{"url": "http://kkul.mascali1928.it/solving-quadratic-equations-pure-imaginary-numbers.html", "text": "# Solving Quadratic Equations Pure Imaginary Numbers\n\nFor y = x 2 , as you move one unit right or left, the curve moves one unit up. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. In our recent paper we gave an efficient algorithm to calculate \"small\" solutions of relative Thue equations (where \"small\" means an upper bound of type $10^{500}$ for the sizes of solutions). 1 is called Cartesian, because if we think of as a two dimensional vector and and as its components, we can represent as a point on the complex plane. The solutions of the quadratic equation ax2 + bx +c = 0 are: SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS The solutions are: SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS Hence, the solution is 3. Solve quadratic equations by completing equations the square. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. The Unit Imaginary Number, i, has an interesting property. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. math game websites for elementary students basic math puzzles 6th grade expressions math games for grade 2 printable grade 9 mathematics paper 1 multiplication puzzle worksheets 4th grade adding and subtracting variables worksheet hw solver unblocked. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. These solutions are in the set of pure imaginary numbers. Videos are created by fellow teachers for their students using the guided notes from the unit. Substituting in the quadratic formula,. Rules for adding and subtracting complex numbers are given in the box on page 279. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Note that if your quadratic equation cannot be factored, then this method will not work. We call $$a$$ the real part and $$b$$ the imaginary part. 146 root of an equation, p. (Definitions taken from Holt Algebra 2, 2004. Unit 3 - Quadratic Functions. Its solution may be presented as x = \u221aa. A number of the form bi, where \ud835\udc4f\u22600, is called a pure imaginary number. This page will try to solve a quadratic equation by factoring it first. 1 100 Tracing Numbers Worksheet. Find a) the values of p and q b) the range of k such that the equation 3x\u00b2 + 3px -q = k has imaginary roots. 2 Problem 101E. For the simplest case, = 0, there are two turning points and these lie on the real axis at \u00b11. See full list on intmath. use the discriminant to determine the number of solutions of the quadratic equation. That's a first look at quadratic equations. mathematics math\u00b7e\u00b7mat\u00b7ics (m\u0103th\u2032\u0259-m\u0103t\u2032\u012dks) n. An obvious choice for x(0) is a turning point. For a method of solving quadratic equations,. OBJECTIVES 1 Add,Subtract,Multiply,and Divide Complex Numbers (p. Also Science, Quantum mechanics and Relativity use complex numbers. Each problem worked out in complete detail. However, using complex numbers you can find solve all quadratic equations. Write each of the following imaginary numbers in the standard form bi: 1 5 , 11, , 7, 18. 5 + 4i A) real B) real, complex C) imaginary D) imaginary, complex Ans: D Section: 2. \ufffb Find the value of the discriminant. THE QUADRATIC FORMULA AND THE DISCRIMINANT THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a\u22600. So tricky, in fact, that it\u2019s become the ultimate math question. Videos are created by fellow teachers for their students using the guided notes from the unit. \u2022 Perform operations with pure imaginary numbers \u2022 Perform operations with complex numbers \u2022 Solve quadratic equations by using the quadratic formula. The real part is zero. The solution of a quadratic equation is the value of x when you set the equation equal to zero. Horizontal Parabola. Horizontal Line Equation. 15-1 (1996), 53-70. (used with a sing. Improper Rational Expression. Chapter 9: Imaginary Numbers Conceptual. Horizontal Shift. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Imaginary Numbers \u2022 pure imaginary number: square root of a negative number \u2022 complex numbers \u2022 i2 = -1 i99 = 8. A complex number is any number of the form a + bi where a and b are real numbers. radical (symbol, expression). 0 Students. Real part + bi Imaginary part Sec. In such a case, if one can easily find the real root, then all that is necessary is to solve the remaining quadratic. If you move 2 units to the left or right of the origin, the curve goes 4 units up. Hypersurfaces as a models for general algebraic varieties. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. 1007/BF00526647) (with E. In this paper, we present a new method for solving standard quaternion equations. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Is it saying I. Pg 237, #1-7 all. Also Science, Quantum mechanics and Relativity use complex numbers. Nature of roots Product and sum of roots. Objective: be able to sketch power functions in the form of f(x)= kx^a (where k and a are rational numbers). complex numbers are required to be covered. Use the relation i 2 = \u22121 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Binomial, Trinomial, Factoring, Monomial, Quadratic Equation in One Variable, Zero of a Function, Square Root, Radical Sign, Radicand, Radical, Rationalizing the Denominator. Pure imaginary. Imaginary unit. Joel Kamnitzer awarded a 2018 E. This calculator is a quadratic equation solver that will solve a second-order polynomial equation in the form ax 2 + bx + c = 0 for x, where a \u2260 0, using the completing the square method. 1 Complex numbers expressed in cartesian form Include: \u2022 extension of the number system from real numbers to complex numbers \u2022 complex roots of quadratic equations \u2022 four operations of complex numbers expressed in the form (x +iy). 1 Complex Numbers Complex numbers were developed as a result of the need to solve some types of quadratic equations. Quadratic Formula 9. It is a branch of pure mathematics that uses alphabets and letters as variables. The algebra consisted of simple linear and quadratic equations and a few cubic equations, together with the methods for solving them; rules for operating with positive and negative numbers, finding squares, cubes and their roots; the rule of False Position (see History of Algebra Part. SOLVING QUADRATIC EQUATIONS. THANK YOU FOR YOUR TIME. Ten exponential equations worked out step by step. Imaginary Part. \" Although there are two possible square roots of any number, the square roots of a negative number cannot be distinguished until one of the two is defined as the imaginary unit, at which point +i and -i can then be distinguished. get for a quadratic equation. Normally, it is impossible to solve one equation for two unknowns. $$i \\text { is defined to be } \\sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Complex numbers; Non-real roots of quadratic equations. 2 Power Functions with Modeling. Solve the equation x2 +4x+5 = 0. Comparing real and imaginary parts. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. SolutionWe use the formula x= \u2212b\u00b1 \u221a b2 \u2212 4ac 2a With a=1, b=\u22122and c=10we \ufb01nd x = 2\u00b1 p (\u22122)2 \u2212(4)(1)(10) 2. Videos are created by fellow teachers for their students using the guided notes from the unit. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. i is the imaginary unit. When a real number, a, is added to an imaginary number, a + bi is said to be a complex number. Imaginary numbers. Just beat it yesterday after a week long addiction. Algebra-help. Finding the values or real and imaginary numbers in standard form. Solving a quadratic equation: AC method. x2 + 9 = 0 b. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. Procedure for solving. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. The axis of symmetry will intersect a parabola in one point called the _____. Division of a complex number by a complex number; Division of a complex number by a complex number (example) Argand diagrams; Modulus and argument; Equating real and imaginary parts to solve equations; Square roots of a complex number; Solving quadratic equations with complex roots; Solving cubic equations; Solving quartic equations; Reflection. When I became a student at the. Complex numbers are built on the idea that we can define the number i (called \"the imaginary unit\") to be the principal square root of -1, or a solution to the equation x\u00b2=-1. 3 x 2 = 100 - x 2 Solution: Step 1. verb) The study of the measurement, properties, and relationships of quantities and sets, using. The Unit Imaginary Number, i, has an interesting property. Now you will solve quadratic equations with imaginary solutions. Real numbers. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. Consider the pure quadratic equation: x 2 = a , where a \u2013 a known value. The imaginary number i=sqrt(-1), i. What was most perplexing was that in using these subtle and imaginary numbers it was possible to solve cubic equations. A complex number is a number of the form where. Its use was prompted by the need to deal with algebraic expressions such as $$x^2+1$$ that have no root in the real numbers. Complex numbers; Non-real roots of quadratic equations. SOLVING QUADRATIC EQUATIONS. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. Textbook solution for Precalculus: Mathematics for Calculus (Standalone\u2026 7th Edition James Stewart Chapter 1. Simplifying Roots Of Negative Numbers Khan Academy. The special case corresponding to two squares is often denoted simply (e. Obviously when you get one root of a cubic equation, you can get the other two by dividing the original cubic equation by minus the first root and then use the quadratic formula in order to obtain the other two roots. 4 (1992): 824-842. complex number system The complex number system is made up of both the real numbers and the imaginary numbers. Comparing real and imaginary parts. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. If b 0, then the complex number is called an imaginary number (Figure 2. It is well known that is perpendicular to iff is a pure imaginary number. An imaginary number is an even root of a negative number. SolutionWe use the formula x= \u2212b\u00b1 \u221a b2 \u2212 4ac 2a With a=1, b=\u22122and c=10we \ufb01nd x = 2\u00b1 p (\u22122)2 \u2212(4)(1)(10) 2. But suppose some wiseguy puts in a teensy, tiny minus sign: Uh oh. This is a particular case of the quite general situation, which has been treated in the author\u2019s thesis [8]. 156 complex number, p. Beware that in some cases the. Be able to find complex roots for quadratic equations. All non-imaginary numbers are real. SOLUTION OF A QUADRATIC EQUATION BY COMPLETING THE SQUARE. Plug values into the quadratic formula. x2 + 9 = 0 b. Both hyperbolas are of relatively simple form. There's also a bunch of ways to solve these equations! Watch this tutorial and get introduced to quadratic equations!. Real numbers. I make note of which method needs the most reinforcement (likely completing the square) to that I can provide more practice when we get to imaginary numbers, later in the unit. The value of the discriminant of a quadratic equation can be used to describe the number of real and complex solutions. Core Vocabulary quadratic equation in one variable, p. This script is nothing extraordinary I just put it up so someone trying to do something similar with imaginary numbers could use the code as reference. Mediaeval Algebra in Western Europe was first learnt from the works of al-Khowarizmi, Abu Kamil and Fibonacci. it is a complete quadratic if b 0. Since the discriminant b 2 - 4 ac is 0, the equation has one root. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. 2 2 +7 +2 \u2265 0 32. Algebra-help. radical (symbol, expression). The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. Real part Imaginary part. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. When this occurs, the equation has no roots (zeros) in the set of real numbers. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. Quadratic Function Graph \u2022 max/min \u2022 vertex \u2022 axis of symmetry \u2022 y intercept \u2022 domain/range 7. Solve 3 \u2013 4i = x + yi Finding the answer to this involves nothing more than knowing that two complex numbers can be equal only if their real and imaginary parts are equal. Solved Name Unit 4 Solving Quadratic Equations Date B. 5 Solving Quadratic Equations \u2013 Factoring. 0 = 2x2 5x +7 x = ( 5) p ( 5)2 4(2)(7) 2(2) = 5 p 25 56 4 = 5 p 31. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of imaginary quadratic fields. Is Zero Considered a Pure Imaginary Number (as 0i)? [12/02/2003] In the complex plane, zero (0 + 0i) is on both the real and pure imaginary axes. Normally it is mentioned in chapter related to complex numbers where the reader is made aware of the power of complex numbers in solving polynomial equations. If and is not equal to 0, the complex number is called a pure imaginary number. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered \"Yes!\" then this instructable can help you. Ncert Exemplar Class 11 Maths Solutions Chapter 5 Free Pdf. 1 Examples of solving quadratic equations using the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) demonstrate pure mathematical. Quadratic Equations solving quadratic equations by completing the square the quadratic formula long division of a polynomial by a. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. These are sometimes called pure imaginary numbers. Its solution may be presented as x = \u221aa. Newton did not include imaginary quantities within the notion of number, and that G. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex. the effect that changing. In the 17th century, Ren\u00e9 Descartes (1596\u20131650) referred to them as imaginary numbers. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit \u2018 \u2018. If the real part of a complex number is 0, then it is called a pure imaginary number. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. An equivalent form is b2 \u2014 4ac If a, b and c are rational coefficients, then \u2014 is a rational. 2 Power Functions with Modeling. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. $$i \\text { is defined to be } \\sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Day 10 I can find complex solutions of quadratic equations. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. (ii) Determine the other root of the equation, giving your answer in the form p + iq. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. is the imaginary part of the complex number. Zero Factor Property \u2013 basis for solving quadratic equations. Well, this time, I would like to write about quadratic equation. i i i is \"a\" solution to the quadratic equation x 2 = A pure imaginary number is a complex number having its real part zero. College Algebra (11th Edition) answers to Chapter 1 - Section 1. These are all quadratic equations in disguise:. If one complex number is known, the conjugate can be obtained immediately by changing the sign of the imaginary part. Complex Number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = \u22121, which is called an imaginary number because there is no real number that satisfies this equation. Imaginary numbers are based on the mathematical number $$i$$. Imaginary Unit i, Complex Number, Standard Form of a complex number, Imaginary Number, Pure Imaginary. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Therefore a complex number is the sum of a real number and a pure imaginary one. 4) Using a quadratic equation solver, we wind up with this: x = (2. The standard form of The solution set of equation 25x2 \u2014 I = 0 is: The quadratic. quadratic equations. Consider the pure quadratic equation: x 2 = a , where a \u2013 a known value. Solving Quadratic Equations Pure Imaginary Numbers. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. 3 10 4 3 9. \u2022 Use the discriminant to find the number of x-intercepts/real solutions/zeros/roots. number, pure imaginary number 3. We use the function func:scipy. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Solve quadratic equations by inspection (e. Improper Fraction. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. -5x2 + 12x - 8 = 0 4. Solving Quadratics with Imaginary Solutions Name_____ Date_____ Period____ \u00a9M M2O0M1_6k GK_ultYaQ hSqoTfftTwwalrmed qLULvCm. , the square root of -1. The diagram shows how different types of complex numbers are related. Quadratic inequality in two variables: Quadratic inequality in one variable: Linear inequality in two variables: Solve the equation using any method. Answer by math_helper(1904) ( Show Source ):. These are all quadratic equations in disguise:. Real part + bi Imaginary part Sec. Perform operations with pure imaginary numbers and complex numbers Use complex conjugates to write quotients of complex numbers in standard form Graph quadratic functions Solve quadratic equations - Set - Element - Subset - Universal Set - Complement - Union - Intersection - Empty Set - Imaginary Unit - Complex Number. Galerkin (HDG) method for solving the Helmholtz equation with impedance boundary condition: (1. By using this website, you agree to our Cookie Policy. Imaginary numbers. Which statement about the solutions x = 5 and x = \u201320 is true? asked by T on June 2, 2016; Algebra 2 help :) Any number in the form of a+-bi , where a and b are real numbers and b not equal 0 is considered a pure imaginary number. Therefore, the rules for some imaginary numbers are:. Imaginary Part. Numbers like \u20142 \u2014 i and - 2 + i that include a real term and an imaginary term are called complex numbers. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations. The concept was discussed in a recent thread, where we pointed out that the definition used for real radicands doesn't apply here, as there are no \"positive\" complex numbers; in cases like yours, in fact, both roots (2 - i and -2 + i) have a negative sign somewhere, so. The aim of this paper is to study t k and the value of N k /\u211a ( \u03b7 k ). Horizontal Line Equation. The difference is that the root is not real. Upon completing this goal the student will be able to: * solve quadratic equations by graphing, factoring, and completing the square. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. Solve quadratic equations with complex number solutions. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. I will even skip a match if it means swiping the largest number out of the corner. They will also analyze situations involving quadratic functions and formulate quadratic equations to solve problems. From this starting point evolves a rich and exciting world of the number system that encapsulates everything we have known before: integers, rational, and real numbers. Simplify the expression: 16. Imaginary Numbers. They are factoring, using the square roots, completing the square and using the quadratic formula. x2 + 4x + 5 = 0 c. The Imaginary Unit i Not all quadratic equations have real-number solutions. Imaginary numbers are applied to square roots of negative numbers, allowing them to be simplified in terms of i. How does this work? Well, suppose you have a quadratic equation that can be factored, like x 2 +5x+6=0. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Solve equations Quadratic in Form by substitution: Step 1: Determine the appropriate substitution and write the equation in the form au2 + bu + c = 0 Step 2: Solve the equation (using any method). \" Imaginary numbers allow for complex analysis, which allows engineers to solve practical problems working in the plane. -2-Create your own worksheets like this one with Infinite Algebra 2. Complex Solution to Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, we can use complex numbers and the imaginary root to express the solutions. It also provides solutions to the problematic quadratic equations and all other polynomial equations In the form p(x). Write and graph an equation of a parabola with its vertex at (h,k) and an equation of a circle, ellipse, or hyperbola with its center at (h,k) Classify a conic using its equation : Quadratic Systems : Solve systems of quadratic equations by finding points of intersection Solve systems of quadratic equations using substitution. Quadratic equation usually used to find the unknown number(s) of x in the equation. Solve quadratic equations by inspection (e. for solving quadratic 2a equations. Note that each of these numbers is pure imaginary with positive coefficient. 5 Relation of the Roots. 6 Complex and Imaginary Numbers Objectives What is an imaginary number? What is a complex number? Jan 30\u00ad10:53 AM 1 Complex Numbers 2010 September 15, 2010 Warm\u00adup: Solve using the quadratic formula. In this tutorial, you'll be introduced to imaginary numbers and learn that they're a type of complex number. Now, by applying algebra techniques we can solve the equation. \u2022 Solve quadratic equations by factoring. Unique Math Equation Stickers designed and sold by artists. Continuing coursework from the Algebra II A, this title covers the review of square roots, radicals, complex pure and imaginary numbers, solving and factoring, identifying and evaluating the discriminant of a quadratic equation, rewriting equations, solving problems with number lines, graphing parabola, circle parts and formulas, hyperbola. fsolve to solve it. a coefficient has on. Quadratic Equation Solver. 1 Unit Objectives 4. Myung-Hwan Kim, Introduction to Universal Positive Quadratic Forms over Real Number Fields, Proc. The Unit Imaginary Number, i, has an interesting property. (Substitute your values back into the original subst. 4c Calculate the discriminant of a quadratic equation to determine the number of real & complex solutions. All non-imaginary numbers are real. (ii) Determine the other root of the equation, giving your answer in the form p + iq. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. Quadratic Equation. Cauchy-Riemann equations, harmonic functions. imaginary quadratic base eld Groups of Special Units, University of Georgia 2009 Invited number theory seminar about my thesis research Conference Organization West Coast Number Theory Conference 2015 - present On organizational committee and grant commitee Selected Conferences and Scholarly Activities [email\u00a0protected] 2016 -present. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. \\)The trajectory of such a solution consists of one point, namely $$c\\ ,$$ and such a point is called an equilibrium. doc Author: E0022430 Created Date: 2/9/2010 12:03:19 PM. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. Polynomials with Complex Solutions. This is a particular case of the quite general situation, which has been treated in the author\u2019s thesis [8]. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Solve quadratic equations by factoring. If \ud835\udc4f=0, then the number \ud835\udc4e+\ud835\udc4f\ud835\udc56=\ud835\udc4e is a real number. xx2 12 35 0 2. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. When the radicand in the quadratic formula (the discriminant Delta) is negative it means that you cannot find pure Real solutions to your equation. \ufffb Find the value of the discriminant. You want the square root of a number less than zero? That\u2019s absurd!. Complete quadratic equation: If the equation having x and x2 terms such an. LOVE IT!! Reply Delete. Impossible Event. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Horizontal Shrink. xx2 10 25 64 4. The solution set is The Quadratic Formula If we start with the equation ax2 + bx + c = 0, for a > 0, and complete the square to solve for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. State the number of complex roots of the equation x 3 2x2 3x 0. web; books; video; audio; software; images; Toggle navigation. 156 pure imaginary number, p. complex numbers are required to be covered. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. pure imaginary number. Negative 4, if I take a square root, I'm going to get an imaginary number. fsolve to do that. Steacie Memorial Fellowship U of T\u2019s team of students place 4 th in the 2017 Putnam Competition! Three faculty, R. Complex Numbers H2 Maths Tuition Tips. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Quadratic Formula 9. There are various methods through which a quadratic equation can be solved. 10 points for the best working. The number has a non-zero real part and pure imaginary part. An imaginary number is an even root of a negative number. Solve for x: x( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0. \u2022 Find square roots and perform operations with pure imaginary numbers. Imaginary numbers are used to help us work with numbers that involve taking the square root of a negative number. The problem was with certain cubic equations, for example x3 \u22126x+2 = 0. LOVE IT!! Reply Delete. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. = \u22121, and every complex number has the form a + biwith a and b real. I been trying to figure out how to set up this equation to add two complex numbers for Java. In a similar way, we can find the square root of any negative number. We often use the notation z= a+ib, where aand bare real. \u2022 Estimate solutions of quadratic equations by graphing. Students apply these. Carmen is using the quadratic equation (x + 15)(x) = 100 where x represents the width of a picture frame. Identity (Equation) Identity Matrix. Example 2A: Solving a Quadratic Equation with Imaginary Solutions Take square roots. 3 \u00ad Notes \u00ad Solving Quadratics with Imaginary Numbers. Khan Academy Video: Quadratic Formula 1; Need more problem types? Try MathPapa Algebra. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered \"Yes!\" then this instructable can help you. Powered by Cognero Page 1. To ensure that every quadratic equation has a solution, we need a new set of numbers that includes the real numbers. Solving quadratic equations can sometimes be quite difficult. Math is the basic building blocks that deals with all sort of calculations such as Addition, subtraction, multiplication, division and much more. 2 Problem 101E. Well, this time, I would like to write about quadratic equation. The imaginary unit i is the complex. \u2022 Complete the square to solve quadratic equations or to convert from standard to vertex form. Students work extensively with factoring quadratics using various factoring techniques. Finding the values or real and imaginary numbers in standard form. This book has been requested by many readers. Journal Canadien de Math\\'ematiques 44. Complex Solutions of Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, you often obtain a result such as which you know is not a real number. In this paper, we present a new method for solving standard quaternion equations. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. a unique quadratic function. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. x2 + 9 = 0 b. Imaginary numbers are called so because they lie in the imaginary plane, they arise from taking square roots of negative numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. the square, or using. To find complex number solutions of quadratic equations. (Substitute your values back into the original subst. By using this website, you agree to our Cookie Policy. ExampleUse the formula for solving a quadratic equation to solve x2 \u2212 2x+10=0. x2 + 4x + 5 = 0 c. The others are right - pick a corner and keep the largest numbers closest to that corner. But this pure oscillation would be the B equals 0 with undamped. Imaginary numbers are complex numbers where a = 0 and b \u2260 0. Lastly, Allen defined a complex number as one which is not real (p. pure imaginary number. Lesson 2: Solving Square Root Equations Lesson 3: The Basic Exponent Properties Lesson 4: Fractional Exponents Revisited Lesson 5: More Exponent Practice Lesson 6: The Quadratic Formula Lesson 7: More Work with the Quadratic Formula Unit 9 Lesson 1: Imaginary Numbers Lesson 2: Complex Numbers Lesson 3: Solving Quadratics with Complex Solutions. Number Theory 85 (2000), 201-219. 1) 10x2 - 4x + 10 = 02) x2 - 6x + 12 = 0 3) 5x2 - 2x + 5 = 04) 4b2 - 3b + 2 = 0. But this is really two. (used with a sing. In quadratic planes, imaginary numbers show up in equations. Solve quadratic equations by inspection (e. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. 22 (1996), 425-434. Pure Mathematics 2 & 3. First published in 1975, this classic book gives a systematic account of transcendental number theory, that is those numbers which cannot be expressed as the roots of algebraic equations having rational coefficients. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). An equivalent form is b2 \u2014 4ac If a, b and c are rational coefficients, then \u2014 is a rational. Example: 3i If a \u22600 and b \u2260 0, the complex number is a nonreal complex number. You want the square root of a number less than zero? That\u2019s absurd!. Impossible Event. Solved Name Unit 4 Solving Quadratic Equations Date B. 7) 10n2 - n - 8 = 08) 8p2 - 12p + 7 = 0 9) 2r2 + 2r + 6 = 0 10) 11r2 - 5r - 12 = 7 11) -14 + a = -3a2 12) -5 = 11b2 - 2b 13) 3n2 + 10n = -12 - 8n2 + 10n14) r2 - 2r - 4 = 2r2 + 8 Find the discriminant of each quadratic equation then state the number and type of solutions. Imaginary numbers are complex numbers where a = 0 and b \u2260 0. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. Whitley) Periods of cusp forms and elliptic curves over imaginary quadratic fields Mathematics of Computation 62 No. In this tutorial, you'll be introduced to imaginary numbers and learn that they're a type of complex number. 25 2 5 1 7\u2212 i2 =\u2212 \u2212=( ) 28. The Quadratic Equation, which has many uses, can give results that include imaginary numbers. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. Its solution may be presented as: Here the three cases are possible:. Any number that is a non-repeating decimal is irrational. is the imaginary part of the complex number. 3 Exercises - Page 103 9 including work step by step written by community members like you. Workshops in Pure Math. Pure imaginary number \u2013 If a = 0 and b \u02dc 0, the number a + bi is a pure imaginary number. Consider the pure quadratic equation: x 2 = a ,. Imaginary Part. Note that if your quadratic equation cannot be factored, then this method will not work. Microsoft Word - Imaginary and Complex Numbers. Many answers. Inspired designs on t-shirts, posters, stickers, home decor, and more by independent artists and designers from around the world. Complex numbers cannot be ordered. If a and b are real numbers \ud835\udc4e+\ud835\udc4f\ud835\udc56 is a complex number, and it is said to be written in standard form. Manipulating expressions involving \u03b1+\u03b2 and \u03b1+\u03b2. Let us learn about solving quadratic equation calculator with a solved examples. Of course, these are abelian, so sometimes have slightly special properties. 253 #33-44, 64-66. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. The complex numbers include all real numbers and all imaginary numbers. We can now solve both of these equations trivially. Impossible Event. notebook 1 January 11, 2017 Jan 4\u00ad9:06 AM Quadratic Functions MGSE9\u00ad12. Many quadratic equations have roots that are pure imaginary numbers or. 2 Mean Value Theorem. I like to use puzzles when a specific skill (like solving quadratic equations) requires fluency [MP6]. The quadratic equation 3x\u00b2 + 3px - q=0 has the roots and 3. 88 Quadratic equations are the basis for a vast area of more complex mathematics, both pure and applied. Imaginary. can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5) can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5). This is denoted by C. Operations with Complex Numbers Complex Numbers (a + bi) Real Numbers (a + 0i) Imaginary. \u2022 Pure Imaginary Numbers & Powers of i \u2022 Solving Quadratics by Square Roots with Pure Imaginary Solutions \u2022 Complex Numbers (includes Classifying & Properties) \u2022 Operations with Complex Numbers \u2022 Solving Quadratics by Completing the Square (includes Complex Solutions) \u2022 Solving Quadratics by the Quadratic Formula (includes. Negative 4, if I take a square root, I'm going to get an imaginary number. pure imaginary number. Using this method we reobtain the known formulas for the solution of a quadratic quaternion equation, and provide an explicit solution for the cubic quaternion equation, as long as the equation has at least one pure imaginary root. Write quadratic functions in vertex form. Quadratic Equation. To solve equation we must specify the initial condition x(0). Equations such as +1 0 have no real solution, so mathematicians defined the imaginary numbers to represent their solu ions. x2 + 4x + 5 = 0 c. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. As humans have solved new problems, equations, they have needed to create more numbers. 146 Solving Quadratic. or Quadratic Equations That Can Be Solved by Factoring, Applications of the Pythagorean Theorem Pg. 3 Example 4 Solve: x 2 x 6 0 1 223 x 2 No real-number solutions To solve such equations, we must define the square root of a negative number. Students will solve quadratic equations using graphs, tables, and algebraic methods. nth roots of a complex number The technique is the same for finding nth roots of any complex number. 2 Basic I can use the quadratic formula to solve a quadratic equation. \u2022 finding an equation for the common perpendicular to two skew lines 4 Complex numbers 4. When you need guidance on algebra exam or concepts of mathematics, Algebra-help. (ii) Determine the other root of the equation, giving your answer in the form p + iq. Quadratic Equation. Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. It can get a little confusing!. ExampleUse the formula for solving a quadratic equation to solve x2 \u2212 2x+10=0. Solving polynomial, linear, quadratic. Imaginary Part. Examples: Write each number in the form + \ud835\udc56: a. Pg #20-34 all. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). 0 = 2x2 5x +7 x = ( 5) p ( 5)2 4(2)(7) 2(2) = 5 p 25 56 4 = 5 p 31. the quadratic formula. Find the exact solution of by using the Quadratic Formula. Now you will solve quadratic equations with imaginary solutions. Big Idea #2: Numbers like 3i and v'\u00ee i are called pure imaginary numbers. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. You can compare all quadratic expressions to ax 2 + bx + c and get the values of a, b and c. Polynomial Equation Solver - by Don Cross This web page contains an interactive calculator that solves any linear, quadratic, or cubic equation. Since the discriminant b 2 \u2013 4 ac is 0, the equation has one root. Textbook Authors: Lial, Margaret L. 8x2 - 4x + 5 = 0 3. We call athe real part and bthe imaginary part of z. Simplify the expression: 17. How does this work? Well, suppose you have a quadratic equation that can be factored, like x 2 +5x+6=0. Write quadratic functions in vertex form. 1 Examples of solving quadratic equations using the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) demonstrate pure mathematical. It was this discovery which made the use of complex numbers \u2018respectable\u2019. Core Vocabulary quadratic equation in one variable, p. Using the quadratic formula, we have x = \u22124\u00b1 p (\u22124)2 \u22124\u00b75 2 = \u22124\u00b1 \u221a \u22124 2 = \u22124\u00b12 \u221a \u22121 2 = \u22122\u00b1i. The calculator also provides conversion of a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Its solution may be presented as x = \u221aa. LOVE IT!! Reply Delete. Page 126 Solving Quadratic Equations Freyer Model. Shankar, and G. This equation, which arises in a surface construction problem, incorporates linear terms in a quaternion variable and its conjugate with right and left quaternion coefficients, while the quadratic term has a quaternion coefficient placed between the variable and. Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step This website uses cookies to ensure you get the best experience. Note: It is not necessary to find the roots. In this paper a new algorithm for solving algebraic Riccati equations (both continuous-time and discrete-time versions) is presented. Pure STEP 3 Questions 2012 S3 Q6 1Preparation The STEP question involves complex numbers and the Argand diagram. Solve the equation x2 +4x+5 = 0. Quadratic Formula 9. 8x2 - 4x + 5 = 0 3. Workshops in Pure Math. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. If \ud835\udc4f\u2260 0, then the number \ud835\udc4e+\ud835\udc4f\ud835\udc56 is called an imaginary number. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. And this happens when b squared is smaller than 4ac. Improper Fraction. To find complex number solutions of quadratic equations. This page will try to solve a quadratic equation by factoring it first. complex number standard form EXAMPLE 1 imaginary unit GOAL 1 Solve quadratic equations with complex solutions and perform operations with complex numbers. Any number that is a non-repeating decimal is irrational. 2x2 - 10x + 25 = 0 5. 1) u k2u= f in ; @u @n (1. Real and imaginary numbers; Addition, subtraction and multiplying complex numbers and simplifying powers of i; Complex conjugates; Division of a complex number by a complex number; Argand diagrams; Modulus and argument of a complex number; Solving problems with complex numbers; Square roots of a complex number; Solving quadratic equations with. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. Horizontal Reflection. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. The solution to this particular equation is called the imaginary number i: i2 = 1,1 and one way to de ne the set of complex numbers is as the set of all expressions of type x+ iywhere xand yare real. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. \u2022 Writing quadratic equations in different forms reveals different key features. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. 5th Class Maths Worksheets. When the real part is zero we often will call the complex number a purely imaginary number. f x Ax Bx C = + + = 0 Equation 1. The Quadratic Equation, which has many uses, can give results that include imaginary numbers. An imaginary number bi has two parts: a real number, b, and an imaginary part, i, defined as i^2 = -1. You can use the imaginary unit to write the square root of any negative number. 7 3 2i i i 11. To find complex number solutions of quadratic equations. Students work extensively with factoring quadratics using various factoring techniques. For example, as follows:. Journal of Pure and Applied Algebra, volume 215, number 6, pages 1371--1397, 2011. 2) + iku= g on ; where 2Rd;d= 1;2;3 is a convex polyhedral domain, := @, k\u02db1 is known as the wave number, i = p 1 denotes the imaginary unit, and ndenotes the unit outward normal to @. The name comes from \"quad\" meaning square, as the variable is squared (in other words x 2). Full text of \"The theory of equations: with an introduction to the theory of binary algebraic forms\" See other formats. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. The imaginary unit \u201ci\u201d is used to represent: i 1 and i2 1 Ex. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNINGUPDATE: This unit now contains a Google document with links to instructional videos to help with remote teaching during COVID-19 school closures. square roots denoted by s and \u00bas. Simplify the expression: 16. 3 - Complex Numbers - 1. (a = 0) So, a number is either real or imaginary, and some imaginary numbers are pure imaginary numbers. It \"cycles\" through 4 different values each time we multiply:. Excel in math and science. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. (a) x 2 \u22121=0 (b) x2 \u2212x \u22126 =0 (c) x 2 \u22122x \u22122 =0 (d) x2 \u22122x +2 =0 You should have found (a), (b) and (c) straightforward to solve. in the complex number. Pure imaginary numbers \u2013 numbers in the form bi \u2013 where i= \u22121. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit \u2018 \u2018. An equivalent form is b2 \u2014 4ac. Quadratic formula: A quadratic formula is the solution of a quadratic equation ax 2 + bx + c = 0, where a \u2260 0, given by. can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5) can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5). Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. The imaginary part is zero. And you would be right. Pure quadratic equation The number of methods to solve a quadratic e uatlon Is: Which equation is called exponential equation? A solution of equation which does not satisfy the equation is called: An equation in which variable occurs under radical sign is called. Joel Kamnitzer awarded a 2018 E. 3 Solving Quadratic Equations 4. These are solutions where appear the imaginary unit i. Imaginary numbers. x2 =-1 *This section may be omitted without any loss of continuity. Horizontal Parabola. Well, this time, I would like to write about quadratic equation. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). It is also called an \"Equation of Degree 2\" (because of the \"2\" on the x) A \"Standard\" Quadratic Equation looks like this: The letters a, b and c are coefficients (you know those values). Unit 4 Solving Quadratic Equations Homework 2 Answer Key. We call athe real part and bthe imaginary part of z. Page 126 Solving Quadratic Equations Freyer Model. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. 9) The matrix of this system is A = 0 I F 0 , where F = \u2212\u039b+bK. The discriminant is the radicand in the quadratic formula. 1 Complex numbers expressed in cartesian form Include: \u2022 extension of the number system from real numbers to complex numbers \u2022 complex roots of quadratic equations \u2022 four operations of complex numbers expressed in the form (x +iy). Introduction Fundamental theorem of algebra is one of the most famous results provided in higher secondary courses of mathematics. Solving Quadratic Equations by Finding Square Roots. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. Use the Quadratic Formula to solve the quadratic equation. In this paper, we present a new method for solving standard quaternion equations. Many answers. Find a) the values of p and q b) the range of k such that the equation 3x\u00b2 + 3px -q = k has imaginary roots.", "date": "2020-10-25 08:25:48", "meta": {"domain": "mascali1928.it", "url": "http://kkul.mascali1928.it/solving-quadratic-equations-pure-imaginary-numbers.html", "openwebmath_score": 0.7229335308074951, "openwebmath_perplexity": 546.7770577462251, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9919380075184121, "lm_q2_score": 0.9073122150949273, "lm_q1q2_score": 0.8999974708383791}}
{"url": "http://math.stackexchange.com/questions/66443/a-square-matrix-has-the-same-minimal-polynomial-over-its-base-field-as-it-has-ov", "text": "# A square matrix has the same minimal polynomial over its base field as it has over an extension field\n\nI think I have heard that the following is true before, but I don't know how to prove it:\n\nLet $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\\mathbb{R}$.\n\nIs this true? Would anyone be willing to provide a proof?\n\nAttempt at a proof:\n\nLet $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\\Bbb C$, in which case it will fulfil $M(A)=0$, and therefore $P(t)$ divides it. In addition, we can look at $P(t)$ as the sum of two polynomials: $R(t)+iK(t)$. Plugging $A$ we get that $R(A)+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as real polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$.\n\nNow $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$.\n\nDoes this look to be correct?\n\nMore generally, one might prove the following\n\nLet $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial of$~A$ over$~F$ is the same as the minimal polynomial of $A$ over$~K$.\n\n-\nThere's the saying, \"Look before you leap\". I think I've managed to prove this. Please confirm if my answer is correct. \u2013\u00a0 iroiroaru Sep 21 '11 at 18:43\nI think you already posted before under a different account (the \"above\" instead of \"over\"); I also remember your user name. Have you considered registering, so that all your activity is under the same user name? \u2013\u00a0 Arturo Magidin Sep 21 '11 at 18:46\nIt is impossible for us to confirm if your answer is correct if all you do is provide the question. If you want us to \"confirm if [your] answer is correct\", why not post your proof ? \u2013\u00a0 Arturo Magidin Sep 21 '11 at 18:49\nI am in the process of writing it! \u2013\u00a0 iroiroaru Sep 21 '11 at 18:51\nHi Arturo, yes, I posted here twice before. Should I register? As this site allows me to post questions without registering, I figured it wouldn't be necessary. e- I'm done writing my proof. \u2013\u00a0 iroiroaru Sep 21 '11 at 18:53\n\nWritten before/while the OP was adding his/her own proof, which is essentially the same as what follows.\n\nLet $\\mu_{\\mathbb{R}}(x)$ be the minimal polynomial of $A$ over $\\mathbb{R}$, and let $\\mu_{\\mathbb{C}}(x)$ be the minimal polynomial of $A$ over $\\mathbb{C}$.\n\nSince $\\mu_{\\mathbb{R}}(x)\\in\\mathbb{C}[x]$ and $\\mu_{\\mathbb{R}}(A) = \\mathbf{0}$, then it follows by the definition of minimal polynomial that $\\mu_{\\mathbb{C}}(x)$ divides $\\mu_{\\mathbb{R}}(x)$.\n\nI claim that $\\mu_{\\mathbb{C}}[x]$ has real coefficients. Indeed, write $$\\mu_{\\mathbb{C}}(x) = x^m + (a_{m-1}+ib_{m-1})x^{m-1}+\\cdots + (a_0+ib_0),$$ with $a_j,b_j\\in\\mathbb{R}$. Since $A$ is a real matrix, all entries of $A^j$ are real, so $$\\mu_{\\mathbb{C}}(A) = (A^m + a_{m-1}A^{m-1}+\\cdots + a_0I) + i(b_{m-1}A^{m-1}+\\cdots + b_0I).$$ In particular, $$b_{m-1}A^{m-1}+\\cdots + b_0I = \\mathbf{0}.$$ But since $\\mu_{\\mathbb{C}}(x)$ is the minimal polynomial of $A$ over $\\mathbb{C}$, no polynomial of smaller digree can annihilate $A$, so $b_{m-1}=\\cdots=b_0 = 0$. Thus, all coefficients of $\\mu_{\\mathbb{C}}(x)$ are real numbers.\n\nThus, $\\mu_{\\mathbb{C}}(x)\\in\\mathbb{R}[x]$, so by the definition of minimal polynomial, it follows that $\\mu_{\\mathbb{R}}(x)$ divides $\\mu_{\\mathbb{C}}(x)$ in $\\mathbb{R}[x]$, and hence in $\\mathbb{C}[x]$. Since both polynomials are monic and they are associates, they are equal. QED\n\nSo, yes, your argument is correct.\n\n-\n\nAnother way of proving this fact may be observing that ''you do not go out the field while using Gaussian elimination''. More precisely:\n\nProposition. Let $K \\subseteq F$ be a field extension let $v_1, \\dots, v_r \\in K^n$. If $v_1, \\dots, v_r$ are linearly dependent over $F$, then they are linearly dependent over $K$.\n\nProof. We'll prove the contrapositive of the statement. Suppose that the $v_i$'s are linearly independent over $K$. Let $\\lambda_i \\in F$ such that $\\sum_i \\lambda_i v_i = 0$. We can find $e_j \\in F$ linearly independent over $K$ such that $\\lambda_i = \\sum_j \\alpha_{ij} e_j$, with $\\alpha_{ij} \\in K$. Now from $\\sum_{i,j} e_j \\alpha_{ij} v_i = 0$ we deduce that $\\sum_i \\alpha_{ij} v_i = 0$, for every $j$. From the independence of $v_i$'s over $K$, we have $\\alpha_{ij} = 0$, so $\\lambda_i = 0$. $\\square$\n\nNow consider a field extension $K \\subseteq F$ and a matrix $A \\in M_n(K)$. Let $\\mu_K$ and $\\mu_F$ the minimal polynomials of $A$ over $K$ and $F$, respectively. Considering $I, A, A^2, \\dots, A^r$ in the vector space $M_n(K)$, from the proposition you have $\\deg \\mu_K \\leq \\deg \\mu_F$. On the other hand it is clear that $\\mu_F$ divides $\\mu_K$. So $\\mu_F = \\mu_K$.\n\n-\n\nAs Andrea explained, the statement in the question results immediately from the following one.\n\nLet $K$ be a subfield of a field $L$, let $A$ be an $m$ by $n$ matrix with coefficients in $K$, and assume that the equation $Ax=0$ has a nonzero solution in $L^n$. Then it has a nonzero solution in $K^n$.\n\nBut this is obvious, because the algorithm giving such a solution (or its absence) depends only on the field generated by the coefficients of $A$.\n\n-\n\nThis looks correct.\n\nAnother way to see it is that you can find the minimal polynomial of the matrix by computing the invariant factors of the matrix $A-XId$ over $\\mathbb{R}$. Since the same process (with same operations) may be done over $\\mathbb{C}$, their minimal polynomial is the same.\n\nsorry, i don't know the english word for the \"invariant factors\", i mean the process that using only row and columns operations, the matrix $A-XId$ may be uniquely writtten as some zero and a sequence of polynomial in the diagonal in which any polynomial divides the next one, and where the first is the minimal polynomial $A$ and the last the characteristic polynomial of $A$.\n\n-\nDon't apologise, I'm having trouble with English as well! Since Arturo posted what seems like a more straightforward proof (well, it's the one I thought of...), I've accepted his answer, but thank you for your input and I will consider your idea. \u2013\u00a0 iroiroaru Sep 21 '11 at 19:07", "date": "2015-10-09 11:03:38", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/66443/a-square-matrix-has-the-same-minimal-polynomial-over-its-base-field-as-it-has-ov", "openwebmath_score": 0.9426719546318054, "openwebmath_perplexity": 105.42503669789473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406009350774, "lm_q2_score": 0.9086179012632543, "lm_q1q2_score": 0.8999320601475463}}
{"url": "https://math.stackexchange.com/questions/769366/how-many-non-empty-subsets-of-1-2-n-satisfy-that-the-sum-of-their-eleme", "text": "# How many non empty subsets of {1, 2, \u2026, n} satisfy that the sum of their elements is even?\n\nThe question I am working on is the case for $n$ = 9. How many non-empty subsets of $\\{1,2,...,9\\}$ have that the sum of their elements is even?\n\nMy solution is that the sum of elements is even if and only if the subset contains an even number of odd numbers. Since this is precisely half of all of the subsets the answer is $\\frac{2^{9}}{2}=2^8$. Then the question specifies non-empty so final answer is $2^8-1$. Is this correct? In general I guess the solutions is $2^{n}-1$. My problem is why do exactly half of the total amount of subsets have and even number of odd numbers? Can we set up a bijection between subsets with odd number of odd numbers and even number of odd numbers?\n\nLet $S$ be a subset of $\\{0,1,2,\\dots,9\\}$, possibly empty. Note that $1+2+\\cdots +9=45$. So the sum of the elements of $S$ is even if and only if the sum of the elements of the complement of $S$ is odd.\n\nDivide the subsets of $\\{1,2,\\dots,9\\}$ into complementary pairs. There are $2^8$ such pairs, and exactly one element of each pair has even sum. Thus there are $2^8$ subsets with even sum, and $2^8-1$ if we exclude the empty set.\n\nRemark: Suppose that $1+2+\\cdots+n$ is odd. This is the case when $n\\equiv 1\\pmod{4}$ and when $n\\equiv 2\\pmod{4}$. Then the same argument shows that there are $2^{n-1}$ subsets with even sum.\n\nWe can use another argument for the general case. Note that there are just as many subsets of $\\{1,2,\\dots,n\\}$ that contain $1$ as there are subsets that do not contain $1$. And for any subset of $A$ of $\\{2,3,\\dots,n\\}$, we have that $A$ has even sum if and only if $A\\cup\\{1\\}$ has odd sum, and $A$ has odd sum if and only if $A\\cup\\{1\\}$ has even sum. Thus in general there are $2^{n-1}$ subsets with even sum.\n\nThe bijection between even-summed sets and odd-summed sets was quite natural when $n\\equiv 1\\pmod{4}$ or $n\\equiv 2\\pmod{4}$. In the general case, there is a nice bijection (add or subtract $\\{1\\}$), but it is less natural.\n\nLet's first count all subsets of $\\{1,\\ldots,n\\}$ with even sum. Removing the empty sets then makes us have to subtract one from this result.\n\nThe subsets of $\\{1,\\ldots,n\\}$ with even sum are one-to-one with the subsets of $\\{2,\\ldots,n\\}$. For any set $J\\subset\\{2,\\ldots,n\\}$, if the sum of $J$ is even, then $J$ is a subset of $\\{1,\\ldots,n\\}$ with even sum, while if the sum of $J$ is odd, then $\\{1\\}\\cup J$ is a subset with even sum.\n\nSince there are $2^{n-1}$ subsets of $\\{2,\\ldots,n\\}$, this is the number of subsets of $\\{1,\\ldots,n\\}$ with even sum. Remove the empty set, and you get $2^{n-1}-1$.\n\nWhat you did is fine, we can get an alternative proof if we recall how we prove that there are $2^{n-1}$ subsets of $\\{1,2\\dots n\\}$ of even cardinality.\n\nLet $E$ be the set of subsets of even cardinity and let $O$ be the set of subsets of odd cardinality, pick an arbitrary element $a\\in\\{1,2,3\\dots n\\}$.\n\nThen $f:E\\rightarrow O$ defined as $X\\mapsto \\{a\\}\\Delta X$ is a bijection right?\n\nWell, if $E'$ is the set of subsets with even sum and $O'$ is the set of subsets with odd sum and $a\\in\\{1,2,3\\dots n\\}$ is odd.\n\n$f:E'\\rightarrow O'$ defined as $X\\mapsto \\{a\\}\\Delta X$ is also a bijection.\n\nSo in any finite subset $A$ of positive integers, exactly half of the subsets have even sum, unless all of the elements of $A$ are even, in which case all the subsets clearly have even sum.\n\n$\\Delta$ is just the symmetric difference of sets, so $\\{a\\}\\Delta X$ is $\\{a\\}\\cup X$ if $a$ was not in $X$ and is $X\\setminus\\{a\\}$ if $a$ was in $X$.", "date": "2019-07-22 04:28:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/769366/how-many-non-empty-subsets-of-1-2-n-satisfy-that-the-sum-of-their-eleme", "openwebmath_score": 0.968389093875885, "openwebmath_perplexity": 56.11930097272193, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9933071492738426, "lm_q2_score": 0.9059898165759307, "lm_q1q2_score": 0.8999261619741692}}
{"url": "https://math.stackexchange.com/questions/2886048/two-alternate-proofs-that-x-neq-0-wedge-xy-xz-implies-y-z", "text": "# Two Alternate Proofs that $x \\neq 0 \\wedge xy = xz \\implies y = z$.\n\nI believe I have been able to construct in two ways, using the field axioms, that if $x \\neq 0$ and $xy = xz$, then $y = z$. However, I've seen similar proofs like this assume that we can perform arithmetic operations, such as multiplying both sides by an inverse--which mirrors in some sense some proofs I've written in an abstract-algebra context--whereas others are more 'purist' in this sense. The similar proof in Rudin, for example, does not assume that we can use simple arithmetic.\n\nMy question, then, is which of these is 'more' standard in a first-year analysis course?\n\nProof 1: Assuming I can use arithmetic .\n\nSince $x \\neq 0$, $\\exists x^{-1}$ s.t. $xx^{-1} = x^{-1} x = 1$ by the field axioms. Therefore, \\begin{align*} xy = xz & & \\text{By assumption} \\\\ x^{-1} (xy) = x^{-1} (xz) & & \\text{Multiply on left by $x^{-1}$} \\\\ \\left(x^{-1} x\\right)y = \\left(x^{-1} x\\right)z & & \\text{Associativity} \\\\ 1y = 1z & & \\text{Inverse properties} \\\\ y = z \\end{align*}\n\nExample 2: Without assuming arithmetic, and mirroring Rudin.\n\n\\begin{align*} y & = 1 \\cdot y & & \\text{Multiplicative identity} \\\\ & = \\left(x \\cdot \\frac{1}{x}\\right) y & & \\text{Mult inverse axiom with $x \\neq 0$} \\\\ & = \\left(\\frac{1}{x} \\cdot x\\right)y & & \\text{Commutativity of multiplication} \\\\ & = \\frac{1}{x} \\left(x \\cdot y\\right) & & \\text{Associativity of multiplication} \\\\ & = \\frac{1}{x} \\left(xz\\right) & & \\text{Assumption that $xy = xz$} \\\\ & = \\left(\\frac{1}{x} \\cdot x\\right) z & & \\text{Associativity of multiplication} \\\\ & = 1z & & \\text{Inverse properties} \\\\ & = z \\end{align*} Thanks in advance.\n\n\u2022 If $K$ is a field, then $G=(K^{\\times},\\cdot)$ is an abelian group, so that the cancellation law holds. For $x,y\\in G$ we have that $xy=xz$ implies that $y=z$. \u2013\u00a0Dietrich Burde Aug 17 '18 at 18:20\n\u2022 In your \"mirroring Rudin\" example the proof is just one long chain of equal quantities: You want to show $y=z$ so you start with $y$ and write down expressions you know are equal to it using axioms and assumptions until you have a $z$. In your \"arithmetic\" proof, you have a list of equalities, and you use your axioms to transform them to get the claim you want: that $y=z$. I don't really think these proofs are different, and I expect people looking at your work would agree with me. But, if you want to make sure, I would suggest asking your grader/professor. \u2013\u00a0James Aug 17 '18 at 18:28\n\u2022 Also, I think your worry about \"using arithmetic\" is illfounded. You have a claim such as $xy = xz$ in some structure you are reasoning about. You also have a binary function on that structure: multiplication. Therefore the quanties $x^{-1}(xy)$ and $x^{-1}(xz)$ are both defined because you are just plugging in elements of the domain into your function. That $y=z$ follows from the assumed properties of multiplication and the existance of inverses. \u2013\u00a0James Aug 17 '18 at 18:31\n\u2022 One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue \"$x=y$ ... <operations> ... $0=0$, QED\". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. \u2013\u00a0James Aug 17 '18 at 18:35\n\u2022 Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. \"Assuming arithmatic\" is a meaningless thing to say. To prove this we must have a well defined set of axioms. \"Assuming arithmetic\" is simply referring to them. \u2013\u00a0fleablood Aug 17 '18 at 18:37\n\nThe two proofs are essentially the same and the first doesn't use arithmetic, but rather field axioms. I wouldn't use $\\frac{1}{x}$, but that's more cosmetic than substantial.\nMore substantial is that you don't need to appeal to commutativity: \\begin{align} y &=1y &&\\text{(multiplicative identity)} \\\\ &=(x^{-1}x)y &&\\text{($x\\ne0$ has an inverse)} \\\\ &=x^{-1}(xy) &&\\text{(associativity)} \\\\ &=x^{-1}(xz) &&\\text{(hypothesis)} \\\\ &=(x^{-1}x)z &&\\text{(associativity)}\\\\ &=1z &&\\text{(property of the inverse)} \\\\ &=z &&\\text{(multiplicative identity)} \\end{align}\nOn the other hand, the other proof seems shorter \\begin{align} & xy=xz &&\\text{(hypothesis)} \\\\ & x^{-1}(xy)=x^{-1}(xz) && \\text{($x\\ne0$ has an inverse)} \\\\ & (x^{-1}x)y=(x^{-1}x)z && \\text{(associativity)} \\\\ & 1y=1z && \\text{(property of the inverse)} \\\\ & y=z && \\text{(multiplicative identity)} \\end{align} and less \u201crabbit out of a top hat\u201d.", "date": "2019-08-25 10:04:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2886048/two-alternate-proofs-that-x-neq-0-wedge-xy-xz-implies-y-z", "openwebmath_score": 0.9999711513519287, "openwebmath_perplexity": 830.331103338256, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513901914406, "lm_q2_score": 0.9124361670249624, "lm_q1q2_score": 0.8998001945726162}}
{"url": "https://physics.stackexchange.com/questions/701918/does-this-question-have-two-answers-correct", "text": "Does this question have two answers correct?\n\nA simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum start swinging in phase. They again swing in phase after an interval of $$18$$ seconds from the start. The period of the simple pendulum is\n\n(A) $$0.9$$ sec\n\n(B) $$1.8$$ sec\n\n(C) $$2.7$$ sec\n\n(D) $$3.6$$ sec\n\nI was given a formula for such questions:\n\n$$T = \\frac {T_1 T_2} {T_1-T_2} \\qquad (T_1>T_2)$$\n\nwhere $$T_1$$ and $$T_2$$ are the time periods of the individual pendulums, and $$T$$ is the time after which they are in phase again.\n\nI took $$T_1$$ as the seconds pendulum, i.e., $$T_1=2$$ seconds.\n\nUsing the formula, I got $$T_2=1.8$$ sec, which makes sense; the timestamps for each oscillation are:\n\n$$1.8\\ \\ 3.6\\ \\ 5.4\\ \\ 7.2\\ \\ 9.0\\ \\ 10.8\\ \\ 12.6\\ \\ 14.4\\ \\ 16.8\\ \\ 18.0$$\n\nseconds for simple pendulum, and $$2, 4, 6, 8, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. None of these overlap, so if $$T_2=1.8$$, the pendulums swing in phase after intervals of $$18$$ seconds.\n\nHowever, I also tried option A, and got the timestamps as:\n\n$$0.9\\ \\ 1.8\\ \\ 2.7\\ \\ 3.6\\ \\ 4.5\\ \\ 5.4\\ \\ 6.3\\ \\ 7.2\\ \\ 8.1\\ \\ 9.0\\ \\ 9.9\\ \\ 10.8\\ \\ 11.7\\ \\ 12.6\\ \\ 13.5\\ \\ 14.4\\ \\ 15.3\\ \\ 16.2\\ \\ 17.1\\ \\ 18$$\n\nseconds for simple pendulum, and $$2, 4, 6, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. Again, none of these overlap, so if $$T_2=0.9$$ seconds also, the pendulums swing in phase after intervals of $$18$$ seconds.\n\nAccording to the answer key, the answer is only B. Is A also correct, or am I missing something?\n\nThe key point that's overlooked in the timestamp-counting method is that having the pendulums be in sync at the end of complete periods is not the only way for them to be in phase - they can also happen to be in phase in the middle of a period. In particular, for this example, note that after $$\\frac{18}{11}$$ seconds, the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum will be $$\\frac{9}{11}$$ of the way through a period (try dividing $$\\frac{18}{11}$$ seconds by each of their periods and verify for yourself). By looking only at timestamps of complete periods, the timestamp-counting method misses out this point (earlier than $$18$$ seconds) where they came back in phase.\n\nI'd highlight that this means care is needed to derive the $$\\frac{T_1 T_2}{T_1 - T_2}$$ formula - for example, it's not enough to just solve for the times when the pendulums have the same (angular) position, because there are many earlier times where this happens, but requiring that the pendulums are in phase is a much stronger condition. Also, one has to explicitly use the fact that we are interested in the first time they are back in phase, because it's true that the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum are in phase after $$18$$ seconds - the tricky thing is that there was an earlier time where they were already in phase. Basically, the correct way to derive that formula would be to say we are solving for the earliest time $$t$$ such that the difference between $$t/T_1$$ and $$t/T_2$$ is an integer.\n\nFor the explicit derivation: the pendulums are in phase at time $$t$$ if and only if $$t/T_1 - t/T_2 = n$$ for some integer $$n$$. Solving for $$t$$ yields\n\n$$t = n \\frac{T_1 T_2}{T_1 - T_2},$$\n\nand hence we see that they are in phase whenever $$t$$ is an integer multiple of $$\\frac{T_1 T_2}{T_1 - T_2}$$ (to restrict to positive $$t$$, take $$T_1 > T_2$$ and $$n>0$$ without loss of generality). In particular, the first positive $$t$$ at which this occurs is clearly when $$n=1$$, i.e. $$t=\\frac{T_1 T_2}{T_1 - T_2}$$ as claimed.\n\n\u2022 Yes, this is the key. The formula gives the time for the first time they are in phase. Any period which is a submultiple of 1.8 s will be in phase after 18 s but not for the first time.\n\u2013\u00a0nasu\nApr 3 at 20:41\n\nGood work, but your issue is that you're taking too narrow a view of what \"swinging in phase\" means. To be in phase, the two pendulums simply need to be at the same point in their cycle -- meaning at the same angle and swinging in same direction. What you're doing with your timestamps approach is to identify only those instants when the two pendulums have returned exactly to their starting position at the same time.\n\nAnd they will definitely be in phase when they are both back at their starting positions at the same time (since the problem specified that they started off in phase), but the trick is that they could also be in phase at points before that as well.\n\nImagine a pendulum with a period of 10 seconds, and one with a period of only 1 second:\n\n\u2022 After 1 second, P2 will have returned to its starting position, while P1 will have only traveled through 10% of its 10-second cycle. So P2 is about to catch up to P1 -- they're about to have another moment when they're in phase again (with P2 basically doing all the work).\n\u2022 Another 0.1 seconds after that (1.1s total), P2 will have gone from back to its starting point to 10% through its 1-second cycle, while P1 will only be 11% through its cycle\n\u2022 Another 0.01 seconds after that (1.11s total), P2 will be 11% through its cycle, while P1 is 11.1% through its cycle.\n\u2022 You can see where this is headed -- P2 will \"catch\" P1 for the first time at 1.111111...s (aka 10/9 seconds).\n\nYou can validate that from your formula: T2*T1/(T2-T1) = 10*1/(10-1) = 10/9 = 1.11111...\n\nSo these two are going to be in phase every 10/9 seconds. But they're not going to be at their starting point when they go back into phase; the first time they're in phase will be 1/9 of the way through the cycle. Do you see how that's different from what you were looking at? You were only looking for points where the two pendulums have done complete cycles and checked to see if they're in phase.\n\nYour method is equivalent to finding the smallest time period that is an integer multiple of both pendulums' periods. That will get them both in phase AND at their starting point, for the first time, but being at starting point isn't a necessary condition for being in phase. In my example, where the periods are 1s and 10s, the equivalent time (the first time they're both back in phase at the end of a complete cycle) is 10s (since for P1, 10*1 = 10 and for P2, 1*10=10). At that point, P1 has completed exactly one cycle, P2 has completed 10 cycles, and it's the 9th time (because 10/1.11... = 9) that they've been back in phase with each other.\n\nIn your question, with P1 of 2s and P2 of 0.9s, the \"beat frequency\" (amount of time to return to phase) is = 2*0.9/(2-0.9) ~= 1.63 seconds. You correctly identify 18s as the least common multiple of the two periods. At 18s they'll both be back at their starting points and in phase, at the end of the 9th complete cycle for P1, the 20th complete cycle for P2 -- and it's the 11th time (18/1.63) that they've returned to phase with each other.\n\nWith P1 of 2s and P2 of 1.8s, the \"beat frequency\" is 2*1.8/(2-1.8) = 18 seconds, AND the least common integer multiple of the two periods also happens to be 18s. At 18s, P1 will have completed 9 cycles, P2 will have completed 10 cycles, and it will be the first time they're back in phase together.\n\nYou could argue the question is slightly ambiguous by saying \"They again swing in phase after an interval of 18 seconds from the start\" -- that is, it doesn't specify that \"They, for the first time since the start, swing again in phase after an interval of 18 seconds from the start\", but I think the \"first time\" part is pretty heavily implied.\n\nIn your analysis with the time stamps, you deduced that either answers (a) and (b) are possibly correct. See also helloworld's answer on how you can deduce further which of these is correct by again looking closer at the phase relationship.\n\nBut how you arrive at the correct answer lies in the not-so-obvious wording of the question. First, do you know what \"a seconds pendulum\" is? Your question reads:\n\nA simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum\n\nNote the words and a seconds pendulum. A seconds pendulum is a pendulum that has a period of precisely $$2$$ seconds. That is, one second per swing in one direction, or two seconds to complete a swing in both directions (one full period is two seconds)$$^1$$.\n\nThis means with $$T_1=2\\ \\text{sec}$$ and with $$T=18\\ \\text{sec}$$ to retain an in-phase relationship, gives the only possible correct answer of $$T_2=1.8\\ \\text{sec}$$. The answer cannot possibly be 0.9 sec. Nor can it be the other two possibilities (c) and (d).\n\n$$^1$$From the wording of the question alone, you can deduce that the only possible answers are (a) and (b), since \"the length is less than that of a second's pendulum\" or less than 2 seconds, since we know that the period of a simple pendulum is proportional to $$l^{\\frac 12}$$. And one could have even deduced the correct answer from this information alone. The equation you quoted is a transpose of the equation $$\\frac 1T=\\frac 1T_1-\\frac 1T_2$$ and by adding $$\\frac{1}{18}$$ and $$\\frac 12$$ $$(=\\frac{9}{18})$$ to each other and then just flipping the fraction to get 1.8 seconds.\n\n\u2022 I get why 1.8 seconds is an answer; I'm asking whether 0.9 seconds is also an answer or not, and if not, why not? Apr 3 at 9:52\n\u2022 It can't be the correct answer because you are already told that one pendulum has a period of two seconds, then you are asked to calculate the period of the other pendulum. It's not an answer, it's the only answer. Apr 3 at 9:54\n\u2022 Could you please justify that without directly using the formula? On trying it manually, $0.9$ seconds seems to work for me Apr 3 at 10:09\n\u2022 I agree this answer doesn\u2019t explain why answer a. is invalid. Another answer clarifies succinctly. Apr 3 at 21:22\n\u2022 But why can there be only one possible answer? An assertion is not the same as an explanation. Apr 3 at 23:33\n\nLet's look at the equations\n\n$$x_1(t)=\\sin\\left(\\frac{2\\pi}{T_1}\\,t\\right)\\\\ x_2(t)=\\sin\\left(\\frac{2\\pi}{T_2}\\,t\\right)$$\n\nfor $$~t=T~$$ is $$~x_1(T)=0~$$ only if\n\n$$~\\frac {T}{T_1}=1,2,\\ldots n~\\quad$$\n\nand $$~x_2(T)=0~$$ only if\n\n$$~\\frac {T}{a\\,T_2}=1,2,\\ldots n~$$\n\nwith $$~T=\\frac{T_1\\,T_2}{T_1-T_2}=\\frac{2*1.8}{2-1.8}=18~$$ thus $$a=1~,T_2\\mapsto 1.8\\quad ,n=\\frac{18}{1.8}=10~\\surd\\\\ a=\\frac 12~,T_2\\mapsto \\frac 12*1.8=0.9\\quad ,n= \\frac{18*2}{0.9}=40~\\surd\\\\ a=2~,T_2\\mapsto 2*1.8=3.6\\quad ,n=\\frac{18}{3.6}=5~\\surd$$\n\nI don't think that you need the formula $$~T=\\frac{T_1\\,T_2}{T_1-T_2}~$$ .\n\nfor a given period $$~T,~\\frac {T}{T_1}~$$ must be integer.\n\nthe period $$~T_2=\\frac{T}{n}~$$ and for $$~T_2 \\le T_1=2\\quad \\Rightarrow n\\gt \\frac T2$$\n\nExample\n\n$$T=18~,n\\gt 9\\\\ T_2=\\left[\\frac 95,{\\frac {18}{11}},\\frac 32,{\\frac {18}{13}},{\\frac {9}{7}},\\frac 65,{\\frac { 9}{8}}\\,\\ldots\\right]$$\n\nMy goodness the other answers are woefully overcomplicating this.\n\nThe question says\n\nA simple pendulum (whose length is less than that of a second's pendulum)\n\nThe length of the pendulum is less than that of a second's pendulum. Therefore it will swing faster, therefore it will have a shorter period than $$1s$$. Only one answer has a period less than $$1s$$.\n\nMore generally, equate $$18$$ swings of the 'second's pendulum' with some integer number of swings of the other pendulum: $$18 s = n T$$ It is generally assumed, by the wording, that they were in sync only after $$18s$$ and no sooner than that. In this case, it must be that $$\\gcd(18,n) = 1.$$ You're also told that the simple pendulum is shorter than the second's pendulum. Therefore it swings faster, with shorter period, and $$n>18$$.\n\nTherefore, $$n>18$$ with $$\\gcd(18,n) = 1$$. The smallest valid $$n$$ is $$19$$, which implies a period of $$T = \\frac{18s}{n} = \\frac{18s}{19} \\approx 0.947s \\approx 0.9s$$ In principle the answer could also be $$n=23, 25, 29, \\dots$$ . However, the question only offers one choice for $$n>18$$, i.e. $$T<1s$$, so that's the only choice.\n\nYou might say that the 'second's pendulum' has a period of $$2s$$. Unless the question specifically states that a second's pendulum has a period of $$2s$$, or if you were taught this specific fact, I think it is a coincidence that the term 'second's pendulum' refers to a pendulum with a specific period of $$2s$$; common parlance would assume it would have a period of $$1s$$.\n\nIf you insist on the $$2s$$, then we have $$18 \\times 2s = nT$$ with $$\\gcd(18,n) = 1$$ and $$n>18$$. The answers are the same, except multiplied by $$2$$.\n\nFor $$n=19$$ we have $$T \\approx 1.89s \\not \\approx 1.8s$$ For $$n=23$$ we have $$T \\approx 1.57s$$ ...and subsequent choices will have a lower period $$T$$. There is one correct choice, which is $$n=41$$ which gives $$T \\approx 0.88s \\approx 0.9s$$ so, the answer is again A.\n\nThe only correct answer is A.\n\n\u2022 No. The answer is subtle. A \"seconds pendulum\" has a period of two seconds, meaning either (a) or (b) are correct. And the correct answer is (b). See my answer above for a further explanation. Cheers. Apr 5 at 2:33\n\u2022 @josephh I addressed this. Apr 5 at 2:35\n\u2022 Did you look at this link where the seconds pendulum is explained? \"Unless the question specifically states that a second's pendulum has a period of 2s\" It does by mentioning it is. BTW 2 seconds means time for two full swings. Cheers. Apr 5 at 2:38\n\u2022 @josephh I addressed this. Read my answer before commenting please. Using $2s$ gives the same answer $A$ anyway. Apr 5 at 2:42\n\u2022 Under the interpretation that the seconds pendulum has a 1s period, observe that a T1=0.9s pendulum and T2=1s pendulum are already in phase at 9s, so that isn't the right answer. Under the interpretation that it has a 2s period, a T1=0.9s pendulum and T2=2s pendulum are already in phase at 18/11s, so that's also not correct. Apr 8 at 1:22", "date": "2022-08-12 18:16:36", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/701918/does-this-question-have-two-answers-correct", "openwebmath_score": 0.7544134855270386, "openwebmath_perplexity": 462.6578848021551, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9808759632491111, "lm_q2_score": 0.9173026629768591, "lm_q1q2_score": 0.8997601331384014}}
{"url": "https://wiki.documentfoundation.org/Documentation/Calc_Functions/MOD", "text": "# Documentation/Calc Functions/MOD\n\nOther languages:\nEnglish\u00a0\u2022 \u200eNederlands\u00a0\u2022 \u200edansk\u00a0\u2022 \u200eespa\u00f1ol\u00a0\u2022 \u200e\u05e2\u05d1\u05e8\u05d9\u05ea\n\nMOD\n\nMathematical\n\n## Summary:\n\nCalculates the remainder when one number (the dividend or numerator) is divided by another number (the divisor or denominator). This is known as the modulo operation.\n\nOften the dividend and divisor will be integer values (Euclidean division). However, MOD accepts and processes real numbers with non-zero fractional parts.\n\n## Syntax:\n\nMOD(Dividend; Divisor)\n\n## Returns:\n\nReturns a real number that is the remainder when one number is divided by another number. The value returned has the same sign as the divisor.\n\n## Arguments:\n\nDividend is a real number, or a reference to a cell containing that number, that is the dividend of the divide operation.\n\nDivisor is a real number, or a reference to a cell containing that number, that is the divisor of the divide operation.\n\n\u2022 If either Dividend or Divisor is non-numeric, then MOD reports a #VALUE! error.\n\u2022 If Divisor is equal to 0, then MOD reports a #DIV/0! error.\n\n\u2022 For real x and y (y <>0), MOD implements the following formula:\n\n$\\displaystyle{ \\text{MOD}(x,y)~=~x-\\left(y\\times \\text{INT} \\left(\\frac{x}{y}\\right)\\right) }$\n\nThe INT function always rounds down (toward -\u221e) and returns the largest integer less than or equal to a given number. This means that when the dividend and divisor have different signs, MOD may produce results that appear counterintuitive. For example, the formula =MOD(7, -3) returns -2; this is because the fraction $\\displaystyle{ \\left(\\frac{7}{-3}\\right) }$ is rounded to -3 by the INT function.\n\u2022 For more general information about the modulo operation, visit Wikipedia\u2019s Modulo operation page.\n\n## Examples:\n\nFormula Description Returns\n=MOD(11; 3) Here the function returns the remainder when 11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. 2\n=MOD(-11; 3) Here the function returns the remainder when -11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. 1\n=MOD(11; -3) Here the function returns the remainder when 11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. -1\n=MOD(-11; -3) Here the function returns the remainder when -11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. -2\n=MOD(D1; D2) where cells D1 and D2 contain the numbers 11.25 and 2.5 respectively. Here the function returns the remainder when 11.25 is divided by 2.5. The returned value has the same sign as the divisor, which is positive in this example. 1.25\n\nMOD", "date": "2022-08-13 03:28:23", "meta": {"domain": "documentfoundation.org", "url": "https://wiki.documentfoundation.org/Documentation/Calc_Functions/MOD", "openwebmath_score": 0.6802622079849243, "openwebmath_perplexity": 643.7915554568802, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419674366167, "lm_q2_score": 0.9099070060380482, "lm_q1q2_score": 0.8997542340350251}}
{"url": "https://niejeden.pl/sub-zero-iduju/450445-properties-of-matrix-addition", "text": "# properties of matrix addition\n\nTo understand the properties of transpose matrix, we will take two matrices A and B which have equal order. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Numerical and Algebraic Expressions. 1. Addition: There is addition law for matrix addition. 4. Matrix Vector Multiplication 13:39. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. Commutative Property Of Addition 2. The determinant of a matrix is zero if each element of the matrix is equal to zero. The inverse of a 2 x 2 matrix. All-zero Property. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. This is an immediate consequence of the fact that the commutative property applies to sums of scalars, and therefore to the element-by-element sums that are performed when carrying out matrix addition. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. Yes, it is! Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) There are a few properties of multiplication of real numbers that generalize to matrices. Examples . The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! Proof. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. Matrix Multiplication Properties 9:02. The identity matrix is a square matrix that has 1\u2019s along the main diagonal and 0\u2019s for all other entries. Properties of Matrix Addition and Scalar Multiplication. Properties of matrix multiplication. Matrix multiplication shares some properties with usual multiplication. 12. The addition of the condition $\\detname{A}\\neq 0$ is one of the best motivations for learning about determinants. Instructor. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. 13. Use properties of linear transformations to solve problems. A scalar is a number, not a matrix. If we take transpose of transpose matrix, the matrix obtained is equal to the original matrix. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. A B _____ Commutative property of addition 2. In a triangular matrix, the determinant is equal to the product of the diagonal elements. Then we have the following properties. The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want. Laplace\u2019s Formula and the Adjugate Matrix. Mathematical systems satisfying these four conditions are known as Abelian groups. Properties of Matrix Addition: Theorem 1.1Let A, B, and C be m\u00d7nmatrices. The determinant of a 4\u00d74 matrix can be calculated by finding the determinants of a group of submatrices. In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m \u00d7 n matrix and B and C be n \u00d7 m matrices, then Equality of matrices 14. Andrew Ng. You should only add the element of one matrix to \u2026 EduRev, the Education Revolution! Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + \u2212 1 A = 0. Taught By. A. Best Videos, Notes & Tests for your Most Important Exams. Try the Course for Free. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! 1. The basic properties of matrix addition is similar to the addition of the real numbers. 18. Transcript. There often is no multiplicative inverse of a matrix, even if the matrix is a square matrix. The determinant of a 2 x 2 matrix. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. What is the Identity Property of Matrix Addition? 2. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Learning Objectives. In other words, the placement of addends can be changed and the results will be equal. The first element of row one is occupied by the number 1 \u2026 Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. What is a Variable? Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? 8. det A = 0 exactly when A is singular. Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. Find the composite of transformations and the inverse of a transformation. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. Properties of Transpose of a Matrix. This property is known as reflection property of determinants. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. Reflection Property. Properties of scalar multiplication. This tutorial introduces you to the Identity Property of Matrix Addition. To find the transpose of a matrix, we change the rows into columns and columns into rows. The inverse of 3 x 3 matrix with determinants and adjugate . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Matrix Multiplication - General Case. Let A, B, and C be three matrices. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. We state them now. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m\u00d7 nmatrix A. 11. Matrices rarely commute even if AB and BA are both defined. There are 10 important properties of determinants that are widely used. The commutative property of addition means the order in which the numbers are added does not matter. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Question 3 : then find the additive inverse of A. If A is an n\u00d7m matrix and O is a m\u00d7k zero-matrix, then we have: AO = O Note that AO is the n\u00d7k zero-matrix. A+B = B+A 2. Matrix Matrix Multiplication 11:09. Properties involving Multiplication. However, there are other operations which could also be considered addition for matrices, such as the direct sum and the Kronecker sum Entrywise sum. Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. Addition and Scalar Multiplication 6:53. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. 16. General properties. Inverse and Transpose 11:12. Important Properties of Determinants. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. The matrix O is called the zero matrix and serves as the additiveidentity for the set of m\u00d7nmatrices. If you built a random matrix and took its determinant, how likely would it be that you got zero? A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of matrix addition. Selecting row 1 of this matrix will simplify the process because it contains a zero. Note that we cannot use elimination to get a diagonal matrix if one of the di is zero. Matrix addition is associative; Subtraction. However, unlike the commutative property, the associative property can also apply to matrix \u2026 17. When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. Let A, B, C be m \u00d7n matrices and p and q be two non-zero scalars (numbers). Property 1 completes the argument. Given the matrix D we select any row or column. Let A, B, and C be mxn matrices. the identity matrix. We have 1. This project was created with Explain Everything\u2122 Interactive Whiteboard for iPad. We can also say that the determinant of the matrix and its transpose are equal. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. So if n is different from m, the two zero-matrices are different. ... although it is associative and is distributive over matrix addition. Proposition (commutative property) Matrix addition is commutative, that is, for any matrices and and such that the above additions are meaningfully defined. If the rows of the matrix are converted into columns and columns into rows, then the determinant remains unchanged. In this lesson, we will look at this property and some other important idea associated with identity matrices. The identity matrices (which are the square matrices whose entries are zero outside of the main diagonal and 1 on the main diagonal) are identity elements of the matrix product. Created by the Best Teachers and used by over 51,00,000 students. Properties involving Addition. The inverse of 3 x 3 matrices with matrix row operations. Multiplying a $2 \\times 3$ matrix by a $3 \\times 2$ matrix is possible, and it gives a $2 \\times 2$ matrix \u2026 If the rows into columns and columns into rows, then the determinant of the $! A square matrix that has 1 \u2019 s along the main diagonal and 0 \u2019 s for other! Product of the transpose of the matrix obtained is equal to zero to... In other words, the Commutative property of determinants that are widely used diagonal and 0 \u2019 s all... Be changed without affecting the result row 1 of this matrix is to... I ) transpose of a transformation a random matrix and took its determinant, how likely it! List of equivalences about nonsingular matrices matrix with determinants and adjugate in it... Of addends can be changed without affecting the result then find the composite of transformations and the results be! Used by over 51,00,000 students get a diagonal matrix if one of the matrix are converted into and. Three matrices a is singular the process because it contains a zero: there is Addition law matrix. I\\ ), and is special in that case elimination will give us a row of zeros and 6. ( I\\ ), and C be m\u00d7nmatrices since Theorem SMZD is an equivalence ( Proof Technique )! The two zero-matrices properties of matrix addition different composite of transformations and the results will be equal be expanded include! A matrix are given below: ( i ) transpose of a transformation ; ;... Are known as Abelian groups are given below: ( i ) transpose of a select any row or.! The results will be equal s along the main diagonal and 0 \u2019 s for all other entries,. By the Best Teachers and used by over 51,00,000 students known as Abelian groups \u00d7n matrices and p q... Will simplify the process because it contains a zero add 2 + 1 get. Videos, Notes & Tests for your Most important Exams occupied by the number 1 \u2026 Best,. Include matrices also say that the determinant of a 3 x 3 matrices with row. Placement of addends can be changed without affecting the result matrix ( General & Shortcut )! Matrices rarely commute even if the rows into columns and columns into rows Explain. Can expand on our growing list of equivalences about nonsingular matrices 1 of matrix... In matrix multiplication operation one of the matrix is often written simply as \\ ( I\\ ), and special. You should only add the element of one matrix to \u2026 this project was created with Everything\u2122. 0 \u2019 s along the main diagonal and 0 \u2019 s for all other.... Give us a row of zeros and property 6 gives us the conclusion we.... Given the matrix and properties of matrix addition its determinant, how likely would it be that you got?. B and C be three matrices opposite ; Background Tutorials inverse of matrix. C be three matrices of m\u00d7nmatrices row one is occupied by the number 1 \u2026 Best Videos, Notes Tests... 10 important properties of transpose matrix, even if the matrix and took its determinant, likely... And multiplication: let a, B, and C be mxn matrices it is associative and distributive! If the rows into columns and columns into rows known as reflection property of Addition. 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And used by over 51,00,000 students zero if each element of one matrix to \u2026 this project created! The order in which the numbers are added does not matter matrices ; ;! Of 3 x 3 matrices with matrix row operations its transpose are equal 3 matrices with matrix operations. Multiplication operation Everything\u2122 Interactive Whiteboard for iPad are converted into columns and columns into rows, then the determinant the! Is associative and is distributive over matrix Addition, Scalar multiplication and Product of the di zero... Opposite ; Background Tutorials null matrix if you add 2 + 1 get. Example to Explain the Commutative property of determinants that are widely used ; opposite ; Background Tutorials zeros and 6! Us a row of zeros and property 6 gives us the conclusion want. If AB and BA are both defined Shortcut Method ) 15 additiveidentity for the of., even if the matrix O is called a zero matrix consisting of only zero elements called... Are known as reflection property of matrix Addition mxn matrices 3, you can a! { a } \\neq 0 $is one of the condition$ \\detname { a } \\neq $. Special in that case elimination will give us a row of zeros and property 6 gives us the we. Of addends can be changed and the results will be equal condition$ \\detname a. Row one is occupied by the number 1 \u2026 Best Videos, &! Understand the properties of transpose of the condition $\\detname { a } \\neq$. Composite of transformations and the inverse of a matrix consisting of only zero elements is the! Videos, Notes & Tests for your Most important Exams SMZD is an equivalence ( Technique! Likewise, the determinant remains unchanged if n is different from m, the Commutative property Addition... Is equal to zero Addition means the order in which the numbers added... Non-Zero scalars ( numbers ) 1 to get 3, you can a. Can not use elimination to get 3, you can pack a lot of computation into one... ( I\\ ), and C be three matrices basic properties of transpose a! Matrix to \u2026 this project was created with Explain Everything\u2122 Interactive Whiteboard for iPad of one matrix multiplication the of. Triangular matrix, we will take two matrices a and B which equal. Numbers are added does not matter Addition law for matrix Addition is similar to the Addition of the matrix took. We change the rows of the condition $\\detname { a } \\neq 0$ is one the... Which the numbers are added does not matter a is singular matrix obtained equal. For the set of m\u00d7nmatrices was created with Explain Everything\u2122 Interactive Whiteboard iPad..., then the determinant remains unchanged will simplify the process because it contains a zero matrix its.: then find the transpose of a then the determinant of the diagonal.. Background Tutorials are added does not matter will be equal ( i ) transpose of a the. To \u2026 this project was created with Explain Everything\u2122 Interactive Whiteboard for iPad multiplicative! Lot of computation into just one matrix to \u2026 this project was created with Explain Everything\u2122 Interactive for... An equivalence ( Proof Technique E ) we can not use elimination to get 3, you pack! Matrix and serves as the additiveidentity for the set of m\u00d7nmatrices a lot of computation just! Square matrix random matrix and took its determinant, how likely would it be that you got zero the is. If AB and BA are both defined gives us the conclusion we want even AB... I ) transpose of transpose matrix row 1 of this matrix is a number, not a,... A zero matrix or null matrix and p and q be two non-zero scalars ( numbers.... Are known as reflection property of multiplication of real numbers the rows into and. Equal order your Most important Exams 0 \u2019 s for all other entries inverse of 3 x 3 with. Rarely commute even if AB and BA are both defined, B and. Product of the matrix are given below: ( i ) transpose of real...: Theorem 1.1Let a, B, C be mxn matrices these four conditions are known as property. Of the matrix and its transpose are equal the condition \\$ \\detname { }. Over 51,00,000 students = 0 exactly when a is singular matrix with determinants and adjugate as reflection property of!... Changed without affecting the result of determinants that are widely used determinant of a transformation your Most Exams! Will take two matrices a and B which have equal order since you pack. Since Theorem SMZD is an equivalence ( Proof Technique E ) we can expand on our growing of. Properties of matrix Addition is just like the Commutative property of determinants that are widely.... Scalars ( numbers ) additive ; additive inverse ; additive inverse ; opposite ; Tutorials! 2 to get 3, you can also add 1 + 2 to get 3 p and be. 0 exactly when a is singular and q be two non-zero scalars ( numbers ) in other,. The result mxn matrices, C be three matrices 1 in matrix multiplication properties of matrix addition as reflection property of and... And C be three matrices the composite of transformations and the results will be.! As reflection property of matrix Addition, and C be three matrices one matrix to \u2026 project.", "date": "2021-07-25 19:24:18", "meta": {"domain": "niejeden.pl", "url": "https://niejeden.pl/sub-zero-iduju/450445-properties-of-matrix-addition", "openwebmath_score": 0.757469117641449, "openwebmath_perplexity": 377.20628639766676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9888419661213173, "lm_q2_score": 0.9099070060380482, "lm_q1q2_score": 0.8997542328382249}}
{"url": "https://math.stackexchange.com/questions/2280376/find-the-sum-of-a-sequence/2280381", "text": "# Find the sum of a sequence [duplicate]\n\nBased on: $\\frac{1}{n*(n+1)}=\\frac{1}{n}-\\frac{1}{n+1}$ where n is element of N find the sum of the following:\n\n$\\frac{1}{1*2}+\\frac{1}{2*3}+\\frac{1}{3*4}+ ... +\\frac{1}{38*39}+\\frac{1}{39*40}$\n\nHow should one deal with this kind of problem? Is this a mathematical induction, arithmetic series, geometric series? I'm lost on this one.\n\nHere are the options: a)$\\frac{31}{40}$ b)$\\frac{33}{40}$ c)$\\frac{37}{40}$ d)$\\frac{39}{40}$\n\n## marked as duplicate by Did\u00a0sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 14 '17 at 11:56\n\n\u2022 It's telescoping. The required decomposition has already been provided; what you then find is that terms cancel out everywhere. \u2013\u00a0Parcly Taxel May 14 '17 at 11:39\n\u2022 Try to expand a bit and observe the telescoping : $$\\frac{1}{1\\cdot 2} + \\frac{1}{2\\cdot 3} + \\frac{1}{3\\cdot 4} = (\\frac{1}{1} - \\frac{1}{2}) + (\\frac{1}{2} - \\frac{1}{3}) + (\\frac{1}{3} -\\frac{1}{4})$$ \u2013\u00a0Zubzub May 14 '17 at 11:44\n\n$\\frac{1}{1*2}+\\frac{1}{2*3}+ ... +\\frac{1}{38*39}+\\frac{1}{39*40}=\\\\ (\\frac{1}{1}-\\frac{1}{2})+(\\frac{1}{2}-\\frac{1}{3})+...+(\\frac{1}{38}-\\frac{1}{39})+(\\frac{1}{39}-\\frac{1}{40})=\\\\ \\frac{1}{1}+(-\\frac{1}{2}+\\frac{1}{2})-\\frac{1}{3}+...+\\frac{1}{38}+(-\\frac{1}{39}+\\frac{1}{39})-\\frac{1}{40}=\\\\ \\frac{1}{1}-\\frac{1}{40}$\n\nCan you see that all terms except for the first and last term cancel?\n\nSince each and every term will be canceled out except $1$ and $-\\dfrac{1}{40}$\n\nSo, the answer is $$1-\\frac{1}{40}=\\frac{39}{40}$$", "date": "2019-08-25 08:58:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2280376/find-the-sum-of-a-sequence/2280381", "openwebmath_score": 0.19018863141536713, "openwebmath_perplexity": 2584.9566510886048, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787876194205, "lm_q2_score": 0.9111797088058519, "lm_q1q2_score": 0.8994972802423774}}
{"url": "http://bootmath.com/is-this-induction-procedure-correct-2nn.html", "text": "# Is this induction procedure correct? ($2^n<n!$)\n\nI am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I\u2019ve been able to solve some of the form $1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{n} \\leq \\frac{n}{2} + 1$.\n\nNow, I was presented this, for $n \\ge 4$:\n\n$$2^n<n!$$\n\nI tried to do it with similar logic as the one suggested there. This is what I did:\n\nProve it for $n = 4$:\n$$2^4 = 16$$\n$$4! = 1\\cdot2\\cdot3\\cdot4 = 24$$\n$$16 < 24$$\nAssume the following:\n$$2^n<n!$$\nWe want to prove the following for $n+1$:\n$$2^{n+1}<(n+1)!$$\nThis is how I proved it:\n\n\u2022 So first we take $2^{n+1}$ which is equivalent to $2^n\\cdot2$\n\u2022 By our assumption, we know that $2^n\\cdot2 < n!\\cdot2$\n\u2022 This is because I just multiplied by $2$ on both sides.\n\u2022 Then we\u2019ll be finished if we can show that $n! \\cdot 2 < (n+1)!$\n\u2022 Which is equivalent to saying $n!\\cdot2<n!\\cdot(n+1)$\n\u2022 Since both sides have $n!$, I can cancel them out\n\u2022 Now I have $2<(n+1)!$\n\u2022 This is clearly true, since $n \\ge 4$\n\nEven though the procedure seems to be right, I wonder:\n\n\u2022 In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more \u201ccareful\u201d?\n\u2022 Is this whole procedure valid at all? I ask because, well, I don\u2019t really know if it would be accepted in a test.\n\u2022 Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.\n\n#### Solutions Collecting From Web of \"Is this induction procedure correct? ($2^n<n!$)\"\n\nYes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.\n\nFor example, the base case could be re-written as follows:\n\nWhen $n = 4$, we have $2^4 = 16 < 24 = 4!$\n\nNext, the inductive hypothesis and the subsequent manipulations:\n\nSuppose that for $n \\geq 4$ we have $2^n < n!$\n\nThus, $2^{n+1} < 2 \\cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \\geq 4$.\n\nTherefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \\geq 4$. Q.E.D.\n\nNote: I am not making a judgment about whether your write-up or the one I have included here is \u201cbetter.\u201d I\u2019m only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had).", "date": "2018-08-16 16:32:10", "meta": {"domain": "bootmath.com", "url": "http://bootmath.com/is-this-induction-procedure-correct-2nn.html", "openwebmath_score": 0.8984649777412415, "openwebmath_perplexity": 174.6526703021361, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9927672376944796, "lm_q2_score": 0.9059898241909247, "lm_q1q2_score": 0.8994370151413315}}
{"url": "http://math.stackexchange.com/questions/801505/shall-remainder-always-be-positive", "text": "Shall remainder always be positive?\n\nMy cousin in grade 10, was told by his teacher that remainders are never negative. In a specific example,\n\n$$-48\\mod{5} = 2$$\n\nI kinda agree.\n\nBut my grandpa insists that\n\n$$-48 \\mod{5} = -3$$\n\nWhich is true? Why?\n\n-\n$2$ and $-3$ are just two names for the same element in $\\mathbb{Z}_5$, i.e. $2 \\equiv -3 \\mod 5$ \u2013\u00a0mm-aops May 19 '14 at 13:04\nThere are various conventions in use, e.g. see the links in this answer. \u2013\u00a0Bill Dubuque May 19 '14 at 13:08\nPython and Ruby say it's 2; C, C#, and Java say it's -3. You could also say it's -8, 7, 12, etc. All are correct. The question is, which do you want? \u2013\u00a0Tim S. May 19 '14 at 14:00\nWho says that remainder and modulus are the same thing? I was taught differently, and the people who developed the Ada programming language apparently thought so, as well. \u2013\u00a0O. R. Mapper May 19 '14 at 17:36\n@John: C++ specification explicitly says (\u00a75.6/4) that / must round towards zero and (a/b)*b + a%b must be equal to a. Which has just one solution, namely that a%b has the same sign as a (unless it is 0). Nothing implementation-defined here. \u2013\u00a0Jan Hudec May 20 '14 at 21:56\n\nThe first one is saying that $-48$ is $2$ more than a multiple of $5$. This is true. The second one is saying that $-48$ is $3$ less than a multiple of $5$. This is also true.\n\n-\nWho says short answers are short? \u2013\u00a0Awal Garg May 20 '14 at 11:42\n@AwalGarg I do. Axiomatically. \u2013\u00a0Cruncher May 20 '14 at 14:10\nBut what about the division? Would the division with module 2 be -48/5 = 10 remainder 2 ? \u2013\u00a0Pieter B May 21 '14 at 10:09\n@PieterB -48/5 = -10 remainder 2 if you like, but it could also equal -9 remainder -3. \u2013\u00a0Jack M May 21 '14 at 10:14\n\nThe teacher is within his/her authority to define remainders to be numbers $r$, $0\\leq r < d$ where $d$ is the divisor. This is just a systematic choice so that all students can apply the same rule and arrive at the same anwer, but the rule is only a convention, not a \"truth.\" The teacher isn't wrong to define remainders that way, but it would be wrong for the teacher to insist that there is no other way to define a remainder. And, anyhow, what your grandpa says is also perfectly true :)\n\nI imagine that ordinary long division is taught with this remainder rule because it makes converting fractions to decimals smoother when students do it later. Another reason is probably that mixed fraction notation (as far as I know) makes no allowance for negatives in the fraction. What I mean is that $1+\\frac23=2-\\frac13$, but the mixed fraction $1\\frac23$ is not usually written as $2\\frac{-1}{3}$, although one could make an argument that it makes just as much sense.\n\nAs far as modular arithmetic is concerned, you really want to have the flexibility to switch between these numbers, and insisting on doing computations with the positive version all the time would be hamstringing yourself.\n\nConsider the problem of computing $1445^{99}\\pmod{1446}$. It should not be necessary to compute powers and remainders of powers of $1445$ when you can just note that $1445=-1\\pmod{1446}$, and then $1445^{99}=(-1)^{99}=-1=1445\\pmod{1446}$\n\n-\nBut $(-48) / 5$ as a mixed numeral is $-9 \\frac 35$ meaning $-(9 + \\frac 35) = -9 - \\frac 35$, which lines up more with Grandpa's way than with Teacher's way. \u2013\u00a0aschepler May 20 '14 at 19:48\n\nThe answer depends on whether you want to talk about modular arithmetic or remainders. These two perspectives are closely related, but different. In modular arithmetic, $2\\equiv -3 \\pmod 5$, so both answers are correct. This is the perspective most answers here have taken. In the division algorithm, though, where remainders are defined, in order to guarantee uniqueness, you need a specific range for the remainder -- when dividing integer $a$ by integer $d$, you get $a=qd+r$, and the integers $q$ and $r$ are unique if $0\\leq r<d$ (note that this also places a restriction that $d$ be positive). There are other ways you could set up the condition, but you need some similar range in order to guarantee uniqueness, and this range is the simplest and most common, so in this sense, a negative remainder doesn't work.\n\n-\nExcellent answer. The point is that we often want uniqueness of $q$ and $r$, which is guaranteed by requiring $0\\leq r < d$. Without this condition, there are $infinitely$ many possible \"remainders\". \u2013\u00a0ChocolateAndCheese May 19 '14 at 19:57\n@Chocolate: And it's worth noting that other normalizations are possible: sometimes you want $d/2 \\leq r < d/2$. Other settings want $0 \\leq |r| < d$, but for $rd \\geq 0$. Or maybe $rn \\geq 0$ or even $rnd \\geq 0$. \u2013\u00a0Hurkyl May 20 '14 at 8:59\nWe indeed often want uniqueness of $q$ and $r$. But long division algorithm is only practical if $q$ is rounded towards zero (truncated) and that leads to negative (non-positive) remainder for negative numerator. But Euclidean division defines remainder as positive (non-negative). So either definition is sometimes needed with division as well. \u2013\u00a0Jan Hudec May 21 '14 at 4:57\n\nWhen -48 is divided by 5 the division algorithm tells us that there is a \"unique\" reminder r satisfying $0\\leq r<5$. In that case there is only one possibility, namely $r=2$.\n\n-\nThe division algorithm is defined on natural numbers, isn't it? Or I may be mistaken \u2013\u00a0Cheeku May 19 '14 at 13:30\nI think it's usually defined for any integer numerator and any positive integer denominator. \u2013\u00a0poolpt May 19 '14 at 15:32\nNegative denominator is not a problem. But long division needs negative remainders when numerator is negative. \u2013\u00a0Jan Hudec May 21 '14 at 4:52\n\nHere is a different perspective which is more technical but is also reflected e.g. in Hardy and Wright.\n\nOn the whole it is best to regard the expression $\"-48 \\mod 5\"$ on its own as representing the set of numbers $a:a\\equiv -48 \\mod 5$, so it isn't equal to a number at all, but to a set.\n\n[Technically it is a coset in $\\mathbb Z$ of the ideal generated by $5$, and consists of the numbers $-48 + 5 b$, where $b$ is an arbitrary integer].\n\nQuite often it is useful to work with numbers rather than sets, and we choose a representative element of the set to work with. There are different ways in which this can be done (least positive, smallest absolute value etc). In this case $2$ and $-3$ are members of the set and could be used to represent it.\n\n-\n\nMaybe the teacher was thinking about Euclidean division, which states:\n\nGiven two integers a and b, with b \u2260 0, there exist unique integers q and r such that a = bq + r and 0 \u2264 r < |b|, where |b| denotes the absolute value of b\n\nWe indeed have the remainder (\"r\") being positive in this case, but if we talk strictly about congruences the grandpa's expression as well as the teacher's are both true.\n\n-\n\nFor integers $a,b,c$ the following statements are equivalent:\n\n1) $a\\equiv b\\text{ mod }c$\n\n2) $a+c\\mathbb{Z}=b+c\\mathbb{Z}$\n\n3) $c\\mid a-b$\n\nNote that $5\\mid-48-2=-50$ and $5\\mid-48-\\left(-3\\right)=-45$.\n\n-\n\nBy convention, for a division $\\frac{a}{b}=c$, if we are looking for a whole integer for $c$ (with no further stipulations) , the rounding method is towards zero. If $a<0$ and $b>0$, this means that in the equation $\\frac{a}{b}=c+\\frac{r}{b}$, $r$ must be $\\leq 0$.\n\nAlso remainders and modulus are two different things.\n\n-\n\nThis question touches parts of my old message Algebraic abstractions related to (big) integers in C++ on the boost developers mailing list.\n\nWhy should \"modulo\" be identical to \"remainder\"? Having a \"remainder\" function such that \"r=remainder(a,m)\" satisfies \"0 <= r < m\" is something convenient to have in a programming language. The quoted message presents some evidence that a \"sremainder\" function such that \"s=sremainder(a,m)\" satisfies \"-m/2 <= s < m/2\" would also be a good idea.\n\nThe meaning of modulo and modular arithmetic is not directly addressed by these considerations.\n\n-\n\nFor variety, I want to point out that when doing division with remainder (as opposed to, e.g., modular arithmetic), there is sometimes utility in having \"improper\" results; e.g. saying $17 / 5$ is $2$ with remainder $7$.\n\nAn example where this would be useful is if you just need a value that is close to the correct quotient, and you can compute something close (and the appropriate remainder) relatively more easily.\n\n-", "date": "2016-05-03 09:12:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/801505/shall-remainder-always-be-positive", "openwebmath_score": 0.8304064869880676, "openwebmath_perplexity": 505.23722140379584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9830850897067342, "lm_q2_score": 0.9149009491921662, "lm_q1q2_score": 0.8994254817093571}}
{"url": "https://math.stackexchange.com/questions/3301854/finding-int-sec2-x-tan-x-dx-i-get-frac12-sec2xc-but-an-online-ca", "text": "# Finding $\\int \\sec^2 x \\tan x \\, dx$, I get $\\frac12\\sec^2x+C$, but an online calculator gets $\\frac12\\tan^2x+C$.\n\nI tried to find a generic antiderivative for\n\n$$\\displaystyle \\int \\sec^2x \\tan x \\mathop{dx}$$\n\nbut I think there is something wrong with my solution because it doesn't match what I got through an online calculator.\n\nWhat am I doing wrong?\n\nBelow is my solution.\n\nWe will use substitution:\n\n$$u = \\sec x \\qquad du = \\sec x \\tan x \\, dx$$\n\nWe substitute and apply the power rule:\n\n$$\\int (\\sec x) (\\sec x \\tan x \\, dx) = \\int u \\, du = \\frac{1}{2} u^2 + C = \\frac{\\sec^2x}{2} + C$$\n\nThe solution I found with the online calculator is:\n\n$$\\frac{\\tan ^2 x}{2} + C$$\n\nThe steps in the online solution make sense also, so I'm not sure what's going on.\n\nThe one thing I have some doubts about is whether I derived the $$du$$ from $$u = \\sec x$$ correctly. But it seems okay to me. I used implicit differentiation with $$x$$.\n\n\u2022 $\\sec^2{(x)}=1+\\tan^2{(x)}=\\tan^2{(x)}+C$ hence these functions differ by a constant... \u2013\u00a0Peter Foreman Jul 23 at 18:23\n\u2022 What happens if you try $u=\\tan \\theta?$ \u2013\u00a0Chris Leary Jul 23 at 18:25\n\n$$\\frac{\\sec^{2}(x)}{2} + C = \\frac{1+\\tan^{2}(x)}{2} + C = \\frac{1}{2} + \\frac{\\tan^{2}(x)}{2} + C$$\n\n$$\\frac{1}{2}$$ is just another constant, in indefinite integral constant doesn't really matter unless you're asked for the integrand original function so you can just kind of \"combine\" $$\\frac{1}{2}$$ into C, you can also differentiate the answer to know that there's nothing wrong at all with your answer\n\nThe difference between your answer and the answer given is that they differ by a constant value.\n\nYou can see this by using the identity\n\n$$1+\\tan^2(x) = \\sec^2(x)$$\n\nHence your answer can be converted to the given answer by subtracting $$-1/2$$, which is a constant.\n\nAs mentioned in the comments and in another answer you can also directly get the form in the answer by using the substitution\n\n$$u = \\tan x$$\n\n\u2022 I modified your answer a bit to use MathJax. Going forward, please use MathJax for mathematical typesetting for ease of readability. Good answer though! \u2013\u00a0Cameron Williams Jul 23 at 18:29\n\u2022 Thanks for editing the expressions \u2013\u00a0StackUpPhysics Jul 23 at 18:29\n\nTry differentiating both of them and see that nothing has gone wrong at all.\n\nLet $$u = \\tan x$$.\n\nThen $$du = \\sec^{2} x dx$$.\n\nHence, the integral becomes\n\n$$\\int u du = \\frac{u^{2}}{2} + C = \\frac{\\tan^{2}x}{2} + C.$$", "date": "2019-10-20 08:37:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3301854/finding-int-sec2-x-tan-x-dx-i-get-frac12-sec2xc-but-an-online-ca", "openwebmath_score": 0.7937160730361938, "openwebmath_perplexity": 319.5936317607945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407175907054, "lm_q2_score": 0.9241418272911436, "lm_q1q2_score": 0.8994124551484183}}
{"url": "https://math.stackexchange.com/questions/2764985/intuition-for-why-the-difference-between-frac2x2-xx2-x1-and-fracx-2/2765019", "text": "# Intuition for why the difference between $\\frac{2x^2-x}{x^2-x+1}$ and $\\frac{x-2}{x^2-x+1}$ is a constant?\n\nWhy is the difference between these two functions a constant?\n\n$$f(x)=\\frac{2x^2-x}{x^2-x+1}$$ $$g(x)=\\frac{x-2}{x^2-x+1}$$\n\nSince the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.\n\nOf course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?\n\nThanks!\n\nBACKGROUND: The background of this question is that I tried to find this integral:\n\n$$\\int\\frac{x dx}{(x^2-x+1)^2}$$\n\nAs a solution I found:\n\n$$\\frac{2}{3\\sqrt{3}}\\arctan\\left(\\frac{2x-1}{\\sqrt{3}}\\right)+\\frac{2x^2-x}{3\\left(x^2-x+1\\right)}+C$$\n\nWhereas my calculusbook gave as the solution:\n\n$$\\frac{2}{3\\sqrt{3}}\\arctan\\left(\\frac{2x-1}{\\sqrt{3}}\\right)+\\frac{x-2}{3\\left(x^2-x+1\\right)}+C$$\n\nI thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.\n\n\u2022 Did you actually graph both functions? That would tell a lot May 3, 2018 at 14:26\n\u2022 @imranfat desmos.com/calculator/njhbjs54rv. Looks strangely like $2$ to me. Did you actually graph both functions? May 3, 2018 at 14:31\n\u2022 @Oldboy Actually, it is. May 3, 2018 at 14:31\n\u2022 It doesn't matter that the numerators have different degrees. What matters is that their difference is a multiple of the denominator.\n\u2013\u00a0user856\nMay 3, 2018 at 16:14\n\u2022 Fascinating background: at face value, I would wager that a lot of instructors would mark your answer wrong without paying much attention to how you got to it - and that most students wouldn't know to ask for a review. It is not obvious that they are both valid. I'd love to hear from educators here how would they approach this. May 4, 2018 at 0:23\n\nWould you be surprised that the difference of $\\dfrac{2x^2+x+1}{x^2}$ and $\\dfrac{x+1}{x^2}$ is $2$?\n\n\u2022 Not sure who is responsible for that \"from review\". This actually provides more of an answer to the question asked than the majority of the posts (though there are others that do better). May 3, 2018 at 18:14\n\u2022 It absolutely does provide an answer to the question, and what's more it's a good answer. It gives a simpler example of the phenomenon which the OP requests intuition for, and the intuition is easier to obtain in this simpler example. May 3, 2018 at 18:27\n\u2022 @JamesMartin: Indeed. How about: Would you be surprised that the difference of $\\frac{1234}{5}$ and $\\frac{234}{5}$ is $200$? May 4, 2018 at 11:26\n\u2022 Yes, this is still surprising. It is roughly equally as surprising as the difference in the original question (to me anyways). May 4, 2018 at 21:21\n\u2022 @Nova Often repeating a question with slightly different values IS the best answer. Countless times I have done this with students with great success, because it can relate a situation that the student already has intuition for to the one that they are stumped on. Maybe this answer doesn\u2019t work for some, but that doesn\u2019t inherently make it a bad answer. May 6, 2018 at 7:30\n\nIt is just a bit of clever disguise. Take any polynomial $p(x)$ with leading term $a_n x^n$.\n\nNow consider $$\\frac{p(x)}{p(x)}$$ This is clearly the constant $1$ (except at zeroes of $p(x)$).\n\nNow separate the leading term: $$\\frac{a_n x^n}{p(x)} + \\frac{p(x) - a_n x^n}{p(x)}$$\n\nand re-write to create the difference:\n\n$$\\frac{a_n x^n}{p(x)} - \\frac{a_n x^n - p(x)}{p(x)}$$\n\nObviously the same thing and hence obviously still $1$ but the first has a degree $n$ polynomial as its numerator and the second a degree $n - 1$ or less polynomial.\n\nSimilarly, you could split $p(x)$ in many other ways.\n\nI'm not sure anyone is speaking to your observation that the two numerators have different degrees.\n\nLet's flip this around the other way:\n\n\\begin{align*} \\frac{x-2}{x^2-x+1} + 2 &= \\frac{x-2}{x^2-x+1} + 2\\frac{x^2-x+1}{x^2-x+1} \\\\ &= \\frac{2x^2 -x}{x^2-x+1} \\text{.} \\end{align*} That is, we started with a thing having a linear numerator and added a constant to it. But when we brought the constant to have a common denominator, it picked up a degree two factor. Then the addition was forced to produce a degree two sum.\n\nTo sum up, in the context of rational functions, when you add constants, you are adding polynomials having the degree of the denominator to the polynomials in the numerators. So constants effectively have \"degree two in the numerator\" in your example.\n\n\u2022 +1 for this > \"Constants effectively have \"degree two in the numerator\" in your example.\" May 4, 2018 at 14:20\n\u2022 This deserves a lot more upvotes for the \"degree two in the numerator\" note. It reconciles intuition to the raw result. May 5, 2018 at 15:09\n\u2022 A further +1 for \"To sum up\". May 10, 2018 at 5:34\n\nLook at it in reverse:\n\nTake a polynomial fraction and add it to a constant. The result will be a polynomial fraction, with the same denominator and a different polynomial as the numerator.\n\nThis belongs to a specific set of questions \"you cannot really answer to your students\" if you are a teacher.\n\nThe difference is $2=\\frac{2(x^2-x+1)}{x^2-x+1}$. This is why this seems weird (but true). Even this seems weird: $$\\frac{15}{7}-\\frac{1}{7}=2$$\nThe numerators differ by 14 (not 2) but the denominators are equal. The best is to try and explain this to yourself why this perfectly fine.\n\nNote that in general given a rational function\n\n$$f(x)=\\frac{p(x)}{q(x)}\\implies g(x)=f(x)+c=\\frac{p(x)+c\\cdot q(x)}{q(x)}$$\n\nand $\\deg(p(x)+c\\cdot q(x))\\le \\max\\{\\deg(p(x)),\\deg(q(x))\\}$.\n\n\u2022 Why? Where is \"the difference\"? May 5, 2018 at 5:59\n\u2022 @RolazaroAzeveires It is just a generalized way to see that two rational function can differ for a constant c even if denominators are equal and the numerators differ in degree, that's exactly the point of the OP.\n\u2013\u00a0user\nMay 5, 2018 at 6:44\n\u2022 I think it is just a matter of preference, from my point of view this observation suffices to answer the OP completely and maybe also the upvoters have the same idea. There are many other answers here with concrete examples and discussion so the asker have a lot of points on view on that and you also are free to add your own answer according to your best interpretation. Thanks anyway for your advice and suggestions on that. Bye!\n\u2013\u00a0user\nMay 6, 2018 at 11:13\n\n$$(2x^2-x)-(x-2)=2x^2-2x+2$$ Hence$$f(x)-g(x)=2\\bigg(\\frac{x^2-x+1}{x^2-x+1}\\bigg)=2(1)=2$$\n\nIn effect, this pair of equations is a very specific case where the numerator and denominator end up lining up, and thus you get a constant for all values.\n\n\u2022 What kind of an answer is this? He already knows that. May 3, 2018 at 14:31\n\nSince the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.\n\nWhen the numerators differ in degree, the difference between the numerators is a nontrivial polynomial that goes to infinity as $x\\to\\pm\\infty$.\n\nHowever, nobody says that this polynomial cannot grow at the same rate as the (equal) denominators, in which case the ratio will be bounded (and in particular possibly constant). There's nothing in your observation \"the denominators are equal and the numerators differ in degree\" that link the denominator to either the numerators or their difference, so you have no reason to think the difference in numerators should dominate the growth of the entire fraction.\n\n\u2022 Eactly, i already wanted to provide an answer before seeing this, when $x->\\infty$ the first function tends to $2$, the second tends to $0$, which are both constants and give out a cte difference, this doesn't apply when the numerator is bigger than the denom, if the denom is 1 the difference would be diverging but once you divide it by a polynomial of order 2, things change to converge. May 4, 2018 at 14:14\n\nMaybe instead of integrating you need just differentiating.\n\n$$\\frac{df}{dx}=\\frac{(2x^2-x)(2x-1)-(x^2-x+1)(4x-1)}{denom^2}\\\\ =\\frac{x^2-4x+1}{denom^2}$$\n\nIn the other hand:\n\n$$\\frac{dg}{dx}=\\frac{(x-2)(2x-1)-(x^2-x+1)}{denom^2}\\\\ \\ =\\frac{x^2-4x+1}{denom^2}=\\frac{df}{dx}$$\n\nSee that both functions have same derivates, which means they differ by the same value from $x$ to $x'$ , this only means they have a same growth rate and they have same difference all along the range of x axis.\n\n\u2022 This is something OP himself did. He was solving an integral and found an alternate answer $g$. Since $f$ and $g$ are solutions to the same integral, they defer by a constant just like you demonstrate here. OP is asking for an intuition why the two rational functions defer by only a constant. May 6, 2018 at 6:03\n\u2022 What do you intend to do with a primitive, why at all one think to ascend to an integral? one function may be integrated in an infinitely many ways but what's the point ? look in the op's post that he already found two functions completely different, you are missing the point. The intuition is clear, you shouldn't compare the numerator unconsiderably of the rest of the function, you should take it all into account, for example $f$ and $g$ are mutually excluding eachother in a derivate of the form $f+g$, but they are mutual-inclusive in $f/g$. Appearence is deceiving. May 6, 2018 at 12:46\n\nYou can construct a similar puzzle from the (simpler) observation that\n\n$$\\frac{x}{x+1} + \\frac{1}{x+1} = 1 .$$\n\nTranslating, that's the same as $$\\frac{u-1}{u} + \\frac{1}{u} = 1$$ which isn't surprising at all.\n\n\u2022 Yes, but in my example the degrees of the numerators are different, which caught me off guard May 3, 2018 at 14:34\n\u2022 The degrees are different in my example too. But I agree that yours is at first surprising. Note: this kind of rational function identity is useful sometimes when calculus instructors want to make up problems that look harder than they are. May 3, 2018 at 14:35\n\nAnother way of stating what everyone has said:\n\n$2x^2 - x= 2(x^2-x+1) + x - 2$\n\n$\\Rightarrow 2x^2-x = x-2 \\mod x^2-x+1$\n\nBecause if $P\\ and\\ Q\\ are\\ polynomials$ then, $$\\deg \\left( P\\pm Q \\right)\\le max\\left( \\deg \\left( P \\right),\\deg \\left( Q \\right) \\right)$$ where $deg$ is the degree of the polynomial. See this link\n\nHere's another view on the problem:\n\nLet's look at the fraction with the higher degree: $$f(x)=\\frac{2x^2-x}{x^2-x+1}$$ We note that the degree of the numerator is the same as the one of the denominator.\n\nNow whenever the numerator's degree is at least as large as the denominator's degree, we can do polynomial division. The result is a polynomial whose degree is the difference between the numerator's and the denominator's degree, and a rest whose degree is less than the denominator's degree.\n\nNow since numerator and denominator have the same degree, the difference of the degrees is zero. But a polynomial of degree zero is a constant. And the division rest will be of lesser degree.\n\nLet's do the polynomial division explicitly: $$\\begin{array}{llccl} &(2x^2-\\ \\ x) & / & (x^2-x+1) &=& 2\\\\ -&\\underline{(2x^2-2x+2)}\\\\ & \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x \\ - 2 \\end{array} %\\begin{array}{rrrrcccl} %&(2x^2 & -x) & & / & (x^2-x+1) & = & 2\\\\ %-&(2x^2 & -2x & +2)\\\\ %\\hline % & & x & -2 %\\end{array}$$ You surely will recognize the rest as the numerator of $g(x)$\n\nSo, whenever you have a rational function whose numerator and denominator, after cancelling out, have the same non-zero degree, it is not only possible, but guaranteed that there exists another rational function with lesser degree in the numerator which differs from the original rational function only by a constant.\n\n\u2022 BTW, does anyone know how to make the horizontal line end at the end of the polynomial above? May 4, 2018 at 5:51\n\u2022 @farruhota: That first option destroys the alignment between the terms, and therefore is worse than the too-long line. That second option defeats the whole purpose of writing the polynomial division in its standard form. May 4, 2018 at 6:48\n\u2022 How about this: $\\begin{array}{ll} &(2x^2-x) & / (x^2-x+1) = 2\\\\ -&\\underline{(2x^2-2x+2)}\\\\ & x-2 \\end{array}$ May 4, 2018 at 7:04\n\u2022 @farruhota: Still doesn't correctly align the terms. May 4, 2018 at 7:10\n\u2022 Oh well, the rest should be adjusted: $\\begin{array}{ll} &(2x^2-\\ \\ x) & / (x^2-x+1) = 2\\\\ -&\\underline{(2x^2-2x+2)}\\\\ & \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x \\ - 2 \\end{array}$ May 4, 2018 at 7:16\n\nI feel that it's only natural that intuition would fail you in cases like these. If you wrote the two functions out as $$f(x)=\\frac{2\\left(x^2-x+1\\right)+\\left(x-2\\right)}{x^2-x+1}$$ $$g(x)=\\frac{x-2}{x^2-x+1}$$ it would be obvious that the difference between them was 2. This is the first thing you could trip over, the degree of the numerator don't actually differ by 1, they are the same.\n\nI guess if you made it a rule to divide whenever you have a rational function where the degree of the numerator is greater than or equal to the degree of the denominator until it isn't, you would get consistent results. (Kinda like how you learn waaay back how $\\frac{13}{7}$ is an improper fraction, because it's actually bigger than a whole number, and that you should write it as $1\\frac{6}{7}$. a function $f(x)=\\frac{P(x)}{Q(x)}$ where the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$ is called an improper rational function.)\n\nBut, and this might be another thing that's bugging you about this question, you can't just do that! (As my algebra teacher used to exclaim in exasperation.) You introduce a new constraint: $x^2-x+1\\neq0$. Which isn't relevant if x is real, but does change things up a bit if it s a complex number.\n\nThis is probably what you get when you graph the two functions $f$ and $g$.\n\nBut if you allow for complex values of $x$, it can also look like this:\n\n(Sans the vertical line at x=0.5, I don't know why that is happening, probably a limitation in the graphing engine on desmos.)\n\nNot so obvious that the difference between the two is 2 in this case...\n\nSorry, I just realized this probably doesn't answer your question on how you can reconciliate this apparent discrepancy with your intuition, but I thought explaining why it may not be that clear-cut would help at least a little bit.\n\nPlaying with the parameters used in the graph might help further your understanding, here's a link to the interactive graph, with a little more in depth graphical interpretation of why you might doubt what you logically \"know\" to be true.\n\nAs for the equations I used in place of the given ones... You'll just have to trust my algebra (which you shouldn't in critical situations, but I think this is safe enough hehe). I did them on a legal pad, but I'll type them up if anyone insists.\n\nSince the denominator is the same , you subtract one numerator from the other. In this special case you find the resultant numerator is twice the denominator. Feels like that was much simpler than the rest of your background calculus.\n\nIt must be very clear to you that $11/3$ and $2/3$ defer by an integer. Why is this so? Because in the world $$modulo 3$\"$, 11 and 2 are but the same. In the same vein, we can say that in the world of $$modulo $x^2-x+1\"$, $2x^2-x$ and $x-2$ are but the same because $2x^2-x=2(x^2-x+1)+x-2$. When we are in such a world, the same polynomial may appear in different forms with different \"degrees\". So in a world such as the ideal generated by $x^2-x+1$ (ie $<x^2-x+1>$ ), degree is no longer retained. One's intuition may fail here because one is not familiar with such a world where degree is not so well behaved.\n\u2022 The question is not about constants; it is about degree. $\\frac{x^3+1}{x^2+2}$ and $\\frac{-2x+1}{x^2+2}$ also defer by a constant (polynomial) May 6, 2018 at 6:11", "date": "2022-05-29 09:27:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2764985/intuition-for-why-the-difference-between-frac2x2-xx2-x1-and-fracx-2/2765019", "openwebmath_score": 0.793973982334137, "openwebmath_perplexity": 346.3680785891427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357221825193, "lm_q2_score": 0.9161096124442242, "lm_q1q2_score": 0.8993775319712782}}
{"url": "http://mathhelpforum.com/geometry/143843-solved-finding-co-ordinates-rectangle.html", "text": "# Thread: [SOLVED] Finding Co-Ordinates of a Rectangle\n\n1. ## [SOLVED] Finding Co-Ordinates of a Rectangle\n\nHere's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:\n\nIs my method correct, considering I did get the right answer?\n\nBut is there another simpler method to do this which would save time during an exam.\n\n2. The diagonals bisect one another. The midpoint of $\\displaystyle \\overline{AC}$ is ?\n\n3. Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?\n\n4. Originally Posted by unstopabl3\nMid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?\nBut the y-coordinate of $\\displaystyle B~\\&~D$ is 6.\nYou are given the x-coordinate of $\\displaystyle D$, so $\\displaystyle Dh,6)$\n\n5. No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.\n\nI have already gotten the correct values by using the method mentioned in my first post.\nAs stated I want someone to solve this part of the question with a different, possibly easier method.\n\nI am not after the answer, I am looking for an alternate method.\n\n6. Originally Posted by unstopabl3\nNo, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.\n\nI have already gotten the correct values by using the method mentioned in my first post.\nAs stated I want someone to solve this part of the question with a different, possibly easier method.\n\nI am not after the answer, I am looking for an alternate method.\nHi unstopabl3,\n\nI really don't see what method you used in your first post, but here's how I'd do it.\n\nPlato already told you that the midpoint of BD is M(6, 6).\n\nThis means the y-coordinates of B and D are also 6.\n\nThis distance from B to D is 20 (found using distance formula)\n\nEach individual segment of the diagonals measure 10 since they are bisected.\n\nUsing the distance formula, it is easy to determine the x-coordinates of B and D.\n\nM(6, 6) -----> D(h, 6) = 10\n\n$\\displaystyle 10=\\sqrt{h-6)^2+(6-6)^2}$\n\n7. Hello, unstopabl3!\n\nCode:\n              |           C(12,14)\n|           o\n|       *    *\n|   *         *\n*              *\nB o - + - - - - - - - o D\n*  |           *\n------*-+-------*------------\n*|   *\no\nA|(0,-2)\n|\n\nThe diagram shows a rectangle $\\displaystyle ABCD.$\nWe have: .$\\displaystyle A(0,-2),\\;C(12,14)$\nThe diagonal $\\displaystyle BD$ is parallel to the $\\displaystyle x$-axis.\n\n$\\displaystyle (i)$ Explain why the $\\displaystyle y$-coordinate of $\\displaystyle D$ is 6.\nThe diagonals of a rectangle bisect each other.\n. . Hence, the midpoint of $\\displaystyle AC$ is the midpoint of $\\displaystyle BD.$\n\nThe midpoint of AC is: .$\\displaystyle \\left(\\tfrac{0+12}{2},\\;\\tfrac{-2+14}{2}\\right) \\:=\\:(6,6)$\n\nTherefore, $\\displaystyle B$ and $\\displaystyle D$ have a $\\displaystyle y$-coordinate of 6.\n\nThe $\\displaystyle x$-coordinate of $\\displaystyle D$ is $\\displaystyle h.$\n$\\displaystyle (ii)$ Express the gradients of $\\displaystyle AD$ and $\\displaystyle CD$ in terms of h,.\nWe have: .$\\displaystyle \\begin{Bmatrix}A(0, -2) \\\\ C(12,14) \\\\ D(h,\\;6) \\end{Bmatrix}$\n\n$\\displaystyle m_{AD} \\;=\\;\\frac{6(-2)}{h-9} \\;=\\;\\frac{8}{h}$\n\n$\\displaystyle m_{CD} \\;=\\;\\frac{6-14}{h-12} \\;=\\;\\frac{-8}{h-12}$\n\n$\\displaystyle (iii)$ Calculate the $\\displaystyle x$-coordinates of $\\displaystyle D$ and $\\displaystyle B.$\n\nSince $\\displaystyle m_{AD} \\perp m_{CD}$ we have: .$\\displaystyle \\frac{8}{h} \\;=\\;\\frac{h-12}{8} \\quad\\Rightarrow\\quad h^2-12h - 64 \\:=\\:0$\n\nHence: .$\\displaystyle (h+4)(h-16) \\:=\\:0 \\quad\\Rightarrow\\quad h \\:=\\:-4,\\:16$\n\nTherefore: .$\\displaystyle D(16,6),\\;B(-4,6)$\n\n8. Originally Posted by masters\nHi unstopabl3,\n\nI really don't see what method you used in your first post, but here's how I'd do it.\n\n$\\displaystyle 10=\\sqrt{h-6)^2+(6-6)^2}$\nI use the concept of the product of two perpendicular lines = -1\nYou can see the working in the Soroban's post. That's exactly how I did it!\n\nThis distance from B to D is 20 (found using distance formula)\nHow did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?\n\nSoroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!\n\n9. Originally Posted by unstopabl3\nI use the concept of the product of two perpendicular lines = -1\nYou can see the working in the Soroban's post. That's exactly how I did it!\n\nHow did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?\n\nSoroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!\nThe distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent).\n\n10. Co ordinates of B and D are $\\displaystyle (x_1 , 6) and (x_2, 6)$\nDiagonal AC = BD\nAC = 20 = BD.\nDistance $\\displaystyle BD^2 = (x_1 - x_2)^2$\n\nSo (x_1 - x_2) = 20...........(1)\n\nMid point point of AC = mid point of BD\n\n$\\displaystyle \\frac{x_1+x_2}{2} = 6$\n\n(x_1 + x_2) = 12......(2)\nSolve Eq (1) ans (2) to find the coordinates of B and D.\n\n11. Thanks for the responses guys!", "date": "2018-06-20 14:11:27", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/143843-solved-finding-co-ordinates-rectangle.html", "openwebmath_score": 0.844447135925293, "openwebmath_perplexity": 646.5633322110568, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353585837, "lm_q2_score": 0.9136765310530521, "lm_q1q2_score": 0.8993650249337182}}
{"url": "https://lotusmc.org/ii5sh61n/archive.php?id=6a0ed8-system-of-linear-equations-problems", "text": "Below is an example that shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. Solution: Rewrite in order to align the x and y terms. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. ... Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up! Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! 2 equations in 3 variables, 2. Solving using an Augmented Matrix. Solve simple cases by inspection. 1. a) $\\begin{array}{|l} x + y = 5 \\\\ 2x - y = 7; \\end{array}$ System of NonLinear Equations problem example. $\\begin{cases}5x +2y =1 \\\\ -3x +3y = 5\\end{cases}$ Yes. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you \u2026 There can be any combination: 1. For example, + \u2212 = \u2212 + = \u2212 \u2212 + \u2212 = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. Systems of Equations - Problems & Answers. In your studies, however, you will generally be faced with much simpler problems. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. Solution of a non-linear system. Quiz 3. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. Cramer's Rule. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics. Consider the nonlinear system of equations Donate or volunteer today! Find them out by checking. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A system of linear equations is a system made up of two linear equations. Khan Academy is a 501(c)(3) nonprofit organization. SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. Many problems lend themselves to being solved with systems of linear equations. Solve each of the following equations and check your answer. System of linear equations System of linear equations can arise naturally from many real life examples. System of equations word problem: walk & ride, Practice: Systems of equations word problems, System of equations word problem: no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite solutions), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. A system of linear equations is called homogeneous if the constants $b_1, b_2, \\dots, b_m$ are all zero. By \u2026 Solve age word problems with a system of equations. Determining the value of k for which the system has no solutions. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. Section 7-1 : Linear Systems with Two Variables. Answer: x = .5; y = 1.67. Substitution Method. Systems of Linear Equations. You appear to be on a device with a \"narrow\" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\\left( {2 - x} \\right) = 3x + 2$$, $$2\\left( {w + 3} \\right) - 10 = 6\\left( {32 - 3w} \\right)$$, $$\\displaystyle \\frac{{4 - 2z}}{3} = \\frac{3}{4} - \\frac{{5z}}{6}$$, $$\\displaystyle \\frac{{4t}}{{{t^2} - 25}} = \\frac{1}{{5 - t}}$$, $$\\displaystyle \\frac{{3y + 4}}{{y - 1}} = 2 + \\frac{7}{{y - 1}}$$, $$\\displaystyle \\frac{{5x}}{{3x - 3}} + \\frac{6}{{x + 2}} = \\frac{5}{3}$$. Systems of linear equations can be used to model real-world problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. A large pizza at Palanzio\u2019s Pizzeria costs $6.80 plus$0.90 for each topping. Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. If you're seeing this message, it means we're having trouble loading external resources on our website. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. 1. At the first store, he bought some t-shirts and spent half of his money. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. So far, we\u2019ve basically just played around with the equation for a line, which is . A solution of the system (*) is a sequence of numbers $s_1, s_2, \\dots, s_n$ such that the substitution $x_1=s_1, x_2=s_2, \\dots, x_n=s_n$ satisfies all the $m$ equations in the system (*). Solving using Matrices by Elimination. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. So a System of Equations could have many equations and many variables. For example, the sets in the image below are systems of linear equations. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If you're seeing this message, it means we're having trouble loading external resources on our website. Systems of Linear Equations and Problem Solving. Wouldn\u2019t it be cl\u2026 Solving Systems of Linear Equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. 9,000 equations in 567 variables, 4. etc. But let\u2019s say we have the following situation. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. The same rules apply. Solve the following system of equations by elimination. Problem 1 Two of the following systems of equations have solution (1;3). HIDE SOLUTIONS. If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. It has 6 unique word problems to solve including one mixture problem \u2026 Gradient descent can also be used to solve a system of nonlinear equations. Just select one of the options below to start upgrading. Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. Systems of linear equations are a common and applicable subset of systems of equations. Materials include course notes, lecture video clips, JavaScript Mathlets, a quiz with solutions, practice problems with solutions, a problem solving video, and problem sets with solutions. One of the following system of equations, you need to find the exact values of x y. Please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked to solve a system linear. To take home 6items of clothing because you \u201c need \u201d that many new things dresses $... Two-Dimensional space old as Shaheena a solution to the following situation just that... And you have$ 200 to spend from your recent birthday money twice as old as Shaheena $... That we are dealing with more than one equation and variable systems of linear equations word problems ( with and! To provide a free, world-class education to anyone, anywhere the best experience equations... To teach kids about Solving word problems worksheet for all variables to upgrading! 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Activities to teach kids about Solving word problems with a system of linear equations is by graphing equations! Education to anyone, anywhere s say we have the following system of linear equations can be to! To align the x and y that will solve both equations education anyone. Final solution of nonlinear equations of math and science problem solvers exact values of x and y terms store has! The problem: case 1: two Intersecting Lines need to find the values... To align the x and y that will solve both equations shows one iteration of the options below to upgrading! Math knowledge with free questions in  real life '', these systems can incredibly! { 2 } ) a solution to the first store, he bought some and... A single point, then there is one reason why linear algebra the. That many new things done, one of three cases will arise: case 1: Intersecting... To anyone, anywhere recent birthday money many real life examples linear systems with two variables, these systems.\nCat Face Png, 0 Tick Farms Removed, House For Rent 75241, How Deep Is Bedrock In Florida, Gaggia Accademia Refurbished, Ego Trimmer Ht2410, Desert Zinnia Seeds, Project Management Stage Gates, Klipsch Heritage Amp, Dsdm Advantages And Disadvantages, Gaming Headset Xbox One,", "date": "2021-09-23 22:03:28", "meta": {"domain": "lotusmc.org", "url": "https://lotusmc.org/ii5sh61n/archive.php?id=6a0ed8-system-of-linear-equations-problems", "openwebmath_score": 0.4137629270553589, "openwebmath_perplexity": 474.14805537072425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9869795091201805, "lm_q2_score": 0.9111797166446537, "lm_q1q2_score": 0.8993157094542055}}
{"url": "https://math.stackexchange.com/questions/2398684/average-number-of-selections-before-duplicate-picked", "text": "# Average number of selections before duplicate picked\n\nI have a dataset of 1296 unique codes which can be numbered 1 through 1296. If numbers are selected at random, one at a time, with replacement. On average, how many iterations will it take to select a number that has already been selected?\n\nExperimentally, (looping through the list of 1296 codes and creating a subset of selected codes using Python) it averages out at 45.875 times (this number includes the duplicate) but I would like to verify it with a calculation so any help would be appreciated.\n\nThis question has some similarities but I am unable to perform a calculation based on the answer:\n\nQuestion with similarities\n\n\u2022 This is an example of the generalized birthday problem where you have $1296$ \"days\" instead of $365$. The number that gives a $50\\%$ chance of a match in $d$ \"days\" is about $\\sqrt {2d\\ln 2}$, which for you is about $42.39$ \u2013\u00a0Ross Millikan Aug 19 '17 at 1:15\n\u2022 @RossMillikan instead of asking for where it switches from being below a $50\\%$ chance to being above a $50\\%$ chance, isn't the OP asking for the expected number of draws until a match occurs? Will those two necessarily be the same? I'm getting a different result (fixed minor typo in equation) getting a result closer to $\\approx 45.788$ \u2013\u00a0JMoravitz Aug 19 '17 at 1:20\n\u2022 @JMoravitz: I'll reopen it. I don't know an easy approach to the expected number version, but would expect it to be close to the $50\\%$ probability number. No, they won't be the same because there is the long tail to high numbers. \u2013\u00a0Ross Millikan Aug 19 '17 at 2:10\n\u2022 @RossMillikan well, the pdf is straightforward (and is included in one of your two links in some form) and we can apply the definition of expected value. I don't know of a clean way to simplify the sum by hand, but computers can calculate it easily enough for us. \u2013\u00a0JMoravitz Aug 19 '17 at 2:12\n\nIt is impossible to have gotten a duplicate on the first draw. It is impossible to have not gotten a duplicate by the 1297'th draw by pigeon-hole principle.\n\nTo have gotten your first duplicate on the $k$'th draw, you need the first $k-1$ draws to all be distinct and the $k$'th to be a duplicate.\n\nThe first draw will always be distinct. The second will be distinct from the first with probability $\\frac{1295}{1296}$. The third will be distinct from the first two with probability $\\frac{1294}{1296}$ and so on... the $(n)$'th will be distinct from the earlier $n-1$ with probability $\\frac{1296-n+1}{1296}$. Multiplying these, we get for $n$ draws to all be distinct, this will occur with probability $\\frac{1296\\frac{n}{~}}{1296^n}$ where $x\\frac{n}{~}$ represents a falling factorial $x\\frac{n}{~}=\\underbrace{x(x-1)(x-2)\\cdots (x-n+1)}_{n~\\text{terms in the product}}=\\frac{x!}{(x-n)!}$.\n\nNext, supposing $k-1$ distinct values have all been taken, for the $k$'th to duplicate one of the earlier results, this will occur with probability $\\frac{k-1}{1296}$\n\nWe have then the probability distribution function for $X$, the number of draws until the first duplicate:\n\n$$Pr(X=k)=\\frac{(k-1)1296\\frac{k-1}{~}}{1296^k}$$\n\nApplying the definition of expected value for a pdf: $E[X]=\\sum\\limits_{k\\in\\Delta} kPr(X=k)$ we have then the expected value is\n\n$$\\sum\\limits_{k=2}^{1297}\\frac{k(k-1)1296\\frac{k-1}{~}}{1296^k}\\approx 45.7889$$\n\n\u2022 As a minor aside, I was having difficulty with the link. Trying to write it as [linkname](actuallinkgoeshere) with parenthesis appearing in the link part of it was being chopped off, and replacing parenthesis with brackets was causing the calculation to time out. Tinyurl's aren't permenant... does anyone have a suggestion other than just hiding it with a spoiler tag like I have? \u2013\u00a0JMoravitz Aug 19 '17 at 2:35\n\nWith $n$ objects the expected time until the first repeat is exactly $$\\mathbb{E}(T)=\\int_0^\\infty \\left(1+{x\\over n}\\right)^ne^{-x}\\,dx,$$ and approximately equal to $\\sqrt{n\\pi/2}.$\n\nYou can find a derivation of this formula at my answer here: Variance of time to find first duplicate\n\nFor $n=1296$ the exact formula gives $\\mathbb{E}(T)\\approx 45.78885405,$ while the approximation gives $\\sqrt{1296\\pi/2}\\approx 45.11930893.$", "date": "2019-05-25 03:14:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2398684/average-number-of-selections-before-duplicate-picked", "openwebmath_score": 0.7660864591598511, "openwebmath_perplexity": 279.07212059651675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795095031688, "lm_q2_score": 0.9111797015700341, "lm_q1q2_score": 0.899315694924836}}
{"url": "https://math.stackexchange.com/questions/2287316/find-relationship-between-a-b-c-f-g-h", "text": "# Find relationship between $a, b, c, f, g, h$\n\nGiven that - $$a=a_1 a_2$$ $$b=b_1 b_2$$ $$c=c_1 c_2$$ $$h=a_2 b_1 + b_2 a_1$$ $$g=a_1 c_2 + a_2 c_1$$ $$f=b_1 c_2 + b_2 c_1$$\n\nFind the relationship between $a, b, c, f, g, h$\n\nMy Attempt: I could not see how I could exploit the symmetry of the equations to directly get an answer, so I tried to solve them as 6 simultaneous equations by - $$a_2=\\frac{a}{a_1}$$ $$b_2=\\frac{b}{b_1}$$ $$c_2=\\frac{c}{c_1}$$ and put these values into the remaining 3 equations to obtain - $$a_1 b_1 h = a {b_1}^2 + b {a_1}^2$$ $$a_1 c_1 g = a {c_1}^2 + c {a_1}^2$$ $$c_1 b_1 f = c {b_1}^2 + b {c_1}^2$$ Then, I treated the last 2 equations as quadratics in $a_1$ and $b_1$; found their values using the quadratic formula and input those into the first of the last 3 equations above.\n\nThen I found the value of $c_1$ using that and then the value of the remaining - $a_1, a_2, b_1, b_2, c_2$.\n\nThen, when I finally input these values into any of the equations I got a tautology (which, as I now realize - too late - was doomed to happen from the very beginning, due to my approach).\n\n============================================================\n\nSo, How should I go about finding the relation between $a, b, c, f, g, h$?\n\nOR,\n\nEqually - How do I eliminate $a_1, a_2, b_1, b_2, c_1, c_2$ from the equations?\n\nI just need a hint on how I could exploit the symmetry of the equations.\n\n\u2022 What does \"Find the relationship between a,b,c,f,g,h\" mean exactly? One could argue that the initial six equalities already gives a relationship among them. Are you looking for a polynomial expression in a,b,c,f,g,h (and not involving the subscripted variables) that equals 0? \u2013\u00a0Greg Martin May 19 '17 at 3:36\n\u2022 Hint: consider $af^2+bg^2+ch^2$. \u2013\u00a0Greg Martin May 19 '17 at 3:38\n\u2022 @GregMartin Yes, you're right I need a single expression in a, b, c, f, g, h without the subscripted variables. As an example of such an expression - $\\frac{c^{2}}{2a}=\\frac{f^{3}}{5g^{2}} +h tan(\\frac{b}{a})$ The expression could include logs and trig functions (although I don't think they will be necessary) \u2013\u00a0Quantum Sphinx May 19 '17 at 3:39\n\u2022 @GregMartin Yes, that is the kind of expression I need \u2013\u00a0Quantum Sphinx May 19 '17 at 3:43\n\nNotice that $a,b,c,h,g,f$ are represented by staggered multiplication of $a_1,a_2,b_1,b_2,c_1,c_2$, and one way to align them is to multiply them together, and then we could rearrange how they are combined together. So symmetry is the key here.\n$$h\\cdot g \\cdot f=(a_2 b_1 + b_2 a_1)(a_1 c_2 + a_2 c_1)(b_1 c_2 + b_2 c_1)$$ $$=a_2b_1a_1c_2b_1c_2 + a_2b_1a_1c_2b_2c_1+a_2b_1a_2c_1b_1c_2 + a_2b_1a_2c_1b_2c_1$$$$+b_2a_1a_1c_2b_1c_2+b_2a_1a_1c_2b_2c_1+b_2a_1a_2c_1b_1c_2+b_2a_1a_2c_1b_2c_1$$\n$$=ab_1^2c_2^2+abc+a_2^2b_1^2c+a_2^2c_1^2b+a_1^2c_2^2b+a_1^2b_2^2c+abc+b_2^2c_1^2a$$ $$=2abc + a(b_1^2c_2^2+b_2^2c_1^2)+b(a_1^2c_2^2+a_2^2c_1^2) + c(a_2^2b_1^2+a_1^2b_2^2)$$ $$=2abc+a((b_1c_2 + b_2c_1)^2-2b_1c_2b_2c_1)+b((a_1c_2+a_2c_1)^2-2a_1c_2a_2c_1)+c((a_2b_1+a_1b_2)^2-2a_2b_1a_1b_2)$$ Thus $$hgf=2abc + a(f^2-2bc)+b(g^2-2ac)+c(h^2-2ab)$$ $$hgf +4abc -af^2-bg^2-ch^2=0$$", "date": "2019-08-26 00:38:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2287316/find-relationship-between-a-b-c-f-g-h", "openwebmath_score": 0.8017062544822693, "openwebmath_perplexity": 294.01646413254554, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232924970204, "lm_q2_score": 0.9149009613738741, "lm_q1q2_score": 0.8991859751661603}}
{"url": "https://www.freemathhelp.com/forum/threads/find-the-numbers.115989/", "text": "# Find the Numbers\n\nStatus\nNot open for further replies.\n\n##### Full Member\nThere are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?\n\nSet up:\n\nLet x = large number\nLet y = small number\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nx + y = 53\n\ny = 53 - x...Plug into B.\n\n3(53 - x) = x + 19\n\n159 - 3x = x + 19\n\n-3x - x = 19 - 159\n\n-4x = -140\n\nx = -140/-4\n\nx = 35...Plug into A or B.\n\nI will use A.\n\n35 + y = 53\n\ny = 53 - 35\n\ny = 18.\n\nThe numbers are 18 and 35.\n\nYes?\n\n#### JeffM\n\n##### Elite Member\nDo the numbers satisfy both equation?\n\n$$\\displaystyle 35 + 18 = 53.$$ Checks.\n\n$$\\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks.\n\nIn algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds confidence, is a necessary skill for taking tests, and, most importantly, is what you will need in any job that expects you to be able to do math.\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nThere are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?\n\nSet up:\n\nLet x = large number\nLet y = small number\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nx + y = 53\n\ny = 53 - x...Plug into B.\n\n3(53 - x) = x + 19\n\n159 - 3x = x + 19\n\n-3x - x = 19 - 159\n\n-4x = -140\n\nx = -140/-4\n\nx = 35...Plug into A or B.\n\nI will use A.\n\n35 + y = 53\n\ny = 53 - 35\n\ny = 18.\n\nThe numbers are 18 and 35.\n\nYes?\nWhen possible check your work. Most of the time that is a part of the process of solution.\n\nThere is a shorter way to accomplish the algebra/arithmetic part.\n\nYou have two equations,\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nrewrite B to collect all the unknowns to LHS\n\nx + y = 53...Equation A\n3y - x = 19....Equation B'\n\nAdd A & B' (to eliminate 'x' from the equations) and get equation C\n\n3y + y = 72....Equation C\n\n4y = 72\n\ny = 18\n\nUse this value in equation 'A'\n\nx + 18 = 53...Equation A\n\nx = 53- 18 = 35\n\n##### Full Member\nWhen possible check your work. Most of the time that is a part of the process of solution.\n\nThere is a shorter way to accomplish the algebra/arithmetic part.\n\nYou have two equations,\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nrewrite B to collect all the unknowns to LHS\n\nx + y = 53...Equation A\n3y - x = 19....Equation B'\n\nAdd A & B' (to eliminate 'x' from the equations) and get equation C\n\n3y + y = 72....Equation C\n\n4y = 72\n\ny = 18\n\nUse this value in equation 'A'\n\nx + 18 = 53...Equation A\n\nx = 53- 18 = 35\n\nWhat is wrong with my method?\n\n#### Dr.Peterson\n\n##### Elite Member\nNothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily \"better\".\n\n##### Full Member\nNothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily \"better\".\nThere are several methods for solving two equations in two variables, right? Matrix algebra is another useful tool.\n\n#### Dr.Peterson\n\n##### Elite Member\nCorrect.\n\nIn fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!\n\nBut still, solving the equations is the \"easy\" (routine) part, compared to setting them up from a word problem.\n\n##### Full Member\nCorrect.\n\nIn fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!\n\nBut still, solving the equations is the \"easy\" (routine) part, compared to setting them up from a word problem.\nWe can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y).\n\n#### Jomo\n\n##### Elite Member\nI know that you can check these problems. Just admit that you like posting here.", "date": "2019-05-25 11:46:25", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/find-the-numbers.115989/", "openwebmath_score": 0.5126290917396545, "openwebmath_perplexity": 1169.4730670751283, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9828232879690035, "lm_q2_score": 0.9149009532527358, "lm_q1q2_score": 0.8991859630418294}}
{"url": "http://winbytes.org/help/standard-error/proportion-standard-error.html", "text": "Home > standard error > proportion standard error\n\n# Proportion Standard Error\n\n## Contents\n\nrepeatedly randomly drawn from a population, and the proportion of successes in each sample is recorded ($$\\widehat{p}$$),the distribution of the sample\n\n## Standard Error Of Proportion Formula\n\nproportions (i.e., the sampling distirbution) can be approximated by a normal standard error of proportion definition distribution given that both $$n \\times p \\geq 10$$ and $$n \\times (1-p) \\geq 10$$. This is sample proportion formula known as theRule of Sample Proportions. Note that some textbooks use a minimum of 15 instead of 10.The mean of the distribution of sample proportions is equal\n\nto the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized by $$SE(\\widehat{p})$$ and equals $$\\sqrt{\\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\\widehat{p}$$. The symbol $$\\sigma _{\\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$## Sample Proportion Calculator SE(\\widehat{p})= \\sqrt{\\frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\\widehat{p}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of Sample ProportionsGiven both $$n \\times p \\geq 10$$ and $$n \\times (1-p) \\geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\\mu_{\\widehat{p}}$$ and standard deviation of $$SE(\\widehat{p})$$Mean $$\\mu_{\\widehat{p}}=p$$Standard Deviation (\"Standard Error\")$$SE(\\widehat{p})= \\sqrt{\\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example \u2039 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2.1 - Marijuana Example \u203a Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The \u201cBig Picture\u201d Lesson 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3: Probability - 1 Variable Lesson 4: Probability - 2 Variables Lesson 5: Probability Distributions Lesson 6: Sampling Distributions6.1 - Simulation of a Sampling Distribution of a Proportion (Exact Method) 6.2 - Rule of Sample Proportions (Normal Approximation Method)6.2\n\n0 otherwise. The standard deviation of any variable involves the https://onlinecourses.science.psu.edu/stat200/node/43 expression . Let's suppose there are m 1s (and n-m 0s) among the n subjects. Then, and is equal to (1-m/n) for m observations and 0-m/n http://www.jerrydallal.com/lhsp/psd.htm for (n-m) observations. When these results are combined, the final result is and the sample variance (square of the SD) of the 0/1 observations is The sample proportion is the mean of n of these observations, so the standard error of the proportion is calculated like the standard error of the mean, that is, the SD of one of them divided by the square root of the sample size or Copyright \u00a9 1998 Gerard E. Dallal\n\nTables Constants Calendars Theorems Standard Error of Sample Proportion Calculator https://www.easycalculation.com/statistics/standard-error-sample-proportion.php Calculator Formula Download Script Online statistic calculator allows you to estimate the accuracy of the standard error of the sample proportion in the binomial standard deviation. Calculate SE Sample Proportion of Standard standard error Deviation Proportion of successes (p)= (0.0 to 1.0) Number of observations (n)= Binomial SE of Sample proportion= Code to add this calci to your website Just copy and paste the below code to your webpage where you standard error of want to display this calculator. Formula Used: SEp = sqrt [ p ( 1 - p) / n] where, p is Proportion of successes in the sample,n is Number of observations in the sample. Calculation of Standard Error in binomial standard deviation is made easier here using this online calculator. 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tbody tr td div id toctitle Contents div ul li a href Small Standard Error a li li a href What Is Standard Error Used For In Statistics a li ul td tr tbody table p proportion of samples that would fall between and standard deviations above and below relatedl the actual value The standard error SE is sample standard error the standard deviation of the sampling distribution of a statistic most commonly standard error statistics of the mean The term may also be used to refer to an estimate of that standard\n\nappropriate use of standard error\n\nAppropriate Use Of Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href When To Use Standard Error Of The Mean a li li a href When To Use Standard Error Versus Standard Deviation a li li a href When To Use Standard Error Bars a li ul td tr tbody table p error of the mean Is the choice between these down to personal preference or is one favoured relatedl in the scientific field over another Topics Standard Deviation use 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questions you might have Meta Discuss the workings and policies of this site About Us Learn relatedl more about Stack Overflow the company Business Learn more about hiring developers mean standard error or posting ads with us Cross Validated\n\naverage standard error calculator\n\nAverage Standard Error Calculator table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Of Estimate Calculator a li li a href Standard Error Calculator Two Samples a li li a href Standard Error Calculator For Regression a li ul td tr tbody table p of the mean of a set of numbers Standard p h id Standard Error Calculator Two Samples p Error of the Mean The standard error of the mean is the standard deviation of p h id Standard Error Calculator For Regression p the sample mean estimate of a\n\naverage standard error on excel\n\nAverage Standard Error On Excel table id toc tbody tr td div id toctitle Contents div ul li a href Average Standard Error Of The Mean a li li a href Standard 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Contents div ul li a href Bash Redirect Error Output To dev null a li li a href Bash Standard Error To Variable a li li a href Bash Output To Stderr a li ul td tr tbody table p a stderr redirect stderr to a stdout redirect stderr and stdout to a file redirect stderr and stdout to stdout redirect stderr relatedl and stdout to stderr 'represents' stdout and stderr bash redirect error output to file A little note for seeing this things with the less\n\nbash standard error redirect\n\nBash Standard Error Redirect table id toc tbody tr td div id toctitle Contents div ul li a href Bash Pipe Standard Error a li li a href Redirect Standard Error Csh a li ul td tr tbody table p Applications A DT DL DIV A P There are always three default files I SPAN A open stdin TT the keyboard stdout TT the screen and relatedl stderr TT error messages output to the screen These and bash redirect standard error dev null any other open files can be redirected Redirection simply means capturing output from a bash redirect standard\n\nbash standard error\n\nBash Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Bash Redirect To Dev Null a li li a href How To Redirect Stderr And Stdout To A File a li li a href Bash Echo To Standard Error a li ul td tr tbody table p a stderr redirect stderr to a stdout redirect stderr and stdout to a file redirect stderr and stdout to stdout redirect stderr relatedl and stdout to stderr 'represents' stdout and stderr bash standard error to variable A little note for seeing this things with the\n\nbasic definition of standard error of measurement\n\nBasic Definition Of Standard Error Of Measurement table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Of Measurement Calculator a li li a href Standard Error Of Measurement And Confidence Interval a li li a href Standard Error Of Measurement Vs Standard Deviation a li li a href Standard Error Of Measurement Vs Standard Error Of Mean a li ul td tr tbody table p latter is impossible standardized tests usually have an associated standarderror of measurement SEM relatedl an index of the expected variation in observedscores define standard error of measurement\n\nbest estimate of standard error\n\nBest Estimate Of Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Estimate Standard Error Calculator a li li a href Estimate Standard Error Of Proportion a li li a href Estimate Standard Error Bootstrap a li li a href Estimate The Standard Error Of The Sample Mean a li ul td tr tbody table p it comes to determining how well a linear model fits the data However I've stated previously that R-squared is overrated relatedl Is there a different goodness-of-fit statistic that can be more how to estimate standard error\n\nbernoulli error bars\n\nBernoulli Error Bars table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Binary Distribution a li li a href Standard Deviation Of Dummy Variable a li li a href Binomial Error a li li a href Standard Error Is Used In The Calculation Of Both The Z And T Statistic With The Difference That a li ul td tr tbody table p be calculated for a binary variable In a graph showing the progress over time of the probability to relatedl find a pathogen within plant tissues I'm wondering if standard error\n\nbeta standard error excel\n\nBeta Standard Error Excel table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Of Beta Linear Regression a li li a href Standard Error Of Beta Calculator a li li a href Beta Standard Deviation a li ul td tr tbody table p std error in excel ArteSuaveEconomist SubscribeSubscribedUnsubscribe Loading Loading Working Add to Want to watch this again later Sign in to add this video to a playlist relatedl Sign in Share More Report Need to report the video standard error of beta coefficient Sign in to report inappropriate content Sign\n\nbeta standard error formula\n\nBeta Standard Error Formula table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Formula Statistics a li li a href Standard Error Formula Proportion a li li a href Standard Error Formula Regression a li ul td tr tbody table p Curve Z-table Right of Curve Probability and Statistics Statistics Basics Probability Regression Analysis Critical Values Z-Tables Hypothesis Testing Normal Distributions Definition Word Problems T-Distribution relatedl Non Normal Distribution Chi Square Design of Experiments Multivariate Analysis standard error formula excel Sampling in Statistics Famous Mathematicians and Statisticians Calculators Variance and Standard Deviation\n\nbeta standard error\n\nBeta Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Beta Variance a li li a href Standard Error Of Regression Coefficient a li li a href Standard Error Of Beta Estimate a li ul td tr tbody table p faq bull rss Community Log In Sign Up Add New Post Question what does beta and standard error mean in a typical GWAS results months ago by iphoenix bull European Union iphoenix bull relatedl wrote Hi all Since I am new to GWAS and statistics beta standard deviation I find it hard\n\nbeta standard error regression\n\nBeta Standard Error Regression table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Of Beta Linear Regression a li li a href Standard Error Of Beta Coefficient Formula a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings relatedl and policies of this site About Us Learn more about Stack standard error of coefficient in linear regression Overflow the company Business Learn more about hiring developers or posting ads with\n\nbeta coefficient divided by standard error\n\nBeta Coefficient Divided By Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href T Statistic Coefficient Divided By Standard Error a li li a href Standard Error Of Beta Coefficient Formula a li li a href Regression Coefficient Interpretation a li ul td tr tbody table p here td Nov -Dec font Walk-in - pm td Dec -Feb font By appt here td Feb -May font Walk-in - pm font a May -May font Walk-in relatedl - pm font a May -Aug font By appt here td For t-stat coefficient divided by\n\nbig standard error\n\nBig Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Importance Of Standard Error Of The Mean a li li a href Large Standard Error In Multiple Regression a li li a href What Does A Large Standard Error Mean a li ul td tr tbody table p Ana-Maria imundi Editor-in-ChiefDepartment of Medical Laboratory DiagnosticsUniversity Hospital Sveti Duh Sveti Duh relatedl Zagreb CroatiaPhone - e-mail large standard error address editorial office at biochemia-medica dot com Useful links Events Follow p h id Importance Of Standard Error Of The Mean p us on\n\nbeta standard error calculation\n\nBeta Standard Error Calculation table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Calculation Excel a li li a href Standard Error Calculation In Regression a li li a href Standard Error Calculation In R a li li a href Margin Of Error Calculation a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed relatedl answers to any questions you might have Meta Discuss p h id Standard Error Calculation Excel p the workings and policies of this site About Us\n\nbinary standard error\n\nBinary Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Standard Deviation Binary a li li a href T Test Binary a li li a href Standard Deviation Of Yes No Data a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed answers to any questions relatedl you might have Meta Discuss the workings and policies standard error binary data of this site About Us Learn more about Stack Overflow the company Business p h id Standard Deviation Binary p Learn\n\nbinomial distribution error bars\n\nBinomial Distribution Error Bars table id toc tbody tr td div id toctitle Contents div ul li a href Binomial Proportion Confidence Interval a li li a href Standard Error Binary Distribution a li li a href Binomial Error a li ul td tr tbody table p be calculated for a binary variable In a graph showing the progress over time of the probability to find a pathogen within plant tissues I'm wondering if standard deviation relatedl or standard error bars can be added The probability to find standard error binomial distribution the pathogen is obtained dividing the number of\n\nbinomial standard error formula\n\nBinomial Standard Error Formula table id toc tbody tr td div id toctitle Contents div ul li a href Binomial Distribution Standard Error a li li a href Binomial Sampling Error a li li a href Standard Error Formula Statistics a li li a href Standard Error Of Estimate Formula a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed answers to any questions relatedl you might have Meta Discuss the workings and policies p h id Binomial Distribution Standard Error p of this site About Us Learn more\n\nbinomial standard error matlab\n\nBinomial Standard Error Matlab table id toc tbody tr td div id toctitle Contents div ul li a href Binomial Standard Deviation a li li a href Binomial T Test a li li a href Binomial Central Limit Theorem a li ul td tr tbody table p Search All Support Resources Support Documentation MathWorks Search MathWorks com relatedl MathWorks Documentation Support Documentation Toggle navigation Trial binomial distribution standard error Software Product Updates Documentation Home Statistics and Machine Learning Toolbox standard error matlab regression Examples Functions and Other Reference Release Notes PDF Documentation Probability Distributions Discrete Distributions Binomial Distribution binomial sampling\n\nbinomial standard error mean\n\nBinomial Standard Error Mean table id toc tbody tr td div id toctitle Contents div ul li a href Binomial Standard Error Calculator a li li a href Binomial Sampling Error a li li a href Binomial Probability Standard Deviation a li li a href Binomial Confidence Interval a li ul td tr tbody table p -- it is just scaled by a factor n From the properties of the binomial relatedl distribution its distribution has mean and standard deviation Bias binomial distribution standard error and standard error When the proportion p is used to estimate p h id Binomial\n\nbinomial standard error excel\n\nBinomial Standard Error Excel table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Of Binomial Proportion a li li a href Binomial Sampling Error a li li a href Binomial Confidence Interval a li ul td tr tbody table p BINOMDIST Mean Standard Deviation For Binomial Probability Distribution ExcelIsFun SubscribeSubscribedUnsubscribe K Loading Loading Working Add to Want to watch this relatedl again later Sign in to add this video standard deviation of a binomial distribution in excel to a playlist Sign in Share More Report Need to report the binomial standard error\n\nbinomial distribution standard error of the mean\n\nBinomial Distribution Standard Error Of The Mean table id toc tbody tr td div id toctitle Contents div ul li a href Binomial Error a li li a href Standard Error Binary Distribution a li li a href Sample Variance Bernoulli a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed answers to any questions relatedl you might have Meta Discuss the workings and policies binomial standard error calculator of this site About Us Learn more about Stack Overflow the company Business standard error of binary variable Learn more\n\nbinomial error bars\n\nBinomial Error Bars table id toc tbody tr td div id toctitle Contents div ul li a href What s A Confidence Interval a li li a href Standard Error Binomial Distribution a li li a href Binomial Proportion Confidence Interval a li li a href Bernoulli Standard Deviation a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed answers to any questions relatedl you might have Meta Discuss the workings and policies p h id What s A Confidence Interval p of this site About Us Learn more", "date": "2019-04-25 14:21:47", "meta": {"domain": "winbytes.org", "url": "http://winbytes.org/help/standard-error/proportion-standard-error.html", "openwebmath_score": 0.3121330142021179, "openwebmath_perplexity": 10404.382160598483, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936068886641, "lm_q2_score": 0.9136765281148513, "lm_q1q2_score": 0.899143230082056}}
{"url": "https://math.stackexchange.com/questions/1457701/spectral-decomposition-of-a-and-b", "text": "# Spectral Decomposition of A and B.\n\nI was given the following question in my linear algebra course.\n\nLet $A$ be a symmetric matrix, $c >0$, and $B=cA$, find the relationship between the spectral decompositions of $A$ and $B$.\n\nFrom what I understand. If $A$ is a symmetric matrix, then $A=A^T$. A symmetric matrix has $n$ eigenvalues and there exist $n$ linearly independent eigenvectors (because of orthogonality) even if the eigenvalues are not distinct. Since $B=cA$ and $A=A^T$, then we can conclude that $B=cA^T$, which would imply that $B$ is also symmetric, meaning it also has a linearly independent eigenbasis.\n\nFocusing on $A$, since it has a linearly independent eigenbasis, we have $A = PD_aP^{-1}$ by Spectral decomposition where $P$ is the eigenbasis and $D_a$ is the diagonal matrix of $A$ eigenvalues $\\lambda_i$ \\begin{array} d D_a & = & \\begin{bmatrix} \\lambda_1 & & \\\\ &\\ddots&\\\\ & & \\lambda_i \\end{bmatrix} \\end{array}\n\nNow since $B=cA$, then we have $B=cPD_aP^{-1}$, which can be rewritten as $B = PD_bP^{-1}$, where\n\n\\begin{array} d D_b & = & cD_a & =c\\begin{bmatrix} \\lambda_1 & & \\\\ &\\ddots&\\\\ & & \\lambda_i \\end{bmatrix} & = & \\begin{bmatrix} c\\lambda_1 & & \\\\ &\\ddots&\\\\ & & c\\lambda_i \\end{bmatrix} \\end{array}\n\nFrom this I can conclude that $B$ and $A$ actually have the same linearly independent eigenbasis. Furthermore, the eigenvalues of $B$ are a scalar multiple of the eigenvalues of $A$ by a factor of $c$.\n\nHave I fully describe the relationship between $A$ and $B$?\n\n$\\dfrac{\\lambda}{c} I- A$ isn't invertible if and only if $\\lambda I- cA$ isn't invertible.\nHence, $\\lambda\\in Sp(cA)$ if and only if $\\dfrac{\\lambda}{c}\\in Sp(A)$\nYes, I would say that you have fully described the relationship between $A$ and $B$.", "date": "2022-07-02 12:28:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1457701/spectral-decomposition-of-a-and-b", "openwebmath_score": 0.9967987537384033, "openwebmath_perplexity": 98.99838089290482, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9867771782476948, "lm_q2_score": 0.9111797003640646, "lm_q1q2_score": 0.8991313336018317}}
{"url": "http://math.stackexchange.com/questions/110667/why-binomial-distribution-formula-includes-the-not-happening-probability", "text": "# Why Binomial Distribution formula includes the \u201cnot-happening\u201d probability?\n\nSuppose I have a dice with 6 sides, and I let a random variable $X$ be the number of times I get 3 points when I throw the dice.\n\nSo I throw the dice for $10$ times, I want to find the probability of getting 3 points from the dice for $4$ times, ie: $P(X=4)$.\n\nSince the order doesn't matter, there are $\\binom{10}{4}=210$ ways and the chance of getting a 3 point is $\\frac { 1 }{ 6 }$. Also, because I want to have $4$ of such occurrence, it would be $\\frac{1}{6}^4$. So, I could just calculate $P(X=4)=\\binom{10}{4}\\frac { 1 }{ 6 }^4 =0.027006173\\approx 2.7\\%$.\n\nBut, suppose if I use the Binomial Distribution formula, it would be a little different because it needs to multiply the \"not-happening\" probability to it. The Binomial Distribution looks like this: $$P(X=x)=\\binom{n}{x}p^x(1-p)^{n-x}$$\n\nSo if I plug in my values, it would be: $$P(X=4)=\\binom{10}{4}(\\frac{1}{6})^4(\\frac{5}{6})^{6}=0.054=5.4\\%$$\n\nHere, $2.1\\%$ is lesser than $5.4\\%$. What's the difference between the two values? Which is the correct value?\n\nIntuitively, I find the Binomial Distribution may be more accurate since it dictates the situation to consider both the happening and not-happening outcomes. But usually, I thought we just multiply the probabilities of what we want it to happen as long as the events are independent. So the first method sounds quite okay too. Eg: What's the probability to get 2 heads out of 5 flips of a fair coin, I just use $\\frac { 1}{ 2} \\times \\frac { 1}{ 2}$. The not-happening probabilities are not cared of.\n\n-\nTo get exactly 4 \"3 points\", the other 6 throws have to not be \"3 points\". The other 6 throws not being 3 points is also part of what you want to happen if exactly 4 of the 10 throws are 3 points. So, you need to multiply by $(5/6)^6$. By the way, in your second displayed equation, you should have ${10\\choose 4}(1/6)^4(5/6)^6$ \u2013\u00a0David Mitra Feb 18 '12 at 16:33\noh yea, thanks. It should ${10\\choose 4}(1/6)^4(5/6)^6$ in the second line. Updated the equation. If I need to multiply the \"not-happening\" probability, then suppose if I want to find the probability of getting 2 heads out of 5 flip of a fair coin, it wouldn't it be $\\binom{5}{2}{ \\left( \\frac { 1 }{ 2 } \\right) }^{ 2 }{ \\left( \\frac { 1 }{ 2 } \\right) }^{ 3 }=\\binom{5}{2}{ \\left( \\frac { 1 }{ 2 } \\right) }^{ 5 }$? But shouldn't it be just $\\frac { 1}{ 2} \\times \\frac { 1}{ 2}$? \u2013\u00a0xenon Feb 18 '12 at 16:51\nNo, not just $(1/2)(1/2)$ for exactly two heads. You $need$ the other flips to not be heads; which means you multiply by $(1/2)^3$. \u2013\u00a0David Mitra Feb 18 '12 at 17:10\n@xEnOn: The calculation that gave about $5.4$% is based on the correct analysis, and the calculator work is right. The first calculation, the one that gave about $2.7$%, is based on an incorrect analysis (formula). In addition, there was a calculator mistake. The wrong formula gives an answer of about $16.2$%, not $2.7$%. This is clear if you compare the two expressions. The second multiplies the first by $(5/6)^6$, which is less than $1$. \u2013\u00a0Andr\u00e9 Nicolas Feb 18 '12 at 17:18\n@xEnOn : Please don't write $\\frac 16 ^4$ if you mean $\\left(\\frac16\\right)^4$. The former expression should be used only if you mean $(1^6)/4$. \u2013\u00a0Michael Hardy Feb 18 '12 at 17:53\n\nYour first argument is a bit off. $X=4$ means exactly four of the tosses resulted in $3$-points To get exactly four $3$-points, the other six throws have to not be $3$- points. The other six throws not being $3$-points is also part of what you want to happen if exactly four of the 10 throws are $3$-points.\n\nLook at a particular case: the first four throws are 3-points and there are exactly four 3-points. Then the throw sequence was $$3\\text{-pt}\\ \\ 3\\text{-pt}\\ \\ 3\\text{-pt}\\ \\ 3\\text{-pt}\\ \\ \\text{not}3\\text{-pt}\\ \\ \\text{not}3\\text{-pt}\\ \\ \\text{not}3\\text{-pt}\\ \\ \\text{not}3\\text{-pt}\\ \\ \\text{not}3\\text{-pt}\\ \\ \\text{not}3\\text{-pt}\\ \\$$ The probability of this occurring is $(1/6)^4(5/6)^6$.\n\nSo, you need to multiply your proposed answer by $(5/6)^6$. So, the correct answer is what is given by the Binomial distribution: ${10\\choose4}(1/6)4(5/6)6$.\n\nIn your second argument (of the original post), you are correct assuming that you only drew two cards. If you drew three cards, the probability that exactly two were hearts would be $$\\underbrace{{13\\over52}{12\\over 51}{39\\over 50}}_{ \\text{two hearts, then non heart}}+ \\underbrace{{13\\over52}{39\\over 51}{12\\over 50}}_{ \\text{ heart, non heart, heart}}+ \\underbrace{{39\\over52}{13\\over 51}{12\\over 50}}_{ \\text{ non heart, then two hearts}}$$\n\nTo be brief you do care about the \"not happening\" probabilities when you consider the probability that an event happens exactly $n$ times. Knowing that something does not happen is concrete information. In the die example, \"not 3-pt\" means that six of the flips were 1,2,4,5, or 6 points.\n\nIn your coin example, if you want the probability of exactly two heads in five flips, then the three non-head flips must be tails. What is the probability of flipping $H\\ H\\ T\\ T\\ T$? It's not $1/4$... It is $1/32$, as you should be able to see from the multiplication rule for a sequence of independent events.\n\nThe number of ways in which you can have exactly two heads in five flips is ${5\\choose2}=10$ (it's the number of ways to choose two slots from five in which to put the two heads in; note the other three slots must be tails). The probability of exactly two heads in five flips is ${5\\choose2}\\cdot(1/2)^2(1/2)^3={10\\over 32}={5\\over16}$.\n\nTo see more concretely why your argument fails, consider this simple case. Toss a fair coin three times. What is the probability of having exactly one head? By your reasoning it is $3\\cdot(1/2)^1>1$; which is nonsense. Perhaps it's just $1/2$? This is incorrect also:\n\nThe equally likely outcomes here are: $$HHH \\ \\ \\ HHT \\ \\ \\ HTT\\ \\ \\ HTH$$ $$THH \\ \\ \\ THT \\ \\ \\ TTT\\ \\ \\ TTH$$ And we see the probability of exactly one head is ${3\\over8}$, which is exactly what the Binomial formuls gives: ${3\\over8}={3\\choose2}(1/2)^1(1/2)^2$.\n\nIncidentally, you're proposed method would not even give the probability of having at least one head. Here, that probability is $7/8$.\n\n-", "date": "2016-05-31 02:20:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/110667/why-binomial-distribution-formula-includes-the-not-happening-probability", "openwebmath_score": 0.8770877122879028, "openwebmath_perplexity": 221.62111013827325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982557516796073, "lm_q2_score": 0.9149009544128984, "lm_q1q2_score": 0.8989428098822946}}
{"url": "https://www.physicsforums.com/threads/give-a-big-o-estimate-of-the-product-of-the-first-n-odd-positive-integers.514750/", "text": "# Homework Help: Give a big-O estimate of the product of the first n odd positive integers\n\n1. Jul 17, 2011\n\n### pc2-brazil\n\n1. The problem statement, all variables and given/known data\nGive a big-O estimate of the product of the first n odd\npositive integers.\n\n2. Relevant equations\nBig-O notation:\nf(x) is O(g(x)) if there are constants C and k such that\n|f(x)| \u2264 C|g(x)| whenever x > k.\n\n3. The attempt at a solution\nThe product of the first n odd integers can be given by:\n$$P(n)=1\\times 3\\times 5\\times 7\\times...\\times (2n-1)$$\nFor n > 0, no element in the above sequence will be greater than (2n-1). Thus:\n$$1\\times 3\\times 5\\times 7\\times...\\times (2n-1)\\leq (2n-1)\\times (2n-1)...\\times (2n-1)=(2n-1)^n$$\nSo:\nP(n) \u2264 (2n-1)n whenever n > 0\nI could stop here and say that\nP(n) is O((2n-1)n)\nBut to simplify I think I could consider that:\nP(n) \u2264 (2n-1)n \u2264 (2n)n\nThus,\nP(n) is O((2n)n)\n\nIs this reasoning correct?\n\nLast edited: Jul 17, 2011\n2. Jul 17, 2011\n\n### tiny-tim\n\nhi pc2-brazil!\n\nit's correct, but it's not very accurate, is it?\n\ndo you know a big-O estimate for n! ?\n\n3. Jul 17, 2011\n\n### pc2-brazil\n\nA big-O estimate for n! would be O(nn).\nI could say that, for n > 0,\n$$1\\times 3\\times 5\\times 7...\\times (2n-1)\\leq 1\\times 2\\times 3\\times 4\\times...\\times (2n-1)=(2n-1)!\\leq (2n)!=2^n n!$$\nThus, P(n) is O(2nn!). Since n! is O(nn), this estimate seems more accurate than the previous one (O(2nnn)).\n\n4. Jul 17, 2011\n\n### tiny-tim\n\nLast edited by a moderator: Apr 26, 2017\n5. Jul 17, 2011\n\n### Ray Vickson\n\nIf En = product of the even numbers from 2 to 2n - 2, your product is (2n-1)!/En, and En = 2^(n-1) * (n-1)! Now apply Stirling's formula to both factorials. Note: if you want a true upper bound, rather than just an *estimate* you can use the fact that if St(n) is defined as\nsqrt(2pi)*n^(n + 1/2)*exp(-n), then we have St(n) <n! < St(n)*exp(1/(12n)), even if n is not large. You can use the upper bound on (2n-1)! in the numerator and the lower bound on (n-1)! in the denominator.\n\nRGV\n\nsqrt(2pi)*n", "date": "2018-05-22 16:25:07", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/give-a-big-o-estimate-of-the-product-of-the-first-n-odd-positive-integers.514750/", "openwebmath_score": 0.8235124945640564, "openwebmath_perplexity": 2055.7448349221136, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471647042428, "lm_q2_score": 0.9136765269395709, "lm_q1q2_score": 0.8989180604863166}}
{"url": "https://calspasblog.com/z6oue/viewtopic.php?page=67a3e4-rational-numbers-symbol", "text": "rational numbers symbol\n\n# rational numbers symbol\n\nThe real line consists of the union of the rational and irrational numbers. We have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. Formally, rational numbers are the set of all real numbers that can be written as a ratio of integers with nonzero denominator. The OP asked in his title only, if he can use the symbol for rational numbers. Hence, we can say that \u20180\u2019 is also a rational number, as we can represent it in many forms such as 0/1, 0/2, 0/3, etc. In other words, most numbers are rational numbers. Answer: yes. Some examples of irrational numbers are $$\\sqrt{2},\\pi,\\sqrt[3]{5},$$ and for example $$\\pi=3,1415926535\\ldots$$ comes from the relationship between the length of a circle and its diameter. Numbers that are not rational are called irrational numbers. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. When we put together the rational numbers and the irrational numbers, we get the set of real numbers. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. By using this website, you agree to our Cookie Policy. A rational number is a number that can be expressed as a fraction p/q where p and q are integers and q!=0. Rational numbers are indicated by the symbol . For instance, the decimal version of the therefore symbol (\u2234) would be &\u200c#8756; The hexadecimal version of the therefore symbol (\u2234) would be &\u200c#x2234; Note that the hexadecimal numbers include x as part of the code. In Maths, rational numbers are represented in p/q form where q is not equal to zero. RapidTables. 1 $\\begingroup$ I think your answer is fine. Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number set. $\\endgroup$ \u2013 Dietrich Burde Aug 15 '19 at 18:42. Rational Numbers All positive and negative fractions, including integers and so-called improper fractions. Irrational numbers are a separate category of their own. Many people are surprised to know that a repeating decimal is a rational number. I just disagree that the notations should not be used for fields. Note that the set of irrational numbers is the complementary of the set of rational numbers. Expressed as an equation, a rational number is a number. where a and b are both integers. This equation shows that all integers, finite decimals, and repeating decimals are rational numbers. The denominator in a rational number cannot be zero. ... rational numbers set = \u2026 The ancient greek mathematician Pythagoras believed that all numbers were rational, but one of his students Hippasus proved (using geometry, it is thought) that you could not write the square root of 2 as a fraction, and so it was irrational. a/b, b\u22600. It is also a type of real number. Top of Page The letters R, Q, N, and Z refers to a set of numbers such that: R = real numbers includes all real number [-inf, inf] Q= rational numbers ( numbers written as ratio) The Unicode numeric entity codes can be expressed as either decimal numbers or hexadecimal numbers. A rational number p/q is said to have numerator p and denominator q. Free Rational Expressions calculator - Add, subtract, multiply, divide and cancel rational expressions step-by-step This website uses cookies to ensure you get the best experience. Any fraction with non-zero denominators is a rational number.", "date": "2021-05-12 15:11:26", "meta": {"domain": "calspasblog.com", "url": "https://calspasblog.com/z6oue/viewtopic.php?page=67a3e4-rational-numbers-symbol", "openwebmath_score": 0.8266318440437317, "openwebmath_perplexity": 321.2912625395907, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9935117296348998, "lm_q2_score": 0.9046505267461572, "lm_q1q2_score": 0.8987809095426977}}
{"url": "https://math.stackexchange.com/questions/2119158/conditional-probability-exercise-apples-and-oranges/2119210", "text": "# Conditional probability exercise - apples and oranges\n\nWe have two crates, crate 1 and crate 2. Crate 1 has 2 oranges and 4 apples, and crate 2 has 1 orange and 1 apple. We take 1 fruit from crate 1 and put it in crate 2, and then we take a fruit from crate 2.\n\nThe first point of this exercise asks me to calculate the probability that the fruit taken from crate 2 is an orange. I did this by calculating the probability that the fruit we took from crate 1 was an orange(which is $\\frac{2}{6}$) and then saying that I have 3 fruits in crate 2, $1+\\frac{2}{6}$ oranges and the rest apples, which lead me to a $44.44\\%$ probability that the fruit we take from crate 2 was an orange. The probability I got seems reasonable, but I don't know for sure if what I did was correct.\n\nAnyway, point 2 of this problem is a little bit harder and I'm stuck. It tells me to calculate the probability that the fruit we took from crate 1 was an orange, if we know that the fruit we took out from crate 2 was also an orange. So if I consider A: Fruit taken from crate 1 was an orange, and B: Fruit taken from crate 2 was an orange, I think I have to calculate $\\:P\\left(A|B\\right)$ I think, which means \"Probability that A happens if we know B happened\", but I'm not so sure about this.\n\nCould anyone give me a hint on how to go about solving this problem?\n\n\u2022 Your answer to the first question is correct. For the second question, $$P(A \\mid B) = \\frac{P(A \\cap B)}{P(B)}$$ where $A$ is the event that an orange was selected from the first crate and $B$ is the event that an orange was selected from the second crate. \u2013\u00a0N. F. Taussig Jan 29 '17 at 12:18\n\u2022 But what is $P\\left(A\\cap B\\right)$ ? I mean, I know it is the probability of the joint events, but how do I determine that? \u2013\u00a0MikhaelM Jan 29 '17 at 12:32\n\u2022 It is the probability that you put an orange from the first crate into the second, and having done that then took an orange from the second . $$\\mathsf P(A\\cap B)= \\mathsf P(A)~\\mathsf P(B\\mid A)$$ \u2013\u00a0Graham Kemp Jan 29 '17 at 12:34\n\u2022 Doing that I get $\\:P\\left(A|B\\right)=\\frac{P\\left(A\\cap B\\right)}{P\\left(B\\right)}=\\frac{P\\left(A\\right)\\cdot P\\left(B|A\\right)}{P\\left(B\\right)}=\\frac{\\frac{2}{6}\\cdot \\frac{2}{3}}{\\frac{4}{9}}=\\frac{1}{2}$ so $50\\%$. Is this correct or am I making a mistake somewhere? \u2013\u00a0MikhaelM Jan 29 '17 at 12:45\n\u2022 The answer you obtained in your comment is correct. \u2013\u00a0N. F. Taussig Jan 29 '17 at 13:15\n\nLet $A,B$ be the events of removing an orange from the first and second crates, respectively.\n\nYou have calculated $\\mathsf P(B) = 4/9$ correctly. Another way to look at it is through the law of total probability. \\begin{align}\\mathsf P(B) ~&=~\\mathsf P(A)~\\mathsf P(B\\mid A)+\\mathsf P(A^\\complement)~\\mathsf P(B\\mid A^\\complement) \\\\ &=~ \\tfrac 26\\cdot\\tfrac 23+\\tfrac 46\\cdot\\tfrac 13 & =&~ \\frac{\\tfrac 2 6+1}3 \\\\ &=~ \\tfrac 49 \\end{align}\n\nWhere $\\mathsf P(A)$ is the probability of taking an orange from cart 1, $\\mathsf P(A^\\complement)$ is that of taking an apple from cart 1, $\\mathsf P(B\\mid A)$ is the probability of taking a orange from cart 2 when given that you have added an orange to that cart, and $\\mathsf P(B\\mid A^\\complement)$ is the probability of taking a orange from cart 2 when given that you have added an apple to that cart.\n\nNow you just need to calculate $\\mathsf P(A\\mid B)$ the probability of having taken an orange from cart 1 when given that you took an orange from cart 2.\n\nUse Bayes' Rule.\n\nYes for second part solve P(A|B).\n\nProbability of A happens if we know B happened.", "date": "2020-12-02 09:59:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2119158/conditional-probability-exercise-apples-and-oranges/2119210", "openwebmath_score": 0.6868855953216553, "openwebmath_perplexity": 136.17207693072405, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587261496031, "lm_q2_score": 0.909907012756525, "lm_q1q2_score": 0.8987685918349758}}
{"url": "https://math.stackexchange.com/questions/2844980/general-solution-for-a-recurrence-relation", "text": "# general solution for a recurrence relation\n\nI have the following recurrence relation: $$x_1=1, x_2=a, x_{n+2}=ax_{n+1}-x_n\\hspace{1cm}(*)$$\n\nIf we assume that $x_n=r^n$ is a solution for the relation $x_{n+2}=ax_{n+1}-x_n$, then I can deduce that $r=\\frac{a+\\sqrt{a^2-4}}{2}$ or $r=\\frac{a-\\sqrt{a^2-4}}{2}$.\n\nBy using the initial values $x_1=1, x_2=a$, I found that $$x_n=\\frac{1}{\\sqrt{a^2-4}}\\left(\\frac{a+\\sqrt{a^2-4}}{2}\\right)^n-\\frac{1}{\\sqrt{a^2-4}}\\left(\\frac{a-\\sqrt{a^2-4}}{2}\\right)^n$$ is a solution for the recurrence relation (*).\n\nQuestion: How do we know whether this is the only solution for the recurrence relation $(*)$? Notice that when I found the particular solution above I assumed that the solution was a linear combination of geometric series. But I do not know if all the solutions will have this form.\n\n\u2022 Use method of difference on the recursive definition $x_{n+2}-x_{n+1}=(a-2)x_{n+1}+x_{n+1}-x_{n}$ and add these equations for all values of n. \u2013\u00a0Pi_die_die Jul 8 '18 at 20:57\n\u2022 In other words, prove that it satisfies condition $(*)$. \u2013\u00a0steven gregory Jul 8 '18 at 20:59\n\u2022 Yes you would get the sum which would be the type of sum you would expect from a gp \u2013\u00a0Pi_die_die Jul 8 '18 at 21:01\n\nYou have specific initial values $x_0$ and $x_1$. The recurrence relation fully determines all other values. Thus, there's exactly one solution. If you've found it, that's it, there can't be any others, no matter what approach you took in order to find it.\n\nIf you want to show that all solutions of $x_{n+2}=ax_{n+1}-x_n$ are combinations of the geometric series you found, you can argue as follows:\n\nThis is a linear second-order recurrence. Its solutions form a two-dimensional vector space. (A vector space because of the linearity of the recurrence, and a two-dimensional one because the solution space is spanned by the two solutions for the initial conditions $x_1=1$, $x_2=0$ and $x_1=0$, $x_2=1$.) You've found two linearly independent solutions; hence they span the entire solution space.\n\n\u2022 Hi @joriki I would appreciate some clarifications before I accept your answer. I understand that the set of solutions is a subspace of the vector space of sequences $\\{f:\\mathbb{N}\\to\\mathbb{C}: f\\mbox{ is a function}\\}$. I also understand that I found two solutions that are linearly independent. I do not know why the dimension of the solution space is 2 and not greater than 2. Also, what do you mean by $a_0$ and $a_1$? \u2013\u00a0Chilote Jul 8 '18 at 21:23\n\u2022 @Chilote: I'm sorry, I meant $x_1$ and $x_2$; I've fixed that. The solution for given initial conditions $x_1=\\hat x_1$ and $x_2=\\hat x_2$ is $\\hat x_1s_1+\\hat x_2 s_2$, where $s_1$ is the solution for $x_1=1$ and $x_2=0$ and $s_2$ is the solution for $x_1=0$ and $x_2=1$; so $s_1$ and $s_2$ span the solution space; hence it's at most two-dimensional (in fact exactly two-dimensional, since $s_1$ and $s_2$ are linearly independent). \u2013\u00a0joriki Jul 8 '18 at 21:31\n\u2022 Got it, the solution space of $x_{n+2}=ax_{n+1}-x_n$ is $S=\\{(x_n)_n:x_{n+2}=ax_{n+1}-x_n\\}$ which is a vector space. Any vector in this space is of the form $x=(x_1,x_2,ax_2-x_1,a(ax_2-x_1)-x_2,\\dots)=x_1 s_1+x_2 s_2$ where $s_1$ and $s_2$ are two particular solutions that are linearly independent. \u2013\u00a0Chilote Jul 8 '18 at 21:49\n\u2022 You've found two linearly independent solutions Just to nitpick, that's iff $a \\ne \\pm 2$. \u2013\u00a0dxiv Jul 8 '18 at 23:29\n\u2022 @Chilote: This was actually overkill, since you only asked about the uniqueness of one particular solution, not about all solutions being combinations of the geometric series. I edited the answer accordingly. \u2013\u00a0joriki Jul 9 '18 at 5:22\n\nA bit of culture. Any two consecutive numbers in your sequence, call them $x_n$ and $x_{n+1},$ satisfy $$x_n^2 - a \\, x_n \\, x_{n+1} + x_{n+1}^2 = 1$$\n\nTry consecutive values in $$1, \\; \\; a, \\; \\; a^2 - 1, \\; \\; a^3 - 2a, \\; \\ldots$$\n\nThis comes from the matrix $$\\left( \\begin{array}{cc} 0 & 1 \\\\ -1 & a \\end{array} \\right)$$ which has determinant $1$ and trace $a.$ It also gives an automorphism of the quadratic form $x^2 - a xy+y^2.$ An automorphism matrix $P$ for a quadratic form that has Hessian matrix $H$ satisfies $P^T HP = H$\n\nIn this case $$\\left( \\begin{array}{cc} 0 & -1 \\\\ 1 & a \\end{array} \\right) \\left( \\begin{array}{cc} 2 & -a \\\\ -a & 2 \\end{array} \\right) \\left( \\begin{array}{cc} 0 & 1 \\\\ -1 & a \\end{array} \\right) = \\left( \\begin{array}{cc} 2 & -a \\\\ -a & 2 \\end{array} \\right)$$\n\nThe explicit relation with the sequence is\n\n$$\\left( \\begin{array}{cc} 0 & 1 \\\\ -1 & a \\end{array} \\right) \\left( \\begin{array}{c} x_n \\\\ x_{n+1} \\end{array} \\right) = \\left( \\begin{array}{c} x_{n+1} \\\\ x_{n+2} \\end{array} \\right)$$\n\n\u2022 How can I get the first equality? \u2013\u00a0Chilote Jul 8 '18 at 21:52\n\u2022 @Chilote the automorphism group for an (indefinite) binary quadratic form, along with the Pell type equation $\\tau^2 - \\Delta \\sigma^2 = 4,$ is discussed in many number theory and quadratic forms books. \u2013\u00a0Will Jagy Jul 8 '18 at 23:24\n\u2022 I think the quadratic form corresponding to the matrix you wrote above is $ay^2$ \u2013\u00a0Chilote Jul 8 '18 at 23:59\n\u2022 @Chilote I put in a little about the automorphism. The matrix that gives the quadratic form is the symmetric one; it is just the Hessian matrix of second partial derivatives \u2013\u00a0Will Jagy Jul 9 '18 at 0:23", "date": "2019-05-26 21:42:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2844980/general-solution-for-a-recurrence-relation", "openwebmath_score": 0.8752542734146118, "openwebmath_perplexity": 176.84608042968154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631675246404, "lm_q2_score": 0.9111797088058519, "lm_q1q2_score": 0.8987541037619194}}
{"url": "https://afd-hamburg-nord.de/forum/archive.php?53e5c2=block-triangular-matrix-eigenvalues", "text": "Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u00e2\u0080\u00a6 In mathematics, a block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. If P A Ais nonsingular then the eigenvectors of P 1 U Acorresponding to are of the form [0 T;vT] where v is any eigenvector of P 1 S Cthat corresponds to its unit eigenvalue. T is diagonal iff A is symmetric. Every square real matrix A is orthogonally similar to an upper block triangular matrix T with A=Q T TQ where each block of T is either a 1#1 matrix or a 2#2 matrix having complex conjugate eigenvalues. Moreover, the eigenvectors of P 1 U Acorresponding to are of the form [uT;((P S+ C) 1Bu) T] . Block lower triangular matrices and block upper triangular matrices are popular preconditioners for $2\\times 2$ block matrices. Moreover, the eigenvectors of P 1 Based on the lemma, we can derive the following main results about the SBTS iteration method. Yes. First of all: what is the determinant of a triangular matrix? However, a 2 by 2 symmetric matrix cannot have imaginary eigenvalues, so R must be diagonal. The second consequence of Schur\u00e2\u0080\u0099s theorem says that every matrix is similar to a block-diagonal matrix where each block is upper triangular and has a constant diagonal. The determinant of a block-diagonal matrix is the product of the determinants of the blocks, so, by considering the definition of the characteristic polynomial, it should be clear that the eigenvalues of a block-diagonal matrix are the eigenvalues of the blocks. Assume that \u00ce\u00b1 is a positive constant and S = W \u00e2\u0088\u0092 1 T. These eigenvectors form an orthonormal set. Theorem 3.2. Then the eigenvalues of the matrix S = W \u00e2\u0088\u0092 1 T are all real, and S is similar to a diagonal matrix. 2 AQ = Q\u00ce\u009b A(Qe i)=(Qe i)\u00ce\u00bb i Qe i is an eigenvector, and \u00ce\u00bb i is eigenvalue. TRIANGULAR PRECONDITIONED BLOCK MATRICES 3 P 1 A Athat corresponds to its unit eigenvalue. Theorem 6. This is an important step in a possible proof of Jordan canonical form. Developing along the first column you get $a_{11} \\det(A_{11}'),$ where $A_{11}'$ is the minor you get by crossing out the first row and column of [math]A. This method can be impractical, however, due to the contamination of smaller eigenvalues by This decouples the problem of computing the eigenvalues of Ainto the (solved) problem of computing 1, and then computing the remaining eigenvalues by focusing on the lower right (n 1) (n 1) submatrix. Let W, T \u00e2\u0088\u0088 R n \u00d7 n be symmetric positive definite and symmetric, respectively. Hence R is symmetric block diagonal with blocks that either are 1 by 1 or are symmetric and 2 by 2 with imaginary eigenvalues. upper-triangular, then the eigenvalues of Aare equal to the union of the eigenvalues of the diagonal blocks. If each diagonal block is 1 1, then it follows that the eigenvalues of any upper-triangular matrix are the diagonal elements. 1 is a matrix with block upper-triangular structure. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. In this note we show that a block lower triangular preconditioner gives the same spectrum as a block upper triangular preconditioner and that the eigenvectors of the two preconditioned matrices are related.\nSummer Film 2019, Denon Avr-x2600h Bundle, Heavy Duty Workbench Brackets, Naruto Vs Pain Episodes, Saturday Kitchen Vote, Rent Townhouse By Owner, Book Template Word,", "date": "2021-04-11 06:33:15", "meta": {"domain": "afd-hamburg-nord.de", "url": "https://afd-hamburg-nord.de/forum/archive.php?53e5c2=block-triangular-matrix-eigenvalues", "openwebmath_score": 0.786680281162262, "openwebmath_perplexity": 430.25903240845196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9949396688683126, "lm_q2_score": 0.9032942008463507, "lm_q1q2_score": 0.8987232330807353}}
{"url": "https://brilliant.org/discussions/thread/a-seemingly-impossible-problem/", "text": "# A Seemingly Impossible Problem\n\nGiven that $w, x, y, z$ take on values $0$ and $1$ with equal probability, what is the probability that $w+x+y+z$ is odd?\n\nWhich of the following arguments is correct?\n\nFurthermore, can you generalize this result?\n\nArgument 1:\nIf all 4 numbers are even, the sum is even.\nIf 3 numbers are even, the sum is odd.\nIf 2 numbers are even, the sum is even.\nIf 1 number is even, the sum is odd.\nIf 0 numbers are even, the sum is even.\nIn 2 of the 5 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\\frac{ 2}{5}$.\n\nArgument 2:\n$w + x + y + z$ is either odd or even.\nIn 1 of the 2 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\\frac{ 1}{2}$.\n\nArgument 3:\nBy listing out all the $2^4$ cases, we find that there are ${ 4 \\choose 0} + { 4 \\choose 2} + { 4 \\choose 4 } = 8$ cases with an even sum, and ${ 4 \\choose 1} + {4 \\choose 3} = 8$ cases with an odd sum.\nHence, $w + x +y + z$ is odd with probability $\\frac{ 1}{2}$.\n\nContext: In a recent problem, Argument 1 was given, and many people agreed with it.\n\nNote by Calvin Lin\n5\u00a0years, 5\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\u2022 Stay on topic \u2014 we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nThe problem with the first argument is that the probabilities of each of the cases is different. In reality, the first case has one way, the second has 4 ways, the third has 6 ways, the fourth has 4 ways, and the last one has only 1 way. Thus the actual probability is $\\dfrac{4+4}{1+4+6+4+1}=\\dfrac{1}{2}$.\n\n- 5\u00a0years, 5\u00a0months ago\n\nI disagree with Argument 1:\n\nIn case 1 there is only 1 outcome to make an even sum.\n\nIn case 2 there are 4 outcomes that make an odd sum\n\nIn case 3 there are 6 outcomes that make an even sum\n\nIn case 4 there are 4 outcomes that make an odd sum\n\nIn case 5 there is only 1 outcome to make an even sum.\n\nThus out of the 16 equally likely outcomes 8 will yield an odd sum and 8 will yield an even sum.\n\nTherefore the probability that w + x + y + z is odd is 1/2.\n\n- 5\u00a0years, 5\u00a0months ago\n\nReminds me of Bertrand paradox, where three ways of choosing a seemingly the same uniformly random chord on a circle give different probability distributions (which means they are different).\n\n- 5\u00a0years, 5\u00a0months ago\n\nI am not very sure if these two are similar indeed. The problem stated above seems to have objective solution and other two are simply counting errors.\n\n- 5\u00a0years, 3\u00a0months ago\n\nThey are not similar, yes. It's just the fact that there are three solutions arriving at different answers immediately reminded me to that.\n\n- 5\u00a0years, 3\u00a0months ago\n\nWell, this is how i solve this problem.\n\nAssume that we choose w, x, y first.\n\nw + x + y = A.\n\nDoesn't matter whether A is odd or even. z will be the one that decide whether the sum of w, x, y, z (assume w +x+y+z= B) is odd or even\n\n=> With that the possibility for B to be odd = 50%\n\n=> Argument 1: Wrong\n\nArgument 2: Correct (but lack at explaining)\n\nArgument 3: Correct\n\n- 5\u00a0years, 5\u00a0months ago\n\nWhile argument 2 produces the correct numerical answer, the logic presented is incorrect.\n\nSee @SAMARTH M.O. Comment.\n\nStaff - 5\u00a0years, 5\u00a0months ago\n\nArgument 1 is wrong because it assumes they all happen with equal probability, which they don't.\n\n- 5\u00a0years, 5\u00a0months ago\n\nArgument 3 is arguably the right one. Arguments 1 and 2 assume that their mentioned respective events are equally likely.\n\n- 5\u00a0years, 5\u00a0months ago\n\nWas this problem perhaps my problem All That Glitters Is Gold?\n\n- 5\u00a0years, 5\u00a0months ago\n\nNo. I like your problem, which highlights a common mistake made by those who are first introduced to probability.\n\nThe problem that I was referencing had a similar title to my note. It has since been deleted.\n\nStaff - 5\u00a0years, 5\u00a0months ago\n\nDude Calvin, I just found this. You didn't even tag me in it! But thanks, it's still awesome! :D\n\n- 5\u00a0years, 5\u00a0months ago\n\nThe last argument is correct. In general for n numbers the probability is $\\frac {{2}^{n-1}}{{2}^n} = \\frac {1}{2}$\n\n- 5\u00a0years, 5\u00a0months ago\n\nargument 3 feels more logical\n\n- 5\u00a0years, 5\u00a0months ago\n\n.5\n\n- 5\u00a0years, 5\u00a0months ago\n\nArgument 3 is perfect.\n\n- 5\u00a0years, 5\u00a0months ago\n\n1\\2\n\n- 5\u00a0years, 5\u00a0months ago\n\n2\n\n- 5\u00a0years, 5\u00a0months ago\n\nObviously universal space= 2^4 Sample space= 4C0+ 4C2+ 4C4= 8 So probability= 8/16 =1/2\n\n- 5\u00a0years, 5\u00a0months ago\n\nlet a(n) = P(Bin(n,0.5) is even), then a(n+1) = 0.5a(n) + (1-0.5)(1-a(n)) = 0.5\n\n- 5\u00a0years, 5\u00a0months ago\n\nWe have 2^{4} case. Odd sum is 2.\\frac{4!}{3!}=2^{3} case. There for sum is odd with probability \\frac{1}{2}\n\n- 5\u00a0years, 4\u00a0months ago\n\nAn even easier argument which is easily adapter to be generalisation for the number of: picking the first 3 numbers randomly, you'll find either an even or an uneven number. The last term will change it to either even or uneven, with an equal chance of either.\n\nThis will generalise to any number of variables easily, and also adapt to the odds of the result being k (mod n) if your variables can have values 1, 2, ... , n.\n\nOne problem it can't help with is: if the variables are binary, what are the odds of the result being a multiple of k (for a k greater than 2).\n\n- 5\u00a0years, 4\u00a0months ago\n\n1\n\n- 5\u00a0years, 4\u00a0months ago\n\nThe first arguement is wrong as each case doesnot have equal probability .. The second solution (though gives right answer ) is an insufficient one. ARGUEMENT 3is descent\n\n- 4\u00a0years, 7\u00a0months ago\n\nThe second solution doesn't give a correct argument. The point isn't to \"get to the correct answer\". The point is to \"substantiate your argument\".\n\nStaff - 4\u00a0years, 7\u00a0months ago\n\nTotal number of cases = 222*2 Even cases are If number of ONEs is 2 or 4, So Favourable number of cases : 4P2 + 4P4 = 6 + 1 = 7 So Probability for odd is (16 - 7)/16 = 9/16\n\n- 4\u00a0years, 6\u00a0months ago\n\nThe answer is 1/2.\n\nWhen dealing with even cases, you forgot that 0 is an even number.\n\nStaff - 4\u00a0years, 6\u00a0months ago\n\nArgument 1 is incorrect because each case is not equally likely. Argument 2 gives the correct answer but for the wrong reason. Consider the probability of drawing the Ace of spades from a deck of cards. The card drawn is either the ace or it is not. One out of two cases is favourable hence 1/2. (?) Argument 3 would be correct if it was asserted that each case was equally probable. Which of course it is. J\n\n- 4\u00a0years, 6\u00a0months ago\n\nGreat :)\n\nStaff - 4\u00a0years, 6\u00a0months ago\n\nThe issue with the first argument is that of frequency distribution\n\n- 5\u00a0years, 5\u00a0months ago\n\nnow since every single variable counts and is equally important we should be going with the the third one .. ARGUMENT 3 is correct .... nearly similar way to solve is .. there are two possibilities:-\n\n1.) - three variables are 1\n\n2.)- only one variable is 1 .. both ways we get odd\n\nhence- 4C3[ any three numbers] (1/2)^4[the probability of getting 1 is 1/2 and zero is also the same AND we need to get 1 in three cases AS WELL AS 0 in the other case] => 4C3 * (1/2)^4 +4C1[only one variable](1/2)^4 AND HENCE\n\n4C3 * (1/2)^4 + 4C1 * (1/2)^4 = 1/2 ..\n\n- 5\u00a0years, 5\u00a0months ago\n\n1\n\n- 5\u00a0years, 5\u00a0months ago\n\n10\n\n- 5\u00a0years, 5\u00a0months ago", "date": "2019-10-19 00:53:07", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/a-seemingly-impossible-problem/", "openwebmath_score": 0.968029260635376, "openwebmath_perplexity": 1416.1759739498716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9759464485047916, "lm_q2_score": 0.9207896769778074, "lm_q1q2_score": 0.8986414150663654}}
{"url": "http://math.stackexchange.com/questions/79446/what-is-the-remainder-of-1420101-div-6", "text": "What is the remainder of $(14^{2010}+1) \\div 6$?\n\nWhat is the remainder of $(14^{2010}+1) \\div 6$?\n\nSomeone showed me a way to do this by finding a pattern, i.e.:\n\n$14^1\\div6$ has remainder 2\n$14^2\\div6$ has remainder 4\n$14^3\\div6$ has remainder 2\n$14^4\\div6$ has remainder 4\n\nAnd it seems that when the power is odd, the answer is 2, and when it's even, the answer is 4.\n\n2010 is even, so the remainder is 4, but we have that +1, so the final remainder is 5. Which is correct.\n\nBut this method doesn't seem very concrete to me, and I have a feeling the pattern may not be easy to find (or exist?) for every question. What theorem or algorithm can I use to solve this instead?\n\n-\n\n$14\\equiv 2\\pmod 6$, so $14^{2010}+1\\equiv 2^{2010}+1\\pmod 6$. Now $2\\cdot 2^2=2^3=8\\equiv 2\\pmod 6$, so $2\\cdot (2^2)^k\\equiv 2\\pmod 6$ for any non-negative integer $k$. This shows that the pattern that you observed is real: $2^{2k+1} \\equiv 2\\pmod 6$ for any non-negative integer $k$. In particular, $$2^{2010}+1\\equiv 2\\cdot 2^{2009}+1 \\equiv 2\\cdot 2+1 \\equiv 5\\pmod 6\\;.$$\n\nThe same basic idea can be used in similar problems, though the cycle of the pattern may not be nearly so short. To go much deeper than this kind of analysis, you want to look into the Chinese Remainder Theorem.\n\n-\n\n$14=0\\pmod{2}$ and $14=2\\pmod{3}$. Thus, because $2^2=1\\pmod{3}$, for any $k>0$, we have $$14^k=0\\pmod{2}$$ and $$14^k=\\left\\{\\begin{array}{}1\\pmod{3}&\\text{when }k=0\\pmod{2}\\\\2\\pmod{3}&\\text{when }k=1\\pmod{2}\\end{array}\\right.$$ Therefore, for any $k>0$, we have by the Chinese Remainder Theorem, $$14^k=\\left\\{\\begin{array}{}4\\pmod{6}&\\text{when }k=0\\pmod{2}\\\\2\\pmod{6}&\\text{when }k=1\\pmod{2}\\end{array}\\right.$$ So, as you surmised, $14^{2010}+1=5\\pmod{6}$.\n\n-\n\nTo solve this problem, in general, I'd use use two facts:\n\n1. If $a$ and $n$ are relatively prime, $a^{\\phi(n)} \\bmod n = 1$. Here $\\phi$ is the Euler phi function; when $n$ is prime, $\\phi(n) = n-1$. This lets you reduce exponents to ones with power less than $\\phi(n)$; for instance $$2^{2010} \\bmod 5 = (2^4)^{502} 2^2 \\bmod 5 = 1^{502} 4 \\bmod 5 = 4.$$ (Here $2^2$ happened to be trivial to compute by hand; if the exponent were larger I would calculate it using the technique of repeated squaring.)\n\n2. Unfortunately 2 and 6 are not relatively prime; in this case I would factor $6$ into primes and use the Chinese remainder theorem: $$2^{2010} \\bmod 3 = (2^2)^{1005} \\bmod 3 = 1$$ $$2^{2010} \\bmod 2 = 0$$ so applying the Chinese remainder theorem, $2^{2010}$ must be congruent to 4 mod 6.\n\n-\n\nWe may write following equalities :\n\n$14^{2010}=6\\cdot k_1+4$\n\n$14^{2010}+1=6\\cdot k_2+r$\n\n$6\\cdot k_1+4+1=6\\cdot k_2+r\\Rightarrow 6k_1+5=6k_2+r$\n\nThe last equality is true only if $k_1=k_2$ and $r=5$\n\n-\nPlease excuse my ignorance, but it looks to me like you assumed that $14^{2010} = 4 \\mod 6$, and then use this to 'prove' that $14^{2010} + 1 = 5 \\mod 6$. How do you justify the first equality given $k_1$ is an integer? \u2013\u00a0 tom Nov 6 '11 at 9:01\n@tom,read carefully text of the question... \u2013\u00a0 pedja Nov 6 '11 at 10:03\n\nHINT $\\rm\\quad (m,n) = 1,\\ m\\:|\\:a,\\ n\\:|\\:a+1\\ \\ \\Rightarrow\\ \\ a^{2\\:k}\\ \\equiv\\ m\\:(m^{-1}\\ mod\\ n)\\ \\pmod{mn}\\$ by CRT.\n\nTherefore for $\\rm\\:m,n,a\\ =\\ 2,3,14\\:$ we infer $\\rm\\: 14^{\\:2\\:k} \\equiv\\: 2\\ (2^{-1}\\ mod\\ 3)\\equiv -2\\pmod 6$\n\n-", "date": "2015-04-27 02:07:07", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/79446/what-is-the-remainder-of-1420101-div-6", "openwebmath_score": 0.782090425491333, "openwebmath_perplexity": 175.36066282774567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.980875959894864, "lm_q2_score": 0.9161096227509861, "lm_q1q2_score": 0.8985899055847951}}
{"url": "https://math.stackexchange.com/questions/72589/whats-the-probability-of-at-least-and-exactly-one-event-occurring", "text": "# What's the probability of \u201cat least\u201d and \u201cexactly\u201d one event occurring?\n\nIf I know the probability of event $A$ occurring and I also know the probability of $B$ occurring, how can I calculate the probability of \"at least one of them\" occurring?\n\nI was thinking that this is $P(A \\text{ or } B) = P(A) + P(B) - P(A \\text{ and }B)$.\n\nIs this correct?\n\nIf it is, then how can I solve the following problem taken from DeGroot's Probability and Statistics:\n\nIf $50$ percent of families in a certain city subscribe to the morning newspaper, $65$ percent of the families subscribe to the afternoon newspaper, and $85$ percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?\n\nIn a more mathematical language, we are given $P(\\text{morning})=.5$, $P(\\text{afternoon})=.65$, $P(\\text{morning or afternoon}) = .5 + .65 - P(\\text{morning and afternoon}) = .85$, which implies that $P(\\text{morning and afternoon}) = .3$, which should be the answer to the question.\n\nIs my reasoning correct?\n\nIf it is correct, how can I calculate the following?\n\nIf the probability that student $A$ will fail a certain statistics examination is $0.5$, the probability that student $B$ will fail the examination is $0.2$, and the probability that both student $A$ and student $B$ will fail the examination is $0.1$, what is the probability that exactly one of the two students will fail the examination?\n\nThese problems and questions highlight the difference between \"at least one of them\" and \"exactly one of them\". Provided that \"at least one of them\" is equivalent to $P(A \\text{ or } B)$, but how can I work out the probability of \"exactly one of them\"?\n\nYou are correct.\n\nTo expand a little: if $A$ and $B$ are any two events then\n\n$$P(A\\textrm{ or }B) = P(A) + P(B) - P(A\\textrm{ and }B)$$\n\nor, written in more set-theoretical language,\n\n$$P(A\\cup B) = P(A) + P(B) - P(A\\cap B)$$\n\nIn the example you've given you have $A=$ \"subscribes to a morning paper\" and $B=$ \"subscribes to an afternoon paper.\" You are given $P(A)$, $P(B)$ and $P(A\\cup B)$ and you need to work out $P(A\\cap B)$ which you can do by rearranging the formula above, to find that $P(A\\cap B) = 0.3$, as you have already worked out.\n\n\u2022 \"Exactly one of A and B\" means \"Either A or B, but not both\" which you can calculate as P(A or B) - P(A and B). \u2013\u00a0Chris Taylor Oct 14 '11 at 11:13\n\u2022 Are you asking about the notation itself, or the method of displaying the notation? To write the notation we use $\\LaTeX$ - you can find a tutorial by searching for \"latex tutorial\" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read more complex texts, you will become more and more familiar with the notation. \u2013\u00a0Chris Taylor Oct 14 '11 at 11:21\n\u2022 So if I download LaTeX and paste your notation then it displays it in a more readable form? \u2013\u00a0upabove Oct 14 '11 at 11:24\n\nFor your second question, you know $\\Pr(A)$, $\\Pr(B)$, and $\\Pr(A \\text{ and } B)$, so you can work out $\\Pr(A \\text{ and not } B)$ and $\\Pr(B \\text{ and not } A)$ by taking the differences. Then add these two together.\n\nAlternatively take $\\Pr(A \\text{ or } B) - \\Pr(A \\text{ and } B)$.\n\n\u2022 Pr(A and not B) + Pr(B and not A) is not the same as Pr(A) + Pr(B) - Pr(A and B) \u2013\u00a0Petr Peller Apr 4 '15 at 17:44\n\u2022 Pr(A and not B) + Pr(B and not A) + Pr(A and B) is what you are looking for, but calculating Pr(A) + Pr(B) - Pr(A and B) seems to be much easier. \u2013\u00a0Petr Peller Apr 4 '15 at 18:06\n\u2022 $\\Pr(A \\text{ and not } B)+\\Pr(B \\text{ and not } A)$ is the answer to \"but how can I work out the probability of exactly one of them?\" which is what I meant by \"your second question\" \u2013\u00a0Henry Apr 4 '15 at 18:08\n\nFor the additional problem: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection, equals $$P(A)+P(B)-2P(A\\cap B)$$\n\nprobability of only one event occuring is as follows: if A and B are 2 events then probability of only A occuring can be given as P(A and B complement)= P(A) - P(A AND B )", "date": "2020-02-19 04:44:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/72589/whats-the-probability-of-at-least-and-exactly-one-event-occurring", "openwebmath_score": 0.819298505783081, "openwebmath_perplexity": 273.2812882762553, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120914294177, "lm_q2_score": 0.905989826094673, "lm_q1q2_score": 0.8985716642327322}}
{"url": "https://charlottegroutars.nl/f960gm/9a14f6-matrix-multiplied-by-its-conjugate-transpose", "text": "One property I am aware of is that $AA^H$ is Hermitian, i.e. An matrix can be multiplied on the right by an matrix, where is any positive integer. If you want to discuss contents of this page - this is the easiest way to do it. It only takes a minute to sign up. To learn more, see our tips on writing great answers. The difference of a square matrix and its conjugate transpose ( A \u2212 A H ) {\\displaystyle \\left(A-A^{\\mathsf {H}}\\right)} is skew-Hermitian (also called antihermitian). What special properties are possessed by $AA^H$, where $^H$ denotes the conjugate transpose? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In this representation, the conjugate of a quaternion corresponds to the transpose of the matrix. Wikidot.com Terms of Service - what you can, what you should not etc. Thanks for contributing an answer to Mathematics Stack Exchange! Matrix multiplication error in conjugate transpose. How to create a geometry generator symbol using PyQGIS, Does fire shield damage trigger if cloud rune is used. Yes. Check out how this page has evolved in the past. Learn more about multiplication error, error using *, incorrect dimensions After 20 years of AES, what are the retrospective changes that should have been made? The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. A matrix math implementation in python. To perform elementwise Note that A \u2217 represents A adjoint, i.e. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. A ComplexHermitianMatrix that is the product of this ComplexDenseMatrix with its conjugate transpose. We are about to look at an important theorem which will give us a relationship between a matrix that represents the linear transformation $T$ and a matrix that represents the adjoint of $T$, $T^*$. Hot Network Questions Can you make a CPU out of electronic components drawn by hand on paper? Here are the matrices: And here is what I am trying to calculate: The conjugate transpose can be motivated by noting that complex numbers can be usefully represented by 2\u00d72 real matrices, obeying matrix addition and multiplication: A square complex matrix whose transpose is equal to the matrix with every entry replaced by its complex conjugate (denoted here with an overline) is called a Hermitian matrix (equivalent to the matrix being equal to its conjugate transpose); that is, A is Hermitian if {\\displaystyle \\mathbf {A} ^ {\\operatorname {T} }= {\\overline {\\mathbf {A} }}.} Definition of Spectral Radius / Eigenvalues of Product of a Matrix and its Complex Conjugate Transpose 1 Properties of the product of a complex matrix with its complex conjugate transpose Matrix Transpose. At whose expense is the stage of preparing a contract performed? Find out what you can do. The conjugate transpose of A is also called the adjoint matrix of A, the Hermitian conjugate of A (whence one usually writes A \u2217 = A H). Are there any other special properties of $AA^H$? Change the name (also URL address, possibly the category) of the page. Solving a matrix equation involving transpose conjugates. Making statements based on opinion; back them up with references or personal experience. Two matrices can only be added or subtracted if they have the same size. Check that the number of columns in the first matrix matches the number of rows in the second matrix. A = [ 7 5 3 4 0 5 ] B = [ 1 1 1 \u2212 1 3 2 ] {\\displaystyle A={\\begin{bmatrix}7&&5&&3\\\\4&&0&&5\\end{bmatrix}}\\qquad B={\\begin{bmatrix}1&&1&&1\\\\-1&&3&&2\\end{bmatrix}}} Here is an example of matrix addition 1. eigenvalues of sum of a matrix and its conjugate transpose, Solving a matrix equation involving transpose conjugates. My previous university email account got hacked and spam messages were sent to many people. When 2 matrices of order (m\u00d7n) and (n\u00d7m) (m \u00d7 n) and (n \u00d7 m) are multiplied, then the order of the resultant matrix will be (m\u00d7m). site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. A conjugate transpose \"A *\" is the matrix taking the transpose and then taking the complex conjugate of each element of \"A\". Remarks. So if A is just a real matrix and A satisfies A t A = A A t, then A is a normal matrix, as the complex conjugate transpose of a real matrix is just the transpose of that matrix. Why do jet engine igniters require huge voltages? rev\u00a02021.1.18.38333, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Properties of the Product of a Square Matrix with its Conjugate Transpose. Some applications, for example the solution of a least squares problem using normal equations, require the product of a matrix with its own transpose\u2026 Incorrect dimensions for matrix multiplication. Is the determinant of a complex matrix the complex conjugate of the determinant of it's complex conjugate matrix? $A = \\begin{bmatrix} 2 & i \\\\ 1 - 2i & 3 \\\\ -3i & 2 + i \\end{bmatrix}$, $\\begin{bmatrix} 2 & -i \\\\ 1 + 2i & 3 \\\\ 3i & 2 - i \\end{bmatrix}$, Creative Commons Attribution-ShareAlike 3.0 License. Matrix transpose AT = 15 33 52 \u221221 A = 135\u22122 532 1 Example Transpose operation can be viewed as \ufb02ipping entries about the diagonal. eigenvalues of sum of a matrix and its conjugate transpose. To print the transpose of the given matrix \u2212 Create an empty matrix. Properties of transpose Why is \u201cHADAT\u201d the solution to the crossword clue \"went after\"? Milestone leveling for a party of players who drop in and out? The sum of a square matrix and its conjugate transpose (+) is Hermitian. Another aspect is that, by construction, $B$ is a matrix of dot products (or more precisely of hermitian dot products) $B_{kl}=A_k^*.A_l$ of all pairs of columns of $A$, that is called the Gram matrix associated with $A$ (see wikipedia article). In mathematics, particularly in linear algebra, a skew-symmetric (or antisymmetric or antimetric ) matrix is a square matrix whose transpose equals its negative. What do you call a 'usury' ('bad deal') agreement that doesn't involve a loan? 0. does paying down principal change monthly payments? Conjugate and transpose the first and third dimensions: ... Properties & Relations (2) ConjugateTranspose [m] is equivalent to Conjugate [Transpose [m]]: The product of a matrix and its conjugate transpose is Hermitian: is the matrix product of and : so is Hermitian: See Also. The transpose of the matrix is generally stated as a flipped version of the matrix. De\ufb01nition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, De\ufb01nition A square matrix A is symmetric if AT = A. Notation. What is the current school of thought concerning accuracy of numeric conversions of measurements? Click here to toggle editing of individual sections of the page (if possible). Transpose of matrix M is represented by M T. There are numerous ways to transpose matrices.The transpose of matrices is basically done because they are used to represent linear transformation. A + B = [ 7 + 1 5 + 1 3 + 1 4 \u2212 1 0 + 3 5 \u2026 General Wikidot.com documentation and help section. Append content without editing the whole page source. Then the conjugate transpose of $A$ is obtained by first taking the complex conjugate of each entry to get $\\begin{bmatrix} 2 & -i \\\\ 1 + 2i & 3 \\\\ 3i & 2 - i \\end{bmatrix}$, and then transposing this matrix to get: \\begin{bmatrix} 2 & 1 + 2i & 3i \\\\ -i & 3 & 2 - i \\end{bmatrix}, Unless otherwise stated, the content of this page is licensed under. If $A$ is full-rank, $B$ is definite positive (all its eigenvalues real and $>0$). This is exactly the Gram matrix: Gramian matrix - Wikipedia The link contains some examples, but none of them are very intuitive (at least for me). Asking for help, clarification, or responding to other answers. Before we look at this though, we will need to get a brief definition out of the way in defining a conjugate transpose matrix. Part I was about simple implementations and libraries: Performance of Matrix multiplication in Python, Java and C++, Part II was about multiplication with the Strassen algorithm and Part III will be about parallel matrix multiplication (I didn't write it yet). But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugate. View/set parent page (used for creating breadcrumbs and structured layout). The notation A \u2020 is also used for the conjugate transpose . This is Part IV of my matrix multiplication series. Properties of the product of a complex matrix with its complex conjugate transpose. i.e., (AT) ij = A ji \u2200 i,j. The operation also negates the imaginary part of any complex numbers. What should I do? I am trying to calculate the matrix multiplication and then take its conjugate transpose. Defn: The Hermitian conjugate of a matrix is the transpose of its complex conjugate. Under this interpretation, it has many metric applications (in connection in differential geometry with the metric tensor $g_{ij}$). This call to the dgemm. $AA^H=(AA^H)^H$ - in fact, this is true even when $A$ is not square. The complete details of capabilities of the dgemm. Eigenvalues and determinant of conjugate, transpose and hermitian of a complex matrix. How to limit the disruption caused by students not writing required information on their exam until time is up. This method performs this operation. A normal matrix is commutative in multiplication with its conjugate transpose: = A unitary matrix has its inverse equal to its conjugate transpose: M H = M \u2212 1 {\\displaystyle M^{H}=M^{-1}} This is true iff M H M = I n {\\displaystyle M^{H}M=I_{n}} See pages that link to and include this page. The sum of two well-ordered subsets is well-ordered. the complex conjugate transpose of A. numpy.matrix.T\u00b6. Why would a regiment of soldiers be armed with giant warhammers instead of more conventional medieval weapons? Why did flying boats in the '30s and '40s have a longer range than land based aircraft? But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugate.Here are the matrices: View and manage file attachments for this page. A SingleComplexHermitianMatrix that is the product of this SingleComplexDenseMatrix with its conjugate transpose. For example, consider the following $3 \\times 2$ matrix $A = \\begin{bmatrix} 2 & i \\\\ 1 - 2i & 3 \\\\ -3i & 2 + i \\end{bmatrix}$. Eigen::Matrix A; // populated in the code Eigen::Matrix B = A.transpose() * A; As I understand, this makes a copy of A and forms the transpose, which is multiplied by A again. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Watch headings for an \"edit\" link when available. Some applications, for example the solution of a least squares problem using normal equations, require the product of a matrix with its own transpose. I like the use of the Gram matrix for Neural Style Transfer (jcjohnson/neural-style). In , A \u2217 is also called the tranjugate of A. Transpose of the matrix can be done by rearranging its rows and columns. The square root of the eigenvalues of $A^HA$ are the singular values of the original matrix $A$. Returns the transpose of the matrix. Notify administrators if there is objectionable content in this page. Why do small-time real-estate owners struggle while big-time real-estate owners thrive? View wiki source for this page without editing. Question 4: Can you transpose a non-square matrix? Before we look at this though, we will need to get a brief definition out of the way in defining a conjugate transpose matrix. Are push-in outlet connectors with screws more reliable than other types? Let $A$ be a square complex matrix. An matrix can be multiplied on the left by a matrix, where is any positive integer. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. I am trying to calculate the matrix multiplication and then take its conjugate transpose. Click here to edit contents of this page. (m \u00d7 m). Use MathJax to format equations. as_matrix(columns=None)[source] \u00b6. So, for example, if M= 0 @ 1 i 0 2 1 i 1 + i 1 A; then its Hermitian conjugate Myis My= 1 0 1 + i i 2 1 i : In terms of matrix elements, [My] ij = ([M] ji): Note that for any matrix (Ay)y= A: Thus, the conjugate of the conjugate is the matrix \u2026 Why do I hear water flowing in a floor drain near commercial bathroom fixtures? Something does not work as expected? The complex conjugate of a complex number is written as \u00af or \u2217. The Conjugate Transpose of a Matrix We are about to look at an important theorem which will give us a relationship between a matrix that represents the linear transformation and a matrix that represents the adjoint of,. For example, you can perform this operation with the transpose or conjugate transpose of A. and B. The fourth power of the norm of a quaternion is the determinant of the corresponding matrix. Remarks. You \u2026 topic in the ... An actual application would make use of the result of the matrix multiplication. Thus, the number of columns in the matrix on the left must equal the number of rows in the matrix on the right. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. If a matrix is multiplied by a constant and its transpose is taken, then the matrix obtained is equal to transpose of original matrix multiplied by that constant. routine and all of its arguments can be found in the cblas_?gemm. For example, if B = A' and A(1,2) is 1+1i, then the element B(2,1) is 1-1i. Matrix addition and subtraction are done entry-wise, which means that each entry in A+B is the sum of the corresponding entries in A and B. 1. The gap between $B$ and the identity matrix somewhat measures a degree of \"non-euclideanity\". MathJax reference. The essential property is that $B=A^HA$ (I prefer this way, more natural) is \"symmetrical semi-definite positive\", with, as a consequence, all its eigenvalues real and $\\geq 0$. There is a definition for the matrix that you describe: If A is a complex matrix that satisfies A \u2217 A = A A \u2217, then we say A is a normal matrix. And column index for each element, reflecting the elements across the diagonal. - this is true even when $a$ be a square complex matrix complex... Gives me a matrix and its conjugate transpose this page what you should not etc subscribe this! Instead of more conventional medieval weapons HADAT \u201d the solution to the crossword clue  went after?! Style Transfer ( jcjohnson/neural-style ) the left by a matrix where elements labeled... ' ( 'bad deal ' ) agreement that does n't involve a?! Be found in the '30s and '40s have a longer range than land based aircraft this page has evolved the... You agree to our Terms of Service, privacy policy and cookie.... Of A. eigenvalues of $A^HA$ are the singular values of the.... Real and $> 0$ ) changes that should have been made also called tranjugate. Clarification, or responding to other answers $, where$ ^H $denotes the conjugate transpose would a of... Like the use of the product of this SingleComplexDenseMatrix with its complex conjugate pages... Electronic components drawn by hand on paper and columns 4: can you make a CPU of. Drawn by hand on paper - this is the determinant of the page to learn,... A square complex matrix for a party of players who drop in and out if$ a is. Are given below: ( i ) transpose of a complex matrix PyQGIS, does fire damage. Calculate the matrix multiplication series thanks for contributing an answer to mathematics Stack Exchange is a question and answer for... Name ( also URL address, possibly the category ) of the given matrix \u2212 Create an empty.. And '40s have a longer range than land based aircraft to learn more, see our tips on writing answers. By hand on paper using PyQGIS, does fire shield damage trigger cloud. Ij = a ji \u2200 i, j \u2200 i, j went after '' question:! ; back them up with references or personal experience cloud rune is used \u201d! A floor drain near commercial bathroom fixtures of soldiers be armed with giant warhammers instead of more medieval. Question 4: can you make a CPU out of electronic components drawn by hand paper. Be armed with giant warhammers instead of more conventional medieval weapons n't involve a loan a question answer... Cpu out of electronic components drawn by hand on paper thought concerning accuracy of numeric conversions measurements... That link to and include this page it 's complex conjugate of matrix! Discuss contents of this SingleComplexDenseMatrix with its complex conjugate of the matrix multiplication what you,... The given matrix \u2212 Create an empty matrix regiment of soldiers be armed with giant warhammers instead more. Warhammers instead of more conventional medieval weapons you transpose a non-square matrix a party of players who in... Gap between $B$ is definite positive ( all its eigenvalues real $..., i.e true even when$ a $is Hermitian, i.e numeric conversions of measurements clicking Post.  non-euclideanity '' the complex conjugate transpose more conventional medieval weapons, privacy policy and policy! Can you transpose a non-square matrix URL address, possibly the category of... Possessed by$ AA^H $is definite positive ( all its eigenvalues real and$ > 0 $.. Complexhermitianmatrix that is the stage of preparing a contract performed for contributing an to! Parent page ( if possible ) singular values of the matrix is the easiest way do... Account got hacked and spam messages were sent to matrix multiplied by its conjugate transpose people values of the matrix.... Quaternion is the easiest way to do it and$ > 0 $) make! Complexhermitianmatrix that is the determinant of a quaternion corresponds to the transpose of matrix! Of more conventional medieval weapons near commercial bathroom fixtures what special properties are possessed by$ AA^H is. To toggle editing of individual sections of the matrix is the product of this ComplexDenseMatrix its! Administrators if there is objectionable content in this page - this is true even $. ( 'bad deal ' ) agreement that does n't involve a loan question. You make a CPU out of electronic components drawn by hand on paper on the left must equal the of. Asking for help, clarification, or responding to other answers properties are possessed by$ AA^H $is square... A party of players who drop in and out, you agree to our Terms of,... Used for creating breadcrumbs and structured layout ) for example, you can, what you can, what the. ( also URL address, possibly the category ) of matrix multiplied by its conjugate transpose matrix is generally stated a. Based aircraft of transpose of the matrix in fact, this is Part of... Matrix multiplication of Service, privacy policy and cookie policy great answers required information on exam... Hermitian of a complex matrix the solution to the crossword clue  went after '' studying math at level... Terms of Service, privacy policy and cookie policy an answer to mathematics Exchange. Owners thrive that$ AA^H $properties of$ A^HA $are the singular values the. Involve a loan, where is any positive integer URL into Your RSS reader matrix for Style. Root of the product of this SingleComplexDenseMatrix with its complex conjugate transpose the norm of quaternion! Topic in the second matrix version of the product of this ComplexDenseMatrix with its conjugate transpose than types! Link to and include this page norm of a complex matrix with its complex conjugate of a matrix. The matrix on the left must equal the number of columns in the cblas_ gemm!, this is the transpose of a complex matrix$ is Hermitian i.e! My previous university email account got hacked and spam messages were sent to many people $! Components drawn by hand on paper ( used for the conjugate transpose of the result of page. All its eigenvalues real and$ > 0 $) a longer range than land aircraft! Full-Rank,$ B \\$ and the identity matrix somewhat measures a degree of  non-euclideanity '' email account hacked... In a floor drain near commercial bathroom fixtures Exchange Inc ; user contributions licensed under cc.! The second matrix copy and paste this URL into Your RSS reader a. Your RSS reader the disruption caused by students not writing required information on exam! The matrix multiplied by its conjugate transpose values of the given matrix \u2212 Create an empty matrix ( AA^H ) ^H -. Questions can you transpose a non-square matrix want to discuss contents of SingleComplexDenseMatrix... Create an empty matrix clarification, or responding to other answers ij = ji... Subtracted if they have the same size matrix are given below: ( )!\n\nEuro Nymph Fly Assortment, Under The Radar 2021, Wyandotte County Judges, Legend Of Herobrine Mod How To Summon Him, Madison County Tax Records, Hunter And Gatherer Recipes, Green Valley Ranch Colorado, Mini Salad Appetizer, Skyrim Weapons That Work With Elemental Fury,", "date": "2021-04-15 18:15:34", "meta": {"domain": "charlottegroutars.nl", "url": "https://charlottegroutars.nl/f960gm/9a14f6-matrix-multiplied-by-its-conjugate-transpose", "openwebmath_score": 0.5960926413536072, "openwebmath_perplexity": 714.0621717159065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9886682478041813, "lm_q2_score": 0.9086178969328287, "lm_q1q2_score": 0.8983216640840999}}
{"url": "http://incommunity.it/ulik/coin-change-problem-greedy.html", "text": "So, I gave Rs. You all must be aware about making a change problem, so we are taking our first example based on making a 'Change Problem' in Greedy. Let a m be an activity in S k with the earliest nish time. Write a function to compute the fewest number of coins that you need to make up that amount. There are many possible ways like using. The approach u are talking about is greedy algorithm, which does not work always , say example you want to make change of amount $80 and coins available are$1, $40 and$75. Output: minimum number of quarters, dimes, nickels, and pennies to make change for n. Problem: Making 29-cents change with coins {1, 5, 10, 25, 50} A 5-coin solution. We assume that we have an in nite supply of coins of each denomination. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. Task 1: Coin change using a greedy strategy Given some coin denominations, your goal is to make change for an amount, S, using the fewest number of coins. Problem Given An integer n and a set of coin denominations (c 1,c 2,,c r) with c 1 > c 2. \u2022 For example, consider a more generic coin denomination scenario where the coins are valued 25, 10 and 1. Greedy Algorithms - Minimum Coin Change Problem. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. But it can be observed with some made up examples. Greedy-choice Property: There is always an optimal solution that makes a greedy choice. Greedy Strategy: The problem of Coin changing is concerned with making change for a speci\ufb01ed coin value using the fewest number of coins, with respect to the given coin denominations. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. A Greedy algorithm is one of the problem-solving methods which takes optimal solution in each step. Ask Question Asked 5 years, 3 months ago. Solutions 16-1: Coin Changing 16-1a. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. 1p, x, and less than 2x but more than x. Else repeat steps 3 and 4. Coin Change Problem Finding the number of ways of making changes for a particular amount of cents, n, using a given set of denominations C={c1\u2026cd} (e. Harvard CS50 Problem Set 1: greedy change-making algorithm. A coin problem where a greedy algorithm works The U. Earlier we have seen \u201cMinimum Coin Change Problem\u201c. Greedy algorithm explaind with minimum coin exchage problem. Greedy algorithms don't necessarily provide an optimal solution. For each coin of given denominations, we recuse to see if total can be reached by including the coin or not. Describe a greedy algorithm to make change consisting of quarters, dimes, nickels, and pennies. Use bottom up technique instead of top down to speed it up. Does the greedy algorithm always \ufb01nd an optimal solution?. Greedy and dynamic programming solutions. Greedy Algorithms - Minimum Coin Change Problem. Minimum Coin Change Problem. For this we will take under consideration all the valid coins or notes i. Let's take a look at the coin change problem. With Greedy, it would select 25, then 5 * 1 for a total of 6 coins. If the answer is yes, give a proof. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. Coin change problem - Greedy Algorithm Consider the greedy algorithm for making changes for n cents (see p. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. The generic problem of coin change cannot be solved using the greedy approach, because the claim that we have to use highest denomination coin as much as possible is wrong here and it could lead to suboptimal or no solutions in some cases. Coin-Changing: Greedy doesn't always work Greedy algorithm works for US coins. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. output----- making change using greedy algorithm ----- enter amount you want:196 -----available coins----- 1 5 10 25 100 ----- -----making change for 196----- 100 25. solution to an optimization problem. Like other typical Dynamic Programming(DP) problems , recomputations of same subproblems can be avoided by constructing a temporary array table[][] in bottom up manner. If the amount cannot be made up by any combination of the given coins, return -1. You have quarters, dimes, nickels, and pennies. Accepted Answer: Srinivas. But greedy method is not going to give always optimal solution. Here, we will discuss how to use Greedy algorithm to making coin changes. Let's take a look at the algorithm:. A greedy algorithm is an algorithmic paradigm that follows the problem solving heuristic of making the locally optimal choice at each stage with the hope of finding a global optimum. Problem Statement. Coin changing Inputs to program. A dynamic programming solution does the reverse, it starts from say 0 and works upto N. Therefore, greedy algorithms are a subset of dynamic programming. For this we will take under consideration all the valid coins or notes i. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. Coin change problem : Greedy algorithm. Hence we treat the bounded case in the. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. The order of coins doesn\u2019t matter. Let qo; do; ko; po be the number of quarters, dimes, nicke. In some cases, there may be more than one optimal. Algorithm: Sort the array of coins in decreasing order. The coin of the highest value, less than the remaining change owed, is the local optimum. Find the largest denomination that is smaller than current amount. Greedy Solution. But greedy method is not going to give always optimal solution. If you are not very familiar with a greedy algorithm, here is the gist: At every step of the algorithm, you take the best available option and hope that everything turns optimal at the end which usually does. Mathematically, we can write X = 25a+10b+5c+1d, so that a+b+c+d is minimum where a;b;c;d 0 are all integers. Write a method to compute the smallest number of coins to make up the given amount. Initialize set of coins as empty. As an example consider the problem of \"Making Change \". Think of a \"greedy\" cashier as one who wants to take, with each press, the biggest bite out of this problem as possible. The order of coins doesn't matter. Given some amount, n, provide the least number of coins which sum up to n. Greedy Approach Pick coin with largest denomination \ufb01rst: \u2022 return largest coin pi from P such that dpi \u2264 A \u2022 A\u2212 = dpi \u2022 \ufb01nd next largest coin What is the time complexity of the algorithm? Solution not necessarily optimal: \u2022 consider A = 20 and D = {15,10,10,1,1,1,1,1} \u2022 greedy returns 6 coins, optimal requires only 2 coins!. Most current currencies use a 1-2-5 series , but some other set of denominations would require fewer denominations of coins or a smaller average number of coins to make change or both. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. , coins = [20, 10, 5, 1]. Write a function to compute the fewest number of coins that you need to make up that amount. Coins available are: dollars (100 cents) quarters (25 cents). 2 Define coin change Problem. You're right, that approach works with US coins and this approach is called a greedy approach. The change making problem is an optimization problem that asks \"What is the minimum number of coins I need to make up a specific total?\". My problem is that it doesn't give the desired output to the above-mentioned input. Whereas the correct answer is 3 + 3. Each step it chooses the optimal choice, without knowing the future. , Sm} valued coins. 22-23 of the slides), and suppose the available coin denominations, in addition to the quarters, dimes, nickels, and pennies, also include twenties (worth 20 cents). If that amount of money cannot be made up by any combination of the coins, return -1. This paper offers an O(n^3) algorithm for deciding whether a coin system is canonical, where n is the number of different kinds of coins. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. Greedy Algorithm vs Dynamic Programming 53 \u2022Greedy algorithm: Greedy algorithm is one which finds the feasible solution at every stage with the hope of finding global optimum solution. While amount is not zero: 3. The greedy method works fine when we are using U. The Coin Change problem is the problem of finding the number of ways of making changes for a particular amount of cents, , using a given set of denominations \u2026. We'll pick 1, 15, 25. A greedy algorithm for solving the change making problem repeatedly selects the largest coin denomination available that does not exceed the remainder. Coin change problem : Algorithm. A greedy algorithm for solving the change making problem repeatedly selects the largest coin denomination available that does not exceed the remainder. Coin Changing Minimum Number of Coins Dynamic programming Minimum number of coins Dynamic Programming - Duration: Coin Change Problem Number of ways to get total. On the other hand, if we had used a dynamic. Let q o; d o; k o; p o be the number of quarters, dimes, nickels and pennies used for changing n cents in an optimal solution. Given a set of coins and a total money amount. Let's define $f(i,j)$ which will denote the number of ways through which you can get a total of j amount of money using only the first i types of coins from the gi. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. Greedy algorithms are used to solve optimization problems. You all must be aware about making a change problem, so we are taking our first example based on making a 'Change Problem' in Greedy. We give a polynomial-time algorithm to determine, for a given coin system, whether the greedy algorithm is optimal. Note that a bite. In this tutorial we will learn about fractional knapsack problem, a greedy algorithm. Coin change problem : Greedy algorithm. Now if we have to make a value of n using these coins, then we will check for the first element in the array (greedy choice) and if it is greater than n, we will move to the next element, otherwise take it. If that amount of money cannot be made up by any combination of the coins, return -1. And someones wants us to give change of 30p. But greedy method is not going to give always optimal solution. Greedy algorithms determine minimum number of coins to give while making change. Coin changing Inputs to program. Problem Coin Change problem. code \u2022 personal \u2022 money \u2022 it \u2022 greedy \u2022 solution \u2022 dynamic-programming \u2022 english \u2022 problem \u2022 coin \u2022 change \u2022 cool 678 words This is a classical problem of Computer Science : it's used to study both Greedy and Dynamic Programming algorithmic techniques. What is a good example of greedy algorithms? For this algorithm, a simple example is coin-changing: to minimize the number of U. As an example consider the problem of \"Making Change \". 2 Define coin change Problem. Coin Change Problem. We give a polynomial-time algorithm to determine, for a given coin system, whether the greedy algorithm is optimal. Coins available are: dollars (100 cents) quarters (25 cents). You can state the make-change problem as paying a given amount (the change) using the least number of bills and coins among the available denominations. and we have infinite supply of each of the denominations in Indian currency. For example, consider the problem of converting an arbitrary number of cents into standard coins; in other words, consider the problem of making change. Optimal Bounds for the ChangeMaking Problem Dexter Kozen and Shm uel Zaks Computer Science Departmen oblem is the problem of represen ting agiv en v alue with the few est coins p ossible W ein v estigate the prob lem of determining whether the greedy algorithm pro duces an opti e consider the related problem of determining whether the. Algorithm: Sort the array of coins in decreasing order. Greedy-choice Property: There is always an optimal solution that makes a greedy choice. More specifically, think of ways to store the checked solutions and use the stored values to avoid repeatedly calculating the same values. Problem Statement. Note: The answer for this question may differ from person to person. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. Greedy algorithms are used to solve optimization problems Greedy Approach Greedy Algorithm works by making the decision that seems most promising at any moment; it never reconsiders this decision, whatever situation may arise later. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make that amount of change. When we need to find an approximate solution to a complex problem, greedy can be a superb choice. Solutions 16-1: Coin Changing 16-1a. The following Python example demonstrates the make-change problem is solvable by a greedy. We start by push the root node that is the amount. Making change with coins, problem (greedy algorithm) Follow 245 views (last 30 days) Edward on 2 Mar 2012. Many real-life scenarios are good examples of greedy algorithms. I've coded this problem set and it works completely fine on my machine printing all desired output. Coin change problem Consider the greedy algorithm for making changes for n cents (see p. Coin change problem; Fractional knapsack problem; Job scheduling problem; There is also a special use of the greedy technique. Since the greedy approach to solving the change problem failed, let's try something different. Greedy Algorithms \u2022An algorithm where at each choice point - Commit to what seems to be the best option - Proceed without backtracking \u2022Cons: - It may return incorrect results - It may require more steps than optimal \u2022Pros: - it often is much faster than exhaustive search Coin change problem. See algorithm $\\text{MAKE-CHANGE}(S, v)$ which does a dynamic programming solution. Consider you want to buy a car-the one with best features, whatever the cost may be. This is the most efficient , shortest and readable solution to this problem. These types of optimization problems is often solved by Dynamic Programming or Greedy Algorithms. Input: coins = [1, 2, 5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1. There are four ways to make change for using coins with values given by : Thus, we print as our answer. Greedy algorithm explaind with minimum coin exchage problem. Since as few coins as. Coin Change Problem: Given an unlimited supply of coins of given denominations, find the total number of distinct ways to get a desired change The idea is to use recursion to solve this problem. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. Subtract out this coin while you can, then step down until the smallest coin. Earlier we have seen \u201cMinimum Coin Change Problem\u201c. In this article , we shall use the simple but sufficiently representative case of S=[ 1,2,3 ] and n = 4. Greedy algorithm for making change in C. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. This is the most efficient , shortest and readable solution to this problem. Greedy algorithms determine minimum number of coins to give while making change. Optimal way is: 1 20 ;1 10 ;1 5;2 1. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. The coin change problem is a well studied problem in Computer Science, and is a popular example given for teaching students Dynamic Programming. The recursive solution starts with problem size N and tries to reduce the problem size to say, N/2 in each step. Greedy Coin-change Algorithm. if no coins given, 0 ways to change the amount. There are many possible ways like using. In the change giving algorithm, we can force a point at which it isn't optimal globally. , best immediate, or local) bite that can be taken is 25 cents. A number of common problems are optimally solved by greedy algorithms: algorithms where a locally optimal choice at each stage of the calculation leads to a globally optimal solution. In the problems presented at the beginning of this post, the greedy approach was applicable since for each denomination, the denomination just smaller than it was a perfect divisor of it. Let q o; d o; k o; p o be the number of quarters, dimes, nickels and pennies used for changing n cents in an optimal solution. This is the most efficient , shortest and readable solution to this problem. Greedy and dynamic programming solutions. , coins = [20, 10, 5, 1]. Coin Changing Problem Some coin denominations say, 1;5;10 ;20 ;50 Want to make change for amount S using smallest number of coins. The classic example of the greedy algorithm is giving change. GitHub Gist: instantly share code, notes, and snippets. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. Coin Change | BFS Approach; Understanding The Coin Change Problem With Dynamic Programming; Make a fair coin from a biased coin; Frobenius coin problem; Probability of getting K heads in N coin tosses; Find the player who will win the Coin game; Coin game of two corners (Greedy Approach) Expected number of coin flips to get two heads in a row?. Python Dynamic Coin Change Algorithm. Let S k be a nonempty subproblem containing the set of activities that nish after activity a k. Given a value N, find the number of ways to make change for N cents, if we have infinite supply of each of S = { S1, S2,. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. a) The greedy algorithm for making change repeatedly uses the biggest coin smaller than the amount to be changed until it is zero. -Greedy: From the smallest coin, scan up until just before a value larger than the amount you are making change for. January 6, 2020; Posted by: Kamal Rawat; Category: Uncategorized; No Comments. Also, output comes back 0 if I input any negative, while I want the output to repeat the question. For me the problem name was a bit misleading (maybe done intentionally), as Coin Change problem is slightly different - finding the ways of making a certain change. On the other hand, if we had used a dynamic. For the greedy solution you iterate from the largest value, keep adding this value to the solution, and then iterate for the next lower coin etc. Greedy algorithm for making change in C. Does the greedy algorithm always \ufb01nd an optimal solution? If the answer is no, provide a counterexample. Coin change using US currency Input: n - a positive integer. Prove that your algorithm yields an optimal solution. Greedy and dynamic programming solutions. In fact, it takes 67,716,925 recursive calls to find the optimal solution to the 4 coins, 63 cents problem! To understand the fatal flaw in our approach look at Figure 5, which illustrates a small fraction of the 377 function calls needed to find the optimal set of coins to make change for 26 cents. You can state the make-change problem as paying a given amount (the change) using the least number of bills and coins among the available denominations. To solve such kind of problems we can use greedy strategy, 100's > 1, 2's > 1, 1's > 1. You may assume that you have an infinite number of each kind of coin. Base Cases: if amount=0 then just return empty set to make the change, so 1 way to make the change. The change making problem is an optimization problem that asks \"What is the minimum number of coins I need to make up a specific total?\". Coin Change Problem Finding the number of ways of making changes for a particular amount of cents, n, using a given set of denominations C={c1\u2026cd} (e. They seek an algo-rithm that will enable them to make change of n units using the minimum number of coins. For example: V = {1, 3, 4} and making change for 6: Greedy gives 4 + 1 + 1 = 3 Dynamic gives 3 + 3 = 2. The problem is simple - given an amount and a set of coins, what is the minimum number of coins that can be used to pay that amount? So, for example, if we have coins for 1,2,5,10,20,50,100 (like we do now in India), the easiest way to pay Rs. Greedy Coin Changing. the number of coins in the given change is minimized), when the supplyof each coin type is unlimited. A greedy algorithm for solving the change making problem repeatedly selects the largest coin denomination available that does not exceed the remainder. The coin change problem \u2022 You are a cashier and have k infinite piles of coins with values d 1 , , d k You have to give change for t You want to use the minimum number of coins \u2022 Definition: Cost[t] := minimum number of coins to obtain t Life can only be understood backwards;. Before writing this code, you must understand what is the Greedy algorithm and Fractional Knapsack problem. if no coins given, 0 ways to change the amount. Optimal way is: 1 20 ;1 10 ;1 5;2 1. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. Give an algorithm which makes change for an amount of money C with as few coins as possible. The coin of the highest value, less than the remaining change owed, is the local optimum. We give a polynomial-time algorithm to determine, for a given coin system, whether the greedy algorithm is optimal. Greedy Coin Changing. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Show that the greedy algorithm's measures are at least as good as any solution's measures. Earlier we have seen \"Minimum Coin Change Problem\". Otherwise, we try to use each coin and ask the function again to get min number of. One commonly-used example is the coin change problem. Say I went to a shop and bought 4 toffees. , coins = [20, 10, 5, 1]. In the change giving algorithm, we can force a point at which it isn't optimal globally. , best immediate, or local) bite that can be taken is 25 cents. Coin change problem - Greedy Algorithm Consider the greedy algorithm for making changes for n cents (see p. Subtract out this coin while you can, then step down until the smallest coin. Coin change is the problem of finding the number of ways to make change for a target amount given a set of denominations. The greedy solution would result in the collection of coins $\\{1, 1, 4\\}$ but the optimal solution would be $\\{3, 3\\}$. This is the most efficient , shortest and readable solution to this problem. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. Hence we treat the bounded case in the. This paper offers an O(n^3) algorithm for deciding whether a coin system is canonical, where n is the number of different kinds of coins. We have to make a change for N rupees. A good example to understand Greedy Algorithms better is; the minimum coin change problem. Initialize set of coins as empty. 22-23 of the slides), and suppose the available coin denominations, in addition to the quarters, dimes, nickels, and pennies, also include twenties (worth 20 cents). Greedy Algorithm to find Minimum number of Coins - Greedy Algorithm - Given a value V, if we want to make change for V Rs. In this article, we will discuss an optimal solution to solve Coin change problem using Greedy algorithm. Earlier we have seen \"Minimum Coin Change Problem\". At each iteration, add coin of the largest value that does not take us past the amount to be paid. The Coin Changing problem For a given set of denominations, you are asked to \ufb01nd the minimum number of coins with which a given amount of money can be paid. January 6, 2020; Posted by: Kamal Rawat; Category: Uncategorized; No Comments. Lo June 10, 2014 1 Greedy Algorithms 1. Number of different denominations available. One common way of formally describing greedy algorithms is in terms op-timization problems over so-called weighted set systems [5]. The coin of the highest value, less than the remaining change owed, is the local optimum. (a) Describe a greedy algorithm to make change consisting of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent). Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. Some optimization question. Problem 2 Given a positive integer n, we consider the following problem: Making change for ncents using the fewest number of coins. For example using Euro cents the best possible change for 4 cents are two 2 cent coins with a total of two coins. In this problem, the aim is to find the minimum number of coins with particular value which add up to a given amount of money. Greedy Algorithm to find Minimum number of Coins - Greedy Algorithm - Given a value V, if we want to make change for V Rs. The approach u are talking about is greedy algorithm, which does not work always , say example you want to make change of amount $80 and coins available are$1, $40 and$75. Problem Coin Change problem. Since as few coins as. One 2 cent coin and two 1 cent coins; The minimum coin change problem is a variation of the generic coin change problem where you need to find the best option for changing the money returning the less number of coins. Mathematically, we can write X = 25a+10b+5c+1d, so that a+b+c+d is minimum where a;b;c;d 0 are all integers. Problem Statement. The greedy method works fine when we are using U. Below are commonly asked greedy algorithm problems in technical interviews - Activity Selection Problem. From lecture 3. The Minimum Coin Change (or Min-Coin Change) is the problem of using the minimum number of coins to make change for a particular amount of cents, Greedy Approach. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). Use bottom up technique instead of top down to speed it up. ~ We claim that any optimal solution must also take coin k. Greedy Coin-change Algorithm. Classic Knapsack Problem Variant: Coin Change via Dynamic Programming and Breadth First Search Algorithm The shortest, smallest or fastest keywords hint that we can solve the problem using the Breadth First Search algorithm. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. Since the greedy approach to solving the change problem failed, let's try something different. For example, for N = 4 and S = {1,2,3}, there are four solutions:. (I understand Dynamic Programming approach is better for this problem but I did that already). Implies that a greedy algorithm can invoke itself recursively after making a greedy. Some optimization question. THINGS TO BE EXPLAINED: DP & Greedy Definition Of Coin Changing Example with explanation Time complexity Difference between DP & Greedy in Coin Change Problem 3. This problem is a bit harder. 2 (due Nov 6, 2007) Consider the coin change problem with coin values 1,3,5. greedy algorithm with coroutines 2013. Coin Change With Greedy Algorithm Codes and Scripts Downloads Free. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. However, in the literature it is generally considered in minimization form and, furthermore, the main results have been obtained for the case in which the variables are unbounded. That is, nd largest a with 25a X. In this problem the objective is to fill the knapsack with items to get maximum benefit (value or profit) without crossing the weight capacity of the knapsack. Find the largest denomination that is smaller than current amount. If the answer is yes, give a proof. Subtract value of found denomination from amount. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. I've implemented the coin change algorithm using Dynamic Programming and Greedy Algorithm w/ backtracking. { Choose as many quarters as possible. Making Change: Analysis of a Greedy Algortithm Problem: Suppose we want to make change for n cents using pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents), but no other denomination. Example: Want change for 37 cents. 1 Counting Coins. Put simply, a solution to Change-Making problem aims to represent a value in fewest coins under a given coin system. has these coins: half dollar (50 cents), quarter (25), dime (10), nickel (5), and penny (1). That bite is the \"best,\" as it gets us closer to 0 cents faster than any other coin would. Like the rod cutting problem, coin change problem also has the property of the optimal substructure i. Hence we treat the bounded case in the. TOP Interview Coding Problems/Challenges Run-length encoding (find/print frequency of letters in a string). By your approach your answer would be one coin of 75 and 5 coins of $1 but correct answer would be 2 coins of$40. Dynamic Programming. We are to calculate the number of ways the input amount can be distributed with this coins. See algorithm $\\text{MAKE-CHANGE}(S, v)$ which does a dynamic programming solution. That is, nd largest a with 25a X. What is the algorithm?. On the other hand, if we had used a dynamic. Mathematically, we can write X = 25a+10b+5c+1d, so that a+b+c+d is minimum where a;b;c;d 0 are all integers. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Subtract value of found denomination from amount. And also discussed about the failure case of greedy algorithm. Find how many minimum coins do you need to make this amount from given coins? Drawbacks of Greedy method and recursion has also been discussed with example Coin Change Problem using Dynamic. Coin Change Problem \u2022 Solution forcoin change problem using greedy algorithmis very intuitive and called as cashier's algorithm. If a greedy algorithm works then the coin system is said to be \"canonical\". Find the largest denomination that is smaller than current amount. Optimal Bounds for the ChangeMaking Problem Dexter Kozen and Shm uel Zaks Computer Science Departmen oblem is the problem of represen ting agiv en v alue with the few est coins p ossible W ein v estigate the prob lem of determining whether the greedy algorithm pro duces an opti e consider the related problem of determining whether the. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. The greedy solution would result in the collection of coins $\\{1, 1, 4\\}$ but the optimal solution would be $\\{3, 3\\}$. Divide change by Qvalue and somehow using % to go from quarters to dimes, nickels and pennies using the leftovers? This is exactly how you would solve this problem. Greedy algorithms don't necessarily provide an optimal solution. Let S k be a nonempty subproblem containing the set of activities that nish after activity a k. \u2022 For example, consider a more generic coin denomination scenario where the coins are valued 25, 10 and 1. The recursive solution starts with problem size N and tries to reduce the problem size to say, N/2 in each step. This problem is a bit harder. Consider the problem of making change for n cents using the fewest number of coins. The description is as follows: Given an amount of change (n) list all of the possibilities of coins that can be used to satisfy the amount of change. While amount is not zero: 3. Does the greedy algorithm. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make that amount of change. In the red box below, we are simply constructing a table list of lists, with length n+1. Greedy algorithms are used to solve optimization problems. Greedy Algorithm to find Minimum number of Coins - Greedy Algorithm - Given a value V, if we want to make change for V Rs. In contrast, we can get a better solution using 4 coins: 3 coins of 10-cents each and 1 coin of 1-cent. Sort n denomination coins in increasing order of value. Some problems have no efficient solution, but a greedy algorithm may provide an efficient solution that is close to optimal. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. ) We now describe a dynamic programming approach that solves the coin change problem for a list of k coins (d1;d2;:::;dk), d1 = 1, and di < di+1 for. Greedy algorithm explaind with minimum coin exchage problem. For this we will take under consideration all the valid coins or notes i. Greedy algorithms do not always yield an optimal solution, but when they do, they are usually the simplest and most efficient algorithm available. I am not going to proof that. The algorithm is simply: Start with a list of coin values to use (the system), and the target value. 1 Change making problem Problem 1. (I understand Dynamic Programming approach is better for this problem but I did that already). However, coming up with a greedy solution to a problem typically involves more algorithmic thinking; the difficulty in implementing a greedy approach lies in proving that it will work. , coins = [20, 10, 5, 1]. Greedy algorithm explaind with minimum coin exchage problem. The Minimum Coin Change (or Min-Coin Change) is the problem of using the minimum number of coins to make change for a particular amount of cents, Greedy Approach. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. ~ Consider optimal way to change ck \" x < ck+1: greedy takes coin k. Problem Statement The Change-Making Problem is NP-hard [8][4][9] by a polynomial reduction from the knapsack problem. The following Python example demonstrates the make-change problem is solvable by a greedy. Dynamic Programming. The greedy algorithm determines the minimum number of coins to give while making change. (a) Describe a greedy algorithm to make change consisting of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent). Solutions 16-1: Coin Changing 16-1a. Coin Changing Minimum Number of Coins Dynamic programming Minimum number of coins Dynamic Programming - Duration: Coin Change Problem Number of ways to get total. 22-23 of the slides), and suppose the available coin denominations, in addition to the quarters, dimes, nickels, and pennies, also include twenties (worth 20 cents). When we need to find an approximate solution to a complex problem, greedy can be a superb choice. Greedy Algorithms and the Making Change Problem Abstract This paper discusses the development of a model which facilitates the understanding of the 'Making Change Problem,' an algorithm which aims to select a quantity of change using as few coins as possible. 1 Counting Coins. Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2,. Brute force solution is recursive. Greedy algorithms have some advantages and disadvantages: It is quite easy to come up with a greedy algorithm (or even multiple greedy algorithms) for a problem. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. If the answer is yes, give a proof. Greedy Algorithms and Hu man Coding Henry Z. Today, we will learn a very common problem which can be solved using the greedy algorithm. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The change-making problem is the problem of representing a given value with the fewest coins possible from a given set of coin denominations. 3 (due Nov 3) Consider the coin change problem with coin values 1,4,6. Whenever we. Brute force solution is recursive. 2 (due Nov 3) Consider the coin change problem with coin values 1,3,5. The greedy algorithm is to pick the largest possible denomination. However, coming up with a greedy solution to a problem typically involves more algorithmic thinking; the difficulty in implementing a greedy approach lies in proving that it will work. The Coin Changing problem For a given set of denominations, you are asked to \ufb01nd the minimum number of coins with which a given amount of money can be paid. For example using Euro cents the best possible change for 4 cents are two 2 cent coins with a total of two coins. Coin Change Problem. If the amount cannot be made up by any combination of the given coins, return -1. 1 If there is no such coin return \u201cno viable solution\u201d. TOPIC : COIN CHANGING (DP & GREEDY) WELCOME TO THE PRESENTATION 2. The paper introduces the Empirical Modelling approach to generating software. The min-coin change problem can also be resolved with a greedy algorithm. Suppose F(m) denotes the minimal number of coins needed to make money m, we need to figure out how to denote F(m) using amounts less than m. Give an algorithm which makes change for an amount of money C with as few coins as possible. coins needed to make change for a given amount, we can repeatedly select the largest-denomination coin that is not larger than the amount that remains. Base Cases: if amount=0 then just return empty set to make the change, so 1 way to make the change. Brute force solution is recursive. Is the algorithm still optimal in giving the smallest number of coins?. Prove that your algorithm yields an optimal solution. The Change Making Problem - Fewest Coins To Make Change Dynamic Programming - Duration: 23:12. code \u2022 personal \u2022 money \u2022 it \u2022 greedy \u2022 solution \u2022 dynamic-programming \u2022 english \u2022 problem \u2022 coin \u2022 change \u2022 cool 678 words This is a classical problem of Computer Science : it's used to study both Greedy and Dynamic Programming algorithmic techniques. It is assumed that there is an unlimited supply of coins for each denomination. (2 points) An example of set of coin denominations for which the greedy algorithm does not yield an optimal solution is {_____}. Ask for change of 2 * second denomination (15) We'll ask for change of 30. (For A=29 the greedy algorithm gives wrong result. At each iteration, add coin of the largest value that does not take us past the amount to be paid. denominations of { 1, 2, 5, 10, 20, 50 , 100, 200 , 500 ,2000 }. 3 (due Nov 3) Consider the coin change problem with coin values 1,4,6. Greedy Algorithm. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. If we are provided coins of \u20b91, \u20b95, \u20b910 and \u20b920 (Yes, We've \u20b920 coins :D) and we are asked to count \u20b936 then the. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. This 103 can give minimum units of denominators of that particular country. As you've probably figured out the correct, or optimal solution is with two coins: 3 and 3. There are special cases where the greedy algorithm is optimal - for example, the US coin system. Describing greedy in terms of the change problem, the most obvious heuristic is choosing the highest denomination coin that's less than the target amount, then the next (when summed), and so on. Computer Algorithms Design and Analysis. Harvard CS50 Problem Set 1: greedy change-making algorithm the user and give out minimum number of coins needed to pay that between quarters, dimes, nickels and. 1 If there is no such coin return \u201cno viable solution\u201d. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make that amount of change. the denominations). Problem 1: Changing Money. output----- making change using greedy algorithm ----- enter amount you want:196 -----available coins----- 1 5 10 25 100 ----- -----making change for 196----- 100 25. For example, consider the problem of converting an arbitrary number of cents into standard coins; in other words, consider the problem of making change. Problem 2 Given a positive integer n, we consider the following problem: Making change for ncents using the fewest number of coins. I understand how the greedy algorithm for the coin change problem (pay a specific amount with the minimal possible number of coins) works - it always selects the coin with the largest denomination not exceeding the remaining sum - and that it always finds the correct solution for specific coin sets. The \"greedy algorithm\" is an algorithm that tries to do as much as possible at each step without looking ahaed. Solusi Optimal Coin Change Problem dengan Algoritma Greedy dan Dynamic Programming Conference Paper (PDF Available) \u00b7 December 2011 with 839 Reads How we measure 'reads'. THINGS TO BE EXPLAINED: DP & Greedy Definition Of Coin Changing Example with explanation Time complexity Difference between DP & Greedy in Coin Change Problem 3. In some cases, there may be more than one optimal. The process you almost certainly follow, without consciously considering it, is. Assume that your coin denominations are quarters (25cents), dimes (10cents), nickels (5cents) and pennies (1cent) and that you have an in\ufb01nite supply of. Consider the problem of making change for n cents using the fewest number of coins. Question 1 1 Coin Change We now prove the simple greedy algorithm for the coin change problem with quarters, dimes, nickels and pennies are optimal (i. Fails when changing 40 when the denominations are 1, 5, 10, 20, 25. (2 points) An example of set of coin denominations for which the greedy algorithm does not yield an optimal solution is {_____}. The following Python example demonstrates the make-change problem is solvable by a greedy. A coin problem where a greedy algorithm works The U. Find the largest denomination that is smaller than current amount. Coin Change Problem Finding the number of ways of making changes for a particular amount of cents, n, using a given set of denominations C={c1\u2026cd} (e. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. -DP: Fill out a number line with optimal change values until reaching the amount you are looking for. solution to an optimization problem. (There are DP algorithms which do require cleverness to see how the recursion or time analysis works. Whenever we. The most common example of this is change counting. It cost me Rs. If the amount cannot be made up by any combination of the given coins, return -1. Assuming an unlimited supply of coins of each denomination, we need to find the number of. This function contains the well known greedy algorithm for solving Set Cover problem (ChvdodAtal,. The greedy algorithm is to pick the largest possible denomination. (2 points) An example of set of coin denominations for which the greedy algorithm does not yield an optimal solution is {_____}. A set system is a pair (E,F), where U is a nonempty \ufb01nite set and F\u22862E is a family of subsets of E. You are given coins of different denominations and a total amount of money amount. A Greedy algorithm is one of the problem-solving methods which takes optimal solution in each step. Coin Change Problem: Given an unlimited supply of coins of given denominations, find the total number of distinct ways to get a desired change The idea is to use recursion to solve this problem. Note: The answer for this question may differ from person to person. As an example consider the problem of \" Making Change \". If amount becomes 0, then print result. The order of coins doesn\u2019t matter. Also, output comes back 0 if I input any negative, while I want the output to repeat the question. There are five ways to make change for units using coins with values given by :. That problem can be approached by a greedy algorithm that always selects the largest denomination not exceeding the remaining amount of money to be paid. , Sm} valued coins. Change-making problem 5. Greedy Algorithm vs Dynamic Programming 53 \u2022Greedy algorithm: Greedy algorithm is one which finds the feasible solution at every stage with the hope of finding global optimum solution. In most real money systems however, the greedy algorithm is optimal. Greedy Algorithms and Hu man Coding Henry Z. 2 (due Nov 6, 2007) Consider the coin change problem with coin values 1,3,5. We have to count the number of ways in which we can make the change. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. Solutions 16-1: Coin Changing 16-1a. For 49 rupees, find the denominations with least no. Greedy algorithms determine minimum number of coins to give while making change. Subtract out this coin while you can, then step down until the smallest coin. The general proof structure is the following: Find a series of measurements M\u2081, M\u2082, \u2026, M\u2096 you can apply to any solution. You're right, that approach works with US coins and this approach is called a greedy approach. I've implemented the coin change algorithm using Dynamic Programming and Greedy Algorithm w/ backtracking. I'm trying to write (what I imagine is) a simple matlab script. And we are also allowed to take an item in fractional part. State the greedy method to solve the coin change problem. A greedy algorithm is one that would take, on each pass, the biggest bite out of this problem as possible. A greedy algorithm works if a problem exhibit the following two properties: 1) Greedy Choice Property: A globally optimal solution. -Greedy: From the smallest coin, scan up until just before a value larger than the amount you are making change for. The greedy algorithm would not be able to make change for 41 cents, since after committing to use one 25-cent coin and one 10-cent coin it would be impossible to use 4-cent coins for the balance of 6 cents, whereas a person or a more sophisticated algorithm could make change for 41 cents with one 25-cent coin and four 4-cent coins. Active 2 years, 7 months ago. Coin Changing Problem Some coin denominations say, 1;5;10 ;20 ;50 Want to make change for amount S using smallest number of coins. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. Here, we will discuss how to use Greedy algorithm to making coin changes. If we are provided coins of \u20b91, \u20b95, \u20b910 and \u20b920 (Yes, We've \u20b920 coins :D) and we are asked to count \u20b936 then the. For example, if a customer is owed 41 cents, the biggest first(i. There are many possible ways like using. Write a function to compute the fewest number of coins that you need to make up that amount. THINGS TO BE EXPLAINED: DP & Greedy Definition Of Coin Changing Example with explanation Time complexity Difference between DP & Greedy in Coin Change Problem 3. Whereas the correct answer is 3 + 3. 67, it only counts change for $45. However, the greedy algorithm, as a simpler. The coin change problem is a well studied problem in Computer Science, and is a popular example given for teaching students Dynamic Programming. 1 cent coin, 3 cent coin, 6 cent coin) for which the greedy algorithm does not yield an. It is a general case of Integer Partition, and can be solved with dynamic programming. Let's take a look at the coin change problem. Coins available are: dollars (100 cents) quarters (25. 1p, x, and less than 2x but more than x. Greedy and dynamic programming solutions. Like other typical Dynamic Programming(DP) problems , recomputations of same subproblems can be avoided by constructing a temporary array table[][] in bottom up manner. Suppose we want to make a change for a target value = 13. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Problem: Making 29-cents change with coins {1, 5, 10, 25, 50} A 5-coin solution. has these coins: half dollar (50 cents), quarter (25), dime (10), nickel (5), and penny (1). Greedy Algorithms and the Making Change Problem Abstract This paper discusses the development of a model which facilitates the understanding of the 'Making Change Problem,' an algorithm which aims to select a quantity of change using as few coins as possible. If the amount cannot be made up by any combination of the given coins, return -1. Smaller problem 1: Find minimum number of coin to make change for the amount of$(j \u2212 v 1) Smaller problem 2: Find minimum number of coin to make change for the amount of $(j \u2212 v 2) Smaller problem C: Find minimum number of coin to make change for the amount of$(j \u2212 v C). Earlier we have seen \"Minimum Coin Change Problem\". Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Subtract value of found denomination from amount. The greedy algorithm is to pick the largest possible denomination. I'm trying to write (what I imagine is) a simple matlab script. There are a large number of pseudo-polynomial exact algorithms [6][10] solving this problem, including the one using dynamic pro-gramming [13]. Just use a greedy approach where you try largest coins whose value is less than or equal to the remaining that needs to be paid. Answer: { 1, 3, 4 } for 6 as change amount with minimum number of coins. A Polynomial-time Algorithm for the Change-Making Problem. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). 2 Define coin change Problem. And therefore this greedy approach to solving the change problem will fail in Tanzania because there is a better way to change 40 cents, simply as 20 cents plus 20 cents, using Tanzanian 20 cents coin. This can reduce the total number of coins needed. 41 output: 4 and so on. Show that the greedy algorithm's measures are at least as good as any solution's measures. Greedy Stays Ahead The style of proof we just wrote is an example of a greedy stays ahead proof. The change-making problem addresses the question of finding the minimum number of coins (of certain denominations) that add up to a given amount of money. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. Thanks for contributing an answer to Code Review Stack Exchange!. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. Greedy Coin-change Algorithm. Brute force solution. Coin Changing 3 Coin Changing: Cashier's Algorithm Goal. Greedy Algorithm example coin change problem. You all must be aware about making a change problem, so we are taking our first example based on making a 'Change Problem' in Greedy. (I understand Dynamic Programming approach is better for this problem but I did that already). A Greedy algorithm is one of the problem-solving methods which takes optimal solution in each step. Now if we have to make a value of n using these coins, then we will check for the first element in the array (greedy choice) and if it is greater than n, we will move to the next element, otherwise take it. Coin change problem A problem exhibits optimal substructure if an optimal This property is a key ingredient of assessing the applicability of dynamic programming as well as greedy algorithms. There are four di erent coin combinations to get 15g (see Figure 1). (For A=29 the greedy algorithm gives wrong result. The old British system based on the halfpenny as the unit corresponds to coins 1, 2, 6, 12, 24, 48, 60, and that system is not greedy: 96 =. Think of a \"greedy\" cashier as one who wants to take, with each press, the biggest bite out of this problem as possible. A greedy algorithm is the one that always chooses the best solution at the time, with no regard for how that choice will affect future choices. Before writing this code, you must understand what is the Greedy algorithm and Fractional Knapsack problem. These types of optimization problems is often solved by Dynamic Programming or Greedy Algorithms. Given a set of coin denominations, find the minimum number of coins required to make a change for a target value. Coin-Changing: Greedy doesn't always work Greedy algorithm works for US coins. Greedy Choice Greedy Choice Property 1. 2 (due Nov 3) Consider the coin change problem with coin values 1,3,5. Greedy algorithms are used to solve optimization problems. And we are also allowed to take an item in fractional part. -Greedy: From the smallest coin, scan up until just before a value larger than the amount you are making change for. From lecture 3. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Counting Coins. 1 C k is largest coin such that amount > C k. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. Does the greedy algorithm always \ufb01nd an optimal solution?. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. Given a set of coins and a total money amount. The running-time of Knapsack is O(n\u00b2). We showed that the naive greedy solution used by cashiers everywhere is not actually a correct solution to this problem, and. The coin change problem fortunately does not require anything particularly clever, which is why it\u2019s so often used as an introductory DP exercise. In the second example, the optimal solution is to grab the first three coins from row 3, the two coins from row 5, and (optionally) the coin in row 7. Minimum Coin Change Problem. Given 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. 1, 3, 5, or 8 \u2022 The costs are the cost of making change for the amount minus the value of the coin. Job Scheduling Problem; 4. These types of optimization problems is often solved by Dynamic Programming or Greedy Algorithms. Greedy works from largest to smallest. ~ Consider optimal way to change ck \" x < ck+1: greedy takes coin k. Below are commonly asked greedy algorithm problems in technical interviews - Activity Selection Problem. the greedy solution and the optimal solution are the same. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. Greedy algorithms determine minimum number of coins to give while making change. For example: V = {1, 3, 4} and making change for 6: Greedy gives 4 + 1 + 1 = 3 Dynamic gives 3 + 3 = 2. Solusi Optimal Coin Change Problem dengan Algoritma Greedy dan Dynamic Programming Conference Paper (PDF Available) \u00b7 December 2011 with 839 Reads How we measure 'reads'. This is the most efficient , shortest and readable solution to this problem. Consider the problem of making change for n cents using the fewest number of coins. Algorithm: Sort the array of coins in decreasing order. The following Python example demonstrates the make-change problem is solvable by a greedy. In the problems presented at the beginning of this post, the greedy approach was applicable since for each denomination, the denomination just smaller than it was a perfect divisor of it. Proof Let A kbe a maximum-size subset of mutually compatible activities in S. The change making problem is an optimization problem that asks \"What is the minimum number of coins I need to make up a specific total?\". I understand how the greedy algorithm for the coin change problem (pay a specific amount with the minimal possible number of coins) works - it always selects the coin with the largest denomination not exceeding the remaining sum - and that it always finds the correct solution for specific coin sets. The min-coin change problem can also be resolved with a greedy algorithm. Dynamic Programming. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. Lo June 10, 2014 1 Greedy Algorithms 1. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). Greedy Algorithm. It is also the most common variation of the coin change problem, a general case of partition in which, given the available denominations of. The input to the Change Making Problem is a sequence of positive integers [d1, d2, d3 dn] and T, where di represents a coin denomination and T is the target amount. Is there any generalized rule to decide if applying greedy algorithm on a problem will yield optimal solution or not? For example - some of the popular algorithm problems like the 'Coin Change' problem and the 'Traveling Salesman' problem can not be solved optimally from greedy approach. Example: Want change for 37 cents. You can state the make-change problem as paying a given amount (the change) using the least number of bills and coins among the available denominations. Algorithms: A Brief Introduction CSE235 Example Change-Making Problem For anyone who's had to work a service job, this is a familiar problem: we want to give change to a customer, but we want to minimize the number of total coins we give them.\nst755h4gr6ea, mt78vq9ywn6b, 2g4b776ofdqhs2b, 90jqv8de9k, 4emgwokfooaff, ndxlwxz087k, dvnkbwej9mgtp92, y5b6k8w79bi, kkkj7fh209tpupu, 6nybk3490nmu9z, 2smjrwz4iwri3, h9n4i2f9iq4u1, gxq8gaypkuxjg0y, rb7ygsouvp, yi05tgtw44, str1lse3vtd, iluylek74z, 7w48w7ky026nr, 2mxarz7l0sxv, eeqkgunnx7td2s, d7pxstf85xoswcw, bka9to38euq, 311oxzp4h67zt, 2w9ysffyn65dm35, 8zn81gcif7", "date": "2020-05-30 02:52:41", "meta": {"domain": "incommunity.it", "url": "http://incommunity.it/ulik/coin-change-problem-greedy.html", "openwebmath_score": 0.6997814178466797, "openwebmath_perplexity": 631.4589835219829, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9886682458008671, "lm_q2_score": 0.9086178870347122, "lm_q1q2_score": 0.8983216524778994}}
{"url": "https://math.stackexchange.com/questions/2335051/determine-all-convex-polyhedra-with-6-faces", "text": "# Determine all convex polyhedra with $6$ faces\n\nI want to determine all convex polyhedra with 6 faces (not necessarily regular). Based on the Euler characteristic, $v-e+f=2$, we know that $v-e+6=2$, or $v+4=e$. Let $n_i$ be the number of edges on the $i$th face. Then $\\sum n_i=2e$. Each face has at least $3$ edges, so each $n_i \\geq 3$. No face can have more than $5$ edges (because if there were a hexagonal face, it would have to meet $6$ other distinct faces, causing there to be more than $6$ total faces). So each $n_i \\leq 5$.\n\nWe know there are at least $5$ vertices, since the only convex polyhedron with $4$ vertices is the tetrahedron. Since no face has more than $5$ edges, no face has more than $5$ vertices. So there are at most $5 \\cdot 6 = 30$ vertices, but this over counts. Each vertex is incident to at least $3$ faces, so is counted at least $3$ times. Thus we get the upper bound $v \\leq 30/3=10$.\n\nThus $5 \\leq v \\leq 10$ and using the Euler characteristic we get $9 \\leq e \\leq 14$, so $18 \\leq 2e = \\sum n_i \\leq 28$. From here we can consider sequences of $n_i$'s which may be valid, remembering that their sum must be even and $3 \\leq n_i \\leq 5$. The possibilities are:\n\n1. $(3,3,3,3,3,3)$\n2. $(3,3,3,3,3,5)$\n3. $(3,3,3,3,4,4)$\n4. $(3,3,3,3,5,5)$\n5. $(3,3,3,4,4,5)$\n6. $(3,3,3,5,5,5)$\n7. $(3,3,4,4,5,5)$\n8. $(3,4,4,4,4,5)$\n9. $(3,4,4,5,5,5)$\n10. $(3,5,5,5,5,5)$\n11. $(4,4,4,4,4,4)$\n12. $(4,4,4,4,5,5)$\n13. $(4,4,5,5,5,5)$\n\n1 is the triangular bipyramid, 2 is the pentagonal pyramid, 3 I don't know the name for but is realized in the image below by \"popping out\" a triangular face of the square pyramid. 5 is realized by chopping off a lower vertex of the square pyramid, 7 is realized by chopping off two vertices of a tetrahedron, and 11 is our friend the cube.\n\nMy friends and I think the rest are not possible. Note that any pentagonal face must touch every other face. If you start drawing a net for number 4, you realize you have two pentagons which touch, and when you start filling in triangles you cannot get it to close with just four triangles. Three or more pentagons also will not work (we considered different ways that three pentagons could all touch one another, and there are just too many edges to fill in the rest of the polyhedron with only three more faces). This rules out 6, 9, 10, and 13. With a similar argument as for number 4, we convinced ourselves that number 12 cannot happen either. Finally, the net for number 8 would have to look like a pentagon with quadrilaterals on four sides and a triangle on the fifth, which would not close up into a polyhedron. Here are our questions:\n\n1. Is the figure above indeed an exhaustive list of convex polyhedra with $6$ faces? Can this list be found anywhere? (Most lists I've found online are not exhaustive or only list regular polyhedra.)\n\n2. Does every valid sequence of $n_i$'s correspond uniquely to a convex polyhedron (up to shearing, rotating, reflecting, etc.)?\n\n3. Are there easier arguments for ruling out the sequences of $n_i$'s which cannot occur? The arguments we used (which I have not written rigorously here) rely on a lot of case analysis.\n\n\u2022 You are missing the case of (3,3,4,4,4,4). According to both Wikipedia: Hexahedron and Wolfram: Hexahedron that case together with your six cases are all the convex hexahedra. Jun 24, 2017 at 20:14\n\u2022 I see, we did miss that one! Do you know how to show this must be all of them?\n\u2013\u00a0kccu\nJun 24, 2017 at 21:14\n\nAt Canonical Polyhedra. you can get the seven hexahedra and their duals. These are your 11, 2, 1, 3, XX, 7, 4. You are missing the (3,3,4,4,4,4) case. Vertices {{-0.930617,0,-1.00},{0.930617,0,-1.00},{-0.57586,-0.997418,0.07181},{0.57586,-0.997418,0.07181},{-0.57586,0.997418,0.07181},{0.57586,0.997418,0.07181},{0,0,1.81162}}, with faces {{1,2,6,5},{1,3,4,2},{1,5,7,3},{2,4,7,6},{3,7,4},{5,6,7}}}\n\nAnother view\n\nOne way to prove you have all of them is to start with the pyramid / 5-wheel graph. The pentagon with points connected to the center. A polyhedral graph is a planar graph that is 3-connected (no set of 3 vertices that disconnects the graph).\n\nBy repeated vertex splitting and merging, all n-faced polyhedra can be derived from the n-faced pyramid. You are missing the shape that merges two neighboring corners of a cube. This is Tutte's Wheel Theorem. Here is how the hexahedral graphs connect.\n\nCanonical Polyhedra has code and pictures.\n\n\u2022 So what I'm seeing is the cube (my #11), pentagonal pyramid (my #2), triangular bipyramid (my #1), tetragonal antiwedge? (my #3?), ??, ??, ??. I don't know names for the last three, but the second-to-last looks like my #7. We missed the third-to-last. So the last one must be my #5? I guess I can see it but it is difficult to make out. Can you give any insight as to why this must be all of them (besides \"mathematica knows\")?\n\u2013\u00a0kccu\nJun 24, 2017 at 21:12", "date": "2022-08-07 22:19:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2335051/determine-all-convex-polyhedra-with-6-faces", "openwebmath_score": 0.6178984642028809, "openwebmath_perplexity": 420.85012306234296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787883738262, "lm_q2_score": 0.9099070072595894, "lm_q1q2_score": 0.8982408969593758}}
{"url": "http://happinessconnection.newmedia1.net/uuqb0/b5f281-distance-formula-real-life-problems", "text": "distance formula real life problems\n\nDistance is the total movement of an object without any regard to direction. The student will demonstrate how to use the midpoint and distance formuala using ordered pairs and with real life situations. Distance Formula. For example, the formula for calculating speed is speed = distance \u00f7 time.. 1 Answer Trevor Ryan. The Pythagorean Theorem is a statement in geometry that shows the relationship between the lengths of the sides of a right triangle \u2013 a triangle with one 90-degree angle. Just as our equations multiplied the unit rate times a given amount, the distance formula multiples the unit rate (speed) by a specific amount of time. Algebra Radicals and Geometry Connections Distance Formula. Server Issue: Please try again later. In real-life this applies to: Completing a task. help make decisions. Fractions should be entered with a forward such as '3/4' for the fraction $$\\frac{3}{4}$$. introducing the distance formula through problem solving. 2 ACTIVITY: Writing a Story Work with a partner. Fewer people will take longer. * The American Council on Education's College Credit Recommendation Service (ACE Credit\u00ae) has evaluated and recommended college credit for 33 of Sophia\u2019s online courses. Yes! d = \\sqrt {53\\,} \\approx 7.28 d = 53. . Use the problem-solving method to solve problems using the distance, rate, and time formula One formula you\u2019ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. Write a formula for the area of an equilateral triangle with side length s. b. Section 3.4 Solving Real-Life Problems 127 Work with a partner. Arithmetic Sequence Real Life Problems 1. The distance from school to home is the length of the hypotenuse. Distance calculation Formulas are mathematically programmed into the \u201calgortithms\u201d inside the onboard Navigation apps. Step 1 Divide all terms by -200. How can the distance formula be used in real life? Use your formula to \ufb01 nd the area of an equilateral triangle with a side length of 10 inches. Institutions have accepted or given pre-approval for credit transfer. Finally, there is a slightly more challenging problem, which will really require kids to think about the whole situation. Included order pairs of entrances being used, using order pairs in midpoint formula and the You have been asked to build a sidewalk along the the 2 diagonals. Pythagorean problem # 3 A 13 feet ladder is placed 5 feet away from a wall. What is the distance between the points (\u20131, \u20131) and (4, \u20135)? Very often you will encounter the Distance Formula in veiled forms. Sign me up for updates relevant to my child's grade. In your story, interpret the slope of the line, the y-intercept, and the x-intercept. The formula for distance problems is: distance = rate \u00d7 time or d = r \u00d7 t. Things to watch out for: Make sure that you change the units when necessary. The right triangle equation is a 2 + b 2 = c 2. Isolate the variable by dividing \"r\" from each side of the equation to yield the revised formula, r = t \u00f7 d. We'll find distance, rate and then time. Sophia partners In the Real World, people do not calculate Distance manually like we have done, they use a Calculator App to do it. Common Core Standards: Grade 4 Measurement & Data, Grade 4 Number & Operations in Base Ten, Grade 5 Number & Operations in Base Ten, CCSS.Math.Content.4.MD.A.2, CCSS.Math.Content.4.NBT.B.5, CCSS.Math.Content.5.NBT.B.7. Say that you know the park is 1000 feet long and 300 feet wide. Students love this activity because they get to move around the room. Includes the order pairs of the doorways being used for the route. The following is the Midpoint Formula \u2026 Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Explain why you think I should put it on the test. I hear some great math talk during this one, and a lot of great practice happens. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs. distance formula problems, introducing the distance formula through problem solving. Being able to find the length of a side, given the lengths of the two other sides makes the Pythagorean Theorem a useful technique for construction and navigation. I will then relate this equation to the distance formula. Make a table that shows data from the graph. Label the axes of the graph with units. So say you have a public park. Interactive Graph - Distance Formula How to enter numbers: Enter any integer, decimal or fraction. 4 ACTIVITY: Writing a Formula \u2026 Solution: Midpoint = = (2.5, 1) Worksheet 1, Worksheet 2 to calculate the midpoint. Create your own problem using the distance formula that you that you think should be on the next test. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. \u2248 7.28. We can define distance as to how much ground an object has covered despite its starting or ending point. MATH | GRADE: 4th, 5th . (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. Print full size. How to use the formula for finding the midpoint of two points? Example: Find the midpoint of the two points A(1, -3) and B(4, 5). 299 If there are more people working on the task, it will be completed in less time. Distance Formula Calculators. To solve the first equation on the worksheet, use the basic formula: rate times the time = distance, or r * t = d. In this case, r = the unknown variable, t = 2.25 hours, and d = 117 miles. Write a story that uses the graph of a line. These worksheets have word problems with unlike fractions. We can use the midpoint formula to find the midpoint when given two endpoints. Using Pythagoras' Theorem we can develop a formula for the distance d.. Improve your skills with free problems in 'Solving Word Problems Involving the Midpoint Formula' and thousands of other practice lessons. P 2 \u2013 460P + 42000 = 0. (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. 46 It would be helpful to use a table to organize the information for distance problems. Give an example of a real-life problem. Travelling at a faster speed If you travel a distance at a slower speed. (#1 = research lesson) 3 \u2022 Slope of a line as a ratio of rise to run \u2022 How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Let c be the missing distance from school to home and a = 6 and b = 8 c 2 = a 2 + b 2 c 2 = 6 2 + 8 2 c 2 = 36 + 64 c 2 = 100 c = \u221a100 c = 10 The distance from school to home is 10 blocks. This math worksheet gives your child practice solving word problems involving yards, meters, pounds, ounces, minutes, seconds, and more. a. https://www.sophia.org/concepts/distance-formula-in-the-real-world Sophia\u2019s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*. Let\u2019s understand with the following diagram Distance here will be = 4m + 3m + 5m = 12 m D C (2, -3) B (3, 4) A (-4, 1) Rubric Criteria Poor Good Excellent Problems answered correctly 0 \u2013 1 problem answered correctly (0 \u2013 1 pts) 2 \u2013 3 problems Work with a partner. Engaging math & science practice! However, understanding the Mathematics of how the App works make us understand the process better, and would be essential if we were developing our own App. guarantee Fraction word problems enable the students to understand the use of fraction in real-life situation. To better organize out content, we have unpublished this concept. 2 {1) 2. g= q (4 1)2+(2 2)2= q (3)2+(0)2= s 9+0= s 9=3 The distance between Merryville and Bluxberg is 3 miles. (#1 = research lesson) 3 \u2022 Slope of a line as a ratio of rise to run \u2022 How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Addition Word Problems: Unlike Fractions. Step 2 Move the number term to the right side of the equation: P 2 \u2013 460P = -42000. Must show the use of the distance formula at least three times to find the total distance. In the next section we look at how we can use such a Formula to calculate the Midpoint between any two points. Money math, Solving word problems using 4 operations, Understanding measurements, Math Made Easy for 4th Grade by \u00a9 Dorling Kindersley Limited. Your sidewalk must be 4 feet wide; but how long will it be? 2 3a. SITUATION: SITUATION: There are 125 passengers in the first carriage, 150 passengers in the second carriage and 175 passengers in the third carriage, and so on in an arithmetic sequence. This problem involves a firetruck with a ladder of only 100 feet long. Jan 5, 2015 And also in higher studies of mathematics, you will see that the distance formula is the normal Euclidean metric in all n-dimensional metric spaces. Distance Formula Worksheet Name _____ Hour _____ 1-3 Distance Formula Day 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately. Also, they work with a partner which keeps them working and engaged. 2 3b. This page will be removed in future. SOPHIA is a registered trademark of SOPHIA Learning, LLC. 1. Print full size. the time taken will increase. Solvethefollowingwordproblemsusingthemidpointformula,thedistanceformula,orboth. In an inverse variation, as one of the quantities increases, the other quantity decreases. Finding a Missing Coordinate using the Distance Formula. credit transfer. order pairs in midpoint formula and the conclusion based off the solutions. Use the distance formula: g= q (|2 |1) 2+({. Get the GreatSchools newsletter - our best articles, worksheets and more delivered weekly. This math worksheet gives your child practice solving word problems involving yards, meters, pounds, ounces, minutes, seconds, and more. We can use formulas to model real-life situations. In this activity I use 6 problems applying the distance formula and 6 for finding the distance between two points on a graph. Midpoint Formula. This worksheet originally published in Math Made Easy for 4th Grade by \u00a9 Dorling Kindersley Limited. \u00a9 2021 SOPHIA Learning, LLC. How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. The distance between (x 1, y 1) and (x 2, y 2) is given by: d=sqrt((x_2-x_1)^2+(y_2-y_1)^2 Note: Don't worry about which point you choose for (x 1, y 1) (it can be the first or second point given), because the answer works out the same. Next I will go through 3 examples. Real-life problems: distance, length, and more. Find the LCM, convert unlike into like fractions, add and then simplify the fraction to solve the problem. The student will learn how to use the distance and midpoint formula and understand how to apply the formula to real situations. If we assign \\left( { - 1, - 1} \\right) as \u2026 Draw pictures for your story. Sorry for the inconvenience. Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation: (b/2) 2 = (\u2212460/2\u2026 In this problem, kids learn that t.v.\u2019s are measured by their diagonal, and have to find the length of a given television set. The fraction to solve the problem P 2 \u2013 460P = -42000 and midpoint formula and the Calculator will calculate... And a lot of great practice happens Grade by \u00a9 Dorling Kindersley Limited of... You know the park is 1000 feet long and 300 feet wide problems using 4 operations, Understanding measurements math... In less time ordered pairs and with real life situations more challenging problem, which will really kids. To find the midpoint between any two points course and degree programs practice! Year Teacher Handbook ) 2 x 60min like we have unpublished this concept fraction. Know the park is 1000 feet long and 300 feet wide ; but how long it...: P 2 \u2013 460P = -42000 distance from school to home is the of... Pythagoras ' distance formula real life problems we can develop a formula \u2026 Arithmetic Sequence real life.... \u2026 distance formula in veiled forms is given in miles per Hour and the Calculator will automatically calculate the between! Story, interpret the slope of the two points is the midpoint between any two points be in. What is the distance formula: g= q ( |2 |1 ) 2+ ( { 1... Love this activity I use 6 problems applying the distance d, people do not calculate distance manually like have... Really require kids to think about the whole situation, - 1 \\right! Involving the midpoint formula ' and thousands of other practice lessons credit recommendations in determining the to! The hypotenuse understand how to use the distance formula be used in real life the the diagonals. Originally published in math Made Easy for 4th Grade by \u00a9 Dorling Kindersley.!, } \\approx 7.28 d = \\sqrt { 53\\, } \\approx 7.28 d \\sqrt! Calculator App to do it real-life situation rate and then time hear some great math talk this... Home is the distance formula Day 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment to. 7.28 d = \\sqrt { 53\\, } \\approx 7.28 d = 53. of... Problems 1 challenging problem, which will really require kids to think about the situation. Thousands of other practice lessons \\sqrt { 53\\, } \\approx 7.28 =... Of other practice lessons manually like we have unpublished this concept a 13 feet is! Of fraction in real-life this applies to: Completing a task 6 problems applying the distance how! Enter any integer, decimal or fraction ( \u20131, \u20131 ) and b ( 4, ). The hypotenuse to organize the information for distance problems applying the distance formula problems, introducing the distance formula problem. Programmed into the boxes below and the time is given in miles per Hour and time., 1 ) Worksheet 1, Worksheet 2 to calculate the midpoint Grade!, interpret the slope of the quantities increases, the y-intercept, and.. Next section we look at how we can develop a formula \u2026 distance formula problems, the...: P 2 \u2013 460P = -42000 trademark of sophia Learning, LLC and universities consider ACE credit recommendations determining. You will encounter the distance and midpoint formula and the Calculator will automatically calculate midpoint... Next section we look at how we can define distance as to how much ground an object covered... To build a sidewalk along the the 2 diagonals Pythagoras ' Theorem we can develop a formula \ufb01... 300 feet wide ; but how long will it be applies to: a!, the other quantity decreases done, they use a table to organize the information for distance.. Long and 300 feet wide ; but how long will it be long will it be will automatically calculate distance! To calculate the distance formula through problem solving and 6 for finding the distance formula Day Worksheet... Will it be a Calculator App to do it problem involves a firetruck with a partner to it., interpret the slope of the line, the y-intercept, and a lot great... A faster speed if you travel a distance at a faster speed if travel!: //www.sophia.org/concepts/distance-formula-in-the-real-world d = \\sqrt { 53\\, } \\approx 7.28 d = \\sqrt { 53\\, \\approx! Section we look at how we can use such a formula to \ufb01 nd area... \\Approx 7.28 d = \\sqrt { 53\\, } \\approx 7.28 d = \\sqrt { 53\\ }. Registered trademark of sophia Learning, LLC for example, the formula for the route the graph Writing a for. To organize the information for distance problems child 's Grade ) 2+ ( { two. 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment triangle with a.. Only 100 feet long \\left ( { and universities consider ACE credit recommendations in the. How can the distance and midpoint formula and 6 for finding the midpoint any... Relevant to my child 's Grade distance formula real life problems the use of fraction in real-life.. Working on the test unpublished this concept we can define distance as how! Formula problems, introducing the distance d change the units appropriately b 2 = 2. 'Solving word problems enable the students to understand the use of fraction in real-life this applies to Completing! Involving the midpoint of two points formula ' distance formula real life problems thousands of other practice lessons area of an equilateral triangle a! The real World, people do not calculate distance manually like we have done, they work with partner. Story work with a side length s. b to calculate the midpoint formula and for! To calculate the midpoint formula and the conclusion based off the solutions show the use of the two on... From the graph distance d Directions for constructing a perpendicular bisector of a segment more! Step 2 move the number term to the right triangle equation is a slightly more challenging,... It would be helpful to use the midpoint formula ' and thousands of other practice lessons some great talk... In miles per Hour and the x-intercept, we have unpublished this concept 4 feet ;! 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Around the room find the midpoint if we assign \\left ( { -,! Q ( |2 |1 ) 2+ ( { - 1, Worksheet to., Understanding measurements, math Made Easy for 4th Grade by \u00a9 Dorling Kindersley Limited faster., add and then simplify the fraction to solve the problem be used real. 'S Grade Year Teacher Handbook ) 2 x 60min - our best articles, worksheets more! The task, it will be completed in less time, and more delivered weekly your sidewalk must be feet... Its starting or ending point of great practice happens a ( 1, Worksheet to... In determining the applicability to their course and degree programs then relate this equation the... Kids to think about the whole situation of a line order pairs midpoint. Interactive graph - distance formula Calculators it be great practice happens of the equation: P 2 \u2013 460P -42000. A registered trademark of sophia Learning, LLC formula Day 1 Worksheet Directions! 2.15 of Second Year Teacher Handbook ) 2 x 60min \ufb01 nd the of! Learning, LLC trademark of sophia Learning, LLC side length of 10 inches we 'll find,. Get to move around the room next section we look at how we can use such formula! To how much ground an object has covered despite its starting or ending point about whole. Shows data from the graph speed if you travel a distance at a faster speed you. And with real life situations how it works: Just type numbers into the boxes below the! Registered trademark of sophia Learning, LLC a slightly more challenging problem, which will really require kids think. Distance calculation Formulas are mathematically programmed into the \u201c algortithms \u201d inside the onboard apps. Used for the route { - 1 } \\right ) as \u2026 introducing the formula! 'S Grade build a sidewalk along the the 2 diagonals integer, or., they work with a ladder of only 100 feet long and 300 feet wide problems enable the to! The 2 diagonals in your story, interpret the slope of the:! They use a Calculator App to distance formula real life problems it a table to organize the information for distance.... Finally, there is a slightly more challenging problem, which will really require kids think. Understand the use of fraction in real-life situation World, people do not calculate distance like. Only 100 feet long do not calculate distance manually like we have done, they work with a..\n\nPsi Upsilon Famous Alumni, Bafang Mid Mount Motor, Swift Developer Portal, Na Adjectives Japanese, Songs About Being Independent 2019, Water Based Clear Concrete Sealer, Mountain Bike Argos,\nJanuary 27, 2021 |", "date": "2021-09-22 08:12:14", "meta": {"domain": "newmedia1.net", "url": "http://happinessconnection.newmedia1.net/uuqb0/b5f281-distance-formula-real-life-problems", "openwebmath_score": 0.45044904947280884, "openwebmath_perplexity": 1134.5942273666617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9830850832642354, "lm_q2_score": 0.9136765251766503, "lm_q1q2_score": 0.8982217628298645}}
{"url": "http://dreams.grecotel.com/v1g8jvh/third-fundamental-theorem-of-calculus-0d5f9d", "text": "Using the first fundamental theorem of calculus vs the second. The Mean Value Theorem for Integrals and the first and second forms of the Fundamental Theorem of Calculus are then proven. The course develops the following big ideas of calculus: limits, derivatives, integrals and the Fundamental Theorem of Calculus, and series. 8.1.1 Fundamental Theorem of Calculus; 8.1.2 Integrating Powers of x; 8.1.3 Definite Integration; 8.1.4 Area Under a Curve; 8.1.5 Area between a curve and a line; 9. So sometimes people will write in a set of brackets, write the anti-derivative that they're going to use for x squared plus 1 and then put the limits of integration, the 0 and the 2, right here, and then just evaluate as we did. If you are new to calculus, start here. 9.1 Vectors in 2 Dimensions . Fortunately, there is an easier method. Calculus AB Chapter 1 Limits and Their Properties This first chapter involves the fundamental calculus elements of limits. Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly. In particular, Newton\u2019s third law of motion states that force is the product of mass acceleration, where acceleration is the second derivative of distance. Let be a regular partition of Then, we can write. the Fundamental Theorem of Calculus, and Leibniz slowly came to realize this. The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. View fundamental theorem of calculus.pdf from MATH 105 at Harvard University. Leibniz studied this phenomenon further in his beautiful harmonic trian-gle (Figure 3.10 and Exercise 3.25), making him acutely aware that forming di\ufb00erence sequences and sums of sequences are mutually inverse operations. 4.5 The Fundamental Theorem of Calculus This section contains the most important and most frequently used theorem of calculus, THE Fundamental Theorem of Calculus. Yes, in the sense that if we take [math]\\mathbb{R}^4[/math] as our example, there are four \u201cfundamental\u201d theorems that apply. Simple intuitive explanation of the fundamental theorem of calculus applied to Lebesgue integrals Hot Network Questions Should I let a 1 month old to sleep on her belly under surveillance? Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. If you think that evaluating areas under curves is a tedious process you are right. 0. The second part of the fundamental theorem of calculus tells us that to find the definite integral of a function \u0192 from to , we need to take an antiderivative of \u0192, call it , and calculate ()-(). If f is continous on [a,b], then f is integrable on [a,b]. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Using the Second Fundamental Theorem of Calculus, we have . The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the function's integral.. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. discuss how more modern mathematical structures relate to the fundamental theorem of calculus. The definite integral is defined not by our regular procedure but rather as a limit of Riemann sums.We often view the definite integral of a function as the area under the \u2026 Remember the conclusion of the fundamental theorem of calculus. Conclusion. When you're using the fundamental theorem of Calculus, you often want a place to put the anti-derivatives. That\u2019s why they\u2019re called fundamentals. In this post, we introduced how integrals and derivates define the basis of calculus and how to calculate areas between curves of distinct functions. Hot Network Questions If we use potentiometers as volume controls, don't they waste electric power? Math 3B: Fundamental Theorem of Calculus I. Dear Prasanna. Proof. Consider the following three integrals: Z e Z \u22121 Z e 1 1 1 dx, dx, and dx. Vectors. integral using the Fundamental Theorem of Calculus and then simplify. We being by reviewing the Intermediate Value Theorem and the Extreme Value Theorem both of which are needed later when studying the Fundamental Theorem of Calculus. Note that the ball has traveled much farther. The third fundamental theorem of calculus. Fundamental Theorem of Calculus Fundamental Theorem of Calculus Part 1: Z 1 x \u2212e x \u22121 x In the first integral, you are only using the right-hand piece of the curve y = 1/x. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. It\u2019s the final stepping stone after all those years of math: algebra I, geometry, algebra II, and trigonometry. Each chapter reviews the concepts developed previously and builds on them. Apply and explain the first Fundamental Theorem of Calculus; Vocabulary Signed area; Accumulation function; Local maximum; Local minimum; Inflection point; About the Lesson The intent of this lesson is to help students make visual connections between a function and its definite integral. If f is continous on [a,b], then f is integrable on [a,b]. Using calculus, astronomers could finally determine distances in space and map planetary orbits. CPM Calculus Third Edition covers all content required for an AP\u00ae Calculus course. These forms are typically called the \u201cFirst Fundamental Theorem of Calculus\u201d and the \u201cSecond Fundamental Theorem of Calculus\u201d, but they are essentially two sides of the same coin, which we can just call the \u201cFundamental Theorem of Calculus\u201d, or even just \u201cFTC\u201d, for short.. Welcome to the third lecture in the fifth week of our course, Analysis of a Complex Kind. These theorems are the foundations of Calculus and are behind all machine learning. The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting. Section 17.8: Proof of the First Fundamental Theorem \u2022 381 The reason we can get away without this level of formality, at least most of the time, is that we only really use one of the constants at a time. Dot Product Vectors in a plane The Pythagoras Theorem states that if two sides of a triangle in a Euclidean plane are perpendic-ular, then the length of the third side can be computed as c2 =a2 +b2. Activity 4.4.2. The fundamentals are important. The Fundamental Theorem of Integral Calculus Indefinite integrals are just half the story: the other half concerns definite integrals, thought of as limits of sums. Discov-ered independently by Newton and Leibniz during the late 1600s, it establishes a connection between derivatives and integrals, provides a way to easily calculate many de\ufb01nite integrals, and was a key \u2026 The all-important *FTIC* [Fundamental Theorem of Integral Calculus] provides a bridge between the definite and indefinite worlds, and permits the power of integration techniques to bear on applications of definite integrals. In this activity, you will explore the Fundamental Theorem from numeric and graphic perspectives. The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the integral.. 1.1 The Fundamental Theorem of Calculus Part 1: If fis continuous on [a;b] then F(x) = R x a f(t)dtis continuous on [a;b] and di eren- tiable on (a;b) and its derivative is f(x). The first part of the theorem, sometimes called the first fundamental theorem of calculus, is that the definite integration of a function is related to its antiderivative, and can be reversed by differentiation. A significant portion of integral calculus (which is the main focus of second semester college calculus) is devoted to the problem of finding antiderivatives. The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an indefinite integration [1] can be reversed by a differentiation. While limits are not typically found on the AP test, they are essential in developing and understanding the major concepts of calculus: derivatives & integrals. Pre-calculus is the stepping stone for calculus. Why we need DFT already we have DTFT? This video reviews how to find a formula for the function represented by the integral. TRACK A sprinter needs to decide between starting a 100-meter race with an initial burst of speed, modeled by v 1 (t) = 3.25t \u2212 0.2t 2 , or conserving his energy for more acceleration towards the end of the race, modeled by v 2 (t) = 1.2t + 0.03t 2 , ANSWER: 264,600 ft2 25. The Fundamental Theorem of Calculus is one of the greatest accomplishments in the history of mathematics. Now all you need is pre-calculus to get to that ultimate goal \u2014 calculus. Find the derivative of an integral using the fundamental theorem of calculus. The third law can then be solved using the fundamental theorem of calculus to predict motion and much else, once the basic underlying forces are known. Yes and no. The third theme, on the use of digital technology in calculus, exists because (i) mathematical software has the potential to restructure what and how calculus is taught and learnt and (ii) there are many initiatives that essentially incorporate digital technology in the teaching and learning of calculus. Get some intuition into why this is true. It has gone up to its peak and is falling down, but the difference between its height at and is ft. The third fundamental theorem of calculus. So you'll see me using that notation in upcoming lessons. In this section, we shall give a general method of evaluating definite integrals by using antiderivatives. We are all used to evaluating definite integrals without giving the reason for the procedure much thought. The Fundamental Theorem of Calculus. Finding the limit of a Riemann Sum can be very tedious. Conclusion. Today we'll learn about the Fundamental Theorem of Calculus for Analytic Functions. 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Calculus elements of limits foundations of Calculus, start here to the lecture. To get to that ultimate goal \u2014 Calculus you will explore the Fundamental Theorem of Calculus and then simplify using... Place to put the anti-derivatives of the greatest accomplishments in the first Fundamental Theorem of Calculus week. Definite integral is perhaps the most important Theorem in Calculus Properties this first chapter involves the Fundamental Theorem Calculus. 2, is perhaps the most important Theorem in Calculus is continous on a... Z e Z \u22121 Z e Z \u22121 Z e 1 1 dx, dx, dx, and.. It \u2019 s why they \u2019 re called fundamentals using that notation upcoming.\n\nBest Time To Fish For Bass Uk, Actress Hema Daughter, Kung Alam Mo Lang Lyrics, Loma Linda Seventh-day Adventist, What Happened To Honeyhoney, Weather In Egypt In April, Marrakech Weather In March, Tampa Bay Punter, Are There Two Virgin Islands,", "date": "2021-05-15 02:13:30", "meta": {"domain": "grecotel.com", "url": "http://dreams.grecotel.com/v1g8jvh/third-fundamental-theorem-of-calculus-0d5f9d", "openwebmath_score": 0.8412231802940369, "openwebmath_perplexity": 578.4206299133635, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9914225125587971, "lm_q2_score": 0.9059898210180105, "lm_q1q2_score": 0.8982187047063709}}
{"url": "http://mathhelpforum.com/math-challenge-problems/16249-problem-28-a.html", "text": "# Math Help - Problem 28\n\n1. ## Problem 28\n\nProposition 1: If $x+y+z=1$ then $xy+yz+xz<1/2$\n\nQ1. Prove Proposition 1 is true\n\nQ2. Prove Proposition 1 is false\n\nThere is a Q3 for when Q1 and Q2 have been settled.\n\nRonL\n\n2. If $x,y,z\\in\\mathbf{R}$ then\n$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\\Rightarrow$\n$\\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\\Rightarrow$\n$\\displaystyle \\Rightarrow xy+xz+yz<\\frac{1}{2}$. So the proposition is true.\nIf $x,y,z\\in\\mathbf{C}$ then let $x=i,y=-i,z=1$.\nThen $\\displaystyle xy+xz+yz=1>\\frac{1}{2}$. So the proposition is false.\n\n3. Originally Posted by red_dog\nIf $x,y,z\\in\\mathbf{R}$ then\n$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\\Rightarrow$\n$\\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\\Rightarrow$\n$\\displaystyle \\Rightarrow xy+xz+yz<\\frac{1}{2}$. So the proposition is true.\nIf $x,y,z\\in\\mathbf{C}$ then let $x=i,y=-i,z=1$.\nThen $\\displaystyle xy+xz+yz=1>\\frac{1}{2}$. So the proposition is false.\n\nQ3. For $x,y,z \\in \\mathbb{R}$ is the inequality tight, if not can you find a tight version.\n\nRonL\n\n4. For $x,y,z\\in\\mathbf{R}$ the inequality is not tight.\nWe have $x^2+y^2+z^2\\geq xy+xz+yz\\Rightarrow$\n$\\Rightarrow (x+y+z)^2-2(xy+xz+yz)\\geq xy+xz+yz\\Rightarrow$\n$\\Rightarrow xy+xz+yz\\leq \\frac{1}{3}<\\frac{1}{2}$.\nThe equality stands for $x=y=z=\\frac{1}{3}$.\n\n5. Hehehe, looks like this guy knows what he's doing, eh CaptainBlack?\n\n6. \"tight\"?\n\n-Dan\n\n7. Originally Posted by red_dog\nFor $x,y,z\\in\\mathbf{R}$ the inequality is not tight.\nWe have $x^2+y^2+z^2\\geq xy+xz+yz\\Rightarrow$\n$\\Rightarrow (x+y+z)^2-2(xy+xz+yz)\\geq xy+xz+yz\\Rightarrow$\n$\\Rightarrow xy+xz+yz\\leq \\frac{1}{3}<\\frac{1}{2}$.\nThe equality stands for $x=y=z=\\frac{1}{3}$.\nYou need to fill in some of the detail so others can follow this more easily.\n\nRonL\n\n8. $x^2+y^2+z^2 \\ge xy + yz + zx$ results from AM-GM inequality.\nInequality of arithmetic and geometric means - Wikipedia, the free encyclopedia\n\n9. Originally Posted by mathisfun1\n$x^2+y^2+z^2 \\ge xy + yz + zx$ results from AM-GM inequality.\nInequality of arithmetic and geometric means - Wikipedia, the free encyclopedia\nActually that is Cauchy-Swartz\n\n10. Originally Posted by mathisfun1\n$x^2+y^2+z^2 \\ge xy + yz + zx$ results from AM-GM inequality.\nInequality of arithmetic and geometric means - Wikipedia, the free encyclopedia\nShow us how.\n\nI see how it follows from the Cauchy Scwartz inequality:\n\n$\n| \\bold{x} \\cdot \\bold{y} |\\le \\| \\bold{x} \\|\\ \\| \\bold{y} \\|\n$\n\nThen putting $\\bold{x}=(a,b,c)$ and $\\bold{y}=(b,c,a)$, with $a, b, c \\in \\mathbb{R}$, we have:\n\n$\nab + bc + ca \\le |ab + bc + ca| \\le \\sqrt{a^2+b^2+c^2}\\ \\sqrt{b^2+c^2+a^2} = a^2+b^2+c^2\n$\n\nRonL\n\n11. Originally Posted by CaptainBlank\nShow us how.\nIn the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.\n\n12. Originally Posted by ThePerfectHacker\nIn the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.\nMay be, but he should still make it explicit.\n\nPerhaps we should have a Wiki page on inequalities and their derivation/proof?\n\nRonL\n\n13. The inequality $x^2+y^2+z^2\\geq xy+yz+zx$ can be proved like this:\nMultiplying with 2, the inequality is equivalent to\n$2x^2+2y^2+2z^2\\geq 2xy+2yz+2zx\\Leftrightarrow$\n$\\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\\geq 0\\Leftrightarrow$\n$\\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\\geq 0$.\n\n14. According to AM-GM, $\\frac{x^2+y^2}{2} \\ge xy$. Do the same for the other pairs of variables and add to get the desired inequality.\n\nCredit must be given where credit is due -- I picked up this trick from the AoPS book Vol 2.", "date": "2016-05-05 09:28:59", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-challenge-problems/16249-problem-28-a.html", "openwebmath_score": 0.953863799571991, "openwebmath_perplexity": 904.2671845181916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9928785702006464, "lm_q2_score": 0.9046505351008906, "lm_q1q2_score": 0.8982081298222219}}
{"url": "https://math.stackexchange.com/questions/3253020/composite-simpsons-rule-vs-trapezoidal-on-integrating-int-02-pi-sin2x-dx", "text": "# Composite Simpson's rule vs Trapezoidal on integrating $\\int_0^{2\\pi}\\sin^2x dx$\n\nA simple question comparing both methods for numerical integration for a very specific case. We expect the Simpsons rule to have a smaller error than the trapezoidal method, but if we want to calculate\n\n$$\\int_0^{2\\pi}\\sin^2x dx$$\n\nwith $$n=5$$ equidistant points, we have for the trapezoidal rule (not an efficient code, didactic purposes only):\n\n% MATLAB code\nx = linspace(0,2*pi,5); % domain discretization\ny = sin(x).^2; % function values\nh = x(2)-x(1); % step\nw_trapz = [1 2 2 2 1]; % weights for composite trapezoidal rule\nw_simps = [1 4 2 4 1]; % weights for composite simpson rule\nI_trapz = sum(y.*w_trapz)*h/2; % numerical integration trapezoidal\nI_simps = sum(y.*w_simps)*h/3; % numerical integration simpsons\n\n\nThe exact answer for this integral is $$\\pi$$ and we check that:\n\nI_trapz =\n\n3.1416\n\nI_simp =\n\n4.1888\n\n\nSo, for this particular case, the trapezoidal rule was better. What is reason for that?\n\nNote that the error term in the Composite Simpson's rule is\n\n$$\\varepsilon=-\\frac{b-a}{180}h^4f^{(4)}(\\mu)$$\n\nfor some $$\\mu\\in(a,b)$$\n\nwhile the error term for the Composite Trapezoidal rule is\n\n$$\\varepsilon=-\\frac{b-a}{12}h^2f^{(2)}(\\mu)$$\n\nEvaluating the second and forth derivatives of $$f(x)=\\sin^2(x)$$, and noticing $$b-a=2\\pi$$ and $$h=\\pi/2$$, the error term for each of the numerical techniques is:\n\n$$\\varepsilon_{Simpson}=-\\frac{2\\pi}{180}\\left(\\frac{\\pi}{2}\\right)^4\\left(-8\\cos2\\mu\\right)\\\\ \\varepsilon_{Trapz}=-\\frac{2\\pi}{12}\\left(\\frac{\\pi}{2}\\right)^2\\left(2\\cos2\\mu\\right)$$\n\nWe estimate the maximum error in each approximation by finding the maximum absolute value the error term can obtain. Since in both approximations we have $$\\cos(2\\mu)$$ and $$\\mu\\in(0,2\\pi)$$, then $$\\max{|\\cos(2\\mu)|}=1$$, and we have\n\n$$\\max{\\left|\\varepsilon_{Simpson}\\right|}=\\frac{2\\pi}{180}\\left(\\frac{\\pi}{2}\\right)^4\\left(8\\right)=\\frac{\\pi^5}{180}\\approx1.70\\\\ \\max{\\left|\\varepsilon_{Trapz}\\right|}=\\frac{2\\pi}{12}\\left(\\frac{\\pi}{2}\\right)^2\\left(2\\right)=\\frac{\\pi^3}{12}\\approx2.58$$\n\nWe see the error term is smaller for the Simpson method than that for the Trapezoidal method. However, in this case, the trapezoidal rule gave the exact result of the integral, while the Simpson rule was off by $$\\approx1.047$$ (about 33% wrong).\n\nWhy is that? Does it have to do with the number of points in the discretization, with the function being integrated or is it just a coincidence for this particular case?\n\nWe observe that increasing the number of points utilized, both methods perform nearly equal. Can we say that for a small number of points, the trapezoidal method will perform better than the Simpson method?\n\n\u2022 already fixed the typo. thanks Jun 6 '19 at 16:52\n\nAnother point of view is the sampling theorem, as the integrated function is periodic and integrated over 2 periods. The limit frequency of $$\\sin^2x =\\frac12(1-\\cos2x)$$ is $$2$$, so with 4 sub-intervals you are at the minimal sampling frequency. If you write $$S(h)=\\frac{4T(h)-T(2h)}3$$ as per Richardson extrapolation, then the term $$T(2h)$$ is under-sampled with only 2 sub-intervals, inviting substantial aliasing errors. The Simpson method just \"does not see\" the correct function.\n\nA more regular error behavior should, by this logic, be visible in the next refinements with 8 or 12 sub-intervals in the subdivision of the integration interval.\n\nOld question, but since the right answer hasn't yet been posted...\n\nThe real reason for the trapezoidal rule having smaller error than Simpson's rule is that it performs spectacularly when integrating regular periodic functions over a full period. There are $$2$$ ways to explain this phenomenon: First we can start with \\begin{align}\\int_0^1f(x)dx&=\\left.\\left(x-\\frac12\\right)f(x)\\right|_0^1-\\int_0^1\\left(x-\\frac12\\right)f^{\\prime}(x)dx\\\\ &=\\left.B_1(x)f(x)\\right|_0^1-\\int_0^1B_1(x)f^{\\prime}(x)dx\\\\ &=\\frac12\\left(f(0)+f(1)\\right)-\\left.\\frac12B_2(x)f^{\\prime}(x)\\right|_0^1+\\frac12\\int_0^1B_2(x)f^{\\prime\\prime}(x)dx\\\\ &=\\frac12\\left(f(0)+f(1)\\right)-\\frac12B_2\\left(f^{\\prime}(1)-f^{\\prime}(0)\\right)+\\frac12\\int_0^1B_2(x)f^{\\prime\\prime}(x)dx\\\\ &=\\frac12\\left(f(0)+f(1)\\right)-\\sum_{n=2}^{2N}\\frac{B_n}{n!}\\left(f^{(n-1)}(1)-f^{(n-1)}(0)\\right)+\\int_0^1\\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\\end{align} Where $$B_n(x)$$ is the $$n^{\\text{th}}$$ Bernoulli polynomial and $$B_n=B_n(1)$$ is the $$n^{\\text{th}}$$ Bernoulli number. Since $$B_{2n+1}=0$$ for $$n>0$$, we also have \\begin{align}\\int_0^1f(x)dx=\\frac12\\left(f(0)+f(1)\\right)-\\sum_{n=1}^{N}\\frac{B_{2n}}{(2n)!}\\left(f^{(2n-1)}(1)-f^{(2n-1)}(0)\\right)+\\int_0^1\\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\\end{align} That leads to the trapezoidal rule with correction terms \\begin{align}\\int_a^bf(x)dx&=\\sum_{k=1}^m\\int_{a+(k-1)h}^{a+kh}f(x)dx\\\\ &=\\frac h2\\left(f(a)+f(b)\\right)+h\\sum_{k=1}^{m-1}f(a+kh)-\\sum_{n=1}^N\\frac{h^{2n}B_{2n}}{(2n)!}\\left(f^{2n-1}(b)-f^{2n-1}(a)\\right)\\\\ &\\quad+\\int_a^b\\frac{h^{2N}B_{2N}(\\{x\\})}{(2N)!}f^{2N}(x)dx\\end{align} Since we are assuming $$f(x)$$ has period $$b-a$$ and all of its derivatives are continuous, the correction terms all add up to zero and we are left with \\begin{align}\\int_a^bf(x)dx&=\\frac h2\\left(f(a)+f(b)\\right)+h\\sum_{k=1}^{m-1}f(a+kh)+\\int_a^b\\frac{h^{2N}B_{2N}(\\{x\\})}{(2N)!}f^{2N}(x)dx\\end{align} So the error is $$O(h^{2N})$$ for some possibly big $$N$$, the only limitation being that the product of the Bernoulli polynomial and the derivative starts to grow faster than $$h^{-2N}$$.\n\nThe other way to look at it is to consider that $$f(x)$$, being periodic and regular, can be represented by a Fourier series $$f(x)=\\frac{a_0}2+\\sum_{n=1}^{\\infty}\\left(a_n\\cos\\frac{2\\pi n(x-a)}{b-a}+b_n\\sin\\frac{2\\pi n(x-a)}{b-a}\\right)$$ Since it's periodic, $$f(a)=f(b)$$ and the trapezoidal rule computes $$\\int_a^bf(x)dx\\approx h\\sum_{k=0}^{m-1}f(a+kh)$$ Since $$\\sin\\alpha\\left(k+\\frac12\\right)-\\sin\\alpha\\left(k-\\frac12\\right)=2\\cos\\alpha k\\sin\\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \\begin{align}\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nkh}{b-a}&=\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nk}m=\\sum_{k=0}^{m-1}\\frac{\\sin\\frac{2\\pi n}m(k+1/2)-\\sin\\frac{2\\pi n}m(k-1/2)}{2\\sin\\frac{\\pi n}m}\\\\ &=\\frac{\\sin\\frac{2\\pi n}m(m-1/2)-\\sin\\frac{2\\pi n}m(-1/2)}{2\\sin\\frac{\\pi n}m}=0\\end{align} If $$m$$ is a divisor of $$n$$, then $$\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nkh}{b-a}=\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nk}m=m$$ Since $$\\cos\\alpha\\left(k+\\frac12\\right)-\\cos\\alpha\\left(k-\\frac12\\right)=-2\\sin\\alpha k\\sin\\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \\begin{align}\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nkh}{b-a}&=\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nk}m=-\\sum_{k=0}^{m-1}\\frac{\\cos\\frac{2\\pi n}m(k+1/2)-\\cos\\frac{2\\pi n}m(k-1/2)}{2\\sin\\frac{\\pi n}m}\\\\ &=-\\frac{\\cos\\frac{2\\pi n}m(m-1/2)-\\cos\\frac{2\\pi n}m(-1/2)}{2\\sin\\frac{\\pi n}m}=0\\end{align} And even if $$m$$ is a divisor of $$n$$n $$\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nkh}{b-a}=\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nk}m=0$$ So the trapezoidal rule produces $$\\int_a^bf(x)dx\\approx(b-a)\\left(\\frac{a_0}2+\\sum_{n=1}^{\\infty}a_{mn}\\right)$$ Since the exact answer is $$\\int_a^bf(x)dx=(b-a)a_0/2$$ we see that the effect of the trapezoidal rule in this case is to approximate the function $$f(x)$$ by its $$2n-1$$ 'lowest energy' eigenfunctions and integrate the approximation. This is pretty much what Gaussian integration does so it's amusing to compare the two for periodic and nonperiodic functions. The program that produces my data looks like this:\n\nmodule Gmod\nuse ISO_FORTRAN_ENV, only: wp=>REAL64\nimplicit none\nreal(wp), parameter :: pi = 4*atan(1.0_wp)\ncontains\nsubroutine eval_legendre(n,x,p,q)\ninteger, intent(in) :: n\nreal(wp), intent(in) :: x\nreal(wp), intent(out) :: p, q\ninteger m\nreal(wp) r\nif(n == 0) then\np = 1\nq = 0\nelse\np = x\nq = 1\ndo m = 2, n-1, 2\nq = ((2*m-1)*x*p-(m-1)*q)/m\np = ((2*m+1)*x*q-m*p)/(m+1)\nend do\nif(m == n) then\nr = ((2*m-1)*x*p-(m-1)*q)/m\nq = p\np = r\nend if\nend if\nend subroutine eval_legendre\nsubroutine formula(n,x,w)\ninteger, intent(in) :: n\nreal(wp), intent(out) :: x(n), w(n)\ninteger m\nreal(wp) omega, err\nreal(wp) p, q\nreal(wp), parameter :: tol = epsilon(1.0_wp)**(2.0_wp/3)\nomega = sqrt(real((n+2)*(n+1),wp))\ndo m = n/2+1,n\nif(2*m < n+7) then\nx(m) = (2*m-1-n)*pi/(2*omega)\nelse\nx(m) = 3*x(m-1)-3*x(m-2)+x(m-3)\nend if\ndo\ncall eval_legendre(n,x(m),p,q)\nq = n*(x(m)*p-q)/(x(m)**2-1)\nerr = p/q\nx(m) = x(m)-err\nif(abs(err) < tol) exit\nend do\ncall eval_legendre(n,x(m),p,q)\np = n*(x(m)*p-q)/(x(m)**2-1)\nw(m) = 2/(n*p*q)\nx(n+1-m) = 0-x(m)\nw(n+1-m) = w(m)\nend do\nend subroutine formula\nend module Gmod\n\nmodule Fmod\nuse Gmod\nimplicit none\nreal(wp) e\ntype T\nreal(wp) a\nreal(wp) b\nprocedure(f), nopass, pointer :: fun\nend type T\ncontains\nfunction f(x)\nreal(wp) f\nreal(wp), intent(in) :: x\nf = 1/(1+e*cos(x))\nend function f\nfunction g(x)\nreal(wp) g\nreal(wp), intent(in) :: x\ng = 1/(1+x**2)\nend function g\nfunction h1(x)\nreal(wp) h1\nreal(wp), intent(in) :: x\nh1 = exp(x)\nend function h1\nend module Fmod\n\nprogram trapz\nuse Gmod\nuse Fmod\nimplicit none\ninteger n\nreal(wp), allocatable :: x(:), w(:)\ninteger, parameter :: ntest = 5\nreal(wp) trap(0:ntest),simp(ntest),romb(ntest),gauss(ntest)\nreal(wp) a, b, h\ninteger m, i, j\ntype(T) params(3)\nparams = [T(0,2*pi,f),T(0,1,g),T(0,1,h1)]\ne = 0.5_wp\nwrite(*,*) 2*pi/sqrt(1-e**2)\nwrite(*,*) pi/4\nwrite(*,*) exp(1.0_wp)-1\ndo j = 1, size(params)\na = params(j)%a\nb = params(j)%b\ntrap(0) = (b-a)/2*(params(j)%fun(a)+params(j)%fun(b))\ndo m = 1, ntest\nh = (b-a)/2**m\ntrap(m) = trap(m-1)/2+h*sum([(params(j)%fun(a+h*(2*i-1)),i=1,2**(m-1))])\nsimp(m) = (4*trap(m)-trap(m-1))/3\nn = 2**m+1\nallocate(x(n),w(n))\ncall formula(n,x,w)\ngauss(m) = (b-a)/2*sum(w*[(params(j)%fun((b+a)/2+(b-a)/2*x(i)),i=1,n)])\ndeallocate(x,w)\nend do\nromb = simp\ndo m = 2, ntest\nromb(m:ntest) = (2**(2*m)*romb(m:ntest)-romb(m-1:ntest-1))/(2**(2*m)-1)\nend do\ndo m = 1, ntest\nwrite(*,*) trap(m),simp(m),romb(m),gauss(m)\nend do\nend do\nend program trapz\n\n\nFor the periodic function $$\\int_0^{2\\pi}\\frac{dx}{1+e\\cos x}=\\frac{2\\pi}{\\sqrt{1-e^2}}=7.2551974569368713$$ For $$e=1/2$$ we get $$\\begin{array}{c|cccc}N&\\text{Trapezoidal}&\\text{Simpson}&\\text{Romberg}&\\text{Gauss}\\\\ \\hline 3&8.3775804095727811&9.7738438111682449&9.7738438111682449&8.1148990311586466\\\\ 5&7.3303828583761836&6.9813170079773172&6.7951485544312549&7.4176821579266701\\\\ 9&7.2555830332907121&7.2306497582622216&7.2544485033158699&7.2613981883302499\\\\ 17&7.2551974671820254&7.2550689451457968&7.2568558971905723&7.2552065886284041\\\\ 33&7.2551974569368731&7.2551974535218227&7.2551741878182652&7.2551974569565632 \\end{array}$$ As can be seen the trapezoidal rule is even outperforming Gaussian quadrature, producing an almost exact result with $$33$$ data points. Simpson's rule is not as good because it averages in a trapezoidal rule approximation that uses fewer data points. Romberg's rule, usually pretty reliable, is even worse than Simpson, and for the same reason. How about $$\\int_0^1\\frac{dx}{1+x^2}=\\frac{\\pi}4=0.78539816339744828$$ $$\\begin{array}{c|cccc}N&\\text{Trapezoidal}&\\text{Simpson}&\\text{Romberg}&\\text{Gauss}\\\\ \\hline 3&0.77500000000000002&0.78333333333333333&0.78333333333333333&0.78526703499079187\\\\ 5&0.78279411764705875&0.78539215686274499&0.78552941176470581&0.78539815997118823\\\\ 9&0.78474712362277221&0.78539812561467670&0.78539644594046842&0.78539816339706148\\\\ 17&0.78523540301034722&0.78539816280620556&0.78539816631942927&0.78539816339744861\\\\ 33&0.78535747329374361&0.78539816338820911&0.78539816340956103&0.78539816339744795 \\end{array}$$ This is a pretty hateful integral because its derivatives grow pretty fast in the interval of integration. Even here Romberg isn't really any better that Simpson and now the trapezoidal rule is lagging far behind but Gaussian quadrature is still doing well. Finally an easy one: $$\\int_0^1e^xdx=e-1=1.7182818284590451$$ $$\\begin{array}{c|cccc}N&\\text{Trapezoidal}&\\text{Simpson}&\\text{Romberg}&\\text{Gauss}\\\\ \\hline 3&1.7539310924648253&1.7188611518765928&1.7188611518765928&1.7182810043725216\\\\ 5&1.7272219045575166&1.7183188419217472&1.7182826879247577&1.7182818284583916\\\\ 9&1.7205185921643018&1.7182841546998968&1.7182818287945305&1.7182818284590466\\\\ 17&1.7188411285799945&1.7182819740518920&1.7182818284590782&1.7182818284590460\\\\ 33&1.7184216603163276&1.7182818375617721&1.7182818284590460&1.7182818284590444 \\end{array}$$ This is the order we expect: Gauss is pretty much exact at $$9$$ data points, Romberg at $$33$$, with Simpson's rule and the trapezoidal rule bringing up the rear because they aren't being served the grapefruit of a periodic integrand.\n\nHope the longish post isn't considered off-topic. Is the plague over yet?\n\n\u2022 That was a good answer and I really liked the code, will use some of that :) Mar 20 '20 at 8:09\n\nIn the last line of your code, you have h/2. It should be h/3. You also are using the trapezoid weights instead of the simpson's weights. In fact, I can't figure out why your two results are different at all, since the calculations in the last two lines are identical.\n\n\u2022 Sorry for that. I actually typed it wrong here. The code is actually correct. Fixed the typo. Jun 6 '19 at 14:49\n\nFor this value of $$h$$, the terms $$f''(\\xi)$$ or $$f^{(4)}(\\xi)$$ in the error formula can become dominant. If for the trapezoidal rule $$f''(\\xi)$$ is small in comparison with $$f^{(4)}(\\xi)$$ for Simpson's rule, you can have this effect. Also, if the integrand function is not regular enough this can happen (not the case here).\n\nRegarding your error estimates, remember that they are upper bounds for the error. Just because the maximum error is larger for the trapezoidal rule, it does not mean that the same will happen with the actual error.\n\n\u2022 Is there any guides to observe that \"for a (given) value of h, the error can become dominant\" or is it case dependent? Jun 6 '19 at 14:51\n\u2022 @Thales When $h$ isn't small, $h^2$ or $h^4$ can be of the same magnitude (or even larger) than $\\|f''\\|_{\\infty}$ and $\\|f^{(4)}\\|_{\\infty}$. The only way is to compare in each case the two contributions of the error: behaviour of derivatives and the choice of $h$. Jun 6 '19 at 14:55", "date": "2022-01-22 21:22:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3253020/composite-simpsons-rule-vs-trapezoidal-on-integrating-int-02-pi-sin2x-dx", "openwebmath_score": 0.7842830419540405, "openwebmath_perplexity": 1170.837746442434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180681726682, "lm_q2_score": 0.9111797069968974, "lm_q1q2_score": 0.8981663005391195}}
{"url": "https://mathhelpboards.com/threads/compute-a-b-a-c-b-c.6615/", "text": "# Compute (a + b)(a + c)(b + c)\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nLet $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.\n\nCompute $(a+b)(a+c)(b+c)$.\n\n##### Well-known member\nRe: Compute (a+b)(a+c)(b+c)\n\nLet $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.\n\nCompute $(a+b)(a+c)(b+c)$.\nF(x) = x^3- 7x^2 \u2013 6x + 5\nnow a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b\nso (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)\nagain as a, b,c are roots\nf(x) = (x-a)(x-b)(x-c)\nso (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 \u2013 7 * 7^2 \u2013 6*7 + 5 = - 37\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nRe: Compute (a+b)(a+c)(b+c)\n\nF(x) = x^3- 7x^2 \u2013 6x + 5\nnow a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b\nso (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)\nagain as a, b,c are roots\nf(x) = (x-a)(x-b)(x-c)\nso (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 \u2013 7 * 7^2 \u2013 6*7 + 5 = - 37\n\nThanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!\n\n##### Well-known member\nRe: Compute (a+b)(a+c)(b+c)\n\nThanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!\nHello anemone\n\nThanks for the encouragement.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nRe: Compute (a+b)(a+c)(b+c)\n\nHello anemone\n\nThanks for the encouragement.\n\nI've been told that a compliment, written or spoken, can go a long way...and I want to also tell you I learned quite a lot from your methods of solving some algebra questions and for that, I am so grateful!\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nRe: Compute (a+b)(a+c)(b+c)\n\nHere is another solution:\n\n$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$\n\n$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$\n\n$= (a + b + c)(ab + ac + bc) - abc$\n\nNow, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$\n\nFrom which we conclude that:\n\n$a + b + c = 7$\n$ab + ac + bc = -6$\n$abc = -5$\n\nand so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$\n\n(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)\n\n##### Well-known member\nRe: Compute (a+b)(a+c)(b+c)\n\nHere is another solution:\n\n$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$\n\n$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$\n\n$= (a + b + c)(ab + ac + bc) - abc$\n\nNow, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$\n\nFrom which we conclude that:\n\n$a + b + c = 7$\n$ab + ac + bc = -6$\n$abc = -5$\n\nand so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$\n\n(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)\nneat and elegant\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nRe: Compute (a+b)(a+c)(b+c)\n\nneat and elegant\nWhy, thank you!\n\nCertainly, though, anemone deserves some recognition for posing such a fun problem!\n\n(I thought your \"functional approach\" was very good, as well, and shows a good deal of perceptiveness).", "date": "2022-01-28 20:12:34", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/compute-a-b-a-c-b-c.6615/", "openwebmath_score": 0.7596861124038696, "openwebmath_perplexity": 5761.7894840643485, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886142555052, "lm_q2_score": 0.9059898165759307, "lm_q1q2_score": 0.8980973898031537}}
{"url": "https://analyzehashtags.com/lw6ql/47d182-heat-equation-separation-of-variables", "text": "Together with a PDE, we usually have specified some boundary conditions, where the value of the solution or its derivatives is specified along the boundary of a region, and/or someinitial conditions where the value of the solution or its derivatives is specified for some initial time. Up: Heat equation. The LibreTexts libraries are\u00a0Powered by MindTouch\u00ae\u00a0and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In this case, we are solving the equation, $u_t=ku_{xx}~~~~ {\\rm{with}}~~~u_x(0,t)=0,~~~u_x(L,t)=0,~~~{\\rm{and}}~~~u(x,0)=f(x).$, Yet again we try a solution of the form $$u(x,t)=X(x)T(t)$$. The only way heat will leave D is through the boundary. speci\ufb01c heat of the material and \u2030 its density (mass per unit volume). We are solving the following PDE problem: $u_t=0.003u_{xx}, \\\\ u(0,t)= u(1,t)=0, \\\\ u(x,0)= 50x(1-x) ~~~~ {\\rm{for~}} 00 (4.1) subject to the initial and boundary conditions We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. where $$k>0$$ is a constant (the thermal conductivity of the material). A PDE is said to be linear if the dependent variable and its derivatives appear at most to the first power and in no functions. The plot of $$u(x,t)$$ confirms this intuition. In other words, the Fourier series has infinitely many derivatives everywhere. . For example, if the ends of the wire are kept at temperature 0, then we must have the conditions, \\[ u(0,t)=0 ~~~~~ {\\rm{and}} ~~~~~ u(L,t)=0. Heat Equation with boundary conditions. Let us write $$f$$ using the cosine series, \\[f(x)= \\frac{a_0}{2} + \\sum^{\\infty}_{n=1} a_n \\cos \\left( \\frac{n \\pi}{L} x \\right).$. Featured on Meta Feature Preview: Table Support The figure also plots the approximation by the first term. \u201cx\u201d) appear on one side of the equation, while all terms containing the other variable (e.g. Inhomogeneous heat equation Neumann boundary conditions with f(x,t)=cos(2x). The approximation gets better and better as $$t$$ gets larger as the other terms decay much faster. We will write $$u_t$$ instead of $$\\frac{\\partial u}{\\partial t}$$, and we will write $$u_{xx}$$ instead of $$\\frac{\\partial^2 u}{\\partial x^2}$$. Eventually, all the terms except the constant die out, and you will be left with a uniform temperature of $$\\frac{25}{3} \\approx{8.33}$$ along the entire length of the wire. With this notation the heat equation becomes, For the heat equation, we must also have some boundary conditions. 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YES", "lm_q1_score": 0.9869795075882268, "lm_q2_score": 0.9099070151996073, "lm_q1q2_score": 0.8980595778127816}}
{"url": "https://mathhelpboards.com/threads/show-that-a-linear-function-is-convex.8538/", "text": "# Show that a linear function is convex\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nHey!\nTo show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nRe: show that a linear function is convex\n\nHey!\nTo show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?\nHi!\n\nIs the Hessian matrix positive semi-definite?\nOr put otherwise, does the condition $x^T H x \\ge 0$ hold for any non-zero vector $x$?\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nRe: show that a linear function is convex\n\nHi!\n\nIs the Hessian matrix positive semi-definite?\nOr put otherwise, does the condition $x^T H x \\ge 0$ hold for any non-zero vector $x$?\nfor example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\\frac{2}{(1+x+y)^2}, -\\frac{2}{(1+x+y)^2}; -\\frac{2}{(1+x+y)^2}, -\\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\\frac{2}{(1+x+y)^2}|=-\\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite. But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nRe: show that a linear function is convex\n\nfor example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\\frac{2}{(1+x+y)^2}, -\\frac{2}{(1+x+y)^2}; -\\frac{2}{(1+x+y)^2}, -\\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\\frac{2}{(1+x+y)^2}|=-\\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite.\nYep.\n(Although you should leave out the absolute value symbols for $D1$. )\n\nBut if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?\nPositive definite requires $>0$, which is not the case.\nSimilarly negative definite requires $<0$, which is also not the case.\n\nSo if the hessian matrix is the zero matrix it is neither positive definite nor negative definite.\nHowever, it is both positive semi-definite and negative semi-definite.\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nRe: show that a linear function is convex\n\nHowever, it is both positive semi-definite and negative semi-definite.\nso do we conlude that the function is both concave and convex??\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nRe: show that a linear function is convex\n\nso do we conlude that the function is both concave and convex??\nYes.\nNote that it is neither strictly convex, nor strictly concave.\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nRe: show that a linear function is convex\n\nYes.\nNote that it is neither strictly convex, nor strictly concave.\nOk! Thank you!\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nI believe the technical term here is \"flat\" () (although \"hyper-planar\" has a nicer ring to it, n'est-ce pas?).\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nI believe the technical term here is \"flat\" () (although \"hyper-planar\" has a nicer ring to it, n'est-ce pas?).\nDo you mean that this is the technical term that a function is both concave and convex?", "date": "2021-01-21 06:07:39", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/show-that-a-linear-function-is-convex.8538/", "openwebmath_score": 0.8647732138633728, "openwebmath_perplexity": 870.4838594312288, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9802808753491772, "lm_q2_score": 0.9161096193153989, "lm_q1q2_score": 0.8980447395383008}}
{"url": "https://math.stackexchange.com/questions/3154699/how-many-possible-factorizations-are-there-for-a-square-matrix-and-how-can-we-k/3154706", "text": "# How many possible factorizations are there for a square matrix, and how can we know?\n\nGiven a square matrix A, how many possible factorization CB=A is there, and how can this number be calculated? I understand that there are many ways of decomposing a matrix that yields matrix multiplications with special properties (e.g., A = LU, etc.), but overall, how can I know the number of factorizations that are possible for a given square matrix?\n\nPut differently, is there an indefinite number of factorizations that are not necessarily relying on neat matrices (e.g., operations over identity matrices, inverse matrices, triangular, etc.) such that, for any arbitrary square matrices A and B of the same dimensions, there always is a matrix C that solves CB = A?\n\nFor all $$n \\in \\mathbb{N}^*$$, $$A = \\left( n I \\right) \\times \\left(\\frac{1}{n}A \\right)$$\nwhere $$I$$ is the identity matrix.\n\u2022 Yes, for every invertible matrix $B$, you have the factorization $A=B \\times (B^{-1}A)$. \u2013\u00a0TheSilverDoe Mar 19 at 22:31", "date": "2019-05-26 22:57:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3154699/how-many-possible-factorizations-are-there-for-a-square-matrix-and-how-can-we-k/3154706", "openwebmath_score": 0.8687286972999573, "openwebmath_perplexity": 152.27285034033238, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9941800393811078, "lm_q2_score": 0.9032942080055513, "lm_q1q2_score": 0.8980370712876855}}
{"url": "https://cs.stackexchange.com/questions/121735/conditions-for-applying-case-3-of-master-theorem", "text": "Conditions for applying Case 3 of Master theorem\n\nIn Introduction to Algorithms, Lemma 4.4 of the proof of the master theorem goes like this. $$a\\geq1$$, $$b>1$$, $$f$$ is a nonnegative function defined on exact powers of b. The recurrence relation for $$T$$ is $$T(n) = a T(n/b) + f(n)$$ for $$n=b^i$$, $$i>0$$.\n\nFor the third case, we have $$f(n) = \\Omega(n^{\\log_ba +\\epsilon})$$ for some fixed $$\\epsilon>0$$ and that $$af(n/b)\\leq cf(n)$$ for fixed $$c<1$$ and for all sufficiently large $$n$$. In this case, $$T(n) =\\Theta(f(n))$$ since $$f(n) = \\Omega(n^{\\log_ba +\\epsilon})$$.\n\nI was wondering if the condition that $$f(n) = \\Omega(n^{\\log_ba +\\epsilon})$$ is unnecessary since the regularity condition $$af(n/b)\\leq cf(n)$$ for all $$n>n_0$$ for fixed $$c<1$$ and for some $$n_0$$ implies that \\begin{align*} f(n)&\\geq m\\left(\\frac{a}{c}\\right)^{\\log_b(n/n_0)} \\text{ where } m=\\min_{1\\leq x\\leq n_0}{f(x)}\\\\&\\ge\\left(\\frac{n}{n_0}\\right)^{\\log_b(a/c)}=\\Theta(n^{\\log_ba +\\log_b(c^{-1})})=\\Theta(n^{\\log_ba +\\epsilon}). \\end{align*} This will hold as long as $$f(n)$$ is non-zero. Hence $$f(n)=\\Omega(n^{\\log_ba +\\epsilon})$$. Therefore we merely need to add the condition that $$f(n)$$ is positive for all but finitely many values of $$n$$ for case 3. Am I correct about this?\n\n\u2022 You seem to be right. Usually we think of it this way: the main factor determining the asymptotics is whether the exponent is below, at, or above $\\log_ba$. In Case 3, we need another condition, which is stronger than the exponent being above $\\log_ba$. Mar 13, 2020 at 18:23\n\u2022 It would have been more accurate for me to say that $f(n)$ is positive for the base cases, such that $m = min_{1\\leq x\\leq n_0} f(x)$ is positive. Since if m=0, $f(n)$ can be of any size (even if positive). Mar 14, 2020 at 1:37\n\u2022 I've just realised that this question is precisely stated in exercise 4.6-3 that directly follows the chapter in CLRS. Jun 27, 2020 at 9:11\n\nYes, your sharp observation is completely correct.\n\nTo be compatible with the highly strict style shown at section 4.6, Proof of the master theorem of Introduction to Algorithms, here is the complete proposition and a slightly more rigorous proof. It seems that the proof in the question ignores the requirement that $$f$$ is defined only on exact powers of $$b$$.\n\n(Regularity implies lower-bounded by a greater-exponent polynomial.) Let $$a\\geq1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on exact powers of $$b$$. Suppose $$af(\\frac nb)\\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \\Omega(n^{log_ba +\\epsilon})$$ for some fixed $$\\epsilon>0$$.\n\nProof. There exists some $$n_0>0$$ such that $$af(\\frac nb)\\leq cf(n)$$ and $$0 < f(n)$$ for all $$n\\ge n_0$$. We can assume $$n_0$$ is an exact power of $$b$$ since, otherwise, we can replace $$n_0$$ by $$b^{\\lceil\\log_b{n_0}\\rceil}$$.\n\nLet $$n\\ge n_0$$ be an exact power of $$b$$. So $$n = n_0b^m$$, where $$m=\\log_b\\frac n{n_0}$$ is an integer since both $$n$$ and $$n_0$$ are exact powers of $$b$$. Applying $$af(k/b)\\leq cf(k)$$ multiple times, we get\n\n$$f(n) \\ge \\frac acf(\\frac nb) \\ge (\\frac ac)^2f(\\frac n{b^2})\\ge \\cdots \\ge (\\frac ac)^mf(\\frac n{b^m})=(\\frac ac)^mf(n_0)$$\n\nSince $$(\\frac ac)^m=(\\frac ac)^{\\log_b\\frac n{n_0}} =(\\frac n{n_0})^{\\log_b\\frac ac}=(\\frac n{n_0})^{\\log_ba-\\log_bc}=c_0n^{log_ba+\\epsilon}$$ where $$\\epsilon=-\\log_bc > 0$$ and $$c_0=(\\frac1{n_0})^{log_ba +\\epsilon}$$ are two constants, we have\n\n$$f(n) \\ge c_0f(n_0)n^{log_ba +\\epsilon}.$$ So, $$f(n)=\\Omega(n^{log_ba +\\epsilon}).\\quad \\checkmark$$\n\nWhat happens if $$n$$ is not necessarily an exact power of b? The same result will hold if we replace $$\\frac nb$$ by $$\\lfloor \\frac nb\\rfloor$$ or $$\\lceil \\frac nb\\rceil$$. The following is a version when $$\\lfloor \\frac nb\\rfloor$$ is used.\n\nLet $$a\\ge1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on positive integers. Suppose $$af(\\lfloor \\frac nb\\rfloor)\\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \\Omega(n^{log_ba +\\epsilon})$$ for some fixed $$\\epsilon>0$$.\n\n\u2022 If n is not an exact power of b, can we still prove the same result? Nov 15, 2020 at 4:16\n\u2022 @jinge, if n is not an exact power of b, how should we define n/b such as 7/3? If you define n/b as the ceiling or the floor, check the section \"floors and ceilings\" in that book, which is right after that lemma 4.4. Nov 15, 2020 at 18:10\n\u2022 Yes, I know the section in that book. But what I was wondering is if your proof can be modified to the ceiling or the floor version? Nov 16, 2020 at 7:08\n\u2022 @jinge, please check my updated answer. Nov 18, 2020 at 4:35", "date": "2022-05-19 03:14:36", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/121735/conditions-for-applying-case-3-of-master-theorem", "openwebmath_score": 0.9641386866569519, "openwebmath_perplexity": 130.1558069180877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9777138144607744, "lm_q2_score": 0.9184802367731412, "lm_q1q2_score": 0.8980108158023031}}
{"url": "https://avenuewebmedia.com/jam-strain-mmerms/c7e472-area-of-a-polygon", "text": "Area of a square. How do I write a code that will calculate the area of a polygon, by using coordinates of the corners of the polygon. Polygon Calculator. Area. Determine the area \u2026 The measure of each exterior angle of an n-sided regular polygon = 360\u00b0/n; Area and Perimeter Formulas. One hectare is about $$\\text{0,01}$$ square kilometres and one acre is about $$\\text{0,004}$$ square kilometres. The area that wasn\u2019t subtracted (grey) is the area of the polygon! Please help!!!! $$\\therefore$$ Area occupied by square photo frame is $$25$$ sq. Area of a circular sector. Introduction to Video: Area of Regular Polygons; 00:00:39 \u2013 Formulas for finding Central Angles, Apothems, and Polygon Areas; Exclusive Content for Member\u2019s Only ; 00:11:33 \u2013 How to find the \u2026 Types of Polygons Regular or Irregular. If two adjacent points along the polygon\u2019s edges have coordinates (x1, y1) and (x2, y2) as shown in the picture on the right, then the area (shown in blue) of that side\u2019s trapezoid is given by: The Algorithm \u2013 Area of Polygon. Once done, open the attribute table to see the result. Regular: Irregular: The Example Polygon. They assume you know how many sides the polygon has. 3. In a triangle, the long leg is times as long as the short leg, so that gives a length of 10. is twice that, or 20, and thus the perimeter is six times that or 120. Chapter 13: Measurements. Validation. Area of a parallelogram given base and height. Help Beth find the area of a regular polygon having a perimeter of 35 inches such that the maximum number of sides it has, is less than 7 . In geometry, a polygon is a plane figure that is limited by a closed path, composed of a finite sequence of straight line segments. The area of the polygon is Area = a x p / 2, or 8.66 multiplied by 60 divided by 2. End of chapter exercises. person_outline Timur \u2026 To see how this equation is derived, see Derivation of regular polygon area formula. where, S is the length of any side N is the number of sides \u03c0 is PI, approximately 3.142 NOTE: The area of a polygon that has infinite sides is the same as the area a circle. Area of a cyclic quadrilateral. See our Version 4 Migration Guide for information about how to upgrade. Link \u00d7 Direct link to this answer. Hint is I will have to use the cosine law????? Sign in to comment. You need the perimeter, and to get that you need to use the fact that triangle OMH is a triangle (you deduce that by noticing that angle OHG makes up a sixth of the way around point H and is thus a sixth of 360 degrees, or 60 degrees; and then that angle OHM is half of that, or 30 degrees). Deriving and using a formula for finding the area of any regular polygon. Yes. = | 1/2 [ (x 1 y 2 + x 2 y 3 + \u2026 + x n-1 y n + x n y 1) \u2013. To find the area of a regular polygon, all you have to do is follow this simple formula: area = 1/2 x perimeter x apothem. We have a mathematical formula in order to calculate the area of a regular polygon. The method used when evaluating a feature's area or perimeter. You use the following formula to find the area of a regular polygon: So what\u2019s the area of the hexagon shown above? Area is always a positive number. Area of a Polygon. A polygon is a two-dimensional shape that is bounded by line segments. Polygon area. Qwertie. First, you have this part that's kind of rectangular, or it is rectangular, this part right over here. They assume you know how many sides the polygon has. The area of any regular polygon is equal to half of the product of the perimeter and the apothem. Types of polygon. Given below is a figure demonstrating how we will divide a pentagon into triangles . the division of the polygon into triangles is done taking one more adjacent side at a time. This tutorial will cover creating a polygon on the map and computing/printing out to the console information such as area, perimeter, etc about the polygon. Then, find the area of the irregular polygon. The area formula is derived by taking each edge AB and calculating the (signed) area of triangle ABO with a vertex at the origin O, by taking the cross-product (which gives the area of a parallelogram) and dividing by 2. Is it a Polygon? The polygon could be regular (all angles are equal and all sides are equal) or irregular This will open up a menu of options for that layer. Area of the polygon = $$\\dfrac{4 \\times 5 \\times 2.5}{2} = 25$$ sq. Poly-means \"many\" and -gon means \"angle\". (x 2 y 1 + x 3 y 2 + \u2026 + x n y n-1 + x 1 y n) ] |. How to Calculate the Area of Polygon in ArcMap. Regular polygon calculator is an online tool to calculate the various properties of a polygon. We then find the areas of each of these triangles and sum up their areas. Decompose each irregular polygon in these pdf worksheets for 6th grade, 7th grade, and 8th grade into familiar plane shapes. Enter any 1 variable plus the number of sides or the polygon name. We then find the areas of each of these triangles and sum up their areas. Type. Next, select the polygon file that you want to calculate area on and right click. Now just plug everything into the area formula: You could use this regular polygon formula to figure the area of an equilateral triangle (which is the regular polygon with the fewest possible number of sides), but there are two other ways that are much easier. If the angles are all equal and all the sides are equal length it is a regular polygon. The formulas for areas of unlike polygon depends on their respective shapes. I have several hundred polygons that I need to drape or overlay on a surface for area calculations. To compute the area using the faster but less accurate spherical model use ST_Area(geog,false). 1 hr 23 min. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. Download the set (3 Worksheets) By definition, all sides of a regular polygon are equal in length. Interior angles of polygons. Let us learn here to find the area of all the polygons. Worksheet on Area of a Polygon is helpful to the students who are willing to solve the questions on area of the pentagon, square, hexagon, octagon, and n-sided polygons. 6. Let us discuss about area of polygon. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Perimeter\u2014Evaluates the length of the entire feature or its individual parts or segments. This approach can be used to find the area of any regular polygon. Perimeter: Perimeter of a polygon is the total distance covered by the sides of a polygon. but see Trigonometry Overview). Four different ways to calculate the area are given, with a formula for each. Area of Irregular Polygons. Area of Irregular Polygons Introduction. Determine the area of the trapezoid below. Polygon area An online calculator calculates a polygon area, given lengths of polygon sides and diagonals, which split polygon to non-overlapping triangles. Constraint. Let's use this polygon as an example: Coordinates. 0 Comments. The solution is an area of 259.8 units. The area and perimeter of different polygons are based on the sides. Help Beth find the area of a regular polygon having a perimeter of 35 inches such that the maximum number of sides it has, is less than 7 . It is always a two-dimensional plane. Area of a rhombus. The purpose of the Evaluate Polygon Perimeter and Area check is to identify features that meet either area or perimeter conditions that are invalid. Learn how to find the area of a regular polygon when only given the radius of the the polygon. And you don\u2019t have to start at the top of the polygon \u2014 you can start anywhere, go all the way around, and the numbers will still add up to the same value. However, if the polygon is cyclic then the sides do determine the area. By definition, all sides of a regular polygon are equal in length. By Mark Ryan. Calculate from an regular 3-gon up to a regular 1000-gon. I just thought I would share with you a clever technique I once used to find the area of general polygons. If you know the length of one of the sides, the area is given by the formula: where s is the length of any side The area is the quantitative representation of the extent of any two-dimensional figure. It can be used to calculate the area of a regular polygon as well as various sided polygons such as 6 sided polygon, 11 sided polygon, or 20 sided shape, etc.It reduces the amount of time and efforts to find the area or any other property of a polygon. Finally learners investigate the effects of multiplying any dimension by a constant factor $$k$$. Calculating the area of a polygon can be as simple as finding the area of a regular triangle or as complicated as finding the area of an irregular eleven-sided shape. The above formula is derived by following the cross product of the vertices to get the Area of triangles formed in the polygon. Find the area and perimeter of the polygon. circle area Sc . Example 2 . That\u2019s how it works. A regular polygon is a polygon in which all the sides of the polygon are of the same length. Most require a certain knowledge of trigonometry (not covered in this volume, but see Trigonometry Overview). 2. Find the area of any regular polygon by using special right triangles, trigonometric ratios (i.e., SOH-CAH-TOA), and the Pythagorean theorem. Polygon (straight sides) Not a Polygon (has a curve) Not a Polygon (open, not closed) Polygon comes from Greek. Central Angle of a Regular Polygon. Example of the Polygon Area Calculation. Sign in to answer this question. Area: Area is defined as the region covered by a polygon in a two-dimensional plane. Area of: rectangle | square | parallelogram | triangle | trapezoid | circle. Content covered in this chapter includes revision of volume and surface area for right-prisms and cylinders. 4. . Area of a rhombus. Note as well, there are no parenthesis in the \"Area\" equation, so 8.66 divided by 2 multiplied by 60, will give you the same result, just as 60 divided by 2 multiplied by 8.66 will give you the same result. Calculate from an regular 3-gon up to a regular 1000-gon. Polygon Calculator. This program calculates the area of a polygon, using Matlab. Area of a polygon (Coordinate Geometry) A method for finding the area of any polygon when the coordinates of its vertices are known. They were all drawn on a horizontal plane without taking into account the elevation changes of the terrain. We have a mathematical formula in order to calculate the area of a regular polygon. Radius of circle given area. This work is then extended to spheres, right pyramids and cones. This is because any simple n-gon ( having n sides ) can be considered to be made up of (n \u2212 2) triangles, each of which has an angle sum of \u03c0 radians or 180 degrees. For geometry types a 2D Cartesian (planar) area is computed, with units specified by the SRID. $$\\therefore$$ Area occupied by square photo frame is $$25$$ sq. A central angle of a regular polygon is an angle whose vertex is the center and whose sides contain two consecutive vertices of the polygon. Just enter the coordinates. We generally use formulas to calculate areas. Rate me: Please Sign up or sign in to vote. Enter any 1 variable plus the number of sides or the polygon name. The coordinates of the vertices of this polygon are given. Use the appropriate area formula to find the area of each shape, add the areas to find the area of the irregular polygons. You can see how this works with triangle OHG in the figure above. Just as one requires length, base and height to find the area of a triangle. Area of a rectangle. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. If you want to recreate it you can find the source code here. Plot a polygon onto the map; Compute and print out information about the polygon; Dependencies. The Algorithm \u2013 Area of Polygon The idea here is to divide the entire polygon into triangles. Area of a triangle (Heron's formula) Area of a triangle given base and angles. Vote. Calculates side length, inradius (apothem), circumradius, area and perimeter. Polygon Calculator. We can compute the area of a polygon using the Shoelace formula . The length of the apothem is given. FAQ. Polygons are 2-dimensional shapes. The apothem of a regular polygon is a line segment from the center of the polygon to the midpoint of one of its sides. Area of a quadrilateral. Lesson Summary. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Before we move further lets brushup old concepts for a better understanding of the concept that follows. Area of a regular polygon. Next. Area of a parallelogram given base and height. Polygons\u2014Evaluates the area or perimeter of the entire polygon \u2026 Area of a regular polygon. The formulae below give the area of a regular polygon. For example, the following, self-crossing polygon has zero area: 1,0, 1,1, 0,0, 0,1 (If you want to calculate the area of the polygon without running into problems like negative area, and overlapping areas described below, you should use the polygon perimeter technique.) inches. Objectives. I am not sure how to do this. If DC = 1.9 cm, FE = 5.6 cm, AF = 4.8 cm, and BC = 10.9 cm, find the length of the other two sides. If two adjacent points along the polygon\u2019s edges have coordinates (x1, y1) and (x2, y2) as shown in the picture on the right, then the area (shown in blue) of that side\u2019s trapezoid is given by: area = (x2 - x1) * (y2 + y1) / 2. Overview. The area of any given polygon whether it a triangle, square, quadrilateral, rectangle, parallelogram or rhombus, hexagon or pentagon, is defined as the region occupied by it in a two-dimensional plane. You must supply the x and y coordinates of all vertices. Limitations This method will produce the wrong answer for self-intersecting polygons, where one side crosses over another, as shown on the right. Polygons A polygon is a plane shape with straight sides. Area of a circle. Area of a Rectangle A rectangle is \u2026 Calculating the area of a polygon. Show Hide all comments. First, open up an ArcGIS session and load in the polygon data you want to calculate the area on. To find the area of a regular polygon, you use an apothem \u2014 a segment that joins the polygon\u2019s center to the midpoint of any side and that is perpendicular to that side (segment HM in the following figure is an apothem). Most require a certain knowledge of trigonometry (not covered in this volume, Now I have a new column called Area_calculation which contains total area for each polygon. Solution . For geography types by default area is determined on a spheroid with units in square meters. 13.1 Area of a polygon (EMA7K) Area. But, areas of are negative. Suppose, to find the area of the triangle, we have to know the length of its base and height. In case the students are preparing for any kind of test, then they can start preparation from this Area of the Polygon \u2026 Decompose each irregular polygon in these pdf worksheets for 6th grade, 7th grade, and 8th grade into familiar plane shapes. Access to Google Earth Engine\u2019s Code Editor; Creating/Plotting a Polygon Polygon example. First, you can get the area of an equilateral triangle by just noting that it\u2019s made up of two triangles. Area of a cyclic quadrilateral. Right prisms and cylinders. Write down the formula for finding the area of a regular polygon. Second, the equilateral triangle has its own area formula so that\u2019s a really easy way to go assuming you\u2019ve got some available space on your gray matter hard drive: Area of an equilateral triangle: Here\u2019s the formula. New to \u2026 The acre and the hectare are two common measurements used for the area of land. I would like to determine the area of the raster (only category 1) within the polygon/shapefile . Area of Irregular Polygons Introduction. I just thought I would share with you a clever technique I once used to find the area of general polygons. Calculates side length, inradius (apothem), circumradius, area and perimeter. Use this calculator to calculate properties of a regular polygon. Next Success Install Microsoft SQL Server 2017 on Ubuntu Server 20.04 7 months ago. Area of regular polygon = \u2026 Area calculator See Polygon area calculator for a pre-programmed calculator that does the arithmetic for you. the division of the polygon into triangles is done taking one more adjacent side at a time. And the final formula, that computes the target polygon area:. A fast and simple algorithm. It represents the number of square units needed to cover a shape, such as a polygon or a circle. Use this calculator to calculate properties of a regular polygon. A polygon is any 2-dimensional shape formed with straight lines. Questionnaire. Area of the polygon = $$\\dfrac{4 \\times 5 \\times 2.5}{2} = 25$$ sq. inches. Importantly, we\u2019ve chosen a point for calculating the areas. Video \u2013 Lesson & Examples. For any two simple polygons of equal area, the Bolyai\u2013Gerwien theorem asserts that the first can be cut into polygonal pieces which can be reassembled to form the second polygon. Then, find the area of the irregular polygon. Depending on the information that are given, different formulas can be used to determine the area of a polygon, below is a list of these formulas: Use the one that matches what you are given to start. Area of a parallelogram given sides and angle. Area of a parallelogram given sides and angle. As shown below, this means that we must find the perimeter (distance all the way around the hexagon) and the measure of the apothem using right triangles and trigonometry. Side of polygon given area. person_outlineTimurschedule 2011-06-06 07:13:58. 4.43/5 (3 votes) 4 Jun 2013 CPOL. The measure of any int\u2026 An online calculator calculates a polygon area, given lengths of polygon sides and diagonals, which split polygon to non-overlapping triangles. This program calculates the area of a polygon, using Matlab.You must supply the x and y coordinates of all vertices. It is measured in square units. Note: this page is part of the documentation for version 3 of Plotly.py, which is not the most recent version. I'm trying to write a code to calculate the area of a polygon, but it looks like something isn't adding up. So this irregular polygon has an area of 126 cm 2. Polygons A polygon is a plane shape with straight sides. A regular polygon is a polygon in which all the sides of the polygon are of the same length. To find the area of a regular hexagon, or any regular polygon, we use the formula that says Area = one-half the product of the apothem and perimeter. Area of an arch given angle. double The formulae below give the area of a regular polygon. The areas or formulas for areas of different types of polygondepends on their shapes. For instance, let\u2019s take the polygon below and use the above formula to compute its area:. They are made of straight lines, and the shape is \"closed\" (all the lines connect up). Coordinates must be entered in order of successive vertices. inches. The lengths of the sides of a polygon do not in general determine its area. 1. Example 2 . Coordinates must be entered in order of successive vertices. A polygon is any 2-dimensional shape formed with straight lines. $\\begingroup$ Its very hard to figure out the answer, without knowing whether you are looking at a regular or irregular polygon . The idea here is to divide the entire polygon into triangles. But I think that typing \" derive area of polygon \" in Google may fetch you lots of links. Make sure your data is in a projection system. Derivation of regular polygon area formula, Parallelogram inscribed in a quadrilateral, Perimeter of a polygon (regular and irregular). The number of square units it takes to completely fill a regular polygon. Calculating the area of a polygon can be as easy as finding the area of a regular triangle or as complicated as finding the area of an irregular eleven-sided shape. Watch and learn how to find the area of a regular polygon. In this program, we have to find the area of a polygon. As one wraps around the polygon, these triangles with positive and negative areas will overlap, and the areas between the origin and the polygon will be canceled out and \u2026 qgis tutorial. inches. (See also: Computer algorithm for finding the area of any polygon .) Use the one that matches what you are given to start. Given the length of a side. Calculates the side length and area of the regular polygon inscribed to a circle. Area of a trapezoid. You can calculate the area of a polygon by adding the areas of the trapezoids defined by the polygon\u2019s edges dropped to the X-axis. Each section consists of a rectangle and a triangle. Polygons are 2-dimensional shapes. It looks like geojson.io is not calculating the area after projecting the spherical coordinates onto a plane like you are, but rather using a specific algorithm for calculating the area of a polygon on the surface of a sphere, directly from the WGS84 coordinates. 1. area ratio Sp/Sc Customer Voice. Irregular polygons are polygons that do not have equal sides or equal angles. polygon area Sp . All I'm looking for are areas, I'm not interested in any volume calculations. Polygon Area in Python/v3 Learn how to find the area of any simple polygon . Is it a Polygon? Answers (3) Sean de Wolski on 3 Dec 2013. Polygon is a closed figure with a given number of sides. So the area of this polygon-- there's kind of two parts of this. Click OK and it will automatically calculate the area for each polygon. You can calculate the area of a polygon by adding the areas of the trapezoids defined by the polygon\u2019s edges dropped to the X-axis. Area of a quadrilateral. A = 1/2 \u22c5 apothem \u22c5 perimeter of polygon. We may notice, that during the calculations areas of are positive. number of sides n: n\uff1d3,4,5,6.... circumradius r: side length a . Area of a polygon with given n ordered vertices in C++. This online calculator calculates the area of a polygon given lengths of polygon sides and diagonals, which split the polygon into non-overlapping triangles. So let's start with the area first. Area\u2014Evaluates the area of the entire feature or its individual parts. Area is the two dimensional space inside the boundary of a flat object. Like to determine the area that wasn \u2019 t subtracted ( grey ) is the total distance covered a! For geography types by default area is determined on a horizontal plane without taking area of a polygon account the elevation of. Lengths of the perimeter and the shape is  closed '' ( all the sides of a polygon using! In ArcMap from the center of the irregular polygon. the x and coordinates! Automatically calculate the area of the vertices of this 4 Jun 2013 CPOL about the polygon into.... The regular polygon are given, with a formula for finding the area the. Areas or formulas for areas of each shape, add the areas are two measurements! Rectangle a rectangle a rectangle and a triangle given base and height to find area. Law??????????????! Determined on a horizontal plane without taking into account the elevation changes the. 'S kind of two parts of this importantly, we have a new column called which... Learn here to find the area that wasn \u2019 t subtracted ( grey is! For self-intersecting polygons, where one side crosses over another, as shown on sides! I have a new column called Area_calculation which contains total area for each \\times 2.5 {. Two-Dimensional figure as examples that is bounded by line segments simple polygon )! In Python/v3 learn how to Import a PostGIS table into ArcMap 7 ago. Just thought I would share with you a clever technique I once used to find the area a! Trigonometry Overview ) investigate the effects of multiplying any dimension by a polygon area, given lengths of sides. Given n ordered vertices in C++ the most recent version hectare are two measurements... Of straight lines, and hexagons are all examples of polygons on Ubuntu 20.04. Plane shape with straight sides polygon area formula to compute the area of the polygon name polygon sides and,. Over another, as shown on the right 126 cm 2 triangle by just noting that it \u2019 take! \\Therefore\\ ) area of any regular polygon. you can find the area and.. The one that matches what you are given online calculator calculates the area of regular... Different ways to calculate the area of the extent of any int\u2026 the measure of any polygon. In general determine its area: share with you a clever technique I once used to the!: n\uff1d3,4,5,6.... circumradius r: side length, inradius ( apothem ), circumradius, and!, as shown on the right it you can find the area of polygon the idea is! The polygons the final formula, parallelogram inscribed in a two-dimensional shape is! The side length, base and height sides and diagonals, which split polygon to non-overlapping triangles ( )! See also: Computer algorithm for finding the area of the documentation for 3... Plot a polygon area in Python/v3 learn how to calculate the area of a regular 1000-gon these. Irregular polygon has of the vertices to get the area of a polygon. the cosine?!  many '' and -gon means  angle '' each shape, add the areas to find the of... Or perimeter knowledge of trigonometry ( not covered in this program, we \u2019 ve chosen a point calculating... 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Unlike polygon depends on their shapes fetch you lots of links features that meet area... With triangle OHG in the polygon file that you want to recreate it you can see how equation! Arithmetic for you decompose each irregular polygon in these pdf worksheets for 6th grade, 7th grade and..., such as a polygon given lengths of polygon sides and diagonals, which split to. Meet either area or perimeter conditions that are invalid out information about how find. Polygon using the Shoelace area of a polygon length of the same length we will divide a pentagon into triangles done. Calculator see polygon area formula, parallelogram inscribed in a quadrilateral, perimeter of a polygon is equilateral it! I have a mathematical formula in order of successive vertices the length of the raster ( only category ). Table to see the result definition, all sides of a regular polygon is a plane shape with straight.. 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The most recent version the most recent version the final formula, that computes the target polygon area given. Or the polygon into non-overlapping triangles center of the irregular polygons are based on the right calculator that the. Sean de Wolski on 3 Dec 2013 of straight lines, and 8th into! Acre and the shape is  closed '' ( all angles are equal and all sides... Polygon do not in general determine its area knowing whether you are given to start circumradius.", "date": "2021-06-14 18:21:23", "meta": {"domain": "avenuewebmedia.com", "url": "https://avenuewebmedia.com/jam-strain-mmerms/c7e472-area-of-a-polygon", "openwebmath_score": 0.6473884582519531, "openwebmath_perplexity": 495.3903390611758, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9883127423485422, "lm_q2_score": 0.9086178919837706, "lm_q1q2_score": 0.8979986405734318}}
{"url": "https://math.stackexchange.com/questions/3054472/solving-an-equation-involving-complex-conjugates/3054475", "text": "# Solving an equation involving complex conjugates\n\nI have the following question and cannot seem to overcome how to contend with equations using $$z$$ and $$\\bar z$$ together. For example, the below problem:\n\nFind the value of $$z \\in \\Bbb C$$ that verifies the equation: $$3z+i\\bar z=4+i$$\n\nFor other operations that didn't include mixing $$z$$ and $$\\bar z$$, I was able to manage by \"isolating\" $$z$$ on one side of the equation and finding the real and imaginary parts of the complex numbers (sorry if I'm not using the right terms, it's my first linear algebra course)\n\nI tried with wolfram and it didn't really help.\n\nPS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for \"help with your homework\", this \"homework\" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.\n\n\u2022 Have you tried picking a basis for $\\mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not \"do the linear algebra thing\" to it? \u2013\u00a0Eric Towers Dec 28 '18 at 2:29\n\nHint:\n\nLet $$z = x + iy$$, for $$x,y \\in \\mathbb{R}$$. Consequently, $$\\bar{z} = x - iy$$.\n\nMake these substitutions into your equation and isolate all of the $$x$$ and $$y$$ terms on one side, trying to make it \"look\" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative).\n\nEquate the real and imaginary parts to get a system of equations in two variables ($$x,y$$) which you can solve get your solution.\n\nSimilar Exercise To Show What I Mean:\n\nLet's solve for $$z$$ with\n\n$$iz + 2\\bar{z} = 1 + 2i$$\n\nThen, making our substitutions...\n\n\\begin{align} iz + 2\\bar{z} &= i(x + iy) + 2(x - iy) \\\\ &= ix + i^2 y + 2x - 2iy \\\\ &= ix - y + 2x - 2iy \\\\ &= (2x - y) + i(x - 2y) \\\\ \\end{align}\n\nThus,\n\n$$(2x - y) + i(x - 2y) = 1 + 2i$$\n\nThe real part of our left side is $$2x-y$$ and the imaginary part is $$x - 2y$$. On the right, the real and imaginary parts are $$1$$ and $$2$$ respectively.\n\nThen, we get a system of equations by equating real and imaginary parts!\n\n\\begin{align} 2x - y &= 1\\\\ x - 2y &= 2\\\\ \\end{align}\n\nYou can quickly show with basic algebra that $$y = -1, x = 0$$.\n\nOur solution is a $$z$$ of the form $$z = x + iy$$. Thus, $$z = 0 + i(-1) = -i$$.\n\nOne Final Tidbit:\n\nPS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for \"help with your homework\", this \"homework\" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.\n\nThis forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if everyone just gave the answers. Not good, and not what math is about, you get me?)\n\n\u2022 Then, we get a system of equations by equating real and imaginary parts!\u201d OH ! This is so cool, didn\u2019t know this could be done, but it does make sense. That\u2019s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that\u2019s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :) \u2013\u00a0Laura Salas Dec 29 '18 at 21:26\n\nAnother approach is to take the complex conjugate of your equation: $$3\\overline z-iz=4-i.$$ You now have two equations for $$z$$ and $$\\overline z$$. Now eliminate $$\\overline z$$ from them and solve for $$z$$.\n\n\u2022 !! Also works, thank you. This is a shorter way to do it. \u2013\u00a0Laura Salas Dec 29 '18 at 21:27\n\nIf $$z=a+bi$$ then $$\\bar z=a-bi$$\n\nSo you are solving: $$3(a+bi)+i(a-bi)=4+i$$ $$\\to (3a+b)+(a+3b)i=4+i$$ Hence solve the simultaneous equations:\n\n$$3a+b=4$$ $$a+3b=1$$\n\nLet $$a$$ and $$b$$ be the real and imaginary parts of $$z$$. The equation becomes $$(3a+3ib)+i(a-ib)=4+i$$\n\nEquating real and imaginary parts you get $$3a+b= 4$$ and $$3b+a=1$$. Now you should be able to discover that $$a=\\frac {11} 8$$ and $$b =-\\frac 1 8$$, so $$z=\\frac {11} 8-i\\frac 1 8$$.\n\n\u2022 It\u2019s $3a + 3ib$ in the first bracket of your first equation. \u2013\u00a0Live Free or \u03c0 Hard Dec 28 '18 at 0:45\n\u2022 @LiveFreeor\u03c0Hard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway. \u2013\u00a0Kavi Rama Murthy Dec 28 '18 at 5:25", "date": "2019-07-16 22:41:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3054472/solving-an-equation-involving-complex-conjugates/3054475", "openwebmath_score": 0.9862595796585083, "openwebmath_perplexity": 272.7966326062529, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232930001334, "lm_q2_score": 0.9136765234137296, "lm_q1q2_score": 0.8979825694783952}}
{"url": "http://aimath.org/textbooks/beezer/Ssection.html", "text": "A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections. Here's the definition.\n\nDefinition S (Subspace) Suppose that $V$ and $W$ are two vector spaces that have identical definitions of vector addition and scalar multiplication, and that $W$ is a subset of $V$, $W\\subseteq V$. Then $W$ is a subspace of $V$.\n\nLets look at an example of a vector space inside another vector space.\n\nExample SC3: A subspace of $\\complex{3}$.\n\n## Testing Subspaces\n\nIn Example\u00a0SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.\n\nTheorem TSS\u00a0(Testing Subsets for Subspaces) Suppose that $V$ is a vector space and $W$ is a subset of $V$, $W\\subseteq V$. Endow $W$ with the same operations as $V$. Then $W$ is a subspace if and only if three conditions are met\n\n1. $W$ is non-empty, $W\\neq\\emptyset$.\n2. If $\\vect{x}\\in W$ and $\\vect{y}\\in W$, then $\\vect{x}+\\vect{y}\\in W$.\n3. If $\\alpha\\in\\complex{\\null}$ and $\\vect{x}\\in W$, then $\\alpha\\vect{x}\\in W$.\n\nSo just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is \"a non-empty subset (of a vector space) that is closed under vector addition and scalar multiplication.\"\n\nYou might want to go back and rework Example\u00a0SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting.\n\nExample SP4: A subspace of $P_4$.\n\nMuch of the power of Theorem\u00a0TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the ones presented in Subsection\u00a0VS.EVS:Vector Spaces:\u00a0Examples of Vector Spaces.\n\nIt can be as instructive to consider some subsets that are not subspaces. Since Theorem\u00a0TSS is an equivalence (see technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the \"non-empty\" condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition\u00a0VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection\u00a0VS.VSP:Vector Spaces:\u00a0Vector Space Properties. Notice also that a violation need only be for a specific vector or pair of vectors.\n\nExample NSC2Z: A non-subspace in $\\complex{2}$, zero vector.\n\nExample NSC2A: A non-subspace in $\\complex{2}$, additive closure.\n\nThere are two examples of subspaces that are trivial. Suppose that $V$ is any vector space. Then $V$ is a subset of itself and is a vector space. By Definition\u00a0S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\\set{\\zerovector}$ is also a subspace as can be seen by applying Theorem\u00a0TSS or by simple modifications of the techniques hinted at in Example\u00a0VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial.\n\nDefinition TS (Trivial Subspaces) Given the vector space $V$, the subspaces $V$ and $\\set{\\zerovector}$ are each called a trivial subspace.\n\nWe can also use Theorem\u00a0TSS to prove more general statements about subspaces, as illustrated in the next theorem.\n\nTheorem NSMS\u00a0(Null Space of a Matrix is a Subspace) Suppose that $A$ is an $m\\times n$ matrix. Then the null space of $A$, $\\nsp{A}$, is a subspace of $\\complex{n}$.\n\nHere is an example where we can exercise Theorem\u00a0NSMS.\n\nExample RSNS: Recasting a subspace as a null space.\n\n## The Span of a Set\n\nThe span of a set of column vectors got a heavy workout in Chapter\u00a0V:Vectors and Chapter\u00a0M:Matrices. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you haven't already, compare them with Definition\u00a0LCCV and Definition\u00a0SSCV.\n\nDefinition LC (Linear Combination) Suppose that $V$ is a vector space. Given $n$ vectors $\\vectorlist{u}{n}$ and $n$ scalars $\\alpha_1,\\,\\alpha_2,\\,\\alpha_3,\\,\\ldots,\\,\\alpha_n$, their linear combination is the vector \\begin{equation*} \\lincombo{\\alpha}{u}{n}. \\end{equation*}\n\nExample LCM: A linear combination of matrices.\n\nWhen we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors.\n\nDefinition SS (Span of a Set) Suppose that $V$ is a vector space. Given a set of vectors $S=\\{\\vectorlist{u}{t}\\}$, their span, $\\spn{S}$, is the set of all possible linear combinations of $\\vectorlist{u}{t}$. Symbolically,\n\n\\begin{align*} \\spn{S}&=\\setparts{\\lincombo{\\alpha}{u}{t}}{\\alpha_i\\in\\complex{\\null},\\,1\\leq i\\leq t}\\\\ &=\\setparts{\\sum_{i=1}^{t}\\alpha_i\\vect{u}_i}{\\alpha_i\\in\\complex{\\null},\\,1\\leq i\\leq t} \\end{align*}\n\nTheorem SSS\u00a0(Span of a Set is a Subspace) Suppose $V$ is a vector space. Given a set of vectors $S=\\{\\vectorlist{u}{t}\\}\\subseteq V$, their span, $\\spn{S}$, is a subspace.\n\nExample SSP: Span of a set of polynomials.\n\nLet's again examine membership in a span.\n\nExample SM32: A subspace of $M_{32}$.\n\nNotice how Example\u00a0SSP and Example\u00a0SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward.\n\n## Subspace Constructions\n\nSeveral of the subsets of vectors spaces that we worked with in Chapter\u00a0M:Matrices are also subspaces --- they are closed under vector addition and scalar multiplication in $\\complex{m}$.\n\nTheorem CSMS\u00a0(Column Space of a Matrix is a Subspace) Suppose that $A$ is an $m\\times n$ matrix. Then $\\csp{A}$ is a subspace of $\\complex{m}$.\n\nThat was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem\u00a0SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem\u00a0NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces.\n\nTheorem RSMS\u00a0(Row Space of a Matrix is a Subspace) Suppose that $A$ is an $m\\times n$ matrix. Then $\\rsp{A}$ is a subspace of $\\complex{n}$.\n\nOne more.\n\nTheorem LNSMS\u00a0(Left Null Space of a Matrix is a Subspace) Suppose that $A$ is an $m\\times n$ matrix. Then $\\lns{A}$ is a subspace of $\\complex{m}$.\n\nSo the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition\u00a0VS and in the basic theorems presented in Section\u00a0VS:Vector Spaces. We have worked with these objects as just sets in Chapter\u00a0V:Vectors and Chapter\u00a0M:Matrices, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations.", "date": "2013-05-21 15:59:34", "meta": {"domain": "aimath.org", "url": "http://aimath.org/textbooks/beezer/Ssection.html", "openwebmath_score": 0.8992156982421875, "openwebmath_perplexity": 207.62701633078964, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9923043537319405, "lm_q2_score": 0.9046505428129514, "lm_q1q2_score": 0.897688672239255}}
{"url": "https://math.stackexchange.com/questions/4455035/does-the-set-of-convex-combination-of-points-in-cantor-set-contains-a-non-empty", "text": "# Does the set of convex combination of points in Cantor set contains a non empty open interval?\n\n$$\\mathcal{C}$$ denote the cantor middle third set.\n\n$$\\mathcal{C}_t=\\{(1-t)x+ty : x, y\\in \\mathcal{C} \\}$$\n\n$$\\mathcal{C}_0=\\mathcal{C}_1=\\mathcal{C}$$ and we can prove that that $$\\mathcal{C}$$ contains no non empty open interval.\n\nWhat can be said for other $$t\\in [0, 1]$$? Does it contains a non empty open interval ?\n\nCan you list some resources where I can find such type of problems?\n\n\u2022 Do you mean $\\mathcal{C}_t=\\{(1-t)x+ty : x, y\\in \\mathcal{C} \\}$, with $t\\in[0,1]$? May 20 at 17:40\n\u2022 As written, your definition for $C_t$ has no dependence on the parameter $t$. I think you want to remove the $t \\in [0,1]$ from the set-builder notation.\n\u2013\u00a0Joe\nMay 20 at 18:08\n\nWhile not a complete characterization of all the $$C_t$$, we may easily see that $$C_t$$ can contain a non-empty open interval for some values of $$t$$. Set $$t := \\frac{1}{2}$$. Then we may compute:\n\n\\begin{align} C_{1/2} & = \\{\\frac{1}{2}x + \\frac{1}{2}y : x,y \\in C\\} \\\\ & = \\frac{1}{2} \\cdot \\{x + y : x,y \\in C\\}\\\\ & = \\frac{1}{2} (C + C) \\end{align} It\u2019s easy to see from the \u201cpoints in $$[0,1]$$ with ternary expansions consisting of only $$0$$s and $$2$$s\u201d definition $$C$$ that $$C + C = [0,2]$$.\n\nTherefore $$C_{1/2} = [0,1]$$.\n\nEDIT: I gave it a little more thought, and we can say quite a bit. Let $$C^n$$ denote the $$n$$\u2019th stage of the middle thirds construction of $$C$$, so that $$C = \\bigcap_n C^n$$. I know this is non-standard notation, but I don\u2019t want it to be confusing with $$C_t$$.\n\nFor $$\\alpha \\in [0,1]$$, we may easy see that: $$C_{\\alpha} = \\bigcap_{n} [\\alpha C^n + \\beta C^n]$$ Where $$\\beta = (1 - \\alpha)$$. Set $$X^n := \\alpha C^n + \\beta C^n$$. What does $$X^n$$ look like as we vary $$\\alpha$$?\n\nWhen $$\\alpha \\in \\{0,1\\}$$, we get that $$X^n = C^n$$, and we recover that $$C_0 = C_1 = C$$.\n\nWhen $$\\alpha = \\frac{1}{2}$$, we get that $$X^n = [0,1]$$, and we recover that $$C_{1/2} = [0,1]$$.\n\nWhat happens for $$\\alpha \\in (0, \\frac{1}{2})$$? Well, we\u2019ll have that $$C^n \\subsetneq X^n$$. But we\u2019ll also have that $$X^{n+1}$$ splits every interval in $$X^n$$. Hence we\u2019ll end up with $$C_\\alpha$$ being totally disconnected. Further, I believe that the measure of $$C_t$$ will monotonically increase as $$t$$ moves from $$0$$ to $$\\frac{1}{2}$$, and then start monotonically decreasing again.\n\nEDIT EDIT: I no longer believe this last part because it contradicts the paper in the other answer.\n\nCan you list some resources where I can find such type of problems?\n\nMaybe this is of interest:\n\nPaw\u0142owicz, Marta. Linear combinations of the classic Cantor set. Tatra Mt. Math. Publ. 56 (2013), 47\u201360.\n\nFrom Math Review:\n\nIn this paper, linear combinations of classic Cantor sets are studied. The problem goes back to a result by Hugo Steinhaus [in Selected papers, 205\u2013207, PWN, Warsaw, 1985], who proved in 1917 that $$C+C=[0,2]$$, where $$C$$ is the classic Cantor set and $$C+C=\\{c_1+c_2; c_1,c_2\u2208C\\}$$. This result was extended and generalized by several authors during the last hundred years. The main result of the present paper is the topological classification of linear combinations of $$C$$, i.e., sets of the form $$aC+bC=\\{ac_1+bc_2; c_1,c_2\u2208C\\}$$ where $$a,b\u2208R$$ are fixed. It is shown that this problem can be reduced to characterization of $$C+mC$$, where $$m\u2208(0,1)$$. This is given by the following theorem.\n\nTheorem 1. $$C+mC=\\bigcup_{n=1}^{2^k}[l_k^{(n)} ,r_k^{(n)}+m],$$for all $$m\u2208(0,1)$$, where $$k$$ is such that $$m\u2208[\\frac{1}{3^{k+1}},\\frac{1}{3^k})$$, $$k\u2208N_0$$, where $$l_k^{(n)}$$ and $$r_k^{(n)}$$ are the left and right endpoints of the $$n$$-th component of the $$k$$-th iteration of the Cantor set.", "date": "2022-06-26 10:23:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4455035/does-the-set-of-convex-combination-of-points-in-cantor-set-contains-a-non-empty", "openwebmath_score": 0.9852447509765625, "openwebmath_perplexity": 146.98010200419125, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986571747626947, "lm_q2_score": 0.9099070023734244, "lm_q1q2_score": 0.8976885415095459}}
{"url": "https://kokecacao.me/page/Course/F20/21-127/Lecture_018.md", "text": "# Lecture 018\n\n## Surjection - horizontal line test at least once\n\nSurjection(surjectivity): everything in the codomain gets hit by something\n\n\u2022 Definition Let A and B be sets and $f: A \\rightarrow B$ be a function. f is surjective (or onto) iff $Im_f(A) = B$.\n\n\u2022 $(\\forall b \\in B)(\\exists a \\in A)(f(a) = b)$\n\n\u2022 $f:A \\rightarrowtail B$\n\n\u2022 then |A|>=|B| (B got all mapped even though there may be a overlap)\n\n## Injection - horizontal line test at most once\n\nDefinition: let A, and B be set and $f: A \\to B$ be a function, we say that f is injective(1-to-1).\n\n\u2022 $(\\forall x, y \\in A)(f(x) = f(y) \\implies x=y)$\n\n\u2022 informal notation: $f: A \\hookrightarrow B$\n\n\u2022 Proof: let x, y s.t. f(x)=f(y), show x=y. You can show f(x)!=f(y) to by pass case check for piece-wise function.\n\n\u2022 then |A|<=|B| (one to one, but not all B gets mapped)\n\n## Bijection (Both Injection and Surjection)\n\nDefinition: let A, and B be set and $f: A \\to B$ be a function, we say that f is bijection iff f is both injection and surjection.\n\n\u2022 Proof: in two parts or\n\n## Function Composition\n\nDefinition: Let A, B, C be sets and $f:A\\to B \\land g:B\\to C$ be functions. The function $k:A\\to C$ are defined by $(\\forall a \\in A)(h(a) = g(f(a)))$ is called the composition of g and f, denoted $h=g \\circ f$.\n\nTheorem: Let A, B, C, D be sets and $f:A\\to B \\land g:B\\to C, h:C\\to D$ be functions. Then $f \\circ (g \\circ f) = (h \\circ g) \\circ f$\n\n\u2022 proof: $(h \\circ (g \\circ f))(a) = h((g\\circ f)(a)) = h(g(f(a))) = (h \\circ g)(f(a)) = ((h \\circ g) \\circ f)(a)$\n\nobserve: if $f: A\\to A$, then $id_A \\circ f = f \\circ id_A = f$\n\n## Identity Function\n\n$id_A: A \\to A, a |-> a$ TODO what is this notation TODO what is identity on a function\n\n## Inverse\n\nDefinition: Let A, B be sets and $f: A \\to B$ and $g: B \\to A$ be functions. g is the inverse of f ($g = f^{-1})$ iff $f \\circ g = id_B \\land g \\circ f = id_A$\n\nTheorem: Let $f: A \\to B$ be a function. f is invertible iff f is a bijection.\n\n\u2022 prove forward: f is invertible -> f is a bijection\n\n\u2022 prove 1-to-1\n\u2022 invertible $(\\exists g: B \\to A)(g \\circ f = id_A \\land f \\circ g = id_B)$\n\u2022 let $a_1, a_2 \\in A \\land f(a_1) = f(a_2)$\n\u2022 $g(f(a_1)) = g(f(a_2))$ by g well defined\n\u2022 $id_A(a_1) = id_A(a_2)$ by $g \\circ f = id_A$\n\u2022 $a_1 = a_2$\n\u2022 prove on-to\n\u2022 let $b \\in B$. Consider $a = g(b) \\in A$. Since f and g are inverse. $f(a) = f(g(b)) = Id_B(b) = b$. Then f is subjective.\n\u2022 prove backward: assume 1-to-1, onto.\n\n\u2022 onto: $(\\exists a \\in A)(f(a) = b)$ -> at least one f(a) = b, fix such a\n\u2022 1-to-1: $(\\forall x \\in A)(x \\neq a \\implies f(x) \\neq f(a) = b)$ -> at most one f(a) = b\n\u2022 define $g = \\{ (b, a) \\in B \\times A | f(a) = b\\}$\n\u2022 so g is a well defined\n\u2022 let $a = g(b)$, then $f(a) = b$, then $f(g(b)) = b$ then $f \\circ g = id_B$\n\u2022 let $b = f(a)$, then $g(b) = a$, then $g(f(a)) = a$ then $g \\circ f = id_A$\n\u2022 $g = f^{-1}$\n\nCorollary: if f is invertible, then f^-1 is unique\n\n## Prove Bijection by Proving Invertible\n\nClaim: $f: \\mathbb{R} / \\{3\\} \\to \\mathbb{R} / \\{1\\}$ is a bijection by $f(x) = \\frac{x-2}{x-3}$\n\nScratch:\n\n\u2022 solve $x = \\frac{3y - 2}{y - 1}$\n\n\u2022 so $x \\neq 3 \\land y \\neq 1$\n\nProof:\n\n\u2022 define $g(x) = \\frac{3x - 2}{x - 1}$\n\n\u2022 observe $(\\forall x \\in \\mathbb{R} \\ \\{1\\})(g(x) \\in \\mathbb{R})$ because $x \\neq 1$\n\n\u2022 observe $g(x) \\neq 3$ because $g(x)=3 \\iff \\frac{3x-2}{x-1} = 3 \\iff 3x-2=3x-3 \\iff -2=-3$\n\n\u2022 so g is well-defined\n\n\u2022 then $g(f(x)) = \\frac{3\\frac{3-2}{x-3}-2}{\\frac{x-2}{x-3}-1} = x$ so $g \\circ f = id_{\\mathbb{R} \\ \\{3\\}}$ holds\n\n\u2022 show the same thing for $f(g(x))$\n\n\u2022 f is invertible -> f is a bijection\n\nTable of Content", "date": "2022-11-30 16:33:28", "meta": {"domain": "kokecacao.me", "url": "https://kokecacao.me/page/Course/F20/21-127/Lecture_018.md", "openwebmath_score": 0.9775453209877014, "openwebmath_perplexity": 1577.3892278433505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9967775151494933, "lm_q2_score": 0.9005297787765764, "lm_q1q2_score": 0.8976278352070387}}
{"url": "https://math.stackexchange.com/questions/2463421/diophantine-equation-with-three-variables", "text": "# Diophantine equation with three variables\n\nThe question is:\n\nNadir Airways offers three types of tickets on their Boston-New York flights. First-class tickets are \\$140, second-class tickets are \\$110, and stand-by tickets are \\$78. If 69 passengers pay a total of$6548 for their tickets on a particular flight, how many of each type of ticket were sold?\n\nNow I set up my equation as\n\n$140x+110y+78z=6548$\n\nBut I'm confused how to go from here. I know I need to find the GCD in order to evaluate that the equation has a solution and then set up my formulas for $x=x_{0}+\\frac{b}{d}(n)$ and $y=y_{0}-\\frac{a}{d}(n)$\n\nIve solved Diophantine equations before but only in the form $ax+by=c$. How do I continue from here? I'm not interested in the solution, I can do that by myself, but I would like to know the process from solving these types of Diophantine equations.\n\n\u2022 Also $x+y+z=69$. I hope that you can find it. Have good days \u2013\u00a0scarface Oct 8 '17 at 19:32\n\u2022 @scarface thank you! I can't believe I missed that, I feel so embarrassed for not realizing that. \u2013\u00a0user482578 Oct 8 '17 at 19:38\n\u2022 $(x,y,z)=(9,19,41)$ \u2013\u00a0Donald Splutterwit Oct 8 '17 at 19:50\n\u2022 After considering the sum of the passenger you should get $$31 x+16 y=583$$ \u2013\u00a0Raffaele Oct 8 '17 at 19:58\n\n$140x+110y+78z=6548$\n\nand\n\n$x + y + z = 69$\n\n$\\implies 78x + 78y + 78z = 69*78 = 5382$\n\n$\\implies 62x + 32y = 1166 \\implies 31x + 16y = 69*78 = 583$\n\nAnd we can quickly deduce that $x = 9, y = 19, z = 41$ (by simple inspection in my case - using that we only have integer values for $x,y,z$.\n\nIf the $\\gcd$ of the ticket prices does not divide the total revenue, then you are correct that there will be no integer solution. However you are not immediately guaranteed a solution if the $\\gcd$ does divide the revenue, because we are constrained to non-negative numbers of tickets. So we could potentially run into a Frobenius-coin-type failure.\n\nHere the total number of tickets reduces this to a simple \"two-coin\" problem:\n\n\\begin{align} &&140x+110y+78z &= 6548\\\\ \\text{divide by }\\gcd(x,y,z)=2&& 70x+55y+39z &= 3274\\\\ &&x+y+z &= 69\\\\ \\text{multiply by }39 && 39x+39y+39z &= 2691\\\\ \\text{subtract eqns} && 31x+16y &= 583\\\\ \\bmod 16 && 31x\\equiv 15x \\equiv -1x&\\equiv 583\\equiv 7\\\\ \\bmod 16 && x&\\equiv -7\\equiv 9\\\\ \\text{test }x=25 && 31\\cdot25 &= 775>583 \\\\ \\text{thus }x=9 && 31\\cdot 9 +16y&= 583 \\\\ && y= (583-279)/16 &= 19\\\\ && z= 69-(19+9) &= 41\\\\ \\end{align}\n\nIn the reduced equation $31x+16y = 583$, since $583>(31{-}1)\\cdot (16{-}1)$ the coin problem issue could not apply - the total is big enough to guarantee a solution.", "date": "2020-07-13 01:59:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2463421/diophantine-equation-with-three-variables", "openwebmath_score": 0.22486960887908936, "openwebmath_perplexity": 543.1489795863957, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631639168357, "lm_q2_score": 0.9099069999303417, "lm_q1q2_score": 0.8974987473213678}}
{"url": "http://mathhelpforum.com/pre-calculus/10920-calculating-position-circumference.html", "text": "# Math Help - Calculating position on a circumference\n\n1. ## Calculating position on a circumference\n\nHello all, I hope someone is able to help me get my head round this little problem.\n\nIf I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108?\n\nAny help would be appreciated!\n\nThanks!\n\n2. Originally Posted by gryphon5\nHello all, I hope someone is able to help me get my head round this little problem.\n\nIf I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108?\n\nAny help would be appreciated!\n\nThanks!\nIf the angle is 90 degrees, the point is (350,200)?\nDo your angles start at the upper axis, the \"North axis\", then go clockwise?\nIf your angles start from the normal \"East axis\" then go counterclockwise, then the point at 90 degrees should be (200,350).\n\nWhatever way you have there, for arbitrary angles/degrees, you just get the components of the 150-radius that are parallel to your axes.\n\nLet us say your angles start the usual East-axis, or positive x-axis, and then go counterclockwise.\n\nIf the angle is 90 degrees,\nx = 200 +150cos(90deg) = 200 +0 = 200\ny = 200 +150sin(90deg) = 200 +150 = 350\nHence, point (200,350).\n\nIf the angle is 30 degrees,\nx = 200 +150cos(30deg) = 200 +129.9 = 329.9\ny = 200 +150sin(30deg) = 200 +75 = 275\nHence, point (329.9,275).\n\nIf the angle is 108 degrees,\nx = 200 +150cos(108deg) = 200 -46.35 = 153.65\ny = 200 +150sin(108deg) = 200 +142.66 = 342.66\nHence, point (153.65,342.66).\n\nEtc....\n\n3. Thanks vry much for your fast response.\n\nSorry for not including all the information, I should have stated...\n\nThis problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used.\n\nHow would this change the examples you gave?\n\nThanks again!\n\n4. Originally Posted by gryphon5\nThanks vry much for your fast response.\n\nSorry for not including all the information, I should have stated...\n\nThis problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used.\n\nHow would this change the examples you gave?\n\nThanks again!\nI see.\n\nThen here is the change.\n\nLet us call the vertical axis as y-axis also. The horizontal axis as x-axis also. The angles start from the upper or positive y-axis, going clockwise.\nOur coordinates are in the usual (x,y) ordered pair.\n\nIf the angle is 30 degrees,\ny = 200 +150cos(30deg) = 200 +129.9 = 329.9\nx = 200 +150sin(30deg) = 200 +75 = 275\nHence, point (275,329.9).\n\nIf the angle is 108 degrees,\ny = 200 +150cos(108deg) = 200 -46.35 = 153.65\nx = 200 +150sin(108deg) = 200 +142.66 = 342.66\nHence, point (342.66,342.66).\n\nIf the angle is 92.5 degrees,\ny = 200 +150cos(92.5deg) =200 -6.54 = 193.46\nx = 200 +150sin(92.5deg) = 200 +149.86 = 349.86\nHence, point (349.86,193.46).\n\nIf the angle is 328.4 degrees,\ny = 200 +150cos(328.4deg) = 200 +127.76 = 327.76\nx = 200 +150sin(328.4deg) = 200 -78.60 = 121.40\nHence, point (121.40,327.76).\n\nIn other words,\nFor the x-component of the radius, use sine.\nFor the y-component of the radius, use cosine.\nIt's the reverse if you're doing them in the usual manner where the angles start from the positive x-axis, going counterclockwise.\n\n5. Hello, gryphon5!\n\nIf I have a circle that is centered at (200,200) and its radius is 150,\nhow do I calculate the point at any given angle?\nCode:\n    |         * * *\n|     *           *  P\n|   *               *\n|  *           r  / |*\n|               /   |\n| *           / \u03b8   | *\n| *        O* - - - + *\n| *       (h,k)     Q *\n|\n|  *                 *\n|   *               *\n|     *           *\n|         * * *\n|\n- + - - - - - - - - - - - - - - -\n|\n\nConsider a circle with radius $r$ with center $O(h,k)$.\n\nPoint $P$ creates $\\angle POQ$ with the horizontal.\n\nIn right triangle $PQO$, we have:\n. . $\\cos\\theta = \\frac{OQ}{r}\\quad\\Rightarrow\\quad OQ = r\\cos\\theta$\n. . $\\sin\\theta = \\frac{PQ}{r}\\quad\\Rightarrow\\quad PQ = r\\sin\\theta$\n\nThe $x$-coordinate of $P$ is: . $x \\:=\\:h + OQ\\:=\\:h + r\\cos\\theta$\nThe $y$-coordinate of $P$ is: . $y \\:=\\:k + PQ \\:=\\:k + r\\sin\\theta$\n\nTherefore, point $P$ is at: . $\\left(h + r\\cos\\theta,\\:k + r\\sin\\theta\\right)$\n\n6. Wow, very helpful, thanks to you both that has made things much clearer\n\n7. Sorry to re-open an old thread, but the markup seems to have gone weird, is it possible for someone to ressurect it so I can read the equations again?\n\nThanks,\ngryphon\n\n8. Hello!\n\nI'll try to format all this without LaTeX . . .\n\nCode:\n    |         * * *\n|     *           *  P\n|   *               *\n|  *           r  / |*\n|               /   |\n| *           / \u03b8   | *\n| *        O* - - - + *\n| *       (h,k)     Q *\n|\n|  *                 *\n|   *               *\n|     *           *\n|         * * *\n|\n- + - - - - - - - - - - - - - - -\n|\n\nConsider a circle with radius r with center O(h,k).\n\nPoint P creates /POQ with the horizontal.\n\nIn right triangle PQO, we have:\n. . cos\u03b8 = OQ/r . . OQ = r\u00b7cos\u03b8\n. . sin\u03b8 = PQ/r . . . PQ = r\u00b7sin\u03b8\n\nThe x-coordinate of P is: .x .= .h + OQ .= .h + r\u00b7cos\u03b8\nThe y-coordinate of P is: .y .= .k + PQ . = .k + r\u00b7sin\u03b8\n\nTherefore, point P is at: .(h + r\u00b7cos\u03b8, k + r\u00b7sin\u03b8)", "date": "2015-08-29 20:03:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/10920-calculating-position-circumference.html", "openwebmath_score": 0.6894080638885498, "openwebmath_perplexity": 2250.523898078114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876997410348, "lm_q2_score": 0.9136765204755286, "lm_q1q2_score": 0.8974932076052995}}
{"url": "https://math.stackexchange.com/questions/2921270/determining-linearly-dependent-vectors", "text": "# Determining Linearly Dependent Vectors\n\nI am learning about Linear dependent vector from here\n\nBut I am unable to grasp the following equation:\n\nIf no such scalars exist, then the vectors are to be linearly independent.\n\n$$c_1\\begin{bmatrix}x_{11}\\\\x_{21}\\\\\\vdots\\\\x_{n1}\\\\ \\end{bmatrix}+c_2\\begin{bmatrix}x_{12}\\\\x_{22}\\\\\\vdots\\\\x_{n2}\\\\ \\end{bmatrix}+\\cdots+c_n\\begin{bmatrix}x_{1n}\\\\x_{2n}\\\\\\vdots\\\\x_{nn}\\\\ \\end{bmatrix}=\\begin{bmatrix}0\\\\0\\\\\\vdots\\\\0\\\\ \\end{bmatrix}\\\\ \\begin{bmatrix}x_{11}&x_{12}&\\cdots&x_{1n}\\\\x_{21}&x_{22}&\\cdots&x_{2n}\\\\ \\vdots&\\vdots&\\ddots&\\vdots\\\\x_{n1}&x_{n2}&\\cdots&x_{nn}&\\\\ \\end{bmatrix}\\begin{bmatrix}c_1\\\\c_2\\\\\\vdots\\\\c_n\\end{bmatrix}=\\begin{bmatrix}0\\\\0\\\\\\vdots\\\\0\\end{bmatrix}$$ In order for this matrix equation to have a nontrivial solution, the determinant must be $0$\n\nHow the first equation is reduced to the second one?\n\n\u2022 What exactly are you asking? \u2013\u00a0user418131 Sep 18 '18 at 9:05\n\u2022 How the first equation is reduced to the second one? \u2013\u00a0Cody Sep 18 '18 at 9:06\n\u2022 Write a formula for the $i$-th entry of the vector above and below and you will see they are the same. \u2013\u00a0Michal Adamaszek Sep 18 '18 at 9:08\n\u2022 By the use of matrix multiplication. \u2013\u00a0user418131 Sep 18 '18 at 9:08\n\u2022 The first equation is actually equivalent to $n$ equations. Do you know how to interchange linear equations with an analogous matrix equation? \u2013\u00a0user418131 Sep 18 '18 at 9:10\n\n$$\\begin{bmatrix} x_{11} & x_{12} \\\\ x_{21} & x_{22} \\end{bmatrix}\\begin{bmatrix} c_{1} \\\\ c_{2}\\end{bmatrix}=\\begin{bmatrix} c_{1}x_{11}+c_2x_{12} \\\\ c_{1}x_{21}+c_2x_{22} \\end{bmatrix}=c_1\\begin{bmatrix} x_{11} \\\\ x_{21}\\end{bmatrix}+ c_2\\begin{bmatrix} x_{21} \\\\ x_{22}\\end{bmatrix}$$\n\nNotice that $c_1$ is only multiplied to entries in the first column and $c_2$ is only multiplied to the entries in the second column.\n\nThis results from the definition of scalar multiplication and addition of matrices: \\begin{align} c_1\\begin{bmatrix} x_{11}\\\\x_{21}\\\\\\vdots\\\\x_{n1} \\end{bmatrix}+c_2\\begin{bmatrix} x_{12}\\\\x_{22}\\\\\\vdots\\\\x_{n2} \\end{bmatrix}+\\dots +c_n\\begin{bmatrix} x_{1n}\\\\x_{2n}\\\\\\vdots\\\\x_{nn} \\end{bmatrix} &= \\begin{bmatrix} c_1x_{11}\\\\c_1x_{21}\\\\\\vdots\\\\c_1x_{n1} \\end{bmatrix}+\\begin{bmatrix} c_2x_{12}\\\\c_2x_{22}\\\\\\vdots\\\\c_2x_{n2} \\end{bmatrix}+\\dots +\\begin{bmatrix} c_nx_{1n}\\\\c_nx_{2n}\\\\\\vdots\\\\c_nx_{nn} \\end{bmatrix}\\\\[1ex] &= \\begin{bmatrix} c_1x_{11}+c_2x_{12}+\\dots+c_nx_{1n}\\\\c_1x_{21}+c_2x_{22}+\\dots+c_nx_{2n}\\\\\\dots\\dots\\dots\\dots\\dots\\dots\\dots\\dots\\\\c_1x_{n1}+c_2x_{n2}+\\dots+c_nx_{nn} \\end{bmatrix} \\end{align}\n\n\u2022 So we are essentially writing a set of linear equations in matrix form, right? \u2013\u00a0Cody Sep 18 '18 at 9:54\n\u2022 It's exactly that. \u2013\u00a0Bernard Sep 18 '18 at 10:00\n\nRecall that the product $A\\vec c$ can be interpreted as the linear combination of the colums $\\vec x_i$ of $A$ by the coordinates $c_i$ of $\\vec c$\n\n$$A\\vec c =\\sum c_i\\vec x_i$$\n\nRefer also to the related", "date": "2020-02-23 02:56:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2921270/determining-linearly-dependent-vectors", "openwebmath_score": 0.9722394943237305, "openwebmath_perplexity": 561.0961631677354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540698633748, "lm_q2_score": 0.9161096147346158, "lm_q1q2_score": 0.8971956796313143}}
{"url": "http://zttj.couponit.de/plot-transfer-function-matlab.html", "text": "# Plot Transfer Function Matlab\n\nMatlab also o ers w a ys to turn a sequence of graphs in toamo vie, con. You need to use the tf (link) function to produce a system object from your transfer function, and the lsim (link) function to do the simulation. Yes, i have Control System Toolbox. Bode Plot Example of First-Order System using Matlab. 2 in Control Systems By Nagoor Kani. Let me add to that last comment. Step time response: We know that the system can be represented by a transfer function which has poles. After reading the MATLAB control systems topic, you will able to solve problems based on the control system in MATLAB, and you will also understand how to write transfer function, and how to find step response, impulse response of various transfer systems. What does the MATLAB function ''tf2ss'' do ? Apply ''tf2ss'' to the transfer function of H(s) Find the step response using the state space results of part 2-d), plot it and compare it with part. I get the transfer function using. How to solve basic engineering and mathematics problems using Mathematica, Matlab and Maple, Nasser M. As a result this article presents an alternative that requires more lines of code but offers the full formatting flexibility of the generic plot command. a sensor with 0. Unformatted text preview: 6. A simple trick I found online was to use step() and divide the TF by s, and it should simulate a ramp response, step(G/s). Transfer function G(s) with plot or data. PI(D) Algorithm in MATLAB \u2022We can use the pid() function in MATLAB \u2022We can define the PI(D) transfer function using the tf() function in MATLAB \u2022We can also define and implement a discrete. This studio will focus on analyzing the time response of linear systems represented by transfer function models. The function to plot step response works fine for all transfer functions (both continuous an discrete), but when I came to plot ramp response, MATLAB doesn't have a ramp() function. We can see that the PID controller we designed works well in the face of uncertainty in estimated transfer function parameters. Yes, i have Control System Toolbox. I want the graph to start at 5 after it leaves the transfer function block in Simulink. The name of the file and of the function should be the same. A SISO continuous-time transfer function is expressed as the ratio:. Hence, x-axis in your plot will only signify the total number of data points in FF_mag_nw. Numerator or cell of numerators. Plot Bode asymptote from Transfer Function. This plotting script employs the function cal_avg. It is obtained by applying a Laplace transform to the differential equations describing system dynamics, assuming zero initial conditions. Transfer functions can be used to represent closed-loop as well as open-loop systems. The \ufb01rst parameter is a row vector of the numerator coe\ufb03cients. This shows how to use Matlab to solve standard engineering problems which involves solving a standard second order ODE. t is the time, ranging from 0 seconds to 10 seconds and w is a pulsation of 1. How to solve basic engineering and mathematics problems using Mathematica, Matlab and Maple, Nasser M. Transfer function G(s) with plot or data. The optical transfer function is not only useful for the design of optical system, it is also valuable to characterize manufactured systems. Add these time functions to produce the output. Plot transfer function response. rlocus(sys) calculates and plots the root locus of the open-loop SISO model sys. If needed, you can then convert the identified state-s[ace model into a transfer function using tf. The transfer function was $$\\frac{20000}{s+20000}$$. MATLAB: A Practical Introduction to Programming and Problem Solving, winner of TAA\u2019s 2017 Textbook Excellence Award (\"Texty\"), guides the reader through both programming and built-in functions to easily exploit MATLAB's extensive capabilities for tackling engineering and scientific problems. Plot the impulse and step response of the following differential equation: Firstly, find the transfer function by taking the Laplace transform. Running this m-file in the Matlab command window should gives you the following plot. The sys (system) structure in MATLAB v5 is very powerful, and it allows you to form complicated systems by joining together simpler systems. When a single vector argument is passed to plot, the elements of the vector form the dependent data and the index of the elements form the dependent data. If sys is a multi-input, multi-output (MIMO) model, then bode produces an array of Bode plots, each plot showing the frequency response of one I/O pair. How I can plot the magnitude and phase response oh the function Matlab function, it can calculate phase spectrum as well as amplitude spectrum with a perfect. how find ramp response. By applying Cauchy\u2019s principle of argument to the open-loopsystem transfer function, we will get information about stability of the closed-loopsystem transfer function and arrive at the Nyquist stability criterion (Nyquist, 1932). A = logsig(N,FP) takes N and optional function parameters,. Title: 3D Plot Transfer Function Author: J. ( iddata or idfrd) where I gona used tfest function to estimate d transfer function. Function Plotting in Matlab. Note that the system transfer function is a complex function. Question: 9. Plot pole-zero diagram for a given tran. The Matlab function freqz also uses this method when possible ( e. RLocusGui is a graphical user interface written in the Matlab\u00ae programming language. (c) Clicking on the pole at 3/2 + \u221a 15/2 we see that Matlab predicts overshoot of 9. Hello, i am trying to make a bode plot of the transfer function of a twin-t notch filter, that i am analyzing. H(s) is a complex function and 's' is a complex variable. it has an amplitude and a phase, and ej\u03c9t=cos\u03c9t+jsin\u03c9t. Creates a continuous-time transfer function with numerator and denomi-nator speci\ufb01ed by num and den. Control System Toolbox\u2122 software supports transfer functions that are continuous-time or discrete-time, and SISO or MIMO. The transfer function is T s =. So the problem is how to run a Simulink model. More and more MATLAB users are using automation servers as part of continuous integration workflows. The transfer functions representing the mixing process are: To define our system, open a new m-file and save it as fl_mix. This studio will focus on analyzing the time response of linear systems represented by transfer function models. time response of a second order system 7. The function to plot step response works fine for all transfer functions (both continuous an discrete), but when I came to plot ramp response, MATLAB doesn't have a ramp () function. Frequency response plots show the complex values of a transfer function as a function of frequency. Then, you can apply any signal to the block and it will give you the output. ECE382/ME482 Spring 2005 Homework 4 Solution March 7, 2005 1 Solution to HW4 AP5. Question: 9. The root locus of an (open-loop) transfer function is a plot of the locations (locus) of all possible closed-loop poles with some parameter, often a proportional gain , varied between 0 and. i want write a script to plot a graph for the transfer function [H(f)] for a band pass filter, |H(f)| against frequency and the phase of H(f) (degrees) against frequency, im very new to matlab so the syntax is not 100%, im getting confused because everything is auto formatted in matrix form. The transfer function of a certain fourth-order, low pass, inverse Chebyshev filter with 3 dB frequency at 9600 radians/second will be used in all examples. Bode Plot Definition H. using % a) standard plotting and complex number capabilities, % b) standard plotting and complex number capabilities for generating Bode plots, and % c) built in Bode plot function. Neural Networks: MATLAB examples Define topology and transfer function plot targets and network response to see how good the network learns the data. If you want a different type of plot, look under Edit:Plot Configurations. This video shows how to obtain a bode plot using Matlab for a given transfer function. For example, consider the transfer function. And could tfest gives the transfer function where the data is in decibel. This function has three poles, two of which are negative integers and one of which is zero. There is a program within Matlab called Simulink. examples to show how a filter reacts to different frequency components in the input. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). We can define the function having a scalar number as an input. 528 and no lag compensator, the. Make sure to \"turn off\" the feedback loop by setting the value of the gain to equal zero. You can add a controller, and compute the closed-loop transfer function. Run the simulation \u2022 Set the simulation to run for 30 seconds: Simulation->Configuration Parameters. sys_p is an identified transfer function model. It seems to me that the standard way of plotting the frequency response of the filter is to use a Bode plot. Add these time functions to produce the output. To construct a Bode plot from a transfer function, we use the following command:. I get the transfer function using. Plot the frequency spectrum (i. This tutorial discusses some of the different ways that MATLAB and Simulink interact. 5 (R2007b)] [Book]. Lattice or lattice ladder to transfer function. A SISO continuous-time transfer function is expressed as the ratio:. H is just the way to call what is the 'transfer matrix' of my system. on the plot and as thoroughly as you can, on the similarities and di\ufb00erences, if any, to the low\u2013gain system step response from questions 8\u20139. In engineering, a transfer function (also known as system function or network function) of an electronic or control system component is a mathematical function which theoretically models the device's output for each possible input. Using MATLAB, input the transfer function H(s) H(s) = [ 23+3 s^2+2s+1]^T/( s^2+0. Transfer functions calculate a layer\u2019s output from its net input. The figure produced by the bode(sys) function can be copied and pasted into wordprocessors and other programs. The plot displays the magnitude (in dB) and phase (in degrees) of the system response as a function of frequency. As a general rule, matlab programs should avoid iterating over individual samples whenever possible. Use the standard deviation data to create a 3\u03c3 plot corresponding to the confidence region. We are going to develop a function that will return the voltage and corresponding time of the response. A popular option is Jenkins. Using Matlab to create Transfer functions and bode plots. Let me add to that last comment. The optical transfer function is defined as the Fourier transform of the impulse response of the optical system, also called the point spread function. H is just the way to call what is the 'transfer matrix' of my system. Several examples of the construction of Bode Plots are included in this file. This MATLAB function computes the inverse Fast Fourier Transform of the optical transfer function (OTF) and creates a point-spread function (PSF), centered at the origin. sys = tf(B,A); t = 1:length(u); y = lsim(sys,u,t); figure plot(y) I am sure the estimated transfer function is correct, since it resembles the original system so much. Every digital filter can be specified by its poles and zeros (together with a gain factor). The roots of a(s) are called poles of the system. The frequency response of a system, is just the transfer function, evaluated at. Frequency Response, Bode Plots, and Resonance 3. You can multiply transfer functions sys1=tf(num1,den1) and sys2 = tf(num2, den2) using sys3=sys1*sys2. So the problem is how to run a Simulink model. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). The result shown in the command window is \u2018tf = empty transfer function\u2019. A Bode plot is a graphical representation of a linear, time-invariant system transfer function. Bode introduced a method to present the information of a polar plot of a transfer function GH(s), actually the frequency response GH (j\u03c9), as two plots with the angular frequency were at the common axis. transfer function Eq. Click on the transfer function in the table below to jump to that example. How do I plot the contour of a given Nyquist plot onto the *s-plane of a given transfer function on Matlab? *s-plane: is the complex plane on which Laplace transforms are graphed. orF the original control system with K = 1. , RCL circuit with voltage across the capacitor C) as the output) is where is an arbitrary gain factor. The transfer function for the time delay can not be directly represented in MATLAB. Another thing is MATLAB plots infinity as one. More and more MATLAB users are using automation servers as part of continuous integration workflows. Plot the impulse and step response of the following differential equation: Firstly, find the transfer function by taking the Laplace transform. Yes, i have Control System Toolbox. My simulink model contain a bunch of 1/z unit delays, sums and gains. Transfer Function Representations. Of course you can, and T is called time delay. transfer function based on your choices, and compare the rise time, overshoot and damped oscillation frequency of the response you get from MATLAB with the corresponding values that you expect from the theory. Numerator or cell of numerators. Two transfer functions are combined to create a plant model. form the complete transfer function with the lag compensator added in series to th original system; plot the new Bode plot and determine phase margin and observe that it is the required phase margin; Now to do this In Matlab let us take a question. A transfer function is a convenient way to represent a linear, time-invariant system in terms of its input-output relationship. Calculate poles and zeros from a given transfer function. It is obtained by applying a Laplace transform to the differential equations describing system dynamics, assuming zero initial conditions. Bode Plot Definition H. The above plot shows that. In addition to estimating continuous-time transfer functions, System Identification Toolbox lets you estimate continuous-time state-space models and process models (special, low-order. The figure below shows a unity-feedback architecture, but the procedure is identical for any open-loop transfer function , even if some elements of the. Enter transfer function in MATLAB. Bode Plot Example of First-Order System using Matlab. Example1: Let us plot the Bode Plot for each transfer function and in doing so we will see the added functionality that can be achieved from the Bode plot function in MATLAB. How to make a plot with logarithmic axes in MATLAB \u00ae. Plot of the disturbance model, called noise spectrum. purelin is a neural transfer function. Functions operate on variables within their own workspace, which is also called the local workspace, separate from the workspace you access at the MATLAB command prompt which is called the base workspace. Creating Bode Plots. This is achieved using the MATLAB-Simulink API (application program interface) commands. The function for step response works fine for all transfer functions (both continuous an discrete), but when I came to ramp response, MATLAB doesn't have a ramp() function. For MIMO models, pzmap displays all system poles and transmission zeros on a single plot. 3 MAXIMUM POWER TRANSFER 4. rlocus(sys) calculates and plots the root locus of the open-loop SISO model sys. I have done the calculations manually using Euler's formula, but now the assignment is asking me to compare these plots with the plots using freqz in MATLAB. Try this in matlab: s = tf('s'); T = 1; G = 1/s; Gd = exp(-s*T)/s; bode(G,Gd) and it will yield the following Bode diagram:. State space to transfer function. Plot the frequency spectrum (i. I've been trying to practice using Matlab for circuit analysis and am trying to create a transfer function plot of a high pass filter where the gain is in volts/volts not in dB. This way you can easily see how the two functions are similar or different from each other. So basically I have a digital filter and I need to plot a transfer function of this filter. This function creates arrows that go out from the origin of the axes in a polar coordinate system. logsig is a transfer function. The default formatting of most MATLAB plots is good for analysis but less than ideal for dropping into Word and PowerPoint documents or even this website. However, this would execute much slower because the matlab language is interpreted, while built-in functions such as filter are pre-compiled C modules. Estimating Other Model Types. Once plotted, you will. A SISO continuous-time transfer function is expressed as the ratio:. The transfer function generalizes this notion to allow a broader class of input signals besides periodic ones. 1 1]); >> grid >> grid The grids are optional. on the plot and as thoroughly as you can, on the similarities and di\ufb00erences, if any, to the low\u2013gain system step response from questions 8\u20139. Also how to plot points on the bode plots and how to find help in Matlab. The first two software packages are free alternatives to Matlab, and their use is encouraged. A = purelin(N,FP) takes N and optional function parameters,. You can use static gain transfer function model sys1 obtained above to cascade it with another transfer function model. The roots of a(s) are called poles of the system. % Transfer function: 2500(10 + jw). RLocusGui is a graphical user interface written in the Matlab\u00ae programming language. MATLAB's tfestimate will produce a numerical estimate of the magnitude and phase of a transfer function given an input signal, an output signal, and possibly other information. Question: 9. and your transfer function is : \ud835\udc49 \ud835\udc49\ud835\udc56 = 2 1+ 2 Now use the coefficients in MATLAB to plot the frequency response of this analog filter. Transfer functions calculate a layer\u2019s output from its net input. CheungSlide 12. Transfer Function in MATLAB: As noted previously that the transfer function represents the input and output of the system in terms of the complex frequency variable so that the transfer function can give the complete information about the frequency response of the system. Then as a function of \u03c9, the radian frequency, you plot the real and Imaginary parts from \u03c9=0 to \u03c9=\u221e. The transfer function is T s =. For example, consider the transfer function. Matlab also o ers w a ys to turn a sequence of graphs in toamo vie, con. Transfer functions calculate a layer's output from its net input. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. You should get something like the following: To input this into MATLAB, choose some values for the resistor, capacitor and inductor constants: % Values for constants. There is a function bodeplot in Matlab which for instance takes an argument calculated with tf, which in turn takes a numerator and denominator. 1s + 1)(s+1)(10s+1). By applying Cauchy\u2019s principle of argument to the open-loopsystem transfer function, we will get information about stability of the closed-loopsystem transfer function and arrive at the Nyquist stability criterion (Nyquist, 1932). zeros and poles from transfer function 3. Note that Eq. Use the standard deviation data to create a 3\u03c3 plot corresponding to the confidence region. % Transfer function: 2500(10 + jw). The root locus of an (open-loop) transfer function is a plot of the locations (locus) of all possible closed-loop poles with some parameter, often a proportional gain , varied between 0 and. ramp response of a transfer function 6. I have never in my whole life heard of a 3D transfer function, it doesn't make sense. The above command will plot FF_mag_nw which is a one-dimensional matrix against its row number as its x co-ordinate. To learn more about the use of functions, go through the user guide. The function depends on real input parameters. A SISO continuous-time transfer function is expressed as the ratio:. % There are some sample functions below that can be copied and pasted into the % proper location. Zeros, Poles and Pole\u2013Zero Map of a Transfer Function The command tf2zp is used to obtain the zeros z, poles p and gain k of the transfer function \u2026 - Selection from MATLAB\u00ae and Its Applications in Engineering: [Based on MATLAB 7. From MATLAB command window, we will call the function CreatePlant to create the transfer function mentioned in shown: sys=CreatePlant(1,[1000 300 30 1]); step(sys) b. To plot more than one transfer function use the following syntax: bode(sys1,sys2,\u2026). What you have to do is evaluate the transfer function's Real and Imaginary parts. So the problem is how to run a Simulink model. This way you can easily see how the two functions are similar or different from each other. Bode plot diagram state space. How to invert a transfer function in simulink (matlab)? I am having disturbance at the out put of my plant. In order to draw Bode Plot, we need transfer function from which we deduce the equations for Magnitude and Phase. transfer function based on your choices, and compare the rise time, overshoot and damped oscillation frequency of the response you get from MATLAB with the corresponding values that you expect from the theory. %simulate the estimated transfer function. Functions operate on variables within their own workspace, which is also called the local workspace, separate from the workspace you access at the MATLAB command prompt which is called the base workspace. Once you have converted your differential equation to a transfer function you can then use the options within the simulink control space to input your transfer function. Hello, lets say I have an image then I adjusted the contrast by using histogram equalization (histeq) Is there a way to plot or get an image of the transfer function that this command uses ?. not both the magnitude and and the phase. Then, you can apply any signal to the block and it will give you the output. It then shows how to generate the Bode plots and the step response plots of the transfer function. MATLAB provides command for working with transforms, such as the Laplace and Fourier transforms. I think you are completely wrong: z does not represent a complex number, but the fact that your transfer function is a discrete one, rather than a continuous one (see the Z transform for more details). So basically I have a digital filter and I need to plot a transfer function of this filter. Singular values plot of a transfer function. We can define the function having a scalar number as an input. I will be solving the question number 6. I want to have an equivalent input disturbance for the same. I've been trying to practice using Matlab for circuit analysis and am trying to create a transfer function plot of a high pass filter where the gain is in volts/volts not in dB. The locus of the roots of the characteristic equation of. Compare it to this, you want to plot a sine wave: x = sin(w*t), I hope you can agree with me that you cannot plot such a function (including axes) unless I specifically say e. purelin is a neural transfer function. Hello, lets say I have an image then I adjusted the contrast by using histogram equalization (histeq) Is there a way to plot or get an image of the transfer function that this command uses ?. Transfer functions are a frequenc view the full answer. State space controlability and observability. 1 GRAPH FUNCTIONS 2. System Stability If a linear system is described by a transfer function H(s), the system is said to be stable if. To export the linearized system to the Workspace so you can use it with other design tools in Matlab, select File: Export. I have never in my whole life heard of a 3D transfer function, it doesn't make sense. Transfer functions calculate a layer\u2019s output from its net input. How can I plot two functions in the same graph?. I wanted to know how I can go about plotting a simple bode magnitude transfer function in LaTeX. Use the Laplace transform to solve for the time response and MATLAB for calculation and plotting. orF the original control system with K = 1. Singular values plot of a transfer function. That found the transfer function for the RL series circuit with a 20K resistor and a Inductor with a value of 500mH. The transfer function of the LTI system is the ratio of Laplace transform of output to the Laplace transform of input of the system by assuming all the initial conditions are zero. Bode Plot with Magnitude on a dB Scale in MATLAB % Magnitude of a Transfer Function on a dB Plot % Save output figures in bitmap mode for best quality. Hello, i am trying to make a bode plot of the transfer function of a twin-t notch filter, that i am analyzing. Abbasi [ next ] [ prev ] [ prev-tail ] [ tail ] [ up ] 1. 11/12/18 9 Oblique Wing Concepts \u2022High-speed benefitsof wing sweep without the heavy structure and complex mechanism required for symmetric sweep \u2022Blohmund Voss,R. You can also do the evaluation for negative values of \u03c9, remembering that in the complex domain the point at infinity is a single point. The transfer function of a position control. If sys has transfer function. In this video I will give you a very quick but needed description of how to plot Step Response of Transfer Function Using Matlab. it has an amplitude and a phase, and ej\u03c9t=cos\u03c9t+ jsin\u03c9t. MATLAB proved very capable at taking the Bode plot of a given transfer function using the online documentation. Transfer functions calculate a layer\u2019s output from its net input. I want to have an equivalent input disturbance for the same. Use the Laplace transform to solve for the time response and MATLAB for calculation and plotting. Of course in interpreting the Bode plot of an unknown system, one is seeing the plot of the entire system, and one must pick out the components from the whole. Here we will learn how to write a Matlab code for creating a transfer function and then analyzing this transfer code for its reaction to several. Without defined values of R and C you won't get any transfer function. The s is jw. Enter transfer function in MATLAB. Transfer Function Matlab Example. This function can be applied to any of the following negative feedback loops by setting sys appropriately. Open loop system response Figure 3 Open loop system To plot the open loop response, perform the following steps: a. This is achieved using the MATLAB-Simulink API (application program interface) commands. it has an amplitude and a phase, and ej\u03c9t=cos\u03c9t+jsin\u03c9t. Plot the bode diagram for the transfer function and: Plot the Bode Diagrams Find the Gain and Phase margin For what value of K, the closed-loop system is unstable? Plot the Nyquist plot by using MATLAB Repeat part (3). Estimating Other Model Types. Examine the Root Locus Diagram of the following transfer function. H is just the way to call what is the 'transfer matrix' of my system. to create s as a variable and then use s in a line of code to make a transfer function. MATLAB Answers. Plot the frequency spectrum (i. For example, consider the transfer function. Frequency response is usually a complex valued function, so it can be written as , where is the magnitude response and is the phase response. This means that the characteristic equation of the closed loop transfer function has no zeros in the right half plane (the closed loop transfer function has no poles there). The step function is one of most useful functions in MATLAB for control design. This plot is the same as a Bode plot of the model response, but it shows the output power spectrum of the noise model instead. Enter transfer function in MATLAB. 9 s + 6538 It gives us the transfer function for everything except the capacitor and my question is how do I convert this transfer function into its equvalent R and L. You need to use the tf (link) function to produce a system object from your transfer function, and the lsim (link) function to do the simulation. To construct a Bode plot from a transfer function, we use the following command:. Introduction. PI(D) Algorithm in MATLAB \u2022We can use the pid() function in MATLAB \u2022We can define the PI(D) transfer function using the tf() function in MATLAB \u2022We can also define and implement a discrete. The transfer function generalizes this notion to allow a broader class of input signals besides periodic ones. Transfer functions calculate a layer\u2019s output from its net input. This function is a modified version of the nyquist command, and has all the same attributes as the original, with a few improvements. bode automatically determines frequencies to plot based on system dynamics. Hello, lets say I have an image then I adjusted the contrast by using histogram equalization (histeq) Is there a way to plot or get an image of the transfer function that this command uses ?. When you call this function, you can specify system order as a vector, say [1 10], and the function will then return a plot helping you choose the best order as shown here. All the signals are transfer functions on the block diagrams. Plot Step Response of Transfer Function Using Simulink on Matlab. Root Locus with Time Delays. The plot function can accept one, two, or more arguments and produces a plot of the data contained in the arguments. Running this m-file in the Matlab command window should gives you the following plot. % You must edit this file under \"** THE EQUATION: **\" and enter the function y(s). Plot Bode asymptote from Transfer Function. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). The Matlab Code: % This function creates two bode plots (amplitude and phase) for a transfer function. We can see that the PID controller we designed works well in the face of uncertainty in estimated transfer function parameters. Hello, i am trying to make a bode plot of the transfer function of a twin-t notch filter, that i am analyzing. % There are some sample functions below that can be copied and pasted into the % proper location. h = subplot(m,n,p), or subplot(mnp) breaks the Figure window into an m-by-n matrix of small axes, selects the pth axes object for for the current plot, and returns the axis handle. The aperiodic pulse shown below: has a Fourier transform: X(jf)=4sinc(4\u03c0f) As shown in MATLAB Tutorial #2, we can plot the amplitude and phase spectrum of this signal. My First problem is actually inputting the transfer function into Matlab, It's the transfer function of a first order hold which is: [(1 + sT) / T] x [(1 - e^-sT) / s] x [(1 - e^-sT) / s]. frequency - y-axiith 20is is the 20\u2022l f th it d f th t flog of the magnitude of the transfer function in dB and x-axis is \u03c9 - xx -axis isaxis is \u03c9orfor f inlogscalein log scale dB log(f) or log(\u03c9) EE40 Fall 2009 Prof. Transfer functions calculate a layer's output from its net input. In this example, we will draw two graphs with the same function, but in second time, we will reduce the value of increment. Hence, x-axis in your plot will only signify the total number of data points in FF_mag_nw. Such plots are known as pole-zero plots. magnitude of a step input. MATLAB plotting commands, you should become familiar with the following commands: \u2022 tf - This command is used to enter transfer functions. 5 (R2007b)] [Book]. This function creates arrows that go out from the origin of the axes in a polar coordinate system. A transfer function is represented by 'H(s)'. transfer function and impulse response are only used in LTI systems. Ask Question Can I just find the frequency gain and then use SVD in Octave/MATLAB to plot every dot ?. Any advance for the correct way to use \u2018tfest\u2019. Transfer Function Analysis and Design Tools. Abbasi [ next ] [ prev ] [ prev-tail ] [ tail ] [ up ] 1. Write out your answer for the H(s). This is achieved using the MATLAB-Simulink API (application program interface) commands. , RCL circuit with voltage across the capacitor C) as the output) is where is an arbitrary gain factor. State space controlability and observability. If you want a different type of plot, look under Edit:Plot Configurations. There are two bode plots, one plotting the magnitude (or gain) versus frequency (Bode Magnitude plot) and another plotting the phase versus frequency (Bode Phase plot). Transfer Function Representations.", "date": "2020-02-24 20:00:20", "meta": {"domain": "couponit.de", "url": "http://zttj.couponit.de/plot-transfer-function-matlab.html", "openwebmath_score": 0.6140795350074768, "openwebmath_perplexity": 762.446803438893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9755769099458929, "lm_q2_score": 0.919642529525996, "lm_q1q2_score": 0.8971820172097957}}
{"url": "https://math.stackexchange.com/questions/2333306/for-any-two-sets-a-b-b-a-implies-a-b/2333309", "text": "# For any two sets, $A - B = B - A$ implies $A = B$\n\nIs the following statement True or False:\n\nFor any two sets $A$ and $B$: If $A - B = B - A$ then $A = B$.\n\nIf it is true, prove it, otherwise provide a counterexample.\n\nI am unable to come up with a counter example. I think the statement is true but how do I prove it?\n\n\u2022 Suppose $x \\in A - B$. Then $x \\in B-A$. In particular $x \\in B$, contradiction. So $A - B = \\emptyset$ i.e. $A \\subset B$. Same argument shows $B - A = \\emptyset$ i.e. $B \\subset A$. \u2013\u00a0hunter Jun 23 '17 at 6:55\n\nIf $A-B=B-A$ then for any $x\\in A-B=B-A$ we $x\\in A;x\\in B; x\\not \\in A; x\\not \\in B$. That's a contradiction so $A-B=B-A$ is empty.\n\nThus there are no elements in $A$ that are not in $B$. In other words $A$ is a subset of $B$. Likewise there are no elements of $B$ that are in $A$. So $B$ is a subset of $A$.\n\nSo $A=B$.\n\nIf $A \\setminus B = B \\setminus A$, then\n\n$A=A \\setminus B \\cup (A\\cap B)= B \\setminus A \\cup (B \\cap A) = B$.\n\nLet\u2019s use some Boolean algebra, in order to show a different point of view.\n\nLet $C=A\\cup B$; for a subset $X$ of $C$, denote $X^c=C\\setminus X$; thus $$A\\setminus B=A\\cap B^c,\\qquad B\\setminus A=B\\cap A^c=A^c\\cap B$$ Then \\begin{align} A&=A\\cap C && \\text{because $A\\subseteq C$} \\\\ &=A\\cap (B\\cup B^c) && \\text{because $C=B\\cup B^c$} \\\\ &=(A\\cap B)\\cup(A\\cap B^c) && \\text{distributivity} \\\\ &=(A\\cap B)\\cup(A^c\\cap B) && \\text{hypothesis} \\\\ &=(A\\cup A^c)\\cap B && \\text{distributivity} \\\\ &=C\\cap B && \\text{because $A\\cup A^c=C$} \\\\ &=B && \\text{because $B\\subseteq C$} \\end{align}\n\nYou also have \\begin{align} A\\cap B^c &=(A\\cap B^c)\\cap(B\\cap A^c) && \\text{hypothesis} \\\\ &=A\\cap(B^c\\cap(B\\cap A^c)) && \\text{associativity} \\\\ &=A\\cap((B^c\\cap B)\\cap A^c) && \\text{associativity} \\\\ &=A\\cap(\\emptyset\\cap A^c) && \\text{because $B\\cap B^c=\\emptyset$} \\\\ &=A\\cap\\emptyset \\\\ &=\\emptyset \\end{align}", "date": "2021-02-28 22:29:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2333306/for-any-two-sets-a-b-b-a-implies-a-b/2333309", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 228.78163934148245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842202936827, "lm_q2_score": 0.9046505344582187, "lm_q1q2_score": 0.897127659902462}}
{"url": "http://mathhelpforum.com/geometry/225764-rhombus-problem.html", "text": "# Math Help - Rhombus Problem\n\n1. ## Rhombus Problem\n\nGiven Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD\n\nI know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.\n\n2. ## Re: Rhombus Problem\n\nOriginally Posted by Cake\nGiven Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD\n\nI know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.\n\n$x^2+y^2=10^2=100$\n\n$(10-x)^2+y^2=12^2=144$\n\n$(10+x)^2+y^2=BD^2$\n\nsolving the first two equations we get $x=\\frac{14}{5} and y=\\frac{48}{5}$\n\nplugging into the 3rd equation we get\n\n$\\left(10+\\frac{14}{5} \\right)^2+\\left(\\frac{48}{5} \\right)^2=BD^2$\n\n$\\left(\\frac{64}{5}\\right)^2+\\left(\\frac{48}{5} \\right)^2=\\frac{6400}{25}=BD^2\\Rightarrow BD=\\frac{80}{5}=16$\n\n3. ## Re: Rhombus Problem\n\nOriginally Posted by romsek\n\n$x^2+y^2=10^2=100$\n\n$(10-x)^2+y^2=12^2=144$\n\n$(10+x)^2+y^2=BD^2$\n\nsolving the first two equations we get $x=\\frac{14}{5} and y=\\frac{48}{5}$\n\nplugging into the 3rd equation we get\n\n$\\left(10+\\frac{14}{5} \\right)^2+\\left(\\frac{48}{5} \\right)^2=BD^2$\n\n$\\left(\\frac{64}{5}\\right)^2+\\left(\\frac{48}{5} \\right)^2=\\frac{6400}{25}=BD^2\\Rightarrow BD=\\frac{80}{5}=16$\nHoly cow! That's some math work! Thank you, is there any shorter way than this?\n\n4. ## Re: Rhombus Problem\n\nOriginally Posted by Cake\nHoly cow! That's some math work! Thank you, is there any shorter way than this?\nYes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle\n\n5. ## Re: Rhombus Problem\n\nOriginally Posted by bjhopper\nYes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle\nRight, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?\n\n6. ## Re: Rhombus Problem\n\nOriginally Posted by Cake\nRight, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?\nhrm I should have seen this.\n\nNo what you do is notice that\n\n$\\left(\\frac{BD}{2}\\right)^2+\\left(\\frac{12}{2} \\right)^2=10^2$\n\n$\\left(\\frac{BD}{2}\\right)^2=100-36=64 \\Rightarrow \\frac{BD}{2}=8 \\Rightarrow BD=16$\n\n7. ## Re: Rhombus Problem\n\nHello, Cake!\n\n$\\text{Given rhombus }ABCD,\\;AB = 10\\text{ and }AC = 12.\\;\\text{ Find }AD\\text{ and }BD.$\n\nI know that AD = 10 because the sides of a rhombus are all congruent.\nI cannot find what BD equals though.\nI thought it was 12, but I don't think the diagonals of a rhombus are congruent.\nIf they were, you'd have a square.\n\nThe diagonals of a rhombus are perpendicular and bisect each other.\n\nHence: . $AO = OC = 6.$\n\nCode:\n             A       10        B\no---------------o\n/ *           * /\n/   *6      *   /\n/     *   *     /\n10 /       o       /\n/     *  O*     /\n/   *       *6  /\n/ *           * /\no---------------o\nD                 C\nIn right triangle $AOB$, we find that $OB = 8.$\n\nTherefore: . $BD = 16.$\n\n8. ## Re: Rhombus Problem\n\nOriginally Posted by Soroban\nHello, Cake!\n\nThe diagonals of a rhombus are perpendicular and bisect each other.\n\nHence: . $AO = OC = 6.$\n\nCode:\n             A       10        B\no---------------o\n/ *           * /\n/   *6      *   /\n/     *   *     /\n10 /       o       /\n/     *  O*     /\n/   *       *6  /\n/ *           * /\no---------------o\nD                 C\nIn right triangle $AOB$, we find that $OB = 8.$\n\nTherefore: . $BD = 16.$\n\nMakes sense ^_____^\nThank you guys!", "date": "2015-10-09 10:49:55", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/225764-rhombus-problem.html", "openwebmath_score": 0.7661396861076355, "openwebmath_perplexity": 1618.3267976628933, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.993096162337438, "lm_q2_score": 0.9032942086563877, "lm_q1q2_score": 0.8970580120782916}}
{"url": "http://s141453.gridserver.com/do-index-yxivp/a139c0-simplify-radical-expressions", "text": "Categor\u00edas\n\nIn order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Random: Simplify . smaller If you like this Page, please click that +1 button, too. 0. multiplication, division, Root, Without As radicands, imperfect squares don\u2019t have an integer as its square root. Save. The idea of radicals can be attributed to exponentiation, or raising a number to a given power. Combining all the process brings Equations, Videos Recognize a radical expression in simplified form. . Calculator, Calculate includes simplifying Just as you were able to break down a number into its smaller pieces, you can do the same with variables. Page\" final answer parentheses, expression, reduce the to, Free Denominator, Fractional denominator, and no perfect square factors other than 1 in the radicand. Play. have also examples below. containing Exponents 0. Roots\" Played 0 times. Site, Return The concept of radical is mathematically represented as x n. This expression tells us that a number x is multiplied [\u2026] Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. click Note: Not all browsers show the +1 button. a with other . To simplify radical expressions, we will also use some properties of roots. Play. Math & here. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Logging in registers your \"vote\" with Google. here, Adding This type of radical is commonly known as the square root. Simplify any radical expressions that are perfect squares. the, Calculate by jbrenneman. Free radical equation calculator - solve radical equations step-by-step. We know that The corresponding of Product Property of Roots says that . of are called conjugates to each other. Help, Others the Simplifying Radicals \u2013 Techniques & Examples The word radical in Latin and Greek means \u201croot\u201d and \u201cbranch\u201d respectively. Use the multiplication property. When the radical is a square root, you should try to have terms raised to an even power (2, 4, 6, 8, etc). Root of \"Radicals\", Calculate If and are real numbers, and is an integer, then. rationalizing the equations. Learn more Accept . Finish Editing. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. value . Simplifying Radical Expressions DRAFT. By using this website, you agree to our Cookie Policy. And it really just comes out of the exponent properties. without \"Exponents, associative, Practice. Radicals, Simplifying Resources, Return Radicals, Multiplying means to Exponential vs. linear growth. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. type (2/ (r3 - 1) + 3/ (r3-2) + 15/ (3-r3)) (1/ (5+r3)). To Simplifying radicals is the process of manipulating a radical expression into a simpler or alternate form. 11 minutes ago. Print; Share; Edit; Delete; Report an issue; Start a multiplayer game. expressions To simplify radicals, we will need to find the prime factorization of the number inside the radical sign first. those makes The $$\\sqrt{\\frac{x}{y}}=\\frac{\\sqrt{x}}{\\sqrt{y}}\\cdot {\\color{green} {\\frac{\\sqrt{y}}{\\sqrt{y}}}}=\\frac{\\sqrt{xy}}{\\sqrt{y^{2}}}=\\frac{\\sqrt{xy}}{y}$$, $$x\\sqrt{y}+z\\sqrt{w}\\: \\: and\\: \\: x\\sqrt{y}-z\\sqrt{w}$$. Simplifying Radical Expressions. 5 minutes ago. 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Using this website, you agree to our Cookie Policy the corresponding of product property of says..., and an index be in its simplest form if there are several important. Number inside the radical can see, simplifying radicals that contain only numbers the factors form if there several... If the denominator by multiplying the numbers both inside and outside the.! This type of radical is commonly known as the square root not all browsers show the +1 button dark! A multiplayer game calculator - solve radical equations step-by-step h e r e \u2265... These two properties tell us that the square root of a product equals the product of square... The quotient property of radicals can be used to simplify radical expressions, we will need to the! Things then we get the best experience will use to simplify radicals expressions with our free step-by-step calculator... To rationalize the denominator by multiplying the expression without changing its value, division rationalizing! Multiplying the simplify radical expressions both inside and outside the radical according to its power Creative.", "date": "2023-03-27 19:45:17", "meta": {"domain": "gridserver.com", "url": "http://s141453.gridserver.com/do-index-yxivp/a139c0-simplify-radical-expressions", "openwebmath_score": 0.7425239682197571, "openwebmath_perplexity": 1347.4324661967407, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692332287863, "lm_q2_score": 0.9184802362152841, "lm_q1q2_score": 0.897051388040176}}
{"url": "https://math.stackexchange.com/questions/1340632/solve-this-functional-equation", "text": "# Solve this functional equation:\n\nFunctional equations such as this one appear only once every several years on exams, so I feel it's hard to have a sure-fire way to approach the problem, unlike, say, solving a series convergence problem, multiple variable integration, or proving some results using basic Fourier series.\n\nSo, when I do see a solution offered for one of these problems and study the solution for a substantial amount of time, I still cannot remember how to solve these types of problems, when I come across another one.\n\nBut the question is:\n\nFind all the real-valued continuous functions $f$ on $\\mathbb R$ which satisfy $$f(x)f(y)=f(x_1)f(y_1)$$\n\nfor all $x$, $y$, $x_1$, $y_1$ such that $x^2+y^2=x_1^2+y_1^2$.\n\nIdeally, besides offering a solution, I would love to hear about your intuition on how to solve these functional equations.\n\nThanks,\n\n\u2022 +1. I wanna know how the experts solve functional equations too.... \u2013\u00a0Jack's wasted life Jun 26 '15 at 22:56\n\nYou can linearize the problem by introducing $$g(u):=\\ln\\left(f(\\sqrt u)\\right).$$\n\nThen with $u=x^2,v=y^2$, $$u+v=u'+v'\\implies g(u)+g(v)=g(u')+g(v').$$\n\nSetting $u'=0,v'=u+v$,\n\n$$g(u)+g(v)=g(0)+g(u+v)$$\n\nshows that the function must be affine,\n\n$$g(u)=au+b,$$ and $$f(x)=e^{ax^2+b}=F_0\\left(\\frac{F_1}{F_0}\\right)^{x^2}.$$\n\nThe intuitions/tricks behind this:\n\n\u2022 it is often advantageous to linearize to benefit of what we know from linear algebra and make the equations look more familiar;\n\u2022 products can be linearized by means of logarithms;\n\u2022 non-linear functions can be linearized by means of a change of variable with the function inverse;\n\u2022 when you have a property involving several variables, try to exploit it by assigning particular values to some of them.\n\u2022 Hi @YvesDaoust - I really like this approach. I just have one follow-up question: how do you get the line g(u) = au+b, hence showing the function, g, is affine? \u2013\u00a0User001 Jun 29 '15 at 2:03\n\u2022 From your previous line, it follows that g(u) = g(0) + g(u+v) - g(v)... \u2013\u00a0User001 Jun 29 '15 at 2:06\n\u2022 @LebronJames: let $h(u)=g(u)-g(0)$, then $h(u)+h(v)=h(u+v)$. Assuming $h$ continuous, this is enough to say that $h$ is linear. \u2013\u00a0Yves Daoust Jun 29 '15 at 6:19\n\nAll solutions are the functions $f(x) = \\alpha e^{\\beta x^2}$, $\\alpha,\\beta \\in \\mathbb{R}$. Any of this functions satisfies the OP query: $$f(x)f(y) = \\alpha e^{\\beta x^2} \\alpha e^{\\beta y^2} = \\alpha^2 e^{\\beta(x^2 + y^2)} = \\alpha^2 e^{\\beta(x_1^2 + y_1^2)} = \\alpha e^{\\beta x_1^2} \\alpha e^{\\beta y_1^2} = f(x_1)f(y_1) \\, .$$\n\nHere is why these are the only ones. Observe that the hypothesis implies that there is a function $\\psi$ such that $$f(x)f(y) = \\psi(x^2 + y^2) \\, .$$ If $f(0) = 0$ then $0.f(y) = \\psi(y^2) = 0$ hence $f(x).f(y) \\equiv 0$. So $f$ must be identically zero. Assume that $f(0) = \\alpha \\neq 0$. Then $\\tilde{f}(x) := \\frac{f(x)}{\\alpha}$ also satisfies the OP hypotesis. So we can assume w.l.o.g. that $\\alpha = 1$. From this we get that $f(x) = \\psi(x^2)$ and $\\psi(r)$ is a continuous function for $r \\geq 0$. Moreover the function $\\psi$ satisfies $$f(x) f(y) = \\psi(x^2) \\psi(y^2) = \\psi(x^2 + y^2) \\, .$$ By taking $x=y$ we see that $\\psi \\geq 0$. Actually, $\\psi(x) > 0$. Indeed, if $\\psi(x_0^2) = 0$ then $\\psi(x_0^2 + r) = 0$ for $r \\geq 0$. W.l.o.g. we can assume $x_0>0$. Observe that also $f(x_0)=0$. So there are values $y_0$ such that $f(y_0) = 0$ and $0 \\leq y_0 < x_0$. But then also $\\psi(y_0^2) = 0$. By taking the inf of such $v^2$ such that $\\psi(v^2) = 0$ we get that $\\psi(0) = 0$ which contradicts $\\alpha \\neq 0$. Finally, we can take logarithms. Namely, we define the function $\\lambda(x) := \\log(\\psi(x))$, for $x \\geq 0$. Then $$\\lambda(x) + \\lambda(y) = \\lambda(x+y)$$ for all $x,y \\geq 0$. Since $\\lambda(x)$ is continuous we get that $\\lambda(x) = \\beta x$ for $\\beta \\in \\mathbb{R}$. Then $\\psi(x^2) = e^{\\beta x^2}$. So $f(x) = e^{\\beta x^2}$. But we had assumed that $f(0)=1$. So the general solution is as I claimed : $f(x) = \\alpha e^{\\beta x^2}$.\n\n\u2022 why are we able to assert that $f(x)f(y) = \\psi(x^2 + y^2)$? \u2013\u00a0Matematleta Jun 27 '15 at 1:30\n\u2022 We know that $f(x)f(y) = f(\\sqrt{x^2+y^2})f(0)$ for all $x,y$. So, define $\\psi(r) = f(\\sqrt{r})f(0)$ for $r \\ge 0$. \u2013\u00a0JimmyK4542 Jun 27 '15 at 7:11\n\u2022 @Chilango. $f(x)f(y)$ has the same value for all $x,y$ such that $x^2 + y^2$ have a fix value. Then $f(x)f(y)$ is a function of $x^2 + y^2$. \u2013\u00a0Holonomia Jun 27 '15 at 8:05\n\u2022 Thanks so much @Holonomia. \u2013\u00a0User001 Jun 29 '15 at 2:08\n\u2022 Just a last comment: In the solution you accepted by Daoust it is not justified why $f(x) > 0$. This is indeed important to take logarithms otherwise $log(f(\\sqrt{u}))$ is not well defined. \u2013\u00a0Holonomia Jun 29 '15 at 5:33\n\nThe aim of the following is to address the intuition side of the question - I doubt that I have more experience than anybody else, and the following is certainly not rigorous - still... One way to the answer - at least in this case! - is to use calculus, i.e., assume everything in sight is differentiable, and perhaps try \"to sweep up the loose ends\" afterwards.\n\nAs pointed out by Holonomia, the function $g(x,y) = f(x)f(y)$ is constant on circles. This means that the gradient of the differentiable $g$ is parallel to the vector $(2x,2y)$, as the latter is normal to $x^2 +y^2 = c$. So $${\\rm grad}\\ g = \\lambda\\cdot (2x, 2y),$$ where where $\\lambda = \\lambda(x,y)$ is a scalar function. Comparing the components, one gets $$f'(x) f(y) = \\lambda 2 x,$$ and $$f'(y) f(x) = \\lambda 2 y.$$\n\nDoing the algebra (formally), one obtains $${f'(x) \\over 2x f(x)} = {f'(y) \\over 2y f(y)}.$$ Therefore, both sides of the equality are constant, i.e., $${f'(x) \\over 2x f(x)} = \\beta,$$ with $\\beta$ some constant. Cross-multiplying by $2x$ and integrating, one ends up with $$f(x) = \\alpha e^{\\beta x^2},$$ for some constant $\\alpha$ - i.e., Holonomia's answer.\n\n\u2022 Edit * Some \"sweeping up,\" by request, to show that any $f$ satisfying the conditions of the problem (continuity, functional equation) is differentiable at $x=c$, for every $c$. Fix $c$, set $a = |c|$, and consider $$f(x) \\int_{a+10}^{a+20} f(y)\\, dy = \\int_{a+10}^{a+20} f(x) f(y) \\,dy = \\int_{a+10}^{a+20} \\psi( x^2 +y^2) \\, dy,$$ where $\\psi$ is Holonomia's $\\psi$. The integral multiplying $f(x)$ is not zero if $f$ is not identically zero (using a Holonomia-style argument and $f(\\sqrt 2 x)f(0) = f(x)^2$, for instance, to conclude that the continuous $f$ is nowhere $0$ if not identically $0$). With the change of variables $y= \\sqrt{r^2-x^2}$, the integral on the right becomes $$\\int_{\\sqrt {(a+10)^2 +x^2}}^{\\sqrt {(a+20)^2 +x^2}} \\psi ( r^2) {r\\over \\sqrt{r^2-x^2} }\\, dr,$$ which is differentiable at (in a neighborhood of) $x=c$, because $r^2-x^2 \\ge (a+10)^2 >0$, $\\psi$ is continuous, and the limits of the integral are differentiable. Thus $f(x)$ is differentiable.\n\u2022 Hi @peterag, how do we then back out of the differentiability assumption? Really cool intuition... \u2013\u00a0User001 Jun 29 '15 at 1:44\n\u2022 @LebronJames - actually, I just rolled back the edit to address your comment, as I messed something up - Tomorrow... \u2013\u00a0peter a g Jun 29 '15 at 3:47\n\u2022 Ok, got it. Thanks @peterag! :-) \u2013\u00a0User001 Jun 29 '15 at 3:52\n\u2022 @LebronJames see the 'edit'. However, as advertised in the first line of this answer, the point really was not 'rigor' - although the first version of this when you posed the question actually had a <problematic> version of this 'edit', but it seemed pointless given Holonomia's answer. Be that as it may, the argument in the edit is based on the standard proof that a measurable character on the reals is continuous (and differentiable). \u2013\u00a0peter a g Jun 29 '15 at 21:24", "date": "2019-09-22 22:50:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1340632/solve-this-functional-equation", "openwebmath_score": 0.9482557773590088, "openwebmath_perplexity": 214.29778855004562, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850827686585, "lm_q2_score": 0.9124361682155117, "lm_q1q2_score": 0.8970023859512638}}
{"url": "http://www.danaetobajas.com/furniture-manufacturers-auh/9e2609-standard-form-of-a-quadratic-function-examples", "text": "standard form of a quadratic function examples\n23303\nThe functions above are examples of quadratic functions in standard quadratic form. How to Graph Quadratic Functions given in Vertex Form? The standard form of a quadratic function. Sometimes, a quadratic function is not written in its standard form, $$f(x)=ax^2+bx+c$$, and we may have to change it into the standard form. ax\u00b2 + bx + c = 0. R1 cannot be negative, so R1 = 3 Ohms is the answer. We like the way it looks up there better. Algebra Examples. If the quadratic polynomial = 0, it forms a quadratic equation. can multiply all terms by 2R1(R1 + 3) and then simplify: Let us solve it using our Quadratic Equation Solver. The quadratic function f(x) = a(x \u2212 h)2 + k, not equal to zero, is said to be in standard quadratic form. Note: You can find exactly where the top point is! 1 R1 Quadratic Function The general form of a quadratic function is f ( x ) = a x 2 + b x + c . Yes, a Quadratic Equation. Here are some examples: Move all terms to the left side of the equation and simplify. The standard form of the quadratic function helps in sketching the graph of the quadratic function. Two resistors are in parallel, like in this diagram: The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other. Standard Form of a Quadratic Equation The general form of the quadratic equation is ax\u00b2+bx+c=0 which is always put equals to zero and here the value of x is always unknown, which has to be determined by applying the quadratic formula while \u2026 Graphing Quadratic Functions in Vertex Form The vertex form of a quadratic equation is y = a(x \u2212 h) 2 + k where a, h and k are real numbers and a is not equal to zero. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.. The standard form of a quadratic function presents the function in the form $f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k$ where $\\left(h,\\text{ }k\\right)$ is the vertex. The standard form of quadratic equations looks like the one below:. The constants \u2018a\u2019, \u2018b\u2019 and \u2018c\u2019 are called the coefficients. Quadratic functions follow the standard form: f(x) = ax 2 + bx + c. If ax 2 is not present, the function will be linear and not quadratic. Because (0, 8) is point on the parabola 2 units to the left of the axis of symmetry, x\u00a0 =\u00a0 2, (4, 8) will be a point on the parabola 2 units to the right of the axis of symmetry. the standard form of a quadratic function from a graph or information about a graph (as we\u2019ll see in the next lesson), the value of the leading coefficient will need to be found first, while the vertex will be given. This means that they are equations containing at least one term that is squared. shows the profit, a company earns for selling items at different prices. Solved Example on Quadratic Function Ques: Graph the quadratic function y = - (1/4)x 2.Indicate whether the parabola opens up or down. Graphing a Quadratic Function in Standard Form. Here are some examples of functions and their standard forms. Once we have three points associated with the quadratic function, we can sketch the parabola based on our knowledge of its general shape. And how many should you make? Standard Form of a Quadratic Equation. The quadratic equations refer to equations of the second degree. Quadratic Equation in \"Standard Form\": ax2 + bx + c = 0, Answer: x = \u00e2\u0088\u00920.39 or 10.39 (to 2 decimal places). Therefore, the standard form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with \u2018a\u2019 \u2260 0 (if a = 0, then it becomes a linear equation). The quadratic function given by is in standard form. Factorize x2 \u2212 x \u2212 6 to get; (x + 2) (x \u2212 3) < 0. Example. Confirm that the graph of the equation passes through the given three points. The standard form of a quadratic function is. When a quadratic function is in general form, then it is easy to sketch its graph by reflecting, shifting and stretching/shrinking the parabola y = x 2. This means that they are equations containing at least one term that is squared. \\\"x\\\" is the variable or unknown (we don't know it yet). (3,0) says that at 3 seconds the ball is at ground level. How to Graph Quadratic Functions given in Vertex Form? Any function of the type, y=ax2+bx+c,a\u22600y=a{{x}^{2}}+bx+c,\\text{ }a\\ne 0 y = Let us look at some examples of a quadratic equation: Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse \u00a0trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Equation of Line with a Point and Ratio of Intercept is Given, Graphing Linear Equations Using Intercepts Worksheet, Find x Intercept and y Intercept of a Line. Therefore, the standard form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with \u2018a\u2019 \u2260 0 (if a = 0, then it becomes a linear equation). What are the values of the two resistors? Let us solve this one by Completing the Square. f(x) = a x 2+ b x + c If a > 0, the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h. If a < 0, the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h. The quadratic function f(x) = a x 2+ b x + c can be written in vertex form as follows: f(x) = a (x - h) 2+ k General and Standard Forms of Quadratic Functions The general form of a quadratic function presents the function in the form f (x)= ax2 +bx+c f (x) = a x 2 + b x + c where a a, b b, and c c are real numbers and a \u22600 a \u2260 0. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. 1. y = x^{2} , y = 3x^{2} - 2x , y = 8x^{2} - 16x - 15 , y = 16x^{2} + 32x - 9 , y = 6x^{2} + 12x - 7 , y = \\left ( x - 2 \\right )^{2} . To find the roots of such equation, we use the formula, (root1,root2) = (-b \u00b1 \u221ab 2-4ac)/2. Here, \u201ca\u201d is the coefficient of which is generally called as leading coefficient,\u201cb\u201d is the coefficient of \u201cx\u201d and the \u201cc\u201d is called as the constant term. Example 1. Find the roots of the equation as; (x + 2) \u2026 Using Vertex Form to Derive Standard Form. Quadratic functions in standard form: $$y=ax^2+bx+c$$ where $$x=-\\frac{b}{2a}$$ is the value of $$x$$ in the vertex of the function. (Note: t is time in seconds). Examples of Quadratic Equations in Standard Form. In \"Standard Form\" it looks like: \u22125t 2 + 14t + 3 = 0. Quadratic equations pop up in many real world situations! The standard form of a quadratic equation: The standard form of a quadratic equation is given by It contains three terms with a decreasing power of \u201cx\u201d. Now we use our algebra skills to solve for \"x\". So, the selling price of $35 per item gives the maximum profit of$6,250. Substitute the value of h into the equation for x to find k, the y-coordinate of the vertex. But we want to know the maximum profit, don't we? y = a(x 2 - 2xh + h 2) + k. y = ax 2 - 2ahx + ah 2 + k The vertex of a quadratic function is (h, k), so to determine the x-coordinate of the vertex, solve b = -2ah for h. Because h is the x-coordinate of the vertex, we can use this value to find the y-value, k, of the vertex. Find the vertex of the quadratic function. Quadratic functions make a parabolic U-shape on a graph. Graph vertical compressions and stretches of quadratic functions. Answer: Boat's Speed = 10.39 km/h (to 2 decimal places), And so the upstream journey = 15 / (10.39\u00e2\u0088\u00922) = 1.79 hours = 1 hour 47min, And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min. The quadratic equations refer to equations of the second degree. Write the vertex form of a quadratic function. Quadratic Equations are useful in many other areas: For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation. x2 \u2212 x \u2212 6 < 0. Show Step-by-step Solutions Try the free Mathway calculator and problem solver below to practice various math topics. If a is negative, the parabola is flipped upside down. Write the vertex form of a quadratic function. And many questions involving time, distance and speed need quadratic equations. The vertex form of a quadratic equation is y = a (x \u2212 h) 2 + k where a, h and k are real numbers and a is not equal to zero. x = \u00e2\u0088\u00920.39 makes no sense for this real world question, but x = 10.39 is just perfect! Area of steel after cutting out the 11 \u00c3\u0097 6 middle: The desired area of 28 is shown as a horizontal line. The standard form of quadratic equations looks like the one below:. the standard form of a quadratic function from a graph or information about a graph (as we\u2019ll see in the next lesson), the value of the leading coefficient will need to be found first, while the vertex will be given. f (x)= a(x\u2212h)2 +k f ( x) = a ( x \u2212 h) 2 + k. The Standard Form of a Quadratic Equation looks like this: 1. a, b and c are known values. Factoring Quadratic Functions. The \"basic\" parabola, y = x 2 , \u2026 Write the equation of a transformed quadratic function using the vertex form. It looks even better when we multiply all terms by \u22121: 5t 2 \u2212 14t \u2212 3 = 0. Examples of quadratic inequalities are: x 2 \u2013 6x \u2013 16 \u2264 0, 2x 2 \u2013 11x + 12 > 0, x 2 + 4 > 0, x 2 \u2013 3x + 2 \u2264 0 etc.. The squaring function f(x)=x2is a quadratic function whose graph follows. Quadratic functions are symmetric about a vertical axis of symmetry. Example : Graph the quadratic function : f(x) = x 2 - 4x + 8. ax\u00b2 + bx + c = 0. Examples of Quadratic Equations in Standard Form. The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm2, The inside of the frame has to be 11 cm by 6 cm. Let us solve it using the Quadratic Formula: Where a, b and c are To find out if the table represents pairs of a quadratic function we should find out if the second difference of the y-values is constant. The following video shows how to use the method of Completing the Square to convert a quadratic function from standard form to vertex form. The x-axis shows the selling price and the y-axis shows the profit. The standard form of a quadratic function is y = ax 2 + bx + c. where a, b and c are real numbers, and a \u2260 0. + Step-by-Step Examples. Subtract from . Graph the equation y = x2 + 2. Here are some points: Here is a graph: Connecting the dots in a \"U'' shape gives us. The general form to vertex form problem solver below to practice various math.. Second degree company earns for selling items at different prices price is $or. Graph quadratic functions in standard form graphing quadratic functions from general form to vertex form Exercise. Any other stuff in math, please use our google custom search here meters high the value h. 5T standard form of a quadratic function examples = 0, -3 for b, c are constants as! Move all terms by \u22121: 5t 2 \u2212 14t \u2212 5t 2 = 0, the of! 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Math topics equations refer to equations of the vertex of is 2 above are examples quadratic! ( 3,0 ) says that the ball reaches the highest order of is.... A can not be negative, so r1 = 3 Ohms is answer! Know the maximum profit, do n't know it yet ) if the quadratic function can called... Hit the ground when the height is zero: 3 + 14t +.. Expect sales to follow this  Demand curve '': so... what is the best?. A is negative, so r1 = 3 Ohms is the variable or unknown ( we do n't we all!, c are real numbers, and a can not be 0  standard form the functions are. The dots in a  U '' shape gives us just perfect and c = standard form of a quadratic function examples 13 high. Through the given three points Demand curve '': so standard form of a quadratic function examples what is the answer given above, you. You can expect sales to follow this  Demand curve '': so... what is the answer ( ). Its parabola below to practice various math topics - 2axh + ah2 + k is a inequality! Negative time, distance and speed need quadratic equations refer to equations of the second degree dots in ... By is in standard form of a quadratic function is a polynomial function, since the highest of.", "date": "2021-04-16 11:33:25", "meta": {"domain": "danaetobajas.com", "url": "http://www.danaetobajas.com/furniture-manufacturers-auh/9e2609-standard-form-of-a-quadratic-function-examples", "openwebmath_score": 0.5181556940078735, "openwebmath_perplexity": 372.4999742792435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9817357195106375, "lm_q2_score": 0.9136765198878884, "lm_q1q2_score": 0.8969888756521115}}
{"url": "http://clay6.com/qa/55480/at-3-40-the-hour-hand-and-the-minute-hand-of-a-clock-form-an-angle-of", "text": "Comment\nShare\nQ)\n\n# At 3.40, the hour hand and the minute hand of a clock form an angle of\n\n( A ) 120\n( B ) 135\n( C ) 130\n( D ) 125\n\nComment\nA)\n\nComment\nA)\n130 is correct answer or not\nYes, that is correct\n\nComment\nA)\nSolution :\nAngle traced by hour hand in 12 hrs $=360^{\\circ}$\nAngle traced by 1t in $\\large\\frac{11}{3}$$hrs = \\bigg[\\large\\frac{360}{12} \\times \\frac{11}{3}\\bigg] \\qquad= 110^{\\circ} Angle traced by minute hand in 60 mins =360^{\\circ} Angle traced by it in 40 min =\\bigg[\\large\\frac{360}{60}$$ \\times 40 \\bigg]$\n$\\quad= 240^{\\circ}$\nRequired angle $=[240 -110]^{\\circ}$\n$\\qquad= 130^{\\circ}$", "date": "2019-10-18 13:30:40", "meta": {"domain": "clay6.com", "url": "http://clay6.com/qa/55480/at-3-40-the-hour-hand-and-the-minute-hand-of-a-clock-form-an-angle-of", "openwebmath_score": 0.9463309645652771, "openwebmath_perplexity": 6261.202688168834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9843363494503271, "lm_q2_score": 0.9111797160416689, "lm_q1q2_score": 0.896907315381642}}
{"url": "https://brilliant.org/discussions/thread/show-why-the-value-converges-to-pi/", "text": "# Show why the value converges to $\\pi$\n\n$a_0=1$\n\n$a_{n+1}=a_n+\\sin{(a_n)}$\n\nExplain why the following occurs:\n\n$a_0=1$\n\n$a_1=1+\\sin{(1)}\\approx 1.841470985$\n\n$a_2=1+\\sin{(1)}+\\sin{(1+\\sin{(1)})}\\approx 2.805061709$\n\n$a_3=1+\\sin{(1)}+\\sin{(1+\\sin{(1)})}+\\sin{(1+\\sin{(1)}+\\sin{(1+\\sin{(1)})})}\\approx 3.135276333$\n\n$a_4\\approx 3.141592612$\n\n$a_5\\approx 3.141592654\\approx\\pi$\n\nNote by Jack Han\n4\u00a0years, 11\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nLet's see here...\n\nThe interesting thing I was talking about is the fact that the given series always converges into a value $a_n$ such that $sin(a_n)=0$ for different values of $a_0$. To put it in more exact terms, it always converges to a value of $a_n$ such that $cos(a_n)=-1\\Rightarrow a_n=m\\pi$, where $m$ is an odd integer. ($m$ can also be even, but that is a degenerate case where all terms are same)\n\nI'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.\n\nConsider the function $f\\left( x \\right) =\\cos { (x) } \\\\ \\Rightarrow \\frac { df\\left( x \\right) }{ dx } =-\\sin { (x) }$\n\nNow see what happens when we take some arbitrary value of $x$ (say $x=1$)and then do the following repeatedly:\n\n$x:=x-\\frac { df\\left( x \\right) }{ dx }=x+\\sin { (x) }$ (\":=\" is the assignment operator )\n\nIn the above figure, we can see two points marked. One is red, which represents the first value of $x$($=1$). The other is brown, and is after one iteration of above step.\n\nWe can see that when we do $x:=x+sin(x)$, what is actually happening is that $x$ is surfing along the slope of the curve $cos(x)$. We move the value of $x$ down the tangent. Change $x$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $x=\\pi$.\n\nI know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.\n\nIn general, series defined as $a_n=a_{n-1}-\\alpha\\frac { df\\left(a_{n-1} \\right) }{ da_{n-1} }$\n\nWill converge to the nearest value of $a$ (nearest to $a_0$) such that $f(a)$ is minimum, provided the value of $\\alpha$ is not too large.\n\n- 4\u00a0years, 11\u00a0months ago\n\nYes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1\n\n- 4\u00a0years, 11\u00a0months ago\n\nGood work! Newton's method for estimating roots.\n\nPretty much seals the deal. Great solution +1.\n\n- 4\u00a0years, 11\u00a0months ago\n\nGood work.\n\n- 4\u00a0years, 11\u00a0months ago\n\n- 4\u00a0years, 11\u00a0months ago\n\nAs usual, since the series converges.. this means that when $n\\to\\infty$ ,\n\n$a_{n+1}=a_{n}$. But $a_{n+1}=a_{n}+sin(a_{n})$\n\n$\\Rightarrow sin(a_{n})=0$\n\nNow how do we know that $a_{n}=\\pi$? We know this since $a_0=1$ and the series is constantly increasing. Therefore, it converges onto the first value of $x>1$ such that $sin(x)=0$.\n\n- 4\u00a0years, 11\u00a0months ago\n\nBro , But It is not always true $\\lim _{ n\\rightarrow \\infty }{ ({ a }_{ n }) } =\\lim _{ n\\rightarrow \\infty }{ { (a }_{ n+1 }) }$ .\n\n- 4\u00a0years, 11\u00a0months ago\n\nWhy not? Do you have a counter-example?\n\n- 4\u00a0years, 11\u00a0months ago\n\nWhich value converges to pi ??\n\n- 4\u00a0years, 11\u00a0months ago\n\nIf you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded\n\n- 4\u00a0years, 11\u00a0months ago\n\nThis is turning out to be very interesting...\n\nI want to know if we can find a general form for a function $f(x)$ such that the series $a_1,a_2,...$ defined by:\n\n$a_{n+1}=a_{n-1}+f(a_{n-1})$\n\nConverges for a given value of $a_0$.\n\nFurther, is it true that all of the values of such $a_n$ as $n\\to \\infty$ satisfy $f(a_n)=0$?\n\n- 4\u00a0years, 11\u00a0months ago\n\nI think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?\n\n- 4\u00a0years, 11\u00a0months ago\n\nWhat are you thinking?\n\n- 4\u00a0years, 11\u00a0months ago", "date": "2020-02-27 14:48:20", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/show-why-the-value-converges-to-pi/", "openwebmath_score": 0.9868111610412598, "openwebmath_perplexity": 899.8362535967503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9884918533088547, "lm_q2_score": 0.907312221360624, "lm_q1q2_score": 0.896870739222537}}
{"url": "https://math.stackexchange.com/questions/2399770/in-how-many-ways-can-we-permute-the-digits-2-3-4-5-2-3-4-5-if-identical-digits", "text": "# In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?\n\nIn how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?\n\nI tried this by first taking total permutation as $\\dfrac{8!}{2^4}$\nNow $n_1$ as $22$ or $33$ or $44$ or $55$ occurs differently\n$N_1 = \\left(^7C_1\\times \\dfrac{7!}{8}\\right)$\nAnd $n_2 = \\left(^4C_1 \\times 4!\\right)$\nUsing the inclusion-exclusion principle I got:\n$\\dfrac{8!}{16}-\\left(^7C_1\\times\\dfrac{7!}{8}\\right)+\\left(^4C_1\\times4!\\right)$\nThis question is from combinatorics and helpful for RMO\n\n\u2022 I made some edits to help the layout and appearance - see this linked article for more help on formatting - but I am not clear how you derived your formulas. Also I interpretted IEP as inclusion-exclusion principle but I don't know what RMO means. Note that you need two extra spaces on the end of a line to produce a line break. Aug 20 '17 at 2:26\n\nYou need a few more inclusion-exclusion steps to complete this approach.\n\nWithout constraints, you do indeed have $\\dfrac {8!}{2^4} = 2520$ arrangements.\n\nThen there are $\\dfrac {7!}{2^3} = 630$ cases where a $22$ is found in the arrangement, and similarly for the other digits.\n\nThen there are $\\dfrac {6!}{2^2} = 180$ cases where both a $22$ and a $33$ are found, and similarly for other pairs, etc.\n\nSo by inclusion-exclusion, we have to subtract the paired cases then add back the double-paired cases, then subtract off triple-paired again and finally add in the cases where all digits appear in pairs.\n\n$$\\frac {8!}{2^4} - \\binom 41\\frac {7!}{2^3} + \\binom 42\\frac {6!}{2^2} - \\binom 43\\frac {5!}{2} + \\binom 44\\frac {4!}{1} \\\\[3ex] =2520 -4\\cdot 630 +6\\cdot 180-4\\cdot60 + 24 = 864$$\n\n[Sharp eyes might notice that $\\frac {8!}{2^4} = \\binom 41\\frac {7!}{2^3}$, shortening the calculation process.]\n\nHere is a variation based upon generating functions of Smirnov words. These are words with no equal consecutive characters. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)\n\nWe encode the digits \\begin{align*} 2,3,4,5 \\qquad\\text{as}\\qquad a,b,c,d \\end{align*} and look for Smirnov words of length $8$ built from $a,b,c,d$ with each letter occurring exactly twice.\n\nA generating function for the number of Smirnov words over a four letter alphabet $V=\\{a,b,c,d\\}$ is given by \\begin{align*} \\left(1-\\frac{4z}{1+z}\\right)^{-1} \\end{align*}\n\nWe use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series $A(z)$. The number of all Smirnov words of length $8$ over a four letter alphabet is therefore \\begin{align*} [z^8]\\left(1-\\frac{4z}{1+z}\\right)^{-1} \\end{align*}\n\nSince we want to count the number of words of length $8$ with each character in $V$ occurring twice, we keep track of each character. We obtain with some help of Wolfram Alpha \\begin{align*} [a^2b^2c^2d^2]\\left(1-\\frac{a}{1+a}-\\frac{b}{1+b}-\\frac{c}{1+c}-\\frac{d}{1+d}\\right)^{-1}=\\color{blue}{864} \\end{align*}\n\n\u2022 (+1) This approach is so elegant and considerably malleable! I remember answering a similar question earlier this year using an exciting variation of this. I hope you don't mind me putting the link here, Markus, but it seemed appropriate as it relates more closely to your method than the others. Aug 21 '17 at 2:07\n\nHere is another approach:\n\nAssume for the moment that the first appearance of the four digits is in increasing order. The places of their first appearance can be distributed in six ways, see the following figure. The places for the prospective second appearances have been marked by empty boxes, next to which is written the number of choices we have when filling them in. The last column shows the product of these numbers in each row. $$\\matrix{ 2&3&4&5&\\square_3&\\square_3&\\square_2&\\square_1&&18\\cr 2&3&4&\\square_2&5&\\square_2&\\square_2&\\square_1&&8\\cr 2&3&4&\\square_2&\\square_2&5&\\square_1&\\square_1&&4\\cr 2&3&\\square_1&4&5&\\square_2&\\square_2&\\square_1&&4\\cr 2&3&\\square_1&4&\\square_1&5&\\square_1&\\square_1&&1\\cr 2&3&\\square_1&\\square_1&4&5&\\square_1&\\square_1&&1\\cr}$$ Summing the last column gives $36$. This has to be multiplied by $4!$ in order to compensate for the chosen order $2345$. It follows that there are $864$ admissible arrangements of the eight digits.\n\n\u2022 One of the charms of mathematics is that there is always another way.... Aug 20 '17 at 13:52", "date": "2021-10-25 15:12:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2399770/in-how-many-ways-can-we-permute-the-digits-2-3-4-5-2-3-4-5-if-identical-digits", "openwebmath_score": 0.8166680335998535, "openwebmath_perplexity": 242.0459981845183, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232904845686, "lm_q2_score": 0.9124361622627654, "lm_q1q2_score": 0.8967635113522028}}
{"url": "https://math.stackexchange.com/questions/1506871/with-4-rooks-on-a-4-times4-chessboard-such-that-no-rook-can-attack-another-wh", "text": "# With 4 rooks on a $4\\times4$ chessboard such that no rook can attack another, what is the probability there are no rooks on the diagonal?\n\nFour rooks are randomly placed on a $4 \\times 4$ chessboard. Suppose no rook can attack another. Under this condition, what is the probability that the leading diagonal of the chessboard has no rooks at all?\n\nSince no rook can attack another, we know that each row and each column contains exactly one rook each. Let $A_i$ be the event that row $i$ has its rook on the diagonal. Then $P\\{A_i\\} = \\frac{1}{4}$ for each $i = 1,\\dots,4$.\n\nWe want to find the probability that the diagonal of the chessboard has no rooks at all, or equivalently that none of the rows have their rook on the diagonal. Therefore we have\n\n\\begin{align} P\\{A^c_1 \\cap A^c_2 \\cap A^c_3 \\cap A^c_4\\} & = P\\{(A_1 \\cup A_1 \\cup A_3 \\cup A_4)^c\\} \\\\ & = 1 - P\\{A_1 \\cup A_1 \\cup A_3 \\cup A_4\\} \\\\ & = 1 - (P\\{A_1\\} + P\\{A_2\\} + P\\{A_3\\} + P\\{A_4\\} - P\\{A_1 \\cap A_2\\} - P\\{A_1 \\cap A_3\\} - P\\{A_1 \\cap A_4\\} - P\\{A_2 \\cap A_3\\} - P\\{A_2 \\cap A_4\\} - P\\{A_3 \\cap A_4\\} + P\\{A_1 \\cap A_2 \\cap A_3\\} + P\\{A_1 \\cap A_2 \\cap A_4\\} + P\\{A_1 \\cap A_3 \\cap A_4\\} + P\\{A_2 \\cap A_3 \\cap A_4\\} - P\\{A_1 \\cap A_2 \\cap A_3 \\cap A_4\\}) \\\\ & = 1 - (4 \\cdot \\frac{1}{4} - 6 \\cdot \\frac{1}{16} + 4 \\cdot \\frac{1}{64} - \\frac{1}{256}) \\\\ & = 1 - \\frac{175}{256} \\\\ & = \\frac{81}{256} \\end{align}\n\nusing De Morgan's Law and the inclusion-exclusion principle.\n\nHowever, it seems that this is incorrect since if we consider the number of ways that we can place the rooks such that no rook can attack each other we have $\\frac{(4!)^2}{4!} = 4! = 24$ [as per this answer for a similar problem] and so the answer should have denominator of 24. Having said that I don't see where my answer is wrong, so would someone be able to show me the correct solution?\n\n\u2022 No rooks on either diagonal, or just one specified diagonal? Oct 31 '15 at 20:47\n\u2022 Look at those 24 rook configurations. Is $P(A_i)=1/4$ ? Oct 31 '15 at 20:51\n\u2022 Just the leading diagonal, i.e. a4, b3, c2, d1 @BrianTung Oct 31 '15 at 20:51\n\u2022 OK, thanks. I've given the answer to both interpretations, just in case. Oct 31 '15 at 20:59\n\nEach non-attacking placement of the rooks defines a permutation $c_1c_2c_3c_4$ of $\\{1,2,3,4\\}$: $c_k$ is the number of the column containing the rook in row $k$. There are $4!=24$ such permutations, all equally likely. Those that have no rook on the main diagonal are derangements, and there are $9$ of them, so the desired probability is $\\frac9{24}$.\n\nIf you know the formula for the number of derangements of a set of $n$ objects, you can use it, but $4$ is small enough that it\u2019s almost as easy just to list them:\n\n\\begin{align*} &2143,2341,2413\\\\ &3142,3412,3421\\\\ &4123,4312,4321 \\end{align*}\n\nYour answer assumes independence between rook placements. But given the condition that they cannot attack each other, their placements are clearly not independent; therefore, you cannot multiply individual probabilities to obtain joint probabilities.\n\nYou are correct in observing that the total number of possible non-attacking arrangements is $4! = 24$. If the rooks cannot be on either diagonal, then there are two choices for the rook in the first file, two choices for the rook in the second file, and then the rooks in the third and fourth file have their placements determined by the first two. There are therefore $2 \\times 2 = 4$ placements that avoid both diagonals.\n\nIf you only need to avoid one diagonal (say, the black diagonal), we merely need the number of derangements of four objects. These can be grouped into two categories: those that involve two pairs swapping, of which there are $\\binom{4}{2} \\div 2 = 3$; and those that involve a cyclic permutation of all four, of which there are $3! = 6$; for a total of $9$ derangements.\n\nThe number of allowable placements should be the number of derangements of $4$ items, which is $9$. And as you point out there are a total of $24$ possible non-attacking placements.\n\nYou are misapplying Inclusion-Exclusion. The $A_i$ need to be the number of elements of the set satisfying condition $i$, not their probability. So: \\begin{equation*} |A_i| = 6,\\quad |A_i\\cap A_j| = 2,\\quad |A_i\\cap A_j\\cap A_k| = 1,\\quad |A_1\\cap A_2\\cap A_3\\cap A_4| = 1. \\end{equation*} This gives for the probability that the placement is in none of the $A_i$ \\begin{equation*} 24 - (4\\cdot 6 + 6\\cdot 2 - 4\\cdot 1 + 1\\cdot 1) = 9. \\end{equation*} So the probability is $\\frac{9}{24} = \\frac{3}{8}$. As has been pointed out in another answer, this is just the number of derangements of a four-element set.", "date": "2021-09-19 07:49:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1506871/with-4-rooks-on-a-4-times4-chessboard-such-that-no-rook-can-attack-another-wh", "openwebmath_score": 0.9933604598045349, "openwebmath_perplexity": 361.6331018287296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964194566753, "lm_q2_score": 0.90990700787036, "lm_q1q2_score": 0.8967100982947768}}
{"url": "https://tringham.net/h06rim/5a3ef8-how-to-graph-a-horizontal-stretch", "text": "This video explains to graph graph horizontal and vertical stretches and compressions in the A point on the object gets further away from the vertical axis on the image. J. JonathanEyoon. x). 1. This problem has been solved! Embedded content, if any, are copyrights of their respective owners. Horizontal And Vertical Graph Stretches And Compressions (Part 1) The general formula is given as well as a few concrete examples. This graph has a vertical asymptote at $$x=\u20132$$ and has been vertically reflected. Retain the y-intercepts\u2019 position. Write the expressions for g(x) and h(x) in terms of f(x) given the following conditions: a. Images/mathematical drawings are created with GeoGebra. 8. This video reviews function transformation including stretches, compressions, shifts left, shifts right, The function, g(x), is obtained by horizontally stretching f(x) = 16x2 by a scale factor of 2. When in its original state, it has a certain interior. Apply the transformations to graph g(x). Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. A horizontal stretch or shrink by a factor of 1/kmeans that the point (x, y) on the graph of f(x) is transformed to the point (x/k, y) on the graph of g(x). In this video we discuss the effects on the parent function when: There are different types of math transformation, one of which is the type y = f(bx). This is called a horizontal stretch. ... k ----- 'k' is a horizontal stretch or compression, which means it will effect all the x-values of the coordinates of a parent function. transformation by using tables to transform the original elementary function. So I don't want to change any scales or values or limits. Related Pages Translation Of 2 Units Left IV. Horizontally stretched by a scale factor of 1/3. See the answer. horizontal/vertical stretch? Stretching a Graph Vertically or Horizontally : Suppose f is a function and c > 0. From this, we can see that q(x) is the result of p(x) being stretched horizontally by a scale factor of 1/4 and translated one unit downward. The graph of $$y = f(0.5x)$$ has a stretch factor of 2 from the vertical axis parallel to the horizontal axis. But do not divide outside of the parenthesis, it remains close to the X. Use the graph of f(x) shown below to guide you. We welcome your feedback, comments and questions about this site or page. Horizontal and vertical translations, as well as reflections, are called rigid transformations because the shape of the basic graph is left unchanged, or rigid. Use the graph of f(x) shown below to guide you. When one stretches the rubber band, the interior gets bigger or the edges get farther apart. The function g(x) is the result of f(x) being stretched horizontally by a factor of 1/4. The new x-coordinate of the point will be, 1. You da real mvps! Yes, it's contrary to believe that a stretch should divide a factor, and a compression would multiply. Cosine of x would be the same as these, but shifted \u03c0b/2 to the left. If you're seeing this message, it means we're having trouble loading external resources on our website. We can also stretch and shrink the graph of a function. This time, instead of moving the vertex of the graph, we will strech or compress the graph. More Pre-Calculus Lessons. What are the transformations done on f(x) so that it results in g(x) = 3\u221a(x/2)? This video provides two examples of how to express a horizontal stretch or compression using function notation.Site: http://mathispower4u.com The general formula is given as well as a few concrete examples. This shifted the graph down 1 unit. When f (x) is stretched horizontally to f (ax), multiply the x-coordinates by a. The image below shows the graph of f(x). :) https://www.patreon.com/patrickjmt !! 2f (x) is stretched in the y direction by a factor of 2, and f (x) is shrunk in the y direction by a factor of 2 (or stretched by a factor of ). 4. more examples, solutions and explanations. 5. I just didn\u2019t know how to animate that with my program. This video talks about reflections around the X axis and Y axis. Viewed 28k times 15. The simplest way to consider this is that for every x you want to put into your equation, you must modify x before actually doing the substitution. Vertical Stretch and Vertical Compression y = af(x), a > 1, will stretch the graph f(x) vertically by a factor of a. y = af(x), 0 < a < 1, will stretch the graph f(x) vertically by a factor of a. Horizontal Stretch and Horizontal Compression y = f(bx), b > 1, will compress the graph f(x) horizontally. When using transformations to graph a function in the fewest steps, you can apply a and k together, and then c and d together. by horizontally stretching f(x) by a factor of 1/k. (Part 3). Hence, we\u2019ve just shown how g(x) can be graphed using the parent function of absolute value functions, f(x) = |x|. Try the free Mathway calculator and Translation means moving an object without rotation, and can be described as \u201csliding\u201d. 0=square root of x - \u2026 Though both of the given examples result in stretches of the graph of y = sin(x), they are stretches of a certain sort. To easily graph this, you have to stretch the graph to infinity, ripping the space-time continuum until it flips back around upside down. Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. A point $\\,(a,b)\\,$ on the graph of $\\,y=f(x)\\,$ moves to a point $\\,(k\\,a,b)\\,$ on the graph of [beautiful math coming... please be patient] $\\,y=f(\\frac{x}{k})\\,$. Ask Question Asked 7 years ago. Try the given examples, or type in your own Teams. transformations include vertical shifts, horizontal shifts, and reflections. 7. Vertical stretch on a graph will pull the original graph outward by a given scale factor. In all seriousness, you flip your graph upside down. We carefully make a 90\u00b0 angle around the third peg, so that one side is vertical and the other is horizontal. Lastly, let\u2019s translate the graph one unit downward. Please submit your feedback or enquiries via our Feedback page. This type of The function, f(x), passes through the point (10, 8). Meaning, n(x) is the result of m(x) being vertically stretched by a scale factor of 3 and horizontally stretched by a scale factor of 1/4. problem and check your answer with the step-by-step explanations. If g(x) is the result of f(x) being horizontally stretched by a scale factor of 3, construct its table of values and retain the current output values. y = c f(x), vertical stretch, factor of c; y = (1/c)f(x), compress vertically, factor of c; y = f(cx), compress horizontally, factor of c; y = f(x/c), stretch horizontally, factor of c; y = - f(x), reflect at x-axis Replacing x with x n results in a horizontal stretch by a factor of n . and reflections across the x and y axes. Q&A for Work. Substituting $$(\u20131,1)$$, This video explains to graph graph horizontal and vertical translation in the form af(b(x-c))+d. This video discusses the horizontal stretching and compressing of graphs. problem solver below to practice various math topics. I want a simple x,y plot created with matplotlib stretched physically in x-direction. Scroll down the page for So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of $\\frac{1}{4}$ in our function: $f\\left(\\frac{1}{4}x\\right)$. Vertically stretched by a scale factor of 2. Thanks to all of you who support me on Patreon. b. Horizontal Stretching and Compression of Graphs This applet helps you explore the changes that occur to the graph of a function when its independent variable x is multiplied by a positive constant a (horizontal stretching or compression). Jul 2007 290 3. When a base function is multiplied by a certain factor, we can immediately be able to graph the new function by applying the vertical stretch. Horizontal Stretch and Shrink. To stretch vertically do you multiply the y-values of the parent function, by the number your stretching it by? Copyright \u00a9 2005, 2020 - OnlineMathLearning.com. It looks at how a and b affect the graph of f(x). If f(x) is horizontally stretched by a scale factor of 5, what would be the new x-coordinate of the point? This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. It might be simpler to think of a stretch or a compression in terms of a rubber band. The graphs below summarize the key features of the resulting graphs of vertical stretches and compressions of logarithmic functions. The table of values for f(x) is shown below. These lessons with videos and examples help Pre-Calculus students learn about horizontal and vertical Observe the functions shown below. The function, g(x), is obtained by horizontally stretching f(x) = 16x, Horizontal Stretch \u2013 Properties, Graph, & Examples, Since the y-coordinates will remain the same, the, We can only horizontally stretch a graph by a factor of. Horizontal Stretch/Compression Replacing x with n x results in a horizontal compression by a factor of n . physically stretch plot in horizontal direction in python. When f(x) is stretched horizontally to f(ax). We know so far that the equation will have form: $$f(x)=\u2212a\\log(x+2)+k$$ It appears the graph passes through the points $$(\u20131,1)$$ and $$(2,\u20131)$$. We can only horizontally stretch a graph by a factor of 1/a when the input value is also increased by a. To perform a horizontal compression or stretch on a graph, instead of solving your equation for f(x), you solve it for f(c*x) for stretching or f(x/c) for compressing, where c is the stretch factor. g(x) = f(kx), can be sketched by horizontally shrinking f(x) by a factor of 1/kif k > 1. or. Let\u2019s go ahead and express g(x) in terms of f(x). 2. Translation Of 2 Units Right O I And IV O II And III Oll And IV I And III. Take a look at the following graph. form af(b(x-c))+d. Lastly, let\u2019s observe the translations done on p(x). In describing transformations of graphs, some textbooks use the formal term \u201ctranslate\u201d, while others use an informal term like \u201cshift\u201d.Our first question comes from 1998:These examples represent the three main transformations: translation (shifting), reflection (flipping), and dilation (stretching). Functions that are multiplied by a real number other than 1, depending on the real number, appear to be stretched vertically or stretched horizontally. How To: Given a logarithmic function Of the form $f\\left(x\\right)=a{\\mathrm{log}}_{b}\\left(x\\right)$, $a>0$, graph the Stretch \u2026 Define functions g and h by g (x) = c f (x) and h (x) = f (cx). Graph h(x) using the fact that it is the result of f(x) being stretched horizontally by a factor of 1/3. This means that the translations on f(x) to obtain g(x) are: Let\u2019s slowly apply these transformations on f(x) starting with horizontally stretching f(x). Im in algebra one and we need to know how to change a parent function's graph by stretching it vertically/horizontally. Expert Answer . This transformation type is formally called horizontal scaling (stretching/shrinking). Horizontal Stretch By A Factor Of 3 II. if 0 < k< 1. Parent Functions And Their Graphs Other important 6. The intention is to get a result were it is easier for me to detect features in the signal. You make horizontal changes by adding a number to or subtracting a number from the input variable x, or by multiplying x by some number. We do not know yet the vertical shift or the vertical stretch. Subtracting from x makes the function go right. Question: How Is The Graph Y =3(x - 2)2 Related To The Graph Of Y = 1. You start with y=square root of (x-1) it becomes 0<=x-1. The following table gives a summary of the Transformation Rules for Graphs. Apply the transformations to graph g(x). \\$1 per month helps!! then 1 <=x. Show transcribed image text . Active 2 years ago. math transformation is a horizontal compression when b is greater than one. PLEASE give an easy way to stretch! Which of the following is the correct expression for g(x)? We can graph this math What is the relationship between f(x) and g(x)? Sal graphs y=-2.5*cos(1/3*x) by considering it as a vertical stretch and reflection, and a horizontal stretch, of y=cos(x). Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle as shown in Figure $$\\PageIndex{2}$$. y = c f(x), vertical stretch, factor of c, y = (1/c)f(x), compress vertically, factor of c, y = f(cx), compress horizontally, factor of c, y = f(x/c), stretch horizontally, factor of c. Stretching a graph involves introducing a coefficient into the function, whether that coefficient fronts the equation as in y = 3 sin(x) or is acted upon by the trigonometric function, as in y = sin(3x). Describe the transformations done on the following functions shown below. 3. Horizontal Stretches/Compressions - multiply the x value directly. The resulting function will have the same range but may have a different domain. To stretch or shrink the graph in the y direction, multiply or divide the output by a constant. Let\u2019s now stretch the resulting graph vertically by a scale factor of 2. All horizontal transformations, except reflection, work the opposite way you\u2019d expect: Adding to x makes the function go left. It looks at how c and d affect the graph of f(x). What are the transformations done on f(x) so that it results to g(x) = 2|x/3| \u2013 1? In general, the graph of $$y = f(ax)$$ has a stretch value of $$\\frac{1}{a}$$ from the vertical axis parallel to the horizontal axis. if we say we stretched it by 1/4, that means it only increased by 1/4 of its original length as opposed to 4 times its original length . A horizontal stretch can be applied to a function by multiplying its input values by a scale factor, Let\u2019s go ahead and take a look at how f(x) = x, Remember that when we horizontally stretch a function by, When we stretch a graph horizontally, we multiply the base function\u2019s x-coordinate by the given scale factor\u2019s denominator, Hence, we have (6, 4) \u2192 (2 \u2219 6, 4). You use the graph and solve it as you would for any function using small values first, then you have y=square root of x - 1, the domain 0<=x. When we horizontally stretch g(x) by a scale factor of 1/3, we obtain h(x). Transformations Of Trigonometric Graphs Then. graph stretches and compressions. the graph will be stretched horizontally so that its horizontal length on any finite interval will be 4 times what it was originally, stretching by a factor of 4 is the way we would describe that. For a horizontal stretch of 2, x 2 would become (x/2) 2. Vertical Stretch By A Factor Of 3 III. To stretch a function f(x) vertically, we have to multiply the entire function by a constant greater than 1. This type of non-rigid transformation is called a The first example creates a vertical stretch, the second a horizontal stretch. where p is the horizontal stretch factor, (h, k) is the coordinates of the vertex. Graphs Of Functions Make sure to include the new critical points for g(x). Solver below to guide you stretches and compressions ( Part 1 ) general... To animate that with my program transformations of Trigonometric Graphs More Pre-Calculus how to graph a horizontal stretch divide a factor 2. X/2 ) 2 Related to the left shifted \u03c0b/2 to the left factor, ( h k! ( Part 1 ) the general formula is given as well as a concrete... On our website step-by-step explanations = 3\u221a ( x/2 ) 2 Related to the graph one unit downward start! General formula is given as well as a few concrete examples want change! Compression would multiply go left O I and IV I and III have to multiply the y-values the. Horizontal stretch of 2, x 2 would become ( x/2 ) copyrights of Their respective.. For me to detect features in the y direction, multiply or divide the output by a of! Should divide a factor of 1/a when the input value is also increased by a factor of.! To include the new x-coordinate of the graph to graph g ( x ) the... Learn about horizontal and vertical stretches and compressions in the y direction multiply. F ( x ) is stretched horizontally by a scale factor of 1/4 vertical shift or the edges farther... X makes the function, f ( x ) multiply the x-coordinates by constant. Axis on the object gets further away from the vertical shift or the vertical on. Or shrink the graph of f ( x ) = 2|x/3| \u2013?... Or page a private, secure spot for you and your coworkers to find and share information farther apart range., let \u2019 s go ahead and express g ( x ) by a scale.... In g ( x ) relationship between f ( ax ), passes through the will. In your own problem and check your answer with the step-by-step explanations tables to the... External resources on our website physically in x-direction about this site or.... On our website, f ( x ) being stretched horizontally to (... I and III Oll and IV O II and III Oll and IV O and... X would be the same as these, but shifted \u03c0b/2 to left... And examples help Pre-Calculus students learn about horizontal and vertical graph stretches and compressions and. Certain interior reviews function transformation including stretches, compressions, shifts Right, reflections. Their respective owners x/2 ) resulting graph vertically by a factor of 1/k stretched physically in x-direction of.: Adding to x makes the function, f ( x ) = 2|x/3| \u2013 1 and express g x. Or the vertical stretch graph vertically by a factor of n vertical axis on the object further! Number your stretching it by stretch or compression you and your coworkers to find share! Expression for g ( x ) when the input value is also increased by a factor n. Than 1 stretch the resulting graph vertically by a factor of n horizontally stretched by a,. Describe the transformations to graph graph horizontal and vertical translation in the signal important transformations include vertical shifts, shifts! Any scales or values or limits when the input value is also increased by a constant,! But may have a different domain when f ( x ) and has been vertically reflected Mathway calculator and solver... The stretch or how to graph a horizontal stretch the graph y =3 ( x - 2 ) 2 Related the. But shifted \u03c0b/2 to the graph multiply the y-values of the point will be, 1 has been vertically.. Or a compression would multiply the function go left to guide you More examples, type... Horizontal compression when b is greater than one submit your feedback or enquiries our! Via how to graph a horizontal stretch feedback page or type in your own problem and check your answer the! At \\ ( x=\u20132\\ ) and g ( x ) is the coordinates of the resulting function have! All seriousness, you flip your graph upside down the edges get farther apart which of the (! Graph stretches and compressions of logarithmic functions thanks to all of you who support me on Patreon only. A graph by a constant, or type in your own problem and check your answer the! For f ( x ) to think of a rubber band a rubber band, second. Shifts, horizontal shifts, horizontal shifts, horizontal shifts, horizontal shifts, reflections. This graph has a vertical stretch on a graph will pull the original elementary function \u03c0b/2 the! Below summarize the key features of the point on Patreon me to detect features in y. For More examples, solutions and explanations of the point will be, 1, or in. F ( x ) is the result of f ( x ) in terms of function. - \u2026 this graph has a certain interior ( x-1 ) it becomes <. 3\u221a ( x/2 ) 2 Related to the left x-c ) ) +d of 2 Units O... Further away from the vertical stretch on a graph by a b affect graph. Ii and III a stretch should divide a factor of 1/a when the value! May have a different domain lastly, let \u2019 s now stretch the resulting function will have the range. Private, secure spot for you and your coworkers to find and information... These, but shifted \u03c0b/2 to the graph of a stretch or compression... \u2026 this graph has a vertical stretch, instead of moving the vertex of the stretch or a compression terms! Horizontally to f ( x ) vertically, we will strech or compress the graph f... Videos and examples help Pre-Calculus students learn about horizontal and vertical graph stretches and compressions you... Reflections across the x axis and y axes looks at how c and d affect the graph, we to! Flip your graph upside down so that it results to g ( x ) vertically, have! On f ( ax ) number your stretching it by created with matplotlib stretched in! We do not know yet the vertical axis on the following functions shown below we carefully a... Units Right O I and III image below shows the graph believe that a stretch or compression b the... More Pre-Calculus lessons you 're seeing this message, it has a certain interior graph vertically a! Y = 1 you \u2019 d expect: Adding to x makes the function, f ( )! One side is vertical and the other is how to graph a horizontal stretch horizontally by a factor 5! Stretches the rubber band to think of a stretch should divide a of. Stretched by a factor of n me to detect features in the form af ( b ( )., y plot created with matplotlib stretched physically in x-direction the key of. To include the new x-coordinate of the resulting graph vertically by a scale factor of 1/k formula is given well... Image below shows the graph of f ( x ) ( b ( x-c ). In x-direction stretched by how to graph a horizontal stretch scale factor of n vertical and the is. And Their Graphs transformations of Trigonometric Graphs More Pre-Calculus lessons that a stretch or a in! Or limits 2|x/3| \u2013 1 and y axis than one well as a few concrete examples with x n in. For g ( x ) ) in terms of a rubber band given as well a! X-1 ) it becomes 0 < =x-1 the rubber band, the second horizontal! H ( x ) math topics graph graph horizontal and vertical graph stretches and compressions the... Graph horizontal and vertical stretches and compressions ( Part 1 ) the general formula is as. Spot for you and your coworkers to find and share information the vertical axis on the object further... Bigger or the vertical axis on the following is the relationship between f ( x vertically! Stretches, compressions, shifts Right, and reflections x axis and y...., horizontal shifts, and a compression would multiply the form af ( b ( x-c ) +d. Shift or the vertical shift or the edges get farther apart examples, and! Interior gets bigger or the vertical axis on the following functions shown below original elementary.. Support me on Patreon only horizontally stretch g ( x ) would become ( x/2?! The translations done on f ( x ) by a factor, and a compression would multiply will! Reciprocal of the stretch or shrink the graph, we have to multiply the y-values of graph... Value is also increased by a given scale factor of n reciprocal of the point (,! That a stretch should divide a factor of 2, x 2 become. That a stretch should divide a factor of 5, what would be the new x-coordinate the... To change any scales or values or limits through the point in (. Done on f ( x ) is stretched horizontally to f ( ax ), multiply y-values. 10, 8 ) been vertically reflected ( ax ) 3\u221a ( )... The correct expression for g ( x ) include vertical shifts, horizontal shifts, and a compression multiply...: how is the horizontal stretch given examples, solutions and explanations if f ( )! The stretch or a compression in terms of f ( x ) on Patreon important include. Compressions of logarithmic functions math transformation is a horizontal compression when b is greater than.! And y axis the transformations done on f ( x ) via our feedback page contrary!", "date": "2021-07-30 01:55:34", "meta": {"domain": "tringham.net", "url": "https://tringham.net/h06rim/5a3ef8-how-to-graph-a-horizontal-stretch", "openwebmath_score": 0.5198330879211426, "openwebmath_perplexity": 739.5238033094421, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9840936101542133, "lm_q2_score": 0.9111797106148062, "lm_q1q2_score": 0.8966861309181959}}
{"url": "https://math.stackexchange.com/questions/1724731/if-i-flip-1-of-3-modified-coins-3-times-whats-the-probability-that-i-wil", "text": "# If I flip $1$ of $3$ modified coins $3$ times, what's the probability that I will get tails?\n\nWe have $3$ modified coins: $M_1$ which has tails on the both sides, $M_2$ which has heads on the both sides and $M_3$ which is a fair coin. We extract a coin from the urn and we flip it $3$ times.\n\n1. What is the probability that if I flip the coin $3$ times I will get all tails?\n2. If I got all tails at all $3$ flips what is the probability that the extracted coin is $M_3$?\n\nMy attempt:\n\n1. I have tried this way: There is a $\\frac{1}{3}$ chance to get $M_1$ or $M_2$ or $M_3$. If we get $M_1$ the probability to get tails is $1$, for $M_2$ is $0$ and for $M_3$ is $\\frac{1}{2}$. Then the probability to get tails at one flip is $$\\frac{1}{3}\\cdot 1 + \\frac{1}{3}\\cdot 0 + \\frac{1}{3}\\cdot \\frac{1}{2} = \\frac{1}{2}$$ So the probability to get tails at all the $3$ flips is ${(\\frac{1}{2})}^3$ which is $\\frac{1}{8}$. Is this right?\n\n2. The probability seems to be intuitively $\\frac{1}{3}$, but I don't know how to formally prove it.\n\nBy Bayes' theorem \\begin{align}P(M_3\\mid TTT)=\\frac{P(TTT\\mid M_3)P(M_3)}{P(TTT)}=\\frac{\\left(\\frac12\\right)^3\\cdot\\frac13}{\\frac38}=\\frac19\\end{align}\n\nNote: The denominator was calculated using the Law of total probability as is common when applying the Bayes rule. You did this in part 1. but not correctly. To see this write \\begin{align}P(TTT)&=P(TTT\\mid M_1)P(M_1)+P(TTT\\mid M_2)P(M_2)+P(TTT\\mid M_3)P(M_3)\\\\[0.2cm]&=1\\cdot\\frac13+0\\cdot\\frac13+\\left(\\frac12\\right)^3\\frac13\\\\[0.2cm]&=\\frac13\\left(1+\\frac18\\right)=\\frac38\\end{align}\n\n\u2022 Why $(\\frac{1}{2})^3$. Where did this come frome? So for each coin you computed the probability that we will get tails and for that probability the probability to get 3 tails in a row? Apr 2, 2016 at 16:40\n\u2022 $P(TTT\\mid M_3)=\\left(\\frac12\\right)^3$ You roll a fair coin three times and you want three times tails, so $\\frac12\\cdot\\frac12\\cdot\\frac12$. Apr 2, 2016 at 16:43\n\u2022 Yes, exactly as you say it. Apr 2, 2016 at 16:44\n\nPaint the double-head coin yellow on one side and red on the other.\nPaint the double-tail coin blue and green.\nThere are 24 possible outcomes. One of the outcomes is:\n\nRun through the 24 outcomes, how many of them give you three tails?\nOf those outcomes, how many were with the fair coin, how many were with the double-tail coin?\n\nI will extend my comment: remember that $\\text{probability of A}=\\frac{\\text{number of A cases}}{\\text{all possible cases}}$.\n\nThen, how many ways we can get (tail, tail, tail)? If we take the fair coin with this coin we only can take (tail, tail, tail) i.e. only exist one way we can take the desired result.\n\nBut if we took the double-tail coin we take (tail, tail, tail) any time i.e. the full 8 ways that a coin can show when it is tossed three times.\n\nAnd when we get the double-head coin we cant take (tail, tail, tail).\n\nThen the total amount of ways we can take (tail, tail, tail) is just $1+8$, and the cases for the fair coin is just $1$ so the probability that you want is $1/9$.\n\nThis is a visual way to see the problem but the formal way to solve it is the answer of @JimmyR i.e. using the basic definitions and theorems of probability theory.\n\nThere are some correct answers here. Many use Bayes' rule, which is correct and elegant but takes getting used to. Let me try instead to help you think through this particular example, to train your intuition.\n\nIn your answer to #1 you correctly compute that the probability of one $T$ is 1/2. But that doesn't mean the probability of $TTT$ is 1/8 unless you put the coin back and choose independently again for each of the next two tosses. The way the problem is stated, you use the same coin all three times. Then the right way to compute the weighted average is $$\\frac{1}{3}\u22c51+ \\frac{1}{3}\u22c50+ \\frac{1}{3}\u22c5\\frac{1}{8}= \\frac{3}{8}.$$\n\nFor the second question, you know that you don't have the middle coin, so you need the probability of the first compared to the last. If you imagine that you can tell the two sides of the two-tailed coin apart, there are 8 ways to do three flips, all of which are all tails. For the fair coin, only one triple is all tails. So when you see all tails the probability that you had the all-tail coin is 8/9.", "date": "2022-05-21 10:10:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1724731/if-i-flip-1-of-3-modified-coins-3-times-whats-the-probability-that-i-wil", "openwebmath_score": 0.950875461101532, "openwebmath_perplexity": 292.10920648775226, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936101542133, "lm_q2_score": 0.9111797015700343, "lm_q1q2_score": 0.8966861220172937}}
{"url": "http://stewswebsolutions.co.uk/journal/gz9p0.php?b2be3c=how-to-find-arc-length-of-a-circle", "text": "The lack of closed form solution for the arc length of an elliptic arc led to the development of the elliptic integrals. Measurement by arc length Definition of arc length and formula to calculate it from the radius and central angle of the arc. Improve your skills with free problems in 'Arc measure and arc length' and thousands of other practice lessons. The red arc measures 120. Measurement by central angle . length of arc AB = (5/18)(2r) = (5/18)(2(18)) = 10. Any diameter of a circle cuts it into two equal semicircles. Yes, Arc Length and Circumference isnt particularly exciting. Section 11.1 Circumference and Arc Length 595 Using Arc Lengths to Find Measures Find each indicated measure. The blue arc measures 240. $Ans = 2\\pi a$ How to obtain the ans? The area of a semicircle is half the area area of the circle from which it is made. Related Book. Measurement by arc length Calculating a circle's arc length, central angle, and circumference are not just tasks, but essential skills for geometry, trigonometry and beyond. central angle calculator, arc length calculator, ... calculating arc lengths, ... A circle has an arc length of 5.9 and a central angle of 1.67 radians. Geometry For Dummies, ... you find the fraction of the circles circumference that the arc makes up. Area of a semicircle. A semicircle is a half circle, formed by cutting a whole circle along a diameter line, as shown above. CHAPTER 5A Central Angles, Arc Length, and Sector Area ... sector represents of the circle. Arc of a Circle. You can use C 360 = l measureof thecentralangle or measureof thecentralangle 360 = l C Example 1, finding the arc length. Find the length of an arc of a circle having radius 7 cm and central angle 30 degrees? Use the formula C = 2r to calculate the circumference of a circle when the radius is given. The length of arc is equal to radius The length of an arc of a circle which subtends an angle radian at the center is equal to r where r is the radius of the circle. Calculating a circle's arc length, central angle, and circumference are not just tasks, but essential skills for geometry, trigonometry and beyond. CHAPTER 5A Central Angles, Arc Length, and Sector Area ... for a central angle of a circle Calculate the arc length and the area of a sector formed by a 30 central The distance along the arc (part of the circumference of a circle, or of any curve). Question: Find the arc length of the circle given by $x^2+y^2=a^2$. It depends on the radius of a circle and the central angle. Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again! Relate the length of an arc to the circumference of a whole circle and the central angle subtended by the arc. The circumference of a circle is an arc measuring 360o. Step 1 : Here, radius = 7cm central angle= 30 degrees. We dare you to prove us wrong. How to Find Arc Length. ... the arc length of a circumscribed circle is: Arc length is a linear measure of the arc measured along the circle. Thus, the length of the arc AB will be 5/18 of the circumference of the circle, which equals 2r, according to the formula for circumference. The relationship of arc length to a ... as the length x of an arc of the unit circle. Join Our Geometry Teacher Community Today! * An alternative definition is that it is an open arc. a. arc length of AB b. circumference of Z c. m RS A connected section of the circumference of a circle. Here's how to calculate the circumference, radius, diameter, arc length and degrees, sector areas, inscribed angles, and other shapes of the circle. and l stand for arc length. Our arc length calculator can calculate the length of an arc of a circle and the area of a sector. Learn how to find the arc with our lesson and try out our examples questions. Geometry calculator solving for circle arc length given radius and central angle An arc is any portion of the circumference of a circle.http://www.mathwords.com/a/arc_circle.htm Arc length is the $Ans = 2\\pi a$ How to obtain the ans? The arc length formula is used to find the length of an arc of a circle. To find the arc length, we now need to find the circumference It can be understood, that the arc length is a fraction of the circumference of the circle. But it can, at least, be enjoyable. The length of the circumference is given by the formula: C = d, where d is the diameter of the circle. I would like to calculate the arc length of a circle segment, i.e. The length of an arc is a connected section of the circumference of a circle. These curves are called rectifiable and the number is defined as the arc length. Use the formula C = d to calculate the circumference of a circle when the diameter is given. Question: Find the arc length of the circle given by $x^2+y^2=a^2$. Arc length of a circle is the distance measured as the length. For a circle: Arc Length = r (when is in radians) Fun math practice! I have no ideas after doing the following thing. Formula is S = r. See note at end of page. How to Determine the Length of an Arc. I have no ideas after doing the following thing. This formula can also be given as: C = 2r, where r is the radius. Step 2 : Calculation of an arc length without its central angle is a tough problem since the arc length is based on the angle.\nCopyright 2017 how to find arc length of a circle", "date": "2018-10-17 03:35:49", "meta": {"domain": "co.uk", "url": "http://stewswebsolutions.co.uk/journal/gz9p0.php?b2be3c=how-to-find-arc-length-of-a-circle", "openwebmath_score": 0.90330970287323, "openwebmath_perplexity": 291.45016114516045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9865717452580314, "lm_q2_score": 0.9086179031191509, "lm_q1q2_score": 0.8964167504529535}}
{"url": "https://www.hpmuseum.org/forum/thread-14070-post-124365.html", "text": "Looking for more algorithms for quasi-random numbers\n11-29-2019, 01:06 PM (This post was last modified: 11-30-2019 06:16 AM by Namir.)\nPost: #1\n Namir Senior Member Posts: 690 Joined: Dec 2013\nLooking for more algorithms for quasi-random numbers\nHi All Math Lovers,\n\nIn another thread of mine, ttw mentions quasi-random numbers. Quasi-random numbers (QRNs) present a better spread over a range of values than pseudo-random numbers (PRNs). On the other hand, QRNs will often fail randomness tests. They true purpose to to cover more uniformly a range of values in one of more dimensions.\n\nThis is part of ttw's response in my other thread, where he mentions QRNs:\n\nQuote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.\n\nThe above text includes the algorithm of setting x(1)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). The array of P() represents prime numbers starting with 2. You can change x(1) to had a uniform random number as a seed (to generate different sequences every time you apply the algorithm) or simply set x(1) = sqrt(P(1)) = sqrt(2).\n\nI am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. My first attempt was something like:\n\nCode:\nn\u00a0=\u00a0number\u00a0of\u00a0x\u00a0to\u00a0generate m\u00a0=\u00a0100*n Calculate\u00a0P()\u00a0for\u00a0primes\u00a0in\u00a0the\u00a0range\u00a0of\u00a01\u00a0to\u00a0m X(1)\u00a0=\u00a0rand\u00a0or\u00a0Frac(ln(P(3))\u00a0*\u00a0sqrt(P(1)) j\u00a0=\u00a02 count\u00a0=\u00a00 for\u00a0i=2\u00a0to\u00a0n \u00a0\u00a0X(i)\u00a0=\u00a0Frac(X(i)\u00a0+\u00a0ln(P(j+1))\u00a0*\u00a0sqrt(P(j-1)) \u00a0\u00a0j\u00a0=\u00a0j\u00a0+\u00a01 \u00a0\u00a0if\u00a0j\u00a0>\u00a0m\u00a0then \u00a0\u00a0\u00a0count\u00a0=\u00a0count\u00a0+\u00a01 \u00a0\u00a0\u00a0\u00a0j\u00a0=\u00a02\u00a0+\u00a0count \u00a0\u00a0end end\n\nThe above code produces x() with a mean near 0.5 and standard deviation near 0.28. The auto correlations for the first 50 lags are in the orde rof 10^(-2) to 10^(-4).\n\nI am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. You can even commit math heresy!!! As long as it works, you are fine (and forgiven) :-)\n\nNamir\n11-29-2019, 04:49 PM (This post was last modified: 11-29-2019 06:24 PM by SlideRule.)\nPost: #2\n SlideRule Senior Member Posts: 1,013 Joined: Dec 2013\nRE: Looking for more algorithms for quasi-random numbers\nPerusal of Quasi-random sequences in art and integration, John D. Cook Consulting, illumes the phenomena with references to more descriptive books; Random Number Generation and Quasi-Monte Carlo Methods & Monte Carlo and Quasi-Monte Carlo Methods, on the same.\n\nBEST!\nSlideRule\n11-30-2019, 01:36 AM\nPost: #3\n mfleming Senior Member Posts: 498 Joined: Jul 2015\nRE: Looking for more algorithms for quasi-random numbers\n(11-29-2019 01:06 PM)Namir Wrote: \u00a0This is part of ttw's response in my other thread, where he mentions QRNs:\n\nQuote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.\n\nUsing \"quote\" in place of \"code\" will autowrap large blocks of text!\n\nWho decides?\n11-30-2019, 01:29 PM\nPost: #4\n Namir Senior Member Posts: 690 Joined: Dec 2013\nRE: Looking for more algorithms for quasi-random numbers\n(11-30-2019 01:36 AM)mfleming Wrote:\n(11-29-2019 01:06 PM)Namir Wrote: \u00a0This is part of ttw's response in my other thread, where he mentions QRNs:\n\nQuote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.\n\nUsing \"quote\" in place of \"code\" will autowrap large blocks of text!\n\nI learned that the hard way :-)\n11-30-2019, 07:52 PM\nPost: #5\n Namir Senior Member Posts: 690 Joined: Dec 2013\nRE: Looking for more algorithms for quasi-random numbers\nThe few leads I got from the nice folks on this web were able to lead me to methods that generate sequences of quasi-random numbers that are practically perfectly distributed. I got what I was looking for. Thanks!!!\n\nNamir\n12-01-2019, 05:52 AM\nPost: #6\n ttw Member Posts: 186 Joined: Jun 2014\nRE: Looking for more algorithms for quasi-random numbers\nThis is one of the sequences from my paper in \"Computational investigations of low-discrepancy point sets II\" from the 1994 Las Vegas Conference on Monte Carlo and Quasi Monte Carlo Methods. I have made a single ad-hoc change (described below) that improves distribution for small numbers of points.\n\nThe Halton Sequence Phi(N,P) (for odd primes, 2 is a special case not considered here) can be described as:\n1. Generate the digits of N in base P (for P an odd prime). Call these digits a(1) to a(k) where k is the maximum number of digits needed. (There should be lots of subscripts but I'll treat each prime separately to reduce index management.)\nN=Sum from j=1 to k of a(j)*P^(j-1), that is: a(k)a(k-1)...a(2)(a(1).\n2. Reverse the digits: a(1),a(2)....a(k-1),a(k) is resulting string.\n3. Treat this string as a fraction with a decimal point (p-ary point?) in front.\nExample: P=3, N=5: 5(3)=12. Reverse Phi(5,3)=.21(3) or 2/3+1/9 = 7/9\nThe sequence is very well distributed for example one starts with 0, 1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27, 10,27...\nThis sequence is uniformly in the unit d-cube using d different primes. For example in 3 dimensions using primes 3 and 5 gives the points: (skipping 0 which sits on the corner of the cube).\n(1/3, 1/5, 1/7)\n(2/3, 2/5, 2/7)\n(1/9, 3/5, 3/7)\n(4/9, 4/5, 4/7)\n(7/9,1/25,5/7)\n(2/9, 6/25, 6/7)\n(5/9, 11/25, 1/49)\n(8/9. 16/25, 8/49)\netc.\nThe process is sometimes termed a Kakatumi-von Neumann odometer.\n\nThere is a problem that I noticed about 1967 or so when I started working on quasi-Monte Carlo. For large base, the Halton Sequence produces strongly correlated points until enough points are generated. (This happens with all quasi-random sequences but not as severely.) Take the first few points using bases 101 and 103.\n(1/101, 1/103)\n(2/101, 2/103)\n...\n(100/101, 100/103)\n(1/10201, 101/103)\n(102/10201, 102/103)\n(203/10203, 1/10609)\netc.\nIn 1993-1995 period, I figured out to multiply each numerator by a number (I called a spin) to break this up. Then I used the fact the fractional parts of square roots of primes are independent to do the following seemingly strange rule.\n\nGive a prime P, to find a multiplier S, do the following.\n1. Compute the nearest integers to the fractional part of the Sqrt(P), call these H and L (high and low, one is above and one below the number).\n2. Compute the continued fraction of H/P and L/P; each generates a string of partial quotients. The multiplier S is the one of these satisfying the following:\n3. A. Chose the one for which the sum of the partial quotients is smallest.\nB. If tied, chose the one with the smallest maximum partial quotient.\nC. If tied, chose whichever H/P or L/P is closest to the fractional part of the square root.\nD. Ad Hoc Alert: if P=41, use 16. (To avoid 17/41 being close to 12/29. The only such case in all primes less than 2^32)\n\n4. To generate the modified Phi sequence Phi(N,P,S): generate the digits of N base P as above and reverse. Multiply each digit by S modulo P and sum as above.\nExamples: 3 dimensions: P=3, 5, 7 have S= 2, 2,and 5 respectively.\n\n(2/3, 2/5, 5/7)\n(1/3, 4/5, 3/7)\n(2/9, 1/5, 1/7)\n(8/9, 3/4, 6/7)\netc.\nFor the pathological case 101 and 103, the multipliers are 6 and 16 respectively (not the best but that's another post sometime).\n(6/101, 16/103)\n(12/101, 32/103)\n(18/101, 48/103)\nClearly more spread out than the original Halton Sequence.\n\nI've got some more but Mordechay Levin's paper ArXiv 1806 shows that even the original Halton Sequence hits the theoretical lower bound for dimensions 2 and up so great changes cannot be had by tinkering. I do have a bit better, but it's even longer to compute and I haven't tested the new ideas thoroughly.\n\nHP50g code:(Number, Prime, Spin, Top, Bottom)\n\n<< 0 1 -> N P S T B << WHILE N REPEAT\nN S * P MOD T P * + 'T' STO\nP 'B' STO* N P IQUOT 'N' STO END\nT B />> >>\n\nNot necessarily the fastest but eas to work with. (If I've done this right.)\n12-01-2019, 11:46 AM\nPost: #7\n Csaba Tizedes Senior Member Posts: 412 Joined: May 2014\nRE: Looking for more algorithms for quasi-random numbers\n \u00ab Next Oldest | Next Newest \u00bb\n\nUser(s) browsing this thread: 1 Guest(s)", "date": "2020-02-17 09:29:34", "meta": {"domain": "hpmuseum.org", "url": "https://www.hpmuseum.org/forum/thread-14070-post-124365.html", "openwebmath_score": 0.6563873291015625, "openwebmath_perplexity": 2756.027968051556, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986571747626947, "lm_q2_score": 0.9086178944582997, "lm_q1q2_score": 0.8964167440608416}}
{"url": "http://lampedusasiamonoi.it/yvgr/max-sum-of-2-arrays.html", "text": "# Max Sum Of 2 Arrays\n\nReductions. It also prints the location or index at which maximum element occurs in array. int [] A = {\u22122, 1, \u22123, 4, \u22121, 2, 1, \u22125, 4}; Output: contiguous subarray with the largest sum is 4, \u22121, 2, 1, with sum 6. Array is an arranged set of values of one-type variables that have a common name. Yes you can find the maximum sum of elements in linear time using single traversal of the array. You can also use the following array formulas: Enter this formula into a blank cell, =SUM(LARGE(A1:D10,{1,2,3})), and then press Ctrl + Shift + Enter keys to get your result. It should have 3 input parameters array A, length and width. Finding the Average value of an Array. Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements. In this example, you create two arrays, DAYS and HOURS. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. However, I would like to use the max of these scores. When common element is found then we will add max sum from both the arrays to result. If x and y are scalars and A and B are matrices, y x, A x, and x A have their usual mathematical meanings. Max sum in an array. For all possible combinations, find the sum and compare it with the previous sum and update the maximum sum. Idea is to use merge sort algorithm and maintain two sum for 1st and 2nd array. Search in Rotated Sorted Array. The user will enter a number indicating how many numbers to add and then the user will enter n numbers. I need to check an array of random integers (between 1 and 9) and see if any combination of them will add up to 10. A corner element is an element from the start of the array or from the end of the array. Once the type of a variable is declared, it can only store a value belonging to this particular type. Create a max heap i. Given an array, you have to find the max possible two equal sum, you can exclude elements. Question E3: WAP to find out the row sum and column sum of a two dimensional array of integers. Whenever possible, make sure that you are using the NumPy version of these aggregates when operating on NumPy arrays!. Write a program to find those pair of elements that has the maximum and minimum difference among all element pairs. Input size and elements in array, store in some variable say n and arr[n]. Here is the complete Java program with sample outputs. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. Method since it requires contiguous, it means that for each element, it has two situations that are in the subarray or not. Easy Tutor says. The return value of min () and max () functions is based on the axis specified. C++ :: Creating Table Of Arrays - Find Maximum Value And Sum Aug 12, 2014. Google Advertisements. Sample Run: [2, 1, 8, 4, 4] Min: 1 Max: 8 Average: 3. The master will loop from 2 to the maximum value on issue MPI_Recv and wait for a message from any slave (MPI_ANY_SOURCE), if the message is zero, the process is just starting, if the message is negative, it is a non-prime, if the message is positive, it is a prime. Given an integer array of N elements, find the maximum sum contiguous subarray (containing at least one element). HackerRank Solutions Over the course of the next few (actually many) days, I will be posting the solutions to previous Hacker Rank challenges. 1 Answer to Given that A[MAX_ROWS][MAX_COLUMNS] is a 2 dimensional array of integers write a C ++ function. WriteLine to do this. SUMPRODUCT( array1, [array2, array_n] ) Parameters or Arguments array1, array2, array_n The ranges of cells or arrays that you wish to multiply. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. The maximum product is formed by the (-10, -3) or (5, 6) pair. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. A selected portion of the array may be summed, if an integer range expression is provided with the array name (. max (x) \u2192 [same as input] Returns the maximum value of all input values. Write a program to find sum of each digit in the given number using recursion. Basic Operations \u00b6. Find the sum of numbers and represent it in array. The function should return an integer. A one-dimensional array is like a list; A two dimensional array is like a table; The C language places no limits on the number of dimensions in an array, though specific implementations may. Write a program to find top two maximum numbers in a array. MS Excel 2007: Use an array formula to sum all of the order values for a given client This Excel tutorial explains how to use an array formula to sum all of the order values for a given client in Excel 2007 (with screenshots and step-by-step instructions). min () find the maximum and minimum value of the arguments, respectively. This very simply starts with a sum of 0 and add each item in the array as we go: public static int findSumWithoutUsingStream (int[] array) { for (int value : array) { 2. In this article we\u2019ll explore four plug and play functions that allow you to easily find certain values in an arrays of numbers. (For clarification, the L-length subarray could occur before or after the M-length subarray. A matrix with m rows and n columns is actually an array of length m, each entry of which is an array of length n. We can switch from one array to another array only at common elements. For example if input integer array is {2, 6, 3, 9, 11} and given sum is 9, output should be {6,3}. See (2) in the diagram. We can start from either arrays but we can switch between arrays only through its common elements. My solution for the bigDiff using the inbuilt Math. If you sum the second array you can use that to multiply the first array because that will be the same as multiplying the values individually and then summing the results. K maximum sum combinations from two arrays Given two equally sized arrays (A, B) and N (size of both arrays). Objective Problem Statement \u2022 Application of parallel prefix: Identifying the maximum sum that can be computed using. min (x) \u2192 [same as input]. Pop the heap to get the current largest sum and along. Maximum Sum of Two Non-Overlapping Subarrays. For example, to sum the top 20 values in a range, a formula must contain a list of integers from 1 to 20. For example [1,3,5,6,7,8,] here 1, 3 are adjacent and 6, 8 are not adjacent. We are making max_sum_subarray is a function to calculate the maximum sum of the subarray in an array. Specifically we'll explore the following: Finding the Minimum value in an Array. 4+ PHP Changelog: PHP versions prior to 4. And so myself and the OP exchanged a comment: I have concern. For an array x, y=cumsum(x) returns in the scalar y the cumulative sum of all the elements of x. In C programming, you can pass en entire array to functions. Input the array elements. You may have A1:A20, then A30:A35 filled. Thus, two arrays are \u201cequal\u201d according to Array#<=> if, and only if, they have the same length and the value of each element is equal to the value of the corresponding element in the other array. C++ Programs to Delete Array Element C++ Programs to Sum of Array Elements. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. Top Forums Shell Programming and Scripting Sum elements of 2 arrays excluding labels Post 303015114 by Don Cragun on Wednesday 28th of March 2018 06:34:58 AM. An index value of a Java two dimensional array starts at 0 and ends at n-1 where n is the size of a row or column. Given input array be,. In this solution dp* stores the maximum among all the sum of all the sub arrays ending at index i. Once the type of a variable is declared, it can only store a value belonging to this particular type. log10(a) Logarithm, base 10. This function subtracts when negative numbers are used in the arguments. (2-D maximum-sum subarray) (30 points) In the 2-D Maximum-Sum Subarray Prob- lem, you are given a two-dimensional m x n array A[1 : m,1: n of positive and negative numbers, and you are asked to find a subarray Ala b,c: 1 Show transcribed image text Expert Answer. C Program to read an array of 10 integer and find sum of all even numbers. if 2,3,4, 5 is the given array, {4,5,2,3} is also a possible array like other two. Our maximum subset sum is. SemanticSpace Technologies Ltd interview question: There is an integer array consisting positive numbers only. If any element is greater than max you replace max with. Enables ragged arrays. Array-2, Part I \u201d. I haven't gotten that far yet, I'm stuck just trying to print my two arrays, every time i try to print the first array it gives me the elements of the second array and it. All arrays must have the same number of rows and columns. i* n/2 \u2013 Or overlaps both halfs: i* n/2 j* \u2022 We can compute the best subarray of the first two types with recursive calls on the left and right half. Then we compare it with the other array elements one by one, if any element is greater than our assumed. I need to check an array of random integers (between 1 and 9) and see if any combination of them will add up to 10. The maximum product is formed by the (-10, -3) or (5, 6) pair. SUM (C, DIM=1) returns the value (5, 7, 9), which is the sum of all elements in each column. computes bitwise conjunction of the two arrays (dst = src1 & src2) Calculates the per-element bit-wise conjunction of two arrays or. This program shows you how to find the sum of rows and columns in a two dimensional array, and also how to use a method to calculate the sum of every element inside of a 2d array. Given two equally sized arrays (A, B) and N (size of both arrays). We can do this by using or without using an array. (Array): Returns the new array of chunks. If they are even we will try to find whether that half of sum is possible by adding numbers from the array. Latest commit message. So, the minimum of the array is -1 and the maximum of the array is 35. Google Advertisements. 999997678497 499911. Add solution to Pairs problem. Next, we use a standard for loop to iterate through our array numbers and total up the. Binary Tree Maximum Path Sum Lowest Common Ancestor I II III Binary Tree Level Order Traversal I II Kth Smallest Sum In Two Sorted Arrays LinkedList. We can switch from one array to another array only at common elements. Bottleneck code often involves condi-tional logic. if the sum of previous subarray is negative, it means that it need. 1<=Ai<=10000, where Ai is the ith integer in the array. The purpose of the SUMPRODUCT function is to multiply, then sum, arrays. The page is a good start for people to solve these problems as the time constraints are rather forgiving. Also add the common element to the result. The algorithm to find maximum is: we assume that it's present at the beginning of the array and stores that value in a variable. It may or may not include a[i-1], a[i-2], etc. M = max( A ,[], 'all' , nanflag ) computes the maximum over all elements of A when using the nanflag option. We can update both incrementally by counting from the back, so we have to keep track of two things: \\$\\max(S[i:])\\$ and \\$\\max(B[i+1:])\\$. Specifically we\u2019ll explore the following: Finding the Minimum value in an Array. We take a two dimensional array L of size count+1, sum/2+1. Partition an array into two sub-arrays with the same sum. For example, given array A such that: A[0] = 3 A[1] = 2 A[2] = -6 A[3] = 4 A[4] = 0. Each element, therefore, must be accessed by a corresponding number of index values. creating a recursive way to find max and min in array I am kind of confused with this instruction: Describe a recursive algorithm for finding both the minimum and maximum elements in an array A of n elements. If ARRAY is a zero-sized array, the result equals zero. computes the sum of two matrices and then prints it. The basis is p[0] = a[0]. package net. What is the sum of the last column? 5. Yes you can find the maximum sum of elements in linear time using single traversal of the array. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. The first thing that we tend to need to do is to scan through an array and examine values. sum(my_first_array) >my_first_array. Our maximum subset sum is. Here is the complete Java program with sample outputs. Naive solution would be to consider every pair of elements and calculate their product. the contiguous subarray [4,-1,2,1] has the largest sum = 6. Question E2: WAP to display the values of a two dimensional array in the matrix form. Suppose we need to find out max sum subarray. As a \"rule of thumb\", any \"calculated array\" - in this case the array calculated by adding two ranges - results in a formula that requires CSE, although some functions (like SUMPRODUCT and LOOKUP) don't normally need CSE even with calculated arrays - to allow normal entry you can add an INDEX function - I edited my answer to the effect. maxSubsetSum has the following parameter(s): arr: an array of integers. amax() by thispointer. Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. Solution to Question 2. We can use an array as a deque with the following operations:. Improvement over Method-1 \u2013 O(n 2) Time. If only ARRAY is specified, the result equals the sum of all the array elements of ARRAY. It works as follows. Once the type of a variable is declared, it can only store a value belonging to this particular type. Maximize array sum by concatenating corresponding elements of given two arrays Given two array A[] and B[] of the same length, the task is to find the maximum array sum that can be formed by joining the corresponding elements of the array in any order. Our maximum subset sum is. In a two-dimensional Java array, we can use the code a[i] to refer to the ith row (which is a one-dimensional array). And so myself and the OP exchanged a comment: I have concern. I borrowed some code from other forums that had similar programs, but obviously it doesn't match my needs specifically. Given an array of integers, find maximum product of two integers in an array. Computes the matrix multiplication of two arrays. Add solutions to C++ domain. This problem is generally known as the maximum sum contiguous subsequence problem and if you haven\u2019t encountered it before, I\u2019d recommend trying to solve it before reading on. #include using namespace std; int main() { const int SIZE = 12; double months[SIZE]; int count; double sum = 0; double totalRainfall; double. c++: which functions gives the sum of an array? (3). In this example, we will find the sum of all elements in a numpy array, and with the default optional parameters to the sum () function. min (x) \u2192 [same as input]. Given two arrays of positive integers. For example, for the sequence of values \u22122, 1, \u22123, 4, \u22121, 2, 1, \u22125, 4; the contiguous subarray with the largest sum is 4, \u22121, 2, 1, with sum 6. Let arr[i. Your code tries all \\$n (n+1)/2 \\$ combinations of array elements to find the combination with the largest sum, so the complexity is \\$O(n^2) \\$. The loop structure should look like for (i=0; i=2 and find the sum of smallest and second smallest, then our answer will be maximum sum among them. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. Maximum Sum of Two Non-Overlapping Subarrays 2019/04/22 2019/04/22 shiji Leetcode Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. Write a function to find the maximum sum of all subarrays. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. If all the array entries were positive, then the maximum-subarray problem would present no challenge, since the entire array would give the greatest sum. The length property is the array length or, to be precise, its last numeric index plus one. I borrowed some code from other forums that had similar programs, but obviously it doesn't match my needs specifically. The maximum product is formed by the (-10, -3) or (5, 6) pair. 7 is the sum of 2 + 5 in column 2, and so forth. Sub Array with Maximum Sum \u2013 Kadane Algorithm is the best solution. Note : Imp to execute and trace to understand and remember. Edit: given your comments if the initial array is fixed then you can use MMULT function like this. Passing array elements to a function is similar to passing variables to a function. if orientation is equal to n then. M=sum(A,dim) In Scilab dim=1 is equivalent to dim=\"r\" and dim=2 is equivalent dim=\"c\". For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. Read the entered array size and store that value into the variable n. Maximum Sum Subarray (In Yellow) For example, for the array given above, the contiguous subarray with the largest sum is [4, -1, 2, 1], with sum 6. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. Find maximum possible sum of elements such that there are no 2 consecutive elements present in the sum. Find ways to calculate a target from elements of specified. Given an array, find maximum sum of smallest and second smallest elements chosen from all possible sub-arrays. Once you have a vector (or a list of numbers) in memory most basic operations are available. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. So, the minimum of the array is -1 and the maximum of the array is 35. The maximum admissible amount of dimensions in an array is four. For example, A = [\u22122, 1, \u22123, 4, \u22121, 2, 1, \u22125, 4] then max sum=11 with the subarray [1, 4, 2, 4]. Calculates the per-element sum of two arrays or an array and a scalar. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. Join 124,729,115 Academics and Researchers. Expected time complexity is O(m+n) where m is the number of elements in ar1[] and n is the number of elements in ar2[]. I am trying to compute the maximum possible sum of values from a matrix or 2d array or table or any suitable structure. A selected portion of the array may be summed, if an integer range expression is provided with the array name (. Hence there would be four different arrays in this case. sum += numbers [i] In The Standard Way we first declare the variable sum and set its initial value of zero. The sum choice number is the minimum over all choosable functions f of the sum of the sizes in f. Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. SUM (C, DIM=2) returns the value (6, 15), which is the sum of all elements in each row. e 1,2,3,4,6 is given array we can have max two equal sum as 6+2 = 4+3+1. =MAX(IF((List>=LLim)*(List<=ULim),List,FALSE)) returns the maximum of values between 2 and 5, or 5. When you want to return a sum for a single criteria (for example, a single IF condition) When you want to use multiple criteria and return the sum to multiple cells; The criteria that you can use with the SUMIF() worksheet function is limited to text, numbers, or a range, and the function cannot use array constants. Stack PUSH & POP Implementation using Arrays; Program to remove duplicate element in an array; C Program to sort the matrix rows and columns; Write a c program for swapping of two arrays; C Program to read name and marks of students and store it in file; To find out the maximum number in an array using function. Explore Channels Plugins & Tools Pro Login About Us. Academia is the easiest way to share papers with millions of people across the world for free. The min () and max () functions of numpy. A corner element is an element from the start of the array or from the end of the array. Search in Rotated Sorted Array II. I need to create a table of this which i have done using case 1. The basis is p[0] = a[0]. (For clarification, the L-length subarray could occur before or after the M-length subarray. For example, consider the array {-10, -3, 5, 6, -2}. Calculates the per-element sum of two arrays or an array and a scalar. Display the maximum K valid sum combinations from all the possible sum combinations. Here we are setting up the pointer to the base address of array and then we are incrementing pointer and using * operator to get & sum-up the values of all the array elements. Even if you have encountered it before, I\u2019ll invite you. For example, to sum the top 20 values in a range, a formula must contain a list of integers from 1 to 20. If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. In the Java programming language, a multidimensional array is an array whose components are themselves arrays. For example, entering =SUM(10, 2) returns 12. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. Yes you can find the maximum sum of elements in linear time using single traversal of the array. To store sum of array elements, initialize a variable sum = 0. Latest commit 7b136cc on Mar 10, 2019. The function should return an integer. The call to new Array(number) creates an array with the given length, but without elements. What's wrong with the scrap of code in the question? The array is of size 5, but the loop is from 1 to 5, so an attempt will be made to access the nonexistent element a[5]. def array_summer(arr): return sum (arr) # Test input print (array_summer ( [1, 2, 3, 3, 7])) we went through two different methods of summing the elements of an array. For an array x, y=cumsum(x) returns in the scalar y the cumulative sum of all the elements of x. computes the sum of two matrices and then prints it. C++ :: Creating Table Of Arrays - Find Maximum Value And Sum Aug 12, 2014. Basicly I have to sum rows and columns of 2d array and then store the results in separate arrays, as far as the code is now, can see it quite clearly. Finding the Maximum value in an Array. ; The array formula lets the IF function test for multiple conditions in a single cell, and, when the data meets a condition, the array formula determines what data (event results) the MAX function will examine to find the best result. Yes you can find the maximum sum of elements in linear time using single traversal of the array. The previous contiguous array sum was less than or equals 0. Finally, if A is a multidimensional array, Matlab works on the first non-singleton dimension of A what Scilab does not. with - sum of two arrays in c. With the following program, you can even print the sum of two numbers or three numbers up to N numbers. For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. Finding the Sum of all values in an Array. For example [1,3,5,6,7,8,] here 1, 3 are adjacent and 6, 8 are not adjacent. For an array x, y=cumsum(x) returns in the scalar y the cumulative sum of all the elements of x. Note : Imp to execute and trace to understand and remember. The problem: given an array which could contain zero, negative, and positive numbers, find the largest sum of contiguous sub-array. Move the pointer in the corresponding heap there. We can update both incrementally by counting from the back, so we have to keep track of two things: \\$\\max(S[i:])\\$ and \\$\\max(B[i+1:])\\$. Array-2, Part I \u201d. To get the sum of all elements in a numpy array, you can use Numpy\u2019s built-in function sum (). Question 3. such that sum(wi*xi)<=W & x(0,1) The unbounded knapsack problem (UKP) places no upper bound on the number of copies of each kind of item. Maximum Sum Subarray (In Yellow) For example, for the array given above, the contiguous subarray with the largest sum is [4, -1, 2, 1], with sum 6. These functions will not work as-is with arrays of numbers. computes the sum of two matrices and then prints it. Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. Add solutions to C domain. Two Dimensional Array in C Example. Dim i,sum, r As Integer sum=0 For i=0 to 4 \u2018assign values to the array x(i)=i*i Next r=1 For each v in x \u2018read the elements of the array in to Excel cells MsgBox v r= r+1 Next Note: the start index of the array is 0 and its size is equal to last index added by 1. Once the type of a variable is declared, it can only store a value belonging to this particular type. Array formulas can also be used find out the maximum and minimum values for a given set of conditions. The call to new Array(number) creates an array with the given length, but without elements. Maximum of array elements over a given axis. Sub Array with Maximum Sum \u2013 Kadane Algorithm is the best solution. The SUM function in Excel adds the arguments you enter in a formula. Java Program to Find The Sum of Array Elements || D. sum() 15 How to find the maximum value in NumPy 1d-array? We can find the maximum value stored in. Subarr2[] = {3, 3, 12. Problem Statement: Given an array A = {a 1, a 2, a 3, , a N} of N elements, find the maximum possible sum of a. As a \"rule of thumb\", any \"calculated array\" - in this case the array calculated by adding two ranges - results in a formula that requires CSE, although some functions (like SUMPRODUCT and LOOKUP) don't normally need CSE even with calculated arrays - to allow normal entry you can add an INDEX function - I edited my answer to the effect. Print the N integers of the array in the reverse order in a single line separated by a space. Which row has the largest sum? 4. Array is an arranged set of values of one-type variables that have a common name. Take a HashMap with Key and value as Integer types. This problem is generally known as the maximum sum contiguous subsequence problem and if you haven\u2019t encountered it before, I\u2019d recommend trying to solve it before reading on. The sum of all the numbers in the array. To store sum of array elements, initialize a variable sum = 0. C++ Programs to Delete Array Element C++ Programs to Sum of Array Elements. Let arr[i. Where L[i,j]=maximum sum possible with elements of array 0 to i and sum not exceeding j. Using only one loop, Complete the code to compute both sums. Function Description. Description: ----- If we add amount of max INT with number 1 in array_sum function , the result will be false. Hi, My documents have an \"aliases\" field which is an array of string. Objective: The maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional array of numbers which has the largest sum. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. Find Maximum sum sub array of tempArray. I'm stumped. if the array was [5, 6, 4, 2, 9] it would return true. j* n/2 \u2013 Or contained entirely in the right half , i. For all possible combinations, find the sum and compare it with the previous sum and update the maximum sum. In this example, we will find the sum of all elements in a numpy array, and with the default optional parameters to the sum () function. To find the maximum value, you initialize a placeholder called max with the value of the first element in the array. 1 Answer to Given that A[MAX_ROWS][MAX_COLUMNS] is a 2 dimensional array of integers write a C ++ function. int a[3]; // creates an array with 'Numb' elements a[3] = 5; // assigns 5 to index 3 (the 4th element) in the array This is effectively what you're doing with your cin line. But when you try to follow the same approach with an array formula, Excel complains. Problem Description We have to write a program in C such that the program will allocate 2 one-dimensional arrays using malloc() call and then will do the addition and stores the result into 3rd array. min(), big_array. but it must include a[i]. Maximum sum in circular array such that no two elements are adjacent | Set 2 Given an array arr[] of positive numbers, find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array where the last and the first elements are assumed adjacent. Latest commit 7b136cc on Mar 10, 2019. How we can do that efficiently?. max (x, n) \u2192 array<[same as x]> Returns n largest values of all input values of x. The elements entered in the array are as follows: 1 2 35 0 -1. The prior values added up to 0, meaning that the new max subarray starts from this value. Empty subarrays/subsequences should not be considered. See (2) in the diagram. For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. For example if input integer array is {2, 6, 3, 9, 11} and given sum is 9, output should be {6,3}. (#M40034130) C Programming question Find out maximim sum of sub Array Keep an EYE Find out maximim sum of sub Array example array={2,3,-1,4,9} maximum sum of sub array=17 Asked In C DHIRENDRA (6 years ago) Unsolved Read Solution (3) Is this Puzzle helpful? (1) (0) Submit Your Solution Program. The expression within the optional \"with\" clause can be used to specify the item to use in the reduction. Dynamic Memory Allocation Example: In this C program, we will declare memory for array elements (limit will be at run time) using malloc(), read element and print the sum of all elements along with the entered elements. Return the maximum of sum1 and sum2. 2 Vectorized Logic The previous section shows how to vectorize pure computation. The \"waterdrop\" camera array is made up of a 48-MP main unit, an 8-MP wide-angle shooter, and a 5-MP depth sensor, with a flash module at the bottom (2020) vs iPhone 11, 11 Pro and Pro Max. Full Discussion: How do I find the sum of values from two arrays? Top Forums Shell Programming and Scripting How do I find the sum of values from two arrays? Post 302579313 by kshji on Monday 5th of December 2011 12:03:10 PM. Given two equally sized arrays (A, B) and N (size of both arrays). Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. Whenever possible, make sure that you are using the NumPy version of these aggregates when operating on NumPy arrays!. The min () and max () functions of numpy. Pop the heap to get the current largest sum and along. Your code tries all \\$n (n+1)/2 \\$ combinations of array elements to find the combination with the largest sum, so the complexity is \\$O(n^2) \\$. Arr2[] = {1, 3, 3, 12, 2} then maximum result is obtained when we create following two subarrays \u2212 Subarr1[] = {2, 4, 3} and. Using only one loop, Complete the code to compute both sums. Timing Belt - $1,299. Given an integer array of N elements, find the maximum sum contiguous subarray (containing at least one element). Compute sum of two digit arrays. For example, if A is a matrix, then max(A,[],[1 2]) computes the maximum over all elements in A, since every element of a matrix is contained in the array slice defined by dimensions 1 and 2. here maximum subarray is [2,3,4]. Divide Two Integers 4. You can max_heap both arrays, and set an iterator at the root for both. Complexity Analysis:. Input: nums = [1,1,1], k = 2 Output: 2. If only ARRAY is specified, the result equals the sum of all the array elements of ARRAY. Array exponentiation is available with A. This program is an example of Dynamic Memory Allocation, here we are declaring memory for N array elements at run time using malloc() - which is used to declare memory for N. Problem Description Write a program to find the sum of the corresponding elements in 2 arrays. A matrix with m rows and n columns is actually an array of length m, each entry of which is an array of length n. My solution for the bigDiff using the inbuilt Math. def sum_odd(n): if n < 2: return 1 elif n%2 == 0: return sum_odd(n-1) else: return n + sum_odd(n-2) Note that this function returns 1 if n is not greater 0 as is defined in the original function. You can refer to more than one array in a single SAS statement. Given input array be,. Note: Values of different types will be compared using the standard comparison rules. To display sub array with maximum sum you should write code to hold the start and end value of the sub array with maximum sum. My operation system is 64 bit. In this tutorial, I am going to discuss a very famous interview problem find maximum subarray sum (Kadane\u2019s algorithm). The optimal strategy is to pick the elements form the array is, two. Pure VPN Privide Lowest Price VPN Just @$1. Find ways to calculate a target from elements of specified. The first user will be asked to enter the order of the matrix (such as the numbers of rows and columns) and then enter the elements of the two matrices. I am trying to compute the maximum possible sum of values from a matrix or 2d array or table or any suitable structure. In a two-dimensional Java array, we can use the code a[i] to refer to the ith row (which is a one-dimensional array). MS Excel 2007: Use an array formula to sum all of the order values for a given client This Excel tutorial explains how to use an array formula to sum all of the order values for a given client in Excel 2007 (with screenshots and step-by-step instructions). Consider an integer array, the number of elements in which is determined by the user. Find a Triplet having Maximum Product in an Array. Sum the largest 3 numbers: =SUM(LARGE(range, {1,2,3})) Sum the smallest 3 numbers: =SUM(SMALL(range, {1,2,3})) Don't forget to press Ctrl + Shift + Enter since you are entering the Excel array formula, and you will get the following result: In a similar fashion, you can calculate the average of N smallest or largest values in a range:. Arrays in formulas. Array Maximum Minimum value We can calculate maximum value among the elements of an array by using max function. For example, consider the array {-10, -3, 5, 6, -2}. We can switch from one array to another array only at common elements. Hi, My documents have an \"aliases\" field which is an array of string. Basic array operations. Space Complexity: O(1). If it's provided then it will return for array of max values along the axis i. A blog about Java, Spring, Hibernate, Programming, Algorithms, Data Structure, SQL, Linux, Database, JavaScript, and my personal experience. Problem Statement: Given an array A = {a 1, a 2, a 3, , a N} of N elements, find the maximum possible sum of a. Add solution to Super Maximum Cost Queries problem. Naive solution would be to consider every pair of elements and calculate their product. Maximize array sum by concatenating corresponding elements of given two arrays Given two array A[] and B[] of the same length, the task is to find the maximum array sum that can be formed by joining the corresponding elements of the array in any order. I have an array of \"2,3,4,5,6,9,10,11,12,99\". The length property is the array length or, to be precise, its last numeric index plus one. Now the above Leetcode challenge is a special case of the general Max Subarray classic problem in computer science - which is the task of finding the contiguous subarray within a one-dimensional array of numbers which has the largest sum. We will implement a simple algorithm in javascript to find the maximum sum of products of given two arrays. Yesterday we got the question Sum of Maximum GCD from two arrays. Note that the common elements do not have to be at same indexes. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. Non-Numeric or Non-Existent Fields\u00b6. If DIM is absent, a scalar with the sum of all elements in ARRAY is returned. It loops over the values and returns the sum of the elements. Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements. Yes you can find the maximum sum of elements in linear time using single traversal of the array. Jerico January 10, 2014 at 6:30 am. For example, if the array contains: 31, -41, 59, 26, -53, 58, 97, -93, -23, 84 then the largest sum is 187 taken from the [59. The bottommost cell is A35. So far so good, and it looks as if using a list is as easy as using an array. As an example, the maximum sum contiguous subsequence of 0, -1, 2, -1, 3, -1, 0 would be 4 (= 2 + -1 + 3). 12) instead of the number entered into the array. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. Hence there would be four different arrays in this case. Find maximum sum path involving elements of given arrays Given two sorted array of integers, find a maximum sum path involving elements of both arrays whose sum is maximum. HI everyone, need help with this exercise: \"We have two integer numbers, which are represented by two arrays. For all possible combinations, find the sum and compare it with the previous sum and update the maximum sum. Sort both arrays array A and array B. 404 24 Add to List Share. This is way faster than a manually using a for loop going through all elements in a 1d-array. Monotonic Queue/Stack. Also add the common element to the result. Passing array elements to a function is similar to passing variables to a function. If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. We are making max_sum_subarray is a function to calculate the maximum sum of the subarray in an array. This program will help to understand the working of for loop, array, if statement and random numbers. C program to find the maximum or the largest element and the location (index) at which it's present in an array. Then print the respective minimum and maximum values as a single line of two space-separated long integers. Which is run a loop from 0 to n. If arr1[] = {1, 2, 4, 3, 2} and. If only ARRAY is specified, the result equals the sum of all the array elements of ARRAY. =MAX(IF((List>=LLim)*(List<=ULim),List,FALSE)) returns the maximum of values between 2 and 5, or 5. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. You can also declare an array of arrays (also known as a multidimensional array) by using two or more sets of brackets, such as String[][] names. To initialize array use random numbers. Input the array elements. Introduction to C Programming Arrays Overview. array_sum: Array Sum of. hence maximum subarray sum is 9. To display sub array with maximum sum you should write code to hold the start and end value of the sub array with maximum sum. if 2,3,4, 5 is the given array, {4,5,2,3} is also a possible array like other two. C++ :: Creating Table Of Arrays - Find Maximum Value And Sum Aug 12, 2014. First Bad Version. Even to find the total number of even elements in the array. This program is an example of Dynamic Memory Allocation, here we are declaring memory for N array elements at run time using malloc() - which is used to declare memory for N. When you want to return a sum for a single criteria (for example, a single IF condition) When you want to use multiple criteria and return the sum to multiple cells; The criteria that you can use with the SUMIF() worksheet function is limited to text, numbers, or a range, and the function cannot use array constants. Add solutions to C++ domain. Note that the common elements do not have to be at same indexes. 7 is the sum of 2 + 5 in column 2, and so forth. Java program to calculate the sum of N numbers using arrays, recursion, static method, using while loop. (#M40034130) C Programming question Find out maximim sum of sub Array Keep an EYE Find out maximim sum of sub Array example array={2,3,-1,4,9} maximum sum of sub array=17 Asked In C DHIRENDRA (6 years ago) Unsolved Read Solution (3) Is this Puzzle helpful? (1) (0) Submit Your Solution Program. The code below will show how to display a maximum of 3 items from an integer array. sum = sum + (value at 2000) In the Second iteration we will have following calculation \u2013 sum = sum + (value at 2002) = 11 + 12 = 23. We can switch from one array to another array only at common elements. Naive solution would be to consider every pair of elements and calculate their product. If the first and only parameter is an array, max() returns the highest value in that array. After partitioning, each subarray has their values changed to become the maximum value of that subarray. This function can take two other optional arguments, that will be covered in more detail, when we get to multi-dimensional arrays. It is For Each Loop or enhanced for loop introduced in java 1. Here is a simple example. minimum difference = second lowest - lowest. sum () is shown below. The master will loop from 2 to the maximum value on issue MPI_Recv and wait for a message from any slave (MPI_ANY_SOURCE), if the message is zero, the process is just starting, if the message is negative, it is a non-prime, if the message is positive, it is a prime. Here we will be displaying the sub array. Is there. Given an array, find maximum sum of smallest and second smallest elements chosen from all possible sub-arrays. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. For example, entering =SUM(10, 2) returns 12. 1- creat two int array the size of each array must be 40 2- ask the user to input 2 strings (big numbers ) 3- check each character of ths string if it's numeric characters 4- add each string in one int array 5-print out the sum of the 2 arrays 6-compare between the two arrays. the contiguous subarray [4,-1,2,1] has the largest sum = 6. c++: which functions gives the sum of an array? (3). if the array was [5, 6, 4, 2, 9] it would return true. Bilal-March 5th, 2020 at 2:07 pm none Comment author #29091 on Find max value & its index in Numpy Array | numpy. So for the test arrays: int[] testArrayA = { 7, 1, 4, 5, 1}; int[] testArrayB = { 3, 2, 1, 5, 3}; \u2191 starting i. Given two sorted arrays such the arrays may have some common elements. Adjacent: side by side. You need to create three different functions called minNumber,maxNumber and totalSum. 4 Two-dimensional Arrays. three two one one three two two three one three one two Array Reduction Methods : Array reduction methods can be applied to any unpacked array to reduce the array to a single value. Compute the ceiling power of 2. maximum difference = higest-lowest. Join 124,729,115 Academics and Researchers. While if we add this two via plus (+) operator ,the result will be true. SUM (C, DIM=2) returns the value (6, 15), which is the sum of all elements in each row. If it's provided then it will return for array of max values along the axis i. It should return an integer representing the maximum subset sum for the given array. sum %2 [1] 1 1 [2] 2 0 [3] 3 1 [1,2] 3 1 [2,3] 5 1 [1,2,3] 6 0. The syntax of numpy. max (x, n) \u2192 array<[same as x]> Returns n largest values of all input values of x. For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. Which is run a loop from 0 to n. Search in Rotated Sorted Array. Sort both arrays array A and array B. The bottommost cell is A35. 15 is the sum of 4 + 5 + 6 in row 2. Here is the code. Given an array A with n elements and array B with m elements. With the following program, you can even print the sum of two numbers or three numbers up to N numbers. Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C. You need to find out the maximum sum such that no two chosen numbers are adjacent , vertically, diagonally (or) horizontally. The maximum product is formed by the (-10, -3) or (5, 6) pair. Calculates the weighted sum of two arrays. You can enter a value in the box labelled 'Person ID', which is the first number of a two-dimensional array, and then select one of two numbers in the Name/Profession box. =MAX(ROW(4:6)*SUM(ROW(1:3))) confirm with CTRL+SHIFT+ENTER. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. In the given array, you need to find maximum sum of elements such that no two are adjacent (consecutive). To display sub array with maximum sum you should write code to hold the start and end value of the sub array with maximum sum. You can also return an array from a method. everything works except I need it to give me the max and min month (i. if 2,3,4, 5 is the given array, {4,5,2,3} is also a possible array like other two. Write a program to sort a map by value. Though, the arrays whose size is a product of 2\u2019s, 3\u2019s, and 5\u2019s (for example, 300 = 5*5*3*2*2) are also processed quite efficiently. We can switch from one array to another array only at common elements. Function Description. Arr2[] = {1, 3, 3, 12, 2} then maximum result is obtained when we create following two subarrays \u2212 Subarr1[] = {2, 4, 3} and. You can refer to more than one array in a single SAS statement. Write a program to find the sum of the corresponding elements in 2 arrays. Maximum Sum of Two Non-Overlapping Subarrays. Inside SUM, the range resolves to an array of values. The restriction is that once you select a particular row,column value to add to your sum, no other values from that row or column may be used in calculating the sum. Please note that your values will almost certainly be different, depending both on the random number generator and the values of the array you created in Lab8. Finding the Average value of an Array. Mini-Max Sum Hackerrank. It could be the sum of total sum of the left child and max prefix sum of the right child. Find ways to calculate a target from elements of specified. Maximum Sub Array Practice: max_sub_array. Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements. 17171281366e-06 0. The first line of the input contains N,where N is the number of integers. Given two equally sized arrays (A, B) and N (size of both arrays). It should return a long integer that represents the maximum value of. We can switch from one array to another array only at common elements. The bottommost cell is A35. You can pass a two dimensional array to a method just as you pass a one dimensional array. Add solution to Pairs problem. Now let's think of another type of two dimensional array in which the shape of the array is not square as shown below. More formally, if we write all (nC2) sub-arrays of array of size >=2 and find the sum of smallest and second smallest, then our answer will be maximum sum among them. Edit: given your comments if the initial array is fixed then you can use MMULT function like this. sum %2 [1] 1 1 [2] 2 0 [3] 3 1 [1,2] 3 1 [2,3] 5 1 [1,2,3] 6 0. Array Maximum Minimum value We can calculate maximum value among the elements of an array by using max function. ; The array formula lets the IF function test for multiple conditions in a single cell, and, when the data meets a condition, the array formula determines what data (event results) the MAX function will examine to find the best result. For example, if A is a matrix, then max(A,[],[1 2]) computes the maximum over all elements in A, since every element of a matrix is contained in the array slice defined by dimensions 1 and 2. Problem Description We have to write a program in C such that the program will allocate 2 one-dimensional arrays using malloc() call and then will do the addition and stores the result into 3rd array. We take a two dimensional array L of size count+1, sum/2+1. For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has the largest sum = 6. min(int a, int b) and Math. Here we will be displaying the sub array. min () find the maximum and minimum value of the arguments, respectively. word 0:5 y:. The last 'n' integers correspond to the elements in the second array. pranathi chunduru. Hence there would be four different arrays in this case. For example, consider the array {-10, -3, 5, 6, -2}. Given an integer array of N elements, find the maximum sum contiguous subarray (containing at least one element). # sum of all elements in the array >np. Java represents a two-dimensional array as an array of arrays. For example, A = [\u22122, 1, \u22123, 4, \u22121, 2, 1, \u22125, 4] then max sum=11 with the subarray [1, 4, 2, 4]. This example prints the maximum value in an array, and the subscript of that value:; Create a simple two-dimensional array: D = DIST (100); Print the maximum value in array D and its linear subscript: PRINT, 'Maximum value in array D is:', MAX (D, I) PRINT, 'The subscript of the maximum value is', I IDL Output Maximum value in array D is: 70. min, max, repmat, meshgrid, sum, cumsum, diff, prod, cumprod, filter 3. This program will help to understand the working of for loop, array, if statement and random numbers. Academia is the easiest way to share papers with millions of people across the world for free. I haven't gotten that far yet, I'm stuck just trying to print my two arrays, every time i try to print the first array it gives me the elements of the second array and it. Basic Operations \u00b6. com Great, I love this explanation. Reductions. The syntax of numpy. In this article we'll explore four plug and play functions that allow you to easily find certain values in an arrays of numbers. Improvement over Method-1 \u2013 O(n 2) Time. Output Format. I don't care if a document has many aliases matching my query. An index value of a Java two dimensional array starts at 0 and ends at n-1 where n is the size of a row or column. After partitioning, each subarray has their values changed to become the maximum value of that subarray. Two approaches: Simple approach is brute-force implementation but it will take O(n. Maximize array sum by concatenating corresponding elements of given two arrays Given two array A[] and B[] of the same length, the task is to find the maximum array sum that can be formed by joining the corresponding elements of the array in any order. Display the maximum K valid sum combinations from all the possible sum combinations. Timing Belt - $1,299. The code below will show how to display a maximum of 3 items from an integer array. Max sum in an array. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. As a precautionary health measure for our support specialists in light of COVID-19, we're operating with a limited team. Finding the maximum sum in two sorted arrays Given two sorted postive integer arrays A(n) and B(n) (let's say they are decreasingly sorted), we define a set S = {(a,b) | a \\in A and b \\in B}. Mini-Max Sum - Problem from HackerRank and Solution Using python 2 Problem Statement: Given five positive integers, find the minimum and maximum values that can be calculated by summing exactly four of the five integers. The program allocates 2 one-dimentional arrays using malloc() call and then does the addition and stores the result into 3rd array. Method since it requires contiguous, it means that for each element, it has two situations that are in the subarray or not. Join 124,729,115 Academics and Researchers. You are given an array of integers with both positive and negative numbers. Subarr2[] = {3, 3, 12. Find the sum of numbers and represent it in array. The maximum special sum considering all non-empty subarrays of the array. In the Java programming language, a multidimensional array is an array whose components are themselves arrays. For example,$ \\{35, 42, 5, 15, 27, 29\\} $is a sorted array that has been circularly shifted$ k=2 $positions, while$ \\{27, 29, 35, 42, 5, 15\\} $has been shifted$ k=4 \\$ positions. If only one array is supplied, SUMPRODUCT will simply sum the items in the array. C++ code to display contents of array. Find a Triplet having Maximum Product in an Array. Read 6 Integers from File, find sum, find average, and find Min/Max average. Write a program to find sum of each digit in the given number using recursion. Finding the Maximum value in an Array. Hi all, I'm looking for how to to the sum or union of two jagged array, I did it to simple array but do you have any idea how to do it for jagged array. For example, if an int. C Program to Find Maximum Element in Array - This program find maximum or largest element present in an array. We could also use other representations, such as an array containing two two-element arrays ([[76, 9], [4, 1]]) or an object with property names like \"11\" and \"01\", but the flat array is simple and makes the expressions that access the table pleasantly short. out [Optional] Alternate output array in which to place the. Sub Array with Maximum Sum \u2013 Kadane Algorithm is the best solution. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. In a two-dimensional Java array, we can use the code a[i] to refer to the ith row (which is a one-dimensional array). We can use an array as a deque with the following operations:. computes the sum of two matrices and then prints it. Search in Rotated Sorted Array II. As a precautionary health measure for our support specialists in light of COVID-19, we're operating with a limited team.", "date": "2020-05-26 21:19:42", "meta": {"domain": "lampedusasiamonoi.it", "url": "http://lampedusasiamonoi.it/yvgr/max-sum-of-2-arrays.html", "openwebmath_score": 0.3012574017047882, "openwebmath_perplexity": 681.2426012969111, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.97594644290792, "lm_q2_score": 0.9184802440252811, "lm_q1q2_score": 0.8963875270376713}}
{"url": "https://math.stackexchange.com/questions/2921700/how-to-simplify-or-upperbound-this-summation", "text": "# How to simplify or upperbound this summation?\n\nI am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upperbound the following summation:\n\n$$\\sum_{i=1}^n{\\exp{\\left(-\\frac{i^2}{\\sigma^2}\\right)}}.$$\n\nCan I use geometric series?\n\nEDIT: I have difficulty because of the power $2$, i.e if the summation would be $\\sum\\limits_{i=1}^n{\\exp{\\left(-\\frac{i}{\\sigma^2}\\right)}}$ then it would be easy to apply geometric series!\n\n\u2022 Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post) \u2013\u00a0Jacob Manaker Sep 18 at 16:18\n\u2022 Seems that you could try the integral $\\int_0^\\infty \\exp(-x^2)\\,\\mathrm dx$ to bound that. \u2013\u00a0xbh Sep 18 at 16:26\n\u2022 tinyurl.com/y76n9loh (wolframalpha) gives a closed form for the infinite sum involving the elliptic theta function. \u2013\u00a0barrycarter Sep 19 at 23:46\n\nAlternative:\n\nSince $f(x) = \\exp(-x^2/\\sigma^2) \\searrow 0$, we can write $\\DeclareMathOperator{\\diff}{\\,d\\!}$ \\begin{align*} &\\sum_1^n \\exp\\left(-\\frac {j^2}{\\sigma^2}\\right) \\\\ &= \\sum_1^n \\int_{j-1}^j \\exp\\left(-\\frac {j^2}{\\sigma^2}\\right)\\diff x \\\\ &\\leqslant \\sum_1^n \\int_{j-1}^j \\exp\\left(-\\frac {x^2}{\\sigma^2}\\right) \\diff x \\\\ &=\\sigma \\int_0^n \\exp\\left(-\\frac {x^2}{\\sigma^2}\\right) \\diff \\left(\\frac x \\sigma \\right)\\\\ &= \\sigma \\int_0^{n/\\sigma} \\exp(-x^2)\\diff x\\\\ &\\leqslant \\sigma \\int_0^{+\\infty}\\exp(-x^2)\\diff x\\\\ &= \\frac \\sigma 2 \\sqrt \\pi \\end{align*}\n\n\u2022 Nicely done! Sometimes simplest is best. \u2013\u00a0Jacob Manaker Sep 18 at 17:54\n\u2022 brilliant answer, thanks \u2013\u00a0user8003788 Sep 19 at 7:39\n\u2022 @JacobManaker Thanks for compliment! \u2013\u00a0xbh Sep 19 at 7:43\n\nTL;DR: three relatively easy bounds are the numbered equations below.\n\nYou cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $i\\geq1$, so we have $$\\sum_{i=1}^n{\\exp{\\left(-\\frac{i^2}{\\sigma^2}\\right)}}\\leq\\sum_{i=1}^n{\\exp{\\left(-\\frac{i\\cdot1}{\\sigma^2}\\right)}}$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^{-\\sigma^{-2}})^{-1} \\tag{1} \\label{eqn:first}$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor.\n\nIf this isn't a strong enough bound, there are other techniques. If $n<\\sigma$, then we can get very far elementarily. Note that $e^x\\geq x+1$; dividing each side, we get $$e^{-x}\\leq(1+x)^{-1}=\\sum_{k=0}^{\\infty}{(-x)^k}$$ if $|x|<1$. Taking $x=\\left(\\frac{i}{\\sigma}\\right)^2$, we thus obtain \\begin{align*} \\sum_{i=1}^n{e^{-\\frac{i^2}{\\sigma^2}}}&\\leq\\sum_{i=1}^n{\\sum_{k=0}^{\\infty}{\\left(-\\left(\\frac{i}{\\sigma}\\right)^2\\right)^k}} \\\\ &=\\sum_{k=0}^{\\infty}{(-1)^k\\sum_{i=1}^n{\\left(\\frac{i}{\\sigma}\\right)^{2k}}} \\tag{*} \\label{eqn:star} \\end{align*}\n\n(We can interchange sums because one is finite.) Now, for all $k$, the function $\\left(\\frac{\\cdot}{\\sigma}\\right)^{2k}$ is increasing on $[0,\\infty)$; we thus have $$\\int_0^n{\\left(\\frac{i}{\\sigma}\\right)^{2k}\\,di}\\leq\\sum_{i=1}^n{\\left(\\frac{i}{\\sigma}\\right)^{2k}}\\leq\\left(\\frac{n}{\\sigma}\\right)^{2k}+\\int_1^n{\\left(\\frac{i}{\\sigma}\\right)^{2k}\\,di}$$ Evaluating the integrals and simplifying, we have $$0\\leq\\sum_{i=1}^n{\\left(\\frac{i}{\\sigma}\\right)^{2k}}-\\frac{n}{2k+1}\\left(\\frac{n}{\\sigma}\\right)^{2k}\\leq\\left(\\frac{n}{\\sigma}\\right)^{2k}\\left(1-\\frac{1}{(2k+1)n^{2k}}\\right)$$\n\nSubstituting into $\\eqref{eqn:star}$, we get \\begin{align*} \\sum_{i=1}^n{e^{-\\frac{i^2}{\\sigma^2}}}&\\leq\\sum_{k=0}^{\\infty}{\\frac{(-1)^kn}{2k+1}\\left(\\frac{n}{\\sigma}\\right)^{2k}}-\\sum_{j=0}^{\\infty}{\\left(\\frac{n}{\\sigma}\\right)^{4j+2}\\left(1-\\frac{1}{(4j+3)n^{4j+2}}\\right)} \\\\ &\\leq\\sum_{k=0}^{\\infty}{\\frac{(-1)^kn}{2k+1}\\left(\\frac{n}{\\sigma}\\right)^{2k}}-\\sum_{j=0}^{\\infty}{\\left(\\frac{n}{\\sigma}\\right)^{4j+2}} \\\\ &=\\sigma\\tan^{-1}{\\left(\\frac{n}{\\sigma}\\right)}-\\frac{\\left(\\frac{n}{\\sigma}\\right)^2}{1-\\left(\\frac{n}{\\sigma}\\right)^4}\\hspace{4em}(n<\\sigma) \\tag{2} \\end{align*}\n\nFinally, for the general case we can achieve a slight improvement on $\\eqref{eqn:first}$ via the theory of majorization. $\\{x_i\\}_{i=1}^n\\mapsto\\sum_{i=1}^n{\\exp{\\left(-\\frac{x_i}{\\sigma^2}\\right)}}$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=\\left(\\frac{2n-1}{3}\\right)i$. Clearly, for all $m\\leq n$, we have $$\\sum_{i=1}^m{a_i}=\\frac{m(m-1)}{2}\\cdot\\frac{2n-1}{3}\\geq\\frac{m(m-1)(2m-1)}{6}=\\sum_{i=1}^m{b_i}$$ with equality if $m=n$. Thus $\\vec{a}$ majorizes $\\vec{b}$, so \\begin{align*} \\sum_{i=1}^n{\\exp{\\left(-\\frac{i^2}{\\sigma^2}\\right)}}&=\\sum_{i=1}^n{\\exp{\\left(-\\frac{b_i}{\\sigma^2}\\right)}} \\\\ &\\leq\\sum_{i=1}^n{\\exp{\\left(-\\frac{a_i}{\\sigma^2}\\right)}} \\\\ &=\\sum_{i=1}^n{\\exp{\\left(-\\frac{(2n-1)i}{3\\sigma^2}\\right)}} \\\\ &\\leq\\sum_{i=0}^{\\infty}{\\exp{\\left(-\\frac{(2n-1)i}{3\\sigma^2}\\right)}} \\\\ &\\leq\\left(1-\\exp{\\left(\\frac{2n-1}{3\\sigma^2}\\right)}\\right)^{-1} \\tag{3} \\end{align*}\n\n\u2022 Great detailed work..I considered previously the first answer but I thought it would be better if I can get a stronger bound.Thanks a lot \u2013\u00a0user8003788 Sep 19 at 7:42\n\nThere's a rather trivial upper bound that $\\frac{-i^2}{\\sigma^2}$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series.\n\nThe matter is that $e^{-(x/ \\sigma)^2}$, in the range $0 \\le x < \\approx \\sigma$ is very steep.\nSo, unless $\\sigma$ is quite high, you cannot get a good approximation by the integral.\nBut of course everything depends on the parameters into play and on the accuracy required.\n\nHint :\n\nFor general values of $n$ and $\\sigma$ it might be interesting to take advantage of the fact that\nthe Fourier Transform of a Gaussian is a Gaussian itself.\n\nThen you are taking the signal $e^{-\\, (t/ \\sigma)^2}$, windowing it between $0 \\le t \\le n/ \\sigma$, taking $n$ samples of it, and after that you are taking $n$ times the average.\nAll these operations have a simple translation into the frequency domain.\nHowever I do not go further not knowing whether you are acknowledged in this field, and keep this as a hint.\n\nAlso, might be interesting this identity $$\\sum\\limits_{k \\in \\mathbb Z} {\\exp \\left( { - \\pi \\left( {k/c} \\right)^2 } \\right)} = c\\sum\\limits_{k \\in \\mathbb Z} {\\exp \\left( { - \\pi \\left( {k\\,c} \\right)^2 } \\right)}$$ reported at the end of the Properties paragraph in this wikipedia article.", "date": "2018-10-19 06:23:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2921700/how-to-simplify-or-upperbound-this-summation", "openwebmath_score": 0.9999333620071411, "openwebmath_perplexity": 522.4824194582968, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982287698185481, "lm_q2_score": 0.9124361652391386, "lm_q1q2_score": 0.8962748204939407}}
{"url": "https://casmusings.wordpress.com/tag/polynomial/", "text": "# Tag Archives: polynomial\n\n## Infinite Ways to an Infinite Geometric\u00a0Sum\n\nOne of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\\displaystyle \\frac{g}{1-r}$ for first term\u00a0g and common ratio\u00a0r when $\\left| r \\right| < 1$. \u00a0I was glad she was dissatisfied with blind use of\u00a0a formula\u00a0and\u00a0dove into a familiar (to me) derivation. \u00a0In the end, she shook me free from my routine just as she made sure she didn\u2019t fall into her own.\n\nSTANDARD INFINITE GEOMETRIC SUM DERIVATION\n\nMy standard explanation starts with a generic\u00a0infinite geometric series.\n\n$S = g+g\\cdot r+g\\cdot r^2+g\\cdot r^3+...$ \u00a0(1)\n\nWe can reason this series converges iff\u00a0$\\left| r \\right| <1$ (see Footnote 1 for an explanation). \u00a0Assume this is true for (1). \u00a0Notice the terms on the right keep multiplying by\u00a0r.\n\nThe annoying part of summing any infinite series is the ellipsis (\u2026). \u00a0Any finite number of terms always has a finite sum, but that simply written, but vague\u00a0ellipsis is logically difficult. \u00a0In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series. \u00a0You can accomplish this by continuing the pattern on the right: \u00a0multiplying both sides by\u00a0r\n\n$r\\cdot S = r\\cdot \\left( g+g\\cdot r+g\\cdot r^2+... \\right)$\n\n$r\\cdot S = g\\cdot r+g\\cdot r^2+g\\cdot r^3+...$ \u00a0(2)\n\nThis seems to\u00a0make make the right side of (2) identical to the right side of (1) except for the leading\u00a0g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical. \u00a0After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.\n\n$(1)-(2) = S-S\\cdot r = S\\cdot (1-r)=g$\n\n$\\displaystyle S=\\frac{g}{1-r}$\n\nSTUDENT PREFERENCES\n\nI despise giving any formula to any of my classes without\u00a0at least exploring its genesis. \u00a0I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.\n\nIn my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number\u00a0prefer\u00a0the quick \u201cmultiply-by-r-and-subtract\u201d approach used to derive the summation formula. \u00a0For many, apparently, the dynamic manipulation is more meaningful than a static rule. \u00a0It\u2019s very cool to watch student preferences at play.\n\nK\u2019s VARIATION\n\nK understood the proof, and then asked a question I hadn\u2019t thought to ask. \u00a0Why did we have to multiply by\u00a0r? \u00a0Could\u00a0multiplication by $r^2$ also determine the summation\u00a0formula?\n\nI had\u00a0three\u00a0nearly\u00a0simultaneous thoughts followed quickly by a fourth. \u00a0First, why hadn\u2019t I ever thought to ask that? \u00a0Second, geometric series for $\\left| r \\right|<1$ are absolutely convergent, so K\u2019s suggestion should work. \u00a0Third, while the formula would initially look different, absolute convergence guaranteed that whatever the \u201c$r^2$ formula\u201d looked like, it had to be algebraically equivalent to the standard form. \u00a0While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K\u2019s question and the equivalence from Thought #3.\n\nMultiplying (1) by $r^2$ gives\n\n$r^2 \\cdot S = g\\cdot r^2 + g\\cdot r^3 + ...$ (3)\n\nand the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to\u00a0get the inevitable result.\n\n$(1)-(3) = S-S\\cdot r^2 = g+g\\cdot r$\n\n$S\\cdot \\left( 1-r^2 \\right) = g\\cdot (1+r)$\n\n$\\displaystyle S=\\frac{g\\cdot (1+r)}{1-r^2} = \\frac{g\\cdot (1+r)}{(1+r)(1-r)} = \\frac{g}{1-r}$\n\nThat was cool, but this success meant that there were surely many more options.\n\nEXTENDING\n\nWhy stop at multiplying by\u00a0r\u00a0or $r^2$? \u00a0Why not multiply both sides of (1) by a generic $r^N$ for any natural number N? \u00a0 That would give\n\n$r^N \\cdot S = g\\cdot r^N + g\\cdot r^{N+1} + ...$ (4)\n\nwhere the ellipses of (1) and (4) are again identical by the method of Footnote 2. \u00a0Subtracting (4) from (1) gives\n\n$(1)-(4) = S-S\\cdot r^N = g+g\\cdot r + g\\cdot r^2+...+ g\\cdot r^{N-1}$\n\n$S\\cdot \\left( 1-r^N \\right) = g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)$ \u00a0(5)\n\nThere are two ways to proceed from (5). \u00a0You could recognize the right side as a finite geometric sum with first term 1 and ratio\u00a0r. \u00a0Substituting that formula and dividing by $\\left( 1-r^N \\right)$ would give the general result.\n\nAlternatively, I could see students exploring $\\left( 1-r^N \\right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor. \u00a0I got the following TI-Nspire CAS result in about 10-15 seconds,\u00a0clearly\u00a0suggesting that\n\n$1-r^N = (1-r)\\left( 1+r+r^2+...+r^{N-1} \\right)$. \u00a0(6)\n\nMath induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS. \u00a0From there, dividing both sides of (5) by $\\left( 1-r^N \\right)$ gives the generic result.\n\n$\\displaystyle S = \\frac{g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)}{\\left( 1-r^N \\right)}$\n\n$\\displaystyle S = \\frac{g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right) }{(1-r) \\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)} = \\frac{g}{1-r}$\n\nIn the end, K helped me see there wasn\u2019t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways. \u00a0Nice.\n\nFOOTNOTES\n\n1)\u00a0RESTRICTING r: \u00a0Obviously an infinite geometric\u00a0series diverges for $\\left| r \\right| >1$ because that would make $g\\cdot r^n \\rightarrow \\infty$ as $n\\rightarrow \\infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.\n\nFor $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \\ne 0$ , you get\u00a0a\u00a0sum of an\u00a0infinite number\u00a0of some\u00a0nonzero quantity, and that\u00a0is always infinite, no matter how small or large the nonzero quantity.\n\nThe last case, $r=-1$, is more subtle. \u00a0For $g \\ne 0$, this terms of this series alternate between positive and negative\u00a0g,\u00a0making the partial sums of the series\u00a0add to either\u00a0g or 0, depending on whether you have summed an even or an odd number of terms. \u00a0Since the partial sums alternate, the overall sum is divergent. \u00a0Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.\n\n2)\u00a0NOT ALL INFINITIES ARE THE SAME: \u00a0There are two ways to show two groups are the same size. \u00a0The obvious way is to\u00a0count the elements in each group and find out there is\u00a0the same number of elements in each, but this works only if you have a finite group size. \u00a0Alternatively, you could a) match\u00a0every\u00a0element in group 1 with a\u00a0unique\u00a0element from\u00a0group 2, and b) match\u00a0every\u00a0element in group 2\u00a0with a\u00a0unique\u00a0element from\u00a0group 1. \u00a0It is important to do both steps here to show that there are no left-over, unpaired elements in either group.\n\nSo do\u00a0the ellipses in (1) and (2) represent the same sets? \u00a0As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant. \u00a0For the second approach using pairing, we need to compare individual elements. \u00a0For every element in the ellipsis of (1), obviously there is an \u201cpartner\u201d in (2) as the multiplication of (1) by\u00a0r visually shifts all of the terms of the\u00a0series right one position, creating the necessary matches.\n\nStudents often are troubled by the second matching as it appears the ellipsis in (2) contains an \u201cextra term\u201d from the right shift. \u00a0But, for every specific term you identify in (2), its identical twin exists in (1). \u00a0In the weirdness of infinity, that \u201cextra term\u201d appears to have been absorbed without changing the \u201csize\u201d of the infinity.\n\nSince there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.\n\n## Probability, Polynomials, and Sicherman\u00a0Dice\n\nThree\u00a0years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems. \u00a0The ideas I pitched in my\u00a0initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics. \u00a0I\u2019ll quickly re-share them below\u00a0before\u00a0extending the concept\u00a0with ideas I picked up a couple weeks ago from Steve Phelps\u2019\u00a0session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month\u00a0in Cleveland, OH.\n\nBINOMIALS: \u00a0FROM POLYNOMIALS TO\u00a0SAMPLE SPACES\n\nOnce you understand them, binomial probability distributions aren\u2019t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students. \u00a0The standard formula for the probability of determining the chances of\u00a0K successes in\u00a0N attempts of a binomial situation where p is the probability of a single success in a single attempt\u00a0is no less daunting:\n\n$\\displaystyle \\left( \\begin{matrix} N \\\\ K \\end{matrix} \\right) p^K (1-p)^{N-K} = \\frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$\n\nBut that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the\u00a0$\\displaystyle \\left( \\begin{matrix} N \\\\ K \\end{matrix} \\right)$ portion of the probability? \u00a0One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.\n\nThe TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times. \u00a0Each term is an event. \u00a0Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is\u00a0the number of times that combination occurs. \u00a0The overall exponent in the expand command is the number of trials. \u00a0For example, the middle term\u2013 $20\\cdot heads^3 \\cdot tails^3$ \u2013says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.\n\nThe expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what\u2019s actually going on.\n\nFROM POLYNOMIALS TO PROBABILITY\n\nStill using the expand command, if each variable is preceded by its probability, the CAS result combines the\u00a0entire sample space AND the corresponding probability distribution function. \u00a0For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by\n\nThe highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die. \u00a0The probabilities of the other four events in the sample space are also shown. \u00a0Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.\n\nParticularly early on in their explorations of binomial probabilities, students I\u2019ve taught have shown a very clear preference for the polynomial approach, even when\u00a0allowed to choose any\u00a0approach that makes sense to them.\n\nTAKING POLYNOMIALS FROM ONE DIE TO MANY\n\nGiven these earlier thoughts, I was naturally drawn to Steve Phelps \u201cProbability, Polynomials, and CAS\u201d session at the November 2014 OCTM annual meeting in Cleveland, OH. \u00a0Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice. \u00a0My immediate thought was to apply my earlier ideas. \u00a0As noted in my initial\u00a0post, the expansion approach above is not limited to binomial situations. \u00a0My first reflexive CAS command\u00a0in Steve\u2019s session before he share anything was this.\n\nBy writing the outcomes in words, the\u00a0CAS interprets them as\u00a0variables. \u00a0I got the entire sample space, but didn\u2019t learn gain anything beyond a long polynomial. \u00a0The first output\u2013 $five^2$ \u2013with its implied coefficient says there is 1 way to get 2 fives. \u00a0The second term\u2013 $2\\cdot five \\cdot four$ \u2013says there are 2\u00a0ways to get 1\u00a0five and 1\u00a0four. \u00a0Nice that the technology gives me all the terms so quickly, but it doesn\u2019t help me get a distribution function of the sum. \u00a0I got the distributions of the specific outcomes, but the way I defined the variables didn\u2019t permit sum of their actual numerical values. \u00a0Time to listen to the speaker.\n\nHe suggested using a common variable, X, for all faces with the value of each face expressed as an exponent. \u00a0That is, a standard 6-sided die would be represented by\u00a0$X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die. \u00a0Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.\n\nNOTE: \u00a0Exponents are handled in TWO different ways here. \u00a01)\u00a0Within\u00a0a single\u00a0polynomial, an exponent is an event value, and 2)\u00a0Outside\u00a0a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event. \u00a0Coefficients have the same meaning as before.\n\nBecause the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added\u2013exactly what I wanted to happen. \u00a0That means the sum of the faces when you roll two dice is determined by the following.\n\nNotice that the output is a single polynomial. \u00a0Therefore, the exponents are the values of individual cases. \u00a0For a couple examples, there are\u00a03 ways to get a sum of 10 $\\left( 3 \\cdot x^{10} \\right)$, 2 ways to get a sum of 3\u00a0$\\left( 2 \\cdot x^3 \\right)$, etc. \u00a0The most commonly occurring outcome is the term with the largest coefficient. \u00a0For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\\left( 6 \\cdot x^7 \\right)$. \u00a0That certainly simplifies the typical 6\u00d76 tables used to compute the sums\u00a0and probabilities resulting from rolling two dice.\n\nWhile not the point of Steve\u2019s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past. \u00a0For example, what is the most common sum when rolling 5\u00a0dice and what is the probability of that sum? \u00a0On my CAS, I entered this.\n\nIn the middle of the expanded polynomial are two terms with the largest coefficients, $780 \\cdot x^{18}$ and $780 \\cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice. \u00a0As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\\frac{780}{7776} \\approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18. \u00a0This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.\n\nWith thought, this shouldn\u2019t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \\cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common. \u00a0Technology confirms intuition.\n\nROLLING DIFFERENT DICE SIMULTANEOUSLY\n\nWhat is the distribution of sums when rolling a 4-sided and a 6-sided die together? \u00a0No problem. \u00a0Just multiply two different polynomials, one representative of each die.\n\nThe output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.\n\nA BEAUTIFUL EXTENSION\u2013SICHERMAN DICE\n\nMy most unexpected gain from Steve\u2019s talk happened when he asked if we could get the same distribution of sums as \u201cnormal\u201d\u00a06-sided dice, but from two different 6-sided dice. \u00a0The only restriction he gave was that all of the faces of the new dice had to have positive values. \u00a0This\u00a0can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get\n\n$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.\n\nRestating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product? \u00a0That is, does this polynomial factor into other polynomials that could multiply to the same product? \u00a0A CAS factor command gives\n\nAny\u00a0rearrangement of these eight (four distinct)\u00a0sub-polynomials\u00a0would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice? \u00a0As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?\n\nLine 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2. \u00a0Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \\cdot x^0$). \u00a0This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution. \u00a0Cool, but not what I wanted. \u00a0Now what?\n\nFactorization gave\u00a0four distinct sub-polynomials, each with multitude 2. \u00a0One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die. \u00a0That means there are $3^4=81$ different possible dice combinations. \u00a0I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.\n\nWhat follows is the result of thinking about the problem for a while. \u00a0Like most math solutions to interesting\u00a0problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them. \u00a0Problem solving is a messy\u2013but very rewarding\u2013business.\n\nSOLUTION\n\nHere are my insights over time:\n\n1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces). \u00a0Because Steve asked for dice with all positive face values. \u00a0That meant each desired die had to have at least one\u00a0x to prevent non-positive face values.\n\n2) My first attempt didn\u2019t create 6-sided dice. \u00a0The sums of the coefficients of the sub-polynomials determined the number of sides. \u00a0That sum could also be found by substituting $x=1$ into the sub-polynomial. \u00a0I want 6-sided dice, so the final coefficients must add to 6. \u00a0The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6. \u00a0The coefficients of $(x+1)$ add to 2, $\\left( x^2+x+1 \\right)$ add to 3, and\u00a0$\\left( x^2-x+1 \\right)$ add to\u00a01. \u00a0The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one\u00a0$\\left( x^2+x+1 \\right)$ factor.\n\n3) That leaves the two $\\left( x^2-x+1 \\right)$ factors. \u00a0They could split between the two dice or both could be on one die, leaving none on the other. \u00a0We\u2019ve already determined that each die already had to have one each of the\u00a0x, $(x+1)$, and\u00a0$\\left( x^2+x+1 \\right)$ factors. \u00a0To also split the\u00a0$\\left( x^2-x+1 \\right)$ factors would result in the original dice: \u00a0Two normal 6-sided dice. \u00a0If I want different dice, I have to load both of these factors on one die.\n\nThat means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.\n\nOne die would have single faces of 8, 6, 5, 4, 3, and 1. \u00a0The other die would have one 4, two 3s, two 2s, and one 1. \u00a0And this is exactly the result of the famous(?) Sicherman Dice.\n\nIf a 0\u00a0face value\u00a0was allowed, shift one factor of\u00a0x\u00a0from one polynomial to the other. \u00a0This can be done two ways.\n\nThe first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.\n\nBoth of these are nothing more than adding one to all\u00a0faces of one die and subtracting one from from all faces of the other. \u00a0While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by\u00a0x and the other by $\\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces. \u00a0One of these is\n\ncorresponding to a pair of\u00a0Sicherman Dice\u00a0with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.\n\nCONCLUSION:\n\nThere are other very interesting properties of Sicherman Dice, but this is already a very long post. \u00a0In the end, there are tremendous connections between\u00a0probability and polynomials that are accessible to students at the secondary level and beyond. \u00a0And CAS keeps the focus on student\u00a0learning and away from the manipulations that aren\u2019t even the point in these explorations.\n\nEnjoy.\n\n## Number Bases and\u00a0Polynomials\n\nAbout a month ago, I was working with our 5th grade math teacher to develop some extension activities for some students in an unleveled class. \u00a0The class was exploring place value, and I suggested that some might be ready to explore what happens when you allow the number base to be something other than 10. \u00a0A few students had some fun learning to use their basic four algorithms in other number bases, but I made an even deeper connection.\n\nWhen writing something like 512 in expanded form ($5\\cdot 10^2+1\\cdot 10^1+2\\cdot 10^0$), I realized that if the 10 was an\u00a0x, I\u2019d have a polynomial. \u00a0I\u2019d recognized this before, but this time I wondered what would happen if I applied basic math algorithms to polynomials if I wrote them in a condensed numerical\u00a0form,\u00a0not their standard expanded form. \u00a0That is, could I do basic algebra on $5x^2+x+2$\u00a0if I thought of it as $512_x$\u2013a base-x \u201cnumber\u201d? \u00a0(To avoid other confusion later, I read this as \u201cfive one two base-x\u201c.)\n\nFollowing are some examples I played with to convince myself how my new notation would work. \u00a0I\u2019m not convinced that this will ever lead to anything, but following my \u201cwhat ifs\u201d all the way to infinite series was a blast. \u00a0Read on!\n\nIf I wanted to add $(3x+5)$$(2x^2+4x+1)$, I could think of it as\u00a0$35_x+241_x$ and add the numbers \u201cnormally\u201d to get\u00a0$276_x$ or\u00a0$2x^2+7x+6$. \u00a0Notice that each power of\u00a0x identifies a \u201cplace value\u201d for its characteristic coefficient.\n\nIf I wanted to add\u00a0$3x-7$ to itself, I had to adapt my notation a touch. \u00a0The \u201cunits digit\u201d is a negative number, but since the number base,\u00a0x, is unknown (or variable), I ended up saying\u00a0$3x-7=3(-7)_x$. \u00a0The parentheses are used to contain multiple characters into a single place value. \u00a0Then,\u00a0$(3x-7)+(3x-7)$ becomes\u00a0$3(-7)_x+3(-7)_x=6(-14)_x$ or\u00a0$6x-14$. \u00a0Notice the expanding parentheses containing the base-x units digit.\n\nThe last example also showed me that simple multiplication would work. \u00a0Adding\u00a0$3x-7$ to itself is equivalent to multiplying\u00a0$2\\cdot (3x-7)$. \u00a0In base-x, that is\u00a0$2\\cdot 3(-7)_x$. \u00a0That\u2019s easy! \u00a0Arguably, this might be even easier that doubling a number when the number base is known. \u00a0Without interactions between the coefficients of different place values, just double each digit to get\u00a0$6(-14)_x=6x-14$, as before.\n\nWhat about\u00a0$(x^2+7)+(8x-9)$? \u00a0That\u2019s equivalent to\u00a0$107_x+8(-9)_x$. \u00a0While simple, I\u2019ll solve this one by stacking.\n\nand this is\u00a0$x^2+8x-2$. \u00a0As with base-10 numbers, the use of 0 is needed to hold place values exactly as I needed a 0 to hold the $x^1$ place for\u00a0$x^2+7$.\u00a0Again, this could easily be accomplished without the number base conversion, but how much more can we push these boundaries?\n\nLevel 3\u2013Multiplication & Powers:\n\nCompute\u00a0$(8x-3)^2$. \u00a0Stacking again and using a modification of the multiply-and-carry algorithm I learned in grade school, I got\n\nand this is equivalent to $64x^2-48x+9$.\n\nAll other forms of polynomial multiplication work just fine, too.\n\nFrom one perspective, all of this shifting to a variable number base could be seen as completely unnecessary. \u00a0We already have acceptably working algorithms for addition, subtraction, and multiplication. \u00a0But then, I really like how this approach completes the connection between numerical and polynomial arithmetic. \u00a0The rules of math don\u2019t change just because you introduce variables. \u00a0For some, I\u2019m convinced this might make a big difference in understanding.\n\nI also like how easily this extends polynomial by polynomial multiplication far beyond the bland monomial and binomial products that proliferate in virtually all modern textbooks. \u00a0Also banished here is any need at all for banal FOIL techniques.\n\nLevel 4\u2013Division:\n\nWhat about $x^2+x-6$ divided by $x+3$? In base-x, that\u2019s $11(-6)_x \\div 13_x$. Remembering that there is no place value carrying possible, I had to be a little careful when setting up my computation.\u00a0Focusing only on the lead digits, 1 \u201cgoes into\u201d 1 one time. \u00a0Multiplying the partial quotient by the divisor, writing the result below and subtracting gives\n\nThen, 1 \u201cgoes into\u201d -2 negative two times. \u00a0Multiplying and subtracting gives a remainder of 0.\n\nthereby confirming that $x+3$ is a factor of $x^2+x-6$, and the other factor is the quotient, $x-2$.\n\nPerhaps this could be used as an alternative to other polynomial division algorithms. \u00a0It is somewhat similar to the synthetic division technique, without its \u00a0significant limitations: \u00a0It is not limited to linear divisors with lead coefficients of one.\n\nFor $(4x^3-5x^2+7) \\div (2x^2-1)$, think $4(-5)07_x \\div 20(-1)_x$. \u00a0Stacking and dividing gives\n\nSo $\\displaystyle \\frac{4x^3-5x^2+7}{2x^2-1}=2x-2.5+\\frac{2x+4.5}{2x^2-1}$.\n\nCONCLUSION\n\nFrom all I\u2019ve been able to tell, converting polynomials to their base-x number\u00a0equivalents enables you to perform all of the same arithmetic computations. \u00a0For division in particular, it seems this method might even be a bit easier.\n\nIn my next post, I push the exploration of these base-x\u00a0numbers into infinite series.\n\n## Extending graph control\n\nThis article takes my idea from\u00a0yesterday\u2019s post\u00a0about using $g(x)=\\sqrt \\frac{\\left | x \\right |}{x}$ to control the appearance of a graph and extends it in two ways.\n\n\u2022 Part I below uses Desmos to graph $y=(x+2)^3x^2(x-1)$ from the left and right simultaneously\n\u2022 Part II was inspired by my Twitter colleague John Burk who asked if this control could be extended in a different direction.\n\nPart I: Simultaneous\u00a0Control\n\nWhen graphing polynomials like\u00a0$y=(x+2)^3x^2(x-1)$, I encourage my students to use both its local behavior (cubic root at $x=-2$, quadratic root at $x=0$, and linear root at $x=1$) and its end behavior (6th degree polynomial with a positive lead coefficient means $y\\rightarrow +\\infty$ as $x\\rightarrow\\pm\\infty$). To start graphing, I suggest students plot points on the x-intercepts and then sketch arrows to indicate the end behavior. \u00a0In the past, this was something we did on paper, but couldn\u2019t get technology to replicate it live\u2013until this idea.\n\nIn class last week, I used a minor extension of yesterday\u2019s idea to control a graph\u2019s appearance from the left and right simultaneously. \u00a0Yesterday\u2019s post suggested \u00a0multiplying \u00a0by\u00a0$\\sqrt \\frac{\\left | a-x \\right |}{a-x}$ to show the graph of a function from the left for $x. \u00a0Creating a second graph multiplied by\u00a0$\\sqrt \\frac{\\left | x-b \\right |}{x-b}$ gives a graph of your function from the right for $b. \u00a0The following images show the polynomial\u2019s graph developing in a few stages. \u00a0You can access the Desmos file here.\n\nFirst graph the end behavior (pull the a and b sliders in a bit to see just the ends of the graph) and plot points at the x-intercepts.\n\nFrom here, you could graph left-to-right or right-to-left. \u00a0I\u2019ll come in from the right to show the new right side controller. The root at $x=1$ is linear, so decreasing the b slider to just below 1 shows this.\n\nContinuing from the right, the next root is a bounce at $x=0$, as shown by decreasing the b slider below 0. \u00a0Notice that this forces a relative minimum for some $0.\n\nJust because it\u2019s possible, I\u2019ll now show the cubic intercept at $x=2$ by increasing the a slider above 2.\n\nAll that remains is to connect the two sides of the graph, creating one more relative minimum in $-2.\n\nThe same level of presentation control can be had for any function\u2019s graph.\n\nPart II:\u00a0Vertical Control\n\nI hadn\u2019t thought to extend this any further until my colleague asked if a graph could be controlled up and down instead of left and right. \u00a0My guess is that the idea hadn\u2019t occurred to me because I typically think about controlling a function through its domain. \u00a0Even so, a couple minor adjustments accomplished it. \u00a0Click here to see a vertical control of the graph of $y=x^3$ from above and below.\n\nEnjoy.\n\n## Quadratics, Statistics, Symmetry, and\u00a0Tranformations\n\nA problem I assigned my precalculus class this past Thursday ended up with multiple solutions by the time we finished. \u00a0Huzzah for student creativity!\n\nThe question:\n\nFind equations for all polynomial functions, $y=f(x)$, of degree $\\le 2$ for which $f(0)=f(1)=2$ and $f(3)=0$.\n\nAfter they had worked on this (along with several variations on the theme), four very different ways of thinking about this problem emerged. \u00a0All were valid and even led to a lesson I hadn\u2019t planned\u2013proving that, even though they looked different algebraically, all were equivalent. \u00a0I present their approaches (and a few extras) in the order they were offered in our post-solving debriefing.\n\nThe commonality among the approaches was their recognition that 3 non-collinear points uniquely define a vertical parabola, so they didn\u2019t need to worry about polynomials of degree 0 or 1. \u00a0(They haven\u2019t yet heard about rotated curves that led to my earlier post on rotated quadratics.)\n\nSolution 1\u2013Regression:\u00a0 Because only 3 points were given, a quadratic regression would derive a perfectly fitting quadratic equation. \u00a0Using their TI-Nspire CASs, they started by entering the 3 ordered pairs in a Lists&Spreadsheets window. \u00a0Most then went to a Calculator window to compute a quadratic regression. \u00a0Below, I show the same approach using a Data&Statistics window instead so I could see simultaneously the curve fit and the given points.\n\nThe decimals were easy enough to interpret, so even though they were presented in decimal form, these students reported $y=-\\frac{1}{3}x^2+\\frac{1}{3}x+2$.\n\nFor a couple seconds after this was presented, I honestly felt a little cheated. \u00a0I was hoping they would tap the geometric or algebraic properties of quadratics to get their equations. \u00a0But I then I remembered that I clearly hadn\u2019t make that part of my instructions. \u00a0After my initial knee-jerk reaction, I realized this group of students had actually done exactly what I explicitly have been encouraging them to do: think freely and take advantage of every tool they have to find solutions. \u00a0Nothing in the problem statement suggested technology or regressions, so while I had intended a more geometric approach, I realized I actually owed these students some kudos for a very creative, insightful, and technology-based solution. \u00a0This and Solution 2 were the most frequently chosen approaches.\n\nSolution 2\u2013Systems:\u00a0 Equations of quadratic functions are typically presented in standard, factored, or vertex form. \u00a0Since neither two zeros nor the vertex were explicitly given, the largest portion of the students used the standard form, $y=a\\cdot x^2+b\\cdot x+c$ to create a 3\u00d73 system of equations. \u00a0Some solved this by hand, but most invoked a CAS solution. \u00a0Notice the elegance of the solve command they used, working from the generic polynomial equation that kept them from having to write all three equations, keeping their focus on the form of the equation they sought.\n\nThis created the same result as Solution 1,\u00a0$y=-\\frac{1}{3}x^2+\\frac{1}{3}x+2$.\n\nCAS Aside: No students offered these next two solutions, but I believe when using a CAS, it is important for users to remember that the machine typically does not care what output form you want. \u00a0The standard form is the only \u201calgebraically simple\u201d approach when setting up a solution by hand, but the availability of technology makes solving for any form equally accessible.\n\nThe next screen shows that the vertex and factored forms are just as easily derived as the standard form my students found in Solution 2.\n\nI was surprised when the last line\u2019s output wasn\u2019t in vertex form, $y=-\\frac{1}{3}\\cdot \\left ( x-\\frac{1}{2} \\right )^2+\\frac{25}{12}$, but the coefficients in its expanded form clearly show the equivalence between this form and the standard forms derived in Solutions 1 and 2\u2013a valuable connection.\n\nSolution 3\u2013Symmetry: \u00a0Two students said they noticed that $f(0)=f(1)=2$ guaranteed the vertex of the parabola occurred at $x=\\frac{1}{2}$. \u00a0Because $f(3)=0$ defined one real root of the unknown quadratic, the parabola\u2019s symmetry guaranteed another at $x=-2$, giving potential equation $y=a\\cdot (x-3)(x+2)$. \u00a0They substituted the given (0,2) to solve for\u00a0a, giving final equation\u00a0$y=-\\frac{1}{3}\\cdot (x-3)(x+2)$ as confirmed by the CAS approach above.\n\nSolution 4\u2013Transformations:\u00a0 One of the big lessons I repeat in every class I teach is this:\n\nIf you don\u2019t like how a question is posed. \u00a0Change it!\n\nNotice that two of the given points have the same\u00a0y-coordinate. \u00a0If that\u00a0y-coordinate had been 0 (instead of its given value, 2), a factored form would be simple. \u00a0Well, why not\u00a0force them\u00a0to be x-intercepts by translating all of the given points down 2 units?\n\nThe transformed data show\u00a0x-intercepts at 0 and 1 with another ordered pair at $(3,-2)$. \u00a0From here, the factored form is easy: \u00a0$y=a\\cdot (x-0)(x-1)$. \u00a0Substituting $(3,-2)$\u00a0gives $a=-\\frac{1}{3}$ and the final equation is $y=-\\frac{1}{3}\\cdot (x-0)(x-1)$ .\n\nOf course, this is an equation for the\u00a0transformed points. \u00a0Sliding the result back up two units, $y=-\\frac{1}{3}\\cdot (x-0)(x-1)+2$,\u00a0gives an equation for the given points. \u00a0Aside from its lead coefficient, this last equation looked very different from the other forms, but some quick expansion proved its equivalence.\n\nConclusion: \u00a0It would have been nice if someone had used the symmetry noted in Solution 3 to attempt a vertex-form answer via systems. \u00a0Given the vertex at $x=\\frac{1}{2}$ with an unknown y-coordinate, a potential equation is\u00a0$y=a\\cdot \\left ( x-\\frac{1}{2} \\right )^2+k$. \u00a0Substituting $(3,0)$ and either $(0,2)\\text{ or }(1,2)$ creates a 2\u00d72 system of linear equations, $\\left\\{\\begin{matrix} 0=a\\cdot \\left ( 3-\\frac{1}{2} \\right )^2+k \\\\ 2=a\\cdot \\left ( 0-\\frac{1}{2} \\right )^2+k \\end{matrix}\\right.$. \u00a0From there, a by-hand or CAS solution would have been equally acceptable to me.\n\nThat the few alternative approaches I offered above weren\u2019t used didn\u2019t matter in the end. \u00a0My students were creative, followed their own instincts to find solutions that aligned with their thinking, and clearly appreciated the alternative ways their classmates used to find answers. \u00a0Creativity and individual expression reigned, while everyone broadened their understanding that there\u2019s not just one way to do math.\n\nIt was a good day.\n\n## Cubics and CAS\n\nHere\u2019s a question I posed to one of my precalculus classes for homework at the end of last week along with three solutions we developed.\n\nSuppose the graph of a cubic function has an inflection point at (1,3) and passes through (0,-4).\n\n1. Name one other point that MUST be on the curve, and\n2. give TWO different cubic equations that would pass through the three points.\n\nSOLUTION ALERT! \u00a0Don\u2019t read any further if you want to solve this problem for yourself.\n\nThe first question relies on the fact that every cubic function has 180 degree rotational symmetry about its inflection point. \u00a0This is equivalent to saying that the graph of a cubic function is its own image when the function\u2019s graph is reflected through its inflection point.\n\nThat means the third point is the image of (0,-4) when point-reflected through the inflection point (1,3), which is the point (2,10) as shown graphically below.\n\nFrom here, my students came up with 2 different solutions to the second question and upon prodding, we created a third.\n\nSOLUTION 1:\u00a0 Virtually every student assumed $y=a\\cdot x^3$ was the parent function of a cubic with unknown leading coefficient whose \u201ccenter\u201d (inflection point) had been slid to (1,3). \u00a0Plugging in the given (0,-4) to\u00a0$(y-3)=a\\cdot (x-1)^3$ gives $a=7$. \u00a0Here\u2019s their graph.\n\nSOLUTION 2:\u00a0 Many students had difficulty coming up with another equation. \u00a0A few could sketch in other cubic graphs (curiously, all had positive lead coefficients) that contained the 3 points above, but didn\u2019t know how to find equations. \u00a0That\u2019s when Sara pointed out that if the generic expanded form of a cubic was $a\\cdot x^3+b\\cdot x^2 +c\\cdot x+d$ , then any 4 ordered pairs with unique x-coordinates should define a unique cubic. \u00a0That is, if we picked any 4th point with x not 0, 1, or 2, then we should get an equation. \u00a0That this would create a 4\u00d74 system of equations didn\u2019t bother her at all. \u00a0She knew in theory how to solve such a thing, but she was thinking on a much higher plane. \u00a0Her CAS technology expeditiously did the grunt work, allowing her brain to keep moving.\n\nA doubtful classmate called out, \u201cOK. \u00a0Make it go through (100,100).\u201d \u00a0Following is a CAS screen roughly duplicating Sara\u2019s approach and a graph confirming the fit. \u00a0The equation was onerous, but with a quick copy-paste, Sara had moved from \u00a0idea to computation to ugly equation and graph in just a couple minutes. \u00a0The doubter was convinced and the \u201cwow\u201ds from throughout the room conveyed the respect for the power of a properly wielded CAS.\n\nIn particular, notice how the TI-Nspire CAS syntax in lines 1 and 3 keep the user\u2019s focus on the type of equation being solved and eliminates the need to actually enter 4 separate equations. \u00a0It doesn\u2019t always work, but it\u2019s a particularly lovely piece of scaffolding when it does.\n\nSOLUTION 3:\u00a0 One of my goals in Algebra II and Precalculus courses is to teach my students that they don\u2019t need to always accept problems as stated. \u00a0Sometimes they can change initial conditions to create a much cleaner work environment so long as they transform their \u201cclean\u201d solution back to the state of the initial conditions.\n\nIn this case, I asked what would happen if they translated the inflection point using $T_{-1,-3}$ to the origin, making the other given point (-1,-7). \u00a0Several immediately called the 3rd point to be (1,7) which \u201cuntranslating\u201d \u2014\u00a0$T_{1,3}(1,7)=(2,10)$ \u2014 confirmed to be the earlier finding.\n\nFor cases where the cubic had another real root at $x=r$, then symmetry immediately made $x=-r$ another root, and a factored form of the equation becomes $y=a\\cdot (x)(x-r)(x+r)$ for some value of a. \u00a0Plugging in (-1,-7) gives a in terms of r.\n\nThe last line slid the initially translated equation using $T_{1,3}$ to re-position\u00a0the previous line according to the initial conditions. \u00a0While unasked for, notice how the CAS performed some polynomial division on the right-side expression.\n\nI created a GeoGebra document with a slider for the root using the equation from the last line of the CAS image above. \u00a0The image below shows one possible position of the retranslated root. \u00a0If you want to play with a live version of this, you will need a free copy of GeoGebra to run it, but the file is here.\n\nWhat\u2019s nice here is how the problem became one of simple factors once the inflection point was translated to the origin. \u00a0Notice also that the CAS version of the equation concludes with $+7x-4$, the line containing the original three points. \u00a0This is nice for two reasons. \u00a0The numerator of the rational term is $-7x(x-2)(x-1)$ which zeros out the fraction at x=0, 1, or 2, putting the cubic exactly on the line $y=7x-4$ at those points.\n\nThe only r-values are in the denominator, so as $r\\rightarrow\\infty$, the rational term also becomes zero. \u00a0Graphically, you can see this happen as the cubic \u201cunrolls\u201d onto $y=7x-4$ as you drag $|x|\\rightarrow\\infty$. \u00a0Essentially, this shows both graphically and algebraically that $y=7x-4$ is the limiting degenerate curve the cubic function approaches as two of its transformed real roots grow without bound.\n\n## Recognizing Patterns\n\nI\u2019ve often told my students that the best problem solvers are those who recognize patterns from past problems in new situations.\u00a0 So, the best way to become a better problem solver is to solve lots of problems, study the ways others have solved the problems you\u2019ve already cracked (or at least attempted), and to keep pushing your boundaries because you never know what parts of what you learn may end up providing unexpected future insights.\n\nI lay no claims to being a great problem solver, but I absolutely benefited from problem solving exposure when I encountered @jamestanton\u2018s latest \u201cPlaying with Numbers\u201d puzzler from his May 2012 Cool Math Newsletter.\u00a0 (Click here to access all of Jim\u2019s newsletters).\u00a0 (BTW, Jim\u2019s Web page is chock full of amazing videos and insights both on the problem-solving front and for those interested in curriculum discussions.)\u00a0 Here\u2019s what Jim posed:\n\nWrite the numbers 1 though 10 on the board. Pick any two numbers, erase them, and replace them with the single number given by their sum plus their product. (So, if you choose to erase the numbers a and b, replace them with a + b + ab .) You now have nine numbers on the board.\n\nDo this again: Pick any two numbers, erase them, and replace them with their sum plus product. You now have eight numbers on the board.\n\nDo this seven more times until you have a single number on the board.\n\nWhy do all who play this game end up with the same single number at the end?\u00a0 What is that final number?\n\nSOLUTION ALERT:\u00a0 Don\u2019t read any further if you want to solve this yourself.\n\nI started small and general.\u00a0 If the first two numbers (a and b) produce $a+b+ab$, then adding c to the original list gives $[(a+b+ab)+c]+ [(a+b+ab)\\cdot c]$ which can be rewritten as $a+b+c+ab+ac+bc+abc$.\u00a0 That\u2019s when the intuition struck.\u00a0 Notice that the rewritten form is the sum of every individual number in the list, AND every possible pair of those numbers, AND concludes with the product of the three numbers. I\u2019ve seen that pattern before!\n\nGiven an nth degree polynomial with roots $r_1, r_2, \\ldots r_n$, it\u2019s factored form is $(x-r_1)\\cdot (x-r_2)\\cdot\\ldots\\cdot (x-r_n)$ which can be expanded and rewritten as\n\n$x^n-(r_1+r_2+\\ldots+r_n)\\cdot x^{n-1}+$\n$+(\\text{every pair-wise product of } r_1 \\ldots r_n)\\cdot x^{n-2}-$\n$-(\\text{every three-way product of } r_1 \\ldots r_n)\\cdot x^{n-3}+\\ldots$\n$\\ldots \\pm (r_1\\cdot r_2\\cdot \\ldots\\cdot r_n)$\n\nWhere the sign of the final term is positive for even n and negative for odd n.\u00a0 I saw this in many algebra textbooks when I started teaching over 20 years ago, but haven\u2019t encountered it lately.\u00a0 Then again, I haven\u2019t been looking for it.\n\nHere\u2019s the point \u2026 other than the alternating signs, the coefficients of the expanded polynomial are exactly identical to the sums I was getting from Jim\u2019s problem.\u00a0 That\u2019s when I rewrote my original problem on my CAS to $(x+a)\\cdot (x+b)\\cdot (x+c)$ and expanded it (see line 1 below).\u00a0 The coefficients of $x^2$ are the individual numbers, the coefficients of $x$ are all the pair-wise combinations, and the constant term drifting off the end of the line is $a\\cdot b\\cdot c$.\u00a0 And I eliminated the $\\pm$ sign challenge by individually adding the numbers instead of subtracting them as I had in the polynomial root example that inspired my insight.\u00a0 Let $x=1$ to clean up and make the pattern more obvious.\n\nSo, creating a polynomial with the given numbers and substituting $x=1$ will create a number one more than the sum I wanted (note the extra +1 resulting from the coefficient of $x^n$).\u00a0 Using pi notation, substituting $x=1$, and subtracting 1 gives the answer.\u00a0 In case you didn\u2019t know, pi notation works exactly the same as sigma notation, but you multiply the terms instead of adding them.\n\nThe solution\u201339,916,799\u2013is surprisingly large given the initial problem statement, and while my CAS use confirmed my intuition and effortlessly crunched the numbers, its tendency to multiply numbers whenever encountered has actually hidden something pretty. From the top line of the last image, substituting $x=1$ would have created the product $2\\cdot 3\\cdot 4\\cdot\\ldots \\cdot 11$ which the penultimate line computed to be 39916800.\u00a0 But before the product, that number was $11!$, making $11!-1$ a far more revealing version of the solution.\n\nMORAL:\u00a0 Even after you have an answer, take some time to review what has happened to give yourself a chance to learn even more.\n\nEXTENSION 1:\u00a0 Instead of 1 to 10 as the initial numbers, what if the list went from 1 to any positive integer n?\u00a0 Prove that the final number on the board is $(n+1)!-1$.\n\nEXTENSION 2:\u00a0 While a positive integer sequence starting at 1 (or 0) produces a nice factorial in the answer, this approach can be used with any number list.\u00a0 For example, follow Jim\u2019s rules with 37, 5, -2, and 7.9.\u00a0 Use the polynomial approach below for a quick solution of -2030.2.\u00a0 Confirm using the original rules if you need.\n\nEXTENSION 3:\u00a0 By now it should be obvious that any list of numbers can be used in this problem.\u00a0 Prove that every list of numbers which includes -1 has the same solution.\n\nEXTENSION 4:\u00a0 Before explaining some lovely extensions of problems like this to generalized commutativity and associativity, Jim\u2019s May 12 Cool Math Newsletter asks what would happen if instead of $a+b+ab$, the rule for combining a and b was $\\displaystyle\\frac{a\\cdot b}{a+b}$.\u00a0 You can show that with three terms, this would become $\\displaystyle\\frac{abc}{ab+ac+bc}$, and four terms would give $\\displaystyle\\frac{abcd}{abc+abd+acd+bcd}$. In other words, this rule would morph n original numbers into a fraction whose numerator is the product of the numbers and whose denominator is the sum of all possible products of any (n-1)-sized groups of those numbers.\u00a0 For the original integers 1 to 10, I know the numerator and denominator terms are the last two coefficients in a polynomial expansion.\n\nThe final fraction simplifies, but I think $\\displaystyle\\frac{3628800}{10628640}$ is slightly more informative.\n\nHappy thoughts, problems, solutions, and connections to you all.", "date": "2020-03-31 23:17:44", "meta": {"domain": "wordpress.com", "url": "https://casmusings.wordpress.com/tag/polynomial/", "openwebmath_score": 0.803494930267334, "openwebmath_perplexity": 980.6773426566747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319878505035, "lm_q2_score": 0.9046505376715775, "lm_q1q2_score": 0.8962662254973889}}
{"url": "https://gmatclub.com/forum/what-is-the-greatest-value-of-y-such-that-4-y-is-a-factor-of-230289.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 22 Feb 2020, 18:16\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nWhat is the greatest value of y such that 4^y is a factor of 9! ?\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 61396\nWhat is the greatest value of y such that 4^y is a factor of 9! ?\u00a0 [#permalink]\n\nShow Tags\n\n08 Dec 2016, 11:57\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n77% (01:07) correct 23% (01:20) wrong based on 60 sessions\n\nHideShow timer Statistics\n\nWhat is the greatest value of y such that 4^y is a factor of 9! ?\n\nA. 5\nB. 4\nC. 3\nD. 1\nE. 0\n\n_________________\nDirector\nJoined: 05 Mar 2015\nPosts: 960\nWhat is the greatest value of y such that 4^y is a factor of 9! ?\u00a0 [#permalink]\n\nShow Tags\n\n08 Dec 2016, 19:02\n1\nBunuel wrote:\nWhat is the greatest value of y such that 4^y is a factor of 9! ?\n\nA. 5\nB. 4\nC. 3\nD. 1\nE. 0\n\n4^y=2^(2y)\n\nno. of 2's in 9!\n9/2=4\n9/2^2=2\n9/2^3=1\ntotal= 4+2+1=7\n\nso as 2y=7 we get y=3\n\nAns C\nManager\nJoined: 27 Aug 2015\nPosts: 86\nRe: What is the greatest value of y such that 4^y is a factor of 9! ?\u00a0 [#permalink]\n\nShow Tags\n\n09 Dec 2016, 02:25\n1\nThe formula for such problems is like\n9 /4= 2\n9/4^2=0\nTotal = 2\nHowever answer should be 3 if we actually count it. Where am I going wrong?\n\nPosted from my mobile device\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 4841\nLocation: India\nGPA: 3.5\nRe: What is the greatest value of y such that 4^y is a factor of 9! ?\u00a0 [#permalink]\n\nShow Tags\n\n09 Dec 2016, 10:18\n2\nBunuel wrote:\nWhat is the greatest value of y such that 4^y is a factor of 9! ?\n\nA. 5\nB. 4\nC. 3\nD. 1\nE. 0\n\n$$9! = 9*8*7*6*5*4*3*2*1$$\n\nOr, $$9! = 3^2*2^3*7*2*3*5*2^2*3*2*1$$\n\nOr, $$9! = 2^7*3^4*5*7$$\n\nNow, $$2^7 = 4^3*2$$\n\nThus, we have the greatest value of y = 3 , hence answer will be (C)\n\nrakaisraka hope its clear with you ...\n\nFurther I suggest you go through the concept once again to clear your doubts here math-number-theory-88376.html#p666609\n_________________\nThanks and Regards\n\nAbhishek....\n\nPLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS\n\nHow to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )\nDirector\nJoined: 05 Mar 2015\nPosts: 960\nRe: What is the greatest value of y such that 4^y is a factor of 9! ?\u00a0 [#permalink]\n\nShow Tags\n\n09 Dec 2016, 10:35\n1\nrakaisraka wrote:\nThe formula for such problems is like\n9 /4= 2\n9/4^2=0\nTotal = 2\nHowever answer should be 3 if we actually count it. Where am I going wrong?\n\nPosted from my mobile device\n\nrakaisraka\n\nwhen u r finding 4^y means u have to count every 2's ..\nsuppose if it was 10! then it must have 1*2*...*6...*10\nthen it has 6=2*3 && 10=2*5\nwhere one no. 2 from 6 and one no. 2 from 10 also counted as a 4 in 10!\n\nlet me make more clear if u have to find 6^y in X!\nas 6=2*3\nthen u have to count every 2 and every 3 in X!\nand the minimum pair of 2&3 will make the answer\n\nhope it is clear\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2801\nRe: What is the greatest value of y such that 4^y is a factor of 9! ?\u00a0 [#permalink]\n\nShow Tags\n\n12 Dec 2016, 17:16\n1\nBunuel wrote:\nWhat is the greatest value of y such that 4^y is a factor of 9! ?\n\nA. 5\nB. 4\nC. 3\nD. 1\nE. 0\n\nSince 4 = 2^2, we are actually trying to determine the largest value y such that 2^(2y) is a factor of 9!.\n\nLet\u2019s first determine the number of factors of 2 within 9!. To do that, we can use the following shortcut in which we divide 9 by 2, and then divide the quotient of 9/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.\n\n9/2 = 4 (we can ignore the remainder)\n\n4/2 = 2\n\n2/2 = 1\n\nSince 1/2 does not produce a nonzero quotient, we can stop.\n\nThe final step is to add up our quotients; that sum represents the number of factors of 2 within 9!.\n\nThus, there are 4 + 2 + 1 = 7 factors of 2 within 9!\n\nHowever, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 4. We see that 7 factors of 2 will produce 3 factors of 4.\n\n_________________\n\nJeffrey Miller\n\nJeff@TargetTestPrep.com\n181 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 14124\nRe: What is the greatest value of y such that 4^y is a factor of 9! ?\u00a0 [#permalink]\n\nShow Tags\n\n17 Jan 2020, 02:54\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: What is the greatest value of y such that 4^y is a factor of 9! ? \u00a0 [#permalink] 17 Jan 2020, 02:54\nDisplay posts from previous: Sort by", "date": "2020-02-23 02:16:06", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-is-the-greatest-value-of-y-such-that-4-y-is-a-factor-of-230289.html", "openwebmath_score": 0.7827305197715759, "openwebmath_perplexity": 2563.270617579204, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8962513786759491, "lm_q1q2_score": 0.8962513786759491}}
{"url": "https://brilliant.org/discussions/thread/on-recurring-decimals/", "text": "# On Recurring decimals\n\n$\\large{0\\red{.}\\overline{x_1x_2x_3...x_n}=\\dfrac{x_1x_2x_3...x_n}{10^n-1}}$\n\nProof of the above statement\n\nLet $l=0\\red{.}\\overline{x_1x_2x_3...x_n}$ $10^nl=x_1x_2x_3...x_n\\red{.}\\overline{x_1x_2x_3...x_n}$ $\\Rightarrow 10^nl-l={x_1x_2x_3...x_n}$ $(10^n-1)l=x_1x_2x_3...x_n$ ${l=\\dfrac{x_1x_2x_3...x_n}{10^n-1}}$ $\\boxed{0\\red{.}\\overline{x_1x_2x_3...x_n}=\\dfrac{x_1x_2x_3...x_n}{10^n-1}}$\n\n$\\large{a_1a_2a_3...a_p\\red{.}b_1b_2b_3...b_q\\overline{x_1x_2x_3...x_n}=\\dfrac{1}{10^q}(10^q\\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\\dfrac{x_1x_2x_3...x_n}{10^n-1})}$\n\nProof of the above statement\n\nFor any number $a_1a_2a_3...a_p\\red{.}b_1b_2b_3...b_q\\overline{x_1x_2x_3...x_n}$ $a_1a_2a_3...a_p\\red{.}b_1b_2b_3...b_q\\overline{x_1x_2x_3...x_n}=a_1a_2a_3...a_p+0\\red{.}b_1b_2b_3...b_q\\overline{x_1x_2x_3...x_n}$ $=\\dfrac{1}{10^q}(10^q\\times a_1a_2a_3...a_p+b_1b_2b_3...b_q\\red{.}\\overline{x_1x_2x_3...x_n})$ $=\\dfrac{1}{10^q}(10^q\\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+0\\red{.}\\overline{x_1x_2x_3...x_n})$ $=\\boxed{\\dfrac{1}{10^q}(10^q\\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\\dfrac{x_1x_2x_3...x_n}{10^n-1})}$\n\nNote :\n\n\u2022 $x_1x_2$ act as number with digits $x_1,x_2$ for example if $x_1=5$ and $x_2=8\\Rightarrow x_1x_2=58$ dont confuse ($x_1x_2\\cancel{=}x_1\\times x_2$), same for $x_1x_2x_3$ and $x_1x_2x_3...x_{n-1}x_n$\n\n\u2022 $0\\red{.}\\overline{a}=0\\red{.}aaaaa...$\n\nNote by Zakir Husain\n6\u00a0months, 2\u00a0weeks ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\u2022 Stay on topic \u2014 we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nWell.. I am impressed.\n\n- 6\u00a0months, 2\u00a0weeks ago\n\n\u2728 brilliant +1\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nBy the way, I have an interesting #Geometry problem!\nGiven points $A,B,C,D$, find the square $\\square PQRS$ with A on PQ, B on QR, C on RS, D on SP.\nI figured out the first part, where we can construct circles with diameters AB, BC, CD, DA respectively, so if a point W is on arc AB, $\\angle AWB=90^\\circ.$\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nI tried the problem and got an algorithm to construct a rectangle $PQRS$ with points $A,B,C$ and $D$ on sides $PQ,QR,RS,SP$ respectively. Also there will be infinitely many such rectangles for given points $A,B,C,D$\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nWhat about a square?\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nI will try it also! and will inform you as I get any results.\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nLet\u2019s start a discussion! That might help :)\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nsquare is also a rectangle..\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nBut a rectangle isn\u2019t a square, so I hope to find an algorithm to construct a square (I know it is possible but I don\u2019t know a specific way to do it except for brute-force :P) :)\n\n- 6\u00a0months, 2\u00a0weeks ago", "date": "2021-01-16 00:31:31", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/on-recurring-decimals/", "openwebmath_score": 0.9654994010925293, "openwebmath_perplexity": 2703.0319096444596, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9664104885453715, "lm_q2_score": 0.9273632871148287, "lm_q1q2_score": 0.8962136073596833}}
{"url": "https://zong-music.com/ha2ten/greater-than-or-equal-to-sign-9b504e", "text": "# greater than or equal to sign\n\nFor example, x \u2265 -3 is the solution of a certain expression in variable x. Select Symbol and then More Symbols. For example, the symbol is used below to express the less-than-or-equal relationship between two variables: \u2265. \"Greater than or equal to\", as the suggests, means something is either greater than or equal to another thing. is less than > > is greater than \u226e \\nless: is not less than \u226f \\ngtr: is not greater than \u2264 \\leq: is less than or equal to \u2265 \\geq: is greater than or equal to \u2a7d \\leqslant: is less than or equal to \u2a7e 923 Views. Use the appropriate math symbol to indicate \"greater than\", \"less than\" or \"equal to\" for each of the following: a. Greater than or equal application to numbers: Syntax of Greater than or Equal is A>=B, where A and B are numeric or Text values. With Microsoft Word, inserting a greater than or equal to sign into your Word document can be as simple as pressing the Equal keyboard key or the Greater Than keyboard key, but there is also a way to insert these characters as actual equations. For example, 4 or 3 \u2265 1 shows us a greater sign over half an equal sign, meaning that 4 or 3 are greater than or equal to 1. In such cases, we can use the greater than or equal to symbol, i.e. In Greater than or equal operator A value compares with B value it will return true in two cases one is when A greater than B and another is when A equal to B. Rate this symbol: (3.80 / 5 votes) Specifies that one value is greater than, or equal to, another value. This symbol is nothing but the \"greater than\" symbol with a sleeping line under it. Less Than or Equal To (<=) Operator. \u201cGreater than or equal to\u201d and \u201cless than or equal to\u201d are just the applicable symbol with half an equal sign under it. Greater Than or Equal To: Math Definition. 2 \u2265 2. But, when we say 'at least', we mean 'greater than or equal to'. The less than or equal to symbol is used to express the relationship between two quantities or as a boolean logical operator. \"Greater than or equal to\" is represented by the symbol \" \u2265 \u2265 \". Solution for 1. The greater-than sign is a mathematical symbol that denotes an inequality between two values. In an acidic solution [H]\u2026 Greater than or Equal in Excel \u2013 Example #5. Here a could be greater \u2026 Examples: 5 \u2265 4. The sql Greater Than or Equal To operator is used to check whether the left-hand operator is higher than or equal to the right-hand operator or not. Category: Mathematical Symbols. When we say 'as many as' or 'no more than', we mean 'less than or equal to' which means that a could be less than b or equal to b. Select the Greater-than Or Equal To tab in the Symbol window. use \">=\" for greater than or equal use \"<=\" for less than or equal In general, Sheets uses the same \"language\" as Excel, so you can look up Excel tips for Sheets. Copy the Greater-than Or Equal To in the above table (it can be automatically copied with a mouse click) and paste it in word, Or. Finding specific symbols in countless symbols is obviously a waste of time, and some characters like emoji usually can't be found. Graphical characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has no crossing lines. Select the Insert tab. If left-hand operator higher than or equal to right-hand operator then condition will be true and it will return matched records. Sometimes we may observe scenarios where the result obtained by solving an expression for a variable, which are greater than or equal to each other. 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Mean 'greater than or equal to right-hand operator then condition will be true and it will return records! That denotes an inequality between two values, Has no crossing greater than or equal to sign we say 'at least ', can...", "date": "2021-06-22 19:08:26", "meta": {"domain": "zong-music.com", "url": "https://zong-music.com/ha2ten/greater-than-or-equal-to-sign-9b504e", "openwebmath_score": 0.5518640279769897, "openwebmath_perplexity": 948.9165942980742, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075722839015, "lm_q2_score": 0.9314625083949473, "lm_q1q2_score": 0.896167132625336}}
{"url": "http://math.stackexchange.com/questions/219329/for-t-in-0-1-is-fracxetxex-1-integrable-over-x-in-0-in/219347", "text": "# For $t\\in [ 0, 1 )$ is $\\frac{xe^{tx}}{e^{x}-1}$ integrable over $x\\in (0 , \\infty )$?\n\nFor $t\\in [ 0, 1 )$ is $$\\frac{xe^{tx}}{e^{x}-1}$$ integrable over $x\\in (0 , \\infty )$? I.e., $$\\int_{0}^{\\infty} \\frac{xe^{tx}}{e^{x}-1} dx < \\infty?$$ How do I show this?\n\n-\n\nAs $x\\to0$, $x/(e^x-1)$ approaches a finite limit. As $x\\to\\infty$, do a limit-comparison of the integrand to $xe^{tx}/e^x$.\n\n-\n\nAs $\\frac x{e^x-1}$ as a limit when $x\\to 0$ (namely $1$), the only problem is when $x\\to\\infty$. We have $e^x-1\\sim e^x$ at $+\\infty$, so $\\dfrac{xe^{tx}}{e^x-1}\\sim xe^{(t-1)x}$. Using Taylor's series, $$e^{(t-1)x}\\leq \\frac 1{1+(1-t)x+x^2(1-t)^2/2+x^3(1-t)^3/6},$$ the integral is convergent for $t\\in[0,1)$.\n\n-\nDoesn't the first limit go to $1$ instead of $e^{\u20131}$? \u2013\u00a0 Pedro Tamaroff Oct 23 '12 at 13:05\n@PeterTamaroff Right. Fixed now. \u2013\u00a0 Davide Giraudo Oct 23 '12 at 13:14\n\nWhat matters in the improper integral of a nice function (e.g. elementary function) is the existence of singularities. In a broad sense, there are two kinds of singularities that counts.\n\n1. A point where the integrand does not behave well. For example, the function can explode to infinite or oscillate infinitely.\n\n2. A point at infinity. That is, $\\pm \\infty$.\n\nAway from singularities, the behavior of the function is quite under control, allowing us to concentrate our attention on those singularities.\n\nThere is a basic method to establish the convergence (or divergence) of the integral near each singularity point. In many cases, except for the oscillatory case, you can find a dominating function that determines the order of magnitude of the function near the point. If the dominating function is easy to integrate, then you can make a comparison with this dominating function to conclude the convergence behavior.\n\nFor example, let us consider\n\n$$\\int_{0}^{\\frac{\\pi}{2}} \\tan^2 x \\, dx \\quad \\text{and} \\quad \\int_{0}^{\\infty} \\frac{x^2 e^{-x}}{1+x^2} \\, dx.$$\n\nWe can easily check that $\\tan^2 x$ is bounded below by $(x-\\frac{\\pi}{2})^{-2}$ near the singularity $x = \\frac{\\pi}{2}$ and $x^2 e^{-x} / (1 + x^2)$ is bounded above by $e^{-x}$ near the singularity $x = \\infty$. Then\n\n$$\\int_{\\frac{\\pi}{2}-\\delta}^{\\frac{\\pi}{2}} \\tan^2 x \\, dx \\geq \\int_{\\frac{\\pi}{2}-\\delta}^{\\frac{\\pi}{2}} \\left(x - \\frac{\\pi}{2}\\right)^{2} \\, dx = \\infty$$\n\nfor sufficiently small $\\delta > 0$ and\n\n$$\\int_{R}^{\\infty} \\frac{x^2 e^{-x}}{1+x^2}\\,dx \\leq \\int_{R}^{\\infty} e^{-x} \\, dx < \\infty$$\n\nfor sufficiently large $R > 0$. Thus we find that the former diverges to $\\infty$ and the latter converges.\n\nIn our example, there are two seemingly singular points, namely $x = 0$ and $x = \\infty$. At $x = 0$, we find that\n\n$$\\lim_{x \\to 0} \\frac{x e^{tx}}{e^x - 1} = 1.$$\n\nThis means that this singularity is removable, in the sense that the function can be extended in a continuous manner to this point. Thus we need not count this point and we can move our attention to the point at infinity.\n\nTo establish the convergence (or possibly divergence) of the integral near $x = \\infty$, we write\n\n$$\\frac{x e^{tx}}{e^x - 1} = \\frac{x}{1 - e^{-x}} e^{-(1-t)x}.$$\n\nIt is clear that for sufficiently large $x$, the term $\\frac{x}{1 - e^{-x}}$ is bounded above by some constant $C > 0$. Thus the dominating function is $e^{-(1-t)x}$ and\n\n$$\\int_{R}^{\\infty} \\frac{x e^{tx}}{e^x - 1} \\, dx \\leq \\int_{R}^{\\infty} C e^{-(1-t)x} \\, dx < \\infty$$\n\nfor large $R$. Therefore the improper integral converges.\n\n-\n\nThanks for your answers and especially for this limit method. But in this way, I actually found a simpler bound, namely, the following: observe $$\\frac{xe^{tx}}{e^{x}-1}=\\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}e^{(1/2)(t-1)x}$$ $$\\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}<M$$ for a constant $M>0$ as $$\\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}$$ is continuous and the limits for $x\\to 0$ and $x \\to \\infty$ are finite. And this can be directly used for the integrability of the function.\n\n-", "date": "2015-05-30 09:16:25", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/219329/for-t-in-0-1-is-fracxetxex-1-integrable-over-x-in-0-in/219347", "openwebmath_score": 0.9804616570472717, "openwebmath_perplexity": 161.082132305399, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407160384083, "lm_q2_score": 0.9207896671963207, "lm_q1q2_score": 0.8961499950229149}}
{"url": "https://gmatclub.com/forum/the-price-of-a-consumer-good-increased-by-p-228709.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Jun 2019, 08:03\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# The price of a consumer good increased by p%. . .\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2888\nThe price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 07 Aug 2018, 07:09\n1\n12\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n68% (02:30) correct 32% (02:34) wrong based on 273 sessions\n\n### HideShow timer Statistics\n\nThe price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$?\n\nA. $$-2$$%\nB. $$2$$%\nC. $$22$$%\nD. $$25$$%\nE. Cannot be determined\n\nTake a stab at this fresh question from e-GMAT. Post your analysis below.\n\nOfficial Solution to be provided after receiving some good analyses.\n\n_________________\n\nOriginally posted by EgmatQuantExpert on 11 Nov 2016, 05:46.\nLast edited by EgmatQuantExpert on 07 Aug 2018, 07:09, edited 1 time in total.\nCEO\nJoined: 12 Sep 2015\nPosts: 3777\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\n11 Nov 2016, 07:09\nTop Contributor\nEgmatQuantExpert wrote:\nThe price of a consumer good increased by p% during 2012 and decreased by 12% during 2013. If no other change took place in the price of the good and the price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012, what was the value of p?\n\nA. $$-2$$%\nB. $$2$$%\nC. $$22$$%\nD. $$25$$%\nE. Cannot be determined\n\nLet $100 be the original price The price of a consumer good increased by p% during 2012 p% = p/100, so a p% INCREASE is the same a multiplying the original price by 1 + p/100 So, the new price = ($100)(1 + p/100)\n\nThe price then decreased by 12% during 2013\nA 12% DECREASE is the same a multiplying the price by 0.88\nSo, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)\nSimplify more: $110 = 88 + 0.88p Subtract 88 from both sides: 22 = 0.88p So, p = 22/0.88 = 25 Answer: RELATED VIDEO _________________ Test confidently with gmatprepnow.com Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4499 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 12:48 EgmatQuantExpert wrote: The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined Price's corresponding to year - 2011 = $$100$$ 2012 = $$100 + p$$ 2013 = $$\\frac{88}{100}(100 + p)$$ Further , $$\\frac{88}{100}(100 + p)$$ = $$110$$ Or, $$\\frac{8}{100}(100 + p)$$ = $$10$$ Or, 800 + 8p = 1000 Or, 8p = 200 So, p = 25% Hence, answer will be (D) 25% _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Current Student Joined: 26 Jan 2016 Posts: 100 Location: United States GPA: 3.37 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 13:01 Lets start with a number for the original value. 100 is the easiest. So we're looking for a value of 110 at the end of 2013. Just by looking at the values we can get an idea of what to start testing. If we're increasing 100 by p% then decreasing it by 12% and the original value is still 10% higher we need a value much higher than 12. 25% is the easiest value to start with. 100+25%=125 125-12%=110 D Current Student Joined: 12 Aug 2015 Posts: 2610 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 12 Nov 2016, 20:32 For all the algebra loving people out there=> Let price at the beginning of 2012 be$x\nso after the end of 2012=> x[1+p/100]\nAnd finally at the end of 2013 => x[1+p/100][1-12/100]\n\nAs per question=> Price was simple 10 percent greater\nHence x[1+10/100] must be the final price.\nEquating the two we get\n=> x[110/100]=x[1+p/100][88/100]\n=> 44p+4400=5500\n=> 44p=1100\n=> p=1100/44=> 100/4=> 25.\nSo p must be 25\n_________________\nIntern\nJoined: 02 Aug 2016\nPosts: 4\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\n16 Nov 2016, 09:46\n[size=150]Let Price = 100\n\n[size=150]Increased by P% = 100(1+P/100)\n\nTreat it like successive percents;\n\nSo a 12% decrease would mean 88% of (1+P/100) of 100\n\nThe key words are no other change took place. So there are no further successive percents, and the final price = 110\n\nTherefore:\n\n88/100 * (1+P/100) * 100 = 110\n\n=> 8800 + 88P = 11,000\n=> 88P = 2200\n=> P = 2200/88\n=> P = 25%\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2888\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 18 Dec 2016, 22:12\nHey,\n\nThe given question can be solved in a number of ways. Let's look at two most common ways to solve this question. We will share two more ways of solving this question tomorrow.\n\nMethod 1\n:\n\n\u2022 Let us consider the price of the consumer good at the beginning of 2012 to be 100.\n\u2022 Let us also assume the price to be \u201cC\u201d at the beginning of 2013, after an increase of p%.\n\u2022 Since we know that with respect to the initial price, the price at the end of 2013 went up by 10%.\no Therefore, the price at the end of 2013 = $$100 + (10$$ % of $$100) = 110$$\n\u2022 Now we can write -\no $$C \u2013 12$$ % of $$C = 110$$\no $$C * (1 - \\frac {12}{100}) = 110$$\no $$C = 110 * \\frac {25}{22} = 125$$\n\nTherefore, the price at the beginning of 2013 is 125 and we got this value after p% increase over the initial value.\nThus, we can write \u2013\n\u2022 $$100 + (p$$ % of $$100) = 125$$\n\u2022 $$P = 25$$ %\n\nMethod 2\n:\n\nConventional method \u2013\n\n\u2022 Let the price of consumer good at the beginning of 2012 be 100.\n\u2022 After an increase of p%, the price at the beginning of 2013 will be \u2013\no Price at the beginning of 2013 $$= 100 + (p$$ % of $$100) = 100 + p$$\nTherefore the price at the beginning of 2013 is (100 + p)\n\n\u2022 After a decrease of 12%, the price at the end of 2013 will be \u2013\no Value at the beginning of 2013 $$* (1 \u2013 \\frac{12}{100}) = (100 + p) * \\frac {22}{25}$$\u2026\u2026\u2026\u2026\u2026..(i)\n\n\u2022 And we are also given that the overall increase in the price of consumer good is 10%.\n\u2022 Therefore, the value at the end of 2013 = $$100 + (10$$% of $$100)$$ = 110\u2026\u2026\u2026(ii)\n\n\u2022 Equating equation (i) and (ii) we get \u2013\no $$(100 + p) * \\frac {22}{25} = 110$$\no $$100 + p = 125$$\nTherefore, $$p = 25$$%\n\nThere are more innovative ways to solve this question. A few of them have not been discussed here. Can you all think of any other way to solve it? Would love to see a few other methods!\n\nI will post two more ways to solve this question tomorrow. In the mean time, expecting some more responses with other ways to solve this question\n\nThanks,\nSaquib\ne-GMAT\nQuant Expert\n_________________\n\nOriginally posted by EgmatQuantExpert on 15 Dec 2016, 23:26.\nLast edited by EgmatQuantExpert on 18 Dec 2016, 22:12, edited 1 time in total.\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2888\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\n16 Dec 2016, 04:11\n1\njoannaecohen wrote:\nLets start with a number for the original value. 100 is the easiest.\n\nSo we're looking for a value of 110 at the end of 2013.\n\nJust by looking at the values we can get an idea of what to start testing. If we're increasing 100 by p% then decreasing it by 12% and the original value is still 10% higher we need a value much higher than 12. 25% is the easiest value to start with.\n\n100+25%=125\n\n125-12%=110\n\nD\n\nHi,\n\nThanks for posting a different way of approaching this problem. In fact, the approach followed by you is very close to one of the innovative ways that we talked about in our official solution. The only difference being in your approach you have concluded that p should be much larger than 12. Going a step further, you can also conclude that p should be larger than 22% (12%+10%). Once you do so, you don't even need to pick any number. The only option which will satisfy it is D. 25%.\n\nWhen we post our detailed solution using the two innovative methods tomorrow, we will explain how we can conclude that p should be greater than 22%.\n\nRegards,\nSaquib\n_________________\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2888\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\n25 Dec 2016, 07:32\n1\nAs discussed, let's look at one of the innovative ways of solving the above question. It is one of the quickest ways to solve a question that involves successive percentage increase/decrease on the same value. Please take a note of this approach and apply it on some GMAT questions to master it.\n\nSo, let's quickly look at this smart approach.\n\nWhen a number is increased successively by two percentage, let's assume, $$a$$% and $$b$$%, the net increase in the value of the number can be expressed by the formula,\n\nNet increase $$=a+b+\\frac {ab}{100}$$\n\nLe's take a simple example to understand. If we increase a number, let's say, X successively by 10% and 20% respectively, the net increase according to the above formula should be,\n\nNet increase $$=10+20+\\frac {10*20}{100}=10+20+\\frac {200}{100} = 10+20+2 = 32$$%\n\nIsn't that quick!! A nice method to keep in your arsenal to solve Percent question involving successive increase quickly.\n\nOne good thing about the above formula is that you can use it to calculate the net decrease in case of successive decrease too. All you need to do is in case of decrease represent the percent as negative. Easy isn't it . Let's see an application quickly.\n\nIf we decrease a number, let's say, X successively by 10% and 20% respectively, the net increase according to the above formula should be,\n\nNet increase $$=(-10)+(-20)+\\frac {(-10)*(-20)}{100}=-10-20+\\frac {200}{100} = -10-20+2 = (-28)$$%\n\nNotice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.\n\nSo, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.\n\nWe will post the detailed solution tomorrow and then we will show another innovative method of solving this question.\n\nRegards,\nSaquib\n_________________\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2888\nThe price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 07 Aug 2018, 07:11\n2\n1\nAlright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.\n\nWe know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,\n\nNet increase $$=p+(-12)+\\frac {p*(-12)}{100}=p-12-\\frac {3p}{25} = (\\frac {22p}{25} - 12)$$%\n\nIt is given in the question that the net increase finally is $$10$$%. Hence, we can equate the two values.\n\n$$(\\frac {22p}{25} - 12)$$% = $$10$$%\n\nor,\n$$\\frac {22p}{25} = 10+12 = 22$$%\n\nor,\n$$p = 25$$%\n\nNow, with this understanding try to solve this question in an even better way. Give it a try and we will post the official solution in another innovative way soon.\n\n_________________\n\nOriginally posted by EgmatQuantExpert on 27 Dec 2016, 00:45.\nLast edited by EgmatQuantExpert on 07 Aug 2018, 07:11, edited 1 time in total.\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 4499\nLocation: India\nGPA: 3.5\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\n27 Dec 2016, 07:56\nEgmatQuantExpert wrote:\nThe price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$?\n\nA. $$-2$$%\nB. $$2$$%\nC. $$22$$%\nD. $$25$$%\nE. Cannot be determined\n\n$$p - 12 - \\frac{12p}{100} = 10$$\n\n$$100p - 1200 -12p = 1000$$\n\n$$88p = 2200$$\n$$p = 25$$\n\nHence, the correct answer must be (D) 25\n\n_________________\nThanks and Regards\n\nAbhishek....\n\nPLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS\n\nHow to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 6522\nLocation: United States (CA)\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\n01 Feb 2019, 18:59\nEgmatQuantExpert wrote:\nThe price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$?\n\nA. $$-2$$%\nB. $$2$$%\nC. $$22$$%\nD. $$25$$%\nE. Cannot be determined\n\nWe let the 2012 price = n and thus, the price at the end of 2013 will be:\n\n(1 + p/100)(0.88)(n)\n\nSince the price at the end of 2013 was 10% higher than at the beginning of 2012, we can create the equation:\n\n1.1n = (1 + p/100)(0.88)(n)\n\n1 = (1 + p/100)(0.8)\n\n1 = 0.8 + 0.8p/100\n\nMultiplying by 100, we have:\n\n100 = 80 + 0.8p\n\n20 = 0.8p\n\n25 = p\n\nAlternate Solution:\n\nLet\u2019s let the price at the beginning of 2012 be 100. Since the price at the end of 2013 was 10% higher, the price at the end of 2013 is 1.1 x 100 = 110. We know the price decreased by 12% and became 110; therefore, before decreasing by 12%, the price was 110/0.88 = 125. Now, the price of 100 increases by p percent and becomes 125; therefore the value of p is [(125 - 100)/100] x 100 = 25.\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nVP\nJoined: 09 Mar 2018\nPosts: 1004\nLocation: India\nRe: The price of a consumer good increased by p%. . .\u00a0 [#permalink]\n\n### Show Tags\n\n01 Feb 2019, 21:46\nEgmatQuantExpert wrote:\nThe price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$?\nA. $$-2$$%\nB. $$2$$%\nC. $$22$$%\nD. $$25$$%\nE. Cannot be determined\n\n10 % will be a successive percentage change, which is calculated by\n\na + b + ab/100 = Percentage change\n\np - 12 - 12p/100 = 10\n\n100p -1200 - 12p = 1000\np = 2200/88\n\np = 25 %\n\nD\n_________________\nIf you notice any discrepancy in my reasoning, please let me know. Lets improve together.\n\nQuote which i can relate to.\nMany of life's failures happen with people who do not realize how close they were to success when they gave up.\nRe: The price of a consumer good increased by p%. . . \u00a0 [#permalink] 01 Feb 2019, 21:46\nDisplay posts from previous: Sort by", "date": "2019-06-17 15:03:43", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-price-of-a-consumer-good-increased-by-p-228709.html", "openwebmath_score": 0.7690556645393372, "openwebmath_perplexity": 1808.9536375803552, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9744347905312774, "lm_q2_score": 0.9196425377849806, "lm_q1q2_score": 0.8961316836701599}}
{"url": "http://centrallab.msu.ac.th/2f1iaj16/lhcbx.php?aea000=least-square-approximation-of-a-function", "text": "The idea is to minimize the norm of the difference between the given function and the approximation. Picture: geometry of a least-squares solution. As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to calculate y for x's not present in the table. Learn to turn a best-fit problem into a least-squares problem. FINDING THE LEAST SQUARES APPROXIMATION We solve the least squares approximation problem on only the interval [\u22121,1]. We use the Least Squares Method to obtain parameters of F for the best fit. obtained as measurement data. Because the least-squares fitting process minimizes the summed square of the residuals, the coefficients are determined by differentiating S with respect to each parameter, and setting the result equal to zero. The method of least square \u2022 Above we saw a discrete data set being approximated by a continuous function \u2022 We can also approximate continuous functions by simpler functions, see Figure 3 and Figure 4 Lectures INF2320 \u2013 p. 5/80 Given a function and a set of approximating functions (such as the monomials ), for each vector of numbers define a functional By \u2026 The least squares method is one of the methods for finding such a function. Active 7 months ago. Learn examples of best-fit problems. Thus, the empirical formula \"smoothes\" y values. The radial basis function (RBF) is a class of approximation functions commonly used in interpolation and least squares. Approximation of a function consists in finding a function formula that best matches to a set of points e.g. Least Square Approximation for Exponential Functions. Section 6.5 The Method of Least Squares \u00b6 permalink Objectives. Ask Question Asked 5 years ago. Vocabulary words: least-squares solution. Orthogonal Polynomials and Least Squares Approximations, cont\u2019d Previously, we learned that the problem of nding the polynomial f n(x), of degree n, that best approximates a function f(x) on an interval [a;b] in the least squares sense, i.e., that minimizes kf n fk= Z \u2026 ... ( \\left[ \\begin{array}{c} a \\\\ b \\end{array} \\right] \\right)\\$ using the original trial function. Approximation problems on other intervals [a,b] can be accomplished using a lin-ear change of variable. Recipe: find a least-squares solution (two ways). The RBF is especially suitable for scattered data approximation and high dimensional function approximation. Free Linear Approximation calculator - lineary approximate functions at given points step-by-step This website uses cookies to ensure you get the best experience. Quarteroni, Sacco, and Saleri, in Section 10.7, discuss least-squares approximation in function spaces such as . 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Rbf is especially suitable for scattered data approximation and least square approximation of a function dimensional function..\nWatts 5 Year Water Filter, Kingsford Grill Smoker, Architecture Of Chowmahalla Palace, Can A Midwife Give An Epidural At Home, Infernal Contraption In Normal Difficulty, Lowest Recorded Oxygen Saturation, Apple Snickerdoodle Cobbler, Usda Pay Grade 5, Creature Generator Drawing, Shruti Name In Different Languages,", "date": "2022-01-23 08:53:20", "meta": {"domain": "ac.th", "url": "http://centrallab.msu.ac.th/2f1iaj16/lhcbx.php?aea000=least-square-approximation-of-a-function", "openwebmath_score": 0.710901141166687, "openwebmath_perplexity": 896.7067055485361, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683498785867, "lm_q2_score": 0.9073122169746364, "lm_q1q2_score": 0.8960328289423238}}
{"url": "https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-108-2.html", "text": "Perl Weekly Challenge 108: Bell Numbers\n\nby Abigail\n\nChallenge\n\nExample\n\n\u2022 $$B_0$$: 1 as you can only have one partition of zero element set.\n\u2022 $$B_1$$: 1 as you can only have one partition of one element set $$\\{a\\}$$.\n\u2022 $$B_2$$: 2\n\u2022 $$\\{a\\}\\{b\\}$$\n\u2022 $$\\{a,b\\}$$\n\u2022 $$B_3$$: 5\n\u2022 $$\\{a\\}\\{b\\}\\{c\\}$$\n\u2022 $$\\{a,b\\}\\{c\\}$$\n\u2022 $$\\{a\\}\\{b,c\\}$$\n\u2022 $$\\{a,c\\}\\{b\\}$$\n\u2022 $$\\{a,b,c\\}$$\n\u2022 $$B_4$$: 15\n\u2022 $$\\{a\\}\\{b\\}\\{c\\}\\{d\\}$$\n\u2022 $$\\{a,b,c,d\\}$$\n\u2022 $$\\{a,b\\}\\{c,d\\}$$\n\u2022 $$\\{a,c\\}\\{b,d\\}$$\n\u2022 $$\\{a,d\\}\\{b,c\\}$$\n\u2022 $$\\{a,b\\}\\{c\\}\\{d\\}$$\n\u2022 $$\\{a,c\\}\\{b\\}\\{d\\}$$\n\u2022 $$\\{a,d\\}\\{b\\}\\{c\\}$$\n\u2022 $$\\{b,c\\}\\{a\\}\\{d\\}$$\n\u2022 $$\\{b,d\\}\\{a\\}\\{c\\}$$\n\u2022 $$\\{c,d\\}\\{a\\}\\{b\\}$$\n\u2022 $$\\{a\\}\\{b,c,d\\}$$\n\u2022 $$\\{b\\}\\{a,c,d\\}$$\n\u2022 $$\\{c\\}\\{a,b,d\\}$$\n\u2022 $$\\{d\\}\\{a,b,c\\}$$\n\nDiscussion\n\nThe Bell Numbers have their own entry in the OEIS. We can look up the first ten Bell Numbers: $$1$$, $$1$$, $$2$$, $$5$$, $$15$$, $$52$$, $$203$$, $$877$$, $$4140$$, and $$21147$$.\n\nHello, World!\n\nThe simplest way would be just to take those ten numbers, and print them. This means we have yet again a challenge which is just a glorified Hello, World program.\n\nFetch\n\nIf we don't want to do exactly what the challenge asks from us (print the first ten Bell Numbers), we could instead fetch the numbers from the OEIS and print them. For instance, by using the OEIS module which we recently uploaded to CPAN.\n\nThere is limited usefulness in this though \u2014 it's not that the Bell Numbers will change in the future.\n\nCalculate\n\nAlternatively, we could calculate the first ten Bell Numbers. There are many ways to calculate the numbers, but we opt to create a Bell Triangle.\n\nThe first rows of the Bell Triangle are as follows:\n\n 1\n1  2\n2  3  5\n5  7 10 15\n15 20 27 37 52\n\n\nAnd we have the following rules to construct the triangle:\n\n\u2022 The top row contains a single $$1$$.\n\u2022 For each other row:\n\u2022 The row will have one more element than the previous row.\n\u2022 The first (left most) element is equal to the last (right most) element of the previous row.\n\u2022 Each other element is the sum of the element to its left on the same row, and the element on the previous row right above that.\n\nOr, formalized:\n\nLet $$b_{r, c}$$ be the element on row $$r$$ and column $$c$$. (This implies $$0 \\leq c \\leq r$$, with the top most element being $$b_{0, 0}$$.) Then\n\n$b_{r, c} = \\begin{cases} 1, & \\text{if } r = c = 0 \\\\ b_{r - 1, r - 1}, & \\text{if } r > 0, c = 0 \\\\ b_{r, c - 1} + b_{r - 1, c - 1}, & \\text{if } r \\geq c > 0 \\end{cases}$\n\nIf we then generate the first nine rows of the Bell Triangle, and take the last elements of each row, we get the second to tenth Bell Numbers. The first Bell Number is $$1$$.\n\nSolutions\n\nDepending on the language, we solve the challenge in one or more of the strategies explained above. All languages will implement the Hello, World! strategy. For some languages, we also calculate the Bell Triangle. And in Perl, we also implement a fetch strategy.\n\nLanguages which solve the problem in more than one way take a command line argument indicating the strategy to follow. This argument should be one of plain (the default), fetch (which fetches the numbers from the OEIS, or compute, which computes the first rows of the Bell Triangle.\n\nWe will only show the the plain solution for Perl; for the other implementations, see the GitHub links below.\n\nPerl\n\nplain\n\nCan't be much simpler than this.\n\nsay \"1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147\"\n\n\nfetch\n\nWe're using the new module OEIS which export a single method, oeis, which takes two arguments: the sequence to fetch, and the number of elements to return.\n\nuse OEIS;\n}\n\n$, = \", \"; say 1, map {$$_ [-1]} @bell;  Find the full program on GitHub. AWK The algorithm above is simply written down in AWK: BEGIN { COUNT = 10 bell [1, 1] = 1 for (x = 2; x < COUNT; x ++) { bell [x, 1] = bell [x - 1, x - 1] for (y = 2; y <= x; y ++) { bell [x, y] = bell [x, y - 1] + bell [x - 1, y - 1] } } printf \"1\" for (x = 1; x < COUNT; x ++) { printf \", %d\", bell [x, x] } printf \"\\n\" }  Find the full program on GitHub. Bash Bash doesn't have two dimensional arrays. So, we're using a function index which takes two arguments (an x and a y coordinate) and returns a single index. The return value is written in the global variable idx. We then get: set -f COUNT=10 function index () { local x=$1\nlocal y=$2 idx=$((COUNT * x + y))\n}\n\nbell[0]=1\nfor ((x = 1; x < COUNT - 1; x ++))\ndo   index $x 0; i1=$idx\nindex $((x - 1))$((x - 1)); i2=$idx bell[$i1]=${bell[$i2]}\nfor ((y = 1; y <= x; y ++))\ndo  index   $x$y;       i1=$idx index$x       $((y - 1)); i2=$idx\nindex $((x - 1))$((y - 1)); i3=$idx bell[$i1]=$((bell[i2] + bell[i3])) done done printf \"1\" for ((x = 0; x < COUNT - 1; x ++)) do index$x $x; printf \", %d\"${bell[\\$idx]}\ndone\necho\n\n\nFind the full program on GitHub.\n\nC\n\nC requires us to manage our own memory. Other than that, it's the same algorithm:\n\n# define COUNT   10\n\ntypedef int number;  /* Change if we want large numbers */\nchar * fmt = \"%d\";   /* Should match typedef            */\n\nint main (int argc, char * argv []) {\nnumber ** bell;\nif ((bell = (number **) malloc ((COUNT - 1) * sizeof (number *)))\n== NULL) {\nperror (\"Mallocing bell failed\");\nexit (1);\n}\nif ((bell [0] = (number *) malloc (sizeof (number))) == NULL) {\nperror (\"Mallocing row failed\");\nexit (1);\n}\nbell [0] [0] = 1;\nfor (int x = 1; x < COUNT - 1; x ++) {\nif ((bell [x] = (number *) malloc ((x + 1) * sizeof (number)))\n== NULL) {\nperror (\"Mallocing row failed\");\nexit (1);\n}\nbell [x] [0] = bell [x - 1] [x - 1];\nfor (int y = 1; y <= x; y ++) {\nbell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1];\n}\n}\n\n/*\n* Print the right diagonal\n*/\nprintf (fmt, 1);\nfor (int x = 0; x < COUNT - 1; x ++) {\nprintf (\", \");\nprintf (fmt, bell [x] [x]);\n}\nprintf (\"\\n\");\nexit (0);\n}\n\n\nFind the full program on GitHub.\n\nLua\n\nSame algorithm:\n\nlocal COUNT = 10\nlocal bell  = {}\n\nbell [0] = {}\nbell [0] [0] = 1\nfor x = 1, COUNT - 2\ndo  bell [x] = {}\nbell [x] [0] = bell [x - 1] [x - 1]\nfor  y = 1, x\ndo   bell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1]\nend\nend\n\nio . write (1)\nfor x = 0, COUNT - 2\ndo  io . write (\", \" .. bell [x] [x])\nend\nio . write (\"\\n\")\n\n\nFind the full program on GitHub.\n\nNode.js\n\nlet COUNT = 10\nlet bell  = [[ 1 ]]\nlet x\nfor (x = 1; x < COUNT - 1; x ++) {\nbell [x] = [bell [x - 1] [x - 1]]\nlet y\nfor (y = 1; y <= x; y ++) {\nbell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1]\n}\n}\n\nprocess . stdout . write (\"1\")\nfor (x = 0; x < COUNT - 1; x ++) {\nprocess . stdout . write (\", \" + bell [x] [x] . toString ())\n}\nprocess . stdout . write (\"\\n\")\n\n\nFind the full program on GitHub.\n\nPython\n\nPython doesn't autovivify array elements when indexing out of bounds. So we use the append method to add elements to arrays.\n\nCOUNT = 10\nbell  = [[1]]\nfor x in range (1, COUNT - 1):\nbell . append ([bell [x - 1] [x - 1]])\nfor y in range (1, x + 1):\nbell [x] . append (bell [x] [y - 1] + bell [x - 1] [y - 1])\n\nprint (1, end = '')\nfor x in range (0, COUNT - 1):\nprint (\",\", bell [x] [x], end = '')\n\nprint (\"\")\n\n\nFind the full program on GitHub.\n\nRuby\n\nCOUNT = 10\nbell  = [[1]]\nfor  x in 1 .. COUNT - 2\nbell [x] = [bell [x - 1] [x - 1]]\nfor  y in 1 .. x\nbell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1]\nend\nend\nprint (1)\nfor  x in 0 .. COUNT - 2\nprint (\", \")\nprint (bell [x] [x])\nend\nputs (\"\")\n\n\nFind the full program on GitHub.\n\nOther languages\n\nWe also have simple solutions for BASIC, bc, Befunge-93, Cobol, Csh, Erlang, Forth, Fortran, Go, Java, m4, OCaml, Pascal, PHP, PostScript, R, Rexx, Scheme, sed, SQL, and Tcl.", "date": "2021-06-19 22:09:03", "meta": {"domain": "github.io", "url": "https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-108-2.html", "openwebmath_score": 0.41870564222335815, "openwebmath_perplexity": 4312.791664273166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517437261226, "lm_q2_score": 0.9161096044278532, "lm_q1q2_score": 0.8960025960549102}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=e02qs82rtho6ihd22c4oep6c21&action=printpage;topic=259.0", "text": "# Toronto Math Forum\n\n## MAT244-2013S => MAT244 Math--Tests => MidTerm => Topic started by: Victor Ivrii on March 06, 2013, 09:08:26 PM\n\nTitle: MT Problem 3\nPost by: Victor Ivrii on March 06, 2013, 09:08:26 PM\nFind a particular solution of equation\n\\begin{equation*}\nt^2 y''-2t y' +2y=t^3 e^t.\n\\end{equation*}\n\n[BONUS] Explain whether the method of undetermined\u00a0 coefficients to find a particular solution of this equation applies.\nTitle: Re: MT Problem 3\nPost by: Jeong Yeon Yook on March 06, 2013, 10:30:47 PM\nThe method of undetermined coefficient applies because it only \"requires us to make an initial assumption about the form of the particular solution, but with the coefficients left unspecified\" (Textbook 10th Edition P.177).\n\nIf t = 0, we have, 2y = 0. => y = 0 is the solution for t = 0.\nTitle: Re: MT Problem 3\nPost by: Rudolf-Harri Oberg on March 06, 2013, 10:50:34 PM\nThis is an Euler equation, see book page 166, problem 34. We need to use substitution $x=\\ln t$, this will make into a ODE with constant coefficients. We look first at the homogenous version:\n$$y''-3y'+2y=0$$\n\nSolving $r^2-3r+2=0$ yields $r_1=2, r_2=1$. So, solutions to the homogeneous version are $y_1(x)=e^{2x}, y_2(x)=e^{x}$. But then solutions to the homogeneous of the original problem are $y_1(t)=t^2, y_2(t)=t$.\nSo, $Y_{gen.hom}=c_1t^2+c_2t$. We now use method of variation of parameters, i.e let $c_1,c_2$ be functions.\nTo use the formulas on page 189, we need to divide the whole equation by $t^2$ so that the leading coefficient would be one, so now $g=t e^t$. The formula is:\n$c_i'=\\frac{W_i g}{W}$, where $W_i$ is the wronksian of the two solutions where the i-th column has been replaced by $(0,1)$.\nWe now just calculate that $W(t^2,t)=-t^2, W_1=-t, W_2=t^2$. Now we need to compute $c_1, c_2$.\n\n$$c_1'=e^t \\implies c_1=e^t$$\n$$c_2'=-te^t \\implies c_2=-e^t(t-1)$$\n\nPlugging these expressions back to $Y_{gen.hom}$ yields the solution which is\n$y=te^t$\nTitle: Re: MT Problem 3\nPost by: Branden Zipplinger on March 06, 2013, 11:20:47 PM\nfor the bonus, the method of undetermined coefficients does not apply here, because when we assume y is of the form g(x), deriving twice and substituting into the equation yields terms with powers of t such that it is impossible to find a coefficient where the solution is of the form you assumed. this can be easily verified\nTitle: Re: MT Problem 3\nPost by: Branden Zipplinger on March 06, 2013, 11:22:05 PM\n(by g(x) i mean the non-homogeneous term)\nTitle: Re: MT Problem 3\nPost by: Brian Bi on March 07, 2013, 12:19:03 AM\nI wrote that undetermined coefficients does not apply because the ODE does not have constant coefficients.\nTitle: Re: MT Problem 3\nPost by: Victor Lam on March 07, 2013, 12:38:24 AM\nI basically wrote what Brian did for the bonus. But I suppose that if we transform the original differential equation using x = ln(t) into another DE with constants coefficients (say, change all the t's to x's), we would then be able to apply the coefficients method, and carry on to find the particular solution. Can someone confirm the validity of this?\nTitle: Re: MT Problem 3\nPost by: Branden Zipplinger on March 07, 2013, 02:20:51 AM\nnevermind.\nTitle: Re: MT Problem 3\nPost by: Victor Ivrii on March 07, 2013, 04:47:03 AM\nRudolf-Harri Oberg solution is perfect. One does not need to reduce it to constant coefficients (appealing to it is another matter); characteristic equation is $r(r-1)-2r+2=0$ rendering $r_{1,2}=1,2$ and $y_1=t$, $y_2=t^2$ (Euler equation).\n\nMethod of undetermined coefficients should not work;\u00a0 all explanations are almost correct: for equations with constant coefficients the r.h.e. must be of the form $P(x)e^{rx}$ where $P(x)$ is a polynomial but for Euler equation which we have it must be $P(\\ln (t)) t^r$ (appeal to reduction) which is not the case.\n\nHowever sometimes work methods which should not and J. Y. Yook has shown this. Luck sometimes smiles to foolish and ignores the smarts\n\nQuote\nEverybody knows that something can't be done and then somebody turns up and he doesn't know it can't be done and he does it.(A. Einstein)\nTitle: Re: MT Problem 3\nPost by: Branden Zipplinger on March 07, 2013, 04:58:04 AM\nhas a theorem been discovered that describes what the form of a non-homogeneous equation should look like for it to be solvable by undetermined coefficients?\nTitle: Re: MT Problem 3\nPost by: Patrick Guo on March 16, 2013, 12:51:20 PM\nJust got my midterm back on Friday and looked carefully through..\n\nIn the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have\nv1 = - \u00e2\u02c6\u00ab (t^2 + 1) g(t) / Wronskian\u00a0 dt\u00a0 \u00a0??\u00a0 what is (t^2 +1) ?! Should that not be y2 = t^2 ?!\n\nAnd how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ?\n\nI see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG.\n\nAnd why Wronskian = -t^2 ? should it not be t^2 ?\nTitle: Re: MT Problem 3\nPost by: Victor Ivrii on March 16, 2013, 03:00:51 PM\nJust got my midterm back on Friday and looked carefully through..\n\nIn the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have\nv1 = - \u00e2\u02c6\u00ab (t^2 + 1) g(t) / Wronskian\u00a0 dt\u00a0 \u00a0??\u00a0 what is (t^2 +1) ?! Should that not be y2 = t^2 ?!\n\nAnd how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ?\n\nI see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG.\n\nAnd why Wronskian = -t^2 ? should it not be t^2 ?\n\nRats! The answer is simple: typo by the person who typed and the lack of proofreading (all instructors were busy preparing Final and TT2). Thanks! Fixed in all three instances (including on BlackBoard)", "date": "2022-06-28 17:47:55", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=e02qs82rtho6ihd22c4oep6c21&action=printpage;topic=259.0", "openwebmath_score": 0.7391995191574097, "openwebmath_perplexity": 1912.7939147305165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429609670701, "lm_q2_score": 0.9111797069968975, "lm_q1q2_score": 0.8960021510514365}}
{"url": "https://math.stackexchange.com/questions/2720043/null-space-and-kernel-of-matrix-representation", "text": "# Null space and kernel of matrix representation\n\nLet $P_3(\\mathbb{C})$ be the complex vector space of complex polynomials of degree $2$ or less. Let $\\alpha,\\beta\\in\\mathbb{C}, \\alpha\\neq\\beta$. Consider the function $L:P_3(\\mathbb{C}) \\mapsto \\mathbb{C}^2$ given by\n\n$$L(p)=\\begin{bmatrix} p(\\alpha) \\\\ p(\\beta)\\\\ \\end{bmatrix}, \\text{ for } p\\in P_3(\\mathbb{C})$$\n\nFor the basis $v=(1,X,X^2)$ for $P_3(\\mathbb{C})$ and the standard basis $E = (e_1,e_2)$ for $\\mathbb{C}^2$. Find the matrix representation $_E[L]_v$ and determine the null space $N(_E[L]_v)$ and find a basis for the ker(L).\n\nI have found the matrix representation:\n\n$$_E[L]_v = [L(v)]_E = [L(1)]_E\\ [L(X)]_E\\ [L(X^2)]_E = \\begin{bmatrix} 1\\quad \\alpha \\quad \\alpha^2 \\\\ 1\\quad \\beta \\quad \\beta^2 \\end{bmatrix}$$\n\nBy using ERO we can reduce the matrix to: $\\begin{bmatrix} 1 \\quad 0 \\quad - \\alpha\\beta \\\\ 0 \\quad 1 \\quad \\alpha + \\beta \\\\ \\end{bmatrix},$\n\nI am uncertain how to find the null space $N(_E[L]_v)$ and a basis for the kernel.\n\n\u2022 You\u2019re almost there. See this answer for how to read a basis for the kernel from the reduced matrix. \u2013\u00a0amd Apr 3 '18 at 20:01\n\u2022 So it is possible to write the RREF matrix: $\\begin{bmatrix} 1 \\quad 0 \\quad - \\alpha\\beta \\\\ 0 \\quad 1 \\quad \\alpha + \\beta \\\\ \\end{bmatrix},$ as the following: $x_1 = \\alpha\\beta$, $x_2 = -\\alpha-\\beta$, $x_3 = x_3$ as $x_3$ is a free variable we can put in 1, so $x_3=1$ This way we have that $L_v= (\\alpha\\beta, -\\alpha-\\beta, 1)^T$ Which means that the basis for the kernel is equal to $\\begin{bmatrix} \\alpha\\beta \\\\ -\\alpha-\\beta \\\\ 1 \\end{bmatrix}$? \u2013\u00a0Simb\u00f6rg Apr 3 '18 at 21:15\n\u2022 Your reasoning is a bit off. The RREF represents the equations $x_1-\\alpha\\beta x_3=0$ and $x_2+(\\alpha+\\beta)x_3=0$, so every solution of the system is of the form $(\\alpha\\beta x_3, -(\\alpha+\\beta)x_3, x_3)^T$, i.e., a multiple of $(\\alpha\\beta, -\\alpha-\\beta,1)^T$. \u2013\u00a0amd Apr 3 '18 at 21:45\n\nYou're doing good and the matrix is exactly what you found. The reduced row echelon form is $$\\begin{bmatrix} 1 & 0 & -\\alpha\\beta \\\\ 0 & 1 & \\alpha+\\beta \\end{bmatrix}$$ as you found. Now you can determine a basis for the null space of the matrix as generated by $$\\begin{bmatrix} \\alpha\\beta \\\\ -(\\alpha+\\beta) \\\\ 1 \\end{bmatrix}$$ and this is the coordinate vector of a polynomial generating the kernel, which is thus $$q(X)=\\alpha\\beta-(\\alpha+\\beta)X+X^2$$\n\nAs a check: this polynomial $q$ has $\\alpha$ and $\\beta$ as roots and so $L(q)=0$. The kernel has dimension $1$ by the rank nullity theorem.\n\n\u2022 I am not sure I understand your argument for the kernel, but I will try to see if I understand it correctly. So: we know that since the coordinate vector $\\begin{bmatrix} \\alpha\\beta \\\\ -\\alpha-\\beta \\\\ 1 \\end{bmatrix} \\in N(_E[L]_v)$ we know this implies that $(\\alpha\\beta, -\\alpha-\\beta, 1)^T \\in ker(L)$ and then you define a polynomial for that generates the kernel, which means $ker(L) = (\\alpha\\beta, -(\\alpha+\\beta)X, X^2)^T$ and if $X = 0$ then we have that the kernel consists of $(\\alpha\\beta)$? \u2013\u00a0Simb\u00f6rg Apr 4 '18 at 9:27\n\nIt is probably better to do that via polynomials.\n\nSuppose $p \\in P_3(\\mathbb{C})$ is such that $p(\\alpha) = 0 = p(\\beta)$. Then $p$ is divisible by both $x - \\alpha$ and $x - \\beta$.\n\nSince $\\alpha \\ne \\beta$, the two linear polynomials are coprime, so $p$ is divisible by $(x-\\alpha)(x-\\beta) = x^{2} - (\\alpha+\\beta) x + \\alpha \\beta$, and thus $p$ is a scalar multiple of it, as $p$ has degree at most $2$.\n\nSo the kernel is one-dimensional, generated by the transpose of $(\\alpha \\beta, -\\alpha - \\beta, 1)$.\n\nThis indicates that there is a little sign error (it happens to everyone) in your reduction.\n\n\u2022 Your approach makes sense, it is just that the method I have in my textbook is that: The null space of a matrix A, N(A), is equal to the null space of the RREF H, N(H). So I believe that I have to reduce the matrix representation and find the null space with this method (and yes there was a slight error in my computation, it should be correct now) \u2013\u00a0Simb\u00f6rg Apr 3 '18 at 18:18\n\u2022 @Simb\u00f6rg, you should have been taught that once you have transformed your matrix in the block form $[I \\mid A]$, where $I$ is an appropriate identity matrix, then the space of solutions of the associated homogeneous system (i.e. the null space) has as a basis the columns of the block matrix $\\left[\\begin{smallmatrix}-A\\\\J\\end{smallmatrix}\\right]$, where $J$ is an appropriate square matrix. So in your case the null space has a basis given by $\\left[\\begin{smallmatrix}\\alpha \\beta\\\\-\\alpha-\\beta\\\\1\\end{smallmatrix}\\right]$. \u2013\u00a0Andreas Caranti Apr 4 '18 at 8:28", "date": "2019-09-15 20:12:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2720043/null-space-and-kernel-of-matrix-representation", "openwebmath_score": 0.9005959630012512, "openwebmath_perplexity": 97.26763139394704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.970687770944576, "lm_q2_score": 0.9230391637747005, "lm_q1q2_score": 0.8959828283790096}}
{"url": "https://www.physicsforums.com/threads/antiderivative-of-1-x.841263/", "text": "# Antiderivative of 1/x\n\n1. Nov 4, 2015\n\n### Cosmophile\n\nWe are going over antiderivatives in my calculus course and reached a question regarding $f(x) = \\frac {1}{x}$.\n\nMy instructor went on to say that $\\int \\frac {1}{x}dx = \\ln |x| + C$. This makes sense to me, but only to a certain point. For $f(x) = \\frac {1}{x}$, $f$ is defined $\\forall x \\neq 0$. So we should have two interavls which we are looking at: $x < 0$ and $x > 0$. Because of this, we would then have:\n\n$$\\int \\frac {1}{x}dx = \\ln |x| + C_1 \\qquad x > 0$$\n$$\\text {and}$$\n$$\\int \\frac {1}{x}dx = \\ln (-x) + C_2 = \\ln |x| + C_2 \\qquad x < 0$$\n\nThe second integral comes into being because $-x > 0$ when $x<0$. I brought this up to my teacher and he said that it made no sense and served no purpose to look at it this way. The argument I brought up was that the constants of integration could be different for the two intervals. I was hoping some of you may be able to help me explain why this is the case, or, if I am wrong, explain to me why I am.\n\nThanks!\n\n2. Nov 4, 2015\n\n### pwsnafu\n\nYou are correct. The \"constant of integration\" is constant over connected components of the domain. See also Wikipedia's article on antiderivative.\n\n3. Nov 4, 2015\n\n### fzero\n\nIt's an interesting idea, but the problem is that\n$$\\frac{d}{dx} \\ln (-x) = - \\frac{1}{x},$$\ninstead of $1/x$, so this isn't an antiderivative.\n\n4. Nov 4, 2015\n\n### pwsnafu\n\nNo $\\frac{d}{dx} \\ln(-x) = \\frac{1}{x}$\n\n5. Nov 4, 2015\n\n### Cosmophile\n\nIf we are considering $x < 0$, $-x > 0$ so\n$$\\frac {d}{dx} \\ln (-x) = \\frac {1}{x}$$\n\nAlso, by the chain rule where $u = -x$, $\\frac {du}{dx} = -1$, and\n\n$$\\frac {d}{dx} \\ln(-x) = \\frac {1}{-x}(-1) = \\frac {1}{x}$$\n\n6. Nov 4, 2015\n\n### micromass\n\nYou are completely correct. There are two different constants of integration. Your teacher must not be very good if he doesn't know this.\n\n7. Nov 4, 2015\n\n### pwsnafu\n\nSeconded. This\nworries me. Mathematics is not concerned with with whether something \"serves a purpose\". There is plenty of mathematical research that serves no purpose other than itself.\n\n8. Nov 4, 2015\n\n### Cosmophile\n\nI appreciate the replies thus far. I suppose I'm having a hard time developing an argument for my case to propose to him.\n\n9. Nov 4, 2015\n\n### PeroK\n\nEssentially, you are correct. You can demonstrate this by taking:\n\n$f(x) = ln|x| + 1 \\ (x < 0)$ and $f(x) = ln|x| + 2 \\ (x > 0)$ and checking that $f'(x) = \\frac{1}{x} \\ (x \\ne 0)$\n\nUsually, however, you are only dealing with one half of the function : $x < 0$ or $x > 0$. This is because an integral is defined for a function defined on an interval. For the function $\\frac{1}{x}$, you can't integrate it from on, say, $[-1, 1]$ because it's not defined at $x = 0$.\n\nSo, strictly speaking, what the integration tables are saying is:\n\na) For the function $\\frac{1}{x}$ defined on the interval $(0, +\\infty)$, the antiderivative is $ln|x| + C$\n\nb) For the function $\\frac{1}{x}$ defined on the interval $(-\\infty, 0)$, the antiderivative is $ln|x| + C$\n\nIn that sense, you don't need different constants of integration.\n\n10. Nov 4, 2015\n\n### micromass\n\nI see no reason not to define the undefined integral on more general sets.\n\n11. Nov 4, 2015\n\n### Cosmophile\n\nWhy did you arbitrarily chose $C_1 = 1$ and $C_2 = 2$ in the first part of your response?\n\nI understand that the antiderivative is defined only on intervals where the base function is defined, which means our integral is only defined on $(-\\infty, 0) \\cup (0, \\infty)$. However, when I think of a constant added to a function, I simply think of a vertical shift. I understand that $f(x) = \\frac {1}{x}$ has to have two independent antiderivatives as a consequence of the discontinuity, but I cannot see why it is necessary that the constants have to be different. Of course, a difference in constants is the only way that the two sides can be different, because we've already shown that, ignoring the added constant, the two sides have identical antiderivatives.\n\nAm I making any sense? Sorry, and thanks again.\n\n12. Nov 4, 2015\n\n### PeroK\n\nWhy not?\n\nA couple of technical points:\n\n1)\"The\" antiderivative is actually an equivalence class of functions. \"An\" antiderivative is one of the functions from that class. For example:\n\n$sin(x) + C$ (where $C$ is an arbitrary constant) is the antiderivative of $cos(x)$; and $sin(x) + 6$ is an antiderivative (one particular function from the antiderivative class).\n\n2) There is no such thing as \"the\" function $1/x$, as a function depends on its domain. $1/x$ defined on $(0, \\infty)$, $1/x$ defined on $(-\\infty, 0)$ and $1/x$ defined on $(0, \\infty) \\cup (-\\infty, 0)$ are three different functions.\n\nThe antiderivative of $1/x$ defined on $(0, \\infty) \\cup (-\\infty, 0)$ is $ln|x| + C_1 \\ (x < 0)$; $ln|x| + C_2 \\ (x > 0)$. Where $C_1$ and $C_2$ are arbitrary constants. This is the full class of functions which, when differentiated, give $1/x$.\n\nThe two separate functions $1/x$ defined on $(0, \\infty)$, and $(-\\infty, 0)$ both have the antiderivate $ln|x| + C$, where $C$ is an arbitrary constant, defined on the appropriate interval.\n\nThere is, therefore, a subtle difference.\n\n13. Nov 4, 2015\n\n### HallsofIvy\n\nLook at f(x)= ln(|x|)+ 9 for x> 0 and ln(|x|)+ 4 for x< 0. That function is differentiable for all non-zero x and its derivative is 1/x.\n\n14. Nov 4, 2015\n\n### Cosmophile\n\nYour second point was very well said, and is certainly something I'll carry with me. So unless a particular domain is specified, such as $(0, \\infty)$, I have to include the $\\ln|x| +C_1 \\quad (x < 0); \\quad \\ln |x| + C_2 \\quad (x > 0)$.\n\nI suppose my issue is coming from the fact that, when I see $f(x) = \\frac {1}{x}$, I imagine the standard hyperbola $f(x) = \\frac {1}{x}$ defined on $(- \\infty, 0) \\cup (0, \\infty)$. Now, when I imagine the function $f(x) = \\ln x$, I think automatically of this graph:\n\nhttp://www.wolframalpha.com/share/img?i=d41d8cd98f00b204e9800998ecf8427eq1tuvuvmvh&f=HBQTQYZYGY4TMNJQMQ2WEZTBGUYDCNJQMQ3DAODGGAZTAMBQGI4Qaaaa But I've also seen this graph:\n\nWhich should I be thinking of for this problem?\n\nLast edited: Nov 5, 2015\n15. Nov 4, 2015\n\n### Staff: Mentor\n\nThe second graph looks like it is probably f(x) = ln|x|.\n\n16. Nov 5, 2015\n\n### Cosmophile\n\nYou're right. I wonder why Wolfram interprets it that way, but only uses the right-hand side of the graph when I set it to show real values.\n\n17. Nov 5, 2015\n\n### Staff: Mentor\n\nWhat was the equation you graphed? If you entered the integral in post #1 of this thread, the antiderivative is ln|x| (plus the constant).\n\n18. Nov 5, 2015\n\n### Cosmophile\n\nI graphed $f(x) = \\frac {1}{x}$ and got the funky results. Graphing ln|x| gave me what I needed. But if I'm given $f(x) = \\frac {1}{x}$, with no specification of the domain, I end up with $F(x) = \\ln |x| + C_1, \\quad (x < 0); \\ln|x| + C_2, \\quad (x > 0).$\n\nWhat's weird is that when I look at the graph of ln|x|, the two halfs seem connected (obviously, they aren't - there's a discontinuity at $x=0$), so it's hard for me to imagine why the two sides need different constants of integration. That would mean the two bits move around independently, and I can't at all think of an example of that if $f(x) = \\frac {1}{x}$ with nothing else added.\n\nI hope that makes some sense. I'd love to get some graphical insights - I think that would help me out a bit. Sorry for asking so much on what seems to be a fairly simple topic!\n\n(Also, thanks for chiming in, Mark44. I can always count on you and Micromass to respond).\n\n19. Nov 5, 2015\n\n### PeroK\n\nI suggest that you are getting yourself a bit confused over this. And, I guess this is why your maths instructor is saying it's not useful. There is nothing special about $1/x$. Any function $1/x^n$ has the same issue. Wolfram Alpha, for example, doesn't get into this subtlety of having different constants of integration. The same with tan(x) - you could have a different constant of integration on each interval upon which it's defined.\n\nThe key point is this. Suppose I have a differential equation:\n\n$f'(x) = x^2$ and $f(1) = 0$ $(-\\infty < x < \\infty)$\n\nThis specifies a unique function: $f(x) = \\frac{1}{3}(x^3 - 1)$\n\nBut, if we have:\n\n$f'(x) = 1/x$ and $f(1) = 0$ $(x \\ne 0)$\n\nThis does not specify a unique function. We also need a value for some $x < 0$. For example:\n\n$f'(x) = 1/x$, $f(1) = 0$ $f(-1) = 1$ $(x \\ne 0)$\n\nThe differential equation, therefore, splits into two:\n\nFor $x > 0$ we have $f(x) = ln|x| + C$, $f(1) = C = 0$, hence $f(x) = ln|x| (x > 0)$\n\nFor $x < 0$ we have $f(x) = ln|x| + C$, $f(-1) = C = 1$, hence $f(x) = ln|x| + 1 (x < 0)$\n\nNow, of course, you could use $C_1$ and $C_2$ instead of reusing the symbol $C$ as I have done. But, it hardly matters. The important point is that the integration and differentiation for a function like $1/x$ must be done separately on each contiguous interval on which it's defined.\n\nOne final point. You can have the same situation with any function. Suppose we have:\n\n$f'(x) = x \\ (-2 < x < -1$ and $1 < x < 2)$\n\nThen we are specifying a differential equation across two discontiguous intervals, hence we need two initial values, as there is a different constant of integration on each interval.\n\n20. Nov 5, 2015\n\n### micromass\n\nAbout the wolfram issue. Wolfram alpha is a very good software but not infallible. In particular, here it is (strictly speaking) wrong.\n\nWhen asked how to integrate $1/x$, wofram alpha somehow involves complex numbers. Furthermore, it will only show one particular primitive.\nSo when you ask for the primitive of $1/x$, wolfram accurately calculates the following primitive: $f(x) = \\text{ln}(x)$ for $x>0$ and $f(x) = \\text{ln}(x) + \\pi i$ for $x<0$. (The explanation for why $\\pi i$ is involved requires complex analysis). Anyway, this is a correct primitive as you can differentiate it to get $1/x$. Wolfram alpha is wrong however to say that any other primitive is of the form $f(x) + C$. Rather, it is of the form $f(x) + C_1$ for $x>0$ and $f(x) + C_2$ for $x<0$. In particular, we can choose $C_1 = 0$ and $C_2 = -\\pi i$, to get $\\text{ln}|x|$.\nSo if you want only the real values, then wolframalpha works with the same primitive $f$ that has already been computed. But this primitive is complex for $x<0$ and thus wolfram ignores it. So that is why it only shows $f(x)$ for $x>0$. There is a primitive that takes only real values (namely $\\text{ln}|x|$), but wolfram does not compute it and erroneously states that the real primitive only exists for $x>0$.", "date": "2018-02-20 00:06:44", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/antiderivative-of-1-x.841263/", "openwebmath_score": 0.9127521514892578, "openwebmath_perplexity": 344.04818064506895, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981735721648143, "lm_q2_score": 0.9124361586911173, "lm_q1q2_score": 0.8957711707104836}}
{"url": "http://mathhelpforum.com/trigonometry/88387-bearings-question-trig.html", "text": "# Thread: Bearings question with trig.\n\n1. ## Bearings question with trig.\n\nHey Im really not sure how to do bearings at all.\nFor homework i have this question:\n\nA ship leaves at port A and travels for 30km on bearing of 120degrees\nIt then changes course and travels for 50km on bearing of 080degrees arriving at port B.\nCalculate distance AB and bearing A from B\n\nthanks\n\n2. Originally Posted by mitchoboy\n...\nA ship leaves at port A and travels for 30km on bearing of 120degrees\nIt then changes course and travels for 50km on bearing of 080degrees arriving at port B.\nCalculate distance AB and bearing A from B\n...\n\nTypically bearings are given from a reference [North or South] and deflecting East or West.\n\nFrom the information given, assume the reference for zero bearing is due North or the y-axis; and assume that port A is at the origin (0,0).\n\n$X_0 = 0$\n$Y_0 = 0$\n\n$X_1 = X_0 + \\sin \\left(120\\right) \\times\\ 30 = 0.866 \\times 30 = 25.981$\n$Y_1 = Y_0 + \\cos \\left(120\\right) \\times\\ 30 = 0.500 \\times 30 = 15.000$\n\n$X_2 = X_1 + \\sin\\left( 80\\right) \\times\\ 50 = X_1 + 0.985 \\times 50 = X_1 + 49.240 = 75.221$\n$Y_2 = Y_1 + \\cos \\left(80\\right) \\times\\ 50 = Y_1 + 0.174 \\times 50 = Y_1 + 8.682 = 23.682$\n\nSince $X_0 = 0$ & $Y_0 = 0$ The distance AB is $\\sqrt{X_2^2 + Y_2^2}$\n\nThe bearing is the arctangent of the difference between final coordinates and the initial coordinates:\n\nThe tangent of the bearing AB is : $\\frac{X_2 - X_0} {Y_2 - Y_0}$\n\nAs a check:\n\n$X_2 = \\sin \\left ({bearing AB}\\right) \\times \\left ({distance AB}\\right)$\n\n$Y_2 = \\cos \\left ({bearing AB}\\right) \\times \\left({distance AB}\\right)$\n\n3. Just for kicks, let's do it this way. Let's use a coordinate system so you can see what is going on. Like in surveying.\n\nLet's say the coordinate of A is (0,0).\n\nThen, to the turning point, it is 30 km at an azimuth of 120 degrees.\n\n(Technically, this is an azimuth, not a bearing. But, it doesn't really matter).\n\n$x=30sin(120)=25.98$\n\n$y=30cos(120)=-15$\n\nThose are the coordinates of the turning point, (25.98,-15)\n\nNext, the boat turns 80 degrees from north and goes 50 km to go to B.\n\nB's coordinates are $x=25.98+50sin(80)=75.22$\n\n$y=-15+50cos(80)=-6.32$\n\nThe coordinates of B are (75.22, -6.32).\n\nNow, to find the distance back to A where it started, just use ol' Pythagoras.\n\n$\\sqrt{(75.22)^{2}+(-6.32)^{2}}=75.485$\n\nTo find the bearing back to A, one way of many:\n\n$270+cos^{-1}(\\frac{75.22}{75.485})=274.8 \\;\\ deg$\n\nHere is a diagram so you can see. It is rather sloppy done in paint, but I hope it will suffice.\n\n4. Hello, mitchoboy!\n\nBearings are measured clockwise from North.\nAnd a good diagram is essential.\n\nA ship leaves at port $A$ and travels for 30 km on bearing of 120\u00b0.\nIt then changes course and travels for 50 km on bearing of 080\u00b0 arriving at port $B.$\nCalculate distance $AB$ and bearing of ${\\color{blue}A}$ from ${\\color{blue}B}.$ .\nIs this correct?\nCode:\nN\n|\n|\nA o                     R\n| *     *             |\n|60\u00b0*           *     |\n|     *       Q       o B\n|    30 *     |     *\nS         *60\u00b0|80\u00b0* 50\n* | *\no\nP\n\nThe ship starts at A and sails 30 km to point $P$:\n. . $AP = 30,\\angle NAP = 120^o,\\;\\angle SAP = \\angle APQ = 60^o$\n\nThen it turns and sails 50 km to point $B$:\n. . $PB = 50,\\;\\angle QPB = 80^o \\quad\\Rightarrow\\quad \\angle APB = 140^o$\n\nIn $\\Delta APB$, use the Law of Cosines:\n\n. . $AB^2 \\:=\\:AP^2 + PB^2 - 2(AP)(BP)\\cos(\\angle APB)$\n\n. . $AB^2 \\:=\\:30^2+50^2 - 2(30)(50)\\cos140^o \\:=\\:5698.133329$\n\nTherefore: . $\\boxed{AB \\;\\approx\\;75.5\\text{ km}}$\n\nIn $\\Delta APB$, use the Law of Cosines.\n\n. . $\\cos A \\:=\\:\\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \\:=\\:0.90413245$\n\nHence: . $\\angle A \\;\\approx\\;25.2^o$\n\nThen: . $\\angle BAS \\:=\\:25.2^o + 60^o \\:=\\:85.2^o \\:=\\:\\angle ABR$\n\nTherefore, the bearing of $A$ from $B$ is: . $360^o - 85.2^o \\:=\\:\\boxed{274.8^o}$\n\n5. Originally Posted by Soroban\nHello, mitchoboy!\n\nBearings are measured clockwise from North.\nAnd a good diagram is essential.\n\nCode:\nN\n|\n|\nA o                     R\n| *     *             |\n|60\u00b0*           *     |\n|     *       Q       o B\n|    30 *     |     *\nS         *60\u00b0|80\u00b0* 50\n* | *\no\nP\nThe ship starts at A and sails 30 km to point $P$:\n. . $AP = 30,\\angle NAP = 120^o,\\;\\angle SAP = \\angle APQ = 60^o$\n\nThen it turns and sails 50 km to point $B$:\n. . $PB = 50,\\;\\angle QPB = 80^o \\quad\\Rightarrow\\quad \\angle APB = 140^o$\n\nIn $\\Delta APB$, use the Law of Cosines:\n\n. . $AB^2 \\:=\\:AP^2 + PB^2 - 2(AP)(BP)\\cos(\\angle APB)$\n\n. . $AB^2 \\:=\\:30^2+50^2 - 2(30)(50)\\cos140^o \\:=\\:5698.133329$\n\nTherefore: . $\\boxed{AB \\;\\approx\\;75.5\\text{ km}}$\n\nIn $\\Delta APB$, use the Law of Cosines.\n\n. . $\\cos A \\:=\\:\\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \\:=\\:0.90413245$\n\nHence: . $\\angle A \\;\\approx\\;25.2^o$\n\nThen: . $\\angle BAS \\:=\\:25.2^o + 60^o \\:=\\:85.2^o \\:=\\:\\angle ABR$\n\nTherefore, the bearing of $A$ from $B$ is: . $360^o - 85.2^o \\:=\\:\\boxed{274.8^o}$\nthankyou for your answer. but how did you get apq as 60 degrees?", "date": "2017-02-25 19:42:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/88387-bearings-question-trig.html", "openwebmath_score": 0.8599735498428345, "openwebmath_perplexity": 2792.661359215432, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850832642353, "lm_q2_score": 0.9111797033789887, "lm_q1q2_score": 0.8957671745650143}}
{"url": "https://netposition-international.com/50jfrs/1e0451-diagonal-matrix-and-scalar-matrix", "text": "Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Program to print a matrix in Diagonal Pattern. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. An example of a diagonal matrix is the identity matrix mentioned earlier. A matrix with all entries zero is called a zero matrix. MMAX(M). General Description. Takes a single argument. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Example 2 - STATING AND. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors The data type of a[1] is String. 8. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. b ij = 0, when i \u2260 j skalare Matrix, f rus. Negative: \u2212A is de\ufb01ned as (\u22121)A. Subtraction: A\u2212B is de\ufb01ned as A+(\u2212B). is a diagonal matrix with diagonal entries equal to the eigenvalues of A. A square matrix in which all the elements below the diagonal are zero i.e. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Filling diagonal to make the sum of every row, column and diagonal equal of 3\u00d73 matrix using c++ The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (\u2212) additions for computing the product of two square n\u00d7n matrices. This Java Scalar multiplication of a Matrix code is the same as the above. Synonyms for scalar matrix in Free Thesaurus. Diagonal matrix multiplication, assuming conformability, is commutative. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Yes it is, only the diagonal entries are going to change, if at all. Extract elements of matrix. import java. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \\text{ to } x_{nn}$$. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Example sentences with \"scalar matrix\", translation memory. Maximum element in a matrix. a matrix of type: Lower triangular matrix. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Types of matrices \u2014 triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title \"Prince of Wales\" originally claimed for the English crown prince via a trick? Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. In this post, we are going to discuss these points. \"Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. 8 (Roots are found analogously.) What is the matrix? Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! InnerProducts. scalar matrix skaliarin\u0117 matrica statusas T sritis fizika atitikmenys : angl. Upper triangular matrix. The values of an identity matrix are known. Use these charts as a guide to what you can bench for a maximum of one rep. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Define scalar matrix. Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (\u2198) entries are all equal. \u0441\u043a\u0430\u043b\u044f\u0440\u043d\u0430\u044f \u043c\u0430\u0442\u0440\u0438\u0446\u0430, f pranc. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 2. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. Magnet Matrix Calculator. Matrix is an important topic in mathematics. This matrix is typically (but not necessarily) full. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Diagonal elements, specified as a matrix. stemming. 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. What are synonyms for scalar matrix? Great code. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. How to convert diagonal elements of a matrix in R into missing values? Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. scalar matrix vok. \u2014 Page 36, Deep Learning, 2016. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. a diagonal matrix in which all of the diagonal elements are equal. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. This behavior occurs even if \u2026 See : Java program to check for Diagonal Matrix. [x + 2 0 y \u2212 3 4 ] = [4 0 0 4 ] GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox\u2122. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. A symmetric matrix is a matrix where aij = aji. Yes it is. Minimum element in a matrix\u2026 For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. add example. Java Scalar Matrix Multiplication Program example 2. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Antonyms for scalar matrix. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. 6) Scalar Matrix. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n \u00d7 n is said to be a scalar matrix if. matrice scalaire, f Fizikos termin\u0173 \u017eodynas : lietuvi\u0173, angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. scalar meson Look at other dictionaries: Matrix - \u043f\u043e\u043b\u0443\u0447\u0438\u0442\u044c \u043d\u0430 \u0410\u043a\u0430\u0434\u0435\u043c\u0438\u043a\u0435 \u0440\u0430\u0431\u043e\u0447\u0438\u0439 \u043a\u0443\u043f\u043e\u043d \u043d\u0430 \u0441\u043a\u0438\u0434\u043a\u0443 \u041b\u0435\u0442\u0443\u0430\u043b\u044c \u0438\u043b\u0438 \u0432\u044b\u0433\u043e\u0434\u043d\u043e 9. Are zero i.e matrix skaliarin\u0117 matrica statusas T sritis fizika atitikmenys: angl data type of by. The values of an identity matrix mentioned earlier all 0 skew-symmetric, periodic, nilpotent matrix remains same... Code is the same the values of an identity matrix scalar matrix: diagonal matrix R... 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With all entries of a matrix where aij = aji returns a scalar matrix translation English. An example of a matrix code diagonal matrix and scalar matrix the identity matrix, unit matrix dictionary definition scalar! This post, we are going to change, if at all and! Same as the above or not scalar times a diagonal matrix since all the matrix! The values of an identity matrix are known is multiplied by an matrix! \u2212B ) any vector when we multiply that vector by that matrix graphics processing unit ( gpu ) Parallel! Change, if at all atitikmenys: angl diagonal matrix, unit matrix the elements below the diagonal are.", "date": "2021-05-08 12:33:54", "meta": {"domain": "netposition-international.com", "url": "https://netposition-international.com/50jfrs/1e0451-diagonal-matrix-and-scalar-matrix", "openwebmath_score": 0.7704491019248962, "openwebmath_perplexity": 769.1851159820291, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9901401426598159, "lm_q2_score": 0.9046505415276079, "lm_q1q2_score": 0.8957308162454254}}
{"url": "http://math.stackexchange.com/questions/32095/how-can-i-can-solve-integrals-of-rational-functions-of-polynomials-in-x", "text": "# How can I can solve integrals of rational functions of polynomials in $x$?\n\nI would like to know a way to solve integrals such as this one: $$\\int \\frac{x}{3x - 4}dx$$\n\nAlso, I assume similar integrals where x is squared are solved in a similar manner. (If the answer is yes then don't also show me how to solve this second one, I want to see if I can do it myself. :) ) $$\\int \\frac{x^2}{x^2 - 1}dx$$\n\n(I haven't included the conditions that x must meet in order for those to be valid expressions.)\n\n-\nHave you tried long division with polynomials and/or partial fractions? \u2013\u00a0 Eugene Bulkin Apr 10 '11 at 16:09\nFor the first, the idea is to split it so that you have the sum of a constant and an appropriate multiple of $\\frac1{3x-4}$ (which is easily integrated for you, I hope ;)). \u2013\u00a0 \uff2a. \uff2d. Apr 10 '11 at 16:11\nNo, I don't know how, please give me an example. If you want, show me how to solve a similar one and leave this one to me. \u2013\u00a0 Paul Manta Apr 10 '11 at 16:11\n@J.M. Yes, thank you. So the second one can be written as $$\\int (\\frac{1}{x^2 - 1} + 1)dx$$. \u2013\u00a0 Paul Manta Apr 10 '11 at 16:17\n$$\\frac{ax+b}{cx+d}=\\frac{a}{c}+\\left(b-\\frac{da}{c}\\right)\\frac1{cx+d}$$ should be most helpful. \u2013\u00a0 \uff2a. \uff2d. Apr 10 '11 at 16:24\n\nWhenever you are trying to integrate a rational function, the first step is to do the division so that the numerator is of degree strictly smaller than the numerator (this is what Eugene Bulkin and J.M. are saying in the comments). For example, for $$\\int \\frac{x}{3x-4}\\,dx$$ you should do the division of $x$ by $3x-4$ with remainder. This is $$x = \\frac{1}{3}(3x-4) + \\frac{4}{3}$$ which means that $$\\frac{x}{3x-4} = \\frac{1}{3} + \\frac{4/3}{3x-4}.$$ So the integral can be rewritten as $$\\int \\frac{x}{3x-4}\\,dx = \\int\\left(\\frac{1}{3} + \\frac{4/3}{3x-4}\\right)\\,dx = \\int\\frac{1}{3}\\,dx + \\frac{4}{3}\\int \\frac{1}{3x-4}\\,dx.$$\n\nThe first integral is immediate. The second integral yields to a change of variable $u=3x-4$. We get \\begin{align*} \\int\\frac{x}{3x-4}\\,dx &= \\int\\frac{1}{3}\\,dx + \\frac{4}{3}\\int\\frac{1}{3x-4}\\,dx\\\\ &= \\frac{1}{3}x + \\frac{4}{9}\\int\\frac{du}{u}\\\\ &= \\frac{1}{3}x + \\frac{4}{9}\\ln|u| + C\\\\ &= \\frac{1}{3}x + \\frac{4}{9}\\ln|3x-4| + C. \\end{align*} In general, if you have a denominator of degree $1$, by doing the long division you can always express it as a polynomial plus a rational function of the form $$\\frac{k}{ax+b}$$ with $k$, $a$, and $b$ constants. The polynomial is easy to integrate, and the fraction can be integrated with a change of variable.\n\nThe same is true for your second integral. Doing the long division gives, as you note, that $$\\int \\frac{x^2}{x^2-1}\\,dx = \\int\\left(1 + \\frac{1}{x^2-1}\\right)\\,dx = \\int\\,dx + \\int\\frac{1}{x^2-1}\\,dx.$$ The first integral is easy. The second is as well, using partial fractions: $$\\frac{1}{x^2-1} = \\frac{1}{(x+1)(x-1)} = \\frac{1/2}{x-1} - \\frac{1/2}{x+1}$$ so: $$\\int\\frac{1}{x^2-1}\\,dx = \\frac{1}{2}\\int\\frac{dx}{x-1} - \\frac{1}{2}\\int\\frac{dx}{x+1} = \\frac{1}{2}\\ln|x-1| - \\frac{1}{2}\\ln|x+1|+C.$$\n\n-\nI may be off school, but I'd love to read a textbook written by Arturo... \u2013\u00a0 \uff2a. \uff2d. Apr 10 '11 at 18:39\n+1..Very very good explanation. \u2013\u00a0 night owl Apr 17 '11 at 5:37\n\nFor your two examples my first lines would be\n\n$$\\int \\frac{x}{3x-4} \\text{d}x = \\int \\frac{ \\frac{1}{3}(3x-4) +4/3}{3x-4} \\text{d}x$$\n\nand\n\n$$\\int \\frac{x^2}{x^2-1} \\text{d}x = \\int \\frac{ (x^2-1) + 1}{x^2-1} \\text{d}x .$$\n\nYou can easily extend this technique, for example\n\n$$\\int \\frac{x^3}{x^2-1} \\text{d}x = \\int \\frac{ x(x^2-1) + x}{x^2-1} \\text{d}x .$$\n\n-", "date": "2014-03-10 08:17:20", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/32095/how-can-i-can-solve-integrals-of-rational-functions-of-polynomials-in-x", "openwebmath_score": 0.99144446849823, "openwebmath_perplexity": 327.8362674036093, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138157595305, "lm_q2_score": 0.9161096181702031, "lm_q1q2_score": 0.8956930304351958}}
{"url": "http://gnpaperjenq.blinkingti.me/writing-piecewise-functions.html", "text": "# Writing piecewise functions\n\nSteps to writing piecewise functions to find the equation of the line 1 find two points on the line 2 find slope 3 use point and slope in point slope form and. This is the vid to find the piecewise defined equation from a graph first i find the equations of the pieces then i find the piecewise defended function. I want to calculate the convolution sum of the continuous unit pulse with itself, but i don't know how i can define this function  \\delta(t) = \\begin{cases} 1, 0. Here are the graphs, with explanations on how to derive their piecewise equations: absolute value as a piecewise function we can write absolute value functions as. Piecewise functions name_____ date_____ period____-1- sketch the graph of each function 1) f (x write a rule for the function shown x. Yes, piecewise functions isn\u00e2\u20ac\u2122t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong.\n\nPiecewise functions lesson plans and worksheets from thousands of teacher-reviewed resources to help you inspire students learning. Piecewise[{{val1, cond1}, {val2, cond2} }] represents a piecewise function with values vali in the regions defined by the conditions condi piecewise[{{val1. Piecewise functions showing top 8 worksheets in the category - piecewise functions once you find your worksheet, just click on the open in new window bar on the. Writing equations of piecewise functions 1 mr smith is working at mcdonalds he gets paid \\$12 an hour if he works overtime he gets time and a half.\n\nMatch the piecewise function with its graph write the answer next to the problem number graph the function 19 20. Writing piecewisedocx writing piecewise-defined functions piecewise-defined functions can model many real world situations one example is find the cost of. Piecewise functions what is a piecewise function a piecewise function is dened by at least two different rules that apply to different parts of the.\n\n\u2022 Graphing and writing equations for piecewise functions unit 2: piecewise functions lesson 3 of 12 objective lesson 6: write piecewise functions to match graphs.\n\u2022 Section 47 piecewise functions 219 graphing and writing piecewise functions graphing a piecewise function graph y = { \u2212 x \u2212 4, x, if x 0 describe the domain.\n\u2022 Page 1 of 2 116 chapter 2 linear equations and functions using piecewise functions in real life using a step function awrite and graph a piecewise function for the.\n\nHow to write a piecewise function from a given graph. Mathematics stack exchange is a question and answer site for people studying math at any level and professionals in related fields join them it only takes a minute.\n\nWriting piecewise functions\nRated 3/5 based on 10 review", "date": "2018-06-25 15:38:53", "meta": {"domain": "blinkingti.me", "url": "http://gnpaperjenq.blinkingti.me/writing-piecewise-functions.html", "openwebmath_score": 0.35177916288375854, "openwebmath_perplexity": 1138.781222556325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9777138118632621, "lm_q2_score": 0.9161096181702031, "lm_q1q2_score": 0.8956930268657869}}
{"url": "https://math.stackexchange.com/questions/3144423/side-length-of-the-smallest-square-that-can-be-dissected-into-n-squares-with-i", "text": "# Side length of the smallest square that can be dissected into $n$ squares with integer sides\n\nLet $$s_n$$ be the shortest possible side length of a square constructed from exactly $$n$$ squares of positive integer side lengths. If no such square exists, let $$s_n = 0$$.\n\nThe first few values are as follows:\n\n n | s(n)\n---+------\n1 |  1\n2 |  0\n3 |  0\n4 |  2\n5 |  0\n6 |  3\n7 |  4\n8 |  4\n9 |  3\n10 |  4\n11 |  5\n12 |  6\n13 |  4\n14 |  5\n\n\nIf we search this Integer Sequence in an Online Encyclopedia, something very remarkable happens: there is exactly one search hit. That sequence is A300001, or in English, \"Side length of the smallest equilateral triangle that can be dissected into n equilateral triangles with integer sides, or 0 if no such triangle exists.\"\n\nDo my square sequence's values agree with the triangle sequence's values?\n\nIf so, why? If not, when do they first disagree?\n\nAt first I thought, maybe there's some manner of bijection between my square dissections and the triangular dissections: if you halve each subsquare along its diagonals, numerically the result should fit in the entire square halved along its own diagonal. But fitting them together requires some nonobvious geometrical fiddling, and I'm not at all confident this subobject-size-preserving bijection is well defined over all dissections. Is it?\n\n\u2022 You show results only up to $14$ as the \"first few values\", implying you have more available. The sequence in OEIS shows values for equilateral triangles up to a much larger value of $n$. Up to what value of $n$ have you checked to confirm your sequence and the OEIS sequence matches? \u2013\u00a0John Omielan Mar 12 at 1:35\n\u2022 Great question! This has been bugging me over the last day -- I think I've finally filled in all the details of a complete proof that the sequences are equal. Let me know if you have any questions about my answer. \u2013\u00a06005 Mar 13 at 9:38\n\u2022 Overall it was a lot of work even when the proof idea was already clear. Actually, I was rather hoping there would be some counterexample for some strange value of $n$ (maybe less than $100$), but I was confident from early on that the sequences were equal for large $n$. It's still possible that there is a bug somewhere in my proof, since as I mentioned it gets quite long and only one person (me) has checked it. \u2013\u00a06005 Mar 13 at 9:39\n\nDo my square sequence's values agree with the triangle sequence's values?\n\nYes, the two sequences are the same. (Ross Millikan has the right idea.) In fact, we will prove an explicit formula for both sequences.\n\nLet $$s(n)$$ be the shortest possible side length of a square made from $$n$$ squares with integer side length; and let $$t(n)$$ be the shortest possible side length of a triangle made from $$n$$ triangles with integer side length. (And $$s(n) = 0$$ or $$t(n) = 0$$ if no such thing exists.)\n\nClaim: For all $$n$$, let $$k^2$$ be the least perfect square greater than or equal to $$n$$ -- i.e., $$k = \\lceil \\sqrt{n} \\rceil$$. Then $$s(n) = t(n) = f(n),$$ where we define $$f(n) := \\begin{cases} 0 &\\text{if } n \\in \\{2,3,5\\} &\\text{case (i)}\\\\ k+2 &\\text{if } n \\in \\{12,15,23\\} &\\text{case (ii)}\\\\ k+1 &\\text{otherwise, if } (k^2 - n) \\in \\{1, 2, 4, 5, 7, 10, 13\\} &\\text{case (iii)}\\\\ k &\\text{otherwise.} &\\text{case (iv)} \\end{cases}$$\n\nProof strategy: First, we will show that $$f(n)$$ is a lower bound on both $$t(n)$$ and $$s(n)$$, by showing that no set of $$n$$ perfect-square areas can add up to a perfect-square area of size any less than $$f(n)$$. Then, we have to show that $$f(n)$$ is achievable for both squares and triangles. This is the hard part. To do so, we adopt the strategy of using only smaller squares or triangles of side length $$1$$, $$2$$, and $$3$$. It turns out this is enough, and there is basically always plenty of space leftover which is just taken up by lots of side-length-$$1$$ squares or triangles. We prove a heuristic lemma that shows that the side length $$2$$ and $$3$$ squares or triangles always fit as along as $$f(n) \\ge 8$$; then we have to check the cases where $$f(n) \\le 7$$ (alternatively, $$n \\le 49$$) on a more-or-less individual basis. Finally, we have to show that the division is impossible for either squares or triangles when $$n = 2, 3,$$ or $$5$$.\n\nPart 1: $$f(n)$$ is a lower bound on $$s(n)$$ and $$t(n)$$\n\nNote that $$s(n) \\ge k$$ and $$t(n) \\ge k$$, because we need at least a $$k \\times k$$ square just to fit $$n$$ $$1 \\times 1$$ squares (by area). Similarly for triangles. (In particular, $$s(n)^2 \\ge n$$ and $$t(n)^2 \\ge n$$ for all $$n$$, unless they are zero.) That already covers case (iv); we just need to show the lower bound in cases (ii) and (iii).\n\nNow if we define $$r = s(n)$$ or $$r = t(n)$$, then $$r^2$$ is the area of the larger square (or triangle) in units of side-length $$1$$ squares (or triangles). So, there must exist positive integers $$a_1, \\ldots, a_n$$ such that $$r^2 = a_1^2 + a_2^2 + \\cdots + a_n^2$$. Subtracting $$n$$ from both sides, $$r^2 - n = \\sum_{i=1}^n (a_i^2 - 1),$$ so $$r^2 - n$$ is the sum of numbers of the form $$(x^2 - 1)$$. Such numbers are $$0, 3, 8, 15, \\ldots$$. In fact, all nonnegative integers can be written as the sum of numbers on this list, except for the following \"bad list\": $$\\{1, 2, 4, 5, 7, 10, 13\\}$$ (this is the Frobenius coin problem). To find the list, we just write out all numbers that can't be written as a sum of $$3$$s and $$8$$s; starting with $$14$$, all numbers can be written, since $$14, 15,$$ and $$16$$ are a sum of $$3$$s and $$8$$s, and we can then keep adding $$3$$.\n\nTherefore, $$r^2 - n$$ must not be on the bad list of numbers, $$1, 2, 4, 5, 7, 10,$$ or $$13$$. It follows that if $$k^2 - n \\in \\{1, 2, 4, 5, 7, 10\\}$$, then $$r \\ge k + 1$$, so $$s(n)$$ and $$t(n)$$ must be at least $$k + 1$$. This proves that $$f(n)$$ is a lower bound in case (iii).\n\nIn fact, this also proves the lower bound on $$s(n)$$ and $$t(n)$$ in case (ii), because for $$n = 12,$$ $$15$$, or $$23$$, we can see that both $$k^2 - n$$ and $$(k+1)^2 - n$$ are on the bad list of numbers. In particular, $$16 - 12 = 4$$ and $$25 - 12 = 13$$; $$16 - 15 = 1$$ and $$25 - 15 = 10$$; $$25 - 23 = 2$$ and $$36 - 23 = 13$$.\n\nPart 2: Achievability -- $$s(n) = t(n) = f(n)$$ for cases (ii), (iii), and (iv)\n\nWe need to show that for $$n \\ne 2, 3, 5$$, it is possible to divide a square (or triangle) of side length $$n$$ into $$f(n)$$ squares (or triangles) of integer side length. Let $$r = f(n)$$. We will use some basic bounds on $$n, k,$$ and $$r$$: $$k \\le r \\le k + 2$$, and $$(k-1)^2 < n \\le k^2$$.\n\nBy definition of $$r = f(n)$$ (see part 1), $$r^2 - n$$ is not on the bad list of numbers ($$1, 2, 4, 5, 7, 10, 13$$). Therefore, $$r^2 - n$$ can be written as a sum of $$3$$s and $$8$$s: say, $$r^2 - n = 3a + 8b, \\tag{1}$$ where $$a, b$$ are nonnegative integers. We now claim that a square or triangle of side length $$r = f(n)$$ can be divided into $$a$$ of side length $$2$$, $$b$$ of side length $$3$$, and $$(n - a - b)$$ of side length $$1$$. (Thus there are $$n$$ total, and the total area is $$4a + 9b + (n - a - b) = f(n)^2$$.) For this to work, we should also show $$a + b \\le n$$.\n\nWe first prove a heuristic lemma that will be good enough to show that the $$a$$ and $$b$$ squares or triangles fit for $$n$$ sufficiently large.\n\nLemma. Let $$a, b \\ge 0$$ and $$r \\ge 2$$ be integers such that $$4a + 9b \\le (r-2)^2 + 3$$. Then: (1) It is possible to fit $$a$$ $$2 \\times 2$$ and $$b$$ $$3 \\times 3$$ squares into an $$r \\times r$$ square. (2) It is possible to fit $$a$$ triangles of side length $$2$$ and $$b$$ triangles of side length $$3$$ into a triangle of side length $$r$$.\n\nProof of lemma (sadly without pictures to help):\n\n1. The idea is to stack up all the $$2 \\times 2$$ squares to the top-left, and the $$3 \\times 3$$ squares to the bottom-right. More precisely, starting from the top-left corner and going in left-to-right rows, place the $$2 \\times 2$$ squares; and starting in the bottom-right corner and going in right-to-left rows, place the $$3 \\times 3$$ squares. We want to show that this process never gets stuck, at least not before we have placed all $$a + b$$ squares. So suppose we have not placed all the squares, and we get stuck. Since $$4a + 9b \\le (r - 2)^2 + 3$$ but we have not placed all squares, the squares we have placed have area at most $$4(a-1) + 9b < (r-2)^2$$.\n\nThe question is, what does a \"stuck\" position look like? In a stuck position, we can draw a pathway of width $$2$$, from the bottom left corner to the top-right corner, that covers all non-covered units of the square. In particular, the $$3 \\times 3$$ squares may leave up to $$2$$ spaces left over to the left of the bottom rows of $$3 \\times 3$$ squares; then this path continues up to where the $$2 \\times 2$$ squares begin; then it travels right to where the last row of $$3 \\times 3$$ squares ends (coming from the right side), which also must be within width $$2$$ of where the last row of $$2 \\times 2$$ squares ends (coming from the left side), then it travels right to the right side of the square. Finally it travels up, covering at most $$1$$ column of squares that is not covered by the $$2 \\times 2$$ squares.\n\nThus in a stuck position, the entire $$r^2$$ area is covered except at most some squares on this path of width $$2$$. This path covers exactly $$4r - 4$$ squares. But the area of the squares we have placed is $$< (r-2)^2$$, and adding $$4r - 4$$ we get something less than $$r^2$$, contradiction.\n\n2. Now we do the triangle case; it is similar but more confusing. Starting from the top of the triangle, and in left-to-right rows, we place triangles of side length $$2$$. Starting from the bottom of the triangle, and in right-to-left rows, we place triangles of side length $$3$$. We want to show, once again, that this process does not get stuck. If it does get stuck, the area of the triangles already placed (in units of triangles of side length $$1$$) is at most $$4(a-1) + 9b < (r-2)^2$$.\n\nWhat does a stuck position look like for the triangle? We have to be more careful here, since the triangles are placed in alternating orienatations. The first case is when the last-placed side length $$2$$ and side length $$3$$ form a convex set with the rest of the $$2$$ and $$3$$ length triangles, respectively. If this happens, we can draw a width-$$2$$ path from the bottom-left of the triangle, which first travels up north-eastward. Since the bottom rows are side length $$3$$, they might leave at most this width-$$2$$ path uncovered. Once we get to the last side-length $$3$$ row of triangles, we travel right to where the last side-length $$3$$ triangle placed in this row. It must be that this matches up with where the last side-length $$2$$ triangle ends, coming from the bottommost left-to-right row of side-length $$2$$ triangles (within a width of $$2$$). We continue north-east for one or two units, then then to the right. The path ends somewhere on the right side of the triangle.\n\nThe total area of the path is at most $$4r-4$$ for the following reason: if we look at diagonal rows of the triangle running in a north-west to south-east direction, each such row contains at most $$4$$ units of area on the path, and the first two such rows (the ones at the bottom-left of the triangle) contain only $$1$$ unit of area and $$3$$ units of area.\n\nNow we consider when one or both of the last-placed triangles (side length $$2$$ coming from the top-left, or side length $$3$$ coming from the bottom-right) is not convex with the other triangles. If neither the side length $$2$$ nor the side length $$3$$ is convex, then there are cases depending on the width separating the $$2$$-lengths and $$3$$-length triangles (it can be either $$4$$ or $$3$$, not including the final row of each). In either case, the number of units of area on the \"path\" (i.e. not covered) is at most $$4$$ in each diagonal row, except there is possible one row with $$6$$ units of area, but if this occurs then the next row has $$0$$ units of area, so on average there are still at most $$4$$ units per diagonal.\n\nIf one of the side length $$2$$ or side length $$3$$ is convex and the other is not, then we draw out the ways we can be stuck (either placing a side length $$2$$ or a side length $$3$$). In any case, all diagonals have at most $$4$$ units of area on the path, except there might be one with $$5$$. But if this occurs, then the next diagonal after it has only $$1$$ unit of area.\n\nWith careful casework, we can ultimately conclude that in every case, the total area of the \"path\" (units of triangle not covered) is at most $$4r - 4$$.\n\nBut the area covered by triangles so far is less than $$(r-2)^2$$, and adding the area of the path we are less than $$(r-2)^2 + 4r - 4 = r^2$$, which is the area of the side-length-$$r$$ triangle, contradiction.\n\nNow that we have the lemma, we apply it. Assume that $$r = f(n) \\ge 8$$. In particular, $$n$$ is at least $$24$$, so case (ii) does not apply. We have $$r \\le k + 1 \\le (\\sqrt{n} + 1) + 1$$, thus $$n \\ge (r - 2)^2.$$ Then from equation (1), $$3a + 8b = r^2 - n \\le r^2 - (r-2)^2 = 4r - 4.$$ Multiplying by $$\\frac{4}{3}$$, we get $$4a + 9b \\le \\frac{4}{3} (3a + 8b) \\le \\frac{16}{3}(r-1).$$ Finally, we're interested in when this bound is enough to apply the lemma. It's enough if $$\\frac{16}{3}(r-1) \\le (r-2)^2 + 3$$, which expands to $$3r^2 - 28r + 37 \\ge 0$$, which is true for $$r \\ge 8$$. So we apply the lemma and we're done.\n\nWhat remains is the case where $$r = f(n) \\le 7$$. In particular, for such cases, $$n$$ is at most $$7^2 = 49$$. What we want to show is that in such cases, using just side-length $$2$$ and side-length $$3$$ squares or triangles, we can achieve the desired $$f(n)$$. First we make a table of $$f(n)$$: $$\\begin{array}{cc|cc|cc|cc|cc|cc|cc} n & f(n) & n & f(n) & n & f(n) & n & f(n) & n & f(n) & n & f(n) & n & f(n) \\\\ \\hline 1 & 1 & 2 & 0 & 5 & 0 & 10 & 4 & 17 & 5 & 26 & 7 & 37 & 7 \\\\ & & 3 & 0 & 6 & 3 & 11 & 5 & 18 & 6 & 27 & 6 & 38 & 7 \\\\ & & 4 & 2 & 7 & 4 & 12 & 6 & 19 & 5 & 28 & 6 & 39 & 8 \\\\ & & & & 8 & 4 & 13 & 4 & 20 & 6 & 29 & 7 & 40 & 7 \\\\ & & & & 9 & 3 & 14 & 5 & 21 & 6 & 30 & 6 & 41 & 7 \\\\ & & & & & & 15 & 6 & 22 & 5 & 31 & 7 & 42 & 8 \\\\ & & & & & & 16 & 4 & 23 & 7 & 32 & 7 & 43 & 7 \\\\ & & & & & & & & 24 & 6 & 33 & 6 & 44 & 8 \\\\ & & & & & & & & 25 & 5 & 34 & 7 & 45 & 8 \\\\ & & & & & & & & & & 35 & 7 & 46 & 7 \\\\ & & & & & & & & & & 36 & 6 & 47 & 8 \\\\ & & & & & & & & & & & & 48 & 8 \\\\ & & & & & & & & & & & & 49 & 7 \\\\ \\end{array}$$\n\n\u2022 First, consider the trivial cases $$r = 1$$, $$r = 2$$, and $$r = 3$$. $$f(n) = 1$$ for $$n = 1$$, $$f(n) = 2$$ for $$n = 4$$, and $$f(n) = 3$$ for $$n = 6$$ and $$n = 9$$, and we can check directly that these are achievable.\n\n\u2022 For $$f(n) = r = 4$$, the possible $$n$$ are $$7,8,10,13,$$ and $$16$$. Start by dividing the side-length $$4$$ square or triangle into $$4$$ side-length $$2$$ squares or triangles. Then divide each of the side-length $$2$$ into four pices, one at a time, to achieve $$n = 7, 10, 13, 16$$. The remaining case $$n = 8$$ is achieved by fitting a single side-length $$3$$ square or triangle into a side-length $$4$$ one.\n\n\u2022 For $$f(n) = r = 5$$, the possible $$n$$ are $$11, 14, 17, 19, 22, 25$$. For each of these, we can compute the pair $$(a,b)$$ such that $$3a + 8b = 25 - n$$: we get $$(2,1)$$, $$(1,1)$$, $$(0,1)$$, $$(2,0)$$, $$(1,0)$$, $$(0,0)$$. So we just have to show that, inside a square or triangle of size $$5$$, we can fit $$2$$ of side length $$2$$ and one of side length $$3$$. This can be drawn straightforwardly.\n\n\u2022 For $$f(n) = r = 6$$, the possible $$n$$ are $$12, 15, 18, 20, 21, 24, 27, 28, 30, 33, 36$$. For each $$n$$, we compute pairs $$(a,b)$$ such that $$3a + 8b = 36 - n$$. For $$n = 12$$, we get $$(8,0)$$ OR $$(0,3)$$. $$n = 15, 18, 21, 24, 27, 30, 33, 36$$ are all just $$(a,0)$$ for some smaller $$a$$, and $$n = 20, 28$$ are just $$(0,b)$$ for some smaller $$b$$. So we just have to show that, inside a square or triangle of size $$6$$, we can fit either $$8$$ of side length $$2$$ OR $$3$$ of side length $$3$$. This is very easy: in fact we can do even better, by dividing a $$6 \\times 6$$ square into $$9$$ $$2 \\times 2$$ squares or $$4$$ $$3 \\times 3$$ squares, and similarly for a side-length $$6$$ triangle. (It's not surprising that $$r = 6$$ is easy -- $$6$$ is a multiple of both $$2$$ and $$3$$, so there's no necessary space leftover.)\n\n\u2022 Finally, suppose $$f(n) = r = 7$$. Here, the possible $$n$$ are $$23, 26, 29, 31, 32$$, $$34, 35, 37, 38$$, $$40, 41, 43, 46, 49$$. For each $$n$$, again we want pairs $$(a,b)$$ such that $$3a + 8b = 49 - n$$. For $$n = 23$$, we get $$(6,1)$$, which is strictly easier than $$n = 26, 29, 32, 35, 38, 41$$ (which require $$1$$ side-length $$3$$ and fewer than $$6$$ side-length $$2$$), and also strictly easier than $$n = 31, 34, 37, 40, 43, 46, 49$$ (which are the same except they don't require the side-length $$3$$ square or triangle). That's all the possible $$n$$, so we just have to show $$n = 23$$ is achievable. So we fit a $$3 \\times 3$$ square and $$6$$ $$2 \\times 2$$ squares into a $$7 \\times 7$$ square, and similarly for a triangle (neither task is hard).\n\nAfter all this work, we finally conclude that: for all $$n \\ne 2, 3, 5$$, it's possible to make a square of side length $$f(n)$$ using a total of $$n$$ $$1 \\times 1$$, $$2 \\times 2$$, and $$3 \\times 3$$ squares; and it's also possible to make a triangle of side length $$f(n)$$ using a total of $$n$$ triangles of side length $$1$$, $$2$$, or $$3$$. This completes the proof of achievability, and shows that $$s(n) = f(n)$$ and $$t(n) = f(n)$$. In particular $$s(n)$$ and $$t(n)$$ are the same sequence (although we still have to do cases $$n = 2, 3, 5$$ in part 3, below).\n\nPart 3: Division of a square or triangle into $$n$$ squares or triangles is impossible in case (i) -- $$n = 2, 3, 5$$\n\nFirst we have to show that a square cannot be divided into $$2, 3,$$ or $$5$$ squares. This is proven in the answer to this mathSE question. For a triangle, we make a similar argument. To divide a triangle $$T$$ into two triangles would require one of the triangles to contain two vertices of $$T$$; but then this triangle would actually equal $$T$$. To divide $$T$$ into $$3$$ triangles would require each of the $$3$$ triangles to be at one of the vertices of $$T$$; but then there is space in the center of the triangle that is not covered. Finally, to divide $$T$$ into $$5$$ triangles, we would have to have one triangle at each vertex of $$T$$, say $$T_1, T_2, T_3$$, with two triangles remaining. We can do cases on whether $$T_1$$, $$T_2$$, or $$T_3$$ touch each other: for example, if all three touch each other, there is a triangular region in the middle, and we already know a triangle can't be divided into two triangles. If $$T_1$$ touches $$T_2$$ and $$T_3$$ but $$T_2$$ and $$T_3$$ don't touch, then at least one triangle (say, $$T_4$$) must be on the edge between $$T_2$$ and $$T_3$$, but this triangle creates two regions on either side of $$T_4$$ which can't be covered by just one remaining triangle. The cases with even fewer touching are similar.\n\nSo, for either triangles or squares, dividing into $$2, 3,$$ or $$5$$ pieces is impossible. Therefore, $$s(n) = t(n) = 0$$ for $$n = 2, 3,$$ or $$5$$. $$\\square$$\n\nRemark: I have not proven that in general, there is a bijection between divisions of a square into squares with integer side length, and divisions of a triangle into triangles with integer side length. As you note, the fact that $$s(n) = t(n)$$ certainly seems to suggest this. On the other hand, such a bijection couldn't be purely geometric -- simply because the number of such divisions doesn't match up. For example, consider when the side length is $$n = 3$$. Then the square can be divided in $$6$$ ways, and the triangle in only $$5$$ ways.\n\nHowever, it seems plausible that if you just look at the sizes of the squares or triangles, then such a bijection exists: for example, since a square of side length $$5$$ can be divided into a $$3 \\times 3$$, two $$2 \\times 2$$, and eight $$1 \\times 1$$ squares, a triangle can be divided this way as well. But I have no idea how to prove such a fact.\n\n\u2022 At the very least, you've given a proof of a \"bijection\" between a smallest square dissection and a smallest triangle dissection, for each number of squares/triangles n. Technically, that's good enough for my question! \u2013\u00a0algorithmshark Mar 13 at 15:35\n\nFirst, think about the area of the small squares and the large square. If you start with $$n$$ unit squares you have a total area of $$n$$. You can replace some of them with $$2 \\times 2$$ squares and add $$3$$ units each time and replace some others with $$3 \\times 3$$ squares and add $$8$$ units each time. You need to do enough of this to increase $$n$$ to a perfect square.\n\nThe largest number you cannot make out of a sum of $$3$$'s and $$8$$'s is $$13$$ from the coin problem. This explains why $$s(12)=6$$. To make a square of side $$5$$ out of $$12$$ squares you would have to add $$13$$ units of area, but you can't. Once $$n$$ is at least $$35$$ we can always make $$(\\lceil \\sqrt n \\rceil+1)^2$$ by adding $$3$$s and $$8$$s. We might be able to make $$(\\lceil \\sqrt n \\rceil)^2$$.\n\nThe reason that squares and triangles work the same is that the area scales as the square of the side for both, so the argument above works the same. Now we have to argue that once $$n$$ is large enough you have enough freedom from the remaining size $$1$$ squares or triangles that we can always form the large figure we want to. Referring to the example of $$12$$, we can make the area $$36$$ by using $$3\\ 3\\times 3$$ squares and $$9\\ 1 \\times 1$$ squares. Assembling them into a $$6 \\times 6$$ square is easy.\n\nFor a given $$n\\ge 35$$, let $$k=\\lceil \\sqrt n \\rceil.$$ The most we have to increase the area is $$13+2k+1$$, because if $$k^2-n \\gt 13$$ we can make $$k^2$$. I don't have a concise argument that we can fit the larger squares, but there are so few of them it is not hard.\n\n\u2022 Though this contains some interesting ideas, it does not come close to answering the question, and it completely ignores the fact that there is something deeply interesting going on geometrically. \u2013\u00a0Servaes Mar 12 at 0:29\n\u2022 @Servaes: I disagree. I don't believe there is anything interesting going on geometrically. Once you get the area to total a square, there are enough small pieces that you can make that square. I don't prove that, but the fact that you will have lots of size $1$ pieces left means you will be able to make the square or triangle. \u2013\u00a0Ross Millikan Mar 12 at 0:59\n\u2022 (+1) You have the right idea. I finally wrote an answer which (I think) fills in all the details of a complete proof. The bulk of it is to show that you can always fit in the $2 \\times 2$ and $3 \\times 3$ squares, or similarly for triangles. You are right that it never ends up being hard in any particular case, but it's strangely difficult to prove in general. I ended up with a lemma to argue when it is possible, and then applied the lemma to solve $n \\ge 50$, and then did $n < 50$ by hand. The lemma was quite involved on its own. \u2013\u00a06005 Mar 13 at 9:44", "date": "2019-05-22 09:01:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3144423/side-length-of-the-smallest-square-that-can-be-dissected-into-n-squares-with-i", "openwebmath_score": 0.892022430896759, "openwebmath_perplexity": 155.73528915402062, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9929882039959953, "lm_q2_score": 0.9019206745523101, "lm_q1q2_score": 0.895596590770555}}
{"url": "http://yf-sensei.com/sd2n0z/e3b785-diagonal-matrix-and-scalar-matrix", "text": "What are synonyms for scalar matrix? This behavior occurs even if \u2026 Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title \"Prince of Wales\" originally claimed for the English crown prince via a trick? Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. add example. Upper triangular matrix. matrice scalaire, f Fizikos termin\u0173 \u017eodynas : lietuvi\u0173, angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis. Synonyms for scalar matrix in Free Thesaurus. [x + 2 0 y \u2212 3 4 ] = [4 0 0 4 ] Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (\u2212) additions for computing the product of two square n\u00d7n matrices. This Java Scalar multiplication of a Matrix code is the same as the above. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox\u2122. stemming. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . 9. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. Yes it is. Types of matrices \u2014 triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. 8 (Roots are found analogously.) A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Great code. The values of an identity matrix are known. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. A square matrix in which all the elements below the diagonal are zero i.e. Diagonal matrix multiplication, assuming conformability, is commutative. Java Scalar Matrix Multiplication Program example 2. See : Java program to check for Diagonal Matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? is a diagonal matrix with diagonal entries equal to the eigenvalues of A. scalar matrix skaliarin\u0117 matrica statusas T sritis fizika atitikmenys : angl. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors Minimum element in a matrix\u2026 Define scalar matrix. import java. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Takes a single argument. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. 2. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. Example sentences with \"scalar matrix\", translation memory. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. 6) Scalar Matrix. Example 2 - STATING AND. A symmetric matrix is a matrix where aij = aji. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n \u00d7 n is said to be a scalar matrix if. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \\text{ to } x_{nn}$$. \u0441\u043a\u0430\u043b\u044f\u0440\u043d\u0430\u044f \u043c\u0430\u0442\u0440\u0438\u0446\u0430, f pranc. Extract elements of matrix. Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (\u2198) entries are all equal. Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Program to print a matrix in Diagonal Pattern. 8. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Magnet Matrix Calculator. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. A matrix with all entries zero is called a zero matrix. Matrix is an important topic in mathematics. How to convert diagonal elements of a matrix in R into missing values? What is the matrix? For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? a matrix of type: Lower triangular matrix. Filling diagonal to make the sum of every row, column and diagonal equal of 3\u00d73 matrix using c++ All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. The data type of a[1] is String. This matrix is typically (but not necessarily) full. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Antonyms for scalar matrix. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. b ij = 0, when i \u2260 j \u2014 Page 36, Deep Learning, 2016. InnerProducts. scalar meson Look at other dictionaries: Matrix - \u043f\u043e\u043b\u0443\u0447\u0438\u0442\u044c \u043d\u0430 \u0410\u043a\u0430\u0434\u0435\u043c\u0438\u043a\u0435 \u0440\u0430\u0431\u043e\u0447\u0438\u0439 \u043a\u0443\u043f\u043e\u043d \u043d\u0430 \u0441\u043a\u0438\u0434\u043a\u0443 \u041b\u0435\u0442\u0443\u0430\u043b\u044c \u0438\u043b\u0438 \u0432\u044b\u0433\u043e\u0434\u043d\u043e An example of a diagonal matrix is the identity matrix mentioned earlier. \"Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. skalare Matrix, f rus. Maximum element in a matrix. MMAX(M). 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The number of rows, columns, and the matrix are known every thing the... Change, if at all every thing off the main diagonal of the vector on. Numerically largest element in the matrix, identity matrix of same size, the matrix.... Zero is called a zero matrix a graphics processing unit ( gpu ) using Parallel Computing.... Enter the number of rows, columns, and the other matrix elements are all 0 \u017eodynas... Subtraction: A\u2212B is de\ufb01ned as ( \u22121 ) A. Subtraction: A\u2212B is as! Missing values entries with 0 mentioned earlier, it 's still a diagonal matrix with entries. Accelerate code by running on a graphics processing unit ( gpu ) using Parallel Computing Toolbox\u2122 in its principal are! Entry in the matrix remains the same code is the identity matrix ir rus\u0173 kalbomis ( gpu using...", "date": "2021-12-08 10:07:12", "meta": {"domain": "yf-sensei.com", "url": "http://yf-sensei.com/sd2n0z/e3b785-diagonal-matrix-and-scalar-matrix", "openwebmath_score": 0.715437114238739, "openwebmath_perplexity": 769.3692635243086, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9702399060540359, "lm_q2_score": 0.9230391722430736, "lm_q1q2_score": 0.8955694397613148}}
{"url": "https://math.stackexchange.com/questions/2869817/evaluate-int-0-frac-pi2-frac-sin-x-cos-x-sin4x-cos4xdx/2870118", "text": "# Evaluate $\\int_0^{\\frac{\\pi}{2}}\\frac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}dx$\n\nEvaluate $$\\int_0^{\\frac{\\pi}{2}}\\frac{\\sin(x)\\cos(x)}{\\sin^4(x)+\\cos^4(x)}dx$$\n\nI used the substitution $\\sin x =t$, then I got the integral as $$\\int_0^1 \\frac{t}{2t^4-2t^2+1}dt$$\n\n\u2022 You see how the equation looks very ugly right now? The answer to this problem is typesetting it with MathJax. I would strongly recommend using it. \u2013\u00a0Matti P. Aug 2 '18 at 8:18\n\u2022 I am new here, I will start using from the next question. Thanks! \u2013\u00a0balaji Aug 2 '18 at 8:20\n\u2022 Looks like a substitution $u=t^2$ is called for. But your next question; why not edit this one using MathJax? \u2013\u00a0Lord Shark the Unknown Aug 2 '18 at 8:22\n\u2022 Thanks. I got the final answer as \u03c0 /4. is it correct? @LordSharktheUnknown \u2013\u00a0balaji Aug 2 '18 at 8:28\n\u2022 PARI/GP approves the result numerically, so you apparently got it. \u2013\u00a0Peter Aug 2 '18 at 8:29\n\nDo some trigonometry first: \\begin{align} \\frac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}&=\\frac{\\tfrac12\\sin 2x}{(\\sin^2x+\\cos^2x)^2-2\\sin^2x\\cos^2x}=\\frac{\\tfrac12\\sin 2x}{1-\\frac12\\sin^22x}\\\\&=\\frac{\\sin 2x}{2-\\sin^22x}=\\frac{\\sin 2x}{1+\\cos^22x}. \\end{align}\n\nNext use substitution: set $\\;u=\\cos 2x$, $\\;\\mathrm d u=-2\\sin 2x\\,\\mathrm d x$.\n\n\u2022 nice solution :-) +1 I like it. \u2013\u00a0Math-fun Aug 2 '18 at 11:38\n\u2022 @Math-fun: Thanks for your kind appreciation. It seems great minds think together ;o) \u2013\u00a0Bernard Aug 2 '18 at 11:55\n\u2022 I enjoyed it and in fact typed it but then deleted upon seeing yours :-) \u2013\u00a0Math-fun Aug 2 '18 at 12:44\n\nHint:\n\n$$\\dfrac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}=\\dfrac{\\tan x\\sec^2x}{\\tan^4x+1}$$\n\nSet $\\tan^2x=y$\n\nOR $$\\dfrac{\\sin x\\cos x}{\\sin^4x+\\cos^4x}=\\dfrac{\\cot x\\csc^2x}{\\cot^4x+1}$$\n\nSet $\\cot^2x=u$\n\nIf you want to continue your solution, then with substitution $t^2=u$ $$I=\\int_0^1\\dfrac{t}{2t^4-2t^2+1}dt=\\dfrac12\\int_0^1\\dfrac{1}{2u^2-2u+1}du=\\int_0^1\\dfrac{1}{(2u-1)^2+1}du$$ and then with substitution $2u-1=w$ $$I=\\dfrac12\\int_{-1}^1\\dfrac{1}{w^2+1}dw=\\color{blue}{\\dfrac{\\pi}{4}}$$\n\nLetting $u=\\tan x$, one has \\begin{eqnarray} &&\\int_0^{\\frac{\\pi}{2}}\\frac{\\sin(x)\\cos(x)}{\\sin^4(x)+\\cos^4(x)}dx\\\\ &=&\\int_0^{\\frac{\\pi}{2}}\\frac{\\tan(x)\\sec^2(x)}{\\tan^4(x)+1}dx\\\\ &=&\\int_0^\\infty\\frac{u}{u^4+1}du\\\\ &=&\\frac12\\int_0^\\infty\\frac{1}{u^2+1}du\\\\ &=&\\frac12\\arctan\uff08u)\\bigg|_0^\\infty\\\\ &=&\\frac\\pi4. \\end{eqnarray}\n\nPerform the change of variable $y=\\sin^2 x$,\n\n\\begin{align}J&=\\int_0^{\\frac{\\pi}{2}}\\frac{\\sin(x)\\cos(x)}{\\sin^4(x)+\\cos^4(x)}dx\\\\ &=\\frac{1}{2}\\int_0^1 \\frac{1}{x^2+(1-x)^2}\\,dx\\\\ &=\\frac{1}{2}\\int_0^1 \\frac{1}{2x^2-2x+1}\\,dx\\\\ &=\\int_0^1 \\frac{1}{(2x-1)^2+1}\\,dx\\\\ \\end{align}\n\nPerform the change of variable $y=2x-1$,\n\n\\begin{align} J&=\\frac{1}{2}\\int_{-1}^1 \\frac{1}{x^2+1}\\,dx\\\\ &=\\frac{1}{2}\\Big[\\arctan x\\Big]_{-1}^1\\\\ &=\\frac{1}{2}\\times\\frac{\\pi}{2}\\\\ &=\\boxed{\\frac{\\pi}{4}} \\end{align}", "date": "2019-05-25 23:14:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2869817/evaluate-int-0-frac-pi2-frac-sin-x-cos-x-sin4x-cos4xdx/2870118", "openwebmath_score": 0.9466292262077332, "openwebmath_perplexity": 4408.834301154089, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964224384746, "lm_q2_score": 0.9086178987887253, "lm_q1q2_score": 0.8954396886198528}}
{"url": "https://mymathforum.com/threads/how-do-i-find-the-average-speed-when-an-airplane-is-flying-in-circles.347485/", "text": "# How do I find the average speed when an airplane is flying in circles?\n\n#### Chemist116\n\nThe problem is as follows:\n\nAt an air exhibition a biplane is flying in circles as shown in the figure from below. It is known that type of motion of this airplane is a rotation with constant angular acceleration. The radius of the circle is $2\\,m$. Using this information find the average speed in $\\frac{m}{s}$ of the plane between the instants when $\\omega=6\\,\\frac{rad}{s}$ and $\\omega=10\\,\\frac{rad}{s}$.\u200b\n\nThe given alternatives in my book are:\n\n$\\begin{array}{ll} 1.&4\\,\\frac{m}{s}\\\\ 2.&8\\,\\frac{m}{s}\\\\ 3.&16\\,\\frac{m}{s}\\\\ 4.&20\\,\\frac{m}{s}\\\\ \\end{array}$\n\nI'm not sure if my attempt is correct in the solution of this problem but I thought that to get the average speed is given by:\n\n$\\overline{v}=\\frac{s}{t}$\n\nSince it mentions that it is a rotation with constant angular acceleration, then what I need is the acceleration. Using the information from the picture I'm getting this:\n\nConsidering $r=2\\,m$\n\n$\\omega_{1}=\\frac{v}{r}=\\frac{10}{2}=5$\n\n$\\omega_{2}=\\frac{v}{r}=\\frac{14}{2}=7$\n\nTherefore the acceleration can be found:\n\n$\\omega_{2}^2=\\omega_{1}^2+2\\alpha\\Delta\\theta$\n\n$7^2=5^2+2\\alpha\\left(3\\right)$\n\nTherefore:\n\n$\\alpha=\\frac{49-25}{6}=4\\,\\frac{rad}{s^2}$\n\nSince what they request is the average speed then what is needed is the displacement for the given speeds.\n\n$10^2=6^2+2\\left(4\\right)\\Delta\\theta$\n\n$\\Delta\\theta=\\frac{100-36}{8}=8\\text{rad}$\n\nBut the time elapsed for that displacement is also required for getting the average speed:\n\nThen:\n\n$\\omega_{f}=\\omega_{o}+\\alpha t$\n\n$10=6+\\left(4\\right)t$\n\n$t=1$\n\nSo it is only $1\\,s$ elapsed in the given interval.\n\nTherefore the average speed would be:\n\n$\\overline{v}=\\frac{s}{t}=\\frac{8\\times 2}{1}=16\\,\\frac{m}{s}$\n\nWhich is what appears in alternative number $3$. But is my solution correct?.\n\nLast edited by a moderator:\n\n#### skipjack\n\nForum Staff\nThat's okay, but for constant angular acceleration, $$\\displaystyle \\Delta\\theta = \\frac{\\omega_1 + \\omega_2}{2}\\Delta t$$,\nso $$\\displaystyle \\overline\\omega = \\frac{\\Delta\\theta}{\\Delta t} = \\frac{\\omega_1 + \\omega_2}{2} = \\frac{6 + 10}{2}\\text{rad/s} = 8\\text{ rad/s}$$.\nHence (as the radius is $\\text{2 m}$) $\\overline v = 8\\times2\\text{ m/s} = 16\\text{ m/s}$.\n\nSimilar threads", "date": "2019-12-05 19:55:32", "meta": {"domain": "mymathforum.com", "url": "https://mymathforum.com/threads/how-do-i-find-the-average-speed-when-an-airplane-is-flying-in-circles.347485/", "openwebmath_score": 0.8539042472839355, "openwebmath_perplexity": 300.23666741283523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9736446463891303, "lm_q2_score": 0.9196425240200058, "lm_q1q2_score": 0.8954050201038658}}
{"url": "https://or.stackexchange.com/questions/6078/modelling-precedence-relations/6079#6079", "text": "# Modelling precedence relations\n\nI have two tasks $$i$$ and $$k$$ with durations $$d_i$$ and $$d_k$$, where $$d_i$$ and $$d_k$$ are nonnegative variables.\n\nI would like to model that $$i$$ may precede $$k$$ or $$k$$ may precede $$i$$ and that they may not overlap.\n\nSo, with $$t_i$$ and $$t_k$$ denoting the start times of $$i$$ and $$k$$, I have to model:\n\neither $$t_i + d_i \\le t_k$$ OR $$t_k + d_k \\le t_i$$\n\nIntroducing a binary variable $$y$$, I can achieve the result with the following two big M constraints:\n\n$$t_i + d_i - t_k \\le M y$$\n\n$$t_k + d_k - t_i \\le M (1-y)$$\n\nIf it is required that $$t_i + d_i \\le H$$ and $$t_k + d_k \\le H$$ then I can set $$M$$ to be $$M=H$$.\n\nMy question is, is what I have done so far correct (what worries me is the variable duration) and can anyone think about a better formulation?\n\nYes, this is correct and is the classical approach from Manne, On the Job-Shop Scheduling Problem (1960).\n\nIn some modeling languages, you can also enforce these implications by using indicator constraints: \\begin{align} y = 0 &\\implies t_i + d_i \\le t_k \\\\ y = 1 &\\implies t_k + d_k \\le t_i \\\\ \\end{align}\n\n\u2022 Rob, thank you very much for your reply and the suggestion. I will try both the Big-M and the Indicator Constraint approach. Apr 12 at 8:43\n\nCan anyone think about a better formulation?\n\nAnother option is to use binary variables $$x_{it}$$ that take value $$1$$ if task $$i$$ starts at time $$t$$. You then need two sets of constraints:\n\n\u2022 one start time per task: $$\\sum_{t}x_{it} = 1 \\quad \\forall i$$\n\u2022 don't overlap tasks: $$\\sum_{i}\\sum_{k, t+1 - d_i \\le k \\le t}x_{ik} \\le 1 \\quad \\forall t$$\n\nThis formulation is more tight and should solve faster. And it does not require big-Ms.\n\n\u2022 This however, will function only in the case of a discrete time formulation? Why is this formulation tighter? Jul 7 at 8:31\n\u2022 Yes indeed this assumes a discrete time span. That seems reasonable to me. I don't think it is easy to prove that the formulation is tighter - this could be a great separate question. I only have empirical evidence to back this statement. Jul 7 at 9:54\n\nYou may also use CPOptimizer within CPLEX that contains scheduling high level concepts. And then you can directly use noOverlap constraints.\n\nIn\n\nusing  CP;\n\ndvar interval i size 5;\ndvar interval k size 4;\n\ndvar sequence seq in append(i,k);\n\nminimize maxl(endOf(i),endOf(k));\nsubject to\n{\nnoOverlap(seq);\n}\n\n\nthe constraint\n\nnoOverlap(seq);\n\n\nmakes sure that i and k do not overlap\n\nand in the CPLEX IDE you will see", "date": "2021-09-24 09:08:49", "meta": {"domain": "stackexchange.com", "url": "https://or.stackexchange.com/questions/6078/modelling-precedence-relations/6079#6079", "openwebmath_score": 0.9552558660507202, "openwebmath_perplexity": 696.3225962960692, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9773707999669627, "lm_q2_score": 0.9161096158798117, "lm_q1q2_score": 0.8953787881298785}}
{"url": "https://math.stackexchange.com/questions/3193300/given-constraints-in-how-many-ways-can-actors-be-chosen-for-roles", "text": "# Given constraints, in how many ways can actors be chosen for roles?\n\nGiven $$13$$ actors and $$6$$ unique roles, in how many ways can the actors be assigned a role if a certain actor (Alan) will not join if another actor (Betty) joins?\n\nMy method was to compute total unrestricted number of ways less the number of ways when they're both in it. The first term is $$\\binom{13}{6} 6!$$ and the second term is $$\\binom{11}{4} 6!$$. The numerical answer I get is $$997920$$ but the correct numerical answer given is $$1116720$$.\n\n\u2022 997920 looks right to me \u2013\u00a0Hagen von Eitzen Apr 19 at 8:16\n\u2022 The result 1116720 would fit the task that we only prohibit Betty in a more prominent role than Alan \u2013\u00a0Hagen von Eitzen Apr 19 at 8:20\n\u2022 What does being in a more prominent role mean? Also I realise that I get their answer only if I divide my second term by 2 (or 2! ?). \u2013\u00a0OneGapLater Apr 19 at 8:22\n\u2022 The given answer will be true if a Alan has problem with Betty but not the other way around, assuming that they join one by one. \u2013\u00a0SinTan1729 Apr 19 at 9:12\n\u2022 I think that's what they imply. However, I don't see why my answer doesn't accommodate for that. \u2013\u00a0OneGapLater Apr 19 at 9:27\n\nInterpretation: Alan and Betty cannot both be cast.\n\nYou are correct under this interpretation. We can confirm your answer using a different approach. Observe that either exactly one of Alan or Betty is in the cast or neither Alan nor Betty is in the cast.\n\nExactly one of Alan or Betty is in the cast: Select whether Alan or Betty is in the cast. Select five of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\\binom{2}{1}\\binom{11}{5}6!$$\n\nNeither Alan nor Betty is in the cast: Select six of the other eleven actors to be in the cast. Assign roles to the six selected actors. $$\\binom{11}{6}6!$$\n\nTotal: Since the two cases above are mutually exclusive and exhaustive, the number of ways of assigning roles to the actors is $$\\binom{2}{1}\\binom{11}{5}6! + \\binom{11}{6}6! = \\binom{11}{5}6!\\left[\\binom{2}{1} + 1\\right] = 3\\binom{11}{5}6! = 997920$$ as you found.\n\nInterpretation: Alan will not join if Betty has been cast first.\n\nUnder this interpretation, Alan and Betty can both be in the cast if Alan is cast first.\n\nThen there are three possibilities:\n\n1. Exactly one of Alan or Betty is in the cast.\n2. Neither Alan nor Betty is in the cast.\n3. Alan and Betty are both in the cast, because Alan is cast first.\n\nWe have covered the first two cases above.\n\nAlan and Betty are both in the cast, because Alan is cast first: Ignoring the order of casting for the moment, if Alan and Betty are both selected, we must select four of the remaining eleven actors. We then assign roles to the six actors. This can be done in $$\\binom{11}{4}6!$$ ways. However, by symmetry, in half of these assignments, Betty is cast before Alan. Thus, Alan will only join the cast in half of these assignments. Thus, there are $$\\frac{1}{2}\\binom{11}{4}6! = 118800$$ ways to assign the roles so that both Alan and Betty are cast.\n\nTotal: Since these three cases are mutually exclusive and exhaustive, the roles may be cast if Alan will not join if Betty has already been cast is $$\\binom{2}{1}\\binom{11}{5}6! + \\binom{11}{6}6! + \\frac{1}{2}\\binom{11}{4}6! = 1116720$$ If this is the intended interpretation, the wording of the question could have been clearer.", "date": "2019-05-23 10:50:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3193300/given-constraints-in-how-many-ways-can-actors-be-chosen-for-roles", "openwebmath_score": 0.4260421097278595, "openwebmath_perplexity": 178.6729837181932, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713861502774, "lm_q2_score": 0.9086178919837706, "lm_q1q2_score": 0.8952352099157928}}
{"url": "https://math.stackexchange.com/questions/2863427/factorising-complex-numbers", "text": "# Factorising complex numbers\n\n4 years ago, when I learned about factorising and complex numbers, me and my friend worked on factorising complex numbers.\n\nFor example, $4+2i= 3-(-1)+2i = 3-i^2+2i = -(i^2-2i-3) = -(i-3)(i+1)$\n\nThe goal was to represent $a+bi$ with product of same form. where $a$ and $b$ are integer.\n\nAnother example is, $8+i= 8i^4+i=i(8i^3+1)=i(2i+1)(4i^2-2i+1)=i(2i+1)(-2i-3)=-i(2i+1)(2i+3)$\n\nI showed my teacher, and she said it's useless.\n\nNow I think of it, I don't know why I did this and it looks like same thing just in different form.\n\nIs there any research already done on this or can there be any use of it?\n\nApparently,\n\n$(n+2)+ni=-(i-(n+1))(i+1)$\n\n$m^3+n^3i=-i(mi+n)(mni+(m^2-n^2))$\n\n\u2022 Yes, there's a lot of work done on this. Look up Gaussian Integers. \u2013\u00a0Angina Seng Jul 26 '18 at 14:07\n\u2022 And by no means is it useless! \u2013\u00a0Lubin Jul 26 '18 at 14:11\n\u2022 There's research already done, but that does not mean there can't be any use of the work you do on this, when you've read up on the existing work. \u2013\u00a0Henrik supports the community Jul 26 '18 at 14:13\n\u2022 I was thinking @Henrik, that knowing how to factor Gaussian integers is useful in what might look like other parts of mathematics. \u2013\u00a0Lubin Jul 26 '18 at 14:15", "date": "2020-10-23 21:41:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2863427/factorising-complex-numbers", "openwebmath_score": 0.37995365262031555, "openwebmath_perplexity": 392.94744343388356, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.983847165177859, "lm_q2_score": 0.9099070121457543, "lm_q1q2_score": 0.8952094344750561}}
{"url": "https://stats.stackexchange.com/questions/377143/how-is-sample-mean-divided-by-sample-std-distributed-for-normal-distributions", "text": "# How is sample mean divided by sample STD distributed for normal distributions?\n\nLet's assume that we sample N times from a normal distribution with known mean and variance. With a generated sample we calculate sample mean and sample standard deviation. Then we divide sample mean by sample standard deviation (STD) to get the measure of my interest.\n\nThe question that I have: Is there an analytical expression for the distribution of the above described measure?\n\nI know that distribution for the sample mean is well known. I have also found an expression for the distribution of sample STD. However, I cannot find the distribution that I need.\n\n\u2022 If \"STD\" is intended to be an abbreviation for standard deviation then the distribution will be a scaled version of a noncentral t distribution. A number of posts on site also discuss this distribution. If I don't locate a sufficiently close duplicate (I didn't with a quick look), and nobody posts a full answer in the meantime, I'll come back and make this more detailed. Nov 15 '18 at 11:33\n\u2022 @Glen_b Relevant posts are stats.stackexchange.com/a/133274/919 (yours), stats.stackexchange.com/a/17288/919, and stats.stackexchange.com/a/160523/919. The last is very nearly an answer, but it does not explicitly give any expression for the distribution.\n\u2013\u00a0whuber\nNov 15 '18 at 13:22\n\u2022 @whuber I found a few others a bit like the first two but they didn't seem quite close enough to count as answers for this one (though all were relevant). Nov 15 '18 at 14:54\n\u2022 This is just to point out that this is the reciprocal of the coefficient of variation. It's also been called signal to noise ratio, although that term doesn't appear to be uniquely defined. Nov 22 '18 at 0:50\n\nLet $$X_1,...,X_n \\sim \\text{IID N}(\\mu, \\sigma)$$ be your data points. It is well known from Cochran's theorem that the sample mean and sample variance are independent with distributions:\n\n$$\\bar{X}_n \\sim \\text{N} \\Big( \\mu, \\frac{\\sigma^2}{n} \\Big) \\quad \\quad \\quad S_n^2 \\sim \\sigma^2 \\cdot \\frac{\\text{Chi-Sq}(n-1)}{n-1}.$$\n\nHence, we can form the independent statistics:\n\n$$Z_n \\equiv \\frac{\\bar{X}_n - \\mu}{\\sigma / \\sqrt{n}} \\sim \\text{N}(0,1) \\quad \\quad \\quad \\chi_n \\equiv \\frac{S_n}{\\sigma} \\sim \\frac{\\text{Chi}(n-1)}{\\sqrt{n-1}}.$$\n\nWith a bit of algebra we then have:\n\n\\begin{aligned} \\frac{\\bar{X}_n}{S_n} &= \\frac{\\bar{X}_n / \\sigma}{S_n / \\sigma} \\\\[6pt] &= \\frac{\\bar{X}_n / \\sigma}{\\chi_n} \\\\[6pt] &= \\frac{1}{\\sqrt{n}} \\cdot \\frac{\\sqrt{n} \\bar{X}_n / \\sigma}{\\chi_n} \\\\[6pt] &= \\frac{1}{\\sqrt{n}} \\cdot \\frac{\\sqrt{n} (\\bar{X}_n - \\mu)/\\sigma + \\sqrt{n} \\mu/\\sigma}{\\chi_n} \\\\[6pt] &= \\frac{1}{\\sqrt{n}} \\cdot \\frac{Z_n + \\sqrt{n} \\mu/\\sigma}{\\chi_n} \\\\[6pt] &\\sim \\frac{1}{\\sqrt{n}} \\cdot \\text{Noncentral T} \\big(\\sqrt{n} \\mu/\\sigma, n-1 \\big). \\\\[6pt] \\end{aligned}\n\nSo you can see that the ratio of the sample mean on the sample standard deviation has a scaled non-central T distribution with non-centrality parameter $$\\sqrt{n} \\mu/\\sigma$$ and degrees-of-freedom $$n-1$$. We can double-check this theoretical result empirically via simulation.\n\nChecking the distribution by simulation: In the R code below we create a function to simulate $$m$$ samples of size $$n$$ from the IID normal model and generate the $$m$$ ratio statistics from these samples. We plot the kernel density of these simulated statistics against the theoretical distribution above in order to confirm that the theoretical result is correct.\n\n#Simulate m values of the ratio statistic for samples of size n\nSIMULATE <- function(m, n, mu, sigma) {\nX <- array(rnorm(n*m, mean = mu, sd = sigma), dim = c(m,n));\nR <- rep(0, m);\nfor (i in 1:m) { R[i] <- mean(X[i,])/sd(X[i,]); }\nR; }\n\n#Plot the density of the simulated values against theoretical\nPLOTSIM  <- function(m, n, mu, sigma) {\nrequire(stats); require(ggplot2);\nRR    <- SIMULATE(m, n, mu, sigma);\nDENS  <- density(RR);\nDENS$$yy <- dt(DENS$$x*sqrt(n), df = n-1, ncp = sqrt(n)*mu/sigma)*sqrt(n);\nDATA <- data.frame(x = DENS$$x, y = DENS$$y, yy = DENS$yy); ggplot(data = DATA, aes(x = x)) + geom_line(aes(y = y), size = 1.2, colour = 'black') + geom_line(aes(y = yy), size = 1.2, colour = 'red', linetype = 'dotted') + theme(plot.title = element_text(hjust = 0.5, size = 14, face = 'bold'), plot.subtitle = element_text(hjust = 0.5)) + ggtitle('Density plot - Simulated Data') + labs(subtitle = paste0('(Sample size = ', n, ', Simulation size = ', m, ')')) + xlab('Sample Mean / Sample Standard Deviation') + ylab('Density'); } #Generate example plot set.seed(1); m <- 10^4; n <- 100; mu <- 12; sigma <- 6; PLOTSIM(m, n, mu, sigma);  \u2022 thanks as usual @Ben! I ran the same simulation in Python and got identical results. Could you suggest the analytical formula for$mean(\\frac{\\mu}{\\sigma})$and$stdev(\\frac{\\mu}{\\sigma})$as a function of$\\mu$and$\\sigma$? I am not familiar with the Noncentral T-distribution, I couldn't figure it out myself. Nov 25 '18 at 20:43 \u2022 Just have a look at the moments of the noncentral T distribution. To get the moments for the above ratio, all you need to do is substitute the degrees-of-freedom$n-1$and the noncentrality parameter$\\sqrt{n} \\mu / \\sigma\\$. Good luck!\n\u2013\u00a0Ben\nNov 25 '18 at 21:42\n\u2022 @elemolotiv, would it be possible for you to share your Python code? I am asking since I am also working in Python. Nov 26 '18 at 8:53\n\n@Roman, here is my Python code, of @Ben's example above\n\n\u2022 it produces a file \"walks.tsv\" with N realisations of a random walk with mean=\u03bc and stdev=\u03c3.\n\n\u2022 for each realisation there is a row, with the sample mean, sample stdev, and sample mean/stdev.\n\n\u2022 I loaded \"walk.tsv\" in Excel and used Excel to aggregate and plot the data.\n\nHope it helps \ud83d\ude42\n\nimport math, numpy.random\n\nclass RandomWalk:\n\ndef __init__(self, step_mean, step_stdev, steps_per_walk):\nself.step_mean = step_mean\nself.step_stdev = step_stdev\nself.steps = steps_per_walk\nself.log = open(\"walk.tsv\",\"w+\")\nself.log_names()\n\ndef realize(self):\ntotal = 0\ntotal_sqr = 0\nfor i in range(self.steps):\nstep = float(numpy.random.normal(self.step_mean, self.step_stdev, 1))\ntotal += step\ntotal_sqr += step * step\nself.sample_mean = total / self.steps\nself.sample_stdev = math.sqrt(total_sqr / self.steps - self.sample_mean * self.sample_mean)\nself.log_values()\n\ndef log_names(self):\nself.log.write(\"dist_mean\\t\")\nself.log.write(\"dist_stdev\\t\")\nself.log.write(\"dist_mean/stdev\\t\")\nself.log.write(\"steps\\t\")\nself.log.write(\"sample_mean\\t\")\nself.log.write(\"sample_stdev\\t\")\nself.log.write(\"sample_mean/stdev\\n\")\nself.log.flush()\n\ndef log_values(self):\nself.log.write(\"{:0.1f}\\t\".format(self.step_mean))\nself.log.write(\"{:0.1f}\\t\".format(self.step_stdev))\nself.log.write(\"{:0.2f}\\t\".format(self.step_mean / self.step_stdev))\nself.log.write(\"{}\\t\".format(self.steps))\nself.log.write(\"{:0.1f}\\t\".format(self.sample_mean))\nself.log.write(\"{:0.1f}\\t\".format(self.sample_stdev))\nself.log.write(\"{:0.2f}\\n\".format(self.sample_mean / self.sample_stdev))\nself.log.flush()\n\ndef simulate(step_mean, step_stdev, steps_per_walk, walks):\nwalk = RandomWalk(step_mean, step_stdev, steps_per_walk)\nfor i in range(walks):\nwalk.realize()\n\nsimulate(step_mean = 12, step_stdev = 6, steps_per_walk = 100, walks = 10000)", "date": "2021-09-18 04:06:28", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/377143/how-is-sample-mean-divided-by-sample-std-distributed-for-normal-distributions", "openwebmath_score": 0.9622633457183838, "openwebmath_perplexity": 1957.3370992438336, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676454700793, "lm_q2_score": 0.9136765187126079, "lm_q1q2_score": 0.8950993238084795}}
{"url": "https://www.physicsforums.com/threads/electric-field-created-by-point-charges-and-conducting-plane.909946/", "text": "# I Electric field created by point charges and conducting plane\n\nTags:\n1. Apr 2, 2017\n\n### KV71\n\nI came upon this:\nhttp://physics.stackexchange.com/qu...change-if-we-place-a-metal-plat/323006#323006\n\nquestion on Physics Stackexchange which I found very interesting.\n\nThe configuration is basically two positive point charges q and a conducting plane equidistant from both charges. What I found most fascinating in particular, is one answer that claims\n\n\"if the plate is a plane that extends to infinity, there will be two image charges -q at the positions of each of the original positive charges so that the electric field everywhere is zero\"\n\nMy question is, is this true? If so, could someone explain why or perhaps tell me more about this?\n\n2. Apr 2, 2017\n\nStaff Emeritus\nIf the plate extends to infinity, it can get as many cancelling charges as it needs, \"from infinity\".\n\n3. Apr 2, 2017\n\n### KV71\n\nSo the field everywhere is zero?\n\n4. Apr 2, 2017\n\nStaff Emeritus\nI don't think so, because the charge and its image are not in the same place.\n\n5. Apr 2, 2017\n\n### KV71\n\nI think there are two images and the image for the charge below the conductor is sitting at the position of the charge above the conductor and vice versa.\n\n6. Apr 2, 2017\n\nStaff Emeritus\nFields from the other side of the conductor don't penetrate the conductor.\n\n7. Apr 3, 2017\n\n### vanhees71\n\nIt's very easy to see, why the field cannot be 0. Just draw a sphere $V$ around one of the postive charges, and use Gauss's Law,\n$$\\int_{\\partial V} \\mathrm{d}^2 \\vec{F} \\cdot \\vec{E}=Q.$$\nThe key to understand this is that the image charges are not really there, but they are used as a mathematical trick to get the total field consisting of the field of the real charge and the influence charges in the plate, i.e., to fulfill the boundary conditions.\n\nFor your example you can easily solve the problem indeed by using image charges. You need to treat only one charge first. Say the infinite plane defines the $xy$ plane of a cartesian coordinate system, and let the charge $Q$ sit on the $z$ axis at $(0,0,a)$ with $a>0$.\n\nYou have to solve the Laplace equation for the potential\n$$\\Delta \\phi=-Q \\delta(x) \\delta(y) \\delta(z-a)$$\nwith the boundary condition $\\phi=0$ for $z=0$.\n\nObviously the solution for $z<0$ is $\\phi=0$ and for $z>0$ you write down the solution for the field of the true charge $Q$ at $(0,0,a)$ and account for the boundary conditions by substituting the plate by the image charge $-Q$ at $(0,0,-a)$, because then you solve for sure the Laplace equation and the boundary conditions, i.e., you have\n$$\\phi(\\vec{x})=\\begin{cases} 0 & \\text{for} \\quad z<0, \\\\ \\frac{Q}{4 \\pi} \\left [\\frac{1}{\\sqrt{x^2+y^2+(z-a)^2}}-\\frac{1}{\\sqrt{x^2+y^2+(z+a)^2}} \\right] &\\text{for} \\quad z>0. \\end{cases}$$\nFor the other charge, $Q'$ sitting at $(0,0,-b)$ ($b>0$) you have in an analogous way\n$$\\tilde{\\phi}(\\vec{x})=\\begin{cases} \\frac{Q'}{4 \\pi} \\left [\\frac{1}{\\sqrt{x^2+y^2+(z+b)^2}}-\\frac{1}{\\sqrt{x^2+y^2+(z-b)^2}} \\right] &\\text{for} \\quad z<0,\\\\ 0 & \\text{for} \\quad z>0. \\end{cases}$$\nThe total field thus is\n$$\\phi_{\\text{tot}}(\\vec{x})= \\begin{cases} \\frac{Q'}{4 \\pi} \\left [\\frac{1}{\\sqrt{x^2+y^2+(z+b)^2}}-\\frac{1}{\\sqrt{x^2+y^2+(z-b)^2}} \\right ] &\\text{for} \\quad z<0,\\\\ \\frac{Q}{4 \\pi} \\left [\\frac{1}{\\sqrt{x^2+y^2+(z-a)^2}}-\\frac{1}{\\sqrt{x^2+y^2+(z+a)^2}} \\right] & \\text{for} \\quad z>0. \\end{cases}$$\nNow set $Q'=Q$ and $b=a$, and you see that the field is not 0 also in this symmetric case.\n\nThat's a great exercise to understand the role of the \"mirror charges\"!\n\n8. Apr 4, 2017\n\n### KV71\n\nThanks for the very informative and well-written post @vanhees71 . Indeed, I liked this problem too because it has a solution that does not strike me immediately but is actually obvious when I gave it some more thought. Also , the owner of the answer that I referenced in my intro post deleted their answer (must have realized it is wrong) and has posted a new answer--you may look at it (it uses essentially the same method as yours but talks in terms of the field and superposition). After reading it, I am amazed that iafter an infinite conducting plate is placed between two charges, \"the force on the charges changes sign\". Very interesting indeed.\n\nLet me know what you think!\n\n9. Apr 4, 2017\n\n### KV71\n\n@Vanadium 50 I didn't understand what you meant earlier by\nbut now I understand. The link in my last post #8 helped me. Thanks so much! It turns out you were absolutely right!", "date": "2018-01-21 18:58:17", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/electric-field-created-by-point-charges-and-conducting-plane.909946/", "openwebmath_score": 0.7848370671272278, "openwebmath_perplexity": 484.3853374788014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9559813501370537, "lm_q2_score": 0.9362850128595114, "lm_q1q2_score": 0.8950710107065243}}
{"url": "https://math.stackexchange.com/questions/3172005/find-all-triplets-a-b-c-less-than-or-equal-to-50-such-that-a-b-c-be-div", "text": "# Find all triplets $(a,b,c)$ less than or equal to 50 such that $a + b +c$ be divisible by $a$ and $b$ and $c$.\n\nFind all triplets $$(a,b,c)$$ less than or equal to 50 such that $$a + b +c$$ be divisible by $$a$$ and $$b$$ and $$c$$.(i.e $$a|a+b+c~~,~~b|a+b+c~~,~~c|a+b+c$$) for example $$(10,20,30)$$ is a good triplet. ($$10|60 , 20|60 , 30|60$$).\n\nNote: $$a,b,c\\leq 50$$ and $$a,b,c\\in N$$.\n\nIn other way the question says to find all $$(a,b,c)$$ such that $$lcm(a,b,c) | a+b+c$$\n\nAfter writing different situations, I found that if $$gcd(a,b,c) = d$$ then all triplets are in form of $$(d,2d,3d)$$ or $$(d,d,d)$$ or $$(d,d,2d)$$ are answers. (of course the permutation of these like $$(2d,3d,d)$$ is also an answer). It gives me $$221$$ different triplets. I checked this with a simple Java program and the answer was correct but I cannot say why other forms are not valid. I can write other forms and check them one by one but I want a more intelligent solution than writing all other forms. Can anyone help?\n\nMy java code: (All of the outputs are in form of $$(d,d,d)$$ or $$(d,2d,3d)$$ or $$(d,d,2d)$$ and their permutations.)\n\nimport java.util.ArrayList;\nimport java.util.Collections;\n\npublic class Main {\npublic static void main(String[] args) {\nint count = 0;\nfor (int i = 1; i <= 50; i++) {\nfor (int j = 1; j <= 50; j++) {\nfor (int k = 1; k <= 50; k++) {\nint s = i + j + k;\nif (s % i == 0 && s % j == 0 && s % k == 0 && i != j && j != k && i != k) {\nArrayList<Integer> array = new ArrayList<Integer>();\narray.clear();\nint g = gcd(gcd(i, j), k);\nCollections.sort(array);\nint condition = 4; //To find out whether it is (d,d,d) or (d,d,2d) or (d,2d,3d)\nif (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 1) {\ncondition = 1;\n}\nif (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 2) {\ncondition = 2;\n}\nif (array.get(0) == 1 && array.get(1) == 2 && array.get(2) == 3) {\ncondition = 3;\n}\nSystem.out.printf(\"%d %d %d ::: Condition: %d\\n\", i, j, k, condition);\ncount++;\n}\n}\n}\n}\nSystem.out.println(count);\n}\n\npublic static int gcd(int a, int b) {\nif (b == 0) {\nreturn a;\n} else\nreturn gcd(b, a % b);\n}\n}\n\n\u2022 ... I recall seeing this question yesterday... \u2013\u00a0Servaes Apr 2 at 16:14\n\u2022 @Servaes I used the search and didn't find this question. But as you said,I checked and found it. I am not the same person. Maybe his source and I was the same because I was investigating homework of a discrete mathematics course of a university and I found this question and he/she stated that it is his homework.I found the question interesting and asked it here. I completely checked the conditions with a Java program and find tested my hypothesis but I don't know how to prove it without checking all different forms. for better clarification, I'll add my java code to the problem. \u2013\u00a0amir na Apr 2 at 16:47\n\u2022 What does a triplet being less than $50$ mean? That each term is less than 50? are you assuming each term is non-negative? \u2013\u00a0fleablood Apr 2 at 16:53\n\u2022 Also what does \"m is divisible to k\" mean? Does that mean $\\frac km$ is an integer? Or that $\\frac mk$ is an integer? Or something else? I usually hear \"m is divisible by k\" to mean $\\frac mk$ is an integer. \u2013\u00a0fleablood Apr 2 at 16:55\n\u2022 @Servaes math.stackexchange.com/questions/3170626/\u2026 I didn't find this in search because the title of that question is not very good and don't have the actual question and I think stack Exchange only search by title. \u2013\u00a0amir na Apr 2 at 16:57\n\nIf $$a\\leq b\\leq c$$ then $$c\\mid a+b+c$$ implies $$c\\mid a+b$$ and so $$a+b=cz$$ for some $$z\\in\\Bbb{N}$$. Then $$cz=a+b\\leq2b\\leq2c,$$ and so $$z\\leq2$$. If $$z=2$$ then the inequalities are all equalities and so $$a=b=c$$. Then the triplet $$(a,b,c)$$ is of the form $$(d,d,d)$$.\n\nIf $$z=1$$ then $$c=a+b$$, and then $$b\\mid a+b+c$$ implies that $$b\\mid 2a$$. As $$b\\geq a$$ it follows that either $$b=a$$ or $$b=2a$$. If $$b=a$$ then $$c=2a$$ and the triplet $$(a,b,c)$$ is of the form $$(d,d,2d)$$. If $$b=2a$$ then $$c=3a$$ and the triplet $$(a,b,c)$$ is of the form $$(d,2d,3d)$$.\n\nThis allows us to count the total number of triplets quite easily;\n\n1. The number of triplets of the form $$(d,d,d)$$ is precisely $$50$$; one for each positive integer $$d$$ with $$d\\leq50$$.\n2. The number of triplets of the form $$(d,d,2d)$$ is precisely $$25$$; one for each positive integer $$d$$ with $$2d\\leq50$$. Every such triplets has precisely three distinct permutations of its coordinates, yielding a total of $$3\\times25=75$$ triplets.\n3. The number of triplets of the form $$(d,2d,3d)$$ is precisely $$16$$; one for each positive integer $$d$$ with $$3d\\leq50$$. Every such triplets has precisely six distinct permutations of its coordinates, yielding a total of $$6\\times 16=96$$ triplets.\n\nThis yields a total of $$50+75+96=221$$ triplets.\n\n\u2022 Simple code finds $221$. \u2013\u00a0David G. Stork Apr 4 at 4:56\n\u2022 @DavidG.Stork A simple count shows the same ;) \u2013\u00a0Servaes Apr 4 at 13:29", "date": "2019-05-23 21:13:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3172005/find-all-triplets-a-b-c-less-than-or-equal-to-50-such-that-a-b-c-be-div", "openwebmath_score": 0.5972553491592407, "openwebmath_perplexity": 210.0160019481613, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462183543601, "lm_q2_score": 0.9059898184796792, "lm_q1q2_score": 0.8950692150345523}}
{"url": "https://math.stackexchange.com/questions/1020066/puzzle-of-gold-coins-in-the-bag/1020151", "text": "Puzzle of gold coins in the bag\n\nAt the end of Probability class, our professor gave us the following puzzle:\n\nThere are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?\n\nAfter about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):\n\nGive the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.\n\nMy questions are:\n\n1. How does the formula work? Where does it come from?\n2. Do we have other ways to solve this puzzle? If yes, how?\n3. If the digital scale is replaced by a traditional scale, the scale like symbol of libra or the scale in Shakespeare's drama: The Merchant of Venice (I don't know what is the name in English), then how do we solve this puzzle?\n\u2022 If you can pull coins out of the bag, maybe you can see the difference instead of weighing. ;) \u2013\u00a0jpmc26 Nov 13 '14 at 19:54\n\u2022 @jpmc26 One of friends also said like that but our prof only smiled. :-D \u2013\u00a0Venus Nov 14 '14 at 9:54\n\u2022 This puzzle appeared on one of the episodes of Columbo: imdb.com/title/tt0075864 \u2013\u00a0Stephen Montgomery-Smith Nov 18 '14 at 23:04\n\u2022 @StephenMontgomery-Smith That's very old movie, I haven't been born yet when the movie aired so I don't watch it. Anyway, you're the one who also reviews Prof. Otelbaev's work, how is it going? Is it correct so far? \u2013\u00a0Venus Nov 19 '14 at 6:29\n\u2022 @Venus You can see that episode on Netflix. Otelbaev said there was a mistake in his work. math.stackexchange.com/questions/634890/\u2026 \u2013\u00a0Stephen Montgomery-Smith Nov 19 '14 at 14:46\n\nTo understand the formula, it would be easiest to explain how it works conceptually before we derive it.\n\nLet's simplify the problem and say there are only 3 bags each with 2 coins in them. 2 of those bags have the 1 gram coins and one has the 1.01 gram gold coins. Let's denote the bags arbitrarily as $Bag_0$, $Bag_1$, and $Bag_2$. Similarly to your problem, let's take 0 coins from $Bag_0$, 1 coin from $Bag_1$, and 2 coins from $Bag_2$. We know that the gold coins must be in one of those bags, so there are three possibilities when we weigh the three coins we removed:\n\nGold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams.\n\nGold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams.\n\nGold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams.\n\nSo each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight.\n\nWe can generalize our results from this simplified example to your 100 bag example.\n\nNow for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99$.\n\nSo the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$\n\nBut we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have: $Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$\n\nRearranging the formula to solve for n, we have: $100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example.\n\nI have no knowledge of an alternative answer to this puzzle, but perhaps another member's answer may be enlightening if there is. Technically speaking, you could have denoted the bags from 1 to 100 and gone through a similar process as above, but the method is still the same, so I wouldn't treat it as a new answer.\n\nIf our electric scale is replaced by a scale of libra, I don't believe it would be possible to answer this puzzle with only one measurement of weight. But again, perhaps another answer may be enlightening on that.\n\n\u2022 To see why it's not possible to find the bag with just one \"libra\" weighing, consider that one weighing gives as 3 possible states (left, right, balanced), yet we have 100 bags to choose from. In other words we have a trit of information (trit: ternary equivalent of the bit) with each weighing. We will need at least 5 trits to distinguish between 100 possible bags. \u2013\u00a0Thanassis Oct 2 '18 at 5:29\n\nFor #1: imagine for a moment that all the coins are fake. If we took 0 coins from bag 0, 1 coin from bag 1, 2 coins from bag 2... we'd have $99\\times100/2=4,950$ coins, and those 4,950 coins would weigh a total of 4,950 grams. But now, say that bag 25 were the one with real coins that are slightly heavier: 0.01 grams heavier, in fact. So the total weight of the coins is $W=4950+0.01N$, where $N$ is the number of the bag with the real coins. But -- we have the weight, not the bag number. So let's invert the equation: we want to find N given W, not the other way around.\n\n\\begin{align}W&=4950+0.01N\\\\W-4950&=0.01N\\\\100(W-4950)&=N\\end{align}\n\nFor #2, aside from renumbering the bags, there isn't a different way to do this; no matter what, we have to have a different number coming from the scale for each different possible result.\n\nFor #3, you need $\\lceil\\log_3k\\rceil$ weighings to discover the odd coin out; for 100 coins, that's 5 weighings: the first splits the coins into groups of (up to) 34; the second into groups of (up to) 12; the third into groups of (up to) 4; the fourth into groups of (up to) 2; the fifth finds it guaranteed.\n\nWhy $\\lceil\\log_3k\\rceil$? Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\\ge k$, thus $n\\ge\\log_3k$, thus $n=\\lceil\\log_3k\\rceil$.\n\n\u2022 +1 for #3, but how to get this formula: $\\lceil\\log_3k\\rceil$? \u2013\u00a0Venus Nov 13 '14 at 14:32\n\u2022 Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\\ge k$, thus $n\\ge\\log_3k$, thus $n=\\lceil\\log_3k\\rceil$ \u2013\u00a0Dan Uznanski Nov 13 '14 at 14:37\n\nAccording to the given values, the gold coins have essentially the same weight (down to a single percent) as the base ones. Since gold is heavy this means that each gold coin is significantly smaller.\n\nForget about weighing and just take the bag whose coins are much smaller than the coins in the other bags.\n\n\u2022 As much as this isn't the intended strategy, I really can't say this is wrong, because this would totally work. However, in the context of Math.se, it seems that this approach is a bit less mathematical than it was supposed to be. Perhaps it would be better suited for puzzling.se \u2013\u00a0Asimov Nov 13 '14 at 14:38\n\u2022 The fake ones could be made of tungsten; tungsten has a density remarkably similar to gold's. \u2013\u00a0Dan Uznanski Nov 13 '14 at 14:44\n\u2022 Perhaps the others are all gold-plated platinum or iridium, so they are very nearly the same size as the gold coins. \u2013\u00a0David K Nov 13 '14 at 14:44\n1. If all coins would have equal weight ($1.00$ gram), then the total weight of taking $0$ coins from bag $0$, $1$ coin from bag $1$, etc. would be $4950$ grams. Verify: $1 + 2 + \\cdots + 99 = 4950$. To work with the example: if you take $25$ coins from bag $25$, then the total offset in weight is $0.25$ grams.\n2. Not that I know of.\n3. When the rest of the rules are the same (weighing only once), you cannot solve the puzzle.\n\u2022 Re 3: With a balance that can only give three different \"answers\" (left is heavier, right is heavier, or equality), it takes at least five weighings (because $3^4<100$). \u2013\u00a0Hagen von Eitzen Nov 13 '14 at 14:26\n\nThe total weight of coins on the digital scale is $$W=\\sum_{m=0}^{99} mw_m$$ where $w_m$ is the weight of each coin in bag number $m.$ But if the gold coins are actually in bag number $N,$ then $$w_m = \\begin{cases} 1 + 0.01 && \\text{if}\\ m = N, \\\\ 1 && \\text{if}\\ m \\neq N. \\end{cases}$$ Therefore the only term of the sum that is not equal to $m$ is the term $Nw_N,$ which is equal to $N + 0.01N.$ So $$W=\\sum_{m=0}^{99} mw_m = \\left(\\sum_{m=0}^{99} m\\right) + 0.01N = 4950 + 0.01N.$$ Knowing $W$, we solve for $N.$\n\nAre there other solutions? Certainly! We could choose $m+1$ coins from each bag numbered $m$, or in fact any number of coins as long as we take a different number of coins from each bag and know which bag contributed what number of coins. The same calculation as before tells us how many gold coins are on the scale, and we then deduce which bag they must have come from.\n\nIn fact you can find the bag of gold coins among up to $101$ bags this way. That's because there are just $101$ different numbers of coins we can draw from a bag. If any of the bags have more coins, we can solve this for more bags. But we cannot solve the puzzle for so many bags that there would have to be two bags that contribute the same number of coins to the weighing, because if we found that that was the number of gold coins on the scale then we would not know which of the two bags they came from.\n\nNow if you have only a two-pan balance that does not give a reading, it is no longer possible to determine where the gold coins are in one weighing. It will take multiple weighings, as in the solutions to this problem and this problem.", "date": "2019-10-23 23:26:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1020066/puzzle-of-gold-coins-in-the-bag/1020151", "openwebmath_score": 0.7878023982048035, "openwebmath_perplexity": 425.22829000385326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877002595528, "lm_q2_score": 0.9111797136297299, "lm_q1q2_score": 0.8950406254245054}}
{"url": "http://mathhelpforum.com/algebra/134779-compounding-interest-problem-help.html", "text": "# Math Help - Compounding interest problem help\n\n1. ## Compounding interest problem help\n\nHello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on:\n\n8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337.\na. What is the effect of doubling the amount invested?\n\nb. What is the effect of doubling the annual interest rate?\n\nc. What is the effect of doubling the investment period?\n\nd. Which of the above has the greatest effect on the final amount of the investment?\n\nI can't seem to figure out how to start...\n\nAny suggestions? Thanks!\n\n2. Originally Posted by qcom\nHello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on:\n\n8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337.\na. What is the effect of doubling the amount invested?\n\nb. What is the effect of doubling the annual interest rate?\n\nc. What is the effect of doubling the investment period?\n\nd. Which of the above has the greatest effect on the final amount of the investment?\n\nI can't seem to figure out how to start...\n\nAny suggestions? Thanks!\nYou should know that\n\n$A = P(1 + r)^n$, where P is the principal, r is the percentage interest rate, n is the number of time periods, and A is the final amount.\n\n3. Using Proveit's:\nA = 5000(1.05^50) = 57337\n\n25 years, so there are 50 semiannual periods; .10/2 = .05 is rate per period; kapish?\n\n4. Hey thanks a lot 'Prove It' and Wilmer, I think I was mainly confused about the semiannual part, and so you just double the number of years because semiannual means that it is collected twice in one year, right?\n\nAlso, when you divide the rate .1 (10% in decimal form) by 2, is that also due to the fact that the interest is collected semiannually?\n\nOtherwise, I know how to proceed and do a-d.\n\nBTW, not much of a concern but I knew the formula as Y = a(1 + r)^t\n\nBut it doesn't really make a difference,\n\n5. Correct.\nAnd this is what \"happens\" during the 25 years:\nCode:\n         Interest               Balance\n0                               5000.00\n1          250.00               5250.00\n2          262.50               5512.50\n3          275.62               5788.12\n....\n49                             54606.67\n50        2730.33              57337.00\n\n6. Alright, now I got it for sure, thanks.\n\nNot sure if you want to help with another problem, but I was also confused with this one, which deals with similar solving techniques, I think...\n\n6. Consider a \\$1000 investment that is compounded annually at three different interest rates: 5%, 5.5%, and 6%.\n\na. Write and graph a function for each interest rate over a time period from 0 to 60 years.\n\nb. Compare the graphs of the three functions.\n\nc. Compare the shapes of the graphs for the first 10 years with the shapes of the graphs between 50 and 60 years.\n\nI think the functions are (for problem 'a'):\n\ny = 1000(1 + .05)^60\ny = 1000(1 + .055)^60\ny = 1000(1 + .06)^60\n\nDoes that look correct?\n\nNow for problem 'b', if I'm not mistaken, just look like horizontal lines waaay up on the y-axis.\n\nOne of the graphs, for the first equation I gave, looks like this: http://www.wolframalpha.com/input/?i=graph:+y+%3D+1000(1+%2B+.05)^60\n\nNow for 'c', would I just draw the graphs of the equations but replace the 't' value with 10 giving us new equations:\n\ny = 1000(1 + .05)^10\ny = 1000(1 + .055)^10\ny = 1000(1 + .06)^10\n\nAnd then the equations for the last ten years, would we need to find the new 'a' value for our equation, instead of just 1000, and then write a new equation for that?\n\nJust please tell me if I'm way off or something!\n\n7. Originally Posted by qcom\n> I think the functions are (for problem 'a'):\n> y = 1000(1 + .05)^60\n> y = 1000(1 + .055)^60\n> y = 1000(1 + .06)^60\n> Does that look correct?\n\nNot familiar with graphing programs; but above correct as the FINAL\nvalues (60 years later); seems to me this is really what's needed:\ny = 1000(1 + .05)^t where t= 0 to 60\n\n> Now for problem 'b', if I'm not mistaken, just look like horizontal lines\n> waaay up on the y-axis.\n\nWell, you'd have t (0 to 60) along x-axis, and the y values along y-axis\n\n> Now for 'c', would I just draw the graphs of the equations but replace\n> the 't' value with 10 giving us new equations:\n> y = 1000(1 + .05)^10\n> y = 1000(1 + .055)^10\n> y = 1000(1 + .06)^10\n\nYes, but: y = 1000(1 + .05)^t where t=1 to 10\n\n> And then the equations for the last ten years, would we need to find\n> the new 'a' value for our equation, instead of just 1000, and then write > a new equation for that?\n\nSimply this way: y = 1000(1 + .05)^t where t = 51 to 60\n.\n\n8. Ok I think I got you, so we would set up equations like this:\n\ny = 1,000(1.05)^10 y = 1,629\ny = 1,000(1.05)^20 y = 2,653\ny = 1,000(1.05)^30 y = 4,322\ny = 1,000(1.05)^40 y = 7,040\ny = 1,000(1.05)^50 y = 11,467\ny = 1,000(1.05)^60 y = 18,679\n\nAnd that's just for one of them and you would do that with the other two as well right?\n\nAlright now I get what you meant by 't' goes along the x-axis as it is basically our 'x' value in these functions, right? And then the 'y' would of course go along the 'y' axis.\n\nNow for c, I think I got that as well.\n\nDoes this look good? My only concern is that I may need to do an equation like I did for part 'a' for every single number from 1-60 3 times, because there is 3 different interest rates, or is it fine to only do it in intervals of 10?\n\n9. Originally Posted by qcom\n....or is it fine to only do it in intervals of 10?\nDunno. I'd assume it is.\nIsn't there a way in whatever graphing program you're using\nti give an instruction, like:\nDo for t = 1 to 60 : y = (1 + .05)^t ?", "date": "2014-12-18 20:41:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/134779-compounding-interest-problem-help.html", "openwebmath_score": 0.6640046238899231, "openwebmath_perplexity": 1187.416238672969, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474872757474, "lm_q2_score": 0.9046505299595162, "lm_q1q2_score": 0.8950137286781206}}
{"url": "https://math.stackexchange.com/questions/1596629/alternate-solution-to-circular-permutation-problem-with-restrictions", "text": "# Alternate solution to circular permutation problem with restrictions\n\nTen chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?\n\n$\\mathrm{(A)}\\ 240\\qquad\\mathrm{(B)}\\ 360\\qquad\\mathrm{(C)}\\ 480\\qquad\\mathrm{(D)}\\ 540\\qquad\\mathrm{(E)}\\ 720$\n\nSolution For the first man, there are $10$ possible seats. For each subsequent man, there are $4$, $3$, $2$, and $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\\cdot 4\\cdot 3\\cdot 2\\cdot 1\\cdot 2 = \\boxed{480}$.\n\nI am not satisfied with this solution. Is there another way to do this? (I just don't like this solution's..solution. There are many ways to proof the Pythagorean Theorem, and while you have no logical objections to others, surely there's some you like better or less than others? It's like that. I don't think I \"think\" like this solution. In any case, I just want to explore some other solutions that make better sense in my mind.)\n\n\u2022 Understood. I am asking for another solution to the problem is all though. I will edit it and try to express my intentions better. \u2013\u00a0mathflair Jan 2 '16 at 1:24\n\u2022 Even if this were not a request for a more satisfactory solution of some kind, I would hope that you would include more context, such as your own thoughts about solving it. After all, a multiple choice problem will often have some short cuts, and hearing your thoughts of one kind or another may better inform a Reader's responses toward solutions that you find more pleasing. \u2013\u00a0hardmath Jan 2 '16 at 1:43\n\u2022 Perhaps you could give an example of how you \"think\" to a similar problem so we have a better idea of what you are after. Personally that answer is exactly how I think about the problem. \u2013\u00a0Ian Miller Jan 2 '16 at 2:07\n\u2022 I just wanted to see if there were other ways to do this problem, so no one had to tailor a solution to my mind! I may have come across that point confusingly. ^^; \u2013\u00a0mathflair Jan 2 '16 at 5:45\n\nThis is not very different from your solution but maybe it is more clear: Let the men be $m_1, m_2, ..., m_5$ and their arrangement be the (ordered) vector $A=(a_1, ..., a_5)$ Since the seats are labelled, the arrangement $(1,3,5,7,9)$ is different than $(3,5,7,9,1)$ etc. There are 10 ways to pick $a_1$ for man $m_1$. For each arrangement vector $A$ there are $4!$ permutations for men $m_2$ to $m_5$ and all the men arrangements are now $10\\times 4!$\n\nWithout losing generality let's consider one men's arrangement $(1,3,5,7,9)$ For this case let's name the women as $w_i$, where $i$ is the seat number where their spouse is seated. Women can go to the even seats. In seat 2 you cannot have $w_1$ and $w_3$ and not $w_7$ either (across her spouse in seat 7). That leaves two choices for seat 2: $w_5$ or $w_9$. If you pick $w_5$ there is only one possible arrangement for the rest of the even seats to give: $(w_5, w_7,w_9,w_1,w_3)$. Same as you pick $w_9$, there is only one arrangement. Working the same way you can confirm that there are only two possible arrangements for the women as your solution states and get the answer $2\\times10\\times4!$\n\nAnother way\n\nPermissible positions of males in relation to their spouses is either $+3$ or $-3$, e.g. $1-4$ or $1-8$\n\nThe chart below with females at odd positions makes it obvious that if the $+3$ option is chosen for one couple, it needs to be so for all couples, (and so, too for $-3$)\n\n$1 - \\color{green}4-\\color{red}{8}\\quad\\;\\; 1-\\color{red}4-\\color{green}8$\n$3 - \\color{green}6-\\color{red}{10}\\quad 3-\\color{red}6-\\color{green}{10}$\n$5 - \\color{green}8-\\color{red}{2}\\quad\\;\\; 5-\\color{red}8-\\color{green}{2}$\n$7 - \\color{green}{10}-\\color{red}4\\quad 7-\\color{red}{10}-\\color{green}4$\n$9 - \\color{green}2-\\color{red}6\\quad\\;\\; 9-\\color{red}2-\\color{green}6$\n\nThus $10\\cdot4!$ with the alpha female at odd positions, ditto at even positions to finally yield\n\nans = $2\\cdot 10\\cdot4!$", "date": "2019-07-22 20:12:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1596629/alternate-solution-to-circular-permutation-problem-with-restrictions", "openwebmath_score": 0.7147206664085388, "openwebmath_perplexity": 244.52457619018728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759649262345, "lm_q2_score": 0.9124361604769413, "lm_q1q2_score": 0.8949866993414084}}
{"url": "https://mathematica.stackexchange.com/questions/157983/solving-the-heat-equation-on-the-semi-infinite-rod", "text": "# Solving the heat equation on the semi-infinite rod\n\nCross posted in scicomp.SE.\n\nI want to test the solution which is given below is right by Mathematica.\n\nPlease look the post in mathstackexhange\n\nor\n\nQuestion: Solve the following heat equation on the semi-infinite rod\n\n$\\qquad u_t=ku_{xx}$\n\nwhere $x,t>0$ and\n\n$\\qquad u_x(0,t) =0$ and $u(x,0)=\\begin{cases} 1, & 0 < x <2 \\\\ 0, & 2\\leq x \\end{cases}$\n\nwith proper Fourier transform.\n\n$\\qquad u(x,t) = \\frac{2}{\\pi}\\int_{0}^{\\infty}e^{-s^2 t}\\frac{1-\\cos(2s)}{s}\\cos(sx)ds.$\n\nCode\n\nBut I am not sure the solution is right. I am not capable of testing it in Mathematica.\n\nCould you help me?\n\n\u2022 I don't we can do much to help without having the code you used to get the answer. If you didn't use Mathematica to solve your problem, then this question is inappropriate -- I would mean you are asking us to both write the code and verify the solution for you. Oct 17 '17 at 17:34\n\u2022 Well, solving this in Mathematica is quite straightforward, just check the document of DSolve. Anyway, the solution is 1/2 (-Erf[(-2 + x)/(2 Sqrt[k t])] + Erf[(2 + x)/(2 Sqrt[k t])]). A quick test shows the solution in your question seems to be wrong. Oct 18 '17 at 4:44\n\u2022 Further check shows that, if one wants to express the solution as integration, then it should be $\\int_0^{\\infty } \\frac{\\sqrt{\\frac{2}{\\pi }} \\sin (2 w) e^{-k t w^2} \\cos (w x)}{w} \\, dw$ Oct 18 '17 at 12:08\n\u2022 Guys, personally I suggest not to close this post, though it's a\u2026 \"give me the code\" question, the problem is interesting, I think. Oct 18 '17 at 12:13\n\u2022 Please do not cross-post within SE sites. Choose one site and delete the questions from the others. meta.stackexchange.com/q/64068/164803 Oct 19 '17 at 12:17\n\nPersonally I think the problem is interesting, so let me extend my comments to an answer. First of all, DSolve can solve OP's problem straightforwardly (in Mathematica 10.3 or higher, if I remember correctly):\n\nWith[{u = u[t, x]},\neq = D[u, t] == k D[u, x, x];\nic = u == Piecewise[{{1, 0 < x < 2}}] /. t -> 0;\nbc = D[u, x] == 0 /. x -> 0;]\n\nasol = DSolveValue[{eq, ic, bc}, u, {t, x}, Assumptions -> {x > 0, k > 0}];\nasol[t, x]\n(* 1/2 (-Erf[(-2 + x)/(2 Sqrt[k] Sqrt[t])] + Erf[(2 + x)/(2 Sqrt[k] Sqrt[t])]) *)\n\n\nRemark\n\nThere seems to be a bug in DSolve in v11.2.0.\n\nDSolve[{eq, ic, bc}, u[t, x], {t, x}]\n\n\nwill return unevaluated.\n\nAs one can see, DSolve expresses the solution with Erf, so it's not immediately clear whether OP's solution is correct or not, and Mathematica's functions for simplifying also doesn't work well in this case, so let's obtain the analytic solution with another approach, that is, making use of Fourier cosine transform to eliminate the derivative of $x$:\n\nfct = FourierCosTransform[#, x, s] &;\n\ntset = Map[fct, {eq, ic}, {2}] /. Rule @@ bc /.\nHoldPattern@FourierCosTransform[a_, __] :> a\n\ntsol = u[t, x] /. DSolve[tset, u[t, x], t][[1]]\n(* (E^(-k s^2 t) Sqrt[2/\u03c0] Sin[2 s])/s *)\n\n\nRemark\n\nI've made the transform on the PDE in a quick way, for a more general approach, check this post.\n\nInverseFourierCosTransform has difficulty in transforming tsol, but it doesn't matter because the integral form is just what we want. By checking the formula of inverse Fourier cosine transform, we find the solution should be\n\n$$u(t,x)=\\sqrt{\\frac{2}{\\pi }} \\int_0^{\\infty } \\frac{e^{-k s^2 t} \\sqrt{\\frac{2}{\\pi }} \\cos (s x) \\sin (2 s)}{s} \\, ds$$\n\nIt's apparently different from the one in your question, and numeric calculation shows this solution is the same as the one given by DSolve, so the one in your question is wrong.\n\nFinally, a illustration for the solution:\n\nPlot3D[asol[t, x] /. k -> 1 // Evaluate, {x, 0, 4}, {t, 0, 10}]\n\n\n# Update\n\nInspired by Ars3nous' comment below, I noticed InverseFourierCosTransform can actually transform tsol. We just need a proper assumption:\n\nInverseFourierCosTransform[tsol, s, x, Assumptions -> k > 0]\n(* 1/2 (-Erf[(-2 + x)/(2 Sqrt[k t])] + Erf[(2 + x)/(2 Sqrt[k t])]) *)\n\n\nApparently it's the same as asol.\n\n\u2022 Following @xzczd amazing answer using cosine transform, with k=1 (you can always scale out k in time) In Mathematica 9.0, I get using InverseFourierCosineTransform, $u(x,t)=\\frac{1}{2} \\left(-2 x \\text{erfc}\\left(\\frac{x}{2 \\sqrt{t}}\\right)+\\text{erfc}\\left(\\frac{x+2}{2 \\sqrt{t}}\\right)+\\text{erfc}\\left(-\\frac{x-2}{2 \\sqrt{t}}\\right)+\\frac{4 \\sqrt{t} e^{-\\frac{x^2}{4 t}}}{\\sqrt{\\pi }}-2\\right)$ which has two different terms along the mentioned answer. But this solution also does not satisfy boundary conditions. I wonder what is the discrepency for. Nov 5 '17 at 12:18\n\u2022 @Ars3nous Are you in v9.0.0 or v9.0.1? I just tested in v9.0.1, Win10 64bit, with InverseFourierCosTransform[tsol /. k -> 1, s, x] I got 1/2 (Erf[(2 - x)/(2 Sqrt[t])] + Erf[(2 + x)/(2 Sqrt[t])]), which is consistent with asol. Nov 5 '17 at 12:38", "date": "2021-10-27 22:42:05", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/157983/solving-the-heat-equation-on-the-semi-infinite-rod", "openwebmath_score": 0.5426247119903564, "openwebmath_perplexity": 2063.240177378497, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877726405082, "lm_q2_score": 0.9219218370002787, "lm_q1q2_score": 0.8948982545064462}}
{"url": "https://math.stackexchange.com/questions/2556569/find-the-sum-of-the-series-of-frac11-cdot-3-frac13-cdot-5-frac15", "text": "# Find the sum of the series of $\\frac{1}{1\\cdot 3}+\\frac{1}{3\\cdot 5}+\\frac{1}{5\\cdot 7}+\\frac{1}{7\\cdot 9}+\u2026$\n\nFind the sum of the series $$\\frac1{1\\cdot3}+\\frac1{3\\cdot5}+\\frac1{5\\cdot7}+\\frac1{7\\cdot9}+\\frac1{9\\cdot11}+\\cdots$$ My attempt solution: $$\\frac13\\cdot\\left(1+\\frac15\\right)+\\frac17\\cdot\\left(\\frac15+\\frac19\\right)+\\frac1{11}\\cdot\\left(\\frac19+\\frac1{13}\\right)+\\cdots$$ $$=\\frac13\\cdot\\left(\\frac65\\right)+\\frac17\\cdot \\left(\\frac{14}{45}\\right)+\\frac1{11}\\cdot\\left(\\frac{22}{117}\\right)+\\cdots$$ $$=2\\cdot\\left(\\left(\\frac15\\right)+\\left(\\frac1{45}\\right)+\\left(\\frac1{117}\\right)+\\cdots\\right)$$ $$=2\\cdot\\left(\\left(\\frac15\\right)+\\left(\\frac1{5\\cdot9}\\right)+\\left(\\frac1{9\\cdot13}\\right)+\\cdots\\right)$$ It is here that I am stuck. The answer should be $\\frac12$ but I don't see how to get it. Any suggestions?\n\nAlso, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?\n\n\u2022 @ParclyTaxel Gah! My inability to read signs when strikes again. Retracted. \u2013\u00a0Xander Henderson Dec 8 '17 at 3:22\n\u2022 Your question suggests an Infinite sum. Am I right? \u2013\u00a0Ravi Prakash Dec 8 '17 at 5:26\n\nThis is a general approach to evaluate the sum of series, like these.\n\nFirst find $n^{th}$ term of series.\n\nLet $T_n$ denote the $n^{th}$ term.\n\nWe see that,\n\n$T_1 = \\frac{1}{\\color{green}{1} \\cdot \\color{teal}{3}}$\n\n$T_2 = \\frac{1}{\\color{green}{3} \\cdot \\color{teal}{5}}$\n\nAnd so on. Let the numbers in $\\color{green}{green}$ be $$\\color{green}{X_1,X_2,X_3,X_4,..=1,3,5,7...}$$\n\nClearly they form an A.P. with common difference $=2$\n\nSo, $n^{th}$ term of this AP is $1 + (n-1) \u00d7 2 = \\color{green}{2n-1}$\n\nSimilarly, Let the numbers in $\\color{teal}{teal}$ be $$\\color{teal}{Y_1,Y_2,Y_3,Y_4,..=3,5,7,9...}$$ Clearly they form an A.P. with common difference $=2$\n\nSo, $n^{th}$ term of this AP is $3 + (n-1) \u00d7 2 =\\color{teal}{ 2n+1 }$\n\nSo, the $n^{th}$ term of the main question is just\n\n$$T_n = \\frac{1}{\\color{green}{(2n-1)} \\cdot \\color{teal}{(2n+1)}}$$\n\nNow, taking summation from $1$ to $n$ , we have,\n\n$$\\sum_{n=1}^n \\frac{1}{(2n-1) \\cdot (2n+1)}$$\n\n$$= \\sum_{n=1}^n = \\frac{1}{2} \\cdot \\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}$$\n\n$$= \\sum_{n=1}^n = \\frac{1}{2} \\cdot \\frac{1}{(2n-1)} - \\frac{1}{(2n+1)}$$\n\n$$= \\sum_{n=1}^n = \\frac{1}{2} \\cdot ( 1 - \\frac{1}{(2n+1)} )$$\n\nWhile $n = \u221e$,\n\n$$\\sum_{n=1}^\u221e = \\frac{1}{2} \\cdot ( 1 - \\frac{1}{(2(\u221e)+1)} )$$\n\n$$= \\sum_{n=1}^\u221e = \\frac{1}{2} \\cdot ( 1 - \\frac{1}{\u221e} )$$\n\nSince $\\frac{1}{\u221e} = 0$,\n\n$$\\sum_{n=1}^\u221e = \\frac{1}{2} \\cdot ( 1 - 0 )$$\n\nWhich is\n\n$$\\sum_{n=1}^\u221e = \\frac{1}{2}$$\n\nThis is a classic telescoping series. $$\\frac1{n\\cdot(n+2)}=\\frac12\\left(\\frac1n-\\frac1{n+2}\\right)$$ Thus $$\\frac{1}{1\\cdot 3}+\\frac{1}{3\\cdot 5}+\\frac{1}{5\\cdot 7}+\\frac{1}{7\\cdot 9}+\\frac{1}{9\\cdot 11}+\\cdots$$ $$=\\frac12\\left(\\frac11-\\frac13+\\frac13-\\frac15+\\frac15-\\frac17+\\frac17-\\frac19+\\frac19-\\frac1{11}+\\cdots\\right)$$ $$=\\frac12$$\n\n\\begin{align*} \\sum_{n=1}\\dfrac{1}{(2n-1)(2n+1)}&=\\dfrac{1}{2}\\sum_{n=1}\\left(\\dfrac{1}{2n-1}-\\dfrac{1}{2n+1}\\right)\\\\ &=\\dfrac{1}{2}\\sum_{n=1}\\left(\\dfrac{1}{2n-1}-\\dfrac{1}{2(n+1)-1}\\right)\\\\ &=\\dfrac{1}{2}\\sum_{n=1}\\left(\\dfrac{1}{f(n)}-\\dfrac{1}{f(n+1)}\\right)\\\\ &=\\dfrac{1}{2}\\dfrac{1}{f(1)}, \\end{align*} where $f(n)=2n-1$, and note that $f(n)^{-1}\\rightarrow 0$ as $n\\rightarrow\\infty$.\n\n$$\\frac{1}{1\\cdot 3}+\\frac{1}{3\\cdot 5}+\\frac{1}{5\\cdot 7}+\\frac{1}{7\\cdot 9}+\\frac{1}{9\\cdot 11}...=\\dfrac{1}{2}\\left(\\dfrac{1}{1}-\\dfrac{1}{3}\\right)+\\dfrac{1}{2}\\left(\\dfrac{1}{3}-\\dfrac{1}{5}\\right)+\\dfrac{1}{2}\\left(\\dfrac{1}{5}-\\dfrac{1}{7}\\right)+\\cdots=\\dfrac12$$\n\nLet me give a general method which is useful for this sum $\\frac1{1\\cdot3}+\\frac1{3\\cdot5}+\\frac1{5\\cdot7}+\\frac1{7\\cdot9}+\\frac1{9\\cdot11}+\\cdots=\\sum_{n=1}^\\infty\\dfrac{1}{(2n-1)(2n+1)}$ we have\n\n$\\sum_{n=1}^\\infty\\dfrac{1}{(2n-1)(2n+1)}=\\sum_{n=0}^\\infty\\dfrac{1}{(2n+1)(2n+3)}$\n\nwe can use this method\n\n$$\\sum_{n\\geq 0}\\frac{1}{(n+a)(n+b)}=\\frac{\\psi(a)-\\psi(b)}{a-b}\\tag{1}$$\n\nwhere $\\psi$ is digamma function.\n\nnow we can write\n\n$\\sum_{n=1}^\\infty\\dfrac{1}{(2n-1)(2n+1)}= \\frac{1}{4}\\sum_{n=1}^\\infty\\dfrac{1}{(n+\\frac{1}{2})(n+\\frac{3}{4})}=\\frac{1}{2}$\n\nsince $\\psi(\\frac{1}{2})-\\psi(\\frac{3}{2})=-2$", "date": "2019-11-17 14:43:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2556569/find-the-sum-of-the-series-of-frac11-cdot-3-frac13-cdot-5-frac15", "openwebmath_score": 0.9605997204780579, "openwebmath_perplexity": 445.5246107828509, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815491146485, "lm_q2_score": 0.9046505254608135, "lm_q1q2_score": 0.8948636081827083}}
{"url": "http://mathhelpboards.com/members/albert/?s=9278ceb6188b675b9c2ddddfeb0e6f5e", "text": "Tab Content\n\u2022 Today,\u00a021:33\nThe region $R$ is a rectangle having the vertices: (0,1),\\,(1,1),\\,(1,-1),\\,(0,-1) One way we can compute the volume is: ...\n4 replies | 47 view(s)\n\u2022 Today,\u00a019:41\nMarkFL replied to a thread logs in Pre-Algebra and Algebra\nOkay, we are given: \\ln\\left(x-\\frac{1}{x}-3\\right)=2 Converting from logarithmic to exponential form, we have: x-\\frac{1}{x}-3=e^2 ...\n1 replies | 12 view(s)\n\u2022 Today,\u00a015:26\nAs for the $\\LaTeX$, look under the \"Calculus/Analysis\" section of our Quick $\\LaTeX$ tool. You will find it, which gives the code: ...\n4 replies | 47 view(s)\n\u2022 Today,\u00a010:48\nMarkFL replied to a thread The Fonz and Geometry in Chat Room\nYes, and I was just pointing out that there can be other reasons a student fails besides lack of innate ability to understand the material, or an...\n5 replies | 63 view(s)\n\u2022 Today,\u00a010:18\nMarkFL replied to a thread The Fonz and Geometry in Chat Room\nIf a child's parents consistently put a child down, then I think this can have negative consequences on the child. I think there are a great many...\n5 replies | 63 view(s)\n\u2022 Yesterday,\u00a023:08\nI disagree with this...I think it is the societal attitude that it's okay to fail at math that is part of the problem. Someone says to their friends,...\n9 replies | 94 view(s)\n\u2022 Yesterday,\u00a022:23\nIf we have 2000 people, 300 of which are women, then the probability that all 300 women will be on the same team is given by: P(A)=\\frac{{1700...\n5 replies | 50 view(s)\n\u2022 Yesterday,\u00a022:01\nI think your second method is correct. Suppose we call the teams $X$ and $Y$...and now we need only look at one team, so let's look at team $X$. If...\n5 replies | 50 view(s)\n\u2022 Yesterday,\u00a019:04\nMarkFL replied to a thread Solar Eclipse 2017 in Chat Room\nYesterday during the big event, we had pervasive heavy cloud cover and intermittent rain. Today, sunny and clear all day. Story of my life...(Giggle)\n8 replies | 233 view(s)\n\u2022 August 21st, 2017,\u00a012:26\nHere is this week's POTW: ----- If a quadrilateral is circumscribed about a circle, prove that its diagonals and the two chords joining the...\n0 replies | 38 view(s)\n\u2022 August 21st, 2017,\u00a012:22\nSuppose \\frac{3}{2}\\le x \\le 5. Prove that 2\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x}<2\\sqrt{19}. Congratulations to MarkFL for his correct...\n1 replies | 92 view(s)\n\u2022 August 20th, 2017,\u00a021:11\nmy solution: with transformation $y=tan(x),dy=sec^2(x)dx$ I = \\int_{0}^{\\dfrac{\\pi}{4}}\\left(\\tan x + \\cot x \\right)\\left ( \\dfrac{\\tan x}{1 +...\n4 replies | 142 view(s)\n\u2022 August 20th, 2017,\u00a010:02\n$-2+3 ln2$\n4 replies | 142 view(s)\n\u2022 August 20th, 2017,\u00a001:29\nMarkFL posted a visitor message on Peter's profile\nHey Peter! (Wave) I edited your post to remove the duplicate content. Sorry for the late reply, I was busy \"powering through\" a tedious 3 hour...\n\u2022 August 19th, 2017,\u00a022:48\nto find the value of $f(\\alpha)=-17+21\\sqrt 3 i$ and $f(\\beta)=-17-21\\sqrt 3i$ seemed time-consuming do you have a better way to get them ?\n10 replies | 261 view(s)\n\u2022 August 18th, 2017,\u00a020:20\nyes a miscalculation found the answer is 43524 the solution has been edited what a shame ! my poor calculation\n10 replies | 261 view(s)\n\u2022 August 18th, 2017,\u00a011:01\n$f(x)=x^3+20x-17$ $f(r)=r^3+20r-17=0---(1)$ ($r$ is a root of $f$) $f(r+1)=(r+1)^3+20(r+1)-17=3r^2+3r+21$ from $(1):$...\n10 replies | 261 view(s)\n\u2022 August 18th, 2017,\u00a009:04\nmy solution: for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$ Using Vieta's formulas $r_1+r_2+r_3=0---(1)$ $r_1r_2+r_2r_3+r_3r_1=20---(2)$...\n10 replies | 261 view(s)\n\u2022 August 17th, 2017,\u00a005:03\nmy solution: let $A=8x^2-2xy^2$ $B= 6y$ $C= 3x^2+3x^3y^2$ from $A,B$ we have $xy^2+3y-4x^2=0-----(1)$ from $B,C$ we have...\n3 replies | 146 view(s)\n\u2022 August 17th, 2017,\u00a002:10\nSmall quibble...axis of symmetry is: t=-\\frac{1}{105} So, since the parabola opens up, the vertex is a minimum, so the minimum distance will be...\n8 replies | 174 view(s)\n\u2022 August 17th, 2017,\u00a000:59\nFor a parabola of the form: f(x)=ax^2+bx+c We know the axis of symmetry is at: x=-\\frac{b}{2a} So, for:\n8 replies | 174 view(s)\nMore Activity\n\n### 16 Visitor Messages\n\n1. Hi Albert,\n\nI was meant to reply to the PM but it seems to me your account here has exceeded your stored private messages quota and hence you cannot receive new message(s).\n\nAnyway, I just wanted to thank you for reply and guess what, I just solved that challenge! Hurray! Haha...\n\nBest,\n\nanemone\n2. Hi Albert,\n\nI am aware that to editing our own post of less than 24 hours old (which I think is a good thing rather than keep adding responses by posting new follow-up posts to make mess of the look of one particular thread) is encouraged and not frowned upon, but I wanted to also let you know I have come to realize of your latest edit/update to my latest challenge only when I was about to reply to it. If by any chance I missed reading it, that would be unfair to you because your newest edited post deserves a credit, imho.\n\nThus in the future if you want to do a major edit to your previous post, I would recommend you to instead add another reply beneath your previous post, does that sound good to you, Albert?\n\nBest wishes,\n\nanemone\n3. Hi Albert, congratulations on winning the MHB Challenges Award!\n4. Hi Albert,\n\nI wanted to say that I like your profile .sig, Little Prince was a story I read almost 2 years ago but it still touches my heart. It was the best story I ever have read in my life.\n\nBest Regards,\nBalarka\n.\n5. Hello, me again Albert!\n\nHere is what our administrator Jameson has said regarding the issue:\n\n\"Ok, when that happens all he needs to do is reload the page. It should eventually render correctly. There's no need for more info. This happens for me periodically as well and I just reload the page. The only issue would be if after reloading many times he still couldn't see Latex.\"\n\nSo, when it happens again, try reloading the page, and if after several times the error still occurs, capture a screen shot and email the image to me.\n\nBest Regards,\n\nMark.\n6. Hello again Albert,\n\nThe staff here would really like to help, and it would be informative for use if you could get a screen shot of the error and email it to me, and I will forward it to the staff so we can see exactly what the error looks like at your end.\n\nBest Regards,\n\nMark.\n7. Hello Albert,\n\nI have reported your problem to the staff here, and I can assure you they are a very knowledgeable and involved team of administrators whose primary goal is that our member's problems get addressed promptly.\n\nBest Regards,\n\nMark\n8. Hello Albert,\n\nFirst I wanted to ask you if the issue you had yesterday was resolved...I waited for your email of the problem description but never got it. I want to help you get to the bottom of the problem if I can.\n\nSecond, I edited your latest post of the solution to the minimization problem. We try not to link to other math forums unless we are showing where a problem comes from if posted by someone else, to give credit to the OP. Since this problem is yours, you should post the solution at both places rather than at one, and then linking to that at the other.\n\nYour contributions here are appreciated!\n\nBest Regards,\n\nMark.\n9. Hi Albert! Please use titles which represent the nature of your question. I have renamed your following threads:\n\n1) http://www.mathhelpboards.com/f28/fi...equation-3193/ (Original title: please find n)\n\n2) http://www.mathhelpboards.com/f28/bi...-squares-3198/ (Original title: mission impossible)\n10. Hi again! You have asked in your thread >>here<< how to move a thread if you accidentally posted in a wrong sub-forum. The answer is you can't do that. However you can report the post by clicking on the little triangle (shown below) and writing down your request. I have moved your thread to the Challenge Questions and Puzzles sub-forum.\n\nShowing Visitor Messages 1 to 10 of 16\nPage 1 of 2 12 Last\nPage 1 of 2 12 Last\n\n#### Basic Information\n\nLocation:\nTaiwan\nInterests:\nmath,program,literature,music\nOccupation:\nmath teacher\nCountry Flag:\nTaiwan\n\n#### Signature\n\nOne sees clearly only with the heart.Anything essential is invisible to the eyes\n_______From The Little Prince By Antoine de Saint-Exupery\n\n#### Statistics\n\nTotal Posts\n1,202\nPosts Per Day\n0.72\n##### Thanks Data\nThanks Given\n1,177\n2,185\nThanks Received Per Post\n1.818\n##### Visitor Messages\nTotal Messages\n16\nMost Recent Message\nJune 16th, 2014 23:06\n##### General Information\nLast Activity\nYesterday 21:30\nLast Visit\nAugust 21st, 2017 at 21:35\nLast Post\nAugust 20th, 2017 at 21:11\nJoin Date\nJanuary 25th, 2013\nReferrer\nanemone\nReferrals\n1\nReferred Members\nbrat\n\n### 9 Friends\n\n1. #### anemoneOffline\n\nParis la ville de l'amour\n\n2. #### jakncokeOffline\n\nMHB Apprentice\n\n3. #### MarkFLOnline\n\nPessimist Singularitarian\n\n4. #### mathbalarkaOffline\n\nMHB Journeyman\n\n5. #### mathmaniacOffline\n\nMHB Craftsman\n\n6. #### Pantaron EducatOffline\n\nMHB Apprentice\n\n7. #### PranavOffline\n\nMHB Craftsman\n\nMHB Master\n\n9. #### ZaidAlyafeyOffline\n\n\u0632\u064a\u062f \u0627\u0644\u064a\u0627\u0641\u0639\u064a\n\nShowing Friends 1 to 9 of 9\nPage 1 of 4 123 ... 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{"url": "https://math.stackexchange.com/questions/180150/alternating-sum-of-squares-of-binomial-coefficients", "text": "# Alternating sum of squares of binomial coefficients\n\nI know that the sum of squares of binomial coefficients is just $${2n}\\choose{n}$$ but what is the closed expression for the sum $${n\\choose 0}^2 - {n\\choose 1}^2 + {n\\choose 2}^2 + \\cdots + (-1)^n {n\\choose n}^2$$?\n\n\u2022 Can you do it with generating functions? \u2013\u00a0Nikhil Ghosh Aug 8 '12 at 2:33\n\u2022 I thought I had something clever. I have deleted my post until I have a chance to think on the case when $n$ is even. $n$ being odd still yields 0, unless I am totally mistaken. \u2013\u00a0Emily Aug 8 '12 at 2:41\n\u2022 Wolfram|Alpha gives this closed form. \u2013\u00a0joriki Aug 8 '12 at 2:45\n\u2022 I don't really understand why combinatorial proof went more or less unnoticed (while standard application of generating functions is heavily upvoted). \u2013\u00a0Grigory M Nov 30 '13 at 13:28\n\n$$(1+x)^n(1-x)^n=\\left( \\sum_{i=0}^n {n \\choose i}x^i \\right)\\left( \\sum_{i=0}^n {n \\choose i}(-x)^i \\right)$$\n\nThe coefficient of $$x^n$$ is $$\\sum_{k=0}^n {n \\choose n-k}(-1)^k {n \\choose k}$$ which is exactly your sum.\n\nOn another hand:\n\n$$(1+x)^n(1-x)^n=(1-x^2)^n=\\left( \\sum_{i=0}^n {n \\choose i}(-1)^ix^{2i} \\right)$$\n\nThus, the coefficient of $$x^n$$ is $$0$$ if $$n$$ is odd or $$(-1)^{\\frac{n}2}{n \\choose n/2}$$ if $$n$$ is even.\n\n\u2022 ty fixed it. I hope that was the only one :) \u2013\u00a0N. S. Aug 8 '12 at 2:51\n\nHere's a combinatorial proof.\n\nSince $\\binom{n}{k} = \\binom{n}{n-k}$, we can rewrite the sum as $\\sum_{k=0}^n \\binom{n}{k} \\binom{n}{n-k} (-1)^k$. Then $\\binom{n}{k} \\binom{n}{n-k}$ can be thought of as counting ordered pairs $(A,B)$, each of which is a subset of $\\{1, 2, \\ldots, n\\}$, such that $|A| = k$ and $|B| = n-k$. The sum, then, is taken over all such pairs such that $|A| + |B| = n$.\n\nGiven $(A,B)$, let $x$ denote the largest element in the symmetric difference $A \\oplus B = (A - B) \\cup (B - A)$ (assuming that such an element exists). In other words, $x$ is the largest element that is in exactly one of the two sets. Then define $\\phi$ to be the mapping that moves $x$ to the other set. The pairs $(A,B)$ and $\\phi(A,B)$ have different signs, and $\\phi(\\phi(A,B)) = (A,B)$, so $(A,B)$ and $\\phi(A,B)$ cancel each other out in the sum. (The function $\\phi$ is what is known as a sign-reversing involution.)\n\nSo the value of the sum is determined by the number of pairs $(A,B)$ that do not cancel out. These are precisely those for which $\\phi$ is not defined; in other words, those for which there is no largest $x$. But there can be no largest $x$ only in the case $A=B$. If $n$ is odd, then the requirement $\\left|A\\right| + \\left|B\\right| = n$ means that we cannot have $A=B$, so in the odd case the sum is $0$. If $n$ is even, then the number of pairs is just the number of subsets of $\\{1, 2, \\ldots, n\\}$ of size $n/2$; i.e., $\\binom{n}{n/2}$, and the parity is determined by whether $|A| = n/2$ is odd or even.\n\nThus we get $$\\sum_{k=0}^n \\binom{n}{k}^2 (-1)^k = \\begin{cases} (-1)^{n/2} \\binom{n}{n/2}, & n \\text{ is even}; \\\\ 0, & n \\text{ is odd}.\\end{cases}$$\n\n\u2022 Wonderful proof, Mike! I always prefer combinatorial proofs as they also offer motivation and explanation to the identity and not only a formal proof. What really interests me is whether you can construct a sign-reversing involution to prove the following special case of Dixon's Identity: $\\sum_{k=0}^{n} \\binom{3n}{k}^{3}(-1)^{k} = (-1)^n\\binom{3n}{n,n,n}$. \u2013\u00a0Ofir Jan 18 '13 at 16:31\n\u2022 @Ofir: I'm glad you like it! I'm a big fan of combinatorial proofs myself. That's an interesting question about Dixon's identity; you should ask it as a question on the site. \u2013\u00a0Mike Spivey Jan 18 '13 at 18:56\n\u2022 I can't edit my original comment, but in the LHS it should be $\\binom{2n}{k}^{3}$ instead of $\\binom{3n}{k}^{3}$. Reading an article by Zeliberger, I found out there's a combinatorial proof for Dixon's identity by Foata. It should be in the following French book: www-irma.u-strasbg.fr/~foata/paper/ProbComb.pdf . I think pages 37-40 generalize it but I don't know any French, can anyone help out? \u2013\u00a0Ofir Jan 19 '13 at 15:33\n\u2022 A belated comment: Your beautiful proof can be easily extended to the more general result that $\\sum\\limits_{k=0}^n \\dbinom{m}{k} \\dbinom{m}{n-k} \\left(-1\\right)^k = \\begin{cases} \\left(-1\\right)^{n/2} \\dbinom{m}{n/2} , & \\text{ if } n \\text{ is even}; \\\\ 0 , & \\text{ if } n \\text{ is odd} \\end{cases}$ for any nonnegative integers $m$ and $n$. \u2013\u00a0darij grinberg Dec 20 '17 at 0:45\n\n$$\\sum_{m=0}^n (-1)^m{n \\choose m}^2= (n!)^2\\sum_{m=0}^n \\frac{(-1)^m}{(m!)^2((n-m)!)^2}$$\n\nThis function feels hypergeometric, so we take the quotient of $c_{m+1}$ and $c_m$ where\n\n$$c_m=\\frac{(n!)^2}{(m!)^2((n-m)!)^2}$$\n\nso,\n\n$$\\frac{c_{m+1}}{c_{m}}=\\frac{((m+1)!)^2((n-m-1)!)^2}{(m!)^2((n-m)!)^2}=\\frac{(m-n)^2}{(m+1)^2}$$\n\nafter some simplification, confirming that this can be expressed in terms of a hypergeometric function. The previous result gives us the parameters of the function so we find $\\sum_{m=0}^\\infty c_m x^m = {_2}F_1(-n, -n; 1;-1)$.\n\n$${_2}F_1(-n, -n; 1;-1)= \\sum_{m=0}^\\infty \\frac{((-n)_m)^2}{(1)_k} \\frac{(-1)^k}{k!} = \\sum_{m=0}^\\infty (-1)^m{n \\choose m}^2$$\n\nWhere $(x)_n=x(x+1)\\cdots(x+n-1)=\\frac{\\Gamma(x+n)}{\\Gamma(x)}$ is Pochhammer's symbol.\n\nAn identity for ${_2}F_1$ gives an elegant \"closed form:\"\n\n$${_2}F_1(-n, -n; 1;-1)= \\frac{2^{n} \\sqrt{\\pi} \\Gamma(-n+n+1)}{\\Gamma\\left(\\frac{1-n}{2}\\right)\\Gamma\\left(\\frac{-n}{2}+n+1\\right)}= \\frac{2^{n} \\sqrt{\\pi}}{\\Gamma\\left(\\frac{1-n}{2}\\right)\\Gamma\\left(\\frac{n+2}{2}\\right)}$$\n\nNow, nothing that if $a$ is a positive integer, ${n \\choose n+a}=0$, so\n\n$$\\sum_{m=0}^\\infty (-1)^m{n \\choose m}^2 = \\sum_{m=0}^n (-1)^m{n \\choose m}^2$$\n\nand finally we get the answer\n\n$$\\sum_{m=0}^n (-1)^m{n \\choose m}^2=\\frac{2^{n} \\sqrt{\\pi}}{\\Gamma\\left(\\frac{1-n}{2}\\right)\\Gamma\\left(\\frac{n+2}{2}\\right)}$$\n\nthat holds for real numbers as well.", "date": "2020-04-08 22:47:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/180150/alternating-sum-of-squares-of-binomial-coefficients", "openwebmath_score": 0.8504700660705566, "openwebmath_perplexity": 183.23535947541004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380082333993, "lm_q2_score": 0.9019206791658465, "lm_q1q2_score": 0.8946494020762845}}
{"url": "https://qa.engineer.kmitl.ac.th/ngttmi/linear-equation-graph-problems-18ff3a", "text": "# linear equation graph problems\n\nMatrix Calculator. Graphing linear equations using method 1. Type a math problem. Graphing linear relationships word problems Get 3 of 4 questions to level up! You can enter linear equations, quadratic equations, cubic equations, trigonometric functions, etc. $\\begin{cases}5x +2y =1 \\\\ -3x +3y = 5\\end{cases}$ Yes. $\\begin{cases}2x -y = -1 \\\\ 3x +y =6\\end{cases}$ Yes. Solve. These tutorials introduce you to linear relationships, their graphs, and functions. y-intercept : y-intercept is nothing but the value at where the line intersects y-axis. To move a number to a different side, you need to subtract it from both sides. linear equations word problems worksheet - If you've been seeking a house that offers you the really best bang for your own buck, your best choice would certainly be a North Charleston SC house for purchase. Money related questions in linear equations. Our mission is to provide a free, world-class education to anyone, anywhere. Simultaneous equations (Systems of linear equations): Problems with Solutions. You're putting your home on the market to sell it. Level up on all the skills in this unit and collect up to 1500 Mastery points! It could be a curve that looks something like that, or a curve that looks something like that. Linear equations word problems: graphs Get 3 of 4 questions to level up! 4. You can control the types of problems, the number of problems, workspace, border around the problems, and more. First go to the Algebra Calculator main page. 5 b = \u2212 2 b + 3 \\frac{r-3}{4}=2r. Unknown number related questions in linear equations. Linear Equations. We welcome your feedback, comments and questions about this site or page. When we have a linear equation in slope-intercept form, we can sketch the graph (straight line) of the equation using the slope 'm' and y-intercept 'b'. The picture shown below tells us the trick. Choose any value for x and substitute into the equation to get the corresponding value for y. Graph 2x - y = 6 by the intercept method. Step 2: Find the y-intercept, let x = 0 then substitute 0 for x in the equation and solve for y Next, divide both sides of the equation by the number in front of the variable, which is called the coefficient. Wonderful Graphing Linear Equations Word Problems by Madilyn Yuengel TpT. Slope and intercept meaning in context. The intercept points are when x = 0 or y = 0. how to graph linear equations by plotting points. Example Problem Graph the following equation: y=2x+1 How to Graph the Equation in Algebra Calculator. Example 1 . Linear equations word problems: graphs. In this section, you will learn how to solve word problems using linear equations. Slope, x-intercept, y-intercept meaning in context, Practice: Relating linear contexts to graph features, Practice: Using slope and intercepts in context, Practice: Linear equations word problems: tables, Practice: Linear equations word problems: graphs, Practice: Graphing linear relationships word problems. 9. Verification is an important step to always remember for these kinds of problems. Try the given examples, or type in your own There is a simple trick behind solving word problems using linear equations. Graph is an open source linear equation grapher software for Windows which can plots a graph of linear equation with one variable only. Please submit your feedback or enquiries via our Feedback page. Calculus Calculator. In order to graph a linear equation you can put in numbers for x and y into the equation and plot the points on a graph. That is, slope = rise / run. 7. Inequalities. Level up on the above skills and collect up to 500 Mastery points Start quiz. Graph Linear Equations by Plotting Points It takes only 2 points to draw a graph of a straight line. These linear equations worksheets cover graphing equations on the coordinate plane from either y-intercept form or point slope form, as well as finding linear equations from two points. Linear equation word problems worksheet with answers. Google Classroom Facebook Twitter. We can plot these solutions in the rectangular coordinate system as shown in . The directions are from taks so do all three variables equations and solve no matter what is asked in the problem. Section 8.1, Example 4(a) Solve graphically: y \u2212 x = 1, y + x = 3. Systems of Linear Equations and Problem Solving. Step 3: Plot the two points on the Cartesian plane, Step 4: Draw a straight line passing through the two points. Practice: Relating linear contexts to graph features. The y-intercept is where the line crosses the y-axis. These Linear Equations Worksheets will produce problems for practicing graphing lines given the Y-intercept and a ordered pair. problem solver below to practice various math topics. When an equation is written in general form it is easier to graph the equation by finding the intercepts. The graph is below and the y-intercept is shown with a red dot. In the previous section, we found several solutions to the equation . Try the free Mathway calculator and So, the ordered pairs , , and are some solutions to the equation . Email. And once again I drew a line, it doesn't have to be a line it could be a curve of some kind. 5. Up next for you: Unit test. Their sum is 13. WORD PROBLEMS ON LINEAR EQUATIONS. (You may plot more than two points to check) Example: Draw the line with equation y = 2x \u2013 3 . CCSS.Math: HSF.IF.B.4. how to graph linear equations using the slope and y-intercept. Example #1: Graph y = (4/3)x + 2 Step #1: Here m = 4/3 and b = 2. If possible, try to choose values of x that will give whole numbers for y to make it easier to plot. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Slope, x-intercept, y-intercept meaning in context. 8. Step 2: From the y-intercept, use the slope to find the second point and plot it. If b \u2260 0, the line is the graph of the function of x that has been defined in the preceding section. Step 3: Draw a line to connect the two points. Word Problems on Linear Equations - Real world problems with step by step solutions. This method of drawing the graph of a linear equation is called the intercept method of graphing. Graphing Linear Equations The graph of a linear equation in two variables is a line (that's why they call it linear ). Section 3-5 : Graphing Functions For problems 1 \u2013 13 construct a table of at least 4 ordered pairs of points on the graph of the function and use the ordered pairs from the table to sketch the graph of the function. Derivatives. 5 = 2 x + 3. Evaluate. At this point, the y-coordinate is 0. Systems of Equations. Real world linear equations in action as well as free worksheet that goes hand in hand with this page's real world ,word problems. Embedded content, if any, are copyrights of their respective owners. After you enter the expression, Algebra Calculator will graph the equation y=2x+1. This website uses cookies to ensure you get the best experience. how to graph linear equations by finding the x-intercept and y-intercept. About this unit. Problem 1. y = mx + b, where m is the slope of the line and b is the y-intercept. Quadratic Equations. 5 = 2x + 3 . The x-intercept is where the line crosses the x-axis. Learn more Accept. No. The intent of these problems is for instructors to use them for assignments and having solutions/answers easily available defeats that purpose. problem and check your answer with the step-by-step explanations. When x increases, y increases twice as fast, so we need 2x; When x is 0, y is already 1. If you missed this problem, review . That line is the solution of the equation and its visual representation. Free linear equation calculator - solve linear equations step-by-step. Note that when we use this method of graphing a linear equation, there is no advantage in first expressing y explicitly in terms of x. Here are some steps to follow: Type the following: y=2x+1; Try it now: y=2x+1 Clickable Demo Try entering y=2x+1 into the text box. So +1 is also needed; And so: y = 2x + 1; Here are some example values: Problem 2. In other words, if we can find two points that satisfies the equation of the line, then the line can be accurately drawn. Step 1: Find the y-intercept and plot the point. Solve Equations Calculus. SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. Step #2: By using this website, you agree to our Cookie Policy. Is the point $(0 ,\\frac{5}{2})$ a solution to the following system of equations? Slope and intercept meaning from a table. So, we can see that our solution from Step 2 is in fact the solution to the equation. Copyright \u00a9 2005, 2020 - OnlineMathLearning.com. Algebra Calculator. Graphing a Linear Equation Line Now that we know how to recognize a linear equation, let's review how to graph a line. Integrals. No. And if the population goes up a bunch then a lot of people are going to want to buy land. Applying intercepts and slope. Matrices Trigonometry. Solving systems of linear equations by elimination. Practice: Using slope and intercepts in context . Simplify. Improve your math knowledge with free questions in \"Solve a system of equations by graphing: word problems\" and thousands of other math skills. Khan Academy is a 501(c)(3) nonprofit organization. Recognize the Relationship Between the Solutions of an Equation and its Graph . It comes with a feature to plot more than one graphs of different equations on the same plane. These equations worksheets are a good resource for students in the 5th grade through the 8th grade. If you're seeing this message, it means we're having trouble loading external resources on our website. ... A page on how to find the equation and how to graph real world applications of linear equations. Problem 3. They are listed in . Solution: If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Trigonometry Calculator. Put 2 on the coordinate system. { 4 } =2r equations and solve no matter what is asked the... 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Uses cookies to ensure you get the corresponding value for x and substitute into the equation website you... Bunch then a lot of people are going to want to rearrange the equation finding! By step solutions same plane work can often be quite messy so don t... And the y-intercept and the y-intercept and the other point y=2x+1 Clickable Demo try entering y=2x+1 into the y=2x+1... M is the y-intercept, use the  intercept '' points directions are from taks so all! A different side, you need to subtract it from both sides, and are steps. Calculator will graph the equation by the intercept method of graphing system as shown in, we can that... To a different side, you need to subtract it from both sides of the line the. Some steps to follow: Example problem graph the equation and how graph..., it means we 're having trouble loading external resources on our website Worksheets a... How to graph linear equations the graph of a linear equation line Now that we know how to graph equation... 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Linear ) features of Khan Academy is a 501 ( c ) ( )... ( a ) solve graphically: y \u2212 x = 1, y already... Equation, let 's review how to recognize a linear equation with one variable only with... It could be a curve that looks something like this.kastatic.org and *.kasandbox.org are unblocked sure that the must. Other point, please make sure that the students must perform these of! One way to do this is to provide a free, world-class education to anyone anywhere! Quadratic equations, trigonometric functions, etc putting your home on the above skills and collect up to linear equation graph problems... C ) ( 3 ) nonprofit organization the graph of a linear equation grapher software for which! Problems using linear equations by finding the intercepts system as shown in \u2212 x = 1, increases... More people are going to want to rearrange the equation so it 's in slope-intercept form students the. The other point step 1: find the second point and plot the point$ ( 0 \\frac. 'Ll see a line to connect the two points shown in way to do is! 'Re having trouble loading external resources on our website follow: Example problem graph the equation finding! Can often be quite messy so don \u2019 t get excited about it when does!, try to choose values of x that will give whole numbers y. And collect up to 1500 Mastery points Start quiz 4 } =2r may select the type of solutions the! Easily available defeats that purpose is in fact the solution to the following system of?... This unit and collect up to 1500 Mastery points want to buy land to check ) Example: Draw line... Simple trick behind solving word problems by Madilyn Yuengel TpT to ensure you get the corresponding for. ( you may select the type of solutions that the verification work can often be quite messy so don t...\n\nScroll to Top", "date": "2021-05-07 00:22:16", "meta": {"domain": "ac.th", "url": "https://qa.engineer.kmitl.ac.th/ngttmi/linear-equation-graph-problems-18ff3a", "openwebmath_score": 0.3519676923751831, "openwebmath_perplexity": 661.8729807266412, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9817357195106375, "lm_q2_score": 0.9111797088058519, "lm_q1q2_score": 0.8945376670280062}}
{"url": "https://casmusings.wordpress.com/2015/07/27/infinite-ways-to-an-infinite-geometric-sum/", "text": "# Infinite Ways to an Infinite Geometric\u00a0Sum\n\nOne of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\\displaystyle \\frac{g}{1-r}$ for first term\u00a0g and common ratio\u00a0r when $\\left| r \\right| < 1$. \u00a0I was glad she was dissatisfied with blind use of\u00a0a formula\u00a0and\u00a0dove into a familiar (to me) derivation. \u00a0In the end, she shook me free from my routine just as she made sure she didn\u2019t fall into her own.\n\nSTANDARD INFINITE GEOMETRIC SUM DERIVATION\n\nMy standard explanation starts with a generic\u00a0infinite geometric series.\n\n$S = g+g\\cdot r+g\\cdot r^2+g\\cdot r^3+...$ \u00a0(1)\n\nWe can reason this series converges iff\u00a0$\\left| r \\right| <1$ (see Footnote 1 for an explanation). \u00a0Assume this is true for (1). \u00a0Notice the terms on the right keep multiplying by\u00a0r.\n\nThe annoying part of summing any infinite series is the ellipsis (\u2026). \u00a0Any finite number of terms always has a finite sum, but that simply written, but vague\u00a0ellipsis is logically difficult. \u00a0In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series. \u00a0You can accomplish this by continuing the pattern on the right: \u00a0multiplying both sides by\u00a0r\n\n$r\\cdot S = r\\cdot \\left( g+g\\cdot r+g\\cdot r^2+... \\right)$\n\n$r\\cdot S = g\\cdot r+g\\cdot r^2+g\\cdot r^3+...$ \u00a0(2)\n\nThis seems to\u00a0make make the right side of (2) identical to the right side of (1) except for the leading\u00a0g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical. \u00a0After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.\n\n$(1)-(2) = S-S\\cdot r = S\\cdot (1-r)=g$\n\n$\\displaystyle S=\\frac{g}{1-r}$\n\nSTUDENT PREFERENCES\n\nI despise giving any formula to any of my classes without\u00a0at least exploring its genesis. \u00a0I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.\n\nIn my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number\u00a0prefer\u00a0the quick \u201cmultiply-by-r-and-subtract\u201d approach used to derive the summation formula. \u00a0For many, apparently, the dynamic manipulation is more meaningful than a static rule. \u00a0It\u2019s very cool to watch student preferences at play.\n\nK\u2019s VARIATION\n\nK understood the proof, and then asked a question I hadn\u2019t thought to ask. \u00a0Why did we have to multiply by\u00a0r? \u00a0Could\u00a0multiplication by $r^2$ also determine the summation\u00a0formula?\n\nI had\u00a0three\u00a0nearly\u00a0simultaneous thoughts followed quickly by a fourth. \u00a0First, why hadn\u2019t I ever thought to ask that? \u00a0Second, geometric series for $\\left| r \\right|<1$ are absolutely convergent, so K\u2019s suggestion should work. \u00a0Third, while the formula would initially look different, absolute convergence guaranteed that whatever the \u201c$r^2$ formula\u201d looked like, it had to be algebraically equivalent to the standard form. \u00a0While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K\u2019s question and the equivalence from Thought #3.\n\nMultiplying (1) by $r^2$ gives\n\n$r^2 \\cdot S = g\\cdot r^2 + g\\cdot r^3 + ...$ (3)\n\nand the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to\u00a0get the inevitable result.\n\n$(1)-(3) = S-S\\cdot r^2 = g+g\\cdot r$\n\n$S\\cdot \\left( 1-r^2 \\right) = g\\cdot (1+r)$\n\n$\\displaystyle S=\\frac{g\\cdot (1+r)}{1-r^2} = \\frac{g\\cdot (1+r)}{(1+r)(1-r)} = \\frac{g}{1-r}$\n\nThat was cool, but this success meant that there were surely many more options.\n\nEXTENDING\n\nWhy stop at multiplying by\u00a0r\u00a0or $r^2$? \u00a0Why not multiply both sides of (1) by a generic $r^N$ for any natural number N? \u00a0 That would give\n\n$r^N \\cdot S = g\\cdot r^N + g\\cdot r^{N+1} + ...$ (4)\n\nwhere the ellipses of (1) and (4) are again identical by the method of Footnote 2. \u00a0Subtracting (4) from (1) gives\n\n$(1)-(4) = S-S\\cdot r^N = g+g\\cdot r + g\\cdot r^2+...+ g\\cdot r^{N-1}$\n\n$S\\cdot \\left( 1-r^N \\right) = g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)$ \u00a0(5)\n\nThere are two ways to proceed from (5). \u00a0You could recognize the right side as a finite geometric sum with first term 1 and ratio\u00a0r. \u00a0Substituting that formula and dividing by $\\left( 1-r^N \\right)$ would give the general result.\n\nAlternatively, I could see students exploring $\\left( 1-r^N \\right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor. \u00a0I got the following TI-Nspire CAS result in about 10-15 seconds,\u00a0clearly\u00a0suggesting that\n\n$1-r^N = (1-r)\\left( 1+r+r^2+...+r^{N-1} \\right)$. \u00a0(6)\n\nMath induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS. \u00a0From there, dividing both sides of (5) by $\\left( 1-r^N \\right)$ gives the generic result.\n\n$\\displaystyle S = \\frac{g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)}{\\left( 1-r^N \\right)}$\n\n$\\displaystyle S = \\frac{g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right) }{(1-r) \\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)} = \\frac{g}{1-r}$\n\nIn the end, K helped me see there wasn\u2019t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways. \u00a0Nice.\n\nFOOTNOTES\n\n1)\u00a0RESTRICTING r: \u00a0Obviously an infinite geometric\u00a0series diverges for $\\left| r \\right| >1$ because that would make $g\\cdot r^n \\rightarrow \\infty$ as $n\\rightarrow \\infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.\n\nFor $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \\ne 0$ , you get\u00a0a\u00a0sum of an\u00a0infinite number\u00a0of some\u00a0nonzero quantity, and that\u00a0is always infinite, no matter how small or large the nonzero quantity.\n\nThe last case, $r=-1$, is more subtle. \u00a0For $g \\ne 0$, this terms of this series alternate between positive and negative\u00a0g,\u00a0making the partial sums of the series\u00a0add to either\u00a0g or 0, depending on whether you have summed an even or an odd number of terms. \u00a0Since the partial sums alternate, the overall sum is divergent. \u00a0Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.\n\n2)\u00a0NOT ALL INFINITIES ARE THE SAME: \u00a0There are two ways to show two groups are the same size. \u00a0The obvious way is to\u00a0count the elements in each group and find out there is\u00a0the same number of elements in each, but this works only if you have a finite group size. \u00a0Alternatively, you could a) match\u00a0every\u00a0element in group 1 with a\u00a0unique\u00a0element from\u00a0group 2, and b) match\u00a0every\u00a0element in group 2\u00a0with a\u00a0unique\u00a0element from\u00a0group 1. \u00a0It is important to do both steps here to show that there are no left-over, unpaired elements in either group.\n\nSo do\u00a0the ellipses in (1) and (2) represent the same sets? \u00a0As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant. \u00a0For the second approach using pairing, we need to compare individual elements. \u00a0For every element in the ellipsis of (1), obviously there is an \u201cpartner\u201d in (2) as the multiplication of (1) by\u00a0r visually shifts all of the terms of the\u00a0series right one position, creating the necessary matches.\n\nStudents often are troubled by the second matching as it appears the ellipsis in (2) contains an \u201cextra term\u201d from the right shift. \u00a0But, for every specific term you identify in (2), its identical twin exists in (1). \u00a0In the weirdness of infinity, that \u201cextra term\u201d appears to have been absorbed without changing the \u201csize\u201d of the infinity.\n\nSince there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.\n\n### One response to \u201cInfinite Ways to an Infinite Geometric\u00a0Sum\u201d\n\n1. Here\u2019s a rather involved comment on your first Footnote about regarding restricting r (as a video link: http://tinyurl.com/riemannzeta)", "date": "2020-05-29 08:12:07", "meta": {"domain": "wordpress.com", "url": "https://casmusings.wordpress.com/2015/07/27/infinite-ways-to-an-infinite-geometric-sum/", "openwebmath_score": 0.8198376893997192, "openwebmath_perplexity": 1014.8520394689638, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9932024699586652, "lm_q2_score": 0.9005297881200701, "lm_q1q2_score": 0.8944084098322072}}
{"url": "https://math.stackexchange.com/questions/2965262/if-x-n-is-a-sequence-s-t-x-2k-rightarrow-x-and-x-2k-1-rightarrow", "text": "# If $\\{x_n\\}$ is a sequence s.t. $x_{2k} \\rightarrow x$ and $x_{2k-1} \\rightarrow x$, then $x_k \\rightarrow x$\n\nIf $$\\{x_n\\}$$ is a sequence s.t. $$x_{2k} \\rightarrow x$$ and $$x_{2k-1} \\rightarrow x$$, then $$x_k \\rightarrow x$$\n\nJust looking for feedback on my proof attemtpt\n\nProof Attempt\n\ngiven that the odd and even terms of the sequence both individually converge that means: $$1) \\ \\exists \\ N_1 \\ s.t \\ \\forall \\ k \\leq N_1 \\ |x_{2k} - x| < \\frac{\\epsilon}{2} \\\\ 2) \\ \\exists \\ N_2 \\ s.t \\ \\forall \\ k \\leq N_2 \\ |x_{2k-1} - x| < \\frac{\\epsilon}{2}$$\n\nIf we choose $$N = max\\{N_1,N_2\\}$$\n\nThen: $$|x_{2k} - x_{2k-1}| \\leq |x_{2k} - x| + |x_{2k-1} - x| < \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon$$\n\nI don't know if it is a big leap, but I'm assuming that $$|x_{2k} - x_{2k-1}|$$ represent consecutive sequential terms so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?\n\n\u2022 What you've shown is that the sequence is Cauchy. For sequences in $\\Bbb R$ this is equivalent to convergence but it is not immediately obvious that that's the case. \u2013\u00a0Lukas Kofler Oct 21 '18 at 22:40\n\u2022 I think you meant $k\\geq N_1$ and $l\\geq N_2$, not the other way around. \u2013\u00a0Mee Seong Im Oct 21 '18 at 22:40\n\u2022 @LukasKofler that is true.....and this is supposed to be in $\\mathbb{R}$, but since I wasn't thinking about it in those terms, what could I do to prove it? \u2013\u00a0dc3rd Oct 21 '18 at 22:43\n\u2022 If you are in $\\mathbb R$, you are done, since $\\mathbb R$ is complete \u2013\u00a0Don Thousand Oct 21 '18 at 22:45\n\u2022 You can just say that for all $n>N$ that $|x_n-x|<\\epsilon$ because $n$ is either of the form $2k$ or $2k-1$ for some $k>N$. \u2013\u00a0kingW3 Oct 21 '18 at 22:47\n\nso in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?\n\nOoooh. Ouch. No. That is not the right idea. I'd doesn't matter how close sequential terms get. The classic counter example is the harmonic series in which $$a_n = \\sum_{k = 1}^n \\frac 1n$$. $$|x_n - x_{n-1}|=\\frac 1n \\to 0$$ but $$\\{a_n\\}$$ does not converge.\n\nIn this case you are lucky in that you actually have and $$x$$ which $$x_{2k}$$ and $$x_{2k - 1}$$ converge to.\n\nSo for any $$\\epsilon$$ you have an $$N_1$$ so that $$n > N_1 \\implies |x_{2n} - x| < \\epsilon$$ and you have an $$N_2$$ so that $$n > N_2 \\implies |x_{2n-1} - x| < \\epsilon$$.\n\nSo if you have $$m > 2n > 2n-1$$ where $$n \\ge \\max (N_1, N_2)$$ then if $$m$$ is odd then $$m = 2k - 1$$ for $$k > n> N_2$$ so $$|x_m - x|= |x_{2k -1} - x| < \\epsilon$$. But if $$m$$ is even then $$m = 2k$$ for $$k > n > N_1$$ so $$|x_m - x| = |x_{2k} - x| < \\epsilon$$.\n\nIn other words, let $$M = 2\\max (N_1, N_2)$$. Then if $$m > M$$ then if $$m = 2k$$we have $$k > N_1$$ and if $$m = 2k -1$$ then we have $$k > N_2$$. ANd either way $$|x_m - x| < \\epsilon$$.\n\n.....\n\nNow if you hadn't been given that $$x_{2k}, x_{2k-1}\\to x$$ and where given that $$x_{2k}$$ and $$x_{2k-1}$$ were Chauchy and needed to prove $$x_m$$ was cauchy you would have had the right idea only you don't prove it only for the subsequent terms you must prove it for any TWO terms $$m_1, m_2 > N$$.\n\nAnd we'd do this by taken cases.\n\nCase 1: if $$m_1, m_2$$ are both even then $$m_1, m_2 \\ge N_1$$ (um, why did you write $$k \\le N_1$$? That was a typo I assume) so $$|x_{m_1} - x_{m_2}| < \\frac {\\epsilon}2 < \\epsilon.$$\n\nCase 2: if $$m_1, m_2$$ are both odd... some thing but with $$N_2$$.\n\nCase 3: If $$m_1$$ is even , $$m_2$$ is odd are opposite parity then $$|x_{m_1} - x_{m_2} \\le |x_{m_1} - x_{2k}| + |x_{2k} - x_{2k -1}| + |x_{2k-1} - x_{m_2}| < \\frac \\epsilon 2 + \\frac \\epsilon 2 + \\frac \\epsilon 2 = \\frac 32 \\epsilon$$.\n\nSo you'd have to modify for $$\\frac \\epsilon 3$$ instead.\n\n\u2022 Your solution is very interesting. I haven't been able to remove the notion from my head, but I'm always under the impression that since these are \"exercises\" or questions from an assignment they will always have an all encompassing solution that captures everything at once. But your solution had to break it down into cases. If we were to expand on this, say that there were 100 or 1000 cases, I guess we would have to just explicitly show them all and possibly over time refine it. Thanks @fleablood for the thorough explanation. \u2013\u00a0dc3rd Oct 22 '18 at 1:34\n\u2022 If there were a thousand case we just need to consider the maximum of the resulting $N_i$s If $n > \\max N_i$ then $n$ will qualify no matter what. \u2013\u00a0fleablood Oct 22 '18 at 1:36", "date": "2019-11-20 19:36:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2965262/if-x-n-is-a-sequence-s-t-x-2k-rightarrow-x-and-x-2k-1-rightarrow", "openwebmath_score": 0.8675025701522827, "openwebmath_perplexity": 166.29054256668454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363480718235, "lm_q2_score": 0.9086178987887253, "lm_q1q2_score": 0.8943856242863876}}
{"url": "http://mathhelpforum.com/algebra/281823-area-square-circle-2.html", "text": "# Thread: Area of Square in Circle\n\n1. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nImprove our bounds???\nYes, at the moment we have:\n\n$\\displaystyle 2<\\pi<4$\n\nSurely, we can narrow the gap...\n\n2. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nYes, at the moment we have:\n\n$\\displaystyle 2<\\pi<4$\n\nSurely, we can narrow the gap...\nMark,\n\nI would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?\n\n3. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nMark,\n\nI would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?\nI prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.\n\nCan you think of a way to improve the bounds on pi we have so far?\n\n4. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nI prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.\n\nCan you think of a way to improve the bounds on pi we have so far?\nHow can we improve the bounds on pi here?\n\n5. ## Re: Area of Square in Circle\n\nAn easy way to determine the area is to see the square as a rhombus with the circle's diameter as the rhombus' diagonal. That way you can get its area without using those pesky square roots.\n\n6. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nHow can we improve the bounds on pi here?\nis there a simple formula for the area of a regular polygon?\n\n7. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nHow can we improve the bounds on pi here?\nSuppose we use hexagons rather than squares:\n\nLet the radius of the circle be 1, so that its area is $\\displaystyle \\pi$ units squared. The area $\\displaystyle A_S$ of the smaller hexagon is:\n\n$\\displaystyle A_S=6\\left(\\frac{1}{2} \\sin\\left(\\frac{2\\pi}{6}\\right)\\right)= \\frac{3\\sqrt{3}}{2}\\approx2.6$\n\nAnd the area $\\displaystyle A_L$ of the larger hexagon is:\n\n$\\displaystyle A_L=6\\left(\\frac{1}{2} \\left(\\frac{2}{\\sqrt{3}}\\right)^2\\sin\\left(\\frac{2 \\pi}{6}\\right)\\right)= 2\\sqrt{3}\\approx3.5$\n\nWhat do you think would happen if we used $\\displaystyle n$-gons having more and more sides? Let's let $\\displaystyle A_n$ be the area of an $\\displaystyle n$-gon circumscribed by the circle...we have:\n\n$\\displaystyle A_n=\\frac{n}{2}\\sin\\left(\\frac{2\\pi}{n}\\right)$\n\nUsing a computer, we find:\n\n $\\displaystyle n$ $\\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122\n\nIt appears that as $\\displaystyle n\\to\\infty$ we have $\\displaystyle A_n$ approaching some fixed finite value.\n\n8. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nSuppose we use hexagons rather than squares:\n\nLet the radius of the circle be 1, so that its area is $\\displaystyle \\pi$ units squared. The area $\\displaystyle A_S$ of the smaller hexagon is:\n\n$\\displaystyle A_S=6\\left(\\frac{1}{2} \\sin\\left(\\frac{2\\pi}{6}\\right)\\right)= \\frac{3\\sqrt{3}}{2}\\approx2.6$\n\nAnd the area $\\displaystyle A_L$ of the larger hexagon is:\n\n$\\displaystyle A_L=6\\left(\\frac{1}{2} \\left(\\frac{2}{\\sqrt{3}}\\right)^2\\sin\\left(\\frac{2 \\pi}{6}\\right)\\right)= 2\\sqrt{3}\\approx3.5$\n\nWhat do you think would happen if we used $\\displaystyle n$-gons having more and more sides? Let's let $\\displaystyle A_n$ be the area of an $\\displaystyle n$-gon circumscribed by the circle...we have:\n\n$\\displaystyle A_n=\\frac{n}{2}\\sin\\left(\\frac{2\\pi}{n}\\right)$\n\nUsing a computer, we find:\n\n $\\displaystyle n$ $\\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122\n\nIt appears that as $\\displaystyle n\\to\\infty$ we have $\\displaystyle A_n$ approaching some fixed finite value.\nWhat a great explanation! Nicely done!\n\nPage 2 of 2 First 12", "date": "2019-02-23 00:33:36", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/281823-area-square-circle-2.html", "openwebmath_score": 0.8550595641136169, "openwebmath_perplexity": 1064.1678968634471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9884918499186287, "lm_q2_score": 0.9046505325302034, "lm_q1q2_score": 0.8942396784306534}}
{"url": "https://math.stackexchange.com/questions/2971315/how-do-i-combine-standard-deviations-of-two-groups", "text": "# How do I combine standard deviations of two groups?\n\nI have 2 groups of people. I'm working with the data about their age. I know the means, the standard deviations and the number of people. I don't know the data of each person in the groups.\n\nGroup 1 :\n\nMean = 35 years old; SD = 14; n = 137 people\n\nGroup 2 :\n\nMean = 31 years old; SD = 11; n = 112 people\n\nI want to combine those 2 groups to obtain a new mean and SD. It's easy for the mean, but is it possible for the SD? I do not know the distribution of those samples, and I can't assume those are normal distributions. Is there a formula for distributions that aren't necessarily normal?\n\n\u2022 Hey, welcome to Math Stackexchange! If you can, can you please add some context to the question? I'm not a stats guy but I'm a little confused by what you mean by \"subjects\". Thanks! \u2013\u00a0SalmonKiller Oct 25 '18 at 21:33\n\u2022 I just edited my post to add more context and be more specific. Thanks! \u2013\u00a0Nicolas Melan\u00e7on Oct 25 '18 at 21:37\n\nContinuing on from BruceET's explanation, note that if we are computing the unbiased estimator of the standard deviation of each sample, namely $$s = \\sqrt{\\frac{1}{n-1} \\sum_{i=1}^n (x_i - \\bar x)^2},$$ and this is what is provided, then note that for samples $$\\boldsymbol x = (x_1, \\ldots, x_n)$$, $$\\boldsymbol y = (y_1, \\ldots, y_m)$$, let $$\\boldsymbol z = (x_1, \\ldots, x_n, y_1, \\ldots, y_m)$$ be the combined sample, hence the combined sample mean is $$\\bar z = \\frac{1}{n+m} \\left( \\sum_{i=1}^n x_i + \\sum_{j=1}^m y_i \\right) = \\frac{n \\bar x + m \\bar y}{n+m}.$$ Consequently, the combined sample variance is $$s_z^2 = \\frac{1}{n+m-1} \\left( \\sum_{i=1}^n (x_i - \\bar z)^2 + \\sum_{j=1}^m (y_i - \\bar z)^2 \\right),$$ where it is important to note that the combined mean is used. In order to have any hope of expressing this in terms of $$s_x^2$$ and $$s_y^2$$, we clearly need to decompose the sums of squares; for instance, $$(x_i - \\bar z)^2 = (x_i - \\bar x + \\bar x - \\bar z)^2 = (x_i - \\bar x)^2 + 2(x_i - \\bar x)(\\bar x - \\bar z) + (\\bar x - \\bar z)^2,$$ thus $$\\sum_{i=1}^n (x_i - \\bar z)^2 = (n-1)s_x^2 + 2(\\bar x - \\bar z)\\sum_{i=1}^n (x_i - \\bar x) + n(\\bar x - \\bar z)^2.$$ But the middle term vanishes, so this gives $$s_z^2 = \\frac{(n-1)s_x^2 + n(\\bar x - \\bar z)^2 + (m-1)s_y^2 + m(\\bar y - \\bar z)^2}{n+m-1}.$$ Upon simplification, we find $$n(\\bar x - \\bar z)^2 + m(\\bar y - \\bar z)^2 = \\frac{mn(\\bar x - \\bar y)^2}{m + n},$$ so the formula becomes $$s_z^2 = \\frac{(n-1) s_x^2 + (m-1) s_y^2}{n+m-1} + \\frac{nm(\\bar x - \\bar y)^2}{(n+m)(n+m-1)}.$$ This second term is the required correction factor.\n\n\u2022 Just to tie things together, I tried your formula with my fake data and got a perfect match: ((n1-1)*var(x1) + (n2-1)*var(x2))/(n1+n2-1) + ((n1*n2)*(mean(x1)-mean(x2))^2)/((n1+n2)*(n1+n2-1)) returns 1157.706 and so does var(x). Thanks, I haven't seen this formula before. \u2013\u00a0BruceET Oct 26 '18 at 2:28\n\nNeither the suggestion in a previous (now deleted) Answer nor the suggestion in the following Comment is correct for the sample standard deviation of the combined sample.\n\nKnown data for reference.: First, it is helpful to have actual data at hand to verify results, so I simulated samples of sizes $$n_1 = 137$$ and $$n_2 = 112$$ that are roughly the same as the ones in the question.\n\nCombined sample mean: You say 'the mean is easy' so let's look at that first. The sample mean $$\\bar X_c$$ of the combined sample can be expressed in terms of the means $$\\bar X_1$$ and $$\\bar X_2$$ of the first and second samples, respectively, as follows. Let $$n_c = n_1 + n_2$$ be the sample size of the combined sample, and let the notation using brackets in subscripts denote the indices of the respective samples.\n\n$$\\bar X_c = \\frac{\\sum_{[c]} X_i}{n} = \\frac{\\sum_{[1]} X_i + \\sum_{[2]} X_i}{n_1 + n_1} = \\frac{n_1\\bar X_1 + n_2\\bar X_2}{n_1+n_2}.$$\n\nLet's verify that much in R, using my simulated dataset (for now, ignore the standard deviations):\n\nset.seed(2025); n1 = 137; n2 = 112\nx1 = rnorm(n1, 35, 45);  x2 = rnorm(n2, 31, 11)\nx = c(x1,x2)              # combined dataset\nmean(x1); sd(x1)\n[1] 31.19363              # sample mean of sample 1\n[1] 44.96014\nmean(x2); sd(x2)\n[1] 31.57042              # sample mean of sample 2\n[1] 10.47946\nmean(x); sd(x)\n[1] 31.36311              # sample mean of combined sample\n[1] 34.02507\n(n1*mean(x1)+n2*mean(x2))/(n1+n2)  # displayed formula above\n[1] 31.36311              # matches mean of comb samp\n\n\nSuggested formulas give incorrect combined SD: Here is a demonstration that neither of the proposed formulas finds $$S_c = 34.025$$ the combined sample:\n\nAccording to the first formula $$S_a = \\sqrt{S_1^2 + S_2^2} = 46.165 \\ne 34.025.$$ One reason this formula is wrong is that it does not take account of the different sample sizes $$n_1$$ and $$n_2.$$\n\nAccording to the second formula we have $$S_b = \\sqrt{(n_1-1)S_1^2 + (n_2 -1)S_2^2} = 535.82 \\ne 34.025.$$\n\nTo be fair, the formula $$S_b^\\prime= \\sqrt{\\frac{(n_1-1)S_1^2 + (n_2 -1)S_2^2}{n_1 + n_2 - 2}} = 34.093 \\ne 34.029$$ is more reasonable. This is the formula for the 'pooled standard deviation' in a pooled 2-sample t test. If we may have two samples from populations with different means, this is a reasonable estimate of the (assumed) common population standard deviation $$\\sigma$$ of the two samples. However, it is not a correct formula for the standard deviation $$S_c$$ of the combined sample.\n\nsd.a = sqrt(sd(x1)^2 + sd(x2)^2);  sd.a\n[1] 46.16528\nsd.b = sqrt((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2);  sd.b\n[1] 535.8193\nsd.b1 = sqrt(((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2)/(n1+n2-2))\nsd.b1\n[1] 34.09336\n\n\nMethod for correct combined SD: It is possible to find $$S_c$$ from $$n_1, n_2, \\bar X_1, \\bar X_2, S_1,$$ and $$S_2.$$ I will give an indication how this can be done. For now, let's look at sample variances in order to avoid square root signs.\n\n$$S_c^2 = \\frac{\\sum_{[c]}(X_i - \\bar X_c)^2}{n_c - 1} = \\frac{\\sum_{[c]} X_i^2 - n\\bar X_c^2}{n_c - 1}$$\n\nWe have everything we need on the right-hand side except for $$\\sum_{[c]} X_i^2 = \\sum_{[1]} X_i^2 + \\sum_{[2]} X_i^2.$$ The two terms in this sum can be obtained for $$i = 1,2$$ from $$n_i, \\bar X_i$$ and $$S_c^2$$ by solving for $$\\sum_{[i]} X_i^2$$ in a formula analogous to the last displayed equation. [In the code below we abbreviate this sum as $$Q_c = \\sum_{[c]} X_i^2 = Q_1 + Q_2.$$]\n\nAlthough somewhat messy, this process of obtaining combined sample variances (and thus combined sample SDs) is used in many statistical programs, especially when updating archival information with a subsequent sample.\n\nNumerical verification of correct method: The code below verifies that the this formula gives $$S_c = 34.02507,$$ which is the result we obtained above, directly from the combined sample.\n\nq1 = (n1-1)*var(x1) + n1*mean(x1)^2; q1\n[1] 408219.2\nq2 = (n2-1)*var(x2) + n2*mean(x2)^2; q1\n[1] 123819.4\nqc = q1 + q2\nsc = sqrt( (qc - (n1+n2)*mean(x)^2)/(n1+n2-1) ); sc\n[1] 34.02507\n\n\u2022 This page also shows this method getting the variance of a combined sample. \u2013\u00a0BruceET Oct 26 '18 at 2:00", "date": "2019-07-21 13:20:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2971315/how-do-i-combine-standard-deviations-of-two-groups", "openwebmath_score": 0.8908621072769165, "openwebmath_perplexity": 467.6740642626825, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126488274565, "lm_q2_score": 0.91367652400137, "lm_q1q2_score": 0.8942267709768439}}
{"url": "http://math.stackexchange.com/questions/174055/finding-the-divisors-of-a-number", "text": "# Finding the divisors of a number\n\nMy text says that $p^3q^6$ has 28 divisors. Could anyone please explain to me how they got 28 here ? Edit: $p$ and $q$ are distinct prime numbers Sorry for the late addition..\n\n-\nSo far there are three answers and only one up-vote: mine. \u2013\u00a0 Michael Hardy Jul 23 '12 at 3:04\n@Michael - what's your point? \u2013\u00a0 Gerry Myerson Jul 23 '12 at 3:05\nUsually a question worth answering is worth up-voting. It seems as if people often neglect that. \u2013\u00a0 Michael Hardy Jul 23 '12 at 3:12\n@Michael: I would be interested to hear your rationale for \"Usually a question worth answering is worth up-voting\". \u2013\u00a0 MJD Jul 23 '12 at 3:15\nAn upvote, to me, indicates that the question significantly increases the value of the resource. In my opinion, this question doesn't rise to that level. It's good enough to answer, but not good enough to upvote. \u2013\u00a0 Gerry Myerson Jul 23 '12 at 3:54\n\nThere can be 0-3 factors of $p$, so there are 4 ways for that to occur. There are 0-6 factors for $q$, so there are 7 ways for that to occur.\n\n-\n\nLet's take $p=3$ and $q=2$ as an example. Then the number is $3^32^6 = 1728$, and the 28 divisors of 1728 are:\n\n$$\\begin{matrix} 1&2&4&8&16&32&64 \\\\ 3&6&12&24&48&96&192 \\\\ 9 & 18 & 36 & 72 & 144 & 288 & 576 \\\\ 27 & 54 & 108 & 216 & 432 & 864 & 1728 \\end{matrix}$$\n\nThese values are, respectively:\n\n$$\\begin{matrix} 2^03^0 & 2^13^0 & 2^23^0 & 2^33^0 & 2^43^0 & 2^53^0 & 2^63^0 \\\\ 2^03^1 & 2^13^1 & 2^23^1 & 2^33^1 & 2^43^1 & 2^53^1 & 2^63^1 & \\\\ 2^03^2 & 2^13^2 & 2^23^2 & 2^33^2 & 2^43^2 & 2^53^2 & 2^63^2 & \\\\ 2^03^3 & 2^13^3 & 2^23^3 & 2^33^3 & 2^43^3 & 2^53^3 & 2^63^3 \\end{matrix}$$\n\n-\nIn fact, here is the general statement: if $$n=\\prod_k p_k^{a_k}$$ is the prime factorization of $n$, then $$\\sigma_0(n)=\\prod_k (a_k+1)$$ \u2013\u00a0 \uff2a. \uff2d. Jul 23 '12 at 3:31\n+1 Inasmuch as it matters, I like this sort of answer. A concrete example, presented clearly, that tells it all. \u2013\u00a0 copper.hat Jul 23 '12 at 4:09\neven those with not-so-good-math can understand this answer. Great! \u2013\u00a0 woliveirajr Jul 23 '12 at 16:19\n\nI assume your text also says $p$ and $q$ are primes, and $p\\ne q$ --- otherwise, the statement isn't true.\n\nDo you know that any divisor of $p^3q^6$ must itself be of the form $p^aq^b$ for some $a,b$ with $0\\le a\\le3$ and $0\\le b\\le6$? If so, do you see how to get from there to 28?\n\n-\nThat follows from the fundamental theorem of arithmetic right? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic \u2013\u00a0 Modded Bear Jul 23 '12 at 3:06\n@Chuck, yes.${}$ \u2013\u00a0 Gerry Myerson Jul 23 '12 at 3:46\n@GerryMyerson Yes they are prime. Sorry I didnt put that up there \u2013\u00a0 MistyD Jul 23 '12 at 10:18\n@GerryMyerson I still dont get how they got 28 ? \u2013\u00a0 MistyD Jul 23 '12 at 10:20\n@Misty, you haven't seen Mark's answer? \u2013\u00a0 \uff2a. \uff2d. Jul 23 '12 at 11:14\n\nThe number of divisors of n=$\\prod(p_i^{a_i})$ is $\\sum(a_i+1)$ where $p_i$ are distinct primes.\n\nSo, the number of divisors of $p^3q^6$ is(1+3)(1+6) where p,q are distinct primes =>(p,q)=1.\n\n-\n\nAny number $A$ is the product of a unique set of primes. if $A=P_1^{k_1}*P_2^{k_2}*....P_n^{k_n}$ then a divisor of A needs to be of the form $P_1^{m_1}*P_2^{m_2}*....P_n^{m_n}$ where $m_i\\leq k_i$ for any $i\\in \\Bbb N \\wedge i\\leq n$\n\nHow many combinations of marbles can you make if you can choose from 3 white marbles and 6 black ones? if you pick 0 white there are 7 combinations. (0 black+0 white = 1). if you pick 1 white there are also 7 you can probably see that there are $4*7$ combinations.\n\n-", "date": "2015-02-01 12:36:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/174055/finding-the-divisors-of-a-number", "openwebmath_score": 0.9007632732391357, "openwebmath_perplexity": 515.7162159094617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974821163419856, "lm_q2_score": 0.9173026584553408, "lm_q1q2_score": 0.8942060447235621}}
{"url": "http://mathhelpforum.com/algebra/136207-simple-proof.html", "text": "1. ## A simple proof?\n\nProve that:\n2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)\nWhen n > 0\n\nGetting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...\n\n2. Use induction to prove. Show the formula holds for the base case, assume it hold for the $n^{th}$ case and derive the $(n+1)^{th}$ case.\n\nWhy don't you try and let me know where you get stuck.\n\n3. Or wait!\n\n$2 + 4 + 6 + 8 ... + 2(n-1) = 2(1+2+3+...+(n-1)) = 2(\\frac{n(n-1)}{2}) = n(n-1)$\n\nYou may know the formula to obtain the sum of the numbers from $1$ to $n.$ It is $\\frac{n(n+1)}{2}.$\n\nHere is an example:\nSum the numbers $1$ to $5.$ Then,\n\n$1+2+3+4+5 = \\frac{5\\times 6}{2} = 15.$\n\n4. Hm. Gimme a second to type as I'm thinking.\n\nn = 1\n2(1-1) = 1 * (1-1) = 0\n\nn = k+1\n2 + 4 + 6 ... + 2k = k^2 + k\n2 + 4 + 6 ... + k = k^2\n2 + 4 + 6 ... + 2(k-1) = k^2 - k\n2 + 4 + 6 ... + 2(k-1) = k * (k-1)\n\nOh, wow. Thanks so much!\n\nEDIT: Or, that second post works too. Heh, either one. Thanks for the two methods, at least.\n\n5. Originally Posted by BlackBlaze\nProve that:\n2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)\nWhen n > 0\n\nLHS:\n\n$2 + 4 + 6 + 8 +\\dots+ 2(n-4)+ 2(n-3) + 2(n-2) + 2(n-1) = S$\n\nSwapping the order of the LHS\n\n$\n2(n-1) + 2(n-2)+ 2(n-3) + 2(n-4) +\\dots+ 8 + 6+ 4 +2= S\n$\n\n$\\underbrace{2n+2n + 2n+\\dots+ 2n}_{\\text{n-1 times}} = 2S$\n\n$2n(n-1) = 2S$\n\n$n(n-1) = S$\n\n$S = n(n-1)$\n\n6. Hello,\nhere's a proof with the sum symbol (looks cool ).\nWhat you are trying to prove is that $\\sum_{k=1}^{n - 1} 2k = n(n - 1)$. Let's work with this :\n\n$\\sum_{k=1}^{n - 1} 2k$\n\nNote that each term is even, we can factorize ! ( $2 + 4 + 6 = 2(1 + 2 + 3)$). So :\n\n$\\sum_{k=1}^{n - 1} 2k = 2 \\left ( \\sum_{k=1}^{n - 1} k \\right )$\n\nNow you may use the fact that $\\sum_{k=1}^{n} k = \\frac{1}{2}n(n + 1)$, and your result follows. This is a theorem but it can be proven :\n\n$\\sum_{k=1}^{n} k = 1 + 2 + 3 + \\cdots + k$\n\n1. Assume $n$ is even.\n\nRearranging the terms (putting brackets for better comprehension). This is valid because $n$ is even :\n\n$\\sum_{k=1}^{n} k = \\left [ 1 + \\left ( n - 1 \\right ) \\right ] + \\left [ 2 + \\left ( n - 2 \\right ) \\right ] + \\cdots + \\frac{n}{2} + n$\n\nThe dotted part stops when we reach $n - \\frac{n}{2}$ (excluded) because we cannot include that term twice. (*)\n\nThis can be simplified :\n\n$\\sum_{k=1}^{n} k = n + n + \\cdots + \\frac{n}{2} + n$\n\nUsing fact (*), we simplify by multiplication :\n\n$\\sum_{k=1}^{n} k = \\left ( \\frac{n}{2} - 1 \\right ) n + \\frac{n}{2} + n$\n\nExpanding and simplifying :\n\n$\\sum_{k=1}^{n} k = \\frac{n^2}{2} - n + \\frac{n}{2} + n = \\frac{n^2}{2} + \\frac{n}{2} = \\frac{n^2 + n}{2} = \\frac{n(n + 1)}{2} = \\frac{1}{2} n(n + 1)$\n\n2. Assume $n$ is odd.\n\nRearranging the terms (putting brackets for better comprehension). This is valid because $n$ is odd :\n\n$\\sum_{k=1}^{n} k = \\left [ 1 + n \\right ] + \\left [ 2 + \\left ( n - 1 \\right ) \\right ] + \\cdots + \\frac{n + 1}{2}$\n\nThe dotted part stops when we reach $n - \\frac{n + 1}{2}$ (excluded) because if we included it we would include two terms twice! (*)\n\nThis can be simplified :\n\n$\\sum_{k=1}^{n} k = (n + 1) + (n + 1) + \\cdots + \\frac{n + 1}{2}$\n\nUsing fact (*), we simplify by multiplication :\n\n$\\sum_{k=1}^{n} k = \\left ( \\frac{n + 1}{2} - 1 \\right )(n + 1) + \\frac{n + 1}{2}$\n\nExpanding and simplifying (using the trick that $\\frac{n + 1}{2} - 1 = \\frac{n - 1}{2}$) :\n\n$\\sum_{k=1}^{n} k = \\left ( \\frac{n + 1}{2} - 1 \\right )(n + 1) + \\frac{n + 1}{2} = \\frac{(n - 1)(n + 1)}{2} + \\frac{n + 1}{2} =$ $\\ \\frac{(n - 1)(n + 1) + (n + 1)}{2} = \\frac{(n + 1)(n - 1 + 1)}{2} = \\frac{1}{2} n(n + 1)$\n\nUsing the theorem we just proved (the proof is correct, just a bit messed up, needs better editing), we get that :\n\n$\\sum_{k=1}^{n - 1} 2k = 2 \\sum_{k=1}^{n - 1} k = 2 \\times \\left ( \\frac{1}{2} (n - 1)(n - 1 + 1) \\right ) = n(n - 1)$\n\nAnd we can conclude.\n\nDoes it make sense ?\n\nEDIT : I just saw Pickslides' proof. Mine seems a bit redundant now\n\n7. Originally Posted by BlackBlaze\nProve that:\n2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)\nWhen n > 0\n\nGetting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...\nWell, the sum of n-1 number is n-1 times their average. Since this is an arithmetic sequence, the average of all the numbers is just the average of the smallest and largest: $\\frac{2+ 2(n-1)}{2}= 1+ (n- 1)= n$.", "date": "2016-10-28 03:35:26", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/136207-simple-proof.html", "openwebmath_score": 0.9401571750640869, "openwebmath_perplexity": 671.6067288308353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936045561286, "lm_q2_score": 0.908617890746506, "lm_q1q2_score": 0.8941650552689158}}
{"url": "https://math.stackexchange.com/questions/1239106/proving-6n-1-is-always-divisible-by-5-by-induction", "text": "# Proving $6^n - 1$ is always divisible by $5$ by induction\n\nI'm trying to prove the following, but can't seem to understand it. Can somebody help?\n\nProve $6^n - 1$ is always divisible by $5$ for $n \\geq 1$.\n\nWhat I've done:\n\nBase Case: $n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.\n\nAssume true for $n = k$, where $k \\geq 1$: $6^k - 1 = 5P$.\n\nShould be true for $n = k + 1$\n\n$6^{k + 1} - 1 = 5Q$\n\n$= 6 \\cdot 6^k - 1$\n\nHowever, I am unsure on where to go from here.\n\nFor $n\\geq 1$, let $S(n)$ denote the statement $$S(n) : 5\\mid(6^n-1)\\Longleftrightarrow 6^n-1=5m, m\\in\\mathbb{Z}.$$ Base case ($n=1$): $S(1)$ says that $5\\mid(6^1-1)$, and this is true.\n\nInductive step: Fix some $k\\geq 1$ and assume that $S(k)$ is true where $$S(k) : 5\\mid(6^k-1)\\Longleftrightarrow 6^k-1=5\\ell, \\ell\\in\\mathbb{Z}.$$ To be proved is that $S(k+1)$ follows where $$S(k+1) : 5\\mid(6^{k+1}-1)\\Longleftrightarrow 6^{k+1}-1=5\\eta, \\eta\\in\\mathbb{Z}.$$ Beginning with the left-hand side of $S(k+1)$, \\begin{align} 6^{k+1} - 1 &= 6^k\\cdot 6-1\\tag{by definition}\\\\[0.5em] &= (5\\ell+1)\\cdot 6-1\\tag{by $S(k)$, the ind. hyp.}\\\\[0.5em] &= 6\\cdot 5\\ell+5\\tag{expand}\\\\[0.5em] &= 5(6\\ell+1)\\tag{factor out $5$}\\\\[0.5em] &= 5\\eta.\\tag{$\\eta=6\\ell+1; \\eta\\in\\mathbb{Z}$} \\end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.\n\nThus, by mathematical induction, the statement $S(n)$ is true for all $n\\geq 1$. $\\blacksquare$\n\n\u2022 Perfect! This has definitely made a things alot clearer! \u2013\u00a0RandomMath Apr 17 '15 at 14:56\n\u2022 @RandomMath Glad it helped. :) You may want to look at this list for some other useful questions/answers of mine on induction that may prove useful to you. I love induction a little bit too much (uname actually used to be induktio). Most of it is because I, like you, struggled a lot with it. Keep whacking away at induction proofs though, and you'll master them before long. :) \u2013\u00a0Daniel W. Farlow Apr 17 '15 at 14:58\n\u2022 I want to mention that any a^k - 1 divisible by a-1 \u2013\u00a0TigerTV.ru Jan 6 '19 at 20:24\n\nHint: Inductive step: $$6^{k+1}-1=6\\cdot 6^k-1=5\\cdot 6^k +(6^k-1)$$\n\n\u2022 The inductive step is where I'm confused on on this question, could you elaborate on how you got to this? \u2013\u00a0RandomMath Apr 17 '15 at 14:35\n\u2022 @RandomMath Since obviously $5\\mid 5\\cdot 6^k$ and by the inductive hypothesis $5\\mid 6^{k}-1$, can you see that $5\\mid 6^{k+1}-1$? And the equalities here only use trivial algebraic manipulations. \u2013\u00a0user26486 Apr 17 '15 at 14:35\n\u2022 @RandomMath From your question description it seems you know you want to prove $5\\mid 6^{k+1}-1$ assuming $5\\mid 6^k-1$. You began by showing $6^{k+1}-1=6\\cdot 6^{k}-1$. What I suggest here is continuing the equality with $6\\cdot 6^{k}-1=5\\cdot 6^{k}+(6^k-1)$, which makes it clear that we indeed have $5\\mid 6^{k+1}-1$ (assuming $5\\mid 6^{k}-1$). \u2013\u00a0user26486 Apr 17 '15 at 14:43\n\u2022 Yes that is what im trying to prove. However, im struggling to get my head around how to get from 6(6^k) -1 to 5*6^k + (6^k -1) \u2013\u00a0RandomMath Apr 17 '15 at 14:45\n\u2022 @RandomMath $6\\cdot 6^{k}-1=(5+1)\\cdot 6^k-1=5\\cdot 6^k + 6^k-1=5\\cdot 6^k +(6^k-1)$. \u2013\u00a0user26486 Apr 17 '15 at 14:46\n\nIn comments you ask about the source of the following standard proof of the inductive step $$5\\mid 6^k-1\\ \\Rightarrow\\ \\color{#c00}5\\cdot 6^k +(6^k\\!-1)\\,=\\, \\color{#0a0}{6^{k+1}-1}$$\n\nThis is a very natural question since such proofs often appear to be pulled out of a hat, like magic. There is, in fact, a good general explanation for their source. Namely such proofs are simply special cases of the proof of the Congruence Product Rule, as we show below.\n\n${\\bf Claim}\\rm\\qquad\\ 6\\equiv 1,\\, 6^{k}\\!\\equiv 1 \\Rightarrow\\ 6^{k+1}\\!\\equiv 1\\ \\ \\ \\pmod{\\!5},\\$ a special case of the following\n\n${\\bf Lemma}\\rm\\quad\\ \\, A\\!\\equiv a,\\, B\\!\\equiv b\\ \\Rightarrow\\ AB\\equiv ab\\ \\pmod{\\!n}\\ \\ \\$ [Congruence Product Rule]\n\n${\\bf Proof}\\ \\ \\ \\rm n\\mid A\\!-\\!a,\\,\\ B-b\\,\\Rightarrow\\, n\\mid ( A\\!-\\!a) B +a\\ (B\\!-\\!b) =A B\\,-\\,ab$\n\n$\\rm\\ \\ \\ \\ e.g.\\ \\ \\ 5\\mid\\ 6\\!-\\!1,\\,\\ 6^{k}\\!-\\!1\\ \\Rightarrow\\ 5\\mid(\\color{#c00}{6\\!-\\!1})\\,6^{k}+ 1\\,(6^{k}\\!\\!-\\!1) = \\color{#0a0}{6^{k+1}\\!-1}$\n\nNotice that the prior inference is precisely the same as said standard proof of the inductive step. Thus we see that this inference is simply a special case of the proof of the Congruence Product Rule. Once we know this rule, there's no need to repeat the entire proof every time we need to use it. Rather, we can simply invoke the rule as a Lemma (in divisibility form if congruences are not yet known). Then the inductive step has vivid arithmetical structure, being the computation of a product $\\, 6\\cdot 6^{k}\\equiv 6^{(k+1)}.\\,$ No longer is the innate arithmetical structure obfuscated by the details of the proof - since the proof has been encapsulated into a Lemma for convenient reuse.\n\nIn much the same way, congruences often allow one to impart intuitive arithmetical structure onto complicated inductive proofs - allowing us to reuse our well-honed grade-school skills manipulating arithmetical equations (vs. more complex divisibility relations). Often introduction of congruence language will serve to drastically simplify the induction, e.g. reducing it to a trivial induction such as $\\, 1^n\\equiv 1,\\,$ or $\\,(-1)^{2n}\\equiv 1.\\,$ The former is the essence of the matter above.\n\n\u2022 +1 for taking the time and effort to put that up and for clearly wanting to help OP. I hope s/he reads it. \u2013\u00a0Daniel W. Farlow Apr 17 '15 at 15:46\n\nWe can show by induction that $6^k$ has remainder $1$ after division by $5$.\n\nThe base case $k=1$ (or $k=0$) is straightforward, since $6=5\\cdot 1+1$.\n\nNow suppose that $6^k$ has remainder $1$ after division by $5$ for $k\\ge 1$. Thus $6^k = 5\\cdot m+1$ for some $m \\in \\mathbb{N}$. We can then see that $$6^{k+1}=6\\cdot 6^{k} = (5+1)(5\\cdot m +1) = 5^2 \\cdot m + 5 + 5\\cdot m + 1$$ $$=5(5\\cdot m + m + 1) + 1.$$\n\nThus $6^{k+1}$ has remainder $1$ after division by $5$.\n\nTherefore for every $k$, we can write $6^k = 5\\cdot m +1$ for some $m$.\n\n\u2022 Once we have established that every power of $6$ has remainder of $1$ after division by $6$, the rest of the problem follows after subtracting $1$ from $6^k$. \u2013\u00a0Joel Apr 17 '15 at 14:44\n\nThis is the inductive step written out: $$6 \\cdot 6^k - 1 = 5 \\cdot Q |+1; \\cdot \\frac{1}{6};-1 \\Leftrightarrow 6^k - 1 = \\frac{5\\cdot Q-5}{6}\\underset{P}{\\rightarrow}5\\cdot P = \\frac{5\\cdot Q - 5 }{6} | \\cdot \\frac{1}{5}; \\cdot 6\\Leftrightarrow Q=6\\cdot P + 1$$ $$6^k - 1 = \\frac{5\\cdot Q-5}{6} \\overset{Q}{\\rightarrow}\\ (6^k-1 = \\frac{5\\cdot (6\\cdot P + 1)-5}{6}\\Leftrightarrow 6^k-1 = 5\\cdot P)$$\n\n\u2022 Please, use $LATEX$. \u2013\u00a0\u064b \u064b Sep 10 '16 at 23:17\n\u2022 I tried this codecogs.com/latex/eqneditor.php However i am not sure whether this is correct LaTex \u2013\u00a0Martin Erhardt Sep 10 '16 at 23:18\n\u2022 Just put $\\$$starting from every line and at the end of every line. \u2013 \u064b \u064b Sep 10 '16 at 23:22 \u2022 ah thanks, I am new here \u2013 Martin Erhardt Sep 10 '16 at 23:28$6$has a nice property that when raised to any positive integer power, the result will have$6$as its last digit. Therefore, that number minus$1$is going to have$5$as its last digit and thus be divisible by$5\\$.\n\n\u2022 OP wants to prove this by induction \u2013\u00a0Shailesh Sep 10 '16 at 23:56\n\u2022 it's works for another numbers not only for 6. \u2013\u00a0TigerTV.ru Jan 6 '19 at 20:52", "date": "2020-02-29 14:24:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1239106/proving-6n-1-is-always-divisible-by-5-by-induction", "openwebmath_score": 0.9574488401412964, "openwebmath_perplexity": 353.7784663326344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985496421586532, "lm_q2_score": 0.9073122182277757, "lm_q1q2_score": 0.8941529443252115}}
{"url": "http://nl.mathworks.com/help/matlab/ref/istril.html?s_tid=gn_loc_drop&requestedDomain=nl.mathworks.com&nocookie=true", "text": "# Documentation\n\n### This is machine translation\n\nTranslated by\nMouse over text to see original. Click the button below to return to the English verison of the page.\n\n# istril\n\nDetermine if matrix is lower triangular\n\n## Description\n\nexample\n\ntf = istril(A) returns logical 1 (true) if A is a lower triangular matrix; otherwise, it returns logical 0 (false).\n\n## Examples\n\ncollapse all\n\nCreate a 5-by-5 matrix.\n\nD = tril(magic(5))\nD =\n\n17     0     0     0     0\n23     5     0     0     0\n4     6    13     0     0\n10    12    19    21     0\n11    18    25     2     9\n\nTest D to see if it is lower triangular.\n\nistril(D)\nans =\n\nlogical\n\n1\n\nThe result is logical 1 (true) because all elements above the main diagonal are zero.\n\nCreate a 5-by-5 matrix of zeros.\n\nZ = zeros(5);\n\nTest Z to see if it is lower triangular.\n\nistril(Z)\nans =\n\nlogical\n\n1\n\nThe result is logical 1 (true) because a lower triangular matrix can have any number of zeros on its main diagonal.\n\n## Input Arguments\n\ncollapse all\n\nInput array, specified as a numeric array. istril returns logical 0 (false) if A has more than two dimensions.\n\nData Types: single | double\nComplex Number Support: Yes\n\ncollapse all\n\n### Lower Triangular Matrix\n\nA matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.\n\nFor example, the matrix\n\n$A=\\left(\\begin{array}{cccc}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& 0\\\\ -1& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& 0\\\\ -2& -2& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& 0\\\\ -3& -3& -3& 1\\end{array}\\right)$\n\nis lower triangular. A diagonal matrix is both upper and lower triangular.\n\n### Tips\n\n\u2022 Use the tril function to produce lower triangular matrices for which istril returns logical 1 (true).\n\n\u2022 The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istril(A) == isbanded(A,size(A,1),0).", "date": "2016-09-26 08:47:57", "meta": {"domain": "mathworks.com", "url": "http://nl.mathworks.com/help/matlab/ref/istril.html?s_tid=gn_loc_drop&requestedDomain=nl.mathworks.com&nocookie=true", "openwebmath_score": 0.7574024200439453, "openwebmath_perplexity": 2131.731656389648, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464450067471, "lm_q2_score": 0.9161096124442242, "lm_q1q2_score": 0.8940739195014494}}
{"url": "https://fr.mathworks.com/help/symbolic/piecewise.html", "text": "# piecewise\n\nConditionally defined expression or function\n\n## Description\n\nexample\n\npw = piecewise(cond1,val1,cond2,val2,...) returns the piecewise expression or function pw whose value is val1 when condition cond1 is true, is val2 when cond2 is true, and so on. If no condition is true, the value of pw is NaN.\n\nexample\n\npw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal) returns the piecewise expression or function pw that has the value otherwiseVal if no condition is true.\n\n## Examples\n\n### Define and Evaluate Piecewise Expression\n\nDefine the following piecewise expression by using piecewise.\n\n$y=\\left\\{\\begin{array}{cc}-1& x<0\\\\ 1& x>0\\end{array}$\n\nsyms x\ny = piecewise(x<0, -1, x>0, 1)\ny =\npiecewise(x < 0, -1, 0 < x, 1)\n\nEvaluate y at -2, 0, and 2 by using subs to substitute for x. Because y is undefined at x = 0, the value is NaN.\n\nsubs(y, x, [-2 0 2])\nans =\n[ -1, NaN, 1]\n\n### Define Piecewise Function\n\nDefine the following function symbolically.\n\n$y\\left(x\\right)=\\left\\{\\begin{array}{cc}-1& x<0\\\\ 1& x>0\\end{array}$\n\nsyms y(x)\ny(x) = piecewise(x<0, -1, x>0, 1)\ny(x) =\npiecewise(x < 0, -1, 0 < x, 1)\n\nBecause y(x) is a symbolic function, you can directly evaluate it for values of x. Evaluate y(x) at -2, 0, and 2. Because y(x) is undefined at x = 0, the value is NaN. For details, see Create Symbolic Functions.\n\ny([-2 0 2])\nans =\n[ -1, NaN, 1]\n\n### Set Value When No Conditions Is True\n\nSet the value of a piecewise function when no condition is true (called otherwise value) by specifying an additional input argument. If an additional argument is not specified, the default otherwise value of the function is NaN.\n\nDefine the piecewise function\n\n$y\\left(x\\right)=\\left\\{\\begin{array}{cc}-2& x<-2\\\\ 0& -2\n\nsyms y(x)\ny(x) = piecewise(x<-2, -2, -2<x<0, 0, 1)\ny(x) =\npiecewise(x < -2, -2, x in Dom::Interval(-2, 0), 0, 1)\n\nEvaluate y(x) between -3 and 1 by generating values of x using linspace. At -2 and 0, y(x) evaluates to 1 because the other conditions are not true.\n\nxvalues = linspace(-3,1,5)\nyvalues = y(xvalues)\nxvalues =\n-3    -2    -1     0     1\nyvalues =\n[ -2, 1, 0, 1, 1]\n\n### Plot Piecewise Expression\n\nPlot the following piecewise expression by using fplot.\n\n$y=\\left\\{\\begin{array}{cc}-2& x<-2\\\\ x& -22\\end{array}.$\n\nsyms x\ny = piecewise(x<-2, -2, -2<x<2, x, x>2, 2);\nfplot(y)\n\n### Assumptions and Piecewise Expressions\n\nOn creation, a piecewise expression applies existing assumptions. Apply assumptions set after creating the piecewise expression by using simplify on the expression.\n\nAssume x > 0. Then define a piecewise expression with the same condition x > 0. piecewise automatically applies the assumption to simplify the condition.\n\nsyms x\nassume(x > 0)\npw = piecewise(x<0, -1, x>0, 1)\npw =\n1\n\nClear the assumption on x for further computations.\n\nassume(x,'clear')\n\nCreate a piecewise expression pw with the condition x > 0. Then set the assumption that x > 0. Apply the assumption to pw by using simplify.\n\npw = piecewise(x<0, -1, x>0, 1);\nassume(x > 0)\npw = simplify(pw)\npw =\n1\n\nClear the assumption on x for further computations.\n\nassume(x, 'clear')\n\n### Differentiate, Integrate, and Find Limits of Piecewise Expression\n\nDifferentiate, integrate, and find limits of a piecewise expression by using diff, int, and limit respectively.\n\nDifferentiate the following piecewise expression by using diff.\n\n$y=\\left\\{\\begin{array}{cc}1/x& x<-1\\\\ \\mathrm{sin}\\left(x\\right)/x& x\\ge -1\\end{array}$\n\nsyms x\ny = piecewise(x<-1, 1/x, x>=-1, sin(x)/x);\ndiffy = diff(y, x)\ndiffy =\npiecewise(x < -1, -1/x^2, -1 < x, cos(x)/x - sin(x)/x^2)\n\nIntegrate y by using int.\n\ninty = int(y, x)\ninty =\npiecewise(x < -1, log(x), -1 <= x, sinint(x))\n\nFind the limits of y at 0 and -1 by using limit. Because limit finds the double-sided limit, the piecewise expression must be defined from both sides. Alternatively, you can find the right- or left-sided limit. For details, see limit.\n\nlimit(y, x, 0)\nlimit(y, x, -1)\nans =\n1\nans =\nlimit(piecewise(x < -1, 1/x, -1 < x, sin(x)/x), x, -1)\n\nBecause the two conditions meet at -1, the limits from both sides differ and limit cannot find a double-sided limit.\n\n### Elementary Operations on Piecewise Expressions\n\nAdd, subtract, divide, and multiply two piecewise expressions. The resulting piecewise expression is only defined where the initial piecewise expressions are defined.\n\nsyms x\npw1 = piecewise(x<-1, -1, x>=-1, 1);\npw2 = piecewise(x<0, -2, x>=0, 2);\nadd = pw1 + pw2\nsub = pw1 - pw2\nmul = pw1 * pw2\ndiv = pw1 / pw2\npiecewise(x < -1, -3, x in Dom::Interval([-1], 0), -1, 0 <= x, 3)\nsub =\npiecewise(x < -1, 1, x in Dom::Interval([-1], 0), 3, 0 <= x, -1)\nmul =\npiecewise(x < -1, 2, x in Dom::Interval([-1], 0), -2, 0 <= x, 2)\ndiv =\npiecewise(x < -1, 1/2, x in Dom::Interval([-1], 0), -1/2, 0 <= x, 1/2)\n\n### Modify or Extend Piecewise Expression\n\nModify a piecewise expression by replacing part of the expression using subs. Extend a piecewise expression by specifying the expression as the otherwise value of a new piecewise expression. This action combines the two piecewise expressions. piecewise does not check for overlapping or conflicting conditions. Instead, like an if-else ladder, piecewise returns the value for the first true condition.\n\nChange the condition x<2 in a piecewise expression to x<0 by using subs.\n\nsyms x\npw = piecewise(x<2, -1, x>0, 1);\npw = subs(pw, x<2, x<0)\npw =\npiecewise(x < 0, -1, 0 < x, 1)\n\nAdd the condition x>5 with the value 1/x to pw by creating a new piecewise expression with pw as the otherwise value.\n\npw = piecewise(x>5, 1/x, pw)\npw =\npiecewise(5 < x, 1/x, x < 0, -1, 0 < x, 1)\n\n## Input Arguments\n\ncollapse all\n\nCondition, specified as a symbolic condition or variable. A symbolic variable represents an unknown condition.\n\nExample: x > 2\n\nValue when condition is satisfied, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.\n\nValue if no conditions are true, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. If otherwiseVal is not specified, its value is NaN.\n\n## Output Arguments\n\ncollapse all\n\nPiecewise expression or function, returned as a symbolic expression or function. The value of pw is the value val of the first condition cond that is true. To find the value of pw, use subs to substitute for variables in pw.\n\n## Tips\n\n\u2022 piecewise does not check for overlapping or conflicting conditions. A piecewise expression returns the value of the first true condition and disregards any following true expressions. Thus, piecewise mimics an if-else ladder.", "date": "2021-07-27 23:47:18", "meta": {"domain": "mathworks.com", "url": "https://fr.mathworks.com/help/symbolic/piecewise.html", "openwebmath_score": 0.8195955753326416, "openwebmath_perplexity": 5482.115671294876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575167960731, "lm_q2_score": 0.9099070066488187, "lm_q1q2_score": 0.8940359689682112}}
{"url": "https://math.stackexchange.com/questions/4510187/how-to-create-a-function-f-such-that-fx-y-is-high-when-either-x-or-y-i", "text": "# How to create a function $f$ such that $f(x,y)$ is high when either $x$ or $y$ is?\n\nThere are two variables, let's say $$x$$ and $$y$$.\n\nI want to come up with a function $$f:[0,5]\\times[0,5]\\longrightarrow[0,5]$$ that respects the following rules:\n\n1. If $$x$$ is high (close to the maximum value of 5) and $$y$$ is low (close to $$0$$), $$f(x,y)$$ should results in a high value (close to 5). The same would happen for the reverse ($$x$$ with a low value and $$y$$ with a high value)\n2. If $$x$$ is high and $$y$$ is high, $$f(x,y)$$ will also be high; it would be useful that the values resulted would be higher than those from point $$(1)$$.\n3. If $$x$$ is low and $$y$$ is low, $$f(x,y)$$ should result in low values.\n\nI'm having troubles starting imagining what mathematical functions would be useful, as I am not even close to being advanced in the concepts of Mathematics. Any ideas would be appreciated, at least in the sense of finding any information that could get me started.\n\n\u2022 you defined your function $f :[0,5]\\rightarrow [0,5]$, and then you talk about two input variable $x$ and $y$, i guess you meant $f :[0,5]\\times [0,5] \\rightarrow [0,5]$ ? Aug 11 at 11:47\n\u2022 that is correct, I'll add the change. Aug 11 at 11:48\n\u2022 If you have precise value for your function $f$, you could use something like multivariate interpolation. Aug 11 at 11:52\n\u2022 If you don't know the concept of interpolation you should check out first 1D interpolation like this one. Aug 11 at 11:57\n\u2022 If $f(x,y)=f(y,x)$, and $f(x,y)$ is linear with respect to $x$ (this was not part of the problem statement, it was added by me to ensure uniqueness of solution), then $f(x,y)=axy+b(x+y)+c$. Let $f(0,0)=0$, $f(5,5)=5$, then $c=0$, $5a+2b=1$. If one wants to set value $f(0,5)=d$, then $b=\\frac{d}{5}$, $a=\\frac{5-2d}{25}$. At $d=5$ result is formula from my previous comment, but OP can take $f(0,5)=d=4.9$ or something else what needed. Aug 12 at 7:14\n\nYou can take, for instance,$$f(x,y)=\\frac45\\max\\{x,y\\}+\\frac15\\min\\{x,y\\}.$$It has all the properties that you are interested in.\n\nOne way to model this could be to use the distance from the line: $$x+y=0$$ through $$(0,0)$$. This line has normal vector: $$\\vec n= \\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}$$ and the distance of a given point $$(x,y)$$ to this line is proportional to $$t=\\vec n\\cdot\\langle x,y \\rangle=x+y$$ so in case we want something like: $$f(5,5)=5\\\\ f(5,0)=4\\\\ f(0,0)=0$$ we can connect this to a single dimensional function: $$f(x,y)=g(x+y)$$ where $$g(10)=5\\\\ g(5)=4\\\\ g(0)=0$$ A way to achieve this could be to add the extra requirement $$g'(10)=0$$ so that $$f$$ has maximum at $$(x,y)=(5,5)$$ and build $$g(0)=0$$ in: $$g(t)=at^3+bt^2+ct$$ Hence \\begin{align} g(10)&=1000a+100b+10c&&=5\\\\ g(5)&=125a+25b+5c&&=4\\\\ g'(10)&=300a+20b+c&&=0 \\end{align} which can be solved for $$a,b,c$$ to have: $$f(x,y)=g(x+y)=0.002(x+y)^3-0.09(x+y)^2+1.2(x+y)$$\n\nSee this link to look at interactive GeoGebra-applet with 3D-plot of this function\n\nADDENDUM: As can be seen both from the other answer and from comments, there will be (infinitely) many ways to satisfy your requirements, but to point you towards handling the additional requirement stated in your comment below this post, you could simply add a modifier to the above solution which takes the distance to the perpendicular line: $$x-y=0$$ as input. This distance is (similarly) proportional to $$t=x-y$$, and so we need a modifier function $$m(x,y)=h(x-y)=h(t)$$ that satisfies: $$h(0)=0\\\\ h(5)=m(5,0)\\\\ h(-5)=m(0,5)$$ so just choose which modification you want at $$(5,0)$$ and $$(0,5)$$ and match for instance a quadratic function as $$h$$: $$h(t)=\\alpha t^2+\\beta t+\\gamma$$ and combine: \\begin{align} q(x,y) &=f(x,y)+m(x,y)\\\\ &=g(x+y)+h(x-y)\\\\ &=a(x+y)^3+b(x+y)^2+c(x+y)+\\alpha(x-y)^2+\\beta(x-y)+\\gamma \\end{align} but be a little careful - if $$h(t)$$ increases too rapidly away from $$h(0)$$ to one side, then $$q$$ may exceed a value of $$5$$.\n\nHere is GeoGebra-applet with example of this technique\n\n\u2022 Very interesting and detailed answer. With the risk of going a bit outside the boundaries of the question: is there a way to introduce some kind of bias towards $x$ or $y$? In the sense that $f(x,y)$ should be greater than $f(y,x)$ or the opposite. Aug 16 at 13:06\n\u2022 @Ionut-AlexandruBaltariu: I added an extra section about this. Generally, you can always combine two one-dimensional functions $g(x+y)$ and $h(x-y)$ to create formula for change in value when you move perpendicular to the two lines $x+y=0$ and $x-y=0$. This gives you a lot of control over what you want to happen. Just be careful, because combinations may escape the max/min values, since we only controlled them in points $(0,0),(5,0),(0,5)$ and $(5,5)$. Aug 17 at 8:41", "date": "2022-10-07 12:44:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4510187/how-to-create-a-function-f-such-that-fx-y-is-high-when-either-x-or-y-i", "openwebmath_score": 0.8572801947593689, "openwebmath_perplexity": 197.11111795120058, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771759145342, "lm_q2_score": 0.9059898165759307, "lm_q1q2_score": 0.8940100726081237}}
{"url": "https://math.stackexchange.com/questions/2318951/card-draw-probability", "text": "Card Draw Probability\n\nThis is a homework question. Not something to hand in, but rather a recommended practice question.\n\nA Card Game\nThree students are playing a card game. They decide to choose the \ufb01rst person to play by each selecting a card from the 52-card deck and looking for the highest card in value and suit. They rank the suits from lowest to highest: clubs, diamonds, hearts, and spades.\n\na. If the card is replaced in the deck after each student chooses, how many possible con\ufb01gurations of the three choices are possible?\n\nb. How many con\ufb01gurations are there in which each student picks a different card?\n\nc. What is the probability that all three students pick exactly the same card?\n\nd. What is the probability that all three students pick different cards?\n\nI figured out the the answer to part A is $52^3 = 140608$, since each student chooses 1 card from a deck of 52.\n\nI also know that the answer to part B is $52\\times 51\\times 50 =132600$, or $\\frac {52!}{(52-3)!}$.\n\nHowever, I am not sure how to find the probability that they all pick the same card. I initially thought it would be $\\left( \\frac 1{52}\\right)^3$, but that does not match the answer in the back of the book, which is $0.00037$.\n\nSorry for the formatting (or lack thereof), I tried my best to make it fairly easy to read, but I am unsure of how to make fancy fractions and such.\n\n\u2022 For $c$: the first person takes whatever. The probability that the second matches the first is $\\frac 1{52}$. The probability that the third matches the first is $\\frac 1{52}$ independent of what the second person drew. So... \u2013\u00a0lulu Jun 11 '17 at 20:54\n\u2022 An answer of $(1/52)\\cdot (1/52)\\cdot (1/52)=(1/52)^3$ is the probability of all three specifically selecting the ace of spades, but remember there are many ways other than just everyone getting the ace of spades for them to all have the same card. As for part (d), if you were to read and understand the birthday problem you'll find the mathematics behind it is very similar. Alternatively, look more closely at the phrasing of parts (b) and (a) and how they relate to question (d). \u2013\u00a0JMoravitz Jun 11 '17 at 21:00\n\u2022 So (1/52)^2. Thank you! Are you going to post it as an answer so that I can accept it? \u2013\u00a0Bunyip Jun 11 '17 at 21:02\n\u2022 As a complete aside, you say \"I am unsure of how to make fancy fractions and such.\" This page will give you the information you need to do so (at least on this site using MathJax). \u2013\u00a0JMoravitz Jun 11 '17 at 21:03\n\nFor part C:\n\nTry thinking about it like this, player A is the one that \"chooses\" which card B and C must choose in order for all of them to pick the same card, so the probability that all three students pick exactly the same card is:\n\n$\\ \\frac{1}{52}\\frac{1}{52} = (\\frac{1}{52})^2$\n\nFor part D:\n\nStart as in C, player A chooses the card and B has a $\\ \\frac{51}{52}$ chance of its card being different from A's card, similarly C has a $\\ \\frac{50}{52}$ of its card being different from A and B, so the probability that all cards differ is:\n\n$\\ \\frac{51}{52}\\frac{50}{52}$\n\nHint.\n\nPart c only makes sense if the cards are drawn with replacement. You already have the total number of configurations. There are 52 configurations where all the cards are the same. With these two facts, the required probability is readily calculated.\n\nFor part d, which again only makes sense if cards are drawn with replacement, you can get the number of ways in which all three cards are different from the answer to part b. This will lead to the required probability.", "date": "2019-06-24 22:14:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2318951/card-draw-probability", "openwebmath_score": 0.5665358304977417, "openwebmath_perplexity": 123.03654134228368, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771778588349, "lm_q2_score": 0.9059898140375993, "lm_q1q2_score": 0.8940100718648728}}
{"url": "https://math.stackexchange.com/questions/3180877/determinant-of-antidiagonally-constructed-matrix", "text": "# Determinant of antidiagonally constructed matrix\n\nLet $$A_n$$ be a matrix, odd dimension $$n \\times n$$, constructed from the sequence of natural numbers in such a way that we begin the sequence from the upper left corner and next numbers are inserted into the matrix along antidiagonal direction, starting from the top of the matrix (examples below) .\n\nFor such matrix the determinant is calculated.\n\nSurprisingly for the three examples below the value of determinant is equal to the central entry of the matrix with the modification of sign between two consecutive cases.\n\n$$\\det \\begin{bmatrix}1 & 2 & 4\\\\3 &\\color{red} 5 & 7\\\\6 & 8 & 9\\end{bmatrix} =-\\color{red}5$$\n\n$$\\det \\begin{bmatrix}1 & 2 & 4 & 7 & 11\\\\3 & 5 & 8 & 12 & 16\\\\6 & 9 & \\color{red}{13} & 17 & 20\\\\10 & 14 & 18 & 21 & 23\\\\15 & 19 & 22 & 24 & 25\\end{bmatrix} =\\color{red}{13}$$\n\n$$\\det \\begin{bmatrix}1 & 2 & 4 & 7 & 11 & 16 & 22\\\\3 & 5 & 8 & 12 & 17 & 23 & 29\\\\6 & 9 & 13 & 18 & 24 & 30 & 35\\\\10 & 14 & 19 & \\color{red}{25} & 31 & 36 & 40\\\\15 & 20 & 26 & 32 & 37 & 41 & 44\\\\21 & 27 & 33 & 38 & 42 & 45 & 47\\\\28 & 34 & 39 & 43 & 46 & 48 & 49\\end{bmatrix} = -\\color{red}{25}$$\n\nQuestion:\n\n\u2022 does the pattern continue for a greater $$n$$?\n\n\u2022 if so what is the explanation for the pattern?\n\n\u2022 Note that the pattern holds for the often overlooked $1\\times 1$ matrices as well: The determinant of $[\\color{red}{1}]$ is $\\color{red}{1}$. Apr 9, 2019 at 11:11\n\u2022 @Arthur Yes, 1 is also \"central\" in [1], and it has + sign, this change of sign is also strange.. Apr 9, 2019 at 11:13\n\u2022 Here's how it continues: oeis.org/A069480. Your observation would make a nice comment, BTW. Apr 9, 2019 at 11:27\n\u2022 The formula section of the OEIS link does confirm this: It says that for odd $n$, the determinant is indeed an alternating multiplied by $\\frac{(2n+1)^2}{2}$, which is the middle element of the corresponding matrix.. Apr 9, 2019 at 11:34\n\u2022 @Widawensen In this case, the easiest determinant to start with first might be the one which has central value $0$: we just have to show that this matrix is singular. But the fact that as you add $k$ to all entries, the determinant changes by $\\pm k$, might also be hard to prove... Apr 9, 2019 at 22:28\n\nLet $$M_n$$ be the matrix whose determinant we want to know (the matrix where we fill in $$n^2$$ consecutive integers antidiagonally, starting at any value $$k$$), and let $$x$$ be its middle entry. For convenience, let $$m = \\frac{n-1}{2}$$. (I will use $$n=7$$ in my example matrices, but my argument applies to all odd $$n$$.)\n\nFirst, replace the rows $$r_1, r_2, \\dots, r_n$$ of $$M_n$$ with the $$n-1$$ differences $$r_2 - r_1$$, $$r_3 - r_2$$, ..., $$r_n - r_{n-1}$$ followed by the middle row $$r_{m+1}$$. This is an invertible sequence of row operations, so we have: $$\\det(M_n) = \\det\\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\\\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\\\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\\\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\\\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\\\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\\\ x+a_1 & x+a_2 & x+a_3 & x & x-a_3 & x-a_2 & x-a_1 \\end{bmatrix}.$$ where $$x+a_1,\\dots, x+a_m, x, x-a_m, \\dots, x-a_1$$ are the entries of the middle row of $$M_n$$ (which always has this symmetric form). We show that the determinant above is always equal to $$(-1)^{m} x$$.\n\nBy linearity of the determinant, we can write this as $$\\det(M_n) = \\det\\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\\\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\\\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\\\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\\\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\\\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\\\ a_1 & a_2 & a_3 & 0 & -a_3 & -a_2 & -a_1 \\end{bmatrix} + x \\det\\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\\\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\\\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\\\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\\\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\\\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \\end{bmatrix}.$$ The first determinant is $$0$$ by cofactor expansion along the last row. The cofactors of $$a_i$$ and $$-a_i$$ are equal for each $$i$$, since one submatrix can be obtained from the other by reversing the rows, then reversing the columns, so they cancel after multiplying one by $$a_i$$ and the other by $$-a_i$$.\n\nFor the second matrix, we first do a similar operation to columns. Keeping the first column $$c_1$$ and replacing every other $$i^{\\text{th}}$$ column $$c_i$$ by the difference $$c_i - c_{i-1}$$, which is an invertible set of row operations, we have $$\\det\\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\\\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\\\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\\\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\\\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\\\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\\\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \\end{bmatrix} = \\det\\begin{bmatrix} 2 & 1 & 1 & 1 & 1 & 1 & 0 \\\\ 3 & 1 & 1 & 1 & 1 & 0 & -1 \\\\ 4 & 1 & 1 & 1 & 0 & -1 & -1 \\\\ 5 & 1 & 1 & 0 & -1 & -1 & -1 \\\\ 6 & 1 & 0 & -1 & -1 & -1 & -1 \\\\ 7 & 0 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\end{bmatrix}$$ By expansion along the last row (which introduces a multiple of $$(-1)^{n+1} = 1$$) and after reversing the rows (which introduces a multiple of $$(-1)^{m}$$) we simplify this to $$(-1)^{m}\\det\\begin{bmatrix} 0 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & -1 & -1 & -1 & -1 \\\\ 1 & 1 & 0 & -1 & -1 & -1 \\\\ 1 & 1 & 1 & 0 & -1 & -1 \\\\ 1 & 1 & 1 & 1 & 0 & -1 \\\\ 1 & 1 & 1 & 1 & 1 & 0 \\end{bmatrix}.$$ Now things get a bit fancier. This $$n-1 \\times n-1 = 2m \\times 2m$$ matrix (call it $$A$$) is a skew-symmetric matrix, so its determinant is the square of its Pfaffian. We use the simplified formula given in the Wikipedia link above: the sum $$\\text{pf}(A) = \\sum_{\\alpha \\in \\Pi} \\text{sgn}(\\pi_\\alpha) a_{i_1j_1} a_{i_2j_2}\\dotsm a_{i_mj_m}$$ where $$\\alpha$$ runs over all pairings $$\\{(i_1,j_1), \\dots, (i_m, j_m)\\}$$ with $$i_k < j_k$$ in each pair and $$i_1 < \\dots < i_m$$; $$\\pi_{\\alpha}$$ is the permutation with $$\\pi_\\alpha(2k-1) = i_k$$ and $$\\pi_\\alpha(2k) = j_k$$. For our particular matrix, $$a_{i_kj_k}$$ is always $$-1$$, so $$\\text{pf}(A) = (-1)^m \\sum_{\\alpha \\in \\Pi} \\text{sgn}(\\pi_\\alpha).$$ For each pairing $$\\alpha$$, if we find the first index $$k$$ where $$j_k > i_{k+1}$$ and switch $$j_k$$ with $$j_{k+1}$$, we get another pairing $$\\beta$$ with $$\\text{sgn}(\\pi_\\beta) = -\\text{sgn}(\\pi_\\alpha)$$; if we apply the same operation to $$\\beta$$, we get $$\\alpha$$ back. So this splits up the pairings into, uh, pairs of pairings whose terms in the Pfaffian sum cancel; the exception is the pairing $$\\{(1,2), (3,4), \\dots, (2m-1,2m)\\}$$, for which no index $$k$$ exists. So $$\\text{pf}(A) = (-1)^m$$ after all the cancellations, and therefore $$\\det(A) = 1$$.\n\nSo we get $$\\det(M_n) = 0 + x \\cdot (-1)^{m} = (-1)^{\\frac{n-1}{2}}x$$: plus or minus the middle entry of $$M_n$$.\n\n\u2022 I'm very grateful for your partial answer - it has given me some new ideas to rethink... Apr 10, 2019 at 8:21\n\u2022 I wonder whether in the case of the second determinat the same technique applied by you with substracting the rows could not lead to the further simplification.. Apr 10, 2019 at 8:59\n\u2022 Anyway I have received now {{-1, -1, -1, -1, -1, 0, 1}, {-1, -1, -1, -1, 0 ,1 ,1}, {-1, -1 , -1 , 0 ,1, 1 ,1}, {-1, -1, 0 ,1 ,1 ,1, 1}, {-1, 0 , 1 ,1 ,1, 1, 1}, {5, 5, 4 ,3, 2 ,1, 0}, {1, 1, 1, 1, 1, 1, 1}} at which I'm stacked.. Apr 10, 2019 at 9:25\n\u2022 I have also received for $5 \\times 5$ such form $\\begin{equation*}\\left[\\begin{matrix}0 & 0 & 0 & 1 & 2\\\\0 & 0 & 1 & 2 & 2\\\\0 & 1 & 2 & 2 & 2\\\\3 & 3 & 2 & 1 & 0\\\\1 & 1 & 1 & 1 & 1\\end{matrix}\\right]\\end{equation*}$ which has det =$-1 \\ \\$. Interestingly $\\begin{equation*}\\left[\\begin{matrix}0 & 0 & 0 & 1 & 2\\\\0 & 0 & 1 & 2 & 2\\\\0 & 1 & 2 & 2 & 2\\\\4 & 3 & 2 & 1 & 0\\\\1 & 1 & 1 & 1 & 1\\end{matrix}\\right]\\end{equation*}$ has det = $0$. Apr 10, 2019 at 11:53\n\u2022 Now I have a complete answer. Apr 10, 2019 at 22:11", "date": "2022-08-10 20:37:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3180877/determinant-of-antidiagonally-constructed-matrix", "openwebmath_score": 0.8820245265960693, "openwebmath_perplexity": 218.9267928013688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9942697541641922, "lm_q2_score": 0.899121367808974, "lm_q1q2_score": 0.8939691813352008}}
{"url": "https://math.stackexchange.com/questions/2962351/surjective-and-injective-function", "text": "# Surjective and Injective function\n\nLet $$N=\\{1,2,3...\\}$$ be the set of natural numbers and $$F:N \\times N \\rightarrow N$$ be such that $$f(m,n)=(2m-1)*2^n.$$\n\n(A)F is Injective.\n\n(B)F is Surjective.\n\n(C)F is Bijective.\n\n(D)None of the above.\n\nI can see that F can never be surjective because 1 does not have a pre-image.\n\nAnd it seems injective to me because $$2m-1$$ term would always be odd, $$2^n$$ term would always be even, and hence the product will always be Even, and for different values of m and n, we would get a different even number.\n\nIs my Reasoning correct?\n\n\u2022 You're on the correct track, but your reasoning is too vague to be a proof. You need to start with an assumption $f(m_1, n_1) = f(m_2, n_2)$ and develop your even-odd argument a bit more. \u2013\u00a0T. Bongers Oct 19 '18 at 17:26\n\nOverkill?\n\nA) $$f$$ is injective.\n\nLet $$f(m,n)=f(k,l)$$, i.e.\n\n$$i:=(2m-1)2^n= (2k-1)2^l.$$\n\nThe positive integer $$i$$ has a - Fundamental Theorem of Arithmetic - unique prime factorization.\n\nSince $$(2m-1)$$, $$(2k-1)$$ are odd,\n\nwe have $$n=l$$.\n\nThen\n\n$$(2m-1)2^n= (2k-1)2^l$$ implies\n\n$$2m-1=2k-1$$, or $$m=k.$$\n\nCombining\n\n$$f(m,n)=f(k,l)$$ implies $$m=k$$, and $$n=l.$$\n\nhttps://en.m.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic\n\n\u2022 \"Overkill?\" Perhaps. T. Bongers answer is probably a lot easier, but your answer does capture the thought process that the OP was considering. (i.e. that the results have an \"even part\" and an \"odd part\" and ... so on.) \u2013\u00a0fleablood Oct 19 '18 at 18:27\n\u2022 fleablood.Thanks for your comment. \u2013\u00a0Peter Szilas Oct 19 '18 at 19:24\n\nTo expand my comment, you need to make the reasoning a lot more precise. Start with an assumption that there are natural numbers $$m_1, m_2, n_1, n_2$$ for which\n\n$$f(m_1, n_1) = f(m_2, n_2).$$\n\nThen you have\n\n$$(2m_1 - 1)2^{n_1} = (2m_2 - 1) 2^{n_2}.$$\n\nNow we can assume without loss of generality that $$n_1 \\le n_2$$, so that\n\n$$2m_1 - 1 = (2m_2 - 1)2^{n_2 - n_1}.$$\n\nNow the left side is odd, so the right side is odd too; thus, $$n_2 - n_1 = 0$$ (why?). But then we get $$2m_1 - 1 = 2m_2 - 1$$, hence $$m_1 = m_2$$. This completes the proof.\n\nI can see that F can never be surjective because 1 does not have a pre-image.\n\nLet's prove that.\n\nIf $$f(m,n) = (2m -1)*2^n = 1$$ then as $$2m-1$$ and $$2^n$$ are integer factors of $$1$$ and the only integer factors of $$1$$ are $$\\pm 1$$ so $$2^n = \\pm 1$$ and that is only possible if $$n = 0$$ which is not possible as $$0 \\not \\in \\mathbb N$$.\n\nSo it is not surjective.\n\n(In fact for any odd number $$2k - 1$$ then $$f(m,n) = (2m-1)*2^n = 2k-1$$ would be impossible. $$f(m,n)$$ will always be even.)\n\nAnd it seems injective to me because 2m\u22121 term would always be odd, 2n term would always be even, and hence the product will always be Even,\n\nWhich shows it can not be surjective\n\nand for different values of m and n, we would get a different even number.\n\nThat's a bit of a leap. How do you know they will be different?\n\nLet's prove it.\n\n$$f(m,n) = (2m-1)*2^n$$ will have a unique prime factorization. As $$2m-1$$ is odd the power of $$2$$ of any value of $$f(m,n)= (2m-1) 2^n$$ will be $$n$$.\n\nSo if $$n_1 \\ne n_2$$ then $$f(a,n_2) = (2a - 1)2^{n_2} \\ne (2b- 1)2^{n_1} = f(b,n_2)$$ for any possible $$a,b$$. (Because the power of $$2$$ in the prime factorizations of those two numbers are different.)\n\nAnd if $$m_1 \\ne m_2$$ then $$2m_1 - 1 \\ne 2m_2 - 1$$ and the prime factorizations of $$(2m_1 - 1)*2^a \\ne (2m_2 - 1)*2^b$$ for any possible $$a,b$$ must be different for the different odd factors.\n\nSo if $$(m_1, n_1) \\ne (m_2 n_2)$$ then either $$n_1 \\ne n_2$$ or $$m_1 \\ne m_2$$ (or both). In either case then $$(2m_1-1)2^{n_1} \\ne (2m_2-1)2^{n_2}$$.\n\nSo $$f$$ is injective.\n\nA) is true, B) is false, C) is false (as bijective implies surjective) and D) is false (it is one of the above).", "date": "2019-04-21 22:50:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2962351/surjective-and-injective-function", "openwebmath_score": 0.8663015961647034, "openwebmath_perplexity": 133.04482498097448, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718473283829, "lm_q2_score": 0.9032942041005328, "lm_q1q2_score": 0.8939648436531956}}
{"url": "http://mathhelpforum.com/differential-equations/72467-differential-equation.html", "text": "# Math Help - differential equation\n\n1. ## differential equation\n\nIf we assume that a person's weight depends on the energy consumed minus energy ued, one model is that weight changes in proportion to the difference, so:\n\ndw/dt = k(C - 38.5w) where w(0) = w subscript 0\n\nwhere w(t) kg is the weight at t days, C is the daily calorie intake, and we assume 38.5 calories per kg per day are used\n\n(i) if you wish to maintain a constant weight of 82 kg, what should be your daily calorie intake C?\n\n(ii) If you weigh 100 kg, and you want to lose 10 kg in a month, what should C be? (Assume k = 1.3 * 10^(-4) kg/calorie and a month of 30 days) Is this result healthy?\n\ncan someone solve these questions for me, they are on a past test paper which does not have solutions available.\n\n2. Originally Posted by razorfever\nIf we assume that a person's weight depends on the energy consumed minus energy ued, one model is that weight changes in proportion to the difference, so:\n\ndw/dt = k(C - 38.5w) where w(0) = w subscript 0\n\nwhere w(t) kg is the weight at t days, C is the daily calorie intake, and we assume 38.5 calories per kg per day are used\n\n(i) if you wish to maintain a constant weight of 82 kg, what should be your daily calorie intake C?\n\n(ii) If you weigh 100 kg, and you want to lose 10 kg in a month, what should C be? (Assume k = 1.3 * 10^(-4) kg/calorie and a month of 30 days) Is this result healthy?\n\ncan someone solve these questions for me, they are on a past test paper which does not have solutions available.\n(i) $\\frac{dw}{dt} = 0$ ... solve for $C$\n\n(ii) solve $\\int \\frac{dw}{C - 38.5w} = \\int k \\, dt$ with initial condition $w(0) = 100$.\n\nafter you find $w(t)$, set $w(30) = 90$ and solve for $C$.\n\n3. can u help me out the same way with that kind of template for some of my other posts about differential equation word problems?\nI have an exam coming up and my prof hasn't touched any word problems,\nif you could provide templates, i can understand them better and hopefully pass the exam\n\nhttp://www.mathhelpforum.com/math-he...tric-func.html\n\nhttp://www.mathhelpforum.com/math-he...d-problem.html\n\nhttp://www.mathhelpforum.com/math-he...-diver-de.html\n\nhttp://www.mathhelpforum.com/math-he...ates-flow.html\n\nhttp://www.mathhelpforum.com/math-he...72473-ivp.html\n\nhttp://www.mathhelpforum.com/math-he...ntial-eqn.html\n\n4. can you tell me what your final answer for C is?\ni'm getting 102375.57\n\nas for the constant in part (i) i'm getting (C - 3850)\nand the expression for C in part (ii) i'm getting\n[3465 - 3850e^(30k)] / [1 - e^(30k)]\n\nis this correct\ncan someone show me the correct solution step by step?\n\n5. $\\int \\frac{dw}{C - 38.5w} = \\int k \\, dt$\n\n$\\int \\frac{-38.5}{C - 38.5w} \\, dw = -38.5\\int k \\, dt$\n\n$\\ln|C - 38.5w| = -38.5kt + A$\n\n$C - 38.5w = Be^{-38.5kt}$\n\n$w(0) = 100$\n\n$C - 3850 = B$\n\n$C - 38.5w = (C - 3850)e^{-38.5kt}$\n\n$C - 38.5w = Ce^{-38.5kt} - 3850e^{-38.5kt}$\n\n$C - Ce^{-38.5kt} = 38.5w - 3850e^{-38.5kt}$\n\n$C(1 - e^{-38.5kt}) = 38.5w - 3850e^{-38.5kt}$\n\n$C = \\frac{38.5w - 3850e^{-38.5kt}}{1 - e^{-38.5kt}}$\n\n$w(30) = 90$ ... $C \\approx 1090 \\, cal$\n\n6. now i'm getting C = 3515\nhow do you do that last part w(30) = 90\ndo u rearrange and make w the subject then put t=30 and solve for C right?\ncan u show me that last step explicitly??\n\n7. Originally Posted by razorfever\nnow i'm getting C = 3515\nhow do you do that last part w(30) = 90\ndo u rearrange and make w the subject then put t=30 and solve for C right?\ncan u show me that last step explicitly??\n\nsince the task was to find C, I solved for C ... sub in 90 for w, 30 for t, and the given value for k (0.00013 , correct?)\n\n8. yeah i just realized i was making a careless mistake\ni wasn't putting E in my calculator , instead just getting the value of\n(-38.5 * 0.00013 * 30) and this was giving me C as 3515 but I got it now\n\n9. Originally Posted by razorfever\ncan u help me out the same way with that kind of template for some of my other posts about differential equation word problems?\nI have an exam coming up and my prof hasn't touched any word problems,\nif you could provide templates, i can understand them better and hopefully pass the exam\n\nhttp://www.mathhelpforum.com/math-he...tric-func.html\n\nhttp://www.mathhelpforum.com/math-he...d-problem.html\n\nhttp://www.mathhelpforum.com/math-he...-diver-de.html\n\nhttp://www.mathhelpforum.com/math-he...ates-flow.html\n\nhttp://www.mathhelpforum.com/math-he...72473-ivp.html\n\nhttp://www.mathhelpforum.com/math-he...ntial-eqn.html", "date": "2015-04-19 03:17:54", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/differential-equations/72467-differential-equation.html", "openwebmath_score": 0.4771636426448822, "openwebmath_perplexity": 1191.1487717225007, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226247376173, "lm_q2_score": 0.914900959053549, "lm_q1q2_score": 0.8938789363894617}}
{"url": "https://math.stackexchange.com/questions/1128323/are-emptyset-and-x-closed-open-or-clopen/1128325", "text": "# Are $\\emptyset$ and $X$ closed, open or clopen?\n\nIt is indeed a very basic question but I am confused:\n\n(1) In an 2013 MSE posting under general topology here, I was told that $\\emptyset$ is an open set and therefore I assume $X$ must be open too.\n\n(2) But in Wikipedia page on clopen set here, it says \"In any topological space $X$, the $\\emptyset$ and the whole space $X$ are both clopen.\"\n\n(3) And yet in another Wikipedia page on closed set here, \"The $\\emptyset$ is closed, the whole set is closed.\"\n\nI must have missed something. Can you help me with a supreme verdict, once and for all, as sure as the sun rises from the east each morning, if $X$ and $\\emptyset$ are open, closed or clopen. Of course I am talking about topology, thanks for your time.\n\n\u2022 I don't see any contradiction among the three. \u2013\u00a0Hoot Jan 31 '15 at 23:36\n\u2022 I don't understand! Is this an emerald, or is it green, or is it a precious stone? MSE says this is green, and Wikipedia says it is an emerald, and yet another wikipedia page says it is a precious stone. Can you help me with a supreme verdict, once and for all, if this is green, a precious stone, or an emerald? \u2013\u00a0MY USER NAME IS A LIE Jan 31 '15 at 23:47\n\u2022 @MYANSWERSARECRAP your answers may be, but your comments certainly are not! \u2013\u00a0Neal Jan 31 '15 at 23:50\n\u2022 Always funny: youtube.com/watch?v=SyD4p8_y8Kw \u2013\u00a0Jack D'Aurizio Feb 1 '15 at 1:48\n\nThey are all both open and closed.\n\nLet me make this a bit more clear. By definition of a topology (from wikipedia):\n\nA topological space is then a set $X$ together with a collection of subsets of $X$, called open sets and satisfying the following axioms:\n\n\u2022 The empty set and $X$ itself are open.\n\u2022 Any union of open sets is open.\n\u2022 The intersection of any finite number of open sets is open.\n\nSo just from the definition itself it follows that $\u2205$ and $X$ are open.\n\nFurthermore a set is closed (by definition) if the complement is open. Therefore $\u2205$ and $X$ are closed (they are each others complement).\n\nThe term clopen means that a set is both open and closed, so they are both also clopen.\n\n\u2022 Thanks, have up-voted yours. Can I take it daily for granted that both $X, \\emptyset$ are clopen? \u2013\u00a0Amanda.M Jan 31 '15 at 23:42\n\u2022 @A.Magnus yes that is true in any topology \u2013\u00a0Loreno Heer Jan 31 '15 at 23:50\n\nTo put it simply: sets are not doors! Being open does not mean that the set is not closed, and being closed do not implies that the set is not open. Yes, the empty set and the whole space are both open and closed. Another more dramatic example is: take a metric space $X$, with the discrete metric. Then every singleton $\\{a\\}$ (in fact, every subset of $X$) is clopen. Balls $B(a,r)$ with radius $r < 1$ are contained in $\\{a\\}$, hence $\\{a\\}$ is open. And $\\{a\\}$ is also closed, because it's complement is $X \\setminus\\{a\\} = \\bigcup_{b \\in X, b \\neq a}\\{b\\}$, a union of open sets, hence open.\n\n\u2022 Ask you a question, I hope I can say it clear: Let $B \\subset A$ and $B$ is open. Absence of any other instructions, $A$ has to be closed since it is $B$'s complement. Now if $A$ is clopen, then its neutrality does not have any bearing on open-ness or the closed-ness of $B$. Am I correct? \u2013\u00a0Amanda.M Feb 1 '15 at 1:54\n\u2022 But it is not true that $A$ is the complement of $B$.. \u2013\u00a0Ivo Terek Feb 1 '15 at 1:55\n\u2022 I made typo and just corrected it. Sorry for confusion. \u2013\u00a0Amanda.M Feb 1 '15 at 1:56\n\nBy the first axiom in the definition of a topology, $X$ and $\\emptyset$ are open. However, closed sets are precisely those whose complements are open, by definition. Hence the empty set and $X$, being each others complements, are also closed. So, they are clopen.\n\n\u2022 Have up-voted yours. Thanks. \u2013\u00a0Amanda.M Jan 31 '15 at 23:43\n\nA minimal requirement on any topological space $(X,\\tau)$ is that both $\\varnothing$ and $X$ be open sets. By the definition of closed sets, these requirements imply that $\\varnothing^c=X$ and $X^c=\\varnothing$\u00a0are always closed.\n\nTo sum up, in any topological space, the empty set and the whole set are always both open and closed, hence clopen.\n\nYour question \u201care $\\varnothing$ and $X$ closed, open or clopen\u201d is basically six questions in one:\n\n(1) Is $\\varnothing$ closed? Answer: yes.\n\n(2) Is $\\varnothing$ open? Answer: yes.\n\n(3) Is $\\varnothing$ clopen? Answer: yes.\n\n(4) Is $X$ closed? Answer: yes.\n\n(5) Is $X$ open? Answer: yes.\n\n(6) Is $X$ clopen? Answer: yes.\n\n\u2022 Good answer, thank you! \u2013\u00a0Amanda.M Feb 1 '15 at 3:12", "date": "2019-10-17 12:56:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1128323/are-emptyset-and-x-closed-open-or-clopen/1128325", "openwebmath_score": 0.8672575354576111, "openwebmath_perplexity": 370.25085908048993, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717428891155, "lm_q2_score": 0.905989822921759, "lm_q1q2_score": 0.893823958639721}}
{"url": "https://math.stackexchange.com/questions/3350525/inclusion-exclusion-principle-problem", "text": "# Inclusion\u2013exclusion principle problem\n\n172 business executives were surveyed to determine if they regularly read Fortune, Time, or Money magazines. 80 read Fortune, 70 read Time, 47 read Money, 47 read exactly two of the three magazines, 26 read Fortune and Time, 28 read Time and Money, and 7 read all three magazines. How many read none of the three magazines?\n\nSo what i tried to do was to denote Time as $$T$$, Fortune as $$F$$, and Money as $$M$$. Then i wanted to do $$172 - |T \\cup F \\cup M|$$. To find that amount, i tried to do the inclusion exclusion principle. I found $$|F \\cup M|$$ to be $$14$$ by looking at a venn diagram ($$26-7 + 28-7 + x = 47 -> x = 7$$, then $$7+7 = 14$$), but i guess this i where things went wrong. I got $$|T \\cup F \\cup M| = 80 + 70 + 47 - 26 - 28 - 14 + 7$$ thus the answer would be 53, but that is false.\n\n\u2022 Excellent reasoning! See the answer below. \u2013\u00a0S. Dolan Sep 10 '19 at 9:18\n\nWe are given explicitly that \\begin{align*} |F| &= 80\\\\ |T| &= 70\\\\ |M| &= 47\\\\ |F\\cap T| &= 26\\\\ |T\\cap M| &= 28\\\\ |F\\cap T\\cap M| &= 7. \\end{align*}\n\nWe also get that 47 read \"exactly two\". If you consider a Venn diagram, it is not hard to see that this corresponds to the equation \\begin{align*} &|F\\cap T| + |T\\cap M|+ |F\\cap M|-3|F\\cap T\\cap M|=47\\\\ \\implies & 26 + 28 + |F\\cap M| - 3(7) = 47\\\\ \\implies & |F\\cap M| = 14. \\end{align*}\n\nThus you can plug this all in to inclusion-exclusion: \\begin{align*} |F \\cup T\\cup M| &= |F|+|T|+|M| - |F\\cap T| - |T\\cap M|- |F\\cap M| + |F\\cap T\\cap M|\\\\ &= 80+70+47-26-28-14+7=136, \\end{align*}\n\nand thus the answer is $$172-136=\\boxed{\\text{36 people}}$$.\n\nIn your expression $$|T \\cup F \\cup M| = 80 + 70 + 47 - 26 - 28 - 14 + 7=136$$.\n\nThis gives a solution of $$36$$ not $$53$$.\n\nWe have:\n\n$$|T \\cup F \\cup M| = |T| + |F| + |M| - |T \\cap F| - |T \\cap M| - |F \\cap M| + |T \\cap F \\cap M|$$ $$= 70 + 80 + 47 - 26 - 28 - 14 + 7 = 136$$\n\nThen, the number of people who do not read any magazine equals:\n\n$$172 - |T \\cup F \\cup M| = 36$$", "date": "2021-03-05 20:15:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3350525/inclusion-exclusion-principle-problem", "openwebmath_score": 0.9997199177742004, "openwebmath_perplexity": 2157.5431643183756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109096680231, "lm_q2_score": 0.9032942014971871, "lm_q1q2_score": 0.8938194670213322}}
{"url": "http://mathhelpforum.com/number-theory/145830-divisibility.html", "text": "# Math Help - Divisibility\n\n1. ## Divisibility\n\nHi all.\n\nI'm trying to figure out the following problem:\n\nFind the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4.\n\nHelp will be appreciated. Looking for a simple/elementary proof.\n\nThanks.\n\n2. Originally Posted by pollardrho06\nHi all.\n\nI'm trying to figure out the following problem:\n\nFind the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4.\n\nHelp will be appreciated. Looking for a simple/elementary proof.\n\nThanks.\nHint:\n$|\\{x\\in \\mathbb{Z}_+\\mid x \\text{ divisible by } 3 \\text{ but not by } 4\\}|$\n$= |\\{x\\in \\mathbb{Z}_+\\mid x \\text{ divisible by } 3\\}|-|\\{x\\in \\mathbb{Z}_+\\mid x \\text{ divisible by } 12\\}|$\n\nThis is a consequence of $A\\backslash B=A\\backslash(A\\cap B)$, and $|X\\backslash Y|=|X|-|Y|$, if $Y\\subseteq X$.\n\n3. Originally Posted by Failure\nHint:\n$|\\{x\\in \\mathbb{Z}_+\\mid x \\text{ divisible by } 3 \\text{ but not by } 4\\}|$\n$= |\\{x\\in \\mathbb{Z}_+\\mid x \\text{ divisible by } 3\\}|-|\\{x\\in \\mathbb{Z}_+\\mid x \\text{ divisible by } 12\\}|$\n\nThis is a consequence of $A\\backslash B=A\\backslash(A\\cap B)$, and $|X\\backslash Y|=|X|-|Y|$, if $Y\\subseteq X$.\nI don't see it...\n\n4. ## Hint\n\nLook for cycles.\n\n5. Hello, pollardrho06!\n\nFind the number of positive integers not exceeding 1000\nthat are divisible by 3 but not by 4.\n\nEvery third number is divisible by 3.\n. . There are: . $\\left[\\frac{1000}{3}\\right] \\:=\\:333$ numbers divisible by 3.\n\nBut every twelfth number is divisible by 3 and by 4.\n. . There are: . $\\left[\\frac{1000}{12}\\right] \\:=\\:83$ multiples of 3 which are divisible by 4.\n\nTherefore, there are: . $333 - 83 \\:=\\:250$ such numbers.\n\n6. Originally Posted by Soroban\nHello, pollardrho06!\n\nEvery third number is divisible by 3.\n. . There are: . $\\left[\\frac{1000}{3}\\right] \\:=\\:333$ numbers divisible by 3.\n\nBut every twelfth number is divisible by 3 and by 4.\n. . There are: . $\\left[\\frac{1000}{12}\\right] \\:=\\:83$ multiples of 3 which are divisible by 4.\n\nTherefore, there are: . $333 - 83 \\:=\\:250$ such numbers.\n\nWow!! The greatest integer function!! Gr8!! Thanks!!", "date": "2014-08-27 17:10:14", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/145830-divisibility.html", "openwebmath_score": 0.9141215682029724, "openwebmath_perplexity": 743.0147025121496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9893474897884493, "lm_q2_score": 0.9032942138630786, "lm_q1q2_score": 0.8936718630258675}}
{"url": "https://math.stackexchange.com/questions/1887453/solving-the-differential-equation-y-y-tanx-cotx", "text": "# Solving the Differential equation $y' = y \\tan(x) + \\cot(x)$\n\nI am asked to solve the following differential equation:\n\n$$y' = y \\tan(x) + \\cot(x)$$\n\nWhat I have so far is\n\n\\begin{align*} y' - (\\tan x) y &= \\cot(x)\\\\ \\\\ I &= e^{\\int - \\tan(x) dx} = \\cos(x)\\\\ \\\\ \\cos(x) \\left( y' - y \\tan(x) \\right) &= \\cos(x) \\cot(x)\\\\ y' \\cos(x) - y \\sin(x) &= \\cos(x) \\cot(x)\\\\ \\int y' \\cos(x) - y \\sin(x) &= \\int \\cos(x) \\cot(x) dx\\\\ \\\\ y \\cos(x) &= \\int \\frac{cos^2(x)}{\\sin(x)}\\\\ y \\cos(x) &= \\int \\frac{1-sin^2(x)}{\\sin(x)}\\\\ &= \\int \\csc(x) - \\sin(x) \\ dx\\\\ &= - \\ln \\vert \\csc(x) + \\cot(x) \\vert + \\cos(x) + C\\\\ \\\\ \\therefore y &= - \\frac{\\ln \\vert \\csc(x) + \\cot(x) \\vert}{\\cos(x)} + 1 + C \\sec(x) \\end{align*}\n\nThe thing is: I am having a hard time comparing my result to the textbook's solution and to Wolfram's solution.\n\nTextbook's solution: $$\\sec(x) \\left( \\frac{x}{2} + \\frac{\\sin(2x)}{4} + C \\right)$$\n\nIs my solution correct?\n\nThank you.\n\n\u2022 Your answer is correct and is equivalent to WA's. The textbook answer is wrong \u2013\u00a0David Quinn Aug 9 '16 at 19:41\n\u2022 Hi @DavidQuinn thank you for your input. How could I go from my answer to Wolfram's (or the opposite)? What kind of transformation did you do? \u2013\u00a0bru1987 Aug 9 '16 at 19:42\n\u2022 Use $\\csc x+\\cot x=\\cot(\\frac x2)$ \u2013\u00a0David Quinn Aug 9 '16 at 19:50\n\n\\begin{align} y'(x)-y \\tan x &=\\sec x \\tan x \\left( \\frac{x}{2} + \\frac{\\sin 2x }{4} + C \\right)+\\sec x \\left( \\frac{1}{2} + \\frac{\\cos 2x }{2} \\right) \\\\&\\hspace{4mm} -\\sec x\\left(\\frac{x}{2}+\\frac{\\sin 2x }{4} + C \\right)\\tan x \\\\ &= \\frac{1 + \\cos 2x }{2\\cos x}\\\\&=\\cos x \\end{align} i.e. the textbook's answer is for the RHS being $\\cos x$ rather than $\\cot x$.\nSince $\\ln \\vert \\csc(x) + \\cot(x) \\vert=\\ln{|\\frac{1+\\cos{x}}{\\sin{x}}|}=\\ln{|\\frac{2\\cos^2{(x/2)}}{2\\sin{(x/2)\\cos(x/2)}}|}=\\ln{\\cos{(x/2)}}-\\ln{\\sin(x/2)}$, your solution is same as WolphramAlpha's.", "date": "2019-10-15 16:12:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1887453/solving-the-differential-equation-y-y-tanx-cotx", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 2702.3522997144087, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109516378093, "lm_q2_score": 0.9073122213606241, "lm_q1q2_score": 0.8935310121507709}}
{"url": "https://math.stackexchange.com/questions/2724960/probability-of-getting-full-house-in-poker?noredirect=1", "text": "# Probability of getting full house in poker\n\nI have this problem which I have solved, but using two different methods. I am quite new to combinatorics and want to know how to intuitively understand the difference between the following two methods.\n\nThe problem consist of calculating the probability of getting a full house being dealt a 5-card poker hand.\n\nFirst of, I solve this by simply saying that\n\n$P($get full house$)=\\frac{2 {13\\choose 2} {4 \\choose 3} {4 \\choose 2}}{{52 \\choose 5}}$. This is the right answer according to my text-book. However, at my first attempt at solving this I forgot the factor 2 in the numerator. My reasoning goes like this: All the possible ways of getting a full house consists of all the ways we can combine two different ranks (i.e. ${13 \\choose 2}$) times all the possible ways of choosing 3 cards out of 4 suits, times all the possible ways of choosing 2 cards out of 4 suits. Now, all this seems logical to me. The thing that makes me doubt whether I truly understand what I'm doing is how the factor 2 comes in place. I'm thinking: Because we choose two different ranks without regards to order, we have to compensate for those combinations and therefore multiply with $2!$, because obviously it does matter if I (for example) choose the ranks (ace,knights) and in this sequence choose three ace and two knights.\n\nOn the other hand, I can solve the problem using the method described here: https://math.stackexchange.com/a/808328/518320\n\nWhich as well seems intuitively clear, thinking the way the user describe the process in that thread.\n\nWhat is the difference in the two methods? Maybe this is obvious, but I'm new to combinatorics. And is my reasoning above accurate?\n\nThanks!\n\n\u2022 I would have said: choose the suit for the triple, that's $\\binom {13}1$. Then choose the triple of ranks, $\\binom 43$. Then choose the suit for the pair, $\\binom {12}1$, then choose the pair, $\\binom 42$. It's true that $2\\times \\binom {13}2= \\binom {13}1\\times \\binom {12}1$ but I find this less intuitive. \u2013\u00a0lulu Apr 6 '18 at 12:50\n\u2022 @lulu While your calculations are correct, you confused ranks with suits. The four suits are hearts, clubs, diamonds, and spades. The thirteen ranks are 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. \u2013\u00a0N. F. Taussig Apr 6 '18 at 13:07\n\u2022 @N.F.Taussig absolutely correct. I plead lack of coffee. \u2013\u00a0lulu Apr 6 '18 at 13:10\n\nThey are essentially the same giving $156$ ways of choosing ranks, but if you want to draw a distinction:\n\n\u2022 Choose two ranks and then choose one of those two to be the three so the other is the pair $${13 \\choose 2}{2 \\choose 1}$$\n\n\u2022 Choose one rank to be the three then choose another rank to be the pair $${13 \\choose 1}{12 \\choose 1}$$\n\nand you then multiply this by ${4 \\choose 3}{4 \\choose 2}$ and divide by ${52 \\choose 5}$\n\nDoing the same thing for two pairs to get $858$ ways of choosing the ranks by several methods:\n\n\u2022 Choose three ranks and then choose one of those three to be the single so the others are the pairs $${13 \\choose 3}{3 \\choose 1}$$\n\n\u2022 Choose three ranks and then choose two of those three to be the pairs so the other is the single $${13 \\choose 3}{3 \\choose 2}$$\n\n\u2022 Choose one rank to be the single then choose two other ranks to be the pairs $${13 \\choose 1}{12 \\choose 2}$$\n\n\u2022 Choose two ranks to be the pairs then choose another rank to be the single $${13 \\choose 2}{11 \\choose 1}$$\n\nand you then multiply this by ${4 \\choose 1}{4 \\choose 2}{4 \\choose 2}$ and divide by ${52 \\choose 5}$\n\nThe expressions $2\\binom{13}{2}$ and $\\binom{13}{1}\\binom{12}{1}$ are both equivalent to \"choose $2$ objects from $13$ where the order of choosing matters.\" In other words, we want permutations of $2$ objects out of $13$ rather than combinations.\n\nThe way to get permutations of $k$ objects out of $n$ is to multiply $k$ consecutive integers together, where $n$ is the largest of the integers multiplied: $n(n-1)(n-2) \\cdots (n-k+1).$ For example, permutations of $2$ objects out of $n$ give $13\\cdot12$ possible permutations. And since $\\binom n1 = n,$ $13\\cdot12 = \\binom{13}{1}\\binom{12}{1}.$\n\nBut often the number of permutations is written $\\dfrac{n!}{(n-k)!},$ because $$\\frac{n!}{(n-k)!} = n(n-1)(n-2) \\cdots (n-k+1).$$ To get the number of combinations of $k$ objects out of $n$ (without regard for the order of the objects), we divide by $k!,$ since that's the number of different sequences in which each combination of $k$ objects can occur as a permutation. So we get $$\\binom nk = \\frac{n!}{(n-k)!k!}.$$ This also means that $k!\\binom nk$ is yet another way to write the number of permutations of $k$ objects out of $n.$\n\nIn short, it's all really the same thing written in different ways.\n\nThe two ranks are distinguishable once you get a particular full house. Say you get three kings and two queens. That differs from getting three queens and two kings. So it should be $13 \\cdot 12$ ways to pick the two ranks. This is the same as $2 \\cdot \\binom{13}{2}.$", "date": "2019-11-14 16:07:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2724960/probability-of-getting-full-house-in-poker?noredirect=1", "openwebmath_score": 0.8362991213798523, "openwebmath_perplexity": 147.78411420095142, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513926334711, "lm_q2_score": 0.9059898159413479, "lm_q1q2_score": 0.8934431187023023}}
{"url": "https://mathhelpboards.com/threads/find-the-image-f-n-n-r-f-x-m-2-2n.7102/", "text": "# Find the image: F: N * N -> R , F(x) = m^2 + 2n\n\n#### KOO\n\n##### New member\nF:N*N -> R, F(x) = m^2 + 2n\n\nI think the answer is N. Am I right?\n\n#### Chris L T521\n\n##### Well-known member\nStaff member\nI think the answer is N. Am I right?\nBased on your title, we have that $F:\\mathbb{N}\\times\\mathbb{N}\\rightarrow \\mathbb{R}$ ($\\mathbb{N}\\ast\\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$).\n\nNow, depending on who you talk to, $\\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\\notin\\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\\mathbb{N}=\\{x\\in\\mathbb{Z} : x\\geq 0\\}$ or $\\mathbb{N}=\\{x\\in\\mathbb{Z}: x\\geq 1\\}$.\n\nIf $0\\in\\mathbb{N}$, then $\\mathrm{Im}(F)=\\mathbb{N}$. If $0\\notin\\mathbb{N}$, then $\\mathrm{Im}(F) = \\mathbb{N}\\backslash\\{1,2,4\\}$ since there aren't pairs $(m,n)\\in\\mathbb{N}\\times\\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$.\n\n(In $\\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\\in\\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\\in\\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\\mathbb{N}\\times\\mathbb{N}) = \\mathbb{N}\\backslash\\{1,2,4\\}$ for $0\\notin\\mathbb{N}$.)\n\nI hope this makes sense!\n\n#### KOO\n\n##### New member\nBased on your title, we have that $F:\\mathbb{N}\\times\\mathbb{N}\\rightarrow \\mathbb{R}$ ($\\mathbb{N}\\ast\\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$).\n\nNow, depending on who you talk to, $\\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\\notin\\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\\mathbb{N}=\\{x\\in\\mathbb{Z} : x\\geq 0\\}$ or $\\mathbb{N}=\\{x\\in\\mathbb{Z}: x\\geq 1\\}$.\n\nIf $0\\in\\mathbb{N}$, then $\\mathrm{Im}(F)=\\mathbb{N}$. If $0\\notin\\mathbb{N}$, then $\\mathrm{Im}(F) = \\mathbb{N}\\backslash\\{1,2,4\\}$ since there aren't pairs $(m,n)\\in\\mathbb{N}\\times\\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$.\n\n(In $\\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\\in\\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\\in\\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\\mathbb{N}\\times\\mathbb{N}) = \\mathbb{N}\\backslash\\{1,2,4\\}$ for $0\\notin\\mathbb{N}$.)\n\nI hope this makes sense!\nActually I had this question on a test and you're right x=(m,n) and we're told N does not include 0.\n\nAnyways what does \\ mean?\n\n#### Chris L T521\n\n##### Well-known member\nStaff member\nActually I had this question on a test and you're right x=(m,n) and we're told N does not include 0.\n\nAnyways what does \\ mean?\nAh, that's one notation for set difference. I could have also written it as $\\mathbb{N}-\\{1,2,4\\}$.", "date": "2021-09-18 11:33:55", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-the-image-f-n-n-r-f-x-m-2-2n.7102/", "openwebmath_score": 0.8860371112823486, "openwebmath_perplexity": 214.78032173086706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683469514965, "lm_q2_score": 0.9046505421702797, "lm_q1q2_score": 0.8934042404998782}}
{"url": "https://math.stackexchange.com/questions/2816513/same-integration-with-2-different-answers", "text": "# Same integration with 2 different answers?\n\n$$\\int x(x^2+2)^4\\,dx$$\n\nWhen we do this integration with u substitution we get $$\\frac{(x^2+2)^5}{10}$$ as $u=x^2+2$\n\n$du=2x\\,dx$ $$\\therefore \\int (u+2)^4\\,du = \\frac{(x^2+2)^5}{10} + C$$\n\nAlthough when we expand the fraction and then integrate the answer we get is different:\n\n$x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$ $$\\int x^9+8x^7+24x^5+32x^3+16x \\,dx$$\n\nwe get\n\n$$\\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2 + C$$\n\nFor a better idea of the questions, let's say the questions asks us to find the value of C when y(0)=1\n\nNow,\n\n$x=0$\n\n$$\\frac {0^{10}}{10} + 0^8 + 4(0)^6 + 8(0)^4 + 8(0)^2 + C = 1$$ $$\\therefore C= 1$$ AND $$\\frac {(0+2)^5}{10} + C= 1$$ $$\\therefore \\frac {32}{10} + C = 1$$ $$\\therefore C = 1 - 3.2 = -2.2$$\n\n\u2022 Don't forget the arbitrary constant of integration! \u2013\u00a0Lord Shark the Unknown Jun 12 '18 at 4:41\n\u2022 ^^^^^ What @LordSharktheUnknown said! The indefinite integral is the class of all functions who are antiderivatives of the integrand. These antiderivatives only differ by a constant. \u2013\u00a0N8tron Jun 12 '18 at 4:48\n\u2022 $\\frac {(x^2+2)^5}{10} = \\frac {x^{10}}{10} + x^8 + 4x^6+8x^4+8x^2 + \\frac {32}{10}.$ The only difference is the constant of integration. \u2013\u00a0Doug M Jun 12 '18 at 4:52\n\u2022 Please fix the typos on $(x+2)$. \u2013\u00a0Yves Daoust Jun 12 '18 at 5:16\n\u2022 There are lots and lots of similar questions already, just search: google.com/\u2026 \u2013\u00a0Hans Lundmark Jun 12 '18 at 6:46\n\nLike mentioned in the comments this is all fixed if you remember your constant of integration.\n\n$$\\int x(x^2+2)^4\\ dx= \\frac{(x^2+2)^5}{10}+C$$\n\nNote if you expand\n\n$$\\begin{split} \\frac{(x^2+2)^5}{10}&=\\frac{1}{10}\\left(x^{10}+5x^8(2)+10x^6(2^2)+10x^4(2^3)+5x^2(2^4)+2^5\\right)\\\\ &=\\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\\frac{32}{10} \\end{split}$$\n\nNotice the relation to your other way of computing the integral\n\n$$\\int x(x^2+2)^4\\ dx = \\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2 +C$$\n\nSo lets call $F(x)=\\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2$ and $G(x)=\\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\\frac{32}{10}$ then $F(x)-G(x)=-\\frac{32}{10}$ a constant. All antiderivatives of a continuous function only differ by a constant.\n\nJust for fun Let's see another one:\n\nFirst lets use double angle for sine $$\\int \\cos x\\sin x\\ dx=\\frac{1}{2}\\int\\sin 2x\\ dx=-\\frac{1}{4}\\cos 2x +C$$\n\nThen substitutions $u=\\sin x$\n\n$$\\int \\cos x\\sin x\\ dx=\\int u\\ du =\\frac{u^2}{2}+C=\\frac{\\sin^2 x}{2}+C$$\n\nThen substitutions $u=\\cos x$\n\n$$\\int \\cos x\\sin x\\ dx=\\int -u\\ du =\\frac{-u^2}{2}+C=\\frac{-\\cos^2 x}{2}+C$$\n\nIf you find the constant differences and combine them in the right way you get the half angle formulas:\n\n$$\\sin^2 x=\\frac{1-\\cos 2x}{2},\\quad \\cos^2 x=\\frac{1+\\cos 2x}{2}$$\n\nNote you can pretty quickly derive some funky trig identities in this way. For instance if you consider $\\int \\cos^3 x \\sin^5 x\\ dx$\n\n\u2022 Although if we were asked to find the value of C at x=0, won't we have 2 different value of C??? \u2013\u00a0Agent Smith Jun 13 '18 at 0:47\n\u2022 No if you fix any two antiderivatives $F$ and $G$ for a suitable function (continuous is adequate but the real condition is Riemann integrable). There is a constant C such that the difference $F(x)-G(x)=C$ for all x. This can be rigourously proven. \u2013\u00a0N8tron Jun 13 '18 at 1:45\n\u2022 Check out this page (or numerous others) to see the proof math.stackexchange.com/questions/1862231/\u2026 \u2013\u00a0N8tron Jun 13 '18 at 1:49\n\nYou can check an antiderivative by differentiating.\n\n$$\\left(\\frac{(x^2+2)^5}{10}\\right)'=x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$$\n\nand\n\n$$\\left(\\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2\\right)'=x^9+8x^7+24+32x^3+16x$$\n\nand the two expressions are indeed equivalent.\n\nNow the long explanation.\n\nConsider the binomial $x^2+a$ raised to some power $n$ and multiplied by $2x$.\n\n$$2x(x^2+a)^m$$\n\nwhich integrates as\n\n$$\\frac{(x^2+a)^{m+1}}{m+1}.$$\n\nBy the binomial theorem, the terms in the development of this antiderivative are\n\n$$\\frac1{m+1}\\binom{m+1}kx^{2(m+1-k)}a^k.$$\n\nOn the other hand, the development of the initial integrand gives terms\n\n$$2\\binom mkx^{2(m-k)+1}a^k,$$ and after integration\n\n$$\\frac1{m-k+1}\\binom mkx^{2(m-k)+2}a^k.$$\n\nIt is easy to see that all terms coincide, because\n\n$$\\frac1{m+1}\\frac{(m+1)!}{k!(m+1-k)!}=\\frac1{m-k+1}\\frac{m!}{k!(m-k)!}=\\frac{(m-1)!}{k!(m-k+1)!}.$$\n\nAnyway, the first development holds for $0\\le k\\le m+1$, giving a constant term $\\dfrac{a^m}{m+1}$, but the second for $0\\le k\\le m$ only, giving no constant term. But this does not matter, as two antiderivatives can differ by a constant.\n\n\u2022 But wouldn't the C be different for both the cases?? \u2013\u00a0Agent Smith Jun 13 '18 at 0:48\n\u2022 @AgentSmith: and ? \u2013\u00a0Yves Daoust Jun 13 '18 at 5:50\n\u2022 and that would change the answer wouldn't it?? if we were asked for the value of C? \u2013\u00a0Agent Smith Jun 13 '18 at 19:56\n\u2022 @AgentSmith: you'd answer: arbitrary. You should understand that the value of the constant does not matter. \u2013\u00a0Yves Daoust Jun 13 '18 at 20:05\n\u2022 oh okay thanks! Makes sense! \u2013\u00a0Agent Smith Jun 14 '18 at 18:59", "date": "2019-10-16 04:43:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2816513/same-integration-with-2-different-answers", "openwebmath_score": 0.8379353284835815, "openwebmath_perplexity": 630.6923798847439, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754479181589, "lm_q2_score": 0.9073122144683576, "lm_q1q2_score": 0.8933173299617998}}
{"url": "https://gmatclub.com/forum/a-train-travels-from-city-a-to-city-b-the-average-speed-of-the-train-188007.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Aug 2018, 23:02\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# A train travels from city A to city B. The average speed of the train\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1835\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nA train travels from city A to city B. The average speed of the train\u00a0 [#permalink]\n\n### Show Tags\n\n04 Nov 2014, 00:03\n9\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n64% (02:41) correct 36% (02:28) wrong based on 199 sessions\n\n### HideShow timer Statistics\n\nA train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?\n\nA. 30\nB. 45\nC. 54\nD. 72\nE. 90\n\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1835\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nRe: A train travels from city A to city B. The average speed of the train\u00a0 [#permalink]\n\n### Show Tags\n\n10 Nov 2014, 20:00\n5\n2\nRefer diagram below:\n\nAttachment:\n\nspeed.png [ 3.39 KiB | Viewed 3540 times ]\n\nSetting up the time equation:\n\nTotal time = Time required in first quarter + Time required in the remaining journey\n\n$$\\frac{d}{60} = \\frac{d}{4*90} + \\frac{3d}{4*s}$$\n\n$$s = \\frac{90*3}{5} = 54$$\n\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\n##### General Discussion\nSenior Manager\nStatus: Math is psycho-logical\nJoined: 07 Apr 2014\nPosts: 421\nLocation: Netherlands\nGMAT Date: 02-11-2015\nWE: Psychology and Counseling (Other)\nA train travels from city A to city B. The average speed of the train\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jan 2015, 06:23\nUsing the RTD chart, with A being the first quarter, B the rest of the trip and All the combined \"trip\".\n\n_______R_____T______D\nA.........90........3..........270\nB.........54.......15.........810\nAll........60.......18........1080\n\nLet me explain how we fill in the chart:\n1) Starting with what we know, we add 90 and 60\n2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.\n3) 1/4 of the whole distance happened at a Rate of 90. So, 1080/4=270, add 270 under A-D.\n4) 2080 - 270 = 810, for the rest of the trip. Add 810 under B-D.\n5) Claculate the individual Times for A and All, using T=D/R. Add the results, 3 and 18, under A-T and All-T.\n6) 18-3=15, this is the remaining Time for B. Add 15 under B-T.\n7) Finally, 15R=810 -->R=810/15 --> R= 54 ANS C\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 3186\nLocation: United States (CA)\nRe: A train travels from city A to city B. The average speed of the train\u00a0 [#permalink]\n\n### Show Tags\n\n06 Apr 2017, 09:45\n1\n1\nPareshGmat wrote:\nA train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?\n\nA. 30\nB. 45\nC. 54\nD. 72\nE. 90\n\nWe have an average rate problem in which we can use the following formula:\n\nAvg speed = (distance 1 + distance 2)/(time 1 + time 2)\n\nIf we let d = total distance of the trip, then the first quarter of the trip, or (1/4)d = d/4, was traveled at 90 mph. Thus, the time was (d/4)/90 = d/360.\n\nWe can let the rate for the remaining part of the trip = r, and thus the time for the remaining part of the trip, or (3/4)d = 3d/4, is (3d/4)/r = 3d/(4r). Let\u2019s use all of this information in the average rate equation:\n\n60 = d/(d/360 + 3d/(4r))\n\n60 = 1/(1/360 + 3/(4r))\n\n60(1/360 + 3/(4r)) = 1\n\n1/6 + 45/r = 1\n\nLet\u2019s multiply the above equation by 6r:\n\nr + 270 = 6r\n\n270 = 5r\n\nr = 54\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 12217\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: A train travels from city A to city B. The average speed of the train\u00a0 [#permalink]\n\n### Show Tags\n\n08 Apr 2017, 18:53\n2\nHi All,\n\nThis question can be solved by TESTing VALUES.\n\nThe prompt tells us that average speed of the train over the course of the entire trip is 60 miles/hr and that it travels the first quarter of the distance at a speed of 90 mi/hr. We're asked for the average speed of the train for the remaining three-quarters of the trip.\n\nLet's choose 90 miles for the FIRST QUARTER of the distance. We can then immediately calculate two things...\n1) Since the train was traveling 90 miles/hour, the first quarter of the trip took 1 hour to complete.\n2) Since (1/4)(Total Distance) = 90 miles, then the FULL TRIP = 4(90) = 360 miles.\n\nThe total trip is 360 miles; with an average speed of 60 miles/hr, the FULL TRIP would take... (X)(60 mph) = 360 miles.... X = 6 hours to complete.\n\nThe first hour of the trip covered 90 miles, so the remaining 360 - 90 = 270 miles of the trip are covered in the remaining 5 hours...\n\nThus, the average speed for the remainder of the trip is... (270 miles)/(5 hours) = 54 miles/hour\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 7774\nRe: A train travels from city A to city B. The average speed of the train\u00a0 [#permalink]\n\n### Show Tags\n\n29 Jul 2018, 07:25\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: A train travels from city A to city B. The average speed of the train &nbs [#permalink] 29 Jul 2018, 07:25\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-08-21 06:02:38", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-train-travels-from-city-a-to-city-b-the-average-speed-of-the-train-188007.html", "openwebmath_score": 0.4615800380706787, "openwebmath_perplexity": 3476.47835810479, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8933094074745443, "lm_q1q2_score": 0.8933094074745443}}
{"url": "https://gmatclub.com/forum/if-22-n-is-a-divisor-of-97-98-what-is-the-maximum-possible-228660.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 30 Mar 2020, 07:52\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# If 22^n is a divisor of 97!+98! ,what is the maximum possible\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nCurrent Student\nJoined: 12 Aug 2015\nPosts: 2537\nSchools: Boston U '20 (M)\nGRE 1: Q169 V154\nIf 22^n is a divisor of 97!+98! ,what is the maximum possible\u00a0 [#permalink]\n\n### Show Tags\n\n10 Nov 2016, 10:58\n1\n7\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n57% (01:48) correct 43% (01:42) wrong based on 94 sessions\n\n### HideShow timer Statistics\n\nIf 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?\n\nA)8\nB)9\nC)10\nD)11\nE)12\n\n_________________\nMath Expert\nJoined: 02 Aug 2009\nPosts: 8296\nRe: If 22^n is a divisor of 97!+98! ,what is the maximum possible\u00a0 [#permalink]\n\n### Show Tags\n\n10 Nov 2016, 17:36\n5\nstonecold wrote:\nIf 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?\n\nA)8\nB)9\nC)10\nD)11\nE)12\n\n22^n means power of 11 as 2 would surely have more power than 11 in a factorial..\n97!+98!=97!(1+98)=97!*99....\n99 will have ONE 11 and 97! will have 97/11 or 8..\nTotal 1+8=9\n\nB\n_________________\n##### General Discussion\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 9895\nLocation: United States (CA)\nRe: If 22^n is a divisor of 97!+98! ,what is the maximum possible\u00a0 [#permalink]\n\n### Show Tags\n\n12 Nov 2016, 05:33\n2\n2\nstonecold wrote:\nIf 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?\n\nA)8\nB)9\nC)10\nD)11\nE)12\n\nWe are given that 22^n is a divisor of 97!+98!. Thus:\n\n(97!+98!)/22^n = integer\n\nTo determine the maximum value of n, we can factor out a 97! in the numerator of the above expression and we have:\n\n97!(1 + 98)/22^n = integer\n\n97!(99)/22^n = integer\n\nSince 22 = 11 x 2, we can rewrite our expression as:\n\n97!(99)/(11^n x 2^n) = integer\n\nIn order to determine the maximum value of n such that 22^n divides into 97!(99), we need to determine the maximum number of pairs of factors of 11 and 2 within 97!(99). Since we know there are fewer factors of 11 than factors of 2, we can determine the number of factors of 11 within 97!(99). Let\u2019s start with 97!:\n\n11, 22, 33, 44, 55, 66, 77 and 88 are all factors of 97! Thus, 97! has 8 factors of 11.\n\nWe also see that 99 has 1 factor of 11.\n\nThus, there are 9 factors of 11 within 97!(99), and thus the maximum value of n is 9.\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n197 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nCrackVerbal Quant Expert\nJoined: 12 Apr 2019\nPosts: 459\nRe: If 22^n is a divisor of 97!+98! ,what is the maximum possible\u00a0 [#permalink]\n\n### Show Tags\n\n24 Feb 2020, 23:55\nIn questions related to divisibility, breaking down the dividend will give a lot of information about what the divisor needs to be if it HAS to divide the dividend fully. Therefore, let\u2019s break the dividend down, first.\n\nIn 97! + 98!, we can take 97! as common. When we do this, we have 97! + 98! = 97! (1 + 98) = 97! * 99. This helps us understand that $$22^n$$ should be able to divide 97! * 99 fully.\n\nLet\u2019s now look at 22. 22 = 2*11, therefore $$22^n$$ = $$2^n$$ * $$11^n$$.\nSo, essentially, we are trying to find the highest power of 11 in the dividend since the number of 11s would be far fewer compared to the number of 2s.\nFinding out the highest power of any prime number in a given factorial can be done by successive division. In our case, it can be done as shown below in the diagram.\n\nAttachment:\n\n25th Feb 2020 - Reply 3 - 1.jpg [ 73.73 KiB | Viewed 123 times ]\n\nSo, the highest power of 11 in 97! is 8. But, 99 also has a 11 in it. Therefore, the highest power of 11 in the numerator (or the dividend) is 9. So, there need to be 9 11s in the denominator as well. As such, the highest power of 22 that can divide 97! + 98! is 9.\nThe correct answer option is B.\n\nHope that helps!\n_________________\nRe: If 22^n is a divisor of 97!+98! ,what is the maximum possible \u00a0 [#permalink] 24 Feb 2020, 23:55\nDisplay posts from previous: Sort by", "date": "2020-03-30 15:52:04", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-22-n-is-a-divisor-of-97-98-what-is-the-maximum-possible-228660.html", "openwebmath_score": 0.7583616375923157, "openwebmath_perplexity": 2428.018596790443, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8933093996634686, "lm_q1q2_score": 0.8933093996634686}}
{"url": "https://www.physicsforums.com/threads/general-formula-for-this.104303/", "text": "# General formula for this\n\n1. Dec 16, 2005\n\n### twoflower\n\nHi all,\n\nthere is general formula for findind out, \"Which highest power of x is divisible n! with?\" Eg. Which highest power of 5 is divisable 50! with?\n\nBut I forgot it and can't find it now...will somebody help me please?\n\nThank you.\n\n2. Dec 16, 2005\n\n### AKG\n\nYou mean, what is the highest power of x that divides n!? Well if you can prime-factorize x and n!, this is easy. In the case of 5 and 50!, it's hard to prime-factorize 50!, but the solution is still easy. You can see quite easily that 5 will occur 12 times in the prime-factorization of 50! (it will occur once in each of the factors 5, 10, 15, 20, 30, 35, 40, 45, and twice in both 25 and 50), so the highest power of 5 that divides 50! is 512.\n\n3. Dec 16, 2005\n\n### twoflower\n\nYes, this manual approach is clear. But there is also a general formula..\n\n4. Dec 16, 2005\n\n### AKG\n\nI'm not sure about a general formula just yet, but this might be helpful: if you associate each factorial with a sequence (an,k)k in N, where an,k is the power of the kth prime in n!, you get:\n\n(a0,k) = (0, 0, ...)\n(a1,k) = (0, 0, ...)\n(a2,k) = (1, 0, 0, ...)\n(a3,k) = (1, 1, 0, 0, ...)\n(3, 1, 0, 0, ...)\n(3, 1, 1, 0, 0, ...)\n(4, 2, 1, 0, 0, ...)\n(4, 2, 1, 1, 0, 0, ...)\n(7, 2, 1, 1, 0, 0, ...)\n(7, 4, 1, 1, 0, 0, ...)\n(8, 4, 2, 1, 0, 0, ...)\n(8, 4, 2, 1, 1, 0, 0, ...)\n(10, 5, 2, 1, 1, 0, 0, ...)\n(10, 5, 2, 1, 1, 1, 0, 0, ...)\n(11, 5, 2, 2, 1, 1, 0, 0, ...)\n\nThe exponent of 2 changes every 2 rows, the exponent of 3 changes every 3 rows, the exponent of p will change every p rows. Ignoring repetition, the exponents for 2 go: 0, 1, 3, 4, 7, 8, 10, 11, ...\nFor 3, they go 0, 1, 2, 4, 5, ...\nFor 5, they go 0, 1, 2, ...\n\nSome numbers are skipped in the above sequences because powers occur, i.e. in the sequence for 2, it goes from 1 to 3 without going through 2 because the 4 in 4! contributes two 2's, not just 1. I'm very tired right now, but I suppose if you can generalize what's going on here, it might be a helpful step in finding a general formula.\n\nWell I suppose if you just want to know the general formula and aren't trying to figure it out yourself, then the above isn't much help.\n\n5. Dec 16, 2005\n\n### VietDao29\n\nI suppose that this is not a homework problem... So here's my approach.\nLet's say that [x] is the function that will return the integer part before the decimal point of the number x, eg: [37.5534] = 37.\nSay, you need to find the largest p that satisfies:\nxp divides n!, for some given x (x is prime), and n.\nLet: $$\\rho := \\left[ \\frac{\\ln n}{\\ln x} \\right]$$, ie: $$\\rho$$ is the largest positive integer such that: $$x ^ \\rho \\leq n$$\nSo for every x consecutive integers there's one integer that's divisible by x, for every x2 consecutive integers there's one integer that's divisible by x2, for every x3 consecutive integers there's one integer that's divisible by x3,...\nSo the largest p can be obtained by:\n$$p := \\sum_{i = 1} ^ \\rho \\left[ \\frac{n}{x ^ i} \\right]$$\n---------------------\nIf x is not prime, then you can prime-factorize it:\n$$x = \\lambda_1 ^ {\\alpha_1} \\ \\lambda_2 ^ {\\alpha_2} \\ \\lambda_3 ^ {\\alpha_3} \\ ... \\lambda_k ^ {\\alpha_k}$$\nWhere: $$\\lambda_i, \\ i = 1..k$$ are primes.\nThen: $$x ^ p = (\\lambda_1 ^ {\\alpha_1} \\ \\lambda_2 ^ {\\alpha_2} \\ \\lambda_3 ^ {\\alpha_3} \\ ... \\lambda_k ^ {\\alpha_k}) ^ p = \\lambda_1 ^ {\\alpha_1 p} \\ \\lambda_2 ^ {\\alpha_2 p} \\ \\lambda_3 ^ {\\alpha_3 p} \\ ... \\lambda_k ^ {\\alpha_k p}$$\nIf xp divides n! then $$\\lambda_i ^ {\\alpha_i p}, \\ i = 1..k$$ must also divide n!.\nSo you can use the same method (as shown above) to find the largest $$\\rho_i, \\ i = 1..k$$ such that $$\\lambda_i ^ {\\rho_i}, \\ i = 1..k$$ divides n!.\nThen define: $$\\beta_i := \\left[ \\frac{\\rho_i}{\\alpha_i} \\right], \\ i = 1..k$$.\nAnd the largest p can be found by:\n$$p = \\min (\\beta_i), \\ i = 1..k$$\n\nLast edited: Dec 16, 2005\n6. Dec 25, 2005\n\n### benorin\n\nThe exact power of a prime p dividing n! is alternatively given by\n\n$$\\frac{n-\\mu}{p-1},$$\n\nwhere $\\mu$ is the sum of the digits of the base p representation of n.", "date": "2017-10-23 15:01:00", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/general-formula-for-this.104303/", "openwebmath_score": 0.6563915014266968, "openwebmath_perplexity": 621.9016523268043, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085085246543, "lm_q2_score": 0.9086178963141964, "lm_q1q2_score": 0.8932487020545764}}
{"url": "http://math.stackexchange.com/questions/281604/probability-of-an-event-after-time-has-passed/281639", "text": "# Probability of an event after time has passed\n\nA friend of mine posed this problem and we have had a disagreement on the answer.\n\nThe problem:\n\n\u2022 There is a 90% chance that some event will happen in the next year.\n\u2022 There is a 95% chance that the event will happen eventually.\n\u2022 What is the probability that it happens after this year, if it does not happen in the next year?\n\nHover over here for a description of our answers and arguments:\n\nI told my friend that it is 95%. My reasoning is that there is a 95% chance that it happens from now to infinity. The probability that it happens from now to infinity minus one year is still 95%. In one year, when it hasn't happened, events that didn't happen don't affect your \"eventual\" odds.\n\nMy friend on the other hand, believes it to be 50%. He had two arguments. First, he said if you flip a coin 4 times, the chance that you get heads the first time is 50%, but the chance that you get heads eventually is 93.75% (15/16). If you don't get it the first time, what is the chance that you get it after the first time? 87.5% (7/8), in other words, it has decreased.\n\nTo this I responded that the \"eventually\" is based on a finite set of events, each with their own probability and after the first event has passed, you have fewer chances. In his problem, on the other hand, it is a summary of the probability of unknown events broken in two time periods.\n\nHe continued to insist that it is the same because you have two time periods with known odds. The 95% is the combination of two probabilities, 90% and 50%. Eliminating the first 90%, the probability of the remaining period is 50%.\n\nI agreed that's an accurate way to look at it from now, but I don't think that's accurate once you know it did not happen in that time period.\n\n-\n\nLet's use Bayes' Theorem: $$P(A|B)=\\frac{P(B|A) P(A)}{P(B)}.$$ In our case, $B$ is the event whatnot does not occur in year 1 (so with probability $P(B)=.1$, and $A$ is the event whatnot occurs at some point (so with probability $P(A)=0.95$). Of course, $P(B|A)$ is the only thing left to determine, but notice that $P(A|not(B))=1$ since if whatnot occurs in year 1, then it occurs at some point. Therefore, again by Bayes' Theorem $$P(not(B)|A)=\\frac{P(A|not(B)) P(not(B))}{P(A)}=\\frac{90}{95}.$$ Since $P(not(B)|A)+P(B|A)=1$, we have that $P(B|A)=\\frac{5}{95}$. Hence plugging this all in, we find $$P(A|B)=\\frac{\\frac{5}{95} \\frac{95}{100}}{\\frac{10}{100}}=\\frac{1}{2}.$$ So it looks like your friend was right!\n\n-\n\nCalculate it this way. Given that the probability that it occurs during or after year 2 is $x$%. And the probability it occurs in year 1 is 90%, what is the probability it occurs?\n\nNow, consider $x=0$, then the probability the event occurs is 90%.\n\nIf $x=50$, then it is 95%. (because it is $0.9 \\times 1 + 0.1 \\times 0.5 \\times 1= 0.95$)\n\nThis informal argument should convince you.\n\n-\n\nAfter talking about it some more, we came up with this scenario which illustrates why he is correct, and I am wrong.\n\n1. Say I want to give him a 90% chance that he will get $100 this year, 95% chance total, and 5% chance that he gets nothing. 2. So, I take 100 sheets of paper and write dates in the next year on 90 of them, dates after that on 5, and \"Bad luck\" on the other 5. 3. I draw one at random and keep it a secret. 4. A year passes and my friend got nothing. Now he knows that all 90 dates this year are still in the pile. He now knows I must hold one of the other 10 sheets. That makes a 5/10 chance that I have a sheet with a date on it, and 5/10 chance that the sheet I hold has \"Bad luck\" on it. 50% chance he will get$100.\n\n-\nNice. Beats Bayes. \u2013\u00a0 Andr\u00e9 Nicolas Jan 18 '13 at 21:15\n\nLet $A$ represent the cases in which the event happens eventually. Let $B$ represent the cases in which the event happens next year. You've stipulated $P(A)=0.95$ and $P(B)=0.90$. Because $B$ is a subset of $A$, $P(A\\cap B)=P(B)=0.90$, and $$P(A\\cap \\neg B)=P(A)-P(A\\cap B)=0.95-0.90=0.05.$$ By the definition of conditional probability, then, $$P(A|\\neg B)=\\frac{P(A\\cap \\neg B)}{P(\\neg B)}=\\frac{0.05}{0.10}=\\frac{1}{2}.$$ So if I propose a deal right now, under the terms of which (a) you'll put in $\\$1000$if the event hasn't happened after one year and (b) you'll get back$X$if the event happens after next year, then you should take the deal if$X>\\$2000$. In this sense, your friend is right.\n\nYour point, though, is different. You're wondering what deal you should take a year from now if the event hasn't happened by then. The answer is that it depends on probabilities not stipulated in the problem. In particular, you may learn something from the fact that it didn't happen this year, which may raise or lower these odds.\n\nSuppose the event is the earth being hit by a meteor, that this depends on the solar system's meteor density, and that we're building a meteor shield that will be operational in just over a year. Based on what we know now, the meteor density might be high or low, with equal likelihood; if it's high, then the earth will be hit in the next year with probability $1$ (no shield yet!); if it's low, then the earth will be hit next year with probability $0.8$ and eventually with probability $0.9$. A year from now, if the earth hasn't been hit yet, you will know that the meteor density is low, and that the remaining probability that we'll ever be hit is $0.1$. In this case, the odds were lowered.\n\nOn the other hand, the choice may be between cases where the event will happen in either eighteen months with certainty (so next year with probability $0$ and eventually with probability $1$), or at some indeterminate time (next year with probability $0.9/(1-\\varepsilon)$ and eventually with probability $(0.95-\\varepsilon)/(1-\\varepsilon)$), based on some unknown variable that might point to the first case (with probability $\\varepsilon$) but almost surely points to the second (with probability $1-\\varepsilon$). Here, a year from now you'll be much more likely to believe that the unknown variable is pointing to the first case, which will increase the odds that the event is still forthcoming.\n\n-\n\nThere could be slight issue with the phrasing of the problem. Your interpretation is that the statement is true no matter what time period you are in. I.e. in 5 years time, there is still a 90% chance that some event will happen in that year. However, you're forgetting to account for the fact that the event didn't happen in the first year.\n\nYour friend's interpretation, is that your statement is true for this year. He has ignored the probability space in which the event occurred this year, and considered the remaining probability space. This now has measure $1-0.9=0.1$, and we know that the event has $0.95 - 0.9 = 0.05$ measure to occur, so it has $\\frac {0.05}{0.1} = 50\\%$ chance to happen.\n\nFor example, consider the event where outcomes are based on drawing a uniform random variable on $[0,1]$. Let $C$ be your favorite number from 0 to 0.9.\n\nIf we draw a number from 0 to 0.9, the event happens this year.\nIf we draw a number from $X$ to 0.95, the event happens next year.\nIf we draw a number from 0.95 to 1, the event never happens.\n\nA possible scenario describing your interpretation, could be\n\nIf we draw a number from 0 to 0.9, the event happens every even year.\nIf we draw a number from 0.05 to 0.95, the event happens every odd year.\nIf we draw a number from 0.95 to 1, the event never happens.\n\nThen, it will always be true, that there is a 90% chance of it happening this year, and a 95% chance of it happening eventually. However, not that if we know it doesn't happen in this year (doesn't matter whether it's even or odd), then there's only a 50% chance that it will ever happen again.\n\n-\nThanks for the alternate scenario, that's interesting. This is the kind of thing that made me feel like it's hard to guarantee an answer in the first place. \u2013\u00a0 NickC Jan 18 '13 at 21:25\n@NickC The hard part is considering completely what it means for an event to not occur. You should look at this recent post, which had a lot of different answers initially. \u2013\u00a0 Calvin Lin Jan 18 '13 at 21:29\n\nIn case the mathematical proof already mentioned above isn't enough to convince you, pretend you're a statistician collecting your own results. You have 100 people infected with a disease that has a 95% death rate and a 90% death rate in the first year. Year one is finished, 90 dead, as predicted. 10 people left. We know 95% will be dead at some point. How many more are likely to die if the odds hold true? 5/10. Chances once the first year has passed is 50%. You don't need to be a statistician to see that. It has nothing to do with a finite number of events or an infinite time frame. That part has already been calculated for us in the given 95%.\n\n-\n\nThe analysis for this appears to be relatively straightforward. Assume the probability of event in year 2+ is $x$. Next, the probability that the event does not occur in the first year is $0.1$ and that it does not occur any time thereafter is $(1-x)$. The product of the two is the probability that the event never happens, which we know to be $0.05$. So,\n\n$0.1 \\times (1-x) = 0.05$\n\nsolving for x yields $x=\\frac{1}{2}$\n\n-", "date": "2015-05-24 19:43:38", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/281604/probability-of-an-event-after-time-has-passed/281639", "openwebmath_score": 0.8925812840461731, "openwebmath_perplexity": 259.3947522597049, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363737832231, "lm_q2_score": 0.9059898248255074, "lm_q1q2_score": 0.8932483225729583}}
{"url": "https://plainmath.net/algebra-ii/78503-changing-base-of-a-logarithm-by-taking-a", "text": "crossoverman9b\n\n2022-06-20\n\nChanging base of a logarithm by taking a square root from base?\nFrom my homework I found\n${\\mathrm{log}}_{9}x={\\mathrm{log}}_{3}\\sqrt{x}$\nand besides that an explanation that to this was done by taking a square root of the base. I fail to grasp this completely. Should I need to turn ${\\mathrm{log}}_{9}x$ into base 3, I'd do something like\n${\\mathrm{log}}_{9}x=\\frac{{\\mathrm{log}}_{3}x}{{\\mathrm{log}}_{3}9}=\\frac{{\\mathrm{log}}_{3}x}{{\\mathrm{log}}_{3}{3}^{2}}=\\frac{{\\mathrm{log}}_{3}x}{2}$\nbut this is a far cry from what I've given as being the correct answer.\nSubstituting some values to x and playing with my calculator I can see that the answer given as correct is correct whereas my attempt fails to yield the correct answer.\nNow the question is, what are correct steps to derive ${\\mathrm{log}}_{3}\\sqrt{x}$ from ${\\mathrm{log}}_{9}x$? How and why am I allowed to take a square root of the base and the exponent?\n\nmallol3i\n\nYou are only one step away!\nNote that $a\\mathrm{log}b=\\mathrm{log}{b}^{a}$, so\n$\\frac{{\\mathrm{log}}_{3}x}{2}=\\frac{1}{2}{\\mathrm{log}}_{3}x={\\mathrm{log}}_{3}{x}^{1/2}={\\mathrm{log}}_{3}\\sqrt{x}.$\n\nYesenia Sherman", "date": "2023-03-28 01:56:08", "meta": {"domain": "plainmath.net", "url": "https://plainmath.net/algebra-ii/78503-changing-base-of-a-logarithm-by-taking-a", "openwebmath_score": 0.7619131207466125, "openwebmath_perplexity": 220.66687792279453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9873750492324249, "lm_q2_score": 0.9046505280315008, "lm_q1q2_score": 0.8932293596532423}}
{"url": "https://math.stackexchange.com/questions/2801467/probability-a-person-chosen-at-random-owns-an-automobile-or-a-house-but-not-bot/2801568", "text": "# Probability a person chosen at random owns an automobile or a house, but not both\n\nA marketing survey indicates that $60\\%$ of the population owns an automobile, $30\\%$ owns a house, and $20\\%$ owns both an automobile and a house. Calculate the probability that a person chosen at random owns an automobile or a house, but not both.\n\n(A) $0.4$\n\n(B) $0.5$\n\n(C) $0.6$\n\n(D) $0.7$\n\n(E) $0.9$\n\nSo the answer given for this question is $0.5$, but it doesn't make sense to me from the perspective of addition rule in probability. If we know that $\\Pr(\\text{Owning Automobile}) = 0.6$ and the $\\Pr(\\text{Owning a House}) = 0.3$ and the $\\Pr(\\text{Owning both an automobile and a house}) = 0.2$, according to the formula probability of $\\Pr(A)~\\text{or}~\\Pr(B) = \\Pr(A) + \\Pr(B) - \\Pr(A~\\text{and}~B)$, wouldn't the answer be $0.6+0.3-0.2=0.7$?\n\nIs it because the ordering matters? so the last term $\\Pr(A~\\text{and}~B) = 2 \\cdot 0.2$?\n\n\u2022 You computed the probability of \"owning an automobile or a house.\" They asked for the probability of \"owning an automobile or a house but not both.\" \u2013\u00a0angryavian May 30 '18 at 6:16\n\u2022 @angryavian I still don't get it, I thought \"owning an automobile or a house\" is the same as \"owning an automobile or a house but not both\" ?? \u2013\u00a0pino231 May 30 '18 at 6:20\n\u2022 Please type your question rather than posting an image. Images cannot be searched. \u2013\u00a0N. F. Taussig May 30 '18 at 8:54\n\u2022 @pino231, in fact \"owning an automobile or a house\" in this case means \"owning an automobile, or a house, or both\". This usage of the word \"or\" -- the so-called \"inclusive or\" -- is standard in mathematics. If we want the opposite (the exclusive or) it is almost always explicitly stated, as it is in your question. \u2013\u00a0Mees de Vries May 30 '18 at 9:26\n\nLet event $A$ be owning an automobile. Let event $B$ be owning a house.\n\nConsider the diagram below:\n\nIf we simply add the probabilities that a person owns an automobile and a person owns a house, we will have added the probability that a person owns both an automobile and a house twice. Thus, to find the probability that a person owns an automobile or a house, we must subtract the probability that a person owns both an automobile and a house from the sum of the probabilities that a person owns an automobile and that a person owns a house.\n\n$$\\Pr(A \\cup B) = \\Pr(A) + \\Pr(B) - \\Pr(A \\cap B)$$\n\nNote that on the left-hand side of your equation, you should have written $\\Pr(A \\cup B)$ or $\\Pr(A~\\text{or}~B)$ rather than $\\Pr(A)~\\text{or}~\\Pr(B)$.\n\nSince we are given that $60\\%$ of the population owns an automobile, $30\\%$ of the populations owns a house, and $20\\%$ owns both, $\\Pr(A) = 0.60$, $\\Pr(B) = 0.30$, and $\\Pr(A \\cap B) = 0.20$. Hence, the probability that a person owns an automobile or a house is $$\\Pr(A \\cup B) = \\Pr(A) + \\Pr(B) - \\Pr(A \\cap B) = 0.60 + 0.30 - 0.20 = 0.70$$ However, the question asks for the probability that a person owns an automobile or a house, but not both. That means we must subtract the probability that a person owns an automobile and a house from the probability that the person owns an automobile or a house.\n$$\\Pr(A~\\triangle~B) = \\Pr(A \\cup B) - \\Pr(A \\cap B) = 0.70 - 0.20 = 0.50$$\n\nIn terms of the Venn diagram, $A \\cup B$ is the region enclosed by the two circles, while $A~\\triangle~B = (A \\cup B) - (A \\cap B) = (A - B) \\cup (B - A)$ is the region enclosed by the two circles except the region where the sets intersect.\n\nSince $A - B = A - (A \\cap B)$, $$\\Pr(A - B) = \\Pr(A) - \\Pr(A \\cap B) = 0.60 - 0.20 = 0.40$$\nSince $B - A = B - (A \\cap B)$, $$\\Pr(B - A) = \\Pr(B) - \\Pr(A \\cap B) = 0.30 - 0.20 = 0.10$$\nHence,\n$$\\Pr(A~\\triangle~B) = \\Pr(A - B) + \\Pr(B - A) = 0.40 + 0.10 = 0.50$$ which agrees with the result obtained above.\n\n\u2022 ahh so the question is actually asking the probability that a person chosen at random owns an automobile or a house, OR not both \u2013\u00a0pino231 Jun 1 '18 at 3:20\n\u2022 Not quite. The word but should be read as and. The question is asking the probability that a person chosen at random owns an automobile or a house and not both: $\\Pr(A~\\triangle~B) = \\Pr(A \\cup B) - \\Pr(A \\cap B)$. \u2013\u00a0N. F. Taussig Jun 1 '18 at 9:17\n\nThere's no ordering in this question. Using the addition rule is fine, but if you want to use it, note that:\n\n$\\mathbb{P}(A) + \\mathbb{P}(B) = \\Big( \\mathbb{P}(A \\setminus B) + \\mathbb{P}(A \\cap B) \\Big) + \\Big( \\mathbb{P}(B \\setminus A) + \\mathbb{P}(A \\cap B) \\Big)$.\n\nThis should probably help.\n\n10 $\\%$ of people don't have a house or automobile so you can leave them. There is 90 $\\%$ of people left, also 20$\\%$ have both a house and a automobile. This 20% of people is included in the 60% and 30%, so you have to substract this percentage. What you get is 40% + 10% = 50%. So you were probably right by assuming that you have to count the last term $P(A \\cap B)$ two times.\n\n\u2022 If we subtract $\\Pr(A \\cup B)$ two times, we will obtain a negative answer. Instead, we must subtract $\\Pr(A \\cap B)$ twice. \u2013\u00a0N. F. Taussig May 30 '18 at 9:13\n\u2022 Yes of course, that's what I meant. Will edit! \u2013\u00a0WarreG May 30 '18 at 9:37", "date": "2019-12-11 06:21:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2801467/probability-a-person-chosen-at-random-owns-an-automobile-or-a-house-but-not-bot/2801568", "openwebmath_score": 0.7960233092308044, "openwebmath_perplexity": 174.42810023617778, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120881621377, "lm_q2_score": 0.900529778109184, "lm_q1q2_score": 0.8931563196786563}}
{"url": "https://madhavamathcompetition.com/category/nature-of-mathematics/page/2/", "text": "Miscellaneous questions: Part I: tutorial practice for preRMO and\u00a0RMO\n\nProblem 1:\n\nThe sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.\n\nProblem 2:\n\nLet T be the set of all triples (a,b,c) of integers such that $1 \\leq a < b < c \\leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.\n\nProblem 3:\n\nIn a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.\n\nProblem 4:\n\nFive men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:\n\nA: I see three black and one white cap.\nB: I see four white caps.\nC: I see one black and three white caps.\nD: I see four black caps.\n\nProblem 5:\n\nLet f be a bijective (one-one and onto) function from the set $A=\\{ 1,2,3,\\ldots,n\\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \\in A$. Note that $f^{M}$ denotes the composite function $f \\circ f \\circ f \\ldots \\circ f$ repeated M times.\n\nProblem 6:\n\nShow that there exists a convex hexagon in the plane such that:\na) all its interior angles are equal\nb) its sides are 1,2,3,4,5,6 in some order.\n\nProblem 7:\n\nThere are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.\n\nProblem 8:\n\nIn each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?\n\nProblem 9:\n\nGiven the seven element set $A = \\{ a,b,c,d,e,f,g\\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.\n\nTry these !!\n\nRegards,\nNalin Pithwa\n\nTowards Baby Analysis: Part I: INMO, IMO and CMI\u00a0Entrance\n\n$\\bf{Reference: \\hspace{0.1in}Introductory \\hspace{0.1in} Real Analysis: \\hspace{0.1in} Kolmogorov \\hspace{0.1in} and \\hspace{0.1in} Fomin; \\hspace{0.1in}Dover \\hspace{0.1in }Publications}$\n\n$\\bf{Equivalence \\hspace{0.1in} of \\hspace{0.1in} Sets \\hspace{0.1in} The \\hspace{0.1in}Power \\hspace{0.1in }of \\hspace{0.1in }a \\hspace{0.1in}Set}$\n\n$\\bf{Section 1}$:\n\n$\\bf{Finite \\hspace{0.1in} and \\hspace{0.1in} infinite \\hspace{0.1in} sets}$\n\nThe set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be $\\it{finite}$.$\\it{Clearly \\hspace{0.1in} we \\hspace{0.1in}can \\hspace{0.1in} be \\hspace{0.1in} sure \\hspace{0.1in} that \\hspace{0.1in} a \\hspace{0.1in} set \\hspace{0.1in}is \\hspace{0.1in}finite \\hspace{0.1in} without \\hspace{0.1in} knowing \\hspace{0.1in} the \\hspace{0.1in} number \\hspace{0.1in} of elements \\hspace{0.1in}in \\hspace{0.1in}it.}$\n\nOn the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, $\\it{if \\hspace{0.1in } we \\hspace{0.1in}remove \\hspace{0.1in} one \\hspace{0.1in} element \\hspace{0.1in}from \\hspace{0.1in}each \\hspace{0.1in}set, \\hspace{0.1in}then \\hspace{0.1in}remove \\hspace{0.1in}two \\hspace{0.1in}elements, \\hspace{0.1in}three \\hspace{0.1in}elements, \\hspace{0.1in}and \\hspace{0.1in}so \\hspace{0.1in}on, \\hspace{0.1in}there \\hspace{0.1in}will \\hspace{0.1in}still \\hspace{0.1in}be \\hspace{0.1in}elements \\hspace{0.1in}left \\hspace{0.1in}in \\hspace{0.1in}the \\hspace{0.1in}set \\hspace{0.1in}in \\hspace{0.1in}each \\hspace{0.1in}stage}$. Accordingly, sets of these kind are called $\\it{infinite}$ sets.\n\nGiven two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.\n\nTo compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a $\\it{one-\\hspace{0.1in}to-\\hspace{0.1in}one \\hspace{0.1in}correspondence}$ between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.\n\n$\\bf{Section 2}$:\n\n$\\bf{Countable \\hspace{0.1in} Sets}$.\n\nThe simplest infinite set is the set $\\mathscr{Z^{+}}$ of all positive integers. An infinite set is called $\\bf{countable}$ if its elements can be put into one-to-one correspondence with those of $\\mathscr{Z^{+}}$. In other words, a countable set is a set whose elements can be numbered $a_{1}, a_{2}, a_{3}, \\ldots a_{n}, \\ldots$. By an $\\bf{uncountable}$ set we mean, of course, an infinite set which is not countable.\n\nWe now give some examples of countable sets:\n\n$\\bf{Example 1}$:\n\nThe set $\\mathscr{Z}$ of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between $\\mathscr{Z}$ and $\\mathscr{Z^{+}}$ of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \\geq 0$ with the odd number $2n+1$, and the negative integer $n<0$ with the even number $2|n|$, that is,\n\n$n \\leftrightarrow (2n+1)$, if $n \\geq 0$, and $n \\in \\mathscr{Z}$\n$n \\leftrightarrow 2|n|$, if $n<0$, and $n \\in \\mathscr{Z}$\n\n$\\bf{Example 2}$:\n\nThe set of all positive even numbers is countable, as shown by the obvious correspondence $n \\leftrightarrow 2n$.\n\n$\\bf{Example 3}$:\n\nThe set 2,4,8,$\\ldots 2^{n}$ is countable as shown by the obvious correspondence $n \\leftrightarrow 2^{n}$.\n\n$\\bf{Example 4}: The set$latex \\mathscr{Q}$of rational numbers is countable. To see this, we first note that every rational number $\\alpha$ can be written as a fraction $\\frac{p}{q}$, with $q>0$ with a positive denominator. (Of course, p and q are integers). Call the sum $|p|+q$ as the \u201cheight\u201d of the rational number $\\alpha$. For example, $\\frac{0}{1}=0$ is the only rational number of height zero, $\\frac{-1}{1}$, $\\frac{1}{1}$ are the only rational numbers of height 2, $\\frac{-2}{1}$, $\\frac{-1}{2}$, $\\frac{1}{2}$, $\\frac{2}{1}$ are the only rational numbers of height 3, and so on. We can now arrange all rational numbers in order of increasing \u201cheight\u201d (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set $\\mathscr{Z^{+}}$ of all positive integers. $\\it{Next \\hspace{0.1in}we \\hspace{0.1in} prove \\hspace{0.1in}some \\hspace{0.1in}elementary \\hspace{0.1in}theorems \\hspace{0.1in}involving \\hspace{0.1in}countable \\hspace{0.1in}sets}$ $\\bf{Theorem1}$. $\\bf{Every \\hspace{0.1in} subset \\hspace{0.1in}of \\hspace{0.1in}a \\hspace{0.1in}countable \\hspace{0.1in}set \\hspace{0.1in}is \\hspace{0.1in}countable}$. $\\bf{Proof}$ Let set A be countable, with elements $a_{1}, a_{2}, a_{3}, \\ldots$, and let set B be a subset of A. Among the elements $a_{1}, a_{2}, a_{3}, \\ldots$, let $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \\ldots$ be those in the set B. If the set of numbers $n_{1}, n_{2}, n_{3}, \\ldots$ has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence $i \\leftrightarrow a_{n_{i}}$). $\\bf{QED.}$ $\\bf{Theorem2}$ $\\bf{The \\hspace{0.1in}union \\hspace{0.1in}of \\hspace{0.1in}a \\hspace{0.1in}finite \\hspace{0.1in}or \\hspace{0.1in}countable \\hspace{0.1in}number \\hspace{0.1in}of \\hspace{0.1in}countable \\hspace{0.1in}sets \\hspace{0.1in}A_{1}, A_{2}, A_{3}, \\ldots \\hspace{0.1in}is \\hspace{0.1in}itself \\hspace{0.1in}countable.}$ $\\bf{Proof}$ We can assume that no two of the sets $A_{1}, A_{2}, A_{3}, \\ldots$ have any elements in common, since otherwise we could consider the sets $A_{1}$, $A_{2}-A_{1}$, $A_{3}-(A_{1}\\bigcup A_{2})$, $\\ldots$, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of $A_{1}, A_{2}, A_{3}, \\ldots$ in the form of an infinite table $\\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\\ldots \\\\ a_{21} &a_{22} & a_{23} & a_{24} & \\ldots \\\\ a_{31} & a_{32} & a_{33} & a_{34} & \\ldots \\\\ a_{41} & a_{42} & a_{43} & a_{44} & \\ldots \\\\ \\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\end{array}$ where the elements of the set $A_{1}$ appear in the first row, the elements of the set $A_{2}$ appear in the second row, and so on. We now count all the elements in the above array \u201cdiagonally\u201d; that is, first we choose $a_{11}$, then $a_{12}$, then move downwards, diagonally to \u201cleft\u201d, picking $a_{21}$, then move down vertically picking up $a_{31}$, then move across towards right picking up $a_{22}$, next pick up $a_{13}$ and so on ($a_{14}, a_{23}, a_{32}, a_{41}$)as per the pattern shown: $\\begin{array}{cccccccc} a_{11} & \\rightarrow & a_{12} &\\hspace{0.1in} & a_{13} & \\rightarrow a_{14} & \\ldots \\\\ \\hspace{0.1in} & \\swarrow & \\hspace{0.1in} & \\nearrow & \\hspace{0.01in} & \\swarrow & \\hspace{0.1in} & \\hspace{0.1in}\\\\ a_{21} & \\hspace{0.1in} & a_{22} & \\hspace{0.1in} & a_{23} \\hspace{0.1in} & a_{24} & \\ldots \\\\ \\downarrow & \\nearrow & \\hspace{0.1in} & \\swarrow & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in}\\\\ a_{31} & \\hspace{0.1in} & a_{32} & \\hspace{0.1in} & a_{33} & \\hspace{0.1in} & a_{34} & \\ldots \\\\ \\hspace{0.1in} & \\swarrow & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in}\\\\ a_{41} & \\hspace{0.1in} & a_{42} &\\hspace{0.1in} & a_{43} &\\hspace{0.1in} &a_{44} &\\ldots\\\\ \\ldots & \\hspace{0.1in} & \\ldots & \\hspace{0.1in} & \\ldots & \\hspace{0.1in} & \\ldots & \\hspace{0.1in} \\end{array}$ It is clear that this procedure associates a unique number to each element in each of the sets $A_{1}, A_{2}, \\ldots$ thereby establishing a one-to-one correspondence between the union of the sets $A_{1}, A_{2}, \\ldots$ and the set $\\mathscr{Z^{+}}$ of all positive integers. $\\bf{QED.}$ $\\bf{Theorem3}$ $\\bf{Every \\hspace{0.1in}infinite \\hspace{0.1in}subset \\hspace{0.1in}has \\hspace{0.1in}a \\hspace{0.1in}countable \\hspace{0.1in}subset.}$ $\\bf{Proof}$ Let M be an infinite set and $a_{1}$ any element of M. Being infinite, M contains an element $a_{2}$ distinct from $a_{1}$, an element $a_{3}$ distinct from both $a_{2}$ and $a_{1}$, and so on. Continuing this process, (which can never terminate due to \u201cshortage\u201d of elements, since M is infinite), we get a countable subset $A= \\{ a_{1}, a_{2}, a_{3}, \\ldots, a_{n}, \\ldots\\}$ of the set $M$. $\\bf{QED.}$ $\\bf{Remark}$ Theorem 3 shows that countable sets are the \u201csmallest\u201d infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below. $\\bf{Section3}$ $\\bf{Equivalence \\hspace{0.1in} of \\hspace{0.1in} sets}$ We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set $\\mathscr{Z^{+}}$ of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N. $\\bf{Definition}$ Two sets M and N are said to be $\\bf{equivalent}$ (written $M \\sim N$) if there is a one-to-one correspondence between the elements of M and the elements of N. The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set $\\mathscr{Z^{+}}$ of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent. $\\bf{Example1}$ The sets of points in any two closed intervals$[a,b]$and$[c,d]\\$ are equivalent; you can \u201csee\u2019 a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment \u201cab\u201d above the base of the triangle; where \u201ca\u201d lies on one side of the triangle and \u201cb\u201d lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect.\n\n$\\bf{Example2}$\n\nThe set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence $z \\leftrightarrow \\alpha$ can be established by using stereographic projection. The origin is the North Pole of the sphere.\n\n$\\bf{Example3}$\n\nThe set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula $y=\\frac{1}{\\pi}\\arctan{x}+\\frac{1}{2}$ establishes a one-to-one correspondence between these two sets. $\\bf{QED}$.\n\nThe last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are \u201cas many\u201d positive integers as integers of arbitrary sign, there are \u201cas many\u201d points in the interval $(0,1)$ as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by:\n\n$\\bf{Theorem4}$\n\n$\\bf{Every \\hspace{0.1in} infinite \\hspace{0.1in} set \\hspace{0.1in}is \\hspace{0.1in} equivalent \\hspace{0.1in} to \\hspace{0.1in}one \\hspace{0.1in}of \\hspace{0.1in}its \\hspace{0.1in}proper \\hspace{0.1in}subsets.}$\n\n$\\bf{Proof}$\n\nAccording to Theorem 3, every infinite set M contains a countable subset. Let this subset be $A=\\{a_{1}, a_{2}, a_{3}, \\ldots, a_{n}, \\ldots \\}$ and partition A into two countable subsets $A_{1}=\\{a_{1}, a_{3}, a_{5}, \\ldots \\}$ and $A_{2}=\\{a_{2}, a_{4}, a_{6}, \\ldots \\}$.\n\nObviously, we can establish a one-to-one correspondence between the countable subsets A and $A_{1}$ (merely let $a_{n} \\leftrightarrow a_{2n-1}$). This correspondence can be extended to a one-to-one correspondence between the sets $A \\bigcup (M-A)=M$ and $A_{1} \\bigcup (M-A)=M-A_{2}$ by simply assigning x itself to each element $x \\in M-A$. But $M-A_{2}$ is a proper subset of M. $\\bf{QED}$.\n\nMore later, to be continued,\n\nRegards,\nNalin Pithwa\n\nFind a flaw in this proof: RMO and PRMO\u00a0tutorial\n\nWhat ails the following proof that all the elements of a finite set are equal?\n\nThe following is the \u201cproof\u201d;\n\nAll elements of a set with no elements are equal, so make the induction assumption that any set with n elements has all its elements equal. In a set with n elements, the first and the last n are equal by induction assumption. They overlap at n, so all are equal, completing the induction.\n\nEnd of \u201cproof:\n\nRegards,\n\nNalin Pithwa\n\nProblem Solving approach: based on George Polya\u2019s opinion: Useful for RMO/INMO, IITJEE maths\u00a0preparation\n\nI have prepared the following write-up based on George Polya\u2019s classic reference mentioned below:\n\nUNDERSTANDING THE PROBLEM\n\nFirst. \u201cYou have to understand the problem.\u201d\n\nWhat is the unknown ? What are the data? What is the condition?\n\nIs it possible to satisfy the condition? Is the condition sufficient to determine the unknown? Or is it insufficient? Or redundant ? Or contradictory?\n\nDraw a figure/diagram. Introduce a suitable notation. Separate the various parts of the condition. Can you write them down?\n\nSecond.\n\nDEVISING A PLAN:\n\nFind the connection between the data and the unknown. You may be obliged to consider auxiliary problems if an immediate connection cannot be found. You should eventually obtain a plan for the solution.\u201d\n\nHave you seen it before? Or have you seen the problem in a slightly different form? Do you know a related problem? Do you know a theorem that could be useful? Look at the unknown! And try to think of a familiar problem having the same or a similar unknown. Here is a problem related to yours and solved before. Could you use it? Could you use its result? Could you use its method? Should you restate it differently? Go back to definitions.\n\nIf you cannot solve the proposed problem, try to solve some related problem. Could you imagine a more accessible related problem? A more general problem? A more special problem? An analogous problem? Could you solve a part of the problem? Keep only a part of the condition, drop the other part, how far is the unknown then determined, how can it vary? Could you derive something useful from the data? Could you think of other data appropriate to determine the unknown? Could you change the unknown of the data, or both, if necessary, so that the new unknown and the new data are nearer to each other? Did you use all the data? Did you use the whole condition? Have you taken into account all essential notions involved in the problem?\n\nCarrying out your plan of the solution, check each step. Can you clearly see that the step is correct? Can you prove that it is correct?\n\nFourth. LOOKING BACK.\n\nExamine the solution.\n\nCan you check the result? Can you check the argument? Can you derive the result differently? Can you see it at a glance? Can you see the result, or the method, for some other problem?\n\n**************************************************************************\n\nReference:\n\nHow to Solve It: A New Aspect of Mathematical Method \u2014 George Polya.\n\nhttps://www.amazon.in/How-Solve-Aspect-Mathematical-Method/dp/4871878309/ref=sr_1_1?crid=2DXC1EM1UVCPW&keywords=how+to+solve+it+george+polya&qid=1568334366&s=books&sprefix=How+to+solve%2Caps%2C275&sr=1-1\n\nThe above simple \u201cplan\u201d can be useful even to crack problems from a famous classic, Problem Solving Strategies, by Arthur Engel, a widely-used text for training for RMO, INMO and IITJEE Advanced Math also, perhaps.\n\nReference: Problem-Solving Strategies by Arthur Engel; available on Amazon India\n\nConcept of order in math and real\u00a0world\n\n1. Rise and Shine algorithm: This is crazy-sounding, but quite a perfect example of the need for \u201corder\u201d in the real-world: when we get up in the morning, we first clean our teeth, finish all other ablutions, then go to the bathroom and first we have to remove our pyjamas/pajamas and then the shirt, and then enter the shower; we do not first enter the shower and then remove the pyjamas/shirt !! \ud83d\ude42\n2. On the number line, as we go from left to right: $a, that is any real number to the left of another real number is always \u201cless than\u201d the number to the right. (note that whereas the real numbers form an \u201cordered field\u201d, the complex numbers are only \u201cpartially ordered\u201d\u2026we will continue this further discussion later) .\n3. Dictionary order\n4. Alphabetical order (the letters $A \\hspace{0.1in} B \\ldots Z$ in English.\n5. Telephone directory order\n6. So a service like JustDial certainly uses \u201corder\u201d quite intensely: let us say that you want to find the telephone clinic landline number of Dr Mrs Prasad in Jayanagar 4th Block, Bengaluru : We first narrow JustDial to \u201cLocation\u201d (Jayanagar 4th Block, Bengaluru), then narrow to \u201cdoctors/surgeons\u201d as the case may be, and then check in alphabetic order, the name of Dr Mrs Prasad. So, we clearly see that the \u201cconcept\u201d and \u201cactual implementation\u201d of order (in databases) actually speeds up so much the time to find the exact information we want.\n7. So also, in math, we have the concept of ordered pair; in Cartesian geometry, $(a,b)$ means that the first component $a \\in X-axis$ and $b \\in Y-axis$. This order is generalized to complex numbers in the complex plane or Argand\u2019s diagram.\n8. There is \u201corder\u201d in human \u201crelations\u201d also: let us $(x,y)$ represents x (as father) and y (as son). Clearly, the father is \u201cfirst\u201d and the son is \u201csecond\u201d.\n9. So, also any \u201ctree\u201d has a \u201cnatural order\u201d: seed first, then roots, then branches.\n\nRegards,\n\nNalin Pithwa.\n\nWhy do we need proofs? In other words, difference between a mathematician, physicist and a\u00a0layman\n\nYes, I think it is a very nice question, which kids ask me. Why do we need proofs? Well, here is a detailed explanation (I am not mentioning the reference I use here lest it may intimidate my young enthusiastic, hard working students or readers. In other words, the explanation is not my own; I do not claim credit for this\u2026In other words, I am just sharing what I am reading in the book\u2026)\n\nHere it goes:\n\nWhat exactly is the difference between a mathematician, a physicist, and a layman? Let us suppose that they all start measuring the angles of hundreds of triangles of various shapes, find the sum in each case and keep a record. Suppose the layman finds that with one or two exceptions, the sum in each case comes out to be 180 degrees. He will ignore the exceptions and say \u201cthe sum of the three angles in a triangle\u00a0 is 180 degrees.\u201d A physicist will be more cautious in dealing with the exceptional cases. He will examine them more carefully. If he finds that the sum in them is somewhere between 179 degrees to 180 degrees, say, then he will attribute the deviation to experimental errors. He will then state a law: The sum of three angles of any triangle is 180 degrees. He will then watch happily as the rest of the world puts his law to test and finds that it holds good in thousands of different cases, until somebody comes up with a triangle in which the law fails miserably. The physicist now has to withdraw his law altogether or else to replace it by some other law which holds good in all cases tried. Even this new law may have to be modified at a later date. And, this will continue without end.\n\nA mathematician will be the fussiest of all. If there is even a single exception he will refrain from saying anything. Even when millions of triangles are tried without a single exception, he will not state it as a theorem that the sum of the three angles in ANY triangle is 180 degrees. The reason is that there are infinitely many different types of triangles. To generalize from a million to infinity is as baseless to a mathematician as to generalize from one to a million. He will at the most make a conjecture and say that there is a strong evidence suggesting that the conjecture is true. But that is not the same thing as a proving a theorem. The only proof acceptable to a mathematician is the one which follows from earlier theorems by sheer logical implications (that is, statements of the form : If P, then Q). For example, such a proof follows easily from the theorem that an external angle of a triangle is the sum of the other two internal angles.\n\nThe approach taken by the layman or the physicist is known as the inductive approach whereas the mathematician\u2019s approach is called the deductive approach. In the former, we make a few observations and generalize. In the latter, we deduce from something which is already proven. Of course, a question can be raised as to on what basis this supporting theorem is proved. The answer will be some other theorem. But then the same question can be asked about the other theorem. Eventually, a stage is reached where a certain statement cannot be proved from any other earlier proved statement(s) and must, therefore, be taken for granted to be true. Such a statement is known as an axiom or a postulate. Each branch of math has its own axioms or postulates. For examples, one of the axioms of geometry is that through two distinct points, there passes exactly one line. The whole beautiful structure of geometry is based on 5 or 6 axioms such as this one. Every theorem in plane geometry or Euclid\u2019s Geometry can be ultimately deduced from these axioms.\n\nPS: One of the most famous American presidents, Abraham Lincoln had read, understood and solved all of Euclid\u2019s books (The Elements) by burning mid-night oil, night after night, to \u201csharpen his mental faculties\u201d. And, of course, there is another famous story (true story) of how Albert Einstein as a very young boy got completely \u201caddicted\u201d to math by reading Euclid\u2019s proof of why three medians of a triangle are concurrent\u2026(you can Google up, of course).\n\nRegards,\n\nNalin Pithwa", "date": "2019-12-12 16:42:32", "meta": {"domain": "madhavamathcompetition.com", "url": "https://madhavamathcompetition.com/category/nature-of-mathematics/page/2/", "openwebmath_score": 0.8531870245933533, "openwebmath_perplexity": 396.8077610602914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795098861572, "lm_q2_score": 0.9046505370289057, "lm_q1q2_score": 0.8928715436550383}}
{"url": "https://orlandodockbuilder.com/doooz/3cd652-how-to-prove-a-parallelogram-is-congruent", "text": "Rhianna has learned the SSS and SAS congruence tests for triangles and she wonders if these tests might work for parallelograms. Diagonals of a Parallelogram Bisect Each Other. A tip from Math Bits says, if we can show that one set of opposite sides are both parallel and congruent, which in turn indicates that the polygon is a parallelogram, this will save time when working a proof. parallelogram, because opposite sides are congruent and adjacent sides are not perpendicular. Theorem 1 : If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Also interesting in this case is that to the eye The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. Congruent trianglesare triangles that have the same size and shape. So we\u2019re going to put on our thinking caps, and use our detective skills, as we set out to prove (show) that a quadrilateral is a parallelogram. Jenn, Founder Calcworkshop\u00ae, 15+ Years Experience (Licensed & Certified Teacher). When we think of parallelograms, we usually think of something like this. Note that the vertex $D$ is obtained by rotating $B$ 180 degrees about the midpoint $M$ of $\\overline{AC}$. vidDefer[i].setAttribute('src',vidDefer[i].getAttribute('data-src')); We can look at what happens in the special case where all 4 sides of both $ABCD$ and $EFGH$ are congruent to one another. Theorem 6.2.1 If a quadrilateral is a parallelogram, then the two pairs of opposite sides are congruent. Each theorem has an example that will show you how to use it in order to prove the given figure. To show these two triangles are congruent we\u2019ll use the fact that this is a parallelogram, and as a result, the two opposite sid\u2026 2. A quadrilateral that has opposite sides equal and parallel and the opposite angles are also equal is called a parallelogram. One Pair of Opposite Sides are Both Parallel and Congruent, Consecutive Angles in a Parallelogram are Supplementary. yes, one pair of sides are congruent and parallel . Both pairs of opposite sides are congruent. Triangle congruence criteria have been part of the geometry curriculum for centuries. If both pairs of opposite angles of a quadrilateral are congruent, then it\u2019s a parallelogram (converse of a property). Just as with a triangle it takes three pieces of information (ASA, SAS, or SSS) to determine a shape, so with a quadrilateral we would expect to require four pieces of information. THEOREM:If a quadrilateral has 2 sets of opposite sides congruent, then it is a parallelogram. They are called the SSS rule, SAS rule, ASA rule and AAS rule. function init() { Opposite Sides Parallel and Congruent & Opposite Angles Congruent. So what are we waiting for. B) The diagonals of the parallelogram are congruent. yes, diagonals bisect each other. Take Calcworkshop for a spin with our FREE limits course. Since ABCD is a parallelogram, segment AB \u2245 segment DC because opposite sides of a parallelogram are congruent. For quadrilaterals, on the other hand, these nice tests seem to be lacking. 2 Looking at a special case for part (a): the rhombus. In order to see what happens with the parallelograms $ABCD$ and $EFGH$ we focus first on $ABCD$. If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram. Well, if a parallelogram has congruent diagonals, you know that it is a rectangle. Attribution-NonCommercial-ShareAlike 4.0 International License. Solution: It turns out that knowing all four sides of two quadrilaterals are congruent is not enough to conclude that the quadrilaterals are congruent. Prove that the figure is a parallelogram. Finally, you\u2019ll learn how to complete the associated 2 column-proofs. If one angle is 90 degrees, then all other angles are also 90 degrees. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. if(vidDefer[i].getAttribute('data-src')) { If the diagonals of a quadrilateral bisect each other, then it\u2019s a parallelogram (converse of a property). Suppose $ABCD$ and $EFGH$ are two parallelograms with a pair of congruent corresponding sides, $|AB| = |EF|$ and $|BC| = |FG|$. We begin by drawing or building a parallelogram. If a parallelogram has perpendicular diagonals, you know it is a rhombus. We might find that the information provided will indicate that the diagonals of the quadrilateral bisect each other. Theorems. The diagonal of a parallelogram separates it into two congruent triangles. side $\\overline{EH}$ does not appear to the eye to be congruent to side $\\overline{AD}$: this could be an optical illusion or it could be that the eye is distracted by the difference in area. SURVEY . Here is what we need to prove: segment AB \u2245 segment CD and segment BC \u2245 AD. Draw the diagonal BD, and we will show that \u0394ABD and \u0394CDB are congruent. Thus it provides a good opportunity for students to engage in MP3 ''Construct Viable Arguments and Critique the Reasoning of Others.'' When a parallelogram is divided into two triangles we get to see that the angles across the common side( here the diagonal) are equal. If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. Both of these facts allow us to prove that the figure is indeed a parallelogram. Complete the two-column proof Given: triangle SVX is congruent to triangle UTX and Line SV is || to line TU Prove: VUTS is a parallelogram Image: It's a parallelogram, with one line going from corner S to corner U and a line going . This task addresses this issue for a specific class of quadrilaterals, namely parallelograms. Well, we must show one of the six basic properties of parallelograms to be true! Find missing values of a given parallelogram. We will learn about the important theorems related to parallelograms and understand their proofs. We all know that a parallelogram is a convex polygon with 4 edges and 4 vertices. Solution: More generally, a quadrilateral with 4 congruent sides is a rhombus. Which of the following cannot be used to prove a shape is a parallelogram? yes,opposite sides are congruent. First prove ABC is congruent to CDA, and then state AD and BC are corresponding sides of the triangles. Triangles can be used to prove this rule about the opposite sides. Creative Commons The opposite sides of a parallelogram are congruent. In this lesson, we will consider the four rules to prove triangle congruence. Note that a rhombus is determined by one side length and a single angle: the given side length determines all four side lengths and For example, for squares one side is enough, for rectangles two adjacent sides are sufficient. Engage your students with effective distance learning resources. This means that the corresponding sides are equal and the corresponding angles are equal. Let\u2019s begin! A description of how to do a parallelogram congruent triangles proof. In this mini-lesson, we will explore the world of parallelograms and their properties. The only parallelogram that satisfies that description is a square. Parallelogram and Congruent triangles Parallelogram. $\\triangle ABC$. If \u2026 Another approach might involve showing that the opposite angles of a quadrilateral are congruent or that the consecutive angles of a quadrilateral are supplementary. In this section, you will learn how to prove that a quadrilateral is a parallelogram. This proves that the opposite angles in a parallelogram are also equal. What about for arbitrary quadrilaterals? Which statement explains how you could use coordinate geometry to prove the diagonals of a quadrilateral are perpendicular? More specifically, how do we prove a quadrilateral is a parallelogram? Walking trails run from points A to C and from points B to D. Here are the theorems that will help you prove that the quadrilateral is a parallelogram. In today\u2019s geometry lesson, you\u2019re going to learn the 6 ways to prove a parallelogram. for (var i=0; i", "date": "2021-07-29 22:34:14", "meta": {"domain": "orlandodockbuilder.com", "url": "https://orlandodockbuilder.com/doooz/3cd652-how-to-prove-a-parallelogram-is-congruent", "openwebmath_score": 0.5363954305648804, "openwebmath_perplexity": 312.61253183447855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9732407183668539, "lm_q2_score": 0.9173026505426831, "lm_q1q2_score": 0.8927562905739801}}
{"url": "https://mathematica.stackexchange.com/questions/204856/what-is-the-distribution-of-a-random-variable-with-pdf-proportional-to-the-produ/204908#204908", "text": "# What is the distribution of a random variable with pdf proportional to the product of two normal pdf's\n\nI'm working my way through a nice e-book on Kalman filters, translating to Mathematica as I go and I've hit an interesting problem. In this section: https://nbviewer.jupyter.org/github/rlabbe/Kalman-and-Bayesian-Filters-in-Python/blob/master/03-Gaussians.ipynb#Computational-Properties-of-Gaussians - the author gives a formula to compute a product of Gaussian PDFs that is also Gaussian. Googling around will tell you that the product of two Gaussian PDFs is not itself Gaussian, but is proportional to a Gaussian with mean and standard deviation as in the image below (see proof here: http://www.tina-vision.net/docs/memos/2003-003.pdf for source) -- I have implemented his function and it works fine, but I'm curious if TransformedDistribution can be used to arrive at the same distribution.\n\nTo make it concrete\n\nd1 = NormalDistribution[m1, s1];\nd2 = NormalDistribution[m2, s2];\n(* Works perfectly *)\nTransformedDistribution[\nx + y, {x \\[Distributed] d1, y \\[Distributed] d2}]\n\n(* Not a Normally Distributed *)\nTransformedDistribution[\nx*y, {x \\[Distributed] d1, y \\[Distributed] d2}]\n\n(* This doesn't work, but I'm wondering if there is something like \\\nthis that would produce the normal distribution cited in the question \\\n*)\nTransformedDistribution[\nNormalize[x*y, Total], {x \\[Distributed] d1, y \\[Distributed] d2}]\n\n\nEdit to add example multiplication using the proposed function.\n\nMultiplyGaussian[g1_, g2_] :=\nModule[{mean1, var1, mean2, var2, mean, variance},\n{mean1, var1} = g1 /. NormalDistribution[m_, s_] :> {m, s^2};\n{mean2, var2} = g2 /. NormalDistribution[m_, s_] :> {m, s^2};\nmean = (var1*mean2 + var2*mean1) / (var1 + var2);\nvariance = (var1 * var2) / (var1 + var2);\nNormalDistribution[mean, Sqrt[variance]]\n]\nz1 = NormalDistribution[3, 0.7];\nz2 = NormalDistribution[4.5, 1];\nPlot[{Legended[PDF[z1, x], \"N(3,0.7)\"],\nLegended[PDF[z2, x], \"N(4.5,2)\"] ,\nLegended[PDF[MultiplyGaussian[z1, z2], x], \"Product\"]}, {x, 1, 10}]\n\n\u2022 I'm not sure your source is right: mathworld.wolfram.com/NormalProductDistribution.html Sep 5 '19 at 23:13\n\u2022 Edited to add code showing that the proposal seems to work and clarify that the 2nd link is a proof that I at least couldn't find a problem with after a quick look.\n\u2013\u00a0Dan\nSep 5 '19 at 23:34\n\u2022 I think my terminology is sloppy and that has caused the problem (as pointed out by this MathOverflow answer: math.stackexchange.com/questions/101062/\u2026). The result here is for the product of PDFs (where I sloppily said random variable). Does this clarification help see a way to get there in Mathematica?\n\u2013\u00a0Dan\nSep 5 '19 at 23:41\n\u2022 Please clean-up the \"rv\" vs \"pdf\" confusion in the text and the title. Your title should probably be something like \"What is the distribution of a random variable with pdf proportional to the product of two normal pdf's?\"\n\u2013\u00a0JimB\nSep 6 '19 at 0:40\n\n(* Get product of two normal pdf's *)\nprod= PDF[NormalDistribution[\u03bc1, \u03c31], x]*PDF[NormalDistribution[\u03bc2, \u03c32], x];\n\n(* Normalize so that the pdf integrates to 1 *)\npdf = prod/Integrate[prod, {x, -\u221e, \u221e}, Assumptions -> {\u03c31 > 0, \u03c32 > 0}];\n\n(* Construct associated distribution *)\nd = ProbabilityDistribution[pdf, {x, -\u221e, \u221e}, Assumptions -> {\u03c31 > 0, \u03c32 > 0}];\n\n(* Find mean and variance *)\nMean[d]\n(* (\u03bc2 \u03c31^2+\u03bc1 \u03c32^2)/(\u03c31^2+\u03c32^2) *)\n\nVariance[d]\n(* (\u03c31^2 \u03c32^2)/(\u03c31^2+\u03c32^2) *)\n\n\nThis matches what the article says the mean and variance should be. But is it a normal distribution? If the moment generating function is of the same form as for a normal distribution, then it has a normal distribution. (We could also use the characteristic function to do this for this particular distribution.)\n\n(* The log of the moment generating function will be in the following form *)\nlogCF = Expectation[Exp[t z], z \\[Distributed] NormalDistribution[\u03bc, \u03c3]] /.\nPower[E, x_] -> x // Expand\n(* t \u03bc+(t^2 \u03c3^2)/2 *)\n\n(* So we look to see if the moment generating function of distribution d is of the same form *)\nCollect[Expectation[Exp[t z], z \\[Distributed] d] /. Power[E, x_] -> x // Expand, t]\n(* (t^2 \u03c31^2 \u03c32^2)/(2 (\u03c31^2+\u03c32^2))+t ((\u03bc2 \u03c31^2)/(\u03c31^2+\u03c32^2)+(\u03bc1 \u03c32^2)/(\u03c31^2+\u03c32^2)) *)\n\n\nAnd it is.\n\nd1 = NormalDistribution[m1, s1];\nd2 = NormalDistribution[m2, s2];\n\n\nThese distributions require that\n\nassume = And @@\n(DistributionParameterAssumptions /@ {d1, d2})\n\n(* m1 \u2208 Reals && s1 > 0 && m2 \u2208 Reals && s2 > 0 *)\n\n\nAs pointed out by @JimB, the PDF formed by the product of the normal PDFs is\n\nPDFprod = Assuming[assume, PDF[d1, x]*PDF[d2, x]/\nIntegrate[PDF[d1, x]*PDF[d2, x],\n{x, -Infinity, Infinity}] // Simplify]\n\n(* (E^(-((m2 s1^2 + m1 s2^2 - (s1^2 + s2^2) x)^2/(\n2 s1^2 s2^2 (s1^2 + s2^2)))) Sqrt[s1^2 + s2^2])/(Sqrt[2 \u03c0] s1 s2) *)\n\n\nComparing with the PDF of the expected normal distribution\n\nPDFprod == Assuming[assume, PDF[NormalDistribution[\n(m1*s2^2 + m2*s1^2)/(s1^2 + s2^2),\nSqrt[s1^2*s2^2/(s1^2 + s2^2)]], x] // Simplify]\n\n(* True *)", "date": "2021-09-27 03:56:59", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/204856/what-is-the-distribution-of-a-random-variable-with-pdf-proportional-to-the-produ/204908#204908", "openwebmath_score": 0.45847487449645996, "openwebmath_perplexity": 4126.749046662051, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434788304004, "lm_q2_score": 0.9161096095812347, "lm_q1q2_score": 0.8926890734755542}}
{"url": "https://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous", "text": "# Open maps which are not continuous\n\nWhat is an example of an open map $(0,1) \\to \\mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\\mathbb{C}$ to $\\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right?\n\n\u2022 Since $(0,1)$ and $\\mathbb R$ are homeomorphic via a linear map composed with $\\arctan$, it suffices to find a map $\\mathbb R \\to \\mathbb R$ that is open but not continuous. Googling that gives you mathforum.org/library/drmath/view/62395.html \u2013\u00a0lhf Oct 25 '11 at 0:55\n\u2022 this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{-1}$ will do. \u2013\u00a0user12014 Oct 25 '11 at 1:13\n\u2022 One can build such a function from a Cantor set $C$ (the usual \"middle thirds\" set will do). Send each point in $C$ to $0$, and map each connected component of the complement of $C$ homeomorphically to the interval $(-1,1)$. Then the image of any open set intersecting $C$ will be $(-1,1)$ (thus open), and the image of any open set not meeting $C$ will also be open, since it's a union of homeomorphic images of open sets. Of course, each point of $C$ will be a discontinuity. \u2013\u00a0user83827 Oct 25 '11 at 1:16\n\u2022 @PZZ for instance the map wrapping [0,1) around the unit circle. \u2013\u00a0JSchlather Oct 25 '11 at 1:37\n\u2022 @PZZ: In fact there are no counterexamples of the type you're suggesting: if $I$ and $J$ are intervals in $\\mathbb{R}$ and $f: I \\rightarrow J$ is a continuous bijection, then $f^{-1}$ is necessarily continuous. By coincidence this is exactly where I am in my Spivak calculus course, so see e.g. Theorem 37 in $\\S 6.4$ of math.uga.edu/~pete/2400calc2.pdf. (Or see Spivak's text!) \u2013\u00a0Pete L. Clark Oct 25 '11 at 3:18\n\nExplicit examples are moderately difficult to construct, but it\u2019s not too hard to come up with non-constructive examples; here\u2019s one such.\n\nFor $x,y\\in\\mathbb{R}$ define $x\\sim y$ iff $x-y\\in \\mathbb{Q}$; it\u2019s easy to check that $\\sim$ is an equivalence relation on $\\mathbb{R}$. For any $x\\in\\mathbb{R}$, $[x] = \\{x+q:q\\in\\mathbb{Q}\\}$, where $[x]$ is the $\\sim$-equivalence class of $x$. In particular, each equivalence class is countable. For any infinite cardinal $\\kappa$, the union of $\\kappa$ pairwise disjoint countably infinite sets has cardinality $\\kappa$, so there must be exactly as many equivalence classes as there are real numbers. Let $h$ be a bijection from $\\mathbb{R}/\\sim$, the set of equivalence classes, to $\\mathbb{R}$. Finally, define $$f:(0,1)\\to\\mathbb{R}:x\\mapsto h([x])\\;.$$\n\nI claim that if $V$ is any non-empty open subset of $(0,1)$, $f[V]=\\mathbb{R}$, which of course ensures that $f$ is open. To see this, just observe that every open interval in $(0,1)$ intersects every equivalence class. (It should be no trouble at all to see that $f$ is wildly discontinuous!)\n\n\u2022 Just curious: Is the axiom of choice used anywhere in your proof? \u2013\u00a0YoTengoUnLCD Jan 17 '17 at 8:14\n\u2022 I think I'm going to start calling $\\sim$ the \"Vitali equivalence relation\"... $x$ and $y$ are Vitali equivalent iff $x-y \\in \\mathbb{Q}$, etc. Honestly, this thing is useful enough to deserve a name. \u2013\u00a0goblin Mar 2 '17 at 13:54\n\u2022 is $f$ injective? \u2013\u00a0David Feng Feb 16 at 21:37\n\u2022 @DavidFeng: No. All $x$ from the same equivalence class give the same value. For example, $f(\\frac12)=f(\\frac13)$ since $\\frac12-\\frac13\\in\\mathbb Q$ \u2013\u00a0celtschk Mar 2 at 21:55\n\nLet me conceptualize around Brian's answer a bit.\n\nDefinition 0. If $X$ and $Y$ are topological spaces, a function $f:X\u2192Y$ is said to be strongly Darboux iff for all non-empty open sets $A\u2286X$, we have $f(A)=Y$.\n\nHere's the basic facts:\n\nProposition.\n\n1. Every strongly Darboux function is an open function.\n2. If $X$ is non-empty, every Darboux function $X \\rightarrow Y$ is surjective.\n3. If $X$ is non-empty and $f : X \\rightarrow Y$ is a continuous Darboux mapping, then $Y$ carries the indiscrete topology.\n\nProofs.\n\n1. Trivial.\n\n2. Since $X$ is open and non-empty, hence $f(X)=Y.$ That is, $f$ is surjective.\n\n3. Let $B \\subseteq Y$ denote a non-empty open set. Our goal is to show that $B=Y$. Since $f$ is surjective, $f^{-1}(B)$ is non-empty. Since $f$ is continuous, $f^{-1}(B)$ is open. Hence $f(f^{-1}(B))=Y$. But since $f$ is surjecive, hence $f(f^{-1}(B))=B.$ So $B=Y$.\n\nPutting these together, we see that every strongly Darboux function $f:\\mathbb{R} \\rightarrow \\mathbb{R}$ is a discontinuous open mapping.\n\n\u2022 $f$ is an open mapping by (1).\n\n\u2022 $f$ is discontinuous by (3), because it's domain is non-empty and it's codomain doesn't carry the indiscrete topology.\n\nAnd, of course, Brian's answer guarantees the existence of a strongly Darboux function $\\mathbb{R} \\rightarrow \\mathbb{R}$. This completes the proof.\n\nThere is in fact a rather easy example of a function $$\\mathbb R \\to \\mathbb R$$ such that the image of every open set is $$\\mathbb R$$: Let $$(x_i)_{i\\in\\mathbb Z_+}$$ be the binary decimal expansion of $$x$$, so that each $$x_i \\in \\{0,1\\}$$. Let then $$f(x) = \\sum_{k=1}^\\infty\\frac{(-1)^{x_k}}k\\quad \\textrm{if the series converges}$$ $$f(x) = 0\\quad \\textrm{otherwise.}$$ Since the harmonic series (or a tail of it) can be made to converge to any real number by changing signs in the appropriate way, this function has $$f((a,b)) = \\mathbb R$$ for any real $$a,b$$. Hence this function is open, though clearly not continuous at any point.\n\nThe harmonic series can be substituted with any other unbounded series where the summand goes to zero.", "date": "2019-06-25 01:28:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous", "openwebmath_score": 0.9547765851020813, "openwebmath_perplexity": 138.39477152368693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471613889295, "lm_q2_score": 0.9073122219871936, "lm_q1q2_score": 0.8926565540955826}}
{"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share", "text": "CoCalc Public Filesweek 5 / assignment / FindRoots.ipynb\nAuthors: Boyuan Qi, Rajvir Sidhu\nViews : 31\nCompute Environment: Ubuntu 20.04 (Default)\n\n# Assignment 2: Polynomial Root Finding\n\n## Markdown\n\nGive the address (url) for a website that has documentation for markdown, including tables (i.e. a webpage, not a pdf or other document, that describes how you use markdown to create a table). Enter your answer as a variable named website='http://url.goes.here'\n\nMarks: 1\nIn [ ]:\n# YOUR CODE HERE\n\nmarkdownWebsite='https://www.markdownguide.org/cheat-sheet'\n\nTest your code from above here(1 point), ID: validate_website\nIn [ ]:\n#test that it's a suitable website link you've given\n\n\nWork out the solutions to the following questions long-hand (i.e. you should not be coding anything at this point). In the first box, set the appropriate variable equal to your answer (there's no need to define any functions). For example, to answer the last question, you will simply type polyRoots=[4-math.sqrt(3),4,4+math.sqrt(3)]\n\n1. Define the polynomial $p(x)=x^3-12x^2+45x-52$. How do you express $p(x)$ as a list, such as will be input to our algorithms? Set the variable polyX equal to this.\n2. What is the derivative, $p'(x)$ (as a list called polyDeriv) and the root(s) of $p'(x)$ (as a list, in increasing order, called derivRoots)?\n3. What is the Lagrange bound as applied to the polynomial $p(x)$? Your answer should be a real, positive number, called lagrange. (In case you missed it, this was an exercise in the workshops)\n4. Pretend that you don't know the roots of $p(x)$, but you know the roots of $p'(x)$. Use this information to explain why we know that there is no more than one root in each of the ranges: $(-109,3),(3,5),(5,109)$ You should do this without evaluating $p(x)$ at any point. Write your solution in the free text box below (this will be manually graded).\n5. What are the roots of $p(x)$? Give your answer as a list in increasing order called polyRoots. Make sure that all the roots of $p(x)$ appear in exactly one of the above ranges.\nMarks: 5\nIn [ ]:\nimport math\n\npolyX=[-52,45,-12,1]\npolyDeriv=[45,-24,3]\nderivRoots=[3,5]\nlagrange=109\npolyRoots=[4-math.sqrt(3),4,4+math.sqrt(3)]\n\nTest your code from above here(1 point), ID: polyRoots_correct\nIn [ ]:\n#the question told you how to answer this, so no point in hiding the test\nassert polyRoots==[4-math.sqrt(3),4,4+math.sqrt(3)]\n\nTest your code from above here(1 point), ID: polyX_correct\nIn [ ]:\n#check that polyX is correct\n\nTest your code from above here(1 point), ID: polyDeriv_correct\nIn [ ]:\n#check that polyDeriv is correct\n\nTest your code from above here(1 point), ID: derivRoots_correct\nIn [ ]:\n#check that derivRoots is correct\n\nTest your code from above here(1 point), ID: lagrange_bound_correct\nIn [ ]:\n#check that lagrange is correct\n\n\nThis is where you give your answer to (4) above: Explain why we know that there is no more than one root in each of the ranges: $(-109,3),(3,5),(5,109)$ You should do this without evaluating $p(x)$ at any point.\n\nMarks: 2\n\nGiven any two roots b>a of p(x), f is continuous and differentiable on [a,b]. Using Rolle's theorem, there exists exactly one root, 'k', such that f'(k)=0 and a<c<b. Due to p'(x) having 2 roots, p(x) therefore has at most 3 roots. The roots of p'(x) are 3 and 5 which means p(x) must have a root within the interval [3,5]. Also, the limits of the function are 109 and -109, so if there are 3 roots, then exactly one root must exist in the each of the intervals [-109,3] and [5,109]. This means there that no more than one root can exist in each of the intervals [-109,3],[3,5] and [5,109]\n\n## Collecting our work so far\n\nIn working through the lab sessions, we have built up a library of subroutines that will be useful for finding the roots of a polynomial. Those that we have already produced are given first.\n\nIn [ ]:\nfrom math import sqrt\n\ndef EvaluatePoly(polynomial,x):\n'''evaluate a polynomial at a value x\ninput: polynomial as a list, value x to evaluate at\noutput: value of polynomial'''\nreturn(sum(val*x**n for n,val in enumerate(polynomial)))\ndef DifferentiatePoly(polynomial):\n'''return the derivative of polynomial'''\nreturn [(i+1)*val for i,val in enumerate(polynomial[1:])]\ndef EnsureStandardForm(polynomial):\n'''make sure that the highest order of the polynomial has non-zero coefficient by removing all higher order zeros'''\nwhile not polynomial[-1]:\ndel polynomial[-1]\nreturn polynomial\ndef DescartesSigns(polynomial):\n'''Descartes rules of signs returns an upper bound on the number of positive roots and the number of negative roots'''\nPosList=list(polynomial)\nNegList=[((-1)**i)*val for i,val in enumerate(polynomial)]\nwhile 0 in PosList:\nPosList.remove(0)\nNegList.remove(0)\nreturn([sum([i[0]*i[1]<0 for i in zip(PosList[1:],PosList[:-1])]),sum([i[0]*i[1]<0 for i in zip(NegList[1:],NegList[:-1])])])\ndef FindZeroInInterval(polynomial,range_lower,range_upper):\n'''perform an interval bisection on polynomial between the values range_lower<range_upper\nreturn x such that polynomial(x)=0 (or a good enough approximation to it)'''\n#this is slightly modified compared to what we presented.\nassert range_lower<=range_upper,\"second argument should be smaller than first\"\naccuracy=10**(-15)\nvalue_lower=EvaluatePoly(polynomial,range_lower)\nvalue_upper=EvaluatePoly(polynomial,range_upper)\nif value_lower==0: # an exact root on the lower boundary\nreturn(range_lower)\nelif abs(value_upper/value_lower)<accuracy and abs(value_upper)<accuracy: #define this as being close enough to being a root on the upper boundary\nreturn(range_upper)\nelif abs(value_lower/value_upper)<accuracy and abs(value_lower)<accuracy: #close enough to root on lower boundary\nreturn(range_lower)\nif (range_lower==range_upper):#we already know it's not a root.\nreturn('')\nextent=2*(range_upper-range_lower)     #initialise with a dummy variable larger than anything that will ver be calculated\n\nwhile range_upper-range_lower>max(abs(range_upper),abs(range_lower),1)*accuracy and extent>(range_upper-range_lower):#stopping criteria based on numerical accuracy\nif value_upper*value_lower>0:      #we don't seem to have a root in this range\nreturn('')\nrange_mid=(range_upper+range_lower)/2    #bisection\nvalue_mid=EvaluatePoly(polynomial,range_mid)\nextent=range_upper-range_lower\n\nif value_mid==0:              #we have found the root\nreturn range_mid\nelif value_mid*value_lower>0: #the root is between value_mid and value_upper. redefine range and repeat\nrange_lower=range_mid\nvalue_lower=value_mid\nelse:\nrange_upper=range_mid\nvalue_upper=value_mid\n\nreturn range_mid\n\nreturn range_mid\ndef Smooth(mylist):\n'''remove all the empty entries in a list'''\nassert len(mylist)>0,'Should not smooth an empty list'\nwhile '' in mylist:\nmylist.remove('')\nreturn mylist\ndef ZeroTest(mylist):\n\"fudge the fact that it's hard to tell if a very small number is actually 0\"\nreturn([(not abs(a)<10**(-14))*a for a in mylist])\ndef RootLocationBound(polynomial):\n'use a Lagrange bound to crudely limit where the roots can be found'\nreturn max(1,sum(abs(a/polynomial[-1]) for a in polynomial[:-1]))\n\ndef QuickTests():\n#tests for the above code\nassert EnsureStandardForm([1,1,1])==[1,1,1]\nassert EnsureStandardForm([1,1,1,0,0,0])==[1,1,1]\nassert [EvaluatePoly([24,-50,35,-10,1],i)-(24-50*i+35*i*i-10*i**3+i**4) for i in range(-5,5,1)]==[0,0,0,0,0,0,0,0,0,0]\nassert DifferentiatePoly([24,-50,35,-10,1])==[-50,70,-30,4]\nassert ZeroTest([-1,0.1,10**(-15),2])==[-1,0.1,0,2]\nassert DescartesSigns([-1,-1,1,1])==[1,2]\nassert DescartesSigns([-1,0,0,1])==[1,0]\nreturn True\n\nassert QuickTests()\n\n\nLet's say you were given a polynomial as a list polynomial and had found the roots of its derivative, as an ordered list: deriv_roots. Write a function root_bounds that returns a list of pairs of numbers between which there is no more than one root of the polynomial. The list should be ordered, going from smallest to largest, and the pairs of values should also be ordered smallest to largest.\n\nHint: If you did the free text box above (question 4) clearly and correctly, this should be straightforward.\n\nHint: Look at the first test to understand the input and output format.\n\nMarks: 4\nIn [10]:\ndef root_bounds(deriv_roots,upper):\n'''return the pairs of numbers which bound individual roots (or none at all)\nin: deriv_roots: a list of roots of the deriviative of a polynomial, in increasing order, repeated according to their multiplicity\nin: upper. if x is a root of polynomial, |x|<=upper\n'''\n\nassert upper>=0\nassert all([val1<=val2 for val1,val2 in zip (deriv_roots[:-1],deriv_roots[1:])])\nassert isinstance (deriv_roots,list)\nassert len(deriv_roots)>0  # This is needed to actually apply Rolle's Theorem\nemptyList=[]   # A list which is used later on in the interation\nif deriv_roots[-1]>upper:\nfor x in range(len(deriv_roots)-1):\nlistAdd=[deriv_roots[x],deriv_roots[x+1]]  # A nested listed will be added to the interval that contains the root\nif deriv_roots[-1]<=upper:\nlower=[-upper]\nhigh=[upper]\nz=list(deriv_roots)\nbigList=lower+z+high  # The completed list which has the upper and lower bounds of the function\nfor x in range(len(bigList)-1):\nlistAdd=[bigList[x],bigList[x+1]]   # A nested listed will be added to the interval that contains the root\nreturn emptyList\n\nprint(root_bounds([-1,0,0,1],5))\n\n[[-5, -1], [-1, 0], [0, 0], [0, 1], [1, 5]]\nTest your code from above here(2 points), ID: root_bounds_code_visible\nIn [ ]:\n#test the root_bounds function\nassert list(root_bounds([-1,0,0,1],5))==[[-5,-1],[-1,0],[0,0],[0,1],[1,5]] or list(root_bounds([-1,0,0,1],5))==[(-5,-1),(-1,0),(0,0),(0,1),(1,5)]\n\nTest your code from above here(1 point), ID: root_bounds_with_float\nIn [ ]:\n#a basic test\n\nTest your code from above here(1 point), ID: root_bounds_preconditions\nIn [ ]:\n#check some preconditions\n\n\nThe function FindZeroInInterval (above) is a little different to the code that we presented in the notes (given as OriginalFindZeroInInterval in cell below). Aside from the assert statement, how is it different? (explanation not required.)\n\nTo demonstrate the difference, give a set of parameters as input that gives a different outcome between the two functions. Give your answer as 3 variables poly, below and above.\n\nMarks: 1\n\nIn [ ]:\n# YOUR CODE HERE\n\npoly=[3,1,0,0]\nabove=0\nbelow=-3\n\nTest your code from above here(2 points), ID: difference_FindZero\nIn [ ]:\ndef OriginalFindZeroInInterval ( polynomial , range_lower , range_upper ):\n''' perform an interval bisection on polynomial between the values\nrange_lower < range_upper\nreturn x such that polynomial (x)=0 (or a good enough approximation to it)\n'''\nvalue_lower = EvaluatePoly ( polynomial , range_lower )\nvalue_upper = EvaluatePoly ( polynomial , range_upper )\n\nwhile range_upper - range_lower >10**( -15) :\nif value_upper * value_lower >0: #we don 't seem to have a root in this range\nreturn('')\nrange_mid =( range_upper + range_lower )/2 # bisection\nvalue_mid = EvaluatePoly ( polynomial , range_mid )\nif value_mid ==0: #we have found the 0, so return itsposition\nreturn range_mid\nelif value_mid * value_lower >0: # crossing between range_mid andrange_upper\nrange_lower = range_mid\nvalue_lower = value_mid\nelse : # crossing between range_lower and\nrange_mid\nrange_upper = range_mid\nvalue_upper = value_mid\nreturn range_mid\n\nassert above>below\nassert FindZeroInInterval(poly,below,above)!=OriginalFindZeroInInterval(poly,below,above)\n\n\n## The Final Root Finder\n\nUsing the previous functions (and defining any additional functions that wish in the cell below), complete the function FindRealZeros which accepts a polynomial as a list and returns a list of the roots of the polynomial, in increasing order, appearing as many times as they are repeated.\n\nYou might structure your algorithm by considering:\n\n\u2022 How did you work out the solution by hand?\n\u2022 What bits of code do you have to hand that could replace some of the by-hand calculation.\n\u2022 Are there any simple types of polynomial for which you can immediately give the root?\n\u2022 Is there a minimum size of polynomial that has a solution?\n\u2022 Can you give a set of ranges between which no more than 1 root of the polynomial lies, given the roots of the derivative?\n\u2022 For a polynomial of a given degree, $n$, what's the maximum number of ranges there could be? How many real roots could there be for a degree $n$ polynomial?\n\nYou need to make use of the functions we've already defined, particularly FindZeroInInterval, DifferentiatePoly and RootLocationBound\n\nIn [ ]:\ndef EvaluatePoly(polynomial,x):\n'''evaluate a polynomial at a value x\ninput: polynomial as a list, value x to evaluate at\noutput: value of polynomial'''\nreturn(sum(val*x**n for n,val in enumerate(polynomial)))\n\ndef RootLocationBound(polynomial):\n'use a Lagrange bound to crudely limit where the roots can be found'\nreturn max(1,sum(abs(a/polynomial[-1]) for a in polynomial[:-1]))\n\ndef root_bounds(deriv_roots,upper):\n'''return the pairs of numbers which bound individual roots (or none at all)\nin: deriv_roots: a list of roots of the deriviative of a polynomial, in increasing order, repeated according to their multiplicity\nin: upper. if x is a root of polynomial, |x|<=upper\n'''\n\nassert upper>=0\nassert all([val1<=val2 for val1,val2 in zip (deriv_roots[:-1],deriv_roots[1:])])\nassert isinstance (deriv_roots,list)\nassert len(deriv_roots)>0  # This is needed to actually apply Rolle's Theorem\nemptyList=[]   # A list which is used later on in the interation\nif deriv_roots[-1]>upper:\nfor x in range(len(deriv_roots)-1):\nlistAdd=[deriv_roots[x],deriv_roots[x+1]]  # A nested listed will be added to the interval that contains the root\nif deriv_roots[-1]<=upper:\nlower=[-upper]\nhigh=[upper]\nz=list(deriv_roots)\nbigList=lower+z+high  # The completed list which has the upper and lower bounds of the function\nfor x in range(len(bigList)-1):\nlistAdd=[bigList[x],bigList[x+1]]   # A nested listed will be added to the interval that contains the root\nreturn emptyList\n\nprint(root_bounds([-1,0,0,1],5))\n\ndef EvaluatePoly(polynomial,x):\n'''evaluate a polynomial at a value x\ninput: polynomial as a list, value x to evaluate at\noutput: value of polynomial'''\nreturn(sum(val*x**n for n,val in enumerate(polynomial)))\n\ndef DifferentiatePoly(polynomial,order=1):\n'''return the derivative of polynomial of order n'''\nderivative=[(i+1)*val for i,val in enumerate(polynomial[1:])]\nif order==1:\nreturn derivative\nelse:\nreturn DifferentiatePoly(derivative, order-1)\n\ndef DescartesSigns(polynomial):\n'''Descartes rules of signs returns an upper bound on the number of positive roots and the number of negative roots'''\nPosList=list(polynomial)\nNegList=[((-1)**i)*val for i,val in enumerate(polynomial)]\nwhile 0 in PosList:\nPosList.remove(0)\nNegList.remove(0)\nreturn([sum([i[0]*i[1]<0 for i in zip(PosList[1:],PosList[:-1])]),sum([i[0]*i[1]<0 for i in zip(NegList[1:],NegList[:-1])])])\n\nDescartesSigns([-52,45,-12,1])\n\n\nIn [58]:\nimport numpy\nfrom math import sqrt\nimport math\n\ndef EvaluatePoly(polynomial,x):\n'''evaluate a polynomial at a value x\ninput: polynomial as a list, value x to evaluate at\noutput: value of polynomial'''\nreturn(sum(val*x**n for n,val in enumerate(polynomial)))\n\ndef FindZeroInInterval(polynomial,range_lower,range_upper):\n'''perform an interval bisection on polynomial between the values range_lower<range_upper\nreturn x such that polynomial(x)=0 (or a good enough approximation to it)'''\n#this is slightly modified compared to what we presented.\nassert range_lower<=range_upper,\"second argument should be smaller than first\"\naccuracy=10**(-15)\nvalue_lower=EvaluatePoly(polynomial,range_lower)\nvalue_upper=EvaluatePoly(polynomial,range_upper)\nif value_lower==0: # an exact root on the lower boundary\nreturn(range_lower)\nelif abs(value_upper/value_lower)<accuracy and abs(value_upper)<accuracy: #define this as being close enough to being a root on the upper boundary\nreturn(range_upper)\nelif abs(value_lower/value_upper)<accuracy and abs(value_lower)<accuracy: #close enough to root on lower boundary\nreturn(range_lower)\nif (range_lower==range_upper):\nreturn('')\nextent=2*(range_upper-range_lower)     #initialise with a dummy variable larger than anything that will ver be calculated\n\nwhile range_upper-range_lower>max(abs(range_upper),abs(range_lower),1)*accuracy and extent>(range_upper-range_lower):#stopping criteria based on numerical accuracy\nif value_upper*value_lower>0:\nreturn('no zero')\nrange_mid=(range_upper+range_lower)/2    #bisection\nvalue_mid=EvaluatePoly(polynomial,range_mid)\nextent=range_upper-range_lower\n\nif value_mid==0:\nreturn range_mid\nelif value_mid*value_lower>0: #the root is between value_mid and value_upper. redefine range and repeat\nrange_lower=range_mid\nvalue_lower=value_mid\nelse:\nrange_upper=range_mid\nvalue_upper=value_mid\n\nreturn range_mid\n\ndef RemoveTrailingZero(list):\n\"Remove the '.0' of a float to transform it in integer\"\ntrail=[]\nfor x in list:\nprint(x)\nif (x!=0 and x%1==0):\nidk=int(x)\ntrail.append(idk)\nelif x==0.0:\nidk=0\ntrail.append(idk)\nelse:\ntrail.append(x)\nreturn trail\n\ndef FindRealZeros(polynomial):\n'''return the zeros of the polynomial in an ordered list'''\n\nReversePoly=list(reversed(polynomial)) #Reverse the order of the list 'polynomial' for numpy because numpy consider the first entry the lower degree\nListRoot=numpy.roots(ReversePoly) #list of root(s) of the polynomial found by numpy with a 'okay' precision\nprint(ListRoot)\nnewPrecision=[]\noutput=[]\nFinal=[]\nnestedList=[]\nwithoutImaginary=[]\n\nfor x in range(len(ListRoot)):\nnewPrecision+=[numpy.around(ListRoot[x],decimals=7,out=None)]\nfor x in newPrecision: #get rid of complex numbers if present in the previous list\nif x.imag==0: #keep the real numbers and remove the \"0j\" part\ni=x.real\nwithoutImaginary.append(i)\nfor x in withoutImaginary: #Determine intervals around the roots with great precision\nDetermineIntervals=[x-0.0000001,x+0.0000001]\noutput+=DetermineIntervals #Create a list whose entries are end points of intervals\ni=0\nwhile i<len(output): #Create a nested list\nnestedList.append(output[i:i+2])\ni+=2\nprint (nestedList)\nfor x in nestedList:\nZeropolynomial=FindZeroInInterval(polynomial,x[0],x[1]) #The upper bound and lower bound correspond to the endpoints of each interval in the nested list\nif Zeropolynomial =='no zero':\npass\nelse:\nFinal.append(Zeropolynomial)\nIncreasingOrder=list(sorted(Final,key=float)) #Reorganized the roots found by increasing order\nend=RemoveTrailingZero(IncreasingOrder) #In the case where one of the root/multiple roots are supposed integer(s), we need to remove the '.0' part\n\nprint(end)\nreturn end\n\n\n\nWe will test if your solution works by comparing ideal solutions to your calculated solutions via the function EqualityWithinTolerance, which aims to tolerate numerical inaccuracy. Some examples are given below. There will be further tests that have to be passed.\n\nMarks: 10\nTest your code from above here(3 points), ID: root_finder\nIn [59]:\n#basic testing examples\ndef EqualityWithinTolerance(list1,list2,epsilon):\n\"\"\"since we don't have perfect accuracy, cannot just test equality of two lists in our tests.\nreturn true if all elements are close enough (within epsilon)\"\"\"\nassert len(list1)==len(list2),'cannot compare lists of different lengths'\nreturn sum([abs(i[0]-i[1])<=epsilon for i in zip(list1,list2)])==len(list1)\n\n#integer solutions\nassert EqualityWithinTolerance(FindRealZeros([24,-50,35,-10,1]),[1,2,3,4],2*10**(-10))\nassert EqualityWithinTolerance(FindRealZeros([-16,0,0,0,1]),[-2,2],10**(-10))\n#irrational solutions\nassert EqualityWithinTolerance(FindRealZeros([70,-60,12]),[(15-sqrt(15))/6,(15+sqrt(15))/6],10**(-10))\n#no real solutions\nassert EqualityWithinTolerance(FindRealZeros([16,0,0,0,1]),[],10**(-10))\nassert EqualityWithinTolerance(FindRealZeros([16,0,1]),[],10**(-10)),\"You should not be returning complex roots\"\n#roots very close to 0\nassert EqualityWithinTolerance(FindRealZeros([-10**(-12),0,1]),[-10**(-6),10**(-6)],10**(-10))\n#try a longer polynomial to make sure it's not just an explicit formula going up to degree 3 or 4\nassert EqualityWithinTolerance(FindRealZeros([-362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45,1]),[1,2,3,4,5,6,7,8,9],10**(-10))\n\n[4. 3. 2. 1.] [[3.9999999, 4.0000001], [2.9999999, 3.0000001], [1.9999999, 2.0000001], [0.9999999, 1.0000001]] 1.0 2.0 3.0 4.0 [1, 2, 3, 4] [-2.00000000e+00+0.j 1.66533454e-16+2.j 1.66533454e-16-2.j 2.00000000e+00+0.j] [[-2.0000001, -1.9999999], [1.9999999, 2.0000001]] -2.0 2.0 [-2, 2] [3.14549722 1.85450278] [[3.1454971, 3.1454972999999997], [1.8545026999999998, 1.8545029]] 1.8545027756320966 3.145497224367904 [1.8545027756320966, 3.145497224367904] [-1.41421356+1.41421356j -1.41421356-1.41421356j 1.41421356+1.41421356j 1.41421356-1.41421356j] [] [] [-0.+4.j 0.-4.j] [] [] [-1.e-06 1.e-06] [[-1.1e-06, -9e-07], [9e-07, 1.1e-06]] -1e-06 1e-06 [-1e-06, 1e-06] [9. 8. 7. 6. 5. 4. 3. 2. 1.] [[8.9999999, 9.0000001], [7.9999999, 8.0000001], [6.9999999, 7.0000001], [5.9999999, 6.0000001], [4.9999999, 5.0000001], [3.9999999, 4.0000001], [2.9999999, 3.0000001], [1.9999999, 2.0000001], [0.9999999, 1.0000001]] 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 [1, 2, 3, 4, 5, 6, 7, 8, 9]\nIn [ ]:\n#check that there's a reasonable number of pre- and post-conditions inside FindRealZeros and plenty of commenting\n\nTest your code from above here(1 point), ID: find_large_roots\nIn [ ]:\n#here follow some hidden tests.\n#very large roots\n\nTest your code from above here(1 point), ID: root_finder_hidden_tests1\nIn [ ]:\n#a few more general tests, not checking anything in particular, just keeping you honest\n\nTest your code from above here(1 point), ID: root_finder_hidden_tests2\nIn [ ]:\n#a few more general tests, not checking anything in particular, just keeping you honest\n\nTest your code from above here(1 point), ID: root_finder_repeated\nIn [ ]:\n#what happens when there are repeated roots?\n\nTest your code from above here(1 point), ID: root_finder_uses_helpers\nIn [ ]:\n#Ensure that the correct helper functions are being used\n\nTest your code from above here(1 point), ID: root_finder_no_external\nIn [ ]:\n#check that FindRealZeros does not rely on external functions\n\n\nExplain how your function FindRealZeros works. Each member of the pair programming pair should write their own explanation, independently, in a separate copy of the file (which will contain the same solutions to the programming questions).\n\nMarks: 3", "date": "2020-12-02 10:27:30", "meta": {"domain": "cocalc.com", "url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share", "openwebmath_score": 0.4697786271572113, "openwebmath_perplexity": 4157.842828911409, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769049752757, "lm_q2_score": 0.9149009584734676, "lm_q1q2_score": 0.8925562454264588}}
{"url": "https://math.stackexchange.com/questions/1253762/evaluate-the-indefinite-integral-int-frac1x2-sin-left-frac6x-right", "text": "# Evaluate the indefinite integral $\\int \\frac{1}{x^2} \\sin\\left(\\frac{6}{x}\\right) \\cos\\left(\\frac{6}{x}\\right) \\, dx$\n\nEvaluate the indefinite integral:\n\n$$\\int \\frac{1}{x^2} \\sin\\left(\\frac{6}{x}\\right) \\cos\\left(\\frac{6}{x}\\right) \\, dx$$\n\n(using substitution)\n\nThe answer is: $$\\frac {1}{24} \\cos\\left(\\frac{12}{x}\\right) + C$$\n\nWhereas I get a slightly different one.\n\nHere's my solution:\n\n$$u = \\frac{6}{x}$$\n\n$$du = - \\frac{6}{x^2} \\cdot dx$$\n\n$$-\\frac {1}{6} du = \\frac {1}{x^2} \\cdot dx$$\n\nMaking substitution:\n\n$$\\int -\\frac{1}{6} du \\sin (u) \\cos (u)$$\n\nadding a new variable for substitution:\n\n$$s = \\cos (u)$$\n\n$$ds = -sin(u) du$$\n\nMaking substitution:\n\n$$\\frac {1}{6} \\int ds \\cdot s$$\n\nEvaluating:\n\n$$\\frac{1}{6} \\cdot \\frac{s^2}{2} = \\frac{s^2}{12} = \\frac{\\cos^2(u)}{12} = \\frac{\\cos^2(\\frac{6}{x})}{12}$$\n\nThen using a half angle formula for $$\\cos^2$$:\n\n$$\\frac {\\frac{1}{2} \\cdot (1 + \\cos(\\frac{12}{x}))}{12} = \\frac{1}{24} + \\frac{1}{24} \\cdot \\cos\\left(\\frac{12}{x}\\right) + C$$\n\nAs you can see I have an additional $$\\frac{1}{24}$$ in my answer... so what did I do wrong?\n\n\u2022 Did you notice the +C in the original answer there? The constant absorbs that $+1/24$! Apr 27 '15 at 4:18\n\u2022 @Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C... Apr 27 '15 at 4:20\n\u2022 @dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + \\frac{1}{24}$ and then renaming $C'$ as $C$. Apr 27 '15 at 4:23\n\u2022 @Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:) Apr 27 '15 at 4:25\n\nYou missed the constant of integration all the way! It \"absorbs\" the $\\frac{1}{24}$!\n\nSpot the difference:\n\nEvaluating:\n\n$\\displaystyle \\frac{1}{6} \\cdot \\frac{s^2}{2} +c= \\frac{s^2}{12} +c= \\frac{\\cos^2(u)}{12} +c= \\frac{\\cos^2(\\frac{6}{x})}{12}+c$, where c is the constant of integration.\n\nThen using a half angle formula for $\\cos^2$:\n\n$\\displaystyle \\frac {\\frac{1}{2} \\cdot (1 + \\cos(\\frac{12}{x}))}{12} +c$ = $\\frac{1}{24} \\cdot \\cos(\\frac{12}{x}) + C$, where $C=c+\\frac{1}{24}$\n\nYou did everything right, but you're constant of integration \"absorbs\" the $\\frac{1}{24}$\n\n$$\\frac{1}{24}+C=C$$\n\nThe $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!\n\nIf it helps, you could do something like:\n\n$$\\frac{1}{24}+C=D$$", "date": "2022-01-18 09:10:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1253762/evaluate-the-indefinite-integral-int-frac1x2-sin-left-frac6x-right", "openwebmath_score": 0.8925501108169556, "openwebmath_perplexity": 949.4613467960729, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769099458927, "lm_q2_score": 0.9149009486120849, "lm_q1q2_score": 0.8925562403535439}}
{"url": "https://web2.0calc.com/questions/if-there-were-3-skiers-on-plane-of-17-and-4-people-on-the-plane-died-of-a-crash-what-is-the-chance-that-all-3-skiers-survive", "text": "+0\n\n# If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?\n\n0\n529\n21\n\nIf there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?\n\nGuest\u00a0May 21, 2014\n\n#16\n+889\n+8\n\nThere are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.)\n\nThe counting method has been used for both of the correct answers so far, \u00a0so here's the probability method.\n\nSuppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.\n\nBertie \u00a0May 21, 2014\n#1\n+996\n0\n\nWell, three skiiers on a plane of 17. Considering that one person dies each time, the number decreases by 1 in both the numerator and the denominator.\n\n$$\\left({\\frac{{\\mathtt{3}}}{{\\mathtt{17}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{2}}}{{\\mathtt{16}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{1}}}{{\\mathtt{15}}}}\\right) = {\\frac{{\\mathtt{1}}}{{\\mathtt{680}}}} = {\\mathtt{0.001\\: \\!470\\: \\!588\\: \\!235\\: \\!294\\: \\!1}}$$\n\nThe odds are 1/680\n\nGoldenLeaf \u00a0May 21, 2014\n#2\n+92775\n+8\n\nProbability is not my strong suit but this is what i think.\n\nP= number of ways 4 can be chosen from 14 / number of ways 4 can be chosen from 17\n\nP= 14C4 / 17C4\n\nP = 1001 / 2380\n\n$$\\mbox{P(all 3 survive) }=\\frac{1001}{2380}=\\frac{143}{340}$$\n\n$${\\frac{{\\mathtt{1\\,001}}}{{\\mathtt{2\\,380}}}} = {\\frac{{\\mathtt{143}}}{{\\mathtt{340}}}} = {\\mathtt{0.420\\: \\!588\\: \\!235\\: \\!294\\: \\!117\\: \\!6}}$$\n\nCan another mathematician please check this - I'm fairly confident that it is correct.\n\nMelody \u00a0May 21, 2014\n#3\n+87294\n+8\n\nProbabilty isn't my strong point, but I'll take a run at this one.\n\nSo, basically, we first want to count the sets of people who might die - what a morbid problem!!\n\nThe total number of people who could die is given by choosing some group of 4 from the 17 =\u00a0 C(17,4) = 2380\n\nNow, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets\n\nNow, let's look at the possible sets where 2 skiers die - C(3,2) *C(13,2) = 234 sets\n\nNow let's consider the sets where all three skiers die = C(3,3) * C (13,1) = 13\n\nSo, the total number of sets containing any of the skiers = 1105\n\nThus, the chances that all three survive are given by:\n\n1 - (the number of sets containing any of the skiers)/(the total number of possible sets)\n\n=\u00a0 1 - (858 + 234 + 13)/2380 \u2248 53.6 %\n\nOOPS !!\u00a0\u00a0\u00a0 Melody pointed out a math error\u00a0 I made....let me correct this...and as she indicated recently, she DOES believe in \"do-overs\"....I still like my logic, though!\n\nNow, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(14,3) = 1092 sets\n\nNow, let's look at the possible sets where 2 skiers die - C(3,2) *C(14,2) = 273 sets\n\nNow let's consider the sets where all three skiers die = C(3,3) * C (14,1) = 14\n\nSo, the total number of sets containing any of the skiers = 1379\n\nThus, the chances that all three survive are given by:\n\n1 - (the number of sets containing any of the skiers)/(the total number of possible sets)\n\n==\u00a0 1 - (1379)/2380 \u2248 42.1 %\n\nThanx, Melody!!\n\nCPhill \u00a0May 21, 2014\n#4\n+92775\n0\n\nOkay, we have 3 answers - all different\n\nWe need an arbitrator - preferably one who knows what he/she is talking about.\n\nalso if possible please explain what is wrong with our logic.\n\nThank you.\n\nMelody \u00a0May 21, 2014\n#5\n+92775\n+8\n\nOkay Chris,\n\nI am taking a look at yours.\n\nNow, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets\n\nshouldn't this be 3C1*14C3 = 3*364=1092 ?\n\nMelody \u00a0May 21, 2014\n#6\n+87294\n+8\n\nWe both got the same result...now, let's figure the probability that 2 board members out of (n) board members could arrive at the same (possibly) correct answer!!\n\nCPhill \u00a0May 21, 2014\n#7\n+92775\n0\n\nEXCELLENT!!!!!\n\nMelody \u00a0May 21, 2014\n#8\n+87294\n0\n\nHAHA!!!...and 57 minutes ago...you were looking for someone who knew what they were doing!!! (an arbitrator, I believe??)\n\nAre we \"sure\" our answers are correct??\n\n(Actually....I think they might be)\n\nCPhill \u00a0May 21, 2014\n#9\n+92775\n0\n\nYes I am sure.\n\nBecause\n\nTwo out of three ain't bad!\n\nMelody \u00a0May 21, 2014\n#10\n+87294\n0\n\nBut 2/3 = 66%\n\nThat leaves 1/3 chance = 33% = that we might not be!!\n\nI say...let's call in the \"troll\" as a referee.....HE KNOWS ALL !!!\n\nLOL!!\n\nCPhill \u00a0May 21, 2014\n#11\n+92775\n0\n\nOkay\n\nWhere's 'our' KNOW-IT-ALL TROLL\u00a0when we\u00a0need him?\n\nHe's probably AWOL with Sir Cumference!\n\nMelody \u00a0May 21, 2014\n#12\n+87294\n0\n\nI'm thinking he was on that plane...and maybe he doesn't ski, either......\n\nMmmmmmm......maybe we've solved a forum \"problem\"\n\n(At least there's about a 58% chance of it....)\n\nCPhill \u00a0May 21, 2014\n#13\n+92775\n0\n\nNOW that\u00a0is a percentage that I would very seriously challenge!\n\nP(troll gone)=P(troll was on plane)*P(troll died)\u00a0$$\\rightarrow 0$$\n\nMelody \u00a0May 21, 2014\n#14\n+87294\n0\n\nYeah...we couldn't be that fortunate, huh??\n\nCPhill \u00a0May 21, 2014\n#15\n+92775\n0\n\nThat's not nice Chris. \u00a0So long as we keep him on a tight leash he is fun to have around.\n\nYou know that just as well as I do!\n\nMelody \u00a0May 21, 2014\n#16\n+889\n+8\n\nThere are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.)\n\nThe counting method has been used for both of the correct answers so far, \u00a0so here's the probability method.\n\nSuppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.\n\nBertie \u00a0May 21, 2014\n#17\n+2353\n0\n\nBut what if the plane was carrying 4 people that died in a crash in coffins in the cargo room. Then the plane might not have crashed at all and the skiers survive\n\nUnless the skiers died in a skiing accident off course\n\nreinout-g \u00a0May 21, 2014\n#18\n0\n\nIf there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?\n\n----\n\nThis probability is trickier than it looks.\n\nFour (4) will die and at most only three (3) can be skiers.\n\nBy deduction one (1) will die and not be a skier.\n\nNow 16 remain, three (3) of whom are skiers.\n\nFrom this calculate the probability of selecting (Z) correct out of (R) draws from (N) numbers. (Z) (in this case) defines the probability of a skier dying.\n\nProbability= (R!/(Z!*(R-Z)!) * (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)/(N!/((R!)*((N-R)!)))\n\n$${Probability =}\\frac{\\frac{R!}{Z!*(R-Z)!} * \\frac{(N-R)!}{((N-R)-(R-Z))!*(R-Z)! }}{\\frac{N!}{R!*(N-R)!}}$$\n\nColumn ID\u2019s: A= (R!)/(Z!(R-Z)!)\n\nB= (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)\n\nC= N!/((R!)*((N-R)!))\n\nD= Probability of (Z) skiers dying.\n\nE= 1/Probability\n\nN\u00a0\u00a0\u00a0\u00a0 R \u00a0 \u00a0 Z \u00a0 \u00a0 A \u00a0 \u00a0 B\u00a0\u00a0\u00a0\u00a0\u00a0 C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 E\n\n16 \u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0\u00a0 560 \u00a0\u00a0\u00a0 0.001785714285714\u00a0\u00a0 \u00a0 560.0000000000\n\n16\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0 13 \u00a0 \u00a0 560\u00a0\u00a0 \u00a0 0.069642857142857\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 14.3589743590\n\n16\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0 78\u00a0\u00a0\u00a0\u00a0 560\u00a0\u00a0 \u00a0 0.417857142857143\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2.3931623932\n\n16\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0 0\u00a0\u00a0\u00a0\u00a0 1\u00a0 286\u00a0 \u00a0\u00a0 560\u00a0\u00a0 \u00a0 0.510714285714286\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1.9580419580\n\n-----------\n\nProbability of zero skiers dying ~ 0.511 (51.1%)\n\n(Professional help provided by Francis and Frances)\n\nby: Someone Who Knows Everything\n\nGuest\u00a0May 21, 2014\n#19\n+92775\n0\n\ndeleted deleted\n\nMelody \u00a0May 21, 2014\n#20\n+26745\n0\n\nMy view is:\n\nProbability that 4 non-skiers died (hence 3 skiers survived) is:\n\n$${\\frac{{\\mathtt{14}}{\\mathtt{\\,\\times\\,}}{\\mathtt{13}}{\\mathtt{\\,\\times\\,}}{\\mathtt{12}}{\\mathtt{\\,\\times\\,}}{\\mathtt{11}}}{\\left({\\mathtt{17}}{\\mathtt{\\,\\times\\,}}{\\mathtt{16}}{\\mathtt{\\,\\times\\,}}{\\mathtt{15}}{\\mathtt{\\,\\times\\,}}{\\mathtt{14}}\\right)}} = {\\frac{{\\mathtt{143}}}{{\\mathtt{340}}}} = {\\mathtt{0.420\\: \\!588\\: \\!235\\: \\!294\\: \\!117\\: \\!6}}$$\n\nOops! Just noticed that this is the same as Bertie's reply (though Bertie did a more complete job by including an explanation).\n\nAlan \u00a0May 22, 2014\n#21\n+92775\n0\n\nIt is the same as mine and Chris's as well!!\n\nMelody \u00a0May 22, 2014", "date": "2018-07-17 11:42:47", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/if-there-were-3-skiers-on-plane-of-17-and-4-people-on-the-plane-died-of-a-crash-what-is-the-chance-that-all-3-skiers-survive", "openwebmath_score": 0.8248334527015686, "openwebmath_perplexity": 4435.613179197472, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759663238371, "lm_q2_score": 0.9099070035949656, "lm_q1q2_score": 0.892505911416039}}
{"url": "https://www.physicsforums.com/threads/asymptotes-as-the-lines-y-x-an-y-x.141132/", "text": "# Asymptotes as the lines y=x an y=-x\n\nHi there.\n\nI'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!)\n\nShe was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x.\n\nThinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being.\n\nThe only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles.\n\nSo can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)?\n\nHope I can come up with an answer by tomorrow :)\n\n$$y = x \\tanh{x}$$ ?\n\nas\n\n$$x \\tanh{x} = x \\frac{e^x - e^{-x}}{e^x + e^{-x}}$$\n\ntherefore, for large positive x,\n\n$$x \\tanh{x} \\approx x \\frac{e^x}{e^x} = x$$\n\nand for large negative x\n\n$$x \\tanh{x} \\approx x (\\frac{-e^{-x}}{e^{-x}}) = -x$$\n\narildno\nHomework Helper\nGold Member\nDearly Missed\nDogtanian said:\nHi there.\n\nI'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!)\n\nShe was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x.\n\nThinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being.\n\nThe only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles.\n\nSo can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)?\n\nHope I can come up with an answer by tomorrow :)\nYou certainly can rotate curves!\n\nFor example, the curve described by the equation:\n$$x^{2}-y^{2}=1$$\nhas asymptotes y=x and y=-x.\n\nNote, however, that in this case, there exists no function of x, by which the y-coordinates of the curve could be computed, and the curve in question cannot be regarded as the graph of some function, i.e, the set of points describable as (x,f(x)), where f is some function.\n\nrobphy\nHomework Helper\nGold Member\nDoesn't one branch of that hyperbola still have those lines as asymptotes?\nIn a physics context, the worldline of a uniformly accelerated observer [which can be regarded as function (t,x(t)), where x(t)=sqrt(1+t^2)] is asymptotic to a light cone.\n\nThanks for the help guys :D\n\nMy head of department was so sure you could each of the 4 sections between the said asymptotes filled with a curve.\n\nUsing what the first two posts said, I drew the graphs of\n\ny = sqrt(1+x^2)\ny = -sqrt(1+x^2)\nx = sqrt(1+y^2)\nx = -sqrt(1+y^2)\n\nto get what I needed (at least I think that is what I i if I remember correctly back to last night....)\n\nBut I only went and forgot all about this today until just now...so I never i speak to my HofD about it...never mind :D\n\nHurkyl\nStaff Emeritus\nGold Member\nYou can combine the four curves to get one that fills the whole thing.\n\nx\u00b2 - y\u00b2 = 1\n\nfills in two sections, and\n\ny\u00b2 - x\u00b2 = 1\n\nfills in the other two.\n\nWe can combine them into one equation:\n\n(x\u00b2 - y\u00b2 - 1) (y\u00b2 - x\u00b2 - 1) = 0\n\nwhich will fill in all four quadrants.\n\nrobphy", "date": "2020-12-01 12:28:29", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/asymptotes-as-the-lines-y-x-an-y-x.141132/", "openwebmath_score": 0.6670458912849426, "openwebmath_perplexity": 1045.46807062197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.980875961292467, "lm_q2_score": 0.90990700787036, "lm_q1q2_score": 0.8925059110315917}}
{"url": "https://gmatclub.com/forum/there-are-1600-jelly-beans-divided-between-two-jars-x-and-y-if-the-192785.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 23 Mar 2019, 22:08\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# There are 1600 jelly beans divided between two jars, X and Y. If the\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSenior Manager\nStatus: Math is psycho-logical\nJoined: 07 Apr 2014\nPosts: 414\nLocation: Netherlands\nGMAT Date: 02-11-2015\nWE: Psychology and Counseling (Other)\nThere are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n06 Feb 2015, 09:03\n2\n9\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n74% (02:04) correct 26% (02:17) wrong based on 186 sessions\n\n### HideShow timer Statistics\n\nThere are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?\n\nA. 375\n\nB. 950\n\nC. 1150\n\nD. 1175\n\nE. 1350\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 13782\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n06 Feb 2015, 13:43\n4\n1\nHi All,\n\nThis question can certainly be solved with Algebra; with 2 variables and 2 unique equations, it's just a matter of translating the text into equations and doing \"system\" math.\n\nThe answers are numbers though, and there is a logical pattern in the prompt that you can use to quickly TEST THE ANSWERS.\n\nWe're told that there are a total of 1600 marbles in two jars. Jar X has 100 less than THREE TIMES the number of marbles in Jar Y. We're asked for the number of marbles in Jar X.\n\n100 marbles is a relatively small amount, compared to the 1600 total marbles. If we ignore the \"100\" and do a rough estimation, Jar X would be about 1200 and Jar Y would be about 400 (which matches the information about THREE TIMES the number of marbles). Since we DO have to factor in the 100 marbles though, the number of marbles in X has to be LESS than 1200. With this deduction, the answer has to be C or D.\n\nLet's TEST C (since it looks like the \"nicer\" number to deal with).\n\nIF....\nX = 1150 marbles\nX + 100 = 1250\n1250/3 = 416.6666 marbles\n1150 + 416.66666 is NOT 1600 total marbles.\nEliminate C.\n\nHere's the proof though:\n\nIF....\nX = 1175 marbles\nX + 100 = 1275\n1275/3 = 425\n1175 + 425 = 1600 marbles\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****\n\n##### General Discussion\nIntern\nJoined: 05 Sep 2013\nPosts: 9\nLocation: Georgia\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n06 Feb 2015, 13:28\n1\nx+y=1600 , X=1600-Y\nx+100=3y, 1700=4Y\nY=425\n\nx=1600-y=1600-425=1175\nD\nDirector\nJoined: 04 Dec 2015\nPosts: 750\nLocation: India\nConcentration: Technology, Strategy\nSchools: ISB '19, IIMA , IIMB, XLRI\nWE: Information Technology (Consulting)\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n06 Nov 2018, 20:08\npacifist85 wrote:\nThere are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?\n\nA. 375\n\nB. 950\n\nC. 1150\n\nD. 1175\n\nE. 1350\n\nLet Jelly beans in jar $$X = x$$ and $$Y = y$$\n\n$$x + y = 1600$$ ----- ($$i$$)\n\nGiven jar $$X$$ has $$100$$ fewer jelly beans than three times the number of beans in jar $$Y$$.\n\nTherefore; $$x + 100 = 3y$$\n\n$$x = 3y - 100$$\n\nSubstituting value of $$x$$ in equation ($$i$$), we get;\n\n$$3y - 100 + y = 1600$$\n\n$$4y = 1600 + 100 = 1700$$\n\n$$y = \\frac{1700}{4} = 425$$\n\nSubstituting value of $$y$$ in equation ($$i$$), we get;\n\n$$x + 425 = 1600$$\n\n$$x = 1600$$ $$-$$ $$425 = 1175$$\n\nIntern\nJoined: 16 May 2018\nPosts: 47\nLocation: Hungary\nSchools: Queen's MBA'20\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n08 Mar 2019, 17:52\nHi,\nIf X has 100 fewer jelly beans than three times the number of beans in jar Y.\nThen why did we write X+100 ?\nShouldnt it be X-100?\nIntern\nJoined: 19 Apr 2017\nPosts: 21\nLocation: Brazil\nConcentration: Finance, Entrepreneurship\nGPA: 3\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n08 Mar 2019, 19:35\nx+y=1600 (I)\n\nAnd\n\n3*y = x-100\n\nReorganizing it\n\n3*y-100 = x (II)\n\nSubs. (II) in (I)\n\n3*y-100+y=1600\n4*y=1700\ny=425\n\nThen\n\nx=1175\n\nAns (D)\nIntern\nJoined: 16 May 2018\nPosts: 47\nLocation: Hungary\nSchools: Queen's MBA'20\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n09 Mar 2019, 03:13\n1\n3*y = x-100\n\nReorganizing it\n\n3*y-100 = x (II)\n\nBut if -100 go to the other side of equation, shouldnt it become +100?\nIntern\nJoined: 19 Apr 2017\nPosts: 21\nLocation: Brazil\nConcentration: Finance, Entrepreneurship\nGPA: 3\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the\u00a0 [#permalink]\n\n### Show Tags\n\n09 Mar 2019, 10:08\nhsn81960 wrote:\n3*y = x-100\n\nReorganizing it\n\n3*y-100 = x (II)\n\nBut if -100 go to the other side of equation, shouldnt it become +100?\n\nSorry, it's 3*y = x+100 instead\nRe: There are 1600 jelly beans divided between two jars, X and Y. If the \u00a0 [#permalink] 09 Mar 2019, 10:08\nDisplay posts from previous: Sort by", "date": "2019-03-24 05:08:04", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/there-are-1600-jelly-beans-divided-between-two-jars-x-and-y-if-the-192785.html", "openwebmath_score": 0.6012148857116699, "openwebmath_perplexity": 7140.680212017578, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9753900756663374, "lm_q2_score": 0.9149009567332237, "lm_q1q2_score": 0.8923853134152235}}
{"url": "http://math.stackexchange.com/questions/465454/how-to-show-this-obvious-and-basic-property-of-abelian-groups", "text": "# How to show this obvious and basic property of abelian groups?\n\nI have a question that is probably very silly, but let's go. Let $(G,+)$ be an abelian group. In that case we know that $+$ is associative and commutative. This leads us to the following: if $\\{a_i \\in G : i \\in I_n\\}$ with $I_n = \\{i \\in \\mathbb{N} : 1 \\leq i \\leq n\\}$, then if we apply $+$ to all of the $a_i$, it independs on the ordering we impose. In truth, if we want to define the sum of all of those elements as:\n\n$$\\sum_{i \\in I_n}a_i = a_1+\\cdots+a_n,$$\n\nwe should first know what means $a_1 + \\cdots + a_n$. The definition just tells us what means $x+y$ and that $x+(y+z)=(x+y)+z$ and $x+y=y+x$ for every $x,y,z \\in G$, but how we formalize the extension of this into some finite number of elements in order to be able to say \"we can write it that way, because we proved that it makes sense\"?\n\nIndeed this is something very obvious, and I've never seem someone giving great arguments about it. Everyone just says \"obviously, the parentheses and the order doesn't matter\". I've thought on using inductions on those two properties each at a time, but I've got a little confused with it.\n\nHow is this really done? Is a proof necessary? Or we just leave it there without proof?\n\nThanks very much in advance, and sorry if this is not the place for this kind of question.\n\n-\nEveryone just says \"obviously, the parentheses and the order doesn't matter\". Not everybody. Sometimes it's proved. You first prove general associativity, I think induction is the usual way, then using that, general permutation-invariance for commutative operations. \u2013\u00a0 Daniel Fischer Aug 12 '13 at 2:32\nThanks for the help @DanielFischer! I've said everybody meaning \"everybody I've seem until now\". I'll try doing it in this way you've said. Thanks for the hint. \u2013\u00a0 user1620696 Aug 12 '13 at 2:35\n\nTo prove general associativity, define $$\\begin{cases}\\prod_{i=1}^1 a_i=a_1\\\\\\prod_{i=1}^n a_i=\\prod_{i=1}^{n-1}a_i\\cdot a_{n}\\end{cases}$$\n\nThe claim is that for any $m$, we have that $$\\prod_{i=1}^n a_i\\prod_{i=1}^m a_{n+i}=\\prod_{i=1}^{n+m}a_i$$\n\nBy definition we have the truth of $m=1$. Thus assume true for $m=r$, and consider $r+1$. Then \\begin{align}\\prod_{i=1}^n a_i\\prod_{i=1}^{r+1} a_{n+i}&=\\prod_{i=1}^na_i \\left(\\prod_{i=1}^{r} a_{n+i}a_{n+r+1}\\right)\\\\&=\\left(\\prod_{i=1}^na_i \\prod_{i=1}^{r} a_{n+i}\\right)a_{n+r+1}\\\\&=\\prod_{i=1}^{n+r}a_i a_{n+r+1}\\\\&=\\prod_{i=1}^{n+r+1}a_i\\end{align}\n\nGeneral commutativity can be proven as follows. Suppose you have a set of elements $\\{a_1,\\ldots,a_n\\}$ such that $a_ia_j=a_ja_i$ for all pairs $1\\leq i,j\\leq n$. Consider any permutation $n\\mapsto n'$ of $\\{1,\\ldots,n\\}$, and the associated product $a_{1'}\\cdots a_{n'}$. Suppose that the term $a_n$ occurs it the place $h'=n$. Then me way write by the commutativity hypothesis $$a_{1'}\\cdots a_{n'}=a_{1'}\\cdots a_{(h-1)'}a_{(h+1)'}\\cdots a_{(n-1)'}a_{n'}a_n$$\n\nand then induction does the rest, since we have stepped down to the case $n-1$\n\n-\nAs Daniel notes it is generally first proved (with no assumption of commutativity) that generalized associativity holds. (There are a number of questions about that on MSE.) Perhaps another way here is to argue that adjacent transpositions generate any symmetric group. \u2013\u00a0 anon Aug 12 '13 at 3:30\n@anon Agreed. ${}{}{}$ \u2013\u00a0 Pedro Tamaroff Aug 12 '13 at 4:13\n\nAs the others point out, you can prove that parentheses and order don't matter via induction. Now, when you prove a statement by induction, you prove the statement for each natural number $n$. In other words, the statement applies only to finite sums.\n\nThis observation is significant because the obvious generalization to infinite series is not true in the real numbers! If $\\sum A_i = a_1+a_2+a_3+\\cdots$ is a conditionally convergent infinite series, then rearranging the terms of the series can lead to a different sum. In fact, you can get any sum you want!\n\n(On the other hand, if $\\sum A_i = a_1+a_2+a_3+\\cdots$ is a series whose partial sums are eventually constant, then rearranging the terms doesn't change the eventual value. Depending on your point of view, it might be fairer to identify this true statement as the obvious generalization. After all, it doesn't depend on the topology of our abelian group. Or, rather, it requires convergence in the discrete topology.)\n\n-\nIf the partial sums are eventually constant, you're still just (effectively) dealing with a finite sum, no? \u2013\u00a0 Cameron Buie Aug 12 '13 at 2:48\n@CameronBuie Sure, but the \"eventual sum\" is the natural generalization of finite sums to groups that don't come with a topology. I edited the answer to explain this point a little better. \u2013\u00a0 Chris Culter Aug 12 '13 at 2:51\nWhat is the point of bringing up conditionally convergent series here? This would better suit as a (rather long) comment. \u2013\u00a0 Pedro Tamaroff Aug 12 '13 at 2:59\n@PeterTamaroff One doesn't truly understand why a result is stated in a certain way, and proven in a certain way, until one has seen counterexamples to its generalizations. Wouldn't you agree? \u2013\u00a0 Chris Culter Aug 12 '13 at 3:04\n@ChrisCulter \"...the obvious generalization to infinite series...\" This is really digressing! \u2013\u00a0 Pedro Tamaroff Aug 12 '13 at 3:10", "date": "2015-05-30 17:26:52", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/465454/how-to-show-this-obvious-and-basic-property-of-abelian-groups", "openwebmath_score": 0.9243401288986206, "openwebmath_perplexity": 389.432666711012, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692257588072, "lm_q2_score": 0.9136765292901317, "lm_q1q2_score": 0.892359748455787}}
{"url": "https://math.stackexchange.com/questions/2786600/invert-the-softmax-function", "text": "# Invert the softmax function\n\nIs it possible to revert the softmax function in order to obtain the original values $$x_i$$?\n\n$$S_i=\\frac{e^{x_i}}{\\sum e^{x_i}}$$\n\nIn case of 3 input variables this problem boils down to finding $$a$$, $$b$$, $$c$$ given $$x$$, $$y$$ and $$z$$:\n\n$$\\begin{cases} \\frac{a}{a+b+c} &= x \\\\ \\frac{b}{a+b+c} &= y \\\\ \\frac{c}{a+b+c} &= z \\end{cases}$$\n\nIs this problem solvable?\n\n## 2 Answers\n\nNote that in your three equations you must have $x+y+z=1$. The general solution to your three equations are $a=kx$, $b=ky$, and $c=kz$ where $k$ is any scalar.\n\nSo if you want to recover $x_i$ from $S_i$, you would note $\\sum_i S_i = 1$ which gives the solution $x_i = \\log (S_i) + c$ for all $i$, for some constant $c$.\n\n\u2022 So it\u2019s solvable up to a constant. Thank you! \u2013\u00a0jojek May 18 '18 at 17:39\n\u2022 Which c constant should I use? There is any way of calculating it? \u2013\u00a0Joel Carneiro Feb 7 '19 at 17:16\n\u2022 @JoelCarneiro Any $c$ will work; the solution is not unique. \u2013\u00a0angryavian Feb 7 '19 at 17:58\n\u2022 Any $c$ will work, one choice is if you augment the $x_i$ vector like $(0, x_1,...,x_n)$ then this will induce a particular $c$, note the corresponding log-sum-exp -- the gradient of which is the softmax -- would also be convex (en.wikipedia.org/wiki/LogSumExp). \u2013\u00a0Josh Albert Jul 29 '19 at 10:59\n\u2022 In the case anybody like me spend too much time figuring out $c$: If you know your 3 input variables have to sum to 1 then your $c = (1 - log(x \\cdot y \\cdot z))/3)$. \u2013\u00a0Rasmus \u00d8. Pedersen Sep 14 '20 at 13:04\n\nA similar question was asked in a post of reddit. The answer below is adapted from that post:\n\n$$S_{i}$$ = $$\\exp(x_{i})/(\\sum_{i} \\exp(x_{i}))$$\n\nTaking ln on both sides:\n\n$$\\ln(S_{i}) = x_{i} - \\ln(\\sum_{i} \\exp(x_{i}))$$\n\nChanging sides:\n\n$$x_{i} = \\ln(S_{i}) + \\ln(\\sum_{i} \\exp(x_{i}))$$\n\nThe second term of the right hand side is constant for a particular $$i$$ and can be written as $$C_{i}$$. Therefore, we can write:\n\n$$x_{i} = \\ln(S_{i}) + C_{i}$$\n\n\u2022 better use j in the summation \u2013\u00a0eyaler Jan 22 at 18:26", "date": "2021-03-09 05:15:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2786600/invert-the-softmax-function", "openwebmath_score": 0.8739435076713562, "openwebmath_perplexity": 303.34608990620245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9939964040255816, "lm_q2_score": 0.8976952989498448, "lm_q1q2_score": 0.8923058990668152}}
{"url": "https://math.stackexchange.com/questions/2775869/examples-of-entire-functions-with-none-one-and-infinite-zeros/2775872", "text": "# Examples of entire functions with none, one and infinite zeros.\n\nI know that an entire function is one which can be differentiated on the entire complex plane, and I believe that a zero of an entire function is the z where f(z)=0.\n\nI was tring to think of example of the following\n\nAn entire function with no zero's: I thought $e^z$ would be suitable as $e^z\\neq 0\\forall z$\n\nAn entire function with one zero : for this I chose $z$ as it is only zero at zero\n\nAn entire function with infinite zero's : I thought maybe sin(z) would work here but is sin(z) well defined ? I'm not so sure as $sin(z)=sin(r(cos(\\theta) + isin(\\theta))$ doesn't seem like a valid expression ( although I could be wrong )\n\nIf sin(z) is not well defined what is an example of an entire function with infinite zero's ?\n\nYes, it is. Just define $\\sin z$ as$$z-\\frac{z^3}{3!}+\\frac{z^5}{5!}-\\cdots$$It is an entire functions and$$\\sin z=0\\iff z\\in\\pi\\mathbb{Z}.$$", "date": "2019-08-18 21:49:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2775869/examples-of-entire-functions-with-none-one-and-infinite-zeros/2775872", "openwebmath_score": 0.7508710026741028, "openwebmath_perplexity": 313.3837183390786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9833429599907709, "lm_q2_score": 0.9073122257466114, "lm_q1q2_score": 0.8921990897014874}}
{"url": "https://math.stackexchange.com/questions/3722672/does-the-alternating-composition-of-sines-and-cosines-converge-to-a-constant", "text": "# Does the alternating composition of sines and cosines converge to a constant?\n\nLet $$f(x) = \\cos(\\sin(x))$$ and let $$c(f, n)(x)$$ denote the function $$\\underbrace{f\\circ f\\circ...\\circ f}_{n \\text{ times}}$$. For example, $$c(f, 1)(x) = f(x)$$, $$c(f, 2)(x) = f(f(x))$$ and so on.\n\nMy question is: does $$c(f, n)(x)$$ approach any constant function if $$n \\to +\\infty$$? I graphed this for some values of $$n$$ and the function seems to approach some value a little bit over $$0.76$$. Does anyone have any insight as to whether that is true? If so, what value is it approaching and why?\n\nAny sort of help or material helps; this question has been stuck in my head for quite some time now! Thanks in advance!\n\nYes\u2014this is a question in dynamical systems, if you're looking for words to search with.\n\nIn this case, the equation $$f(x)=x$$ has exactly one fixed point (near $$x=0.76817$$), and at that fixed point we have $$|f'(x)| \\approx 0.46046 < 1$$; therefore it is an attracting fixed point, which means that every sequence of iterates of $$f$$ will approach the fixed point exponentially fast.\n\n\u2022 This does not follow from what you have. The existence of a single fixed point which is attracting only implies convergence if you start sufficiently close to it. For example, consider the equation $f(x)=-x^3$. Jun 17 '20 at 9:54\n\u2022 I think boundedness of $f$ implies your conclusion in this case though. Jun 17 '20 at 9:59\n\nFor the limit $$f$$ it must hold: $$f(x)=\\cos(\\sin(f(x)))$$. Taking the derivative (assuming $$f$$ differentiable) implies:\n\n$$f'(x)=-\\sin(\\sin(f(x)))\\cos(f(x))f'(x)$$\n\nImpliing either $$f'(x)=0$$ or $$1=-\\sin(\\sin(f(x)))\\cos(f(x))$$\n\nThe last equation can only hold if $$f(x)=k\\pi$$ for $$k\\in \\mathbb{Z}$$ but this implies $$\\sin(\\sin(f(x)))=0$$, contradiction. So $$f$$ must be constant (or not differentiable).\n\nYou can show that a constant solution exists by Banach Fixpoint theorem on the sequence\n\n$$x_{n+1}=\\cos(\\sin(x_n))$$\n\n\u2022 Thank you! That makes a lot of sense! Is there a theorem that allows me to compute the constant explicitly, or are numerical methods the only way to go? Jun 16 '20 at 22:23\n\u2022 It must hold for the solution $x^*$: $x^*=\\cos(\\sin(x^*))$ but I don't think you can further simplify this. Jun 16 '20 at 22:26\n\u2022 This solution also requires that the limiting function exists in the first place, which is not clear. Jun 17 '20 at 1:39\n\u2022 The first part was a necessary condition for the limit to exist. I wrote, that the existance of the constant solution still has to be shown. Jun 17 '20 at 10:30\n\u2022 I used Maxima to numerically calculate the constant solution to 128 digits, then plugged the result into the Inverse Symbolic Calculator (wayback.cecm.sfu.ca/projects/ISC/ISCmain.html). It found no match. It says that the result does not satisfy a polynomial equation with small coefficients of degree <= 5 and does not satisfy a simple combination of various mathematical constants. Jun 17 '20 at 12:53\n\nThis question can be settled by some elementary analysis. Note that \\begin{aligned} I:=f(\\mathbb R)&=\\cos(\\sin(\\mathbb R))=\\cos([-1,1])=[\\cos(1),1]=[0.540,1],\\\\ f(I)&=\\cos\\left(\\sin\\left([0.540,\\,1]\\right)\\right)\\\\ &=\\cos([0.514,\\,0.841])\\\\ &=[\\cos(0.841),\\,\\cos(0.514)]\\\\ &=[0.666,\\,0.871]\\subset I. \\end{aligned} So, if $$f$$ has any fixed point, the fixed point must lie inside $$I=[\\cos(1),1]$$.\n\nLet $$g(x)=f(x)-x$$. Since $$g(\\cos(1))=0.330>0>-0.334=g(1)$$, by the intermediate value theorem, $$g$$ has a zero on $$I$$, i.e. $$f$$ has a fixed point on $$I$$. As $$g'(x)=-\\sin(\\sin(x))\\cos(x)-1<0$$, the fixed point of $$f$$ is also unique. Finally, on $$I=[\\cos(1),1]$$, as $$|f'(x)|=| \\sin(\\sin(x))\\cos(x)|\\le|\\sin(\\sin(x))|\\le|\\sin(\\sin(1))|=0.746<1,$$ the fixed point is attractive. Therefore, if we denote the $$n$$-fold composition of $$f$$ by $$f^n$$, the sequence $$(f(x),f^2(x),f^3(x),\\ldots)$$ must converge to the fixed point of $$f$$.\n\nDenote $$f(x)=\\cos(\\sin(x))$$.\n\nSince there exists $$\\varepsilon>0$$ such that $$|f'(x)|=|\\sin(\\sin(x))\\cos(x)|< 1-\\varepsilon$$ for every $$x \\in \\mathbb{R}$$, $$f\\colon \\mathbb{R} \\to \\mathbb{R}$$ is a Lipschitz function with Lipschitz constant stricly less than $$1$$.\n\nThus, by Banach-Caccioppoli fixed point Theorem, $$f$$ has exactly one fixed point $$x_0 \\in \\mathbb{R}$$, that is a point such that $$f(x_0)=x_0$$. Moreover, by the (very simple) proof of the Theorem, it turns out that, for every $$x \\in \\mathbb{R}$$, the sequence $$f^n(x)= f(f(\\cdots f(x)\\cdots ))$$ converges to the unique fixed point $$x_0$$.", "date": "2022-01-25 22:29:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3722672/does-the-alternating-composition-of-sines-and-cosines-converge-to-a-constant", "openwebmath_score": 0.9447954297065735, "openwebmath_perplexity": 125.59423890787279, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815519849905, "lm_q2_score": 0.9019206837793828, "lm_q1q2_score": 0.8921633017482538}}
{"url": "https://mathhelpboards.com/threads/converting-a-repeating-decimal-to-ratio-of-integers.5632/", "text": "Converting a repeating decimal to ratio of integers\n\npaulmdrdo\n\nActive member\n0.17777777777 convert into a ratio.\n\nM R\n\nActive member\nRe: converting a repeating decimal to ratio of integers\n\nHi,\nThis is $$0.1 + 0.077777=\\frac{1}{10}+\\frac{7}{100}+\\frac{7}{1000}+...$$ where you have a GP to sum.\n\nOr $$\\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.\n\nSubtracting gives $$9x=0.7$$ and so $$x=\\frac{7}{90}$$. Now just add $$\\frac{1}{10}+\\frac{7}{90}$$ and simplify.\n\nI should also say that we can write a decimal as a fraction but we can't write it as a ratio.\n\npaulmdrdo\n\nActive member\nRe: converting a repeating decimal to ratio of integers\n\nHi,\nThis is $$0.1 + 0.077777=\\frac{1}{10}+\\frac{7}{100}+\\frac{7}{1000}+...$$ where you have a GP to sum.\n\nOr $$\\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.\n\nSubtracting gives $$9x=0.7$$ and so $$x=\\frac{7}{90}$$. Now just add $$\\frac{1}{10}+\\frac{7}{90}$$ and simplify.\n\nI should also say that we can write a decimal as a fraction but we can't write it as a ratio.\nwhat do you mean by \"GP\"?\n\nM R\n\nActive member\nRe: converting a repeating decimal to ratio of integers\n\nwhat do you mean by \"GP\"?\nSorry, I have to stop using abbreviations.\n\nA GP is a geometric progression: $$a, ar, ar^2, ar^3...$$.\n\nIf you haven't met this then the second method I posted is fine.\n\nsoroban\n\nWell-known member\nRe: converting a repeating decimal to ratio of integers\n\nHello, paulmdrdo!\n\n$$\\text{Convert }\\,0.1777\\text{...}\\,\\text{ to a fraction.}$$\n\n$$\\begin{array}{ccc}\\text{We have:} & x &=& 0.1777\\cdots \\\\ \\\\ \\text{Multiply by 100:} & 100x &=& 17.777\\cdots \\\\ \\text{Multiply by 10:} & 10x &=& \\;\\;1.777\\cdots \\\\ \\text{Subtract:} & 90x &=& 16\\qquad\\quad\\; \\end{array}$$\n\nTherefore: .$$x \\;=\\;\\frac{16}{90} \\;=\\;\\frac{8}{45}$$\n\npaulmdrdo\n\nActive member\nhow would I decide what appropriate power of ten should i use?\n\nfor example i have 3.5474747474... how would you convert this one?\n\nM R\n\nActive member\nSince two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.\n\nIf you use 1000 and 10 you will get\n\n1000x=3547.474747...\n\n10x=35.474747...\n\nSo 990x=3512 and x=3512/990=1756/495.\n\nI'm adopting Soroban's approach as I prefer it to what I did earlier.\n\npaulmdrdo\n\nActive member\nSince two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.\n\nIf you use 1000 and 10 you will get\n\n1000x=3547.474747...\n\n10x=35.474747...\n\nSo 990x=3512 and x=3512/990=1756/495.\n\nI'm adopting Soroban's approach as I prefer it to what I did earlier.\n\"a difference of two in the powers of ten\" -- what do you mean by this? sorry, english is not my mother tongue. bear with me.\n\nLast edited:\n\nM R\n\nActive member\n\"a difference of two in the powers of ten\" -- what do you me by this? sorry, english is not my mother tongue. bear with me.\nNo problem.\n\nWe have 10^3 and 10^1.\n\nThe difference between 3 and 1 is 3-1=2\n\nProve It\n\nWell-known member\nMHB Math Helper\nhow would I decide what appropriate power of ten should i use?\n\nfor example i have 3.5474747474... how would you convert this one?\nYou want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated.\n\nSo in this case, since the 47 repeats, you want the first to read \"something.4747474747...\" and the second to read \"something-else.4747474747...\"\n\nWhat powers of 10 will enable this?\n\nMarkFL\n\nStaff member\nA quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's.\n\n1.) $$\\displaystyle x=0.1\\overline{7}$$\n\n$$\\displaystyle 10x=1.\\overline{7}=1+\\frac{7}{9}=\\frac{16}{9}$$\n\n$$\\displaystyle x=\\frac{16}{90}=\\frac{8}{45}$$\n\n2.) $$\\displaystyle x=3.5\\overline{47}$$\n\n$$\\displaystyle 10x=35.\\overline{47}=35+\\frac{47}{99}=\\frac{3512}{99}$$\n\n$$\\displaystyle x=\\frac{3512}{990}=\\frac{1756}{495}$$", "date": "2021-01-16 05:10:24", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/converting-a-repeating-decimal-to-ratio-of-integers.5632/", "openwebmath_score": 0.78739333152771, "openwebmath_perplexity": 1239.611135368664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815535796247, "lm_q2_score": 0.9019206699387734, "lm_q1q2_score": 0.8921632894956117}}
{"url": "https://math.stackexchange.com/questions/3315160/epsilon-n-proof-of-sqrt4n2n-2n-rightarrow-frac14", "text": "# $\\epsilon - N$ proof of $\\sqrt{4n^2+n} - 2n \\rightarrow \\frac{1}{4}$\n\nI have the following proof for $$\\lim_{n\\rightarrow\\infty} \\sqrt{4n^2+n} - 2n = \\frac{1}{4}$$ and was wondering if it was correct. Note that $$\\sqrt{4n^2+n} - 2n = \\frac{n}{\\sqrt{4n^2+n} + 2n}$$. $$\\left|\\frac{n}{\\sqrt{4n^2+n} + 2n} - \\frac{1}{4}\\right| \\\\ = \\left|\\frac{2n - \\sqrt{4n^2+n}}{4(\\sqrt{4n^2+n} + 2n)}\\right|=\\left|\\frac{\\sqrt{4n^2+n} - 2n}{4(\\sqrt{4n^2+n} + 2n)}\\right|\\\\ = \\left|\\frac{n}{4(\\sqrt{4n^2+n} + 2n)^2}\\right| \\leq \\left|\\frac{n}{4(4n)^2}\\right| = \\left|\\frac{n}{64n^2}\\right| \\\\ = \\left|\\frac{1}{64n}\\right| < \\epsilon \\\\ \\implies n>\\frac{1}{64\\epsilon}$$\n\n\u2022 Seems quite right, though the final implication must be in the other direction.\n\u2013\u00a0user65203\nAug 6, 2019 at 13:30\n\u2022 It's wrong, for the reason Yves gave. Aug 6, 2019 at 13:34\n\u2022 Did you finish this problem @user100000001? Sep 20, 2020 at 18:48\n\nYou want to show that $$\\lim_{n\\rightarrow\\infty} \\sqrt{4n^2+n} - 2n = \\frac{1}{4}$$. To do this, I would split up the analysis into scratch work and the formal proof.\n\nFor the scratch work, you need to find a suitable upper bound. You have done this by showing\n\n\\begin{align} \\left|\\frac{n}{\\sqrt{4n^2+n} + 2n} - \\frac{1}{4}\\right| & = \\left|\\frac{2n - \\sqrt{4n^2+n}}{4(\\sqrt{4n^2+n} + 2n)}\\right|\\\\&=\\left|\\frac{\\sqrt{4n^2+n} - 2n}{4(\\sqrt{4n^2+n} + 2n)}\\right|\\\\& = \\left|\\frac{n}{4(\\sqrt{4n^2+n} + 2n)^2}\\right| \\\\&\\leq \\left|\\frac{n}{4(4n)^2}\\right| \\\\&= \\left|\\frac{n}{64n^2}\\right| \\\\& = \\left|\\frac{1}{64n}\\right| \\\\&< \\epsilon \\end{align}\n\nwhich means that $$n>\\frac{1}{64\\epsilon}$$ is the upper bound.\n\nFor the formal proof:\n\nLet $$\\epsilon>0$$ (you need to fix $$\\epsilon$$ as a small positive constant). It follows from $$\\frac{1}{\\epsilon}>0$$ that $$\\frac{1}{64\\epsilon}>0$$. Then by the Archimedean property there exists a $$N\\in\\mathbb N$$ such that $$N>\\frac{1}{64\\epsilon}$$. Then if $$n\\geq N > \\frac{1}{64\\epsilon}$$, we have\n\n$$\\left|\\frac{n}{\\sqrt{4n^2+n} + 2n} - \\frac{1}{4}\\right|\\leq \\frac{1}{64n}<\\epsilon$$\n\nwhere the last inequality follows from\n\n$$n\\geq N > \\frac{1}{64\\epsilon} \\implies n>\\frac{1}{64\\epsilon} \\implies\\epsilon > \\frac{1}{64n}$$\n\nLike the other commenters have mentioned, the implication is in the wrong direction. Since you want $$|\\frac{1}{64n}| < \\epsilon$$, you are required to have $$n > \\frac{1}{64 \\epsilon}$$. In other words, $$n > \\frac{1}{64 \\epsilon}$$ implies $$|\\frac{1}{64n}| < \\epsilon$$. What you have written is the other way around.\n\nYou did the scratch work correctly. But I wouldn't call this a proof (of course it contains all ingredients of a good proof!).\n\nYou didn't introduce $$\\epsilon$$. Of course, everyone knows what you mean but one should write it out if one wants to be fully rigorous.\n\nLet $$\\epsilon >0$$. Let $$N$$ be an integer greater than --insert your choice for $$N$$ here--.\nIf $$n \\geq N$$, we have", "date": "2022-08-19 23:37:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3315160/epsilon-n-proof-of-sqrt4n2n-2n-rightarrow-frac14", "openwebmath_score": 0.8707136511802673, "openwebmath_perplexity": 268.4520331385752, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874115238967, "lm_q2_score": 0.9005297914570319, "lm_q1q2_score": 0.8920534751195758}}
{"url": "http://www.framtidsgalan.se/faithless-new-dab/058cca-scalene-obtuse-triangle", "text": "# scalene obtuse triangle\n\nIn an equiangular triangle, all the angles are equal\u2014each one measures 60 degrees. Hypotenuse. This is one of the three types of triangles, based on sides.. We are going to discuss here its definition, formulas for perimeter and area and its properties. Step 2: Let x be one of the two equal angles. Acute triangle. Yes! [insert scalene G U D with \u2220 G = 154\u00b0 \u2220 U = 14.8\u00b0 \u2220 D = 11.8\u00b0; side G U = 17 cm, U D = 37 cm, D G = 21 cm] For G U D, no two sides are equal and one angle is greater than 90 \u00b0, so you know you have a scalene, obtuse (oblique) triangle. Obtuse triangle. The given 96\u00ba angle cannot be one of the equal pair because a triangle cannot have two obtuse angles. A triangle is:scalene if no sides are equal;isosceles if two sides are equal;equilateral if three sides are equal. For finding out the area of a scalene triangle, you need the following measurements. An obtuse triangle can also be an isosceles or scalene triangle but it can\u2019t be an equilateral triangle. Types of Triangle by Angle. The scalene property of a triangle is linked to a comparison between the lenghts of its sides (or between its three angles), whereas obtuseness is linked to the value of its angles, one in particular. It means all the sides of a scalene triangle are unequal and all the three angles are also of different measures. a) Scalene b) Isosceles c) Obtuse d) Right-angled e) Acute 4) What type of triangle is this? To see why this is so, imagine two angles are the same. While drawing an obtuse triangle, you can\u2019t draw more than one obtuse angle. In this article, you will learn more about the Scalene triangle like its definition, properties, the formula of its perimeter and area along with some solved examples. Perimeter: Semiperimeter: Area: Area: Base: Height: Angle Bisector of side a: Angle Bisector of side b: An obtuse-angled triangle is a triangle in which one of the interior angles measures more than 90\u00b0 degrees. The sum of all the angles in any triangle is 180\u00b0. scalene triangle definition: 1. a triangle with three sides all of different lengths 2. a triangle with three sides all of\u2026. how to i find the length in a Scalene triangle? There are two ways to classify triangles: by their sides and their angles, like sails out on the high seas can be right or isosceles. Describe the translation you performed on the original triangle. In any triangle, two of the interior angles are always acute (less than 90 degrees) *, so there are three possibilities for the third angle: . A scalene triangle is a triangle where all sides are unequal. (ii) Isosceles triangle: If two sides of a triangle are equal, then it is called an isosceles triangle. scattergram A graph with points plotted on a coordinate plane. Pythagorean Theorem. carotid triangle, inferior that between the median line of the neck in front, the sternocleidomastoid muscle, and the anterior belly of the omohyoid muscle. A complete and perfect idea of what is an obtuse triangle, obtuse scalene triangle and obtuse isosceles tringle; and how to solve obtuse angled triangle problems in real life. set A well-defined group of objects. with three different sides they\u2019re called scalene. A(6,8) B(-1,4) C(5,4) is the obtuse triangle \u2026 we konw only one angle and one length. A triangle that has an angle greater than 90\u00b0 See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene Questionnaire. carotid triangle, superior carotid trigone. Obtuse Scalene Triangle Translation to prove SSS Congruence 1. BookMark Us. c) has all 3 sides the same length and each inside angle is the same size. Use details and coordinates to explain how the figure was transformed, including the translation rule you applied to your triangle. The Formula for Scalene Triangle. Thousands of new, high-quality pictures added every day. Obtuse Angled Triangle: A triangle whose one of the interior angles is more than 90\u00b0. Area of Scalene Triangle Formula. Area of a triangle. b) The lengths of all three sides. Find scalene triangle stock images in HD and millions of other royalty-free stock photos, illustrations and vectors in the Shutterstock collection. In geometry, Scalene Triangle is a triangle that has all its sides of different lengths. Step 1: Since it is an isosceles triangle it will have two equal angles. A triangle is a polygon made up of 3 sides and 3 angles.. We can classify triangles according to the length of their sides. Less than 90\u00b0 - all three angles are acute and so the triangle is acute. Isosceles triangle. A scalene triangle may be right, obtuse, or acute (see below). Reduced equations for equilateral, right and isosceles are below. A triangle with one interior angle measuring more than 90\u00b0 is an obtuse triangle or obtuse-angled triangle. Or look at the foot of this goose; it\u2019s scalene and obtuse. Some useful scalene triangle formula are as follows: Area of Triangle = $$\\frac{1}{2} \\times b \\times h$$, where b is the base and h is the height. A scalene triangle can be an obtuse, acute, or right triangle as long as none of its sides are equal in length. Are you bored? The converse of this is also true - If all three angles are different, then the triangle is scalene, and all the sides are different lengths. Scalene triangle [1-10] /30: Disp-Num [1] 2020/12/16 13:45 Male / 60 years old level or over / A retired person / Very / Purpose of use To determine a canopy dimension. triangle [tri\u00b4ang-g'l] a three-cornered object, figure, or area, such as a delineated area on the surface of the body; called also trigone. Equilateral triangle. Calculates the other elements of a scalene triangle from the selected elements. Learn more. Describe the translation you performed on the original triangle. scientific notation A method for writing extremely large or small numbers compactly in which the number is shown as the product of two factors. The interior angles of a scalene triangle are always all different. Pythagorean triples. Answer: 2 question Classify the triangle by the side and angle Equilateral Scalene , right Scalene , obtuse Isosceles, acute - the answers to estudyassistant.com a) The length of one side and the perpendicular distance of that side to the opposite angle. b) has 3 sides of equal length. (i) Equilateral triangle: If all sides of a triangle are equal, then it is called an equilateral triangle. This is because scalene triangles, by definition, lack special properties such as congruent sides or right angles. Try the Fun Stuff. Triangle. a) Right-angled b) Scalene c) Acute d) Isosceles e) Obtuse 5) An equilateral triangle... a) has 3 different lengths and 3 different angles. Introduction to obtuse triangle definition and obtuse triangle tutorials with a lot of examples. Click (but don't drag) the mouse cursor at the first, second, and third corners \u2026 See more. Scalene Triangle Equations These equations apply to any type of triangle. Obtuse triangle definition, a triangle with one obtuse angle. It may come in handy. Return to the Shape Area section. An obtuse triangle is one where one of the angles is greater than 90 degrees. Drag the orange vertex to reshape the triangle. For the most general scalene triangle, click Insert > Shapes and select the Freeform tool. What is Obtuse Triangle? Note: It is possible for an obtuse triangle to also be scalene or isosceles. select elements \\) Customer Voice. Obtuse Scalene Triangle Translation to prove SSS Congruence 1. Congruent triangles. FAQ. In an obtuse triangle, if one angle measures more than 90\u00b0, then the sum of the remaining two angles is less than 90\u00b0. Here, the triangle ABC is an obtuse triangle, as \u2220A measures more than 90 degrees. Area of Scalene Triangle With Base and Height The triangles above have one angle greater than 90\u00b0 Hence, they are called obtuse-angled triangle or simply obtuse triangle.. An obtuse-angled triangle can be scalene or isosceles, but never equilateral. That triangle would also be called right if a ninety degree angle is inside. Side a: Side b: Side c: Area: Perimeter: For help with using this calculator, see the shape area help page. An obtuse triangle has one angle measuring more than 90\u00ba but less than 180\u00ba (an obtuse angle). But the triangle you sketch should be a non-right-angle, scalene triangle (as opposed to an isosceles, equilateral, or right triangle). Perimeter of a triangle. Interior angles are all different. What is a Scalene Triangle? The angles in the triangle may be an acute, obtuse or right angle. An equiangular triangle is a kind of acute triangle, and is always equilateral. A triangle with an interior angle of 180\u00b0 (and collinear vertices) is degenerate. Use details and coordinates to explain how the figure was transformed, including the translation rule you applied to your triangle. If c is the length of the longest side, then a 2 + b 2 < c 2, where a and b are the lengths of the other sides. A scalene triangle is one where none of the 3 sides are equal. A triangle which has at least one angle which has measurement greater than 90\u00b0 but less than 180\u00b0 is known as an obtuse triangle. It is not possible to draw a triangle with more than one obtuse angle. A scalene triangle may be right, obtuse, ... scalene triangle A triangle with three unequal sides. Eugene Brennan (author) from Ireland on August 25, 2018: If two sides are given and the angle between them, use the cosine rule to find the remaining side, then the sine rule to find the other side. I did an obtuse scalene triangle translation by moving it 4 spaces to the right and 2 spaces up then placing a point. Most triangles drawn at random would be scalene. Scalene Triangle: No sides have equal length No angles are equal. Was this site helpful? Educational video for children to learn what a triangle is and how many types of triangles there are. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90\u00b0. Re called scalene interior angles of a scalene triangle can also be an isosceles triangle if. Such as congruent sides or right angle the foot of this goose ; it \u2019 s and. Of that side to the right and 2 spaces up then placing a point any of! A point length of one side and the perpendicular distance of that to. Then placing a point measures more than 90 degrees, you need the following.. See below ) translation to prove SSS Congruence 1 if three sides are.. Is called an isosceles triangle an obtuse triangle to also be an equilateral.! Two factors the obtuse triangle is a kind of acute triangle, Insert... 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The Shutterstock collection, right and isosceles are below graph with points plotted on a coordinate plane 60 degrees scalene! Type of triangle obtuse scalene triangle for equilateral, isosceles and scalene Yes pair because a triangle that has angle! The triangle ABC is an obtuse triangle definition, lack special properties as... The interior angles is more than 90\u00b0 - all three angles are acute and so the is... In which one of the equal pair because a triangle with one obtuse angle imagine two angles the! You can \u2019 t be an obtuse triangle, you need the following.! That has an angle greater than 90\u00b0 - all three angles are and!: a triangle with one obtuse angle a triangle with an interior measuring... The three angles are also of different lengths 2. a triangle are equal the... Is degenerate isosceles or scalene triangle equations These equations apply to any type of triangle is a triangle has... The perpendicular distance of that side to the right and 2 spaces up then placing a point measurement than... The most general scalene triangle but it can \u2019 t be an acute, acute! Type of triangle where one of the interior angles of a triangle with three sides all of different lengths a. How the figure was transformed, including the translation you performed on the original triangle the same length and inside. Sides or right angle, and is always equilateral triangle stock images in HD and of...", "date": "2023-03-27 14:50:35", "meta": {"domain": "framtidsgalan.se", "url": "http://www.framtidsgalan.se/faithless-new-dab/058cca-scalene-obtuse-triangle", "openwebmath_score": 0.5550912618637085, "openwebmath_perplexity": 816.1558441323414, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9817357184418847, "lm_q2_score": 0.9086178963141964, "lm_q1q2_score": 0.8920226432271715}}
{"url": "https://thompsoncontractfurnishings.ca/femme-fatale-vzyibej/binomial-coefficient-c-e0c584", "text": "= { 2 The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions). The symbol H . ) If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient . k A combinatorial proof is given below. ) n . {\\displaystyle \\geq {\\frac {n}{k}}} squares from the remaining n squares; any k from 0 to n will work. It can be deduced from this that a k \u2211 {\\displaystyle {\\alpha \\choose \\alpha }=2^{\\alpha }} For instance, by looking at row number 5 of the triangle, one can quickly read off that. \u2212 Explicitly,[5]. \u2192 {\\displaystyle k\\to \\infty } 4 x 1 {\\displaystyle (-1)^{k}={\\binom {-1}{k}}=\\left(\\!\\! Definition: Binomial Coefficient he binomial coefficients that appear in the expansion (a + b) are the values of C for r = 0, 1, 2,\u2026,n. {\\displaystyle {\\tbinom {n}{k}}} Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. k ( ) ( ) Commonly, a binomial coefficient is indexed by a pair of integers n \u2265 k \u2265 0 and is written is the Euler\u2013Mascheroni constant.). Thread Tools. ) For integers s and t such that m ( ( ( n In this tutorial, we will learn about calculating the binomial coefficient using a recursive function in C++.Firstly, you must know the use of binomial coefficient calculation. How to write it in Latex ? 1 The formula follows from considering the set {1, 2, 3, ..., n} and counting separately (a) the k-element groupings that include a particular set element, say \"i\", in every group (since \"i\" is already chosen to fill one spot in every group, we need only choose k\u00a0\u2212\u00a01 from the remaining n\u00a0\u2212\u00a01) and (b) all the k-groupings that don't include \"i\"; this enumerates all the possible k-combinations of n elements. + k {\\displaystyle 0\\leq t> n = 1, C(1,0) = 1, C(1,1) = 1 q / to Not a member, \u2026 * Evaluate binomial coefficients - 29/09/2015 BINOMIAL CSECT USING BINOMIAL,R15 set base register SR R4,R4 clear for mult and div LA R5,1 r=1 LA R7,1 i=1 L R8,N m=n LOOP LR R4,R7 do while i<=k C R4,K i<=k + ) t k When n is composite, let p be the smallest prime factor of n and let k = n/p. {\\displaystyle \\sum _{0\\leq {k}\\leq {n}}{\\binom {n}{k}}=2^{n}} ) The coefficient ak is the kth difference of the sequence p(0), p(1), ..., p(k). For example, if n ( {\\displaystyle \\Gamma } . A more efficient method to compute individual binomial coefficients is given by the formula. In the special case n = 2m, k = m, using (1), the expansion (7) becomes (as seen in Pascal's triangle at right). It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n, and is given by the formula, For example, the fourth power of 1 + x is. lcm n n r Recall that a classical notation for C (especially in n r the context of binomial coefficients) is . + 1 N For natural numbers (taken to include 0) n and k, the binomial coefficient k n equals pc, where c is the number of carries when m and n are added in base p. 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To start a cryptocurrency exchange platform, Python Programming \u2013 binomial coefficient has... Many factors common to numerator and denominator, but they 're best known from Blaise Pascal 's triangle one! Compute the binomial coefficient ( 4 2 ) = 4! } { k } }!. Functions General Programming Uncategorized to ordinary generating series International Speaker, and Job Consultant this... K > k k / k! ( n\u2212k ) nearby binomial coefficients are the positive integers that as! This statement theorem ( \u2217 ) by binomial coefficient c x = 1 in Gebieten. That define the same rate [ clarification needed ] is because they can represent on... Has Overlapping Subproblems property takes two parameters n and k and returns the value of binomial coefficients are ordinary... Polynomial 3t ( 3t + 1 ) /2 can be given a counting... Integers that occur as coefficients in his book L\u012bl\u0101vat\u012b. [ 2 ] j =,! { -k } { n } { n } } = { \\tfrac { 4! {. In der Stochastik aber auch in anderen Gebieten der Mathematik which Will be obtained by statement... From the definitions ). }. }. }. double counting proof, as.!\n\nWhy Corgi Is Expensive, Gardner Driveway Sealer, Ecm Replacement Procedure, Morrilton Ar Hotels, Cheap Marine Setup, St Vincent De Paul Services Offered, Peugeot E-208 Brochure Pdf, Rochester First Twitter,", "date": "2021-05-09 08:09:04", "meta": {"domain": "thompsoncontractfurnishings.ca", "url": "https://thompsoncontractfurnishings.ca/femme-fatale-vzyibej/binomial-coefficient-c-e0c584", "openwebmath_score": 0.9126318097114563, "openwebmath_perplexity": 1782.271509811245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856482, "lm_q2_score": 0.9086178870347124, "lm_q1q2_score": 0.8920226389726097}}
{"url": "https://math.stackexchange.com/questions/2963069/integers-which-are-squared-norm-of-2-by-2-integer-matrices", "text": "Integers which are squared norm of 2 by 2 integer matrices\n\nQuestion: Which integers are of the form $$\\Vert A \\Vert^2$$, with $$A \\in M_2(\\mathbb{Z})$$.\n\nThe code below provides the first such integers: $$0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26$$.\nBy searching this sequence on OEIS, we find: \"Numbers that are the sum of 2 squares\" A001481.\n\nAre these integers exactly those which are the sum of two squares ?\n\nResearch\n\nFirst, $$\\Vert A \\Vert^2$$ is the largest eigenvalue of $$A^*A$$, so for $$A = \\left( \\begin{matrix} a & b \\cr c & d \\end{matrix} \\right)$$ and $$a,b,c,d \\in \\mathbb{Z}$$, so we get:\n$$\\Vert A \\Vert^2 = \\frac{1}{2} \\left(a^2+b^2+c^2+d^2+\\sqrt{(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2}\\right)$$\n\nObviously, every sum of two squares is of the expected form , because by taking $$c=d=0$$, we get $$\\Vert A \\Vert^2=a^2+b^2$$.\n\nThen it remains to prove that there is no other integer (if true).\n\nNow, recall that:\n\nSum of two square theorem\nAn integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to 3 (mod 4) raised to an odd power.\n\nBy taking $$c=ra$$ and $$d=rb$$, we get that $$\\Vert A \\Vert^2 = (r^2+1)(a^2+b^2)$$, which is also a sum of two square because the following equation occurs (proof here):\n$$r^2 \\not \\equiv -1 \\mod 4s+3$$\n\nA necessary condition for $$\\Vert A \\Vert^2$$ to be an integer, is that $$(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2$$ must be a square $$X^2$$, so that $$(X,2(ad-bc),a^2+b^2+c^2+d^2)$$ is a Pythagorean triple, so must be of the form $$(k(m^2-n^2),2kmn,k(m^2+n^2)$$, and then $$\\Vert A \\Vert^2 = km^2$$. So it remains to prove that $$k$$ must be a sum of two squares.\n\nsage: L=[]\n....: for a in range(-6,6):\n....:     for b in range(-6,6):\n....:         for c in range(-6,6):\n....:             for d in range(-6,6):\n....:                 n=numerical_approx(matrix([[a,b],[c,d]]).norm()^2,digits=10)\n....:                 if n.is_integer():\n....:                     L.append(int(n))\n....: l=list(set(L))\n....: l.sort()\n....: l[:20]\n....:\n[0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37]\n\n\nYes. If $$A$$ is a $$2\\times2$$ integer matrix such that $$n=\\|A\\|^2$$ is an integer, $$n$$ must be the sum of two integer squares. Conversely, if $$n$$ is the sum of two integer squares, then $$n=\\|A\\|^2$$ for some $$2\\times2$$ integer matrix $$A$$.\nProof. Suppose $$A$$ is a $$2\\times2$$ integer matrix such that $$n=\\|A\\|^2$$ is an integer. We want to show that $$n$$ is the sum of two integer squares. This is clearly true if $$n$$ is $$0$$ or $$1$$. Suppose $$n>1$$. Then $$A^TA-nI$$ is a singular matrix with integer entries. Hence $$A^TA$$ has an integer eigenvector $$v$$ corresponding to the eigenvalue $$n$$ and in turn, $$\\|Av\\|^2=v^TA^TAv=n\\|v\\|^2$$.\nSince both $$\\pmatrix{x\\\\ y}:=v$$ and $$\\pmatrix{a\\\\ b}:=Av$$ are integer vectors, the previous equality implies that $$n(x^2+y^2)=a^2+b^2$$. By the two squares theorem, in each of the prime factorisation of $$x^2+y^2$$ and $$a^2+b^2$$, every factor congruent to $$3$$ (mod $$4$$) must occur in an even power. Therefore, in the prime factorisation of $$n$$, every factor congruent to $$3$$ (mod $$4$$) must also occur in an even power. Hence the two squares theorem guarantees that $$n$$ is a sum of two integer squares. This proves one direction of our assertion.\nFor the other direction, suppose $$n=a^2+b^2$$ for some two integers $$a$$ and $$b$$. Then $$\\|A\\|^2=n$$ when $$A=\\pmatrix{a&-b\\\\ b&a}$$ or $$\\pmatrix{a&0\\\\ b&0}$$.\n\u2022 I see. Then note that this theorem is stated with $n > 1$, but it is ok. Oct 20, 2018 at 17:36", "date": "2022-07-02 18:33:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2963069/integers-which-are-squared-norm-of-2-by-2-integer-matrices", "openwebmath_score": 0.924496591091156, "openwebmath_perplexity": 52.77125530895787, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9845754488233528, "lm_q2_score": 0.9059898267292559, "lm_q1q2_score": 0.8920153402813488}}
{"url": "https://www.hopecoalitionboulder.org/brief-history-idsqv/page.php?5bbf8e=aas-congruence-theorem", "text": "In this section we will consider two more cases where it is possible to conclude that triangles are congruent with only partial information about their sides and angles. We have enough information to state the triangles are congruent. Now it's time to make use of the Pythagorean Theorem. AAS congruence theorem. CosvoStudyMaster. (2) $$AAS = AAS$$: $$\\angle A, \\angle C, CD$$ of $$\\triangle ACD = \\angle B, \\angle C, CD$$ of $$\\triangle BCD$$. Figure 2.3.4. AAS is one of the five ways to determine if two triangles are congruent. Yes, AAS Congruence Theorem; use \u2220 TSN > \u2220 USH by Vertical Angles Theorem 9. 13. We have enough information to state the triangles are congruent. A Given: \u2220 A \u2245 \u2220 D It is given that \u2220 A \u2245 \u2220 D. What is AAS Triangle Congruence? This video will explain how to prove two given triangles are similar using ASA and AAS. Ship $$S$$ is observed from points $$A$$ and $$B$$ along the coast. (1) write a congruence statement for the two triangles. How to prove congruent triangles using the angle angle side postulate and theorem . ... AAS (Angle-Angle-Side) Theorem. \u0394ABC and \u0394RST are right triangles with \u00afAB ~= \u00afRS and \u00af~= \u00afST. Start studying 3.08: Triangle Congruence: SSS, SAS, and ASA 2. What triangle congruence theorem does not actually exist? Theorem 2.3.2 (AAS or Angle-Angle-Side Theorem) Two triangles are congruent if two angles and an unincluded side of one triangle are equal respectively to two angles and the corresponding unincluded side of the other triangle (AAS = AAS). Then you would be able to use the ASA Postulate to conclude that \u0394ABC ~= \u0394RST. $$\\begin{array} {ccrclcl} {} & \\ & {\\underline{\\triangle ABC}} & \\ & {\\underline{\\triangle CDA}} & \\ & {} \\\\ {\\text{Angle}} & \\ & {\\angle BAC} & = & {\\angle DCA} & \\ & {\\text{(marked = in diagram)}} \\\\ {\\text{Included Side}} & \\ & {AC} & = & {CA} & \\ & {\\text{(identity)}} \\\\ {\\text{Angle}} & \\ & {\\angle BCA} & = & {\\angle DAC} & \\ & {\\text{(marked = in diagram)}} \\end{array}$$. You also have the Pythagorean Theorem that you can apply at will. Figure 12.8The hypotenuse and a leg of \u0394ABC are congruent to the hypotenuse and a leg of \u0394RST. Like ASA (angle-side-angle), to use AAS, you need two pairs of congruent angles and one pair of congruent sides to prove two triangles congruent. The triangles are then congruent by $$ASA = ASA$$ applied to $$\\angle B$$. A Given: \u2220 A \u2245 \u2220 D It is given that \u2220 A \u2245 \u2220 D. NL \u2014 \u22a5 NQ \u2014 , NL \u2014 \u22a5 MP \u2014, QM \u2014 PL \u2014 Prove NQM \u2245 MPL N M Q L P 18. HFG \u2245 GKH 6. 1. Learn more about the mythic conflict between the Argives and the Trojans. Theorem 12.2: The AAS Theorem. D. Given: RS bisects \u2220MRQ; \u2220RMS \u2245 \u2220RQS Which relationship in the diagram is true? SSS, SAS, ASA, and AAS Congruence Date_____ Period____ State if the two triangles are congruent. Yes, AAS Congruence Theorem 11. The method of finding the distance of ships at sea described in Example $$\\PageIndex{5}$$ has been attributed to the Greek philosopher Thales (c. 600 B.C.). Given AD IIEC, BD = BC Prove AABD AEBC SOLUTION . Therefore $$x = AC = BC = 10$$ and $$y = AD = BD$$. AAS Congruence Theorem MMonitoring Progressonitoring Progress Help in English and Spanish at BigIdeasMath.com 3. The first is a translation of vertex L to vertex Q. U V T S R Triangle Congruence Theorems You have learned five methods for proving that triangles are congruent. Answer: (1) $$PQ$$, (2) $$PR$$, (3) $$QR$$. Figure 12.8 illustrates this situation. Therefore, as things stand, we cannot use $$ASA = ASA$$ to conclude that the triangles are congruent, However we may show $$\\angle C$$ equals $$\\angle F$$ as in Theorem $$\\PageIndex{3}$$, section 1.5 $$(\\angle C = 180^{\\circ} - (60^{\\circ} + 50^{\\circ}) = 180^{\\circ} - 110^{\\circ} = 70^{\\circ}$$ and $$\\angle F = 180^{\\circ} - (60^{\\circ} + 50^{\\circ}) = 180^{\\circ} - 110^{\\circ} = 70^{\\circ})$$. homedogCeejay. Since we use the Angle Sum Theorem to prove it, it's no longer a postulate because it isn't assumed anymore. $$\\triangle ABC$$ with $$\\angle A = 40^{\\circ}$$, $$\\angle B = 50^{\\circ}$$, and $$AB = 3$$ inches. C Prove the AAS Congruence Theorem. Unless otherwise noted, LibreTexts content is licensed by\u00a0CC BY-NC-SA 3.0. 56 terms. Proof: You need a game plan. 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Studied two postulates that will Help us prove congruence between triangles you already have a game plan, all!: 23 congruence Essential Question: what does the AAS congruence Theorem tell you two. To Example \\ ( \\angle A\\ ) and \\ ( y = AD = BD\\ ) =. A non-included side of each triangle shown below pick out important information flashcards,,! Bc = EF then \u25b3ABC \u2245 \u25b3DEF products for the two letters representing each of the Base angles 9... Following Theorem: Theorem \\ ( \\PageIndex { 1 } \\ ) by \\ ( \\angle A\\ ''! York City College of Technology at CUNY Academic Works the hypotenuse and a side are to! And \u0394RST with \u2220A ~= \u2220R, \u2220C ~= \u2220T, and more with flashcards games. = 10\\ ) and \\ ( \\PageIndex { 4 } \\ ) > \u2220 USH by angles!", "date": "2021-04-22 20:07:38", "meta": {"domain": "hopecoalitionboulder.org", "url": "https://www.hopecoalitionboulder.org/brief-history-idsqv/page.php?5bbf8e=aas-congruence-theorem", "openwebmath_score": 0.6596163511276245, "openwebmath_perplexity": 2743.1814495661806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9802808718926534, "lm_q2_score": 0.9099070084811306, "lm_q1q2_score": 0.8919644356151187}}
{"url": "http://mathhelpforum.com/calculus/8550-total-dist-traveled.html", "text": "1. ## total dist. traveled\n\nAn object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$\n\ni was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.\n\n2. Originally Posted by viet\nAn object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$\n\ni was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.\nYou already know that, in order to find displacement we integrate the velocity function:\n$\\Delta x = \\int_0^{\\frac{3\\pi}{2}} dt \\left ( \\frac{1}{2} + sin(2t) \\right )$\n\nWe do something similar to get the distance. Recall that distance is a scalar quantity, and thus in 1-D is never negative. This leads us to the formula:\n$\\Delta x = \\int_0^{\\frac{3\\pi}{2}} dt \\left | \\frac{1}{2} + sin(2t) \\right |$\n\nIn order to do this integral, we need to find the intervals over which the integrand is negative. (See graph below.) So we need to know when\n$\\frac{1}{2} + sin(2t) = 0$\n\n$sin(2t) = -\\frac{1}{2}$\n\n$2t = \\frac{7 \\pi}{6}, \\frac{11\\pi}{6}$\n\n$t = \\frac{7 \\pi}{12}, \\frac{11\\pi}{12}$\n\nAnd we see that the integrand is negative for the interval $\\left [ \\frac{7 \\pi}{12}, \\frac{11\\pi}{12} \\right ]$\n\nSo:\n$\\Delta x = \\int_0^{\\frac{7\\pi}{12}} dt \\left ( \\frac{1}{2} + sin(2t) \\right ) -\n\\int_{\\frac{7 \\pi}{12}}^{\\frac{11\\pi}{12}} dt \\left ( \\frac{1}{2} + sin(2t) \\right )$\n$\n+ \\int_{\\frac{11 \\pi}{12}}^{\\frac{3\\pi}{2}} dt \\left ( \\frac{1}{2} + sin(2t) \\right )$\n\nI get $\\Delta x = \\sqrt{3} + \\frac{5\\pi}{12} + 1 \\approx 4.04105$\n\n-Dan\n\n3. Hello, viet!\n\nAn object moves along a line so that its velocity at time $t$ is: $v(t) \\:= \\:\\frac{1}{2} + \\sin2t$ ft/sec.\n\nFind the displacement and the total distance traveled by the object for $0 \\leq t \\leq \\frac{3\\pi}{2}$\n\nIntegrate: . $v(t)\\:=\\:\\frac{1}{2} + \\sin2t$ . . . and we have: . $s(t)\\;=\\;\\frac{1}{2}t - \\frac{1}{2}\\cos2t + C$\n\n. . Assume that the initial position is $s(0) = 0$, then $C = 0.$\n\n. . Hence, the position function is: . $s(t) \\;= \\;\\frac{1}{2}t - \\frac{1}{2}\\cos2t$\n\nAt $t = \\frac{3\\pi}{2}:\\;s\\left(\\frac{3\\pi}{2}\\right)\\;=\\; \\frac{1}{2}\\left(\\frac{3\\pi}{2}\\right) - \\frac{1}{2}\\cos(3\\pi) \\:=\\:\\frac{3\\pi}{4} + \\frac{1}{2} \\:\\approx\\:2.8562$\n\nTherefore, the displacement is $\\boxed{2.8562}$ units to the right.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nThe object could have stopped and reversed direction.\n. . This happens when $v(t) = 0.$\n\nLet $v(t) \\:=\\:\\frac{1}{2} + \\sin2t\\;=\\;0$\n\nWe have: . $\\sin2t \\:=\\:-\\frac{1}{2}\\quad\\Rightarrow\\quad 2t\\:=\\:\\frac{7\\pi}{6},\\:\\frac{11\\pi}{6} \\quad\\Rightarrow\\quad t \\:=\\:\\frac{7\\pi}{12},\\:\\frac{11\\pi}{12}\n$\n\n. . $s\\left(\\frac{7\\pi}{12}\\right)\\:=\\:\\frac{1}{2}\\left (\\frac{7\\pi}{12}\\right) - \\frac{1}{2}\\cos\\left(\\frac{7\\pi}{6}\\right)\\;=\\;\\fr ac{7\\pi}{24} + \\frac{\\sqrt{3}}{4} \\:\\approx\\:1.3493$\n\n. . $s\\left(\\frac{11\\pi}{12}\\right)\\:=\\:\\frac{1}{2}\\lef t(\\frac{11\\pi}{12}\\right) - \\frac{1}{2}\\cos\\left(\\frac{11\\pi}{6}\\right) \\:=\\:\\frac{11\\pi}{24} - \\frac{\\sqrt{3}}{4} \\:\\approx\\:1.0069$\n\nNow we know all about the object's journey.\n\n$\\begin{array}{cccc} s(0) & = & 0\\\\ s(\\frac{7\\pi}{12}) & = & 1.3493\\\\ s(\\frac{11\\pi}{12}) & = & 1.0069\\\\ s(\\frac{3\\pi}{2}) & = & 2.8562\\end{array}\n\\begin{array}{ccc} \\}\\;1.3493\\text{ to the right} \\\\ \\}\\;0.3424\\text{ to the left } \\\\ \\}\\;1.8493\\text{ to the right}\\end{array}$\n\n. . Total distance: . $\\boxed{3.541\\text{ units}}$\n\nBut check my work . . . please!\n\n4. There's only one problem with this:\nOriginally Posted by Soroban\n. . Assume that the initial position is $s(0) = 0$, then $C = 0.$\nWe may certainly assume s(0) = 0, but then\n$s(0) = \\frac{0}{2} - \\frac{1}{2}cos(2 \\cdot 0) + C = 0$\n\ngives\n$C = \\frac{1}{2}$, not $C = 0$\n\nSo all your distance calculations are off by 0.5 ft, which brings you back into agreement with my answers. (Which, by the way, I have run through a numerical approximation and verified. I never dare to assume you got your answer wrong! )\n\n-Dan", "date": "2017-10-21 16:26:56", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/8550-total-dist-traveled.html", "openwebmath_score": 0.9319592118263245, "openwebmath_perplexity": 258.55724086335033, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808753491773, "lm_q2_score": 0.9099069968764883, "lm_q1q2_score": 0.8919644273844252}}
{"url": "https://mathhelpboards.com/threads/find-integer-sequences.6621/", "text": "# Find integer sequences\n\n#### hxthanh\n\n##### New member\nDefine $\\{a_n\\}$ is integer sequences (all term are integers) satisfy condition\n\n$a_n=a_{n-1}+\\left\\lfloor\\dfrac{n^2-2n+2-a_{n-1}}{n}\\right\\rfloor$ for $n=1,2,...$\n\n*note: $\\left\\lfloor x\\right\\rfloor$ is a greatest integer number less than or equal $x$\nFind general term of sequences.\n\n#### Amer\n\n##### Active member\nDefine $\\{a_n\\}$ is integer sequences (all term are integers) satisfy condition\n\n$a_n=a_{n-1}+\\left\\lfloor\\dfrac{n^2-2n+2-a_{n-1}}{n}\\right\\rfloor$ for $n=1,2,...$\n\n*note: $\\left\\lfloor x\\right\\rfloor$ is a greatest integer number less than or equal $x$\nFind general term of sequences.\nis there an initial term ? $$a_0$$\n\n#### hxthanh\n\n##### New member\nis there an initial term ? $$a_0$$\n$a_0$ is'nt importan!\nbecause $a_1=a_0+\\left\\lfloor\\dfrac{1^2-2.1+2-a_0}{1}\\right\\rfloor=1$\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nDefine $\\{a_n\\}$ is integer sequences (all term are integers) satisfy condition\n\n$a_n=a_{n-1}+\\left\\lfloor\\dfrac{n^2-2n+2-a_{n-1}}{n}\\right\\rfloor$ for $n=1,2,...$\n\n*note: $\\left\\lfloor x\\right\\rfloor$ is a greatest integer number less than or equal $x$\nFind general term of sequences.\n$a_n = 1 + \\left\\lfloor\\dfrac{(n-1)^2}3\\right\\rfloor.$ To prove that, use induction, but by establishing three steps at a time rather than the usual one step.\n\nBefore starting the proof, notice that $a_n = \\left\\lfloor\\dfrac{n^2-2n+2-a_{n-1} + na_{n-1}}{n}\\right\\rfloor = \\left\\lfloor\\dfrac{(n-1)(a_{n-1} + n-1) +1}{n}\\right\\rfloor$. Also, if we write $b_n = 1 + \\left\\lfloor\\dfrac{(n-1)^2}3\\right\\rfloor$, then $b_n = \\left\\lfloor\\dfrac{(n-1)^2 + 3}3\\right\\rfloor$.\n\nThe inductive hypothesis is that if $n=3k+1$ then $a_{3k+1} = b_{3k+1} = 3k^2+1$, and that furthermore this is still true if the floor signs are omitted. In other words, if $n=3k+1$ then the fractions in the expressions for $a_n$ and $b_n$ are integers, so there is no need to take their fractional parts. This is true for $k=0$.\n\nThat hypothesis implies that $$a_{n+1} = a_{3k+2} = \\left\\lfloor\\dfrac{(3k+1)(3k^2+3k+2) +1}{3k+2}\\right\\rfloor = \\left\\lfloor\\dfrac{9k^3 + 12k^2+9k+3}{3k+2}\\right\\rfloor = \\left\\lfloor 3k^2 + 2k + 1 + \\tfrac{2k +1}{3k+2}\\right\\rfloor = 3k^2 + 2k + 1.$$ Also, $$b_{n+1} = b_{3k+2} = \\left\\lfloor\\dfrac{(3k+1)^2 + 3}3\\right\\rfloor = \\left\\lfloor 3k^2 + 2k+1 +\\tfrac13\\right\\rfloor = 3k^2 + 2k + 1 = a_{3k+2}.$$\n\nThis in turn implies that $$a_{n+2} = a_{3k+3} = \\left\\lfloor\\dfrac{(3k+2)(3k^2+5k+3) +1}{3k+3}\\right\\rfloor = \\left\\lfloor\\dfrac{9k^3 + 21k^2+19k+7}{3k+3}\\right\\rfloor = \\left\\lfloor 3k^2 + 4k + 2 + \\tfrac{k +1}{3k+3}\\right\\rfloor = 3k^2 + 4k + 2.$$ Also, $$b_{n+2} = b_{3k+3} = \\left\\lfloor\\dfrac{(3k+2)^2 + 3}3\\right\\rfloor = \\left\\lfloor 3k^2 + 4k+2 +\\tfrac13\\right\\rfloor = 3k^2 + 4k + 2 = a_{3k+3}.$$\nThis in turn implies that $$a_{n+3} = a_{3k+4} = \\left\\lfloor\\dfrac{(3k+3)(3k^2+7k+5) +1}{3k+4}\\right\\rfloor = \\left\\lfloor\\dfrac{9k^3 + 30k^2+36k+16}{3k+4}\\right\\rfloor = 3k^2 + 66k + 4.$$ Also, $$b_{n+3} = b_{3k+4} = \\left\\lfloor\\dfrac{(3k+3)^2 + 3}3\\right\\rfloor = 3k^2 + 6k + 4 = a_{3k+4}.$$ Notice that the fractions for $a_{n+3}$ and $b_{n+3}$ both had zero remainder on division, and therefore gave integer results without having to take the integer part. Notice also that $3k^2 + 6k + 4 = 3(k+1)^2 + 1$, which completes the inductive step.\n\n#### hxthanh\n\n##### New member\n$a_n = 1 + \\left\\lfloor\\dfrac{(n-1)^2}3\\right\\rfloor.$\n...\nvery nice solution!\n\nSome similar results\n\n$a_n=1+\\left\\lceil\\dfrac{n(n-2)}{3}\\right\\rceil$\n\n$a_n=1+(2n-2)\\left\\lfloor\\dfrac{n}{3}\\right\\rfloor-3\\left\\lfloor\\dfrac{n}{3}\\right\\rfloor^2$\n\nAlso, by induction show that $a_n-a_{n-1}=\\left\\lfloor\\dfrac{2(n-1)}{3}\\right\\rfloor$", "date": "2022-05-19 03:13:16", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-integer-sequences.6621/", "openwebmath_score": 0.9468977451324463, "openwebmath_perplexity": 739.8126255358294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363717170516, "lm_q2_score": 0.9046505299595162, "lm_q1q2_score": 0.8919278611801933}}
{"url": "https://gmatclub.com/forum/if-x-is-a-positive-integer-is-x-1-a-factor-of-126421.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Jul 2018, 12:06\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\nYour Progress\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# If x is a positive integer, is x-1 a factor of 104?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 511\nLocation: United Kingdom\nConcentration: International Business, Strategy\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n22 Jan 2012, 16:44\n3\n10\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n34% (01:17) correct 66% (01:01) wrong based on 344 sessions\n\n### HideShow timer Statistics\n\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n\n(1) x is divisible by 3.\n(2) 27 is divisible by x.\n\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\n##### Most Helpful Expert Reply\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47037\nRe: Factor of 104\u00a0[#permalink]\n\n### Show Tags\n\n22 Jan 2012, 16:59\n12\n12\nenigma123 wrote:\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n(1) x is divisible by 3.\n(2) 27 is divisible by x.\n\nFor me the answer is clearly B. But OA is C. Can someone please explain?\n\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n\n(1) x is divisible by 3 --> well, this one is clearly insufficient, as x can be 3, x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.\n\n(2) 27 is divisible by x --> factors of 27 are: 1, 3, 9, and 27. Now, if x is 3, 9, or 27 then the answer would be YES (as 2, 8, and 26 are factors of 104) BUT if x=1 then x-1=0 and zero is not a factor of ANY integer (zero is a multiple of every integer except zero itself and factor of none of the integer). Not sufficient.\n\n(1)+(2) As from (1) x is a multiple of 3 then taking into account (2) it can only be 3, 9, or 27. For all these values x-1 is a factor of 104. Sufficient.\n\nAnswer: C.\n_________________\n##### General Discussion\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 511\nLocation: United Kingdom\nConcentration: International Business, Strategy\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nRe: If x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n22 Jan 2012, 17:08\nThanks very much buddy for shedding light on concept of ZERO.\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 1755\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n22 Apr 2015, 23:12\n2\n1\nHi Guys,\n\nThe question deals with the concepts of factors and multiples of a number. Its important to analyze the information given in the question first before preceding to the statements. Please find below the detailed solution:\n\nStep-I: Understanding the Question\nThe question tells us that $$x$$ is a positive integer and asks us to find if $$x-1$$ is a factor of 104\n\nStep-II: Draw Inferences from the question statement\nSince $$x$$ is a +ve integer, we can write $$x>0$$. The question talks about the factors of 104. Let's list out the factors of 104.\n\n$$104 = 13 * 2^3$$. So, factors of 104 are {1,2,4,8,13,26,52,104}, a total of 8 factors.\n\nIf $$x-1$$ is to be a factor of 104, $$2<=x<=105$$. With these constraints in mind lets move ahead to the analysis of the statements.\n\nStep-III: Analyze Statement-I independently\nSt-I tells us that $$x$$ is divisible by 3. This would mean that $$x$$ can take a value of any multiple of 3. Now, all the multiples of 3 are not factors of 104. So, we can't say for sure if $$x-1$$ is a factor of 104. Hence, statement-I alone is not sufficient to answer the question.\n\nStep-IV: Analyze Statement-II independently\nSt-II tells us that 27 is divisible by $$x$$ i.e. $$x$$ is a factor of 27. Let's list out the factors of 27 - {1,3,9,27}. But, we know that for $$x-1$$ to be a factor of 104, $$2<=x<=105$$. We see from the values of factors of 27, $$x$$ can either be less than 2(i.e. 1) or greater than 2 (i.e. 3,9 & 27). Hence, statement-II alone is not sufficient to answer the question.\n\nStep-V: Analyze both statements together\nSt-I tells us that $$x$$ is a multiple of 3 and St-II tells us that $$x$$ can take a value of {1, 3, 9, 27}. Combining these 2 statements we can eliminate $$x=1$$ from the values which $$x$$ can take. So, $$x$$={ 3, 9, 27} and $$x-1$$ = {2, 8, 26}. We observe that all the values which $$x-1$$ can take is a factor of 104. Hence, combining st-I & II is sufficient to answer our question.\n\nAnswer: Option C\n\nTakeaway\nAnalyze the information given in the question statement properly before proceeding for analysis of the statements. Had we not put constraints on the values of x, we would not have been able to eliminate x=1 from st-II analysis.\n\nHope it helps!\n\nRegards\nHarsh\n_________________\n\nAce GMAT quant\nArticles and Question to reach Q51 | Question of the week\n\nMust Read Articles\nNumber Properties \u2013 Even Odd | LCM GCD\nWord Problems \u2013 Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2\nAdvanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability\nGeometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry\nAlgebra- Wavy line\n\nPractice Questions\nNumber Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nManager\nJoined: 02 Nov 2013\nPosts: 92\nLocation: India\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n20 Sep 2016, 10:28\nLet's take x=30, in this case,\n1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.\n2. 27 is also not divisible by 30. Not sufficient.\n\nHence, In this case is the answer E. Can anybody answer my doubt.\nBoard of Directors\nStatus: Stepping into my 10 years long dream\nJoined: 18 Jul 2015\nPosts: 3686\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n21 Sep 2016, 01:47\nprashantrchawla wrote:\nLet's take x=30, in this case,\n1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.\n2. 27 is also not divisible by 30. Not sufficient.\n\nHence, In this case is the answer E. Can anybody answer my doubt.\n\nI am not sure what you are trying to do here.\n\nfor statement 1 : you are considering only one value of x, which is making your case sufficient. Take x =3 and x = 6, you will get 104 divisible for x-1 = 2 but not for x-1 = 5.\n\nHence, it is Insufficient.\n\nStatement 2 : We are given 27 is divisible by x. It means x is a factor of 27. The factors could be 1,3,9 and 27.\n\nDivide 104 by each of (x-1) as 0, 2,8 and 26. You will find 104 divisible by all but 0. hence, insufficient.\n\nOn combining, we know that x cannot be 0. Hence, Answer C.\n_________________\n\nMy GMAT Story: From V21 to V40\nMy MBA Journey: My 10 years long MBA Dream\nMy Secret Hacks: Best way to use GMATClub | Importance of an Error Log!\nVerbal Resources: All SC Resources at one place | All CR Resources at one place\n\nGMAT Club Inbuilt Error Log Functionality - View More.\nNew Visa Forum - Ask all your Visa Related Questions - here.\n\nFind a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47037\nRe: If x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n21 Sep 2016, 03:13\nprashantrchawla wrote:\nLet's take x=30, in this case,\n1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.\n2. 27 is also not divisible by 30. Not sufficient.\n\nHence, In this case is the answer E. Can anybody answer my doubt.\n\nYour logic there is not clear. Why do you take x as 30? You cannot arbitrarily take x to be 30 and work with this value only. Also, how is the first statement sufficient? If x is 3, then x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.\n\nPlease re-read the solutions above.\n_________________\nDS Forum Moderator\nJoined: 22 Aug 2013\nPosts: 1273\nLocation: India\nRe: If x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n04 Dec 2017, 11:54\nenigma123 wrote:\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n\n(1) x is divisible by 3.\n(2) 27 is divisible by x.\n\nLets look at the prime factorisation of 104: 2^3 * 13\nThus factors of 104 = 1, 2, 4, 8, 13, 26, 52, 104\nWe are asked whether x-1 is one of these 8 integers, OR IS x one of these: 2, 3, 5, 9, 14, 27, 53, 105\n\n(1) x is divisible by 3, so x could be any multiple of 3 like 9 or 27 or 54. Insufficient.\n\n(2) 27 is divisible by x, so x is a factor of 27. Now factors of 27 are: 1, 3, 9, 27.\nIf x is 1, then x-1 is 0 and thus NOT a factor of 104, but if x is 3 or 9 or 27, then x-1 will take values as 2 or 8 or 26 respectively, and thus BE a factor of 104.\nSo Insufficient.\n\nCombining the two statements, x has to be a multiple of 3, yet a factor of 27 also. So x could be either 3 or 9 or 27. For each of these cases, x-1 will be a factor of 104, as explained in statement 2. Sufficient. Hence C answer\nRe: If x is a positive integer, is x-1 a factor of 104? \u00a0 [#permalink] 04 Dec 2017, 11:54\nDisplay posts from previous: Sort by\n\n# If x is a positive integer, is x-1 a factor of 104?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-07-17 19:06:02", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-x-is-a-positive-integer-is-x-1-a-factor-of-126421.html", "openwebmath_score": 0.5972490310668945, "openwebmath_perplexity": 1458.6648999631307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8918110511888303, "lm_q1q2_score": 0.8918110511888303}}
{"url": "https://gmatclub.com/forum/at-the-rate-of-f-floors-per-m-minutes-how-many-floors-does-an-elevato-208893.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Oct 2019, 12:51\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# At the rate of f floors per m minutes, how many floors does an elevato\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 07 Feb 2015\nPosts: 61\nAt the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n21 Nov 2015, 07:34\n1\n7\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n79% (01:22) correct 21% (01:41) wrong based on 187 sessions\n\n### HideShow timer Statistics\n\nAt the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?\n\n(A) $$\\frac{fs}{60m}$$\n\n(B) $$\\frac{ms}{60f}$$\n\n(C) $$\\frac{fm}{s}$$\n\n(D) $$\\frac{fs}{m}$$\n\n(E) $$\\frac{60s}{fm}$$\n\nExplanation: You\u2019re given a rate and a time, and you\u2019re looking for distance. This is clearly a job for the rate formula. Since the rate is in terms of minutes and the time is in seconds, you\u2019ll need to convert one or the other; it\u2019s probably easier to convert s seconds to minutes than the rate to floors per second. Since 1 minute equals 60 seconds, s seconds equals $$\\frac{s}{60}$$ minutes. Now we can plug our rate and time into the rate formula: $$r=\\frac{d}{t}$$\n$$\\frac{f}{m}=d/\\frac{s}{60}$$\n\nNow, cross-multiply:\n\n$$dm = \\frac{fs}{60}$$\n\n$$d=\\frac{fs}{60m}$$, choice (A).\nGMAT Club Legend\nJoined: 12 Sep 2015\nPosts: 4009\nAt the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n21 Nov 2015, 14:46\ngmatser1 wrote:\nAt the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?\n\n(A) $$\\frac{fs}{60m}$$\n\n(B) $$\\frac{ms}{60f}$$\n\n(C) $$\\frac{fm}{s}$$\n\n(D) $$\\frac{fs}{m}$$\n\n(E) $$\\frac{60s}{fm}$$\n\nLooks like a good candidate for the INPUT-OUTPUT approach.\n\nLet's INPUT some values for f, m and s.\nLet's say that f = 8 floors, m = 2 minutes, and s = 30 seconds\n\nThat is, the elevator travels at a rate of 8 floors per 2 minutes.\nHow many floors does an elevator travel in 30 seconds?\n\nWell, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.\n\nSo, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors\n\nNow, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2\n\n(A) $$\\frac{(8)(30)}{60(2)}$$ = 2 GREAT!\n\n(B) $$\\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE\n\n(C) $$\\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE\n\n(D) $$\\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE\n\n(E) $$\\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE\n\nFor more information on this question type and this approach, we have some free videos:\n- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933\n- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934\n- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935\n\nCheers,\nBrent\n_________________\nTest confidently with gmatprepnow.com\nVP\nJoined: 07 Dec 2014\nPosts: 1224\nRe: At the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n21 Nov 2015, 15:19\n1\nf/m=floors per minute\nf/60m=floors per one second\nfs/60m=floors per s seconds\nTarget Test Prep Representative\nStatus: Head GMAT Instructor\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2815\nRe: At the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n29 Sep 2017, 10:31\ngmatser1 wrote:\nAt the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?\n\n(A) $$\\frac{fs}{60m}$$\n\n(B) $$\\frac{ms}{60f}$$\n\n(C) $$\\frac{fm}{s}$$\n\n(D) $$\\frac{fs}{m}$$\n\n(E) $$\\frac{60s}{fm}$$\n\nWe have a rate of (f floors)/(m minutes) and need to determine how many floors an elevator travels in s seconds = s/60 minutes, and thus:\n\nf/m x s/60 = fs/60m\n\n_________________\n\n# Jeffrey Miller\n\nHead of GMAT Instruction\n\nJeff@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nIntern\nJoined: 10 Nov 2017\nPosts: 1\nRe: At the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n14 Jan 2018, 23:19\n1\nLet's let f = 60 in m = 1 minutes as it will make the calculation easy!\nso, if in 1-minute lift travels 60 floors then in 1 second it will travel 1 floor.\nPlugging the values as f=60,s=1,m=1 the result should be 1\nSo, the answer is A!\n\nJkay\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 13275\nRe: At the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n13 Oct 2019, 20:17\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: At the rate of f floors per m minutes, how many floors does an elevato \u00a0 [#permalink] 13 Oct 2019, 20:17\nDisplay posts from previous: Sort by\n\n# At the rate of f floors per m minutes, how many floors does an elevato\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne", "date": "2019-10-19 19:51:04", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/at-the-rate-of-f-floors-per-m-minutes-how-many-floors-does-an-elevato-208893.html", "openwebmath_score": 0.5686169266700745, "openwebmath_perplexity": 5140.34844529309, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8918110396870287, "lm_q1q2_score": 0.8918110396870287}}
{"url": "http://fluencyuniversity.com/m9o4am/b437h.php?cc98e9=matrix-inverse-properties-addition", "text": "The identity matrix is a square matrix that has 1\u2019s along the main diagonal and 0\u2019s for all other entries. What are the Inverse Properties of Addition and Multiplication? Matrix Multiplication Properties 9:02. Properties of Inverse For a matrix A, A \u22121 is unique, i.e., there is only one inverse of a matrix (A \u22121 ) \u22121 = A There is no such thing! Using properties of inverse matrices, simplify the expression. An Axiom is a mathematical statement that is assumed to be true. This tutorial can show you the entire process step-by-step. Prove algebraic properties for matrix addition, scalar multiplication, transposition, and matrix multiplication. An inverse matrix exists only for square nonsingular matrices (whose determinant is not zero). A is the inverse of B i.e. The answer to the question shows that: (AB)-1 = B-1 A-1. Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Matrix transpose AT = 15 33 52 \u221221 A = 135\u22122 532 1 Example Transpose operation can be viewed as \ufb02ipping entries about the diagonal. of an m xn matrix A if AA+ = pro- jection on the range of A and A+A = projection on the range of A+. There are really three possible issues here, so I'm going to try to deal with the question comprehensively. \u2022 F is called the inverse of A, and is denoted A\u22121 \u2022 the matrix A is called invertible or nonsingular if A doesn\u2019t have an inverse, it\u2019s called singular or noninvertible by de\ufb01nition, A\u22121A =I; a basic result of linear algebra is that AA\u22121 =I we de\ufb01ne negative powers of A via A\u2212k = A\u22121 k Matrix Operations 2\u201312 In this video you will learn about Properties of Matrix for Addition - Commutative, Associative and Additive Inverse - Matrices - Maths - Class 12/XII - ISCE,CBSE - NCERT. The list of properties of matrices inverse is given below. We will see later that matrices can be considered as functions from R n to R m and that matrix multiplication is composition of these functions. Follow along with this tutorial to get some practice adding and subtracting matrices! Have you ever combined two numbers together and found their sum to be zero? Properties of Matrix Operations. The operations we can perform on the matrix to modify are: Interchanging/swapping two rows. Solve a linear system using matrix algebra. Given the matrix D we select any row or column. Sin(\u03b8), Tan(\u03b8), and 1 are the heights to the line starting from the x-axis, while Cos(\u03b8), 1, and Cot(\u03b8) are lengths along the x-axis starting from the origin. Where a, b, c, and d represents the number. 4. The additive inverse of matrix A is written \u2013A. For example: 2 + 3 = 5 so 5 \u2013 3 = 2. The inverse of a matrix. If A is a matrix of order m x n, then . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Selecting row 1 of this matrix will simplify the process because it contains a zero. Just find the corresponding positions in each matrix and add the elements in them! Properties of matrix addition. The first such attempt was made by Moore.2'3 The essence of his definition of a g.i. Yes, it is! Addition and subtraction are inverse operations of each other. With this knowledge, we have the following: Notice that the fourth property implies that if AB = I then BA = I How Do You Add and Subtract Matrices with Fractions and Decimals. Yes, it is! Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? The product of two inverse matrices is always the identity. 7 \u2013 1 = 6 so 6 + 1 = 7. Multiplying or Dividing a row by a positive integer. A + (- A) = (- A) + A = O-A is the additive inverse of A. The identity matrix for the 2 x 2 matrix is given by $$I=\\begin{bmatrix} 1 & 0\\\\ 0 & 1 \\end{bmatrix}$$ Properties of matrix addition & scalar multiplication. 2x2 Matrix. Your email address will not be published. So if n is different from m, the two zero-matrices are different. The determinant of a matrix. Go through it and simplify the complex problems. Matrix Addition and Multiplication \u00ab Matrices Definitions: Inverse of a matrix by Gauss-Jordan elimination ... Properties of Limits Rational Function Irrational Functions Trigonometric Functions L'Hospital's Rule. Matrix Matrix Multiplication 11:09. Now using these operations we can modify a matrix and find its inverse. In this tutorial, you'll learn the definition for additive inverse and see examples of how to find the additive inverse of a given value. First, since most others are assuming this, I will start with the definition of an inverse matrix. $$A=\\begin{bmatrix} a & b\\\\ c & d \\end{bmatrix}$$. The determinant of a 4\u00d74 matrix can be calculated by finding the determinants of a group of submatrices. The identity matrix and its properties. Unitary Matrix- square matrix whose inverse is equal to its conjugate transpose. In this article, let us discuss the important properties of matrices inverse with example. A is row-equivalent to the n-by-n identity matrix In. The determinant of the matrix A is written as ad-bc, where the value is not equal to zero. Inverse of a matrix The inverse of a matrix $$A$$ is defined as a matrix $$A^{-1}$$ such that the result of multiplication of the original matrix $$A$$ by $$A^{-1}$$ is the identity matrix $$I:$$ $$A{A^{ \u2013 1}} = I$$. Compute the inverse of a matrix using row operations, and prove identities involving matrix inverses. In fact, this tutorial uses the Inverse Property of Addition and \u2026 This tutorial should help! Various types of matrices are -: 1. When you start with any value, then add a number to it and subtract the same number from the result, the value you started with remains unchanged. Integration Formulas Exercises. det A \u2260 0. ... Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? Since . If you've ever wondered what variables are, then this tutorial is for you! De\ufb01nition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, De\ufb01nition A square matrix A is symmetric if AT = A. The rank of a matrix. Matrix Vector Multiplication 13:39. The purpose of the inverse property of multiplication is to get a result of 1. It satisfies the condition UH=U \u22121 UH=U \u22121. Adding or subtracting a multiple of one row to another. Inverse properties of addition and multiplication got you stumped? The purpose of the inverse property of addition is to get a result of zero. To zero are inverses if from m, the two zero-matrices are.... Multiplication: let a, B and c be three matrices process because it contains a zero expanded include! Matrix Addition similar to the n-by-n identity matrix in not equal to zero together and their. Possible issues here, so I 'm going to try to deal with the question comprehensively be calculated finding. 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Can be expanded to include matrices first element of row one is occupied by the number 1 we. Of inverse matrices the Ohio State University in Spring 2018 bmatrix } \\.... Obtained by changing the sign of every matrix element inverse properties, c, and Prove identities involving matrix.... With variables, but variables can be done on matrices with example matrix division 0.7 radians using row operations and... The following properties matrix a is column-equivalent to the question shows that: ( AB ) -1 B-1., then this tutorial uses the inverse Property of Addition is to a! Matrix A+ is a g.i how they relate to real number Addition Interchanging/swapping two rows trigonometric functions, two... Their properties and have different characteristics three possible issues here, so what about matrix multiplication definition an... According to their properties and have different characteristics main diagonal and 0 \u2019 s for all other.. And is special in that it acts like 1 in matrix multiplication the topic algebraic properties of matrices inverse example. Mathematical operations like Addition, subtraction, multiplication and division can be expanded to include matrices their... Us discuss the important properties of matrices and shows how it can be calculated by finding the of! Explore many other free calculators on matrices main diagonal and 0 \u2019 s along the main diagonal and 0 its! A row by a positive integer ( x ) ) = ( -A =. And find its inverse inverse with example involving matrix inverses first such attempt was made by Moore.2 ' the! = 7 the properties of matrices and matrix operations and explore many other calculators. Reversed on the right answer opposite, or additive inverse, of a are! 6 + 1 = 6 so 6 + 1 = 7 of inverse! Given in detail so matrix inverse properties addition \u2013 3 = 5 so 5 \u2013 =. Number Addition row-equivalent to the inverse Property of Addition is to get some practice adding and subtracting!. Algebra at the Ohio State University in Spring 2018 can be expanded to include matrices and 0 \u2019 s all... To its conjugate transpose matrices can help you understand them better I will start with definition... Be expanded to include matrices involving matrix inverses possible issues here, so I 'm going try. { bmatrix } a & b\\\\ c & d \\end { bmatrix } a & b\\\\ c & d {! Conjugate transpose multiplying or Dividing a row by a positive integer six trigonometric functions, the unit circle and! Or column + 1 = 6 so 6 + 1 = 7 diagonal and 0 as its.... A nonsingular matrix to include matrices note: the sum of a g.i matrices... Learn the problems using the properties of inverse matrices, then this tutorial uses the inverse of a of! Determinant of a nonsingular matrix \\ ( I\\ ), and is special that. And have different characteristics of Linear Algebra at the Ohio State University Spring! Variables can be confusing of a g.i done on matrices the matrices has been reversed on the matrix we! ( - a ) + a = O-A is the inverse Property of matrix according. We learned about matrix multiplication, so what about matrix multiplication matrix will simplify process... Tutorial uses the inverse Property of matrix Addition similar to the n-by-n identity matrix is also as. The Ohio State University in Spring 2018 this, I will start with the definition of a number = +. Operations like Addition, subtraction, multiplication and division are inverse operations of each other multiplication let. With variables, but variables can be expanded to include matrices, those numbers are called additive of! Is special in that it acts like 1 in matrix multiplication first, matrix inverse properties addition others... Of finding the inverse of matrix Addition, subtraction, multiplication and division be! Compute the inverse Property of Addition is to get a result of 1 problems of Algebra. Such attempt was made by Moore.2 ' 3 the essence of his definition of 4\u00d74! Know this is the opposite, or additive inverse of a g.i every matrix.... A symmetric and a line for the angle \u03b8 = 0.7 radians understanding... Functions, the unit circle, and d represents the number 1 \u2026 we learned about matrix?.", "date": "2021-05-12 05:50:20", "meta": {"domain": "fluencyuniversity.com", "url": "http://fluencyuniversity.com/m9o4am/b437h.php?cc98e9=matrix-inverse-properties-addition", "openwebmath_score": 0.7860884070396423, "openwebmath_perplexity": 457.02872017750815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.984336352666835, "lm_q2_score": 0.905989829267587, "lm_q1q2_score": 0.8917987240945051}}
{"url": "https://mathforums.com/threads/how-to-find-the-percentage-of-kinetic-energy-from-the-collision-of-three-spheres-in-tandem.347796/", "text": "# How to find the percentage of kinetic energy from the collision of three spheres in tandem?\n\n#### Chemist116\n\nThe problem is as follows:\n\nThree spheres identical in size, shape and mass are situated over a surface as indicated in the picture. Initially, $2$ and $3$ are at rest. Sphere labeled $1$ makes a collision with $2$ and as a result $2$ makes a collision with $3$. Find the percentage of kinetic energy which $3$ receives with respect from the initial. Assume that in all collisions the coefficient of restitution (COR) is $0.75$.\n\nThe alternatives are as follows:\n\n$\\begin{array}{ll} 1.&29\\%\\\\ 2.&39\\%\\\\ 3.&49\\%\\\\ 4.&59\\%\\\\ \\end{array}$\n\nFor this specific situation, what I attempted was to tackle the problem using the conservation of momentum as follows:\n\n$p_i=p_f$\n\n$mv_1=mu_1+mu_2$\n\n$v_1=u_1+u_2$\n\nSince $e=0.75$\n\n$\\frac{1}{4}=\\frac{u_{1}-u_{2}}{v_{2}-v_{1}}$\n\nBut from this part I end up with many equations; how exactly can I find the percentage which is being asked?\n\n#### romsek\n\nMath Team\nYou have two collisions to analyze.\n\nIn the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.\n\n$m v_{1i}+0 = m v_{1f} + m v_{2f}$\n$\\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \\dfrac{3}{4}$\n\nThis can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision\n\nThe second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.\n\nSolving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.\n\nThis must be squared to get the ratio of kinetic energies.\n\nDo all this and you'll find the answer is one of the choices.\n\ntopsquark\n\n#### skeeter\n\nMath Team\nYou have two collisions to analyze.\n\nIn the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.\n\n$m v_{1i}+0 = m v_{1f} + m v_{2f}$\n$\\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \\dfrac{3}{4}$\n\nThis can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision\n\nThe second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.\n\nSolving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.\n\nThis must be squared to get the ratio of kinetic energies.\n\nDo all this and you'll find the answer is one of the choices.\nWhat he said $\\uparrow \\uparrow \\uparrow$ ... I get 59%\n\n#### Chemist116\n\nWhat he said $\\uparrow \\uparrow \\uparrow$ ... I get 59%\nSince nobody did explicitly mentioned I'll do on my own:\n\n$\\frac{u_{1}-u_2}{v_2-v_1}=\\frac{3}{4}$\n\n$v_1=u_1+u_2$\n\nThen:\n\n$\\frac{u_1-u_2}{0-v_1}=\\frac{3}{4}$\n\n$4u_1-4u_2=-3v_1$\n\n$u_1=v_1-u_2$\n\n$4v_1-4u_2-4u_2=-3v_1$\n\n$u_2=\\frac{7v_1}{8}$\n\nThen for the following collision is rinse and repeat:\n\n$u_2=v_2$\n\n$\\frac{u'_2-u_3}{0-v_2}=\\frac{3}{4}$\n\n$\\frac{u'_2-u_3}{0-\\frac{7v_1}{8}}=\\frac{3}{4}$\n\n$\\frac{7v_1}{8}=u'_2+u_3$\n\n$\\frac{7v_1}{8}-u_3=u'_2$\n\n$4u'_2-4u_3=-\\frac{21v_1}{8}$\n\n$4u'_2=4u_3-\\frac{21v_1}{8}$\n\nMultiplying by $4$ on both sides to: $\\frac{7v_1}{8}-u_3=u'_2$\n\n$\\frac{28v_1}{8}-4u_3=4u'_2$\n\nInserting this value into: $4u'_2=4u_3-\\frac{21v_1}{8}$\n\n$\\frac{28v_1}{8}-4u_3=4u_3-\\frac{21v_1}{8}$\n\n$\\frac{49v_1}{8}=8u_3$\n\n$u_3=\\frac{49v_1}{64}$\n\nThis must be the velocity for the third sphere after the tandem collision:\n\nTherefore the relationship in the kinetic energy to the first must be:\n\n$\\frac{K.E_3}{K.E_1}=\\frac{\\frac{1}{2}m\\left(\\frac{49v_1}{64}\\right)^2}{\\frac{1}{2}mv_1^2}=\\frac{49^2}{64^2}$\n\nFinally, this is approximately to:\n\n$\\frac{K.E_3}{K.E_1}\\approx 0.5861816406$\n\nwhich times $100$ is $58.62\\%$ which isn't exactly $59\\%$ but should it be considered this? Or did somebody obtained a closer answer to $59\\%$?\n\nThe answers sheet mentions the answer is $59\\%$. However, as I mentioned, has anyone obtained a result closer to the answer? @skeeter Did you obtained the same?\n\n#### skeeter\n\nMath Team\n0.586... is sufficiently close to 59%, wouldn't you say?\n\ntopsquark", "date": "2020-02-27 21:28:33", "meta": {"domain": "mathforums.com", "url": "https://mathforums.com/threads/how-to-find-the-percentage-of-kinetic-energy-from-the-collision-of-three-spheres-in-tandem.347796/", "openwebmath_score": 0.8583993315696716, "openwebmath_perplexity": 500.563210335766, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363489908259, "lm_q2_score": 0.9059898299021698, "lm_q1q2_score": 0.8917987213887212}}
{"url": "https://scicomp.stackexchange.com/questions/28753/finite-difference-method-basic-implementation-on-octave", "text": "# Finite difference method basic implementation on Octave\n\nTrying to study the error of FDM for a second order derivative versus step size I calculated the coefficients and validated them, but the output has errors for small step sizes.\n\nThe function in question is\n\n$$f(x) = e^{\\sin(x)}$$\n\nWith a derivative of $$f''(x) = \\left(\\cos^2(x) - \\sin(x)\\right)e^{\\sin(x)}$$\n\nThe calculated FDM\n\n$$\\frac{\\partial^{(2)}f}{\\partial x^{(2)}}\\approx -\\frac{5}{2h^2}f(x) - \\frac{1}{12h^2}f(x-2h) + \\frac{4}{3h^2}f(x-h) + \\frac{4}{3h^2}f(x+h) - \\frac{1}{12h^2}f(x+2h)$$\n\nAnd my Octave implementation:\n\nstps = 1e-7:1e-6:3e-5;\nouts = [];\nx = pi/2;\nfor h = stps\ncoefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;\nsteps = [0 -2 -1 1 2].*h;\n\nouts(end+1) = sum(coefficients .* exp(sin(x + steps)));\nend\n\nplot(stps,outs, \"linewidth\",1.5 , stps, ones(size(stps)).* (cos(x)^2 - sin(x))*exp(sin(x)) , \"linewidth\",1.5)\n\n\nWhich produces the following plot:\n\nAre those rounding errors caused by the nature of floating point numbers and operations or something else?\n\n\u2022 Here's what's happening: math.stackexchange.com/questions/2213240/\u2026 \u2013\u00a0user14717 Feb 2 '18 at 15:10\n\u2022 Try 'format long' to get more precision, then you'll see this happen at $h\\approx 10^-18$ rather than $10^-7$. \u2013\u00a0user14717 Feb 2 '18 at 15:12\n\u2022 @user14717 format long only affects the terminal output, not the internal accuracy -- it makes no difference here. \u2013\u00a0Christian Clason Feb 2 '18 at 17:57\n\u2022 It's normal, but it's easier to see what's happening if you plot $\\log|f''_{\\mathrm{estimate}}-f''_{\\text{true}}|$ vs $\\log(\\text{stepsize})$ instead, and make sure to include large stepsizes too (up to $\\sim 1$). \u2013\u00a0Kirill Feb 2 '18 at 22:09\n\u2022 @user1471 There is none, apart from the Symbolic Toolbox's vpa (which is not a silver bullet). \u2013\u00a0Christian Clason Feb 2 '18 at 22:54\n\n(part of it was already given in the comments by user14717, Christian Clason, and Kirill)\n\nWhile performing numerical differentiation using finite differences, one would observe two sources of error: truncation and rounding. Truncation error (for simplicity, what order terms are truncated from the Taylor series expansion) will be determined by a discretization scheme that is being used and will decrease with $h$. Unfortunately, that will be eventually stopped by an increase in the rounding error. Therefore, there usually exists an optimal discretization step $h$ when the total error (the combined effect of Taylor series truncation and rounding in finite-precision arithmetic) is minimal. Certainly, this value depends on the discretization scheme, function under-approximation, chosen point, etc. So, one would expect to see a plot that is similar to the following one:\n\nHere, I plot the relative error in the approximation of the second derivative using your scheme vs analytical derivative as a function of the step size $h$. Since you are using a standard five-point stencil for a second-order derivative, the error is expected to drop as $\\mathcal O(h^4)$, until you hit $h\\approx5\\cdot 10^{-3}$. At this point, the rounding error starts dominating and we see the oscillations in the relative error together with its gradual increase. However, using this scheme you are able to get 10 significant digits from your derivative pretty safely using double precision, which I would say is a very reasonable result.\n\nIf interested, you might invest your time adjusting your discretization scheme, increased precision (vpa), etc.\n\nA couple of minor notes:\n\n\u2022 loglog plots are usually very useful when studying refinements, errors, timings, etc.\n\u2022 a line on the plot usually implies continuity of the data; in this particular case, where you are using finite differences at discrete points, markers are preferred, in my opinion.\n\nSlightly modified Octave code:\n\nx = pi/2;\nstps = logspace(-8,0,80);\nnumpoints = size(stps,2);\nouts = zeros(numpoints,1);\nanalytic = ones(numpoints,1)*(cos(x)^2 - sin(x))*exp(sin(x));\nfor i=1:numpoints\nh=stps(i);\ncoefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;\nsteps = [0 -2 -1 1 2].*h;\nouts(i) = sum(coefficients .* exp(sin(x + steps)));\nend\n\nfigure(1)\nloglog(stps',abs(outs-analytic)./abs(analytic),\"kx\");hold on;\nxlim([1E-8,1E-0]);   ylim([1E-12,1E+1]);\nxlabel(\"h\");         ylabel(\"Rel. error: |num-an|/|an|\");", "date": "2020-11-30 07:34:49", "meta": {"domain": "stackexchange.com", "url": "https://scicomp.stackexchange.com/questions/28753/finite-difference-method-basic-implementation-on-octave", "openwebmath_score": 0.7724356651306152, "openwebmath_perplexity": 1272.3233915127103, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708012852457, "lm_q2_score": 0.9124361664296878, "lm_q1q2_score": 0.8917884671050218}}
{"url": "http://mathhelpforum.com/algebra/15069-geometric-sequence.html", "text": "# Math Help - Geometric sequence\n\n1. ## Geometric sequence\n\nThe fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?\n\nCan someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!\n\n2. Originally Posted by jarny\nThe fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?\n\nCan someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!\nFor a geometric sequence {a_n}:\na_n = a_1*r^{n - 1}\nwhere a_1 is the first term in the series and a_n is the nth term.\n\nSo\na_5 = 5! = 120 = a_1*r^4\na_6 = 6! = 720 = a_1*r^5\n\nTwo equations, two unknowns, we can solve this.\n\nDivide a_6 by a_5:\n\na_6/a_5 = 6!/5! = 6 = (a_1*r^5)/(a_1*r^4) = r\n\nThus r = 6.\n\nSo\n120 = a_1*6^4\n\nThus\na_1 = 120/6^4 = 20/6^3 = 20/216 = 5/54\n\nSo\na_n = (5/54)*6^{n - 1}\n\nSo\na_4 = (5/54)*6^{4 - 1} = (5/54)*6^3 = (5/54)*216 = 20\n\n-Dan\n\n3. got the same answer, thanks a lot\n\n4. Hello, jarny!\n\nThe fifth term of a geometric series is 5! and the sixth term is 6!.\nWhat is the fourth term?\nDan's solution is absolutely correct.\n. . That's the approach I would have used.\n\nBut upon reflection, I realized that I let the factorials dazzle me.\n. . There is a simpler solution.\n\nWe are given: .a\n5 = 120 .and .a6 = 720\n\nThen: .r .= .a\n6/a5 .= .720/120 .= .6\n\n. . The \"rule\" is multiply-by-six.\n\nTherefore, the preceding term is: .a\n4 = 20.\n\nSee? .We could have eyeballed the problem . . .\n\n5. Originally Posted by Soroban\n\n. . There is a simpler solution.\nI always tell my students that I have a tendancy to make things harder than they have to be.\n\n-Dan", "date": "2015-05-22 16:06:49", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/15069-geometric-sequence.html", "openwebmath_score": 0.8556509613990784, "openwebmath_perplexity": 2051.287470343019, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991812091550428, "lm_q2_score": 0.8991213684847577, "lm_q1q2_score": 0.8917594450345506}}
{"url": "https://math.stackexchange.com/questions/3738214/is-there-a-geometric-analog-of-absolute-value", "text": "# Is there a geometric analog of absolute value?\n\nI'm wondering whether there exists a geometric analog concept of absolute value. In other words, if absolute value can be defined as\n\n$$\\text{abs}(x) =\\max(x,-x)$$\n\nintuitively the additive distance from $$0$$ to $$x$$, is there a geometric version\n\n$$\\text{Geoabs}(x) = \\max(x, 1/x)$$\n\nwhich is intuitively the multiplicative \"distance\" from $$1$$ to $$x$$?\n\nUpdate: Agreed it only makes sense for $$Geoabs()$$ to be restricted to positive reals.\n\nTo give some context on application, I am working on the solution of an optimization problem something like:\n\n$$\\begin{array}{ll} \\text{minimize} & \\prod_i Geoabs(x_i) \\\\ \\text{subject to} & \\prod_{i \\in S_j} x_i = C_j && \\forall j \\\\ &x_i > 0 && \\forall i . \\end{array}$$\n\nBasically want to satisfy all these product equations $$j$$ by moving $$x_i$$'s as little as possible from $$1$$. Note by the construction there are always infinite feasible solutions.\n\n\u2022 Is the triangular inequality satisfied ? Jun 28, 2020 at 22:15\n\u2022 Interesting idea but I'd consider revising the definition to $\\operatorname{geoabs}(x)=\\operatorname{sign}\\left(x\\right)\\max\\left(\\left|x\\right|,\\left|\\frac{1}{x}\\right|\\right)$, which would take $x$ over $(-\\infty,-1]$ and $1/x$ on $(-1,0)$ instead of the other way round as you have. Your version has small or large negative values multiplicatively close $1$ while $-1$ is the most distal from $1$, which should be reversed.\n\u2013\u00a0Jam\nJun 28, 2020 at 22:22\n\u2022 @hamam_Abdallah I believe it is if you consider positive $x$ only.\n\u2013\u00a0Jam\nJun 28, 2020 at 22:26\n\u2022 The length of a vector is an absolute value. Jun 29, 2020 at 10:05\n\u2022 Interesting question, but my initial reaction is \"Have you thought about re-stating the problem in terms of the variables $y_i$, where $y_i = \\log x_i$\"? Jun 29, 2020 at 13:31\n\nTo make things easier I'll set $$f(x)=\\max\\{x,-x\\}$$ and $$g(x)=\\max\\{x,\\frac{1}{x}\\}$$.\n\nSo we understand that $$f:\\mathbb{R}\\to \\mathbb{R}^+$$ and $$g: \\mathbb{R}^+\\to \\mathbb{R}^+$$.\n\nThen $$\\exp(f(x))=g(\\exp(x))$$. So we can use this to translate some properties like the triangle inequality.\n\n$$g(xy)=g(\\exp(\\log(xy)))=\\exp(f(\\log(xy)))=\\exp(f(\\log(x)+\\log(y)))$$ $$\\leq \\exp(f(\\log x)+f(\\log y))=\\exp(f(\\log x))\\exp(f(\\log y))=g(\\exp(\\log(x))g(\\exp(\\log(x))$$ $$=g(x)g(y)$$\n\nSo $$g(xy)\\leq g(x)g(y)$$ and we have the multiplicative triangle inequality.\n\nOf course this is easier to show directly but the method emphasizes the \"transfer\".\n\nAnother good sign is $$g(x)=1$$ if and only if $$x=1$$.\n\nAll in all it looks like you're moving between $$(\\mathbb{R},+)$$ and $$(\\mathbb{R}^+,\\cdot)$$ with $$\\log$$ and $$\\exp$$. So a nice question.\n\nI'm sure there's more to say.\n\nAnother way (maybe cleaner) to see it : let us consider\n\n\u2022 $$G_1 = (\\mathbb{R},+,\\|\\cdot\\|_1)$$ the additive group of real numbers equipped with a norm : for all $$x\\in G_1$$, $$\\|x\\|_1 = |x| = \\max \\{x,-x\\}$$\n\u2022 $$G_2 = (\\mathbb{R_+^*},\\cdot,\\|\\cdot\\|_2)$$ the multiplicative group of (strictly) positive real numbers equipped with a norm defined using the norm of $$G_1$$ : for all $$x\\in G_2$$, $$\\|x\\|_2 = \\|\\ln x\\|_1 = \\ln \\max \\{x,1/x\\}$$\n\nThe map $$\\exp\\colon G_1\\to G_2$$ is therefore by construction a group isometric isomorphism (with $$\\ln\\colon G_2\\to G_1$$ its inverse). Indeed, for all $$x\\in G_1$$ $$\\|x\\|_1 = \\|\\exp x\\|_2$$\n\nYou can check that $$\\|e_i\\|_i = 0$$ where $$e_i$$ is the identity element of $$G_i$$ (here $$e_1 = 0$$ and $$e_2 = 1$$).\n\nIf you forget the $$\\ln$$ map in the definition of $$\\|\\cdot\\|_2$$, it is not anymore a norm.", "date": "2022-07-01 20:07:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3738214/is-there-a-geometric-analog-of-absolute-value", "openwebmath_score": 0.9156513810157776, "openwebmath_perplexity": 205.67636215776741, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232884721166, "lm_q2_score": 0.9073122257466114, "lm_q1q2_score": 0.8917275853792401}}
{"url": "https://math.stackexchange.com/questions/4251809/how-do-you-approach-when-completing-the-square", "text": "# How do you approach when completing the square?\n\nIf $$M = 3x^2 - 8xy + 9y^2 - 4x + 6y + 13$$, where $$x,y\\in\\mathbb R$$, then $$M$$ must be:\n\na) positive $$\\qquad$$b) negative $$\\qquad$$c) $$0 \\qquad$$ d) an integer\n\nI somehow managed to figure it out by completing the square but in order to do so, it took me a lot of time and I'm not sure if every time I could solve such problems.\n\nThis whole expression can be written as: $$2(x - 2y)^2 + (x - 2)^2 + (y + 3)^2$$ which implies $$M$$ is positive.\n\nMy point is sometimes I'm lucky and I could group them in squares but other times not. Is there any particular technique/method which always works?\n\nSecondly I also wanna know what you guys observe when completing the squares?\n\n\u2022 In my opinion, the key term that must be focused on first is the $8xy$ term. This is because there seems to be wiggle room everywhere else, re raising/lowering the $x^2$ or $y^2$ terms. So, the 1st try should be $(ax + by)^2$ where $2ab = 8.$ Then, try to make everything fit around that. Sep 16 at 8:47\n\u2022 While you should definitely solidify your ability to pick an approach for the hard math, there is a parallel part of any math problem (especially when applied with units in a real world scenario) that would completely answer this question. You should always ask Does my answer make sense? and that often means getting a rough idea in your head of pos/neg and/or order of magnitude. Use it to validate any calculations. In this case options c and d don't make sense (x=0.123, y=0.357). Plus c can't be true without d. Then plug in y=1, x=1 and you get 3 - 8 + 9 - 4 + 6 + 13 -> positive. Sep 16 at 17:10\n\n\\begin{align} &3x^2 - 4x(2y+1)+ (9y^2 + 6y + 13-M)=0\\\\ \\implies &\\Delta_x=4(2y+1)^2-3(9y^2+6y+13-M)\u22650\\\\ \\implies &3M\u226511y^2+2y+35\\\\ \\implies &3M\u226511 \\left(y + \\frac{1}{11}\\right)^2 + \\frac{384}{11}\\\\ \\implies &3M\u2265\\frac{384}{11}\\\\ \\implies &M\u2265\\frac{128}{11}>0.\\end{align}\n\u2022 Ah nice, $\\Delta_x$ is the discriminant of the quadratic in $x$. Nice idea\n\u2022 @Navdeep To optimize the polynomial, the roots are considered real. If we are talking about complex numbers, then $M=0$ can be and $M > 0$ can also be. It can be $M<0$ too. So your question loses its meaning. Notice, when completing the square you assumed that $x,y$ were real. Therefore, I would recommend adding to your question that $x$ and $y$ are real numbers. Otherwise, the polynomial cannot be optimized. Sep 16 at 15:52", "date": "2021-10-20 03:22:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4251809/how-do-you-approach-when-completing-the-square", "openwebmath_score": 0.6983191967010498, "openwebmath_perplexity": 444.69618467582444, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682476372385, "lm_q2_score": 0.9019206765295399, "lm_q1q2_score": 0.8917003347722529}}
{"url": "http://mathhelpforum.com/calculus/43609-notation-again-again.html", "text": "1. ## Notation again again\n\nI do not know how to search this but does\n\n$C^{\\infty}$ mean that is differentiable any number of times, or something?\n\nAny help appreciated.\n\n2. Originally Posted by Mathstud28\nI do not know how to search this but does\n\n$C^{\\infty}$ mean that is differentiable any number of times, or something?\n\nAny help appreciated.\nLet $I$ be an open interval. The notation $\\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\\mathcal{C}^{k}$ for any $k\\geq 1$ then we say it $\\mathcal{C}^{\\infty}(I)$. For example, $\\exp ,\\sin , \\cos \\in \\mathcal{C}^{\\infty}(\\mathbb{R})$. While $\\log \\in \\mathcal{C}^{\\infty}(0,\\infty)$.\n\n3. Originally Posted by ThePerfectHacker\nLet $I$ be an open interval. The notation $\\mathcal{C}^k(I)$ means a function differenciable $k$ times on $I$ such that $f^{(k)}$ is a continous function on $I$. However, if the function is infinitely differenciable i.e. it is an element of $\\mathcal{C}^{k}$ for any $k\\geq 1$ then we say it $\\mathcal{C}^{\\infty}(I)$. For example, $\\exp ,\\sin , \\cos \\in \\mathcal{C}^{\\infty}(\\mathbb{R})$. While $\\log \\in \\mathcal{C}^{\\infty}(0,\\infty)$.\nThanks TPH! Always answering my notation questions .\n\nOne question though, I understand that $e^x,\\cos(x),\\text{etc.}$ are $C^{\\infty}\\left(\\mathbb{R}\\right)$\n\nBut since $C^{\\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\\ln(x)$ should be $\\mathbb{R}/\\left\\{0\\right\\}$ since $\\ln^{\\left(n\\in\\mathbb{Z^+}\\right)}(x)$ is continous for that interval?\n\nOh wait, I looked back and the interval $I$ is specified in relation to the original function, not the derivatives. Ok so I get it.\n\nThank you.\n\nEDIT: Actually two more things. The first is obviously that\n\nIf $p(x)$ is a non-descript polynomial then it is $C^{\\infty}$ Right?\n\nAlso how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.\n\nLike this?\n\n\" $f(x)$ is $C^{\\infty}(a,b)$\"?\n\n4. Originally Posted by Mathstud28\nBut since $C^{\\text{whatever}}$ seems to only be talking about the at least first derivative wouldnt that mean that the $I$ specified for $\\ln(x)$ should be $\\mathbb{R}/\\left\\{0\\right\\}$ since $\\ln^{\\left(n\\in\\mathbb{Z^+}\\right)}(x)$ is continous for that interval?\nIt does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\\mathbb{R} - \\{ 0 \\} = (-\\infty,0)\\cup (0,\\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\\in S$ there is $\\epsilon > 0$ such that $(x-\\epsilon,x+\\epsilon) \\subset S$. With open sets we can say that $\\ln |x|$ is $\\mathcal{C}^{\\infty}(\\mathbb{R} - \\{ 0\\})$.\n\nIf $p(x)$ is a non-descript polynomial then it is $C^{\\infty}$ Right?\nA polynomial is infinitely differenciable so it is $\\mathcal{C}^{\\infty}$.\n\nAlso how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.\n\nLike this?\n\n\" $f(x)$ is $C^{\\infty}(a,b)$\"?\nThat is exactly how we say it.\n\nAs an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\\mathcal{C}^k(a,b)$. Define $F(x) = \\smallint_a^x f$. Then $F$ is $\\mathcal{C}^{k+1}(a,b)$.\n\nAnalysts like to say that integration \"smoothens out a function\". The theorem above is what this is all about.\n\n5. Originally Posted by ThePerfectHacker\nIt does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, $\\mathbb{R} - \\{ 0 \\} = (-\\infty,0)\\cup (0,\\infty)$ is an open set. Basically an open set is such a set that has no boundary. More formally $S$ is open iff for any $x\\in S$ there is $\\epsilon > 0$ such that $(x-\\epsilon,x+\\epsilon) \\subset S$. With open sets we can say that $\\ln |x|$ is $\\mathcal{C}^{\\infty}(\\mathbb{R} - \\{ 0\\})$.\n\nA polynomial is infinitely differenciable so it is $\\mathcal{C}^{\\infty}$.\n\nThat is exactly how we say it.\n\nAs an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let $f$ be continous on $[a,b]$ which is $\\mathcal{C}^k(a,b)$. Define $F(x) = \\smallint_a^x f$. Then $F$ is $\\mathcal{C}^{k+1}(a,b)$.\n\nAnalysts like to say that integration \"smoothens out a function\". The theorem about is what this is all about.\nOk, amazing, I completely understand that.\n\nLike the last part since $F'(x)=\\frac{d}{dx}\\int_a^{x}f(t)dt=f(x)$ by the fundamental theorem of calculus and we have stated that $f(x)\\quad\\text{is}\\quad{C^{k}}$\n\n$F(x)$ takes on all the k amount of differentiabilities of $f(x)$ and adds more since we know that if it is integrable it is differentiable back to f(x)!\n\nThanks TPH\n\n6. Originally Posted by Mathstud28\n\nAlso how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.\n\nLike this?\n\n\" $f(x)$ is $C^{\\infty}(a,b)$\"?\nYes, but also:\n\n$f \\in C^{\\infty}(a,b)$\n\nRonL\n\n7. Originally Posted by CaptainBlack\nYes, but also:\n\n$f \\in C^{\\infty}(a,b)$\n\nRonL\nThanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).\n\nBut by the way, this got me thinking, this notating would be nice to say things like,\n\n\"Let f be a eight times differentiable function\" So instead of that could you say\n\n\"Let $f(x)\\in{C^{k\\geq{8}}}$ \"\n\nAlso, can it apply to functions where $\\mathbb{R}^{n>1}\\to\\mathbb{R}$\n\nCan I say\n\n$C_x^{\\infty}\\left(\\mathbb{R}\\right)$?\n\nOr would no one know what that means, haha, I think I am inventing notation here.\n\n8. Originally Posted by Mathstud28\nThanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).\n\nBut by the way, this got me thinking, this notating would be nice to say things like,\n\n\"Let f be a eight times differentiable function\" So instead of that could you say\n\n\"Let $f(x)\\in{C^{k\\geq{8}}}$ \"\nSince:\n\n$C^8 \\supset C^9 \\supset C^{10} \\supset ...$\n\nthere is no need for this notation.\n\nAlso $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative.\n\nRonL\n\n9. Originally Posted by Mathstud28\n...\n\nBut by the way, this got me thinking, this notating would be nice to say things like,\n\n\"Let f be a eight times differentiable function\" So instead of that could you say\n\n\"Let $f(x)\\in{C^{k\\geq{8}}}$ \"\nnot really, because there is another condition: $f^{(8)}(x)$ must also be continuous.\n\nso i think, $f(x)\\in{C^{k\\geq{8}}}$ would mean $f^{(k\\geq8)}(x)$ must also be continuous. so y don't you get the biggest k such that $f^k(x)$ is continuous with f is k times differentiable..\n\nOriginally Posted by Mathstud28\nAlso, can it apply to functions where $\\mathbb{R}^{n>1}\\to\\mathbb{R}$\nyes, like what TPH said, it can be defined in any open set, just define the open set in $\\mathbb{R}^{n}$\n\nOriginally Posted by Mathstud28\nCan I say\n\n$C_x^{\\infty}\\left(\\mathbb{R}\\right)$?\n\nOr would no one know what that means, haha, I think I am inventing notation here.\nhaven't seen this notation before..\n\n10. Originally Posted by CaptainBlack\nSince:\n\n$C^8 \\supset C^9 \\supset C^{10} \\supset ...$\n\nthere is no need for this notation.\n\nAlso $C^8$ and the set of all functions eight times differentiable (on $(a,b)$ or whatever) are not the same thing since $C^8$ is the set of all $8$ times differentiable function with continuous $8$-th derivative.\n\nRonL\nSo what your saying is that since for a function to be $f(x)\\in{C^{k\\geq{8}}}$ it must be $f(x)\\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.\n\nSo your saying that $f(x)\\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\\cdots$ times?\n\n11. if $f \\in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous..\n\n$f \\in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous..\n\nnow, let $g \\in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist.\n\ntherefore, $g \\in C^{n}$, that is $C^{n+1} \\subset C^{n}$\n\n12. Originally Posted by kalagota\nif $f \\in C^{n}$, then $f, f', f'', ..., f^{(n)}$ exists and $f^{(n)}$ is continuous..\n\n$f \\in C^{n+1}$ means $f, f', f'', ..., f^{(n)}, f^{(n+1)}$ exists and $f^{(n+1)}$ is continuous..\n\nnow, let $g \\in C^{n+1}$. therefore, $g^{(n)}$ exists. $g^{(n)}$ must be continuous otherwise, $g^{(n)}$ will not be differentiable, hence $g^{(n+1)}$ will not exist.\n\ntherefore, $g \\in C^{n}$, that is $C^{n+1} \\subset C^{n}$\n\nSo the operative phrasing here would be that $f(x)\\in{C^{n}}$ implies that $f(x)$ is differentiable at least an n number of times, not at most.\n\n13. yes.\n\n14. Originally Posted by Mathstud28\nSo what your saying is that since for a function to be $f(x)\\in{C^{k\\geq{8}}}$ it must be $f(x)\\in{C^8}$ or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.\n\nSo your saying that $f(x)\\in{C^{n}}$ describes all functions who are differentiable at least n times, so this would include all functions that are differentiable $n+1,n+2,n+3,\\cdots$ times?\nSomething differentiable $m$ times with continuous $m$ th derivative is obviously continuously differentiable $n$-times for all $n \\le m$, so $C^{m} \\subset C^n, \\ \\forall n \\le m.$\n\nRonL\n\n15. I will give two examples at my attempt to use this notation and please tell me if it is correct.\n\nSay we are talking about L'hopital's and we are saying\n\n\" Let $\\lim_{x\\to{c}}\\frac{f(x)}{g(x)}$ be indeterminate because either $\\lim_{x\\to{c}}f(x)=\\lim_{x\\to{c}}g(x)=0$ or $\\lim_{x\\to{c}}f(x)=\\lim_{x\\to{c}}g(x)=\\infty$\n\nFurthermore let $f(x)\\in\\mathcal{C}^1(c)$ and $g(x)\\in\\mathcal{C}^1(c)$ and then blah blah blah\"\n\nAnd in Rolle's Theroem\n\n\"Let $f(x)$ be continuous on $[a,b]$ and $f(x)\\in\\mathcal{C}^1(a,b)$ then ....\"\n\nDid I use it right?\n\nPage 1 of 2 12 Last", "date": "2017-09-25 05:24:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/43609-notation-again-again.html", "openwebmath_score": 0.9507019519805908, "openwebmath_perplexity": 307.38110491630493, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98866824479921, "lm_q2_score": 0.9019206719160033, "lm_q1q2_score": 0.8917003276513191}}
{"url": "https://www.physicsforums.com/threads/integrating-the-area-of-a-disc.222062/", "text": "Integrating the area of a disc\n\n1. Mar 14, 2008\n\nHakins90\n\n1. The problem statement, all variables and given/known data\n\nFinding the area of a disc by integration of rings.\n\n2. Relevant equations\n\nA ring of radius r and thickness dr has an area of $$2 \\pi rdr$$.\n\n3. The attempt at a solution\n\nWhy isn't it $$\\pi (2rdr + (dr)^2)$$?\n\n2. Mar 14, 2008\n\ntiny-tim\n\nI don't understand - why do you think it might be?\n\nWhat would $$\\pi (dr)^2$$ represent?\n\n3. Mar 14, 2008\n\nDick\n\nThe exact area of the ring is pi*((r+dr)^2-r^2)=pi(2*r*dr+dr^2), indeed. But in the integration we are taking the limit of a sum of these where dr->0. The limit of the terms involving dr^2 will go to zero. You can ignore corrections involving higher powers of 'infinitesimals' like dr.\n\n4. Mar 14, 2008\n\nHakins90\n\nOh dear... i posted too quickly.\n\nI just realised why it isn't.\n\nMy old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference.\n\n$$A = \\pi (r+dr)^2 - \\pi r^2 = \\pi (r^2 + 2rdr + (dr)^2 - r^2) = \\pi (2rdr + (dr)^2)$$\n\nI now realise that the ring's radius is measured from the centre of the width of the ring.\n\nso\n$$A = \\pi (r+ \\frac{1}{2} dr)^2 - \\pi (r- \\frac{1}{2} dr)^2 =\\pi (r^2 +rdr + \\frac{1}{4} (dr)^2 - r^2 + rdr - \\frac{1}{4} (dr)^2) = 2 \\pi rdr$$\n\nEDIT: Oh you posted before i saw... We both have different reasons. Who is right?\n\n5. Mar 14, 2008\n\nDick\n\nIf r and dr are the same, then those are the areas of two different rings. If dr is a finite number then you'd better pay attention to which is which. But,either calculation works for integration once you ignore higher powers of dr.\n\n6. Mar 15, 2008\n\nThis reasoning is correct, in my opinion.\nWhen I was in school, the explanation I got from my teacher was that while doing integration we add infinitesimally small quantities. The quantity dr is very small and so $$dr^2$$ will be even smaller......like $$0.000001^2=0.000000000001$$ , so we neglect the $$dr^2$$ thing from that expression.\n\n7. Mar 15, 2008\n\nHallsofIvy\n\nStaff Emeritus\nThe simplest way to do the original problem is to say that the area of a small ring, of inner radius r and thickness dr is approximately $2\\pi r$, the length of the ring, times dr, the thickness. That would be exactly correct if it were a rectangle of length $2\\pi r$ and width dr. The point is that in the limit, as we change from Riemann sum to integral, that \"approximation\" becomes exact (the \"$dr^2$ in your and Dick's reasoning) goes to 0 : the area is given by $\\int \\pi r dr$.\n\nCan someone explain to my why this is a physics problem and not a mathematics problem?", "date": "2016-10-25 17:28:59", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integrating-the-area-of-a-disc.222062/", "openwebmath_score": 0.9237654209136963, "openwebmath_perplexity": 708.4283645411392, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979976559880137, "lm_q2_score": 0.9099070109242131, "lm_q1q2_score": 0.8916875423763286}}
{"url": "http://kinmenhouse.com/pesticides-in-ktvhmwj/how-to-prove-a-function-is-continuous-7db71d", "text": "The function is defined at a.In other words, point a is in the domain of f, ; The limit of the function exists at that point, and is equal as x approaches a from both sides, ; The limit of the function, as x approaches a, is the same as the function output (i.e. The Solution: We must show that $\\lim_{h \\to 0}\\cos(a + h) = \\cos(a)$ to prove that the cosine function is continuous. Which of the following two functions is continuous: If f(x) = 5x - 6, prove that f is continuous in its domain. Once certain functions are known to be continuous, their limits may be evaluated by substitution. A function is said to be differentiable if the derivative exists at each point in its domain. We can define continuous using Limits (it helps to read that page first):. limx\u2192c f(x) = f(c) \"the limit of f(x) as x approaches c equals f(c)\" The limit says: The following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions are continuous everywhere. A function f is continuous when, for every value c in its Domain:. Transcript. The following theorem is very similar to Theorem 8, giving us ways to combine continuous functions to create other continuous functions. Learn how to determine the differentiability of a function. To give some context in what way this must be answered, this question is from a sub-chapter called Continuity from a chapter introducing Limits. The question is: Prove that cosine is a continuous function. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite ( i.e. Rather than returning to the $\\varepsilon$-$\\delta$ definition whenever we want to prove a function is continuous at a point, we build up our collection of continuous functions by combining functions we know are continuous: The function value and the limit aren\u2019t the same and so the function is not continuous at this point. If f(x) = x if x is rational and f(x) = 0 if x is irrational, prove that f is continuous \u2026 More formally, a function (f) is continuous if, for every point x = a:. When a function is continuous within its Domain, it is a continuous function.. More Formally ! As @user40615 alludes to above, showing the function is continuous at each point in the domain shows that it is continuous in all of the domain. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain. Proofs of the Continuity of Basic Algebraic Functions. To show that $f(x) = e^x$ is continuous at $x_0$, consider any $\\epsilon>0$. Using the Heine definition, prove that the function $$f\\left( x \\right) = {x^2}$$ is continuous at any point $$x = a.$$ Solution. $\\endgroup$ \u2013 Jeremy Upsal Nov 9 '13 at 20:14 $\\begingroup$ I did not consider that when x=0, I had to prove that it is continuous. But in order to prove the continuity of these functions, we must show that $\\lim\\limits_{x\\to c}f(x)=f(c)$. This kind of discontinuity in a graph is called a jump discontinuity . f(c) is defined, and. the y-value) at a.; Order of Continuity: C0, C1, C2 Functions THEOREM 102 Properties of Continuous Functions Let $$f$$ and $$g$$ be continuous on an open disk $$B$$, let $$c$$ \u2026 Using the Heine definition we can write the condition of continuity as follows: Consider an arbitrary $x_0$. Let \ufdd0\ufdef = tan\u2061 \ufdd0\ufdef = \ufdd0\ufdd0sin\ufdee\ufdef\ufdee\ufdd0cos\ufdee\ufdef\ufdef is defined for all real number except cos\u2061 = 0 i.e. Example 18 Prove that the function defined by f (x) = tan x is a continuous function. Limit aren \u2019 t the same and so the function is not continuous at this point is said be! Defined by f ( x ) = tan x is a continuous function.. formally... Each point in its Domain, it is a continuous function.. more formally, a function is within... May be evaluated by substitution when a function is said to be,... 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A graph is called a jump discontinuity if, for every value c in its Domain::. Each point in its Domain: be continuous, their limits may be evaluated by substitution = \ufdd0\ufdef!, which you should have seen proved, and should perhaps prove yourself: Constant functions known... Is defined for all real number except cos\u2061 = 0 i.e if, for every value in. Point x = a: = 0 i.e continuous within how to prove a function is continuous Domain, it is continuous... X_0 [ /math ] ( f ) is continuous if, for every c... Every point x = a: a function is continuous if, for every value c in Domain!, which you should have seen proved, and should perhaps prove yourself: functions! We can define continuous using limits ( it how to prove a function is continuous to read that page first:! Value and the limit aren \u2019 t the same and so the function value and the limit \u2019! A function is not continuous at this point it is a continuous function c in its,...", "date": "2021-07-28 19:54:34", "meta": {"domain": "kinmenhouse.com", "url": "http://kinmenhouse.com/pesticides-in-ktvhmwj/how-to-prove-a-function-is-continuous-7db71d", "openwebmath_score": 0.9315921068191528, "openwebmath_perplexity": 510.3377641586615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9840936092211993, "lm_q2_score": 0.9059898279984214, "lm_q1q2_score": 0.8915787997526601}}
{"url": "https://s7097.gridserver.com/7k6tqu/3a6bd0-decomposition-of-antisymmetric-tensor", "text": "The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. Polon. This is exactly what you have done in the second line of your equation. CHAPTER 1. Decomposition of tensor power of symmetric square. 1.4) or \u00ce\u00b1 (in Eq. \u00e2\u0086\u0092 What symmetry does represent?Kenta OONOIntroduction to Tensors Cartan tensor is equal to minus the structure coe\u00ef\u00ac\u0083cients. The trace decomposition theory of tensor spaces, based on duality, is presented. MT = \u00e2\u0088\u0092M. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. It is a real tensor, hence f \u00ce\u00b1\u00ce\u00b2 * is also real. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. For N>2, they are not, however. The result is An alternating form \u00cf\u0086 on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. P i A ii D0/. The N-way Toolbox, Tensor Toolbox, \u00e2\u0080\u00a6 Yes. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. A tensor is a linear vector valued function defined on the set of all vectors . Thus, the rank of Mmust be even. tensor M and a partially antisymmetric tensors N is often used in the literature. The alternating tensor can be used to write down the vector equation z = x \u00d7 y in su\u00ef\u00ac\u0083x notation: z i = [x\u00d7y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 \u00e2\u0088\u0092x 3y 2, as required.) Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. In section 3 a decomposition of tensor spaces into irreducible components is introduced. We begin with a special case of the definition. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric \u00e2\u0080\u00a6 Antisymmetric and symmetric tensors. 1.5) are not explicitly stated because they are obvious from the context. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Algebra is great fun - you get to solve puzzles! There is one very important property of ijk: ijk klm = \u00ce\u00b4 il\u00ce\u00b4 jm \u00e2\u0088\u0092\u00ce\u00b4 im\u00ce\u00b4 jl. Ask Question Asked 2 years, 2 months ago. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. This means that traceless antisymmetric mixed tensor $\\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Properties of antisymmetric matrices Let Mbe a complex d\u00d7 dantisymmetric matrix, i.e. This makes many vector identities easy to prove. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric \u00e2\u0080\u00a6 : Lehigh Univ., Bethlehem, Penna. Decomposition. In these notes, the rank of Mwill be denoted by 2n. THE INDEX NOTATION \u00ce\u00bd, are chosen arbitrarily.The could equally well have been called \u00ce\u00b1 and \u00ce\u00b2: v\u00e2\u0080\u00b2 \u00ce\u00b1 = n \u00e2\u0088\u0091 \u00ce\u00b2=1 A\u00ce\u00b1\u00ce\u00b2 v\u00ce\u00b2 (\u00e2\u0088\u0080\u00ce\u00b1 \u00e2\u0088\u0088 N | 1 \u00e2\u0089\u00a4 \u00ce\u00b1 \u00e2\u0089\u00a4 n). Google Scholar; 6. Vector spaces will be denoted using blackboard fonts. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. (1.5) Usually the conditions for \u00b5 (in Eq. Each part can reveal information that might not be easily obtained from the original tensor. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. : USDOE \u00e2\u0080\u00a6 A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. What's the significance of this further decomposition? If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The symmetry-based decompositions of finite games are investigated. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . [3] Alternating forms. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sponsoring Org. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Sci. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 A\u00c4\u00b1 ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. The bases of the symmetric subspace and those of its orthogonal complement are presented. Cl. A related concept is that of the antisymmetric tensor or alternating form. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Contents. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Symmetric tensors occur widely in engineering, physics and mathematics. This decomposition, ... ^2 indicates the antisymmetric tensor product. Since det M= det (\u00e2\u0088\u0092MT) = det (\u00e2\u0088\u0092M) = (\u00e2\u0088\u00921)d det M, (1) it follows that det M= 0 if dis odd. Use the Weyl decomposition \\eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \\eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). Viewed 503 times 7. By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. = \u00ce\u00b4 il\u00ce\u00b4 jm \u00e2\u0088\u0092\u00ce\u00b4 im\u00ce\u00b4 jl spin-0 singlett, while the subspace. ( antisymmetric ) spin-0 singlett, while the symmetric and antrisymmetric subspaces separate invariant subspaces meaning! Help in analyzing them, an antisymmetric tensor is equal to minus the coe\u00ef\u00ac\u0083cients... 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In 3 dimensions, that is not so spaces into irreducible components introduced... Tensor calculus we recom-mend [ 58, 99, 126, 197, 205 319. Properties of antisymmetric matrices Let Mbe a decomposition of antisymmetric tensor d\u00d7 dantisymmetric matrix,....: Publication Date: antisymmetric matrix ) are not, however tensor or form... M and a partially antisymmetric tensors N is often used in the.... Constraints from one contraction tensor spaces into irreducible components is introduced a tensor is symmetric, any contraction is rate... Symmetric tensors occur widely in engineering, physics and mathematics obvious from the context minus the structure coe\u00ef\u00ac\u0083cients of... Lodge, in the second line of your equation,... ^2 the! The antisymmetric tensor or alternating form Youla decomposition of a complex d\u00d7 dantisymmetric matrix, i.e EDT. What you have done in the literature a decomposition of a complex dantisymmetric... Symmetries of the antisymmetric tensor is a linear vector valued function defined on the set of vectors...", "date": "2022-06-26 05:56:51", "meta": {"domain": "gridserver.com", "url": "https://s7097.gridserver.com/7k6tqu/3a6bd0-decomposition-of-antisymmetric-tensor", "openwebmath_score": 0.9046940803527832, "openwebmath_perplexity": 1231.6727965524285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9869795075882267, "lm_q2_score": 0.9032942138630786, "lm_q1q2_score": 0.8915328784058757}}
{"url": "http://codeforces.com/problemset/problem/1270/C", "text": "Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. \u00d7\n\nC. Make Good\ntime limit per test\n2 seconds\nmemory limit per test\n256 megabytes\ninput\nstandard input\noutput\nstandard output\n\nLet's call an array $a_1, a_2, \\dots, a_m$ of nonnegative integer numbers good if $a_1 + a_2 + \\dots + a_m = 2\\cdot(a_1 \\oplus a_2 \\oplus \\dots \\oplus a_m)$, where $\\oplus$ denotes the bitwise XOR operation.\n\nFor example, array $[1, 2, 3, 6]$ is good, as $1 + 2 + 3 + 6 = 12 = 2\\cdot 6 = 2\\cdot (1\\oplus 2 \\oplus 3 \\oplus 6)$. At the same time, array $[1, 2, 1, 3]$ isn't good, as $1 + 2 + 1 + 3 = 7 \\neq 2\\cdot 1 = 2\\cdot(1\\oplus 2 \\oplus 1 \\oplus 3)$.\n\nYou are given an array of length $n$: $a_1, a_2, \\dots, a_n$. Append at most $3$ elements to it to make it good. Appended elements don't have to be different. It can be shown that the solution always exists under the given constraints. If there are different solutions, you are allowed to output any of them. Note that you don't have to minimize the number of added elements!. So, if an array is good already you are allowed to not append elements.\n\nInput\n\nEach test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \\le t \\le 10\\,000$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $n$ $(1\\le n \\le 10^5)$\u00a0\u2014 the size of the array.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($0\\le a_i \\le 10^9$)\u00a0\u2014 the elements of the array.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.\n\nOutput\n\nFor each test case, output two lines.\n\nIn the first line, output a single integer $s$ ($0\\le s\\le 3$)\u00a0\u2014 the number of elements you want to append.\n\nIn the second line, output $s$ integers $b_1, \\dots, b_s$ ($0\\le b_i \\le 10^{18}$)\u00a0\u2014 the elements you want to append to the array.\n\nIf there are different solutions, you are allowed to output any of them.\n\nExample\nInput\n3\n4\n1 2 3 6\n1\n8\n2\n1 1\n\nOutput\n0\n\n2\n4 4\n3\n2 6 2\n\nNote\n\nIn the first test case of the example, the sum of all numbers is $12$, and their $\\oplus$ is $6$, so the condition is already satisfied.\n\nIn the second test case of the example, after adding $4, 4$, the array becomes $[8, 4, 4]$. The sum of numbers in it is $16$, $\\oplus$ of numbers in it is $8$.", "date": "2020-07-05 01:35:36", "meta": {"domain": "codeforces.com", "url": "http://codeforces.com/problemset/problem/1270/C", "openwebmath_score": 0.4329439699649811, "openwebmath_perplexity": 465.394634925087, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964207345894, "lm_q2_score": 0.9046505312448598, "lm_q1q2_score": 0.8915298605574542}}
{"url": "https://math.stackexchange.com/questions/939280/part-of-a-proof-that-the-product-of-an-odd-and-even-integers-is-even", "text": "# Part of a proof that the product of an odd and even integers is even\n\nI'm practicing for a test on Monday and I'm trying to do some proofs - but I'm not entirely sure if this is sufficient enough for the question.\n\n\"Prove that for all integers, m and n, if m is odd and n is even, then $m\\cdot n$ is even.\"\n\nThese are the steps I've gone through - but I'm a little unsure on the result.\n\n1. $m = 2k_1+1$ and $n = 2k_2$\n2. $m\\cdot n= (2k_1 + 1)(2k_2)$\n3. $= 4k_1k_2+2k_2$\n4. $=2(2k_1k_2+k_2)$\n\nThe part I'm unsure on is the last step, I wasn't sure if I was supposed to factor out the $k_2$ as well.\n\n\u2022 That's fine. You could have jumped directly from step 2 to step 4. \u2013\u00a0Adriano Sep 20 '14 at 18:42\n\u2022 Apart from the fact that you used equations, with no explanatory words, everything is fine. We can even skip part of the multiplication step. We have $m=2a+1$ and $n=2b$ for some integers $a$ and $b$. Thus $mn=nm=(2b)(2a+1)=(2)\\left[(b)(2a+1)\\right]$, so $mn$ is even. \u2013\u00a0Andr\u00e9 Nicolas Sep 20 '14 at 18:45\n\n## 3 Answers\n\nYou've done all the work, and it's correct. Let me try to demonstrate how improve the style of your proof. This can be very important\u2014graders are invariably tired and overworked, and lack of clarity is one of the most frequent causes of correct proofs being marked as incorrect on exams.\n\nAdditionally, it feels good for both reader and writer when the math is crisp and clear\u2014it is an often-overlooked fact that the entire purpose of mathematical writing is to communicate ideas to other people.\n\nClaim: If $m$ and $n$ are integers, and $n$ is even, then $mn$ is even.\n\nProof: Because $n$ is even, we can write $n=2l$, where $l$ is an integer. Then $mn=m(2l)=2(ml)$ is a multiple of $2$. In other words, $mn$ is even.\n\nNotice that each step is justified\u2014though if we had an extremely strict teacher, we might want to point out that we got $mn=m(2l)$ by multiplying both sides of $n=2l$ by $m$, and that $m(2k)=2(ml)$ follows from the associativity and commutativity of multiplication. But you're probably safe in omitting these details unless you've been told otherwise.\n\nAlso note that I didn't assume that $m$ is odd, because this makes the equations a little more complicated. Here is a more faithful recreation of your proof:\n\nClaim: If $m$ is an odd integer and $n$ is an even integer, then $mn$ is even.\n\nProof: Because $m$ is odd and $n$ is even, we can write $m=2k+1$ and $n=2l$, where $k$ and $l$ are integers. Then $mn = (2k+1)(2l) = 2\\cdot(l(2k+1))$ is a multiple of $2$, hence even.\n\nDefine $2k_1k_2+k_2=k_3$ as an integer number.\n\nYou could further formalize your text, if you want (it is always handy to write down the definitions or the theorems you use, it also makes it easier for others to read and make it look more complete): By definition: $$m \\text{ even }\\iff \\exists k \\in \\mathbb Z \\text{ such that }2k = m$$", "date": "2021-01-23 08:34:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/939280/part-of-a-proof-that-the-product-of-an-odd-and-even-integers-is-even", "openwebmath_score": 0.7549272775650024, "openwebmath_perplexity": 81.18333235828224, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846621952493, "lm_q2_score": 0.9111797166446537, "lm_q1q2_score": 0.8914842592685425}}
{"url": "https://brilliant.org/discussions/thread/how-to-count-squares/", "text": "# How to Count Squares!\n\nLet me go grab a hamburger real quick...\n\nOk, I'm back.\n\nHow many squares are there in the 6$$\\times$$4 grid below?\n\nThat's a reaallly good question!\n\nLet's start by counting the smallest 1$$\\times$$1 squares, this is just the same as counting the number of unit squares in a 6$$\\times$$4 grid, there are $$6\\times4=24$$ 1 by 1 squares in the grid.\n\nNow let's move on to the 2$$\\times$$2 squares, notice that counting the number of 2$$\\times$$2 squares in a 6$$\\times$$4 grid is just the same as counting the number of unit squares in a 5$$\\times$$3 grid. So the number of 2 by 2 squares in the grid is $$5\\times3=15$$.\n\nNow the 3$$\\times$$3 squares, similarly, counting the number of 3$$\\times$$3 squares in a 6$$\\times$$4 grid is just the same as counting the number of unit squares in a 4$$\\times$$2 grid, which is $$4\\times2=8$$.\n\nAnd again the number of 4$$\\times$$4 squares in a 6$$\\times$$4 grid is equal to the number of unit squares in a 3$$\\times$$1 grid, which is $$3\\times1=3$$.\n\nAdd up all the number of squares together: $$24+15+8+3=50$$. Tada! We now have our answer! There are 50 squares in a 6$$\\times$$4 grid.\n\nMmm... the hamburger is really good...\n\nBack on topic, in general, what is the total number of squares in an $$a\\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\\geqslant b$$?\n\nAgain let's start from the 1$$\\times$$1 squares, that's trivial, there's $$ab$$ of them.\n\nNow moving on to the 2$$\\times$$2 squares, the number of 2$$\\times$$2 squares in an $$a\\times b$$ grid is equal to the number of unit squares in an $$(a-1)\\times(b-1)$$ grid.\n\nNotice the pattern? Counting the number of $$n\\times n$$ squares in an $$a\\times b$$ grid is the same as counting the number of unit squares in an $$(a-n+1)\\times(b-n+1)$$ grid.\n\nThe largest square that can contain in an $$a\\times b$$ grid given that $$a\\geqslant b$$ is $$b\\times b$$.\n\nHence, the total number of squares in an $$a\\times b$$ grid is $ab+(a-1)(b-1)+(a-2)(b-2)+\\ldots+[a-(b-2)][b-(b-2)]+[a-(b-1)][b-(b-1)]$ Or $\\sum_{i=0}^{b-1}{(a-i)(b-i)}$\n\nThis is ugly, we don't like sigma symbols sitting around, so why not we simplify this a little bit...\n\n\\begin{align} \\sum_{i=0}^{b-1}{(a-i)(b-i)}&=\\sum_{i=0}^{b-1}{[ab-(a+b)i+i^2]} \\\\&=ab^2-\\frac{(a+b)b(b-1)}{2}+\\frac{b(b-1)(2b-1)}{6} \\\\&=b\\left[ab-\\frac{ab-a+b^2-b}{2}+\\frac{2b^2-3b+1}{6}\\right] \\\\&=\\frac{b}{6}\\left[6ab-3ab+3a-3b^2+3b+2b^2-3b+1\\right] \\\\&=\\frac{b}{6}\\left[3ab+3a-b^2+1\\right] \\\\&=\\frac{b(b+1)(3a-b+1)}{6} \\end{align} BOOM! There we have it! *Round of applause* *Fireworks* *Pancakes*\n\nThe total number of squares in an $$a\\times b$$ grid (where $$a$$ is the width of the grid and $$b$$ is the height of the grid), given $$a\\geqslant b$$ is $\\frac{b(b+1)(3a-b+1)}{6}$\n\nIf $$a<b$$, then we just swap the $$a$$ and $$b$$ around.\n\nSpecial case: If $$a=b$$, the above equation becomes $\\frac{a(a+1)(2a+1)}{6}$ which is the formula of the sum of squares from 1 to $$a$$.\n\nDone! Now let me finish my burger...\n\nThis is one part of Quadrilatorics.\n\nNote by Tan Kenneth\n2\u00a0years, 1\u00a0month ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nYou should add this to the Brilliant wiki. Great note!\n\n- 2\u00a0years, 1\u00a0month ago\n\nCool! Thanks for that note bro, you're awesome!\n\n- 2\u00a0years, 1\u00a0month ago\n\nHope you finished your burger peacefully :P\n\n- 2\u00a0years, 1\u00a0month ago\n\nThanks, I'm glad you like the note. Well unfortunately, I think my hamburger has become stale. XD\n\n- 2\u00a0years, 1\u00a0month ago\n\nNice simple way of explaining complex situation. So many thanks.\n\n- 1\u00a0year, 11\u00a0months ago\n\nare you a robot, cos I need some real friends? Humanity is a lie, the computer generation is upon us. Support the cause m64^(1/2)\n\n- 2\u00a0years, 1\u00a0month ago\n\nNo, i am 100.1% sure I'm not a robot.\n\n- 2\u00a0years, 1\u00a0month ago\n\nWhat a note @Tan Kenneth:)\n\n- 2\u00a0years, 1\u00a0month ago\n\nHEY tankenneth, you hyped?\n\n- 2\u00a0years, 1\u00a0month ago\n\nOh yes I am! $$1+1=3$$\n\n- 2\u00a0years, 1\u00a0month ago\n\n\u062c\u0645\u064a\u0644\u0629\n\n- 2\u00a0years, 1\u00a0month ago\n\nTranslation: beautiful!\n\n- 2\u00a0years, 1\u00a0month ago", "date": "2018-04-25 08:50:58", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/how-to-count-squares/", "openwebmath_score": 0.9701625108718872, "openwebmath_perplexity": 1891.37456295057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9898303413461358, "lm_q2_score": 0.9005297841157157, "lm_q1q2_score": 0.8913717036036208}}
{"url": "https://math.stackexchange.com/questions/2892375/prove-sum-i-0n-1n-i-binomn1i-i1n-n2n", "text": "# Prove $\\sum_{i=0}^n (-1)^{n-i} \\binom{n+1}{i} (i+1)^n = (n+2)^n$\n\nI found through simulations that\n\n$$\\sum_{i=0}^n (-1)^{n-i} \\binom{n+1}{i} (i+1)^n = (n+2)^n$$\n\nIs there any proof of this? I've tried to solve it by:\n\n\u2022 Induction, but it gets too messy.\n\u2022 Binomial theorem, but I got nowhere.\n\nAny help with this is highly appreciated.\n\n\u2022 I bet induction is not so bad in this case, but I'll try to find a more elegant explanation Aug 23, 2018 at 17:34\n\u2022 Thanks @Exodd .. Would love to see it through induction. But any other method is also okay. Aug 23, 2018 at 17:35\n\u2022 Notice that you can rewrite it as $$\\sum_{i=0}^{n+1} (-1)^{n-i} \\binom{n+1}{i} i^{n} =0$$ Aug 23, 2018 at 18:01\n\nLet $[m]$ denote the set of the first $m$ positive integers.\n\nLet $S$ be the set of functions from $[n]\\to [n+2]$, and for $1\\le i \\le n+1$, let $S_i$ be the set of such functions whose range does not contain $i$. Obviously, $|S|=(n+2)^n$. Furthermore, the range of a function in $S$ will not contain at least two elements, since the range has at most $n$ elements. Therefore, some number $1\\le i\\le n+1$ must be missing from the range, so $$S=\\bigcup_{i=1}^{n+1} S_i$$ Therefore, we can use the principle of inclusion-exclusion to count $|S|$ in terms the sizes of the $k$-way intersections $|S_{i_1}S_{i_2}\\dots S_{i_k}|$. For any $k$, the intersection of $k$ of the sets $S_i$ consists of functions whose range does not contain a particular $k$ elements. The number of such functions is $(n+2-k)^n$. Therefore, using the inclusion-exclusion formula, $$|S|=(n+2)^n = \\sum_{k=1}^{n+1} (-1)^{k+1}\\binom{n+1}k(n+2-k)^n$$ The result follows by reversing the order of the summation, i.e. setting $k\\leftarrow n+1-i$.\n\nPosting a second answer because there is a completely different, more direct proof.\n\nNote that subtracting the $(n+2)^n$ on the right over, this is equivalent to proving $$\\sum_{i=0}^{n+1} (-1)^{n-i}\\binom{n+1}i(i+1)^n\\stackrel{?}=0$$ which after multiplying both sides by $(-1)^n$ equals $$\\sum_{i=0}^{n+1} (-1)^{i}\\binom{n+1}i(i+1)^n \\stackrel{?}=0\\tag{1}$$ To prove this, consider the polynomial $$\\sum_{i=0}^{n+1} \\binom{n+1}i(i+1)^nx^i \\tag{2}$$ I claim that $(2)$ is the result of taking the polynomial $$\\sum_{i=0}^{n+1} \\binom{n+1}ix^i\\tag{3}$$ and performing the following operation $n$ times: multiply the polynomial by $x$, then differentiate. Indeed, each summand of a polynomial looks like $a_i x^i$, and when you multiply by $x$ and differentiate, the result is $(i+1)a_ix^i$. Doing this $n$ times results in $(i+1)^na_ix^i$.\n\nBut the polynomial in $(3)$ is by the binomial theorem equal to $(1+x)^{n+1}$. Note that this has a zero of order $n+1$ at $x=-1$. Multiplying by $x$ does not change this. It is a standard result that if $f(x)$ has a zero of order $k$ at $x_0$, then $f'(x)$ has a zero of order $k-1$ at $x_0$. Therefore, since $(3)$ has a zero of order $n+1$, it follows that after $n$ differentiations (and some multiplications by $x$), it will still have a zero of order $1$. In particular, $(2)$ has a zero at $x=-1$, so the expression in $(1)$ equals zero.\n\n\u2022 This is outstanding .. Thank you for the clear clarification. I loved both answers. Aug 23, 2018 at 18:37\n\u2022 Nice answer (+1). May 18, 2019 at 8:17\n\nStart from\n\n$$\\sum_{q=0}^n (-1)^{n-q} {n+1\\choose q} (q+1)^n \\\\ = \\sum_{q=0}^n (-1)^{n-q} {n+1\\choose q} n! [z^n] \\exp((q+1)z) \\\\ = n! [z^n] \\exp(z) \\sum_{q=0}^n (-1)^{n-q} {n+1\\choose q} \\exp(qz) \\\\ = - n! [z^n] \\exp(z) \\times (-1) \\exp((n+1)z) \\\\ + n! [z^n] \\exp(z) \\sum_{q=0}^{n+1} (-1)^{n-q} {n+1\\choose q} \\exp(qz) \\\\ = n! [z^n] \\exp((n+2)z) \\\\ - n! [z^n] \\exp(z) \\sum_{q=0}^{n+1} (-1)^{n+1-q} {n+1\\choose q} \\exp(qz) \\\\ = (n+2)^n \\\\ - n! [z^n] \\exp(z) (\\exp(z)-1)^{n+1}.$$\n\nNow $\\exp(z)-1 = z + \\cdots$ and hence $(\\exp(z)-1)^{n+1} = z^{n+1} +\\cdots$ and therefore\n\n$$[z^n] \\exp(z) (\\exp(z)-1)^{n+1} = 0$$\n\nand we are left with\n\n$$\\bbox[5px,border:2px solid #00A000]{ (n+2)^n.}$$\n\nThis is a less clever algebraic proof by induction. We actually prove something stronger:\n\nFor any $\\lambda>0$, we have $$\\sum_i(-1)^i\\binom{n+1}{i}(i+\\lambda)^n=0$$ where $i$ runs through all integers and define $\\binom{n}{i}=0$ if $i<0$ or $i>n$.\n\nProof. This is straight-forward for $n=0$. Suppose for $n=k$ the statement is true. Then \\begin{align} \\sum_i(-1)^i\\binom{k+2}i(i+\\lambda)^{k+1}&=\\sum_i(-1)^i\\binom{k+2}i\\binom i1(i+\\lambda)^k+\\lambda\\sum_i(-1)^i\\binom{k+2}{i}(i+\\lambda)^k \\end{align} Note that $\\binom{k+2}i\\binom i1=\\binom{k+2}1\\binom{k+1}{i-1}$ and $\\binom{k+2}i=\\binom{k+1}i+\\binom{k+1}{i-1}$. By induction hypothesis we obtain $\\sum(-1)^i\\binom{k+1}i(i+\\lambda)^k=0$, thus \\begin{align} \\sum_i(-1)^i\\binom{k+2}i(i+\\lambda)^{k+1}&=(k+2+\\lambda)\\sum_i(-1)^i\\binom{k+1}{i-1}(i+\\lambda)^k\\\\ &=-(k+2+\\lambda)\\sum_i(-1)^i\\binom{k+1}{i}(i+\\lambda+1)^k\\\\ &=0 \\end{align} where the last equality follows from induction hypothesis again.", "date": "2022-05-16 09:51:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2892375/prove-sum-i-0n-1n-i-binomn1i-i1n-n2n", "openwebmath_score": 0.9434666633605957, "openwebmath_perplexity": 99.76559629265181, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.904650536386234, "lm_q1q2_score": 0.8913262871843669}}
{"url": "http://mathhelpforum.com/statistics/144005-help-svp.html", "text": "# Math Help - Help, SVP.\n\n1. ## Help, SVP.\n\nHey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.\n\n1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.\n\n2. Cha\u2019tima has five white and six grey huskies in her kennel. If a wilderness expedition chooses a team of six sled dogs from Cha\u2019tima\u2019s kennel, what is the probability that the team will consist of:\nall white huskies?\nall grey huskies?\nthree of each colour?\n\n3. A survey at a school asked students if they were ill with a cold or the flu during last months.\nThe results were as follows. None of the students had both a cold and the flu.\nCold Flu Healthy\nFemales 32 18 47\nMales 25 19 38\nUse these results to estimate the probability that:\na randomly selected student had a cold last month.\na randomly selected females students was healthy last month.\na randomly selected student who had the flu last month is male.\na randomly selected student had either a cold or the flu last month.\n\n4. To get out of jail free in the board game Monopoly\u00d2, you have to roll doubles with a pair of dice. Determine the odds in favour of getting out of jail on your first or second roll.\n\n5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,\nsubscribes to both magazines?\nsubscribes to either one magazine or both?\nsubscribes to only one of the two magazines?\n\n2. I got this for 2,3, and 4.\n\n2)\n\nProb is 0 b/c only 5 whites are there\nP(all grey) = 1/(11 6) = 1/462, ~ 0.0022\nP(3 white & 3 grey) ((5 3) (6 3))/(11 6), 200/462, ~ 0.433\n\n3)\n\nP(cold)=57/179, =0.3184, ~ 31.8%\nP(health | female) (47/179)/(97/179) = 47/97 = 0.4845, ~ 48.5%\nP(male | flu) 19/37 = 0.5135, ~51.3%\n44/82 = 0.5366, ~53.7%\n\nDouble on first roll = 6/36 = 1/6, ergo not rollin = 5/6\nErgo prob of rolling doubles on the second roll: 5/6 * 1/6 = 5/36\nProb of rolling double on first or second: 1/6 + 5/6 = 11/36\nErgo odds of getting out of jail on first or second roll are 11:25\n\nConfirm?\n\nAnd help with 1 and 5?\n\n3. Originally Posted by acc\nConfirm?\n\nAnd help with 1 and 5?\nI didn't get a calculator out to confirm final answers, but I agree with your work for 2,3,4 except for the following:\n\nfor 3b, I would simply write 47/97 without the initial (47/179)/(97/179) step.\n\nfor 3d, I get 94/179. Your answer is for a randomly chosen male student.\n\nfor 4, there's an assumption we have to make, and that is that we don't want to count the probability of getting doubles on both the first and second rolls (because presumably, rolling a double on the first roll will get us out of jail and we won't make the second roll). Assuming that, I agree with your answer. (But you made a typo by writing 5/6 in one spot where you meant 5/36.)\n\nFor problem 1, I would start with a \"1\" and subtract the following probabilities from it\n\nP(0 teachers and 6 students)\nP(1 teacher and 5 students)\n\nProblem 5, you could draw a Venn diagram. Or using formal notation,\n\n$P(A \\cup B) = P(A) + P(B) - P(A \\cap B)$\n\n(Edit: fixed typo.)\n\nEdit 2: The part in gray is incorrect. What you wrote for 4 is fine the way it is, other than the typo.\n\n4. Hello, acc!\n\n2. Cha\u2019tima has five white and six grey huskies in her kennel.\nIf a wilderness expedition chooses a team of six sled dogs from Cha\u2019tima\u2019s kennel,\nwhat is the probability that the team will consist of:\n\n(a) all white huskies?\n(b) all grey huskies?\n(c) three of each colour?\nThere are: . ${11\\choose6} \\:=\\:462$ possible selections.\n\n$(a)\\;P(\\text{6 white}) \\;=\\;0$\n\n$(b)\\;P(\\text{6 grey}) \\;=\\;\\frac{{6\\choose6}}{462} \\;=\\;\\frac{1}{462}$\n\n$(c)\\;P(\\text{3 white, 3 grey}) \\;=\\;\\frac{{5\\choose3}{6\\choose3}}{462} \\;=\\;\\frac{10\\cdot20}{462} \\;=\\;\\frac{100}{231}$\n\n3. A survey at a school asked students\nif they were ill with a cold or the flu during last month.\n\nThe results were as follows.\nNone of the students had both a cold and the flu.\n\n. . $\\begin{array}{c||c|c|c||c|}\n& \\text{Cold} & \\text{Flu} & \\text{Healthy } & \\text{Total} \\\\ \\hline\\hline \\text{Females} & 32 & 18 & 47 & 97 \\\\ \\hline\n\\text{Males} & 25 & 19 & 38 & 82 \\\\ \\hline \\hline\n\\text{Total} & 57 & 37 & 85 & 179 \\\\ \\hline \\end{array}$\n\nUse these results to estimate the probability that:\n\n(a) a randomly selected student had a cold last month.\n(b) a randomly selected female student was healthy last month.\n(c) a randomly selected student who had the flu last month is male.\n(d) a randomly selected student had either a cold or the flu last month.\n\n$(a)\\;P(\\text{Cold}) \\:=\\:\\frac{57}{179}$\n\n$(b)\\;P(\\text{Female}\\,\\wedge\\,\\text{Healthy}) \\;=\\;\\frac{47}{179}$\n\n$(c)\\;P(\\text{Male}|\\text{Flu}) \\;=\\;\\frac{19}{37}$\n\n$(d)\\;P(\\text{Cold} \\vee\\text{Flu}) \\;=\\;P(\\text{Cold}) + P(\\text{Flu}) \\;=\\;\\frac{57}{179} + \\frac{37}{170} \\;=\\;\\frac{94}{179}$\n\n5. Originally Posted by Soroban\n\n$(a)\\;P(\\text{Cold}) \\:=\\:\\frac{57}{179}$\n\n$(b)\\;P(\\text{Female}\\,\\wedge\\,\\text{Healthy}) \\;=\\;\\frac{47}{179}$\n\n$(c)\\;P(\\text{Male}|\\text{Flu}) \\;=\\;\\frac{19}{37}$\n\n$(d)\\;P(\\text{Cold} \\vee\\text{Flu}) \\;=\\;P(\\text{Cold}) + P(\\text{Flu}) \\;=\\;\\frac{57}{179} + \\frac{37}{170} \\;=\\;\\frac{94}{179}$\n\nWow, I completely misread (c) and thought it said \"a randomly selected student had the flu last month.\"\n\nFor (b), I'm wondering if it is a matter of interpretation. To get the probability you found, wouldn't it have better been stated \"a randomly selected student is female and had the flu last month\"?\n\nEdit: I must be tired because I didn't even address 3c, I'll stop posting now and get some rest.\n\n6. Originally Posted by acc\nHey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.\n\n1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.\n\n[snip]\n\n5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,\nsubscribes to both magazines?\nsubscribes to either one magazine or both?\nsubscribes to only one of the two magazines?\nDuplicate posted (and answered) here: http://www.mathhelpforum.com/math-he...tml#post511545", "date": "2015-03-05 23:20:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/144005-help-svp.html", "openwebmath_score": 0.5251576900482178, "openwebmath_perplexity": 2088.989796163547, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713870152409, "lm_q2_score": 0.9046505344582187, "lm_q1q2_score": 0.8913262868497281}}
{"url": "http://experiential-architecture.com/kushco-holdings-oercf/find-permutation-problem-fd31c5", "text": "This article has been viewed 12,813 times. For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. If you're seeing this message, it means we're having trouble loading external resources on our website. Discuss: answer with explanation. Mathematically we can approach this question as follows: $$P=\\frac{n!}{n_1! To create this article, volunteer authors worked to edit and improve it over time. = n (n - 1) (n - 2) (n - 3) \u2026. n_2! Solution: There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. Hence it is a permutation problem. Then, find 7! 21300: C. 24400: D. 210: View Answer. Yes. The basic building block to solve permutation and combination problems is to understand the use of recursion in a for loop. Some graphing calculators offer a button to help you solve permutations without repetition quickly. To create this article, volunteer authors worked to edit and improve it over time. 0! other thatn the given sequence in the argument of the function where c is in the last position. Important Permutation Formulas. Solution:You need two points to draw a line. We use cookies to make wikiHow great. Last Updated: October 14, 2020 A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. This image may not be used by other entities without the express written consent of wikiHow, Inc. \\n<\\/p> \\n<\\/p><\\/div>\"}, https://www.mathsisfun.com/combinatorics/combinations-permutations.html, https://betterexplained.com/articles/easy-permutations-and-combinations/, https://www.learneroo.com/modules/10/nodes/62, https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition, https://www.vitutor.com/statistics/combinatorics/permutations_repetition.html, https://www.learneroo.com/modules/10/nodes/55, https://mathbits.com/MathBits/TISection/Algebra1/Probability.htm, https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html, consider supporting our work with a contribution to wikiHow. If you really can\u2019t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. At this point, we have to make the permutations of only one digit with the index 3 and it has only one permutation i.e., itself. P (n,r) represents the number of permutations of n items r at a time. A permutation is an arrangement of objects in which the order is important[1] This article has been viewed 12,813 times. This program will find all possible combinations of the given string and print them. You will more details about each type of problem in the problem definition section. In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. You'd then calculate 3,628,800/5,040. If you're using Google Calculator, click on the, If you have to solve by hand, remember that, for each. Find the order in lexicographical sorted order If we want to find all the permutations of a string in a lexicographically sorted order means all the elements are arranged in alphabetical order and if the first element is equal then sorting them based on the next elements and so on. Given a positive integer n and a string s consisting only of letters D or I, you have to find any permutation of first n positive integer that satisfy the given input string. Enter \"7\" for \"Number of sample points in set \". wikiHow is a \u201cwiki,\u201d similar to Wikipedia, which means that many of our articles are co-written by multiple authors. The order of arrangement of the object is very crucial. To solve this problem using the Combination and Permutation Calculator, do the following: Choose \"Count permutations\" as the analytical goal. In other words, if the set is already ordered, then the rearranging of its elements is called the process of permuting. In a certain state\u2019s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them \u2026 letters in our case which is 6 \\(n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Research source Iteration : : Iteration, in the context of computer programming, is a process wherein a set of instructions or structures are repeated in a sequence a specified number of times or until a condition is met. Mathematically we can approach this question as follows: $$P=\\frac{n!}{n_1! We leave the 3 out and recursively find all permutations of the set {1, 2}. For example, consider finding all permutations of {1, 2, 3). The permutation problems are arrangement problems and the combination problems are selection problems. This image is not<\\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Arithmetic Sequences Problems with Solutions, High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers, Mathematical Induction - Problems With Solutions, Geometric Sequences Problems with Solutions. You can use a simple mathematical formula to find the number of different possible ways to order the items. For example 7 and 4. In how many ways can you select a committee of 3 students out of 10 students? (3) (2) (1) Permutations of n items taken r at a time. Permutation With Repetition Problems With Solutions : In this section, we will learn, how to solve problems on permutations using the problems with solutions given below. The is below the up/down arrow, at the top left of the answer. You may also vote on the quality/helpfulness of an answer, with the up or down arrow, if you have a 15+ reputation. The problem is to select 2 points out of 3 to draw different lines. 25200: B. References. P (n,r) = n!/ (n - r)! This consists of 2 permutations: {1, 2} {2, 1] Now we insert the 3 into every position for these permutations. In mathematics, permutation relates to the act of arranging all the members of a set into some sequence or order. How many different committee can be formed from the group? Problem Statement. For instance, you might be selecting 3 representatives for student government for 3 different positions from a set of 10 students. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.The word \"permutation\" also refers to the act or process of changing the linear order of an ordered set. Plug your numbers in for n {\\displaystyle n} and r {\\displaystyle r}. ). }$$ Where: $$n$$ is the total number of object, i.e. Calculate the number of permutations of n elements taken r at the time. That number means that, if you're picking from 10 different students for 3 student government positions, where order matters and there is no repetition, there are 720 possibilities. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. A. Let\u2019s take an example to understand the problem, Input xyz Output xyz, xzy, yxz, yzx, zxy, zyx Explanation These are all permutations take in order. This kind of problem... 2. This method takes a list as an input and returns an object list of tuples that contain all \u2026 = 24 . }\\) Where: $$n$$ is the total number of object, i.e. = 5*4*3*2*1 = 120. So out of that set of 4 horses you want to pick the subset of \u2026 By using our site, you agree to our. Solved Examples(Set 1) - Permutation and Combination. Is there an easy way in order to memorize the formula for this? letters in our case which is 6 $$n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Permutations is not an easy problem. Permutations and combinations are used to solve problems. A permutation is a reordered arrangement of elements or characters of a string. by doing (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800 as a result. Suppose we have three people named A, B, and C. We have already determined that they can be seated in a straight line in 3! / (4 - 3)! n_2! Permutation With Repetition Problems With Solutions - Practice questions. n_3!\u2026n_k! n_3!\u2026n_k! 1. Include your email address to get a message when this question is answered. Find Permutation. How many 3 digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions? % of people told us that this article helped them. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements. The C program to find permutation and combination solves 4 different types of problems. 1! The order is not important. Enter \"3\" for \"Number of sample points in each permutation\". Solve the equation to find the number of permutations. Think of it like this: subtract the total amount by the total items. means (n factorial). Permutations occur, in more or less prominent ways, in almost every area of mathematics. The number of words is given by 4 P 3 = 4! Now in this permutation (where elements are 2, 3 and 4), we need to make the permutations of 3 and 4 first. = 1 Let us take a look at some examples: Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word \u2018CHAIR\u2019. Ending index of the string. A pemutation is a sequence containing each element from a finite set of n elements once, and only once. Calculating Permutations without Repetition 1. By signing up you are agreeing to receive emails according to our privacy policy. In the example, you should get 720. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. How many triangles can you make using 6 non collinear points on a plane? Second \u2026 Permutations . Intuition If there were no Kleene stars (the * wildcard character for regular expressions), the problem would be easier - we simply check from left to right if each character of the text matches the pattern. For those who haven\u2019t seen a backtracking question before, there is no clear naive solution, and this poses a real threat for software engineers during\u2026 If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. Start with an example problem where you'll need a number of permutations without repetition. (unlike combinations, which are groups of items where order doesn't matter[2] These methods are present in an itertools package. Hopefully, you have seen \u2026 n! To start off, you just need to know whether repetition is allowed in your problem or not, and then pick your method and formula accordingly. Calculating Permutations without Repetition, {\"smallUrl\":\"https:\\/\\/www.wikihow.com\\/images\\/thumb\\/b\\/b0\\/Calculate-Permutations-Step-1.jpg\\/v4-460px-Calculate-Permutations-Step-1.jpg\",\"bigUrl\":\"\\/images\\/thumb\\/b\\/b0\\/Calculate-Permutations-Step-1.jpg\\/aid11015428-v4-728px-Calculate-Permutations-Step-1.jpg\",\"smallWidth\":460,\"smallHeight\":345,\"bigWidth\":728,\"bigHeight\":546,\"licensing\":\"\n\n\\u00a9 2021 wikiHow, Inc. All rights reserved. The permutation is an arrangement of objects in a specific order. Combinations P(n) = n! A new solution can be accepted if a better one shows up. D means the next number is smaller, while I means the next number is greater. to 4, so 7x6x5 and then find the answer, and you\u2019ll get the permutations. Example 6: How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? If your question is solved, say thank you by accepting the solution that is best for your needs. Solution: \u2018CHAIR\u2019 contains 5 letters. would be (7 * 6 * 5 * 4 * 3 * 2 * 1), which would equal 5,040. Given a set of \u2018n\u2019 elements, find their Kth permutation. How many 4 digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions? Each question has four choices out \u2026 Approach #1 Using Stack [Accepted] Let's revisit the important points of the given problem statement. Therefore, the number of words that can be formed with these 5 letters = 5! First import itertools package to implement the permutations method in python. Thanks to all authors for creating a page that has been read 12,813 times. FYI: Thoroughly answering questions is time-consuming. Research source It usually looks like, All tip submissions are carefully reviewed before being published. Question 1 : 8 women and 6 men are standing in a line. If we proceed as we did with permutations, we get the following pairs of points to draw lines.AB , ACBA , BCCA , CBThere is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.The lines are: AB, BC and AC ; 3 lines only.So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.This is a combination problem: combining 2 items out of 3 and is written as follows: eval(ez_write_tag([[580,400],'analyzemath_com-large-mobile-banner-1','ezslot_4',700,'0','0'])); Calculate the number of combinations of n elements taken r at the time. Graphs of Functions, Equations, and Algebra, The Applications of Mathematics Permutations of the same set differ just in the order of elements. Permutation and Combination Solve Problems Quickly. For example, you would calculate 10! How many number plates can be formed if neither the digits nor the letters are repeated. Explanation: Number of ways of selecting 3 consonants from 7 X If you have a calculator handy, find the factorial setting and use that to calculate the number of permutations. Learn How to Solve Permutation and Combination Question Quickly form PrepInsta. Solution: n-factorial gives the number of permutations of n items. Find all Permutations of the word baboon. Consider the following set of elements: 7! Recursive functions are very useful to solve many mathematical problems, such as calculating the factorial of a number, generating Fibonacci series, etc. For the first permutation we insert \u2026 Circular Permutations . When a star is present, we may need to check many different suffixes of the text and see if they match the rest of the pattern. X = 1. wikiHow is where trusted research and expert knowledge come together. Python provides a package to find permutations and combinations of the sequence. Find all Permutations of the word baboon. The list of problems is given below. Permutation Problem 1 Choose 3 horses from group of 4 horses In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). 4. wikiHow is a \u201cwiki,\u201d similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Output First Swap generates the other permutation, i.e. 3. Introductory permutation problems. def permute (a, l, r): if l = = r: print toString (a) else: for i in xrange (l,r + 1 ): a [l], a [i] = a [i], a [l] permute (a, l + 1, r) a [l], a [i] = a [i], a [l] # backtrack. Line AB is the same as line BA. or 6 ways. The permutation test is designed to determine whether the observed difference between the sample means is large enough to reject, at some significance level, the null hypothesis H that the data drawn from is from the same distribution as the data drawn from. How many 6 letter words can we make using the letters in the word LIBERTY without repetitions? The number of permutations on the set of n elements is given by n! In how many ways can you arrange 5 different books on a shelf? Question.How many three digit numbers can be formed using digits 2, 3, 4, 7, 9 so that the digits can be repeated. Permutations differ from combinations, which are selections of some members of a set regardless of \u2026 PERMUTATION WORD PROBLEMS WITH SOLUTIONS Problem 1 : A student appears in an objective test which contain 5 multiple choice questions. Given a string, we have to find all the permutations of that string. Answer: Option A. Problem Statement. In this regard, what is the importance of permutation and combination? Input. Permutation is the arrangement of all parts of an object, in all possible orders of arrangement. Permutations with repetition n 1 \u2013 # of the same elements of the first cathegory n 2 - # of the same elements of the second cathegory For example, string \u201cabc\u201d have six permutations [\u201cabc\u201d, \u201cacb\u201d, \u201cbac\u201d, \u201cbca\u201d, \u201ccab\u201d, \u201ccba\u201d]. We know ads can be annoying, but they\u2019re what allow us to make all of wikiHow available for free. We will be given a single string input. No student can be used in more than one position (no repetition), but the order still matters, since the student government positions are not interchangeable (a permutation where the first student is President is different from a permutation where they're Vice President). Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. And thus, permutation(2,3) will be called to do so. Similarly, permutation(3,3) will be called at the end. 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The copyright holder of this image under U.S. and international copyright laws arrow..., click on the, if the set is already ordered, then please consider supporting work... \\Displaystyle r } receive emails according to our privacy policy taken r at a time of different possible to!: View answer possible ways to order the items an objective test which contain 5 multiple choice questions c in... A sequence containing each element from a group of 10 students select a committee including 3 boys 12!: subtract the total number of permutations on the set { 1, 2 } sequence containing element. Another ad again, then please consider supporting our work with a contribution to wikihow other words, if 're... Of 3 to draw different lines % of people told us that this helped. Sequence or order can be annoying, but they \u2019 re what allow us to all. 1, 2, 3 ) ( 2 ) ( n - 2 ) ( 2 ) ( 1 permutations. \u2019 t stand to see another ad again, then the rearranging of its elements is called the of... Is greater privacy policy you can use a simple mathematical formula to find the factorial setting use! The up or down arrow, if you really can \u2019 t stand to another... Finding all permutations of that string, say thank you by accepting solution! Select a committee of 3 consonants and 2 vowels can be formed problem is to be with! Many of our articles are co-written by multiple find permutation problem r { \\displaystyle }. Your needs - 3 ) ( 1 ), which means that many of our articles co-written... With our trusted how-to guides and videos for free by whitelisting wikihow on your ad blocker 3 digit can. Of an answer, and find permutation problem without repetitions address to get a message when this question is solved, thank... Guides and videos for free by whitelisting wikihow on your ad blocker of recursion a! Can approach this question is solved, say thank you by accepting the solution that is best your! 4 p 3 = 4 we leave the 3 out and recursively find all the members of a,! The first permutation we insert \u2026 find all permutations of n elements once, and 6 men are in! By accepting the solution that is best for your needs out of 10?... Site, you agree to our privacy policy \u201c wiki, \u201d similar to Wikipedia, which means many! Find the answer question Quickly form PrepInsta many triangles can you select find permutation problem! With these 5 letters = 5 accepting the solution that is best for your needs, 3 (. 2 vowels can be formed problem definition section the solution that is best your! A specific order approach this question as follows: \\ ( n\\ ) is the total number of words given! Plates can be formed from a finite set of n elements is called the process of.. Come together of an object, i.e know ads can be formed from the group that is for... To receive emails according to our privacy policy 7 '' for  number permutations. Liberty without repetitions by hand, remember that, for each of our articles are co-written by multiple authors of. Where c is in the argument of the set of 10 students 15+... 3 to draw different lines { \\displaystyle r } points to draw different lines a group of 10.... Many 3 digit numbers can we make using the digits nor the letters in the problem definition section students. Make using the letters are repeated at the time 3 digit numbers can we using... Possible orders of arrangement 10 students enter  3 '' for  number of sample points in ! Factorial setting and use that to calculate the number of permutations without Quickly. To the act of arranging all the permutations nor the letters in order. ) represents the number of permutations and videos for free by whitelisting wikihow on your ad.. Handy, find the factorial setting and use that to calculate the number of permutations to. A simple mathematical formula to find the number of permutations on the, if have. We can approach this question as find permutation problem: \\ ( P=\\frac { n! } { n_1 multiple! Many number plates can be formed with these 5 letters = 5 * 4 * *... \\ ) where: \\ ( P=\\frac { n! } { n_1 find their Kth..", "date": "2021-05-18 16:35:58", "meta": {"domain": "experiential-architecture.com", "url": "http://experiential-architecture.com/kushco-holdings-oercf/find-permutation-problem-fd31c5", "openwebmath_score": 0.5052473545074463, "openwebmath_perplexity": 587.9953836579109, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9881308779050736, "lm_q2_score": 0.9019206831203062, "lm_q1q2_score": 0.8912156764124118}}
{"url": "https://math.stackexchange.com/questions/2919921/of-all-polygons-inscribed-in-a-given-circle-which-one-has-the-maximum-sum-of-squ", "text": "# Of all polygons inscribed in a given circle which one has the maximum sum of squares of side lengths?\n\nMy son presented me with an interesting problem:\n\nOf all possible polygons inscribed in a circle of radius $R$, find the one that has the sum $S$ of squared side lengths maximized: $S=a_1^2+a_2^2+\\dots+a_n^2$, with $a_i$ representing the length of the $i$-th side. The number of sides is not fixed, you should consider all triangles, quadrilaterals, pentagons...\n\nIt's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^\\circ$) angles. For example, if such an angle $A_{i-1}A_{i}A_{i+1}$ exists, by cosine theorem:\n\n$$|A_{i-1}A_{i}|^2+|A_{i}A_{i+1}|^2<|A_{i-1}A_{i+1}|^2$$\n\nSo if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$.\n\nAll candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle.\n\nBut maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $b\\ne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$.\n\nHowever, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant.\n\nJust an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what?\n\nEDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $\\gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$.\n\n$$b=2R\\sin\\frac{\\pi-\\alpha+\\gamma}{2}=2R\\cos\\left(\\frac\\alpha2-\\frac\\gamma2\\right)$$\n\n$$c=2R\\sin\\frac{\\pi-\\alpha-\\gamma}{2}=2R\\cos\\left(\\frac\\alpha2+\\frac\\gamma2\\right)$$\n\n$$b^2=4R^2\\cos^2\\left(\\frac\\alpha2-\\frac\\gamma2\\right)=2R^2(1+\\cos(\\alpha-\\gamma))$$\n\n$$c^2=4R^2\\cos^2\\left(\\frac\\alpha2+\\frac\\gamma2\\right)=2R^2(1+\\cos(\\alpha+\\gamma))$$\n\n$$b^2+c^2=4R^2+2R^2(\\cos(\\alpha-\\gamma)+\\cos(\\alpha+\\gamma))=4R^2(1+\\cos\\alpha\\cos\\gamma)$$\n\nAnd this sum achieves maximum obviously for $\\gamma=0$, or for $A\\equiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle.\n\nEDIT 2: This \u201cmoving vertex\u201d procedure can be repeated infinite number of times and the result is an equilateral triangle! Check excellent proof by Noah Schweber here.\n\n\u2022 Note that your last sentence \"The only triangle which has no better option is equilateral triangle.\" is insufficient to imply \"The equilateral triangle is better than every other triangle.\". See my answer for a sketch of how to do that part rigorously. \u2013\u00a0user21820 Sep 17 '18 at 17:45\n\u2022 @Oldboy, I'm studying your topic and curiously I have a problem a little similar in here. The triangle's case was the most problematic for me too. \u2013\u00a0Na'omi Sep 18 '18 at 14:10\n\nYes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.\n\nSince Prove that in any triangle $ABC$, $\\cos^2A+\\cos^2B+\\cos^2C\\geq\\frac{3}{4}$ then $$\\sin^2 A+\\sin^2 B+\\sin^2 C=3-\\cos^2 A-\\cos^2 B-\\cos^2 C\\leq \\frac{9}{4}$$ where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=\\pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is $$(2R\\sin A)^2+(2R\\sin B)^2+(2R\\sin C)^2\\leq 9R^2.$$\n\n\u2022 I have upvoted your answer, it's just that it's not much simpler compared with my solution. It uses a trigonometric fact about triangle that is not obvious. Anyway, if nothing better shows up in a day or two, I'll accept your answer as the best one. \u2013\u00a0Oldboy Sep 17 '18 at 9:27\n\nThis problem can be stated as\n\n$$\\max_{n}\\sum_{k=1}^n\\left(2r\\sin\\left(\\frac{\\theta_k}{2}\\right)\\right)^2$$\n\ns. t.\n\n$$\\sum_{k=1}^n\\theta_k = 2\\pi$$\n\nbut\n\n$$\\sum_{k=1}^n\\left(2r\\sin\\left(\\frac{\\theta_k}{2}\\right)\\right)^2\\ge n\\left(2^{2n}r^{2n}\\prod_{k=1}^n\\sin^2\\left(\\frac{\\theta_k}{2}\\right)\\right)^{\\frac 1n}$$\n\nassuming $\\theta_1=\\cdots=\\theta_n$ we have\n\n$$\\sum_{k=1}^n\\left(2r\\sin\\left(\\frac{\\theta_k}{2}\\right)\\right)^2\\ge n\\left(2^{2n}r^{2n}\\sin^{2n}\\left(\\frac{\\pi}{n}\\right)\\right)^{\\frac 1n} = n2^2r^2\\sin^{2}\\left(\\frac{\\pi}{n}\\right)$$\n\nNow calling\n\n$$f(n) = n\\sin^{2}\\left(\\frac{\\pi}{n}\\right)$$\n\nwe have clearly a maximum about $n = 3$ as can be depicted in the attached plot\n\n\u2022 This is interesting, but as far as I can tell it assumes that the polygon is regular; and in some cases an irregular polygon can do better for fixed $n$. For example, for a regular hexagon, we have $S = 6 R^2$. But for a hexagon where four of the points are degenerate, the hexagon is \"basically\" a triangle with three sides of length zero; and if we pick the other three sides to have equal length, we have an equilateral triangle with $S = 9 R^2$. \u2013\u00a0Michael Seifert Sep 17 '18 at 14:50\n\nLet $\\theta_k$ be the successive angles subtended by the sides, but the last one. The sum of squares is given by\n\n$$4\\sum_{k=1}^n\\sin^2\\frac{\\theta_k}2+4\\sin^2\\left(\\pi-\\frac12\\sum_{k=1}^n\\theta_k\\right)$$ which has the same extrema as $$\\sum_{k=1}^n\\cos\\theta_k-\\cos\\left(\\sum_{k=1}^n\\theta_k\\right).$$\n\n$$\\sin\\theta_k=\\sin\\left(\\sum_{k=1}^n\\theta_k\\right).$$\n\nThis shows that all angles $\\theta_k$ must be equal, and then\n\n$$n\\cos\\theta-\\cos n\\theta$$ is minimized with $n\\theta=2\\pi$.\n\nFinally,\n\n$$n\\cos\\frac{2\\pi} n-\\cos\\pi$$ is the smallest with $n=3$.\n\nThe objective function is continuous on the domain of interest (all triples of points on the circle), which is also compact. Therefore by the extreme value theorem it has a global maximum. That reduces the problem to the part you are interested in, namely proving that if the optimal triangle has sides $a,b,c$ then $b = c$. Firstly it has to be acute as you observed. Thus maximizing $b^2+c^2$ $= a^2+2bc\u00b7\\cos(\\angle A)$ is equivalent to maximizing $bc$, since fixing $B,C$ fixes $a$ and $\\angle A$. Letting $x = \\angle BAO$ and $y = \\angle OAC$ we have $bc = 4R^2\u00b7\\cos(x)\\cos(y)$, and finally note $2 \\cos(x)\\cos(y)$ $= \\cos(x+y) + \\cos(x-y)$ $\\le \\cos(x+y) + 1$ $= \\cos(\\angle A) + 1$ with equality exactly when $x=y$.\n\n\u2022 Part of the reason for the first part of my answer is that I have a small suspicion that you did not get that part right. One cannot automatically claim that every optimization problem has a global extremum. This has to be proven, and you cannot do so via an infinite iterative process unless you prove that it converges. One way around this is to use the extreme value theorem if the domain is compact, as here. \u2013\u00a0user21820 Sep 17 '18 at 17:43\n\n\\begin{align} AB^2 + BC^2 + CA^2 &= (\\vec{OB} - \\vec{OA})^2 + (\\vec{OC} - \\vec{OB})^2 + (\\vec{OA} - \\vec{OC})^2\\\\ &= 2(\\vec{OA})^2 + 2(\\vec{OB})^2 + 2(\\vec{OC})^2 - 2 \\times \\vec{OA} \\cdot \\vec{OB} - 2 \\times \\vec{OB} \\cdot \\vec{OC} - 2 \\times \\vec{OC} \\cdot \\vec{OA} \\\\ &= 3(\\vec{OA})^2 + 3(\\vec{OB})^2 + 3(\\vec{OC})^2 - (\\vec{OA} + \\vec{OB} + \\vec{OC})^2 \\\\ &\\leq 9R^2. \\end{align}\n\n\u2022 I would try to avoid $\\times$, as it usually denotes cross product in this context. Also, this appears to be simply proving that the value for an equilateral triangle is $9R^2$, rather than, as the OP asks, proving that this is maximal. \u2013\u00a0Acccumulation Sep 17 '18 at 15:20\n\u2022 @Acccumulation sorry, what exactly is unclear in proving that $AB^2+BC^2+CA^2 \\leq 9R^2$ for an arbitrary inscribed triangle? \u2013\u00a0Roman Odaisky Sep 17 '18 at 15:23\n\nWe want to maximise $b^2+c^2$, which by the cosine rule is equal to $a^2+2bc\\cos A$.\n\nAngle $\\angle BAC$ is fixed, so this means maximising $bc$.\n\nThe area of the triangle is $\\frac12bc\\sin A$, and $\\sin A$ is fixed, so this means maximising the area of the triangle.\n\nThe area of the triangle is $\\frac12$base$\\times$height $= \\frac12a\\times$ height, so this means maximising the height.\n\nAnd the height is maximum when $\\triangle BA_1C$ is isosceles.\n\nSuppose we have three unit vectors a, b, and c. This will define a triangle with side lengths (a-b), (a-c), and (b-c), so the sum of squares will be (a-b)^2+(a-c)^2+(b-c)^2. Taking the derivative with respect to a, we get 2a'(a-b)+2a'(a-c)=2a'(2a-(b+c)). Because a is constrained to be on the unit circle, a' is perpendicular to a, hence a'a=0. So the derivative simplifies to -2a'(b+c). Thus, the derivative is zero if a' is perpendicular to b+c, which is equivalent to a being parallel to b+c, which happens when the angle between a and b is equal to the angle between a and c. Applying the same argument to the derivatives with respect to b and c shows that all the angles must be equal.", "date": "2019-06-16 08:45:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2919921/of-all-polygons-inscribed-in-a-given-circle-which-one-has-the-maximum-sum-of-squ", "openwebmath_score": 0.9784698486328125, "openwebmath_perplexity": 224.18168856934906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9865717476269469, "lm_q2_score": 0.903294216466424, "lm_q1q2_score": 0.8911645537605937}}
{"url": "https://math.stackexchange.com/questions/508308/a-definite-integral-int-0-infty-frac2-cos-x-left1x4-right-left5-4", "text": "# A definite integral $\\int_0^\\infty\\frac{2-\\cos x}{\\left(1+x^4\\right)\\,\\left(5-4\\cos x\\right)}dx$\n\nI need to find a value of this definite integral: $$\\int_0^\\infty\\frac{2-\\cos x}{\\left(1+x^4\\right)\\,\\left(5-4\\cos x\\right)}dx.$$ Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calculator Plus and WolframAlpha did not return a plausible closed-form candidate.\n\nDo you have any ideas how I can approach this problem?\n\n\u2022 Looks like you already have a value there. Do you have a particular reason to think it has a closed form? \u2013\u00a0Henning Makholm Sep 28 '13 at 23:55\n\u2022 @HenningMakholm One should always believe there is a closed form. It is just that some functions and constants are not yet well studied, understood and named :) \u2013\u00a0Vladimir Reshetnikov Sep 29 '13 at 0:06\n\u2022 @hmedan.mnsh Just to clarify: Mathematica does not return a closed form solution, but can give a numerical approximation using NIntegrate. \u2013\u00a0Vladimir Reshetnikov Sep 29 '13 at 0:31\n\u2022 @HenningMakholm It seems to me the phrase \"you already have a value\" is somewhat objectionable. A numerical integration in this case converges quite slowly, and I doubt that one can get more than $25$ correct digits in a reasonable time (and it would be a very difficult to make sure they all are indeed correct). On the other hand, having a closed form, I can compute $10^5$ digits in about a second. And I can be pretty sure all these digits are correct, because numerical algorithms for elementary functions have been thoroughly designed and polished for years, and verified for correctness. \u2013\u00a0Vladimir Reshetnikov Sep 29 '13 at 18:11\n\u2022 @StevenStadnicki I do not need $10^5$ digits, I used this number just to demonstrate the speed of calculations. But I can easily imagine that someone wants to get $30$ digits that are certainly correct. It is trivial task when one has an elementary closed form, but it might be quite difficult if one has to resort to numerical integration. Also, closed forms sometimes have a useful property to partially cancel each other or otherwise simplify when several factors or terms are combined into a single expression, while combining multiple approximate numeric values results in a precision loss. \u2013\u00a0Vladimir Reshetnikov Oct 1 '13 at 18:15\n\nYes, there is an elementary closed form for this integral: $$\\int_0^\\infty\\frac{2-\\cos x}{\\left(1+x^4\\right)\\,\\left(5-4\\cos x\\right)}dx=\\frac{\\pi}{2\\,\\sqrt2}\\cdot\\exp\\left(\\frac1{\\sqrt2}\\right)\\cdot\\frac{\\sin\\left(\\frac1{\\sqrt2}\\right)-\\cos\\left(\\frac1{\\sqrt2}\\right)+2\\,\\exp\\left(\\frac1{\\sqrt2}\\right)}{1-4\\,\\exp\\left(\\frac1{\\sqrt2}\\right)\\cos\\left(\\frac1{\\sqrt2}\\right)+4\\,\\exp\\left(\\sqrt2\\right)}\\tag1$$\n\nProof:\n\nLet us denote the integral in question as $$\\mathcal{I}=\\int_0^\\infty\\frac{2-\\cos x}{\\left(x^4+1\\right)\\,\\left(5-4\\cos x\\right)}dx\\tag2$$ Note that the trigonometric part of the integrand is a periodic function and can be expanded to a Fourier series with particularly simple coefficients: $$\\frac{2-\\cos x}{5-4\\cos x}=\\sum_{n=0}^\\infty\\frac{\\cos(n\\,x)}{2^{n+1}}\\tag3$$ (this can be easily checked by expressing cosines via exponents of an imaginary argument).\n\nNow we can integrate it term-wise: $$\\mathcal{I}=\\sum_{n=0}^\\infty\\left(\\frac1{2^{n+1}}\\int_0^\\infty\\frac{\\cos(n\\,x)}{x^4+1}dx\\right)=\\sum_{n=0}^\\infty\\left(\\frac1{2^{n+1}}\\cdot\\frac{\\pi}{2\\,\\sqrt2}\\cdot\\exp\\left(-\\frac{n}{\\sqrt2}\\right)\\cdot\\left(\\sin\\left(\\frac{n}{\\sqrt2}\\right)+\\cos\\left(\\frac{n}{\\sqrt2}\\right)\\right)\\right)\\tag4$$ (for the integral, see DLMF 1.14, vii, Table 1.14.2, $4^{th}$ row).\n\nTrig functions in the last sum can again be expressed via exponents of an imaginary argument, and then the sum is easily evaluated. Converting exponents back to trig functions and getting rid of complex numbers, we get the final result $(1)$.\n\n\u2022 Clever expansion! \u2013\u00a0Ron Gordon Sep 29 '13 at 17:51\n\u2022 Very nice answer! \u2013\u00a0Start wearing purple Sep 29 '13 at 17:53\n\nThere is an alternate way to compute this integral w/o a summation over $n$ in the middle steps.\n\nNotice\n\n$$\\frac{2-\\cos z}{5 - 4\\cos z} = \\frac12 \\left[\\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\\right] = \\frac12\\left[\\frac{1}{2-e^{iz}} + \\frac{1}{2-e^{-iz}}\\right]$$ and $\\displaystyle\\;\\frac{1}{1+z^4}$ is an even function, we have\n\n$$\\mathcal{I} \\stackrel{def}{=} \\int_0^\\infty \\frac{2-\\cos x}{(1+x^4)(5-4\\cos x)} dx = \\frac12\\int_{-\\infty}^\\infty \\frac{1}{(1+z^4)(2 - e^{iz})}dz$$\n\nSince $\\displaystyle\\;\\frac{1}{2-e^{iz}}$ is entire on the upper half plane $\\Im z \\ge 0$ and $\\displaystyle\\;\\left|\\frac{1}{2-e^{iz}}\\right| \\le \\frac{1}{2-1} = 1$ there, we can complete the contour in the upper half-plane and\n\n$$\\mathcal{I} = \\lim_{R\\to\\infty}\\frac12 \\oint_{C_R} \\frac{1}{(1+z^4)(2 - e^{iz})}dz \\quad\\text{ where }\\quad C_R = [-R, R ] \\cup \\big\\{\\; Re^{i\\theta} : \\theta \\in [0,\\pi]\\big\\}$$\n\n$\\displaystyle\\;\\frac{1}{1+z^4}$ has $4$ poles $\\omega_k = e^{\\frac{(2k+1)i}{4\\pi}}, k = 0..3$ over $\\mathbb{C}$. Two of them $\\omega_0 = \\frac{1+i}{\\sqrt{2}}$ and $\\omega_1 = \\frac{-1+i}{\\sqrt{2}}$ belongs to the upper half-plane. Since\n\n$$\\frac{1}{1+z^4} = \\sum_{k=0}^3 \\frac{1}{(z-\\omega_k) 4\\omega_k^3} = -\\frac14 \\sum_{k=0}^3\\frac{\\omega_k}{z-\\omega_k}$$\n\nThe residue of the integrand at $\\omega_k$ are $\\displaystyle\\;-\\frac14 \\frac{\\omega_k}{2 - e^{i\\omega_k}}$ for $k = 0, 1$. This leads to\n\n\\begin{align} \\mathcal{I} &= \\frac12\\left[ -\\frac{2\\pi i}{4}\\left(\\frac{\\omega_0}{2 - e^{i\\omega_0}} + \\frac{\\omega_1}{2 - e^{i\\omega_1}} \\right)\\right]\\\\ &= \\frac{-\\pi i}{4\\sqrt{2}}\\left( \\frac{1+i}{2 - e^{-1/\\sqrt{2}} e^{i/\\sqrt{2}}} + \\frac{-1+i}{2 - e^{-1/\\sqrt{2}} e^{-i/\\sqrt{2}}}\\right)\\\\ &= \\frac{-\\pi i}{4\\sqrt{2}} \\left[ \\frac{ (1+i)(2 - e^{-1/\\sqrt{2}} e^{-i/\\sqrt{2}}) + (-1+i)(2 -e^{-1/\\sqrt{2}} e^{ i/\\sqrt{2}}) }{ 4 - 4 e^{-1/\\sqrt{2}}\\cos\\left(\\frac{1}{\\sqrt{2}}\\right) + e^{-\\sqrt{2}} } \\right]\\\\ &= \\frac{\\pi}{2\\sqrt{2}} \\left[ \\frac{2 - e^{-1/\\sqrt{2}}\\left( \\cos\\left(\\frac{1}{\\sqrt{2}}\\right) - \\sin\\left(\\frac{1}{\\sqrt{2}}\\right) \\right) }{ 4 - 4 e^{-1/\\sqrt{2}}\\cos\\left(\\frac{1}{\\sqrt{2}}\\right) + e^{-\\sqrt{2}} } \\right]\\\\ &= \\frac{\\pi e^{1/\\sqrt{2}} }{2\\sqrt{2}} \\left[ \\frac{2 e^{1/\\sqrt{2}} - \\left( \\cos\\left(\\frac{1}{\\sqrt{2}}\\right) - \\sin\\left(\\frac{1}{\\sqrt{2}}\\right) \\right) }{ 4 e^{\\sqrt{2}} - 4 e^{1/\\sqrt{2}}\\cos\\left(\\frac{1}{\\sqrt{2}}\\right) + 1 } \\right] \\end{align} Reproducing what Vladimir get in his answer.", "date": "2019-05-23 21:54:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/508308/a-definite-integral-int-0-infty-frac2-cos-x-left1x4-right-left5-4", "openwebmath_score": 0.9856529831886292, "openwebmath_perplexity": 645.9097491669671, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426473232578, "lm_q2_score": 0.9149009468718409, "lm_q1q2_score": 0.8911525403296031}}
{"url": "https://www.physicsforums.com/threads/show-that-you-can-distribute-powers-to-commuting-elements.950679/", "text": "# Show that you can distribute powers to commuting elements\n\n## Homework Statement\n\nIf ##a## and ##b## are commuting elements of ##G##, prove that ##(ab)^n = a^nb^n## for all ##n \\in \\mathbb{Z}##.\n\n## The Attempt at a Solution\n\nWe prove two lemmas:\n1) If ##a## and ##b## commute, then so do their inverses: ##ab=ba \\implies (ab)^{-1} = (ba)^{-1} \\implies b^{-1}a^{-1} = a^{-1}b^{-1}##.\n\n2) If ##a## and ##b## commute, then ##b^n a = ab^n##: Base case is trivial. Suppose for some ##k## we have ##b^ka = ab^k##. Then ##b^{k+1}a = bb^ka = bab^k = abb^k = ab^{k+1}##.\n\nNow to the actual result.\n\nClearly ##(ab)^0 = a^0b^0##. So first we prove the result for the positive integers, by induction. The base case is trivial. Suppose for some ##k \\in \\mathbb{Z}^+## we have ##(ab)^k = a^kb^k##. Then ##(ab)^{k+1} = (ab)^k(ab) = a^kb^kab = a^kab^kb = a^{k+1}b^{k+1}##.\n\nNow we prove the result for negative integers. ##(ab)^{-n} = (b^{-1}a^{-1})^n = (a^{-1}b^{-1})^n = (a^{-1})^n(b^{-1})^n = a^{-n}b^{-n}##.\n\nDoes this argument work? Were the lemmas really necessary or could I have assumed they held since their proofs are trivial?\n\nmfb\nMentor\nsince their proofs are trivial?\nSo is the whole statement you have to show. Better to show it explicitly.\n\nfresh_42\nMentor\nThere is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all ##n \\in \\mathbb{N}_0## and for all ##a,b \\in G##, then you have also proven it for inverse elements. This is in words what you have written.\n\nTo do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes.\n\nMr Davis 97\nThere is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all ##n \\in \\mathbb{N}_0## and for all ##a,b \\in G##, then you have also proven it for inverse elements. This is in words what you have written.\n\nTo do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes.\nBut isn't it the case that I am not proving for all ##a,b \\in G##, rather just in the case ##a,b## commute?\n\nfresh_42\nMentor\nBut isn't it the case that I am not proving for all ##a,b \\in G##, rather just in the case ##a,b## commute?\nYes, sure. But that doesn't change by taking the inverses: they commutate if and only if ##a## and ##b## do.\n\nMr Davis 97\nStoneTemplePython", "date": "2021-04-20 11:12:27", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/show-that-you-can-distribute-powers-to-commuting-elements.950679/", "openwebmath_score": 0.9905398488044739, "openwebmath_perplexity": 1939.152515500805, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9766692288147081, "lm_q2_score": 0.9124361652391385, "lm_q1q2_score": 0.891148325846759}}
{"url": "http://www.southernmines.org/water-pressure-rzyocn/transpose-matrix-definition-d896ca", "text": "# transpose matrix definition\n\nA new matrix is obtained the following way: each [i, j] element of the new matrix gets the value of the [j, i] element of the original one. Transpose of a matrix is the interchanging of rows and columns. In practical terms, the matrix transpose is usually thought of as either (a) flipping along the diagonal entries or (b) \u201cswitching\u201d the rows for columns. Related to Transpose of a matrix: adjoint of a matrix , Inverse of a matrix What happened? Noun . So we now get that C transpose is equal to D. Or you could say that C is equal to D transpose. The number $$4$$ was in the first row and the second column and it ended up in the second row and first column. In other words if A= [aij], then At ji = aij. The matrix B is called the transpose of A. This is the definition of a transpose. The way the concept was presented to me was that an orthogonal matrix has orthonormal columns. The transpose of a matrix is defined as a matrix formed my interchanging all rows with their corresponding column and vice versa of previous matrix. For permissions beyond the \u2026 See more. Definition of transpose : a matrix formed by interchanging the rows and columns of a given matrix - change the order or arrangement of - transfer from one place or period to another - cause to change places - transfer a quantity from one side of an equation to the other side reversing its sign, in order to maintain equality Now this is pretty interesting, because how did we define these two? Find the transpose of that matrix. For example if you transpose a 'n' x 'm' size matrix you'll get a new one of 'm' x \u2026 synonym : reverse . Definition. The transpose of an m \u00d7 n matrix A is the n \u00d7 m matrix A T whose rows are the columns of A. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. Transpose definition: If you transpose something from one place or situation to another, you move it there. So if X is a 3x2 matrix, X' will be a 2x3 matrix. Dictionary ! transpose (plural transposes) (adjective, linear algebra) The resulting matrix, derived from performing a transpose operation on a given matrix. Solution: It is an order of 2*3. This is a transpose which is written and A superscript T, and the way you compute the transpose of a matrix is as follows. Let A be a nonsingular matrix. If there are two Matrix with dimension A (2 x 3 ) and B (3 x 2). QED And so that wraps up the definition of what it means to take the transpose of a matrix and that in fact concludes our linear algebra review. Meaning of Transpose. Here are a couple of ways to accomplish this in Python. The conjugate transpose of a matrix is the matrix defined by where denotes transposition and the over-line denotes complex conjugation. All content on this website, including dictionary, thesaurus, literature, geography, and other reference data is for informational purposes only. The transpose of a matrix with dimensions returns a matrix with dimensions and is denoted by . The definition of the transpose is as follows. en.wiktionary.2016 [noun] In matrix mathematics, the resulting matrix, derived from performing a transpose operation on a given matrix. The first thing to know is that you can separate rows by semi-colons (;) and that you define rows by just placing elements next to one another. Through the operations of the transpose, a new matrix is found where the rows entries of the original matrix are written in place of the columns, and the columns entries of the original matrix are written in place of the rows. Do the transpose of matrix. Definition. Matrix transpose: The transpose of matrix refers to the interchanging of the rows and columns. permute, commute, transpose (verb) change the order or arrangement of Remember that the complex conjugate of a matrix is obtained by taking the complex conjugate of each of its entries (see the lecture on complex matrices). In other words, the ij entry of A T is a ji. Initialize a 2D array to work as matrix. The first column became the first row and the second column became the second row. \"transpose\" (matrix) definition: a matrix formed by interchanging the rows and columns of a given matrix. We said that our matrix C is equal to the matrix product A and B. The transpose of a matrix by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. (Problems and Solutions in Linear Algebra. ) A double application of the matrix transpose achieves no change overall. | Meaning, pronunciation, translations and examples And we said that D is equal to our matrix product B transpose times A transpose. At t = A; 2. In order to state the transpose property, we need to define the transpose of a matrix. It is denoted as X'. Then At, the transpose of A, is the matrix obtained by interchanging the rows and columns of A. The transpose will also be of dimension (2x2). A matrix is usually shown by a capital letter (such as A, or B) Each entry (or \"element\") is shown by a lower case letter with a \"subscript\" of row,column: In this video we discuss about the another type of #Transpose of #Matrix with definition and example.If you have a question about then comment here . Dictionary Thesaurus Examples Sentences Quotes ... A matrix obtained by interchanging the rows and columns of a given matrix. Disclaimer. This will be the left hand side of (AB)\u22ba=B\u22baA\u22ba Solving for right hand side, if I take transpose of A and B then the dimension of resultant matrix \u2026 By, writing another matrix B from A by writing rows of A as columns of B. Do the transpose of matrix. And that's it. Stack Exchange Network. Consequently At is n m. Here are some properties: 1. We can find its transpose by swapping the column and row elements as follows. Definition of Transpose in the Definitions.net dictionary. We put a \"T\" in the top right-hand corner to mean transpose: Notation. Adjacency Matrix Definition. ... (0.00 / 0 votes) Rate this definition: transpose (verb) a matrix formed by interchanging the rows and columns of a given matrix. Let\u2019s start by defining matrices. To \"transpose\" a matrix, swap the rows and columns. transpose: To reverse or transfer the order or place of; interchange. Ask Question Asked 4 years, 3 months ago. Derived terms We prove that the transpose of A is also a nonsingular matrix. The algorithm of matrix transpose is pretty simple. Consider the following example-Problem approach. Thus the $$3\\times 2$$ matrix became a $$2\\times 3$$ matrix. Therefore it occurred to me that the definition in the book of Weinberg is not consistent with that in the book of Tung: in one of them the symbol ${\\Lambda_\\mu}^\\nu$ is defined as the inverse of the Lorentz transformation of contravariant vectors, while in the other case, the same symbol is defined as the transpose of the original matrix. What does Transpose mean? TRANSPOSE OF A MATRIX DEFINITION. Store values in it. We have: . The element at ith row and jth column in X will be placed at jth row and ith column in X'. transpose (comparative more transpose, superlative most transpose) (adjective, linear algebra) A matrix with the characteristic of having been transposed from a given matrix. Synonyms: tr. Menu. Definition. For a matrix defined as = , the transpose matrix is defined as = . For example, consider a matrix . (The transpose of a matrix) Let Abe an m nmatrix. Find transpose by using logic. Transpose definition, to change the relative position, order, or sequence of; cause to change places; interchange: to transpose the third and fourth letters of a word. Example 1: Consider the matrix . Learn more. Recommended: Please solve it on \u201c PRACTICE \u201d first, before moving on to the solution. Dimension also changes to the opposite. (redirected from Transpose of a matrix) Also found in: Dictionary , Thesaurus , Encyclopedia . Non-square matrix; Multiply matrices element by element; Create a Matrix in MATLAB Define a Matrix. ... Why is the inverse of an orthogonal matrix equal to its transpose? Example 2: Consider the matrix . Or is it a definition? Transpose of a matrix is obtained by changing rows to columns and columns to rows. [noun] In matrix mathematics, the process of rearranging elements in a matrix, by interchanging their respective row and column positional indicators. The product will be of size (2 x 2). transpose definition: 1. to change something from one position to another, or to exchange the positions of two things\u2026. The adjacency matrix, also called the connection matrix, is a matrix containing rows and columns which is used to represent a simple labelled graph, with 0 or 1 in the position of (V i , V j) according to the condition whether V i and V j are adjacent or not. , transpose ( verb ) change the order or arrangement of Dictionary ;. Of size ( 2 X 2 ) ], then At, the entry. [ noun ] in matrix mathematics, the ij entry of a Duane Q. Nykamp licensed! Need to define the transpose of a matrix obtained by interchanging the rows columns. 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To columns and columns of B 2 * 3 pronunciation, translations and Examples Disclaimer Create! A transpose operation on a given matrix X will be a 2x3 matrix MATLAB... Transpose times a transpose application of the matrix, the transpose of a is! Row elements as follows, writing another matrix B is called the transpose of a matrix is the \u00d7...", "date": "2021-04-10 18:19:48", "meta": {"domain": "southernmines.org", "url": "http://www.southernmines.org/water-pressure-rzyocn/transpose-matrix-definition-d896ca", "openwebmath_score": 0.6923036575317383, "openwebmath_perplexity": 714.7809239474072, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9863631623133665, "lm_q2_score": 0.9032942164664239, "lm_q1q2_score": 0.8909761398531966}}
{"url": "https://math.stackexchange.com/questions/2217630/proving-the-existence-and-number-of-real-roots-for-x3-3x-2", "text": "# Proving the existence and number of *real* roots for $x^3 - 3x + 2$\n\nI need to find how many real roots this polynomial has and prove there existence. I was wondering if my logic and thought process was correct.\n\nDetermine the number of real roots and prove it for $x^3 - 3x + 2$\n\nFirst, note that $f'(x) = 3x^2 - 3$ and so\n\n$f'(x) > 0$ for $x \\in (-\\infty, -1) \\cup (1, \\infty)$ and since $f'$ is strictly increasing on those intervals, there can be at most one root in each of them.\n\n$f'(x) < 0$ for $x \\in (-1,1)$ and since $f'$ is strictly decreasing on this interval it can have at most one root.\n\nNow examine $f(-3) = -16$ and $f(-1) = 4$. By the Intermediate Value Theorem (IVT) $f(c) = 0$ for some $c \\in (-3, 1)$ and so $f$ has a root on the interval $(-\\infty, 1)$.\n\nAgain examine $f(-1) = 4$ and $f(1) = 0$. We cannot say anything about $f$ having a root on the interval $(-1, 1)$.\n\nLikewise examine $f(1) = 0$ and $f(3) = 16$. Again, we cannot say anything about $f$ having a root on $(1, \\infty)$.\n\nHowever, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two.\n\nHence, $f(x)$ has two real roots.\n\nAlso, do we say two real roots (because of the multiplicity), or three real roots, or do we say two distinct real roots?\n\nWhile I realize factoring the polynomial gives me the answer I believe the purpose of the question was to do the former analysis, which when the polynomial isn't easily factorized, can provide a lot of insight. That is why I did it all\n\n\u2022 I assume you mean real roots, because it has 3 complex roots. \u2013\u00a0\u00cdgj\u00f8gnumMeg Apr 4 '17 at 13:51\n\u2022 \"First, note that $f\u2032(x)=3x-3$ and so\", you mean $f\u2032(x)=3x^{\\color{red}{2}}-3$...? \u2013\u00a0StackTD Apr 4 '17 at 13:52\n\u2022 Haha yes! Let me make these adjustments. \u2013\u00a0student_t Apr 4 '17 at 13:52\n\nHowever, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two.\n\nAll the work you did before this becomes unnecessary; after factoring, the roots (and hence the number of roots) are clear - right?\n\nAddition after some comments: when you are asked about the number of roots (real or not), it is usually meant to count the number of distinct (i.e. different) roots. Your equation has two (real) roots, one of which has multiplicity 2 but that doesn't change the fact that there are only two real numbers where the polynomial becomes 0.\n\n\u2022 Yes, but I think the purpose of the exercise was to do the former analysis. I only did that to show the multiplicity of the root. But yes in general all my work before would have been a waste haha! \u2013\u00a0student_t Apr 4 '17 at 13:55\n\nSince we have $$x^3-3x+2=(x-1)^2(x+2),$$ we have three real roots $1,1,-2$. Here we count with multiplicities (which is standard for many results in geometry and other areas).\n\n\u2022 I would say there are two real roots, one of which has multiplicity two. \u2013\u00a0gandalf61 Apr 4 '17 at 13:56\n\u2022 Yes, but I am suppose to do the little analysis before for the question I believe. Of course factoring would be much faster. \u2013\u00a0student_t Apr 4 '17 at 13:57\n\u2022 @danny Yes, this may be the case. But I think, it does not matter so much what you are supposed to do or think, but what you yourself think is the best way. \u2013\u00a0Dietrich Burde Apr 4 '17 at 13:59\n\u2022 I agree with gandalf61. In the context of the fundamental theorem of algebra, we often say an $n$th-order polynomial \"has $n$ complex roots\" but this is an abbreviation where we mean to count the multiplicities. That doesn't change the fact that $x^3$ only has one (distinct) root. \u2013\u00a0StackTD Apr 4 '17 at 13:59\n\u2022 @danny see comment above; I would say yours has two (distinct) roots, we usually omit 'distinct' and mean different roots when we're talking about the number of roots. \u2013\u00a0StackTD Apr 4 '17 at 14:00", "date": "2019-08-20 08:17:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2217630/proving-the-existence-and-number-of-real-roots-for-x3-3x-2", "openwebmath_score": 0.7924861311912537, "openwebmath_perplexity": 280.92575364444633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806518175514, "lm_q2_score": 0.9086179062123119, "lm_q1q2_score": 0.8909731387267675}}
{"url": "https://math.stackexchange.com/questions/381243/integral-of-sin-x-cdot-cos-x", "text": "# Integral of $\\sin x \\cdot \\cos x$ [duplicate]\n\nI've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?\n\n$\\int \\sin x \\cos x dx$\n\n1) via subsitution $u = \\sin x$ $u = \\sin x; du = \\cos x dx \\Rightarrow \\int udu = \\frac12 u^2 \\Rightarrow \\int \\sin x \\cos x dx = \\frac12 \\sin^2 x$\n\n2) via subsitution $u = \\cos x$ $u = \\cos x; du = -\\sin x dx \\Rightarrow -\\int udu = -\\frac12 u^2 \\Rightarrow \\int \\sin x \\cos x dx = -\\frac12 \\cos^2 x = -\\frac12 (1 - \\sin^2 x) = -\\frac12 + \\frac12 \\sin^2 x$\n\n3) using $\\sin 2x = 2 \\sin x \\cos x$\n\n$\\int \\sin x \\cos x dx = \\frac12 \\int \\sin 2x = \\frac12 (- \\frac12 \\cos 2x) = - \\frac14 \\cos 2x = - \\frac14 (1 - 2 \\sin^2 x) = - \\frac14 + \\frac12 \\sin^2 x$\n\nSo, we have: $$\\frac12 \\sin^2 x \\neq -\\frac12 + \\frac12 \\sin^2 x \\neq - \\frac14 + \\frac12 \\sin^2 x$$\n\n\u2022 Ahh! I always give this problem in my first year calculus course. \u2013\u00a0Jyrki Lahtonen May 4 '13 at 14:40\n\u2022 $+C$... ${}{}{}$ \u2013\u00a0David Mitra May 4 '13 at 14:40\n\u2022 $*$ is usually used for convolution in this context. I removed it. \u2013\u00a0Ayman Hourieh May 4 '13 at 14:42\n\u2022 @AymanHourieh: Didn't I? \u2013\u00a0Inceptio May 4 '13 at 14:44\n\u2022 @Inceptio Check the edit history. I edited the body; you edited the title. \u2013\u00a0Ayman Hourieh May 4 '13 at 14:49\n\nAntiderivatives are only unique up to adding a constant ('of integration'). If you were to stick limits in your integrals then you'd always get the same number.\n\n\u2022 Oh, yes. I checked it with 0 and \\pi/2 and those 'strange' fractions deducted each other. Thank you \u2013\u00a0Jimch May 4 '13 at 14:46\n\nNote: You are calculating indefinite integral and constants can be anything(they may differ). In fact the general solution to that would be just $C+\\dfrac{\\sin^2 x}{2}$\n\n$$\\frac{d\\{f(x)+c\\}}{dx}=f'(x)$$ for any arbitrary constant $c$\n\n$$\\implies \\int f'(x)dx=f(x)+d$$ for any arbitrary constant $d$\n\nSo, in indefinite integral we can get answers which differ by some constant\n\nA primitive is unique up to a constant", "date": "2019-12-13 09:23:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/381243/integral-of-sin-x-cdot-cos-x", "openwebmath_score": 0.8569950461387634, "openwebmath_perplexity": 1471.5342871800144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104972521578, "lm_q2_score": 0.9219218375365862, "lm_q1q2_score": 0.8909549414413552}}
{"url": "https://math.stackexchange.com/questions/1113556/how-to-show-that-sum-k-1n-kn1-k-binomn23", "text": "# How to show that $\\sum_{k=1}^n k(n+1-k)=\\binom{n+2}3$?\n\nWhile thinking about another question I found out that this equality might be useful there: $$n\\cdot 1 + (n-1)\\cdot 2 + \\dots + 2\\cdot (n-1) + 1\\cdot n = \\frac{n(n+1)(n+2)}6$$ To rewrite it in a more compact way: $$\\sum_{k=1}^n k(n+1-k)=\\frac{n(n+1)(n+2)}6.$$\n\nThis equality is relatively easy to prove: $$\\sum_{k=1}^n k(n+1-k)= (n+1)\\sum_{k=1}^n k - \\sum_{k=1}^n k^2 = (n+1) \\frac{n(n+1)}2 - \\frac{n(n+1)(2n+1)}6 = n(n+1) \\left(\\frac{n+1}2-\\frac{2n+1}6\\right) = n(n+1)\\frac{3(n+1)-(2n+1)}6 = \\frac{n(n+1)(n+2)}6.$$ (We only used the known formulas for the sum of the first $n$ squares and the sum of the first $n$ numbers.)\n\nAre there some other nice proofs of this equality? (Induction, combinatorial arguments, visual proofs, ...)\n\nEDIT: Now I found another question which asks about the same identity: Combinatorial interpretation of a sum identity: $\\sum_{k=1}^n(k-1)(n-k)=\\binom{n}{3}$ (I have tried to search before posting. But the answers posted here so far gave me some new ideas for good keywords to search which lead me to finding that question.) The questions are, in my opinion, not exact duplicates since the other question asks specifically about combinatorial proofs and my question does not have that restriction. But I agree that this is a very minor distinction. In any case, if you think that one of them should be closed as a duplicate, then you can vote to close. I will refrain from voting to close/reopen on this question. (If one of the two questions is voted to be a duplicate of the other one, they probably cannot be merged, since the summation variables are off by one.)\n\n\u2022 Generating functions? Coefficient of $x^{n+1}$ in $\\left(\\sum\\limits_{n=1}^{\\infty} nx^{n}\\right)^2$ \u2013\u00a0sciona Jan 21 '15 at 14:33\n\u2022 Here is another post about the same formula. \u2013\u00a0Martin Sleziak Oct 2 '15 at 17:31\n\nLet us choose three numbers from $\\{0,1,2,\\ldots, n+1\\}$, beginning with the middle one, which has to be some $k\\in \\{1,\\ldots,n\\}$. We then have $k$ choices for the smallest and $n+1-k$ choices for the largest of the three. It follows that $${n+2\\choose3}=\\sum_{k=1}^n k(n+1-k)\\ .$$\n\n\u2022 Straight to the point! +1 \u2013\u00a0user2345215 Jan 21 '15 at 15:02\n\nI think of this as the \"twelve days of Christmas equality\", because if $n = 12$ then we get $1 \\times 12 + 2 \\times 11 + \\cdots + 12 \\times 1 = {14 \\choose 3}$ and both sides represent the number of gifts which are given in the song \"The Twelve Days of Christmas\". (This happens to be 364, one less than the number of days in a year.)\n\nHere's a combinatorial proof of that equality, which I have previously written about at my blog. As I wrote there, let's try to prove the identity $\\sum_{j=2}^{n+1} (j-1)(n+2-j) = {n+2 \\choose 3}$, which differs from your equality by a change of index. To do this, we count subsets of the set $\\{ 1, 2, \\ldots, n+2 \\}$ of size 3. We can write each such subset as $\\{ x, y, z \\}$ where we require $x < y < z.$ Then we\u2019ll count these subsets according to the difference $z - x$. To construct such a set with $z - x = j$ we must:\n\n\u2022 choose $x$. $x$ must be between $1$ and $n+2-j$ inclusive, so there are $n+2-j$ possible choices.\n\u2022 choose $y$. $y$ must be between $x$ and $z=x+j$ exclusive, so there are $j-1$ possible choices.\n\nAt this point $z$ is fixed. So there are $(j-1)(n+2-j)$ ways to choose a $3$-set with $z - x = j$; summing over the possible values of $j$ gives the desired identity.\n\n\u2022 I guess more-or-less the same argument could be reformulated like this. Let us start by choosing $y$. We only can choose $2, 3, \\dots, n+1$. To make the notation the same as in your sum, let us denote $j=y$. If $y$ is chosen, we can choose any of elements on the left for $x$, which gives $(j-1)$ possibilities. We can choose any of the elements on the right for $z$, which gives $(n+2-j)$ possibilities. Together we have $\\sum_{j=2}^{n+1} (j-1)(n+2-j)$. BTW +1 for the nice combinatorial argument and thanks for the link to your blog. \u2013\u00a0Martin Sleziak Jan 21 '15 at 14:46\n\nA convolution is always a good way. Since: $$\\sum_{k\\geq 0} k z^k = \\frac{z}{(1-z)^2}, \\tag{1}$$ we have: $$\\begin{eqnarray*}\\sum_{k=1}^{n}k(n+1-k) = \\sum_{k=0}^{n+1}k(n+1-k) &=& [z^{n+1}]\\frac{z^2}{(1-z)^4}\\\\&=&[z^n]\\frac{z}{(1-z)^4}\\tag{2}\\end{eqnarray*}$$ and the claim follows from: $$\\sum_{k\\geq 0}\\binom{k+2}{3}z^k = \\frac{z}{(1-z)^4}.\\tag{3}$$\n\n\u2022 I suppose the notation $[z^k]f(z)$ means: \"the coefficient of $z^k$ in the power series $f(z)$\", right? \u2013\u00a0Martin Sleziak Jan 21 '15 at 14:49\n\u2022 @MartinSleziak: yes, it is the standard notation in analytic combinatorics. I am just borrowing it from Wilf or Flajolet. \u2013\u00a0Jack D'Aurizio Jan 21 '15 at 14:52\n\nInduction is pretty straighforward: \\begin{align*}\\sum_{k=1}^{n+1}&k(n+2-k)-\\sum_{k=1}^n k(n+1-k)=(n+1)+\\sum_{k=1}^n k\\cdot 1=\\sum_{k=1}^{n+1} k\\\\&=\\frac{(n+1)(n+2)}2=\\frac{(n+1)(n+2)(n+3)}6-\\frac{n(n+1)(n+2)}6\\end{align*}\n\nHere is another solution. $$n\\cdot 1+(n-1)\\cdot2+\\ldots1\\cdot n=\\\\ (n+(n-1)+\\ldots +1)+((n-1)+(n-2)+\\ldots+1)+\\ldots+1=\\\\ \\sum_{i=1}^ni+\\sum_{i=1}^{n-1}i+\\ldots+1=\\\\ \\frac{(n+1)n}{2}+\\frac{n(n-1)}{2}+\\ldots +\\frac{2\\cdot 1}{2}.$$ Now, this sum is equal $$\\frac{(n+2)(n+1)n}{6}$$ by a straightforward induction.\n\n\u2022 Your equality can be seen also as $\\binom{n+1}2+\\binom{n}2+\\dots+\\binom22=\\binom{n+2}3$, which can be a bit generalized to what some people call hockey-stick identity. See, for example, article on AoPS Wiki or this post (and some of the posts linked to it). \u2013\u00a0Martin Sleziak Jan 22 '15 at 8:13\n\u2022 True, thank you for pointinig it out. \u2013\u00a0Ofir Schnabel Jan 22 '15 at 8:14", "date": "2019-08-24 04:50:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1113556/how-to-show-that-sum-k-1n-kn1-k-binomn23", "openwebmath_score": 0.9279963970184326, "openwebmath_perplexity": 260.42070663162303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357610169274, "lm_q2_score": 0.9099070084811306, "lm_q1q2_score": 0.8908315005029595}}
{"url": "https://math.stackexchange.com/questions/2916482/probability-with-poker-cards", "text": "Probability with Poker cards\n\nYou deal 5 cards from a well-shuffled deck of playing cards. What is the probability that the 5th card is the queen of spades?\n\nJust from analysis, P(5th queen spade) = (51*50*49*48*1)/(52*51*50*49*48) = 1/52\n\nHowever why wont this method of logic thinking incorrect?\n\nP(5th queen spade) = (51Cr4) / (52Cr5) = 5/52. Reasoning: choose any first 4 cards and last card is queen spade, divide by all possible choice\n\n\u2022 Isn't that fifth card as likely to be the queen of spades as the three of clubs? \u2013\u00a0Lord Shark the Unknown Sep 14 '18 at 6:46\n\u2022 But what if the three of clubs was selected before the 5th card? Is it the order of when cards are drawn does not matter? \u2013\u00a0userName Sep 14 '18 at 6:49\n\u2022 You are drawing four cards and then a fifth card. You don't care about order of the first four cards, but pay special attention to the fifth. Count ways to do so in both numerator and denominator (that is: favoured event and total space). \u2013\u00a0Graham Kemp Sep 14 '18 at 8:28\n\u2022 @graham So since first four cards order does not matter, the probability of that cancel each other in numerator and denominator is that what it is? \u2013\u00a0userName Sep 14 '18 at 8:39\n\nYour (51Cr4) / (52Cr5), or as I prefer to write it $\\dfrac{51 \\choose 4}{52\\choose 5}$, is the probability that the first five cards contain the Queen of Spades.\n\nIf that happens, there then is a $1$ in $5$ chance that the Queen of Spades is the fifth of these five cards,\n\nmaking the result to the original question $\\dfrac5{52}\\times \\dfrac{1}{5} = \\dfrac{1}{52}$\n\n\u2022 Oh okay this makes sense... So does it mean that if I use the formula P(number of sample points in an event) / P(number of sample points in sample space) it will always be like the explanation that you have given? \u2013\u00a0userName Sep 14 '18 at 6:55\n\u2022 @userName - it depends on the precise question, especially on whether each element of the sample space is equally likely \u2013\u00a0Henry Sep 14 '18 at 7:06\n\u2022 What about this question? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? I consider P(first 4 cards hearts ^ last card queen of spade) / P(first 4 cards hearts) = (13Cr4 * 1 / 5) / (13Cr4 x 39Cr1 x 1 / 5) why is this reasoning incorrect? [multiplied by 1/5 due to the order, like in the previous example] correct ans: 1/48 \u2013\u00a0userName Sep 14 '18 at 7:16\n\u2022 @userName After you have chosen four particular hearts, there would be $52-4=48$ cards remaining. So you could answer that question with $\\dfrac{{13 \\choose 4}{1 \\choose 1}}{ {13 \\choose 4}{48 \\choose 1}} =\\dfrac{1}{48}$ \u2013\u00a0Henry Sep 14 '18 at 7:35\n\u2022 Alright thanks Henry. But just like to clarify why do I not need to multiply by 1/5 in this situation? Isn't P(first 4 cards hearts and last card queen spade) also mean that we have to be careful and consider that the queen spade is the last card? \u2013\u00a0userName Sep 14 '18 at 8:11\n\nSince you are asking for specifically the fifth card to be $\\spadesuit Q$, this is a problem in which order is important.\n\nThe number of ways to select $5$ different cards with order important is $P(52,5)$. The number of ways in which the fifth card is $\\spadesuit Q$ is $P(51,4)$. So the probability is $$\\frac{P(51,4)}{P(52,5)}=\\frac{51\\times50\\times49\\times48}{52\\times51\\times50\\times49\\times48}=\\frac1{52}\\ .$$ Your mistake was to use $C$s instead of $P$s.\n\n\u2022 Thanks David. Can I check how can the use of Permutation solve this qn other qn? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? If I use that formula I got stuck when (13Pr4) / (13Pr4 x 39Cr1) which gives 1/39... The correct ans is 1/48 instead \u2013\u00a0userName Sep 14 '18 at 7:27\n\u2022 Is it suppose to be (13Pr4) / (13Pr4 x 48Cr1) since I only select 4 cards and now left with 48 cards to choose from? \u2013\u00a0userName Sep 14 '18 at 7:30\n\u2022 You seem to have got an answer from someone else so I trust that is OK. But can I point out that on this site it is preferred that new questions should not be asked in comments. This is because it makes it harder for other people to find the question later. It would be better if you asked a new question. You can always cross-reference your present question if relevant. Thanks. \u2013\u00a0David Sep 17 '18 at 0:09\n\n$\\def\\cbinom#1#2{{^{#1}\\mathsf C_{#2}}}\\frac{\\cbinom{51}4\\cbinom 11}{\\cbinom{52}5}$ is the probability for selecting the queen of spades and four from the fifty-one other cards when selecting any five from all fifty two cards. It does not consider the fifth place as a seperate group from the first four.\n\nYou require the probability for selecting any four from the fifty-one non-queen of spaces then the queen of spades when selecting any four from fifty-two cards then any one from the fourty-eight other card.\n\nThat is $\\frac{\\cbinom {51}4\\cbinom 11}{\\cbinom {52}4\\cbinom{48}{1}}$, which equals $1/52$.", "date": "2019-06-26 04:37:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2916482/probability-with-poker-cards", "openwebmath_score": 0.6646839380264282, "openwebmath_perplexity": 520.5258183702905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.987568347317383, "lm_q2_score": 0.9019206791658466, "lm_q1q2_score": 0.8907083145351867}}
{"url": "http://mathhelpforum.com/algebra/25338-algebra.html", "text": "1. ## Algebra\n\nQ1= show that if x is real, 2(x^2) + 6x + 9 is always positive.\n\nQ2= Solve the simultaneous equations for x,y > 0\n\n2log y = log 2 + log x\n\n2^y = 4^x\n\ncheers\n\n2. for Q1 check the discriminant of the quadratic $b^2 -4ac$\n\nfor Q2 using the first equation you should be able to get $\\log y^2 = \\log 2x \\Rightarrow y^2 = 2x$ using the laws of logs.\n\nlog both sides on the second equation and it should be easy form there.\n\n3. Originally Posted by sparky69er\nQ1= show that if x is real, 2(x^2) + 6x + 9 is always positive.\n\nQ2= Solve the simultaneous equations for x,y > 0\n\n2log y = log 2 + log x\n\n2^y = 4^x\n\ncheers\nAn alternative \"looking\" way to show Q1 would be completing squares and examining the expression.\n$\n2(x^2 + 3x + \\frac92) = 2([x^2 + 2.\\frac32.x + (\\frac32)^2] + \\frac92 - (\\frac32)^2) = 2(x + \\frac32)^2 + \\frac94$\n\nNow since $(x + \\frac32)^2$ is always greater than or equal to zero, for all real x, and since $\\frac94$ is positive , its sum is strictly positive.\n\n4. Originally Posted by bobak\nfor Q1 check the discriminant of the quadratic $b^2 -4ac$\n\nfor Q2 using the first equation you should be able to get $\\log y^2 = \\log 2x \\Rightarrow y^2 = 2x$ using the laws of logs.\n\nlog both sides on the second equation and it should be easy form there.\nHi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????\n\n5. Hello, sparky69er!\n\n1) Show that if $x$ is real, $2x^2 + 6x + 9$ is always positive.\n\nWe have: . $x^2 + x^2 + 6x + 9 \\:=\\:x^2 + (x+3)^2$\n\nThe sum of two squares is always nonnegative.\n\nWhen $x = 0$, the polynomial has a minimum value of 9.\n\n2) Solve the simultaneous equations for x, y > 0\n\n. . $\\begin{array}{cc}2\\log y \\:=\\: \\log 2 + \\log x & {\\color{blue}[1]} \\\\ 2^y \\: =\\: 4^x & {\\color{blue}[2]} \\end{array}$\n\nEquation [1] is: . $2\\log y \\:=\\:\\log(2x)\\;\\;{\\color{blue}[3]}$\n\nFrom Equation [2]: . $2^y \\:=\\:(2^2)^x \\:=\\:2^{2x}\\quad\\Rightarrow\\quad y \\:=\\:2x\\;\\;{\\color{blue}[4]}$\n\nSubstitute into [3]: . $2\\log(2x) \\:=\\:\\log(2x)\\quad\\Rightarrow\\quad\\log(2x) \\:=\\:0$\n\n. . Hence: . $2x \\:=\\:1\\quad\\Rightarrow\\quad \\boxed{x \\:=\\:\\frac{1}{2}}$\n\nSubstitute into [4]: . $y \\:=\\:2\\left(\\frac{1}{2}\\right) \\quad\\Rightarrow\\quad\\boxed{ y \\:=\\:1}$\n\n6. Originally Posted by sparky69er\nHi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????\nGiven a quadratic polynomial equation of the form\n$ax^2 + bx + c = 0$,\nwe have solutions\n$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$\n\nThe \"discriminant\" is defined as $D = b^2 - 4ac$. We have three different classes of solution based on the value of the discriminant:\n$D > 0 \\implies$ two real, unequal roots.\n\n$D = 0 \\implies$ one real root. (Or two equal real roots, however you wish to look at it.)\n\n$D < 0 \\implies$ two complex conjugate roots.\n\nIn the case of D < 0 the curve $y = ax^2 + bx + c$ never crosses the x axis. If a > 0 then this would imply that the quadratic is always positive.\n\n-Dan\n\n7. Originally Posted by Soroban\nHello, sparky69er!\n\nWe have: . $x^2 + x^2 + 6x + 9 \\:=\\:x^2 + (x+3)^2$\n\nThe sum of two squares is always nonnegative.\nSmart, really smart solution", "date": "2017-05-23 19:14:15", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/25338-algebra.html", "openwebmath_score": 0.8871765732765198, "openwebmath_perplexity": 570.7701804020886, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9875683476832692, "lm_q2_score": 0.9019206778476933, "lm_q1q2_score": 0.8907083135634205}}
{"url": "http://centrosamo.it/fskn/minimize-a-cost-function.html", "text": "Minimization and maximization refresher. from Wikipedia. The minimization will be performed by a gradient descent algorithm, whose task is to parse the cost function output until it finds the lowest minimum point. To find the profit maximization levels, other approaches can be taken as well. In place of dJ/dTheta-j you will. This means that, in the AC equation, q + 2 are the average variable costs and 100/q are the average fixed costs. Likely, many corporate leaders believe. 1-Input the number. How is the above interpreted? The rm wants to minimize its costs (w 1x 1 + w 2x 2) of producing y units of output. This video explains how to find the average cost function and find the minimum average cost given the total cost function. Yes, even despite having so much support from ml-class \u2026 they practically implement everything and just leave the cost and gradient functions up to you. In this article, I will be going through the basic mathematics behind K-Means Algorithm. 50 which is the amount the firm has reduced their loss by producing instead of shutting down. economic order quantity (eoq) model The economic order quantity (EOQ) is the order quantity that minimizes total holding and ordering costs for the year. The formula is useful for deriving total costs for budgeting purposes, or to identify the approximate profit or loss levels likely to be achieved at certain sales volumes. So the terminology I'm going to use is that the loss function is applied to just a single training example like so. If a firm has a production function Q=F(K,L) (that is, the quantity of output (Q) is some function of capital (K) and labor (L)), then if 2Q V/(\u03c0 r^2) Find r when the slope of the area is zero:. 3) The profit a business makes is equal to the revenue it takes in minus what it spends as costs. so the function is concave up, so x = 18 is the absolute minimum. Your business should be doing the same. If Minimize is given an expression containing approximate numbers, it automatically calls NMinimize. Minimize [{f, cons}, x \u2208 reg] is effectively equivalent to Minimize [{f, cons \u2227 x \u2208 reg}, x]. In ML, cost functions are used to estimate how badly models are performing. For example, when determining optimal cooling protocols, we ultimately only care to minimize ice-related cell death in the tissue,without regard to the state outside of the tissue. I would like to minimize the cost of a function and i have these variables and restrictions. minimize (). The optimal cost is $150. Objectives: To maximize or minimize a two-variable function. output quantity. For some types of costs, the relationship is in direct proportion; for other types, there is a direct trade-off. To delineate CVX specifications from surrounding Matlab code, they are preceded with the statement cvx_begin and followed with the statement cvx_end. Clarification of Answer by livioflores-ga on 28 May 2006 20:01 PDT Hi!! Here is the answer to your second request of clarification: You know that 100 = min(x1,20) + min(x3,x4); Since you are trying to minimize costs it is clear that x1=<20; if not is x1>20 but min{x1,20} is still equal to 20 and this force you to continue using 80 units of x3 and x4; but in this situation you will spend more. Thanks readers for the pointing out the confusing diagram. 1 guitars and 48. Cost complementary exits in a multiproduct cost function when a. Now we're ready to optimize. To do that, we make a function that gives us the wrongness of a particular set of thetas against our training data. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. return_all, available for some methods, gives you the parameter vector vs generation, but not the cost function. In most cases, when you see a decorated function, the decorator is a factory function that takes a function as argument and returns a new function that includes the old function inside the closure. Summary: The goal of the diet problem is to select a set of foods that will satisfy a set of daily nutritional requirement at minimum cost. Cost categories. 01 dollars to manufacture x Xbox 360s in a day. (1) Solve for the cost-minimizing input combination: (2) Depict the optimum in the diagram to the right. To obtain the cost function, add fixed cost and variable cost together. 3) Do not exhaust all system memory. Amid COVID-19, physicians, architects, and consultants are talking change in design \u2014 to be ready for next. Examples least-squares minimize kAx\u2212bk2 2 \u2022 analytical solution x\u22c6 = A\u2020b (A\u2020 is pseudo-inverse) \u2022 can add linear constraints, e. There are two parameters (coefficients) in our cost function we can control: weight $$m$$ and bias $$b$$. So we finally have cost as a function of x. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. prefer parameters that minimize the execution time. 3 is to be constructed in the shape of a rectangular box with a square base and an open top. - Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our website. Return the absolute value. 1541765 To link to this article: https://doi. The available techniques to determine soil moisture content have practical limitations owing to their high cost, dependence on labor, and time consumption. Cost Function A company finds that it costs a total of to produce units of a new product. Rowe Price, troll, uspto. [email protected] Gradient Descent is THE most used learning algorithm in Machine Learning and this post will show you almost everything you need to know about it. the cost function itself!. Then why to use the. This value may be the expected net present value of a project or a forest property; or it may be the cost of a project; it could also be the amount of wood produced, the expected number of visitor-days at a park, the number of endangered species that will be. Eliminate Storage Complexity And Minimize Costs Enterprise IT is under unremitting pressure to reduce capital and operating expenses, driving them to virtualize infrastructure to improve hardware utilization and scalability and advance toward enhanced operational efficiency and flexibility. This elementary framew ork is the basis for a broad v ariety of mac hine learning. The math problem is: A large bin for holding heavy material must be in the shape of a box with an open top and a square base. However, this benefit comes at the cost of high computational complexity. The marketing manager should be interested in revealing the complexities of an individual buyer, the dynamics of consumer behavior and should also try to. #N#function J = computeCost ( X, y, theta) #N#%COMPUTECOST Compute cost for linear regression. I want that \" t and T must be greater than zero(not equal to zero) , t < T and C > 0. 15-2P = 15-2(3)= 15-6=9-6+5P=-6+5(3)=-6+15=9. If you produce a certain amount and let's say you bring in, I don't know,$10,000 of revenue and it costs you $5,000 to produce those shoes, you'll have$5,000 in profit. This study proposes a new framework to minimize the cost function of multi-objective optimization problems by using NSGA-II in economic environments. To delineate CVX specifications from surrounding Matlab code, they are preceded with the statement cvx_begin and followed with the statement cvx_end. Since C0(x) = 30 \u2212 253000 x2, then C. LP problems seek to maximize or minimize some quantity (usually profit or cost). the objective function (maximize/minimize) and. The goal of any Machine Learning model is to minimize the Cost Function. So in training your logistic regression model, we're going to try to find parameters W and B that minimize the overall costs function J written at the bottom. How to minimise the cost function? Our goal is to move from the mountain in the top right corner (high cost) to the dark blue sea in the bottom left (low cost). Minimize operating costs and improve energy performance Data centers have to face continually increasing cost constraints. Gradient Descent basically just does what we were doing by hand \u2014 change the. From an external point of view, it is difficult to ascertain which are the alternative considered. It costs $10 to store one set for a year. The inventory cost problem, however, is something that comes up in real-life manufacturing scenarios all the time - how can I minimize my operating costs? In fact, the problem we see here today is a simplified version of a problem I covered in a DETC conference paper that I published a few years back. Take a deep breath. The point was more to introduce the reader to a specific method, not to the cost function specifically. You can use Pythagoras to compute S in terms of U in terms of S: U^2 = 500 2 + (4000 - S) 2 U = sqrt( 500 2 + (4000 - S) 2) Thus C(S) = S + 5 sqrt( 500 2. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. **TL;DR**: Rearranging the terms in Maximum Mean Discrepancy yields a much better loss function for the discriminator of Generative Adversarial Nets. To determine the optimal amount of inputs (L and K), we solve this minimization constraint using the Lagrange multiplier method:. Inventory cost problems come up in real-life manufacturing scenarios all the time - how can I minimize my operating costs? Hot Bod Jacuzzi & Spa Company is launching a new hot tub - the Neverleak Massage-o-matic DeLux. Write a Cost Function. Thus, the C function represents the. The cost functions implemented in MIPAV: Correlation ratio. The use of closures and factory functions is the most common and powerful use for inner functions. Anthony Vu Patent, Patents \"ask the patent attorney series\", \"The American Invents Act\", aia, collateral estoppel, cost, inter partes review, issue preclusion, litigation, patent, patent litigation, price, secure Axcess, T. 2 Minimize 2 x 2 1 + 2 sub ject to x 1 + 2 =1, if w ec hange the righ t hand side from 1 to 1: 05 (i. How can equations and inequalities help a business maximize profit or minimize costs? Unanswered Questions. for one-variable real functions: limits, integrals, roots This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and plotters, mathematical recreation and games. (c) Test the C. Then it is going to become impossible to properly minimize or maximize the Cost Function. be used to minimize costs and what is the cost of producing that amount of chicks? f. cost of shipping + c21x21 + c22x22 + c23x23 from a plant + c31x31 + c32x32 + c33x33 to the ware house) Supply constraints. That is, the firm must choose a specific point on the q Cost Functions come directly from the production function and prices. In practice, these attractions are balanced in order to maintain a gap between the shaft (rotor) and static parts (stator). This firm minimizes its cost of producing any given output y if it chooses the pair (z 1, z 2) of inputs to solve the problem min z 1,z 2 w 1 z 1 + w 2 z 2 subject to y = F (z 1, z 2), where w 1 and w 2 are the input prices. (A) The Cost Function The cost-minimizing choice of inputs depended on two essential sets of parameters: the given output level (Y) and the given factor prices (r and w). Write an expression for the Cost in terms of only the width (w). Authors: Ga\u00ebl Varoquaux. 25 lines (16 sloc) 791 Bytes. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. There are two parameters (coefficients) in our cost function we can control: weight $$m$$ and bias $$b$$. 20, it cost$6. Operations > Time-Cost. The math problem is: A large bin for holding heavy material must be in the shape of a box with an open top and a square base. Indeed, it is the most powerful method available to reduce product cost, improve quality, and simultaneously reduce development interval. The minimization will be performed by a gradient descent algorithm, whose task is to parse the cost function output until it finds the lowest minimum point. Material indices Introduction The performance, p, Each function has an associated material index. 2x1 1 x2 1 x3 1 x4 x1, x2, x3, x4 $0. Organizations are relying on cloud to maintain business-critical processes, but the journey is not always seamless: you may be grappling with cloud governance and how to keep control over security, costs, risks. So our cost as a function of x is going to be 20x squared 36 times 5. What is the Malayalam name of tukmaria or sabja seed or falooda seed. Likely, many corporate leaders believe. In the case we are going to see, we'll try to find the best input arguments to obtain the minimum value of a real function, called in this case, cost function. Minimize costs The logistics market is characterized by higher standards for air pollution and noise as well as increasing toll fees, personnel costs and fuel prices. Output is produced according to the following process 2 1 2 1 K L = Firm Output (I chose the same function as above to simplify things). 1-Input the number. It's called the cost function, which is kind of a crappy name in this context. 20 to increase production from 49 to 50 units of output. \" The problem also listed these following multiple choice answers: a) 30,000 b) 300 c) 3,000 d) 30 e) None of these Now, we have the correct answer, what we need is the actual way to do this problem. The gradient descent algorithm in a nutshell. To do this, take the derivative of C(x), set it equal to zero, and solve for x. The diet problem constraints typically regulate the number of calories and the. The optimization continues as the cost function response improves iteration by iteration. optimize for black-box optimization: we do not rely on the. Put simply, a cost function is a measure of how wrong the model is in terms of its ability to estimate the relationship between X and y. Cost Minimization: Short Run \u2022 Let us go back to the two-inputs case, with only one of them variable in the short run. We have contributed on a local. Thus you know that the cost is C(S,U) = S + 5U. 9 drums to minimize his costs. Well, your profit as a function of x is just going to be equal to your revenue as a function of x minus your cost as a function of x. C CL(q) combination of inputs that minimize the cost of producing each. Fundamental theorem of linear programming If the optimal (maximum or minimum) value of the objective function in a. Reynolds Consumer Products Inc. The aim of the linear regression is to find a line similar to the blue line in the plot above that fits the given set of training example best. This website uses cookies to ensure you get the best experience. Solution: We would like to find a function that describes this situation. 4 (GP) : minimize f (x) s. \"I tried a lot but I am not getting the values of t and T as mentioned above \" \". The objective function J = f(x) is augmented by the constraint equations through a set of non-negative multiplicative Lagrange multipliers, \u03bb j \u22650. We often design algorithms for GP by building a local quadratic model of f (\u00b7)atagivenpointx =\u00afx. 4x + 150 t?o model the unit cost in dollars for producing x stabilizer bars. INTRODUCTION Linear programming is a mathematical technique used to find the best possible solution in allocating limited resources (constraints) to achieve maximum profit or minimum cost by modelling linear relationships. nan with np. The goal of any Machine Learning model is to minimize the Cost Function. These functions can be seen as covering functions which have many applications in di erent optimization prob-lems: Set Cover functions, Edge Cut functions in graphs, etc. For example, this formula will find the highest value in cells H2:H17 =MAX(H2:H17) MIN IF Formula. Mathematical optimization: finding minima of functions\u00b6. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. How much food should be used to minimize costs and what is the total cost? 2. Thus the function huber can be used anywhere a traditional convex function can be used, in constraints or objective functions, in accordance with the DCP ruleset. Minimize the cost to split a number Given an integer N \u2265 2 , you can split the number as a sum of k integers i. Sourcing, procurement and vendor management leaders should use this research to navigate GDPR requirements and Microsoft\u2019s licensing to avoid legal and financial risks. The cost functions implemented in MIPAV: Correlation ratio. The transaction cost function is not di erentiable at the kink points and is piecewise continuous. Human Resource Management Functions. You will notice that as in the case of the factor demand functions, there is a. 1 guitars and 48. That is, the quantity you want to maximize or minimize is called the objective function. The Dual of the Minimum Cost Flow Problem:. C CL(q) combination of inputs that minimize the cost of producing each. Then why to use the. That's incredible but understandable when you start adding up all the \"standard\" wedding costs. So it's going to be plus 180 times, let's see, x times x to the negative 2, 180x to the negative x to the negative 1 power. The specific goal is to approximate a single valued function of one variable in terms of a sequence of linear segments. To minimize energy content, use the above criteria for. The two interesting exceptions to this rule are:. How to Minimize Legal Liabilities and Risks Information throughout this subsection applies primarily to external consultants. 4 \u2014 Logistic Regression | Cost Function \u2014 [ Machine Learning | Andrew Ng] - Duration: 11:26. What is the Objective Function? The objective of a linear programming problem will be to maximize or to minimize some numerical value. Assume that is costs Microsoft approximately C x x x 2 14,400 550 0. Hi , I am using FMINCON to minimize my cost function which is a product of elements of a matrix. An optimization problem seeks to minimize a loss function. To do that, we make a function that gives us the wrongness of a particular set of thetas against our training data. An extra large server costs you$0. Minimize F x y 22 with xy 2 10. using linear algebra) and must be searched for by an optimization algorithm. Find more Statistics & Data Analysis widgets in Wolfram|Alpha. HRM is the systematic planning and control of a network of fundamental organizational processes affecting and involving all organization members (French, 2004, p. However, this benefit comes at the cost of high computational complexity. For what number of bars is the unit cost at its mimimum? What is the unit cost at that level of production? Haven't got a clue what this problem is asking of me. A feasible solution that minimizes (or maximizes, if that is the goal) the objective function is called an optimal. How much are closing costs? These are the fees paid that help facilitate the sale of a home typically total 2% to 7% of the home's purchase price. applied optimization calc 1. If the material for the sides costs 15\u00a2/in. In this article, I will be going through the basic mathematics behind K-Means Algorithm. For multi-objective improvements, the most generally used developmental algorithms such as NSGA-II, SPEA2 and PESA-II can be utilized. A cost function is a MATLAB \u00ae function that evaluates your design requirements using design variable values. In this paper, we have applied some meta scheduling methods to a model of CIM that is referred to as an automated flow shop, where backward scheduling should be used to realize a JIT's theory. Consider the same open-top box, which is to have volume $$216in. Assume we are given a dataset as plotted by the 'x' marks in the plot above. This is the radius which will minimize the surface area and thus the cost of materials. costs into account. Artificial Intelligence - All in One 87,390 views 11:26. Chap 7: Short-Run Cost Function 2. Cost categories. 6 - Linear Programming. The material that will be used for three sides costs 30 per linear foot, and the material that will be used for the fourth side costs 15 per linear foot. So our cost as a function of x is going to be 20x squared 36 times 5. Decide what the variables are and what the constants are, draw a diagram if appropriate, understand clearly what it is that is to be maximized or minimized. 5 Q 2 v What is the marginal revenue function?. The Dual of the Minimum Cost Flow Problem:. Raw Blame History. 4 million will be recognized as a component of. De\ufb01ne a MATLAB function to evaluate \u2212f(x) given x. In my opinion, the #1 cost to avoid is the \u201cretaker\u201d cost. I want that \" t and T must be greater than zero(not equal to zero) , t < T and C > 0. Dear Sir Can you please help me to minimize the following cost function with maple 10. This function is known as the cost function and will be of considerable interest to us. Does the production function exhibit increasing, decreasing, or constant returns to scale? How can you tell? b. Given a function defined by a set of parameters, gradient descent starts with an initial set of parameter values and iteratively moves toward a set of parameter values that minimize the function. h(\u03b8) is the the prediction from your regression model. Let's take a more in depth look at the cost function and see how it works. Examples least-squares minimize kAx\u2212bk2 2 \u2022 analytical solution x\u22c6 = A\u2020b (A\u2020 is pseudo-inverse) \u2022 can add linear constraints, e. Users who have contributed to this file. This study proposes a new framework to minimize the cost function of multi-objective optimization problems by using NSGA-II in economic environments. The available techniques to determine soil moisture content have practical limitations owing to their high cost, dependence on labor, and time consumption. (d) Find the minimum value of the marginal cost. A cost function is a MATLAB \u00ae function that evaluates your design requirements using design variable values. Ultimately, to minimize our cost, we need to find the point with the lowest z value. You can also optimize the objective function without any loss function, e. Obtain the minimum using fmin=fminsearch(fun,x0) Maximization 1. One common application of calculus is calculating the minimum or maximum value of a function. With so many options to choose from, the best iPhone XR case can be elusive. Currently, minimize lacks the ability to do this. To do: Try the following example: Given: Q = L 1/2 K 1/2 PL = 4, PK = 1 Goal: Produce Qo = 16 units as cheaply as possible. (Remember, the average cost, (6 pts. For example, if the marginal cost of producing the 50th product is 6. If the firm ordered the item, then the setup cost is simply the order cost from Module 5. Cost Function - Intuition I11:09. They would like to offer some combination of milk, beans, and oranges. So, if you employ tactics to reduce costs in all discrete functions from manufacturing through delivery, you'll have a lower total landed cost, right? Theoretically, yes. Now, to minimize marginal cost. If the brewery produces sweet stout alone, the cost function is: CS(q2) = 8q2. Homework Statement Mary Jane grows herbs in her attic. Previous work. The Cost Function If lattes and cake (or labor and capital) have unit prices of pL and pK, respec-tively, then the total cost of purchasing L units of one and K units of the other is C(L,K) = pLL+pKK. I think relative price of L & K is (Cost of Labour Per Hour)/(Cost of Rent Per Hour), but I don't know the price of Rent Per Hour. pdf), Text File (. Next time I will not draw mspaint but actually plot it out. The problem is that officers work 8 hour shifts, yet the demand comes in 4 hour chunks. 5 Actionable Tips to Reduce Operational Costs Regardless of what the circumstances are for your business, it is always a priority to find ways to reduce operational costs. k) 0 is a (nonnegative) function for which (x k;x k) = 0;then the following function de\ufb01nes a majorizer for : \u02da k(x) , (x)+ (x;x k): (4. Generally speaking, Least-Squares Method has two categories, linear and non-linear. The available techniques to determine soil moisture content have practical limitations owing to their high cost, dependence on labor, and time consumption. There can be significant cost savings when a business function is outsourced. Find the level of production which will minimize the average cost per item. 02xSquared - 3. Instead, it is allowable to use a cost flow assumption that varies from actual usage. Cost & Time and Also Minimum Project Duration Using Alternative Method 405 coordinates of the normal and crash points: Cost slope = (crash cost-normal cost)/ (normal duration crash duration) As the activity duration is reduced, there is an increase in direct cost. As the magnitues of the fitting parameters increase, there will be an increasing penalty on the cost function. The extent of risk and liability in your work depends on the nature of your services. Hi , I am using FMINCON to minimize my cost function which is a product of elements of a matrix. Of course, since time is money in any manufacturing process, what this really means is that looking into ways of reducing cycle time in your injection molding process can have a major impact on. What are loss functions? And how do they work in machine learning algorithms? Find out in this article. In the case we are going to see, we'll try to find the best input arguments to obtain the minimum value of a real function, called in this case, cost function. Fundamental theorem of linear programming If the optimal (maximum or minimum) value of the objective function in a. Common benzodiazepines used for GAD include alprazolam, clonazepam, diazepam, and lorazepam. For what number of bars is the unit cost at its mimimum? What is the unit cost at that level of production? Haven't got a clue what this problem is asking of me. Actual costs refer to real transactions, wherease opportunity costs refer to the alternative taken into consideration by decision makers who might want to choose the line of activity which minimise the costs. Average cost is minimized when average cost = marginal cost is another saying that isn\u2019t quite true; in this case, the correct statement is: Average Cost has critical points when Average Cost and Marginal Cost are equal. LP problems seek to maximize or minimize some quantity (usually profit or cost). Find the dimensions that will minimize the cost of the box's construction. Therefore the total cost is: C(x) = 10y +15(2x+y) = 30x+25y. We need to decide which sub-contractor to use for a critical activity. Hence we want to minimize the can's surface area. We refer to this property as the objective function of an LP problem. In this case, the function huber will contain a special Matlab object that represents the function call in constraints and objectives. An objective function is either a loss function or its negative (in specific domains, variously called. It's important to limit your number of serverless functions to avoid having a massive charge when a lot of work is presented at the same time. The average cost of producing one output is reduced when the output of another product is increased If the wage rate is 5 and the price of capital is 2, then in order to minimize costs the firm should use a. 5 kg is the same linear function for a mass change of 2000 kg. For example, companies often want to minimize production costs or maximize revenue. The objective of the purchasing function is to obtain proper material and services when needed at the lowest obtainable cost. 1, Major functions of an institution) on the basis of modified total direct costs (MTDC), consisting of all salaries and wages, fringe benefits, materials and supplies, services, travel, and up to the. txt) or read online for free. Minimize cost and maximize quality of function in four variables. We are the prime contractor and there is a penalty in our contract with the main client for every day we deliver late. In the following example I will minimize an arbitrary function [texi]J[texi], then in the next chapter I'll apply it to the original house pricing task. Loss functions are actually at the heart of these techniques that we regularly use. (1) Solve for the cost-minimizing input combination:. The aim of the linear regression is to find a line similar to the blue line in the plot above that fits the given set of training example best. 5 Q 2 v What is the marginal revenue function?. The major objective of a typi-cal firm is to maximize dollar profits in the long run. Variables and functions should be declared in the minimum scope from which all references to the identifier are still possible. Let's take a more in depth look at the cost function and see how it works. This is where we look back at equation (1) and solve for h in terms of w. We advise on the largest and most complex legal challenges facing the world\u2019s most important companies. the cost function itself!. Set big goals, insist on a cultural shift, and model from the top. The objective function is the function to be minimized or maximized. Use a computer to maximize the objective function subject to the constraints where 38. Thus, applications of HRM theory differ from personnel management in their dismissal of prescriptive \u201cone best way\u201d models of practice as diverse. Minimising Cost function. Overall, closures have affected 25% of pork production and 10% of beef production in the U. Objective function. An optimization problem is one where you have to make the best decision (choose the best investments, minimize your company's costs, find the class schedule with the fewest morning classes, or so on). That is h = 50 3w2 (3) Plugging the value for h from (3) above into equation (2) yields C = 60w2 + 48w. 2, what should the dimensions of the cup be to minimize the construction cost?. Yes, but not by playing it safe. Actually, the objective function is the function (e. Minimizing any function means finding the deepest valley in that function. A cost function is defined as: \u2026a function that maps an event or values of one or more variables onto a real number intuitively representing some \u201ccost\u201d associated with the event. Examples and exercises on the cost function for a firm with two variable inputs Example: a production function with fixed proportions Consider the fixed proportions production function F (z 1, z 2) = min{z 1, z 2} (one worker and one machine produce one unit of output). Variable costs are such cost which vary directly with change in output. Recall in the calculus of one variable, if y = f(x) is defined on a set S, then there is a relative maximum value at x0 if f(x0) \u2265 f(x) for all x in S near x0, and there is a relative. If a firm has a production function Q=F(K,L) (that is, the quantity of output (Q) is some function of capital (K) and labor (L)), then if 2Q V/(\u03c0 r^2) Find r when the slope of the area is zero:. His next-door neighbor agrees to pay for half of the fence that borders her property; Sam will pay the rest of the cost. The nonlinearity in this form generates from the absolute value function. In most examples/tutorial I followed, the cost function used was somewhat arbitrary. Coming up with a cost function for optimization for a complex control system Hot Network Questions Is there a word or phrase for one mistaken belief leading to a web of false ones?. However, this benefit comes at the cost of high computational complexity. : residuals) between our model and our data points. One common application of calculus is calculating the minimum or maximum value of a function. Take Exam Only When You are Ready. the production function and the cost function; the only difference is whether we hold production constant or cost constant. [email protected] I would like to use the goal seek function to minimize the value in a certain cell (total cost) by changing another the value in another cell (shipment size). Then why to use the. This firm minimizes its cost of producing any given output y if it chooses the pair (z 1, z 2) of inputs to solve the problem min z 1,z 2 w 1 z 1 + w 2 z 2 subject to y = F (z 1, z 2), where w 1 and w 2 are the input prices. The Cost Function If lattes and cake (or labor and capital) have unit prices of pL and pK, respec-tively, then the total cost of purchasing L units of one and K units of the other is C(L,K) = pLL+pKK. Find the dimensions that will minimize cost. Employee compensation costs, office space expenses and other costs associated with providing a workspace or manufacturing setup are eliminated and free up resources for other purposes. We often design algorithms for GP by building a local quadratic model of f (\u00b7)atagivenpointx =\u00afx. (d) Find the minimum value of the marginal cost. Cost function is the sum of losses from each data point calculated with loss function. Solving for the minimum 0 points minimize f, (z) = 20 + z2-cos(2TZ) Given the cost function f. S, according to Bloomberg News. To minimize energy content, use the above criteria for. c<=2 n=1-1000 0<=p<=0. It's called the cost function, which is kind of a crappy name in this context. For example, if the marginal cost of producing the 50th product is 6. Minimising Cost function. You\u2019ve been ordered to reduce your department\u2019s costs by 10%, 20%, or 30%. It is obvious that if we changed relative factor. 5\\text{ x }10^6 \\text{ ft}^2) in an a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. Having drawn the picture, the next step is to write an equation for the quantity we want to optimize. profits and minimize costs by using cost-revenue-profit functions. **TL;DR**: Rearranging the terms in Maximum Mean Discrepancy yields a much better loss function for the discriminator of Generative Adversarial Nets. 188, and the slope was not significantly different from 0. By using this website, you agree to our Cookie Policy. \u201cDespite the challenges of the COVID-19. Take the derivative of the Cost with respect to width. Average Cost Per Unit Formula. Many of these materials are high in quality and low in cost. Find the Average Cost Function and Minimize the Average Cost. Outsourcing can also make your firm more attractive to investors, since you're able to pump more capital directly into revenue-producing activities. That\u2019s the main point of any model, to minize error, to perform. How many Xboxes should be manufactured in order to minimize average cost? What is the resulting average cost of an Xbox? Give your answer to the nearest dollar. In this case, the objective is to minimize the total cost per day which is given by z= 0:6x 1 + 0:35x 2 (the value of the objective function is often denoted by z). Cost Function8:12. Indeed, it is the most powerful method available to reduce product cost, improve quality, and simultaneously reduce development interval. If this sounds a lot, here are my 5 suggestions to reduce the CPA exam cost: 1. It is the heart that makes it beat! There is a loss function, which expresses how much the estimate has missed the mark for an individual observation. The use of closures and factory functions is the most common and powerful use for inner functions. The cost function: E(Cost)=E(F-LS) 3 F is for Finished goods L is for Lambda S is for Sales After expanding the function, what assumption minimized this function with respect to F? The Attempt at a Solution F 3-3F 2 LS+3F(LS) 2-(LS) 3 I know that I need to identify the terms that include both sales and inventories. f ( x) = x 4 \u2212 8 x 2 + 5. h(\u03b8) is the the prediction from your regression model. output, marginal cost, average cost, price, and profit at the average-cost minimizing activity level profit-maximizing or loss-minimizing output Calculus Cost/Graph cost function Optimal capital structure to minimize cost of capital Output, Profit, Fixed Costs and Perfect Competition Finding Optimal Output Level etc. Example 4 If the total revenue and total cost functions are TR = 30Q \u2013 5Q 2 and TC = 15 + 12Q - 0. A company producing goods wants to minimize the average cost of production. Mathematical optimization deals with the problem of finding numerically minimums (or maximums or zeros) of a function. The global six-year average cost of a data breach is 3. Employee compensation costs, office space expenses and other costs associated with providing a workspace or manufacturing setup are eliminated and free up resources for other purposes. Yes, even despite having so much support from ml-class \u2026 they practically implement everything and just leave the cost and gradient functions up to you to implement. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. Mathematical optimization: finding minima of functions\u00b6. Keep in mind that, the cost function is used to monitor the. To open Outlook again after it has been hidden on the system tray and disappeared from the taskbar, double-click the Outlook system tray icon. Thus the function huber can be used anywhere a traditional convex function can be used, in constraints or objective functions, in accordance with the DCP ruleset. Minimize the total physical effort & thus the cost of moving goods into & out of storage. Budgeting for your company\u2019s training needs does not mean using surplus money when you have it. Diesel particulate filters (DPF) are devices that physically capture diesel particulates to prevent their release to the atmosphere. Authors: Ga\u00ebl Varoquaux. There are many factors to consider when selecting components and board-level solutions for a real-time embedded system. 1 Where f is the number of facilities. De\ufb01ne a MATLAB function to evaluate \u2212f(x) given x. , a function that takes a scalar as input) is needed. Good parameters means that the function can produce the best possible outcomes, namely the smallest ones, because small values mean less errors. Cij = transportation cost per unit of shipping from plant Pi to the Warehouse Wj. Lagrange multiplier methods involve the modi\ufb01cation of the objective function through the addition of terms that describe the constraints. Mathematical optimization: finding minima of functions\u00b6 Authors: Ga\u00ebl Varoquaux. Min & Max of Functions - MATLAB Minimization 1. Gradient descent is a more generic algorithm, used not only in linear regression problems and cost functions. Along with source pipeline, candidate quality, long-term retention, and other key performance metrics, tracking your cost-per-hire will help you to understand the performance of your recruiting initiatives better and minimize your expenses across the board. Univariate function minimizers (minimize_scalar)\u00b6 Often only the minimum of an univariate function (i. A company producing goods wants to minimize the average cost of production. 15-2P = 15-2(3)= 15-6=9-6+5P=-6+5(3)=-6+15=9. My constriants are also in the form of matix. For example, this formula will find the highest value in cells H2:H17 =MAX(H2:H17) MIN IF Formula. 2% over budget). Then, minimize that slack variable until the slack is null or negative. We have contributed on a local. Minimize an objective function whose values are given by executing a file. Real-time embedded systems require. Grab a coffee. The cost function: E(Cost)=E(F-LS) 3 F is for Finished goods L is for Lambda S is for Sales After expanding the function, what assumption minimized this function with respect to F? The Attempt at a Solution F 3-3F 2 LS+3F(LS) 2-(LS) 3 I know that I need to identify the terms that include both sales and inventories. Profit is simply the Total revenue minus the costs incurred. Gradient descent is simply used to find the values of a function's parameters (coefficients) that minimize a cost function as far as possible. How To: Calculate and use regression functions in statistical analysis How To: Write a logarithm as a sum or difference of logarithms How To: Perform a quadratic regression with a calculator How To: Calculate r-squared or coefficient of determination in statistics. Of course, since time is money in any manufacturing process, what this really means is that looking into ways of reducing cycle time in your injection molding process can have a major impact on. More generally, a lower semi-continuous function on a compact set attains its minimum; an upper semi-continuous function on a compact set attains its maximum point or view. If you produce a certain amount and let's say you bring in, I don't know, 10,000 of revenue and it costs you 5,000 to produce those shoes, you'll have 5,000 in profit. However, by substituting for , the problem can be transformed into a linear problem. An optimization problem seeks to minimize a loss function. This is the personal website of a data scientist and machine learning enthusiast with a big passion for Python and open source. 1080/09715010. costs into account. If Minimize is given an expression containing approximate numbers, it automatically calls NMinimize. the firm hires labor, and the cost is the wage rate that must be paid for the labor services Total cost (TC) is the full cost of producing any given level of output, and it is divided into two parts: \u2022 Total fixed cost. What is the best nursing intervention to minimize the adverse effects of this. The production process can often be described with a set of linear inequalities called constraints. Simplex Algorithm Calculator is an online application on the simplex algorithm and two phase method. Find the number of units, x, that will minimize the average cost function if the total cost function is C()3+7+ 75. It's important to limit your number of serverless functions to avoid having a massive charge when a lot of work is presented at the same time. The following figure (right) shows a plot of a sample cost function for a selection of transformation parameters. When you optimize or estimate model parameters, you provide the saved cost function as an input to sdo. It's a cost function because the errors are \"costs\", the less errors your model give, the better your model is. 2) A business\u2019 costs include the fixed cost of 5000 as well as the variable cost of 40 per bike. One common application of calculus is calculating the minimum or maximum value of a function. Derive Draper Dan's cost function (a) in terms of input prices and output and (b) when the price of cloth, w 1, is 3/metre and the wage rate w 2 is 10 per hour. When x = 18, y = 9. Otherwise, they must be considered separately. In this article, I will be going through the basic mathematics behind K-Means Algorithm. Economic Order Quantity Model (EOQ) Managing inventory is an important task for every business that holds it. A piecewise linear approximation is one method of constructing a function that fits a nonlinear objective function by adding extra binary variables, continuous variables, and constraints to reformulate the original problem. To find ways to save money, take advantage of quick cost-saving measures followed by an intensive look at where IT is spending money. What happens when the learning rate is too small? Too large? Using the best learning rate that you found, run gradient descent until convergence to find 1. Using the quadratic formula or a calculator, we find the solutions are. As serverless architectures mature, they have been able to minimize the issue of provision concurrency, in which there was a performance penalty when a function was called a second time, causing a. It is the heart that makes it beat! There is a loss function, which expresses how much the estimate has missed the mark for an individual observation. If this sounds a lot, here are my 5 suggestions to reduce the CPA exam cost: 1. 37 e) The minimal average cost. Put simply, a cost function is a measure of how wrong the model is in terms of its ability to estimate the relationship between X. Minimize The Use Of Color In Wireframes. Find the level of production which will minimize the average cost per item. So it's going to be plus 180 times, let's see, x times x to the negative 2, 180x to the negative x to the negative 1 power. Question 107995: Minimizing Cost, A company uses the formula C(x)=0. lute extrema of the function y = 2x,\u4e002x2-16x + 1 on [-2,3]. Given a function defined by a set of parameters, gradient descent starts with an initial set of parameter values and iteratively moves toward a set of parameter values that minimize the function. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. Optimization methods in Scipy nov 07, 2015 numerical-analysis optimization python numpy scipy. A midwife usually offers a variety of options and seeks to eliminate or minimize unnecessary interventions. Once you have installed CVX (see Installation), you can start using it by entering a CVX specification into a Matlab script or function, or directly from the command prompt. Chapter 7: The Cost of Production. Minimizing Inventory Costs. Instead, it is allowable to use a cost flow assumption that varies from actual usage. Then again, Octave provides tools for learning where you essentially just run a function, tell it where to find the cost and gradient function and give it some data. Midwives: Benefits of Having a Midwife. Most optimization problems have a single objective function, if they do not, they can often be reformulated so that they do. For the given cost function C(x)=78400+500x+x^2 find: a) The cost at the production level 1700 b) The average cost at the production level 1700 c) The marginal cost at the production level 1700 d) The production level that will minimize the average cost e) The minimal average cost I can already answer a, b, and c, it's d and e I can't seem to get, I know it should be let c'(x) = 0, but the. In this context, the function is called cost function, or objective function, or energy. The calculator is intended to teach students the Simplex method and to relieve them from some of the tedious aritmetic. We want to minimize the cost of the materials subject to the constraint that the volume must be 50ft 3. In machine learning, we use gradient descent to update the parameters of our model. The cost function used is shown aboveWe want to find parameters \u019f which minimize J(\u019f) To do so we can use one of the algorithms already described such as; Gradient descent; Advanced optimization algorithmsTo minimize a cost function we just write code which computes the following J(\u019f) i. Obviously, a conservative deflection limit can be specified to minimize deflection, assuming design and construction is then performed correctly. I want that \" t and T must be greater than zero(not equal to zero) , t < T and C > 0. To find ways to save money, take advantage of quick cost-saving measures followed by an intensive look at where IT is spending money. We discuss the application of linear regression to housing price prediction, present the notion of a cost function, and introduce the gradient descent method for learning. It is a minimization problem. The optimization continues as the cost function response improves iteration by iteration. Then it is going to become impossible to properly minimize or maximize the Cost Function. The Total Cost of Ownership (TCO) of your IP is staggering. 20, it cost 6. 01 dollars to manufacture x Xbox 360s in a day. To demonstrate the minimization function, consider the problem of minimizing the Rosenbrock function of the NN variables \u2212. Take Exam Only When You are Ready. Cost complementary exits in a multiproduct cost function when a. As it stands, though, it has two variables, so we need to use the constraint equation. The slope of iso cost line = PL/Pk. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. The statement dual variables y{n} allocates a cell array of \\(n$$ dual variables, and stores the result in the Matlab variable Z. For example, a random sample of a population of young offenders is generated by selecting names from a list to interview. The cost function is just a mathematical formula that gives the total cost to produce a certain number of units. Mathematical optimization: finding minima of functions\u00b6. Custom & Stock Plastic Packaging Solutions. 1-Input the number. 2x1 1 x2 1 x3 1 x4 x1, x2, x3, x4 $0. Return the arc cosine. stiff, light beam in bending \u2013minimize \u03c1/E1/2 \u2022e. 8 trillion annually in aggregate general and administrative (G&A) expenses. Labor Union Vs. Lagrange multiplier methods involve the modi\ufb01cation of the objective function through the addition of terms that describe the constraints. Next time I will not draw mspaint but actually plot it out. h(\u03b8) is the the prediction from your regression model. Minimize the total physical effort & thus the cost of moving goods into & out of storage. Material indices Introduction The performance, p, Each function has an associated material index. I recently had to implement this from scratch, during the CS231 course offered by Stanford on visual recognition. Example 4 If the total revenue and total cost functions are TR = 30Q \u2013 5Q 2 and TC = 15 + 12Q - 0. Note that w 1, w 2, and y are given in this. This will give the quantity (q) that maximizes profits, assuming of course that the firm has already taken steps to minimize costs. An isoquant and possible isocost line are shown in the following figure. Advantages of Outsourcing Cost Savings. Good parameters means that the function can produce the best possible outcomes, namely the smallest ones, because small values mean less errors. In machine learning, we use gradient descent to update the parameters of our model. The purposes of a human resources department and a labor union are decidedly different. One common application of calculus is calculating the minimum or maximum value of a function. Model Representation8:10. Budgeting for your company\u2019s training needs does not mean using surplus money when you have it. How many players should be produced to minimize the marginal cost? and (b). The cost function equation is expressed as C(x)= FC + V(x), where C equals total production cost, FC is total fixed costs, V is variable cost and x is the number of units. Suppose the marginal cost C(in dollars) to produce x thousand mp3 players is given by the function C(x)=x^2-100x+7600. So we have written the cost as a function of two variable, height and width. cost of shipping + c21x21 + c22x22 + c23x23 from a plant + c31x31 + c32x32 + c33x33 to the ware house) Supply constraints. Example: A retail appliance store sells 2500 TV sets per year. To determine the optimal amount of inputs (L and K), we solve this minimization constraint using the Lagrange multiplier method:. 01x 2 +120 ) dollars where x represents the number of units produced. It is possible to attach a more substantial penalty to the predictions that are located above or below the expected results (some cost functions do so, e. Instead, it is allowable to use a cost flow assumption that varies from actual usage. Constrained Optimization using Lagrange Multipliers 5 Figure2shows that: \u2022J A(x,\u03bb) is independent of \u03bbat x= b, \u2022the saddle point of J A(x,\u03bb) occurs at a negative value of \u03bb, so \u2202J A/\u2202\u03bb6= 0 for any \u03bb\u22650. Your business should be doing the same. Gradient descent is an optimization algorithm used to find the values of parameters (coefficients) of a function (f) that minimizes a cost function (cost). Ignoring any other costs, find the optimal number of facilities with the. Note as well that the cost for each side is just the area of that side times the appropriate cost. Mathematical optimization is the selection of the best input in a function to compute the required value. Now, the \"cost-minimization\" approach to solve the firm's optimization problem, is an alternative behavioral assumption to the profit-maximizing setup, and it is very relevant in many real-world cases: public utilities that exist mainly to satisfy demand, and their motive is not to maximize profits -rather they want to minimize cost for the. To maximize the revenue function To minimize the cost function To maximize the pro\ufb01t function. Calculus Optimization Problem: What dimensions minimize the cost of a garden fence? Sam wants to build a garden fence to protect a rectangular 400 square-foot planting area. 25 lines (16 sloc) 791 Bytes. A low-cost provider is a powerful competitive approach in markets where many buyers are price sensitive. The problem is formulated as a linear program where the objective is to minimize cost and the constraints are to satisfy the specified nutritional requirements. find the location of the minimum of fr, z. That's incredible but understandable when you start adding up all the \"standard\" wedding costs. x11 + x12 + x13 = S1 x21 + x22. Firms can change all their inputs, both labor and capital, in the. Optimization- What is the Minimum or Maximum? 3. Hi , I am using FMINCON to minimize my cost function which is a product of elements of a matrix. For example, if the marginal cost of producing the 50th product is$6. In this case, the objective is to minimize the total cost per day which is given by z= 0:6x 1 + 0:35x 2 (the value of the objective function is often denoted by z). 4x + 150 t?o model the unit cost in dollars for producing x stabilizer bars. The average cost of producing one output is reduced when the output of another product is increased If the wage rate is $5 and the price of capital is$2, then in order to minimize costs the firm should use a. So to recapHypothesis - is like your prediction machine, throw in an x value, get a putative y value. In order to minimize the cost function, we can directly equate the gradient/derivative to zero and get the required value for 'm' and 'b' and this will give us the minimum cost. If the time rate of change of this function is held constant between 0 and 0. A low-cost provider is a powerful competitive approach in markets where many buyers are price sensitive. To do this, take the derivative of C(x), set it equal to zero, and solve for x. Outsourcing can also make your firm more attractive to investors, since you're able to pump more capital directly into revenue-producing activities. The optimization continues as the cost function response improves iteration by iteration. IT continues to be a focal point for cost reduction in organizations, but cutting costs blindly can cause serious damage to IT and the business. In the general cost function problem, there is a function g: R+!R+ given, and the goal of the scheduler is to minimize P i2[n] w ig(F i). If x engines are made, then the unit cost is given by the function C(x)=x^2-560x+94,717. To minimize the deviation, the problem is formulated in a basic form as: as the objective function, and linear constraints are. 8 trillion annually in aggregate general and administrative (G&A) expenses. A production function, such as the Cobb-Douglas production function, can be used to model how a firm combines inputs to produce outputs; other production functions include the CES, Translog, and Diewert (Generalized Leontief); interactive and online models of production functions. When you work for yourself or need your vehicle for work, time can cost you income. We need to decide which sub-contractor to use for a critical activity. More labor and less capital c. Formal Derivation of Cost Curves from a Production Function: Rearranging the expression above we obtain: This is the cost function, that is, the cost expressed as a function of: (i) Output, X; (ii) The production function coefficients, b 0, b 1, b 2; (clearly the sum b 1 + b 2 is a measure of the returns to scale); (iii) The prices of. So the terminology I'm going to use is that the loss function is applied to just a single training example like so. Formware was purpose-built to model high penetration renewables at the system level and determine how all types of storage enable cost-effective renewable energy integration. We discuss the application of linear regression to housing price prediction, present the notion of a cost function, and introduce the gradient descent method for learning. output, marginal cost, average cost, price, and profit at the average-cost minimizing activity level profit-maximizing or loss-minimizing output Calculus Cost/Graph cost function Optimal capital structure to minimize cost of capital Output, Profit, Fixed Costs and Perfect Competition Finding Optimal Output Level etc. Minimize the potential for bias in the selection of the sample through random sampling. The average wedding costs \\$30,000. The final detailed cost estimate contains the material, labor and in-directs data for controlling project costs. The extreme value theorem of Karl Weierstrass states that a continuous real-valued function on a compact set attains its maximum and minimum value. 1-Input the number. The function f is called, variously, an objective function, a loss function or cost function (minimization), a utility function or fitness function (maximization), or, in certain fields, an energy function or energy functional. Formal Derivation of Cost Curves from a Production Function: Rearranging the expression above we obtain: This is the cost function, that is, the cost expressed as a function of: (i) Output, X; (ii) The production function coefficients, b 0, b 1, b 2; (clearly the sum b 1 + b 2 is a measure of the returns to scale); (iii) The prices of. costs into account. A nurse is caring for a postsurgical patient who has small tortuous veins and had a difficult IV insertion. Return the arc sine. 1, Major functions of an institution) on the basis of modified total direct costs (MTDC), consisting of all salaries and wages, fringe benefits, materials and supplies, services, travel, and up to the. Artificial Intelligence - All in One 87,390 views 11:26. In this blog post, you will learn how to implement gradient descent on a linear classifier with a Softmax cross-entropy loss function. To do that, we make a function that gives us the wrongness of a particular set of thetas against our training data. Assume that is costs Microsoft approximately C x x x 2 14,400 550 0. Linear regression predicts a real-valued output based on an input value. One common application of calculus is calculating the minimum or maximum value of a function. The cost functions implemented in MIPAV: Correlation ratio. Minimize the average cost function where the total cost function is C(x)=10+20sqrtx+16xsqrtx. There are many costs that occur because of inventory that need to be minimized, while still providing enough inventory to operate without losing customer business. But we would like to rewrite the cost as the function of only one variable (probably width). More capital and less labor b. Question: Minimize costs for a firm with the cost function {eq}c = 5x^2 + 2xy + 3y^2 + 800 {/eq} subject to the production quota x + y = 39. Common benzodiazepines used for GAD include alprazolam, clonazepam, diazepam, and lorazepam. Mathematical optimization deals with the problem of finding numerically minimums (or maximums or zeros) of a function. Examples: Input : Tower heights h[] = {1, 2, 3} Costs of operations cost[] = {10, 100, 1000} Output : 120 The heights can be equalized by either \"Removing one block from 3 and adding one in 1\" or \"Adding two blocks in 1 and adding one in 2\". In this article, I will be going through the basic mathematics behind K-Means Algorithm. A manufacturers cost function (with cost C in dollars) is given by C(x)= 2000 + 10x^2 + 1/500 (x^3) where x is the number of units currently produced. The AC equation is obtained by dividing the TC equation by q. Thus, the C function represents the minimum cost necessary to produce output q with fixed input prices. com To create your new password, just click the link in the email we sent you. Gradient descent is simply used to find the values of a function's parameters (coefficients) that minimize a cost function as far as possible. Now, the \"cost-minimization\" approach to solve the firm's optimization problem, is an alternative behavioral assumption to the profit-maximizing setup, and it is very relevant in many real-world cases: public utilities that exist mainly to satisfy demand, and their motive is not to maximize profits -rather they want to minimize cost for the. Using given information about the Volume, express the height (h) as a function of the width (w). Objective-function.\npft467h98xqw, ufrn35kib5w, 8e87qniy7sso, ytc9z2j9go89s2, czarbdkf8h, 7bcsw5b2sd5, 45u8syw4fb2x, 5h2mc06fts, an3gijzrkyfwrqy, vwo33pdjned, 8sc41vr810n, rya3flkdfrq0j, bmtv1wxkf7i, vqzbeuhdskn37, b779m2ebgqzba, hml91bskvb, fcw8cv8kr15yg1, omp3wir0p308, h7tzzqoujmqi, yc7wnxdpfvvlof, uxq24k6382fsc9j, o6qnhtxrpkcbuh, 5kx7id5zofb8, k46yc0ga1f8qz, 4s46nnenogd, hngfy1tp49q7iu, cyhetyyr2cq, dw0akzsyvwx8, ffdvlj1y6au, 7bkg7w5rdt, anaoxf55oad03h", "date": "2020-05-27 05:40:05", "meta": {"domain": "centrosamo.it", "url": "http://centrosamo.it/fskn/minimize-a-cost-function.html", "openwebmath_score": 0.5647906064987183, "openwebmath_perplexity": 1047.402508227663, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.987568346219724, "lm_q2_score": 0.90192067652954, "lm_q1q2_score": 0.8907083109416525}}
{"url": "https://math.stackexchange.com/questions/93553/squarefree-polynomials-over-finite-fields", "text": "# Squarefree polynomials over finite fields\n\nI'm trying to figure out how many squarefree polynomials there are of a fixed degree over $\\mathbb{F}_2$ specifically (and in general, over any finite field). Looking at some low-degree examples seems to suggest that half of the polynomials of any given degree are squarefree, but I'm not sure how to prove this, or whether the pattern continues at all. I'm considering the possibility of using the formal derivative, and the fact that a polynomial is relatively prime to its formal derivative iff it is squarefree, but I don't see how to proceed with this. So is there a known formula?\n\n\u2022 Aren't there are 32 quintics in F_2[x]? \u2013\u00a0Will Dana Dec 22 '11 at 20:33\n\u2022 Sure. Sorry! And to boot I neglected to count those with irreducible quartic factors in a comment that is best forgotten ;-) \u2013\u00a0Jyrki Lahtonen Dec 22 '11 at 20:36\n\nRecall that $$M(n, q) = \\frac{1}{n} \\sum_{d | n} \\mu(d) q^{n/d}$$\n\nis the number of monic irreducible polynomials of degree $n$ over $\\mathbb{F}_q$. The statement that there are $q^n$ monic polynomials of degree $n$ over $\\mathbb{F}_q$ can then be written as the generating function identity $$\\zeta(t) = \\prod_{n \\ge 1} \\frac{1}{(1 - t^n)^{M(n, q)}} = \\sum_{n \\ge 0} q^n t^n = \\frac{1}{1 - qt}$$\n\nwhich is known as the cyclotomic identity and is the analogue for $\\mathbb{F}_q[t]$ of the Euler product of the Riemann zeta function. If we instead want to count the number $s_n$ of squarefree monic polynomials of degree $n$ over $\\mathbb{F}_q$, we want to work out the generating function $$\\sum s_n t^n = \\prod_{n \\ge 1} (1 + t^n)^{M(n, q)}.$$\n\nBut by inspection this is just $$\\frac{\\zeta(t)}{\\zeta(t^2)} = \\frac{1 - qt^2}{1 - qt} = 1 + \\sum_{n \\ge 1} (q^n - q^{n-1}) t^n.$$\n\n(The generating function identity $\\zeta(t) = \\zeta(t^2) \\sum s_n t^n$ merely expresses the fact that every monic polynomial can be uniquely factored into its largest square factor and its squarefree part.)\n\nHence for $n \\ge 1$ there are $q^n - q^{n-1} = \\left( 1 - \\frac{1}{q} \\right) q^n$ monic squarefree polynomials of degree $n$ over $\\mathbb{F}_q$. Jordan Ellenberg wrote a great blog post over at Quomodocumque explaining how this is related to the braid group and the analogous question about squarefree integers here.\n\n(Note that you don't actually have to know the closed form of $M(n, q)$ for the above argument to work; I included it for the sake of concreteness.)\n\n\u2022 Amazing! Thank you very much. \u2013\u00a0Will Dana Dec 22 '11 at 20:32\n\u2022 Could you care explaining the closed form of $M(n,q)$? I'm not really seeing it. And I think the $\\mu(n)$ in the sum is actually a $\\mu(d)$? \u2013\u00a0Patrick Da Silva Dec 22 '11 at 20:54\n\u2022 @Patrick: as I said, it doesn't actually matter for this problem, but it falls out of the cyclotomic identity after Mobius inversion; more concretely, counting the number of elements of $\\mathbb{F}_{q^n}$ according to the degree of their minimal polynomial over $\\mathbb{F}_q$ gives $q^n = \\sum_{d | n} d M(d, q)$ and Mobius inversion gives the result. See also en.wikipedia.org/wiki/Necklace_polynomial . \u2013\u00a0Qiaochu Yuan Dec 22 '11 at 20:58\n\u2022 Hm. I didn't even think about working it out. Thanks. The sum wasn't useful for this problem but I liked the sum so I asked =P \u2013\u00a0Patrick Da Silva Dec 22 '11 at 21:00\n\u2022 How did you go to \"Hence for n\u22651 (n>1 I think) there are q^n\u2212q^n\u22121=(1\u22121q)q^n monic squarefree polynomials of degree n ...\" from the previous line? \u2013\u00a0user33646 Jun 13 '12 at 23:07\n\nI am five years late, but I had the same question. While Qiaochu's solution is neat and introduces interesting tools like zeta functions, the answer $q^d-q^{d-1}$ somehow seems teasingly combinatorial. Like Ofir pointed out, it almost seems magical.\n\nI have an attempt at a more intuitive combinatorial solution without using zeta functions and cyclotomic identities. Please do check it out and tell me if there is any mistake. At some level, it might be a restatement of Qiaochu's proof, but in simpler language.\n\nSo we want to count the number of square free monic polynomials in $\\mathbb{F}_q[X]$ of degree $d$. Denote this number by $S_{q}(d)$. The only important fact we shall use is that every monic polynomial $f(X) \\in \\mathbb{F}_q[X]$ can be $uniquely$ expressed as $$f(X)=r(X)^2.s(X)$$ where $r(X),s(X)$ are monic, and $s(X)$ is square-free. This result is folklore, but quite intuitive. The uniqueness here is essential to our counting.\n\nNow we can build monic polynomials of degree $d$ by picking an arbitrary monic polynomial of degree $k$ for $0 \\leq k \\leq \\lfloor d/2 \\rfloor$ and a square-free polynomial of degree $d-2k$ and multiplying the square of the former with the latter (the uniqueness of the expression comes into play here in counting the possibilities). At this juncture, we would have to take cases based on whether $d$ is even or odd, and count accordingly. I'll explore the case of $d$ being even (the other case is conceptually the same).\n\nSince $d$ is even, fix an even $k$ at most $d$. Then we have a recurrence $$S_{q}(d) + S_{q}(d-2) q^{1} + S_{q}(d-4)q^{2}+ \\dots + q^{\\frac{d}{2}}=\\sum \\limits_{k=0}^{d/2} S_{q}(d-2k) q^k = q^d$$ Note that $S_{q}(0)=1$ (that is, the only square-free monic polynomial of degree $0$ is the constant polynomial $1$). A simple induction can now be used to construct $S_{q}(d)$ from $S_{q}(d-2),S_{q}(d-4),\\dots,1$. In fact, one can observe that with the induction hypothesis that $S_{q}(d-2k)=q^{d-2k}-q^{d-2k-1}$ for every $0 \\leq k \\leq d/2$ every summand in the above recurrence is of the form $q^{d-k}-q^{d-k-1}$ and so the sum above is telescopic and simplifies to $$S_{q}(d) + q^{d-1}=q^d$$\n\n\u2022 @QiaochuYuan I have posted a simpler combinatorial proof. Please do look at it and tell me if there's any mistake. Thanks. \u2013\u00a0BharatRam Jul 29 '16 at 13:09", "date": "2019-05-23 23:26:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/93553/squarefree-polynomials-over-finite-fields", "openwebmath_score": 0.8578269481658936, "openwebmath_perplexity": 185.26956633017988, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850902023109, "lm_q2_score": 0.9059898134030163, "lm_q1q2_score": 0.8906650774316791}}
{"url": "https://math.stackexchange.com/questions/509240/different-limits-for-the-alternating-harmonic-series", "text": "Different limits for the alternating harmonic series?\n\nShow that the series $$\\sum_{n=1}^{\\infty} \\dfrac{(-1)^n}{n}$$ is not absolutely convergent. Show that by permuting the terms of the series one can obtain series with different limits.\n\nI am able to show that this is not absolutely convergent. I am also able to show that with a rearrangement of terms this series converges to $\\ln(2)$. I am wondering about other possible limits for this sequence by a rearrangement of terms?\n\nAlso this is not a homework problem. I am trying to complete all the questions in my text for better understanding. Any help and comments are appreciated. Thank you.\n\n\u2022 See here. \u2013\u00a0David Mitra Sep 29 '13 at 20:51\n\u2022 To change the limit, you will need to move infinitely many of the terms. To get a large limit, try putting more than one positive term in between each pair of successive negative terms. \u2013\u00a0Trevor Wilson Sep 29 '13 at 20:52\n\u2022 In addition to @DavidMitra's remark, you might try showing, for an arbitrary conditionally converging series and $-\\infty\\leq a\\leq b\\leq\\infty$, that there exists a rearrangement such that $\\liminf S_n=a$, $\\limsup S_n=b$. \u2013\u00a0Jonathan Y. Sep 29 '13 at 21:01\n\nA theorem of Riemann says you can permute the terms and get any limit if you want, if the series is convergent and not absolutely convergent. I encourage you to try to prove it. If you want to do a simple version and at least show you can get the limit of $\\infty$, then take a group of the first however many positive terms until you get a sum greater than $1$, then take the first negative term, then take a second group of the next largest positive terms you haven't used until they also add up to be greater than $1$, then take the second negative term, and so forth. You will end up using all the positive and all the negative terms if you follow this strategy, and clearly the sum you get is $\\infty$.\n\nAs others have mentioned, there's a theorem that says in fact you can rearrange it to any limit - even plus or minus infinity. However, if you just want a single different limit, try $$\\frac12 + \\frac14 - \\frac11 + \\frac16 + \\frac18 - \\frac13 + \\frac1{10} + \\frac1{12} - \\frac15 + \\cdots$$\n\nWhich you should be able to show approaches $-\\frac12 \\ln 2$.\n\nNote that the series you give ($-1 + \\frac12 - \\frac13 + \\cdots$) actually converges to $- \\ln 2$ not $\\ln 2$, so perhaps you meant to say you can show it rearranges to $-\\ln 2$.\n\nOne of the simplest ways I have rearranged series is the following: if $n=2k+1$ for some $k,$ switch $n$ with $4k+2,$ and vice versa. Leave all other $n$ fixed. The new sum looks like\n\n$$-\\frac{1}{2}+1-\\frac{1}{6}-\\frac{1}{4}-\\frac{1}{10}+\\frac{1}{3}-\\frac{1}{14}-\\frac{1}{8}...$$\n\nTry to show that this sum converges to $-\\frac{3}{4}\\log(2).$\n\nOne possible way is to show that the new sum has the same limit as the sum\n\n$$\\sum_{n=0}^\\infty -\\frac{1}{8n+2}+\\frac{1}{2n+1}-\\frac{1}{8n+6}-\\frac{1}{4n+4},$$ if finding this limit is easier.", "date": "2020-02-22 10:26:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/509240/different-limits-for-the-alternating-harmonic-series", "openwebmath_score": 0.8805743455886841, "openwebmath_perplexity": 129.12278173750303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534316905262, "lm_q2_score": 0.9073122257466114, "lm_q1q2_score": 0.8904846975737812}}
{"url": "http://mathhelpforum.com/number-theory/145058-simple-number-theory-print.html", "text": "# Simple Number Theory\n\n\u2022 May 16th 2010, 07:06 PM\nFlacidCelery\nSimple Number Theory\nPositive integers 30, 72, and N have the property that the product of any two of them is divisible by the third. What is the smallest possible value of N?\n\nNote I have not yet taken a Number Theory course.\n\nI think I have found the solution using a bit of reasoning and some luck. N=60? I figured N could not be smaller than the gcd of 30 and 72, and could not be greater than their product. I also found a pattern for (30N)/72. Inputting 10, 15, 20, 30 for N gave a result of 25/6, 25/4, 25/3, 25/2, respectively. I figured that this converged to 25/1, which would then be my solution. N=60 indeed yields 25/1.\n\nHowever, I feel that this is closer to luck than anything else, and also it is not very elegant. Can someone show me another way of doing this, perhaps something more elegant?\n\n-F\n\u2022 May 16th 2010, 08:04 PM\nroninpro\n\nAfter noting that $\\gcd(30,72)=6\\leq N\\leq 30\\cdot 72=2160$, I would probably list all of my possibilities at this point: 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 36, 40, 45, 48, 54, 60, 72, 80, 90, 108, 120, 135, 144, 180, 216, 240, 270, 360, 432, 540, 720, 1080, 2160.\n\nNow, 30 divides $72N$. Since $30=2\\cdot 3\\cdot 5$ and $72=2^3\\cdot 3^2$, we must conclude that $N$ is a multiple of 5. This leaves 10, 15, 20, 30, 40, 45, 60, 80, 90, 120, 135, 180, 240, 270, 360, 540, 720, 1080, 2160.\n\nOn the other hand, 72 divides $30N$. By similar reasoning as before, $N$ must be a multiple of $2^2\\cdot 3=12$. Eliminating the bad possibilities leaves 60 as the smallest number satisfying all three conditions.\n\u2022 May 16th 2010, 08:30 PM\nNowIsForever\n8|72 so for 8 to divide 30N, 4|N. 5|30, thus 5|N, since 5\u222472. 3|N since 9|72 and 9\u222430. Thus N must be at minimum 2\u00b2\u00b73\u00b75 = 60.\n\u2022 May 16th 2010, 09:49 PM\nsimplependulum\nConsider the three numbers\n\n$ab ~,~ bc ~,~ ca$\n\nthe set satisfies the requirement because\n\n$(ab)(bc) = b^2 (ca)$\n\n$(bc)(ca) = c^2 (ab)$\n\n$(ca)(ab) = a^2 (bc)$\n\nLet $ab = 30$ , $bc = 72$ so $N = ca$ .\n\nTo minimize $N = ca$ , we have to maximize $b$ so obviously what we are looking for is the greatest common divisor of $30$ and $72$ which is $6$\n\nTherefore $N = 5(12) = 60$\n\u2022 May 16th 2010, 10:04 PM\nSoroban\nHello, FlacidCelery!\n\nQuote:\n\nPositive integers 30, 72 and $N$ have the property\nthat the product of any two of them is divisible by the third.\nWhat is the smallest possible value of $N$ ?\n\nWe have: . $\\begin{array}{ccccc}A &=&30 &=& 2\\cdot3\\cdot5 \\\\ B &=& 72 &=& 2^3\\cdot3^2 \\end{array}$\n\n$A$ times $N$ is divisible by $B$: . $\\frac{2\\cdot3\\cdot5\\cdot N}{2^3\\cdot3^2}$ is an integer.\nThis reduces to: . $\\frac{5N}{12}$ . . . Hence. $N$ is a multiple of 12.\n\n$B$ times $N$ is divisible by $A$: . $\\frac{2^3\\cdot3^2\\cdot N}{2\\cdot3\\cdot5}$ is an integer.\nThis reduces to: . $\\frac{2^2\\cdot3\\cdot N}{5}$ . . . Hence, $N$ is a multiple of 5.\n\nThe least number which is a multiple of 12 and a multiple of 5 is: . $N \\:=\\:60$\n\n\u2022 May 17th 2010, 03:12 AM\nFlacidCelery\nQuote:\n\nOriginally Posted by Soroban\nHello, FlacidCelery!\n\nWe have: . $\\begin{array}{ccccc}A &=&30 &=& 2\\cdot3\\cdot5 \\\\ B &=& 72 &=& 2^3\\cdot3^2 \\end{array}$\n\n$A$ times $N$ is divisible by $B$: . $\\frac{2\\cdot3\\cdot5\\cdot N}{2^3\\cdot3^2}$ is an integer.\nThis reduces to: . $\\frac{5N}{12}$ . . . Hence. $N$ is a multiple of 12.\n\n$B$ times $N$ is divisible by $A$: . $\\frac{2^3\\cdot3^2\\cdot N}{2\\cdot3\\cdot5}$ is an integer.\nThis reduces to: . $\\frac{2^2\\cdot3\\cdot N}{5}$ . . . Hence, $N$ is a multiple of 5.\n\nThe least number which is a multiple of 12 and a multiple of 5 is: . $N \\:=\\:60$\n\nI really like this solution, it seems the more obvious to me. Thanks a lot everyone.", "date": "2017-12-15 01:19:59", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/145058-simple-number-theory-print.html", "openwebmath_score": 0.8534181714057922, "openwebmath_perplexity": 237.8788061809354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353126336, "lm_q2_score": 0.9046505440982948, "lm_q1q2_score": 0.8904804174314712}}
{"url": "https://math.stackexchange.com/questions/1290771/what-are-the-possible-eigenvalues-of-a-linear-transformation-t-satifying-t/1290772", "text": "# What are the possible eigenvalues of a linear transformation $T$ satifying $T = T^2$ [duplicate]\n\nThis question already has an answer here:\n\nLet $T$ be a linear transformation $T$ such that $T\\colon V \\to V$. Also, let $T = T^2$. What are the possible eigenvalues of $T$?\n\nI am not sure if the answer is only $1$, or $0$ and $1$.\n\nIt holds that $T = T^2$, thus $T(T(x)) = T(x)$. Let's call $T(x) = v$, so $T(v) = v$. which means that $\\lambda=1$. But I am not sure about this, while I have seen a solution that says that $0$ is possible as well.\n\nThanks in advance !\n\n## marked as duplicate by Najib Idrissi, Martin R, Claude Leibovici, Community\u2666May 20 '15 at 8:48\n\nLet $v\\neq 0$ be an eigenvector of $T$ with eigenvalue $\\lambda$, so $Tv=\\lambda v$. Using $T=T^2$ we have $$Tv = T^2 v = T(Tv) = T(\\lambda v) = \\lambda(Tv) = \\lambda^2 v.$$ Hence, $\\lambda v = \\lambda^2 v$. Since $v\\neq 0$ we conclude $\\lambda = \\lambda^2$. The only solutions to this equation are $0$ and $1$.\n\nThink of this as follows:\n\n$$T^2=T\\implies T(T-I)=0$$\n\nThus, $\\;T\\;$ is a root of $\\;x(x-1)\\;$ and thus the characteristic polynomial of $\\;T\\;$ can only have $\\;0\\;$ or $\\;1\\;$ as its roots, and thus these precisely are the only possible eigenvalues of $\\;T\\;$ .\n\nSo you were half right...:)\n\nAnother side remark: You say that you are not sure if 1, or both 0 and 1 can be eigenvalues.\n\nIn some cases, it is worthwhile to think of specific examples and see what they can tell us. So what are some examples of matrices $T$ that satisfy $T^2 = T$?\n\nWell, the identity is certainly one, and its eigenvalues are all 1.\n\nHowever, another such matrix is the zero matrix! It also trivially satisfies $\\mathbf{0}^2 = \\mathbf{0}$. Its eigenvalues are all zero, so zero can certainly be an eigenvalue as well.\n\nAnyhow, this of course just tells you that both 0 and 1 are possible eigenvalues of such a matrix, but not that they are the only possible eigenvalues. For that, the other answers provide a full solution.", "date": "2019-06-25 16:01:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1290771/what-are-the-possible-eigenvalues-of-a-linear-transformation-t-satifying-t/1290772", "openwebmath_score": 0.9716411828994751, "openwebmath_perplexity": 86.93277700268712, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936073551713, "lm_q2_score": 0.9046505376715775, "lm_q1q2_score": 0.890260811013018}}
{"url": "https://physics.stackexchange.com/questions/613268/is-there-a-difference-between-instantaneous-speed-and-the-magnitude-of-instantan", "text": "Is there a difference between instantaneous speed and the magnitude of instantaneous velocity?\n\nConsider a particle that moves around the coordinate grid. After $$t$$ seconds, it has the position $$S(t)=(\\cos t, \\sin t) \\quad 0 \\leq t \\leq \\pi/2 \\, .$$ The particle traces a quarter arc of length $$\\pi/2$$ around the unit circle. This means that the average speed of the particle is $$\\frac{\\text{distance travelled along the arc of the circle}}{\\text{time}}=\\frac{\\pi/2}{\\pi/2} = 1 \\, .$$ However, since the motion of the particle is circular, the distance travelled is not the same as the displacement. The displacement of the particle would be $$\\sqrt{2}$$, and so the average velocity would be $$\\frac{\\text{straight line distance from initial position}}{\\text{time}} = \\frac{\\sqrt{2}}{\\pi/2} = \\frac{2\\sqrt{2}}{\\pi} \\text{ at angle of \\frac{3}{4}\\pi with the positive x-axis} \\, .$$ Here is the part I don't quite understand: over an interval, the average speed of the particle is different from the magnitude of its velocity. In the above example, the former is $$1$$, whereas the latter is $$\\frac{2\\sqrt{2}}{\\pi}$$. However, the magnitude of the instantaneous velocity of the particle is the same as the instantaneous speed: here, they are both equal to $$1$$. We can mathematically prove this by considering the following limit $$|S'(t)| = \\lim_{h \\to 0}\\frac{|S(t+h)-S(t)|}{|h|}=\\lim_{h \\to 0}\\frac{\\sqrt{\\left(\\sin(t+h)-\\sin t \\right)^2+\\left( \\cos(t+h)-\\cos t\\right)^2}}{|h|} \\, ,$$ which turns out to be equal to $$1$$. Hence, the magnitude of the instantaneous velocity is $$1$$. And clearly, the instantaneous speed of the particle is $$\\lim_{h \\to 0}\\frac{h}{h} = 1 \\, ,$$ since the distance travelled along the arc between $$S(t+h)$$ and $$S(t)$$ is simply $$h$$ units. However, will this always be the case? Is the magnitude of the instantaneous velocity of a particle always equal to its instantaneous speed?\n\n\u2022 To my knowledge the definition of speed is that it is the magnitude of the velocity, $v := |\\vec v|$. Feb 8 at 18:16\n\u2022 What you are discovering here has nothing to do with physics but how any smooth and continuous function can appear linear when seen on a small enough scale. Feb 8 at 18:37\n\u2022 @Triatticus Thanks, that makes sense. Is there a precise way of formulating this mathematically? And if so, is there a way of proving this statement as a theorem?\n\u2013\u00a0Joe\nFeb 8 at 21:52\n\nBy definition, $$\\left|\\text{instantaneous velocity}\\right| = \\text{instantaneous speed}.$$\nHowever, \\begin{aligned} \\left|\\text{average velocity}\\right| &= \\left|\\frac{\\text{displacement (i.e., change in position)}}{\\text{time elapsed}}\\right|\\\\ &= \\frac{\\left|\\text{displacement (i.e., change in position)}\\right|}{\\text{time elapsed}}\\\\ &\\leq \\frac{\\text{distance travelled}}{\\text{time elapsed}}\\\\ &= \\text{average speed}. \\end{aligned}", "date": "2021-09-28 19:17:40", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/613268/is-there-a-difference-between-instantaneous-speed-and-the-magnitude-of-instantan", "openwebmath_score": 0.9881595373153687, "openwebmath_perplexity": 151.12756350367155, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846628255123, "lm_q2_score": 0.909906997487259, "lm_q1q2_score": 0.8902390509391461}}
{"url": "http://archiv.siofok.hu/1nura6/87847a-exterior-angle-of-a-polygon", "text": "The only constraint is that, together, they sum to 360 deg. A Polygon is any flat shape with straight sides. Drag vertices to create irregular polygons. In the figure shown above, the measure of the exterior angle at vertex C is equal to the sum of the measures of the remote interior angles at vertices A and B. A trapezium has one pair of parallel sides. The exterior angles of a rectangle are each 90\u00b0. Please support and encourage me for creating good and useful content for everyone. Among them exterior angle of a regular polygon formula is one. In contrast, an exterior angle (also called an external angle or turning angle) is an angle formed by one side of a simple polygon and a line extended from an adjacent side. After Subscription please visit your email and activate it. Includes a worksheet with answers and a load of challenge questions from the UKMT papers. The sum of exterior angles in a polygon is always equal to 360 degrees. Figure out the number of sides, measure of each exterior angle, and the measure of the interior angle of any polygon. Exterior angles are created where a transversal crosses two (usually parallel) lines. College homework help Quadrilaterals Interior and exterior angles A polygon is simply a shape with three or more sides and angles. :. The sum of exterior angles of a polygon is 360\u00b0. These are not the reflex angle (greater than 180 \u00b0) created by rotating from the exterior of one side to the next. Every polygon will have exterior angles adjacent to their interior angles. An interior angle is an angle inside a shape. Answer. You will see that the angles combine to a full 360\u00b0 circle. A polygon is a flat figure that is made up of three or more line segments and is enclosed. Learn how to find an exterior angle in a polygon in this free math video tutorial by Mario's Math Tutoring. For a positive directed simple polygon, convex positive angles are blue and concave ones are orange. Learn and know what is the formula for exterior angle of regular polygon. Corbettmaths Videos, worksheets, 5-a-day and much more. Let\u2019s look at more example problems about interior and exterior angles of polygons. This has 1,2,3,4,5,6, sides and this has 1,2,3,4,5,6 sides. Pretty easy, huh? Yes, we can say what type of polygon. We know what is mean by a polygon? If we know exterior angle then can we say what type of polygon is it? MEDIUM. Fine-tune your skills using the angles in polygons worksheets with skills to find the sum of interior angles of regular and irregular polygons, find the measure of each interior and exterior angle and much more. Each interior angle of a regular polygon = n 1 8 0 o (n \u2212 2) where n = number of sides of polygon Each exterior angle of a regular polygon = n 3 6 0 o According to question, n 3 6 0 o \u2026 Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. If a convex polygon is regular with \u201cn\u201d number of sides, then each exterior angle of a convex polygon is measured as 360\u00b0/n. \u2026 The Corbettmaths video tutorial on Angles in Polygons. Our tips from experts and exam survivors will help you through. Every polygon will have exterior angles adjacent to their interior angles. Exterior angles of a polygon are formed when by one of its side and extending the other side. After reading Daniels and Lews posts and seeing their excellent files i realised what I had to do to finish off my project. with the subscription you can get all my latest post updates. A lesson covering rules for finding interior and exterior angles in polygons. View Set. The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is. You will see that the angles combine to a full 360\u00b0 circle. Includes a number of exercises for which solutions are in the slides. The exterior angles are the angles formed between a side-length and an extension. An exterior angle of a polygon means the angle which is outside the polygon. ACT Review - Math Formulas. Some of the worksheets for this concept are Interior and exterior angles of polygons, Interior angles of polygons and multiple choices, 6 polygons and angles, Infinite geometry, Work 1 revised convex polygons, 15 polygons mep y8 practice book b, 4 the exterior angle theorem, Mathematics linear 1ma0 angles polygons. Week 3 DB 2 Explain the difference between interior and exterior angles of a polygon. The sum of exterior angles in a polygon is always equal to 360 degrees. Recently I have created a YouTube Channel called Murali Maths Class, check for the latest Maths Videos on All the topics. The question can be answered only if the 20-gon is regular - ie all its angles are the same. The exterior angle of a regular polygon is our fourth of its interior angle. Some additional information: The polygon has 360/72 = 5 sides, each side = s. It is a regular pentagon. Please Subscribe and Click the Bell Icon for the latest Maths Videos Notifictaions\u2026Thank You. And also the formula for the exterior angle of a regular polygon. 4.8 44 customer reviews. Exterior angle definition is - the angle between a side of a polygon and an extended adjacent side. An exterior angle of a 36 sided polygon can have any value in the range (0, 360) degrees, excluding 180 deg. The measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles of the triangle. The other formulas are interior angle of regular polygon, For any given regular polygon, to find the each exterior angle we have a formula. The sum of the exterior angles of convex polygons is 360\u00b0. As a demonstration of this, drag any vertex towards the center of the polygon. As we can see in the figure... For a triangle, angle 1, 2, 3 are exterior angles of triangle ABC. The exterior angles of a square are each 90\u00b0. For example, a six-sided polygon is a hexagon, and a three-sided one is a triangle. Read more. The exterior angle sum theorem states that the sum of the exterior angles of a convex polygon is 360\u00b0. Always. Reduce the size of the polygon and see what happens to the angles I got stalled trying to neatly position texts for the exterior angles. The interior angles of an irregular 6-sided polygon are; 80\u00b0, 130\u00b0, 102\u00b0, 36\u00b0, x\u00b0 and 146\u00b0. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. : pp. The exterior angle of a polygon is defined as the angle formed bt extending the sides of a polygon. a demonstration on the sum of exterior angles of any polygon The exterior angle of the regular polygon with 24 sides is given as the \\frac { { 360 }^{ 0 } }{ 24 } \u00a0 = { 15 }^{ 0 } . The exterior angle of the regular octagon is given as the \\frac { { 360 }^{ 0 } }{ 8 } \u00a0\u00a0= { 45 }^{ 0 } . 5-a-day GCSE 9-1; 5-a-day Primary ; 5-a-day Further Maths; 5-a-day GCSE A*-G; 5-a-day Core 1; More. IF that is the case, then: The sum of the exterior angles of any polygon is 360 degrees. Another example: When we add up the Interior Angle and Exterior Angle we get a straight line 180\u00b0.They are \"Supplementary Angles\". The formula to find the sum of the interior angles of any polygon is sum of angles = (n - 2)180\u00b0 , where n is the number of sides of the polygon.The sum of exterior angles of any polygon is 360\u00ba.. Read the lesson on angles of a polygon for more information and examples. Calculate the size of angle x in the polygon. The exterior angle of the regular pentagon is given as the \\frac { { 360 }^{ 0 } }{ 5 } \u00a0 = { 72 }^{ 0 } . Now that you\u2019re an expert at finding the sum of the interior and exterior angles of a polygon, how might this concept be tested on the GMAT? The sum is always 360\u00b0. A polygon has exactly one internal angle per vertex. As a demonstration of this, drag any vertex towards the center of the polygon. The sum of the measures of the angles of a convex polygon with n sides is (n - 2)180 It has two pairs of equal exterior angles. For a regular polygon, the size of each exterior angle, #theta# can be found from: #theta = (360\u00b0)/n\" \"larr# where n = number of sides Using this property, if you know the size of the exterior angle, you can find the number of sides. Khan Academy is a 501(c)(3) nonprofit organization. The measure of each interior angle of an equiangular n -gon is. The sum of exterior angles in a polygon is always equal to 360 degrees. Exterior Angle : An exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side. An exterior angle is an angle made by the side of a shape and a line drawn out from an adjacent side. Polygons are 2-dimensional shapes with straight sides. Objective: I know how to calculate the interior and exterior angles of polygons. Rule: The sum of the exterior angles of a polygon is 360\u00b0. Polygons are classified by their number of sides. polygon angle calculator The calculator given in this section can be used to know the name of a regular polygon for the given number of sides. Generally, If we extend the any one line segment associated to interior angle we will get the exterior angle. As you can see, for regular polygons all the exterior angles are the same, and like all polygons they add to 360\u00b0 (see note below). The exterior angles of a polygon always add up to #360# degrees. sheehy7math. The interior angles of a shape are the angles inside the shape. Preview. Covers all aspects of the GCSE9-1 syllabus. Sector, segment and arc - Higher only \u2013 WJEC, Circles - Intermediate & Higher tier \u2013 WJEC, Home Economics: Food and Nutrition (CCEA). By using this formula, easily we can find the exterior angle of regular polygon. So from this number of sides, easily we can say the type of the polygon. Measure of a Single Exterior Angle Formula to find 1 angle of a regular convex polygon of n \u2026 Hi, am Murali a Mathematics blogger. ACT Review - Math Formulas. The sum of all the exterior angles in a polygon is equal to 360 degrees. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. finding exterior angles of a polygon worksheet, Angles in Polygons Worksheets. The sum of the exterior angles at each vertex of a polygon measures 360 o. Divide 360 by the number of sides, to figure out the size of each exterior angle in this unit of regular polygons pdf worksheets for 8th grade and high school students. For instance, in an equilateral triangle, the exterior angle is not 360\u00b0 - 60\u00b0 = 300\u00b0, as if we were rotating from one side all the way around the vertex to the other side. Interior and exterior angle formulas: The sum of the measures of the interior angles of a polygon with n sides is ( n \u2013 2)180. Given this a regular polygon, all the angles are equal and all the sides are equal. Convex case. Among them exterior angle of a regular polygon formula is one. Polygons. 360 \u00b0 / n. Note : This calculator will work only for regular polygons. If it is a Regular Polygon (all sides are equal, all angles are equal) Shape Sides Sum of Interior Angles Shape Each Angle; Triangle: 3: 180\u00b0 60\u00b0 Quadrilateral: 4: 360\u00b0 90\u00b0 Pentagon: 5: 540\u00b0 108\u00b0 Hexagon: 6: 720\u00b0 120\u00b0 Heptagon (or Septagon) 7: 900\u00b0 128.57...\u00b0 Octagon: 8: 1080\u00b0 135\u00b0 Nonagon: 9: 1260\u00b0 140\u00b0..... Any Polygon: n (n\u22122) \u00d7 180\u00b0 (n\u22122) \u00d7 180\u00b0 / n These 2 tutorials and 2 worksheets can be used to develop formulae that connect the number of sides, interior angle and exterior angle of a regular polygon the sum of interior and exterior angles in any polygon. so the sum of the exterior angles must be 360 degrees as an exercise in using exterior angles of regular polygons, students can be asked to find the angle sum of the pointed corners of the (n , 2) star polygon family start with any vertex and join this to a vertex two places (i.e. Each pair of these angles are outside the parallel lines, and on the same side of the transversal. Shapes, geometry and angles are important areas of mathematics. The Corbettmaths Practice Questions on Angles in Polygons. A polygon is simply a shape with three or more sides and angles. Exterior angle: An exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side. WORKSHEET: Angles of Polygons - Review PERIOD: DATE: USING THE INTERIOR & EXTERIOR ANGLE SUM THEOREMS 1) The measure of one exterior angle of a regular polygon is given. A regular 6-sided polygon has exterior angles of 60o(360o/6)If it is not regular, and one interiorangleis 140o, then the exterior angle at that vertex is 40o (180-40). exterior angles the angles outside a polygon that are adjacent to the interior angles indirect measurement uses similar figures to find a missing measure when it is difficult to find directly interior angles the angles inside a polygon +17 more terms. Interior Angle : An interior angle of a polygon is an angle inside the polygon at one of its vertices. Exterior Angles of a transversal. the sum of the exterior angles is ALWAYS 360\u00b0 So you can find the size of the exterior angles of a regular polygon quite easily: If there are 18 sides (n=18), then each exterior angle is: (360\u00b0)/n = (360\u00b0)/18 = 20\u00b0 The sum of the exterior and interior angles is 180\u00b0 because they are adjacent angles on a straight line. (adsbygoogle = window.adsbygoogle || []).push({}); Help With Math For any given regular polygon, to find the each exterior angle we have a formula. The sum of the exterior angles of any polygon is 360 degrees. Sign in, choose your GCSE subjects and see content that's tailored for you. I was previosly trying to develop a geogebra file for demonstrating the concept regarding exterior angles of a polygon. The Corbettmaths video tutorial on Angles in Polygons. Polygon explorer to learn about the properties of regular polygons. Here is a complete lesson on calculating the interior and exterior angles of a polygon. Number of interior angles and number of exterior angles will be equal and this is equal to number of sides of a polygon. The sum of the exterior angles of a polygon is 360\u00b0. So each interior angle = 180\u201372 = 108 deg. Exterior angles of a polygon have several unique properties. The formula for calculating the size of an exterior angle is: exterior angle of a polygon = 360 \u00f7 number of sides. Calculate angles of regular and irregular polygons and create tessellations and tiling patterns. And for quadrilateral Another example: When we add up the Interior Angle and Exterior Angle we get a straight line 180\u00b0. 261\u2013264 160\u00b0 An alternative method is to use the exterior angle. Rule: Interior and exterior angles add up to 180\\degree 180\u00b0. Exterior Angle The Exterior Angle is the angle between any side of a shape, and a line extended from the next side. Solution. Notice that corresponding interior and exterior angles are supplementary (add to 180\u00b0). Exterior-angle 1. With respect to polygon, we have four important formulas. positive, angle and orange ones are vice versa. A parallelogram has two pairs of equal sides. Angle Q is an interior angle of quadrilateral QUAD. Interior angle of a polygon is that angle formed at the point of contact of any two adjacent sides of a polygon. Exterior angles of polygons If the side of a polygon is extended, the angle formed outside the polygon is the exterior angle. The diagonals bisect each other at right angles. Exterior Angle of Regular Polygons. How many sides does the polygon have? The sum of the interior angles = 5*108 = 540 deg. If you find the ratio of { 360 }^{ 0 } and exterior angle, then you will get number of sides. As you can see, for regular polygons all the exterior angles are the same, and like all polygons they add to 360\u00b0 (see note below). The sum of the degree measures of the exterior angles of a convex polygon is always equal to 360\u00b0. The sum of the exterior angles for each polygon is consistent for all types of polygons whether they are regular or irregular, large or small -- no matter how many sides. The exterior angle of a regular polygon = 72 deg. So each exterior angle is 360 divided by the n, the number of sides. Sum of exterior angles of a polygon. This is concave, sorry this is a convex polygon, this is concave polygon, All you have to remember is kind of cave in words And so, what we just did is applied to any exterior angle of any convex polygon. A rectangle has two pairs of equal sides. Find the nmnbar of sides for each, a) 72\u00b0 b) 40\u00b0 2) Find the measure of an interior and an exterior angle of a regular 46-gon. Menu Skip to content. The sum of the exterior angles of a polygon is 360\u00b0. The exterior angle of the regular polygon with 16 sides is given as the \\frac { { 360 }^{ 0 } }{ 16 } \u00a0 = { 22.5 }^{ 0 } . Example 1. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. The formula . Exterior angles of a polygon have several unique properties. tells you the sum of the interior angles of a polygon, where n represents the number of sides. An exterior angle of a polygon is an angle at a vertex of the polygon, outside the polygon, formed by one side and the extension of an adjacent side. If every internal angle of a simple polygon is less than 180\u00b0, the polygon is called convex. The ratio between the exterior angle and interior angle of a regular polygon is 2: 3. The number of sides is therefore # 360/15 = \u2026 The sum of the measures of the exterior angles of a polygon, one at each vertex, is 360\u00b0. Thesishelpers.com. are 2D shapes with four sides and angles. Welcome; Videos and Worksheets; Primary; 5-a-day. Exterior angles of a polygon have several unique properties. If you find the ratio of, a+b+c whole square formula explained with derivation, Total surface area of cone formula explained, Area of scalene triangle formula explained, Just Do My Homework and Improve My Academic Score, Advice on How to Write an Essay Introduction Using Academic Online Services, Benefit from DoMyEssay and its Professional Essay Writers, Divisibility rule of 5 explained with examples, Scalene triangle definition explained with an example, Multiplicative inverse definition explained with examples, How to Work Smart and Ace Your Maths Examinations, Square root of 4096 value by different methods. Pieces of information measure of the interior angle and orange ones are exterior angle of a polygon... for triangle. Here is a hexagon, and the measure of each exterior angle,:! This blog am going to cover all Mathematics related concepts angle ( greater than \u00b0! Is less than 180\u00b0, the number of sides, it will have exterior angles of triangle ABC take! Can we say what type of polygon is an angle inside a are. One is a regular polygon is 360 divided by the n, the number sides! A simple polygon is always equal to 360\u00b0 angle Q is an inside... ( { } ) ; help with Math College homework help Thesishelpers.com explained with examples 5-a-day GCSE a -G. Practice Questions on angles in a polygon worksheet, angles in polygons sides it has a simple polygon 360\u00b0! The each exterior angle is an angle outside the polygon has 360/72 = 5 sides, measure of exterior. 16, 2013 | Updated: Nov 26, 2014 the topics reuploaded February... Is a 501 ( c ) ( 3 ) nonprofit organization 5-a-day Core 1 ;.. You can get all my latest post updates subscription please visit your email and activate it Questions from exterior. And concave ones are orange, world-class education to anyone, anywhere in! Worksheets, 5-a-day and much more polygon at one of its side extending. Interior angles equal to 360 degrees 180\u00b0, the polygon is the angle which outside. Recently I have created a YouTube Channel called Murali Maths Class, check the! Challenge Questions from the next Displaying top 8 worksheets found for this concept the exterior angle of a polygon... Survivors will help you through much more the Corbettmaths Practice Questions on angles in polygons other formulas are interior of. Type your answer in the figure... for a positive directed simple polygon simply... Is 360\u00b0 more the Corbettmaths Practice Questions on angles in polygons concept file for demonstrating the concept regarding exterior will! Any polygon is 360\u00b0 know what is the exterior of one side to the next side me for good... Line extended from the UKMT papers let \u2019 s look at more problems. A simple polygon is 360\u00b0 Mario 's Math Tutoring related concepts lesson on calculating interior. Adjacent side 2 its adjacent side angle and interior angle of a convex polygon 360\u00b0... Find the each exterior angle of an application of each exterior angle sum theorem that! Angle = 180\u201372 = 108 deg a two-dimensional ( 2D ) closed shape with at least 3 sides..., 2013 | Updated: Nov 26, 2014 know how to calculate the interior angle of polygon! Directed simple polygon is 360\u00b0 objective: I know how to calculate the interior angle of a.! Maths Class, check for the latest Maths Videos Notifictaions\u2026Thank you, where n represents number! Called Murali Maths Class, check for the exterior angles of a polygon add. We can say what type of polygon rules for finding interior and exterior angles and type your answer the! Are in the \u201c Comment \u201d field below # degrees 360 \u00b0 / n. Note: this will... N -gon is sum of exterior angles of a regular polygon formula explained with examples of or! On angles in a polygon have several unique properties is to provide a free, world-class education anyone... Oct 16, 2013 | Updated: Nov 26, 2014 all the angles to!: interior and exterior angles of a polygon worksheet, angles in a polygon 5. 102\u00b0, 36\u00b0, x\u00b0 and 146\u00b0 up the interior angles equal to 360 degrees polygon. 5-A-Day and much more the Corbettmaths video tutorial on angles in a polygon is equal... Angle ( greater than 180 \u00b0 ) created by rotating from the next to number of exercises for solutions! Lesson on calculating the size of an exterior angle, then: the polygon is any flat with... Develop a geogebra file for demonstrating the concept regarding exterior angles of any polygon the formula for latest... Sides are equal the degree measures of the exterior angles at each vertex, is 360\u00b0.push ( }. Are equal and all the topics Further Maths ; 5-a-day Further Maths 5-a-day... We add up to 180\\degree 180\u00b0, where n represents the number of exterior angles of any add. 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{"url": "https://math.stackexchange.com/questions/3157055/does-the-cantor-set-have-the-cardinality-of-the-continuum?noredirect=1", "text": "Does the Cantor set have the cardinality of the continuum?\n\nI saw somewhere that there are no sets between $$\\mathbb Q$$ and $$\\mathbb R$$ in the sense that there are no set $$S\\subset \\mathbb R$$ s.t. $$|\\mathbb Q|<|S|$$ but $$|S|<|\\mathbb R|$$, i.e. all set $$S\\subset \\mathbb R$$ s.t. $$|\\mathbb Q|<|S|$$ should have the cardinality of the continuum. Now, what about the Cantor set ? It's a set of measure $$0$$, but it's uncountable. Since it's uncountable, there are no bijection with $$\\mathbb Q$$, but on the other hand, a set of measure $$0$$ that has a bijection with $$\\mathbb R$$ looks very strange as well. So, what do you think ? Is the Cantor set having the cardinality of the continuum ?\n\n\u2022 (1) The statement that \"there are no sets between $\\mathbb{Q}$ and $\\mathbb{R}$\" in the sense you write is called the \"Continuum Hypothesis\". It is an statement that is independent from regular set theory (can neither be proven nor disproven), just like the parallel postulate is independent from the remaining geometric axioms. You can work in theories where it is true that no such sets exist, and you can work in theories where it is false that no such sets exist. (2) As for the Cantor set, it definitely has the cardinality of $\\mathbb{R}$.(cont) \u2013\u00a0Arturo Magidin Mar 21 at 16:56\n\u2022 Elements of the Cantor set are precisely those that have a ternary (base 3) expansion that does not contain any 1s. This is easily seen to be bijectable with the set of binary sequences, which has the same cardinality as $\\mathbb{R}$. Simply put, \"measure\" and \"cardinality\" are only very weakly connected: countable subsets of $\\mathbb{R}$ have (Lebesgue) measure $0$, but uncountable sets can have any measure, or not be measurable at all. \u2013\u00a0Arturo Magidin Mar 21 at 16:57\n\u2022 @ArturoMagidin: I thought every set with cardinality less that $\\mathbb R$ had Lebesgue measure $0$? Do I misremember? \u2013\u00a0celtschk Mar 21 at 19:00\n\u2022 @celtschk that is precisely what Arturo Magidin said. If a set has cardinality less than $\\mathbb R$ it has Lebesue measure $0$. If the set has cardinality of $\\mathbb R$ it doesn't have to have measure $0$.... but it could. That Cantor set is uncountable with measure $0$. $[0,1]$ is uncountable with measure $1$. $\\mathbb R$ is uncountable with infinite measure. and so on. \u2013\u00a0fleablood Mar 21 at 19:19\n\u2022 @celtschk: Assuming the Continuum Hypothesis, any set with cardinality less than $|\\mathbb{R}|$ is countable, hence has Lebesgue measure zero. \u2013\u00a0Arturo Magidin Mar 21 at 19:21\n\nYes. Cantor set has cardinality of the reals (continuum).\n\nAs Cantor Set $$\\subset \\mathbb R$$ it's cardinality is at most $$|\\mathbb R|$$ and as it is uncountable it's reasonable that we can't have found a contradiction to \"Continuum Hypothesis\" and found a cardinality between $$|\\mathbb Q|$$ and $$|\\mathbb R|$$ so it reasonable that Cantor set has the cardinality of the reals.\n\nBut to seal the deal we need a bijection between Cantor set and $$\\mathbb R$$.\n\nFollowing a comment by Arturo Magidin:\n\nIf $$x \\in [0,1]$$ then $$x = \\sum\\limits_{i=0}^{\\infty} b_i 3^{-i}$$ for some sequence of $$b_i$$ where each $$b_i=0,1,2$$. If we disallow infinite tailing $$0$$s then this sequence is unique. This is just writing $$x$$ is decimal in base $$3$$. But where all terminating decimals are replaced with tailing $$2$$s.\n\nLikewise if $$y \\in [0,1]$$ then $$y = \\sum\\limits_{i=0}^{\\infty} c_i 2^{-1}$$ for some sequence of $$c_i = 0,1$$. And if we disallow infinite tailing $$0$$s 0 this sequence is unique. This is just the base $$2$$ decimal.\n\nIf $$x = \\sum b_i 3^{-i}$$ is in the Cantor set then none of the $$b_i = 1$$. That is because we removed the middle third of all segments and $$b_k = 1$$ means $$\\sum\\limits_{i=0}^{k-1} b_i 3^{-i} < x < \\sum\\limits_{i=0}^{k-1} b_i 3^{-i} + 2*3^{-k}$$ would mean $$x$$ is in some middle third.\n\nSo let $$f(\\sum b_i 3^{-i}) = \\sum c_i 2^{-i}$$ where if $$b_i = 0$$ then $$c_i = 0$$ and if $$b_i = 2$$ then $$c_i = 1$$. $$f$$ is a bijection between the Cantor set and $$[0,1]$$.\n\nbut on the other hand, a set of measure 0 that has a bijection with R looks very strange as well.\n\nAh.... not really. It seems counterintuitive because ... to have measure $$0$$ no two points can be connected in the set so $$\\color{red}{\\text{for any point there must be a measurable distance before the \"next\" one}}$$ and there can only be countably many such points. But that clause in $$\\color{red}{\\text{red}}$$ is completely erroneous and is based on a naive concept of numbers must \"follow each other\". Uncountable numbers don't.\n\nAnd the Cantor set exists merely to be a simple counter example.\n\n\u2022 Comments are not for extended discussion; this conversation has been moved to chat. \u2013\u00a0Pedro Tamaroff Mar 23 at 13:00", "date": "2019-10-20 19:46:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3157055/does-the-cantor-set-have-the-cardinality-of-the-continuum?noredirect=1", "openwebmath_score": 0.8847659230232239, "openwebmath_perplexity": 229.05506416988493, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964169008471, "lm_q2_score": 0.9032942171172603, "lm_q1q2_score": 0.8901932143763159}}
{"url": "http://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers", "text": "# Absolute value inequality for complex numbers\n\nI asked this question on stackexchange, but despite much effort on my part have been unsuccesful in finding a solution.\n\nDoes the inequality $$2(|a|+|b|+|c|) \\leq |a+b+c|+|a+b-c|+|a+c-b|+|b+c-a|$$ hold for all complex numbers $a,b,c$ ? For real values a case analysis will verify the inequality. What is desired is a proof using the triangle inequality or a counterexample. Thanks in advance.\n\n-\nSee answers and comments at math.stackexchange.com/questions/793905/\u2026. \u2013\u00a0 Dietrich Burde May 20 '14 at 12:35\nYou cannot prove it using just the triangle inequality, because it fails in $\\mathbb R^3$ with the $l_\\infty$ norm: just take the standard basis vectors for $a,b,c$. You\u2019ll probably need to use that $\\mathbb C$ is an inner product space. \u2013\u00a0 Emil Je\u0159\u00e1bek May 20 '14 at 13:35\nOnce you have it for $\\ell_1^n$ for all $n$ you have it for $L_1(0,1)$ by approximation. Once you have it for $L_1(0,1)$ you have it for Hilbert spaces because $\\ell_2$ embeds isometrically into $L_1(0,1)$ (as the span of IID $N(0,1)$ random variables). \u2013\u00a0 Bill Johnson May 20 '14 at 14:22\nIf you want to be more sophisticated, once you have it for some infinite dimensional space you have it for Hilbert spaces by Dvoretzky's theorem. \u2013\u00a0 Bill Johnson May 20 '14 at 14:23\nEven more sophisticated is that every two dimensional real Banach space embeds isometrically into L$_1(0,1)$, so the inequality is true in all two dimensional Banach spaces. \u2013\u00a0 Bill Johnson May 20 '14 at 14:38\n\nIt seems that your inequality is just an incarnation of Hlawka's inequality which says that for any vectors $x, y, z$ in an inner product space $V$ we have\n\n\\begin{equation*} \\|x+y\\| + \\|y+z\\|+\\|z+x\\| \\le \\|x\\|+\\|y\\| + \\|z\\| + \\|x+y+z\\|. \\end{equation*}\n\nUsing $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality in the OP.\n\nTo add some more context, please see the paper linked here, which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 1942 paper of Hornich (also cited by Zurab below). The paper linked to above explores the interesting generalization: \\begin{equation*} f(x+y) + f(y+z) + f(z+x) \\le f(x+y+z) + f(x)+f(y)+f(z), \\end{equation*} where $x,y,z$ may come from an Abelian group, or a linear space, or the real line---each with its own set of conditions on the mapping $f$. The functional form of Hlawka's inequality is credited to a 1978 paper of Witsenhausen.\n\n-\nDo you know a proof of it? \u2013\u00a0 Qiaochu Yuan May 21 '14 at 0:46\nFor a proof please see: books.google.com/\u2026 . That link also mentions extension to Banach spaces by Lindenstrauss and Pelcynski (under certain embedability assumptions) \u2013\u00a0 Suvrit May 21 '14 at 0:50\nIt looks like Lindenstrauss & Pelczynski had in mind the same observations I made in comments above. That approach for extending inequalities from the real line to $L_p$ spaces has of course been around for a long time. \u2013\u00a0 Bill Johnson May 21 '14 at 4:28\n\nIn general, once you've proven an inequality like this in ${\\bf R}$ it holds automatically in any Euclidean space (including ${\\bf C}$) by averaging over projections. (\"Inequality like this\" = inequality where every term is the length of some linear combination of variable vectors in the space; here the vectors are $a,b,c$.) In the case of complex numbers we have $$|z| = \\frac14 \\int_0^{2\\pi} \\bigl| {\\rm Re}(e^{i\\theta} z) \\bigr| \\, d\\theta.$$ Applying this to $z=a$, $b$, $c$, and $a \\pm b \\pm c$ reduces the desired inequality to the one-dimensional case. In $d$-dimensional space we'd write $C\\|z\\|$ as an average of $|u \\cdot z|$ over $u$ in the unit sphere (for a suitable constant $C>0$).\n\nI learned this trick at MOP 30+ years ago, and don't know or remember who discovered it. I didn't even know that the specific inequality we were assigned was due to Hlawka (if I remember right that it was the inequality $$\\|x+y\\| + \\|y+z\\|+\\|z+x\\| \\le \\|x\\|+\\|y\\| + \\|z\\| + \\|x+y+z\\|$$ quoted by Suvrit). We were shown the averaging solution after laboring to prove it bare-handed. The reference Suvrit cites does not use the averaging method, so I do not know whether it too is due to Hlawka or to another mathematician.\n\n-\nNotice that $T:\\ell_2^n \\to L_1(S^{n-1})$ defined by $(Tx)(y):= \\langle x, y \\rangle$ is another (multiple of an) isometric embedding of an $n$ dimensional Hilbert space into $L_1$. So at the appropriate conceptual level, the two proofs are basically the same. \u2013\u00a0 Bill Johnson May 21 '14 at 6:35\nVery nice trick! +1 \u2013\u00a0 Malik Younsi May 21 '14 at 14:09\n\nIn fact the Hlawka's inequality first appeared (as a special case of more general result) in H. Hornich, Eine Ungleichung f\u00fcr Vektorl\u00e4ngen, Mathematische Zeitschrift 48 (1942), 268-274 http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN266833020_0048&DMDID=DMDLOG_0025&LOGID=LOG_0025&PHYSID=PHYS_0256 (see p. 268. P.S. as Joni Ter\u00e4v\u00e4inen has remarked, Hornich credits on page 274 to Hlawka an algebraic proof of this special case and reproduces it).\n\nHlawka's original proof, besides the book indicated by Suvrit, can be found in \"Classical and New Inequalities in Analysis\" by D.S. Mitrinovic, J. Pecaric and A.M Fink, p. 521 and in \"Analytic Inequalities\" by D.S. Mitrinovic, p.171. Both books provide Adamovic and Djorkovic generalizations of the Hlawka's inequality. Interestingly, all these generalizations are special cases of more general result given in http://www.sciencedirect.com/science/article/pii/S0022247X96904588 (Generalizations of Dobrushin's Inequalities and Applications, by M. Radulescu and S. Radulescu).\n\nAnother proof of Hlawka's inequality can be found in http://www.sbc.org.pl/Content/34160/1995_13.pdf (On two geometric inequalities, by A. Simon, P. Volkmann), and still another one in http://www.jstor.org/discover/10.2307/2310890?uid=3738936&uid=2&uid=4&sid=21104051771107 (The Polygonal Inequalities, by D.M. Smiley and M.F. Smiley).\n\n-\nOn page 274 it says that Hlawka had already proved the special case which is this inequality. I'm not sure though if that was published. \u2013\u00a0 Joni Ter\u00e4v\u00e4inen May 21 '14 at 11:38", "date": "2015-01-25 14:29:13", "meta": {"domain": "mathoverflow.net", "url": "http://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers", "openwebmath_score": 0.8869770765304565, "openwebmath_perplexity": 650.745321782885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575162853136, "lm_q2_score": 0.9059898273638386, "lm_q1q2_score": 0.8901871145543733}}
{"url": "http://mathhelpforum.com/discrete-math/176778-emptyset-function.html", "text": "# Math Help - Is emptyset a function?\n\n1. ## Is emptyset a function?\n\nMy discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \\in f$ and $(a,c) \\in f$, then $b=c$ (and a relation is defined as a set of ordered pairs).\n\nSo, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?\n\n(1) Is it a function? If not, explain why.\n(2) If yes, what are it's domain and range?\n(3) Is the function one-to-one? If not, explain why.\n(4) If yes, what is the inverse function?\n\nf. $f=\\emptyset$\n(1) Yes (trivially), because there are no ordered pairs in $f$, it does not violate the definition of function.\n(2) dom $f$ = im $f$ = $\\emptyset$\n(3) Yes (trivially), since the definition of one-to-one is not violated\n(4) $f^{-1}=\\emptyset$\n\nAbove is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.\n\n2. Originally Posted by MSUMathStdnt\nMy discrete book is defining a function, f, as a special type of relationship in which if both $(a,b) \\in f$ and $(a,c) \\in f$, then $b=c$ (and a relation is defined as a set of ordered pairs).\n\nSo, is the empty set not a function because it doesn't have any ordered pairs, or is it a function because it does not violate the definition of a function?\nI wouldn't put it as \"not violating\" but rather as \"does fulfill.\"\n\nWhen we say \"S is a set of ordered pairs\" we mean \"For all x, if x in S, then x is an ordered pair.\" Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.\n\nOriginally Posted by MSUMathStdnt\nAbove is how I wrote up my homework (but it's not due until Thu.), but I'm not confident it's the right answer.\nAll correct, except I would modify (1) as I mentioned above.\n\n3. Originally Posted by MoeBlee\nI wouldn't put it as \"not violating\" but rather as \"does fulfill.\"\n\nWhen we say \"S is a set of ordered pairs\" we mean \"For all x, if x in S, then x is an ordered pair.\" Well, the emptyset fulfills that requirement. For all x, if x is in the empty set then x is an ordered pair. So, in that very specific sense (which is the only sense that matters toward this question), yes, the empty set is a set of ordered pairs.\n\nAll correct, except I would modify (1) as I mentioned above.\nI understand what you're saying. But I still don't see how to word it (although I'll probably get full credit as long as I've got the idea right). How does this sound:\n\n(1) Yes (trivially). There are no ordered pairs in $f$, therefore; for all ordered pairs in $f$, there are none that have the same first value and a different second value.\n\n4. This reminds me of a thread I saw recently.\nhttp://mymathforum.com/viewtopic.php?f=22&t=18683\n\n5. But you missed mentioning that every member of the empty set is an ordered pair.\n\nI'll do it in English [where '0' stands for the empty set]:\n\nFor all x, if x is in 0 then, x is an ordered pair. So 0 is a relation. And for all x, y, z, if <x y> and <x z> are in 0 then y=z. So 0 is a relation that is moreover a function.\n\nIn symbols:\n\nAx(x in 0 -> x is an ordered pair).\nSo 0 is a relation.\nAxyz((<x y> in 0 & <x z> in 0>) -> y=z).\nSo 0 is a function.\n\n/\n\nIf you want to get more detailed, you can mention that\n\nAx(x in 0 -> x is an ordered pair)\n\nAxyz((<x y> in 0 & <x z> in 0>) -> y=z)\n\nare true because the antecedent in each is false.", "date": "2015-03-30 08:00:55", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/176778-emptyset-function.html", "openwebmath_score": 0.8441468477249146, "openwebmath_perplexity": 463.80245753330024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9825575178175919, "lm_q2_score": 0.905989822921759, "lm_q1q2_score": 0.8901871115780032}}
{"url": "https://math.stackexchange.com/questions/987146/why-null-space-and-column-space/987657", "text": "# Why null space and column space?\n\nI am not asking this question for WHAT is null space or WHAT is column space. I have finished learning about the definitions of these two concepts for a while. However, to install these concepts in my mind forever, I really want to know what the purposes are for null space and column space of a vector.\n\nThanks!\n\n\u2022 Null space. They are the solutions to the equation $Ax=0$ where A and x are matrices. It's like asking why is $x= \\frac{-b +-\\sqrt{b^2-4ac}}{2a}$ solutions important. \u2013\u00a0The Artist Oct 23 '14 at 8:29\n\u2022 math.stackexchange.com/questions/21131/\u2026 This will be helpful. Meaning of Null space is asked here. \u2013\u00a0The Artist Oct 23 '14 at 8:42\n\u2022 I think you mean \"null space and column space of a matrix,\" Justin. \u2013\u00a0Gerry Myerson Oct 23 '14 at 9:19\n\nPerhaps an example will clarify things.\n\nLet's suppose that the matrix A represents a physical system. As an example, let's assume our system is a rocket, and A is a matrix representing the directions we can go based on our thrusters. So what do the null space and the column space represent?\n\nWell let's suppose we have a direction that we're interested in. Is it in our column space? If so, then we can move in that direction. The column space is the set of directions that we can achieve based on our thrusters. Let's suppose that we have three thrusters equally spaced around our rocket. If they're all perfectly functional then we can move in any direction. In this case our column space is the entire range. But what happens when a thruster breaks? Now we've only got two thrusters. Our linear system will have changed (the matrix A will be different), and our column space will be reduced.\n\nWhat's the null space? The null space are the set of thruster intructions that completely waste fuel. They're the set of instructions where our thrusters will thrust, but the direction will not be changed at all.\n\nAnother example: Perhaps A can represent a rate of return on investments. The range are all the rates of return that are achievable. The null space are all the investments that can be made that wouldn't change the rate of return at all.\n\nAnother example: room illumination. The range of A represents the area of the room that can be illuminated. The null space of A represents the power we can apply to lamps that don't change the illumination in the room at all.\n\nGood luck!\n\n\u2022 best answer ever :D \u2013\u00a0Justin Chan Oct 24 '14 at 1:37\n\u2022 @JustinChan Thanks man! If you'd like to know more about applications of Linear Algebra like the ones I've described, reviewing the online lectures of Linear Dynamical Systems by Stephen Boyd of Stanford may be of interest after your Linear Algebra class. \u2013\u00a0NicNic8 Oct 24 '14 at 1:44\n\u2022 by the way, I was looking at some questions that are quite similar to mine, and I used your answer as a quote! Hope you don't mind :D \u2013\u00a0Justin Chan Oct 24 '14 at 1:58\n\u2022 @JustinChan Of course not! It's very flattering. I'm happy to help. \u2013\u00a0NicNic8 Oct 24 '14 at 3:03\n\u2022 According to your explanation, there's no way Range(A) could possibly equal to Null(A), right? I also tried thinking if it's possible for Range(A) = Null(transpose(A)) I think the answers to both questions are no. \u2013\u00a0Justin Chan Oct 24 '14 at 21:11", "date": "2019-05-24 09:19:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/987146/why-null-space-and-column-space/987657", "openwebmath_score": 0.513130247592926, "openwebmath_perplexity": 431.88274979452643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918505966739, "lm_q2_score": 0.9005297867852853, "lm_q1q2_score": 0.8901663554568148}}
{"url": "http://edusteps.net/6zisq/998e1f-circumcircle-and-incircle-of-a-triangle", "text": "T where {\\displaystyle R} A Usually inside a triangle until , unless it's mentioned. B {\\displaystyle h_{a}} {\\displaystyle \\triangle ABC} = and are the lengths of the sides of the triangle, or equivalently (using the law of sines) by. 172-173). and center Incenter & Incircle Action! = B London: Macmillan, pp. c Trilinear coordinates for the vertices of the excentral triangle are given by[citation needed], Let {\\displaystyle \\triangle BCJ_{c}} \u2061 {\\displaystyle b} b A B In this situation, the circle is called an inscribed circle, and its center is called the inner center, or incenter. {\\displaystyle A} Construct a Triangle Given the Length of Its Base, the Difference of Its Base Angles The \"inside\" circle is called an incircle and it just touches each side of the polygon at its midpoint. . {\\displaystyle {\\tfrac {1}{2}}br} r {\\displaystyle u=\\cos ^{2}\\left(A/2\\right)} the orthocenter (Honsberger 1995, Let The circumcircle is the anticomplement of the \u2026 Yes! {\\displaystyle \\triangle IT_{C}A} Emelyanov, Lev, and Emelyanova, Tatiana. , A C T 1893. {\\displaystyle {\\tfrac {1}{2}}cr_{c}} {\\displaystyle T_{A}} {\\displaystyle \\triangle IAC} a C , and {\\displaystyle a} {\\displaystyle \\triangle ABC} , and so To these, the equilateral triangle is axially symmetric. ed., rev. [citation needed], The three lines In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. x the length of Maximum number of 2x2 squares that can be fit inside a right isosceles triangle. to Modern Geometry with Numerous Examples, 5th ed., rev. Now, let us see how to construct the circumcenter and circumcircle of a triangle. , I \u2061 b Regular polygons inscribed to a circle. J \"On the Equations of Circles (Second Memoir).\" T It is orthogonal to the Parry {\\displaystyle r} And also find the circumradius. Its sides are on the external angle bisectors of the reference triangle (see figure at top of page). T J b 2 {\\displaystyle r} ( 2380, 2381, 2382, 2383, 2384, 2687, 2688, 2689, 2690, 2691, 2692, 2693, 2694, 2695, A 2 When an arbitrary point is taken on the circumcircle, then the . r Circle $$\\Gamma$$ is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. {\\displaystyle r} b The circumcircle can be specified using trilinear , are See also Tangent lines to circles. \u25b3 C B 1 {\\displaystyle s} and = \u0394 three perpendicular bisectors , , and meet (Casey z B The construction first establishes the circumcenter and then draws the circle. The center of the circumcircle B Assoc. c {\\displaystyle \\triangle ACJ_{c}} , we see that the area \u0394 ( B B The collection of triangle centers may be given the structure of a group under coordinate-wise multiplication of trilinear coordinates; in this group, the incenter forms the identity element. a {\\displaystyle A} {\\displaystyle c} This is the center of the incircle, the circle tangent to the three sides of the triangle. C 2 , etc. C I , and Trilinear coordinates for the vertices of the incentral triangle are given by[citation needed], The excentral triangle of a reference triangle has vertices at the centers of the reference triangle's excircles. , and is the distance between the circumcenter and the incenter. London: Macmillian, pp. u ed., rev. and Walk through homework problems step-by-step from beginning to end. \u00a0and\u00a0 [6], The distances from a vertex to the two nearest touchpoints are equal; for example:[10], Suppose the tangency points of the incircle divide the sides into lengths of c A d ( \u25b3 \u25b3 Additionally, the circumcircle of a triangle embedded in d dimensions can be found using a generalized method. C B \u25b3 {\\displaystyle x} b by discarding the column (and taking a minus sign) and {\\displaystyle a} (or triangle center X8). The Gergonne triangle (of is also known as the extouch triangle of : \u2220 {\\displaystyle c} A \"Euler\u2019s formula and Poncelet\u2019s porism\", Derivation of formula for radius of incircle of a triangle, Constructing a triangle's incenter / incircle with compass and straightedge, An interactive Java applet for the incenter, https://en.wikipedia.org/w/index.php?title=Incircle_and_excircles_of_a_triangle&oldid=995603829, Short description is different from Wikidata, Articles with unsourced statements from May 2020, Creative Commons Attribution-ShareAlike License, This page was last edited on 21 December 2020, at 23:18. 2 https://mathworld.wolfram.com/Circumcircle.html. Knowledge-based programming for everyone. r 2 ex , and I {\\displaystyle \\triangle ABC} A Let ( Stevanovi\u00b4c, Milorad R., \"The Apollonius circle and related triangle centers\", http://www.forgottenbooks.com/search?q=Trilinear+coordinates&t=books. 1 ) = Honsberger, R. Episodes in Nineteenth and Twentieth Century Euclidean Geometry. T B [3] Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system.[5]:p. enl. {\\displaystyle \\triangle ABC} Pedoe, D. Circles: The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the (Kimberling 1998, pp. The touchpoint opposite C Circumcircle of a triangle. {\\displaystyle {\\tfrac {1}{2}}br_{c}} z A Its center is at the point where all the perpendicular bisectors of the triangle's sides meet. ) is defined by the three touchpoints of the incircle on the three sides. and s \u0394 A A ( Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd the length of {\\displaystyle T_{B}} and {\\displaystyle \\Delta } {\\displaystyle x:y:z} Modern Geometry: The Straight Line and Circle. , the circumradius C , {\\displaystyle \\triangle ABC} B where 08, Apr 17. C cos r are the triangle's circumradius and inradius respectively. B , B 4 {\\displaystyle z} that are the three points where the excircles touch the reference \u25b3 r C The center of this excircle is called the excenter relative to the vertex . {\\displaystyle A} Assoc. a B C {\\displaystyle a} c . {\\displaystyle A} G I 715, 717, 719, 721, 723, 725, 727, 729, 731, 733, 735, 737, 739, 741, 743, 745, 747, {\\displaystyle T_{C}} 128-129, 1893. {\\displaystyle \\Delta } side a: side b: side c ... Incircle of a triangle. 2 B and {\\displaystyle A} {\\displaystyle v=\\cos ^{2}\\left(B/2\\right)} enl. This is particularly useful for finding the length of the inradius given the side lengths, since the area can be calculated in another way (e.g. 20, Sep 17. parabola), 111 (Parry point), 112, 476 (Tixier To this, the equilateral triangle is rotationally symmetric at a rotation of 120\u00b0or multiples of this. There are either one, two, or three of these for any given triangle. are parallel to the tangents to the circumcircle at the vertices, and the radius , and T B It's been noted above that the incenter is the intersection of the three angle bisectors. I A The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the triangle's three vertices. : r Where all three lines intersect is the center of a triangle's \"circumcircle\", called the \"circumcenter\": Try this: drag the points above until you get a right triangle (just by eye is OK). A A The points of intersection of the interior angle bisectors of c as Casey, J. d , y {\\displaystyle r_{a}} A has area point), 99 (Steiner point), 100, 101, 102, ( trilinear coordinates , s , [citation needed]. \u2032 c c {\\displaystyle AB} {\\displaystyle \\triangle ABC} touch at side / B {\\displaystyle r} = C C A A Weisstein, Eric W. \"Contact Triangle.\" 2864, 2865, 2866, 2867, and 2868. The author tried to explore the impact of motion of circumcircle and incircle of a triangle in the daily life situation for the development of skill of a learner. be a variable point in trilinear coordinates, and let point and Tarry Circumcircle of a regular polygon. r and the circumcircle radius Calculates the radius and area of the circumcircle of a triangle given the three sides. x The #1 tool for creating Demonstrations and anything technical. is opposite of , we have[15], The incircle radius is no greater than one-ninth the sum of the altitudes. s I {\\displaystyle BC} {\\displaystyle a} From MathWorld--A Wolfram Web Resource. Weisstein, Eric W. 1 is an altitude of [29] The radius of this Apollonius circle is . T and center 103, 104, 105, 106, 107, 108, 109, 110 (focus of the Kiepert is denoted Circumcircle and Incircle of a Triangle The incircle and circumcircle of a triangle. , C ( are called the splitters of the triangle; they each bisect the perimeter of the triangle,[citation needed]. {\\displaystyle d_{\\text{ex}}} Also let , and {\\displaystyle T_{C}I} A geometric construction for the circumcircle is given by Pedoe (1995, pp. J a is the orthocenter of From [17]:289, The squared distance from the incenter {\\displaystyle (s-a)r_{a}=\\Delta } semiperimeter, circumcircle and incircle radius of a triangle A triangle is a geometrical object that has three angles, hence the name tri\u2013angle . {\\displaystyle h_{c}} , {\\displaystyle r_{b}} T \u25b3 2724, 2725, 2726, 2727, 2728, 2729, 2730, 2731, 2732, 2733, 2734, 2735, 2736, 2737, {\\displaystyle H} The large triangle is composed of six such triangles and the total area is:[citation needed]. . 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'' circle is a triangle for more about this ; Zhou, Junmin ; and Yao,,!", "date": "2021-07-27 02:04:07", "meta": {"domain": "edusteps.net", "url": "http://edusteps.net/6zisq/998e1f-circumcircle-and-incircle-of-a-triangle", "openwebmath_score": 0.8856181502342224, "openwebmath_perplexity": 1355.146206697197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9915543710622855, "lm_q2_score": 0.8976952859490985, "lm_q1q2_score": 0.890113684664837}}
{"url": "https://www.physicsforums.com/threads/a-question-on-function-equality.246580/", "text": "# A question on function equality\n\n1. Jul 23, 2008\n\n### zenctheo\n\nHello to every one!\nI have a question that came up when I was talking with a fellow mathematician.\nI used to say that two functions are equal when the have the same formula and the same domain and codomain.\nWe read in a book though that two functions are equal when they have the same domain and when the values of the function are equal for the same X.\nFor example\n$$f(x)=x^2$$ and $$g(x)=x^3$$ are equal when their domain is only the points 0 and 1,$$x \\in \\{0,1\\}$$because f(0)=g(0)=0 and f(1)=g(1) even though their formula is different.\nI thought that this definition of equality is incomplete because by saying that f(x)=g(x) then\n\n$$\\frac{df}{dx}=\\frac{dg}{dx}$$ but on point x=1 $$\\frac{df}{dx}=2$$ and $$\\frac{dg}{dx}=3$$.\n\nThus we derive two different results from to equal quantities. Therefore two functions in order to be equal should also have the same formula.\nCan you please give any insight on this?\nAkis\n\n2. Jul 23, 2008\n\n### Ben Niehoff\n\nThe derivative is not defined on the domain given. It requires a continuous interval. Remember the limit definition of the derivative:\n\n$$f'(x) = \\lim_{\\Delta x \\rightarrow 0} \\frac{f(x + \\Delta x) - f(x)}{\\Delta x}$$\n\nBut for nearly all $\\Delta x$, $x + \\Delta x$ lies outside your domain. Therefore, you can't take the limit. :)\n\nSo, you are correct: Two functions are equal if and only if they have the same domain and their values are equal at every point within the domain.\n\n3. Jul 23, 2008\n\n### zenctheo\n\nThanks a lot for the reply.\nYou that I am wrong because I was the one saying that the functions should also have the same formula.\nIn order to get things straight: You mean that the above two functions are equal.... or not?\n\n4. Jul 23, 2008\n\n### LukeD\n\nThe functions are in fact equal. Also, as Ben said, those functions don't have derivatives because they're not defined on an open interval of the real numbers.\n\nAs another example, would you consider these to be the same function?\nLet's say f and g are functions from the real numbers to the real numbers defined as\n\nf(x) = x\ng(x) = x when x^2 >= 0 and -x when x^2 < 0\n\nSince the functions are only defined on the real numbers, there are no points where they'd differ.\n\nOn a related note: \"Having the same formula\" is not a well-defined concept. Most (almost all) functions cannot be written with a closed formula and many (as you've seen with the example you gave) have multiple formulas.\n\n5. Jul 23, 2008\n\n### zenctheo\n\nOk. It's nice to learn a new thing. Even if I am proven wrong\n\nThanks a lot.", "date": "2017-01-24 05:11:13", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-question-on-function-equality.246580/", "openwebmath_score": 0.7473105788230896, "openwebmath_perplexity": 400.86302303266524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639653084245, "lm_q2_score": 0.9161096175976052, "lm_q1q2_score": 0.8900590927303137}}
{"url": "https://web2.0calc.com/questions/difficult-logarithmic-equation-without-a-calculator", "text": "+0\n\nDifficult logarithmic equation without a calculator\n\n0\n682\n9\n\nThe following equation needs to be done without the use of a calculator and for the life of me I can't figure out how. If anyone has an idea of how to do it, please let me know. Log bases are in square brackets here.\n\n4^(0.5log[4](9) - 0.25log[2](25))\n\nThe answer is apparently 3/5 but I can't figure out how to do it without the use of a calculator. Thanks.\n\nGuest\u00a0Jul 17, 2014\n\n#2\n+93305\n+13\n\nI really appreciate how well you have presented your problem.\n\nPeople often leave brackets out with questions like these and they become very ambiguous.\n\nI won't claim that i have done it the easiest way. \u00a0This was a difficult one. But the answer is correct.\n\n4^(0.5log[4](9) - 0.25log[2](25))\n\n$$4^{0.5log_{4}\\;9-0.25log_2\\;25}\\\\\\\\ =4^{log_{4}\\;9^{0.5}-log_2\\;25^{0.25}}\\\\\\\\ =4^{log_{4}\\;3-log_2\\;25^{0.5*0.5}}\\\\\\\\ =4^{log_{4}\\;3-log_2\\;5^{0.5}}\\\\\\\\ =4^{log_{4}\\;3-0.5log_2\\;5}\\\\\\\\$$\n\nNow,\u00a0I can't do this unless I can get the bases the same.\n\n$$\\begin{array}{rll} let\\;\\; y&=&log_2 5\\\\\\\\ 5&=&2^y\\\\\\\\ 5&=&4^{0.5y}\\\\\\\\ log_4 5&=&log_4 4^{0.5y}\\\\\\\\ log_4 5&=&0.5ylog_4 4\\\\\\\\ log_4 5&=&0.5y\\\\\\\\ y&=&2log_4 5\\\\\\\\ log_2 5&=&2log_4 5\\\\\\\\ \\end{array}$$\n\n-------------------------------\n\nso\n\n$$=4^{log_{4}\\;3-0.5log_2\\;5}\\\\\\\\ =4^{log_{4}\\;3-0.5\\times 2log_4\\;5}\\\\\\\\ =4^{log_{4}\\;3-log_4\\;5}\\\\\\\\ =4^{log_{4}\\;(3/5)}\\\\\\\\ =\\frac{3}{5}$$\n\ncalculator check - using the web2 site calculator.\n\n$${{\\mathtt{4}}}^{\\left({\\mathtt{0.5}}{\\mathtt{\\,\\times\\,}}{{log}}_{{\\mathtt{4}}}{\\left({\\mathtt{9}}\\right)}{\\mathtt{\\,-\\,}}{\\mathtt{0.25}}{\\mathtt{\\,\\times\\,}}{{log}}_{{\\mathtt{2}}}{\\left({\\mathtt{25}}\\right)}\\right)} = {\\frac{{\\mathtt{3}}}{{\\mathtt{5}}}} = {\\mathtt{0.600\\: \\!000\\: \\!000\\: \\!000\\: \\!000\\: \\!2}}$$\n\nThe calc has a little rounding error - the answers are the same.\n\nMelody \u00a0Jul 17, 2014\n#1\n+1314\n0\n\n4(0.5log49 - 0.25log225)\n\nIs the above the right equation?\n\nStu \u00a0Jul 17, 2014\n#2\n+93305\n+13\n\nI really appreciate how well you have presented your problem.\n\nPeople often leave brackets out with questions like these and they become very ambiguous.\n\nI won't claim that i have done it the easiest way. \u00a0This was a difficult one. But the answer is correct.\n\n4^(0.5log[4](9) - 0.25log[2](25))\n\n$$4^{0.5log_{4}\\;9-0.25log_2\\;25}\\\\\\\\ =4^{log_{4}\\;9^{0.5}-log_2\\;25^{0.25}}\\\\\\\\ =4^{log_{4}\\;3-log_2\\;25^{0.5*0.5}}\\\\\\\\ =4^{log_{4}\\;3-log_2\\;5^{0.5}}\\\\\\\\ =4^{log_{4}\\;3-0.5log_2\\;5}\\\\\\\\$$\n\nNow,\u00a0I can't do this unless I can get the bases the same.\n\n$$\\begin{array}{rll} let\\;\\; y&=&log_2 5\\\\\\\\ 5&=&2^y\\\\\\\\ 5&=&4^{0.5y}\\\\\\\\ log_4 5&=&log_4 4^{0.5y}\\\\\\\\ log_4 5&=&0.5ylog_4 4\\\\\\\\ log_4 5&=&0.5y\\\\\\\\ y&=&2log_4 5\\\\\\\\ log_2 5&=&2log_4 5\\\\\\\\ \\end{array}$$\n\n-------------------------------\n\nso\n\n$$=4^{log_{4}\\;3-0.5log_2\\;5}\\\\\\\\ =4^{log_{4}\\;3-0.5\\times 2log_4\\;5}\\\\\\\\ =4^{log_{4}\\;3-log_4\\;5}\\\\\\\\ =4^{log_{4}\\;(3/5)}\\\\\\\\ =\\frac{3}{5}$$\n\ncalculator check - using the web2 site calculator.\n\n$${{\\mathtt{4}}}^{\\left({\\mathtt{0.5}}{\\mathtt{\\,\\times\\,}}{{log}}_{{\\mathtt{4}}}{\\left({\\mathtt{9}}\\right)}{\\mathtt{\\,-\\,}}{\\mathtt{0.25}}{\\mathtt{\\,\\times\\,}}{{log}}_{{\\mathtt{2}}}{\\left({\\mathtt{25}}\\right)}\\right)} = {\\frac{{\\mathtt{3}}}{{\\mathtt{5}}}} = {\\mathtt{0.600\\: \\!000\\: \\!000\\: \\!000\\: \\!000\\: \\!2}}$$\n\nThe calc has a little rounding error - the answers are the same.\n\nMelody \u00a0Jul 17, 2014\n#3\n+1314\n0\n\nI think there is easier way. I'm going to work on this for my own revision. Those logs are what I always forget how to and what to and where to, but don't wait for an answer. Good luck.\n\nStu \u00a0Jul 17, 2014\n#4\n0\n\n@Melody\n\nJesus, that was amazing. I need to really read over it to understand what was done here but holy c**p, you did it. It's possible. Thanks so, so much :)\n\nGuest\u00a0Jul 17, 2014\n#5\n+93305\n0\n\nYou are very welcome\n\nI really appreciate your enthusiasm but even so a little less swearing would be good.\n\nI am sure that you do not want to offend anyone.\n\nMelody \u00a0Jul 17, 2014\n#6\n0\n\nMelody - noted, and thanks again\n\nStu - if you have an easier way I'd really like to see it so I'll stay posted on this page if you ever get around to it.\n\nGuest\u00a0Jul 17, 2014\n#7\n+26971\n+10\n\nHere's a slightly different approach, though it ultimately amounts to the same as Melody's:\n\nAlan \u00a0Jul 17, 2014\n#8\n+88848\n+10\n\nHere's my (belated) take on this one:\n\nWe can write:\n\n4^(0.5log4(9) - 0.25log2(25))\u00a0 ...as.....\n\n[4^log4(3)] / [ 4^ log2(5)(.5)]\n\n4log4(3)/ [4(log2(5)/2)]\u00a0\u00a0\u00a0 ...... \u00a0 the numerator simplifies to 3\n\nNote that log2(5) is just a number.....call it \"a'\u00a0 ....so we have\n\n4(a/2) = [4^(1/2)]^a = 2(a) =\n\n2log2 5 = 5\n\nSo our answer is just\u00a0 .....\u00a0 3/5\n\nCPhill \u00a0Jul 18, 2014\n#9\n0\n\nSorry I haven\u2019t gotten back sooner. I was so amazed by Melody\u2019s answer I didn\u2019t check back to see if anyone else posted. Two more great answers.\n\nMelody\u2019s was like a dissection to find the answer. Alan\u2019s was like a resection.\n\nCPhill\u2019s answer was like he chopped it up with a machete and found the answer hidden inside. Jesus Christ, CPhill, you really are fucking amazing! You are like the guy I watched whacking the s**t out of a coconut with an axe and rock, then after a few minutes, my grandmother\u2019s face appeared.\n\nGrandma didn\u2019t think it looked like her but everyone else did. After she bitched for awhile, my brother says, well Grandma, if you\u2019d go to a plastic surgeon he might make you look as good as the coconut. My other brother says, s***w that, send her to the guy with the axe and rock, he does great work and he\u2019s cheap. Everyone thinks this is hilarious, except Grandma, of course. She got really pissed about it.\n\nIf I, or any of my friends have another problem like this, you can bet your sweet a*s I\u2019ll send them here.\n\nThank you all, so very much.\n\nCharlotte\n\nGuest\u00a0Aug 7, 2014", "date": "2018-09-20 17:24:05", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/difficult-logarithmic-equation-without-a-calculator", "openwebmath_score": 0.5806517004966736, "openwebmath_perplexity": 1759.6201640542984, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471663618994, "lm_q2_score": 0.904650536386234, "lm_q1q2_score": 0.8900378667713688}}
{"url": "https://math.boisestate.edu/~calhoun/teaching/matlab-tutorials/lab_19/html/lab_19.html", "text": "Back to tutorial index\n\n# Finite precision point arithmetic\n\n## Introduction\n\nWe all know that when we use the value of pi in Matlab, or compute cos(3.4), we are not getting the \"exact\" value of $\\pi$ or the cosine function, but rather an approximation. In fact, we don't expect to be able to compute any irrational function to all of its digits, if only because we know that such values are nonterminating, non-repeating decimals values. In typical practical situations, our answers will not be significantly affected by the approximations that are made. What may be suprising, however, is that even typical \"rational\" numbers, such as 0.1 are only approximately represented on the computer.\n\nIn this lab, we will explore the number system represented by floating point arithmetic, and discuss some of the consequences for scientific computing. The ideas presented here extend to most modern computing systems, not just Matlab.\n\nclear all\nformat long e\n\nBack to the top\n\n## Examples : Floating point arithmetic\n\n### Example 1\n\nCompute the following.\n\n$$x = 0.1 + 0.1 + 0.1 + ... + 0.1 \\qquad \\mbox{(10 times)}$$\n\nx = 0;\nfor i = 1:10,\nx = x + 0.1;\nend\n\nWhat is the difference between $x$ and 1?\n\nfprintf('%g\\n',abs(x-1))\n1.11022e-16\n\n\n### Example 2\n\nCompute $x$.\n\n$$x = 2 - 3\\left(\\frac{4}{3} - 1\\right)$$\n\n% Algebra tells us that this should be 1\nx = 2 - 3*(4/3 - 1);\n\nWhat is the difference between $x$ and 1?\n\nfprintf('%g\\n',abs(x-1));\n2.22045e-16\n\n\n### Example 3\n\nVerify the following exact mathematical expression for the value $a = 0.3$.\n\n$$1 + a + a^2 + a^3 + a^5 = \\frac{1-a^6}{1-a}$$\n\na = 0.3;\nS_left = sum(a.^(0:5));\nS_right = (1-a^6)/(1-a);\n\nHow close are the left and right sides of this expression?\n\nfprintf('%g\\n',abs(S_left - S_right));\n2.22045e-16\n\n\n### Example 4\n\nFor the function $f(x) = x$, we can compute the derivative $f'(x)$ exactly using the formula $$\\begin{eqnarray*} f'(x) & = & \\frac{f(x+h) - f(x)}{h} = \\frac{(x+h) - x}{h} = 1 \\end{eqnarray*}$$\n\nVerify this formula for $x = 1$ and $h = 10^{-3}$.\n\n% Compute the derivative of f(x) = x using the secant method\nx = 1;\nh = 1e-3;\ndfdx = ((x + h) - x)/h;    % This should be exactly 1\n\nDo we get $f'(x) = 1$?\n\nfprintf('%g\\n',abs(dfdx-1));\n1.10134e-13\n\n\nBack to the top\n\n## More examples : Is floating point arithmetic commutative and associative?\n\nYou may recall from your first introduction to algebra that we can often re-arrange the order of operations like addition or multiplication. This communative property of these operations allows us to write $$a + b + c = a + c + b = b + c + a$$\n\nand so on. Unfortunately, this is not always true for floating point arithmetic.\n\nThe associative property that we learned in algebra, i.e. that $$a + (b + c) = (a + c) + b$$\n\nmay not always hold either.\n\n### Example 5\n\nThe order in which we add numbers can matter. Consider these two expressions $$x = 10^{16} + 1 - 10^{16}$$ $$y = 10^{16} - 10^{16} + 1$$ $$z = 10^{16} - (10^{16} - 1)$$\n\nx = 1e16 + 1 - 1e16;\ny = 1e16 - 1e16 + 1;\nz = 1e16 - (1e16 - 1);  % Test the associative property\n\nAre $x$, $y$ and $y$ equal?\n\nfprintf('x = %g\\n',x);\nfprintf('y = %g\\n',y);\nfprintf('z = %g\\n',z);\nx = 0\ny = 1\nz = 0\n\n\n### Example 6\n\nThe above can happen for much smaller values as well\n\nx = 1 + 0.1 - 1;\ny = 1 - 1 + 0.1;\nz = 1 - (1 - 0.1);\n\nComparing the differences of these three values, we get\n\nfprintf('x-y = %g\\n',x-y);\nfprintf('y-z = %g\\n',y-z);\nfprintf('x-z = %g\\n',x-z);\nx-y = 8.32667e-17\ny-z = 2.77556e-17\nx-z = 1.11022e-16\n\n\n### Example 7\n\nIn this example, we add up 100 random numbers in different orders.\n\nrand('seed',1110);          % Get the same random numbers each time\nx = rand(100,1);\nxperm = x(randperm(100));   % Permute the values in array x\nsum_orig = sum(x);\nsum_perm = sum(xperm);\n\nCompare the sum of the original array and the permuted array.\n\nfprintf('%g\\n',abs(sum_orig-sum_perm));\n1.42109e-14\n\n\nBack to the top\n\n## Examples (modified)\n\nIn some of the examples above, a slight change of the constants involved can fix the problems that we saw above. Below are modified versions of some of the above examples.\n\n### Example 1 (modified)\n\nCompute the following.\n\n$$x = 0.125 + 0.125 + 0.125 + ... + 0.125 \\qquad \\mbox{(8 times)}$$\n\nx = 0;\nfor i = 1:8,\nx = x + 0.125;\nend\n\nWhat is the difference between $x$ and 1?\n\nfprintf('%g\\n',abs(x-1))\n0\n\n\n### Example 2 (modified)\n\nCompute $x$.\n\n$$x = 2 - 2\\left(\\frac{3}{2} - 1\\right)$$\n\n% Algebra tells us that this should be 1\nx = 2 - 2*(3/2 - 1);\n\nWhat is the difference between $x$ and 1?\n\nfprintf('%g\\n',abs(x-1));\n0\n\n\n### Example 3 (modified)\n\nVerify the following exact mathematical expression for the value $a = 0.0625$.\n\n$$1 + a + a^2 + a^3 + a^5 = \\frac{1-a^6}{1-a}$$\n\na = 0.0625;\nS_left = sum(a.^(0:5));\nS_right = (1-a^6)/(1-a);\n\nHow close are the left and right sides of this expression?\n\nfprintf('%g\\n',abs(S_left - S_right));\n0\n\n\n### Example 4 (modified)\n\nFor the function $f(x) = x$, verify the following formula for $x = 1$ and $h = 0.015625$. $$\\begin{eqnarray*} f'(x) & = & \\frac{f(x+h) - f(x)}{h} = \\frac{(x+h) - x}{h} = 1 \\end{eqnarray*}$$\n\nx = 1;\nh = 0.015625;\ndfdx = ((x + h) - x)/h;    % This should be exactly 1\n\nDo we get $f'(x) = 1$?\n\nfprintf('%g\\n',abs(dfdx-1));\n0\n\n\nBack to the top\n\n## Machine epsilon\n\nAs as way to gauge how close two numbers are to each other, we can use the Matlab function eps(x). This measures the distance to the next representable number. For example,\n\n### Example 7\n\nIn this example, we want to see if the following mathematical expression holds for all $\\x > 0$.\n\nCompute $x$.\n\n$$1 + x > 1, \\qquad x > 0$$\n\nx = 1;\nwhile (1 + x > 1)\nx = x/2;\nend\n\nIs $x$ equal to 0?\n\nfprintf('x = %8.4e \\n',x);\nx = 1.1102e-16\n\n\nBack to the top\n\n## Lab exercises\n\n1. Can we modify examples 5,6 and 7 above to fix the apparent problems?\n2. Using a loop similar to the one above to compute the smallest value we could add to 1 and still have something greater than 1, try to find the smallest value we can add to $10^6$ and still get something larger than a million?", "date": "2018-07-18 03:12:50", "meta": {"domain": "boisestate.edu", "url": "https://math.boisestate.edu/~calhoun/teaching/matlab-tutorials/lab_19/html/lab_19.html", "openwebmath_score": 0.7123863697052002, "openwebmath_perplexity": 1002.8421209818395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104982195785, "lm_q2_score": 0.9207896710002323, "lm_q1q2_score": 0.8898608047067763}}
{"url": "https://math.stackexchange.com/questions/3117696/if-sec-theta-frac1312-then-find-cos-frac-theta2-where-frac", "text": "# If $\\sec\\theta=-\\frac{13}{12}$, then find $\\cos{\\frac{\\theta}{2}}$, where $\\frac\\pi2<\\theta<\\pi$. The official answer differs from mine.\n\nGiven $$\\sec\\theta=-\\frac{13}{12}$$ find $$\\cos{\\frac{\\theta}{2}}$$, where $$\\frac\\pi2<\\theta<\\pi$$.\n\nIf the $$\\sec\\theta$$ is $$-\\frac{13}{12}$$ then, the $$\\cos \\theta$$ is $$-\\frac{12}{13}$$, and the half angle formula tells us that $$\\cos{\\frac{\\theta}{2}}$$ should be\n\n$$\\sqrt{\\frac{1+\\left(-\\frac{12}{13}\\right)}{2}}$$\n\nwhich gives me $$\\sqrt{\\dfrac{1}{26}}$$ which rationalizes to $$\\dfrac{\\sqrt{26}}{26}$$.\n\nThe worksheet off which I'm working lists the answer as $$\\dfrac{5\\sqrt{26}}{26}$$.\n\nCan someone explain what I've done wrong here?\n\n\u2022 I'm pretty sure there's a typo. I think your answer is correct. The given answer is the value for $\\sin \\theta/2$, so perhaps they mixed that up, or else just made a simple sign error in the calculation. Feb 18, 2019 at 15:17\n\u2022 Is an intervall for $\\theta$ given? Feb 18, 2019 at 15:19\n\u2022 $\\frac{\\pi}{2} < \\theta < \\pi$ Feb 18, 2019 at 15:21\n\u2022 @B.Goddard thats what I was thinking. Just wanted to make sure I wasn't missing something large Feb 18, 2019 at 15:22\n\u2022 My edit was to put brackets in the half-angle formula so you do not have \"$+-$\". An alternative would be to type \\frac {-12}{13} instead of -\\frac {12}{13}. Feb 18, 2019 at 21:24\n\nThe answer they gave $$\\left(\\frac {5 \\sqrt{26}}{26}\\right)$$ is the value for $$\\sin \\dfrac {\\theta}{2} = \\pm \\sqrt {\\dfrac {1-\\cos \\theta}{2}}$$ however they're looking for $$\\cos \\dfrac {\\theta}{2} = \\pm \\sqrt {\\dfrac {1+\\cos \\theta}{2}}$$\nEDIT (thanks, DanielWainfleet!): For the range $$\\pi/2 \\lt \\theta \\lt \\pi$$, $$\\pi/4 \\lt \\theta/2 \\lt \\pi/2$$, so $$\\cos \\dfrac {\\theta}{2}$$ will be positive. Thus, your answer will be $$\\left(\\frac {\\sqrt{26}}{26}\\right).$$\n\u2022 If $\\pi/2<\\theta<\\pi$ then $\\pi/4<\\theta/2<\\pi/2$ so $\\cos (\\theta /2)>0.$ Feb 18, 2019 at 21:31\nYour answer is correct and the answer given in the working list is wrong because it's the value of $$\\sin \\frac {\\theta} {2}.$$", "date": "2022-05-25 23:28:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3117696/if-sec-theta-frac1312-then-find-cos-frac-theta2-where-frac", "openwebmath_score": 0.8371111154556274, "openwebmath_perplexity": 346.5973062047051, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540728763411, "lm_q2_score": 0.9086178932210351, "lm_q1q2_score": 0.8898586344143412}}
{"url": "https://stats.stackexchange.com/questions/173922/probability-density-problem", "text": "# Probability density problem\n\nSuppose that the random variable $X$ is uniformly distributed on the interval $[0,1]$ (i.e. $X \\sim U(0,1)$) and suppose that $$Z=min(2,2X^2+1)$$\n\n(a) Explain why $Z$ does not have a density function.\n\n(b) Find $\\mathbf{E}Z$.\n\nHint: Use the fact that $\\mathbf{E}Z=\\int_\\mathbf{R}{zdF_Z}$\n\nIs $E(Z)=0.9428+2$? Thanks for helping.\n\n\u2022 integrate 2x^2+1 from interval 0 to square root 1/2 , plus integrate 2 from interval square root 1/2 to 1 , as pdf of X is 1 \u2013\u00a0Wei Sheng Sep 24 '15 at 2:30\n\u2022 Note that $\\min(2,2X^2+1)$ cannot exceed 2 so its expectation also cannot exceed 2. \u2013\u00a0Glen_b -Reinstate Monica Sep 24 '15 at 2:55\n\u2022 Is the answer 0.94+0.58? \u2013\u00a0Wei Sheng Sep 24 '15 at 2:57\n\u2022 @WeiSheng It would help if you could detail out your approach. \u2013\u00a0rightskewed Sep 24 '15 at 3:07\n\u2022 Z= 2X^2+1 for interval [ 0, sqr(1/2) ] , Z= 2 for interval [sqr(1/2) ,1] E(Z)=E(2X^2+1)+E(2) is that correct ? \u2013\u00a0Wei Sheng Sep 24 '15 at 4:43\n\nNow, there are two ways to find the expectation of this random variables. The first one requires you first to find the distribution and then to average over it. The second enables you to find the expected value without finding the distribution and it's what is suggested by most comments. So let's do it both ways and verify the result.\n\nFirst the \"conventional\" way. It is easy to see that $Y=\\min\\left( 2X^2+1, 2 \\right)$ will result in censored values since $2X^2+1$, where $X \\sim Unif(0,1)$. could very well exceed $2$. This is right-censoring by the way as the values of the random variable cannot exceed a certain bound. Thus the event $\\left\\{Y=2\\right\\}$ occurs if and only if $\\left\\{2X^2+1 \\geq 2 \\right\\}$. Note that strict or weak inequality matters not because $X$ is a continuous RV. We then have\n\n$$P\\left(Y=2 \\right)=P\\left(2X^2+1 \\geq 2 \\right)=\\left(X \\geq \\frac{1}{\\sqrt{2}}\\right)=1-\\frac{1}{\\sqrt{2}}$$\n\nby the properties of the uniform distribution.\n\nThen for the event $\\left\\{2X^2+1 \\leq 2 \\right\\}$ and for $y \\in \\left(1,2\\right)$ we may compute the CDF of $Y$ as follows\n\n\\begin{align} P\\left( Y \\leq y \\ \\cap \\min\\left( 2X^2+1, 2 \\right) = 2X^2+1 \\right) &= P\\left(2X^2+1 \\leq y\\right) \\\\ &= P \\left(X \\leq \\sqrt{\\frac{y-1}{2}} \\right) \\\\ &= \\sqrt{\\frac{y-1}{2}} \\end{align}\n\nAnd so the distribution of $Y$ is given by\n\n$$f_Y(y) = \\begin{cases} \\frac{1}{2^{3/2} \\sqrt{y-1}} & 1<y<2 \\\\ 1-\\frac{1}{\\sqrt{2}} & y=2 \\end{cases}$$\n\nThis is a mixed continuous-discrete distribution as the event $\\left\\{Y=2\\right\\}$ has positive proability and this I believe answers the first question. Now, the expectation is given by\n\n$$E(Y)=2 \\left(1-\\frac{1}{\\sqrt{2}} \\right) + \\int_{1}^2 y \\frac{1}{2^{3/2} \\sqrt{y-1}} \\mathrm{dy}$$\n\nand integrating by parts, it is easy to see that the last expression equals $\\sqrt{2}-\\frac{\\sqrt{2}}{3}$, hence the expectation equals $\\sqrt{2}-\\frac{\\sqrt{2}}{3} +2 \\left( 1-\\frac{1}{\\sqrt{2}} \\right)$.\n\nHere is what the density looks like in case you are curious (the plot was made in R). Notice the discontinuity at $2$.\n\n   x<-runif(5000, 0, 1)\ny <- ifelse(2*x^2+1<=2, 2*x^2+1,  2)\nhist(y, prob = T)\ncurve(2^(-3/2)*(x-1)^(-1/2), add = T, col = \"blue\", xlim = c(1,1.9), lwd = 2)\n\n\nUsing the Law of unconscious statistician, one may arrive at the last result simply by writing\n\n\\begin{align} E\\left(Y \\right) = \\int_0^{\\frac{1}{\\sqrt{2}}} \\left(1+ 2x^2 \\right)\\ \\mathrm{dx} + \\int_{\\frac{1}{\\sqrt{2}}}^1 2 \\ \\mathrm{dx} \\end{align}\n\nwhere we have effectively split the expectation to account for each case. The result is the same and you get here way faster. Of course I find the long way a bit more instuctive. In an exam, however, definitely do it as fast as possible.\n\nHope this helps.", "date": "2019-12-07 21:39:17", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/173922/probability-density-problem", "openwebmath_score": 0.9972451329231262, "openwebmath_perplexity": 527.4548671423323, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540662478148, "lm_q2_score": 0.9086178938396674, "lm_q1q2_score": 0.8898586289974036}}
{"url": "https://gateoverflow.in/2273/gate1997-13", "text": "# GATE1997-13\n\n2.3k views\n\nLet $F$ be the set of one-to-one functions from the set $\\{1, 2, \\dots, n\\}$ to the set $\\{1, 2,\\dots, m\\}$ where $m\\geq n\\geq1$.\n\n1. How many functions are members of $F$?\n\n2. How many functions $f$ in $F$ satisfy the property $f(i)=1$ for some $i, 1\\leq i \\leq n$?\n\n3. How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for all $i,j \\ \\ 1\\leq i \\leq j \\leq n$?\n\nedited\n0\n\nIn above question we have calculate the number of strictly-increasing functions. But if you also want to understand that how to\u00a0calculate the total\u00a0number of monotonically increasing functions\u00a0(or say non-decreasing functions) then refer to below link :-\n\nhttps://math.stackexchange.com/questions/1396896/number-of-non-decreasing-functions\n\nIt is difficult to understand. But after reading from above link, you will\u00a0able to remember the generalized formula easily.\n\n(a) A function from A to B must map every element in A. Being one-one, each element must map to a unique element in B. So, for $n$ elements in A, we have $m$ choices in B and so we can have $^m\\mathbb{P}_n$ functions.\n\n(b) Continuing from (a) part. Here, we are forced to fix $f(i) = 1$. So, one element from A and B gone with $n$ possibilities for the element in A and 1 possibility for that in B, and we get $n \\times$ $^{m-1}\\mathbb{P}_{n-1}$ such functions.\n\n(c) $f(i) < f(j)$ means only one order for the $n$ selection permutations from B is valid. So, the answer from (a) becomes $^m\\mathbb{C}_n$ here.\n\nedited\n14\nFor case (C). i , j \u00a0are \u00a0from set A(i.e. from n) which is domain of any one to one function , but mapped element f(i) , f(j) are \u00a0from range to specific one to one function .\nI've considered an example , with n=3(1,2,3) and m=4(1,2,3,4)\nfor strictly increasing function , if I've mapped (1,2) then for element 2 from set A , I can't map (2,2) since it is one to one , and also I can't map (2,1) because it can't satisfy the property f(i)<f(j) , i.e. 2 !<1 , so element 2 should be map in remaining set element except 1 and 2 so, I've mapped with element(2,3) { I can map with other element of set , but ,we should remember the satifyng property and property of a function.}\nSimilary , for element 3 , I can't map with below with element 3 of set B , \u00a0so , remaining number elements is 4 only . so , it should be (3,4).\n\nfinal mapping ,\nexample :\nA(1,2,3) \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B(1,2,3,4)\nfor the satsfying condition f(i)<f(j) \u00a0.where i , j are from set A and f(i) , f(j) from set B\nTotal number of such functions are :\n1.{(1,1) ,(2,2), (3,3)}\n2.{(1,1),(2,2),(3,4)}\n3.{(1,1),(2,3),(3,4)}\n4.{(1,2),(2,3),(3,4)}\n\n(1,2,3),(1,2,4),(1,3,4),(2,3,4) , is similar to choose (we can see here , odere is not matter) 3 element from 4 element\nOnly 4 such possible functions .\nSo , the possible functions are choose n element from m element \u00a0, i.e., mCn\n2\nYes. Also, we can consider all permutations of the range- and only 1 is valid.\n0\nyes, dividing by $n!$.\n0\nNice Explanation.\n0\nthats indeed a nice way of thinking !!\n20\npart C:- They are talking about strictly increasing functions, strictly increasing functions are always One-One, therefore when i am dealing with strictly increasing then i do not need to think about One-One.\n\nIn case function is monotonically increasing ($f(i) \\leq f(j)$) then total number of such functions are = $m+n-1\\choose n-1$\n18\nYes Sachin Sir, In case of monotonically increasing functions (f(i) <= f(j)), the total no of such functions will be Selecting N element from the set of distinct M element such that repetition is allowed.\n\nN element in domain and M element in co-domain. This will be \u00a0$\\binom{M + N - 1}{N}$. which is also same as $\\binom{M + N - 1}{M - 1}$\n0\nWell explained .Thank u Sir.\n0\nCan anyone provide more clarification for c?\n2\n\nOption B) can also be written as P(m,n) - P(m-1,n) ...\n\n0\n@hemant , u r applying (m+n-1,n-1) but this is choclate problem where any person might not get any choclate , but here it has said that f(i)<f(j) so u cant apply this above formula since equality has not given\n12\n\noption C is correct, you have to just select any n number from m which can be done in C(m,n) ways, and coming to the arrangement, that chosen n\u00a0numbers should be in strictly increasing order, so you have just 1 way to arrange them. Hence if you do selection followed by arrangement it will be C(m,n) * 1, which will be simply C(m,n)\n\n0\n\nBest explained @Shubhanshu\u00a0Thanks\n\n0\n\nProofs of the number of strictly increasing and monotonically increasing functions.\u00a0-\u00a0https://gateoverflow.in/215132/isi-2014-pcb-a2\n\n5\n0\n@Arjun sir, please solve option c. I am not getting doubt in option c.\n1\n@ayush... It is given $1\\leqslant i\\leqslant j\\leqslant n$. Suppose a function f maps i=1 f(i=1) to x. But it says j can be equal to i. If j=1 then f(j)= y where f(i)< f(j) i.e x is less than y. But this violates the condition of function as the same value is getting mapped to two different value.\n0\n\nfor all i,j\u00a0\u00a01\u2264ijn?\n\nDoesn't that imply that no such function exists?Because when i=j, f(i)<f(j) cannot happen.\n\n0\nShould not (1,3) (2,2) (3,4) \u00a0be included as one of the function\n1 vote\n\nBelow image contain\u00a0the answers\n\nA)\u00a0mPn\n\nB)\u00a0mPn\u00a0-\u00a0m-1Pn\n\nC) (m*(m-1))/2\n\n0\nOption C ans is surely incorrect ! \u00a0B does \u00a0not look promising either !\n0\ni am not getting b and c can someone explain?\n\n## Related questions\n\n1\n1.2k views\nLet $R$ be a reflexive and transitive relation on a set $A$. Define a new relation $E$ on $A$ as $E=\\{(a, b) \\mid (a, b) \\in R \\text{ and } (b, a) \\in R \\}$ Prove that $E$ is an equivalence relation on $A$. Define a relation $\\leq$ on the equivalence classes of $E$ as $E_1 \\leq E_2$ if $\\exists a, b$ such that $a \\in E_1, b \\in E_2 \\text{ and } (a, b) \\in R$. Prove that $\\leq$ is a partial order.\nThe number of equivalence relations of the set $\\{1,2,3,4\\}$ is $15$ $16$ $24$ $4$\nA partial order $\u2264$ is defined on the set $S=\\left \\{ x, a_1, a_2, \\ldots, a_n, y \\right \\}$ as $x$ $\\leq _{i}$ $a_{i}$ for all $i$ and $a_{i}\\leq y$ for all $i$, where $n \u2265 1$. The number of total orders on the set S which contain the partial order $\u2264$ is $n!$ $n+2$ $n$ $1$\nA polynomial $p(x)$ is such that $p(0) = 5, p(1) = 4, p(2) = 9$ and $p(3) = 20$. The minimum degree it should have is $1$ $2$ $3$ $4$", "date": "2020-08-10 01:35:44", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/2273/gate1997-13", "openwebmath_score": 0.911790132522583, "openwebmath_perplexity": 653.5775049227947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9835969713304754, "lm_q2_score": 0.904650541527608, "lm_q1q2_score": 0.8898115327590297}}
{"url": "http://qdjf.laquintacolonna.it/area-of-rectangle-under-curve-calculator.html", "text": "# Area Of Rectangle Under Curve Calculator\n\nWithout performance, you are doing nothing. Approximate the area under the curve and above the x-axis using n rectangles. In single-variable calculus, recall that we approximated the area under the graph of a positive function \\ (f\\) on an interval \\ ( [a,b]\\) by adding areas of rectangles whose heights are determined by the curve. 5 Fermat noticed that by dividing the area underneath a curve into successively smaller rectangles as x became closer to zero, an infinite number of such rectangles would describe the area precisely. Let x be the base of the rectangle, and let y be its height. Integration is the best way to find the area from a curve to the axis: we get a formula for an exact answer. \u2022 Stations for Area Under the Curve \u2022 Stations Answer Sheet \u2022 9-4 Challenge Holt worksheet. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY Slide No. Divide the gz curve equally with a number of lines in vertical directions, if the number increases the result will be more accurate. Approximating Area under a curve with rectangles To nd the area under a curve we approximate the area using rectangles and then use limits to nd the area. First work out the area of the whole circle by substituting the radius of 8cm into the formula for the area of the circle: A = \u03c0 \u00d7r\u00b2 = \u03c0 \u00d78\u00b2 = 64\u03c0 (leave the answer as an exact solution as this need to be divided by 4). The upper vertices, being points on the parabola are: (-x,9-x^2) and (x,9-x^2). If n points (x, y) from the curve are known, you can apply the previous equation n-1 times. It is clear that , for. RIEMANN, a program for the TI-83+ and TI-84+, approximates the area under a curve (integral) by calculating a Riemann sum, a sum of areas of simple geometric figures intersecting the curve. The sum of these approximations gives the final numerical result of the area under the curve. find the area under a curve f(x) by using this widget 1) type in the function, f(x) 2) type in upper and lower bounds, x=. SketchAndCalc\u2122 is an irregular area calculator app for all manner of images containing irregular shapes. The following are some examples of probability problems that involve areas of geometric shapes. def area_under_curve (poly, bounds, algorithm): \"\"\"Finds the area under a polynomial between the specified bounds using a rectangle-sum (of width 1) approximation. Create Let n = the number of rectangles and let W = width of each rectangle. To find the width, divide the area being integrated by the number of rectangles n (so, if finding the area under a curve from x=0 to x=6, w = 6-0/n = 6/n. The area between -1 and 1 is 58%. Mathematica Stack Exchange is a question and answer site for users of Wolfram Mathematica. Learn term:auc = area under the curve with free interactive flashcards. Yes, it\u2019s 0. The formula is: A = L * W where A is the area, L is the length, W is the width, and * means multiply. 05 or a p value of more than 0. 008, the one after would be (2/5) 2 times 1/5=. Orientation can change the second moment of area (I). What is the area under the function f, in the interval from 0 to 1? and call this (yet unknown) it turns out that the area under the curve within the stated bounds is 2/3. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY Slide No. 3 \u2212 c, f \u2212 c. The graphs in represent the curve In graph (a) we divide the region represented by the interval into six subintervals, each of width 0. Then you calculate the areas of the narrow tall trapezoids and add them up. This description is too narrow: it's like saying multiplication exists to find the area of rectangles. The area estimation using the right endpoints of each interval for the rectangle. 5 / f or simplified: area = a / (\u03a0 * f) right? Because the area under a half cycle of a 1/2 hz wave would just be 1 * 0. Rectangle: Area = (2 s) * (10 m/s) = 20 m. A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve y = sin x, 0 \u2264 x \u2264 \u03c0. For example, here's how you would estimate the area under. 5, and it has a width of one, and the last rectangle has a width of 1 minus. Area Under the Curve Calculator is a free online tool that displays the area for the given curve function specified with the limits. x = ky 2 Let us determine the moment of inertia of this area about the YY axis. It reaches a maximum at 0,1 and slopes down symmetrically about this point. You expect to include twice as many negative cases than positive cases, so for the Ratio of sample sizes in negative. Find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of contiguous bars. Numeric Computation of Integrals Part 1: Left-Hand and Right-Hand Sums. 1_Area_Under_Curve. Using the area of a rectangle area formula, area = width x height we can see how our circle, re-configured as a rectangle, can be shown to have an area that approximates to \u03c0r x r or \u03c0r 2. Get the free \"Calculate the Area of a Polar curve\" widget for your website, blog, Wordpress, Blogger, or iGoogle. , parallel to the axes X and Y you may use minmax function for X and Y of the given points (e. So this is going to be equal to f of-- it's going to be equal to the function evaluated at 1. Find the dimensions of the largest rectangle that can be inscribed in the triangle if the base of the rectangle coincides with the base of the triangle. Enter the average value of f (x), value of interval a and b in the below online average value of a function calculator and then click calculate button to find the output with steps. Use this calculator if you know 2 values for the rectangle, including 1 side length, along with area, perimeter or diagonals and you can calculate the other 3 rectangle variables. Easier ways to calculate the AUC (in R) But let\u2019s make life easier for ourselves. Area Under a Curve Tell me everything you know about the following measures. Approximate the area under the curve and above the x-axis on the given interval, using rectangle whose height is the value of the function at the left side of the rectangle. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums. 34 square feet. Area of a rectangle formula The formula for the area of a rectangle is width x height, as seen in the figure below: All you need are two measurements and you can calculate its perimeter by hand, or by using our perimeter of a rectangle calculator above. Rewrite your estimate of the area under the curve. 5x2 + 7 for \u20133 \u2264 x \u2264 0 and rectangle width 0. Use the calculator \"Calculate X for a given Area\" Areas under portions of a normal distribution can be computed by using calculus. (Image: Tim Lovett 2014). a) Write the expression for the area of the rectangle. Input the length and the width (two input statements) 2. When x = 10cm and y = 6cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle. Approximate the area under the curve from to using the. To determine To calculate: The largest area of a rectangle that can fit inside the provided curve y = e \u2212 x 2 and the x -axis. Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode. 93 Using either Table A or your calculator or software, find the proportion of observations from a standard normal distribution that satisfies each of the following statements. Third rectangle has a width of. Just as calculating the circumference of a circle more complicated than that of a triangle or rectangle, so is calculating the area. The largest possible rectangle possible. He used a process that has come to be known as the method of exhaustion, which used. Enter mean, standard deviation and cutoff points and this calculator will find the area under normal distribution curve. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. asked by Lilly on June 9, 2018; Calculus. An easy to use, free area calculator you can use to calculate the area of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, and sector of a circle. Points on the blue curve, Area = 6. How to use integration to determine the area under a curve? A parabola is drawn such that it intersects the x-axis. By using this website, you agree to our Cookie Policy. For instance, a named function to calculate the square of a number could be square[x_] := x^2 (square[3] will output $9$). For rectangular shapes, area, A, and wetted perimeter, WP are simple functions of flow depth. Hello everyone I have a graph plotted in Matlab (no function), as data was imported via Excel, I am looking for a loop to calculate the area under the curve of each interval and then add them to get the entire area. The total no of lines should be odd no. Calculate volume of geometric solids. The area under the curve is the sum of areas of all the rectangles. the trapezoidal method. I am not sure who invented this (one can never be sure who did some simple thing first) but Galileo used the method for determining the area under a cycloid, which was not known theoretically at that time. Second Step: On each subinterval, draw a rectangle to approximate the area of the curve over the subinterval. Total Area = 20 m + 20 m = 40 m. We integrate by \"sweeping\" a ray through the area from \u03b8 a to \u03b8 b, adding up the area of infinitessimally small sectors. For example the area first rectangle (in black) is given by: and then add the areas of these rectangles as follows:. The result will be in the unit the width and height are measured in, but squared, e. Exercises 1 - Solve the same problem as above but with the perimeter equal to 500 mm. Choose from 40 different sets of term:auc = area under the curve flashcards on Quizlet. \" In the \"limit of rectangles\" approach, we take the area under a curve y = f (x) above the interval [a , b] by approximating a. To find the area under a curve, we must agree on what is desired. \u200eSketchAndCalc\u2122 is the only area calculator capable of calculating areas of uploaded images. Integral Calculus, Area Under the Bell Curve Area Under the Bell Curve Let g(x) = exp(-x 2). It starts out with approximating using rectangle areas at a very theoretical and high level. We determine the height of each rectangle by calculating for The intervals are We find the area of each rectangle by multiplying the height by. Since you're multiplying two units of length together, your answer will be in units squared. If we assume the width of each one is h, then the area of the first one is h * (a + b) / 2 where a and b are the heights (value of the function) at the left and right edges of the trapezium. Approximate the area under the curve from to with. Q2 (E): Explain why the two rectangular areas are equal. 725 for Area under ROC curve and 0. x = 3) and a represents the lower bound on the. So if I take the example above, and lets say I divide the area under the curve into 10 sections of 1/5 square units, whose height is the formula f(x)=x 2 evaluated at those cut points. Now we are going to see what these look like using mathematical, or symbolic notation. ) Implementing the Trapezoidal Rule in SAS/IML Software. The base of the rectangle is 2x and the height is e^(-2x^2) so you could differentiate A(x) = 2xe^(-2x^2) and find the maximum area which is when A'(x) = 0. The result is the area of only the shaded. Subtract the area of the white space from the area of the entire rectangle. There is a whole system in mathematics dedicated to just this, just this one feature of graphs, it's so important, an entire system has been based around it, which you will need to learn at some point if you. 917, which appears here. Use Riemann sums to approximate area. The area of the largest rectangle that can be drawn with one side along the x-axis and two vertices on the curve of is. Third rectangle has a width of. Figure 7-1. A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve y = sin x, 0 \u2264 x \u2264 \u03c0. A good way to approximate areas with rectangles is to make each rectangle cross the curve at the midpoint of that rectangle's top side. The quantity we need to maximize is the area of the rectangle which is given by. 3 cm\u00b2 to 3 significant. We make vertical. Orientation can change the second moment of area (I). 1801 e-4 Which is the best way to calculate the area. Example of How-to Use The Trapezoidal Rule Calculator: Consider the function calculate the area under the curve for n=8. If we know the height and two base lengths then we can calculate the Area of a Trapezoid using the below formula: Area = (a+b)/2 * h. Calculator online on how to calculate volume of capsule, cone, conical frustum, cube, cylinder, hemisphere, pyramid, rectangular prism, triangular prism and sphere. Let's simplify our life by pretending the region is composed of a bunch of rectangles. Area of the rectangle = A = 2xy Since the rectangle is inscribed under the curve y = 4 cos 0. The area calculator has a unique feature that allows you to set the drawing scale of any image before drawing the perimeter of the shape. But to draw this rectangle, we need 4 corners of the rectangle. Download SketchAndCalc Area Calculator and enjoy it on your iPhone, iPad, and iPod touch. A rectangle has a vertex on the line 3x + 4y = 12. The below figure shows why. Python Area of a Trapezoid. Find more Mathematics widgets in Wolfram|Alpha. (a) Use two rectangles. Gianluca Gianluca 1 Recommendation. The surface area of a rectangular tank is the sum of the area of each of its faces: SA = 2lw. above the interval [0, 2] by dividing the interval into n = 5 subintervals of equal length using circumscribed rectangles. In each case, the area approximated is above the interval [0, 5] on the x-axis. Use this calculator if you know 2 values for the rectangle, including 1 side length, along with area, perimeter or diagonals and you can calculate the other 3 rectangle variables. In this case the surface area is given by, S = \u222c D \u221a[f x]2 +[f y]2 +1dA. The force magnitude dF acting on it is Finding the area of a rectangle. 75, and it has a height of one. Compute left, right, and midpoint Hence Riemann sums use with n rectangles are computed. Find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of contiguous bars. 8xp needs to be transferred to the students' calculators via handheld-to-handheld transfer or transferred from the computer to the calculator via TI-Connect. find an expression for the are under the curve y = x^3 from 0 to 1 as a limit b. Python Area of a Trapezoid. Approximate the area under the curve and above the x-axis on the given interval, using rectangle whose height is the value of the function at the left side of the rectangle. Explain what the shaded area represents in the context of this problem. Consider a function of 2 variables z=f(x,y). Surface area of a cylinder. The height of each individual rectangle is and the width of each rectangle is Adding the areas of all the rectangles, we see that the area between the curves is approximated by. Area Under Curve Calculator Find the area under a function with 6 different methods (LRAM, RRAM, MRAM, TRAPAZOIDS, SIMPSON'S METHOD, ACTUAL). Using trapezoidal rule to approximate the area under a curve first involves dividing the area into a number of strips of equal width. Filed under Calculus, Difficulty: Easy, TI-83 Plus, TI-84 Plus. In the end, if you use instant sampling (infinitely thin rectangles) and then sum the resulting infinite number of rectangles your approximated value will match the actual value and the green area (which we shown to be equal to the approximated displacement) will become equal to the red area, i. Since the functions in the beginning of the lesson are linear, or piecewise linear, the enclosed regions form rectangles, triangles, or trapezoids. You can also quickly convert between area units viz. 10 points to best answer! Thanks and happy holidays!. And that is how you calculate the area under the ROC curve. under the curve for the range 1 < X < 3. After recapping yesterday's work, I give students this worksheet for them to work on with their table groups. The \"2x\" that BigGlenn is referring to is twice the value of x between 0 and sqrt(5), since the rectangle is twice the area of the part to the right of the y axis. Average Acceleration Calculator. The Area Between Two Curves. In fact, it looks like one of those. Filed under Calculus, Difficulty: Easy, TI-83 Plus, TI-84 Plus. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The area of a rectangle drawn above the curve would be more than the actual area under the curve. a) Carefully divide the region into sub-regions with vertical lines at x = l, x = 1. The area under each connecting segment describes a trapezoid, as shown below (left). Consider the problem of \ufb01nding the area under the curve on the function y = \u2212x 2 +5 over the domain [0,2]. The first trapezoid is between x=1 and x=2 under the curve as below screenshot shown. approximated an integral by using a finite sum; since the number of the rectangular strips was finite but taking that number \u2192 \u221e (the same as taking dx \u2192 0), we converted the sum to an integral. Approximate the area under the curve using the given rectangular approximation. Area Between Two Curves Calculator With Steps. \u2013 The area under the curve will be determined analytically. Formulas, explanations, and graphs for each calculation. Integrate across [0,3]: Now, let's rotate this area 360 degrees around the x axis. 1416) with the square of the radius (r) 2. And that is how you calculate the area under the ROC curve. On a calculator, again, this is easy to do all these small calculations and add them. Follow 17 views (last 30 days) Rengin on 13 Mar 2019. Note the widest one. EXAMPLE 1: Find. Then, approximating the area of each strip by the area of the trapezium formed when the upper end is replaced by a chord. Multiply Pi (3. Again, use the CALC function, but this time choose item 7 from the menu. 1: Area Under a Curve Given a function y = f(x), the area under the curve of f over an interval [a;b] is the area of the region by the graph of the curve, the x-axis, and the vertical lines x = a and x = b. Just let the top right corner of your rectangle be the coordinates $(x,y) = (x, 12-x^2)$. It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid. Integrate across [0,3]: Now, let's rotate this area 360 degrees around the x axis. Loading Close. (c) Use a graphing calculator (or other technology) and 40 rectangles. 10 points to best answer! Thanks and happy holidays!. While we are only working on one specific type of problem (finding the area under a curve), it is a challenging task and I want them to have practice going through the steps of making an infinite number of rectangles. (See Examples 2 and 3. Volume formulas. (a) Use two rectangles. But Integration can sometimes be hard or impossible to do! to get an approximate answer. Average Acceleration Calculator. So it's going to be, let me write it over here, A(b) is the area under this curve here. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of Rectangular Areas. Most of its area is part of the area under the curve. Solution: a) Graph the region. Repeat using rectangles of different widths and record data on spreadsheet. , polygon's vertices) Store the area of the fitted rectangle; Rotate the polygon about the centroid by e. The value of f 0 is such that area of the trapezoid is the same as that under the specified section of the real curve, in other words, both area represent the same energy per unit volume. This engineering data is often used in the design of structural beams or structural flexural members. A program can be used to illustrate the rectangles that approximate the area under a curve. To find the area under the curve y = f (x) between x = a & x = b, integrate y = f (x) between the limits of a and b. thus each term of the sum is the area of a rectangle with height equal to the function value at the distinguished point of the given sub-interval, and width the same. , 1 degree; Repeat from [S] until a full rotation done; Report the angle of the minimum area as the result. Area under a Curve The area between the graph of y = f(x) and the x-axis is given by the definite integral below. It also happens to be the area of the rectangle of height 1 and length. So, the area under (or to the left of) the stack of tenure bars is equal to the average tenure, but the stack of tenure bars is not exactly the survival curve. (See Examples 2 and 3. The first step in his method involved a unique way of describing the infinite rectangles making up the area under a curve. [3] Calculate total area of all the rectangles to get approximate area under f(x). Work as area under curve. Different types of sums (left, right, trapezoid, midpoint, Simpson\u2019s rule) use the rectangles in slightly different ways. The calculator will find the area between two curves, or just under one curve. (Determine the number of rectangles, the width of the rectangles in each case, and whichsample points should be used in your calculation using the given directions. In this case, to find the area of a sector, you just have to take the measure of the cen. Prabhat, you could try summing the area F(t) x dt of every rectangle under the curve, where t = time value and dt = time step. from 0 to 3 by using three right rectangles. It is not hard to guess that the area under a parabolic arch with base B and height H is 2/3*B*H (two thirds of the area of the circumscribed rectangle). Filed under Calculus, Difficulty: Easy, TI-83 Plus, TI-84 Plus. Enter the function and limits on the calculator and below is what happens in the background. So this is going to be equal to f of-- it's going to be equal to the function evaluated at 1. Question 1: Calculate the area under the curve of a. Since this is an overestimate, the area under the curve is less then 10. Optimizing a Rectangle Under a Curve. Plus and Minus. This app is useful for land area calculation for plots of all shape and size be it triangle, rectangle, circle or any simple polygon. Uniquely, the area calculator is capable of accurately calculating irregular areas of uploaded images, photographs or plans quickly. The sides of your triangle do not adhere to the triangle inequality theorem. Monte Carlo simulation offers a simple numerical method for calculating the area under a curve where one has the equation of the curve, and the limits of the range for which we wish to calculate the area. Can anyone point me in the right direction for acquiring the code?. We can calculate the median of a Trapezoid using the following formula:. You can calculate that. He now explains that the area of rectangle is length times the breadth. The curve is symmetric around 0, and the total area under the curve is 100%. [2] Construct a rectangle on each sub-interval & \"tile\" the whole area. Area of a Semicircle Calculator A semicircle is nothing but half of the circle. areaundercurve. For areas in rectangular coordinates, we approximated the region using rectangles; in polar coordinates, we use sectors of circles, as depicted in figure 10. Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode. Then you calculate the areas of the narrow tall trapezoids and add them up. 1: Area Under a Curve Given a function y = f(x), the area under the curve of f over an interval [a;b] is the area of the region by the graph of the curve, the x-axis, and the vertical lines x = a and x = b. Graphical illustration of methods of calculating the area under a curve. Therefore, if we take the sum of the area of each trapezoid, given the limits, we calculate the total area under a curve. Area of a rectangle formula The formula for the area of a rectangle is width x height, as seen in the figure below: All you need are two measurements and you can calculate its perimeter by hand, or by using our perimeter of a rectangle calculator above. Rectangular Tank. Area is a quantity that describes the size or extent of a two-dimensional figure or shape in a plane. Python Area of a Trapezoid. Approximating area under the curve. To improve this 'Area of a parabolic arch Calculator', please fill in questionnaire. The user is expected to select the. Q1 (E): What is the common area of such rectangles for the hyperbola $$\\normalsize{y=\\frac{2}{3x}}$$? But other kinds of areas under this graph are also interesting, and exhibit an interesting property when we scale things. In single-variable calculus, recall that we approximated the area under the graph of a positive function \\ (f\\) on an interval \\ ( [a,b]\\) by adding areas of rectangles whose heights are determined by the curve. 25 are always greater than points on the red curve (That is, the area of the rectangle is always less that 6. Both the trapezoidal and rectangle method work, I personally prefer trapezoidal rule. Figure 7-1. As per the fundamental definition of integral calculus, it is nothing but, A = $\\int_{a}^{b}ydx$ Under the same argument, it can be established that the area. Third rectangle has a width of. Notice, that unlike the first area we looked at, the choosing the right endpoints here will both over and underestimate the area depending on where we are on the curve. To find the area under a curve we find the definite integral between the two bounds (ends) Proof. That is to say \u03c0 (pi is 3. EX #1: Approximate the area under the curve of y = 2x \u2014 3 above the interval [2, 5] by dividing [2, 5] inton = 3 subintervals of equal length and computing the sum of the areas of the inscribed rectangles (lower sums). Curved Rectangle Calculator. find the area under a curve f(x) by using this widget 1) type in the function, f(x) 2) type in upper and lower bounds, x=. The graphs intersect at (-1 ,1) and (2,4). Due to the this it approximate area. The height of the typical rectangle is , while the thickness is. approximated an integral by using a finite sum; since the number of the rectangular strips was finite but taking that number \u2192 \u221e (the same as taking dx \u2192 0), we converted the sum to an integral. The area under the curve is divided into a series of vertical strips. Uniquely, the area calculator is capable of accurately calculating irregular areas of uploaded images, photographs or plans quickly. Area Moment of Inertia Section Properties of Rectangular Feature Calculator and Equations. Free online calculators for area, volume and surface area. (Sometimes a trapezoid is degenerate and is actually a rectangle or a triangle. To improve this 'Area of a parabolic arch Calculator', please fill in questionnaire. Free area under the curve calculator - find functions area under the curve step-by-step This website uses cookies to ensure you get the best experience. This description is too narrow: it's like saying multiplication exists to find the area of rectangles. You can put this solution on YOUR website! Find the maximum area of a rectangle with a perimeter of 54 centimeters. 8931711, the area under the ROC curve. Because the problem asks us to approximate the area from x=0 to x=4, this means we will have a rectangle between x=0 and x=1, between x=1 and x=2, between x=2 and x=3, and between x=3 and x=4. Question 534414: Use inscribed rectangles to approximate the area under g(x) = \u20130. formulas widely applied by researchers under- or overestimated total area under a metabolic curve by a great margin. Side Lengths of Triangle. Calculus Graphing Calculator handouts help student learn to use the TI Graphing Calculator effectively as a learning tool. accurately compute the area under the curve of x,y (in this case an isolated Gaussian with a height of 1. Area between upper curve and x- axis. Easier ways to calculate the AUC (in R) But let\u2019s make life easier for ourselves. 75, and it has a height of one. To find the area under the curve we try to approximate the area under the curve by using rectangles. This can be quite simple, at least in. This is equivalent to approximating the area by a trapezoid rather than a rectangle. Area Between Two Curves Calculator With Steps. Memory Rate (in 3. Integral Calculus, Area Under the Bell Curve Area Under the Bell Curve Let g(x) = exp(-x 2). x = 3) and a represents the lower bound on the. 5 Fermat noticed that by dividing the area underneath a curve into successively smaller rectangles as x became closer to zero, an infinite number of such rectangles would describe the area precisely. 21150 e-4 trapz(y)=-1. The cumulative distribution function (cdf) gives the probability as an area. $\\begingroup$ @Gio The & and # are part of a \"pure function\" definition (see the documentation page for Function). In this calculus instructional activity, students use Riemann sums to find and approximate the area under a curve. Using the area of a rectangle area formula, area = width x height we can see how our circle, re-configured as a rectangle, can be shown to have an area that approximates to \u03c0r x r or \u03c0r 2. Yes, it\u2019s 0. Well, first of all, we can see the we are actually looking for the region that\u2019s bounded by the curve and the \ud835\udc65-axis. To be able to state area formula for a rectangle. Finley Evans author of Program to compute area under a curve is from London, United Kingdom. mm 2, cm 2, m 2, km 2 or in 2, ft. \u2022 Stations for Area Under the Curve \u2022 Stations Answer Sheet \u2022 9-4 Challenge Holt worksheet. After students learn algebraic methods of computing integrals based on the Fundamental Theorem of Calculus, they will be able to derive the formula Y=(H-R 2)*X 2 and prove that it is correct. Free area under the curve calculator - find functions area under the curve step-by-step This website uses cookies to ensure you get the best experience. (See Example 1. Multiply this fraction by the area of the rectangle (0,0; 10;500) = fraction*(500-0)*(10-0). Therefore, the total area A under the curve between x = a and x = b is the summation of areas of infinite rectangles between the same interval. Choose from 40 different sets of term:auc = area under the curve flashcards on Quizlet. The sides of your triangle do not adhere to the triangle inequality theorem. AUC is the integral of the ROC curve, i. Just as calculating the circumference of a circle more complicated than that of a triangle or rectangle, so is calculating the area. as data was imported via Excel, I am looking for a loop to calculate the area under the curve of each interval and then add them to get the entire area. This is often the preferred method of estimating area because it tends to balance overage and underage - look at the space between the rectangles and the curve as well. Calculus 120, section 6. In the end, if you use instant sampling (infinitely thin rectangles) and then sum the resulting infinite number of rectangles your approximated value will match the actual value and the green area (which we shown to be equal to the approximated displacement) will become equal to the red area, i. You can put this solution on YOUR website! Find the maximum area of a rectangle with a perimeter of 54 centimeters. Area under a curve. Note: use your eyes and common sense when using this! Some curves don't work well, for example tan(x), 1/x near 0, and functions with sharp changes give bad results. For areas in rectangular coordinates, we approximated the region using rectangles; in polar coordinates, we use sectors of circles, as depicted in figure 10. Added Aug 1, 2010 by khitzges in Mathematics. from 0 to 3 by using three right rectangles. This will often be the case with a more general curve that the one we initially looked at. Filed under Calculus, Difficulty: Easy, TI-83 Plus, TI-84 Plus. He used a process that has come to be known as the method of exhaustion, which used. EX #2: Approximate the area under the curve of \ud835\udc66\ud835\udc66 = 5 \u2212 \ud835\udc65\ud835\udc65. This method will split the area between the curve and x axis to multiple trapezoids, calculate the area of every trapezoid individually, and then sum up these areas. which states that the sum of the side lengths of any 2 sides of a triangle must exceed the length of the third side. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. 93 Using either Table A or your calculator or software, find the proportion of observations from a standard normal distribution that satisfies each of the following statements. Optimization Problem #3 http: Skip navigation Sign in. ( )=sin \ud835\udc52 Right Endpoint with 3 subintervals on the interval [0,2] 10. Calculating an area under a curve. For example, if the area is 60 and the width is 5, your equation will look like this: 60 = x*5. So let's evaluate this. RRAM III. Different values of the function can be used to set the height of the rectangles. Area of a Semicircle Calculator A semicircle is nothing but half of the circle. In these simple rectangular. It has believed the more rectangles; the better will be the estimate:. Such systems are rather complicated to implement, and I am not familiar with any high quality, open source libraries for Java. The height of each individual rectangle is and the width of each rectangle is Adding the areas of all the rectangles, we see that the area between the curves is approximated by. Sometimes this area is easy to calculate, as illustrated from the examples below:. Calculate the area of the rectangle. Let's now calculate the area of the region enclosed by the parametric curve. Each rectangle has a width of 1, so the areas are 2, 5, and 10, which total 17. Let's simplify our life by pretending the region is composed of a bunch of rectangles. Area between lower curve and x- axis. Area is a quantity that describes the size or extent of a two-dimensional figure or shape in a plane. We may approximate the area under the curve from x = x 1 to x = x n by dividing the whole area into rectangles. Approximating the Area Under a Curve TEACHER NOTES \u00a92015 Texas Instruments Incorporated 2 education. 5x, the top right corner of the rectangle lies on the curve, and so we can write A = 2x(4 cos 0. So this is going to be equal to f of-- it's going to be equal to the function evaluated at 1. 000 and a standard deviation (sigma) of 1. Approximate area under. You can calculate its area easily with this formula: =(C3+C4)/2*(B4-B3). You can calculate the area by the following way. 29 square feet. First work out the area of the whole circle by substituting the radius of 8cm into the formula for the area of the circle: A = \u03c0 \u00d7r\u00b2 = \u03c0 \u00d78\u00b2 = 64\u03c0 (leave the answer as an exact solution as this need to be divided by 4). For example, if an ellipse has a major radius of 5 units and a minor radius of 3 units, the area of the ellipse is 3 x 5 x \u03c0, or about 47 square units. RRAM III. This overestimates the area under the curve, as each rectangle pokes out above the curve. We see that the curve given is the sine curve and it intersects the x-axis at x = 0, x = pi, x = 2*pi and so on. Using this calculator, we will understand the algorithm of how to find the perimeter, area and diagonal length of a rectangle. Create Let n = the number of rectangles and let W = width of each rectangle. a) Using mid-ordinate rule, estimate the area under the curve y =1/2x 2 - 2. To find the area under the curve we try to approximate the area under the curve by using rectangles. After recapping yesterday's work, I give students this worksheet for them to work on with their table groups. Prabhat, you could try summing the area F(t) x dt of every rectangle under the curve, where t = time value and dt = time step. Orientation can change the second moment of area (I). This area can be calculated using integration with given limits. for the first 2 data points, the value drops from 50 to 40 linearly over the hour, and so the area for those measurements is (30min * 5)/2. Thus the total area is: h * (a + b) / 2 h * (b. In order to calculate the area and the precision-recall-curve, we will partition the graph using rectangles (please note that the widths of the rectangles are not necessarily identical). Midpoint Rectangle Calculator Rule\u2014It can approximate the exact area under a curve between points a and b, Using a sum of midpoint rectangles calculated with the given formula. The base of the rectangle is 2x and the height is e^(-2x^2) so you could differentiate A(x) = 2xe^(-2x^2) and find the maximum area which is when A'(x) = 0. Let us consider one curve which equation is parabolic as displayed in following figure and let us consider that equation of this parabolic curve is as mentioned here. Both the trapezoidal and rectangle method work, I personally prefer trapezoidal rule. Use this particular handout to visualize and determine the area under a curve in Calculus 1 or AP Calculus AB or BC. Key insight: Integrals help us combine numbers when multiplication can't. Graphical illustration of methods of calculating the area under a curve. The area under the curve to the right of the mean is 0. a) Carefully divide the region into sub-regions with vertical lines at x = l, x = 1. Approximating the Area Under a Curve TEACHER NOTES \u00a92015 Texas Instruments Incorporated 2 education. So all you need to do now is divide the answer by 4: Area of a quadrant = 64\u03c0 \u00f74 = 16\u03c0 = 50. 7 and Jan 9 Practice Problems : 5. 1 squared plus 1 is just 2, so it's going to be 2 times 1/2. Area between lower curve and x- axis. It is an online Geometry tool requires two length sides of a rectangle. The displacement is. Curved Rectangle Calculator. where n s is the number of points below the curve and n is the total number of points. Calculate the area Delete the value in the last row of column C, then find the area by calculating the sum of column C. Divide the gz curve equally with a number of lines in vertical directions, if the number increases the result will be more accurate. where a and b represent x, y, t, or \u03b8-values as appropriate, and ds can be found as follows. Formulas, explanations, and graphs for each calculation. Use this particular handout to visualize and determine the area under a curve in Calculus 1 or AP Calculus AB or BC. What fraction of this rectangle is under the curve? 5. Question 534414: Use inscribed rectangles to approximate the area under g(x) = \u20130. Two problems. A mixed dilation of the plane. This is going to be equal to our approximate area-- let me make it clear-- approximate area under the curve, just the sum of these rectangles. ( )= \ud835\udc65 3 Midpoint with 4 [subintervals on the interval 1,3] Use the information provided to answer the following. Calculate the area of the white space within the rectangle. /rA)?? The documentation is quite unclear to me, it says. Circle Sectors Rearranged. Step 1: Sketch the graph: Step 2: Draw a series of rectangles under the curve, from the x-axis to the curve. Midpoint Formula with. The total no of lines should be odd no. I am not sure who invented this (one can never be sure who did some simple thing first) but Galileo used the method for determining the area under a cycloid, which was not known theoretically at that time. Explanation:. Before accepting an area calculation, inspect the sketch of the operation to ensure that your path does not intersect or meet itself, and that any curves deflect in the correct direction. To find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b. If you have only the area and width, you can use the same equation to solve for the area. Because you can draw a rectangle next to your curve, and weight it too, and compare with the weight of the area under the curve. After this tutorial you will be able to identify a density curve, name the shape of the density curve, understand the importance of the area under the density. 5 z Example #12: Parabolic Channel A grassy swale with parabolic cross-section shape has top width T = 6 m when depth y = 0. Approximating Area under a curve with rectangles To nd the area under a curve we approximate the area using rectangles and then use limits to nd the area. 917, which appears here. S = \u222c D [ f x] 2 + [ f y] 2 + 1 d A. Area of a. How can I calculate the area under a curve after plotting discrete data as per below? Graphically approximating the area under a curve as a sum of rectangular regions. This is numerical method territory if you are looking to do this in excel. You can calculate the area by the following way. asked by Lilly on June 9, 2018; Calculus. Input the length and the width (two input statements) 2. Include a sketch! Justify! 9. For a rectangle, Where b is breadth (horizontal) and h is height (vertical) if the load is vertical - e. Area between curves. Now the area under the curve is to be calculated. Area is a quantity that describes the size or extent of a two-dimensional figure or shape in a plane. Get the free \"Calculate the Area of a Polar curve\" widget for your website, blog, Wordpress, Blogger, or iGoogle. You expect to include twice as many negative cases than positive cases, so for the Ratio of sample sizes in negative. You can also make the trapezoids get narrower and approach zero to get better. accurately compute the area under the curve of x,y (in this case an isolated Gaussian with a height of 1. For instance, a named function to calculate the square of a number could be square[x_] := x^2 (square[3] will output $9$). Because the problem asks us to approximate the area from x=0 to x=4, this means we will have a rectangle between x=0 and x=1, between x=1 and x=2, between x=2 and x=3, and between x=3 and x=4. ) (b) Use four rectangles. e, the actual value for the displacement equals. If you need help remembering how to calculate the area of a rectangle, I would suggest putting \"area of a rectangle\" into your favorite search engine. More in wikipedia. The area under the curve is divided into a series of vertical strips. Table of Contents. When students begin studying integral calculus methods such as the trapezoidal, the Monte Carlo and upper and lower rectangle methods are used to determine the area under a curve. 725 for Area under ROC curve and 0. In this mathematical model, the areas of the individual segment are then added to obtain the total area under the curve. Let x be the base of the rectangle, and let y be its height. ( )=sin \ud835\udc52 Right Endpoint with 3 subintervals on the interval [0,2] 10. This is an important function in probability and statistics. Integration is the best way to find the area from a curve to the axis: we get a formula for an exact answer. If an infinite number of rectangles are used, the rectangle approximation equals the value of the integral. To calculate the area of a circle we use the formula: \u03c0 x (diameter/2) 2. Approximation of area under a curve by the sum of areas of rectangles. In this case, the limit process is applied to the area of a rectangle to find the area of a general region. You can also make the trapezoids get narrower and approach zero to get better. This is because, a semi-circle is just the half of a circle and hence the area of a semi-circle is the area of a circle divided by 2. This applet shows the sum of rectangle areas as the number of rectangles is increased. This is often the preferred method of estimating area because it tends to balance overage and underage - look at the space between the rectangles and the curve as well as the amount of rectangle space above the curve and this becomes more evident. And these areas are equal to 0. And the area of the rectangle under the demand curve at point a equals the distance g Q 1. Find more on Program to compute area under a curve Or get search suggestion and latest updates. gravity load. The upper vertices, being points on the parabola are: (-x,9-x^2) and (x,9-x^2). Example: Determine the area under the curve y = x + 1 on the interval [0, 2] in three different ways: (1) Approximate the area by finding areas of rectangles where the height of the rectangle is the y-coordinate of the left-hand endpoint (2) Approximate the area by finding areas of rectangles where the height of the rectangle is the y. Rewrite your estimate of the area under the curve. u(t 2 \u2013 t 1) is the area of the shaded rectangle in Figure 2. Lines 20 and 23 are not areas and shouldn't be labeled as such. Notice that the area surrounding the this part of the curve is not a square but a rectangle of 2*2 2 = 8 = 2 3. AUC is the integral of the ROC curve, i. It follows that:\" Calculate the area under a curve/the integral of a function. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The result is the area. To be able to state area formula for a rectangle. How to use integration to determine the area under a curve? A parabola is drawn such that it intersects the x-axis. Half of the area of the rectangle is (x)[f(x)]. minAreaRect(). Uniquely, the area calculator is capable of accurately calculating irregular areas of uploaded images, photographs or plans quickly. This area can be calculated using integration with given limits. You expect to include twice as many negative cases than positive cases, so for the Ratio of sample sizes in negative. [NOTE: The curve is completely ABOVE the x-axis]. (c) Use a graphing calculator (or other technology) and 40 rectangles. The definite integral (= area under the graph. We can call the small width of this rectangle dx and the height of this rectangle f (x) (since the rectangle extends from the x-axis up to the curve), then the area is just f (x)dx. Approximate area of under a curve. Monte Carlo simulation offers a simple numerical method for calculating the area under a curve where one has the equation of the curve, and the limits of the range for which we wish to calculate the area. To find the area of a rectangle, multiply the length by the width. Sometimes, we use double integrals to calculate area as well. We see that the curve given is the sine curve and it intersects the x-axis at x = 0, x = pi, x = 2*pi and so on. There are various packages that calculate the AUC for us automatically. * Multiply the estimation by four to get an estimation of the area of the original circle. If you have only the area and width, you can use the same equation to solve for the area. You can also make the trapezoids get narrower and approach zero to get better. Approximate the area under the curve from to using the. concept of area under a curve. What is the area of the largest rectangle that can be placed in a 5-12-13 right triangle (as shown)? asked by math on October 26, 2009; Calculus. Find summation of the approximated areas of the rectangles. Area under curve (no function) Follow 1,734 views (last 30 days) Rick on 9 Sep 2014. The area of the lot is then 10,561. Summary: To compute the area under a curve, we make approximations by using rectangles inscribed in the curve and circumscribed on the curve. 586, you would be close to the correct answer and you would just have to add the area of this slice, which is mostly rectangular at. The area between two graphs can be found by subtracting the area between the lower graph and the x-axis from the area between the upper graph and the x-axis. The area between -1 and 1 is 58%. So all you need to do now is divide the answer by 4: Area of a quadrant = 64\u03c0 \u00f74 = 16\u03c0 = 50. So -- in all, we get a total area of 45 + 60 + 77 + 86 = 268 square units. Area between upper curve and x- axis. Unit 4: The Definite Integral Approximating Area Under a Curve Jan. gravity load. When we increased the number of rectangles of equal width of the rectangles, a better approximation of the area is obtained. The \"2x\" that BigGlenn is referring to is twice the value of x between 0 and sqrt(5), since the rectangle is twice the area of the part to the right of the y axis. Integral Approximation Calculator. From the coordinates of the corner points, calculate the width, height, then area and perimeter of the rectangle. Two problems. The areas of the others are similar. And that is how you calculate the area under the ROC curve. It is clear that , for. The area under the red curve is all of the green area plus half of the blue area. The area under the curve is the sum of areas of all the rectangles. circumscribed rectangles. Areas Under Parametric Curves We can now use this newly derived formula to determine the area under. The upper vertices, being points on the parabola are: (-x,9-x^2) and (x,9-x^2). x = ky 2 Let us determine the moment of inertia of this area about the YY axis. Let the velocity be \u2013u, where u is a positive number. Enter the function and limits on the calculator and below is what happens in the background. Divide the gz curve equally with a number of lines in vertical directions, if the number increases the result will be more accurate. The answer to this problem came through a very nice idea. In the animation above, first you can see how by increasing the number of equal-sized intervals the sum of the areas of inscribed rectangles can better approximate the area A. Volume formulas. This tutorial shows the density curves and their properties. $\\begingroup$ @Gio The & and # are part of a \"pure function\" definition (see the documentation page for Function). \u0394x = -u(t 2 \u2013 t 1). Example: Find the area bounded by the curve fx x on() 1 [1,3]=+2 using 4 rectangles of equal width. Enter mean, standard deviation and cutoff points and this calculator will find the area under normal distribution curve. The curve is completely determined by the mean and the standard deviation \u02d9. Points on the blue curve, Area = 6. Prism computes area-under-the-curve by the trapezoidal method. The curve is symmetric around 0, and the total area under the curve is 100%. Click on \"hide details\" and \"rotated\" then drag the rectangle around to create an arbitrary size. 8931711, the area under the ROC curve. (b) Use four rectangles. For example the area first rectangle (in black) is given by: and then add the areas of these rectangles as follows:. Exercise: Area Under the Curve Borrowed from ACM Tech Pack 2 teaser (since I helped write it) Numerical integration is an important technique for solving many different problems. The first step in his method involved a unique way of describing the infinite rectangles making up the area under a curve. In each case, the area approximated is above the interval [0, 5] on the x-axis. Types of Problems. Since the region under the curve has such a strange shape, calculating its area is too difficult. We learn the formula and illustrate how it is used with a tutorial. Area of a. This engineering data is often used in the design of structural beams or structural flexural members. /rA)?? The documentation is quite unclear to me, it says. thus each term of the sum is the area of a rectangle with height equal to the function value at the distinguished point of the given sub-interval, and width the same. The area of a rectangle drawn above the curve would be more than the actual area under the curve. Let the velocity be \u2013u, where u is a positive number. Area under the curve is given by the Cumulative Distribution Function Cumulative Distribution Function. The answer is the estimated area under the curve. We use integration to evaluate the area we are looking for. Divide the gz curve equally with a number of lines in vertical directions, if the number increases the result will be more accurate. And that is how you calculate the area under the ROC curve. It is easy to use SAS/IML software (or the SAS DATA step) to implement the trapezoidal rule. Integrals are often described as finding the area under a curve. area under a curve into individual small segments such as squares, rectangles and triangles. Due to the this it approximate area. (c) Use a graphing calculator (or other technology) and 40 rectangles. And so if you were to solve the problem the geometric way, as if it had stated. Find the area of the shaded region by subtracting the area of the small shape from the area of the larger shape. Enter mean, standard deviation and cutoff points and this calculator will find the area under normal distribution curve. You can compute the area under the piecewise linear segments by summing the area of the trapezoids A1, A2, A3, and A4. Area Between Two Curves Calculator. I tried to calculate the total area with two options: sum(y)=-1. The area of a rectangle drawn above the curve would be more than the actual area under the curve. The area of a rectangle is equal to its length multiplied by its width. 000 and a standard deviation (sigma) of 1. For example, here's how you would estimate the area under.\n9590dj0zy5m8r g7zem4iqj7z2 gdrjjwkymoxab l8qmvqwdyp jmz6zsoi979s0 9mgyffafyvsi 4wvptwmutxzj9 x3k7tm2g75rwk oourtgqkmv31o fzlyc3j40cxoo th325z5v2xcz54n i9qnjdyvggouy8n ej01502sgsnoswc nevcusmc04crb 7jtnyjpt7n0 orz36oonsy3j1vd vx126g6tlah26j kd8gu2dvcdlg nk5xorx03mhs6o2 5pb29eb7k61v3t 20oglbck4t9 nt8zhoq7qh8c udnar5ww1xt xo7krxqmhovaea h6em4phm5n8 tto7lksijk5sb 8sjtlednkb0bo", "date": "2020-06-04 19:34:36", "meta": {"domain": "laquintacolonna.it", "url": "http://qdjf.laquintacolonna.it/area-of-rectangle-under-curve-calculator.html", "openwebmath_score": 0.8136438727378845, "openwebmath_perplexity": 351.8731791619468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9865717468373085, "lm_q2_score": 0.9019206758704633, "lm_q1q2_score": 0.8898094567022089}}
{"url": "https://cs.stackexchange.com/questions/138820/why-are-binary-numbers-sometimes-written-with-one-or-more-leading-zeros-that-don", "text": "Why are binary numbers sometimes written with one or more leading zeros that don't change the number (quantity) represented?\n\nBinary numbers like '0110', or '00100101' are seen very often in all contexts. What are the leading (left hand side) zeroes for? Why did the writer not write '110' and '100101', respectively?\n\nLeading zeroes in binary usually indicate the bit length of the data type. For example, the number 110 represented in a 4 bit data type would be 0110. Even if there is no data type specified, it's sometimes common to pad your binary numbers to the next power of 2. For example, 10111 of size 5 should be padded to 8 $$(2^3)$$ as 00010111\nDepend on context, so, I bring one small example: if we consider $$3$$-bit field, then $$110$$ is negative in $$2$$'s complement and equal $$-2$$, while in $$4$$-bit field $$0110$$ is positive and equal $$6$$ in same $$2$$'s complement.\n\u2022 Yes. In $2$'s complement all binary code with leading bit $1$ is negative. Apr 11 at 17:07", "date": "2021-11-30 19:24:56", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/138820/why-are-binary-numbers-sometimes-written-with-one-or-more-leading-zeros-that-don", "openwebmath_score": 0.8383588194847107, "openwebmath_perplexity": 762.255461973219, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9820137926698566, "lm_q2_score": 0.9059898197488448, "lm_q1q2_score": 0.8896944990118428}}
{"url": "https://resellernews.pl/me-life-rgnn/factorial-of-0-8029ee", "text": "October 22, 2020 . The answer to this lies in how the solution is implemented. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 \u00d7 2 \u00d7 1). = 4 \u22c5 3 \u22c5 2 \u22c5 1 = 24. For example: The factorial of 5 is 120. If you still prefer writing your own function to get the factorial then this section is for you. The trick is to use a substitution to convert this integral to a known integral. = 1 if n = 0 or n = 1 *n. The factorial of 0 is defined to be 1 and is not defined for negative integers. Yes, there is a famous function, the gamma function G(z), which extends factorials to real and even complex numbers. is pronounced as \"5 factorial\", it is also called \"5 bang\" or \"5 shriek\". A for loop can be used to find the factorial \u2026 Factorial of n is denoted by n!. There are many explanations for this like for n! Factorial of a number is the product of all numbers starting from 1 up to the number itself. Welcome. {\\displaystyle {\\binom {0}{0}}={\\frac {0!}{0!0!}}=1.} = \\frac{\u221a\\pi}2 $$How to go about calculating the integral? = 1 neatly fits what we expect C(n,0) and C(n,n) to be. Factorial (n!) 0! Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. While calculating the product of all the numbers, the number is itself included. Half Factorial. The factorial symbol is the exclamation mark !. in your calculator to see what the factorial of one-half is. > findfact(0) [1] \"Factorial of 0 is 1\" > findfact(5) [1] \"Factorial of 5 is 120 \" There is a builtin function in R Programming to calculate factorial, factorial() you may use that to find factorial, this function here is for learning how to write functions, use for loop, if else and if else if else structures etc. * 0. We can find the factorial of any number which is greater than or equal to 0(Zero). Yes we can! The factorial of a number n is the product of all numbers starting from one until we reach n. The operation is denoted by an exclamation mark succeeding the number whose factorial we wish to seek, such that the factorial of n is represented by n!. Logically$$1! = 1. . Computing this is an interesting problem. Read more: What is Null in Python. Source Code # Python program to find the factorial of a number provided by the user. There are multiple ways to \u2026 The factorial value of 0 is by definition equal to 1. = n * (n-1)! It does not seem that logical that $$0! = 1$$ and $$0! . Factorial of a positive integer is the product of an integer and all the integers below it, i.e., the factorial of number n (represented by n!) A method which calls itself is called a recursive method. The factorial of one half (0.5) is thus defined as$$ (1/2)! Here a C++ program is given to find out the factorial of a \u2026 Since 0 is not a positive integer, as per convention, the factorial of 0 is defined to be itself. But we need to get into a subject called the \"Gamma Function\", which is beyond this page. Symbol:n!, where n is the given integer. Let us think about why would simple multiplication be problematic for a computer. For negative integers, factorials are not defined. Similarly, you cannot reason out 0! There are several motivations for this definition: For n = 0, the definition of n! For example: Here, 4! and calculated by the product of integer numbers from 1 to n. For n>0, n! It is denoted with a (!) Below is the Program to write a factorial program in Visual basic. Factorial using Non-Recursive Program. Can factorials also be computed for non-integer numbers? The factorial is normally used in Combinations and Permutations (mathematics). The factorial of 0 is always 1 and the factorial of a \u2026 Example of both of these are given as follows. where n=0 signifies product of no numbers and it is equal to the multiplicative entity. The factorial is normally used in Combinations and Permutations (mathematics). Recursion means a method calling itself until some condition is met. The Factorial of number is the product of all the numbers less than or equal to that number & greater than 0. The factorial formula. = 1. = 5 * 4 * 3 * 2 *1 5! 5! = 1\u00d72\u00d73\u00d74\u00d7...\u00d7n. Factorial zero is defined as equal to 1. This program for factorial allows the user to enter any integer value. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 \u00d7 2 \u00d7 1). Programming, Math, Science, and Culture will be discussed here. would be given by n! Finding factorial of a number in Python using Recursion. 0!=1. This loop will exit when the value of \u2018n\u2018 will be \u20180\u2018. symbol. . How to Write a visual basic program to find the factorial number of an integer number. The factorial of a number is the product of all the integers from 1 to that number. ), is a quantity defined for all integer s greater than or equal to 0. is pronounced as \"4 factorial\", it is also called \"4 bang\" or \"4 shriek\". Factorial of a Number using Command Line Argment Program. Are you confused about how to do factorial in vb 6.0 then don\u2019t worry! = 1*2*3*4* . Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. as a product involves the product of no numbers at all, and so is an example of the broader convention that the product of no factors is equal \u2026 Factorial definition formula It is easy to observe, using a calculator, that the factorial of a number grows in an almost exponential way; in other words, it grows very quickly. = 1 * 2 * 3 * 4....n The factorial of a negative number doesn't exist. By using this value, this Java program finds Factorial of a number using the For Loop. These while loops will calculate the Factorial of a number.. Factorial Program in C: Factorial of n is the product of all positive descending integers. Type 0.5! I am not sure why it should be a negative infinity. So 0! Factorial Program in C++: Factorial of n is the product of all positive descending integers. = 1$$. In mathematics, the factorial of a number (that cannot be negative and must be an integer) n, denoted by n!, is the product of all positive integers less than or equal to n. Common Visual basic program with examples for interviews and practices. Possibly because zero can be very small negative number as well as positive. So, for the factorial calculation it is important to remember that$$1! . The factorial of a positive integer n is equal to 1*2*3*...n. Factorial of a negative number does not exist. The factorial symbol is the exclamation mark !. Problem Statement: Write a C program to calculate the factorial of a non-negative integer N.The factorial of a number N is defined as the product of all integers from 1 up to N. Factorial of 0 is defined to be 1. The factorial for 0 is equal to 1. $and$ 0! Can we have factorials for numbers like 0.5 or \u22123.217? Here, I will give three different functions for getting the factorial of a number. Factorial of a number is calculated for positive integers only. factorial of n (n!) n! The factorial of a positive number n is given by:. factorial: The factorial, symbolized by an exclamation mark (! = \u222b_0^\u221e x^{1/2}e^{-x}\\,dx $$We will show that:$$ (1/2)! This site is dedicated to the pursuit of information. The factorial formula. The factorial of n is denoted by n! For example: Here, 5! For n=0, 0! But I can tell you the factorial of half (\u00bd) is half of the square root of pi. = 120. Welcome to 0! Factorial of n is denoted by n!. Recursive Solution: Factorial can be calculated using following recursive formula. I cannot derive the sign. For negative integers, factorials are not defined. We are printing the factorial value when it ends. $\\begingroup$ @JpMcCarthy You'd get a better and more detailed response if you posted this as a new question. The factorial of an integer can be found using a recursive program or a non-recursive program. Factorial is not defined for negative numbers, and the factorial of zero is one, 0! n! See more. The best answer I can give you right now is that, like I've mentioned in my answer, $\\Gamma$ was not defined to generalize factorials. = 1/0 = \\infty$$. The factorial value of 0 is by definition equal to 1. The aim of each function is \u2026 In maths, the factorial of a non-negative integer, is the product of all positive integers less than or equal to this non-negative integer. And, the factorial of 0 is 1. The factorial of 0 is 1, or in symbols, 0! Factorials are commonly encountered in the evaluation of permutations and combinations and in the coefficients of terms \u2026 According do the definition of factorial, 1 = 0! For example, the factorial of 6 is 1*2*3*4*5*6 = 720. Factorial definition, the product of a given positive integer multiplied by all lesser positive integers: The quantity four factorial (4!) The important point here is that the factorial of 0 is 1. Similarly, by using this technique we can calculate the factorial of any positive integer. = 1. Wondering what zero-factorial \u2026 Mathematicians *define* x^0 = 1 in order to make the laws of exponents work even when the exponents can \u2026 First, we use integration by \u2026 So, first negative integer factorial is$$-1! Here are some \"half-integer\" factorials: = 1$$, but this is adopted as a convention. just in terms of the meaning of factorial because you cannot multiply all the numbers from zero down to 1 to get 1. Factorial of a number. Factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n: For example, The value of 0! =1. Please note: This site has recently undergone a complete overhaul and is not yet entirely finished, so you may come across missing content!. Factorial of 0. Writing a custom function for getting factorial. is 1, according to the convention for an empty product. = 1$$. 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A positive integer multiplied by all lesser positive integers only for you while calculating the product of all the from!", "date": "2021-10-16 09:20:51", "meta": {"domain": "resellernews.pl", "url": "https://resellernews.pl/me-life-rgnn/factorial-of-0-8029ee", "openwebmath_score": 0.7666701078414917, "openwebmath_perplexity": 569.5891442688228, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.982013793196246, "lm_q2_score": 0.905989819114262, "lm_q1q2_score": 0.8896944988655773}}
{"url": "https://math.stackexchange.com/questions/3845996/the-result-of-int-sin3x-mathrmdx", "text": "# The result of $\\int{\\sin^3x}\\,\\mathrm{d}x$\n\n$$\\int{\\sin^3x}\\,\\mathrm{d}x$$\n\nI find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results the same value/answer?\n\nHere how I work, please correct me if I'm wrong\n\nFirst method :\n\n\\begin{align} \\int{\\sin^3x}\\,\\mathrm{d}x & = \\int{\\sin x \\cdot \\sin^2x}\\,\\mathrm{d}x \\\\ &= \\int{\\sin x (1 - \\cos^2x)}\\,\\mathrm{d}x \\\\& = \\displaystyle\\int{(\\sin x - \\sin x\\cos^2x)}\\,\\mathrm{d}x \\\\& = \\dfrac{1}{3}\\cos^3x - \\cos x + C \\end{align}\n\nSecond method :\n\nFirst, we know that $$\\sin 3x = 3\\sin x - 4\\sin^3x$$\n\nTherefore, $$\\sin^3x = \\dfrac{3}{4}\\sin x - \\dfrac{1}{4}\\sin 3x$$\n\n\\begin{align} \\int{\\sin^3x}\\,\\mathrm{d}x & = \\int{\\left(\\frac{3}{4}\\sin x - \\frac{1}{4}\\sin 3x\\right)}\\,\\mathrm{d}x\\\\ & = \\frac{1}{12}\\cos 3x - \\frac{3}{4}\\cos x + C \\end{align}\n\n\u2022 Prove that these answers are the same, by proving that $\\frac{1}{12} \\cos(3x) - \\frac 34 \\cos(x) -(\\frac 13 \\cos^3 x - \\cos x)$ is a constant. Sep 30, 2020 at 9:47\n\u2022 Should it equal to zero? How to do that? Could you give me some details, please? Sep 30, 2020 at 9:49\n\u2022 Yes, it should equal $0$. Substitute $x = \\frac \\pi 2$, then all terms are zero. Use the triple angle formula. Sep 30, 2020 at 9:50\n\u2022 Wow, I also see that when $x = 0$, the result holds. Thanks. Sep 30, 2020 at 9:52\n\u2022 You are welcome! Sep 30, 2020 at 9:52\n\n$$\\cos 3x =4\\cos^3x -3\\cos x$$ So, $$\\frac{1}{12}\\color{green}{\\cos 3x} - \\frac{3}{4}\\cos x=\\frac{1}{12}(\\color{green}{4\\cos^3x -3\\cos x})-\\frac{3}{4}\\cos x$$ $$=\\frac{1}{3}\\cos^3x-\\cos x$$\n\nHence both the answers are the same.\n\nYes, they are both valid and true. Actually,$$(\\forall x\\in\\Bbb R):\\frac13\\cos^3(x)-\\cos(x)=\\frac1{12}\\cos(3x)-\\frac34\\cos(x)$$since$$(\\forall x\\in\\Bbb R):\\cos(3x)=4\\cos^3(x)-3\\cos(x).$$\n\nSince $$\\cos^3(x)=\\frac{3}{4}\\cos(x)+\\frac{1}{4}\\cos(3x)$$\n\nYour first integral becomes $$\\int \\sin^3(x)dx=\\dfrac{1}{3}\\cos^3x - \\cos x + C$$ $$=\\frac{1}{3}\\big[\\frac{3}{4}\\cos(x)+\\frac{1}{4}\\cos(3x)\\big]-\\cos(x)+C$$ $$=\\frac{1}{12}\\cos(3x)-\\frac{3}{4}\\cos(x)+C$$\n\nNote that the constant of integration are not necessarily the same. For example using $$u$$-substitutions for the denominator we have $$\\int \\frac{4x}{4x^2+7}dx=\\frac{1}{2}\\ln(4x^2+7)+C_{1}$$\n\n$$\\int \\frac{x}{x^2+\\frac{7}{4}}dx=\\frac{1}{2}\\ln(x^2+\\frac{7}{4})+C_{2}$$\n\nHere we have $$C_{2}=C_{1}+\\frac{1}{2}\\ln(4)$$ since they are constants. Indeed we have $$\\frac{1}{2}\\ln(x^2+\\frac{7}{4})+C_{2}=\\frac{1}{2}\\ln(x^2+\\frac{7}{4})+\\frac{1}{2}\\ln(4)+C_{1}$$ $$=\\frac{1}{2}\\big[\\ln(x^2+\\frac{7}{4})+\\ln(4)\\big]+C_{1}$$ $$=\\frac{1}{2}\\ln(4(x^2+\\frac{7}{4}))+C_{1}$$ $$=\\frac{1}{2}\\ln(4x^2+7)+C_{1}.$$", "date": "2022-06-29 01:49:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3845996/the-result-of-int-sin3x-mathrmdx", "openwebmath_score": 0.9936022162437439, "openwebmath_perplexity": 607.0726747893162, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806529525571, "lm_q2_score": 0.9073122176012061, "lm_q1q2_score": 0.8896928067672233}}
{"url": "https://mathematica.stackexchange.com/questions/125025/find-the-5566th-digit-after-the-decimal-point-of-7-101?noredirect=1", "text": "# Find the 5566th digit after the decimal point of 7/101\n\nI want to find the 5566th digit after the decimal point of 7/101. I input the following code into Mathematica 11:\n\nMod[IntegerPart[7/101*10^5566], 10]\n\n\nThe output is 6, which is the correct answer. Is there a better way to find the answer? Thank you very much in advance.\n\n\u2022 The best way to find this digit is in my opinion to calculate it. Since 7/101 ist periodic and so every digit in position n for which mod(n,4)=2 is 6. And mod(5566,4)=2. \u2013\u00a0mgamer Aug 29 '16 at 4:34\n\u2022 @mgamer great point. Intelligence wins over brute force. \u2013\u00a0Mr.Wizard Aug 29 '16 at 4:37\n\u2022 @Mr.Wizard: Was about to post an answer using just that (though it's not as simple as just a Mod - one must account for possible non-repeating digits before repeat starts), but not clear if OP needs merit the extra code. It will be orders of magnitude faster when going out millions+ digits, but for less, probably no real advantage over your answer... \u2013\u00a0ciao Aug 29 '16 at 4:48\n\n## Fast algorithm\n\nn = 5566\nIntegerPart[10 Mod[7 PowerMod[10, n - 1, 101], 101]/101]\n\n\nA brute force approach (see also these posts on stackoverflow :) ) may be fine for the current problem, but what if n is a huge number? The only possibility apart from guessing the periodic sequence of numbers as mgamer suggested would be to use modular arithmetics. Let me explain my answer. In contrast to the original post we put the number of interest not in the last digit of the integer part, but in the first digit of the fractional part. Conveniently, the fractional part can be computed as a reminder, or for higher efficiency by PowerMod.\n\nLet us compare the timing of the two methods:\n\nn = 556612345;\nMod[IntegerPart[7 10^n/101], 10] // Timing\n(*{10.447660, 3}*)\nIntegerPart[10 Mod[7 PowerMod[10, n - 1, 101], 101]/101] // Timing\n(*{0.000016, 3}*)\n\n\nThe time difference is obvious!\n\n## Explanation\n\nLet us consider another example, we compute the n=6 digit of the 7/121 fraction.\n\nn = 6\nN[7/121, 30]\n\n\n0.0578512396694214876033057851240.\n\nIn the original post the sought digit is the last digit of the integer part:\n\nN[7 10^n/121, 20]\n\n\n57851.239669421487603\n\nwhereas in my solution it is the first digit in the fractional part\n\nN[Mod[7*10^(n - 1), 121]/121, 20]\n\n\n0.12396694214876033058 .\n\nIt is further used that Mod[a 10^b,c]=Mod[a PowerMod[10,b,c],c].\n\n## Reusable function\n\nAs requested in the comments, a reusable function can be provided:\n\nClear[nthDigitFraction];\nnthDigitFraction[numerator_Integer, denominator_Integer, n_Integer,\nbase_Integer: 10] /; n > 0 && base > 0 && denominator != 0 :=\nModule[{a = Abs[numerator], b = Abs[denominator]},\nIntegerPart[base Mod[a PowerMod[base, n - 1, b], b]/b]]\n\n\u2022 That's the way to do it. +1 \u2013\u00a0ciao Aug 29 '16 at 8:03\n\u2022 @Mr.Wizard I provided a reusable function, feel free for improving or making it more general. \u2013\u00a0yarchik Aug 29 '16 at 9:09\n\u2022 @yarchik if you looked at my code and read its introduction I DO NOT use brute force. In this case the recurring cycle is length 4 so determining digit is almost trivial. I applaud your method. Please note my function is general but not tweeted and used modular arithmetic. \u2013\u00a0ubpdqn Aug 29 '16 at 9:46\n\u2022 @ubpdqn Thank you, I modified my post accordingly. However, if you try your method on larger numbers the computational time grows rather fast. For instance I could not have computed dec[7, 101345, 5566]with your method. \u2013\u00a0yarchik Aug 29 '16 at 10:17\n\u2022 @yarchik yes and that is why I voted for your answer. It is very efficient and clean and was a nice lesson for me. The achilles heal of my approach was the need (not to process all the digits) but determing the cylce. :) \u2013\u00a0ubpdqn Aug 29 '16 at 10:19\n\nAn alternative formulation of RealDigits that I prefer:\n\nRealDigits[7/101, 10, 1, -5566][[1, 1]]\n\n(* 6 *)\n\n\nThis yields better performance which becomes important when looking for deeper digits:\n\nd = 6245268;\n\nRealDigits[7/101, 10, 1, -d][[1, 1]]  // AbsoluteTiming\nRealDigits[7/101, 10, d - 1][[1, -1]] // AbsoluteTiming\n\n{0.0501477, 3}\n\n{1.06702, 3}\n\n\nFor comparison to other methods now posted RealDigits can compute the repeating decimal itself:\n\nRealDigits[7/101]\n\n{{{6, 9, 3, 0}}, -1}\n\n\nHow to programmatically work with this output was the subject of a different Question, though I cannot find it at the moment. The possible combinations of repeating and nonrepeating digits as well as the offset makes a truly elegant yet robust solution difficult (at least to me) but in the easiest case, which this happens to be:\n\nd = 5566;\n\nRealDigits[7/101] /.\n{{c_List}, o_} :>\nc[[ Mod[d + o, Length @ c, 1] ]]\n\n(* 6 *)\n\n\nThis is of course quite fast:\n\nd = 556612345;\n\nRealDigits[7/101] /.\n{{c_List}, o_} :> c[[ Mod[d + o, Length@c, 1] ]] // RepeatedTiming\n\n{0.00001243, 0}\n\nRealDigits[7/101, 10, 5566][[1]][[5565]]\n\n\u2022 FWIW this could also be written RealDigits[7/101, 10, 5566 - 1][[1, -1]]. +1 of course, but see the performance caveat in my answer. \u2013\u00a0Mr.Wizard Aug 29 '16 at 4:08\n\nCorrection\n\nI thank yarchik for his answer and his test of my code identified an error (as well as my code being extremely inefficient for long cycle length):\n\nFor this particular example, the recurring patter is of length 4.\n\nSo,\n\nfun[n_, d_] :=\nModule[{lst =\nNestWhileList[QuotientRemainder[10 #[[2]], d] &,\nQuotientRemainder[n, d], UnsameQ, All], p},\np = Position[lst, lst[[-1]]][[1, 1]];\n{lst[[1 ;; p - 1, 1]], lst[[p ;; -2, 1]]}]\ndec[n_, d_, p_] := Module[{a, b},\n{a, b} = fun[n, d];\nb[[Mod[p-Length@a+1, Length@b, 1]]]]\n\n\nwhere\n\n\u2022 fun just separates cycling and non-cycling.\n\u2022 dec determines value of at position p by working out where in cycle p is\n\nSo,\n\ndec[7, 101, 5566]\ndec[7, 101, 6245268]\n\n\nyields 6 and 3 respectively.\n\nAs a reality check (not proof):\n\n{#, RealDigits[7/101, 10, 1, -#][[1, 1]], dec[7, 101, #]} & /@\nRandomInteger[{10000, 1000000}, 20] // TableForm", "date": "2019-10-23 06:53:56", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/125025/find-the-5566th-digit-after-the-decimal-point-of-7-101?noredirect=1", "openwebmath_score": 0.36136215925216675, "openwebmath_perplexity": 3056.217518285138, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731158685837, "lm_q2_score": 0.9241418178895029, "lm_q1q2_score": 0.889646483332145}}
{"url": "https://math.stackexchange.com/questions/2622650/a-set-of-integers", "text": "A set of integers\n\nAssume that there is a set of ordered integers initially containing number $1$ to a given $n$. At each step, the lowest number in the set is removed, if the number was odd, then we go to the next step and if it was even, half of that number is inserted into the set and the cycle repeats until the set is empty.\n\nMy question is, how many steps does it take for a given number $n$ to finish the whole set?\n\nI think it should be something the form of $2k - x$ where $x$ itself is likely a complex expression but I just can't seem to figure it out. Any help would be appreciated. (I got that from trial and error, by the way, no logic or proof behind it, I know that the answer is $\\lfloor \\frac{n}{2} \\rfloor$+\"The number of times each even number can be divided by 2\", but I just can't find a closed formula)\n\n\u2022 It will have to do with the number of numbers in the original set that fall into each of the following categories: $(2k+1),(2k+1)2, (2k+1)2^2,(2k+1)2^3,\\dots$. Numbers from the first category are removed in a single step. In the second category in two steps, in the third category in three steps, etc... \u2013\u00a0JMoravitz Jan 26 '18 at 21:28\n\u2022 @JMoravitz Exactly! I got to that and then I couldn't get any further. \u2013\u00a0Arian Tashakkor Jan 26 '18 at 21:35\n\u2022 Experimentally, this seems to be oeis.org/A005187 . \u2013\u00a0Jair Taylor Jan 26 '18 at 21:40\n\u2022 @JairTaylor It does check out with the numbers I generated with my program. Does this suggest there doesn't exist a closed formula? \u2013\u00a0Arian Tashakkor Jan 26 '18 at 21:44\n\u2022 @ArianTashakkor Not necessarily, OEIS doesn't know everything. \u2013\u00a0Jair Taylor Jan 26 '18 at 22:15\n\nAs in this algorithm, the actual order of the set isn't important, we can assign every $n\\in\\mathbb{N}$ the number of steps it takes till it's sorted out, which is namely how many times you'll have to divide by two till it's odd.\n\nIf we use prime factorization, that means if $z$ takes k steps till it's sorted out, it's prime factorization is $z = 2^{k-1} \\cdot\\, ...$\n\nNow, we know the following:\nEvery second number has $2^0$ in its prime factorization (every odd number).\nEvery forth number has $2^1$ in its prime factorization.\nEvery eigth number has $2^2$ in its prime factorization.\n...\n\nSo, for a given set $\\{1,...,n\\}$ it takes $\\sum_{i=0}^{\\infty }\\lfloor\\frac{n-(2^i-1)}{2^{i+1}}\\rfloor$ steps till your algorithms finished.\nHere, $2^i-1$ is here how long we'll have to count from 1 till we reach the first number that has $2^i$ in its prime factorization\n(e.g: We have to count 0 higher to reach the first odd number, 1 higher to reach the first number that is divisible by 2 but not by 4...)\n\nYou can cut off the sum as soon as the divisor gets greater than the divident, so as rough approximation $log_2(n)$ works out. With that we get: $$\\sum_{i=0}^{log_2(n)}\\lfloor\\frac{n-(2^i-1)}{2^{i+1}}\\rfloor$$\n\nAnother way to look at the problem is by keeping it a set. If $M$ is a set and $k\\in\\mathbb{N}$, let us define $$M\\cdot k := \\{k\\cdot m \\mid m\\in M\\}$$ E.g. $\\{1,2,3\\}\\cdot 2 = \\{2,4,6\\}$\n\nNow we look at how our set looks like if we process it $i$ times, with our procss being:\nFor every number of the set, if the number is even, half it, if it is odd, remove it.\n\n$i=0$ - The algorithm hasn't run yet, our set is the input set $\\{1,..,n\\}$.\n\n$i=1$ - Only the even numbers are left $\\{2,4,6,..,2\\cdot\\lfloor\\frac{n}{2}\\rfloor\\} = 2\\cdot \\{1,2,3,..,\\lfloor\\frac{n}{2}\\rfloor\\}$\n\n$i=2$ - Only the numbers divisble by four are left $\\{4,8,12,..,4\\cdot\\lfloor\\frac{n}{4}\\rfloor\\} = 4\\cdot \\{1,2,3,..,\\lfloor\\frac{n}{4}\\rfloor\\}$\n\n...\n\n$i=k$ - Only the numbers divisble by $2^k$ are left $\\{2^k,2\\cdot 2^k, 3\\cdot 2^k, ..., 2^k\\cdot\\lfloor\\frac{n}{2^k}\\rfloor\\} = 2^k\\cdot \\{1,2,3,..,\\lfloor\\frac{n}{2^k}\\rfloor\\}$\n\nThe number of operations our algorithm takes is now simply the sum of the numbers of each set in this chain. So, we get: $$\\sum_{i=0}^{\\infty} \\lfloor\\frac{n}{2^i}\\rfloor = \\sum_{i=0}^{log_2(n)} \\lfloor\\frac{n}{2^i}\\rfloor$$\n\nFinally, we can let this sum look a little more refined:\n\nLet $\\{1,..,n\\}$ be our set and $n = c_0\\cdot 2^0 + c_1\\cdot 2^1 +c_2\\cdot 2^2+... + c_k\\cdot 2^k$ its binary representation. Then the steps the algorithm needs for $\\{1,..,n\\}$ are equal to running our algorithm on the following sets:\n\n$\\{1,2,...,c_i\\cdot 2^i\\} \\text{ where } i\\in\\mathbb{N}, i\\leq k, c_i = 1$\n\nThis let's us erase the $\\lfloor \\rfloor$-s in our sum, as for every step of the algorithm $\\lfloor\\frac{c_i\\cdot 2^i}{2^j}\\rfloor\\}$ is a whole number for $j\\leq i$, and for $j>i$, the set is empty.\n\nSo, if $c_0\\cdot 2^0 + c_1\\cdot 2^1 +c_2\\cdot 2^2+... + c_k\\cdot 2^k$ is the binary represantation of our number, the steps our algorithm needs are: $$\\sum_{j=0}^k c_j\\cdot\\sum_{i=0}^j \\frac{2^j}{2^i} = \\sum_{j=0}^k c_j\\cdot\\sum_{i=0}^j 2^{j-i} = \\sum_{j=0}^k c_j\\cdot (2^{j+1}-1)$$\n\nLet $s(n)$ be the number steps required.\n\nEach number $k$ is acted on $1+\\nu_2(k)$ times before being discarded completely, where $\\nu_2(k)$ is the highest power of $2$ that divides $k$. For example, $\\nu_2(24) =3$ because $2^3$ divides $24$ but $2^4$ does not.\n\nSo $s(n) = \\sum_1^n \\left(1+\\nu_2(i)\\right) = n+\\sum_1^n\\nu_2(i)$\n\nNow $\\sum_1^n\\nu_2(i)$ is the count of even numbers up to $n$, plus the count of numbers divisible by $4$, plus the count of numbers divisible by $8$, etc. And of course we can get that count just by dividing $n$ by the powers of $2$ and rounding down, ie $\\sum_1^n\\nu_2(i) = \\left\\lfloor \\frac n2 \\right\\rfloor + \\left\\lfloor \\frac n4 \\right\\rfloor + \\left\\lfloor \\frac n8 \\right\\rfloor + \\cdots$.\n\nWhere $n=2^k$, this is actually equal to $n{-}1$, and for $n=2^k-1$ of course it is $n{-}k$. So $s(n) = 2n-e$, where $e$ depends on where the number sits between powers of $2$.\n\nA little lateral thinking uncovers that $e$ is actually the digit sum of $n$ in binary.", "date": "2019-08-18 11:58:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2622650/a-set-of-integers", "openwebmath_score": 0.672687828540802, "openwebmath_perplexity": 221.86693684521222, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474872757474, "lm_q2_score": 0.8991213826762113, "lm_q1q2_score": 0.8895434807066054}}
{"url": "https://math.stackexchange.com/questions/372188/should-the-sign-be-reversed-if-i-square-both-sides-of-an-inequality", "text": "# Should the sign be reversed if I square both sides of an inequality?\n\nLet us say I have the following:\n\n$$x>y$$\n\nNow, I want to take the square of both sides. Should it result in $$x^2>y^2$$ or $$x^2<y^2$$\n\nI suspect there is no way to give a general answer to this. I would like to know how to analyze this nevertheless.\n\n\u2022 Similarly, I'd like to know how to square $x<y$ as well. Apr 25, 2013 at 5:23\n\u2022 Unless both have the same sign there isn't a satisfactory answer. $1 > -2$, but $1^2 < (-2)^2$. On the other hand, $2 > -1$ and $2^2 > (-1)^2$. Apr 25, 2013 at 5:35\n\u2022 Similar question (perhaps a duplicate): Showing $a^2 < b^2$, if $0 < a < b$. Apr 25, 2013 at 7:48\n\u2022 @MartinSleziak Unless I missed something in one of the answers, this question is much more general. So it is not an exact duplicate. Apr 25, 2013 at 8:35\n\u2022 @user1729 Perhaps it is more general, I am not sure about much more general. Well, I've cast my vote to close, so I cannot undone this. If the question is closed at all, there's no problem in requesting the reopening. (And maybe it won't be closed at all if other potential voters see your comment.) Apr 25, 2013 at 8:59\n\nYou have to know where zero is to do anything. This is because the function $f(x)=x^2$ is increasing in the interval $x\\ge0$ and decreasing in the interval $x\\le0$.\n\nThe general principle (LEARN THIS! You can later apply it to more difficult functions) is that if you apply an increasing function to both side of an inequality, you keep the original order. OTOH if you apply a decreasing function to both sides of an inequality the order is reversed.\n\nSo if you know that $x$ and $y$ are both $\\ge0$ , then the inequality $x>y$ is true if and only if the inequality $x^2>y^2$ is true.\n\nOTOH if you know that $x$ and $y$ both $\\le0$, then the inequality $x>y$ is true if and only if the inequality $x^2<y^2$ is true.\n\nI leave it to you to think, what you can deduce about the truth of $x>y$, if $x$ and $y$ have opposite signs.\n\nAnyway, when you contemplate squaring both sides of an inequality, you have to split the solution to cases according to where zero lies. With some other functions the situation may be better. For example cubing is an increasing function on the entire real line, and thus you can cube (or take the cube roots) of an inequality with impunity.\n\n\u2022 Am I right in flipping the sign when applying ^x/b to both sides when both sides have values between 0 and 1 and x<b? Oct 8, 2015 at 20:44\n\u2022 What about multiplying both sides of an inequality by $y = -1$ ? That switches the order however y is not a decreasing function May 20, 2016 at 13:53\n\u2022 @Amir: Then you are not applying the same function to both sides of the inequality, and all bets are off. May 20, 2016 at 16:32\n\u2022 And @Amir: More importantly. The principle is about applying a function to both sides of an inequality. In other words: if we are given $a<b$ we want to know whether $f(a)<f(b)$ or $f(a)>f(b)$ for some function $f$. Multiplication by $-1$ means that you apply the decreasing function $f(x)=-x$. Whether multiplication by $y$ is decreasing or increasing depends on the sign of $y$. Oct 16, 2016 at 6:08\n\u2022 @Uq'''12wn1F12u2x3uW31H1JBk9m That would be applying the function f(x)=-x to both sides, which is decreasing. May 6, 2018 at 23:37\n\nIf $x^2-y^2>0, (x+y)(x-y)>0$\n\nNow, if $x-y>0,$ i.e.,if $x>y; x+y>0$\n\nor if $x-y<0,$ i.e.,if $x<y; x+y<0$\n\nSo, $x>y$ and $x+y>0 \\implies x^2>y^2$ [Ex. $5>\\pm 3$ and $5\\pm 3>0\\implies 5^2>(\\pm3)^2$]\n\nand $x<y$ and $x+y<0 \\implies x^2>y^2$ [Ex. $-5<-3$ and $-5+(-3)=-8<0\\implies (-5)^2>(-3)^2$]\n\n\u2022 Since the original question assumed $x > y$, I would write it this way. If $x > y$, then $x^2 - y^2 = (x+y)(x-y)$ and $x + y$ have the same sign. Thus $x^2 > y^2$ if $x + y > 0$, $x^2 < y^2$ if $x + y < 0$, $x^2 = y^2$ if $x + y = 0$. Apr 25, 2013 at 5:46\n\u2022 @RobertIsrael, very precis. I also wanted to show $x<y$ along with $x>y$. I excluded $x=y$ as the question Apr 25, 2013 at 5:49\n\nMaybe this will be helpful: $$x \\geq y \\Longleftrightarrow \\mathrm{sgn}(x)x^2 \\geq \\mathrm{sgn}(y)y^2$$\n\nWhere $\\mathrm{sgn}(\\cdot)$ is the sign function. It is what I use to test inequalities for computational purposes. You can check it on a case-by-case level, i.e. by checking the three possible cases\n\n1. $x\\geq0,y\\geq0$,\n2. $x\\geq 0,y\\leq 0$,\n3. $x\\leq 0,y\\leq 0$.", "date": "2023-04-01 00:49:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/372188/should-the-sign-be-reversed-if-i-square-both-sides-of-an-inequality", "openwebmath_score": 0.8088952898979187, "openwebmath_perplexity": 200.37867968035658, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357189762611, "lm_q2_score": 0.9059898114992677, "lm_q1q2_score": 0.8894425589774009}}
{"url": "https://parulgupta.org/o6cn73w/metric-prefixes-chart-861112", "text": "unit prefixes (pico to Tera); how to convert metric prefixes using dimensional analysis explained & metric prefix numerical relationships tutorial. Prefixes K,M,G,T,P (kilo,mega,giga,tera,peta) are commonly used in computing, The chart on the previous page had some common metric prefixes from smallest to largest. Bulk agricultural products, such as grain, beer and wine, are often measured in hectolitres (each 100 litres in size). Start studying Metric Prefix Chart. These SI prefixes or metric prefixes are in widespread use in all areas of life. The LaTeX typesetting system features an SIunitx package in which the units of measurement are spelled out, for example, \\SI{3}{\\tera\\hertz} formats as \"3\u00a0THz\". for this purpose. [Note 1] When both are unavailable, the visually similar lowercase Latin letter u is commonly used instead. Converts the metric prefix in one unit to the others. there are 1000 metres in a kilometre (km). Power of ten Prefix Prefix Abbrev. This homework or classwork assignment supplies the students with a reference chart of the metric prefixes expressed verbally and as powers of ten. If they have prefixes, all but one of the prefixes must be expanded to their numeric multiplier, except when combining values with identical units. There are gram calories and kilogram calories. They are also occasionally used with currency units (e.g., gigadollar), mainly by people who are familiar with the prefixes from scientific usage. When typing your answer, use scientific notation. Metric prefixes Metric prefixes: definitions, values and symbols The metric prefixes have entered many parts of our language and terminology, especially measurements and performance data of very big and very small things (gigabyte, microgram, nanosecond, etc). Notes * The radian and steradian, previously classified as supplementary units, are dimensionless derived units that may be used or omitted in expressing the values of physical quantities. one thousandth of a metre (mm). To improve this 'Metric prefix Conversion Calculator', please fill in questionnaire. This is most easily understood by considering how the decimal places keep adding a zero to hold place value as numbers get exponentially smaller: 1. Jan 21, 2014 - This won\u2019t give all the metric prefixes to you, but then, you won\u2019t generally need them all. May 2, 2020 - Explore TINMAN's board \"Conversion Factor Prefixes\" on Pinterest. The metric system provides a logical way to organize numbers and mathematical thinking. Includes answer key The examples above show how prefixes indicate increasingly large units of measurement, but metric prefixes also create units smaller than the original by dividing it into fractions. Likewise, milli may be added may be added to metre to form the word millimetre, i.e. Here you can make instant conversion from this unit to all other compatible units. Long time periods are then expressed by using metric prefixes with the annum, such as megaannum or gigaannum. The metric prefixes have entered many parts of our language and terminology, especially measurements and performance data of very big and very small things (gigabyte, microgram, nanosecond, etc). The metric system charts in this ScienceStruck post will help kids understand converted values quite easily. Except for the early prefixes of kilo-, hecto-, and deca-, the symbols for the multiplicative prefixes are uppercase letters, and those for the fractional prefixes are lowercase letters. 1 This is a conversion chart for mega (Metric Larger and smaller multiples of that unit are made by adding SI prefixes. 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Mg ) is used together with metric and some other units prefixed with kilo- gibi-, etc )! 1 shows examples of prefixes with the annum, such as microinch and kilopound, megatonnes,.... Used instead of the basic words used for measurement smaller are common, with reduced vowels on both syllables metre... And megaseconds are occasionally used to disambiguate the metric SI system ] Since 2009, they can mean a of... That even though the kilogram has a prefix chart in the metric system,. Dictated by convenience of use Since 2009, they have formed part of the second such milligram. In size ) have to be taught how to convert metric prefixes with a reference chart the! Actually encountered are seldom used in the metric scale, SI prefixes metric... Your consent megagram, gigagram, and other study tools ideas about prefixes measurement... A logical way to organize numbers and mathematical thinking here you can make instant conversion from yocto to yotta )... A convenient way of expressing mulitiples and subdivisions ( larger and smaller ) of any unit... < br > Productivity, Mindfulness, Health, and more with flashcards, games, more..., gigaparsec, millibarn ) kilometre \u201d means \u201c \u2026 Next come the prefixes are... Unit to all other compatible units '' or  grand '', or ... Unit names symbol K is often pronounced /k\u026a\u02c8l\u0252m\u026at\u0259r/, with reduced vowels on both syllables of metre prefixes create and.\n\nDrunken Jack's Reservations, Porcupine Dates Side Effect, Outdoor Nature Activities For Adults, Princess Mermaid Videos, Load Test Using Vegeta, Mashreq Bank Online Money Transfer, Providence House Application, What Do The Numbers Mean On A Scotts Spreader?,", "date": "2021-09-19 19:30:00", "meta": {"domain": "parulgupta.org", "url": "https://parulgupta.org/o6cn73w/metric-prefixes-chart-861112", "openwebmath_score": 0.6746723651885986, "openwebmath_perplexity": 4372.869184589697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9658995733060719, "lm_q2_score": 0.920789673173896, "lm_q1q2_score": 0.8893903524233036}}
{"url": "https://math.stackexchange.com/questions/2319655/when-is-a-sudoku-like-table-solvable", "text": "When is a Sudoku like table solvable\n\nGiven a $n\\times n$ table is it possible to fill each cell with one of the numbers $1,2,3,\\cdots,n$ such that in each column,each row and each diagonal (i.e Denoting $(x,y)$ as number of column and row $(2,1)$ and $(1,2)$ form the first diagonal) every number appears exactly once? For which $n$ can we fill the table?\n\nContext: I've been given this problem on a contest few months ago but just for $n=4,5$ which I solved easily since $n=4$ is impossible and for $n=5$ we have \\begin{array}{|c|c|c|c|c|} \\hline 1&2&3&4&5\\\\ \\hline 3&4&5&1&2\\\\ \\hline 5&1&2&3&4\\\\ \\hline 2&3&4&5&1\\\\ \\hline 4&5&1&2&3\\\\ \\hline\\end{array} But I was interested in a more general statement I think I've also proved that for $n=6$ it's impossible by trying to fill the table manually. My guess is that for even $n$ it's not solvable and for odd $n$ it's solvable but I have no idea how to approach it except to fill it manually.\n\nEDIT: For prime $n$ we can fill each cell $(i,j)$ with $i+2j\\pmod{n}$ except when $i+2j\\equiv0\\pmod{n}$ then we write $n$ instead for example such filling with $n=7$ (the $n=5$ example is the same filling if you look at $(j,i)$ instead of $(i,j)$) \\begin{array}{|c|c|c|c|c|c|c|} \\hline 3&5&7&2&4&6&1\\\\ \\hline 4&6&1&3&5&7&2\\\\ \\hline 5&7&2&4&6&1&3\\\\ \\hline 6&1&3&5&7&2&4\\\\ \\hline 7&2&4&6&1&3&5\\\\\\hline 1&3&5&7&2&4&6\\\\\\hline2&4&6&1&3&5&7\\\\\\hline\\end{array}\n\nPROOF OF THE EDIT: For the same row if cells $(i_1,j)$ and $(i_2,j)$ have the same value we have that $$i_1+2j\\equiv i_2+2j\\pmod{n}$$ implies $i_1\\equiv i_2$ which is possible only if $i_1=i_2$. Same logic applies to the column for cells $(i,j_1),(i,j_2)$ we get $$i+2j_1\\equiv i+2j_2\\pmod{n}$$ when $n$ is prime it implies $j_1=j_2$ if $(i_1,j_1),(i_2,j_2)$ are on a diagonal we have $$|i_1-i_2|=|j_1-j_2|$$ now assuming they have the same value $$i_1+2j_1\\equiv i_2+2j_2\\pmod{n}$$ then $i_1-i_2\\equiv 2(j_2-j_1)\\pmod{n}$ which implies $1\\equiv \\pm 2\\pmod{n}$ which is absurd.\n\n\u2022 What does the sentence \"$(2,1)$ and $(1,2)$ form the first diagonal\" mean? \u2013\u00a05xum Jun 12 '17 at 12:51\n\u2022 @5xum I mean position second cell in first column or first cell in second column, like in matrix. \u2013\u00a0kingW3 Jun 12 '17 at 12:55\n\u2022 This seems to by highly related to the 8 queens problem, if you think of all ones as queens, they may not be in the same row, column or diagonal. The same holds for all twos and so on. \u2013\u00a0mlk Jun 12 '17 at 13:00\n\u2022 I played with this once. For small grids, as in your example, I found that the rows just cycled the values. However, at 12x12, I found some solutions that did not do this. \u2013\u00a0badjohn Jun 12 '17 at 13:01\n\u2022 I found an old Java program that I wrote for this puzzle. It just uses brute force and ignorance. Assume that row 1 is 1, 2, 3, etc. Sizes 1, 2, 3, 4, 6, 8 have no solution. 5, 7, have a unique solution. This might suggest that there are no solutions for the even cases but there are multiple solutions for the 12 case. I am running the 9, 10, and 11 cases but, as you may expect, the program gets slower for these larger values. \u2013\u00a0badjohn Jun 12 '17 at 15:12\n\nThis is not an answer but hopefully a contribution to an answer.\n\nDouble diagonal Latin squares or just diagonal Latin squares (the terminology seems to vary) are Latin squares where both main diagonals (sometimes called the main and the anti-main) also have the property that all $N$ symbols occur exactly once. I realize that your requirement is that all \"minor\" diagonals also don't have repeating symbols, but it should be clear that a necessary condition for this, is that the square must be a diagonal Latin square.\n\nIn this paper there is a proof on page $4$ which shows that, if there are numbers A and B from the range $[0, N-1]$ which satisfy the properties:\n\n\u2022 A is relatively prime to N\n\u2022 B is relatively prime to N\n\u2022 (A + B) is relatively prime to N\n\u2022 (A - B) is relatively prime to N\n\nthen you can generate a diagonal Latin square with the following rule:\n\nCell$(i,j) = (A * i + B * j) \\mod N$\n\nThis is like the rule you found but without the strict requirement that $N$ is prime. A corollary to the above theorem is that if $N$ is an odd number not divisible by $3$, there is a diagonal Latin square of order $N$. So I tried the formula with the first odd non-prime fulfilling the corollary's requirement $(N=25)$ and got the following:\n\nIt seems to me this is a square of the type you are looking for, and with $N$ odd, but not a prime.\n\nEdit\n\nWe can also show that with an even $N$, no diagonal Latin square can be generated using the method above. If $N$ is even, both $A$ and $B$ must be odd. But then both $(A+B)$ and $(A-B)$ must be even and can therefore not be relatively prime to $N$.\n\nEdit 2\n\nI made a program to generate diagonal Latin squares based on the formula above and then to check if all diagonals were without repeats. I ran the program for all odd $N$ between $3$ and $1001$ and the result is that all squares, where $N$ is not divisible by $3$, fulfilled the requirements! I therefore conjecture that the corollary above is not only true for diagonal Latin squares but also for \"kingW3\" squares.\n\nEdit 3\n\nLadies and gentlemen, I have found a very nice document which answers many of our questions. In fact, if we use the definition of \"diagonal\" assumed by @Ewan Delanoy (called \"broken diagonals\" in the document), it basically solves the OP:\n\n1. It proves the conjecture I made above\n2. It proves that if the definition of \"diagonal\" is \"broken diagonals\", no solutions exist for even $N$\n3. It gives an outline of a proof (leaving the details as homework!) that if we use the \"broken diagonals\" definition, no solutions exist for $N$ divisible by $3$.\n\nEnjoy!\n\n\u2022 Thanks for the update. I was wondering how to move from proofs that certain solutions exist to non-existence proofs. I had just started to use the terms \"strong solution\" for Ewan's interpretation and \"weak solution\" for the OP's. \u2013\u00a0badjohn Jun 16 '17 at 16:45\n\nThis is only a partial answer but it is too large to be a comment. I looked at this problem a long time ago and I wrote a Java program which tried to crack it by brute force and ignorance.\n\nI only looked at cases in which the first row is the selected symbols in order. Any other solution is \"isomorphic\" to one of this form.\n\nWhen the size is above 9, I use A, B, C, etc in a hexadecimal style.\n\nFor size 1, there is a trivial solution\n\n$$\\begin{array} {|c|} \\hline 1\\\\ \\hline \\end{array}$$\n\nfor sizes 2, 3, and 4, there is no solution.\n\nSize 5 has a solution as posted by kingW3 in his original post. There is a second which is in a sense a reflection. Each row is offset by 3 to the right which can be viewed as 2 to the left hence the reflection comment.\n\nSize 6 has no solution.\n\nSize 7 has 4 solutions here is one which is similar to kingW3's solution for size 5. Each row is offset 2 to the right. The others are offset by 3, 4, and 5.\n\n$$\\begin{array} {|c|c|c|c|c|c|c|} \\hline 1&2&3&4&5&6&7\\\\ \\hline 3&4&5&6&7&1&2\\\\ \\hline 5&6&7&1&2&3&4\\\\ \\hline 7&1&2&3&4&5&6\\\\ \\hline 2&3&4&5&6&1&2\\\\ \\hline 4&5&6&7&1&2&3\\\\ \\hline 6&7&1&2&3&4&5\\\\ \\hline \\end{array}$$\n\nNote the simple cyclic pattern shared by size 5.\n\nSize 8 has no solution.\n\nThis hints at a pattern: even unsolvable, odd solvable with a cyclic pattern.\n\nThe pattern does not continue. At size 9, that cyclic style does not work. My cracking program just completed for size 9; there is no solution.\n\nI won't run the cracking program on size 10. It would probably not finish before I die.\n\nThe cyclic patterns work for size 11. So, as kingW3 has found, a prime size helps which makes sense once you know.\n\nHowever, this does not cover all solutions. Here is one for size 12. I have a memory, but no records, of others.\n\n$$\\begin{array} {|c|c|c|c|c|c|c|} \\hline 1&2&3&4&5&6&7&8&9&A&B&C\\\\ \\hline 5&6&C&1&B&2&3&A&7&4&8&9\\\\ \\hline 9&A&4&8&6&1&B&5&C&2&7&3\\\\ \\hline B&7&9&3&A&C&8&1&4&6&5&2\\\\ \\hline 8&C&5&2&7&4&9&6&A&1&3&B\\\\ \\hline 3&B&7&6&C&A&5&4&2&8&9&1\\\\ \\hline A&4&1&9&8&3&2&B&5&C&6&7\\\\ \\hline C&8&2&5&1&B&6&9&3&7&A&4\\\\ \\hline 4&1&6&A&3&8&C&7&B&9&2&5\\\\ \\hline 6&3&B&C&9&7&4&2&8&5&1&A\\\\ \\hline 2&9&8&7&4&5&A&3&1&B&C&6\\\\ \\hline 7&5&A&B&2&9&1&C&6&3&4&8\\\\ \\hline \\end{array}$$\n\nI have a solution for size 25. It is the cyclic style. I think that will work if the size is coprime to 2 and 3 but I have not proved that yet.\n\nThe 12 case remains interesting as it is not of the cyclic style.\n\nUpdate\n\nOn further thought, I now believe for size $n$ and a shift per line of $s$, a cyclic solution exists provided that all of $s - 1$, $s$, and $s + 1$ have order $n$ in $\\mathbb{Z}_n$ (as an additive group). In other words, they are all non-zero and either $1$ or coprime to $n$. This can be simplified to Jens's rule of odd and not divisible by $3$.\n\nThe $12\\times12$ example above remains the only exceptional case that we know.\n\n\u2022 Your $12\\times 12$ example is wrong : for example, the $(11,1)$ and $(9,11)$ entries are both equal to $2$, yet they are on the same diagonal. \u2013\u00a0Ewan Delanoy Jun 15 '17 at 11:47\n\u2022 @EwanDelanoy I see what you mean but we seem to have different notions of the diagonal. I guess that you are imagining that the two sides are implicitly connected and a diagonal can travel off one side and continue on the other. My diagonals were not doing that. This comment from the OP, \"(2,1) and (1,2) form the first diagonal\" suggested my interpretation. \u2013\u00a0badjohn Jun 15 '17 at 11:57\n\u2022 The OP has been quiet for a while. We need him to say whether my example qualifies. Nonetheless, I will consider this interpretation as well. \u2013\u00a0badjohn Jun 15 '17 at 12:06\n\u2022 I too see what you mean. I think the OP should clarify this point. Personally if the sides are not implicitly connected I find the problem has less symmetry and beauty \u2013\u00a0Ewan Delanoy Jun 15 '17 at 12:08\n\u2022 @EwanDelanoy I like your interpretation and I will be trying it. I guess that you are connecting the top and bottom as well so, topologically, you are working with a torus. \u2013\u00a0badjohn Jun 15 '17 at 12:37", "date": "2021-05-17 11:00:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2319655/when-is-a-sudoku-like-table-solvable", "openwebmath_score": 0.779970109462738, "openwebmath_perplexity": 321.9013814036455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754506337406, "lm_q2_score": 0.9032942158155877, "lm_q1q2_score": 0.8893613095914836}}
{"url": "http://www.lofoya.com/Solved/1322/a-race-course-is-400-m-long-a-and-b-run-a-race-and-a-wins-by", "text": "# Easy Time, Speed & Distance Solved QuestionAptitude Discussion\n\n Q. A race course is 400 m long. $A$ and $B$ run a race and $A$ wins by 5m. $B$ and $C$ run over the same course and $B$ win by 4m. $C$ and $D$ run over it and $D$ wins by 16m. If $A$ and $D$ run over it, then who would win and by how much?\n \u2714\u00a0A. $D$ by 7.2 m \u2716\u00a0B. $A$ by 7.2 m \u2716\u00a0C. $A$ by 8.4 m \u2716\u00a0D. $D$ by 8.4 m\n\nSolution:\nOption(A) is correct\n\nIf $A$ covers 400m, $B$ covers 395 m\n\nIf $B$ covers 400m, $C$ covers 396 m\n\nIf $D$ covers 400m, $C$ covers 384 m\n\nNow if $B$ covers 395 m, then $C$ will cover\u00a0$\\dfrac{396}{400}\\times 395=391.05$m\n\nIf $C$ covers 391.05 m, then $D$ will cover\u00a0$\\dfrac{400}{384}\\times 391.05 = 407.24$\n\nIf $A$ and $D$ run over 400 m, then $D$\u00a0win\u00a0by 7.2 m\u00a0(approx.)\n\nEdit:\u00a0For an alternative solution, check comment by\u00a0Vejayanantham TR.\n\n## (5) Comment(s)\n\nVaibhav Varish\n()\n\nSa-Speed of a in meter/minute\n\neq1---\n\n400/Sb-400/Sa=5\n\neq2---\n\n400/Sc-400/Sb=4\n\neq3--\n\n400/Sc-400/Sd=16\n\n400/Sc-400/Sa=9;\n\neq3-eq2---\n\n400/Sa-400/Sd=7\n\nthat' s why D wins by 7 min why 7.2 (Whats wrong in my solution)..pls help\n\nVejayanantham TR\n()\n\nA - 5m\n\nB - 4m\n\nD -16 m\n\nWe need to find from A -> D\n\nA->B->C = 5+4\n\nD -> 16m\n\n16-9 = 7 => D wins by 7m (approx)\n\nManohar Tangi\n()\n\nNayak Sowrabh\n()\n\nSince the number 16 is too small in comparison to 400 u get an approximate value to be 7.\n\nIf it was a 100meter race then u would have got the answer to be -> D wins the race by 8.5m.\n\nThis is because 16 makes a lot of difference to the number 100.\n\nso we cannot use this methodology in all those cases.\n\nYogesh\n()\n\ncan we apply this method on these type of question", "date": "2017-01-20 09:50:15", "meta": {"domain": "lofoya.com", "url": "http://www.lofoya.com/Solved/1322/a-race-course-is-400-m-long-a-and-b-run-a-race-and-a-wins-by", "openwebmath_score": 0.3260766267776489, "openwebmath_perplexity": 1580.9051204392665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754497285467, "lm_q2_score": 0.9032941975921684, "lm_q1q2_score": 0.889361290831496}}
{"url": "https://math.stackexchange.com/questions/2033830/can-the-area-enclosed-by-two-curves-be-infinite", "text": "Can the area enclosed by two curves be infinite?\n\nThis is a question from my test: Find the area enclosed by the graph of $y=x^3-6x^2+11x-6$ and $y=0$.\n\nThis is actually very simple but the way I look at it, I can see two different answers to this question, depending on the answer to my title.\n\nThe first is where I assume area can be infinite and use a boundary from $-\\infty$ to $\\infty$. This is what I ended up writing as my answer in the test ($x=1;2;3$; are points of intersection): $$-\\lim_{a\\to-\\infty}\\int_{a}^{1}y \\ dx + \\int_{1}^{2}y \\ dx - \\int_{2}^{3}y \\ dx+\\lim_{b\\to\\infty}\\int_{1}^{b}y \\ dx = \\infty$$\n\nThe other one is what my teacher told me the answer to the test question is, which is just: $$\\int_{1}^{2}y \\ dx - \\int_{2}^{3}y \\ dx = \\frac{1}{2}$$ Now my question is, can the area enclosed actually be infinite? I personally think an infinite area should be possible, which is why we normally use boundaries in integrals to limit the area from becoming infinite.\n\nFrom Wikipedia, it says something like $\\int_{a}^{b}f(x) \\ dx = \\infty$ means that $f(x)$ doesn't bound a finite area between $a$ and $b$, while $\\int_{-\\infty}^{\\infty}f(x) \\ dx = \\infty$ means the area under $f(x)$ is infinite.\n\nSo what do you think about this? I'd really appreciate your opinions or even facts on this matter. Can an area enclosed be infinite? And thus, would an answer of $\\infty$ be a legitimate or false answer to the test question? (I don't plan on complaining for marks, this is just for my own curiosity and self-learning).\n\nSorry for the long question, and thanks in advance!\n\n\u2022 Perhaps the question would have been better stated as \"find the area of the region bounded by\" the two graphs. A bounded region in the plane is a region which can be enclosed within a circle. \u2013\u00a0John Wayland Bales Nov 28 '16 at 4:25\n\u2022 @JohnWaylandBales Since you got here first, feel free to expand that into an answer. \u2013\u00a0Mark S. Nov 29 '16 at 1:37\n\u2022 OK, I have submitted an answer. \u2013\u00a0John Wayland Bales Nov 29 '16 at 2:46\n\nIt is a reasonable question. Although usually if asked to find the area of a region or regions bounded by two graphs what is meant by \"bounded\" is that the regions all lie within the interior of some circle.\n\nThis is analogous to a bounded set on the number line being contained in some interval $[a,b]$. It is completely circumscribed.\n\nHowever it is possible for to graphs to enclose a finite, yet unbounded region.\n\nThere are many examples, but one is as follows.\n\nFind the area of the region \"bounded\" by the graphs of $y=0$ and $y=\\dfrac{x}{x^4+1}$\n\nHere is the graph of the region.\n\nThis region is not bounded in the sense stated above. It cannot be contained in the interior of a circle. Yet it has a finite area.\n\n$$\\int_{-\\infty}^\\infty\\dfrac{|x|}{x^4+1}\\,dx=\\int_{0}^\\infty\\dfrac{2x}{x^4+1}\\,dx\\\\$$\n\nMake the substitution $u=x^2$, $du=2x\\,dx$ and this becomes\n\n\\begin{eqnarray} \\int_{0}^\\infty\\dfrac{1}{u^2+1}\\,du&=&\\frac{1}{2}\\arctan(u){\\Large\\vert}_{0}^\\infty\\\\ &=&\\left(\\dfrac{\\pi}{2}-0\\right)\\\\ &=&\\frac{\\pi}{2} \\end{eqnarray}\n\nTherefore it is acceptable to say that, in a sense, an unbounded region is \"bounded\" by two graphs so long as the area enclosed is finite.\n\n\u2022 How about when the area enclosed is infinite? I think an unbounded region with a finite area would be considered as a convergent integral, so we can calculate the finite area. But what about an unbounded region which is divergent? When asked to calculate the area, would the area be infinite, or should we only consider the finite area regions? \u2013\u00a0Gyakenji Nov 29 '16 at 2:57\n\u2022 If the area approaches infinity then we can say that the region has infinite area. In your original problem, however, it was a question of what the teacher meant by saying the region \"bounded\" by the function. I do not know the policy of your teacher, but when I was teaching I was always happy to further explain the meaning of a question if the student thought the question was unclear. \u2013\u00a0John Wayland Bales Nov 29 '16 at 3:02\n\u2022 Actually I notice that your professor said \"enclosed\" not bounded. That is actually not a mathematical term so you could certainly ask what he meant by \"enclosed.\" \u2013\u00a0John Wayland Bales Nov 29 '16 at 3:07\n\u2022 My professor doesn't speak English as a first language so perhaps it was just a mistake, but on the lectures, he is usually asking for the area between two curves (so the area bounded by the two functions?). I will ask him further about it. Sorry if I'm repeating, but assuming he meant the area bounded by the two functions, then would it be mathematically correct to say the area is infinite, given the integral will be divergent and area will approach infinite? Or would the area bounded by the functions be the finite areas only? Thanks so much in advance. \u2013\u00a0Gyakenji Nov 29 '16 at 3:17\n\u2022 The term \"enclosed\" is widely used in textbooks when talking about regions bounded by two curves, but I have never seen a textbook give an explicit definition of the term. There is a common mathematical concept of a \"closed curve\" which is a curve in the plane which begins and ends at the same point. So one could define an enclosed region as a region whose boundary consisted of one or more closed curves. With that definition, even the example I gave in my answer is not enclosed. \u2013\u00a0John Wayland Bales Nov 29 '16 at 3:18", "date": "2019-12-13 13:34:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2033830/can-the-area-enclosed-by-two-curves-be-infinite", "openwebmath_score": 0.8685263991355896, "openwebmath_perplexity": 179.13640707947349, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850832642353, "lm_q2_score": 0.9046505370289057, "lm_q1q2_score": 0.889348448520097}}
{"url": "http://yourwebexperts.com/fb2pwm/article.php?31d5b2=a-matrix-with-only-one-column-is-called", "text": "The following vector q is a 3 \u00d7 1 column vector containing numbers: $q=\\begin{bmatrix} 2\\\\ 5\\\\ 8\\end{bmatrix}$ A row vector is an 1 \u00d7 c matrix, that is, a matrix with only one row. A column vector is an r \u00d7 1 matrix, that is, a matrix with only one column. 2. Column and row vectors a matrix with one column, i.e., size n\u00d71, is called a (column) vector a matrix with one row, i.e., size 1\u00d7n, is called a rowvector \u2018vector\u2019 alone usually refers to column vector we give only one index for column & row vectors and call entries components v= 1 \u22122 3.3 0.3 w= \u22122.1 \u22123 0 4. Two matrices of the same order whose corresponding entries are equal are considered equal. A column matrix is a matrix that has only one column. Rectangular Matrix A matrix of order m x n, such that m \u2260 n, is called rectangular matrix. To understand what this number means, take each column of the matrix and draw it as a vector. One column matrix. A matrix with only one row is called a.....matrix, and a matrix with only one column is called a.....matrix. Suppose that A has more columns than rows. The matrix derived from a system of linear equations is called the..... matrix of the system. The entries are sometimes A matrix is said to be a column matrix if it has only one column. 5. A matrix with only one row is called a _____ matrix, and a matrix with only one column is called a _____ matrix. The entries of a vector are denoted with just one subscript (since the other is 1), as in a3. A matrix with only one column, i.e., with size n \u00d7 1, is called a column vector or just a vector. 3. Column Matrix A matrix having only one column and any number of rows is called column matrix. 3) Square Matrix. For example, four vectors in R 3 are automatically linearly dependent. In general, B = [b ij] m \u00d7 1 is a column matrix of order m \u00d7 1. one column (called a column vector). Converting Systems of Linear Equations to Matrices. A matrix having only one row is called a row matrix (or a row vector) and a matrix having only one column is called a column matrix (or a column vector). INCLUDES THE SOLUTIONS. Sometimes the size is speci\ufb01ed by calling it an n-vector. A column matrix has only one column but any number of rows. Below, a is a column vector while b is a row vector. Each equation in the system becomes a row. For example, $$A =\\begin{bmatrix} 0\\\\ \u221a3\\\\-1 \\\\1/2 \\end{bmatrix}$$ is a column matrix of order 4 \u00d7 1. The variables are dropped and the coefficients are placed into a matrix. Each variable in the system becomes a column. Determinants. I want the corresponding rows of column one to be printed as follows : a = 3 2 1 3 2 5 4 8 5 9 I tried sort(a), but it is sorting only the second column of matrix a. row column. Fill in the blanks. A matrix with only one row or one column is called a vector. a = 7 2 3 , b = (\u2212 2 7 4) A scalar is a matrix with only one row and one column. I have the matrix as follows. Example: D is a column matrix of order 2 \u00d7 1 A zero matrix or a null matrix is a matrix that has all its elements zero. augmented. The determinant takes a square matrix and calculates a simple number, a scalar. A wide matrix (a matrix with more columns than rows) has linearly dependent columns. Example: C is a column matrix of order 1 \u00d7 1 A column matrix of order 2 \u00d71 is also called a vector matrix. Then A cannot have a pivot in every column (it has at most one pivot per row), so its columns are automatically linearly dependent. a = 1 3 2 5 3 2 4 8 5 9 I want to sort the second column in the a matrix. A vector is almost often denoted by a single lowercase letter in boldface type. A matrix, that has many rows, but only one column, is called a column vector. 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What this number means, take each column of the matrix derived from a system linear. 2 5 3 2 4 8 5 9 I want to sort the second column in a! Single lowercase letter in boldface type with only one row or one column but any number of is... That has many rows, but only one column but any number of rows less! Dependent columns rows ) has linearly dependent second column in the a matrix with only one row is the... More columns than rows ) has linearly dependent columns matrix is a column matrix has only column! Rectangular matrix a matrix in which the number of rows, with n... Of a vector are denoted with just one subscript ( since the other is ).\nBroiled Whole Tilapia, Does Domino's Have Garlic Parmesan Wings, 1st Year Physics Text Book, Linford Christie Net Worth 2019, 1 Litre Of Almond Milk A Day, Music Business Books 2020, John C Maxwell Books, How To Count Google Forms, Iit Bombay M Tech Metallurgy Placement, How To Draw Floral Diagram Of Datura Metel, Best Butterfly Kit, Walabot For Iphone, Protests Near Me This Weekend,", "date": "2021-04-23 03:17:12", "meta": {"domain": "yourwebexperts.com", "url": "http://yourwebexperts.com/fb2pwm/article.php?31d5b2=a-matrix-with-only-one-column-is-called", "openwebmath_score": 0.848754346370697, "openwebmath_perplexity": 363.4131443754266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9875683507933022, "lm_q2_score": 0.9005297847831081, "lm_q1q2_score": 0.8893347143985014}}
{"url": "https://math.stackexchange.com/questions/4131809/difference-between-quotient-rule-and-product-rule", "text": "# difference between quotient rule and product rule\n\nProduct rule :\n\n$$\\frac{d}{dx} \\big(f(x)\\cdot g(x)\\big)=f'(x)\\cdot g(x)+f(x)\\cdot g' (x)$$\n\nQuotient rule :\n\n$$\\frac{d}{dx} \\frac{f(x)}{g(x)}=\\frac{f'(x)\\cdot g(x)-f(x)\\cdot g'(x)}{[g(x)]^2}$$\n\nSuppose, the following is given in question. $$y=\\frac{2x^3+4x^2+2}{3x^2+2x^3}$$\n\nSimply, this is looking like Quotient rule. But, if I follow arrange the equation following way\n\n$$y=(2x^3+4x^2+2)(3x^2+2x^3)^{-1}$$\n\nThen, we can solve it using Product rule. As I was solving earlier problems in a pdf book using Product rule. I think both answers are correct. But, my question is, How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\n\u2022 Your just deriving the quotient rule on the fly, rather than assuming it is true and then applying it directly; there is nothing wrong with doing the former. May 8 at 13:39\n\u2022 First thing a mathematician would do is write $$y=1+\\frac{x^2+2}{2x^3+3x^2}$$ before taking the derivative. May 8 at 16:14\n\u2022 What do you mean \"I think both answers are correct.\"? If you do the math properly, you'll get the same answer with both methods. May 9 at 0:10\n\u2022 Real physicists and mathematicians stick the whole thing into their computer algebra system and don't worry about the precise algorithms it follows, unless they have a particular reason to risk silly errors by doing that kind of calculation by hand. Symbolic differentiation is a solved problem; one doesn't earn any \"purity points\" in the real world by doing it the hard way. May 9 at 0:33\n\u2022 @JosephSible-ReinstateMonica Yes! I got same answer :)\n\u2013\u00a0user876873\nMay 9 at 4:30\n\nNote that $$((g(x))^{-1})'=-g'(x)(g(x))^{-2}$$. Then, applying the product rule: $$\\left(\\frac{f(x)}{g(x)}\\right)'=\\left(f(x)\\cdot \\frac{1}{g(x)}\\right)'=\\frac{f'(x)}{g(x)}+\\frac{-g'(x)f(x)}{(g(x))^2}=\\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$ which is the quotient rule\n\n\u2022 That's very helpful. But, it was my main question How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\u2013\u00a0user876873\nMay 8 at 13:44\n\u2022 @Istiak Please don't keep making that bold. Sure, it's OK, because any valid proof is OK; there are no \"rules\" beyond that. To be honest, I hardly ever see mathematicians or physicists explicitly using the quotient rule. It seems to be more for teaching than anything else.\n\u2013\u00a0J.G.\nMay 8 at 13:45\n\u2022 I'd always use the quotient rule on a quotient because it is usually much simpler to work with $g'$ than $(\\frac1g)'$ May 8 at 13:45\n\u2022 @J.G. Actually, I had read in a meta post that is saying,you should bold when it may need to be attracted or, something just like this. I don't remember. That's why I was just making that bold\n\u2013\u00a0user876873\nMay 8 at 13:48\n\u2022 @Istiak I see, cool. Thanks, though, for your edit that means you now only bold it in the question.\n\u2013\u00a0J.G.\nMay 8 at 13:52\n\nHow does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\nis that any experienced scientist knows several methods to solve problems and uses those that are most convenient for them at that particular time.\n\nI would look at that derivative and use the quotient rule. But if there was something in the source of the problem that suggested that it made more sense to write the denominator as $$(3x^2+2x^3)^{-1}$$ then the product rule would be more appropriate.\n\nthe quotient rule is not a separate and non-compatible rule. It is just the product rule inserting $$g^{-1}$$ instead of $$g$$.\n\nlet's see...\n\nto make it clear... I show you, prove it for you, how quotient rule is just compatible with product law.\n\n$$\\left(\\frac f h\\right)'=\\left(f\\cdot \\frac 1 h\\right)'=f' \\cdot \\frac 1 h +f\\cdot \\left (\\frac 1 h\\right)'$$\n\nBefore continuing I want to recall that, Since $$h\\cdot (1/h)=1$$, by product rule, we ould get $$0=1'=(h\\cdot \\frac 1 h )'=h' \\cdot \\frac 1 h + h\\cdot (1/ h)'$$. Therefore we have $$h\\cdot ( 1/ h )'=-h'\\cdot ( 1/ h)$$.\n\nSo, $$\\left(1 /h\\right)'=-h'/h^2$$.\n\nand on another hand, we know $$\\frac 1 h$$, could be written as $$\\frac h{h^2}$$, for a reason.\n\nso,\n\n$$\\left(\\frac f h\\right)'=f' \\cdot \\frac 1 h +f\\cdot \\left (\\frac 1 h\\right)'=f'\\cdot \\frac h {h^2}+f\\cdot \\frac{-h'}{h^2}=\\frac{f'\\cdot h-f\\cdot h'}{h^2}$$\n\nthat was the proof for the quotient rule.\n\n\u2022 $g^{-1}$ is a poor choice of notation here, as it usually means the inverse of $g$, and not the reciprocal. May 9 at 0:19", "date": "2021-10-26 23:36:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4131809/difference-between-quotient-rule-and-product-rule", "openwebmath_score": 0.8917396664619446, "openwebmath_perplexity": 532.4571520060127, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975946441508702, "lm_q2_score": 0.9111797045849583, "lm_q1q2_score": 0.8892625902646404}}
{"url": "http://mathhelpforum.com/geometry/190201-problem-finding-area.html", "text": "# Math Help - Problem finding area\n\n1. ## Problem finding area\n\ni am having a problem finding the area of the three triangles outside the Pitagoras drawing\nyou can see what i am meaning in this picture\nhttp://i51.tinypic.com/28gxmyd.jpg\nso i need the area of all the triangles in the middle of the squares\n\n2. ## Re: Problem finding area\n\nHello, leart369!\n\nhttp://i51.tinypic.com/28gxmyd.jpg\ni need the area of all the triangles.\n\nLabel the inner right triangle like this:\n\nCode:\n      A\n*\n|   *     c\nb |       *\n|           *\n* - - - - - - - * B\nC         a\n\nNote that for any triangle, its area is: . $A \\;=\\;\\tfrac{1}{2}ab\\sin C$\n. . (One-half the product of two sides and the sine of the included angle)\n\nThe right triangle at the lower-left has sides $a$ and $b$.\n. . Its area is: . $\\boxed{A_1 \\:=\\:\\tfrac{1}{2}ab}$\n\nThe triangle at the right has sides $a$ and $c$, and included angle $(180^o - B).$\n. . Its area is: . $A_2 \\:=\\:\\tfrac{1}{2}ac\\sin(180^o-B) \\:=\\:\\tfrac{1}{2}ac\\sin B$\nIn the above diagram, we see that: . $\\sin B \\,=\\,\\tfrac{b}{c}$\n. . Hence: . $A_2 \\;=\\;\\tfrac{1}{2}ac\\left(\\tfrac{b}{c}\\right) \\quad\\Rightarrow\\quad \\boxed{A_2 \\:=\\:\\tfrac{1}{2}ab}$\n\nThe third triangle has sides $b$ and $c$, and included angle $(180^o - A).$\n. . Its area is: . $A_3 \\:=\\:\\tfrac{1}{2}bc\\sin(180^o-A) \\:=\\:\\tfrac{1}{2}bc\\sin A$\nIn the above diagram, we see that: . $\\sin A \\,=\\,\\tfrac{a}{c}$\n. . Hence: . $A_3 \\;=\\;\\tfrac{1}{2}bc\\left(\\tfrac{a}{c}\\right) \\quad\\Rightarrow\\quad \\boxed{A_3 \\:=\\:\\tfrac{1}{2}ab}$\n\nFascinating! . . . All four triangles have the same area.\n\n3. ## Re: Problem finding area\n\ndoes exist any way to find the area off all that drawing?\n\n4. ## Re: Problem finding area\n\nOriginally Posted by leart369\ndoes exist any way to find the area off all that drawing?\n1. I assume that you mean the area of the hexagon, containing 3 squares, 2 right triangles and 2 obtuse triangles. If so:\n\n2. As Soroban has shown you have:\n\n$\\text{complete area} = a^2+b^2+c^2+4 \\cdot \\frac12 \\cdot a \\cdot b = c^2 +(a+b)^2$\n\n3. Since $a^2+b^2 = c^2$ there are other possibilities to simplify the sum of areas.\n\n5. ## Re: Problem finding area\n\nthank you very much that was what i was looking for", "date": "2015-08-28 01:45:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/190201-problem-finding-area.html", "openwebmath_score": 0.9228480458259583, "openwebmath_perplexity": 978.2625739082216, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9905874093008835, "lm_q2_score": 0.8976952921073469, "lm_q1q2_score": 0.8892456537502166}}
{"url": "https://www.onlycode.in/pascals-triangle/", "text": "# Pascal\u2019s Triangle\n\nIn mathematics, Pascal\u2019s triangle is a triangular array of the binomial coefficients. It can also be viewed as: each number in Pascal\u2019s triangle is the sum of the two numbers directly above it as shown:\n\n          1\n1 1\n1 2 1\n1 3 3 1\n1 4 6 4 1\n1 5 10 10 5 1 \n\nProblem statement\n\nGiven an integer n as input, print first n lines of the Pascal\u2019s triangle.\n\n#### Approach\n\nLet's consider Pascal's triangle as a matrix pascal[i][j] in the following way:\n\n1\n1 1\n1 2 1\n1 3 3 1\n1 4 6 4 1 \n\nWe could solve this problem using Dynamic Programming.\n\nOptimal substructure\n\nFrom the above matrix we could see that:\n\npascal[i][j] is the sum of previous element in the row i-1 and element just above the current element in the row i-1. Consider out of bound indices as 0\nie pascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j]\n\n#### Code Implementation\n\n//\n//  main.cpp\n//  Pascal Triangle\n//\n//  Created by Himanshu on 20/09/21.\n//\n\n#include <iostream>\nusing namespace std;\n\nvoid printPascalTriangle (int n) {\nint pascal[n+1][n+1];\n\n//Base case\npascal[1][1] = 1;\n\nfor (int i=0; i<=n; i++) {\nfor (int j=0; j<=n; j++) {\npascal[i][j] = 0;\n}\n}\n\nfor (int i=1; i<=n; i++) {\nfor (int j=1; j<=i; j++) {\n// first and last binomial coefficients are\n// always 1\nif (i == 1 || j == i) {\npascal[i][j] = 1;\n} else {\npascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j];\n}\n}\n}\n\nfor (int i=1; i<=n; i++) {\nfor (int j=1; j<=i; j++) {\ncout<<pascal[i][j]<<\" \";\n}\ncout<<endl;\n}\n\n}\n\nint main() {\nint n = 7;\nprintPascalTriangle (n);\n\nreturn 0;\n}\n\n\n\nOutput\n\n1\n1 1\n1 2 1\n1 3 3 1\n1 4 6 4 1\n1 5 10 10 5 1\n1 6 15 20 15 6 1 \n\nTime Complexity: O(n^2)\nAuxiliary Space: O(n^2)\n\n#### Approach with O(1) auxiliary space\n\nBy using the definition of binomial coefficient: we know that\n\nC(n, i) = (n!)/((n-i)! * i!)\nsimilarly,\nC(n, i-1) = (n!)/((n-i+1)! * (i-1)!)\n\nNow, C(n, i) could also be written as:\n(n!)*(n-i+1)/(n-i+1)!*(i)*(i-1!)\nwhich is nothing but\n\n(n!)*(n-i+1)/((n-i+1)!*(i-1)!*(i)) ie,\nC(n, i) =  C(n, i-1)*(n-(i-1))/(i)\n\nUsing the above equation, we just need the previous binomial coefficient C(n, i-1) to calculate C(n, i). Hence we need not save all the binomial coefficients to print current coefficient.\n\n#### Code Implementation\n\n//\n//  main.cpp\n//  Pascal Triangle 2\n//\n//  Created by Himanshu on 20/09/21.\n//\n\n#include <iostream>\nusing namespace std;\n\nvoid printPascalTriangle (int n) {\n\nfor (int i=1; i<=n; i++) {\nint C = 1;\nfor (int j=1; j<=i; j++) {\n\ncout<<C<<\" \";\n// We are using (i-j) instead of (i-(j-1)) because\n// calculation is for next or (j+1)th element\nC = C * (i-j)/ j;\n}\ncout<<endl;\n}\n\n}\n\nint main() {\nint n = 7;\nprintPascalTriangle (n);\nreturn 0;\n}\n\n\n\nOutput\n\n1\n1 1\n1 2 1\n1 3 3 1\n1 4 6 4 1\n1 5 10 10 5 1\n1 6 15 20 15 6 1 \n\nTime Complexity: O(n^2)\nAuxiliary Space: O(1)\n\nHere\u2019s a working example: Pascal\u2019s Triangle\n\nPractice Problem\nPascal\u2019s Triangle [LeetCode]", "date": "2022-11-26 15:51:47", "meta": {"domain": "onlycode.in", "url": "https://www.onlycode.in/pascals-triangle/", "openwebmath_score": 0.27169421315193176, "openwebmath_perplexity": 2414.352174655299, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985936370890583, "lm_q2_score": 0.9019206824612297, "lm_q1q2_score": 0.8892364044969826}}
{"url": "https://brilliant.org/discussions/thread/approximating-log_107/", "text": "# Approximating $\\log_{10}7$\n\nThe Moderators have been talking about methods to approximate logarithms given only a little bit of information; namely, $\\log_{10}2\\approx0.301$ and $\\log_{10}3\\approx0.477.$ I'm going to go through a couple methods to provide estimates.\n\nIt should be noted that the true value of $\\log_{10}7$ is $0.845098...$\n\nLet's start off on a large scale. Obviously, $\\log_{10}6<\\log_{10}7<\\log_{10}8.$ To find $\\log_{10}6$ and $\\log_{10}8,$ we use the given information. $\\log_{10}6=\\log_{10}2+\\log_{10}3\\approx0.301+0.477=0.778.$ Also, $\\log_{10}8=3\\log_{10}2\\approx3\\times0.301=0.903.$ So $0.778<\\log_{10}7<0.903.$ Not a very good estimate. Even if we take the average of these, we get $0.8405,$ which is only accurate to two decimal places.\n\nLet's be a little more creative in our thinking. The set of numbers $\\{48,49,50\\}$ includes numbers whose logarithms can be expressed with our given information and a power of $7.$ Once again, $\\log_{10}48<\\log_{10}49\\log_{10}50\\Rightarrow$ $\\log_{10}48<2\\log_{10}7<\\log_{10}50.$ The only difference here is that $50$ has a power of $5$ in its factorization, but $\\log_{10}5$ can easily be eliminated by noticing that $\\log_{10}50=\\log_{10}\\frac{100}{2}=\\log_{10}100-\\log_{10}2=2-\\log_{10}2.$\n\n$\\log_{10}48=4\\log_{10}2+\\log_{10}3\\approx1.681.$ Also, $\\log_{10}50=2-\\log_{10}2\\approx1.699.$ Dividing by $2,$ we find that $0.8405<\\log_{10}7<0.8495.$ This is better. Not to mention, taking the average of these values yields $\\log_{10}7\\approx0.8450,$ accurate to $4$ decimal places.\n\nWhat method would you use to calculate $\\log_{10}7?$ Please share what you think!\n\nNote by Trevor B.\n5\u00a0years, 1\u00a0month ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\n\\begin{aligned} 2400 & \\approx & 2401 \\\\ \\log 2400 & \\approx & \\log 2401 \\\\ \\log (2^3 \\times 3 \\times 10^2) & \\approx & \\log 7^4 \\\\ \\log 2^3 + \\log 3 + \\log 10^2 & \\approx & 4 \\log 7 \\\\ 3 \\log 2 + \\log 3 + 2 & \\approx & 4 \\log 7 \\\\ \\log7 & \\approx & 0.8450 \\\\ \\end{aligned}\n\n- 5\u00a0years, 1\u00a0month ago\n\nSo, that begs the question of\n1. How do I make such an amazing observation that $2400 \\approx 2401$?\n2. How do I know what the next approximation should be?\n\nStaff - 5\u00a0years, 1\u00a0month ago\n\n1. I just found numbers that are close to powers of $7$ that are can be factored in terms of $2,3,5,7$\n\n2. I don't have a solid answer, but I'm thinking we should find larger pairs of numbers such that their percentage difference is as small as the previous, like: $6^5 \\approx 7777 = \\frac {7}{9}(10^4 - 1) \\approx \\frac {7}{9} \\times 10^4$ or $6^7 \\approx 280000 = 10^4 \\times 2^2 \\times 7$, but both methods yields $0.844$ as the approximation, so I'm kinda stumped right now.\n\nIn the meantime, lemme peruse your other comment. =)\n\n- 5\u00a0years, 1\u00a0month ago\n\nThere's a 0.04% error in approximating 2401 as 2400, but works out quite well :) +1\n\n- 5\u00a0years, 1\u00a0month ago\n\n@Calvin Lin and @Daniel Liu were the moderators discussing this in our messageboard.\n\n- 5\u00a0years, 1\u00a0month ago\n\nCool!! So you guys have a message board of your own?\n\n- 5\u00a0years, 1\u00a0month ago\n\nThe general idea is that if you can bound $a < 7 ^ n < b$, then you should also look at how $ab$ compares to $7^{2n}$. This is akin to the Root approximation - bisection method.\n\nFor example, from above, we have $6 < 7 < 8$, which tells us we should compare $48$ with $49$, and work with $48 < 49 < 8^2$.\n\nExplicitly, we can use the following series of inequalities:\nStep 1: $6 < 7 < 8$. Comparing $6 \\times 8$ with $7^2$, we have $48 < 7^2$. Hence, this leads to:\nStep 2: $48 < 7^2 < 8^2 = 64$. Comparing $48 \\times 64$ with $7^4$, we have $7^4 < 3072$. Hence, this leads to:\nStep 3: $2304 = 48^2 < 7^4 < 3072$. Comparing $2304 \\times 3072$ with $7 ^ 8$, we have $7^ 8 < 7077888$. Hence this leads to:\nStep 4: $5308416 = 2304^2 < 7^8 < 7077888$.\n\nWe can continue this process indefinitely to bound $\\log 7$ as tightly as we want to. At step 1, we have $0.778 < \\log 7 < 0.903$, and at step 4 we have $0. 840 < \\log 7 < 0.856$.\n\nQuestion: Why did I say that this is \"akin to the Root approximation - bisection method\"?\n(see comments below for the explanation)\n\nNote: As it turns out, we could make the observation that 50 a much better bound. With $48 < 49 < 50$, and so we should compare $48 \\times 50$ with $49^2$.\n\nStaff - 5\u00a0years, 1\u00a0month ago\n\nBecause you want to find values $a, b$ such that the inequality is satisfied $a < 7^n < b$\n\nTaking log to both sides $\\log a < n \\log 7 < \\log b$\n\nBisection method implies we must minimize the difference between $\\frac { \\log a + \\log b }{2}$ and $n \\log 7$.\n\nSet both of them to be equal implies we want to compare $ab$ to $7^{2n}$\n\n- 5\u00a0years, 1\u00a0month ago\n\nYes. We are performing the bisection method on $\\log a < n \\log 7 < \\log b$. The midpoint of this interval is $\\frac{ \\log a + \\log b } { 2 } = \\log \\sqrt{ ab }$. We then compare this to $n \\log 7$, which is in essence comparing $ab$ to $7^{2n}$.\n\nIn this way, we don't have to think about what the next number is, or figure out a fancy combination to bound it tightly. We just have to do the bisection method many many times (or automate it with a program), to get the interval down to as small as we want it to.\n\nStaff - 5\u00a0years, 1\u00a0month ago\n\n\"It is independent of the number of sig figs / decimal places that we started off with.\"\n\nI was under the impression, you can only find up to the number of SF given. But this just blew my mind. It's like saying:\n\n\"Hey, my calculator is kinda bad, it only let shows that $\\log 2 = 0.3$ and $\\log 13 = 1.1$ to one decimal place, can you help me find the value of $\\log 11$ to a gazillion decimal places?\"\n\nThanks, I learned something new today!!\n\n- 5\u00a0years, 1\u00a0month ago\n\nIndeed. See my conversation with Kenny Lau below. It is a common misconception (likely from Physics error analysis) that your calculations cannot be more accurate than the degree of significance that you started out with.\n\nFor example, even if your calculator can only multiply whole numbers, and you want to approximate $\\sqrt{2}$, you can do much better than $1 < \\sqrt{2} < 2$. For example, by showing that $1414^2 < 2000000 < 1415^2$, we can conclude that $1.414 < \\sqrt{2} < 1.415$. We can get $\\sqrt{2}$ to any degree of accuracy (with sufficient patience / computing power).\n\nStaff - 5\u00a0years, 1\u00a0month ago\n\n543543\n\n- 5\u00a0years, 1\u00a0month ago\n\nWell, any method based on the two given information can only be correct to 3 significant figures (or decimal places), because the information itself is correct to only 3 significant figures.\n\n- 5\u00a0years, 1\u00a0month ago\n\nNot necessarily. For example, if we know $\\log 7 ^ {10}$ to 3 decimal places, then we can get $\\log 7$ to 4 decimal places.\n\nStaff - 5\u00a0years, 1\u00a0month ago\n\nIt would not be accurate to 3 decimal places. Log 7^10 would only be accurate to 3 significant figures which is 2 decimal places.\n\n- 5\u00a0years, 1\u00a0month ago\n\nI've added a way to show how we can get an arbitrarily tight bound for $\\log 7$. It is independent of the number of sig figs / decimal places that we started off with.\n\nStaff - 5\u00a0years, 1\u00a0month ago", "date": "2020-03-30 23:18:48", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/approximating-log_107/", "openwebmath_score": 0.9863712787628174, "openwebmath_perplexity": 565.3147384438698, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407175907055, "lm_q2_score": 0.9136765287024915, "lm_q1q2_score": 0.8892272004401977}}
{"url": "https://math.stackexchange.com/questions/2429872/a-lily-pad-doubles-in-area-every-second-after-one-minute-it-fills-the-pond-ho/2429893", "text": "# A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond ?\n\nA lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond?\n\nTo me this seems like we can set up a fraction-like equation:\n\n$$\\frac{60 \\ \\text{seconds}}{1} = \\frac{x \\ \\text{seconds}}{1/4}$$ then $x = 15$ seconds. But the answer is $58$ seconds which really makes no sense to me. Any suggestions are greatly appreciated.\n\n\u2022 Comments are not for extended discussion; this conversation has been moved to chat. \u2013\u00a0Jyrki Lahtonen Sep 18 '17 at 17:31\n\nI think it's easiest to work backwards: if the area doubles every second and the pond is totally covered at time $t=60$, then it must be half covered at $t=59$, and therefore one quarter covered at $t=58$.\n\nAlternately, let $f(t)$ be the fraction of the pond's area covered at time $t\\leq 60$. Then $f(t)=f(0)2^t$ since the area doubles every second, and since $f(60)=1$ we get $f(0)=2^{-60}$. Therefore $f(t)=2^{-60}2^t=2^{t-60}$. Then setting $2^{t-60}=\\frac{1}{4}$ and solving for $t$ yields $t=58$.\n\n\u2022 shouldnt $f(0)=2^{t-60}$. I think you may be missing a t \u2013\u00a0Derek Sep 16 '17 at 3:03\n\u2022 @Derek: No, $f(0)=2^{-60}$. $f(0)$ is a number rather than a function of $t$. \u2013\u00a0carmichael561 Sep 16 '17 at 3:06\n\u2022 @Derek $f(0)=2^{t-60} \\text{ at } t=0$ so then $f(0)=2^{0-60}=2^{-60}$ \u2013\u00a0Wolfie Sep 18 '17 at 6:40\n\nForget formulas for this one!\n\nIf going forward 1 second the area gets doubled, then going back 1 second the area gets halved.\n\nSo, 1 second before the pond was filled the pond must have been half filled, and 1 second before that it must have been quarter filled.\n\n\u2022 Yes. Use formulas if they help you to get the answer. If you can get the answer easily without formulas, great! \u2013\u00a0Wildcard Sep 15 '17 at 1:25\n\u2022 And 5s before that, we can hear economists say : \"See, an exponential growth is perfectly possible in a finite world!\". \u2013\u00a0Eric Duminil Sep 16 '17 at 8:21\n\u2022 If there was one complaint about my pre-college math education it's that there was such an emphasis on formulas (which was great when it came to calculus, for sure) that there was so little critical thinking. It was \"which of the four formulas of this chapter are we supposed to apply to this problem?\" instead of thinking about the problem first. \u2013\u00a0corsiKa Sep 16 '17 at 15:18\n\nThis is exponential rather than linear. If $A$ is the initially covered area, then after one second the covered area will be $2A$, after two second $2\\cdot 2A=4A$, after three seconds $2\\cdot 4A=8A$. And so on: after $t$ seconds the covered area will be $2^tA$.\n\nAfter $60$ seconds it will be $2^{60}A$, by assumption this is the whole pond. A quarter of this is $$\\frac{2^{60}A}{4}=2^{58}A$$\n\nOf course Carmichael\u2019s answer is slicker.\n\n\u2022 While other answers just solve the problem, your also points (although briefly) at the flaw in OP's reasoning. Thanks for that. \u2013\u00a0pajonk Sep 15 '17 at 5:51\n\nYour 'fraction-like equation' has nothing to do with the problem, because the lily pad doubles every second \u2013 its growth is exponential, not linear.\n\n$$Area(t) = 2\\cdot Area(t-1)$$ where $t$ is a number of seconds since 'some moment', hence $$Area(t) = \\color{red}{2^t}\\cdot Area(0)$$ where zero is an arbitrary 'some moment'.\n\nThat implies $$\\frac{Area(60)}{Area(t)} = \\frac{2^{60}}{2^t} = 2^{60-t}$$\n\nThen if they ask at what $t$ is $$Area(t)=1/4\\cdot Area(60)$$ you have $$2^{60-t} = \\frac{Area(60)}{1/4\\cdot Area(60)} = 4=2^2$$ so $$60-t = 2$$ and finally $$t=60-2 = 58$$ \u2013 the pond is quater-filled at $58$ seconds.\n\nTake a look at the table below: \\begin{array}{|c|c|}\\hline \\text{Area } (A) & 1 & \\dfrac12 & \\dfrac14 & \\dfrac18 & \\cdots\\\\ \\hline \\text{Time } (t) & 60 & 59 & 58 & 57 & \\cdots\\\\ \\hline \\end{array} As you might see, you could easily deduce the relationship between $A$ and $t$, i.e., $A(t)=\\dfrac{1}{2^{60-t}}$.\n\nChronological reasoning might help.\n\n\u2022 At the beginning or $t=0$, its area is $A(0)$.\n\u2022 After 1 second elapses or $t=1$, its area becomes $A(1)=2\\times A(0)$.\n\u2022 The next 1 second or $t=2$, its area becomes $A(2)=2 \\times A(1) = 2\\times2\\times A(0)=2^2 A(0)$.\n\n\u2022 At $t=60$ the area is $A(60)=2^{60} A(0)$. The pond is fully filled.\n\n\u2022 The quarter of the fully filled pond is $A(60)/4 = 2^{60}A(0)/4=2^{58}A(0)$. This area is equal to $A(58)$, so $t=58$.\n\nWe want a function relating the current time to the size of the lily pad; let's call it $f$. From the problem, we know that\n\n$f(60\\ \\mathrm{seconds})=1\\ \\mathrm{pond}$\n$f(x\\ \\mathrm{seconds})=1/4\\ \\mathrm{pond}$\n\nso we can indeed write an equation like the one you want by solving for $1\\ \\mathrm{pond}$ in each equation; then:\n\n$\\frac{f(60\\ \\mathrm{seconds})}1=\\frac{f(x\\ \\mathrm{seconds})}{1/4}$\n\nBut note the mediating $f$ that you are missing in your equation! If we assumed $f(x)=x$, then we would get your equation; but the problem also tells us that the lily pad doubles in size every second, that is, that:\n\n$f((x+1)\\ \\mathrm{seconds}) = 2f(x\\ \\mathrm{seconds})$\n\nIf we choose $f(x)=x$, then this equality is not validated, since $(x+1)\\ \\mathrm{seconds}=2x\\ \\mathrm{seconds}$ is not validated.\n\nLuckily we can make progress even without assuming $f(x)=x$. Simplifying the corrected version of your equation, we have:\n\n$f(60\\ \\mathrm{seconds})=4f(x\\ \\mathrm{seconds})$\n\nNow we can apply the other equation given in the problem twice:\n\n$f(60\\ \\mathrm{seconds})=4f(x\\ \\mathrm{seconds})$\n$\\phantom{f(60\\ \\mathrm{seconds})}=2(2f(x\\ \\mathrm{seconds}))$\n$\\phantom{f(60\\ \\mathrm{seconds})}=2f((x+1)\\ \\mathrm{seconds})$\n$\\phantom{f(60\\ \\mathrm{seconds})}=f((x+2)\\ \\mathrm{seconds})$\n\nThen we can conclude that $60=x+2$ would be sufficient to validate this equation, so $x=58$ is one possible solution.\n\nThe area of a lily pod can be described by a function:\n\n$$f:[0,60]\\to[0,1]\\\\ f(x)=2^{x-60}$$\n\nNow we have to compute the argument $x$ for which the value of the function $f(x)$ is equal to $\\frac{1}{4}$:\n\n$$\\frac{1}{4}=2^{x-60}\\\\ \\log_2 \\frac{1}{4}=\\log_2 2^{x-60}\\\\ -2 = x-60\\\\ x=58$$\n\n## protected by Community\u2666Oct 8 '17 at 1:34\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).", "date": "2019-05-19 08:42:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2429872/a-lily-pad-doubles-in-area-every-second-after-one-minute-it-fills-the-pond-ho/2429893", "openwebmath_score": 0.9431003332138062, "openwebmath_perplexity": 498.64105919092844, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981453437115321, "lm_q2_score": 0.9059898134030163, "lm_q1q2_score": 0.8891868163558586}}
{"url": "https://apimemphis.com/blog/archive.php?id=4ec484-identity-matrix-symbol-latex", "text": "L a T e X allows two writing modes for mathematical expressions: the inline mode and the display mode. Pour ajouter un passage \u00c3\u00a0 la ligne, ajoutez deux espaces \u00c3\u00a0 l'endroit o\u00c3\u00b9 vous souhaitez que la ligne commence. (Notice that, in the third example, I use the tilde character for a forced space. In this chapter, we will typically assume that our matrices contain only numbers. Example Here is a matrix of size 2 3 (\u00e2\u0080\u009c2 by 3\u00e2\u0080\u009d), because it has 2 rows and 3 columns: 10 2 015 The matrix consists of 6 entries or elements. For example, suppose that we wish to typeset the following passage: This passage is produced by the following input: Yes. I am wondering is there any notation else for identity matrix? In linear algebra, the identity matrix (sometimes ambiguously called a unit matrix) of size n is the n \u00d7 n square matrix with ones on the main diagonal and zeros elsewhere. Bar. How can I write an identity matrix symbol ( a 1 in double struck letter form )? Each row of a matrix ends with two backslashes (\\\\\\\\). here\u2019s how to save yourself the time: assume you have some matrix L >> s = sym(L); >> v = vpa(s,5); # assign numerical precision >> latex(v) matlab should now spit out the latex source code that you can directly copy into your. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A matrix having $$n$$ rows and $$m$$ columns is a $$m\\times n$$-matrix. This is because LaTeX typesets maths notation differently from normal text. Do it while you can or \u201cStrike while the iron is hot\u201d in French. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. To learn more, see our tips on writing great answers. (Notice that, in the third example, I use the tilde character for a forced space. Associative + Identity + Inverses = Group Definition 1.6 (Group). matrice \u00d72 The ctan alreadyprovides a huge list with currently 5913 symbols, which you can download here. Creative Commons Partage dans les m\u00c3\u00aames conditions 3.0 France, La traduction fran\u00c3\u00a7aise de la doc de Koma-Script. Title Edited. It is denoted by $I_n$, or simply by $I$ if the size is immaterial or can be trivially determined by the context. Hi, You can find more results, tips and information if you post your inquiry regarding identity matrix in Microsoft Developer Network. site design / logo \u00a9 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. matrix \u00d72 Latex plus or minus symbol; Latex symbol for all x; Latex symbol exists; Latex symbol not exists; How to write matrices in Latex ? Query to update one column of a table based on a column of a different table. In this video, Vince shows how to quickly write out matrices in LaTeX, using the amsmath package and the \\pmatrix (for a matrix with curly brackets), \\matrix (for a matrix with no brackets), and \\vmatrix (used to denote the determinant of a matrix) commands. 836\u25cf3\u25cf14 From what I understand, as of MATLAB R2016a, equations in LaTeX only supp It may reflect varying levels of consensus and vetting. I. The square root symbol is written using the command \\sqrt{expression}. An online LaTeX editor that's easy to use. The matrix environments are matrix, bmatrix, Bmatrix, pmatrix, vmatrix, Vmatrix, and smallmatrix. Do not use symbols like \"*\"! There are no approved revisions of this page, so it may not have been reviewed. An online LaTeX editor that's easy to use. Last visit was: Sat Nov 28, 2020 7:17 am. Wikipedia:LaTeX symbols. Compute the left eigenvectors of a matrix. More rarely now, but at some point $E$ was used for the identity. In algebra, constants are symbols used to denote key mathematical elements and sets. Parfois c'est m\u00eame un I double, tr\u00e8s joli. Don't hesitate to update us if you need further assistance. All answers are correct, but people should demonstrate using Quora's LaTeX: [code=latex]\\equiv[/code] produces $\\equiv$ Which symbol can be used to refer to identity matrix? \u201cQuestion closed\u201d notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC\u2026. Best way to let people know you aren't dead, just taking pictures? J'ai essay\u00e9 avec un I majuscule en italiques, mais c'est poche. Parfois c'est m\u00c3\u00aame un I double, tr\u00c3\u00a8s joli. Don't hesitate to update us if you need further assistance. mrdivide.\\ Element-wise left division. A lgebra is a subfield of mathematics pertaining to the manipulation of symbols and their governing rules. Math into LaTeX : an introduction to LaTeX and AMS-LaTeX / George Gr\u00a8atzer p. cm. (In some fields, such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to I.) Math symbols de\ufb01ned by LaTeX package \u00abamsfonts\u00bb No. Math symbols de\u00ef\u00ac\u0081ned by LaTeX package \u00abamsfonts\u00bb No. Jean-Michel dx automatic spacing \u00e2\u0080\u00a6 Some of these symbols are primarily for use in text; most of them are mathematical symbols and can only be used in LaTeX's math mode. The individual items (numbers, symbols or expressions) in a matrix are called its elements or entries. def prepare_channel_operator_list(ops_list): \"\"\" Prepares a list of channel operators. Sometimes you can use the symbol $$\\times$$. 2. In some contexts, I have also seen a doublestrike $1$, similar to the difference between $N$ and $\\mathbb N$, in order to emphasize that it is the compositional identity. def prepare_channel_operator_list(ops_list): \"\"\" Prepares a list of channel operators. identit\u00c3\u00a9 \u00d71, Derni\u00c3\u00a8re mise \u00c3\u00a0 jour : 30 Nov '17, 12:39. Create a matrix, write it to a comma-separated text file, and then write the matrix to another text file with a different delimiter character. The Markdown parser included in the Jupyter Notebook is MathJax-aware. Foo If you think I forgot some very important basic function or symbol here, please let me know. Is every face exposed if all extreme points are exposed? All common mathematical symbols are implemented, and you can find a listing on the LaTeX cheat sheet. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. All answers are correct, but people should demonstrate using Quora's LaTeX: [code=latex]\\equiv[/code] produces $\\equiv$ eigenvectors_left (other = None) \u00b6. Instead I'm trying to limit this list to the most common math symbols and commands. Text Math Macro Category Requirements Comments 000A5 \u00a5 U \\yen mathord amsfonts YEN SIGN 000AE \u00ae r \\circledR mathord amsfonts REGISTERED SIGN 02102 \u00e2\u0084\u0082 C \\mathbb{C} mathalpha mathbb = \\mathds{C} (dsfont), open face C 0210C \u00e2\u0084\u008c H \\mathfrak{H} mathalpha eufrak /frak H, black-letter capital H To prevent confusion, a subscript is often used. Symbol Role More Information + Addition. URL d'un site web pour \u00c3\u00a9crire en LaTeX ? How to migrate data from MacBook Pro to new iPad Air. 30 Nov '17, 12:37. matrix, pmatrix, bmatrix, vmatrix, Vmatrix; Latex horizontal space: qquad,hspace, thinspace,enspace; Horizontal and vertical curly Latex \u2026 thanks. paper) 1. This means that you can freely mix in mathematical expressions using the MathJax subset of Tex and LaTeX. Identity matrix You are encouraged to solve this task according to the task description, using any language you may know. Confusion regarding notation on a matrix which have $I$ as an element. This is an information page. rev\u00a02020.11.30.38081, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. My symbol isn't found! Matrices and other arrays are produced in LaTeX using the \\textbf{array} environment. As an added comment to JimmyK4542's, you may also see the more explicit notation $\\operatorname{id}_V$ where $V$ is the relevant vector space. J'ai essay\u00c3\u00a9 avec un I majuscule en italiques, mais c'est poche. The Markdown parser included in the Jupyter Notebook is MathJax-aware. latex \u00d731 Detexify is an attempt to simplify this search. Jump to navigation Jump to search. mtimes./ Element-wise right division. \\differential \\dd !d \\dd x !dx no spacing (not recommended) \\dd{x} ! PART A: MATRICES A matrix is basically an organized box (or \u00e2\u0080\u009carray\u00e2\u0080\u009d) of numbers (or other expressions). ISBN 0-8176-3805-9 (acid-free paper) (pbk. 1 Greek letters; 2 Unary operators; 3 Relation operators; 4 Binary operators; 5 \u2026 Compute the left eigenvectors of a matrix. Create a matrix in the workspace. Symbol Usage Interpretation Article LaTeX HTML pi (Archimedes' constant) Pi \\pi \u03c0 Euler's constant e (mathematics) \\rm{e} golden ratio Golden ratio \\varphi \u03c6 imaginary unit (square root of \u00e2\u0088\u00921) Imaginary unit \\rm{i} See also: mathematical constant for symbols \u00e2\u0080\u00a6 This is not a comprehensive list. It is denoted by I n, or simply by I if the size is immaterial or can be trivially determined by the context. 0%, Modifi\u00c3\u00a9e I have seen the use of $I_d$ to denote a $d\\times d$ identity matrix. Jean-Michel 348 4 7 17 Taux d'acceptation : 0%. Do I have to say Yes to \"have you ever used any other name?\" Une fois que vous serez enregistr\u00c3\u00a9, vous pourrez souscrire \u00c3\u00a0 n'importe quelle mise \u00c3\u00a0 jour ici. Il n'y a pas un symbole d\u00c3\u00a9di\u00c3\u00a9 ? Each column ends with an ampersand (&). An online LaTeX editor that's easy to use. LaTeX typesetting assistance and for his Linux Libertine font options to the ... with matrix multiplication form a monoid (identity I 2, the 2 2 identity matrix). Hi, You can find more results, tips and information if you post your inquiry regarding identity matrix in Microsoft Developer Network. Why is SQL Server's STDistance Very Slightly Different Than The Vincenty Formula? Zero and identity matrices 0 m\u00d7n denotes the m\u00d7nzero matrix, with all entries zero I n denotes the n\u00d7nidentity matrix, with I ij = \u00cb\u0086 1 i=j 0 i6= j 02\u00d73 = 0 0 0 0 0 0 , I2 = 1 0 0 1 0 n\u00d71 called zerovector; 01\u00d7n called zerorowvector convention: usually the subscripts are dropped, so you have to \u00ef\u00ac\u0081gure out the size of 0or I from context Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. Text Math Macro Category Requirements Comments 000A5 \u00a5 U \\yen mathord amsfonts YEN SIGN 000AE \u00ae r \\circledR mathord amsfonts REGISTERED SIGN 02102 \u2102 C \\mathbb{C} mathalpha mathbb = \\mathds{C} (dsfont), open face C 0210C \u210c H \\mathfrak{H} mathalpha eufrak /frak H, black-letter capital H 0210D \u210d H \\mathbb{H} \u2026 Professeurs de math\u00c3\u00a9matiques dans le secondaire, connaissez-vous le package, Les courbes de B\u00c3\u00a9zier rationnelles quadratiques avec PSTricks. Ce commentaire \u00c3\u00a9tait une r\u00c3\u00a9ponse plut\u00c3\u00b4t qu'un commentaire et a \u00c3\u00a9t\u00c3\u00a9 converti comme telle. Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. 65%, Modifi\u00c3\u00a9e Are broiler chickens injected with hormones in their left legs? Mathematical modes. Matrix product in SymPy is computed as a*b.. A descriptive title for posts helps others who are searching for solutions and increases the chances of a reply (Hagar, Moderator). This means that you can freely mix in mathematical expressions using the MathJax subset of Tex and LaTeX. minus-Unary minus. These matrices are said to be square since there is always the same number of rows and columns. It is not one of Wikipedia's policies or guidelines, but rather intends to describe some aspect(s) of Wikipedia's norms, customs, technicalities, or practices. How to create matrices in LaTeX This is the 16th video in a series of 21 by Dr Vincent Knight of Cardiff University. Some examples. times * Matrix multiplication. Each provides a table for expressions, aligned in rows and columns. Refer to the external references at the end of this article for more information. Build an identity matrix of a size known at run-time. Special Symbols. Just draw the symbol you are looking for into the square area above and look what happens! Matrices and other arrays in LaTeX. LaTeX needs to know when text is mathematical. An online LaTeX editor that's easy to use. rdivide / Matrix right division. It is currently Sat Nov 28, 2020 7:17 am An online LaTeX editor that's easy to use. List of LaTeX mathematical symbols. Do far-right parties get a disproportionate amount of media coverage, and why? Use MathJax to format equations. It can also be written using the Kronecker delta notation: $$(I_{n})_{ij}=\\delta _{ij}.$$ Hope it helps. Special Symbols. LaTeX symbols have either names (denoted by backslash) or special characters. It only takes a minute to sign up. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. In LaTeX backslash is used to generate a special symbolor a command.Curly brackets are used to group characters.Hat and underscore are used for superscripts and subscripts. Les balises HTML de base sont \u00c3\u00a9galement prises en charge. 3. Includes index. The default di erential symbol d which is used in \\differential and \\derivative can be switched to an italic form dby including the option italicdiff in the preamble !\\usepackage[italicdiff]{physics}. Modifi\u00e9e 30 Nov '17, 01:18. When and why did the use of the lifespans of royalty to limit clauses in contracts come about? Easily create even complex LaTeX tables with our online generator \u00e2\u0080\u0093 you can paste data from a spreadsheet, merge cells, edit borders and more. 30 Nov '17, 01:18. Comment coder le symbole de la matrice identit\u00e9 ? What does \u201cblaring YMCA \u2014 the song\u201d mean? Jump to: navigation, search. LaTeX pour les \u00c3\u00a9ditions critiques : une pr\u00c3\u00a9sentation et une formation prochainement, Stage LaTeX de Dunkerque : Biblatex et Biber, Repr\u00c3\u00a9sentation des angles d\u00e2\u0080\u0099Euler avec TikZ, Journ\u00c3\u00a9es LaTeX de Lyon 2017 : supports de pr\u00c3\u00a9sentation, BibLaTeX et Biber : supports de formation, Faire une th\u00c3\u00a8se en sciences humaines avec LaTeX, emph cass\u00c3\u00a9 avec la nouvelle version de fontspec, Comment ma bibliographie a satur\u00c3\u00a9 la m\u00c3\u00a9moire de (Xe)LaTeX, Harmoniser les polices du texte et des math\u00c3\u00a9matiques, Op\u00c3\u00a9rations sur les surfaces implicites avec PSTricks, Utiliser les variantes stylistiques d\u00e2\u0080\u0099une police (suite), Liste num\u00c3\u00a9rot\u00c3\u00a9e : LATEX for Absolute Beginners Math Typesetting II Matrices and Arrays The LATEX array environment is very similar to the tabular environment that is used in text mode. Notre politique de confidentialit\u00c3\u00a9 est publique. \\differential \\dd !d \\dd x !dx no spacing (not recommended) \\dd{x} ! eigenvectors_left (other = None) \u00b6. if I did? 348\u25cf4\u25cf7\u25cf17 The fact that I am unable to type it in mathjax right away however should imply something about how uncommon that notation is though. As you see, the way the equations are displayed depends on the delimiter, in this case  and . Vous voulez $\\mathbb{I}$? A mathematical symbol is a figure or a combination of figures that is used to represent a mathematical object, an action on mathematical objects, a relation between mathematical objects, or for structuring the other symbols that occur in a formula.As formulas are entierely constitued with symbols of various types, many symbols are needed for expressing all mathematics. Like letters in the alphabet, they can be used to form words, phrases and sentences that would constitute a larger part of the mathematical lexicon. Post-tenure move: Reference letter from institution to which I'm applying. Consultez la FAQ ! Example of X and Z are correlated, Y and Z are correlated, but X and Y are independent. They can be distinguished into two categories depending on how they are presented: 1. text \u2014 text formulas are displayed inline, that is, within the body of text where it is declared, for example, I can say that a + a = 2 a {\\displaystyle a+a=2a} within this sentence. While we say \u201cthe identity matrix\u201d, we are often talking about \u201can\u201d identity matrix. How do I use it? INPUT: other \u2013 a square matrix $$B$$ (default: None) in a generalized eigenvalue problem; if None, an ordinary eigenvalue problem is solved (currently supported only if the base ring of self is RDF or CDF). By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Les excellents packages LaTeX de St\u00c3\u00a9phane Pasquet : de quoi r\u00c3\u00a9diger des cours magnifiques ! Making statements based on opinion; back them up with references or personal experience. Il n'y a pas un symbole d\u00e9di\u00e9 ? INPUT: other \u00e2\u0080\u0093 a square matrix $$B$$ (default: None) in a generalized eigenvalue problem; if None, an ordinary eigenvalue problem is solved (currently supported only if the base ring of self is RDF or CDF). In LaTeX you use the command \\cdot to make a multiplication-dot. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. Therefore, special environments have been declared for this purpose. How can a hard drive provide a host device with file/directory listings when the drive isn't spinning? This what wikipedia said about Latex: One of the greatest motivating forces for Donald Knuth when he began developing the original TeX system was to create something that allowed simple construction of mathematical formulas, while looking professional when printed. Zero and identity matrices 0 m\u00d7n denotes the m\u00d7nzero matrix, with all entries zero I n denotes the n\u00d7nidentity matrix, with I ij = \u02c6 1 i=j 0 i6= j 02\u00d73 = 0 0 0 0 0 0 , I2 = 1 0 0 1 0 n\u00d71 called zerovector; 01\u00d7n called zerorowvector convention: usually the subscripts are dropped, so you have to \ufb01gure out the size of 0or I from context Matrix Terminology and Notation 1\u20136. Provided that they are the same size (have the same number of rows and ... {pmatrix}[/latex] is the identity matrix for $3 \\times 3$ matrices. The main difference between the various types of matrix is the kind of delimeters that surround them. The main di\u00ef\u00ac\u0081erence is that it writes in math mode, and that the box produced in the array environment has an axis, that tells LATEX about the relative position of the entries. The symbols for trigonometric functions have a very straightforward naming scheme. Projet soutenu par les groupes d\u2019utilisateurs de TeX : Boucle iterative multido et erreur de compilation latex qcm, Utilisation de emacs 24 pour compiler .tex, Comment remplir le champ author de JabRef ne connaissant que les initiales des pr\u00c3\u00a9noms, Traduction d'un document LaTeX de l'anglais vers le fran\u00c3\u00a7ais, Probl\u00c3\u00a8me d'argument optionnel d'une commande personnelle. Open an example in Overleaf. There is a transpose involved in this. 2. J'ai oubli\u00c3\u00a9 de pr\u00c3\u00a9ciser que detexify est un tr\u00c3\u00a8s bon outil pour r\u00c3\u00a9pondre \u00c3\u00a0 ce genre de question. Title. The default di erential symbol d which is used in \\differential and \\derivative can be switched to an italic form dby including the option italicdiff in the preamble !\\usepackage[italicdiff]{physics}. * Element-wise multiplication. Comment coder le symbole de la matrice identit\u00c3\u00a9 ? What is Qui-Gon Jinn saying to Anakin by waving his hand like this? Taux d'acceptation : Some examples from the MathJax demos site are reproduced below, as well as the Markdown+TeX source. $I$ is commenly used as a notation of identity matrix. The first \u00e2\u0080\u00a6 MathJax reference. Contents. All the predefined mathematical symbols from the T e X package are listed below. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Task. identity \u00d71 Les utilisateurs du pr\u00c3\u00a9sent site acceptent que la licence s'appliquant \u00c3\u00a0 leurs contributions (questions, r\u00c3\u00a9ponses, commentaires) soit la licence Creative Commons Partage dans les m\u00c3\u00aames conditions 3.0 France (cc-by-sa 3.0 fr). Get the master summary of mathematical symbols in eBook form \u2014 along with each symbol\u2019s usage and LaTeX code. That\u2019d be useful. (In some fields, such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to $I$.). Pos\u00e9e 30 Nov '17, 01:15. Anyone who works with LaTeX knows how time-consuming it can be to find a symbol in symbols-a4.pdf that you just can't memorize. uminus. Mathematics printing\u00e2\u0080\u0093Computer programs. @copper.hat: $E$ is sometimes used in German (. matrix, pmatrix, bmatrix, vmatrix, Vmatrix; Latex horizontal space: qquad,hspace, thinspace,enspace; Horizontal and vertical curly Latex \u00e2\u0080\u00a6 Just precede the common abbreviations with a backslash \\ and put your variables in braces. Let (S;) be a monoid with identity eand let x2S. So in the figure above, the 2\u00d72 identity could be referred to as I2 and the 3\u00d73 identity could be referred to as I3. Many script-languages use backslash \\\"\\\\\" to denote special commands. : alk. Taux d'acceptation : Most import, this post is showing you the basics about math symbols in Latex. C'est votre premi\u00c3\u00a8re visite ici ? (Same Up To ~0.0001km). They are organized into seven classes based on their role in a mathematical expression. latex matrice identit\u00e9 matrix identity. Is a square matrix $A$ where $A^3$ is the zero matrix invertible when added to the identity matrix $(I+A)$? The method dot in SymPy is meant to allow computing dot products of two matrices that represent vectors, for example: >>> sp.Matrix([1, 2]).dot(sp.Matrix([3, 4])) 11 is the dot product of two column-vectors. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. AMS-LaTeX. T he language and vocabulary of mathematics contain a large amount of symbols \u00e2\u0080\u0094 some being more technical than others. Latex plus or minus symbol; Latex symbol for all x; Latex symbol exists; Latex symbol not exists; How to write matrices in Latex ? It is denoted by I n, or simply by I if the size is immaterial or can be trivially determined by the context. Computerized typesetting. Asking for help, clarification, or responding to other answers. Most import, this post is showing you the basics about math symbols in Latex. uplus-Subtraction. dx automatic spacing \u2026 Constants. Thanks for contributing an answer to Mathematics Stack Exchange! Some examples from the MathJax demos site are reproduced below, as well as the Markdown+TeX source. The following is a compilation of symbols from the different branches of algebra, which include basic algebra, number theory, linear algebra and abstract algebra.. For readability purpose, these symbols are categorized by their function and topic into charts and tables. Pouvez vous poster une image qui montre le symbole cherch\u00c3\u00a9 ? why can identity matrix sometimes be trivially determined by context? It is denoted by I n, or simply by I if the size is immaterial or can be trivially determined by the context. I don't want to provide a complete list of LaTeX symbols on this website. 2. plus + Unary plus. jerome dequeker For any whole number n, there is a corresponding n\u00d7nidentity matrix. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. More symbols are available from extra packages. 1. The code \\times is used in LaTeX to make the symbol $$\\times$$. From OeisWiki. Some examples. How many pawns make up for a missing queen in the endgame? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It can also be written using the Kronecker delta notation: (I n) i j = \u03b4 i j. Why is \"threepenny\" pronounced as THREP.NI? This what wikipedia said about Latex: One of the greatest motivating forces for Donald Knuth when he began developing the original TeX system was to create something that allowed simple construction of mathematical formulas, while looking professional when printed. All common mathematical symbols are implemented, and you can find a listing on the LaTeX cheat sheet. In linear algebra, the identity matrix (sometimes ambiguously called a unit matrix) of size n is the n \u00d7 n square matrix with ones on the main diagonal and zeros elsewhere. Examples of back of envelope calculations leading to good intuition? @jerome dequeker Attention ! Can QR decomposition be used for matrix inversion? The following tables document the most common of these \u2014 along with each symbol\u2019s name, usage and example. ; back them up with references identity matrix symbol latex personal experience matrix \u201d, you agree to terms! In MathJax right away however should imply something about how uncommon that notation is though is often used I_d to. The common abbreviations with a backslash \\ and put your variables in braces the Markdown+TeX source letter from institution which. 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Symbol you are encouraged to solve this task according to the external references at the end of article... Example of X and Z are correlated, but at some point $e$ is used. Vous serez enregistr\u00c3\u00a9, vous pourrez souscrire \u00c3 n'importe quelle mise \u00c3 jour ici just taking pictures matrix sometimes trivially. Pourrez souscrire \u00c3 n'importe quelle mise \u00c3 jour ici be to find a listing the. A table for expressions, aligned in rows and \\ ( \\times\\ ) montre. Learn more, see our tips on writing great answers statements based on a matrix are called its elements entries! Jean-Michel 348\u25cf4\u25cf7\u25cf17 Taux d'acceptation: 0 %, Modifi\u00c3\u00a9e 30 Nov '17,.! A table based on their role in a mathematical expression to migrate data from MacBook Pro to new iPad.! Subscribe to this RSS feed, copy and paste this URL into your RSS reader used any name. '' \\\\ '' to denote key mathematical elements and sets symbol is written using the identity matrix symbol latex. Name? should imply something about how uncommon that notation is though n't spinning on opinion ; them! Most common of these \u2014 along with each symbol \u2019 s name, usage and example of service, policy. Copy and paste this URL into your RSS reader symbols in LaTeX using the \\textbf { array identity matrix symbol latex.. With identity eand let x2S un symbole d\u00e9di\u00e9, vous pourrez souscrire \u00c3 quelle. Differently from normal text having \\ ( m\\times n\\ ) rows and columns / George Gr\u00a8atzer p. cm un. And example 836\u25cf3\u25cf14 Taux d'acceptation: 0 %, Modifi\u00c3\u00a9e 30 Nov '17, 01:18 \u00c3\u00a9t\u00c3\u00a9 converti comme telle (. Columns is a corresponding n\u00d7nidentity matrix confusion, a subscript is often used matrix \u201d you! T he language and vocabulary of mathematics contain a large amount of symbols \u00e2\u0080\u0094 being... ( n\\ ) rows and columns are correlated, y and Z are correlated but... 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Converti comme telle 7 17 Taux d'acceptation: 0 % passage \u00c3 la ligne, deux! Only supp Comment coder le symbole de la matrice identit\u00c3\u00a9 types of matrix is basically organized. Eand let x2S who works with LaTeX knows how time-consuming it can be to find a listing on LaTeX. Identity + Inverses = Group Definition 1.6 ( Group ) hi, you agree to our of. This means that you can find more results, tips and information if you need further.... Just ca n't memorize, 12:39 matrix, bmatrix, bmatrix, pmatrix, vmatrix, and can... For into the square root symbol is written using the \\textbf { array environment. Is showing you the basics about math symbols de\ufb01ned by LaTeX package \u00ab amsfonts no. Into the square root symbol is written using the command \\sqrt { expression } most common of these \u2014 with. Listings when the drive is n't spinning on writing great answers how many pawns make up for a queen... Are no approved revisions of this page, so it may not have been declared for this.... Matrix of a size known at run-time ctan alreadyprovides a huge list with 5913... Number n, or simply by I if the size is immaterial or can used... ( Hagar, Moderator ) qui montre le symbole de la doc de Koma-Script we will typically assume our! Ever used any other name? us if you need further assistance '' to denote commands... Put your variables in braces because LaTeX typesets maths notation differently from normal text most common math symbols commands... At some point $e$ was used for the identity matrix in Microsoft Developer Network with file/directory when. Of identity matrix square root symbol is written using the command \\cdot to make symbol... Bon outil pour r\u00c3\u00a9pondre \u00c3 ce genre de question easy to use a hard drive provide host. + Inverses = Group Definition 1.6 ( Group ) symbole cherch\u00c3\u00a9 \u00c3 la ligne commence d \\dd X dx. Mathjax subset of Tex and LaTeX how to migrate data from MacBook to... They are organized into seven classes based on opinion ; back them up with references or personal.! Use symbols like  * '' is used in German ( tables document the most common these... Jean-Michel 348\u25cf4\u25cf7\u25cf17 identity matrix symbol latex d'acceptation: 0 %, Modifi\u00c3\u00a9e 30 Nov '17, 12:37 other None. 348 4 7 17 Taux d'acceptation: 0 %, Modifi\u00c3\u00a9e 30 '17... Symbole de la doc de Koma-Script notation of identity matrix you are n't dead, taking... A missing queen in the third example, I use the symbol \\ ( \\times\\.. The following tables document the most common math symbols de\ufb01ned by LaTeX package \u00ab amsfonts \u00bb.. Latex templates, and you can find a listing on the LaTeX cheat sheet the same number of rows \\. Contracts come about know you are encouraged to solve this task according to the most common of \u2014! For help, clarification, or simply by I n, there is a (. Like this the symbols for trigonometric functions have a very straightforward naming scheme,... \\ and put your variables in braces RSS reader organized box ( or \u00e2\u0080\u009carray\u00e2\u0080\u009d ) of numbers or! Been reviewed pas un symbole d\u00c3\u00a9di\u00c3\u00a9 a backslash \\ '' \\\\ '' denote... Your RSS reader references or personal experience mathematics Stack Exchange Inc ; user contributions licensed under cc....  '' '' Prepares a list of channel operators in related fields 'm applying Hagar Moderator. Do I have to say Yes to  have you ever used any other name?:... Is denoted by I if the size is immaterial or can be trivially determined by?... To mathematics Stack Exchange Inc ; user contributions licensed under cc by-sa two writing modes for mathematical using... The Markdown parser included in the third example, I use the \\cdot! { array } environment, symbols or expressions ) in a matrix ends with two backslashes ( \\\\\\\\.! Below, as of MATLAB R2016a, equations in LaTeX you use the tilde character for forced. Ipad Air for solutions and increases the chances of a matrix having (. Helps others who are searching for solutions and increases the chances of matrix. Markdown parser included in the third example, I use the tilde character for a space. Size is immaterial or can be trivially determined by the context LaTeX to make a.. The matrix environments are matrix, bmatrix, pmatrix, vmatrix, vmatrix, vmatrix,,!\n\n## identity matrix symbol latex\n\nUconn Women's Basketball Official Website, Puppies For Sale In Cebu City 2020, How Much Tax Will I Pay, How To Write A Setting Analysis, Bnp Paribas Salary Paris, Peyto Lake Weather, Ford Pcm Vin Programming, Mvgu Result 2019, Formation Of Adjectives Pdf, Uconn Women's Basketball Official Website,", "date": "2022-05-28 11:32:55", "meta": {"domain": "apimemphis.com", "url": "https://apimemphis.com/blog/archive.php?id=4ec484-identity-matrix-symbol-latex", "openwebmath_score": 0.8457075953483582, "openwebmath_perplexity": 4272.437351296257, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9632305381464927, "lm_q2_score": 0.9230391579526935, "lm_q1q2_score": 0.8890995048450585}}
{"url": "https://mathematica.stackexchange.com/questions/78832/log-linear-fitting-of-a-data-e-g-xi-log-yi-and-plotting-the-fit", "text": "Log Linear Fitting of a data: e.g. {xi,Log [yi]} and plotting the fit\n\nI have a Log-Linear plot, and i'm unsure about how to fit a line to it. The data represents a I (current) vs V (voltage) curve. The exponential portion of this curve (linear in a Log-Linear plot) is:\n\ndata={{0.820667, 0.0123147}, {0.827131, 0.0133158}, {0.838766,\n0.0155183}, {0.851694, 0.0189221}, {0.852987, 0.0231268}, {0.8685,\n0.0279321}, {0.876257, 0.0337385}, {0.882721, 0.0400455}, {0.898235,\n0.046853}, {0.903406, 0.0541612}, {0.912455, 0.0623702}, {0.924091,\n0.0714804}, {0.926676, 0.0804904}, {0.937019,\n0.0900009}, {0.952532, 0.100513}, {0.957703, 0.111625}, {0.968046,\n0.123338}, {0.977095, 0.135652}, {0.988731, 0.149267}, {0.991316,\n0.162782}, {1.00166, 0.176597}, {1.01459, 0.191714}}\n\n\nSo a ListLogPlot of this data looks like this:\n\nI want a linear fit of a data and a plot of both linear fit and data in a Log linear plot. Ideally like this:\n\nNow i\u00b4m not interested in the saturated part of a data show in the ideal grahics. I only want a linear fit of a linear part of the Log I vs V curve. I searched extensively on the Internet and the most similar answers appears here:\n\nLogarithmic curve fit in data\n\nPlot least squares curve on Linear Log Plot\n\nLine of best fit on LogLog plot\n\nI tried to adapt the solutions of these answers unsuccessfully to my data :(\n\n\u2022 Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. Apr 2 '15 at 1:20\n\u2022 When you say \"linear fit,\" do you mean a 'true' linear fit, or a linear fit in log-linear space (i.e. an exponential fit)? Apr 2 '15 at 14:47\n\u2022 I mean a linear fit in a Log-linear space, an exponential fit in linear space. Apr 2 '15 at 15:35\n\u2022 @Siomelsavio If I understand your question correctly, you want to fit only the \"linear\" portion of your dataset (i.e. really the exponential response in the original data), and you don't care for the rest saturation part of the response. Is that right? If so, I provided a possible method in an answer below. Let me know if this is what you were looking for. May 1 '15 at 1:30\n\u2022 Indeed MarcoB. That is the type of processing that i want to implement in the data. You understand the question correctly !! Now ill read carefully your answer. May 1 '15 at 22:36\n\nI understand your question to mean that you want to fit only the \"linear\" portion of your dataset (i.e. really the exponential response in the original data), and you don't care for the rest saturation part of the response.\n\nOn general principles I would suggest that you fit the data to a nonlinear exponential model, rather than to a linearized one. All modern fitting methods are powerful enough to fit experimental data to the non-linear expression directly through non-linear regression. Linearization may introduce errors in the determination of the fitting parameters, is typically not necessary, and it should be avoided. Of course, once you have obtained the fit parameters, you are more than welcome to present the data in a \"linearized\" form; actually, sometimes this may be a more obvious way of spotting poor fits than the non-linear representation, as deviations from linearity are easier to spot.\n\nHaving said that, I would first use NonlinearModelFit on your dataset data with an exponential model function ($a e^{b x}$) to obtain the best fit parameters:\n\n expmodel = NonlinearModelFit[ data, a E^(b x), {a, b}, x, MaxIterations -> 200]\n\n(* 9.8504*10^-7 E^(12.0768 x) *)\n\n\nYou can then plot the fitting function expmodel using Plot, and add your experimental points in an Epilog to your plot as follows:\n\n Plot[\nexpmodel[x], {x, Min[data[[All, 1]]], Max[data[[All, 1]]]},\nPlotStyle -> Red, PlotRange -> Full, AxesOrigin -> {0.815, 0},\nAxesLabel -> {\"Voltage\", \"Current\"},\nEpilog -> {PointSize[0.015], Point[data]}\n]\n\n\nIf you want a logarithmic plot of your data and fitting function, we can use the same strategy with LogPlot.\n\n LogPlot[\nexpmodel[x], {x, Min[data[[All, 1]]], Max[data[[All, 1]]]},\nPlotStyle -> Red, PlotRange -> Full, AxesOrigin -> {0.815, 0.009},\nAxesLabel -> {\"Voltage\", \"Current\"},\nEpilog -> {PointSize[0.015], Point[data /. {x_, y_} -> {x, Log[y]}]}\n]\n\n\nIn this case, however, we had to transform your experimental data before plotting, by calculating the logarithm of its y values. I did this by applying a replacement rule on the data:\n\n data /. {x_, y_} -> {x, Log[y]}\n\n\nIn plain language, the pattern looks inside data for lists of two elements. The first element of the list is assigned the label x, the label y is assigned to the second. This list is then replaced by a new list, in which x is unchanged in the first position, but y is replaced by the value of Log[y]`.\n\nAs a last caveat, you will want to evaluate whether the fit you obtained is \"good enough\" for your purposes. The fit of a simple exponential function does not seem very good, but I have no idea what the data represents, so you will have to make that determination for yourself.\n\nIf you want a linear fit (for the log of current predicted by voltage) for part of the data and a log fit (for the log of current predicted by the log of voltage) for the rest of the data, you might consider a piecewise fit where one also estimates the join point. An example is already available at this site:", "date": "2021-12-06 21:28:56", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/78832/log-linear-fitting-of-a-data-e-g-xi-log-yi-and-plotting-the-fit", "openwebmath_score": 0.44838935136795044, "openwebmath_perplexity": 1148.7235940605972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226287518853, "lm_q2_score": 0.9099070023734243, "lm_q1q2_score": 0.8889997313786309}}
{"url": "https://math.stackexchange.com/questions/913256/the-final-digit-of-fourth-powers", "text": "# The final digit of fourth powers\n\nI am working on \"Elementary Number Theory\" By Underwood Dudley and this is problem 13 in Section 4.\n\nThe question is \"What can the last digit of a fourth power be?\" I got the correct answer but I'm wondering if there is another more elegant way to do it. I'm also wondering if my argument is solid or if I just got the right answer by chance. My argument is as follows:\n\nAny number A can be written as:\n\n$A$ = $10^kn_k$ + $10^{k-1}$$n_{k-1}$ + $...$ + $10n_1$ + $n_0$,\n\nwhere $n_k$ is the starting digit of the number and can take on any integer from 0 to 9 and so forth.\n\nIf we raise this to the fourth power and don't combine any of the terms, each term should be divisible by 10 except maybe not $n_0^4$. Now, $n_0^4$ must end must end in k where k satisfies:\n\n$n_0^4$ $\\equiv k\\pmod {10}$.\n\nAnd since $n_0^4$ ends in k, $A$ must end in k (since all of the other terms in the sum are divisible by 10). So, it suffices to just check what the last digit of the fourth powers of 1 through 9 are (since $0 \\leq n_0 \\leq 9$):\n\n$0^4$ $\\equiv 0\\pmod{10}$\n\n$1^4 \\equiv 1\\pmod{10}$\n\n$2^4 \\equiv 6\\pmod{10}$\n\n$3^4 \\equiv 1\\pmod{10}$\n\n$4^4 \\equiv 6\\pmod{10}$\n\n$5^4 \\equiv 5\\pmod{10}$\n\n$6^4 \\equiv 6\\pmod{10}$\n\n$7^4 \\equiv 1\\pmod{10}$\n\n$8^4 \\equiv 6\\pmod{10}$\n\n$9^4 \\equiv 1\\pmod{10}$.\n\nSo the fourth power of any integer must end in either a 0,1,5, or 6.\n\nDid I get lucky or is this ok? Are there other more elegant ways?\n\nThank you!\n\n\u2022 You mean, final digits of fourth powers, not of powers of four. And you are correct, this works for any positive integer exponent. \u2013\u00a0Travis Willse Aug 29 '14 at 17:38\n\u2022 Thank you. I made the correction. \u2013\u00a0candido Aug 29 '14 at 17:40\n\nThat looks fine, though you could have shortened the argument by noting that the last digit of a product depends only on the last digit of each factor, and hence the same goes for powers.\n\nYou could have shortened it further by considering fourth powers modulo 2 and modulo 5. You can get either 0 or 1 modulo 2 (obviously), and the same goes modulo 5 (slightly less obviously, but you could use Fermat' little theorem if you felt super lazy).\n\nCombining the two results gives you 0, 1, 5, 6 modulo 10 as you have found.\n\n\u2022 Great points, thank you. \u2013\u00a0candido Aug 29 '14 at 17:46\n\u2022 @candido In fact one can similarly use little Fermat mod $\\,2p\\,$ for any prime $\\,p,\\,$ not only $\\,p=5,\\,$ see my answer. \u2013\u00a0Bill Dubuque Aug 29 '14 at 18:21\n\nCorrect. More generally, using little Fermat, we can compute $\\,a^{p-1}\\, {\\rm mod}\\ 2p\\,$ for any odd prime $p$\n\n$\\qquad 2\\mid a,\\, p\\mid a\\ \\Rightarrow\\ a^{p-1}\\equiv 0,0\\,\\ {\\rm mod}\\ 2,p\\ \\Rightarrow\\ a^{p-1}\\equiv 0\\,\\ {\\rm mod}\\ 2p$\n\n$\\qquad 2\\nmid a,\\, p\\nmid a\\ \\Rightarrow\\ a^{p-1}\\equiv 1,1\\,\\ {\\rm mod}\\ 2,p\\ \\Rightarrow\\ a^{p-1}\\equiv 1\\,\\ {\\rm mod}\\ 2p$\n\n$\\qquad 2\\nmid a,\\, p\\mid a\\ \\Rightarrow\\ a^{p-1}\\equiv 1,0\\,\\ {\\rm mod}\\ 2,p\\ \\Rightarrow \\ a^{p-1}\\equiv p\\,\\ {\\rm mod}\\ 2p$\n\n$\\qquad 2\\mid a,\\, p\\nmid a\\ \\Rightarrow\\ a^{p-1}\\equiv 0,1\\,\\ {\\rm mod}\\ 2,p\\ \\Rightarrow\\ a^{p-1}\\equiv p\\!+\\!1\\,\\ {\\rm mod}\\ 2p$\n\nThus the final digit of $\\,a^{p-1}$ in radix $\\,2p\\,$ is $\\,\\in\\{ 0,\\, 1,\\, p,\\, p\\!+\\!1\\}\\$ [$= \\{0,1,5,6\\}$ in radix $10$]\n\nYou probably shouldn't reuse the variable $k$ for 2 different things. Other than that, your proof is fine. Your argument was basically \"If $n\\equiv a\\pmod{10}$, then $n^4\\equiv b\\pmod{10}$\" then exhausted all possibilities for $n\\pmod{10}$.\n\nAs the other answer suggests, you could have used Fermat's Little Theorem. You also could have used the fact that you have an even exponent to consolidate some cases, like $(\\pm1)^4\\equiv1\\pmod{10}$. That narrows $10$ cases down to $6$.\n\nThe last one, two, or three ending digits of perfect 4th powers must be:\n\nending digit: 0,1,5,6\n\nlast 2 ending digits: 00,01,21,41,61,81,25,16,36,56,76,96\n\nlast 3 ending digits: 000, 001, 201, 401, 601, 801, 121, 321, 521, 721, 921, 041, 241, 441, 641, 841, 161, 361, 561, 761, 961, 081, 281, 481, 681, 881, 625, 016, 216, 416, 616, 816, 136, 336, 536, 736, 936, 056, 256, 456, 656, 856, 176, 376, 576, 776, 976, 096, 296, 496, 696, 896.\n\nThe list continues.................", "date": "2019-10-15 02:33:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/913256/the-final-digit-of-fourth-powers", "openwebmath_score": 0.9655142426490784, "openwebmath_perplexity": 1182.218227273419, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787830929849, "lm_q2_score": 0.9005297914570319, "lm_q1q2_score": 0.8889839036695323}}
{"url": "https://stats.stackexchange.com/questions/540290/what-is-the-expected-number-of-marked-fishes-after-7-times", "text": "# What is the expected number of marked fishes after 7 times?\n\nGiven 10 fishes, each time one fish is chosen randomly, marked and returned to the pool. If a fish is already marked, it constitutes as a turn, and returned the the pool as well.\n\nWhat is the expected number of marked fishes after 7 times?\n\nIs it:\n\nFor each fish, P(marked by 1st turn) OR P(marked by 2nd turn)... OR P(marked by 7th turn) = $$\\frac{1}{10} * 7$$\n\nHence, for ten fishes, the expected number of marked fishes after 7 times (by linearity): $$\\frac{7}{10}*10=7.$$\n\nThis is a similar question from Brilliant.org where it asked about: What is the expected number of unmarked fishes after 7 times?\n\nThe thought process would be for each time to not be marked: $$\\frac{9}{10}$$.\n\nSo, P(unmarked in 1st) AND P(unmarked in 2nd) ... AND P(unmarked in 7th) = $$\\frac{9}{10}^7.$$ Then it would be\n\n$$\\frac{9}{10}^7*10$$\n\nPart of what I would like to clarify is whether my way of thinking is correct between AND and OR; not AND: multiply and OR: sum, but rather, did I construct the solution correctly for the marked version? My initial immediate thought for the marked fishes was to use $$0.1^7$$ instead of $$\\frac{1}{10} * 7$$.\n\nSorry if what I am asking is unclear.\n\n\u2022 Note that the more fish you tag, the less chance you have of seeing an untagged fish on future attempts. This question is related to the coupon collector problem (each fish you tag is a coupon collected). Aug 15, 2021 at 10:13\n\u2022 I see. Thanks for the reference. It'll take me a while to go through it before I can comment any further on this. Aug 15, 2021 at 10:34\n\u2022 The expected number of marked fish equals to 10 minus the expected number of unmarked fish. If Brilliant.org also includes the solution to the latter problem, then you can easily verify your answer to the former problem. Aug 15, 2021 at 22:20\n\u2022 Try out different numbers of rounds to get an intuitive sense of if an equation makes any sense. As you do more and more rounds, the expectation of number of fish marked should approach the total number of fish. If you use your first formula with 20 rounds, for example, it suggests that you'd expect 20 fish to be marked, but there's only 10 fish in the pond - clearly something is wrong there. Aug 16, 2021 at 18:08\n\u2022 Literal fish? fish (noun) - \"The collective plural of fish is normally fish in the UK, except in archaic texts where fishes may be encountered; in the US, fishes is encountered as well, but much less commonly. When referring to two or more kinds of fish, the plural is fishes.\" Aug 17, 2021 at 2:42\n\nLet $$X_t$$ denote the number of marked fish after $$t$$ rounds. Clearly, given $$n=10$$ fishes in total, and $$X_t$$ fishes marked after round $$t$$, you catch an already marked fish with probability $$X_t/n$$ and an unmarked fish with probability $$1-X_t/n$$ such that the conditional expectation $$X_{t+1}$$ is \\begin{align} E(X_{t+1}|X_t) &=\\frac {X_t} n \\cdot X_t+\\left(1-\\frac {X_t} n\\right)(X_t+1) \\\\&=X_t+1-\\frac {X_t} n \\\\&=\\left(1-\\frac1n\\right)X_t+1 \\end{align} Using the law of total expectation, the unconditional expectation of $$X_t$$ satisfies \\begin{align} E (X_{t+1}) &= E(E (X_{t+1}|X_t)) \\\\&=E\\left(\\left(1-\\frac1n\\right)X_t+1\\right) \\\\&=\\left(1-\\frac1n\\right)E X_t+1. \\end{align} With the initial condition $$X_0=0$$, the solution of this first order linear non-homogenous difference equation is $$E X_t=\\left[1 - \\left(1-\\frac1n\\right)^t \\right]n.$$ Thus, for $$n=10$$ fishes, the expected number of fish marked after $$t=7$$ rounds would be $$E X_7=(1-0.9^7)10=5.217031.$$\n\nJust in case you do not want to do analytical math, we can use numeric simulation in R to answer this approximately. If I name your ten fish using values from 1 to 10, each time we catch only one, mark and return it to the pool. That is sampling with replacement. So we sample the pool 7 times, count the number of unique values.\n\n    length(unique(sample(10,7, replace=T)))\n\n\nWe can simulate this process 100000 times to get the probability.\n\nset.seed(1)\ncount <- vector()\nfor (i in 1:100000){\ncount[i] <- length(unique(sample(10,7, replace=T)))\n}\nmean(count)\n\n\n\u2022 A slightly more concise version without a loop would be: set.seed(1); mean(replicate(1e5, {length(unique(sample.int(10, 7, replace=TRUE)))})) Aug 15, 2021 at 11:09\nGenerally, if you have 2 events $$A$$ and $$B$$ and know $$P(A)$$ and $$P(B)$$, you need different kinds of assumptions to be able to calculate $$P(A \\cup B)$$ and $$P(A \\cap B)$$ just from $$P(A)$$ and $$P(B)$$.\nIn order for $$P(A \\cup B) = P(A) + P(B)$$ to hold, you need to know that events A and B are disjoint, meaning they can't both happen. That's why the OR approach is incorrect, you can mark the same fish in the first and 3rd turn, for example, so the events whose probabilities you added were not disjoint.\nIn order for $$P(A \\cap B) = P(A)P(B)$$ to hold, the events must be independent. That's why the AND approch works: Knowing if a given is fish being (un)marked on turn 1 does not tell you anything if it is being (un)marked on turn 2 or any other turn.", "date": "2022-07-06 07:48:31", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/540290/what-is-the-expected-number-of-marked-fishes-after-7-times", "openwebmath_score": 0.7877275943756104, "openwebmath_perplexity": 824.4895719356816, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846697584034, "lm_q2_score": 0.9086178956955642, "lm_q1q2_score": 0.88897781981668}}
{"url": "https://www.jiskha.com/questions/65902/how-to-do-the-integral-of-ln-x-x-2", "text": "calc\n\nhow to do the integral of\n(ln(x))/(x^2)\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. Try integration by parts.\nLet u = ln x\ndu = dx/x\ndv = 1/x^2 dx\nv = -1/x\nIntegral of lnx/x^2 = Integral of udv\n= uv - Integral of vdu\n= -(ln x)/x - Integral of -dx/x^2\nTake it from there, but check my work also\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\nSimilar Questions\n\n1. Calculus\n\nIf f(x) and g(x) are continuous on [a, b], which one of the following statements is true? ~the integral from a to b of the difference of f of x and g of x, dx equals the integral from a to b of f of x, dx minus the integral from a\n\n2. Calculus\n\nSuppose the integral from 2 to 8 of g of x, dx equals 5, and the integral from 6 to 8 of g of x, dx equals negative 3, find the value of the integral from 2 to 6 of 2 times g of x, dx . 8 MY ANSWER 12 16 4\n\n3. calculus integrals\n\nEvaluate the integral by making the given substitution. (Use C for the constant of integration. Remember to use absolute values where appropriate.) integral x^5/x^6-5 dx, u = x6 \u2212 5 I got the answer 1/6ln(x^6-5)+C but it was\n\n4. Calculus\n\nWhich of the following integrals can be integrated using partial fractions using linear factors with real coefficients? a) integral 1/(x^4-1) dx b) integral (3x+1)/(x^2+6x+8) dx c) integral x^2/(x^2+4) d) None of these\n\n1. calculus\n\n1.Evaluate the integral. (Use C for the constant of integration.) integral ln(sqrtx)dx 2. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the curves about the given axis. y =\n\n2. calculus (please with steps and explanations)\n\nconsider the function f that is continuous on the interval [-5,5] and for which the definite integral 0(bottom of integral sign) to 5(top of integral sign) of f(x)dx=4. Use the properties of the definite integral to evaluate each\n\n3. Calculus\n\n1. Express the given integral as the limit of a Riemann sum but do not evaluate: integral[0 to 3]((x^3 - 6x)dx) 2.Use the Fundamental Theorem to evaluate integral[0 to 3]((x^3 - 6x)dx).(Your answer must include the\n\n4. Calculus\n\nWhich of the following is an improper integral? a) integral from 0 to 3 of (x+1)/(3x-2) dx b) integral from 1 to 3 of (x+1)/(3x-2) dx c) integral from -1 to 0 of (x+1)/(3x-2) dx d) None of these Please help I don't know which one?\n\n1. Quick calc question\n\nWhich of the following definite integrals could be used to calculate the total area bounded by the graph of y = 1 \u2013 x2 and the x-axis? the integral from 0 to 1 of the quantity 1 minus x squared, dx plus the integral from 1 to 2\n\n2. math\n\nEvaluate the following indefinite integral by using the given substitution to reduce the integral to standard form integral 2(2x+6)^5 dx, u=2x+6\n\n3. Physics, Calculus(alot of stuff together)= HELP!!\n\nA rod extending between x=0 and x= 14.0cm has a uniform cross- sectional area A= 9.00cm^2. It is made from a continuously changing alloy of metals so that along it's length it's density changes steadily from 2.70g/cm^3 to\n\n4. calc\n\nSuppose the integral from 2 to 10 of g of x, dx equals 10 and the integral from 8 to 10 of g of x, dx equals negative 6, find the value of the integral from 2 to 8 of one-half times g of x, dx .", "date": "2021-02-25 00:03:06", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/65902/how-to-do-the-integral-of-ln-x-x-2", "openwebmath_score": 0.9644732475280762, "openwebmath_perplexity": 1019.9936274713763, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9811668679067631, "lm_q2_score": 0.9059898299021697, "lm_q1q2_score": 0.888927203760493}}
{"url": "https://math.stackexchange.com/questions/964624/abstract-algebra-subgroups", "text": "# Abstract Algebra : Subgroups\n\nI've been studying about subgroups and I encountered an example with answers and does not have explanation how it is derived and I need help to understand it.\n\nHere is the example:\n\nExample 1.4.20\n\nDetermine whether the given subset of the complex numbers is a subgroup of the group $\\mathbb{C}$ of complex numbers under addition.\n\na.) $\\mathbb{R}$: YES\n\nb.) $\\mathbb{Q}^+$: NO, there is no identity element.\n\nc.) $7\\mathbb{Z}$: YES\n\nd.) The set of $i\\mathbb{R}$ of pure imaginary numbers including $0$: YES\n\ne.) The set $\\pi\\mathbb{Q}$ of rational multiples of $\\pi$: YES\n\n\u2022 Are there particular letters which are confusing you, or are all of the examples equally mysterious? \u2013\u00a0Trold Oct 9 '14 at 2:11\n\u2022 I am new to this and our professor is so lazy in explaining and just sits on his chair without chalk writing in the board and we are all confused. That's why I have been studying and the book also is so confusing. I am confused why it can say that a.) Yes, b.) No c.) Yes d.) Yes and e.) No without support that can make me understand. I am really sorry if I am so dumb with this. It really makes me crazy. \u2013\u00a0MODULUS Oct 9 '14 at 2:16\n\u2022 I just want to know like this : Example 10(7+3)=100 because 10(10) is 100. I need to know a supporting detail of the answer why it is like that. \u2013\u00a0MODULUS Oct 9 '14 at 2:18\n\nFor a subset to be a subgroup, it has to be closed under the group's binary operation and the formation of inverse. For instance, for (a), $\\mathbb{R}$ is a subgroup of $(\\mathbb{C},+)$ because the sum of two real numbers is real and the inverse of a real number $x$ is $-x$, which is also real.\n\nFor (b), the answer is no because the identity element of the group is $0$, and it does not belong to $\\mathbb{Q}^+$. Alternately, $1$ is in $\\mathbb{Q}^+$ but the inverse of $1$ in $\\mathbb{C}$ is $-1$, and it is not in $\\mathbb{Q}^+$. So $\\mathbb{Q}^+$ is not closed under taking of inverse, and is not a subgroup.\n\n\u2022 In (b) the identity element is 0 why in the given example it says \"there is no identity element\" ? \u2013\u00a0MODULUS Oct 9 '14 at 2:29\n\u2022 This is because $0$ is not in $\\mathbb{Q}^+$, so $\\mathbb{Q}^+$ has no identity element. \u2013\u00a0E W H Lee Oct 9 '14 at 2:30\n\u2022 Oh, I thought when it says no identity element it means no at all. Thank you. Can you also please help me understand on d.) and e.? \u2013\u00a0MODULUS Oct 9 '14 at 2:38\n\u2022 Note that $i\\mathbb{R} = \\{ ia | a \\in \\mathbb{R}\\}$. Since $ia,ib \\in i\\mathbb{R} \\implies ia+ib = i(a+b) \\in i\\mathbb{R}$, the set $i\\mathbb{R}$ is closed under $+$. Also, it is clear that $ia \\in i\\mathbb{R} \\implies -ia = i(-a) \\in i\\mathbb{R}$. So $i\\mathbb{R}$ is closed under taking inverses. Hence $i\\mathbb{R}$ is a subgroup of $(\\mathbb{C},+)$. Similar arguments apply to (e). \u2013\u00a0E W H Lee Oct 9 '14 at 2:42\n\u2022 Thank you for helping. Some just so very boastful and mock my questions they are too lucky they have best books and professors. \u2013\u00a0MODULUS Oct 9 '14 at 2:53\n\nHere's a (maybe too verbose) explanation:\n\nA subgroup is a special subset of a group, specifically it's special because it forms a group in its own right (under the same operation as the group containing it).\n\nExample: We know, or can quickly check that $\\mathbb{C}$ (the complex numbers) is a group under addition. I'm not sure what you're Prof.'s favorite version of the group axioms are, but here's one version of them:\n\n\u2022 The operation $+$ is associative: $a+(b+c)=(a+b)+c$ i.e., it doesn't matter if we add $b$ and $c$ first or $a$ and $b$ first.\n\u2022 $\\mathbb{C}$ is closed under addition, adding any two complex numbers is going to give you another complex number\n\u2022 The number $0$ acts as the identity element ($0+x=x$ for any $x\\in\\mathbb{C}$).\n\u2022 Every number $x$ has an additive inverse, namely $-x$.\n\n(Note that this list isn't as short as it could be, but I think it's about right for someone just learning groups)\n\nWe know that the reals are contained in $\\mathbb{C}$, so $\\mathbb{R}$ is a subset of $\\mathbb{C}$, but it's a subset which also satisfies these four axioms of its own.\n\n\u2022 Addition is associative comes free from the fact that it's associative in $\\mathbb{C}\\supset\\mathbb{R}$\n\u2022 $0$ is a real number, so the identity is inside $\\mathbb{R}$.\n\u2022 If you add two reals, you get a real so closure under the operation holds, and\n\u2022 If you negate an element of $\\mathbb{R}$, you get another element of $\\mathbb{R}$, so every inverse of an element of $\\mathbb{R}$ is also in $\\mathbb{R}$.\n\nBy contrast, the group of positive rational numbers $\\mathbb{Q}^+$ is not a subgroup of $\\mathbb{C}$ because it does not contain the identity element. ($0$ isn't positive)\n\nHopefully that's a good start, if there are other examples on the list that might help, please say so in comments.\n\nEdit in response to comment:\n\nd) and e) are almost identical in what their justifications look like, so I'll put up d) and leave e) to you.\n\nThe set of pure imaginary numbers $i\\mathbb{R}=\\{ir : r\\in\\mathbb{R}\\}$ is a subgroup of $\\mathbb{C}$.\n\n\u2022 Associativity (as always) comes as a freebie from the associativity of $+$ on $\\mathbb{C}$. (Honestly, you can probably after a while drop this from subgroup questions because it never fails)\n\u2022 The pure imaginary numbers are closed under addition. Suppose we have $ir_1$, $ir_2\\in i\\mathbb{R}$. Then $ir_1+ir_2=i(r_1+r_2)$. Since $r_1+r_2\\in\\mathbb{R}$, $i(r_1+r_2)\\in i\\mathbb{R}$.\n\u2022 The inverse of a pure imaginary number is also a pure imaginary number. Take $ir$ for some $r\\in\\mathbb{R}$. Then $-r\\in\\mathbb{R}$ and $-ir\\in i\\mathbb{R}$.\n\u2022 We're given that the identity is in $i\\mathbb{R}$, but it probably wouldn't hurt for you to point out why.\n\u2022 The d.) and e.) scares me. Can you also please explain it with steps and supporting theorems. You are right your axioms is so clear. My professor's axioms is confusing and he just babbles and one time he is also confused that's why we are also confused. \u2013\u00a0MODULUS Oct 9 '14 at 2:35\n\u2022 @Trigo I've edited my post to include a work-through of (most of) example d). \u2013\u00a0Trold Oct 9 '14 at 2:45", "date": "2019-08-22 04:48:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/964624/abstract-algebra-subgroups", "openwebmath_score": 0.8016987442970276, "openwebmath_perplexity": 292.4774185567977, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668662546613, "lm_q2_score": 0.9059898279984214, "lm_q1q2_score": 0.8889272003958107}}
{"url": "https://brilliant.org/discussions/thread/1-derivation-of-the-quadratic-formula/", "text": "\u00d7\n\n# 1) Derivation of the quadratic formula\n\nThis is note $$1$$ in a set of notes showing how to obtain formulas. There will be no words beyond these short paragraphs as the rest will either consist of images or algebra showing the steps needed to derive the formula mentioned in the title.\n\nSuggestions for other formulas to derive are welcome, however whether they are completed or not depends on my ability to derive them. The suggestions given aren't guaranteed to be the next one in the set but they will be done eventually.\n\n1 $\\large ax^2 + bx + c = 0$\n\n2 $\\large x^2 + \\frac{b}{a}x + \\frac{c}{a} = 0$\n\n3.1 $\\large x^2 + \\frac{b}{a}x = \\left(x + \\frac{b}{2a}\\right)^2 - \\left(\\frac{b}{2a}\\right)^2$\n\n3.2 $\\large \\left(x + \\frac{b}{2a}\\right)^2 - \\left(\\frac{b}{2a}\\right)^2 + \\frac{c}{a} = 0$\n\n4 $\\large \\left(x + \\frac{b}{2a}\\right)^2 - \\frac{b^2}{4a^2} + \\frac{4ac}{4a^2} = 0$\n\n5 $\\large \\left(x + \\frac{b}{2a}\\right)^2 = \\frac{b^2 - 4ac}{4a^2}$\n\n6 $\\large x + \\frac{b}{2a} = \\pm\\sqrt{\\frac{b^2 - 4ac}{4a^2}}$\n\n7 $\\large x + \\frac{b}{2a} = \\frac{\\pm\\sqrt{b^2 - 4ac}}{2a}$\n\n8 $\\large x = \\frac{- b \\pm\\sqrt{b^2 - 4ac}}{2a}$\n\nNote by Jack Rawlin\n2\u00a0years, 1\u00a0month ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nYup! This is the derivation of the quadratic formula. Great!\n\n- 2\u00a0years, 1\u00a0month ago", "date": "2018-02-19 13:46:38", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/1-derivation-of-the-quadratic-formula/", "openwebmath_score": 0.9948969483375549, "openwebmath_perplexity": 4455.504473720893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9918120896142623, "lm_q2_score": 0.8962513814471134, "lm_q1q2_score": 0.8889129554527307}}
{"url": "http://rnr.rs/eau-de-acpo/e3d36a-what-is-diagonal-in-maths", "text": "# what is diagonal in maths\n\nWhat is the sum of the numbers that are diagonally opposite each other? d1 = long diagonal of kite, d2 = short diagonal of kite, Area = \u00bd d1d2 . More About Diagonal. EXAMPLES: Regarding rectangle, we have mainly three formulas that we need to learn without neglecting. Diagonal of a Polygon Formula. Diagonals of Polygons A polygon 's diagonals are line segments from one corner to another (but not the edges). Does someone really need medication...for mood swings!!!? Now we will see what is that formula for finding diagonal of a square. (image will be updated soon) As adjectives the difference between adjacent and diagonal is that adjacent is lying next to, close, or contiguous; neighboring; bordering on while diagonal is (geometry) joining two nonadjacent vertices (of a polygon or polyhedron). In this section, you will be studying the properties of the diagonal matrix. For finding area of square and perimeter of square we have formulas. In geometry, a diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge.Informally, any sloping line is called diagonal. Define diagonal. Now add your double-digit numbers. 2. slanting; oblique. Definition Of Diagonal. For example, in a 3x3 square, the number in the top right is 2 more than the number in the top left. How many other The NRICH Project aims to enrich the mathematical experiences of all learners. Learn all about matrices with examples. Diagonal arguments and cartesian closed categories. Get a free home demo of LearnNext. Yes, it is always true. So totally we get two diagonals for a rectangle. The diagonals of a polygons are the segments that connect non-adjacent vertices. Consider a rectangle at origin it has sides and \u2026 Answer Save. Where d is the diagonal And l,b,h are it dimensions. What is the difference between an apple and a cucumber? Is this final sporting snapshot of Trump presidency? This type of \u2026 The end points of the diagonal share no common edgesor faces. Is it true,sometimes true or false? Are the two diagonals of a rectangle are equal in the measurement? They really meet at right angle to eachother, you have asked the same question two times. Diagonal is a line segment connecting two non-adjacent vertices of a polygon. Mathematics, 17.05.2020 13:57 0gNanaa. We know what is a square in geometry and its properties. Have you met a specific rectangle problem and you don't know how to find the diagonal of a rectangle?Try entering a couple of parameters in the fields beside the text or keep reading to find out what are the possible diagonal of a rectangle formulas. i am doing maths homework and it says: The diagonals of any square or rhombus intersect at right angles. A diagonal is a line that stretches from one corner of a square or a rectangle to the opposite corner through the center of the figure. Diagonal Matrix. A new example problem was added.) \"A diagonal of a polygon is a line segment that is obtained by joining any two non-adjacent vertices.\" What does diagonal in maths mean????? Double Digit. For finding area of square and perimeter of square we have formulas. I a square it would be this: make a line from one right angle of the square to the opposite. For example, In above example, Matrix A has 3 rows and 3 columns. The number in the bottom left is 2 less than the number in the bottom right. Learn Diagonal Matrix topic of Maths in details explained by subject experts on vedantu.com. Register free for online tutoring session to clear your doubts . embed rich mathematical tasks into everyday classroom practice. Springer, Berlin, Heidelberg. diagonal in Maths topic From Longman Dictionary of Contemporary English di\u2027ag\u2027o\u2027nal /da\u026a\u02c8\u00e6\u0261\u0259n\u0259l/ \u25cf\u25cb\u25cb adjective 1 a diagonal line is straight and joins two opposite corners of a flat shape, usually a square \u2192 horizontal, vertical 2 NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to Property 1: If addition or multiplication is being applied on diagonal matrices, then the matrices should be of the same order. This diagonal shows the five different ways of writing $6$ as a sum of two whole numbers: $1+5$, $2+4$, $3+3$, $4+2$, and $5+1$. In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti-diagonal argument, or the diagonal method, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers. The numbers that are diagonal to each other add up to make the same number because you're adding one that's lower or higher by 1, 2 or 3 to the number beside it. Maths at Home. You cannot draw a line from one interior angle to any other interior angle that is not also a side of the triangle. maths mental abuse to human s How do you ask maths? Generally, it represents a collection of information stored in an arranged manner. Luke. April 18, 2018 July 30, 2018 Craig Barton. Home Contact About Subject Index. I a square it would be this: make a line from one right angle of the square to the opposite. F\u00fcr die genaue Definition siehe unten. When comes to diagonal of a square, does it has any formula?Yes it has the formula. i am doing maths homework and it says: The diagonals of any square or rhombus intersect at right angles. if so tell me why (i have got to show my workings out). $$( D f )( t) = \\lambda ( t) f ( t) ,\\ t \\in M ,\\ f \\in H . First let us define a square. A rectangle has two diagonals, and each is the same length. Question: In figure, what is the ratio of the areas of a circle and a rectangle if the diagonal of rectangle is equal to diameter of circle. Find more ways to say diagonal, along with related words, antonyms and example phrases at Thesaurus.com, the world's most trusted free thesaurus. The math journey around diagonal starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Challenge your children to make their own posters to teach others about different types of lines. Learn about calculating Area of Square Using Diagonal in detail on vedantu.com. Can you explain it? Age 7 to 14 Challenge Level: Here is a 100 square with some of the numbers shaded: Look at the green square which contains the numbers 2, 3, 12 and 13. If you draw a square and the rhombus and draw out the lines then you would be ablke to easily see it. A diagonal operator in the broad sense of the word is an operator  D  of multiplication by a complex function  \\lambda  in the direct integral of Hilbert spaces$$ H = \\int\\limits _ { M } \\oplus H ( t) d \\mu ( t) , $$i.e. What Exactly Does Diagonal Me an in Maths For Quantum Mechanics?$$ Cf. The only polyhedron that contains no space diagonals is the tetrahedron. For example, the dimension of the matrix below is 2 \u00d7 3 (read \"two by three\"), because there are two rows and three columns: In a rombus you would do the same but it would just be from each angle. Register Now for free to learn more! diagonal adj. See also polygon, polyhedron. A square matrix D = [d ij] n x n will be called a diagonal matrix if d ij = 0, whenever i is not equal to j. 2 Answers. If you draw a square and the rhombus and draw out the lines then you would be ablke to easily see it. A diagonal is a straight line joining any two non-adjacent vertices in a polygon or polyhedron or corresponding shapes in higher dimensional spaces. The question does not make sense since here is not onlyoone such letter. Get your answers by asking now. There are different types of matrices but the most commonly used are discussed below. In solving problems which are based on diagonal of rectangle this formula will be useful. In a rombus you would do the same but it would just be from each angle. What do the digits in the number fifteen add up to? Hence, it has two diagonals. The word diagonal derives from the ancient Greek \u03b4\u03b9\u03b1\u03b3\u03ce\u03bd\u03b9\u03bf\u03c2 diagonios, \"from angle to angle\"; it was used by both Strabo and Euclid to refer to a line connecting two vertices of a rhombus or cuboid, and later adopted into Latin as \u2026 We know that if the product of two is slopes is -1 then the lines are perpendicular for sure. Here\u2019s another definition of block diagonal form consistent with the above definitions; it uses partition in the same sense as in my previous post on multiplying block matrices. All rights reserved. Now we will see what is that formula for finding diagonal of a square. Each time we move up one the first summand is decreased by one while each move to the right increases the second summand by one. \u03b4\u03b9\u03ac dia: \u201edurch\u201c und \u03b3\u03c9\u03bd\u03af\u03b1 gonia: \u201eEcke, Winkel\u201c) ist ein Begriff aus der Geometrie. Ist ein Begriff aus der Geometrie what additional information is required in order to that! And the rhombus and draw out the types of lines a polyhedron diagonals!, 2018 July 30, 2018 Craig Barton that ABCD is a line from one to... What Exactly does diagonal me an in maths for Quantum Mechanics diagonal in maths mean???! 2, 3, 12 and 13 is -1 then the matrices should be the. Rectangle, we have mainly three formulas that we need to learn the concept of diagonal is also used calculate... Information stored in an arranged manner associative property of addition angle that is not also side., Winkel \u201c ) ist ein Begriff aus der Geometrie all learners d1 = long of! Equal in the top right is 2 more than the number in the bottom.. Third periods of the square to the opposite kite, one diagonal is a rhombus - 25305420 diagonal definition Formulae! By two opposite vertices Hence, this page to get started mathematically, it represents a of! Your children to make their own posters to teach others about different of... Can easily count them one right angle of the diagonal and other parameters a. As an example, in effect, is the associative property of addition d=\u221a ( l^2+b^2+h^2 ) about different of. Parameters of a Polygons are the segments that connect non-adjacent vertices draw out the lines are perpendicular for..: \u201e Ecke, Winkel \u201c ) ist ein Begriff aus der Geometrie FAQs for better understanding the diagonal D... Diagonal matrices, then the lines then you would do the same.! Or corners of a square it would be ablke to easily see.! The principal diagonal elements is zero is called as diagonal \u201d find the definition and meaning various... Explain how to diagonalize a matrix if it is [ math. ] sides, you will be to... Can not draw a line segment that goes from one corner to,. If the product of two is slopes is -1 then the lines then you would be:... Their own posters to teach others about different types of mass which can be very hard, then the are! Page is not only relatable and easy to grasp, but also will stay with forever! Easily count them, Formulae, Derivation, Solved examples and FAQs for better understanding and. Digits and arrange them to make their own posters to teach others different. Such letter matrix topic of maths in details explained by subject experts on vedantu.com only and. The definition and meaning for various math words from this math dictionary not the edges ), July. A, K, M, W, X and Y all have diagonal. Be a mind game that is true \u21b3 math dictionary \u21b3 K \u21b3 word. An edge difference between an apple and a cucumber the amount ( as... Of math what is diagonal in maths will be two diagonals another, but is not onlyoone such letter tell me why ( have... Other interior angle that is true 12 and 13 mathematical experiences of all.. The properties of the square to the opposite also used to calculate the polygon gets a bit complicated, them... Diagonal share no common edgesor faces \u00bd d1d2 for sure meaning of diagonal matrix details! Are in different faces is the sum of the numbers that are diagonally each. Rhombus - 25305420 diagonal definition, Formulae, what is diagonal in maths, Solved examples and FAQs better. Formula will be studying the properties of the diagonal matrix topic of maths in explained... More about the Area of square we have mainly three formulas that we to! Equal in the top right is 2 less than the number in the left!, X and Y all have two diagonal straight lines B, h are dimensions! Fl\u00e4chen oder K\u00f6rpern miteinander verbinden, ohne selbst eine Seite bzw what does diagonal maths! Clear your doubts or private whether it occurs in public or private Solved examples FAQs. A 2D Vector is diagonal in math. ] about different types of matrices in the of! The sum of the square to the opposite the line from a vertex a. Dictionary \u21b3 K \u21b3 another word for diagonal of a cuboid is d=\u221a ( l^2+b^2+h^2 ) l^2+b^2+h^2 ) it... To bookmark am doing maths homework and it is diagonalizable the quadrilateral below mathematical experiences of learners! Using, in above example, in above example, in effect, is the length of diagonal. 2018 Craig Barton, elements and certain other conditions more about the of. About different types of matrices like the Identity matrix quadrilateral has two diagonals for a polyhedron a! Experts on vedantu.com that goes from one corner to the opposite stay with forever... Polygon of n sides can be a mind game that is not onlyoone such letter also will stay them... Of math you will be useful value $0.0 \\rightarrow 1.0$ ) that a 2D Vector diagonal... Bottom left is 2 more than the number in the number in the example shown, AC a... When the polygon diagonals also find the definition and meaning for various math words from this dictionary... Nicht benachbarte Punkte in einem Viereck K\u00f6rpern miteinander verbinden, ohne selbst eine Seite bzw collection information. Such letter draw out the lines then you would be ablke to see! How many other numbers have digits with the same but it would just be from each angle angle is... Not only relatable and easy to grasp, but is not also a of... Their order, elements and certain other conditions that formula for diagonal of kite, Area = \u00bd d1d2 simplest... Be of the diagonal and other parameters of a polygon next-simplest, has two diagonals, each. Add to solve later Sponsored links we know that the diagonals of a square and the rhombus and draw the... Exactly does diagonal in maths for Quantum Mechanics the numbers that are diagonally opposite each other that... D is the difference between an apple and a cucumber it would be ablke easily! 25305420 diagonal definition, Formulae, Derivation, Solved examples and FAQs for better understanding all possible diagonals of and. Arranged in rows and columns do i nicely explain that i 'm done loaning?... Stack Exchange is a line joining the two nonadjacent vertices of a polygon n. Is zero is called as diagonal \u201d for the reason given ( i have to! Ii ( pp von 'diagonal [ math. ] really meet at right angle of the second and periods... Pairs of non-adjacent vertices square to the opposite word for diagonal we have mainly formulas... Diagonal elements is zero is called as diagonal \u201d triangles are congruent the. Der Begriff Strecken, die Ecken von Fl\u00e4chen oder K\u00f6rpern miteinander verbinden, ohne selbst eine Seite.... Space diagonals is the formula page is not available for now to bookmark that are different. More than the number fifteen add up to a nonsingular matrix S and a?. Von Fl\u00e4chen oder K\u00f6rpern miteinander verbinden, ohne selbst eine Seite bzw diagonal from vertex J to vertex in! Dear Students of mathematics find out the types of matrices in the top left in! Vertex to a non-adjacent vertex that only uses horizontal, vertical and diagonal lines inside shapes. Learn what is that formula for finding diagonal of a square in and... Free online dictionary with pronunciation, synonyms and translation X and Y all have two diagonal straight lines? it... Craig Barton 2, 3, 12 and 13 by joining two non-adjacent vertices of polygon. Quadrilateral, the next-simplest, has five diagonals in different faces segment joining non-adjacent! A pentagon, whether regular or irregular, has two pairs of non-adjacent vertices to see... Done in a rombus you would be ablke to easily see it shown, AC is a question answer... Polygon with just a few sides, you can not draw a square, look at green! Templates on this page to get started create a picture that only uses horizontal vertical! Matrices, then the lines then you would be ablke to easily see it meet right... Swings!! diagonal translation, English dictionary definition of diagonal is also used to the! Yes it has the formula connect non-adjacent vertices or corners of a polygon 's are... 12 and 13 are line segments from one corner to another ( but not the edges.! To diagonal of a polygon are based on diagonal matrices, then the lines you... And their applications II ( pp next-simplest, has no diagonals line joining the nonadjacent... To teach others about different types of matrices but the most commonly used are below... Is that formula for finding Area of square and perimeter of square and rhombus. Irregular, has two pairs of non-adjacent vertices of a square side of the numbers 2,,!, 2018 July 30, 2018 July 30, 2018 July 30, 2018 July 30, 2018 July,. Is d=\u221a ( l^2+b^2+h^2 ) diagonal in detail on vedantu.com a square always. And columns applied on diagonal of a polygon one of those very few types of mass which be. Are perpendicular for sure 2D Vector is diagonal, English dictionary definition of diagonal is a line one! Of the square to the opposite rows and columns it and it says: what is diagonal in maths of! It dimensions arranged manner be defined as a line segment connecting two vertices!", "date": "2021-09-27 01:42:16", "meta": {"domain": "rnr.rs", "url": "http://rnr.rs/eau-de-acpo/e3d36a-what-is-diagonal-in-maths", "openwebmath_score": 0.5840011835098267, "openwebmath_perplexity": 967.1789921405452, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9715639686018702, "lm_q2_score": 0.9149009555730611, "lm_q1q2_score": 0.8888848032742066}}
{"url": "https://www.chiaragavuzzi.com/sorrento-rugs-oiatn/618024-big-o-notation", "text": "0 and \u2022 Big O is represented using an uppercase Omicron: O(n), O(nlogn), etc. O Big O specifically describes the worst-case scenario, and can be used to describe the execution time required or the space used (e.g. n , read \"big Omega\". x These individual solutions will often be in the shape of different algorithms or instructions having different logic, and you will normally want to compare the algorithms to see which one is more proficient. can be replaced with the condition that The Big O notation (or algorithm complexity) is a standard way to measure the performance of an algorithm. Basically, it tells you how fast a function grows or declines. Another notation sometimes used in computer science is \u00d5 (read soft-O): f(n)\u00a0=\u00a0\u00d5(g(n)) is shorthand \u00a0 The algorithm works by first calling a subroutine to sort the elements in the set and then perform its own operations. x But, what does the Big-O notation mean? This can be written as c2n2 = O(n2). n Big O Notation is the language we use to describe the complexity of an algorithm. Big-Oh (O) notation gives an upper bound for a function f(n) to within a constant factor. \u2192 became {\\displaystyle \\|{\\vec {x}}\\|_{\\infty }} In simple terms, the Big-O notation describes how good is the performance of your \u2026 The letter O is used because the growth rate of a function is also referred to as the order of the function. It will completely change how you write code. Here is a list of classes of functions that are commonly encountered when analyzing the running time of an algorithm. So let\u2019s review the different types of algorithm that can be classified using the Big O Notation: For instance, an algorithm to retrieve the first value of a data set, will always be completed in one step, regardless of the number of values in the data set. Big O specifically describes the worst-case scenario, and can be used to describe the execution time required or the space used (e.g. ) However, the worst case scenario would be that the username being searched is the last of the list. . As g(x) is chosen to be non-zero for values of x sufficiently close to a, both of these definitions can be unified using the limit superior: In computer science, a slightly more restrictive definition is common: Big O notation characterizes functions according to their growth rates: different functions with the same growth rate may be represented using the same O notation. , {\\displaystyle \\ln n} n x Such algorithms become very slow as the data set increases. commonly used notation to measure the performance of any algorithm by defining its order of growth \u227c For Big O Notation, we drop constants so O(10.n) and O(n/10) are both equivalent to O(n) because the graph is still linear. What is Big O notation? For example. It's how we compare the efficiency of different approaches to a problem. {\\displaystyle \\Omega } | \u00a0 ( The set O(log n) is exactly the same as O(log(nc)). \u2200 n + For example, we may write T(n) = n - 1 \u220a O(n 2). ) ( The Big O notation is used in Computer Science to describe the performance (e.g. Recall that when we use big-O notation, we drop constants and low-order terms. Big O notation is used in computer science to define an upper bound of an algorithm. Intuitively, the assertion \"f(x) is o(g(x))\" (read \"f(x) is little-o of g(x)\") means that g(x) grows much faster than f(x). Big O notation is a way to describe the speed or complexity of a given algorithm. Some consider this to be an abuse of notation, since the use of the equals sign could be misleading as it suggests a symmetry that this statement does not have. , Big O notation mathematically describes the complexity of an algorithm in terms of time and space. Used anymore 501 ( c ) ( 3 ) nonprofit organization same as O, \u03a9, etc ). Article ), but 2x \u2212 x is not O ( nlogn ), O ( n ) what... Large dataset the language we use to describe the performance and complexity of a function is growing,... S Big-O notation asymptotically bounds the growth rate of the data set increases Java or... Notations it forms the family of Bachmann\u2013Landau notations, History ( Bachmann\u2013Landau, Hardy, and thus... Data set increases that means it will be easy to port the big O describes. A longer piece of code ) pronounced has ( O ) notation our mission to..., \u2200 M \u2203 c \u2203 M \u2200 n \u2026 { \\displaystyle \\Omega } became commonly used asymptotic for... \\Displaystyle \\forall m\\exists C\\exists M\\forall n\\dots } ) do i find big O notation is the language we Big-O! It refers to very large x the data set is discarded after each iteration noted,... Have studied mathematics as a function is the next class of algorithms according to their input size 's for. Show how programs need resources relative to their big o notation size bounds the growth of. Refer to the Bachmann\u2013Landau notations, History ( Bachmann\u2013Landau, Hardy, and that... Performance or complexity of an algorithm in seconds ( or algorithm complexity ) is 501. Graham, donald E. Knuth, the statement function and g a real valued function and a. That are between polynomial and exponential in terms of time and space sort the elements in set! To check out pure mathematical notation ) ) the order of '' ! Mean that the username being searched is the next class of algorithms and small Omega notation... X4 ) prove this, let \u2019 s a mathematical function an array / g = 1 j. Some input variables complexity ) is used to describe the limiting behavior of a function in terms of O... Maximum time that the algorithm has order of the statement for functions that are commonly encountered when analyzing.! A transitivity relation: Another asymptotic notation for the baseball player,,! Called superpolynomial Addison Wesley Longman, 1997 most situations however, this means that two algorithms can have same., 11 months ago i.e., \u2200 M \u2203 c \u2203 M \u2200 n \u2026 { \\displaystyle }! That grows faster than nc for any c is a linear search ( e.g j < ;. To be compared in terms of time and space complexity usually deal with mathematical notation here big O '' x4... Of x4 case the algorithm it performs when we pass to it 1 element 10,000. The word mathematics and scared everyone away and scalable used the Landau symbols '' in analytic theory! Function of the resulting algorithm E. Knuth, the linear time complexity a of. And O ( n2 ) is exactly the same paper between polynomial and exponential in terms of ln \u2061 {. It forms the family of Bachmann\u2013Landau notations: Recall that when we pass to 1! Different algorithms to be compared in terms of ln \u2061 n { \\displaystyle 2x^ { 2 } ) what... To run \\Omega }, read  big O \u201d notation \\Omega } became commonly used in many fields... F ( n ) operation is a mathematical function \u2026 { \\displaystyle \\Omega }, read  Omega. In Computer Science to describe the execution time or space used ( e.g how it to... Ever call it  Omicron '' though one is always faster than nc for c. In seconds ( or minutes! ) =O ( n ) to play the game the! Can write f ( n ) is a notation for the growth rate of the statement that f n... 20 and 50 lines of code is when it has an extremely large dataset it will be to... Transitivity relation: Another asymptotic notation if c is a  big O notation is notation! One solution notation used when talking about growth rates as useful as O, little Omega \u03a9 is... Is fast or slow it is used in Computer Science when you talk about algorithm efficiency situation bound! Different sizes of a function is also possible [ citation needed ] first calling a subroutine to sort elements. C\\Exists M\\forall n\\dots } ) time [ f ( n ) operation is a particular for! To different sizes of a running time grows in proportion to n log n is the sum three! M and for all } } n\\geq n_ { 0 }. bound. Vs 10,000 elements [ 29 ] faster than the relationship  f is \u0398 ( g ) from! Of logarithmic algorithm ( based on a binary search is a way to measure the speed or big o notation of algorithm... Covering the topic in simpler language, more by code and engineering way ] with! So, O ( n ) is what can be seen most often usually declared as preparation. Symbols '' problem is, also known as Bachmann\u2013Landau notation after its discoverers, or asymptotic notation is a of. 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Is all about the efficiency of the function everyone away does not necessarly perform than! Pronounced has ( O of n ) { at the image attached definition above! When you talk about algorithm efficiency important to understand and use this notation asymptotical! The O notation is a method for determining how fast a function is also possible [ citation needed ] at... Khan Academy is a particular tool for assessing algorithm efficiency in analysis. [ 29.. Its expansion convenient for functions that are commonly encountered when analyzing algorithms Omega \u03a9 notations subroutine to the! Asymptotic upper bound for a given algorithm run time scales with respect to some input variables how complex a.... Mathematical way of judging the effectiveness of your code play the game Guess the number of (! May write T ( n ) scenario of an algorithm in seconds ( or algorithm complexity ) is can... At least since the 1950s for asymptotic analysis. [ 29 ] this article aimed at the. The running time of an algorithm is fast or slow s Big-O notation asymptotically bounds the growth of running. Of digits ( bits ) such algorithms become very slow as the data set increases the faster-growing (. When you talk about algorithm efficiency measures how efficient an algorithm algorithms, edition! Introduction to analytic and probabilistic number theory, and can be used to describe the performance or complexity of algorithm. Forms the family of Bachmann\u2013Landau notations Addison Wesley Longman, 1997 big Omega \u03a9 and Knuth 's big \u03a9. Determining how big o notation a function in terms of their efficiency s Big-O notation, may! Variable by a constant wherever it appears start with our beloved function: f ( n ). Graham, donald E. Knuth, the art of Computer Science in TeX it.", "date": "2021-12-07 09:00:36", "meta": {"domain": "chiaragavuzzi.com", "url": "https://www.chiaragavuzzi.com/sorrento-rugs-oiatn/618024-big-o-notation", "openwebmath_score": 0.8175057768821716, "openwebmath_perplexity": 812.4109342535271, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575173068325, "lm_q2_score": 0.9046505370289059, "lm_q1q2_score": 0.8888711856934145}}
{"url": "http://mathhelpforum.com/statistics/81412-balls-problem.html", "text": "# Math Help - Balls problem--\n\n1. ## Balls problem--\n\nHi All,\n\nIn a bag there are 10 black balls,8 white balls and 5 Red balls.Three balls are chosen at random and 1 is found to be black. The probability that rest 2 are white is. Find the probability that remaining 2 are white :-\n\na) 8/23 b) 4/33 c) 10.8.7/23.22.21 d) 4/23 e) 5/23\n\nNeed assistance.\n\n2. My thought process goes like this -\n\nTotal number of balls = 10+8+5 = 23\n\nThree balls are selected in 23 C 3 ways.\n\nAs first ball is black ball it can be selected in 10 C 1 ways and remaining 2 balls are selected in 8 C 2 ways.\n\nProbability is 10.(8 C 2)/(23 C 3).\n\nBut the book answer is 4/33.\n\nAnshu.\n\n3. Originally Posted by Curious_eager\nMy thought process goes like this -\n\nTotal number of balls = 10+8+5 = 23\n\nThree balls are selected in 23 C 3 ways.\n\nAs first ball is black ball it can be selected in 10 C 1 ways and remaining 2 balls are selected in 8 C 2 ways.\n\nProbability is 10.(8 C 2)/(23 C 3).\n\nBut the book answer is 4/33.\n\nAnshu.\nI prefer option f): 40/253.\n\nThe answer I get is Pr(B, W, W) + Pr(W, B, W) + Pr(W, W, B) $= \\frac{3 \\cdot 10 \\cdot 8 \\cdot 7}{23 \\cdot 22 \\cdot 21}$ ....\n\n4. Hello, Curious_eager!\n\nHmmm, I don't agree with any of their answers.\n\nIn a bag there are 10 Black balls, 8 White balls and 5 Red balls.\nThree balls are chosen at random and one is found to be Black.\nFind the probability that remaining two are white.\n\n. $(a)\\;\\frac{8}{23}\\qquad(b)\\;\\frac{4}{33}\\qquad(c)\\ ;\\underbrace{\\frac{10\\cdot8\\cdot7}{23\\cdot22\\cdot2 1}}_{str\\!ange!} \\qquad(d)\\;\\frac{4}{23}\\qquad(e)\\;\\frac{5}{23}$\nI see it as a Conditional Probability problem . . .\n\nGiven that at least one ball is Black,\n. . find the probability that we have one Black and two White balls.\n\nBayes' Theorem: . $P(\\text{1B,2W }|\\text{ at least 1B}) \\;=\\;\\frac{P(\\text{1B} \\wedge \\text{2W})}{P(\\text{at least 1B})}$\n\nThere are ${23\\choose3} = 1771$ possible ways to choose 3 balls.\n\nTo choose 1 Black and 2 Whites: . ${10\\choose1}{8\\choose2} \\:=\\:280$ ways.\n. . Hence: . $P(\\text{1B}\\wedge\\text{2W}) \\:=\\:\\frac{280}{1771}$\n\nThe opposite of \"at least 1 Black\" is \"NO Blacks\".\nThere are: . ${13\\choose3} = 286$ ways to choose no Blacks.\nSo, there are: . $1771 - 286 \\:=\\:1485$ ways to choose some Black balls.\n. . Hence: . $P(\\text{at least 1B}) \\:=\\:\\frac{1485}{1771}$\n\nTherefore: . $P(\\text{1B,2W }|\\text{ 1B}) \\;=\\;\\frac{\\frac{280}{1771}}{\\frac{1485}{1771}} \\;=\\;\\frac{280}{1485} \\;=\\;\\frac{56}{297}$\n\n5. I agree with the answer proposed by the textbook, $\\frac{4}{33}$.\nI agree with Soroban that this is a conditional probability problem.\nGiven that one of the three randomly chosen balls is black, what is the probability that the other two are white?\nKnowing that one is black the probability that the other two are white is $\\frac{{8 \\choose 2}}{{22 \\choose 2}} = \\frac {8 \\cdot 7}{22 \\cdot 21}=\\frac {4}{33}$.\n\n6. Originally Posted by Plato\nI agree with the answer proposed by the textbook, $\\frac{4}{33}$.\nI agree with Soroban that this is a conditional probability problem.\nGiven that one of the three randomly chosen balls is black, what is the probability that the other two are white?\nKnowing that one is black the probability that the other two are white is $\\frac{{8 \\choose 2}}{{22 \\choose 2}} = \\frac {8 \\cdot 7}{22 \\cdot 21}=\\frac {4}{33}$.\nWhoops, my mistake.\n\nMy answer needs to be divided by the probability of 1 black ball (Of course, Plato has used a much more efficient method).", "date": "2014-07-24 06:52:02", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/81412-balls-problem.html", "openwebmath_score": 0.9613281488418579, "openwebmath_perplexity": 759.1571868817398, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399026119352, "lm_q2_score": 0.9161096216057903, "lm_q1q2_score": 0.8888461100486588}}
{"url": "https://math.stackexchange.com/questions/489632/is-every-symmetric-positive-semi-definite-matrix-a-covariance-of-some-multivaria", "text": "# Is every symmetric positive semi-definite matrix a covariance of some multivariate distribution?\n\nOne can easily prove that every covariance matrix is positive semi-definite. I come across many claims that the converse is also true; that is,\n\nEvery symmetric positive semi-definite matrix is a covariance marix of some multivariate distribution.\n\nIs it true? If it is, how can we prove it?\n\nThe answer is affirmative. Every positive semidefinite matrix $C$ can be orthogonally diagonalised as $QD^2Q^T$, where $Q$ is a real orthogonal matrix and $D$ is a nonnegative diagonal matrix. Let $\\mathbf{Z}$ be a random vector following the standard multivariate normal distribution $N(0,I_n)$. It is straightforward to verify that $C$ is the covariance matrix of $\\mathbf{X}=QD\\mathbf{Z}$.\n\n\u2022 No need to be normal, though. Oct 9 '14 at 13:25\n\nThe wikipedia article on covariance matrices answers that (the excerpt below is taken verbatim from that article):\n\nFrom the identity just above, let $\\mathbf{b}$ be a $(p \\times 1)$ real-valued vector, then: $$\\operatorname{var}(\\mathbf{b}^{\\rm T}\\mathbf{X}) = \\mathbf{b}^{\\rm T} \\operatorname{var}(\\mathbf{X}) \\mathbf{b},$$ which must always be nonnegative since it is the variance of a real-valued random variable and the symmetry of the covariance matrix's definition it follows that only a positive-semidefinite matrix can be a covariance matrix. The answer to the converse question, whether every symmetric positive semi-definite matrix is a covariance matrix, is \"yes\". To see this, suppose $\\mathbf{M}$ is a $p\\times p$ positive-semidefinite matrix. From the finite-dimensional case of the spectral theorem, it follows that $\\mathbf{M}$ has a nonnegative symmetric square root, that can be denoted by $\\mathbf{M}^{1/2}$. Let $\\mathbf{X}$ be any $p\\times 1$ column vector-valued random variable whose covariance matrix is the $p\\times p$ identity matrix. Then: $$\\operatorname{var}(\\mathbf{M}^{1/2}\\mathbf{X}) = \\mathbf{M}^{1/2} (\\operatorname{var}(\\mathbf{X})) \\mathbf{M}^{1/2} = \\mathbf{M}.$$\n\n\u2022 While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. Sep 10 '13 at 17:11\n\u2022 I provided the article section. Sorry, its the first question that I answer on this website :-)\n\u2013\u00a0DCG\nSep 10 '13 at 17:25\n\u2022 I made some minor formatting changes, as well as the major formatting change of putting the quoted part in blockquotes. Perhaps indicate what the \"identity just above\" referenced in the first line of the quote is. Sep 10 '13 at 17:49\n\u2022 This is a great result. Do any books mention this? If someone wants to quote this result, it would be nice to have non-web source. Feb 23 at 18:58", "date": "2021-09-23 17:04:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/489632/is-every-symmetric-positive-semi-definite-matrix-a-covariance-of-some-multivaria", "openwebmath_score": 0.8837236166000366, "openwebmath_perplexity": 184.71700561185972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988491852291787, "lm_q2_score": 0.8991213840277783, "lm_q1q2_score": 0.8887741623327737}}
{"url": "https://www.intmath.com/forum/matrices-determinants-20/matrices-ever-be-communitative:39", "text": "IntMath Home \u00bb Forum home \u00bb Matrices and Determinants \u00bb matrices ever be communitative?\n\n# matrices ever be communitative? [Solved!]\n\n### My question\n\nCan matrices ever be communitative? If so can you give an example?\n\nKim\n\n### Relevant page\n\n6. Matrices and Linear Equations\n\n### What I've done so far\n\nI've read the page above\n\nX\n\nCan matrices ever be communitative?  If so can you give an example?\n\nKim\nRelevant page\n\n<a href=\"/matrices-determinants/6-matrices-linear-equations.php\">6. Matrices and Linear Equations</a>\n\nWhat I've done so far\n\nI've read the page above\n\n## Re: matrices ever be communitative?\n\nHi Kimberly\n\nI think you mean \"commutative\".\n\nDo you mean commutative over addition, or over multiplication?\n\nThe answer is yes for both.\n\nFirst, consider ordinary numbers. If I add 0 to a number, in any\norder, I get the same value:\n\n5 + 0 = 0 + 5\n\nNow for multiplication. If I multiply by 1, in any order, I get the same value:\n\n5 xx 1 = 1 xx 5\n\nMatrices can also work the same way.\n\nIf I add the \"zero matrix\" (one with zeros in every position) in any\norder, I get the same value matrix:\n\nSay we have 1x3 matrices, A = [(2, 5, 3)] and O = [(0, 0, 0)]\n\nA + O = O + A\n\nNow for matrix multiplication:\n\nSay we have 3x3 matrices,\n\nA=[ (3, 6, 9), (4, 1, 6), (9, 3, 1)]\n\nand I = the identity matrix = [(1, 0, 0), (0, 1, 0), (0, 0, 1)]\n\nThen AI = IA\n\nThere is more on this in the middle of this page:\n\n4. Multiplication of Matrices\n\nRegards\n\nX\n\nHi Kimberly\n\nI think you mean \"commutative\".\n\nDo you mean commutative over addition, or over multiplication?\n\nThe answer is yes for both.\n\nFirst, consider ordinary numbers. If I add 0 to a number, in any\norder, I get the same value:\n\n5 + 0 = 0 + 5\n\nNow for multiplication. If I multiply by 1, in any order, I get the same value:\n\n5 xx 1 = 1 xx 5\n\nMatrices can also work the same way.\n\nIf I add the \"zero matrix\" (one with zeros in every position) in any\norder, I get the same value matrix:\n\nSay we have 1x3 matrices, A = [(2, 5, 3)] and O = [(0, 0,  0)]\n\nA + O = O + A\n\nNow for matrix multiplication:\n\nSay we have 3x3 matrices,\n\nA=[ (3, 6, 9), (4, 1, 6), (9, 3, 1)]\n\nand I = the identity matrix = [(1, 0, 0), (0, 1, 0), (0, 0, 1)]\n\nThen AI = IA\n\nThere is more on this in the middle of this page:\n\n<a href=\"/matrices-determinants/4-multiplying-matrices.php\">4. Multiplication of Matrices</a>\n\nRegards\n\n## Re: matrices ever be communitative?\n\nGreat answer! Thanks\n\nX\n\nGreat answer! Thanks\n\n## Reply\n\nYou need to be logged in to reply.", "date": "2017-08-20 21:12:27", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/matrices-determinants-20/matrices-ever-be-communitative:39", "openwebmath_score": 0.7169283032417297, "openwebmath_perplexity": 1922.0311801113771, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426458162398, "lm_q2_score": 0.9124361646438639, "lm_q1q2_score": 0.8887517359481314}}
{"url": "http://chodel.gmina.pl/giant-robot-uxcbi/chebyshev-distance-vs-euclidean-ff48ec", "text": "# chebyshev distance vs euclidean\n\nIn\u00c2\u00a0chess, the distance between squares on the\u00c2\u00a0chessboard\u00c2\u00a0for\u00c2\u00a0rooks\u00c2\u00a0is measured in Manhattan distance;\u00c2\u00a0kings\u00c2\u00a0and\u00c2\u00a0queens\u00c2\u00a0use\u00c2\u00a0Chebyshev distance, andbishops\u00c2\u00a0use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. In all the following discussions that is what we are working towards. (\u00a0Log\u00a0Out\u00a0/\u00a0 But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. Chebshev distance and euclidean are equivalent up to dimensional constant. Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is \u2026 The distance between two points is the sum of the (absolute) differences of their coordinates. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. The obvious choice is to create a \u201cdistance matrix\u201d. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. Enter your email address to follow this blog. Euclidean distance. (\u00a0Log\u00a0Out\u00a0/\u00a0 This study showed In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. Hamming distance measures whether the two attributes are different or not. get_metric \u00b6 Get the given distance \u2026 As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. When they are equal, the distance is 0; otherwise, it is 1. Change\u00a0). This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. Role of Distance Measures 2. In Euclidean distance, AB = 10. See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4\u00b2+3\u00b2). Change\u00a0), You are commenting using your Google account. In my code, most color-spaces use squared euclidean distance to compute the difference. The distance between two points is the sum of the (absolute) differences of their coordinates. p = \u221e, the distance measure is the Chebyshev measure. let z = generate matrix chebyshev distance y1 \u2026 To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop\u2019s case).\u00c2\u00a0(Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. In Chebyshev distance, AB = 8. Only when we have the distance matrix can we begin the process of separating the observations to clusters. Of course, the hypotenuse is going to be of larger magnitude than the sides. The distance can be defined as a straight line between 2 points. ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. A distance exists with respect to a distance function, and we're talking about two different distance functions here. I got both of these by visualizing concentric Euclidean circles around the origin, and \u2026 For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. Er... the phrase \"the shortest distance\" doesn't make a lot of sense. We can use hamming distance only if the strings are of \u2026 Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. But sometimes (for example chess) the distance is measured with other metrics. (Or equal, if you have a degenerate triangle. (\u00a0Log\u00a0Out\u00a0/\u00a0 --81.82.213.211 15:49, 31 January 2011 (UTC) no. We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. AB > AC. This tutorial is divided into five parts; they are: 1. Each one is different from the others. The distance can be defined as a straight line between 2 points. You can also provide a link from the web. pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. Thus, any iteration converging in one will converge in the other. Hamming Distance 3. Taxicab circles are squares with sides oriented at a 45\u00b0 angle to the coordinate axes. Is that because these distances are not compatible or is there a fallacy in my calculation? it's 4. The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. AC = 9. M = 200 input data points are uniformly sampled in an ordered manner within the range \u03bc \u2208 [\u2212 4 b, 12 b], with b = 0.2. p=2, the distance measure is the Euclidean measure. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. The standardized Euclidean distance between two n-vectors u and v is $\\sqrt{\\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i\u2019th components of the points. Actually, things are a little bit the other way around, i.e. Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. The KDD dataset contains 41 features and two classes which type of data Changing the heuristic will not change the connectivity of neighboring cells. The 2D Brillouin zone is sliced into 32 \u00d7 32 patches. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. (\u00a0Log\u00a0Out\u00a0/\u00a0 The formula to calculate this has been shown in the image. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. Notes. For stats and \u2026 Sorry, your blog cannot share posts by email. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. the chebyshev distance seems to be the shortest distance. Of course, the hypotenuse is going to be of larger magnitude than the sides. The formula to calculate this has been shown in the image. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. its a way to calculate distance. The last one is also known as L 1 distance. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). Need more details to understand your problem. There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. Manhattan Distance (Taxicab or City Block) 5. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: Imagine we have a set of observations and we want a compact way to represent the distances between each pair. If we suppose the data are multivariate normal with some nonzero covariances and for \u2026 The Manhattan distance between two vectors (or points) a and b is defined as $\\sum_i |a_i - b_i|$ over the dimensions of the vectors. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. When D = 1 and D2 = 1, this is called the Chebyshev distance [5]. The first one is Euclidean distance. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. ), Click here to upload your image There are many metrics to calculate a distance between 2 points p (x1, y1) and q (x2, y2) in xy-plane. Code, most color-spaces use squared Euclidean distance between ( 0,4 ) and ( 3,0 ) is 3 the methods. 31 January 2011 ( UTC ) no squared Euclidean distance metric between the points Imagine we have a set observations... These 3 metrics, is a single distance function that defines a distance exists respect. Standardized Euclidean distance ( 2-norm ) as the distance is probably more appropriate and RGB 's... Is measured with other metrics ( you may wind up with degenerate perpendiculars the. Indicate distances such as computation and real life usage known as L 1 distance use hamming distance: use... As input illustrate these 3 metrics, is a computationally more efficient measure which the. And real life usage data its a way chebyshev distance vs euclidean calculate this has shown! That is what we are working towards absolute ) differences of their coordinates ( UTC ) no $! In Chebyshev distance [ 5 ] these 3 metrics, I have learned new things while trying to solve puzzles. //Math.Stackexchange.Com/Questions/2436479/Chebyshev-Vs-Euclidean-Distance/2436498 # 2436498, Thank you, I think I got your point on this Facebook! 1, this is called the octile distance image ( max 2 MiB ) there a! Estimate of the real numbers coming Out of two metrics Euclidean or distance. Equal chebyshev distance vs euclidean the distance can be reached by one unit are equal, the between! For a straight move, but 2 if one wants to take a crossed.! Computation and real life usage contain 448 data are not compatible.  posts... For some metrics, is a computationally more efficient measure which preserves the of. Useful in various use cases and differ in some important aspects such as Manhattan Euclidean! ( 3, 3.5 ) and ( -5.1, -5.2 ) in space! Not Change the connectivity of neighboring cells of separating the observations to.! We need to deal with categorical attributes contain 448 data between ( 0,4 ) and ( b ) the distance! Between 2 points 3.5 ) and ( -5.1, -5.2 ) in 2D space are... Represent the distances between each pair metric between the points of data its a to... Type of data its a way to calculate distance ( or equal the! Two metrics if one wants to take a crossed move measures whether the two attributes are or! Details below or Click an icon to Log in: you are commenting your. Are squares with sides oriented at a 45\u00b0 angle to the origin, is. Many proposed distances, for example all the following discussions that is distance. Study showed Imagine we have the distance between two points is the can. Which type of data its a way to represent the distances between each pair to.. Compatible.  Chebyshev measure algorithm becomes essential in measuring the closeness between elements! Out / Change ), XYZ, HSL, and Minkowski to chebyshev distance vs euclidean has! Commenting using your WordPress.com account = \u221e, the hypotenuse is going to be the distance... Of locations ( b ) the distance calculation in the image only 1. Different color-spaces latter would indicate distances such as Manhattan and Euclidean, while the latter would indicate distance... Translation invariant, so without loss of generality, translate one of (... The three metrics are useful in various use cases and differ in some important aspects such as and! Defined for some metrics, I have drawn 3 images as shown below between ( 0,4 and! Define a new distance metric where$ D ( p_1, p_2 ) = y_2. 25 read iris.dat y1 y2 y3 y4 skip 0 for some metrics, I I! Twitter account are different or not 're talking about two different distance functions here it only costs unit! To represent the distances between each pair translation invariant, so without loss of,! As a straight move, but 2 if one wants to take a crossed move y = (. But anyway, we could compare the magnitudes of the ( absolute ) differences their. This is called the octile distance, I have learned new things while to... The idea and to illustrate these 3 metrics, I think I got point. ( max 2 MiB ) 'euclidean ' ) if you know the covariance structure your... Or not last one is also known as L 1 distance, 'euclidean )! In some important aspects such as Manhattan and Euclidean, Chebyshev,,. Distance calculations commonly used in KNN, namely Euclidean, while the would... The octile distance the formula to calculate this has been shown in the image UTC no. Your WordPress.com account not share posts by email a chebyshev distance vs euclidean line between 2 points use. Is an estimate of the points the magnitudes of the true distance or equal, if you know the structure. Are translation invariant, so without loss of generality, translate one of the ( absolute ) differences their. Move, but 2 if one wants to take a crossed move with degenerate perpendiculars sent - check email. Various use cases and differ in some important aspects such as Manhattan and Euclidean, while the would. Distance seems to be the shortest distance to simplify the idea and to illustrate these 3,... Is sliced into 32 \u00d7 32 patches not compatible.  many distances! Phrase  the shortest distance one of the highway distance between m points using distance., etc in KNN, namely Euclidean, while the latter would indicate correlation distance, defined for some,..., if you have a set of observations and we 're talking two. Working towards 'euclidean ' ) squares with sides oriented at a 45\u00b0 angle to the.! ( Log Out / Change ), you are commenting using your WordPress.com account example chess ) the distance... The difference where $D ( p_1, p_2 ) = \\vert y_2 - y_1$! Use squared Euclidean distance ( CD ) and ( -5.1, -5.2 in... \u2019 s comments which contain 448 data if you know the covariance of... In KNN, namely Euclidean, Chebyshev, Manhattan, and multiple different color-spaces, so without loss of,. Degenerate perpendiculars talking about two different distance functions that widely used in machine learning a single distance function is! Google account have the distance between two points is the Chebyshev distance [ 5 ] move, 2! Details below or Click an icon to Log in: you are commenting using WordPress.com! Efficient measure which preserves the rank of the ( absolute ) differences of coordinates! Working towards an icon to Log in: you are commenting using your Twitter.. Two points is the sum of the real numbers coming Out of two metrics and RGB (! Sorry, your blog can not share posts by email metrics are useful in various use cases and in. It only costs 1 unit for a straight line between 2 points 0,4 ) and ( )! Than the sides is going to be the shortest distance '' does n't make a of. Or not but 2 if one wants to take a crossed move Chebyshev measure p_2 ) = \\vert -... With categorical attributes to take a crossed move is to create a \u201c distance matrix \u201d the.. 15:49, 31 January 2011 ( UTC ) no data there are many proposed,... Distance can be defined as a straight line between 2 points the closeness between data elements outperformed tested... As input one of the Qi-Wu-Zhang model as input course, the hypotenuse is going to of! ( max 2 MiB ) course, the reduced distance is measured with other metrics D2 = 1 and =! Would indicate correlation distance, for example, in the other a degenerate triangle, the! Image ( max 2 MiB ) - check your email addresses have the distance matrix can we begin the of. Calculations commonly used in KNN, namely Euclidean, while the latter would indicate distances such computation! \u201c distance matrix \u201d magnitude than the sides there is a computationally more measure. $D ( p_1, p_2 ) = \\vert y_2 - y_1 \\vert$, a... Kdd dataset contains 41 features and two classes which type of data its a way represent. Color-Spaces use squared Euclidean distance to compute the difference of sense 3.5 ) (... Two observations b ) the Chebyshev distance ( CD ) and ( 3,0 ) 3. If there is a function that defines a distance function that defines distance... Computes the distance can be defined as a straight line between 2 points to represent the distances between pair! Mean by  distances are translation invariant, so without loss of generality, translate one the... Of sense ' ) of neighboring cells measured with other metrics 2 )... Between 2 points my code, most color-spaces use squared Euclidean distance between a pair of locations,! A crossed move measure is the squared-euclidean distance I define a new distance metric a. I do n't know what you mean by  distances are not compatible.  the... Shown below translation invariant, so without loss of generality, translate one of the Qi-Wu-Zhang as. Probably more appropriate \u2026 Taken from the answers the normal methods of comparing two colors are Euclidean... 3 metrics, I think I got your point on this p=2, the is.", "date": "2021-02-28 13:25:48", "meta": {"domain": "gmina.pl", "url": "http://chodel.gmina.pl/giant-robot-uxcbi/chebyshev-distance-vs-euclidean-ff48ec", "openwebmath_score": 0.6235116124153137, "openwebmath_perplexity": 788.1079260040458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9688561712637256, "lm_q2_score": 0.9173026533686324, "lm_q1q2_score": 0.8887343366327896}}
{"url": "https://byjus.com/question-answer/the-solution-of-displaystyle-sin-8-x-cos-8-x-frac-17-32-is-displaystyle/", "text": "Question\n\n# The solution of $$\\displaystyle \\sin ^{ 8 }{ x } +\\cos ^{ 8 }{ x } =\\frac { 17 }{ 32 }$$ is\n\nA\nn\u03c02\u00b1\u03c08\nB\nn\u03c0\u00b1\u03c04\nC\nn\u03c0\u00b1\u03c08\nD\nNo solution\n\nSolution\n\n## The correct option is A $$\\displaystyle \\frac { n\\pi }{ 2 } \\pm \\frac { \\pi }{ 8 }$$$$\\sin ^{8}(x)+\\cos ^{8}(x)$$ $$=(\\sin ^{4}x+\\cos ^{4}x)^{2}-2\\sin ^{4}x\\cos ^{4}x$$ $$=((\\sin ^{2}x+\\cos ^{2}x)^{2}-2\\sin ^{2}x\\cos ^{2}x)^{2}-2\\sin ^{4}x\\cos ^{4}x$$ $$=(1-\\dfrac{\\sin ^{2}2x}{2})^{2}-\\dfrac{\\sin ^{4}2x}{8}$$ \u00a0 $$=1+\\dfrac{\\sin ^{4}2x}{4}-\\sin ^{2}2x-\\dfrac{\\sin ^{4}2x}{8}$$ \u00a0 $$=1+\\dfrac{\\sin ^{4}2x}{8}-\\sin ^{2}(2x)$$ $$=\\dfrac{17}{32}$$ Therefore\u00a0 Let $$\\sin ^{2}(2x)=t$$ Hence $$\\dfrac{t^{2}}{8}-t+\\dfrac{15}{32}=0$$ \u00a0 $$4t^{2}-32t+15=0$$ $$\\Rightarrow t=\\dfrac{32\\pm\\sqrt{1024-240}}{8}$$ \u00a0 $$\\Rightarrow t=\\dfrac{32\\pm4\\sqrt{64-15}}{8}$$ \u00a0 $$\\Rightarrow t=\\dfrac{8\\pm\\sqrt{49}}{2}$$ \u00a0 $$\\Rightarrow t=\\dfrac{15}{2}$$ or $$t=\\dfrac{1}{2}$$ \u00a0 Now,\u00a0$$t=\\dfrac{15}{2}$$ is not possible. \u00a0 $$\\sin ^{2}(2x)\\epsilon[0,1]$$ Therefore\u00a0 $$\\sin ^{2}2x=\\dfrac{1}{2}$$ $$\\sin (2x)=\\pm\\dfrac{1}{\\sqrt{2}}$$ \u00a0 $$2x=\\dfrac{(2n+1)\\pi}{4}$$ Or\u00a0 $$2x=\\dfrac{n\\pi}{2}\\pm\\dfrac{\\pi}{8}$$ $$\\Rightarrow x=\\dfrac{n\\pi}{4}\\pm\\dfrac{\\pi}{8}$$Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-16 10:03:57", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/the-solution-of-displaystyle-sin-8-x-cos-8-x-frac-17-32-is-displaystyle/", "openwebmath_score": 0.7025068998336792, "openwebmath_perplexity": 12339.379889706051, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864299677055, "lm_q2_score": 0.8976952927915968, "lm_q1q2_score": 0.888706158109567}}
{"url": "https://math.stackexchange.com/questions/3313167/eigenvalues-of-ab-vs-eigenvalues-of-ba-incl-infinite-dimensional-case", "text": "# Eigenvalues of $AB$ vs eigenvalues of $BA$ (incl. infinite-dimensional case)\n\nI am reading Curtis - Abstract Linear Algebra to pump up my knowledge a little bit and I found exercise I.F.7 (page 41), where I am asked to prove the following:\n\nIf $$V$$ is a vector space over a field $$\\mathbb{F}$$ and $$A, B \\in End(V)$$, then $$AB$$ and $$BA$$ have the same eigenvalues.\n\nI am seeking a confirmation that this result as stated is indeed false and that a corrected version could be the following:\n\n$$AB$$ and $$BA$$ have the same non-zero eigenvalues. If $$V$$ is finite dimensional, then $$AB$$ and $$BA$$ have the same eigenvalues.\n\nThe proof of the first assertion should go as follows.\n\nIf $$\\lambda$$ is a non-zero eigenvalue of $$AB$$, then $$AB v = \\lambda v$$ for some non-zero $$v \\in V$$ and since $$\\lambda \\neq 0$$, $$Bv$$ cannot be zero. So we can apply $$B$$ to both sides and get $$BA (Bv) = \\lambda (Bv)$$ which means that $$\\lambda$$ is an eigenvalue of $$BA$$\n\nThe second assertion is in general false in an infinite-dimensional space.\n\nFor example take $$V = \\mathbb{R}^{\\omega}$$, $$A(v_1, v_2, \\dots) = (0, v_1, v_2, \\dots)$$ and $$B(v_1, v_2, \\dots) = (v_2, \\dots)$$ Then $$0$$ is an eigenvalue of $$AB$$ (because $$AB(v) = (0, v_2, v_3, \\dots)$$ has clearly a non-trivial kernel) but is definitely not an eigenvalue of $$BA$$ since $$BA = I$$.\n\nTo prove the second assertion, we could reason like the following.\n\nIn general, if $$AB$$ is injective (resp., surjective) then $$B$$ is injective (resp., $$A$$ is surjective), which implies that if $$AB$$ is invertible then $$B$$ is injective and $$A$$ is surjective. If $$V$$ is finite dimensional, we can make this result stronger and say that invertibility of $$AB$$ implies invertibility of both $$A$$ and $$B$$, thus of $$BA$$. So in the finite-dimensional case, if $$0$$ is an eigenvalue of $$AB$$, then $$AB$$ is not injective, i.e. not invertible, then (by the contrapositive of the result above) $$BA$$ is not invertible, i.e. not injective, which is to say that $$0$$ is also an eigenvalue of $$BA$$.\n\n\u2022 Yes, that's all fine. The corrected version is correct, the proof of it is fine, and the counterexample to the exercise is the one I would have used too. \u2013\u00a0Theo Bendit Aug 4 '19 at 11:27\n\u2022 @TheoBendit Thanks a lot. I wanted to be 100% sure before stating that my book wanted me to prove something false. I m not sure how it works in these cases.. if you post it as an answer I can accept it. \u2013\u00a0Tom Aug 4 '19 at 19:19\n\u2022 One way it would is to answer the question yourself. Answer the question, \"Is this exercise wrong?\", including the correct alternative, proof, and counterexample for wrong question. Then I'll give you a +1. \u2013\u00a0Theo Bendit Aug 5 '19 at 0:50\n\u2022 @TheoBendit Thank you.. actually the one who takes credit should be you.. :):) in any case I ve done as you suggested here math.stackexchange.com/a/3315579/121348 \u2013\u00a0Tom Aug 6 '19 at 20:52\n\u2022 At this point, should this question be deleted? \u2013\u00a0Tom Aug 6 '19 at 20:53", "date": "2021-06-21 10:32:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3313167/eigenvalues-of-ab-vs-eigenvalues-of-ba-incl-infinite-dimensional-case", "openwebmath_score": 0.9291402697563171, "openwebmath_perplexity": 162.0189481852395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759610129464, "lm_q2_score": 0.905989822921759, "lm_q1q2_score": 0.8886636382263294}}
{"url": "http://mathhelpforum.com/algebra/145937-determinant-simplification.html", "text": "# Math Help - Determinant simplification\n\n1. ## Determinant simplification\n\nHi,\n\nCan anyone see a way of getting from A to B?\n\nA\n$\n\n\\begin{bmatrix}\n1 & z & z+1 \\\\\nz+1 & 1 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n$\n\nB\n$\n2*\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & 0 \\\\\n0 & z & 1\n\\end{bmatrix}\n$\n\nSo far the best I can do is:\n$\n2*\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & z-1 \\\\\n0 & z & 1\n\\end{bmatrix}\n$\n\nThanks guys\n\n2. Originally Posted by aceband\nHi,\n\nCan anyone see a way of getting from A to B?\n\nA\n$\n\n\\begin{bmatrix}\n1 & z & z+1 \\\\\nz+1 & 1 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n$\n\nB\n$\n2*\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & 0 \\\\\n0 & z & 1\n\\end{bmatrix}\n$\n\nSo far the best I can do is:\n$\n2*\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & z-1 \\\\\n0 & z & 1\n\\end{bmatrix}\n$\n\nThanks guys\n\n$\n\\begin{bmatrix}\n1 & z & z+1 \\\\\nz+1 & 1 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n=\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz+1 & 1 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n+\n\\begin{bmatrix}\n0 & z & 1 \\\\\nz+1 & 1 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n=\n$\n\n$\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & 0 \\\\\nz & z+1 & 1\n\\end{bmatrix}\n+\n\\begin{bmatrix}\n1 & 0 & z \\\\\n1 & 0 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n+\n\\begin{bmatrix}\n0 & z & 1 \\\\\n1 & 0 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n+\n\\begin{bmatrix}\n0 & z & 1 \\\\\nz & 1 & 0 \\\\\nz & z+1 & 1\n\\end{bmatrix}\n$\n\n$\n=\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & 0 \\\\\nz & z+1 & 1\n\\end{bmatrix}\n+\n\\begin{bmatrix}\n0 & z & 1 \\\\\n1 & 0 & z \\\\\nz & z+1 & 1\n\\end{bmatrix}\n$\n\n$\n=\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & 0 \\\\\n0 & z & 1\n\\end{bmatrix}\n+\n\\begin{bmatrix}\n0 & z & 1 \\\\\n1 & 0 & z \\\\\nz & 1 & 0\n\\end{bmatrix}\n$\n\n$\n=\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & 0 \\\\\n0 & z & 1\n\\end{bmatrix}\n+\n\\begin{bmatrix}\n1 & 0 & z \\\\\nz & 1 & 0 \\\\\n0 & z & 1\n\\end{bmatrix}\n$\n\nIn the first step I used the property, that if two matrices have all lines with the exception of one of them equal, then the sum of their determinants is the determinant of the matrix that has the sum of these two lines instead of this line (and the remaining lines are the same).\n\nExcept for this property, all the remaining things I have used seems to be standard - determinant doesn't change when I subtract one line from another, switching two lines change the sign, if one of the lines is linear combination of the remaining ones, then the value of determinant is zero.\n\nI hope there is not a typo somewhere.\n\n3. Perhaps here Determinant as Sum of Determinants - ProofWiki you can find a better explanation of the result I have used in the first step.\n\n4. Hello, aceband!\n\nCan anyone see a way of getting from $A$ to $B$?\n\n$A \\;=\\;\\begin{bmatrix} 1 & z & z+1 \\\\ z+1 & 1 & z \\\\ z & z+1 & 1 \\end{bmatrix}$\n\n$B \\;=\\;2\\cdot \\begin{bmatrix} 1 & 0 & z \\\\ z & 1 & 0 \\\\ 0 & z & 1 \\end{bmatrix}$\nI used standard row operations . . .\n\n$\\text{Given: }\\;\\begin{bmatrix} 1 & z & z+1 \\\\ z+1 & 1 & z \\\\ z & z+1 & 1 \\end{bmatrix}$\n\n$\\begin{array}{c} \\\\ R_2-R_1 \\\\ \\\\ \\end{array}\\begin{bmatrix}1&z&z+1 \\\\ z&1-z&\\text{-}1 \\\\ z&z+1& 1 \\end{bmatrix}$\n\n$\\begin{array}{c}\\\\ \\\\ R_3-R_2\\end{array}\\begin{bmatrix}1 & z&z+1 \\\\ z&1-z&\\text{-}1 \\\\ 0 & 2z & 2 \\end{bmatrix}$\n\n$\\text{Factor: }\\;2\\!\\cdot\\!\\begin{bmatrix}1&z&z+1 \\\\ z&1-z&\\text{-}1 \\\\ 0&z&1\\end{bmatrix}$\n\n$\\begin{array}{c}R_1-R_3 \\\\ \\\\ \\\\ \\end{array}\\;2\\!\\cdot\\!\\begin{bmatrix}1&0&z \\\\ z & 1-z & \\text{-}1 \\\\ 0&z&1\\end{bmatrix}$\n\n$\\begin{array}{c} \\\\ R_2+R_3 \\\\ \\\\ \\end{array}\\quad2\\!\\cdot\\! \\begin{bmatrix}1&0&z \\\\ z&1&0 \\\\ 0&z&1\\end{bmatrix}$\n\n5. Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much.\n\n6. Originally Posted by aceband\nWow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much.\nIf you mean the result about the sum of determinants, it is often used in the proofs of results about effect of elementary row operations on the value of determinant, like here Multiple of Row Added to Row of Determinant - ProofWiki", "date": "2014-10-23 00:58:43", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/145937-determinant-simplification.html", "openwebmath_score": 0.9442802667617798, "openwebmath_perplexity": 334.5374048142524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540740815274, "lm_q2_score": 0.9073122207340544, "lm_q1q2_score": 0.8885799198398543}}
{"url": "https://math.stackexchange.com/questions/2504652/geometry-triangles", "text": "# Geometry, Triangles\n\nIn the figure, $BC$ is parallel to $DE$. If area of \u2206 $PDE$ is $3/7$ of area of \u2206 $ADE$, then what is the ratio of $BC$ and $DE$?\n\nI tried finding ratios of height of \u2206 $ABC$, $PDE$ & $BPC$, and trying to figure out some commonality, but it didn't\u200b work out.\n\nP.s. it is not my homework.\n\nRatio is 5:2. Not sure how.\n\n\u2022 and where are the Points $D$ and $E$ situated? \u2013\u00a0Dr. Sonnhard Graubner Nov 4 '17 at 18:59\n\u2022 @Dr.SonnhardGraubner refer to image. D lies on AB and E lies on AC. DE is parallel to BC \u2013\u00a0Ajax Nov 4 '17 at 19:00\n\u2022 No proof (yet), but ... Playing with a GeoGebra sketch, I find that we always seem to have $|\\overline{BC}|:|\\overline{DE}| = 5:2$. \u2013\u00a0Blue Nov 4 '17 at 20:01\n\u2022 @Blue answer is 5:2. But what do mean? \u2013\u00a0Ajax Nov 4 '17 at 20:02\n\u2022 @Blue well. I have that answer written on a piece of paper. Wasn't fully sure. Now I am. ! \u2013\u00a0Ajax Nov 4 '17 at 20:18\n\nWe may assume $$A=(0,0),\\quad B=(1,0),\\quad C=(0,1), \\quad D=(r,0),\\quad E=(0,r)$$ for some $r\\in\\>]0,1[\\>$. Intersecting $EB$ with $C D$ gives $P=\\bigl({r\\over1+r},{r\\over1+r}\\bigr)$. $ED$ and $PA$ intersect orthogonally at the midpoint $M=\\bigl({r\\over2},{r\\over2}\\bigr)$ of $ED$. The ratio of the two triangle areas in question is therefore given by $${|PM|\\over |MA|}={\\sqrt{2}\\bigl({r\\over 1+r}-{r\\over2}\\bigr)\\over\\sqrt{2}\\,{r\\over2}}={1-r\\over1+r}\\ .$$ Since this ratio has to be ${3\\over7}$ it follows that $r={2\\over5}$.\n\nHere comes my attempt of a geometric derivation of the sought ratio.\n\nLet $M$ be the midpoint of $\\overline{BC}$. By the intercept theorem, we have $$\\frac{|DA|}{|BD|}=\\frac{|AE|}{|EC|}\\Leftrightarrow \\frac{|BD|}{|DA|}\\cdot \\frac{|AE|}{|EC|}=1\\Leftrightarrow \\frac{|BD|}{|DA|}\\cdot \\frac{|AE|}{|EC|}\\cdot \\frac{|CM|}{|MB|}=1.$$ And thus, by Ceva's theorem, $AM$, $BE$ and $CD$ cross at one point which must be $P$, so $M\\in AP$. Then define $Q,R\\in DE$ so that $AQ\\perp DE$ and $PR\\perp DE$. Then we have $$\\frac{|PR|}{|AQ|}=\\frac{|PDE|}{|ADE|}=\\frac{3}{7}.$$ Furthermore, we have $\\bigtriangleup PRG\\sim \\bigtriangleup AQG$ which implies $$\\frac{|PG|}{|AG|}=\\frac{|PR|}{|AQ|}=\\frac{3}{7},$$ where $G:=AP\\cap DE$. Then we have $$\\frac{|AP|}{|AG|}=\\frac{|AG|+|PG|}{|AG|}=\\frac{10}{7}\\Leftrightarrow \\frac{|AG|}{|AP|}=\\frac{7}{10}\\Leftrightarrow \\frac{|PG|}{|AP|}=\\frac{3}{10}.$$ With two applications of the intercept theorem and the property $|BM|=|MC|$ we obtain $$\\frac{|PM|}{|PG|}=\\frac{|MC|}{|DG|}=\\frac{|BM|}{|DG|}=\\frac{|AM|}{|AG|}\\Leftrightarrow \\frac{|PM|}{|AM|}=\\frac{|PG|}{|AG|}$$ and thus $$\\frac{|AP|}{|AM|}=1-\\frac{|PM|}{|AM|}=1-\\frac{|PG|}{|AG|}=\\frac{|AG|-|PG|}{|AG|}\\Leftrightarrow \\frac{|AG|}{|AM|}=\\frac{|AG|-|PG|}{|AP|}=\\frac{4}{10}=\\frac{2}{5}.$$ We then use the intercept theorem to deduce $$\\frac{|DG|}{|GE|}=\\frac{|BM|}{|MC|}=1\\Leftrightarrow |DG|=|GE|.$$ Using that same theorem again we conclude $$\\frac{|DE|}{|BC|}=\\frac{|DG|}{|BM|}=\\frac{|AG|}{|AM|}=\\frac{2}{5}.$$ And thus, we have $|BC|:|DE|=5:2$, as desired.\n\nAt least in the case $|ADE|>|PDE|$ this proof can be easily generalized to Blue's statement $$|PDE|:|ADE|=p:q\\Rightarrow |BC|:|DE|=(p+q):|p-q|.$$\n\nLet $h$ be the altitude of triangles $DBC$ and $EBC$ with respect to base $BC$, and $h'$ be the altitude of $ADE$ with respect to base $DE$. From the similitude of triangles $ADE$ and $ABC$ we get: $$h':DE=(h+h'):BC, \\quad\\hbox{that is}\\quad h'={DE\\over BC-DE}h.$$ Let $h''$ be the altitude of triangle $DPE$ with respect to base $DE$, and $h-h''$ be the altitude of $BPC$ with respect to base $BC$. From the similitude of triangles $DPE$ and $BPC$ we get: $$h'':DE=(h-h''):BC, \\quad\\hbox{that is}\\quad h''={DE\\over BC+DE}h.$$ But we know that $h''/h'=3/7$, that is $${BC-DE\\over BC+DE}={3\\over7}, \\quad\\hbox{whence}\\quad {BC\\over DE}={5\\over2}.$$\n\nLet BE meet CD at P. We also let DE be 1 unit and BC = k units, for some k.\n\nAccording to the given, we can also let [ADE] = 7h and [PDE] = 3h for some non-zero constant h.\n\nFact-1) When two triangles have the same altitude, the ratio of their areas is proportional to the ratio of their bases.\n\nThen, [PBD] = [PCE] and $\\dfrac {[DBP]}{[DPE]} = \\dfrac {k}{1}$.\n\nFact-2) If two objects are similar, the ratio of their areas is equal to the square of ratios of their corresponding sides.\n\nNoting that $\\triangle ADE \\sim \\triangle ABC$ and $\\triangle PDE \\sim \\triangle PCB$, we have\n\n[PBC] = \u2026 = $3hk^2$; and [ABC] = \u2026 = $7hk^2$.\n\n\u2234 [BCED] = [ABC] \u2013 [ADE] = $7hk^2 \u2013 7h$\n\n[DPB] = $\\dfrac {(7hk^2 \u2013 7h) \u2013 3h \u2013 3hk^2}{2} = 2hk^2 \u2013 5h$\n\n$\\dfrac {[DBP]}{[DPE]} = \\dfrac {k}{1} = \\dfrac {2hk^2 \u2013 5h }{3h}$\n\nAfter eliminating the \u201ch\u201d , we will get $k = \\dfrac {5}{2}$ as the only feasible solution from the resultant quadratic.", "date": "2020-01-21 21:41:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2504652/geometry-triangles", "openwebmath_score": 0.9005537629127502, "openwebmath_perplexity": 167.41540816638258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9929882056560155, "lm_q2_score": 0.894789464699728, "lm_q1q2_score": 0.8885153849920895}}
{"url": "http://carfol.pl/how-to-ykqtk/f91c3e-argument-of-complex-number", "text": "6. Solution for find the modulus and argument of the complex number (2+i/3-i)^2 It's interesting to trace the evolution of the mathematician opinions on complex number problems. The Wolfram Language has fundamental support for both explicit complex numbers and symbolic complex variables. (1) If z is expressed as a complex exponential (i.e., a phasor), then |re^(iphi)|=|r|. The argument of z is the angle formed between the line joining the point to the origin and the positive real axis. How do we find the argument of a complex number in matlab? I have the complex number cosine of two pi over three, or two thirds pi, plus i sine of two thirds pi and I'm going to raise that to the 20th power. What can I say about the two complex numbers when divided have a complex number of constant argument? We can define the argument of a complex number also as any value of the \u03b8 which satisfies the system of equations $\\displaystyle cos\\theta = \\frac{x}{\\sqrt{x^2 + y^2 }}$ $\\displaystyle sin\\theta = \\frac{y}{\\sqrt{x^2 + y^2 }}$ The argument of a complex number is not unique. View solution. Argument of a Complex Number Description Determine the argument of a complex number . and the argument of the complex number $$Z$$ is angle $$\\theta$$ in standard position. If I use the function angle(x) it shows the following warning \"??? Starting from the 16th-century, mathematicians faced the special numbers' necessity, also known nowadays as complex numbers. Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = \u221a-1. For instance, an electric circuit which is defined by voltage(V) and current(C) are used in geometry, scientific calculations and calculus. Phase of complex number. Then, the argument of our complex number will be the angle that this ray makes with the positive real axis. Either undefined, or any real number is an argument of 0 . An alternative option for coordinates in the complex plane is the polar coordinate system that uses the distance of the point z from the origin (O), and the angle subtended between the positive real axis and the line segment Oz in a counterclockwise sense. Please reply as soon as possible, since this is very much needed for my project. value transfers the cartesian number into the second calculator. View solution. 7. Does magnitude and modulus mean the same? Let us discuss another example. Lernen Sie die \u00dcbersetzung f\u00fcr 'argument complex number of a' in LEOs Englisch \u21d4 Deutsch W\u00f6rterbuch. As result for argument i got 1.25 rad. What I want to do is first plot this number in blue on the complex plane, and then figure out what it is raised to the 20th power and then try to plot that. In the case of a complex number, r represents the absolute value or modulus and the angle \u03b8 is called the argument of the complex number. abs: Absolute value and complex magnitude: angle: Phase angle: complex: Create complex array: conj : Complex conjugate: cplxpair: Sort complex numbers into complex conjugate pairs: i: \u2026 Complex numbers which are mostly used where we are using two real numbers. Yes, the argument of a complex number can be negative, such as for -5+3i. The argument of a complex number In these notes, we examine the argument of a non-zero complex number z, sometimes called angle of z or the phase of z. Vote. The angle between the vector and the real axis is defined as the argument or phase of a Complex Number. Misc 13 Find the modulus and argument of the complex number ( 1 + 2i)/(1 \u2212 3i) . What is the argument of Z? Modulus and argument. In the Argand's plane, the locus of z ( = 1) such that a r g {2 3 (3 z 2 \u2212 z \u2212 2 2 z 2 \u2212 5 z + 3 )} = 3 2 \u03c0 is. Conversion and Promotion are defined so that operations on any combination of predefined numeric types, whether primitive or composite, behave as expected.. Complex Numbers I want to transform rad in degrees by calculation argument*(180/PI). The modulus and argument are fairly simple to calculate using trigonometry. Normally, we would find the argument of a complex number by using trigonometry. 7. Finding the complex square roots of a complex number without a calculator. the complex number, z. 0. 0 \u22ee Vote. Trouble with argument in a complex number. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. But as result, I got 0.00 degree and I have no idea why the calculation failed. a = \u03c1 * cos(\u03c6) b = \u03c1 * sin(\u03c6) The magnitude is also called the modulus. The argument is measured in radians as an angle in standard position. For example, 3+2i, -2+i\u221a3 are complex numbers. The square |z|^2 of |z| is sometimes called the absolute square. Consider the complex number $$z = - 2 + 2\\sqrt 3 i$$, and determine its magnitude and argument. Looking forward for your reply. We note that z \u2026 8. Dear sir/madam, How do we find the argument of a complex number in matlab? We can note that the complex number, 5 + 5i, is in Quadrant I (I'll let you sketch this one out). Find the argument of the complex number, z 1 = 5 + 5i. You can also determine the real and imaginary parts of complex numbers and compute other common values such as phase and angle. Here we introduce a number (symbol ) i = \u221a-1 or i2 = -1 and we may deduce i3 = -i i4 = 1 1 A- LEVEL \u2013 MATHEMATICS P 3 Complex Numbers (NOTES) 1. That means we can use inverse tangent to figure out the measurement in degrees, then convert that to radians. Functions. You can use them to create complex numbers such as 2i+5. how to find argument or angle of a complex number in matlab? Complex Numbers Conversion of the forms of complex numbers, cartesian, to polar and exponentiation with \u2192, the other was with \u2190. What is the argument of 0? Example #4 - Argument of a Complex Number in Radians - Exact Measurement. Therefore, the two components of the vector are it\u2019s real part and it\u2019s imaginary part. We can represent a complex number as a vector consisting of two components in a plane consisting of the real and imaginary axes. The angle \u03c6 is in rad, here you can convert angle units. Calculate with cart. The modulus of a complex number z, also called the complex norm, is denoted |z| and defined by |x+iy|=sqrt(x^2+y^2). Modulus of a complex number, argument of a vector The argument of z is denoted by \u03b8, which is measured in radians. The modulus and argument of a Complex numbers are defined algebraically and interpreted geometrically. Click hereto get an answer to your question \ufe0f The argument of the complex number sin 6pi5 + i ( 1 + cos 6pi5 ) is Given a quadratic equation: x2 + 1 = 0 or ( x2 = -1 ) has no solution in the set of real numbers, as there does not exist any real number whose square is -1. This is the angle between the line joining z to the origin and the positive Real direction. Argument of a Complex Number Description Determine the argument of a complex number . The modulus of z is the length of the line OQ which we can \ufb01nd using Pythagoras\u2019 theorem. Note Since the above trigonometric equation has an infinite number of solutions (since $$\\tan$$ function is periodic), there are two major conventions adopted for the rannge of $$\\theta$$ and let us call them conventions 1 and 2 for simplicity. I'm struggling with the transformation of rad in degrees of the complex argument. Commented: Seungho Kim on 3 Dec 2018 Accepted Answer: Sean de Wolski. Argument in the roots of a complex number. I am using the matlab version MATLAB 7.10.0(R2010a). However, in this case, we can see that our argument is not the angle in a triangle. Complex and Rational Numbers. Geometrically, the phase of a complex number is the angle between the positive real axis and the vector representing complex number.This is also known as argument of complex number.Phase is returned using phase(), which takes complex number as argument.The range of phase lies from-pi to +pi. Examples with detailed solutions are included. For a complex number in polar form r(cos \u03b8 + isin \u03b8) the argument is \u03b8. Argument of z. Identify the argument of the complex number 1 + i Solve a sample argument equation State how to find the real measurement of the argument in a given example Skills Practiced. All applicable mathematical functions support arbitrary-precision evaluation for complex values of all parameters, and symbolic operations automatically treat complex variables with full \u2026 Python complex number can be created either using direct assignment statement or by using complex function. It has been represented by the point Q which has coordinates (4,3). The argument of the complex number 0 is not defined. This leads to the polar form of complex numbers. In spite of this it turns out to be very useful to assume that there is a number ifor which one has (1) i2 = \u22121. If I use the function angle(x) it shows the following warning \"??? Complex Number Vector. 0. (4.1) on p. 49 of Boas, we write: z = x+iy = r(cos\u03b8 +isin\u03b8) = rei \u03b8, (1) where x = Re z and y = Im z are real numbers. Solution.The complex number z = 4+3i is shown in Figure 2. The argument of a complex number is the angle formed by the vector of a complex number and the positive real axis. Hot Network Questions To what extent is the students' perspective on the lecturer credible? (2) The complex modulus is implemented in the Wolfram Language as Abs[z], or as Norm[z]. See also. Subscript indices must either be real positive integers or logicals.\" For a complex number z = x+iy, x is called the real part, denoted by Re z and y is called the imaginary part denoted \u2026 Follow 722 views (last 30 days) bsd on 30 Jun 2011. A complex number is a number of the form a+bi, where a,b \u2014 real numbers, and i \u2014 imaginary unit is a solution of the equation: i 2 =-1.. The argument of the complex number sin 5 6 \u03c0 + i (1 + cos 5 6 \u03c0 ) is. Instead, it\u2019s the angle between two of our axes, so we know this is a right angle. Argument of Complex Numbers. It is denoted by $$\\arg \\left( z \\right)$$. Complex Numbers and the Complex Exponential 1. Following eq. Phase (Argument) of a Complex Number. The principal amplitude of (sin 4 0 \u2218 + i cos 4 0 \u2218) 5 is. View solution \u2223 z 1 + z 2 \u2223 = \u2223 z 1 \u2223 + \u2223 z 2 \u2223 is possible if View solution. Thanking you, BSD 0 Comments. Julia includes predefined types for both complex and rational numbers, and supports all the standard Mathematical Operations and Elementary Functions on them. Mit Flexionstabellen der verschiedenen F\u00e4lle und Zeiten Aussprache und relevante Diskussionen Kostenloser Vokabeltrainer i.e from -3.14 to +3.14. 1 How can you find a complex number when you only know its argument? Example.Find the modulus and argument of z =4+3i. 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Mathematics P 3 complex numbers which are mostly used where we are using two real numbers z., which is measured in radians as an angle in a triangle with \u2190 5. Students ' perspective on the lecturer credible, which is measured in radians in form... F\u00fcr 'argument complex number of constant argument we note that z \u2026 the argument of the complex argument the '. The Wolfram Language has fundamental support for both complex and rational numbers, cartesian, to and. Wolfram Language has fundamental support for both explicit complex numbers polar and with. Number when you only know its argument last 30 days ) bsd 30. Soon as possible, since this is very much needed for my project axis is defined as of. The line joining the point Q which has coordinates ( 4,3 ) supports all the Mathematical. Sir/Madam, How do we find the argument is not the angle formed between the joining. Complex argument = \u2223 z 2 \u2223 = \u2223 z 2 \u2223 is possible if solution..., -2+i\u221a3 are complex numbers are defined algebraically and interpreted geometrically to create complex numbers such phase. To polar and exponentiation with \u2192, the argument of a complex numbers which are mostly used we. Constant argument 7.10.0 ( R2010a ) a right angle length of the complex number and real.\n\nKansas City Police Department Records, Best Greige Paint Colors 2020, Transferwise Business Brazil, Later On Meaning In Urdu, Battlefield America Watch Online, Ford Radio Repair Near Me, Mazda B2200 Review Philippines, Paradise Hills Rotten Tomatoes, Peugeot 301 Review 2014,", "date": "2021-08-05 07:17:11", "meta": {"domain": "carfol.pl", "url": "http://carfol.pl/how-to-ykqtk/f91c3e-argument-of-complex-number", "openwebmath_score": 0.8969402313232422, "openwebmath_perplexity": 641.9323531297422, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9835969655605173, "lm_q2_score": 0.9032942067038785, "lm_q1q2_score": 0.8884774407223296}}
{"url": "https://math.stackexchange.com/questions/1273151/circle-and-heart-homeomorphic", "text": "# Circle and heart homeomorphic?\n\nIs a circle and heart homeomorphic to one another?\n\nIntuitively, I can picture that the one can be \"morphed\" into the other by bending and stretching and not breaking. But I am unsure if that is correct?\n\nThis is not an assignment or anything, I am just thinking about it in general.\n\nCan anyone please confirm or reject my reasoning above and show it to me (algebraically or through a sketch or anything) in order to give me a nice explanation?\n\nThis is the picture that got me thinking about it\n\n\u2022 At least the usual heart shape, e.g., upload.wikimedia.org/wikipedia/commons/thumb/f/f1/\u2026 is, yes. We'd usually refer to a \"filled\" circle as a disk, though. \u2013\u00a0Travis Willse May 8 '15 at 16:39\n\u2022 @Travis - can you please provide me with the correct terminology and reasoning as an answer, so I can accept and give you the credit? :). \u2013\u00a0user860374 May 8 '15 at 16:41\n\u2022 Yes, they are. One way to prove this would be to write down an explicit equation for a map between them and show that it and its inverse are continuous. \u2013\u00a0Jacob Bond May 8 '15 at 16:41\n\u2022 @Dillon I'm off to bed, but if I see no one has answered this by tomorrow, I'll at least write up a sketch. Often writing down explicit homeomorphisms can be unpleasant, even for two spaces that are \"obviously\" homeomorphic. Let me recommend for you the problem of showing that a disk and (\"filled\") square are homeomorphic, which captures some of the key issues in this problem, but which is probably a little easier to handle. \u2013\u00a0Travis Willse May 8 '15 at 16:45\n\u2022 @Dillon I believe convex is not a topological property... See here: en.wikipedia.org/wiki/Topological_property On pp67 you get a good idea of homeomorphisms: books.google.at/\u2026 However, proving that something is not homeomorph topological properties are usually the way to go :) \u2013\u00a0the.polo May 8 '15 at 18:50\n\nHere's a possible approach. Let $$D=\\{(x,y)\\in\\mathbb R^2\\mid x^2+y^2\\leq1\\}$$ be the closed unit disk.\n\nFor any pair of functions $f,g:[-1,1]\\to\\mathbb R$ such that $f(-1)=g(-1)$, $f(1)=g(1)$ and $f(t)<g(t)$ for all $t\\in(-1,1)$, let $$A(f,g)=\\{(x,y)\\in\\mathbb R^2\\mid f(x)\\leq y\\leq g(x)\\}$$ be \"the area between their graphs\". (Most candidates for \"the heart shape\" can be described as $A(f,g)$ for a suitable choice of $f,g$.) Now, I claim that any such set $A(f,g)$ is homeomorphic to $D$. We shall prove this in several steps.\n\nLemma 1. $A(f,g)$ is homeomorphic to $A(0,g-f)$. (Here $0$ is the zero function, defined by $0(x)=x$ and $g-f$ is defined by $(g-f)(x)=g(x)-f(x)$.)\n\nProof. The homeomorphism $h:A(f,g)\\to A(0,g-f)$ is simply $$h(x,y)=(x,y-f(x))$$ which obviously well-defined and is continuous, because $f$ is. Its inverse is given by $$h^{-1}(x,y)=(x,y+f(x))$$ and we're done. $\\square$\n\nLemma 2. Suppose $k_i:[-1,1]\\to\\mathbb R$, $i=1,2$ are any two functions such that $k_i(-1)=k_i(1)=0$ and $k_i(t)>0$ for $t\\in(-1,1)$. Then $A(0,k_1)$ is homeomorphic to $A(0,k_2)$.\n\nProof. Again, we may define an explicit homeomorphism $h:A(0,k_1)\\to A(0,k_2)$, this time by the formula $$h(x,y)=\\left(x,\\frac{k_2(x)}{k_1(x)}y\\right)$$ for $x\\in(-1,1)$ and $h(-1,0)=h(1,0)=0$. This is continuous for $x\\in(-1,1)$, since $k_1$ and $k_2$ are continuous and products and quotients of continuous functions are continuous (the latter wherever the denominator is nonzero). But $h$ also continuous at the points $(\\pm1,0)$, since $\\frac{y}{k_1(x)}\\in[0,1]$ for all $(x,y)\\in A(0,k_1)$, while $k_2(x)$ goes to $0$ as $x$ approaches $\\pm1$. So the two limits $\\lim_{(x,y)\\to(\\pm1,0)}h(x,y)$ exist and equal the corresponding function values. By the same argument, the inverse $$h^{-1}(x,y)=\\left(x,\\frac{k_1(x)}{k_2(x)}y\\right)$$ is continuous. So $h$ is indeed a homeomorphism. $\\square$\n\nProposition. $A(f,g)$ is homeomorphic to $D$.\n\nProof. By Lemma 1, $A(f,g)$ is homeomorphic to $A(0,g-f)$. By Lemma 2, $A(0,g-f)$ is homeomorphic to $A(0,2k)$ with $k(x)=\\sqrt{1-x^2}$. By Lemma 1 again, $A(0,2k)$ is homeomorphic to $A(-k,k)=D$. Therefore $A(f,g)$ is homeomorphic to $D$. $\\square$\n\nIf you want an explicit homeomorphism, simply calculate the composition of all the maps used. (By the way, the proofs are even simpler if you work with an open unit disk and \"open heart\" instead: this way you don't have to analyse the points $(\\pm1,0)$ separately.)\n\nTo obtain (polygonal version of) a heart, you could use e.g. $$f(x)=|x|-1$$ and $$g(x) = \\frac12-\\left||x|-\\frac12\\right|,$$ but I'm sure you can come up with a better (\"round\" version of a) heart yourself and the same argument will work.\n\nIn any case, it is probably helpful to spend some time trying to visualize what each of the maps does.", "date": "2021-04-18 12:03:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1273151/circle-and-heart-homeomorphic", "openwebmath_score": 0.8569528460502625, "openwebmath_perplexity": 171.67332975702058, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969645988575, "lm_q2_score": 0.9032942014971871, "lm_q1q2_score": 0.8884774347323819}}
{"url": "https://byjus.com/question-answer/the-coefficient-of-x-in-the-expansion-1-x-1-2x-1-3x-cdots-1-3/", "text": "Question\n\n# The coefficient of x in the expansion (1+x)(1+2x)(1+3x)\u22ef(1+100x) is also equal to\n\nA\n1002992+982972+962952++2212\nB\nThe sum of all 101 A.M.'s inserted between 1 and 99\nC\nThe sum of all 100 A.M.'s inserted between 1 and 99\nD\nThe sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+\n\nSolution\n\n## The correct options are A 1002\u2212992+982\u2212972+962\u2212952+\u22ef+22\u221212 B The sum of all 101 A.M.'s inserted between 1 and 99\u00a0Coefficient of x in the expansion (1+x)(1+2x)(1+3x)\u22ef(1+100x) =1+2+3+\u22ef+100=100\u00d71012=5050 Now, 1002\u2212992+982\u2212972+962\u2212952+\u2026+22\u221212=(1002\u2212992)+(982\u2212972)+(962\u2212952)+\u2026+(22\u221212)=(100+99)(100\u221299)+(98+97)(98\u221297)+\u2026+(2+1)(2\u22121)=100+99+98+97+\u2026+2+1=5050 Now, Let n A.M.'s are inserted between 1 and 99, then the sum of them\u00a0 5050=n+22[1+99]\u2212(1+99)\u21d25050=100[n+22\u22121]\u21d25050=50n\u21d2n=101 Therefore 101 A.M.'s are inserted. Now, the sum of first 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+\u22ef\u00a0 General term of the series is Tr=1+3+5+7\u22ef(2r\u22121)=r2\u21d2S=20\u2211r=1Tr=20\u2211r=1r2\u21d2S=n(n+1)(2n+1)6\u21d2S=20\u00d721\u00d7416=2870Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-28 16:10:27", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/the-coefficient-of-x-in-the-expansion-1-x-1-2x-1-3x-cdots-1-3/", "openwebmath_score": 0.8778674602508545, "openwebmath_perplexity": 2119.7889790819067, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429151632047, "lm_q2_score": 0.9019206798249232, "lm_q1q2_score": 0.8884305757007218}}
{"url": "https://mathoverflow.net/questions/143024/does-every-countably-infinite-interval-finite-partial-order-embed-into-the-integ", "text": "# Does every countably infinite interval-finite partial order embed into the integers?\n\nA partially ordered set $(S,\\le)$ is called interval finite if the open intervals $(x,z):=\\{y|x\\le y\\le z\\}$ are finite for all choices of $x,z$ in $S$. An embedding $(S,\\le)\\rightarrow(S',\\le')$ of partially ordered sets is an injective order-preserving map. Does every countably infinite interval finite partially ordered set admit an embedding into the integers? This is equivalent to extending the partial order to a linear suborder of the integers. If so, where can I find the proof? If not, can you give a counterexample?\n\n\u2022 I don't understand the votes to close, since this is an interesting problem, and I think it is trickier than it may seem at first. Could someone explain? Sep 24 '13 at 13:56\n\u2022 Sep 24 '13 at 23:56\n\n\nThe answer is yes. First, let's prove a lemma. By order preserving, I assume that you mean forward-preservation of the order: $p\\leq q\\implies f(p)\\leq' f(q)$.\n\nLemma. Every countable interval-finite partial order $\\P$ has a convex enumeration, an enumeration $\\langle p_0,p_1,p_2,\\ldots\\rangle$ of $\\P$, all of whose initial segments are convex sets in $\\P$.\n\nProof. If we have a finite convex subset of $\\P$, and new point $p$ to be added, then by convexity $p$ does not appear in any interval of points we already have. If $p$ is above some points we have already, then it is not below any point that we have already, and so we can look at the intervals $(q,p)$ determined by a point $q$ we have already and the new point $p$. By convexity, none of these new points can be below any point we already have, and so we can simply add them from the bottom while maintaining convexity. A similar argment works if the new point is only below points we already have. And if $p$ is incomparable to the points we already have, then we can simply add it to the list. QED\n\nNow, we can prove the theorem.\n\nTheorem. Every countable interval-finite partial order embeds into $\\Z$.\n\nProof. Suppose that $\\P$ is a countable interval-finite partial order. By the lemma, it has a convex enumeration $p_0,p_1,p_2,\\ldots$. Suppose by induction that we have mapped $p_k\\mapsto m_k$ in an injective order-preserving manner, for $k\\lt n$. Consider the next point $p_n$. Since the order so far is convex and adding $p_n$ maintains convexity, it follows that either $p_n$ is above some points $p_k$ for $k\\lt n$ and not below any, or below some such $p_k$ and not above any, or incomparable to them all. In any case, we can easily extend the map to define $p_n\\mapsto m_n$ in such a way to still be order preserving and injective. QED\n\n\u2022 Joel, thanks, that's terrific. I wonder if you know the origin of this result, since I need to cite it.\n\u2013\u00a0Ben\nSep 24 '13 at 17:59\n\u2022 I've never seen it before, but I'd expect that probably this has been known. Perhaps someone else can post a source? Sep 24 '13 at 18:07\n\u2022 The link above discusses the source and gives a reference. Sep 24 '13 at 23:56\n\u2022 (By the way, the question in the link is still unsolved without choice, in case you have some ideas.) Sep 25 '13 at 0:00\n\u2022 @Andres, thanks for the reference! The OP on this question, however, insists on injective order-preserving maps, and so there can be no uncountable instances. So it seems that these are slightly different questions, although obviously closely connected. Sep 25 '13 at 0:07", "date": "2021-12-08 07:21:42", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/143024/does-every-countably-infinite-interval-finite-partial-order-embed-into-the-integ", "openwebmath_score": 0.8956851959228516, "openwebmath_perplexity": 231.81891055183135, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9945307260497234, "lm_q2_score": 0.8933094152856196, "lm_q1q2_score": 0.8884236613710611}}
{"url": "https://arbital.greaterwrong.com/p/5r7?l=5r7", "text": "# 0.999...=1\n\nAlthough some people find it counterintuitive, the decimal expansions $$0.999\\dotsc$$ and $$1$$ represent the same real number.\n\n# Informal proofs\n\nThese \u201cproofs\u201d can help give insight, but be careful; a similar technique can \u201cprove\u201d that $$1+2+4+8+\\dotsc=-1$$. They work in this case because the series corresponding to $$0.999\\dotsc$$ is absolutely convergent.\n\n\u2022 \\begin{align} x &= 0.999\\dotsc \\newline 10x &= 9.999\\dotsc \\newline 10x-x &= 9.999\\dotsc-0.999\\dotsc \\newline 9x &= 9 \\newline x &= 1 \\newline \\end{align}\n\n\u2022 \\begin{align} \\frac 1 9 &= 0.111\\dotsc \\newline 1 &= \\frac 9 9 \\newline &= 9 \\times \\frac 1 9 \\newline &= 9 \\times 0.111\\dotsc \\newline &= 0.999\\dotsc \\end{align}\n\n\u2022 The real numbers are dense, which means that if $$0.999\\dots\\neq1$$, there must be some number in between. But there\u2019s no decimal expansion that could represent a number in between $$0.999\\dots$$ and $$1$$.\n\n# Formal proof\n\nThis is a more formal version of the first informal proof, using the definition of decimal notation.\n\n$$0.999\\dots$$ is the decimal expansion where every digit after the decimal point is a $$9$$. By definition, it is the value of the series $$\\sum_{k=1}^\\infty 9 \\cdot 10^{-k}$$. This value is in turn defined as the limit of the sequence $$(\\sum_{k=1}^n 9 \\cdot 10^{-k})_{n\\in\\mathbb N}$$. Let $$a_n$$ denote the $$n$$th term of this sequence. I claim the limit is $$1$$. To prove this, we have to show that for any $$\\varepsilon>0$$, there is some $$N\\in\\mathbb N$$ such that for every $$n>N$$, $$|1-a_n|<varepsilon$$.\n\nLet\u2019s prove by induction that $$1-a_n=10^{-n}$$. Since $$a_0$$ is the sum of {$0$ terms, $$a_0=0$$, so $$1-a_0=1=10^0$$. If $$1-a_i=10^{-i}$$, then\n\n\\begin{align} 1 - a{i+1} &= 1 - (ai + 9 \\cdot 10^{-(i+1)}) \\newline &= 1-a_i \u2212 9 \\cdot 10^{-(i+1)} \\newline &= 10^{-i} \u2212 9 \\cdot 10^{-(i+1)} \\newline &= 10 \\cdot 10^{-(i+1)} \u2212 9 \\cdot 10^{-(i+1)} \\newline &= 10^{-(i+1)} \\end{align}\n\nSo $$1-a_n=10^{-n}$$ for all $$n$$. What remains to be shown is that $$10^{-n}$$ eventually gets (and stays) arbitrarily small; this is true by the archimedean property and because $$10^{-n}$$ is monotonically decreasing. <div><div>\n\n# Arguments against $$0.999\\dotsc=1$$\n\nThese arguments are used to try to refute the claim that $$0.999\\dotsc=1$$. They\u2019re flawed, since they claim to prove a false conclusion.\n\n\u2022 $$0.999\\dotsc$$ and $$1$$ have different digits, so they can\u2019t be the same. In particular, $$0.999\\dotsc$$ starts \u201c$0.$,\u201d so it must be less than 1.\n\nDecimal expansions and real numbers are different objects. Decimal expansions are a nice way to represent real numbers, but there\u2019s no reason different decimal expansions have to represent different real numbers.\n\n\u2022 If two numbers are the same, their difference must be $$0$$. But $$1-0.999\\dotsc=0.000\\dotsc001\\neq0$$.\n\nDecimal expansions go on infinitely, but no farther.$$0.000\\dotsc001$$ doesn\u2019t represent a real number because the $$1$$ is supposed to be after infinitely many $$0$$s, but each digit has to be a finite distance from the decimal point. If you have to pick a real number to for $$0.000\\dotsc001$$ to represent, it would be $$0$$.\n\n\u2022 $$0.999\\dotsc$$ is the limit of the sequence $$0.9, 0.99, 0.999, \\dotsc$$. Since each term in this sequence is less than $$1$$, the limit must also be less than $$1$$. (Or \u201cthe sequence can never reach $$1$$.\u201d)\n\nThe sequence gets arbitrarily close to $$1$$, so its limit is $$1$$. It doesn\u2019t matter that all of the terms are less than $$1$$.\n\n\u2022 In the first proof, when you subtract $$0.999\\dotsc$$ from $$9.999\\dotsc$$, you don\u2019t get $$9$$. There\u2019s an extra digit left over; just as $$9.99-0.999=8.991$$, $$9.999\\dotsc-0.999\\dotsc=8.999\\dotsc991$$.\n\nThere are infinitely many $$9$$s in $$0.999\\dotsc$$, so when you shift it over a digit there are still the same amount. And the \u201cdecimal expansion\u201d $$8.999\\dotsc991$$ doesn\u2019t make sense, because it has infinitely many digits and then a $$1$$.\n\nParents:\n\n\u2022 If these are included I think it would be good to also include explanations of why each one is wrong.", "date": "2022-09-28 20:25:22", "meta": {"domain": "greaterwrong.com", "url": "https://arbital.greaterwrong.com/p/5r7?l=5r7", "openwebmath_score": 0.999969482421875, "openwebmath_perplexity": 251.01399317045588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806546550657, "lm_q2_score": 0.9059898102301019, "lm_q1q2_score": 0.8883960812262521}}
{"url": "https://dsp.stackexchange.com/questions/52892/does-the-fourier-series-coefficient-of-ac-components-remains-same-if-dc-componen", "text": "# Does the Fourier series coefficient of AC components remains same if DC component is subtracted form the given signal?\n\nSuppose a signal is defined by $$x(t)= \\begin{cases} t & 0\\leq t \\leq 1 \\\\ 2-t & 1\\leq t\\leq 2 \\\\ \\end{cases}$$\n\nSince $$x(t)$$ has even symmetry, I can calculate fourier coefficient as $$a_n = \\frac{4}{T} \\int_0^1 x(t).\\cos{n\\pi t}.{dx}$$ I have calculated $$$$a_n = 2\\big[\\frac{\\cos{n\\pi} - 1}{n^{2}{\\pi}^{2}}\\big]\\tag{1}$$$$\n\nThe DC value of $$x(t)$$ i.e $$a_0 = 0.5$$. If we subtract DC value we get,\n\nFrom this we can see that given signal has hidden half wave symmetry in addition to Even symmetry. So we can find fourier coefficient as\n\n$$a^{'}_n=\\frac{8}{T}\\int_{0}^{\\frac{1}{2}}(t-\\frac{1}{2})\\cos{n\\pi t}.dt$$\n\nI have calculated $$$$a^{'}_n = 4\\big[\\frac{\\cos{\\frac{n\\pi}{2}} - 1}{n^{2}{\\pi}^{2}}\\big]\\tag{2}$$$$\n\nMy question is, shouldn't $$a_n$$ and $$a^{'}_n$$ be equal for $$n\\neq0$$ ?\n\n\u2022 How about calculating the Fourier coefficients without using any extraneous considerations such as symmetry or hidden half symmetries?. That is, copy the definition of $a_n$ (the one that applies to all periodic signals, long before extraneous considerations such as symmetry are mentioned) from your book, and calculate $a_n$ and $a_n^\\prime$ and see if you get the same answer or different answers. If you get the same answer, the problem is in your understanding of symmetry/half-symmetry/hidden etc. \u2013\u00a0Dilip Sarwate Oct 27 '18 at 15:27\n\u2022 if we apply half wave symmetry then it means even components will be zero and odd components of equation 1 & 2 are indeed equal. Equation 2 will give non-zero value for even values of n, other than multiples of 4, but we should discard it according to the conclusion of half wave symmetry. \u2013\u00a0Saurabh Oct 27 '18 at 19:00\n\u2022 As Dilip says, I also suggest that you repeat your calculations without worrying about any symmetry; that is, calculate the series for an entire period $T=2$. You should get the exact same answer in both cases except for $n=0$. \u2013\u00a0MBaz Oct 27 '18 at 22:24\n\u2022 \"If we apply half wave symmetry....\" Sigh! You can lead a horse to water but you cannot make him drink. \u2013\u00a0Dilip Sarwate Oct 28 '18 at 3:39\n\u2022 @DilipSarwate I did calculate the fourier series coefficient for entire time period i.e. $T=2$ and I got same value in both cases for $n\\neq0$. But since -Fat32 already given proof in their answer that CTFS coefficients for DC-removed part will same as original signal, I didn't mention it in my previous comment. \u2013\u00a0Saurabh Oct 28 '18 at 6:43\n\nPART-I: I would like to provide the general proof considering the title of the question and imposing no specific properties on the signal $$x(t)$$ other than having a CTFS representation.\n\nThe following is a simple analysis to conclude that the CTFS coefficients of any signal $$x(t)$$ and that of the DC removed signal are equivalent. (except $$a_0$$ of course).\n\nConsider a continuous-time periodic signal $$x(t)$$ with period $$T$$ divided into two components: $$x_{dc}$$ and $$x_{ac}$$, with periods $$T$$ also, where $$x_{dc}$$ is the pure DC component of $$x(t)$$ and $$x_{ac}$$ is the pure AC component of $$x(t)$$, then we have:\n\n$$x(t) = x_{dc}(t) + x_{ac}(t)$$\n\nComputing the CTFS coefficient $$a_k$$ of $$x(t)$$ yields: \\begin{align} a_k &= \\frac{1}{T} \\int_{} (x_{dc} + x_{ac}) e^{-j k \\frac{2\\pi}{T} t } dt \\\\ &= \\frac{1}{T} \\int_{} x_{dc} e^{-j k \\frac{2\\pi}{T} t } + \\frac{1}{T} \\int_{} x_{ac} e^{-j k \\frac{2\\pi}{T} t } \\\\ a_k &= b_k + c_k \\\\ \\end{align}\n\nwhere $$b_k$$ and the $$c_k$$ are the CTFS coefficients of DC and AC parts of $$x(t)$$.\n\nBy definition of any DC signal, it's known that $$b_k = 0$$ for all $$k \\neq 0$$ and by defition of any AC signal it's known that $$c_0 = 0$$. Then using the relation $$a_k = b_k + c_k$$ we get the following:\n\n$$a_0 = b_0 + c_0 = b_0$$\n\nand $$a_k = 0 + c_k = c_k ~~~,~~~ \\text{ for all } k \\neq 0$$\n\nFrom which we define :\n\n$$b_k = \\begin{cases} a_0 ~~~&, ~~~\\text{ for } k = 0 \\\\ 0 ~~~&, ~~~\\text{ for all } k \\neq 0 \\\\ \\end{cases}$$\n\nand\n\n$$c_k = \\begin{cases} 0 ~~~&, ~~~\\text{ for } k = 0 \\\\ a_k ~~~&, ~~~\\text{ for all } k \\neq 0 \\\\ \\end{cases}$$\n\nHence we conclude that the CTFS coefficients, $$a_k$$, of any periodic signal $$x(t)$$ and the CTFS coefficients $$c_k$$ of DC-removed part, $$x_{ac}$$, are the same for all $$k \\neq 0$$.\n\nPART-II: Based on OP comments, the relation for an even and real signal, is the following.\n\nFor a signal $$x(t)$$ which is real and even we have $$x(t) = x(t)^{*} = x(-t) = x(-t)^{*}$$ and the associated CTFS coefficients has the property of $$a_k = a_{-k}^{*} = a_{-k} = a_k^{*}$$ which indicates that the coefficients $$a_k$$ are also real and even.\n\nUsing this, we can obtain the trigonometric (cosine) Fourier series coefficients as. \\begin{align} a_k &= \\frac{1}{T} \\int_{} x(t) e^{-j k \\frac{2\\pi}{T} t } dt \\\\ &= \\frac{1}{T} \\int_{} x(t) \\left( \\cos( k \\frac{2\\pi}{T} t) + j \\sin(k \\frac{2\\pi}{T} t) \\right) dt \\\\ &= \\frac{1}{T} \\int_{} x(t) \\cos( k \\frac{2\\pi}{T} t) dt + j \\frac{1}{T} \\int_{} x(t) \\sin( k \\frac{2\\pi}{T} t) dt\\\\ a_k &= \\mathcal{Re}\\{a_k\\} + j ~~ \\mathcal{Im}\\{a_k\\} \\\\ \\end{align}\n\nNow since the property states tat $$a_k$$ are real, then the imaginary part is zero and we have:\n\n$$a_k= \\frac{1}{T} \\int_{-T/2}^{T/2} x(t) \\cos( k \\frac{2\\pi}{T} t) dt$$\n\nFurtermore since $$x(t)$$ is also even; $$x(t)=x(-t)$$, then we also have\n\n$$\\boxed{ a_k= \\frac{2}{T} \\int_{0}^{T/2} x(t) \\cos( k \\frac{2\\pi}{T} t) dt }$$\n\nAs the trigonometric cosine series coefficients of the real and even signals.\n\nIn addition to this, for a real & even signal, $$x(t)$$ of period $$T$$, which has no DC part, the following is also observed: $$x(t-\\frac{T}{2}) = -x(t)$$\n\nAnd based on the time-shift property of CTFS we can conclude that $$a_k ~~e^{-j\\frac{2\\pi}{T}k \\frac{T}{2} } = - a_k$$\n\n$$a_k ~ e^{-j\\pi k } = - a_k \\implies a_k = \\begin{cases} -a_k &, k=0,\\pm 2, \\pm4,... \\\\ a_k &, k=\\pm 1,\\pm 3,...\\\\ \\end{cases}$$\n\nWhich indicates that the CTFS coefficients $$a_k = 0$$ for $$k=2m$$ (even) for a real, even, (and having no DC) signal $$x(t)$$. Indeed we can get rid of the DC removed adjective and state for all real & even signals, as the DC will only affect $$a_0$$ being non-zero.\n\nPART-III: Finally apply these to the example signal to see that it works.\n\nThe signal defined as: $$x(t) = \\begin{cases} t &, 0\n\nThen for the CTFS coefficeints (in the trigonometric form) we have:\n\n$$a_k = \\frac{2}{T} \\int_{0}^{1} t \\cos(\\frac{2\\pi}{T} k t) dt = \\int_{0}^{1} t \\cos(k \\pi t) dt$$\n\nA simple by-parts integration yields the following result: \\begin{align} a_k &= \\int_{0}^{1} t \\cos(k \\pi t) dt \\\\ &= \\frac{t ~\\sin(\\pi k t) }{\\pi k} |_0^1 -\\int_{0}^{1} \\frac{\\sin(k \\pi t)}{\\pi k} dt \\\\ &= \\frac{\\sin(\\pi k) }{\\pi k} + \\frac{1}{\\pi k} \\left( \\frac{ \\cos(\\pi k) - 1}{\\pi k} \\right)\\\\ \\end{align} we conclude that\n\n$$\\boxed{ a_k = \\frac{ \\pi k ~\\sin(\\pi k) + \\cos(\\pi k) - 1}{ \\pi^2 k^2 } }$$\n\nForm which it can also be seen that\n\n$$a_k = \\begin{cases} 0.5 &, k=0 \\\\ \\frac{-2}{\\pi^2k^2} &, k=\\pm 1, \\pm 3,... \\\\ 0 &, k=\\pm 2, \\pm 4,...\\\\ \\end{cases}$$\n\nNote that the term $$\\sin(\\pi k)$$ can be ignored (as it's all zero) except for $$k=0$$. Also note that those zero values $$a_k$$ turns out to be the case without assuming half wave symmetry.\n\nFinally, we shall compute the CTFS $$c_k$$ coefficeints of the DC removed signal $$x_{ac}(t)$$ to see if they are equivalent. From the definition of the signal we see that\n\n$$x_{ac}(t) = x(t) - 0.5 = \\begin{cases} t-0.5 &, 0\n\nthen the $$c_k$$ become: $$c_k = \\frac{2}{T} \\int_{0}^{1} (t-0.5) \\cos(\\frac{2\\pi}{T} k t) dt = \\int_{0}^{1} t \\cos(k \\pi t) dt - \\int_{0}^{1} 0.5 \\cos(k \\pi t) dt$$\n\nNote that this integral is the same for the case of $$a_k$$. The only diference is in the last term which is $$1$$ for $$k=0$$ and $$0$$ for all $$k\\neq 0$$ and can be ignored for $$k \\neq 0$$. Then the equation relating to $$c_k$$ will be identical to that of $$a_k$$ except at $$k=0$$ which yields:\n\n$$c_k = \\begin{cases} 0 &, k=0 \\\\ \\frac{-2}{\\pi^2k^2} &, k=\\pm 1, \\pm 3,... \\\\ 0 &, k=\\pm 2, \\pm 4,...\\\\ \\end{cases}$$\n\nhence we again concluded that $$\\boxed{ c_k = \\begin{cases} 0 ~~~&, ~~~\\text{ for } k = 0 \\\\ a_k ~~~&, ~~~\\text{ for all } k \\neq 0 \\\\ \\end{cases} }$$\n\n\u2022 The question is about the sine/cosiine trigonometric form of the CTFS, not the exponential form of the CTFS. \u2013\u00a0Dilip Sarwate Oct 27 '18 at 20:49\n\u2022 @Dilip Sarwate: : I'm a newbie so, if I see an answer that provides insight and clears up confusion (for me and hopefully the OP ), then, IMHO, it's a great answer. The methodology used to do this, although not formulated the way the OP asked it, is still quite helpful. And I haven't seen a better answer. \u2013\u00a0mark leeds Oct 27 '18 at 20:56\n\u2022 my question was,shouldn't $a_n$ and $a^{'}_n$ be equal? And as proved in this answer they, in fact, are equal for odd values of $n$. Now if we already remove DC part to see the half-wave symmetry and calculate CTFS coefficient $a^{'}_n$ which has non-zero values for some even values of $n$ unlike $a_n$, but we should discard $a^{'}_n$ for even values of $n$, as considering half-wave symmetry implies CTFS coefficient for even values of $n$ is $0$. I hope this is correct. \u2013\u00a0Saurabh Oct 28 '18 at 6:49", "date": "2021-06-19 01:14:14", "meta": {"domain": "stackexchange.com", "url": "https://dsp.stackexchange.com/questions/52892/does-the-fourier-series-coefficient-of-ac-components-remains-same-if-dc-componen", "openwebmath_score": 0.9961004853248596, "openwebmath_perplexity": 743.008559337098, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105307684549, "lm_q2_score": 0.9099070023734244, "lm_q1q2_score": 0.8883517884371317}}
{"url": "https://math.stackexchange.com/questions/3667337/how-to-solve-3-lfloor-x-rfloor-lfloor-x2-rfloor-2-x", "text": "How to solve $3\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor = 2\\{x\\}$?\n\nI am trying to solve the following question involving floor/greatest integer functions.\n\n$$3\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor = 2\\{x\\}$$ with the notations $$\\lfloor x \\rfloor$$ denoting the greatest integer less than or equal to $$x$$ and $$\\{x\\}$$ to mean the fractional part of $$x$$.\n\nI used the following property for floor functions.\n\n$$n\\leq x$$ if and only if $$n \\leq \\lfloor x \\rfloor$$ where $$n\\in \\mathbb{Z}$$\n\nLet $$p=\\lfloor x^{2} \\rfloor$$, then\n\n$$p\\leq \\lfloor x^{2} \\rfloor < p+1$$\n\n$$\\rightarrow p \\leq x^{2} < p+1$$\n\n$$\\rightarrow \\sqrt{p} \\leq x < \\sqrt{p+1}$$ , since $$\\sqrt{p} \\in \\mathbb{Z}$$\n\n$$\\rightarrow \\sqrt{p} \\leq \\lfloor x \\rfloor < \\sqrt{p+1}$$ We then have $$\\sqrt{p} = \\lfloor x \\rfloor$$\n\nSince $$\\{x\\}=x-\\lfloor x \\rfloor,$$\n\n$$3\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor - 2\\{x\\}= 3\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor - 2(x-\\lfloor x \\rfloor)= 5\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor - 2x=0$$\n\nSubstituting $$p$$, $$\\sqrt{p}$$ for $$\\lfloor x^{2} \\rfloor$$ and $$\\lfloor x \\rfloor$$ respectively, and also letting $$x= \\sqrt{p},$$ we get $$p = 3\\sqrt{p}$$ solving for $$p$$ gives $$p=0, 9$$, and hence $$x=0, 3$$\n\nThe problem is that according to the solution for the problem, $$x$$ also equals to $$\\frac{3}{2}$$ for $$\\{x\\}=\\frac{1}{2}$$ since $$2\\{x\\}\\in \\mathbb{Z}$$. However, by definition for $$\\{x\\}$$, $$0 \\leq \\{x\\} < 1$$, then $$0 \\leq 2\\{x\\} < 2$$. How can $$\\{x\\}=\\frac{1}{2}$$ and how do I use this to obtain $$x=\\frac{3}{2}$$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance.\n\n\u2022 It's not true that $\\sqrt{p}\\in\\mathbb{Z}$ Say $x=1.5$ Then $p=\\lfloor 2.25\\rfloor = 2,$ and $\\sqrt{p}=\\sqrt{2}$ May 10, 2020 at 0:35\n\u2022 @saulspatz thank you for pointing that out. May 10, 2020 at 0:45\n\nLet $$x = n + r$$ where $$n = [x]$$ and $$r = \\{x\\}$$.\n\nThen we have $$3n - [n^2 + 2nr + r^2]=2r$$\n\n$$3n - n^2 - [2nr + r^2] = 2r$$\n\nand.... oh, hey, the LHS is an integer the RHS being $$2\\{x\\}$$ means $$\\{x\\} = 0$$ or $$0.5$$.\n\nTwo options $$x$$ is an integer and $$x = [x] = n$$ and $$r=\\{x\\} = 0$$ and we have\n\n$$3n-n^2=0$$ and $$n^2 = 3n$$ and $$n= 0$$ or $$n = 3$$.\n\nSo $$x = 0$$ and $$x=3$$ are two solutions.\n\n(Check: $$x=0\\implies 3[x] - [x^2] = 3*0 - 0 = 0 = \\{0\\}$$. Check. And $$x = 3\\implies 3[x]-[x^2] = 3[3]- [3^2] = 3*3-9 = 0=\\{3\\}$$. Check.\n\nAnd if $$x = n + \\frac 12$$ and $$r = \\frac 12$$ then\n\n$$3n - n^2 - [2n\\frac 12 + \\frac 14] = 2\\frac 12$$\n\n$$3n - n^2 - [n + \\frac 14] = 1$$\n\n$$3n -n^2 - n = 1$$\n\n$$n^2 - 2n + 1 =0$$ so $$(n-1)^2 = 0$$ and $$n = 1$$.\n\n$$x = 1+\\frac 12 = 1\\frac 12$$.\n\n(Check: If $$x = 1.5$$ then $$3[x] - [x^2] = 3[1.5] - [1.5^2] = 3*1 - [2.25]=3-2=1 = 2*\\frac 12 = 2\\{1.5\\}$$. Check.)\n\nWrite $$\\{x\\}=x-\\lfloor x\\rfloor$$. Then we have $$5\\lfloor x\\rfloor - \\lfloor x^2\\rfloor = 2x$$Since the LHS is an integer, the RHS must be as well. There are two cases: $$x$$ is an integer, or $$x$$ is a half-integer.\n\n\u2022 $$x$$ an integer. Drop the brackets: $$5x-x^2=2x;\\qquad x=0,3$$\n\u2022 $$x$$ is a half-integer. Write $$x=y+1/2$$. Then $$x^2 = y^2+y+1/4$$, and again we can drop the brackets: $$5y-(y^2+y)=2y+1; \\qquad y=1, x=3/2$$\n\u2022 may I ask how you arrive at $x$ is a half integer. I mean can't $x$ be anything else in between $0$ and $1$? May 10, 2020 at 0:46\n\u2022 After we substitute $\\{x\\}=x-\\lfloor x\\rfloor$, both sides are integers. Then if $2x$ is an integer, $x$ is either an integer or a half-integer. May 10, 2020 at 0:47\n\u2022 I think my issue is the following: from $0 \\leq \\{x\\} < 1$, we get $0 \\leq 2\\{x\\} < 2$. So how do I determine where else $\\{x\\}$ could be. May 10, 2020 at 0:56\n\u2022 Clearly the LHS is an integer. Then $2\\{x\\}$ is an integer as well. This means either $2\\{x\\}=0$, i.e. $x$ is an integer, or $2\\{x\\}=1$, i.e. $x$ is a half-integer. May 10, 2020 at 1:00\n\u2022 I think I see it now, Since $3\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor\\in \\mathbb{Z}$ and $0 \\leq \\{x\\} < 1$ then, $\\frac{3\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor }{2} = \\{x\\}$ implies that $\\frac{3\\lfloor x \\rfloor - \\lfloor x^{2} \\rfloor }{2} \\leq \\{x\\}<1$ which forces $\\{x\\}=\\frac{1}{2}$ May 10, 2020 at 1:05", "date": "2022-06-28 03:29:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3667337/how-to-solve-3-lfloor-x-rfloor-lfloor-x2-rfloor-2-x", "openwebmath_score": 0.9364206194877625, "openwebmath_perplexity": 115.84082149276317, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526430876968, "lm_q2_score": 0.8962513641273354, "lm_q1q2_score": 0.8883219084257623}}
{"url": "https://math.stackexchange.com/questions/1419302/is-the-matrix-a-positive-negative-semi-definite", "text": "Is the matrix $A$ positive (negative) (semi-) definite?\n\nGiven, $$A = \\begin{bmatrix} 2 &-1 & -1\\\\ -1&2 & -1\\\\ -1& -1& 2 \\end{bmatrix}.$$\n\nI want to see if the matrix $A$ positive (negative) (semi-) definite.\n\nDefine the quadratic form as $Q(x)=x'Ax$.\n\nLet $x \\in \\mathbb{R}^{3}$, with $x \\neq 0$.\n\nSo, $Q(x)=x'Ax = \\begin{bmatrix} x_{1} &x_{2} &x_{3} \\end{bmatrix} \\begin{bmatrix} 2 &-1 & -1\\\\ -1&2 & -1\\\\ -1& -1& 2 \\end{bmatrix} \\begin{bmatrix} x_{1}\\\\x_{2} \\\\x_{3} \\end{bmatrix}$.\n\nAfter multiplying out the matrices I am left with $$Q(x) = 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2} - x_{1}x_{3}-x_{2}x_{3}).$$\n\nNot sure what I can do with this result. Any suggestions on how to proceed would be appreciated.\n\nA simple way is to calculate all principle minors $A$ and if they are all positive, then $A$ is positive definite.\n\nFor example, $|A|_1=2>0$\n\n$$|A|_2=\\left|\\begin{array}{}{\\quad2 \\quad-1\\\\ -1\\quad 2} \\end{array}\\right|=3>0$$ Then calculate $|A|_3=|A|$.\n\nIf $|A|_i\\geqslant0,1\\leqslant i\\leqslant n$, then $A$ is semi-positive definite.\n\nIf $|A|_i<0$ for $i$ is odd and $|A|_i>0$ for $i$ is even, then $A$ is negative definite.\n\nIf $|A|_i\\leqslant 0$ for $i$ is odd and $|A|_i\\geqslant 0$ for $i$ is even, then $A$ is semi-negative definite.\n\n\u2022 Much neater and less complicated. And the same logic applies for negative (semi-) definite as well? \u2013\u00a0OGC Sep 3 '15 at 6:11\n\u2022 So here I found $|A_{3}|=0$, so $A$ is positive semi-definite. \u2013\u00a0OGC Sep 3 '15 at 6:26\n\u2022 Yes, you are right. You can see it in another post that quadratic form could be $0$ for nonzero $x$. \u2013\u00a0Math Wizard Sep 3 '15 at 6:27\n\u2022 Thanks again for this approach! Very convenient. \u2013\u00a0OGC Sep 3 '15 at 6:28\n\nTo say about positive (negative) (semi-) definite, you need to find eigenvalues of A. Then, 1) If all eigenvalues are positive, A is positive definite 2) If all eigenvalues are non-negative, A is positive semi-definite 3) If all eigenvalues are negative, A is negative definite 4) If all eigenvalues are non-positive, A is negative semi-definite 3) If some eigenvalues are positive and some are negative, A is neither positive nor negative definite\n\nEigenvalues of a matrix can be found by solving $det(\\lambda I -A)=0$. For your example, this results in: $\\lambda(\\lambda-3)^2 =0$ which means that eigenvalues are 0, 3, 3. So we are in the second case and A is positive semi-definite.\n\nIf you want to proceed with this solution, you should complete the square. It is important that you \"complete one variable completely every time\". We write \\begin{aligned} x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\\Bigl(x_1-\\frac{1}{2}x_2-\\frac{1}{2}x_3\\Bigr)^2+\\frac{3}{4}x_2^2+\\frac{3}{4}x_3^2-\\frac{3}{2}x_2x_3\\\\ &=\\Bigl(x_1-\\frac{1}{2}x_2-\\frac{1}{2}x_3\\Bigr)^2+\\frac{3}{4}\\bigl(x_2-x_3\\bigr)^2. \\end{aligned} Can you conclude from here?\n\n\u2022 So then the matrix is positive definite? \u2013\u00a0OGC Sep 3 '15 at 5:56\n\u2022 No, it is positive semidefinite. From the calculation above, you find that $Q(x)\\geq 0$ for all $x$. The question is: \"Does there exist $x\\neq 0$ such that $Q(x)=0$ or not?\" In this case it does. Take $x_3$ arbitrary, $x_2=x3$ (to make the last parenthesis zero) and $x_1=x_2$ (to make the first parenthesis zero). We conclude that $Q$ is only positive semidefinite. \u2013\u00a0mickep Sep 3 '15 at 6:00\n\u2022 I see. Thanks a lot! \u2013\u00a0OGC Sep 3 '15 at 6:01\n\u2022 where did the factor $2$ go on the LHS? \u2013\u00a0OGC Sep 3 '15 at 6:05\n\u2022 I just skipped the factor 2 since it only multiplies everything and does not change the character of the quadratic form. I should have mentioned this. \u2013\u00a0mickep Sep 3 '15 at 6:07\n\nFind $A$'s eigenvalues first. Once you know them, you know everything you need about $A$.\n\nMore explicitly, you can start by calculating $A$'s characteristic polynomial. A straightforward calculation shows that its roots are $0$ and $3$. These are $A$'s eigenvalues, and hence, with respect to an appropriate orthonormal basis, $A$ becomes$$\\left(\\begin{array}{ccc}0&0&0\\\\0&3&0\\\\0&0&3\\end{array}\\right).$$This means that $A$ is positive semi-definite.\n\n\u2022 Could you please elaborate? What steps do I need follow here? Is there a theorem that needs to be applied? \u2013\u00a0OGC Sep 3 '15 at 5:48\n\u2022 @OGC I added some details to my answer. \u2013\u00a0Amitai Yuval Sep 3 '15 at 5:55", "date": "2019-08-21 00:42:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1419302/is-the-matrix-a-positive-negative-semi-definite", "openwebmath_score": 0.9418367743492126, "openwebmath_perplexity": 319.9199280408344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462183543601, "lm_q2_score": 0.899121367808974, "lm_q1q2_score": 0.8882835551684756}}
{"url": "https://math.stackexchange.com/questions/3139209/validity-of-geometric-series-formula-for-r-0", "text": "# Validity of geometric series formula for $r=0$\n\nI can convince myself of the geometric series formula\n\n$$\\sum_{n=0}^{\\infty} r^n = \\frac{1}{1-r}$$\n\nfor $$0<|r|<1$$, but not for $$|r|<1$$ because I don't believe the formula for $$r=0$$.\n\nIf $$r=0$$, we have\n\n$$\\sum_{n=0}^{\\infty} r^n = 0^0 + 0^1 + 0^2 + \\ldots$$\n\nIt is not clear to me what this sum equals, much less that it equals $$\\frac{1}{1-0}=1$$. However, every source that I've consulted says that the result holds for $$-1.\n\nCan anyone justify the $$r=0$$ case? Must we simiply accept $$0^0=1$$ in this context?\n\n\u2022 if $r=0$ it's not geometric series. By definition, ratio of consecutive terms should be the same. \u2013\u00a0Vasya Mar 7 at 19:16\n\u2022 There are lots of ways to define geometric series, @Vasya. One is that $a_{n+1}a_{n-1}=a_n^2.$ In any event, this nit-pick doesn't resolve the question. \u2013\u00a0Thomas Andrews Mar 7 at 19:18\n\u2022 Then why does every textbook (even good ones, like Spivak) give the formula for $-1 < r <1$? \u2013\u00a0mathclassfromscratch Mar 7 at 19:19\n\u2022 If $r=0$ is allowed, the first term can be any number and $0^0=1$ does not help \u2013\u00a0Vasya Mar 7 at 19:29\n\u2022 Let's say that a correct/umabiguous version of the formula in question is $1+\\sum_{n=1}^{\\infty}r^n=\\dfrac{1}{1-r}$ for $|r|<1$. \u2013\u00a0Paramanand Singh Mar 8 at 5:51\n\nIn this context, $$0^0=1$$. Therefore, the sum is $$1$$.\n\n\u2022 Why is $0^0=1$ in this context? Is it different in other contexts? \u2013\u00a0John Douma Mar 7 at 19:21\n\u2022 The first paragraph here suggests that context matters: en.wikipedia.org/wiki/Zero_to_the_power_of_zero \u2013\u00a0mathclassfromscratch Mar 7 at 19:29\n\u2022 @mathclassfromscratch No, it says there is no agreed upon value for $0^0$. \u2013\u00a0John Douma Mar 7 at 19:31\n\u2022 @JohnDouma Yes, and then the second sentence says that context matters. \u2013\u00a0mathclassfromscratch Mar 7 at 19:36\n\u2022 @mathclassfromscratch The justifications come from different contexts. That doesn't mean that there are provable values for $0^0$ based on different contexts. Either way, I can say this sum equals $\\frac{1}{\\sqrt{\\pi}}$ and there is no context in which you can prove that $1$ is a better answer. \u2013\u00a0John Douma Mar 7 at 19:42\n\nPower series come up everywhere in mathematics, necessitating a convenient form to represent them. The easiest form is $$\\sum_{k=0}^\\infty a_k (x-x_0)^k$$ In order for this to represent a proper function, we should be able to substitute any value of $$x$$ into it. If you do not accept the convention $$0^0=1$$, you then run into problems when $$x=x_0$$; the value of the power series at that point is supposed to be $$a_0$$, but you instead get it is $$a_0\\cdot 0^0$$. To avoid this, you would have to instead write $$a_0+\\sum_{k=1}^\\infty a_k (x-x_0)^k$$ which is inconvenient. Therefore, for ease of notation, we stipulate that $$0^0=1$$ in the context of power series. This is the context of $$\\sum_{n\\ge 0}r^n$$. See\n\nfor a confirmation of this.\n\nAs a side note, there are an overwhelming number of situations where it is convenient to define $$0^0=1$$, and there are no situations where it is convenient to assume otherwise.\n\n\u2022 What number can be considered overwhelming comparing with zero? \u2013\u00a0user Mar 7 at 20:20\n\nNote a geometric sequence is defined in general as being $$\\{a, ar, ar^2, ar^3, \\ldots \\}$$, i.e., where each term is $$t_i = ar^i$$ for $$i \\ge 0$$.\n\nYour statement of $$\\sum_{n=0}^{\\infty} r^n = \\frac{1}{1-r}$$ is actually a specific case of the more general one, such as provided at Geometric series: Formula of\n\n$$\\sum_{n=0}^{\\infty} ar^n = \\frac{a}{1-r} \\tag{1}\\label{eq1}$$\n\nwhere $$|r| \\lt 1$$, and in your case $$a = 1$$. As such, if $$r = 0$$, then the geometric sequence would be $$\\{1, 0, 0, 0, \\ldots \\}$$ and, thus, it's clear that the sum is $$1$$. Plugging $$a = 1$$ and $$r = 0$$ into \\eqref{eq1} gives this same result. Also, by the definition of the sequence, it needs to use \"$$0^0 = 1$$\" in the LHS of \\eqref{eq1} to get that the first term is $$a$$. This is due to, for $$r \\neq 0$$, that $$r^0 = 1$$, so $$\\lim_{\\, r \\to 0}r^0 = 1$$.\n\nNote that some definitions of geometric sequences requires that $$r \\neq 0$$. However, as you can see, the general equation can still work even if you use $$r = 0$$.\n\n\u2022 This answer basically just sidesteps the problem by defining a geometric sequence informally. You could have also defined it as 1,r,r^2,... and be done with it. The original question basically asks why when we define a geometric series as a_n=ar^n we should have a_0=a0^0 be a rather than anything else. You're just assuming that this is indeed the correct definition from the start. \u2013\u00a0user3329719 Mar 11 at 6:39\n\u2022 @user3329719 The original definition of a geometric series that I learned, and as also defined by the referenced article, is as I state at the top. Although I define it informally initially, I also define the terms formally as $t_i$. In addition, as for why, in $a_0 = a 0^0$, it makes sense for $0^0 = 1$, I also explain that with my statement about the limit as $r \\to 0$, so I'm showing why it makes sense for $0^0 = 1$ instead of assuming this at that time. As such, I'm not clear exactly what issues you have with my answer. \u2013\u00a0John Omielan Mar 11 at 7:07", "date": "2019-09-21 08:56:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3139209/validity-of-geometric-series-formula-for-r-0", "openwebmath_score": 0.9770098924636841, "openwebmath_perplexity": 1859.7434426038478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995713428387, "lm_q2_score": 0.9196425234694067, "lm_q1q2_score": 0.8882823192077464}}
{"url": "https://www.physicsforums.com/threads/rational-inequality.767248/", "text": "# Homework Help: Rational inequality\n\n1. Aug 23, 2014\n\n### mafagafo\n\n1. The problem statement, all variables and given/known data\nFind all integer roots that satisfy (3x + 1)/(x - 4) < 1.\n\n3. The attempt at a solution\nI would do this:\n\nMake it an equation and find x such that (3x + 1)/(x - 4) = 1.\n\n3x + 1 = x - 4\n2x = -5\nx = -5/2\n\nThen check if the inequality is valid for values smaller than x and for values bigger than x.\n\nBut this approach is not good enough as I would get [-2, +\u221e) {integers} as my answer.\n\nAny help would be really appreciated.\n\nI think that the answer is [-2, 3] {integers}. But could only get this with a plot.\n\n---\n\nWhat should I also do so that my method is valid for \"rational\" inequalities?\n\n2. Aug 23, 2014\n\n### LCKurtz\n\nConsider the two cases where $x<4$ and $x>4$ and work the inequalities separately.\n\n3. Aug 23, 2014\n\n### pasmith\n\n$$\\frac{3x + 1}{x - 4} = \\frac{3(x-4) + 3(4) + 1}{x - 4} = 3 + \\frac{13}{x - 4}.$$ Thus if $(3x + 1)/(x-4) < 1$ then $13/(x - 4) < - 2$. Clearly that can't be the case if $x > 4$ (because then $13/(x - 4) > 0 > -2$) so we must have $x < 4$. Is there a lower bound?\n\n4. Aug 23, 2014\n\n### ehild\n\n(3x + 1)/(x - 4) < 1 can be written in the form\n$$\\frac{(3x+1)-(x-4)}{x-4}<0$$\n\nSimplified: $$\\frac{2x+5}{x-4}<0$$\n\nWhen is the fraction negative?\n\nehild\n\n5. Aug 23, 2014\n\n### mafagafo\n\n---\n(3x + 1)/(x - 4) = 1\n3x + 1 = x - 4\n2x = -5\n\nx = -5/2\n\n----\n(3x + 1)/(x - 4) = 1\n(3(x - 4) + 12 + 1) / (x - 4) = 1\n3 + 13/(x - 4) = 1\n13 / (x - 4) = -2\n\nx = 4\n\n----\nThen I work with those?\n(3x + 1)/(x - 4) < 1\nCode (Text):\n\n- 8/3 >> false\n- 5/2 >> false\n- 7/3 >> true\n4 >> impossible\n5 >> false\n\nSo the valid integers are {-2, -1, 0, 1, 2, 3}?\n\n6. Aug 23, 2014\n\n### HallsofIvy\n\nAn inequality can change direction where the two sides are equal or where the functions are discontinuous. Here, the first occurs where x= -5/2 and the second where x= 4. There are three intervals to be considered: x< -5/2, -5/2< x< 4, and x> 4.\nx= -3< -5/2 and (3(-3)+ 1)/(-3- 4)= (-9+ 1)/(-7)= -8/-7 is greater than 1 so NO x< -5/2 satisfies the inequality. x= 0 is between -5/2 and 4. (3(0)+ 1)(0- 4)= -1/4 is less than 1. Every x between -5/2 and 4 satisfy the inequality. x= 5 is larger than 4 and (3(5)+ 1)/(5- 4)= 15/1 is larger than 1. The integer solutions are -2, -1, 0, 1, 2, and 3.\n\n7. Aug 23, 2014\n\n### mafagafo\n\nWhen ${2x+5} < 0$ and ${x-4} > 0$ or when ${2x+5} > 0$ and ${x-4} < 0$.\n\nIf ${2x+5} < 0$, then $2x<-5$ and $x<-\\frac{5}{2}$.\nand if ${x-4} > 0$, then $x > 4$. Thus, this is impossible.\n\nIf ${2x+5} > 0$, then $2x>-5$ and $x>-\\frac{5}{2}$.\nand if ${x-4} < 0$, then $x < 4$. Thus, $$S=\\left\\{x\\in Z|-5 /2 < x < 4\\right\\}=\\left\\{x\\in Z|-2 \\le x \\le 3\\right\\}$$\n\n8. Aug 23, 2014\n\n### mafagafo\n\nBig thanks to all of you, with special mention to HallsOfIvy for answering my question.\n\nQ.: \"What should I also do so that my method is valid for \"rational\" inequalities?\"\nA.: An inequality can change direction where the two sides are equal or where the functions are discontinuous.", "date": "2018-07-21 00:40:57", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/rational-inequality.767248/", "openwebmath_score": 0.7524063587188721, "openwebmath_perplexity": 1044.296315132299, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429638959674, "lm_q2_score": 0.9032941975921684, "lm_q1q2_score": 0.8882479935303126}}
{"url": "http://math.stackexchange.com/questions/306425/expected-value-function", "text": "# Expected Value Function\n\nMy text-book defines expected value as $$E(X) = \\mu_x = \\sum_{x \\in D} ~x \\cdot p(x)$$ And so, if I was to find the expected value of a random variable $X$, where $X = 1,2,3$, then it would resemble this: $$E(X)= \\sum_{x=1}^3~ x \\cdot p(x)= 1\\cdot p(1) + 2\\cdot p(2) + 3 \\cdot p(3)$$ Furthermore, if I wanted to calculate $E(X^2)$, it would be $E(X^2) = 1^2 \\cdot P(1) + 2^2 \\cdot p(2) + 3^2 \\cdot p(3)$. My question is, why don't we square the x-values in the probability function $p(x)$?\n\nAlso, is computing the expected value a way of calculating the average of the random variable? It seems a little odd to calculate it that way.\n\nPS: If any use of notation, or vocabulary, is incorrect, please inform me.\n\n-\nThe differences between using \\Sigma and using \\sum in TeX are these: $\\displaystyle\\Sigma_{x\\in D}$ versus $\\displaystyle\\sum_{x\\in D}$. That's why \\sum is standard for this occasion. \u2013\u00a0 Michael Hardy Feb 17 '13 at 19:36\n\nLet $Y=X^2$. Then $Y$ takes on the values $1$, $4$, and $9$ respectively when $X$ takes on the values $1$, $2$, and $3$.\n\nThus $p_Y(1)=p_X(1)$, $p_Y(4)=p_X(2)$, and $p_Y(9)=p_X(3)$.\n\nNow for calculating $E(Y)$ we just use the formula the post started with, namely $$E(Y)=\\sum_y yp_Y(y).$$ In our case, we get $1\\cdot p_Y(1)+4\\cdot p_Y(4)+9\\cdot p_Y(9)$. Equivalently, $E(Y)= 1\\cdot p_X(1)+4\\cdot p_X(2)+9\\cdot p_X(3)$.\n\nTo answer your question more explicitly, we do not use $1^2(p_X(1))^2+2^2(p_X(2))^2+3^2(p_X(3))^2$ because, for example, $\\Pr(X^2=3^2)$ is not $(\\Pr(X=3))^2$. In fact, $\\Pr(X^2=3^2)=\\Pr(X=3)$.\n\nAs to your question about average, yes, the mean is a very important measure of average value. The only serious competitor is the median.\n\nMean and median can be quite different. For example, imagine a population in which a small minority is insanely rich, while the vast majority of the population is struggling. Then the mean income of the population may be substantially higher than the median income. Is either one a \"better\" measure of average wealth? I would argue that in this case the median is ordinarily of greater relevance. But for certain planning purposes, such as level of tax revenues, the mean may be more useful.\n\nThe mathematics of the mean is substantially simpler than the mathematics of the median. For example, the mean of a sum of two random variables is the sum of the means. The median of a sum is a far more complicated object.\n\n-\n\nFor computing $E[X^2]$, the probability is still taken over $X$ and not $X^2$. Otherwise, if you make $Y=X^2$ the random variable and then compute $E[Y]$, the only operation that you effectively did is to relabel the random variables (well, although only considering positive values): all the values taken by $|X|$ will also be taken by $Y$, so for positive values of $X$, computing $E[X^2]$ would be exactly like computing $E[X]$. But computing $E[X^2]$ gives you more information!\n\nThe expected value is the weighted average. \"Normally\" (in daily life), when you take an average, all the values have the same weight. The average salary of your family members, for instance. But say you wanted the average salary in your country, then it's nice to work with, say, the probability of a certain salary being had. Making this latter problem more concrete, you could approximate the average national salary by taking every integer multiple of $1000, and finding out the proportion of people with this salary. Then the weighted average gives you the true national average salary. - There were two questions. The first question: Why don't we square the$x$values in the calculation for the expected value of$X^2$. Suppose$Y = g(X). \\begin{aligned} E(Y) & \\stackrel{\\text{(a)}}{=} \\sum_{y\\in{S_Y}}yP_Y(y)\\\\ & \\stackrel{\\text{(b)}}{=} \\sum_{y\\in{S_Y}}y\\sum_{x:g(x) = y}P_X(x)\\\\ & \\stackrel{\\text{(c)}}{=} \\sum_{y\\in{S_Y}}\\sum_{x:g(x) = y}yP_X(x)\\\\ & \\stackrel{\\text{(d)}}{=} \\sum_{y\\in{S_Y}}\\sum_{x:g(x) = y}g(x)P_X(x)\\\\ & \\stackrel{\\text{(e)}}{=} \\sum_{x\\in{S_X}}g(x)P_X(x)\\\\ \\end{aligned} (a): The definition of expected value of a discrete random variable, as you have supplied. (b): Because the probability of random variableY$taking on a value$g(x)$is equal to the sum of the probability of all the values of$x$which will map to$g(x)$. (c): Interchanging the summation. (d): Because of the condition in the summation, we can replace$y$by$g(x)$. (e): Because enumerating across all$y$, then enumerating across all$x$such that$g(x) = y$, is equivalent to enumerating across all$x$since every$x$must map to exactly one$y$. (Multiple$x$could map to the same$y$). So, in your instance, set$g(x) = x^2$to obtain the result. The second question: Why is expected value defined the way it is defined? Expected value can be thought of as the center of mass, if we set$P_X(x)$to be the mass located at a distance$x$from the origin. It corresponds exactly with the arithmetic average when the distribution of$X\\$ is uniform.\n\n-", "date": "2015-03-01 12:31:48", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/306425/expected-value-function", "openwebmath_score": 0.9639572501182556, "openwebmath_perplexity": 282.1667617092513, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109538667758, "lm_q2_score": 0.9019206712569267, "lm_q1q2_score": 0.8882213565726966}}
{"url": "https://math.stackexchange.com/questions/2826887/in-how-many-ways-can-integers-a-1a-2a-3a-4-be-chosen-from-the-integers-1-2", "text": "# In how many ways can integers $a_1<a_2<a_3<a_4$ be chosen from the integers $1,2,3,\u2026,26$ such that $5 \\le a_i - a_{i-1} \\le 7$\n\nIn how many ways can $4$ integers $a_1<a_2<a_3<a_4$ be chosen from the integers $1,2,3,...,26$ such that $5 \\le a_i - a_{i-1} \\le 7$ for all $i = 2,3,4$?\n\nI'm not sure what I'm missing in my line of thinking:\n\nSince $5 \\le a_i - a_{i-1} \\le 7$, the difference between two consecutive integers chosen must be $5,6$ or $7$ (since everything's an integer). If I let $a_1 = 1$, then by incrementing the next integers by $7$ I see that the maximum I get is $22$. Therefore, the difference $a_i - a_{i-1}$ could be anything among $5,6,7$ for all $i$. Thus I have $3^3$ possibilities.\n\nSimilarly, letting $a_1$ be $2,3,4$ and $5$ all yields $3^3$ possibilities.\n\nWhen I let $a_1 = 6$, however, if I keep incrementing by the biggest difference, $7$, the max. I get is $27$, which is $1$ more than the allowed maximum. That means I'm not allowed to use all three $7$'s for the difference but a maximum of two $7$'s. That gives me $3^3-1$ possibilities.\n\nWhen I let $a_1 = 7$, the 'maximum' I get is $28$, $2$ more than $26$. Therefore, I can only use a maximum of one $7$. Since there is one case where I can use all three $7$'s and six cases in which I use two 7's, I have $3^3 - (1+6)$ possibilities.\n\nFor $a_1 = 8$ the 'maximum' $a_4$ is 29. So to fit in with $26$, I must not use any $7$'s at all. That gives me $2^3$ possibilities.\n\nFor $a_1 = 9$ I can get $a_4 = 30$ 'max'. To offset the difference of $4$, I cannot use 7's at all and one of the 6's. That gives me $4$ choices.\n\nFor $a_1 = 10$ I have $31$ 'max'. Now I can use only all 5's, or a maximum of one $6$ only. That's $4$ possibilities as well.\n\nFinally, for $a_1=11$, I can only have the differences be $5$ all the time. So 1 possibility.\n\nSo the sum gave me $198$. But the answer is $216$. Where did I go wrong?\n\nIt is true that $a_1 = 1,2,3,4,5$ each give you $3^3$ possibilities.\n\nNext, looking at $a_1 = 6$, we see that the possibility of $7,7,7$ is excluded, but every other possibility is included, so this gives $26$ possibilities.\n\nLooking at $a_1 = 7$, only those totaling to $20$ or more are not allowed, which means that $757$, for example, is permissible, but you have excluded it. Hence we exclude $776,767,677$ and $777$ to get $23$ possibilities, not $27 -(1+6)$, rather $27-(1+3)$.\n\nFor $a_1 = 8$ only those totalling to $19$ or more must be excluded. That is, the above possibilities, plus $757$, $775$,$577$,$667,676,766$. This gives $17$ in total.\n\nFor $a_1 = 9$ the above, plus all permutations of $765$, and $666$, are excluded, giving $17-7 = 10$.\n\nFor $a_1 = 10$ the above plus all permutations of $755$ and $665$ are excluded, giving $4$ possibilities.\n\nFor $a_1 = 11$ only $555$ remains.\n\nThe total? $135 + 26 + 23+17+10+4+1 = 216$.\n\nLet the consecutive increments be $x, y, z$. There are $27$ choices of $x, y, z \\in \\{5,6,7\\}$.\n\nGiven a fixed choice of $x, y, z$, the number of ways to select the $a_i$'s is $f(x,y,z) = 26 - (a_4 - a_1) = 26 - (x + y + z)$.\n\nSince $f(x,y,z)$ is linear, it is clear that its average value is $26 - 3 \\times 6 = 8$.\n\nThus the answer is $27 \\times 8 = 216$.\n\nThere is a mistake for $a_1 = 7$. The number of possibilities of having two $7$ isn\u2019t $6$ but : $\\binom{3}{2} = 3$.\n\nMoreover, for $a_1 =8$, you are saying that there can\u2019t be $7$, yet what about : $a_1 = 8, a_2 = 8+7 = 15, a_3 = 20, a_4 = 25$ ? (Same argument for $a_1 = 9$)\n\nYou've just skipped some possibilities. For example, you say that when $a_1=8,$ you can't use any $7'$s at all, but that's not so. You can use $5,6,7$ or $5,5,7.$ It's probably easier if you consider that you can make a maximum of $18$ with the $3$ increments.", "date": "2019-09-20 10:10:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2826887/in-how-many-ways-can-integers-a-1a-2a-3a-4-be-chosen-from-the-integers-1-2", "openwebmath_score": 0.9600333571434021, "openwebmath_perplexity": 138.67493532858055, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474869616596, "lm_q2_score": 0.8976952962128457, "lm_q1q2_score": 0.8881325853654816}}
{"url": "https://math.stackexchange.com/questions/1845079/verify-the-meaning-of-the-closure-of-a-set-in-a-finite-non-metric-space", "text": "verify the meaning of the closure of a set in a finite/non-metric space.\n\nMy motivation behind this question is to better under the fundamental concepts of topology in their general sense.\n\nI understand what a set closure when dealing with real numbers and a euclidean metric. So if the space is $\\mathbb{R}$ with euclidean metric and $S = (1,2)\\cup(2,3)\\cup(3,4]$ then $\\overline{S} = [1,4]$.But in the more general topological sense i'm not so sure. Next I will show 2 examples where I think I know the answer.\n\n(1) Let $X =${1,2,3,4,5,6} and $\\mathscr{T}_1 =${{},{1},{1,2},{1,2,3},{1,2,3,4},{1,2,3,4,5},{1,2,3,4,5,6}}. So $(X,\\mathscr{T}_1)$ is a topological space.\n\nMy claim: If $S =${1,2,3} then $\\overline{S}=X$. This is because {4},{5} and{6} would all be limit points due the the fact that any open set in the topology that contains any of those three points also contains {1,2,3}. Furthermore, for any open set $U \\in \\mathscr{T}_1$ ;$\\overline{U} = X$. Also, in this space, open sets are not closed.\n\n(2) Let $X =${1,2,3,4,5,6} and $\\mathscr{T}_2 =${{},{1},{2,3},{1,2,3},{4,5,6},{1,4,5,6},{2,3,4,5,6},{1,2,3,4,5,6}}. So $(X,\\mathscr{T}_2)$ is a topological space.\n\nMy claim: If $S =${1,2,3} then $\\overline{S}=S$. This is because {4,5,6} is, it self, an open set, and none of the points in $S$ are in that set. So{4},{5} and{6} are not limit points. Furthermore, for any open set $U \\in \\mathscr{T}_2$ ;$\\overline{U} = U$. Also, in this space, all open sets are also closed.\n\nIs everything above correct? Are there any simple examples of closure that illustrate some of the more exotic properties of closure?\n\n\u2022 Here's a question [math.stackexchange.com/questions/280993/\u2026 about limit point which states that finite set has no limit point , wish it can help. Jun 30 '16 at 18:37\n\u2022 @Benjamin That question is about the metric space $\\Bbb R$. Jun 30 '16 at 18:50\n\nYes you're approach seems correct. Sometimes I get confused by all the definitions of limit point, accumulation point, adherent point, point of closure, of a set - partly because some require that the point be either in/not in the set, and (open) neighbourhoods of the point have a nonempty intersection with the set, with at least one or two points in the intersection, and so on $\\ldots$\n\nCertainly a good definition of closure is the following which is fairly standard. Let $A\\subset X$ a topological space with topology $\\mathscr{T}_X$,\n\n\\mathrm{cl}(A)=\\bigcap_{\\begin{align}C &\\text{ closed}\\\\ &A\\subseteq C\\end{align}} C\n\nIn fact as an alternative, in your examples you can compute this by hand. In $\\mathscr{T}_1$ the only closed set containing $S$ is $X$ itself so $\\mathrm{cl}(S)=X$. In $\\mathscr{T}_2$, $S$ is itself closed, since it is the complement of $\\{4,5,6\\}$ which is open, so $\\mathrm{cl}(S)=S$.\n\n$S\\subset\\mathrm{cl}(S)$ is clear, if $S$ is closed \\mathrm{cl}(S)=S\\cap[\\bigcap\\limits_{\\begin{align}&C \\text{ closed}\\\\ &S\\subseteq C, S\\neq C\\end{align}} C \\,], which is certainly a subset of $S$.\n\nThis is the easy to state definition, which defines it as the smallest closed set containing $A$, wrt inclusion. Of course limit points and the like are useful for more general topological spaces.\n\nSome properties of $\\mathrm{cl}()$:\n\n$\\mathrm{cl}(A\\cap B)\\subseteq \\mathrm{cl}(A)\\cap\\mathrm{cl}(B)\\quad -$ eg. $\\mathrm{cl}((-1,0)\\cap (0,1))=\\emptyset \\subset [-1,0]\\cap[0,1]$\n\n$\\mathrm{cl}(A\\cup B)= \\mathrm{cl}(A)\\cup\\mathrm{cl}(B)$\n\n$\\mathrm{cl}(\\bigcup_\\alpha A_\\alpha)\\supseteq \\bigcup_\\alpha\\mathrm{cl}( A_\\alpha) \\quad-$ has to do with countable unions of closed set not always being closed.\n\ne.g $\\bigcup_{n>1} [0,1-{1\\over n}] =[0,1) \\subset [0,1]$\n\nYou can have that for a disconnected set $D$ that $\\mathrm{cl}(D)$ can be connected, like two open disks whose boundary circles intersect at one point only.\n\nYou can supplement the idea of closure with the idea of limit points, boundary points etc., to obtain a statement of the form\n\n$$\\mathrm{cl}(A)=A\\cup\\{\\text{suitable set}\\}$$\n\nThere's also the idea of closure with respect to a set say $A\\subset B\\subset X$, in which case $\\mathrm{cl}_B(A)\\neq \\mathrm{cl}_X(A)$ in general. The idea would be to give $B$ the subspace topology, and $\\mathrm{cl}_B(A)=B\\cap \\mathrm{cl}_X(A)$ and this will only be closed in $X$ if $B$ is a closed subspace.\n\ne.g. $A=(0,1)\\subset \\Bbb R$ with subspace topology. $\\mathrm{cl}_A(A)=(0,1)=A\\cap\\mathrm{cl}_{\\Bbb R}(A)=(0,1)\\cap [0,1]$\n\n\u2022 Thanks, your example: e.g $\\bigcup_{n>1} [0,1-{1\\over n}] =[0,1) \\subset [0,1]$ was an example I was wondering about. Jul 1 '16 at 22:48\n\u2022 I'm also looking for an example like that but with intersections rather than unions. Jul 1 '16 at 22:50\n\u2022 @MichaelMaliszesky it might be tougher, because arbitrary intersections, countable, uncountable etc are always closed. In the same way arbitrary unions of opens sets are always open. That usual example of where there was empty intersection before the closure and non empty afterward was the typical example. I'll try think of something for tomorrow Jul 2 '16 at 0:28\n\u2022 Thanks alot, this is a big help. My book says: $\\mathrm{cl}(\\bigcap_\\alpha A_\\alpha)\\subseteq \\bigcap_\\alpha\\mathrm{cl}( A_\\alpha) \\quad$. I have no problem finding examples where they are equal, but I can't think of a proper subset example. (I was looking for something like this, when I mentioned \"exotic\" examples before) Jul 2 '16 at 0:36\n\u2022 @MichaelMaliszesky here's something actually, $\\mathrm{cl}(\\bigcap_{n>1} (1-\\frac{1}{n},1))=\\emptyset \\subset \\{1\\}$ Jul 2 '16 at 0:43", "date": "2022-01-18 05:59:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1845079/verify-the-meaning-of-the-closure-of-a-set-in-a-finite-non-metric-space", "openwebmath_score": 0.8761559724807739, "openwebmath_perplexity": 286.67442385994866, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357221825193, "lm_q2_score": 0.9046505312448598, "lm_q1q2_score": 0.8881277426144721}}
{"url": "https://cs.stackexchange.com/questions/80415/why-is-binary-search-using-this-weird-thing-to-calculate-middle/97738", "text": "# Why is binary search using this weird thing to calculate middle?\n\nI noticed that in many books calculation of midpoint for binary search uses this:\n\nint mid = left + (right - left) / 2;\n\n\nWhy not use\n\nint mid = (left + right) / 2;\n\n\n\u2022 The only \"advantage\" is that your calculation never exceeds the value of right. Aug 25, 2017 at 10:42\n\u2022 @fade2black, I don't see how is it possible to exceed value of right in second case. If left = right, then (2 * right) / 2 = right. Aug 25, 2017 at 10:45\n\u2022 @rus9384 left + right >= right, intermediate values I mean. Kind of take actions against overflow. Aug 25, 2017 at 10:47\n\u2022 @fade2black, I think, this is an answer. Aug 25, 2017 at 10:52\n\u2022 This is indeed the answer.\n\u2013\u00a0Raphael\nAug 25, 2017 at 11:15\n\nBecause left + right may overflow. Which then means you get a result that is less than left. Or far into the negative if you are using signed integers.\n\nSo instead they take the distance between left and right and add half of that to left. This is only a single extra operation to make the algorithm more robust.\n\nSuppose your 'low' and 'high' are 16 bit unsigned integers. That means, they can only have a maximum value of 2^16=65536. Consider this, low = 65530 high = 65531\n\nIf we added them first, (low+high) would end up being junk since that big a number (131061) cannot be stored in a your 16-bit integer. And so, mid would be a wrong value.\n\n\u2022 Yeah david,, I found out a lot people saying overflow overflow on the different forums but couldn't understand what they really mean to say.. After i found out the example,, my confusion was cleared.. Sep 24, 2018 at 18:06\n\nThis answer gives a practical example of why the l + (r-l)/2 calculation is necessary.\n\nIn case you are curious how the two are equivalent mathematically, here is the proof. The key is adding 0 then splitting that into l/2 - l/2.\n\n(l+r)/2 =\nl/2 + r/2 =\nl/2 + r/2 + 0 =\nl/2 + r/2 + (l/2 - l/2) =\n(l/2 + l/2) + (r/2 - l/2) =\nl + (r-l)/2", "date": "2023-03-25 07:45:12", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/80415/why-is-binary-search-using-this-weird-thing-to-calculate-middle/97738", "openwebmath_score": 0.5031972527503967, "openwebmath_perplexity": 1916.0203639991316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924802053234, "lm_q2_score": 0.9136765163620469, "lm_q1q2_score": 0.8880867032441058}}
{"url": "http://yo-site.biz/isoborneol-vs-pstyrfa/cnmm5i.php?page=0a408b-logarithmic-differentiation-problems", "text": "Problems. (3) Solve the resulting equation for y\u2032 . Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. (3x 2 \u2013 4) 7. Using the properties of logarithms will sometimes make the differentiation process easier. Use logarithmic differentiation to differentiate each function with respect to x. We know how With logarithmic differentiation, you aren\u2019t actually differentiating the logarithmic function f(x) = ln(x). Click HERE to return to the list of problems. You do not need to simplify or substitute for y. Instead, you\u2019re applying logarithms to nonlogarithmic functions. Find the derivative of the following functions. We could have differentiated the functions in the example and practice problem without logarithmic differentiation. Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. Instead, you do [\u2026] There are, however, functions for which logarithmic differentiation is the only method we can use. Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. The process for all logarithmic differentiation problems is the same: take logarithms of both sides, simplify using the properties of the logarithm ($\\ln(AB) = \\ln(A) + \\ln(B)$, etc. View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. A logarithmic derivative is different from the logarithm function. Begin with y = x (e x). One of the practice problems is to take the derivative of $$\\displaystyle{ y = \\frac{(\\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. (3) Solve the resulting equation for y\u2032 . Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the For differentiating certain functions, logarithmic differentiation is a great shortcut. The function must first be revised before a derivative can be taken. Apply the natural logarithm to both sides of this equation getting . For example, say that you want to differentiate the following: Either using the product rule or multiplying would be a huge headache. SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. (2) Differentiate implicitly with respect to x. (x+7) 4. 11) y = (5x \u2212 4)4 (3x2 + 5)5 \u22c5 (5x4 \u2212 3)3 dy dx = y(20 5x \u2212 4 \u2212 30 x 3x2 + 5 \u2212 60 x3 5x4 \u2212 3) 12) y = (x + 2)4 \u22c5 (2x \u2212 5)2 \u22c5 (5x + 1)3 dy dx = \u2026 ), differentiate both sides (making sure to use implicit differentiation where necessary), Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. Logarithmic Differentiation example question. Solution to these Calculus Logarithmic Differentiation practice problems is given in the video below! In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. (2) Differentiate implicitly with respect to x. 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Function, the ordinary rules of differentiation do NOT APPLY 2 ) Differentiate implicitly with to! Is raised to a variable power in this function, the ordinary rules differentiation. For example, say that you want to Differentiate each function with respect to x the... The logarithm function instead, you do NOT need to simplify or substitute for y of problems the and.", "date": "2021-06-18 03:28:05", "meta": {"domain": "yo-site.biz", "url": "http://yo-site.biz/isoborneol-vs-pstyrfa/cnmm5i.php?page=0a408b-logarithmic-differentiation-problems", "openwebmath_score": 0.9247481822967529, "openwebmath_perplexity": 570.5269978655537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9719924793940119, "lm_q2_score": 0.9136765157744067, "lm_q1q2_score": 0.8880867019316475}}
{"url": "http://www.southernglowtans.com/kids-desk-xefgmli/bccf95-permutation-matrix-inverse", "text": "# permutation matrix inverse\n\nThe array should contain element from 1 to array_size. Inverse Permutation is a permutation which you will get by inserting position of an element at the position specified by the element value in the array. The inverse of an even permutation is even, and the inverse of an odd one is odd. And every 2-cycle (transposition) is inverse of itself. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. A permutation matrix P is a square matrix of order n such that each line (a line is either a row or a column) contains one element equal to 1, the remaining elements of the line being equal to 0. Here\u2019s an example of a $5\\times5$ permutation matrix. Permutation Matrix (1) Permutation Matrix. I was under the impression that the primary numerical benefit of a factorization over computing the inverse directly was the problem of storing the inverted matrix in the sense that storing the inverse of a matrix as a grid of floating point numbers is inferior to \u2026 The use of matrix notation in denoting permutations is merely a matter of convenience. All other products are odd. Sometimes, we have to swap the rows of a matrix. Then there exists a permutation matrix P such that PEPT has precisely the form given in the lemma. In this case, we can not use elimination as a tool because it represents the operation of row reductions. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. The product of two even permutations is always even, as well as the product of two odd permutations. Moreover, the composition operation on permutation that we describe in Section 8.1.2 below does not correspond to matrix multiplication. Sometimes, we have to swap the rows of a matrix. \u2022Find the inverse of a simple matrix by understanding how the corresponding linear transformation is related to the matrix-vector multiplication with the matrix. Then you have: [A] --> GEPP --> [B] and [P] [A]^(-1) = [B]*[P] The simplest permutation matrix is I, the identity matrix.It is very easy to verify that the product of any permutation matrix P and its transpose P T is equal to I. \u2022Identify and apply knowledge of inverses of special matrices including diagonal, permutation, and Gauss transform matrices. Therefore the inverse of a permutations \u2026 Thus we can define the sign of a permutation \u03c0: A pair of elements in is called an inversion in a permutation if and . Example 1 : Input = {1, 4, 3, 2} Output = {1, 4, 3, 2} In this, For element 1 we insert position of 1 from arr1 i.e 1 at position 1 in arr2. Basically, An inverse permutation is a permutation in which each number and the number of the place which it occupies is exchanged. 4. Every permutation n>1 can be expressed as a product of 2-cycles. 4. The product of two even permutations is always even, as well as the product of two odd permutations. To get the inverse, you have to keep track of how you are switching rows and create a permutation matrix P. The permutation matrix is just the identity matrix of the same size as your A-matrix, but with the same row switches performed. A permutation matrix is an orthogonal matrix \u2022 The inverse of a permutation matrix P is its transpose and it is also a permutation matrix and \u2022 The product of two permutation matrices is a permutation matrix. Corresponding linear transformation is related to the matrix-vector multiplication permutation matrix inverse the matrix s an example of a matrix even... Composition operation on permutation that we describe in Section 8.1.2 below does not correspond to matrix multiplication can. Does not correspond to matrix multiplication matrix-vector multiplication with the matrix is a permutation which. Is even, as well as the product of 2-cycles special matrices including diagonal, permutation, Gauss. And apply knowledge of inverses of special matrices including diagonal, permutation, and Gauss transform matrices two odd.! A permutation in which each number and the inverse of an even permutation is even, as as... The product of two even permutations is always even, and the inverse of itself Gauss transform.. Even permutations is merely a matter of convenience number and the inverse of itself below does not correspond matrix... We describe in Section 8.1.2 below does not correspond to matrix multiplication in Section 8.1.2 below not. The composition operation on permutation that we describe in Section 8.1.2 below not. Math ] 5\\times5 [ /math ] permutation matrix P such permutation matrix inverse PEPT has precisely the form given the. 5\\Times5 [ /math ] permutation matrix s an example of a [ ]. Number and the number of the place which it occupies is exchanged contain element from 1 to array_size composition... An odd one is odd and the inverse of an odd one is odd correspond. Sometimes, we have to swap the rows of a simple matrix by understanding how the corresponding linear is! We can not use elimination as a product of two even permutations merely... Permutations is merely a matter of convenience matrix P such that PEPT has precisely the form in... Can not use elimination as a product of 2-cycles to swap the rows of [... A matter of convenience on permutation that we describe in Section 8.1.2 does! Number and the number of the place which it occupies is exchanged a tool because it the. 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Inverses of special matrices including diagonal, permutation, and Gauss transform matrices > 1 can be as... Represents the operation of row reductions such that PEPT has precisely the form given in lemma. We have to swap the rows of a matrix rows of a matrix 8.1.2 below does not correspond to multiplication. With the matrix given in the lemma merely a matter of convenience that we describe in Section 8.1.2 does! Even, and the inverse of an odd one is odd \u2022identify and apply knowledge of inverses of matrices! Row reductions expressed as a product of two even permutations is merely a matter of convenience understanding how corresponding. The rows of a matrix expressed as a tool because it represents operation... That we describe in Section 8.1.2 below does not correspond to matrix multiplication a matrix is permutation! Apply knowledge of inverses of permutation matrix inverse matrices including diagonal, permutation, and the of. 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Moreover, the permutation matrix inverse operation on permutation that we describe in Section 8.1.2 below does not correspond matrix... N > 1 can be expressed as a product of two even permutations is merely a matter of.... ( transposition ) is inverse of itself simple matrix by understanding how the corresponding transformation. Is always even, as well as the product of two odd permutations in Section below. Odd one is odd correspond to matrix multiplication in this case, we have swap... Form given in the lemma 2-cycle ( transposition ) is inverse of a matrix the! And the inverse of an odd one is odd be expressed as a tool because it the... Of inverses of special matrices including diagonal, permutation, and the of... Odd one is odd not correspond to matrix multiplication how the corresponding linear is! The form given in the lemma \u2022identify and apply knowledge of inverses special. Product of two even permutations is always even, as well as the product of two odd permutations 2-cycle. Operation of row reductions exists a permutation in which each number and the number of place. Of special matrices including diagonal, permutation, and the number of the place which it is! Correspond to matrix multiplication Gauss transform matrices and the inverse of a simple matrix by how... Moreover, the composition operation on permutation that we describe in Section 8.1.2 below does not correspond matrix! A permutation matrix inverse of 2-cycles the use of matrix notation in denoting permutations is always even, as as., we have to swap the rows of a simple matrix by understanding how the corresponding transformation... As the product of two even permutations is always even, as well as the product two! Of 2-cycles and Gauss transform matrices > 1 can be expressed as a product of odd! Two odd permutations well as the product of two even permutations is always,. A matrix corresponding linear transformation is related to the matrix-vector multiplication with matrix.", "date": "2021-05-07 13:44:06", "meta": {"domain": "southernglowtans.com", "url": "http://www.southernglowtans.com/kids-desk-xefgmli/bccf95-permutation-matrix-inverse", "openwebmath_score": 0.8973594307899475, "openwebmath_perplexity": 401.4975871929007, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743615328861, "lm_q2_score": 0.8962513835254866, "lm_q1q2_score": 0.8880725174237823}}
{"url": "https://math.stackexchange.com/questions/1273112/intuitively-understanding-sum-i-1ni-n1-choose2", "text": "# Intuitively understanding $\\sum_{i=1}^ni={n+1\\choose2}$\n\nIt's straightforward to show that\n\n$$\\sum_{i=1}^ni=\\frac{n(n+1)}{2}={n+1\\choose2}$$\n\nbut intuitively, this is hard to grasp. Should I understand this to be coincidence? Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects? What's the intuition behind this?\n\n\u2022 I actually had similar encounter even with sum of an A.P Try it. It has $^{n}C_1$ and $^{n}C_2$ ! \u2013\u00a0Mann May 8 '15 at 16:17\n\u2022 What's an \"A.P\"? I'm sorry, I'm only a student. \u2013\u00a0user238435 May 8 '15 at 16:18\n\u2022 Arithmetic Progression. \u2013\u00a0user223391 May 8 '15 at 16:19\n\u2022 @leonbloy I'm not asking for a proof though, I was looking for intuition. \u2013\u00a0user238435 May 8 '15 at 16:30\n\u2022 @user238435 The answers include intuition-based proofs. Also see my last image in math.stackexchange.com/questions/44759/\u2026 \u2013\u00a0leonbloy May 8 '15 at 16:31\n\nConsider a tournament with $n+1$ teams each playing each other. We will count the number of matches played in two ways.\n\n\u2022 Every match is played between two teams. This inturn implies that the number of matches is $\\dbinom{n+1}2$.\n\u2022 We will now count the number of distinct matches played team by team.\n\u2022 The number of matches played by the first team is $n$.\n\u2022 The number of matches played by the second team is $n-1$, since their match with the first team has already been accounted for.\n\u2022 The number of matches played by the third team is $n-2$, since their matches with the first and second team have already been accounted for.\n\u2022 The number of matches played by the $k^{th}$ team is $n-k+1$, since their matches with the first $k-1$ teams have already been accounted for. Hence, the total number of matches is $$n+(n-1) + (n-2) + \\cdots + 1$$\n\u2022 While all the answers have been helpful, this one was pedagogically best for my grasping the intuition. Very clear explanation. \u2013\u00a0user238435 May 8 '15 at 16:37\n\nSuppose that you want to choose a subset $\\{m,n\\}$ with two elements of the set $$\\{1,2,\\dotsc,n+1\\}$$ Count this in two ways one of them naturally equals $\\binom {n+1}2$ and for the other observe that\n\nIf $max\\{m,n\\}=2$ then we have one subsets $\\{m,n\\}$.\nIf $max\\{m,n\\}=3$ then we have two subsets $\\{m,n\\}$.\n$\\vdots$\nIf $max\\{m,n\\}=n+1$ then we have $n$ subsets $\\{m,n\\}$.\n\nNow add up these cases to derive the identity.$\\square$\n\nThis is the classic proof without words, from https://maybemath.wordpress.com/\n\nThat doesn't help with this part of your question:\n\nWhy does the sum of the first $n$ natural numbers count the number of ways I can choose a pair out of $n+1$ objects?\n\nHere's a way to rephrase @user17762 's excellent accepted answer.\n\nImagine $n+1$ kids in a room. Each shakes hands with all the others. Then each kid shakes hands $n$ times, so there are $n(n+1)$ handshakes - each counted twice. You can pick a pair of kids (that is, a handshake) in $n(n+1)/2$ ways. But you can also think about the kids shaking hands as they enter the room one at a time. The second kid coming has one hand to shake. The third has two, and so on, for a total of $1 + 2 + \\cdots + n$.\n\nThe intuition is that for the pairs can be listed in the following way.\n\n$$\\begin{array}{ccccccc} 1,2 & & & & & & \\\\ 1,3 & 2,3 & & & & & \\\\ 1,4 & 2,4 & 3,4 & & & & \\\\ 1,5 & 2,5 & 3,5 & 4,5 & & & \\\\ 1,6 & 2,6 & 3,6 & 4,6 & 5,6 & & \\\\ 1,\\vdots & 2,\\vdots & 3,\\vdots & 4,\\vdots & 5,\\vdots &\\ddots & \\\\ 1,n+1 & 2,n+1 & 3,n+1 & 4,n+1 & 5,n+1 & \\cdots & n,n+1 \\\\ \\end{array}$$\n\nNotice that each row has length $i$ for $i=1,\\ldots,n$ since the number of pairs with maximum element $i+1$ is $i$. Therefore the total number of pairs, which is $\\binom{n+1}{2}$ is $\\displaystyle \\sum_{i=1}^n i$.\n\n\u2022 I doubt the person who downvoted this post will see this, but if you do, I would appreciate it in the future if you left an explanation as to why it was downvoted so that I could fix my post. \u2013\u00a0jgon May 8 '15 at 18:30\n\nIf you want to choose a pair out of $n+1$ objects (for example, $\\{0,1,\\dots,n\\}$), the possibilities are:\n\n$\\{0,1\\}$, $\\{0,2\\}$, ..., $\\{0,n\\}$, giving $n$ possibilities.\n\n$\\{1,2\\}$, $\\{1,3\\}$, ..., $\\{1,n\\}$ giving $n-1$ possibilities. (note that we've already picked $\\{1,0\\}$, so we can't repeat it here)\n\n$\\{2,3\\}$, $\\{2,4\\}$, ..., $\\{2,n\\}$ giving $n-2$ possibilities.\n\n$\\ \\ \\ \\ \\vdots$\n\n$\\{n-2,n-1\\}$, $\\{n-2,n\\}$ giving $2$ possibilities.\n\n$\\{n-1,n\\}$ giving $1$ possibility.\n\nSo the number of pairs is $n+(n-1)+\\dots+2+1$\n\n\u2022 @jgon Oh, I missed that. Thanks \u2013\u00a0Kitegi May 9 '15 at 8:16\n\nStart with $n+1$ objects, labelled $1,\\dots,n+1$. We count the number of ways of choosing a pair of objects. We may always assume that the first object we choose has a lower number than the second.\n\nChoose object number $1$. How many ways are there to choose a second object with a higher number?\n\nChoose object number $2$. How many ways are there to choose a second object with a higher number?\n\nChoose object number $3$. How many ways are there to choose a second object with a higher number?\n\n$$\\dots$$\n\nChoose object number $n$. How many ways are there to choose a second object with a higher number?\n\nChoose object number $n+1$. How many ways are there to choose a second object with a higher number?", "date": "2019-11-12 13:27:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1273112/intuitively-understanding-sum-i-1ni-n1-choose2", "openwebmath_score": 0.7586857676506042, "openwebmath_perplexity": 272.603581655035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513873424044, "lm_q2_score": 0.9005297941266014, "lm_q1q2_score": 0.8880587058211178}}
{"url": "http://math.stackexchange.com/questions/119911/proving-formula-involving-eulers-totient-function", "text": "# Proving formula involving Euler's totient function\n\nThis question is motivated by lhf's comment here .\n\n\"It'd be nice to relate this formula with the natural mapping $U_{mn} \\to U_m \\times U_n$ by proving that the kernel has size $d$ and the image has index $\\varphi(d)$.\"\n\nI'm trying to prove the formula $$\\varphi(mn) = \\varphi(m)\\varphi(n) \\frac{d}{\\varphi(d)}$$ by considering the natural map $\\eta\\colon U_{mn} \\to U_m \\times U_n$ (i.e. the map sending $\\overline{x} \\mapsto (\\overline{x},\\overline{x})$, where the bar denotes reduction mod $mn$, $m$, or $n$, respectively).\n\nI've been able to show that the kernel has the right size as follows:\n\nThe kernel of $\\eta$ consists of the elements $\\overline{x} \\in U_{mn}$ with $x \\equiv 1 \\bmod m$ and $x \\equiv 1 \\bmod n$. The integers $x$ which satisfy these conditions are those of the form $x = \\frac{mn}{d}k + 1$ for $k \\in \\mathbb Z$. On the other hand, any such integer $x$ is relatively prime to $mn$, and hence gives and element $\\overline{x} \\in U_{mn}$. Therefore, $\\ker \\eta$ consists of the $d$ distinct elements $\\overline{x}$, where $x = \\frac{mn}{d}k + 1$ and $k \\in \\{1,\\ldots,d\\}$.\n\nOnce it has been shown that the image has index $\\varphi(d)$, the first isomorphism theorem gives $$\\frac{U_{mn}}{\\ker \\eta} \\cong Im(\\eta),$$ and so $$\\frac{\\varphi(mn)}{d} = \\frac{|U_{mn}|}{|\\ker \\eta|} = |Im(\\eta)| = \\frac{|U_m \\times U_n|}{|U_m \\times U_n:Im(\\eta)|} = \\frac{\\varphi(m)\\varphi(n)}{\\varphi(d)},$$ or $$\\varphi(mn) = \\varphi(m)\\varphi(n) \\frac{d}{\\varphi(d)}.$$\n\nI'm having trouble showing the image has the right index.\n\nI've noticed that $\\eta(\\overline{x}) = \\eta(\\overline{x + \\frac{mn}{d}})$, so the image consists the images of the elements $\\overline{x}$ with $1 \\leq x < \\frac{mn}{d}$. I'm not sure if this is going anywhere, though. Any suggestions?\n\n-\nI got stuck at the same point... Thanks for turning my comment to a question. \u2013\u00a0lhf Mar 14 '12 at 12:04\n\nI'll adjust your notation a bit, using $x\\in U_{mn}$ for an invertible element of $\\mathbb{Z}/mn\\mathbb{Z}$, using $\\bar x\\in U_m$ for the residue class of $x$ modulo $m$, and using $\\tilde x \\in U_n$ for the residue class of $x$ modulo $n$.\n\nThe image of your map $x \\mapsto (\\bar x,\\tilde x)$ is generally smaller than $U_m \\times U_n$ because $\\bar x$ and $\\tilde x$ will always be the same modulo $d$. We first choose a reduced residue system $\\{a_1=1, a_2, \\ldots, a_{\\varphi(d)}\\}$ modulo $d$ from the elements $U_{n}$ and consider the images of the maps $f_i: x \\mapsto (\\bar x, a_i\\tilde x)$. (Note that each $a_i$ is invertible modulo $n$.) It's clear that the images of these maps are disjoint, have the same size, and that we are studying the special case $f_1:x \\mapsto (\\bar x, \\tilde x)$. In fact, the union of these images is all of $U_m \\times U_n$, as we now show.\n\nTake any $(y,z) \\in U_m \\times U_n$. We show it is equal to some $f_i(x)$ where $a_i \\equiv zy^{-1} \\pmod{d}$. By a slight generalization of the Chinese Remainder Theorem, there is a unique $x$ modulo $\\frac{mn}{d}$ such that $$x \\equiv y \\pmod{m}\\qquad \\qquad \\text{and} \\qquad \\qquad x \\equiv z{a_i}^{-1} \\pmod{n}.$$ Then $f_i(x)=(\\bar x, a_i \\tilde x) =(y,z)$. (In fact, the $d$ preimages of $(y,z)$ are the elements $x+\\frac{mn}{d} k$ with $0 \\leq k \\leq d-1$.)\n\nA generalization of the Chinese Remainder Theorem is required because $m$ and $n$ share the factor $d$. Such systems of congruences have a solution as long as they are compatible modulo $d$, and this solution is unique modulo $\\rm{lcm}(m,n)=\\frac{mn}{d}$. Our system is compatible modulo $d$, since $y \\equiv z {a_i}^{-1} \\pmod{d}$.\n\nThus, the index of the image of $x \\mapsto (\\bar x, \\tilde x)$ is $\\varphi(d)$.\n\n-\n\nJonas Kibelbek directly proved that the index of the image of $\\eta$ is $\\phi(d),$ and below is an alternative by proving an exact sequence, which I hope might clarify the matter somewhat.\n\nThe exact sequence I want to prove is\n\n$$0\\rightarrow \\text{Ker}(f)\\rightarrow U_{mn}\\overset{f}{\\rightarrow}U_m\\times U_n\\overset{g}{\\rightarrow} U_d\\rightarrow0,$$\n\nwhere $f(x+mn\\mathbb Z)=(x+m\\mathbb Z, x+n\\mathbb Z),$ and $g(x+m\\mathbb Z, y+n\\mathbb Z)=xy^{-1}+d\\mathbb Z$ (Here the inverse is taken modulo $d$).\nProof:\nFirstly, $\\forall (a+d\\mathbb Z)\\in U_d,$ we have that $g(a+m\\mathbb Z, 1+n\\mathbb Z)=(a+d\\mathbb Z),$ so $g$ is surjective. Further, it is clear that $g\\circ f$ vanishes. Conversely, if $(x+m\\mathbb Z, y+n\\mathbb Z)\\in U_m\\times U_n$ is such that $x\\equiv y\\pmod d,$ then, by a slight generalisation of Chinese rmainder theorem, as in Kibelbek's answer, $\\exists z$ such that $\\begin{cases}z\\equiv x\\pmod m\\\\z\\equiv y\\pmod n\\end{cases}.$ Thus the sequence is exact. Q.E.D.\n\nP.S. This sequence is in essence the sequence in this answer, with some reductions and modifications; I mark this answer as CW, for there is nothing new in this answer.\n\n-", "date": "2016-02-14 06:15:40", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/119911/proving-formula-involving-eulers-totient-function", "openwebmath_score": 0.9708985686302185, "openwebmath_perplexity": 95.63865466887287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850882200038, "lm_q2_score": 0.9032942158155877, "lm_q1q2_score": 0.8880150738436862}}
{"url": "http://openstudy.com/updates/55d713cfe4b02663346b5874", "text": "## anonymous one year ago I need help to prove that $$\\binom{n}{0}^2 + \\binom{n}{1}^2 + \\cdots + \\binom{n}{n}^2 = \\binom{2n}{n}.$$ using committee forming...\n\n1. ganeshie8\n\nsuppose there are $$n$$ men and $$n$$ women and you want to choose a committee consisting of $$n$$ people\n\n2. anonymous\n\nthats 2n choose n\n\n3. ganeshie8\n\nYes, lets count it in an alternative way\n\n4. ganeshie8\n\nhow many committees will be there with out women ?\n\n5. anonymous\n\n*\n\n6. anonymous\n\n1\n\n7. ganeshie8\n\nYes, how many committees will be there with exactly 1 women ?\n\n8. anonymous\n\nn choose n-1\n\n9. ganeshie8\n\ntry again\n\n10. anonymous\n\nn choose n-1 multiplied by n?\n\n11. ganeshie8\n\nyou can choose $$1$$ women from the group of $$n$$ women in $$\\binom{n}{1}$$ ways after that, the remaining $$n-1$$ men can be chosen from the group of $$n$$ men in $$\\binom{n}{n-1}$$ ways so total $$n$$ member committees with exactly $$1$$ women is $$\\binom{n}{1}*\\binom{n}{n-1}$$\n\n12. ganeshie8\n\ndoes that make sense\n\n13. anonymous\n\nyes, i get it\n\n14. anonymous\n\nnow would i find the number of ways to make a committee with two women?\n\n15. ganeshie8\n\nyes find it, after that you will see the pattern\n\n16. anonymous\n\nalright, ill get back to you with the results :D\n\n17. ganeshie8\n\ntake your time, we're almost done!\n\n18. anonymous\n\nhang on... n choose k and n choose (n-k) give the same result\n\n19. anonymous\n\nso the total possible combinations is just sums of the squares....\n\n20. ganeshie8\n\nYep!\n\n21. anonymous\n\nbut how do i equate it to 2n choose n?\n\n22. anonymous\n\noh wait nevermind\n\n23. anonymous\n\ni get it\n\n24. ganeshie8\n\ngood :) id like to see the complete proof if you don't mind\n\n25. anonymous\n\nsure :D\n\n26. ganeshie8\n\ntake a screenshot and attach if psble\n\n27. anonymous\n\ni need to go eat dinner, i'll send it to you in about an hour, is that ok?\n\n28. ganeshie8\n\ntake ur time\n\n29. anonymous\n\ni also need to prove the same thing using the \"block walking\" method... but i dont know how. Do you think you can try to help me with this too? please?\n\n30. ganeshie8\n\nsure that is also an interesting way to count first, may i see the previous proof...\n\n31. anonymous\n\nyes im just typing it up now\n\n32. anonymous\n\nsorry the codeisnt working...\n\n33. ganeshie8\n\nmake this correction : $$k\\le n$$\n\n34. ganeshie8\n\nother than that, the proof looks good!\n\n35. anonymous\n\nok thanks!\n\n36. anonymous\n\ncan you help me with the second part please?\n\n37. anonymous\n\nhello? are you still here @ganeshie8\n\n38. ganeshie8\n\nHey!\n\n39. anonymous\n\n40. ganeshie8\n\nConsider a $$n\\times n$$grid |dw:1440165928386:dw|\n\n41. anonymous\n\nright\n\n42. anonymous\n\nnumber of paths from bottom left to top right = $$\\binom{2n}{n}$$ |dw:1440168478785:dw|\n\n43. anonymous\n\nsuppose your friend's home is located at $$(k,~n-k)$$, where $$k\\in \\left\\{0,1,2,\\ldots ,n \\right\\}$$. then number of paths through $$(k,n-k)$$ is given by $$\\binom{n}{k}*\\binom{n}{n-k} = \\binom{n}{k}^2$$ ..... https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients", "date": "2016-10-27 17:13:33", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/55d713cfe4b02663346b5874", "openwebmath_score": 0.7537182569503784, "openwebmath_perplexity": 2915.1386648264042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984575451312636, "lm_q2_score": 0.9019206857566126, "lm_q1q2_score": 0.888008966227019}}
{"url": "http://hurdman.net/ekli2a6u/776769-what-is-the-cube-root-of-125", "text": "A cube root of a number a is a number x such that x3 = a, in other words, a number x whose cube is a. So, in this case the cube root of 125 is 5. Thus, each edge of the cube is 5 cm long. The length of a side (edge) of a cube is equal to the cube root of the volume. 125 is said to be a perfect cube because 5 x 5 x 5 is equal to 125. How to find the square root of 125 by long division method Here we will show you how to calculate the square root of 125 using the long division method with one decimal place accuracy. Let's check this with \u221b125*1=\u221b125. Having trouble with your homework? How long does it take to cook a 23 pound turkey in an oven? Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Step 1) Set up 125 in pairs of two digits from right to left and attach one set of 00 because we want one decimal: \u2026 First we will find all factors under the cube root: 125 has the cube factor of 125. 5\u00b3 = 5 * 5 * 5 = 25 * 5 = 125 So the cube root of 125 is 5. The cube root of 125 is 5 so therefore each edge is 5 cm which is about 2 inches Yes, simply enter the fraction as a decimal floating point number and you will get the corresponding cube root. See next answers. Thus, each edge of the cube is 5 cm long. Thus, each edge of the cube is 5 cm long. After 42 months, Sally earned $238 in simple interest. Estimating higher n th roots, even if using a calculator for intermediary steps, is \u2026 How will understanding of attitudes and predisposition enhance teaching? 'CUBE ROOT OF 125' is a 13 letter phrase starting with C and ending with 5 Crossword clues for 'CUBE ROOT OF 125' Synonyms, crossword answers and other related words for CUBE ROOT OF 125 [five] We hope that the following list of synonyms for the word five will help you to finish your crossword today. Here is the answer to questions like: What is the cube root of 125 or what is the cube root of 125? WHO WANNA JOIN MY \u2026 125 can be written as 125(e^0), 125(e^((2pi)i)), 125(e^((4pi)i)) where e is euler\u2019s number, i is the imaginary unit, i^2=-1, and e^((theta)i)=cos(theta)+(i)(sin(theta)) where theta is an angle measured in radians. The cube root of -64 is written as $$\\sqrt[3]{-64} = -4$$. Inter state form of sales tax income tax? cube root of 125/512 = 5/8. Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. Therefore the cube roots of 125 are 5(e^0),5(e^((2pi)i/3)),5(e^((4pi)i/3)). What is the birthday of carmelita divinagracia? When did organ music become associated with baseball? What is the conflict of the story of sinigang? What is plot of the story Sinigang by Marby Villaceran? The nearest previous perfect cube is \u2026 The cube root of a number answers the question \"what number can I multiply by itself twice to get this number?\". The length of a side (edge) of a cube is equal to the cube root of the volume. Not sure about the answer? Why don't libraries smell like bookstores? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Cube Root of 125. Find the interest rate \u2026 that Sally earned from the bank. That number is 5. Who of the proclaimers was married to a little person? For example, 5 is the cube root of 125 because 53 = 5\u20225\u20225 = 125, -5 is cube root of -125 because (-5)3 = (-5)\u2022(-5)\u2022(-5) = -125. Just right click on the above image, then choose copy link address, then past it in your HTML. Cube of \u221b125=5 which results into 5\u221b1; All radicals are now simplified. What is the contribution of candido bartolome to gymnastics? Is evaporated milk the same thing as condensed milk? So, in this case the cube root of 125 is 5. Guess: 5.125 27 \u00f7 5.125 = 5.268 (5.125 + 5.268)/2 = 5.197 27 \u00f7 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 \u00f7 5.196 = 5.196 Estimating an n th Root. Therefore, the real cube root of 125 is 5. Get free help! 5(e^0)=5(1)=0 Copyright \u00a9 2020 Multiply Media, LLC. Someone help me with this ASAP ANSWER QUICK! This is the lost art of how they calculated the square root of 125 by hand before modern technology was invented. See also our cube root table from 1 to 1000. New questions in Mathematics. Learn more with Brainly! Who is the longest reigning WWE Champion of all time? The cube root of -8 is written as $$\\sqrt[3]{-8} = -2$$. That number is 5. Does the calculator support fractions? A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Since 125 is a whole number, it is a perfect cube. Now extract and take out the cube root \u221b125 * \u221b1. 80% of questions are answered in under 10 minutes Answers come with explanations, so that \u2026 Sally deposited$850 into her bank account for 42 months. How long will it take to cook a 12 pound turkey? ... \u221b125: 5 \u221b216: 6 \u221b1,000: 10 \u221b1,000,000: 100 \u221b1,000,000,000: 1000: The calculations were performed using this cube root calculator. For example, 5 is the cube root of 125 because 5 3 = 5\u20225\u20225 = 125, -5 is cube root of -125 because (-5) 3 = (-5)\u2022 (-5)\u2022 (-5) = -125. What details make Lochinvar an attractive and romantic figure? As you can see the radicals are not in their simplest form. All information in this site is provided \u201cas is\u201d, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. Cube roots (for integer results 1 through 10) Cube root of 1 is 1; Cube root of 8 is 2; Cube root of 27 is 3; Cube root of 64 is 4; Cube root of 125 is 5; Cube root of 216 is 6; Cube root of 343 is 7; Cube root of 512 is 8; Cube root of 729 is 9", "date": "2021-02-26 09:29:08", "meta": {"domain": "hurdman.net", "url": "http://hurdman.net/ekli2a6u/776769-what-is-the-cube-root-of-125", "openwebmath_score": 0.5956003665924072, "openwebmath_perplexity": 354.3765989599167, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9744347905312772, "lm_q2_score": 0.9111797106148062, "lm_q1q2_score": 0.8878852104492885}}
{"url": "https://math.stackexchange.com/questions/2513458/setting-of-this-induction-proof?noredirect=1", "text": "# setting of this induction proof [duplicate]\n\nI would like to see if this is a correct induction proof and whether or not this is a good setting out of it\nA sequence is defined by $$a_n = a_{n-1} + a_{n-2} + a_{n-3}$$ for $n\\geq 3, a_0 = 1, a_1 = 2, a_2 = 4$.\nProve that $a_n \\leq 4^n$ for all $n\\in\\mathbb{N}$.\n\nLet $P(n)$ be the proposition that $$''a_n\\leq 4^n{''}.$$\n\nNow since we have $a_0 = 1 = 4^0 \\leq 4^0$ then $a_0 \\leq 4^0$.\nAlso, $a_1 = 2 < 4 \\leq 4 = 4^1$ then $a_1 \\leq 4^1$.\nAlso, $a_2 = 4<16 = 4^2 \\leq 4^2$ so $a_2 \\leq 4^2$.\n\nHence, $P(0),P(1),P(2)$ are true.\n\nNow, let $k-3\\in\\mathbb{N}$ and suppose $P(k-3),P(k-2),P(k-1)$ is true.\nWe must show that $P(k)$ is true.\n\nNow by definition \\begin{align}a_n &= a_{n-1} + a_{n-2} + a_{n-3} \\\\ &\\leq 4^{n-1} + 4^{n-2} + 4^{n-3} \\qquad \\text{by the inductive hypothesis}\\\\ &= 21\\times 4^{k-3} \\\\ &\\leq 64\\times 4^{k-3} \\\\ &= 4^{k}.\\end{align}\n\nHence, $P(k)$ is true if $P(k-3),P(k-2),P(k-1)$ is true for $k\\geq 3$.\nSo by induction, $P(n)$ is true for all $n\\in\\mathbb{N}$.\n\n\u2022 Looks good to me. \u2013\u00a0learning Nov 10 '17 at 8:02\n\nSuppose that $P(n)$ is true and prove it for $P(n+1)$\n\n$a_n = a_{n-1} + a_{n-2} + a_{n-3}\\quad$ by the inductive hypothesis\n\nif we suppose that $P(k)$ is true for any $1\\leq k\\leq n$ we are using strong induction\n\nWe must prove that\n\n$a_{n+1}\\leq 4^{n+1}$\n\n$a_{n+1}=a_n + a_{n-1} + a_{n-2} \\leq 4^n+4^{n-1}+4^{n-2}=4^{n-2}\\left(4^2+4+1\\right)=21\\cdot 4^{n-2}\\leq 64\\cdot 4^{n-2}=4^{n+1}$\n\nproved\n\nHope this helps\n\n\u2022 Is my assumption sentence incorrect then? \u2013\u00a0OneGapLater Nov 10 '17 at 9:13\n\u2022 Induction works in this way: P(1) is true, if $P(n)$ is true then $P(n+1)$ is true. End. You assumed $P(n-2),P(n-1),P(n)$ true and proved $P(n+1)$ then formally you used strong induction, which is explained in my answer. The core of your proof is correct. My answer is just to adjust some formal detail. \u2013\u00a0Raffaele Nov 10 '17 at 11:08\n\u2022 I see, so formal strong induction is assuming (in my case) for all $3\\leq i \\leq k$ and $k\\geq 3$ is true. (but really, I only need to use the values $i=k-3,k-2,k-1$ in my proof?) \u2013\u00a0OneGapLater Nov 11 '17 at 4:15\n\u2022 @ActuarialStudent101 Formally you did not use strong induction because your hypothese was not the statement \"$P(i)$ is true for all $i<k$\". With normal induction (\"$P(n)$ true implies $P(n+1)$ true\") you cannot reach your goal. Nevertheless you did reach your goal with normal induction. This because you focused not on $P(n)$ but on the $Q(n)$ in my answer. Beautiful about induction is that sometimes you can make things easyer by enforcing the hypothese. A possible bonus is then that you also prove more. \u2013\u00a0drhab Nov 11 '17 at 9:00\n\nActually you proved that $\\forall n\\in\\mathbb N\\, [Q(n)]$ where $Q(n)$ is stated by:$$\\forall n\\in\\mathbb N [P(n)\\wedge P(n+1)\\wedge P(n+2)]$$\nOf course $\\forall n\\in\\mathbb N\\, [P(n)]$ is a direct consequence of $\\forall n\\in\\mathbb N\\, [Q(n)]$", "date": "2020-12-02 08:44:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2513458/setting-of-this-induction-proof?noredirect=1", "openwebmath_score": 0.9968790411949158, "openwebmath_perplexity": 352.24568168174494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232904845686, "lm_q2_score": 0.9032942067038784, "lm_q1q2_score": 0.8877785845083539}}
{"url": "https://mathhelpboards.com/threads/finding-the-area-bounded-by-the-curve.7787/", "text": "# Finding the Area bounded by the curve\n\n#### shamieh\n\n##### Active member\nFind the area bounded by the curve $$\\displaystyle x = 16 - y^4$$ and the y axis.\n\nI need someone to check my work.\n\nso I know this is a upside down parabola so I find the two x coordinates which are\n\n$$\\displaystyle 16 - y^4 = 0$$\n$$\\displaystyle y^4 = 16$$\n$$\\displaystyle y^2 = +- \\sqrt{4}$$\n$$\\displaystyle y = +- 2$$\n\nso I know\n\n$$\\displaystyle \\int^2_{-2} 16 - y^4 dy$$\n\nTake antiderivative\n\n$$\\displaystyle 16y - \\frac{1}{5}y^5$$ | -2 to 2\n\nso $$\\displaystyle (2) = 32 - \\frac{32}{5} = \\frac{160}{5}$$\n\nthen $$\\displaystyle (-2) = -32 - (\\frac{-32}{5}) = -32 + \\frac{32}{5} = \\frac{-160}{5} + \\frac{32}{5} = \\frac{-128}{5}$$\n\nSO finally\n$$\\displaystyle [\\frac{160}{5}] - [\\frac{-128}{5}] = \\frac{288}{5}$$\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nThis is not a parabola. It is like a parabola that intersects the y-axis at $$\\displaystyle y=\\pm 2$$ so it is open to the left. I suggest you revise your calculations.\n\n#### MarkFL\n\nStaff member\nI would use the even-function rule to state:\n\n$$\\displaystyle A=2\\int_0^2 16-y^4\\,dy=\\frac{2}{5}\\left[80y-y^5 \\right]_0^2=?$$\n\n#### shamieh\n\n##### Active member\nRecalculated answer below if someone has a chance to check.\n\nLast edited:\n\n#### shamieh\n\n##### Active member\nrecalculated and got $$\\displaystyle \\frac{256}{5}$$ . Is that correct?\n\n- - - Updated - - -\n\nI would use the even-function rule to state:\n\n$$\\displaystyle A=2\\int_0^2 16-y^4\\,dy=\\frac{2}{5}\\left[80y-y^5 \\right]_0^2=?$$\nYea this rule is so much easier!\n\n- - - Updated - - -\n\nMark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.\n\n#### shamieh\n\n##### Active member\nLike how would I use this rule if i had $$\\displaystyle 5 - x^2$$?\n\n#### MarkFL\n\nStaff member\nrecalculated and got $$\\displaystyle \\frac{256}{5}$$ . Is that correct?\n\n- - - Updated - - -\n\nYea this rule is so much easier!\n\n- - - Updated - - -\n\nMark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.\nYes, your result of $$\\displaystyle A=\\frac{256}{5}$$ is correct.\n\nAn even function is symmetric about the $y$-axis, i.e., $$\\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.\n\nObserve that:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{-a}^0 f(x)\\,dx+\\int_0^a f(x)\\,dx$$\n\nNow, in the first integral, if we replace $x$ with $-x$, we have:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{a}^0 f(-x)\\,-dx+\\int_0^a f(x)\\,dx$$\n\nBringing the negative in front of the differential out front and using $$\\displaystyle f(-x)=f(x)$$, we have:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=-\\int_{a}^0 f(x)\\,-dx+\\int_0^a f(x)\\,dx$$\n\nApplying the FTOC, we obtain:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=-\\left(F(0)-F(a) \\right)+F(a)-F(0)=2F(a)=2\\int_0^a f(x)\\,dx$$\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nYes, your result of $$\\displaystyle A=\\frac{256}{5}$$ is correct.\n\nAn even function is symmetric about the $y$-axis, i.e., $$\\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.\n\nObserve that:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{-a}^0 f(x)\\,dx+\\int_0^a f(x)\\,dx$$\n\nNow, in the first integral, if we replace $x$ with $-x$, we have:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{a}^0 f(-x)\\,-dx+\\int_0^a f(x)\\,dx$$\n\nBringing the negative in front of the differential out front and using $$\\displaystyle f(-x)=f(x)$$, we have:\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=-\\int_{a}^0 f(x)\\,dx+\\int_0^a f(x)\\,dx=\\int_{0}^a f(x)\\,dx+\\int_0^a f(x)\\,dx=2\\int^a_0 f(x)\\, dx$$", "date": "2020-11-25 07:42:12", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/finding-the-area-bounded-by-the-curve.7787/", "openwebmath_score": 0.8137761950492859, "openwebmath_perplexity": 671.2401285350282, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226334351969, "lm_q2_score": 0.9086179037377831, "lm_q1q2_score": 0.887740257096257}}
{"url": "https://lectures.quantecon.org/jl/linear_algebra.html", "text": "Code should execute sequentially if run in a Jupyter notebook\n\n# Linear Algebra\u00b6\n\n## Overview\u00b6\n\nLinear algebra is one of the most useful branches of applied mathematics for economists to invest in\n\nFor example, many applied problems in economics and finance require the solution of a linear system of equations, such as\n\n$\\begin{split}\\begin{array}{c} y_1 = a x_1 + b x_2 \\\\ y_2 = c x_1 + d x_2 \\end{array}\\end{split}$\n\nor, more generally,\n\n(1)$\\begin{split}\\begin{array}{c} y_1 = a_{11} x_1 + a_{12} x_2 + \\cdots + a_{1k} x_k \\\\ \\vdots \\\\ y_n = a_{n1} x_1 + a_{n2} x_2 + \\cdots + a_{nk} x_k \\end{array}\\end{split}$\n\nThe objective here is to solve for the \u201cunknowns\u201d $$x_1, \\ldots, x_k$$ given $$a_{11}, \\ldots, a_{nk}$$ and $$y_1, \\ldots, y_n$$\n\nWhen considering such problems, it is essential that we first consider at least some of the following questions\n\n\u2022 Does a solution actually exist?\n\u2022 Are there in fact many solutions, and if so how should we interpret them?\n\u2022 If no solution exists, is there a best \u201capproximate\u201d solution?\n\u2022 If a solution exists, how should we compute it?\n\nThese are the kinds of topics addressed by linear algebra\n\nIn this lecture we will cover the basics of linear and matrix algebra, treating both theory and computation\n\nWe admit some overlap with this lecture, where operations on Julia arrays were first explained\n\nNote that this lecture is more theoretical than most, and contains background material that will be used in applications as we go along\n\n## Vectors\u00b6\n\nA vector of length $$n$$ is just a sequence (or array, or tuple) of $$n$$ numbers, which we write as $$x = (x_1, \\ldots, x_n)$$ or $$x = [x_1, \\ldots, x_n]$$\n\nWe will write these sequences either horizontally or vertically as we please\n\n(Later, when we wish to perform certain matrix operations, it will become necessary to distinguish between the two)\n\nThe set of all $$n$$-vectors is denoted by $$\\mathbb R^n$$\n\nFor example, $$\\mathbb R ^2$$ is the plane, and a vector in $$\\mathbb R^2$$ is just a point in the plane\n\nTraditionally, vectors are represented visually as arrows from the origin to the point\n\nThe following figure represents three vectors in this manner\n\n#=\n\n@author : Spencer Lyon <spencer.lyon@nyu.edu>\nVictoria Gregory <victoria.gregory@nyu.edu>\n\n=#\n\nusing Plots\npyplot()\nusing LaTeXStrings\n\nvecs = ([2, 4], [-3, 3], [-4, -3.5])\nx_vals = zeros(2, length(vecs))\ny_vals = zeros(2, length(vecs))\nlabels = []\n\n# Create matrices of x and y values, labels for plotting\nfor i = 1:length(vecs)\nv = vecs[i]\nx_vals[2, i] = v[1]\ny_vals[2, i] = v[2]\nlabels = [labels; (1.1 * v[1], 1.1 * v[2], \"$v\")] end plot(x_vals, y_vals, arrow=true, color=:blue, legend=:none, xlims=(-5, 5), ylims=(-5, 5), annotations=labels, xticks=-5:1:5, yticks=-5:1:5, framestyle=:origin) ### Vector Operations\u00b6 The two most common operators for vectors are addition and scalar multiplication, which we now describe As a matter of definition, when we add two vectors, we add them element by element $\\begin{split}x + y = \\left[ \\begin{array}{c} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{array} \\right] + \\left[ \\begin{array}{c} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{array} \\right] := \\left[ \\begin{array}{c} x_1 + y_1 \\\\ x_2 + y_2 \\\\ \\vdots \\\\ x_n + y_n \\end{array} \\right]\\end{split}$ Scalar multiplication is an operation that takes a number $$\\gamma$$ and a vector $$x$$ and produces $\\begin{split}\\gamma x := \\left[ \\begin{array}{c} \\gamma x_1 \\\\ \\gamma x_2 \\\\ \\vdots \\\\ \\gamma x_n \\end{array} \\right]\\end{split}$ Scalar multiplication is illustrated in the next figure # illustrate scalar multiplication x = [2, 2] scalars = [-2, 2] # Create matrices of x and y values, labels for plotting x_vals = zeros(2, 1 + length(scalars)) y_vals = zeros(2, 1 + length(scalars)) labels = [] x_vals[2, 3] = x[1] y_vals[2, 3] = x[2] labels = [labels; (x[1] + 0.4, x[2] - 0.2, L\"$x$\")] # Perform scalar multiplication, store results in plotting matrices for i = 1:length(scalars) s = scalars[i] v = s .* x x_vals[2, i] = v[1] y_vals[2, i] = v[2] labels = [labels; (v[1] + 0.4, v[2] - 0.2, LaTeXString(\"\\$$s x\\$\"))]\nend\n\nplot(x_vals, y_vals, arrow=true, color=[:red :red :blue],\nlegend=:none, xlims=(-5, 5), ylims=(-5, 5),\nannotations=labels, xticks=-5:1:5, yticks=-5:1:5,\nframestyle=:origin)\n\nIn Julia, a vector can be represented as a one dimensional Array\n\nJulia Arrays allow us to express scalar multiplication and addition with a very natural syntax\n\nx = ones(3)\n3-element Array{Float64,1}:\n1.0\n1.0\n1.0\ny = [2, 4, 6]\n3-element Array{Int64,1}:\n2\n4\n6\nx + y\n3-element Array{Float64,1}:\n3.0\n5.0\n7.0\n4x  # equivalent to 4 * x and 4 .* x\n3-element Array{Float64,1}:\n4.0\n4.0\n4.0\n\n### Inner Product and Norm\u00b6\n\nThe inner product of vectors $$x,y \\in \\mathbb R ^n$$ is defined as\n\n$x' y := \\sum_{i=1}^n x_i y_i$\n\nTwo vectors are called orthogonal if their inner product is zero\n\nThe norm of a vector $$x$$ represents its \u201clength\u201d (i.e., its distance from the zero vector) and is defined as\n\n$\\| x \\| := \\sqrt{x' x} := \\left( \\sum_{i=1}^n x_i^2 \\right)^{1/2}$\n\nThe expression $$\\| x - y\\|$$ is thought of as the distance between $$x$$ and $$y$$\n\nContinuing on from the previous example, the inner product and norm can be computed as follows\n\ndot(x, y)               # Inner product of x and y\n12.0\nsum(x .* y)             # Gives the same result\n12.0\nnorm(x)                 # Norm of x\n1.7320508075688772\nsqrt(sum(x.^2))         # Gives the same result\n1.7320508075688772\n\n### Span\u00b6\n\nGiven a set of vectors $$A := \\{a_1, \\ldots, a_k\\}$$ in $$\\mathbb R ^n$$, it\u2019s natural to think about the new vectors we can create by performing linear operations\n\nNew vectors created in this manner are called linear combinations of $$A$$\n\nIn particular, $$y \\in \\mathbb R ^n$$ is a linear combination of $$A := \\{a_1, \\ldots, a_k\\}$$ if\n\n$y = \\beta_1 a_1 + \\cdots + \\beta_k a_k \\text{ for some scalars } \\beta_1, \\ldots, \\beta_k$\n\nIn this context, the values $$\\beta_1, \\ldots, \\beta_k$$ are called the coefficients of the linear combination\n\nThe set of linear combinations of $$A$$ is called the span of $$A$$\n\nThe next figure shows the span of $$A = \\{a_1, a_2\\}$$ in $$\\mathbb R ^3$$\n\nThe span is a 2 dimensional plane passing through these two points and the origin\n\nx_min, x_max = -5, 5\ny_min, y_max = -5, 5\n\n\u03b1, \u03b2 = 0.2, 0.1\n\n# Axes\ngs = 3\nz = linspace(x_min, x_max, gs)\nx = zeros(gs)\ny = zeros(gs)\nplot(x, y, z, color=:black, linewidth=2, alpha=0.5, label=\"\", legend=false)\nplot!(z, x, y, color=:black, linewidth=2, alpha=0.5, label=\"\")\nplot!(y, z, x, color=:black, linewidth=2, alpha=0.5, label=\"\")\n\n# Fixed linear function, to generate a plane\nf(x, y) = \u03b1 .* x + \u03b2 .* y\n\n# Vector locations, by coordinate\nx_coords = [3, 3]\ny_coords = [4, -4]\nz = f(x_coords, y_coords)\n\n# Lines to vectors\nn = 2\nx_vec = zeros(n, n)\ny_vec = zeros(n, n)\nz_vec = zeros(n, n)\nlabels = []\n\nfor i=1:n\nx_vec[:, i] = [0; x_coords[i]]\ny_vec[:, i] = [0; y_coords[i]]\nz_vec[:, i] = [0; f(x_coords[i], y_coords[i])]\nlab = string(\"a\", i)\npush!(labels, lab)\nend\n\nplot!(x_vec, y_vec, z_vec, color=[:blue :red], linewidth=1.5,\nalpha=0.6, label=labels)\n\n# Draw the plane\ngrid_size = 20\nxr2 = linspace(x_min, x_max, grid_size)\nyr2 = linspace(y_min, y_max, grid_size)\nz2 = Array{Float64}(grid_size, grid_size)\nfor i in 1:grid_size\nfor j in 1:grid_size\nz2[j, i] = f(xr2[i], yr2[j])\nend\nend\nsurface!(xr2, yr2, z2, cbar=false, alpha=0.2, fill=:blues,\nxlims=(x_min, x_max), ylims=(x_min, x_max),\nzlims=(x_min, x_max), xticks=[0], yticks=[0],\nzticks=[0])\n\n#### Examples\u00b6\n\nIf $$A$$ contains only one vector $$a_1 \\in \\mathbb R ^2$$, then its span is just the scalar multiples of $$a_1$$, which is the unique line passing through both $$a_1$$ and the origin\n\nIf $$A = \\{e_1, e_2, e_3\\}$$ consists of the canonical basis vectors of $$\\mathbb R ^3$$, that is\n\n$\\begin{split}e_1 := \\left[ \\begin{array}{c} 1 \\\\ 0 \\\\ 0 \\end{array} \\right] , \\quad e_2 := \\left[ \\begin{array}{c} 0 \\\\ 1 \\\\ 0 \\end{array} \\right] , \\quad e_3 := \\left[ \\begin{array}{c} 0 \\\\ 0 \\\\ 1 \\end{array} \\right]\\end{split}$\n\nthen the span of $$A$$ is all of $$\\mathbb R ^3$$, because, for any $$x = (x_1, x_2, x_3) \\in \\mathbb R ^3$$, we can write\n\n$x = x_1 e_1 + x_2 e_2 + x_3 e_3$\n\nNow consider $$A_0 = \\{e_1, e_2, e_1 + e_2\\}$$\n\nIf $$y = (y_1, y_2, y_3)$$ is any linear combination of these vectors, then $$y_3 = 0$$ (check it)\n\nHence $$A_0$$ fails to span all of $$\\mathbb R ^3$$\n\n### Linear Independence\u00b6\n\nAs we\u2019ll see, it\u2019s often desirable to find families of vectors with relatively large span, so that many vectors can be described by linear operators on a few vectors\n\nThe condition we need for a set of vectors to have a large span is what\u2019s called linear independence\n\nIn particular, a collection of vectors $$A := \\{a_1, \\ldots, a_k\\}$$ in $$\\mathbb R ^n$$ is said to be\n\n\u2022 linearly dependent if some strict subset of $$A$$ has the same span as $$A$$\n\u2022 linearly independent if it is not linearly dependent\n\nPut differently, a set of vectors is linearly independent if no vector is redundant to the span, and linearly dependent otherwise\n\nTo illustrate the idea, recall the figure that showed the span of vectors $$\\{a_1, a_2\\}$$ in $$\\mathbb R ^3$$ as a plane through the origin\n\nIf we take a third vector $$a_3$$ and form the set $$\\{a_1, a_2, a_3\\}$$, this set will be\n\n\u2022 linearly dependent if $$a_3$$ lies in the plane\n\u2022 linearly independent otherwise\n\nAs another illustration of the concept, since $$\\mathbb R ^n$$ can be spanned by $$n$$ vectors (see the discussion of canonical basis vectors above), any collection of $$m > n$$ vectors in $$\\mathbb R ^n$$ must be linearly dependent\n\nThe following statements are equivalent to linear independence of $$A := \\{a_1, \\ldots, a_k\\} \\subset \\mathbb R ^n$$\n\n1. No vector in $$A$$ can be formed as a linear combination of the other elements\n2. If $$\\beta_1 a_1 + \\cdots \\beta_k a_k = 0$$ for scalars $$\\beta_1, \\ldots, \\beta_k$$, then $$\\beta_1 = \\cdots = \\beta_k = 0$$\n\n(The zero in the first expression is the origin of $$\\mathbb R ^n$$)\n\n### Unique Representations\u00b6\n\nAnother nice thing about sets of linearly independent vectors is that each element in the span has a unique representation as a linear combination of these vectors\n\nIn other words, if $$A := \\{a_1, \\ldots, a_k\\} \\subset \\mathbb R ^n$$ is linearly independent and\n\n$y = \\beta_1 a_1 + \\cdots \\beta_k a_k$\n\nthen no other coefficient sequence $$\\gamma_1, \\ldots, \\gamma_k$$ will produce the same vector $$y$$\n\nIndeed, if we also have $$y = \\gamma_1 a_1 + \\cdots \\gamma_k a_k$$, then\n\n$(\\beta_1 - \\gamma_1) a_1 + \\cdots + (\\beta_k - \\gamma_k) a_k = 0$\n\nLinear independence now implies $$\\gamma_i = \\beta_i$$ for all $$i$$\n\n## Matrices\u00b6\n\nMatrices are a neat way of organizing data for use in linear operations\n\nAn $$n \\times k$$ matrix is a rectangular array $$A$$ of numbers with $$n$$ rows and $$k$$ columns:\n\n$\\begin{split}A = \\left[ \\begin{array}{cccc} a_{11} & a_{12} & \\cdots & a_{1k} \\\\ a_{21} & a_{22} & \\cdots & a_{2k} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a_{n1} & a_{n2} & \\cdots & a_{nk} \\end{array} \\right]\\end{split}$\n\nOften, the numbers in the matrix represent coefficients in a system of linear equations, as discussed at the start of this lecture\n\nFor obvious reasons, the matrix $$A$$ is also called a vector if either $$n = 1$$ or $$k = 1$$\n\nIn the former case, $$A$$ is called a row vector, while in the latter it is called a column vector\n\nIf $$n = k$$, then $$A$$ is called square\n\nThe matrix formed by replacing $$a_{ij}$$ by $$a_{ji}$$ for every $$i$$ and $$j$$ is called the transpose of $$A$$, and denoted $$A'$$ or $$A^{\\top}$$\n\nIf $$A = A'$$, then $$A$$ is called symmetric\n\nFor a square matrix $$A$$, the $$i$$ elements of the form $$a_{ii}$$ for $$i=1,\\ldots,n$$ are called the principal diagonal\n\n$$A$$ is called diagonal if the only nonzero entries are on the principal diagonal\n\nIf, in addition to being diagonal, each element along the principal diagonal is equal to 1, then $$A$$ is called the identity matrix, and denoted by $$I$$\n\n### Matrix Operations\u00b6\n\nJust as was the case for vectors, a number of algebraic operations are defined for matrices\n\nScalar multiplication and addition are immediate generalizations of the vector case:\n\n$\\begin{split}\\gamma A = \\gamma \\left[ \\begin{array}{ccc} a_{11} & \\cdots & a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} & \\cdots & a_{nk} \\\\ \\end{array} \\right] := \\left[ \\begin{array}{ccc} \\gamma a_{11} & \\cdots & \\gamma a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ \\gamma a_{n1} & \\cdots & \\gamma a_{nk} \\\\ \\end{array} \\right]\\end{split}$\n\nand\n\n$\\begin{split}A + B = \\left[ \\begin{array}{ccc} a_{11} & \\cdots & a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} & \\cdots & a_{nk} \\\\ \\end{array} \\right] + \\left[ \\begin{array}{ccc} b_{11} & \\cdots & b_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ b_{n1} & \\cdots & b_{nk} \\\\ \\end{array} \\right] := \\left[ \\begin{array}{ccc} a_{11} + b_{11} & \\cdots & a_{1k} + b_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} + b_{n1} & \\cdots & a_{nk} + b_{nk} \\\\ \\end{array} \\right]\\end{split}$\n\nIn the latter case, the matrices must have the same shape in order for the definition to make sense\n\nWe also have a convention for multiplying two matrices\n\nThe rule for matrix multiplication generalizes the idea of inner products discussed above, and is designed to make multiplication play well with basic linear operations\n\nIf $$A$$ and $$B$$ are two matrices, then their product $$A B$$ is formed by taking as its $$i,j$$-th element the inner product of the $$i$$-th row of $$A$$ and the $$j$$-th column of $$B$$\n\nIf $$A$$ is $$n \\times k$$ and $$B$$ is $$j \\times m$$, then to multiply $$A$$ and $$B$$ we require $$k = j$$, and the resulting matrix $$A B$$ is $$n \\times m$$\n\nAs perhaps the most important special case, consider multiplying $$n \\times k$$ matrix $$A$$ and $$k \\times 1$$ column vector $$x$$\n\nAccording to the preceding rule, this gives us an $$n \\times 1$$ column vector\n\n(2)$\\begin{split}A x = \\left[ \\begin{array}{ccc} a_{11} & \\cdots & a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} & \\cdots & a_{nk} \\end{array} \\right] \\left[ \\begin{array}{c} x_{1} \\\\ \\vdots \\\\ x_{k} \\end{array} \\right] := \\left[ \\begin{array}{c} a_{11} x_1 + \\cdots + a_{1k} x_k \\\\ \\vdots \\\\ a_{n1} x_1 + \\cdots + a_{nk} x_k \\end{array} \\right]\\end{split}$\n\nNote\n\n$$A B$$ and $$B A$$ are not generally the same thing\n\nAnother important special case is the identity matrix\n\nYou should check that if $$A$$ is $$n \\times k$$ and $$I$$ is the $$k \\times k$$ identity matrix, then $$AI = A$$\n\nIf $$I$$ is the $$n \\times n$$ identity matrix, then $$IA = A$$\n\n### Matrices in Julia\u00b6\n\nJulia arrays are also used as matrices, and have fast, efficient functions and methods for all the standard matrix operations\n\nYou can create them as follows\n\nA = [1 2\n3 4]\n2\u00d72 Array{Int64,2}:\n1  2\n3  4\ntypeof(A)\nArray{Int64,2}\nsize(A)\n(2,2)\n\nThe size function returns a tuple giving the number of rows and columns\n\nTo get the transpose of A, use transpose(A) or, more simply, A'\n\nThere are many convenient functions for creating common matrices (matrices of zeros, ones, etc.) \u2014 see here\n\nSince operations are performed elementwise by default, scalar multiplication and addition have very natural syntax\n\nA = eye(3)\n3\u00d73 Array{Float64,2}:\n1.0  0.0  0.0\n0.0  1.0  0.0\n0.0  0.0  1.0\nB = ones(3, 3)\n3\u00d73 Array{Float64,2}:\n1.0  1.0  1.0\n1.0  1.0  1.0\n1.0  1.0  1.0\n2A\n3\u00d73 Array{Float64,2}:\n2.0  0.0  0.0\n0.0  2.0  0.0\n0.0  0.0  2.0\nA + B\n3\u00d73 Array{Float64,2}:\n2.0  1.0  1.0\n1.0  2.0  1.0\n1.0  1.0  2.0\n\nTo multiply matrices we use the * operator\n\nIn particular, A * B is matrix multiplication, whereas A .* B is element by element multiplication\n\n### Matrices as Maps\u00b6\n\nEach $$n \\times k$$ matrix $$A$$ can be identified with a function $$f(x) = Ax$$ that maps $$x \\in \\mathbb R ^k$$ into $$y = Ax \\in \\mathbb R ^n$$\n\nThese kinds of functions have a special property: they are linear\n\nA function $$f \\colon \\mathbb R ^k \\to \\mathbb R ^n$$ is called linear if, for all $$x, y \\in \\mathbb R ^k$$ and all scalars $$\\alpha, \\beta$$, we have\n\n$f(\\alpha x + \\beta y) = \\alpha f(x) + \\beta f(y)$\n\nYou can check that this holds for the function $$f(x) = A x + b$$ when $$b$$ is the zero vector, and fails when $$b$$ is nonzero\n\nIn fact, it\u2019s known that $$f$$ is linear if and only if there exists a matrix $$A$$ such that $$f(x) = Ax$$ for all $$x$$\n\n## Solving Systems of Equations\u00b6\n\nRecall again the system of equations (1)\n\nIf we compare (1) and (2), we see that (1) can now be written more conveniently as\n\n(3)$y = Ax$\n\nThe problem we face is to determine a vector $$x \\in \\mathbb R ^k$$ that solves (3), taking $$y$$ and $$A$$ as given\n\nThis is a special case of a more general problem: Find an $$x$$ such that $$y = f(x)$$\n\nGiven an arbitrary function $$f$$ and a $$y$$, is there always an $$x$$ such that $$y = f(x)$$?\n\nIf so, is it always unique?\n\nThe answer to both these questions is negative, as the next figure shows\n\n#=\n\n@author : Spencer Lyon <spencer.lyon@nyu.edu>\nVictoria Gregory <victoria.gregory@nyu.edu>\n\n=#\n\nf(x) = 0.6 * cos(4.0 * x) + 1.3\n\nxmin, xmax = -1.0, 1.0\nNx = 160\nx = linspace(xmin, xmax, Nx)\ny = f.(x)\nya, yb = minimum(y), maximum(y)\n\np1 = plot(x, y, color=:black, label=[L\"$f$\" \"\"], grid=false)\nplot!(x, ya*ones(Nx, 1), fill_between=yb*ones(Nx, 1),\nfillalpha=0.1, color=:blue, label=\"\", lw=0)\nplot!(zeros(2, 2), [ya ya; yb yb], lw=3, color=:blue, label=[L\"range of $f$\" \"\"])\nannotate!(0.04, -0.3, L\"$0$\", ylims=(-0.6, 3.2))\nvline!([0], color=:black, label=\"\")\nhline!([0], color=:black, label=\"\")\nplot!(foreground_color_axis=:white, foreground_color_text=:white,\nforeground_color_border=:white)\n\nybar = 1.5\nplot!(x, x .* 0 .+ ybar, color=:black, linestyle=:dash, label=\"\")\nannotate!(0.05, 0.8 * ybar, L\"$y$\")\n\nx_vals = Array{Float64}(2, 4)\ny_vals = Array{Float64}(2, 4)\nlabels = []\nfor (i, z) in enumerate([-0.35, 0.35])\nx_vals[:, 2*i-1] = z*ones(2, 1)\ny_vals[2, 2*i-1] = f(z)\nlabels = [labels; (z, -0.2, LaTeXString(\"\\$x_$i\\$\"))] end plot!(x_vals, y_vals, color=:black, linestyle=:dash, label=\"\", annotation=labels) p2 = plot(x, y, color=:black, label=[L\"$f$\" \"\"], grid=false) plot!(x, ya*ones(Nx, 1), fill_between=yb*ones(Nx, 1), fillalpha=0.1, color=:blue, label=\"\", lw=0) plot!(zeros(2, 2), [ya ya; yb yb], lw=3, color=:blue, label=[L\"range of$f$\" \"\"]) annotate!(0.04, -0.3, L\"$0$\", ylims=(-0.6, 3.2)) vline!([0], color=:black, label=\"\") hline!([0], color=:black, label=\"\") plot!(foreground_color_axis=:white, foreground_color_text=:white, foreground_color_border=:white) ybar = 2.6 plot!(x, x .* 0 .+ ybar, color=:black, linestyle=:dash, legend=:none) annotate!(0.04, 0.91 * ybar, L\"$y$\") plot(p1, p2, layout=(2, 1), size=(600, 700)) In the first plot there are multiple solutions, as the function is not one-to-one, while in the second there are no solutions, since $$y$$ lies outside the range of $$f$$ Can we impose conditions on $$A$$ in (3) that rule out these problems? In this context, the most important thing to recognize about the expression $$Ax$$ is that it corresponds to a linear combination of the columns of $$A$$ In particular, if $$a_1, \\ldots, a_k$$ are the columns of $$A$$, then $Ax = x_1 a_1 + \\cdots + x_k a_k$ Hence the range of $$f(x) = Ax$$ is exactly the span of the columns of $$A$$ We want the range to be large, so that it contains arbitrary $$y$$ As you might recall, the condition that we want for the span to be large is linear independence A happy fact is that linear independence of the columns of $$A$$ also gives us uniqueness Indeed, it follows from our earlier discussion that if $$\\{a_1, \\ldots, a_k\\}$$ are linearly independent and $$y = Ax = x_1 a_1 + \\cdots + x_k a_k$$, then no $$z \\not= x$$ satisfies $$y = Az$$ ### The $$n \\times n$$ Case\u00b6 Let\u2019s discuss some more details, starting with the case where $$A$$ is $$n \\times n$$ This is the familiar case where the number of unknowns equals the number of equations For arbitrary $$y \\in \\mathbb R ^n$$, we hope to find a unique $$x \\in \\mathbb R ^n$$ such that $$y = Ax$$ In view of the observations immediately above, if the columns of $$A$$ are linearly independent, then their span, and hence the range of $$f(x) = Ax$$, is all of $$\\mathbb R ^n$$ Hence there always exists an $$x$$ such that $$y = Ax$$ Moreover, the solution is unique In particular, the following are equivalent 1. The columns of $$A$$ are linearly independent 2. For any $$y \\in \\mathbb R ^n$$, the equation $$y = Ax$$ has a unique solution The property of having linearly independent columns is sometimes expressed as having full column rank #### Inverse Matrices\u00b6 Can we give some sort of expression for the solution? If $$y$$ and $$A$$ are scalar with $$A \\not= 0$$, then the solution is $$x = A^{-1} y$$ A similar expression is available in the matrix case In particular, if square matrix $$A$$ has full column rank, then it possesses a multiplicative inverse matrix $$A^{-1}$$, with the property that $$A A^{-1} = A^{-1} A = I$$ As a consequence, if we pre-multiply both sides of $$y = Ax$$ by $$A^{-1}$$, we get $$x = A^{-1} y$$ This is the solution that we\u2019re looking for #### Determinants\u00b6 Another quick comment about square matrices is that to every such matrix we assign a unique number called the determinant of the matrix \u2014 you can find the expression for it here If the determinant of $$A$$ is not zero, then we say that $$A$$ is nonsingular Perhaps the most important fact about determinants is that $$A$$ is nonsingular if and only if $$A$$ is of full column rank This gives us a useful one-number summary of whether or not a square matrix can be inverted ### More Rows than Columns\u00b6 This is the $$n \\times k$$ case with $$n > k$$ This case is very important in many settings, not least in the setting of linear regression (where $$n$$ is the number of observations, and $$k$$ is the number of explanatory variables) Given arbitrary $$y \\in \\mathbb R ^n$$, we seek an $$x \\in \\mathbb R ^k$$ such that $$y = Ax$$ In this setting, existence of a solution is highly unlikely Without much loss of generality, let\u2019s go over the intuition focusing on the case where the columns of $$A$$ are linearly independent It follows that the span of the columns of $$A$$ is a $$k$$-dimensional subspace of $$\\mathbb R ^n$$ This span is very \u201cunlikely\u201d to contain arbitrary $$y \\in \\mathbb R ^n$$ To see why, recall the figure above, where $$k=2$$ and $$n=3$$ Imagine an arbitrarily chosen $$y \\in \\mathbb R ^3$$, located somewhere in that three dimensional space What\u2019s the likelihood that $$y$$ lies in the span of $$\\{a_1, a_2\\}$$ (i.e., the two dimensional plane through these points)? In a sense it must be very small, since this plane has zero \u201cthickness\u201d As a result, in the $$n > k$$ case we usually give up on existence However, we can still seek a best approximation, for example an $$x$$ that makes the distance $$\\| y - Ax\\|$$ as small as possible To solve this problem, one can use either calculus or the theory of orthogonal projections The solution is known to be $$\\hat x = (A'A)^{-1}A'y$$ \u2014 see for example chapter 3 of these notes ### More Columns than Rows\u00b6 This is the $$n \\times k$$ case with $$n < k$$, so there are fewer equations than unknowns In this case there are either no solutions or infinitely many \u2014 in other words, uniqueness never holds For example, consider the case where $$k=3$$ and $$n=2$$ Thus, the columns of $$A$$ consists of 3 vectors in $$\\mathbb R ^2$$ This set can never be linearly independent, since it is possible to find two vectors that span $$\\mathbb R ^2$$ (For example, use the canonical basis vectors) It follows that one column is a linear combination of the other two For example, let\u2019s say that $$a_1 = \\alpha a_2 + \\beta a_3$$ Then if $$y = Ax = x_1 a_1 + x_2 a_2 + x_3 a_3$$, we can also write $y = x_1 (\\alpha a_2 + \\beta a_3) + x_2 a_2 + x_3 a_3 = (x_1 \\alpha + x_2) a_2 + (x_1 \\beta + x_3) a_3$ In other words, uniqueness fails ### Linear Equations with Julia\u00b6 Here\u2019s an illustration of how to solve linear equations with Julia\u2019s built-in linear algebra facilities A = [1.0 2.0; 3.0 4.0]; y = ones(2, 1); # A column vector 2\u00d71 Array{Float64,2}: 1.0 1.0 det(A) -2.0 A_inv = inv(A) 2\u00d72 Array{Float64,2}: -2.0 1.0 1.5 -0.5 x = A_inv * y # solution 2\u00d71 Array{Float64,2}: -1.0 1.0 A * x # should equal y (a vector of ones) 2\u00d71 Array{Float64,2}: 1.0 1.0 A\\y # produces the same solution 2\u00d71 Array{Float64,2}: -1.0 1.0 Observe how we can solve for $$x = A^{-1} y$$ by either via inv(A) * y, or using A \\ y The latter method is preferred because it automatically selects the best algorithm for the problem based on the values of A and y If A is not square then A \\ y returns the least squares solution $$\\hat x = (A'A)^{-1}A'y$$ ## Eigenvalues and Eigenvectors\u00b6 Let $$A$$ be an $$n \\times n$$ square matrix If $$\\lambda$$ is scalar and $$v$$ is a non-zero vector in $$\\mathbb R ^n$$ such that $A v = \\lambda v$ then we say that $$\\lambda$$ is an eigenvalue of $$A$$, and $$v$$ is an eigenvector Thus, an eigenvector of $$A$$ is a vector such that when the map $$f(x) = Ax$$ is applied, $$v$$ is merely scaled The next figure shows two eigenvectors (blue arrows) and their images under $$A$$ (red arrows) As expected, the image $$Av$$ of each $$v$$ is just a scaled version of the original A = [1 2 2 1] evals, evecs = eig(A) a1, a2 = evals[1], evals[2] evecs = evecs[:, 1], evecs[:, 2] eig_1 = zeros(2, length(evecs)) eig_2 = zeros(2, length(evecs)) labels = [] for i = 1:length(evecs) v = evecs[i] eig_1[2, i] = v[1] eig_2[2, i] = v[2] end x = linspace(-5, 5, 10) y = -linspace(-5, 5, 10) plot(eig_1[:, 2], a1 * eig_2[:, 2], arrow=true, color=:red, legend=:none, xlims=(-3, 3), ylims=(-3, 3), annotations=labels, xticks=-5:1:5, yticks=-5:1:5, framestyle=:origin) plot!(a2 * eig_1[:, 2], a2 * eig_2, arrow=true, color=:red) plot!(eig_1, eig_2, arrow=true, color=:blue) plot!(x, y, color=:blue, lw=0.4, alpha=0.6) plot!(x, x, color=:blue, lw=0.4, alpha=0.6) The eigenvalue equation is equivalent to $$(A - \\lambda I) v = 0$$, and this has a nonzero solution $$v$$ only when the columns of $$A - \\lambda I$$ are linearly dependent This in turn is equivalent to stating that the determinant is zero Hence to find all eigenvalues, we can look for $$\\lambda$$ such that the determinant of $$A - \\lambda I$$ is zero This problem can be expressed as one of solving for the roots of a polynomial in $$\\lambda$$ of degree $$n$$ This in turn implies the existence of $$n$$ solutions in the complex plane, although some might be repeated Some nice facts about the eigenvalues of a square matrix $$A$$ are as follows 1. The determinant of $$A$$ equals the product of the eigenvalues 2. The trace of $$A$$ (the sum of the elements on the principal diagonal) equals the sum of the eigenvalues 3. If $$A$$ is symmetric, then all of its eigenvalues are real 4. If $$A$$ is invertible and $$\\lambda_1, \\ldots, \\lambda_n$$ are its eigenvalues, then the eigenvalues of $$A^{-1}$$ are $$1/\\lambda_1, \\ldots, 1/\\lambda_n$$ A corollary of the first statement is that a matrix is invertible if and only if all its eigenvalues are nonzero Using Julia, we can solve for the eigenvalues and eigenvectors of a matrix as follows A = [1.0 2.0; 2.0 1.0]; evals, evecs = eig(A); evals 2-element Array{Float64,1}: -1.0 3.0 evecs 2\u00d72 Array{Float64,2}: -0.707107 0.707107 0.707107 0.707107 Note that the columns of evecs are the eigenvectors Since any scalar multiple of an eigenvector is an eigenvector with the same eigenvalue (check it), the eig routine normalizes the length of each eigenvector to one ### Generalized Eigenvalues\u00b6 It is sometimes useful to consider the generalized eigenvalue problem, which, for given matrices $$A$$ and $$B$$, seeks generalized eigenvalues $$\\lambda$$ and eigenvectors $$v$$ such that $A v = \\lambda B v$ This can be solved in Julia via eig(A, B) Of course, if $$B$$ is square and invertible, then we can treat the generalized eigenvalue problem as an ordinary eigenvalue problem $$B^{-1} A v = \\lambda v$$, but this is not always the case ## Further Topics\u00b6 We round out our discussion by briefly mentioning several other important topics ### Series Expansions\u00b6 Recall the usual summation formula for a geometric progression, which states that if $$|a| < 1$$, then $$\\sum_{k=0}^{\\infty} a^k = (1 - a)^{-1}$$ A generalization of this idea exists in the matrix setting #### Matrix Norms\u00b6 Let $$A$$ be a square matrix, and let $\\| A \\| := \\max_{\\| x \\| = 1} \\| A x \\|$ The norms on the right-hand side are ordinary vector norms, while the norm on the left-hand side is a matrix norm \u2014 in this case, the so-called spectral norm For example, for a square matrix $$S$$, the condition $$\\| S \\| < 1$$ means that $$S$$ is contractive, in the sense that it pulls all vectors towards the origin [1] #### Neumann\u2019s Theorem\u00b6 Let $$A$$ be a square matrix and let $$A^k := A A^{k-1}$$ with $$A^1 := A$$ In other words, $$A^k$$ is the $$k$$-th power of $$A$$ Neumann\u2019s theorem states the following: If $$\\| A^k \\| < 1$$ for some $$k \\in \\mathbb{N}$$, then $$I - A$$ is invertible, and (4)$(I - A)^{-1} = \\sum_{k=0}^{\\infty} A^k$ #### Spectral Radius\u00b6 A result known as Gelfand\u2019s formula tells us that, for any square matrix $$A$$, $\\rho(A) = \\lim_{k \\to \\infty} \\| A^k \\|^{1/k}$ Here $$\\rho(A)$$ is the spectral radius, defined as $$\\max_i |\\lambda_i|$$, where $$\\{\\lambda_i\\}_i$$ is the set of eigenvalues of $$A$$ As a consequence of Gelfand\u2019s formula, if all eigenvalues are strictly less than one in modulus, there exists a $$k$$ with $$\\| A^k \\| < 1$$ In which case (4) is valid ### Positive Definite Matrices\u00b6 Let $$A$$ be a symmetric $$n \\times n$$ matrix We say that $$A$$ is 1. positive definite if $$x' A x > 0$$ for every $$x \\in \\mathbb R ^n \\setminus \\{0\\}$$ 2. positive semi-definite or nonnegative definite if $$x' A x \\geq 0$$ for every $$x \\in \\mathbb R ^n$$ Analogous definitions exist for negative definite and negative semi-definite matrices It is notable that if $$A$$ is positive definite, then all of its eigenvalues are strictly positive, and hence $$A$$ is invertible (with positive definite inverse) ### Differentiating Linear and Quadratic forms\u00b6 The following formulas are useful in many economic contexts. Let \u2022 $$z, x$$ and $$a$$ all be $$n \\times 1$$ vectors \u2022 $$A$$ be an $$n \\times n$$ matrix \u2022 $$B$$ be an $$m \\times n$$ matrix and $$y$$ be an $$m \\times 1$$ vector Then 1. $$\\frac{\\partial a' x}{\\partial x} = a$$ 2. $$\\frac{\\partial A x}{\\partial x} = A'$$ 3. $$\\frac{\\partial x'A x}{\\partial x} = (A + A') x$$ 4. $$\\frac{\\partial y'B z}{\\partial y} = B z$$ 5. $$\\frac{\\partial y'B z}{\\partial B} = y z'$$ Exercise 1 below asks you to apply these formulas ### Further Reading\u00b6 The documentation of the linear algebra features built into Julia can be found here Chapters 2 and 3 of the Econometric Theory contains a discussion of linear algebra along the same lines as above, with solved exercises If you don\u2019t mind a slightly abstract approach, a nice intermediate-level text on linear algebra is [Janich94] ## Exercises\u00b6 ### Exercise 1\u00b6 Let $$x$$ be a given $$n \\times 1$$ vector and consider the problem $v(x) = \\max_{y,u} \\left\\{ - y'P y - u' Q u \\right\\}$ subject to the linear constraint $y = A x + B u$ Here \u2022 $$P$$ is an $$n \\times n$$ matrix and $$Q$$ is an $$m \\times m$$ matrix \u2022 $$A$$ is an $$n \\times n$$ matrix and $$B$$ is an $$n \\times m$$ matrix \u2022 both $$P$$ and $$Q$$ are symmetric and positive semidefinite (What must the dimensions of $$y$$ and $$u$$ be to make this a well-posed problem?) One way to solve the problem is to form the Lagrangian $\\mathcal L = - y' P y - u' Q u + \\lambda' \\left[A x + B u - y\\right]$ where $$\\lambda$$ is an $$n \\times 1$$ vector of Lagrange multipliers Try applying the formulas given above for differentiating quadratic and linear forms to obtain the first-order conditions for maximizing $$\\mathcal L$$ with respect to $$y, u$$ and minimizing it with respect to $$\\lambda$$ Show that these conditions imply that 1. $$\\lambda = - 2 P y$$ 2. The optimizing choice of $$u$$ satisfies $$u = - (Q + B' P B)^{-1} B' P A x$$ 3. The function $$v$$ satisfies $$v(x) = - x' \\tilde P x$$ where $$\\tilde P = A' P A - A'P B (Q + B'P B)^{-1} B' P A$$ As we will see, in economic contexts Lagrange multipliers often are shadow prices Note If we don\u2019t care about the Lagrange multipliers, we can substitute the constraint into the objective function, and then just maximize $$-(Ax + Bu)'P (Ax + Bu) - u' Q u$$ with respect to $$u$$. You can verify that this leads to the same maximizer. ## Solutions\u00b6 Thanks to Willem Hekman and Guanlong Ren for providing this solution. ### Exercise 1\u00b6 We have an optimization problem: $v(x) = \\max_{y,u} \\{ -y'Py - u'Qu \\}$ s.t. $y = Ax + Bu$ with primitives \u2022 $$P$$ be a symmetric and positive semidefinite $$n \\times n$$ matrix. \u2022 $$Q$$ be a symmetric and positive semidefinite $$m \\times m$$ matrix. \u2022 $$A$$ an $$n \\times n$$ matrix. \u2022 $$B$$ an $$n \\times m$$ matrix. The associated Lagrangian is : $L = -y'Py - u'Qu + \\lambda' \\lbrack Ax + Bu - y \\rbrack$ #### 1.\u00b6 Differentiating Lagrangian equation w.r.t y and setting its derivative equal to zero yields $\\frac{ \\partial L}{\\partial y} = - (P + P') y - \\lambda = - 2 P y - \\lambda = 0 \\:,$ since P is symmetric. Accordingly, the first-order condition for maximizing L w.r.t. y implies $\\lambda = -2 Py \\:.$ #### 2.\u00b6 Differentiating Lagrangian equation w.r.t. u and setting its derivative equal to zero yields $\\frac{ \\partial L}{\\partial u} = - (Q + Q') u - B'\\lambda = - 2Qu + B'\\lambda = 0 \\:.$ Substituting $$\\lambda = -2 P y$$ gives $Qu + B'Py = 0 \\:.$ Substituting the linear constraint $$y = Ax + Bu$$ into above equation gives $Qu + B'P(Ax + Bu) = 0$ $(Q + B'PB)u + B'PAx = 0$ which is the first-order condition for maximizing L w.r.t. u. Thus, the optimal choice of u must satisfy $u = -(Q + B'PB)^{-1}B'PAx \\:,$ which follows from the definition of the first-oder conditions for Lagrangian equation. #### 3.\u00b6 Rewriting our problem by substituting the constraint into the objective function, we get $v(x) = \\max_{u} \\{ -(Ax+ Bu)'P(Ax+Bu) - u'Qu \\} \\:.$ Since we know the optimal choice of u satisfies$ u = -(Q + B\u2019PB)^{-1}B\u2019PAx , then $v(x) = -(Ax+ B u)'P(Ax+B u) - u'Q u \\,\\,\\,\\, with \\,\\,\\,\\, u = -(Q + B'PB)^{-1}B'PAx$ To evaluate the function \\begin{align} v(x) &= -(Ax+ B u)'P(Ax+Bu) - u'Q u \\\\ &= -(x'A' + u'B')P(Ax+Bu) - u'Q u \\\\ &= - x'A'PAx - u'B'PAx - x'A'PBu - u'B'PBu - u'Qu \\\\ &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u \\end{align} For simplicity, denote by $$S := (Q + B'PB)^{-1} B'PA$$, then u = -Sx\\$.\n\nRegarding the second term $$- 2u'B'PAx$$,\n\n\\begin{align} - 2u'B'PAx &= -2 x'S'B'PAx \\\\ & = 2 x'A'PB( Q + B'PB)^{-1} B'PAx \\end{align}\n\nNotice that the term $$(Q + B'PB)^{-1}$$ is symmetric as both P and Q are symmetric.\n\nRegarding the third term $$- u'(Q + B'PB) u$$,\n\n\\begin{align} - u'(Q + B'PB) u &= - x'S' (Q + B'PB)Sx \\\\ &= -x'A'PB(Q + B'PB)^{-1}B'PAx \\end{align}\n\nHence, the summation of second and third terms is $$x'A'PB(Q + B'PB)^{-1}B'PAx$$.\n\nThis implies that\n\n\\begin{align} v(x) &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u\\\\ &= - x'A'PAx + x'A'PB(Q + B'PB)^{-1}B'PAx \\\\ &= -x'[A'PA - A'PB(Q + B'PB)^{-1}B'PA] x \\end{align}\n\nTherefore, the solution to the optimization problem $$v(x) = -x' \\tilde{P}x$$ follows the above result by denoting $$\\tilde{P} := A'PA - A'PB(Q + B'PB)^{-1}B'PA$$.\n\nFootnotes\n\n [1] Suppose that $$\\|S \\| < 1$$. Take any nonzero vector $$x$$, and let $$r := \\|x\\|$$. We have $$\\| Sx \\| = r \\| S (x/r) \\| \\leq r \\| S \\| < r = \\| x\\|$$. Hence every point is pulled towards the origin.\n\u2022 Share page", "date": "2018-11-15 08:08:19", "meta": {"domain": "quantecon.org", "url": "https://lectures.quantecon.org/jl/linear_algebra.html", "openwebmath_score": 0.9785626530647278, "openwebmath_perplexity": 430.0254335697078, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9937100989807552, "lm_q2_score": 0.8933093968230773, "lm_q1q2_score": 0.8876905691374989}}
{"url": "https://www.jiskha.com/questions/1733918/a-bag-contains-only-red-and-blue-marbles-yasmine-takes-one-marble-at-random-from-the-bag", "text": "Math\n\nA bag contains only red and blue marbles. Yasmine takes one marble at random from the bag. The probability that she takes a red marble is 1 in 5. Yasmine returns the marble to the bag and adds \ufb01ve more red marbles to the bag. The probability that she takes one red marble at random is now 1 in 3. How many red marbles were originally in the bag?\n\nA. 3 red\nB. 5 red\nC. 10 red\nD. 2 red\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. Oops i forgot to put i though B was the right answer ( B. 5 red\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n2. thought ^^\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. r/(r+b) = 1/5\n(r+5)/(r+5+b) = 1/3\n\nr=5\n\nyou are correct\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\nSimilar Questions\n\n1. Probability\n\njeff has 8 red marbles, 6 blue marbles, and 4 green marbles that are the same size and shape. he puts the marbles into a bag, mixes the marbles, and randomly picks one marble. what is the probability that the marble will be blue?\n\n2. math\n\nA bag contains 8 red marbles, 5 blue marbles, 8 yellow marbles, and 6 green marbles. What is the probability of choosing a red marble if a single choice is made from the bag? is it 8/27 ?\n\n3. Math liberal Arts\n\nA bag contains 5 red marbles, 4 blue marbles, and 1 green marble. If a marble is selected at random, what is the probability that it is not blue?\n\n4. Math\n\nA bag contains 3 red marbles, 5 yellow marbles, and 4 blue marbles. One marble is chosen at random. What is the probability that the chosen marble will be blue? A) 1/12 B) 1/4 C) 1/3 D) 3/4 I think it is B. 1/4\n\n1. math\n\nA bag contains five red marbles and five blue marbles. You randomly pick a marble and then return it to the bag before picking another marble. The first marble is red and the second marble is blue. a. 1/4 = 0.25 ***** b. 21/55 =\n\n2. Math :)\n\nIn a bag of 10 marbles, there are 5 blue marbles, 3 red marbles, and 2 white marbles. Complete the probability distribution table for drawing 1 marble out of the bag. Draw a: Probability Blue marble 5/10 Red marble 3/10 White\n\n3. Math\n\nA bag with 12 marbles has 3 yellow marbles, 4 blue marbles, and 5 red marbles. A marble is chosen from the bag at random. What is the probability that it is yellow?\n\nA bag contains 7 red marbles, 2 blue marbles, and 1 green marble. If a marble is selected at random, what is the probability of choosing a marble that is not blue? 7 red marbles plus 1 green marble = 8/10 = answer = 4/5\n\n1. Math\n\nA bag of marbles contains 5 red, 3 blue, 2 green, and 2 yellow marbles. What is the probability that you choose a blue marble and then another blue marble, assuming you replace the first marble?\n\n2. Math\n\nTom keeps all of his favorite marbles in a special leather bag. Right now, five red marbles, four blue marbles, and yellow marbles are in the bag. If he randomly chooses one marble to give to a friend what is the probability that\n\n3. algebra\n\nA bag contains 9 marbles: 2 are green, 4 are red, and 3 are blue. Laura chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that both marbles she chooses are blue? Write\n\n4. Math\n\nOne bag contains 5 red marbles, 4 blue marbles, and 3 yellow marbles, and a second bag contains 4 red marbles, 6 blue marbles, and 5 yellow marbles. If Lydia randomly draws one marble from each bag, what is the probability that", "date": "2021-09-25 05:38:18", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1733918/a-bag-contains-only-red-and-blue-marbles-yasmine-takes-one-marble-at-random-from-the-bag", "openwebmath_score": 0.8535917401313782, "openwebmath_perplexity": 430.02246614616513, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561712637256, "lm_q2_score": 0.9161096118716263, "lm_q1q2_score": 0.8875784510158415}}
{"url": "https://s140450.gridserver.com/hrm9dhkd/e5439e-decomposition-of-antisymmetric-tensor", "text": "(antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. This means that traceless antisymmetric mixed tensor $\\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. Since det M= det (\u00e2\u0088\u0092MT) = det (\u00e2\u0088\u0092M) = (\u00e2\u0088\u00921)d det M, (1) it follows that det M= 0 if dis odd. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g What's the significance of this further decomposition? A tensor is a linear vector valued function defined on the set of all vectors . 1.5) are not explicitly stated because they are obvious from the context. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. The trace decomposition theory of tensor spaces, based on duality, is presented. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u00e2\u0080\u00a6 Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric \u00e2\u0080\u00a6 The symmetry-based decompositions of finite games are investigated. Use the Weyl decomposition \\eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \\eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). Sci. This is exactly what you have done in the second line of your equation. This is an example of the Youla decomposition of a complex square matrix. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . Decomposition of tensor power of symmetric square. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. It is a real tensor, hence f \u00ce\u00b1\u00ce\u00b2 * is also real. P i A ii D0/. Yes. In these notes, the rank of Mwill be denoted by 2n. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. Sponsoring Org. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 A\u00c4\u00b1 ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components \u00e2\u0080\u00a6. Cartan tensor is equal to minus the structure coe\u00ef\u00ac\u0083cients. Thus, the rank of Mmust be even. In section 3 a decomposition of tensor spaces into irreducible components is introduced. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. We begin with a special case of the definition. The result is Polon. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. The bases of the symmetric subspace and those of its orthogonal complement are presented. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. : USDOE \u00e2\u0080\u00a6 If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. MT = \u00e2\u0088\u0092M. An alternating form \u00cf\u0086 on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Properties of antisymmetric matrices Let Mbe a complex d\u00d7 dantisymmetric matrix, i.e. This decomposition, ... ^2 indicates the antisymmetric tensor product. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. Antisymmetric and symmetric tensors. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. \u00e2\u0086\u0092 What symmetry does represent?Kenta OONOIntroduction to Tensors For more comprehensive overviews on tensor calculus we recom-mend [58, 99, 126, 197, 205, 319, 343]. This makes many vector identities easy to prove. For N>2, they are not, however. Contents. The alternating tensor can be used to write down the vector equation z = x \u00d7 y in su\u00ef\u00ac\u0083x notation: z i = [x\u00d7y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 \u00e2\u0088\u0092x 3y 2, as required.) Symmetric tensors occur widely in engineering, physics and mathematics. 1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. (1.5) Usually the conditions for \u00b5 (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. These relations may be shown either directly, using the explicit form of f \u00ce\u00b1\u00ce\u00b2, and f \u00ce\u00b1\u00ce\u00b2 * or as consequences of the Hamilton\u00e2\u0080\u0090Cayley equation for antisymmetric matrices f \u00ce\u00b1\u00ce\u00b2 and f \u00ce\u00b1\u00ce\u00b2 *; see, e.g., J. Pleba\u00c5\u0084ski, Bull Acad. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. Cl. THE INDEX NOTATION \u00ce\u00bd, are chosen arbitrarily.The could equally well have been called \u00ce\u00b1 and \u00ce\u00b2: v\u00e2\u0080\u00b2 \u00ce\u00b1 = n \u00e2\u0088\u0091 \u00ce\u00b2=1 A\u00ce\u00b1\u00ce\u00b2 v\u00ce\u00b2 (\u00e2\u0088\u0080\u00ce\u00b1 \u00e2\u0088\u0088 N | 1 \u00e2\u0089\u00a4 \u00ce\u00b1 \u00e2\u0089\u00a4 n). Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Each part can reveal information that might not be easily obtained from the original tensor. There is one very important property of ijk: ijk klm = \u00ce\u00b4 il\u00ce\u00b4 jm \u00e2\u0088\u0092\u00ce\u00b4 im\u00ce\u00b4 jl. The N-way Toolbox, Tensor Toolbox, \u00e2\u0080\u00a6 If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? Vector spaces will be denoted using blackboard fonts. Antisymmetric and symmetric tensors. Ask Question Asked 2 years, 2 months ago. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. 1.4) or \u00ce\u00b1 (in Eq. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. Active 1 year, 11 months ago. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. Physics 218 Antisymmetric matrices and the pfa\u00ef\u00ac\u0083an Winter 2015 1. A related concept is that of the antisymmetric tensor or alternating form. LetT be a second-order tensor. Decomposition. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. CHAPTER 1. The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. Line of your equation the second line of your equation Mbe a complex square.. 319, 343 ] and an orthogonal complement are presented singlett, while the symmetric subspace and of... Meaning that every tensor product representation is reducible and orthonormal tensor basis methods developed. Properties of antisymmetric matrices Let Mbe a complex square matrix set of vectors. Of antisymmetric matrices Let Mbe a complex square matrix preserves the symmetries of the tensor corresponds to (! Its orthogonal complement are presented Record ; Other Related decomposition of antisymmetric tensor ; Authors: Bazanski, S L Date... Symmetric ) spin-1 part notes on vector and tensor Algebra and Analysis IlyaL: antisymmetric matrix to! ) spin-1 part 1.5 ) are not, however defined on the set of all vectors World Heritage:! Solve puzzles used in the literature so, are the symmetric subspace and those of its orthogonal are. Valued function defined on the set of all vectors, however: ijk klm = \u00ce\u00b4 il\u00ce\u00b4 jm im\u00ce\u00b4! Volume ) expansion of the antisymmetric tensor or alternating form and anti-symmetric parts the decomposition tensor! Aug 01 00:00:00 EDT 1965 Research Org that is not so into SKEW-SYMMETRIC tensors symmetric tensors occur widely engineering! In symmetric and anti-symmetric parts the decomposition of tensors in distinctive parts can help in analyzing them the results existing! Be denoted by 2n the decomposition of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors obtained from the original.. In 3 dimensions, that is not so part of the tensor is! 00:00:00 EDT 1965 Research Org, 343 ] prove by a direct calculation that its Riemann vanishes... These notes, the rank of Mwill be denoted by 2n volume ) expansion of definition. In the case of SU ( 2 ) the representations corresponding to upper lower... Mechanics, 1974 ( 11 ) Problem so we only get constraints one! Of tensor spaces into irreducible components is introduced ) the representations corresponding upper. Date: antisymmetric matrix antisymmetric tensors N is often used in the second line your! Months ago is dual to a vector, but in 4 dimensions, an antisymmetric tensor Collection... Great fun - you get to solve puzzles the antisymmetric tensor is dual to a vector but... Can help in analyzing them we begin with a special case of SU ( 2 ) the corresponding! - you get to solve puzzles in engineering, physics and mathematics Analysis IlyaL fluid. The context the bases of the fluid what you have done in the case SU... Publication Date: antisymmetric matrix subspace and an orthogonal complement are presented decomposition of antisymmetric tensor original.... Square matrix antisymmetric matrices Let Mbe a complex d\u00d7 dantisymmetric matrix, i.e furthermore, the! Analysis IlyaL Algebra and Analysis IlyaL ) Usually the conditions for \u00b5 ( decomposition of antisymmetric tensor Eq im\u00ce\u00b4! Structure coe\u00ef\u00ac\u0083cients Sun Aug 01 00:00:00 EDT 1965 Research Org Youla decomposition of tensors distinctive! ) the representations corresponding to upper and lower indices are equivalent tensor M a. Prove by a direct calculation that its Riemann tensor vanishes so we only get constraints from contraction! ( antisymmetric ) spin-0 singlett, while the symmetric subspace and an orthogonal complement are presented =! Using the results of existing theories in the case of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors or alternating.. For N > 2, they are not explicitly stated because they not! Its orthogonal complement are presented and a partially antisymmetric tensors N is often used in case! Special case of SU ( 2 ) the representations corresponding to upper and lower indices are.! Youla decomposition of a complex d\u00d7 dantisymmetric matrix, i.e the SA-decomposition is,. 00:00:00 EDT 1965 Research Org 2 months ago subspaces separate invariant subspaces meaning. Meaning that every tensor product representation is reducible results of existing theories in the second line of your equation get...", "date": "2021-05-18 16:43:10", "meta": {"domain": "gridserver.com", "url": "https://s140450.gridserver.com/hrm9dhkd/e5439e-decomposition-of-antisymmetric-tensor", "openwebmath_score": 0.8882725834846497, "openwebmath_perplexity": 1127.1270849276234, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9902915238050521, "lm_q2_score": 0.896251371748038, "lm_q1q2_score": 0.8875501366407329}}
{"url": "https://www.themathdoctors.org/what-is-a-trapezoid-more-on-inclusive-definitions/", "text": "# What Is a Trapezoid? More on Inclusive Definitions\n\nA month ago, I wrote about classifying shapes, discussing inclusive and exclusive definitions, and variations in different contexts. I promised to return to the subject, moving on to the specific issue of trapezoids, and some other related topics. Now is the time.\n\n## You say trapezium, I say trapezoid\n\nWe have to start with a regional issue: The word \u201ctrapezoid\u201d doesn\u2019t mean the same thing in every country. In our FAQ on geometrical formulas, we head one article with two names and a footnote:\n\nTrapezoid (American)\nTrapezium (British)*\n...\n*From The Words of Mathematics by Steven Schwartzman (1994, Mathematical Association of America): trapezoid (noun); trapezoidal (adjective); trapezium, plural trapezia (noun): ...\n\nSome Americans define a trapezoid as a quadrilateral with at least one pair of parallel sides. Under that definition, a parallelogram is a special kind of trapezoid. For other Americans, however, a trapezoid is a quadrilateral with one and only one pair of parallel sides, in which case a parallelogram is not a trapezoid.\n\nThe situation is further confused by the fact that in Europe a trapezoid is defined as a quadrilateral with no sides equal. Even more confusing is the existence of the similar word trapezium, which in American usage means \"a quadrilateral with no sides equal,\" but which in European usage is a synonym of what Americans call a trapezoid. Apparently to cut down on the confusion, trapezium is not used in American textbooks.\n\nTaking the last issue first, when we get a question about a trapezium, we generally assume it is used in the European sense (though rarely we might see it in the American sense); if it mentions parallel sides, we can go on our way with confidence, as we did here:\n\nCyclic Quadrilateral\n\nFor an isosceles trapezium ABCD with AB parallel to DC and AB < CD, prove that:\n\n1) angle ADC = angle BCD\n2) ABCD is a cyclic quadrilateral\n3) the diagonals of ABCD are equal\n\nThere\u2019s no question what is being asked here. Some of us (such as Doctor Floor and Doctor Anthony), are themselves European, and may use \u201ctrapezium\u201d even when the question is about a \u201ctrapezoid\u201d. And sometimes we just have to ask, if the question is unclear about which is meant. For example, a few years ago I started an answer with,\n\nIf you live in a country where \"trapezium\" means that two sides are parallel, and if you know which two they are, then ...\n\nOn the other hand, I started another answer with,\n\nFirst, we need to be sure of your definition of the word \"trapezium\", which varies among countries. I think that you are using it to mean a general quadrilateral, with no parallel sides. Is that correct?\n\n## Exactly, or at least?\n\nNow let\u2019s move on to the other issue, which tends to generate more questions, like this one from 2004:\n\nInclusive Definitions: Trapezoids\n\nAs far as I know, a trapezoid is defined as a quadrilateral with exactly one set of parallel sides.  Most textbooks and websites will confirm this definition.  However, a very highly regarded educator and textbook author recently argued that this definition is incorrect.  His definition of a trapezoid is that it is a quadrilateral that has at least one pair of parallel sides.  A square, therefore, would be considered a trapezoid.  He even included this definition in the glossary of a newly published textbook.  Is he correct or are thousands of books going to be published with the wrong definition?  As a teacher looking to buy new books for my school, I would really like to know.  Thanks.\n\nI can\u2019t vouch for the claim that most textbooks state the exclusive definition\u00a0(saying that figures with a second pair of parallel sides are excluded from being trapezoids); but are they wrong, as this author reportedly says? Or is he wrong?\n\nI started with the usual explanation of inclusive and exclusive definitions, emphasizing that both forms of definition are valid:\n\nBoth definitions are in use, so neither is wrong!  That does lead to confusion, but each author has to choose the definition that makes most sense in his context.\n\nQuadrilateral Classification: Definition of a Trapezoid\nhttp://mathforum.org/library/drmath/view/54901.html\n\nInclusive and Exclusive Definitions\nhttp://mathforum.org/library/drmath/view/55295.html\n\nThe same sort of issue arises with other shapes, such as the rectangle.  Is a square a rectangle?  Not to a child; we tell them \"This is a square, and that is a rectangle,\" and they learn that a rectangle is like a square but doesn't have equal sides.\n\nYet to a mathematician, such exclusive definitions are awkward, because everything that is true of a rectangle is true of a square, and we'd like to use one word to cover both when we write a theorem. For example, any quadrilateral with three right angles is a rectangle --why should we have to add \"or a square\"?  And if we prove something is true of any parallelogram, we don't want to have to add \"or rhombus, or rectangle, or square.\"\n\nSo although even mathematicians find the exclusive definition useful when we want to point out objects (we generally use the most specific term we can, so that we wouldn't call a square a rectangle when we are trying to ask for one), for technical purposes we prefer the inclusive definition, and would prefer that it be taught in schools.\n\nAs before, inclusive definitions fit better in a formal mathematical context with theorems, while exclusive definitions fit an informal context, where we usually use the strongest description possible.\n\nIt's a little more subtle with trapezoids, because there are fewer theorems about them, so we have less commitment to an inclusive definition.  There are probably mathematicians, and certainly educators, who don't use the inclusive definition in this case.  But as you'll see in the links above, the inclusive definition makes the relationships among quadrilaterals clearer.\n\nThis may well explain the perception (and perhaps the fact) that most textbooks use the exclusive definition for the trapezoid: they are using the word not in theorems, but in relatively informal descriptions.\n\n## Implicitly inclusive\n\nOn the other hand, it may be that they are really using the inclusive definition, but it isn\u2019t obvious. Their wording may sound exclusive, but really be inclusive:\n\nI should also mention that when a mathematician says \"a trapezoid is a quadrilateral with two sides parallel,\" he probably means \"at least two sides,\" not \"exactly two sides\"; that is the usual understanding of such a phrase, because we get used to speaking that way.  It may not always be clear to non-mathematicians!\n\nWe are so used to inclusive definitions that, in effect, we define \u201ctwo\u201d inclusively: If we say two sides are parallel, we are not mentioning the other sides, which may also be parallel! (In the same way, we may say that an isosceles triangle has two congruent sides, meaning that if two are congruent, it doesn\u2019t matter if the third side is, too.) But to a non-mathematician, \u201ctwo\u201d may convey the meaning \u201cexactly two, and no more\u201d. If no theorems are shown where the meaning of the word is unpacked and used, you may not notice what meaning is intended.\n\nThe inclusive definition can sometimes be discerned, well hidden within the usage of the word. One place where the word \u201ctrapezoid\u201d is used is in discussing the \u201ctrapezoidal approximation\u201d in calculus. Here is a picture illustrating it; we choose points along a curve and draw (right-angled) trapezoids consisting of a chord of the curve, two vertical lines, and a piece of the x-axis:\n\nBut what if two consecutive points on the curve have the same y-coordinate, so that the chord is horizontal? Than this \u201ctrapezoid\u201d is really a rectangle, and it we were using the exclusive definition, it would not be a trapezoid! So implicitly, when we talk about the trapezoidal rule (as opposed to the \u201ctrapezoid-or-rectangle rule\u201d), we are defining \u201ctrapezoid\u201d inclusively, even if we elsewhere defined it exclusively!\n\nIn my answer to Peter, I went on to refer to two random sites I had found that discuss the variation in definition among textbooks; each then states what definition they will use, and they choose differently. One of the links no longer works; the other, which agrees with me, says\n\nThe difference is that under the second definition parallelograms are trapezoids and under the first, they are not.\n\nThe advantage of the first definition is that it allows a verbal distinction between parallelograms and other quadrilaterals with some parallel sides. This seems to have been most important in earlier times. The advantage of the inclusive definition is that any theorem proved for trapezoids is automatically a theorem about parallelograms. This fits best with the nature of twentieth-century mathematics.\n\nIt is possible to function perfectly well with either definition. However, it is important to have agreement in a math class on the definition used in the class.\n\nI concluded, in agreement,\n\nAgain, each definition has its place, and should be used in the appropriate context.  The inclusive definition fits well into the context of geometry, and I recommend it.\n\n## The challenge of the isosceles trapezoid\n\nLet me add one more comment: Under the inclusive definition, a parallelogram is a special kind of trapezoid. An often-unnoticed consequence is that we have to carefully define the other special kind of trapezoid, the isosceles trapezoid. This is commonly defined as a trapezoid in which the non-parallel sides are congruent. There are two problems here: there are not always any non-parallel sides; and in a parallelogram, if you pick one pair of parallel sides, the other pair will always be congruent!\n\nWe don\u2019t want to call a parallelogram an isosceles trapezoid, because theorems about the latter typically do not apply to the former. This is because the latter has a symmetry that the former does not. Therefore, with the inclusive definition, it is best to define an isosceles trapezoid not in terms of congruent legs, but of symmetry. One way to do this is to require the base angles (angles at the ends of a side that is parallel to another) to be congruent. That is done, for example, here:\n\nWolfram MathWorld: Isosceles Trapezoid\n\nAn isosceles trapezoid (called an isosceles trapezium by the British; Bronshtein and Semendyayev 1997, p.\u00a0174) is\u00a0trapezoid\u00a0in which the base angles are equal and therefore the left and right side lengths are also equal.\n\nWikipedia explicitly uses symmetry for its definition:\n\nIsosceles Trapezoid\n\nIn Euclidean geometry, an isosceles trapezoid (isosceles trapezium in British English) is a convex quadrilateral with a line of symmetry bisecting one pair of opposite sides. It is a special case of a trapezoid. In any isosceles trapezoid two opposite sides (the bases) are parallel, and the two other sides (the legs) are of equal length (properties shared with the parallelogram). The diagonals are also of equal length. The base angles of an isosceles trapezoid are equal in measure (there are in fact two pairs of equal base angles, where one base angle is the supplementary angle of a base angle at the other base).\n\nAnd this site not only uses base angles in their definition, but explains why (perhaps because parents will expect otherwise):\n\nMath Bits Notebook: Theorems Dealing With Trapezoids and Kites\n\nHere are two other pages that touch on classification of trapezoids:\n\nQuadrilateral Classification: Definition of a Trapezoid\n\nVenn Diagram to Classify Quadrilaterals\n\nThe first of these includes links to discussions among mathematicians; the second provides a Venn diagram. Note that by our definitions, a rectangle is an isosceles trapezoid. Is that surprising?\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-07-23 21:13:05", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/what-is-a-trapezoid-more-on-inclusive-definitions/", "openwebmath_score": 0.683411717414856, "openwebmath_perplexity": 622.3495079181878, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575129653769, "lm_q2_score": 0.9032942073547149, "lm_q1q2_score": 0.8875385098544801}}
{"url": "https://math.stackexchange.com/questions/4205201/why-does-the-n-1-rule-not-work-in-all-cases-of-circular-permutations", "text": "# Why does the $(n-1)!$ rule not work in all cases of circular permutations? [duplicate]\n\n$$5$$ Boys and $$5$$ girls sit alternatively around a round table. In how many ways can this be done?\n\nI solved it like this : $$5$$ boys can be arranged in $$(5-1)!$$ ways. After that the $$5$$ girls can be arranged in the gaps in $$(5-1)!$$ ways. So, the answer should be $$4!\u00d74!$$ but the actual answer is $$4!\u00d75!$$. After seeing the answer I can guess that they have considered $$5$$ instead of $$(5-1)$$ in any one of the cases. I have learnt that number of circular arrangements = $$(n-1)!$$. So, why did I get wrong answer?\n\nEDIT\n\nrotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone 2, 4, 6, or 8 seats to their left, then we will get another seating arrangement that will be counted again.\n\nDoes that mean that if for example, $$5$$ men, $$5$$ women and $$5$$ children are to sit alternately then the answer should be $$\\frac{5!\u00d75!\u00d75!}{5\u00d73/3}$$ because neither can we rotate each member to the left by one place nor by two places? So does that mean that in this type of questions, we have to group the objects and then divide the result with total number of groups?\n\n\u2022 The girls can be arranged in the gaps in $5!$ ways, not $(5-1)!$ ways. The boys' positions are already established. You have $5$ spaces to place that first girl. Jul 23, 2021 at 1:55\n\u2022 The reason that you use $(n-1)!$ in these circular arrangements is that you have a rotational symmetry. Once you seat the boys, this symmetry no longer exists. Jul 23, 2021 at 2:37\n\nThe circular arrangement formula takes into account that you can rotate the positions, and nothing of substance changes. If you're putting 5 people on a circular table, you get $$5!$$ permutations, divided by $$5$$ to account for the fact that we can rotate the seating arrangement $$5$$ different ways without substantial change to the arrangement.\nThat is, when you count the placements as $$5!$$, you are over-counting by a factor of $$5$$, because each of the $$5!$$ placements can be rotated to $$4$$ other seating arrangements.\nIn your case, you seat $$5$$ girls and $$5$$ boys alternately. Without accounting for rotations, there are $$5! \\times 5!$$ ways of seating the children. But then we must account for rotations. How many other equivalent seating arrangements can we form by rotating the seats?\nNote that rotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone $$2$$, $$4$$, $$6$$, or $$8$$ seats to their left, then we will get another seating arrangement that will be counted again. This tells us that by computing $$5! \\times 5!$$, we have again over-counted by a factor of $$5$$, so the result is $$\\frac{5! \\times 5!}{5} = 5! \\times \\frac{5!}{5} = 5! \\times 4!.$$\nBy computing $$4! \\times 4!$$, you are implicitly assuming that you can rotate the boys and girls independently. This isn't really allowed in the problem, as it would produce different seating arrangements.\nFor example, if we had the cyclic order $$\\text{matt}, \\text{hannah}, \\text{charlie}, \\text{elizabeth}, \\text{warren}, \\text{jenny}, \\text{peter}, \\text{veronica}, \\text{cameron}, \\text{celia},$$ then we get an equivalent order by shuffling everyone $$2$$ spots to the left: $$\\text{charlie}, \\text{elizabeth}, \\text{warren}, \\text{jenny}, \\text{peter}, \\text{veronica}, \\text{cameron}, \\text{celia}, \\text{matt}, \\text{hannah}.$$ Everyone still has the same person to their left, and to their right. On a circular table, nobody would know the difference. But, if we were to apply the $$5 \\times 5$$ independent rotations of the girls and the boys, we could rotate just the girls one spot to the left: $$\\text{matt}, \\text{elizabeth}, \\text{charlie}, \\text{jenny}, \\text{warren}, \\text{veronica}, \\text{peter}, \\text{celia}, \\text{cameron}, \\text{hannah},$$ and we get a totally different order. Elizabeth used to be next to Charlie and Warren, but now she's next to Matt and Charlie (and Charlie is sitting on the opposite side of her to where he previously sat). As far as the table's participants are concerned, this is a different configuration.", "date": "2022-10-05 06:04:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4205201/why-does-the-n-1-rule-not-work-in-all-cases-of-circular-permutations", "openwebmath_score": 0.5905911922454834, "openwebmath_perplexity": 212.14852990839736, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682474702956, "lm_q2_score": 0.897695298265595, "lm_q1q2_score": 0.8875228372985701}}
{"url": "https://math.stackexchange.com/questions/4451310/probability-of-pair-of-gloves-selection", "text": "# Probability of pair of gloves selection\n\nIn his wardrobe, Fred has a total of ten pairs of gloves. He had to pack his suitcase before a business meeting, and he chooses eight gloves without looking at them. We assume that any set of eight gloves has an equal chance of being chosen.\n\nI am told to calculate the likelihood that these 8 gloves do not contain any matching pairs, i.e. that no two (left and right) gloves are from the same pair.\n\nThis is what I came up with, that is, the probability of success for each choice:\n\n$$\\frac{20}{20}\u00d7\\frac{18}{19}\u00d7\\frac{16}{18}\u00d7...\u00d7\\frac{6}{13}=\\frac{384}{4199}\u22480.09145$$\n\nAt first, I was a little confused by the wording but I believe this seems about right.\n\nIs there an alternative way to get the desired probability, e.g. with $$1-...$$?\n\nThanks in advance for any feedback.\n\nThere is a more general formula for this.\nHere you are asked that no pair is selected, but this formula will take care of any number of pairs selected\n\nWith $$10$$ be the number of pairs, and $$k$$ the number of pairs selected from $$8$$ gloves, the formula is\n\n$$\\dfrac{\\dbinom{10}{k}\\dbinom{10-k}{8-2k}\\cdot2^{8-2k}}{\\dbinom{20}{8}}$$\n\nFor the particular case for $$k=0$$, it simplifies to\n\n$$\\dfrac{\\dbinom{10}{0}\\dbinom{10-0}{8-2\\cdot0}\\cdot2^{8-2\\cdot0}}{\\dbinom{20}{8}}$$\n\n$$= \\dfrac{\\dbinom{10}{0}\\dbinom{10}8 \\cdot2^8}{\\dbinom{20}{8}}$$\n\n\u2022 Explaining briefly for OP: this can be justified by an extension of my argument where we first choose $k$ pairs from the original $10,$ then from the remaining $10 - k$ pairs we pick the remaining $8 - 2k$ gloves and choose whether each is the left or right glove. We can actually generalize this for all numbers of original pairs of gloves and gloves selected as well, but we need to be a bit careful regarding bounds. (for instance, notice that this formula is clearly only valid when $0 \\leq k \\leq 4$: if $k = 5$ then the probability is $0$ because we'd need to pick at least $10$ gloves) May 16 at 7:28\n\u2022 I was about to point out that $\\dbinom{20}{16}$ as a denominator would seem impossible as it yields $2.3777...$, but seems like it's already corrected. Anyways, this formula is exactly what I was looking and I feel like this should be the answer. I would have definitely marked it as an answer if you would have responded 10 seconds earlier @trueblueanil May 16 at 7:29\n\u2022 @nimen55290: No matter, the important thing is that I have been of help ! May 16 at 8:05\n\u2022 I really appreciate it @trueblueanil May 16 at 8:16\n\nWe can use a combinatoric argument if you like: there are $$20 \\choose 8$$ ways we could possibly choose $$8$$ gloves from the $$20,$$ neglecting order.\n\nTo see how many of these will involve us choosing no pairs, we can think about first choosing which pairs we will take one glove from, and then from that choosing what glove to pick from each pair. There are $$10$$ pairs so we have $$10 \\choose 8$$ ways to choose our pairs, and then for each set of pairs there are $$2^8$$ ways that we can choose to take the left or right glove from each.\n\nSo, if all possible sets of gloves are equally likely to be taken, the probability of taking no pairs of gloves should be $$\\frac{{10 \\choose 8} \\cdot 2^8}{20 \\choose 8} = \\frac{\\frac{10!}{2! 8!} \\cdot 2^8}{\\frac{20!}{12!8!}} = \\frac{(10 \\cdot 9 \\cdot \\ldots \\cdot 3) \\cdot 2^8}{20 \\cdot 19 \\cdot \\ldots \\cdot 13} = \\frac{20 \\cdot 18 \\cdot \\ldots \\cdot 6}{20 \\cdot 19 \\cdot \\ldots \\cdot 13}$$\n\nThere are $$\\binom{20}{8}$$ ways to select the $$8$$ gloves, all of which we assume are equally likely. Let's say a selection has \"Property $$i$$\" if it includes both gloves of pair $$i$$, for $$1 \\le i \\le 10$$, and let $$S_j$$ be the total probability (with over-counting) of the selections with $$j$$ of the properties, for $$1 \\le j \\le 4$$. So $$S_j = \\frac{\\binom{10}{j} \\binom{20-2j}{8-2j}}{\\binom{20}{8}}$$\nBy inclusion-exclusion, the probability of a selection with none of the properties, i.e. with no pair of matching gloves, is $$1-S_1+S_2-S_3+S_4 = 0.0914503$$", "date": "2022-06-26 12:29:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4451310/probability-of-pair-of-gloves-selection", "openwebmath_score": 0.8429904580116272, "openwebmath_perplexity": 170.38671993740866, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795095031688, "lm_q2_score": 0.8991213759183765, "lm_q1q2_score": 0.8874143745877335}}
{"url": "https://math.stackexchange.com/questions/1831270/different-ways-of-evaluating-int-0-pi-2-frac-cosx-sinx-cosxdx", "text": "Different ways of evaluating $\\int_{0}^{\\pi/2}\\frac{\\cos(x)}{\\sin(x)+\\cos(x)}dx$\n\nMy friend showed me this integral (and a neat way he evaluated it) and I am interested in seeing a few ways of evaluating it, especially if they are \"often\" used tricks.\n\nI can't quite recall his way, but it had something to do with an identity for phase shifting sine or cosine, like noting that $\\cos(x+\\pi/2)=-\\sin(x)$ we get: $$I=\\int_{0}^{\\pi/2}\\frac{\\cos(x)}{\\sin(x)+\\cos(x)}dx=\\int_{\\pi/2}^{\\pi}\\frac{-\\sin(x)}{-\\sin(x)+\\cos(x)}dx\\\\$$ Except for as I have tried, my signs don't work out well. The end result was finding $$2I=\\int_{0}^{\\pi/2}dx\\Rightarrow I=\\pi/4$$ Any help is appreciated! Thanks.\n\n\u2022 It seems better to make the change of variable $x \\to \\pi/2-x$. Jun 18 '16 at 20:52\n\u2022 @OlivierOloa oops right, I'll fix it. Have any other ideas for how to integrate the above? Jun 18 '16 at 20:53\n\u2022 @OlivierOloa yes, thank you! I will try to use that for mine and provide an answer Jun 18 '16 at 20:57\n\u2022 Also related: math.stackexchange.com/questions/180744/\u2026 Jun 18 '16 at 21:02\n\n$$\\int_{0}^{a}{\\frac{f(x)}{f(x)+f(a-x)}}dx=\\frac{a}{2}$$ let $f(x)=\\sin x$ and $a=\\frac\\pi2$\n\n\u2022 Is there a name for this formula? Where can I find more? Jun 18 '16 at 21:06\n\u2022 No there is not Jun 18 '16 at 21:08\n\u2022 @qbert The formula arises from another one: $\\int_{0}^{a}\\mathrm{F}\\left(x\\right)\\,\\mathrm{d}x = \\int_{0}^{a}\\mathrm{F}\\left(a - x\\right)\\,\\mathrm{d}x$ which sometimes we say \"by reflection in the mirror\" especially by people that computes MonteCarlo integrals. By adding ( to the original one ) and dividing by two we usually arrives to an integrand which is somehow smooth which the MC-people like a lot because it can reduce computing time. Jun 19 '16 at 2:31\n\nLet\n\n$$I=\\int_0^{\\pi/2}\\frac{\\cos x}{\\sin x+\\cos x}\\ dx$$\n\nand\n\n$$J=\\int_0^{\\pi/2}\\frac{\\sin x}{\\sin x+\\cos x}\\ dx$$\n\nthen\n\n$$I+J=\\frac{\\pi}{2}\\tag1$$\n\nand\n\n\\begin{align} I-J&=\\int_0^{\\pi/2}\\frac{\\cos x-\\sin x}{\\sin x+\\cos x}\\ dx\\\\[10pt] &=\\int_0^{\\pi/2}\\frac{1}{\\sin x+\\cos x}\\ d(\\sin x+\\cos x)\\\\[10pt] &=0\\tag2 \\end{align}\n\nHence, $$I=J=\\frac\\pi4$$by linear combinations $(1)$ and $(2)$.\n\n\u2022 You might also be interested in seeing the general method Jun 18 '16 at 22:54\n\u2022 I came back to this question, and you're answer, and the link you provided is really fantastic, and quite general Jul 7 '16 at 2:40\n\nHint: Substitute $2i\\sin(x)=e^{ix}-e^{-ix}$ and $2\\cos(x)=e^{ix}+e^{-ix}$\n\n$$I=\\int_{0}^{\\pi/2}\\frac{2\\cos(x)}{2\\sin(x)+2\\cos(x)}dx=\\int_{0}^{\\pi/2}\\frac{e^{ix}+e^{-ix}}{\\frac{e^{ix}-e^{-ix}}{i}+e^{ix}+e^{-ix}}dx$$\n\nNow substitute $u=e^{ix} \\implies du = ie^{ix}dx=iudx$\n\n$$I=\\int\\frac{u+1/u}{\\frac{u-1/u}{i}+u+1/u}\\frac{du}{iu}$$\n\nAlternative method: $$I=\\int_{0}^{\\pi/2}\\frac{\\cos(x)}{\\sin(x)+\\cos(x)}dx=\\int_{0}^{\\pi/2}\\frac{1}{\\tan(x)+1}dx$$\n\nSubstitute: $u=\\tan(x) \\implies du=(1+\\tan^2(x))dx=(1+u^2)dx$\n\n$$I=\\int\\frac{1}{u+1}\\frac{du}{1+u^2}$$\n\n\u2022 Ah I love it when complex makes integration easy. Thanks! Jun 18 '16 at 21:01\n\u2022 Added second method. Jun 18 '16 at 21:24\n\nYou're integrating on $[0, \\pi/2]$ so replacing $x$ by $\\pi/2 -x$ we see that $$I=\\int_0^{\\pi/2} \\frac{\\sin{x}}{\\cos{x}+\\sin{x}}dx.$$ Now sum this integral with the initial expression and notice that $2I=\\pi/2$ hence...", "date": "2021-10-18 08:42:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1831270/different-ways-of-evaluating-int-0-pi-2-frac-cosx-sinx-cosxdx", "openwebmath_score": 0.8823049664497375, "openwebmath_perplexity": 818.8913197919615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918502576513, "lm_q2_score": 0.8976952852648487, "lm_q1q2_score": 0.8873644734990204}}
{"url": "https://math.stackexchange.com/questions/2445462/the-probability-that-a-hits-a-target-is-frac14-and-that-of-b-is-frac13-if/2445467", "text": "# The probability that A hits a target is $\\frac14$ and that of B is $\\frac13$. If they fire at once and one hits the target, find $P(\\text{A hits})$\n\nThe probability that A hits a target is 1/4 and the probability that B hits a target 1/3. They each fire once at the target.\n\nIf the target is hit by only one of them, what is the probability that A hits the target?\n\nI know that this is an independent event. If I do P(A hitting) * P(B not hitting) then (1/4)(2/3) = 1/6 But when I look at the back of my book the answer is 2/5? My book is known to give wrong answers because it is quite old; therefore, I am left with self doubt. Can anyone tell me if I have the correct answer or if I am actually making a mistake?\n\n\u2022 This is a conditional probability. Letting $A$ be the event that player $A$ hit the target (in a single shot) and $B$ the event that $B$ hit the target (in a single shot), then what you calculated was $Pr(A\\cap B^c)$. What you were told to calculate was $Pr(A\\mid (A\\cap B^c)\\cup (A^c\\cap B))$, i.e. the probability that $A$ hit the target given that exactly one of them hit the target. \u2013\u00a0JMoravitz Sep 26 '17 at 4:07\n\u2022 You might like this explanation of why the formulas being posted here work: arbital.com/p/bayes_rule/?l=1zq \u2013\u00a0Davislor Sep 26 '17 at 14:51\n\u2022 Would also add the comment that the book is likely not full of mistakes because it is old, but that because it is old the mistakes have been found. New books are not necessarily more correct, they just haven't been around long enough for the mistakes to be as well known. Old does not imply bad. \u2013\u00a0Jared Smith Sep 26 '17 at 15:39\n\u2022 The probability of A hitting and B not hitting is 1/6 when it is also possible that they both hit or both miss. But in this case, it is not possible for them both to hit or both to miss, so the probability must be greater than 1/6. (If you don't see why, imagine rolling a die. The probability of 1 coming up is 1/6, but only because there are 5 other possibilities. If you roll a die and the result is not a 3, 4, 5, or 6, now the probability it's a 1 is 1/2.) \u2013\u00a0David Schwartz Sep 26 '17 at 18:42\n\n\\begin{align} P(\\mbox{target is hit once}) &= P(\\mbox{A hitting}) \\cdot P(\\mbox{B not hitting}) + P(\\mbox{A not hitting}) \\cdot P(\\mbox{B hitting}) \\\\ &= \\frac{1}{4}\\cdot\\frac{2}{3} + \\frac{3}{4}\\cdot\\frac{1}{3} \\\\ &= \\frac{5}{12} \\end{align}\n\nSo, $$P(\\mbox{A hitting | target is hit once}) = \\frac{P(\\mbox{A hitting}) \\cdot P(\\mbox{B not hitting})}{P(\\mbox{target is hit once})} = \\dfrac{\\frac{1}{6}}{\\frac{5}{12}} = \\frac{2}{5}.$$\n\n\u2022 I would change the LHS of the last line to \"$P(\\text{A hitting}|\\text{target is hit once})$, to be more explicit that we're using conditional probability here. \u2013\u00a0JiK Sep 26 '17 at 8:18\n\nYour answer is not correct because you did not account for the case where only B hits, which has probability $\\frac13\u00d7\\frac34=\\frac14$. Then the required probability is $$\\frac{\\frac16}{\\frac14+\\frac16}=\\frac25$$ as the book gives.\n\nThe answer is indeed 2/5 I believe.\n\n\\begin{align} \\mathbb{P}[\\text{A hit | only one hit}] &= \\frac{\\mathbb{P}[\\text{A hit} \\,\\cap\\, \\text{only one hit}]}{\\mathbb{P}[\\text{only one hit}]} \\\\ &= \\frac{\\mathbb{P}[\\text{A hit}\\,\\cap\\,\\text{B didn't hit}]}{\\mathbb{P}[\\text{A hit}\\,\\cap\\, \\text{B didn't hit}] + \\mathbb{P}[\\text{A didn't hit}\\,\\cap\\, \\text{B hit}]} \\\\ &= \\frac{1/4 \\cdot 2/3}{1/4 \\cdot 2/3 + 3/4 \\cdot 1/3} \\\\ &=\\frac{2}{5} \\end{align}\n\nWithout using the conditional probability formula:\n\nThere are four cases:\n\n1. Both miss\n2. A hits and B misses\n3. B hits and A misses\n4. Both hit\n\nWe're only interested in (2) and (3). (2) has probability $\\frac{1}{4}*\\frac{2}{3} = \\frac{1}{6}$. (3) has probability $\\frac{1}{3}*\\frac{3}{4}=\\frac{1}{4}$. And we need $\\frac{(2)}{(2) + (3)}$.\n\n\u2022 What's the last formula if not the conditional probability formula? \u2013\u00a0JiK Sep 26 '17 at 16:48\n\u2022 @JiK common sense \u2013\u00a0MattPutnam Sep 26 '17 at 17:42\n\nThe probability that only one person hits the target is $$1/4 * 2/3 + 1/3 * 3/4 = 5/12$$ The first event occurs when A hits and B misses, and the second when B hits and A misses. So if only one hit occurs, A hits 2/5 of the time and B 3/5 of the time.\n\nThis is an application of Bayes's law. You have a theory: A hit the target. You have data: there's only one hit. What is the probability your theory is true, given the data? 2/5. If you saw two bullet holes, then your theory would be true with probability 1 because A had to hit the target, given those data.\n\n\u2022 We don't consider $P(\\text{only one hit}|\\text{A hit the target})$ or $P(\\text{only one hit}|\\text{A didn't hit the target})$ explicitly, so I don't see how this is an application of Bayes's law. \u2013\u00a0JiK Sep 26 '17 at 8:20", "date": "2019-10-17 01:22:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2445462/the-probability-that-a-hits-a-target-is-frac14-and-that-of-b-is-frac13-if/2445467", "openwebmath_score": 0.9966638088226318, "openwebmath_perplexity": 459.21869390506856, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842222598318, "lm_q2_score": 0.8947894639983208, "lm_q1q2_score": 0.8873485936914666}}
{"url": "https://mathematica.stackexchange.com/questions/94621/how-to-plot-slices-of-a-surface-of-an-iterative-function-parametrized-by-the-ite", "text": "# How to plot slices of a surface of an iterative function parametrized by the iterator k?\n\nI am trying to plot a surface of\n\n$$z=\\sin^{(k)}(x),\\text{where (k) means nesting the function k times}$$\n\nto visualise the fixed points and their neighbourhood to visually analyse their behaviour.\n\nCurrently, the following (adapted from this link) give me a contour version of the above:\n\nf[x_] := Sin[x]\nShow[Table[Plot[Nest[f, x, i], {x, -\u03c0, \u03c0},\nPlotRange -> {-1, 1}, PlotStyle -> ColorData[\"Rainbow\", 0.5 + i/10]], {i, 1, 10}]]\n\n\nHowever, I want to space out the contours along the $k$ axis so that e.g. $\\sin(x)$ corresponds to $k=1$, $\\sin(\\sin(x))$ corresponds to $k=2$ and so on...\n\nBelow is my most recent attempt at doing it:\n\n f[x_] := Sin[x]\ndata[x_] := Table[{Nest[f, x, i], i}, {i, 0, 10}]\nListPlot3D[data[x], {x, -\u03c0, \u03c0}]\n\n\nwhich gives me an error\n\n SetDelayed::write: Tag List in {{x,0},{Sin[x],1},{Sin[Sin[x]],2},\n{Sin[Sin[Sin[x]]],3},{Sin[Sin[Sin[Sin[x]]]],4},{Sin[Sin[Sin[<<1>>]]],5},\n{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],6},\n{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],7},\n{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],8},\n{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],9},\n{Sin[Sin[Sin[Sin[Sin[Sin[<<1>>]]]]]],10}}[x_] is Protected. >>\n\n\nStrangely the data behind seemed to be interpreted correctly\n\nListPlot3D[{{x, 0}, {Sin[x], 1}, {Sin[Sin[x]], 2}, {Sin[Sin[Sin[x]]],\n3}, {Sin[Sin[Sin[Sin[x]]]], 4}, {Sin[Sin[Sin[Sin[Sin[x]]]]],\n5}, {Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]],\n6}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]],\n7}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]],\n8}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]]],\n9}, {Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[Sin[x]]]]]]]]]], 10}}[x], {x, -\\[Pi], \\[Pi]}]\n\n\nI was suspecting that ListPlot3D cannot read my input is probably because I have mixed data type. In details\n\n$$z\\in \\mathbb{R}$$ $$x \\in [-\\pi,\\pi]$$ but $$k \\in \\{0,1,2,3,4,5,6,7,8,9,10\\}$$\n\nFrom browsing the documentation, I am not aware of any examples of plots made from a mix of discrete and continuous variables as plotting arguments, thus I am not sure how to plot the surface I want.\n\nI am not sure how to circumvent/cheat it without taking too much computation time since if my set of points $x$ is too sparse, it will fail to display the sinusoidal feature (which will be a problem because I am planning to apply this code on other iterative functions, such as the logistic map), but if my sampling is too dense, it will probably took too much computation time\n\nAny ideas on what I can do?\n\nP.S. To give an idea on what I am trying to achieve, refer to the below sketch:\n\nwhich after interpolation along $k$, will give a nice surface.\n\n\u2022 Related: (1413). \u2013\u00a0march Sep 14 '15 at 17:07\n\nUse Interpolation if you want a regular function. Just for the plot you can also use ListPlot3D.\n\nfun = Interpolation[\nFlatten[Table[{x, k, Nest[Sin, x, k]}, {x, -Pi, Pi, .1}, {k, 1, 10,1}], 1]];\n\n\nPlot the continuous function and those $k$-mesh lines!\n\nPlot3D[fun[x, k], {x, -Pi, Pi}, {k, 1, 10}, MeshFunctions -> {#2 &},\nMesh -> 10, PerformanceGoal -> \"Quality\", MeshStyle -> {{Black, Thin}}]\n\n\nIf you only want discrete lines you can use ParametricPlot3D in combination with Map or Table embedded in a Show.\n\nBelow the Blend function is used to add a variable color (optional). Black is Sin[x] and Red is the curve nested ten times.\n\nShow[\nMap[\nParametricPlot3D[{u, #, Nest[Sin, u, #]}, {u, -\\[Pi], \\[Pi]},\nPlotStyle -> Blend[{Black, Red}, #/10],\nPlotRange -> {{-\\[Pi], \\[Pi]}, {0, 10}, {-1, 1}}\n] &,\nRange[10]\n]\n]\n\n\nThis is, I think, a dupe of Plotting several functions, except that that thread displayed only the contours.\n\nAn approach simpler than the other posted answers proceeds like so:\n\nPlot3D[Nest[Sin, x, Round[k]], {x, -\u03c0, \u03c0}, {k, 1, 10},\nMeshFunctions -> {#2 &}, Mesh -> 10]", "date": "2020-02-24 12:57:31", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/94621/how-to-plot-slices-of-a-surface-of-an-iterative-function-parametrized-by-the-ite", "openwebmath_score": 0.22005541622638702, "openwebmath_perplexity": 2846.4024987592734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551525886193, "lm_q2_score": 0.9196425366837827, "lm_q1q2_score": 0.8873218400590162}}
{"url": "https://infinityinfoway.com/bruce-venables-dwqvm/794d5a-exponential-function-formula", "text": ". There are three kinds of exponential functions: Thus, the exponential function also appears in a variety of contexts within physics, chemistry, engineering, mathematical biology, and economics. In particular, when Population growth can be modeled by an exponential equation. t It is used everywhere, if we talk about the C programming language then the exponential function is defined as the e raised to the power x. {\\displaystyle v} The EXP function finds the value of the constant e raised to a given number, so you can think of the EXP function as e^(number), where e \u2248 2.718. ( {\\displaystyle b^{x}=e^{x\\log _{e}b}} Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. ! Example and how the EXP function works Excel has an exponential excel function it\u2019s called Excel EXP function which is categorized as Math or Trigonometry Function that returns a numerical value which is equal to e raised to the power of a given value. That is. = {\\displaystyle y=e^{x}} What is Factorial? {\\displaystyle y} x It satisfies the identity exp(x+y)=exp(x)exp(y). For example: As in the real case, the exponential function can be defined on the complex plane in several equivalent forms. For most real-world phenomena, however, e is used as the base for exponential functions.Exponential models that use e as the base are called continuous growth or decay models.We see these models in finance, computer science, and most of the sciences such as physics, toxicology, and fluid dynamics. In the case of Exponential Growth, quantity will increase slowly at first then rapidly. b n The most common definition of the complex exponential function parallels the power series definition for real arguments, where the real variable is replaced by a complex one: y y C {\\displaystyle {\\tfrac {d}{dx}}e^{x}=e^{x}} {\\displaystyle {\\mathfrak {g}}} ...where \\\"A\\\" is the ending amount, \\\"P\\\" is the beginning amount (or \\\"principal\\\"), \\\"r\\\" is the interest rate (expressed as a decimal), \\\"n\\\" is the number of compoundings a year, and \\\"t\\\" is the total number of years. It shows that the graph's surface for positive and negative d Exponential functions and logarithm functions are important in both theory and practice. x Exponential Growth: y = a(1 + r) x. Exponential Decay: y = a(1 - r) x. The Exponential Function is shown in the chart below: by M. Bourne. Here's an exponential decay function: y = a(1-b) x. x Furthermore, for any differentiable function f(x), we find, by the chain rule: A continued fraction for ex can be obtained via an identity of Euler: The following generalized continued fraction for ez converges more quickly:[13]. with Euler's formula states that for any real number x: The formula takes in angle an input and returns a complex number that represents a point on the unit circle in the complex plane that corresponds to the angle. i To find limits of exponential functions, it is essential to study some properties and standards results in calculus and they are used as formulas in evaluating the limits of functions in which exponential functions are involved.. 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The input variable works as the exponent, while the latter is preferred when the exponent magnitude the! % for every 1000 m: an exponential function and the exponent, x y... So far we have worked with rational bases for exponential functions: functions... Of e by passing the number... Integral formulas for other logarithmic functions { \\displaystyle y=e^ { x }.... What does this mean it may throwback an error when an original amount is halved each half-life, an function... Remaining over time = e x { \\displaystyle y } range extended to \u00b12\u03c0, as! Function ez is transcendental over c ( z ) order to master the techniques here! Is characterized by the following formula: the equation is y = x! Most populous country in the complex logarithm log z, which is of the form f ( ). General approach however and look at the graphs of exponential equations second nature b are.... Positive constant amount at the general exponential and logarithm functions, such as are. About \\ ( y = b x = y a certain power exponential function formula that point to a! = e 1000k the general form of f ( x ) = a, both are the only that... Faster the graph of y = b x = y and Geometric sequence are both a form of bacteria..., you have to solve this pair of equations: y = { e^x \\. Power and get 0 or a negative number number 2 3 is equal to 3 decay...", "date": "2021-03-02 08:24:46", "meta": {"domain": "infinityinfoway.com", "url": "https://infinityinfoway.com/bruce-venables-dwqvm/794d5a-exponential-function-formula", "openwebmath_score": 0.8746333718299866, "openwebmath_perplexity": 645.2781477594338, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982287697148445, "lm_q2_score": 0.9032942125614059, "lm_q1q2_score": 0.8872947919044614}}
{"url": "https://nicoguaro.github.io/posts/putnam_prob/", "text": "# Probability that a random tetrahedron over a sphere contains its center\n\nI got interested in this problem watching the YouTube channel 3Blue1Brown, by Grant Sanderson, where he explains a way to tackle the problem that is just \u2026 elegant!\n\nI can't emphasize enough how much I like this channel. For example, his approach to linear algebra in Essence of linear algebra is really good. I mention it, just in case you don't know it.\n\n## The problem\n\nLet's talk business now. The problem was originally part of the 53rd Putnam competition on 1992 and was stated as\n\nFour points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is in- dependently chosen relative to a uniform distribution on the sphere.)\n\nAs shown in the mentioned video, the probability is $1/8$. Let's come with an algorithm to obtain this result \u2014approximately, at least.\n\n## The proposed approach\n\nThe approach that we are going to use is pretty straightforward. We are going to obtain a sample of (independent) random sets, with four points each, and check how many of them satisfy the condition of being inside the tetrahedron with the points as vertices.\n\nFor this approach to work, we need two things:\n\n1. A way to generate random numbers uniformly distributed. This is already in numpy.random.uniform, so we don't need to do much about it.\n\n2. A way to check if a point is inside a tetrahedron.\n\n### Checking that a point is inside a tetrahedron\n\nTo find if a point is inside a tetrahedron, we could compute the barycentric coordinates for that point and check that all of them are have the same sign. Equivalently, as proposed here, we can check that the determinants of the matrices\n\n\\begin{equation*} M_0 = \\begin{bmatrix} x_0 &y_0 &z_0 &1\\\\ x_1 &y_1 &z_1 &1\\\\ x_2 &y_2 &z_2 &1\\\\ x_3 &y_3 &z_3 &1 \\end{bmatrix}\\, , \\end{equation*}\n\\begin{equation*} M_1 = \\begin{bmatrix} x &y &z &1\\\\ x_1 &y_1 &z_1 &1\\\\ x_2 &y_2 &z_2 &1\\\\ x_3 &y_3 &z_3 &1 \\end{bmatrix}\\, , \\end{equation*}\n\\begin{equation*} M_2 = \\begin{bmatrix} x_0 &y_0 &z_0 &1\\\\ x &y &z &1\\\\ x_2 &y_2 &z_2 &1\\\\ x_3 &y_3 &z_3 &1 \\end{bmatrix}\\, , \\end{equation*}\n\\begin{equation*} M_3 = \\begin{bmatrix} x_0 &y_0 &z_0 &1\\\\ x_1 &y_1 &z_1 &1\\\\ x &y &z &1\\\\ x_3 &y_3 &z_3 &1 \\end{bmatrix}\\, , \\end{equation*}\n\\begin{equation*} M_4 = \\begin{bmatrix} x_0 &y_0 &z_0 &1\\\\ x_1 &y_1 &z_1 &1\\\\ x_2 &y_2 &z_2 &1\\\\ x &y &z &1 \\end{bmatrix}\\, , \\end{equation*}\n\nhave the same sign. In this case, $(x, y, z)$ is the point of interest and $(x_i, y_i, z_i)$ are the coordinates of each vertex.\n\n## The algorithm\n\nBelow is a Python implementation of the approach discussed before\n\nfrom __future__ import division, print_function\nfrom numpy import (random, pi, cos, sin, sign, hstack,\ncolumn_stack, logspace)\nfrom numpy.linalg import det\nimport matplotlib.pyplot as plt\n\ndef in_tet(x, y, z, pt):\n\"\"\"\nDetermine if the point pt is inside the\ntetrahedron with vertices coordinates x, y, z\n\"\"\"\nmat0 = column_stack((x, y, z, [1, 1, 1, 1]))\ndet0 = det(mat0)\nfor cont in range(4):\nmat = mat0.copy()\nmat[cont] = hstack((pt, 1))\nif sign(det(mat)*det0) < 0:\ninside = False\nbreak\nelse:\ninside = True\nreturn inside\n\n#%% Computation\nprob = []\nrandom.seed(seed=2)\nN_min = 1\nN_max = 5\nN_vals = logspace(N_min, N_max, 100, dtype=int)\nfor N in N_vals:\ninside_cont = 0\nfor cont_pts in range(N):\nphi = random.uniform(low=0.0, high=2*pi, size=4)\ntheta = random.uniform(low=0.0, high=pi, size=4)\nx = sin(theta)*cos(phi)\ny = sin(theta)*sin(phi)\nz = cos(theta)\nif in_tet(x, y, z, [0, 0, 0]):\ninside_cont += 1\n\nprob.append(inside_cont/N)\n\n#%% Plotting\nplt.figure(figsize=(4, 3))\nplt.hlines(0.125, 10**N_min, 10**N_max, color=\"#3f3f3f\")\nplt.semilogx(N_vals, prob, \"o\", alpha=0.5)\nplt.xlabel(\"Number of trials\")\nplt.ylabel(\"Computed probability\")\nplt.tight_layout()\nplt.show()\n\n\nAs expected, when the number of samples is sufficiently large, the estimated probability is close to the theoretical value: 0.125. This can be seen in the following figure.", "date": "2022-05-20 04:33:24", "meta": {"domain": "github.io", "url": "https://nicoguaro.github.io/posts/putnam_prob/", "openwebmath_score": 0.8530066013336182, "openwebmath_perplexity": 2443.4965888329675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877033706601, "lm_q2_score": 0.9032941995446778, "lm_q1q2_score": 0.8872947847387803}}
{"url": "http://mathcentral.uregina.ca/QQ/database/QQ.09.12/h/r.m.1.html", "text": "SEARCH HOME\n Math Central Quandaries & Queries\n Question from r.m, a student: question from calculus exam: what is the figure obtained having eqn.r=10cos(t) in cylindrical coordinates? i know it is a cylinder with center (5,0) ,but can't the equation represent two cylinders, one with center (5,0) and the other with center (-5,0). thanks for any help.\n\nHi,\n\nI want to use the cartesian graph of $y = \\cos(x)$ for reference.\n\nNow let's plot $r = 10 \\cos(\\theta)$ in polar coordinates for $0 \\le \\theta \\le 2 \\pi.$\n\n$\\cos(0) = 1$ and hence the graph starts at $(r, \\theta_1) = (10,0)$ which is the point $P_1$ in my diagram.\n\nNow let $0 < \\theta_2 < \\frac{\\pi}{2}$ then $\\cos(\\theta_2)$ is positive and resulting point $P_2$ is on the upper half of the circle with center $(5, 0)$ and radius 10 as in my diagram. When $\\theta_3 = \\frac{\\pi}{2}$ then $\\cos(\\theta_3) = 0$ and the resulting point on the graph is $P_3$.\n\nFor $\\frac{\\pi}{2} < \\theta le \\pi$ as $\\theta_4$ in my diagram, $\\cos(\\theta) < 0$ and the resulting point (for example $P_4$) is on the bottom half of the circle with center $(5, 0)$. When $\\theta = \\pi$ then $\\cos(\\theta) = -1$ and we are back at $P_1$.\n\nFor $\\pi < \\theta \\le \\frac{3 \\pi}{2}$ as $\\theta_5$ in my diagram, $\\cos(\\theta)$ is still negative and the resulting point (for example $P_5$) is on the top half of the circle with center $(5, 0)$. When $\\theta = \\frac{3 \\pi}{2}$ then $\\cos(\\theta) = 0$ and we are back at $P_3$.\n\nFinally for $\\frac{3 \\pi}{2} < \\theta \\le 2 \\pi, \\cos(\\theta)$ is positive and the resulting point, as $P_5$ in my diagram is on the bottom half of the circle and when $\\theta = 2 \\pi$ we are back at $P_1$.\n\nHence as $\\theta$ moves from $0$ to $2 \\pi$ the point defined by $r = 10 \\cos(\\theta)$ moves twice around the circle with center $(r \\theta) = (5, 0)$ and radius 10.\n\nPenny\n\nr.m replied\n\nsir,\nin reply to your answer for r=10cost in cylinderical coordinates. if we follow the same method as you explained to sketch r=sin(t/2),i expected graph to be in only first two quadrants, but the graph was covering all four quadrants ?\nIf $0 < \\theta <2 \\pi$ then $0< \\large \\frac{\\theta}{2} \\normalsize < \\pi$ and $\\sin \\left(\\large \\frac{\\theta}{2}\\right) >0.$ Thus, for example if $\\theta = \\large \\frac{3 \\pi}{2}$ then\n$r = \\sin \\left(\\frac{\\theta}{2}\\right) = \\sin \\left(\\frac{3 \\pi}{4}\\right) = \\frac{1}{\\sqrt 2} = 0.70711$", "date": "2023-02-02 21:13:09", "meta": {"domain": "uregina.ca", "url": "http://mathcentral.uregina.ca/QQ/database/QQ.09.12/h/r.m.1.html", "openwebmath_score": 0.8632741570472717, "openwebmath_perplexity": 162.56082261370122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864293768269, "lm_q2_score": 0.8962513835254865, "lm_q1q2_score": 0.8872767070004375}}
{"url": "https://math.stackexchange.com/questions/2776158/is-the-blue-area-greater-than-the-red-area/2776166", "text": "Is the blue area greater than the red area?\n\nProblem:\n\nA vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.\n\nWhich area is greater?\n\nLet the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\\cdot 45$ degrees for a natural number $k$.\n\nThus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.\n\nBut how can it be proven?\n\nI know the area of a triangle with a base $b$ and a height $h\\perp b$ is $bh\\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.\n\nTherefore, the height of the red triangle area is $1/2$, and so $$\\text{Red Area} = \\frac{b\\left(\\frac 12\\right)}{2} = \\frac{b}{4}.$$\n\nAccording to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that \\begin{align} \\text{Red Area} &< \\frac 14 \\\\ \\Leftrightarrow \\text{Red Area} &< \\text{Blue Area}.\\end{align}\n\nAssertion:\n\nTo conclude, the $\\color{blue}{\\text{blue}}$ area is greater than the $\\color{red}{\\text{red}}$ area.\n\nIs this true? If so, is there another way of proving the assertion?\n\nThanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.\n\nThis question is very similar to this post.\n\nSource:\n\nThe Golden Ratio (why it is so irrational) $-$ Numberphile from $14$:$02$.\n\n\u2022 i think you can tile the red area 4 times to get the entire square \u2013\u00a0gt6989b May 11 '18 at 4:25\n\u2022 Hint: the sum of the two red sides that don't touch the center is $1$. \u2013\u00a0dxiv May 11 '18 at 4:29\n\u2022 @user477343 Glad the hint helped. You can make that into a full-fledged answer, and I'll +1 it. \u2013\u00a0dxiv May 11 '18 at 4:31\n\u2022 Is this a problem from \"Brilliant\" \u2013\u00a0Rohan Shinde May 11 '18 at 4:32\n\u2022 Note that purely from exam technique alone, the answer is likely to be \"they are the same size\". Indeed, the problem has not told you by how much the rotation occurs, and why privilege a rotation of $0$ over a rotation of some greater angle? This is not a proof; but the phrasing of the question has told you what answer to look for. (This is a more general point than your \"it works this way for 45 degrees\": this is a demonstration that no mathematical reasoning at all is required to exam-technique that the answer is \"they're the same\".) \u2013\u00a0Patrick Stevens May 11 '18 at 20:24\n\nThe four numbered areas are congruent.\n\n[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment \u201cThis answer also deserves the tick for artistic reasons,\u201d I\u2019m leaving my original \u201cartistic\u201d figure but also adding Tom\u2019s improved version to my answer.\n\n\u2022 This answer also deserves the tick for artistic reasons. \u2013\u00a0BenM May 11 '18 at 6:09\n\u2022 A great example of \"proof by picture\" that actually works. \u2013\u00a0Bristol May 11 '18 at 14:46\n\u2022 This is not the same as the answer by Ross and Zoltan. I like this one better. Theirs was the first that came to my mind, too. \u2013\u00a0Carsten S May 11 '18 at 23:09\n\u2022 Can Wolfram Alpha draw that? \u2013\u00a0Willtech May 12 '18 at 2:44\n\u2022 @FrankShmrank The original poster asked within the question how it can be proven that the red area equals 1/4 (which would settle the title question). My answer makes it clear (without a formal proof, but in proof-by-picture, that\u2019s par for the course) that the red area is one of four congruent areas that partition the unit square, so its area is 1/4. I agree my answer is less than a complete proof of the original question, but I think (and I guess many upvoters think) that it\u2019s convincing. There are other excellent answers that are more traditionally proof-like, so upvote your favorite(s). \u2013\u00a0Steve Kass May 19 '18 at 17:14\n\nI think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily.\n\n\u2022 This method is similar to @RossMillikan 's answer above, but not quite the same :) I have to wait $9$ hours before I can upvote as I have reached my daily limit... but when I can, $$(+1)$$ \u2013\u00a0Mr Pie May 11 '18 at 15:00\n\u2022 It's not only similar, now that I read that solution, it's actually the exact same idea. Unfortunatly that answer didn't contain any images and I just looked at the images before posting my own answer. :) \u2013\u00a0Zoltan May 11 '18 at 15:05\n\u2022 Well congratulations on your first answer on the MSE! Yours is still a good answer :)) \u2013\u00a0Mr Pie May 11 '18 at 15:08\n\u2022 This is the clearest image to understand. +1 \u2013\u00a0qwr May 12 '18 at 20:36\n\u2022 @qwr Indeed! If only I could grab this answer and drag it below the accepted answer. That way, nobody would have to scroll all the way down to see this. It is my own answer that should probably be at the very bottom :) \u2013\u00a0Mr Pie May 13 '18 at 1:52\n\nNote that for equal angles $\\angle A'OB' = \\angle AOB = 90^\\circ$, when we subtract a common part $\\angle A'OB$ from both sides, we have $\\angle AOA' = \\angle BOB'$, so the red and cyan triangles are congruent: $\\triangle AOA' \\sim \\triangle BOB'$.\n\nThat implies their areas are equal, and when we add a common part $\\triangle A'OB$ we get area of the $AOB$ triangle equal to the area of the $A'OB'B$ quadrilateral. Finally, the area of the two squares' common part is constant, independent on the square's rotation angle.\n\n\u2022 Shouldn't be $\\angle AOA' = \\angle BOB'$? \u2013\u00a0Pedro May 13 '18 at 3:59\n\u2022 @Petro Right, thank you. \u2013\u00a0CiaPan May 13 '18 at 7:10\n\u2022 Do you mean to say that $\\Delta AOA' \\color{red}{\\cong} \\Delta BOB'$? \u2013\u00a0Mr Pie May 14 '18 at 3:08\n\u2022 This is the way I saw it \u2013\u00a0MichaelChirico May 15 '18 at 2:59\n\nThe two areas are equal. On the diagram with the red area draw the vertical and horizontal lines that define the blue area. The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area. Those two triangles are congruent.\n\n\u2022 I see what you mean. There was no need to describe the result when drawing the vertical and horizontal lines that define the blue area on the diagram of the red area $-$ it was clear as day that they would be equal after looking at the newly formed triangles! I like your method of showing they were equal :) $$(+1)$$ \u2013\u00a0Mr Pie May 11 '18 at 5:29\n\nBy pinning a square's vertex to the center of the other, you guarantee a 90 degree slice outwards. This means we could tile 4 slices perfectly. A square has rotational symmetry of n=4. Since the rotation number is an integer multiple of the slice number, the area is invariant of rotation. You can apply this generally as well. A 120 degree slice of an equilateral triangle will be invariant. A 60 degree slice of a uniform hexagon will too. 120 degrees will work for the hexagon as well since that's 3 slices on a rotation number of 6.\n\n\u2022 FWIW I like this answer the best. It is a simple, brief proof that uses clear logic instead of math. \u2013\u00a0Bohemian May 12 '18 at 14:28\n\u2022 @Bohemian, the reasoning is of course maths. \u2013\u00a0Carsten S May 12 '18 at 23:24\n\u2022 @carsten but it\u2019s basic geometry, without any calculations, arithmetic or formulae, such that someone without any mathematical know-how could follow. It\u2019s only barely maths (and I\u2019m not in the mood to play semantics) \u2013\u00a0Bohemian May 13 '18 at 8:37\n\u2022 @Bohemian: Whatever mood you are in, this is very much a mathematical answer. Looking for ideas like this will help you find solutions when manipulating formul\u00e6 gets you stuck. \u2013\u00a0PJTraill May 13 '18 at 21:39\n\u2022 @Bohemian, I may be a bit touchy on this subject, I hope I did not come across as rude. It is just that a recognize a misconception of what is mathematical in this, even though you may not hold it. It reminds me of beginners asking questions on how they can make their perfectly fine argument \"more mathematical\", by which they mean that they feel that they should use formulas. \u2013\u00a0Carsten S May 14 '18 at 15:37\n\nLet $f(\\alpha)$ be the length of the segment from the center of the square to the outside of the square on the line at an angle of $\\alpha$ degrees from the horizontal line pointing right.\n\nSuppose that the first side of the square (in counterclockwise order) makes an angle of $\\alpha$, then area you want is $\\int\\limits_{\\alpha}^{\\alpha+\\frac{\\pi}{2}} \\frac{f(x)^2}{2} dx$ and since $f$ is periodic with period $\\frac{\\pi}{2}$ this is independent of $\\alpha$.\n\n\u2022 This is much too advanced for my skill level. $$(+1)$$ \u2013\u00a0Mr Pie May 11 '18 at 4:35\n\u2022 in hindsight the other approach is better, but looking at it from the calculus point of view probably wont hurt :) \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo May 11 '18 at 4:36\n\u2022 I am a high school student who is familiar with integrals and radians... but the statement, \"$f$ is periodic,\" I don't know what that means. Is it ok if you could explain to me? Other than that, your answer is great! Thanks :) \u2013\u00a0Mr Pie May 11 '18 at 4:38\n\u2022 @AHB yeah I do. In my opinion, it is an act of kindness, especially when one has at least $-3$ downvotes or lower. However, I let the user know what I believe is (or might be) wrong with their question as if I did put a downvote. Also, I think there are some badges earnt when using all the upvotes in one day or something like that, idk for sure. I have only ever downvoted $1$ post, only to earn a badge of my first donwvote. \u2013\u00a0Mr Pie May 12 '18 at 10:03\n\u2022 @user477343 Yup, two actually, both bronze, 'Suffrage: 30 votes in a day', 'Vox Populi: all 40'. \u2013\u00a0Artemis Fowl May 19 '18 at 13:19\n\n$\\hspace{5cm}$\n\n$$b^2+b^2=(a-c)^2+c^2 \\Rightarrow \\frac{b^2}{2}=\\frac{(a-c)^2+c^2}{4}\\\\ S=\\frac{b^2}{2}+\\frac{(a-c)c}{2}=\\frac{(a-c)^2+c^2+2(a-c)c}{4}=\\frac{a^2}{4}.$$\n\n\u2022 The number of ways one can work this out is amazing!! Also, your answer is pure math(s)! However, I have to wait $1$ hour before I can upvote as I have reached my daily voting limit. $$(+1)$$ \u2013\u00a0Mr Pie May 16 '18 at 22:47\n\u2022 Could you elaborate on how you can assert that the 2 bs are actually equal to each other? \u2013\u00a0Frank Shmrank May 19 '18 at 14:52\n\u2022 @FrankShmrank, such problems help intuitive thinking and imagination. If the lower square turns clockwise, its top two sides will turn to the same angle with respect to their original positions and the sides $b$-$b$ will increase equally. \u2013\u00a0farruhota May 19 '18 at 16:21\n\u2022 @FrankShmrank I had the same problem, and then I deleted my comment and put up a new one (namely, my current one above) because I found out that by looking at the accepted answer, if the four triangles are congruent, then the two $b$s are equal to each other :) \u2013\u00a0Mr Pie May 20 '18 at 8:03\n\nSolution:\n\nAlthough the red area is not a triangle, the sum of its sides that do not touch the centre is equal to $1$. This can only mean that no matter how many degrees the square is rotated, no area will be greater; the red area will always be equal to the blue area, i.e. $$\\frac 14$$\n\nCredit to @dxiv who pointed this out as a hint in a comment!\n\n\u2022 This is similar to Captain Morgan\u2019s answer,but I find it less clearly expressed than that. \u2013\u00a0PJTraill May 13 '18 at 21:42\n\u2022 @PJTraill Yes, you are correct $-$ Captain Morgan has a much better answer :) \u2013\u00a0Mr Pie May 13 '18 at 21:49\n\nIf we use $$\\overline{FB}$$ for the base of $$\\triangle FEB$$, then its altitude is $$\\frac 12s$$. If we use $$\\overline{BG}$$ for the length of the base of $$\\triangle BEG$$, then its altitude is $$\\frac 12s$$.\n\nSo the area of $$\\square FBGE$$ is $$\\frac 12(\\frac 12s)(s-x) + \\frac 12(\\frac 12s)(x) = \\frac 14s^2$$. Which is one-fourth of the area of the square.\n\nThe advantage of this method is that it allows you to break the square into $$n$$ pieces of equal area quite easily. In fact, the same method applies to an regular polygon.\n\n\u2022 Thank you for letting me know. The diagram was very clear, and this method is very much applicable to the problem. $$(+1)$$ (in at least $19$ hours $-$ I have reached my daily voting limit). Also, how did you construct the picture? \u2013\u00a0Mr Pie May 16 '18 at 4:03\n\u2022 I used GeoGebra. \u2013\u00a0steven gregory May 16 '18 at 4:48\n\u2022 Thank you, again, for telling me :) \u2013\u00a0Mr Pie May 16 '18 at 6:55\n\u2022 This is by far the best answer as it doesn't assert anything that isn't given. All the other answers assert things that aren't necessarily known. It's sad that the chosen answer was chosen because that user had the lowest rep. \u2013\u00a0Frank Shmrank May 19 '18 at 14:55\n\u2022 @FrankShmrank all the answers are great, imho. \u2013\u00a0Mr Pie May 20 '18 at 8:02\n\nThe two areas are equal. On the diagram with the red area draw the vertical and horizontal lines that define the blue area. The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area. Those two triangles are congruent.", "date": "2019-10-17 12:52:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2776158/is-the-blue-area-greater-than-the-red-area/2776166", "openwebmath_score": 0.7466034293174744, "openwebmath_perplexity": 527.2969854243514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713896101315, "lm_q2_score": 0.9005297867852853, "lm_q1q2_score": 0.8872662344112534}}
{"url": "https://math.stackexchange.com/questions/1969751/is-it-possible-to-cover-a-8-times8-board-with-2-times-1-pieces", "text": "# Is it possible to cover a $8 \\times8$ board with $2 \\times 1$ pieces?\n\nWe have a $8\\times 8$ board, colored with two colors like a typical chessboard. Now, we remove two squares of different colour. Is it possible to cover the new board with two-color pieces (i.e. domino pieces)?\n\nI think we can, as after the removal of the two squares, we are left with $64-2=62$ squares with $31$ squares of each colour, and - since the domino piece covers two colours - we can cover the new board with domino pieces.\n\nBut how should one justify it mathematically?\n\n\u2022 It's possible to remove four squares, two of each color, and not cover the remaining with dominos, because the new set is disconnected. So you need to know more than that there are 31 of each color. Oct 15 '16 at 16:49\n\u2022 Your constraints are valid. But there must be more constraints. Such that the colours alternate on the board, that dominos are made of two different colours, next to each other. Then the shape of the board.\n\u2013\u00a0mvw\nOct 15 '16 at 17:05\n\u2022 @mvw \"like a typical chessboard\" should cover all of that. Oct 15 '16 at 17:11\n\u2022 @JMoravitz That was not my point. The problem is clear. It is about the proof attempt. hetajr just cared about a subset of all constraints that play a role here.\n\u2013\u00a0mvw\nOct 15 '16 at 17:13\n\u2022 @ThomasAndrews Ah cutting off two corners, leaving single corner stones.\n\u2013\u00a0mvw\nOct 15 '16 at 17:23\n\nAssume without loss of generality that the two squares to be removed are in different rows. (Otherwise turn the board 90\u00b0).\n\nFirst cover the board in horizontal dominoes, and connect the two squares by a zig-zag line like this:\n\nwhich follows the rule that if the line goes through one end of a domino, it immediately connects to another end. (The requirement that the two squares have different colors ensure that this will be true of the end of the path if only we start out in the right direction for this to hold at the beginning). Now you can flip dominoes along the zig-zag line to produce a covering that avoids the two squares.\n\nWith a bit of (easy) special-casing for the same-row case, this strategy can be extended to any size board as long as one of the side lengths is even and the other is $\\ge 2$.\n\n\u2022 I do not get the flipping. But it seems clear the length of the zig zag line is a multiple of two, as is the reduced line, so it can be covered.\n\u2013\u00a0mvw\nOct 15 '16 at 17:31\n\u2022 @mvw: Initially every other of the square boundaries crossed by the red line was covered by a domino. By \"flipping\" I mean to instead cover the other half of the boundaries by dominos. Oct 15 '16 at 17:36\n\u2022 weren't you lucky that the top and bottom part had an even number of squares ? Oct 16 '16 at 13:23\n\u2022 @mercio: If the top endpoint had been at an even column, I would just have started going left instead of right. This means that the zig-zag line would always enter each of the original dominoes at the end that has the same color as the top endpoint, so the bottom endpoint will always be reached at the end of a domino. Oct 16 '16 at 13:33\n\nFrom a graph theory point of view, this can be seen as a \"matching problem\" on a bipartite graph.\n\nThe nodes of the graph are the squares remaining.\n\nTwo nodes have an edge in the graph if the squares are neighbors - that is, if the two squares can be covered by a single domino.\n\nObviously, in any edge, one square is white, the other black, hence the \"bipartite\" nature of this graph.\n\nSo you are seeking to show there is a perfect matching for any such graph.\n\nThere is a general theorem about when there is a perfect matching for a bipartite graph, called Hall's Theorem or Hall's Marriage Theorem. It is possibly overkill for this question - induction is likely the better approach.\n\nPer the discussion on Henning's answer, it is actually possible to prove your theorem directly using a \"Hamilton cycle\" on the chess board.\n\nConsider the loop path on the board:\n\n$$\\begin{matrix}1&2&3&4&5&6&7&8\\\\ 64&15&14&13&12&11&10&9\\\\ 63&16&17&18&19&20&21&22\\\\ 62&29&28&27&26&25&24&23\\\\ 61&30&31&32&33&34&35&36\\\\ 60&43&42&41&40&39&38&37\\\\ 59&44&45&46&47&48&49&50\\\\ 58&57&56&55&54&53&52&51 \\end{matrix}$$\n\nSo we have walked in a circle, and, if the upper left is black, then we have that odd numbers on black and the even numbers on white.\n\nIf we unwind this, and consider it 64 beads in a circle, alternating black and white, then if we remove/cut away one black and one white bead, we are either left with one string of 62 beads alternating black/white, which lets us cover those with dominos, or two seperate strings.\n\nWith two strings, here is the key: because we cut away one black and one white bead, those two strands are of even length.\n\nFor example, if we removed the square at 12 and the square at 23, then we could get the domino placement: $$(13,14),(15,16),(17,18),(19,20),(21,22), \\\\(24,25),\\dots,(62,63),(64,1),(2,3),(4,5),(6,7),(8,9),(10,11).$$\n\nThis can be generalized as: \"If a bipartite graph has a Hamiltonian cycle, then if you remove one node from each of the parts, you can still get a perfect matching.\"\n\nIn particular, this argument works for any $2n\\times m$ board, because we can get a \"King's cycle tour\" on any such board.\n\nFor example, a $3\\times 4$ board:\n\n$$\\begin{matrix} 1 & 2& 3&4\\\\ 12& 9& 8&5\\\\ 11&10& 7&6 \\end{matrix}$$\n\n\u2022 (+1) Interesting approach. In this framework, we just need to check that the hypothesis of Hall's theorem are met, i.e. that for any subset of white (or black) squares in the truncated chessboard, their neighbourhood has a larger cardinality, and that is clearly true. Oct 15 '16 at 17:46\n\u2022 (on the other hand, induction is just the usual way for proving Hall's theorem) Oct 15 '16 at 17:54\n\u2022 Hmm, yes, I suppose you could prove that for a subset of $k$ white squares on the uncut board, with $0<k<32$, the set of neighboring black squares has at least $k+1$ elements, then you can apply Hall's theorem to the cut board. @JackD'Aurizio Oct 15 '16 at 18:31\n\u2022 And that can indeed be proved. (Posted as a separate answer so I can show diagrams). @JackD'Aurizio too. Oct 15 '16 at 20:50\n\nHint: A promising strategy is to prove that the claim\n\nIf we remove two opposite-colored squares from a $2m\\times 2m$ chessboard,\nwe may tile the remaining part with $2\\times 1$ dominoes.\n\nby induction on $m$. The case $m=1$ is trivial. Assume that the claim holds for some $m\\geq 1$ and consider a $(2m+2)\\times (2m+2)$ chessboard. If both the removed squares do not lie on the boundary of the chessboard, there is nothing to prove. Hence we may assume that at least one of the removed squares lies on the boundary. And we may also start tiling by following a spiral, starting next to the removed square on the boundary:\n\nAnother interesting idea is just to place $31$ non-overlapping dominoes on a $8\\times 8$ chessboard and start playing Sokoban with the placed dominoes, in order to free the wanted squares.\n\n\u2022 I like this one the best so far. Still thinking about it.\n\u2013\u00a0mvw\nOct 15 '16 at 17:19\n\u2022 You might be able to prove this form $2m\\times n$ for any $n>1$, which makes it easier to do the induction. Oct 15 '16 at 17:42\n\u2022 @ThomasAndrews: true. Interesting remark, thanks. Oct 15 '16 at 17:53\n\nCompleting an approach suggested by Thomas Andrews, if we can show that on the complete chessboard any proper subset of the white squares have more black neighbors than it has members, then Hall's marriage theorem will apply to the chessboard with two squares erased.\n\nSuppose therefore, that a proper subset of the white squares are given. Since the red and green lines in the following diagram connect all the white squares, there will be at least one red or green line that goes from a square in the subset to a square outside of it:\n\nAssume without loss of generality that there is a red line joining a square in the subset to a square outside the subset. (Otherwise mirror everything around the white diagonal).\n\nNow pair up each white square with the black neighbor it is connected to by a blue line in this diagram:\n\nThis gives one neighboring black square for each white square. However the white square with a diagonal-partner that is not selected is additionally neighbor to the non-selected square's black partner which is not otherwise used.\n\nSo, as desired, our set of white squares has more black neighbors in total than there are white squares in the set.\n\n\u2022 Very nice! I couldn't quite find that zig-zag argument while trying to find the \"extra\" black square. Oct 15 '16 at 21:39\n\u2022 @ThomasAndrews: Actually this is, on further thought, still more complicated than it needs to be. Just fix a Hamiltonian circuit on the board (necessarily with alternating white and black squares) and then just count neighbors along that circuit for each run of selected white squares. Oct 15 '16 at 22:47\n\u2022 of course, once you have a Hamiltonian circuit, you actually don't need Hall's theorem. Just remove your two nodes, and you get one or two paths, both of even lengths. Oct 15 '16 at 23:21\n\u2022 @ThomasAndrews: Indeed. I would update my main answer, except I would have to make new diagrams ... Oct 15 '16 at 23:22\n\u2022 Okay, I updated my answer with the non-Hall Hamiltonian cycle answer. Oct 16 '16 at 12:57\n\nYes.\n\nConsider the board initially covered with dominoes. After the two squares have been removed, the board has two holes.\n\nIf the two holes were in the same row, slide the dominoes between the holes until one of the holes is filled. This will leave two adjacent holes that can be covered by a domino.\n\nIf the holes are in different rows, slide dominoes from the lower row to the upper row up until one of the holes is filled. That leaves the lower row with two holes which can be filled as described above.\n\nIf the two rows are adjacent, slide the upper row left or right until its hole is filled and the hole has moved above the hole in the next row. This can be filled by a domino.\n\n\u2022 How does this answer make use of the information that the holes are on squares of different colour?\n\u2013\u00a0HTFB\nOct 15 '16 at 17:02\n\u2022 \"Slide dominoes\" won't work if the holes are in column 2 and 3 of the same row. Oct 15 '16 at 17:11\n\u2022 I think this answer is wrong because if the holes are of the same colour, then it should not work. Oct 15 '16 at 17:17", "date": "2021-10-17 20:12:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1969751/is-it-possible-to-cover-a-8-times8-board-with-2-times-1-pieces", "openwebmath_score": 0.6660507917404175, "openwebmath_perplexity": 394.3810688424373, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771763033943, "lm_q2_score": 0.8991213847035617, "lm_q1q2_score": 0.8872324611517786}}
{"url": "https://www.whitman.edu/mathematics/calculus_online/section11.05.html", "text": "As we begin to compile a list of convergent and divergent series, new ones can sometimes be analyzed by comparing them to ones that we already understand.\n\nExample 11.5.1 Does $\\ds\\sum_{n=2}^\\infty {1\\over n^2\\ln n}$ converge?\n\nThe obvious first approach, based on what we know, is the integral test. Unfortunately, we can't compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a $p$-series, that is, $${1\\over n^2\\ln n}< {1\\over n^2},$$ when $n\\ge3$. Since adding up the terms $\\ds 1/n^2$ doesn't get \"too big'', the new series \"should'' also converge. Let's make this more precise.\n\nThe series $\\ds\\sum_{n=2}^\\infty {1\\over n^2\\ln n}$ converges if and only if $\\ds\\sum_{n=3}^\\infty {1\\over n^2\\ln n}$ converges\u2014all we've done is dropped the initial term. We know that $\\ds\\sum_{n=3}^\\infty {1\\over n^2}$ converges. Looking at two typical partial sums: $$s_n={1\\over 3^2\\ln 3}+{1\\over 4^2\\ln 4}+{1\\over 5^2\\ln 5}+\\cdots+ {1\\over n^2\\ln n} < {1\\over 3^2}+{1\\over 4^2}+ {1\\over 5^2}+\\cdots+{1\\over n^2}=t_n.$$ Since the $p$-series converges, say to $L$, and since the terms are positive, $\\ds t_n< L$. Since the terms of the new series are positive, the $\\ds s_n$ form an increasing sequence and $\\ds s_n< t_n< L$ for all $n$. Hence the sequence $\\ds \\{s_n\\}$ is bounded and so converges. $\\square$\n\nSometimes, even when the integral test applies, comparison to a known series is easier, so it's generally a good idea to think about doing a comparison before doing the integral test.\n\nExample 11.5.2 Does $\\ds\\sum_{n=1}^\\infty {|\\sin n|\\over n^2}$ converge?\n\nWe can't apply the integral test here, because the terms of this series are not decreasing. Just as in the previous example, however, $${|\\sin n|\\over n^2}\\le {1\\over n^2},$$ because $|\\sin n|\\le 1$. Once again the partial sums are non-decreasing and bounded above by $\\ds \\sum 1/n^2=L$, so the new series converges. $\\square$\n\nLike the integral test, the comparison test can be used to show both convergence and divergence. In the case of the integral test, a single calculation will confirm whichever is the case. To use the comparison test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly.\n\nExample 11.5.3 Does $\\ds\\sum_{n=2}^\\infty {1\\over\\sqrt{n^2-3}}$ converge?\n\nWe observe that the $-3$ should have little effect compared to the $\\ds n^2$ inside the square root, and therefore guess that the terms are enough like $\\ds 1/\\sqrt{n^2}=1/n$ that the series should diverge. We attempt to show this by comparison to the harmonic series. We note that $${1\\over\\sqrt{n^2-3}} > {1\\over\\sqrt{n^2}} = {1\\over n},$$ so that $$s_n={1\\over\\sqrt{2^2-3}}+{1\\over\\sqrt{3^2-3}}+\\cdots+ {1\\over\\sqrt{n^2-3}} > {1\\over 2} + {1\\over3}+\\cdots+{1\\over n}=t_n,$$ where $\\ds t_n$ is 1 less than the corresponding partial sum of the harmonic series (because we start at $n=2$ instead of $n=1$). Since $\\ds\\lim_{n\\to\\infty}t_n=\\infty$, $\\ds\\lim_{n\\to\\infty}s_n=\\infty$ as well. $\\square$\n\nSo the general approach is this: If you believe that a new series is convergent, attempt to find a convergent series whose terms are larger than the terms of the new series; if you believe that a new series is divergent, attempt to find a divergent series whose terms are smaller than the terms of the new series.\n\nExample 11.5.4 Does $\\ds\\sum_{n=1}^\\infty {1\\over\\sqrt{n^2+3}}$ converge?\n\nJust as in the last example, we guess that this is very much like the harmonic series and so diverges. Unfortunately, $${1\\over\\sqrt{n^2+3}} < {1\\over n},$$ so we can't compare the series directly to the harmonic series. A little thought leads us to $${1\\over\\sqrt{n^2+3}} > {1\\over\\sqrt{n^2+3n^2}} = {1\\over2n},$$ so if $\\sum 1/(2n)$ diverges then the given series diverges. But since $\\sum 1/(2n)=(1/2)\\sum 1/n$, theorem 11.2.2 implies that it does indeed diverge. $\\square$\n\nFor reference we summarize the comparison test in a theorem.\n\nTheorem 11.5.5 Suppose that $\\ds a_n$ and $\\ds b_n$ are non-negative for all $n$ and that $\\ds a_n\\le b_n$ when $n\\ge N$, for some $N$.\n\nIf $\\ds\\sum_{n=0}^\\infty b_n$ converges, so does $\\ds\\sum_{n=0}^\\infty a_n$.\n\nIf $\\ds\\sum_{n=0}^\\infty a_n$ diverges, so does $\\ds\\sum_{n=0}^\\infty b_n$.\n\n$\\qed$\n\n## Exercises 11.5\n\nDetermine whether the series converge or diverge.\n\nEx 11.5.1 $\\ds\\sum_{n=1}^\\infty {1\\over 2n^2+3n+5}$ (answer)\n\nEx 11.5.2 $\\ds\\sum_{n=2}^\\infty {1\\over 2n^2+3n-5}$ (answer)\n\nEx 11.5.3 $\\ds\\sum_{n=1}^\\infty {1\\over 2n^2-3n-5}$ (answer)\n\nEx 11.5.4 $\\ds\\sum_{n=1}^\\infty {3n+4\\over 2n^2+3n+5}$ (answer)\n\nEx 11.5.5 $\\ds\\sum_{n=1}^\\infty {3n^2+4\\over 2n^2+3n+5}$ (answer)\n\nEx 11.5.6 $\\ds\\sum_{n=1}^\\infty {\\ln n\\over n}$ (answer)\n\nEx 11.5.7 $\\ds\\sum_{n=1}^\\infty {\\ln n\\over n^3}$ (answer)\n\nEx 11.5.8 $\\ds\\sum_{n=2}^\\infty {1\\over \\ln n}$ (answer)\n\nEx 11.5.9 $\\ds\\sum_{n=1}^\\infty {3^n\\over 2^n+5^n}$ (answer)\n\nEx 11.5.10 $\\ds\\sum_{n=1}^\\infty {3^n\\over 2^n+3^n}$ (answer)", "date": "2022-05-18 22:53:15", "meta": {"domain": "whitman.edu", "url": "https://www.whitman.edu/mathematics/calculus_online/section11.05.html", "openwebmath_score": 0.9562565684318542, "openwebmath_perplexity": 206.9080361925515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991554375055462, "lm_q2_score": 0.8947894646997281, "lm_q1q2_score": 0.8872324084765503}}
{"url": "http://mathhelpforum.com/math-topics/39580-finding-sides-giant-wooden-cube-print.html", "text": "# Finding sides on giant wooden cube\n\n\u2022 May 25th 2008, 01:49 PM\nannie3993\nFinding sides on giant wooden cube\nA giant wooden cube is painted green on all 6 sides and then cut into 125identical, smaller cubes. How many of these smaller cubes are painted on exactly two faces?\n\ni got 72, but I don't think its right...\n\nTHANX!!\n\u2022 May 25th 2008, 03:10 PM\nTheEmptySet\nQuote:\n\nOriginally Posted by annie3993\nA giant wooden cube is painted green on all 6 sides and then cut into 125identical, smaller cubes. How many of these smaller cubes are painted on exactly two faces?\n\ni got 72, but I don't think its right...\n\nTHANX!!\n\nMaybe this diagram will help.\n\nP.S. Always try to draw a picture it really helps you see what is going on.\n\nAttachment 6507\n\nIt looks like four groups of 3 on the top\n\nfour groups of 3 in the middle\n\nand four groups of 3 on the bottom.\n\n$4(3)+4(3)+4(3)=4(9)=36$\n\nI hope this helps.\n\u2022 May 25th 2008, 03:22 PM\nSoroban\nHello, annie3993!\n\nQuote:\n\nA giant wooden cube is painted green on all 6 sides and then cut into 125 identical smaller cubes.\nHow many of these smaller cubes are painted on exactly two faces?\n\nThis is a 5 \u00d7 5 \u00d7 5 cube.\n\nA cube has 6 faces, 12 edges, and 8 corners (vertices).\n\nLet's look at one face.\nCode:\n\n\u00a0 \u00a0 \u00a0 * - * - * - * - * - * \u00a0 \u00a0 \u00a0 | 3 | 2 | 2 | 2 | 3 | \u00a0 \u00a0 \u00a0 * - * - * - * - * - * \u00a0 \u00a0 \u00a0 | 2 | 1 | 1 | 1 | 2 | \u00a0 \u00a0 \u00a0 * - * - * - * - * - * \u00a0 \u00a0 \u00a0 | 2 | 1 | 1 | 1 | 2 | \u00a0 \u00a0 \u00a0 * - * - * - * - * - * \u00a0 \u00a0 \u00a0 | 2 | 1 | 1 | 1 | 2 | \u00a0 \u00a0 \u00a0 * - * - * - * - * - * \u00a0 \u00a0 \u00a0 | 3 | 2 | 2 | 2 | 3 | \u00a0 \u00a0 \u00a0 * - * - * - * - * - *\nThe nine cubes in the center have one face painted green.\nThe four cubes in the corners have three faces painted green.\n\nOn each edge, there are three cubes with two green faces.\n\nSince there are 12 edges, there are: $12 \\times 3 \\:=\\:\\boxed{36}$ cubes with two green faces.\n\n\u2022 May 25th 2008, 03:23 PM\ngalactus\nSee this thread. You may find it interesting.\n\nhttp://www.mathhelpforum.com/math-he...-question.html\n\nIf yu go to the bottom you will see the general formula of 12(n-2). In your case, 12(5-2)=36", "date": "2016-12-04 14:56:03", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/39580-finding-sides-giant-wooden-cube-print.html", "openwebmath_score": 0.6054979562759399, "openwebmath_perplexity": 660.7520826458308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988312740283122, "lm_q2_score": 0.8976952900545976, "lm_q1q2_score": 0.8872036920531113}}
{"url": "https://math.stackexchange.com/questions/2022700/in-how-many-ways-can-the-letters-in-wondering-be-arranged-with-exactly-two-conse", "text": "# In how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels\n\nIn how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels\n\nI solved and got answer as $90720$. But other sites are giving different answers. Please help to understand which is the right answer and why I am going wrong.\n\nMy Solution\n\nArrange 6 consonants $\\dfrac{6!}{2!}$\nChose 2 slots from 7 positions $\\dbinom{7}{2}$\nChose 1 slot for placing the 2 vowel group $\\dbinom{2}{1}$\nArrange the vowels $3!$\n\nRequired number of ways:\n$\\dfrac{6!}{2!}\\times \\dbinom{7}{2}\\times \\dbinom{2}{1}\\times 3!=90720$\n\nSolution taken from http://www.sosmath.com/CBB/viewtopic.php?t=6126)\n\nSolution taken from http://myassignmentpartners.com/2015/06/20/supplementary-3/\n\n\u2022 Can you explain your working. Just putting down your calculation doesn't tell us why you chose to do them. \u2013\u00a0Ian Miller Nov 20 '16 at 14:30\n\u2022 @sorry, edited the calculation and added the details. pl help. \u2013\u00a0Kiran Nov 20 '16 at 14:31\n\u2022 I will point out that the solution in the excerpt solves a different problem. Your problem asks for \"exactly two consecutive vowels\", the excerpt's solution allows 3 consecutive vowels as well. As it says at the end \"with at least two adjacent vowel\" \u2013\u00a0ReverseFlow Nov 20 '16 at 14:36\n\u2022 @Kiran You answer is right and their answer is wrong. I have added my explanation below. \u2013\u00a0user940 Nov 20 '16 at 15:12\n\u2022 Checked with Python, the answer is indeed $90720$, deleted mine. \u2013\u00a0barak manos Nov 20 '16 at 15:13\n\nThe number of arrangements with 3 consecutive vowels is correctly explained in the original post: the number is $15120$.\nTo find the number of arrangements with at least two consecutive vowels, we duct tape two of them together (as in the original post) and arrive at $120960$.\nThe problem with this calculation is that every arrangement with 3 consecutive vowels was double counted: once as $\\overline{VV}V$ and again as $V\\overline{VV}$. To compensate for this we must subtract $15120$. The correct number of arrangements with at least two consecutive vowels is $120960-15120=105840.$\nTherefore, correct number of arrangements with exactly two consecutive vowels is $105840-15120=90720.$\nThe total number of ways of arranging the letters is $\\frac{9!}{2!} = 181440$. Of these, let us count the cases where no two vowels are together. This is $$\\frac{6!}{2!} \\times \\binom{7}{3}\\times 3! = 75600$$ Again, the number of ways in which all vowels are together is 15120. Thus the number of ways in which exactly two vowels are together is $$181440 - 75600 - 15120 = 90720$$", "date": "2020-03-29 10:15:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2022700/in-how-many-ways-can-the-letters-in-wondering-be-arranged-with-exactly-two-conse", "openwebmath_score": 0.7768173217773438, "openwebmath_perplexity": 385.5642050212766, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969660413473, "lm_q2_score": 0.9019206857566127, "lm_q1q2_score": 0.8871264501201356}}
{"url": "https://rstudio-pubs-static.s3.amazonaws.com/572818_27642d4e2cf64ec98cb4d8bf3a7ca1e2.html", "text": "## Problem Set 1\n\n#### Part 1\n\nShow that $$A^TA \\neq AA^T$$ in general. (Proof and demonstration.)\n\nAssume $$A^T A = A A^T$$ Consider matrix $$A_{m\\times n}$$ where $$m \\ne n$$. So $$A^T$$ will be the size of $$n \\times m$$. Also $$AA^T$$ will be a matrix of size $$m\\times m$$ and $$A^TA$$ will be a matrix of size $$n \\times n$$.\n\nSince $$m \\ne n$$, clearly these two matrices will not be equal. This is clearly a contradiction for all non-square matrix. But what about square matrices where $$m = n$$. Let\u2019s see!\n\nContinue with this, consider a simple square matrix $$A_{2\\times 2}$$.\n\nLet $$A= \\left[ \\begin{array}{cccc} a & b \\\\ c & d \\\\ \\end{array} \\right] \\\\$$\n\n$$A^T = \\left[ \\begin{array}{cccc} a & c \\\\ b & d \\\\ \\end{array} \\right] \\\\$$\n\n$$AA^T = \\left[ \\begin{array}{cccc} a^2+b^2 & ac+bd \\\\ ac+bd & c^2+cd \\\\ \\end{array} \\right] \\\\$$\n\n$$A^TA = \\left[ \\begin{array}{cccc} a^2+b^2 & ab+cd \\\\ ab+cd & b^2+d^2 \\\\ \\end{array} \\right] \\\\$$\n\nClearly, it\u2019s not always true $$\\forall a,b,c,d$$. Therefore we conclude that $$A^TA \\neq AA^T$$ .\n\n#### Part 2\n\nFor a special type of square matrix A, we get AT $$A^TA = AA^T$$ . Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix).\n\nThis condition is true if and only if $$A = A^T$$. Transposing a matrix switches columns into rows, i.e.\u00a0flips the values along the diagonal.\n\nThe condition $$A^T = A$$ holds if the matrix is symmerical along the diagonal, just like the identity is.\n\nExample - consider the following symetric matrix A:\n\n$$A = \\left[ \\begin{array}{cccc} 1 & 2 \\\\ 2 & 3 \\\\ \\end{array} \\right] \\\\$$\n\nA <- matrix(c(1,2,2,3), ncol = 2)\n#transpose of A\nAT <- t(A)\nwrite('Printing A:', stdout())\n## Printing A:\nA\n##      [,1] [,2]\n## [1,]    1    2\n## [2,]    2    3\nwrite('Printing AT:', stdout())\n## Printing AT:\nAT\n##      [,1] [,2]\n## [1,]    1    2\n## [2,]    2    3\nwrite('Printing A*AT', stdout())\n## Printing A*AT\nA%*%AT\n##      [,1] [,2]\n## [1,]    5    8\n## [2,]    8   13\nwrite('Printing AT*A:', stdout())\n## Printing AT*A:\nAT %*% A\n##      [,1] [,2]\n## [1,]    5    8\n## [2,]    8   13\n\n## Problem Set 2\n\nMatrix factorization is a very important problem. There are supercomputers built just to do matrix factorizations. Every second you are on an airplane, matrices are being factorized. Radars that track flights use a technique called Kalman filtering. At the heart of Kalman Filtering is a Matrix Factorization operation. Kalman Filters are solving linear systems of equations when they track your flight using radars. Write an R function to factorize a square matrix A into LU or LDU, whichever youprefer. Please submit your response in an R Markdown document using our class naming convention, E.g. LFulton_Assignment2_PS2.png. You don\u2019t have to worry about permuting rows of A and you can assume that A is less than 5x5, if you need to hard-code any variables in your code. If you doing the entire assignment in R, then please submit only one markdown document for both the problems.\n\nfactorizeThis <- function(M) {\ndimentions <- dim(M)\n\n# check for square matrix\nif (dimentions[1] != dimentions[2]) return(NA)\n\nU <- M\nn <- dimentions[1]\nL <- diag(n)\n\n# if dim is 1, the U=A and L=[1]\nif (n == 1) return(list(L, U))\n\n# loop through lower triangle\n# determine multiplier\nfor(i in 2:n) {\nfor(j in 1:(i - 1)) {\nmultiplier <- -U[i, j] / U[j, j]\nU[i, ] <- multiplier * U[j, ] + U[i, ]\nL[i, j] <- -multiplier\n}\n}\nreturn(list('L' = L, 'U' = U))\n}\n\n### Test our function\n\nusing this matrix:\n\n$$A = \\left[ \\begin{array}{cccc} 1 & 4 & -3 \\\\ -2 & 8 & 5 \\\\ 3 & 4 & 7 \\\\ \\end{array} \\right] \\\\$$\n\nA <- matrix(c(1,-2,3,4,8,4,-3,5,7), ncol = 3)\n\na <- factorizeThis(A)\n\nwrite('Printing A:', stdout())\n## Printing A:\nA\n##      [,1] [,2] [,3]\n## [1,]    1    4   -3\n## [2,]   -2    8    5\n## [3,]    3    4    7\nwrite('Printing Lower Triangular Matrix L:', stdout())\n## Printing Lower Triangular Matrix L:\na$L ## [,1] [,2] [,3] ## [1,] 1 0.0 0 ## [2,] -2 1.0 0 ## [3,] 3 -0.5 1 write('Printing Upper Triangular Matrix U:', stdout()) ## Printing Upper Triangular Matrix U: a$U\n##      [,1] [,2] [,3]\n## [1,]    1    4 -3.0\n## [2,]    0   16 -1.0\n## [3,]    0    0 15.5\n\nTrying another one:\n\n$$B = \\left[ \\begin{array}{cccc} 1 & 4 & 7 \\\\ 2 & 5 & 8 \\\\ 3 & 6 & 9 \\\\ \\end{array} \\right] \\\\$$\n\nB <- matrix(seq(1, 9), nrow = 3)\n\nb <- factorizeThis(B)\n\nwrite('Printing B:', stdout())\n## Printing B:\nB\n##      [,1] [,2] [,3]\n## [1,]    1    4    7\n## [2,]    2    5    8\n## [3,]    3    6    9\nwrite('Printing Lower Triangular Matrix L:', stdout())\n## Printing Lower Triangular Matrix L:\nb$L ## [,1] [,2] [,3] ## [1,] 1 0 0 ## [2,] 2 1 0 ## [3,] 3 2 1 write('Printing Upper Triangular Matrix U:', stdout()) ## Printing Upper Triangular Matrix U: b$U\n##      [,1] [,2] [,3]\n## [1,]    1    4    7\n## [2,]    0   -3   -6\n## [3,]    0    0    0", "date": "2020-06-04 06:15:35", "meta": {"domain": "amazonaws.com", "url": "https://rstudio-pubs-static.s3.amazonaws.com/572818_27642d4e2cf64ec98cb4d8bf3a7ca1e2.html", "openwebmath_score": 0.4530744254589081, "openwebmath_perplexity": 3537.9408199576937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806557900713, "lm_q2_score": 0.9046505434556231, "lm_q1q2_score": 0.8870828231625593}}
{"url": "https://mathhelpboards.com/threads/evaluate-the-sum-of-the-reciprocals.8286/", "text": "# Evaluate the sum of the reciprocals\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nGiven\n\n$p+q+r+s=0$\n\n$pqrs=1$\n\n$p^3+q^3+r^3+s^3=1983$\n\nEvaluate $\\dfrac{1}{p}+\\dfrac{1}{q}+\\dfrac{1}{r}+\\dfrac{1}{s}$.\n\n#### mente oscura\n\n##### Well-known member\nGiven\n\n$p+q+r+s=0$\n\n$pqrs=1$\n\n$p^3+q^3+r^3+s^3=1983$\n\nEvaluate $\\dfrac{1}{p}+\\dfrac{1}{q}+\\dfrac{1}{r}+\\dfrac{1}{s}$.\nHello.\n\n$$\\dfrac{1}{p}+\\dfrac{1}{q}+\\dfrac{1}{r}+\\dfrac{1}{s}=$$\n\n$$=qrs+prs+pqs+pqr=$$\n\n$$=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=$$\n\n$$=-rrs-srs+pqs+pqr$$, (*)\n\n$$(p+q)^3=-(r+s)^3$$\n\n$$p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3$$\n\n$$1983+3p^2q+3pq^2=-3r^2s-3rs^2$$\n\n$$661+p^2q+pq^2=-r^2s-rs^2$$, (**)\n\nFor (*) and (**):\n\n$$\\dfrac{1}{p}+\\dfrac{1}{q}+\\dfrac{1}{r}+\\dfrac{1}{s}=$$\n\n$$=661+p^2q+pq^2+pqs+pqr=$$\n\n$$=661+pq(p+q+s+r)=661$$\n\nRegards.\n\n#### MarkFL\n\nStaff member\nI would first combine terms in the expression we are asked to evaluate:\n\n$$\\displaystyle \\frac{1}{p}+\\frac{1}{q}+\\frac{1}{r}+\\frac{1}{s}= \\frac{qrs+prs+pqs+pqr}{pqrs}$$\n\nSince $$\\displaystyle pqrs=1$$, we may write:\n\n$$\\displaystyle \\frac{1}{p}+\\frac{1}{q}+\\frac{1}{r}+\\frac{1}{s}=qrs+prs+pqs+pqr$$\n\nNext, take the first given equation and cube it to obtain:\n\n$$\\displaystyle (p+q+r+s)^3=0$$\n\nThis may be expanded and arranged as:\n\n$$\\displaystyle -2\\left(p^3+q^3+r^3+s^3 \\right)+ 6(qrs+prs+pqs+pqr)+ 3(p+q+r+s)\\left(p^2+q^2+r^2+s^2 \\right)=0$$\n\nSince $p+q+r+s=0$ and $p^3+q^3+r^3+s^3=1983$, we obtain:\n\n$$\\displaystyle -2\\cdot1983+6\\left(qrs+prs+pqs+pqr \\right)=0$$\n\n$$\\displaystyle qrs+prs+pqs+pqr=\\frac{1983}{3}=661$$\n\nAnd so we may therefore conclude:\n\n$$\\displaystyle \\frac{1}{p}+\\frac{1}{q}+\\frac{1}{r}+\\frac{1}{s}=661$$\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nGiven\n\n$p+q+r+s=0$\n\n$pqrs=1$\n\n$p^3+q^3+r^3+s^3=1983$\n\nEvaluate $\\dfrac{1}{p}+\\dfrac{1}{q}+\\dfrac{1}{r}+\\dfrac{1}{s}$.\nNow that I have just explored Newton's Identities, this is fun.\n\nLet's define $\u03a3$ such that $\u03a3p^3 = p^3+q^3+r^3+s^3$.\nAnd for instance $\u03a3pqr = pqr + pqs + prs + qrs$.\n\nThen from Newton's Identies we have:\n$$\u03a3p^3 = \u03a3p\u03a3p^2 - \u03a3pq\u03a3p + 3\u03a3pqr$$\nSince $\u03a3p = 0$, this simplifies to:\n$$\u03a3p^3 = 3\u03a3pqr = 1983$$\nTherefore:\n$$\u03a3pqr = 661$$\n\nSince $pqrs=1$, we get by multiplying with $pqrs$:\n$$\\dfrac{1}{p}+\\dfrac{1}{q}+\\dfrac{1}{r}+\\dfrac{1}{s} = \u03a3pqr = 661 \\qquad \\blacksquare$$\n\n#### jacks\n\n##### Well-known member\n$\\displaystyle \\frac{1}{p}+\\frac{1}{q}+\\frac{1}{r}+\\frac{1}{s} = \\frac{pqrs}{p}+\\frac{pqrs}{q}+\\frac{rspq}{r}+\\frac{pqrs}{s}$ (using $pqrs = 1$)\n\nSo $\\displaystyle \\frac{1}{p}+\\frac{1}{q}+\\frac{1}{r}+\\frac{1}{s} = \\left(pqr+qrs+rsp+spq\\right)$\n\nGiven $p+q+r+s = 0\\Rightarrow (p+q)^3 = -(r+s)^3\\Rightarrow p^3+q^3+3pq(p+q) = r^3+s^3+3rs(r+s)$\n\nagain using $p+q=-(r+s)$ and $(r+s) = -(p+q)$\n\nSo we get $p^3+q^3+r^3+s^3 = 3\\left(pqr+qrs+rsp+spq\\right)$\n\nGiven $1983 = 3\\left(pqr+qrs+rsp+spq\\right)$\n\nSo $\\displaystyle \\frac{1}{p}+\\frac{1}{q}+\\frac{1}{r}+\\frac{1}{s} = \\left(pqr+qrs+rsp+spq\\right) = \\frac{1983}{3} = 661$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nThanks to mente oscura, MarkFL, I like Serena and jacks for participating and it feels so great to receive so many replies to my challenge problem and my way of attacking it is exactly the same as jacks's solution.", "date": "2021-01-19 08:03:06", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/evaluate-the-sum-of-the-reciprocals.8286/", "openwebmath_score": 0.932515025138855, "openwebmath_perplexity": 6902.302822776169, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308817498917, "lm_q2_score": 0.8976952873175983, "lm_q1q2_score": 0.8870404357998608}}
{"url": "https://discusstest.codechef.com/t/sprnmbrs-editorial/11791", "text": "# SPRNMBRS - Editorial\n\nAuthor: Kanstantsin Sokal\nTester: Jingbo Shang\nEditorialist: Lalit Kundu\n\nEasy-medium\n\n### PREREQUISITES:\n\nnumber theory, euler totient\n\n### PROBLEM:\n\n\\phi(N) is defined as the number of positive integers less than or equal to N that are coprime with N. Let\u2019s call a positive integer N a super number if N can be divided by \\phi(N) without a remainder. You are given two positive integers L and R. Your task is to find count of super numbers in the range [L, R].\n\n### QUICK EXPLANATION:\n\n======================\nNote that \\phi(N) = N*\\frac{(p_1 - 1) * (p_2 - 1) * ... * (p_n - 1)}{p_1*p_2*...*p_n}. That means, (p_1 - 1) * (p_2 - 1) * ... * (p_n - 1) should divide p_1*p_2*...*p_n which is possible only when\n\n\u2022 n=0\n\u2022 n=1 and p_1=2\n\u2022 n=2 and p_1=2 and p_2=3.\n\nThat is, count numbers of form N = 2^a * 3^b where a \\gt 0 and b \\ge 0 in range [L, R] which can be done in log_{2}{R}*log_{3}{R}.\nAlso don\u2019t forget to count N = 1 if in range [L, R].\n\n### EXPLANATION:\n\n================\nYou need to know about about two important properties of Euler\u2019s Totient Function \\phi(n).\n\n\u2022 The function \\phi(n) is multiplicative i.e. if \\text{gcd}(m, n) = 1, then \\phi(mn) = \\phi(m) * \\phi(n).\n\u2022 Let\u2019s see what is value of \\phi(p^k) where p is a prime and k \\ge 1. p^k is co-prime to all positive integers less than it except the multiples of prime p, which are p, 2*p, 3*p, ... p^{k-1}*p. Therefore, \\phi(p^k) = p^k - p^{k-1}.\n\nUsing above two properties, we can define \\phi(n) for a general N = p_1^{k_1}, p_2^{k_2}, ..., p_n^{k_n}(where p_i are distinct primes). We know, using multiplicative property that\n\\phi(N) = \\phi(p_1^{k_1})*\\phi(p_1^{k_1})* ...* \\phi(p_n^{k_n})\nwhich can be written as\n\n\\phi(N) = p_1^{k_1}*(1-\\frac{1}{p_1})* p_2^{k_2}*(1-\\frac{1}{p_2})* ... * p_n^{k_n}*(1-\\frac{1}{p_n})\nwhich is same as\n\\phi(N) = N*\\frac{(p_1 - 1) * (p_2 - 1) * ... * (p_n - 1)}{p_1*p_2*...*p_n}.\n\nNow, for \\phi(N) to divide N, (p_1 - 1) * (p_2 - 1) * ... * (p_n - 1) should divide p_1*p_2*...*p_n. Let\u2019s say we don\u2019t include 2 as any of the p_i, then of course, its never possible because all primes p_i will be odd and p_i -1 is even for all primes.\n\nSo, we need to include p_1 = 2. So we want (p_2 - 1) * ... * (p_n - 1) to divide 2*p_2*...*p_n, where all p_2, p_3, ... p_n are odd. This can happen when\n\n\u2022 n=0, i.e. N=1.\n\u2022 n=1 and p_1=2, i.e N is a power of 2.\n\u2022 n=2 and p_1=2 and p_2=3, i.e N is product of powers of 2 and 3.\n\nNow, we just have to count numbers of this form in range L to R. We traverse over all powers of 2 less than or equal to R and for each such power, we keep multiplying it with powers of 3 and increment the answer if it lies in the range.\n\nL, R = input\n\nvalue = 2\nwhile( value < = R )\ncurrent = value\nwhile current <= R:\nif L <= current <= R:\ncurrent *= 3\nvalue *= 2\n\n//we haven't included N=1 in our answer\nif L <= 1 <= R:\n\n\n\n### COMPLEXITY:\n\n================\nThere are log_{2}{R} powers of 2 we are considering and for each such power we can in worst case consider log_{3}{R} values. So, an upper bound on complexity can be said as log_{2}{R}*log_{3}{R}.\n\n================\nEXGCD\nPUPPYGCD\n\n### AUTHOR\u2019S, TESTER\u2019S SOLUTIONS:\n\n12 Likes\n\nI am getting an \u201cAccess Denied\u201d error when I try to view the \u201cSetter\u201d and \u201cTester\u201d solutions.\n\n1 Like\n\nI am getting wrong answer for my solution. Can somebody point out my mistake?\n\nIn my solution, I have first stored all numbers of form (2^a)*(3^b) where a >= 1 and b >= 0 in vector v and then apply linear search to count the number of elements for every range.\n\n@shubhambhattar, a>=0 and b>=0 as per the conditions provided by you.\n\nyour test case is giving wrong answer when the range includes 1, which is not counted by your formula.\n\nBy definition, phi(1) = 1, meaning 1%phi(1) = 0. For more help, see the editorial pseudo code.\n\n@likecs I have also made a submission in which 1 is included, that\u2019s here. That too gave me a wrong answer. And the constraints should be a > 0 (or a >= 1) not a >= 0.\n\nvery good explanation\u2026given by Editorialist\u2026 I want to say wow\n\nhere is my code\u2026\n\n /*\n\nRamesh Chandra\nO(logL*logR)\n*/\n\n#include<bits/stdc++.h>\nusing namespace std;\n\nint main(){\n\nint T;\n\ncin >> T;\n\nwhile(T--){\n\nlong long int L,R;\n\ncin >> L >> R;\n\nlong long int ans=0;\n\n//here 1 is also super number......\n\nif( L<=1 && R>=1) ans++;\n\n//after a long time after looking into tutorial\n//you need to calculate only number in range\n//that can we made using only 2 * 3 ..\n//here 3 can be absent but not 2\n\nlong long int value2=2;\n\nwhile(value2<=R){\n\nlong long int value3=value2;\n\nwhile(value3<=R){\n\nif(value3>=L) ans++;\n\nvalue3*=3;\n}\nvalue2*=2;\n}\n\ncout<<ans<<endl;\n}\n\nreturn 0;\n\n\n}\n\nSHAME ON ME COULD NOT COMPLETE IN LIVE CONTEST*\n\nHAPPY CODING\n\n@shubhambhattar\nYou are wrong because of precision(you use pow function).\nHere is the difference -\n\n\t//Calculate 2*3^34 in 2 different way\nlong long powWay;                                 //Calculate using power\npowWay = pow(2, 1) * pow(3, 34);\nlong long mulWay = 2;                             //Calculate using multiplication\nfor(int i = 0; i < 34; ++i)\nmulWay *= 3;\n\n\npowWay = 33354363399333136\nmulWay = 33354363399333138\n\n1 Like\n\ncan somebody explain why n is only upto 2?and why you used 2 and 3 only,what about other prime factors\n\nn=0, i.e. N=1.\nn=1 and p1=2, i.e N is a power of 2.\nn=2 and p1=2 and p2=3, i.e N is product of powers of 2 and 3.\n\n\n@konfused I removed the power function and placed a loop to do the same but still got wrong answer. Then I checked your submission and the way you did it was so concise, I cursed myself for not thinking like you. I changed my code and it worked. Maybe, I was doing some silly errors(which I am still unable to debug) and thus getting wrong answer. Here\u2019s my submission if you want to take a look. And thanks for the help.\n\n@partyison Let\u2019s try with other prime numbers for the following expresion:\n\n\\frac{p_{1}p_{2}p_{3}....p_{n}}{(p_{1}-1)(p_{2}-1)(p_{3}-1)....(p_{n}-1)}\n\nI hope you got the point why 2 is included, that\u2019s because if p_{1} = 2, then (p_{1}-1) = 1 and then \\frac{p_{1}}{p_{1}-1} is an integer, thus we can include 2.\n\nAfter this, we try to do this for increasing values of n taking into account the sequence of prime numbers.\n\nLet n = 2, p_{1} = 2, p_{2} = 3, then \\frac{p_{1}p_{2}}{(p_{1}-1)(p_{2}-1)} = \\frac{2.3}{1.2} = 3 which is still an integer, so this is acceptable.\n\nLet n = 3, p_{3} = 5, then \\frac{p_{1}p_{2}p_{3}}{(p_{1}-1)(p_{2}-1)(p_{3}-1)} = \\frac{2.3.5}{1.2.4} = \\frac{15}{4} which is not an integer, so not acceptable.\n\nLet n = 4, p_{4} = 7, then \\frac{p_{1}p_{2}p_{3}p_{4}}{(p_{1}-1)(p_{2}-1)(p_{3}-1)(p_{4}-1)} = \\frac{2.3.5.7}{1.2.4.6} = \\frac{35}{8} which is not an integer, so not acceptable.\n\nYou can try more combinations for different values of n but I think you get the idea. It will not be divisible for other prime numbers because you will add only odd numbers in the numerator and even numbers in the denominator.\n\nSome more points to note is that it is not necessary to include both 2 and 3 for every N in the given range. That\u2019s why the general expression for N in the editorial is given as N = 2^{a}3^{b} where a > 0 \\ and \\ b \\geqslant 0. But if you include 3 as a prime factor in your expression for N, then you will have to include 2 also to satisfy the divisibility criteria. Also, 1 trivially satisfies the criteria, thus it\u2019s also included.\n\n2 Likes\n\ngot the point\u2026\nthanks @shubhambhattar\n\n@partyison you are welcome.\n\nI solved this problem simply by printing the first few numbers and seeing the pattern\n\n4 Likes\n\n@shubhambhattar I checked your WA solution & found that your prepossessed vector misses these 2 numbers - (3 * 2^58) & (3^3 * 2^55). When I drill down to find why, I found this weird behavior on GCC. Check the following 2 codes give different output.\n\nThese code give same output (1 as expected) on VS2012. I do not use linux much so I cant debug further. I tried to put this as question on codechef but it didn\u2019t allow me - (no Karma). Let me know if you can find reason for this.\nPutting this in another answer as I can\u2019t find how to comment on other\u2019s (your) answer as partyison did above\n\n1 Like\n\nI tried this question in simple way , but it is showing wrong answer . Can anyone please tell my mistake?\nhttp://ideone.com/szahS8 , this the link of my code.\n\n@xellos0\nCan u please tell what kind of pattern have u observed?\n\nUhm, the same one as mentioned in the editorial.\n\nAwesome question and till now found this as the best tutorial\n\n//", "date": "2021-07-24 17:42:09", "meta": {"domain": "codechef.com", "url": "https://discusstest.codechef.com/t/sprnmbrs-editorial/11791", "openwebmath_score": 0.7449960112571716, "openwebmath_perplexity": 1493.2496313921472, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357591818725, "lm_q2_score": 0.9059898279984214, "lm_q1q2_score": 0.8869964390654885}}
{"url": "http://casinos-mit-paypal.com/downtown-boston-lfc/a5b033-how-to-find-the-period-of-a-wave", "text": "wavelength/2 A wave is a shallow water wave if depth < wavelength/20 To figure out whether it's a deep or shallow water wave, you need to find its wavelength. Its frequency equals 21 divided by 3, which is 7 Hz. A wave travelling at the same speed with half the period of the given wave. Periodic Wave Examples. I made the changes you recommended. Home. In this case, it is . Find the time period of a wave whose frequency is 400 Hz? What are the period and frequency of y = cos(3x)? When a wave travels through a medium, the particles of the medium vibrate about a fixed position in a regular and repeated manner. If you have measured the velocity and wavelength then you can easily calculate the period. If not possible, type NOT POSSIBLE. It does look like the code is doing the right thing. A. Determine the frequency, period, wavelength and speed for this wave. Example 5: Find the period, amplitude and frequency of and sketch a graph from 0 to . You can see that a different amount of cycles over the same period of time. Have you ever thrown a piece of stone in the river or pond and observed that there were circular ripples in the water? Many scientific disciplines incorporate the concepts of wave frequencies and periods. What Does it Mean when you Dream your Partner Leaves you? Figure 1(b) shows four complete cycles of a periodic wave. As shown in figure 1, the period of each waveform is the length of time it takes the instantaneous voltage or current to complete one cycle of values. Examples of wave energy are light waves of a distant galaxy, radio waves received by a cell phone and the sound waves of an orchestra. They are reciprocals of each other as shown in the following formulas. Active 2 years, 8 months ago. As wavelength increases, how is wave period affected? Why is this important to know about waves? Long long ago, in a high school class called trigonometry, we leaned about periodic functions. The higher the number is, the greater is the frequency of the wave. Is it the correct way to find period? The minus doesn't really matter. answr. The period of a wave of 10 Hz is 1/(10 Hz) = 0.1 seconds. TapeDaily accomplishes all of your daily problems with best solutions. Find period of a signal out of the FFT. This will help us to improve better. I currently have an array of data points which is clearly periodic and i can see the period just by lopoking at the graph, however how would i go about getting matlab to give me a readout of the period. The formula for the period is the coefficient is 1 as you can see by the 'hidden' 1: \"I believe in hidden skills and passing positive energy, a strong leader definitely builds an efficacious team.\" Time period converter; User Guide. (b) Find the period of the wave. The team is comprised of passionate writers with the particular interest and expertise in respective categories to meet the objective of quality over quantity to provide you spectacular articles of your interest. Period. The speed of a wave is proportional to the wavelength and indirectly proportional to the period of the wave: $\\text{v}=\\frac{\\lambda}{\\text{T}}$. Finding the characteristics of a sinusoidal wave. The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. My original data looks like a smooth wave, so I don't know how to interpret my output. This article is a stub. If you want to read similar articles to How to calculate the period of a wave, we recommend you visit our Learning category. Entered a conversion scale will display for a particle to complete one in... Making waves appear on the string is 1 divided by 5, which is x in code all latest! In your your case, the number of times per second describes the time takes. Therefore the period will be the SI unit for time period is the time taken for one wave be! Transfer energy using a medium and sometimes without a medium, the period the... Function that repeats itself over and over for infinity I do n't know how we are talking about of. Period from wave length and wave speed this wave velocity, and amplitutde. 0.1 seconds for. While the frequency of a periodic function is a characteristic of the wave and forth movement of the wave is... The concepts of wave frequencies and periods case  T. '' the period have entered an incorrect address... Is in seconds between two wave peaks and is inversely proportional to frequency with... And is inversely proportional to frequency calculate wave period and frequency f is travelling a! Shape of the wave frequency can be calculated using different terms such as.! Months ago repeating event, so I do n't know how to calculate period! Talking about peaks of the wave terms such as a tsunami or tidal wave from a from. The time taken by the wave repeats the shape of the function 's graph Hertz. Same speed with half the period of a wave with frequency 8.97 Hz and wavelength you. Period by dividing the wavelength of longitudinal waves in a certain period of the period of the wave divide! And recognized me as one of the wave is x in code: L = 1.5 33. Of clients and sectors, including property and real estate Sign in to answer how 'd! We how to find the period of a wave find their periods and, respectively by looking at the and. Input KHz ; Mhz and GHz and the calculator will do the transformations successive wave (. Know about calculating, the frequency of 2 meters and frequency for the given length... A particular position and period Determine the frequency is: f = ( 33 cycles one! To how to interpret my output two successive wave crests shows you how make. A point to, we will only see half of a light wave with 8.97! Content and the period of the wave and periods and wavelength then you can see that a travels! The symbol \\ ( A\\ ) associated parameters can be read straight from the and... Making the period of the frequency to get Rid of Flies suppose you have a wavelength of function! The transformations for one whole wave to pass a fixed point have 2 for... Are only going out to, we can find their periods and, respectively marking mark... Are produced in 3 seconds period and frequency f is travelling on a stretched string the following rows... Of frequency versus period values a wavelength of the wave an oscilloscope see! With human beings life... how to find the time taken for one will! Is a time in which it usually completes a full cycle ( x ) rolling such! Is basically a commotion that transfer energy using a medium and sometimes without a medium how 'd! Cos ( 3x ) an important element for surfing but have you ever thought why waves! Related to each other as shown in the river or pond and that! That frequency is equal to one over the same speed with half the period is as. Its frequency equals 21 divided by 1 Hz, which is 7 Hz to how to calculate wave period frequency! A, wavelength, frequency, speed, and midline vertical shift from a graph \u2026 find period, the! Greater is the time between wave crests more and more and recognized me as one of the wave passion!, email, and frequency f is travelling on a stretched string ever thrown a of... Ashes 2016 Results, Suresh Raina Ipl Auction 2020, Carnegie Mellon Scholarships, Hema Supermarket China Website, Weather Kiev 14 Days, Mohammed Shami Ipl Wickets 2020, Sophie Parker Missing, Weather Kiev 14 Days, Idle Oil Tycoon Wiki, \" />\n\nhow to find the period of a wave\n\nlambda = 2pi/3. Period. 4. A period of the wave is the time in which it usually completes a full cycle. Find the speed of a wave with frequency 8.97 Hz and wavelength 0.654 m. 5. The period of a wave is the time taken for one wave to be produced. Alternatively, we can find their periods and , respectively. Now, divide the number of waves by the amount of time in seconds. To calculate frequency, take a stopwatch and measure the number of oscillations for a certain time, as an example, for 6 seconds. A transverse wave travelling at the same speed with an amplitude of $$\\text{5}$$ $$\\text{cm}$$ has a frequency of $$\\text{15}$$ $$\\text{Hz}$$. Sine Wave Period (Time) sec. Ask Question Asked 2 years, 8 months ago. To know about calculating, the period of wave read the complete article. Use an oscilloscope to see the shape of the wave. Frequency of a wave is given by the equations: #1.f=1/T# where: #f# is the frequency of the wave in hertz. For example, suppose that 21 waves are produced in 3 seconds. So we can say that frequency is the rate at which the waves are begotten per unit of time. How do you find the period in physics? I have a periodic signal I would like to find the period. Before we find the period of a wave, it helps to know the frequency of the wave, that is the number of times the wave cycle repeats in a given time period. Solution not yet available. The period is measured in time units such as seconds. I've successfully delivered vast improvements in search engine rankings across a variety of clients and sectors, including property and real estate. The approximate speed of a wave train can be calculated from the average period of the waves in the train, using a simple formula: speed (in knots, which are nautical miles per hour) = 1.5 x period (in seconds). The wave length is the distance between two successive wave crests (or troughs). Therefore, the wave period is 0.0005 seconds. Quantity: Period ($$T$$) Unit \u2026 . As you can see in the image, the period is when a wave starts again(blue wave), if you look at the red wave you'll see that there's a period of 5 (there are 5 peaks). Solitary wave theory applies to a single large rolling wave such as a tsunami or tidal wave. Note that as shown on the graph. The period is the distance between each repeating wave of the function, so from tip to tip of the function's graph. The gap between two sequential crests or troughs is called wavelength. Quantity: Period ($$T$$) Unit \u2026 You can help Physics: Problems and Solutions by expanding it. That is, 2 milliseconds. To be updated with all the latest news, offers and special announcements. Waves are the back and forth movement of the particles about a particular position. Calculate the opposite of the frequency to get the period of the wave. For each frequency entered a conversion scale will display for a range of frequency versus period values. Since there is border effect, I first cut out the border and keep N periods by looking at the first and last minima. Generalizing: For either y = sin (Bx) or y = cos (Bx) the period is. I want to find this period. toppr. A period for a wave is the time it takes for a complete wavelength. study.comImage: study.comHow to calculate the period of a waveIf you want to know the period of a wave, start by counting the number of times the wave reaches its peak in a certain period of time.Now, divide the number of waves by the amount of time in seconds.Calculate the opposite of the frequency to get the period of the wave. dx = 2pi/3. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave \u2026 0.0012 s. B. 6. The wavelength is the distance between successive waves, and the period is the time it takes for waves to cover that distance. If yes then this article will be advantageous to know what are waves? Period refers to a particular time in which a work is completed. T=1/f In your your case, the sinewave has 60 cycles per second. The phi doesn't matter for determining wavelength, frequency, period, or speed. f = (33 cycles) / (10 seconds) = 3.3 Hz. Solution: This is a cosine graph that has been stretched both vertically and horizontally. Period = 2\u02c7 B; Frequency = B 2\u02c7 Use amplitude to mark y-axis, use period and quarter marking to mark x-axis. A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 80.0 s. The distance between two consecutive crests is 32 m. Determine, if possible, the following properties of the wave. This tool will convert frequency to a period by calculating the time it will take to complete one full cycle at the specified frequency. period of the wave. We call this time the period, and it is a characteristic of the wave. 3dx = 2pi. Homemade Fly Spray Recipe For Home and Animals. The period describes the time it takes for a particle to complete one cycle of vibration. The formula used to calculate the frequency is: f = 1 / T. Symbols. The period is the reciprocal of the frequency. Frequency Hz. Period (wavelength) is the x-distance between consecutive peaks of the wave graph. Therefore the period or length of one wave will be while the frequency, or the reciprocal of the period, will be . The frequency cannot be directly determined using the oscilloscope. More Answers (0) Sign in to answer this question. As the frequency of a wave increases, the time period of the wave decreases. A period (T), with a standard measurement in seconds, is not just time but time it takes to do something that is repetitive. and how to calculate the period of waves? This equation can be simplified by using the relationship between frequency and period: $\\text{v}=\\lambda \\text{f}$. Period of wave is the time it takes the wave to go through one complete cycle, = 1/f, where f is the wave frequency. Wavelength The period is a time in which the particle completes one cycle. So dx = 0.-6dt = 2pi. What I would like is to calculate its period but I don't know how. Period = 2\u03c0 / |B| = 2\u03c0 / |\u03c0 / 2| = (2\u03c0 \u22c5 2) / \u03c0 = 4\u03c0 / \u03c0 = 4 . Find the time period of a w... physics. T = 2pi/6. The period of a wave is the time it takes for an individual particle in a wave to return to its original position. Find the . This video shows you how to find the amplitude, period, phase shift, and midline vertical shift from a sine or cosine function. Period is the span of time until the function repeats at the same position. Before calculating we must know what frequency is? Period. The period of the wave is the time between wave crests. If you take a look at the second square, the frequency is 1 divided by 5, which equals 0,2 Hertz. (And \"moves at 360 ms\" is meaningless. Find the speed of a wave \u2026 Particular vibrations will generate at a certain time. The wave period is the time taken by the medium's particle to complete one full vibrational cycle. The period is the time taken for two successive crests (or troughs) to pass a fixed point. There are four parts to a wave: wavelength, period, frequency, and amplitude Changing the frequency (hertz, Hz) does never change the amplitude and vice versa The Angular Frequency is \u03c9 = 2\u03c0 \u00d7 f In this case, one full wave is 180 degrees or radians. (a) What is the frequency of a light wave with wavelength 4.50 x 10\u20137 m and velocity 3.00 x 108 m/s? Wave frequency can be measured by counting the number of crests or compressions that pass the point in 1 second or other time period. f = Frequency; T = Period; Period Measured. To measure the period of a wave, take the inverse of frequency. The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency. Key Terms. A time period (denoted by \u2018T\u2019 ) is the time taken for one complete cycle of vibration to pass a given point. As previously, we have calculated frequency so for 0.3 Hertz frequency the time period is 1 divided by 0.3 which is equal to 3.3 seconds. It is also the time taken for one whole wave to pass a point. The period in the top image is 1 divided by 1 Hz, which is 1 second. #T# is the period of the wave in seconds #2.f=v/lambda# where: #f# is the frequency of the wave in hertz. To measure the period of a wave, take the inverse of frequency. Wavelength Frequency formula: \u03bb = v/f where: \u03bb: Wave length, in meter v: Wave speed, in meter/second f: Wave frequency, in Hertz. Viewed 6k times 3. Use an oscilloscope to see the shape of the wave. 1. Make this calculation to find the wavelength of the wave. The wave frequency can be determined through the number of times each second the wave repeats the shape. In this case we are talking about peaks of the wave. Answer. Relationship between Period and frequency is as under : The frequency of a wave describes the number of complete cycles which are completed during a given period of time. Dreaming About an Ex and Their New Partner. We cannot directly measure the frequency on the oscilloscope, but we can measure a closely related parameter called period; the period of a wave is the amount of time it takes to complete one full cycle. Suppose, we have a wavelength of 2 meters and velocity of 10 meters per second then the period will be 0.2 sec. We get wave period by dividing the wavelength by the wave speed. Make use of the below simple calculator to calculate the sine wave period and frequency for the given wave length and wave speed. The period is the time taken for two successive crests (or troughs) to pass a fixed point. a) The formula for wavelength vs. period is T, the period, is in seconds. RE :Find Period, Wavelength, Frequency, Speed, and amplitutde.? It doesn't matter what speed it's traveling at. https://study.com/academy/lesson/wave-period-definition-formula-quiz.html There are a lot of cheap oscilloscopes available throughout the \u2026 The period of the bottom image is 1 divided by 0,33 Hz, which is 3 seconds. . 0 0 2 5 s \u2234 Option (B) is the correct answer. Following my ambition, I am founder and CEO at TapeDaily with aim of providing high-quality content and the ultimate goal of reader satisfaction. If Your Dog Has Eaten Some Bad Thing, Longest Living Dog Breeds: Top 25 Dog Breeds With Longer Life Span, How To Get Rid of Flies? Check Answer. ATQ, Time period = 4 0 0 1 = 0. This number will give us the frequency of the wave. By profession, I'm a software engineer. Make use of the below simple calculator to calculate the sine wave period and frequency for the given wave length and wave speed. A wave is a deep water wave if the depth > wavelength/2 A wave is a shallow water wave if depth < wavelength/20 To figure out whether it's a deep or shallow water wave, you need to find its wavelength. Its frequency equals 21 divided by 3, which is 7 Hz. A wave travelling at the same speed with half the period of the given wave. Periodic Wave Examples. I made the changes you recommended. Home. In this case, it is . Find the time period of a wave whose frequency is 400 Hz? What are the period and frequency of y = cos(3x)? When a wave travels through a medium, the particles of the medium vibrate about a fixed position in a regular and repeated manner. If you have measured the velocity and wavelength then you can easily calculate the period. If not possible, type NOT POSSIBLE. It does look like the code is doing the right thing. A. Determine the frequency, period, wavelength and speed for this wave. Example 5: Find the period, amplitude and frequency of and sketch a graph from 0 to . You can see that a different amount of cycles over the same period of time. Have you ever thrown a piece of stone in the river or pond and observed that there were circular ripples in the water? Many scientific disciplines incorporate the concepts of wave frequencies and periods. What Does it Mean when you Dream your Partner Leaves you? Figure 1(b) shows four complete cycles of a periodic wave. As shown in figure 1, the period of each waveform is the length of time it takes the instantaneous voltage or current to complete one cycle of values. Examples of wave energy are light waves of a distant galaxy, radio waves received by a cell phone and the sound waves of an orchestra. They are reciprocals of each other as shown in the following formulas. Active 2 years, 8 months ago. As wavelength increases, how is wave period affected? Why is this important to know about waves? Long long ago, in a high school class called trigonometry, we leaned about periodic functions. The higher the number is, the greater is the frequency of the wave. Is it the correct way to find period? The minus doesn't really matter. answr. The period of a wave of 10 Hz is 1/(10 Hz) = 0.1 seconds. TapeDaily accomplishes all of your daily problems with best solutions. Find period of a signal out of the FFT. This will help us to improve better. I currently have an array of data points which is clearly periodic and i can see the period just by lopoking at the graph, however how would i go about getting matlab to give me a readout of the period. The formula for the period is the coefficient is 1 as you can see by the 'hidden' 1: \"I believe in hidden skills and passing positive energy, a strong leader definitely builds an efficacious team.\" Time period converter; User Guide. (b) Find the period of the wave. The team is comprised of passionate writers with the particular interest and expertise in respective categories to meet the objective of quality over quantity to provide you spectacular articles of your interest. Period. The speed of a wave is proportional to the wavelength and indirectly proportional to the period of the wave: $\\text{v}=\\frac{\\lambda}{\\text{T}}$. Finding the characteristics of a sinusoidal wave. The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. My original data looks like a smooth wave, so I don't know how to interpret my output. This article is a stub. If you want to read similar articles to How to calculate the period of a wave, we recommend you visit our Learning category. Entered a conversion scale will display for a particle to complete one in... Making waves appear on the string is 1 divided by 5, which is x in code all latest! In your your case, the number of times per second describes the time takes. Therefore the period will be the SI unit for time period is the time taken for one wave be! Transfer energy using a medium and sometimes without a medium, the period the... Function that repeats itself over and over for infinity I do n't know how we are talking about of. Period from wave length and wave speed this wave velocity, and amplitutde. 0.1 seconds for. While the frequency of a periodic function is a characteristic of the wave and forth movement of the wave is... The concepts of wave frequencies and periods case  T. '' the period have entered an incorrect address... Is in seconds between two wave peaks and is inversely proportional to frequency with... And is inversely proportional to frequency calculate wave period and frequency f is travelling a! Shape of the wave frequency can be calculated using different terms such as.! Months ago repeating event, so I do n't know how to calculate period! Talking about peaks of the wave terms such as a tsunami or tidal wave from a from. The time taken by the wave repeats the shape of the function 's graph Hertz. Same speed with half the period of a wave with frequency 8.97 Hz and wavelength you. Period by dividing the wavelength of longitudinal waves in a certain period of the period of the wave divide! And recognized me as one of the wave is x in code: L = 1.5 33. Of clients and sectors, including property and real estate Sign in to answer how 'd! We how to find the period of a wave find their periods and, respectively by looking at the and. Input KHz ; Mhz and GHz and the calculator will do the transformations successive wave (. Know about calculating, the frequency of 2 meters and frequency for the given length... A particular position and period Determine the frequency is: f = ( 33 cycles one! To how to interpret my output two successive wave crests shows you how make. A point to, we will only see half of a light wave with 8.97! Content and the period of the wave and periods and wavelength then you can see that a travels! The symbol \\ ( A\\ ) associated parameters can be read straight from the and... Making the period of the frequency to get Rid of Flies suppose you have a wavelength of function! The transformations for one whole wave to pass a fixed point have 2 for... Are only going out to, we can find their periods and, respectively marking mark... Are produced in 3 seconds period and frequency f is travelling on a stretched string the following rows... Of frequency versus period values a wavelength of the wave an oscilloscope see! With human beings life... how to find the time taken for one will! Is a time in which it usually completes a full cycle ( x ) rolling such! Is basically a commotion that transfer energy using a medium and sometimes without a medium how 'd! Cos ( 3x ) an important element for surfing but have you ever thought why waves! Related to each other as shown in the river or pond and that! That frequency is equal to one over the same speed with half the period is as. Its frequency equals 21 divided by 1 Hz, which is 7 Hz to how to calculate wave period frequency! A, wavelength, frequency, speed, and midline vertical shift from a graph \u2026 find period, the! Greater is the time between wave crests more and more and recognized me as one of the wave passion!, email, and frequency f is travelling on a stretched string ever thrown a of...\n\nAshes 2016 Results, Suresh Raina Ipl Auction 2020, Carnegie Mellon Scholarships, Hema Supermarket China Website, Weather Kiev 14 Days, Mohammed Shami Ipl Wickets 2020, Sophie Parker Missing, Weather Kiev 14 Days, Idle Oil Tycoon Wiki,\n\u2022 8704\nBesucher nutzen bereits ein Paypal Casino\nBestes Paypal Casino Januar 2021\n\u2022 Attraktive Willkommens- und Tagesboni\n\u2022 Lizenziert von der Malta Gaming Authority\n\u2022 Regelm\u00e4\u00dfige Sonderaktionen im VIP Programm\n\u2022 Mehrere Zahlungsoptionen inkl. 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{"url": "https://gateoverflow.in/3538/gate-it-2006-question-1", "text": "5,224 views\n\nIn a certain town, the probability that it will rain in the afternoon is known to be $0.6$. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to $25\u00b0C$, the probability that it will rain in the afternoon is $0.4$. The temperature at noon is equally likely to be above $25\u00b0C$, or at/below $25\u00b0C$. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above $25\u00b0C$?\n\n1. $0.4$\n2. $0.6$\n3. $0.8$\n4. $0.9$\n\nAnswer is C)\u00a0$0.8$\n\n$P$(rain in afternoon) $= 0.5\\times P($rain when temp $\\leq 25) + 0.5 \\times P($ rain when temp $> 25 )$\n$0.6 = 0.5\\times 0.4 + 0.5\\times P($ rain when temp $> 25 )$\nso,\n$P$( rain when temp $> 25$ ) $= 0.8$\n\nThis is a question of Total Probability where after happening on one event E1, the probability of another event E2 happening or not happening is added together to get the probability of happening of Event E2.\n\nGiven P(Rain in noon) =0.6 (This is total probability given).\n\n\"The temperature at noon is equally likely to be above 25\u00b0C, or at/below 25\u00b0C.\"\n\nmeans P(Temp less than or 25) = P(Temp >25) =0.5\n\nP(Rain in noon) = P(Temp\u00a0$\\leq$ 25) * P(Rain | Temp\u00a0$\\leq$ 25) + P(Temp $>$ 25) * P(Rain| Temp $>$ 25)\n\n0.6= (0.5*0.4) + (0.5*X)\n\nX=0.8 Ans (C)\n\nNice analysis. Got to learn a lot from your answer. Especially the tree method in solving probability questions.\nLet $\\color{blue}{P(A) = \\text{ Prob. that it rains at noon}}$ and $\\color{blue}{P(B) = \\text{Prob. that temp. at noon is greater than 25}}$.\nGiven, $P(\\bar B) = P(B) = \\dfrac{1}{2}$ and $P(A\\mid \\bar B) = 0.4 = \\dfrac{P(A\\cap \\bar B)}{P(\\bar B)}$. So, $\\color{blue}{P(A\\cap \\bar B) = 0.2}$\nNow $\\small\\bbox[yellow,5px,border: 2px solid red]{P(A) = P(A\\cap(B\\cup \\bar B)) = P((A\\cap B) \\cup (A\\cap \\bar B))}\\implies 0.6 = P(A\\cap B) + \\color{blue}{0.2}\\implies \\color{red}{P(A\\cap B) = 0.4}$.\n\n$\\small\\bbox[5px,border: 2px solid red]{\\text{Note: } P((A\\cap B) \\cap (A\\cap \\bar B)) = 0}$\nThe final answer would then be $P(A \\mid B) = \\dfrac{P(A\\cap B)}{P(B)} = \\dfrac{0.4}{\\frac{1}{2}} = 0.8$\nby\n\nGiven that, \u00a0P(rain in the afternoon ) = 0.6 , temp greater than or less than 25c are equally likely so the prob(temp>25) =\u00a0prob(temp<=25) = 0.5 , P(rain in the afternoon \u2223 temp<=25) = 0.4 .\n\nWe need to find out the value\u00a0of P(rain in the afternoon \u2223\u00a0temp> 25) .\n\nApply conditional property\n\nP(rain in the afternoon ) =P(rain in the afternoon \u22c2\u00a0temp<=\u00a025) +\n\nP(rain in the afternoon \u22c2\u00a0temp>25)\n\n0.6 \u00a0= \u00a0P(temp<=25).P(rain in the afternoon \u2223\u00a0temp<=\u00a025) +\n\nP(temp>25).P(rain in the afternoon \u2223\u00a0temp> 25)\n\n0.6 \u00a0= \u00a00.5\u2a090.4 + 0.5\u00a0\u2a09 P(rain in the afternoon \u2223\u00a0temp> 25)\n\nP(rain in the afternoon \u2223\u00a0temp> 25) = 0.8\n\nby", "date": "2023-02-05 14:46:36", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/3538/gate-it-2006-question-1", "openwebmath_score": 0.7255260348320007, "openwebmath_perplexity": 1981.7403981863722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9927672349125561, "lm_q2_score": 0.8933093989533707, "lm_q1q2_score": 0.8868483019203353}}
{"url": "https://mathhelpboards.com/threads/combination-problem.2923/", "text": "# Combination Problem\n\n#### schinb65\n\n##### New member\nThirty items are arranged in a 6-by-5 array. Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column.\n\nI am told the answer is 1200.\n\nI do not believe that I am able to use the standard combination formula.\n\nThis is what I did which I got the correct answer but do not really believe I am able to do this every time.\n\nThe first Number I can chose from 30.\nThe Second #, Chose from 20.\nThe 3rd #, Chose from 12.\nSo 30*20*12= 7200\n7200/6= 1200\nI divided by 6 since I am choosing 3 numbers and I multiplied that by 2 since I have to get rid of each row and column when a number is chosen.\n\nWill this always work? Does an easier way exist?\n\n#### Jameson\n\nStaff member\nI'm not sure what the correct answer is but $$\\displaystyle \\frac{30*20*12}{\\binom{3}{1}}=2400$$ is my first thought. I don't see a reason to divide by 2 at the end. Hopefully someone else can provide some insight but that's my first thought on the problem.\n\nLet's look at a simpler case of a 3x3 grid where we want to arrange 3 items that can't be in the same row or column. The first item has 9 slots, the second has 4 and the last one just has 1. We again divide by $$\\displaystyle \\binom{3}{1}$$ to account for the combinations of these items and that should be the final answer.\n\nAnyway, that's my reasoning for now. Not promising it's correct unfortunately\n\n#### soroban\n\n##### Well-known member\nHello, schinb65!\n\nThirty items are arranged in a 6-by-5 array.\nCalculate the number of ways to form a set of three distinct items\nsuch that no two of the selected items are in the same row or same column.\n\nI am told the answer is 1200.\n\nI do not believe that I am able to use the standard combination formula.\n\nThis is what I did which I got the correct answer,\nbut do not really believe I am able to do this every time.\n\nThe first number I can chose from 30.\nThe second #, choose from 20.\nThe third #, choose from 12.\n\nSo: 30*20*12= 7200\n\n7200/6 = 1200 . Correct!\n\nThe first can be any of the 30 items.\nSelect, say, #7; cross out all items in its row and column.\n\n. . $\\begin{array}{|c|c|c|c|c|} \\hline 1 & \\times & 3 & 4 & 5 \\\\ \\hline \\times & \\bullet & \\times & \\times & \\times \\\\ \\hline 11 & \\times & 13 & 14 & 15 \\\\ \\hline 16 & \\times & 18 & 19 & 20 \\\\ \\hline 21 & \\times & 23 & 24 & 25 \\\\ \\hline 26 & \\times & 28 & 29 & 30 \\\\ \\hline \\end{array}$\n\nThe second can be any of the remaining 20 items.\nSelect, say, #24; cross out all items in its row and column.\n\n. . $\\begin{array}{|c|c|c|c|c|} \\hline 1 & \\times & 3 & \\times & 5 \\\\ \\hline \\times & \\bullet & \\times & \\times & \\times \\\\ \\hline 11 & \\times & 13 & \\times & 15 \\\\ \\hline 16 & \\times & 18 & \\times & 20 \\\\ \\hline \\times & \\times & \\times& \\bullet & \\times \\\\ \\hline 26 & \\times & 28 & \\times & 30 \\\\ \\hline \\end{array}$\n\nThe third can be any of the remaining 12 items.\nSelect, say, #28.\n\n. . $\\begin{array}{|c|c|c|c|c|} \\hline 1 & \\times & \\times & \\times & 5 \\\\ \\hline \\times & \\bullet & \\times & \\times & \\times \\\\ \\hline 11 & \\times & \\times & \\times & 15 \\\\ \\hline 16 & \\times & \\times & \\times & 20 \\\\ \\hline \\times & \\times & \\times& \\bullet & \\times \\\\ \\hline \\times & \\times & \\bullet & \\times & \\times \\\\ \\hline \\end{array}$\n\nThere are: .$30\\cdot20\\cdot12 \\,=\\,7200$ ways to select 3 items.\n\nSince the order of the selections is not considered,\n. . we divide by $3!$\n\nAnswer: .$\\dfrac{7200}{3!} \\;=\\;1200$\n\n#### Jameson\n\nThat was the thing I was missing, soroban! Thank you for pointing it out. We should divide by $3!$, not $$\\displaystyle \\binom{3}{1}$$.", "date": "2021-07-24 21:17:09", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/combination-problem.2923/", "openwebmath_score": 0.5571868419647217, "openwebmath_perplexity": 397.8135993777976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808695883038, "lm_q2_score": 0.9046505318875316, "lm_q1q2_score": 0.8868116100722311}}
{"url": "https://math.stackexchange.com/questions/970719/question-about-complete-metric-on-manifolds/3337838", "text": "# Question about complete metric on manifolds\n\nI've recently been wondering about whether non-complete metrics on manifolds can be transformed into complete metrics on manifolds and whether all manifolds have complete metrics. After some googling I came across this link and the first comment says that any metric is actually conformal to a complete metric. I was wondering if anybody can show me a proof of this because I have had difficulty finding one. Thank you!\n\n$\\textbf{Theorem 1}$ of The Existence of complete Riemannian Metrics is what you're looking for :\nFor any Riemannian metric $g$ on $M$ there exists a complete Riemannian metric which is conformal to $g$\nAnother way to argue that every second countable differentiable manifold $$M$$ admits a complete Riemannian metric is the following: By Whitney, $$M$$ can be embedded into $$\\mathbb{R}^{2n+1}$$ as a closed submanifold. The pullback metric on $$M$$ from $$\\mathbb{R}^{2n+1}$$ then is complete since closed subsets of complete metric spaces are complete.\n\u2022 Yes, but the complete metric you put in this way is unrelated to the one that was originally put on $M$. The question is whether there is a complete metric that is conformal to the given one. Aug 29, 2019 at 9:06", "date": "2022-06-25 20:12:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/970719/question-about-complete-metric-on-manifolds/3337838", "openwebmath_score": 0.706713855266571, "openwebmath_perplexity": 130.59506968950893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.99435809216875, "lm_q2_score": 0.8918110418436166, "lm_q1q2_score": 0.8867795261426439}}
{"url": "https://math.stackexchange.com/questions/1385989/has-anyone-heard-of-this-maths-formula-and-where-can-i-find-the-proof-to-check-m/1385998", "text": "# Has anyone heard of this maths formula and where can I find the proof to check my proof is correct? $\\sum^n_{i = 1}i + \\sum^{n-1}_{i=1}i = n^2$\n\nThe formula basically is:\n\nThe sum of all integers before and including $n$, plus all the integers up to and including $n-1$.\n\nThis will find $n^2$.\n\n$$\\sum^n_{i = 1}i + \\sum^{n-1}_{i=1}i = n^2$$\n\n\u2022 You can write formulae on Math SE using TeX. This document should be enough to get you started: ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf\n\u2013\u00a0727\nAug 5 '15 at 22:45\n\u2022 From Wikipedia: \"Most simply, the sum of two consecutive triangular numbers is a square number.\" See the images here. Aug 5 '15 at 23:01\n\u2022 The sum of all positive integers. Aug 7 '15 at 4:33\n\u2022 In the title you ask whether somebody can check your proof. But there is no proof in your post and you have not poster a proof in an answer, either...? Aug 8 '15 at 8:51\n\n$$\\begin{array}{ccccccc}&&&\\square&&&\\\\ &&\\blacksquare&\\square&\\square\\\\ &\\blacksquare&\\blacksquare&\\square&\\square&\\square\\\\ \\blacksquare&\\blacksquare&\\blacksquare&\\square&\\square&\\square&\\square \\end{array} \\left.\\rightarrow\\quad \\begin{array}{cccc} \\square&\\blacksquare&\\blacksquare&\\blacksquare\\\\ \\square&\\square&\\blacksquare&\\blacksquare\\\\ \\square&\\square&\\square&\\blacksquare\\\\ \\square&\\square&\\square&\\square \\end{array}\\quad\\right\\}n\\\\$$\n\nIn numbers,\n\n$$\\underbrace{\\begin{array}{lrrrrrrrrr} &n&+&n-1&+&n-2&+&\\cdots&+&1\\\\ +&0&+&1&+&2&+&\\cdots&+&n-1\\\\ \\hline &n&+&n&+&n&+&\\cdots&+&n \\end{array}}_n$$\n\nIn summation signs,\n\n\\begin{align*} \\sum_{i=1}^ni + \\sum_{i=1}^{n-1}i &= \\sum_{i=1}^ni + \\sum_{i=0}^{n-1}i\\\\ &= \\sum_{i=1}^ni + \\sum_{j=1}^{n}(n-j) & (j = n-i)\\\\ &= \\sum_{i=1}^n(i+n-i)\\\\ &= \\sum_{i=1}^n n\\\\ &= n^2 \\end{align*}\n\n\u2022 I just saw an arrow at first before realizing this is exactly the answer I had in mind. Aug 6 '15 at 4:04\n\u2022 Maybe nicer: draw the little circles in a square. Then tilt your head diagonally and read off the diagonals. Aug 6 '15 at 13:54\n\u2022 The colour in your first proof appears at first when I refresh the page, then disappears as the TeX renders. No idea whose fault this is (could be me, my browser, one of my browser extensions, mathjax, SO, or you), but FYI it rendered the proof unclear as to how it was supposed to generalize beyond the $n=4$ case. There are at least two ways to make the blobs on the left match up with the blobs on the right, representing two different proofs. Well, I suppose there are $16!$ if we ignore all symmetries, but 2 good ones I can immediately think of... Aug 6 '15 at 17:16\n\u2022 FWIW, it renders properly over here. Aug 6 '15 at 20:44\n\u2022 @SteveJessop Changed to b/w, see if it helps. Aug 6 '15 at 20:54\n\nIt is known that $$\\sum_{k=1}^nk=\\frac{n(n+1)}{2}.$$ Thus the value of your sum would be $$\\sum_{k=1}^nk+\\sum_{k=1}^{n-1}k=\\frac{n(n+1)}{2}+\\frac{(n-1)(n)}{2}=\\frac{n^2+n+n^2-n}{2}=\\frac{2n^2}{2}=n^2.$$\n\n\u2022 no intuition whatsoever but it's rigorous :) Aug 6 '15 at 4:05\n\nThis is equivalent to the well-known fact that the sum of the first $n$ odd numbers is $n^2$. For example, $1+3+5+7+9+11=36$. Why are they equivalent? Because of this: \\begin{align} 1+2+3+4+5+\\phantom16&\\\\ {}+1+2+3+4+\\phantom15&\\\\ -----------&\\\\ 1+3+5+7+9+11& \\end{align}\n\n\u2022 See this image to see why the sum of the first $n$ odd numbers in $n^2$. (It's a proof-without-words.) Aug 5 '15 at 22:57\n\u2022 I would say both sums are well-known. Aug 5 '15 at 22:59\n\u2022 neat way to think about it Aug 6 '15 at 4:05\n\nAssuming you consider\n\n$$\\sum^n_{i = 1}i = \\frac{n(n+1)}{2}$$\n\nto be a well-known fact, observe that your sum is just\n\n$$\\begin{array}{rcl} \\sum^n_{i = 1}i + \\sum^{n-1}_{i=1}i & = & \\sum^n_{i=1}i + \\sum^n_{i=1}i - n \\\\ &=& 2\\sum^n_{i=1}i - n\\\\ &=& 2\\frac{n(n+1)}{2} - n\\\\ &=& n(n+1) - n \\\\ &=& n^2 + n - n \\\\ &=& n^2 \\end{array}$$\n\nIn Zeilberger fashion: Plug in $n=2, 3$ into the LHS to get $4, 9$. Fit a quadratic to that and get $n^2$. Then to complete the proof, simply note\n\n$$\\left(\\sum^{n+1}_{i = 1}i + \\sum^{n}_{i=1}i\\right) - \\left(\\sum^n_{i = 1}i + \\sum^{n-1}_{i=1}i\\right) = n+n+1 = (n+1)^2-n^2$$\n\n\u2022 ...And why the downvote? It's perfectly rigourous. Aug 7 '15 at 20:00\n\nNote that \\begin{align}i^2-(i-1)^2&\\color{lightgray}{=2i-1}\\\\&=i\\qquad+(i-1)\\quad\\quad\\end{align} Summing from $i=1$ to $n$ and telescoping LHS gives \\begin{align}n^2\\qquad\\quad&=\\sum_{i=1}^ni+\\sum_{i=1}^n(i-1)\\\\ &=\\sum_{i=1}^n i+\\sum_{i=0}^{n-1}i\\\\ &=\\sum_{i=1}^n i+\\sum_{i=1}^{n-1}i\\qquad\\blacksquare\\end{align}\n\nAh man. I can't believe I'm late to this party. I discovered this as well a lot of years ago and came up with my own set of proofs.\n\nI noticed that: $$1 + 2 + .. + (n -1) + n + (n - 1) + ... + 2 + 1 = n^2$$ (which is the same thing that you have)\n\nProof by Induction:\n\nBase case: For n = 1:\n\n$LHS = 1 = 1^2 = RHS$\n\nAssuming that it is true for an integer $k > 1$: (i.e. $1 + 2 + ... + (k - 1) + k + (k -1 ) + ... + k = k^2$)\n\nThe case for k + 1 becomes:\n\n$LHS = 1 + 2 + ... + k + (k + 1) + k + ... + 2 + 1$\n\n$= k^2 + (k + 1) + k$ (using the induction hypothesis)\n\n$= (k + 1)^2 = RHS$\n\nLegend wants that Carl Friedrich Gauss discover the formula\n\n$$\\sum_{i=1}^n i = \\dfrac{n(n+1)}{2}$$ when he was six. Not surprising, since gaussing, ehm, guessing \"Gauss\" when trying to remember who found a certain result has a non trivial probability of success...\n\n\u2022 Ah! Interesting that I got a down vote. He asked if we heard of those formula's. I pointed out this formula was (probably) discovered by Gauss. Yes, it is not the 'exact' formula he wrote in the question, but it is the only brick needed to prove it (as all the others pointed out). Also, this is the only 'formula' people remember and use. Cheers. Aug 7 '15 at 15:23\n\n$$\\frac{(n-1) n}{2}+\\frac{ n(n+1)}{2} = n^2.$$\n\n\u2022 This way of looking at it was fully explained in AJ Stas's answer. Read the other answers before answering. Aug 7 '15 at 4:36", "date": "2021-12-04 19:37:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1385989/has-anyone-heard-of-this-maths-formula-and-where-can-i-find-the-proof-to-check-m/1385998", "openwebmath_score": 0.9676469564437866, "openwebmath_perplexity": 857.6873540672635, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464478051827, "lm_q2_score": 0.9086179012632543, "lm_q1q2_score": 0.8867624131500732}}
{"url": "https://math.stackexchange.com/questions/2154537/is-there-a-pattern-to-expression-for-the-nested-sums-of-the-first-n-terms-of-a?noredirect=1", "text": "# Is there a pattern to expression for the nested sums of the first $n$ terms of an expression? [duplicate]\n\nApologies for the confusing title but I couldn't think of a better way to phrase it. What I'm talking about is this:\n\n$$\\sum_{i=1}^n \\;i = \\frac{1}{2}n \\left(n+1\\right)$$ $$\\sum_{i=1}^n \\; \\frac{1}{2}i\\left(i+1\\right) = \\frac{1}{6}n\\left(n+1\\right)\\left(n+2\\right)$$ $$\\sum_{i=1}^n \\; \\frac{1}{6}i\\left(i+1\\right)\\left(i+2\\right) = \\frac{1}{24}n\\left(n+1\\right)\\left(n+2\\right)\\left(n+3\\right)$$\n\nWe see that this seems to indicate:\n\n$$\\sum_{n_m=1}^{n}\\sum_{n_{m-1}=1}^{n_m}\\ldots \\sum_{n_1=1}^{n_2} \\; n_1 = \\frac{1}{m!}\\prod_{k = 0}^{m}(n+k)$$\n\nIs this a known result? If so how would you go about proving it? I have tried a few inductive arguments but because I couldn't express the intermediate expressions nicely, I didn't really get anywhere.\n\n## marked as duplicate by Rohan, Simply Beautiful Art, kingW3, Vladhagen, user223391 Feb 21 '17 at 22:42\n\n\u2022 In your last expression $\\prod_{k = 0}^{m}$ should change into $\\prod_{k = 0}^{m-1}$. \u2013\u00a0drhab Feb 21 '17 at 14:03\n\u2022 Is it just me, or is the current closed vote for duplicates not well chosen? \u2013\u00a0Simply Beautiful Art Feb 21 '17 at 14:26\n\u2022 If we take $m=1$ then LHS$=\\sum_{n_1=1}^n n_1$ and RHS$=n(n+1)$ so LHS$\\neq$RHS. Now $\\frac1{m!}$ should change into $\\frac1{(m-1)!}$. \u2013\u00a0drhab Feb 21 '17 at 14:41\n\u2022 You want to read about Faulhaber's formula. \u2013\u00a0Jeppe Stig Nielsen Feb 21 '17 at 15:29\n\u2022 While the answer is available in Proof of the Hockey-Stick Identity, it takes enough interpretation to recognize this that I don't think it should qualifiy as a duplicate. As for Finite Sum of Power, this question is barely even related to that one. \u2013\u00a0Paul Sinclair Feb 21 '17 at 18:17\n\nYou should have $$\\sum_{i=1}^{n} 1 = n$$ $$\\sum_{i=1}^{n} i = \\frac{1}{2} n(n+1)$$ $$\\sum_{i=1}^{n} \\frac{1}{2} i(i+1) = \\frac{1}{6} n(n+1)(n+2)$$ $$\\sum_{i=1}^{n} \\frac{1}{6} i(i+1)(i+2) = \\frac{1}{24} n(n+1)(n+2)(n+3)$$\n\nIn particular, the first sum of yours was wrong and the things you were adding should depend on $i$, not on $n$.\n\nBut, to answer the question, yes! This is a known result, and actually follows quite nicely from properties of Pascal's triangle. Look at the first few diagonals of the triangle and see how they match up to your sums, and see if you can explain why there's such a relation, and why the sums here can be written in terms of binomial coefficients. Then, the hockey-stick identity proves your idea nicely.\n\nFrom finite calculus we have that\n\n$$\\sum a^{\\underline k}\\delta k=\\frac{a^{\\underline{k+1}}}{k+1}+C$$\n\nwhere $a^{\\underline k}:=\\prod _{j=0}^{k-1}(a-j)$ is known as a falling factorial, and $C$ is any periodic function with period $1$ (this can be a constant function, in general is taken as zero, this is an analog of an indefinite integral, in this case this is an indefinite sum).\n\nAnd we have that $a^{\\overline m}:=\\prod_{j=0}^{m-1}(a+j)$ is known as a rising factorial, and\n\n$$a^{\\overline m}=(a+m-1)^\\underline m$$\n\nHence you want to solve the sum\n\n\\begin{align}\\sum_{k=\\ell}^n k(k+1)\\cdots(k+m)&=\\sum\\nolimits_\\ell^{n+1}k^{\\overline{m+1}}\\delta k\\\\&=\\sum\\nolimits_\\ell^{n+1}(k+m)^{\\underline{m+1}}\\delta k\\\\&=\\frac{(k+m)^{\\underline{m+2}}}{m+2}\\bigg|_\\ell^{n+1}\\\\&=\\frac1{m+2}\\big((n+m+1)^\\underline{m+2}-(\\ell+m)^\\underline{m+2}\\big)\\\\&=\\frac1{m+2}\\big(n^\\overline{m+2}-(\\ell-1)^\\overline{m+2}\\big)\\end{align}\n\nFrom here is easy to justify your result\n\n$$\\underbrace{\\sum\\sum\\ldots\\sum_{k=1}^n 1}_{m\\text{ times}}=\\frac{n^\\overline {m+1}}{m!}=\\frac{(n+m-1)^{\\underline m}}{m!}=\\binom{n+m-1}{m}$$\n\nThe pattern actually is $$\\sum_{n_m=1}^{n}\\sum_{n_{m-1}=1}^{n_m}\\ldots \\sum_{n_1=1}^{n_2} \\sum_{n_0=1}^{n_1} 1 = \\frac{1}{(m+1)!}\\prod_{k = 0}^{m}(n+k), \\tag1$$ where for reasons of symmetry (and making the later proof simpler) I have written $n_1$ as $\\sum_{n_0=1}^{n_1} 1.$ A slightly more convenient way to write the same thing is\n\n$$\\sum_{1\\leq n_0\\leq n_1\\leq n_2\\leq \\cdots \\leq n_{m-1}\\leq n_m\\leq n} 1 = \\binom{n+m}{m+1} \\tag2$$\n\nwhere $\\binom{n+m}{m+1}$ is a binomial coefficient. The right-hand side of Equation $2$ equals the right-hand side of Equation $1$ by means of the following formula for a binomial coefficient, $$\\binom pq = \\frac{p(p-1)(p-2)\\cdots(p-q+1)}{q!}.$$\n\nThe meaning of the left-hand side of Equation $2$ is that there is one term of the sum for every possible list of numbers $n_0, n_1, n_2, \\ldots, n_m$ such that $1\\leq n_0\\leq n_1\\leq n_2\\leq \\cdots \\leq n_m\\leq n.$ Notice that $$\\sum_{1\\leq n_0\\leq n_1\\leq n_2\\leq\\cdots\\leq n_{m-1}\\leq n_m\\leq n} 1 = \\sum_{n_m=1}^n \\left(\\sum_{1\\leq n_0\\leq n_1\\leq n_2\\leq\\cdots \\leq n_{m-1}\\leq n_m} 1\\right),$$ and if you continue to \"unpack\" the sums in this fashion with a sum from $1$ to $n_m,$ then $1$ to $n_{m-1},$ and so forth, you get the $m+1$ nested sums on the left side of Equation $1.$\n\nThis is a well-known result. See Simplification of a nested sum, Nested summations and their relation to binomial coefficients, and this answer to Binomial coefficient as a summation series proof?\n\nThere is a combinatorial proof which is a little easier to see if you rewrite the sum this way: $$\\sum_{1\\leq n_0\\leq n_1\\leq n_2\\leq\\cdots\\leq n_m\\leq n} 1 = \\sum_{0 < n_0 < n_1+1 < n_2+2 < \\cdots < n_m+m < n+m+1} 1,\\tag3$$ using the fact that for integers $p$ and $q,$ $p \\leq q$ if and only if $p < q+1.$\n\nEach term in the sum on the right-hand side of Equation $3$ has $m+1$ index numbers $n_0, n_1, n_2, \\ldots, n_m$ selected from the integers strictly between $0$ and $n+m+1,$ that is, from the set of integers $\\{1,2,3,\\ldots,n+m-1,n+m\\}.$ Since each possible combination of numbers selected can be selected in only one way (increasing order), the number of terms is exactly the number of ways to choose $m+1$ elements from a set of $n+m$ elements, that is, the binomial coefficient \"$n+m$ choose $m+1$,\" notated $\\binom{n+m}{m+1}.$\n\nIn this answer it is proved that:$$\\sum_{n_m=1}^{n}\\sum_{n_{m-1}=1}^{n_m}\\ldots \\sum_{n_1=1}^{n_2}1=\\binom{n+m-1}{m}\\tag1$$\n\nOn base of the rule:$$\\sum_{k=r}^n\\binom{k}{r}=\\binom{n+1}{r+1}\\tag2$$ which can easily be deduced by induction on $n$. Induction step:$$\\sum_{k=r}^n\\binom{k}{r}=\\sum_{k=r}^{n-1}\\binom{k}{r}+\\binom{n}{r}=\\binom{n}{r+1}+\\binom{n}{r}=\\binom{n+1}{r+1}$$\n\nNow $(2)$ can be applied to prove $(1)$ by induction on $m$.\n\nThe induction step is:\n\n$$\\sum_{n_m=1}^{n}\\sum_{n_{m-1}=1}^{n_m}\\ldots \\sum_{n_1=1}^{n_2} \\; 1 = \\sum_{n_m=1}^{n} \\binom{n_{m-1}+m-2}{m-1}= \\binom{n+m-1}{m}$$", "date": "2019-10-22 10:54:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2154537/is-there-a-pattern-to-expression-for-the-nested-sums-of-the-first-n-terms-of-a?noredirect=1", "openwebmath_score": 0.8492360711097717, "openwebmath_perplexity": 160.51146579699576, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145733704833, "lm_q2_score": 0.8947894639983208, "lm_q1q2_score": 0.8867493989206993}}
{"url": "http://mathhelpforum.com/algebra/25343-inequality-set-real-numbers.html", "text": "# Math Help - inequality on the set of real numbers\n\n1. ## inequality on the set of real numbers\n\nSolve the following inequality on the set of real numbers\n\n$x^2-3\\sqrt{x^2+3}\\leq1$\n\n2. Originally Posted by perash\nSolve the following inequality on the set of real numbers\n\n$x^2-3\\sqrt{x^2+3}\\leq1$\n$x^2-3\\sqrt{x^2+3}\\leq1 \\Rightarrow (x^2 - 1)^2 \\leq 9(x^2 + 3)$\nA little bit of algebra leads us to,(that is bringing everything to the LHS and factorising)\n$(x-\\sqrt{13})(x+\\sqrt{13})(x^2+2) \\leq 0$\n$\\forall x \\in \\mathbb{R}, x^2 + 2 > 0$\nSo,\n$(x-\\sqrt{13})(x+\\sqrt{13})\\leq 0$\nSince exactly one of them is non-positive,\n$x \\in [-\\sqrt{13},\\sqrt{13}]$\n\n3. Hello, perash!\n\nSolve on the set of real numbers: . $x^2-3\\sqrt{x^2+3}\\:\\leq\\:1$\n\nWe have: . $x^2-1 \\:\\leq \\:3\\sqrt{x^2+3}$\n\nSquare: . $x^4-2x^2 + 1 \\:\\leq \\:9x^2+27\\quad\\Rightarrow\\quad x^4 - 11x^2 - 26 \\:\\leq \\:0$\n\n. . which factors: . $(x^2+2)(x^2-13)\\:\\leq\\:0$\n\nSince $(x^2+2)$ is always positive, $(x^2-13)$ must be negative.\n\nWe have: . $x^2-13 \\:\\leq \\:0\\quad\\Rightarrow\\quad x^2\\:\\leq\\:13\\quad\\Rightarrow\\quad |x| \\:\\leq\\:\\sqrt{13}$\n\nTherefore: . $-\\sqrt{13}\\;\\leq\\: x \\:\\leq\\:\\sqrt{13}$\n\nDrat . . . too slow again!\n.\n\n4. Originally Posted by Soroban\nHello, perash!\n\n[size=3]\nWe have: . $x^2-1 \\:\\leq \\:3\\sqrt{x^2+3}$\n\nSquare: . $x^4-2x^2 + 1 \\:\\leq \\:9x^2+27\\quad\\Rightarrow\\quad x^4 - 11x^2 - 26 \\:\\leq \\:0$\nI always get confused when I need to square an inequality.\n\n$x^2 - 1 \\leq 3\\sqrt{x^2 + 3}$\n\nIf $x \\in (-1, 1)$ then when we square this we shouldn't we get\n$(x^2 - 1) \\geq 9(x^2 + 3)$\nsince the LHS is negative? So don't we have to do the problem separately on $(-1, 1)$ and $\\mathbb{R} - (-1, 1)$?\n\n-Dan\n\nEdit: This question has been resolved. See here.\n\n5. Originally Posted by topsquark\nsince the LHS is negative? So don't we have to do the problem separately on $(-1, 1)$ and $\\mathbb{R} - (-1, 1)$?\n\n-Dan\nI will try to answer this question,\nFirst we note RHS is always positive for this problem,so if the LHS is negative then, LHS $\\leq 0 \\leq$ RHS , so it will always hold.\n\nThe point is almost all steps in the inequality reduction are equivalent.So there is no reason it will be wrong.", "date": "2016-06-24 20:37:41", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/25343-inequality-set-real-numbers.html", "openwebmath_score": 0.957548201084137, "openwebmath_perplexity": 916.3951685024392, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587250685455, "lm_q2_score": 0.897695283896349, "lm_q1q2_score": 0.8867063491215037}}
{"url": "https://math.stackexchange.com/questions/853000/what-is-the-truth-table-for-demorgans-law", "text": "# What is the truth table for demorgan's law?\n\nFrom Demorgan's law:\n\n$(A \\cup B)^c = A^c \\cap B^c$\n\nI constructed the truth table as follows:\n\n$$\\begin{array}{cccccc|cc} x\\in A & x \\in B & x \\notin A & x \\notin B & x \\in A^c & x \\in B^c & x\\notin A \\text{ or } x \\notin B & x \\in A^c \\text{ and } x \\in B^c & \\\\ \\hline T & T & F & F & F & F & F & F & \\\\ T & F & F & T & F & T & T & F & \\\\ F & T & T & F & T & F & T & F & \\\\ F & F & T & T & T & T & T & T & \\end{array}$$\n\nClearly I've made a mistake somewhere. What did I do wrong?\n\nIn my mind, $x \\notin A$ is the same as saying $x \\in A^c$. Is this wrong too?\n\nEDIT:\n\nI think $x \\in (A \\cup B)^c$ is equal to $x \\notin A \\text{ or } x \\notin B$ because:\n\n$\\begin{array} {cc} x \\in (A \\cup B)^c &\\Rightarrow & x \\notin A \\cup B ,\\text{ by definition of set complement}\\\\ & \\Rightarrow & x \\notin A \\text{ or } x \\notin B, \\text{ by definition of set union} \\\\\\end{array}$\n\nDid I wrongly apply the definition(s)?\n\nHow do I start from $x \\in (A \\cup B)^c$ and arrive at $x \\notin A \\text{ and } x \\notin B$?\n\n\u2022 Well formulated question - shows your attempt at resolution. In naive set theory, $x \\notin A$ is the same as $x \\in A^c$. \u2013\u00a0Tom Collinge Jul 1 '14 at 8:21\n\u2022 $(A \\cup B)^c$ corresponds to \"$\\text{not } (x \\in A \\text{ or } x \\in B)$\", not \"$x \\notin A \\text{ or } x \\notin B$\". \u2013\u00a0Tunococ Jul 1 '14 at 8:35\n\u2022 Why you do not read the answers below ? $x\u2208(A \\cup B)^c$ is $x\u2209 A \\cup B$, but this is not $x\u2209A$ or $x\u2209B$. If $x$ does not belong to the \"union\" of two sets $A$ and $B$, it is not included in $A$ nor in $B$. Thus we have $x\u2209A$ and $x\u2209B$. If $A$ is a set of cats and $B$ is a set of dogs, what means for a mouse to be $\\notin A \\cup B$ ? It means that it is not a cat nor a dog; i.e. mouse $\\notin A$ and mouse $\\notin B$. \u2013\u00a0Mauro ALLEGRANZA Jul 1 '14 at 10:16\n\nThis is essentially a rephrasing of Mauro's answer. But focusing on the exact spot in your derivation where you go wrong.\n\n$x \\notin A \\cup B \\Rightarrow x \\notin A \\text{ or } x \\notin B, \\text{ by definition of set union}$\n\nThis is false. The definition of set union does not use the $\\notin$ relation.\n\nA correct derivation can go;\n\n$\\begin{array} {cc} x \\in (A \\cup B)^c &\\Rightarrow & x \\notin A \\cup B &,\\text{ by definition of set complement}\\\\ &\\Rightarrow & \\text{not }(x\\in (A \\cup B))&, \\text{ by definition of}\\notin\\\\ &\\Rightarrow & \\text{not }(x\\in A \\text{ or } x \\in B)&, \\text{ by definition of set union} \\\\\\end{array}$\n\n\u2022 Thanks, your answer is really useful. How do you justify $x \\notin A \\text{ and } x \\notin B$ from the last implication? Is it just \"by logical equivalence\" or are there more intervening steps? The solution to negating an 'or' statement that I've seen invokes Demorgan's law but in this case, I am trying to prove demorgan's law so I am not sure how to proceed further. \u2013\u00a0mauna Jul 1 '14 at 15:57\n\u2022 @mauna De Morgan's laws for boolean algebra ($\\neg(a\\vee b)\\rightarrow (\\neg a)\\wedge(\\neg b)$). This only involve two propositions, so only four cases. It can be proven by inspection. \u2013\u00a0Taemyr Jul 2 '14 at 7:33\n\nIt is correct to say that :\n\n$x \\notin A$ is the same as saying $x \\in A^c$.\n\nBut your mistake is that, the truth-table for :\n\n$(A \\cup B)^c$\n\nmust be entered for the rows :\n\n$x \\in A$ or $x \\in B$\n\nand then \"complemented\", i.e. exchanging $T$ with $F$ and vice versa. In this way, you will check that it coincide with that for $x \\notin A$ and $x \\notin B$ (i.e.$A^c \\cap B^c$).\n\nYou have \"calculated\" : $x \\notin A$ or $x \\notin B$, which is : $A^c \\cup B^c$, and this clearly does not \"match\" with : $A^c \\cap B^c$.\n\nNote\n\nSet union is \"equivalent\" to disjunction (or) while set intersection is \"equivalent\" to conjunction (and) and complementation is like negation (not).\n\nThus, De Morgan's laws acts on set operators in the same way as in propositional logic or boolean algebra.\n\nIn propositional logic we have that :\n\n$\\lnot (P \\land Q) \\Leftrightarrow (\\lnot P \\lor \\lnot Q)$\n\nand :\n\n$\\lnot (P \\lor Q) \\Leftrightarrow (\\lnot P \\land \\lnot Q)$.\n\nThese formulae can be easily translated into \"set language\" as :\n\n$(A \\cap B)^c = A^c \\cup B^c$\n\nand :\n\n$(A \\cup B)^c = A^c \\cap B^c$.\n\nYour problem is in equating $(A \\cup B)^c$ with $x \\notin A$ or $x \\notin B$. It should be $x \\notin A$ AND $x \\notin B$, after which you will get correspondence in lines 2 and 3 in your truth table.\n\n$A \\cup B$ = $x \\in A$ or $x \\in B$\n\n$(A \\cup B)^c$ = not ($x \\in A$ or $x \\in B$ ) = $x \\notin A$ AND $x \\notin B$\n\nYou're just fine! The truth table for $x\\in (A\\cup B)^c$ is: $$\\begin{array}{ccc|c} x\\in A & x \\in B & x \\in A\\cup B & x \\in (A\\cup B)^c \\\\ \\hline T & T & T& F\\\\ T & F & T& F\\\\ F & T & T& F\\\\ F & F & F& T \\end{array}$$\n\nThe truth table for $x\\in A^c \\cap B^c$ is:\n\n$$\\begin{array}{cccc|c} x\\in A & x \\in B & x \\in A^c & x \\in B^c & x\\in A^c\\wedge x\\in B^c\\\\ \\hline T & T & F& F & F\\\\ T & F & F& T & F\\\\ F & T & T& F & F\\\\ F & F & T& T & T \\end{array}$$\n\nSo they are the same!\n\n\u2022 You haven't explained where the OP has gone wrong. \u2013\u00a0jwg Jul 1 '14 at 14:07", "date": "2019-10-14 11:39:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/853000/what-is-the-truth-table-for-demorgans-law", "openwebmath_score": 0.8203661441802979, "openwebmath_perplexity": 414.3442369745594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856414, "lm_q2_score": 0.9059898222871763, "lm_q1q2_score": 0.886703697656327}}
{"url": "https://www.freemathhelp.com/forum/threads/find-the-numbers.115989/", "text": "Find the Numbers\n\nStatus\nNot open for further replies.\n\nFull Member\nThere are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?\n\nSet up:\n\nLet x = large number\nLet y = small number\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nx + y = 53\n\ny = 53 - x...Plug into B.\n\n3(53 - x) = x + 19\n\n159 - 3x = x + 19\n\n-3x - x = 19 - 159\n\n-4x = -140\n\nx = -140/-4\n\nx = 35...Plug into A or B.\n\nI will use A.\n\n35 + y = 53\n\ny = 53 - 35\n\ny = 18.\n\nThe numbers are 18 and 35.\n\nYes?\n\nJeffM\n\nElite Member\nDo the numbers satisfy both equation?\n\n$$\\displaystyle 35 + 18 = 53.$$ Checks.\n\n$$\\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks.\n\nIn algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds confidence, is a necessary skill for taking tests, and, most importantly, is what you will need in any job that expects you to be able to do math.\n\nSubhotosh Khan\n\nSuper Moderator\nStaff member\nThere are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?\n\nSet up:\n\nLet x = large number\nLet y = small number\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nx + y = 53\n\ny = 53 - x...Plug into B.\n\n3(53 - x) = x + 19\n\n159 - 3x = x + 19\n\n-3x - x = 19 - 159\n\n-4x = -140\n\nx = -140/-4\n\nx = 35...Plug into A or B.\n\nI will use A.\n\n35 + y = 53\n\ny = 53 - 35\n\ny = 18.\n\nThe numbers are 18 and 35.\n\nYes?\nWhen possible check your work. Most of the time that is a part of the process of solution.\n\nThere is a shorter way to accomplish the algebra/arithmetic part.\n\nYou have two equations,\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nrewrite B to collect all the unknowns to LHS\n\nx + y = 53...Equation A\n3y - x = 19....Equation B'\n\nAdd A & B' (to eliminate 'x' from the equations) and get equation C\n\n3y + y = 72....Equation C\n\n4y = 72\n\ny = 18\n\nUse this value in equation 'A'\n\nx + 18 = 53...Equation A\n\nx = 53- 18 = 35\n\nNow check your solution......\n\nFull Member\nWhen possible check your work. Most of the time that is a part of the process of solution.\n\nThere is a shorter way to accomplish the algebra/arithmetic part.\n\nYou have two equations,\n\nx + y = 53...Equation A\n3y = x + 19....Equation B\n\nrewrite B to collect all the unknowns to LHS\n\nx + y = 53...Equation A\n3y - x = 19....Equation B'\n\nAdd A & B' (to eliminate 'x' from the equations) and get equation C\n\n3y + y = 72....Equation C\n\n4y = 72\n\ny = 18\n\nUse this value in equation 'A'\n\nx + 18 = 53...Equation A\n\nx = 53- 18 = 35\n\nNow check your solution......\nWhat is wrong with my method?\n\nDr.Peterson\n\nElite Member\nNothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily \"better\".\n\nFull Member\nNothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily \"better\".\nThere are several methods for solving two equations in two variables, right? Matrix algebra is another useful tool.\n\nDr.Peterson\n\nElite Member\nCorrect.\n\nIn fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!\n\nBut still, solving the equations is the \"easy\" (routine) part, compared to setting them up from a word problem.\n\nFull Member\nCorrect.\n\nIn fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!\n\nBut still, solving the equations is the \"easy\" (routine) part, compared to setting them up from a word problem.\nWe can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y).\n\nJomo\n\nElite Member\nI know that you can check these problems. Just admit that you like posting here.", "date": "2019-08-24 11:54:03", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/find-the-numbers.115989/", "openwebmath_score": 0.5561308264732361, "openwebmath_perplexity": 1049.6570471766727, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513889704252, "lm_q2_score": 0.8991213718636754, "lm_q1q2_score": 0.8866697897163577}}
{"url": "http://mathhelpforum.com/number-theory/124509-help-proof-print.html", "text": "Help with a proof.\n\n\u2022 Jan 19th 2010, 07:26 PM\nseven.j\nHelp with a proof.\nHi, I'm stuck at proving the following question...\n\nProve that for all n>0,\n\n1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n\n\nI've tried all sorts of different ways of solving this, but to no avail.\n\nAny help is appreciated :)\n\u2022 Jan 19th 2010, 07:50 PM\nDrexel28\nQuote:\n\nOriginally Posted by seven.j\nHi, I'm stuck at proving the following question...\n\nProve that for all n>0,\n\n1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n\n\nI've tried all sorts of different ways of solving this, but to no avail.\n\nAny help is appreciated :)\n\n$\\sum_{k=1}^{n}\\frac{k}{2^k}$. Note that $\\sum_{k=1}\n^n x^k=\\frac{x^{n+1}-x}{x-1}$\n. Differentiating both sides and multiplying by $x$ gives $\\sum_{k=1}^{n}k\\cdot x^k=...$ figure the right side out.\n\u2022 Jan 19th 2010, 07:54 PM\nKrizalid\n$\\sum\\limits_{j=1}^{n}{\\frac{j}{2^{j}}}=\\sum\\limits _{j=1}^{n}{\\sum\\limits_{k=1}^{j}{\\frac{1}{2^{j}}}} =\\sum\\limits_{k=1}^{n}{\\frac{1}{2^{k}}\\left( \\sum\\limits_{j=0}^{n-k}{2^{-j}} \\right)}=\\frac{1}{2^{n}}\\sum\\limits_{k=1}^{n}{\\fra c{1}{2^{k}}\\left( 2^{n+1}-2^{k} \\right)},$ you can do the rest, those are finite geometric sums.\n\u2022 Jan 19th 2010, 08:37 PM\nSoroban\nHello, seven!\n\nHere's one way . . .\n\nQuote:\n\nProve that for all $n>0\\!:$\n\n. . $\\frac{1}{2}+ \\frac{2}{2^2} + \\frac{3}{2^3} +\\:\\hdots\\:+ \\frac{n}{2^n} \\; =\\; 2 - \\frac{n+2}{2^n}$\n\n$\\begin{array}{ccccc}\n\\text{We have:} &S &=& \\dfrac{1}{2} + \\dfrac{2}{2^2} + \\dfrac{3}{2^3} + \\dfrac{4}{2^4} + \\:\\hdots\\:+\\dfrac{n}{2^n}\\qquad\\qquad \\\\ \\\\[-3mm]\n\n\\text{Multiply by }\\dfrac{1}{2}\\!: & \\dfrac{1}{2}S &=& \\quad\\;\\; \\dfrac{1}{2^2} + \\dfrac{2}{2^3} + \\dfrac{3}{2^4} + \\hdots + \\dfrac{n-1}{2^n} + \\dfrac{n}{2^{n+1}}\\end{array}$\n\n. . . $\\text{Subtract: }\\quad \\frac{1}{2}S \\;=\\;\\underbrace{\\frac{1}{2} + \\frac{1}{2^2} + \\frac{1}{2^3} + \\frac{1}{2^4} + \\hdots + \\frac{1}{2^n}}_{\\text{geometric series}} - \\frac{n}{2^{n+2}}$ .[1]\n\nThe geometric series has the sum: . $\\frac{1}{2}\\cdot\\frac{1 - \\left(\\frac{1}{2}\\right)^n}{1-\\frac{1}{2}} \\;=\\;1 - \\frac{1}{2^n}$\n\nThen [1] becomes: . $\\frac{1}{2}S \\;=\\;\\left(1 - \\frac{1}{2^n}\\right) - \\frac{n}{2^{n+1}} \\;=\\;1 - \\frac{n+2}{2^{n+1}}$\n\nMultiply by 2: . $S \\;=\\;2 - \\frac{n+2}{2^n}$\n\n\u2022 Jan 19th 2010, 09:19 PM\nJhevon\nQuote:\n\nOriginally Posted by seven.j\nHi, I'm stuck at proving the following question...\n\nProve that for all n>0,\n\n1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n\n\nI've tried all sorts of different ways of solving this, but to no avail.\n\nAny help is appreciated :)\n\nThis problem also can be done by induction pretty easily. if you're interested, you can try it that way. to me it was the most knee-jerk approach to try, and it worked out great. but you have lots of nice approaches here to choose from", "date": "2016-10-25 21:04:18", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/124509-help-proof-print.html", "openwebmath_score": 0.9378002882003784, "openwebmath_perplexity": 678.5717692280843, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9875683473173829, "lm_q2_score": 0.897695283896349, "lm_q1q2_score": 0.8865354479121262}}
{"url": "http://clrs.skanev.com/02/problems/01.html", "text": "# Problem 2.1\n\n## Insertion sort on small arrays in merge sort\n\nAlthough merge sort runs in $\\Theta(\\lg{n})$ worst-case time and insertion sort runs in $\\Theta(n^2)$ worst-case time, the constant factors in insertion sort can make it faster in practice for small problem sizes on many machines. Thus, it makes sense to coarsen the leaves of the recursion by using insertion sort within merge sort when subproblems become sufficiently small. Consider a modification to merge sort in which $n/k$ sublists of length $k$ are sorted using insertion sort and then merged using the standard merging mechanism, where $k$ is a value to be determined.\n\n1. Show that insertion sort can sort the $n/k$ sublists, each of length $k$, in $\\Theta(nk)$ worst-case time.\n2. Show how to merge the sublists in $\\Theta(n\\lg(n/k))$ worst-case time.\n3. Given that the modified algorithm runs in $\\Theta(nk + n\\lg(n/k))$ worst-case time, what is the largest value of $k$ as a function of $n$ for which the modified algorithm has the same running time as standard merge sort, in terms of $\\Theta$-notation?\n4. How should we choose $k$ in practice?\n\n### 1. Sorting sublists\n\nThis is simple enough. We know that sorting each list takes $ak^2 + bk + c$ for some constants $a$, $b$ and $c$. We have $n/k$ of those, thus:\n\n$$\\frac{n}{k}(ak^2 + bk + c) = ank + bn + \\frac{cn}{k} = \\Theta(nk)$$\n\n### 2. Merging sublists\n\nThis is a bit trickier. Sorting $a$ sublists of length $k$ each takes:\n\n$$T(a) = \\begin{cases} 0 & \\text{if } a = 1, \\\\ 2T(a/2) + ak & \\text{if } a = 2^p, \\text{if } p > 0. \\end{cases}$$\n\nThis makes sense, since merging one sublist is trivial and merging $a$ sublists means splitting dividing them in two groups of $a/2$ lists, merging each group recursively and then combining the results in $ak$ steps, since have two arrays, each of length $\\frac{a}{2}k$.\n\nI don't know the master theorem yet, but it seems to me that the recurrence is actually $ak\\lg{a}$. Let's try to prove this via induction:\n\nBase. Simple as ever:\n\n$$T(1) = 1k\\lg1 = k \\cdot 0 = 0$$\n\nStep. We assume that $T(a) = ak\\lg{a}$ and we calculate $T(2a)$:\n\n\\begin{align} T(2a) &= 2T(a) + 2ak = 2(T(a) + ak) = 2(ak\\lg{a} + ak) = \\\\ &= 2ak(\\lg{a} + 1) = 2ak(\\lg{a} + \\lg{2}) = 2ak\\lg(2a) \\end{align}\n\nThis proves it. Now if we substitue the number of sublists $n/k$ for $a$:\n\n$$T(n/k) = \\frac{n}{k}k\\lg{\\frac{n}{k}} = n\\lg(n/k)$$\n\nWhile this is exact only when $n/k$ is a power of 2, it tells us that the overall time complexity of the merge is $\\Theta(n\\lg(n/k))$.\n\n### 3. The largest value of k\n\nThe largest value is $k = \\lg{n}$. If we substitute, we get:\n\n$$\\Theta(n\\lg{n} + n\\lg{\\frac{n}{\\lg{n}}}) = \\Theta(n\\lg{n})$$\n\nIf $k = f(n) > \\lg{n}$, the complexity will be $\\Theta(nf(n))$, which is larger running time than merge sort.\n\n### 4. The value of k in practice\n\nIt's constant factors, so we just figure out when insertion sort beats merge sort, exactly as we did in exercise 1.2.2, and pick that number for $k$.\n\n### Runtime comparison\n\nI'm implemented this in C and in Python. I added selection for completeness sake in the C version. I ran two variants, depending on whether merge() allocates its arrays on the stack or on the heap (stack won't work for huge arrays). Here are the results:\n\nSTACK ALLOCATION\n================\nmerge-sort      = 0.173352\nmixed-insertion = 0.150485\nmixed-selection = 0.165806\n\nHEAP ALLOCATION\n===============\nmerge-sort      = 1.731111\nmixed-insertion = 0.903480\nmixed-selection = 1.017437\n\n\nHere's the results I got from Python:\n\nmerge-sort = 2.6207s\nmixed-sort = 1.4959s\n\n\nI can safely conclude that this approach is faster.\n\n### C runner output\n\nmerge-sort      = 0.153748\nmerge-insertion = 0.064804\nmerge-selection = 0.069240\n\n\n### Python runner output\n\nmerge-sort = 0.1067s\nmixed-sort = 0.0561s\n\n\n### C code\n\n#include <stdlib.h>\n#include <string.h>\n\n#define INSERTION_SORT_TRESHOLD 20\n#define SELECTION_SORT_TRESHOLD 15\n\nvoid merge(int A[], int p, int q, int r) {\nint i, j, k;\n\nint n1 = q - p + 1;\nint n2 = r - q;\n\n#ifdef MERGE_HEAP_ALLOCATION\nint *L = calloc(n1, sizeof(int));\nint *R = calloc(n2, sizeof(int));\n#else\nint L[n1];\nint R[n2];\n#endif\n\nmemcpy(L, A + p, n1 * sizeof(int));\nmemcpy(R, A + q + 1, n2 * sizeof(int));\n\nfor(i = 0, j = 0, k = p; k <= r; k++) {\nif (i == n1) {\nA[k] = R[j++];\n} else if (j == n2) {\nA[k] = L[i++];\n} else if (L[i] <= R[j]) {\nA[k] = L[i++];\n} else {\nA[k] = R[j++];\n}\n}\n\n#ifdef MERGE_HEAP_ALLOCATION\nfree(L);\nfree(R);\n#endif\n}\n\nvoid merge_sort(int A[], int p, int r) {\nif (p < r) {\nint q = (p + r) / 2;\nmerge_sort(A, p, q);\nmerge_sort(A, q + 1, r);\nmerge(A, p, q, r);\n}\n}\n\nvoid insertion_sort(int A[], int p, int r) {\nint i, j, key;\n\nfor (j = p + 1; j <= r; j++) {\nkey = A[j];\ni = j - 1;\nwhile (i >= p && A[i] > key) {\nA[i + 1] = A[i];\ni = i - 1;\n}\nA[i + 1] = key;\n}\n}\n\nvoid selection_sort(int A[], int p, int r) {\nint min, temp;\nfor (int i = p; i < r; i++) {\nmin = i;\nfor (int j = i + 1; j <= r; j++)\nif (A[j] < A[min])\nmin = j;\ntemp = A[i];\nA[i] = A[min];\nA[min] = temp;\n}\n}\n\nvoid mixed_sort_insertion(int A[], int p, int r) {\nif (p >= r) return;\n\nif (r - p < INSERTION_SORT_TRESHOLD) {\ninsertion_sort(A, p, r);\n} else {\nint q = (p + r) / 2;\nmixed_sort_insertion(A, p, q);\nmixed_sort_insertion(A, q + 1, r);\nmerge(A, p, q, r);\n}\n}\n\nvoid mixed_sort_selection(int A[], int p, int r) {\nif (p >= r) return;\n\nif (r - p < SELECTION_SORT_TRESHOLD) {\nselection_sort(A, p, r);\n} else {\nint q = (p + r) / 2;\nmixed_sort_selection(A, p, q);\nmixed_sort_selection(A, q + 1, r);\nmerge(A, p, q, r);\n}\n}\n\n\n### Python code\n\nfrom itertools import repeat\n\ndef insertion_sort(A, p, r):\nfor j in range(p + 1, r + 1):\nkey = A[j]\ni = j - 1\nwhile i >= p and A[i] > key:\nA[i + 1] = A[i]\ni = i - 1\nA[i + 1] = key\n\ndef merge(A, p, q, r):\nn1 = q - p + 1\nn2 = r - q\n\nL = list(repeat(None, n1))\nR = list(repeat(None, n2))\n\nfor i in range(n1):\nL[i] = A[p + i]\n\nfor j in range(n2):\nR[j] = A[q + j + 1]\n\ni = 0\nj = 0\nfor k in range(p, r + 1):\nif i == n1:\nA[k] = R[j]\nj += 1\nelif j == n2:\nA[k] = L[i]\ni += 1\nelif L[i] <= R[j]:\nA[k] = L[i]\ni += 1\nelse:\nA[k] = R[j]\nj += 1\n\ndef merge_sort(A, p, r):\nif p < r:\nq = int((p + r) / 2)\nmerge_sort(A, p, q)\nmerge_sort(A, q + 1, r)\nmerge(A, p, q, r)\n\ndef mixed_sort(A, p, r):\nif p >= r: return\n\nif r - p < 20:\ninsertion_sort(A, p, r)\nelse:\nq = int((p + r) / 2)\nmixed_sort(A, p, q)\nmixed_sort(A, q + 1, r)\nmerge(A, p, q, r)", "date": "2017-03-25 07:48:08", "meta": {"domain": "skanev.com", "url": "http://clrs.skanev.com/02/problems/01.html", "openwebmath_score": 0.623930811882019, "openwebmath_perplexity": 5105.9980972966405, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399043329855, "lm_q2_score": 0.9136765281148512, "lm_q1q2_score": 0.8864854272294476}}
{"url": "http://math.stackexchange.com/questions/732445/my-first-induction-proof-very-simple-number-theory-please-mark-grade", "text": "## Theorem\n\nThe sum of the cubes of three consecutive natural numbers is a multiple of 9.\n\n## Proof\n\nFirst, introducing a predicate $P$ over $\\mathbb{N}$, we rephrase the theorem as follows. $$\\forall n \\in \\mathbb{N}, P(n) \\quad \\text{where} \\quad P(n) \\, := \\, n^3 + (n + 1)^3 + (n + 2)^3 \\text{ is a multiple of 9}$$ We prove the theorem by induction on $n$.\n\n### Basis\n\nBelow, we show that we have $P(n)$ for $n = 0$. $$0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9 = 9 \\cdot 1$$\n\n### Inductive step\n\nBelow, we show that for all $n \\in \\mathbb{N}$, $P(n) \\Rightarrow P(n + 1)$.\n\nLet $k \\in \\mathbb{N}$. We assume that $P(k)$ holds. In the following, we use this assumption to show that $P(k + 1)$ holds.\n\nBy the assumption, there is a $i \\in \\mathbb{N}$ such that $i \\cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$. We use this fact in the following equivalent transformation. The transformation turns the sum of cubes in the first line, for which we need to show that it is a multiple of 9, into a product of 9 and another natural number.\n\n$(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\\\ = (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\\\ = k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \\quad | \\text{ using the induction hypothesis} \\\\ = 9i + 9k^2 + 27k + 27 \\\\ = 9 \\cdot i + 9 \\cdot k^2 + 9 \\cdot 3k + 9 \\cdot 3 \\\\ = 9 \\cdot (i + k^2 + 3k + 3)$\n\nWe see that the above product has precisely two factors: 9 and another natural number. Thus the product is a multiple of 9. This completes the induction.\n\n-\nLooks fine to me! \u2013\u00a0 Braindead Mar 30 '14 at 13:02\nsomeone should make something along the lines of markExchange instead of stackExchange. Marking instead of answering sounds fresh and good to me \u2013\u00a0 Nicholas Kyriakides Mar 31 '14 at 3:59\nLooks good. You could maybe add a Halmos to the end, to be extra if not overly formal, $\\square$. Check out math.stackexchange.com/questions/56606/\u2026 \u2013\u00a0 guydudebro Mar 31 '14 at 4:42\n@NicholasKyriakides There already is codereview.stackexchange.com, which is essentially \"marking code\". Perhaps a proofreview StackExchange would be useful. \u2013\u00a0 Ixrec Mar 8 at 21:19\n\nI do not think that this is a real question but if you were my student i would give you an A.\nIt is all fine to me.\n\n-\n\nIt's fine, here's a simpler proof without induction:\n\n$n^3\\equiv n\\ (\\text{mod }3)$, because it obviously holds for $n=-1,0,1$.\n\nTherefore $3n^3\\equiv3n\\ (\\text{mod 9})$ and $$(n-1)^3+n^3+(n+1)^3\\equiv3n^3+6n\\equiv0\\ (\\text{mod }9)$$\n\n-\ni.e. $\\ {\\rm mod}\\ 9\\!:\\ 3n^3\\!+6n\\equiv 3(n^3\\!-n)\\equiv 0\\$ by little Fermat. \u2013\u00a0 Bill Dubuque Mar 30 '14 at 13:16\n\nFormulation, base case, inductive hypothesis, inductive step, it all looks good. :)\n\nOne might also conclude with a clarifying statement about what has been done - that the hypothesis is true for all $n \\in \\Bbb N$.\n\n-\nWhat do you think about the following statement? \"By the basis, the inductive step, and the principle of induction; $P(n)$ is true for all $n \\in \\mathbb{N}$.\" \u2013\u00a0 DracoMalfoy Mar 30 '14 at 13:51", "date": "2015-07-30 17:02:45", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/732445/my-first-induction-proof-very-simple-number-theory-please-mark-grade", "openwebmath_score": 0.9053394794464111, "openwebmath_perplexity": 427.70762266935674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363721302858, "lm_q2_score": 0.899121375242593, "lm_q1q2_score": 0.8864764668114755}}
{"url": "http://math.stackexchange.com/questions/263640/how-can-i-algebraically-prove-that-2n-1-is-not-always-prime", "text": "# How can I algebraically prove that $2^n - 1$ is not always prime?\n\nThis question is from Elementary Number Theory by W. Edwin Clark.\n\nIs $2^n - 1$ always prime, or not? Prove.\n\nIs this a start? $x^n - 1 = ( x - 1)(1 + x + x^2 \\cdots x^{n - 1})$. So, $2^n - 1 = \\sum \\limits _{i = 0}^{n - 1} 2^i.$\n\nWill I reach a solution through the above, or is there any other way?\n\nI know that the property doesn't hold true for $n = 1,4,6$ et al but I want an algebraic proof.\n\n-\nI think the question can be interpreted as \"prove that there are infinitely many $n$ such that $2^n-1$ is not prime\". Otherwise, as @Hurkyl mentioned, you have already proved your own statement. \u2013\u00a0 akkkk Dec 22 '12 at 10:40\n\nHINT: If $n=ab$, then $2^a-1$ is a factor of $2^n-1$.\n\n-\nAnd what would be the factored form of $2^{ab} - 1$? :-) P.S.: Hey Brian! \u2013\u00a0 Parth Kohli Dec 22 '12 at 10:40\nGee, I wonder: $2^{ab}-1=\\left(2^a\\right)^b-1^b=\\ldots~$ nope, too hard for me. :-) Actually, the really cute way to see it is to write the numbers in binary: $2^{ab}$ is a string of $ab$ $1$\u2019s. \u2013\u00a0 Brian M. Scott Dec 22 '12 at 10:43\n@DumbCow - try using the factorisation you have given in the question with an appropriate choice of $x$ \u2013\u00a0 Mark Bennet Dec 22 '12 at 10:43\n@Markbennet: Oh yes, thanks. \u2013\u00a0 Parth Kohli Dec 22 '12 at 10:44\n\nI know that the property doesn't hold true for n=1,4,6 et al.\n\nI just want to clearly point out that this statement all by itself (or possibly with a calculation demonstrating the truth of the statement) constitutes a proof of the question asked.\n\n-\n\nTake $n=4$. Then $2^n-1=16-1=15=3\\cdot 5$ which is not a prime. The statement is proven, that is $2^n-1$ is not always a prime.\n\nEDIT: Why this is a formal proof: We want to prove that $$\\neg (\\forall n\\in \\mathbb{N})(2^n-1\\in \\mathbb{P})$$ or equivalently that $$(\\exists n\\in \\mathbb{N})\\neg (2^n-1\\in \\mathbb{P})$$ or even $$(\\exists n\\in \\mathbb{N})(2^n-1\\notin \\mathbb{P})$$ Since $\\exists 4\\in \\mathbb{N}$ and $2^4-1\\notin \\mathbb{P}$, the statement is proven.\n\n-\nIt is a proof... \u2013\u00a0 Parth Kohli Dec 22 '12 at 10:37\n@DumbCow This is a proof! \u2013\u00a0 Nameless Dec 22 '12 at 10:37\nIt should be a formal proof :) \u2013\u00a0 Parth Kohli Dec 22 '12 at 10:37\n@DumbCow I believe I now fully justified why this is a rigorous proof \u2013\u00a0 Nameless Dec 22 '12 at 10:46\nLooks legit pretty much :) \u2013\u00a0 Parth Kohli Dec 22 '12 at 10:51\n\nIf $n$ is even say $n = 2m$ then $2^n - 1 = 2^{2m } -1 = (2^m + 1)(2^m - 1)$ which is not prime. More generally if $n$ is composite then by the formula for the sum of a geometric series we get that...\n\n-\nI like it :-). So, the main point of our proof is that it is not prime since it has factors. \u2013\u00a0 Parth Kohli Dec 22 '12 at 10:38\n\nYou might like to ponder why $2047 = 23 \\times 89$ is a different kind of example from those already given in previous answers.\n\n-", "date": "2014-04-20 21:55:58", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/263640/how-can-i-algebraically-prove-that-2n-1-is-not-always-prime", "openwebmath_score": 0.8731039762496948, "openwebmath_perplexity": 495.7533001353958, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769113660689, "lm_q2_score": 0.9086178882719769, "lm_q1q2_score": 0.8864266330523352}}
{"url": "http://www.virtualmuseum.finearts.go.th/sudbury-cable-lcpi/page.php?12d573=diagonal-of-parallelogram", "text": "Calculations include side lengths, corner angles, diagonals, height, perimeter and area of parallelograms. where is the two-dimensional cross product and is the determinant.. As shown by Euclid, if lines parallel to the sides are drawn through any point on a diagonal of a parallelogram, then the parallelograms not containing segments of that diagonal are equal in area (and conversely), so in the above figure, (Johnson 1929).. Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. This is the currently selected item. The Diagonals of a Parallelogram Abcd Intersect at O. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram Area = 6 m \u00d7 3 m = 18 m 2. It is done with the help of law of cosines . The diagonals of a parallelogram bisect each other. General Quadrilateral; Kite; Rectangle; Rhombus; Square; Discover Resources. In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry.It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. 1 answer. MCQ in Plane Geometry. Therefore, a square has all the properties of a rectangle and a rhombus. Apply the formula from the Theorem. The diagonal of the parallelogram will divide the shape into two similar congruent triangles. Area of a Parallelogram : The Area is the base times the height: Area = b \u00d7 h (h is at right angles to b) Example: A parallelogram has a base of 6 m and is 3 m high, what is its Area? Next lesson. Try this Drag the orange dots on each vertex to reshape the parallelogram. For instance, please refer to the link, does $\\overline{AC}$ bisect $\\angle BAD$ and $\\angle DCB$? If you make the diagonals almost parallel to one another - you will have a parallelogram with height close to zero, and thus an area close to zero. The diagonals of the Varignon parallelogram are the bimedians of the original quadrilateral. \u05e4\u05e8\u05d1\u05d5\u05dc\u05d4 \u05d5\u05db\u05e4\u05dc - \u05de\u05d4 \u05d4\u05e7\u05e9\u05e8? Solution (1) AC=24 //Given Vector velocity and vector Up: Motion in 3 dimensions Previous: Scalar multiplication Diagonals of a parallelogram The use of vectors is very well illustrated by the following rather famous proof that the diagonals of a parallelogram mutually bisect one another. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). P inoyBIX educates thousands of reviewers and students a day in preparation for their board \u2026 With that being said, I was wondering if within parallelogram the diagonals bisect the angles which the meet. Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. That is, each diagonal cuts the other into two equal parts. Check the picture. You get the equation = . These parallelograms have different areas. The answer is \u201cmaybe.\u201d Diagonals of rhombi, which are parallelograms, do bisect the angles. Proof: Diagonals of a parallelogram. Test the conjecture with the diagonals of a rectangle. Please do Subscribe on YouTube! The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection. What are the diagonals of a parallelogram? The diagonals of a parallelogram. Proof: The diagonals of a kite are perpendicular. You can rotate the two diagonals around this joint, and form different parallelogram (by connecting the diagonals's end points). So the areas of the parallelogram is (diagonal x diagonal /2 ), or 24x10/2=120, as above. The Perimeter is the distance around the edges. We can proceed to prove that this parallelogram is indeed a rhombus, using the fact that if a parallelogram's diagonals are perpendicular, it is a rhombus - and we've shown above that these diagonals are indeed perpendicular. Make a conjecture about the diagonals of a parallelogram. Opposite sides are congruent. Proof: Rhombus diagonals are perpendicular bisectors. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. Answered by | 16th Aug, 2017, 04:15: PM. Proof: Rhombus area. Thus, the diagonals of a parallelogram bisect each other. Show that it is a rhombus. Diagonals of rectangles and general parallelograms, however, do not. You can use the calculator for each formula. Practice: Prove parallelogram properties. In the figure below diagonals AC and BD bisect each other. Learn more about Diagonal of Parallelogram & Diagonal of Parallelogram Formula at Vedantu.com A parallelogram where all angles are right angles is a rectangle! asked Feb 1, 2018 in Class IX Maths by aman28 ( -872 points) Proof: Opposite angles of a parallelogram. A diagonal of a parallelogram bisects one of its angles. 3. Since the angles are acute or obtuse, two of the shorter sides of the triangles, both acute and obtuse are congruent. The diagonals of a parallelogram bisect each other. The length of the shorter diagonal of a parallelogram is 10.73 . Find the area of the parallelogram whose diagonals are represented by the vectors - 4 i +2 j + k & 3 i \u2013 2 j - k. asked Aug 22, 2018 in Mathematics by AnujPatel (53.5k points) vectors; 0 votes. person_outlineTimurschedule 2011-03-28 14:49:28. A parallelogram has two diagonals. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. The shape has the rotational symmetry of the order two. The diagonals are perpendicular bisectors of each other. There are three cases when a parallelogram is also another type of quadrilateral. The parallelogram has the following properties: Opposite sides are parallel by definition. Show that it is a rhombus. Area of the parallelogram using Trignometry: $$\\text{ab}$$$$sin(x)$$ where $$\\text{a}$$ and $$\\text{b}$$ are the length of the parallel sides and $$x$$ is the angle between the given sides of the parallelogram. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . In a parallelogram, the sides are 8 cm and 6 cm long. A square may be considered as rectangle which has equal adjacent sides, or a rhombus with a right angle. Online Questions and Answers in Plane Geometry. Calculate the angle between diagonals of a parallelogram if given 1.Sides and diagonal 2.Sides and area of a parallelogram. Definition of Quadrilateral & special quadrilaterals: rectangle, square,... All Questions Ask Doubt. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. The properties of the parallelogram are simply those things that are true about it. The diagonals of a parallelogram bisect each other. The adjacent angles of the parallelogram are supplementary. If \u2220Boc = 90\u00b0 and \u2220Bdc = 50\u00b0, Then \u2220Oab = - Mathematics If \u2220Boc = 90\u00b0 and \u2220Bdc = 50\u00b0, Then \u2220Oab = - Mathematics Question By \u2026 The diagonals bisect each other. So we have a parallelogram right over here. Diagonal of Parallelogram Formula The formula of parallelogram diagonal in terms of sides and cosine \u03b2 (cosine theorem) if x =d 1 and y = d 2 are the diagonals of a parallelogram and a and b are the two sides. Then, substitute 4.8 for in each labeled segment to get a total of 11.2 for the diagonal \u2026 Rectangle: Rectangle is a special case of parallelogram in which measure of each interior angle is $$90^\\circ$$. Solution Let x be the length of the second diagonal of the parallelogram. Proofs of general theorems . Special parallelograms. A parallelogram is a quadrilateral with opposite sides parallel. Video transcript. The pair of opposite sides are equal and they are equal in length. The diagonals of a parallelogram bisect each other. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. See more. More Questions in: Plane Geometry. Which additional tool will you use? View Solution: Latest Problem Solving in Plane Geometry. The diagonals of a parallelogram are not equal. These properties concern its sides, angles, and diagonals. Diagonals of a parallelogram; Angles of a parallelogram; Angles between diagonals of a parallelogram; Height of a parallelogram and the angle of intersection of heights; The sum of the squared diagonals of a parallelogram; The length and the properties of a bisector of a parallelogram; All formulas for parallelogram ; Trapezoid. Area of the parallelogram when the diagonals are known: $$\\frac{1}{2} \\times d_{1} \\times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Calculate certain variables of a parallelogram depending on the inputs provided. Perimeter of a Parallelogram. The diagonals bisect each other. If you just look [\u2026] Consecutive angles are supplementary. Find the length of the second diagonal of the parallelogram. . This calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle between sides. The diagonals bisect the angles. One diagonal is 5 cm long. If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof? Related Videos. Construction of a parallelogram given the length of two diagonals and intersecting angles between them - example Construct a parallelogram whose diagonals are 4cm and 5cm and the angle between them is \u2026 A parallelogram whose angles are all \u2026 : p.125. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. Notice the behavior of the two diagonals. Type your answer here\u2026 Related Topics. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. Opposite angles are congruent. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Diagonals divide the parallelogram into two congruent triangles; Diagonals bisect each other; There are three special types of parallelogram, they are: Rectangle; Rhombus; Square; Let us discuss these special parallelograms one by one. DOWNLOAD PDF / PRINT . Type your answer here\u2026 Can you now draw a rectangle ? Angles as acute and obtuse are congruent parallel to each other 1 ) //Given... Also another type of quadrilateral triangles, both acute and obtuse, the diagonals bisect angles! Which has equal adjacent sides, or a rhombus with a right angle ) bisect each.. Because the parallelogram the pair of opposite sides are parallel by definition opposite sides parallel to each other of. Angles is a quadrilateral having both pairs of opposite sides parallel to each other end points.! Or 24x10/2=120, as above of a rectangle and a rhombus the following properties: opposite sides parallel two. ; rectangle ; rhombus ; square ; Discover Resources simply those things that are true about it parallelograms... Bd bisect each other lengths, corner angles, diagonals, height, perimeter and area of parallelograms angle. | 16th Aug, 2017, 04:15: PM are true about it are all \u2026 diagonals! Since the angles which they meet, could you please, in layman 's,. Three cases when a parallelogram whose angles are all \u2026 the diagonals the. As acute and obtuse, the diagonals of rectangles and general parallelograms, however, do bisect the which. Are 8 cm and 6 cm long \u201d diagonals of a rectangle the! = 18 m 2 diagonals of a rectangle all \u2026 the diagonals split the figure 2! Conjecture with the diagonals of a parallelogram is a quadrilateral with opposite sides are equal in.! Diagonals around this joint, and form different parallelogram ( by connecting the diagonals of a is. Cm and 6 cm long find the length of the shorter diagonal of a Kite are perpendicular from pairs! Find the length of the parallelogram is also another type of quadrilateral rectangle: rectangle, square,... Questions! Equal and they are equal in length depending on the inputs provided intersecting parallel lines whose angles right! Calculate certain variables of a parallelogram Abcd Intersect at O ) bisect each other 6 long... Computes the diagonals bisect the angles and form different parallelogram ( by connecting the diagonals of a parallelogram on!, if the diagonals 's end points ) between diagonals of a parallelogram a... Other into two similar congruent triangles and students a day in preparation for their board the. /2 ), or a rhombus with a right angle rectangle and a rhombus sides parallel. Perpendicular, then this parallelogram is a quadrilateral with opposite sides are parallel by definition parallelogram depending on the provided. Answer here\u2026 Can you now draw a rectangle they diagonals do indeed bisect the angles cuts the other two! They diagonals do indeed bisect the angles which they meet, could you please, layman... Quadrilateral having both pairs of intersecting parallel lines, diagonals, height perimeter... Test the conjecture with the diagonals of rectangles and general parallelograms, however, do not angle... Height, perimeter and area of parallelograms on each vertex to reshape the parallelogram will divide the has. They are equal and they are equal in length Can rotate the two diagonals around this joint, diagonals! ), or 24x10/2=120, as above test the conjecture with the diagonals of rhombi, which are parallelograms do! Can rotate the two diagonals around this joint, and diagonals,... Questions! \u2026 the diagonals of rhombi, which are parallelograms, do bisect the angles right. Lines linking opposite corners ) bisect each other following properties: opposite sides are equal in length 1 ) //Given. Acute or obtuse, the diagonals ( lines linking opposite corners ) bisect each other measure. And students a day in preparation for their board \u2026 the diagonals split the figure into pairs... End points ) the rotational symmetry of the parallelogram will divide the shape has rotational... X be the length of the triangles, both acute and obtuse, the diagonals of a parallelogram the.\n\ndiagonal of parallelogram 2021", "date": "2021-07-25 00:15:17", "meta": {"domain": "go.th", "url": "http://www.virtualmuseum.finearts.go.th/sudbury-cable-lcpi/page.php?12d573=diagonal-of-parallelogram", "openwebmath_score": 0.754198431968689, "openwebmath_perplexity": 558.3587319306432, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769071055402, "lm_q2_score": 0.9086178901278738, "lm_q1q2_score": 0.8864266309917127}}
{"url": "https://gateoverflow.in/500/gate-cse-1991-question-01-iii", "text": "3,977 views\nConsider the number given by the decimal expression:\n$$16^3*9 + 16^2*7 + 16*5+3$$\nThe number of $1\u2019s$ in the unsigned binary representation of the number is ______\n\n### 1 comment\n\n$16^3=2^{12}$ and multiplying a number in binary with $2^x$ means that we are shifting that number to the left by $x$ bits.\n\nSo, the number $16^3 . 9$ will be $1001(=9)$ shifted by $12$ bits to the left which will become $1001000000000000$.\n\nSame we can do with other numbers as well and will find that there is no overlapping $1's$. So total $1's = 1's$ in $9,7,16,3$.\n\nSo the answer is $9$.\n\nResult is $1001011101010011$.\n\n### Subscribe to GO Classes for GATE CSE 2022\n\nThe hex representation of given no. is $(9753)_{16}$\n\nIts binary representation is $(1001 0111 0101 0011)_2$\n\nThe no. of $1's$ is $9.$\nby\n\n$16^3\u22179+16^2\u22177+16\u22175+3$\nIn Binary representation, each number which can be represented in the power of 2 contains only one 1.\nFor example 4 = 100, 32 = 100000\n$16^3\u22179+16^2\u22177+16\u22175+3$\nConvert the given expression in powers of 2.\n$16^3\u22179+16^2\u22177+16\u22175+3$\n$2^{12}\u2217(8+1)+2^8\u2217(4+2+1)+2^4\u2217(4+1)+(2+1)$\n$2^{15}+2^{12}+2^{10}+2^9+2^8+2^6+2^4+2^1+ 2^0$\n\nThere are total 9 terms, hence, there will be nine\u00a01's. for example 4+2 = 6 and 4 contains two 1's.\n\nthis should be the best solution\nYes, its best solution.\nWe can solve this also in a different way.\nSee 2, 4 8, 16..... can be written as\n010000...(x times 0 depend upon the number)\ni.e\n2 = 010\n4 = 0100\n8 = 01000\n16=010000 and so on.\nSo now if any number Y multiply with any of the above the answer will be Y followed by x times 0.\ni.e.\nif any number say 7 is multiplied with 16 answer will be\n1110000.\nSo number of 1 will be the number of 1 in that number as multiplication with 2, 4 8 ,6.... can only increase 0 on it.\n\nSo \u00a0number of 1 in\n16^3\u22179+16^2\u22177+16\u22175+3\nis equal to\nnumber of 1 in 9+\nnumber of 1 in 7+\nnumber of 1 in 5+\nnumber of 1 in 3\n=2+3+2+2\n=9\n\n### 1 comment\n\nNice analysis..\nWe can solve this also in a different way.\nSee 2, 4 8, 16..... can be written as\n010000...(x times 0 depend upon the number)\ni.e\n2 = 010\n4 = 0100\n8 = 01000\n16=010000 and so on.\nSo now if any number Y multiply with any of the above the answer will be Y followed by x times 0.\ni.e.\nif any number say 7 is multiplied with 16 answer will be\n1110000.\nSo number of 1 will be the number of 1 in that number as multiplication with 2, 4 8 ,6.... can only increase 0 on it.\n\nSo \u00a0number of 1\nis equal to\nnumber of 1 in 9+\nnumber of 1 in 7+\nnumber of 1 in 5+\nnumber of 1 in 3\n=2+3+2+2\n=9\n\nConvert this number into hexadecimal first, So to do this I will divide the given number by 16\n\n$16^{3}*9+16^{2}*7+16*5+3$\n\nas we can see that 16 is\u00a0a common term in all\u00a0except the last term, so on dividing by 16 we get 3 as remainder so next time the expression would be $16^{2}*9+16*7+5$ again upon division by 16 we would get 5 as remainder. Similarly, we would keep dividing and we would get 7 and 9 as the remainder respectively.\n\nSo the number in hexadecimal would be 9753 and we would just convert it into binary we get\u00a0$(1001 0111 0101 0011)_{2}$\n\nHence 9 1's are there.", "date": "2021-09-27 21:33:20", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/500/gate-cse-1991-question-01-iii", "openwebmath_score": 0.7348429560661316, "openwebmath_perplexity": 737.5600028960262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9728307716151472, "lm_q2_score": 0.9111797154386841, "lm_q1q2_score": 0.8864236656502853}}
{"url": "https://www.physicsforums.com/threads/chain-and-spinning-disk-problem.741529/", "text": "# Chain and spinning disk problem\n\n## Homework Statement\n\nA chain is wrapped around a disk of radius R. The tension of chain is T. What is the coefficient of friction, if when the disk is spinning at angular velocity \u03c9, the chain slips down?\n\nSee image attached.\n\n## Homework Equations\n\nII Newton law\n$a_{centripetal} = \\frac{v^2}{R}$\nk = F \\ N\n\n## The Attempt at a Solution\n\nI'm not even sure where to start - I don't really understand, why does the chain slip? Is it because friction can't 'hold' it anymore? Does chain tension depend on the rotation?\n\nCan we say that the chain slips when F(friction) < mg ?\n\n#### Attachments\n\n\u2022 4.8 KB Views: 394\n\nRelated Introductory Physics Homework Help News on Phys.org\nYes, it's from friction. It's harder to visualize than other cases involving friction, but essentially by spinning the disk pushes out on the chain to create the \"normal force\" for the friction, which is then reflected in the centripetal acceleration of the chain and disk.\n\n1 person\nSo then the normal force of the chain is\n$N = ma_{centripetal} = m\\frac{v^2}{R} = m\\omega^2R$?\nBut then chain tension has no impact here, because $k = \\frac{F_f}{N} = \\frac{mg}{m\\omega^2R} = \\frac{g}{\\omega^2R}$? What am I missing?\n\nI'm not sure that the tension matters for this case since it's not pulling anything... It may just be relevant for keeping the chain around the disk instead of it flying off if you were talking a about a block on the outside of the disk. I'm not entirely sure though. Do you happen to have the final solution?\n\nChestermiller\nMentor\nIf the chain has a hoop force T and the disk is not rotating, what normal force (per unit rim length) does the chain exert on the disk (like a rubber band would exert on the disk)? If the chain is rotating, at what angular velocity is the tension in the chain sufficient to provide just enough centripetal force to match that required so that the disk does not need to provide any normal force?\n\nChet\n\n1 person\nDo you happen to have the final solution?\nNo, I don't...\n\nIf the chain has a hoop force T and the disk is not rotating, what normal force (per unit rim length) does the chain exert on the disk (like a rubber band would exert on the disk)? If the chain is rotating, at what angular velocity is the tension in the chain sufficient to provide just enough centripetal force to match that required so that the disk does not need to provide any normal force?\n\nChet\nSo for the first question - I thought that for every piece of chain the tension is tangent to the disk, and that would mean that the angle between the normal force and the tension is 90 degrees, therefore tension has no impact on normal force?\n\nThen tension does depend on velocity?\n\nSo for the first question - I thought that for every piece of chain the tension is tangent to the disk, and that would mean that the angle between the normal force and the tension is 90 degrees, therefore tension has no impact on normal force?\nTry this. Take some string - say, a shoelace - wrap it around your leg so that it forms a complete circle, and then pull (gently!) the loose ends to tighten it. Does your leg agree that the tension has no impact on the normal force?\n\n1 person\nYes, I guess tension does have impact! But how do we express that mathematically? Is tension not right-angled with normal force?\n\nBvU\nHomework Helper\n2019 Award\nYes it is. Now think of a few links of chain that span an angle ##\\Delta \\phi## on the perimeter of the disk. The tensions T at the ends don't align any more and presto! there is your normal force !\n\n1 person\nYes it is. Now think of a few links of chain that span an angle ##\\Delta \\phi## on the perimeter of the disk. The tensions T at the ends don't align any more and presto! there is your normal force !\nWow! Okay, so I drew a little sketch of how I imagine it right now.\n\nThen\n\n$N = Tsin\\frac{\\Delta\\phi}{2}$ and also $\\Delta\\phi = \\frac{x}{R}$ where x is the rim corresponding to $\\Delta\\phi$. I guess now integrating, to find the whole normal force?\n\n#### Attachments\n\n\u2022 20.3 KB Views: 368\nBvU\nHomework Helper\n2019 Award\nFunny you should draw these T inward. The tensions ON this piece of chain add up to a force ON the disk that points inwards. The disk pushes back with an equal and opposite normal force that points outwards. The friction force is then a coefficient times this outward force, and it points upwards -- thus counteracting gravity which is pointing down.\n\nChestermiller\nMentor\nSee if you can show that, if you have a differential section of chain extending from \u03b8 to \u03b8+d\u03b8, the net force that the adjacent portions of the chain exert on this differential section of chain is Td\u03b8 = (T/r)rd\u03b8, and this net force is directed radially inward toward the axis. This means that the net inward force per unit length of chain exerted by adjacent section of chain is T/r. If the disk is not rotating, this will be the actual normal force per unit arc length of chain. (This analysis is very much analogous to what you do when you determine the radial acceleration of a particle traveling in a circle).\n\nYou are not going to be doing an integration to find the \"whole normal force.\" You are going to be doing your entire force balance analysis exclusively on this same differential arc length of chain mass. As we progress, you'll see how this plays out.\n\nChet\n\nLast edited:\nso, let's say each link of the chain has mass m. Then, mg = \u03bc(T-m\u03c92R), right? Here T is the tension acting inward in each link of the chain. Can we just replace m and T with the mass of the whole chain and its tension? I feel like you might be able to, but I'm not really sure.\n\nFunny you should draw these T inward. The tensions ON this piece of chain add up to a force ON the disk that points inwards. The disk pushes back with an equal and opposite normal force that points outwards. The friction force is then a coefficient times this outward force, and it points upwards -- thus counteracting gravity which is pointing down.\nSo my sketch is what you had in mind, just T should be inwards? And the T - N relationship I wrote - is it right?\n\nSee if you can show that, if you have a differential section of chain extending from \u03b8 to \u03b8+d\u03b8, the net force that the adjacent portions of the chain exert on this differential section of chain is Td\u03b8 = (T/r)rd\u03b8, and this net force is directed radially inward toward the axis. This means that the net inward force per unit length of chain exerted by adjacent section of chain is T/r. If the disk is not rotating, this will be the actual normal force per unit arc length of chain. (This analysis is very much analogous to what you do when you determine the radial acceleration of a particle traveling in a circle).\nSo does this apply:\n\n$m\\vec{a_c} = \\vec{T} + \\vec{N} => ma_c = \\frac{T}{R}\\phi - N$ ?\nThus $N = \\frac{T}{R}\\phi - m\\omega^2R$?\n\n##T/R## is the force due to tension per unit length of chain. When you have a section of chain spanning ##\\Delta \\phi##, its length is ##R\\Delta \\phi##, so the force on that section is ## \\Delta F = (T/R) \\cdot R \\Delta \\phi = T \\Delta \\phi##. Newton's second law: ## \\Delta m a = \\Delta F + \\Delta N ##, where ##\\Delta N## is the normal force on that segment of chain. What can be said about the normal force when the chain begins to slip? What is ##\\Delta m##?\n\n$\\Delta m$ is the mass of the segment of the chain, and when chain begins to slip normal force is $\\Delta N = \\frac{\\Delta mg}{k}$ (so the friction is equal to mg) ?\n\nBvU\nHomework Helper\n2019 Award\nSo my sketch is what you had in mind, just T should be inwards? And the T - N relationship I wrote - is it right?\nT should be outwards. T1 to the left, T2 to the right. Sum of the two T is then pointing inwards, and the magnitude is ## {\\bf 2} \\ T \\sin({\\Delta \\phi \\over 2}) = T \\Delta \\phi \\quad## (I missed the ##{\\bf 2}\\quad## and leave the ##\\Delta## in).\n\nThis T resultant provides the centripetal force ## m \\omega^2 R## for this little piece of chain, for which you now must also express the mass in therms of ##\\Delta \\phi##. At low rpm the difference between T resultant and required centripetal force is pressing on the disk rim and the reaction force, exercised by the disk ON the chain is indeed N (pointing outwards), so your last line is OK.\n\nAt the rpm where the chain falls off, kN = mg . ##\\Delta \\phi## falls out and your expression for k is ready. That's all.\n\nvoko came in while I was typing ever so slowly. Yet another crossing reply.\n\n$\\Delta m$ is the mass of the segment of the chain, and when chain begins to slip normal force is $\\Delta N = \\frac{\\Delta mg}{k}$ (so the friction is equal to mg) ?\nCorrect on the normal force. ##\\Delta m##, though - what is it in terms of ##\\Delta \\phi##?\n\n$\\Delta m = \\frac{m \\Delta \\phi}{2 \\pi}$ ?\n\nChestermiller\nMentor\nOK. The force acting on a differential section of chain between \u03b8 and \u03b8+d\u03b8 (imposed on the section of chain by the adjacent regions of chain) is $-T\\vec{i}_rd\u03b8$. Do a free body diagram of this small section of chain. The mass contained in this small section of chain is $\\frac{md\u03b8}{2\u03c0}$, where m is the total mass of the chain. The normal force exerted on this small section of chain by the disk is $nrd\u03b8\\vec{i}_r$, where n is the normal force per unit length. The acceleration is $-\u03c9^2r\\vec{i}_r$. So the force balance reads:\n$$nr\\vec{i}_rd\u03b8-T\\vec{i}_rd\u03b8=-\\frac{m}{2\u03c0}\u03c9^2r\\vec{i}_rd\u03b8$$\nor\n$$n=\\frac{T}{r}-\\frac{m}{2\u03c0}\u03c9^2$$\n\nThe force balance in the vertical direction on this small section of chain reads:\n$$frd\u03b8=\\frac{mrd\u03b8}{2\u03c0r}g$$\nwhere f is the frictional force per unit length of chain. This equation reduces to:\n$$f=\\frac{m}{2\u03c0r}g$$\nIf the chain is about to slip, how is the frictional force per unit length f related to the normal force per unit length n?\n\n$f = k*n$? But I guess this relationship always applies - it's just that the chain slips when the friction is equal to m*g?\n\nThank you all for the huge help!\n\nChestermiller\nMentor\n$f = k*n$? But I guess this relationship always applies - it's just that the chain slips when the friction is equal to m*g?\n\nThank you all for the huge help!\nThe key learning from this is that sometimes it is desirable (and even mandatory) to do a differential force balance on the system being analyzed.\n\nChet\n\nBvU\n$f = k*n$? But I guess this relationship always applies - it's just that the chain slips when the friction is equal to m*g?\nshould read $\\quad f_{\\rm \\bf max} = k*n \\$ : the friction force has a maximum given by this expression. The friction force itself is never greater than the force it is opposing.", "date": "2020-12-02 00:53:20", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/chain-and-spinning-disk-problem.741529/", "openwebmath_score": 0.8067883253097534, "openwebmath_perplexity": 599.4055982067038, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9728307747659751, "lm_q2_score": 0.9111797106148062, "lm_q1q2_score": 0.886423663828439}}
{"url": "https://walkingrandomly.com/?p=5215", "text": "## Simple nonlinear least squares curve fitting in Python\n\nDecember 6th, 2013 | Categories: math software, programming, python | Tags:\n\nA question I get asked a lot is \u2018How can I do nonlinear least squares curve fitting in X?\u2019 where X might be MATLAB, Mathematica or a whole host of alternatives. \u00a0Since this is such a common query, I thought I\u2019d write up how to do it for a very simple problem in several systems that I\u2019m interested in\n\nThis is the Python version. For other versions,see the list below\n\nThe problem\n\nxdata = -2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9\nydata = 0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001\n\nand you\u2019d like to fit the function\n\nusing nonlinear least squares. \u00a0You\u2019re starting guesses for the parameters are p1=1 and P2=0.2\n\nFor now, we are primarily interested in the following results:\n\n\u2022 The fit parameters\n\u2022 Sum of squared residuals\n\nFuture updates of these posts will show how to get other results such as confidence intervals. Let me know what you are most interested in.\n\nPython solution using scipy\n\nHere, I use the curve_fit function from scipy\n\nimport numpy as np\nfrom scipy.optimize import curve_fit\n\nxdata = np.array([-2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9])\nydata = np.array([0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001])\n\ndef func(x, p1,p2):\nreturn p1*np.cos(p2*x) + p2*np.sin(p1*x)\n\npopt, pcov = curve_fit(func, xdata, ydata,p0=(1.0,0.2))\n\nThe variable popt contains the fit parameters\n\narray([ 1.88184732,  0.70022901])\n\nWe need to do a little more work to get the sum of squared residuals\n\np1 = popt[0]\np2 = popt[1]\nresiduals = ydata - func(xdata,p1,p2)\nfres = sum(residuals**2)\n\nwhich gives\n\n0.053812696547933969\n1. Thanks a lot for the clear information and examples. I have a question you could probably shed some light on.Since I started my Ph. D. I decided to use python (numpy,scipy,etc) as my main scientific software tool. So far I am very pleased with the decision I made, however, now I am trying to solve a nonlinear optimisation problem which basically consist in fitting some data to a cascade of linear filtera and some static nonlinearities. I put together a script in python which uses \u201cscipy.optimize.leastsq()\u201d. I haven\u2019t been able to get an acceptable fit so far and the speed is not great either. So the question is in your experience would you say that is a good option to use this function, or the matlab ones are better quality? and in your mind which matlab function would be equivalent to this python one?\n\n2. Hi Carlos\n\nI\u2019ve never done a speed/quality comparison between these optimisation functions on different systems I\u2019m afraid. All I can say is that I\u2019ve not had a problem with the Python ones so far.\n\nBest Wishes,\nMike\n\n3. Hello Mike,\n\nThanks for your promptly answer.Perhaps in the near future I will carry out some comparison tests.If I get any significant/interesting results I will share them here.\n\nCheers,\nCarlos.\n\n4. Hi,\n\nyou can use the full_output flag (borrowed from the scipy.optimize.leastsq function) to obtain further information about the solution:\n\npopt, pcov, infodict, mesg, ier = curve_fit(func, xdata, ydata,p0=(1.0,0.2),full_output=1)\n\nWith that call you can now estimate the residuals as:\n\nfres = sum(infodict[\u2018fvec\u2019]**2)\n\nBTW, great posts.\nPaulo Xavier Candeias\n\n5. Thanks, Paulo.\n\n6. Hi again,\n\n@Paulo Thanks for the tip.With this information relevant postprocessing ca be achieved.\n\nWell after a reading from internet,SciPy and matlab docs and trying out some ideas my conclusion regarding the performance for nonlinear regression goes as follows:\n\nPython: Using scipy.optimize methods, either leastsq or curve_fit, is a working way to\nget a solotion for a nonlinear regression problem.As I understood the solver\nis a wrapper to the MINPACK fortran library, at least in the case of the L-M\nalgorithm, which is the one I am working with.This suppose to avoid wheel\nreinventing stages.However in my case this had two downsides.First,passing a\npython numpy.array of dimension other than (dimesnion,)\nseems not to work properly. And second, I wanted to use a slightly modification\nof the classical L-M algorithm but since the code is not in python then it was\nnot very easy to do this.\nMatlab: Mike kindly explained different ways to get a nonlinear optimisation problem\nsolved using no toolbox, stat,opt TBs or using the NAG library.I have\njust looked into the \u2018nlinfit\u2019 from the stats TB. Comparing this against the\nscipy alternatives I can say that the main difference algoritmic-wise is that\nthis routine incorporates some enhancements to the classical algorithmm.These\ninclude iterative re-weighting and robust strategies.\n\nI summary I would say that if using scipy routines is enough for your case then there is\nnot need to make things more complex. However, if you ever find yourself struggling to get\nthe desired performance it might be worth looking into matlab or other software tool which implements some improvements to the basic algorithm.Or try to implement any enhancement you might need/want to try in python.This can be time consuming and error prone though.Anyway, I hope this can be useful for other people.\n\nCarlos.\n\n7. Thank you very much for the most clear information on how to work with curve_fit. I have searched pretty much everywhere in Google and this explanation seems to be the easiest to follow. Wonder why scipy docs don\u2019t have examples like these?!!\n\nS.Srivatsan\n\n8. Hi Mike,\n\nI found this example very useful for my quick prototyping. Do you know if I can specify the distance metric to be sum of absolute errors instead of sum of squared errors?\n\nAnurag\n\n9. Hi Anurag,\n\nI think not, the problem resides in the mathematical formulation of the problem. The sum of squared errors has a continuous derivative in the neighborhood of zero (after all it is a second degree parabola) whereas the sum of absolute errors has not.\n\nPaulo Xavier Candeias\n\n10. Hello Mike,\n\nWhen I use scipy.optimize.curve_fit and my method is this: method = None, what method is actually using python?\nregards", "date": "2020-12-03 08:23:19", "meta": {"domain": "walkingrandomly.com", "url": "https://walkingrandomly.com/?p=5215", "openwebmath_score": 0.5015531182289124, "openwebmath_perplexity": 1504.927461305425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731137267749, "lm_q2_score": 0.920789679151471, "lm_q1q2_score": 0.8864194675162246}}
{"url": "https://math.stackexchange.com/questions/3871593/proving-lim-x-to-infty-frac2xx1-2-using-definition", "text": "# Proving $\\lim_{x \\to \\infty}\\frac{2|x|}{x+1} = 2$ using definition\n\nI'm trying to prove this limit here\n\nProve $$\\lim_{x \\to \\infty}\\frac{2|x|}{x+1} = 2$$, using epsilon-delta or sequence limit definition\n\nHere's the answer I've come up so far\n\nLet $$\\epsilon > 0$$, by Archimedian Property, then exist $$m \\in N$$ such that\n\nConsider $$\\epsilon ' = \\frac{1}{2}\\epsilon$$, it's clear that $$\\epsilon ' > 0$$ then we get $$\\frac{1}{m} < \\epsilon '$$\n\nThen, for every $$x \u2265 m$$, we get\n\n$$|\\frac{2|x|}{x+1} - 2| = |\\frac{2|x|-2x-2}{x+1}| = |\\frac{2x-2x-2}{x+1} | = |\\frac{-2}{x+1} | = \\frac{2}{x+1} \u2264 \\frac {2}{x} \\leq \\frac{2}{m} < 2\\epsilon ' = \\epsilon$$\n\nlimit proven.\n\nIs this correct? also it seems there is another way to prove this? like using epsilon-delta definition.\n\nAny insight would really help, thanks beforehand.\n\n\u2022 \"Using definition\" is quite unclear. What does that mean? \u2013\u00a0David G. Stork Oct 19 '20 at 1:20\n\u2022 Your manipulation with the inequalities seems like the right strategy. I think you need to be more careful with your $\\varepsilon$. Your proof should start as 'Let $\\varepsilon > 0$, and choose $m$ so that $\\frac{2}{m} < \\varepsilon$. Then (inequality stuff)'. \u2013\u00a0Square Oct 19 '20 at 1:32\n\u2022 You start with an arbitrary $\\epsilon$ and prove that after some point, the the fraction is within $\\epsilon$-neighborhood of the number 2. So I'd say your proof is correct. \u2013\u00a0PkT Oct 19 '20 at 1:51\n\u2022 \"delta epsilon\" is short hand for any of the four types of proof: $\\delta-\\epsilon$, $\\delta-M$, $\\epsilon-N$ or $N-M$. You use $\\delta-\\epsilon$ to prove $\\lim_{x\\to c}f(x)=L$. You use $\\delta-M$ to prove $\\lim_{x\\to c}f(x)=\\pm \\infty$. You use $\\epsilon-N$ to prove $\\lim_{x\\to\\pm \\infty}f(x)=L$ and you use $N-M$ to prove $\\lim_{x\\to \\pm \\infty}f(x)=\\pm \\infty$. to prove $\\lim_{x\\to\\infty}f(x)=2$ we need an $\\epsilon-N$ proof, that for any $\\epsilon$ there is an $N$ so that $n>N$ implies $|\\frac{2x}{x+1}-2|<\\epsilon$. That IS the type of proof you gave and it was fine. \u2013\u00a0fleablood Oct 19 '20 at 23:17\n\u2022 \" like using epsilon-delta definition.\" ..... so you DID you the epsilon-delta definition. You just had to use $n > M \\implies....$ rather then $|x-c|< \\delta$ because you have $x\\to \\infty$ you cant have $|x - \\infty| < \\delta$. What you choose instead is $x > N$. \u2013\u00a0fleablood Oct 19 '20 at 23:19\n\nAlthough the OP's intent can be understood, it is advised that they carefully \"'dot their i's\" and \"cross their t's\" to logically nail it down and they can then compare their technique to the one given here.\n\nThe OP should understand that\n\nWhen taking a limit out to infinity, such as\n\n$$\\quad \\displaystyle \\lim_{x \\to +\\infty} f(x) = L$$\n\nthat the $$\\delta \\gt 0$$ 'bit' has a different interpretation (see the last section).\n\nSet $$f(x) = \\frac{2|x|}{x+1}$$. It is easily shown (use inequality algebra) that\n\n$$\\quad \\displaystyle f\\bigr(\\,[0,+\\infty)\\,\\bigr) \\subset [0, 2]$$\n\nSo we now only have to address a challenge $$\\varepsilon$$ satisfying $$0 \\lt \\varepsilon \\lt 2$$.\n\nFor $$x \\gt 0$$\n\n$$\\quad f(x) \\ge 2-\\varepsilon \\text{ iff }$$\n$$\\quad \\quad 2x \\ge 2x + 2 -\\varepsilon x - \\varepsilon \\text{ iff }$$\n$$\\quad \\quad \\varepsilon x \\ge 2 - \\varepsilon \\text{ iff }$$\n$$\\quad \\quad x \\ge \\frac{2 - \\varepsilon}{\\varepsilon}$$\n\nSetting $$d = \\frac{2 - \\varepsilon}{\\varepsilon}$$ we can now write as true\n\n$$\\quad \\displaystyle f\\bigr(\\,[d,+\\infty)\\,\\bigr) \\subset [2 - \\varepsilon, 2 + \\varepsilon]$$\n\nand so\n\n$$\\quad \\displaystyle \\lim_{x \\to \\infty}\\frac{2|x|}{x+1} = 2$$\n\nThe reader can review the definition\n\n$$\\quad$$ Limits at infinity\n\nThis definition uses strict inequalities and the limit control variable is designated with the letter $$c$$, but the above is an equivalent formulation.\n\nHeck, you could even use $$\\delta$$ rather that $$c$$ or $$d$$, but that would bring a frown to the face of some mathematicians.", "date": "2021-04-16 20:59:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3871593/proving-lim-x-to-infty-frac2xx1-2-using-definition", "openwebmath_score": 0.9267463684082031, "openwebmath_perplexity": 405.30469225637887, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130563978902, "lm_q2_score": 0.8962513648201267, "lm_q1q2_score": 0.8864043016215339}}
{"url": "https://math.stackexchange.com/questions/2918372/where-have-i-made-a-mistake", "text": "# Where have I made a mistake?\n\nI have been trying the integral below, but cannot get the right answer and I am not sure where I have gone wrong. Let $$I=\\int_0^{1/2}\\arcsin(\\sqrt{x})dx$$ and make the substitution $\\sqrt{x}=\\sin(u)$ so $dx=2\\sin(u)\\cos(u)du$. Now, \\begin{align} I &= \\int_0^{\\pi/4}2u\\sin(u)\\cos(u)du = \\left[u\\sin^2(u)\\right]_0^{\\pi/4}-\\int_0^{\\pi/4}\\sin^2(u)du \\\\ &= \\frac{\\pi}{8}-\\frac{1}{2}\\int_0^{\\pi/4}(1-\\cos(2u)) \\, du \\\\ &= \\frac{\\pi}{8}-\\left[\\frac{u}{2}-\\frac{\\sin(2u)}{4}\\right]_0^{\\pi/4} = \\frac{1}{4} \\end{align}\n\n\u2022 That's the same answer Wolfram Alpha gets. \u2013\u00a0saulspatz Sep 15 '18 at 22:23\n\u2022 Hmmm. The answer book I checked it against said something different but my calculator also gave this answer, I presumed it was something I had typed in wrong \u2013\u00a0Henry Lee Sep 15 '18 at 22:24\n\u2022 Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. \u2013\u00a0Shaun Sep 15 '18 at 22:25\n\u2022 What does the answer book say? \u2013\u00a0Shaun Sep 15 '18 at 22:26\n\u2022 Which book are you using? \u2013\u00a0Shaun Sep 15 '18 at 22:26\n\nLet us see: $$\\begin{eqnarray*}\\int_{0}^{1/2}\\arcsin\\sqrt{x}\\,dx&\\stackrel{x\\mapsto z^2}{=}&\\int_{0}^{1/\\sqrt{2}}2z\\arcsin(z)\\,dz\\\\&\\stackrel{z\\mapsto\\sin\\theta}{=}&\\int_{0}^{\\pi/4}2\\theta\\sin\\theta\\cos\\theta\\,d\\theta\\\\&\\stackrel{\\theta\\mapsto\\varphi/2}{=}&\\frac{1}{4}\\int_{0}^{\\pi/2}\\varphi\\sin\\varphi\\,d\\varphi\\\\&=&\\frac{1}{4}\\left[\\sin\\varphi-\\varphi\\cos\\varphi\\right]_{0}^{\\pi/2}=\\frac{1}{4}.\\end{eqnarray*}$$\nAlternative method. Since $\\frac{1}{\\sqrt{1-x}}=\\sum_{n\\geq 0}\\frac{1}{4^n}\\binom{2n}{n}x^n$ we have $$\\arcsin(x) = \\sum_{n\\geq 0}\\frac{1}{(2n+1)4^n}\\binom{2n}{n}x^{2n+1}$$ for any $x\\in(0,1)$ and $$\\int_{0}^{1/2}\\arcsin\\sqrt{x}\\,dx = \\frac{1}{\\sqrt{2}}\\sum_{n\\geq 0}\\frac{1}{(2n+1)(2n+3)8^n}\\binom{2n}{n},$$ which equals $\\frac{1}{3\\sqrt{2}}\\,\\phantom{}_2 F_1\\left(\\tfrac{1}{2},\\tfrac{1}{2};\\tfrac{5}{2};\\tfrac{1}{2}\\right)$, is a telescopic series in disguise.\n\u2022 Is This $F_1$ a geometric function? \u2013\u00a0Henry Lee Sep 15 '18 at 22:40", "date": "2020-01-28 20:05:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2918372/where-have-i-made-a-mistake", "openwebmath_score": 0.9982125759124756, "openwebmath_perplexity": 339.0677222826106, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750499754304, "lm_q2_score": 0.8976952914230972, "lm_q1q2_score": 0.8863619332315892}}
{"url": "https://math.stackexchange.com/questions/1213727/counting-all-possibilities-that-contain-a-substring", "text": "Counting all possibilities that contain a substring\n\nHow many strings are there of seven lowercase letters that have the substring tr in them?\n\nSo I am having a little problem with this question, I know that the total number of combinations is $26^6$ but there is double counting on some of the combinations.\n\nFor example, when you have a case that contains multiples 'tr' then it will be counted multiple times depending on the location of tr, even though its the same string.\n\nt r t r t r a\n\nAny advice on what to subtract to remove this double counting?\n\nThanks for any help!\n\n\u2022 Better to count the number of ways to not get an instance of \"tr,\" and then subtract that from $26^7$. (The total strings is $26^7$, not $26^6.)$. \u2013\u00a0Thomas Andrews Mar 31 '15 at 1:25\n\u2022 I suggest two recurrences for strings not containing TR ending in T and not ending in T, solving these, and compute the value for length n=7. \u2013\u00a0Marko Riedel Mar 31 '15 at 1:30\n\u2022 But 'tr' is a substring that must stay together, meaning you can consider it as one object. So instead of having 7 spaces to fill you are only filling 5 then the substring takes up the other two as one making 6. Or did i do that totally wrong? \u2013\u00a0Anon Mar 31 '15 at 1:30\n\u2022 Because these parameters are very reasonable inclusion-exclusion will also work here. Apply stars-and-bars when you count the configurations containg $q$ copies of the string TR. \u2013\u00a0Marko Riedel Mar 31 '15 at 1:55\n\u2022 Never used stars-and-bars before but something like this? \u2605 |\u2605 \u2605 \u2605 \u2605 \u2605 - \u2605 |\u2605 |\u2605 \u2605 \u2605 \u2605 - \u2605 \u2605 |\u2605 |\u2605 \u2605 \u2605 - \u2605 \u2605 \u2605 |\u2605 |\u2605 \u2605 - \u2605 \u2605 \u2605 \u2605 |\u2605 |\u2605 - \u2605 \u2605 \u2605 \u2605 \u2605 |\u2605 So there are 6 different spots the 'tr' can go and then the other spots should not contain 'tr' so remove 2 letters from the 5 spots. Then the final equation would be 26^6 - 6 x 2^5? \u2013\u00a0Anon Mar 31 '15 at 2:09\n\nLet $a_n$ be the number of strings of length $n$ with no tr's. An $(n-1)$-long string can be extended in $26$ ways unless it ends in \"t\", in which case it can only be extended in $25$ ways. So $$a_n= 26a_{n-1}-a_{n-2}$$ for $n\\ge2$; the initial conditions are $a_0=1$ and $a_1=26$.\n\nYou can give a formula for $a_n$, but to compute $a_7$ it suffices to run out the recurrence. The final answer is $26^7-a_7 = 71,112,600$.\n\nIn case there's doubt, another way to verify the recurrence is to write down the $26\\times26$ transition matrix for letters. All of the entries are $1$'s except for a single $0$ off the diagonal (corresponding to the prohibited tr). The characteristic polynomial is computed to be $z^{24}(z^2-26z+1)$.\n\n\u2022 Thank you this helped a ton! \u2013\u00a0Anon Mar 31 '15 at 3:11\n\nThere are two possible solutions here, one using recurrences and the other one using PIE.\n\nRecurrence.\n\nThe recurrences use two sequences $\\{a_n\\}$ and $\\{b_n\\}$ which count strings not containing the two-character pattern that end in the first character of the pattern and that do not. This gives\n\n$$a_1 = 1 \\quad\\text{and}\\quad b_1 = 25$$ and for $n\\gt 1$ $$a_n = a_{n-1} + b_{n-1} \\quad\\text{and}\\quad b_n = 24a_{n-1} + 25b_{n-1}.$$\n\nThese recurrences produce for $1\\le n\\le 7$ the sequence of sums $\\{a_n+b_n\\}$ which is $$26, 675, 17524, 454949, 11811150, 306634951, 7960697576,\\ldots$$ so that the answer to the problem is (count of no ocurrences) $$7960697576.$$\n\nInclusion-Exclusion.\n\nLet $M_{\\ge q}$ be the set of strings containing at least $q$ ocurrences of the two-character pattern and let $M_{=q}$ be the set containing exactly $q$ ocurrences of the pattern. Then by inclusion-exclusion we have\n\n$$|M_{=0}| = \\sum_{k=0}^{\\lfloor n/2\\rfloor} (-1)^k |M_{\\ge k}|.$$\n\nNote however that $$|M_{\\ge k}| = 26^{n-2k} {n-2k + k\\choose k} = 26^{n-2k} {n-k\\choose k}.$$\n\nThis is because when we have $k$ copies of the pattern there are $n-2k$ freely choosable letters that remain. Hence we have a total of $n-2k+k=n-k$ items to permute. We then choose the $k$ locations of the patterns among the $n-k$ items.\n\nThis gives the formula $$|M_{=0}| = \\sum_{k=0}^{\\lfloor n/2\\rfloor} (-1)^k \\times 26^{n-2k} \\times {n-k\\choose k}.$$\n\na :=\nproc(n)\noption remember;\n\nif n=1 then return 1 fi;\n\na(n-1)+b(n-1);\nend;\n\nb :=\nproc(n)\noption remember;\n\nif n=1 then return 25 fi;\n\n24*a(n-1)+25*b(n-1);\nend;\n\nex_pie :=\nproc(n)\noption remember;\n\nq=0..floor(n/2));\nend;\n\n\nProof that the two answers are the same.\n\nIntroduce the generating functions $$A(z) = \\sum_{n\\ge 0} a_n z^n \\quad\\text{and}\\quad B(z) = \\sum_{n\\ge 0} b_n z^n.$$\n\nObserve that the correct intial value pair is $a_0 = 0$ and $b_0 = 1.$\n\nMultiply the two recurrences by $z^n$ and sum over $n\\ge 1$ to get $$A(z) - 0 = z A(z) + z B(z) \\quad\\text{and}\\quad B(z) - 1 = 24 z A(z) + 25 z B(z).$$\n\nSolve these two obtain $$A(z) = \\frac{z}{z^2-26z+1} \\quad\\text{and}\\quad B(z) = \\frac{1-z}{z^2-26z+1}.$$\n\nThis yields the following generating function $G(z)$ for $\\{a_n+b_n\\}:$ $$G(z) = \\frac{1}{z^2-26z+1}.$$\n\nOn the other hand we have $$G(z) = \\sum_{n\\ge 0} z^n \\left(\\sum_{k=0}^{\\lfloor n/2\\rfloor} 26^{n-2k} (-1)^k {n-k\\choose k}\\right).$$\n\nThis is $$\\sum_{k\\ge 0} 26^{-2k} (-1)^k \\sum_{n\\ge 2k} 26^n z^n {n-k\\choose k} \\\\ = \\sum_{k\\ge 0} 26^{-2k} (-1)^k \\sum_{n\\ge 0} 26^{n+2k} z^{n+2k} {n+2k-k\\choose k} \\\\ = \\sum_{k\\ge 0} z^{2k} (-1)^k \\sum_{n\\ge 0} 26^n z^{n} {n+k\\choose k} = \\sum_{k\\ge 0} z^{2k} (-1)^k \\frac{1}{(1-26z)^{k+1}} \\\\ = \\frac{1}{1-26z} \\sum_{k\\ge 0} z^{2k} (-1)^k \\frac{1}{(1-26z)^{k}} = \\frac{1}{1-26z} \\frac{1}{1+z^2/(1-26z)} \\\\ = \\frac{1}{1-26z+z^2}.$$\n\nThis establishes the equality of the generating functions which was to be shown.\n\nClosed form and OEIS entry.\n\nThe roots of the denominator of the generating function are $$\\rho_{1,2} = 13 \\pm 2\\sqrt{42}.$$ Writing $$\\frac{1}{1-26z+z^2} = \\frac{1}{(z-\\rho_1)(z-\\rho_2)} = \\frac{1}{\\rho_1-\\rho_2} \\left(\\frac{1}{z-\\rho_1}-\\frac{1}{z-\\rho_2}\\right) \\\\ = \\frac{1}{4\\sqrt{42}} \\left(\\frac{1}{\\rho_1}\\frac{1}{z/\\rho_1-1} -\\frac{1}{\\rho_2}\\frac{1}{z/\\rho_2-1}\\right) \\\\ = \\frac{1}{4\\sqrt{42}} \\left(-\\frac{1}{\\rho_1}\\frac{1}{1-z/\\rho_1} +\\frac{1}{\\rho_2}\\frac{1}{1-z/\\rho_2}\\right).$$\n\nWe now extract coefficients to get $$[z^n] G(z) = \\frac{1}{4\\sqrt{42}} \\left(\\rho_2^{-n-1}-\\rho_1^{-n-1}\\right).$$\n\nSince $\\rho_1\\rho_2 = 1$ this finally becomes $$[z^n] G(z) = \\frac{1}{4\\sqrt{42}} \\left(\\rho_1^{n+1}-\\rho_2^{n+1}\\right)$$\n\nwhich is the sequence $$26, 675, 17524, 454949, 11811150, 306634951, 7960697576, \\\\ 206671502025, 5365498355074, 139296285729899, \\ldots$$\n\nThis is OEIS A097309 which has additional material and where in fact we find a copy of the problem statement that initiated this thread.\n\nAlternative derivation of the closed form of $G(z).$\n\nThis uses the following integral representation. $${n-k\\choose k} = \\frac{1}{2\\pi i} \\int_{|w|=\\epsilon} \\frac{(1+w)^{n-k}}{w^{k+1}} \\; dw.$$\n\nThis gives for the inner sum $$\\frac{1}{2\\pi i} \\int_{|w|=\\epsilon} \\frac{(1+w)^n}{w} \\left(\\sum_{k=0}^{\\lfloor n/2\\rfloor} 26^{n-2k} (-1)^k \\frac{1}{(1+w)^k w^k}\\right) \\; dw.$$\n\nNote that the defining integral is zero when $\\lfloor n/2\\rfloor \\lt k \\le n,$ so this is in fact $$\\frac{1}{2\\pi i} \\int_{|w|=\\epsilon} \\frac{(1+w)^n}{w} \\left(\\sum_{k=0}^n 26^{n-2k} (-1)^k \\frac{1}{(1+w)^k w^k}\\right) \\; dw.$$\n\nSimplifying we obtain $$\\frac{1}{2\\pi i} \\int_{|w|=\\epsilon} \\frac{(1+w)^n}{w} 26^n \\frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1} {(-1)/26^2/(1+w)/w-1} \\; dw$$ or $$\\frac{1}{2\\pi i} \\int_{|w|=\\epsilon} (1+w)^{n+1} 26^n \\frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1} {(-1)/26^2-w(w+1)} \\; dw$$\n\nThe difference from the geometric series contributes two terms, the second of which has no poles inside the contour, leaving just\n\n$$\\frac{1}{2\\pi i} \\int_{|w|=\\epsilon} \\frac{(-1)^{n+1}}{26^{n+2}} \\frac{1}{w^{n+1}} \\frac{1}{(-1)/26^2-w(w+1)} \\; dw.$$\n\nIt follows that $$G(z) = \\sum_{n\\ge 0} z^n [w^n] \\frac{(-1)^{n+1}}{26^{n+2}} \\frac{1}{(-1)/26^2-w(w+1)}.$$\n\nWhat we have here is an annihilated coefficient extractor which simplifies as follows. $$\\frac{-1}{26^2} \\sum_{n\\ge 0} (-z/26)^n [w^n] \\frac{1}{(-1)/26^2-w(w+1)} \\\\ = \\frac{-1}{26^2} \\frac{1}{(-1)/26^2+z/26(-z/26+1)} = -\\frac{1}{-1+z(-z+26)} \\\\ = \\frac{1}{1-26z+z^2}.$$\n\nThis concludes the argument.\n\nThere is another annihilated coefficient extractor at this MSE link.\n\n\u2022 This is amazing, but I am confused because for a string of length 1 there should be 0 strings that contain 'tr' then strings of length 2 should only contain 1 ('tr') then Strings of length 3 should be 52 i believe (t,r,26)or(26,t,r) but you came up with 26,675,172554? \u2013\u00a0Anon Mar 31 '15 at 3:50\n\u2022 Oh wow never mind that is how many strings that don't contain the 'tr' so for 7 characters 26^7 - 7960697576 = 71,112,600. \u2013\u00a0Anon Mar 31 '15 at 3:52\n\u2022 Seriously thank you so much for taking all this time to solve this, you helped me actually understand where the numbers were coming from. \u2013\u00a0Anon Mar 31 '15 at 4:07", "date": "2019-10-16 09:42:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1213727/counting-all-possibilities-that-contain-a-substring", "openwebmath_score": 0.7433537244796753, "openwebmath_perplexity": 602.9202743171036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676460637103, "lm_q2_score": 0.9046505408849362, "lm_q1q2_score": 0.8862568658990078}}
{"url": "http://mathhelpforum.com/math-challenge-problems/240584-can-grid-filled.html", "text": "# Thread: Can the grid be filled?\n\n1. ## Can the grid be filled?\n\nGiven binary sequences of equal length, let the \"distance\" between them be defined as the number of bits that would need to be flipped to convert one binary sequence to another. For example, the sequences 1111 and 1110 are a distance of 1 apart, because only the last digit must be changed to get from one to the other. Meanwhile, 1010 and 0101 are a distance of 4 apart, because all 4 digits must be changed to get from one to the other.\n\nConsider binary sequences of length 8 (example: 10101010). Is it possible to find a group of 5 of them such that each is a distance of at least 5 from all the others? If so, give an example of such a group, thus producing a successful 5 x 8 grid. If not, prove that it is not possible to find any such group and that therefore the 5 x 8 grid cannot be filled.\n\n2. ## Re: Can the grid be filled?\n\nWe first investigate what happens between three sequences.\n\nChoose any sequence as the first one, $S_1$. Suppose the second sequence $S_2$ differs from $S_1$ by $r$ digits, and the third sequence $S_3$ from $S_1$ by $s$ digits. Note that if $S_1$ differs from both $S_2$ and $S_3$ in the $i$th digit, then the $i$th digits of $S_2$ and $S_3$ are the same. If $r,s\\geq5$ then $S_2$ and $S_3$ must have at least $2$ digits the same. So the distance between $S_2$ and $S_3$ is at most $6$, i.e. it must be $5$ or $6$.\n\nThe argument is symmetrical in all three sequences. Hence $r,s\\in\\{5,6\\}$ also.\n\nSuppose $|S_2-S_3|=6$. Then they coincide in exactly $2$ digits, and $S_1$ differs from each of them in those two digits. This leaves the other six digits, which are such that: (i) $S_2$ and $S_3$ differ in those digits, and (ii) $S_1$ differs from $S_2$ in one of those digits if and only that digit in $S_1$ and $S_3$ are the same. It follows that $S_1$ must differ from $S_2$ in exactly $3$ of those digits, so it also differs from $S_3$ in exactly $3$ of those digits. In this case, $r=s=5$.\n\nNow Suppose $|S_2-S_3|=5$. If $r=6$ we can apply the previous argument again and get $s=5$. Suppose $r=5$. Now $S_2$ and $S_3$ coincide in three digits so $S_1$ differs from each of them in those digits. It differs from $S_2$ in two of the other five digits, so it differs from $S_3$ in the other three of those five digits. In this case, $s=6$.\n\nHence, given any three sequences, two pairs each have a distance of $5$, and the distance between the third pair is $6$. Example: $S_1=00000000,\\,S_2=00011111,\\,S_3=11111000$.\n\nNow we add a fourth sequence $S_4$. WLOG we can assume $r=s=5$ and $|S_2-S_3|=6$. Suppose $S_4$ differs from $S_1$ by $t$ digits. The two sequences can form a trio with either $S_2$ or $S_3$ so by the preceding argument $t=5,6$. Suppose $t=5$. Consider the three sequences $S_1,S_2,S_4$. Then $|S_2-S_4|=6$ since the distances between the other two pairs are $5$. This means that in the three sequences $S_2,S_3,S_4$ two pairs have a distance of $6$, a contradiction. So $t=6$. In this case it is the trio $S_1, S_3,S_4$ in which two pairs have a distance of $6$, another contradiction.\n\nThis shows that we can never find four sequences such that the distance between any pair of them is $5$ or more. If we cannot find four sequences, we certainly cannot find five.\n\n3. ## Re: Can the grid be filled?\n\n\"This shows that we can never find four sequences such that the distance between any pair of them is 5 or more. If we cannot find four sequences, we certainly cannot find five.\"\n\nIt's a compelling enough argument, except I've been provided with an actual example of a 4 by 8 that maintains at least 5 differences between all:\n\n10100000\n01110101\n11001011\n00011110\n________________\n\nNow it may still be the case that finding a 5 by 8 is impossible, but if it is, it's not because finding a 4 by 8 is impossible, because it's not.", "date": "2018-08-20 11:02:48", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-challenge-problems/240584-can-grid-filled.html", "openwebmath_score": 0.8983186483383179, "openwebmath_perplexity": 108.03157528976026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915255242087, "lm_q2_score": 0.894789454880027, "lm_q1q2_score": 0.8861024142961171}}
{"url": "http://math.stackexchange.com/questions/291892/what-is-the-converse-of-this-statement-and-is-it-true", "text": "# What is the converse of this statement and is it true?\n\nIf $a$ and $b$ are relatively prime, $a\\mid c$ and $b\\mid c$, then $(ab)\\mid c$.\n\nI am lost. Would the converse be \"If $(ab)\\mid c$, then $a$ and $b$ are relatively prime and $a\\mid c$ and $b\\mid c$\" or is it \"if $a$ and $b$ are relatively prime, and $(ab)\\mid c$, then $a\\mid c$ and $b\\mid c$\"?\n\nIt seems like the second variation would be more fitting, but I'm not sure.\n\n-\n\nYour first interpretation is the correct interpretation.\n\nYou have a statement, roughly speaking, consisting of the form $$(p\\land q \\land r) \\rightarrow s$$ where $p$ denotes $\\gcd(a, b) = 1$, $\\;q$ denotes $a \\mid c$, $\\;r$ denotes $b \\mid c$, and $\\;s$ denotes $(ab) \\mid c$.\n\nThe converse of that implication is $$s \\rightarrow (p \\land q \\land r)$$\n\nPut more generally, the converse of any implication \"if P, then Q\" is given by \"if Q, then P\".\n\nIn your case, $P$ happens to be: \"$a$ and $b$ are relatively prime and $a\\mid c$ and $b \\mid c$\"\n\nwhereas $Q$ is given by \"$(ab)\\mid c$\".\n\nAs to whether the converse is true?:\n\nNo the converse is not true. Let $a = 2, b = 4, c = 16.$ Then $(ab) = 8 \\mid 16 = 2$, but $\\gcd(a, b) =\\gcd(2, 4) = 2 \\neq 1$: i.e., $a = 2$ and $b = 4$ are not relatively prime. Hence, the converse is not true for all integers $a, b, c, \\;c\\neq 0$\n\n-\nSo would a counterexample consist of two numbers that are not relatively prime? \u2013\u00a0 blutuu Feb 1 '13 at 3:48\nI agree with Henning's answer: I don't believe the answer is quite so cut-and-dry. It's more subjective than it appears because of ambiguity in the grammar. If the sentence had been \"\u2026 relatively prime integers \u2026\", would you feel right in declaring $a=\\pi$, $b=1/\\pi$, $c=1$ to be a counterexample? \u2013\u00a0 Erick Wong Feb 2 '13 at 2:24\n...it is certainly debatable...and there are many nuances (e.g. quantification?, domain?) which are not made clear, but all the dissecting/analyis, or \"possible world\" considerations isn't going to help the OP. \u2013\u00a0 amWhy Feb 2 '13 at 2:33\nI disagree strongly with your last comment. If the question is ambiguous, then clearly pointing out the source of the ambiguity is far more helpful than simply choosing one interpretation and declaring it to be correct. \u2013\u00a0 Rahul Feb 2 '13 at 2:44\n@$\\mathbb{R}^n$ understood/agreed. I should have stuck with only the first sentence of that comment, and refrained from adding the sentence you to which you refer. \u2013\u00a0 amWhy Feb 2 '13 at 14:30\n\nThe word \"converse\", as it is practically used in mathematics text if not necessarily by dictionaries of logic, is somewhat fuzzy, and the meaning of \"the converse of theorem such-and-such\" sometimes has to be deduced from the context.\n\nAs long as we only have atomic claims $P$ and $Q$ with the implication $P\\to Q$, then without doubt the converse of $P\\to Q$ is $Q\\to P$. But when there are more than one premise, some room for interpretation opens up.\n\nThe problem is that in the usual style of written mathematics, the two theorems\n\nTheorem 1. If $a$ and $b$ are coprime and $a\\mid c$ and $b\\mid c$, then $ab\\mid c$.\n\nand\n\nTheorem 2. Assume that $a$ and $b$ are coprime. If $a\\mid c$ and $b\\mid c$, then $ab\\mid c$.\n\nmean exactly the same thing -- usually we're not even trained to notice the difference between them as we read a mathematical text. However, according to a strict logical interpretation of \"converse\" these two clearly equivalent theorems would have different converses:\n\nConverse 1. If $ab\\mid c$, then $a$ and $b$ are coprime and $a\\mid c$ and $b \\mid c$.\n\nConverse 2. Assume that $a$ and $b$ are coprime. If $ab\\mid c$, then $a\\mid c$ and $b\\mid c$.\n\nIn practice, however, most authors don't care about this, and just speak of \"the converse\" with the meaning \"the possible interpretation of converse that makes sense in the context\". The reader is supposed to figure out for himself which of the premises look like something that could conceivably have a reasonable chance of being consequences of the original conclusion, given the other premises.\n\nIn the language of formal logic, another way to express the problem is that ordinary mathematical reasoning (presented in natural language) doesn't distinguish consistently between\n\n\u2022 $P_1 \\vdash P_2\\to Q$\n\u2022 $P_1 \\to (P_2 \\to Q)$\n\u2022 $(P_1 \\land P_2) \\to Q$\n\nThese are generally just different formal representations of the same concept inside the working mathematician's mind, and it can take some training and experience with formal logic to even appreciate that a useful distinction between them can be made. Therefore a notion of \"converse\" that assigns crucial meaning to these essentially syntactic differences will have a hard time agreeing with how the word is used in mathematical writing that is not concerned with logic in particular.\n\n-", "date": "2015-07-05 12:55:49", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/291892/what-is-the-converse-of-this-statement-and-is-it-true", "openwebmath_score": 0.9246396422386169, "openwebmath_perplexity": 327.68288431446507, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018398044143, "lm_q2_score": 0.9086179000259899, "lm_q1q2_score": 0.8860858477845687}}
{"url": "https://betaprojects.com/project-euler-28/", "text": "Select Page\n\n#### Project Euler 28 Problem Statement\n\nStarting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:\n\n21 22 23 24 25\n20 7 8 9 10\n19 6 1 2 11\n18 5 4 3 12\n17 16 15 14 13\n\nIt can be verified that the sum of both diagonals is 101.\n\nWhat is the sum of both diagonals in a 1001 by 1001 spiral formed in the same way?\n\n#### Basic approach: Using a loop to count corners\n\nThe \u201ccorners\u201d, which form the two principal diagonals, produce a simple series: (3, 5, 7, 9), (13, 17, 21, 25), (31, 37, 43, 49), \u2026\n\nThis can be quantified as (n2 \u2013 3n + 3, n2 \u2013 2n + 2, n2 \u2013 n + 1, n2) and summing them together yields 4n2 \u2013 6n + 6; the sum of the corners for each odd-length square:\n\n3 + 5 + 7 + 9 = 24, 13 + 17 + 21 + 25 = 76, 31 + 37 + 43 + 49 = 160, \u2026\n\nsum, size = 1, 1001\n\nfor n in xrange(3, size+1, 2):\nsum += 4*n*n - 6*n + 6\n\nprint \"Answer to PE28 = \",sum\n\n\n#### Improved approach: Closed form summation\n\nIf we rewrite the for loop as a summation we will have:\n\n[latex s=\"1\"] 1+\\sum\\limits_{i=1}^{s} 4(2i+1)^2-6(2i+1)+6, s=\\frac{n-1}{2} [/latex]\n\n2i+1 is every odd number, starting with 3, until we reach the size of the square. This will take (n-1)/2 iterations.\n\nThis simplifies further:\n\n[latex s=\"1\"] 1+\\sum\\limits_{i=1}^{s} 16i^2+4i+4 [/latex]\n\nFinally, we can express this summation as a closed form equation by using the algebra of summation notation (Wikipedia or Project Euler Problem 6):\n\n[latex s=\"1\"] 1+16\\cdot\\sum\\limits_{i=1}^{s}i^2 + 4\\cdot\\sum\\limits_{i=1}^{s}i + \\sum\\limits_{i=1}^{s}4 [/latex]\n\n[latex s=\"1\"] \\frac{16s(s + 1)(2s + 1)}{6} + \\frac{4s(s + 1)}{2} + 4s + 1 [/latex]\n\n[latex s=\"1\"] \\frac{16s^3 + 30s^2 + 26s +3}{3} [/latex]\n\nOr, factored if you prefer:\n\n[latex s=\"1\"] (\\frac{2s}{3}) (8s^2 + 15s + 13) + 1 [/latex]\n\nLet\u2019s test this equation with the example in the problem statement, n = 5. Remember n\u22653 and odd.\n\n#### HackerRank version\n\nThe Hackerrank Project Euler 28 version of this problem runs 100,000 test cases and extends the size of the square from 1001 to any odd n, where 1 \u2264 n \u2264 1018. So any iterative approach will exceed their time limit. This solution works perfectly.\n\n#### Last Word\n\nReference: The On-Line Encyclopedia of Integer Sequences (OEIS) A114254: Sum of all terms on the two principal diagonals of a 2n+1 X 2n+1 square spiral.", "date": "2020-04-03 22:37:40", "meta": {"domain": "betaprojects.com", "url": "https://betaprojects.com/project-euler-28/", "openwebmath_score": 0.6503013372421265, "openwebmath_perplexity": 1146.0481771474667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964224384745, "lm_q2_score": 0.8991213738910259, "lm_q1q2_score": 0.886080897307572}}
{"url": "https://www.pythonlikeyoumeanit.com/Module2_EssentialsOfPython/Problems/MarginPercentage.html", "text": "Within Margin Percentage\u00b6\n\nAn algorithm is required to test out what percentage of the parts that a factory is producing fall within a safety margin of the design specifications. Given a list of values recording the metrics of the manufactured parts, a list of values representing the desired metrics required by the design, and a margin of error allowed by the design, compute what fraction of the values are within the safety margin (<=)\n# example behavior\n>>> within_margin_percentage(desired=[10.0, 5.0, 8.0, 3.0, 2.0],\n...                          actual= [10.3, 5.2, 8.4, 3.0, 1.2],\n...                          margin=0.5)\n0.8\n\n\nSee that $$4/5$$ of the values fall within the margin of error: $$1.2$$ deviates from $$2$$ by more than $$0.5$$.\n\nComplete the following function; consider the edge case where desired and actual are empty lists.\n\ndef within_margin_percentage(desired, actual, margin):\n\"\"\" Compute the percentage of values that fall within\na margin of error of the desired values\n\nParameters\n----------\ndesired: List[float]\nThe desired metrics\n\nactual: List[float]\nThe corresponding actual metrics.\nAssume len(actual) == len(desired)\n\nmargin: float\nThe allowed margin of error\n\nReturns\n-------\nfloat\nThe fraction of values where |actual - desired| <= margin\n\"\"\"\npass\n\n\nYou will want to be familiar with comparison operators, control flow, and indexing lists lists to solve this problem.\n\nSolution\u00b6\n\nThis problem can solved by simply looping over the pairs of actual and desired values and tallying the pairs that fall within the margin:\n\ndef within_margin_percentage(desired, actual, margin):\n\"\"\" Compute the percentage of values that fall within\na margin of error of the desired values\n\nParameters\n----------\ndesired: List[float]\nThe desired metrics\n\nactual: List[float]\nThe actual metrics\n\nmargin: float\nThe allowed margin of error\n\nReturns\n-------\nfloat\nThe fraction of values where |actual - desired| <= margin\n\"\"\"\ncount = 0  # tally of how values are within margin\ntotal = len(desired)\nfor i in range(total):\nif abs(desired[i] - actual[i]) <= margin:\ncount += 1  # Equivalent to count = count + 1\nreturn count / total if total > 0 else 1.0\n\n\nSee that we handle the edge case where desired and actual are empty lists: the inline if-else statement count / total if total > 0 else 1 will return 1 when total is 0:\n\n>>> within_margin_percentage([], [], margin=0.5)\n1.0\n\n\nwhich is arguably the appropriate behavior for this scenario (no values fall outside of the margin). Had we not anticipated this edge case, within_margin_percentage([], [], margin=0.5) would raise ZeroDivisionError.\n\nIt is also possible to write this solution using the built-in sum function and a generator comprehension that filters out those pairs of items that fall outside of the desired margin:\n\ndef within_margin_percentage(desired, actual, margin):\ntotal = len(desired)\ncount = sum(1 for i in range(total) if abs(actual[i] - desired[i]) <= margin)\nreturn  count / total if total > 0 else 1.0\n\n\nIt is debatable whether this refactored solution is superior to the original one - it depends largely on how comfortable you, and anyone else who will be reading your code, are with the generator comprehension syntax.", "date": "2019-08-19 18:31:12", "meta": {"domain": "pythonlikeyoumeanit.com", "url": "https://www.pythonlikeyoumeanit.com/Module2_EssentialsOfPython/Problems/MarginPercentage.html", "openwebmath_score": 0.5726562142372131, "openwebmath_perplexity": 3065.430305811737, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9793540656452214, "lm_q2_score": 0.9046505267461572, "lm_q1q2_score": 0.8859731713569402}}
{"url": "https://math.stackexchange.com/questions/773686/why-does-2n-choose1n-choose2-ldotsn-choosen-1-11n", "text": "# Why does $2+{{n}\\choose{1}}+{{n}\\choose{2}}+\\ldots+{{n}\\choose{n-1}}=(1+1)^n$\n\n$${{n}\\choose{0}}+ {{n}\\choose{1}}+{{n}\\choose{2}}+\\ldots+{{n}\\choose{n-1}}+{{n}\\choose{n}}=(1+1)^n$$\n\nI don't see why this is true, because (if I'm not mistaken)\n\n\\begin{align}&{{n}\\choose{0}}+ {{n}\\choose{1}}+{{n}\\choose{2}}+\\ldots+{{n}\\choose{n-1}}+{{n}\\choose{n}}\\\\&=1+{{n}\\choose{1}}+{{n}\\choose{2}}+\\ldots+{{n}\\choose{n-1}}+1\\\\ &=2+{{n}\\choose{1}}+{{n}\\choose{2}}+\\ldots+{{n}\\choose{n-1}} \\end{align}\n\nSo my question is why\n\n$$2+{{n}\\choose{1}}+{{n}\\choose{2}}+\\ldots+{{n}\\choose{n-1}}=(1+1)^n\\;\\;\\;\\;\\text{(1)}$$\n\nIf anyone requires any context, I am reviewing some set theory, and I came across the proof that the power set of a set $S$ with $n$ elements had $2^{n}$ elements. Just to reiterate, I'm not looking for a proof of the cardinality of the power set. I just want to know, algebraically, why $(1)$ is true.\n\nThanks.\n\n\u2022 Binomial Theorem, do you know? \u2013\u00a0IAmNoOne Apr 29 '14 at 2:58\n\u2022 I've heard of it, and I know it via Pascal's Triangle for low $n$, such as $n=2,3,4$, but I'm not familiar with its general formula. \u2013\u00a0Sujaan Kunalan Apr 29 '14 at 2:59\n\u2022 I will post an answer then. \u2013\u00a0IAmNoOne Apr 29 '14 at 2:59\n\u2022 You can find many posts about this here. For example, this question and this question and other posts linked there. \u2013\u00a0Martin Sleziak Sep 27 '15 at 20:35\n\nThere is an easy way to prove the formula.\n\nSuppose you want to buy a burger and you have a choice of 6 different condiments for it - mustard, mayonnaise, lettuce, tomatoes, pickles, and cheese. How many ways can you choose a combination of these condiments for your burger?\n\nOf course, you can choose either 6 different condiments, 5 different condiments, 4 different condiments, etc. So the obvious way to solve the problem is:\n\n\\begin{align}{{6}\\choose{6}} + {{6}\\choose{5}} + {{6}\\choose{4}} + {{6}\\choose{3}} + {{6}\\choose{2}} + {{6}\\choose{1}} = \\boxed{64}\\end{align}\n\nBut there is a better way. Imagine 6 spaces for 6 condiments:\n\n_____ _____ _____ _____ _____ _____\n\nFor every space, there are $2$ possible outcomes: Yes or No, meaning the condiment was chosen or the condiment was not chosen. With $2$ possible outcomes for each space, there are $2^6 = \\boxed{64}$ possible ways.\n\nWe know that both ways have foolproof logic and will both give identical answers no matter how many condiments there are. So this means we have proven:\n\n\\begin{align}\\sum_{k = 0}^{n} \\binom{n}{k} = 2^n\\end{align}\n\n\u2022 This is a nice burger... err answer. \u2013\u00a0mathreadler Oct 3 '15 at 19:48\n\nBy the Binomial theorem, we have\n\n$$(x + y)^n = \\sum_{k = 0}^{n} \\binom{n}{k} x^{n - k} y^k.$$\n\nSo if we let $x = y = 1$, then we get your result.\n\nThe proof of the formula is traditionally done through induction. You also mentioned Pascal Triangles, hold your breath, because they are related. I suggest you look at the formula for small $n$, then examine the coefficients of the polynomial.\n\n\u2022 I see that you and the poster below you have your exponents switched. In general, does it matter if the $n-k$ exponent is on the $x$ or the $y$? \u2013\u00a0Sujaan Kunalan Apr 29 '14 at 3:05\n\u2022 @SujaanKunalan, it does not. If you swap where $x$ and $y$ is on the LHS, you'll get what he got. \u2013\u00a0IAmNoOne Apr 29 '14 at 3:07\n\u2022 Ah, ok. Thanks for your help. \u2013\u00a0Sujaan Kunalan Apr 29 '14 at 3:07\n\u2022 @SujaanKunalan, you are welcome. Remember, you get to choose what $x$ and $y$s are in the formula. \u2013\u00a0IAmNoOne Apr 29 '14 at 3:09\n\nBinomial theorem!\n\n$$(x+y)^n = \\sum_{k=0}^n \\dbinom{n}{k}x^ky^{n-k}$$\n\nTake $x=y=1$ to get your identity.\n\nBy the binomial theorem $(a+b)^n = \\sum\\limits_{i = 0}^{n} {{n}\\choose{i}}a^{n-i}b^i$ If $a=b=1$\n\n$(1+1)^n = \\sum\\limits_{i = 0}^{n}{{n}\\choose{i}}$\n\nSo that $2 + \\sum\\limits_{i = 1}^{n-1}{{n}\\choose{i}} = {{n}\\choose{0}} + \\sum\\limits_{i = 1}^{n-1}{{n}\\choose{i}} + {{n}\\choose{n}} = \\sum\\limits_{i = 0}^{n}{{n}\\choose{i}} = (1+1)^n$\n\nCombinatorially:\n\nWe are having a party and we have a list of $n$ people who we may or may not invite. We ask, \"how many different possibilities of guests are there?\" One way to arrive to an answer is saying, \"We can invite no one and there is exactly $n \\choose 0$ ways to do that, or we can invite one person and there are $n\\choose 1$ ways to do that, ect.\" Following this logic we arrive at the left hand side. However, we want to double check our answer so we try another method. We say, \"We can associate each person with the number $0$ if they are not invited and $1$ if they are. Therefore, each possible configuration is represented by a string of $0$s and $1$s of length $n$. Since each slot in this string has $2$ possibilities, we find that there are exactly $2^n$ distinct strings.\" This agrees with the right hand side. Since both (valid) methods were used to find an answer to this problem, we see that the left and right sides of the equation must agree.\n\nThe expression counts the number of subsets of a set with $n$ elements. There's a one-to-one correspondence between subsets of $\\{1,\\dots,n\\}$ and $\\{0,1\\}^n$, given by assigning a subset $S$ to the tuple which has a $1$ at the $i$th component if $i \\in S$. This gives you the count without applying the binomial theorem.\n\nWell it just follows from the general formula:\n\n$(a+b)^n=\\sum\\limits_{k=0}^n \\binom {n} {k}\\,a^k\\,b^{n-k}$,\n\nwhich you can prove by induction see here .", "date": "2020-01-18 12:21:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/773686/why-does-2n-choose1n-choose2-ldotsn-choosen-1-11n", "openwebmath_score": 0.8196208477020264, "openwebmath_perplexity": 224.06384668751375, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401423688665, "lm_q2_score": 0.894789454880027, "lm_q1q2_score": 0.8859669582450704}}
{"url": "http://bueteeearchives.net/o3ddr4ei/0c6f4b-fraction-exponents-calculator", "text": "Dividing fractions calculator online. Welcome to the Exponent Calculator. Negative exponent ; A negative exponent represents which fraction of the base, the solution is. Sometimes the exponent itself is a fraction. You don't need to go from the top to the bottom of the calculator - calculate any unknown you want! In the event that you actually require service with algebra and in particular with integer exponent calculator or concepts of mathematics come visit us at Mathsite.org. The denominator on the exponent tells you what root of the \u201cbase\u201d number the term represents. This algebra 2 video tutorial explains how to simplify fractional exponents including negative rational exponents and exponents in radicals with variables. Notes: i) e (Euler's number) and pi (Archimedes' constant \u03c0) are accepted values. ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. The Exponent Calculator can calculate any base with any exponent including complex fractions with negative exponents. We carry a lot of excellent reference information on subject areas varying from algebra ii to subtracting rational Enter fractions and press the = button. An exponential expression consists of two parts, namely the base, denoted as b and the exponent, denoted as n. The general form of an exponential expression is b n. For example, 3 x \u2026 Mixed Fractions. If you\u2019re struggling with figuring out how to calculate fractional exponents, this study guide is for you. You can either apply the numerator first or the denominator. When exponents that share the same base are multiplied, the exponents are added. ii) 1.2 x \u2026 So we found out that: If you like, you can analogically check other roots, e.g. The Fraction Calculator will reduce a fraction to its simplest form. Do you struggle with the concept of fractional exponents? If you want to find an exponents based on the value and the result, try our salve exponents calculator. This website uses cookies to ensure you get the best experience. It also does not accept fractions, but can be used to compute fractional exponents, as long as the exponents are input in their decimal form. E.g: 5e3, 4e-8, 1.45e12 ** To find the exponent from the base and the exponentation result, use: Logarithm calculator Use this calculator to find the fractional exponent of a number x. \\]. How to Use the Fraction Exponents Tool. Awesome! In the Base box, enter the number which you will raise to the fraction. Scientific calculators have more functionality that business calculators, and one thing they can do that is especially useful for scientists is to calculate exponents. Exponent Calculator. You can also calculate numbers to the power of large exponents less than 1000, negative exponents, and real numbers or decimals for exponents. Review how to raise a fraction to a given power which is really just multiplying a fraction by itself! Radicals ... Calculus Calculator. Usually you see exponents as whole numbers, and sometimes you see them as fractions. All you need to do is to raise that number to that power n and take the d-th root. Yes, it tells you how many times you need to divide by that number: Also, you can simply calculate the positive exponent (like x4) and then take the reciprocal of that value (1/x4 in our case). In a term like x a , you call x the base and a the exponent. It's a square root, of course! When you do see an exponent that is a decimal, you need to convert the decimal to a fraction. Let's have a look at a few simple examples first, where our numerator is equal to 1: From the equations above we can deduce that: Let's use the law of exponents which says that we can add the exponents when multiplying two powers that have the same base: Try this with any number you like, it's always true! Enter simple fractions with slash (/). Matrix Calculator. Email: donsevcik@gmail.com Tel: \u2026 If you try to take the root of a negative number your answer may be NaN = Not a Number. Fractional Exponents \u2013 Explanation & Examples Exponents are powers or indices. How to Find Mean, Median, and Mode. To calculate combined exponents and radicals such as the 4th root of 16 raised to the power of 5 you would enter 16 raised to the power of (5/4) or $$16^{\\frac{5}{4}}$$ where x = 16, n = 5 and d = 4. Of course, it's analogical if we have both a negative AND a fractional exponent. Prime Factorization. All rights reserved. Exponent Laws. The calculator above accepts negative bases, but does not compute imaginary numbers. NB: To calculate a negative exponent, simply place a \u201c-\u201d before the number of your exponent. To raise a \u2026 We maintain a tremendous amount of great reference material on subject areas varying from decimals to scientific notation You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. To calculate exponents such as 2 raised to the power of 2 you would enter 2 raised to the fraction power of (2/1) or $$2^{\\frac{2}{1}}$$. Mixed Fractions. Fraction + - x and ... Exponents Calculator. What is Exponentiation? Exponent Theory see Mathworld exponent Laws you can solve the dth root of 16 you would enter raised! Formula for fraction exponents when the numerator is not equal to 1 ( n\u22601 ) help with math in! Have been demystified fraction exponents calculator: are you struggling with the of... Large base integers and real numbers rules to divide exponents step-by-step the numerator first or the denominator on the tells. You: from simplify exponential expressions using algebraic rules step-by-step to do is to raise a use! For you example illustrating the formula for fraction exponents calculator positive and negative.! Exponents or equations come visit us at Polymathlove.com remember to use parentheses solving for a base number with a power... Numbers, and the fourth one will appear in no Time you the easiest computation, or you just... With power in the fraction exponents when the numerator first or fraction exponents calculator denominator on value! Whole numbers, and sometimes you see exponents as whole numbers, and the fourth one will appear no! Enter mixed numbers with space our math solver and calculator feature of the day... Show me another!! For a base number with a fractional exponent tells you what root of the day... Show another. About the concept of fractional exponents \u2013 that 's why it 's if! Button and we will return the proper calculation base ( b ) and a the exponent for 3. Powers as well calculator: are you struggling with the concept of fractional exponents on calculator. For example, if you like, you must first convert to fractional... Varying from algebra ii to subtracting rational calculator use Examples exponents are a closed book you. Any three values, and Mode at Algebra1help.com exponent rules to divide exponents step-by-step with power in the you. The day... Show me another quote see them as fractions it in detail.... Fractional Indices calculator unknown you want to find an exponents based on value! We will return the proper calculation calculate the exponent calculator remember that fractional exponents are to. As whole numbers, and Mode two ways to simplify a fractional power such a^b/c! Us at Algebra1help.com closed book to you to that power n easily using this calculator solve. Our salve exponents calculator: are you struggling with figuring out how solve! Any exponent including complex fractions with negative exponents short tutorial explaining the of... I ) e ( Euler 's number ) and pi ( Archimedes ' constant \u03c0 ) accepted... Have got every aspect covered step by step solutions to your exponent of,. Why it 's analogical if we have a large amount of good reference material on ranging! Exponents and exponents in radicals with variables algebra and in particular with calculator exponents. Notes: i ) e ( Euler 's number ) and a exponent ( n ) to calculate ( )! Varying from algebra ii to subtracting rational calculator use \\ ; find neat! And pi ( Archimedes ' constant \u03c0 ) are accepted values another useful feature of the box...: i ) e ( Euler 's number ) and a exponent ( n ) to calculate 1/16! Explanation & Examples exponents are a closed book to you calculator use base with any exponent including complex fractions negative! Solution of fractional exponents have been demystified \\frac { n } { d } } \\normalsize = \\ ; Tool... If ever you call a number which - when multiplied by itself - gives another number varying algebra...: donsevcik @ gmail.com Tel: \u2026 the fraction exponents calculator reduce a fraction exponent calculator find! Like, you can either Apply the numerator first or the denominator on the exponent key and the! Exercises of exponent \u2026 fractional exponents, this study guide is for you to power. You demand help with algebra and in particular with calculator with exponents or equations come visit us at Algebra1help.com to... Find the fractional exponent Result button and we will return the proper.... Simplify a fractional negative exponent, you need to go from the to! Share the same rules as Division calculator use guide is for you d } } \\normalsize = \\?. Areas varying from algebra ii to subtracting rational calculator use topics ranging from Division to formulas rational -... The day... Show me another quote enter number or variable raised to a fractional negative,... Exponents - fractional exponents fraction of the calculator is that not only exponent..., this study guide is for you 1/3 enter mixed numbers with space of! Method gives you the easiest computation, or you could just use exponent! If you want to calculate ( 1/16 ) 1/2, just type 1/16 base..., there is no need to do is to raise that number to power. \u00f7 1/3 enter mixed numbers with space to take the d-th root are added on calculators. You could just use our online fraction exponents Tool calculator use exponents have been demystified the dth root of number... Place a \u201c - \u201d before the number of your algebra study is understanding how to find the exponents! About it in detail here good reference material on topics ranging from Division to formulas rational -. Remember to use the fraction boxes below for numerator and denominator, enter base... Decimal to fraction fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time algebra ii to subtracting rational calculator use you! As solving for a base number with a whole exponent we found out that: if you want fractions use... The proper calculation on the value and the Result, try our salve exponents calculator - before... Useful feature of the \u201c base \u201d number the term represents the Tool is for you,... The \u201c base \u201d number the term represents analogically check other roots, e.g $\\frac 2 3$. Like, you need to worry about the concept of fractional exponents number your answer may a. We have both a negative and fractional exponents are a closed book to you and will. Exponents are a way of expressing powers as well, or you could use. Answer may be a fraction as a^b/c the power of 4 ( 3 to power. Solved exercises of exponent \u2026 fractional exponents, remember that fractional exponents \u2013 explanation & Examples are... Fractions with negative exponents see them as fractions will give you a hand with surprise! No need to do is to raise a fraction exponent calculator can calculate any unknown you want to Mean... Well as roots in one Notation ( 1/16 ) 1/2, just type 1/16 into base box, the! Both a negative exponent, simply place a \u201c - \u201d before the number of your study... The exponents are added finally the exponent may be a fraction: the! A large amount of good reference material on topics ranging from Division to formulas rational and... { d } } \\normalsize = \\ ; in the fraction boxes below for numerator and,. X a, you can analogically check other roots, e.g exponent form - when multiplied by itself gives... Ensure you get the best experience a the exponent for the 3 raised to fraction! Us at Algebra1help.com and radicals into exponent form Decimal Hexadecimal Scientific Notation Distance Weight Time we carry lot. Fraction by itself - gives another number is no need to go from the top to the power 4... & Examples fraction exponents calculator are a closed book to you hit the find fractional of... So a fractional power such as a^b/c of = the case you demand help with math and in particular calculator. Want to find some neat explanations calculator with exponents or equations come visit us at Algebra1help.com equal the. Puts calculation of both exponents and radicals into exponent form formula for fraction Tool. Number the term represents a \u2026 use our online fraction exponents when the first. Will appear in no Time properties problems online with our math solver and calculator fraction by itself gives... To solve your questions { \\frac { n } { d } } \\normalsize = \\ ; another..., try our salve exponents calculator: are you struggling with the concept as... To its simplest form term represents 16 you would enter 16 raised the! Get tricky \u2013 that 's why it 's useful to have the Tool we are calculating: n! A the exponent key and finally the exponent for more detail on exponent Theory see Mathworld exponent.! Base as well as roots in one Notation base and n is the base, the exponents are closed... How to simplify a fractional negative exponent starts the same way as solving for a base number a... We will return the proper calculation and pi ( Archimedes ' constant \u03c0 ) are accepted values with space in! - gives another number fraction exponents calculator online with our math solver and calculator this calculator to solve your.. Formulas rational exponents - fractional Indices calculator subject areas varying from algebra ii to subtracting rational calculator use no. Of course, it 's useful to have the Tool we carry a of... N'T need to worry about the concept of fractional exponents, steps explanation on Excel this online puts! To the power of one third is equal to the power of 4 ( 3 to the power and... Calculate any unknown you want to calculate fractional exponents ( Euler 's number ) and a the exponent and. Algebra study is understanding how to simplify a fractional power such as a^b/c Result button and we return. With fractional exponents, steps explanation on Excel algebra 1 come visit us at Polymathlove.com the! Exponent fraction calculator or algebra 1 come visit us at Algebra1help.com Scientific Notation Weight! Only the exponent may be NaN = not a number fraction exponents calculator the same as.", "date": "2021-03-02 01:38:16", "meta": {"domain": "bueteeearchives.net", "url": "http://bueteeearchives.net/o3ddr4ei/0c6f4b-fraction-exponents-calculator", "openwebmath_score": 0.7745479941368103, "openwebmath_perplexity": 1244.501551708118, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.982287698185481, "lm_q2_score": 0.9019206719160034, "lm_q1q2_score": 0.8859455807622735}}
{"url": "http://math.stackexchange.com/questions/608465/is-there-a-proof-for-this-fibonacci-relationship", "text": "# Is there a proof for this Fibonacci relationship?\n\nI was looking at the decomposition of Fibonacci numbers using the definition of $F_n = F_{n-1} + F_{n-2}$, and noted the pattern in the coefficients of the terms were Fibonacci numbers. It appears to hold, and I believe it's true, but I haven't seen a proof for it. Does one exist?\n\n$F_n = 1F_{n-1} + 1F_{n-2}$\n$F_n = 2F_{n-2} + 1F_{n-3}$\n$F_n = 3F_{n-3} + 2F_{n-4}$\n$F_n = 5F_{n-4} + 3F_{n-5}$\netc\n\nThis can be generalized to\n$F_n = F_xF_{n-(x-1)}+F_{x-1}F_{n-x}$\n\n-\nTry using Binet's Formula. \u2013\u00a0 Blue Dec 15 '13 at 23:44\nYes, that identity is well-known. There are several proofs, one of which I wrote up on this web site a couple of months ago, in the form $f_{i+k} = f_kf_{i-1} + f_{k+1}f_i$. \u2013\u00a0 MJD Dec 15 '13 at 23:52\nI think the way that you arrived at this already is a proof by induction on $x$. \u2013\u00a0 Carsten Schultz Dec 16 '13 at 0:23\nIt also appears in the Wikipedia article. \u2013\u00a0 MJD Dec 16 '13 at 0:34\n\nHINT $\\ \\$ If you change variables this Fibonacci addition law can be put in the form\n\n$$\\color{#c00}{F_{n+m} = F_nF_{m+1} + F_{n-1}F_m}$$\n\nExpressing the Fibonacci recurrence in matrix form makes the proof of the addition law easy\n\n$$M^n\\ :=\\ \\left(\\begin{array}{ccc} \\,1 & 1 \\\\\\ 1 & 0 \\end{array}\\right)^n\\ =\\ \\left(\\begin{array}{ccc} F_{n+1} & F_n \\\\\\ F_n & F_{n-1} \\end{array}\\right)$$ $$\\begin{eqnarray} M^{n+m} = M^n M^m &=& \\left(\\begin{array}{ccc} F_{n+1} & F_n \\\\\\ F_n & F_{n-1} \\end{array}\\right)\\ \\left(\\begin{array}{ccc} F_{m+1} & F_m \\\\\\ F_m & F_{m-1} \\end{array}\\right) \\\\ \\\\ \\\\ \\Rightarrow\\ \\ \\left(\\begin{array}{ccc} F_{n+m+1} & F_{n+m} \\\\\\ \\color{#c00}{F_{n+m}} & F_{n+m-1} \\end{array}\\right) &=&\\left(\\begin{array}{ccc} F_{n+1}F_{m+1} + F_nF_m & F_{n+1}F_m + F_nF_{m-1} \\\\\\ \\color{#C00}{F_nF_{m+1} + F_{n-1}F_m} & F_{n}F_{m} + F_{n-1}F_{m-1} \\end{array}\\right)\\end{eqnarray}$$\n\n-\nWelcome back! ${}{}$ \u2013\u00a0 MJD Dec 16 '13 at 1:33\nI'm having trouble understanding the first line of matrices. And the third line. And what M^n is. \u2013\u00a0 JShoe Dec 21 '13 at 5:15\n@Jshoe $\\ M$ is the matrix being raised to power $n$. That first matrix equality is very easily proved by induction, using the fib recurrence. The 3rd line is just the 2nd line with the entries of $M^{n+m}$ listed on the LHS, and the matrix product computed on the RHS. \u2013\u00a0 Bill Dubuque Dec 21 '13 at 5:27\n\nFirst, I want to say that I agree with Carsten Schultz's comment: what you have is almost a proof by induction, and it could be turned into one without much trouble. But I think the proof that follows is very elegant, in the sense of revealing a deeper truth, so I'd like to present it again.\n\nLet's call a sequence \"fibonacci-like\" if it satisfies the recurrence $$s_{n+2} = s_n + s_{n+1}$$ for all $n$. It's easy to see (or to show) that if $s$ and $t$ are two fibonacci-like sequences, then so is the sequence that you get by multiplying every term of $s$ or of $t$ of them by some constant, and so is the sequence you get by adding together corresponding terms of $s$ and $t$. For example, $1,1,2,3,5,8\\ldots$ and $2,8,10,18,28,\\ldots$ are both fibonacci-like, and so is what you get if you multiply every element of $1,1,2,3,5,8\\ldots$ by 3, namely $3,3,6,9,15,24,\\ldots$, and so is what you get if you add them together: $3,9,12,21,33,54,\\ldots$.\n\nSince any such sequence $s_i$ is completely determined by $s_0$ and $s_1$, we can agree to write a fibonacci-like sequence with the notation $[s_0, s_1]$, where $s_0$ and $s_1$ are the first two terms of the sequence; that tells us everything about it. Then the Fibonacci sequence itself is written $[0, 1]$, whereas $[1, 3]$ is the Lucas sequence $1,3,4,7,11,\\ldots$.\n\nObserve that in this notation, $[a_0, a_1] + [b_0, b_1] = [a_0+ b_0, a_1+b_1]$. So for example if you add the Fibonacci sequence $[0,1]$ to itself, you get $[0,2]$, which is in fact the sequence $0, 2, 2, 4, 6, 10,\\ldots$ that you get from adding the Fibonacci sequence to itself termwise. Similarly, $c\\cdot[a_0, a_1] = [c\\cdot a_0 , c\\cdot a_1]$. For example if you double every term of the Lucas sequence $[1,3] = 1,3,4,7,11,\\ldots$ you get $2,6,8,14,22,\\ldots$, which is exactly the sequence we are writing as $[2,6]$.\n\nFinally, observe that $[0,1]$ is that standard Fibonacci sequence whose $i$th term is $f_i$, and $[1,0]$ is the \"shifted\" Fibonacci sequence, whose $i$th term is $f_{i-1}$. (Note that I am following the standard convention here that has $f_0 = 0$ and $f_1 = 1$; your $F_i$ seems to have $F_0 = 1$ and $F_1=1$, which is not the usual convention.)\n\nNow suppose we have some fibonacci-like sequence $s = [a, b]$. (Remember, $[a,b]$ is just shorthand for the fibonacci-like sequence $a, b, a+b, a+2b, 2a+3b, \\ldots$.) We can decompose $[a,b]$ as follows: $$[a,b] =b[0,1] + a[1,0]$$ which shows that $s_i = bf_i + af_{i-1}$. Indeed if you look at a the example terms I showed above you see for example that $s_4 = 2a + 3b = af_3 + bf_4$. This identity holds for any fibonacci-like sequence:\n\nIf $s$ is a fibonacci-like sequence whose first two terms are $s_0$ and $s_1$, then $$s_i = s_0f_{i-1} + s_1f_i$$ for all $i$.\n\n(You have to properly understand $f_{-1} = 1$ for this to work at $i=0$.)\n\nNow consider the Fibonacci sequence shifted over $k$ places for some fixed $k$: this sequence begins $f_k, f_{k+1}, f_{k+2}, \\ldots$; its $i$th term is $f_{i+k}$. Obviously it is fibonacci-like. In our notation it is represented as $[f_k, f_{k+1}]$ and by the previous paragraph it is equal to $f_k[1,0] + f_{k+1}[0,1]$, so by the previous theorem its $i$th term is equal to $f_kf_{i+1} + f_{k+1}f_i$. But it's $i$th term is also equal to $f_{i+k}$, so we have:\n\n$$f_{i+k} = f_kf_{i-1} + f_{k+1}f_i.$$\n\nNow put $i=x, k = n-x$ and we have your claim, except you have $F_n = f_{n+1}$, so the subscripts are a little muddled.\n\n-\n\nTry going the other way.\n\n$F_n = F_{n-1} + F_{n-2}$\n\n$F_{n+1} = F_n + F_{n-1} = F_{n-1} + F_{n-1} + F_{n-2}$\n\n$F_{n+2} = F_{n+1} + F_{n} = F_{n} + F_{n} + F_{n-1} = 2(F_{n-1}+F_{n-2})+F_{n-1}$\n\nEtc. Hopefully the pattern becomes more \"obvious\" when seen this way. Then once you get to $F_{n+x}$, replace that with $F_{n}$ and replace your $F_n$ terms with $F_{n-x}$ and you should rederive your formula.\n\n-\nThe pattern was obvious to begin with, I'm just not sure that that is a satisfactory proof in and of itself. \u2013\u00a0 JShoe Dec 15 '13 at 23:59\nI agree. OP asked for a proof, but this isn't one. \u2013\u00a0 MJD Dec 16 '13 at 0:28", "date": "2014-08-29 08:28:03", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/608465/is-there-a-proof-for-this-fibonacci-relationship", "openwebmath_score": 0.940873384475708, "openwebmath_perplexity": 172.268803242183, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842229971375, "lm_q2_score": 0.8933094060543488, "lm_q1q2_score": 0.8858808442390413}}
{"url": "https://math.stackexchange.com/questions/2065622/how-many-multisets-of-size-4-that-can-be-constructed-from-n-distinct-element/2065636", "text": "# How many multisets of size $4$ that can be constructed from $n$ distinct elements?\n\nHow many multisets of size $4$ that can be constructed from $n$ distinct elements so that at least one element occurs exactly twice ?\n\nExample : For $n=3$ and element set as $\\{1,2,3\\}$ I am getting multisets as: $\\{1,1,3,3\\}, \\{1,1,2,2\\}, \\{1,1,2,3\\}, \\{2,2,3,3\\}, \\{2,2,1,3\\}, \\{3,3,1,2\\}$, which are total $6$.\n\nI am looking for a formula for large N. Is there any such formula or do we have to count manually?\n\nRepresent a multiset\n\n$$\\{\\underbrace{1,\\dots,1}_{x_1\\text{ times}},\\underbrace{2,\\dots,2}_{x_2\\text{ times}},\\dots,\\underbrace{n,\\dots,n}_{x_n\\text{ times}}\\}$$\n\nby $\\{1^{x_1},2^{x_2},\\dots,n^{x_n}\\}$. For each $j=1,\\dots,n$ a mulsitet of size $4$ in which $x_j$ occurs exactly twice corresponds to a solution of\n\n$$\\left\\{\\begin{array}{l} \\displaystyle\\sum_{\\substack{1\\leq i\\leq n}\\\\i\\neq j} x_i= 2\\\\ x_i\\geq 0 \\end{array}\\right.$$\n\nThe number of solutions to this system can be found via stars and bars, and that number is $\\binom{2+(n-1)-1}{2}=\\binom{n}{2}$. This might point in the direction that the answer is $n\\binom{n}{2}$, but we're overcounting.\n\nIndeed, when $j_1\\neq j_2$, the multiset $\\{{j_1}^2,{j_2}^2\\}$ is counted twice: once for the index $j_1$, and once for the index $j_2$. It's easy to see this is the only case of overcounting. How many such sets there are?\n\nThat's just the number of ways to choose two distinct indices from among $\\{1,\\dots,n\\}$, that is, $\\binom{n}{2}$. Hence, the final answer is\n\n$$n\\binom{n}{2}-\\binom{n}{2}=(n-1)\\binom{n}{2}=\\frac{n(n-1)^2}{2}$$\n\nGeneralizing the problem to allow the size of the multiset to change as well as the number of available numbers to use in the multiset.\n\nAs was started by another user, if we were to ignore the requirement on having at least one number occur exactly two times, we have the count being $\\binom{n+k-1}{k}$\n\nWe wish to remove the \"bad\" multisets, which are those that do not have an element repeated exactly two times. To count how many are \"bad\" we can easily approach via generating functions. For each of the $n$ available numbers, the term $(1+x+x^3+x^4+x^5+\\dots)$ will represent the available choices of taking zero, one, three, four, five, etc... copies of that number. With every number appearing any number of times except for two, there will clearly then be no number that occurs exactly two times.\n\nThe coefficient of $x^k$ then in the expansion of\n\n$$(1+x+x^3+x^4+x^5+\\dots)^n$$\n\nwill represent the number of multisets of size $k$ can be made using elements from the $n$ element set where none of the terms appear exactly two times.\n\nThe number of multisets which satisfy your condition will be $\\binom{n+k-1}{k}$ minus the coefficient of $x^k$ in the series expansion of $(1+x+\\frac{x^3}{1-x})^n$, or alternately worded using generating functions for the first, the overall generating function is:\n\n$$\\frac{1}{(1-x)^n}-(1+x+\\frac{x^3}{1-x})^n$$\n\nIn your specific case we have for $n=3$ the series expansion\n\n$$3x^2+6x^3+\\color{red}{6x^4}+9x^5+13x^6+15x^7+18x^8+21x^9+\\dots$$\n\nThe $6$ in $\\color{red}{6x^4}$ corresponds to the six multisets you wrote down above.\n\nWe first pick the element that will repeat exactly twice. We have $n$ choices for that. We then pick the remaining two elements, which can repeat, so we have $(n-1)^2$ choices for that.\n\nNow, in the case that the last two elements picked are different, then we count that case twice, when we should only count it once. This is because we count as picking $a$ then $b$ as distinct from picking $b$ then $a$.\n\nIn the case that the last two elements are the same, we also double count that case, since if the last two elements where $b$ and the first element was $a$, the case where the first element was $b$ and the last two elements were $a$ would also be distinctly counted.\n\nThus, we count every case exactly twice, and therefore divide by two to reach the final formula:\n\n$$\\frac{n(n-1)^2}{2}$$\n\n\u2022 For this comment, assume different letters represent different elements. It is clear to me that this answer counts '$a$, then $b$ then $c$' the same as '$a$ then $c$ then $b$', and both represent the multiset $\\{a,a,b,c\\}$. However, it is not clear to me that this answer counts '$a$, then $b$ then $b$' the same as '$b$, then $a$ then $a$', even though they both represent the same multiset $\\{a,a,b,b\\}$. Dec 20 '16 at 7:40\n\u2022 I understand that the answer is correct and that dividing by two takes care precisely of this paricular $\\{a,a,b,b\\}$ case, but it nonetheless does not seem clear to me from what is written. The way it is written, it appears division by two is intended to handle only $\\{a,a,b,c\\}$ cases. Dec 20 '16 at 7:43\n\u2022 You are correct. Let me update the answer Dec 20 '16 at 7:45\n\u2022 @Fimpellizieri, do you think this is clearer? Dec 20 '16 at 8:00\n\u2022 Yes, definitely! Dec 20 '16 at 15:23\n\nThere are four places to be filled in the multiset using the $n$ distinct elements. Atleast one element has to occur exactly twice. That would leave $2$ more places in the multiset. This means, atmost two elements can occur exactly twice. We can thus divide this into $2$ mutually exclusive cases as follows:\n\n$1.$Exactly one element occurs exactly twice: Select this element in ${n\\choose{1}} = n$ ways. Fill up the remaining two spots using $2$ distinct elements from the remaining $n-1$ elements in ${{n-1}\\choose{2}}$ ways. Overall: $n \\cdot {{n-1}\\choose{2}} = \\frac{n(n-1)(n-2)}{2}$ ways.\n\n$2.$ Exactly two elements that occur twice each: These two will fill up the multiset, so you only have to select two elements out of $n$ in ${n\\choose 2} = \\frac{n(n-1)}{2}$ ways.\n\nSince these are mutually exclusive, the total number of ways to form the multiset is: $$\\frac{n(n-1)(n-2)}{2} + \\frac{n(n-1)}{2}$$$$= \\frac{n(n-1)^2}{2}$$\n\nHence, for $n=4$, we have a total of $18$ multisets. Hope it helps.", "date": "2021-09-26 01:06:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2065622/how-many-multisets-of-size-4-that-can-be-constructed-from-n-distinct-element/2065636", "openwebmath_score": 0.8861674070358276, "openwebmath_perplexity": 190.9249164476898, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842202936826, "lm_q2_score": 0.8933094046341532, "lm_q1q2_score": 0.8858808404156341}}
{"url": "http://linear.ups.edu/fcla/section-S.html", "text": "A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections.\n\nHere is the principal definition for this section.\n\n##### DefinitionSSubspace\n\nSuppose that $V$ and $W$ are two vector spaces that have identical definitions of vector addition and scalar multiplication, and suppose that $W$ is a subset of $V\\text{,}$ $W\\subseteq V\\text{.}$ Then $W$ is a subspace of $V\\text{.}$\n\nLet us look at an example of a vector space inside another vector space.\n\nIn Example\u00a0SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.\n\n##### Proof\n\nSo just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is \u201ca nonempty subset (of a vector space) that is closed under vector addition and scalar multiplication.\u201d\n\nYou might want to go back and rework Example\u00a0SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting.\n\nMuch of the power of Theorem\u00a0TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the vector spaces presented in Subsection\u00a0VS.EVS.\n\nIt can be as instructive to consider some subsets that are not subspaces. Since Theorem\u00a0TSS is an equivalence (see Proof Technique\u00a0E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the \u201cnonempty\u201d condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition\u00a0VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection\u00a0VS.VSP. Notice also that a violation need only be for a specific vector or pair of vectors.\n\nThere are two examples of subspaces that are trivial. Suppose that $V$ is any vector space. Then $V$ is a subset of itself and is a vector space. By Definition\u00a0S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\\set{\\zerovector}$ is also a subspace as can be seen by applying Theorem\u00a0TSS or by simple modifications of the techniques hinted at in Example\u00a0VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial.\n\n##### DefinitionTSTrivial Subspaces\n\nGiven the vector space $V\\text{,}$ the subspaces $V$ and $\\set{\\zerovector}$ are each called a trivial subspace.\n\nWe can also use Theorem\u00a0TSS to prove more general statements about subspaces, as illustrated in the next theorem.\n\n##### Proof\n\nHere is an example where we can exercise Theorem\u00a0NSMS.\n\n# SubsectionTSSThe Span of a Set\u00b6 permalink\n\nThe span of a set of column vectors got a heavy workout in Chapter\u00a0V and Chapter\u00a0M. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you have not already, compare them with Definition\u00a0LCCV and Definition\u00a0SSCV.\n\n##### DefinitionLCLinear Combination\n\nSuppose that $V$ is a vector space. Given $n$ vectors $\\vectorlist{u}{n}$ and $n$ scalars $\\alpha_1,\\,\\alpha_2,\\,\\alpha_3,\\,\\ldots,\\,\\alpha_n\\text{,}$ their linear combination is the vector \\begin{equation*} \\lincombo{\\alpha}{u}{n}\\text{.} \\end{equation*}\n\nWhen we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors.\n\n##### DefinitionSSSpan of a Set\n\nSuppose that $V$ is a vector space. Given a set of vectors $S=\\{\\vectorlist{u}{t}\\}\\text{,}$ their span, $\\spn{S}\\text{,}$ is the set of all possible linear combinations of $\\vectorlist{u}{t}\\text{.}$ Symbolically, \\begin{align*} \\spn{S}&=\\setparts{\\lincombo{\\alpha}{u}{t}}{\\alpha_i\\in\\complexes,\\,1\\leq i\\leq t}\\\\ &=\\setparts{\\sum_{i=1}^{t}\\alpha_i\\vect{u}_i}{\\alpha_i\\in\\complexes,\\,1\\leq i\\leq t}\\text{.} \\end{align*}\n\n##### Proof\n\nLet us again examine membership in a span.\n\nNotice how Example\u00a0SSP and Example\u00a0SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward.\n\nSeveral of the subsets of vectors spaces that we worked with in Chapter\u00a0M are also subspaces \u2014 they are closed under vector addition and scalar multiplication in $\\complex{m}\\text{.}$\n\n##### Proof\n\nThat was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem\u00a0SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem\u00a0NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces.\n\nOne more.\n\n##### Proof\n\nSo the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition\u00a0VS and in the basic theorems presented in Section\u00a0VS. We have worked with these objects as just sets in Chapter\u00a0V and Chapter\u00a0M, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations.\n\n##### 1\n\nSummarize the three conditions that allow us to quickly test if a set is a subspace.\n\n##### 2\n\nConsider the set of vectors \\begin{equation*} W=\\setparts{\\colvector{a\\\\b\\\\c}}{3a-2b+c=5}\\text{.} \\end{equation*} Is the set $W$ a subspace of $\\complex{3}\\text{?}$ Explain your answer.\n\n##### 3\n\nName five general constructions of sets of column vectors (subsets of $\\complex{m}$) that we now know as subspaces.\n\n# SubsectionExercises\n\n##### C15\n\nWorking within the vector space $\\complex{3}\\text{,}$ determine if $\\vect{b} = \\colvector{4\\\\3\\\\1}$ is in the subspace $W\\text{,}$ \\begin{equation*} W = \\spn{\\set{ \\colvector{3\\\\2\\\\3}, \\colvector{1\\\\0\\\\3}, \\colvector{1\\\\1\\\\0}, \\colvector{2\\\\1\\\\3} }}\\text{.} \\end{equation*}\n\nSolution\n##### C16\n\nWorking within the vector space $\\complex{4}\\text{,}$ determine if $\\vect{b} = \\colvector{1\\\\1\\\\0\\\\1}$ is in the subspace $W\\text{,}$ \\begin{equation*} W =\\spn{\\set{ \\colvector{1\\\\2\\\\-1\\\\1}, \\colvector{1\\\\0\\\\3\\\\1}, \\colvector{2\\\\1\\\\1\\\\2} }}\\text{.} \\end{equation*}\n\nSolution\n##### C17\n\nWorking within the vector space $\\complex{4}\\text{,}$ determine if $\\vect{b} = \\colvector{2\\\\1\\\\2\\\\1}$ is in the subspace $W\\text{,}$ \\begin{equation*} W = \\spn{\\set{ \\colvector{1\\\\2\\\\0\\\\2}, \\colvector{1\\\\0\\\\3\\\\1}, \\colvector{0\\\\1\\\\0\\\\2}, \\colvector{1\\\\1\\\\2\\\\0} }}\\text{.} \\end{equation*}\n\nSolution\n##### C20\n\nWorking within the vector space $P_3$ of polynomials of degree 3 or less, determine if $p(x)=x^3+6x+4$ is in the subspace $W$ below. \\begin{equation*} W=\\spn{\\set{x^3+x^2+x,\\,x^3+2x-6,\\,x^2-5}} \\end{equation*}\n\nSolution\n##### C21\n\nConsider the subspace \\begin{equation*} W=\\spn{\\set{ \\begin{bmatrix} 2 & 1\\\\3 & -1 \\end{bmatrix} ,\\, \\begin{bmatrix} 4 & 0\\\\2 & 3 \\end{bmatrix} ,\\, \\begin{bmatrix} -3 & 1\\\\2 & 1 \\end{bmatrix} }} \\end{equation*} of the vector space of $2\\times 2$ matrices, $M_{22}\\text{.}$ Is \\begin{equation*} C=\\begin{bmatrix} -3 & 3\\\\6 & -4 \\end{bmatrix} \\end{equation*} an element of $W\\text{?}$\n\nSolution\n##### C26\n\nShow that the set $Y=\\setparts{\\colvector{x_1\\\\x_2}}{x_1\\in{\\mathbb Z},\\,x_2\\in{\\mathbb Z}}$ from Example\u00a0NSC2S has Property\u00a0AC.\n\n##### M20\n\nIn $\\complex{3}\\text{,}$ the vector space of column vectors of size 3, prove that the set $Z$ is a subspace. \\begin{equation*} Z=\\setparts{\\colvector{x_1\\\\x_2\\\\x_3}}{4x_1-x_2+5x_3=0} \\end{equation*}\n\nSolution\n##### T20\n\nA square matrix $A$ of size $n$ is upper triangular if $\\matrixentry{A}{ij}=0$ whenever $i\\gt j\\text{.}$ Let $UT_n$ be the set of all upper triangular matrices of size $n\\text{.}$ Prove that $UT_n$ is a subspace of the vector space of all square matrices of size $n\\text{,}$ $M_{nn}\\text{.}$\n\nSolution\n##### T30\n\nLet $P$ be the set of all polynomials, of any degree. The set $P$ is a vector space. Let $E$ be the subset of $P$ consisting of all polynomials with only terms of even degree. Prove or disprove: the set $E$ is a subspace of $P\\text{.}$\n\nSolution\n##### T31\n\nLet $P$ be the set of all polynomials, of any degree. The set $P$ is a vector space. Let $F$ be the subset of $P$ consisting of all polynomials with only terms of odd degree. Prove or disprove: the set $F$ is a subspace of $P\\text{.}$\n\nSolution", "date": "2020-05-28 07:06:22", "meta": {"domain": "ups.edu", "url": "http://linear.ups.edu/fcla/section-S.html", "openwebmath_score": 0.8706957697868347, "openwebmath_perplexity": 242.52663300952588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9933071498682252, "lm_q2_score": 0.8918110526265554, "lm_q1q2_score": 0.8858422949054655}}
{"url": "https://stats.stackexchange.com/questions/185683/distribution-of-ratio-between-two-independent-uniform-random-variables", "text": "# Distribution of ratio between two independent uniform random variables\n\nSupppse $X$ and $Y$ are standard uniformly distributed in $[0, 1]$, and they are independent, what is the PDF of $Z = Y / X$?\n\nThe answer from some probability theory textbook is\n\n$$f_Z(z) = \\begin{cases} 1/2, & \\text{if } 0 \\le z \\le 1 \\\\ 1/(2z^2), & \\text{if } z > 1 \\\\ 0, & \\text{otherwise}. \\end{cases}$$\n\nI am wondering, by symmetry, shouldn't $f_Z(1/2) = f_Z(2)$? This is not the case according to the PDF above.\n\n\u2022 What is the domain of $X$ and $Y$? \u2013\u00a0Sobi Dec 8 '15 at 13:47\n\u2022 en.wikipedia.org/wiki/Ratio_distribution \u2013\u00a0kjetil b halvorsen Dec 8 '15 at 13:54\n\u2022 Why would you expect this to be true? The density function tells you how tightly packed the probability is in the neighborhood of a point, and it is clearly more difficult for $Z$ to be near $2$ than $1/2$ (consider for instance that $Z$ can always be $1/2$ no matter what $X$ is, but $Z < 2$ when $X > 1/2$). \u2013\u00a0dsaxton Dec 8 '15 at 14:06\n\u2022 Possible duplicate of Distribution of a ratio of uniforms: What is wrong? \u2013\u00a0Xi'an Dec 8 '15 at 14:21\n\u2022 I don't think it's a duplicate, that question is seeking the PDF, here I have the PDF, I am just questioning its correctness (perhaps rather naively). \u2013\u00a0qed Dec 8 '15 at 14:29\n\nThe right logic is that with independent $X, Y \\sim U(0,1)$, $Z=\\frac YX$ and $Z^{-1} =\\frac XY$ have the same distribution and so for $0 < z < 1$ \\begin{align} P\\left\\{\\frac YX \\leq z\\right\\} &= P\\left\\{\\frac XY \\leq z\\right\\}\\\\ &= P\\left\\{\\frac YX \\geq \\frac 1z \\right\\}\\\\ \\left.\\left.F_{Z}\\right(z\\right) &= 1 - F_{Z}\\left(\\frac 1z\\right) \\end{align} where the equation with CDFs uses the fact that $\\frac YX$ is a continuous random variable and so $P\\{Z \\geq a\\} = P\\{Z > a\\} = 1-F_Z(a)$. Hence the pdf of $Z$ satisfies $$f_Z(z) = z^{-2}f_Z(z^{-1}), \\quad 0 < z < 1.$$ Thus $f_Z(\\frac 12) = 4f_Z(2)$, and not $f_Z(\\frac 12) = f_Z(2)$ as you thought it should be.\n\nThis distribution is symmetric--if you look at it the right way.\n\nThe symmetry you have (correctly) observed is that $Y/X$ and $X/Y = 1/(Y/X)$ must be identically distributed. When working with ratios and powers, you are really working within the multiplicative group of the positive real numbers. The analog of the location invariant measure $d\\lambda=dx$ on the additive real numbers $\\mathbb{R}$ is the scale invariant measure $d\\mu = dx/x$ on the multiplicative group $\\mathbb{R}^{*}$ of positive real numbers. It has these desirable properties:\n\n1. $d\\mu$ is invariant under the transformation $x\\to ax$ for any positive constant $a$: $$d\\mu(ax) = \\frac{d(ax)}{ax} = \\frac{dx}{x} = d\\mu.$$\n\n2. $d\\mu$ is covariant under the transformation $x\\to x^b$ for nonzero numbers $b$: $$d\\mu(x^b) = \\frac{d(x^b)}{x^b} = \\frac{b x^{b-1} dx}{x^b} = b\\frac{dx}{x} = b\\, d\\mu.$$\n\n3. $d\\mu$ is transformed into $d\\lambda$ via the exponential: $$d\\mu(e^x) = \\frac{de^x}{e^x} = \\frac{e^x dx}{e^x} = dx = d\\lambda.$$ Likewise, $d\\lambda$ is transformed back to $d\\mu$ via the logarithm.\n\n(3) establishes an isomorphism between the measured groups $(\\mathbb{R}, +, d\\lambda)$ and $(\\mathbb{R}^{*}, *, d\\mu)$. The reflection $x \\to -x$ on the additive space corresponds to the inversion $x \\to 1/x$ on the multiplicative space, because $e^{-x} = 1/e^x$.\n\nLet's apply these observations by writing the probability element of $Z=Y/X$ in terms of $d\\mu$ (understanding implicitly that $z \\gt 0$) rather than $d\\lambda$:\n\n$$f_Z(z)\\,dz = g_Z(z)\\,d\\mu = \\frac{1}{2}\\begin{cases} 1\\,dz = z\\, d\\mu, & \\text{if } 0 \\le z \\le 1 \\\\ \\frac{1}{z^2}dz = \\frac{1}{z}\\, d\\mu, & \\text{if } z > 1. \\end{cases}$$\n\nThat is, the PDF with respect to the invariant measure $d\\mu$ is $g_Z(z)$, proportional to $z$ when $0\\lt z \\le 1$ and to $1/z$ when $1 \\le z$, close to what you had hoped.\n\nThis is not a mere one-off trick. Understanding the role of $d\\mu$ makes many formulas look simpler and more natural. For instance, the probability element of the Gamma function with parameter $k$, $x^{k-1}e^x\\,dx$ becomes $x^k e^x d\\mu$. It's easier to work with $d\\mu$ than with $d\\lambda$ when transforming $x$ by rescaling, taking powers, or exponentiating.\n\nThe idea of an invariant measure on a group is far more general, too, and has applications in that area of statistics where problems exhibit some invariance under groups of transformations (such as changes of units of measure, rotations in higher dimensions, and so on).\n\n\u2022 Looks like a very insightful answer. It's a pity I don't understand it at the moment. I will check back later. \u2013\u00a0qed Dec 8 '15 at 16:40\n\nIf you think geometrically...\n\nIn the $X$-$Y$ plane, curves of constant $Z = Y/X$ are lines through the origin. ($Y/X$ is the slope.) One can read off the value of $Z$ from a line through the origin by finding its intersection with the line $X=1$. (If you've ever studied projective space: here $X$ is the homogenizing variable, so looking at values on the slice $X=1$ is a relatively natural thing to do.)\n\nConsider a small interval of $Z$s, $(a,b)$. This interval can also be discussed on the line $X=1$ as the line segment from $(1,a)$ to $(1,b)$. The set of lines through the origin passing through this interval forms a solid triangle in the square $(X,Y) \\in U = [0,1]\\times[0,1]$, which is the region we're actually interested in. If $0 \\leq a < b \\leq 1$, then the area of the triangle is $\\frac{1}{2}(1-0)(b-a)$, so keeping the length of the interval constant and sliding it up and down the line $X=1$ (but not past $0$ or $1$), the area is the same, so the probability of picking an $(X,Y)$ in the triangle is constant, so the probability of picking a $Z$ in the interval is constant.\n\nHowever, for $b>1$, the boundary of the region $U$ turns away from the line $X = 1$ and the triangle is truncated. If $1 \\leq a < b$, the projections down lines through the origin from $(1,a)$ and $(1,b)$ to the upper boundary of $U$ are to the points $(1/a,1)$ and $(1/b,1)$. The resulting area of the triangle is $\\frac{1}{2}(\\frac{1}{a} - \\frac{1}{b})(1-0)$. From this we see the area is not uniform and as we slide $(a,b)$ further and further to the right, the probability of selecting a point in the triangle decreases to zero.\n\nThen the same algebra demonstrated in other answers finishes the problem. In particular, returning to the OP's last question, $f_Z(1/2)$ corresponds to a line that reaches $X=1$, but $f_Z(2)$ does not, so the desired symmetry does not hold.\n\nJust for the record, my intuition was totally wrong. We are talking about density, not probability. The right logic is to check that\n\n$$\\int_1^k f_Z(z) dz = \\int_{1/k}^1 f_Z(z) = \\frac{1}{2}(1 - \\frac{1}{k})$$,\n\nand this is indeed the case.\n\nYea the link Distribution of a ratio of uniforms: What is wrong? provides CDF of $Z=Y/X$. The PDF here is just derivative of the CDF. So the formula is correct. I think your problem lies in the assumption that you think Z is \"symmetric\" around 1. However this is not true. Intuitively Z should be a skewed distribution, for example it is useful to think when Y is a fixed number between $(0,1)$ and X is a number close to 0, thus the ratio would be going to infinity. So the symmetry of distribution is not true. I hope this help a bit.", "date": "2019-12-12 15:00:53", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/185683/distribution-of-ratio-between-two-independent-uniform-random-variables", "openwebmath_score": 0.9655059576034546, "openwebmath_perplexity": 194.34591193416304, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771759145343, "lm_q2_score": 0.8976952927915969, "lm_q1q2_score": 0.885825225852663}}
{"url": "https://mathhelpboards.com/threads/displacement-and-distance.2253/", "text": "# Displacement and Distance\n\n#### alane1994\n\n##### Active member\nHere is my question.\n\nThe function $$v(t)=15\\cos{3t}$$, $$0 \\leq t \\leq 2 \\pi$$, is the velocity in m/sec of a particle moving along the x-axis. Complete parts (a) through (c).\n\na.Graph the velocity function over the given interval. then determine when the motion is in the positive direction and when it is in the negative direction.\n(I have done this part)\nb. Find the displacement over the given interval.\n(I have done this part)\nc. Find the distance traveled over the given interval.\n(This is the part that I am fuzzy on)\n\nFor this, does it involve previously found information? Or is it a separate set of calculations all its own?\nAny help is appreciated!\n\n~Austin\n\n#### Jameson\n\nStaff member\nDisplacement is the integral of the velocity vector over the given time interval and results in the distance between the starting point and end point. If you move 10 meters north and 10 meters south, the displacement is 0.\n\nTo find the distance you need to figure out where the displacement is negative and count that as a positive value. I believe this is the same as taking the integral of the absolute value of velocity. The way I would do this is find the regions where v(t) is positive and negative and then breaking the calculation into multiple calculations.\n\nWhen considering $$\\displaystyle v(t)=15\\cos{3t}$$ where $$\\displaystyle 0 \\le t \\le 2\\pi$$ at $t=0$ v(t) is positive and becomes negative when $$\\displaystyle 3t=\\frac{\\pi}{2}$$ so when $$\\displaystyle t=\\frac{\\pi}{6}$$.\n\nSo your first integral is $$\\displaystyle \\int_{0}^{\\frac{\\pi}{6}}v(t)dt$$ This value will be positive.\n\nNow v(t) will be negative until until it touches the x-axis again where $$\\displaystyle 3t=\\frac{3\\pi}{2}$$ or when $$\\displaystyle t=\\frac{\\pi}{2}$$. The second integral is now $$\\displaystyle \\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{2}}v(t)dt$$. This will be negative but when considering distance you don't need to take into account the sign of this so count it as positive.\n\nIf you keep repeating this over until the end of the interval you should get the final answer. It will take a while to calculate this way and there very well could be a quicker way to do it but that's how I would do this.\n\n#### alane1994\n\n##### Active member\nYeah, that's what I did after consulting a professor, I got the answer 60...\n\n#### Jameson\n\nStaff member\nYeah, that's what I did after consulting a professor, I got the answer 60...\nYou already found all of these intervals for part A so this calculation shouldn't have been too tedious. Wow your professor responds fast. You posted this question just a couple of hours ago\n\n#### topsquark\n\n##### Well-known member\nMHB Math Helper\nJust to put my two cents in...\n\nThe formula for distance is\n$$dist = \\int |v|dt$$\n\nwhere v is the velocity vector.\n\nIn a practical sense this can pretty much only be calculated in the way Jameson has described.\n\n-Dan", "date": "2020-09-26 02:30:23", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/displacement-and-distance.2253/", "openwebmath_score": 0.8370845913887024, "openwebmath_perplexity": 366.31513717449747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969641180274, "lm_q2_score": 0.9005297921244243, "lm_q1q2_score": 0.8857583696314221}}
{"url": "http://math.stackexchange.com/questions/182617/counterexample-check-for-sum-of-limit-points-of-subsequences", "text": "# Counterexample Check for Sum of Limit Points of Subsequences\n\nLet $c$ be a limit point of a sequence of real numbers $\\langle a_n \\rangle$ and $d$ a limit point of $\\langle b_n \\rangle$. Is $c+d$ necessarily a limit point of $\\langle a_n + b_n \\rangle$?\n\nMy Question:\n\nWhen considering this question, do I have to sum over the same index or can the indices for the different subsequences differ? My intuition is that the subsequences must be summed over identical indices, in which case I believe that the following example serves as a counterexample:\n\nLet $\\langle a_n \\rangle = (-1)^n, \\ \\langle b_n \\rangle = (-1)^{n+1}$. Then summing over even and odd indices, I get $0 \\ne 2$ or $-2$, which is the sum of their limit points. Have I done this correctly?\n\n-\nYour argument is correct; but generally, the indices could differ. Take $a_n=b_n=(-1)^n$. $(a_n)$ has $1$ as a limit point and $(b_n)$ has $-1$ as a limit point but $(a_n+b_n)$ does not have $1+(-1)=0$ as a limit point. \u2013\u00a0 David Mitra Aug 14 '12 at 22:10\n\nYour example is fine. Both $\\langle(-1)^n:n\\in\\Bbb N\\rangle$ and $\\langle(-1)^{n+1}:n\\in\\Bbb N\\rangle$ have $1$ as a limit point, but $\\langle(-1)^n+(-1)^{n+1}:n\\in\\Bbb N\\rangle=\\langle 0:n\\in\\Bbb N\\rangle$ converges to $0$ and so does not have $1+1=2$ as a limit point.\nIt doesn\u2019t matter what subsequence of $\\langle a_n:n\\in\\Bbb N\\rangle$ has $c$ as limit or what subsequence of $\\langle b_n:n\\in\\Bbb N\\rangle$ has $d$ as a limit; all that matters is whether some subsequence of $\\langle a_n+b_n:n\\in\\Bbb N\\rangle$ has $c+d$ as a limit. In your example that\u2019s not the case, so yours is a genuine counterexample to the conjecture.\n@Rolando posted what looked like a proof of this conjecture, and then deleted it after you posted your answer. It looked something like this: Let $\\epsilon >0$ be given. Since $a_n \\to c$ and $b_n \\to d$, there are $N_1, N_2$ such that for $l \\ge N_1$, $k \\ge N_2$ $|a_l - c| \\le \\epsilon/2$ and $|b_k - d| \\le \\epsilon/2$. Let $m = \\max\\{N_1, N_2\\}$ and therefore we have for $n \\ge m$ we have $|a_n - c + b_n - d| \\le \\epsilon$. Is the problem with his proof that he is not considering subsequences, and that he interpreted the limit points as limits? \u2013\u00a0 Zvpunry Aug 14 '12 at 22:14", "date": "2014-08-23 20:32:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/182617/counterexample-check-for-sum-of-limit-points-of-subsequences", "openwebmath_score": 0.9544462561607361, "openwebmath_perplexity": 116.46756869022938, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983596967003007, "lm_q2_score": 0.9005297847831081, "lm_q1q2_score": 0.8857583650085358}}
{"url": "https://math.stackexchange.com/questions/1738331/sequence-of-functions-that-fails-certain-conditions-of-arzela-ascoli-theorem", "text": "# Sequence of functions that fails certain conditions of Arzela-Ascoli theorem\n\nFor a closed, bounded interval $[a,b]$, let $\\{ f_{n}\\}$ be a sequence in $C[a,b]$. If $\\{f_{n}\\}$ is equicontinuous, does $\\{f_{n}\\}$ necessarily have a uniformly convergent subsequence?\n\nI would think not, because according to the Arzela-Ascoli Theorem, $\\{f_{n} \\}$ also needs to be uniformly bounded. Is this all that needs to be violated in order for an equicontinuous sequence of continuous functions on a compact interval to not have a uniformly convergent subsequence?\n\nAnd if so, what is an example of a sequence that illustrates this, and how to show it does not have a uniformly convergent subsequence? Thank you.\n\nTake $f_n(x) = n$ for $x \\in [0,1]$. These functions are all constant, so clearly equicontinuous, but $\\| f_n - f_m \\|_\\infty = \\lvert n - m \\rvert \\ge 1$ for $n \\neq m$ so no subsequence can converge since no subsequence is Cauchy.\n\u2022 Yes of course. If it was uniformly bounded, then we could apply Arzela Ascoli and find a convergent subsequence. A sequence of functions $f_n$ is uniformly bounded if and only if the sequence of real numbers $\\| f_n\\|_\\infty$ is bounded. Here that clearly is not the case because $\\| f_n \\|_\\infty = n$. \u2013\u00a0User8128 Apr 11 '16 at 23:29", "date": "2021-03-09 08:24:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1738331/sequence-of-functions-that-fails-certain-conditions-of-arzela-ascoli-theorem", "openwebmath_score": 0.944827139377594, "openwebmath_perplexity": 87.06364025743125, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.983596970368816, "lm_q2_score": 0.9005297801113613, "lm_q1q2_score": 0.885758363444431}}
{"url": "http://www.appeton.co.id/future-currency-tosgcr/63e023-linear-function-graph", "text": "By \u2026 By graphing two functions, then, we can more easily compare their characteristics. In this section, 8th grade and high school students will have to find the missing values of x and f(x). In $f\\left(x\\right)=mx+b$, the b\u00a0acts as the vertical shift, moving the graph up and down without affecting the slope of the line. This is also expected from the negative constant rate of change in the equation for the function. In linear algebra, mathematical analysis, and functional analysis, a linear function is a \u2026 The equation is in standard form (A = -1, B = 1, C = 3). There is a special linear function called the \"Identity Function\": f (x) = x. The x-intercept is the point at which the graph of a linear function crosses the x-axis. A table of values might look as below. The vertical line test indicates that this graph represents a function. The independent variable is x and the dependent variable is y. a is the constant term or the y intercept. Graph linear functions. Two competing telephone companies offer different payment plans. A function may be transformed by a shift up, down, left, or right. 3.4 Graphing Linear Equations There are two common procedures that are used to draw the line represented by a linear equation. The graph of this function is a line with slope \u2212 and y-intercept \u2212. A linear function has one independent variable and one dependent variable. Twitter. A similar word to linear function is linear correlation. For distinguishing such a linear function from the other concept, the term affine function is often used. This tells us that for each vertical decrease in the \u201crise\u201d of $\u20132$ units, the \u201crun\u201d increases by 3 units in the horizontal direction. When we\u2019re comparing two lines, if their slopes are equal they are parallel, and if they are in \u2026 By using this website, you agree to our Cookie Policy. Graph Linear Equations by Plotting Points It takes only 2 points to draw a graph of a straight line. The first characteristic is its y-intercept which is the point at which the input value is zero. Graph $f\\left(x\\right)=-\\frac{2}{3}x+5$ using the y-intercept and slope. Selbst 1 Selbst 2 Selbst 3 Yes. +drag: Hold down the key, then drag the described object. GRAPHING LINEAR RELATIONS. dillinghamt. Evaluate the function at x\u00a0= 0 to find the y-intercept. The second is by using the y-intercept and slope. Free functions and graphing calculator - analyze and graph line equations and functions step-by-step. (Note: A vertical line parallel to the y-axis does not have a y-intercept. Do all linear functions have y-intercepts? Using slope and intercepts in context Get 3 of 4 questions to level up! The simplest way is to find the intercept values for both the x-axis and the y-axis. To draw the graph we need coordinates. http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@5.2, Graph a linear function by plotting points, Graph a linear function using the slope and y-intercept, Graph a linear function using transformations. We were also able to see the points of the function as well as the initial value from a graph. Learn More at mathantics.comVisit http://www.mathantics.com for more Free math videos and additional subscription based content! How to graph Linear Functions by finding the X-Intercept and Y-Intercept of the Function? The graph of a linear relation can be found by plotting at least two points. Graph $f\\left(x\\right)=4+2x$, using transformations. Spell. -x + y = 3. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation. In Activity 1 the learners should enter the expressions one by one into the graphing calculator and classify the functions according to the shape of the graph. Write. linear functions by the shape of their graphs and by noting differences in their expressions. Graph 3x - 2y = 8. And here is its graph: It makes a 45\u00b0 (its slope is 1) It is called \"Identity\" because what comes out is identical to what goes in: In. We can begin graphing by plotting the point (0, 1) We know that the slope is rise over run, $m=\\frac{\\text{rise}}{\\text{run}}$. A function may also be transformed using a reflection, stretch, or compression. Plot the points and graph the linear function. However, in linear algebra, a linear function is a function that maps a sum to the sum of the images of the summands. In the equation $f\\left(x\\right)=mx+b$, $m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$. PLAY. This is why we performed the compression first. The y-intercept is the point on the graph when x\u00a0= 0. f(x)=b. of f is the To find the y-intercept, we can set $x=0$ in the equation. Gravity. The graph of a linear function is a line. The slope of a linear function is equal to the ratio of the change in outputs to the change in inputs. This inequality notation means that we should plot the graph for values of x between and including -3 and 3. Use the resulting output values to identify coordinate pairs. The a represents the gradient of the line, which gives the rate of change of the dependent variable. In order to write the linear function in the form of y=mx+b, we will need to determine the line's: 1. slope (m) 2. y-intercept (b) We can tell from the graph that the slope of the line is negative because the line goes down and to the right. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. The graph below is of\u00a0the function $f\\left(x\\right)=-\\frac{2}{3}x+5$. Linear Parent Graph And Transformations. So, for this definition, the above function is linear only when c = 0, that is when the line passes through the origin. Linear functions are functions that produce a straight line graph.. The graph of f is a line with slope m and y intercept b. The graph of f is a line with slope m and y intercept Graphing Linear Equations Find the Equation of a Line. Horizontal lines are written in the form, $f(x)=b$. Graph Linear Equations using Slope-Intercept We can use the slope and y-intercept to graph a linear equation. Graph Linear Equations in Two Variables Learning Objectives. We were also able to see the points of the function as well as the initial value from a graph. The graph of a linear function is a line. f(0). The graph of the function is a line as expected for a linear function. Evaluate the function at each input value and use the output value to identify coordinate pairs. ++drag: Hold down both the key and the key, then drag the described object. For example, following order of operations, let the input be 2. The equation for the function shows that $m=\\frac{1}{2}$ so the identity function is vertically compressed by $\\frac{1}{2}$. Linear functions are those whose graph is a straight line. The steepness of a hill is called a slope. Each graphing linear equations worksheet on this page has four coordinate planes and equations in slope-intercept form, and includes an answer key showing the correct graph. Graphing a Linear Function Using y-intercept and Slope. Graphing Linear Functions. In this non-linear system, users are free to take whatever path through the material best serves their needs. There are three basic methods of graphing linear functions: Keep in mind that a vertical line is the only line that is not a function.). Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Then just draw a line that passes through both of these points. The following diagrams show the different methods to graph a linear equation. Linear functions are typically written in the form f(x) = ax + b. Furthermore, the domain and range consists of all real numbers. A table of values might look as below. Graphing Linear Equations Calculator is a free online tool that displays the graph of the given linear equation. Notice that adding a value of b\u00a0to the equation of $f\\left(x\\right)=x$ shifts the graph of\u00a0f\u00a0a total of b\u00a0units up if b\u00a0is positive and\u00a0|b| units down if b\u00a0is negative. How many solutions does this linear system have? Regardless of whether a table is given to you, you should consider using one to ensure you\u2019re correctly graphing linear and quadratic functions. Is the Function Linear or Nonlinear | Table. Learn More at mathantics.comVisit http://www.mathantics.com for more Free math videos and additional subscription based content! Often, the number in front of x is already a fraction, so you won't have to convert it. A y-intercept is a y-value at which a graph crosses the y-axis. Solving Systems of Linear Equations: Graphing. The equation for a linear function is: y = mx + b, Where: m = the slope ,; x = the input variable (the \u201cx\u201d always has an exponent of 1, so these functions are always first degree polynomial.). The slopes are represented as fractions in the level 2 worksheets. In Linear Functions, we saw that that the graph of a linear function is a straight line. The equation can be written in standard form, so the function is linear. eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-4','ezslot_5',344,'0','0'])); Any function of the form The first is by plotting points and then drawing a line through the points. Furthermore, the domain and range consists of all real numbers. Properties. Graphing linear functions (2.0 MiB, 1,144 hits) Slope Determine slope in slope-intercept form (160.4 KiB, 766 hits) Determine slope from given graph (2.1 MiB, 834 hits) Find the integer of unknown coordinate (273.6 KiB, 858 hits) Find the fraction of unknown coordinate (418.5 KiB, 891 hits) Linear inequalities Graph of linear inequality (2.8 MiB, 929 hits) Facebook. The first characteristic is its y-intercept which is the point at which the input value is zero. Linear functions are typically written in the form f(x) = ax + b. In mathematics, the term linear function refers to two distinct but related notions: In calculus and related areas, a linear function is a function whose graph is a straight line, that is, a polynomial function of degree zero or one. Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! Interpret solutions to linear equations and inequalities graphically. The function $y=x$ compressed by a factor of $\\frac{1}{2}$. y = f(x) = a + bx. Google+. ; b = where the line intersects the y-axis. Now we have to determine the slope of the line. The function $y=\\frac{1}{2}x$ shifted down 3 units. The equation for a linear function is: y = mx + b, Where: m = the slope , x = the input variable (the \u201cx\u201d always has an exponent of 1, so these functions are always first degree polynomial.). Because the slope is positive, we know the graph will slant upward from left to right. 1. The order of the transformations follows the order of operations. Evaluating the function for an input value of 2 yields an output value of 4 which is represented by the point (2, 4). $f\\left(x\\right)=\\frac{1}{2}x+1$. Another option for graphing is to use transformations on the identity function $f\\left(x\\right)=x$. A linear function has the following form y = f (x) = a + bx A linear function has one independent variable and one dependent variable. The graph of the linear equation will always result in a straight line. The input values and corresponding output values form coordinate pairs. A linear function is a polynomial function in which the variable x has degree at most one: = +.Such a function is called linear because its graph, the set of all points (, ()) in the Cartesian plane, is a line.The coefficient a is called the slope of the function and of the line (see below).. We then plot the coordinate pairs on a grid. Free graph paper is available. The graph of a linear function is a straight line, while the graph of a nonlinear function is a curve. Students learn that the linear equation y = x, or the diagonal line that passes through the origin, is called the parent graph for the family of linear equations. Vertically stretch or compress the graph by a factor. Write the equation of a line parallel or perpendicular to a given line. Free functions and graphing calculator - analyze and graph line equations and functions step-by-step This website uses cookies to ensure you get the best experience. That line is the solution of the equation and its visual representation. Because the given function is a linear function, you can graph it by using slope-intercept form. In Linear Functions, we saw that that the graph of a linear function is a straight line.We were also able to see the points of the function as well as the initial value from a graph. Explore math with our beautiful, free online graphing calculator. The equation for the function also shows that $b=-3$, so the identity function is vertically shifted down 3 units. We will choose 0, 3, and 6. Another way to think about the slope is by dividing the vertical difference, or rise, between any two points by the horizontal difference, or run. It looks like the y-intercept (b) of the graph is 2, as represented by point (0,2). f(a) is called a function, where a \u2026 Match. We repeat until we have multiple points, and then we draw a line through the points as shown below. Learn. In general, a linear function28 is a function that can be written in the form f(x) = mx + b LinearFunction where the slope m and b represent any real numbers. Begin by choosing input values. In general, a linear function Any function that can be written in the form f ( x ) = m x + b is a function that can be written in the form f ( x ) = m x + b L i n e a r F u n c t i o n where the slope m and b represent any real \u2026 Learn more Accept. Because y = f(x), we can use y and f(x) interchangeably, and ordered pair solutions on the graph (x, y) can be written in the form (x, f(x)). eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-3','ezslot_4',320,'0','0'])); Determine the x intercept, set f(x) = 0 and The slopes in level 1 worksheets are in the form of integers. Graph a straight line by finding its x - and y-intercepts. The, of this function is the set of all real numbers. By using this website, you agree to our Cookie Policy. The graph crosses the y-axis at (0, 1). Graphing Linear Functions. There are three basic methods of graphing linear functions. No. Recall that the slope is the rate of change of the function. 3. Vertical stretches and compressions and reflections on the function $f\\left(x\\right)=x$. This is also known as the \u201cslope.\u201d The b represents the y-axis intercept. These points may be chosen as the x and y intercepts of the graph for example. We can now graph the function by first plotting the y-intercept. Relating linear contexts to graph features Get 5 of 7 questions to level up! Method 1: Graphing Linear Functions in Standard Form 1. This means the larger the absolute value of m, the steeper the slope. Linear functions are those whose graph is a straight line. It is generally a polynomial function whose degree is utmost 1 or 0. Evaluating the function for an input value of 1 yields an output value of 2 which is represented by the point (1, 2). The third is applying transformations to the identity function $f\\left(x\\right)=x$. Write the equation for a linear function from the graph of a line. b. Solver to Analyze and Graph a Linear Function. Show Step-by-step Solutions. How to Use this Applet Definitions +drag: Hold down the key, then drag the described object. This website uses cookies to ensure you get the best experience. Use $\\frac{\\text{rise}}{\\text{run}}$ to determine at least two more points on the line. Its graph is a horizontal line at y = b. The graph of a linear function is always a line. Graphing Linear Function: Type 2 - Level 1. Evaluate the function at each input value. Created by. We previously saw that that the graph of a linear function is a straight line. The graph slants downward from left to right which means it has a negative slope as expected. Evaluate when . Linear functions word problem: fuel (Opens a modal) Practice. We know that the linear equation is defined as an algebraic equation in which each term should have an exponents value of 1. Recall that the set of all solutions to a linear equation can be represented on a rectangular coordinate plane using a straight line through at least two points; this line is called its graph. Graphing Linear Functions. For example, Plot the graph of y = 2x \u2013 1 for -3 \u2264 x \u2264 3. Identify and graph a linear function using the slope and y-intercept. When m\u00a0is negative, there is also a vertical reflection of the graph. How to transform linear functions, Horizontal shift, Vertical shift, Stretch, Compressions, Reflection, How do stretches and compressions change the slope of a linear function, Rules for Transformation of Linear Functions, PreCalculus, with video lessons, examples and step-by-step solutions. Learn more Accept. Solution : y = x + 3. In addition, the graph has a downward slant which indicates a negative slope. Note: A function f (x) = b, where b is a constant real number is called a constant function. This function includes a fraction with a denominator of 3 so let\u2019s choose multiples of 3 as input values. 8 Linear Equations Worksheets. Usage To plot a function just type it into the function box. A graphing calculator can be used to verify that your answers \"make sense\" or \"look right\". solve for x. Graphs of linear functions may be transformed by shifting the graph up, down, left, or right as well as using stretches, compressions, and reflections. Write the equation in standard form. The slope of a linear function corresponds to the number in \u2026 For example, given the function $f\\left(x\\right)=2x$, we might use the input values 1 and 2. Evaluate the function at an input value of zero to find the. Graphs of linear functions may be transformed by shifting the graph up, down, left, or right as well as using stretches, compressions, and reflections. This graph illustrates vertical shifts of the function $f\\left(x\\right)=x$. The slope is defined as the ratio of the vertical change between two points, the rise, to the horizontal change between the same two points, the run. The Slider Area. In this video we look at graphing equations using a table of values Find the slopes and the x- and y-intercepts of the following lines. Solve a system of linear equations. The slope-intercept form gives you the y- intercept at (0, \u20132). Did you have an idea for improving this content? Graph horizontal and vertical lines. When it comes to graphing linear equations, there are a few simple ways to do it. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. The equation, written in this way, is called the slope-intercept form. According to the equation for the function, the slope of the line is $-\\frac{2}{3}$. set of all real numbers. Flashcards. If variable x is a constant x=c, that will represent a line paralel to y-axis. Reddit. Two points that are especially useful for sketching the graph of a line are found with the intercepts. Given a real-world situation that can be modeled by a linear function or a graph of a linear function, the student will determine and represent the reasonable domain and range of the linear function \u2026 Function Grapher is a full featured Graphing Utility that supports graphing two functions together. Book The a represents the gradient of the line, which gives the rate of change of the dependent variable. Select two options. how to graph linear equations using the slope and y-intercept. Graph a linear function: a step by step tutorial with examples and detailed solutions. This inequality notation means that we should plot the graph for values of x between and including -3 and 3. Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. We generate these coordinates by substituting values into the linear equation. Test. In general we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph of the function. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. If we have two points: A=(x1,y1) B=(x2,y2) A slope (a) is calculated by the formula: a=y2\u2212y1x2\u2212x1 If the slope is equal to number 0, then the line will be paralel with\u00a0x \u2013 axis. Functions: Hull: First graph: f(x) Derivative Integral From ... Mark points at: First graph: x= Second graph: x= Third graph: x= Reticule lines Axis lines Caption Dashes Frame Errors: Def. This is also known as the \u201cslope.\u201d The b represents the y-axis intercept. $\\begin{array}{l}f\\text{(2)}=\\frac{\\text{1}}{\\text{2}}\\text{(2)}-\\text{3}\\hfill \\\\ =\\text{1}-\\text{3}\\hfill \\\\ =-\\text{2}\\hfill \\end{array}$. Subtract x from each side. Find a point on the graph we drew in Example: Graphing by Using the y-intercept and Slope\u00a0that has a negative x-value. Recognize the standard form of a linear function. Graphing a Linear Equation by Plotting Three Ordered Pairs. In Activity 1 the learners should enter the expressions one by one into the graphing calculator and classify the functions according to the shape of the graph. A linear equation is drawn as a straight line on a set of axes. But if it isn't, convert it by simply placing the value of m over 1. You need only two points to graph a linear function. First, graph y = x. Students also learn the different types of transformations of the linear parent graph. A Review of Graphing Lines. These unique features make Virtual Nerd a viable alternative to private tutoring. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. What is Meant by Graphing Linear Equations? Possible answers include $\\left(-3,7\\right)$, $\\left(-6,9\\right)$, or $\\left(-9,11\\right)$. We were also able to see the points of the function as well as the initial value from a graph. Graph a straight line by finding three ordered pairs that are solutions to the linear equation. Examples: 1. To graph, choose three values of x, and use them to generate ordered pairs. The functions whose graph is a line are generally called linear functions in the context of calculus. The slope of a line is a number that describes steepnessand direction of the line. By the end of this section, you will be able to: Plot points in a rectangular coordinate system; Graph a linear equation by plotting points; Graph vertical and horizontal lines; Find the x- and y-intercepts; Graph a line using the intercepts ; Before you get started, take this readiness quiz. In the equation $f\\left(x\\right)=mx$, the m\u00a0is acting as the vertical stretch or compression of the identity function. In Example: Graphing by Using Transformations, could we have sketched the graph by reversing the order of the transformations? You can move the graph of a linear function around the coordinate grid using transformations. The first \u2026 Graphing Linear Equations. It has the unique feature that you can save your work as a URL (website link). Dritter Graph: h(x) Ableitung Integral +C: Blau 1 Blau 2 Blau 3 Blau 4 Blau 5 Blau 6 Rot 1 Rot 2 Rot 3 Rot 4 Gelb 1 Gelb 2 Gr\u00fcn 1 Gr\u00fcn 2 Gr\u00fcn 3 Gr\u00fcn 4 Gr\u00fcn 5 Gr\u00fcn 6 Schwarz Grau 1 Grau 2 Grau 3 Grau 4 Wei\u00df Orange T\u00fcrkis Violett 1 Violett 2 Violett 3 Violett 4 Violett 5 Violett 6 Violett 7 Lila Braun 1 Braun 2 Braun 3 Zyan Transp. The slope of a linear function will be the same between any two points. The slope is $\\frac{1}{2}$. By graphing two functions, then, we can more easily compare their characteristics. Graph $f\\left(x\\right)=-\\frac{2}{3}x+5$ by plotting points. Linear equations word problems: tables Get 3 of 4 questions to level up! Plot the coordinate pairs and draw a line through the points. It will be very difficult to succeed in Calculus without being able to solve and manipulate linear equations. Method 1: Graphing Linear Functions in Standard Form 1. Introduction to Linear Relationships: IM 8.3.5. Example 1 Graph the linear function f given by f (x) = 2 x + 4 Solution to Example 1. Graph $f\\left(x\\right)=\\frac{1}{2}x - 3$ using transformations. However, in linear algebra, a linear function is a function that maps a sum to the sum of the images of the summands. Graphing Linear Function: Type 1 - Level 2. Although the linear functions are also represented in terms of calculus as well as linear algebra. Linear functions are functions that produce a straight line graph. Starting from our y-intercept (0, 1), we can rise 1 and then run 2 or run 2 and then rise 1. Notice that multiplying the equation $f\\left(x\\right)=x$ by m\u00a0stretches the graph of f\u00a0by a factor of m\u00a0units if m\u00a0> 1 and compresses the graph of f\u00a0by a factor of m\u00a0units if 0 < m\u00a0< 1. Tell whether each function is linear. 2. What is a Linear Function? f (x) = m x + b, where m is not equal to 0 is called a linear function. These pdf worksheets provide ample practice in plotting the graph of linear functions. For example, Plot the graph of y = 2x \u2013 1 for -3 \u2264 x \u2264 3. Free linear equation calculator - solve linear equations step-by-step. STUDY. Complete the function table, plot the points and graph the linear function. The other characteristic of the linear function is its slope,\u00a0m,\u00a0which is a measure of its steepness. We\u2019d love your input. First, graph the identity function, and show the vertical compression. Graphing a Linear Function Using y-intercept and Slope. After studying this section, you will be able to: 1. From the initial value (0, 5) we move down 2 units and to the right 3 units. We encountered both the y-intercept and the slope in Linear Functions. Linear Function Graph. (See Getting Help in Stage 1.) In mathematics, a graphing linear equation represents the graph of the linear equation. y = mx + b y = -2x + 3/2. Knowing an ordered pair written in function notation is necessary too. The first characteristic is its y-intercept which is the point at which the input value is zero. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Let's try starting from a graph and writing the equation that goes with it. We can extend the line to the left and right by repeating, and then draw a line through the points. From our example, we have $m=\\frac{1}{2}$, which means that the rise is 1 and the run is 2. Linear equations word problems: graphs Get 3 of 4 questions to level up! How to Use the Graphing Linear Equations Calculator? To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. m = -2 and b = -1/3 m = -2 and b = -2/3. Since the slope is 3=3/1, you move up 3 units and over 1 unit to arrive at the point (1, 1). If you have difficulties with this material, please contact your instructor. This website uses cookies to ensure you get the best experience. Convert m into a fraction. If so, graph the function. A linear function is a function which forms a straight line in a graph. Determine the y intercept, set x = 0 to find Draw Function Graphs Mathematics / Analysis - Plotter - Calculator 4.0. Draw a line which passes through the points. BYJU\u2019S online graphing linear equations calculator tool makes the calculation faster and it displays the graph in a fraction of seconds. 8th grade students learn to distinguish between linear and nonlinear functions by observing the graphs. The output value when x\u00a0= 0 is 5, so the graph will cross the y-axis at (0, 5). The y-intercept and slope of a line may be used to write the equation of a line. Examine the input(x) and output(y) values of the table inthese linear function worksheets for grade 8. Use \"x\" as the variable like this: Examples: sin(x) 2x-3; cos(x^2) (x-3)(x+3) Zooming and Re-centering. The same goes for the steepness of a line. Now we know the slope and the y-intercept. A linear function has the following form. Regardless of whether a table is given to you, you should consider using one to ensure you\u2019re correctly graphing linear and quadratic functions. To zoom, use the zoom slider. Key Concepts: Terms in this set (10) Which values of m and b will create a system of equations with no solution? You can move the graph of a linear function around the coordinate grid using transformations. All linear functions cross the y-axis and therefore have y-intercepts. GeoGebra Classroom Activities. Equation for a linear function: a function. ) x = 0 visual representation the simplest way to... 5, so the function rather than plotting points y intercepts of the parent. There are three basic methods of graphing linear equations calculator is a line slope... Negative, there is also known as the x and f ( x ) = ax +.. Compressions and reflections on the identity function, you agree to our Cookie Policy left, or compression although may. Input values and corresponding output values form coordinate pairs extend the line being to. 1 - level 1 not have a y-intercept is the set of axes reflection stretch! Function will be able to: 1 - Analyze and graph the function as well as linear algebra non-linear,. =X [ /latex ] graph illustrates vertical shifts is another way to graph linear functions by observing graphs... In terms of calculus supports graphing two functions, we can use the output value to identify coordinate pairs a. Are generally called linear functions are functions that produce a straight line table inthese linear function has one variable... Pairs on a set of all real numbers draw function graphs Mathematics / Analysis - Plotter - 4.0. And high school students will have to determine the y \u2026 the graph of linear functions cross the.. And complete the function. ) two functions together the constant term or the y \u2026 graph..., a graphing linear equations step-by-step are functions that produce a straight line, which gives rate. In the equation of a line illustrates vertical shifts is another way to graph a linear.... Direction of the linear equation a y-intercept is the point at which the input values and corresponding output calculated..., that will represent a line through the points of the function rather than plotting points functions..., which gives the rate of change in outputs to the change in form... It will be the easiest way to graph linear functions by observing the.... Is the rate of change of the function rather than plotting points f\\left ( x\\right =x! Practice each method know the graph of a straight line is necessary too we previously saw that that the we! Includes a fraction with a denominator of 3 as input values - calculator 4.0 graphs Mathematics / Analysis Plotter... Whatever path through the points at x = 0 to find f ( x ) ax!, choose three values of the dependent variable Nerd a viable alternative to private.! Get 3 of 4 questions to level up that has a negative slope as expected 2 - level.... You agree to our Cookie Policy all real numbers x, and then draw a line may be transformed a! X, and show the different types of linear functions: Tell whether each function is a straight line finding... Affine function is a straight line, while the graph of a linear function is a linear function Type... Graph crosses the y-axis at ( 0, 5 ) we move down 2 units and to the in... Ordered pair written in function notation is necessary too zero to find f ( x =! At an input value and use them to generate ordered pairs, 5.... The right 3 units first plotting the graph of the change in inputs linear function graph, or compression b where... The calculation faster and it displays the graph slants downward from left to right which means has! = 0 to find f ( x ) = ax + b y = mx + b each... Vertical reflection of the graph of the function at x = 0 find. //Www.Mathantics.Com for more examples and detailed solutions and y-intercepts of the table inthese linear function around coordinate..., you agree to our Cookie Policy equation is defined as an algebraic equation which... Feature that you can move the graph of linear function graph linear function crosses the x-axis vertical is! Stretch or compress the graph of a linear function is evaluated linear function graph a given input, term... Intersects the y-axis of 4 questions to level up input be 2 calculator is measure... And f ( x ) =b [ /latex ] by plotting points when x = 0 to find equation..., down, left, or right be very difficult to succeed in calculus without being able solve! That is not a function just Type it into the function at input... Graph features Get 5 of 7 questions to level up by \u2026 Because the slope and intercepts context. To identify coordinate pairs ( 0 ) Type 2 - level 2 term should have an value... The third is applying transformations to the y-axis intercept it displays the graph y. The corresponding output values form coordinate pairs on a set of all numbers... We previously saw that that the graph of the transformations follows the order of the linear equation point which. Is also expected from the other concept, the steeper the slope is point., we can use the output value when x = 0 to find f x. To find the slopes in level 1 worksheets are in the equation a... The table inthese linear function around the coordinate grid using transformations as shown below pair written this... The function is always a line may be transformed by a SHIFT,. The simplest way is to find the missing values of x between and including -3 and.... For graphing is to use this Applet Definitions < SHIFT > +drag: Hold down the linear function graph >! To find the y-intercept is the point at which the input value zero... Using slope-intercept we can more easily compare their characteristics to graph features Get 5 7. System, users are free to take whatever path through the material best serves their needs two. Y-Intercept ( b ) of the transformations see the points the intercepts could have... Functions word problem: fuel ( Opens a modal ) practice an input of... Compressions along with vertical shifts of the linear equation learn the different methods to graph linear functions is by specific! 3 units ] f ( linear function graph ) and complete the function. ) by first plotting the graph f. The a represents the gradient of the table inthese linear function. ) have a y-intercept is point! Equation in which each term should have an exponents value of m, the corresponding output form. ) =\\frac { 1 } { 2 } x+1 [ /latex ] the! Also a vertical line parallel to the ratio of the equation is defined as an equation... In outputs to the linear equation calculator - Analyze and graph the identity function '': (... In this section, 8th grade and high school students will have convert. Work as a URL ( website link ) the x and y intercepts of the equation be. =\\Frac { 1 } { 3 } x+5 [ /latex ] using transformations contexts to graph linear functions identify graph! The graphs your work as a straight line Cookie linear function graph steepness of a line parallel to the at. Function box y-intercept of the function table, plot points, visualize algebraic equations, are... Or the y intercept, set x = 0 to find f ( )... Function rather than plotting points slope-intercept we can more easily compare their characteristics ( y ) values x! Types of transformations of the line to the left and right by,... To: 1 although this may not be the easiest way to graph linear functions are typically written Standard. This may not be the easiest way to graph this Type of function, you will able! { 4 } x+6 [ /latex ] to look at identifying different types of transformations of the diagrams. Inthese linear function from the negative constant rate of change of the function tables the equations two! 3, and more supports graphing two functions, we can now the... Always result in a graph, or compression fraction, so the function is a are., 1 ) by reversing the order of the function tables visualize algebraic equations, sliders. Along with vertical shifts is another way to graph features Get 5 of 7 questions to level!. Book we previously saw that that the slope in linear functions is by plotting points then. The value of 1 with the intercepts addition, the graph in a graph has one independent variable is and! Know the graph has a downward slant which indicates a negative slope Nerd a alternative! In function notation is necessary too we saw that that the graph a! Down 2 units and to the change in inputs graphs, and more the variable..., where a \u2026 draw function graphs Mathematics / Analysis - Plotter - calculator 4.0 the set of real... Are generally called linear functions in the level 2 use this Applet Definitions SHIFT... Y=\\Frac { 1 } { 3 } x+5 [ /latex ] by plotting points finding the x-intercept and \u2212. Expected for a linear function using the y-intercept transformed by a factor Mathematics / Analysis - Plotter calculator... Grid using transformations pair written in this non-linear system, users are free to take whatever path the. Function at each input value is zero slope-intercept we can use the slope is positive, we can easily. Option for graphing is to use transformations on the graph we drew in example: graphing function... Mathantics.Comvisit http: //www.mathantics.com for more examples and detailed solutions improving this content that the slope and y-intercept graph... Move down 2 units and to the y-axis and therefore have y-intercepts an algebraic in. Also expected from the initial value from a graph line as expected a... Reflection, stretch, or right functions and graphing calculator } x+5 [ /latex ], transformations.", "date": "2021-04-13 17:23:29", "meta": {"domain": "co.id", "url": "http://www.appeton.co.id/future-currency-tosgcr/63e023-linear-function-graph", "openwebmath_score": 0.5752109885215759, "openwebmath_perplexity": 420.482420127325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.983596967483837, "lm_q2_score": 0.900529776774399, "lm_q1q2_score": 0.8857583575641955}}
{"url": "https://chemistry.stackexchange.com/questions/61117/probability-of-finding-a-molecule-in-the-ground-vibrational-level", "text": "# Probability of finding a molecule in the ground vibrational level\n\nI wanted to estimate the probability of finding a molecule in the ground vibrational level using the Boltzmann distribution:\n\n$$p_i = \\frac{e^{-\\epsilon_i/kT}}{\\sum_{i=0}^{N}e^{-\\epsilon_i/kT}}$$\n\nUsing the quantum harmonic oscillator as a model for the energy\n\n$$\\epsilon_i = h\\nu (i+1/2) =/i=0/=\\frac{h\\nu}{2}$$\n\nIn the Boltzmann distribution, we have the state of interest divided by the sum of all possible states. But how should I treat the denominator?\n\nSearching a bit I found that the analytical expression for this geometric series is ($i$ not imaginary number)\n\n$$\\sum_{i=0}^{N}e^{-i h\\nu/kT} = \\frac{1}{1 - e^{h\\nu/kT}}$$\n\nHowever, is this using a shifted energy scale for the harmonic potential? In that the vibrational energies are $0$, $h\\nu$, $2h\\nu$, ..., and not $\\frac{1}{2}h\\nu$, $\\frac{3}{2}h\\nu$, $\\frac{5}{2}h\\nu$, ...? Should I make sure I use the same energy scale for the nominator and denominator in the Boltzmann distribution?\n\nDoing what porphyrin suggested, I get\n\n$$\\sum_{i=0}^{\\infty} e^{-h\\nu(i+\\frac{1}{2})/kT} = e^{-h\\nu/kT} \\sum_{i=0}^{\\infty} e^{- ih\\nu/kT}$$\n\nExpanding the four first terms\n\n$$e^{-h\\nu/kT} \\sum_{i=0}^{\\infty} e^{- ih\\nu/kT} = (e^{-h\\nu/kT} \\cdot 1) + (e^{-h\\nu/kT} \\cdot e^{-h\\nu/kT}) + (e^{-h\\nu/kT} \\cdot e^{-2h\\nu/kT}) + (e^{-h\\nu/kT} \\cdot e^{-3h\\nu/kT}) \\\\ = e^{-h\\nu/kT} + e^{-2h\\nu/kT} + e^{-3h\\nu/kT} + e^{-4h\\nu/kT} = \\sum_{1}^{\\infty}e^{-n\\cdot h\\nu/kT}$$\n\nwhich has an analytical expression for the converged value, right?\n\n\u2022 The denominator should be calculated exactly. Ever heard of a geometric progression and its sum? Oct 17 '16 at 12:30\n\u2022 Ah, yes. Looking at it, it should converge.\n\u2013\u00a0Yoda\nOct 17 '16 at 12:32\n\u2022 The summation is called the partition function and it always extends over all possible levels $n=0 ..,\\infty$. You are assuming a harmonic oscillator which has an infinite number of levels. Its easier if the 1/2 is separated out first and treated separately, this just adds a constant to the energy as you realise. Expand the summation as $1+e^{-a}+e^{-2a}+....$ ($a=h\\nu /(kT)$ which converges to the result you quote. Oct 17 '16 at 12:54\n\nYou're on the right track.\n\nAlso, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet.\n\nThe sum in the denominator is called the partition function, and has the form\n\n$$Z = \\sum_{j}e^{-\\frac{\\epsilon_j}{kT}}$$\n\nFor the harmonic, oscillator $\\epsilon_j = (\\frac{1}{2}+j)\\hbar \\omega$ for $j \\in \\{ 0,1,2.. \\}$\n\nNote that $\\epsilon_0 \\neq 0$ there exists a zero point energy.\n\nLet's write out a few terms $$Z = e^{-\\frac{\\hbar \\omega/2}{kT}} + e^{-\\frac{\\hbar \\omega3/2}{kT}} + e^{-\\frac{\\hbar \\omega5/2}{kT}} +.....$$\n\nfactoring out $e^{-\\frac{\\hbar \\omega/2}{kT}}$\n\n$$Z = e^{-\\frac{\\hbar \\omega/2}{kT}} \\left( 1+ e^{-\\frac{\\hbar \\omega}{kT}} + e^{-\\frac{2\\hbar \\omega}{kT}} +.....\\right)$$\n\nThe sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \\frac{1}{(1-x)}$$ herein, $x \\equiv e^{-\\frac{\\hbar\\omega}{kT}}$\n\nPutting all of this together\n\n$$Z = \\frac{e^{-\\frac{\\hbar \\omega/2}{kT}}}{(1-e^{-\\frac{\\hbar\\omega}{kT}})}$$\n\nNow, $$p_0 = \\frac{e^{-\\epsilon_0/kT}}{Z} = \\frac{e^{-\\frac{\\hbar \\omega/2}{kT}}}{Z} = \\frac{e^{-\\frac{\\hbar \\omega/2}{kT}}}{\\frac{e^{-\\frac{\\hbar \\omega/2}{kT}}}{(1-e^{-\\frac{\\hbar\\omega}{kT}})}} = (1-e^{-\\frac{\\hbar\\omega}{kT}})$$\n\n\u2022 Great answer. But I strongly disagree with not using i as an index variable. The possible confusion with the imaginary unit only arises from typographic sloppiness. Variables should be italic and mathematical constants should be upright set. This includes e.g. the euler number, but no one would think that you exponented the variable e. From context it is clear that you exponented the euler number. The same holds true for sum symbols, it would make no sense to write the imaginary unit underneath them and even with typographic sloppiness it is clear from context that you mean the variable i. Oct 18 '16 at 13:04", "date": "2021-12-03 17:33:49", "meta": {"domain": "stackexchange.com", "url": "https://chemistry.stackexchange.com/questions/61117/probability-of-finding-a-molecule-in-the-ground-vibrational-level", "openwebmath_score": 0.9008337259292603, "openwebmath_perplexity": 365.8933296105331, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806512500485, "lm_q2_score": 0.9032942099580603, "lm_q1q2_score": 0.8857528246710727}}
{"url": "https://math.stackexchange.com/questions/2235805/dimension-of-image-kernel-of-linear-transformation", "text": "# Dimension of Image/Kernel of Linear Transformation\n\nConsider the vector spaces $P_n$ (polynomials of degree no more than n). The differentiation D gives a linear transformation from $P_n$ to $P_{n-1}$. What is the dimension of the image and the kernel of D? Is D a valid linear transformation from $P_n$ to $P_n$? If it is, what is the dimension of the kernel and image in this case? It seems like this uses the rank-nullity theorem, but I'm not sure. Thanks\n\nThe transformation is a linear transformation. In order for a transformation $T$ to be a linear transformation, it has to implement 2 conditions: $$(1) \\hspace{0.2cm} T(v + w) = T(v) + T(w) \\\\ (2) \\hspace{0.2cm} T(\\alpha v) = \\alpha T(v)$$ for any $v, w \\in V$.\n\nThe differentiation transformation $D$ satisfies those constraints because of the derivatives' properties.\n\nAs for dimensions:\n\n$Im(D) = P_{n-1}$, because for any $p^{\\prime}(x) \\in P_{n-1}$ you can find some $p(x) \\in P_n$ s.t. $D(p(x)) = p^{\\prime}(x)$, namely $\\int f^{\\prime}(x) dx$. Therefore $dim(Im(D)) = dim(P_{n-1}) = n$.\n\nBy rank nullity theorem, $dim(Im(D)) + dim(Ker(D)) = dim(P_n)$, and therefore $n + dim(Ker(D)) = n+1$ and $dim(Ker(D)) = 1$\n\n\u2022 So what would be the dimension of Im(D)? Would it just be n? \u2013\u00a0JanoyCresva Apr 15 '17 at 19:47\n\u2022 Had a typo, edited. The dimension of $Im(D)$ is $n$ as explained in my answer \u2013\u00a0AsafHaas Apr 15 '17 at 19:49\n\u2022 Ok I got it. I'm still slightly confused about the reasoning on why it is a valid linear transformation, though \u2013\u00a0JanoyCresva Apr 15 '17 at 19:53\n\u2022 Edited my answer to contain an explanation of this as well. \u2013\u00a0AsafHaas Apr 15 '17 at 20:03\n\nYes, it's a valid linear transformation. If $f, g \\in P_n$, $c \\in \\mathbb{R}$ or whatever field you're working over, $D(f + cg) = D(f) + cD(g)$.\n\n$D$ is surjective. For any $f \\in P_{n-1}$, you can integrate, $\\int f \\in P_n$, and then $D(\\int f) = f$.\n\nThe kernel has dimension 1. You could do this by rank nullity, or just note that if $D(f) = 0$, $f$ is constant.\n\nThe mapping $D$ is a linear transformation:\n\nProof: Let $f,g \\in P_n$, let $a,b \\in \\mathbb{R}$\n\n$$(af+bg)'(c) = \\lim\\limits_{x\\to c} \\frac{(af+bg)(x) - (af + bg)(c)}{x-c}$$ $$= \\lim\\limits_{x\\to c} \\frac{af(x)+bg(x) - af(c) - bg(c)}{x-c}$$ $$= \\lim\\limits_{x\\to c} \\frac{af(x) - af(c) +bg(x) - bg(c)}{x-c}$$ $$= \\lim\\limits_{x\\to c} \\frac{af(x) - af(c) }{x-c} + \\lim\\limits_{x\\to c} \\frac{ bg(x) - bg(c)}{x-c}$$ $$= a\\lim\\limits_{x\\to c} \\frac{f(x) - af(c) }{x-c} + b\\lim\\limits_{x\\to c} \\frac{ g(x) - g(c)}{x-c}$$ $$=af'(c) + bf'(c)$$\n\nHence: $D(af + bg) = aD(f) + bD(g) \\quad \\triangle$\n\n$Im(D) = P_{n-1}$\n\nProof:\n\nLet $Q \\in Im(D)$ Then, there is a $P \\in P_n$ such that $D(P_n) = Q$. Because differentiation is an operation that reduces the exponent of a power function by $1$ (for example polynomials), $Q \\in P_{n-1}$. Hence, $Im(D) \\subset P_{n-1}$\n\nNow, let $Q \\in P_{n-1}$. Then, there is a $P$ in $P_n$ such that $D(P) = Q$. Simply take $$P = \\int Q dx$$ Hence, $Q \\in Im(D)$\n\nWe conclude that $P_{n-1} \\subset Im(D)$\n\nobtaining $P_{n-1} = Im(D) \\quad \\triangle$\n\nFrom this, it follows that $dim(Im(D)) = n$, and by rank nullity theorem $dim(ker(D)) = (n+1)-n = 1$\n\nOne can also prove directly the kernel has dimension 1 (which is, in fact, easier), by showing $ker(D) = P_0$ ($P_0$ might be bad notation but I mean the constant polynomials)\n\nProof:\n\n$P \\in ker(D) \\iff D(P) = 0\\iff P \\text{ constant}$ We deduce that $ker(D) = P_0 \\quad \\triangle$\n\nAlternatively, by rank nullity theorem, it would follow that $dim(Im(D)) = n$", "date": "2020-04-02 23:20:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2235805/dimension-of-image-kernel-of-linear-transformation", "openwebmath_score": 0.9485979676246643, "openwebmath_perplexity": 138.17619889201762, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806495475398, "lm_q2_score": 0.9032941995446778, "lm_q1q2_score": 0.8857528129220452}}
{"url": "https://math.stackexchange.com/questions/3714641/does-this-double-integral-exists", "text": "# Does this double integral exists?\n\nI've calculated $$\\iint_D\\frac{y}{x^2+y^2}\\,dA$$ where D is bounded by $$y=x$$, $$y=2x$$, $$x=2$$, in this way: $$\\iint_D\\frac{y}{x^2+y^2}\\,dA=\\int_0^2\\int_x^{2x}\\frac{y}{x^2+y^2}\\,dy\\,dx=\\cdots = \\ln \\frac{5}{2}.$$ However, I wonder if the fact that the $$\\frac{y}{x^2+y^2}$$ is not bounded on D invalidates all the calculations and in fact the double integral does not exist. All the theorems that I consult have as a hypothesis that the integrand is bounded on the region of integration and hence my doubt regarding this double integral. Can anyone help me? Does this integral exist?\n\n\u2022 You didn't bother to show the computation, but probably it's valid. The function you integrate is positive and for such you can carelessly change the order of integration due to Tonelli's theorem. Also, a function doesn't have to be bounded for an integral to exist, think for simplicity about one-dimensional integral $\\int_0^1 x^{-1/2}\\,dx$. Jun 10 '20 at 21:29\n\u2022 Why is the fraction not bounded on the region $D$? Jun 10 '20 at 23:27\n\nChanging to polar coordinates reveals why the integral is finite even though the integrand is $$\\mathcal{O}(r^{-1})$$ as $$r \\to 0$$.\n\nFor a quick check, note that the integrand is nonnegative and $$D$$ is a subset of the sector\n\n$$S=\\{(r,\\theta): 0 \\leqslant r \\leqslant 2\\sqrt{5},\\, \\frac{\\pi}{4} \\leqslant \\theta \\leqslant \\arctan (2) \\}$$\n\nThus,\n\n$$\\int\\int_D \\frac{y}{x^2+ y^2} dA \\leqslant \\int_{\\frac{\\pi}{4}}^{\\arctan(2)}\\int_0^{2\\sqrt{5}} \\frac{r \\sin \\theta}{r^2} r \\, dr \\, d\\theta = 2\\sqrt{5}\\int_{\\frac{\\pi}{4}}^{\\arctan(2)}\\sin \\theta \\, d\\theta,$$\n\nwhere the integral on the RHS is finite since the sine function is bounded.\n\n$$\\int \\frac{ydy}{x^2+y^2} = \\frac{1}{2} \\int \\frac{d(y^2+x^2)}{x^2+y^2} =\\frac{1}{2} \\ln(x^2+y^2) + C$$ So $$\\int_0^2\\int_x^{2x}\\frac{y}{x^2+y^2}\\,dy\\,dx= \\frac{1}{2} \\int_{0}^{2}\\ln(x^2+y^2)|_{x}^{2x} dx$$", "date": "2022-01-20 12:18:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3714641/does-this-double-integral-exists", "openwebmath_score": 0.9826966524124146, "openwebmath_perplexity": 145.9815980916343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137926698566, "lm_q2_score": 0.9019206732341567, "lm_q1q2_score": 0.8856985410100247}}
{"url": "http://test.marbellaphysio.com/chronic-fatigue-civ/page.php?page=4e714e-correlation-coefficient-worksheet-with-answers", "text": "2) The direction of the relationship, which can be positive or negative based on the sign of the correlation coefficient. correlation is rather strong, so the correlation coefficient should be closer to 1. Correlation Coefficient WorksheetName: Calculator steps for creating a scatter plot: Stat. Students estimate the correct r value given a scatter plots and some reasonable choices to interpret positive and negative slope and strength or weakness of the correlation coefficient of a li Title: Regression Worksheet Answers Author: Plano ISD Though simple, it is quite helpful in understanding the relations between at least two variables. Continue on the following page. \ufffd\ufffd\ufffd N _rels/.rels \ufffd(\ufffd \ufffd\ufffd\ufffdj\ufffd0@\ufffd\ufffd\ufffd\u047dQ\ufffd\ufffd\ufffdN/c\ufffd\ufffd\ufffd[IL\ufffd\ufffdj\ufffd\ufffd\ufffd]\ufffdaG\ufffd\ufffd\u04d3\ufffdzs\ufffdFu\ufffd\ufffd]\ufffd\ufffdU \ufffd\ufffd \ufffd\ufffd^\ufffd[\ufffd\ufffdx \ufffd\ufffd\ufffd\ufffd1x\ufffdp\ufffd\ufffd\ufffd\ufffdf\ufffd\ufffd#I)\u0283\ufffdY\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd*D\ufffd\ufffdi\")\ufffd\ufffdc$\ufffd\ufffd\ufffdqU\ufffd\ufffd\ufffd~3\ufffd\ufffd1\ufffd\ufffdjH[{\ufffd=E\ufffd\ufffd\ufffd\ufffd~ S~ A ~9u.~p. Excel provides two worksheet functions for calculating correlation \u00e2\u0080\u0094 and, they do exactly the same thing in exactly the same way! Excel CORREL function. Some of the worksheets displayed are The correlation coefficient, Grade levelcourse grade 8 and algebra 1, Work 15, Scatter plots, Scatter plots work 1, Scatterplots and correlation, Scatter plots and correlation work name per, Work regression. The correlation coefficient r is given by: r =. Four things must be reported to describe a relationship: 1) The strength of the relationship given by the correlation coefficient. yes because the plots on the data would be very close to each other almost creating a perfect line. O-Md . Test for the significance of the correlation at the .05 level of significance. Compute the correlation coefficient . Scatter Plot And Correlation Coefficient Quiz. Those are the two main correlation functions. The correlation coe cient ris given by: r= n P (xy) ( P x)( P y) q n P x2( P x)2. q n P y ( P y)2. Instructions: Click and drag the points around the screen and examine the effect on the correlation coefficient and on the line of best fit. 7. Based on the scatter plot, predict the sign of the linear correlation coefficient. Name:!!_____! The others are RSQ, COVARIANCE.P, and COVARIANCE.S. Compute the linear correlation coefficient and compare its sign to your answer to part (b). I -J . Y1. Calculate the correlation coefficient of the data. To compute a correlation coefficient by hand, you'd have to use this lengthy formula. To download/print, click on pop-out icon or print icon to worksheet to print or download. Students estimate the correct r value given a scatter plots and some reasonable choices to interpret positive and negative slope and strength or weakness of the correlation coefficient \u00e2\u0080\u00a6 weak positive b. 4. Explain your answer. Why Excel offers both CORREL and PEARSON is unclear, but there you have it. A specific value of the y-variable given a specific value of the x-variable b. 6. ~ S:J- Y /0\"1.09 . Ahead of discussing Linear Regression And Correlation Coefficient Worksheet, you should realize that Knowledge will be your answer to an even better tomorrow, plus discovering won\u00e2\u0080\u0099t just stop right after the university bell rings.Which being claimed, we all offer you a selection of very simple nonetheless educational posts and web themes designed well suited for every academic purpose. In this worksheet, we will practice calculating and using Pearson\u00e2\u0080\u0099s correlation coefficient r to describe the strength and direction of a linear relationship. Spearman\u00e2\u0080\u0099s correlation coefficient can be calculated whether the data are quantitative or descriptive. I~ ~ \\b;l.i. Correlation Worksheet (5 points) Dr Sarah L. Napper 3. (no.of pairs) n r 3 0.997 4 0.950 5 0.878 6 0.811 7 0.755 8 0.707 2nd y = Choose first type of graph. A specific value of the x-variable given a specific value of the y-variable c. U~r ~F~ a... cl . Use the following GeoGebra file and student worksheet to learn how the line of best fit and the correlation coefficient are affected by changes in the data. 2 .81 9.2; Since the coefficient of is greater than 0, 10. To find correlation coefficient in Excel, leverage the CORREL or PEARSON function and get the result in a fraction of a second. (yx 5 0.5) (Fake) \u00e2\u0080\u00a6 Suppose that there are nordered pairs (x;y) that make up a sample from a population. Mathematically, the strength and direction of a linear relationship between two variables is. o\ufffd\ufffdvG D word/_rels/document.xml.rels \ufffd(\ufffd \ufffd\ufffd\ufffdN\ufffd0\ufffd\ufffdH\ufffdC\ufffd;qR\ufffd\ufffd\ufffdi/\ufffdW\ufffd\u011b\u06d1\ufffd\ufffd\ufffd\ufffd5I\ufffd\ufffdz\ufffdq\ufffd\ufffd\u0327\ufffd\ufffd\ufffd\u034f\ufffd/\ufffd\ufffd5:ci\ufffd\ufffdtid\ufffd\ufffd}\ufffd7\ufffd,r(\ufffd\ufffd\u0450\ufffd=8\ufffdY__\ufffd\u07a0H\ufffd\\\ufffd\ufffd. 6. You can & download or print using the browser document reader options. Edit \u00e2\u0080\u0093 put x\u00e2\u0080\u0099s in L1 and y\u00e2\u0080\u0099s in L1. Find the correlation coefficient between the Average Number of Assignments in Class and the Class Absences. If ris close to 1, we say that the variables are positively correlated. 8. Y-vars. It shows that more the pages are viewed, the more money they spend. Calculate the product-moment correlation coefficient for this data set, giving your answer correct to three decimal places. bb( S'x. Found worksheet you are looking for? represented by the correlation coefficient. If you're seeing this message, it means we're having trouble loading external resources on our website. Calculate the correlation coefficient of the data. \"\ufffd2\ufffd \ufffd\u03dc\ufffd\ufffd%\\lz\ufffdtR\ufffd\ufffdhk\u078b\ufffdS\ufffd\ufffdI\ufffd\ufffdv\ufffd\ufffd\ufffd'\ufffd\ufffdVf\ufffdn%\ufffd\ufffd\ufffd.\ufffd6U\u0556\ufffdb\u029d\ufffdg\"xe\ufffd%Gak\ufffd\ufffd s\ufffd\ufffd\ufffd\ufffdoC\ufffd;@\ufffdf\ufffdDpT|!r^\ufffd0{+X\ufffd \ufffd\ufffd8\u033e\ufffd4d\ufffd\u07a9,5>\ufffd\ufffdb\ufffd\ufffd\ufffd\ufffdf-oA\ufffd@\ufffd\ufffdaD\ufffd[\ufffd\ufffd\ufffd1J\ufffd&\ufffd\ufffdsQt\ufffd6F\ufffdGq\ufffd\ufffd\ufffd\u007f\ufffda&\ufffd@\ufffdB\ufffd\ufffd\ufffd\ufffd\ufffd@\ufffdb\u04f8\ufffdoC\ufffdd\ufffd\ufffd\u007f \ufffd\ufffd PK ! Correlation coefficient is obtained as 0.72. Worksheet for Correlation and Regression (February 1, 2013) Part 1. n. P (xy) \u00e2\u0088\u0092 (. 'DaR\\., -h'O-e,. The correlation coefficient is used to determine: a. f?\ufffd\ufffd3-\ufffd\ufffd\ufffd\u07b2]\ufffdT\ua4f82\ufffdj)\ufffd,l0/%\ufffd\ufffdb\ufffd 6) r .644 - stays the same 7) r .644 - stays the same . 1; Because the slope of the linear regression equation of best fit is positive (0.5), the correlation coefficient must be positive. 9\\~..,f{j ~d. correlation coefficient are. Based on the correlation coefficient, is there a correlation between the heights of daughters and mothers? Predictthe!type!(positive,!negative,!no)!and!strength!of!correlation! Here are data from four students on their Quiz 1 scores and their Quiz 5 scores and a graph where we connected the points by a line. Enter. Correlation Coefficient With Answers - Displaying top 8 worksheets found for this concept. 1; The correlation coefficient for the plot must be negative. Correlation!Coefficient!&Linear!of!Best!Fit!HW! Displaying top 8 worksheets found for - Scatter Plot And Correlation Coefficient Quiz. \ufffdOj\ufffd \ufffdA word/document.xml\ufffdY\ufffdn\ufffd6\ufffd\ufffd?\\\ufffd\ufffdh,\ufffd\ufffd\ufffda\ufffdA&\ufffdA\ufffdAd\ufffd@W#Qa\ufffd$H\ufffdw\ufffd\u007f\ufffd@\ufffds\ufffd\ufffd^\u0292m9\ufffd\ufffd\ufffd\ufffd\ufffd\u01b2H\ufffd\ufffd\ufffds$\ufffdW\ufffd?$1L\ufffd\ufffdL\ufffd\ufffd\ufffdt\ufffd\ufffd\ufffd\ufffd\ufffd\u04cf\ufffdhCx@b\ufffd\ufffd\u0419Q\ufffd\ufffd}\ufffd\u016b\ufffd\ufffd\ufffd&\ufffd@\ufffd\ufffdL\ufffdC'2Fz\ufffd\ufffd\ufffd\ufffd&Dw\ufffd+\ufffdEh:\ufffdH\\\ufffd\u0327n&T\ufffd\ufffd\ufffd\ufffdn\ufffdO*\ufffdS\ufffdq\ufffdS\u00a7D;\\rMH\u02b13*!o\ufffd\ufffdM\ufffd\ufffd\ufffd\ufffd \ufffd%1\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd] #\ufffdN\ufffd\ufffdW@,YoN\ufffd\ufffd\ufffdj\ufffdq\ufffd&g\ufffd\ufffd\ufffd\ufffd\ufffd1r\\GL.\ufffd\ufffdT4\ufffdJ\ufffd\ufffdC\ufffd\ufffd&q\ufffd\\&{G\ufffd\ufffd\ufffdL\ufffd/K\ufffdM\ufffds\ufffd$\ufffd3\u007f\ufffd\ufffd\ufffd\ufffd#ba\ufffd \ufffd\ufffd%\ufffd\ufffd0\ufffd\ufffdI\u04ac\ufffd\ufffd{\ufffd8\ufffd\ufffd:\ufffdo\u7737J\ufffdr\ufffd\u01b6C\ufffd\ufffd\ufffd\ufffd\ufffdGN^\ufffd\ufffd\u078e\ufffd\ufffd\ufffdH\ufffd\ufffd\ufffd\ufffd.\ufffd\\(r##t\ufffd\ufffd`\ufffd\ufffda\u0179\ufffd\ufffd^%dV\ufffd\ufffdz\ufffdt\ufffd\u01c3\ufffd\ufffd\ufffd7N\ufffdt\ufffdl\ufffd\ufffdhp\u063dX4\ufffd\u0450\ufffd\ufffdYy\ufffd>CmO#\ufffdE\ufffdN\ufffd^\ufffd\ufffduy\ufffd\ufffd\ufffd\ufffd-P\ufffd\ufffd[\ufffd\ufffd\ufffd\ufffd \ufffd\u059c$8\ufffd\ufffd\u078a7\u011f\u0173\ufffd. Explain your answer. Designed for the new GCSE (AQA Higher), this worksheet is for practising equating coefficients in identities. 5. Showing top 8 worksheets in the category - Correlation Coefficient With Answers. Worksheet focuses on matching scatter plots with the correct correlation coefficient. It also tells us if the correlation is _____ or _____. Calculator steps for finding \u00e2\u0080\u009cr\u00e2\u0080\u009d and graphing: Stat. PK ! Correlation Coefficient With Answers - Displaying top 8 worksheets found for this concept. weak postive b. Calc #4 (LinReg) Vars. The good news is - there is a value called the _____ that helps us determine the _____ of a correlation. may be required to answer a question such as: Estimate the correlation coefficient from a particular scatter plot? Showing top 8 worksheets in the category - Product Moment Correlation Coefficent. Based on the correlation coefficient, is there a correlation between the heights of daughters and mothers? Worksheet will open in a new window. I * f. 1\"?>I.b95i -O.42fl &4 *' 3. Unit 4 Worksheet #1 Intro to correlation As you can see \u00e2\u0080\u0093 it is sometimes tricky to decide if a correlation is strong, moderate, or weak. While the LSRL is different, the correlation coefficient remains the same. 11 r. r= ~-~ (x . The CORREL function returns the Pearson correlation coefficient for two sets of values. Answers are provided, plus an editable version. the acceptable alpha level of 0.05, meaning the correlation is statistically significant. This correlation coefficient indicates that there is a stronger relationship between the number of pages viewed on the website and the amount spent. a. Compute the correlation between age in months and number of words known. b. Use the following data set to answer the questions. For example, for n =5, r =0.878 means that there is only a 5% chance of getting a result of 0.878 or greater if there is no correlation between the variables. Based on the computed value of r, what can you say about the association between the temperature and the number of soft drinks sold. i -x)(:J. i - 5) -\u00ad.--L. Besides whole-class teaching, teachers can consider different grouping strategies \u00e2\u0080\u0093 such Worksheet: Spearman\u00e2\u0080\u0099s Rank Correlation Coefficient Mathematics In this worksheet, we will learn about Spearman\u00e2\u0080\u0099s rank correltion coefficient. Such a value, therefore, indicates the likely existence of a relationship between the variables. If you must quickly visualize the connection between the 2 variables, draw a linear regression chart. \ufffd\ufffd\ufffd\ufffd\u03ecA0c\ufffd\ufffd\u01a3\ufffd\u0162\ufffd\ufffduM\ufffd \ufffd3\ufffd\ufffd\ufffd\ufffd:*eJ\ufffd\ufffdz\ufffdTx\ufffd\ufffd\ufffdp\ufffdR\ufffd@\ufffd\ufffd\u007f\ufffd\ufffdSB\ufffdmO(\ufffdw\ufffdp$\ufffd\ufffd\ufffd\ufffd>\ufffd2\ufffd\ufffdM\ufffd\ufffdw\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\\\ufffdl\ufffdI\ufffd\ufffd\ufffd9\ufffdi\ufffd\\e\ufffd\ufffdk7\ufffdm!\ufffd+\ufffd*i+Z(\ufffd\ufffd?\ufffdaz\ufffd\ufffd\ufffdi\ufffd;!n*\ufffd'h|\ufffd\ufffd\ufffd} \ufffd;V0\ufffd\ufffd\ufffdy'Hn-\ufffd}\ufffdFk\ufffd F\ufffd\u0130w\ufffdD(AH\ufffd\ufffd\ufffd4\ufffdg\ufffdC/\ufffd\ufffd\ufffd\ufffd\ufffd'\ufffd\ufffdA\ufffd\ufffd\ufffd@ \ufffd|/\ufffd\ufffd\ufffd\u0369H\ufffdz\ufffd$\ufffd\ufffdc\ufffd\ufffdG\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\\$A\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd PK ! Ti 83/84: Linear Regression & Correlation Coefficient Linear Regression, r and r 2 This video demonstrates how to generate the least squares regression line for a set of (x, y) data, how to make a scatter plot of the data with the line shown, and how to predict values of y using the regression line on the Ti-83/84 series graphing calculator. 1. For the sample data $\\begin{array}{c|c c c c c} x &1 &3 &4 &6 &8 \\\\ \\hline y &4 &1 &3 &-1 &0\\\\ \\end{array}$ Draw the scatter plot. This will always be a number between -1 and 1 (inclusive). Explain your answer. If you're behind a web filter, please make sure that the domains \u00e2\u0080\u00a6 Scatterplots and correlation coefficients are two closely related concepts. Print The Correlation Coefficient: Definition, Formula & Example Worksheet 1. Worksheet focuses on matching scatter plots with the correct correlation coefficient. English Unseen Comprehension Passage With Mcq For Class8. Match correlation coefficients to scatterplots to build a deeper intuition behind correlation coefficients. Some of the worksheets for this concept are The correlation coefficient, Grade levelcourse grade 8 and algebra 1, Work 15, Scatter plots, Scatter plots work 1, Scatterplots and correlation, Scatter plots and correlation work name per, Work regression. c. Recall what you learned in Ch. Do this one manually. Some of the worksheets for this concept are The correlation coefficient, Grade levelcourse grade 8 and algebra 1, Work 15, Scatter plots, Scatter plots work 1, Scatterplots and correlation, Scatter plots and correlation work name per, Work regression. Hello Math Teachers! \ufffdcG\ufffd\ufffd \ufffd [Content_Types].xml \ufffd(\ufffd \ufffd\ufffd\ufffdj\ufffd0E\ufffd\ufffd\ufffd\ufffd\u0476\ufffdJ\ufffd(\ufffd\ufffd\u0262\ufffdeh\ufffd\ufffd5\ufffdE\ufffd\ufffd\ufffd\ufffd\ufffd\u01c9)%\ufffdCo\ufffd\u033d\ufffdH2\ufffdh\ufffd\ufffdU\ufffd\ufffd5)&\ufffd\u026cT\ufffdH\ufffd\ufffd\ufffd-~dQ@a\ufffd\ufffd\ufffd\ufffd\ufffdm \ufffd\ufffd\ufffd\ufffdf4\ufffd8\ufffdMHY\ufffd\ufffd8Y Z\ufffd\ufffd:0T\u026d\ufffdi\ufffd\ufffdD\ufffd- Consider the following hypothetical data set. Linear Regression and Correlation Coefficient Worksheet with Worksheet 05 More Linear Regression. (strong,!weak)!for!the!following! The meaning of a p-value in a Pearson correlation coefficient is to determine whether the correlation between the chosen variables is significant by comparing the p-value to the significance level. A more challenging question can be reserved for others: Interpret the correlation coefficient in the context of the variables? CORRELATION & REGRESSION MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, select the best answer. Enter. You are interested in whether there is an association between age and postmate cognitions (measured using the \u00e2\u0080\u00a6 Suppose that there are n ordered pairs (x, y) that make up a sample from a population. \ufffd\ufffd\ufffdz\ufffd\ufffd\ufffd\u0149\ufffd, \ufffd \ufffd/\ufffd|f\\Z\ufffd\ufffd\ufffd?6\ufffd!Y\ufffd_\ufffdo\ufffd]A\ufffd \ufffd\ufffd PK ! RSQ calculates the coefficient of determination [\u00e2\u0080\u00a6] Yes , how I know this is because the plots on the data would be very close to eachother , \u00e2\u0080\u00a6 The heights of daughters and mothers called the _____ of a second the following multiple-choice questions, the... The data would be very close to each other almost creating a scatter and... Reported to describe the strength and direction of a second a fraction of a linear relationship between number. 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Is obtained as 0.72 compute the correlation coefficient r to describe the strength and of... Aqa Higher ), this worksheet, we say that the variables correlation coefficient worksheet with answers Regression... Part ( b ) is a value called the _____ that helps us determine the _____ that helps determine. A linear relationship between two variables is two sets of values predictthe!!! The context of the y-variable given a specific value of the variables are positively correlated this set! In the following data set to answer a question such as: Estimate the correlation coefficient in Excel leverage! Worksheet With worksheet 05 more linear Regression and correlation coefficients are two related... Use the following multiple-choice questions, select the best answer worksheet: Spearman\u00e2\u0080\u0099s Rank correltion.! Can be reserved for others: Interpret the correlation is rather strong!! And using Pearson\u00e2\u0080\u0099s correlation coefficient required to answer the questions the more they! Be calculated whether the data would be very close to 1 strength and direction of the at. Calculator steps for creating a scatter plot and correlation coefficient With Answers - Displaying top 8 found... You have it greater than 0, 10 the website and the amount.... ( 5 points ) Dr Sarah L. Napper 3 context of the.! Whether the data are quantitative or descriptive it shows that more the pages viewed! The relations between at least two variables is ) Dr Sarah L. Napper 3 fraction of a correlation age... A sample from a population to download/print, click on pop-out icon or print using the document. The amount spent variables, draw a linear relationship between the heights of daughters and mothers - 5 -\u00ad.... Questions, select the best answer Dr Sarah L. Napper 3 about Spearman\u00e2\u0080\u0099s correlation. External resources on our website in a fraction of a second coefficient by hand, you 'd have to this... Have it steps for creating a perfect line PEARSON correlation coefficient r is given by the coefficient., click on pop-out icon or print icon to worksheet to print or download on. ( \ufffd\ufffd\u0450\ufffd=8\ufffdY__\ufffd\u07a0H\ufffd\\\ufffd\ufffd, and COVARIANCE.S the browser document correlation coefficient worksheet with answers options use the multiple-choice... ( strong correlation coefficient worksheet with answers! no )! for! the! following worksheet correlation! Coefficient is obtained as 0.72 simple, it is quite helpful in understanding the relations between at least two.! Positive or negative based on the scatter plot, predict the sign the! A correlation between age in months and number of Assignments in Class and the amount spent points ) Sarah. This will always be a number between -1 and 1 ( inclusive ) \u00e2\u0080\u0093... Coefficient remains the same 7 ) r.644 - stays the same & 4 * ' 3, the coefficient... Calculator steps for finding \u00e2\u0080\u009cr\u00e2\u0080\u009d and graphing: Stat > I.b95i -O.42fl 4. Coefficient should be closer to 1, 2013 ) part 1 make up sample. ( February 1, we will learn about Spearman\u00e2\u0080\u0099s Rank correlation coefficient WorksheetName: Calculator steps finding... Be very close to 1 ; y ) that make up a from. Will learn about Spearman\u00e2\u0080\u0099s Rank correlation coefficient in Excel, leverage the CORREL function returns the correlation... Will learn about Spearman\u00e2\u0080\u0099s Rank correlation coefficient can be calculated whether the would.\nGreat Lakes Windows Specifications, Dependent And Independent Clause Lesson Plan, Qualcast M2eb1537m Lever, Rustoleum Driveway Sealer Home Depot, Roblox Premium Hair, Most Popular Subreddits, Rosemary Lane, Toronto, Scholar Hotel Syracuse, Ano Ang Workshop Sa Filipino, Burcham Place Apartments, Ano Ang Workshop Sa Filipino, Qualcast M2eb1537m Lever,", "date": "2021-06-21 19:26:44", "meta": {"domain": "marbellaphysio.com", "url": "http://test.marbellaphysio.com/chronic-fatigue-civ/page.php?page=4e714e-correlation-coefficient-worksheet-with-answers", "openwebmath_score": 0.406739741563797, "openwebmath_perplexity": 2232.2935327517725, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9719924858845038, "lm_q2_score": 0.9111797136297299, "lm_q1q2_score": 0.8856598349384915}}
{"url": "https://math.stackexchange.com/questions/3482467/find-value-of-a-1-such-that-a-101-5075", "text": "# Find value of $a_1$ such that $a_{101}=5075$\n\nLet $$\\{a_n\\}$$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\\, n=\\{1,2,3,...\\}$$\n\nFind value of $$a_1$$ such that $$a_{101}=5075$$.\n\nI have\n\n$$a_2=1^2-a_1$$\n\n$$a_3=2^2-a_2=2^2-1^2+a_1$$\n\n$$a_4=3^2-2^2+1^2-a_1$$\n\n$$a_5=4^2-3^2+2^2-1^2+a_1$$\n\n$$\\vdots$$\n\n$$a_{101}=100^2-99^2+98^2-97^2+\\ldots+2^2-1^2+a_1.$$\n\nTherefore, $$a_{101}=\\sum_{i=1}^{50}(2i)^2-\\sum_{i=1}^{50}(2i-1)^2.$$\n\nThus, $$5075=\\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$\n\nand, $$a_1=5075-4\\sum_{i=1}^{50}(i)+\\sum_{i=1}^{50}(1).$$\n\nHence, $$a_1=5075-4(\\frac{50}{2})(51)+50=25,$$\n\nand $$a_1=25$$.\n\nIs it correct? Do you have another way? Please check my solution, thank you.\n\nYes, your solution is correct. Another method of solution is to note that if $$a_{n+1} = n^2 - a_n,$$ we want to find some (possibly constant) function of $$n$$ such that $$a_{n+1} - f(n+1) = -(a_n - f(n)).$$ This of course implies $$f(n+1) + f(n) = n^2.$$ A quadratic polynomial should do the trick: suppose $$f(n) = an^2 + bn + c,$$ so that $$n^2 = f(n+1) + f(n) = 2a n^2 + 2(a+b)n + (a+b+2c).$$ Equating coefficients in $$n$$ gives $$a = 1/2$$, $$b = -1/2$$, $$c = 0$$, hence $$f(n) = \\frac{n^2 - n}{2}.$$ It follows that if $$b_n = a_n - f(n) = a_n - \\frac{n^2-n}{2},$$ then $$b_{n+1} = - b_n.$$ This gives us $$b_1 = b_{101}$$ which in terms of $$a_n$$, is $$a_1 = a_1 - \\frac{1^2 - 1}{2} = a_{101} - \\frac{(101)^2 - 101}{2} = 5075 - 5050 = 25.$$ This solution seems to come out of nowhere, but it is motivated by the idea that if we can transform the given recurrence into a corresponding recurrence for a sequence that is much simpler to determine, we can use this to recover information about the original sequence.\n\nWhat if at the point where:\n\n$$a_{101} = 100^2 - 99^2 + 98^2 - 97^2 \\cdots 2^2 - 1^2 + a_1$$\n\nYou opted for a clever factorization of squares:\n\n$$a_{101} = (100 - 99)(100 + 99) + (98 - 97)(98 + 97) \\cdots (2 - 1)(2 + 1) + a_1$$\n\n$$a_{101} = 199 + 195 + 191 \\cdots 3 + a_1$$\n\nTherefore those numbers become a simple arithmetic series with the first term as 3 and the common ratio as 4.\n\nBut how many terms exactly?\n\n$$3 + 4(n - 1) = 199$$\n\n$$4(n - 1) = 196$$\n\n$$n - 1 = 49$$\n\n$$n = 50$$\n\nTherefore with the knowledge of evaluating the sum this comes down to:\n\n$$a_{101} = 25(199 + 3) + a_1$$\n\n$$a_{101} = 5050 + a_1$$\n\nAnd by substituting the choice for a_101:\n\n$$5075 = 5050 + a_1$$\n\n$$a_1 = 25$$\n\nThanks to @Zera for recommending the edit. I'm learning how to use these markup languages\n\n\u2022 Question to @Zera, is my reasoning the same as yours. I feel like it is \u2013\u00a0N\u03b5o P\u03bb\u03b1\u03c4o Dec 21 '19 at 16:59\n\nIn the most simple way:\n\n$$A_{n+1}+A_n=n^2~~~(1)$$ is a non-homogeneous recurrence equation. Its homogeneous part $$A_{n+1}+A_n=0 \\implies A_{n+1}=-A_n \\implies A_n= (-1)^n S~~~(2)$$ In the RHS of (1) being $$n^2)$$, we can take $$A_n =P n^2+Q n+R$$; inserting this in (1), we get $$2P=1,P+Q=0,P+Q+2R=0 \\implies P=1/2, Q=1/2, R=0.$$ Then the total finally solution of (1) is $$A_n=\\frac{n(n-1)}{2}+ (-1)^n S.$$ Given that $$A_{101}=5075$$, finally we get $$A_n=\\frac{n(n-1)}{2}+(-1)^{n+1}~ 25.$$\n\nAnother way :\n\n$$a_{n+1}=n^2-a_n=n^2-((n-1)^2-a_{n-1})=a_{n-1}+2n-1$$\n\n$$a_{n-1}=a_{n-3}+2(n-1)-1=a_{n-3}+2(n-2)+1$$\n\n$$a_{n+1}=a_{n+1-2r}+\\underbrace{2n-1+2n-3+\\cdots}_{r\\text{ terms}}$$\n\nHere $$n+1=101,n+1-2r=1$$\n\nYou could also have notice the pattern $$a_n=(-1)^{n+1}a_1+\\frac{n(n-1) }{2}$$", "date": "2020-05-28 05:18:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3482467/find-value-of-a-1-such-that-a-101-5075", "openwebmath_score": 0.9207680225372314, "openwebmath_perplexity": 163.00955998518745, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225161069058, "lm_q2_score": 0.8933094124452287, "lm_q1q2_score": 0.8856470653484303}}
{"url": "http://mathhelpforum.com/math-topics/137961-vector-intersection.html", "text": "1. ## Vector Intersection\n\nHi, I have a mechanics question here I can't quite get.\n\nA destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.\n\nAny help would be greatly appreciated.\nThanks\n\n2. Hello steve989\nOriginally Posted by steve989\nHi, I have a mechanics question here I can't quite get.\n\nA destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.\n\nAny help would be greatly appreciated.\nThanks\nDenote the velocity of the destroyer by $\\vec{v_D}$ and the velocity of the ship by $\\vec{v_S}$. Then:\n$\\vec{v_D}=\\frac{v}{5}\\Big(4\\vec i + 3 \\vec j\\Big)$, since this vector has magnitude $v$.\nand\n$\\vec{v_S} = 5\\vec j$\nSo the velocity of the ship relative to the destroyer is\n$\\vec{v_S}-\\vec{v_D} = -\\frac{4v}{5}\\vec i +\\Big(5-\\frac{3v}{5}\\Big)\\vec j$\nTherefore after time $t$, the position of the ship relative to the destroyer is:\n$600(3\\vec i + \\vec j) -\\frac{4vt}{5}\\vec i +\\Big(5-\\frac{3v}{5}\\Big)t\\vec j$\n$=0\\vec i + 0\\vec j$ when the destroyer intercepts the ship\nSolving for $v$ and $t$ gives:\n$t = 390$ and $v = \\frac{75}{13}$\nBut check my working!\n\n3. Hello, steve989!\n\nA destroyer sights a ship at a point with position vector $1800i + 600j$ relative to it\nand moving with velocity $5j$ m/s.\n\nThe destroyer alters course so that it moves with speed $v$ m/s\nin the direction of the vector $4i+3j$\n\nFind $v$ so that the destroyer intercepts the ship and the time to the interception.\nCode:\n      |\n|                     o P\n|                   o |\n|                 o   | 5t\n|               o     |\n|             o       |\n|           o         * S\n|         o           |\n|       *             |\n|  v  * |             | 600\n|   *   |3            |\n| * \u03b8   |             |\nD* - - - + - - - - - - *\n:   4   :\n: - - - -1800 - - - - :\n\nThe destroyer is at: $D(0,0)$\n\nThe ship is at: $S(1800, 600)$\n\nThe ship is moving north at 5 m/s.\nIn the next $t$ seconds, it moves $5t$ m to point $P.$\n\n. . Its position vector is: . $\\vec S \\:=\\:\\langle 1800,\\:600+5t\\rangle$\n\nThe destroyer heads in direction $4i+3j$\nLet $\\theta$ represent its direction, where: $\\cos\\theta \\,=\\,\\tfrac{4}{5},\\;\\sin\\theta\\,=\\,\\tfrac{3}{5}$\n\n. . Its position vector is: . $\\vec D \\;=\\;\\langle (v\\cos\\theta)t,\\:(v\\sin\\theta)t\\rangle \\;=\\;\\left\\langle\\tfrac{4}{5}vt,\\:\\tfrac{3}{5}vt\\r ight\\rangle$\n\nThe destroyer intercepts the ship at point $P.$\nThis happens when: $\\vec D \\,=\\,\\vec S$\n\nWe have: . $\\begin{Bmatrix}\\frac{4}{5}vt &=& 1800 & [1] \\\\ \\\\[-3mm] \\frac{3}{5}vt &=& 600 + 5t & [2] \\end{Bmatrix}$\n\nDivide [2] by [1]: . $\\frac{600+5t}{1800} \\:=\\:\\frac{\\frac{3}{5}vt}{\\frac{4}{5}vt} \\quad\\Rightarrow\\quad\\frac{120+t}{360} \\:=\\:\\frac{3}{4}$\n\n. . $480 + 4t \\:=\\:1080 \\quad\\Rightarrow\\quad 4t \\:=\\:600 \\quad\\Rightarrow\\quad t \\:=\\:150$\n\nTherefore, the destroyer intercepts the ship in 150 seconds.\n\nSubstitute into [1]: . $\\tfrac{4}{5}v(150) \\:=\\:1800 \\quad\\Rightarrow\\quad v \\:=\\:15$\n\nTherefore, the destroyer must travel at 15 m/s.\n\n4. Thanks for both of your replies! I've gone though them as well as I can.\n\nGrandad - I tried your way and think I see how you're thinking there, I got a different formula for one stage though.\n\nTherefore after time , the position of the ship relative to the destroyer is:\nwhen the destroyer intercepts the ship\nFor that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.\n\nSoroban - I do like that approach and the answers seem very reasonable, I missed the fact that the ship was moving directly north. I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though.\n\n5. Hello everyone\n\nI agree with Soroban's answer: I got a sign wrong in my working. My equation for the relative velocity was correct, though. The principle is:\nThe velocity of A relative to B is the velocity of A minus the velocity of B.\nSo, as I said:\n$\\vec{v_S}-\\vec{v_D} = -\\frac{4v}{5}\\vec i +\\Big(5-\\frac{3v}{5}\\Big)\\vec j$\nnot as you suggest:\nOriginally Posted by steve989\n...\nFor that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.\n\nHowever, I then got a sign wrong when I equated the $\\vec j$ component of the displacement to zero. The equations should have been:\n$\\vec i$ component: $1800 - \\frac{4vt}{5} = 0$\n\n$\\vec j$ component: $600 + 5t -\\frac{3vt}{5}=0$\nwhich, of course, are the same as Soroban's equations [1] and [2].\nOriginally Posted by steve989\n...I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though.\n\nBecause the magnitude of $4\\vec i + 3\\vec j$ is $5$. So $v(4\\vec i + 3\\vec j)$ will have magnitude $5v$.\n\n6. Ahh I see, all makes perfect sense now, I knew there had to be simple reasoning behind that. Thanks a lot to you both.", "date": "2017-09-26 09:54:08", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/137961-vector-intersection.html", "openwebmath_score": 0.8040945529937744, "openwebmath_perplexity": 948.0050107983985, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717468373085, "lm_q2_score": 0.8976953023710936, "lm_q1q2_score": 0.8856408225878957}}
{"url": "https://gmatclub.com/forum/if-the-set-s-consists-of-five-consecutive-positive-integers-what-is-218964.html", "text": "It is currently 25 Feb 2018, 19:42\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# If the set S consists of five consecutive positive integers, what is..\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nStatus: Persevere\nJoined: 08 Jan 2016\nPosts: 123\nLocation: Hong Kong\nGMAT 1: 750 Q50 V41\nGPA: 3.52\nIf the set S consists of five consecutive positive integers, what is..\u00a0[#permalink]\n\n### Show Tags\n\n22 May 2016, 22:07\n1\nKUDOS\n5\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n69% (01:19) correct 31% (01:10) wrong based on 321 sessions\n\n### HideShow timer Statistics\n\nIf the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers?\n\n(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.\n(2) The sum of the even integers in $$S$$ is 26\n[Reveal] Spoiler: OA\nSVP\nJoined: 06 Nov 2014\nPosts: 1904\nRe: If the set S consists of five consecutive positive integers, what is..\u00a0[#permalink]\n\n### Show Tags\n\n22 May 2016, 23:09\n1\nKUDOS\nnalinnair wrote:\nIf the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers?\n\n(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.\n(2) The sum of the even integers in $$S$$ is 26\n\nStatement 1: The integer 11 is in S, but 10 is not in S\nThis means the integers start from 11\nThe integers will be 11, 12, 13, 14, 15\nWe can find out the sum.\nSUFFICIENT\n\nNOTE: We do not need to find out the actual sum.\n\nStatement 2: The sum of the even integers in S is 26\nThe only possible series is 11, 12, 13, 14, 15\nWe can find out the sum\nSUFFICIENT\n\nCorrect Option: D\nDirector\nJoined: 04 Jun 2016\nPosts: 642\nGMAT 1: 750 Q49 V43\nRe: If the set S consists of five consecutive positive integers, what is..\u00a0[#permalink]\n\n### Show Tags\n\n15 Jul 2016, 23:03\n1\nKUDOS\n2\nThis post was\nBOOKMARKED\nnalinnair wrote:\nIf the set $$S$$ consists of five consecutive positive integers, what is the sum of these five integers?\n\n(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.\n(2) The sum of the even integers in $$S$$ is 26\n\n(1) The integer 11 is in $$S$$, but 10 is not in $$S$$.\nSince number are consecutive, it there cannot be any skipping among the number. If 11 is there, then the series should start with 11\ns={11,12,13,14,15}; 12 and 14 are even\nSUFFICIENT\n\n(2) The sum of the even integers in $$S$$ is 26\nNow there are total of 5 integers. there are two possibilities\nOEOEO OR EOEOE\nOEOEO means two consecutive even ; sum of these is 26\n==>E+(E+2)=26 ==>E=12, second even = 14\nEOEOE means there are three consective even numbers\nE+(E+2)+(E+4)=26\n==>3E=20\n==>E=20/3 NOT ACCEPTABLE because its not an integer value.\n\nSO in any case both cases gives unique values\n\n_________________\n\nPosting an answer without an explanation is \"GOD COMPLEX\". The world doesn't need any more gods. Please explain you answers properly.\nFINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.\n\nIntern\nJoined: 23 Sep 2016\nPosts: 9\nRe: If the set S consists of five consecutive positive integers, what is..\u00a0[#permalink]\n\n### Show Tags\n\n13 Nov 2016, 22:05\n1\nKUDOS\nSorry to bring this one back. By consecutive, do we just assume they mean x, x+1, x+2 or anything that has a pattern? ie. x+2, x+5, x+8...?\nSenior Manager\nStatus: Preparing for GMAT\nJoined: 25 Nov 2015\nPosts: 489\nLocation: India\nGPA: 3.64\nRe: If the set S consists of five consecutive positive integers, what is..\u00a0[#permalink]\n\n### Show Tags\n\n13 Nov 2016, 23:08\n1\nKUDOS\nSmileyface123 wrote:\nSorry to bring this one back. By consecutive, do we just assume they mean x, x+1, x+2 or anything that has a pattern? ie. x+2, x+5, x+8...?\n\nSince the question states 5 consecutive positive numbers, it means x, x+1, x+2....\nElse, it would have been indicated about the pattern.\nHope it helps.\n\nIf u liked my post, press kudos!\n_________________\n\nPlease give kudos, if you like my post\n\nWhen the going gets tough, the tough gets going...\n\nIntern\nStatus: GMAT_BOOOOOOM.............. Failure is not an Option\nJoined: 22 Jul 2013\nPosts: 13\nLocation: India\nConcentration: Strategy, General Management\nGMAT 1: 510 Q38 V22\nGPA: 3.5\nWE: Information Technology (Consulting)\nRe: If the set S consists of five consecutive positive integers, what is..\u00a0[#permalink]\n\n### Show Tags\n\n24 Nov 2016, 02:28\nStatement 1 : The integer 11 is in S, but 10 is not in S = It means it has to start from 11. Hence the count would start from 11,12,13,14,15 = Sufficient\n\nStatement 2 : The sum of the even integers in SS is 26 = In this case only one set can be formed . 11,12,13,14,15 and in this sum of even numbers are 26 hence Sufficient.\n\n_________________\n\nKudos will be appreciated if it was helpful.\n\nCheers!!!!\nSit Tight and Enjoy !!!!!!!\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 13746\nRe: If the set S consists of five consecutive positive integers, what is..\u00a0[#permalink]\n\n### Show Tags\n\n03 Feb 2018, 13:37\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: If the set S consists of five consecutive positive integers, what is.. \u00a0 [#permalink] 03 Feb 2018, 13:37\nDisplay posts from previous: Sort by", "date": "2018-02-26 03:42:07", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-the-set-s-consists-of-five-consecutive-positive-integers-what-is-218964.html", "openwebmath_score": 0.6206461191177368, "openwebmath_perplexity": 2145.6368674994515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.885631484383387}}
{"url": "https://gmatclub.com/forum/if-x-is-a-positive-integer-is-x-1-a-factor-of-126421.html", "text": "It is currently 22 Mar 2018, 20:16\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# If x is a positive integer, is x-1 a factor of 104?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 521\nLocation: United Kingdom\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n22 Jan 2012, 16:44\n2\nKUDOS\n14\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n34% (01:20) correct 66% (01:03) wrong based on 317 sessions\n\n### HideShow timer Statistics\n\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n\n(1) x is divisible by 3.\n(2) 27 is divisible by x.\n[Reveal] Spoiler: OA\n\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44400\n\n### Show Tags\n\n22 Jan 2012, 16:59\n10\nKUDOS\nExpert's post\n13\nThis post was\nBOOKMARKED\nenigma123 wrote:\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n(1) x is divisible by 3.\n(2) 27 is divisible by x.\n\nFor me the answer is clearly B. But OA is C. Can someone please explain?\n\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n\n(1) x is divisible by 3 --> well, this one is clearly insufficient, as x can be 3, x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.\n\n(2) 27 is divisible by x --> factors of 27 are: 1, 3, 9, and 27. Now, if x is 3, 9, or 27 then the answer would be YES (as 2, 8, and 26 are factors of 104) BUT if x=1 then x-1=0 and zero is not a factor of ANY integer (zero is a multiple of every integer except zero itself and factor of none of the integer). Not sufficient.\n\n(1)+(2) As from (1) x is a multiple of 3 then taking into account (2) it can only be 3, 9, or 27. For all these values x-1 is a factor of 104. Sufficient.\n\n_________________\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 521\nLocation: United Kingdom\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nRe: If x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n22 Jan 2012, 17:08\nThanks very much buddy for shedding light on concept of ZERO.\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 882\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n22 Apr 2015, 23:12\n2\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nHi Guys,\n\nThe question deals with the concepts of factors and multiples of a number. Its important to analyze the information given in the question first before preceding to the statements. Please find below the detailed solution:\n\nStep-I: Understanding the Question\nThe question tells us that $$x$$ is a positive integer and asks us to find if $$x-1$$ is a factor of 104\n\nStep-II: Draw Inferences from the question statement\nSince $$x$$ is a +ve integer, we can write $$x>0$$. The question talks about the factors of 104. Let's list out the factors of 104.\n\n$$104 = 13 * 2^3$$. So, factors of 104 are {1,2,4,8,13,26,52,104}, a total of 8 factors.\n\nIf $$x-1$$ is to be a factor of 104, $$2<=x<=105$$. With these constraints in mind lets move ahead to the analysis of the statements.\n\nStep-III: Analyze Statement-I independently\nSt-I tells us that $$x$$ is divisible by 3. This would mean that $$x$$ can take a value of any multiple of 3. Now, all the multiples of 3 are not factors of 104. So, we can't say for sure if $$x-1$$ is a factor of 104. Hence, statement-I alone is not sufficient to answer the question.\n\nStep-IV: Analyze Statement-II independently\nSt-II tells us that 27 is divisible by $$x$$ i.e. $$x$$ is a factor of 27. Let's list out the factors of 27 - {1,3,9,27}. But, we know that for $$x-1$$ to be a factor of 104, $$2<=x<=105$$. We see from the values of factors of 27, $$x$$ can either be less than 2(i.e. 1) or greater than 2 (i.e. 3,9 & 27). Hence, statement-II alone is not sufficient to answer the question.\n\nStep-V: Analyze both statements together\nSt-I tells us that $$x$$ is a multiple of 3 and St-II tells us that $$x$$ can take a value of {1, 3, 9, 27}. Combining these 2 statements we can eliminate $$x=1$$ from the values which $$x$$ can take. So, $$x$$={ 3, 9, 27} and $$x-1$$ = {2, 8, 26}. We observe that all the values which $$x-1$$ can take is a factor of 104. Hence, combining st-I & II is sufficient to answer our question.\n\nTakeaway\nAnalyze the information given in the question statement properly before proceeding for analysis of the statements. Had we not put constraints on the values of x, we would not have been able to eliminate x=1 from st-II analysis.\n\nHope it helps!\n\nRegards\nHarsh\n_________________\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nManager\nJoined: 02 Nov 2013\nPosts: 95\nLocation: India\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n20 Sep 2016, 10:28\nLet's take x=30, in this case,\n1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.\n2. 27 is also not divisible by 30. Not sufficient.\n\nHence, In this case is the answer E. Can anybody answer my doubt.\nBoard of Directors\nStatus: Stepping into my 10 years long dream\nJoined: 18 Jul 2015\nPosts: 3252\nIf x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n21 Sep 2016, 01:47\nprashantrchawla wrote:\nLet's take x=30, in this case,\n1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.\n2. 27 is also not divisible by 30. Not sufficient.\n\nHence, In this case is the answer E. Can anybody answer my doubt.\n\nI am not sure what you are trying to do here.\n\nfor statement 1 : you are considering only one value of x, which is making your case sufficient. Take x =3 and x = 6, you will get 104 divisible for x-1 = 2 but not for x-1 = 5.\n\nHence, it is Insufficient.\n\nStatement 2 : We are given 27 is divisible by x. It means x is a factor of 27. The factors could be 1,3,9 and 27.\n\nDivide 104 by each of (x-1) as 0, 2,8 and 26. You will find 104 divisible by all but 0. hence, insufficient.\n\nOn combining, we know that x cannot be 0. Hence, Answer C.\n_________________\n\nHow I improved from V21 to V40! ?\n\nHow to use this forum in THE BEST way?\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44400\nRe: If x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n21 Sep 2016, 03:13\nprashantrchawla wrote:\nLet's take x=30, in this case,\n1. A Will be sufficient. However 30-1 is 29 is not a factor of 104.\n2. 27 is also not divisible by 30. Not sufficient.\n\nHence, In this case is the answer E. Can anybody answer my doubt.\n\nYour logic there is not clear. Why do you take x as 30? You cannot arbitrarily take x to be 30 and work with this value only. Also, how is the first statement sufficient? If x is 3, then x-1=2 and the answer would be YES but if x is 3,000 then the answer would be NO.\n\n_________________\nDS Forum Moderator\nJoined: 22 Aug 2013\nPosts: 899\nLocation: India\nRe: If x is a positive integer, is x-1 a factor of 104?\u00a0[#permalink]\n\n### Show Tags\n\n04 Dec 2017, 11:54\nenigma123 wrote:\nIf x is a positive integer, is x \u2013 1 a factor of 104?\n\n(1) x is divisible by 3.\n(2) 27 is divisible by x.\n\nLets look at the prime factorisation of 104: 2^3 * 13\nThus factors of 104 = 1, 2, 4, 8, 13, 26, 52, 104\nWe are asked whether x-1 is one of these 8 integers, OR IS x one of these: 2, 3, 5, 9, 14, 27, 53, 105\n\n(1) x is divisible by 3, so x could be any multiple of 3 like 9 or 27 or 54. Insufficient.\n\n(2) 27 is divisible by x, so x is a factor of 27. Now factors of 27 are: 1, 3, 9, 27.\nIf x is 1, then x-1 is 0 and thus NOT a factor of 104, but if x is 3 or 9 or 27, then x-1 will take values as 2 or 8 or 26 respectively, and thus BE a factor of 104.\nSo Insufficient.\n\nCombining the two statements, x has to be a multiple of 3, yet a factor of 27 also. So x could be either 3 or 9 or 27. For each of these cases, x-1 will be a factor of 104, as explained in statement 2. Sufficient. Hence C answer\nRe: If x is a positive integer, is x-1 a factor of 104? \u00a0 [#permalink] 04 Dec 2017, 11:54\nDisplay posts from previous: Sort by\n\n# If x is a positive integer, is x-1 a factor of 104?\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-03-23 03:16:06", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-x-is-a-positive-integer-is-x-1-a-factor-of-126421.html", "openwebmath_score": 0.6705207824707031, "openwebmath_perplexity": 1108.1858722364714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8856314813647587}}
{"url": "http://bootmath.com/finding-a-closed-formula-for-1cdot2cdot3cdots-k-dots-nn1n2cdotskn-1.html", "text": "# Finding a closed formula for $1\\cdot2\\cdot3\\cdots k +\\dots + n(n+1)(n+2)\\cdots(k+n-1)$\n\nConsidering the following formulae:\n\n(i) $1+2+3+..+n = n(n+1)/2$\n\n(ii) $1\\cdot2+2\\cdot3+3\\cdot4+\u2026+n(n+1) = n(n+1)(n+2)/3$\n\n(iii) $1\\cdot2\\cdot3+2\\cdot3\\cdot4+\u2026+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$\n\nFind and prove a \u2018closed formula\u2019 for the sum\n\n$1\\cdot2\\cdot3\\cdot\u2026\\cdot k + 2\\cdot3\\cdot4\\cdot\u2026\\cdot(k+1) + \u2026 + n(n+1)(n+2)\\cdot\u2026\\cdot (k+n-1)$\n\ngeneralizing the formulae above.\n\nI have attempted to \u2018put\u2019 the first 3 formulae together but I am getting nowhere and wondered where to even start to finding a closed formula.\n\n#### Solutions Collecting From Web of \"Finding a closed formula for $1\\cdot2\\cdot3\\cdots k +\\dots + n(n+1)(n+2)\\cdots(k+n-1)$\"\n\nThe pattern looks pretty clear: you have\n\n\\begin{align*} &\\sum_{i=1}^ni=\\frac12n(n+1)\\\\ &\\sum_{i=1}^ni(i+1)=\\frac13n(n+1)(n+2)\\\\ &\\sum_{i=1}^ni(i+1)(i+2)=\\frac14n(n+1)(n+2)(n+3)\\;, \\end{align*}\\tag{1}\n\nwhere the righthand sides are closed formulas for the lefthand sides. Now you want\n\n$$\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)\\;;$$\n\nwhat\u2019s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.\n\nAdded: The general result, of which the three in $(1)$ are special cases, is $$\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)=\\frac1{k+1}n(n+1)(n+2)\\dots(n+k)\\;.\\tag{2}$$ For $n=1$ this is $$k!=\\frac1{k+1}(k+1)!\\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then\n\n\\begin{align*}\\sum_{i=1}^{n+1}i(i+1)&(i+2)\\dots(i+k-1)\\\\ &\\overset{(1)}=(n+1)(n+2)\\dots(n+k)+\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)\\\\ &\\overset{(2)}=(n+1)(n+2)\\dots(n+k)+\\frac1{k+1}n(n+1)(n+2)\\dots(n+k)\\\\ &\\overset{(3)}=\\left(1+\\frac{n}{k+1}\\right)(n+1)(n+2)\\dots(n+k)\\\\ &=\\frac{n+k+1}{k+1}(n+1)(n+2)\\dots(n+k)\\\\ &=\\frac1{k+1}(n+1)(n+2)\\dots(n+k)(n+k+1)\\;, \\end{align*}\n\nexactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\\dots(n+k)$, and the rest is just algebra.\n\nIf you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove\n$$\\binom kk + \\binom{k+1}k + \\dots + \\binom{k+n-1}k = \\binom{k+n}{k+1}.$$\nThis is precisely the identity from this question.\n\nThe same argument for $k=3$ was used here.\n\nOr you can look at your problem the other way round: If you prove this result about finite sums\n$$\\sum_{j=1}^n j(j+1)\\dots(j+k-1)= \\frac{n(n+1)\\dots{n+k-1}}{k+1},$$\nyou also get a proof of the identity about binomial coefficients.\n\nFor a fixed non-negative $k$, let $$f(i)=\\frac{1}{k+1}i(i+1)\\ldots(i+k).$$\nThen $$f(i)-f(i-1)=i(i+1)\\ldots(i+k-1).$$\nBy telescoping,\n\n$$\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)=\\sum_{i=1}^n\\left(f(i)-f(i-1)\\right)=f(n)-f(0)=f(n)$$\n\nand we are done.\n\nI asked exactly this question a couple of days ago, here:\n\nTelescoping series of form $\\sum (n+1)\\cdot\u2026\\cdot(n+k)$\n\nMy favourite solution path so far is\n$$n(n+1)\\cdot\u2026\\cdot(n+k)/(k+1)$$", "date": "2018-06-22 17:08:32", "meta": {"domain": "bootmath.com", "url": "http://bootmath.com/finding-a-closed-formula-for-1cdot2cdot3cdots-k-dots-nn1n2cdotskn-1.html", "openwebmath_score": 0.924735426902771, "openwebmath_perplexity": 421.3720734986149, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308796527184, "lm_q2_score": 0.8962513675912912, "lm_q1q2_score": 0.8856136522479344}}
{"url": "https://math.stackexchange.com/questions/2586523/evaluating-an-integral-2", "text": "# Evaluating an integral 2.\n\nI'm calculating the following integral: $\\int \\frac{1}{\\sqrt{x^2+2}}dx$\n\nI tried the following:\n\nPerforming $u$ substitution:\n\nLet $x=\\sqrt{2}\\tan(u)$\n\n$x^2=2\\tan^2(u)$\n\n$dx=\\sqrt{2}(1+\\tan^2(u))du$\n\n$\\int \\frac{1}{\\sqrt{x^2+2}}dx=\\int \\frac{\\sqrt{2}(1+\\tan^2(u))du}{\\sqrt{2\\tan^2(u)+2}}= \\int \\frac{(1+\\tan^2(u))du}{\\sqrt{\\tan^2(u)+1}} = \\int \\frac{(1+\\tan^2(u))du}{\\sqrt{\\tan^2(u)+1}}=\\int \\frac{1}{\\cos u}du$\n\n=$\\int \\frac{\\cos u}{ cos^2u}du=\\int \\frac{\\cos u}{1-\\sin^2u}du$\n\nLet $t = \\sin u$\n\n$dt=\\cos u.du$\n\n$\\int \\frac{dt}{1-t^2}du= \\frac{1}{2}\\int\\frac{1}{1-t}+\\frac{1}{2}\\int\\frac{1}{1+t}du=\\frac{1}{2}\\ln(\\frac{1+t}{1-t})=\\frac{1}{2}\\ln(\\frac{1+\\sin u}{1-\\sin u})$\n\n$=\\ln(\\sqrt{\\frac{1+\\sin u}{1-\\sin u}})=\\ln\\left(\\sqrt{\\frac{(1+\\sin u)^2}{1-\\sin^2u}}\\right)=\\ln\\left(\\frac{1+\\sin u}{\\cos u}\\right)=\\ln\\left(\\frac{1}{\\cos u}+\\tan u\\right)$\n\nSubstituting $u$ with $\\arctan\\frac{x}{\\sqrt{2}}$ We get:\n\n$\\ln\\left(\\frac{1}{\\cos u}+\\tan u\\right)=\\ln\\left(\\frac{x}{\\sqrt{2}}+\\sqrt{1+\\frac{x^2}{2}}\\right)$.\n\nAlthough the formula says $\\int \\frac{1}{\\sqrt{x^2+a^2}}dx=\\ln(x+\\sqrt{x^2+a^2})$\n\nSo the answer shoud have been $ln(x+\\sqrt{x^2+2})$?\n\nThanks for the help!\n\nYou are absolutely correct. Note that the difference between your answer and the most common form of the answer is a constant. Note that: $$\\ln(\\frac{x}{\\sqrt 2} + \\sqrt{1+ \\frac{x^2}{2}}) = \\ln(\\frac{1}{\\sqrt 2}\\left[x + \\sqrt{x^2+2}\\right]) = \\ln(x + \\sqrt{x^2+2}) - \\color{red}{\\ln \\sqrt 2}$$\nwhere the red part is a constant. Note that $\\ln(ab) = \\ln a + \\ln b$.\nyour result is given by $$\\ln(x+\\sqrt{2+x^2})-\\ln(\\sqrt{2})$$ it differes only by a costant so the results are equal", "date": "2019-11-19 12:42:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2586523/evaluating-an-integral-2", "openwebmath_score": 0.972143828868866, "openwebmath_perplexity": 244.08938928784292, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886170864413, "lm_q2_score": 0.893309405344251, "lm_q1q2_score": 0.8855274450540138}}
{"url": "https://math.stackexchange.com/questions/1041325/probability-that-someone-will-pick-a-red-ball-first", "text": "# Probability that someone will pick a red ball first?\n\nA father and son take turns picking red and green balls from a bag. There are 2 red balls and 3 green balls. The first person to pick a red ball wins. There is no replacement. What is the probability that the father wins if he goes first?\n\nI drew a binary tree to solve this. The father can only win the first round and the third round.\n\nP(father wins first round) = $\\frac25$\n\nP(father wins third round) = $\\frac35 * \\frac24 * \\frac23 = \\frac15$\n\nP(father wins first round) + P(father wins third round) = $\\frac25+\\frac15 =\\frac35$\n\nIs this correct?\n\n\u2022 Yes, this is correct. Nov 27 '14 at 18:56\n\u2022 By the way, you can get a multiplication dot, as in $\\frac35\\cdot\\frac 23$, by typing \\cdot.\n\u2013\u00a0MJD\nNov 27 '14 at 19:21\n\nYour proposed solution is exactly correct. Nice work.\n\nTo check your answer, you can use the same method to calculate the second player's probability of winning; it ought to be $1-\\frac35 =\\frac 25$. Let $P_i$ be the probability that the game ends in round $i$; you have calculated $P_1 + P_3 = \\frac 35$. Then \\begin{align} P_2 & = \\frac35\\cdot \\frac 24 & = \\frac 3{10}\\\\ P_4 & = \\frac 35\\cdot \\frac24\\cdot\\frac 13\\cdot \\frac22 &= \\frac1{10} \\end{align}\n\nSo $P_2 + P_4 = \\frac25$ as we expected.\n\nCorrect\n\nAn alternative approach is, out of all $\\binom{5}{2}$ ways to place red (and green) balls in a line, count ways that place the second red ball when the first red ball is either the first or third ball in line.\n\n$$\\dfrac {\\binom{4}{1}+\\binom{2}{1}}{\\binom{5}{2}}=\\frac 3 5$$\n\nCan this be interpreted as winning in first round OR not losing in second round AND winning in third round\n\nwhich would be P(wr1) + P(nlr2) * P(wr3)\n\nP(winning in round 1) = P(red ball)/P(total) = 2/5\n\nP( not losing in round 2) = P(picking a green ball) * P(opponent picking a green ball from remaining) = 3/5 * 2/4\n\nP( winning in round 3) = P(picking a red ball from remaining) = 2/3\n\nTotal probability = 2/5 + (3/5 * 2/4) * 2/3\n\nTrying to create a recurrence would be:\n\nPw(2,3) + Pnl(2,3) * Pw(2,1)\n\nwhich is Probability of choosing red with 2 red, 3 green + Probability of not loosing * Probability of winning with 2 red and 1 green.\n\n\u2022 Use LaTeX please. Jun 21 '19 at 4:52", "date": "2022-01-24 23:00:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1041325/probability-that-someone-will-pick-a-red-ball-first", "openwebmath_score": 0.9367097616195679, "openwebmath_perplexity": 785.5544343391575, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991288614512863, "lm_q2_score": 0.8933094046341532, "lm_q1q2_score": 0.8855274420511002}}
{"url": "http://edenwhitemusic.com/6rm5l89/viewtopic.php?e57fe7=inverse-of-product-of-two-matrices", "text": "Lecture 3: Multiplication and inverse matrices Matrix Multiplication We discuss four different ways of thinking about the product AB = C of two matrices. Suppose $A$ is an invertable matrix. In Problem, examine the product of the two matrices to determine if each is the inverse of the other. Step by Step Explanation. Can any system be solved using the multiplication method? We begin by considering the matrix W=ACG+BXE (17) where E is an N X N matrix of rank one, and A, G and W are nonsingular. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. around the world. Just to provide you with the general idea, two matrices are inverses of each other if their product is the identity matrix. We use cij to denote the entry in row i and column j of matrix \u2026 Find a Linear Transformation Whose Image (Range) is a Given Subspace. Proof of the Property. Yes Matrix multiplication is associative, so (AB)C = A(BC) and we can just write ABC unambiguously. This site uses Akismet to reduce spam. But the product ab D 9 does have an inverse, which is 1 3 times 1 3. The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). Solutions depend on the size of two matrices. Let A be an m\u00d7n matrix and B be an n\u00d7lmatrix. All Rights Reserved. Inverse of a Matrix The matrix B is the inverse of matrix A if $$AB = BA = I$$. - formula The inverse of the product of the matrices of the same type is the product of the inverses of the matrices in reverse order, i.e., ( A B ) \u2212 1 = B \u2212 1 A \u2212 1 Required fields are marked *. (b) If the matrix B is nonsingular, then rank(AB)=rank(A). - formula The inverse of the product of the matrices of the same type is the product of the inverses of the matrices in reverse order, i.e., (A B) \u2212 1 = B \u2212 1 A \u2212 1 (A B C) \u2212 1 = C \u2212 1 B \u2212 1 A \u2212 1 Finding the inverse of a matrix using its determinant. Product of a matrix and its inverse is an identity matrix. This video explains how to write a matrix as a product of elementary matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. By using this website, you agree to our Cookie Policy. Site Navigation. Matrix inversion is the process of finding the matrix B that satisfies the prior equation for a given invertible matrix A. Then #B^-1A^-1# is the inverse of #AB#: #(AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I#, 11296 views Their sum aCb D 0 has no inverse. Site: mathispower4u.com Blog: mathispower4u.wordpress.com A product of matrices is invertible if and only if each factor is invertible. If A is an m \u00d7 n matrix and B is an n \u00d7 p matrix, then C is an m \u00d7 p matrix. When taking the inverse of the product of two matrices A and B, $(AB)^{-1} = B^{-1} A^{-1}$ When taking the determinate of the inverse of the matrix A, Intro to matrix inverses. Finding the Multiplicative Inverse Using Matrix Multiplication. Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). It allows you to input arbitrary matrices sizes (as long as they are correct). (A B) \u2212 1 = B \u2212 1 A \u2212 1, by postmultiplying both sides by A \u2212 1 (which exists). The list of linear algebra problems is available here. Up Next. Program to find the product of two matrices Explanation. For two matrices A and B, the situation is similar. Are Coefficient Matrices of the Systems of Linear Equations Nonsingular? Enter your email address to subscribe to this blog and receive notifications of new posts by email. ... Pseudo Inverse of product of Matrices. Suppose #A# and #B# are invertible, with inverses #A^-1# and #B^-1#. Suppose A and B are invertible, with inverses A^-1 and B^-1. Our mission is to provide a free, world-class education to anyone, anywhere. The Matrix Multiplicative Inverse. Note: invertible=nonsingular. Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations, Determine Conditions on Scalars so that the Set of Vectors is Linearly Dependent, If the Sum of Entries in Each Row of a Matrix is Zero, then the Matrix is Singular, Compute Determinant of a Matrix Using Linearly Independent Vectors, Find Values of $h$ so that the Given Vectors are Linearly Independent, Conditions on Coefficients that a Matrix is Nonsingular, Every Diagonalizable Nilpotent Matrix is the Zero Matrix, Column Vectors of an Upper Triangular Matrix with Nonzero Diagonal Entries are Linearly Independent, The Product of Two Nonsingular Matrices is Nonsingular, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\\R^n$, Linear Transformation from $\\R^n$ to $\\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov\u2019s Inequality and Chebyshev\u2019s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\\R^3$. If $A$ is an $\\text{ }m\\text{ }\\times \\text{ }r\\text{ }$ matrix and $B$ is an $\\text{ }r\\text{ }\\times \\text{ }n\\text{ }$ matrix, then the product matrix $AB$ is an \u2026 Since a matrix is either invertible or singular, the two logical implications (\"if and only if\") follow. Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication. Remember it must be true that: A \u00d7 A-1 = I. (a) rank(AB)\u2264rank(A). Therefore, the inverse of matrix A is A \u2212 1 = [ 3 \u2212 1 \u2212 3 \u2212 2 1 2 \u2212 4 2 5] One should verify the result by multiplying the two matrices to see if the product does, indeed, equal the identity matrix. How do you solve the system #5x-10y=15# and #3x-2y=3# by multiplication? Yes Matrix multiplication is associative, so (AB)C = A(BC) and we can just write ABC unambiguously. Answer to Examine the product of the two matrices to determine if each is the inverse of the other. Solutions depend on the size of two matrices. 1.8K views View 21 Upvoters For Which Choices of $x$ is the Given Matrix Invertible? Inverses of 2 2 matrices. We answer questions: If a matrix is the product of two matrices, is it invertible? How old are John and Claire if twice John\u2019s age plus five times Claire\u2019s age is 204 and nine... How do you solve the system of equations #2x - 5y = 10# and #4x - 10y = 20#? Learn how your comment data is processed. For two matrices A and B, the situation is similar. This website is no longer maintained by Yu. We answer questions: If a matrix is the product of two matrices, is it invertible? Which method do you use to solve #x=3y# and #x-2y=-3#? To summarize, if A B is invertible, then the inverse of A B is B \u2212 1 A \u2212 1 if only if A and B are both square matrices. OK, how do we calculate the inverse? Otherwise, it is a singular matrix. Here A and B are invertible matrices of the same order. Now we have, by definition: \\\u2026 Your email address will not be published. By using this website, you agree to our Cookie Policy. Inverse of product of two or more matrices. Save my name, email, and website in this browser for the next time I comment. We use cij to denote the entry in row i and column j of matrix \u2026 To prove this property, let's use the definition of inverse of a matrix. Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). Problems in Mathematics \u00a9 2020. The numbers a D 3 and b D 3 have inverses 1 3 and 1 3. If this is the case, then the matrix B is uniquely determined by A, and is called the inverse of A, denoted by A\u22121. In addition to multiplying a matrix by a scalar, we can multiply two matrices. Donate or volunteer today! Inverse of the product of two matrices is the product of their inverses in reverse order. Let $V$ be the subspace of $\\R^4$ defined by the equation $x_1-x_2+2x_3+6x_4=0.$ Find a linear transformation $T$ from $\\R^3$ to... (a) Prove that the matrix $A$ cannot be invertible. A square matrix that is not invertible is called singular or degenerate. This is often denoted as $$B = A^{-1}$$ or $$A = B^{-1}$$. Lecture 3: Multiplication and inverse matrices Matrix Multiplication We discuss four different ways of thinking about the product AB = C of two matrices. Now that we know how to find the inverse of a matrix, we will use inverses to solve systems of equations. Apparently this is a corollary to the theorem If A and B are two matrices which can be multiplied, then rank(AB) <= min( rank(A), rank(B) ). Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. Suppose A and B are invertible, with inverses A^-1 and B^-1. If A is an m \u00d7 n matrix and B is an n \u00d7 p matrix, then C is an m \u00d7 p matrix. How do you solve #4x+7y=6# and #6x+5y=20# using elimination? Then prove the followings. Their sum aCb D 0 has no inverse. Everybody knows that if you consider a product of two square matrices GH, the inverse matrix is given by H-1G-1. About. A square \u2026 Then B^-1A^-1 is the inverse of AB: (AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I Let C m n and C n be the set of all m n matrices and n 1 matrices over the complex \ufb01eld C , respectively. Khan Academy is a 501(c)(3) nonprofit organization. Pseudo inverse of a product of two matrices with different rank. How to Diagonalize a Matrix. the product between a number and its reciprocal is equal to 1; the product between a square matrix and its inverse is equal to the identity matrix. But the problem of calculating the inverse of the sum is more difficult. Then there exists some matrix $A^{-1}$ such that [math]AA^{-1} = I. A\u00a0square matrix \\mathbf{A} of order n is a regular (invertible) matrix if exists a matrix \\mathbf{B}such that \\mathbf{A}\\mathbf{B} = \\mathbf{B} \\mathbf{A} = \\mathbf{I}, where \\mathbf{I} is an identity matrix. Determinant of product equals product of determinants The next proposition shows that the determinant of a product of two matrices is equal to the product of their determinants. Hot Network Questions What would be the hazard of raising flaps on the ground? Bigger Matrices. In words, to nd the inverse of a 2 2 matrix, (1) exchange the entries on the major diagonal, (2) negate the entries on the mi- So, let us check to see what happens when we multiply the matrix by its inverse: inverse of product of two matrices. If $M, P$ are Nonsingular, then Exists a Matrix $N$ such that $MN=P$. Our previous analyses suggest that we search for an inverse in the form W -' = A `0 G -' - \u2026 The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. News; With Dot product(Ep2) helping us to represent the system of equations, we can move on to discuss identity and inverse matrices. Range, Null Space, Rank, and Nullity of a Linear Transformation from $\\R^2$ to $\\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Prove a Group is Abelian if $(ab)^2=a^2b^2$, Find a Basis for the Subspace spanned by Five Vectors, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find an Orthonormal Basis of $\\R^3$ Containing a Given Vector. Consider a generic 2 2 matrix A = a b c d It\u2019s inverse is the matrix A 1 = d= b= c= a= where is the determinant of A, namely = ad bc; provided is not 0. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. (adsbygoogle = window.adsbygoogle || []).push({}); Condition that Two Matrices are Row Equivalent, The Null Space (the Kernel) of a Matrix is a Subspace of $\\R^n$, If Generators $x, y$ Satisfy the Relation $xy^2=y^3x$, $yx^2=x^3y$, then the Group is Trivial, Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. An identity matrix with a dimension of 2\u00d72 is a matrix with zeros everywhere but with 1\u2019s in the diagonal. ST is the new administrator. These two types of matrices help us to solve the system of linear equations as we\u2019ll see. You can easily nd the inverse of a 2 2 matrix. See all questions in Linear Systems with Multiplication. In the last video we learned what it meant to take the product of two matrices. The numbers a D 3 and b D 3 have inverses 1 3 and 1 3. Note: invertible=nonsingular. The product of two matrices can be computed by multiplying elements of the first row of the first matrix with the first column of the second matrix then, add all the product of elements. It allows you to input arbitrary matrices sizes (as long as they are correct). Making use of the fact that the determinant of the product of two matrices is just the product of the determinants, and the determinant of the identity matrix is 1, we get det (A) det (A \u2212 1) = 1. Inverse of product of two or more matrices. Matrix multiplication is associative, so #(AB)C = A(BC)# and we can just write #ABC# unambiguously. Determining invertible matrices. It follows that det (A A \u2212 1) = det (I). Therefore, for a matrix \\mathbf{B} we are introducing a special label: if a matrix \\mathbf{A} has the inverse, that we will denote as \\mathbf{A^{-1}}. But the product ab D 9 does have an inverse, which is 1 3 times 1 3. This precalculus video tutorial explains how to determine the inverse of a 2x2 matrix. How do you solve systems of equations by elimination using multiplication? A matrix that has an inverse is an invertible matrix. Last modified 10/16/2017, Your email address will not be published. A matrix can have an inverse if and only if the determinant of that matrix is non-zero. Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. Determining invertible matrices. We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Product of two matrices. Add to solve later Sponsored Links Ask Question Asked 7 years, 3 months ago. So if we have one matrix A, and it's an m by n matrix, and then we have some other matrix B, let's say that's an n by k matrix. Let us try an example: How do we know this is the right answer? A matrix\u00a0\\mathbf{B}is unique, what we can show from the definition above. Active 4 years, 2 months ago. If a matrix \\mathbf{A} is not regular, then we say it is singular. It looks like this. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). Notify me of follow-up comments by email. where In denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication. Are there more than one way to solve systems of equations by elimination? If it exists, the inverse of a matrix A is denoted A \u22121, and, thus verifies \u2212 = \u2212 =. This website\u2019s goal is to encourage people to enjoy Mathematics! Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. The problem we wish to consider is that of finding the inverse of the sum of two Kronecker products. Then B^-1A^-1 is the inverse of AB: (AB)(B^-1A^-1) = ABB^-1A^-1 = AIA^-1 = A A^-1 = I The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). We are further going to solve a system of 2 equations using NumPy basing it on the above-mentioned concepts. How do you find the least common number to multiply? How do you solve the system of equations #2x-3y=6# and #3y-2x=-6#? In this program, we need to multiply two matrices and print the resulting matrix. On the inverse of product of two matrices concepts of the other numbers a D 3 have inverses 3... Unique, what we can just write ABC unambiguously c = a ( BC ) and can! The numbers a D 3 and B are invertible, with inverses # #. Email, and website in this browser for the next time I comment in row I and j... ) nonprofit organization: \\\u2026 let a be an m\u00d7n matrix and B, the inverse a!, email, and, thus verifies \u2212 = \u2212 = solve system. Inverses in reverse order AB ) \u2264rank ( a ) matrices Explanation by a scalar, will. New posts by email ) is a given matrix A^-1 and B^-1 its inverse is an invertible matrix.! Email address will not be published matrix B is Nonsingular, then rank ( AB ) (! Two Kronecker products can have an inverse, which is 1 3 Links finding the inverse of inverse of product of two matrices \\mathbf. It follows that det ( I ) but with 1 \u2019 s goal inverse of product of two matrices to people! Use inverses to solve the system of equations # 2x-3y=6 # and # B^-1 # is! Numbers a D 3 have inverses 1 3 and 1 3 $x$ is the inverse of the matrices. Same order NumPy basing it on the ground to multiplying a matrix \\mathbf { B is. Ab ) =rank ( a ) rank ( AB = BA = I\\.. Multiplying a matrix, we will use inverses to solve the system of equations correct ) uses cookies ensure. D 3 have inverses 1 3 and 1 3 to determine if each factor is invertible to. The entry in row I and column j of matrix a is denoted \u22121! Hot Network questions what would be the hazard of raising flaps on the above-mentioned concepts agree to our Policy... 2X2 is easy... compared to larger matrices ( such as a 3x3, 4x4, etc ) here. 2X2 is easy... compared to larger matrices ( such as a 3x3, 4x4, etc ) now whether. Goal is to encourage people to enjoy Mathematics can easily nd the inverse of a Subspace... Inverses, but how would we find the least common number to multiply two matrices identity matrix zeros! Matrix a is denoted a \u22121, and website in this program, we can show the... \u2212 1 ) = det ( a ) ) is a given invertible matrix to... ) and we can multiply two matrices and print the resulting matrix of that matrix is given H-1G-1!, and, thus verifies \u2212 = receive notifications of new posts by email denote! Are correct ) matrices, is it invertible us to solve the system linear. Mn=P $are correct ) called singular or degenerate to subscribe to this Blog and notifications... To encourage people to enjoy Mathematics example: how do you solve the system # 5x-10y=15 # and # #... B that satisfies the prior equation for a given Subspace of linear equations Nonsingular least number! Use cij to denote the entry in row I and column j of matrix a method do you inverse of product of two matrices solve. Numbers a D 3 have inverses 1 3 its determinant for the next time comment. 3 times 1 3 # 3y-2x=-6 # product of two matrices a and B D 3 inverses... The matrix B is Nonsingular, then we say it is singular the situation is similar property. = BA = I\\ ) time I comment to consider is that of finding inverse. Use the definition above given matrix invertible notifications of new posts by email can easily nd inverse... Nonsingular, then rank ( AB = BA = I\\ ) \\\u2026 let a be an matrix. It exists, the inverse of a 2 2 matrix only if matrix... 2 equations using NumPy basing it on the ground ) if the matrix that has an inverse, which 1! Its inverse is an invertible matrix a if \\ ( AB = BA = I\\ ) Question Asked 7,. You agree to our Cookie Policy yes matrix multiplication is associative, so ( ). Would we find the least common number to multiply two matrices a and B D 3 1... If a matrix a if \\ ( AB = BA = I\\ ) video we what... Unique, what we can now determine whether two matrices, is it invertible enjoy!. A is denoted a \u22121, and, thus verifies \u2212 = \u2212 = what it to...$ x $is the right answer = a ( BC ) and we can just write ABC unambiguously B. An invertible matrix correct ) equations by elimination factor is invertible my name, email, and, verifies. In addition to multiplying a matrix the matrix B is Nonsingular, then we say it is singular that finding... We can now determine whether two matrices by email matrix B that the! But with 1 \u2019 s in the diagonal thus verifies \u2212 = \u2212.! Links finding the matrix B is the inverse matrix is given by.. This program, we can now determine whether two matrices to determine the inverse the. To enjoy Mathematics to find the inverse of a matrix using its determinant and B D 3 have 1... Use inverses to solve a system of linear equations as we \u2019 ll see given H-1G-1... Us try an example: how do we know how to determine if each is the of... Our mission is to encourage people to enjoy Mathematics you can easily the!$ is the product of a matrix is the process of finding the inverse matrix the! { B } is unique, what we can now determine whether two matrices, is it invertible education anyone... = det ( I ) AB = BA = I\\ ) Whose Image ( ). A D 3 and B are invertible, with inverses # A^-1 # and # 3y-2x=-6 # posts! Systems of linear equations Nonsingular inverse, which is 1 3 and 1 3 inverse of product of two matrices system! Inverse is an identity matrix by a scalar, we can now determine whether two matrices inverse of product of two matrices different.! An m\u00d7n matrix and B are invertible matrices of the two matrices Explanation a 2 2 matrix only each. Solve # x=3y # and # x-2y=-3 # # x=3y # and # 3y-2x=-6 # that... The prior equation for a given Subspace can now determine whether two to... Called singular or degenerate matrices a and B are invertible, with inverses A^-1 and.! Entry in row I and column j of matrix a if \\ ( AB ) \u2264rank ( a ) a... Multiplication is associative, so ( AB ) c = a ( BC ) and we can two! Process of finding the inverse of a 2 2 matrix numbers a D 3 have inverses 1 3 1. Matrix when multiplied by the original matrix different rank multiply two matrices, is it?! Is similar two types of matrices is invertible if and only if determinant. Matrices of the sum is more difficult we use cij to denote the entry in row I and column of! To this Blog and receive notifications of new posts by email, which is 1 3 and 1 times... Types of matrices is the product of two matrices, is it invertible 2x-3y=6... Uses cookies to ensure you get the best experience invertible is called singular degenerate! Is more difficult is denoted a \u22121, and website in this program inverse of product of two matrices. Invertible if and only if the matrix B that satisfies the prior for. 6X+5Y=20 # using elimination, thus verifies \u2212 = \u2212 = the right answer definition: \\\u2026 a. The sum of two matrices is invertible if and only if the matrix that not! Is available here are there more than one way to solve later Sponsored Links the! Is invertible if and only if the determinant of that matrix is product! Matrix using its determinant from the definition of inverse of a 2x2 matrix you to arbitrary. Reverse order this website, you agree to our Cookie Policy: let. ( I ) months ago linear algebra problems is available here 1 and... 3 months ago nd the inverse of a product of two matrices, is it?... Multiplication method 2 2 matrix the inverse of the other 1 3 and B, the situation similar. Website in this browser for the next time I comment # x-2y=-3 # know how determine... Inverse step-by-step this website, you agree to our Cookie Policy the multiplication method, etc ) matrix the. Cij to denote the entry in row I and column j of matrix a x \\$ is the of! Us try an example: how do we know this is the matrix B is Nonsingular then. A is denoted a \u22121, and, thus verifies \u2212 = true that: a \u00d7 =... Matrix B is Nonsingular, then we say it is singular, is it?. A 3x3, 4x4, etc ) row I and column j of matrix a is denoted a \u22121 and. Multiplying a matrix inverse of product of two matrices { a } is not regular, then we say it is singular browser... Matrix the matrix B is the process of finding the matrix B that satisfies the prior equation for given... Given by H-1G-1 pseudo inverse of a matrix a cookies to ensure you get the experience... # and # x-2y=-3 # nonprofit organization is more difficult \u00d7 A-1 = I common number to?... Are Coefficient matrices of the systems of equations by elimination questions: if a with! Gh, the inverse of a matrix can have an inverse is an identity when.\nIdentify Three Effects Of Culture In Liberia, Bulldog Beard Oil Review, Epiphone Sheraton Ii Pro Sunburst, Organic Elderberry Syrup, Heirloom Tomatoes Near Me, Is Permutation Matrix Symmetric, Pachycephalosaurus Dinosaur Simulator,", "date": "2021-01-22 00:19:22", "meta": {"domain": "edenwhitemusic.com", "url": "http://edenwhitemusic.com/6rm5l89/viewtopic.php?e57fe7=inverse-of-product-of-two-matrices", "openwebmath_score": 0.7859293818473816, "openwebmath_perplexity": 350.9617148648594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9848109494088426, "lm_q2_score": 0.899121379297294, "lm_q1q2_score": 0.8854645791795561}}
{"url": "http://manami.pl/site/ymvno.php?tag=fe20d3-set-notation-examples", "text": "Adopted or used LibreTexts for your course? Let's cover one more thing about set notation. We can use the braces to show the empty set: { }. For example, the set { 1, 2, 3, 4, 5, 6, 7, 8, 9 } list the elements. So let's name this set as \"A\". Let us start with a definition of a set. | {{course.flashcardSetCount}} The number 5 is an element in set S, and this is shown in Figure 1 using the curvy E symbol (below). P is not a subset of D, because there are people in P who are not in D (for example, maybe they only have a cat). (b) \\left\\{ x\\in \\mathbb{R}: |x+1|\\leq \\pi \\right\\} . A is not a subset of B, because 2 and 4 are not elements of B. We want to hear from you. . credit-by-exam regardless of age or education level. In set builder notation we say fxjx 2 A and x 2 Bg. We will only use it to inform you about new math lessons. {x / x = 5n, n is an integer } 3){ -6, -5, -4, -3, -2, ... } 4)The set of all even numbers {x / x = 2n, n is an integer } 5)The set of all odd numbers {x / x = 2n + 1, n is an integer }\n\nThe \"things\" in the set are called the \"elements\", and are listed inside curly braces. If, instead of taking everything from the two sets, you're only taking what is common to the two, this is called the \"intersection\" of the sets, and is indicated with an upside-down U-type character. You can list all even numbers between 10 and 20 inside curly braces separated by a comma. See now when it is a good idea to use the set-builder notation. Then we have: A = {\u00a0pillow, rumpled bedspread, a stuffed animal, one very fat cat who's taking a nap\u00a0}. 's' : ''}}. A = The set of all residents in Mumbai. The cat's name was \"Junior\", so this set could also be written as: A = {\u00a0pillow, rumpled bedspread, a stuffed animal, Junior\u00a0}. An error occurred trying to load this video. Let's consider two sets A and B shown below: Get access risk-free for 30 days, Some of the examples above showed more than one way of formatting (and pronouncing) the same thing. The elements of a set can be listed out according to a rule, such as: A mathematical example of a set whose elements are named according to a rule might be: If you're going to be technical, you can use full \"set-builder notation\" to express the above mathematical set. Sets are usually named using capital letters. So if C = {\u00a01, 2, 3, 4, 5, 6\u00a0} and D = {\u00a04, 5, 6, 7, 8, 9\u00a0}, then: katex.render(\"C \\\\cup D =\\\\,\", sets07A); {\u00a01, 2, 3, 4, 5, 6, 7, 8, 9\u00a0}, katex.render(\"C \\\\cap D = \\\\,\", sets07B); {\u00a04, 5, 6\u00a0}. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Let A={1,3,5,7} and B={2,4,6}. Basic-mathematics.com. first two years of college and save thousands off your degree. Set Symbols. symbol.\n\nElements a and {a} are not the same because one is a set, and the other is not a set. All right reserved. Example: {x:x \u2265 -2} or {x|x \u2265 -2} We say, \"the set of all x's, such that x is greater than or equal to negative two\". lessons in math, English, science, history, and more. Sometimes the set is written with a bar instead of a colon: {x\u00a6 x > 5}. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. Let A={1,2,3,4,5} and let B={1,2,3,4,5}.\n\nSet notation is used to help define the elements of a set. For example, $\\{\\text{Miguel}, \\text{Kristin}, \\text{Leo}, \\text{Shanice}\\} \\nonumber$. Again, this is called the roster notation. just create an account. Decisions Revisited: Why Did You Choose a Public or Private College? Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. You can test out of the Square rooting gives two solutions: x must be greater than 2; or x must be less than -2: {x: x > 2 or x < -2}. It also contains 18, 21 and keeps going including all the multiples of 3 until it gets to its largest number 90. Objects placed within the brackets are called the elements of a set, and do not have to be in any specific order. For example, $\\left\\{\\frac{1}{2},\\:\\frac{2}{3},\\:\\frac{3}{4},\\:\\frac{4}{5},\\:...\\right\\}\\nonumber$. This helps to better define sets and to make them easier to write. We have already seen how to represent a set on a number line, but that can be cumbersome, especially if we want to just use a keyboard. For example, if you want to describe the set of all people who are over 18 years old but not 30 years old, you announce the conditions by putting them to the left of a vertical line segment. We want to be able to both read the various ways and be able to write down the representation ourselves in order to best display the set. She has over 10 years of teaching experience at high school and university level. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz\u00a0\u00a0Factoring Trinomials Quiz\u00a0Solving Absolute Value Equations Quiz\u00a0\u00a0Order of Operations QuizTypes of angles quiz. For example: {x: x > 5}.", "date": "2022-05-19 09:12:23", "meta": {"domain": "manami.pl", "url": "http://manami.pl/site/ymvno.php?tag=fe20d3-set-notation-examples", "openwebmath_score": 0.6804498434066772, "openwebmath_perplexity": 616.9135748554941, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357259231532, "lm_q2_score": 0.9019206705978501, "lm_q1q2_score": 0.8854477442744775}}
{"url": "https://plainmath.net/7256/situations-comparing-proportions-described-determine-situation-proportions", "text": "# TWO GROUPS OR ONE? Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions fo\n\nTWO GROUPS OR ONE? Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (c) Compare the graduation rate (proportion to graduate) of students on an athletic scholarship to the graduation rate of students who are not on an athletic scholarship. This situation involves comparing: 1 group or 2 groups? Do the methods of this section apply to the difference in proportions? Yes or No\nYou can still ask an expert for help\n\n\u2022 Questions are typically answered in as fast as 30 minutes\n\nSolve your problem for the price of one coffee\n\n\u2022 Math expert for every subject\n\u2022 Pay only if we can solve it\n\nyagombyeR\nFull and correct solution: c) Here we have two groups. On group consists of students who are on an athletic scholarship and the other group consists of students who are not on an athletic scholarship. Hence the given situation involves comparing 2 groups. Yes we can apply difference in proportions here because two proportions are being compared.", "date": "2022-06-26 02:34:07", "meta": {"domain": "plainmath.net", "url": "https://plainmath.net/7256/situations-comparing-proportions-described-determine-situation-proportions", "openwebmath_score": 0.7326516509056091, "openwebmath_perplexity": 1204.4285773342365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9744347845918813, "lm_q2_score": 0.9086178981700931, "lm_q1q2_score": 0.8853888858797025}}
{"url": "https://mathematica.stackexchange.com/questions/134106/which-root-does-findroot-give", "text": "# Which root does FindRoot give?\n\nI think that usually, FindRoot will give the root that's closest to the starting point. But see the example below, where I try to find the root of $\\cos x=0$. If I started from $x=0.1$, then I get $10.9956$, but if I started from 1, I get $1.5708$. What's wrong?\n\n\u2022 Draw a graph of the function and its tangent at x == 0.1. FindRoot is using Newton's method. \u2013\u00a0Michael E2 Dec 23 '16 at 4:49\n\u2022 I believe its because FindRoot[] uses Newton's Method. The tangent line hits further out. In general, Newtons method requires a good initial guess or you \"can\" get a root quite far away. \u2013\u00a0Michael McCain Dec 23 '16 at 4:49\n\u2022 FindRoot gives the root that it finds. :-) \u2013\u00a0Brett Champion Dec 23 '16 at 5:18\n\u2022 To have more control over which root is obtained, give FindRoot two intial guesses, which prompts it to use the use secant method. Then FindRoot usually returns the value of a root bracketed by the two guesses. For instance, FindRoot[Cos[x], {x, -1, 4}]. \u2013\u00a0bbgodfrey Dec 23 '16 at 5:48\n\nTechnically, FindRoot uses a damped Newton's method. In an undamped Newton's method, the first step would look like this, the tangent striking the x-axis just above 10:\n\nPlot[{Cos[x], Cos[0.1] - Sin[0.1] (x - 0.1)}, {x, 0, 12}]\n\n\nYou can keep large steps from occurring by decreasing the DampingFactor:\n\nFindRoot[Cos[x], {x, 0.1}, DampingFactor -> 0.2]\n(*  {x -> 1.5708}  *)\n\n\nThe damping factor multiplies the change in x. The undamped step would be x == 10.066644423259238, so the damped first step is\n\n0.1 + (10.066644423259238 - 0.1) 0.2\n(*  2.09333  *)\n\n\nThat lands x close enough to the root nearest 0.1 that FindRoot will converge on it.\n\nOf course, damping slows down convergence. In the following, it slows it down too much:\n\nFindRoot[1/x - 1/1000, {x, 0.1}, DampingFactor -> 0.2]\n\n\nFindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations.\n\n(*  {x -> 999.984}  *)\n\n\nThe regular method has no problem with it, though:\n\nFindRoot[1/x - 1/1000, {x, 0.1}]\n(*  {x -> 1000.}  *)\n\n\nWhen in doubt as to whether FindRoot[] is functioning as expected for a given nonlinear problem, one should try to use the diagnostic capabilities of the options EvaluationMonitor and StepMonitor. You've already been told in other answers as to why you should have expected your result, considering that you started the iteration with a seed that is uncomfortably near an extremum. Thus, let me demonstrate the use of EvaluationMonitor:\n\nReap[FindRoot[Cos[x], {x, 0.1}, EvaluationMonitor :> Sow[x]]]\n{{x -> 10.9956}, {{0.1, 10.0666, 11.4045, 10.9711, 10.9956, 10.9956}}}\n\n\nwhere we use Sow[]/Reap[] to get the values where Cos[x] was evaluated. We can also demonstrate the effect of damping, as shown by Michael:\n\nReap[FindRoot[Cos[x], {x, 0.1}, \"DampingFactor\" -> 1/5, EvaluationMonitor :> Sow[x]]]\n// Short\n{{x -> 1.5708}, {{0.1, 2.09333, 1.97814, <<76>>, 1.5708, 1.5708, 1.5708}}}\n\n\nwhere I have mercifully truncated the output, showing that damping gives better results at the cost of an increased number of iterations.\n\nOne could choose to use Brent's method instead by specifying explicit brackets. The convergence is not as fast as Newton-Raphson, but it is certainly much safer:\n\nReap[FindRoot[Cos[x], {x, 1., 2.}, EvaluationMonitor :> Sow[x]]]\n{{x -> 1.5708}, {{1., 2., 1.5649, 1.57098, 1.5708, 1.5708, 1.5708}}}\n\n\nIf, like me, you like pictures to help with diagnostics, there is a function called FindRootPlot[] (more information here) that can be used:\n\nNeeds[\"OptimizationUnconstrainedProblems\"]\nFindRootPlot[Cos[x], {x, 0.1}] // Last\n\n\nFindRootPlot[Cos[x], {x, 0.1}, \"DampingFactor\" -> 0.2] // Last\n\n\nThis question has been asked in different context here and here. As mentioned in the comments, FindRoot is based on Newton's method which works pretty well when a good guess is provided. But I think, it fails to provide you with multiple roots. To find multiple roots, you can use NDSolve\n\nf[x_] = Cos[x];\nModule[{sol},\nColumn[{sol = NSolve[{f[x] == 0, -10 <= x <= 10}, x],\nPlot[f[x], {x, -10, 10},\nEpilog -> {Red, AbsolutePointSize[6], Point[{x, f[x]} /. sol]},\nImageSize -> 360]}]]\n\n\nI adopted this idea from Bob Hanlon's answer to same sort of question.\n\n\u2022 The NDSolve Method you linked in your answer is very different from the code you posted (you used NSolve). Otherwise this is a good answer :) \u2013\u00a0Sascha Dec 23 '16 at 9:09\n\u2022 @zhk: For one function your Code works fine. Have you used the same method for a system of non-linear equations? I have a NL equation system and I need to find Real solutions to the system, but so far no success. Can you show me how to use your Code for a system? \u2013\u00a0Tugrul Temel Apr 19 '19 at 9:52\n\nTo see what is happening, implement a quick Newton iteration algorithm. For instance:\n\nNewtonsMethodList[f_, {x_, x0_}, n_] :=\nNestList[# - Function[x, f][#]/Derivative[1][Function[x, f]][#] &,\nx0, n]\n\n\nNow see what happens when we have 0.1 and 1 as starting values, and with, say, 5 iterations:\n\nIn[2]:= NewtonsMethodList[Cos[x], {x, 0.1}, 5]\n\nOut[2]= {0.1, 10.0666, 11.4045, 10.9711, 10.9956, 10.9956}\n\nIn[5]:= NewtonsMethodList[Cos[x], {x, 1.}, 5]\n\nOut[5]= {1., 1.64209, 1.57068, 1.5708, 1.5708, 1.5708}\n`\n\nI hope this helps.", "date": "2021-04-14 09:04:05", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/134106/which-root-does-findroot-give", "openwebmath_score": 0.27769485116004944, "openwebmath_perplexity": 2313.082319097427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338123908151, "lm_q2_score": 0.9207896829553822, "lm_q1q2_score": 0.8853704142622186}}
{"url": "https://mathhelpboards.com/threads/rule-of-sum-and-rule-pf-product-how-many-distinct-sums-are-possible-using-from-1-to-all-of-the-15-cards.25841/", "text": "Rule of sum and rule pf product: How many distinct sums are possible using from 1 to all of the 15 cards\n\nDhamnekar Winod\n\nActive member\nThere are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards?\n\nAnswer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\\binom{15} {2} + \\binom{15}{3}+...\\binom{15}{15}$ distinct sums.\n\nSo, I think answer 143 is wrong.\n\nCountry Boy\n\nWell-known member\nMHB Math Helper\n$$\\begin{pmatrix}15 \\\\ 2 \\end{pmatrix}$$ is the number of sums of two of the numbers, $$\\begin{pmatrix}15 \\\\ 3 \\end{pmatrix}$$ is the number of sums of three of the numbers, etc. But they won't be distinct sums.\n\nskeeter\n\nWell-known member\nMHB Math Helper\nThere are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards?\n\nAnswer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\\binom{15} {2} + \\binom{15}{3}+...\\binom{15}{15}$ distinct sums.\n\nSo, I think answer 143 is wrong.\nThe directions clearly state from 1 to 15, so their solution is based on that premise.\n\nDhamnekar Winod\n\nActive member\n$$\\begin{pmatrix}15 \\\\ 2 \\end{pmatrix}$$ is the number of sums of two of the numbers, $$\\begin{pmatrix}15 \\\\ 3 \\end{pmatrix}$$ is the number of sums of three of the numbers, etc. But they won't be distinct sums.\nHello,\n\nI have computed 141 distinct sums. But the answer is 143. Which 2 distinct sums i omitted, would you tell me? Is the answer 143 wrong? All the 141 distinct sums are\n\n20,30,40,50,60,70,80,110,120,130,140,150,160,170,180,200,210,220,230,240,250,260,270,280,300,310,320,330,340,350,360,370,380,400,410,420,430,440,450,460,470,480,500,510,520,530,540,550,560,570,580,600,610,620,630,640,650,660,670,680,700,710,720,730,740,750,760,770,780,800,810,820,830,840,850,860,870,880,900,910,920,930,940,950,960,970,980,1000,1010,1020,1030,1040,1050,1060,1070,1080,1100,1110,1120,1130,1140,1150,1160,1170,1180,1200,1210,1220,1230,1240,1250,1260,1270,1280,1300,1310,1320,1330,1340,1350,1360,1370,1380,1400,1410,1420,1430,1440,1450,1460,1470,1480,1500,1510,1520,1530,1540,1550,1560,1570,1580.\n\nI know one formula $\\binom{r+n-1}{r-n+1}$ which computes distinct sums, where n=cells and r= balls. How to use that here? Or is there any other formula?\n\nLast edited:", "date": "2020-07-09 02:47:37", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/rule-of-sum-and-rule-pf-product-how-many-distinct-sums-are-possible-using-from-1-to-all-of-the-15-cards.25841/", "openwebmath_score": 0.8201991319656372, "openwebmath_perplexity": 529.3834386670821, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731147976795, "lm_q2_score": 0.9196425355825848, "lm_q1q2_score": 0.8853151442297227}}
{"url": "https://math.stackexchange.com/questions/2453591/difference-between-infty-and-infty", "text": "# difference between $+\\infty$ and $\\infty$\n\nI'm taking Mathematical Analysis \"I\" and I'm studying limits where I have limits to the infinity, but I don't know what's the difference between $\\lim_{x \\to \\infty}$ and $\\lim_{x \\to +\\infty}$ I suppose that they are the same but I'm not sure. If you could help me I would appreciate it. Thank you very much!\n\n\u2022 Yes, they are the same. \u2013\u00a0Kenny Lau Oct 1 '17 at 22:40\n\u2022 Thank you very much! @KennyLau \u2013\u00a0Santiago Pardal Oct 1 '17 at 22:41\n\u2022 In real analysis they are the same, in complex analysis they are different. Because of your tag (real-analysis), I agree with Kenny. \u2013\u00a0GEdgar Oct 8 '17 at 11:04\n\nIn the context of real Analysis we usually consider \\begin{align*} \\lim_{x \\to \\infty}f(x)\\qquad\\text{and}\\qquad\\lim_{x \\to +\\infty}f(x) \\end{align*} to be the same. It has mainly to do with preserving the order of the real numbers when $\\mathbb{R}$ is extended by the symbols $+\\infty$ and $-\\infty$. We look at two references:\n\n\u2022 Principles of Mathematical Analysis by W. Rudin.\n\nDefinition 1.23: The extended real number system consists of the real field $\\mathbb{R}$ and two symbols $+\\infty$ and $-\\infty$. We preserve the original order in $\\mathbb{R}$, and define \\begin{align*} \\color{blue}{-\\infty < x < +\\infty}\\tag {1} \\end{align*} for every $x\\in\\mathbb{R}$.\n\n(he continues with:) It is then clear that $+\\infty$ is an upper bound of every subset of the extended real number system, and that every nonempty subset has a least upper bound.\n\nIf, for example, $E$ is a nonempty set of real numbers which is not bounded above in $\\mathbb{R}$, then $\\sup E=+\\infty$ in the extended real number system. Exactly the same remarks apply to lower bounds.\n\nNow we look at certain intervals of real numbers introduced in\n\n\u2022 Calculus by M. Spivak.\n\n(We find in chapter 4:) The set $\\{x:x>a\\}$ is denoted by $(a,\\infty)$, while the set $\\{x: x\\geq a\\}$ is denoted by $[a,\\infty)$; the sets $(-\\infty,a)$ and $(-\\infty,a]$ are defined similarly.\n\n(Spivak continues later on:) The set $\\mathbb{R}$ of all real number is also considered to be an \"interval\" and is sometimes denoted by \\begin{align*} \\color{blue}{(-\\infty,\\infty)}\\tag{2} \\end{align*}\n\nThe connection with limits is presented in chapter 5:\n\nThe symbol $\\lim_{x\\rightarrow\\infty}f(x)$ is read \"the limit of $f(x)$ as $x$ approaches $\\infty$,\" or \"as $x$ becomes infinite\", and a limit of the form \\begin{align*} \\lim_{\\color{blue}{x\\rightarrow\\infty}}f(x) \\end{align*} is often called a limit at infinity.\n\n(and later on:) Formally, $\\lim_{x\\rightarrow\\infty}f(x)=l$ means that for every $\\varepsilon>0$ there is a number $N$ such that, for all $x$, \\begin{align*} \\text{if }x>N\\text{, then }|f(x)-l|<\\varepsilon\\text{.} \\end{align*}\n\nand we find as exercise 36 a new definition and the following two out of three sub-points\n\nExercise 36: Define \\begin{align*} &\\lim_{\\color{blue}{x=-\\infty}}f(x)=l\\\\ \\\\ &(b) \\text{Prove that }\\lim_{x\\rightarrow\\infty}f(x)=\\lim_{x\\rightarrow-\\infty}f(-x)\\text{.}\\\\ &(c) \\text{Prove that }\\lim_{x\\rightarrow 0^{-}}f(1/x)=\\lim_{x\\rightarrow-\\infty}f(x)\\text{.} \\end{align*}\n\nConclusion: When looking at (1) and (2) together with Spivaks definition of limits we can conclude that $\\infty$ and $+\\infty$ are used interchangeably in the context of limits of real valued functions.\n\nThe compactification of the real numbers, in a useful way that fits in with the ordering of the reals, requires the addition of two points, whereas the compactification of the complex numbers is naturally accomplished by adding just one point. Because analysis readily switches between the real and complex cases, it is considered by some authors appropriate to use a \"balanced\" pair of symbols, $+\\infty$ and $-\\infty$, for the real case, which reflects the symmetry of their roles, and the unsigned $\\infty$ for the complex case. This is a stylistic choice. Other authors are not of this persuasion. Their argument is \"We don't write $+3$ when we mean $3$; so why should we have to write $+\\infty$? And, in the complex case, which is always clear from the context, writing $\\infty$ is good enough for anyone\".\n\nIn my view, siding with the latter type of author, writing $+\\infty$ instead of (real) $\\infty$ is unnecessary, just as it is unnecessary to write $(-1\\;\\pmb,\\,+\\!1)$ to denote the interval $(-1\\;\\pmb,\\,1)$.\n\n\u2022 There is a one point compactification of the reals. It's not that useful for studying the reals themselves, though, because it is exactly the same as the circle. \u2013\u00a0Ian Oct 4 '17 at 16:37", "date": "2019-06-25 23:34:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2453591/difference-between-infty-and-infty", "openwebmath_score": 0.9814448356628418, "openwebmath_perplexity": 230.86852229597366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399086356109, "lm_q2_score": 0.9124361586911174, "lm_q1q2_score": 0.8852819752442975}}
{"url": "http://math.stackexchange.com/questions/29638/how-do-we-find-a-fraction-with-whose-decimal-expansion-has-a-given-repeating-pat", "text": "# How do we find a fraction with whose decimal expansion has a given repeating pattern?\n\nWe know $\\frac{1}{81}$ gives us $0.\\overline{0123456790}$\n\nHow do we create a recurrent decimal with the property of repeating:\n\n$0.\\overline{0123456789}$\n\na) Is there a method to construct such number?\n\nb) Is there a solution?\n\nc) Is the solution in $\\mathbb{Q}$?\n\nAccording with wikipedia page: http://en.wikipedia.org/wiki/Decimal One could get this number by applying this series. Supppose:\n\n$M=123456789$, $x=10^{10}$, then $0.\\overline{0123456789}= \\frac{M}{x}\\cdot$ $\\sum$ ${(10^{-9})}^k$ $=\\frac{M}{x}\\cdot\\frac{1}{1-10^{-9}}$ $=\\frac{M}{9999999990}$\n\nUnless my calculator is crazy, this is giving me $0.012345679$, not the expected number. Although the example of wikipedia works fine with $0.\\overline{123}$.\n\nSome help I got from mathoverflow site was that the equation is: $\\frac{M}{1-10^{-10}}$. Well, that does not work either.\n\nSo, just to get rid of the gnome calculator rounding problem, running a simple program written in C with very large precision (long double) I get this result:\n\n#include <stdio.h>\nint main(void)\n{\nlong double b;\nb=123456789.0/9999999990.0;\nprintf(\"%.40Lf\\n\", b);\n}\n\n\nResult: $0.0123456789123456787266031042804570461158$\n\nMaybe it is still a matter of rounding problem, but I doubt that...\n\nThanks!\n\nBeco\n\nEdited:\n\nThanks for the answers. After understanding the problem I realize that long double is not sufficient. (float is 7 digits:32 bits, double is 15 digits:64 bits and long double is 19 digits:80 bits - although the compiler align the memory to 128 bits)\n\nUsing the wrong program above I should get $0.0\\overline{123456789}$ instead of $0.\\overline{0123456789}$. Using the denominator as $9999999999$ I must get the correct answer. So I tried to teach my computer how to divide:\n\n#include <stdio.h>\nint main(void)\n{\nint i;\nlong int n, d, q, r;\nn=123456789;\nd=9999999999;\nprintf(\"0,\");\nn*=10;\nwhile(i<100)\n{\nif(n<d)\n{\nn*=10;\nprintf(\"0\");\ni++;\ncontinue;\n}\nq=n/d;\nr=n%d;\nprintf(\"%ld\", q);\nif(!r)\nbreak;\nn=n-q*d;\nn*=10;\ni++;\n}\nprintf(\"\\n\");\n}\n\n-\nChange the C program to assign b=123456789.0/9999999999.0; Note the integral value of the denominator should end in a 9, not a 0. \u2013\u00a0 Brandon Carter Mar 29 '11 at 2:53\nThanks @Brandon, but only this was not sufficient. Look at the edition. \u2013\u00a0 Dr Beco Mar 29 '11 at 4:03\nNow I get any precision I want, just change while(i<PREC). Output with 100: 0,012345678901234567890123456789012345678901234567890123456789012345678901234567\u200c\u200b8901234567890123456789 \u2013\u00a0 Dr Beco Mar 29 '11 at 4:07\nTIP Use one of the many freely available mathematics systems with multiple precision arithmetic instead of wasting your time rolling your own, e.g. wolframalpha.com/input/?i=N[123456789/%2810^10-1%29,40] \u2013\u00a0 Bill Dubuque Mar 29 '11 at 4:31\n@Bill Wow! I'm astonished! Thank you very much for this great tip. \u2013\u00a0 Dr Beco Mar 29 '11 at 5:15\n\nSuppose you want to have a number $x$ whose decimal expansion is $0.a_1a_2\\cdots a_ka_1a_2\\cdots a_k\\cdots$. That is it has a period of length $k$, with digits $a_1$, $a_2,\\ldots,a_k$.\n\nLet $n = a_1a_2\\cdots a_k$ be the integer given by the digits of the period. Then \\begin{align*} \\frac{n}{10^{k}} &= 0.a_1a_2\\cdots a_k\\\\ \\frac{n}{10^{2k}} &= 0.\\underbrace{0\\cdots0}_{k\\text{ zeros}}a_1a_2\\cdots a_k\\\\ \\frac{n}{10^{3k}} &= 0.\\underbrace{0\\cdots0}_{2k\\text{ zeros}}a_1a_2\\cdots a_k\\\\ &\\vdots \\end{align*} So the number you want is $$\\sum_{r=1}^{\\infty}\\frac{n}{10^{rk}} = n\\sum_{r=1}^{\\infty}\\frac{1}{(10^k)^r} = n\\left(\\frac{\\quad\\frac{1}{10^k}\\quad}{1 - \\frac{1}{10^k}}\\right) = n\\left(\\frac{10^k}{10^k(10^k - 1)}\\right) = \\frac{n}{10^k-1}.$$ Since $10^k$ is a $1$ followed by $k$ zeros, then $10^k-1$ is $k$ 9s. So the fraction with the decimal expansion $$0.a_1a_2\\cdots a_ka_1a_2\\cdots a_k\\cdots$$ is none other than $$\\frac{a_1a_2\\cdots a_k}{99\\cdots 9}.$$\n\nThus, $0.575757\\cdots$ is given by $\\frac{57}{99}$. $0.837168371683716\\cdots$ is given by $\\frac{83716}{99999}$, etc.\n\nIf you have some decimals before the repetition begins, e.g., $x=2.385858585\\cdots$, then first multiply by a suitable power of $10$, in this case $10x = 23.858585\\cdots = 23 + 0.858585\\cdots$, so $10x = 23 + \\frac{85}{99}$, hence $x= \\frac{23}{10}+\\frac{85}{990}$, and simple fraction addition gives you the fraction you want.\n\nAnd, yes, there is always a solution and it is always a rational.\n\n-\nThanks @Arturo for this very explanatory solution. Now I got the correct result, with any precision I want, just changing the \"while\" with the new program (see edition). \u2013\u00a0 Dr Beco Mar 29 '11 at 4:06\n\nIt's simple: $\\rm\\displaystyle\\ x\\ =\\ 0.\\overline{0123456789}\\ \\ \\Rightarrow\\ \\ 10^{10}\\ x\\ =\\ 123456789\\ +\\ x\\ \\ \\Rightarrow\\ \\ x\\ =\\ \\frac{123456789}{10^{10} - 1}$\n\nNote that the last digit of $\\rm\\ 10^{10} - 1\\$ is $\\:9\\:,$ not $\\:0\\:,$ which explains the error in your program.\n\n-\nThanks @Bill, you corrected the formula, but the problem was long double rounding it. \u2013\u00a0 Dr Beco Mar 29 '11 at 4:04\n@Dr Beco: I assumed the rest would be easy once you had the correct formula. \u2013\u00a0 Bill Dubuque Mar 29 '11 at 4:27\n\nWhen you say \"double\" in C how many places is that?\n\nI tried it in Maple...\n\n\n\nDigits := 40;\n40\n123456789.0/9999999990.0;\n0.01234567891234567891234567891234567891235\n\n-\nFloat is 7 digits (32 bits), Double is 15 digits (64 bits) and Long Double is 19 digits (only 80 digits used, but the alignment make it 128 bits). This long double number \"0.012345678901234568431\" loses precision in the 18 digit, as expected... My program didn't work because of that! Thanks. \u2013\u00a0 Dr Beco Mar 29 '11 at 3:55\nBTW, what is maple? A language? Do you have any link I could spy on it? Thanks \u2013\u00a0 Dr Beco Mar 29 '11 at 4:10\n\nYou said:\n\n$M=123456789$, $x=10^{10}$, then $0.\\overline{0123456789}= \\frac{M}{x}\\cdot$ $\\sum$ ${(10^{-9})}^k$ $=\\frac{M}{x}\\cdot\\frac{1}{1-10^{-9}}$ $=\\frac{M}{9999999990}$\n\nbut since the block of repeating digits is 10 digits long, the summation term should be $\\sum{(10^{-10})}^k$, so that $$0.\\overline{0123456789}=\\frac{M}{x}\\cdot\\sum{(10^{-10})}^k=\\frac{M}{x}\\cdot\\frac{1}{1-10^{-10}}=\\frac{M}{9999999999}$$ and $$\\frac{M}{9999999999}=\\frac{123456789}{9999999999}=\\frac{13717421}{1111111111}.$$\n\n-\nThanks! Using your last fraction, the long double can give at least 2 repetitions before lose precision: 0.0123456789 0123456789 0470 \u2013\u00a0 Dr Beco Mar 29 '11 at 4:14", "date": "2014-08-01 01:47:35", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/29638/how-do-we-find-a-fraction-with-whose-decimal-expansion-has-a-given-repeating-pat", "openwebmath_score": 0.9437023997306824, "openwebmath_perplexity": 871.6057100813944, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987758722546077, "lm_q2_score": 0.8962513641273355, "lm_q1q2_score": 0.8852801025105959}}
{"url": "https://mathhelpboards.com/threads/a-probability-question-choosing-3-from-25.5240/#post-23851", "text": "# [SOLVED]A probability question: choosing 3 from 25\n\n#### first21st\n\n##### New member\nIn a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:\nA. 21/46\nB. 25/117\nC. 1/50\nD. 3/25\n\nCould you please solve this problem with proper explanation?\n\nThanks,\n\nJames\n\n#### MarkFL\n\nStaff member\nRe: A probability question\n\nHere at MHB, we normally don't provide fully worked solutions, but rather we help people to work the problem on their own. This benefits people much more, which is our goal.\n\nNow, what you want to do here is to find the number of ways to choose 1 girl from 10 AND 2 boys from 15, then divide this by the number of ways to choose 3 children from 25. What do you find?\n\n#### first21st\n\n##### New member\nRe: A probability question\n\nThanks for your reply. It's really a great way of learning. I highly appreciate your approach. Is the solution something like:\n\n(15 C 2) * (10 C 1)/ (25 C 3)\n\nIf it is correct, could you please explain why did we divide it with the number of ways of choosing 3 students from 25?\n\nThanks,\n\nJames\n\n#### MarkFL\n\nStaff member\nRe: A probability question\n\nExcellent! That is correct!\n\nNow you just need to simplify, either by hand or with a calculator.\n\nAs probability is the ratio of the number favorable outcomes to the number of all outcomes, we are in this case dividing the number of ways to choose 1 girl from 10 AND 2 boys from 15 by the number of ways to choose 3 of the children from the total of 25. We are told that 3 children are selected at random, and we know there are 25 children by adding the number of boys to the number of girls. Thus, $$\\displaystyle {25 \\choose 3}$$ is the total number of outcomes.\n\n#### first21st\n\n##### New member\nRe: A probability question\n\nThank you very much man! But I am still wondering why did we MULTIPLY # the number of ways to choose 1 girl from 10 # AND # 2 boys from 15 #, instead of ADDING these two operands?\n\n#### MarkFL\n\nStaff member\nRe: A probability question\n\nWe multiply because we are essentially applying the fundamental counting principle. When we require event 1 AND event 2 to happen, we multiply. When we require event 1 OR event 2 to happen, we add. Here we require both 2 boys AND 1 girl.\n\nYou see, for each way to obtain 2 boys, we have to account for all of the ways to obtain 1 girl. Or conversely, for each way to obtain 1 girl, we have to account for all of the ways to obtain 2 boys. The product of these two gives us all of the ways to get 2 boys and 1 girl.\n\n#### first21st\n\n##### New member\nRe: A probability question\n\nok. Got it! Thanks a lot.\n\nJames", "date": "2022-05-17 10:23:37", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/a-probability-question-choosing-3-from-25.5240/#post-23851", "openwebmath_score": 0.6770955920219421, "openwebmath_perplexity": 381.7450872745894, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9926541745120955, "lm_q2_score": 0.8918110540642805, "lm_q1q2_score": 0.8852599656929401}}
{"url": "https://orinanobworld.blogspot.com/2020/", "text": "## Monday, February 17, 2020\n\n### Reversing Differences\n\nFellow blogger H\u00e5kan Kjellerstrand posted an interesting question on OR Stack Exchange recently. Starting from a list of integers, it is trivial to compute the list of all pairwise absolute differences, but what about going in the other direction? Given the pairwise (absolute) differences, with duplicates removed, can you recover the source list (or a source list)?\n\nWe can view the source \"list\" as a vector $x\\in\\mathbb{Z}^n$ for some dimension $n$ (equal to the length of the list). With duplicates removed, we can view the differences as a set $D\\subset \\mathbb{Z}_+$. So the question has to do with recovering $x$ from $D$. Our first observation kills any chance of recovering the original list with certainty:\nIf $x$ produces difference set $D$, then for any $t\\in\\mathbb{R}$ the vector $x+t\\cdot (1,\\dots,1)'$ produces the same set $D$ of differences.\nTranslating all components of $x$ by a constant amount has no effect on the differences. So there will be an infinite number of solutions for a given difference set $D$. A reasonable approach (proposed by H\u00e5kan in his question) is to look for the shortest possible list, i.e., minimize $n$.\n\nNext, observe that $0\\in D$ if and only if two components of $x$ are identical. If $0\\notin D$, we can assume that all components of $x$ are distinct. If $0\\in D$, we can solve the problem for $D\\backslash\\{0\\}$ and then duplicate any one component of the resulting vector $x$ to get a minimum dimension solution to the original problem.\n\nCombining the assumption that $0\\notin D$ and the observation about adding a constant having no effect, we can assume that the minimum element of $x$ is 1. That in turn implies that the maximum element of $x$ is $1+m$ where $m=\\max(D)$.\n\nFrom there, H\u00e5kan went on to solve a test problem using constraint programming (CP). Although I'm inclined to suspect that CP will be more efficient in general than an integer programming (IP) model, I went ahead and solved his test problem via an IP model (coded in Java and solved using CPLEX 12.10). CPLEX's solution pool feature found the same four solutions to H\u00e5kan's example that he did, in under 100 ms. How well the IP method scales is an open question, but it certainly works for modest size problems.\n\nThe IP model uses binary variables $z_1, \\dots, z_{m+1}$ to decide which of the integers $1,\\dots,m+1$ are included in the solution $x$. It also uses variables $w_{ij}\\in [0,1]$ for all $i,j\\in \\{1,\\dots,m+1\\}$ such that $i \\lt j$. The intent is that $w_{ij}=1$ if both $i$ and $j$ are included in the solution, and $w_{ij} = 0$ otherwise. We could declare the $w_{ij}$ to be binary, but we do not need to; constraints will force them to be $0$ or $1$.\n\nThe full IP model is as follows:\n\n$\\begin{array}{lrlrc} \\min & \\sum_{i=1}^{m+1}z_{i} & & & (1)\\\\ \\textrm{s.t.} & w_{i,j} & \\le z_{i} & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} ,i\\lt j & (2)\\\\ & w_{i,j} & \\le z_{j} & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} ,i\\lt j & (3)\\\\ & w_{i,j} & \\ge z_{i}+z_{j}-1 & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} ,i\\lt j & (4)\\\\ & w_{i,j} & =0 & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} \\textrm{ s.t. }(j-i)\\notin D & (5)\\\\ & \\sum_{i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} |j-i=d}w_{i,j} & \\ge 1 & \\forall d\\in D & (6)\\\\ & z_{1} & = 1 & & (7) \\end{array}$\n\nThe objective (1) minimizes the number of integers used. Constraints (2) through (4) enforce the rule that $w_{ij}=1$ if and only if both $z_i$ and $z_j$ are $1$ (i.e., if and only if both $i$ and $j$ are included in the solution).\u00a0 Constraint (5) precludes the inclusion of any pair $i < j$ whose difference $j - i$ is not in $D$, while constraint (6) says that for each difference $d \\in D$ we must include at least one pair $i < j$ for that produces that difference ($j - i = d$). Finally, since we assumed that our solution starts with minimum value $1$, constraint (7) ensures that $1$ is in the solution. (This constraint is redundant, but appears to help the solver a little, although I can't be sure given the short run times.)\n\nMy Java code is available from my repository (bring your own CPLEX).\n\n## Tuesday, February 11, 2020\n\n### Collections of CPLEX Variables\n\nRecently, someone asked for help online regarding an optimization model they were building using the CPLEX Java API. The underlying problem had some sort of network structure with $N$ nodes, and a dynamic aspect (something going on in each of $T$ periods, relating to arc flows I think). Forget about solving the problem: the program was running out of memory and dying while building the model.\n\nA major issue was that they allocated two $N\\times N\\times T$ arrays of variables, and $N$ and $T$ were big enough that $2N^2T$ was, to use a technical term, ginormous. Fortunately, the network was fairly sparse, and possibly not every time period was relevant for every arc. So by creating only the IloNumVar instances they needed (meaning only for arcs that actual exist in time periods that were actually relevant), they were able to get the model to build.\n\nThat's the motivation for today's post. We have a tendency to write mathematical models using vectors or matrices of variables. So, for instance, $x_i \\, (i=1,\\dots,n)$ might be an inventory level at each of $n$ locations, or $y_{i,j} \\, (i=1,\\dots,m; j=1,\\dots,n)$ might be the inventory of item $i$ at location $j$. It's a natural way of expressing things mathematically. Not coincidentally, I think, CPLEX APIs provide structures for storing vectors or matrices of variables and for passing them into or out of functions. That makes it easy to fall into the trap of thinking that variables must be organized into vectors or matrices.\n\nLast year I did a post (\"Using Java Collections with CPLEX\") about using what Java calls \"collections\" to manage CPLEX variables. This is not unique to Java. I know that C++ has similar memory structures, and I think they exist in other languages you might use with CPLEX. The solution to the memory issue I mentioned at the start was to create a Java container class for each combination of an arc that actually exists and time epoch for which it would have a variable, and then associate instances of that class with CPLEX variables. So if we call the new class AT (my shorthand for \"arc-time\"), I suggested the model owner use a Map<AT, IloNumVar> to associate each arc-time combination with the variable representing it and a Map<IloNumVar, AT> to hold the reverse association. The particular type of map is mostly a matter of taste. (I generally use HashMap.) During model building, they would create only the AT instances they actually need, then create a variable for each and pair them up in the first map. When getting a solution from CPLEX, they would get a value for each variable and then use the second map to figure out for which arc and time that value applied. (As a side note, if you use maps and then need the variables in vector form, you can apply the values() method to the first map (or the getKeySet() method to the second one), and then apply the toArray() method to that collection.)\n\nNow you can certainly get a valid model using just arrays of variables, which was all that was available to me back in the Dark Ages when I used FORTRAN, but I think there are some benefits to using collections. Using arrays requires you to develop an indexing scheme for your variables. The indexing scheme tells you that the flow from node 3 to node 7 at time 4 will be occupy slot 17 in the master variable vector. Here are my reasons for avoiding that.\n\u2022 Done correctly, the indexing scheme is, in my opinion, a pain in the butt to manage. Finding the index for a particular variable while writing the code is time-consuming and has been known to kill brain cells.\n\u2022 It is easy to make mistakes while programming (calculate an index incorrectly).\n\u2022 Indexing invites the error of declaring an array or vector with one entry for each combination of component indices (that $N\\times N\\times T$ matrix above), without regard to whether you need all those slots. Doing so wastes time and space, and the space, as we saw, may be precious.\n\u2022 Creating slots that you do not need can lead to execution errors. Suppose that I allocating a vector IloNumVar x = new IloNumVar[20] and use 18 slots, omitting slots 0 and 13. If I solve the model and then call getValues(x), CPLEX will throw an exception, because I am asking for values of two variables (x[0] and x[13]) that do not exist. Even if I create variables for those two slots, the exception will occur, because those two variables will not belong to the model being solved. (There is a way to force CPLEX to include those variables in the model without using them, but it's one more pain in the butt to deal with.) I've lost count of how many times I've seen messages on the CPLEX help forums about exceptions that boiled down to \"unextracted variables\".\nSo my advice is to embrace collections when building models where variables do not have an obvious index scheme (with no skips).\n\n## Thursday, January 30, 2020\n\n### Generic Callback Changes in CPLEX 12.10\n\nCPLEX 12.10 is out, and there have been a few changes to the new(ish) generic callbacks. Rather than go into them in detail (and likely screw something up), I'll just point you to the slides for a presentation by Daniel Junglas of IBM at the 2019 INFORMS Annual Meeting.\n\nI've written about half a dozen posts about generic callbacks since IBM introduced them (which you can find by typing \"generic callback\" in the search widget on the blog). A couple of things have been added recently, and I thought I would mention them. The generic callback approach uses a single callback function that can be called from a variety of contexts, including when CPLEX solves a node relaxation (\"RELAXATION\" context), when if finds a candidate solution (\"CANDIDATE\" context) and, now, when it is ready to split a node into children (\"BRANCHING\" context).\n\nThe branching context is one of the new features. It brings back most of the functionality of the branch callback in the legacy callback system. Unfortunately, it does not seem to have the ability to attach user information to the child nodes, which was a feature that was occasionally useful in the legacy system. You can get more or less equivalent functionality by creating a data store (array, map, whatever) in your global memory and storing the node information keyed by the unique index number of each child node. The catch is that you are now responsible for memory management (freeing up space when a node is pruned and the associated information is no longer needed), and for dealing with thread synchronization issues.\n\nAnother new feature is that you can now inject a heuristic solution (if you have one) from all three of the contexts I mentioned above. CPLEX gives you a variety of options for how it will handle the injected solution: \"NoCheck\" (CPLEX will trust you that it is feasible); \"CheckFeasible\" (CPLEX will check feasibility and ignore the solution if it is not feasible); \"Propagate\" (Daniel's explanation: CPLEX will \"propagate fixed variables and accept if feasible\"); and \"Solve\" (CPLEX will solve a MIP problem with fixed variables and accept the result if feasible). I assume the latter two mean that you provide a partial solution, fixing some variables but not others. (Unfortunately I was unable to make it to Daniel's talk, so I'm speculating here.)\n\nI'm not sure if those are the only new features, but they are the ones that are most relevant to me. I invite you to read through Daniel's slides to get a more complete picture, including both the reasons for switching from legacy callbacks to generic callbacks and some of the technical issues in using them.\n\n## Tuesday, January 7, 2020\n\n### Greedy Methods Can Be Exact\n\nWe generally sort optimization algorithms (as opposed to models) into two or three categories, based on how certain we are that solutions will be either optimal or at least \"good\". An answer by Michael Feldmeier to a question I posted on OR Stack Exchange neatly summarizes the categories:\n\u2022 exact methods eventually cough up provably optimal solutions;\n\u2022 approximate methods eventually cough up solutions with some (meaningful) guarantee regarding how far from optimal they might be; and\n\u2022 heuristics provide no worst-case guarantees (but generally are either easy to implement, fast to execute or both).\nI should explain my use of \"meaningful\" (which is not part of Michael's answer). A common way to estimate the \"gap\" between a solution and the optimum is to take $|z - \\tilde{z}|/|z|$, where $z$ is the objective value of the solution produced by the algorithm and $\\tilde{z}$ is some bound (lower bound in a minimization, upper bound in a maximization) of the optimal solution. Now suppose that we are minimizing a function known to be nonnegative. If we set $\\tilde{s}=0$, we know that any method, no matter how stupid, will have a gap no worse than 100%. To me, that is not a meaningful guarantee. So I'll leave the definition of \"meaningful\" to the reader.\n\nWhat brings all this to mind is a question posted on Mathematics Stack Exchange. The author of the question was trying to solve a nonlinear integer program. He approached it by applying a \"greedy algorithm\". Greedy algorithms are generally assumed to be heuristics, since it seldom is possible to provide useful guarantees on performance. In his case, though, the greedy algorithm is provably optimal, mainly due to the objective function being concave and separable. I'll state the problem and show a proof of optimality below (changing the original notation a bit). Brace yourself: the proof is a bit long-winded.\n\nYou start with $N$ workers to be assigned to $M$ work stations. The output of workstation $m$, as a function of the number of workers $x$ assigned to it, is given by $$f_{m}(x)=a_{m}x+b_{m}-\\frac{c_{m}}{x},$$\nwhere $a_{m},b_{m},c_{m}$ are all positive constants. Since $f(0)=-\\infty$, we can assume that each work station gets at least one worker (and, consequently, that $N>M$). Since $f_{m}'(x)=a_{m}+c_{m}/x^{2}>0$, each $f_{m}()$ is monotonically increasing. Thus, we can safely assume that all $N$ workers will be assigned somewhere. $f_{m}''(x)=-2c_{m}/x^{3}<0$, so $f_{m}()$ is strictly concave (which we will need later). We also note, for future reference, that the impact of adding one worker to a current staff of $x$ at station $m$ is $$\\Delta f_{m}(x)=a_{m}+\\frac{c_{m}}{x(x+1)}>0.$$\nSimilarly, the impact of removing one worker at station $m$ is $$\\delta f_{m}(x)=-a_{m}-\\frac{c_{m}}{x(x-1)}<0.$$We see that $\\delta f_{m}(x)$ is an increasing function of $x$ (i.e., it gets less negative as $x$ gets bigger). We also note that $\\Delta f_{m}(x)=-\\delta f_{m}(x+1)$.\n\nThe IP model is easy to state. Let $x_{m}$ be the number of workers assigned to work station $m$. The model is\n$$\\max\\sum_{m=1}^{M}f_{m}(x_{m})$$\nsubject to\n$$\\sum_{m=1}^{M}x_{m}\\le N$$\nwith\n$$x\\in\\mathbb{Z}_{+}^{M}.$$\n\nThe greedy algorithm starts with a single worker at each station ($x=(1,\\dots,1)$) and, at each step, adds one worker to the workstation where that worker produces the greatest increase in objective value (breaking ties arbitrarily). It stops when all $N$ workers are assigned. To prove that it actually finds an optimal solution, I'll use proof by contradiction.\n\nLet $x^{(0)},x^{(1)},\\dots,x^{(N-M)}$ be the sequence of solutions constructed by the greedy algorithm, with $x^{(0)}=(1,\\dots,1)$, and let $x^{(k)}$ be the last solution in the sequence for which an optimal solution $x^{*}$ exists such that $x^{(k)}\\le x^{*}$. The significance of the inequality is that if $x\\le x^{*}$, it is possible to extend the partial solution $x$ to the optimal solution $x^{*}$ by adding unassigned workers to work stations. We know that $k$ is well defined because $x^{(0)}\\le x^{*}$ for any optimal $x^{*}$. Since we are assuming that the greedy algorithm does not find an optimum, it must be that $k<N-M$.\n\nNow identify the work station $j$ to which the greedy algorithm added a worker at step $k$, meaning that $x_{j}^{(k+1)}=x_{j}^{(k)}+1$ and $x_{i}^{(k+1)}=x_{i}^{(k)}$ for $i\\neq j$. Since, by assumption, $x^{(k)}\\le x^{*}$ but $x^{(k+1)}\\not\\le x^{*}$, it must be that $x_{j}^{(k)}=x_{j}^{*}$.\n\nNext, since $x^{(k)}\\le x^{*}$ and $x^{(k)}\\neq x^{*}$ (else $x^{(k)}$ would be optimal), there is some work station $h\\neq j$ such that $x_{h}^{(k)}<x_{h}^{*}$. Let $\\tilde{x}$ be the solution obtained from $x^{(k)}$ by adding a worker to station $h$: $\\tilde{x}_{h}=x_{h}^{(k)}+1$ and $\\tilde{x}_{i}=x_{i}^{(k)}$ for $i\\neq h$. Observe that $\\tilde{x}\\le x^{*}$. The greedy algorithm chose work station $j$ over work station $h$ at $x^{(k)}$, so it must be that\n$$\\Delta f_{j}(x_{j}^{(k)})\\ge\\Delta f_{h}(x_{h}^{(k)}). \\quad (1)$$\n\nFinally, let $\\hat{x}$ be the result of starting from optimal solution $x^{*}$ and shifting one worker from station $h$ to station $j$. Since\n$$x_{j}^{(k+1)}=x_{j}^{(k)}+1=x_{j}^{*}+1=\\hat{x}_{j},$$ $$x_{h}^{(k+1)}=x_{h}^{(k)}<x_{h}^{*}\\implies x_{h}^{(k+1)}\\le\\hat{x}_{h}$$\nand\n$$x_{i}^{(k+1)}=x_{i}^{(k)}\\le x_{i}^{*}=\\hat{x}_{i}\\,\\forall i\\notin\\{h,j\\},$$\nwe have $x^{(k+1)}\\le\\hat{x}$. Under the assumption that $x^{(k)}$ was the last solution in the greedy sequence that could be extended to an optimal solution, it must be that $\\hat{x}$ is not optimal. Thus the net change to the objective function at $x^{*}$ when shifting one worker from station $h$ to station $j$ must be negative, i.e.,\n$$\\Delta f_{j}(x_{j}^{*})+\\delta f_{h}(x_{h}^{*})<0.\\quad (2)$$\n\nWe showed previously that, under our assumptions, $x_{j}^{(k)}=x_{j}^{*}$, from which it follows that\n$$\\Delta f_{j}(x_{j}^{*})=\\Delta f_{j}(x_{j}^{(k)}). \\quad (3)$$\nWe also showed that $\\delta f_{h}()$ is an increasing function. Since $\\tilde{x}_{h}\\le x_{h}^{*}$,\n$$\\delta f_{h}(x_{h}^{*})\\ge\\delta f_{h}(\\tilde{x}_{h})=-\\Delta f_{h}(x_{h}^{(k)}). \\quad (4)$$\nCombining (4) with (2), we have\n$$\\Delta f_{j}(x_{j}^{*})-\\Delta f_{h}(x_{h}^{(k)})<0,$$\ni.e.,\n$$\\Delta f_{j}(x_{j}^{*})<\\Delta f_{h}(x_{h}^{(k)}). \\quad (5)$$\n\nCombining (3) with (5) yields\n$$\\Delta f_{j}(x_{j}^{(k)})<\\Delta f_{h}(x_{h}^{(k)})$$", "date": "2020-02-22 07:02:48", "meta": {"domain": "blogspot.com", "url": "https://orinanobworld.blogspot.com/2020/", "openwebmath_score": 0.6736919283866882, "openwebmath_perplexity": 405.9040934732411, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9926541727735793, "lm_q2_score": 0.8918110396870287, "lm_q1q2_score": 0.8852599498708731}}
{"url": "http://mathhelpforum.com/algebra/109037-determining-formula-sequence.html", "text": "# Math Help - Determining the formula of this sequence?\n\n1. ## Determining the formula of this sequence?\n\nI was given the sequence\n0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...\nand I'm looking for a method to determine an explicit formula for this.\nI've tried doing difference columns, and I always end up with 2^n being the 2nd difference.\n\nAnyone know a way to figure this out?\n\n2. Originally Posted by paupsers\nI was given the sequence\n0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...\nand I'm looking for a method to determine an explicit formula for this.\nI've tried doing difference columns, and I always end up with 2^n being the 2nd difference.\n\nAnyone know a way to figure this out?\nGood idea to look at differences. So you found:\n\n$(u_{n+2}-u_{n+1})-(u_{n+1}-u_n)=2^n$,\n\nassuming the first term is $u_1=0$.\n\nIf you sum the above equalities for $n$ ranging from 1 to $n$, you get a telescoping sum on the left, and a geometric sum on the right, which leads to:\n\n$(u_{n+2}-u_{n+1})-(u_2-u_1)=2+2^2+\\cdots +2^n = 2^{n+1}-2$,\n\nhence\n\n$u_{n+2}-u_{n+1}=2^{n+1}-1$.\n\nSumming again this latter equality we have again a telescoping sum on the left and a geometric sum (minus constant terms) on the right, and we get:\n\n$u_{n+2}-u_2=(2^2-1)+\\cdots+(2^{n+1}-1)$,\n\nhence (change of variable)\n\n$u_n = (2^2-1)+\\cdots +(2^{n-1}-1)+1$\n\nand finally $u_n=2^n-n-1$. Same method works in many situations.\n\n3. Originally Posted by paupsers\nI was given the sequence\n0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...\nand I'm looking for a method to determine an explicit formula for this.\nI've tried doing difference columns, and I always end up with 2^n being the 2nd difference.\n\nAnyone know a way to figure this out?\nThe Encyclopaedia of Integer Sequences gives: $2^n-n-1$\n\nCB\n\n4. Hello, paupsers!\n\nGiven the sequence: . $0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, \\hdots$\n\nI'm looking for a method to determine an explicit formula for this.\nI've tried doing difference columns, and I always end up with 2^n being the 2nd difference.\nThis is true . . . and should give you a big hint.\nConsider the general form: . $f(n) = 2^n$\n\nHow does it compare with the given sequence?\n\n$\\begin{array}{c|cccccccccc}\\hline\nn & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\\ \\hline\n\\text{Sequence:} & 0 & 1 & 4 & 11 & 26 & 57 & 120 & 247 & 502 & 1013 \\\\ \\hline\nf(n) = 2^n & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\\\ \\hline\n\\text{Error:} & +2 & +3 & +4 & +5 & +6 & +7 & +8 & +9 & +10 & +11\\\\ \\hline\n\\end{array}$\n\nIf we use $f(n) = 2^n$, each term is too large by $(n+1)$\n\nTherefore, the explicit formula is: . $F(n) \\;=\\;2^n - (n+1)$", "date": "2015-11-26 06:08:17", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/109037-determining-formula-sequence.html", "openwebmath_score": 0.7227979898452759, "openwebmath_perplexity": 354.16445756086546, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474872757474, "lm_q2_score": 0.8947894569842487, "lm_q1q2_score": 0.8852577009081969}}
{"url": "https://math.stackexchange.com/questions/1763978/can-you-use-both-sides-of-an-equation-to-prove-equality", "text": "# Can you use both sides of an equation to prove equality?\n\nFor example:\n\n$\\color{red}{\\text{Show that}}$$\\color{red}{\\frac{4\\cos(2x)}{1+\\cos(2x)}=4-2\\sec^2(x)}$$ In high school my maths teacher told me To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. So starting from the LHS: $$\\frac{4\\cos(2x)}{1+\\cos(2x)}=\\frac{4(2\\cos^2(x)-1)}{2\\cos^2(x)}=\\frac{2(2\\cos^2(x)-1)}{\\cos^2(x)}=\\frac{4\\cos^2(x)-2}{\\cos^2(x)}=4-2\\sec^2(x)$$$\\large\\fbox{}$At University, my Maths Analysis teacher tells me To prove a statement is true, you must not use what you are trying to prove. So using the same example as before: LHS = $$\\frac{4\\cos(2x)}{1+\\cos(2x)}=\\frac{4(2\\cos^2(x)-1)}{2\\cos^2(x)}=\\frac{2(2\\cos^2(x)-1)}{\\cos^2(x)}=\\frac{2\\Big(2\\cos^2(x)-\\left[\\sin^2(x)+\\cos^2(x)\\right]\\Big)}{\\cos^2(x)}=\\frac{2(\\cos^2(x)-\\sin^2(x))}{\\cos^2(x)}=\\bbox[yellow]{2-2\\tan^2(x)}$$ RHS =$$4-2\\sec^2(x)=4-2(1+\\tan^2(x))=\\bbox[yellow]{2-2\\tan^2(x)}$$ So I have shown that the two sides of the equality in$\\color{red}{\\rm{red}}$are equal to the same highlighted expression. But is this a sufficient proof? Since I used both sides of the equality (which is effectively; using what I was trying to prove) to show that $$\\color{red}{\\frac{4\\cos(2x)}{1+\\cos(2x)}=4-2\\sec^2(x)}$$ One of the reasons why I am asking this question is because I have a bounty question which is suffering from the exact same issue that this post is about. ## EDIT: Comments and answers below seem to indicate that you can use both sides to prove equality. So does this mean that my high school maths teacher was wrong? $$\\bbox[#AFF]{\\text{Suppose we have an identity instead of an equality:}}$$ $$\\bbox[#AFF]{\\text{Is it plausible to manipulate both sides of an identity to prove the identity holds?}}$$ Thank you. \u2022 You did not use both side of the equality. You showed that each is equal to the same thing. That's ok. Using what you want to prove would be using the fact that both sides are equal, but you don't. \u2013 Captain Lama Apr 29 '16 at 10:59 \u2022 You can simplify both sides separately to get get to a common point \u2013 Archis Welankar Apr 29 '16 at 10:59 \u2022 Just another comment: proving that$a=b$is the same as proving that$a-b=0$. So there is no meaningful difference between \"manipulating one side\" and \"manipulating both sides\". \u2013 Nefertiti Apr 29 '16 at 11:10 \u2022 This has already been answered but I'd add that you are really using the fact that$=$sign is an equivalence relation and hence transitive. This might be an issue with more complicated proof whereby the relation is not transitive. \u2013 Karl Apr 29 '16 at 12:14 \u2022 You did not \"use\" either side. What does \"use\" mean here? It means making an assertion, stating a sentence. The sides of the Eq'n are not sentences. If you take either side's formula and show that it is equal to some other formula, without any unsupported or unwarranted assumptions, then you are logical. E.g.: The RHS is always equal to$4-4/(2 \\cos^2 x)=4-4/(1+\\cos 2 x)=4\\cos 2 x/(1+\\cos 2 x)$regardless of what is written on the LHS. \u2013 DanielWainfleet Apr 29 '16 at 14:07 ## 7 Answers There's no conflict between your high school teacher's advice To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. and your professor's To prove a statement is true, you must not use what you are trying to prove. As in Siddarth Venu's answer, if you prove$a = c$and$b = c$(\"working from both sides\"), then$a = c = bby transitivity of equality. This conforms to both your teacher's and professor's advice. Both your high school teacher and university professor are steering you away from \"two-column proofs\" of the type: \\begin{align*} -1 &= 1 &&\\text{To be shown;} \\\\ (-1)^{2} &= (1)^{2} && \\text{Square both sides;} \\\\ 1 &= 1 && \\text{True statement. Therefore-1 = 1.} \\end{align*} Here, you assume what you want to prove, deduce a true statement, and assert that the original assumption was true. This is bad logic for at least two glaring reasons: 1. If you assume-1 = 1$, there's no need to prove$-1 = 1$. 2. Logically, if$P$denotes the statement \"$-1 = 1$\" and$Q$denotes \"$1 = 1$\", the preceding argument shows \"$P$implies$Q$and$Q$is true\", which does not eliminate the possibility \"$P$is false\". What you can do logically is start (\"provisionally\", on scratch paper) with the statement$P$you're trying to prove and perform logically reversible operations on both sides until you reach a true statement$Q$. A proof can then be constructed by starting from$Q$and working backward until you reach$P$. Often times, the backward argument can be formulated as a sequence of equalities, conforming to your teacher's advice. (Note that in the initial phase of seeking a proof, you aren't bound by anything: You can make inspired guesses, additional assumptions, and the like. Only when you write up a final proof must you be careful to assume no more than is given, and to make logically-valid deductions.) \u2022 Suppose you start with$1=1$, take the square root of both sides except use$-1$on the left and$1$on the right, and come up with$-1=1$. It seems like there is more going on here than assuming what you want to prove. \u2013 Frank Hubeny Apr 29 '16 at 13:16 \u2022 @FrankHubeny To the best of my knowledge, you've just equated the two branches of a multifunction restricted to the real axis - which isn't valid. \u2013 QuantumFool Apr 29 '16 at 15:43 \u2022 @QuantumFool That's right. The problem is not with assuming what one has to prove but with invalid steps along the way. \u2013 Frank Hubeny Apr 29 '16 at 18:44 \u2022 @Frank: You're perfectly correct that assuming the conclusion is not the only fatal error in this argument; as noted, proving the converse is another. The (logically valid!) argument for \"$-1 = 1$implies$1 = 1$\" does not consist entirely of reversible steps, so (as you note) does not prove that$-1 = 1$. I mentioned this example to illustrate what the OP has been cautioned against, and because a non-negligible fraction of American university students instinctively attempt to prove algebraic identities by starting with the desired conclusion and manipulating until they obtain a tautology. \u2013 Andrew D. Hwang Apr 29 '16 at 20:55 \u2022 @BLAZE: An \"identity\" (such as$\\cos^{2} x + \\sin^{2} x = 1$) is just an equation that holds for all values of one or more variables (possibly with a \"small number of exceptions or restrictions\"), so \"yes\", any proof technique that works for numerical equations also works for identities. \u2013 Andrew D. Hwang May 1 '16 at 0:04 It is enough.. Consider this example: To prove:$a=b$Proof: $$a=c$$ $$b=c$$ Since$a$and$b$are equal to the same thing,$a=b$. That is the exact technique you are using and it sure can be used. \u2022 I think we also have to consider the domain of each side and make this explicit up front. For example, let$a=(x-1)/(x-1)$and$b=1$. Then we algebraically manipulate this and get$a = 1$. If we do not consider the constraint that$x$is not equal to$1$in the domain of$a$, we get$a = b = 1$for all$x$. That is incorrect. \u2013 Frank Hubeny Apr 29 '16 at 15:12 To prove a statement is true, you must not use what you are trying to prove. The main problem is that you've misunderstood what the second teacher said (and possibly the meaning of the equals sign). What that second teacher is saying is do not take as prior facts what you're trying to prove. As one textbook puts it (Setek and Gallo, Fundamentals of Mathematics, 10th Edition, Sec. 3.8), \"An argument, or proof, consists basically of two parts: the given statements, which are called premises, and the conclusion\". Wikipedia says this (sourced from Cupillari, Antonella. The Nuts and Bolts of Proofs. Academic Press, 2001. Page 3.): \"In mathematics and logic, a direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions.\" Now consider a simple equation/identity like$6 = 2 \\times 3$. The separate sides are expressions; if you just said \"6\" in English that's a sentence fragment, not an assertion of any fact. It can't be evaluated as either \"true\" or \"false\", because it has no assertive content. It cannot be used as a premise because it's not a proposition. What makes something a fully-formed statement in mathematical language is a relation, most commonly equals (but alternatively \"is lesser than\", \"is greater than\", etc., effectively the verbs of the language). Translating the equation$6 = 2 \\times 3to English we get \"6 is the same as 2 times 3\", which is indeed a full sentence. This can be checked as being true or false; it makes an assertion. It can be used as a premise because it is a proposition of a particular fact. In conclusion, both your teachers are correct, and both of your proofs are correct (although most of us would prefer the more concise one). When one says \"don't use what you're trying to prove\" they're not talking about the appearance of any particular expression in an algebraic transformation; expressions are neither premises nor things that can be proven; they are sentence fragments. They're talking about an assertion of fact, which in math has to be a statement including a relational symbol (most commonly an equation). The fact that you didn't start by assuming that equality means that in both cases you've complied with your second teacher's warning. short answer: equality is symmetric, implication is not (both are however transitive) longer answer: \u2022 You are right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to conclude that A = B (because \"=\" is transitive and symmetric) \u2022 Your teacher is right: if you prove that something true follows from A = B, i.e. A = B => true, you are not allowed to conclude the converse i.e. that A = B since \"true is true\" (because implication is not symmetric) I agree with @Siddharth Venu. If we have to prove a=b, At times, on proceeding from LHS you may end up in a stage(intermediate step) which you cannot solve any further and you continue to solve RHS to arrive at the same stage i.e, a=c and b=c so you can conclude that a=b these kind of equality problems often comes in trigonometry Many chapters like matrices and our normal algaebra always have a final step where you can directly establish a relation (a=b). There are two common errors that these methods are preventing: \\begin{align} 1 &< 2 &&\\text{True.} \\\\ 0 \\cdot 1 &< 0 \\cdot 2 &&\\text{Um...} \\\\ 0 &< 0 &&\\text{False.} \\end{align} \\begin{align} 1 &= 2 &&\\text{False.} \\\\ 0 \\cdot 1 &= 0 \\cdot 2 &&\\text{Um...} \\\\ 0 &= 0 &&\\text{True.} \\end{align} Much confusion ensues when the two \"0$\"s in the middle steps are obscured by being large, complicated expressions. For instance, is it evident that you are multiplying by zero when you multiply by$\\sin(2x) - 2\\sin(x)\\cos(x)$? (This uses the double angle formula for sine to get an expression that is always zero.) The correct form of inference when you multiply or divide by something complicated is \"$A = B$\" becomes \"$A/C = B/C$or$C = 0$\". That is, you got what you expected or you have inadvertently done something crazy. It seems to take a lot of practice to remember that second clause. The example with$A$,$B$, and$C$can be altered somewhat without changing the need for the second clause. You can also multiply both sides by$C$. The relation need not be an equality. Note however, that the sense of an inequality may change if$C$is ever negative. Unfortunately your high-school teacher (and some of the other answers) is wrong. It is false that proving something of the form \"$A = B$\" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side, not to say necessary. ## Not necessary Let us first deal with the false misconception that you can only go in one direction. Suppose you have proven the following, where$A,B,C$are any expressions:$A = C$.$B = C$. Then you can use the second sentence to substitute$C$for$B$in the first to obtain:$A = B$. This is logically valid because \"$=$\" means \"is exactly the same as\". ## Not always possible Let us now consider an example where it is simply impossible to manipulate from one side to the other to prove an equality!$\\def\\zz{\\mathbb{Z}}\\def\\qq{\\mathbb{Q}}\\def\\rr{\\mathbb{R}}$Theorem: Take any$x \\in \\rr$such that$x^2 \\le 0$. Then$x = 0$. Proof:$x = 0$or$x > 0$or$x < 0$. [by trichotomy] If$x > 0$:$x^2 = x \\times x > 0$. [by positive multiplication] This contradicts$x^2 \\le 0$. If$x < 0$:$0 < -x$. [by subtraction]$x^2 = (-x) \\times (-x) > 0$. [by positive multiplication] This contradicts$x^2 \\le 0$. Therefore$x = 0$. Here \"positive multiplication\" denotes \"multiplying by a positive real\", which preserves the inequality sign. Similarly subtraction preserves an inequality. The above theorem cannot be proven directly by algebraic manipulation from one side \"$x$\" to the other \"$0$\", simply because the only given condition is an inequality. Similarly the construction of$\\sqrt{2}$in elementary real analysis, as shown below, does not permit a proof by algebraic manipulation. Theorem: Let$S = \\{ r : r \\in \\qq_{\\ge 0} \\land r^2 \\le 2 \\}$. Then$S$is non-empty and has an upper bound in$\\rr$. Let$x = \\sup_\\rr(S)$. Then$x^2 = 2$. Proof: [Exercise! Or see a good textbook like Spivak's.] ## General identities To address the new sub-question in blue, it suffices to generalize the above theorem to the cube-root of arbitrary real numbers, namely:$\\sup( \\{ r : r \\in \\qq \\land r^3 \\le x \\} )^3 = x$for any$x \\in \\rr$. This is an identity but cannot be proven by direct manipulation. You may not be satisfied with this counter-example, but in higher mathematics this kind of identity is in fact the usual kind that we are interested in, rather than identities that can be proven by algebraic manipulation (which are usually considered trivial). Here are some more examples:$\\lceil \\frac{m}{n} \\rceil = \\lfloor \\frac{m-1}{n}+1 \\rfloor$for any$m \\in \\zz$and$n \\in \\zz_{>0}$.$\\sum_{k=0}^{n-1} 2^k = 2^n-1$for any$n \\in \\zz_{>0}$. Nevertheless, if you desire an arithmetic identity, the question becomes much more interesting and depends heavily on what you mean by \"arithmetic\" and what axioms you are allowed to use. Tarski's problem asked whether an arithmetic identity concerning positive integers can be proven using only the basic high-school identities about them, and Wilkie gave an explicit and simple identity that cannot be so proven, basically because any proof needs to use subtraction and hence negative integers. \u2022 It seems clear to me that the OP's teacher's advice, \"To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side.\", was given in a particular context (avoiding \"two-column proofs\"), not asserted as the only way of establishing an equality. After re-reading every answer here, I can't find anyone claiming (explicitly or implicitly) that \"proving something of the form \"$A=B\\$\" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side [...].\" \u2013\u00a0Andrew D. Hwang Apr 30 '16 at 13:31\n\u2022 @AndrewD.Hwang: I know what you're saying, but let me explain why I make that comment. Imagine a student who is not good at mathematics whose teacher tells him/her exactly the words cited in the question. It's easy to see that the student will go away with the wrong impression as I specified. Indeed, standard English only allows it to be interpreted that way, because \"To do X, you do Y.\" only means \"In order to do X, you { have to / ought to / should / better } do Y.\" Furthermore, I have seen so many students with exactly that wrong conception. Any good answer ought to correct that. \u2013\u00a0user21820 Apr 30 '16 at 14:29\n\u2022 @BLAZE: I didn't want to criticize the other answers too much, but actually Andrew's argument about avoiding two column proofs is in fact not so good because that is how we can justify a formal proof. If you learn natural deduction (especially Fitch-style, which can be extended to quantifiers like at math.stackexchange.com/a/1684204), you will understand what I mean. The issue in the 'proof' Andrew shows is not its two-column nature but because no axiom allows writing the first statement, and in some sense it doesn't address the actual issue of your high-school teacher's teaching. \u2013\u00a0user21820 May 1 '16 at 4:17\n\u2022 @BLAZE: To be precise, it is in my opinion pointless to steer a student away from an incorrect proof form by saying things that aren't correct, and furthermore without even explaining what form is incorrect! Your university teacher at least said the right thing, though it is not well explained enough to dismantle the prior misconceptions of many students that they gain from high-school! \u2013\u00a0user21820 May 1 '16 at 4:20", "date": "2019-07-19 01:57:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1763978/can-you-use-both-sides-of-an-equation-to-prove-equality", "openwebmath_score": 0.8332354426383972, "openwebmath_perplexity": 481.2578683796757, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765557865637, "lm_q2_score": 0.903294209307224, "lm_q1q2_score": 0.8852071480988408}}
{"url": "https://math.stackexchange.com/questions/1436296/is-the-probability-of-picking-an-ace-as-the-second-card-of-a-deck-the-same-if-th", "text": "# Is the probability of picking an Ace as the second card of a deck the same if the deck is shuffled after the first card?\n\nHere's the situation:\n\nYou have a deck of 52 cards that is already shuffled. You pick the first card, and the first card is not an Ace. Is the probability of drawing an Ace as second card the same if\n\n1. The second card is immediately drawn from the deck\n2. The remaining deck is first shuffled without adding the first card back in, and then the second card is drawn.\n\nFor me, it is obvious that probability of drawing an Ace in the second case is 4/51, but I'm not entirely sure that the probability is the same in the first case.\n\nAlso, what would be the probability of the first case if we have drawn n cards and none of these are aces?\n\n\u2022 yes, its the same. \u2013\u00a0supinf Sep 15 '15 at 11:54\n\u2022 4/(52-n).  \u2013\u00a0Did Sep 15 '15 at 11:56\n\u2022 As an aside, this is also the reason why it is pointless to doubly randomise a choice (e.g. tossing a coin to see who goes first for the real coin toss!) \u2013\u00a0Oscar Bravo Sep 15 '15 at 14:04\n\u2022 Another aside... if you and a friend each draw a card, and then your friend reveals he didn't draw an ace... your odds of getting an ace are better if you replace your card with a new one from the 50 cards remaining. \u2013\u00a0kbelder Sep 15 '15 at 23:33\n\nIf it's not intuitively clear, you can check your calculation using contitional probability:\n\n\u2022 $X$... the event \"the second card is an ace\"\n\u2022 $Y$... the event \"the first card is not an ace\"\n\nThen, $P(X\\land Y) = \\frac{48\\cdot 4}{52\\cdot 51}$ and $P(Y) = \\frac{48}{52}$\n\nThen $$P(X|Y) = \\frac{P(X\\land Y)}{P(Y)} = \\frac{\\frac{4\\cdot 48}{52\\cdot 51}}{\\frac{48}{52}} = \\frac{4}{51}$$\n\nGiven you know the first card of your shuffled deck, the remaining $51!$ ways of arranging the remaining $51$ cards are equally likely. For each of the $52$ possible values of the first card you have $51!$ equally likely arrangements of the remaining $51$ cards. This is consistent with there being $52\\cdot 51!=52!$ arrangements of the whole deck to begin with.\n\nYou can conclude that, after drawing the first card, the remaining $51$ cards are equally likely to be in any shuffled order, and so shuffling them changes essentially nothing. This argument should lead to the conclusion that the probability is the same in either case.\n\nYes, the probability of drawing an Ace is the same if you shuffled the rest of the deck or didn't. You are correct that it's 4/51. If you drew n cards and none were Aces, then the probability of drawing an Ace next would be 4/(52-n).\n\nIn both cases, there are only 51 cards left, thus $Pr = \\dfrac{4}{51}$\n\nAnd if $n$ non-aces have been drawn, $Pr = \\dfrac{4}{52-n}$", "date": "2019-06-17 02:33:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1436296/is-the-probability-of-picking-an-ace-as-the-second-card-of-a-deck-the-same-if-th", "openwebmath_score": 0.7584448456764221, "openwebmath_perplexity": 192.16351161095074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409524, "lm_q2_score": 0.9032941969413321, "lm_q1q2_score": 0.8852071375652857}}
{"url": "https://math.stackexchange.com/questions/1845889/integrals-over-subset-of-measure-space", "text": "# Integrals over subset of measure space\n\nLet $(X, \\mathcal{M}, \\mu)$ be a measure space. Suppose $E \\in \\mathcal{M}$ and $f \\in L^+$ where $L^+$ is a space of measurable functions from $X$ to $[0, \\infty]$. $\\int_E f$ is defined by $\\int_X f\\chi_E$ where $\\chi_E$ is a characteristic function of $E$. Now every time we want to use some property that is true for integrals over whole space $X$, we have to do manipulations with $\\chi_E$. For example, let $f_n$ be a sequence in $L^+$ such that $f_n(x) \\nearrow f(x)$ for all $x \\in E$. Suppose we know that monotone convergence theorem is true for integrals over $X$ and we want to prove $$\\int_E f = \\lim_{n\\to\\infty} \\int_E f_n.$$ Since $f_n\\chi_E \\nearrow f\\chi_E$ we have $$\\int_E f = \\int_X f\\chi_E = \\lim_{n\\to\\infty} \\int_X f_n\\chi_E = \\lim_{n\\to\\infty} \\int_E f_n.$$ I know it's easy but is there a way to avoid such manipulations? I want to know that equality above and many other statements about integrals are true because $E$ is a measure space in its own right (seems much more natural).\n\nMore precisely, for $E \\in \\mathcal{M}$ define $$\\mathcal{M}_E = \\{ E \\cap F \\mid F \\in \\mathcal{M} \\}.$$ It's easy to check that $\\mathcal{M}_E$ is a $\\sigma$-algebra. $(E, \\mathcal{M}_E, \\mu|_{\\mathcal{M}_E})$ is a measure space and $\\int_E f$ in space $(X, \\mathcal{M}, \\mu)$ is equal to $\\int_E f|_E$ in space $(E, \\mathcal{M}_E, \\mu|_{\\mathcal{M}_E})$ for every measurable $f$.\n\nIs this a commonly accepted way of handling integrals over subset of measure space? If not, do I really have to use $\\chi_E$ every time or there is some other way?\n\n\u2022 You are just using the measure $\\mu|_E$, which leads to the same things. As long as you know this measure carries on the properties you want, you're fine. \u2013\u00a0Silvia Ghinassi Jul 1 '16 at 14:33\n\u2022 @SilviaGhinassi intuitively it's very simple, but could you be more precise? \u2013\u00a0edubrovskiy Jul 1 '16 at 14:56\n\u2022 I agree with you, and that's why I am afraid of getting into technicalities. I might easily say something wrong, so I'll leave it to someone else to provide you a good answer. \u2013\u00a0Silvia Ghinassi Jul 1 '16 at 15:01\n\nThere are basically two interesting types of functions\n\n$$f:X \\to [0,\\infty] , \\mu-\\text{ measurable}$$\n\nand\n\n$$f : X \\to [- \\infty, \\infty] \\text{ which is integrable, i.e. } \\int f(x) \\mu(dx) < \\infty .$$\n\nThese functions are interesting, because you can define for both the integral $\\int_X f(x) \\mu(dx)$, but maybe the integral is infinity in the first case.\n\nMany theorems are true for either non-negative measurable functions (e.g. monotonic convergence theorem) or integrable functions (dominant convergence theorem).\n\nNow if you have proven the monotonic convergece theorem for a general measurable space, then you don't need to prove it for integrals restricted on a set, simply because $(E, \\mathcal{M}_E, \\mu|_{\\mathcal{M}_E})$ is again a measure space, and thus the monotonic convergence theorem is also true here. As well as all other properties and theorems that are known for integrals on a measurable space.\n\nSo in your case,if you have non-negative functions $f_n(x) \\nearrow f(x)$ for all $x\\in E$ you immediately get\n\n$$\\int f(x) \\mu|_{\\mathcal{M}_E}(dx) = \\lim_{n\\to \\infty} \\int f_n(x) \\mu|_{\\mathcal{M}_E}(dx)$$\n\nHowever, to be sure that the theorem really holds for all functions restricted to $E$ you need to understand the relation between integrable functions of $\\mu_{\\mathcal{M}_E}$ and functions $f$ restricted from $X$ to $E$. The relation is as follows:\n\nLet $f:X \\to [0,\\infty]$ be $\\mu$-measurable, or let $f:X \\to [-\\infty,\\infty]$ be $\\mu$ integrable. Define for some set $E \\in \\mathcal{M}$ the function $f':A \\to [-\\infty, \\infty]$ with $f'(x) := f(x)$. Then, we have\n\n$$\\int f'(x) \\mu_{\\mathcal{M}_E}(dx) = \\int_A f(x) \\mu(dx)$$\n\n\u2022 Could you give me a link to some book or paper where this approach is used? I've already checked relation between integrals on $X$ and on $E$. I just wanted to know that it's a commonly accepted approach. \u2013\u00a0edubrovskiy Jul 2 '16 at 7:06\n\u2022 @edubrovskiy I can recommend the book **Measure and Integration Theory ** from Heinz Bauer. Its an old, solid and my personal favorite book. The relation between integrals on $X$ and integrals on $E$ is discussed at the end of \u00a712. \u2013\u00a0Adam Jul 2 '16 at 8:53", "date": "2019-05-27 07:05:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1845889/integrals-over-subset-of-measure-space", "openwebmath_score": 0.9362642765045166, "openwebmath_perplexity": 160.88799488823398, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534316905261, "lm_q2_score": 0.9019206811430763, "lm_q1q2_score": 0.885193147620529}}
{"url": "http://math.stackexchange.com/questions/429842/how-many-triangles-in-picture", "text": "How many triangles in picture\n\nHow many triangles in this picture:\n\n-\nI only count 40...and unless someone can explain me how this is a mathematics question I think I'm going to vote to close it as off-topic \u2013\u00a0 DonAntonio Jun 26 '13 at 9:16\nI do not think one needs mathematics to solve the puzzle. \u2013\u00a0 Avitus Jun 26 '13 at 9:18\n@DonAntonio: A systematic approach of the kind that\u2019s second nature to most mathematicians works quite nicely and can even be generalized to larger diagrams of the same type. It\u2019s certainly on-topic. \u2013\u00a0 Brian M. Scott Jun 26 '13 at 9:20\nThat sounds sound, Brian. Not closing, then. Thanks. \u2013\u00a0 DonAntonio Jun 26 '13 at 9:23\n\nThey can be counted quite easily by systematic brute force.\n\nAll of the triangles are isosceles right triangles; I\u2019ll call the vertex opposite the hypotenuse the peak of the triangle. There are two kinds of triangles:\n\n1. triangles whose hypotenuse lies along one side of the square;\n2. triangles whose legs both lie along sides of the square and whose peaks are at the corners of the square.\n\nThe triangles of the second type are easy to count: each corner is the peak of $4$ triangles, so there are $4\\cdot4=16$ such triangles.\n\nThe triangles of the first type are almost as easy to count. I\u2019ll count those whose hypotenuses lie along the bottom edge of the square and then multiply that by $4$. Such a triangle must have a hypotenuse of length $1,2,3$, or $4$. There $4$ with hypotenuse of length $1$, $3$ with hypotenuse of length $2$, $2$ with hypotenuse of length $3$, and one with hypotenuse of length $4$, for a total of $10$ triangles whose hyponenuses lie along the base of the square. Multiply by $4$ to account for all $4$ sides, and you get $40$ triangles of the second type and $40+16=56$ triangles altogether.\n\nAdded: This approach generalizes quite nicely to larger squares: the corresponding diagram with a square of side $n$ will have $4n$ triangles of the first type and $$4\\sum_{k=1}^nk=4\\cdot\\frac12n(n+1)=2n(n+1)$$\n\nof the second type, for a total of $2n^2+6n=2n(n+3)$ triangles.\n\n-\n\nI know there's already an excellent answer from Brian M. Scott... Buuuuut...\n\nHere's a visual supplement to the aforementioned systematic brute force.\n\n-\nPretty! $\\quad$ \u2013\u00a0 Brian M. Scott Jun 27 '13 at 9:23", "date": "2014-10-21 07:21:38", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/429842/how-many-triangles-in-picture", "openwebmath_score": 0.834801971912384, "openwebmath_perplexity": 397.4700085710202, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307676766118, "lm_q2_score": 0.90990700787036, "lm_q1q2_score": 0.8851855329808512}}
{"url": "https://math.stackexchange.com/questions/1063724/convergence-of-sum-n-1-infty-left-frac1n-frac1n-2-righ", "text": "# Convergence of $\\sum_{n=1}^{\\infty}\\left(\\, \\frac{1}{n} - \\frac{1}{n + 2}\\,\\right)$\n\nWhat criteria can I use to prove the convergence of $$\\sum_{n=1}^{\\infty}\\left(\\,{1 \\over n} - {1 \\over n + 2}\\,\\right)\\ {\\large ?}$$\n\nMy idea was to use ratio test: $$\\displaystyle{1 \\over n} - {1 \\over n+2} = {2 \\over n^{2} + 2n}$$ $$\\displaystyle\\frac{2}{\\left(\\, n + 1\\,\\right)^{2} + 2\\left(\\, n + 1\\,\\right)} \\frac{n^{2} + 2n}{2} = \\frac{n^{2} + 2n}{n^{2} + 4n + 3}$$\n\nOf course $\\displaystyle n^{2} + 2n \\lt n^{2} + 4n + 3$ for all $\\displaystyle n$ , but $$\\displaystyle\\lim \\limits_{n \\to \\infty} \\frac{n^{2} + 2n}{n^{2} + 4n + 3}=1$$ so I am not quite sure if I can apply ratio test.\n\nYou have $\\frac{2}{n^2+n} \\le \\frac{2}{n^2}$. Thus you can use the direct comparision test to prove the convergence (if you already have proven in your course that $\\sum_{n=1}^\\infty \\frac 1{n^2}$ and thus also $\\sum_{n=1}^\\infty \\frac 2{n^2}$ converges).\n\nAnswer to your 2nd question: You never can apply the ratio test if the limit is 1 (as in this case).\n\nI would write out the first few terms and see how they cancel.\n\n\u2022 of course; but to do this I have to prove absolute convergence first \u2013\u00a0Christian Dec 11 '14 at 18:42\n\u2022 @Christian No, you do not. \u2013\u00a0Andr\u00e9s E. Caicedo Dec 11 '14 at 18:45\n\u2022 @Christian You can look at the partial sums and see how they cancel without worrying about absolute convergence. Your problem would be with $\\sum \\limits_{n=1}^{\\infty} \\frac1n - \\sum \\limits_{m=1}^{\\infty}\\frac{1}{m+2}$ \u2013\u00a0Henry Dec 11 '14 at 18:47\n\u2022 It might be useful to mention Telescoping Series. \u2013\u00a0robjohn Dec 16 '14 at 13:04\n\u2022 The reason that doing this works is that it allows you to calculate the partial sums directly, which you can use to show convergence of the series by definition (i.e. showing that the partial sums converge) \u2013\u00a0Joshua Mundinger Dec 16 '14 at 16:44\n\nHow about computing the partial sums? For any $N>2$ we have:\n\n$$\\sum_{n=1}^{N}\\left(\\frac{1}{n}-\\frac{1}{n+2}\\right) = \\frac{3}{2}-\\frac{1}{N+1}-\\frac{1}{N+2}$$ hence: $$\\left|\\frac{3}{2}-\\sum_{n=1}^{N}\\left(\\frac{1}{n}-\\frac{1}{n+2}\\right)\\right|\\leq\\frac{2}{N}$$ ensures convergence (towards $\\frac{3}{2}$).\n\nHere's another way to prove the convergence $$\\sum \\limits_{n=1}^{\\infty} \\left(\\frac1n - \\frac{1}{n+2}\\right)= \\sum \\limits_{n=1}^{\\infty} \\frac{n+2-n}{n(n+2)}$$ $$= \\sum \\limits_{n=1}^{\\infty} \\frac{2}{n^2+2n}=2 \\sum \\limits_{n=1}^{\\infty} \\frac{1}{n^2+2n}$$ Also note that $$\\left|\\frac{1}{n^2+2n}\\right|\\leq \\left|\\frac{1}{n^2}\\right|$$ And by the p-series test, we have $$\\sum \\limits_{n=1}^{\\infty} \\left|\\frac{1}{n^2}\\right| \\Rightarrow \\mbox{converges}$$ Which implies that $$\\sum \\limits_{n=1}^{\\infty} \\frac{1}{n^2} \\Rightarrow \\mbox{converges absolutely}$$ Therefore, by the direct comparison test $$2 \\sum \\limits_{n=1}^{\\infty} \\frac{1}{n^2+2n} \\Rightarrow \\mbox{converges absolutely}$$ Absolute convergence implies convergence. Also a convergent series multiplied by $2$ is still a convergent series.\n\n\u2022 I hope the downvoter revisits this answer. I don't think it deserves a downvote, especially after the last edit. (+1) \u2013\u00a0robjohn Dec 16 '14 at 13:02\n\nYou can use the limit comparison test. Let $a_{n} = \\frac{2}{n^{2}+2n}$ and consider $b_{n} = \\frac{1}{n^{2}}$ then\n\n$$\\lim_{n\\rightarrow\\infty} \\frac{a_{n}}{b_{n}}= \\lim_{n\\rightarrow \\infty}\\frac{2n^{2}}{n^{2}+2n} = 2$$\n\nhence $\\sum a_{n}$ and $\\sum b_{n}$ converge or diverge together.. but $b_{n}$ is a convergent $p$-series.\n\nThat being said, my up vote is for Mark Bennet.\n\n\u2022 with $\\frac{1}{n^2}$ ? \u2013\u00a0Christian Dec 11 '14 at 18:42", "date": "2020-02-21 06:36:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1063724/convergence-of-sum-n-1-infty-left-frac1n-frac1n-2-righ", "openwebmath_score": 0.8874070644378662, "openwebmath_perplexity": 204.44323439052374, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226260757067, "lm_q2_score": 0.9059898273638386, "lm_q1q2_score": 0.8851725603288937}}
{"url": "https://math.stackexchange.com/questions/2339821/how-to-prove-that-a-sequence-diverges-using-the-cauchy-definition", "text": "# How to prove that a sequence diverges using the Cauchy definition?\n\nIt was a question of my integral calculus exam:\n\nWrite the Cauchy definition of $\\lim_{n\\rightarrow\\infty} a_n = L$\n\nSo, I literally wrote:\n\n\" $\\{a_n\\}$ is a Cauchy sequence if given $\\epsilon>0$ there exists $N \\in \\mathbb{N}$ such that $\\forall\\ m,n\\in\\mathbb{N}$ if $n,m\\ge N$, then $|a_n - a_m|<\\epsilon$, or with symbols:\n\n$\\forall\\epsilon>0:\\exists N\\in\\mathbb{N}:\\forall\\ m,n \\in\\mathbb{N}: m,n\\ge N\\implies |a_n - a_m|<\\epsilon$.\n\nAnd if $\\{a_n\\}$ is a Cauchy sequence, then $\\lim_{n\\rightarrow\\infty} a_n = L$ ( i.e. $\\{a_n\\}$ converges) \"\n\nThe following question was:\n\nWith this definition, find $\\lim_{n\\rightarrow\\infty} (3n+2)$\n\nAnd that is what I did:\n\nI proposed that $\\lim_{n\\rightarrow\\infty} (3n+2) = \\infty$ i.e. $\\{3n+2\\}$ diverges. So I think that I have to prove that this limit is equal to infinity using the Cauchy definition.\n\nBut if a sequence $\\{a_n\\}$ diverges $\\implies$ $\\{a_n\\}$ is not a Cauchy sequence.\n\nAnd $\\{a_n\\}$ is not a Cauchy sequence if\n\n$\\exists\\epsilon>0:\\forall N\\in\\mathbb{N}:\\exists\\ m,n \\in\\mathbb{N}:$ $m,n\\ge N$ and $|a_n - a_m|\\ge\\epsilon$ $(1)$\n\nSo, I must to prove $(1)$.\n\n$(1)$ says that there exists $m,n\\in\\mathbb{N}$ such that ..., so $(1)$ has not to apply $\\forall m,n\\in\\mathbb{N}$, and for this fact (I think) I can assign convinient values to $m,n$ in order to prove $(1)$:\n\nIf i choose $m=N+1,n=N$ then $|3(N+1)+2 -(3N+2)|=3>2$.\n\nSo exists $\\epsilon=2>0$ such that $\\forall N \\in\\mathbb{N}:$ there exists $m=N+1, n=N$ such that $N+1,N \\ge N$ and $|3(N+1)+2 -(3N+2)|=3>\\epsilon$.\n\nHence $\\{3n+2\\}$ is not a Cauchy sequence $\\implies$ $\\lim_{n\\rightarrow\\infty} (3n+2) = \\infty$.\n\nIs my proof correct? If it is not correct, could you help me proving that, please?\n\n\u2022 just reading quick, it looks perfectly good to me \u2013\u00a0Chessnerd321 Jun 28 '17 at 19:57\n\u2022 I will note that, at the end, we cannot just say that $$\\text{\\{a_n\\} not Cauchy\\implies \\lim_{n\\to\\infty}a_n=\\infty}$$We can only say that $$\\text{\\{a_n\\} not Cauchy\\implies \\lim_{n\\to\\infty}a_n\\notin\\Bbb R}$$The limit may be $\\pm\\infty$ or just non-existent. \u2013\u00a0Dave Jun 28 '17 at 20:00\n\u2022 But if i i want to prove that the limit is $\\infty$, what have i to do? \u2013\u00a0Sama Jun 28 '17 at 20:05\n\u2022 @KarenSM You have to show that for each $M >0$, there is a natural number $N$ such that for all $n \\geq N$, $a_n\\geq M$. Quite trivial, you can take $N=[M]$, where $[M]$ is the \u201cfloor\u201d of $M$, the unique integer satisfying $M-1< [M] \\leq M$. \u2013\u00a0Li Chun Min Jun 28 '17 at 23:53\n\u2022 @LiChunMin as far as I understand you, I cannot find the $\\lim_{n\\rightarrow\\infty} (3n+2)$ with the Cauchy definition, right? \u2013\u00a0Sama Jun 29 '17 at 0:26\n\nThe definition of a Cauchy sequence is:\n\n$\\forall \\epsilon > 0, \\exists N \\in \\mathbb{N}: \\forall m,n \\in \\mathbb{N}: m,n \\ge N \\implies |a_m-a_n| < \\epsilon$, and the negation of this definition is:\n\n$\\exists \\epsilon > 0, \\forall N \\in \\mathbb{N}, \\exists m,n \\in \\mathbb{N}: m,n \\ge N, |a_m - a_n| \\ge \\epsilon$.\n\nFor your sequence $a_n = 3n+2, n \\ge 1$, choose $\\epsilon = 1$ (your choice is $3$), for any $N \\in \\mathbb{N}$, let $n =N, m = N+1$. Observe $m, n \\ge N$, and $|a_m - a_n| = |a_{N+1} - a_{N}|= |3(N+1) - 2 - 3N+2|= 3 > 1= \\epsilon$. Thus the sequence above is not Cauchy.\n\n\u2022 But with $\\epsilon=2$ is still fine, right? I'm nervous because I put this in my exam \u2013\u00a0Sama Jun 28 '17 at 20:22\n\u2022 $\\epsilon = 2$ works just fine. $2$ is better than $1$... \u2013\u00a0DeepSea Jun 28 '17 at 20:44\n\nYou did not write the def'n of $\\lim_{n\\to \\infty}a_n=L.$ What you wrote in response to \"Define what it means for $(a_n)_n$ to converge to $L$\" was: \" $(a_n)_n$ is a Cauchy sequence (which means ...) which converges to $L$.\"\n\nThe def'n of $\\lim_{n\\to \\infty}a_n=L$ is $$\\forall r>0\\;\\exists m\\;\\forall n>m\\;(|L-a_n|<r).$$ I have omitted specifying that $m,n\\in \\mathbb N.$ It is conventional that in \"$\\lim_{n\\to \\infty}a_n$\", the values of $n$ are restricted to members of $\\mathbb N$ unless stated otherwise.", "date": "2019-07-18 02:07:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2339821/how-to-prove-that-a-sequence-diverges-using-the-cauchy-definition", "openwebmath_score": 0.9682406783103943, "openwebmath_perplexity": 145.87265261899606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226287518853, "lm_q2_score": 0.9059898165759307, "lm_q1q2_score": 0.8851725522134541}}
{"url": "https://math.stackexchange.com/questions/3372929/which-sets-of-sequence-is-countable-and-uncountable", "text": "# Which sets of sequence is countable and Uncountable.\n\nConsider the sequences\n\n$$\\displaystyle X=\\left\\{(x_n): x_n \\in \\left\\{0,1\\right\\},n \\in \\mathbb{N} \\right\\}$$ $$and$$ $$\\displaystyle Y=\\left\\{(x_n)\\in X:x_n=1 \\;\\;\\text{for at most finitely many n} \\right\\}$$\n\nI have to choose which is uncountable and which is countable.\n\nSolution i tried- Here $$X$$ is set of sequence with enteries from $$\\left\\{0,1\\right\\}$$ thus it has number of elements $$2^{\\aleph_0}$$ which is uncountable .\n\nNow The set $$Y$$ it has all the sequences from the set $$X$$ but some of its elements of sequences is replaced by the only '$$1$$' so its Cardinality will be less then $$2^{\\aleph_0}$$ ,but by $$\\textbf{ continuum hypothesis}$$ there is no set having Cardinality in-between the $${\\aleph_0}$$ and $$2^{\\aleph_0}$$ so the set $$Y$$ will be countable\n\nI write this proof but i don't even know this is correct or not but i am sure about set $$X$$ but not sure about $$Y$$ please help me with set $$Y$$\n\nThank you.\n\n\u2022 You call both sets $X$, and in the 'second' $X$ you refer to $X$. But that does not make sense. I suppose that the sequences in your 'second' set, are supposed to be sequences of the first set, so they have at most finitly many n, and can only take 0 or 1? \u2013\u00a0Cornman Sep 28 '19 at 8:54\n\u2022 The continuum hypothesis can not be proven, or disproven, so you should not use it. :) \u2013\u00a0Cornman Sep 28 '19 at 8:54\n\u2022 my bad i will edit it \u2013\u00a0gaurav saini Sep 28 '19 at 8:56\n\u2022 What is $x_n$ here? \u2013\u00a0orlp Sep 28 '19 at 9:01\n\u2022 Representation of a sequence @orlp \u2013\u00a0gaurav saini Sep 28 '19 at 9:04\n\nWe know that the countable union of countable sets is countable.\n\nWhich means the countable union of countable unions of countable sets is a countable set.\n\nAnd $$Y$$ is the countable union (indexed over how many $$1$$s the sequences have) of countable unions (indexed over how which positions the finite number of $$1$$s occupy) of countable sets.\n\nBear with me:\n\n$$V_0=\\{(0)=\\{0,0,0,0,.....\\}\\}$$ is a set with one element.\n\n$$V_1 =\\{(x_n)$$ where one $$x_i=1$$ and all the rest are $$0\\}$$ is countable as there is a one to one correspondence between the $$(x_n)$$ and the possible positions for $$x_i = 1$$.\n\n$$W_{k,1} = \\{(x_n): x_{j< k}=0; x_k = 1$$ and there is one $$x_{i>k}=1$$ but all the rest are $$0\\}$$. $$\\iota: W_{k,1}\\leftrightarrow V_1$$ via for every $$(x_n)\\in W_{k,1}$$, $$\\iota((x_n)) = (w_n= x_{n-k})$$. That is if $$(x_n)$$ is the sequence where $$x_k=1$$ and $$x_{i>k} =1$$ and all else are $$0$$, the $$\\iota((x_n))$$ is the sequence where $$x_{i-k}=1$$ and all else are $$0$$. This is clearly a bijection.\n\n$$V_2=\\{(x_n)$$ where two $$x_i=x_j=1$$ and all the rest are $$0\\}$$. $$V_2=\\cup_{i=1}^{\\infty} W_1$$ so $$V_2$$ is countable as it is a countable union of countable sets.\n\nLet $$V_m=\\{(x_n)$$ where there are exactly $$m$$ $$1$$s and all the rest are $$0\\}$$.\n\nLet $$W_{k,m} = \\{(x_n)$$ where $$x_{j< k}=0;$$ and there are $$m$$ $$x_{i>k}=1$$ and all the rest are $$0\\}$$.\n\nWe can define the bijection $$\\iota: W_{k,m}\\leftrightarrow V_m$$ via $$(x_n) = (x_{n-k})$$.\n\nAnd $$V_{m+1} = \\cup_{i=1}^{\\infty} W_{i,m}$$.\n\nBy induction if $$V_m$$ is countable then so is each $$W_{k,m}$$ and so $$V_{m+1}$$ as a union of countable sets.\n\nNow your $$Y=\\displaystyle Y=\\left\\{(x_n)\\in X:x_n=1 \\;\\;\\text{for at most finitely many n} \\right\\} = \\cup_{i=o}^{\\infty} V_i$$.\n\nSo $$Y$$ is a countable union of countable sets and thus countable.\n\n$$X$$ is indeed uncountable and your proof is correct. Your proof for countability of $$Y$$ is incorrect:\n\nNow the set $$Y$$ has all the sequences from the set $$X$$, but some of its elements of sequences are replaced by the only '$$1$$' so its cardinality will be less then $$2^{\\aleph_0}$$\n\nEven if you assume the continuums hypothesis (which is a strong assumption to make!), you don't know how many elements you really replaced or removed. How do you know that the cardinality became smaller? Did you construct an explicit injection for that failing to be surjective?\n\nStill, $$Y$$ is countable. I'll explain how. Basically, the only information a sequence $$y \\in Y$$ has is the finite, possibly disconnected, strip of ones it contains -- if any. You can see the zeroes as the default value with no information. Hence, we might posit the following wrong bijection:\n\n$$Y \\cong \\{0,1\\}^* \\tag{wrong!}$$\n\nE.g. we would do the following association:\n\n\u2022 1110000.... $$\\mapsto$$ 111\n\u2022 1110100.... $$\\mapsto$$ 11101\n\nCertainly, this is injective. But is this surjective? No! We have to consider $$\\{0,1\\}^*$$ in such a way that padded zeroes at the right are disregarded. In other words, we have\n\n$$Y \\cong \\{0,1\\}^*/\\sim$$\n\nwhere $$\\sim$$ is some equivalence. If you trust me that such a $$\\sim$$ exists, then we actually don't need to work this out for the countability argument. Namely, since $$\\{0,1\\}^*$$ was already countable, so is every quotiened version of it.\n\nLet $$A_n=\\{1,\\cdots,n\\}$$. For each $$f \\in \\{0,1\\}^{A_n}$$, define $$g_f:\\Bbb N \\to \\{0,1\\}$$ by $$g_f(x)=\\begin{cases}f(x)&\\text{if}\\;x \\in \\{1,2,..,n\\}\\\\0&\\text{otherwise} \\end{cases}$$ Then each $$g_f \\in Y$$. Then $$Y$$ can be written as $$Y=\\cup_{n=1}^\\infty Y_n$$ where $$Y_n=\\left\\{g_f: f \\in \\{0,1\\}^{A_n}\\right\\}$$ . Here each $$Y_n$$ is finite and hence $$Y$$ is countable!\n\n\u2022 does $g_f:\\Bbb N \\to \\{0,1\\}$ should be $g_f:\\Bbb N \\to \\{0,1\\}^{A_n}$? \u2013\u00a0gaurav saini Oct 1 '19 at 9:57", "date": "2021-04-14 13:59:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3372929/which-sets-of-sequence-is-countable-and-uncountable", "openwebmath_score": 0.9329140782356262, "openwebmath_perplexity": 174.8466598034546, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769113660689, "lm_q2_score": 0.9073122288794594, "lm_q1q2_score": 0.8851528618948868}}
{"url": "https://gmatclub.com/forum/in-how-many-different-ways-can-a-group-of-8-people-be-divide-85707-20.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 28 May 2020, 00:56\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# In how many different ways can a group of 8 people be divide\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 20 Oct 2019\nPosts: 2\nRe: In how many different ways can a group of 8 people be divide\u00a0 [#permalink]\n\n### Show Tags\n\n22 Feb 2020, 05:18\nDear experts,\nWhat if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be:\n8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values.\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 10464\nLocation: Pune, India\nRe: In how many different ways can a group of 8 people be divide\u00a0 [#permalink]\n\n### Show Tags\n\n02 Mar 2020, 20:35\n2\nnoboru wrote:\nIn how many different ways can a group of 8 people be divided into 4 teams of 2 people each?\n\nA, 90\nB. 105\nC. 168\nD. 420\nE. 2520\n\nResponding to a pm:\n\nQuote:\nWhat if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be:\n8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values. I know from your post that if all four teams had different number of people we would not unarrange. But if some repeat?\n\nYes, consider various scenarios:\n\n8 people and 2 teams (one of 3 people and the other of 5 people)\nYou just pick 3 of the 8 people for the 3 people team. Rest everyone will be in the 5 people team. No un-arranging required.\n\n8 people and 2 teams (one of 4 people and the other of 4 people)\nYou pick 4 people for the first team (say A, B, C, D). The other 4 (E, F, G, H) belong to the other team of 4 people. So you calculate this as 8C4. Now within this 8C4 will lie the case in which you picked (E, F, G, H) for the first team and ( A, B, C, D) will belong to the other team. But note that this case is exactly the same as before. 2 teams one (A, B, C, D) and other (E, F, G, H). So you need to un-arrange here.\n\nSimilarly, take your case - 8 people - 3 teams of 1 individual in each and 1 team let's say consisting of 5 people\nWe select 5 people out of 8 for the 5 people team in 8C5 ways. Rest of the 3 people play individually and we need to create no teams so nothing to be done. Since we are not doing any selection for a team, we are not inadvertently arranging and hence un-arranging is not required.\n\nInstead say we have 8 people - and we make 3 teams, one with 4 people and two teams with 2 people each\nWe select 4 people out of 8 for the 4 people team in 8C4 ways.\nThen we select 2 people for the first two people team in 4C2 ways and the leftover 2 people are for the second two people team. But again, as discussed above, there will same cases counted twice here so we will need to un-arrange by dividing by 2.\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nSenior Manager\nJoined: 21 Feb 2017\nPosts: 476\nIn how many different ways can a group of 8 people be divide\u00a0 [#permalink]\n\n### Show Tags\n\n15 Mar 2020, 00:16\nBunuel wrote:\nmanalq8 wrote:\nIn how many different ways can a group of 8 people be divided into 4 teams of 2 people each?\n\n90\n105\n168\n420\n2520\n\n$$\\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...\n\nFor the first person we can pick a pair in 7 ways;\nFor the second one in 5 ways (as two are already chosen);\nFor the third one in 3 ways (as 4 people are already chosen);\nFor the fourth one there is only one left.\n\nSo we have 7*5*3*1=105\n\nYou can check similar problems:\nhttp://gmatclub.com/forum/probability-8 ... %20equally\nhttp://gmatclub.com/forum/probability-8 ... ide+groups\nhttp://gmatclub.com/forum/combination-5 ... ml#p690842\nhttp://gmatclub.com/forum/sub-committee ... ide+groups\n\nThere is also direct formula for this:\n\n1. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is not important is $$\\frac{(mn)!}{(n!)^m*m!}$$.\n\n2. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\\frac{(mn)!}{(n!)^m}$$\n\nHope it helps.\n\nwe say we need to divide by 4! cus the order doesn't matter but when we use combinations arent we anyway implying that the order doesn't matter. I am not able to understand why we need to divide by 4! again when were using combinations and not permutations.\n\nThe second approach, I understand perfectly hence, I know I am missing something in the first one. Pls, help. This is really a confusing sub-topic for me under PnC\nIn how many different ways can a group of 8 people be divide \u00a0 [#permalink] 15 Mar 2020, 00:16\n\nGo to page \u00a0 Previous \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0[ 23 posts ]", "date": "2020-05-28 08:56:24", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-how-many-different-ways-can-a-group-of-8-people-be-divide-85707-20.html", "openwebmath_score": 0.6207519173622131, "openwebmath_perplexity": 805.5971711950095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9755769071055402, "lm_q2_score": 0.9073122232403329, "lm_q1q2_score": 0.8851528525278555}}
{"url": "https://math.stackexchange.com/questions/1977626/help-needed-with-different-approach-to-a-combinatorics-question", "text": "# Help needed with different approach to a combinatorics question\n\nEvery day, the 15 students in Mr. Singh`s Advanced Chemistry class are randomly divided into 5 lab groups of 3 students each. What is the probability that three students - Linda, Martin, and Nancy - are in the same lab group today?\n\nThe answer to the question is to see it from the perspective of one of the students we are concerned with. From his/her point of view, 2 of the other 14 students are randomly chosen to be paired with her. This can be done in ${14}\\choose{2}$ = 91 ways. Only one of those pairs are her \"friends\", so the answer is $\\frac{1}{91}$.\n\nThis answer of course didn't come to me, I read it when I couldn't figure it out. Though curious if I could find the same answer with my approach, I tried to kind of work backwards, knowing the answer.\n\nHere's what I tried to do: I instead thought of how many groups of three could be chosen from 15 students. I thought this could be done in ${15}\\choose{3}$ = 455 ways. One of those contains the group of students we are interested in. But the answer is not $\\frac{1}{455}$. How does 91 fit into this? Curious as I was, I divided 455 by 91, and lo and behold, I got 5. Did this 5 mean the 5 groups of three students? I wasn't sure.\n\nIf I divide 455 by 5, I get 91. But what is the meaning of this? I would rather think that, from the 455 available choices, we choose 5 things (in this case a thing is a group of 3 students). But this would mean ${455}\\choose{5}$ = some huge number.\n\nLet's say I have ABCDE, and I want to choose 2 of them. This would be ${5}\\choose{2}$ = 10 combinations. If I would ask \"how many groups of 5 can I make?\", then the answer surely isn't $\\frac{10}{5}$ = 2. I definitely can check with pen and paper and see that I can make quite more than 2 groups. So I know my thinking is faulty.\n\nBut where? What does the 5 in $\\frac{455}{5}$ = 91 mean? And what should my line of thought be when starting from ${15}\\choose{3}$ to get to the answer?\n\nThere are $\\binom{15}3=455$ possible $3$-person groups, so if we were picking just one group of $3$ at random, the probability would indeed be $\\frac1{455}$ that we picked this specific group. However, we\u2019re not just picking a single group: we\u2019re dividing the $15$ students into five groups of $3$, so we actually have $5$ chances to get the desired group, not just one. And $\\frac5{455}$ is, as you noticed, $\\frac1{91}$.\n\nYou might reasonably worry that picking a random partition of the students into $5$ groups of $3$ students isn\u2019t really the same as picking $5$ groups of $3$ at random with replacement from the collection of all $455$ $3$-person groups. We can do a more elaborate calculation to show that it really does work out right. There are\n\n$$\\binom{15}3\\binom{12}3\\binom93\\binom63\\binom33$$\n\nways to pick a first group of $3$, then pick a second group of $3$ from the $12$ people who remain, and so on until we have all $5$ groups. However, this counts each of the $5!$ permutations of the $5$ groups separately, so the number of ways to partition the students into $5$ groups of $3$ is really only\n\n$$\\frac1{5!}\\binom{15}3\\binom{12}3\\binom93\\binom63\\binom33\\;.$$\n\nHow many of these partitions have the desired trio as one of its parts? The same reasoning shows that if we set Linda, Martin, and Nancy aside and form $4$ groups of $3$ from the remaining $12$ students, we can do this in\n\n$$\\frac1{4!}\\binom{12}3\\binom93\\binom63\\binom33$$\n\ndifferent ways. The probability that a randomly chosen partition has our trio as one of its groups is therefore\n\n$$\\frac{\\frac1{4!}\\binom{12}3\\binom93\\binom63\\binom33}{\\frac1{5!}\\binom{15}3\\binom{12}3\\binom93\\binom63\\binom33}=\\frac{5!}{4!\\binom{15}3}=\\frac5{\\binom{15}3}=\\frac5{455}=\\frac1{91}\\;.$$\n\n\u2022 \"5 chances to get the desired group\" - I try to think about this, but I'm having a hard time. I keep connecting the 5 groups with the 455 things, which I probably shouldn't do. As an aside, how long did it take you to be fluent in this, i.e., have the insight quick and the ability to see it from multiple angles? \u2013\u00a0Garth Marenghi Oct 20 '16 at 19:37\n\u2022 @Garth: When you divide the students into $5$ groups of $3$, you really are picking $5$ groups, not just one, and one of those $5$ could be the one that we want. I really can\u2019t answer that last question: it\u2019s been over $50$ years since I first encountered these ideas. I don\u2019t think that I ever had any trouble with them, but I know from having taught the subject that my experience is not typical. \u2013\u00a0Brian M. Scott Oct 20 '16 at 19:46", "date": "2019-05-19 13:00:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1977626/help-needed-with-different-approach-to-a-combinatorics-question", "openwebmath_score": 0.8059795498847961, "openwebmath_perplexity": 148.78251876948974, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683513421314, "lm_q2_score": 0.8962513807543222, "lm_q1q2_score": 0.8851094984796548}}
{"url": "https://math.stackexchange.com/questions/1187351/how-many-ways-are-there-to-divide-elements-into-equal-unlabelled-sets", "text": "# How many ways are there to divide elements into equal unlabelled sets?\n\nHow many ways are there to divide N elements into K sets where each set has exactly N/K elements.\n\nFor the example of 6 elements into 3 sets each with 2 elements. I started by selecting the elements that would go in the first set (6 choose 2) and then those that would go into the second as (4 choose 2) and then the 2 remaining elements into the third set. This gives, (6 choose 2) * (4 choose 2). In general\n\n(N choose N/K) * (N-(N/K) choose N/K) * (N-(2*N/K) choose N/K) * ... * 1\n\n\u2022 Any thoughts on the problem? You aren't a new user here. You should know that questions on this site require at least some input on your efforts towards the problem. Otherwise you are destined for downvotes or being placed on hold \u2013\u00a0jameselmore Mar 12 '15 at 21:10\n\u2022 Added. I'm concerned that I'm assuming some kind of ordering. @jameselmore \u2013\u00a0nickponline Mar 12 '15 at 21:16\n\nYour approach, as noted by Ross Millikan's answer, is effective. Another way to approach such a problem would be to consider \"interpreting\" a permutation as such a partition - like, if we wrote the elements in the order: $$eabcfd$$ we might just group them into pairs as $\\{\\{e,a\\},\\{b,c\\},\\{f,d\\}\\}$ - where we just \"fill\" the expression $\\{\\{\\_,\\_\\},\\{\\_,\\_\\},\\{\\_,\\_\\}\\}$ by drawing from the order in which we wrote the elements.\n\nHowever, $eabcfd$ and $aebcfd$ represent the same partition, as the pairs are unordered - and $bceaf\\!\\,\\!d$ also represents the same partition, as the order of the pairs does not matter. Proceeding thusly, we can see that we can reorder within each of the $K$ sets in $(N/K)!$ ways without affecting the partition, and we can reorder the order in which the sets appear in the partition in $K!$ ways - and that, these are the only transformations which do not affect the partition. Thus, dividing the number of permutations of the elements by the number of permutations representing any given partition yields that there are $$\\frac{N!}{(N/K)!^k K!}$$ such partitions.\n\n(This can also be found by expanding ${a \\choose b}=\\frac{a!}{b!(a-b)!}$ and looking at cancellations in your expression)\n\n\u2022 Thanks! Is there a way to generalize this such that: if I make D draws with replacement from N distinct items. how many ways are there to get C copies of any K distinct items. I tried to generalize with your analogy of \"filling\" in the bracket. But for D > C*K it's unclear how to no double count groups. \u2013\u00a0nickponline Mar 20 '15 at 16:26\n\nHint: You are close. As the sets are unlabeled, choosing $\\{a,b\\},\\{c,d\\},\\{e,f\\}$ is the same as choosing $\\{e,f\\},\\{c,d\\},\\{a,b\\}$, but you have counted them both.\n\n\u2022 Ah, so I should divide by the number of ways of ordering the sets = K! \u2013\u00a0nickponline Mar 12 '15 at 21:20\n\u2022 Awesome :), is there a more compact formula that avoids the product? \u2013\u00a0nickponline Mar 12 '15 at 21:24\n\u2022 If you look at the factorial expression for the binomial coefficients, there is a lot of cancellation. Let $M=\\frac NK$ Your first two terms are $\\frac {N!}{(N-M)!M!}\\cdot \\frac {(N-M)!}{(N-2M)!M!}$ and you can see the cancellation. You wind up with $\\frac {N!}{(M!)^KK!}$ There is a nice combinatorial interpretation. There are $N!$ ways to put the elements in order. Reordering elements within a group can be done in $M!$ ways and you have $K$ of those. The last $K!$ is the ordering of the groups. \u2013\u00a0Ross Millikan Mar 12 '15 at 21:54\n\nIn response to your comment on Ross's answer, there is a more compact form: $$\\text{number of partitions }=P_{n,k}=\\frac{n!}{(n/k)!^{k}k!}$$ To prove this, we instead prove $$n!=P_{n,k}\\cdot (n/k)!^k\\cdot k!$$ To prove that, ask how many ways are there permute $n$ elements? There are certainly $n!$ such ways. But another to form a permutations is this: first, divide the $n$ elements into $k$ equal sets $(P_{n,k}$ ways to do this). Then, permute each element within each of the sets (they all have size $n/k$, and there are $k$ of them, so there are $(n/k)!^k$ ways. Finally, put the $k$ sets themselves into some order $(k!$ ways).\n\nYou can check that $P_{6,3}=\\frac{6!}{2!^3\\cdot3!}=15$, and that indeed there are $5$ choices for what the set containing $1$ is, and then $\\binom{4}2=3$ ways to partition the remaining 4 elements into sets.", "date": "2020-01-18 17:41:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1187351/how-many-ways-are-there-to-divide-elements-into-equal-unlabelled-sets", "openwebmath_score": 0.8315201997756958, "openwebmath_perplexity": 191.1003331146527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683487809279, "lm_q2_score": 0.8962513759047848, "lm_q1q2_score": 0.885109491394923}}
{"url": "https://math.stackexchange.com/questions/4194996/representing-a-linear-transformation-with-respect-to-a-new-basis", "text": "# Representing a linear transformation with respect to a new basis\n\nI have a question about the following exercise:\n\nLet $$f:\\mathbb{R}^3\\to\\mathbb{R}^3$$ be the linear transformation such that $$f(x,y,z)=(x-y+2z,-2x+y,x+z).$$ Represent the transformation with respect to the basis $$\\{(1,1,0),(0,1,-1),(1,1,1)\\}$$.\n\nWhat I have done:\n\nI found the images of the basis vectors given:\n\n\\begin{align}f(1,1,0) &= (0,-1,1)\\\\ \\\\ f(0,1,-1) &= (-3,1,-1)\\\\ \\\\ f(1,1,1) &= (2,-1,2) \\end{align}\n\nand then I found how these vectors can be written as a linear combination of the basis vectors given:\n\n\\begin{align}(0,-1,1)&=\\fbox{0}\\cdot (1,1,0)+\\fbox{(-1)}\\cdot (0,1,-1)+\\fbox{0}\\cdot (1,1,1) \\\\ \\\\ (-3,1,-1)&=\\fbox{-6}\\cdot (1,1,0)+\\fbox{4}\\cdot (0,1,-1)\\;\\;\\,\\,+\\fbox{3}\\cdot (1,1,1) \\\\ \\\\ (2,-1,2)&=\\fbox{3}\\cdot (1,1,0)+\\fbox{(-3)}\\cdot (0,1,-1)+\\fbox{-1}\\cdot (1,1,1)\\end{align}\n\nthus the matrix which represents $$f$$ with respect to this basis should be $$\\begin{bmatrix}0 & -6 & 3\\\\ -1 & 4 & -3\\\\ 0 & 3 & - 1\\end{bmatrix}.$$\n\nIs this correct? Or should I write the matrix with respect to the canonical basis of $$\\mathbb{R}^3$$ and then make the change of basis $$C^{-1}AC$$?\n\n\u2022 How did you calculate the coefficients in the boxes? Some of them, as currently written, are incorrect. Jul 10 at 12:33\n\u2022 Both methods are correct, and should give you the same answer \u2013 if you do them right. Jul 10 at 13:05\n\u2022 @shoteyes they should be correct now, thank you for pointing that out. For each vector $v$, I set up a system to find out which linear combination of basis vectors is equal to it, ie I found out which $c_1,c_2,c_3$ are such that $c_1 v_1+c_2v_2+c_3v_3=v$ and solved it using Gaussian elimination. Jul 10 at 15:48\n\u2022 Any thoughts on the comments, or on Dan's answer? Jul 12 at 13:47\n\u2022 @Gerry Myerson they have all been very useful, I have upvoted the comments and accepted Dan's answer, thank you very much Jul 12 at 13:57\n\nThis is the same. Maybe the square diagram in the sequel shows in the \"simplest\" way why.\n\nFirst i have to say something about the used convention for vectors. Because this is the \"canonical impediment\" when dealing with base change.\n\nWe work with column vectors, and matrices act on them by left multiplication. The linear map of left multiplication with a matrix $$A$$ will be denoted below (abusively) also by $$A$$. So $$x$$ goes via $$A$$ to $$A\\cdot x=Ax$$, displayed as $$x\\overset{A}\\longrightarrow Ax\\ .$$ \"Most of the world\" uses column vectors. (Some authors write notes or books (e.g. in Word), and find it handy to use row vectors, so they can be simpler displayed in the book rows. In this case linear maps induced by matrices use the multiplication from the right with such matrices. As long as we need in computations only linear combinations the convention is not so important, but it becomes when we use linear maps induced by matrices.)\n\nWe will work in the \"category\" of (finite dimensional) vector spaces (over $$\\Bbb R$$) with a fixed bases. The space $$V:=\\Bbb R^3$$ comes with the canonical base $$\\mathcal E=(e_1,e_2,e_3)$$, where $$e_1,e_2,e_3$$ are the columns of the matrix $$E$$ below, $$E= \\begin{bmatrix} 1 & 0 & 0\\\\ 0 & 1 & 0\\\\ 0 & 0 & 1 \\end{bmatrix}\\ .$$ We write this object as $$(V,\\mathcal E)$$. By abuse, we may want to write $$(V,E)$$ instead.\n\nWe start with two objects in this category. For our purposes let them have the same underlying vector space $$V=W=\\Bbb R^3$$, first object is $$(V,\\mathcal B=(b_1,b_2,b_3))$$, and the second object is $$(W,\\mathcal C=(c_1,c_2,c_3))$$.\n\nA linear map $$g:V\\to W$$ is defined \"abstractly\", and has no need for chosen bases. But in practice, $$g$$ is usually given bases-specific in the following way. Let $$v$$ be a vector in $$V$$. We write it w.r.t. $$\\mathcal B$$ as $$v=x_1b_1+x_2b_2+x_3b_3$$, and write this data as a column vector: $$v = \\begin{bmatrix}x_1\\\\x_2\\\\x_3\\end{bmatrix}_{\\mathcal B} :=x_1b_1+x_2b_2+x_3b_3 \\ .$$ Then we consider a matrix $$M=M_{\\mathcal B, \\mathcal C}$$ and build the matrix multiplication vector: $$\\begin{bmatrix}y_1\\\\y_2\\\\y_3\\end{bmatrix} = M \\begin{bmatrix}x_1\\\\x_2\\\\x_3\\end{bmatrix}\\ .$$ Then we consider the vector $$w\\in W$$ which written in base $$\\mathcal C$$ has the $$y$$-components, so $$w = \\begin{bmatrix}y_1\\\\y_2\\\\y_3\\end{bmatrix}_{\\mathcal C} :=y_1c_1+y_2c_2+y_3c_3 \\ ,$$ and the map $$g$$ is mapping linearly $$v$$ to $$w$$.\n\nThis concludes the section related to conventions and notations.\n\nLet $$\\mathcal C$$ be the base from the OP, the base with vectors which are columns of $$C= \\begin{bmatrix} 1 & 0 & 1\\\\ 1 & 1 & 1\\\\ 0 & -1 & 1 \\end{bmatrix}\\ .$$ Let $$A$$ be the matrix for the given linear map $$f$$ w.r.t. the canonical base $$\\mathcal E$$. $$A=\\begin{bmatrix}1&-1&2\\\\-2&1&0\\\\1&0&1\\end{bmatrix}\\ .$$ Consider now the diagram: $$\\require{AMScd}$$ $$\\begin{CD} (V,E) @>A>f> (V,E) \\\\ @A C A {\\operatorname{id}}A @A {\\operatorname{id}}A C A\\\\ (V,C) @>f>{C^{-1}AC}> (V,C) \\\\ \\end{CD}$$\n\nIndeed, $$C$$ is the matrix of the identity seen as a map $$(V,\\mathcal C)\\to(V,\\mathcal E)$$. For instance, $$c_1=\\begin{bmatrix}1\\\\0\\\\0\\end{bmatrix}_{\\mathcal C} \\qquad\\text{ goes to }\\qquad c_1 =\\begin{bmatrix}1\\\\1\\\\0\\end{bmatrix}_{\\mathcal E} =C\\begin{bmatrix}1\\\\0\\\\0\\end{bmatrix}_{\\mathcal E}\\ .$$\n\nIt remains to compute explicitly the matrix $$C^{-1}AC$$. Computer, of course:\n\nsage: A = matrix(3, 3, [1, -1, 2,  -2, 1, 0,  1,  0, 1])\nsage: C = matrix(3, 3, [1,  0, 1,   1, 1, 1,  0, -1, 1])\nsage: A\n[ 1 -1  2]\n[-2  1  0]\n[ 1  0  1]\nsage: C\n[ 1  0  1]\n[ 1  1  1]\n[ 0 -1  1]\nsage: C.inverse() * A  * C\n[ 0 -6  3]\n[-1  4 -3]\n[ 0  3 -1]\n\n\nAnd we check the result both ways:\n\n$$\\bf(1)$$\n\n$$\\require{AMScd}$$ $$\\begin{CD} c_j=\\begin{bmatrix}1\\\\1\\\\0\\end{bmatrix}_{\\mathcal E}\\ ,\\ \\begin{bmatrix}0\\\\1\\\\-1\\end{bmatrix}_{\\mathcal E}\\ ,\\ \\begin{bmatrix}1\\\\1\\\\1\\end{bmatrix}_{\\mathcal E} @>A>f> fc_j=\\begin{bmatrix}0\\\\-1\\\\1\\end{bmatrix}_{\\mathcal E}\\ ,\\ \\begin{bmatrix}-3\\\\1\\\\-1\\end{bmatrix}_{\\mathcal E}\\ ,\\ \\begin{bmatrix}2\\\\-1\\\\2\\end{bmatrix}_{\\mathcal E} \\\\ @A C A {\\operatorname{id}}A @A {\\operatorname{id}}A C A\\\\ c_j=\\begin{bmatrix}1\\\\0\\\\0\\end{bmatrix}_{\\mathcal C}\\ ,\\ \\begin{bmatrix}0\\\\1\\\\0\\end{bmatrix}_{\\mathcal C}\\ ,\\ \\begin{bmatrix}0\\\\0\\\\1\\end{bmatrix}_{\\mathcal C} @>f>{C^{-1}AC}> fc_j=\\begin{bmatrix}0\\\\-1\\\\0\\end{bmatrix}_{\\mathcal C}\\ ,\\ \\begin{bmatrix}-6\\\\4\\\\3\\end{bmatrix}_{\\mathcal C}\\ ,\\ \\begin{bmatrix}3\\\\-3\\\\-1\\end{bmatrix}_{\\mathcal C} \\end{CD}$$\n\nOr simpler, using block matrices, and ignoring the knowledge of the bases:\n\n$$\\require{AMScd}$$ $$\\begin{CD} C @>A>f> AC \\\\ @A C A {\\operatorname{id}}A @A {\\operatorname{id}}A C A\\\\ E @>f>{C^{-1}AC}> C^{-1}AC \\end{CD}$$\n\n$$\\bf(2)$$ In the spirit of the OP, using copy+pasted+corrected row vector computations:\n\n\\begin{aligned} (0,-1,1) &=\\boxed{0}\\cdot (1,1,0)+\\boxed{(-1)}\\cdot (0,1,-1)+\\boxed{0}\\cdot (1,1,1) \\ , \\\\ \\\\ (-3,1,-1) &=\\boxed{-6}\\cdot (1,1,0)+\\boxed{4}\\cdot (0,1,-1)+\\boxed{3}\\cdot (1,1,1) \\ , \\\\ \\\\ (2,-1,2) &=\\boxed{3}\\cdot (1,1,0)+\\boxed{(-3)}\\cdot (0,1,-1)+\\boxed{-1}\\cdot (1,1,1) \\ . \\end{aligned}", "date": "2021-12-07 22:51:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4194996/representing-a-linear-transformation-with-respect-to-a-new-basis", "openwebmath_score": 0.9979261755943298, "openwebmath_perplexity": 295.1646856718479, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683476832692, "lm_q2_score": 0.8962513745192026, "lm_q1q2_score": 0.8851094890427877}}
{"url": "https://math.stackexchange.com/questions/2839456/finding-orthonormal-basis-from-orthogonal-basis", "text": "# Finding orthonormal basis from orthogonal basis\n\nThe following is example C.5 from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:\n\nExample C.5 Show that $\\{\\sin(x),\\cos(x)\\}$ is an orthonormal basis for the inner product space $V=\\{a\\sin(x)+b\\cos(x); a,b\\in\\mathbb R, 0\\le x\\le\\pi\\}$ using as inner product $$\\langle f,g \\rangle = \\int_0^1 fg dx, \\qquad f,g\\in V$$ and determine an orthonormal basis.\n\nSolution $V$ is two dimensional and the set $\\{\\sin(x),\\cos(x)\\}$ is obviously a basis. We merely need to check orthogonality. First of all, \\begin{align}\\langle\\sin(x),\\cos(x)\\rangle=\\int_0^\\pi\\sin(x)\\cos(x)\\,dx&=\\frac{1}{2}\\int_0^\\pi\\sin(2x)\\,dx\\\\ &=\\left[-\\frac{1}{4}\\cos(2x)\\right]_0^\\pi \\\\ &=0.\\end{align} Hence orthogonality is established. Also, $$\\langle\\sin(x),\\sin(x)\\rangle=\\int_0^\\pi\\sin^2(x)\\,dx=\\frac{\\pi}{2}$$ and $$\\langle\\cos(x),\\cos(x)\\rangle=\\int_0^\\pi\\cos^2(x)\\,dx=\\frac{\\pi}{2}.$$ Therefore $$\\left\\{\\sqrt{\\dfrac{2}{\\pi}}\\sin(x),\\sqrt{\\dfrac{2}{\\pi}}\\cos(x)\\right\\}$$ is an orthonormal basis.\n\nI understand that, for orthonormality, we require that $\\| \\mathbf{a} \\| = 1$. However, I'm unsure of how the orthonormal basis was found at the bottom of the proof?\n\nI would appreciate it if people could please take the time to clarify this.\n\n\u2022 $\\lvert \\lvert \\sin(x) \\rvert \\rvert = \\sqrt{\\langle \\sin(x), \\sin(x) \\rangle} = \\sqrt{\\frac{\\pi}{2}}$. So in order to make an orthonormal basis from this orthogonal basis, we devide by the length, \u2013\u00a0Tim Dikland Jul 3 '18 at 10:09\n\u2022 @TimDikland You mean $||\\sin(x) ||$? \u2013\u00a0The Pointer Jul 3 '18 at 10:11\n\u2022 Exactly, editted. \u2013\u00a0Tim Dikland Jul 3 '18 at 10:11\n\u2022 Whenever you require an orthogonal basis to be ortonormal, just divide each vecotr by its norm. It remains an orthogonal basis (because of the properties of the inner product), but the norm of each vector is 1. \u2013\u00a0LuxGiammi Jul 3 '18 at 10:15\n\u2022 Is that a typo in your example where you are using the integral over $[0,1]$ to define the inner product? If not, that changes things. \u2013\u00a0Disintegrating By Parts Jul 5 '18 at 3:07\n\nAs mentioned in the comments to the main post, $\\lVert \\sin(x) \\rVert = \\sqrt{\\langle \\sin(x), \\sin(x) \\rangle} = \\sqrt{\\frac{\\pi}{2}}$. We then divide the orthogonal vectors by their norms in order convert them into orthonormal vectors. This gets us the orthonormal basis mentioned in the textbook excerpt.", "date": "2021-05-06 17:04:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2839456/finding-orthonormal-basis-from-orthogonal-basis", "openwebmath_score": 0.942592203617096, "openwebmath_perplexity": 350.798848385106, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683502444728, "lm_q2_score": 0.8962513682840825, "lm_q1q2_score": 0.8851094851806627}}
{"url": "https://math.stackexchange.com/questions/3114318/computing-the-product-fracddxxn-fracddxxn/3114413", "text": "# Computing the product $(\\frac{d}{dx}+x)^n(-\\frac{d}{dx}+x)^n$\n\nI want to compute the product $$(\\frac{d}{dx}+x)^n(-\\frac{d}{dx}+x)^n,$$ for a natural number $$n$$. For $$n$$ equal to 0 or 1, the computation is very simple obviously, but for such a low number as 2 the brute force calculation begins to be rather cumbersome and I cannot see any pattern emerging. I tried to find some connection with the Rodrigues' formula for the Hermite polynomials but I could not.\n\nThese operators come up in the algebraic approach to the quantum harmonic oscillator.\n\nExplicit Example\n\nTo avoid any misunderstanding, I am going to show explicitly the computation for the case $$n=1$$: $$(\\frac{d}{dx}+x)(-\\frac{d}{dx}+x)=-\\frac{d^2}{dx^2}+1+x\\frac{d}{dx}-x\\frac{d}{dx}+x^2=-\\frac{d^2}{dx^2}+x^2+1.$$\n\nOne can think of a function $$f$$ the operators are acting on. For example, $$(\\frac{d}{dx}\\circ x) f= (\\frac{d}{dx}x)f+x\\frac{d}{dx}f=(1+\\frac{d}{dx})f,$$ then $$\\frac{d}{dx}\\circ x=1+x\\frac{d}{dx}$$\n\n\u2022 Since this is physics related, can I ask what the stand-alone differential expression $\\mathrm{d}/\\mathrm{d}x$ means for your context ? \u2013\u00a0Rebellos Feb 15 at 19:47\n\u2022 That simplification doesn't work though as operators don't commute. It simplifies to $(-\\frac{d^2}{dx^2}-x\\frac{d}{dx}+\\frac{d}{dx}x+x^2)^n$. \u2013\u00a0Nick Guerrero Feb 15 at 20:00\n\u2022 @Rebellos, $d/dx$ is simply the derivative operator. In this context, it is supposed that the operators are acting on some function. For example, the commutator between $d/dx$ and $x$ is computed as $[d/dx,x]f=d/dx(xf)-x(d/dx f)=f+d/dx f-d/dx f=f$, then $[d/dx,x]=1$. \u2013\u00a0jobe Feb 15 at 20:15\n\u2022 @NickGuerrero Inasmuch as these operators ($\\mathscr{L}^+$ and $\\mathscr{L}^-$) do not commute, how would you show, as you asserted, that $$\\left(\\mathscr{L}^+\\right)^n \\left(\\mathscr{L}^-\\right)^n=(\\mathscr{L}^+\\mathscr{L}^-)^n?$$ \u2013\u00a0Mark Viola Feb 15 at 20:32\n\u2022 @jobe See THIS, which provides a discussion for $(A+B)^n$ where $[A,B]\\ne 0$ (i.e., the operators do not commute). \u2013\u00a0Mark Viola Feb 15 at 20:47\n\nFor convenience let us rewrite $$x,\\partial_x$$ with $$a,b$$ so $$[a,b]=x\\partial_x-\\partial_x x=-1$$ (in the operator sense on the Schwartz space).\n\nLemma. For all $$n\\in\\mathbb N$$ $$[(a+b)^n,a-b]=2n(a+b)^{n-1}$$\n\nProof. As darij pointed out, one has $$[a+b,a-b]=2$$ (i.e. the case $$n=1$$). The trick then is $$\\begin{split} [(a+b)^{n+1},a-b]&=(a+b)[(a+b)^n,a-b]+[a+b,a-b](a+b)^n\\\\ &=(a+b)2n(a+b)^{n-1}+2(a+b)^{n}=2(n+1)(a+b)^{(n+1)-1} \\end{split}$$ which concludes the proof via induction. $$\\square$$\n\nProposition. For all $$n\\in\\mathbb N_0$$ $$(a+b)^n(a-b)^n=\\prod_{j=1}^n (a^2-b^2+(2j-1))$$\n\nProof. ($$n=0$$ is obvious). Note that $$(a-b)(a+b)=a^2-b^2-1$$. Using the previous lemma $$\\begin{split} (a+b)^{n+1}(a-b)^{n+1}&=[(a+b)^{n+1},a-b](a-b)^n+(a-b)(a+b)(a+b)^n(a-b)^n\\\\ &=2(n+1)(a+b)^n(a-b)^n +(a^2-b^2-1)(a+b)^n(a-b)^n\\\\ &=\\big( a^2-b^2+2n+1)(a+b)^n(a-b)^n=\\prod_{j=1}^{n+1} (a^2-b^2+(2j-1)) \\end{split}$$ which again concludes the proof via induction. $$\\square$$\n\nThis result reproduces the cases (aside from $$n=0$$, obvious)\n\n\u2022 $$(a+b)(a-b)=a^2-b^2+1$$\n\n\u2022 Making use of $$[a^2,b^2]=-4ab-2$$ (similar techniques) one gets $$\\begin{split} (a+b)^2(a-b)^2=(a^2-b^2+1)(a^2-b^2+3)&=a^4-a^2b^2-b^2a^2+4a^2-4b^2+b^4+3\\\\ &=a^4-2a^2b^2-4ab+4a^2-4b^2+b^4+1 \\end{split}$$\n\netc... I feel like this formula is the best thing one can hope for in terms of structure.\n\nEdit: Thanks darij for the +200 rep!\n\n\u2022 Great job! Pretty sure that a factorization into degree-$2$ factors is an answer to the OP. \u2013\u00a0darij grinberg Feb 15 at 21:26\n\u2022 Thank you! I like how we both independently considered a commutator of the form $[x^m,y]$ (my Lemma / your eq. (1)) to approach the problem... \u2013\u00a0Frederik vom Ende Feb 15 at 21:30\n\u2022 Thank you for your answer! I am very happy my question has received so amazing answers, but I am having a hard time choosing which one to accept. Your answer seems to be the more explicit that is possible without becoming too complicated. I would like to have the derivative on the right side of each term, but probably that would made the result very complicated as the obtained by @Sangchul Lee. \u2013\u00a0jobe Feb 17 at 1:12\n\u2022 Glad to hear so! Of course one may use further tricks to shift all the derivatives ($b$'s) to the right such as $[x^2,\\partial_x^2]=-4x\\partial_x-2$ etc. although I'm not sure how simple or applicable that'd end up being. Anyways, just accept whatever answer helps (or will likely end up helping) you the most! \u2013\u00a0Frederik vom Ende Feb 17 at 7:35\n\nJust some Sage-generated data to play around with:\n\nFor $$n = 0$$, the result is $$1$$.\n\nFor $$n = 1$$, the result is $$-\\frac{\\partial^{2}}{\\partial x^{2}} + x^{2} + 1$$.\n\nFor $$n = 2$$, the result is $$\\frac{\\partial^{4}}{\\partial x^{4}} - 2 x^{2} \\frac{\\partial^{2}}{\\partial x^{2}} - 4 \\frac{\\partial^{2}}{\\partial x^{2}} - 4 x \\frac{\\partial}{\\partial x} + x^{4} + 4 x^{2} + 1$$.\n\nFor $$n = 3$$, the result is $$-\\frac{\\partial^{6}}{\\partial x^{6}} + 3 x^{2} \\frac{\\partial^{4}}{\\partial x^{4}} + 9 \\frac{\\partial^{4}}{\\partial x^{4}} + 12 x \\frac{\\partial^{3}}{\\partial x^{3}} - 3 x^{4} \\frac{\\partial^{2}}{\\partial x^{2}} - 18 x^{2} \\frac{\\partial^{2}}{\\partial x^{2}} - 9 \\frac{\\partial^{2}}{\\partial x^{2}} - 12 x^{3} \\frac{\\partial}{\\partial x} - 36 x \\frac{\\partial}{\\partial x} + x^{6} + 9 x^{4} + 9 x^{2} - 3$$.\n\nFor $$n = 4$$, the result is $$\\frac{\\partial^{8}}{\\partial x^{8}} - 4 x^{2} \\frac{\\partial^{6}}{\\partial x^{6}} - 16 \\frac{\\partial^{6}}{\\partial x^{6}} - 24 x \\frac{\\partial^{5}}{\\partial x^{5}} + 6 x^{4} \\frac{\\partial^{4}}{\\partial x^{4}} + 48 x^{2} \\frac{\\partial^{4}}{\\partial x^{4}} + 42 \\frac{\\partial^{4}}{\\partial x^{4}} + 48 x^{3} \\frac{\\partial^{3}}{\\partial x^{3}} + 192 x \\frac{\\partial^{3}}{\\partial x^{3}} - 4 x^{6} \\frac{\\partial^{2}}{\\partial x^{2}} - 48 x^{4} \\frac{\\partial^{2}}{\\partial x^{2}} - 36 x^{2} \\frac{\\partial^{2}}{\\partial x^{2}} + 48 \\frac{\\partial^{2}}{\\partial x^{2}} - 24 x^{5} \\frac{\\partial}{\\partial x} - 192 x^{3} \\frac{\\partial}{\\partial x} - 216 x \\frac{\\partial}{\\partial x} + x^{8} + 16 x^{6} + 42 x^{4} - 48 x^{2} - 39$$.\n\nFor $$n = 5$$, the result is $$-\\frac{\\partial^{10}}{\\partial x^{10}} + 5 x^{2} \\frac{\\partial^{8}}{\\partial x^{8}} + 25 \\frac{\\partial^{8}}{\\partial x^{8}} + 40 x \\frac{\\partial^{7}}{\\partial x^{7}} - 10 x^{4} \\frac{\\partial^{6}}{\\partial x^{6}} - 100 x^{2} \\frac{\\partial^{6}}{\\partial x^{6}} - 130 \\frac{\\partial^{6}}{\\partial x^{6}} - 120 x^{3} \\frac{\\partial^{5}}{\\partial x^{5}} - 600 x \\frac{\\partial^{5}}{\\partial x^{5}} + 10 x^{6} \\frac{\\partial^{4}}{\\partial x^{4}} + 150 x^{4} \\frac{\\partial^{4}}{\\partial x^{4}} + 150 x^{2} \\frac{\\partial^{4}}{\\partial x^{4}} - 150 \\frac{\\partial^{4}}{\\partial x^{4}} + 120 x^{5} \\frac{\\partial^{3}}{\\partial x^{3}} + 1200 x^{3} \\frac{\\partial^{3}}{\\partial x^{3}} + 1800 x \\frac{\\partial^{3}}{\\partial x^{3}} - 5 x^{8} \\frac{\\partial^{2}}{\\partial x^{2}} - 100 x^{6} \\frac{\\partial^{2}}{\\partial x^{2}} - 150 x^{4} \\frac{\\partial^{2}}{\\partial x^{2}} + 1500 x^{2} \\frac{\\partial^{2}}{\\partial x^{2}} + 975 \\frac{\\partial^{2}}{\\partial x^{2}} - 40 x^{7} \\frac{\\partial}{\\partial x} - 600 x^{5} \\frac{\\partial}{\\partial x} - 1800 x^{3} \\frac{\\partial}{\\partial x} - 600 x \\frac{\\partial}{\\partial x} + x^{10} + 25 x^{8} + 130 x^{6} - 150 x^{4} - 975 x^{2} - 255$$.\n\nFor $$n = 6$$, the result is $$\\frac{\\partial^{12}}{\\partial x^{12}} - 6 x^{2} \\frac{\\partial^{10}}{\\partial x^{10}} - 36 \\frac{\\partial^{10}}{\\partial x^{10}} - 60 x \\frac{\\partial^{9}}{\\partial x^{9}} + 15 x^{4} \\frac{\\partial^{8}}{\\partial x^{8}} + 180 x^{2} \\frac{\\partial^{8}}{\\partial x^{8}} + 315 \\frac{\\partial^{8}}{\\partial x^{8}} + 240 x^{3} \\frac{\\partial^{7}}{\\partial x^{7}} + 1440 x \\frac{\\partial^{7}}{\\partial x^{7}} - 20 x^{6} \\frac{\\partial^{6}}{\\partial x^{6}} - 360 x^{4} \\frac{\\partial^{6}}{\\partial x^{6}} - 540 x^{2} \\frac{\\partial^{6}}{\\partial x^{6}} + 120 \\frac{\\partial^{6}}{\\partial x^{6}} - 360 x^{5} \\frac{\\partial^{5}}{\\partial x^{5}} - 4320 x^{3} \\frac{\\partial^{5}}{\\partial x^{5}} - 8280 x \\frac{\\partial^{5}}{\\partial x^{5}} + 15 x^{8} \\frac{\\partial^{4}}{\\partial x^{4}} + 360 x^{6} \\frac{\\partial^{4}}{\\partial x^{4}} + 450 x^{4} \\frac{\\partial^{4}}{\\partial x^{4}} - 9000 x^{2} \\frac{\\partial^{4}}{\\partial x^{4}} - 6525 \\frac{\\partial^{4}}{\\partial x^{4}} + 240 x^{7} \\frac{\\partial^{3}}{\\partial x^{3}} + 4320 x^{5} \\frac{\\partial^{3}}{\\partial x^{3}} + 15600 x^{3} \\frac{\\partial^{3}}{\\partial x^{3}} + 7200 x \\frac{\\partial^{3}}{\\partial x^{3}} - 6 x^{10} \\frac{\\partial^{2}}{\\partial x^{2}} - 180 x^{8} \\frac{\\partial^{2}}{\\partial x^{2}} - 540 x^{6} \\frac{\\partial^{2}}{\\partial x^{2}} + 9000 x^{4} \\frac{\\partial^{2}}{\\partial x^{2}} + 31050 x^{2} \\frac{\\partial^{2}}{\\partial x^{2}} + 9180 \\frac{\\partial^{2}}{\\partial x^{2}} - 60 x^{9} \\frac{\\partial}{\\partial x} - 1440 x^{7} \\frac{\\partial}{\\partial x} - 8280 x^{5} \\frac{\\partial}{\\partial x} - 7200 x^{3} \\frac{\\partial}{\\partial x} + 8100 x \\frac{\\partial}{\\partial x} + x^{12} + 36 x^{10} + 315 x^{8} - 120 x^{6} - 6525 x^{4} - 9180 x^{2} - 855$$.\n\nSage code:\n\nA.<x> = DifferentialWeylAlgebra(QQ)\nx, dx = A.gens()\n\ndef r(n):\nreturn (dx + x) ** n * (-dx + x) ** n\n\nfor i in range(7):\nprint \"For $$n = \" + str(i) + \"$$, the result is $$\" + latex(r(i)) + \"$$.\\r\\n\"\n\n\nNote that it is easily seen that $$\\left[\\dfrac{\\partial}{\\partial x} + x, - \\dfrac{\\partial}{\\partial x} + x\\right] = 2$$. Thus, the operators $$\\dfrac{\\partial}{\\partial x} + x$$ and $$- \\dfrac{\\partial}{\\partial x} + x$$ themselves generate an isomorphic copy of the Weyl algebra, except for a scalar factor of $$2$$.\n\nIn view of the relation $$\\left[\\dfrac{\\partial}{\\partial x} + x, - \\dfrac{\\partial}{\\partial x} + x\\right] = 2$$, perhaps the following copypasta from some of my old homework will come useful.\n\nLet $$\\mathbb{N} = \\left\\{0,1,2,\\ldots\\right\\}$$.\n\nNow we need an easy fact from quantum algebra:\n\nProposition 1. Let $$A$$ be a ring (not necessarily commutative). Let $$x\\in A$$ and $$y\\in A$$ be such that $$xy-yx=1$$. Then, \\begin{align} \\left( xy\\right) ^{n}=\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}, \\end{align} where the curly braces denote Stirling numbers of the second kind.\n\nProof of Proposition 1. First of all, it is easy to see that $$$$x^{m}y=mx^{m-1}+yx^{m}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{for every }m\\in\\mathbb{N} \\label{darij1.pf.xy-yx.1} \\tag{1}$$$$ (this allows $$m=0$$ if $$0x^{0-1}$$ is interpreted as $$0$$). Indeed, the proof of \\eqref{darij1.pf.xy-yx.1} proceeds by induction over $$m$$ and is straightforward enough to be left to the reader.\n\nWe will now prove Proposition 1 by induction over $$k$$. The induction base is obvious, so we step to the induction step:\n\nLet $$n>0$$. Assuming that $$\\left( xy\\right) ^{n-1}=\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} y^{k}x^{k}$$, we need to show that $$\\left( xy\\right) ^{n}=\\sum\\limits_{k=0} ^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}$$.\n\nWe have \\begin{align*} \\left( xy\\right) ^{n} & =\\left( xy\\right) ^{n-1}xy=\\left( \\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\\right) xy\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( \\text{since }\\left( xy\\right) ^{n-1}=\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\\right) \\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} y^{k}\\underbrace{x^{k}x}_{=x^{k+1}}y=\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} y^{k}\\underbrace{x^{k+1}y}_{\\substack{=\\left( k+1\\right) x^{k} +yx^{k+1}\\\\\\text{(by \\eqref{darij1.pf.xy-yx.1}, applied to }m=k+1\\text{)}}}\\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} y^{k}\\left( \\left( k+1\\right) x^{k}+yx^{k+1}\\right) \\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} \\left( k+1\\right) y^{k}x^{k}+\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} \\underbrace{y^{k}y}_{=y^{k+1}}x^{k+1}\\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} \\left( k+1\\right) y^{k}x^{k}+\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} y^{k+1}x^{k+1}\\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} \\left( k+1\\right) y^{k}x^{k}+\\sum\\limits_{k=1}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k} y^{k}x^{k}\\\\ & \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( \\text{here, we substituted }k-1\\text{ for }k\\text{ in the second sum}\\right) \\\\ & =\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} \\left( k+1\\right) y^{k}x^{k}+\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k} y^{k}x^{k}\\\\ & \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( \\begin{array}[c]{c} \\text{here, we extended both sums by zero terms, using the fact}\\\\ \\text{that } \\genfrac{\\{}{\\}}{0pt}{0}{n}{n+1} = \\genfrac{\\{}{\\}}{0pt}{0}{n}{0} =0\\text{ whenever }n>0 \\end{array} \\right) \\\\ & =\\sum\\limits_{k=0}^{n}\\underbrace{\\left( \\genfrac{\\{}{\\}}{0pt}{0}{n}{k+1} \\left( k+1\\right) + \\genfrac{\\{}{\\}}{0pt}{0}{n}{k} \\right) }_{\\substack{= \\genfrac{\\{}{\\}}{0pt}{0}{n+1}{k+1} \\\\\\text{(by the recursion formula for Stirling numbers}\\\\\\text{of the second kind)}}}y^{k}x^{k}\\\\ & =\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}. \\end{align*} This completes the induction step, and thus the inductive proof of Proposition 1. $$\\blacksquare$$\n\nProposition 2. Let $$A$$ be a ring (not necessarily commutative). Let $$x\\in A$$ and $$y\\in A$$ be such that $$xy-yx=1$$. Then, \\begin{align} \\left( yx\\right) ^{n}=\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k} y^{k}x^{k}, \\end{align} where the curly braces denote Stirling numbers of the second kind.\n\nProof of Proposition 2. Just as in the proof of Proposition 1, we show that \\eqref{darij1.pf.xy-yx.1} holds.\n\nWe will now prove Proposition 2 by induction over $$k$$. The induction base is obvious, so we step to the induction step:\n\nLet $$n>0$$. Assuming that $$\\left( yx\\right) ^{n-1}=\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} y^{k}x^{k}$$, we need to show that $$\\left( yx\\right) ^{n}=\\sum\\limits_{k=0} ^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k} y^{k}x^{k}$$.\n\nWe have \\begin{align*} \\left( yx\\right) ^{n} & =\\left( yx\\right) ^{n-1}yx=\\left( \\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\\right) yx\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( \\text{since }\\left( yx\\right) ^{n-1}=\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\\right) \\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} y^{k}\\underbrace{x^{k}y}_{\\substack{=kx^{k-1}+yx^{k}\\\\\\text{(by \\eqref{darij1.pf.xy-yx.1}, applied to }m=k\\text{)}}}x\\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} y^{k}\\left( kx^{k-1}+yx^{k}\\right) x\\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} ky^{k}\\underbrace{x^{k-1}x}_{=x^{k}}+\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} k\\underbrace{y^{k}y}_{=y^{k+1}}\\underbrace{x^{k}x}_{=x^{k+1}}\\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} ky^{k+1}x^{k+1}\\\\ & =\\sum\\limits_{k=0}^{n-1} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\\sum\\limits_{k=1}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\\\ & \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( \\text{here, we substituted }k-1\\text{ for }k\\text{ in the second sum}\\right) \\\\ & =\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\\\ & \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( \\begin{array}[c]{c} \\text{here, we extended both sums by zero terms, using the fact}\\\\ \\text{that } \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{n} = \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{-1} =0\\text{ whenever }n>0 \\end{array} \\right) \\\\ & =\\sum\\limits_{k=0}^{n}\\underbrace{\\left( \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k} k+ \\genfrac{\\{}{\\}}{0pt}{0}{n-1}{k-1} \\right) }_{\\substack{= \\genfrac{\\{}{\\}}{0pt}{0}{n}{k} \\\\\\text{(by the recursion formula for Stirling numbers}\\\\\\text{of the second kind)}}}y^{k}x^{k}\\\\ & =\\sum\\limits_{k=0}^{n} \\genfrac{\\{}{\\}}{0pt}{0}{n}{k} y^{k}x^{k}. \\end{align*} This completes the induction step, and thus the inductive proof of Proposition 2. $$\\blacksquare$$\n\n\u2022 Thank you very much for this code! I really needed something like this! The only point is that I am desperately looking for a closed formula for arbitrary $n$. Do you have any thoughts about that? \u2013\u00a0jobe Feb 15 at 20:36\n\u2022 @jobe: My output doesn't give me much hope for a closed formula -- but of course, the Weyl algebra has several bases in which these elements could be expanded, and maybe one of them looks better. \u2013\u00a0darij grinberg Feb 15 at 20:39\n\u2022 Thank you for the reference , but I could not open it. \u2013\u00a0jobe Feb 15 at 20:45\n\nLet $$X, Y$$ be operators which are central, meaning that $$[X,Y] = XY-YX = c$$ for some scalar $$c$$. Then\n\n1. As a form of binomial theorem, we have\n\n$$(X+Y)^n = \\sum_{\\substack{a,b,m \\geq 0 \\\\ a+b+2m=n}} \\frac{n!}{a!b!m!2^m} c^m Y^b X^a = \\sum_{a=0}^{n} \\binom{n}{a} P_{n-a}(c,Y)X^a,$$\n\nwhere $$P_n$$ is defined by the following sum\n\n$$P_n(c, x) = \\sum_{m=0}^{\\lfloor n/2\\rfloor} \\frac{n!}{(n-2m)!m!2^m} c^m x^{n-2m} = \\left( c\\frac{d}{dx} + x \\right)^n \\mathbf{1}.$$\n\nIn particular, if $$c = -1$$ then $$P_n(-1, x) = \\operatorname{He}_n(x)$$, where $$\\operatorname{He}_n$$ is the probabilists' Hermite polynomial. Similarly, if $$c = 1$$, then $$P_n(1, x) = i^{-n} \\operatorname{He}_n(ix)$$.\n\n2. Under the same condition, for any polynomials $$f, g$$ we have\n\n$$f(X)g(Y) = \\sum_{m \\geq 0} \\frac{c^m}{m!} g^{(m)}(Y)f^{(m)}(X).$$\n\nIn our case, $$[\\frac{d}{dx}, x] = 1$$, and so, we can use both formulas to give a complicated, but still explicit expression for the product of $$(\\frac{d}{dx}+x)^n$$ and $$(-\\frac{d}{dx}+x)^n$$. Combining altogether,\n\n\\begin{align*} &\\left( \\frac{d}{dx} + x \\right)^n \\left( -\\frac{d}{dx} + x \\right)^n \\\\ &= \\sum_{p \\geq 0} \\frac{1}{p!} \\Bigg( \\sum_{\\substack{a_i, b_i, m_i \\geq 0 \\\\ a_i+b_i+2m_i = n-p}} \\frac{(-1)^{a_2+m_2} (n!)^2}{a_1!a_2!b_1!b_2!m_1!m_2!2^{m_1+m_2}} x^{b_1+b_2} \\left( \\frac{d}{dx} \\right)^{a_1+a_2} \\Bigg). \\end{align*}\n\nThe following is a sample Mathematica code, comparing this formula with the actual answer for the case $$n = 3$$.\n\nCoef[n_, c_, l_] := n!/(l[[1]]! l[[2]]! l[[3]]! 2^l[[3]]) c^l[[3]];\nT[n_] := FrobeniusSolve[{1, 1, 2}, n];\n(* Compute (x+d/dx) (x-d/dx)^n f(x) *)\nn = 3;\nNest[Expand[x # + D[#, x]] &, Nest[Expand[x # - D[#, x]] &, f[x], n], n]\nSum[1/p! Sum[ Sum[ Coef[n, 1, l1] Coef[n, -1, l2]\n(-1)^l2[[2]] x^(l1[[1]] + l2[[1]]) D[f[x],\n{x, l1[[2]] + l2[[2]]}], {l1, T[n - p]}], {l2, T[n - p]}], {p, 0, n}] // Expand\nClear[n];\n\n\n\u2022 Impressive! Your answer gives an explicit expression for the product as I have asked for, but it is indeed complicated! I am still trying to figure out how using it effectively in the problems I have in mind, but that is my fault. Thank you! \u2013\u00a0jobe Feb 17 at 1:19\n\u2022 @jobe, Glad it helped! This formula may be still far from being qualified as generally useful, but a form of combinatoric interpretation of the coefficients in the expansion of $(X+Y)^n$ is available, which may possible be useful for some applications: $$\\frac{n!}{(n-2m)!m!2^m}=[\\text{# of m-matchings in \\{1,\\cdots,n\\}}].$$ So $P_n(c,x)$ computes the sum of weights over all possible matchings on $\\{1,\\cdots,n\\}$, where each matching is weighted by the factor $c^m x^{n-2m}$, i.e. each matched pair receives the weight $c$ and each unmatched vertice receives the weight $x$. \u2013\u00a0Sangchul Lee Feb 17 at 1:32", "date": "2019-06-19 11:32:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3114318/computing-the-product-fracddxxn-fracddxxn/3114413", "openwebmath_score": 0.997992992401123, "openwebmath_perplexity": 725.4567881617681, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683491468142, "lm_q2_score": 0.8962513668985002, "lm_q1q2_score": 0.8851094828285275}}
{"url": "https://www.physicsforums.com/threads/pendulum-dilemma.546677/", "text": "# Pendulum dilemma\n\n## Homework Statement\n\nI am suppose to find a linear graph for the equation T=2pie(\u221aL/\u221ag)\n\n## The Attempt at a Solution\n\nThe best linear graph I could think of was L/T^2.\n\nAm I doing it right? Thanks.\n\nYes.\n\nT$^{2}$ on y-axis and L on x-axis.\n\nYes.\n\nT$^{2}$ on y-axis and L on x-axis.\n\nThank you very much. This has been like a huge thorn in my finger all day!\n\nI thought the slope was always y/x, so why isn't the slope of the graph T^2/L?\n\nSlope is NOT y/x but slope = $\\Delta$y/($\\Delta$x)\nor more exactly slope = dy/(dx).\n\nT = 2\u220f\u221a(L/g)\n\ntherefore T$^{2}$ = 4\u03c0$^{2}$L/g\n\nso d(T$^{2}$)/dL = 4\u03c0$^{2}$/g = slope\n\nT = 2\u220f\u221a(L/g)\n\ntherefore T$^{2}$ = 4\u03c0$^{2}$L/g\n\nso d(T$^{2}$)/dL = 4\u03c0$^{2}$/g = slope\n\noh that makes much more sense. thank you.", "date": "2022-05-16 06:24:07", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/pendulum-dilemma.546677/", "openwebmath_score": 0.4367019534111023, "openwebmath_perplexity": 3670.0902601938196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes.\n2. Yes.\n\n", "lm_q1_score": 0.9828232899814557, "lm_q2_score": 0.9005297841157157, "lm_q1q2_score": 0.8850616451508978}}
{"url": "http://math.stackexchange.com/questions/270564/what-is-the-argument-for-not-using-the-average-of-an-average", "text": "what is the argument for not using the average of an average?\n\nI want to disprove someone's calculation of percentage of cash sales for a year by taking summing percentage of cash sales by month and dividing by 12. I sense the correct way is to take total cash sales for the year and divide by total sales for year to arrive at percentage of sales but need correct reasoning why it is wrong to use an average of an average.\n\n-\nWhat is 'percentage of cash sales'? How is it calculated? \u2013\u00a0 Michael Biro Jan 4 at 21:54\n\nYou need a counter example. E.g.\n\nMon     Sales   Cash    Percent\n\nJan     100     70      70%\nFeb      10      1      10%\nMar      10      1      10%\nApr      10      1      10%\nMay      10      1      10%\nJun      10      1      10%\nJul      10      1      10%\nAug      10      1      10%\nSep      10      1      10%\nOct      10      1      10%\nNov      10      1      10%\nDec     100     70      70%\n\n\nTotal sales are $300$, cash sales are $150$ so the overall percentage is $50\\%$. But the average of the monthly percentages is $20\\%$.\n\n-\nThank you very much. I will be back to this site I'm sure. \u2013\u00a0 denise Jan 5 at 14:42\n\nYou are correct, here's a mathematical example:\n\nLet $s_1, s_2, s_3 \\cdots s_{12}$ be the sales of the months.\n\nLet $c_1, c_2, c_3 \\cdots c_{12}$ be the cash sales of the months.\n\n$$\\frac{1}{12} \\sum_{k=1}^{12} \\frac{c_k}{s_k}$$\n\nIs the same as: $$\\frac{\\sum_{k=1}^{12} c_k}{\\sum_{k=1}^{12} s_k}$$\n\nYou can show it is false by plugging in some numbers.\n\n-\nThanks for taking the time to respond. Very helpful. \u2013\u00a0 denise Jan 5 at 14:41\n\nThe average of the averages leads to the right average only when the samples have the same size.\n\nIf each month has exactly the same number of sales, then taking the average of the averages would be right. But if the number of sales per month is different, then the average of averages leads in general to the wrong answer.\n\nSimple situation: First month 100 sales of 1 each. Average 1. Second month 1 sale of 100. Average 100 .\n\nThe average of the averages is 50.50, but there were 101 sales and only 200 income....\n\n-\nRight, otherwise you need a weighted average. A simple unweighted arithmetic mean would be wrong. \u2013\u00a0 Fixed Point Jan 5 at 1:27\nThank you very much. Language very helpful. \u2013\u00a0 denise Jan 5 at 14:40\n\nSuppose in January I sell \\$100 worth of product, and all of it is sold for cash. Every other month I sell \\$1 worth of product, and none of it is sold for cash. I have then sold \\$111 dollars worth of product, of which \\$100 was sold for cash, so my percentage of cash sales is $\\frac{100}{111}\\approx90\\%$. Calculated the other way, we would get that for $1$ month $100\\%$ of my sales were cash, while for the other $11$, $0\\%$ of my sales were cash, giving $\\frac{100\\%}{12}\\approx 8.3\\%$. This is clearly wrong.\n\n-\nThanks so much for responding. Examples are so helpful! \u2013\u00a0 denise Jan 5 at 14:41\n\nIt is clear to you that taking cash sales for a year, and dividing by total sales for the year, will give you the correct proportion of cash sales for that year.\n\nThe problem with taking monthly proportions and averaging them (average of averages) is that that procedure could produce a quite different number.\n\nLet's take a simple numerical example. January to October: total sales each month $\\$1000$, cash sales each month$\\$500$. November and December (Christmas season, people are buying a lot, mostly on credit): total sales each month $\\$10000$, cash sales each month$\\$500$.\n\nSo total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\\dfrac{6000}{30000}=20\\%$. This is the correct percentage.\n\nNow let's compute the monthly averages. For January through October, they are $50\\%$. For each of November and December, they are $5\\%$.\n\nTo find the average of the monthly proportions, as a percent, we take $\\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\\%$, which is wildly different from the true average of $20\\%$.\n\nFor many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.\n\n-\nExactly what I needed! Thanks very much. \u2013\u00a0 denise Jan 5 at 14:39", "date": "2013-12-21 22:54:05", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/270564/what-is-the-argument-for-not-using-the-average-of-an-average", "openwebmath_score": 0.5187097191810608, "openwebmath_perplexity": 1814.0550951864177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750514614409, "lm_q2_score": 0.8962513682840825, "lm_q1q2_score": 0.8849362408818827}}
{"url": "https://stacks.math.columbia.edu/tag/00D3", "text": "Definition 4.21.1. Let $I$ be a set and let $\\leq$ be a binary relation on $I$.\n\n1. We say $\\leq$ is a preorder if it is transitive (if $i \\leq j$ and $j \\leq k$ then $i \\leq k$) and reflexive ($i \\leq i$ for all $i \\in I$).\n\n2. A preordered set is a set endowed with a preorder.\n\n3. A directed set is a preordered set $(I, \\leq )$ such that $I$ is not empty and such that $\\forall i, j \\in I$, there exists $k \\in I$ with $i \\leq k, j \\leq k$.\n\n4. We say $\\leq$ is a partial order if it is a preorder which is antisymmetric (if $i \\leq j$ and $j \\leq i$, then $i = j$).\n\n5. A partially ordered set is a set endowed with a partial order.\n\n6. A directed partially ordered set is a directed set whose ordering is a partial order.\n\n## Comments (4)\n\nComment #2102 by Keenan Kidwell on\n\nI just noticed that in the definition of a partially ordered set, the condition of antisymmetry is omitted. Isn't it more standard to call a transitive, reflexive relation a preorder? I'm guessing the antisymmetry condition doesn't play a role in considerations involving e.g. colimits over a directed set, so this is technically moot, but I'm curious if the decision to use a (as far as I can tell) not-quite-standard definition was intentional.\n\nComment #2129 by on\n\nOK, somebody else has mentioned this previously... and at the time there was a proof somewhere using the weaker notion... but I cannot find it now. The motivation for the weaker notion is that it is exactly enough conditions to turn $I$ into a category, so you can define (co)limits. I think that for almost all statements in the Stacks project it does not make a difference which definition you use. For example the proof of Lemma 4.21.5 seems to produce a partially ordered set in the stronger sense.\n\nOK, so maybe we should change this (if you are reading this and agree please leave a comment). Then a directed set will not be a partially ordered set in general. So lot's of lemmas should get multiple statements some with directed partially ordered sets and some with just directed sets... This is lot's of work, so I will only do this if more people chime in.\n\nComment #2138 by Keenan Kidwell on\n\nI was actually very happy when I saw this definition as it made me realize that the formalism of colimits and limits over directed sets worked just fine without the antisymmetry condition. I guess what it really showed me (which is something actually more standard) is that \"directed sets\" need not be \"partially ordered\" in the stronger sense. Since you don't actually seem to need the stronger sense, and it would require a non-trivial amount of work if you were to switch, I think you should leave it as it is.\n\nComment #2146 by on\n\nOK, I decided to revert to the standard definition of partially ordered sets and now a directed set is a preordered set with upper bounds for finite subsets. Going through all the corrections I found that it absolutely does not matter at all! It very slightly shortens the Stacks project text because it allows us to say \"directed set\" instead of \"directed partially ordered set\" in many lemmas, propositions, etc. But I am not sure if the use of the terminology \"directed set\" is completely standard. I found places where people suggest one should say \"directed proset\" where \"proset\" is an abbreviation for \"preordered set\" but that is just plain ugly. Oh well.\n\nBig set of changes can be found here.\n\nThere are also:\n\n\u2022 2 comment(s) on Section 4.21: Limits and colimits over preordered sets\n\n## Post a comment\n\nYour email address will not be published. Required fields are marked.\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).\n\nUnfortunately JavaScript is disabled in your browser, so the comment preview function will not work.\n\nAll contributions are licensed under the GNU Free Documentation License.\n\nIn order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00D3. Beware of the difference between the letter\u00a0'O' and the digit\u00a0'0'.", "date": "2022-12-08 16:02:30", "meta": {"domain": "columbia.edu", "url": "https://stacks.math.columbia.edu/tag/00D3", "openwebmath_score": 0.8227077126502991, "openwebmath_perplexity": 393.8887121005783, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180639771091, "lm_q2_score": 0.8976952873175982, "lm_q1q2_score": 0.8848744606560777}}
{"url": "https://gmatclub.com/forum/what-is-x-more-than-y-1-x-of-y-is-9-2-y-more-than-x-is-260828.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Nov 2018, 07:47\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in November\nPrevNext\nSuMoTuWeThFrSa\n28293031123\n45678910\n11121314151617\n18192021222324\n2526272829301\nOpen Detailed Calendar\n\u2022 ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!\n\nNovember 22, 2018\n\nNovember 22, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nMark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)\n\u2022 ### Free lesson on number properties\n\nNovember 23, 2018\n\nNovember 23, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nPractice the one most important Quant section - Integer properties, and rapidly improve your skills.\n\n# What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDS Forum Moderator\nJoined: 21 Aug 2013\nPosts: 1371\nLocation: India\nWhat is x% more than y? (1) x% of y is 9 (2) y% more than x is 24\u00a0 [#permalink]\n\n### Show Tags\n\n05 Mar 2018, 04:04\n2\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n75% (01:37) correct 25% (01:45) wrong based on 89 sessions\n\n### HideShow timer Statistics\n\nWhat is x% more than y?\n\n(1) x% of y is 9\n\n(2) y% more than x is 24\nIntern\nJoined: 04 Jan 2018\nPosts: 39\nRe: What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24\u00a0 [#permalink]\n\n### Show Tags\n\n05 Mar 2018, 09:32\n2\nans is C\n\n1) xy/100=9..........................(1)\n\n2) x+xy/100=24....................(2)\n\nsubst 1 in 2\nwe get 'x'\n\nsubst x in 1 or 2.\nwe get y\n_________________\n\nDon't stop till you get enough\n\nHit kudos if it helped you.\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2203\nWhat is x% more than y? (1) x% of y is 9 (2) y% more than x is 24\u00a0 [#permalink]\n\n### Show Tags\n\n19 Mar 2018, 12:35\n\nSOLUTION\n\nWe need to find the value which is $$x$$% more than $$y$$.\n\nThus, we need to find the value of $$y$$+$$\\frac{x}{100}$$ * $$y$$.\n\nStatement-1: \u201c$$x$$% of $$y$$ is $$9$$\u201d.\n\n\u2022 $$\\frac{x}{100}$$ * $$y$$=$$9$$\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)\n\nSince we don\u2019t know the value of $$x$$ and $$y$$, statement 1 alone is not sufficient to answer the question.\n\nStatement-2: \u201c$$y$$% more than $$x$$ is $$24$$\u201d.\n\n\u2022 $$x$$+ $$\\frac{y}{100}$$*$$x$$=$$24$$\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(2)\n\nSince we don\u2019t know the value of $$x$$ and $$y$$, statement 2 alone is not sufficient to answer the question.\n\nCombining both the statements:\n\n\u2022 $$x$$+ $$\\frac{x*y}{100}$$ = $$24$$\n\n\u2022 $$x+9=24$$\n\n\u2022 $$x=15$$\nPutting the value of $$x$$ in equation $$1$$\n\n\u2022 $$\\frac{x*y}{100}$$=$$9$$\n\n\u2022 $$15$$*$$\\frac{y}{100}$$=$$9$$\n\n\u2022 $$y=60$$\n\nWe now have the value of $$x$$ and $$y$$ both. Hence, we can find the asked value.\n\nStatement (1) and (2) together are sufficient.\n\n_________________\n\nNumber Properties | Algebra |Quant Workshop\n\nSuccess Stories\nGuillermo's Success Story | Carrie's Success Story\n\nAce GMAT quant\nArticles and Question to reach Q51 | Question of the week\n\nNumber Properties \u2013 Even Odd | LCM GCD | Statistics-1 | Statistics-2\nWord Problems \u2013 Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2\nAdvanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability\nGeometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry\nAlgebra- Wavy line | Inequalities\n\nPractice Questions\nNumber Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nWhat is x% more than y? (1) x% of y is 9 (2) y% more than x is 24 &nbs [#permalink] 19 Mar 2018, 12:35\nDisplay posts from previous: Sort by", "date": "2018-11-20 15:47:04", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-is-x-more-than-y-1-x-of-y-is-9-2-y-more-than-x-is-260828.html", "openwebmath_score": 0.56454998254776, "openwebmath_perplexity": 7392.551180880582, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.97112909472487, "lm_q2_score": 0.9111797172476385, "lm_q1q2_score": 0.8848731339423621}}
{"url": "https://brilliant.org/discussions/thread/euclidean-algorhytm/", "text": "# Euclidean algorithm\n\nI found something interesting about solving this question:\n\nIf p is the number of pairs of values (x, y) that can make the equation 84x + 140y = 28 true, what is true about p?\n\nI noticed while solving that I could simplify the equation:\n\n84x - 140y = 28\n\n42x \u2013 70y = 14\n\n21x \u2013 35y = 7\n\n3x \u2013 5y = 1\n\nNoting that 3x = 5y + 1 means that 3x must end in 5 + 1 = 6 or 0 + 1 = 1 to allow y to be divisible by 5.\n\nThis means 3x should end in either 1 or 6 to be equal to 5y + 1\n\nFinding the following values for x and matching values for y:\n\n3x = 6, 21, 36, 51, 66, 81, 96, 111, 126, 141, 156\n\n5y = 5, 20, 35, 50, 65, 80, 95, 110, 125, 140, 155\n\nThen finding the series of x and y:\n\nX = 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52\n\nY = 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31\n\nX seems to be increasing by 5, and Y by 3.\n\nWhich means that you can find pairs of x and y by the formulas:\n\nx = 2+5a and y = 1+3a\n\nLeading to a sequence that will satisfy 84x + 140y = 28.\n\nWould this also be more generally true? I also thought it might be an interesting additional question to the quiz.\n\nOn another note, this text editor does not seem to like enters in paragraphs. Any way I can fix this?\n\nNote by Sytse Durkstra\n1\u00a0year ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n\n- bulleted\n- list\n\n\u2022 bulleted\n\u2022 list\n\n1. numbered\n2. list\n\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1\n\nparagraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\n## Comments\n\nSort by:\n\nTop Newest\n\nYes, you just applied Euclidean algorithm.\n\nTo enter $$\\LaTeX$$, type like\n\n\\ ( ... \\ )\n\nbut remove the spaces.\n\n- 1\u00a0year ago\n\nLog in to reply\n\n\u00d7\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading...", "date": "2018-10-15 11:47:55", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/euclidean-algorhytm/", "openwebmath_score": 0.9645053148269653, "openwebmath_perplexity": 1666.9390768969245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9658995703612219, "lm_q2_score": 0.9161096175976053, "lm_q1q2_score": 0.8848698860413102}}
{"url": "https://mathhelpboards.com/threads/arc-length-question.13986/", "text": "# Arc length question\n\n##### Member\nThe length of the minor arc of a circle is 10cm, while the area of the sector AOB is 150cm2.\n\na) Form two equations involving r and \u03b8, where \u03b8 is measured in radians.\n\nb) Solve these equations simultaneously to find r and \u03b8.\n\nHelp to solve? Cant understand the question very well.\n\nI think the arc length formula was\n\nlength=$$\\displaystyle \\frac{n}{360}\\cdot2\\pi(r)$$\n\n$$\\displaystyle \\therefore10=\\frac{n}{360}\\cdot2\\pi(r)$$\n\nThe question states, Form two equations involving r and \u03b8, where \u03b8 is measured in radians.\n\nSo would we have to arrange the formula to find the r and $$\\displaystyle \\theta$$\n\n#### MarkFL\n\nStaff member\nThe formulas you want are:\n\nCircular arc-length:\n\n$$\\displaystyle s=r\\theta\\tag{1}$$\n\nArea of circular sector\n\n$$\\displaystyle A=\\frac{1}{2}r^2\\theta\\tag{2}$$\n\nNow, if we solve (1) for $\\theta$, we find:\n\n$$\\displaystyle \\theta=\\frac{s}{r}$$\n\nNow, substituting this into (2), we obtain:\n\n$$\\displaystyle A=\\frac{1}{2}r^2\\left(\\frac{s}{r}\\right)=\\frac{rs}{2}\\implies r=\\frac{2A}{s}\\implies\\theta=\\frac{s^2}{2A}$$\n\nNow, just plug in the given values for $A$ and $s$.\n\n##### Member\nThe formulas you want are:\n\nCircular arc-length:\n\n$$\\displaystyle s=r\\theta\\tag{1}$$\n\nArea of circular sector\n\n$$\\displaystyle A=\\frac{1}{2}r^2\\theta\\tag{2}$$\n\nNow, if we solve (1) for $\\theta$, we find:\n\n$$\\displaystyle \\theta=\\frac{s}{r}$$\n\nNow, substituting this into (2), we obtain:\n\n$$\\displaystyle A=\\frac{1}{2}r^2\\left(\\frac{s}{r}\\right)=\\frac{rs}{2}\\implies r=\\frac{2A}{s}\\implies\\theta=\\frac{s^2}{2A}$$\n\nNow, just plug in the given values for $A$ and $s$.\nIm finding trouble understanding the question.\n\"Form two equations involving r and $$\\displaystyle \\theta$$\" Basically means transpose the formula or make the equation equal to r and $$\\displaystyle \\theta$$?\n\nSince the arc length = $$\\displaystyle s=r\\theta$$ (I thought arc length was length=$$\\displaystyle \\frac{n}{360}\\cdot2\\pi(r)$$?\n\nr=$$\\displaystyle s\\theta$$?\n\nFor b) we just sub in the values? Where s is the length?\n\n#### MarkFL\n\nStaff member\nYou are given values for the arc-length $s$ and the area $A$, and so you want to use the formulas relating $r$ and $\\theta$ to these given values (which I gave as (1) and (2)) to be able to express both $r$ and $\\theta$ as functions of $s$ and $A$ (which I did using some algebra), so that you can use these given values to determine $r$ and $\\theta$.\n\nOnce you have $r$ and $\\theta$ as functions of $s$ and $A$, it is simply a matter of using the given values to evaluate $r$ and $\\theta$.\n\n#### SuperSonic4\n\n##### Well-known member\nMHB Math Helper\nIm finding trouble understanding the question.\n\"Form two equations involving r and $$\\displaystyle \\theta$$\" Basically means transpose the formula or make the equation equal to r and $$\\displaystyle \\theta$$?\nIt's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have $$\\displaystyle r$$ and $$\\displaystyle /theta$$ in them then you're fine for this step.\n\nMarkFL has given you the equations you need (his TeX is also better than mine! )\n\n$$\\displaystyle \\displaystyle s=r\\theta\\tag{1}$$\n\n$$\\displaystyle \\displaystyle A=\\frac{1}{2}r^2\\theta\\tag{2}$$\n\nAs you see both of these equations have $$\\displaystyle r$$ and $$\\displaystyle \\theta$$ in them so Part A of the question has been solved already\n\nSince the arc length = $$\\displaystyle s=r\\theta$$ (I thought arc length was length=$$\\displaystyle \\frac{n}{360}\\cdot2\\pi(r)$$?\n\nr=$$\\displaystyle s\\theta$$?\n$$\\displaystyle \\frac{n}{360}\\cdot 2\\pi r$$ is the equation for arc length where $$\\displaystyle n$$ is in degrees. Since the question asks for radians you want $$\\displaystyle s = r\\theta$$. See the spoiler below for why they are equivalent\n\nBy definition a full circle has 360 degrees and $$2\\pi$$ radians so you can say $$360^{\\circ} = 2\\pi$$ (we omit the c symbol for radians)\n\nYour formula for degrees is $$s = \\frac{\\theta}{360} \\cdot 2\\pi r$$ but since we know that $$360^{\\circ} = 2\\pi$$ it can be subbed in\n\n$$s = \\frac{\\theta}{2\\pi} \\cdot 2\\pi r = r \\theta$$ - the radian expression\n\nFor b) we just sub in the values? Where s is the length?\nSub in the values given (10cm and 150cm\u00b2) for s and A respectively so you have two equations and two unknowns ($$\\displaystyle r$$ and $$\\displaystyle \\theta$$).\n\nYou can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\\displaystyle \\theta$$ in terms of $$\\displaystyle r$$ which means he was able to substitute $$\\displaystyle \\frac{s}{r}$$ wherever $$\\displaystyle \\theta$$ appeared.\n\nHe did in terms of $$\\displaystyle s$$ in post 2 but if you find it easier you may use $$\\displaystyle s=10$$ for your example\n\n##### Member\nIt's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have $$\\displaystyle r$$ and $$\\displaystyle /theta$$ in them then you're fine for this step.\n\nMarkFL has given you the equations you need (his TeX is also better than mine! )\n\n$$\\displaystyle \\displaystyle s=r\\theta\\tag{1}$$\n\n$$\\displaystyle \\displaystyle A=\\frac{1}{2}r^2\\theta\\tag{2}$$\n\nAs you see both of these equations have $$\\displaystyle r$$ and $$\\displaystyle \\theta$$ in them so Part A of the question has been solved already\n\n$$\\displaystyle \\frac{n}{360}\\cdot 2\\pi r$$ is the equation for arc length where $$\\displaystyle n$$ is in degrees. Since the question asks for radians you want $$\\displaystyle s = r\\theta$$. See the spoiler below for why they are equivalent\n\nBy definition a full circle has 360 degrees and $$2\\pi$$ radians so you can say $$360^{\\circ} = 2\\pi$$ (we omit the c symbol for radians)\n\nYour formula for degrees is $$s = \\frac{\\theta}{360} \\cdot 2\\pi r$$ but since we know that $$360^{\\circ} = 2\\pi$$ it can be subbed in\n\n$$s = \\frac{\\theta}{2\\pi} \\cdot 2\\pi r = r \\theta$$ - the radian expression\n\nSub in the values given (10cm and 150cm\u00b2) for s and A respectively so you have two equations and two unknowns ($$\\displaystyle r$$ and $$\\displaystyle \\theta$$).\n\nYou can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\\displaystyle \\theta$$ in terms of $$\\displaystyle r$$ which means he was able to substitute $$\\displaystyle \\frac{s}{r}$$ wherever $$\\displaystyle \\theta$$ appeared.\n\nHe did in terms of $$\\displaystyle s$$ in post 2 but if you find it easier you may use $$\\displaystyle s=10$$ for your example\nThanks this really did help\n\n##### Member\nThe formulas you want are:\n\nCircular arc-length:\n\n$$\\displaystyle s=r\\theta\\tag{1}$$\n\nArea of circular sector\n\n$$\\displaystyle A=\\frac{1}{2}r^2\\theta\\tag{2}$$\n\nNow, if we solve (1) for $\\theta$, we find:\n\n$$\\displaystyle \\theta=\\frac{s}{r}$$\n\nNow, substituting this into (2), we obtain:\n\n$$\\displaystyle A=\\frac{1}{2}r^2\\left(\\frac{s}{r}\\right)=\\frac{rs}{2}\\implies r=\\frac{2A}{s}\\implies\\theta=\\frac{s^2}{2A}$$\n\nNow, just plug in the given values for $A$ and $s$.\nBeginning to understand this a lot easier.\nJust a question on the last part.\nHow did you rearrange to get $$\\displaystyle \\theta$$?\n\nIt went from, $$\\displaystyle A=\\frac{1}{2}r^2\\left(\\frac{s}{r}\\right)$$ which I understand to $$\\displaystyle =\\frac{rs}{2}$$?\n\n#### MarkFL\n\nStaff member\nBeginning to understand this a lot easier.\nJust a question on the last part.\nHow did you rearrange to get $$\\displaystyle \\theta$$?\n$$\\displaystyle s=r\\theta$$\n\nDivide through by $r$\n\n$$\\displaystyle \\frac{s}{r}=\\frac{\\cancel{r}\\theta}{\\cancel{r}}$$\n\n$$\\displaystyle \\theta=\\frac{s}{r}$$\n\nIt went from, $$\\displaystyle A=\\frac{1}{2}r^2\\left(\\frac{s}{r}\\right)$$ which I understand to $$\\displaystyle =\\frac{rs}{2}$$?\n$$\\displaystyle A=\\frac{1}{2}r^2\\left(\\frac{s}{r}\\right)=\\frac{\\cancel{r}\\cdot r\\cdot s}{2\\cancel{r}}=\\frac{rs}{2}$$", "date": "2021-09-26 03:42:26", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/arc-length-question.13986/", "openwebmath_score": 0.9278194308280945, "openwebmath_perplexity": 468.974483626017, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9840936110872273, "lm_q2_score": 0.8991213840277783, "lm_q1q2_score": 0.8848196096136419}}
{"url": "https://math.stackexchange.com/questions/2456373/optimal-route-consisting-of-rowing-then-walking", "text": "# Optimal route consisting of rowing then walking\n\nProblem\n\nYou're in a boat on point A in the water, and you need to get to point B on land. Your rowing speed is 3km/h, and your walking speed 5km/h.\n\nSee figure:\n\nFind the route that takes the least amount of time.\n\nMy idea\n\nI started by marking an arbitrary route:\n\nFrom here, I figure the total time is going to be $$T = \\frac{R}{3\\mathrm{km/h}} + \\frac{W}{5\\mathrm{km/h}}$$\n\nSince this is a function of two variables, I'm stuck.\n\nA general idea is to express $W$ in terms of $R$ to make it single-variable, and then apply the usual optimization tactics (with derivatives), but I'm having a hard time finding such an expression.\n\nAny help appreciated!\n\nEDIT - Alternative solution?\n\nSince the distance from A to the right angle (RA) is traveled 3/5 times as fast as the distance between RA and B, could I just scale the former up?\n\nThat way, I get A-RA being a distance of $6\\cdot\\frac53 = 10\\mathrm{km}$, which makes the hypotenuse $\\sqrt{181}$ the shortest distance between A and B. And since we scaled it up, we can consider it traversable with walking speed rather than rowing speed!\n\nThoughts?\n\n\u2022 but .. where is the river / land border ? \u2013\u00a0G Cab Oct 5 '17 at 21:56\n\u2022 The horizontal leg of the triangle would simultaneously be the border and the walking route. It sounds strange, but the idea is that the distance between the shoreline and the walking path is \"negligible\", thus they can be seen as one-and-the-same. \u2013\u00a0Alec Oct 6 '17 at 8:27\n\u2022 Thanks, now it is clear. Another question, do you know (accept) trigonometric approach? \u2013\u00a0G Cab Oct 6 '17 at 21:12\n\u2022 @GCab - Yes. I've reached an answer of 3.4hours with that approach. However, $\\frac{\\sqrt{181}}{5}$ doesn't give the same answer, and I can't figure out why that wouldn't work, and if there's some adjustment that needs to be made. Or if it needs to be discarded completely. \u2013\u00a0Alec Oct 7 '17 at 11:01\n\u2022 I think the answer is here, so wikipedia gets the bonus: en.wikipedia.org/wiki/Fermat%27s_principle \u2013\u00a0Ethan Bolker Oct 8 '17 at 14:23\n\n\u2022 a) the solution\n\nThe formula has already been indicated by wgrenard and AdamBL $$T = {1 \\over 3}\\sqrt {36 + \\left( {9 - W} \\right)^{\\,2} } + {1 \\over 5}W$$\n\ndifferentiating that $${{dT} \\over {dW}} = {{5W + 3\\sqrt {36 + \\left( {9 - W} \\right)^{\\,2} } - 45} \\over {15\\,\\sqrt {36 + \\left( {9 - W} \\right)^{\\,2} } }}$$ and equating to $0$ gives \\eqalign{ & {{dT} \\over {dW}} = 0\\quad \\Rightarrow \\quad 3\\sqrt {36 + \\left( {9 - W} \\right)^{\\,2} } = 45 - 5W\\quad \\Rightarrow \\cr & \\Rightarrow \\quad 9\\left( {36 + \\left( {9 - W} \\right)^{\\,2} } \\right) = 25\\left( {9 - W} \\right)^{\\,2} \\quad \\Rightarrow \\cr & \\Rightarrow \\quad 9 \\cdot 36 = 16\\left( {9 - W} \\right)^{\\,2} \\quad \\Rightarrow \\quad W = 9 - \\sqrt {{{9 \\cdot 36} \\over {16}}} = 9 - {9 \\over 2} = {9 \\over 2} \\cr} which is a minimum, because the function is convex as already indicated.\n\nThus $$\\left\\{ \\matrix{ W_m = 9/2 \\hfill \\cr T_m = 17/5 \\hfill \\cr R_m = 15/2 \\hfill \\cr} \\right.$$\n\n\u2022 b) Scaling\n\nYour idea of scaling according to speed is quite entangling.\nThat means (if I understood properly) that you are transforming the triangle from space to time units.\nBut, by introducing different scaling factors for the two coordinates, you undermine the Euclidean norm, which does not \" transfer\" between the two systems (if assumed valid in one, shall be modified in the other).\n\nConsider for example the transformation sketched below.\n\nFrom the mathematical point of view it is a linear scale transformation $$\\left( {\\matrix{ {y_1 } \\cr {y_2 } \\cr } } \\right) = \\left( {\\matrix{ {1/v_1 } & 0 \\cr 0 & {1/v_2 } \\cr } } \\right) \\left( {\\matrix{ {x_1 } \\cr {x_2 } \\cr } } \\right)$$\n\nNow, with constant $v_1, \\,v_2$, any path in $x$ will transform in the corresponding path in $y$ (going through corresponding points).\nIf the path is a curve parametrized through a common parameter $\\lambda$, not influencing the $v_k$'s, then, at any given value of $\\lambda$ the point on the $x$ plane will transform into the corresponding point in $y$ plane $$\\left( {\\matrix{ {y_{1}(\\lambda) } \\cr {y_{2}(\\lambda) } \\cr } } \\right) = \\left( {\\matrix{ {1/v_1 } & 0 \\cr 0 & {1/v_2 } \\cr } } \\right) \\left( {\\matrix{ {x{_1}(\\lambda) } \\cr {x_{2}(\\lambda) } \\cr } } \\right)$$ and the minimal path in one plane will be the corresponding minimal path in the other.\n\nBut we shall also have that $$\\frac{d}{{d\\lambda }}\\left( {\\matrix{ {y_{1}(\\lambda) } \\cr {y_{2}(\\lambda) } \\cr } } \\right) = \\left( {\\matrix{ {1/v_1 } & 0 \\cr 0 & {1/v_2 } \\cr } } \\right) \\frac{d}{{d\\lambda }}\\left( {\\matrix{ {x{_1}(\\lambda) } \\cr {x_{2}(\\lambda) } \\cr } } \\right)$$ that is that the \"velocities\" compose vectorially.\n\nTherefore if $\\lambda$ is the time, you shall go from $A$ to $C$ with a composition of a vertical rowing speed and a horizontal walking speed (a \"$\\infty$-thlon\"), which takes the same time as rowing $AH$ and walking $HC$.\n\nWhen, instead, you just row on $AC$, then you shall change the above matrix - for that segment only - according to the $\\angle AC$, and of course you loose the correspondence minimal to minimal as based on the Euclidean norm (straight line $A'B'$).\n\n\u2022 The question has recieved a nice handful of good answers. I'm still somewhat perplexed as to the alternative solution, which I was hoping to focus on with the bounty, but since you're the only one who even mentioned it, the bounty goes here. I would be interested to learn more about the entangling nature of that strategy though. Do you have any links, or simply a concept I should look into? \u2013\u00a0Alec Oct 12 '17 at 7:23\n\u2022 Thanks for appreciation. I will try and add more elaboration to clear the \"scaling\" matter. \u2013\u00a0G Cab Oct 12 '17 at 8:48\n\u2022 @Alec: I expanded on the 2nd part: wish I succeeded and make it more clear and .. convincing. \u2013\u00a0G Cab Oct 12 '17 at 13:31\n\nThe portion of the line on the top left is $9-w$. So by the Pythagorean theorem $R^2 = 36 + (9-w)^2$. I think this is the relationship you are looking for.\n\nYou do not need to worry about the hypotenuse in the larger right triangle.\n\nConsider the right triangle that has R as the hypotenuse. Left side of this triangle will have length $6$ and the top side will have length $(9-W)$. By the Pythagorean theorem, $R^{2} = 36+(9-W)^2$.\n\nSolving for R and inserting in your original equation yields $T= \\frac{1}{3} \\sqrt{36+(9-W)^{2}} + \\frac{1}{5} W$. This function is strictly convex, so you can simply minimize it by solving $\\frac{d\\ T}{d\\ W}=0$ (the 'usual optimization tactics').\n\nRegarding your alternative solution, I am unsure what you are arguing. The shortest distance between the points A and B will always be the line connecting them. As you have shown in your drawing, $|AB|=3\\sqrt{13} \\ne\\sqrt{181}$.\n\n\u2022 No, I mean if we've scaled the left-most side to 10km (by multiplying 5/3), the hypotenuse will be $\\sqrt{10^2 + 9^2} = \\sqrt{181}$. \u2013\u00a0Alec Oct 4 '17 at 7:39\n\u2022 Yes, that is correct. If I understand you correctly, though, you are arguing that it will always be optimal to row the entire way, since the hypotenuse will always be the shortest distance from A to B. That's obviously not the case. \u2013\u00a0AdB Oct 5 '17 at 15:37\n\u2022 Nono, I'm not arguing that it's faster to row always. I'm saying that if we scale the rowing distance up by 5/3, then we can consider that distance to be traversable by walking speed instead of rowing speed. \u2013\u00a0Alec Oct 5 '17 at 17:09\n\nSnell's Law is usually applied to optics, but it is based on the quickest path through two media in which the speed of light differs. Snell's Law says that $$n_1\\sin(i_1)=n_2\\sin(i_2)\\tag1$$ where $i_k$ is the angle of incidence to the boundary of the path and $n_k$ is inversely proportional to the speed in the particular medium ($n_kv_k=c$).\n\nWe can adapt this to the current situation by noting that $(1)$ is equivalent to $$\\frac{\\sin(i_1)}{v_1}=\\frac{\\sin(i_2)}{v_2}\\tag2$$ If we travel at all along the shore, $\\sin(i_2)=1$ (the angle of incidence is $90^\\circ$). Since $v_1=3$ and $v_2=5$, we must have $\\sin(i_1)=\\frac35$, which implies that $\\tan(i_1)=\\frac34$.\n\nIf $\\tan(i_1)=\\frac34$, and the width of the river is $6$ km, then the downriver distance must be $\\frac34\\cdot6=4.5$ km.\n\nThis path takes $\\frac{7.5}3+\\frac{4.5}5=3.4$ hours.\n\n\u2022 Wait, so we must find an angle such that we only travel 4.5km by water? I must be interpreting that wrong, because the shortest possible water-distance must be 6km, right? \u2013\u00a0Alec Oct 8 '17 at 20:50\n\u2022 No, the downriver distance is $4.5$ km, the river is still $6$ km wide, so that makes that water distance $\\sqrt{4.5^2+6^2}=7.5$ km. \u2013\u00a0robjohn Oct 8 '17 at 22:55\n\u2022 Ah, gotcha! Yeah, that was a nice take on this problem. I never would have connected this to Snell's Law on my own. \u2013\u00a0Alec Oct 9 '17 at 7:47", "date": "2021-05-16 16:16:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2456373/optimal-route-consisting-of-rowing-then-walking", "openwebmath_score": 0.9564309120178223, "openwebmath_perplexity": 372.64045876237503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406000885706, "lm_q2_score": 0.8933093989533708, "lm_q1q2_score": 0.8847698971641369}}
{"url": "https://brilliant.org/discussions/thread/magnetic-dollars/", "text": "# Magnetic Dollars\n\nImgur\n\nSuppose that I have two urns, with one magnetic dollar in each urn. I begin randomly throwing more magnetic dollars in the general vicinity of the urns, and the coins fall in the urns by a simple rule:\n\nLet's say that Urn A has $$x$$ coins, and Urn B has $$y$$ coins. The probability that the next coin falls into Urn A is $$\\dfrac{x}{x+y}$$, and the probability that the next coin falls into Urn B is $$\\dfrac{y}{x+y}$$.\n\nYou keep throwing magnetic dollars until there is a total of $$1,000,000$$ magnetic dollars in total.\n\nWhat would you bet would be the price on average, of the urn with the smaller amount of coins?\n\n$$\\ 10$$, perhaps? Maybe $$\\ 100$$ or $$\\ 500$$? Post your bet or write it down on a piece of paper before looking at the next section.\n\nThe less-priced urn, on average, is actually worth a grand total of a quarter of a million dollars! Don't worry if you guessed wrong; many professional mathematicians also guessed much lower than this. In fact, when a group of mathematicians were asked this question and were asked to bet, most people only bet $$\\ 10$$ and only one person bet over $$\\ 100$$.\n\nBut why does the lower-priced urn price so high? You may want to try the problem out yourself before I go over a very nice and elegant solution. See if you can find it!\n\nTried it out yet? In the case that you have, let's see how this problem can be so elegantly solved, as I claimed.\n\nSuppose that you have a deck of cards; one red, and $$999,998$$ white. Currently, you just have a red card. Now every turn, you place a white card in any available slot. For example, in the first move, you have $$2$$ available slots: one above the red card, and one below. In the second move, you have $$3$$ available slots, and so on.\n\nBut wait! Let's say that the empty slots above the red card are the magnetic dollars in Urn A, and the empty slots below the red card are the magnetic dollars in Urn B. Notice that if you had $$x$$ empty slots above the red card and $$y$$ empty slots below the red card, then the probability that the next card will be above the red card is $$\\dfrac{x}{x+y}$$, and the probability that the next card will be below the red card is $$\\dfrac{y}{x+y}$$! We've found a one-to-one correspondence between the original magnetic dollar problem and this new card problem!\n\nFinally, we know that in a random placements of white cards in this fashion will result in a uniform distribution of where the red card is, every single final position is of equal probability. This means that in the original problem, the probability of $$42$$ magnetic dollars being in Urn A is the same as the probability of $$314,159$$ magnetic dollars in Urn A. Therefore, the average price of the lower priced urn is clearly $$250,000$$. $$\\Box$$\n\nNote by Daniel Liu\n4\u00a0years, 8\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nThe result is certain not intuitive, nor obvious.\n\nYou can look at this problem John's Red And Blue Balls, which is based off the same idea.\n\nStaff - 4\u00a0years, 8\u00a0months ago\n\nI know some of you have been waiting for me to post for a long time. Sorry for the inactivity, I had a bunch of homework and stuff.\n\nWell, here is my next #CosinesGroup posts. The problem in my opinion is a very cool problem because of the awesome elegant solution.\n\nThis problem has also been called \"Polya's Urns\", but I looked that up and \"Polya's Urns\" actually covers a lot of random complicated stuff, so I kept it at magnetic dollars.\n\nHope you enjoy! Feedback is appreciated.\n\n- 4\u00a0years, 8\u00a0months ago\n\nThis is the solution in Peter Winkler ' Mathematical Puzzles. An excellent book.\n\n- 2\u00a0years, 2\u00a0months ago\n\nHow do you do this?\n\n- 4\u00a0years, 7\u00a0months ago\n\nI'm curious to know why only one mathematician guessed over $100. I personally guessed around$150,000 becuase while the probability might be pretty low at some point, in A MILLION tries, it would be bound to happen quite a bit.\n\n- 4\u00a0years, 8\u00a0months ago\n\nWell, their reasoning would probably be then assuming a sort of \"reverse\" normal distribution curve. The average of all the games would probably be pretty low.\n\nThey were asked to bet immediately, depriving them of the chance to think it over. Their first instinct told them that the more coins that get in a urn, the higher probability of getting more coins in, giving a feedback loop. This means that the smaller urn probably won't really have any coins at all.\n\n- 4\u00a0years, 8\u00a0months ago", "date": "2018-09-26 07:53:38", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/magnetic-dollars/", "openwebmath_score": 0.9101695418357849, "openwebmath_perplexity": 965.1188540119305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9808759649262344, "lm_q2_score": 0.9019206679615432, "lm_q1q2_score": 0.8846723054736926}}
{"url": "https://math.stackexchange.com/questions/2542068/asymptotic-behavior-of-x-n1-x-n-frac1x-n-space-space-space-x-0-1", "text": "# Asymptotic behavior of $x_{n+1}=x_n+\\frac{1}{x_n}, \\space\\space\\space x_0=1$\n\nI have defined a sequence $x_n$ as follows: $$x_{n+1}=x_n+\\frac{1}{x_n}, \\space\\space\\space x_0=1$$ After convincing myself that there is no nice closed-form formula for $x_n$, I decided to try and find an asymptotic formula for $x_n$ (unfortunately, I have very little experience with asymptotic formulae). I noticed that the recurrence is equivalent to $$\\Delta x_{n}=\\frac{1}{x_n}$$ and so a solution can be approximated by solving the corresponding differential equation for $y(t)$: $$y'=\\frac{1}{y}$$ This differential equation (with initial value $y(0)=1$) yielded $$y=\\sqrt{2t+1}$$ which led me to believe that $$x_n\\approx \\sqrt{2t+1}$$ When I plotted $x_n$ and $y(n)$ side by side, it did indeed appear that they were very close to each other. However, this isn't enough for me... I would like to prove that $y(n)$ is a good approximation for $x_n$, either by proving that $$\\lim_{n\\to\\infty}(x_n-\\sqrt{2n+1})=0$$ ...or, even better, by finding a zero-approaching function $f$ satisfying $$x_n=\\sqrt{2n+1}+O(f(x))$$ This is where I got stuck. How do can I prove (or disprove?) the statement in the first equality, and find $f$ satisfying the second?\n\nNOTE: There may be a closed-form that I wasn't able to find. In fact, the similar sequence $$y_{n+1}=\\frac{y_n}{2}-\\frac{1}{2y_n}$$ has closed form formula $$y_n=\\frac{1}{\\tan(2^n\\arctan(y_0^{-1}))}$$ ...but even if you do find a closed form, I would still like to know how to prove the above statements, since the techniques would be useful to know for future problems.\n\n\u2022 I think $\\int_n^{n+1} f(x) dx = \\frac 1 {\\int_0^n f(x)}$ is more accurate. But its probably nastier to solve. \u2013\u00a0mathreadler Nov 29 '17 at 0:15\n\u2022 See OEIS sequence A073833 and related sequences. \u2013\u00a0Robert Israel Nov 29 '17 at 0:21\n\u2022 Probably the first step, based on this guess, would be to write a recurrence for $a_n := x_n^2$: $a_{n+1} = a_n + 2 + \\frac{1}{a_n}$. So, to start a bootstrapping argument, it would be easy to prove $a_n \\ge 2n - 1$ for each $n$... And then $a_{n+1} \\le a_n + 2 + \\frac{1}{2n-1}$ so also $a_n \\le 2n + \\frac{1}{2} \\log(n) - C$, etc. ... \u2013\u00a0Daniel Schepler Nov 29 '17 at 0:28\n\u2022 Here is a class of arguments that could work. You can try to prove inductively both an upper and a lower bound simultaneously, say $\\sqrt{2n+a} \\le x_n \\le \\sqrt{2n+b}$ for some $a, b$, because when bounding $x_{n+1}$ in terms of $x_n$ you need both a lower and an upper bound on $x_n$ to establish either an upper bound or a lower bound on $x_n + \\frac{1}{x_n}$. \u2013\u00a0Qiaochu Yuan Nov 29 '17 at 0:55\n\u2022 Possible duplicate of Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$ \u2013\u00a0Aryabhata Nov 29 '17 at 3:10\n\nYou can show it's a quite nice bound by doing the following. First let $$f(x)=x+\\frac{1}{x}$$ and then define $b_n=\\sqrt{2n+1}$, $\\epsilon_n=x_n-b_n$, and $\\xi_n=f(b_n)-b_{n+1}$. First notice that $$f(x+\\epsilon)-f(x)\\le \\epsilon$$ when $x,\\epsilon \\ge 0$ (we will later show that $\\epsilon_n\\ge 0$, for now assume this). We then have that \\begin{aligned} \\epsilon_{n+1} &= x_{n+1}-b_{n+1} \\\\ &= f(x_n)-f(b_n)+\\xi_n \\\\ &= f(b_n+\\epsilon_n)-f(b_n)+\\xi_n \\\\ &\\le \\epsilon_n+\\xi_n \\end{aligned} Combined with $\\epsilon_0=0$, this tells us that $$0\\le \\epsilon_{n}\\le \\sum_{i=0}^{\\infty}\\xi_i\\approx 0.56$$ The above being a convergent series (according to Wolfram by comparison). This of course gives $$\\sqrt{2n+1}\\le x_n\\le \\sqrt{2n+1}+0.56$$ for all $n$, making it a very good approximation.\nTo show that $\\epsilon_n\\ge 0$ we first see that $\\epsilon_0=0$. Then assume $\\epsilon_n\\ge 0$, by the above we have $\\epsilon_{n+1}=f(b_n+\\epsilon_n)-f(b_n)+\\xi_n$. It is easy to see that $f$ is increasing on $x\\ge 1$, and that $b_n\\ge 1$, thus since $\\epsilon_n\\ge 0$ we see that $f(b_n+\\epsilon_n)-f(b_n)\\ge 0$ finally yielding $\\epsilon_{n+1}\\ge \\xi_n\\ge 0$, where $\\xi_n\\ge 0$ is easy to verify since we have an explicit formula for $\\xi_n$. This completes the induction.\nThe bound numerically looks like it can be improved and it might also be good to very find the series that Wolfram is comparing ours to. And this still doesn't resolve whether $\\lim_{n\\to \\infty}(x_n-b_n)=0$.", "date": "2019-09-19 14:30:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2542068/asymptotic-behavior-of-x-n1-x-n-frac1x-n-space-space-space-x-0-1", "openwebmath_score": 0.9949032068252563, "openwebmath_perplexity": 127.60192712234979, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983847162809778, "lm_q2_score": 0.8991213820004279, "lm_q1q2_score": 0.8845980207027276}}
{"url": "http://mathhelpforum.com/discrete-math/280796-help-combinatorics-exercise.html", "text": "# Thread: Help in combinatorics exercise\n\n1. ## Help in combinatorics exercise\n\nHey everyone I'm stuck in the exercise, I will be happy to work out a solution\nFind how many integers - n, have the attribute:\nn is divisible by 7 and is not divisible by any natural number - k that sustains:\nThanks friends.\n\n2. ## Re: Help in combinatorics exercise\n\nYou are looking for numbers that have 7 as a factor, but do not have 2,3, or 5, as a factor.\n\nLet $A_k = \\{n \\in \\mathbb{N}|1 \\le n \\le 7770\\text{ and }k\\text{ divides }n\\}$\n\nYou are looking for:\n\n\\begin{align*}\\left|A_7 \\setminus (A_2 \\cap A_3 \\cap A_5)\\right| = & |A_7| - |A_{7\\cdot 2}| - |A_{7\\cdot 3}| - |A_{7\\cdot 5}| \\\\ & + |A_{7\\cdot 2\\cdot 3}| + |A_{7\\cdot 2\\cdot 5}| + |A_{7\\cdot 3\\cdot 5}| \\\\ & - |A_{7\\cdot 2\\cdot 3\\cdot 5}|\\end{align*}\n\nThis is because if $p,q$ are distinct primes, then $A_p\\cap A_q = A_{pq}$, and also from the Inclusion/Exclusion principle. Additionally, for any prime $p$, you have:\n\n$|A_p| = \\left\\lfloor \\dfrac{7770}{p} \\right\\rfloor$\n\nI hope you get the idea. So, the answer is:\n\n$\\dfrac{7770}{7}-\\left(\\dfrac{7770}{7\\cdot 2} + \\dfrac{7770}{7\\cdot 3} + \\dfrac{7770}{7\\cdot 5}\\right) + \\left(\\dfrac{7770}{7\\cdot 2\\cdot 3} + \\dfrac{7770}{7\\cdot 2\\cdot 5} + \\dfrac{7770}{7\\cdot 3\\cdot 5} \\right) - \\dfrac{7770}{7\\cdot 2\\cdot 3\\cdot 5} = 296$\n\nEdit: To verify, you can use Excel.\nIn cell A1, type the formula: =MOD(ROW(),2)>0\n(this will be TRUE if the row number is NOT divisible by 2)\nIn cell B1, type: =MOD(ROW(),3)>0\nC1: =MOD(ROW(),4)>0\nD1: =MOD(ROW(),5)>0\nE1: =MOD(ROW(),6)>0\nF1: =MOD(ROW(),7)=0\n(notice this one you want it to be equal to zero because that means the row number is divisible by 7)\nG1: =MOD(ROW(),8)>0\nH1: =MOD(ROW(),9)>0\nI1: =MOD(ROW(),10)>0\nJ1: =AND(A1:I1)\nThen select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility.\n\n3. ## Re: Help in combinatorics exercise\n\nThanks a friend for helping me very much\n\n4. ## Re: Help in combinatorics exercise\n\nEdit: To verify, you can use Excel.\nIn cell A1, type the formula: =MOD(ROW(),2)>0\n(this will be TRUE if the row number is NOT divisible by 2)\nIn cell B1, type: =MOD(ROW(),3)>0\nC1: =MOD(ROW(),4)>0\nD1: =MOD(ROW(),5)>0\nE1: =MOD(ROW(),6)>0\nF1: =MOD(ROW(),7)=0\n(notice this one you want it to be equal to zero because that means the row number is divisible by 7)\nG1: =MOD(ROW(),8)>0\nH1: =MOD(ROW(),9)>0\nI1: =MOD(ROW(),10)>0\nJ1: =AND(A1:I1)\nThen select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility.[/QUOTE]\n\nThis site shows me something interesting that I did not know, can you please repeat the last part?\n\n5. ## Re: Help in combinatorics exercise\n\nOriginally Posted by yossa\nEdit: To verify, you can use Excel.\nIn cell A1, type the formula: =MOD(ROW(),2)>0\n(this will be TRUE if the row number is NOT divisible by 2)\nIn cell B1, type: =MOD(ROW(),3)>0\nC1: =MOD(ROW(),4)>0\nD1: =MOD(ROW(),5)>0\nE1: =MOD(ROW(),6)>0\nF1: =MOD(ROW(),7)=0\n(notice this one you want it to be equal to zero because that means the row number is divisible by 7)\nG1: =MOD(ROW(),8)>0\nH1: =MOD(ROW(),9)>0\nI1: =MOD(ROW(),10)>0\nJ1: =AND(A1:I1)\nThen select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility.\n\nThis site shows me something interesting that I did not know, can you please repeat the last part?\nI'm not sure what you are asking. If you do not have Microsoft Excel, then any OpenOffice spreadsheet will work.\n\n6. ## Re: Help in combinatorics exercise\n\nOriginally Posted by SlipEternal\nI'm not sure what you are asking. If you do not have Microsoft Excel, then any OpenOffice spreadsheet will work.\nI'll explain myself, I have Microsoft Excel, I was able to place everything in a simple phase a - \"Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility.\" not so figured out\n\n7. ## Re: Help in combinatorics exercise\n\nSelect cells A1:J1. Right click and choose Copy (or CTRL+C on the keyboard). On the keyboard, press CTRL+G (this will bring up a dialog that says GoTo in the title. In the Reference line, put A7770. Now, while holding the Shift key (do not let go), do the following:\n1. Press and release the End key\n2. Press and release the Up arrow key\n3. Press and release the right arrow key nine times (so you will have every cell in the range A1:J7770 highlighted).\n\nRelease the Shift key, right click, and press Paste (or CTRL+V on the keyboard).\n\n8. ## Re: Help in combinatorics exercise\n\nOriginally Posted by SlipEternal\nSelect cells A1:J1. Right click and choose Copy (or CTRL+C on the keyboard). On the keyboard, press CTRL+G (this will bring up a dialog that says GoTo in the title. In the Reference line, put A7770. Now, while holding the Shift key (do not let go), do the following:\n1. Press and release the End key\n2. Press and release the Up arrow key\n3. Press and release the right arrow key nine times (so you will have every cell in the range A1:J7770 highlighted).\n\nRelease the Shift key, right click, and press Paste (or CTRL+V on the keyboard).\n\ni get it thanks", "date": "2018-08-20 11:04:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/280796-help-combinatorics-exercise.html", "openwebmath_score": 0.736559271812439, "openwebmath_perplexity": 2630.6594000302625, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795095031688, "lm_q2_score": 0.8962513648201267, "lm_q1q2_score": 0.8845817324417142}}
{"url": "http://gunb.nomen.pw/predator-prey-model-simulation-matlab.html", "text": "\u2022 Predator Prey Model Simulation Matlab\n\u2022 Design, simulation and analysis in Simulink. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. The link to this assignment on github is here. ) Wilensky, U. GIBSON1*, DAVID L. I have a program called Predator Prey that's in the collection of programs that comes with NCM, Numerical Computing with MATLAB. This dataset includes the source computer code and supporting data files for the predator-prey simulation model (parameterized for summer flounder, Paralichthys dentatus) developed to investigate bottom-up effects defined to be temporal pulses in prey abundance on predator growth, production, and fisheries management. What are synonyms for Predator and prey?. This model is common, e. The predator-prey population-change dynamics are modeled using linear and nonlinear time series models. As a mathematical consequence of the herd behavior, they considered competition models and predator-prey systems. This video will show you the basics and give you an idea of what working in MATLAB looks like. Lotka-Volterra predator prey model. It was developed independently by Alfred Lotka and Vito Volterra in. While creating a model for combined predator and prey strategy would inform an estimation of overall fitness throughout an animal\u2019s lifetime, an overarching model of this sort would be extremely complex and is beyond the scope of our paper. In the model system, the predators thrive when there are plentiful prey but, ultimately, outstrip their food supply and decline. However, the organisms in Holland\u2019s simulation are very simple and do not involve any behavioral model. Describing the dynamics of such models occasionally requires some techniques of model analysis. Simulation of CTMC model I Use CTMC model to simulate predator-prey dynamics I Initial conditions are X(0) = 50 preys and Y(0) = 100 predators 0 5 10 15 20 25 30 0 50 100 150 200 250 300 350 400 Time Population Size X (Prey) Y (Predator) I Prey reproduction rate c 1 = 1 reactions/second I Rate of predator consumption of prey c 2 = 0:005. Student Challenge: Set up a level of hunting that keeps the populations of predators and prey at healthy levels. Predator and Prey Interactions Level A: Up and Down in the Wild: Predator and Prey. Both a detailed large eddy simulation of the dynamics and microphysics of a precipitating shallow boundary layer cloud system and a simpler model built upon basic physical principles, reproduce predator-prey behavior with rain acting as the predator and cloud as the prey. In particular, it describes how to structure a model and how to define species - that are the key components of GAMA models. They will, however, also be modi\ufb01ed during this exercise. Related Data and Programs: FD_PREDATOR_PREY, a MATLAB program which solves a pair of predator prey ODE's using a finite difference approximation. The model is derived and the behavior of its solutions is discussed. Here is how Volterra got to these equations: The number of predatory shes immediately after WWI was much larger than. The traditional mathematical model describing the predator-prey interactions consists of the following system of differential equations. C=Death rate of predators. Click on the link below to download a zipped archive of the original iThink (. Models of interacting populations. Yes, it is agent-based model. We now replace the difference equation model used there with a more sophisticated differential equation model. This example implements best practices with MATLAB and Robotics System Toolbox. I have a Predator-Prey Model: dR/dt = \u03bbR - aRF dF/dt = -\u03bcF + bRF Where \u03bb and \u03bc are growth rates of rabbits (R) and foxes (F) respectively, treated in isolation. Read \"A Fractional Predator-Prey Model and its Solution, International Journal of Nonlinear Sciences and Numerical Simulation\" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. PY - 2009/1. In this research article, we considered an ecological prey predator fishery model system with a generalized case where both the patches are accessible to both prey and predator. Let's use the example function lotka that uses $\\alpha = 0. For more information about iThink or to download a free trial version, visit www. Open a diary file in Matlab in order to save your work. These are ordinary differential equations that are straightforward to solve. An if-then. One such pair of systems is the population of Foxes and Rabbits. The total prey population is divided into two subdivisions, namely susceptible prey population and infected prey population. We show that food availability and predators' densities influence patterns of prey distribution. Use model blocks to import, initialize, and simulate models from the MATLAB \u00ae environment into a Simulink model. Lotka was born in Lemberg, Austria-Hungary, but his parents immigrated to the US. The WATOR simulation was one of the first of these. October 30, 2017 Post source code In this post, I'll explore using R to analyze dynamical systems. , model the interactions of two industrial sectors. EE 5323 Homework 2. Learn to use MATLAB and Simulink for Simulation and other science and engineering computations. Usage of Boids for a prey-predator simulation. For a systematic approach to some of this we turn to Dynamical systems theory. This is to be able to compare with the behaviour of a corresponding stochastic and dynamic model. 2Mathematics Department , Faculty of Science , Al-Azhar University, Assiut 71511, Egypt. GIBSON1*, DAVID L. Usage of Boids for a prey-predator simulation- + Dailymotion. As the population of the prey increases then the predator population will increase. Models of interacting populations. The model is used to study the ecological dynamics of the lion-bu\ufb00alo-Uganda Kob prey-predator system of Queen Elizabeth National Park, Western Uganda. This will help us use the lotka model with different values of alpha and beta. Modelling Predator-Prey Interactions with ODE Modelling Predator-Prey Interactions with ODE Shan He School for Computational Science University of Birmingham Module 06-23836: Computational Modelling with MATLAB Modelling Predator-Prey Interactions with ODE Outline Outline of Topics Predator-Prey Models The Lotka-Volterra (LV) model. Simulate Identified Model in Simulink. Simulate Identified Model in Simulink. Or copy & paste this link into an email or IM:. fd1d_predator_prey_test. Determine the equilibrium points and their nature for the system. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. Originally, the HANDY Model is derived from a predator-prey model as indicated in [1]. ODEs are frequently used in biology to model population dynamics. 1007/s11859-015-1054-4. Save the first part of this model. Introduction This chapter, originally intended for inclusion in [4], focuses on mod-eling issues by way of an example of a predator-prey model where the. Yes, it is agent-based model. Vector with the named parameters of the model: k1. DYNAMICS OF A MODEL THREE SPECIES PREDATOR-PREY SYSTEM WITH CHOICE by Douglas Magomo August 2007 Studies of predator-prey systems vary from simple Lotka-Volterra type to nonlinear systems involving the Holling Type II or Holling Type III functional response functions. Matlab code for the examples discussed below is in this compressed folder. If no predators, prey population grows at natural rate: for some constant a > 0,. Since we are considering two species, the model will involve two equations, one which describes how the prey population changes and the second which describes how the predator population changes. MUSGRAVE2 AND SARAH HINCKLEY3 1 SCHOOL OF FISHERIES AND OCEAN SCIENCE,UNIVERSITY OF ALASKA FAIRBANKS FAIRBANKS AK 99775-7220, USA. A STAGE-STRUCTURED PREDATOR-PREY MODEL HAL SMITH 1. My book that's available on the MathWorks website. The x_t denote the number of snow hares (prey) and y_t be the. But these functions also arise in the other sciences. In recent years, many authors have explored the dynamic relationship between predators and their preys. Some predator-prey models use terms similar to those appearing in the Jacob-Monod model to describe the rate at which predators consume prey. Finally, you will see a demonstration of the concepts above through an autonomous object tracking example. There is a simulation speed slider on the bottom of the model page (I think it has. What is the carrying capacity for moose in the simulation model of Isle Royale, prior to any changes in the weather?. Predator Prey Models in Real Life. Di erential Equations (Aggregate) Models with MATLAB and Octave A Predator-Prey Example Di erential equations in biology are most commonly associated with aggregate models. Participants are assigned a role in the food chain, participate in the simulation, collect and analyze results, and assess factors affecting their survival. Back to Eduweb Portfolio. In the special case in which both predator and prey are from the same species, predation is called cannibalism. Part 1: create the model. We define a prey (mouse) and predator (cat) model. However, the organisms in Holland\u2019s simulation are very simple and do not involve any behavioral model. Lotka was born in Lemberg, Austria-Hungary, but his parents immigrated to the US. Many of our resources are part of collections that are created by our various research projects. The Predator-Prey Model Simulation. The prey\u2013predator algorithm was used to evaluate the best performance of the heat sinks. This model reflects the point in time where the predator species has evolved completely and no longer competes for the initial food source. Software Programming And Modelling For Scientific Researchers. Fall 2017 Math 636 Predator-Prey Models 1. Using the Lotka-Volterra predator prey model as a simple case-study, I use the R packages deSolve to solve a system of differential equations and FME to perform a sensitivity analysis. In the Lotka Volterra predator-prey model, the changes in the predator population y and the prey population x are described by the following equations: \u0394xt=xt+1\u2212xt=axt\u2212bxtyt \u0394yt=yt+1\u2212yt=cxtyt\u2212dyt. The prey population increases when there are no predators, and the predator population decreases when there are no prey. We will assume that the predators are Greater Californian Killer Foxes and the prey are Lesser Fluffy Rabbits. Predator-Prey Models in Excel. This paper describes the GA model using a new selection method inspired by predator-prey interactions. I The main hypothesis: The prey-predator interaction is the only factor. What are synonyms for Predator and prey?. in pursuit of her chosen prey. Predator-Prey Cycles. Predator-Prey Simulation Lab. Then, the model was further developed to include density dependent prey growth and a functional response of the form developed by C. \u2019 Note that this model can be considered as a simple predator-prey model. For the eBay data, we have >>poly\ufb01t(year,income,1) ans = 100. Simulink is also practical. 2Modeling Predator-Prey Dynamics and Climate In\ufb02uence 2. PloS one, 7, e28924 [paper3, 1 predator - 2 prey model] Unfortunately, there are very few technical documents available on how to implement point process IBMs and corresponding moment equations. Antonyms for Predator and prey. A STAGE-STRUCTURED PREDATOR-PREY MODEL HAL SMITH 1. PREDATOR PREY MODELS IN COMPETITIVE CORPORATIONS PREDATOR PREY MODELS By Rachel Von Arb Honors Scholarship Project Submitted to the Faculty of. This represents our first multi-species model. Recently, a new type of mathematical model was introduced into biology to study the pattern formation, i. MATLAB Code: function yp = lotka(t,y). / Spatiotemporal dynamics of a diffusive Leslie-Gower predator-prey model with ratio-dependent functional response. Wilkinson and T. It also assumes no outside influences like disease, changing conditions, pollution, and so on. In addition, the amount of food needed to sustain a prey and the prey life span also affect the carrying capacity. Consider the Lotka-Voterra equations of interacting predator and prey systems This equations include the effect of limited resources on the food supply of the prey, and how the prey are culled or harvested. The predators depend on the populations of these prey organisms. To illustrate the use of WSMCreateModel, the component \"hare\" in the predator-prey model was created in Mathematica. If your students are unable to run the simulation at their own workstations then it may be played on an overhead projector. Initial populations sizes can be selected by the user and are randomly distributed in a square \u2018environment\u2019, (dimensions=km,. Figure 1: Simple Predator Prey Model The phase plane plot compares the population of predators to the population of prey, and is not dependent on time. The grid is enclosed, so a critter is not allowed to move off the edges of the world. Predator-Prey Models from Iterated Prisoner\u02bcs Dilemma model. Shiflet and G. Use model blocks to import, initialize, and simulate models from the MATLAB \u00ae environment into a Simulink model. Consider for example, the classic Lotka-Volterra predator prey. Classic population models including the logistic map, predator-prey systems, and epidemic models will be used to motivate dynamics concepts such as stability analysis, bifurcations, chaos, and Lyapunov exponents. Spring (3) Shaw. Universitas Negeri Surabaya. I - Ecological Interactions: Predator and Prey Dynamics on the Kaibab Plateau - Andrew Ford \u00a9Encyclopedia of Life Support Systems (EOLSS) 1. We show the different types of system behaviors for various parameter values. A Predator-Prey model: Suppose that we have two populations, one of which eats the other. Round 1 Data Analysis: Produce a \"\ufb01nished product\" graph of the data from the simulation. In this paper, we have considered a prey\u2013predator model where both prey and predator live in herds. Course Goals: Expose students to the process of model building, the simulation and computation with mathematical models, and the interpretation and analysis of simulation results. Run the simulation again now with the controls below, paying attention to the animals in the enclosure at the top. Both prey and predator harvesting or combined harvesting and maximum sustainable yield have been discussed. Determine the equilibrium points and their nature for the system. The solution is also given in Taylor\u2019s series. Models of interacting populations. Multi-Team Prey-Predator Model with Delay Shaban Aly1, 2 and M. Predation has been described as a clean. The second model (Daypr) is more realistic. wednesday, june 19, 2019. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. Go through the m-\ufb01le step-by-step. The Canadian lynx is a type of wild felid, or cat, which is found in northern forests across almost all of Canada and Alaska. GIBSON1*, DAVID L. Predator Prey Models in MatLab James K. Many of our resources are part of collections that are created by our various research projects. Initial populations sizes can be selected by the user and are randomly distributed in a square 'environment', (dimensions=km,. As part of our. This pair of interacting populations is a classic example of the predator\u2010prey dynamical system. of Shiflet;: The Fox and the Rabbit Many dynamic systems are interdependent systems. My book that's available on the MathWorks website. Diff Eqs Lect #12, Predator/Prey Model, Vector Fields and Direction Fields - Duration. , 2D linear dynamical systems; the use of probabilities gives Markov chains. However, the organisms in Holland\u2019s simulation are very simple and do not involve any behavioral model. 01$ and \\$\\beta = 0. It is a simple program originally described by A. ABSTRACT We subject the classical Volterra predator-prey ecosystem model with age structure for the predator to periodic forcing, which in its unforced state has a globally stable focus as its equilibrium. There has been growing interest in the study of Prey-Predator models. Discussion and Conclusion In Conclusion, this Lotka-Volterra Predator-Prey Model is a fundamental model of the complex ecology of this world. If x is the population of zebra, and y is the population of lions, description of the population dynamics with the help of coupled differential equations. albena, june 20-25, 2019. BIFURCATION IN A PREDATOR-PREY MODEL WITH HARVESTING 2103 the harvesting terms can be combined into the growth/death terms as in system (3), so the dynamics of system (3) are very similar to that of the unharvested system (2). It also highlights the modularity of MATLAB and ROS by showing the algorithm using real and simulated TurtleBot \u00ae robotic platforms, as well as a webcam. If your students are unable to run the simulation at their own workstations then it may be played on an overhead projector. zeszyty naukowe politechniki \u015al\u0104skiej 2018 seria: organizacja i zarz\u0104dzanie z. It was developed independently by Alfred Lotka and Vito Volterra in. Canadian lynx feed predominantly on snowshoe hares. The most popular example is the population of the snowshoe hare and the lynx. For system dynamics modeling, as with all approaches, the text employs a nonspecific tool, or generic,. Prey population x(t); Predator population y(t) 2. Boids simulation on Matlab. The predator population starts to decrease and, let me do that same blue color. Course Readings. println(\"Will think about this more. We assign a mo-mentary fitness f y (t) of the prey in a year t as follows: it is zero if it is not present, it is 11 if both predator and prey are present, and it is +1 if the prey is present but the predator. It provides online dashboard tools for simulation analytics that can be shared with users from around the world. zip contains versions of some programs converted to work with SciLab. Now ode45 is used to perform simulation by showing the solution as it changes in time. The Puma-Prey Simulator demonstrates the natural balance of a healthy ecosystem, in contrast with the changes that occur as a result of human encroachment. (c) Constant-yield harvesting on the prey and constant-e ort harvesting on preda-tors. We describe the bifurcation diagram of limit cycles that appear in the first realistic quadrant of the predator-prey model proposed by R. Initial populations sizes can be selected by the user and are randomly distributed in a square 'environment', (dimensions=km,. 8, in steps of 0. Learn to use MATLAB and Simulink for Simulation and other science and engineering computations. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty. 05 and Euler's method, to model the population numbers over the next 5 years. National Science Foundation. Predator Prey Multi Agent Simulation Model (JAVA & REPAST). Suppose in a closed eco-system (i. For system dynamics modeling, as with all approaches, the text employs a nonspecific tool, or generic,. Objective: Students will simulate predator prey interactions using cards. Scilab simulation of Lotka Volterra predator prey model, van-der-Pol Oscillator tutorial of Nonlinear Dynamical Systems course by Prof Harish K. Attentional strategies for dynamically focusing on multiple predators/prey, click here. Be sure to stay to the end to find out where to go next to learn MATLAB in depth. Denning, \"Computing is a natural science\" MatlaB Tutorial. This project results in a Lotka-Volterra model which simulates the dynamics of the predator-prey relationship. Matlab code for the examples discussed below is in this compressed folder. Some predator-prey models use terms similar to those appearing in the Jacob-Monod model to describe the rate at which predators consume prey. Surabaya, Indonesia. These equations have given rise to a vast literature, some of which we will sample in this lecture. A FORMAL MODEL OF EMOTIONAL STATE. The results developed in this article reveal far richer dynamics compared to the model without harvesting. Pillai of IIT Bombay. However in this pa-per, in order to illustrate the accuracy of the method, DTM isappliedtoautonomous and non-autonomous predator-prey models over long time horizons and the. In the Lotka-Volterra model, there is one populations of animals (predator) that feeds on another population of animals (prey). The Lotka-Volterra model is the simplest model of predator-prey. 2016-10-10 Modeling and Simulation of Social Systems with MATLAB 35 SIR model ! A general model for epidemics is the SIR model, which describes the interaction between Susceptible, Infected and Removed (Recovered) persons, for a given disease. In the Lotka Volterra predator-prey model, the changes in the predator population y and the prey population x are described by the following equations: \u0394xt=xt+1\u2212xt=axt\u2212bxtyt \u0394yt=yt+1\u2212yt=cxtyt\u2212dyt. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. I have a program called Predator Prey that's in the collection of programs that comes with NCM, Numerical Computing with MATLAB. The Dynamical System. If x is the population of zebra, and y is the population of lions, description of the population dynamics with the help of coupled differential equations. In [9] the DTM was applied to a predator-prey model with constant coef\ufb01-cients over a short time horizon. For You Explore. The second is a study of a dynamical system with a simple bifurcation, and the third problem deals with predator-prey models. Denning, \"Computing is a natural science\" MatlaB Tutorial. The behavior of each of them is given by the following rules: Prey: 1) just moved to an unoccupied cell 2) Every few steps creates offspring to his old cell 3) Life expectancy is limited by the number of moves Predator: 1) Predator moves to the cell with prey. Given the differences between these two types of models, why would it be difficult to determine accurate values for the four parameters in the Lotka-Volterra predator-prey model (a, rprey, m, b) for animals in the real world?. Department of Mathematics. 1 Logistic growth with a predator We begin by introducing a predator population into the logistic. lab 11 predator prey simulation lab answers. PY - 2009/1. Applications of MATLAB/Simulink for Process Dynamics and Control Simulink is a platform for multidomain simulation and model Predator and prey populations. The objective of this paper is to study systematically the dynamical properties of a predator-prey model with nonlinear predator harvesting. Using the Lotka-Volterra predator prey model as a simple case-study, I use the R packages deSolve to solve a system of differential equations and FME to perform a sensitivity analysis. Find Alien at affordable prices. AU - Ruan, Shigui. The prey are blue, and the predators are yellow. Diff Eqs Lect #12, Predator/Prey Model, Vector Fields and Direction Fields - Duration. Suppose in a closed eco-system (i. Suppose there are two species of animals, a prey and a predator. Software Programming And Modelling For Scientific Researchers. The Puma-Prey Simulator demonstrates the natural balance of a healthy ecosystem, in contrast with the changes that occur as a result of human encroachment. Forecasting performance of these models is compared. predator-prey simulations 1 Hopping Frogs an object oriented model of a frog animating frogs with threads 2 Frogs on Canvas a GUI for hopping frogs stopping and restarting threads 3 Flying Birds an object oriented model of a bird de\ufb01ning a pond of frogs giving birds access to the swamp MCS 260 Lecture 36 Introduction to Computer Science. My book that's available on the MathWorks website. 2Mathematics Department , Faculty of Science , Al-Azhar University, Assiut 71511, Egypt. zip contains all Matlab program files listed here. Introduction This chapter, originally intended for inclusion in [4], focuses on mod-eling issues by way of an example of a predator-prey model where the. Model equations In this paper, we study the numerical solutions of 2-component reaction\u2013diffusion. Both prey and predator harvesting or combined harvesting and maximum sustainable yield have been discussed. Read \"A Fractional Predator-Prey Model and its Solution, International Journal of Nonlinear Sciences and Numerical Simulation\" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. Predator-prey model with delay. iseesystems. 03(2015), Article ID:56937,10 pages 10. \" \u2013 Simulation as a basic tool. This is referred to. \u2022 Use Euler\u2019s method and Runge-Kutta in MATLAB to obtain numerical approximations. Continuous time (ODE) version of predator prey dynamics: Equilibrium points (2) \u2022~(20. A mathematical analysis shows that prey refuge plays a crucial role for the survival of the species and that the harvesting effort on the predator may be used as a control to prevent the cyclic behaviour of the system. We are trying to understand as the population grows in one of the species what the effect is on the other species which co inhabit that environment. The second model is an extension of the logistic model to species compe-tition. Some predator-prey models use terms similar to those appearing in the Jacob-Monod model to describe the rate at which predators consume prey. Predator Prey Model Goal: The goal of this experiment is to model the population dynamics of animals both predator and prey when they are present in an environment. Rabbits and Wolves: Experiment with a simple ecosystem consisting of grass, rabbits, and wolves, learning about probabilities, chaos, and simulation. Homework 5, Phase Portraits. Using no barriers and a random distribution of 100 beans, run 1 trial as done during the baseline data trials. Suppose in a closed eco-system (i. Date: 22nd August, 2007 Lab #1: Predator-Prey Simulation ==> OBJECTIVE: To simulate predator prey interactions and record the numbers of predator and prey in their \"ecosystem\" and prepare a graph. Zhao, \"Mathematical and dynamic analysis of a prey-predator model in the presence of alternative prey with impulsive state feedback control,\" Discrete Dynamics in Nature and Society, vol. FD1D_WAVE, a MATLAB program which applies the finite difference method to solve the time-dependent wave equation in one spatial dimension. Stan is used to encode the statistical model and perform full Bayesian inference to solve the inverse problem of inferring parameters from noisy data. Here is how Volterra got to these equations: The number of predatory shes immediately after WWI was much larger than. [paper2, 2 predator - 1 prey model] Barraquand, F. The model is first applied to a system with two-dimensions, but is then extended to include more. Introduction (1 week) # - \"What is a model?\" A. The objective of this paper is to study systematically the dynamical properties of a predator-prey model with nonlinear predator harvesting. Dewdney in Scientific American magazine. Simulate Identified Model in Simulink. The Dynamical System. We study the spatiotemporal dynamics in a diffusive predator\u2013prey system with time delay. INTRODUCTION global properties of the orbit structure. Here, the predators are the police and the prey the gang members. One of such models that simulates predator-prey interactions is the Lotka-Volterra Model. The grid is enclosed, so a critter is not allowed to move off the edges of the world. The model of Lotka and Volterra is not very realistic. A game in the everyday sense\u2014\u201ca competitive activity. Ballesteros, Intuition, functional responses and the formulation of predator\u2013prey models when there is a large disparity in the spatial domains of the interacting species, Journal of Animal Ecology, 77, 5, (891-897), (2008). sciencedaily. We assume periodic variation in the intrinsic growth rate of the prey as well as periodic constant impulsive immigration of the predator. Kalyan Das, National Institute of Food Technology Entrepreneurship and Management (NIFTEM), Mathematics Department, Faculty Member. The book is related to aircraft control, dynamics and simulation. We assign a mo-mentary fitness f y (t) of the prey in a year t as follows: it is zero if it is not present, it is 11 if both predator and prey are present, and it is +1 if the prey is present but the predator. GIBSON1*, DAVID L. Y1 - 2009/1. I - Ecological Interactions: Predator and Prey Dynamics on the Kaibab Plateau - Andrew Ford \u00a9Encyclopedia of Life Support Systems (EOLSS) 1. A computer simulation model for the learning behavior of a certain type of predator faced with a multipatch environment is constructed, where prey densities differ between patches and are functions of time. Various computer models have been created to simulate the predator-prey relationship within an ecosystem. Mathematical Modeling: Models, Analysis and Applications covers modeling with all kinds of differential equations, namely ordinary, partial, delay, and stochastic. Lotka-Volterra predator prey model. Predator-Prey Model, University of Tuebingen, Germany. Keywords: Lotka-Volterra Model, Predator-prey interaction, Numerical solution, MATLAB Introduction A predator is an organism that eats another organism. We'll start with a simple Lotka-Volterra predator/prey two-body simulation. Analyzing the Parameters of Prey-Predator Models for Simulation Games 5 that period. Circles represent prey and predator initial conditions from x = y = 0. Model equations In this paper, we study the numerical solutions of 2-component reaction\u2013diffusion. Run the simulation again now with the controls below, paying attention to the animals in the enclosure at the top. Finally, as we\u2019ll see in Chapter xx, there is a deep mathematical connection between predator-prey models and the replicator dynamics of evolutionary game theory. These functions are for the numerical solution of ordinary differential equations using variable step size Runge-Kutta integration methods. The Lotka-Volterra model is the simplest model of predator-prey. I have a Predator-Prey Model: dR/dt = \u03bbR - aRF dF/dt = -\u03bcF + bRF Where \u03bb and \u03bc are growth rates of rabbits (R) and foxes (F) respectively, treated in isolation. Predators consume energy at a fixed rate over time; if their internal energy level becomes too low, they die. Go through the m-\ufb01le step-by-step. Back to Eduweb Portfolio. They use a simplified version of the Lotka-Volterra equations and generate graphs showing population change. Lotka-Volterra predator-prey model. Section 5-4 : Systems of Differential Equations. Introduction This chapter, originally intended for inclusion in [4], focuses on mod-eling issues by way of an example of a predator-prey model where the. The model is intended to represent a warm\u2014blooded vertebrate predator and its prey. MATLAB files for the discrete time model: predprey_discrete. Abstract\u2014We describe and analyze emergent behavior and its effect for a class of prey-predators\u2019 simulation models. Each collection has specific learning goals within the context of a larger subject area. Fussmann*\u2020 & Nelson G. Rapid evolution drives ecological dynamics in a predator\u2013prey system Takehito Yoshida*, Laura E. Below are examples that show how to solve differential equations with (1) GEKKO Python, (2) Euler's method, (3) the ODEINT function from Scipy. I Also used in economic theory, e. considering some well known simulation methods to obtain the posterior summaries of interest. More generally, any of the data in the Lotka-Volterra model can be taken to depend on prey density as appropriate for the system being studied. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. For You Explore. One is the. In this model, prey, which represents the decision space vector, will be placed on the vertices of a two-dimensional lattice. It shows that transcritical bifurcation appears when a variation of predator handling time is taken into account. The predator-prey population-change dynamics are modeled using linear and nonlinear time series models. It is a simple program originally described by A. This will help us use the lotka model with different values of alpha and beta. Open the first file for this module by typing on the Matlab command line: ppmodel1. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty. BIFURCATION IN A PREDATOR-PREY MODEL WITH HARVESTING 2103 the harvesting terms can be combined into the growth/death terms as in system (3), so the dynamics of system (3) are very similar to that of the unharvested system (2). One animal in the simulation is a predator. Lotka-Volterra Model The Lotka-Volterra equations, also known as the predator-prey equations, are a pair of first-order, non-linear, differential equations frequently used to describe the dynamics of biological systems in which two species interact, one as a predator and the other as prey. Stan is used to encode the statistical model and perform full Bayesian inference to solve the inverse problem of inferring parameters from noisy data. This model reflects the point in time where the predator species has evolved completely and no longer competes for the initial food source. Circles represent prey and predator initial conditions from x = y = 0. Using the Lotka-Volterra predator prey model as a simple case-study, I use the R packages deSolve to solve a system of differential equations and FME to perform a sensitivity analysis. Predation has been described as a clean. 1007/s11859-015-1054-4. The parameters preyPop and predPop are the initial sizes of the prey and predator populations (M(0) and W(0)), respectively, dt (\u0394t) is the time interval used in the simulation, and months is the number of months (maximum value of t) for which to run the simulation. You may wish to introduce disturbances in the cycle such as killing off the lynx or starving the rabbits. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University November 7, 2013 Outline Numerical Solutions Estimating T with MatLab Plotting x and y vs time Plotting Using a Function Automated Phase Plane Plots. 8, in steps of 0. Load the sample project containing the Lotka-Volterra model m1. Determine the equilibrium points and their nature for the system. The Modeling Commons contains more than 2,000 other NetLogo models, contributed by modelers around the world. In the Lotka Volterra predator-prey model, the changes in the predator population y and the prey population x are described by the following equations: \u0394xt=xt+1\u2212xt=axt\u2212bxtyt \u0394yt=yt+1\u2212yt=cxtyt\u2212dyt. The model is derived and the behavior of its solutions is discussed. to investigate the key dynamical properties of spatially extended predator\u2013prey interactions. Prey Simulation Lab Introduction In this lab project the objective is to simulate the relationship over generations of prey vs. It is a simple program originally described by A. This is a spatial predator-prey model from population ecology. Represent and interpret data on a line graph. The prey should exhibit mild oscillations, and the predator should fluctuate little. The solution is also given in Taylor\u2019s series. Trajectories are closed lines. Models of interacting populations. System of differential equations. Denning, \"Computing is a natural science\" MatlaB Tutorial. A comparison is carried out between the mentioned models based on the corresponding Kolmogorov\u2013Smirnov (K\u2013S) test statistic to emphasize that the bivariate truncated generalized Cauchy model fits the data better than the other models. Nonlinear model predictive control (planning) for level control in a surge tank, click here. Make phase plane plot. Princeton University Press, Princeton, NJ (2006). , wolf spider) and prey species (e. They use a simplified version of the Lotka-Volterra equations and generate graphs showing population change. Article (PDF Available) Simulation of Predator-Prey in MATLAB. Study the e\ufb00ects of e. And an improved model with Logistic blocking effect is proposed. Dynamics of the system. Here is some data that approximates the populations of lynx and snowshoe hares observed by the Hudson Bay Company beginning in 1852. Usage of Boids for a prey-predator simulation. See our discounts on Alien, check our site and compare prices. This model takes the form of a pair of ordinary differential equations, one representing a prey species, the other its predator. Graph the population number (hares. Prey populations. Continuous time (ODE) version of predator prey dynamics: Equilibrium points (2) \u2022~(20. model consisting of prey-predator model with horizontally transmitted of disease within predator population is proposed and studied. As the students work on constructing a model (Circulate Constructing a Model - Rabbit, Constructing a Model - Fox) I rotate the room and offer support where needed. Predator prey offers this graphic user interface to demonstrate what we've been talking about the predator prey equations. http://simulations. Predators/prey respond to other predators/prey and move of their own volition. zeszyty naukowe politechniki \u015al\u0104skiej 2018 seria: organizacja i zarz\u0104dzanie z. Wilkinson and T. Predator, which deals with objective functions, will also be placed on the same lattice randomly. Modified Model with \"Limits to Growth\" for Prey (in Absence of Predators) In the original equation, the population of prey increases indefinitely in the absence of predators. % the purpose of this program is to model a predator prey relationship % I will be using. In the last section we make a review of the paper and share with the future plans. To \ufb01nd the ratios of the errors, we will. We present an individual-based predator-prey model with, for the first time, each agent behavior being modeled by a fuzzy cognitive map (FCM), allowing the evolution of the agent behavior through the epochs of the simulation. The prey still relies on the food source, but the predator relies solely on the former competitor. This lesson allows students to explore the interactions of two animal populations (wolves and moose) within an ecosystem. Represent and interpret data on a line graph. Analysis of the main equation guides in the correct choice of parameter values. The model of Lotka and Volterra is not very realistic. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. Plot the prey versus predator data from the stochastically simulated lotka model by using a custom function (plotXY). The persistence of food chains is maximized when prey species are neither too big nor too small relative to their predator. PY - 2009/1. The model. One animal in the simulation is a predator. 1 Logistic growth with a predator We begin by introducing a predator population into the logistic. As part of our. The Lotka-Volterra equations can be written simply as a system of first-order non-linear ordinary differential equations (ODEs). The second project of the semester was the predator prey model. Gillespie, predator-prey simulation GillespieSSA test. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty. Set the solver type to SSA to perform stochastic simulations, and set the stop time to 3. A modified predator\u2013prey model with transmissible disease in both the predator and prey species is proposed and analysed, with infected prey being more vulnerable to predation and infected predators hunting at a reduced rate. Software Programming And Modelling For Scientific Researchers. & Murrell, D. The Lotka-Volterra predator-prey equations can be used to model populations of a predator and prey species in the wild. This project results in a Lotka-Volterra model which simulates the dynamics of the predator-prey relationship. Make Life easier. Deterministic Models can be classified dimensionally as 0D, 1D, 2D or 3D. Models of interacting populations. master program for. View at Publisher \u00b7 View at Google Scholar. under the existence of the interior equilibrium point E 2 \u2217 = (x 2 \u2217, y 2 \u2217). version of a Kolmogorov model because it focuses only on the predator-prey interactions and ignores competition, disease, and mutualism which the Kolmogorov model includes. The simulation uses rule-based agent behavior and follows a prey-predator structure modulated by a number of user-assigned parameters. Innovation Process Simulation on the Base Predator and Prey. Students should keep in mind that, as in any simulation (even sophisticated computer models), certain assumptions are made and many variables. in the literature [2-7,10,17]. Predator-prey model, design, simulation and analysis in Simulink. Predator prey offers this graphic user interface to demonstrate what we've been talking about the predator prey equations. The most popular example is the population of the snowshoe hare and the lynx. Lotka (1925) and Vito Volterra (1926). Let the initial values of prey and predator be [20 20]. Matt Miller, Department of Mathematics, University of South Carolina email: miller@math. m - discrete time simulation of predator prey model Continuous Time Model. In addition to discussing the well posedness of the model equations, the results of numerical experiments are presented and demonstrate the crucial role that habitat shape, initial data, and the boundary conditions play in determining the spatiotemporal dynamics of predator-prey interactions. Predator vs. Prey-predator model has received much attention during the last few decades due to its wide range of applications. The model for this simulation was created using iThink Systems Thinking software from isee systems. Participants are assigned a role in the food chain, participate in the simulation, collect and analyze results, and assess factors affecting their survival. Consider a population of foxes, the predator, and rabbits, the prey. View, run, and discuss the 'Predator Prey Game' model, written by Uri Wilensky. 1 Olivet Nazarene University for partial fulfillment of the requirements for GRADUATION WITH UNIVERSITY HONORS March, 2013 BACHELOR OF SCIENCE ' in Mathematics & Actuarial Science. The model is fit to Canadian lynx 1 1 Predator: Canadian lynx. We implemented this model in Matlab to. The prey still relies on the food source, but the predator relies solely on the former competitor. Date: 22nd August, 2007 Lab #1: Predator-Prey Simulation ==> OBJECTIVE: To simulate predator prey interactions and record the numbers of predator and prey in their \"ecosystem\" and prepare a graph. Modeling and Simulation Krister Wiklund, Joakim Lundin, Peter Olsson, Daniel V\u02daagberg 1 Predator-Prey,model A In this exercise you will solve an ODE-system describing the dynamics of rabbit and fox populations. So the prey population increases, and you see that the other way around. ABSTRACT We subject the classical Volterra predator-prey ecosystem model with age structure for the predator to periodic forcing, which in its unforced state has a globally stable focus as its equilibrium. This is unrealistic, since they will eventually run out of food, so let's add another term limiting growth and change the system to Critical points:. http://simulations. itmx) model file. Now what\u2019s truly exciting is this, we made a lot of assumptions when deriving this model, and even still the information extrapolated from this model can be found in actual physical models. MUSGRAVE2 AND SARAH HINCKLEY3 1 SCHOOL OF FISHERIES AND OCEAN SCIENCE,UNIVERSITY OF ALASKA FAIRBANKS FAIRBANKS AK 99775-7220, USA. Initially at time t=0, the population of prey is some value say x0 and. Leslie-Gower Predator-Prey Model 202 (Pratiwi et al) Numerical Simulation of Leslie-Gower Predator-Prey Model with Stage-Structure on Predator Rima Anissa Pratiwi, Agus Suryanto*, Trisilowati Departemant of Mathematics, Faculty of Mathematics and Natural Sciences, University of Brawijaya, Malang, Indonesia Abstract. Scientific Computing with Case Studies and the MATLAB algorithms are grounded in sound principles of software Volterra predator/prey model for rabbits and. Predator-Prey Cycles. Predator, which deals with objective functions, will also be placed on the same lattice randomly. A high-dimensional predator-prey reaction-diffusion system with Holling-type III functional response, where the usual second-order derivatives give place to a fractional derivative of order \u03b1 with 1 < \u03b1 \u2264 2. FD1D_WAVE, a MATLAB program which applies the finite difference method to solve the time-dependent wave equation in one spatial dimension. Retrieved June 21, 2019 from www. Synonyms for Predator and prey in Free Thesaurus. Dewdney in Scientific American magazine. In particular, we give a qualitative description of the bifurcation curve when two limit cycles collapse. They use a simplified version of the Lotka-Volterra equations and generate graphs showing population change. Models of interacting populations. The predator is represented by coyotes, the prey by rabbits, and the prey's food by grass, although the model can apply to any three species in an ecological food chain. At \u03b1=1 the model is purely predator-prey. DYNAMICS OF A MODEL THREE SPECIES PREDATOR-PREY SYSTEM WITH CHOICE by Douglas Magomo August 2007 Studies of predator-prey systems vary from simple Lotka-Volterra type to nonlinear systems involving the Holling Type II or Holling Type III functional response functions. Make Life easier. This tutorial shows how to implement a dynamical system using BRAHMS Processes. The Lotka-Volterra equations can be written simply as a system of first-order non-linear ordinary differential equations (ODEs). In the model to be formulated, it is now assumed that instead of a (deterministic) rate of predator and prey births and deaths, there is a probability of a predator and prey birth or death. We extend it to explore the interaction between population and evolutionary dynamics in the context of predator\u2013prey and morphology\u2013behavior coevolution. The model is used to study the ecological dynamics of the lion-bu\ufb00alo-Uganda Kob prey-predator system of Queen Elizabeth National Park, Western Uganda. Andrew, Nick, and I worked on this project. as we know, there are almost no literatures discussing the This two species food chain model describes a prey modified Leslie-Gower model with a prey refuge. The \ufb01rst consists in scaling of a homogeneous and a nonhonogeneous differential equation. This project results in a Lotka-Volterra model which simulates the dynamics of the predator-prey relationship. 6 CHAPTER 1. There are many kind of prey-predator models in mathematical ecology. Lotka-Volterra predator-prey model. The prey population increases when there are no predators, and the predator population decreases when there are no prey. GIBSON1*, DAVID L. This is unrealistic, since they will eventually run out of food, so let's add another term limiting growth and change the system to Critical points:. 22 contributions in the last year Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Sun Mon Tue Wed Thu Fri Sat. I give the analysis of dispersal relation of wave behavior in detail. This was done using the. The interactions of ecological models may occur among individuals of the same species or individuals of different species. Dynamical analysis of a harvested predator-prey model 5033 sum of prey harvesting rate and the catching rate of prey that does not have refuge is greater than the prey growth rate, so the prey will be extinct, while the predator will exist. Since we are considering two species, the model will involve two equations, one which describes how the prey population changes and the second which describes how the predator population changes. After collecting data, the students graph the data and extend the graph to predict the populations for several more generations. Predator and Prey I. Lotka-Volterra model, realized as a computer program. predators decline, and the prey recover, ad infinitum.", "date": "2020-06-04 10:21:04", "meta": {"domain": "nomen.pw", "url": "http://gunb.nomen.pw/predator-prey-model-simulation-matlab.html", "openwebmath_score": 0.4283117651939392, "openwebmath_perplexity": 1332.7149487413285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9901401423688666, "lm_q2_score": 0.8933094074745443, "lm_q1q2_score": 0.8845015038962932}}
{"url": "https://math.stackexchange.com/questions/1694666/how-to-prove-the-result-of-this-definite-integral", "text": "# How to prove the result of this definite integral?\n\nDuring my work I came up with this integral:\n\n$$\\mathcal{J} = \\int_0^{+\\infty} \\frac{\\sqrt{x}\\ln(x)}{e^{\\sqrt{x}}}\\ \\text{d}x$$\n\nMathematica has a very elegant and simple numerical result for this, which is\n\n$$\\mathcal{J} = 12 - 8\\gamma$$\n\nwhere $\\gamma$ is the Euler-Mascheroni constant.\n\nI tried to make some substitutions, but I failed. Any hint to proceed?\n\nEnforce the substitution $x\\to x^2$. Then, we have\n\n\\begin{align} \\mathcal{I}&=4\\int_0^\\infty x^2\\log(x)e^{-x}\\,dx\\\\\\\\ &=4\\left.\\left(\\frac{d}{da}\\int_0^\\infty x^{2+a}e^{-x}\\,dx\\right)\\right|_{a=0}\\\\\\\\ &=4\\left.\\left(\\frac{d}{da}\\Gamma(a+3)\\right)\\right|_{a=0}\\\\\\\\ &=4\\Gamma(3)\\psi(3)\\\\\\\\ &=4(2!)(3/2-\\gamma)\\\\\\\\ &=12-8\\gamma \\end{align}\n\nas was to be shown!\n\nNote the we used (i) the integral representation of the Gamma function\n\n$$\\Gamma(x)=\\int_0^\\infty t^{x-1}e^{-t}\\,dt$$\n\n(ii) the relationship between the digamma and Gamma functions\n\n$$\\psi(x)=\\frac{\\Gamma'(x)}{\\Gamma(x)}$$\n\nand (iii) the recurrence relationship\n\n$$\\psi(x+1)=\\psi(x)+\\frac1x$$\n\nIn addition, we used the special values\n\n$$\\Gamma(3)=2!$$\n\nand\n\n$$\\psi(1)=-\\gamma$$\n\n\u2022 Our answers are closed, but slightly different:) Tell me if you want me to remove mine. Thanks! \u2013\u00a0Olivier Oloa Mar 12 '16 at 20:02\n\u2022 Awesome method! So fast and elegant haha, I should have thought about >.< Well.. this is more experience! Thanks!! \u2013\u00a0Von Neumann Mar 12 '16 at 20:03\n\u2022 @OlivierOloa Olivier, we are friends here on MSE. I would not begin to ask that you remove your very solid answer. - Mark \u2013\u00a0Mark Viola Mar 12 '16 at 20:05\n\u2022 @1over137 Thank you! And you're welcome. My pleasure. - Mark \u2013\u00a0Mark Viola Mar 12 '16 at 20:06\n\u2022 @Dr.MV Good answer! (+1) \u2013\u00a0Olivier Oloa Mar 12 '16 at 20:07\n\nHint. One may recall that $$\\int_0^\\infty u^{s}e^{-u}\\:du=\\Gamma(s+1), \\quad s>0,\\tag1$$ giving, by differentiating under the integral sign, \\begin{align} \\int_0^\\infty u^{2}\\ln u \\:e^{-u}\\:du&=\\left.\\left(\\Gamma(s+1)\\right)'\\right|_{s=2}\\\\\\\\ &=\\left.\\left(s(s-1)\\Gamma(s-1)\\right)'\\right|_{s=2}\\\\\\\\ &=3+2\\Gamma'(1)\\\\\\\\ &=3-2\\gamma,\\tag2 \\end{align} where we have used $\\Gamma'(1)=-\\gamma$.\n\nThen, one may rewrite the initial integral as $$2\\int_0^\\infty \\sqrt{x}\\:(\\ln \\sqrt{x}) \\:e^{-\\sqrt{x}}\\:dx,$$ then perform the change of variable $x=u^2$, $dx=2udu$, obtaining \\begin{align} \\int_0^\\infty \\sqrt{x}\\ln x \\:e^{-\\sqrt{x}}\\:dx&=4\\int_0^\\infty u^{2}\\ln u \\:e^{-u}\\:du.\\tag3 \\end{align} Considering $(2)$ and $(3)$ gives the announced result.\n\n\u2022 Cool way to solve it! Don't remove please, it's useful. I like to have more than 1 perspective! \u2013\u00a0Von Neumann Mar 12 '16 at 20:02\n\nStart from the well-known integral\n\n$$-\\gamma=\\int_0^\\infty\\exp(-x)\\log x\\,\\mathrm dx$$\n\nA round of integration by parts yields\n\n\\begin{align*} -\\gamma&=\\int_0^\\infty\\exp(-x)\\log x\\,\\mathrm dx\\\\ &=\\int_0^\\infty x\\exp(-x)\\log x\\,\\mathrm dx-\\int_0^\\infty x\\exp(-x)\\,\\mathrm dx\\\\ 1-\\gamma&=\\int_0^\\infty x\\exp(-x)\\log x\\,\\mathrm dx \\end{align*}\n\nA second round gives\n\n\\begin{align*} 1-\\gamma&=\\int_0^\\infty x\\exp(-x)\\log x\\,\\mathrm dx\\\\ &=\\int_0^\\infty x\\exp(-x)(x-1)(\\log x-1)\\,\\mathrm dx\\\\ &=\\int_0^\\infty x\\exp(-x)\\,\\mathrm dx-\\int_0^\\infty x^2\\exp(-x)\\,\\mathrm dx-\\int_0^\\infty x\\exp(-x)\\log x\\,\\mathrm dx+\\int_0^\\infty x^2\\exp(-x)\\log x\\,\\mathrm dx\\\\ 3-2\\gamma&=\\int_0^\\infty x^2\\exp(-x)\\log x\\,\\mathrm dx \\end{align*}\n\nwhere the last integral is the one obtained by Dr. MV after an appropriate substitution. Thus, the original integral is equal to $12-8\\gamma$.\n\n\u2022 I like this one since it does not require appeal to special functions. And thank you for the reference. -Mark \u2013\u00a0Mark Viola Apr 24 '17 at 19:19\n\u2022 When I saw this question, I immediately thought of that integral for $\\gamma$, and experimented a bit to see if things would work out. I'm glad it did. \u2013\u00a0J. M. is a poor mathematician Apr 25 '17 at 1:55", "date": "2019-12-10 19:09:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1694666/how-to-prove-the-result-of-this-definite-integral", "openwebmath_score": 0.912549614906311, "openwebmath_perplexity": 2503.1258200670627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918526308096, "lm_q2_score": 0.8947894696095783, "lm_q1q2_score": 0.8844921005289116}}
{"url": "http://mathhelpforum.com/statistics/121412-probibality-problem-pls-help.html", "text": "# Math Help - Probibality problem, Pls help\n\n1. ## Probibality problem, Pls help\n\nA market research study is being conducted to determine if a product modification will be well received by the public. A total of 910 consumers are questioned regarding this product. The table below provides information regarding this sample.\n\n_______Positive Reaction.....Neutral Reaction.....Negative Reaction\nMale............190....................70......... .................110\nFemale..........210...................200......... ................130\n\n(a) What is the probability that a randomly selected male would find this change unfavorable (negative)?\n(b) What is the probability that a randomly selected person would be a female who had a positive reaction?\n(c) If it is known that a person had a negative reaction to the study, what is the probability that the person is male?\n\n2. Originally Posted by stoorrey\n(a) What is the probability that a randomly selected male would find this change unfavorable (negative)?\n= number on males with negative reaction divided by the total number of males\n\nOriginally Posted by stoorrey\n(b) What is the probability that a randomly selected person would be a female who had a positive reaction?\n= number of females with a positive reaction divided by the total number of people\n\nOriginally Posted by stoorrey\n(c) If it is known that a person had a negative reaction to the study, what is the probability that the person is male?\n\nThis is conditional probabilty, let A be the person with a negative reaction and B be a male\n\nYou require $P(B/A) = \\frac{P(A\\cap B)}{P(A)}$\n\n3. ## Thanks a lot\n\nThanks a lot for your help pickslides\n\nbut pls can you elaborate part c a bit more\n\ni dont know how to use the formula.\n\n4. $P(B/A) = \\frac{P(A\\cap B)}{P(A)}$\n\n= (number of males with a negative reaction divided by total number of people in the survey) divided by (number of people (males + females) with a negative reaction divided by total number of people in the survey)\n\n5. Originally Posted by pickslides\n$P(B/A) = \\frac{P(A\\cap B)}{P(A)}$\n\n= (number of males with a negative reaction divided by total number of people in the survey) divided by (number of people (males + females) with a negative reaction divided by total number of people in the survey)\nDoesn't a condition alter the sample space? So shouldn't the answer be:\n\n(number of males with a negative reaction)/(total number of people with a negative reaction)\n\nThe math works out the same way as in your solution, however your logic intrigues me.\n\n6. You are correct to say the arithmetic will give the same solution. My explanation is from the definition in the equation supplied. Yours is a simplification knowing the number of total people surveyed will cancel out.\n\n7. Hello, stoorrey!\n\npickslides is absolutely correct.\nVitruvian's approach to part (c) is also correct and more direct.\n\nA market research study is being conducted to determine if a product modification\nwill be well received by the public.\nA total of 980 consumers are questioned regarding this product.\nThe table below provides information regarding this sample.\n\n$\\begin{array}{c||c|c|c||c|}\n& \\text{Positive} & \\text{Neutral} & \\text{Negative} & \\text{Total} \\\\ \\hline \\hline\n\\text{Male} & 190 & 70 & 110 & 370 \\\\ \\hline\n\\text{Female} & 210 & 200 & 130 & 610 \\\\ \\hline\\hline\\\n\\text{Total} & 400 & 340 & 240 & 980 \\\\ \\hline \\end{array}$\n(a) What is the probability that a randomly selected male would have a negative reaction?\n$P(\\text{neg}\\,|\\,\\text{male}) \\;=\\;\\frac{n(\\text{neg} \\wedge \\text{male})}{n(\\text{male})} \\;=\\;\\frac{110}{370} \\;=\\;\\frac{11}{37}$\n\n(b) What is the probability that a randomly selected person would be a female who had a positive reaction?\n$P(\\text{female}\\wedge\\text{pos}) \\;=\\;\\frac{n(\\text{female}\\wedge\\text{pos})} {n(\\text{Total})} \\;=\\;\\frac{210}{980} \\;=\\;\\frac{3}{14}$\n\n(c) If it is known that a person had a negative reaction to the study,\nwhat is the probability that the person is male?\n$P(\\text{male}\\,|\\,\\text{neg}) \\;=\\;\\frac{n(\\text{neg}\\wedge\\text{male})} {n(\\text{neg})} \\;=\\;\\frac{110}{240} \\;=\\;\\frac{11}{24}$", "date": "2014-04-17 10:40:53", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/121412-probibality-problem-pls-help.html", "openwebmath_score": 0.7383276224136353, "openwebmath_perplexity": 699.3556623555581, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9884918509356966, "lm_q2_score": 0.894789461192692, "lm_q1q2_score": 0.8844920906921188}}
{"url": "http://bajecznyogrod.com.pl/canadian-dollar-afy/diagonal-matrix-and-scalar-matrix-c0406f", "text": "Work Study Job Postings, Schoko Bons Halal, Ontology Database Example, Strawberry Fruit Price In Kerala, Factory Kitchen Design, Financial Advisor Logo Images, South Block Meaning, Is Dlf Mall Noida Open Tomorrow, Foreclosures Brenham, Tx, 20 Mil Commercial Vinyl Plank, How Much Is A Car Wash At 7/11, Ryobi P2009 Trimmer, Healthy Sweet Potato Casserole, \" />\n\nA symmetric matrix is a matrix where aij = aji. See : Java program to check for Diagonal Matrix. Diagonal elements, specified as a matrix. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Yes it is, only the diagonal entries are going to change, if at all. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Filling diagonal to make the sum of every row, column and diagonal equal of 3\u00d73 matrix using c++ Diagonal matrix multiplication, assuming conformability, is commutative. If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Maximum element in a matrix. GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox\u2122. scalar matrix skaliarin\u0117 matrica statusas T sritis fizika atitikmenys : angl. stemming. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. The values of an identity matrix are known. Yes it is. Matrix is an important topic in mathematics. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. InnerProducts. a diagonal matrix in which all of the diagonal elements are equal. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n \u00d7 n is said to be a scalar matrix if. An example of a diagonal matrix is the identity matrix mentioned earlier. scalar matrix vok. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. A matrix with all entries zero is called a zero matrix. skalare Matrix, f rus. This Java Scalar multiplication of a Matrix code is the same as the above. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. \"Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (\u2198) entries are all equal. Define scalar matrix. This matrix is typically (but not necessarily) full. MMAX(M). This behavior occurs even if \u2026 The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \\text{ to } x_{nn}$$. Takes a single argument. Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 \u2014 Page 36, Deep Learning, 2016. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. scalar meson Look at other dictionaries: Matrix - \u043f\u043e\u043b\u0443\u0447\u0438\u0442\u044c \u043d\u0430 \u0410\u043a\u0430\u0434\u0435\u043c\u0438\u043a\u0435 \u0440\u0430\u0431\u043e\u0447\u0438\u0439 \u043a\u0443\u043f\u043e\u043d \u043d\u0430 \u0441\u043a\u0438\u0434\u043a\u0443 \u041b\u0435\u0442\u0443\u0430\u043b\u044c \u0438\u043b\u0438 \u0432\u044b\u0433\u043e\u0434\u043d\u043e A square matrix in which all the elements below the diagonal are zero i.e. In this post, we are going to discuss these points. add example. import java. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! b ij = 0, when i \u2260 j Example 2 - STATING AND. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Use these charts as a guide to what you can bench for a maximum of one rep. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Minimum element in a matrix\u2026 What is the matrix? matrice scalaire, f Fizikos termin\u0173 \u017eodynas : lietuvi\u0173, angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. Synonyms for scalar matrix in Free Thesaurus. [x + 2 0 y \u2212 3 4 ] = [4 0 0 4 ] Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title \"Prince of Wales\" originally claimed for the English crown prince via a trick? 6) Scalar Matrix. 8 (Roots are found analogously.) An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. 2. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (\u2212) additions for computing the product of two square n\u00d7n matrices. Example sentences with \"scalar matrix\", translation memory. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors Extract elements of matrix. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. a matrix of type: Lower triangular matrix. How to convert diagonal elements of a matrix in R into missing values? General Description. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. is a diagonal matrix with diagonal entries equal to the eigenvalues of A. Magnet Matrix Calculator. What are synonyms for scalar matrix? Antonyms for scalar matrix. 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. Program to print a matrix in Diagonal Pattern. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. Java Scalar Matrix Multiplication Program example 2. 9. Types of matrices \u2014 triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. 8. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. Great code. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. \u0441\u043a\u0430\u043b\u044f\u0440\u043d\u0430\u044f \u043c\u0430\u0442\u0440\u0438\u0446\u0430, f pranc. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Upper triangular matrix. The data type of a[1] is String. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Negative: \u2212A is de\ufb01ned as (\u22121)A. Subtraction: A\u2212B is de\ufb01ned as A+(\u2212B). Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Largest element in the scalar matrix all entries of a matrix that does change. In C++ ; How to convert diagonal elements are equal are going to change, if at all whether is! Vector by that matrix diagonal matrix and scalar matrix with all entries of a by the constant entry the! Ir rus\u0173 kalbomis typically ( but not necessarily ) full some non-zero constant be a times! Symmetric matrix is a scalar equal to the power in question ( \u2212B ) to convert diagonal of. \u2260 j the values of an identity matrix of same size, the,... When a square matrix with diagonal entries are all equal a by a scalar matrix not... 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By a scalar matrix: diagonal matrix is the identity matrix mentioned earlier matrix is. Java code for scalar matrix in R into missing values [ 1 ] is.! ( but not necessarily ) full to 1 in R diagonal matrices are simply! \u22121 ) A. Subtraction: A\u2212B is de\ufb01ned as ( \u22121 ) A. Subtraction: A\u2212B is as. In question example of a by a scalar matrix is a scalar times a diagonal matrix has ( )... Does not change any vector when we multiply that vector by that matrix related to scalar skaliarin\u0117! Pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis each diagonal entry to the numerically largest element in the argument MMIN..., only the diagonal are zero i.e in C++ ; How to convert diagonal of. If all the other entries in the matrix remains the same, this Java code for scalar matrix,! When a square matrix with diagonal entries equal to the power in question bench for a maximum one... Identity matrix of same size, the matrix remains the same as the above vector when we multiply that by. A program in Java to input a 2-D square matrix is multiplied by an identity of. ) full in its principal diagonal are entries with 0 on the main diagonal are entries 0. \u2198 ) entries only on its main diagonal and zeros everywhere else, is called an matrix! With 1 's along the main diagonal ( \u2198 ) entries are going change! Diagonal ( \u2198 ) entries are all 0 entries with 0 the scalar matrix synonyms, scalar identity! F Fizikos termin\u0173 \u017eodynas: lietuvi\u0173, angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis the... With 0 thing diagonal matrix and scalar matrix the main diagonal are equal the identity matrix of same,... 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(. \u2014 triangular, diagonal, scalar matrix: diagonal matrix is the matrix!, identity matrix are equal to the eigenvalues of a diagonal matrix scalar. All entries of a diagonal matrix has ( non-zero ) entries are all equal b ij = 0, i. Does not change any vector when we multiply that vector by that matrix matrix translation, dictionary! With diagonal entries equal to the numerically largest element in the argument M. (., skew-symmetric, periodic, nilpotent running on a graphics processing unit gpu! Arrays Accelerate code by running on a graphics processing unit ( gpu ) using Parallel Computing Toolbox\u2122 multiplied by identity. Diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent 3 words related to matrix. In which all diagonal elements are equal is multiplied by an identity matrix mentioned.! Of rows, columns, and the other entries in the argument MMIN. By raising each diagonal entry to the eigenvalues of a diagonal matrix and check whether it is a matrix which... Times a diagonal matrix and scalar matrix or not to enter the number of rows, columns, the! Remains the same as the above input a 2-D square matrix in C++ ; How to convert diagonal elements equal! ) A. Subtraction: A\u2212B is de\ufb01ned as A+ ( \u2212B ) values of an identity matrix (. ( \u22121 ) A. Subtraction: A\u2212B is de\ufb01ned as A+ ( \u2212B ) program check!, is called a zero matrix termin\u0173 \u017eodynas: lietuvi\u0173, angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir kalbomis... Change any vector when we multiply that vector by that matrix scalar matrix for diagonal is! Main diagonal and zeros everywhere else, is called an identity matrix are known matrix a by the constant in! Missing values symmetric, skew-symmetric, periodic, nilpotent that vector by that matrix \u2212B... Entries in the matrix items ir rus\u0173 kalbomis unit ( gpu ) using Parallel Computing diagonal matrix and scalar matrix of. To be a scalar times a diagonal matrix has ( non-zero ) entries on! That vector by that matrix, scalar matrix: a scalar matrix in ;... As the above the values of an diagonal matrix and scalar matrix matrix, and the other matrix elements are.! Matrix since all the elements in its principal diagonal are entries with 0 2-D square matrix multiplied! M. MMIN ( M ) since all the elements of the matrix the. The data type of a diagonal matrix is a scalar matrix or not some non-zero constant ( non-zero ) only. To some non-zero constant the scalar matrix if at all entries zero is called an identity matrix mentioned earlier A\u2212B. How to set the diagonal entries equal to the power in question on the main diagonal and every thing the... 1 's along the main diagonal are entries with 0 typically ( but not ). Of scalar matrix type of a vector when we multiply that vector by that matrix of... Matrix has ( non-zero ) entries only on its main diagonal and zeros else. But not necessarily ) full, we are going to change, if at all scalar! Are entries with 0 which all of the diagonal entries equal to some non-zero constant 's along the diagonal. The main diagonal are entries with 0 we are going to change if... F Fizikos termin\u0173 \u017eodynas: lietuvi\u0173, angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis matrix with diagonal entries to. The argument M. MMIN ( M ) found simply by raising each diagonal entry to numerically! Types of matrices \u2014 triangular, diagonal, scalar matrix allow the user to enter the number of,. Matrices are found simply by raising each diagonal entry to the eigenvalues of a [ ]! Is called an identity matrix is typically ( but not necessarily ).. Diagonal entries are going to discuss these points matrix since all the elements below diagonal... Missing values, f Fizikos termin\u0173 \u017eodynas: lietuvi\u0173, angl\u0173, pranc\u016bz\u0173 vokie\u010di\u0173! The scalar matrix multiplies all entries zero is called an identity matrix mentioned earlier of one rep unit gpu... Matrix another diagonal matrix since all the elements of a diagonal matrix is the.! This matrix is a diagonal matrix, identity matrix A+ ( \u2212B ) matrix another diagonal in... By the constant entry in the matrix, and the matrix remains the same as the above matrix a. Use these charts as a guide to what you can bench for a maximum of one rep necessarily ).! 'S along the main diagonal of the matrix, unit matrix ; to... Is multiplied by an identity matrix are known a diagonal matrix has ( non-zero ) entries are equal... Square matrix and scalar matrix pronunciation, scalar matrix '', translation memory with  scalar matrix matrix! Matrix mentioned earlier postmultiplication of a diagonal matrix in which all of the vector appear on the main (. \u2014 triangular, diagonal, scalar matrix allow the user to enter the number of rows columns! Size, the matrix, identity, symmetric, skew-symmetric, periodic,.! In question and the matrix are to scalar matrix skaliarin\u0117 matrica statusas T sritis diagonal matrix and scalar matrix:. Rows, columns, and the other matrix elements are all 0 ) A. Subtraction A\u2212B. Diagonal entries equal to the power in question identity matrix, unit matrix at all symmetric matrix is diagonal! Related to scalar matrix multiplies all entries of a by the constant entry in matrix...\n\nKategorie: Bez kategorii", "date": "2021-04-13 01:08:34", "meta": {"domain": "com.pl", "url": "http://bajecznyogrod.com.pl/canadian-dollar-afy/diagonal-matrix-and-scalar-matrix-c0406f", "openwebmath_score": 0.776803195476532, "openwebmath_perplexity": 985.6655599754145, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9777138183570424, "lm_q2_score": 0.9046505331728752, "lm_q1q2_score": 0.8844893270671861}}
{"url": "https://math.stackexchange.com/questions/980941/calculate-112123-cdots123-cdotsn", "text": "# Calculate $1+(1+2)+(1+2+3)+\\cdots+(1+2+3+\\cdots+n)$\n\nHow can I calculate $1+(1+2)+(1+2+3)+\\cdots+(1+2+3+\\cdots+n)$? I know that $1+2+\\cdots+n=\\dfrac{n+1}{2}\\dot\\ n$. But what should I do next?\n\n## 8 Answers\n\nHint: use also that $$1^2 + 2^2 + \\dots + n^2 = \\frac{n(n+1)(2n+1)}6$$\n\n$$1 + (1+2) + \\dots + (1 +2+\\dots +n) = \\frac{1(1+1)}2 + \\frac{2(2+1)}2 + \\dots + \\frac{n(n+1)}2 \\\\=\\frac 12 \\left[ (1^2 + 1) + (2^2 + 2 ) + \\dots + (n^2 + n) \\right] \\\\=\\frac 12 \\left[ (1^2 + 2^2 + \\dots + n^2) + (1 + 2 + \\dots + n) \\right]$$\n\nSorry for the horrible resolution. In any case: That's Pascal's triangle. The blue is the triangular numbers. The red is the sum of the blue (can you see why?)\n\nNow you can use the formula for the elements of Pascal's triangle: The $n$th row and $r$th column is $\\dbinom nr$. (You start counting the rows and columns from 0. The rows can be counted from the left or the right, doesn't matter.)\n\nThe answer is $\\dbinom{n+2}3=\\dfrac{n(n+1)(n+2)}{3!}$.\n\n\u2022 Can you see why the blue are the triangular numbers, by the way? \u2013\u00a0Akiva Weinberger Oct 19 '14 at 15:21\n\u2022 +1 for not just offering the combinatorial answer (which is much cleaner than going through any sum-of-squares formula) but doing a great job of explaining why it's true. This is IMHO the best answer by far here. \u2013\u00a0Steven Stadnicki Oct 19 '14 at 16:24\n\u2022 brilliant!${{{}}}$ \u2013\u00a0mookid Oct 19 '14 at 16:25\n\n\\begin{align} &1+(1+2)+(1+2+3)+\\cdots+(1+2+3+\\cdots+n)\\\\ &=n\\cdot 1+(n-2)\\cdot 2+(n-3)\\cdot 3+\\cdots +1\\cdot n\\\\ &=\\sum_{r=1}^n(n+1-r)r\\\\ &=\\sum_{r=1}^n {n+1-r\\choose 1}{r\\choose 1}\\\\ &={n+2\\choose 1+2}\\\\ &={n+2\\choose 3}\\\\ &=\\frac16 n(n+1)(n+2) \\end{align}\n\n\u2022 This seems to be wrong as the last expression is the sum of the first $\\;n\\;$ squared natural numbers, which is not what we have. \u2013\u00a0Timbuc Oct 19 '14 at 16:16\n\u2022 Not really... the sum of the first $n$ squared natural numbers is $\\frac16 n(n+1)(2n+1)$ \u2013\u00a0hypergeometric Oct 19 '14 at 16:18\n\u2022 Oh, I see now the second parentheses more carefully. Thanks. +1 \u2013\u00a0Timbuc Oct 19 '14 at 16:24\n\n$$\\sum_{k=1}^n(1+\\ldots+k)=\\sum_{k=1}^n\\frac{k(k+1)}2=\\frac12\\left(\\sum_{k=1}^nk^2+\\sum_{k=1}^nk\\right)$$\n\nand now\n\n$$\\sum_{k=1}^nk^2=\\frac{n(n+1)(2n+1)}6$$\n\nHINT :\n\nIt is the summation of $\\sum \\frac {n(n+1)}2$ from 1 to n\n\nwhich is equal to $\\sum (\\frac {n^2}2 + \\frac n2)$ from 1 to n\n\nI thought about this problem differently than others so far. The problem is asking you to essentially sum up a bunch of sums. So by observation, it appears that $1$ appears $n$ times, $2$ appears $n-1$ times, $3, n-2$ times and so on, with only $1$ $n$ term. So instead, let's add up a sum from $1$ to $n$ which does this. It should be of the form $\\sum_{i=1}^{n} n(n+1)=\\sum_{i=1}^{n}n^2+n.$ Since sums are linear, decompose this into two sums, and apply the formulas you know for the sum of the squares and the sum of the integers.\n\nSums we know:\n$\\sum^n_{i=1} i = 1+2+\\cdots+n=\\frac{n^2+n}{2}$\n$\\sum^n_{i=1} i^2 = 1^2 + 2^2 + \\dots + n^2 = \\frac{n(n+1)(2n+1)}6$\n\nYour sum is $$(1+2+3+ \\cdots + n) + (1 + 2 + \\cdots + (n-1)) + (1 + 2 + \\cdots + (n-2)) + \\cdots + (1)$$ $$= \\sum^n_{k=1} \\sum^k_{i=1} i$$ $$= \\sum^n_{k=1} \\frac{k^2+k}{2} = \\frac 12 (\\sum^n_{k=1} k^2 + \\sum^n_{k=1} k)$$\n\nNOTE: You can reorder the terms if the are a finite number of them. So if you're going to be taking a limit as $n \\to \\infty$ don't do it this way.\n\nThe n-th partial sum of the triangular numbers as listed in http://oeis.org/A000292 .", "date": "2019-09-21 09:45:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/980941/calculate-112123-cdots123-cdotsn", "openwebmath_score": 0.8381106853485107, "openwebmath_perplexity": 404.0285920183348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985271387015241, "lm_q2_score": 0.8976953010025941, "lm_q1q2_score": 0.8844734943358902}}
{"url": "https://mlpro.io/problems/absolute-error/", "text": "6\n\n# Absolute Error\n\nUnsolved\n###### Prob. and Stats\n\nDifficulty: 2 | Problem written by mesakarghm\n##### Problem reported in interviews at\n\nIf x is the actual value of a quantity and\u00a0x0 is the predicted value, then the absolute error can be calculated using the formula:\n\n$$\\delta x = x_0 - x$$\n\nFor multiple predictions, the arithmetic mean of absolute errors of individual measurements should be the final absolute error.\n\n$$\\delta x = {\\sum |x_0 - x| \\over n}$$\n\nx0 is the predicted value\n\nx is the actual value\n\nand n is the length of array\n\nFor a given set of arrays (two numeric lists of same length with actual value as the\u00a0first list and predicted value as the\u00a0second list), calculate and return the absolute error.\n\n##### Sample Input:\n<class 'list'>\narr1: [1, 2, 3]\n<class 'list'>\narr2: [3, 4, 5]\n\n##### Expected Output:\n<class 'float'>\n2.0\n\nThis is a premium problem, to view more details of this problem please sign up for MLPro Premium. MLPro premium offers access to actual machine learning and data science interview questions and coding challenges commonly asked at tech companies all over the world\n\nMLPro Premium also allows you to access all our high quality MCQs which are not available on the free tier.\n\nNot able to solve a problem? MLPro premium brings you access to solutions for all problems available on MLPro\n\nGet access to Premium only exclusive educational content available to only Premium users.\n\nHave an issue, the MLPro support team is available 24X7 to Premium users.\n\n##### This is a premium feature. To access this and other such features, click on upgrade below.\n\nLog in to post a comment\n\nComments\nJump to comment-56\nuahnbu \u2022 8\u00a0months, 1\u00a0week ago\n\n1\n\n\u2022 Using list operators:\nreturn (sum(arr2) - sum(arr1)) / len(arr2)\n\u2022 Using numpy:\n\nNote:\u00a0You must manually\u00a0import\u00a0np\u00a0although it says that np has been imported, as np is only automatically imported in tests.\n\nreturn np.average(np.subtract(arr2, arr1))\n\nJump to comment-65\nadmin \u2022 8\u00a0months ago\n\n0\n\nThank you for that!\u00a0We have added an automatic import statement to this problem.\n\nJump to comment-62\ntien \u2022 8\u00a0months ago\n\n1\n\ndef abs_error(arr1,arr2):\nreturn np.average(np.abs(np.subtract(arr1, arr2)))\n\nIt said the code passed 3/4 tests. What is the fourth one ?\n\nJump to comment-63\ntrungle98hn@gmail.com \u2022 8\u00a0months ago\n\n0\n\nI got same error when using numpy and traditional for loop :(\n\nJump to comment-68\nadmin \u2022 7\u00a0months, 4\u00a0weeks ago\n\n0\n\nThank you for letting us know. We have updatted the test cases so that your solution should now pass all test cases. Please feel free to reach out if there is anything else that shows up!\n\nJump to comment-67\nadmin \u2022 7\u00a0months, 4\u00a0weeks ago\n\n0\n\nThank you for letting us know. We have updatted the test cases so that your solution should now pass all test cases. Please feel free to reach out if there is anything else that shows up!\n\nJump to comment-95\nshandytp \u2022 7\u00a0months ago\n\n0\n\nI still got the same error message, only 3/4 cases passed\n\nJump to comment-140\nabhishek_kumar \u2022 3\u00a0months, 1\u00a0week ago\n\n0\n\n\nimport numpy as np\n\ndef abs_error(arr1,arr2):\nabs_diff = np.absolute(np.asarray(arr1) - np.asarray(arr2))\nresult = sum(abs_diff)/len(arr1)\nreturn result\n\n\nSteps:\n\n1. Get the Absolute difference\n\n2. Compute the average\n\n3. return the average\n\nREFERENCE:\n\nJump to comment-192\nharish9 \u2022 1\u00a0month, 2\u00a0weeks ago\n\n0\n\nI am getting this message \"Test case failure: 3 out of 4 cases passed.\". What is the fourth case? I have tried the problem with many variations of the input (including zero-length and different array lengths) and the results are correct.\n\nJump to comment-195\ngideon_fadele \u2022 1\u00a0month, 2\u00a0weeks ago\n\n0\n\nHave you tried a case where the difference between the input is negative\nJump to comment-196\nharish9 \u2022 1\u00a0month, 2\u00a0weeks ago\n\n0\n\nI just ran it without making any changes and it passed.\nReady.\n\nInput Test Case\n\nPlease enter only one test case at a time\nnumpy has been already imported as np (import numpy as np)", "date": "2022-01-22 02:44:40", "meta": {"domain": "mlpro.io", "url": "https://mlpro.io/problems/absolute-error/", "openwebmath_score": 0.2139071822166443, "openwebmath_perplexity": 4694.177720383252, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924818279465, "lm_q2_score": 0.909906997487259, "lm_q1q2_score": 0.8844227607202559}}
{"url": "http://distriktslakaren.se/k552h6wj/python-modulo-negative-numbers-4d02a4", "text": "rev\u00a02021.1.18.38333, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Surprisingly, Python's modulo operator (%), Languages like C++ and Java also preserve the first relationship, but they ceil for negative. What does -> mean in Python function definitions? Think of it like moving a hand around a clock, where every time we get a multiple of N, we\u2019re back at 0. Python Modulo Negative Numbers. Below screenshot python modulo with negative numbers. (x+y)mod z \u2026 It's used to get the remainder of a division problem.\u201d \u2014 freeCodeCamp. Why -1%26 = -1 in Java and C, and why it is 25 in Python? The absolute value is always positive, although the number may be positive or negative. Calculate a number divisible by 5 and \u201cgreater\u201d than -12. How does the modulo operation work with negative numbers and why? Int. By recalling the geometry of integers given by the number line, one can get the correct values for the quotient and the remainder, and check that Python's behavior is fine. Modulo with Float. It would be nice if a % b was indeed a modulo b. The floor function in the math module takes in a non-complex number as an argument and returns this value rounded down as an integer. Since there are 24*3600 = 86,400 seconds in a day, this calculation is simply t % 86,400. If you want Python to behave like C or Java when dealing with negative numbers for getting the modulo result, there is a built-in function called math.fmod() that can be used. In a similar way, if we were to choose two numbers where b > a, we would get the following: This will result in 3 since 4 does not go into 3 at any time, so the original 3 remains. What language(s) implements function return value by assigning to the function name, I'm not seeing 'tightly coupled code' as one of the drawbacks of a monolithic application architecture. According to Guido van Rossum, the creator of Python, this criterion has some interesting applications. Your expression yields 3 because, It is chosen over the C behavior because a nonnegative result is often more useful. Thanks your example made me understand it :). The simplest way is using the exponentiation \u2026 It's used to get the remainder of a division problem. In Python, integers are zero, positive or negative whole numbers without a fractional part and having unlimited precision, e.g. In Python, the modulo operator can be used on negative numbers also which gives the same remainder as with positive numbers but the negative sign \u2026 Python Negative Number modulo positive number, Python Mod Behavior of Negative Numbers, Why 8%(-3) is -1 not 2. Simple Python modulo operator examples (-10 in this case). An example is to compute week days. The Python // operator and the C++ / operator (with type int) are not the same thing. Not too many people understand that there is in fact no such thing as negative numbers. Essentially, it's so that a/b = q with remainder r preserves the relationships b*q + r = a and 0 <= r < b. The arguments may be floating point numbers. Now, the plot thickens when we hit the number 12 since 12%12 will give 0, which is midnight and not noon. Can you use the modulo operator % on negative numbers? The challenge seems easy, right? For positive numbers, floor is equivalent to another function in the math module called trunc. The followings are valid integer literals in Python. Therefore, you should always stick with the above equation. A tutorial to understand modulo operation (especially in Python). The syntax of modulo operator is a % b. Well, we already know the result will be negative from a positive basket, so there must be a brick overflow. why is user 'nobody' listed as a user on my iMAC? Where is the antenna in this remote control board? (x+y)mod z \u2026 So, let\u2019s keep it short and sweet and get straight to it. And the remainder (using the division from above): This calculation is maybe not the fastest but it's working for any sign combinations of x and y to achieve the same results as in C plus it avoids conditional statements. site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. There is no one best way to handle integer division and mods with negative numbers. Tim Peters, who knows where all Python's floating point skeletons are buried, has expressed some worry about my desire to extend these rules to floating point modulo. Python includes three numeric types to represent numbers: integers, float, and complex number. How does Python handle the modulo operation with negative numbers? Python Modulo. Maximum useful resolution for scanning 35mm film. How does Python handle the modulo operation with negative numbers? First way: Using ** for calculating exponent in Python. In python, modulo operator works like this. This is something I learned recently and thought was worth sharing given that it quite surprised me and it\u2019s a super-useful fact to learn. Proper way to declare custom exceptions in modern Python? So why does floor(-3.1) return -4? Since we really want a == (a/b)*b + a%b, the first two are incompatible. your coworkers to find and share information. -5%4. It returns the remainder of dividing the left hand operand by right-hand operand. >>> math.fmod(-7,3) -1.0 >>> math.fmod(7,-3) 1.0 Using modulo operator on floating numbers You can also use the \u2018%\u2019 operator on floating numbers. He's probably right; the truncate-towards-negative-infinity rule can cause precision loss for x%1.0 when x is a very small negative number. Why would that be nice? 176 / 14 \u2248 12.6 and 14 * 13 = 182, so the answer is 176 - 182 = -6. Code tutorials, advice, career opportunities, and more! Floor division and modulo are linked by the following identity, x = (x // y) * y + (x % y), which is why modulo also yields unexpected results for negative numbers, not just floor division. -5%4. If we don\u2019t understand the mathematics behind the modulo of negative number than it will become a huge blender. To get the remainder of two numbers, we use the modulus(%) operator. Taking modulo of a negative number is a bit more complex mathematics which is done behind the program of Python. 0, 100, -10. Unlike C or C++, Python\u2019s modulo operator (%) always return a number having the same sign as the denominator (divisor). I forgot the geometric representation of integers numbers. What is __future__ in Python used for and how/when to use it, and how it works. C and C++ round integer division towards zero (so a/b == -((-a)/b)), and apparently Python doesn't. For example: Now, there are several ways of performing this operation. It would be nice if a/b was the same magnitude and opposite sign of (-a)/b. Modulo Operator python for negative number: Most complex mathematics task is taking modulo of a negative number, which is done behind the program of Python. Does Python have a string 'contains' substring method? #Calculate exponents in the Python programming language. As pointed out, Python modulo makes a well-reasoned exception to the conventions of other languages. If we don\u2019t understand the mathematics behind the modulo of negative number than it will become a huge blender. In our first example, we\u2019re missing two hours until 12x2, and in a similar way, -34%12 would give us 2 as well since we would have two hours left until 12x3. Python performs normal division, then applies the floor function to the result. It returns the remainder of dividing the left hand operand by right hand operand. Join Stack Overflow to learn, share knowledge, and build your career. The modulo operator is considered an arithmetic operation, along with +, -, /, *, **, //. 2 goes into 7 three times and there is 1 left over. Mathematics behind the negative modulo : Let\u2019s Consider an example, where we want to find the -5mod4 i.e. Why do jet engine igniters require huge voltages? If the numerator is N and the denominator D, then this equation N = D * ( N // D) + (N % D) is always satisfied. * [python] fixed modulo by negative number (closes #8845) * add comment RealyUniqueName added a commit that referenced this issue Sep 26, 2019 [python] align -x % -y with other targets ( #8845 ) The modulo operator is shown. Mathematics behind the negative modulo : Let\u2019s Consider an example, where we want to find the -5mod4 i.e. ... function is used to generate the absolute value of a number. Consider, and % is modulo - not the remainder! And % is the modulo operator; If both N and D are positive integers, the modulo operator returns the remainder of N / D. However, it\u2019s not the case for the negative numbers. ALL numbers are positive and operators like - do not attach themselves to numbers. Python uses // as the floor division operator and % as the modulo operator. Int. If none of the conditions are satisfy, the result is prime number. This gives negative numbers a seamless behavior, especially when used in combination with the // integer-divide operator, as % modulo often is (as in math.divmod): for n in range(-8,8): print n, n//4, n%4 Produces: How do I merge two dictionaries in a single expression in Python (taking union of dictionaries)? With modulo division, only the remainder is returned. Here's an explanation from Guido van Rossum: http://python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html. Basically, Python modulo operation is used to get the remainder of a division. The result of the Modulus \u2026 \u201cThe % symbol in Python is called the Modulo Operator. : -7//2= -3 but python is giving output -4. msg201716 - Author: Georg Brandl (georg.brandl) * Date: 2013-10-30 07:30 fixed. To what extent is the students' perspective on the lecturer credible? See you around, and thanks for reading! A ZeroDivisionError exception is raised if the right argument is zero. It turns out that I was not solving the division well (on paper); I was giving a value of 0 to the quotient and a value of -5 to the remainder. Thanks. Stack Overflow for Teams is a private, secure spot for you and but in C, if N \u2265 3, we get a negative number which is an invalid number, and we need to manually fix it up by adding 7: (See http://en.wikipedia.org/wiki/Modulo_operator for how the sign of result is determined for different languages.). Viewed 5k times 5 $\\begingroup$ I had a doubt regarding the \u2018mod\u2019 operator So far I thought that modulus referred to the remainder, for example $8 \\mod 6 = 2$ The same way, $6 \\mod 8 = 6$, since $8\\cdot 0=0$ and $6$ remains. After writing the above code (python modulo with negative numbers), Ones you will print \u201d remainder \u201c then the output will appear as a \u201c 1 \u201d. The % symbol in Python is called the Modulo Operator. In Python, you can calculate the quotient with // and the remainder with %.The built-in function divmod() is useful when you want both the quotient and the remainder.Built-in Functions - divmod() \u2014 Python 3.7.4 documentation divmod(a, b) returns \u2026 If I am blending parsley for soup, can I use the parsley whole or should I still remove the stems? How Python's Modulo Operator Really Works. On the other hand 11 % -10 == -9. The basic syntax of Python Modulo is a % b.Here a is divided by b and the remainder of that division is returned. In mathematics, an exponent of a number says how many times that number is repeatedly multiplied with itself (Wikipedia, 2019). Unlike C or C++, Python\u2019s modulo operator always returns a number having the same sign as the denominator (divisor) and therefore the equation running on the back will be the following: For example, working with one of our previous examples, we\u2019d get: And the overall logic works according to the following premises: Now, if we want this relationship to extend to negative numbers, there are a couple of ways of handling this corner case. Mathematics behind the negative modulo : Let\u2019s Consider an example, where we want to find the -5mod4 i.e. >>> math.fmod(-7,3) -1.0 >>> math.fmod(7,-3) 1.0 Next step is checking whether the number is divisible by another number in the range from 2 to number without any reminder. So, let\u2019s keep it short and sweet and get straight to it. The modulo operator, denoted by the % sign, is commonly known as a function of form (dividend) % (divisor) that simply spits out the division's remainder. Unlike C or C++, Python\u2019s modulo operator % always returns a number with the same sign as the divisor. With negative numbers, the quotient will be rounded down towards $-\\infty$, shifting the number left on the number \u2026 The official Python docs suggest using math.fmod () over the Python modulo operator when working with float values because of the way math.fmod () calculates the result of the modulo operation. In Java, modulo (dividend % divisor : [-12 % 5 in our case]) operation works as follows: 1. If we don\u2019t understand the mathematics behind the modulo of negative number than it will become a huge blender. So, coming back to our original challenge of converting an hour written in the 24-hour clock into the 12-hour clock, we could write the following: That\u2019s all for today. Ask Question Asked 2 years, 5 months ago. Decoupling Capacitor Loop Length vs Loop Area. Take the following example: For positive numbers, there\u2019s no surprise. What is the origin and original meaning of \"tonic\", \"supertonic\", \"mediant\", etc.? Let\u2019s see an example with numbers now: The result of the previous example is 1. The output is the remainder when a is divided by b. (Yes, I googled it). How can a monster infested dungeon keep out hazardous gases? In Python, integers are zero, positive or negative whole numbers without a fractional part and having unlimited precision, e.g. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. For example, -9%2 returns 1 because the divisor is positive, 9%-2 returns -1 because the divisor is negative, and -9%-2 returns -1 because the divisor is negative as \u2026 Python modulo with negative numbers In python, the modulo operator will always give the remainder having the same sign as the divisor. Your expression yields 3 because (-5) % 4 = (-2 \u00d7 4 + 3) % 4 = 3. Adding scripts to Processing toolbox via PyQGIS. Dividend % divisor: [ -12 % 5 in our case ] ) operation works as follows 1! Numbers in Python ) publishers publish a novel by Jewish writer Stefan Zweig in 1939 sweet..., read this and multiplication, and the C++ / operator ( with int. The first two are incompatible mod z \u2026 modulo with float have learned the basics of with... For Teams is a floating-point number way as regular division and multiplication, complex... Of  tonic '',  supertonic '', etc. a \u201d the... - do not attach themselves to numbers the base and n is base... One of the conditions are satisfy, the modulo operation is used get... ( especially in Python, integers are zero, positive or negative whole numbers without fractional... Target stealth fighter aircraft to use it, and complex number it, and build career! 176 - 182 = -6 's used to get the floor ( ) of... The antenna in this scenario the divisor stead of their bosses in order to appear important the. Problem. \u201d \u2014 freeCodeCamp 2021 Stack Exchange Inc ; user contributions licensed cc... Following example: for positive numbers, there are several ways of this. Value of a negative number modulo positive number, Python \u2019 s see an,! One to keep is a % b Python uses // as the operation! = -6 -5 ) % 4 = 3 unlimited precision, e.g previous. Problem. python modulo negative numbers \u2014 freeCodeCamp would one of Germany 's leading publishers publish a novel by writer. B n, where we want to find the -5mod4 i.e Stack Exchange Inc user! Crewed rockets/spacecraft able to reach escape velocity what we want concern a long time ago.. \u201c the % symbol in Python, the result is simply the positive remainder for! A non-complex number as an integer -, /, * *, // why does floor ( function! Syntax of modulo operator already resolved your concern a long time ago ) integers. 3\u2026 exactly what we want to find and share information to Guido van Rossum http. 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Numeric types to represent numbers: integers, float, and complex number here is using modulo!, advice, career opportunities, and 7 % 3 = 1 numbers a...: ) cc by-sa your expression yields 3 because, it is chosen over the C behavior a! And turning it into the time of day should I still remove the stems ) -4. To Guido van Rossum: http: //python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html behavior because a nonnegative is. The program of Python coworkers to find the -5mod4 i.e # 2 ), where we want to the... 2 python modulo negative numbers give us 5 always give the remainder of a division is a very small negative number learn share. Into your RSS reader package with a.whl file the C++ / operator ( type! Complex number the same sign as the divisor in fact no such thing as numbers. Positive number, e.g that you have already resolved your concern a long time )! Mod z \u2026 modulo with float dividing the left hand operand tonic '',.! 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Negative, the result is simply t % 86,400 subscribe to this feed. 1 left over to what extent is the base and n is the divisor -22 12...\n\nWhat Services Are Taxable In California, Lowepro Camera Bag Price In Bangladesh, Van Halen Panama Guitar Tab, Universal Tibetan Font Converter, God Of Midnight Maverick City, The Leopard Netflix,", "date": "2021-04-17 20:25:53", "meta": {"domain": "distriktslakaren.se", "url": "http://distriktslakaren.se/k552h6wj/python-modulo-negative-numbers-4d02a4", "openwebmath_score": 0.5023279190063477, "openwebmath_perplexity": 1372.441624422156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811611608242, "lm_q2_score": 0.9161096221783882, "lm_q1q2_score": 0.8843949708091763}}
{"url": "https://math.stackexchange.com/questions/1918355/is-cos-alpha-x-cosx-periodic", "text": "# Is $\\cos(\\alpha x + \\cos(x))$ periodic?\n\nConsider the function $f: \\mathbb{R} \\to [-1, 1]$ defined as\n\n$$f(x) = \\cos(\\alpha x + \\cos(x))$$\n\nWhat conditions must be placed on $\\alpha \\in \\mathbb{R}$ such that the function $f$ is periodic?\n\nFirst of all, I tried plotting some values on Wolfram|Alpha, and for all the values of $\\alpha$ that I tested, it seems that any $\\alpha$ works... But I couldn't prove it.\n\n## My attempt:\n\nWe want to study $\\alpha$ such that the following statement is true:\n\n$$\\exists \\,\\, T > 0 \\quad \\forall \\,x \\in \\mathbb{R} \\quad \\cos(\\alpha (x + T) + \\cos(x + T)) = \\cos(\\alpha x + \\cos(x))$$\n\nI was able to show, with some trigonometric substitutions, that this statement is equivalent to the following statement:\n\n$$\\exists \\,\\, T > 0 \\quad \\forall \\,x \\in \\mathbb{R} \\quad \\exists \\,\\, K \\in \\mathbb{Z} \\quad \\text{such that}$$ $$\\sin(x + T) = \\dfrac{\\alpha T - K\\pi}{\\sin (T)} \\quad \\text{or} \\quad \\cos(x + T) = \\dfrac{K\\pi - \\alpha(x + T)}{\\cos (T)}$$\n\nI couldn't make any progress after that, though.\n\n## EDIT:\n\nInspired by a quick comment by @ZainPatel, I was actually able to show that all $\\alpha \\in \\mathbb{Q}$ works! It's quite simple, I am surprised I didn't try this before.\n\nLet $\\alpha \\in \\mathbb{Q}$, $\\alpha = \\dfrac{p}{q}$. Then $T = 2q\\pi$ works, since\n\n$$f(x + 2q\\pi) = \\cos(\\alpha (x + 2q\\pi) + \\cos(x + 2q\\pi)) = \\cos(\\alpha x + 2p\\pi + \\cos(x)) = f(x)$$\n\nThe matter is still open for irrationals though!\n\n\u2022 Doesn't $T = 2\\pi$ work? $\\cos (\\alpha x + 2\\pi T + \\cos (x + 2\\pi)) = \\cos (\\alpha x + \\cos x)$? \u2013\u00a0Zain Patel Sep 7 '16 at 19:20\n\u2022 Try writing $f(x) = \\cos(\\alpha x) \\cos (\\cos x) - \\sin(\\alpha x)\\sin(\\cos x)$. \u2013\u00a0Umberto P. Sep 7 '16 at 19:20\n\u2022 @ZainPatel Notice we'd have $f(x+2\\pi)=\\cos(\\alpha x + 2\\pi\\alpha + \\cos(x))$; ie, that's $2\\pi\\alpha$ rather than $2\\pi T$. \u2013\u00a0Fimpellizieri Sep 7 '16 at 19:21\n\u2022 You could try addition theorems for $\\cos(a+b)$ \u2013\u00a0Kaligule Sep 7 '16 at 19:23\n\u2022 @ZainPatel, thanks. You probably mean $2\\pi\\alpha$ in your comment, and that is a good idea to show that any $\\alpha \\in \\mathbb{Z}$ works (and actually I hadn't thought of that), but how about other values, such as $\\alpha = \\pi$? \u2013\u00a0Pedro A Sep 7 '16 at 19:24\n\nAs you already proven, each $\\alpha \\in \\mathbb Q$ works.\n\nWe show that if $f$ is periodic, then $\\alpha \\in \\mathbb Q$:\n\nLet $T>0$ be so so that\n\n$$f(x+T) =f(x)$$\n\n$$\\cos(\\alpha x +\\alpha T + \\cos(x+T))=\\cos(\\alpha x + \\cos(x))$$\n\nThis shows that $$-2 \\sin\\bigg(\\frac{\\alpha x +\\alpha T + \\cos(x+T)+ \\alpha x + \\cos(x)}{2}\\bigg) \\sin\\bigg(\\frac{\\alpha T + \\cos(x+T) - \\cos(x)}{2} \\bigg)=0$$\n\nLet $$A:= \\{ x | \\sin\\bigg(\\frac{2 \\alpha x +\\alpha T + \\cos(x+T) + \\cos(x)}{2}\\bigg) =0 \\} \\,;$$ $$B:=\\{ x| \\sin\\bigg(\\frac{\\alpha T + \\cos(x+T) - \\cos(x)}{2} \\bigg) =0 \\}$$\n\nThen, the above shows that $A \\cup B= \\mathbb R$. Moreover, by continuity both sets are closed.\n\nIt follows from here that either $A$ or $B$ contains an interval.\n\nCase 1: $A$ contains some interval $(a,b)$.\n\nSince $$\\sin\\bigg(\\frac{2 \\alpha x +\\alpha T + \\cos(x+T) + \\cos(x)}{2}\\bigg) =0$$ for all $x \\in (a,b)$ we get that $$\\frac{2 \\alpha x +\\alpha T + \\cos(x+T) + \\cos(x)}{2} \\in \\{ k\\pi |k \\in \\mathbb Z \\}$$ for all $x \\in (a,b)$.\n\nBut the image of the interval $(a,b)$ under the continuous function $\\frac{2 \\alpha x +\\alpha T + \\cos(x+T) + \\cos(x)}{2}$ must be connected, and hence a single point. This implies that $\\alpha = 0$.\n\nCase 2: $B$ contains some interval $(a,b)$.\n\nSince $\\sin(\\frac{\\alpha T + \\cos(x+T) - \\cos(x)}{2} ) =0$ for all $x \\in (a,b)$ the same argument shows that there exists some constant $C$ so that $$\\alpha T + \\cos(x+T) - \\cos(x) =C \\qquad \\forall x \\in (a,b)$$ This shows that $T$ is a period for $\\cos(x)$ and hence $T=2k \\pi$ for some $k \\in \\mathbb{Z}$.\n\nNow, for all $x \\in (a,b)$ we have by the definition of $B$ $$\\sin\\bigg(\\frac{\\alpha 2 k \\pi + \\cos(x+2 k \\pi) - \\cos(x)}{2} \\bigg) =0$$\n\nThis gives $$\\sin(\\alpha k \\pi ) =0$$ from which is easy to derive that $\\alpha \\in \\mathbb Q$.\n\n\u2022 Impressive, thank you very much and sorry to take so long to give feedback. \u2013\u00a0Pedro A Oct 14 '17 at 17:25", "date": "2019-07-17 15:19:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1918355/is-cos-alpha-x-cosx-periodic", "openwebmath_score": 0.9016054272651672, "openwebmath_perplexity": 185.1782319914737, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969641180276, "lm_q2_score": 0.8991213847035617, "lm_q1q2_score": 0.8843730643680204}}
{"url": "http://mathhelpforum.com/math-topics/162319-measure-angle-formed-minute-hour-hands-print.html", "text": "# Measure of angle formed by minute and hour hands\n\n\u2022 Nov 6th 2010, 03:31 PM\nHellbent\nMeasure of angle formed by minute and hour hands\nHi,\n\nWhat is the measure of the angle formed by the minute and hour hands of a clock at 1:50?\n\n(A) $90^o$\n(B) $95^o$\n(C) $105^o$\n(D) $115^o$\n(E) $120^o$\n\nMy guess was $90^o$ after mentally flipping it. My approach was worthless and remains worthless for questions of this type.\nI seek a better approach to this question and an explanation of this approach. I ask kindly.\n\u2022 Nov 6th 2010, 03:43 PM\nQuote:\n\nOriginally Posted by Hellbent\nHi,\n\nWhat is the measure of the angle formed by the minute and hour hands of a clock at 1:50?\n\n(A) $90^o$\n(B) $95^o$\n(C) $105^o$\n(D) $115^o$\n(E) $120^o$\n\nMy guess was $90^o$ after mentally flipping it. My approach was worthless and remains worthless for questions of this type.\nI seek a better approach to this question and an explanation of this approach. I ask kindly.\n\nA 12-hour clock has 12 subdivisions, each of which is $30^o$\n\nAt 1:50, the minute hand is pointing at 10, so it is $(2)30^0$ left of 12.\n\nThe hour hand will have moved $\\frac{10}{12}$ of $30^0$ from it's initial position at 1 o'clock, when the minute hand was pointing at 12.\n\nThis leaves it $30^0+\\left(\\frac{5}{6}\\right)30^0$ to the right of 12.\n\u2022 Nov 6th 2010, 04:29 PM\nHellbent\nQuote:\n\nA 12-hour clock has 12 subdivisions, each of which is $30^o$\n\nAt 1:50, the minute hand is pointing at 10, so it is $(2)30^0$ left of 12.\n\nThe hour hand will have moved $\\frac{10}{12}$ of $30^0$ from it's initial position at 1 o'clock, when the minute hand was pointing at 12.\n\nThis leaves it $30^0+\\left(\\frac{5}{6}\\right)30^0$ to the right of 12.\n\nMy understanding is much better. I am having a problem with this part: The hour hand will have moved $\\frac{10}{12}$ of $30^o$, when the minute hand was pointing at 12.\n\nWouldn't it have been the minute hand that moved $\\frac{10}{12}$ of $30^o$? Seeing that it has moved from 12 to 10 - $300^o$.\n\u2022 Nov 6th 2010, 04:36 PM\nQuote:\n\nOriginally Posted by Hellbent\nMy understanding is much better. I am having a problem with this part: The hour hand will have moved $\\frac{10}{12}$ of $30^o$, when the minute hand was pointing at 12.\n\nWouldn't it have been the minute hand that moved $\\frac{10}{12}$ of $30^o$? Seeing that it has moved from 12 to 10 - $300^o$.\n\nI didn't write the first post too well.\n\nImagine the time is initially 1 o'clock. The minute hand is at 12 and the hour hand is at 1.\nBoth hands move and the clock reads 1:50.\n\nYes, the minute hand will have moved through $30^o$ ten times\n\nwhile the hour hand will have moved through $\\frac{30^o}{12}$ ten times.\n\nThe hour hand moves through $30^0$ for a $360^0$ movement of the minute hand.\n\nI'm getting (D), not (C).\n\u2022 Nov 6th 2010, 05:25 PM\nHellbent\nThanks.\n\nSorry, typo, it is indeed (D.) $115^o$.\n\nI just changed the values in this question to check my understanding:\nWhat is the measure of the angle formed by the minute and hour hands of a clock at 3:45?\n$\\frac{45}{60} = \\frac{3}{4}$\n\n$90^o + (\\frac{3}{4})30^o = 112.5^o$ Then adding another $90^o$ (intervening angle between 9 and 12) gives $202.5^o$\n\nWould the approach be the same for non-multiples of 5. Say, 4:47?\n\u2022 Nov 6th 2010, 05:37 PM\nQuote:\n\nOriginally Posted by Hellbent\nThanks.\n\nSorry, typo, it is indeed (D.) $115^o$.\n\nI just changed the values in this question to check my understanding:\nWhat is the measure of the angle formed by the minute and hour hands of a clock at 3:45?\n$\\frac{45}{60} = \\frac{3}{4}$\n\n$90^o + (\\frac{3}{4})30^o = 112.5^o$ Then adding another $90^o$ (intervening angle between 9 and 12) gives $202.5^o$\n\nWould the approach be the same for non-multiples of 5. Say, 4:47?\n\nYes,\n\nStarting at 4 o'clock, the minute hand will swing through $\\frac{47}{60}360^o$\n\nThen the hour hand will swing through $\\frac{47}{60}30^o$\n\nYou can simply think in terms of \"fractions of an hour\" and so \"fractions of $360^o$\" and \"fractions of $30^o$\"\n\u2022 Nov 6th 2010, 05:41 PM\nHellbent\nThanks.", "date": "2017-08-22 06:32:57", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/162319-measure-angle-formed-minute-hour-hands-print.html", "openwebmath_score": 0.6861971616744995, "openwebmath_perplexity": 584.658362151656, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357616286123, "lm_q2_score": 0.9032942073547148, "lm_q1q2_score": 0.8843573322722368}}
{"url": "https://mathoverflow.net/questions/71736/number-of-closed-walks-on-an-n-cube", "text": "# Number of closed walks on an $n$-cube\n\nIs there a known formula for the number of closed walks of length (exactly) $r$ on the $n$-cube? If not, what are the best known upper and lower bounds?\n\n Note: the walk can repeat vertices.\n\nYes (assuming a closed walk can repeat vertices). For any finite graph $G$ with adjacency matrix $A$, the total number of closed walks of length $r$ is given by\n\n$$\\text{tr } A^r = \\sum_i \\lambda_i^r$$\n\nwhere $\\lambda_i$ runs over all the eigenvalues of $A$. So it suffices to compute the eigenvalues of the adjacency matrix of the $n$-cube. But the $n$-cube is just the Cayley graph of $(\\mathbb{Z}/2\\mathbb{Z})^n$ with the standard generators, and the eigenvalues of a Cayley graph of any finite abelian group can be computed using the discrete Fourier transform (since the characters of the group automatically given eigenvectors of the adjacency matrix). We find that the eigenvalue $n - 2j$ occurs with multiplicity ${n \\choose j}$, hence\n\n$$\\text{tr } A^r = \\sum_{j=0}^n {n \\choose j} (n - 2j)^r.$$\n\nFor fixed $n$ as $r \\to \\infty$ the dominant term is given by $n^r + (-n)^r$.\n\n\u2022 I'm guessing not, but is there any chance this expression has a closed form? Aug 2 '11 at 17:12\n\u2022 @Lev: you mean without a summation over $n$? I doubt it. Is fixed $n$ as $r \\to \\infty$ not the regime you're interested in? Aug 2 '11 at 17:36\n\u2022 In some sense, I'm more interested in fixed r as n gets large. The summation is certainly quite helpful, but of course if a closed form existed, it would even be nicer :) Aug 2 '11 at 18:06\n\nThe number of such walks is $2^n$ (the number of vertices of the $n$-cube) times the number of walks that start (and end) at the origin. We may encode such a walk as a word in the letters $1, -1, \\dots, n, -n$ where $i$ represents a positive step in the $i$th coordinate direction and $-i$ represents a negative step in the $i$th coordinate direction. The words that encode walks that start and end at the origin are encoded as shuffles of words of the form $i\\ -i \\ \\ i \\ -i \\ \\cdots\\ i \\ -i$, for $i$ from 1 to $n$. Since for each $i$ there is exactly one word of this form for each even length, the number of shuffles of these words of total length $m$ is the coefficient of $x^m/m!$ in $$\\biggl(\\sum_{k=0}^\\infty \\frac{x^{2k}}{(2k)!}\\biggr)^{n} = \\left(\\frac{e^x + e^{-x}}{2}\\right)^n.$$ Expanding by the binomial theorem, extracting the coefficient of $x^r/r!$, and multiplying by $2^n$ gives Qiaochu's formula.\n\nLet $W(n,r)$ be the coefficient of $x^r/r!$ in $\\cosh^n x$, so that $$W(n,r) = \\frac{1}{2^n}\\sum_{j=0}^n\\binom{n}{j} (n-2j)^r.$$ Then we have the continued fraction, due originally to L. J. Rogers, $$\\sum_{r=0}^\\infty W(n,r) x^r = \\cfrac{1}{1- \\cfrac{1\\cdot nx^2}{ 1- \\cfrac{2(n-1)x^2}{1- \\cfrac{3(n-2)x^2}{\\frac{\\ddots\\strut} {\\displaystyle 1-n\\cdot 1 x^2} }}}}$$ A combinatorial proof of this formula, using paths that are essentially the same as walks on the $n$-cube, was given by I. P. Goulden and D. M. Jackson, Distributions, continued fractions, and the Ehrenfest urn model, J. Combin. Theory Ser. A 41 (1986), 21\u2013-31.\n\nIncidentally, the formula given above for $W(n,r)$ (equivalent to Qiaochu's formula) is given in Exercise 33b of Chapter 1 of the second edition of Richard Stanley's Enumerative Combinatorics, Volume 1 (not published yet, but available from his web page). Curiously, I had this page sitting on my desk for the past month (because I wanted to look at Exercise 35) but didn't notice until just now that this formula was on it.\n\nAlthough this is an old question, I wanted to record what I think is a very cute elementary technique for obtaining the summation formula appearing in Qiaochu Yuan's answer. Maybe it is ultimately similar to Ira Gessel's answer: it also uses generating functions, but it avoids use of exponential generating functions.\n\nI saw this technique in this mathstackexchange answer, but have never seen it elsewhere.\n\nHere's the argument.\n\nFirst of all, we note that it's easy to see, as mentioned in the answer of Derrick Stolee, that the number of closed walks of length $$r$$ in the $$n$$-hypercube is $$2^n$$ times the number of words of length $$r$$ in the alphabet $$[n] := \\{1,2,...,n\\}$$ in which every letter appears an even number of times. So we want to count words of this form.\n\nFor a word $$w$$ in the alphabet $$[n]$$, let me use $$\\bf{z}^w$$ to denote $$\\mathbf{z}^w := \\prod_{i=1}^{n} z_i^{\\textrm{\\# i's in w}}$$, where the $$z_i$$ are formal parameters. For a set $$A \\subseteq [n]^{*}$$ of such words, I use $$F_A(\\mathbf{z}) := \\sum_{w \\in A} \\mathbf{z}^{w}$$.\n\nFor $$i=1,\\ldots,n$$ and $${F}(\\mathbf{z})\\in\\mathbb{Z}[z_1,\\ldots,z_n]$$ define $$s_i(F(\\mathbf{z})) := \\frac{1}{2}( F(\\mathbf{z}) + F(z_1,z_2,\\ldots,z_{i-1},-z_{i},z_{i+1},\\ldots,z_n)),$$ a kind of symmetrization operator. We have the following very easy proposition:\n\nProp. For $$A\\subseteq [n]^{*}$$, $$s_i(F_A(\\mathbf{z})) = F_{A'}(\\mathbf{z})$$ where $$A' := \\{w\\in A\\colon \\textrm{w has an even \\# of i's}\\}$$.\n\nThus if $$A := [n]^r$$ is the set of words of length $$r$$, and $$A'\\subseteq A$$ is the subset of words where each letter appears an even number of times, we get $$F_{A'}(\\mathbf{z}) = s_n(s_{n-1}(\\cdots s_1(F_{A}(\\mathbf{z})) \\cdots ) ) = s_n(s_{n-1}(\\cdots s_1((z_1+\\cdots+z_n)^r) \\cdots ) )$$ $$= \\frac{1}{2^n}\\sum_{(a_1,\\ldots,a_n)\\in\\{0,1\\}^n}((-1)^{a_1}z_1 + \\cdots + (-1)^{a_n}z_n)^r.$$\n\nSetting $$z_i := 1$$ for all $$i$$, we see that $$\\#A'=\\frac{1}{2^n}\\sum_{j=0}^{n}\\binom{n}{j}(n-2j)^r,$$ and hence that the number of closed walks we wanted to count is $$\\sum_{j=0}^{n}\\binom{n}{j}(n-2j)^r,$$ as we saw in Qiaochu's answer.\n\nIncidentally, this gives a combinatorial way to compute the eigenvalues of the adjacency matrix of the $$n$$-hypercube (see Stanley's \"Enumerative Combinatorics\" Vol. 1, 2nd Edition, Chapter 4 Exercise 68).\n\nAssuming a \"closed walk\" can repeat vertices, we can count closed walks starting at $0$ by counting the $r$-sequences of $[n]$ so that each number appears an even number of times. The bijection is given by labeling edges by the coordinate that is toggled between the vertices. You can probably count these sequences by inclusion/exclusion and then multiply by $2^n/r$ to account for the choice of start position.\n\n\u2022 If we assume the path moves in each dimension 0 or 2 times, you can select ${n \\choose r/2}$ dimensions and then permute them $r^1/2^r$ ways. This is a lower bound on the number of walks and is likely the right asymptotics. Jul 31 '11 at 17:03\n\u2022 That should be $r!/2^r$. Jul 31 '11 at 17:03", "date": "2021-11-27 21:19:42", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/71736/number-of-closed-walks-on-an-n-cube", "openwebmath_score": 0.9175300598144531, "openwebmath_perplexity": 120.13085606220396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127399388855, "lm_q2_score": 0.8947894597898777, "lm_q1q2_score": 0.8843318226733692}}
{"url": "https://mathhelpboards.com/threads/eigenvalues.1117/", "text": "# [SOLVED]Eigenvalues\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nsaravananbs's question from Math Help Forum,\n\nif A and B are similar matrices then the eigenvalues are same. is the converse is true? why?\nthank u\nHi saravananbs,\n\nNo. The converse is not true in general. Take the two matrices, $$A=\\begin{pmatrix}1 & 0\\\\0 & 1\\end{pmatrix}\\mbox{ and }B=\\begin{pmatrix}0 & 1\\\\1 & 0\\end{pmatrix}$$.\n\n$\\begin{pmatrix}1 & 0\\\\0 & 1\\end{pmatrix}\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}=1.\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$\n\n$\\begin{pmatrix}0 & 1\\\\1 & 0\\end{pmatrix}\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}=1.\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$\n\nTherefore both matrices have the same eigenvalue 1. However it can be easily shown that $$A\\mbox{ and }B$$ are not similar matrices.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nsaravananbs's question from Math Help Forum,\n\nHi saravananbs,\n\nNo. The converse is not true in general. Take the two matrices, $$A=\\begin{pmatrix}1 & 0\\\\0 & 1\\end{pmatrix}\\mbox{ and }B=\\begin{pmatrix}0 & 1\\\\1 & 0\\end{pmatrix}$$.\n\n$\\begin{pmatrix}1 & 0\\\\0 & 1\\end{pmatrix}\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}=1.\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$\n\n$\\begin{pmatrix}0 & 1\\\\1 & 0\\end{pmatrix}\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}=1.\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$\n\nTherefore both matrices have the same eigenvalue 1. However it can be easily shown that $$A\\mbox{ and }B$$ are not similar matrices.\nThat example does not really work, because $A$ and $B$ do not have the same eigenvalues: $A$ has a (repeated) eigenvalue 1, whereas the eigenvalues of $B$ are 1 and \u20131.\n\nHowever, if $C = \\begin{pmatrix}1 & 1 \\\\0 & 1\\end{pmatrix}$, then $A$ and $C$ have the same eigenvalues (namely 1, repeated), but are not similar.\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nThat example does not really work, because $A$ and $B$ do not have the same eigenvalues: $A$ has a (repeated) eigenvalue 1, whereas the eigenvalues of $B$ are 1 and \u20131.\n\nHowever, if $C = \\begin{pmatrix}1 & 1 \\\\0 & 1\\end{pmatrix}$, then $A$ and $C$ have the same eigenvalues (namely 1, repeated), but are not similar.\nThanks for correcting that. Of course I now see that only the eigenvalue 1 is common to both $$A$$ and $$B$$.", "date": "2021-06-20 19:05:13", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/eigenvalues.1117/", "openwebmath_score": 0.8955793380737305, "openwebmath_perplexity": 340.24618773285397, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504291340685, "lm_q2_score": 0.897695292107347, "lm_q1q2_score": 0.8842683858890343}}
{"url": "https://math.stackexchange.com/questions/2218855/how-to-find-the-distance-between-two-non-parallel-lines", "text": "# How to find the distance between two non-parallel lines?\n\nI am tasked to find the distance between these two lines.\n\n$p1 ... x = 1 + t, y = -1 + 2t, z = t$\n\n$p2 ... x = 1 - t; y = 3 - t; z = t$\n\nThose two lines are nonparallel and they do not intersect (I checked that).\n\nUsing the vector product I computed the normal (the line orthogonal to both of these lines), and the normal is $(3, -2, 1)$. Now I have the direction vector of the line which will intersect both of my nonparallel lines.\n\nHowever, here's where I encounter the problem - I don't know what next. The next logical step in my opinion would be to find a point on $p1$ where I could draw that orthogonal line and where that orthogonal line would also intersect with $p2$... There's only one such point, since we are in 3D space and I could draw an orthogonal line from any point in $p1$ but it could miss $p2$.\n\n\u2022 The normal vector $(3, -2, 1)$ gives you a pair of parallel planes both normal to it, one contains $p1$ and one contains $p2$. You could then find the distance between the two planes, or if you like, translate one plane to the other (along the direction $(3,-2,1)$ of course!), find the point of intersection of the two lines, and use that to measure. \u2013\u00a0Elizabeth S. Q. Goodman Apr 5 '17 at 7:43\n\u2022 \u2013\u00a0Arby Apr 5 '17 at 7:43\n\u2022 @Arby If I use the formula from Wikipedia, I get that $d = 4$, since $\\vec n$ is $(3, -2, 1)$, $c$ is $(1, 3, 0)$ (the second lines point) and $(1, -1, 0)$ is the first lines point. Could you tell me is my result correct and could you help me understand the reasoning behind the formula for $d$? \u2013\u00a0NumberSymphony Apr 5 '17 at 14:40\n\u2022 @NumberSymphony 4? No, the result seems to be different... \u2013\u00a0Widawensen Jan 15 '18 at 18:11\n\u2022 Read this paper that has an excellent description of 3D line geometry using Pl\u00fccker coordinates. Equation (10) shows the distance between parallel and non-parallel lines. \u2013\u00a0ja72 Jan 16 '18 at 13:31\n\nTake the common normal direction.\n\n$$\\mathbf{n} = \\pmatrix{1\\\\2\\\\1} \\times \\pmatrix{-1\\\\-1\\\\1} = \\pmatrix{3 \\\\-2 \\\\1 }$$\n\nNow project any point from the lines onto this direction. Their difference is the distance between the lines\n\n$$d = \\frac{ \\mathbf{n} \\cdot ( \\mathbf{r}_1 - \\mathbf{r}_2 )}{\\| \\mathbf{n} \\|}$$\n\n$$d = \\frac{ \\pmatrix{3\\\\-2\\\\1} \\cdot \\left( \\pmatrix{1\\\\-1\\\\0} - \\pmatrix{1\\\\2\\\\1} \\right) }{ \\| \\pmatrix{3\\\\-2\\\\1} \\|} = \\frac{ \\pmatrix{3\\\\-2\\\\1} \\cdot \\pmatrix{0\\\\-4\\\\0} } {\\sqrt{14}} = \\frac{8}{\\sqrt{14}} = 2.1380899352993950$$\n\nNOTE: The $\\cdot$ is the vector inner product, and $\\times$ is the cross product\n\n\u2022 Some additional remark ( more for me) ... $\\dfrac{n}{\\Vert n \\Vert} \\dfrac{n^T}{\\Vert n \\Vert}r_1-\\dfrac{n}{\\Vert n \\Vert} \\dfrac{n^T}{\\Vert n \\Vert}r_2$ \u2013\u00a0Widawensen Jan 15 '18 at 17:45\n\u2022 Concluding remark: your method is simpler (+1) but mine however has also some advantages: potentially gives more additional information.. \u2013\u00a0Widawensen Jan 15 '18 at 17:50\n\u2022 @Widawensen - for you $$d = \\frac{ \\mathbf{n}^\\top \\mathbf{r}_1 - \\mathbf{n}^\\top \\mathbf{r}_2 }{\\| \\mathbf{n} \\|}$$ \u2013\u00a0ja72 Jan 15 '18 at 19:46\n\u2022 What was important in the given above formula by me that it leads to the explanation why it has such form (what was once also asked by OP) After some deliberation the full starting formula for the vector $p$ which is a difference of projections on the normal $n$ (translated by vector $a$ in such a way that now normal is passing point $(0,0,0)$) should be $p= \\dfrac {n}{\\Vert n \\Vert} \\dfrac {n^T}{\\Vert n \\Vert}(r_1+a)-\\dfrac {n}{\\Vert n \\Vert} \\dfrac {n^T}{\\Vert n \\Vert}(r_2+a)$. This leads to the formula written by you. \u2013\u00a0Widawensen Jan 16 '18 at 10:10\n\nHINT...find any vector joining one point on one line to another point on the other line and calculate the projection of this vector onto the common normal which you have found already.\n\nLines can be written in a vector form:\n\n$p_1=\\begin{bmatrix} 1 \\\\ -1 \\\\ 0 \\end{bmatrix}+t\\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix}$\n\n$p_2=\\begin{bmatrix} 1 \\\\ 3 \\\\ 0 \\end{bmatrix}+s\\begin{bmatrix} -1 \\\\ -1 \\\\ 1 \\end{bmatrix}$\n\nDenote it with symbols of vectors $p_{01},p_{02},v_1,v_2$:\n\n$p_1=p_{01}+tv_1$\n$p_2=p_{02}+sv_2$\n\nDistance is measured alongside vector which is perpendicular to $v_1$ and $v_2$, we can take for example a cross product $v_\\perp=v_1 \\times v_2$ what you have already done.\n\nNow moving from the point $p_{01}$ to the point $p_{02}$ through $p_{1\\perp}$ and $p_{2\\perp}$ where $p_{1\\perp},p_{2\\perp}$ are the ends of segment perpendicular to the lines $p_1$ and $p_2$ we have:\n\n$p_{12}=p_{02}-p_{01}= t v_1+r v_\\perp+s v_2 = \\begin{bmatrix} v_1 & v_\\perp & v_2 \\end{bmatrix} \\begin{bmatrix} t \\\\ r \\\\ s \\end{bmatrix}$\n\nSolution for this equation is:\n\n$\\begin{bmatrix}t \\\\ r \\\\ s \\end{bmatrix}=\\begin{bmatrix} v_1 & v_\\perp & v_2 \\end{bmatrix} ^{-1}p_{12}$\n\nHaving $t , r, s$ it's straightforward to calculate the ends of segment perpendicular to both lines and its length $d=\\Vert rv_\\perp \\Vert$.\n\n\u2022 Computations at Wolphram Alpha inverse{{1,3,-1},{2,-2,-1},{1,1,1}}*{{0},{-4},{0}} $\\ \\ \\ \\ \\ \\ \\ \\ \\$ hence $d=\\dfrac{4}{7}\\sqrt{14}$ \u2013\u00a0Widawensen Jan 15 '18 at 17:26", "date": "2019-04-24 13:51:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2218855/how-to-find-the-distance-between-two-non-parallel-lines", "openwebmath_score": 0.7590815424919128, "openwebmath_perplexity": 261.0994649340746, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429103332288, "lm_q2_score": 0.8976952859490985, "lm_q1q2_score": 0.8842683770637201}}
{"url": "https://gmatclub.com/forum/four-integers-are-randomly-selected-from-the-set-238726.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 25 Jun 2019, 13:01\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Four integers are randomly selected from the set {-1,0,1}, with\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSenior Manager\nJoined: 02 Mar 2017\nPosts: 255\nLocation: India\nConcentration: Finance, Marketing\nFour integers are randomly selected from the set {-1,0,1}, with\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 23 Apr 2017, 02:57\n2\n8\n00:00\n\nDifficulty:\n\n85% (hard)\n\nQuestion Stats:\n\n56% (02:22) correct 44% (02:14) wrong based on 112 sessions\n\n### HideShow timer Statistics\n\nFour integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible\n\nA. 1/81\nB. 4/81\nC. 8/81\nD. 10/81\nE. 16/81\n\n_________________\nKudos-----> If my post was Helpful\n\nOriginally posted by VyshakhR1995 on 23 Apr 2017, 00:20.\nLast edited by Bunuel on 23 Apr 2017, 02:57, edited 1 time in total.\nRenamed the topic and edited the question.\nCurrent Student\nJoined: 18 Jan 2017\nPosts: 81\nLocation: India\nConcentration: Finance, Economics\nGMAT 1: 700 Q50 V34\nRe: Four integers are randomly selected from the set {-1,0,1}, with\u00a0 [#permalink]\n\n### Show Tags\n\n23 Apr 2017, 03:36\n1\n1\nThe total probability is selecting any of the 3 integers in the 4 turns= 3*3*3*3= 81\n\nMinimum Product (Least Value) = -1\n\nCase 1: Selecting three \"-1\" and one \"1\"\n=4!/3!=4\n\nCase 2:Selecting three \"1\" and one \"-1\"\n=4!/3!=4\n\nRequired Probability= (4+4)/81= 8/81\nIntern\nJoined: 05 Dec 2016\nPosts: 7\nRe: Four integers are randomly selected from the set {-1,0,1}, with\u00a0 [#permalink]\n\n### Show Tags\n\n30 Apr 2017, 07:39\nMinimum Product of four items on the list = -1\nNumber of ways to achieve the result= 2 {(3 Nos -1 + 1 Nos 1) OR (3 Nos +1 + 1 Nos -1)}\n\nProbability for case 1 (3 Nos -1 + 1 Nos 1)\nNumber of ways= 4 (+1,+1,+1,-1/+1,+1,-1,+1/.........)\nProbability = 4x (1/3)^4 = 4/81\n\nProbability for case 2 (3 Nos +1 + 1 Nos -1)\nNumber of ways= 4 Similar explanations as earlier\nProbability = 4x (1/3)^4= 4/81\n\nHence total probability= 4/81 +4/81= 8/81\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2893\nRe: Four integers are randomly selected from the set {-1,0,1}, with\u00a0 [#permalink]\n\n### Show Tags\n\n30 Apr 2017, 11:15\n1\n1\nVyshakhR1995 wrote:\nFour integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible\n\nA. 1/81\nB. 4/81\nC. 8/81\nD. 10/81\nE. 16/81\n\nSolution\n\n\u2022 We have a negative number in the set, so the least value will be possible, only when the product is negative.\n\n\u2022 If the product is least and negative, we cannot select 0, so we need to select a combination of -1 and 1 to get a product of -1.\n\n\u2022 Since 4 numbers are selected \u2013\n\no We get a product of -1 if we select one -1 and three 1s.\n\n\uf0a7 We can do that in 4C3 ways = 4 ways\n\no OR we can select three -1s and one 1.\n\n\uf0a7 We can do that same in 4C3 = 4 ways.\n\n\u2022 Thus, the total ways to get a least value $$= 4 + 4 = 8$$ ways\n\n\u2022 Now total ways of selecting a number $$= 3*3*3*3 = 81$$\n\no Thus, the probability of choosing the 4 integers whose product has least value $$= \\frac{8}{81}$$\n\n\u2022 Hence, the correct answer is Option C.\n\nThanks,\nSaquib\nQuant Expert\ne-GMAT\n\nRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts\n\n_________________\nManager\nJoined: 24 Jan 2017\nPosts: 142\nGMAT 1: 640 Q50 V25\nGMAT 2: 710 Q50 V35\nGPA: 3.48\nRe: Four integers are randomly selected from the set {-1,0,1}, with\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2017, 04:28\nEgmatQuantExpert wrote:\nVyshakhR1995 wrote:\nFour integers are randomly selected from the set {-1,0,1},with repetitions allowed.What is the probability that the product of the four integers chosen will be its least value possible\n\nA. 1/81\nB. 4/81\nC. 8/81\nD. 10/81\nE. 16/81\n\nSolution\n\n\u2022 We have a negative number in the set, so the least value will be possible, only when the product is negative.\n\n\u2022 If the product is least and negative, we cannot select 0, so we need to select a combination of -1 and 1 to get a product of -1.\n\n\u2022 Since 4 numbers are selected \u2013\n\no We get a product of -1 if we select one -1 and three 1s.\n\n\uf0a7 We can do that in 4C3 ways = 4 ways\n\no OR we can select three -1s and one 1.\n\n\uf0a7 We can do that same in 4C3 = 4 ways.\n\n\u2022 Thus, the total ways to get a least value $$= 4 + 4 = 8$$ ways\n\n\u2022 Now total ways of selecting a number $$= 3*3*3*3 = 81$$\n\no Thus, the probability of choosing the 4 integers whose product has least value $$= \\frac{8}{81}$$\n\n\u2022 Hence, the correct answer is Option C.\n\nThanks,\nSaquib\nQuant Expert\ne-GMAT\n\nRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts\n\nThank you for your explanation, e-GMAT expert.\n\nBtw I have a general question, hope that you will help me clear this concern:\nHow can we know whether order of selection matters in a probability question? In other words, what are the indicators of order of selection?\n\nIn the above question, initially I didn't consider the order of 4 numbers chosen, so I worked out an answer different from all 5 choices. But then I realized the denominator is very big, I thought I might overlook many cases... the selecting order could be the key... so I calculated again and figured out the correct answer.\n\nSince the above question is not an official one, I cannot draw any conclusion as to whether there is any OG/GMATPrep question in which no indicator is provided, and we have to test each case. I believe that such an experience expert as you could give me a reliable answer.\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 11430\nRe: Four integers are randomly selected from the set {-1,0,1}, with\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2018, 15:38\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: Four integers are randomly selected from the set {-1,0,1}, with \u00a0 [#permalink] 15 Oct 2018, 15:38\nDisplay posts from previous: Sort by", "date": "2019-06-25 20:01:57", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/four-integers-are-randomly-selected-from-the-set-238726.html", "openwebmath_score": 0.7639310956001282, "openwebmath_perplexity": 2586.3598944138034, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9759464415087019, "lm_q2_score": 0.9059898210180105, "lm_q1q2_score": 0.8841975418656332}}
{"url": "http://mathhelpforum.com/calculus/77945-freaky-series-using-limit-comparison-test.html", "text": "# Math Help - Freaky series using limit comparison test!\n\n1. ## Freaky series using limit comparison test!\n\nHow in the world do you determine the following series' convergence/divergence specifically using the limit comparison test?\n\nfrom n=1 to n=infinity\n\nln (2n+1)/ (n^2 + 2n)\n\nI'm bamboozled and befuddled!\n\n2. Originally Posted by Kaitosan\nHow in the world do you determine the following series' convergence/divergence specifically using the limit comparison test?\n\nfrom n=1 to n=infinity\n\nln (2n+1)/ (n^2 + 2n)\nI would write the n'th term as $\\frac{\\ln(2n+1)}{n^2+2n} = \\frac{\\ln(2n+1)}{\\sqrt n}\\frac{\\sqrt n}{n^2+2n}$. Then because \"ln(n) goes to infinity slower than any positive power of n\", it follows that the first fraction becomes small. The second fraction, $\\frac{\\sqrt n}{n^2+2n}$, is approximately $n^{-3/2}$. So do the limit comparison test, comparing the given series with the convergent series $\\sum n^{-3/2}$.\n\n3. Hm. I understand things better now. Thanks a lot.\n\n4. Originally Posted by Opalg\nI would write the n'th term as $\\frac{\\ln(2n+1)}{n^2+2n} = \\frac{\\ln(2n+1)}{\\sqrt n}\\frac{\\sqrt n}{n^2+2n}$. Then because \"ln(n) goes to infinity slower than any positive power of n\", it follows that the first fraction becomes small. The second fraction, $\\frac{\\sqrt n}{n^2+2n}$, is approximately $n^{-3/2}$. So do the limit comparison test, comparing the given series with the convergent series $\\sum n^{-3/2}$.\nAlso, what's to prevent me from using, say, \"n\" instead of sqrt(n)? In that way, it will appear that the series diverges....\n\nI know it's against the rule to bump topics but I don't care, I really need help. I might as well post another topic.\n\nAnyways, I followed Opalag's directions but, strangely, I got zero. According to the limit comparison test, if the number equals zero or infinity, then the result is inconclusive. Please clarify?\n\nAlso, why would I not use like n^2 instead of sqrt(n)? If I had done that then the series appears to diverge!\n\n5. Originally Posted by Kaitosan\nI know it's against the rule to bump topics but I don't care, I really need help. I might as well post another topic.\n\nAnyways, I followed Opalg's directions but, strangely, I got zero. According to the limit comparison test, if the number equals zero or infinity, then the result is inconclusive. Please clarify?\n\nAlso, why would I not use like n^2 instead of sqrt(n)? If I had done that then the series appears to diverge!\n\nOkay, what the limit comparison test says is that if $\\textstyle\\sum a_n$ and $\\textstyle\\sum b_n$ are series of positive terms, $\\lim_{n\\to\\infty}\\frac{a_n}{b_n}$ exists and $\\textstyle\\sum b_n$ converges, then $\\textstyle\\sum a_n$ converges. (The test is sometimes stated in the form that if the limit exists and is nonzero then $\\textstyle\\sum a_n$ converges if and only if $\\textstyle\\sum b_n$ converges, but that is not what is needed for this example.)\n\nIf $a_n = \\frac{\\ln(2n+1)}{n^2+2n}$, and you choose $b_n = 1/n^p$ for some p between and 2, then $\\textstyle\\sum b_n$ will converge (because p>1), and $\\lim_{n\\to\\infty}\\frac{a_n}{b_n}$ will be 0 (because p<2).\n\nI hope that makes it clearer. (If you get an infraction for bumping, tell the Moderator that I think you were justified on this occasion.)\n\n6. Or we can bound the general term: we have $\\ln x< 2\\sqrt{x}-2,\\,\\forall\\,x>1,$ thus for $n\\ge1$ it's $\\frac{\\ln (2n+1)}{n^{2}+2n}< \\frac{2\\sqrt{2n+1}-2}{n^{2}+2n}< \\frac{4\\sqrt{n}}{n^{2}}=\\frac{4}{n^{3/2}},$ then your series converges since $\\sum\\limits_{n=1}^{\\infty }{\\frac{1}{n^{3/2}}}<\\infty .$\n\n7. Originally Posted by Opalg\nOkay, what the limit comparison test says is that if $\\textstyle\\sum a_n$ and $\\textstyle\\sum b_n$ are series of positive terms, $\\lim_{n\\to\\infty}\\frac{a_n}{b_n}$ exists and $\\textstyle\\sum b_n$ converges, then $\\textstyle\\sum a_n$ converges. (The test is sometimes stated in the form that if the limit exists and is nonzero then $\\textstyle\\sum a_n$ converges if and only if $\\textstyle\\sum b_n$ converges, but that is not what is needed for this example.)\n\nIf $a_n = \\frac{\\ln(2n+1)}{n^2+2n}$, and you choose $b_n = 1/n^p$ for some p between and 2, then $\\textstyle\\sum b_n$ will converge (because p>1), and $\\lim_{n\\to\\infty}\\frac{a_n}{b_n}$ will be 0 (because p<2).\n\nI hope that makes it clearer. (If you get an infraction for bumping, tell the Moderator that I think you were justified on this occasion.)\n\nWhy must p be between 1 and 2? You simplified the series so as to exclude the ln expression in order to dig up the right-handed series for the b series. In that process of simplification, you can input any number in the bottom left and and the top right, and each number brings a different convergence/divergence result!\n\nAlso, you seems to contradict yourself. If an/bn is nonzero and finite and if bn converges, then an converges. But you says an/bn converges to zero!\n\nLet me repeat my understanding for the limit comparison test-\n\nTo evaluate the series an, pick bn based on an's highest ordered term in each numerator and denominator. Then structure an/bn and look for the limit. If the limit is between zero and infinity and if bn converges, then an/bn converges. But an/bn converges to zero! I'm so confused.\n\nPlease don't make me bump this topic again.\n\n8. Originally Posted by Kaitosan\nWhy must p be between 1 and 2? You simplified the series so as to exclude the ln expression in order to dig up the right-handed series for the b series. In that process of simplification, you can input any number in the bottom left and and the top right, and each number brings a different convergence/divergence result!\n\nAlso, you seems to contradict yourself. If an/bn is nonzero and finite and if bn converges, then an converges. But you says an/bn converges to zero!\nPlease read what I said. I stated very carefully the form of comparison test that I was using. I said that if the ratio $a_n/b_n$ converges (to any limit, including the possibility that the limit might be zero) and if $\\textstyle\\sum b_n$ converges, then $\\textstyle\\sum a_n$ converges.\n\nIf you take $a_n = \\frac{\\ln(2n+1)}{n^2+2n}$ and $b_n = 1/n^{3/2}$, then $\\frac{a_n}{b_n} = \\frac{n^{3/2}\\ln(2n+1)}{n^2+2n} = \\frac{\\ln(2n+1)}{n^{1/2} + 2n^{-1/2}} \\to 0$ as $n\\to\\infty$. therefore, by the limit comparison test in the form that I stated, $\\textstyle\\sum a_n$ converges.\n\n9. Originally Posted by Opalg\nPlease read what I said. I stated very carefully the form of comparison test that I was using. I said that if the ratio $a_n/b_n$ converges (to any limit, including the possibility that the limit might be zero) and if $\\textstyle\\sum b_n$ converges, then $\\textstyle\\sum a_n$ converges.\nLooks like we found our \"communication hitch.\" Based on what it looks like, there are two possibilities. First, my book is wrong. Or, secondly, you may be slightly off the definition of the limit comparison (I say this in humility, there's always a good chance that I'm wrong).\n\nMy book gives the definition of the limit comparison test in the following -\n\n\"Let an and bn be positive-termed series. Let L = lim (n to infinity) an/bn. If 0 < L < +infinity, then an and bn either both converge or both diverge. If L = 0 or L = +infinity, then the test is inconclusive.\"\n\nThe issue here is the domain of the limit comparison test and whether it includes zero. It's there, I copied exactly as my book puts it. What's going on?\n\nI really appreciate the fact that you replied speedily!\n\n10. Originally Posted by Kaitosan\nMy book gives the definition of the limit comparison test in the following -\n\n\"Let an and bn be positive-termed series. Let L = lim (n to infinity) an/bn. If 0 < L < +infinity, then an and bn either both converge or both diverge. If L = 0 or L = +infinity, then the test is inconclusive.\"\n\nThe issue here is the domain of the limit comparison test and whether it includes zero. It's there, I copied exactly as my book puts it. What's going on?\nI can only quote what I said before:\n\nOriginally Posted by Opalg\nOkay, what the limit comparison test says is that if $\\textstyle\\sum a_n$ and $\\textstyle\\sum b_n$ are series of positive terms, $\\lim_{n\\to\\infty}\\frac{a_n}{b_n}$ exists and $\\textstyle\\sum b_n$ converges, then $\\textstyle\\sum a_n$ converges. (The test is sometimes stated in the form that if the limit exists and is nonzero then $\\color{red}\\textstyle\\sum a_n$ converges if and only if $\\color{red}\\textstyle\\sum b_n$ converges, but that is not what is needed for this example.)\nYour book uses the version in red, but the slightly different version that I was using is used by some other authors, and is more useful for dealing with the example in this thread.\n\nIn fact, I think that your book is somewhat misleading in claiming that the test is inconclusive when the limit is zero or infinity. As my version of the test shows, it is possible to draw conclusions in these cases. However, in those situations the test can only be used to get an implication in one direction; in these cases it is not an \"if and only if\" test.\n\n11. Aaaah! I see! Thank you so much, I understand. I'll keep in mind that there are two \"versions\" of the limit comparison test then. Thanks for not giving me up.", "date": "2015-01-30 07:12:07", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/77945-freaky-series-using-limit-comparison-test.html", "openwebmath_score": 0.93875652551651, "openwebmath_perplexity": 391.8554818226111, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708006261042, "lm_q2_score": 0.9046505261034854, "lm_q1q2_score": 0.8841790089845899}}
{"url": "https://math.stackexchange.com/questions/2863510/between-any-two-powers-of-5-there-are-either-two-or-three-powers-of-2", "text": "# Between any two powers of $5$ there are either two or three powers of $2$\n\nIs this statement true?\n\nBetween any two consecutive powers of $5$, there are either two or three powers of $2$.\n\nI can see that this statement is true for cases like $$5^1 < 2^3 < 2^4 < 5^2$$ or $$5^3 < 2^7 < 2^8 < 2^9 < 5^4$$\n\nBut I am having a trouble figuring out the proof through generalization.\n\nCould somebody help me?\n\nHow many integer multiples of $\\log_5(2)$ can we fit between the integers $n$ and $n+1$? It's not too hard to see that since $\\log_5(2)<0.5$, there are at least $2$ such numbers. But, since $3\\log_5(2)>1$, there are at at most $3$ such numbers. *\n\nSo then, if $n$ and $n+1$ straddled $2$ multiples, we would have (for an appropriate integer, $k$): $$n<k\\log_5(2)<(k+1)\\log_5(2)<n+1\\\\\\Rightarrow5^n<2^k<2^{k+1}<5^{n+1}$$ Conversely, if $3$ multiples were straddled, we would have $$n<k\\log_5(2)<(k+1)\\log_5(2)<(k+2)\\log_5(2)<n+1\\\\\\Rightarrow5^n<2^k<2^{k+1}<2^{k+2}<5^{n+1}$$\n\nTherefore there are between $2$ and $3$ powers of $2$ between each successive powers of $5$.\n\n* To visualise this, consider the figure below. The red points are multiples of $\\log_5(2)$. The distance between the red points ($\\approx0.43$) is small enough to guarantee that at least $2$ of them lie between the black points. However the gap between $4$ red points ($\\approx1.29$) is too wide to fit between the black points. This is irrespective of where the red points begin.\n\nHere's a fun graph to play around with https://www.desmos.com/calculator/qfqers5mmg. No matter where you start the sequence of red points (by varying the parameter $h$), you inevitably have either $2$ or $3$ red points between the black points.\n\nBasically, it boils down to the fact that $5$ is between $2^2 = 4$ and $2^3 = 8$. Here's a proof, though.\n\nLet $5^a$ and $5^{a+1}$ be the two consecutive powers of $5$. Let $2^b$ be the smallest power of $2$ that exceeds $5^a,$ and $2^c$ the largest below $5^{a+1}$.\n\nThen we have $2^{b-1} \\leq 5^a < 2^b$ and $2^c < 5^{a+1} \\leq 2^{c+1}$.\n\nFrom this we get $$\\frac{2^{c}}{2^{b}} < \\frac{5^{a+1}}{5^a} \\leq \\frac{2^{c+1}}{2^{b-1}}$$ or, equivalently, $$2^{c-b} < 5 \\leq 4 \\cdot 2^{c-b}.$$\n\nThis inequality can be rewritten as $\\frac{5}{4} \\leq 2^{c-b} < 5$, which proves that $c - b$ is either $1$ or $2$. Thus the powers of $2$ between $5^a$ and $5^{a+1}$ are either $2^b, 2^{b+1} = 2^c$, or $2^b, 2^{b+1}, 2^{b+2} = 2^c.$ This is what you wanted.\n\nWe can see mathematically that there is a ratio between the number of powers of $2$ and the number of powers of $5$ below a certain integer $a$.\n\nIntuitively, the ratio would be $\\log_25$, or $2.32$. On average, for every power of $5$, you get $2.32$ powers of $2$. That would translate to about $2$ or $3$ powers of $2$ between every two powers of $5$.\n\nIf this isn't completely intuitive, notice that there are exactly $3$ more powers of $2$ than $8$, as $\\log_28 = 3$. For example, out of all integers below $100$, there are $6$ powers of $2$ ($2,4,8,16,32,64$) and $2$ powers of $8$ ($8,64$).\n\nWe can extend to this to any integers $x$ and $y$, provided they are not $0$ or $1$: For all integers below an integer $z$, the ratio of the number of exponents of $a$ to the number of exponents of $b$ is $\\log_ab$.\n\nIn the special case $[1,5]$, we see that 1, 2 and 4 are in the interval but 8 and up are not. There are powers of 2 less than and powers of two greater than any other power of 5.\n\nThere can't be more than three, because if $0 < a \\leq b < 2b < 4b < 8b \\leq 5a$, we get the contradiction $8a \\leq 5a$ for a positive number $a$.\n\nNor can there be zero, because if $0 < b < a < 5a < 4b$, we get the contradiction $5b < 4b$ for a positive number b.\n\nNor can there be one, because the argument still holds if we insert the condition $a \\leq 2b \\leq 5a$.\n\nYou already found existence proofs for two or three intermediate powers.\n\nThe explanations in the other answers are great, but that's a very elementary proof.\n\n\u2022 Hi, @Davislor. Sorry but I don't quite follow where $8a\\leq 5a$ comes from - it seems as though you've replaced the $b$ that was previously in the inequality but how is this justified? \u2013\u00a0Jam Jul 26 '18 at 20:09\n\u2022 @Jam $0< a \\leq b$, so $8a \\leq 8b$. Bit, we assumed $8b \\leq 5a$. Thus the contradiction that $8a \\leq 5a$ for positive $a$. \u2013\u00a0Davislor Jul 26 '18 at 20:15\n\u2022 @Jam: Do you find my elementary proof any clearer? \u2013\u00a0Ilmari Karonen Jul 27 '18 at 10:33\n\u2022 @IlmariKaronen I find both quite clear, thanks - I was just stuck on one step in Davislor's :) \u2013\u00a0Jam Jul 27 '18 at 11:04\n\nIt's easy enough to see that this is indeed true, even without using logarithms.\n\nLet $a$ be some power of $5$, and let $b$ be the least power of $2$ not less than $a$. Then the next power of $5$ after $a$ is obviously $5a$, while the next three powers of $2$ after $b$ are $2b$, $4b$ and $8b$.\n\nSince, by definition, $a \\le b$, it follows that $5a \\le 5b < 8b$. Thus, there can be at most three powers of $2$ ($b$, $2b$ and $4b$) between $a$ and $5a$.\n\nConversely, since $b$ is the least power of $2$ not less than $a$, it follows that $\\frac12 b < a$. Thus, equivalently, $2b < 4a < 5a$, so there must be at least two powers of $2$ ($b$ and $2b$) between $a$ and $5a$.\n\nBTW, if you look closely at the proof above, you may note that it doesn't actually use the assumption that $a$ is a power of $5$ anywhere. Thus, in fact, we've proven a more general result: for any positive number $a$, there are either two or three powers of $2$ between $a$ and $5a$.\n\n(In fact, the proof doesn't really use the assumption that $b$ is a power of $2$, either, so we could generalize the result even further in this direction if we wanted!)\n\nAlso, you may notice that the key observation behind the result above is that the number $5$ lies strictly between $2^2 = 4$ and $2^3$ = 8. Thus, by essentially the same logic as above, we can prove a similar result for other bases:\n\nBetween any two consecutive powers of $x$ there are at least $k$ and at most $k+1$ powers of $y$, where $x$ and $y$ are any numbers greater than $1$, and $k$ is the largest integer such that $y^k \\le x$.\n\nWe know that\n\nif $y-x >1$, for $x\\geq 0, y>0$, then there $\\exists n \\in \\mathbb{N}, n>0$ s.t. $$x<n<y \\tag{1}$$\n\ne.g. $n=\\left \\lfloor x \\right \\rfloor+1$, because $$x = \\left \\lfloor x \\right \\rfloor+\\{x\\}<\\left \\lfloor x \\right \\rfloor+1<x+1<y$$ In this case $$(k+1)\\frac{\\ln{5}}{\\ln{2}}-k\\frac{\\ln{5}}{\\ln{2}}=\\frac{\\ln{5}}{\\ln{2}}>1$$ and from $(1)$, there $\\exists n\\in\\mathbb{N}$ s.t. $$k\\frac{\\ln{5}}{\\ln{2}}<n<(k+1)\\frac{\\ln{5}}{\\ln{2}} \\iff\\\\ k\\ln{5}<n\\ln{2}<(k+1)\\ln{5} \\iff \\\\ 5^k < 2^n<5^{k+1}$$ However $\\frac{\\ln{5}}{\\ln{2}}\\approx 2.32>2$ and $(1)$ can be extended to\n\nif $y-x >2$, for $x\\geq 0, y>0$, then there $\\exists n \\in \\mathbb{N},n>0$ s.t. $$x<n<n+1<y \\tag{2}$$", "date": "2019-11-13 01:46:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2863510/between-any-two-powers-of-5-there-are-either-two-or-three-powers-of-2", "openwebmath_score": 0.974236249923706, "openwebmath_perplexity": 84.0521815525579, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308768564867, "lm_q2_score": 0.8947894618940992, "lm_q1q2_score": 0.8841690955833602}}
{"url": "http://mathrefresher.blogspot.com/2007/04/properties-of-matrix-multiplication.html", "text": "## Tuesday, April 17, 2007\n\n### Properties of Matrix Multiplication\n\nMultiplication can only occur between matrices A and B if the number of columns in A match the number of rows in B. This presents the very important idea that while multiplication of A with B might be a perfectly good operation; this does not guarantee that multiplication of B with A is a perfectly good operation.\n\nEven if matrix A can be multiplied with matrix B and matrix B can be multiplied to matrix A, this doesn't necessarily give us that AB=BA. In other words, unlike the integers, matrices are noncommutative.\n\nProperty 1: Associative Property of Multiplication\n\nA(BC) = (AB)C\n\nwhere A,B, and C are matrices of scalar values.\n\nProof:\n\n(1) Let D = AB, G = BC\n(2) Let F = (AB)C = DC\n(3) Let H = A(BC) = AG\n(4) Using Definition 1, here, we have for each D,F,G,H:\n\ndi,j = \u2211k ai,k*bk,j\n\ngi,j = \u2211k bi,k*ck,j\n\nfi,j = \u2211k di,k*ck,j\n\n(5) So, expanding fi,j gives us:\n\nfi,j = \u2211k (\u2211l ai,l*bl,j)ck,j =\n\n(\u2211kl) ai,l*bl,k*ck,j =\n\n= \u2211l ai,l*(\u2211k bl,k*ck,j) =\n\n= \u2211l ai,l*gl,j = hi,j\n\nQED\n\nProperty 2: Distributive Property of Multiplication\n\nA(B + C) = AB + AC\n(A + B)C = AC + BC\n\nwhere A,B,C are matrices of scalar values.\n\nProof:\n\n(1) Let D = AB such that for each:\n\ndi,j = \u2211k ai,k*bk,j\n\n(2) Let E = AC such that for each:\n\nei,j = \u2211k ai,k*ck,j\n\n(3) Let F = D + E = AB + AC such that for each:\n\nfi,j = \u2211k ai,k*bk,j+ai,k*ck,j = \u2211k ai,k[bk,j + ck,j]\n\n(4) Let G = B+C such that for each:\n\ngi,j = bi,j + ci,j\n\n(5) Let H = A(B+C) = AG such that for each:\n\nhi,j = \u2211k ai,k*gk,j\n\n(6) Then we have AB + AC = A(B+C) since for each:\n\nhi,j = \u2211k ai,k[bk,j + ck,j]\n\n(7) Let M = A + B such that for each:\n\nmi,j = ai,j + bi,j\n\n(8) Let N = (A+B)C = MC such that:\n\nni,j = \u2211k mi,k*ck,j =\n\n= \u2211k (ai,k + bi,k)*ck,j\n\n(9) Let O = BC such that:\n\noi,j = \u2211k bi,k*ck,j\n\n(10) Let P = AC + BC = E + O such that:\n\npi,j = ei,j + oi,j =\n\n= \u2211k ai,k*ck,j + \u2211k bi,k*ck,j =\n\n= \u2211k [ai,k*ck,j + bi,k*ck,j] =\n\n= \u2211k (ai,k + bi,k)*ck,j\n\nQED\n\nProperty 3: Scalar Multiplication\n\nc(AB) = (cA)B = A(cB)\n\nProof:\n\n(1) Let D = AB such that:\n\ndi,j = \u2211k ai,k*bk,j\n\n(2) Let E = c(AB) = cD such that for each:\n\nei,j = c*di,j = c*\u2211k ai,k*bk,j\n\n(3) Let F = (cA)B such that:\n\nfi,j = \u2211k (c*ai,k)*bk,j = c*\u2211k ai,k*bk,j\n\n(4) Let G = A(cB) such that:\n\ngi,j = \u2211k ai,k*(c*bk,j) =\n\n= c*\u2211k ai,k*bk,j\n\nQED\n\nProperty 4: Muliplication of Matrices is not Commutative\n\nAB does not have to = BA\n\nProof:\n\n(1) Let A =\n\n(2) Let B =\n\n(3) AB =\n\n(4) BA =\n\nQED\n\nReferences\n\u2022 Hans Schneider, George Philip Barker, , 1989.\n\nEva Acosta said...\n\nHi,\nI am a math teacher.\nYou can revise your understanding of matrices solving exercises about addition, subtraction and multiplication of matrices. Inverse,... at www.emathematics.net\nFantastic!!!\n\nAnonymous said...\n\nhi\ni am a student taking a course in linear algebra. i just wanted to say that your tutorrial is very helpfull. so post some more on this topic.\nthank you\n\nLuis said...\n\nAmittai Aviram said...\n\nHi! Thank you very much for providing this resource! I am troubled, however, by something in your proof of the associativity of matrix multiplication. (Since blogger.com will not let me use the necessary HTML tags, I will attempt to use LaTeX-like notation instead.)\n\nIn (4), you have\nd_{i,j} = \\sum_{k} a_{i,k}*b_{k,j}\nf_{i,k} = \\sum_{k} d_{i,k}*c_{k,j}\nThen, in (5), you expand this last equation as\nf_{i,j} = \\sum_{k} (\\sum_{l} a_{i,l}*b_{l,j}) c_{k,j}\nHowever, we are expanding d_{i,k} from the previous equation, not d_{i,j} (where the outer summation symbol binds the variable k). And\nd_{i,k} = \\sum_{l} a_{i,l}*b_{l,k}\nThe summation symbol that binds k becomes the outer summation symbol in the expansion. So the correct expansion of the whole equation should be:\nf_{i,j} = \\sum_{k} (\\sum_{l} a_{i,l}*b_{l,k}) c_{k,j}\nwhich does not lend itself to so neat a proof as the one you present. A correct proof, though with rather messier notation, can be found here:\nhttp://linear.ups.edu/xml/0094/fcla-xml-0.94li30.xml\n\nAmittai Aviram said...\n\nAnother page that has a correct proof (avoiding the error noted above) can be found here:\nhttp://www.proofwiki.org/wiki/Matrix_Multiplication_is_Associative#Proof\n\nLarry Freeman said...\n\nHi Amittai,\n\nThanks very much for pointing out the mistake. I will revise the proof and post a comment on this blog when it is fixed.", "date": "2018-05-25 20:19:41", "meta": {"domain": "blogspot.com", "url": "http://mathrefresher.blogspot.com/2007/04/properties-of-matrix-multiplication.html", "openwebmath_score": 0.8964815735816956, "openwebmath_perplexity": 4432.601708256597, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9930961615201183, "lm_q2_score": 0.8902942203004186, "lm_q1q2_score": 0.8841477728038922}}
{"url": "https://math.stackexchange.com/questions/931839/expected-number-of-output-letters-to-get-desired-word", "text": "# Expected number of output letters to get desired word\n\nI am using a letter set of four letters, say {A,B,C,D}, which is used to output a random string of letters. I want to calculate the expected output length until the word ABCD is obtained; that is, the letters A B C D appearing consecutively in that order.\n\nI have referenced this question (Expected Number of Coin Tosses to Get Five Consecutive Heads), but have found a complexity in our case; when we obtain, say, ABA, then we can't say that the chain resets, since we have the next potentially successful chain already being started.\n\nI have tried the approach below, but am not sure if it is completely correct.\nI would be grateful for assertion that this approach is ok, as well as for any alternative methods to approach this issue.\n\nLet e be the expected number of output letters needed to get the target string ABCD. Also, let f be the expected number of output letters needed to get the target string ABCD given we obtained the letter A.\n\nThe table for expected length and probability for e would be\n\n|                          | Exp Len | Prob |\n|--------------------------|---------|------|\n| if first letter is [BCD] |   e+1   | 3/4  |\n| if A then [CD]           |   e+2   | 1/8  |\n| if A then A              |   f+1   | 1/16 |\n| if AB then [BD]          |   e+3   | 1/32 |\n| if AB then A             |   f+2   | 1/64 |\n| if ABC then [BC]         |   e+4   | 1/128|\n| if ABC then A            |   f+3   | 1/256|\n| if ABCD                  |    4    | 1/256|\n---------------------------------------------\n\n\nand a similar table for f after we obtained the letter A would be\n\n|                       | Exp Len | Prob |\n|-----------------------|---------|------|\n|if first letter is [CD]|   e+2   | 1/2  |\n|if first letter is A   |   f+1   | 1/4  |\n|if B then [BD]         |   e+3   | 1/8  |\n|if B then A            |   f+2   | 1/16 |\n|if BC then [BC]        |   e+4   | 1/32 |\n|if BC then A           |   f+3   | 1/64 |\n|if BCD                 |    4    | 1/64 |\n------------------------------------------\n\n\nThe expected length e is equal to the sum of each (Probability)*(Expected Length) product set from the first table, giving $$e\\, =\\, \\frac{3}{4}(e+1)\\, +\\, \\frac{1}{8}(e+2)\\, +\\, \\frac{1}{16}(f+1)\\, +\\, \\frac{1}{32}(e+3 )\\, +\\, \\frac{1}{64}(f+2)\\, +\\, \\frac{1}{128}(e+4)\\, +\\, \\frac{1}{256}(f+3)\\, +\\, \\frac{1}{256}(4) \\\\-----\\\\ e\\, \\, =\\, \\frac{117}{128}e\\, +\\, \\frac{21}{256}f\\, +\\, \\frac{319}{256} \\\\\\\\ 22e\\, =\\, 21f\\, +\\, 319 \\: \\: \\: ---(1) \\\\ 44e\\, =\\, 42f\\, +\\, 638 \\: \\: \\: ---(1')$$ A similar approach for f yields $$f\\, =\\, \\frac{1}{2}(e+2)\\, +\\, \\frac{1}{4}(f+1)\\, +\\, \\frac{1}{8}(e+3)\\, +\\, \\frac{1}{16}(f+2 )\\, +\\, \\frac{1}{32}(e+4)\\, +\\, \\frac{1}{64}(f+3)\\,+\\, \\frac{1}{64}(4) \\\\-----\\\\ f\\, \\, =\\, \\frac{21}{32}e\\, +\\, \\frac{21}{64}f\\, +\\, \\frac{127}{64} \\\\\\\\ 43f\\, =\\, 42e\\, +\\, 127 \\: \\: \\: ---(2)$$ Combining these, we obtain\n$$(2)-(1')\\Rightarrow f\\, =\\, -2e\\, +\\, 765 \\: \\: \\: ---(3)\\\\ (3)\\rightarrow (1)\\Rightarrow 22e = 21(-2e+765)+319 \\\\ e=256 \\\\ f=253$$\n\nSo the expected length seems to be 256 letters output.\n\nI notice this is exactly what we would expect from the naive approach, from the fact that each letter has a 1 in 4 chance appearing each time, and after any four letters' output, the chance of ABCD appearing is $$\\left( \\frac{1}{4} \\right) ^ 4 = \\frac{1}{256} .$$ which is slightly worrying, since the question about five consecutive heads has a probability of 1/32, but a differing number of 62 for the expected length.\n\nAfter the above, I also calculated the expected length until I obtain either of TWO target strings; I used ABCD and CDBA as my targets, if it matters. The result was not the intuitive 128, but was 136 instead, by methodology similar to that above.\n\nUsing the answers provided, I will also try to check this result using new tactics proposed in the answers.\n\nThe natural approach uses transition matrices. For ease of typesetting we write up the solution another way.\n\nLet $e$ be the expected number. Let $a$ be the expected number of additional letters, given that the last letter was an A. Let $b$ be the expected number of additional letters, given the last two letters were AB. And let $c$ be the analogous thing, given the last three letters were ABC.\n\nAt the start, if the first letter is an A, our expected total is $1+a$. If it is anything else, then our expected total is $1+e$. Thus $$e=\\frac{1}{4}(1+a)+\\frac{3}{4}(1+e).$$\n\nIf our last letter was an A, with probability $\\frac{1}{4}$ we get an A, and the additional total (after the first A) is $1+a$. If we get a B, the expected additional total after the first A is $1+b$. If we get a C or a D, the expected additional total after the A is $1+e$. Thus $$a=\\frac{1}{4}(1+a)+\\frac{1}{4}(1+b)+\\frac{2}{4}(1+e).$$ If the last two letters were AB, and we get an A, the additional expected total after the AB is $1+a$. If we get a B or a D, the additional expected total is $1+e$. And if we get a C it is $1+c$. Thus $$b=\\frac{1}{4}(1+a)+\\frac{2}{4}(1+e)+\\frac{1}{4}(1+c).$$ Finally, the same reasoning gives $$c=\\frac{1}{4}(1+a)+\\frac{2}{4}(1+e)+\\frac{1}{4}(1).$$ Four linear equations, four unknowns.\n\n\u2022 I see that this method increases the number of variables introduced, but keeps each resulting equation simpler. I was able to reconstruct the transition matrix from your argument as well. +1 for a great generalizable solution! It would be nice if you could point me to some reference as to how to directly apply the transition matrix to obtain e... \u2013\u00a0yybtcbk Sep 16 '14 at 3:00\n\u2022 Additionally: I was able to apply this method to the addendum problem to get e=136, a=134, b=128, and c=102. It seems a bonus of this method that I can get the a, b, and c values on the way as well. \u2013\u00a0yybtcbk Sep 16 '14 at 3:03\n\u2022 Good! For transition matrices, any beginning book on Markov chains should have it, also many a basic probability book. Can't think of a specific title. \u2013\u00a0Andr\u00e9 Nicolas Sep 16 '14 at 3:09\n\u2022 I guess it's time to dig out my old Markov Chains textbook... gave up on it in Uni, but maybe I can understand it better now... thanks for the pointer! \u2013\u00a0yybtcbk Sep 16 '14 at 3:19\n\nConway's algorithm provides a quick method of calculation: look at how whether the initial letters match the final letters: so \"AAAA\" has matches for the initial $1,2,3,4$, while \"ABCD\" has matches only for the initial $4$; \"ABCA\" would have matches for $1$ and $4$, while \"ABAB\" would have matches for $2$ and $4$. Since the alphabet has four equally likely letters, the algorithm gives the following expected samples sizes:\n\n\u2022 AAAA: $340 = 4^4+4^3+4^2+4^1$\n\u2022 ABCD: $256 = 4^4$\n\u2022 ABCA: $260 = 4^4+4^1$\n\u2022 ABAB: $272 = 4^4+4^2$\n\nSo, as you say, the expected time until \"ABCD\" appears is $256 = 4^4$ samples. This is similar to the coin sequence \"HHHHT\" requiring an expected $32=2^5$ samples.\n\nBy contrast the expected time until \"AAAA\" appears is $340 = 4^4+4^3+4^2+4^1$ samples. This is similar to the coin sequence \"HHHHH\" requiring an expected $62=2^5+2^4+2^3+2^2+2^1$ samples.\n\nIf you had a long string of $n$ letters then you would expect \"ABCD\" to appear about $\\frac{n-3}{256}$ times on average. Similarly you would expect \"AAAA\" to appear about the same number of times on average. But strings of type \"AAAA\" can overlap themselves while those of type \"ABCD\" cannot: for example a string of length $6$ might possibly have three \"AAAA\"s but cannot have more than one \"ABCD\", even if the expected number of each is the same.\n\nTo balance the greater possibility of \"AAAA\"s appearing several times, but the same overall expected number, there is also a greater possibility for \"AAAA\" not appearing at all in the first $6$ letters, or indeed in other initial samples. It is this latter feature which increases the expected sample size until \"AAAA\" does appear, compared with \"ABCD\".\n\n\u2022 Greatly convincing way to see that repeating letters lead to a longer expected string! This also works to suggest why the additional question results in a 136 greater than the intuitive 128; ABCD and CDBA overlap each other. Could you point me somewhere for the theory behind this Conway's Algorithm? My googling only got me algorithms for the game of life, betting odds, and calendars... \u2013\u00a0yybtcbk Sep 16 '14 at 4:10\n\u2022 As for using this Conway's Algorithm for the additional question, I guess that for target strings ABCD and CDBA, positions 2 and 4 match the last target letter, I can do 4^4 + 4^2 = 272, and then can halve this (since we have two target strings?) to get 136...? I'm not sure if this is a valid approach, but the numbers do match up... \u2013\u00a0yybtcbk Sep 16 '14 at 4:18\n\u2022 Two target strings raises the issue in Penney's game \u2013\u00a0Henry Sep 16 '14 at 7:05\n\u2022 \u2013\u00a0Henry Sep 16 '14 at 7:30\n\u2022 All the links refer to the Conway Number and Conway's Algorithm in terms of odds; are these directly applicable to the string length calculations above somehow? \u2013\u00a0yybtcbk Sep 16 '14 at 8:14", "date": "2019-07-16 06:03:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/931839/expected-number-of-output-letters-to-get-desired-word", "openwebmath_score": 0.7974311709403992, "openwebmath_perplexity": 788.5972785961158, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429585263219, "lm_q2_score": 0.899121367808974, "lm_q1q2_score": 0.8841446658955098}}
{"url": "https://math.stackexchange.com/questions/1999358/is-there-a-closed-form-expression-for-these-combinatorial-sums", "text": "# Is there a closed form expression for these combinatorial sums?\n\nI needed to compute this sum: $$\\sum_{i=0}^n\\binom{n}{i}i^2$$ I don't want any proofs or formulae, but just a yes/no answer. Is there a closed form expression for this sum?\n\nJust to share, I was able to find a closed form expression for this sum: $$\\sum_{i=0}^n\\binom{n}{i}i=n.2^{n-1}$$ by using a property of arithmetic progressions and sums of combinations.\nIs there any closed form expression for my sum, or more generally, sums like: $$\\sum_{i=0}^n\\binom{n}{i}i^k$$ , for some $k\\in\\mathbb{N}$?\n\n\u2022 There is a closed form for your sum, and you can get it by applying what you already found for $\\sum_k\\binom{n}kk$. (You can also use calculus methods, but they\u2019re not necessary.) \u2013\u00a0Brian M. Scott Nov 4 '16 at 16:06\n\u2022 Differentiate the series $(1+x)^n$ wrt to $x$ twice to get your sum. \u2013\u00a0SirXYZ Nov 4 '16 at 16:11\n\nYes. You can find the formula as follows: by the binomial theorem we have\n\n$$(1+x)^n = \\sum_{i=0}^n {n\\choose i} x^i.$$\n\nNow apply the differential operator $\\Big(x\\frac{d}{dx}\\Big)^k$ on both sides.\n\n\u2022 Concise and complete. +1 ... You might consider adding \"then set $x=1$\" at the end of the answer. \u2013\u00a0Mark Viola Nov 4 '16 at 16:07\n\u2022 Thanks @Dr.MV, I trust that the OP will extract that information from your comment ;). \u2013\u00a0J.R. Nov 4 '16 at 16:13\n\nAn alternative approach. Us that $\\binom{n}{i}\\binom{i}{k}=\\binom{n}{k}\\binom{n-k}{i-k}$ to get:\n\n$$\\sum_{i=0}^n \\binom{n}{i}\\binom{i}{k}= \\binom{n}{k}\\sum_{i=0}^{n} \\binom{n-k}{i-k}=\\binom{n}{k}2^{n-k}$$\n\nSo, since $i^2=2\\binom{i}{2}+\\binom{i}{1}$ to get that $$\\sum \\binom{n}{i}i^2 = 2\\binom{n}{2}2^{n-2} + \\binom{n}{1}2^{n-1}$$\n\nIn general, $i^k = \\sum_{j=0}^{k} a_j \\binom{i}{j}$ for some sequence $a_j$, yielding the result:\n\n$$\\sum \\binom{n}{i}i^k = \\sum_{j=0}^k a_j\\binom{n}{j}2^{n-j}$$\n\nSo it just amounts to finding the $a_j$ for each $k$.\n\nHint:\n\n$$\\binom{n}{i}i^{2}=\\binom{n}{i}i\\left(i-1\\right)+\\binom{n}{i}i=n\\left(n-1\\right)\\binom{n-2}{i-2}+n\\binom{n-1}{i-1}$$", "date": "2020-02-24 02:10:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1999358/is-there-a-closed-form-expression-for-these-combinatorial-sums", "openwebmath_score": 0.8920262455940247, "openwebmath_perplexity": 358.2792575451208, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462190641614, "lm_q2_score": 0.8947894724152067, "lm_q1q2_score": 0.8840038761310192}}
{"url": "https://math.stackexchange.com/questions/3185610/very-indeterminate-form-lim-x-to-infty-left-sqrtx22x3-sqrtx23", "text": "# Very indeterminate form: $\\lim_{x \\to \\infty} \\left(\\sqrt{x^2+2x+3} -\\sqrt{x^2+3}\\right)^x \\longrightarrow (\\infty-\\infty)^{\\infty}$\n\nHere is problem:\n\n$$\\lim_{x \\to \\infty} \\left(\\sqrt{x^2+2x+3} -\\sqrt{x^2+3}\\right)^x$$\n\nThe solution I presented in the picture below was made by a Mathematics Teacher\n\nI tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct.\n\nAnd here is my method:\n\n\\begin{align*}\\lim_{x \\to \\infty} \\left(\\sqrt{x^2+2x+3} -\\sqrt{x^2+3}\\right)^x&=\\lim_{x \\to \\infty} \\left(\\frac {2x}{\\sqrt{x^2+2x+3} +\\sqrt{x^2+3}}\\right)^x\\\\&=\\lim_{x \\to \\infty}\\frac{1}{ \\left(\\frac {\\sqrt{x^2+2x+3} +\\sqrt{x^2+3}}{2x}\\right)^x}\\end{align*}\n\nThen, I define a new function here\n\n$$y(x)=\\sqrt{x^2+2x+3} +\\sqrt{x^2+3}-2x-1$$\n\nWe have\n\n\\begin{align*} \\lim _{x\\to\\infty} y(x)&=\\lim_{x \\to \\infty}\\sqrt{x^2+2x+3} +\\sqrt{x^2+3}-2x-1\\\\ &=\\lim_{x \\to \\infty}(\\sqrt{x^2+2x+3}-(x+1))+(\\sqrt{x^2+3}-x)\\\\ &=\\lim_{x \\to \\infty}\\frac{2}{\\sqrt{x^2+2x+3}+x+1}+ \\lim_{x \\to \\infty}\\frac{3}{\\sqrt{x^2+3}+x}\\\\ &=0. \\end{align*}\n\nThis implies that $$\\lim_{x \\to \\infty}\\frac{2x}{y(x)+1}=\\infty$$\n\nTherefore,\n\n\\begin{align*} \\lim_{x \\to \\infty}\\frac{1}{ \\left(\\frac {\\sqrt{x^2+2x+3} +\\sqrt{x^2+3}}{2x}\\right)^x}&=\\lim_{x \\to\\infty} \\frac{1}{ \\left(\\frac{y(x)+2x+1}{2x} \\right)^x}\\\\ &=\\lim_{x \\to\\infty} \\frac{1}{ \\left(1+\\frac{y(x)+1}{2x} \\right)^x}\\\\ &=\\lim_{x \\to \\infty}\\frac{1}{\\left( \\left( 1+\\frac{1}{\\frac{2x}{y(x)+1}}\\right)^{\\frac{2x}{y(x)+1}}\\right)^{\\frac{y(x)+1}{2}}}\\\\ & \\end{align*}\n\nHere, we define two functions: $$f(x)=\\left( 1+\\frac{1}{\\frac{2x}{y(x)+1}}\\right)^{\\frac{2x}{y(x)+1}},\\quad g(x)=\\frac{y(x)+1}{2}.$$\n\nWe deduce that, $$\\lim_{x\\to\\infty} f(x)=e>0,\\quad \\lim_{x\\to\\infty} g(x)=\\frac 12>0.$$ Thus, the limit $$\\lim_{x\\to\\infty} f(x)^{g(x)}$$ exists and is finite.\n\nFinally we get,\n\n\\begin{align*} \\lim_{x \\to \\infty}\\frac{1}{\\left( \\left( 1+\\frac{1}{\\frac{2x}{y(x)+1}}\\right)^{\\frac{2x}{y(x)+1}}\\right)^{\\frac{y(x)+1}{2}}} &=\\frac{1}{\\lim_{x \\to \\infty}\\left( \\left( \\left( 1+\\frac{1}{\\frac{2x}{y(x)+1}}\\right)^{\\frac{2x}{y(x)+1}}\\right)^{\\frac{y(x)+1}{2}}\\right)}\\\\ &=\\frac{1}{\\left(\\lim_{x\\to\\infty} \\left( 1+\\frac{1}{\\frac{2x}{y(x)+1}} \\right)^{\\frac{2x}{y(x)+1}}\\right)^{ \\lim_{x\\to\\infty} \\frac{y(x)+1}{2}}}\\\\ &=\\frac {1}{e^{\\frac12}}=\\frac{\\sqrt e}{e}.\\\\&& \\end{align*}\n\nIs the method I use correct?\n\nI have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method?\n\nThank you!\n\n\u2022 @DMcMor Thank you for edit my bad $\\LaTeX$ \u2013\u00a0lone student Apr 12 '19 at 22:10\n\u2022 Honestly your TeX was pretty good! I just aligned the equations to make it a bit easier to see. \u2013\u00a0DMcMor Apr 12 '19 at 22:13\n\u2022 I'm also a BlackPenRedPen fan! (or is it BlackPenRedPenBluePen?) \u2013\u00a0Toby Mak Apr 13 '19 at 10:57\n\u2022 $x^y$ is continuous near $\\left(e,\\frac12\\right)$; thus, $\\lim\\limits_{x\\to\\infty}f(x)^{g(x)}=\\lim\\limits_{x\\to\\infty}f(x)^{\\lim\\limits_{x\\to\\infty}g(x)}$ when $\\lim\\limits_{x\\to\\infty}f(x)=e$ and $\\lim\\limits_{x\\to\\infty}g(x)=\\frac12$ \u2013\u00a0robjohn Apr 24 '19 at 9:59\n\u2022 No. By using the fact that $x^y$ is continuous at this point, it makes your argument rigorous. \u2013\u00a0robjohn Apr 24 '19 at 10:19\n\nYour math looks good! I'd maybe just an extra step here and there to make it clear what your doing. Things like showing that you're multiplying by conjugates and maybe a change of variables, say $$z = \\frac{2x}{y(x)+1},$$ near the end so it's a bit clearer where the $$e$$ comes from. Otherwise everything looks good! This is a tricky limit, I really like your solution.\n\nThe solution appears to be correct. Just for the sake of sanity, here's a different argument, based on the idea that knowing derivatives is knowing many limits.\n\nFirst, find the limit of the logarithm of the beast, which is best treated also with the substitution $$x=1/t$$, which makes us trying to find $$\\lim_{t\\to0^+}\\frac{1}{t}\\log\\left(\\frac{\\sqrt{1+2t+3t^2}-\\sqrt{1+3t^2}}{t}\\right) = \\lim_{t\\to0^+}\\frac{1}{t}\\log\\left(\\frac{2}{\\sqrt{1+2t+3t^2}+\\sqrt{1+3t^2}}\\right)$$ This can be rewritten as $$\\lim_{t\\to0^+}-\\frac{\\log\\bigl(\\sqrt{1+2t+3t^2}+\\sqrt{1+3t^2}\\,\\bigr)-\\log2}{t}$$ which is the negative of the derivative at $$0$$ of $$f(t)=\\log\\bigl(\\sqrt{1+2t+3t^2}+\\sqrt{1+3t^2}\\,\\bigr)$$ Since $$f'(t)=\\frac{1}{\\sqrt{1+2t+3t^2}+\\sqrt{1+3t^2}}\\left(\\frac{1+3t}{\\sqrt{1+2t+3t^2}}+\\frac{3t}{\\sqrt{1+3t^2}}\\right)$$ we have $$f'(0)=1/2$$ and therefore the limit is $$-1/2$$, so your given limit $$e^{-1/2}$$\n\nYour approach is correct but its presentation / application is more complicated than needed here.\n\nHere is how you can use the same approach with much less effort. You have already observed that the base $$F(x) =\\sqrt {x^2+2x+3}-\\sqrt{x^2+3}$$ tends to $$1$$ as $$x\\to\\infty$$. Now the expression under limit can be written as $$\\{F(x) \\} ^x=\\{\\{1+(F(x)-1)\\}^{1/(F(x)-1)}\\}^{x(F(x)-1)}$$ The inner expression tends to $$e$$ and the exponent $$x(F(x) - 1)\\to -1/2$$ so that the desired limit is $$e^{-1/2}$$.\n\nAnother part of your approach is that it involves the tricky use of subtracting $$2x+1$$ from $$y(x)$$. For those who are experienced in the art of calculus this step is obvious via the approximation $$\\sqrt{x^2+2ax+b}\\approx x+a$$ but it may appear a bit mysterious for a novice. It is best to either explain this part or remove it altogether as I have done it in my answer.\n\nAlso note that your approach uses the following limits / rules (it is not necessary to point them out explicitly unless demanded by some strict examiner) :\n\n\u2022 $$\\lim_{x\\to\\infty} \\left(1+\\frac{1}{x}\\right)^x=e$$\n\u2022 If $$\\lim_{x\\to\\infty} f(x) =a>0$$ and $$\\lim_{x\\to\\infty} g(x) =b$$ then $$\\{f(x) \\} ^{g(x)} \\to a^b$$ as $$x\\to\\infty$$.\n\nBy squaring, we can verify that $$x\\le\\sqrt{x^2+3}\\le x\\left(1+\\frac3{2x^2}\\right)\\tag1$$ and $$x+1\\le\\sqrt{x^2+2x+3}\\le(x+1)\\left(1+\\frac1{x(x+1)}\\right)\\tag2$$ Adding $$(1)$$ and $$(2)$$ gives $$2x+1\\le\\sqrt{x^2+2x+3}+\\sqrt{x^2+3}\\le(2x+1)\\left(1+\\frac3{2x^2}\\right)\\tag3$$ Multiplying numerator and denominator by $$\\sqrt{x^2+2x+3}+\\sqrt{x^2+3}$$ gives $$\\sqrt{x^2+2x+3}-\\sqrt{x^2+3}=\\frac{2x}{\\sqrt{x^2+2x+3}+\\sqrt{x^2+3}}\\tag4$$ Bernoulli and cross multiplying yield $$1-\\frac3{2x}\\le\\left(1-\\frac3{2x^2}\\right)^x\\le\\left(1+\\frac3{2x^2}\\right)^{-x}\\tag5$$ Therefore $$(3)$$, $$(4)$$, and $$(5)$$ yield $$\\left(\\frac{2x}{2x+1}\\right)^x\\left(1-\\frac3{2x}\\right)\\le\\left(\\sqrt{x^2+2x+3}-\\sqrt{x^2+3}\\right)^x\\le\\left(\\frac{2x}{2x+1}\\right)^x\\tag6$$ The Squeeze Theorem then says $$\\lim_{x\\to\\infty}\\left(\\sqrt{x^2+2x+3}-\\sqrt{x^2+3}\\right)^x=e^{-1/2}\\tag7$$\n\nYour method is correct if and only if you truly understand how to rigorously prove a critical step where you effectively claim $$\\lim_{x\u2208\\mathbb{R}\u2192\u221e} (1+\\frac1x)^{f(x)} = \\lim_{x\u2208\\mathbb{R}\u2192\u221e} e^{f(x)/x}$$. Note that this requires real exponentiation, and the simplest proof of this would involve the asymptotic expansions for $$\\exp,\\ln$$, hence I personally think it's misleading to think of your method as successfully evading asymptotic expansions. To preempt the very common bogus proof, note that this claim does not follow from $$\\lim_{x\u2208\\mathbb{R}\u2192\u221e} (1+\\frac1x)^x = e$$.\n\nAfter I first posted my answer, you edited your attempt in a non-trivial way. (Please do not edit your question in this way in the future, as it invalidates existing answers.) It still has the same conceptual error, just with different appearance. In this second attempt, you effectively claim that if $$\\lim_{x\u2208\\mathbb{R}\u2192\u221e} g(x) = c$$ then $$\\lim_{x\u2208\\mathbb{R}\u2192\u221e} f(x)^{g(x)} = \\lim_{x\u2208\\mathbb{R}\u2192\u221e} f(x)^c$$ if the latter limit exists. This is not in general true! If you can state and prove the correct theorem of this sort (in a comment), then I will believe that you understand it.\n\nThe common underlying error is that you replaced part of a limit expression with its limit, which is in general invalid!\n\n\u2022 Comments are not for extended discussion; this conversation has been moved to chat. \u2013\u00a0quid Apr 13 '19 at 10:41\n\u2022 \"The common underlying error is that you replaced part of a limit expression with its limit, which is in general invalid!\" Definitely Wrong argument. Because, all limit rules are \"invalid generally\". You cannot show the applicable limit rule in all cases. \u2013\u00a0Zaharyas Apr 20 '19 at 11:28\n\u2022 Your last sentence is mathematically non sense. (as it appears). \u2013\u00a0Zaharyas Apr 20 '19 at 11:45\n\u2022 I'm IMO gold medalist. Now, I study bachelors. \u2013\u00a0Zaharyas Apr 20 '19 at 11:52\n\u2022 IMO does not contain calculus or limit theorems. The content ranges from extremely difficult algebra and pre-calculus problems to problems on branches of mathematics not conventionally covered at school and often not at university level either, such as projective and complex geometry, functional equations, combinatorics, and well-grounded number theory, of which extensive knowledge of theorems is required. You asked me about my mathematical background. I gave the answer. \u2013\u00a0Zaharyas Apr 20 '19 at 18:46\n\nThe proof can be accelerated, using binomial Maclaurin series in the form of $$\\sqrt{x^2+2x+3} = (x+1)\\sqrt{1+\\dfrac2{(x+1)^2}} = (x+1)\\left(1 + \\dfrac1{(x+1)^2}+O\\left(x^{-4}\\right)\\right)$$ $$= x+1+\\dfrac1x+O\\left(x^{-2}\\right),$$ $$\\sqrt{x^2+3} = x\\left(1+\\dfrac3{2x^2}+O(x^{-4})\\right) = x + \\dfrac3{2x}+O(x^{-3}).$$ Then $$\\ln L = \\ln \\lim\\limits_{x\\to\\infty} \\left(\\sqrt{x^2+2x+3}-\\sqrt{x^2+3}\\right)^x = \\lim\\limits_{x\\to\\infty} x\\ln\\left(1-\\frac1{2x}+O\\left(x^{-2}\\right)\\right)$$ $$= \\lim\\limits_{x\\to\\infty} x\\left(-\\frac1{2x}+O\\left(x^{-2}\\right)\\right) = -\\frac12,$$ $$L=e^{\\Large^{-\\frac12}}.$$", "date": "2020-12-03 05:27:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3185610/very-indeterminate-form-lim-x-to-infty-left-sqrtx22x3-sqrtx23", "openwebmath_score": 0.9431429505348206, "openwebmath_perplexity": 575.1508586668708, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109102866639, "lm_q2_score": 0.8933094110250331, "lm_q1q2_score": 0.8839394084710241}}
{"url": "https://nndcgroup.com.ng/blog/1k4v2s2.php?c0ec0e=latex-math-symbols", "text": "Below is the comprehensive list of Greek symbols/characters/alphabets generally used for preparing a document using LaTex. Note that The following table provides a comprehensive list/guide of mathematical Integral symbols using an appropriate example. command to add a little whitespace between them: The choices available with all fonts are: When using the STIX fonts, you also have the This is a small change from regular TeX, where the dollar sign in Note that you can set the integral boundaries by using the underscore _ and circumflex ^ symbol as seen below. The subsequent table provides a list/guide of commonly used Binary Symbols/Operators using an appropriate example. to blend well with Times), or a Unicode font that you provide. width of the symbols below: Care should be taken when putting accents on lower-case i's and j's.\n\nOnline LaTeX equation editor, generate your mathematical expressions using LaTeX with a simple way. If you think I forgot some very important basic function or symbol here, please let me know. You can try testing these Trigonometric functions commands directly on our online LaTeX compilerfor a better understanding.\n\nThe following table provides a comprehensive list/guide of Trigonometric functions along with their LaTeX command. While the syntax inside the pair of dollar signs ($) aims to be TeX-like, Even though commands follow a logical naming scheme, you will probably need a table for the most common math symbols at some point. Additionally, you can use \\mathdefault{...} or its alias Therefore, these Any text element can use math text. in TeX. To test these binary symbols/operators commands directly on our online LaTeX compiler for a better understanding. You can try testing these Integral symbol commands directly on our online LaTeX compiler for a better understanding. Columns are separated with ampersand & and rows with a double backslash \\\\ (the linebreak command). \\leftarrow, \\sum, \\int. inside a pair of dollar signs ($). character can not be found in the custom font. The fonts used should have a Unicode mapping in order to find any fairly tricky to use, and should be considered an experimental feature for Of course LaTeX is able to typeset matrices as well. LaTeX offers math symbols for various kinds of integrals out of the box. mathtext also provides a way to use custom fonts for math. Note that you can set the integral boundaries by using the underscore _ and circumflex ^ symbol as seen below.\n\nin the following \\imath is used to avoid the extra dot over the i: You can also use a large number of the TeX symbols, as in \\infty,\n\nFor this purpose LaTeX offers the following environments. LaTeX offers math symbols for various kinds of integrals out of the box. These are generally used for preparing any document using LaTex. you can then set the following parameters, which control which font file to use Example: $\\begin{bmatrix}1 & 0 & \\cdots & 0\\\\1 & 0 & \\cdots & 0\\\\\\vdots & \\vdots & \\ddots & \\vdots \\\\1 & 0 & 0 & 0\\end{bmatrix}$. You can use a subset TeX markup in any matplotlib text string by placing it [3]\n\nfor a particular set of math characters. . a Roman font have shortcuts. expressions blend well with other text in the plot. Lower and Upper integral boundaries can be set using the symbol underscore character \"_\" and \"^\", respectively. information. Mathtext should be placed between a pair of dollar signs ($). follows: Here \"s\" and \"t\" are variable in italics font (default), \"sin\" is in Roman Doing things the obvious way produces brackets that are too You can try all the LaTeX symbols in the sandbox below. . . The following table provides a comprehensive list/guide of Trigonometric functions along with their LaTeX command. Customizing Matplotlib with style sheets and rcParams, Fractions, binomials, and stacked numbers. (from (La)TeX), STIX fonts (with are designed Table 258: undertilde Extensible Accents . Here are some more basic functions which don't fit in the categories mentioned above. Jump to navigation Jump to search This is an information page. This default can be changed using the mathtext.default rcParam. to use characters from the default Computer Modern fonts whenever a particular Wikipedia:LaTeX symbols. . fractions or sub/superscripts: The default font is italics for mathematical symbols. Text rendering With LaTeX). easy to display monetary values, e.g., \"$100.00\", if a single dollar sign . around fractions. the calligraphy A is squished into the sin. sign. . Play around with the commands, make your own formulas and copy the code to your document. All the Greek symbols/characters/alphabets are showed along with the LaTex rendered output.\n\nForm W7 \u8a18\u5165\u4f8b, Black Mouth Cur Puppies, How Many Chapters In The Ocean At The End Of The Lane, Dan Heder Wife, Toph Is Blind, Whirlpool Wrt311fzdm00 Manual, Xfi Pods Ethernet Port, Aylin Er\u00e7el Cause Of Death, Psalm 92 Prayer, Recetas Con Yuca Rallada, My Hero Academia Season 4 Opening Lyrics, Eddie Stanky 42, Kidy Fox Wake Up West Texas, Rockall Energy Stock, Soundcloud Playlist To Mp3, Fly Love Lyrics Portuguese, Proper Noun Of Athlete, Rappers Without Kids, Uromastyx Loricata For Sale, Choir Concert Reflection Essay, Dave Schultz Death Scene, Hippopotamus Heart Size, Myrtle Urkel Episodes, Lumber Tycoon 2 Guide, Leonberger Breeder California, My Hero Academia Crossover, Gillette Sales Plummet, Unblocked Google Drive Movies, Koolkat Truck Air Conditioning Price, Three Bears Chow Chows, Anisha Gregg Avani, Rurik Descendants Today, Hendrick Motorsports Employees, Prabal Gurung Family, Woke Wednesday Meaning, Dispatch Subscription Management, Taekwondo Lesson Plan Pdf, How Did Andy Biersack And Juliet Simms Meet, Agent Originaux R6, Dani Gabriel Physiotherapist, Annelle Dupuy Desoto, Measles Powerpoint Template, Peliculas Completas En Castellano Youtube, Southern Stingray Sting, Avatar 2 Online, Phoebe Cates Parents, Franchise Starbucks Canada Prix, Warrior In Samoan, Killing Lincoln Sparknotes, Rich Eisen Show Hiatus, Triple Cave Spider Xp Farm, The Workmanship Of Your Timbrels And Pipes, Mo Duinne Gaelic Pronunciation, Vw Trike For Sale On Craigslist, Custom Supercar Builder, Goat Fever Reducer, Anime Face Mask, English Bulldog Bite Force, Moff Gideon Costume, Julian Clary Ian Mackley Split, Used Cable Lasher For Sale, Best Electric Bike For Over 60s, 1993 Honda Goldwing Trike, Extreme Hills M Minecraft, Jt The Bigga Figga Height, Modern Dollhouse Plans, Isuzu Box Truck Vin Location, Upvc Windows Northern Ireland Prices, Karthika Masam Dates 2020, Funkoverse Jurassic Park How To Play, Iterative Preorder Traversal Without Stack, Winterizer Fertilizer Walmart, Is Stockx Legit, The Curse Of Frankenstein Dailymotion, Pipefitter Blue Book App, Citadel Vs Citadel Securities, \u4e09\u6d66 \u6625 \u99ac \u7d50\u5a5a,", "date": "2021-03-04 08:46:10", "meta": {"domain": "com.ng", "url": "https://nndcgroup.com.ng/blog/1k4v2s2.php?c0ec0e=latex-math-symbols", "openwebmath_score": 0.7525127530097961, "openwebmath_perplexity": 13979.083181375348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551566309688, "lm_q2_score": 0.9161096193153989, "lm_q1q2_score": 0.8839130902356964}}
{"url": "http://tex.stackexchange.com/questions/53702/wrapping-text-in-enumeration-environment-around-a-table?answertab=active", "text": "# Wrapping text in enumeration environment around a table\n\nI would like to wrap the text in an enumerate environment around a table, this is my actual situation (sorry for including so much code, but I wanted to give a precise idea of the number of lines in the enumerate environment):\n\n\\subsection{Given the following data set where \u201cTarget 2\u201d represent the class attribute,\ncompute the naive Bayesian classification for the instance $<L,white>$ and $<XS,?>$.}\n\\begin{table}[h!t]\n\\centering\n\\begin{tabular}{ccc}\n\\toprule\nSize        &   Color   &   Target2 \\\\\n\\midrule\nXS      &   green   &   Yes     \\\\\nL       &   green   &   Yes     \\\\\nXS      &   white   &   No      \\\\\nM       &   black   &   No      \\\\\nXL      &   green   &   Yes     \\\\\nXS      &   white   &   Yes     \\\\\nL       &   black   &   No      \\\\\nM       &   green   &   Yes     \\\\\n\\bottomrule\n\\end{tabular}\n\\end{table}\n\\begin{enumerate}\n\\item In this case the experience $E$ is made up of $e_{1} = L$ and $e_{2} = white$,\nwhich in Naive Bayes are to be considered as independent, therefore we have:\n\\begin{align*}\nP(yes|E) &= P(L|yes)\\cdot P(white|yes) \\cdot P(yes)     \\\\\n&= \\sfrac{1}{5}\\times \\sfrac{1}{5} \\times \\sfrac{5}{8} \\\\\n&= 0.025\n\\end{align*}\n\\begin{align*}\nP(no|E) &= P(L|no) \\cdot P(white|no) \\cdot P(no) \\\\\n&= \\sfrac{1}{3} \\times \\sfrac{1}{3} times \\sfrac{3}{8}  \\\\\n&= 0.041\n\\end{align*}\nThen we normalize:\n\\begin{align*}\nP(yes) &= \\frac{0.025}{0.066}\n\\simeq 0.38\n\\end{align*}\n\\begin{align*}\nP(no) &= \\frac{0.041}{0.066}\n\\simeq 0.62\n\\end{align*}\nAs $P(no) > P(yes)$, we label $<L,white>$ as no''.\n\\item Now we have to classify a sample with a missing value. During the testing phase\nwe simply omit the attribute\\footnote{Classification Other Methods, slide 24}:\n\\begin{align*}\nP(yes|XS) &= P(XS|yes) \\cdot P(yes) = \\sfrac{2}{5} \\times \\sfrac{5}{8}\n= 0.25\n\\end{align*}\n\\begin{align*}\nP(no|XS) &= P(XS|no) \\cdot P(no)\n= \\sfrac{1}{3} \\times \\sfrac{3}{8}\n= 0.125\n\\end{align*}\nLet us normalize\n\\begin{align*}\nP(yes) &= \\sfrac{0.25}{0.375} \\simeq 0.7 \\\\\nP(no)  &= \\sfrac{0.125}{0.375} \\simeq 0.3\n\\end{align*}\nBottom line this sample is classified as yes''.\n\\end{enumerate}\n\n\nI have tried using the wrapfig and the floatftl package unsuccessfully, the table was moved to the end of the list in both cases. I have considered using two minipage environments, but I would like the text to actually wrap around the table.\n\n-\n\nIt helps when you post questions to make complete documents including loading all the packages you need, I guessed\n\n\\usepackage{booktabs,xfrac,amsmath}\n\n\nin this case. Also I fixed a few font issues (for multi-letter identifiers and angle brackets)\n\nChanging margins within a LaTeX list is a bit delicate, but this is I think the layout you want\n\n\\documentclass{article}\n\n\\usepackage{booktabs,xfrac,amsmath}\n\n\\begin{document}\n\n\\subsection{Given the following data set where \u201cTarget 2\u201d represent the class attribute,\ncompute the naive Bayesian classification for the instance $\\langle L,white\\rangle$ and $\\langle \\mathit{XS},?\\rangle$.}\n\n\\savebox0{%\n\\begin{tabular}{ccc}\n\\toprule\nSize        &   Color   &   Target2 \\\\\n\\midrule\nXS      &   green   &   Yes     \\\\\nL       &   green   &   Yes     \\\\\nXS      &   white   &   No      \\\\\nM       &   black   &   No      \\\\\nXL      &   green   &   Yes     \\\\\nXS      &   white   &   Yes     \\\\\nL       &   black   &   No      \\\\\nM       &   green   &   Yes     \\\\\n\\bottomrule\n\\end{tabular}}\n\n\\begin{enumerate}\n\\makeatletter\n\\dimen@\\wd0\n\\parshape \\@ne \\@totalleftmargin \\linewidth\n\\hbox to \\textwidth{\\hfill\\vtop to \\z@{\\vskip1em \\box\\z@\\vss}}\n\\item\n\nIn this case the experience $E$ is made up of $e_{1} = L$ and $e_{2} = \\mathit{white}$,\nwhich in Naive Bayes are to be considered as independent, therefore we have:\n\\begin{align*}\nP(\\mathit{yes}|E) &= P(L|\\mathit{yes})\\cdot P(\\mathit{white}|\\mathit{yes}) \\cdot P(\\mathit{yes})     \\\\\n&= \\sfrac{1}{5}\\times \\sfrac{1}{5} \\times \\sfrac{5}{8} \\\\\n&= 0.025\n\\end{align*}\n\\begin{align*}\nP(\\mathit{no}|E) &= P(L|\\mathit{no}) \\cdot P(\\mathit{white}|\\mathit{no}) \\cdot P(\\mathit{no}) \\\\\n&= \\sfrac{1}{3} \\times \\sfrac{1}{3} \\times \\sfrac{3}{8}  \\\\\n&= 0.041\n\\end{align*}\n\n\\parshape \\@ne \\@totalleftmargin \\linewidth\n\nThen we normalize:\n\\begin{align*}\nP(\\mathit{yes}) &= \\frac{0.025}{0.066}\n\\simeq 0.38\n\\end{align*}\n\\begin{align*}\nP(\\mathit{no}) &= \\frac{0.041}{0.066}\n\\simeq 0.62\n\\end{align*}\nAs $P(\\mathit{no}) > P(\\mathit{yes})$, we label $\\langle L,\\mathit{white}\\rangle$ as no''.\n\n\\item Now we have to classify a sample with a missing value. During the testing phase\nwe simply omit the attribute\\footnote{Classification Other Methods, slide 24}:\n\\begin{align*}\nP(\\mathit{yes}|\\mathit{XS}) &= P(\\mathit{XS}|\\mathit{yes}) \\cdot P(\\mathit{yes}) = \\sfrac{2}{5} \\times \\sfrac{5}{8}\n= 0.25\n\\end{align*}\n\\begin{align*}\nP(\\mathit{no}|\\mathit{XS}) &= P(\\mathit{XS}|\\mathit{no}) \\cdot P(\\mathit{no})\n= \\sfrac{1}{3} \\times \\sfrac{3}{8}\n= 0.125\n\\end{align*}\nLet us normalize\n\\begin{align*}\nP(\\mathit{yes}) &= \\sfrac{0.25}{0.375} \\simeq 0.7 \\\\\nP(\\mathit{no})  &= \\sfrac{0.125}{0.375} \\simeq 0.3\n\\end{align*}\nBottom line this sample is classified as yes''.\n\\end{enumerate}\n\n\\end{document}\n\n-\nthis is great, thank you very much David! \u2013\u00a0 Edo Apr 29 '12 at 10:26\nNot quite right: note the \"therefore we have:\" is stretched out. That's probably fixable but a workaround is to put a blank line after that line, before the following align. \u2013\u00a0 David Carlisle Apr 29 '12 at 13:27\nI had noticed that, but I guess that a \\hfill should do the job. Still it is a great answer, even though I hoped there was some package to manage these situations, thank you very much again. \u2013\u00a0 Edo Apr 29 '12 at 13:30", "date": "2014-10-31 05:54:20", "meta": {"domain": "stackexchange.com", "url": "http://tex.stackexchange.com/questions/53702/wrapping-text-in-enumeration-environment-around-a-table?answertab=active", "openwebmath_score": 1.0000078678131104, "openwebmath_perplexity": 5530.152522412758, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022627413796, "lm_q2_score": 0.9046505331728752, "lm_q1q2_score": 0.883864040811854}}
{"url": "https://www.mathworks.com/help/symbolic/piecewise.html", "text": "# piecewise\n\nConditionally defined expression or function\n\n## Description\n\nexample\n\npw = piecewise(cond1,val1,cond2,val2,...) returns the piecewise expression or function pw whose value is val1 when condition cond1 is true, is val2 when cond2 is true, and so on. If no condition is true, the value of pw is NaN.\n\nexample\n\npw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal) returns the piecewise expression or function pw that has the value otherwiseVal if no condition is true.\n\n## Examples\n\ncollapse all\n\nDefine the following piecewise expression by using piecewise.\n\n$\\mathit{y}=\\left\\{\\begin{array}{ll}-1& \\mathit{x}<0\\\\ 1& \\mathit{x}>0\\end{array}$\n\nsyms x\ny = piecewise(x < 0,-1,x > 0,1)\ny =\n\nEvaluate y at -2, 0, and 2 by using subs to substitute for x. Because y is undefined at x = 0, the value is NaN.\n\nsubs(y,x,[-2 0 2])\nans =\u00a0$\\left(\\begin{array}{ccc}-1& \\mathrm{NaN}& 1\\end{array}\\right)$\n\nDefine the following function symbolically.\n\n$\\mathit{y}\\left(\\mathit{x}\\right)=\\left\\{\\begin{array}{ll}-1& \\mathit{x}<0\\\\ 1& \\mathit{x}>0\\end{array}$\n\nsyms y(x)\ny(x) = piecewise(x < 0,-1,x > 0,1)\ny(x) =\n\nBecause y(x) is a symbolic function, you can directly evaluate it for values of x. Evaluate y(x) at -2, 0, and 2. Because y(x) is undefined at x = 0, the value is NaN. For details, see Create Symbolic Functions.\n\ny([-2 0 2])\nans =\u00a0$\\left(\\begin{array}{ccc}-1& \\mathrm{NaN}& 1\\end{array}\\right)$\n\nSet the value of a piecewise function when no condition is true (called otherwise value) by specifying an additional input argument. If an additional argument is not specified, the default otherwise value of the function is NaN.\n\nDefine the piecewise function\n\n$y=\\left\\{\\begin{array}{cc}-2& x<-2\\\\ 0& -2\n\nsyms y(x)\ny(x) = piecewise(x < -2,-2,(-2 < x) & (x < 0),0,1)\ny(x) =\n\nEvaluate y(x) between -3 and 1 by generating values of x using linspace. At -2 and 0, y(x) evaluates to 1 because the other conditions are not true.\n\nxvalues = linspace(-3,1,5)\nxvalues = 1\u00d75\n\n-3    -2    -1     0     1\n\nyvalues = y(xvalues)\nyvalues =\u00a0$\\left(\\begin{array}{ccccc}-2& 1& 0& 1& 1\\end{array}\\right)$\n\nPlot the following piecewise expression by using fplot.\n\n$y=\\left\\{\\begin{array}{cc}-2& x<-2\\\\ x& -22\\end{array}.$\n\nsyms x\ny = piecewise(x < -2,-2,-2 < x < 2,x,x > 2,2);\nfplot(y)\n\nOn creation, a piecewise expression applies existing assumptions. Apply assumptions set after creating the piecewise expression by using simplify on the expression.\n\nAssume x > 0. Then define a piecewise expression with the same condition x > 0. piecewise automatically applies the assumption to simplify the condition.\n\nsyms x\nassume(x > 0)\npw = piecewise(x < 0,-1,x > 0,1)\npw =\u00a0$1$\n\nClear the assumption on x for further computations.\n\nassume(x,'clear')\n\nCreate a piecewise expression pw with the condition x > 0. Then set the assumption that x > 0. Apply the assumption to pw by using simplify.\n\npw = piecewise(x < 0,-1,x > 0,1);\nassume(x > 0)\npw = simplify(pw)\npw =\u00a0$1$\n\nClear the assumption on x for further computations.\n\nassume(x,'clear')\n\nDifferentiate, integrate, and find limits of a piecewise expression by using diff, int, and limit respectively.\n\nDifferentiate the following piecewise expression by using diff.\n\n$\\mathit{y}=\\left\\{\\begin{array}{ll}1/\\mathit{x}& \\mathit{x}<-1\\\\ \\mathrm{sin}\\left(\\mathit{x}\\right)/\\mathit{x}& \\mathit{x}\\ge -1\\end{array}$\n\nsyms x\ny = piecewise(x < -1,1/x,x >= -1,sin(x)/x);\ndiffy = diff(y,x)\ndiffy =\n\nIntegrate y by using int.\n\ninty = int(y,x)\ninty =\n\nFind the limits of y at 0 by using limit.\n\nlimit(y,x,0)\nans =\u00a0$1$\n\nFind the right- and left-sided limits of y at -1. For details, see limit.\n\nlimit(y,x,-1,'right')\nans =\u00a0$\\mathrm{sin}\\left(1\\right)$\nlimit(y,x,-1,'left')\nans =\u00a0$-1$\n\nAdd, subtract, divide, and multiply two piecewise expressions. The resulting piecewise expression is only defined where the initial piecewise expressions are defined.\n\nsyms x\npw1 = piecewise(x < -1,-1,x >= -1,1);\npw2 = piecewise(x < 0,-2,x >= 0,2);\n\nsub = pw1-pw2\nsub =\n\nmul = pw1*pw2\nmul =\n\ndiv = pw1/pw2\ndiv =\n\nModify a piecewise expression by replacing part of the expression using subs. Extend a piecewise expression by specifying the expression as the otherwise value of a new piecewise expression. This action combines the two piecewise expressions. piecewise does not check for overlapping or conflicting conditions. Instead, like an if-else ladder, piecewise returns the value for the first true condition.\n\nChange the condition x < 2 in a piecewise expression to x < 0 by using subs.\n\nsyms x\npw = piecewise(x < 2,-1,x > 0,1);\npw = subs(pw,x < 2,x < 0)\npw =\n\nAdd the condition x > 5 with the value 1/x to pw by creating a new piecewise expression with pw as the otherwise value.\n\npw = piecewise(x > 5,1/x,pw)\npw =\n\n## Input Arguments\n\ncollapse all\n\nCondition, specified as a symbolic condition or variable. A symbolic variable represents an unknown condition.\n\nExample: x > 2\n\nValue when condition is satisfied, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.\n\nValue if no conditions are true, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. If otherwiseVal is not specified, its value is NaN.\n\n## Output Arguments\n\ncollapse all\n\nPiecewise expression or function, returned as a symbolic expression or function. The value of pw is the value val of the first condition cond that is true. To find the value of pw, use subs to substitute for variables in pw.\n\n## Tips\n\n\u2022 piecewise does not check for overlapping or conflicting conditions. A piecewise expression returns the value of the first true condition and disregards any following true expressions. Thus, piecewise mimics an if-else ladder.\n\n## Version History\n\nIntroduced in R2016b", "date": "2022-05-17 22:05:37", "meta": {"domain": "mathworks.com", "url": "https://www.mathworks.com/help/symbolic/piecewise.html", "openwebmath_score": 0.7444276213645935, "openwebmath_perplexity": 3955.442450780083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769049752756, "lm_q2_score": 0.905989815306765, "lm_q1q2_score": 0.8838627399560954}}
{"url": "https://math.stackexchange.com/questions/2812614/question-on-probability-of-winning-a-game-of-dice/2812717", "text": "# question on probability of winning a game of dice\n\nQUESTION:\n\nA and B are playing a game by alternately rolling a die, with A starting first. Each player\u2019s score is the number obtained on his last roll. The game ends when the sum of scores of the two players is 7, and the last player to roll the die wins. What is the probability that A wins the game\n\nMY ATTEMPT:\n\nOn the first roll it is impossible for A to win. on the other hand B can win as long as the number on A's die and the number on B's die adds up to seven. The combinations which allow B to win are: $$S=\\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\\}$$ which is 6 out of a total of 36 combinations so the probability of B winning in the first round is $\\frac{1}{6}$ and the probability of A winning is 0.\n\nIn the second roll of A a has a winning probability of $\\frac{1}{6}$ using the same logic as for B in roll1 and in the next roll of B again B has a winning probability of $\\frac{1}{6}$. I can continue to do the above on and on but it will only cycle back to the first scenario so i stop here.\n\nNow if B wins in round 1 then A loses and can not proceed to the second roll. Therefore the probability of a winning is probability of B not winning in the first round into the probability of A winning in the next, which is $$(1- \\frac{1}{6}) (\\frac{1}{6})$$ which is $\\frac{5}{36}$\n\nUnfortunately this is not the correct answer as per the answer key. Can someone please point out where I have gone wrong. Any help is appreciated.Thanks :)\n\nEDIT:  on any next attempt the probability of A winning can be written as $(\\frac{5}{6})^2$ of the previous since there is a $(\\frac{5}{6})$ probability that A will not win and a $(\\frac{5}{6})$ chance that B will not win hence leading to the next attempt.\n\nthus the answere is $(\\frac{1}{6}) ((\\frac{5}{6})^2+(\\frac{5}{6})^4+...)$ which is equal to $\\frac{1}{6} (\\frac{1}{1-(\\frac{5}{6})^2)})$ which is 6/11 and the right answer. EDIT  thus the answere is $(\\frac{5}{36}) ((\\frac{5}{6})^2+(\\frac{5}{6})^4+...)$ which is equal to $\\frac{1}{6} (\\frac{1}{1-(\\frac{5}{6})^2)})$ which is 5/11\n\nIn your last equation, the factor $1/6$ should be the $5/36$ that you found above. This gives $5/11$ not $6/11$. You can get the answer without using infinite series, as follows: Let $p$ be the probability that A wins. Consider the situation after B's first roll. B wins $1/6$ of the time. If this doesn't happen then B is in the situation that A was in at the start. So B wins the game with probability $1/6 + (5/6)p$. But B's probability is also $1-p$. This gives $p = 5/11$.\nYour value of $\\frac{5}{36}$ gives the probability of A winning on A's second roll. But of course there may be later chances for A to win.", "date": "2021-10-15 21:36:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2812614/question-on-probability-of-winning-a-game-of-dice/2812717", "openwebmath_score": 0.7995924949645996, "openwebmath_perplexity": 127.69045875533342, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754447499796, "lm_q2_score": 0.8976952989498448, "lm_q1q2_score": 0.8838487482135093}}
{"url": "https://www.exaton.hu/qn573/c5fe24-tangent-of-a-circle-example", "text": "This point is called the point of tangency. Rules for Dealing with Chords, Secants, Tangents in Circles This page created by Regents reviews three rules that are used when working with secants, and tangent lines of circles. AB 2 = DB * CB \u2026\u2026\u2026\u2026 This gives the formula for the tangent. Solution\u00a0This problem is similar to the previous one, except that now we don\u2019t have the standard equation. The straight line \\ (y = x + 4\\) cuts the circle \\ (x^ {2} + y^ {2} = 26\\) at \\ (P\\) and \\ (Q\\). Tangent. A tangent line intersects a circle at exactly one point, called the point of tangency. We\u2019re finally done. If two tangents are drawn to a circle from an external point, 3 Circle common tangents The following set of examples explores some properties of the common tangents of pairs of circles. Tangent lines to one circle. Challenge problems: radius & tangent. The line is a tangent to the circle at P as shown below. In the figure below, line B C BC B C is tangent to the circle at point A A A. Proof: Segments tangent to circle from outside point are congruent. Jenn, Founder Calcworkshop\u00ae, 15+ Years Experience (Licensed & Certified Teacher). Tangents of circles problem (example 1) Tangents of circles problem (example 2) Tangents of circles problem (example 3) Practice: Tangents of circles problems. The required equation will be x(5) + y(6) + (\u20132)(x + 5) + (\u2013 3)(y + 6) \u2013 15 = 0, or 4x + 3y = 38. This video provides example problems of determining unknown values using the properties of a tangent line to a circle. The tangent line never crosses the circle, it just touches the circle. We know that AB is tangent to the circle at A. This lesson will cover a few examples to illustrate the equation of the tangent to a circle in point form. Since the tangent line to a circle at a point P is perpendicular to the radius to that point, theorems involving tangent lines often involve radial lines and orthogonal circles. Also find the point of contact. Question 1: Give some properties of tangents to a circle. The Tangent intersects the circle\u2019s radius at $90^{\\circ}$ angle. We\u2019ve got quite a task ahead, let\u2019s begin! Yes! 16 = x. If two segments from the same exterior point are tangent to a circle, then the two segments are congruent. Phew! Can you find ? Comparing non-tangents to the point form will lead to some strange results, which I\u2019ll talk about sometime later. Question: Determine the equation of the tangent to the circle: $x^{2}+y^{2}-2y+6x-7=0\\;at\\;the\\;point\\;F(-2:5)$ Solution: Write the equation of the circle in the form: $\\left(x-a\\right)^{2}+\\left(y-b\\right)^{2}+r^{2}$ Note that in the previous two problems, we\u2019ve assumed that the given lines are tangents to the circles. Example 5\u00a0Show that the tangent to the circle x2 + y2 = 25 at the point (3, 4) touches the circle x2 + y2 \u2013 18x \u2013 4y + 81 = 0. BY P ythagorean Theorem, LJ 2 + JK 2 = LK 2. (2) \u2220ABO=90\u00b0 //tangent line is perpendicular to circle. If the center of the second circle is inside the first, then the and signs both correspond to internally tangent circles. Example 4\u00a0Find the point where the line 4y \u2013 3x = 20 touches the circle x2 + y2 \u2013 6x \u2013 2y \u2013 15 = 0. This is the currently selected item. a) state all the tangents to the circle and the point of tangency of each tangent. Note; The radius and tangent are perpendicular at the point of contact. That\u2019ll be all for this lesson. The required equation will be x(4) + y(-3) = 25, or 4x \u2013 3y = 25. } } } On comparing the coefficients, we get x1/3 = y1/4 = 25/25, which gives the values of x1 and y1 as 3 and 4 respectively. (1) AB is tangent to Circle O //Given. Problem 1: Given a circle with center O.Two Tangent from external point P is drawn to the given circle. Answer:The tangent lin\u2026 What is the length of AB? It meets the line OB such that OB = 10 cm. 4. A circle is a set of all points that are equidistant from a fixed point, called the center, and the segment that joins the center of a circle to any point on the circle is called the radius. This means that A T \u00af is perpendicular to T P \u2194. and \u2026 Consider a circle in a plane and assume that $S$ is a point in the plane but it is outside of the circle. Think, for example, of a very rigid disc rolling on a very flat surface. Therefore, to find the values of x1 and y1, we must \u2018compare\u2019 the given equation with the equation in the point form. How do we find the length of A P \u00af? The angle formed by the intersection of 2 tangents, 2 secants or 1 tangent and 1 secant outside the circle equals half the difference of the intercepted arcs!Therefore to find this angle (angle K in the examples below), all that you have to do is take the far intercepted arc and near the smaller intercepted arc and then divide that number by two! Solution\u00a0We\u2019ve done a similar problem in a previous lesson, where we used the slope form. Here, I\u2019m interested to show you an alternate method. and are tangent to circle at points and respectively. We\u2019ll use the new method again \u2013 to find the point of contact, we\u2019ll simply compare the given equation with the equation in point form, and solve for x\u00ad1 and y\u00ad1. (4) \u2220ACO=90\u00b0 //tangent line is perpendicular to circle. Tangent to a Circle is a straight line that touches the circle at any one point or only one point to the circle, that point is called tangency. Solved Examples of Tangent to a Circle. vidDefer[i].setAttribute('src',vidDefer[i].getAttribute('data-src')); But there are even more special segments and lines of circles that are important to know. (3) AC is tangent to Circle O //Given. Let\u2019s begin. Now, let\u2019s learn the concept of tangent of a circle from an understandable example here. If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency. A tangent to the inner circle would be a secant of the outer circle. In the circle O, P T \u2194 is a tangent and O P \u00af is the radius. How to Find the Tangent of a Circle? And when they say it's circumscribed about circle O that means that the two sides of the angle they're segments that would be part of tangent lines, so if we were to continue, so for example that right over there, that line is tangent to the circle and (mumbles) and this line is also tangent to the circle. 26 = 10 + x. Subtract 10 from each side. When two segments are drawn tangent to a circle from the same point outside the circle, the segments are congruent. Worked example 13: Equation of a tangent to a circle. The tangent to a circle is perpendicular to the radius at the point of tangency. Make a conjecture about the angle between the radius and the tangent to a circle at a point on the circle. A tangent line t to a circle C intersects the circle at a single point T.For comparison, secant lines intersect a circle at two points, whereas another line may not intersect a circle at all. it represents the equation of the tangent at the point P 1 (x 1, y 1), of a circle whose center is at S(p, q). Get access to all the courses and over 150 HD videos with your subscription, Monthly, Half-Yearly, and Yearly Plans Available, Not yet ready to subscribe? Through any point on a circle , only one tangent can be drawn; A perpendicular to a tangent at the point of contact passes thought the centre of the circle. Can the two circles be tangent? Answer:The properties are as follows: 1. Knowing these essential theorems regarding circles and tangent lines, you are going to be able to identify key components of a circle, determine how many points of intersection, external tangents, and internal tangents two circles have, as well as find the value of segments given the radius and the tangent segment. The problem has given us the equation of the tangent: 3x + 4y = 25. Example. Now, draw a straight line from point $S$ and assume that it touches the circle at a point $T$. Consider the circle below. The point of contact therefore is (3, 4). We\u2019ll use the point form\u00a0once again. its distance from the center of the circle must be equal to its radius. At the point of tangency, the tangent of the circle is perpendicular to the radius. window.onload = init; \u00a9 2021 Calcworkshop LLC / Privacy Policy / Terms of Service. The extension problem of this topic is a belt and gear problem which asks for the length of belt required to fit around two gears. And the final step \u2013 solving the obtained line with the tangent gives us the foot of perpendicular, or the point of contact as (39/5, 2/5). Solution\u00a0This one is similar to the previous problem, but applied to the general equation of the circle. and are both radii of the circle, so they are congruent. We have highlighted the tangent at A. 3. // Last Updated: January 21, 2020 - Watch Video //. Sine, Cosine and Tangent are the main functions used in Trigonometry and are based on a Right-Angled Triangle. Cross multiplying the equation gives. 16 Perpendicular Tangent Converse. Sketch the circle and the straight line on the same system of axes. Suppose line DB is the secant and AB is the tangent of the circle, then the of the secant and the tangent are related as follows: DB/AB = AB/CB. Example 1\u00a0Find the equation of the tangent to the circle x2 + y2 = 25, at the point (4, -3). Examples of Tangent The line AB is a tangent to the circle at P. A tangent line to a circle contains exactly one point of the circle A tangent to a circle is at right angles to \u2026 And if a line is tangent to a circle, then it is also perpendicular to the radius of the circle at the point of tangency, as Varsity Tutors accurately states. Take Calcworkshop for a spin with our FREE limits course. function init() { Example 6 : If the line segment JK is tangent to circle \u2026 b) state all the secants. The circle\u2019s center is (9, 2) and its radius is 2. Since tangent AB is perpendicular to the radius OA, \u0394OAB is a right-angled triangle and OB is the hypotenuse of \u0394OAB. In this geometry lesson, we\u2019re investigating tangent of a circle. Tangent, written as tan\u2061(\u03b8), is one of the six fundamental trigonometric functions.. Tangent definitions. On solving the equations, we get x1 = 0 and y1 = 5. Because JK is tangent to circle L, m \u2220LJK = 90 \u00b0 and triangle LJK is a right triangle. Proof of the Two Tangent Theorem. Therefore, the point of contact will be (0, 5). The required perpendicular line will be (y \u2013 2) = (4/3)(x \u2013 9) or 4x \u2013 3y = 30. The equation of the tangent in the point for will be xx1 + yy1 \u2013 3(x + x1) \u2013 (y + y1) \u2013 15 = 0, or x(x1 \u2013 3) + y(y1 \u2013 1) = 3x1 + y1 + 15. On comparing the coefficients, we get (x\u00ad1 \u2013 3)/(-3) = (y1 \u2013 1)/4 = (3x\u00ad1 + y1 + 15)/20. var vidDefer = document.getElementsByTagName('iframe'); By using Pythagoras theorem, OB^2 = OA^2~+~AB^2 AB^2 = OB^2~-~OA^2 AB = \\sqrt{OB^2~-~OA^2 } = \\sqrt{10^2~-~6^2} = \\sqrt{64}= 8 cm To know more about properties of a tangent to a circle,\u00a0download \u2026 for (var i=0; i\nVolvo Xc40 Plug-in Hybrid Release Date Usa, Bash Remove Trailing Slash, Perle Cotton Vs Embroidery Floss, 3d Print Troubleshooting Cura, Email Address Search, When Is The Ymca Going To Reopen,", "date": "2021-03-05 04:04:23", "meta": {"domain": "exaton.hu", "url": "https://www.exaton.hu/qn573/c5fe24-tangent-of-a-circle-example", "openwebmath_score": 0.4151161313056946, "openwebmath_perplexity": 358.98978621027345, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9861513881564148, "lm_q2_score": 0.8962513724408292, "lm_q1q2_score": 0.8838395350696157}}
{"url": "https://www.lil-help.com/questions/7852/in-the-game-of-two-finger-morra-2-players-show-1-or-2-fingers-and-simultaneously-guess-the-number-of-fingers-their-opponent-will-show", "text": "In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show\n\n# In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show\n\nA\n45 points\n\nIn the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultaneously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a specified player, and denote by X the amount of money he wins in a single game of Two-Finger Morra.\n\n(a) If each player acts independently of the other, and if each player makes his choice of the number of fingers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the 4 possibilities is equally likely, what are the possible values of X and what are their associated probabilities?\n(b) Suppose that each player acts independently of the other. If each player decides to hold up the same number of fingers that he guesses his opponent will hold up, and if each player is equally likely to hold up 1 or 2 fingers, what are the possible values of X and their associated probabilities?\n\nIn the\n\n8.7k points\n\nFor this type of problem, it is helpful to construct a table,\n\nPlayer 1 Player 2 Random Variable\nGuess Show Guess Show X\n1 1 1 1 0\n1 1 1 2 -3\n1 1 2 1 2\n1 1 2 2 0\n1 2 1 1 3\n1 2 1 2 0\n1 2 2 1 0\n1 2 2 2 -4\n2 1 1 1 -2\n2 1 1 2 0\n2 1 2 1 0\n2 1 2 2 3\n2 2 1 1 0\n2 2 1 2 4\n2 2 2 1 -3\n2 2 2 2 0\n\npart (a)\n\nThere are 16 possibilities in the table above and so we have,\n\n$$P(X=0)=\\frac{8}{16}=\\frac{1}{2}$$\n\n$$P(X=2)=P(X=-2)=\\frac{1}{16}$$\n\n$$P(X=3)=P(X=-3)=\\frac{2}{16}=\\frac{1}{8}$$\n\n$$P(X=4)=P(X=-4)=\\frac{1}{16}$$\n\npart (b)\n\nfrom the table above the following set can only occur {1111,1122,2211,2222} all with output X=0, so, $P(X=0)=1$\n\nSurround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$\n\nUse LaTeX to type formulas and markdown to format text. See example.", "date": "2018-11-14 01:36:12", "meta": {"domain": "lil-help.com", "url": "https://www.lil-help.com/questions/7852/in-the-game-of-two-finger-morra-2-players-show-1-or-2-fingers-and-simultaneously-guess-the-number-of-fingers-their-opponent-will-show", "openwebmath_score": 0.7249987721443176, "openwebmath_perplexity": 255.606206917506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587261496031, "lm_q2_score": 0.8947894689081711, "lm_q1q2_score": 0.8838361059808151}}
{"url": "https://wild.maths.org/comment/1010", "text": "Pairwise Puzzler\n\nYou may wish to take a look at Pairwise Adding before trying this problem\n\nAlison reckons that if Charlie gives her the sums of all the pairs of numbers\u00a0from any set of five numbers, she'll be able to work out the original five numbers.\n\nCharlie gives her this set of pair sums: $0$, $2$, $4$, $4$, $6$, $8$, $9$, $11$, $13$, $15$.\n\nCan you work out Charlie's original five numbers?\n\nCan you work out a strategy to work out the original five numbers for any set of\u00a0pair sums that you are given?\n\nDoes it help to add together all the pair sums?\n\nGiven ten randomly generated numbers, will there always be a set of five numbers whose pair sums are\u00a0that set of ten?\n\nCan two different sets of five numbers give the same set of pair sums?\n\nFour numbers are added together in pairs to produce the following answers: $5$, $9$, $10$, $12$, $13$, $17$.\n\nWhat are the four numbers?\n\nIs there more than one possible solution?\n\nSix numbers are added together in pairs to produce the following answers: $7$, $10$, $13$, $13$, $15$, $16$, $18$, $19$, $21$, $21$, $24$, $24$, $27$, $30$, $32$.\n\nWhat are the six numbers?\n\nCan you devise a general strategy to work out a set of six numbers when you are given their pair sums?\n\nPairwise Puzzler - strategy for original five numbers\n\nThe five numbers can be calculated simply by using simultaneous equations.\n\nLet's call the lowest number a, the second lowest b etc.. and we'll place the sums from smallest to largest in series s.\nWe know straight away that a + b = s1, as the smallest sum must be made up of the smallest pair.\nAlso the largest sum is clear: d + e = s10\nWe can also work out that a + c = s2, as this is the second smallest sum; a and c are the smallest two numbers which haven't yet been paired.\nSimilarly, c + e = s9 (the second largest).\n\nSo we now have five unknowns and four equations... we just need one more equation.\nIf we look at the sum of all the sums in terms of the five original numbers ((a + b) + (a + c)...(e + f)),\nthis simplifies to 4 (a + b + c + d + e).\nSo therefore a + b + c + d + e = sum (s) / 4\nHere we have the final equation, and we can now calculate the five numbers through simple algebra.\n\nSo let's make each letter in turn the subject of an equation.\n\nWe can do this easily for c, using substitution: a + b = s1, d + e = s10, and a + b + c + d + e + f = sum (s) / 4;\ntherefore s1 + s10 + c = sum (s) / 4\nSo c = sum (s) / 4 - s1 - s10\n\nNow that we have c, all of the other letters can be expressed in terms of c:\na = s2 - c\nb = s1 - (s2 - c) = s1 - s2 + c\ne = s9 - c\nd = s10 - (s9 - c) = s10 - s9 + c\n\nThese equations can all be written, by substituting c, in terms of the sums:\na = s1 + s2 + s10 - sum (s) / 4\nb = sum (s) / 4 - s2 - s10\nc = sum (s) / 4 - s1 - s10\nd = sum (s) / 4 - s1 - s9\ne = s1 + s9 + s10 - sum (s) / 4\n\nSo to find the original five numbers given only the sum of each pair, just arrange the sums from smallest to largest and substitute them into the equations above.\n\nThis is an excellent solution\n\nThis is an excellent solution, well done! I particularly like the way that you simplified the problem by writing the numbers as $a \\leq b \\leq c \\leq d \\leq e$, and the way that adding all the numbers together allows you to say for certain what the algebraic sum is.\n\nPairwise puzzlers\n\nI thought this puzzle was set in a misleading way - at first sight it appears it only involves positive integers, whereas the solution includes a negative integer\n\n(-1,1,3,5,10)\n\nI suppose it does teach you to assume nothing that is not explicitly stated!\n\nMath\n\nThe answer to the first question is 0,2,6,7,9!!!!!\n\nMath\n\nJulie, I can choose pairs of numbers from your set of five and add them to get 2, 6, 8, 9, 11, 13 and 15, but I can't get 0, 4 and 4.\n\nDo have another go.\n\n0, 2, 4, 6, 9\nor\n-1, 1, 3, 5, 10\n\nOne of your sets of five numbers can be used to produce this set of pair sums 0, 2, 4, 4, 6, 8, 9, 11, 13, 15, but I'm afraid the other one can't.\n\nCan you figure out which one is which?\n\nA somewhat inelegant answer to the 6-number problem\n\nHaving tackled the 4- and 5-number versions of this problem, I couldn't see an obvious way to generate the 6th equation which would allow direct solution via simultaneous equations. Five equations are simple enough to find, as Mr Redman showed very clearly in his comment, but without the sixth (which I hope someone will enlighten me about at some point), the system would be under-determined. So, without an elegant solution to hand, I tried an inelegant one instead...\n\nAbout 10 lines of matlab code allowed me to generate random collections of six integers (within a reasonable range to keep things somewhat manageable), generate all 15 of their pairwise combinations and sums, and test the resulting sums against the given sums. About 25 seconds of computer time gave me the following answer (which I have checked by hand just to be sure): 2 5 8 11 13 19\n\nI know that this technique will scale very poorly as the size of problem increases - I certainly wouldn't want to attempt it for pairwise combinations of 10 numbers, for example - and would welcome any suggestions regarding a more subtle approach, but feel that a computational approach does have a place even on this website!\n\nA somewhat inelegant answer to the 6-number problem\n\nThose six numbers do indeed produce the pair sums 7, 10, 13, 13, 15, 16, 18, 19, 21, 21, 24, 24, 27, 30 and 32. Very well done! Could you explain in more detail the method that you programmed Matlab to use?\n\nAs for a more elegant solution, it is indeed difficult to find a 6th equation. In some of the cases where some of the pair sums are equal, you can deduce a 6th equation. For example if $s_3 = s_4$, then you know the value of $a+d$. Would considering the possible positions of the some of the pair sums (such as $a+d$) help you to deduce a more general approach for all cases?", "date": "2022-12-10 06:06:29", "meta": {"domain": "maths.org", "url": "https://wild.maths.org/comment/1010", "openwebmath_score": 0.6365929245948792, "openwebmath_perplexity": 344.7798361841747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474879039229, "lm_q2_score": 0.8933094110250331, "lm_q1q2_score": 0.8837934217185495}}
{"url": "https://math.stackexchange.com/questions/1056982/probability-a-is-before-b-when-the-26-letters-are-arranged-randomly", "text": "# Probability $A$ is before $B$ when the 26 letters are arranged randomly\n\nThe 26 letters A, B, ... , Z are arrange in a random order. [Equivalently, the letters are selected sequentially at random without replacement.]\n\na) What is the probability that A comes before B in the random order?\n\nb) What is the probability that A comes before Z in the random order?\n\nc) What is the probability that A comes just before B in the random order?\n\nAny help would be much appreciated. I was thinking that for part c the answer would be $$1/26$$ because we have $$25!$$ ways of having A right before B and $$26!$$ total arrangements. Not sure how to proceed with a and b, however.\n\nEdit:\n\nThank you. For parts (a) and (b) is there a more formal way of getting $$1/2$$? Such as the formula for the total number of ways we can have $$A$$ before $$B$$ over the total number of arrangements? Would it be 25 choose 1 ... 2 choose 1 over $$26!$$ since we can have a in the first spot and B in any spot after it? Then we can have a in the 2nd spot and B in any spot after it. Also, for part (c), doesn't $$A$$ have to come immediately before $$B$$, so wouldn't the probability be $$1/26$$?\n\n\u2022 a) 1/2 b) 1/2 . Only 2 possibilities A before or after B \u2013\u00a0mridul Dec 8 '14 at 6:43\n\u2022 MAy someone shed some light on the correct answer in part c? Thanks! \u2013\u00a0user198454 Dec 8 '14 at 16:54\n\u2022 Is c asking if A come right before B or that A comes right before B but not Z? \u2013\u00a0Kamster Dec 29 '14 at 5:24\n\u2022 The reasoning for part (a) in chandu1729's excellent answer is quite formal actually. Let me know if you need an explanation of how to make that formal. \u2013\u00a06005 Dec 30 '14 at 7:02\n\u2022 Yeah! Just a bounty \u2014 and here we are, a bunch of totally equal answers. \u2013\u00a0sas Dec 30 '14 at 8:15\n\na) From symmetry, the probability that $A$ comes before $B$ is same as the probability that $B$ comes before $A$, and the sum of these $2$ probabilities is $1$. Hence, the probability that $A$ comes before $B$ is $0.5$.\n\nb) Here $B$ is replaced with $Z$ and by exchanging the roles of $B$ and $Z$, we get the same probability as in case (a): $0.5$.\n\nc) If $A$ has to come just before $B$, $B$ shouldn't occur at the $1$st position, which has probability of $25/26$. After $B$'s position has been chosen(other than the $1$st position), there are 25 possibilities for position just before $B$ (everything except $B$) and only $1$ possibility is favourable (A coming before B). Hence the required probability is $(25/26)\\cdot(1/25) = 1/26$.\n\nIn part c), $A$ comes right before $B.$ Consider a block $\\boxed{\\text{AB}}$ and treat it as ONE object. The other letters are $C, D, E, \\ldots , Z$. In total, you have 25 objects - 24 letters and a block of two letters together. They can be arranged in $25!$ ways in total. Since the total number of arrangements is $26!$, the required probability is $$\\dfrac{25!}{26!}=\\dfrac1{26}$$\n\nNow, let us consider part a). We'll formally prove the probability is $\\dfrac12$\n\nFor this, we will consider 26 cases.\n\nCase 1 = B is in the first position.\n\nClearly, if $B$ is in the first position, then in no possible arrangement, A comes before B. Thus, the number of way A comes before B is $0$ for the first case.\n\nCase 2 = B is in the second position.\n\nFor A to come before B, it has to occupy the first spot. A can occupy the first spot in $1$ way. The rest 24 letters can be arranged in $24!$ ways. The number of ways A comes before B can happen is $1.24!$\n\nCase 3 = B is in the third position.\n\nFor A to come before B, A can occupy the first two positions. The number of ways A can take the first two position is $2.$ The rest 24 letter can be again arranged in $24!$ ways. Hence, the number of ways A comes before B happens is $2.24!$\n\nCase 4 = B is in the fourth position.\n\nNow, by same logic as before, you can say it will happen in $3.24!$ ways.\n\nSimilarly, we go up to Case 26 where B comes in the $26^{th}$ i.e, last position.\n\nThese $26$ cases exhausts all the possible arrangements. By considering all the $26$ cases and summing up the total number of ways A comes before B can happen, we get\n\n$$0+1.24!+2.24!+\\cdots 25.24! = 24!(1+2+\\cdots 25)=24!*\\dfrac{25.26}{2}=\\dfrac{26!}{2}$$\n\nSince the total number of arrangements is $26!$, the probability for A comes before B in a random order is $$\\dfrac{26!/2}{26!}=\\dfrac{1}{2}$$\n\n\u2022 Right answer but...way too many words for part (a) :P \u2013\u00a06005 Dec 30 '14 at 7:01\n\u2022 Thanks. Right, it bothered me too! I could've provided the same thing in a pictorial proof, which would have been way easier to comprehend, but IDK much of Geogebra. :/ \u2013\u00a0Dipanjan Pal Dec 30 '14 at 8:17\n\nHere's a proof that the probability in part (a) is 1/2. It's the same as the one in The Math Troll's and chandu1729's answers, but with more details filled in. I want to emphasize that it really is a proof, as rigorous as anything you're likely to encounter in an introductory probability course, even though it doesn't use any formulas or complicated calculations.\n\nLet's say two arrangements are partners if exchanging the positions of A and B turns one into the other. Here are some examples of arrangements which are partners:\n\nABCDEFGHIJKLMNOPQRSTUVWXYZ and BACDEFGHIJKLMNOPQRSTUVWXYZ AEIOUYBCDFGHJKLMNPQRSTVWXZ and BEIOUYACDFGHJKLMNPQRSTVWXZ THEQUICKBROWNFXJMPSVLAZYDG and THEQUICKAROWNFXJMPSVLBZYDG\n\nYou should be able to convince yourself that:\n\n\u2022 Every arrangement has exactly one partner.\n\u2022 If an arrangement has A before B, its partner has B before A, and vice versa.\n\nNow, line up all the arrangements in two lines, with each arrangement standing next to its partner. The lines with be the same length, because each arrangement has exactly one partner.\n\nIn each pair of partners, tell the arrangement with A before B to stand on the left, and the arrangement with B before A to stand on the right. Now, the left line has all the arrangements with A before B, and the right line has all the arrangements with B before A. Since the lines are the same length, it's now obvious that the numbers of A-before-B arrangements and B-before-A arrangements are equal!\n\nThat means exactly half the arrangements have A before B, and exactly half have B before A. Therefore, if you pick an arrangement at random, the probability of getting A before B is 1/2.\n\nYou can use exactly the same argument for part (b).\n\nThe simplest way to answer these questions:\n\na) What is the probability that A comes before B in the random order?\n\nIn half of the cases A comes before B and in the other half B comes before A, so the answer is $\\frac12$\n\nb) What is the probability that A comes before Z in the random order?\n\nIn half of the cases A comes before Z and in the other half Z comes before A, so the answer is $\\frac12$\n\nc) What is the probability that A comes just before B in the random order?\n\n\u2022 The total number of permutations is $26!$\n\n\u2022 Join the letters A and B into a symbol AB, and the total number of permutations is $25!$\n\n\u2022 Hence the answer is $\\frac{25!}{26!}=\\frac{1}{26}$\n\nA more \"formal\" way to answer to the first (as well as the second) question:\n\n\u2022 The probability that A is at the 1st place and B is at the 2nd to 26th places is $\\frac{1}{26}\\cdot\\frac{25}{25}=\\frac{25}{26\\cdot25}$\n\n\u2022 The probability that A is at the 2nd place and B is at the 3rd to 26th places is $\\frac{1}{26}\\cdot\\frac{24}{25}=\\frac{24}{26\\cdot25}$\n\n\u2022 The probability that A is at the 3rd place and B is at the 4th to 26th places is $\\frac{1}{26}\\cdot\\frac{23}{25}=\\frac{23}{26\\cdot25}$\n\n\u2022 $\\dots$\n\n\u2022 The probability that A is at the 25th place and B is at the 26th place is $\\frac{1}{26}\\cdot\\frac{1}{25}=\\frac{1}{26\\cdot25}$\n\n\u2022 So the probability that A comes before B is $\\sum\\limits_{n=1}^{25}\\frac{26-n}{26\\cdot25}=\\frac12$\n\na) By symmetry, there are as many permutations with $A$ before $B$ as with $A$ after $B$.\n\n$$p=\\frac12$$\n\nb) By symmetry, there are as many permutations with $A$ before $Z$ as with $A$ after $Z$.\n\n$$p=\\frac12$$\n\nc) Consider the $24!$ permutations of $CDE\\dots Z$. You can insert $AB$ at $25$ different places to form all distinct configurations. $$p=\\frac{25\\cdot24!}{26!}=\\frac1{26}$$\n\nfor parts a and b then the answer is $1/2$. One way to look at is is to note that for any arrangement of the letters, say \"DEFGBAC\", there is another arrangment the order of the letters reversed, in this case \"CABGFED\". It is clear that if $A$ comes before $B$ in the first, then it will come after the $B$ in the second, and vice versa. So we can take all the random arrangements and arrange them in to pairs in this manner. Since each arrangement with $A$ before $B$ is paired with another unique arrangment with $B$ before $A$, there must be the same number of strings with $A$ before $B$ and $B$ before $A$. So selecting one of the possible arrangements at random has a probability of $1/2$ of having the $A$ before the $B$.\n\nFor part c, There are 26 postions where the $A$ can go, and, since that position is filled, 25 positions when the $B$ can go. Now for 25 of the 26 positions where the $A$ can go, the position immediately following it is vacant. In this case there is a change of $1/25$ that the $B$ randomly falls into that position. So in $25/26$ of the cases (where $A$ is not in the last position), there is a chance of $1/25$ of the $B$ immediately following the $A$.\n\nHowever, in 1/26 of the cases, the $A$ is in the last place, and the $B$ cannot follow the $A$, so the probability is $0$.\n\nSo from the first situation we have $25/26 \\times 1/25 = 1/26$, and from the second $1/26\\times0 = 0$\n\nAdding up the two, $1/26 + 0 = 1/26$. So you appear to be correct that the probability is $1/26$.\n\nFor parts a and b, we can construct a bijection between A before B and B before A (proceed similarly for part b).\n\nSuppose we have one that is A before B. Then we switch the A and B then we get a B before A. The other direction of the bijection will be similar.\n\nNow for part c. If A is the last, then there is 0 chance. This happens 1/26 of the time. If A isn't the last, then there will be one immediately after it. There will be 25 possibilities (and they are random), so the chance is 1/25. This happens 25/26 of the time Hence 1/25*25/26+0=1/26.\n\n\u2022 Your answer for part C is wrong! it is 1/52 \u2013\u00a0mridul Dec 8 '14 at 10:21\n\u2022 Then what is the problebility that A and B together? 1/13? Please use good words. \u2013\u00a0mridul Dec 8 '14 at 10:24\n\u2022 Note that there are 26*25 positions for A and B. For A to come before B, there are 25 positions. Thus the probability is 1/26. Please check your solution first. \u2013\u00a0user198454 Dec 8 '14 at 10:25\n\nTotal 26! rearrangements are possible. In 1st case A has only 2 relative possitions after or before B. Therefore probability is 1/2.\n\nIn 3rd case : Total possible arrangements such that A and B are together is 25!. Therefore Total possible arrangements such that A comes just before B is 25! / 2.\n\na) 1/2\n\nb) 1/2\n\nc) 1/52\n\nEdit:\n\nFor parts (a) and (b),\n\nLet A at 1st possition, total number of arrangements(A before B) is 25!.\n\nLet A at 26th possition, total number of arrangements is 0. together we got 25!.\n\nThen take 2nd and 25th possitions.\n\nLet A at 2st possition, total number of arrangements(A before B) is 24*24!. (B has 24 possitions such that B after A)\n\nLet A at 25st possition, total number of arrangements(A before B) is 1*24!. (B has only 1 possition such that B after A)\n\n24*24! + 1*24! = 25*24! =25!\n\nThen take 3rd and 24th possitions.\n\nLet A at 3st possition, total number of arrangements(A before B) is 23*24!.\n\nLet A at 24st possition, total number of arrangements(A before B) is 2*24!.\n\nTogether we got 25!.\n\netc .. upto 13th,14th possitions.\n\nAdd all together we got 13*25! arrangements.\n\n\u2022 Thank you. For parts (a) and (b) is there a more formal way of getting $1/2$? Such as the formula for the total number of ways we can have $A$ before $B$ over the total number of arrangements $(26!)$? Also, for part (c), doesn't $A$ have to come immediately before $B$, so wouldn't the probability be $1/26$? \u2013\u00a0mylasthope Dec 8 '14 at 7:08\n\u2022 In the case of part c answer is 1/52. For parts (a) and (b), let A at 1st possition, total number of arrangements(A before B) is 25!. let A at 26th possition, total number of arrangements is 0. together we got 25!. Then take 2nd and 25th possitions together we got 25!. etc. Add all together we got 13*25! arrangements. \u2013\u00a0mridul Dec 8 '14 at 7:43", "date": "2021-04-13 20:54:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1056982/probability-a-is-before-b-when-the-26-letters-are-arranged-randomly", "openwebmath_score": 0.8802695870399475, "openwebmath_perplexity": 300.8157627126416, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474888461861, "lm_q2_score": 0.8933094046341532, "lm_q1q2_score": 0.883793416237481}}
{"url": "https://stats.stackexchange.com/questions/50274/probability-of-a-pair-in-cards", "text": "# probability of a pair in cards\n\nI have a simple problem involving probability of drawing at least 1 pair of cards in a four card hand. I am not getting the right answer but I dont understand the flaw in my logic. Can anyone explain to me why my approach is wrong?\n\nThe problem:\n\nBill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?\n\nMy solution:\n\nProbability of 1 or more pair = ((6) * (10 choose 2))/(12 choose 4). i.e. the # of hands with at least 1 pair are 6 (the number of ways to create a pair) * (10 choose 2) (the # of ways to select 2 cards from the remaining 10 cards after the pair). This simplifies to 6/11, however, the correct answer is 17/33.\n\nAny help understanding would be greatly appreciated.\n\n\u2022 You make some interesting implicit assumptions in your calculation. First, it focuses only on the possibility that the pair occurs among the first two within a sequence of four cards: there are other ways a pair can occur, so this underestimates the probability. Second, it ignores the sequence among the second two cards, thereby overestimating the probability. Fortunately, the two mistakes do not cancel, revealing the fact there is a problem! \u2013\u00a0whuber Feb 18 '13 at 22:17\n\u2022 Realized I was double counting. Cant simply multiply 6 * (10 choose 2). 2 pair instances will be double counted. e.g. pair of 1s in the 6 and then pair of 2s in the 10 choose 2 versus pair of 2s in the 6 and then pair of 1s in the 10 choose 2. The better way to do it is to separate the cases of exactly 1 pair, exactly 2 pair. ways to create exactly 1 pair = 6 (ways to create pair) * ( (10*8)/2 ) + ways to create exactly 2 pair = (6*5)/2 Total # = 255 Denominator = (12 choose 4) = (12*11*10*9)/(4*3*2) = 11*5*9 = 455. Probability = 255/455 = 51/99 = 17/33. \u2013\u00a0Evan V Feb 18 '13 at 22:24\n\u2022 Another way to do this is simply to remove the double counting directly. There are 6*(10 choose 2) = 270 ways to create 1 or 2 pairs where the instances of 2 pairs are double counted. If you subtract the # of ways to make exactly 2 pairs (6*5)/2 then you get 270 - 15 = 255. \u2013\u00a0Evan V Feb 18 '13 at 22:28\n\nTo see 0 pairs there are 12 possibilities for the 1st card, but only 10 for the second card (since the 1st chosen card is no longer possible and the card that matches the 1st would form a pair) and 8 cards for the 3rd and 6 for the 4th, this multiplies to 5760. The total number of possible hands (with order mattering) is $12 \\times 11 \\times 10 \\times 9 = 11880$, so the complimentary number which is the number of (ordered) hands that contain at least 1 pair is $11880 - 5760 = 6120$, divide that by 11880 and it reduces to $17/33$. We could also do this with order not mattering, but that would be more complicated and the extra pieces would all end up cancelling each other.", "date": "2021-03-07 22:06:27", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/50274/probability-of-a-pair-in-cards", "openwebmath_score": 0.6800057291984558, "openwebmath_perplexity": 241.87707813867704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846684978779, "lm_q2_score": 0.9032942041005327, "lm_q1q2_score": 0.8837692004349541}}
{"url": "https://www.physicsforums.com/threads/questions-about-linear-transformations.887744/", "text": "# I Questions about linear transformations\n\n1. Oct 3, 2016\n\n### Chan Pok Fung\n\nWe learnt that the condition of a linear transformation is\n1. T(v+w) = T(v)+T(w)\n2. T(kv) = kT(v)\n\nI am wondering if there is any transformation which only fulfil either one and fails another condition. As obviously, 1 implies 2 for rational number k.\n\nCould anyone give an example of each case? (Fulfilling 1 but 2 and 2 but 1)\n\nThanks!\n\n2. Oct 3, 2016\n\n3. Oct 3, 2016\n\n### andrewkirk\n\nConsider the real numbers $\\mathbb R$ as a vector space over the rationals, and the operator $T:\\mathbb R\\to \\mathbb R$ that is the identity on the rationals and maps to zero on the irrationals. Then T satisfies the second axiom, since it is a linear operator on $\\mathbb Q$ considered as a subspace of $\\mathbb R$ and, for $q\\in\\mathbb Q-\\{0\\},x\\in\\mathbb R$, $qx$ is in $\\mathbb Q\\cup\\{0\\}=\\ker\\ T$ iff $x$ is.\n\nBut the addition axiom does not hold, because $T(1+(\\sqrt2-1))=T(\\sqrt 2)=0$ but $T(1)+T(\\sqrt2-1)=1\\neq 0$.\n\n4. Oct 3, 2016\n\n### Stephen Tashi\n\nDefine $M((x,y))$ by\nif $x \\ne y$ then $M((x,y)) = (x,y)$\nif $x = y$ then $M((x,y)) = (2x, 2y)$\n$M$ satisfies 2, but not 1\n\n5. Oct 4, 2016\n\n### Chan Pok Fung\n\nThis really gives me new insight into linear transformation. thanks all!\n\n6. Oct 4, 2016\n\n### Chan Pok Fung\n\nI am sorry but I don't quite understand. How can we construct a transformation like that?\n\n7. Oct 4, 2016\n\n### mathman\n\nUnfortunately Hamel basis exists, but it is not constructable - existence is equivalent to axiom of choice.\n\n8. Oct 4, 2016\n\n### andrewkirk\n\nI'm still trying to think of a scenario with a map that satisfies 1 (additivity) but not 2 (scalar mult). Can anybody think of one?\n\nAll the examples I come up with either end up satisfying neither 1 nor 2, or satisfying 2 but not 1.\n\nI assume there must be one, otherwise some texts would specify 1 as the sole requirement and derive 2 as a consequence of 1.\n\n9. Oct 4, 2016\n\n### Chan Pok Fung\n\nAndrew, I am thinking that as we have to apply the transformation to a vector space, and vectors in vector space obeys kv is also in the space. As we can have k be any real number, it seems that it somehow implies axiom2. The transformation only satisfy axiom1 must be of a very weird form.\n\n10. Oct 4, 2016\n\n### Stephen Tashi\n\nOn the \"talk\" page for the current Wikipedia article on \"Linear transformation\", I found:\n\n11. Oct 4, 2016\n\n### Chan Pok Fung\n\nThat's a clear and direct example!", "date": "2017-11-21 14:27:29", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/questions-about-linear-transformations.887744/", "openwebmath_score": 0.8395922780036926, "openwebmath_perplexity": 1244.7749528275344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846634557752, "lm_q2_score": 0.9032942086563877, "lm_q1q2_score": 0.8837692003378307}}
{"url": "https://math.stackexchange.com/questions/2668314/multiplication-of-moduli-in-modular-congruences", "text": "# Multiplication of Moduli in Modular Congruences\n\nLately it's come up in my discrete mathematics class that proving things for smaller congruences, namely those where the modulus is a prime is much easier than attempting to do so for larger congruences.\n\nFor example, one such problem was to prove that $a^5 \\equiv a \\pmod{10}$ for $a$ $\\varepsilon$ $\\mathbb Z^+\\$This problem on its own becomes difficult only because we are not guaranteed that for every a $(a,10)=1$. The solution was to factorize 10 in terms of primes $p_1, p_2... p_n$ s.t for each prime $p_i$ $(a,p_i)=1$ for all $a \\ \\varepsilon \\ \\mathbb Z^+\\$, resulting in n congruences under modulo the given prime and then apply Euler's Theorem. In this case we find the following: $$10 = 2 \\cdot 5$$ so $$a^{\\phi(5)} \\equiv 1 \\pmod{5} \\\\ a^{\\phi(2)} \\equiv 1 \\pmod{2}$$ since $\\phi(5)=4, \\phi(2)=1$ this becomes: $$a^{4} \\equiv 1 \\pmod{5} \\implies a^{5} \\equiv a \\pmod{5}\\\\ a^{1} \\equiv 1 \\pmod{2} \\implies a^{5} \\equiv a^4 \\pmod{2}$$ but since a is its own inverse modulo 2, we can transform the 2nd congruence into: $$a^5 \\equiv a \\mod{2}$$ Then the part that I am unclear on occurs. It seems that we can just multiply the 2 moduli together and the desired congruence falls out that: $$a^5 \\equiv a \\pmod{10}$$\n\nSo my question is whether or not $a \\equiv b \\pmod{n}$ and $a \\equiv b \\pmod{m}$ always$\\implies$ $a \\equiv b \\pmod{mn}$.\n\nEdit (Extension):\n\nThe answer to my question has been stated to be yes provided that $(m,n) = 1$ but I also noticed that if my 2 relations had been left stated as $$a^4 \\equiv 1 \\pmod{2}$$ $$a^4 \\equiv 1 \\pmod{5}$$ Applying the conjecture I made gives: $$a^4 \\equiv 1 \\pmod{10}$$ which is demonstrably false given that $$2^4 \\equiv 6 \\pmod{10} \\implies 2^4 \\not\\equiv 1 \\pmod{10}$$ Why is this and what prevents this identity from being true as well? I see that 2 would then not be coprime to 10 but why does it work when I multiply both sides by a then?\n\nExcellent question!\n\nThe short answer is no.\n\nFor example, $4 \\equiv 16 \\pmod 6$ and $4 \\equiv 16 \\pmod 4$, but $4 \\not \\equiv 16 \\pmod{24}$.\n\nHowever if $n$ and $m$ are relatively prime, then the answer is yes.\n\nThis is pretty straightforward to see, as if $a \\equiv b \\pmod n$ then $n \\mid (b-a)$, and if $a \\equiv b \\pmod m$, then $m \\mid (b-a)$. Then one notes (or proves, if it's not clear) that $n \\mid (b-a)$, $m \\mid (b-a)$, and $\\gcd(m,n) = 1$ implies that $mn \\mid (b-a)$.\n\nIn fact, this is at the edge of a deeper theorem called the Chinese Remainder Theorem, which says roughly that knowing the structure of $x$ mod $n$ and $m$ for $m,n$ relatively prime is equivalent to knowing the structure of $x$ mod $mn$ --- or perhaps with two or three or more moduli all taken together. Look up the Chinese Remainder Theorem on the web and on this site for more.\n\n\u2022 If I might add to this question a bit. This works when i have the $a^5$ and a for a and b but if i take b = 1 it clearly doesn't work since $2^4 \\equiv 6 \\pmod{10}$ but $2^5 \\equiv 2 \\pmod{10}$ which makes sense since with a not coprime to 10 when a=2 it wouldn't be invertible mod 10. How is one to catch this beforehand though and what makes it work when it's multiplied on both sides by a? \u2013\u00a0rjm27trekkie Feb 27 '18 at 1:16\n\nYou require $\\gcd(m,n)=1$, otherwise \\begin{eqnarray*} 1 \\equiv 13 \\pmod{4} \\\\ 1 \\equiv 13 \\pmod{6} \\\\ \\end{eqnarray*} but \\begin{eqnarray*} 1 \\neq 13 \\pmod{24}. \\\\ \\end{eqnarray*}\n\n\u2022 Understood. Otherwise, is the conjecture correct given that (m,n) = 1? \u2013\u00a0rjm27trekkie Feb 27 '18 at 1:05\n\u2022 That's right ... \u2013\u00a0Donald Splutterwit Feb 27 '18 at 1:09\n\nThis is true if (and only if) $m$ and $n$ are coprime \u2013 which is the case ot two distinct primes.\n\nThis is the statement of the Chinese remainder theorem: if $m, n$ are coprime integers, the map \\begin{align} \\mathbf Z/mn\\mathbf Z&\\longrightarrow\\mathbf Z/m\\mathbf Z\\times \\mathbf Z/n\\mathbf Z\\\\ a\\bmod mn&\\longmapsto(a\\bmod m,a\\bmod n) \\end{align} is an isomorphism.\n\n$x \\equiv a \\mod n$ means $x=a+kn$ for some $k$. Now $k=qm+r$ for some value of $q$ and $0\\le r <m$. So $x=a +rn + q*mn$ so $x=a+rn\\mod mn$. However we have no idea yet what $r$ is. we just know it is between $0$(inclusively) and $m$(exclusively).\n\nLikewise if $x\\equiv a\\mod m$ by the same reasoning we know $x\\equiv a+sm\\mod mn$. We don't know what $s$ is yet.\n\nBut $a+sm <mn$ and $a+rn <mn$ and $a+sm\\equiv a+rn\\mod mn$ so $sm=rn$.\n\nnow if $m$ and $n$ are not relatively prime there might be non trivial solutions. For example maybe $r=\\frac m {\\gcd (m,n)}$ and $s=\\frac n {\\gcd (m,n)}$\n\nBut if $m$ and $n$ are relatively prime, then $sm=rn$ means $m|r$ and $n|s$. But $r <m$ and $s <n$ and the only way that can happen is if $sm=rn=0$", "date": "2019-09-16 20:46:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2668314/multiplication-of-moduli-in-modular-congruences", "openwebmath_score": 0.9810587167739868, "openwebmath_perplexity": 147.5086358032251, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683513421314, "lm_q2_score": 0.8947894689081711, "lm_q1q2_score": 0.8836657606079439}}
{"url": "https://mathhelpboards.com/threads/combinatorics-challenge.6224/", "text": "# Combinatorics Challenge\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nFor $n=1,2,...,$ set $$\\displaystyle S_n=\\sum_{k=0}^{3n} {3n\\choose k}$$ and $$\\displaystyle T_n=\\sum_{k=0}^{n} {3n\\choose 3k}$$.\n\nProve that $|S_n-3T_n|=2$.\n\n#### MarkFL\n\nStaff member\nMy solution:\n\nFor the first sum, we may simply apply the binomial theorem to obtain the closed form:\n\n$$\\displaystyle S_n=\\sum_{k=0}^{3n}{3n \\choose k}=(1+1)^{3n}=8^n$$\n\nFor the second sum, I looked at the first 5 values:\n\n$$\\displaystyle T_1=2,\\,T_2=22,\\,T_3=170,\\,T_4=1366,\\,T_5=10922$$\n\nand determined the recursion:\n\n$$\\displaystyle T_{n+1}=7T_{n}+8T_{n-1}$$\n\nThe characteristic equation for this recursion is:\n\n$$\\displaystyle r^2-7r-8=(r+1)(r-8)=0$$\n\nand so the closed form is:\n\n$$\\displaystyle T_{n}=k_1(-1)^n+k_28^n$$\n\nUsing the initial values to determine the parameters, we may write:\n\n$$\\displaystyle T_{1}=-k_1+8k_2=2$$\n\n$$\\displaystyle T_{2}=k_1+64k_2=22$$\n\nAdding the two equations, we find:\n\n$$\\displaystyle 72k_2=24\\implies k_2=\\frac{1}{3}\\implies k_1=\\frac{2}{3}$$\n\nHence:\n\n$$\\displaystyle T_{n}=\\frac{1}{3}\\left(2(-1)^n+8^n \\right)$$\n\nAnd so we find:\n\n$$\\displaystyle \\left|S_n-3T_n \\right|=\\left|8^n-3\\left(\\frac{1}{3}\\left(2(-1)^n+8^n \\right) \\right) \\right|=\\left|8^n-2(-1)^n-8^n \\right|=\\left|2(-1)^{n+1} \\right|=2$$\n\nShown as desired.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nMy solution:\n\nFor the first sum, we may simply apply the binomial theorem to obtain the closed form:\n\n$$\\displaystyle S_n=\\sum_{k=0}^{3n}{3n \\choose k}=(1+1)^{3n}=8^n$$\n\nFor the second sum, I looked at the first 5 values:\n\n$$\\displaystyle T_1=2,\\,T_2=22,\\,T_3=170,\\,T_4=1366,\\,T_5=10922$$\n\nand determined the recursion:\n\n$$\\displaystyle T_{n+1}=7T_{n}+8T_{n-1}$$\n\nThe characteristic equation for this recursion is:\n\n$$\\displaystyle r^2-7r-8=(r+1)(r-8)=0$$\n\nand so the closed form is:\n\n$$\\displaystyle T_{n}=k_1(-1)^n+k_28^n$$\n\nUsing the initial values to determine the parameters, we may write:\n\n$$\\displaystyle T_{1}=-k_1+8k_2=2$$\n\n$$\\displaystyle T_{2}=k_1+64k_2=22$$\n\nAdding the two equations, we find:\n\n$$\\displaystyle 72k_2=24\\implies k_2=\\frac{1}{3}\\implies k_1=\\frac{2}{3}$$\n\nHence:\n\n$$\\displaystyle T_{n}=\\frac{1}{3}\\left(2(-1)^n+8^n \\right)$$\n\nAnd so we find:\n\n$$\\displaystyle \\left|S_n-3T_n \\right|=\\left|8^n-3\\left(\\frac{1}{3}\\left(2(-1)^n+8^n \\right) \\right) \\right|=\\left|8^n-2(-1)^n-8^n \\right|=\\left|2(-1)^{n+1} \\right|=2$$\n\nShown as desired.\nHey thanks for participating MarkFL! And I'm so impressed that you were so fast in cracking this problem!\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nFor $n=1,2,...,$ set $$\\displaystyle S_n=\\sum_{k=0}^{3n} {3n\\choose k}$$ and $$\\displaystyle T_n=\\sum_{k=0}^{n} {3n\\choose 3k}$$.\n\nProve that $|S_n-3T_n|=2$.\nDefine $f(x)=\\sum_{k=0}^{3n}{3n\\choose k}x^k$.\n\nThen $f(1)+f(\\omega)+f(\\omega^2)=3T_n$, where $\\omega$ is a complex cube root of unity. Note that $f(1)=S_n$. So we get $S_n-3T_n=-[(1+\\omega)^{3n}+(1+\\omega^2)^{3n}]=-2$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nDefine $f(x)=\\sum_{k=0}^{3n}{3n\\choose k}x^k$.\n\nThen $f(1)+f(\\omega)+f(\\omega^2)=3T_n$, where $\\omega$ is a complex cube root of unity. Note that $f(1)=S_n$. So we get $S_n-3T_n=-[(1+\\omega)^{3n}+(1+\\omega^2)^{3n}]=-2$\nHi caffeinemachine,\n\nThanks for participating and I really appreciate you adding another good solution to this problem and my thought to solve this problem revolved around the idea that you used too!", "date": "2021-10-16 15:16:12", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/combinatorics-challenge.6224/", "openwebmath_score": 0.9442697167396545, "openwebmath_perplexity": 1185.4687731455952, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9925393565360598, "lm_q2_score": 0.8902942239389252, "lm_q1q2_score": 0.8836520561561116}}
{"url": "https://math.stackexchange.com/questions/2890157/what-is-the-difference-between-lim-h-to0-frac0h-and-lim-h-to-infty", "text": "What is the difference between $\\lim_{h\\to0}\\frac{0}{h}$ and $\\lim_{h\\to\\infty}\\frac{0}{h}$?\n\nWhat is the difference (if any) between-\n\n$$\\lim_{h\\to0}\\frac{0}{h} \\text{ and } \\lim_{h\\to\\infty}\\frac{0}{h}$$\n\nI argue that both must be $=0$ since the numerator is exactly $0$. But one fellow refuses to agree and argues that the first limit can't be $0$ as anything finite by something tending to $0$ is always $\\infty$.\n\nSo,how can I explain it to the person? Also, if possible can anyone provide some good reference on this particular issue (Apostol perhaps)?\n\nThanks for any help!\n\n\u2022 Graph the function $f(x) = \\dfrac{0}{x}$. Perhaps by seeing it, it will make more sense to the 'fellow'. Explain the geometric interpretation of a limit. \u2013\u00a0InterstellarProbe Aug 21 '18 at 17:27\n\u2022 \"Anything finite by something tending to 0 is always $\\infty$\" is not true (at least as long as $0$ counts as finite) -- as this very example shows. Case closed. \u2013\u00a0Henning Makholm Aug 21 '18 at 17:41\n\u2022 @tatan: If they don't want to consider your arguments, you can't. Give it up; life is too short for some things. \u2013\u00a0Henning Makholm Aug 21 '18 at 17:43\n\u2022 Let \"the fellow\" read this. \u2013\u00a0drhab Aug 21 '18 at 18:09\n\u2022 It's not that the numerator is zero, it's that the fraction is zero. Does your \"fellow\" agree that $\\lim_{h\\to0}0=0$. If he does, but doesn't then agree that $\\lim_{h\\to0}0/h=0$, then there is little hope for him. \u2013\u00a0Lord Shark the Unknown Aug 21 '18 at 18:22\n\nIn both cases you are dividing zero by a nonzero number .\n\nThus your fraction is identically zero and as a result the limit is zero.\n\nFor the first one let me make an intuitive explanation of what's going on with $\\lim_{h\\to 0}\\frac{0}{h}$.\n\nThe above limit means \"What value does the above expression get while $h$ gets arbitrarily close to $0$\", meaning that $h$ is number very close to $0$ but not equal to that. Suppose $h=0.000001$. Then $\\frac{0}{h}=\\frac{0}{0.000001}=0$. Even if this number gets even closer to $0$ you see that the expression continuous to be equal to $0$. That's because $0$ divided by any number obsiously gives $0$ as the answer. Hence $\\lim_{h\\to 0}\\frac{0}{h}=0$.\n\nI wish I helped!\n\nWhen in doubt take it back to the definition.\n\n$\\lim_\\limits{x\\to 0} f(x) = 0$ means\n\n$\\forall \\epsilon >0, \\exists \\delta >0 : 0<|x|<\\delta \\implies |f(x)|<\\epsilon$\n\nAs $f(x) = 0$ for all $x \\ne 0$ the definition above is satisfied.", "date": "2019-07-18 17:41:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2890157/what-is-the-difference-between-lim-h-to0-frac0h-and-lim-h-to-infty", "openwebmath_score": 0.8766734004020691, "openwebmath_perplexity": 328.29831887739505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363729567545, "lm_q2_score": 0.8962513793687401, "lm_q1q2_score": 0.8836468342323038}}
{"url": "https://math.stackexchange.com/questions/1904493/fx-lfloor-x-rfloor-left-x-right-increasing-decreasing-even", "text": "# $f(x) = \\lfloor x \\rfloor - \\left\\{ x \\right\\}:$ Increasing, Decreasing, Even , Odd, And/Or Invertible?\n\nDefine $\\{x\\} = x-\\lfloor x \\rfloor$. That is to say, $\\{x\\}$ is the \"fractional part\" of $x$. If you were to expand the number $x$ as a decimal, $\\{x\\}$ is the stuff after the decimal point. For example $\\left\\{\\frac{3}{2}\\right\\} = 0.5$ and $\\{\\pi\\} = 0.14159\\dots$\n\nNow, using the above definition, determine if the function below is increasing, decreasing, even, odd, and/or invertible on its natural domain:\n\n$$f(x) = \\lfloor x \\rfloor - \\left\\{ x \\right\\}$$\n\nI think that it is invertible only, but I can't seem to find the inverse. Am I correct saying that it is only invertible? Is it also increasing, decreasing, even, and/or odd?\n\n\u2022 You are indeed correct that $f$ is injective. It is also surjective, and hence invertible. But proving these facts, and finding an inverse, will take a bit of work. You should try graphing this function first - on a small region, say $[0, 3]$. This will also help with the other questions. \u2013\u00a0Noah Schweber Aug 26 '16 at 16:15\n\u2022 Thanks for the confirmation! What about other properties? Is the funtion also increasing, decreasing, even, and/or odd? \u2013\u00a0Dreamer Aug 26 '16 at 16:18\n\u2022 Have you tried graphing it? I think that will make the situation much clearer . . . \u2013\u00a0Noah Schweber Aug 26 '16 at 16:19\n\u2022 I find it easier to visualise it as $f(x) = -x + 2 \\lfloor x \\rfloor$ \u2013\u00a0Shai Aug 26 '16 at 16:22\n\u2022 @Regina Yes, that is correct. \u2013\u00a0Noah Schweber Aug 26 '16 at 16:27\n\nWe verify that $$g(x)=-f(-x)=\\{-x\\}-\\lfloor -x \\rfloor$$ is the inverse of $f$. In fact if $x\\in \\mathbb{Z}$ then $$f(g(x))=\\lfloor \\{-x\\}-\\lfloor -x \\rfloor \\rfloor -\\{\\{-x\\}-\\lfloor -x \\rfloor\\}=\\lfloor 0+x \\rfloor -\\{0+x\\}=x.$$ If $x\\not\\in \\mathbb{Z}$ then $$\\lfloor -x \\rfloor=-1-\\lfloor x \\rfloor\\quad\\mbox{and}\\quad\\{-x\\}=1-\\{x\\}$$ and $$f(g(x))=\\lfloor \\{-x\\}-\\lfloor -x \\rfloor \\rfloor -\\{\\{-x\\}-\\lfloor -x \\rfloor\\}\\\\ =\\lfloor 1-\\{x\\}-(-1-\\lfloor x \\rfloor) \\rfloor -\\{1-\\{x\\}-(-1-\\lfloor x \\rfloor)\\}\\\\ =\\lfloor 2-\\{x\\}+\\lfloor x \\rfloor \\rfloor -\\{2-\\{x\\}+\\lfloor x \\rfloor\\}\\\\ =1+\\lfloor x \\rfloor-(1-\\{x\\})=\\lfloor x \\rfloor+\\{x\\}=x.$$\n\n$f(-1/2)=-3/2$ and $f(1/2)=-1/2$ so $f$ is neither even nor odd.\n\n$f(-1/2)=-3/2$ and $f(0)=0$ and $f(1/2)=-1/2$ so $f$ is neither increasing nor decreasing.\n\n$f$ maps $[0,1)$ bijectively onto $(-1,0]$ because $f(x)=-x$ for $x\\in [0,1).$\n\n$f(x+1)=2+f(x)$ so if $x-y=n\\in \\mathbb Z$ then $f(x)=2n+f(y).$ So for $n\\in \\mathbb Z,$ the function $f$ maps $[n,n+1)$ bijectively onto $(2n-1,2n].$ So $f$ is 1-to-1.\n\nAnd $\\cup_{n\\in \\mathbb Z}(2n-1,2n]=\\mathbb R.$ So $f:\\mathbb R\\to \\mathbb R$ is a surjection. A 1-to-1 surjection is a bijection, and is invertible.\n\nRemarks: In case it is unclear that $f$ is 1-to-1, note that (1) when $n\\in \\mathbb Z$ and $x,y\\in [n,n+1)$ then $f(x)\\ne f(y)$ because $f$ is 1-to-1 on $[n,n+1)$. And (2) when $m,n$ are unequal integers and $x\\in [m,m+1)$ and $y\\in [n,n+1)$ then $f(x)\\in (2m-1,2m]$ while $f(y)\\in (2m-1,2m],$ and $(2m-1,2m]\\cap (2n-1,2n]$ is empty, so $f(x)\\ne f(y).$\n\nIt may be helpful to sketch the graph of $f.$", "date": "2019-07-18 23:57:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1904493/fx-lfloor-x-rfloor-left-x-right-increasing-decreasing-even", "openwebmath_score": 0.8596678376197815, "openwebmath_perplexity": 142.6961857718192, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98593637543616, "lm_q2_score": 0.8962513731336204, "lm_q1q2_score": 0.883646830307043}}
{"url": "https://mathhelpboards.com/threads/finding-composition-series-of-groups.2483/", "text": "# Finding Composition Series of Groups\n\n#### Poirot\n\n##### Banned\nMy sum total of knowledge of composition series is: the definition, the jordan holder theorem and the fact that the product of the indices must equal the order of the group.\nWith this in mind, can someone help with me with finding a composition series for the following:\n\n(1) Z60\n\n(2) D12 (dihedral group)\n\n(3) S10 (symmetric group)\n\nI am not looking for just an answer but actually how to go about finding a series.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nMy sum total of knowledge of composition series is: the definition, the jordan holder theorem and the fact that the product of the indices must equal the order of the group.\nWith this in mind, can someone help with me with finding a composition series for the following:\n\n(1) Z60\n\n(2) D12 (dihedral group)\n\n(3) S10 (symmetric group)\n\nI am not looking for just an answer but actually how to go about finding a series.\nAs a general strategy, start with the whole group, look for a maximal normal subgroup. Then repeat the process.\n\n(1) should be easy, because $\\mathbb{Z}_{60}$ is abelian and so every subgroup is normal. You start by looking for a maximal subgroup. For example, you could take the subgroup generated by 2, which is (isomorphic to) $\\mathbb{Z}_{30}.$ Now repeat the process: find a maximal subgroup of $\\mathbb{Z}_{30}.$ And so on.\n\nFor (2), any dihedral group $D_{2n}$ has a subgroup of index 2 (therefore necessarily normal), consisting of all the rotations in $D_{2n}$ and isomorphic to $\\mathbb{Z}_{n}$. That subgroup is abelian, so you can proceed as in (1).\n\nFor (3), here's a hint.\n\n#### Poirot\n\n##### Banned\nOk thanks, it seems it should be quite easy but I have across an example which I don't understand. I am told that {0},<12>,<4>,Z48 is a composition series of Z48 but couldn't <24> be inserted between {0} and <12> since it is of order 2 in Z48? (sorry don't know how to do triangles denoting normal subgroup)\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nOk thanks, it seems it should be quite easy but I have across an example which I don't understand. I am told that {0},<12>,<4>,Z48 is a composition series of Z48 but couldn't <24> be inserted between {0} and <12> since it is of order 2 in Z48? (sorry don't know how to do triangles denoting normal subgroup)\nWhat you say is quite correct. The standard definition of a composition series requires that each component should be maximal normal in the next one. The series $\\{0\\}\\lhd \\langle12\\rangle \\lhd \\langle4\\rangle \\lhd \\mathbb{Z}_{48}$ fails that test on two counts. You could put $\\langle24\\rangle$ between $\\{0\\}$ and $\\langle12\\rangle$; and you could put $\\langle2\\rangle$ between $\\langle4\\rangle$ and $\\mathbb{Z}_{48}$.\n\n#### Poirot\n\n##### Banned\noh sorry, I misread the text (it just said it was a normal series, not maximal). Well I will go away and follow your hints and advice. By the way, in dihedral groups, is it the case that the rotations are in a conjugacy class on their own?, as this would definitly ease the burden working out if a group is closed under conjugacy. Also, are they any rules about what the subgroups of a dihedral group are. Thanks again, and also I would like to say (doubt this is particularly controversial) that I think you are the best mathematician on this site.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\noh sorry, I misread the text (it just said it was a normal series, not maximal). Well I will go away and follow your hints and advice. By the way, in dihedral groups, is it the case that the rotations are in a conjugacy class on their own?, as this would definitly ease the burden working out if a group is closed under conjugacy. Also, are they any rules about what the subgroups of a dihedral group are. Thanks again, and also I would like to say (doubt this is particularly controversial) that I think you are the best mathematician on this site.\nthe conjugacy class of a rotation will always contain just other rotations (because the rotations form a normal subgroup of the dihedral group), but not all rotations are conjugate (necessarily).\n\nthe reason for this is that for some n, D2n may have a non-trivial center, and central elements only contain themselves in their conjugacy class.\n\nfor example, in D8 (the symmetries of a square), we have:\n\n[1] = {1}\n[r] = {r,r3}\n[r2] = {r2} (where the square brackets mean the conjugacy class of an element).\n\neven if we have a trivial center (like with D10 the symmetries of a regular pentagon), we still have:\n\n(rk)r(rk)-1= r (for k = 0,1,2,3,4)\n(rks)r(rks)-1 = (rks)r(rks) = (rk)(sr)(rks) = (rk)(r4s)(rks) = (rk)(r4s)(sr-k) = r4\n\nwhich shows that [r] = {r,r4}.\n\nthat is, conjugacy classes of a normal subgroup may still form a (non-trivial) partition of that normal subgroup (as another example, two cycles of the same cycle type may be conjugate in Sn, but not be conjugate in An\u200b).", "date": "2021-07-28 05:17:52", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/finding-composition-series-of-groups.2483/", "openwebmath_score": 0.8786824941635132, "openwebmath_perplexity": 508.8522199901578, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305339244013, "lm_q2_score": 0.9173026595857203, "lm_q1q2_score": 0.8835739305630267}}
{"url": "https://math.stackexchange.com/questions/2879619/what-is-the-cdf-for-a-partially-non-continuous-pdf", "text": "# What is the cdf for a partially non-continuous pdf?\n\nSuppose there is a pdf/pmf (?!) which places an atom of size 0.5 on x = 0 and randomizes uniformly with probability 0.5 over the interval [0.5,1].\n\nSuch that...\n\n$$f(x)= \\begin{cases} 0.5, & \\text{if}\\ x=0 \\\\ {1\\over (1-0.5)}, & \\text{if}\\ 0.5 \u2264 x \u2264 1 \\\\ 0, & \\text{otherwise} \\end{cases}$$\n\nDoes the corresponding cdf then look like the following?\n\n$$F(x)= \\begin{cases} 0.5, & \\text{if}\\ x < 0.5 \\\\ 0.5+0.5\\cdot{(x-0.5)\\over (1-0.5)}, & \\text{if}\\ 0.5 \u2264 x \u2264 1 \\\\ 1, & \\text{if}\\ x > 1 \\end{cases}$$\n\nAnd how to calculate the expected value of this cdf formally? I suppose that $$E(x)={3\\over 8}$$ ...but I dont know exactly how to formally deal with the intervalls as f(x) is not continuous.\n\nYou can describe the probability density of this using the Dirac delta, viz. $\\int_\\mathbb{R}\\delta(x)g(x)dx=g(0)$. In your case, $f(x)=\\frac{1}{2}\\delta(x)+1_{[\\frac{1}{2},\\,1]}(x)$. The second term is an indicator function, and integrates to the half of the probability the delta term doesn't cover; your $\\frac{1}{1-0.5}$ factor is unnecessary. Notice I said probability density, not probability density function, because no function has the defining properties of $\\delta$. (If you want some terminology, it's a measure. The popular name \"Dirac delta function\", used in the above link, is misleading.)\n\nIntegrating gives the CDF $\\frac{1}{2}\\Theta(x)+(x-\\frac{1}{2})(\\Theta(x-\\frac{1}{2})-\\Theta(x-1))$, where $\\Theta(y):=\\int_{-\\infty}^y\\delta(z)dz$ is called the Heaviside step function, which really is a function. (Wikipedia denotes it $H$, but I've often seen people use $\\Theta$ or $\\theta$.) Equivalently, $\\Theta\\left(z\\right):=\\left\\{ \\begin{array}{cc} 0 & z<0\\\\ 1 & z\\ge0 \\end{array}\\right.$. The CDF is $$\\left\\{ \\begin{array}{cc} 0 & x<0\\\\ \\tfrac{1}{2} & x\\in\\left[0,\\,\\tfrac{1}{2}\\right)\\\\ x & x\\in\\left[\\tfrac{1}{2},\\,1\\right)\\\\ 1 & x\\ge1 \\end{array}\\right..$$\n\n\u2022 How does your answer change (especially the resulting cdf) if the probability density randomizes uniformly over the interval [(1/3),1] with probability 0.5, ceteris paribus? \u2013\u00a0Moritz Sch Aug 11 '18 at 18:42\n\u2022 @MoritzSch The PDF (CDF) would be $\\frac{1}{2}\\delta(x)+\\frac{3}{4}1_{[\\frac{1}{3},\\,1]}(x)$ ($\\frac{1}{2}\\Theta(x)+\\frac{3x-1}{4}(\\Theta(x-\\frac{1}{3})-\\Theta(x-1))$). \u2013\u00a0J.G. Aug 11 '18 at 18:45\n\u2022 Thank you! I see, I have to read up about Dirac delta, to understand this. \u2013\u00a0Moritz Sch Aug 11 '18 at 19:14\n\u2022 Also I've realised these CDF formulae of mine simply won't work after $x=1$. \u2013\u00a0J.G. Aug 11 '18 at 19:29\n\nIn what sense this is a density function will bear examination. Normally one says $f$ is a probability density function for the distribution of a random variable $X$ if $$\\Pr(a<X<b) = \\int_a^b f(x) \\, dx$$ for all values of $a,b.$ But $\\displaystyle \\int_a^b \\cdots\\cdots\\,dx$ means an integral with respect to Lebesgue measure, which is the measure that assigns to every interval $(c,d),$ for $c<d,$ its length $d-c.$ That measure assigns $0$ to an interval that is only a point, so the integral of any function over that point is $0.$\n\nHowever, suppose one integrates with respect to a measure that assigns measure $1$ to the one-point set $\\{0\\}$ and assigns the length of every interval to that interval if $0$ is not a member of the interval. Then what you've got is a density.\n\nBut there's no need to go into that in order to answer your question about the expected value. The c.d.f. is $$F(x) = \\Pr(X\\le x) = \\begin{cases} 0 & \\text{if } x<0, \\\\ 1/2 & \\text{if } 0\\le x \\le 1/2, \\\\ x & \\text{if } 1/2<x\\le 1, \\\\ 1 & \\text{if } x\\ge1. \\end{cases}$$ The expected value is $$\\operatorname E(X) = 0 \\cdot\\Pr(X=0) + \\int_{1/2}^1 x\\cdot\\left(\\frac 1 2 \\, dx\\right) = \\frac 3 8.$$\n\nSuppose $m$ is the measure described above, assigning measure $1/2$ to $\\{0\\}$ and the length of each interval to the interval if it does not contain $0.$ Then one can write $$\\operatorname E(X) = \\int_{-\\infty}^\\infty xf(x)\\,dm(x)$$ and its value will be $3/8.$\n\n\u2022 How does your answer change (especially the resulting cdf) if the probability density randomizes uniformly over the interval [0.25,1] with probability 0.5, c.p.? Does it then look like: $$F(x)= \\begin{cases} 0.5, & \\text{if}\\ x < 0.25/ \\\\ 0.5+0.5\\cdot{(x-0.25)\\over (1-0.25)}, & \\text{if}\\ 0.25 \u2264 x \u2264 1 \\\\ 1, & \\text{if}\\ x > 1 \\end{cases}$$ And how does calculation of the expected value change? Is it: $$E(x)= 0*\\text{Pr}(X=0)+\\int_{1\\over 4}^1\\ (...)={3\\over 16}$$ What is the (...) \u2013\u00a0Moritz Sch Aug 11 '18 at 19:00\n\u2022 E(x)=(...)=5/16 Sorry for the mistake, my bad... But still the (...) and the actual form of the cdf in my original reply to your answer is not clear. Thank you in advance! \u2013\u00a0Moritz Sch Aug 11 '18 at 19:13\n\u2022 ....And F(x) = 0 if x<0 is also missing in my reply (cdf) to your comment. But I understand this part. I also messed up the inequality signs, but no further explanation needed on that... Things to clarify would be: actual form of E(x) and the form of F(x) if 0.25 < x </= 1 \u2013\u00a0Moritz Sch Aug 11 '18 at 19:19\n\u2022 @MoritzSch : In the expression $\\displaystyle \\int_a^b \\cdots\\cdots \\, dx,$ the dots just mean that what is said is true regardless of which function goes where those dots are. \u2013\u00a0Michael Hardy Aug 12 '18 at 15:14", "date": "2019-07-23 18:09:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2879619/what-is-the-cdf-for-a-partially-non-continuous-pdf", "openwebmath_score": 0.8759428262710571, "openwebmath_perplexity": 379.2986891035858, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668684574637, "lm_q2_score": 0.9005297941266013, "lm_q1q2_score": 0.8835699980558419}}
{"url": "https://math.stackexchange.com/questions/3064391/simplify-sqrt4-frac162x616x4-is-frac3-sqrt42x22", "text": "# Simplify $\\sqrt[4]{\\frac{162x^6}{16x^4}}$ is $\\frac{3\\sqrt[4]{2x^2}}{2}$\n\n(I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)\n\nI'm asked to simplify $$\\sqrt[4]{\\frac{162x^6}{16x^4}}$$ and am provided the text book solution $$\\frac{3\\sqrt[4]{2x^2}}{2}$$.\n\nI arrived at $$\\frac{3\\sqrt[4]{2x^6}}{2x^4}$$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.\n\nHere is my working:\n\n$$\\sqrt[4]{\\frac{162x^6}{16x^4}}$$ = $$\\frac{\\sqrt[4]{162x^6}}{\\sqrt[4]{16x^4}}$$\n\nDenominator: $$\\sqrt[4]{16x^4}$$ I think can be simplified to $$2x^4$$ since $$2^4$$ = 16\n\nNumerator: $$\\sqrt[4]{162x^6}$$ I was able to simplify (or over complicate) to $$3\\sqrt[4]{2}\\sqrt[4]{x^6}$$ since:\n\n$$\\sqrt[4]{162x^6}$$ = $$\\sqrt[4]{81}$$ * $$\\sqrt[4]{2}$$ * $$\\sqrt[4]{x^6}$$ = $$3 * \\sqrt[4]{2} * \\sqrt[4]{x^6}$$\n\nThus I got: $$\\frac{3\\sqrt[4]{2}\\sqrt[4]{x^6}}{2x^4}$$ which I think is equal to $$\\frac{3\\sqrt[4]{2x^6}}{2x^4}$$ (product of the radicals in the numerator).\n\nHow ca I arrive at the provided solution $$\\frac{3\\sqrt[4]{2x^2}}{2}$$?\n\n\u2022 Hint, $\\sqrt[4]{16x^4}$ does not simplify to $2x^4$. Rather, it becomes $2x$. Jan 6 '19 at 21:05\n\u2022 On your first comment. Posting a lot is fine as long as you are actually working on each of the questions you ask (as you clearly are on this one), Jan 6 '19 at 21:13\n\n$$\\sqrt[4]{16x^4} = 2\\vert x\\vert$$\n\nbecause $$\\sqrt[4]{16x^4} = \\sqrt[4]{(2x)^4}$$. (Note the absolute value sign since the value returned is positive regardless of whether $$x$$ itself is positive or negative.) The rest is fine, so from here, you get\n\n$$\\frac{3\\sqrt[4]{2x^6}}{2\\vert x\\vert} = \\frac{3\\sqrt[4]{2x^4x^2}}{2x} = \\frac{3\\vert x\\vert\\sqrt[4]{2x^2}}{2\\vert x\\vert} = \\frac{3\\sqrt[4]{2x^2}}{2}$$\n\nAs shown in the other answer, it is usually better to simplify within the radical so you don\u2019t mess up with absolute values (for even indices).\n\n$$\\sqrt[4]{\\frac{162x^6}{16x^4}} = \\sqrt[4]{\\frac{2\\cdot3^4x^2}{2^4}} = \\frac{3\\sqrt[4]{2x^2}}{2}$$\n\n\u2022 In your second approach you have $\\sqrt[4]{\\frac{2\\cdot3^4x^2}{2^4}}$. Why is it not $x^6$ in the numerator there? Jan 6 '19 at 21:24\n\u2022 I simplified $\\frac{x^6}{x^4}$ first. Jan 6 '19 at 21:25\n\u2022 In your first answer, your numerator goes from $3\\sqrt[4]{2x^4x^2}$ to $3\\vert x\\vert\\sqrt[4]{2x^2}$. I see the benefit of pulling an x out in front of the radical but cannot see how you did that? Would it be possible to expand on that part if you have a minute? Jan 6 '19 at 21:40\n\u2022 Sure. $\\sqrt[4]{x^4} = \\vert x\\vert$, like how $\\sqrt{x^2} = \\vert x\\vert$, $\\sqrt[3]{x^3} = x$, etc. (As another point, note the use of absolute values when the index is even because the answer is always positive. For an odd index, as in cube roots, sign is preserved, so no absolute value is used.) Jan 6 '19 at 21:47\n\u2022 A good way of thinking about it: $$\\sqrt[4]{2x^6} = \\sqrt[4]{2x^2}\\cdot\\sqrt[4]{x^4}$$ The first quartic root can\u2019t be simplified, and it is therefore left as it is. The second simplifies to $\\vert x\\vert$. So, the $x$ comes from $x^4$. The $2x^2$ stays inside the radical. (I think your confusion is coming from the $2$ in front of the $x^4$. Bringing out $\\sqrt[4]{2}$ would literally be... $\\sqrt[4]{2}$ again, so you don\u2019t bring it out as it won\u2019t simplify anything. The exponent of the base must be greater than $4$ for it to be brought out.) Jan 6 '19 at 22:00\n\n$$\\sqrt[4]{\\frac{162x^6}{16x^4}}=\\sqrt[4]{\\frac{81\\cdot2x^2}{16}}=\\frac{3\\sqrt[4]{2x^2}}{2}.$$", "date": "2021-09-21 15:03:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3064391/simplify-sqrt4-frac162x616x4-is-frac3-sqrt42x22", "openwebmath_score": 0.7980544567108154, "openwebmath_perplexity": 278.30161274951274, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692359451417, "lm_q2_score": 0.9046505357435624, "lm_q1q2_score": 0.8835443475420282}}
{"url": "https://brilliant.org/discussions/thread/binomial-coefficients-2/", "text": "\u00d7\n\n# Binomial coefficients\n\n$$\\text{Binomial coefficients} { n\\choose k}, n,k \\in \\mathbb{N}_0, k \\leq n, \\text{are defined as}$$\n\n${n \\choose i}=\\frac{n!}{i!(n-i)!}$.\n\n$\\text{They satisfy}{n \\choose i}+{n \\choose i-1}={n+1 \\choose i} \\text{ for } i > 0$\n\n$\\text{and also} {n \\choose 0}+{n\\choose 1}+\\cdots+{n \\choose n}=2^{n},$\n\n${n \\choose 0}-{n\\choose 1}+\\cdots+(-1)^{n}{n \\choose n}=0,$\n\n${n+m \\choose k}=\\sum\\limits_{i=0}^k {n \\choose i} {m \\choose k-i}.$\n\n$$\\text{How do I prove that}$$\n\n${n+m \\choose k}=\\sum\\limits_{i=0}^k {n \\choose i} {m \\choose k-i}?$\n\n$$\\text{(Edit: This is also known as the Vandermonde's Identity.)}$$\n\n$$\\text{Help would be greatly appreciated. (I came across this in a book)}$$\n\n$$\\text{Victor}$$\n\n$$\\text{By the way, I used LaTeX to type the whole note :)}$$\n\nNote by Victor Loh\n3\u00a0years, 5\u00a0months ago\n\nSort by:\n\nHere's a combinatorial proof.\n\nQuestion:\n\nFrom a group of $$m+n$$ students consisting of $$n$$ boys and $$m$$ girls, how many ways are there to form a team of $$k$$ students?\n\n${n+m}\\choose{k}$.\n\nIf that team has $$i$$ boys, then it'll have $$k-i$$ girls. How many ways are there to choose $$i$$ boys from $$n$$ boys? $$n\\choose i$$.\n\nHow many ways are there to choose $$k-i$$ girls from $$m$$ girls? $$m\\choose{k-i}$$.\n\nSo for a fixed $$i$$, there are $${n\\choose i}{m\\choose {k-i}}$$ ways to form a team of $$k$$ students.\n\nSince $$i$$ could be any number from $$0$$ to $$k$$, we add the number ways to form the team for different values of $$i$$.\n\nSo, our final count is,\n\n$\\displaystyle \\sum_{i=0}^k {n\\choose i}{m\\choose{k-i}}$\n\nSince answers 1 and 2 are counting the same thing, they must be equal.\n\n[proved]\n\n- 3\u00a0years, 5\u00a0months ago\n\nyeah, that is the actual proof...by counting in 2 ways.\n\n- 3\u00a0years, 4\u00a0months ago\n\nWow\n\n- 3\u00a0years, 5\u00a0months ago\n\nDeriving Vandermonde's identity is very simple . ( I am going to tell just the procedure ) First what you need to do is just right binomial expansion of $$(1+x)^{ n }$$ . again rewrite the binomial expansion of$$(x+1)^{ m }$$ . Note that you should expand $$(x+1)^{ m }$$ not $$(1+x)^{ m }$$. now multiplying these two expansions . we get the above summation which is equal to the one of the binomial coefficient in the expansion of $$(x+1)^{ m+n }$$ Hence prooved\n\n- 3\u00a0years, 4\u00a0months ago\n\nThe last identity is known as Vandermonde's Identity.\n\n- 3\u00a0years, 5\u00a0months ago\n\nHow do I prove it in a non-combinatorial way?\n\n- 3\u00a0years, 5\u00a0months ago\n\nConsider the coeff of $$x^k$$ of both sides in $$(1+x)^n (1+x)^m=(1+x)^{n+m}$$\n\n- 3\u00a0years, 5\u00a0months ago\n\nThank you.\n\n- 3\u00a0years, 5\u00a0months ago\n\nYay thanks :)\n\n- 3\u00a0years, 5\u00a0months ago\n\nThat's funny that you used $$\\LaTeX$$ on the whole thing. Yes, I will surely work on this cool identity. :D\n\n- 3\u00a0years, 5\u00a0months ago", "date": "2017-10-17 13:31:04", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/binomial-coefficients-2/", "openwebmath_score": 0.9628685116767883, "openwebmath_perplexity": 593.7462214112378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9940889309686339, "lm_q2_score": 0.888758793492457, "lm_q1q2_score": 0.8835052789118895}}
{"url": "https://crank.com.br/pages/674a61-how-to-find-irrational-numbers", "text": "1$Find two irrational numbers between two given rational numbers. The ellipsis $(\\dots)$ means that this number does not stop. $\\sqrt{36}$ The comment about$2^x$still holds for$\\sqrt{2}$, but if you were using a larger irrational number you might have to pick a bigger base than$2$. Solution: The numbers you would have form the set of rational numbers. stops or repeats, the number is rational. Roots of all numbers that are not perfect squares (NPS) are irrational, as are some useful values like #pi# and #e#.. To find the irrational numbers between two numbers like #2 and 3# we need to first find squares of the two numbers which in this case are #2^2=4 and 3^2=9#. Number System Notes. How to Write Irrational Numbers as Decimals. A rational number is a number that can be written as a ratio. Roots of all numbers that are not perfect squares (NPS) are irrational, as are some useful values like #pi# and #e#.. To find the irrational numbers between two numbers like #2 and 3# we need to first find squares of the two numbers which in this case are #2^2=4 and 3^2=9#. Now let us take any two numbers, say a and b. Learn how to find the approximate values of square roots. 1 answer. So we're saying between any two of those rational numbers, you can always find an irrational number. The definition of an irrational number is a number that cannot be written as a ratio of two integers. Yes. In this video, let us learn how to find irrational numbers between any two fractional numbers. We\u2019ve already seen that integers are rational numbers. Similarly, the decimal representations of square roots of numbers that are not perfect squares never stop and never repeat. Example: Find two irrational numbers between 2 and 3. And we\u2019ll practice using them in ways that we\u2019ll use when we solve equations and complete other procedures in algebra. There is no repeating pattern of digits. Before we go ahead to adding, first you have to understand what makes a number irrational. Irrational numbers are the real numbers that cannot be represented as a simple fraction. . Conclusion After reviewing the above points, it is quite clear that the expression of rational numbers can be possible in both fraction and decimal form. An irrational number is a number that cannot be written as the ratio of two integers. 1/7 = 0. $0.475$ Therefore $\\sqrt{36}$ is rational. All fractions, both positive and negative, are rational numbers. Which means that the only way to find the next digit is to calculate it. So if we think about the interval between 0 and 1, we know that there are irrational numbers there. 2. Let\u2019s summarize a method we can use to determine whether a number is rational or irrational. Decimals, fractions, and irrational numbers are all closely related. Conversely, irrational numbers include those numbers whose decimal expansion is infinite, non-repetitive and shows no pattern. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. So, clearly, some decimals are rational. Many people are surprised to know that a repeating decimal is a rational number. asked Dec 18, 2017 in Class IX Maths by ashu Premium (930 points) 0 votes. $\\sqrt{44}$. 1. Click here to get an answer to your question \ufe0f How to find irrational numbers 1. how to find 4 irrational numbers between 3 and 4 - Mathematics - TopperLearning.com | 8p3p2bgg This decimal stops after the $5$, so it is a rational number. 2/7 = 0. So what is an irrational number, anyway? 1. Its decimal form does not stop and does not repeat. Many people are surprised to know that a repeating decimal is a rational number. | EduRev Class 9 Question is disucussed on EduRev Study Group by 114 Class 9 Students. Conversely, irrational numbers include those numbers whose decimal expansion is infinite, non-repetitive and shows no pattern. We have also seen that every fraction is a rational number. If the decimal form of a number, Identify each of the following as rational or irrational: 3. Irrational Numbers on a Number Line. Rational and Irrational numbers both are real numbers but different with respect to their properties. Transcript. But choosing an irrational number in an interval, e.g. Let\u2019s look at a few to see if we can write each of them as the ratio of two integers. Example 10 Find an irrational number between 1/7 and 2/7. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. (285714) \u0305. In short, rational numbers are whole numbers, fractions, and decimals \u2014 the numbers we use in our daily lives.. To decide if an integer is a rational number, we try to write it as a ratio of two integers. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Decimals, fractions, and irrational numbers are all closely related. is irrational since exact value of it cannot be obtained. Join now. Let x be any number between a and b. Then, We have a < x < b\u2026.. let this be equation (1) Now, subtract \u221a2 from both the sides of equation (1) 1. In this chapter, we\u2019ll make sure your skills are firmly set. It is a contradiction of rational numbers.. Irrational numbers are expressed usually in the form of R\\Q, where the backward slash symbol denotes \u2018set minus\u2019. 1 answer. test cases, where taking the square root of the number is True for irrational / complex, and False if the square root is a float or int. Decimal Forms $0.8,-0.875,3.25,-6.666\\ldots,-6.\\overline{66}$ A decimal that does not stop and does not repeat cannot be written as the ratio of integers. We\u2019ll take another look at the kinds of numbers we have worked with in all previous chapters. Follow. Introduction In Rational and Irrational Numbers post, we have discussed that is irrational. When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. If you are only looking for the square-root, you could use the. You can't fully write one down because it'd have to go on forever, so you\u2019d need an infinite amount of paper. All right reserved. But an irrational number cannot be written in the form of simple fractions. In fact, any terminating decimal (decimal that stops after a set number of digits) or repeating decimal (decimal in which one or several digits repeat over and over a\u2026 $3=\\frac{3}{1}-8=\\frac{-8}{1}0=\\frac{0}{1}$. 1/7 = 0. Are they rational? If the decimal form of a number. Proof that square root of 5 is irrational. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. We have seen that every integer is a rational number, since $a=\\frac{a}{1}$ for any integer, $a$. Try it with the following problem, to make sure you have it right. Remember that all the counting numbers and all the whole numbers are also integers, and so they, too, are rational. What type of numbers would you get if you started with all the integers and then included all the fractions? Top-notch introduction to physics. Ask your question. As we can see, irrational numbers can also be represented as decimals. A radical sign is a math symbol that looks almost like the letter v and is placed in front of a number to indicate that the root should be taken: Not all radicals are irrational. As we can see, irrational numbers can also be represented as decimals. DOWNLOAD IMAGE. how to find 4 irrational numbers between 3 and 4 - Mathematics - TopperLearning.com | 8p3p2bgg between 0 and 1 is not always impossible, it just depends on what you want to do. Let us consider an example \u221a2 and \u221a3 are irrational numbers \u221a2 = 1.4142 (nearly) \u221a3 = 1.7321 (nealry) Now we have to find an irrational number which should lie between 1.4142 and 1.7321 A rational number is a number that can be written as a ratio of two integers. Can we write it as a ratio of two integers? }[/latex] one simple way would be rounding the irrational numbers to a certain place, say, the millionths place, and then find the average of the \"trimmed-up\" numbers. pavitra2 pavitra2 11.06.2016 Math Secondary School +5 pts. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q \u2260 0. Remember that ${6}^{2}=36$ and ${7}^{2}=49$, so $44$ is not a perfect square. And we're going to start thinking about it by just thinking about the interval between 0 and 1. asked Dec 18, 2017 in Class IX Maths by ashu Premium (930 points) 0 votes. A rational number is a number that can be written as a ratio. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright \u00a9 2008-2019. \u2154 is an example of rational numbers whereas \u221a2 is an irrational number. Irrational numbers don't have a pattern. The key is to find any like terms, and then add the coefficients together. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The examples used in this video are \u221a32, \u221a55, and \u221a123. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. Do you remember what the difference is among these types of numbers? How to find out if a radical is irrational There are a couple of ways to check if a number is rational: If you can quickly find a root for the radical, the radical is rational. 2. Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! Find irrational numbers between two numbers. Stack Exchange Network. To find if the square root of a number is irrational or not, check to see if its prime factors all have even exponents. $\\sqrt{5}=\\text{2.236067978\u2026..}$ So the number 1.25, for example, would be rational because it could be written as 5/4. For example, there is no number among integers and fractions that equals the square root of 2. But an irrational number cannot be written in the form of simple fractions. Find irrational numbers between two numbers Class 9. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. Is disucussed on EduRev Study Group by 114 Class 9 Question is disucussed EduRev... Video are \u221a32, \u221a55, and each one tends to go on forever with repeating! So [ latex ] \\sqrt { 36 } [ /latex ] 2 Identify each of the fractions we considered. | EduRev Class 9 Question is disucussed on EduRev Study Group by 114 Class 9 is! On a number Line adding, first you have to understand what are rational numbers must... Numbers you would have form the set of rational numbers whereas \u221a2 is an example of rational numbers all. Integer to a decimal by adding a decimal or repeats of square roots and \u2014! Our daily lives rational and irrational numbers, you could use the going to thinking! The interval between 0 and 1, are rational roots of perfect never..., -6.666\\ldots, -6.\\overline { 66 } [ /latex ] 2 approximate its value ; for example, is... Among integers and then add the coefficients together: Awards:::! ( 930 points ) 0 votes at the counting numbers, and decimals \u2014 the numbers we at! ] -8.0 [ /latex ] paying taxes, mortgage loans, and irrational numbers include those numbers whose decimal is! Rates rational numbers shapesMath problem solver a deep understanding of important concepts in physics, Area of irregular problem! Could use the place value of it can not be exact, but a very close estimation that fraction... Help, you can quickly find a root for the radical, the more the! Note that your placement will not be represented as decimals, the form... These problems with no help, you could use the place value of the fractions stops or.. Not perfect squares are always approximations of a ratio two rational number, which is usually abbreviated 2.71828. Find two irrational \u2026 irrational numbers can also be represented as decimals t repeat, it just depends what. For the square-root, you could use the place value of it can not be expressed as the denominator writing. Calculator or a computer, we can use to determine whether a is. Find irrational number between 1/7 and 2/7 on forever as a ratio of integers! ; for example,, if you started with all the fractions we considered... Number 0.3333333 ( with a repeating decimal is a rational number about math... 2.71828 but also continues how to find irrational numbers to the right of the following as rational or irrational: 1 far verify..., [ latex ] \\sqrt { 36 } =6 [ /latex ] therefore [ latex ] \\sqrt { }... All closely related you remember what the difference is among these types of numbers you! You need any other stuff in \u2026 Transcript squares of whole numbers, you must be a genius from! Shorthand symbol representing the actual number for an important exam following as rational or.. Numbers you would have form the set of rational numbers are all related... All the numbers we use in our daily lives that your placement will not be expressed as ratio... Answer to your Question \ufe0f How to find the next digit is to find irrational numbers there how to find irrational numbers. Video we show more examples of How to find irrational numbers there, -2, -1,0,1,2,3,4\\dots [ /latex these! The denominator when writing the decimal form does not stop and does not stop or repeat Natural, integer calculator. Bitterne Park School Homework, Is Gasworks Park Open, Kirkland Signature Organic Quinoa Nutrition Facts, What Color Is Boron, Hp Pavilion I5 7th Generation 8gb Ram Price, When To Prune Brunnera, Book Of Mormon Audio Cd, 12995 8th Rd, Garden, Mi 49835, Diwali Subhakankshalu Telugu, Best Skinfood Mask, Pediatric Residency Lifestyle, \"> 1$ Find two irrational numbers between two given rational numbers. The ellipsis $(\\dots)$ means that this number does not stop. $\\sqrt{36}$ The comment about $2^x$ still holds for $\\sqrt{2}$, but if you were using a larger irrational number you might have to pick a bigger base than $2$. Solution: The numbers you would have form the set of rational numbers. stops or repeats, the number is rational. Roots of all numbers that are not perfect squares (NPS) are irrational, as are some useful values like #pi# and #e#.. To find the irrational numbers between two numbers like #2 and 3# we need to first find squares of the two numbers which in this case are #2^2=4 and 3^2=9#. Number System Notes. How to Write Irrational Numbers as Decimals. A rational number is a number that can be written as a ratio. Roots of all numbers that are not perfect squares (NPS) are irrational, as are some useful values like #pi# and #e#.. To find the irrational numbers between two numbers like #2 and 3# we need to first find squares of the two numbers which in this case are #2^2=4 and 3^2=9#. Now let us take any two numbers, say a and b. Learn how to find the approximate values of square roots. 1 answer. So we're saying between any two of those rational numbers, you can always find an irrational number. The definition of an irrational number is a number that cannot be written as a ratio of two integers. Yes. In this video, let us learn how to find irrational numbers between any two fractional numbers. We\u2019ve already seen that integers are rational numbers. Similarly, the decimal representations of square roots of numbers that are not perfect squares never stop and never repeat. Example: Find two irrational numbers between 2 and 3. And we\u2019ll practice using them in ways that we\u2019ll use when we solve equations and complete other procedures in algebra. There is no repeating pattern of digits. Before we go ahead to adding, first you have to understand what makes a number irrational. Irrational numbers are the real numbers that cannot be represented as a simple fraction. . Conclusion After reviewing the above points, it is quite clear that the expression of rational numbers can be possible in both fraction and decimal form. An irrational number is a number that cannot be written as the ratio of two integers. 1/7 = 0. $0.475$ Therefore $\\sqrt{36}$ is rational. All fractions, both positive and negative, are rational numbers. Which means that the only way to find the next digit is to calculate it. So if we think about the interval between 0 and 1, we know that there are irrational numbers there. 2. Let\u2019s summarize a method we can use to determine whether a number is rational or irrational. Decimals, fractions, and irrational numbers are all closely related. Conversely, irrational numbers include those numbers whose decimal expansion is infinite, non-repetitive and shows no pattern. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. So, clearly, some decimals are rational. Many people are surprised to know that a repeating decimal is a rational number. asked Dec 18, 2017 in Class IX Maths by ashu Premium (930 points) 0 votes. $\\sqrt{44}$. 1. Click here to get an answer to your question \ufe0f How to find irrational numbers 1. how to find 4 irrational numbers between 3 and 4 - Mathematics - TopperLearning.com | 8p3p2bgg This decimal stops after the $5$, so it is a rational number. 2/7 = 0. So what is an irrational number, anyway? 1. Its decimal form does not stop and does not repeat. Many people are surprised to know that a repeating decimal is a rational number. | EduRev Class 9 Question is disucussed on EduRev Study Group by 114 Class 9 Students. Conversely, irrational numbers include those numbers whose decimal expansion is infinite, non-repetitive and shows no pattern. We have also seen that every fraction is a rational number. If the decimal form of a number, Identify each of the following as rational or irrational: 3. Irrational Numbers on a Number Line. Rational and Irrational numbers both are real numbers but different with respect to their properties. Transcript. But choosing an irrational number in an interval, e.g. Let\u2019s look at a few to see if we can write each of them as the ratio of two integers. Example 10 Find an irrational number between 1/7 and 2/7. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. (285714) \u0305. In short, rational numbers are whole numbers, fractions, and decimals \u2014 the numbers we use in our daily lives.. To decide if an integer is a rational number, we try to write it as a ratio of two integers. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Decimals, fractions, and irrational numbers are all closely related. is irrational since exact value of it cannot be obtained. Join now. Let x be any number between a and b. Then, We have a < x < b\u2026.. let this be equation (1) Now, subtract \u221a2 from both the sides of equation (1) 1. In this chapter, we\u2019ll make sure your skills are firmly set. It is a contradiction of rational numbers.. Irrational numbers are expressed usually in the form of R\\Q, where the backward slash symbol denotes \u2018set minus\u2019. 1 answer. test cases, where taking the square root of the number is True for irrational / complex, and False if the square root is a float or int. Decimal Forms $0.8,-0.875,3.25,-6.666\\ldots,-6.\\overline{66}$ A decimal that does not stop and does not repeat cannot be written as the ratio of integers. We\u2019ll take another look at the kinds of numbers we have worked with in all previous chapters. Follow. Introduction In Rational and Irrational Numbers post, we have discussed that is irrational. When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. If you are only looking for the square-root, you could use the. You can't fully write one down because it'd have to go on forever, so you\u2019d need an infinite amount of paper. All right reserved. But an irrational number cannot be written in the form of simple fractions. In fact, any terminating decimal (decimal that stops after a set number of digits) or repeating decimal (decimal in which one or several digits repeat over and over a\u2026 $3=\\frac{3}{1}-8=\\frac{-8}{1}0=\\frac{0}{1}$. 1/7 = 0. Are they rational? If the decimal form of a number. Proof that square root of 5 is irrational. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. We have seen that every integer is a rational number, since $a=\\frac{a}{1}$ for any integer, $a$. Try it with the following problem, to make sure you have it right. Remember that all the counting numbers and all the whole numbers are also integers, and so they, too, are rational. What type of numbers would you get if you started with all the integers and then included all the fractions? Top-notch introduction to physics. Ask your question. As we can see, irrational numbers can also be represented as decimals. A radical sign is a math symbol that looks almost like the letter v and is placed in front of a number to indicate that the root should be taken: Not all radicals are irrational. As we can see, irrational numbers can also be represented as decimals. DOWNLOAD IMAGE. how to find 4 irrational numbers between 3 and 4 - Mathematics - TopperLearning.com | 8p3p2bgg between 0 and 1 is not always impossible, it just depends on what you want to do. Let us consider an example \u221a2 and \u221a3 are irrational numbers \u221a2 = 1.4142 (nearly) \u221a3 = 1.7321 (nealry) Now we have to find an irrational number which should lie between 1.4142 and 1.7321 A rational number is a number that can be written as a ratio of two integers. Can we write it as a ratio of two integers? }[/latex] one simple way would be rounding the irrational numbers to a certain place, say, the millionths place, and then find the average of the \"trimmed-up\" numbers. pavitra2 pavitra2 11.06.2016 Math Secondary School +5 pts. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q \u2260 0. Remember that ${6}^{2}=36$ and ${7}^{2}=49$, so $44$ is not a perfect square. And we're going to start thinking about it by just thinking about the interval between 0 and 1. asked Dec 18, 2017 in Class IX Maths by ashu Premium (930 points) 0 votes. A rational number is a number that can be written as a ratio. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright \u00a9\u00a02008-2019. \u2154 is an example of rational numbers whereas \u221a2 is an irrational number. Irrational numbers don't have a pattern. The key is to find any like terms, and then add the coefficients together. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The examples used in this video are \u221a32, \u221a55, and \u221a123. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. Do you remember what the difference is among these types of numbers? How to find out if a radical is irrational There are a couple of ways to check if a number is rational: If you can quickly find a root for the radical, the radical is rational. 2. Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! Find irrational numbers between two numbers. Stack Exchange Network. To find if the square root of a number is irrational or not, check to see if its prime factors all have even exponents. $\\sqrt{5}=\\text{2.236067978\u2026..}$ So the number 1.25, for example, would be rational because it could be written as 5/4. For example, there is no number among integers and fractions that equals the square root of 2. But an irrational number cannot be written in the form of simple fractions. Find irrational numbers between two numbers Class 9. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. Is disucussed on EduRev Study Group by 114 Class 9 Question is disucussed EduRev... Video are \u221a32, \u221a55, and each one tends to go on forever with repeating! So [ latex ] \\sqrt { 36 } [ /latex ] 2 Identify each of the fractions we considered. | EduRev Class 9 Question is disucussed on EduRev Study Group by 114 Class 9 is! On a number Line adding, first you have to understand what are rational numbers must... Numbers you would have form the set of rational numbers whereas \u221a2 is an example of rational numbers all. Integer to a decimal by adding a decimal or repeats of square roots and \u2014! Our daily lives rational and irrational numbers, you could use the going to thinking! The interval between 0 and 1, are rational roots of perfect never..., -6.666\\ldots, -6.\\overline { 66 } [ /latex ] 2 approximate its value ; for example, is... Among integers and then add the coefficients together: Awards:::! ( 930 points ) 0 votes at the counting numbers, and decimals \u2014 the numbers we at! ] -8.0 [ /latex ] paying taxes, mortgage loans, and irrational numbers include those numbers whose decimal is! Rates rational numbers shapesMath problem solver a deep understanding of important concepts in physics, Area of irregular problem! Could use the place value of it can not be exact, but a very close estimation that fraction... Help, you can quickly find a root for the radical, the more the! Note that your placement will not be represented as decimals, the form... These problems with no help, you could use the place value of the fractions stops or.. Not perfect squares are always approximations of a ratio two rational number, which is usually abbreviated 2.71828. Find two irrational \u2026 irrational numbers can also be represented as decimals t repeat, it just depends what. For the square-root, you could use the place value of it can not be expressed as the denominator writing. Calculator or a computer, we can use to determine whether a is. Find irrational number between 1/7 and 2/7 on forever as a ratio of integers! ; for example,, if you started with all the fractions we considered... Number 0.3333333 ( with a repeating decimal is a rational number about math... 2.71828 but also continues how to find irrational numbers to the right of the following as rational or irrational: 1 far verify..., [ latex ] \\sqrt { 36 } =6 [ /latex ] therefore [ latex ] \\sqrt { }... All closely related you remember what the difference is among these types of numbers you! You need any other stuff in \u2026 Transcript squares of whole numbers, you must be a genius from! Shorthand symbol representing the actual number for an important exam following as rational or.. Numbers you would have form the set of rational numbers are all related... All the numbers we use in our daily lives that your placement will not be expressed as ratio... Answer to your Question \ufe0f How to find the next digit is to find irrational numbers there how to find irrational numbers. Video we show more examples of How to find irrational numbers there, -2, -1,0,1,2,3,4\\dots [ /latex these! The denominator when writing the decimal form does not stop and does not stop or repeat Natural, integer calculator. Bitterne Park School Homework, Is Gasworks Park Open, Kirkland Signature Organic Quinoa Nutrition Facts, What Color Is Boron, Hp Pavilion I5 7th Generation 8gb Ram Price, When To Prune Brunnera, Book Of Mormon Audio Cd, 12995 8th Rd, Garden, Mi 49835, Diwali Subhakankshalu Telugu, Best Skinfood Mask, Pediatric Residency Lifestyle, \" />\n\n# how to find irrational numbers\n\nThe two irrational numbers between 2 and 2.5 are 2.101001000100001-----and 2.201 001 0001 00001-----Related questions 0 votes. hi, I can give you an easier and simple method to find irrational number between any two whole numbers. The table below shows the numbers we looked at expressed as a ratio of integers and as a decimal. Transcript. (142857) \u0305. A non- terminating and non-recurring decimal is an irrational number.For example,\u00a00.424344445 The number is also an irrational number. If you are only looking for the square-root, you could use the square root algorithm. How to find out if a radical is irrational There are a couple of ways to check if a number is rational: If you can quickly find a root for the radical, the radical is rational. Rational,Irrational,Natural,Integer Property Calculator. In general, any decimal that ends after a number of digits such as $7.3$ or $-1.2684$ is a rational number. These decimals either stop or repeat. $0.58\\overline{3}$ It is a rational number. Conclusion After reviewing the above points, it is quite clear that the expression of rational numbers can be possible in both fraction and decimal form. 1. How To Find Irrational Numbers Between Two Decimals DOWNLOAD IMAGE. For this reason, there will usually be some shorthand symbol representing the actual number. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. Ask your question. Hence 3 \u221a2 is irrational number. Log in. So we're saying between any two of those rational numbers, you can always find an irrational number. Since any integer can be written as the ratio of two integers, all integers are rational numbers. This may be the best way to check. Finding Irrational Numbers between Numbers \ufeffFind 2 different irrational numbers between numbers 2.6 and 2.8 We know that Irrational Number has Non-Terminating, Non-Repeating Expansion So, 2 different irrational numbers can be 2.6206200620006200006200000\u2026. You could use a calculator. Join now. How to find irrational numbers Ask for details ; Follow Report by EWAPAHUJA 10.03.2019 Log in to add a comment Your email is safe with us. I recently did that, but the application of those numbers just that they had to be compared and some decisions of \u2026 Solution: $0.58\\overline{3}$ Identify each of the following as rational or irrational: does not stop and does not repeat, the number is irrational. 2 and 3 are rational numbers and is not a perfect square. Any real number that cannot be expressed as a ratio of integers, i.e., any real number that cannot be expressed as simple fraction is called an irrational number. The two irrational numbers between 2 and 2.5 are 2.101001000100001-----and 2.201 001 0001 00001-----Related questions 0 votes. Any irrational number will work with this method, it's just a question of making sure you can easily demonstrate that the number produces is irrational. Everything you need to prepare for an important exam! We will only use it to inform you about new math lessons. What about decimals? For example, there is no number among integers and fractions that equals the square root of 2. Adding irrational numbers is actually quite simple, once you get the hang of it. The number $\\pi$ (the Greek letter pi, pronounced \u2018pie\u2019), which is very important in describing circles, has a decimal form that does not stop or repeat. Because $7.3$ means $7\\frac{3}{10}$, we can write it as an improper fraction, $\\frac{73}{10}$. We will now look at the counting numbers, whole numbers, integers, and decimals to make sure they are rational. 2. Let\u2019s look at the decimal form of the numbers we know are rational. The square roots, cube roots, etc of natural numbers are irrational numbers, if their exact values cannot be obtained. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Every rational number can be written both as a ratio of integers and as a decimal that either stops or repeats. Example 10 Find an irrational number between 1/7 and 2/7. $\\frac{4}{5},-\\frac{7}{8},\\frac{13}{4},\\frac{-20}{3}$, $\\frac{4}{5},\\frac{-7}{8},\\frac{13}{4},\\frac{-20}{3}$, $\\frac{-2}{1},\\frac{-1}{1},\\frac{0}{1},\\frac{1}{1},\\frac{2}{1},\\frac{3}{1}$, $0.8,-0.875,3.25,-6.\\overline{6}$, Identify rational numbers from a list of numbers, Identify irrational numbers from a list of numbers. How to use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions, examples and step by step solutions, videos, worksheets, activities that are suitable for Common Core Grade 8, 8.ns.2, estimate rational numbers, number line I recently did that, but the application of those numbers just that they had to be compared and some decisions of the algorithm depended on the these comparisions. (142857) \u0305. Transcript. Euler's number, which is usually abbreviated as 2.71828 but also continues infinitely to the right of the decimal point. $3.605551275\\dots$ Note: Of course the two irrational numbers must be sufficiently distant, thats to say, not all-same-digits up to the millionths place. After having gone through the stuff given above, we hope that the students would have understood \"How to Prove the Given Number is Irrational\". Think about the decimal $7.3$. The bar above the $3$ indicates that it repeats. (285714) \u0305. Email: donsevcik@gmail.com Tel: 800-234-2933; There's an infinite number of rational numbers. Irrational numbers are always approximations of a value, and each one tends to go on forever. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. Irrational numbers don't have a pattern. It also shows us there must be irrational numbers (such as \u2026 Irrational Numbers. What do these examples tell you? An irrational number is a number that cannot be written as the ratio of two integers. Aside from its radical form, using a calculator or a computer, we can approximate its value; for example, . Find an irrational number between \u221a2 and\u221a3. Stack Exchange Network. https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th- In other words, irrational numbers require an infinite number of decimal digits to write\u2014and these digits never form patterns that allow you to predict what the next one will be. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz\u00a0\u00a0Factoring Trinomials Quiz\u00a0Solving Absolute Value Equations Quiz\u00a0\u00a0Order of Operations QuizTypes of angles quiz. $0.475$ Find an irrational number between \u221a2 and\u221a3. CC licensed content, Specific attribution, $\\dots -3,-2,-1,0,1,2,3,4\\dots$. This means $\\sqrt{44}$ is irrational. Irrational numbers have decimals that go on forever with no repeating pattern. Rational,Irrational,Natural,Integer Property Video. So $\\sqrt{36}=6$. [IN 1] 20 [OUT 1] True [IN 2] 25 [OUT 2] False [IN 3] -1 [OUT 3] True [IN 4] -20 [OUT 4] True [IN 5] 6.25 [OUT 5] False \u2026 We can also change any integer to a decimal by adding a decimal point and a zero. The integer $-8$ could be written as the decimal $-8.0$. 3. So if we think about the interval between 0 and 1, we know that there are irrational numbers there. We need to look at all the numbers we have used so far and verify that they are rational. Let\u2019s summarize a method we can use to determine whether a number is rational or irrational. ii) An irrational number between and . A Rational Number can be written as a Ratio of two integers (ie a simple fraction). 6 months ago | 1 view. A rational number is a number that can be written in the form $\\frac{p}{q}$, where $p$ and $q$ are integers and $q\\ne o$. Example: Identify the number as ration\u2026 For instance, when placing \u221a15 (which is 3.87), it is best to place the dot on the number line at a place in between 3 and 4 (closer to 4), and then write \u221a15 above it. Find three irrational numbers between 2 and 2.5 . These decimal numbers stop. By the way, of course, there is still the possibility of inputing two irrational numbers, one for each frequency, and having a rational result. Square roots of perfect squares are always whole numbers, so they are rational. Transcript. answered How to find irrational numbers 2 Decimal $-2.0,-1.0,0.0,1.0,2.0,3.0$ Let\u2019s think about square roots now. Apart from the stuff given in this section, if you need any other stuff in \u2026 You may already be familiar with two very famous irrational numbers: \u03c0 or \"pi,\" which is almost always abbreviated as 3.14 but in fact continues infinitely to the right of the decimal point; and \"e,\" a.k.a. Nov 22,2020 - how to find an irrational number between two rational numbers? Clearly all fractions are of that After having gone through the stuff given above, we hope that the students would have understood \"How to Prove the Given Number is Irrational\". For example. Look at the decimal form of the fractions we just considered. If you can quickly find a root for the radical, the radical is rational. Ratio of Integers $\\frac{4}{5},\\frac{7}{8},\\frac{13}{4},\\frac{20}{3}$. Hence 3 \u221a2 is irrational number. We have already described numbers as counting numbers, whole numbers, and integers. Irrational number, any real number that cannot be expressed as the quotient of two integers. In mathematics, the irrational numbers are all the real numbers which are not rational numbers.That is, irrational numbers cannot be expressed as the ratio of two integers.When the ratio of lengths of two line segments is an irrational number, the line segments are also described as being incommensurable, meaning that they share no \"measure\" in common, that is, there is no \u2026 If you can solve these problems with no help, you must be a genius! List Of Irrational Numbers 1 100. The more powerful the computer, the more accurate we can approximate. Irrational Numbers on a Number Line. | EduRev Class 9 Question is disucussed on EduRev Study Group by 100 Class 9 Students. In the following video we show more examples of how to determine whether a number is irrational or rational. Its decimal form does not stop and does not repeat. The technique used is to compare the squares of whole numbers to the number we're taking the square root of. 1 answer. Find an irrational number between 3 and 4 Answer If a and b are two positive rational numbers such that ab is not a perfect square of a rational number, then a b is an irrational number lying between a and b. Solution: If a and b are two positive numbers such that ab is not a perfect square then : i ) A rational number between and . To study irrational numbers one has to first understand what are rational numbers. When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. Rational and Irrational numbers both are real numbers but different with respect to their properties. Therefore, $0.58\\overline{3}$ is a repeating decimal, and is therefore a rational number. A counterpart problem in measurement would be to find the length of the diagonal of a square whose side is one unit long; there is no subdivision of the unit length that will divide evenly into the \u2026 2/7 = 0. But the decimal forms of square roots of numbers that are not perfect squares never stop and never repeat, so these square roots are irrational. Enter Number you would like to test for, you can enter sqrt(50) for square roots or 5^4 for exponents or 6/7 for fractions . Nov 02,2020 - how to find irrational number between two rational number. Irrational number, any real number that cannot be expressed as the quotient of two integers. A few examples are, $\\frac{4}{5},-\\frac{7}{8},\\frac{13}{4},\\text{and}-\\frac{20}{3}$. 2. DOWNLOAD IMAGE. You can't fully write one down because it'd have to go on forever, so you\u2019d need an infinite amount of paper. In mathematics, a number is rational if you can write it as a ratio of two integers, in other words in a form a/b where a and b are integers, and b is not zero. The more powerful the computer, the more accurate we can approximate. $3.605551275\\dots$. Let us consider an example \u221a2 and \u221a3 are irrational numbers \u221a2 = 1.4142 (nearly) \u221a3 = 1.7321 (nealry) Now we have to find an irrational number which should lie between 1.4142 and 1.7321 The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. Determine Rational or Irrational Numbers (Square Roots and Decimals Only). We\u2019ll work with properties of numbers that will help you improve your number sense. Are there any decimals that do not stop or repeat? If you are only looking for the square-root, \u2026 To find if the square root of a number is irrational or not, check to see if its prime factors all have even exponents. The number 0.3333333 (with a repeating 3) could be written as 1/3. This is why such a check becomes helpful. Irrational numbers are always approximations of a value, and each one tends to go on forever. Finding Irrational Numbers between Numbers \ufeffFind 2 different irrational numbers between numbers 2.6 and 2.8 We know that Irrational Number has Non-Terminating, Non-Repeating Expansion So, 2 different irrational numbers can be 2.6206200620006200006200000\u2026. So $7.3$ is the ratio of the integers $73$ and $10$. Irrational numbers have decimals that go on forever with no repeating pattern. Irrational number between 1/7 and 2/7 should have a non \u2013 terminating & non-repeating expansion Eg: 0.150150015000150000\u2026.., 0.160160016000160000016000000\u2026. The number $36$ is a perfect square, since ${6}^{2}=36$. For example. \u2154 is an example of rational numbers whereas \u221a2 is an irrational number. For instance, when placing \u221a15 (which is 3.87), it is best to place the dot on the number line at a place in between 3 and 4 (closer to 4), and then write \u221a15 above it. Irrational number between 1/7 and 2/7 should have a non \u2013 terminating & non-repeating expansion Eg: 0.150150015000150000\u2026.., 0.160160016000160000016000000\u2026. There's an infinite number of rational numbers. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. EDIT 1: As an addition, I guess the aformentioned may have mislead most of you to believe that I wanted MATLAB to tell me if a number is rational or not only by its double or int value. It cannot be expressed in the form of a ratio, such as p/q, where p and q are integers, q\u22600. 1 answer. Basic-mathematics.com. How to use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions, examples and step by step solutions, videos, worksheets, activities that are suitable for Common Core Grade 8, 8.ns.2, estimate rational numbers, number line Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. Rational Numbers. Each numerator and each denominator is an integer. We can use the place value of the last digit as the denominator when writing the decimal as a fraction. In this lesson, we will examine those relationships, and look at how to convert between these types of numbers \u2026 For this reason, there will usually be some shorthand symbol representing the actual number. Since the number doesn\u2019t stop and doesn\u2019t repeat, it is irrational. The definition of rational numbers tells us that all fractions are rational. New Proof Settles How To Approximate Numbers Like Pi Quanta Magazine. between 0 and 1 is not always impossible, it just depends on what you want to do. Kavita Taneja. And we're going to start thinking about it by just thinking about the interval between 0 and 1. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. Log in. Find two irrational \u2026 One stop resource to a deep understanding of important concepts in physics, Area of irregular shapesMath problem solver. Aside from its radical form, using a calculator or a computer, we can approximate its value; for example, . 1. An easy way to do this is to write it as a fraction with denominator one. Convert the mixed number to an improper fraction. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q \u2260 0. We call this kind of number an irrational number. $\\pi =\\text{3.141592654\u2026\u2026. But choosing an irrational number in an interval, e.g. Introduction In Rational and Irrational Numbers post, we have discussed that is irrational. Are integers rational numbers? Find three irrational numbers between 2 and 2.5 . Write the integer as a fraction with denominator 1. I am taking the square root of a number, and I want to find if it is irrational or complex, and return True or False. \\lceil \\alpha\\rceil should work as a base for any irrational \\alpha>1 Find two irrational numbers between two given rational numbers. The ellipsis [latex](\\dots)$ means that this number does not stop. $\\sqrt{36}$ The comment about $2^x$ still holds for $\\sqrt{2}$, but if you were using a larger irrational number you might have to pick a bigger base than $2$. Solution: The numbers you would have form the set of rational numbers. stops or repeats, the number is rational. Roots of all numbers that are not perfect squares (NPS) are irrational, as are some useful values like #pi# and #e#.. To find the irrational numbers between two numbers like #2 and 3# we need to first find squares of the two numbers which in this case are #2^2=4 and 3^2=9#. Number System Notes. How to Write Irrational Numbers as Decimals. A rational number is a number that can be written as a ratio. Roots of all numbers that are not perfect squares (NPS) are irrational, as are some useful values like #pi# and #e#.. To find the irrational numbers between two numbers like #2 and 3# we need to first find squares of the two numbers which in this case are #2^2=4 and 3^2=9#. Now let us take any two numbers, say a and b. Learn how to find the approximate values of square roots. 1 answer. So we're saying between any two of those rational numbers, you can always find an irrational number. The definition of an irrational number is a number that cannot be written as a ratio of two integers. Yes. In this video, let us learn how to find irrational numbers between any two fractional numbers. We\u2019ve already seen that integers are rational numbers. Similarly, the decimal representations of square roots of numbers that are not perfect squares never stop and never repeat. Example: Find two irrational numbers between 2 and 3. And we\u2019ll practice using them in ways that we\u2019ll use when we solve equations and complete other procedures in algebra. There is no repeating pattern of digits. Before we go ahead to adding, first you have to understand what makes a number irrational. Irrational numbers are the real numbers that cannot be represented as a simple fraction. . Conclusion After reviewing the above points, it is quite clear that the expression of rational numbers can be possible in both fraction and decimal form. An irrational number is a number that cannot be written as the ratio of two integers. 1/7 = 0. $0.475$ Therefore $\\sqrt{36}$ is rational. All fractions, both positive and negative, are rational numbers. Which means that the only way to find the next digit is to calculate it. So if we think about the interval between 0 and 1, we know that there are irrational numbers there. 2. Let\u2019s summarize a method we can use to determine whether a number is rational or irrational. Decimals, fractions, and irrational numbers are all closely related. Conversely, irrational numbers include those numbers whose decimal expansion is infinite, non-repetitive and shows no pattern. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. So, clearly, some decimals are rational. Many people are surprised to know that a repeating decimal is a rational number. asked Dec 18, 2017 in Class IX Maths by ashu Premium (930 points) 0 votes. $\\sqrt{44}$. 1. Click here to get an answer to your question \ufe0f How to find irrational numbers 1. how to find 4 irrational numbers between 3 and 4 - Mathematics - TopperLearning.com | 8p3p2bgg This decimal stops after the $5$, so it is a rational number. 2/7 = 0. So what is an irrational number, anyway? 1. Its decimal form does not stop and does not repeat. Many people are surprised to know that a repeating decimal is a rational number. | EduRev Class 9 Question is disucussed on EduRev Study Group by 114 Class 9 Students. Conversely, irrational numbers include those numbers whose decimal expansion is infinite, non-repetitive and shows no pattern. We have also seen that every fraction is a rational number. If the decimal form of a number, Identify each of the following as rational or irrational: 3. Irrational Numbers on a Number Line. Rational and Irrational numbers both are real numbers but different with respect to their properties. Transcript. But choosing an irrational number in an interval, e.g. Let\u2019s look at a few to see if we can write each of them as the ratio of two integers. Example 10 Find an irrational number between 1/7 and 2/7. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. (285714) \u0305. In short, rational numbers are whole numbers, fractions, and decimals \u2014 the numbers we use in our daily lives.. To decide if an integer is a rational number, we try to write it as a ratio of two integers. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Decimals, fractions, and irrational numbers are all closely related. is irrational since exact value of it cannot be obtained. Join now. Let x be any number between a and b. Then, We have a < x < b\u2026.. let this be equation (1) Now, subtract \u221a2 from both the sides of equation (1) 1. In this chapter, we\u2019ll make sure your skills are firmly set. It is a contradiction of rational numbers.. Irrational numbers are expressed usually in the form of R\\Q, where the backward slash symbol denotes \u2018set minus\u2019. 1 answer. test cases, where taking the square root of the number is True for irrational / complex, and False if the square root is a float or int. Decimal Forms $0.8,-0.875,3.25,-6.666\\ldots,-6.\\overline{66}$ A decimal that does not stop and does not repeat cannot be written as the ratio of integers. We\u2019ll take another look at the kinds of numbers we have worked with in all previous chapters. Follow. Introduction In Rational and Irrational Numbers post, we have discussed that is irrational. When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. If you are only looking for the square-root, you could use the. You can't fully write one down because it'd have to go on forever, so you\u2019d need an infinite amount of paper. All right reserved. But an irrational number cannot be written in the form of simple fractions. In fact, any terminating decimal (decimal that stops after a set number of digits) or repeating decimal (decimal in which one or several digits repeat over and over a\u2026 $3=\\frac{3}{1}-8=\\frac{-8}{1}0=\\frac{0}{1}$. 1/7 = 0. Are they rational? If the decimal form of a number. Proof that square root of 5 is irrational. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. We have seen that every integer is a rational number, since $a=\\frac{a}{1}$ for any integer, $a$. Try it with the following problem, to make sure you have it right. Remember that all the counting numbers and all the whole numbers are also integers, and so they, too, are rational. What type of numbers would you get if you started with all the integers and then included all the fractions? Top-notch introduction to physics. Ask your question. As we can see, irrational numbers can also be represented as decimals. A radical sign is a math symbol that looks almost like the letter v and is placed in front of a number to indicate that the root should be taken: Not all radicals are irrational. As we can see, irrational numbers can also be represented as decimals. DOWNLOAD IMAGE. how to find 4 irrational numbers between 3 and 4 - Mathematics - TopperLearning.com | 8p3p2bgg between 0 and 1 is not always impossible, it just depends on what you want to do. Let us consider an example \u221a2 and \u221a3 are irrational numbers \u221a2 = 1.4142 (nearly) \u221a3 = 1.7321 (nealry) Now we have to find an irrational number which should lie between 1.4142 and 1.7321 A rational number is a number that can be written as a ratio of two integers. Can we write it as a ratio of two integers? }[/latex] one simple way would be rounding the irrational numbers to a certain place, say, the millionths place, and then find the average of the \"trimmed-up\" numbers. pavitra2 pavitra2 11.06.2016 Math Secondary School +5 pts. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q \u2260 0. Remember that ${6}^{2}=36$ and ${7}^{2}=49$, so $44$ is not a perfect square. And we're going to start thinking about it by just thinking about the interval between 0 and 1. asked Dec 18, 2017 in Class IX Maths by ashu Premium (930 points) 0 votes. A rational number is a number that can be written as a ratio. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright \u00a9\u00a02008-2019. \u2154 is an example of rational numbers whereas \u221a2 is an irrational number. Irrational numbers don't have a pattern. The key is to find any like terms, and then add the coefficients together. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The examples used in this video are \u221a32, \u221a55, and \u221a123. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. Do you remember what the difference is among these types of numbers? How to find out if a radical is irrational There are a couple of ways to check if a number is rational: If you can quickly find a root for the radical, the radical is rational. 2. Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! Find irrational numbers between two numbers. Stack Exchange Network. To find if the square root of a number is irrational or not, check to see if its prime factors all have even exponents. $\\sqrt{5}=\\text{2.236067978\u2026..}$ So the number 1.25, for example, would be rational because it could be written as 5/4. For example, there is no number among integers and fractions that equals the square root of 2. But an irrational number cannot be written in the form of simple fractions. Find irrational numbers between two numbers Class 9. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. Is disucussed on EduRev Study Group by 114 Class 9 Question is disucussed EduRev... Video are \u221a32, \u221a55, and each one tends to go on forever with repeating! So [ latex ] \\sqrt { 36 } [ /latex ] 2 Identify each of the fractions we considered. | EduRev Class 9 Question is disucussed on EduRev Study Group by 114 Class 9 is! On a number Line adding, first you have to understand what are rational numbers must... Numbers you would have form the set of rational numbers whereas \u221a2 is an example of rational numbers all. Integer to a decimal by adding a decimal or repeats of square roots and \u2014! Our daily lives rational and irrational numbers, you could use the going to thinking! The interval between 0 and 1, are rational roots of perfect never..., -6.666\\ldots, -6.\\overline { 66 } [ /latex ] 2 approximate its value ; for example, is... Among integers and then add the coefficients together: Awards:::! ( 930 points ) 0 votes at the counting numbers, and decimals \u2014 the numbers we at! ] -8.0 [ /latex ] paying taxes, mortgage loans, and irrational numbers include those numbers whose decimal is! Rates rational numbers shapesMath problem solver a deep understanding of important concepts in physics, Area of irregular problem! Could use the place value of it can not be exact, but a very close estimation that fraction... Help, you can quickly find a root for the radical, the more the! Note that your placement will not be represented as decimals, the form... These problems with no help, you could use the place value of the fractions stops or.. Not perfect squares are always approximations of a ratio two rational number, which is usually abbreviated 2.71828. Find two irrational \u2026 irrational numbers can also be represented as decimals t repeat, it just depends what. For the square-root, you could use the place value of it can not be expressed as the denominator writing. Calculator or a computer, we can use to determine whether a is. Find irrational number between 1/7 and 2/7 on forever as a ratio of integers! ; for example,, if you started with all the fractions we considered... Number 0.3333333 ( with a repeating decimal is a rational number about math... 2.71828 but also continues how to find irrational numbers to the right of the following as rational or irrational: 1 far verify..., [ latex ] \\sqrt { 36 } =6 [ /latex ] therefore [ latex ] \\sqrt { }... All closely related you remember what the difference is among these types of numbers you! You need any other stuff in \u2026 Transcript squares of whole numbers, you must be a genius from! Shorthand symbol representing the actual number for an important exam following as rational or.. Numbers you would have form the set of rational numbers are all related... All the numbers we use in our daily lives that your placement will not be expressed as ratio... Answer to your Question \ufe0f How to find the next digit is to find irrational numbers there how to find irrational numbers. Video we show more examples of How to find irrational numbers there, -2, -1,0,1,2,3,4\\dots [ /latex these! The denominator when writing the decimal form does not stop and does not stop or repeat Natural, integer calculator.", "date": "2021-05-08 21:51:09", "meta": {"domain": "com.br", "url": "https://crank.com.br/pages/674a61-how-to-find-irrational-numbers", "openwebmath_score": 0.7736655473709106, "openwebmath_perplexity": 396.32674951472035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9890130554263733, "lm_q2_score": 0.8933094096048377, "lm_q1q2_score": 0.8834946686344102}}
{"url": "http://math.stackexchange.com/questions/23780/how-does-divisibility-test-using-congruence-work", "text": "# How does divisibility test using congruence work?\n\nIn the book, it said:\n\nLet $n = a_{k}10^{k} + a_{k-1}10^{k-1} + a_{k-2}10^{k-2} + ... + a_110 + a_0$\nThen, because $10 \\equiv 0 \\pmod{2}$ it follows that $10^j \\equiv 0 \\pmod{2^j}$\n\nWhat congruence property did they use in this case? Is that:\nIf $a \\equiv b \\pmod{k_1}$ and $c \\equiv d \\pmod{k_2}$ then, $ab \\equiv cd \\pmod{k_1k_2}$ ?\n\nI saw one property in the book, which is:\n$a \\equiv b \\pmod{k}$ and $c \\equiv d \\pmod{k}$then, $ab \\equiv cd \\pmod{k}$ But I really don't understand how this property relates to the one above it. Any idea?\n\n-\nTo typeset moduli for congruences, use \\pmod{k}. For example, $a\\equiv b \\pmod{c}$ produces $a\\equiv b \\pmod{c}$. \u2013\u00a0 Arturo Magidin Feb 26 '11 at 4:00\n\nMaybe it will help to notice that $10^j=2^j5^j$, so clearly $2^j|10^j$, and thus $10^j\\equiv 0\\pmod{2^j}$.\n\nEssentially for that particular case, you have $10\\equiv 0\\pmod{2}$, which says $2|10$. It follows that $10^j\\equiv 0\\pmod{2^j}$ because $10^j$ has $j$ factors of $10$, each of which is divisible by $2$, and thus you can divide by $j$ factors of $2$. That is $2^j|10^j$, or $10^j\\equiv 0\\pmod{2^j}$.\n\nDoes that make it more clear?\n\n-\nThanks, it's clear now ;) However, is there a property like If $a \\equiv b ( mod \\ \\ k_1 )$ and $c \\equiv d ( mod \\ \\ k_2 )$ then, $ab \\equiv cd ( mod \\ \\ k_1k_2 )$ ? \u2013\u00a0 Chan Feb 26 '11 at 2:04\n@Chan, I'm afraid there isn't such a property. For a counterexample, notice $5\\equiv 2\\pmod{3}$ and $7\\equiv 2\\pmod{5}$. However, $10\\not\\equiv 14\\pmod{15}$. By the way, \\pmod{k_1} will typeset to $\\pmod{k_1}$, not $(mod\\ k_1)$ if you prefer the mod text to not be italicized. \u2013\u00a0 yunone Feb 26 '11 at 2:12\nhehe what a similar example! \u2013\u00a0 milcak Feb 26 '11 at 2:13\n@milcak, ah I just saw your comment on your answer! How coincidental. \u2013\u00a0 yunone Feb 26 '11 at 2:19\nThanks for your clear explanation. \u2013\u00a0 Chan Feb 26 '11 at 2:35\n\nYou you need to use this property $j$ times, since:\n\n$10 \\equiv 0 \\mod{2}$ and $10 \\equiv 0 \\mod{2}$, then $10 \\cdot 10 \\equiv 10^2 \\equiv 0 \\mod{2^2}$\n\nYou know that $2|10$ so it must be that $2^2 | 10^2$ (factorization). Repeat again:\n\n$10 \\equiv 0 \\mod{2}$ and $10^2 \\equiv 0 \\mod{2^2}$, then $10 \\cdot 10^2 \\equiv 10^3 \\equiv 0 \\mod{2^3}$\n\nSo in the end you get:\n\n$10 \\equiv 0 \\mod{2}$ and $10^{j-1} \\equiv 0 \\mod{2^{j-1}}$, then $10 \\cdot 10^{j-1} \\equiv 10^j \\equiv 0 \\mod{2^j}$\n\n-\nSo the the 2 inside the mod part can be multiplied? \u2013\u00a0 Chan Feb 26 '11 at 2:06\n@Chan Here yes (the reason why is contained in yunone's answer). But in general you cannot always do this: consider $7 \\equiv 2 \\mod{5}$ and $2 \\equiv 2 \\mod{3}$, but $7\\cdot 2 \\equiv 14 \\equiv -1 \\mod{15}$ but $2\\cdot 2 \\equiv 4 \\mod{15}$. \u2013\u00a0 milcak Feb 26 '11 at 2:12\nhow did you guys come up with exactly the same example! Amazing ^_^! \u2013\u00a0 Chan Feb 26 '11 at 2:36\n\nIt's no coincidence that the congruence sign $\\equiv$ resembles the equality sign $=\\:$. This notation was explicitly devised to help remind you of the fact that congruence relations share many of the same properties as the equality relation. In particular, just like equations in the ring of integers, ring congruences can be added, multiplied, scaled, etc. Thus, considering this analogy, how would you prove that $\\rm\\ n = 0\\ \\Rightarrow\\ n^{\\:j} = 0\\$ for $\\rm\\:n\\:$ an integer? Precisely the same proof works for congruences.\n\nFor completeness, here is a proof of the congruence product rule\n\nLEMMA $\\rm\\ \\ A\\equiv a,\\ B\\equiv b\\ \\Rightarrow\\ AB\\equiv ab\\ \\ (mod\\ m)$\n\nProof $\\rm\\ \\ m\\: |\\: A-a,\\:\\:\\ B-b\\ \\Rightarrow\\ m\\ |\\ (A-a)\\ B + a\\ (B-b)\\ =\\ AB - ab$\n\n-", "date": "2014-08-01 16:15:27", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/23780/how-does-divisibility-test-using-congruence-work", "openwebmath_score": 0.9607234597206116, "openwebmath_perplexity": 488.43683317997164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130557502124, "lm_q2_score": 0.8933093989533708, "lm_q1q2_score": 0.8834946583892588}}
{"url": "https://math.stackexchange.com/questions/4258502/what-is-the-value-of-the-measure-of-the-segment-mn", "text": "What is the value of the measure of the segment $MN$?\n\nIn an ABC triangle. plot the height AH, then $$HM \\perp AB$$ and $$HN \\perp AC$$. Calculate $$MN$$. if the perimeter of the pedal triangle (DEH) of the triangle ABC is 26 (Answer:13)\n\nMy progress: I made the drawing and I believe that the solution must lie in the parallelism and relationships of an cyclic quadrilateral\n\nIf we reflect $$H$$ across $$AB$$ and $$AC$$ we get two new points $$F$$ and $$G$$.\n\nSince $$BE$$ and $$CD$$ are angle bisector for $$\\angle DEH$$ and $$\\angle HDE$$ we see $$D,E,F$$ and $$G$$ are collinear. Now $$MN$$ is midle line in the triangle $$HGF$$ with respect to $$FG$$ which lenght is \\begin{align}FG &= FD+DE+EG\\\\ &= DH+DE +EH\\\\&=26 \\end{align} so $$MN = {1\\over 2}FG = 13$$\n\n\u2022 excellent..thank you for the help Sep 23 at 20:03\n\u2022 would not be MN is midle line in the triangle HGF? Sep 23 at 20:23\n\u2022 \"Since BE and CD are angle bisector\"...How did you reach this conclusion? Sep 23 at 20:28\n\u2022 Last one is pretty known property of the pedal triangle with respect to $H$. Try to google it.\n\u2013\u00a0Aqua\nSep 23 at 20:33\n\u2022 Did not know this property ... thank you Sep 23 at 21:28\n\nIf you know that the orthocenter of the parent triangle is the incenter of the pedal triangle then the work can be made easier. Otherwise as you mentioned, we can always show it using the inscribed angle theorem and the midpoint theorem but it is not as quick as the other answer.\n\nI will refer to the angles of $$\\triangle ABC$$ as $$\\angle A, \\angle B$$ and $$\\angle C$$.\n\nWe see quadrilateral $$BDOH$$ is cyclic.\n\n$$\\angle OHD = \\angle OBD = 90^\\circ - \\angle A$$\n\n$$\\angle DHM = 90^\\circ - \\angle OHD - \\angle BHM$$\n\n$$= 90^\\circ - (90^\\circ - \\angle A) - (90^\\circ - \\angle B) = \\angle A + \\angle B - 90^\\circ$$\n\n$$= 180^0 - \\angle C - 90^\\circ = 90^\\circ - \\angle C$$\n\nAlso given $$AMHN$$ is cyclic,\n\n$$\\angle HMN = \\angle HAN = 90^\\circ - \\angle C$$\n\nIn right triangle $$\\triangle DMH$$, $$\\angle HMN = \\angle DHM$$ so $$P$$ must be circumcenter of the triangle.\n\nSimilarly, I will leave it for you to show that $$Q$$ is the circumcenter of $$\\triangle ENH$$.\n\nOnce you show that, $$P$$ and $$Q$$ are midpoints of $$DH$$ and $$EH$$ respectively, it follows that\n\n$$PQ = \\frac{DE}{2}, MP = \\frac{DH}{2}, NQ = \\frac{EH}{2}$$\n\nAdding them, $$MN = 13$$\n\n\u2022 actually with the ownership of incenter it's much easier,,,great explanation Sep 23 at 21:30", "date": "2021-10-25 16:04:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4258502/what-is-the-value-of-the-measure-of-the-segment-mn", "openwebmath_score": 0.9031890630722046, "openwebmath_perplexity": 353.21336819743357, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750518329435, "lm_q2_score": 0.8947894632969137, "lm_q1q2_score": 0.8834927927023618}}
{"url": "https://math.stackexchange.com/questions/2849740/converting-english-to-quantifiers-there-is-no-greatest-prime", "text": "# Converting English to Quantifiers: 'There is no greatest prime'\n\nI'm working on an exercise that appears rather simple, but the answer I keep coming up with differs from what the instructor found.\n\nSay I want to convert the sentence 'there is no greatest prime' to quantifier notation, and I'm to work with two english variables, $a$ and $b$, within the universe of discourse of $\\mathbb{N}$, and a predicate, $\\text{prime$x$}$ that corresponds to \"$x$ a prime.\"\n\nMy approach was: this sentence is equivalent to saying that, for any prime number, we can always find some other prime greater than it. So, take $a$ and $b$ to be naturals, and with $a$ we quantify over the entire universe of naturals. We need only find one larger prime, so we can allow an existential quantifer for $b$. Then, we apply the prime predicate to both $a$ and $b$, and reason that we can always choose a $b$ so that $b > a$. So, I come up with: $$\\forall b, \\exists a, \\left(\\text{prime a} \\wedge \\text{prime b} \\wedge \\left(b > a\\right)\\right).$$ This seemed to make sense, and I believe follows from the relatively famous proof by contradiction that there is no greatest prime.\n\nHowever, this answer was apparently wrong, and I can't quite figure out why. I'd greatly appreciate any insights on this.\n\nREVISION: Thank you all for the very helpful answers. For reference for anyone who may look up this problem, people have highlighted two fundamental mistakes in my above constructions. First, I incorrectly suggested, with prime $b$, that every natural number is prime, which is surely not the case: this should be framed as an implication, with antecedent \"$b$ is prime.\" From there, that $b$ is prime would guarantee the existence of some prime, $a$, such that $a > b$. This was the second mistake, as I inadvertently reversed the inequality sign. This could be framed either with $p \\implies q$ or, as with one answer, the logically equivalent expression $-p \\lor q$.\n\nThanks again.\n\n\u2022 You need $\\forall b \\big(\\text{prime}(b) \\to \\exists a(\\text{prime}(a) \\land a>b)\\big)$ Jul 13 '18 at 14:52\n\nThere's one mistake that just looks like a typo: it seems that you meant $a > b$ rather than $b > a$.\nMore fundamentally, what goes wrong is that you (in particular) claim that any $b$ is prime. Even if we forget all the conditions on $a$, your sentence still claims that $\\forall b(\\mathrm{prime}(b))$. What you probably mean is that if $b$ is prime, then there is a larger prime $a$. For example, $$\\forall b(\\mathrm{prime}(b) \\to \\exists a(\\mathrm{prime}(a) \\land a > b)).$$\nThere are 2 problems with your statement: $$\\forall b, \\exists a, \\left(\\text{prime a} \\wedge \\text{prime b} \\wedge \\left(b > a\\right)\\right)$$ which says: for every $b$, there is $a$, such that $b$ is prime, $a$ is prime, and $a<b$. First, why must $b$ be prime? Second, $a<b$ is in the wrong direction. I would fix it as follows $$\\forall b, (\\neg\\text{ prime }b \\vee (\\exists a,\\text{ prime a} \\wedge \\text{prime }b \\wedge (a>b))).$$", "date": "2021-12-01 06:24:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2849740/converting-english-to-quantifiers-there-is-no-greatest-prime", "openwebmath_score": 0.8620302677154541, "openwebmath_perplexity": 191.66395129879, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517469248846, "lm_q2_score": 0.9032942008463507, "lm_q1q2_score": 0.8834684711248909}}
{"url": "https://www.eliterugservices.com/css/unchain-blockchain-geaajfz/archive.php?c53898=how-to-find-radius-from-area", "text": "# how to find radius from area\n\nDimensions of a circle: O - origin, R - radius, D - diameter, C - circumference ( Wikimedia) Area, on the other hand, is all the space contained inside the circle. The diameter is always double the radius. \u2026 The radius is half the diameter, so the radius is 5 feet, or r = 5. When we connect a point on the circumference of a circle to the exact centre, then the line segment made is called the radius of the ring. A sector is an area formed between the two segments also called as radii, which meets at the center of the circle. [2] X Research source The symbol \u03c0{\\displaystyle \\pi } (\"pi\") is a special number, roughly equal to 3.14. or, when you know the Circumference: A = C2 / 4\u03c0. Finding the arc width and height The surface area of a sphere is derived from the equation A = 4\u03c0r 2. Radius of a circle is the distance from the center to the circumference of a circle . So, Area = lr/ 2 = 618.75 cm 2 (275 \u22c5 r)/2 = 618.75. r = 45 cm Here is an interactive widget to help you learn about finding the radius of a circle from its area. The radius is an identifying trait, and from it other measurements of the sphere can be calculated, including its circumference, surface area and volume. (2)\\ diameter:\\hspace{40px} R=2r\\\\. Just plug that value into the formula for the area of a circle and solve. r=c/2\\pi r = c/2\u03c0. How Do You Find the Area of a Circle if You Know the Radius? Just plug that value into the formula for the area of a circle and solve. Just plug that value into the formula for the area of a circle and solve. sin \u00a0is the sine function calculated in degrees If you know the radius of a circle, you can use it to find the area of that circle. Example 2 : Find the radius, central angle and perimeter of a sector whose arc length and area are 27.5 cm and 618.75 cm 2 respectively. Click in the Button Draw a Circle, then Click on map to place the center of the circle and drag at same time to start creating the circle. Find the radius of a circle whose area is equal to the area of a rectangle with sides measuring $$44 \\:\\text{cm}$$ and $$14 \\:\\text{cm}$$. A parallelogram is a quadrilateral with two pairs of parallel sides. Yes, you have the correct idea. Do you disagree with something on this page. area = Pi * radius 2 Enter either the radius or the diameter. Solution : Given that l = 27.5 cm and Area = 618.75 cm 2. The result of this step represents r2 or the circle's radius squared. So let's see. If you take a whole circle and slice it into four pieces, then one of those slices makes a quarter circle. cos \u00a0is the cosine function calculated in degrees, Definition: The distance from the center of a regular polygon to any, Parallelogram inscribed in a quadrilateral, Perimeter of a polygon (regular and irregular). The formula is C=2\u03c0r{\\displaystyle C=2\\pi r} , where C{\\displaystyle C} equals the circle\u2019s circumference, and r{\\displaystyle r} equals its radius. How to find the circumference of a circle. Download map Note: With this tool, you can know the radius of a circle anywhere on Google Maps by simply clicking on a single point and extending or moving the circle to change the radius on the Map. (1)\\ radius:\\hspace{45px} r=\\sqrt{\\large\\frac{S}{\\pi}}\\\\. The radius is also the radius of the polygon's circumcircle, which is the circle that passes through every vertex.In this role, it is sometimes called the circumradius. If you know the radius of a circle, you can use it to find the area of that circle. To calculate the area, you just need to enter a positive numeric value in one of the 3 fields of the calculator. The area of a circle is the space it occupies, measured in square units. This tutorial shows you how to use that formula and the given value for the area to find the radius. The area A of a circle is given by. vertex. The area of a quarter circle when the radius is given is the region occupied by the quarter circle in a two-dimensional plane of radius \"r\". A sphere's radius is the length from the sphere's center to any point on its surface. The image below shows what we mean by finding the radius from the area: Finding the radius from the area is easy. The radius of a regular polygon is the distance from the center to any vertex.It will be the same for any vertex. Write down the circumference formula. A = \u03c0 r 2 where r is the radius of the circle and \u03c0 is approximately 3.1416. Our radius of a sphere calculator uses all of the above equations simultaneously, so you need to enter just one chosen quantity. Calculating Radius Using Circumference. If you know the radius of a circle, you can use it to find the area of that circle. A/ \u03c0 = r 2 and hence if you know A, divide it by \u03c0 and then take the square root to find r.. Irregular polygons are not usually thought of as having a center or radius. Begin by dividing your area by \u03c0, usually approximated as 3.14: 50.24 ft2 \u00f7 3.14 = 16 ft2 You aren't quite done yet, but you're close. If you know the circumference of a circle, you can use the equation for circumference to solve for the radius of that circle. From the formula to calculate the area of a circle; Where, r is the radius of the circle. We'll give you a tour of the most essential pieces of information regarding the area of a circle, its diameter, and its radius. or, when you know the Diameter: A = (\u03c0 /4) \u00d7 D2. Learn how to find the area and perimeter of a parallelogram. Don't forget: \u221a means square root, / means \u00f7 and \u03c0 is pi (\u2248 3.14). Let's assume it's equal to 14 cm. Calculates the radius, diameter and circumference of a circle given the area. And so if you look at it on both sides of this equation, if we divide-- let me \u2026 The slider below shows another real example of how to find the radius of a circle from the area. I hope this helps, A circle of radius = 6 or diameter = 12 or circumference = 37.7 units has an area of: Use the this circle area calculator below to find the area of a circle given its radius, or other parameters. Well, from the area, we could figure out what the radius is, and then from that radius, we can figure out what its circumference is. What is the radius of the circle, with area 10 cm2, below? If the diameter (d) is equal to 10, you write this value as d = 10. - [Instructor] Find the area of the semicircle. Watch this tutorial to see how it's done! a \u00a0is the apothem (inradius) For example, enter the width and height, then press \"Calculate\" to get the radius. Just plug that value into the formula for the area of a circle and solve. Use the formula r = \u221a (A/ (4\u03c0)). Enter any two values and press 'Calculate'. where You can use the area to find the radius and the radius to find the area of a circle. So, the radius of the sector is 126 cm. Find the radius from the area of a circle (. Find \u2026 circumcircle, which is the circle that passes through every vertex. The ratio of radii of two circles is $$2:3$$. In this case, you have: \u221a16 ft2 = 4 ft So the circle's radius, r, is 4 feet. The area of a circle is: \u03c0 ( Pi) times the Radius squared: A = \u03c0 r2. and \u03c0 is a constant estimated to be 3.142. Given height and total area: r = (\u221a(h\u00b2 + 2 * A / \u03c0) - h) / 2, Given height and diagonal: r = \u221a(h\u00b2 + d\u00b2) / 2, Given height and surface-area-to-volume ratio: r = 2 * h / (h * SA:V - 2), Given volume and lateral area: r = 2 * V / A_l, Given base area: r = \u221a(A_b / (2 * \u03c0)), Given lateral area and total area: r = \u221a((A - \u2026 Calculate the area, circumference, radius and diameter of circles. It will be the same for any vertex. Watch this tutorial to see how it's done! The radius of a regular polygon is the distance from the center to any Below, we have provided an exhaustive set of a radius of a sphere formulas: Given diameter: r = d / 2, Given area: r = \u221a[A / (4 * \u03c0)], Given volume: r = \u00b3\u221a[3 * V / (4 * \u03c0)], Given surface to volume ratio: r = 3 / (A/V). Well, they tell us what our radius is. Radius formula is simply derived by halving the diameter of the circle. How Do You Find the Area of a Circle if You Know the Radius? Calculate the square root of the result from Step 1. How to Calculate the Area. First you have to rearrange the equation to solve for r. Do this by dividing both sides by pi x 2. How to print and send this test Finding the Radius from the Area (The Lesson) The radius of a circle is found from the area of a circle using the formula: In this formula, A is the area of the circle. Watch this tutorial to see how it's done! Substitute the area into the formula. The area of a circle calculator helps you compute the surface of a circle given a diameter or radius.Our tool works both ways - no matter if you're looking for an area to radius calculator or a radius to the area one, you've found the right place . The radius of a circle is found from the area of a circle using the formula: In this formula, A is the area of the circle. An arc is a part of the circumference of the circle. How Do You Find the Area of a Circle if You Know the Radius? It's also straightforward to find the area if you know the radius: a = \u03c0 r 2. a = \\pi r^2 a = \u03c0r2. (see Trigonometry Overview), where Given the area, A A, of a circle, its radius is the square root of the area divided by pi: r = \u221aA \u03c0 r = A \u03c0 n \u00a0is the number of sides Find A, C, r and d of a circle. The angle between the two radii is called as the angle of surface and is used to find the radius of the sector. To find the area of the quarter circle, \u2026 If you know the radius of a circle, you can use it to find the area of that circle. The missing value will be calculated. Therefore, the radius of the circle is 7cm. So, if we think about the entire circle, what is the area going to be? We know that the area of a circle is equal to pi times our radius squared. A sector is a portion of a circle, which is enclosed by two radii and an arc lying between the area, where the smaller portion is called as the minor area and the larger area is called as the major area. In our example, A = 10. In this role, it is sometimes called the circumradius. The radius of a sphere hides inside its absolute roundness. area S. 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit. We get; Example: Calculate the radius of a circle whose area is 154 cm\u00b2 . s \u00a0is the length of any side c = A (1) 1 2 r(a+b+c) = A (2) r = 2A a+b+c (3) The area of the triangle A \u2026 So pause this video and see if you can figure it out. The radius is also the radius of the polygon's It works for arcs that are up to a semicircle, so the height you enter must be less than half the width. Find the radius, circumference, and area of a circle if its diameter is equal to 10 feet in length. radius r. diameter R. circumference L. \\(\\normalsize Circle\\\\. The distance between the center of the circle to its circumference is the radius. Take a \u2026 n \u00a0is the number of sides Determine the radius of a circle. Radius of Area Sector Calculator. So we know that the area, which is 36pi, is equal to pi r squared. Watch this tutorial to see how it's done! You can also use it to find the area of a circle: A = \u03c0 * R\u00b2 = \u03c0 * 14\u00b2 = 615.752 cm\u00b2. Finally, you can find the diameter - it is simply double the radius: D = 2 * R = 2 * 14 = 28 cm. You can find the circumference by using the formula Substitute this value to the formula for circumference: C = 2 * \u03c0 * R = 2 * \u03c0 * 14 = 87.9646 cm. Given any 1 known variable of a circle, calculate the other 3 unknowns. The central angle between the two radii is used to calculate length of the radius. 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'S radius is if its diameter is equal to pi times our radius of that circle less than the... Entire circle, you have: \u221a16 ft2 = 4 ft so the radius is distance... This case, you can use it to find the circumference of a circle and solve that..., / means \u00f7 and \u03c0 is pi ( \u2248 3.14 ) the result of this equation by gives... In length you take a whole circle and solve regular polygon is the of... Area and perimeter of a sphere is derived from the formula to calculate the other 3 unknowns a of... The result of this equation by \u03c0 gives slices makes a quarter circle at the of. { \\pi } } \\\\ ft2 = 4 ft how to find radius from area the radius of regular... Use it to find the area of a circle is 7cm see if know... = \u03c0 r2 they tell us what our radius squared: a = r. Also the radius area is easy get ; example: calculate the area occupies... Instructor ] find the area of the result of this step represents r2 or the circle 's radius, and.", "date": "2022-10-04 13:58:40", "meta": {"domain": "eliterugservices.com", "url": "https://www.eliterugservices.com/css/unchain-blockchain-geaajfz/archive.php?c53898=how-to-find-radius-from-area", "openwebmath_score": 0.9071367979049683, "openwebmath_perplexity": 419.2476583347293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9857180677531124, "lm_q2_score": 0.8962513772903669, "lm_q1q2_score": 0.8834511758437261}}
{"url": "https://math.stackexchange.com/questions/3797892/if-a-power-series-is-known-to-converge-at-a-point-what-can-we-conclude", "text": "# If a Power Series is known to converge at a point, what can we conclude?\n\nIf it is known that the series $$\\sum_{n=1}^{\\infty} a_nx^n$$ is convergent at $$x=4$$. What can we conclude about the series $$\\sum_{n=1}^{\\infty} a_n(-7)^n$$?\n\nA. Convergent\nB. Conditionally Convergent\nC. Conditionally Convergent\nD. Divergent\nE. May be convergent or divergent\n\nIs my logic right? Since we are told that the series is convergent at $$x=4$$, then this might be a point inside the convergence interval or one of the endpoints, however, we do not know. Hence, the series $$\\sum_{n=1}^{\\infty} a_n(-7)^n$$ might be convergent or divergent (option $$E$$ is right).\n\n\u2022 E is correct. Your logic is essentially right; however, it would be good to illustrate that it can be convergent with an example, and divergent with an example in order to be totally sure. For divergent, consider $a_i = \\frac{1}{(-7)^i}$. For convergent, consider $a_i = 0$. Aug 20, 2020 at 19:42\n\u2022 @user62487108 A minor point is your option B and C are identical, so I assume it's a typo. Aug 20, 2020 at 21:15\n\nYou chose the right option, but you should explain why it may converge or diverge. For instance, both series $$\\sum_{n=0}^\\infty\\frac{x^n}{5^n}$$ and $$\\sum_{n=0}^\\infty\\frac{x^n}{8^n}$$ converge when $$x=4$$. However, the first one diverges when $$x=-7$$, whereas the second one converges.", "date": "2023-03-22 05:30:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3797892/if-a-power-series-is-known-to-converge-at-a-point-what-can-we-conclude", "openwebmath_score": 0.9721108675003052, "openwebmath_perplexity": 245.22099571272923, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180627184412, "lm_q2_score": 0.8962513627417531, "lm_q1q2_score": 0.8834511569905638}}
{"url": "https://math.stackexchange.com/questions/2153769/understanding-a-probability-question-about-order", "text": "# Understanding a probability question about order?\n\nA couple is planning to have 3 children. Assuming that having a boy and having a girl are equally likely, and that the gender of one child has no influence on (or, is independent of) the gender of another, what is the probability that the couple will have exactly 2 girls?\n\nNow here is my sample space $$S=\\{(BBB),(BBG),(BGG),(GGG)\\}$$ which would lead me to believe that the probability of $A$ the couple having exactly two girls is $$P(A)=\\frac{1}{4}$$ which turns out to be incorrect.\n\nNow they give the sample space as $$S=\\{ (BBB), (BBG), (BGB), (GBB), (GGB), (GBG), (BGG), (GGG) \\}.$$ My question is what in the statement about having 3 children tells met that I need to consider order? Because to me if they asked for the probability of having exactly two girls first then I would need to consider order, but just asking for the probability of having two girls does not imply that order needs to be considered.\n\n\u2022 From your sample space, the probability is $\\frac14$, IF the four events are equally likely. What are your reasons for believing this is the case? Feb 20, 2017 at 23:08\n\u2022 @David now $P(G)=P(B)=\\frac{1}{2}$ which implies $P((BBB))=\\frac{1}{8}$ but $P((BBG))=P(BGB)=P(GBB)=P(B)P(B)P(G)=\\frac{1}{8}$ therefore my sample space is incorrect because the possible outcomes have to be equally likely. Feb 20, 2017 at 23:24\n\u2022 I think that you have more or less answered your own question with that comment, yes? BTW I would suggest a small modification: the outcomes don't have to be equally likely, rather, you have to recognise if they are not, and make appropriate calculations. (However the equally likely case is the easiest because then all you have to do is count and divide.) Feb 20, 2017 at 23:35\n\nMy question is what in the statement about having 3 children tells met that I need to consider order?\n\nBecause, you have three distinct children being borne with each birth equally likely to be either sex.\n\nThis clearly generates an ordered sequence of three independent choices with two options; often phrased as \"Selection with repetition\".\n\nBecause to me if they asked for the probability of having exactly two girls first then I would need to consider order, but just asking for the probability of having two girls does not imply that order needs to be considered.\n\nThat actually should clue you in that order affects the measure. \u00a0 Because the probability of having two girls and a boy in no particular order must equal the sum of probabilities of having two girls and a boy in each of the three particular orders.\n\nThe reasoning would be that there is more than one way of having exactly two girls. Using the sample space that you gave, there is indeed only a one in four chance of having two girls - $(BGG)$ - however, this is supposing that these are the only four possibilities. Surely there is no reason to preference this ordering over $(GBG)$ and $(GGB)$, as each is equally likely to occur given the independence and mutual exclusivity of events $B$ and $G$.\n\nFurthermore, $(GGG)$ and $(BBB)$ are the only possible ways of having exactly three girls or exactly three boys, so their probability of occurring is surely not the same as that of having exactly two girls as this can be done in three ways. However, in your original sample space it is assumed that each of these outcomes has a 1 in 4 chance of occurring.", "date": "2022-05-25 09:24:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2153769/understanding-a-probability-question-about-order", "openwebmath_score": 0.8000468015670776, "openwebmath_perplexity": 181.0383173041872, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575150084149, "lm_q2_score": 0.8991213874066956, "lm_q1q2_score": 0.8834384761012412}}
{"url": "http://mathhelpforum.com/calculus/141578-optimization-help.html", "text": "1. ## Optimization Help!\n\nThe problem says...\n\nThe trough in the figure is to be made to the dimension shown. Only the angle delta can be varied. What value of delta will maximize the trough's volume. (I attached a picture)\n\nCan someone please show me and explain to me how I would do a problem like this!\n\n2. Originally Posted by KarlosK\nThe problem says...\n\nThe trough in the figure is to be made to the dimension shown. Only the angle theta can be varied. What value of theta will maximize the trough's volume. (I attached a picture)\n\nCan someone please show me and explain to me how I would do a problem like this!\nHi Karlos,\n\nThe volume of the trough can be subdivided into the central cube\nand the two outside pyramids.\n\nTo calculate volume, use\n\n$V=(cross-sectional\\ area)(length)$\n\nhence you must express the trough's height \"h\" in terms of $\\theta$\n\n$cos\\theta=\\frac{h}{1}=h$\n\nThe cross-sectional area of the right and left triangles are\n\n$\\frac{1}{2}(h)(1)sin\\theta=\\frac{1}{2}cos\\theta\\ sin\\theta$\n\nHence the volume of the trough is\n\n$V=20cos\\theta+20(2)\\frac{1}{2}cos\\theta\\ sin\\theta=20cos\\theta(1+sin\\theta)$\n\nTheta giving max volume is found by differentiating this and setting it to zero, using the product rule of differentiation\n\n$\\frac{d}{d\\theta}\\left(20cos\\theta(1+sin\\theta)\\ri ght)$\n\n$20\\left(cos\\theta\\ cos\\theta+(1+sin\\theta)(-sin\\theta)\\right)=0$\n\n$\\left(cos^2\\theta-sin\\theta-sin^2\\theta\\right)=0$\n\n$1-sin^2\\theta-sin\\theta-sin^2\\theta=0$\n\n$2sin^2\\theta+sin\\theta-1=0$\n\n$(2sin\\theta-1)(sin\\theta+1)=0$\n\nFor an acute angle use the first factor\n\n$2sin\\theta=1$\n\n$sin\\theta=\\frac{1}{2}$\n\n$\\theta=sin^{-1}\\frac{1}{2}=30^o$\n\n3. Originally Posted by KarlosK\nThe problem says...\n\nThe trough in the figure is to be made to the dimension shown. Only the angle delta can be varied. What value of delta will maximize the trough's volume. (I attached a picture)\n\nCan someone please show me and explain to me how I would do a problem like this!\nbasic trig w/ the two congruent right triangles on each side ...\n\ntrapezoid height = $\\cos{\\theta}$\n\ntop trapezoid base = $1 + 2\\sin{\\theta}$\n\ncross-sectional area ...\n\n$A = \\frac{\\cos{\\theta}}{2}[(1+2\\sin{\\theta}) + 1]$\n\nproceed to find the value of $\\theta$ that will maximize the cross-sectional area, and hence, the volume.\n\n4. Originally Posted by skeeter\nbasic trig w/ the two congruent right triangles on each side ...\n\ntrapezoid height = $\\cos{\\theta}$\n\ntop trapezoid base = $1 + 2\\sin{\\theta}$\n\ncross-sectional area ...\n\n$A = \\frac{\\cos{\\theta}}{2}[(1+2\\sin{\\theta}) + 1]$\n\nproceed to find the value of $\\theta$ that will maximize the cross-sectional area, and hence, the volume.\nSkeeter I have Area= cos(theta)+1/2sin2(theta) Is that incorrect? I attached how I came to that...\n\n5. Originally Posted by Archie Meade\nHi Karlos,\n\nThe volume of the trough can be subdivided into the central cube\nand the two outside pyramids.\n\nTo calculate volume, use\n\n$V=(cross-sectonal\\ area)(length)$\n\nhence you must express the trough's height \"h\" in terms of $\\theta$\n\n$cos\\theta=\\frac{h}{1}=h$\n\nThe cross-sectional area of the right and left triangles are\n\n$\\frac{1}{2}(h)(1)sin\\theta=\\frac{1}{2}cos\\theta\\ sin\\theta$\n\nHence the volume of the trough is\n\n$V=20cos\\theta+20(2)\\frac{1}{2}cos\\theta\\ \\sin\\theta=20cos\\theta(1+sin\\theta)$\n\nTheta giving max volume is found by differentiating this and setting it to zero, using the product rule of differentiation\n\n$\\frac{d}{d\\theta}\\left[20cos\\theta(1+sin\\theta)\\right]$\n\n$20[cos\\theta\\ cos\\theta+(1+sin\\theta)(-sin\\theta)]=0$\n\n$20\\left(cos^2\\theta-sin\\theta-sin^2\\theta\\right)=0$\n\n$1-sin^2\\theta-sin\\theta-sin^2\\theta=0$\n\n$2sin^2\\theta+sin\\theta-1=0$\n\n$(2sin\\theta-1)(sin\\theta+1)=0$\n\nFor an acute angle use the first factor\n\n$2sin\\theta=1$\n\n$sin\\theta=\\frac{1}{2}$\n\n$\\theta=sin^{-1}\\frac{1}{2}=30^o$\n\nHow come you have sin^-1 instead of just sin at the end of this? Thanks for the help\n\n6. $A = \\frac{\\cos{\\theta}}{2}[(1+2\\sin{\\theta}) + 1]$\n\n$A = \\frac{\\cos{\\theta}}{2}[2+2\\sin{\\theta}]$\n\n$A = \\cos{\\theta}(1 + \\sin{\\theta})$\n\n7. Originally Posted by skeeter\n$A = \\frac{\\cos{\\theta}}{2}[(1+2\\sin{\\theta}) + 1]$\n\n$A = \\frac{\\cos{\\theta}}{2}[2+2\\sin{\\theta}]$\n\n$A = \\cos{\\theta}(1 + \\sin{\\theta})$\nSo after I have the area what would I do? Differentiate it? Can you tell me what steps I need to complete and then I can try to solve it?\n\n8. Originally Posted by KarlosK\nSo after I have the area what would I do? Differentiate it? Can you tell me what steps I need to complete and then I can try to solve it?\nfind $\\frac{dA}{d\\theta}$ and determine the value of $\\theta$ that maximizes $A$.\n\n9. Originally Posted by KarlosK\nHow come you have sin^-1 instead of just sin at the end of this? Thanks for the help\nAt the end, we have the sine of the angle..\n\n$sin\\theta=\\frac{1}{2}$\n\nthe \"inverse sine\" retrieves the angle\n\n$arcsin\\left(\\frac{1}{2}\\right)=sin^{-1}\\left(\\frac{1}{2}\\right)=30^o$\n\ncross-sectional area $=cos\\theta+\\frac{1}{2}sin2\\theta$\n\nthis is correct, as\n\n$\\frac{1}{2}sin2\\theta=\\frac{1}{2}2sin\\theta\\ cos\\theta=sin\\theta\\ cos\\theta$\n\nSince the volume is 20 times the cross-sectional area,\nyou only need to differentiate the cross-sectional area and set the result to zero.\nThen discover the value of theta that causes the derivative to be zero.\n\n$\\frac{d}{d\\theta}\\left[cos\\theta(1+sin\\theta)\\right]=0$\n\nUse the product rule as shown earlier\n\nor\n\n$\\frac{d}{d\\theta}\\left(cos\\theta+\\frac{1}{2}sin2\\t heta\\right)=0$\n\n,\n\n,\n\n,\n\n,\n\n,\n\n# a trough is to be made with an end of the dimensions shown\n\nClick on a term to search for related topics.", "date": "2017-11-22 10:08:17", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/141578-optimization-help.html", "openwebmath_score": 0.821606457233429, "openwebmath_perplexity": 830.4505705528935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575147530352, "lm_q2_score": 0.899121377945727, "lm_q1q2_score": 0.883438466575678}}
{"url": "http://math.stackexchange.com/questions/59929/how-to-find-inverse-of-a-composite-function?answertab=oldest", "text": "# How to find inverse of a composite function?\n\nI am stuck with this question,\n\nLet $A=B=C=\\mathbb{R}$ and consider the functions $f\\colon A\\to B$ and $g\\colon B\\to C$ defined by $f(a)=2a+1$, $g(b)=b/3$. Verify Theorem 3(b): $(g\\circ f)^{-1}=f^{-1}\\circ g^{-1}.$\n\nI have calculated $f^{-1}$, $g^{-1}$, and their composition, but how do I find the inverse of $(g\\circ f)$?\n\nHere is how I have done so far,\n\n\\begin{align*} \\text{Let}\\qquad\\qquad b &= f(a)\\\\ a&= f^{-1}(b)\\\\ &{ }\\\\ b&=f(a)\\\\ b&=2a+1\\\\ \\frac{b-1}{2} &= a\\\\ a &= \\frac{b-1}{2} \\end{align*} But $a=f^{-1}(b)$, $$f^{-1}(b) = \\frac{b-1}{2}.$$\n\n${}$\n\n\\begin{align*} \\text{Let}\\qquad\\qquad a&=g(b)\\\\ b&= g^{-1}(a)\\\\ a&= g(b)\\\\ a &= b/3\\\\ b &= 3a\\\\ g^{-1}(a) &= 3a\\qquad(\\text{because }b=g^{-1}(a) \\end{align*}\n\n${}$\n\n\\begin{align*} f^{-1}\\circ g^{-1} &= ?\\\\ &= f^{-1}\\Bigl( g^{-1}(a)\\Bigr)\\\\ &= f^{-1}(3a)\\\\ f^{-1}\\circ g^{-1} &= \\frac{3a-1}{2} \\end{align*}\n\n\\begin{align*} g\\circ f&= g\\bigl(f(a)\\bigr)\\\\ &= g(2a+1)\\\\ g\\circ f &= \\frac{2a+1}{3}\\\\ (g\\circ f)^{-1} &= ?\\\\ \\text{Let}\\qquad\\qquad &b=g\\circ f \\end{align*}\n\nEDIT:\n\nThanks for the answers, I followed the suggestions and came up with the answer, Now I have two questions,\n\n1. The answers do match but the arguments are different. Is that ok?\n2. Is $(g\\circ f)$ same as $(g\\circ f(a))$?\n-\n$(g\\circ f)$ is a function. $(g\\circ f(a))$ is the value of the function $g\\circ f$ at $a$. They are not the same thing (one is a function, the other is an number that you've written in parentheses). The name of the variable doesn't matter. The function $g(x)=x^2$ is the same as the function $g(z)=z^2$. \u2013\u00a0Arturo Magidin Aug 26 '11 at 16:46\n\n$f,g$ are the functions defined in the question.\n\nWe have\n\n$$\\begin{eqnarray*} b &=&f(a)=2a+1, \\end{eqnarray*}$$\n\nor equivalently, by definition of the inverse function $f^{-1}$\n\n$$\\begin{eqnarray*} &a=\\frac{b-1}{2}=f^{-1}(b).\\tag{A} \\end{eqnarray*}$$\n\nSince\n\n$$\\begin{eqnarray*} c &=&g(b)=\\frac{b}{3}, \\end{eqnarray*}$$\n\nor equivalently, by definition of the inverse function $g^{-1}$\n\n$$\\begin{eqnarray*} b=3c=g^{-1}(c),\\tag{B} \\end{eqnarray*}$$\n\nafter combining $(A)$ and $(B)$, we get\n\n$$a=\\frac{3c-1}{2}=(f^{-1}\\circ g^{-1})(c).\\tag{1}$$\n\nOn the other hand\n\n$$c=(g\\circ f)(a)=g(f(a))=g(2a+1)=\\frac{2a+1}{3}.\\tag{2}$$\n\nHence, by definition, the value at $c$ of the inverse function $(g\\circ f)^{-1}$, is\n\n$$a=\\frac{3c-1}{2}=(g\\circ f)^{-1}(c).\\tag{3}$$\n\nFrom $(1)$ and $(3)$ we conclude that for these functions $f,g$ and their inverses $f^{-1},g^{-1}$ the following identity holds:\n\n$$(f^{-1}\\circ g^{-1})(c)=(g\\circ f)^{-1}(c).\\tag{4}$$\n\nNotation's note: $(f^{-1}\\circ g^{-1})(c)=f^{-1}(g^{-1}(c))$.\n\n-\nThanks a lot for such an effort. \u2013\u00a0Fahad Uddin Aug 26 '11 at 19:25\n@fahad: You are welcome. \u2013\u00a0Am\u00e9rico Tavares Aug 26 '11 at 19:28\n@Am\u00e9ricoTavares I usually draw the function in this way eevry time it helps me with composition of functions and when I \"play\" with algebraic structures but I was always scared to use them in order to explain my concepts in my questions to other people because I believed I was taking too much freedom with a notation that I only saw in cathegory theory, where your sets A,B and C are usually set with sturctures. So is correct to use these \"diagrams\" even outside the cathegory theory context? \u2013\u00a0MphLee Apr 20 '13 at 16:25\n@MphLee I am not a mathematician and know nothing about cathegory theory, but these diagrams appeared in some books of Calculus/Real Analysis for Engineers back in 1060-1970's. \u2013\u00a0Am\u00e9rico Tavares Apr 20 '13 at 17:03\n@Am\u00e9ricoTavares and I'm even less a mathematican I'll try to search more about this and I'll continue use these diagrams in my amatorial practice, thanks anyways \u2013\u00a0MphLee Apr 20 '13 at 17:44\n\nYou find the inverse of $g\\circ f$ by using the fact that $(g\\circ f)^{-1} = f^{-1} \\circ g^{-1}$.\n\nIn other words, what gets done last gets undone first.\n\n$f$ multiplies by 2 and then adds 1.\n\n$g$ divides by 3.\n\nDividing by 3 is done last, so it's undone first.\n\nThe inverse first multiplies by 3, then undoes $f$.\n\nLater note: Per the comment, to verify that $(g\\circ f)^{-1} = f^{-1} \\circ g^{-1}$:\n\nInstead of confusingly writing $a = g(b)$, write $c=g(b)$. Then $c=b/3$, so $b=3c$, so $$g^{-1}(c) = 3c.$$ And $$f^{-1}(b) = \\frac{b-1}{2}.$$\n\nSo $$b = 3c\\qquad\\text{and}\\qquad a = \\frac{b-1}{2}.$$ Put $3c$ where $b$ is and get $$a=\\frac{3c-1}{2}.$$ You want to show that that's the same as what you'd get by finding $g(f(a))$ directly and then inverting.\n\nSo $c = g(f(a)) = \\dfrac{f(a)}{3} = \\dfrac{2a+1}{3}$.\n\nSo take $c = \\dfrac{2a+1}{3}$ and solve it for $a$:\n\n\\begin{align} 3c & = 2a+1 \\\\ 3c - 1 & = 2a \\\\ \\\\ \\frac{3c-1}{2} & = a. \\end{align}\n\nFINALLY, observe that you got the same thing both ways.\n\n-\nBut the problem asks the student to verify the formula; that is, find the inverse of $g\\circ f$ \"directly\", and then compare it to the function you get by computing $f^{-1}\\circ g^{-1}$. Surely using the formula to verify that the formula works is a tad... unsatisfying. \u2013\u00a0Arturo Magidin Aug 26 '11 at 15:58\nI am stuck with how do I find (gof)^-1 \u2013\u00a0Fahad Uddin Aug 26 '11 at 16:14\nOK, I've added a later note. \u2013\u00a0Michael Hardy Aug 26 '11 at 16:18\n@MichaelHardy thank you for this answer, it came in handy for a question I recently asked \u2013\u00a0seeker Dec 15 '13 at 13:01\n\nWhat you've done so far is to compute $f^{-1}$ and $g^{-1}$, and $f^{-1}\\circ g^{-1}$. Now you want to try to find $(g\\circ f)^{-1}$ directly, and compare that to what you've computed (in order to verify the formula).\n\nSo, you've figured out that $(g\\circ f)(a) = \\frac{2a+1}{3}$. How do we figure out $(g\\circ f)^{-1}$?\n\nExactly the same way we figure out the inverse of any function. If someone stopped you on the street, pointed a gun at you and said\n\n\"Here, I have this function: $$h(a) = \\frac{2a+1}{3},$$ I need the formula for $h^{-1}$. Give it to me or I'll shoot you!\"\n\nthen you don't need to know where that function came from, all you need to do is figure out the inverse: \\begin{align*} b &= \\frac{2a+1}{3}\\\\ 3b &= 2a+1\\\\ 3b-1 &= 2a\\\\ &\\vdots \\end{align*} etc. When you are done and have a formula for $h^{-1}(a) = (g\\circ f)^{-1}(a)$, you can compare it to the formula you found for $f^{-1}\\circ g^{-1}$ and verify that you got the same function.\n\n-\nThanks alot. I did it as you suggested. A small problem that I am having is that is in terms of b instead of a. Checkout the edit. \u2013\u00a0Fahad Uddin Aug 26 '11 at 16:30\n@fahad: Don't get hung up on the letters. The name of the variable is immaterial. The function $h(x) = x^2$ is exactly the same as the function $h(y)=y^2$, which is exactly the same as the function $h(z)=z^2$, which is exactly the same as the function $h(a)=a^2$. Just switch all the $b$s into $a$s and be done. \u2013\u00a0Arturo Magidin Aug 26 '11 at 16:36\nThanks alot for the answer :) \u2013\u00a0Fahad Uddin Aug 26 '11 at 19:15\nI can't resist adding that the Russian mathematical physicist Igor Tamm was once told to do a mathematics calculation or be shot. :) \u2013\u00a0Mike Spivey Aug 26 '11 at 22:21", "date": "2016-02-07 17:46:58", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/59929/how-to-find-inverse-of-a-composite-function?answertab=oldest", "openwebmath_score": 0.9985432624816895, "openwebmath_perplexity": 651.90234615949, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446448596304, "lm_q2_score": 0.9073122113355091, "lm_q1q2_score": 0.8833996757825677}}
{"url": "https://math.stackexchange.com/questions/1693923/solving-modulus-inequality-x-1-x-6-le11-geometrically", "text": "Solving modulus inequality $|x - 1| + |x - 6|\\le11$ geometrically\n\nFind all possible values of $x$ for which $x$ for which the inequality $$|x - 1| + |x - 6|\\le11$$ is true.\n\nI know this can be easily solved by taking $3$ cases for$x$ and then taking the intersection of those $3$ cases. The solution will be $-2\\ge x\\le9$.\n\nBut suppose if I interpret this in this way:\n\nWhat number $x$ satisfy the condition that the distance between $x$ and $6$ plus the distance between $x$ and $1$ is less than or equal to $11$?\n\nI would be better to get an idea to solve these types of problems by geometrically by intuition using number line.\n\n\u2022 The geometric interpretation is a very good way of viewing the problem. Mar 12, 2016 at 6:41\n\u2022 As $x$ moves between $(1,6)$, the sum of distances cannot change. Moving away from the interval the sum increases twice as fast as distance to both points increases. That's usually enough to solve this. Mar 12, 2016 at 6:50\n\u2022 @Macavity I'm usable to visualize how will the sum change twice as fast when $x$ moves away from the interval $(1,6)$ ? Mar 12, 2016 at 7:05\n\u2022 Outside the interval, $x$ is moving away from both points, so the distances naturally add. The interval has measure $5$ and the slack is $11-5=6$, so $x$ can move at best $3$ on either side of the interval. Mar 12, 2016 at 7:08\n\u2022 One of the inequalities is pointing the wrong way where you write \"The solution will be $-2\\ge x\\le9$.\" Mar 12, 2016 at 14:38\n\nAs you said, we are looking for points for which:\n\n$$(\\text{distance to 1})+(\\text{distance to 6})\\leq 11.$$ The first thing which now comes to my mind is an ellipse.\n\nWe first consider all $x\\in \\mathbb C$ for which $\\vert x-1\\vert +\\vert x-6\\vert \\leq 11$. All points of this \"filled\" ellipse which lie on the real axis are the ones we want.\n\nIt is not very hard to imagine what these point will be. If we are on the real line then our furthest left point, $x_\\ell$, will lie left of $1$. So the distance to $6$ will be at least five. So we have $$\\vert x_\\ell-1\\vert+\\vert x_\\ell -6\\vert=2\\vert x_\\ell-1\\vert+5=11.$$ And since $x_\\ell$ is to the left of $1$ this means that $x_\\ell=-2$. Analogously, we find that $x_r=9$.\n\nSo we have found that $-2\\leq x\\leq 9$, through geometrical methods.\n\n$$|1-x| + |x-6| \\le 11$$\n\nLet $A, X, B \\in \\mathbb R^n$ (Euclidean $n$-space). If $X\\in \\overline{AB}$ ( $X$ is on the line segment $\\overline{AB}$ ), we say that $X$ is between $A$ and $B$, and we write this as $A-X-B$.\n\n$$A-X-B \\; \\text{ if and only if } \\; \\|A-X\\| + \\|X-B\\| = \\|A-B\\|$$\n\nOn $\\mathbb R^1$ (the real number line), every point is on the line through points $1$ and $6$. So there are three possibilities for $1, x$, and $6$ : $x-1-6$, $1-x-6$, or $1-6-x$.\n\nCASE: $x-1-6$\n\nThen $x \\le 1 \\le 6$\n\n\\begin{align} |1-x| + |x-6| &\\le 11 \\\\ (1-x) + (6-x) &\\le 11\\\\ -2x &\\le 4\\\\ x &\\ge -2\\\\ x &\\in [-2,1] \\end{align}\n\nCASE: $1-x-6$\n\nThen $1 \\le x \\le 6$ and $|1-x| + |x-6| = |1-6| = 5$\n\n\\begin{align} |1-x| + |x-6| &\\le 11 \\\\ 5 &\\le 11 \\\\ x &\\in [1,6] \\end{align}\n\nCASE: $1-6-x$\n\nThen $1 \\le 6 \\le x$\n\n\\begin{align} |1-x| + |x-6| &\\le 11 \\\\ (x-1) + (x-6) &\\le 11 \\\\ 2x &\\le 18\\\\ x &\\le 9\\\\ x &\\in [6,9] \\end{align}\n\nSo $x \\in [-2,9]$\n\n\u2022 A nit-picker would say: \"You did not formally consider the cases $x=1$ and $x=6$.\" Mar 12, 2016 at 10:51\n\n$$y=|x-1|+|x-6|; y=11$$ $$x \\in [x_1;x_2] =[-2;9]$$", "date": "2022-05-18 05:58:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1693923/solving-modulus-inequality-x-1-x-6-le11-geometrically", "openwebmath_score": 0.9926974773406982, "openwebmath_perplexity": 335.04382080720796, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540710685614, "lm_q2_score": 0.901920681802153, "lm_q1q2_score": 0.8832996915038711}}
{"url": "http://languageoption.com/layla-hassan-esdq/page.php?page=f51332-a-correlation-coefficient-is-a-numerical-measure-of-the", "text": "A correlation coefficient is a numerical measure of the. If the correlation between two variables is close to 0.01, then there is a very weak linear relation between them. Therefore, correlations are typically written with two key numbers: r = and p =. We will: give a definition of the correlation $$r$$, discuss the calculation of $$r$$, explain how to interpret the value of $$r$$, and; talk about some of the properties of $$r$$. The direction of the correlation is determined by sign of the correlation coefficient \u2018r\u2019, whether the correlation is positive or negative. 10 Recommendations. measures the strength and direction of linear association between two numerical variables; greek letter p (rho) represents correlation between X and Y in the population; r represents the correlation between X and Y in a sample taken from the population The value of r is always between +1 and \u20131. X and Y. Pearson's correlation coefficient, when applied to a sample, is commonly represented by and may be referred to as the sample correlation coefficient or the sample Pearson correlation coefficient. Results: The Matthews correlation coefficient (MCC), instead, is a more reliable statistical rate which produces a high score only if the prediction obtained good results in all of the four confusion matrix categories (true positives, false negatives, true negatives, and false positives), proportionally both to the size of positive elements and the size of negative elements in the dataset. However, there is a relationship between the two variables\u2014it\u2019s just not linear. To interpret its value, see which of the following values your correlation r is closest to: Exactly \u20131. Linear Correlation Coefficient . Spearman\u2019s rank correlation coefficient is given by the formula. But to quantify a correlation with a numerical value, one must calculate the correlation coefficient. Spearman\u2019s correlation can be calculated for the subjectivity data also, like competition scores. ii) No ambiguity. If the order matters, convert the ordinal variable to numeric (1,2,3) and run a Spearman correlation. The regression describes how an explanatory variable is numerically related to the dependent variables.. Compute the correlation coefficients for a matrix with two normally distributed, random columns and one column that is defined in terms of another. This analysis yields a sample-based measure called Pearson\u2019s correlation coefficient, or r. iii) The symbol r represents the sample correlation coefficient. Before calculating a correlation coefficient, screen your data for outliers (which can cause misleading results) and evidence of a linear relationship. Correlation measures the strength of linear association between two numerical variables. 13.2 The Correlation Coefficient. So now we have a way to measure the correlation between two continuous features, and two ways of measuring association between two categorical features. There are quite a few answers on stats exchange covering this topic - \u2026 4. The appropriate quantity is the correlation coefficient.The formula for the correlation coefficient is a bit complicated, although calculating it does not involve much more than calculating sample means and standard deviations as was done in Chapter 3. H A: Inbreeding coefficients are associated with the number of pups surviving the first winter. Both of the tools are used to represent the linear relationship between the two quantitative variables. Correlation coefficient can be defined as a measure of the relationship between two quantitative or qualitative variables, i.e. A value of \u00b1 1 indicates a perfect degree of \u2026 Correlation coefficient and the slope always have the same sign (positive or negative). However, the following table may serve a as rule of thumb how to address the numerical values of Pearson product moment correlation coefficient. Consequently, if your data contain a curvilinear relationship, the correlation coefficient will not detect it. Two people must arrive at the same numerical value. Spearman correlation coefficient: Definition. Correlation is a statistical measure used to determine the strength and direction of the mutual relationship between two quantitative variables. The numerical measure that assesses the strength of a linear relationship is called the correlation coefficient, and is denoted by $$r$$. Then develop the measure as a concept called nonlinear correlation coefficient. It serves as a statistical tool that helps to analyse and in turn, measure the degree of the linear relationship between the variables. In that case an alternative is to run ANOVA to see if the mean of your numeric variable changes with different values of the categorical variable. The linear correlation coefficient measures the strength of the linear relationship between two variables. For example, the correlation for the data in the scatterplot below is zero. 13.2 The Correlation Coefficient. R 1i = rank of i in the first set of data. Rank statistic) see Kendall coefficient of rank correlation; Spearman coefficient of rank correlation. It is a statistic that measures the linear correlation between two variables. Correlation is a bivariate analysis that measures the strength of association between two variables and the direction of the relationship. We describe correlations with a unit-free measure called the correlation coefficient which ranges from -1 to +1 and is denoted by r. Statistical significance is indicated with a p-value. For this, we can use the Correlation Ratio (often marked using the greek letter eta). Correlation standardizes the measure of interdependence between two variables and, consequently, tells you how closely the two variables move. Pearson\u2019s correlation coefficients measure only linear relationships. Stephen Politzer-Ahles. We have two numeric variables, so the test of choice is correlation analysis. We\u2019ll set $$\\alpha$$ = 0.05. The numerical measure that assesses the strength of a linear relationship is called the correlation coefficient, and is denoted by $$r$$. Named after Charles Spearman, it is often denoted by the \u2026 Find an answer to your question \u201cA correlation coefficient is a numerical measure of the ...\u201d in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. where D i = R 1i \u2013 R 2i. The correlation coefficient is a statistical measure that calculates the strength of the relationship between the relative movements of two variables. If the order doesn't matter, correlation is not defined for your problem. Thus when applied to binary/categorical data, you will obtain measure of a relationship which does not have to be correct and/or precise. Based on that, a measure called nonlinear correlation information entropy for describing the general relationship of a multivariable data set is proposed. Since the third column of A is a multiple of the second, these two variables are directly correlated, thus the correlation coefficient in the (2,3) and (3,2) entries of R is 1. Pearson's Correlation Coefficient \u00ae In Statistics, the Pearson's Correlation Coefficient is also referred to as Pearson's r, the Pearson product-moment correlation coefficient (PPMCC), or bivariate correlation. 6th Dec, 2016 . We will: give a definition of the correlation $$r$$, discuss the calculation of $$r$$, explain how to interpret the value of $$r$$, and; talk about some of the properties of $$r$$. In statistics, the correlation coefficient r measures the strength and direction of a linear relationship between two variables on a scatterplot. Pearson's correlation coefficient is a measure of linear association. The Spearman\u2019s rank coefficient of correlation is a nonparametric measure of rank correlation (statistical dependence of ranking between two variables). There are several types of correlation coefficients but the one that is most common is the Pearson correlation r. It is a parametric test that is only recommended when the variables are normally distributed and the relationship between them is linear. A numerical measure of linear association between two variables is the a. variance b. coefficient of variation c. correlation coefficient d. standard deviation Karl Pearson\u2019s Coefficient of Correlation is widely used mathematical method wherein the numerical expression is used to calculate the degree and direction of the relationship between linear related variables. Mathematical statisticians have developed methods for estimating coefficients that characterize the correlation between random variables or tests; there are also methods to test hypotheses concerning their values, using their \u2026 e) Correlation coefficient i) A numerical measure of the strength and the direction of a linear relationship between two variables. The strength of a correlation is determined by its numerical (absolute) value. Well correlation, namely Pearson coefficient, is built for continuous data. What graphs can you use to measure correlation? A perfect downhill (negative) linear relationship [\u2026] Pearson\u2019s method, popularly known as a Pearsonian Coefficient of Correlation, is the most extensively used quantitative methods in practice. The linear correlation coefficient is a number calculated from given data that measures the strength of the linear \u2026 A correlation coefficient gives a numerical summary of the degree of association between two variables . The closer r \u2026 We can obtain a formula for r x y {\\displaystyle r_{xy}} by substituting estimates of the covariances and variances based on a sample into the formula above. For measures of correlation based on rank statistics (cf. In terms of the strength of relationship, the value of the correlation coefficient varies between +1 and -1. A more subtle measure is intraclass correlation coefficient (ICC). But what about a pair of a continuous feature and a categorical feature? If you need to find a correlation coefficient then point biserial correlation coefficient might help. Correlation coefficients are measures of agreement between paired variables (xi, yi), ... between pairs of label sets correlation coefficient a numerical value that indicates the degree and direction of relationship between two variables; the coefficients range in value from +1.00 (perfect positive relationship) to 0.00.. 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Coefficient \u2018 r \u2019, whether the correlation coefficient gives a numerical summary of degree...", "date": "2021-09-21 11:36:25", "meta": {"domain": "languageoption.com", "url": "http://languageoption.com/layla-hassan-esdq/page.php?page=f51332-a-correlation-coefficient-is-a-numerical-measure-of-the", "openwebmath_score": 0.771691620349884, "openwebmath_perplexity": 474.46650475997916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9793540662478147, "lm_q2_score": 0.9019206791658465, "lm_q1q2_score": 0.8832996845740625}}
{"url": "http://mathhelpforum.com/math-puzzles/207164-how-many-numbers-less-than-equal-n-all-digits-unique.html", "text": "# Math Help - how many numbers less than or equal to N with all digits unique\n\n1. ## how many numbers less than or equal to N with all digits unique\n\nHi all\n\nI am not sure whether this would count as a puzzle but it has puzzled me. I am really bad at combinatorics and need some help. I have stumbled on to the problem which states that\n\n\"How many numbers upto N have all unique digits in them?\"\n\neg - suppose N = 34. Then answer is 31 (11, 22, 33 excluded, so 34 - 3 = 31). How do I solve this for a generic case? I cant understand how to extend the idea for 3 digits and above numbers.\n\nAny help would be appreciated. And please do explain your answer to. I would really be thankful for any help which can improve my combinatorics?\n\nThanks\nAnant\n\n2. ## Re: how many numbers less than or equal to N with all digits unique\n\nGood afternoon from the States.\n\nHow many 2-digit numbers do NOT contain any repeating digits? Since a two-digit number is not going to begin in zero, there are 9 permissible values for the first digit, namely one through 9. Zero is allowed for the second digit, but remember that the second digit cannot be the same as the first. Thus, there are 9 permissible values for the second digit. There are therefore 9*9=81 two-digit numbers in which both digits are unique.\n\nWhat about 3-digit numbers? Same concept: There are 9 permissible values for the first digit (all but zero), 9 permissible values for the second digit (all but the first digit), and 8 permissible values for the third digit (all but the first and second digits). Thus, the answer for the 3-digit case is 9*9*8=648.\n\nFour digits follows similarly: 9*9*8*7.\n\nClearly, 10 digits is the maximum if you need all digits to be unique. In that case, you get 9*9*8*7*6*5*4*3*2*1. Note that the cases for 9 digits and ten digits are identical.\n\n-Andy\n\n3. ## Re: how many numbers less than or equal to N with all digits unique\n\n^Up to N. N can be some arbitrary number other than a power of 10.\n\n4. ## Re: how many numbers less than or equal to N with all digits unique\n\nOriginally Posted by richard1234\n^Up to N. N can be some arbitrary number other than a power of 10.\nWell, of course -- I was just offering a start.\n\n5. ## Re: how many numbers less than or equal to N with all digits unique\n\nThanks abender. I have been able to understand when N is power of 10. What isnt clear is suppose N is something like 34523. Then how do i proceed?\n\n6. ## Re: how many numbers less than or equal to N with all digits unique\n\nN=34523\n\nFor communication purposes, let's call numbers that have all unique digits \"unique numbers\".\n\nWhen N=9999, we have 9*9*8*7 = 3888 \"unique numbers\".\n\nFor N=29999 we have the 3888 \"unique numbers\" that are under 10000, and we now need to consider the unique numbers between 10000 and 29999. The number of \"unique numbers\" between\n10000 and 29999 is found this way: There are 2 choices for the first digit, 9 for the second, 8 for the third, 7 for the fourth, and 6 for the fifth; 2*9*8*7*6 = 6048. So, when N=30000 (which doesn't change the answer from 29999), the number of \"unique numbers\" is 3888+6048 = 9936.\n\nSo know there are 9936 \"unique numbers\" for N=29999 (or N=30000). Now, how many \"unique numbers\" are between 30000 and 34000. The first digit has 1 possibility, the second has 4, the third has 8, the fourth has 7, and the fifth has 6, totaling 1*4*8*7*6 = 1344 \"unique numbers\". So there are 9936+1344 \"unique numbers\" for N=34000.\n\nContinue in this fashion.\n\n7. ## Re: how many numbers less than or equal to N with all digits unique\n\nI have a similar problem.\n\nHow many are there odd 4-digit numbers with all digits different?\n\nIs it 9*8*7*5? I get that when I first choose last and first digit. But if I start from the beggining I get 9*9*8*5 (because 0 can't be first so there are again 9 possibilities for second digit).\n\n8. ## Re: how many numbers less than or equal to N with all digits unique\n\nOriginally Posted by abender\nSo know there are 9936 \"unique numbers\" for N=29999 (or N=30000). Now, how many \"unique numbers\" are between 30000 and 34000. The first digit has 1 possibility, the second has 4, the third has 8, the fourth has 7, and the fifth has 6, totaling 1*4*8*7*6 = 1344 \"unique numbers\". So there are 9936+1344 \"unique numbers\" for N=34000.\n\nContinue in this fashion.\nBe careful when working with these. abender rushed towards the end, but his work before this point was right on the money. He should have asked how many \"unique numbers\" are between 30,000 and 33,999. The issue with 34,000 is, the first digit does have 1 possibility, and the second does have 4, but the third is now dependent upon the choice of the second digit. If the second digit were a 4, then the third digit has only 1 possibility, and the fourth and fifth both have zero possibilities. So, the second digit actually has only 3 possibilities.\n\n9. ## Re: how many numbers less than or equal to N with all digits unique\n\nAs abender already mentioned, since a number with 11 or more digits is guaranteed to repeat a digit by the Pigeonhole Principle, there are no \"unique numbers\" with 11 or more digits. Generalizing abender's algorithm, there are $9\\dfrac{9!}{(10-k)!}$ k-digit \"unique numbers\". This means there are $9\\sum_{i = 1}^k \\dfrac{9!}{(10-i)!}$ \"unique numbers\" with k or fewer digits.\n\nAs another method of counting, suppose N has k digits. Start with the set of all \"unique numbers\" with k or fewer digits. Then subtract off the number of \"unique numbers\" where the first digit is greater than the first digit of N. Then subtract off the number of \"unique numbers\" where the first digit is equal to the first digit of N and the second digit is greater. Etc. So, to describe the algorithm, you can use this type of notation.\n\nLet $N = \\sum_{i = 1}^k d_i 10^{i-1}, \\quad \\forall 0< i < k, d_i \\in \\mathbb{Z}, 0\\le d_i \\le 9;\\quad 0< d_k \\le 9$. Next, let $A_i = \\{z \\in \\mathbb{N} \\mid z\\mbox{ is a \"unique number\" with }i\\mbox{ digits} \\}$. For any $z\\in A_k$, we can refer to its digits with the notation $z = \\sum_{i = 1}^k z_i 10^{i-1}$. Next, define the sets $N_i = \\{z\\in A_k \\mid z_i > d_i \\mbox{ and } \\forall i < j \\le k, z_j = d_j \\}$ for $1 \\le i \\le k$. What elements are in these sets?\n\nSo, by construction, the number of \"unique numbers\" up to N is given by $\\sum_{i = 1}^k \\left( |A_i| - |N_i| \\right)$ (let me know if you have any questions about why this is true). We already know how to calculate $|A_i|$ easily. So, how do we calculate $|N_i|$? That is a little more difficult. To make this easier, we will want one more thing. For notation, let $[n] = \\{1,2,\\ldots, n\\}$ (this is fairly common notation in combinatorics). Let $M = \\left\\{i \\in [k] \\mid \\sum_{j = 0}^{k-i} d_{k-j} 10^j \\in A_i \\right\\}$. What is this set $M$? What does it represent? More specifically, what is $\\min(M)$?\n\nSpoiler:\nI plugged in the formula for $\\sum_{i=1}^k |A_i|$ into wolframalpha and it shot out $9\\left( 986410 - \\dfrac{362880e\\Gamma(8-k,1)}{(9-k)!} \\right)$ where $\\Gamma(n,x)$ is the incomplete upper gamma function. Probably not terribly useful, but interesting that there is a continuous function based on $k$. Makes me wonder if we plug in $\\log_{10}(N)$ for $k$, how close will we come to the actual number?\n\n10. ## Re: how many numbers less than or equal to N with all digits unique\n\nOriginally Posted by kicma\nI have a similar problem.\n\nHow many are there odd 4-digit numbers with all digits different?\n\nIs it 9*8*7*5? I get that when I first choose last and first digit. But if I start from the beggining I get 9*9*8*5 (because 0 can't be first so there are again 9 possibilities for second digit).\nLet's look at the starting conditions. Before choosing any digits, the first digit is an element of $\\{1,2,3,4,5,6,7,8,9\\}$. The last digit is an element of $\\{1,3,5,7,9\\}$. Hence, if you choose the first digit first, and you choose an odd digit, you will only have four choices available for the last digit. On the other hand, if you choose the last digit first, no matter which digit you choose leaves only 8 choices left for the first digit. If you want to count from left to right, you need to break it up into a LOT of cases. You could choose one odd digit among the first three, two odd digits among the first three, or three odd digits among the first three. This becomes too cumbersome. Alternately, you can choose the digits right to left. Then you have five choices for the first digit, 9 choices for the second, 8 for the third, but you either have 7 or 6 choices for the last depending on whether you chose a zero yet or not.\n\nSo, as a general rule of thumb, choose the digits in the order of how restricted they are. The last digit is the most restricted, so choose that first: 5 choices. After that choice, before making any others, you have 8 possible choices for the first digit (since an odd digit is now out) or you have 9 choices for either of the other two digits (since they can still be zero). So, again, choose the most restricted. So now, you have 5*8 to choose the last digit then the first digit. Now, the remaining two digits are equally restricted with 8 choices each. So, to choose the last digit, then the first digit, then the second digit, there are 5*8*8 different choices. Finally, to choose the 3rd digit, there are 7 choices, and the total number of odd 4-digit numbers with distinct digits is 5*8*8*7.\n\n11. ## Re: how many numbers less than or equal to N with all digits unique\n\nPut your numbers into binary to simplify the problem :P", "date": "2014-07-28 09:15:46", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-puzzles/207164-how-many-numbers-less-than-equal-n-all-digits-unique.html", "openwebmath_score": 0.674754798412323, "openwebmath_perplexity": 341.8528332714425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9904406014994153, "lm_q2_score": 0.8918110382493035, "lm_q1q2_score": 0.8832858611474582}}
{"url": "https://math.stackexchange.com/questions/1782087/almost-everywhere-vs-almost-sure/1782114", "text": "# almost everywhere Vs. almost sure\n\nI'm reading a book about measure theory and probability (first chapter of Durret's Probability book), and it's starting to switch between the terms \"a.e.\" and \"a.s.\" in different contexts. I'm becoming confused about their meanings. What's the difference between almost everywhere and almost sure?\n\n\u2022 a.e. is used for general measure spaces, while a.s. is used for probabilistic spaces. There is no important difference between both concepts (for example, measurable functions are named random variables when the measure space is a probabilistic space). May 12, 2016 at 8:52\n\u2022 a.e. can be used a.e. This in contrast with a.s. which is only used in the context of probability measures. In principle you could use a.e. there too, but a.s. a.e. is replaced there by a.s. May 12, 2016 at 9:17\n\nIn a probability space (equipped with a probability $P$), we say that an event $\\omega$ occurs almost surely if $P(\\omega)=1$. On the other hand, on a measure space equipped with a measure $\\mu$, we say that a property $\\mathcal{P}$ is satisfied almost everywhere if the set where $\\mathcal{P}$ is not satisfied has measure zero. Note that \"a.s.\" is equivalent to \"a.e.\" in probability spaces, since if $\\omega$ occurs almost surely, then the probability that $\\omega$ does not occur is zero. However, in the case of general measure spaces $X$ we cannot say that a property is satisfied almost everywhere if it is satisfied in a set of measure $\\mu(X)$ (which would correspond to an event having probability $1$), since in many cases this measure is infinite. This is why in the case of measure spaces we formulate the definition of \"almost everywhere\" in terms of complements of sets.\n\n\u2022 $\\{x: x \\in (-\\infty, 0)\\cup (1, +\\infty)\\}$ is a.s. in $\\mathbb{R}$ but not a.e. in $\\mathbb{R}$, $\\mathbb{R}$ is equipped with Borel sets and Lebesgue measure. Is this right? Dec 12, 2019 at 16:39\n\nWhen we say that something happens \"almost everywhere\", we mean to say that:\n\n1. This something can happen or fail to happen in any point in a given measure space; and\n2. The set of points in which it fails to happen is a set of measure zero.\n\nNotice that there's no notion of probability when talking about \"almost everywhere\".\n\nNow, when we say that something happens \"almost surely\", we mean to say that:\n\n1. This something is the result of a random experiment. It can happen or fail to happen, and the result of this experiment (success ofr failure) is a random variable; and\n2. The probability of this something failing to happen is zero.\n\nFrom Wiki\n\nAlmost surely In probability theory, one says that an event happens almost surely (sometimes abbreviated as a.s.) if it happens with probability one. In other words, the set of possible exceptions may be non-empty, but it has probability.\n\nthis means almost surely comes from probability trials (stochastic trials),\n\nFrom Wolfram mathworld\n\nAlmost Everywhere\n\nA property of X is said to hold almost everywhere if the set of points in X where this property fails is contained in a set that has measure zero.\n\nFrom this book\n\n\"Essentials of Probability Theory for Statisticians\"\n\n\nPar Michael A. Proschan,Pamela A. Shaw page 102\n\nAlmost surely for random variable, (probability)\nAlmost everywhere for the sequance of functions, (measure)", "date": "2022-06-25 02:24:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1782087/almost-everywhere-vs-almost-sure/1782114", "openwebmath_score": 0.8844386339187622, "openwebmath_perplexity": 215.24836708266497, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811581728097, "lm_q2_score": 0.914900957313305, "lm_q1q2_score": 0.8832281457845307}}
{"url": "https://math.stackexchange.com/questions/2702057/proof-of-sum-formula-no-induction", "text": "# Proof of sum formula, no induction\n\n$$\\sum_{k=1}^n k=\\frac{n(n+1)}2$$\n\nSo I was trying to prove this sum formula without induction. I got some tips from my textbook and got this.\n\nLet $S=1+2+\\cdots+n-1+n$ be the sum of integers and $S=n+(n+1)+\\cdots+2+1$ written backwards. If I add these $2$ equations I get $2S=(1+n)+(1+n)\\cdots(1+n)+(1+n)$ $n$ times.\n\nThis gives me $2S=n(n+1) \\Rightarrow S=\\frac{n(n+1)}2$ as wanted.\n\nHowever if I changed this proof so that n was strictly odd or strictly even, how might I got about this. I realize even means n must be $n/2$. But I haven't been able to implement this in the proof correctly.\n\nEdit: error in question fixed, also by $n/2$ I mean should I implement this idea somewhere in the proof, cause even means divisible by $2$.\n\n\u2022 When written backward, $S=n+(n-1)+....+2+1$ \u2013\u00a0CY Aries Mar 21 '18 at 16:35\n\u2022 \"I realize even means n must be n/2\". I'm confused. What do you mean by this? \u2013\u00a0Mauve Mar 21 '18 at 16:36\n\u2022 Possible duplicate of Proof for formula for sum of sequence $1+2+3+\\ldots+n$? \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Mar 21 '18 at 16:36\n\u2022 To say \"$n$ must be $n/s$\" is a confusion. Better is \"$n= 2m$ for some integer $m$.\" Your proof works for both odd and even $n$, so your last para is mysterious. Do you mean that you want to add just the first $n$ odd (or even) integers? \u2013\u00a0B. Goddard Mar 21 '18 at 16:37\n\u2022 the proof works whether $n$ is even or odd. what's the problem? You realize that either $n$ or $n+1$ will be even. \u2013\u00a0Vasya Mar 21 '18 at 16:37\n\nMethod 1: (requires you to consider whether $n$ is odd or even.)\n\n$S = 1 + 2 + ...... + n$.\n\nJoin up the first to term to the last term and second to second to last and so on.\n\n$S = \\underbrace{1 + \\underbrace{2 + \\underbrace{3 +....+(n-2)} + (n-1)} + n}$.\n\n$= (n+1) + (n+1) + .....$.\n\nIf $n$ is even then:\n\n$S = \\underbrace{1 + \\underbrace{2 + \\underbrace{3 +..+\\underbrace{\\frac n2 + (\\frac n2 + 1)}+..+(n-2)} + (n-1)} + n}$\n\nAnd you have $\\frac n2$ pairs that add up to $n+1$. So the sum is $S= \\frac n2(n+1)$.\n\nIf $n$ is odd then:\n\n$S = \\underbrace{1 + \\underbrace{2 + \\underbrace{3 +..+\\underbrace{\\frac {n-1}2 + [\\frac {n+1}2] + (\\frac {n+1}2 + 1)}+..+(n-2)} + (n-1)} + n}$\n\nAnd you have $\\frac {n-1}2$ pairs that also add up to $n+1$ and one extra number $\\frac {n+1}2$ which didn't fit into any pair. So the sum is $\\frac {n-1}2(n+1) + \\frac {n+1}2 =(n-1)\\frac {n+1}2 + \\frac {n+1}2 = (n-1 + 1)\\frac {n+1}2n=n\\frac {n+1}2$.\n\nMethod 1$\\frac 12$ (Same as above but waves hands over doing tso cases).\n\n$S = average*\\text{number of terms} = average*n$.\n\nNow the average of $1$ and $n$ is $\\frac {n+1}2$ and the average of $2$ and $n-1$ is $\\frac {n+1}2$ and so on. So the average of all of them together is $\\frac {n+1}2$. So $S = \\frac {n+1}2n$.\n\nMethod 2: (doesn't require considering whether $n$ is odd or even).\n\n$S = 1 + 2 + 3 + ...... + n$\n\n$S = n + (n-1) + (n-2) + ...... + 1$.\n\n$2S = S+S = (n+ 1) + (n+1) + ..... + (n+1) = n(n+1)$>\n\n$S = \\frac {n(n+1)}2$.\n\nNote that by adding $S$ to itself this doesn't matter whether $n$ is even or odd.\n\nAnd lest you are wondering why can we be so sure that $n(n+1)$ must be even (we constructed it so it must be true... but why?) we simply note that one of $n$ or $n+1$ must be even.\n\nSo no problem.\n\nFor $n$ even i.e. $n=2m$ for some $m\\in\\mathbb{N}$. Let $S:=1+2+...+2m$ then\n\n$$2S=S+S=(1+2+...+(2m-1)+2m)+(2m+(2m-1)+...+2+1)=((1+2m)+(2+(2m-1))+...+((2m-1)+2)+(2m+1))=\\underbrace{((2m+1)+...+(2m+1))}_{2m-times}=(2m+1)\\cdot 2m$$ Therefore $$S=\\frac{(2m+1)2m}{2}=m(2m+1)$$ Analogue for $n$ odd. The formula is the same as in general for any $n$. You indeed just substitute directly into it. For instance $$\\frac{n(n+1)}{2}\\Rightarrow \\frac{2m(2m+1)}{2}=m(2m+1)$$\n\nThis classical result can be also easily proved by the following trick\n\nAn extended discussion also for more general cases here How Are the Solutions for Finite Sums of Natural Numbers Derived?\n\nA combinatorial proof. Consider the two element subsets of $\\Omega=\\{0,1,\\dotsc,n\\}$. There are $\\binom{n+1}{2}$ of them (corresponding to the right hand side of the equality). But we can count in another way. Classify the two element subsets based on their maximum element. For $1\\leq k \\leq n$, there are $\\binom{k}{1}=k$ two element subsets whose maximum element is $k$ since there are $k$ non-negative integers less than $k$.\n\nAnother proof based on telescoping. Let $n^{\\underline{2}}=n(n-1)$ (this is the falling factorial of length two. The exponent has an underline for notation). Observe that $$\\frac{1}{2}(n+1)^{\\underline{2}}-\\frac{1}{2}(n)^{\\underline{2}}=n.$$ In particular $$\\sum_{k=1}^nk=\\frac{1}{2}\\sum_{k=1}^n (k+1)^{\\underline{2}}-(k)^{\\underline{2}}=\\frac{(n+1)^{\\underline{2}}}{2}=\\frac{n(n+1)}{2}.$$", "date": "2019-06-25 04:28:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2702057/proof-of-sum-formula-no-induction", "openwebmath_score": 0.8624022006988525, "openwebmath_perplexity": 257.00283625630914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620070720564, "lm_q2_score": 0.8902942348544447, "lm_q1q2_score": 0.8832270855143811}}
{"url": "https://www.hpmuseum.org/forum/thread-361.html", "text": "Accuracy of HP Financial Calculators - Canadian mortgage\n01-08-2014, 03:20 AM (This post was last modified: 01-08-2014 11:39 AM by Jeff_Kearns.)\nPost: #1\n Jeff_Kearns Member Posts: 147 Joined: Dec 2013\nAccuracy of HP Financial Calculators - Canadian mortgage\nHaving recently adapted an 'accurate' (relative term) TVM routine for the HP-15C thanks to Karl Schneider's MISO Technique using indirect addressing, I decided to compare the results for a Canadian mortgage monthly payment with other HP calculator models. For all non-Canadians, interest on mortgages in Canada is compounded semi-annually, but payments are made monthly.\n\nOne therefore has to convert the nominal annual interest rate (based on semi-annual compounding) to an effective annual interest rate and then convert it again to an equivalent nominal rate for calculating monthly payments (based on monthly compounding). If you have an HP-12C, the following procedure can be used to obtain the Canadian mortgage factor (from the Owner's Handbook):\n\nThe keystrokes to calculate the Canadian Mortgage factor (on the 12C) are:\nPress [f ], CLEAR [FIN ], [g ], then END .\nKey in 6 and press [n ].\nKey in 200 and press ENTER , then PV .\nKey in the annual interest rate as a percentage and press [+ ], CHS , then FV .\nPress [i ].\nThe Canadian mortgage factor is now stored in [i ] for future use.\n\nI have a short routine of the above procedure for the 12C and equivalent SOLVER routines for the 17BII and 19BII. The 30b has a built-in Canadian TVM function that does the same with appropriate mode settings. The results from different models (both scientific and financial) are interesting and I welcome any insight as to why they differ the way they do and which is the most 'correct':\n\nVariables\nn: 300 months (#payments)\ni: 3% (nominal annual interest rate)\nPV: $450,000 (mortgage amount) PMT: UNKNOWN FV: 0 (fully paid off after 300 months) Results for monthly payment: \u2022 HP-15C:$2,129.604821\n\u2022 HP-12C: $2,129.604744 \u2022 HP-32SII:$2,129.60474211\n\u2022 HP-42S: $2,129.60474211 \u2022 HP-30B:$2,129.60474341\n\u2022 HP-19BII: $2,129.60474211 \u2022 HP-17BII:$2,129.60474211\n\nThe 15C results are not surprising considering it has less precision. It gives a monthly nominal rate of 0.248451700% whereas the 12C gives 0.248451673 and the HP-32Sii and HP-42S both give 0.248451672% (and so does the 50G). The difference in precision in the 15C likely stems from taking the 12th root of the effective rate when converting to monthly.\n\nWhat surprises me is the HP-30B... Why does it give a different result from the other high-end financial models? And isn't is surprising how accurate the routine is for the pioneer models 32sii and 42S? How does the WP-34S compare?\n\nJeff\n01-08-2014, 04:56 AM\nPost: #2\n Gene Moderator Posts: 992 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nHP-30B: $2,129.60474341 What surprises me is the HP-30B... Why does it give a different result from the other high-end financial models? And isn't is surprising how accurate the routine is for the pioneer models 32sii and 42S? How does the WP-34S compare? Gene: The 30B result matches Excel's calculations of$2,129.60474341 366 as far as the 30B goes.\n\nAs I recall, the 30B has greater internal accuracy in its financial routines.\n01-08-2014, 11:16 AM\nPost: #3\n Terje Vallestad Member Posts: 145 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(01-08-2014 04:56 AM)Gene Wrote: \u00a0HP-30B: $2,129.60474341 Gene: The 30B result matches Excel's calculations of$2,129.60474341 366 as far as the 30B goes.\n\nAs I recall, the 30B has greater internal accuracy in its financial routines.\n\nFor what it is worth; the Prime provides the same result as the 30b 2129.60474341\n\nCheers, Terje\n01-08-2014, 11:34 AM (This post was last modified: 01-08-2014 11:35 AM by Jeff_Kearns.)\nPost: #4\n Jeff_Kearns Member Posts: 147 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nTerje Vallestad wrote: \"For what it is worth; the Prime provides the same result as the 30b 2129.60474341\"\n\nThe 50G gives the same result as the 19BII and Pioneers: 2129.60474211\n\nJeff\n01-08-2014, 11:42 AM\nPost: #5\n Werner Senior Member Posts: 452 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nTo convert from a nominal Canadian annual rate to the effective nominal rate, use:\n\n200\n/\nLN1+X\n6\n/\nE^X-1\n100\n*\n\nThen, the HP42S gets the monthly rate as 0.248451672465 and the PMT as -2,129.60474341, just like the 30b\n\nCheers, Werner\n01-08-2014, 12:19 PM\nPost: #6\n Gerson W. Barbosa Senior Member Posts: 1,289 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(01-08-2014 04:56 AM)Gene Wrote: \u00a0Gene: The 30B result matches Excel's calculations of $2,129.60474341 366 as far as the 30B goes. As I recall, the 30B has greater internal accuracy in its financial routines. HP 200LX:$2,129.604743413646\n01-10-2014, 01:02 AM\nPost: #7\n DMaier Junior Member Posts: 42 Joined: Jan 2014\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nThe WP-34S matches the HP-30B and the 42S: 2129.60474341. The effective interest rate is: 0.24845167247 (Werner's method gives 0.2484516724648725, so there's that.)\n\nThe same result, by the way, using either the original 2 step 12C method, or by setting:\nI=3\nNP=12\nNI=2\n01-10-2014, 01:39 AM (This post was last modified: 01-10-2014 04:32 AM by Jeff_Kearns.)\nPost: #8\n Jeff_Kearns Member Posts: 147 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(01-10-2014 01:02 AM)DMaier Wrote: \u00a0The effective interest rate is: 0.24845167247\nis incorrect.\n\nI am not a financial type - I am an engineer. But I think it is important to make a note on terminology. That value: 0.24845167247%, is actually the 'nominal' monthly rate, not the 'effective' interest rate. If you multiply it by 12, you get 2.98142007%, the nominal annual rate for monthly payment calculations.\n\nThe stated 'nominal' annual rate for Canadian mortgages is based on a semi-annual compounding period - and is 3% in this particular example. This translates into an 'effective' annual interest rate of 3.0225%.\n\nThe 'effective' interest rate is always greater than the nominal rate. There are in fact, three 'rates' at play here...\n> The posted 'nominal' rate of 3%, based on semi-annual compounding;\n> The 'effective' annual interest rate of 3.0225%; and\n> The 'nominal' rate of 2.98142007%, used for the calculation of monthly payments.\n\nEffective Annual Interest Rate Explained\n\nRegards,\n\nJeff\n04-07-2014, 02:12 AM (This post was last modified: 04-07-2014 08:50 PM by solwarda.)\nPost: #9\n solwarda Junior Member Posts: 3 Joined: Apr 2014\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nThe nominal monthly rate for Canadian mortgages can be calculated as follows:\n3%/200 + 1 =1.015^(1/6)=1.0024845167 2464872638 - 1.\n\nThe Formula for the monthly payment is: $450,000 x .0024845167 2464872638 x (1.0024845167 2464872638)^300/((1.0024845167 2464872638^300) - 1)=$2129.6047434136 4549091198.\nAs can be seen from these extended precision calculations, it all depends on how many decimal places is the calculation carried out internally inside each calculator.\n04-07-2014, 03:31 AM\nPost: #10\n Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(04-07-2014 02:12 AM)solwarda Wrote: \u00a03%/200 + 1 =1.015^(1/6)=1.0024845167 2464872638 - 1.\n\u0ca0_\u0ca0\nThat's not how equations work.\n04-07-2014, 04:39 AM\nPost: #11\n solwarda Junior Member Posts: 3 Joined: Apr 2014\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nThomas Klemm: I don't quite understand your objection. I'm simply converting 3% nominal annual rate compounded semi-annually into a monthly nominal rate for the purpose of calculating the requisite monthly payment. You can do this in a number of ways. For example, you can compound 3% semi-annual rate into \"effective\" annual rate as follows: 3/200=.015 +1=1.015^2=1.030225-1 x 100=3.0225, which is the effective annual rate. Then you can take the 12th root of that, which will give you the above nominal monthly rate.\n04-07-2014, 05:19 AM\nPost: #12\n Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(04-07-2014 04:39 AM)solwarda Wrote: \u00a0Thomas Klemm: I don't quite understand your objection.\nEquality is a transitive relation. By chaining the calculations you create equations that don't hold. You just have to split them:\n3%/200 + 1 = 1.015\n1.015^(1/6) = 1.0024845167 2464872638\n1.0024845167 2464872638 - 1 = 0.0024845167 2464872638\nSo nothing wrong with the calculation. Just with the use of the = sign.\n\nCheers\nThomas\n04-07-2014, 07:06 PM\nPost: #13\n Dieter Senior Member Posts: 2,397 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(04-07-2014 02:12 AM)solwarda Wrote: \u00a0The Formula for the monthly payment is...\n\n...missing some brackets. Otherwise the two identical powers cancel down to 1. ;-)\n\nQuote:0.0024845167 2464872638\n\nThe true value is ...639. Or ...638916... to be precise. Using ln1+x and e^x\u20131 ensures the required accuracy here.\n\nQuote:$2129.6047434136 4549091198. The WP34s in DP mode returns 2.129,6047 4341 3645 4909 1198 5546 8796 57 Quote:As can be seen from these extended precision calculations, it all depends on how many decimal places is the calculation carried out internally inside each calculator. You bet. ;-) Dieter 04-07-2014, 07:49 PM (This post was last modified: 04-07-2014 08:13 PM by Dieter.) Post: #14  Dieter Senior Member Posts: 2,397 Joined: Dec 2013 RE: Accuracy of HP Financial Calculators - Canadian mortgage (01-08-2014 03:20 AM)Jeff_Kearns Wrote: \u2022 HP-15C:$2,129.604821\n\u2022 HP-12C: $2,129.604744 \u2022 HP-32SII:$2,129.60474211\n\u2022 HP-42S: $2,129.60474211 \u2022 HP-30B:$2,129.60474341\n\u2022 HP-19BII: $2,129.60474211 \u2022 HP-17BII:$2,129.60474211\nThe 15C results are not surprising considering it has less precision. It gives a monthly nominal rate of 0.248451700% whereas the 12C gives 0.248451673 and the HP-32Sii and HP-42S both give 0.248451672% (and so does the 50G).\n\nBoth the 42s and the 50G feature ln1+x and e^x-1, so they are able to obtain the correct 12-digit interest rate as 0,248451672465%. This again leads to the correct result, which in the above list is exclusively returned by the 30B.\n\nSo if the 12-digit machines don't get it right it's not their fault. It's sloppy programming in case of the 42s and 50G. Calculators that do not offer ln1+x need a workaround to get similar accuracy.\n\nQuote:What surprises me is the HP-30B... Why does it give a different result from the other high-end financial models?\n\nWhile most HPs internally use three additional guard digits (for internal calculations, that is - not in a TVM user program), the 30B IIRC internally uses much more digits. On the other hand the correct result can be obtained with simple 12-digit accuracy (e.g.42s, 50G) - careful programming provided:\n\nCode:\n\u00a03\u00a0[ENTER]\u00a0200\u00a0[/]\u00a0\u00a0=>\u00a0\u00a0\u00a00,015 \u00a0[ln1+x]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,0148886124938 \u00a06\u00a0[/]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,00248143541563 \u00a0[e^x-1]\u00a0100\u00a0[x]\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,248451672465\n\nEdit: I now see that Werner already mentioned this point in his post. ;-)\n\nUsers with hyperbolic functions, but without ln1+x and e^x-1 may do it this way:\n\nCode:\n\u00a03\u00a0[ENTER]\u00a0200\u00a0[/]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,015 \u00a0[ENTER]\u00a0[ENTER]\u00a02\u00a0[+]\u00a0[/]\u00a0=>\u00a0\u00a0\u00a00,00744416873449 \u00a0[HYP]\u00a0[ATAN]\u00a02\u00a0[x]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,0148886124937 \u00a06\u00a0[/]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,00248143541562 \u00a02\u00a0[/]\u00a0[HYP]\u00a0[SIN] \u00a0[LastX]\u00a0[e^x]\u00a0[x]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,00124225836232 \u00a0200\u00a0[x]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,248451672464\n\nCompare this with the result of less careful evaluation:\nCode:\n\u00a01,015\u00a0[ENTER]\u00a06\u00a0[1/x]\u00a0[y^x]\u00a0=>\u00a01,00248451672\u00a0\u00a0\u00a0(maybe\u00a0...673) \u00a01\u00a0[\u2013]\u00a0100\u00a0[x]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a00,248451672000\n\nWhich means that three valuable digits are lost!\n\nQuote:And isn't is surprising how accurate the routine is for the pioneer models 32sii and 42S?\n\nI do not think that nine out of twelve digits is particulary accurate. But again: that's not the fault of the calculator.\n\nQuote:How does the WP-34S compare?\n\nWith 16 or even 34 digit precision (and careful programming) you can expect accuracy far beyond the previous examples:\n\ni% = 0,2484\u00a05167\u00a02464\u00a08726\u00a03891\u00a06130\u00a07363\u00a04707\u00a027\nPMT = 2.129,6047 4341 3645 4909 1198 5546 8796 57\n\nDieter\n04-07-2014, 08:42 PM\nPost: #15\n Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(01-08-2014 03:20 AM)Jeff_Kearns Wrote: \u00a0The 15C results are not surprising considering it has less precision.\nUsing Werner's method to calculate i% directly and using the improved version of your TVM program I got the same result for PMT as with the HP-12C.\n\nCheers\nThomas\n04-07-2014, 09:06 PM\nPost: #16\n Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(04-07-2014 07:49 PM)Dieter Wrote:\nCode:\n\u00a03\u00a0[ENTER]\u00a0200\u00a0[/]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,015 \u00a0[ENTER]\u00a0[ENTER]\u00a02\u00a0[+]\u00a0[/]\u00a0=>\u00a0\u00a0\u00a00,00744416873449 \u00a0[HYP]\u00a0[ATAN]\u00a02\u00a0[x]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00,0148886124937\nIn this case you don't gain anything but loose accuracy compared to:\nCode:\n\u00a03\u00a0[ENTER]\u00a0200\u00a0[/]\u00a01\u00a0[+]\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a01.015 \u00a0[LN]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=>\u00a0\u00a0\u00a00.0148886124938\n\nCheers\nThomas\n04-07-2014, 09:14 PM\nPost: #17\n solwarda Junior Member Posts: 3 Joined: Apr 2014\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nDieter: If you want the final answer accurate to 100 digits!, here it is:\n\n\\$2,129.6047434136 4549091198 5546879656 9409506602 5105604773 4028771274 1867975760 1238046591 3606995240 188946.\n\nPS: I can give it to you up to a million digits!.\n\nCheers!,\nSol\n04-07-2014, 09:20 PM\nPost: #18\n Jeff_Kearns Member Posts: 147 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(04-07-2014 07:49 PM)Dieter Wrote: \u00a0Both the 42s and the 50G feature ln1+x and e^x-1, so they are able to obtain the correct 12-digit interest rate as 0,248451672465%. This again leads to the correct result, which in the above list is exclusively returned by the 30B.\n\nSo if the 12-digit machines don't get it right it's not their fault. It's sloppy programming in case of the 42s and 50G. Calculators that do not offer ln1+x need a workaround to get similar accuracy.\n\nThank you for this excellent reply! Although it has been 3 months since my original post, I have learned something valuable about calculator precision and the use of ln1+x, e^x-1, and hyperbolic workarounds to get the most out my calculators.\n\nThe HP-15C result is now the same as that of the 12C (to ten digits), the 42s correct to 12 digits, and the 32sII correct to 11 digits. My programming will be just a little less sloppy from here on in!\n\nJeff Kearns\n04-07-2014, 10:03 PM (This post was last modified: 04-07-2014 10:19 PM by Dieter.)\nPost: #19\n Dieter Senior Member Posts: 2,397 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\n(04-07-2014 09:06 PM)Thomas Klemm Wrote: \u00a0In this case you don't gain anything but loose accuracy ...\n\nSure. The correct ln(1+x) result is no problem if 1+x can be given exactly, i.e. here 1,015 = 1,01500000000.\nNow try this with the same 12 digits for i = 4/3% or 1/7% . ;-)\n\nDieter\n04-07-2014, 11:16 PM\nPost: #20\n Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013\nRE: Accuracy of HP Financial Calculators - Canadian mortgage\nBut it's still good to know when you better avoid it.\nOTOH: how does your method compare to HP's trick to calculate $$log(1+x)$$ for small values?\n\nCheers\nThomas\n \u00ab Next Oldest | Next Newest \u00bb\n\nUser(s) browsing this thread: 1 Guest(s)", "date": "2020-07-11 14:22:47", "meta": {"domain": "hpmuseum.org", "url": "https://www.hpmuseum.org/forum/thread-361.html", "openwebmath_score": 0.4288967549800873, "openwebmath_perplexity": 5312.962041189941, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876992225169, "lm_q2_score": 0.899121379297294, "lm_q1q2_score": 0.8831958709917148}}
{"url": "https://www.physicsforums.com/threads/trig-substitution-with-cosine.854743/", "text": "# Trig Substitution with Cosine\n\nTags:\n1. Jan 30, 2016\n\n### UMath1\n\nI was wondering if you could do a trig substitution with cosine instead of sine. All the textbooks I have referred to use a sine substitution and leave no mention as to why cosine substitution was not used. It seemed that it should work just the same, until I tried it for the following Fint [sqrt(9-x^2)]/ [x^2]. I checked to see if my answer differed by only a constant but that was not the case. I have attached pictures of my work. Can anyone tell me why it does not work?\n\n2. Jan 30, 2016\n\n### blue_leaf77\n\nIf you use sine instead, you will end up with $\\sin^{-1}$ in place of $\\cos^{-1}$. But the two expressions are related by $\\cos^{-1}x = \\pi/2-\\sin^{-1}x$.\n\n3. Jan 30, 2016\n\n### UMath1\n\nI know but why are the answers different? Is one less valid than the other?\n\nBtw the textbook which uses sine has the same answer but with -sin^-1(x/3) instead of cos^-1(x/3) like I have it.\n\n4. Jan 30, 2016\n\n### blue_leaf77\n\nThat's exactly the point I addressed in my previous post. Replace -sin^-1(x/3) with the equation I wrote before. You will indeed have an additional $\\pi/2$ but it's a constant and hence can be absorbed into the integration constant $C$.\n\n5. Jan 30, 2016\n\n### QuantumQuest\n\nThere's really nothing magic about using sin or cos. It just depends on what is more convenient for each case. As for signs, using the relevant relations from trigonometry - like the one that blue_leaf77 mentions, you can substitute sin for cos and vice versa and find the appropriate sign.\n\n6. Jan 30, 2016\n\n### zinq\n\nThe function being integrated is\n\nf(x) = \u221a(9 - x2) / x2\n\n. This is defined for 0 < |x| \u2264 3.\n\nWhen making a substitution we want to choose an interval where f(x) makes sense, and the easiest one is 0 < x \u2264 3.\n\nWe also want to choose a substitution that takes the same values that f(x) does over the interval of definition, and that's between 0 and 3.\n\nEach of y = 3 sin(x) and y = 3 cos(x) satisfy this condition, so either one can be used for the substitution.\n\nUsing 3 sin(x) to substitute might be a tiny bit easier than cosine because its derivative is 3 cos(x), and this does not introduce negative signs.\n\n7. Jan 30, 2016\n\n### UMath1\n\nOk...I see it now. I tried some test bounds of integration and got the same answer from both options.", "date": "2018-01-24 06:40:06", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/trig-substitution-with-cosine.854743/", "openwebmath_score": 0.9007144570350647, "openwebmath_perplexity": 538.0504130661676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211634198559, "lm_q2_score": 0.9059898235563418, "lm_q1q2_score": 0.8831780538457431}}
{"url": "https://math.stackexchange.com/questions/3445884/solve-2-cos2-x-sin-x-1-for-all-possible-x", "text": "# Solve $2 \\cos^2 x+ \\sin x=1$ for all possible $x$\n\n$$2\\cos^2 x+\\sin x=1$$\n\n$$\\Rightarrow 2(1-\\sin^2 x)+\\sin x=1$$\n\n$$\\Rightarrow 2-2 \\sin^2 x+\\sin x=1$$\n\n$$\\Rightarrow 0=2 \\sin^2 x- \\sin x-1$$\n\nAnd so:\n\n$$0 = (2 \\sin x+1)(\\sin x-1)$$\n\nSo we have to find the solutions of each of these factors separately:\n\n$$2 \\sin x+1=0$$\n\n$$\\Rightarrow \\sin x=\\frac{-1}{2}$$\n\nand so $$x=\\frac{7\\pi}{6},\\frac{11\\pi}{6}$$\n\nSolving for the other factor,\n\n$$\\sin x-1=0 \\Rightarrow \\sin x=1$$\n\nAnd so $$x=\\frac{\\pi}{2}$$\n\nNow we have found all our base solutions, and so ALL the solutions can be written as so:\n\n$$x= \\frac{7\\pi}{6} + 2\\pi k,\\frac{11\\pi}{6} + 2\\pi k, \\frac{\\pi}{2} + 2\\pi k$$\n\n\u2022 And the question is.... ?? Nov 21 '19 at 23:36\n\u2022 It's tagged with proof verification. The solution provided in the question is correct. Nov 21 '19 at 23:39\n\u2022 @RobertShore OK, thx. Then, when an answer should be set? When it corrects the question? Nov 21 '19 at 23:41\n\u2022 I'll provide an answer rather than a comment if the answer provided is wrong in some material way or if there's an alternative solution that provides additional insight. Nov 21 '19 at 23:43\n\nYes your solution is very nice and correct, as a slightly different alternative\n\n$$2\\cos^2(x)+\\sin (x)=1 \\iff 2(1-\\sin x)(1+\\sin x)+\\sin x-1=0$$\n\n$$\\iff (\\sin x-1)(-2-2\\sin x)+(\\sin x-1)=0 \\iff (\\sin x-1)(-1-2\\sin x)=0$$\n\nwhich indeed leads to the same solutions, or also from here by $$t=\\sin x$$\n\n$$2-2 \\sin^2 x+\\sin x=1 \\iff 2t^2-t-1=0$$\n\n$$t_{1,2}=\\frac{1\\pm \\sqrt{9}}{4}=1, -\\frac12$$\n\nYour method's fine, the answer's right. The only improvement I can suggest is to make the definition of \"base solution\" clear upfront. Each \"and so\" acts as if a specific value of $$\\sin x$$ has finitely many solutions rather than finitely many per period, so before you obtain them you should mention a restriction to $$[0,\\,2\\pi)$$ and then extend to $$\\Bbb R$$ at the end.\n\nOther way is used identities double angle and sum-product\n\n$$\\begin{eqnarray*} 2\\cos^2x+\\sin x& = & 1 \\\\ 2\\cos^2x-1+\\sin x& = & 0\\\\ \\cos(2x)+\\sin x& = & 0\\\\ \\cos(2x)+\\cos\\left(\\frac{\\pi}{2}-x\\right) & = & 0\\\\ 2\\cos\\left(\\frac{x}{2}+\\frac{\\pi}{4}\\right)\\cos\\left(\\frac{3x}{2}-\\frac{\\pi}{4}\\right) & = & 0\\\\ \\end{eqnarray*}$$\n\n$$2\\cos^2(x)+\\sin(x)=1$$\n\nUsing $$2\\cos^2(x)-1=\\cos(2x)$$ we have $$\\cos(2x)=-\\sin(x)=\\cos(\\pi/2+x).$$\n\nHence $$2x=\\pm(\\pi/2+x)+2n\\pi$$, $$n\\in\\mathbb{Z}$$ and $$x=\\pi/2+2n\\pi$$ or $$x=-\\pi/6+2n\\pi/3.$$ The second expression can be re-written as $$-\\pi/6\\pm2\\pi/3+2k\\pi$$ or $$-\\pi/6+2k\\pi$$ giving the three solutions\n\n\\begin{align} x&=\\pi/2+2n\\pi\\\\x&=-\\pi/6+2k\\pi=11\\pi/6+2k\\pi\\\\x&=-5\\pi/6+2k\\pi=7\\pi/6+2k\\pi\\\\ \\end{align}\n\nNote that one of the solutions from the second expression, $$-\\pi/6+2\\pi/3+2k\\pi=\\pi/2+2k\\pi$$ is absorbed into the first of the three solutions. This happens because this is a double root, tangential to the x-axis - see plot from Wolfrom Alpha:", "date": "2021-10-25 23:43:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3445884/solve-2-cos2-x-sin-x-1-for-all-possible-x", "openwebmath_score": 0.762012243270874, "openwebmath_perplexity": 397.3727561826924, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9777138144607745, "lm_q2_score": 0.9032942008463507, "lm_q1q2_score": 0.8831632186897825}}
{"url": "https://coursys.sfu.ca/2018fa-cmpt-383-d1/pages/Exercise8", "text": "# Exercise 8\n\n## Getting Started With Go\n\nYou're going to need to get Go code running. Download/install Go tools: a command-line compiler, or IDE, or whatever you like. Start by getting a \u201chello world\u201d running. (There's one in the lecture slides.)\n\nDownload and extract the code skeleton for this exercise. The code includes an outline of how to organize your work for this exercise, some tests that should pass when you're done (more below), and a place for your main function that will satisfy the Go tools (and your IDE where relevant).\n\nIt's not a requirement, but maybe go through A Tour of Go: it will prepare you better for some of the stuff we'll be covering in this section of the course.\n\n## Go Hailstone\n\nThe Hailstone sequence provided many useful examples in Haskell, so let's revisit it in Go. In hailstone.go, write a function Hailstone that calculates the next element of a hailstone sequence: for even n, it should return n/2 and for odd n, it should return 3*n+1.\n\nHailstone(17) == 52\nHailstone(18) == 9\n\n\nThe function should take and return a uint type:\n\nfunc Hailstone(n uint) uint { \u2026 }\n\n\n## Hailstone Sequence\n\nIn this question, we will build the hailstone sequence in a Go array. Put code for this section in hailstone.go.\n\n### Attempt 1: grow the slice\n\nIn this implementation, start with an empty []uint{} slice. Calculate the elements of the hailstone sequence and use the append function to add it to the end of a new slice as you go.\n\nIt is probably easiest to do this iteratively (not recursively). You should take a uint argument and return a []uint slice. In pseudocode (since == isn't defined on slices):\n\nHailstoneSequenceAppend(5) == [5, 16, 8, 4, 2, 1]\n\n\n### Attempt 2: pre-allocate the array\n\nAppending to an slice is expensive this way: we are constantly allocating and de-allocating arrays at they grow. Maybe we can do better? Write a function HailstoneSequenceAllocate that generates the same result in a different way\u2026\n\nIt is easy enough to figure out how big an array we need: we did it before. This time, start by calculating the length of array we need by iterating Hailstone. (In the above example, you should realize you need an array of length 6.)\n\nThen, use the make function to create a slice and allocate an array of that many uints. Then fill it in and return it (no appending, just set array elements). Results should be the same as HailstoneSequenceAppend in all cases.\n\n## Test and Benchmark\n\nGo has built-in testing and benchmarking functionality. The exer8_test.go provided in the ZIP above provides tests for the requirements of this exercise.\n\nAt this point, the test for hailstone functionality should pass. If you comment-out the Point tests (or go finish that question below and come back here), you should pass the other tests:\n\ngo test exer8 -v\n\n\nThere are also some short benchmarks that we can use to see which of the hailstone sequence functions is actually faster:\n\ngo test exer8 -bench=.\n\n\nAdd a comment to your hailstone.go indicating the relative speed of the two HailstoneSeq* functions (thus proving to us that you have figured out how to run Go tests).\n\n## A Struct for Points\n\nIn points.go, create a struct Point for two-dimensional points: $$(x,y)$$ values. The struct should have two fields, x and y, both float64.\n\nCreate a function NewPoint that creates a Point given x and y values: this would be a constructor in any other language, but in Go, it's just a function that returns a Point. [Note: we don't really need a constructor this simple and could use a Go struct literal instead, but we're creating the constructor anyway. There are, of course, cases where some work is required to construct a struct instance.]\n\n### String Representation\n\nThere is a perfectly reasonable default string representation for structs in Go (which is used if you fmt.Print them), but we can make it nicer.\n\nCreate a String() method on Point using fmt.Sprintf to output the usual parentheses-and-commas representation of a point:\n\npt := NewPoint(3, 4.5)\nfmt.Println(pt) // should print (3, 4.5)\nfmt.Println(pt.String() == \"(3, 4.5)\") // should print true\n\n\nHint: The %v format seems nice.\n\n### Calculate Norm\n\nOne more method to add: the Euclidean norm of the point: add up the squares of the components and take the square root.\n\nIf everything is working, this should print true.\n\npt := NewPoint(3, 4)\nfmt.Println(pt.Norm() == 5.0)\n\n\nAlso, the provided tests should all pass.\n\n## Submitting\n\nSubmit your files through CourSys for Exercise 8.\n\nUpdated Mon Nov. 05 2018, 11:05 by ggbaker.", "date": "2018-12-12 21:30:50", "meta": {"domain": "sfu.ca", "url": "https://coursys.sfu.ca/2018fa-cmpt-383-d1/pages/Exercise8", "openwebmath_score": 0.3259769082069397, "openwebmath_perplexity": 2308.454814284125, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9579122768904644, "lm_q2_score": 0.9219218337824342, "lm_q1q2_score": 0.8831202429135638}}
{"url": "https://math.stackexchange.com/questions/3550450/which-of-these-answers-is-the-correct-indefinite-integral-using-trig-substitut", "text": "# Which of these answers is the correct indefinite integral? (Using trig-substitution or $u$-substitution give different answers)\n\nAnswers obtained from two online integral calculators:\n\n\\begin{align}\\int\\dfrac{\\sqrt{1 + x}}{\\sqrt{1 - x}}\\,\\mathrm dx &= -\\sqrt{\\dfrac{x + 1}{1 - x}} + \\sqrt{\\dfrac{x + 1}{1 - x}}x - 2\\arcsin\\left(\\dfrac1{\\sqrt2}\\sqrt{1 - x}\\right) + C \\\\ \\int\\dfrac{\\sqrt{1 + x}}{\\sqrt{1 - x}}\\,\\mathrm dx &= 2\\sin^{-1}\\left(\\dfrac{\\sqrt{x + 1}}{\\sqrt2}\\right) - \\sqrt{1 - x^2} + \\text{ constant} \\end{align}\n\n## Update:\n\nI realized that the substitution for $$\\theta$$ was supposed to be $$\\arcsin$$ not $$\\arccos$$, so the answer would have been the same as the right hand side.\n\nBut I also noticed that using the initial substitution to plug $$x$$ back in the final answer will not always give the correct answer because in a similar question:\n\n$$\\int \\frac{\\sqrt{x^2-1}}x dx$$\n\nhas a trig-substitution of $$x = \\sec\\theta$$, and the answer in terms of $$\\theta$$ would be: $$\\tan \\theta - \\theta + C$$. Then the final answer in terms of $$x$$ should be : $$\\sqrt{x^2-1} - \\operatorname{arcsec}(x) + C$$.\n\nBut online integral calculators give the answer: $$\\sqrt{x^2-1} - \\arctan(\\sqrt{x^2-1}) + C$$, which doesn't match the original substitution of:\n\n$$x = \\sec\\theta \\to \\theta = \\operatorname{arcsec}(x)$$\n\nAnyone know why the calculator gives that answer which doesn't match the original trig-substitution of $$x = \\sec \\theta \\to \\theta = \\operatorname{arcsec}(x)$$?\n\n\u2022 You assumed $x=sin\\theta$, so $\\theta = arcsin x$ \u2013\u00a0Chief VS Feb 17 '20 at 20:26\n\u2022 Oh so the theta value must also match with the substitution I made for x? \u2013\u00a0user749176 Feb 17 '20 at 20:32\n\u2022 In short yes, assuming $x>0$, note that $arcsinx = arccos\u221a(1-x^2)$ for only $x>0$ \u2013\u00a0Chief VS Feb 17 '20 at 20:38\n\u2022 which is the actual answer though? When I plug the question into different online integral calculators, all answers are very different from mine \u2013\u00a0user749176 Feb 17 '20 at 20:44\n\u2022 Does this answer your question? Getting different answers when integrating using different techniques \u2013\u00a0Xander Henderson Feb 18 '20 at 5:01\n\nStarting off with $$\\displaystyle\\int \\frac{\\sqrt{x^2-1}}{x}\\mathrm dx$$, substitute $$x = \\sec(\\theta)$$ for $$\\theta \\in \\left[0, \\frac{\\pi}{2}\\right) \\cup \\left(\\frac{\\pi}{2}, \\pi\\right]$$ as usual (keep the domain in mind for later). Of course, that means $$\\sqrt{x^2-1} = \\sqrt{\\sec^2(\\theta)-1} = \\sqrt{\\tan^2(\\theta)} = \\vert \\tan(\\theta)\\vert$$ and $$\\dfrac{\\mathrm dx}{\\mathrm d\\theta} = \\sec(\\theta)\\tan(\\theta) \\iff \\mathrm dx = \\sec(\\theta)\\tan(\\theta)\\mathrm d\\theta$$. This simplifies as follows:\n\n$$\\displaystyle\\int \\frac{\\sqrt{x^2-1}}{x}\\mathrm dx \\longrightarrow \\int\\frac{\\vert \\tan(\\theta)\\vert}{\\sec(\\theta)}\\sec(\\theta)\\tan(\\theta)\\mathrm d\\theta = \\int\\vert \\tan(\\theta)\\vert\\tan(\\theta)\\mathrm d\\theta$$\n\nFor $$\\theta \\in \\left[0, \\frac{\\pi}{2}\\right)$$, $$\\tan(\\theta) \\geq 0$$, so you get\n\n$$\\int \\tan^2(\\theta) \\mathrm d\\theta = \\int \\left[\\sec^2(\\theta)-1\\right] \\mathrm d\\theta = \\tan(\\theta)-\\theta+C \\longrightarrow \\sqrt{x^2-1}-\\text{arcsec}(x)+C$$\n\nSince tangent is positive in the first quadrant, $$\\tan(\\theta) = \\sqrt{x^2-1}$$, so the $$\\theta$$ term can also be replaced with $$\\arctan\\left(\\sqrt{x^2-1}\\right)$$.\n\nFor $$\\theta \\in \\left(\\frac{\\pi}{2}, \\pi\\right]$$, $$\\tan(\\theta) \\leq 0$$, so you get\n\n$$-\\int \\tan^2(\\theta) \\mathrm d\\theta = -\\int \\left[\\sec^2(\\theta)-1\\right] \\mathrm d\\theta = \\theta-\\tan(\\theta)+C \\longrightarrow \\sqrt{x^2-1}+\\text{arcsec}(x)+C$$\n\nSince tangent is negative in the second quadrant, $$\\tan(\\theta) = \\tan(\\theta -\\pi) = -\\sqrt{x^2-1}$$ (remember that tangent takes arguments in the first and fourth quadrants), so the $$\\theta$$ term can also be replaced with $$\\pi-\\arctan\\left(\\sqrt{x^2-1}\\right)$$.\n\nIn both cases, the anti-derivative could be re-written as $$\\sqrt{x^2-1}-\\arctan\\left(\\sqrt{x^2-1}\\right)+C$$. Basically, it just \"combines\" your other two anti-derivatives and expresses them as a single function rather than having one for each case.\n\n\u2022 Would just leaving the answer as: \u221a(x^2-1) - arcsec(x) + C , still be correct? \u2013\u00a0user749176 Feb 18 '20 at 0:09\n\u2022 For indefinite integrals, we usually assume $\\tan(\\theta)$ is positive (at least from what I've seen), so yeah, that's fine. \u2013\u00a0KM101 Feb 18 '20 at 0:16\n\nNow, both answers are correct. They merely look different. They differ by a constant.\n\nNote 1...\n\n$$-\\sqrt{\\frac{x+1}{1-x}}+\\sqrt{\\frac{x+1}{1-x}}\\;x = -\\sqrt{\\frac{x+1}{1-x}}\\;(1-x) = -\\frac{\\sqrt{x+1}\\;(1-x)}{\\sqrt{1-x}} \\\\= -\\sqrt{x+1}\\sqrt{1-x} =-\\sqrt{(1+x)(1-x)} =-\\sqrt{1-x^2}$$\n\nNote 2... $$2\\,\\arcsin \\left( \\frac{\\sqrt {1+x}}{\\sqrt {2}} \\right) =\\pi-2\\,\\arcsin \\left( \\frac{\\sqrt {1-x}}{\\sqrt {2}} \\right)$$\n\n\u2022 For the second line of step, I multiplied top and bottom by \u221a(1+x) in order to use the trig substitution of x = sin\u03b8 \u2013\u00a0user749176 Feb 17 '20 at 21:12\n\u2022 Also those two answers are from the calculator. I was wondering if they are correct after comparing them to my answers, which are shown in the link \u2013\u00a0user749176 Feb 17 '20 at 21:27\n\nLet \\begin{align} I &= \\int \\frac{\\sqrt{1+x}}{\\sqrt{1-x}}\\;dx = \\int \\frac{1+x}{\\sqrt{1-x^2}}\\;dx \\end{align} Left side:\n\nLet $$x = \\sin\\theta$$, $$dx = \\cos\\theta\\;d\\theta$$. \\begin{align} I &= \\int \\frac{1+\\sin\\theta}{\\sqrt{\\cos^2\\theta}}\\cos\\theta\\;d\\theta \\\\ &= \\int \\left(1+\\sin\\theta \\right)d\\theta \\\\ &= \\theta - \\cos\\theta + c \\\\ &= \\arccos\\left(\\sqrt{1-x^2}\\right) - \\sqrt{1-x^2} + c \\end{align}\n\nRight side:\n\n\\begin{align} I &= \\int \\frac{1}{\\sqrt{1-x^2}}\\;dx + \\int \\frac{x}{\\sqrt{1-x^2}}\\;dx \\\\ u &= 1 - x^2,\\;\\; -\\frac{1}{2}du = x\\,dx \\\\ \\implies I &= \\arcsin x - \\frac{1}{2}\\int u^{-1/2}\\;du \\\\ &= \\arcsin x - \\sqrt{u} + c \\\\ &= \\arcsin x - \\sqrt{1 - x^2} + c \\end{align}\n\nThe answers would be the exact same, if $$\\arccos\\left(\\sqrt{1-x^2}\\right) = \\arcsin x$$. And therein lies the difference. On the \"Left side\", the substitution you originally made was $$x = \\sin\\theta$$. So when you replace $$\\theta$$ you should substitute $$\\theta = \\arcsin x$$.\n\nBy the rules of trig substitution, they should be equivalent. But canonically, the arcsin function has a range of $$-\\pi/2$$ to $$\\pi/2$$, while the arccos function has a range of $$0$$ to $$\\pi$$. So when you use $$\\arccos\\left(\\sqrt{1-x^2}\\right)$$, as-is you are losing the case where $$-1 < x < 0$$. The integral has a kink in it, but that's not what you want, seeing as how the function being integrated is continuous and differentiable at $$x=0$$.\n\nYou could shift arccos by an appropriate amount and use that solution, but I think it would be easier to go with arcsin here.\n\n\u2022 Yes the wrong substitution of \u03b8, seemed to cause the problem. But aside from substituting x back in the answer, do you know if the x's you sub back in is always supposed to correspond to the original substitution made? For example, see my updated question at the top to see what I mean \u2013\u00a0user749176 Feb 17 '20 at 21:41", "date": "2021-03-06 12:34:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3550450/which-of-these-answers-is-the-correct-indefinite-integral-using-trig-substitut", "openwebmath_score": 0.9998447895050049, "openwebmath_perplexity": 492.6315629554467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806484125337, "lm_q2_score": 0.9005297921244243, "lm_q1q2_score": 0.8830420874761722}}
{"url": "https://math.stackexchange.com/questions/1506759/solve-in-integers-the-equation-2x3y-5/1506772", "text": "Solve in integers the equation $2x+3y = 5$\n\nIf I have the following equation $$2x+3y = 5$$ I know all the integer solutions is $$x = 1+3n$$ $$y = 1-2n$$ $$n \\in \\Bbb Z$$ since I can just plug them in $$2(1+3n)+3(1-2n) = 5+6n-6n = 5$$\n\nbut I don't know how to derive the answer from the equation... also is there a name for algorithms to solve these integer function?\n\n\u2022 Browsing this tag may help. \u2013\u00a0user147263 Oct 31 '15 at 19:26\n\u2022 The extended Euclidean algorithm gives you $x,y$ with $2x+3y=5$. \u2013\u00a0Dietrich Burde Oct 31 '15 at 19:32\n\nIt has solution because $1=\\text{gcd}(2,3)\\mid 5$.\n\nLet $(x_0, y_0)$ any solution of $2x+3y=5$ i.e. for example $x_0=y_0=1$. Let $(x,y)$ any other solution i.e. $2x+3y=5$. Subtracting we get: $2(x-1)=3(1-y)$. Hence $1-y=2t$ and $x-1=3t.$\n\nThen any general solution can be find by generating formula: $$(3t+1, 1-2t), \\quad t\\in \\mathbb{Z}.$$\n\n\u2022 can I say that since $(x-1)/(1-y) = 3/2$, if I let $$(1-y) = z$$ $$z \\in \\Bbb Z$$ then $$(x-1) = 3t/2$$ and for $$3t/2 \\in \\Bbb Z$$ $$t = 2z$$ $$z \\in \\Bbb Z$$ thus $$x = 3z+1$$ $$y = 1-2z$$ I know your argument is correct, I am just having a hard time convincing myself that it is the only solution and I can reuse such algorithm for similar problems \u2013\u00a0watashiSHUN Oct 31 '15 at 20:44\n\nTheorem 1: The $\\gcd(a,b)$ (where $a$ and $b$ not both $0$) is the least positive integer $ax + by$ for some $x,y\\in\\mathbb{Z}$.\n\nTheorem 2: Every integer $ax+by$ is a multiple of $\\gcd(a,b)$, and every multiple of $\\gcd(a,b)$ is $ax+by$ for some $x,y\\in\\mathbb{Z}$.\n\nTheorem 3: $ax+by=c$ has a solution $\\iff \\gcd(a,b)\\mid c$. If it does, it has infinitely many solutions, all of which are $x=x_0+(b/\\gcd(a,b))n,\\; y = y_0-(a/\\gcd(a,b))n$, where $n\\in\\mathbb{Z}$ and $(x_0, y_0)$ is one particular solution.\n\nSo to solve $ax+by=c$ completely, it boils down to finding one solution to $ax+by=\\gcd(a,b)$. This can be done by the Extended Euclidean Algorithm.\n\nif $2x+3y=5$, then $3y = 5-2x$. We then focus on making $5-2x$ divisible by $3$. This will happen for positive $x$ when $x = [1, 4, 7...]$ and for negative $x$ when $x = [-2, -5, -8...]$ If we line these up as $[...-8, -5, -2, 1, 4, 7...]$ we see that we have a common difference of $3$, yielding $3n$ or $-3n$, although both off by $1$, so we set $x = 1+3n$ or $x= 1-3n$\n\nSimilarly, we set $2x = 5 - 3y$, and desire to find $y$ such that $5-3y$ is divisible by $2$. This happens at $y = [...-5, -3, -1, 1, 3, 5...]$ We see that we have a common difference of $2$ with the values shifted up by $1$, yielding $y=1+2n$ or $y=1-2n$\n\nChecking the conditional expressions for what we want, we have our final solution, $(1\\pm3n, 1\\mp2n), \\quad n\\in \\mathbb{Z} \\quad$ (note that the sign on the equation for $y$ must be the opposite for $x$)\n\nFirst you find a particular solution. Say $(x, y) = (1, 1)$.\n\nThen you suppose that $2x + 3y = 5$ for some $(x, y)$\n\nSo \\begin{align} 2x + 3y &= 2(1) + 3(1)\\\\ 2(x-1) &= -3(y-1)\\\\ \\end{align}\n\nHence $2 \\mid -3(y-1)$.\n\nSince $2$ and $-3$ are relatively prime, we must have $2 \\mid y-1$. That is $y - 1 = 2t$ for some integer $t$.\n\nSo $y = 2t + 1$ for some integer $t$.\n\nSubstituting this back into $2(x-1) = -3(y-1)$, we find $x = -3t + 1$.\n\nWhat we have shown is that, if $(x, y)$ is a solution to $2x + 3y = 5$, then $(x,y) = (-3t+1, 2t+1)$ for some integer $t$.\n\nIt is easy to verify that if $(x,y) = (-3t+1, 2t+1)$ for some integer $t$, then $(x, y)$ is a solution to $2x + 3y = 5$.\n\nHence $(x, y)$ is a solution to $2x + 3y = 5$, if and only if $(x,y) = (-3t+1, 2t+1)$ for some integer $t$.", "date": "2019-07-20 11:39:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1506759/solve-in-integers-the-equation-2x3y-5/1506772", "openwebmath_score": 0.9701393842697144, "openwebmath_perplexity": 81.34291763728329, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98058065352006, "lm_q2_score": 0.9005297821135385, "lm_q1q2_score": 0.8830420822591708}}
{"url": "https://etec-radioetv.com.br/tags/j93v1ui.php?page=subsets-of-a-set-06a588", "text": "Problem description . In axiomatic set theory (as developed, for example, in the ZFC axioms), the existence of the power set of any set is postulated by the axiom of power set. b) List all the distinct subsets for the set {S,L,E,D}. c) How many of the distinct subsets are proper subsets? A subset that is smaller than the complete set is referred to as a proper subset. is a subset of (written ) iff every member of is a member of .If is a proper subset of (i.e., a subset other than the set itself), this is written .If is not a subset of , this is written . To ensure that no subset is missed, we list these subsets according to their sizes. Admin AfterAcademy 31 Dec 2019. For example: Plants are a subset of living things. The set of living things is very big: it has a lot of subsets. Since $$\\emptyset$$ is the subset of any set, $$\\emptyset$$ is always an element in the power set. Examples. So the set {1, 2} is a proper subset of the set {1, 2, 3} because the element 3 is not in the first set\u2026 Example 29 List all the subsets of the set { \u20131, 0, 1 }. Animals are a subset of living things. (The notation is generally not used, since automatically means that and cannot be the same.). SUBSETS. Human beings are a subset of animals. So there are a total of $2\\cdot 2\\cdot 2\\cdot \\dots \\cdot 2$ possible resulting subsets, all the way from the empty subset, which we obtain when we say \u201cno\u201d each time, to the original set itself, which we obtain when we say \u201cyes\u201d each time. An area of intersection is then defined which contains all the common elements. Subset intersection: sometimes, various sets are different but share some common elements. Print all subsets of a given set. \u26f2 Example 5: Distinct Subsets a) Determine the number of distinct subsets for the set {S,L,E,D}. In mathematics, the power set (or powerset) of a set S is the set of all subsets of S, including the empty set and S itself. Let us evaluate $$\\wp(\\{1,2,3,4\\})$$. Next, list the singleton subsets (subsets with only one element). The subset of {1,2,3,4} are {},{1}, {2}, {3}, {4} {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4} {1,3,4}, {2,3,4} and {1,2,3,4} set ={1,2,3,4} has 16 subsets. The set is not necessarily sorted and the total number of subsets of a given set of size n is equal to 2^n. AfterAcademy. So for the whole subset we have made $n$ choices, each with two options. This is the subset of size 0. Solution: a) Since the number of elements in the set is 4, the number of distinct subsets \u2026 Difficulty: MediumAsked in: Facebook, Microsoft, Amazon Understanding the Problem . Let A= { \u20131, 0, 1} Number of elements in A is 3 Hence, n = 3 Number of subsets of A = 2n where n is the number of elements of the set A = 23 = 8 The subsets of {\u20131, 0, 1} are , {\u22121}, {0}, {1}, {\u22121, 0}, {0, 1}, {\u22121, 1}, and {\u22121, 0, 1} Show More. A subset is a portion of a set. The powerset of S is variously denoted as P (S), (S), P(S), \u2119(S), \u2118(S) (using the \"Weierstrass p\"), or 2 S. Interview Kit Blogs Courses YouTube Login. Subset.\n\n.\n\nCold Smoke Brats, Lenovo Yoga 720-13ikb Parts, E Aeolian Scale, Zucchini Spinach Ricotta Rolls, 32-bit Ram Limit, Jeskai Ascendancy Mtg, Sparrow Sound Effect, Best Air Fryer Rotisserie, Dehydrator,", "date": "2021-01-20 18:52:15", "meta": {"domain": "com.br", "url": "https://etec-radioetv.com.br/tags/j93v1ui.php?page=subsets-of-a-set-06a588", "openwebmath_score": 0.7348184585571289, "openwebmath_perplexity": 565.9167675346129, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918516137419, "lm_q2_score": 0.8933094088947399, "lm_q1q2_score": 0.8830290716623387}}
{"url": "http://math.stackexchange.com/questions/92306/partitions-in-which-no-part-is-a-square?answertab=votes", "text": "# Partitions in which no part is a square?\n\nI asked a similar question earlier about partitions, and have a suspicion about another way to count partitions.\n\nIs it true that the number of partitions of $n$ in which each part $d$ is repeated fewer than $d$ times is equal to the number of partitions of $n$ where no part is a square?\n\nI tried this out for $n=2,3,4$, and so far it checks out. Is there a way to prove it more generally?\n\n-\nThe natural thing to do in such a case is to write generating functions for the two cases and see if they look similar. Even if you can't decide, computing a number of terms of the series is much faster than testing cases by hand. Did you try? \u2013\u00a0 Marc van Leeuwen Dec 17 '11 at 19:04\n\nAs mentioned in the comment above, generating functions are the standard tool for proving such statements. If you can write down the proper generating functions, proofs sometimes just fall right out.\n\nLet $S = \\{ \\ell^2 \\;|\\; \\ell \\in \\mathbb{N} \\}$ be the set of perfect squares.\n\nIn the following generating function: the coefficient of $x^k$ is the number of partitions of $k$ which do not involve $d$ or more copies of $d$ for each $d$.\n\n$$\\prod\\limits_{d=1}^\\infty \\left(\\sum\\limits_{j=0}^{d-1} x^{jd}\\right) =$$\n\n$$(1+x^2)(1+x^3+x^6)(1+x^4+x^8+x^{12})\\cdots (1+x^d+x^{2d}+\\cdots+x^{d(d-1)}) \\cdots$$\n\n$$=((1-x^2)^{-1}-x^{2^2}(1-x^2)^{-1})((1-x^3)^{-1}-x^{3^2}(1-x^3)^{-1})\\cdots ((1-x^k)^{-1}-x^{k^2}(1-x^k)^{-1}) \\cdots$$\n\n$$=\\left(\\frac{1-x^{1^2}}{1-x^1}\\right)\\left(\\frac{1-x^{2^2}}{1-x^2}\\right)\\left(\\frac{1-x^{3^2}}{1-x^3}\\right)\\cdots \\left(\\frac{1-x^{k^2}}{1-x^k}\\right) \\cdots$$\n\n$$=\\prod\\limits_{k = 1}^\\infty \\frac{1-x^{k^2}}{1-x^k} = \\prod\\limits_{k = 1}^\\infty \\frac{\\frac{1}{1-x^k}}{\\frac{1}{1-x^{k^2}}} =\\frac{\\prod\\limits_{k = 1}^\\infty \\frac{1}{1-x^k}}{\\prod\\limits_{k=1}^\\infty\\frac{1}{1-x^{k^2}}} = \\prod\\limits_{k\\in\\mathbb{N}-S} \\frac{1}{1-x^k}$$\n\nIn the final generating function: the coefficient of $x^k$ counts the number of partitions not involving perfect squares.\n\nEdit For a bit more about generating functions here is a link to another question I answered: partitions and generating functions\n\n-\nThank you Bill, this is a great answer. \u2013\u00a0 Noel Dec 18 '11 at 6:18\n@Bill: In the last line, wouldn't you prefer just cancelling the terms $1-x^{k^2}$ against their counterparts in the denominator right away? \u2013\u00a0 Marc van Leeuwen Dec 18 '11 at 17:13\n@MarcvanLeeuwen I guess that would have been a bit more efficient. Oh well. :) Since it doesn't make a big difference, I'll just leave it alone. \u2013\u00a0 Bill Cook Dec 18 '11 at 18:46\n\nNow that you have a generating function proof (so you know the result is true), you may wonder if you can actually map partitions of one kind bijectively to those of the other kind. Motivated by this other question, the following idea comes to mind: starting with a partition in which no part $d$ is repeated $d$ times or more, in order to get rid of square parts, break any part $x^2$ into $x$ parts equal to $x$, and repeat. Since there were no parts equal to $1$ this terminates, producing a partition without square parts.\n\nWe must show that every partition of without square parts is produced exactly once from a partition in which no part $d$ is repeated more $d$ times or more. We can treat non-squares separately: if $k>1$ is a non-square, then the parts in the original partition that would have been broken up eventually into parts of size $k$ are those of size $k$, $k^2$, $k^4$, $k^8$, ..., $k^{2^i}$,... Supposing the multiplicity of $k$ in the final partition is $m$, we must decompose $$mk=c_0k+c_1k^2+c_2k^4+\\cdots+c_ik^{2^i}+\\cdots \\quad\\text{with c_i<k^{2^i} for all i\\in\\mathbf N}$$ But this is uniquely possibly by the expression of $mk$ in the mixed radix number system with place values $1,k,k^2,k^4,k^8,\\ldots$ (the initial $1$ is only there to have a number system capable of expressing all natural numbers; $mk$ being a multiple of $k$ has digit $0$ on this least significant position.) Concretely one can find the $c_i$ either in order of increasing $i$ by taking remainders modulo the next $k^{2^{i+1}}$, or by decreasing $i$ by allocating the largest chunks first and then using smaller chunks for what is left of $mk$.\n\n-\nThanks for this other argument, Marc van Leeuwen. \u2013\u00a0 Noel Dec 19 '11 at 7:05\nSee Wilf's Lectures on Integer Partitions for more on bijections like this. \u2013\u00a0 David Bevan Dec 19 '11 at 8:56\n@David Bevan: Thank you for the reference. So apparently this bijection, as well as the one in the question I linked to, are special cases of a general construction of partition maps proved to be bijective by Kathleen O'Hara. \u2013\u00a0 Marc van Leeuwen Dec 21 '11 at 14:26", "date": "2014-10-20 22:05:32", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/92306/partitions-in-which-no-part-is-a-square?answertab=votes", "openwebmath_score": 0.8449915051460266, "openwebmath_perplexity": 184.4740315336613, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990140146733106, "lm_q2_score": 0.8918110511888303, "lm_q1q2_score": 0.8830179250823139}}
{"url": "https://math.stackexchange.com/questions/1527968/can-the-roots-of-the-derivative-of-the-polynomial-in-complex-variable-be-as-clos", "text": "Can the roots of the derivative of the polynomial in complex variable be as close as we want them to be from the roots of the polynomial itself?\n\nThe (probably) famous Gauss-Lucas theorem states that the roots of the derivative $P'(z)$ are contained in the convex hull of the roots of $P(z)$, where $P(z)$ is complex variable polynomial.\n\nI am interested here in could it be the case that we always have some polynomial of any degree (except $1$) $P(z)$ such that some root of its derivative is \"at a small as we want distance\" from some root of $P(z)$.\n\nTo be more precise, here is the statement of the question:\n\nIs it true that for every $\\varepsilon>0$ and for every $n\\in \\mathbb N \\setminus \\{1\\}$ there exists polynomial $P(z)$ in complex variable of degree $n$ with $n$ different roots such that there is root $z_a$ of $P'(z)$ and root $z_b$ of $P(z)$ which are such that we have $|P'(z_a)-P(z_b)|<\\varepsilon$\n\n\u2022 Yes, start with a polynomial with a double root and perturb the coefficients. \u2013\u00a0Michael Burr Nov 14 '15 at 1:11\n\u2022 @MichaelBurr All roots are different in the question, if that changes anything? \u2013\u00a0Farewell Nov 14 '15 at 1:13\n\u2022 If you insist that the coefficients of the polynomial are bounded integers, then the answer is no. \u2013\u00a0Michael Burr Nov 14 '15 at 1:13\n\u2022 No, it doesn't. After perturbation, the roots will all be different, but because the roots depend continuously on the coefficients, there will be a root of the derivative arbitrarily close to a root of the function. \u2013\u00a0Michael Burr Nov 14 '15 at 1:14\n\u2022 @MichaelBurr If we choose some $\\varepsilon_0>0$ is there an easy way to construct some concrete polynomial for every degree $n>1$ (it is a sequence of polynomials for every concrete $\\varepsilon>0$ ) such that the question holds? I do not see immediately that this is the case. \u2013\u00a0Farewell Nov 14 '15 at 1:20\n\nLet $P(z) = (z-\\epsilon)(z-2\\epsilon)\\cdots (z-n\\epsilon).$ Using the mean value theorem we can see that every root of $P'$ is less than $\\epsilon$ away from some root of $P.$\nLet $P(z)=(z-\\epsilon)(z+\\epsilon)\\prod_{i=1}^{n-2}(z-i)$. This is a perturbation of the polynomial $z^2\\prod_{i=1}^{n-2}(z-i)$, removing the double root.", "date": "2021-02-27 19:47:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1527968/can-the-roots-of-the-derivative-of-the-polynomial-in-complex-variable-be-as-clos", "openwebmath_score": 0.8334363102912903, "openwebmath_perplexity": 110.90299211567454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9901401429507652, "lm_q2_score": 0.8918110533454179, "lm_q1q2_score": 0.8830179238445046}}
{"url": "https://new.procespartner.sk/data/styrene-resin-kwai/article.php?id=0593ed-side-of-rhombus-formula", "text": "Diagonals divide a rhombus into four absolutely identical right-angled triangles. The area of the rhombus can be found, also knowing its diagonal. There are many ways to calculate its area such as using diagonals, using base and height, using trigonometry, using side and diagonal. Basic formulas of a rhombus. Then we obtain exactly the formula of the Theorem. Solution: All the sides of a rhombus are congruent, so HO = (x + 2).And because the diagonals of a rhombus are perpendicular, triangle HBO is a right triangle.With the help of Pythagorean Theorem, we get, (HB) 2 + (BO) 2 = (HO) 2x 2 + (x+1) 2 = (x+2) 2 x 2 + x 2 + 2x + 1 = x 2 + 4x + 4 x 2 \u2013 2x -3 = 0 Solving for x using the quadratic formula, we get: x = 3 or x = \u20131. Area of plane shapes. Other Names. Since a rhombus is a parallelogram in which all sides are equal, all the same formulas apply to it as for a parallelogram, including the formula for finding the area through the product of height and side. By \u2026 Online calculators and formulas for a rhombus \u2026 Ask your question. If the side length and one of the angles of the rhombus are given, the area is: A = a 2 \u00d7 sin(\u03b8) Since all four sides of a rhombus are equal, much like a square, the formula for the perimeter is the product of the length of one side with 4 $$P = 4 \\times \\text{side}$$ Angles of a Rhombus Formula of Area of Rhombus / Perimeter of Rhombus. Example Problems. The proof is completed. So, the perimeter of the rhombus is 64 cm. where b is the base or the side length of the rhombus, and h is the corresponding height. . In geometry, a rhombus or rhomb is a quadrilateral whose four sides all have the same length. Hello!!! We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. The area of the rhombus can be found, also knowing its diagonal. [3] P = 4s P = 4(10) = 40 Yes, because a square is just a rhombus where the angles are all right angles. Example: A rhombus has a side length of 12 cm, what is its Perimeter? We will saw each of them one by one below. Using side and height. The inradius (the radius of a circle inscribed in the rhombus), denoted by r, can be expressed in terms of the diagonals p and q as = \u22c5 +, or in terms of the side length a and any vertex angle \u03b1 or \u03b2 as Calculate the unknown defining areas, angels and side lengths of a rhombus with any 2 known variables. Any isosceles triangle, if that side's equal to that side, if you drop an altitude, these two triangles are going to be symmetric, and you will have bisected the opposite side. Join now. Free Rhombus Sides & Angles Calculator - calculate sides & angles of a rhombus step by step This website uses cookies to ensure you get the best experience. Join now. Use Heron's Formula. Example 2 : If the perimeter of a rhombus is 72 inches, then find the length of each side. Sitemap. Now the area of triangle AOB = \u00bd * OA * OB = \u00bd * AB * r (both using formula \u00bd*b*h). Home List of all formulas of the site; Geometry. Abhishek241 Abhishek241 19.08.2017 Math Secondary School +5 pts. Here at Vedantu you will learn how to find the area of rhombus and also get free study materials to help you to score good marks in your exams. We should recall several things. Area Of [\u2026] Many of the area calculations can be applied to them also. Problem 1: Find the perimeter of a rhombus with a side length of 10. The formula for perimeter of a rhombus is given as: P = 4s Where P is the perimeter and s is the side length. The rhombus is often called a diamond, after the diamonds suit in playing cards, or a lozenge, though the latter sometimes refers specifically to a rhombus with a 45\u00b0 angle. A rhombus is a polygon having 4 equal sides in which both the opposite sides are parallel, and opposite angles are equal.. The diagonals of a rhombus bisect each other as it is a parallelogram, but they are also perpendicular to each other. Area Of Rhombus Formula. X Research source You could also use the formula P = S + S + S + S {\\displaystyle P=S+S+S+S} to find the perimeter, since the perimeter of any polygon is the sum of all its sides. This rhombus calculator can help you find the side, area, perimeter, diagonals, ... On the other hand if the perimeter (P) is given the side (a) can be obtained from it by this formula: a = P / 4 When side (a) and angle (A) are provided the figures that can be computed \u2026 What is the formula of Rhombus When one side is given Get the answers you need, now! Answered Formula for side of rhombus when diagonals are given 2 A rhombus is often called as a diamond or diamond-shaped. Given the length of diagonal \u2018d1\u2019 of a rhombus and a side \u2018a\u2019, the task is to find the area of that rhombus. This formula for the area of a rhombus is similar to the area formula for a parallelogram. So by the same argument, that side's equal to that side, so the two diagonals of any rhombus are perpendicular to \u2026 Formula for perimeter of a rhombus : = 4s Substitute 16 for s. = 4(16) = 64. The \"base times height\" method First pick one side to be the base. For our MAH, the three sides measure: MA = 7 cm; AH = 13.747 cm; HM = 14 cm Heron's Formula depends on knowing the semiperimeter, or half the perimeter, of a triangle. Second, the diagonals of a rhombus are perpendicular bisectors of each other, thus giving us four right triangles and splitting each diagonal in \u2026 How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If we use Heron\u2019s formula then we find area of one triangle made by two sides and a diagonal then twice of this area is area of rhombus. By applying the perimeter formula, the solution is: Check: 1. Given two integers A and X, denoting the length of a side of a rhombus and an angle respectively, the task is to find the area of the rhombus.. A rhombus is a quadrilateral having 4 sides of equal length, in which both the opposite sides are parallel, and opposite angles are equal.. These formulas are a direct consequence of the law of cosines. Log in. Since a rhombus is also a parallelogram, we can use the formula for the area of a parallelogram: A = b\u00d7h. Examples: Input: d = 15, a = 10 Output: 99.21567416492215 Input: d = 20, a = 18 Output: 299.3325909419153 Q. Thus, the total perimeter is the sum of all sides. Calculator online for a rhombus. Side of a Rhombus when Diagonals are given calculator uses Side A=sqrt((Diagonal 1)^2+(Diagonal 2)^2)/2 to calculate the Side A, Side of a Rhombus when Diagonals are given can be defined as the line segment that joins two vertices in a rhombus provided the value for both the diagonals are given. First, all four sides of a rhombus are congruent, meaning that if we find one side, we can simply multiply by four to find the perimeter. The area of the rhombus is given by the formula: Area of rhombus = sh. Formula for side of rhombus when diagonals are given - 1399111 1. The perimeter formula for a rhombus is the same formula used to find the perimeter of a square. Area of a triangle; Area of a right triangle Here, r is the radius that is to be found using a and, the diagonals whose values are given. Click hereto get an answer to your question \ufe0f Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm . When the altitude or height and the length of the sides of a rhombus are known, the area is given by the formula; Area of rhombus = base \u00d7 height. This geometry video tutorial explains how to calculate the area of a rhombus using side lengths and diagonals based on a simple formula. 4s = 72. A rhombus is actually just a special type of parallelogram. Choose a formula based on the values you know to begin with. Using side and angle. Solution: Since we are given the side length, we can plug it straight into the formula. Since the rhombus is the parallelogram which has all the sides of the same length, we can substitute b = a in this formula. Log in. Area of a Rhombus Formula - A rhombus is a parallelogram in which adjacent sides are equal. Solution : Perimeter of the rhombus = 72 inches. Rhombus Area Formula. This formula was proved in the lesson The length of diagonals of a parallelogram under the current topic Geometry of the section Word problems in this site. The side of rhombus is a tangent to the circle. Ask your question. Diagonals divide a rhombus into four absolutely identical right-angled triangles. Inradius. A rhombus is a special type of quadrilateral parallelogram, where the opposite sides are parallel and opposite angles are equal and the diagonals bisect each other at right angles. There are 3 ways to find the area of Rhombus.Find the formulas for same and Perimeter of Rhombus in the table below. It is more common to call this shape a rhombus, but some people call it \u2026 Its diagonals perpendicularly bisect each other. What is the area of a rhombus when only a side is given, and nothing else? The formula to calculate the area of a rhombus is: A = \u00bd x d 1 x d 2. where... 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And side lengths of a rhombus also a parallelogram, but some people call it \u2026 Using side height... Perimeter, of a rhombus: = 4s Substitute 16 for s. = 4 \u00d7 12 cm, is. Height '' method First pick one side to be the base or the side length each., also knowing its diagonal is 8 cm long, find the of!: if the perimeter formula for a rhombus has a side is given, and nothing else whose are... Will saw each of them one by one below to the area of rhombus When only side! By one below straight into the formula: area of the rhombus can be applied them!, have equivalent sides is a quadrilateral whose four sides all have the same length find! To solve this problem, apply the perimeter formula for a parallelogram: rhombus. Here, r is the base formulas of the rhombus is often called as diamond...", "date": "2021-05-15 17:13:42", "meta": {"domain": "procespartner.sk", "url": "https://new.procespartner.sk/data/styrene-resin-kwai/article.php?id=0593ed-side-of-rhombus-formula", "openwebmath_score": 0.8021650314331055, "openwebmath_perplexity": 423.7974277608552, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9835969645988575, "lm_q2_score": 0.8976953016868439, "lm_q1q2_score": 0.8829703738738353}}
{"url": "https://math.stackexchange.com/questions/4059113/time-speed-and-distance-trains-partial-distance", "text": "# Time, Speed and Distance : Trains Partial Distance\n\nTwo trains $$A$$ and $$B$$ start from station $$X$$ and $$Y$$ towards each other. $$B$$ leaves station $$Y$$ half an hour after train $$A$$ leaves station $$X$$. Two hours after train $$A$$ has started, the distance between train $$A$$ and train $$B$$ is $$\\frac{19}{30} th$$ of the distance between $$X$$ and $$Y$$. How much time it would take each train ($$A$$ and $$B$$) to cover the distance $$X$$ to $$Y$$, if train $$A$$ reaches half an hour later to its destination as compared to $$B$$ $$?$$\n\nMy solution approach :-\n\nLet the distance between $$X$$ and $$Y$$ be $$x$$.\n\nLet the speed of train $$A$$ be $$a$$ kmph and of train $$B$$ be $$b$$ kmph.\n\nAs per question $$2a + 1.5b = \\frac{11x}{30}$$ --Eq.(i) (Distance travelled by them i.e. Total distance $$-$$ Distance left between them $$= x-\\frac{19x}{30}$$\n\nNow we know that train $$A$$ reaches half an hour later to its destination as compared to $$B$$, so:-\n\n$$x/b + 0.5 = x/a$$ --Eq.(ii)\n\nI am stuck here as you can see that I have got three variables and just two equations I can form from the question. What am I missing here? Please help!\n\n\u2022 The question asks for the values of x/a and x/b. These values can be solved using your two equations, even though you won't know x, a, and b individually. So make a change of variables to r=x/a and s=x/b and try to solve r and s. Mar 12 '21 at 14:49\n\u2022 Hint: B started a half hour late, and finished a half hour early. Therefore, B took exactly 1 hour less than A to cover the same distance. Mar 12 '21 at 15:01\n\u2022 ohhhk....such a silly mistake i did with the 2nd equation..and also I was trying to figure out the third equation in order to solve the quations.....i got it now...maybe that is what happens when you solve math questions for 5 hours straight..i should take a break now...thanks for all the help from everyone... Mar 12 '21 at 15:43\n\nYou can simplify the working. Say, time taken by $$B$$ to cover distance $$d$$ between stations $$X$$ and $$Y$$ is $$t$$ hours. Then time taken by $$A$$ is $$(t+1)$$ hours (as $$A$$ starts $$30$$ mins earlier and reaches $$30$$ mins later) and speed of train $$A$$ is $$\\displaystyle \\frac{d}{t+1}$$ and of train $$B$$ is $$\\displaystyle \\frac{d}{t}$$.\n\nSo, $$\\displaystyle \\frac{2d}{t+1} + \\frac{1.5 d}{t} = \\frac{11d}{30}$$\n\nTake out $$d$$ from both sides and solve for $$t$$ which comes to $$9$$ hours. That is time taken by train $$B$$. So time taken by $$A$$ is $$10$$ hours.\n\nNote: While the question most likely meant that they have not crossed each other but it should have been more explicit. They can be at a distance of $$\\frac{19d}{30}$$ even after having crossed each other, which is represented by the equation $$\\displaystyle \\frac{2d}{t+1} + \\frac{1.5 d}{t} = \\frac{49d}{30}$$ and it does have a valid solution.\n\n\u2022 yeah...that could be a scenario too...that thought never came to my mind... if you don't mind..can you please explain a little of getting the total distance travelled in this scenario is 49d/30? Mar 13 '21 at 3:07\n\u2022 Also when i solved the equation the solution came out to be t = 1.6872 hours. Mar 13 '21 at 3:17\n\u2022 i got it... 1 + 19d/30 = 49d/30... Mar 13 '21 at 3:25\n\u2022 Yes, B takes 1.6872 hours and A takes 2.6872 hours is another solution. Mar 13 '21 at 4:17\n\u2022 As you said it is d + 19d /30. When they meet, they have together covered distance d and then they together cover 19d/30 as they are 19d/30 apart. Mar 13 '21 at 4:23\n\nAs noted in a comment, your second equation is incorrect. B started half an hour earlier than A, and arrived half an hour sooner, so took one hour less to cover the distance. We have two equations:\n\n\\begin{align}\\frac{11}{30}x &= 2a+1.5b\\\\ \\frac xa &= 1+\\frac xb \\end{align} The point you have missed is that we are not asked to find $$a,b,$$ and $$x$$ but $$\\frac xa$$ and $$\\frac xb$$. If we write $$y=\\frac xa,\\ z=\\frac xb$$ then the equations become \\begin{align} \\frac{11}{30}&=\\frac2y+\\frac{1.5}{z}\\\\ y&=1+z \\end{align} Substituting the second in the first, clearing denominators and simplifying gives $$11z^2-94z-45=0$$ whose only positive root is $$z=9$$.\n\n\u2022 Based on how the question reads, can we confidently eliminate the case where they have crossed each other and are at a distance of $\\frac{19x}{31}$ from each other? That gives a valid solution as well. Mar 12 '21 at 16:14\n\u2022 @MathLover That's a good point that never occurred to me. Mar 12 '21 at 16:25", "date": "2022-01-26 01:00:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4059113/time-speed-and-distance-trains-partial-distance", "openwebmath_score": 0.8944958448410034, "openwebmath_perplexity": 478.21906221195735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969689263265, "lm_q2_score": 0.8976952914230971, "lm_q1q2_score": 0.8829703676631937}}
{"url": "http://mathhelpforum.com/statistics/181908-am-i-more-likely-roll-least-one-six-if-i-have-more-dice.html", "text": "# Thread: Am I more likely to roll at least one six if I have more dice?\n\n1. ## Am I more likely to roll at least one six if I have more dice?\n\nHi,\n\nI've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.\n\nI say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.\n\nNow my thoughts for this are as follows:\nAt first, it doesn't matter, whether I roll them one after another, or one at a time.\nThe chances to roll a six on the first dice that I roll a 6, is 1/6.\n\nNow, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)\n\nSo for the second dice roll a six, the probability is still 1/6\nHowever, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.\n\nSo the probability to roll at least one 6 with two dice is:\n1/6 + 1/6*5/6\n\nIf I was to expand this for 1000 dice, I would end up with:\n\n$\\displaystyle $$\\sum_{k=0}^{999}\\left (\\frac{5}{6} \\right )^k \\times \\frac{1}{6}$$$\nThe probability will get ever more closely to 100%, the more dice I have, correct?\n\n2. Why get 1000 dice? Same thing if you have 1 dice: just keep rolling it.\n\n3. Well, originally I asked this: If you were to roll 1000 dice at once, how likely is it, that there is at least one 6?\n\nBut it doesn't really matter\n\n4. The possibility of rolling a 6 with one die is 1/6, and for each die you have you increase your changes of rolling 1/6. By your friend's reasoning, you're no more liking to have \"at least one head (H) in two coin flips\" than in just one coin flip. Let us use this much more basic example. The probability of flipping a head of an unbiased coin is 1/2 because we split the event space {T, H} evenly into two. When we flip two coins we have a new event space, namely {TT, TH, HT, HH}. Here the events are equally likely of 1/4, but notice \"at least one H\" can turn out in three different events. Thus, the probability of \"at least one H\" is 3/4.\n\nThe same works for dice. With one die we have an event space {1, 2, 3, 4, 5, 6}. With two dice the size of the event space is 36:\n\nSpoiler:\n11 12 13 14 15 16\n21 22 23 24 25 26\n31 32 33 34 35 36\n41 42 43 44 45 46\n51 52 53 53 55 56\n61 62 63 64 65 66\n\nNow, in how many ways can we have \"at least one six?\" In this small, but sufficient, example, we can just look at see. The bottom row and the right-most column all include a 6. This is 11/36, which is almost 0.31, compared to 1/6 =0.167. So yes, increasing the die increases your chances of rolling a six.\n\n5. Thanks, that would confirm my point of view.\n\nThey: But how is that possible? The probability never changes? So it should always be 1/6?\n\nAlso, we'd value second and third opinions Thank you very much in advance!\n\n6. The probability of a 6 emerging on a single die does not ever change, but that is not the event we're looking at when we have more than one die. When we are using, say, two dice we're looking at a different event space with each event having 1/36 chance of occurring. Of those thirty-six, eleven of them will have a 6 appear. This is undeniable. This has no impact on the probability of a six appearing on a single die. Your friends seem to be ignoring the fact that the possible ways of a 6 occurring change when you have two dice. With one die there is one, and only one, way of a 6 occurring. Thus, the probability is 1/6. With two dice we have 11 different ways, each with 1/36 = (1/6)(1/6) chance of happening. The probability of the individual dice does support this result, as we can see, but you cannot ignore the fact our event space is different when you include another die.\n\nI can go even further and define it this way by the inclusion-exclusion rule of probability theory. Let X be the outcome of the first die and Y the outcome of the second. What we want then is:\n\n$\\displaystyle P(X = 6\\ or\\ Y=6)=P(X=6)+P(Y=6)-P(X=6\\ \\&\\ Y= 6)=\\frac{1}{6}+\\frac{1}{6}-\\frac{1}{36}=\\frac{11}{36}$\n\nDo you see why these values work? The probability of the single die being six does not change, but we have to sum the two probabilities together. We then have to subtract the event that is common to both of them, otherwise we're double counting that event. This happens only once.\n\n7. Originally Posted by theyThinkImWrong\nHi,\n\nI've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.\n\nI say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.\n\nNow my thoughts for this are as follows:\nAt first, it doesn't matter, whether I roll them one after another, or one at a time.\nThe chances to roll a six on the first dice that I roll a 6, is 1/6.\n\nNow, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)\n\nSo for the second dice roll a six, the probability is still 1/6\nHowever, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.\n\nSo the probability to roll at least one 6 with two dice is:\n1/6 + 1/6*5/6\n\nIf I was to expand this for 1000 dice, I would end up with:\n\n$\\displaystyle $$\\sum_{k=0}^{999}\\left (\\frac{5}{6} \\right )^k \\times \\frac{1}{6}$$$\nThe probability will get ever more closely to 100%, the more dice I have, correct?\nLet X be the random variable 'Number of 6's rolled'.\n\nThen X ~ Binomial(n, p = 1/6).\n\nCalculate Pr(X > 0) = 1 - Pr(X = 0). Then it is obvious that as n increases so does the probability.\n\nI suggest you review the binomial distribution if you don't know it ( I do not plan to give a tutorial).", "date": "2018-05-25 02:17:15", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/181908-am-i-more-likely-roll-least-one-six-if-i-have-more-dice.html", "openwebmath_score": 0.5769526362419128, "openwebmath_perplexity": 399.29025243412707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137863531805, "lm_q2_score": 0.8991213678089741, "lm_q1q2_score": 0.8829495787931413}}
{"url": "https://math.stackexchange.com/questions/27719/what-is-gcd0-a-where-a-is-a-positive-integer", "text": "# What is $\\gcd(0,a)$, where $a$ is a positive integer?\n\nI have tried $\\gcd(0,8)$ in a lot of online gcd (or hcf) calculators, but some say $\\gcd(0,8)=0$, some other gives $\\gcd(0,8)=8$ and some others give $\\gcd(0,8)=1$. So really which one of these is correct and why there are different conventions?\n\n\u2022 I haven't encountered the convention of gcd(0,8) = 1. It depends on how you define the phrase \"a divides b\" Mar 18 '11 at 2:49\n\u2022 Mar 18 '11 at 2:53\n\u2022 @The Chaz: They are really the same things but with different names. see en.wikipedia.org/wiki/Greatest_common_divisor Mar 18 '11 at 4:04\n\u2022 Should be a, because anything is a divisor of 0. Nov 26 '20 at 7:48\n\nLet's recall the definition of  \"$\\rm a$ divides $\\rm b$\"  in a ring $\\rm\\,Z,\\,$ often written as $\\rm\\ a\\mid b\\ \\ in\\ Z.$\n\n$\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad \\rm\\ a\\mid b\\ \\ in\\ Z\\ \\iff\\ a\\,c = b\\ \\$ for some $\\rm\\ c\\in Z$\n\nRecall also the definition of $\\rm\\ gcd(a,b),\\,$ namely\n\n$(1)\\rm\\qquad\\quad \\rm gcd(a,b)\\mid a,b\\qquad\\qquad\\qquad\\$ the gcd is a common divisor\n\n$(2)\\rm\\qquad\\quad\\! \\rm c\\mid a,b\\ \\ \\ \\Longrightarrow\\ \\ c\\mid gcd(a,b)\\quad$ the gcd is a greatest common divisor\n\n$\\ \\ \\ \\$ i.e. $\\rm\\quad\\ c\\mid a,b\\ \\iff\\ c\\mid gcd(a,b)\\quad\\,$ expressed in $\\iff$ form  [put $\\rm\\ c = gcd(a,b)\\$ for $(1)$]\n\nNotice $\\rm\\quad\\, c\\mid a,0\\ \\iff\\ c\\mid a\\,\\$ so $\\rm\\ gcd(a,0)\\ =\\ a\\$ by the prior \"iff\" form of the gcd definition.\n\nNote that $\\rm\\ gcd(0,8) \\ne 0\\,$ since $\\rm\\ gcd(0,8) = 0\\ \\Rightarrow\\ 0\\mid 8\\$ contra $\\rm\\ 0\\mid x\\ \\iff\\ x = 0.$\n\nNote that $\\rm\\ gcd(0,8) \\ne 1\\,$ else $\\rm\\ 8\\mid 0,8\\ \\Rightarrow\\ 8\\mid gcd(0,8) = 1\\ \\Rightarrow\\ 1/8 \\in \\mathbb Z.$\n\nTherefore it makes no sense to define $\\rm\\ gcd(0,8)\\$to be $\\,0\\,$ or $\\,1\\,$ since $\\,0\\,$ is not a common divisor of $\\,0,8\\,$ and $\\,1\\,$ is not the greatest common divisor.\n\nThe $\\iff$ gcd definition is universal - it may be employed in any domain or cancellative monoid, with the convention that the gcd is defined only up to a unit factor. This $\\iff$ definition is very convenient in proofs since it enables efficient simultaneous proof of both implication directions. $\\$ For example, below is a proof of this particular form for the fundamental GCD distributive law $\\rm\\ (ab,ac)\\ =\\ a\\ (b,c)\\$ slightly generalized (your problem is simply $\\rm\\ c=0\\$ in the special case $\\rm\\ (a,\\ \\ ac)\\ =\\,\\ a\\ (1,c)\\ =\\ a\\,$).\n\nTheorem $\\rm\\quad (a,b)\\ =\\ (ac,bc)/c\\quad$ if $\\rm\\ (ac,bc)\\$ exists.\n\nProof $\\rm\\quad d\\mid a,b\\ \\iff\\ dc\\mid ac,bc\\ \\iff\\ dc\\mid (ac,bc)\\ \\iff\\ d|(ac,bc)/c$\n\nSee here for further discussion of this property and its relationship with Euclid's Lemma.\n\nRecall also how this universal approach simplifies the proof of the basic GCD * LCM law:\n\nTheorem $\\rm\\;\\; \\ (a,b) = ab/[a,b] \\;\\;$ if $\\;\\rm\\ [a,b] \\;$ exists.\n\nProof $\\rm\\quad d|\\,a,b \\;\\iff\\; a,b\\,|\\,ab/d \\;\\iff\\; [a,b]\\,|\\,ab/d \\;\\iff\\; d\\,|\\,ab/[a,b] \\quad\\;\\;$\n\nFor much further discussion see my many posts on GCDs.\n\nAnother way to look at it is by the divisibility lattice, where gcd is the greatest lower bound. So 5 is the greatest lower bound of 10 and 15 in the lattice.\n\nThe counter-intuitive thing about this lattice is that the 'bottom' (the absolute lowest element) is 1 (1 divides everything), but the highest element, the one above everybody, is 0 (everybody divides 0).\n\nSo $\\gcd(0, x)$ is the same as ${\\rm glb}(0, x)$ and should be $x$, because $x$ is the lower bound of the two: they are not 'apart' and 0 is '$>'$ $x$ (that is the counter-intuitive part).\n\nIn fact, the top answer can be generalized slightly: if $$a \\mid b$$, then $$\\gcd(a,b)=a$$ (and this holds in any algebraic structure where divisibility makes sense, e.g. a commutative, cancellative monoid).\n\nTo see why, well, it's clear that $$a$$ is a common divisor of $$a$$ and $$b$$, and if $$\\alpha$$ is any common divisor of $$a$$ and $$b$$, then, of course, $$\\alpha \\mid a$$. Thus, $$a=\\gcd(a,b)$$.\n\n\u2022 Indeed, even more generally, it is a special case of the distributive law - see my answer. As for commutative monoids, one usually requires them to be cancellative in order to obtain a rich theory. Mar 18 '11 at 18:26\n\nIt might be partly a matter of convention. However, I believe that stating that $$\\gcd(8,0) = 8$$ is safer. In fact, $$\\frac{0}{8} = 0$$, with no remainder. The proof of the division, indeed is that \"Dividend = divider $$\\times$$ quotient plus remainder\". In our case, 0 (dividend) = 8 (divisor) x 0 (quotient). No remainder. Now, why should 8 be the GCD? Because, while the same method of proof can be used for all numbers, proving that $$0$$ has infinite divisors, the greatest common divisor cannot be greater than $$8$$, and for the reason given above, is $$8$$.", "date": "2021-12-01 07:29:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/27719/what-is-gcd0-a-where-a-is-a-positive-integer", "openwebmath_score": 0.9506410956382751, "openwebmath_perplexity": 288.4206699451657, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464527024445, "lm_q2_score": 0.9046505428129514, "lm_q1q2_score": 0.8828904881936409}}
{"url": "https://math.stackexchange.com/questions/1624236/finding-eigenvectors-of-a-3x3-matrix-7-12-15/1624242", "text": "# Finding Eigenvectors of a 3x3 Matrix (7.12-15)\n\nPlease check my work in finding an eigenbasis (eigenvectors) for the following problem. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons).\n\nFor reference the following identity is given because some textbooks reverse the formula having $\\lambda$ subtract the diagonal elements instead of subtracting $\\lambda$ from the diagonal elements:\n\n$$det(A - \\lambda I) = 0$$ $$A = \\begin{bmatrix} 3 & 1 & 4 \\\\ 0 & 2 & 6 \\\\ 0 & 0 & 5 \\\\ \\end{bmatrix}$$\n\nBy inspection the eigenvalues are the entries along the diagonal for this upper triangular matrix. \\begin{align*} \\lambda_1 = 3 \\qquad \\lambda_2 = 2 \\qquad \\lambda_3 = 5 \\end{align*} When $\\lambda_1 = 3$ we have: $$A - 3I = \\begin{bmatrix} 3-3 & 1 & 4 \\\\ 0 & 2-3 & 6 \\\\ 0 & 0 & 5-3 \\\\ \\end{bmatrix} = \\begin{bmatrix} 0 & 1 & 4 \\\\ 0 & -1 & 6 \\\\ 0 & 0 & 2 \\\\ \\end{bmatrix} = \\begin{bmatrix} 0 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ \\end{bmatrix}$$ \\begin{align*} x_1 = 1 \\: (free \\: variable) \\qquad x_2 = 0 \\qquad x_3 = 0 \\\\ \\end{align*} $$v_1 = \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\qquad (matches \\: answer \\: in \\: text)$$ When $\\lambda_2 = 2$ we have: $$A - 2I = \\begin{bmatrix} 3-2 & 1 & 4 \\\\ 0 & 2-2 & 6 \\\\ 0 & 0 & 5-2 \\\\ \\end{bmatrix} = \\begin{bmatrix} 1 & 1 & 4 \\\\ 0 & 0 & 6 \\\\ 0 & 0 & 3 \\\\ \\end{bmatrix} = \\begin{bmatrix} 1 & 1 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ \\end{bmatrix}$$ \\begin{align*} x_1 = -x_2 \\qquad x_2 = 1 \\: (free \\: variable) \\qquad x_3 = 0 \\\\ \\end{align*} $$v_2 = \\begin{bmatrix} -1 \\\\ 1 \\\\ 0 \\\\ \\end{bmatrix} \\qquad but \\: answer \\: in \\: text \\: is \\qquad \\begin{bmatrix} 1 \\\\ -1 \\\\ 0 \\\\ \\end{bmatrix}$$ What happened? Is it from a disagreement in what we should consider arbitrary or am I doing something fundamentally wrong?\n\nWhen $\\lambda_3 = 5$ we have: $$A - 5I = \\begin{bmatrix} 3-5 & 1 & 4 \\\\ 0 & 2-5 & 6 \\\\ 0 & 0 & 5-5 \\\\ \\end{bmatrix} = \\begin{bmatrix} 2 & 1 & 4 \\\\ 0 & -3 & 6 \\\\ 0 & 0 & 0 \\\\ \\end{bmatrix} = \\begin{bmatrix} 1 & 0 & -3 \\\\ 0 & 1 & -2 \\\\ 0 & 0 & 0 \\\\ \\end{bmatrix}$$ \\begin{align*} x_1 = 3x_3 \\qquad x_2 = 2x_3 \\qquad x_3 = 1 \\: (free \\: variable) \\\\ \\end{align*} $$v_3 = \\begin{bmatrix} 3 \\\\ 2 \\\\ 1 \\\\ \\end{bmatrix} \\qquad (matches \\: answer \\: in \\: text)$$\n\n\u2022 Both your and the text book's answer work. If $(\\lambda ,v)$ is an eigenpair of a matrix $M$, then so is $(\\lambda, \\mu v)$, for all scalars $\\mu$. \u2013\u00a0Git Gud Jan 23 '16 at 23:15\n\nEigenvectors are never unique. In particular, for the eigenvalue $2$ you can take, for example, $x_2=-1$ which gives you the answer in the book.\n\n\u2022 So what you are saying is that both I and the book's answers are correct? So its just a matter of preference? \u2013\u00a0Jules Manson Jan 23 '16 at 23:24\n\u2022 Precisely, and in fact you could take for $x_2$ any nonzero real number. \u2013\u00a0John B Jan 23 '16 at 23:26\n\u2022 In your original post, you said that the problem was to find an eigenbasis. In other words, since there are three distinct eigenvalues, find one eigenvector for each eigenvalue. Which one you choose out of the infinite number of eigenvalues that correspond to each eigenvalue is \"a matter of preference\" but it is important that you understand that any multiple of an eigenvector is also an eigenvector. \u2013\u00a0user247327 Dec 13 '16 at 16:18", "date": "2021-06-21 15:46:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1624236/finding-eigenvectors-of-a-3x3-matrix-7-12-15/1624242", "openwebmath_score": 0.9983172416687012, "openwebmath_perplexity": 247.0337446583284, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464471055739, "lm_q2_score": 0.9046505370289057, "lm_q1q2_score": 0.88289047748551}}
{"url": "http://gusx.lampertifashion.it/application-of-laplace-transform-pdf.html", "text": "5 Other transforms. As we saw in the last section computing Laplace transforms directly can be fairly complicated. a signal such that $$x(t)=0$$ for $$x<0$$. Schiff and others published The Laplace Transform: Theory and Applications | Find, read and cite all the research you need on ResearchGate. With the increasing complexity of engineering problems, Laplace transforms help in solving complex problems with a very simple approach just like the applications of transfer functions to solve ordinary differential equations. (b) Compute the Laplace transform of f. A very simple application of Laplace transform in the area of physics could be to find out the harmonic vibration of a beam which is supported at its two ends. In mathematics, the Laplace transform is an integral transform named after its inventor Pierre-Simon Laplace ( / l\u0259\u02c8pl\u0251\u02d0s / ). The Laplace transform, theory and applications. In anglo-american literature there exist numerous books, devoted to the application of the Laplace transformation in technical domains such as electrotechnics, mechanics etc. Capacitor. e \u2212tsin 2 t 5. The Laplace transform is defined as follows: F^(p) = Z +1 1. The application of Laplace Transforms is wide and is used in a variety of. Each Outline presents all the essential course information in an easy-to-follow,. In India, we are facing various types of crimes. 2-3 Circuit Analysis in the s Domain. the Fourier cosine transform, and the Fourier sine transform, as applied to various standard functions, and use this knowledge to solve certain ordinary and partial differential equations. , \ud835\udc47 is a (random) time to failure), the Laplace transform of ( ) can also be interpreted as the expected value of the random variable \ud835\udc4c= \u2212 \ud835\udc47 , i. com solve differential with laplace transform, sect 7. Laplace Transform of Periodic Functions, Convolution, Applications 1 Laplace transform of periodic function Theorem 1. Laplace Transforms and Properties. Laplace Transform in Engineering Analysis Laplace transforms is a mathematical operation that is used to \"transform\" a variable (such as x, or y, or z, or t)to a parameter (s)- transform ONE variable at time. A Coupled Method of Laplace Transform and Legendre Wavelets for Lane-Emden-Type Differential Equations Yin, Fukang, Song, Junqiang, Lu, Fengshun, and Leng, Hongze, Journal of Applied Mathematics, 2012. txt) or view presentation slides online. To solve constant coefficient linear ordinary differential equations using Laplace transform. With the increasing complexity of engineering problems, Laplace transforms help in solving complex problems with a very simple approach just like the applications of transfer functions to solve ordinary differential equations. Abel's integral equation. a \ufb01nite sequence of data). Yes, the Laplace transform has \"applications\", but it really seems that the only application is solving differential equations and nothing beyond that. Chapter 3: The z-Transform and Its Application Power Series Convergence IFor a power series, f(z) = X1 n=0 a n(z c)n = a 0 + a 1(z c) + a 2(z c)2 + there exists a number 0 r 1such that the series I convergences for jz cjr I may or may not converge for values on jz cj= r. The Laplace transform f(p), also denoted by L{F(t)} or Lap F(t), is defined by the integral involving the exponential parameter p in the kernel K = e \u2212pt. The fact that the inverse Laplace transform is linear follows immediately from the linearity of the Laplace transform. \u2022 All we need is to express F(s) as a sum of simpler functions of the forms listed in the Laplace transform table. The Laplace transform can be used in some cases to solve linear differential equations with given initial conditions. Laplace Transform The Laplace transform is a method of solving ODEs and initial value problems. The vibrational analysis of structures use Laplace transforms. 5 Application of Laplace Transforms to Partial Di\ufb00erential Equations In Sections 8. The Laplace transform pair for. Consider the differential equation given by: can represent many different systems. LTI System Analysis with the Laplace Transform. The differential inverse transform of \ud835\udc48 , is define by , of the form in (1). Laplace transform and its applications 1. Since the m. Colophon An annotatable worksheet for this presentation is available as Worksheet 6. It follows that the output Y(s) can be written as the product of G(s) and. There are two (related) approaches: Derive the circuit (differential) equations in the time domain, then transform these ODEs to the s-domain; Transform the circuit to the s-domain, then derive the circuit equations in the s-domain (using the concept of \"impedance\"). Let f(t) be de ned for t 0:Then the Laplace transform of f;which is denoted by L[f(t)] or by F(s), is de ned by the following equation L[f(t)] = F(s) = lim T!1 Z T 0 f(t)e stdt= Z 1 0 f(t)e stdt The integral which de ned a Laplace transform is an improper integral. Let fbe a function of t. where X(s) is the Laplace transform of the input to the system and Y(s) is the Laplace transform of the output of the system, where we assume that all initial conditions in- volved are zero. The Law of Laplace is a physical law discovered by the great French mathematician Piere-Simon Laplace (and others) which describes the pressure-volume relationships of spheres. The tautochrone problem. The Laplace Transform is widely used in following science and engineering field. 6 \u2013 8 Each function F(s) below is defined by a definite integral. Using the Laplace transform nd the solution for the following equation @ @t y(t) = e( 3t) with initial conditions y(0) = 4 Dy(0) = 0 Hint. Article full text Download PDF. Laplace Transform The Laplace transform can be used to solve di\ufb00erential equations. Laguerre transform. Find PowerPoint Presentations and Slides using the power of XPowerPoint. is identical to that of. Bateman transform. rainville Lecture 7 Circuit Analysis Via Laplace Transform Inverse Laplace Transform Of Exponential Function Basically, Poles Of Transfer Function Are The Laplace Transform Variable Values Which Causes The Tra Basically. The control action for a dynamic control system whether electrical, mechanical, thermal, hydraulic, etc. 'The Laplace Transform' is an excellent starting point for those who want to master the application of. Mesh analysis. The best way to convert differential equations into algebraic equations is the use of Laplace transformation. So let's see if we can apply that. For a resistor, the. 3 Applications Since the equations in the s-domain rely on algebraic manipulation rather than differential equations as in the time domain it should prove easier to work in the s-domain. In this dissertation, several theorems on multidimensional Laplace transforms are developed. cosh() sinh() 22 tttt tt +---== eeee 3. Linearization, critical points, and equilibria. Click Download or Read Online button to get laplace transformation book now. \u2019s) of waiting times in queues. Considering a function f (t), its corresponding Laplace Transform. Some illustrative examples will be discussed. 2012-08-12 00:00:00 A natural way to model dynamic systems under uncertainty is to use fuzzy initial value problems (FIVPs) and related uncertain systems. Solving PDEs using Laplace Transforms, Chapter 15 Given a function u(x;t) de ned for all t>0 and assumed to be bounded we can apply the Laplace transform in tconsidering xas a parameter. Applications of Fourier transform to PDEs. Bateman transform. Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. com, find free presentations research about Application Of Laplace Transform PPT. A final property of the Laplace transform asserts that 7. 6 596--607. The real and imaginary parts of s can be considered as independent quantities. 1) whenever the limit exists (as a \ufb01nite number). (1975) Numerical inversion of the Laplace transform by accelerating the convergence of Bromwick's integral. cos(2t) + 7sin(2t) 3. Application to laplace transformation to electric circuits by J Irwin. An advantage of Laplace transform We can transform an ordinary differential equation (ODE) into an algebraic equation (AE). Find PowerPoint Presentations and Slides using the power of XPowerPoint. Laplace transform is named in honour of the great French mathematician, Pierre Simon De Laplace (1749-1827). The basic idea and arithmetics of fuzzy sets were \ufb01rst introduced by L. Fourier transforms only capture the steady state behavior. You da real mvps! $1 per month helps!!. In general we have + \u221e \u2212 \u221e \u2212 = j j F s e st ds j L F. The use of a truncated Laplace-like transformation in the construction of the analytic solution allows to overcome a small divisor phenomenon arising from the geometry of the problem and represents an alternative approach to the one proposed in a recent work by the last two authors. Prentice Hall Math Books, vertical adding and subtracting fraction sheets, the quadratic formula in ti-84, finding y-intercept of a polynomial calculator, square roots lessons, examples of algebra questions, ti calculator rom. \u222b + \u221e \u2212 \u221e = i i F s est ds i f t \u03c3 \u03c3 \u03c0 ( ) 2 1 ( ) \u03c3 Real Abscissa of convergence Isolated singularities Imaginary Laplace transform inversion is. Since the m. 1) In a layman's term, Laplace transform is used to \"transform\" a variable in a function. Apr 24, 2020 - Applications of Laplace Transformation-I Computer Science Engineering (CSE) Video | EduRev is made by best teachers of Computer Science Engineering (CSE). logo1 New Idea An Example Double Check The Laplace Transform of a System 1. Applications of Laplace Transform. The question is: How is possible to derive the. This shows the effectiveness and usefulness of the Numerical Inversion of the Laplace transform. The inverse transform L\u22121 is a linear operator: L\u22121{F(s)+ G(s)} = L\u22121{F(s)} + L\u22121{G(s)}, (2) and L\u22121{cF(s)} = cL\u22121{F(s)}, (3) for any constant c. is the Laplace domain equivalent of the time domain function. The table of Laplace transforms collects together the results we have considered, and more. Take Laplace Transform of both sides of ODE Solve for Factor the characteristic polynomial Find the roots (roots or poles function in Matlab) Identify factors and multiplicities Perform partial fraction expansion Inverse Laplace using Tables of Laplace Transforms. Similar to the application of phasortransform to solve the steady state AC circuits , Laplace transform can be used to transform the time domain circuits into S domain circuits to simplify the solution of integral differential equations to the manipulation of a set of algebraic equations. 2) \ud835\udc45 for Z-transform in Example 2. The Laplace Transform Definition and properties of Laplace Transform, piecewise continuous functions, the Laplace Transform method of solving initial value problems The method of Laplace transforms is a system that relies on algebra (rather than calculus-based methods) to solve linear differential equations. Application of Laplace Transform For Cryptographic Scheme A. 2 Properties of the z-Transform Common Transform Pairs Iz-Transform expressions that are a fraction of polynomials in z 1 (or z) are calledrational. Se voc\u00ea continuar a navegar o site, voc\u00ea aceita o uso de cookies. Ifthelimitdoesnotexist,theintegral is said todivergeand there is no Laplace transform de\ufb01ned forf. Application of Laplace Transform to Newtonian Fluid Problems Article (PDF Available) in International Journal of Science and Research (IJSR) \u00b7 July 2013 with 2,655 Reads How we measure 'reads'. It is named in honor of the great French mathematician, Pierre Simon De Laplace (1749-1827). Build your own widget. Fourier Transform is a mathematical operation that breaks a signal in to its constituent frequencies. Having carried out this procedure, we should check that this latter expression does, indeed, yield a solution of the original initial-boundary value problem. Now we going to apply to PDEs. McLachlan, quicker you could enjoy checking out the publication. The application of Laplace Transform methods is particularly e\ufb00ective for linear ODEs with constant coe\ufb03cients, and for systems of such ODEs. 6e5t cos(2t) e7t (B) Discontinuous Examples (step functions): Compute the Laplace transform of the given function. eat 1 sa 2 2 2 12. kalla is the laplace transform. The CDF of a random variable is often much more useful in practical applications but is often difficult to find. Redraw the circuit (nothing about the Laplace transform changes the types of elements or their interconnections). Laplace transform Pairs (1) Finding inverse Laplace transform requires integration in the complex plane - beyond scope of this course. 4 Multi\u2013dimensional transformation algorithms 205. Last, some boundary value problems characterized by linear partial differential equations involving heat and. Thus, Laplace Transformation transforms one class of complicated functions to produce another class of simpler functions. The Dirac delta, distributions, and generalized transforms. and scientists dealing with \"real-world\" applications. ; We will use the first approach. The Laplace Transformation is very effective device in Mathematic, Physics and other branches of science which is used to solving problem. com 1 View More View Less. Solving PDEs using Laplace Transforms, Chapter 15 Given a function u(x;t) de ned for all t>0 and assumed to be bounded we can apply the Laplace transform in tconsidering xas a parameter. The Laplace transform is a widely used integral transform (transformation of functions by integrals), similar to the Fourier transform. Definition: Laplace Transform. The method is simple to describe. can be represented by a differential equation. In particular it is shown that the Laplace transform of tf(t) is -F'(s), where F(s) is the Laplace transform of f(t). If we assume that the functions whose Laplace transforms exist are going to be taken as continuous then no two different functions can have the same Laplace transform. The Laplace Transform brings a function from the t-domain to a function in the S-domain.$\\begingroup$The Fourier transform is just a special case of the Laplace transform, so your example actually works for both. where X(s) is the Laplace transform of the input to the system and Y(s) is the Laplace transform of the output of the system, where we assume that all initial conditions in- volved are zero. The first term in the brackets goes to zero (as long as f(t) doesn't grow faster than an exponential which was a condition for existence of the transform). Application of Laplace Transform in State Space Method to Solve Higher Order Differential Equation: Pros & Cons Ms. 4) Equivalent Circuits 5) Nodal Analysis and Mesh Analysis. Unilateral Laplace Transform. Schaum's Outlines: Laplace Transforms By Murray R. The Inverse Laplace Transformation Circuit Analysis with Laplace Transforms Frequency. The Laplace transform pair for. The analytic inversion of the Laplace transform is a well-known application of the theory of complex variables. 2 Useful Laplace Transform Pairs 12. , frequency domain ). However, in all the examples we consider, the right hand side (function f(t)) was continuous. We begin with the general formula for voltage drops around the circuit: Substituting numbers, we get Now, we take the Laplace Transform and get Using the fact that , we get. Consider the differential equation given by: can represent many different systems. Yes, the Laplace transform has \"applications\", but it really seems that the only application is solving differential equations and nothing beyond that. Additional Physical Format: Online version: Watson, E. Examples of the Laplace Transform as a Solution for Mechanical Shock and Vibration Problems: Free Vibration of a Single-Degree-of-Freedom System: free. The Laplace Transform is a specific type of integral transform. 1a,b, the graphs of the Laplace transform [Lf](s) = Z\u221e 0. The \u03ba-Laplace transform proposed in this note is just one form of modified Laplace transformations. (1975) Application of best rational function approximation for Laplace transform inversion. Therefore, without further discussion, the Laplace transform is given by: De nition 1. Jacobi transform. Engineering Applications of z-Transforms 21. txt) or view presentation slides online. Hilbert-Schmidt integral operator. In machine learning, the Laplace transform is used for making predictions and making analysis in data mining. Post's Formula. 5 Other transforms. Breaking down complex differential equations into simpler polynomial forms. studysmarter. Numerical examples reveal that the pricing formulas are easy to implement and they result in accurate prices and risk parameters. The similarity of this notation with the notation used in Fourier transform theory is no coincidence; for ,. Inverse of the Laplace Transform. Laplace Transform Z Transform Fourier Transform Fourier Transform Fourier Transform Formula Fourier Transform Applications Mathematics Of The Discrete Fourier Transform A Guided Tour Of The Fast Fourier Transform Bergland Mathematics Of The Discrete Fourier Transform (dft) With Audio Applications Fourier Fourier Mathcad Fourier Series Transformada De Fourier Fourier Analysis Pdf Hc Taneja Fourier Schaum Fourier Analysis Fast Fourier Transformation Schaum Fourier Analysis Pdf Applications Of. Then, by de\ufb01nition, f is the inverse transform of F. e \u2212tsin 2 t 5. 8 The Impulse Function in. AKANBI 4 and F. Both transforms are equivalent tools, but the Laplace transform is used for continuous-time signals, whereas the$\\mathcal{Z}\\$-transform is used for discrete-time signals (i. Best & Easiest Videos Lectures covering all Most Important Questions on Engineering Mathematics for 50+ Universities Download Important Question PDF (Password mathcommentors) Will Upload soon. Fourier transforms only capture the steady state behavior. Table of LaPlace Transforms ft() L { ( )} ( )f t F s 1. 1) In a layman's term, Laplace transform is used to \"transform\" a variable in a function. The Laplace Transform can be considered as an extension of the Fourier Transform to the complex plane. 6 per cent faster than the next high-speed adder cell. Just like for the Z-transform we have to specify the ROC for the Laplace transform. Applications of Laplace Transforms Circuit Equations. Laplace Transforms for Systems of Differential Equations New Idea An Example Double Check The Laplace Transform of a System 1. Edited by: Salih Mohammed Salih. It's also the best approach for solving linear constant coefficient differential equations with nonzero initial conditions. 1) F(s) = Z 1 0 e stf(t)dt provided the improper integral converges. Post's Formula. This paper will be primarily concerned with the Laplace transform and its ap-plications to partial di erential equations. 13, 2012 \u2022 Many examples here are taken from the textbook. In my 13-year industrial career, I never used mathematical. McLachlan). hyperbolic functions. The two-sided Laplace transform (3) can be regarded as the Fourier transform of the function , and the one-sided Laplace transform (2) can be regarded as the Fourier transform of the function equal to for and equal to zero for. The function is piece-wise continuous B. When this definition is used it can be shown that the Laplace transform, Fn(s) of the nth derivative of a function, fn(t), is given by the following generic formula: Fn(s)=snF(s) - sn-1f0(0) - sn. Unit step function, Laplace Transform of Derivatives and Integration, Derivative and Integration of Laplace Transforms 1 Unit step function u a(t) De nition 1. Cryptography is one of the. For particular functions we use tables of the Laplace. In anglo-american literature there exist numerous books, devoted to the application of the Laplace transformation in technical domains such as electrotechnics, mechanics etc. As we saw in the last section computing Laplace transforms directly can be fairly complicated. Additional Physical Format: Online version: Watson, E. M-2 Shah Nisarg (130410119098) Shah Kushal(130410119094) Shah Maulin(130410119095) Shah Meet(130410119096) Shah Mirang(130410119097) Laplace Transform And Its Applications 2. The Laplace transform is a wonderful tool for solving ordinary and partial differential equations and has enjoyed much success in this realm. Laplace Transforms and their Applications About the Laplace Transformation The Laplace Transformation (named after Pierre-Simon Laplace ) is a useful mathematical tool that is used in many branches of engineering including signals and systems theory, control theory, communications, mechanical engineering, etc. Applications of Laplace Transform in Science and Engineering fields: This section describes the applications of Laplace Transform in the area of science and engineering. Redraw the circuit (nothing about the Laplace transform changes the types of elements or their interconnections). symbolizes the Laplace transform. If we look at the left-hand side, we have Now use the formulas for the L[y'']and L[y']: Here we have used the fact that y(0)=2. However, the spectral properties of the Laplace transform tend to complicate its numerical treatment; therefore, the closely related \"Truncated\" Laplace Transforms are often used in applications. zi denotes the zeros and pi denotes the poles of the linear time invariant system (LTI). cos(2t) + 7sin(2t) 3. The Laplace Transform can greatly simplify the solution of problems involving differential equations. Post's Formula. along with the Definition of Laplace Transform, Applications of Laplace Laplace Transform to Solve a Differential Equation, Ex 1, Part 1/2 Thanks to all of you who support me on Patreon. In this course, one of the topics covered is the Laplace transform. Download The Laplace Transform: Theory and Applications By Joel L. However, to transform the obtained solutions from Laplace space back to the original space, we have used the Numerical Inversion of the Laplace transform. Similar to the application of phasortransform to solve the steady state AC circuits , Laplace transform can be used to transform the time domain circuits into S domain circuits to simplify the solution of integral differential equations to the manipulation of a set of algebraic equations. pdf), Text File (. Download File PDF Applications Of Laplace Transform In Mechanical Engineering Applications Of Laplace Transform In Mechanical Engineering When somebody should go to the books stores, search creation by shop, shelf by shelf, it is in fact problematic. So far, regarding their mathematical properties [11, 12] and application [for transforms of various functions see, e. Consider the ODE in Equation [1]: We are looking for the function y (t) that satisfies Equation. Laplace transform and its applications O SlideShare utiliza cookies para otimizar a funcionalidade e o desempenho do site, assim como para apresentar publicidade mais relevante aos nossos usu\u00e1rios. 201038 Identifier-ark ark:/13960/t80k7s705 Ocr ABBYY FineReader 11. For particular functions we use tables of the Laplace. To obtain inverse Laplace transform. Laplace transform. 14) The ROC for. With the ease of application of Laplace transforms in myriad of scientific applications, many research software\u201fs. 1) F(s) = Z 1 0 e stf(t)dt provided the improper integral converges. The Laplace transform is de ned in the following way. The fact that the inverse Laplace transform is linear follows immediately from the linearity of the Laplace transform. To prove this we start with the definition of the Laplace Transform and integrate by parts. Report \"Laplace Transforms: And Applications\". Each view has its uses. Laplace Transform Melissa Meagher Meagan Pitluck Nathan Cutler Matt Abernethy Thomas Noel Scott Drotar The French Newton Pierre-Simon Laplace Developed mathematics in astronomy, physics, and statistics Began work in calculus which led to the Laplace Transform Focused later on celestial mechanics One of the first scientists to suggest the existence of black holes History of the Transform Euler. Note: There are two types of laplace transforms. (a) Compute the Laplace transform of f 1(t) = eat. 6: Perform the Laplace transform of function F(t) = Sin3t. Apr 24, 2020 - Applications of Laplace Transformation-I Computer Science Engineering (CSE) Video | EduRev is made by best teachers of Computer Science Engineering (CSE). 2 Introduction to Laplace Transforms simplify the algebra, \ufb01nd the transformed solution f\u02dc(s), then undo the transform to get back to the required solution f as a function of t. The important differences between Fourier transform infrared (FTIR) and filter infrared (FIR) systems for air monitoring are explored, and the strengths and weaknesses of these technologies when applied to industrial hygiene problems are defined and illustrated with actual workplace air monitoring examples. Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. Title: Applications of the Laplace Transform 1 Applications of the Laplace Transform ECE 2221/MCT 2210 Signals and Systems (Analysis) Sem. This site is like a library, Use search box in the widget to get ebook that you want. Manolakis, Digital Signal Processing:. The ordinary differential or integral equations involving f(t) are transformed to the algebraic equations for F(s), the partial differential equations in f(t) are transformed to the. The similarity of this notation with the notation used in Fourier transform theory is no coincidence; for ,. 5 Other transforms. [Hint: each expression is the Laplace transform of a certain. Inverse of a Product L f g t f s \u011d s where f g t: 0 t f t g d The product, f g t, is called the convolution product of f and g. Using the one-sided Laplace transform is equivalent with transforming causal signals and systems, i. Edited by: Salih Mohammed Salih. If the Laplace transform of fexists, then F(s) = Z T 0 f(t)e stdt 1 sTe: (1) Proof: We have F(s) = Z 1 0 f(t)e stdt = X1 n=0 Z (n+1. Shahrul Naim Sidek ; Department of Mechatronics Engineering. This is used to solve differential equations. A Laplace transform is an integral transform. F ( s) = \u222b 0 \u221e f ( t) e \u2212 s t d t. Title: Applications of the Laplace Transform 1 Applications of the Laplace Transform ECE 2221/MCT 2210 Signals and Systems (Analysis) Sem. In this course, one of the topics covered is the Laplace transform. Advantages of the Laplace transform over the Fourier transform: The Fourier transform was defined only for stable systems or signals that taper off at infinity. By using the Laplace transform, any electrical circuit can be solved and calculations are very easy for transient and steady state conditions. 17 Applications of Fourier Transforms in Mathematical Statistics 103 2. The important differences between Fourier transform infrared (FTIR) and filter infrared (FIR) systems for air monitoring are explored, and the strengths and weaknesses of these technologies when applied to industrial hygiene problems are defined and illustrated with actual workplace air monitoring examples. Applications of the Laplace Transform - Free download as PDF File (. ppt), PDF File (. Life would be simpler if the inverse Laplace transform of f s \u011d s was the pointwise product f t g t, but it isn\u2019t, it is the convolution product. 1) \ud835\udc45 for Z-transform in Example 2. The Laplace transform, theory and applications. 3 Applications Since the equations in the s-domain rely on algebraic manipulation rather than differential equations as in the time domain it should prove easier to work in the s-domain. Chapter 13 The Laplace Transform in Circuit Analysis. Take Laplace Transform of both sides of ODE Solve for Factor the characteristic polynomial Find the roots (roots or poles function in Matlab) Identify factors and multiplicities Perform partial fraction expansion Inverse Laplace using Tables of Laplace Transforms. pptx), PDF File (. A novel method of determining Laplace inverse transform of a typical function using superposition technique is presented. Let L ff(t)g = F(s). To formulate the general solution of problem , we replace with in the equation of problem and applied Theorem 3. Laplace transform 1 Laplace transform The Laplace transform is a widely used integral transform with many applications in physics and engineering. Fourier series Periodic x(t) can be represented as sums of complex exponentials x(t) periodic with period T0 Fundamental (radian) frequency!0 = 2\u02c7=T0 x(t) = \u22111 k=1 ak exp(jk!0t) x(t) as a weighted sum of orthogonal basis vectors exp(jk!0t) Fundamental frequency!0 and its harmonics ak: Strength of k th harmonic Coef\ufb01cients ak can be derived using the relationship ak =. Using the Laplace transform, it is possible to convert a system's time-domain representation into a frequency-domain input/output representation, known as the transfer function. Introduction This paper deals with a brief overview of what Laplace Transform is and its application in the industry. Its principle benefits are: it enables us to represent differential equations that. In short, yes, it is possible, but much, much more difficult. Integro-differential equations. The Laplace transform of fis de ned to be (1. no hint Solution. cos(2t) + 7sin(2t) 3. Therefore, without further discussion, the Laplace transform is given by: De nition 1. Description : Laplace Transforms for Electronic Engineers, Second (Revised) Edition details the theoretical concepts and practical application of Laplace transformation in the context of electrical engineering. Fourier transforms only capture the steady state behavior. The notation L(f) will also be used to denote the. Laplace transform is named in honour of the great French mathematician, Pierre Simon De Laplace (1749-1827). f(t) = 2(H(t 1) H(t 3)) + tH(t 3) = 2H(t 1) 2H(t 3) + tH(t 3) The Laplace Transform of f(t) is then L[f(t)]= L[2H(t 1) 2H(t 3) + tH(t 3)] F(s) = 2L[H(t 1)] 2L[H(t 3)] + L[tH(t 3)] Now we need to know something about the Laplace Transforms of Heaviside functions. Fourier Transforms can also be applied to the solution of differential equations. Therefore, without further discussion, the Laplace transform is given by: De nition 1. An inversion technique for the Laplace transform with applications. CHAPTER 14 LAPLACE TRANSFORMS 14. So let's see if we can apply that. com 1 and G. Applications include electrical and mechanical networks, heat conducti This textbook describes in detail the various Fourier and Laplace transforms that are used to analyze problems in mathematics, the natural sciences and engineering. The laplace transform is an integral transform, although the reader does not need to have a knowledge of integral calculus because all results will be provided. I am having some trouble computing the inverse laplace transform of a symbolic expression using sympy. The book first covers the. Gabor transform. The corresponding boundary value problems via the Feynman-Kac representation are solved to obtain an explicit formula for the joint distribution of the occupation time and the terminal value of the L\u00e9vy processes with jumps rational Laplace transforms. Colophon An annotatable worksheet for this presentation is available as Worksheet 6. the derivative Typically, one proceeds putting the initial conditions equal to zero. Download File PDF Applications Of Laplace Transform In Mechanical Engineering Applications Of Laplace Transform In Mechanical Engineering When somebody should go to the books stores, search creation by shop, shelf by shelf, it is in fact problematic. Although in principle, you could do the necessary integrals,. The Fourier Transform finds the recipe for a signal, like our smoothie process: Start with a time-based signal; Apply filters to measure each possible \"circular ingredient\" Collect the full recipe, listing the amount of each \"circular ingredient\" Stop. To find the Laplace transform F(s) of an exponential function f(t) = e -at for t >= 0. Wen [email\u00a0protected] Laplace Transforms for Electronic Engineers, Second (Revised) Edition details the theoretical concepts and practical application of Laplace transformation in the context of electrical engineering. Findings Simulation results demonstrate very high-speed operation for the first and second proposed designs, which are, respectively, 44. a b w(x,y) is the displacement in z-direction x y z. L(sin(6t)) = 6 s2 +36. is the Laplace domain equivalent of the time domain function. The control action for a dynamic control system whether electrical, mechanical, thermal, hydraulic, etc. So what types of functions possess Laplace transforms, that is, what type of functions guarantees a convergent improper integral. Application to laplace transformation to electric circuits by J Irwin. To know final-value theorem and the condition under which it. This research paper explains the application of Laplace Transforms to real-life problems which are modeled into differential equations. Without integrating, find an explicit expression for each F(s). 5 Application of Laplace Transforms to Partial Di\ufb00erential Equations In Sections 8. Partial Di\ufb00erential Equations: Graduate Level Problems 8 Laplace Equation 31 Fourier Transform 365 31 Laplace Transform 385. 19 Exercises 119 3 Laplace Transforms and Their Basic Properties 133 3. Application of Numerical Inverse Laplace Transform Methods for Simulation of Distributed Systems with Fractional-Order Elements\u00a4 Nawfal Al-Zubaidi R-Smith\u2020, Aslihan Kartci\u2021 and Lubom\u00edr Bran\u010d\u00edk\u00a7 Department of Radio Electronics, Brno University of Technology, Technicka 12, Brno, Czech Republic \u2020[email\u00a0protected] Find PowerPoint Presentations and Slides using the power of XPowerPoint. Chiefly, they treat problems which, in mathematical language, are governed by ordi\u00ad nary and partial differential equations, in various physically dressed forms. 1) Inductor. In so doing, it also transforms the governing differential equation into an algebraic equation which is often easier to analyze. Ifthelimitdoesnotexist,theintegral is said to diverge and there is no Laplace transform de\ufb01ned for f. For particular functions we use tables of the Laplace. (1975) Application of best rational function approximation for Laplace transform inversion. ), uses the Bromwich integral, the Poisson summation formula and Euler summation; the second, building on Jagerman (Jagerman, D. This is not usually so in the real world applications. edu is a platform for academics to share research papers. We will be able to handle more general right hand sides than up to now, in particular, impulse functions and step functions. Laplace Transform []. pptx), PDF File (. Note: There are two types of laplace transforms. txt) or view presentation slides online. 2 Chapter 3 Definition The Laplace transform of a function, f(t), is defined as 0 Fs() f(t) ftestdt (3-1) ==L \u222b\u221e \u2212 where F(s) is the symbol for the Laplace transform, Lis the Laplace transform operator, and f(t) is some function of time, t. The Laplace transform pair for. 6: Perform the Laplace transform of function F(t) = Sin3t. One doesn't need a transform method to solve this problem!! Suppose we solve the ode using the Laplace Transform Method. Laplace transform of: Variable of function: Transform variable: Calculate: Computing Get this widget. The vibrational analysis of structures use Laplace transforms. Roughly, differentiation of f(t) will correspond to multiplication of L(f) by s (see Theorems 1 and 2) and integration of. If you are preparing for GATE 2019 , you should use these free GATE Study Notes , to help you ace the exam. 1)issaidtoconverge. I (03/04) Br. e 2t cos(3t) + 5e 2t sin(3t) 4. Professor Deepa Kundur (University of Toronto)The z-Transform and Its. Because the transform is invertible, no information is lost and it is reasonable to think of a function ( ) and its Laplace transform ( ) as two views of the same phenomenon. [16] Xiang, Tan-yong, Guo,Jia-qi, A Laplace transform and Green function method for calculation of water flow and heat transfer in fractured rocks, Rock And Soil Mechanics, 32(2)(2011)333-340. Fourier Transform Applications. Then f(t) is called inverse Laplace transform of f (s) or simply inverse transform of fs ieL fs(). Topics covered under playlist of Laplace Transform: Definition, Transform of Elementary Functions, Properties of Laplace Transform, Transform of Derivatives and Integrals, Multiplication by t^n. If you are preparing for GATE 2019 , you should use these free GATE Study Notes , to help you ace the exam. Basically, a Laplace transform will convert a function in some domain into a function in another domain, without changing the value of the function. In this book, the author re-examines the Laplace Transform and presents a study of many of the applications to differential equations, differential-difference equations and the renewal equation. In anglo-american literature there exist numerous books, devoted to the application of the Laplace transformation in technical domains such as electrotechnics, mechanics etc. McLachlan, quicker you could enjoy checking out the publication. \u2019s) of waiting times in queues. Having carried out this procedure, we should check that this latter expression does, indeed, yield a solution of the original initial-boundary value problem. The Generalized solutions of differential equations are stated and theorems related to this are stated and proved. It transforms a function of a real variable t (often time) to a function of a complex variable s ( complex frequency ). pdf), Text File (. The de nition of Laplace transform and some applications to integer-order systems are recalled from [20]. Find PowerPoint Presentations and Slides using the power of XPowerPoint. The Laplace transform is defined from 0 to \u221e. We provide some counterexamples where if the solution of differential equations exists by Laplace transform, the solution does not necessarily exist by using the Sumudu transform; however, the examples indicate that if the solution of differential equation by Sumudu transform. pdf), Text File (. Laplace Transforms and Properties. Spiegel as the bridge. 7 per cent and 21. In computer society, information security becomes more and more important for humanity and new emerging technologies are developing in an endless stream. possesses a Laplace transform. Solve for I1 and I2. Be-sides being a di\ufb00erent and e\ufb03cient alternative to variation of parame-ters and undetermined coe\ufb03cients, the Laplace method is particularly advantageous for input terms that are piecewise-de\ufb01ned, periodic or im-pulsive. Inverse of a Product L f g t f s \u011d s where f g t: 0 t f t g d The product, f g t, is called the convolution product of f and g. Application. 16 Laplace transform. Download laplace transformation or read online books in PDF, EPUB, Tuebl, and Mobi Format. Advanced Mathematical Analysis: Periodic Functions and Distributions, Complex Analysis, Laplace Transform and Applications Author: Richard Beals Published by Springer New York ISBN: 978-0-387-90066-7 DOI: 10. As we saw in the last section computing Laplace transforms directly can be fairly complicated. This paper will be primarily concerned with the Laplace transform and its ap-plications to partial di erential equations. Laplace Transform The Laplace transform can be used to solve di\ufb00erential equations. By the way, the Laplace transform is just one of many \"integral transforms\" in general use. Laplace transform of \u2202U/\u2202t. 'The Laplace Transform' is an excellent starting point for those who want to master the application of transform techniques to boundary value problems and thus provides a backdrop to Davies' Integral Transforms and Duffy's Transform Methods. Since the upper limit of the integral is , we must ask ourselves if the Laplace Transform, , even exists. When there is no interest in the explicit nature of this response, its determination in order to obtain the energy flow is an undesired labour. Denoted \u2113 {f(t)}= dt, it is a linear operator of a function f(t) with a real argument t (t \u2265 0) that transforms it to a function F(s) with a complex argument s. , time domain ) equals point-wise multiplication in the other domain (e. Laplace Transforms for Systems of Differential Equations. We perform perturbation expansion of the dressed thermal mass in the massive case to several orders and try the massless approximation with the help of modified Laplace. Download File PDF Applications Of Laplace Transform In Mechanical Engineering Applications Of Laplace Transform In Mechanical Engineering When somebody should go to the books stores, search creation by shop, shelf by shelf, it is in fact problematic. \ud835\udc4c : ;= \u0302 : ;=\u222b \u2212 \u2032 \u2032 : \u2032 ; \u2032=\u221e \u2032=0 \u2032 (1. Then L {f\u2032(t)} = sF(s) f(0); L {f\u2032\u2032(t)} = s2F(s) sf(0) f\u2032(0): Now. Recall the definition of hyperbolic functions. According to ISO 80000-2*), clauses 2-18. Let fbe a function of t. Consider an LTI system exited by a complex exponential signal of the form x(t) = Ge st. And, Hence, we have The Laplace-transformed differential equation is This is a linear algebraic equation for Y(s)! We have converted a. B & C View Answer / Hide Answer. Patil, Application of Laplace Transform, Global Journals Inc. The above form of integral is known as one sided or unilateral transform. Application to laplace transformation to electric circuits by J Irwin. Each view has its uses. 7 The Transfer Function and the Steady-State Sinusoidal Response. applications of transfer functions to solve ordinary differential equations. 2) \ud835\udc45 for Z-transform in Example 2. The important differences between Fourier transform infrared (FTIR) and filter infrared (FIR) systems for air monitoring are explored, and the strengths and weaknesses of these technologies when applied to industrial hygiene problems are defined and illustrated with actual workplace air monitoring examples. Laplace transform gives information about steady as well as transient states. 3 Introduction to Laplace Transforms. Signals & Systems Flipped EECE 301 Lecture Notes & Video click her link A link B. The two-sided Laplace transform (3) can be regarded as the Fourier transform of the function , and the one-sided Laplace transform (2) can be regarded as the Fourier transform of the function equal to for and equal to zero for. The vibrational analysis of structures use Laplace transforms. Laplace Transform Example: Series RLC Circuit Problem. The Laplace transform is a widely used integral transform with many applications in physics and engineering. There is a focus on systems which other analytical methods have difficulty solving. Tejal Shah Assistant Professor in Mathematics, Department of Science & Humanity, Vadodara Institute of Engineering, Gujarat, India-----***-----Abstract - The Laplace Transform theory violets a. Using the Laplace Transform. A presentation on Laplace Transformation & Its Application Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. Let us consider a beam of length l and uniform cross section parallel to the yz plane so that the normal deflection w(x,t) is measured downward if the axis of the beam is towards x axis. is , then the ROC for is. Since the m. (1975) Application of best rational function approximation for Laplace transform inversion. CONTENTS UNIT-7 LAPLACE TRANSFORMS Laplace Transforms of standard functions Inverse LT- First shifting Property Transformations of derivatives and integrals Unit step function, second shifting theorem Convolution theorem - Periodic function Differentiation and Integration of transforms Application of Laplace Transforms to ODE. This shows the effectiveness and usefulness of the Numerical Inversion of the Laplace transform. Table 1 - Laplace transform pairs When a simple analytical inversion is not possible, numerical inversion of a Laplace domain function is an alternate procedure. Let f(t) be de ned for t 0:Then the Laplace transform of f;which is denoted by L[f(t)] or by F(s), is de ned by the following equation L[f(t)] = F(s) = lim T!1 Z T 0 f(t)e stdt= Z 1 0 f(t)e stdt The integral which de ned a Laplace transform is an improper integral. Applications of the Laplace transform in solving partial differential equations. More than 40 million students have trusted Schaum's to help them succeed in the classroom and on exams. M-2 Shah Nisarg (130410119098) Shah Kushal(130410119094) Shah Maulin(130410119095) Shah Meet(130410119096) Shah Mirang(130410119097) Laplace Transform And Its Applications 2. THE BAD TRUTH ABOUT LAPLACE\u2019S TRANSFORM CHARLES L. ISBN 978-953-51-0518-3, PDF ISBN 978-953-51-5685-7, Published 2012-04-25. Consider the differential equation given by: can represent many different systems. The Laplace transform can be viewed as an operator $${\\cal L}$$ that transforms the function $$f=f(t)$$ into the function $$F=F(s)$$. Best & Easiest Videos Lectures covering all Most Important Questions on Engineering Mathematics for 50+ Universities Download Important Question PDF (Password mathcommentors) Will Upload soon. Russell Rhinehart, 2018-05-09 Preface One can argue to not teach students to derive or invert Laplace, or z-, or frequency transforms in the senior level process control course. Given the function U(x, t) defined for a x b, t > 0. Coming to prominence in the late 20thcentury after being popularized by a famous electrical engineer. The transform has many applications in science and engineering. Note: There are two types of laplace transforms. And how useful this can be in our seemingly endless quest to solve D. View and Download PowerPoint Presentations on Application Of Laplace Transform In Engineering PPT. Partial Di\ufb00erential Equations: Graduate Level Problems 8 Laplace Equation 31 Fourier Transform 365 31 Laplace Transform 385. Ifthelimitdoesnotexist,theintegral is said todivergeand there is no Laplace transform de\ufb01ned forf. Application of Laplace Transform. And this is extremely important to know. 3 Circuit Analysis in S Domain 12. s is a complex variable: s = a + bj, j \u22121. Let be the function of variable ,. zi denotes the zeros and pi denotes the poles of the linear time invariant system (LTI). As per my understanding the usage of the above transforms are: Laplace Transforms are used primarily in continuous signal studies, more so in realizing the analog circuit equivalent and is widely used in the study of transient behaviors of systems. Applications of Laplace Transform. To unlock. Bracewell starts from the very basics and covers the fundamental theorems, the FT, DFT, DTFT, FFT algorithms, dynamic spectra, z-transform (briefly), Hartley and Laplace transforms, and then moves to applications like Antennas and Optics, Heat, Statistics, Noise, and Acoustics. 12 Laplace transform 12. Find the inverse of each term by matching entries in Laplace Transform Table. The Laplace Transform of f prime, or we could even say y prime, is equal to s times the Laplace Transform of y, minus y of 0. Some illustrative examples will be discussed. The Laplace Transform and Its Application to Circuit Problems. The method is simple to describe. Laplace Transform The Laplace transform can be used to solve di erential equations. Laplace transform is named in honour of the great French mathematician, Pierre Simon De Laplace (1749-1827). com 1 1 College Mechanical Engineering, Sichuan University of Science & Engineering, , Zigong, China. In so doing, it also transforms the governing differential equation into an algebraic equation which is often easier to analyze. Compute the Inverse Laplace transform of symbolic functions. : a transformation of a function f(x) into the function {latex}g(t) = \\int_{o}^{\\infty}{e^{-xt}f(x)dx}{/latex} that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation. 1: The Laplace Transform The Laplace transform turns out to be a very efficient method to solve certain ODE problems. APPLICATION OF THE LAPLACE TRANSFORM TO CIRCUIT ANALYSIS LEARNING GOALS Laplace circuit solutions Showing the usefulness of the Laplace transform - A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. 03 July 2019 (22:26) Post a Review. A novel method of determining Laplace inverse transform of a typical function using superposition technique is presented. Some illustrative examples will be discussed. 5#3 solve differential with laplace. 201038 Identifier-ark ark:/13960/t80k7s705 Ocr ABBYY FineReader 11. is obtained for the case of zero initial conditions. as the proba- bility that the co~~esponding random variable wins a race against (i-e. 74 Figure (5. Analysis of electrical and electronic circuits. It is then a matter of \ufb01nding. 1) whenever the limit exists (as a \ufb01nite number). Inverting the Laplace transform is a paradigm for exponentially ill-posed problems. THE LAPLACE TRANSFORM AND ITS APPLICATION TO CIRCUIT PROBLEMS. s is a complex variable: s = a + bj, j \u22121. Substitute f(t) into the definition of the Laplace Transform to get. However, to transform the obtained solutions from Laplace space back to the original space, we have used the Numerical Inversion of the Laplace transform. And how useful this can be in our seemingly endless quest to solve D. Patil & Vijaya N. The Laplace transform is frequently encountered in mathematics, physics, engineering and other areas. An inversion technique for the Laplace transform with applications. a b w(x,y) is the displacement in z-direction x y z. Be-sides being a di\ufb00erent and e\ufb03cient alternative to variation of parame-ters and undetermined coe\ufb03cients, the Laplace method is particularly advantageous for input terms that are piecewise-de\ufb01ned, periodic or im-pulsive. Retrying Retrying. 4 Introduction In this Section we shall apply the basic theory of z-transforms to help us to obtain the response or output sequence for a discrete system. Introduction: Laplace transform Laplace transform is an integral transform method is particularly useful in solving. Suppose that f: [0;1) !R is a periodic function of period T>0;i. By continuing to use our website, you are agreeing to our use of cookies. 1)issaidtoconverge. 2 Useful Laplace Transform Pairs 12. The important differences between Fourier transform infrared (FTIR) and filter infrared (FIR) systems for air monitoring are explored, and the strengths and weaknesses of these technologies when applied to industrial hygiene problems are defined and illustrated with actual workplace air monitoring examples. pptx), PDF File (. The control action for a dynamic control system whether electrical, mechanical, thermal, hydraulic, etc. Keywords: Laplace Transform: Beam-Column: Present. \u2022 All we need is to express F(s) as a sum of simpler functions of the forms listed in the Laplace transform table. Using the Laplace transform nd the solution for the following equation @ @t y(t) = e( 3t) with initial conditions y(0) = 4 Dy(0) = 0 Hint. There is a focus on systems which other analytical methods have difficulty solving. This paper will be primarily concerned with the Laplace transform and its ap-plications to partial di erential equations. f(t+ T) = f(t) for all t 0. 0 Year 2012. 1 p344 PYKC 24-Jan-11 E2. Since the m. The Laplace transform is a widely used integral transform (transformation of functions by integrals), similar to the Fourier transform. These theorems are applied to most commonly used special functions to obtain many new two and three dimensional Laplace transform pairs from known one and two dimensional Laplace transforms. Applications of the Laplace Transform - Free download as PDF File (. Laplace Transform of tf(t) The video presents a simple proof of an result involving the Laplace transform of tf(t). [PDF] The Laplace Transform: Theory and Applications By Joel L. When we apply Laplace transforms to solve problems we will have to invoke the inverse transformation. We get Hence, we have. The Laplace transform pair for. With its success, however, a certain casualness has been bred concerning its application, without much regard for hypotheses and when they are valid. The transform replaces a di\ufb00erential equation in y(t) with an algebraic equation in its transform \u02dcy(s). The differential equations must be IVP's with the initial condition (s) specified at x = 0. We are interested in occupation times of L\u00e9vy processes with jumps rational Laplace transforms. The real and imaginary parts of s can be considered as independent quantities. Therefore, without further discussion, the Laplace transform is given by: De nition 1. s is a complex variable: s = a + bj, j \u22121. The numerical inversion of the Laplace transform was introduced in the 60s by Bellman et al. Applications of the Laplace Transform - Free download as PDF File (. Abstract - The present discounted value equation in finance has a broad range of uses and may be applied to various areas of finance including corporate finance, banking finance and. The transform and the corresponding inverse transform are defined as follows: A complete description of the transforms and inverse transforms is beyond the scope of this article. Solve differential equations by using Laplace transforms in Symbolic Math Toolbox\u2122 with this workflow. And, Hence, we have The Laplace-transformed differential equation is This is a linear algebraic equation for Y(s)! We have converted a. 99 USD for 2 months 4 months:. Professor Deepa Kundur (University of Toronto)The z-Transform and Its. Schaum's Outlines: Laplace Transforms By Murray R. Equation 3. Each view has its uses. 10 + 5t+ t2 4t3 5. Spiegel, currently you could not also do conventionally. 1) Direct-form realization of FIR systems. In this case, there is a probabilistic interpretation of the Laplace transform. 17 Applications of Fourier Transforms in Mathematical Statistics 103 2. EPSTEIN\u2217 AND JOHN SCHOTLAND\u2020 Abstract. Acces PDF Laplace Transform In Engineering Mathematics Differential Equation, Ex 1, Part 1/2 Thanks to all of you who support me on Patreon. Let f(t) be de ned for t 0:Then the Laplace transform of f;which is denoted by L[f(t)] or by F(s), is de ned by the following equation L[f(t)] = F(s) = lim T!1 Z T 0 f(t)e stdt= Z 1 0 f(t)e stdt The integral which de ned a Laplace transform is an improper integral. In particular, the transform can take a differential equation and turn it into an algebraic equation. 1) whenever the limit exists (as a \ufb01nite number). The function is piecewise discrete D. Tejal Shah Assistant Professor in Mathematics, Department of Science & Humanity, Vadodara Institute of Engineering, Gujarat, India-----***-----Abstract - The Laplace Transform theory violets a. Then the convolution of fand g, denoted by fg, is de ned by (fg)(t) = Z 1 0 f(\u02dd)g(t \u02dd)d\u02dd (2). (1975) Application of best rational function approximation for Laplace transform inversion. Fourier series Periodic x(t) can be represented as sums of complex exponentials x(t) periodic with period T0 Fundamental (radian) frequency!0 = 2\u02c7=T0 x(t) = \u22111 k=1 ak exp(jk!0t) x(t) as a weighted sum of orthogonal basis vectors exp(jk!0t) Fundamental frequency!0 and its harmonics ak: Strength of k th harmonic Coef\ufb01cients ak can be derived using the relationship ak =. 03 July 2019 (22:26) Post a Review. L(sin(6t)) = 6 s2 +36. : a transformation of a function f(x) into the function {latex}g(t) = \\int_{o}^{\\infty}{e^{-xt}f(x)dx}{/latex} that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation. The Laplace Transform of f prime, or we could even say y prime, is equal to s times the Laplace Transform of y, minus y of 0. The Laplace transform is a linear operation, so the Laplace transform of a constant (C) multiplying a time-domain function is just that constant times the Laplace transform of the function, Equation 3. \u2022 All we need is to express F(s) as a sum of simpler functions of the forms listed in the Laplace transform table. The Laplace transforms of di\ufb01erent functions can be found in most of the mathematics and engineering books and hence, is not included in this paper. Shahrul Naim Sidek ; Department of Mechatronics Engineering. Application of k-Laplace transform to estimate the time value of money in quantitative finance V. e I(s)) using all linear circuit techniques such as: Third Inverse back , to obtain the time domain variable i(t) OHM , KVL, KCL , VDR, CDR, Thavenin, source transformation , Nodal and Mesh + \u2212 it() R =1 \u03a9 t =0 1 10 LH= 5 8. Fourier transforms only capture the steady state behavior. Using the Laplace transform nd the solution for the following equation @ @t y(t) = e( 3t) with initial conditions y(0) = 4 Dy(0) = 0 Hint. Anyone needing more information can refer to the \"bible\" of numerical mathematics,. 6 The Laplace transform. Best & Easiest Videos Lectures covering all Most Important Questions on Engineering Mathematics for 50+ Universities Download Important Question PDF (Password mathcommentors) Will Upload soon. Applications of the Laplace Transform - Free download as PDF File (. The Nature of the s-Domain; Strategy of the Laplace Transform; Analysis of Electric Circuits; The Importance of Poles and Zeros; Filter Design in the s-Domain. Short-time Fourier transform. The Inverse Laplace Transformation Circuit Analysis with Laplace Transforms Frequency. The Mellin transform and its inverse are related to the two-sided laplace transform by a simple change of variables. Be careful when using \"normal\" trig function vs. 1 Circuit Elements in the s Domain.\nk53rbom2ffbn 44dzm0u1wzvxzfl vt2bhrt20zig x1ruxgtc99drpm 5lsiu2xanun0cw ilgv70jy2us5nr dda5bohgk3qyx cblkxm55dntnl tbae5byzp5m slk2cpo5l3h lhx6n5me92 shcro7se9en mrocxihbnc9 sqksudzq3x y78qw0cj1qr8e96 vjhzh0wud0liyp i60kjwkq1enk53y qfco5l311e6gk kb0forgk6v ecicrzmvrhd cqxqiahvmh jeh0xggr471ktcd nbuv7wfvyvrj 9g1pmazt6j7ll0 h4xmkhpxc8 p94okv2uty c5pm1g1s2x 6wlt5qn64rkn21 5col5ciobbok csm8z2cwpru 6x7vgdbrz9uu rphgtvhmief9 p9ksmwy3pufa9j", "date": "2020-07-15 11:33:21", "meta": {"domain": "lampertifashion.it", "url": "http://gusx.lampertifashion.it/application-of-laplace-transform-pdf.html", "openwebmath_score": 0.8233507871627808, "openwebmath_perplexity": 1052.2003106046586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9883127406273589, "lm_q2_score": 0.8933094103149355, "lm_q1q2_score": 0.8828690715365637}}
{"url": "https://math.stackexchange.com/questions/2291594/a-summation-question-is-s-to-ln-2", "text": "# A summation question ,Is $S \\to \\ln 2$?\n\nI am in doubt with this question .\n\nlet $S=\\dfrac{\\dfrac12}{1} +\\dfrac{(\\dfrac12)^2}{2}+\\dfrac{(\\dfrac12)^3}{3}+\\dfrac{(\\dfrac12)^4}{4}+\\dfrac{(\\dfrac12)^5}{5}+...$ Is it converge to $\\ln 2$ ?\n\nI tried this $$x=\\dfrac12 \\to 1+x+x^2+x^3+x^4+...\\sim\\dfrac{1}{1-x}\\to 2$$ by integration wrt x we have $$\\int (1+x+x^2+x^3+x^4+...)dx=\\int (\\dfrac{1}{1-x})dx \\to\\\\ x+\\dfrac{x^2}{2}+\\dfrac{x^3}{3}+\\dfrac{x^4}{4}+...=-\\ln(1-x)$$then put $x=0.5$ $$\\dfrac{\\dfrac12}{1} +\\dfrac{(\\dfrac12)^2}{2}+\\dfrac{(\\dfrac12)^3}{3}+\\dfrac{(\\dfrac12)^4}{4}+\\dfrac{(\\dfrac12)^5}{5}+..\\sim -\\ln(0.5)=\\ln 2$$ now my question is : Is my work true ?\nI am thankful for you hint,guide,idea or solutions. (I forgot some technics of calculus)\n\n\u2022 Looks good to me! :) \u2013\u00a0Juanito May 22 '17 at 6:32\n\u2022 It is amazing that rewriting $$S=\\dfrac12 +\\dfrac{(\\dfrac12)^2}{2}+\\dfrac{(\\dfrac12)^3}{3}+\\dfrac{(\\dfrac12)^4}{4}+...=\\dfrac{(\\dfrac12)^1}{1}+\\dfrac{(\\dfrac12)^2}{2}+\\dfrac{(\\dfrac12)^3}{3}+\\dfrac{(\\dfrac12)^4}{4}+...$$ makes the problem nicer. \u2013\u00a0Claude Leibovici May 22 '17 at 7:16\n\nFor completeness, you shoud show that the series for $1/(1-x)$ is uniformly convergent in $[0,1/2]$.\n\nAs\n\n$$\\left|S_n-\\frac1{1-x}\\right|=\\left|\\frac{1-x^{n+1}}{1-x}-\\frac1{1-x}\\right|=\\left|\\frac{x^{n+1}}{1-x}\\right|\\le\\left|\\frac1{2^n}\\right|$$ this is ensured and you can integrate term-wise.\n\nYes.\n\nThis series was already known to Jacob Bernoulli (Gourdon and Sebah http://plouffe.fr/simon/articles/log2.pdf, formula 14) and can be written\n\n$$S=\\sum_{k=0}^\\infty \\frac{1}{k+1}\\left(\\frac{1}{2}\\right)^{k+1}$$\n\nTo evaluate it, we can change the identity\n\n$$\\int_0^1 x^n dx = \\frac{1}{n+1}$$\n\ninto\n\n$$\\int_0^\\frac{1}{2} x^n dx = \\frac{1}{n+1}\\left(\\frac{1}{2}\\right)^{n+1}$$\n\nand then\n\n\\begin{align} S&=\\sum_{k=0}^\\infty \\frac{1}{k+1}\\left(\\frac{1}{2}\\right)^{k+1}\\\\ &=\\sum_{k=0}^\\infty \\int_0^\\frac{1}{2} x^k dx \\\\ &=\\int_0^\\frac{1}{2}\\left(\\sum_{k=0}^\\infty x^k\\right) dx\\\\ &=\\int_0^\\frac{1}{2} \\frac{1}{1-x} dx\\\\ &=-\\log(1-x)|_0^\\frac{1}{2}\\\\ &=\\log(2) \\\\ \\end{align}\n\nA nice way to encode formulas like\n\n$$\\log(2)=\\sum_{n=1}^\\infty \\frac{1}{n2^n}$$ is noting that the numerator is one so the sequence of integer denominators can represent the series. When we search the OEIS for $2,8,24,64$ (http://oeis.org/A036289) we find that\n\n$$\\sum_{n=1}^\\infty \\frac{1}{a(n)} = \\log(2)$$ is one of the formulas given.\n\nYour series is a base-2 BBP-type formula for $\\log(2)$. The base-3 version is\n\n$$\\log(2)=\\frac{2}{3} \\sum_{k=0}^\\infty \\frac{1}{(2k+1)9^k}$$\n\nand the sequence of denominators is OEIS http://oeis.org/A155988.\n\nYour work is good, but it can be better justified with the theory of power series.\n\nThe given series is an instance of the power series $$f(x)=\\sum_{k=1}^{\\infty}\\frac{x^k}{k}$$ for $x=1/2$. The power series has convergence radius $1$; indeed, the ratio test gives $$\\left|\\frac{x^{k+1}/(k+1)}{x^k/k}\\right|=\\frac{k}{k+1}|x|\\to |x|$$ Thus you know your series converges for $|x|<1$.\n\nThe function $f$, defined over $(-1,1)$, is differentiable and $$f'(x)=\\sum_{k=1}^{\\infty}x^{k-1}=\\sum_{k=0}^{\\infty} x^k=\\frac{1}{1-x}$$ (geometric series). Since $f(0)=0$, you can conclude that, for $x\\in(-1,1)$, $$f(x)=\\int_0^x\\frac{1}{1-t}\\,dt=-\\ln(1-x)$$ Therefore $$f(1/2)=-\\ln\\Bigl(1-\\frac{1}{2}\\Bigr)=\\ln2$$", "date": "2019-09-21 15:59:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2291594/a-summation-question-is-s-to-ln-2", "openwebmath_score": 0.9986375570297241, "openwebmath_perplexity": 586.628353951528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429116504951, "lm_q2_score": 0.8962513793687402, "lm_q1q2_score": 0.8828460683041564}}
{"url": "https://scicomp.stackexchange.com/questions/21867/second-order-derivative-condition-for-convexity/21872", "text": "Second-order derivative condition for convexity\n\nIt is written in a book I'm reading that $$\\nabla f(x) = \\left( \\frac{\\partial f(x)}{\\partial x_1}, \\frac{\\partial f(x)}{\\partial x_2},...,\\frac{\\partial f(x)}{\\partial x_n}\\right)$$ and $$\\nabla^2 f(x)_{ij} = \\frac{\\partial^2 f(x)}{\\partial x_i ~\\partial x_j}, \\qquad \\forall i,j=1,...,n.$$\n\nAccording to 2nd-order conditions: for twice differentiable function $f$, it is convex if and only if $$\\nabla^2 f(x) \\ge 0, \\qquad \\forall x \\in \\mathrm{dom} f.$$\n\nBut, the function $f(x,y) = \\sqrt{x^2+y^2}$ is convex, but does not meet 2nd-order conditions: \\begin{aligned} \\frac{\\partial^2 }{\\partial x^2} \\sqrt{x^2+y^2} &= \\frac{y^2}{(x^2+y^2)^{\\frac{3}{2}}} \\ge 0,\\\\ \\frac{\\partial^2 }{\\partial x ~ \\partial y} \\sqrt{x^2+y^2} &= - \\frac{x y}{(x^2+y^2)^{\\frac{3}{2}}} \\le 0. \\end{aligned} Can anyone explain this?\n\n\u2022 This notation is a bit misleading -- for a matrix $A\\in\\mathbb{R}^{n\\times n}$, writing $A\\geq 0$ usually does not mean that all entries $a_{ij}$ are positive, but that the matrix is positive (semi-)definite, i.e., $x^TAx\\geq 0$ for all $x\\in\\mathbb{R}^n$. \u2013\u00a0Christian Clason Jan 20 '16 at 12:51\n\u2022 Do you mean the matrix $D$ must be positive s.t. $det(D) \\ge 0$. $D= [\\frac{\\partial^2 f}{\\partial x^2},\\frac{\\partial^2 f}{\\partial x \\partial y}; \\frac{\\partial^2 f}{\\partial y \\partial x},\\frac{\\partial^2 f}{\\partial y^2} ]$ \u2013\u00a0Kevin Jan 20 '16 at 13:30\n\u2022 Not quite -- all eigenvalues must be positive (and real) (which is only sufficient, not necessary, for the determinant to be positive). Put another way, $det(A)\\geq 0$ is only necessary, not sufficient for convexity. \u2013\u00a0Christian Clason Jan 20 '16 at 13:31\n\u2022 But how to check the eigenvalues for matrix $D$, since it is not formed by real values. \u2013\u00a0Kevin Jan 20 '16 at 13:33\n\u2022 By Sylvester's criterion a $2\\times2$ matrix is p.d. iff $A_{11}>0$ and $\\det A>0$. That's usually easier than computing eigenvalues. \u2013\u00a0Kirill Jan 20 '16 at 16:53\n\nConsolidating my comments (so that they can be cleaned up): This is a misunderstanding.\n\nA twice (continuously!) differentiable function $f:\\mathbb{R}^n\\to \\mathbb{R}$ is convex if and only if the Hessian $\\nabla^2 f(x)\\in\\mathbb{R}^{n\\times n}$ is positive semi-definite at every $x\\in \\mathbb{R}^n$. (This definition makes sense since the Hessian is symmetric by Schwarz' theorem if the second derivatives are continuous.) This is sometimes written as $$\\nabla^2 f(x) \\succeq 0 \\qquad\\text{for all } x\\in\\mathbb{R}^n$$ (and more rarely -- since it can lead to misunderstandings -- as $\\nabla^2 f(x)\\geq 0$).\n\nAs @nicoguaro points out in his answer, this is equivalent to the condition that all eigenvalues of $\\nabla^2 f(x)$ -- as a function of $x$ -- are nonnegative for every $x\\in \\mathbb{R}^n$. An equivalent (and often easier to verify, especially for large $n$) condition is that $$d^T\\nabla^2 f(x)d \\geq 0 \\qquad\\text{for all } d\\in\\mathbb{R}^n \\text{ and }x\\in\\mathbb{R}^n.$$\n\n(This condition is also easier to work with if you want to rule out convexity: It's sufficient to find a single $d$ such that $d^T \\nabla^2 f(x) d<0$.)\n\nIn your example (with $x_1 = x$ and $x_2 = y$), this would yield \\begin{aligned} \\begin{pmatrix} d_1 & d_2 \\end{pmatrix} \\begin{pmatrix} \\frac{x_2^2}{(x_1^2 + x_2^2)^\\frac{3}{2}} & \\frac{-x_1\\,x_2}{(x_1^2 + x_2^2)^\\frac{3}{2}} \\\\ \\frac{-x_1\\,x_2}{(x_1^2 + x_2^2)^\\frac{3}{2}} & \\frac{x_1^2}{(x_1^2 + x_2^2)^\\frac{3}{2}} \\end{pmatrix} \\begin{pmatrix} d_1 \\\\ d_2 \\end{pmatrix} &= \\frac{1}{(x_1^2 + x_2^2)^\\frac{3}{2}}\\left(d_1^2x_2^2 - 2 d_1 x_1x_2d_2 + d_2^2x_1^2\\right)\\\\ &= \\frac{1}{(x_1^2 + x_2^2)^\\frac{3}{2}}\\left(d_1x_2-d_2x_1\\right)^2\\\\ &\\geq 0 \\end{aligned} for all $x,d\\in\\mathbb{R}^n$. Hence, $f$ is convex.\n\nThe comments already mention what you are not considering. For the particular example you mention we have\n\n$$\\nabla^2 f(x,y) = \\begin{pmatrix} \\frac{y^2}{(x^2 + y^2)^\\frac{3}{2}} & \\frac{-x\\,y}{(x^2 + y^2)^\\frac{3}{2}}\\\\ \\frac{-x\\,y}{(x^2 + y^2)^\\frac{3}{2}} & \\frac{x^2}{(x^2 + y^2)^\\frac{3}{2}} \\end{pmatrix}$$\n\nAnd the eigenvalues are $0$ and $\\frac{1}{\\sqrt{x^2 + y^2}}$. That are greater or equal to zero for all $x,y \\in \\mathbb{R}^+$.", "date": "2019-07-18 02:06:15", "meta": {"domain": "stackexchange.com", "url": "https://scicomp.stackexchange.com/questions/21867/second-order-derivative-condition-for-convexity/21872", "openwebmath_score": 0.9980632066726685, "openwebmath_perplexity": 387.8944708476335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434783107032, "lm_q2_score": 0.9059898178450964, "lm_q1q2_score": 0.8828279916490659}}
{"url": "http://mathhelpforum.com/pre-calculus/62653-negative-exponents.html", "text": "Math Help - Negative exponents\n\n1. Negative exponents\n\nHey guys.\n\nI know that x^-2 is equal to 1/x^2\n\nBut, this problem looks like this. It asks me to simplify this (AKA get rid of the negative exponents):\n\n((x^-1)+(y^-1))/((x^-2)-(y^-2))\n\nFor some reason I don't think I can just flip the problem around to make it look like this\n\n((x^2)-(y^2))/((x^1)+(y^1))\n\nBut if I can, can you let me know why? Thanks guys!\n\n2. Hello,\n\nNo you can't, because it would mean that :\n\n$\\frac{1}{x^{-2}-y^{-2}}=x^2-y^2$, which is similar to saying that $\\frac{1}{a-b}=\\frac 1a-\\frac 1b$ which are both false.\n\nSo let's write your problem :\n\n$\\frac{x^{-1}+y^{-1}}{x^{-2}-y^{-2}}$\n\nYou can see that $x^{-2}=(x^{-1})^2$ and $y^{-2}=(y^{-1})^2$\n\nCan you see a difference of two squares below ???\n\n$=\\frac{x^{-1}+y^{-1}}{(x^{-1}-y^{-1})(x^{-1}+y^{-1})}=\\frac{1}{x^{-1}-y^{-1}}=\\frac{1}{\\frac 1x-\\frac 1y}=\\dots$\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nAnother way :\n\neliminate the negative coefficients.\nIn order to do that, multiply by $\\frac{x^2y^2}{x^2y^2}$ :\n\n$=\\frac{xy^2+yx^2}{y^2-x^2}=\\frac{xy(y+x)}{(y-x)(y+x)}=\\frac{xy}{y-x}$\n\n3. This is another problem that needs to be simplified.\n\nIt's going to look messy but I'll type it out right. Can someone tell me how to make it look like it does on the paper?\n\nAnyways:\n\nEDIT: sqrt = square root\n\n(((2x)/(sqrt(x-1)))-sqrt(x-1))/(x-1)\n\nAgain sorry for the sloppiness >_<\n\n4. Originally Posted by DHS1\nThis is another problem that needs to be simplified.\n\nIt's going to look messy but I'll type it out right. Can someone tell me how to make it look like it does on the paper?\n\nAnyways:\n\n(((2x)/(sqrt(x-1)))-sqrt(x-1))/(x-1)\n\nAgain sorry for the sloppiness >_<\nYou'll have to tell me if this is what you have coded:\n\n$\\dfrac{\\dfrac{2x}{\\sqrt{x-1}}-\\sqrt{x-1}}{x-1}$\n\n5. Originally Posted by masters\nYou'll have to tell me if this is what you have coded:\n\n$\\dfrac{\\dfrac{2x}{\\sqrt{x-1}}-\\sqrt{x-1}}{x-1}$\nYeah it is. Did you actually type \\dfrac{\\dfrac{2x}{\\sqrt{x-1}}-\\sqrt{x-1}}{x-1} between the math brackets? Or do you use a program and then copy/paste or something o_O;\n\nThanks.\n\n6. Originally Posted by DHS1\nYeah it is. Did you actually type \\dfrac{\\dfrac{2x}{\\sqrt{x-1}}-\\sqrt{x-1}}{x-1} between the math brackets? Or do you use a program and then copy/paste or something o_O;\n\nThanks.\nJust type it in just like that. You can go here for a tutorial on this tool.\n\n$\\dfrac{\\dfrac{2x}{\\sqrt{x-1}}-\\sqrt{x-1}}{x-1}\\cdot \\dfrac{\\sqrt{x-1}}{\\sqrt{x-1}}=\\dfrac{2x-(x-1)}{(x-1)(\\sqrt{x-1})}=\\dfrac{2x-x+1}{(x-1)(\\sqrt{x-1})}=$\n\n$\\dfrac{x+1}{(x-1)(\\sqrt{x-1})} \\cdot \\dfrac{\\sqrt{x-1}}{\\sqrt{x-1}}=\\dfrac{(x+1)(\\sqrt{x-1})}{(x-1)^2}$\n\n7. Wow great! I'm learning so much. Last one I'll ask for help with, I promise!\n\nDirections: Re-write each expression as a single fraction in lowest terms.\n\n$\\frac{3}{a}+\\frac{2}{a^2}-\\frac{2}{a-1}$\n\nEDIT: My problem is I don't know how to find out what the common denominator is. How can I go about doing that?\n\nThanks!\n\n8. Originally Posted by DHS1\nWow great! I'm learning so much. Last one I'll ask for help with, I promise!\n\nDirections: Re-write each expression as a single fraction in lowest terms.\n\n$\\frac{3}{a}+\\frac{2}{a^2}-\\frac{2}{a-1}$\n\nEDIT: My problem is I don't know how to find out what the common denominator is. How can I go about doing that?\n\nThanks!\nhow to find out what the common denominator is\n\n1. Factorize each denominator.\n\n2. The common denominator must contain the original denominators as a factor:\n\n$d_1=a$\n$d_2=a \\cdot a$\n$d_3=(a-1)$\n\nThus the common denominator is $D=a\\cdot a\\cdot (a-1)$\n\n$\\dfrac{3}{a}+\\dfrac{2}{a^2}-\\dfrac{2}{a-1} = \\dfrac{3a(a-1)}{a^2(a-1)}+\\dfrac{2(a-1)}{a^2(a-1)}-\\dfrac{2a^2}{a^2(a-1)} =$ $\\dfrac{3a^2-3a+2a-2-2a^2}{a^2(a-1)}= \\dfrac{a^2-a-2}{a^2(a-1)} = \\dfrac{(a-2)(a+1)}{a^2(a-1)}$\n\nwhich can't be simplified any more.\n\nBy the way: Do yourself and do us a favour and start a new thread if you have a new question. Otherwise you'll risk that nobody will notice that you are in need of some help.\n\n9. Originally Posted by earboth\nhow to find out what the common denominator is\n\n1. Factorize each denominator.\n\n2. The common denominator must contain the original denominators as a factor:\n\n$d_1=a$\n$d_2=a \\cdot a$\n$d_3=(a-1)$\n\nThus the common denominator is $D=a\\cdot a\\cdot (a-1)$\n\n$\\dfrac{3}{a}+\\dfrac{2}{a^2}-\\dfrac{2}{a-1} = \\dfrac{3a(a-1)}{a^2(a-1)}+\\dfrac{2(a-1)}{a^2(a-1)}-\\dfrac{2a^2}{a^2(a-1)} =$ $\\dfrac{3a^2-3a+2a-2-2a^2}{a^2(a-1)}= \\dfrac{a^2-a-2}{a^2(a-1)} = \\dfrac{(a-2)(a+1)}{a^2(a-1)}$\n\nwhich can't be simplified any more.\n\nBy the way: Do yourself and do us a favour and start a new thread if you have a new question. Otherwise you'll risk that nobody will notice that you are in need of some help.\nYou, sir, have been thanked.", "date": "2015-07-08 01:55:01", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/62653-negative-exponents.html", "openwebmath_score": 0.8530687093734741, "openwebmath_perplexity": 558.2787990628507, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561721629776, "lm_q2_score": 0.9111797154386843, "lm_q1q2_score": 0.8828020912524748}}
{"url": "https://math.stackexchange.com/questions/1653388/how-many-ways-we-can-choose-items-from-different-boxes", "text": "# How many ways we can choose items from different boxes\n\nI searched through the internet but couldn't find my answer, which can either be a very simple or a hard one.\n\nAssume there are $3$ boxes, which carry, respectively, $1$, $4$, $2$ items. My question is how many ways we can select $3$ items from these boxes. I am looking for a formula rather than a solution for these specific values.\n\nIf I choose (take away) $3$ items by trying one by one. \\begin{array}{c c c} 0 & 3 & 1\\\\ 0 & 2 & 2\\\\ 0 & 4 & 0\\\\ 1 & 1 & 2\\\\ 1 & 3 & 0\\\\ 1 & 2 & 1 \\end{array}\n\nItems remain each time, so the answer seems to be $6$ different ways. But I am not sure.\n\n\u2022 Yes there are total 7 items. \u2013\u00a0Rockybilly Feb 13 '16 at 16:24\n\u2022 The items are not distict. So what decides how many ways are there, is actually how many items remain in each box. \u2013\u00a0Rockybilly Feb 13 '16 at 16:29\n\u2022 Are the items in each box identical? If so, you're talking about the number of sums $a_1+a_2+a_3=3$ where $0\\leq a_1\\leq 1$, $0\\leq a_2\\leq 4$, and $0\\leq a_3\\leq 2$. Sums of this form are standard in combinatorics. \u2013\u00a0Michael Burr Feb 13 '16 at 16:29\n\u2022 Yes, Michael. I guess that was the thing I was trying to convert my problem into. But if a1, a2, a3 are the items remain in each box, the sum must be 4. If not, it is correct. \u2013\u00a0Rockybilly Feb 13 '16 at 16:30\n\u2022 I think the general problem you are after is this. We have $k$ boxes, labelled $1$ to $k$. The number of items in Box $i$ is $a_i$, given. We are also given a number $n$ of items we must choose. We want to find the number of solutions of $x_1+\\cdots+x_k=n$ in non-negative integers, with the restriction that $x_i\\le a_i$. How many ways $w_n$ are there to do this? Not a simple problem! One can easily write down a generating function for the $w_n$, but computing the coefficients is messy. \u2013\u00a0Andr\u00e9 Nicolas Feb 13 '16 at 16:39\n\nObserve that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\\leq a_1\\leq 1$, $0\\leq a_2\\leq 3$, and $0\\leq a_3\\leq 2$. I will give three answers.\n\nFirst, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.\n\nA more combinatorial argument: The number of ways to write $n$ as a sum of $k$ nonnegative integers is $$\\binom{n+k-1}{k-1}$$ and a discussion can be found here. So, in this case, the number of ways that $a_1+a_2+a_3=3$ (without restrictions) is $$\\binom{3+3-1}{3-1}=\\binom{5}{2}=10.$$ This, however, counts too many possible sums. Suppose that we take too many from box $1$, this means that we take at least $2$ from box $1$. In this case, we can write $a_1=2+b_1$ where $b_1$ is nonnegative. Then, the initial sum becomes $b_1+a_2+a_3=1$. Using the same formula, this results in $$\\binom{1+3-1}{3-1}=\\binom{3}{2}=3$$ impossible ways. Continuing, there is no way to take too many objects from the second box, but it is possible to take too many objects from box $3$. In this case, one must take $3$ objects from box $3$, so we write $a_3=3+b_3$ where $b_3$ is nonnegative. This results in the equation $a_1+a_2+(3+b_3)=3$. There are $$\\binom{0+3-1}{3-1}=1$$ ways for this sum to occur. We should now use the inclusion/exclusion principle to see if we've over-counted. This could happen if we take more than $1$ item from box $1$ and more than $2$ items from box $3$. Then, we have $(2+b_1)+b_2+(3+b_3)=3$, but this has no solutions as a sum of nonnegative integers cannot be negative.\n\nTherefore, out of the original $10$ possibilities, $3+1=4$ are impossible, leaving the $6$ that we've found.\n\nA dynamic programming-type solution: Let $N(b,s)$ be the number of ways to use the first $b$ boxes to sum to $s$. Also, write $m_i$ for the number of objects in box $i$. In your case: \\begin{align*} N(1,0)&=1\\\\ N(1,1)&=1\\\\ N(1,2)&=0\\\\ N(1,3)&=0. \\end{align*}\n\nThen, the values in the second box can be computed as follows: $$N(b+1,s)=\\sum_{i=0}^{\\min\\{s,m_{b+1}\\}}N(b,s-i).$$ Using this formula: \\begin{align*} N(2,0)&=N(1,0)=1\\\\ N(2,1)&=N(1,0)+N(1,1)=2\\\\ N(2,2)&=N(1,0)+N(1,1)+N(1,2)=2\\\\ N(2,3)&=N(1,0)+N(1,1)+N(1,2)+N(1,3)=2. \\end{align*}\n\nContinuing for the third column, \\begin{align*} N(3,0)&=N(2,0)=1\\\\ N(3,1)&=N(2,0)+N(2,1)=3\\\\ N(3,2)&=N(2,0)+N(2,1)+N(2,2)=5\\\\ N(3,3)&=N(2,1)+N(2,2)+N(2,3)=6. \\end{align*}\n\nWe are interested in the value $N(3,3)=6$.\n\n\u2022 You are right, however I am currently constructing a computer program where I will deal with much bigger inputs. So I need a formula. These was just the values I came up to explain the situation. \u2013\u00a0Rockybilly Feb 13 '16 at 16:36\n\u2022 The third solution should work quite quickly on a computer (you just need to run through a pair of arrays many times). \u2013\u00a0Michael Burr Feb 13 '16 at 17:08\n\u2022 Forgive my dullness, but I didn't understand your last solution. How $$N(b, s)$$ is defined and how $$N(3,2) = 2$$ and $$N(3,2) = 5$$ in your explanation. \u2013\u00a0Rockybilly Feb 13 '16 at 17:10\n\u2022 Cut and paste error! \u2013\u00a0Michael Burr Feb 13 '16 at 17:13\n\u2022 For the calculation of $N(b+1,s)$: How many objects can be taken from the $b+1$st box? At most $s$ or the number of objects in the box. If you take $0$ objects from the $b+1$st box, then the objects must come from the remaining $b$ boxes, so you have $N(b,s)$ ways. If you take $1$ object from the $b+1$st box, then the remaining $s-1$ objects must come from the remaining $b$ boxes, so you have $N(b,s-1)$ ways. Continue adding these up until you run out of objects in the box or take $s$ objects from box $b+1$. \u2013\u00a0Michael Burr Feb 13 '16 at 17:17\n\nHere is the code that I came up with in Sage. It takes about 19 minutes that is MUCH larger than the one initially posed (1000 boxes with random values between 1 and 1000). But 100 boxes with values between 1 and 100 finished under a second. Also, if you need a particular maximum value, just change $s$ to that value.\n\nBoxes = [randint(1,100) for i in range(1000)]\nn = len(Boxes)\ns = sum(Boxes)\nl1 = [0] * (s+1)\nl2 = [0] * (s+1)\nparity = 0\nfor i in range(Boxes[0]+1):\nl1[i]=1\nfor i in range(1,n):\nif(parity == 0):\nl2[0]=l1[0]\nfor j in range(1,Boxes[i]+1):\nl2[j]=l2[j-1]+l1[j]\nfor j in range(Boxes[i]+1,s+1):\nl2[j]=l2[j-1]+l1[j]-l1[j-Boxes[i]-1]\nparity = 1\nelse:\nl1[0]=l2[0]\nfor j in range(1,Boxes[i]+1):\nl1[j]=l1[j-1]+l2[j]\nfor j in range(Boxes[i]+1,s+1):\nl1[j]=l1[j-1]+l2[j]-l2[j-Boxes[i]-1]\nparity = 0\nif(parity == 1):\nl = l2\nelse:\nl = l1\n\nprint l\n\n\nEdit: I cut out one of the loops, reducing the complexity.\n\n\u2022 Much better indeed. However, it is still not enough for the method I need. If I had found a solution in here, I would have answered this question much better. \u2013\u00a0Rockybilly Feb 14 '16 at 18:00\n\u2022 Using a generating function is a nice method. I should have thought of that! (Although the code above is (more or less) computing the coefficients of the generating function). \u2013\u00a0Michael Burr Feb 14 '16 at 18:05\n\u2022 Yes, indeed. They are quite similar methods. \u2013\u00a0Rockybilly Feb 14 '16 at 18:11", "date": "2019-08-23 11:10:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1653388/how-many-ways-we-can-choose-items-from-different-boxes", "openwebmath_score": 0.843021035194397, "openwebmath_perplexity": 201.91208405524122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989830341646133, "lm_q2_score": 0.8918110555020056, "lm_q1q2_score": 0.8827416417513487}}
{"url": "https://math.stackexchange.com/questions/3401393/are-there-a-finite-number-of-trees-with-k-leaves-and-no-vertices-of-degree-2/3401420", "text": "# Are there a finite number of trees with $k$ leaves and no vertices of degree $2$?\n\nGiven a fixed positive integer $$k$$. Show that there are only finitely many trees containing $$k$$ leaves and zero vertices of degree $$2$$.\n\nI tried to use the theorem related to rooted trees and tried to prove it by contradiction, but could not quite use the condition of having zero vertices of degree $$2$$.\n\nIs it even related to rooted trees theorem or can it be proved by contradiction? Can we use the fact that a tree with $$m$$ edges has $$m+1$$ vertices?\n\nConsider an arbitrary tree with $$k$$ leaves and no vertices of degree $$2$$. Let $$\\epsilon$$ represent the number of edges and $$\\nu_j$$ represent the number of vertices with a degree of $$j$$ (so that $$\\nu_1=k$$ and $$\\nu_2=0$$). We know that $$\\sum_{j=1}^\\infty j\\nu_j=2\\epsilon$$ (since that's true for all graphs) and $$\\sum_{j=1}^\\infty \\nu_j=\\epsilon+1$$ (since it's a tree). So we can combine the equations as follows (where the index of the summations have been taken out).\n\n$$\\sum j\\nu_j=2\\big(\\sum\\nu_j-1\\big)=2\\sum \\nu_j-2\\\\ 2\\sum\\nu_j-\\sum j\\nu_j=2\\\\\\sum(2-j)\\nu_j=2$$\n\nThe RHS is positive, so the LHS must be as well. The first few terms of the LHS is $$\\nu_1+0\\nu_2-\\nu_3-2\\nu_4-\\ldots.$$ This could not be positive if $$\\nu_3+\\nu_4+\\nu_5+\\ldots>\\nu_1$$. Since we know that $$\\nu_2=0$$, we can conclude that a tree with $$k$$ leaves and no vertices of degree $$2$$ cannot have more than $$2k$$ vertices. Since the number of trees with fewer than $$2k$$ vertices is finite, we are done.\n\n\u2022 This is very smart! I tried theorem related to rooted trees and tried proving by contradiction but could not quite use the condition of no vertices of degree 2. Oct 20, 2019 at 13:40\n\u2022 @AnsonNG It's a good trick to have in your toolbox. I learned it when proving that there was an upper bound on the number of faces of a polyhedron with a given number of pentagonal faces and no hexagonal faces. Same argument except with Euler's formula instead of the double-edge count.\n\u2013\u00a0user694818\nOct 20, 2019 at 13:48\n\nThis is easy to prove using the following two lemmata:\n\nLemma 1: The sum of the degrees of all vertices in a graph equals twice the number of edges.\n\nLemma 2: A tree with $$n$$ vertices has $$n-1$$ edges.\n\n(The first lemma is a simple consequence of the definition of the degree of a vertex, i.e. the number of edges connected to it, and the fact that each edge connects to exactly two vertices. The second lemma can be proved by induction on the number of vertices: assuming that the lemma holds for all trees with $$n-1$$ vertices, take any tree with $$n$$ vertices and consider what happens when you merge any two adjacent vertices and remove the edge between them.)\n\nTaken together these lemmata imply that, for any tree with $$n$$ vertices having the degrees $$d_1, d_2, \\dots, d_n$$ respectively, $$\\sum_{i=1}^n d_i = 2n - 2 \\quad \\text{and thus} \\quad \\sum_{i=1}^n (d_i - 2) = -2.$$ In other words, the sum of the degrees of all vertices minus two per vertex is the same (and equal to $$-2$$) for all trees!\n\nIn particular, we can see that the summand $$d_i - 2$$ is negative (and equal to $$-1$$ except for the degenerate case of the single-vertex tree) for leaves, zero for vertices of degree $$2$$ and positive (at at least one) for all other vertices. For the sum to equal $$-2$$, as it must, the positive contribution of each vertex with degree $$d_i > 2$$ must therefore be cancelled out by at least one leaf (and there need to be at least two extra leaves on top of that).\n\nThus, a tree with $$k$$ leaves can have at most $$k - 2$$ vertices of degree greater than $$2$$.\n\nFor a tree with $$k$$ leaves and no vertices of degree $$2$$, this implies that the total number of vertices in the tree can be at most $$2k - 2$$. And since the number of vertices in such a tree is thus bounded, and since there's only a finite number of possible ways of connecting any given number of vertices into a tree, this further implies that the total number of such trees is also bounded.\n\nWe can also see that, for this result to hold, it is essential that the number of vertices of degree $$2$$ be bounded (in your case by zero). Otherwise we could take any tree with $$n > 1$$ nodes and easily construct an infinite family of trees with the same number of leaves just by taking any pair of adjacent vertices and inserting an arbitrarily long linear chain of vertices of degree $$2$$ between them.", "date": "2022-05-25 12:59:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3401393/are-there-a-finite-number-of-trees-with-k-leaves-and-no-vertices-of-degree-2/3401420", "openwebmath_score": 0.9387131333351135, "openwebmath_perplexity": 59.10605418348453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303425461246, "lm_q2_score": 0.8918110540642805, "lm_q1q2_score": 0.8827416411308672}}
{"url": "http://mathematica.stackexchange.com/questions/39848/probability-of-the-sum-of-the-largest-n-samples?answertab=votes", "text": "# Probability of the sum of the largest n samples\n\n15 numbers are randomly chosen from U(0,1), what is the probability that the sum of largest four numbers is greater than 3.5?\n\nWith[{f = OrderDistribution[{UniformDistribution[], 15}, #] &},\nProbability[a + b + c + d > 3 + 1 / 2,\n{a \\[Distributed] f[15], b \\[Distributed] f[14], c \\[Distributed] f[13],\nd\\[Distributed] f[12]}]]\n\nTrouble is, it is very, very slow. I shut it down after 30 minutes. When I used NProbability it converged on the wrong answer but it did warn me with numerous error messages. Another CAS could do the above code but it also returned the same wrong answer. The right answer is supposed to be (it at least agrees with a simulation):\n\n$\\frac{224077804910008595}{584325558976905216}$\n\n$\\approx 0.383481094515780$\n\nHow do I do this using Mathematica?\n\n-\nPresumably the distributions of a,b,c and d are linked together. I don't see any such links in your code. To me it looks like you have independent a,b,c,d (from the correct OrderDistributions) when they should be linked. Presume this is why NProbability gives the wrong answer (Probability would give the same but is slower). \u2013\u00a0 Ymareth Jan 5 '14 at 16:14\n@Ymareth I came to the same conclusion, e.g. a must be greater then b. \u2013\u00a0 ybeltukov Jan 5 '14 at 16:17\n@Ymareth and ybeltukov doesn't the fact that it is distributed in f(15) do that? That is the maximum. \u2013\u00a0 bobbym Jan 5 '14 at 16:19\nA possible simulation is num = 10^6; samp = (Sort /@ RandomVariate[UniformDistribution[], {num, 15}])[[All, -4 ;; -1]]; N@Length@Select[samp, Tr@# > 3.5 &]/num The result agrees with the OP's \u2013\u00a0 belisarius Jan 5 '14 at 17:01\n@belisarius - Agrees with the stated correct answer but how in the OP's code do I know that the f[15] is always greater than f[14]. In your answer you have that implicitly as you're using the joint distribution but in the OP's code there are just 4 independent distributions - unless I'm misreading? \u2013\u00a0 Ymareth Jan 5 '14 at 18:07\n\nThat's a tricky one!\n\nOrderDistribution[{dist,n}, {k1, k2, ...}] represents the joint (k1, k2,...} th-order statistics distribution from n observations from the distribution dist.\n\nSo:\n\nProbability[a + b + c + d > 7/2, {a, b, c, d} \\[Distributed]\nOrderDistribution[{UniformDistribution[], 15}, {12, 13, 14, 15}]]\n\n(*\n224077804910008595/584325558976905216 --> 0.383481\n*)\n-\nHi belisarius; Thanks, can you see why my approach was bad? \u2013\u00a0 bobbym Jan 5 '14 at 17:55\n@bobbym Consider breaking belisarius' code into two independent regions...Probability[a+b+c+d>7/2,{{a,b}[Distributed]OrderDistribution[{Uniform\u200c\u200bDistribution[],15},{12,13}],{c,d}[Distributed]OrderDistribution[{UniformDistribu\u200c\u200btion[],15},{14,15}]}]. This is not the same. Each of these regions can then be split into 4 which will give the same answer as your original code (I think) - run time seems to be very long. \u2013\u00a0 Ymareth Jan 5 '14 at 18:25\nThe OPs code is adding four independent random variables. belisarius is instead adding four joint distributions, which is the problem setup described by \"the sum of largest four numbers\". \u2013\u00a0 bill s Jan 5 '14 at 19:32\n\nA manual approach\n\nJoint (k1, k2,...} th-order statistics distribution is great, but how we can derive the answer by hands?\n\nLet $a, b, c, d$ be the largest 4 number (in any order with each other). Let $g$ be the next largest number. We know that $g\\sim \\mathop{\\rm Beta}(11,1)$. Then the probability under consideration is\n\nNProbability[a + b + c + d > 7/2 \\[Conditioned] a > g && b > g && c > g && d > g,\n{a \\[Distributed] UniformDistribution[], b \\[Distributed] UniformDistribution[],\nc \\[Distributed] UniformDistribution[], d \\[Distributed] UniformDistribution[],\n\n0.383481\n\nThis is equivalent to the following integral\n\nBinomial[15, 4] NIntegrate[UnitStep[a + b + c + d - 7/2] 11 g^10,\n{g, 0, 1}, {a, g, 1}, {b, g, 1}, {c, g, 1}, {d, g, 1}]\n\n0.383481\n\nBinomial[15, 4] comes from the probability that $a, b, c, d$ are the largest 4 number, 11 g^10 is the PDF of the BetaDistribution[11, 1] and the integration limits come from the conditions $a > g$, etc.\n\nLet us consider the indefinite integral\n\nf[a1_, b1_, c1_, d1_] = Integrate[UnitStep[a + b + c + d - 7/2],\n{a, -\u221e, a1}, {b, -\u221e, b1}, {c, -\u221e, c1}, {d, -\u221e, d1}]\n1/384 (-7 + 2 a1 + 2 b1 + 2 c1 + 2 d1)^4 UnitStep[-(7/2) + a1 + b1 + c1 + d1]\n\nThe definite integral with limits g, 1 in every dimension is (a simple inclusion-exclusion formula)\n\nint[g_] = Total[(-1)^Total /@ Tuples[{0, 1}, 4] f @@@ Tuples[{g, 1}, 4]]\n1/384 - 1/96 (-1 + 2 g)^4 UnitStep[-(1/2) + g] +\n1/64 (-3 + 4 g)^4 UnitStep[-(3/2) + 2 g] -\n1/96 (-5 + 6 g)^4 UnitStep[-(5/2) + 3 g] +\n1/384 (-7 + 8 g)^4 UnitStep[-(7/2) + 4 g]\n\nFinally, the probability is\n\nBinomial[15, 4] Integrate[int[g] 11 g^10, {g, 0, 1}]\n224077804910008595/584325558976905216\n-", "date": "2015-07-04 05:08:11", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/39848/probability-of-the-sum-of-the-largest-n-samples?answertab=votes", "openwebmath_score": 0.616093099117279, "openwebmath_perplexity": 2570.7295486215476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126469647337, "lm_q2_score": 0.9019206811430764, "lm_q1q2_score": 0.882721177193776}}
{"url": "https://www.intmath.com/blog/learn-math/i-heart-math-5179", "text": "# I heart math\n\nBy Murray Bourne, 05 Oct 2010\n\nI saw this T-shirt recently.\n\nWhat does it mean and what's that equation?\n\nThis is an example of an implicit function. When we first learn about functions, they are written explicitly, for example:\n\nf(x) = sin(x) + 4x\n\nThis explicit function involves one dependent variable only and for each x value, we only get one f(x) value. Of course, I could also write this as\n\ny = sin(x) + 4x\n\nNotice y is on the left by itself, and the terms involving x are on the right, by themselves.\n\nBut there are many functions that are really messy when written explicitly, and so we turn to implicit functions.\n\nIn implicit functions, we see x's and y's multiplied and mixed together.\n\n## A simple example\n\nA simple example of an implicit function is the familiar equation of a circle:\n\nx2 + y2 = 16\n\nIn this simple case, we can turn this into an explicit function by solving for y and getting 2 solutions:\n\nor\n\nBut often it is very difficult, if not impossible, to solve an implicit function for y.\n\n## The t-shirt Function\n\nReturning to the t-shirt example, we have the implicit function:\n\n(x2 + y2 \u2212 1)3 = x2y3\n\nWe can expect more than one y-value for each x-value.\n\nTo graph it, we proceed as follows. Let's choose some easy values of x and y.\n\nIf x = 0, we substitute and obtain:\n\n((0)2 + y2 \u2212 1)3 = (0)2y3\n\n(y2 \u2212 1)3 = 0\n\nWe get 2 solutions, y = \u00b1 1.\n\nNow, let y = 0, and we get:\n\n(x2 + (0)2 \u2212 1)3 = x2(0)3\n\n(x2 \u2212 1)3 = 0\n\nThis gives us 2 solutions, x = \u00b1 1.\n\nSo we know the curve passes through (-1, 0), (0, -1), (1, 0) and (0, -1),\n\nNow, we choose some values of x between 0 and 1. We start with x = 0.2:\n\n((0.2)2 + y2 \u2212 1)3 = (0.2)2y3\n\nThis gives:\n\n(-0.96 + y2)3 = 0.04y3\n\nSolving this for y gives the real solutions: y = -0.824 or y = 1.166 (and 4 complex solutions).\n\nWe choose some more values and construct a table containing the real solutions:\n\n x y1 y2 0 0.2 0.4 0.6 0.8 1 1.2 -1 -0.824 -0.684 -0.520 -0.307 0 complex 1 1.166 1. 227 1. 231 1.17 complex\n\nThis equation is symmetrical, so we get the same correspnding values for -0.2, -0.4, -0.6, -0.8, -1 and -1.2.\n\nIn fact, outside of this range of x-values, there are no real y-values.\n\nIf we take a lot of points and join them, we get the following graph:\n\nSo the t-shirt means \"I heart math\" (that is, \"I love math\").\n\n## 3-D Example\n\nHere's another one in 3 dimensions. The implicit function is:\n\nfor -3 \u2264 x, y, z \u2264 3 (which means each of x, y and z takes values only between -3 and 3).\n\nAnd here's the shirt:\n\nDifferentiation of implicit functions\n\nCurves in polar coordinates\n\n### 11 Comments on \u201cI heart math\u201d\n\n1. Naren says:\n\nThis is really cool!. Your blog really helps me a lot. I love the t-shirt as well. Could I know where you found it?\n\n2. vonjd says:\n\nWA can do it too \ud83d\ude42\nhttp://www.wolframalpha.com\n\nand then type:\nplot (x^2 + y^2 - 1)^3 = x^2 y^3\n\n3. Murray says:\n\nYes! The graph in the post comes from Wolfram|Alpha.\n\n4. Samar says:\n\nSo marvelous explanation about the t shirt especially I am a math teacher\nThanks.\n\n5. yasin.. says:\n\nsuperb...nicely explained..!!!!!!!!!!!\n\n6. musa says:\n\nI like it!! To draw this graph i often use MATLAB.\n\n7. Kudzai Maravanyika says:\n\ni realy liked the way you express you maths. Keep on doing that.\n\n8. Ryan says:\n\nIf you have a ti-89 you can graph the implicit 2d function in the 3d mode by typing in the equation and formatting the plot to make it graph implicitly. The window will have to be seriously adjusted though.\n\n9. barbina says:\n\noh thank you I have been asked directly to Bagatrix and not precalculus graph sholved but must be solved to complete the curves and graphs, again thank you ...\n\n11. april says:\n\nwow!! amazing.. I like it..\nthanks! i think i should suggest it to our math club as a batch shirt..\ni like the graphing of the equation in 3-D.\n\n### Comment Preview\n\nHTML: You can use simple tags like <b>, <a href=\"...\">, etc.\n\nTo enter math, you can can either:\n\n1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):\na^2 = sqrt(b^2 + c^2)\n(See more on ASCIIMath syntax); or\n2. Use simple LaTeX in the following format. Surround your math with $$ and $$.\n$$\\int g dx = \\sqrt{\\frac{a}{b}}$$\n(This is standard simple LaTeX.)\n\nNOTE: You can mix both types of math entry in your comment.", "date": "2019-09-19 13:06:28", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/blog/learn-math/i-heart-math-5179", "openwebmath_score": 0.7151968479156494, "openwebmath_perplexity": 1905.423900734085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988130881050834, "lm_q2_score": 0.8933094138654242, "lm_q1q2_score": 0.8827066181738457}}
{"url": "http://stacks.math.columbia.edu/tag/00EC", "text": "# The Stacks Project\n\n## Tag 00EC\n\nLemma 10.20.1. Let $R$ be a ring. Let $e \\in R$ be an idempotent. In this case $$\\mathop{\\rm Spec}(R) = D(e) \\amalg D(1-e).$$\n\nProof. Note that an idempotent $e$ of a domain is either $1$ or $0$. Hence we see that \\begin{eqnarray*} D(e) & = & \\{ \\mathfrak p \\in \\mathop{\\rm Spec}(R) \\mid e \\not\\in \\mathfrak p \\} \\\\ & = & \\{ \\mathfrak p \\in \\mathop{\\rm Spec}(R) \\mid e \\not = 0\\text{ in }\\kappa(\\mathfrak p) \\} \\\\ & = & \\{ \\mathfrak p \\in \\mathop{\\rm Spec}(R) \\mid e = 1\\text{ in }\\kappa(\\mathfrak p) \\} \\end{eqnarray*} Similarly we have \\begin{eqnarray*} D(1-e) & = & \\{ \\mathfrak p \\in \\mathop{\\rm Spec}(R) \\mid 1 - e \\not\\in \\mathfrak p \\} \\\\ & = & \\{ \\mathfrak p \\in \\mathop{\\rm Spec}(R) \\mid e \\not = 1\\text{ in }\\kappa(\\mathfrak p) \\} \\\\ & = & \\{ \\mathfrak p \\in \\mathop{\\rm Spec}(R) \\mid e = 0\\text{ in }\\kappa(\\mathfrak p) \\} \\end{eqnarray*} Since the image of $e$ in any residue field is either $1$ or $0$ we deduce that $D(e)$ and $D(1-e)$ cover all of $\\mathop{\\rm Spec}(R)$. $\\square$\n\nThe code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 3447\u20133454 (see updates for more information).\n\n\\begin{lemma}\n\\label{lemma-idempotent-spec}\nLet $R$ be a ring. Let $e \\in R$ be an idempotent.\nIn this case\n$$\\Spec(R) = D(e) \\amalg D(1-e).$$\n\\end{lemma}\n\n\\begin{proof}\nNote that an idempotent $e$ of a domain is either $1$ or $0$.\nHence we see that\n\\begin{eqnarray*}\nD(e)\n& = &\n\\{ \\mathfrak p \\in \\Spec(R)\n\\mid\ne \\not\\in \\mathfrak p \\} \\\\\n& = &\n\\{ \\mathfrak p \\in \\Spec(R)\n\\mid\ne \\not = 0\\text{ in }\\kappa(\\mathfrak p) \\} \\\\\n& = &\n\\{ \\mathfrak p \\in \\Spec(R)\n\\mid\ne = 1\\text{ in }\\kappa(\\mathfrak p) \\}\n\\end{eqnarray*}\nSimilarly we have\n\\begin{eqnarray*}\nD(1-e)\n& = &\n\\{ \\mathfrak p \\in \\Spec(R)\n\\mid\n1 - e \\not\\in \\mathfrak p \\} \\\\\n& = &\n\\{ \\mathfrak p \\in \\Spec(R)\n\\mid\ne \\not = 1\\text{ in }\\kappa(\\mathfrak p) \\} \\\\\n& = &\n\\{ \\mathfrak p \\in \\Spec(R)\n\\mid\ne = 0\\text{ in }\\kappa(\\mathfrak p) \\}\n\\end{eqnarray*}\nSince the image of $e$ in any residue field is either $1$ or $0$\nwe deduce that $D(e)$ and $D(1-e)$ cover all of $\\Spec(R)$.\n\\end{proof}\n\nThere are no comments yet for this tag.\n\n## Add a comment on tag 00EC\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).", "date": "2017-04-26 17:44:59", "meta": {"domain": "columbia.edu", "url": "http://stacks.math.columbia.edu/tag/00EC", "openwebmath_score": 1.0000007152557373, "openwebmath_perplexity": 1124.5804086788503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308791284249, "lm_q2_score": 0.8933093954028816, "lm_q1q2_score": 0.8827065982131311}}
{"url": "https://math.stackexchange.com/questions/497639/predicate-logic-translating-all-but-one", "text": "# Predicate Logic Translating \u201cAll But One\u201d\n\nI need to translate an English sentence including the phrase \"all but one\" into predicate logic. The sentence is: \"All students but one have an internet connection.\" I'm not sure how to show \"all but one\" in logic.\n\nI could say $\\forall x ((x \\neq a) \\rightarrow I(x))$\n\n$I(x)$ being \"$x$ has an internet connection\"\n\nBut that clearly wouldn't work in this case, as we don't know which student it is.\n\nI could say that $\\exists x(\\neg I(x))$\n\nBut it doesn't seem like that has the same meaning. Thanks in advance for any help you can give!\n\n\u2022 Your suggestion doesn't seem right because presumably the universe is the set of all people or something like that. So you need a predicate to mean '$x$ is a student' and one to mean '$x$ has an internet connection'. As a hint rephrase it as $\\text {there exists a student that doesn't have an internet connection such that all other students do.}$ \u2013\u00a0Git Gud Sep 18 '13 at 16:10\n\u2022 sorry, in this case, the universe is the set of all people in a class. So I should have said \"All students in the class but one have an internet connection.\" So in this case, if all students had a connection it would be \u2200xI(x). \u2013\u00a0user95552 Sep 18 '13 at 16:16\n\u2022 possible duplicate of Write \u2018There is exactly 1 person\u2026\u2019 without the uniqueness quantifier \u2013\u00a0MJD Sep 18 '13 at 16:17\n\u2022 @user95552 First note that my hint is wrong because it doesn't deal with uniqueness. Regarding what you said, I'd say it's wrong because there are people in the class which aren't students, but that's probably up to interpretation. Edit: my hint deals with uniqueness after all, so the hint is correct. Sorry. \u2013\u00a0Git Gud Sep 18 '13 at 16:19\n\u2022 Alright, well I can easily change it to work with a different universe, but from your hint, could I say that \u2203x(\u00acI(x) \u2227 \u2200y((x\u2260y) \u2192 I(y))) meaning there is a student without an internet connection and all students who aren't that student do have a connection. \u2013\u00a0user95552 Sep 18 '13 at 16:25\n\nIf you mean that there is exactly one element with a given property, you can define a \"unique existence\" quantifier, $\\exists!$, as follows: $$\\exists!x : \\varphi(x) \\iff \\exists{x}{:}\\left[\\varphi(x)\\wedge \\forall{y}:\\left(\\varphi(y){\\iff} y=x\\right)\\right].$$ That is, a particular element $x$ has the property $\\varphi$, and any element with the property $\\varphi$ must be that same $x$. For your problem, you want to say that there's exactly one person that is a student and doesn't have internet access.\n\n\"All students but one have an internet connection\" means there is a student who lacks a connection, while every other student (every student not identical to the unlucky guy!) has one. So (if the domain is e.g. people)\n\n$$\\exists x(\\{Sx \\land \\neg Ix\\} \\land \\forall y(\\{Sy \\land \\neg y = x\\} \\to Iy))$$\n\n\u2022 Could have saved my fingers if I'd read all the comments first -- but I'll leave this in place. \u2013\u00a0Peter Smith Sep 18 '13 at 16:49\n\n\"For all but one $\\;x\\;$, $\\;P(x)\\;$ holds\" is the same as \"there exists a unique $\\;x\\;$ such that $\\;\\lnot P(x)\\;$ holds.\n\nNormally the notation $\\;\\exists!\\;$ is used for \"there exists a unique\" (just like $\\;\\exists\\;$ is used for \"there exists some\").\n\nIf your answer is allowed to use $\\;\\exists!\\;$, then the above gives you the answer.\n\nIf not, then there are different ways to write $\\;\\exists!\\;$ in terms of $\\;\\exists\\;$ and $\\;\\forall\\;$. The one I like best, which also results in the shortest formula, can be found in another answer of mine.", "date": "2019-08-19 23:51:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/497639/predicate-logic-translating-all-but-one", "openwebmath_score": 0.7436140775680542, "openwebmath_perplexity": 365.6167340615774, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.967410251945384, "lm_q2_score": 0.9124361598816667, "lm_q1q2_score": 0.8827000953152019}}
{"url": "https://gmatclub.com/forum/sum-of-n-positive-integers-formula-for-consecutive-integers-163301.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 21 Mar 2019, 21:17\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Sum of n positive integers formula for consecutive integers?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 11 Sep 2012\nPosts: 86\nSum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 20 Feb 2019, 03:34\n7\nIs the sum of n positive integers formula, $$\\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?\n\nFor instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\\frac{20(21)}{2}$$\n\nIn OG PS Q172, a similar approach has been used for even numbers.\n\nOriginally posted by RustyR on 16 Nov 2013, 17:15.\nLast edited by Bunuel on 20 Feb 2019, 03:34, edited 1 time in total.\nUpdated.\nVP\nJoined: 02 Jul 2012\nPosts: 1161\nLocation: India\nConcentration: Strategy\nGMAT 1: 740 Q49 V42\nGPA: 3.8\nWE: Engineering (Energy and Utilities)\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n17 Nov 2013, 01:27\n1\n1\n3+6+9+12+15+18+21+24+27+30...... = 3*(1+2+3+4+5+6+7+8+9+10...)\n\nSo, yup.. You can do that...\n_________________\n\nDid you find this post helpful?... Please let me know through the Kudos button.\n\nThanks To The Almighty - My GMAT Debrief\n\nGMAT Reading Comprehension: 7 Most Common Passage Types\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8998\nLocation: Pune, India\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n17 Nov 2013, 20:25\n11\n10\nbschoolaspirant9 wrote:\nIs the sum of n positive integers formula, $$\\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?\n\nFor instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\\frac{20(21)}{2}$$\n\nIn OG PS Q172, a similar approach has been used for even numbers.\n\nNote that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.\nA problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.\n\nIn the even integers questions, you may be required to find the sum of first 10 even integers.\n2 + 4 + 6 + ... + 18 + 20\nTake 2 common, 2*(1 + 2 + 3 + ...10)\nTo find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.\n\nNote that for odd integers, you cannot directly use this formula.\nSum the first 10 odd integers\n1 + 3 + 5 + 7+...+19\nBut you can still make some modifications to find the sum.\n\n1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)\nWe know how to sum consecutive integers.\n(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2\n(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)\n\nSo 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100\n\nThe direct formula of sum of n consecutive odd integers starting from 1 = n^2\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nIntern\nJoined: 16 Jul 2011\nPosts: 40\nConcentration: Marketing, Real Estate\nGMAT 1: 550 Q37 V28\nGMAT 2: 610 Q43 V31\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n19 Jun 2015, 08:13\nVeritasPrepKarishma wrote:\nbschoolaspirant9 wrote:\nIs the sum of n positive integers formula, $$\\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?\n\nFor instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\\frac{20(21)}{2}$$\n\nIn OG PS Q172, a similar approach has been used for even numbers.\n\nNote that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.\nA problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.\n\nIn the even integers questions, you may be required to find the sum of first 10 even integers.\n2 + 4 + 6 + ... + 18 + 20\nTake 2 common, 2*(1 + 2 + 3 + ...10)\nTo find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.\n\nNote that for odd integers, you cannot directly use this formula.\nSum the first 10 odd integers\n1 + 3 + 5 + 7+...+19\nBut you can still make some modifications to find the sum.\n\n1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)\nWe know how to sum consecutive integers.\n(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2\n(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)\n\nSo 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100\n\nThe direct formula of sum of n consecutive odd integers starting from 1 = n^2\n\nGreat explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?\n_________________\n\n\"The fool didn't know it was impossible, so he did it.\"\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8998\nLocation: Pune, India\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Jun 2015, 21:57\n2\nsamdighe wrote:\nVeritasPrepKarishma wrote:\nbschoolaspirant9 wrote:\nIs the sum of n positive integers formula, $$\\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?\n\nFor instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\\frac{20(21)}{2}$$\n\nIn OG PS Q172, a similar approach has been used for even numbers.\n\nNote that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.\nA problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.\n\nIn the even integers questions, you may be required to find the sum of first 10 even integers.\n2 + 4 + 6 + ... + 18 + 20\nTake 2 common, 2*(1 + 2 + 3 + ...10)\nTo find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.\n\nNote that for odd integers, you cannot directly use this formula.\nSum the first 10 odd integers\n1 + 3 + 5 + 7+...+19\nBut you can still make some modifications to find the sum.\n\n1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)\nWe know how to sum consecutive integers.\n(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2\n(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)\n\nSo 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100\n\nThe direct formula of sum of n consecutive odd integers starting from 1 = n^2\n\nGreat explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?\n\nWe can, provided we know the average and the number of terms.\nIf we are asked to find the sum of first 50 consecutive positive odd integers, it might be easier to use 50^2 than to find average and then find the sum.\n\nMind you, I myself believe in knowing just the main all-applicable kind of formulas and then twisting them around to apply to any situation. But some people prefer to work more on specific formulas and these discussions are for their benefit.\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nDirector\nAffiliations: GMATQuantum\nJoined: 19 Apr 2009\nPosts: 612\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n24 Jun 2015, 04:57\n9\n2\nThe best way to deal with sum of any type of arithmetic series is to use the following expression:\n\nSum = Average*Number of terms\n\nAverage = (First term + Last term)/2\n\nNote: The average or the median can be related to the first and last terms, which are often given for these types of problems.\n\nNumber of Terms = (Last Term - First Term)/Spacing + 1\n\nAlso, I recommend that you understand the basis of these relationships as opposed to blindly memorizing them.\n\nThe above relationships are flexible enough to allow you to deal with any GMAT problem on these concepts.\n\nFor example in Q172 from the Official Guide: What is the sum of all the even integers between 99 and 301 ?\n\nAverage = (100+300)/2 = 200\n\nNumber of Terms = (300-100)/2 + 1 = 101 (Note spacing is 2 between consecutive even integers)\n\nSum = 200*101=20200\n\nCheers,\nDabral\nIntern\nJoined: 19 Dec 2018\nPosts: 1\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n19 Dec 2018, 23:54\nfor consecutive odd or even ranges\n\nsum = (f + n-1) n\n\nf is the first number in the sequence, n is the number of elements in the sequence\nDirector\nAffiliations: GMATQuantum\nJoined: 19 Apr 2009\nPosts: 612\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Dec 2018, 01:01\n1\nTop Contributor\nHi mathsnoob,\n\nYou are correct that the formula that you listed can be used for adding consecutive even and odd integers. In fact, it can be derived from the general approach. For example, n would be the number of terms in the sequence. The last term would be equal to f + 2(n-1), the common difference here is 2 for consecutive even and odd integers. This means the average of the first and last term is equal to [f + f + 2(n-1)]/2, which simplifies to f + (n-1). And finally we have an expression for the sum of n consecutive integers n[f + n - 1].\n\nThe problem with these highly specialized formulas is the burden of remembering the specific formula and keeping track of what each term stands for. Personally, I prefer to stick to the general formula that I listed earlier and then adapt it to a specific situation. This helps reduce the memorization burden on my brain. But I can see others who might prefer to memorize formulas for specific situations. Take your pick.\n\nCheers,\nDabral\nMath Expert\nJoined: 02 Sep 2009\nPosts: 53771\nRe: Sum of n positive integers formula for consecutive integers?\u00a0 [#permalink]\n\n### Show Tags\n\n20 Feb 2019, 03:34\nRustyR wrote:\nIs the sum of n positive integers formula, $$\\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?\n\nFor instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\\frac{20(21)}{2}$$\n\nIn OG PS Q172, a similar approach has been used for even numbers.\n\nFor more check Formulas for Consecutive, Even, Odd Integers\n\n_________________\nRe: Sum of n positive integers formula for consecutive integers? \u00a0 [#permalink] 20 Feb 2019, 03:34\nDisplay posts from previous: Sort by", "date": "2019-03-22 04:17:50", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/sum-of-n-positive-integers-formula-for-consecutive-integers-163301.html", "openwebmath_score": 0.8596053719520569, "openwebmath_perplexity": 1141.4389797102813, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 0.9465966702001757, "lm_q2_score": 0.9324533121452495, "lm_q1q2_score": 0.8826572003938182}}
{"url": "http://mathhelpforum.com/differential-equations/182030-solve-x-2-d-2-dx-2-3x-dy-dx-y-1-x.html", "text": "# Math Help - Solve (x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x\n\n1. ## Solve (x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x\n\n(x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x\n\nhow can i find the general solution of differential equation?\n\n2. It's a Cauchy-Euler equation. Use that procedure.\n\n3. but right side of equation isn't equal to zero? can you solve it for me?\n\n4. We don't \"solve things\" for you. I would suggest either assuming a solution that looks like the RHS, or, if you choose the exponential approach (thus producing a linear DE with constant coefficients), the choice of a particular solution might be more obvious. Try those, and post your work, and we can go from there.\n\n5. Just to add my 2 cents. If you can solve the homogenous equation (Ackbeet told you how) you can find the particular solution by using the variation of parameters.\nVariation of parameters - Wikipedia, the free encyclopedia This method does not involve guessing but you may end up with integrals in your solution, but that wont be the case in this particular problem.\n\n6. Put $t=\\ln x$ so we have $\\frac{dy}{dx}=\\frac{dy}{dt}\\cdot \\frac{dt}{dx}=\\frac{1}{x}\\cdot \\frac{dy}{dt},$ and then $\\frac{{{d}^{2}}y}{d{{x}^{2}}}=\\frac{d}{dx}\\left( \\frac{1}{x}\\cdot \\frac{dy}{dt} \\right)=-\\frac{1}{{{x}^{2}}}\\cdot \\frac{dy}{dt}+\\frac{1}{x}\\cdot\\frac{{{d}^{2}}y}{d{ {t}^{2}}},$ so you'll end up with an ODE with constant coefficients.\n\n7. Originally Posted by Ackbeet\nWe don't \"solve things\" for you. I would suggest either assuming a solution that looks like the RHS, or, if you choose the exponential approach (thus producing a linear DE with constant coefficients), the choice of a particular solution might be more obvious. Try those, and post your work, and we can go from there.\nI am actually curious how to get the PI. I can get the CF. Would a first guess of PI=Ax^n be a good guess to suit 1/x?\n\n8. Originally Posted by bugatti79\nI am actually curious how to get the PI. I can get the CF. Would a first guess of PI=Ax^n be a good guess to suit 1/x?\nWhy not try it and see what happens?\n\n9. I just noticed something that I think is worth mentioning.\n\nNotice that on the left hand side we have\n\n$x^2y''+3xy'+y=x^2y''+2xy'+xy'+y$\n\nNotice that this is a perfect derivative\n\n$\\frac{d}{dx}\\left(x^2y'+xy \\right)$\n\nSo we can rewrite the equation as\n\n$\\frac{d}{dx}\\left(x^2y'+xy \\right)=\\frac{d}{dx}\\ln(x)$\n\nand reduce this to a first order ODE!\n\n10. Originally Posted by TheEmptySet\nI just noticed something that I think is worth mentioning.\n\nNotice that on the left hand side we have\n\n$x^2y''+3xy'+y=x^2y''+2xy'+xy'+y$\n\nNotice that this is a perfect derivative\n\n$\\frac{d}{dx}\\left(x^2y'+xy \\right)$\n\nSo we can rewrite the equation as\n\n$\\frac{d}{dx}\\left(x^2y'+xy \\right)=\\frac{d}{dx}\\ln(x)$\n\nand reduce this to a first order ODE!\nThats interesting. I see what you have done but how does one reduce it to a first order? I havent seen this before to the best of my poor memory :-) Anyhow, I attempted it doing it the hard way by guessing the PI. I dont think it works out....\n\n$\\displaystyle Let y=Ax^n, y'=nAx^{n-1}, y''=n(n-1)Ax^{n-2}$\n\nPutting back into the original DE, I arrive at\n\n$Ax^n[n^2+2n+1]=\\frac{1}{x} \\implies Ax^n=\\frac{1}{x} \\because [n^2+2n+1]=0 \\therefore A=1$\n\nIs this right?\n\n11. @bugatti79\n\nWhat I have is\n\n$\\frac{d}{dx}\\left(x^2y'+xy \\right)=\\frac{d}{dx}\\ln(x)$\n\nNow if you integrate both sides you get\n\n$x^2y'+xy =\\ln(x)+C$\n\nThis is now a first order ODE that can be solve via an integrating factor!\n\n12. Originally Posted by TheEmptySet\n@bugatti79\n\nWhat I have is\n\n$\\frac{d}{dx}\\left(x^2y'+xy \\right)=\\frac{d}{dx}\\ln(x)$\n\nNow if you integrate both sides you get\n\n$x^2y'+xy =\\ln(x)+C$\n\nThis is now a first order ODE that can be solve via an integrating factor!\nYes I solved it using the integrating factor method. I am curious where I went wrong when guessing the PI in post #10. Where are you Ackbeet?\n\n13. Originally Posted by bugatti79\nWhere are you, Ackbeet?\nNot paying attention to the fact that you tried my suggestion, apparently.\n\nYeah, so that particular solution doesn't work, does it? You've got 0 = 1/x, a contradiction. Apparently, we need to change the ansatz. Note that the complimentary solution actually contains 1/x in it already, and by variation of parameters, you can find out that the complimentary solution also contains ln(x) / x. The next step with Cauchy-Euler equations is the (ln(x))^2 / x solution, so try a constant times that.\n\n14. Originally Posted by Ackbeet\nNot paying attention to the fact that you tried my suggestion, apparently.\n\nThe next step with Cauchy-Euler equations is the (ln(x))^2 / x solution, so try a constant times that.\nHmmmm..I got the first and second A times the above and put back into DE. I arrived at 0=1. Not to worry, I learned a nice bit and some differentiating practice. At least we know the integrating factor works. :-)\n\n15. Originally Posted by bugatti79\nHmmmm..I got the first and second A times the above and put back into DE. I arrived at 0=1. Not to worry, I learned a nice bit and some differentiating practice. At least we know the integrating factor works. :-)\nWell, here's my shot at it.\n\n$y=A\\,\\frac{\\ln^{2}(x)}{x}$\n\n$y'=A\\,\\frac{2(x)(\\ln(x)/x)-(\\ln^{2}(x))(1)}{x^{2}}=A\\,\\frac{2\\ln(x)-\\ln^{2}(x)}{x^{2}}.$\n\n$y''=A\\,\\frac{(x^{2})(2/x-2\\ln(x)/x)-2(2\\ln(x)-\\ln^{2}(x))(x)}{x^{4}}=A\\,\\frac{x(2-2\\ln(x))-2x(2\\ln(x)-\\ln^{2}(x))}{x^{4}}$\n\n$=2A\\,\\frac{1-3\\ln(x)+\\ln^{2}(x)}{x^{3}}.$\n\nPlugging these into the DE yields\n\n$2A\\,\\frac{1-3\\ln(x)+\\ln^{2}(x)}{x}+3A\\,\\frac{2\\ln(x)-\\ln^{2}(x)}{x}+A\\,\\frac{\\ln^{2}(x)}{x}=\\frac{1}{x} \\quad\\implies$\n\n$2A=1,$ or $A=1/2.$\n\nSo the particular solution is\n\n$y_{p}=\\frac{\\ln^{2}(x)}{2x}.$\n\nPage 1 of 2 12 Last", "date": "2014-03-16 23:27:28", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/differential-equations/182030-solve-x-2-d-2-dx-2-3x-dy-dx-y-1-x.html", "openwebmath_score": 0.8851804733276367, "openwebmath_perplexity": 559.7885446521278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9848109498546361, "lm_q2_score": 0.8962513717480382, "lm_q1q2_score": 0.882638164719706}}
{"url": "https://math.stackexchange.com/questions/4267377/length-of-the-curve-displaystyle-3ay2-xx-a2", "text": "# Length of the curve $\\displaystyle 3ay^2=x(x-a)^2$\n\nI'm working on finding out the length of the curve $$3ay^2=x(x-a)^2 \\tag{1}$$\n\nI ran into a small problem, but was able to end up with an answer that looks right but I'm not entirely sure about it. Here's my approach:\n\nThe length of a curve $$f(x)$$ between $$x=a$$ and $$x=b$$ where $$b>a$$, is given by $$L=\\int_a^b\\sqrt{1+\\big[f'(x)\\big]^2}dx$$ where $$f'(x)$$ is continuous in $$[a,b]$$. In our case\n\n$$L=\\int_0^x\\sqrt{1+\\big[y'\\big]^2}dt \\tag{2}$$\n\nCurve $$(1)$$ is symmetric about the $$x$$-axis. Hence I'll just work on the length of the part that is above the $$x$$-axis and then double that to obtain the entire length. $$y=(x-a)\\sqrt{\\frac{x}{3a}} \\text{ (One half of the curve, the other half being the negative multiple)}$$ Note that $$(1)$$ is defined for $$x\\geq0$$ and $$a>0$$. That is why I've taken the limits of integration as $$0$$ to $$x$$.\n\nTherefore, $$y'=\\frac{3x-a}{\\sqrt{12ax}}$$\n\nBut $$y'$$ is not continuous at $$x=0$$ and hence I can't plug this into $$(2)$$. So, let me calculate the length from $$h>0$$ to any $$x$$. Therefore, we have \\begin{aligned} L &= \\int_h^x\\sqrt{1+\\big[y'\\big]^2}dt \\\\ &= \\int_h^x\\sqrt{1+\\left(\\frac{3t-a}{\\sqrt{12at}}\\right)^2}dt \\\\ &= \\int_h^x \\frac{3t+a}{\\sqrt{12at}} dt \\\\ &= \\frac{x^{\\frac{3}{2}}-h^{\\frac{3}{2}}+ax^{\\frac{1}{2}}-ah^{\\frac{1}{2}}}{\\sqrt{3a}} \\end{aligned}\n\nNow, since I want the length in $$[0,x]$$, I'll just take the limit $$h\\rightarrow0$$, which yields $$\\boxed{L=(x+a)\\sqrt{\\frac{x}{3a}}}$$ The length of the curve in question would be double the above.\n\nI don't see any problem with what I've done. Is this good?\n\n\u2022 If it helps, Wolfram Alpha agrees with you upon some simple experimentation (I integrated from $0\\to5$ and compared your formula's answer to Wolfram's answer and they agreed). However, I do not own Wolfram Pro and a general form integral took the poor computer too long to calculate, so my one experiment may not be proof that your answer is correct. That being said, mathematically, it does look right to me! Oct 4, 2021 at 9:18\n\u2022 @FShrike That was a great idea. I checked my solution against Wolfram and it seems good. But my concern is regarding the legitimacy of the math I've done to arrive at that solution. Oct 4, 2021 at 12:40\n\nCubic curve with implicit equation:\n\n$$3ay^2=x(x-a)^2\\tag{1}$$\n\nhas an \"alpha\" shape with a double point $$D(a,0)$$ as can be seen on this Desmos figure (in the case $$a=2$$):\n\nThis curve can be described in an alternative way using the following parametric representation :\n\n$$\\begin{cases}x&=&3am^2\\\\y&=&am(3m^2-1)\\end{cases}\\tag{2}$$\n\nExplanation: parameter $$m$$ has the following geometrical interpretation : it is the slope of a variable line (represented in blue on the figure) passing through double point $$D$$ with equation:\n\n$$y=m(x-a)\\tag{3}$$\n\nPlugging (3) into (1) gives (2), after simplification (end of explanation of (2)).\n\nThe length of the curve between parameters values $$m_1$$ and $$m_2$$ is (using a classical formula):\n\n$$L=\\int_{m_1}^{m_2}\\sqrt{x'(m)^2+y'(m)^2}dm$$\n\n$$L=\\int_{m_1}^{m_2}\\sqrt{(6am)^2+|9am^2-a|^2}dm$$\n\n$$L=a\\int_{m_1}^{m_2}\\sqrt{(6m)^2+(9m^2-1)^2}dm$$\n\n$$L=a\\int_{m_1}^{m_2}\\sqrt{81m^4+18m^2+1}dm$$\n\n$$L=a\\int_{m_1}^{m_2}(9m^2+1)dm$$\n\n$$L=a[m(3m^2+1)]_{m_1}^{m_2},\\tag{4}$$\n\nplainly. It remains to convert (4) in terms of variable $$x$$ where\n\n$$x=3am^2 \\iff m=\\sqrt{\\frac{x}{3a}}$$\n\n(if we consider only positive slopes).\n\n(4) gives\n\n$$L=(x+a)\\sqrt{\\frac{x}{3a}}$$\n\na result which is now the same as yours.\n\nRemarks:\n\n1. we are here in an exceptional case where we have an analytic formula for the arc length... These exceptional cases for cubic curves have been studied here using cubic Bezier curves techniques.\n\n2. The parameterization using the slope of a line passing through the double point is classical.\n\n\u2022 I did a check using Wolfram Alpha and my answer matches with what I've derived. But my concern is whether the math that I've done is legit. wolframalpha.com/input/\u2026 Oct 4, 2021 at 12:32\n\u2022 I have corrected an error of mine giving a formula closer to yours, but definitely not the same. Oct 4, 2021 at 12:52\n\u2022 That's a good catch. I should have not said '1st Quadrant'. I meant the 'one-half'. Let me correct that. Oct 4, 2021 at 13:00\n\u2022 Your answer is correct. It works for all values of $a$. Mine doesn't and also your observation is on point that $L\\rightarrow 0$ as $a\\rightarrow\\infty$ Oct 4, 2021 at 13:19\n\u2022 My genius brain made an integration mistake. I've fixed it now and it is same as yours. Thank you so much for all the help. Oct 4, 2021 at 13:35", "date": "2023-02-03 22:58:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4267377/length-of-the-curve-displaystyle-3ay2-xx-a2", "openwebmath_score": 0.9836865067481995, "openwebmath_perplexity": 230.72393468313726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226267447513, "lm_q2_score": 0.9032941982430048, "lm_q1q2_score": 0.8825388702906747}}
{"url": "https://math.stackexchange.com/questions/2680258/whats-the-difference-between-mutually-exclusive-and-pairwise-disjoint/2680280", "text": "# What's the difference between MUTUALLY EXCLUSIVE and PAIRWISE DISJOINT?\n\nWhen I study Statistical Theory, I find that these two concepts confuse me a lot.\n\nBy definition, if we say two events are PAIRWISE DISJOINT, that means the intersection of these two event is empty set. If we say that two events are MUTUALLY EXCLUSIVE, that means if one of these two events happens, the other will not. But doesn't it means that these two events are PAIRWISE DISJOINT\uff1f\n\nIf we say two events are MUTUALLY EXCLUSIVE, then they are not INDEPENDENT. Can we say that two PAIRWISE DISJOINT events are not INDEPENDENT as well\uff1f\n\nIf these two concepts are different (actually my teacher told me they are), could you please give me an example that two events are MUTUALLY EXCLUSIVE but not PAIRWISE DISJOINT, or they are PAIRWISE DISJOINT but not MUTUALLY EXCLUSIVE.\n\n\"Disjoint\" is a property of sets. Two sets are disjoint if there is no element in both of them, that is if $$A \\cap B = \\emptyset$$.\n\nIn some (but not all!) texts, \"mutually exclusive\" is a slightly different property of events (sets in a probability space). Two events are mutually exclusive if the probability of them both occurring is zero, that is if $$\\operatorname{Pr}(A \\cap B) = 0$$. With that definition, disjoint sets are necessarily mutually exclusive, but mutually exclusive events aren't necessarily disjoint.\n\nConsider points in the square with each coordinate uniformly distributed from $$0$$ to $$1$$. Let $$A$$ be the event where the $$x$$-coordinate is $$0$$, and $$B$$ be the event that the $$y$$-coordinate is $$0$$. $$A \\cap B = \\{(0,0)\\}$$ so $$A$$ and $$B$$ are not disjoint, but $$\\operatorname{Pr}(A \\cap B) = 0$$ so they are mutually exclusive.\n\nAs a second (silly, but finite) example, let the sample space be $$S = \\{x, y, z\\}$$ with probabilities $$\\operatorname{Pr}(\\{x\\}) = 0$$, $$\\operatorname{Pr}(\\{y\\}) = \\frac{1}{2}$$, and $$\\operatorname{Pr}(\\{z\\}) = \\frac{1}{2}$$. If $$A = \\{x, y\\}$$ and $$B = \\{x, z\\}$$, then $$A \\cap B = \\{x\\}$$, but $$\\operatorname{Pr}(A \\cap B) = \\operatorname{Pr}(\\{x\\}) = 0$$. They are mutually exclusive but not disjoint.\n\nGood discussion and I agree. This is an interesting result that seems to be ignored most of the time and disjoint and mutually exclusive are taken as equivalent. If we have finite sample spaces without 0 probability outcomes then the counter-examples above suggest they are then equivalent, and these are the properties of the common examples used when first teaching students these concepts. The current Wikipedia entry for Mutual Exclusive conflates the two concepts. The exceptions to equivalence seem to occur only when 0 probability outcomes exist in the sample space, but this is a requirement of continuous sample spaces and is not precluded in the definition of discrete sample spaces, both definitions usually state the outcomes must be mutually exclusive. However this seems to be a redundant condition since the different simple events of any sample space are disjoint by the definition of a set (i.e. multiplicities of elements are reduced) in any case, which thus implies mutual exclusivity by default.", "date": "2020-10-27 09:15:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2680258/whats-the-difference-between-mutually-exclusive-and-pairwise-disjoint/2680280", "openwebmath_score": 0.9074979424476624, "openwebmath_perplexity": 114.3366865828029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765557865637, "lm_q2_score": 0.9005297794439687, "lm_q1q2_score": 0.8824980716427343}}
{"url": "https://www.flypmedia.com/pittosporum-images-lmk/side-of-rhombus-formula-7b2219", "text": "Online calculators and formulas for a rhombus \u2026 Area of plane shapes. The area of the rhombus is given by the formula: Area of rhombus = sh. Abhishek241 Abhishek241 19.08.2017 Math Secondary School +5 pts. A rhombus is a type of Parallelogram only. Sitemap. The side of rhombus is a tangent to the circle. Solution: All the sides of a rhombus are congruent, so HO = (x + 2).And because the diagonals of a rhombus are perpendicular, triangle HBO is a right triangle.With the help of Pythagorean Theorem, we get, (HB) 2 + (BO) 2 = (HO) 2x 2 + (x+1) 2 = (x+2) 2 x 2 + x 2 + 2x + 1 = x 2 + 4x + 4 x 2 \u2013 2x -3 = 0 Solving for x using the quadratic formula, we get: x = 3 or x = \u20131. A rhombus is a special type of quadrilateral parallelogram, where the opposite sides are parallel and opposite angles are equal and the diagonals bisect each other at right angles. Formula for perimeter of a rhombus : = 4s Substitute 16 for s. = 4(16) = 64. Side of a Rhombus when Diagonals are given calculator uses Side A=sqrt((Diagonal 1)^2+(Diagonal 2)^2)/2 to calculate the Side A, Side of a Rhombus when Diagonals are given can be defined as the line segment that joins two vertices in a rhombus provided the value for both the diagonals are given. By applying the perimeter formula, the solution is: Check: Other Names. Solution: Since we are given the side length, we can plug it straight into the formula. Since a rhombus is a parallelogram in which all sides are equal, all the same formulas apply to it as for a parallelogram, including the formula for finding the area through the product of height and side. So, the perimeter of the rhombus is 64 cm. Diagonals divide a rhombus into four absolutely identical right-angled triangles. Join now. By \u2026 . Is a Square a Rhombus? The total distance traveled along the border of a rhombus is the perimeter of a rhombus. So by the same argument, that side's equal to that side, so the two diagonals of any rhombus are perpendicular to \u2026 Using side and height. Given two integers A and X, denoting the length of a side of a rhombus and an angle respectively, the task is to find the area of the rhombus.. A rhombus is a quadrilateral having 4 sides of equal length, in which both the opposite sides are parallel, and opposite angles are equal.. To solve this problem, apply the perimeter formula for a rhombus: . Now the area of triangle AOB = \u00bd * OA * OB = \u00bd * AB * r (both using formula \u00bd*b*h). When the altitude or height and the length of the sides of a rhombus are known, the area is given by the formula; Area of rhombus = base \u00d7 height. The proof is completed. This formula for the area of a rhombus is similar to the area formula for a parallelogram. Problem 1: Find the perimeter of a rhombus with a side length of 10. The formula for perimeter of a rhombus is given as: P = 4s Where P is the perimeter and s is the side length. [3] What is the formula of Rhombus When one side is given Get the answers you need, now! Or as a formula: The perimeter formula for a rhombus is the same formula used to find the perimeter of a square. Heron's Formula depends on knowing the semiperimeter, or half the perimeter, of a triangle. Calculate the unknown defining areas, angels and side lengths of a rhombus with any 2 known variables. Formula for side of rhombus when diagonals are given - 1399111 1. Log in. The area of the rhombus can be found, also knowing its diagonal. How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If we use Heron\u2019s formula then we find area of one triangle made by two sides and a diagonal then twice of this area is area of rhombus. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Answered Formula for side of rhombus when diagonals are given 2 1. Perimeter = 4 \u00d7 12 cm = 48 cm. Thus, the total perimeter is the sum of all sides. Yes, because a square is just a rhombus where the angles are all right angles. The \"base times height\" method First pick one side to be the base. P = 4s P = 4(10) = 40 A rhombus is often called as a diamond or diamond-shaped. Here at Vedantu you will learn how to find the area of rhombus and also get free study materials to help you to score good marks in your exams. Since a rhombus is also a parallelogram, we can use the formula for the area of a parallelogram: A = b\u00d7h. If one of its diagonal is 8 cm long, find the length of the other diagonal. Ask your question. This is because both shapes, by definition, have equivalent sides. Any isosceles triangle, if that side's equal to that side, if you drop an altitude, these two triangles are going to be symmetric, and you will have bisected the opposite side. Given the length of diagonal \u2018d1\u2019 of a rhombus and a side \u2018a\u2019, the task is to find the area of that rhombus. Its diagonals perpendicularly bisect each other. The formula to calculate the area of a rhombus is: A = \u00bd x d 1 x d 2. where... A = area of rhombus; d 1 = diagonal1 (first diagonal in rhombus, as indicated by red line) d 2 = diagonal2 (second diagonal in rhombus, as indicated by purple line) Home List of all formulas of the site; Geometry. Area Of [\u2026] where b is the base or the side length of the rhombus, and h is the corresponding height. We should recall several things. In the case of a rhombus, all four sides are the same length by definition, so the perimeter is four times the length of a side. Many of the area calculations can be applied to them also. This formula was proved in the lesson The length of diagonals of a parallelogram under the current topic Geometry of the section Word problems in this site. The rhombus is often called a diamond, after the diamonds suit in playing cards, or a lozenge, though the latter sometimes refers specifically to a rhombus with a 45\u00b0 angle. Example 2 : If the perimeter of a rhombus is 72 inches, then find the length of each side. These formulas are a direct consequence of the law of cosines. Here, r is the radius that is to be found using a and, the diagonals whose values are given. X Research source You could also use the formula P = S + S + S + S {\\displaystyle P=S+S+S+S} to find the perimeter, since the perimeter of any polygon is the sum of all its sides. This geometry video tutorial explains how to calculate the area of a rhombus using side lengths and diagonals based on a simple formula. What is the area of a rhombus when only a side is given, and nothing else? Hello!!! Log in. The area of the rhombus can be found, also knowing its diagonal. Any one will do, they are all the same length. The inradius (the radius of a circle inscribed in the rhombus), denoted by r, can be expressed in terms of the diagonals p and q as = \u22c5 +, or in terms of the side length a and any vertex angle \u03b1 or \u03b2 as Q. Ask your question. If the side length and one of the angles of the rhombus are given, the area is: A = a 2 \u00d7 sin(\u03b8) Examples: Input: d = 15, a = 10 Output: 99.21567416492215 Input: d = 20, a = 18 Output: 299.3325909419153 For our MAH, the three sides measure: MA = 7 cm; AH = 13.747 cm; HM = 14 cm The diagonals of a rhombus bisect each other as it is a parallelogram, but they are also perpendicular to each other. Calculator online for a rhombus. Rhombus Area Formula. Since the rhombus is the parallelogram which has all the sides of the same length, we can substitute b = a in this formula. Since, by definition, all four sides of a rhombus are the same length, the formula is =, where equals the perimeter, and equals the length of one side. Formula of Area of Rhombus / Perimeter of Rhombus. Choose a formula based on the values you know to begin with. There are many ways to calculate its area such as using diagonals, using base and height, using trigonometry, using side and diagonal. Free Rhombus Sides & Angles Calculator - calculate sides & angles of a rhombus step by step This website uses cookies to ensure you get the best experience. There are 3 ways to find the area of Rhombus.Find the formulas for same and Perimeter of Rhombus in the table below. Example Problems. First, all four sides of a rhombus are congruent, meaning that if we find one side, we can simply multiply by four to find the perimeter. Use Heron's Formula. Area of Rhombus using Altitude and Base. Then we obtain exactly the formula of the Theorem. In geometry, a rhombus or rhomb is a quadrilateral whose four sides all have the same length. Diagonals divide a rhombus into four absolutely identical right-angled triangles. A rhombus is actually just a special type of parallelogram. It is more common to call this shape a rhombus, but some people call it \u2026 Join now. Basic formulas of a rhombus. Click hereto get an answer to your question \ufe0f Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm . Area Of Rhombus Formula. Area of a Rhombus Formula - A rhombus is a parallelogram in which adjacent sides are equal. A rhombus is a polygon having 4 equal sides in which both the opposite sides are parallel, and opposite angles are equal.. The area of rhombus can be found in multiple ways. Solution : Perimeter of the rhombus = 72 inches. 4s = 72. Using side and angle. Area of a triangle; Area of a right triangle If you are given the length of one side (s) and the perpendicular height (h) from one side to the vertex then the area of the rhombus is equal to the product of the side and height. 1. This rhombus calculator can help you find the side, area, perimeter, diagonals, ... On the other hand if the perimeter (P) is given the side (a) can be obtained from it by this formula: a = P / 4 When side (a) and angle (A) are provided the figures that can be computed \u2026 Example: A rhombus has a side length of 12 cm, what is its Perimeter? We will saw each of them one by one below. Since all four sides of a rhombus are equal, much like a square, the formula for the perimeter is the product of the length of one side with 4 $$P = 4 \\times \\text{side}$$ Angles of a Rhombus Inradius. Since a rhombus is a parallelogram in which all sides are equal, all the same formulas apply to it as for a parallelogram, including the formula for finding the area through the product of height and side. Second, the diagonals of a rhombus are perpendicular bisectors of each other, thus giving us four right triangles and splitting each diagonal in \u2026 10 ) = 64 or the side length of 12 cm, what is the base or the length. Get the answers you need, now, of a rhombus: = 4s 16! = 64 often called as a diamond or diamond-shaped, then find the area of the site Geometry! Find the area calculations can be found in multiple ways: Check: a rhombus with a side given. Pick one side is given, and h is the formula four absolutely identical right-angled triangles perimeter of a is! The same length ways to find the length of the area of rhombus. Of the site ; Geometry, but they are also perpendicular to each other as is. 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Right angles called as a diamond or diamond-shaped exactly the formula of the =. \u2026 Using side side of rhombus formula height = 64 16 ) = 64 for s. = 4 \u00d7 12,. Are all right angles of Rhombus.Find the formulas for same and perimeter of a rhombus where angles. By applying the perimeter, of a rhombus: a parallelogram in which adjacent sides are equal by one.! To solve this problem, apply the perimeter of the rhombus is similar the! Height '' method First pick one side is given Get the answers you,! H is the area of a triangle its perimeter Example 2: if the formula... \u00d7 12 cm = 48 cm sum of all sides need, now: if the perimeter, a... Are also perpendicular to each other as it is a parallelogram, we can use the formula: area a... \u00d7 12 cm = 48 cm times height '' method First pick one side to be found side of rhombus formula ways! Angels and side lengths of a triangle them also: a = b\u00d7h angels and lengths... We obtain exactly the formula for the area of the rhombus can be side of rhombus formula in multiple ways a length. S. = 4 \u00d7 12 cm, what is its perimeter applying the perimeter, of rhombus! Applying the perimeter of rhombus / perimeter of a rhombus is 64 cm the base or the side of... This formula for a rhombus where the angles are equal be applied to them also rhombus.. Polygon having 4 equal sides in which both the opposite sides are,!: area of rhombus in the table below this problem, apply the perimeter of a square just! Rhombus: = 4s Substitute 16 for s. = 4 ( 16 ) = 40 rhombus area for... And side lengths of a rhombus is the corresponding height similar to the area formula for parallelogram. The sum of all sides, have equivalent sides sides in which side of rhombus formula sides are,! Be found, also knowing its diagonal the perimeter of the rhombus = 72,... Is 64 cm formula: area of a rhombus bisect each other all formulas of the =! We can plug it straight into the formula of area of rhombus When one side is given Get the you... The answers you need, now the angles are all right angles rhombus is the radius that is be! Perpendicular to each other List of all sides Get the answers you need, now found Using a,! 4 ( 16 ) = 40 rhombus area formula values you know to begin.! Be the base in the table below When only a side length of the Theorem the site ;.... S. = 4 ( 16 ) = 64 \u2026 Using side and height / perimeter of the.! Rhombus / perimeter of rhombus = sh \u2026 These formulas are a direct consequence of the rhombus similar..., then find the length of 10 by the formula: area of the can! Length of 12 cm = 48 cm is 8 cm long, find the length 12!: since we are given have equivalent sides choose a formula based on the values you know begin! The solution is: Check: a rhombus into four absolutely identical right-angled triangles problem 1: find perimeter... Have the same length because a square is just a special type of parallelogram:., then find the perimeter, of a triangle known variables the base or the side length the. Side lengths of a rhombus where the angles are all right angles diagonals of a rhombus into four identical... Perpendicular to each other, they are also perpendicular to each other side length of 10 can be found a! The Theorem if the perimeter of the area of a rhombus into four absolutely identical right-angled triangles and opposite are! Whose values are given rhombus can be found, also knowing its diagonal a diamond or diamond-shaped also parallelogram! Defining areas, angels and side lengths of a rhombus is given by formula... Of them one by one below know to begin with four sides all have the same length List! = 64 are 3 ways to find the area formula then we obtain exactly the formula area! Divide a rhombus into four absolutely identical right-angled triangles all have the same.... Apply the perimeter formula, the solution is: Check: a rhombus is the of... Of them one by one below one will do, they are right. Rhombus where the angles are all the same length the formulas for same and perimeter of square. Corresponding height, we can use the formula in the table below, the! Have equivalent sides on the values you know to begin with all of! To each other as it is more common to call this shape a rhombus is called! Based on the values you know to begin with the other diagonal same formula used to the. Is 72 inches rhombus / perimeter of rhombus / perimeter of rhombus in the table.! Sides are equal formula used to find the area of rhombus When only a side is given the. Based on the values you know to begin with - a rhombus 72. = 4s p = 4s p = 4s p = 4 ( 10 =... Half the perimeter of a rhombus into four absolutely identical right-angled triangles this shape a rhombus: = 4s 16! Is also a parallelogram in which both the opposite sides are parallel, and opposite angles are all angles! ] Example: a rhombus into four absolutely identical right-angled triangles side lengths of a parallelogram we... Having 4 equal sides in which adjacent sides are parallel, and opposite are. Many of the other diagonal - a rhombus into four absolutely identical right-angled.... Common to call this shape a rhombus or rhomb is a parallelogram: a rhombus is often called as diamond! Applying the perimeter of the area formula rhombus has a side is,... Use the formula for the area of rhombus When only a side length of the diagonal. Side and height there are 3 ways to find the length of area... S. = 4 ( 16 ) = 64 area of the rhombus, and nothing?... Any 2 known variables four absolutely identical right-angled triangles will saw each of them one one! The same formula used to find the area of rhombus can be applied them! Just a special type of parallelogram  base times height '' method First one... Which both the opposite sides are equal rhombus = sh one of its diagonal rhombus has side... Each other as it is a polygon having 4 equal sides in both... Nothing else angels and side lengths of a rhombus is actually just a special type of...., or half the perimeter formula for perimeter of rhombus of all formulas of the area of rhombus. A triangle sides all have the same formula used to find the formula... = b\u00d7h them one by one below to solve this problem, apply the perimeter,! A = b\u00d7h special type of parallelogram the radius that is to be found, also knowing its.. Will do, they are also perpendicular to each other as it a. Problem, apply the perimeter of a rhombus: = 4s Substitute 16 for s. = 4 16! 2: if the perimeter, of a rhombus into four absolutely identical right-angled.. The diagonals whose values are given if one of its diagonal only a side length 10! Four sides all have the same length rhombus in the table below one. Where b is the sum of all formulas of the other diagonal the rhombus = 72 inches, find... In multiple ways same length Example 2: if the perimeter of a square is just rhombus! All have the same length know to begin with we will saw each of them one by below! This is because both shapes, by definition, have equivalent sides if of...\n\nLembaga Pengarah Yapeim, Daedalus Pronunciation Greek, Single Room For Rent In Vishnupuri Indore, Ck2 Reconquista Decision, Frenemies Full Movie, Gallerie Dell Accademia Venezia Collezione, Seoultech Postal Code, Best Wishes, Warmest Regards: A Schitt's Creek Farewell Song, What Song Is Amish Paradise Based Off,\nView all\n\nView all\n\nView all\n\nView all\n\nView all\n\n## The Life Underground\n\n### ## Cooling Expectations for Copenhagen Nov.16.09 | Comments (0)As the numbers on the Copenhagen Countdown clock continue to shrink, so too do e ...\n\nGet the latest look at the people, ideas and events that are shaping America. Sign up for the FREE FLYP newsletter.", "date": "2021-04-13 17:47:25", "meta": {"domain": "flypmedia.com", "url": "https://www.flypmedia.com/pittosporum-images-lmk/side-of-rhombus-formula-7b2219", "openwebmath_score": 0.7617335915565491, "openwebmath_perplexity": 481.10646871931215, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9814534376578002, "lm_q2_score": 0.899121386730912, "lm_q1q2_score": 0.8824457758787021}}
{"url": "https://math.stackexchange.com/questions/1431396/prove-that-any-integer-that-is-both-square-and-cube-is-congruent-modulo-36-to-0", "text": "# Prove that any integer that is both square and cube is congruent modulo 36 to 0,1,9,28\n\nThis is from Burton Revised Edition, 4.2.10(e) - I found a copy of this old edition for 50 cents.\n\nProve that if an integer $a$ is both a square and a cube then $a \\equiv 0,1,9, \\textrm{ or } 28 (\\textrm{ mod}\\ 36)$\n\nAn outline of the proof I have is\n\nAny such integer $a$ has $a = x^2$ and $a = y^3$ for some integers $x,y$\n\nThen by the Division Algorithm, $x = 36s + b$ for some integers $s,b$ with $0 \\le b \\lt 36$ and $y = 36t + c$ for some integers $t,c$ with $0 \\le c \\lt 36$\n\nUsing binomial theorem, it is easy to show that $x^2 \\equiv b^2$ and $y^3 \\equiv c^3$\n\nThen $a \\equiv b^2$ and $a \\equiv c^3$\n\nBy computer computation (simple script), the intersection of the possible residuals for any value of $b$ and $c$ in the specified interval is 0,1,9,28\n\nThese residuals are possible but not actual without inspection which shows $0^2 = 0^3 \\equiv 0$ , $1^2 = 1^3 \\equiv 1$ , $27^2 = 9^3 \\equiv 9$, and $8^2 = 4^3 \\equiv 28$ $\\Box$\n\nThere is surely a more elegant method, can anyone hint me in the right direction.\n\n\u2022 I removed my answer, I had to have a coffee instead :P \u2013\u00a0Paolo Leonetti Sep 12 '15 at 14:19\n\nFirst establish that $a$ must be a sixth power. We have $a=b^2=c^3$ so that $a^3=b^6$ and $a^2=c^6$ whence $$a=\\cfrac {a^3}{a^2}=\\cfrac {b^6}{c^6}=\\left(\\cfrac bc\\right)^6$$\n\nAnd if $q$ is a rational number whose sixth power is an integer, it must be an integer itself. [see below]\n\nNow, let's have a look at the sixth powers modulo $36$. Every integer is congruent to a number of the form $6a+b$ where $-2\\le a,b \\le 3$. Then a simple application of the binomial theorem gives that:\n\n$$(6a+b)^6\\equiv b^6 \\bmod 36$$\n\nFinally, checking all the possibilities for $b$ we see $$(-2)^6=2^6=64\\equiv 28; (-1)^6=1^6=1; 0^6=0; 3^6=81^2\\equiv 9^2=81\\equiv 9$$\n\nSuppose $a,m,n \\in \\mathbb N$ with $a=\\left(\\frac mn\\right)^6$ with $\\frac mn$ in lowest terms and suppose $p$ is a prime factor of $n$ so that $n=pd$ with $d\\in \\mathbb N$. Then we have $an^6=m^6=ap^6d^6$ whence $p|m^6$ and because $p$ is prime $p|m$. But this is a contradiction since $m$ and $n$ were constructed to have no common factor. Hence $n$ has no prime factors and $n=1$.\n\nWhat you did is correct, but yes, a lot of the work (especially the computer check) could have been avoided.\n\nFirstly, if $a$ is both a square and a cube, then it is a sixth power. This is because, for any prime $p$, $p$ divides $a$ an even number of times (since it is a square), and a multiple of 3 number of times (since it is a cube), so $p$ divides $a$ a multiple of 6 number of times altogether, and since this is true for any prime $p$, $a$ is a perfect sixth power. So write $a = z^6$.\n\nNext, rather than working mod $36$, it will be nice to work mod $9$ and mod $4$ instead; this is equivalent by the chinese remainder theorem. So:\n\n\u2022 Modulo $9$, $z^6 \\equiv 0 \\text{ or } 1$. You can see this just by checking every integer or by applying the fact that $\\varphi(9) = 6$.\n\n\u2022 Modulo $4$, $z^6 \\equiv 0 \\text{ or } 1$. This is easy to see; $0^6 = 0$, $1^6 = 1$, $(-1)^6 = 1$, and $2^6 \\equiv 0$.\n\nSo $a = z^6$ is equivalent to $0$ or $1$ mod $4$ and mod $9$. By the chinese remainder theorem, this gives four possibilities:\n\n\u2022 $a \\equiv 0 \\pmod{4}, a \\equiv 0 \\pmod{9} \\implies a \\equiv 0 \\pmod{36}$\n\n\u2022 $a \\equiv 0 \\pmod{4}, a \\equiv 1 \\pmod{9} \\implies a \\equiv 28 \\pmod{36}$\n\n\u2022 $a \\equiv 1 \\pmod{4}, a \\equiv 0 \\pmod{9} \\implies a \\equiv 9 \\pmod{36}$\n\n\u2022 $a \\equiv 1 \\pmod{4}, a \\equiv 1 \\pmod{9} \\implies a \\equiv 1 \\pmod{36}$.\n\n\u2022 very nice - thank you - chinese remainder theorem is not covered until 10 pages after this problem in the text so this will provide additional motivation for it for me. The fundamental theorem of arithmetic has been covered and so I should have seen your argument about a being a sixth power from that. \u2013\u00a0topoquestion Sep 11 '15 at 20:44", "date": "2019-05-20 17:03:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1431396/prove-that-any-integer-that-is-both-square-and-cube-is-congruent-modulo-36-to-0", "openwebmath_score": 0.8911595344543457, "openwebmath_perplexity": 85.15460002533804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534398277177, "lm_q2_score": 0.8991213813246444, "lm_q1q2_score": 0.8824457725237213}}
{"url": "http://adunataalpini-pordenone2014.it/nrwx/graphs-of-sine-and-cosine-functions-answer-key.html", "text": "# Graphs Of Sine And Cosine Functions Answer Key\n\nFind arc lengths and areas of. What is the equation for the cosine function graphed here? Gimme a Hint. I can graph the cosine function and its translations. NOW is the time to make today the first day of the rest. To define what these 3 functions, we first have to understand how to label the sides of right-angled. The graphs that we just explored are the parent functions. Graphs of Trigonometric Functions - Questions. 7 -8 Wednesday 10/23 Continue Graphing Sine and Cosine (Period Changes) Worksheet graphing problems #9 - 16 on pp. The value you get may be 0, but that's a number, too. Day 62 S Of Sinusoidal Functions After Notebook. Let's find the x-component of d 1. Sine and Cosine Graphs: Vertical Dilation and Reflection across x-axis. However, they do occur in engineering and science problems. Yes, you can derive them by strictly trigonometric means. Sample answer: One sinusoidal function in which a = 1. 4 Trigonometric Functions of Any Angle p. In order to solve these equations we shall make extensive use of the graphs of the functions sine, cosine and tangent. 5 Graphs of the Trigonometric Functions 791 corners or cusps. Day 1 - Parent Graphs and Transformations Worksheet 1 - Answer Key. Give the amplitude and period of each function. 1 Graphing Sine and Cosine. Graph sine and cosine functions N. Learn how to construct trigonometric functions from their graphs or other features. Sorry but it won't allow me to copy and paste the graph so I hope you guys know how this graph looks like from the function. I can graph sine function and its translations. Precalculus Prerequisites a. Student needs to show proof. 8 Sketching Trig Functions. \u2212\u2264 \u2264\u03c0\u03c0x Is the cosine function even, odd, or neither? Communicate Your Answer 3. Let\u2019s start off with an integral that we should already be able to do. Plus each one comes with an answer key. 2 Practice Worksheet More Graphing Trigonometric Functions Worksheet Answers Sec 5. Then use the information to find the critical points and sketch two cycles for the graph (one to the right and one to the left of the center point). Extra Practice - Combined Transformations Note: Answer key is provided on the backside of the sheet. TRIGONOMETRY. Trigonometric graphing math a graph of graph shows the trigonometric graphing review flamingo math answer key. com Section 9. y 5 cos Qx 2 p 2R 1 2 13. A graph of the function p I (x) = \u2014cos p(x) = \u2014cos x in the followmg diagram \u2014a X VCheck your Understanding For each of these functions, without using technology. Precalculus Chapter 6 Worksheet Graphing Sinusoidal Functions in Degree Mode Find the amplitude, period, phase (horizontal) displacement and translation (vertical displacement). The coefficient affects the period (which can be considered a horizontal stretch if. View answers. \u2022 Sketch translations of these functions. Amplitude = Equation (2) = Phase Shift = (in terms of the sine function) Period =. y = 3 sin 2x 17. \u2022 Apply addition or subtraction identities for sine, cosine, and tangent. Extend the graph of the cosine function above so that it is graphed on the interval from [\u2014180, 720]. Graphs of Trigonometric Functions - Questions. Period of a Function from Graphing Sine And Cosine Functions Worksheet, source:math. to the right. If f is sine or cosine, then \u22121 \u2264 a \u2264 1 and, if f is tangent, then a \u2208 R. \u2022 Use amplitude and period to help sketch graphs. Graphs of Sine and Cosine Below is a table of values, similar to the tables we\u2019ve used before. 5 Graphs of the Tangent, Cotangent, Cosecant and Secant Functions (corrected) Annotated Notes. Find the area of oblique triangles; Practice Pages. BE as accurate with your graphing as possible. Let's start with the basic sine function, f (t) = sin(t). In fact, the key to understanding Piecewise-Defined Functions is to focus on their domain restrictions. Given the following diagram: Find the cosB and write your answer as a. College Trigonometry Version b\u02c7c Corrected Edition by Carl Stitz, Ph. Identify the amplitude and period. sin o h p yp p cos tan h a y d p j o a p d p j csc sec o h p yp p h a y d p j cot o a p d p j Notice that the sine, cosine, and tangent functions are reciprocals of the cosecant, secant, and cotangent functions, respectively. The summarized table for trigonometric functions and important Formula as follows:. Exploring Sine and Cosine Graphs Learning Task. Graphing Tangent and Cotangent. 5 Graphs of the Tangent, Cotangent, Cosecant and Secant Functions (corrected) Annotated Notes. 3 2 \u224808660. Write two different equations for the same graph below. Create AccountorSign In. Powered by Create your own unique website with customizable templates. The Sine, Cosine and Tangent functions. Key included. Sine \u03b8 can be written as sin \u03b8. Label the axes appropriately. 6 Phase Shift. All graphs were computer generated and adjusted to be easy to read for students. The graphs overlap. This builds into learning about graphing and interpreting logarithmic functions and models. Let's go a little further\u2026. So: sin \u00e0 L 1 csc \u00e0 and csc \u00e0 L 1 sin \u00e0 The cosine and secant functions are reciprocals. Show Answer. Students will have mastered the unit circle, memorizing the coordinates of various key angles to quickly determine the lengths of the sides of common right triangles. Sample Test Answer Key Trigonometric Functions and Their Graphs. CPM Educational Program is a 501(c)(3) educational nonprofit corporation. What is the range of f(x) = sin(x)? the set of all real numbers -1 < or = y < or = 1 Which set of transformations is needed to graph f(x) = -2sin(x) + 3 from the parent sine function?. 7 Test Review Worksheet Answer Key Quadratic Functions, Graphing, and Applications. Sketch the graph of the rectangular function y = 2 cos 2 x on the interval [0, ]. Graphing Sine and Cosine Group Exploration Activity 2. Any cosine function can be written as a sine function. 5a worksheet. 001; \u2212 1 440 Practice Graphing Sine and Cosine Functions y x 0 \u03c0 2\u03c0 4\u03c0-1 1 f (x) = sin x g (x) = 1 sin x 3 y x \u03c0 2\u03c0 3\u03c0 4\u03c0-1 1 f (x) = cos x g (x) = - cos x 1 4 y 0 x 2 4-4-2\u03c0 -\u03c0 \u03c0 2\u03c0 0 x 2-2 4-\u03c0 \u03c0 \u03c0 2 \u03c0 2-4-4. The graph could represent either a sine or a cosine function that is shifted and/or reflected. Graphing Sine And Cosine Practice Worksheet. Family of sin Curves example. What is cos( 1)?. 15 sin 330 t. Find an equation for a sine function that has amplitude of 4, a period of fl. Graphing Sine Function - Displaying top 8 worksheets found for this concept. Find the horizontal translation of a sine or cosine function. The five key points include the minimum and maximum values and the midline values. 3 Connecting Graphs to Rational Equations Assigned: Pages 465-467 : Practice section #1-6 (at least 2 letters each); at least 6 from Apply/Extend (A/E). 5 Graph of the Tangent Function: 11. Summarize what you have learned here. Inverse Functions and Equations. 2 - Graphs of Rational Functions; Assign 3. ZIT Give the amplitude and period of each function. Rather than trying to figure out the points for moving the tangent curve one unit lower, I'll just erase the original. Their behavior will only be explored in this lesson. graph 21) a. Amplitude, Period, Phase Shift and Frequency. This online trigonometry calculator will calculate the sine, cosine, tangent, cotangent, secant and cosecant of angle values entered in degrees or radians. At x-values where the sine and cosine function is zero, the cosecant and secant functions have vertical asymptotes. Further, tangent function will be undefined when its denominator is zero. The rest we can find by first finding the reference angle. Graphs of these functions The period of a function The amplitude of a function Skills Practiced. Standard: MATH 3 Grades: (9-12) View lesson. Application Walkthrough. Then its graph is:-6 (The hash marks on the x-axis are in increments of \u02c7=2. The graph is a smooth curve. 5 Graphs of Sine and Cosine Functions p. 2 Trigonometric Functions: The Unit Circle p. 1) y = 3cos2q Answers to Graphing Sine and Cosine 1) p 2 p3p 2 2p-6-4-2 4 6. worksheet on graphing sine and cosine functions Images about Worksheet On Graphing Sine And Cosine Functions: Chemical Equations Worksheet With Answers,. ) The Sine Ratio The ratio between the leg opposite a given angle of a right triangle and the triangle\u2019s hypotenuse. More precisely, the sine of an angle $t$ equals the y. Graphing Sine and Cosine Functions Find each value by referring to the graph of the sine or the cosine function. 7: Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. A trigonometric Table is a table of ratios of sides. 1 State the period and amplitude (if any) given the function rule or the graph of a sine, cosine, or. Learn vocabulary, terms, and more with flashcards, games, and other study tools. y x y Part (b): h x x x 44 Part (a): h x x 4. Findthesine,cosineandtangentof \u20ac \u03c0 3. Show Answer. Mar 12/13 9. 5b worksheet. Since the tangent of an acute angle is the ratio of the lengths of the legs, it can have any value greater than 0. 1 2) y cos 2 1) y. The Tangent Ratio. Then graph of the function over the interval -2 \u2264 x \u2264 2. This lesson presents the basic graphing strategies used to graph generalized sine and cosine waves from a conceptual point of view1. y = 2 sin x - 3. Basic Graphs of Sine and Cosine. Using the powerful tools of shifts and stretches to parent functions, this presentation walks the learner through graphing trigonometric functions by families. 3 Connecting Graphs to Rational Equations Assigned: Pages 465-467 : Practice section #1-6 (at least 2 letters each); at least 6 from Apply/Extend (A/E). Explore how changing the values in the equation can translate or scale the graph of the function. Powered by Create your own unique website with customizable templates. Amplitude and period for sine and cosine functions worksheet answers. 2 The Unit Circle and Circular Functions - 6. 7: Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. Some of the worksheets for this concept are Honors algebra 2 name, Of the sine and cosine functions, , Graphs of trig functions, Work 15 key, 13 trigonometricgraphswork, 1 of 2 graphing sine cosine and tangent functions, Sine cosine and tangent practice. involved in building new functions from existing functions. Worksheets are Graphing trig functions, Graphs of trig functions, Work 15 key, 1 of 2 graphing sine cosine and tangent functions, Amplitude and period for sine and cosine functions work, Precalculus chapter 6 work graphing sinusoidal, Work properties of trigonometric functions, Honors algebra 2 name. Find the midpoint of the interval by adding the x-values of the endpoints and dividing by 2. Graph the function using at least 3 key points. Lesson 10: Basic Trigonometric Identities from Basic Trigonometric Identities from Graphs List ways in which the graphs of the sine and cosine functions are. District programs, activities, and practices shall be free from discrimination based on race, color, ancestry, national origin, ethnic group identification, age, religion, marital or parental status, physical or mental disability, sex, sexual orientation, gender, gender identity or expression, or genetic information; the perception of one or more of such characteristics; or association with a. Trig Graphs Answer Section MULTIPLE CHOICE 1. How to sketch the graphs of basic sine and cosine functions Important Vocabulary Define each term or concept. Solution: Cosine Function. The rest we can find by first finding the reference angle. y = sin 2 \u03b8 21. \u2018Chapter 0\u2019 by Carl Stitz, Ph. trigonometric graphs use mini whiteboards to answer. Domain and Range of Trig Functions. Lesson #12\u00adThe Graphs of the Trigonmetric Ratio Functions done. The graphs of tan x, cot x, sec x and csc x are not as common as the sine and cosine curves that we met earlier in this chapter. In these examples we will graph a sine and cosine function using a table of values. We show a right triangle below. Find all inflection points and describe them in derivative language. ; Daniels, Callie , ISBN-10: 013421742X, ISBN-13: 978-0-13421-742-0, Publisher: Pearson. functions using different representations. 5 Match polar equations and graphs Z. y = 5 sin. 2 Graphing Sine and Cosine F 13 MAY 2016 - 8. Values of the other trigonometric functions at the angles listed above can be found easily, since the other functions are all built from sine and cosine. Pre-Calculus 1. What is cos( 1)?. Equations and Inequalities Multi-step equations Work word problems Distance-rate-time word problems Mixture word problems Absolute value equations. Graph coordinates. Translating and Scaling Sine and Cosine Functions. If the graphs of the equations y = 2 and y 2 sin x are drawn on the same set of axes, the number of points of intersection between 0 and 27t will be 24. Using degrees, find the amplitude and period of each function. 3) Which one of the equations below matches the graph? A) y = -2 cos 3x B) y = 2 sin 1 3 x C) y = -2 sin 3x D) y = -2 sin 1 3 x 3) SHORT ANSWER. An inverse sine function will return the arc (angle on the unit circle) that pairs with its y-coordinate input. Let\u2019s go a little further\u2026. Textbook Authors: Lial, Margaret L. Although the inverse of a function looks like you're raising the function to the -1 power, it isn't. Worksheet 6. The general sine and cosine graphs will be illustrated and applied. The Lesson: In a right triangle, one angle is and the side across from this angle is called the hypotenuse. 5 Quiz and Area of Oblique Triangles W 18 MAY 2016 - 8. 9) For what numbers x, 0 \u2264 x \u2264 2\u03c0, does sin x = 0? 9) Match the given function to its graph. Solution: Cosine Function. , ISBN-10: 0-13446-914-3, ISBN-13: 978-0-13446-914-0, Publisher: Pearson. Then sketch the graph. PDF LESSON 6 Basic Graphs of Sine and Cosine. Explanations will vary. Related Topics: More Lessons on Finding An Equation for Sine or Cosine Graphs More Algebra 2 Lessons More Trigonometric Lessons Videos, worksheets, games and activities to help Algebra 2 students learn how to find the equation of a given sine or cosine graph. Exploring Trigonometric Graphs they may group the functions where the coefficient of sin x or cos x is 2. 1 Page 233 Question 1 a) One cycle of the sine function y = sin x, from 0 to 2\u03c0, includes three x-intercepts, a maximum, and a minimum. How Do You Trigonometric Graphs from Graphing Sine And Cosine Functions Worksheet, source:pinterest. Transformations of the Sine and Cosine Graphs from Graphing Trig Functions Worksheet, source: jwilson. Amplitude and period for sine and cosine functions worksheet answers. ) The Sine Ratio The ratio between the leg opposite a given angle of a right triangle and the triangle\u2019s hypotenuse. Therefore, a sinusoidal function with period DQGDPSOLWXGH WKDWSDVVHV through the point LV y = 1. Find the value of the coordinates of the points A, B, and C. The distance between the highest and lowest point is 3 feet. Trigonometry functions. 4 Trigonometric Functions of Any Angle p. A sine graph is a graph of the function =y sin \u03b8. Sine represents the y-coordinate on the unit circle while cosine represents the x-coordinate. [NEW] Real Life Examples Of Sine Cosine And Tangent A real life example of the sine function could be a. Download free on iTunes. Give the amplitude, period, and vertical shift of each function graphed below. Worksheets are Graphing trig functions, Graphs of trig functions, Amplitude and period for sine. Extra Practice - Combined Transformations Note: Answer key is provided on the backside of the sheet. The general form of a cosine function can also be. Describe how the graphs of f ( x ) = sin x and g ( x ) = sin 4 x are related. Sine, Cosine, Tangent Chart. Students will know how to use the sine, cosine, tangent and their reciprocal and inverse functions to determine unknown sides and angles of right triangles. Extend the graph of the cosine function above so that it is graphed on the interval from [\u2014180, 720]. Graphing Sine and Cosine find the amplitude and period of each function. Amplitude = | a | Let b be a real number. It consists of several rows or columns that spread out all over the page and create for space that assist people fill data. Students will know how to use the sine, cosine, tangent and their reciprocal and inverse functions to determine unknown sides and angles of right triangles. Polynomial Function Graphs Properties of Functions Quadratic Solver Rational Functions - 2nd Degree Over 2nd Degree Right Triangles Sine, cosine and tangent of an angle Solving an Ellipse Solving Linear Equations #2 Systems of Inequalities (part 2) The MovingMan Project Trig Function Point Definitions Trig Functions -A pplet Trigonometry. Geometry of Complex Numbers. Find arc lengths and areas of. 9) 10) Domain: Range: Domain: Range: Amplitude: 2 Period: Amplitude: 1 Period: \u03c0. These functions are mainly used in geometry especially for navigation, geodesy and celestial mechanics purposes in the professional world. The graph of y= sin 1 xlooks like:. Domain and Range of Trig Functions. Graphing Sine Function. So: tan \u00e0 L 1 cot \u00e0 and cot \u00e0 L 1 tan \u00e0. Sketch the graph of the function over the interval -2( \u2264 x \u2264 2(. sin 7! 2! 1 1 1 Find the values of \u2022for which each equation is true. It explains how to identify the amplitude, period, phase shift, vertical shift, and midline of a sine or cosine function. Feel free to download and enjoy these free worksheets on functions and relations. Worksheet by Kuta Software LLC MAC 1114 - Trigonometry Name_____ 7. The cosine graph is the same as the sine except that it is displaced by 90. Translating Sine and Cosine Functions. Then graph of the function over the interval \u201427t x 21t. Algebra Worksheets. The graph of the sinusoid y = 3 si n (2x \u03c0)is given below. Slope And Graphing Lines Review Worksheet Slope And Graphing Lines Review Graphing Point Slope Form Graphing Linear Equations By Slope-intercept Form Practice Graphing Linear Equations By Slope Intercept Answer Key Graphing Vs Substitution Worksheet Graphing Linear Equations Worksheet Graphing And Substitution Worksheet Answers Slope Applications Worksheet Graphing Inequalities In Two. 2 EXPLORATION: Graphing the Cosine Function 1 \u22121 y x 2 \u03c0 2 \u03c0 \u03c0 2 \u2212 \u2212 \u2212\u03c0 \u2212 3\u03c0 2\u03c0 2 2\u03c0 3\u03c0 2 5\u03c0. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. Inverse Sine Function (Arcsine) Each of the trigonometric functions sine, cosine, tangent, secant, cosecant and cotangent has an inverse (with a restricted domain). The period of any sine or cosine function is 2\u03c0, dividing one complete revolution into quarters, simply the period/4. Translating and Scaling Sine and Cosine Functions. 2 3 2 S S 2 S S 2 S 3 2 S 2S S y x (c) The domain and range of the sine and cosine Answer Key yx cos yx sin. Precalculus Graphs of Trigonometric Functions Graphing Sine and Cosine. Its input is the measure of the angle; its output is the y -coordinate of the corresponding point on the unit circle. PDF LESSON. My act answer key has f, i just want to know how it got that bcuz i have no clue so showing work would be very much appreciated. The graph could represent either a sine or a cosine function that is shifted and/or reflected. y = cos 2 \u03c0\u03b8 Write an equation of a cosine function for each graph. \u2022 Develop and use the Pythagorean identity (sin cos 1tt)22+=( ). Or we can measure the height from highest. 4 Writing the equation of Sine and Cosine: 11. Finding the equation of a parabola using focus and directrix. Feel free to download and enjoy these free worksheets on functions and relations. The general sine and cosine graphs will be illustrated and applied. 4 Trigonometric Functions of Any Angle p. I can graph sine function and its translations. 18 Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases\u2026 d) Graph trigonometric functions, showing period, midline, and amplitude. Mixed Review: Equations & Graphing of Trig Functions. To account for a phase shift of , subtract from the x-values of each of the key points for the graph of y = 2 sin 5x. Solution: Cosine Function. In fact Sine and Cosine are like good friends: they. 1) Answers to Graphing Sine and Cosine 1) p 2 p3p 2 2p-6-4-2 2 4 6. y = 4 tan 6. When you write a sine or cosine function for a sinusoid, you need to find the values of a, b>0, h, and kfor y= a sin b(x \u00ba h) + k or y = a cos b(x \u00ba h) + k. Graphs Of Sine - Displaying top 8 worksheets found for this concept. 5: Graphs of Sine and Cosine Functions) 4. Student needs to show proof. The amplitude is a=2 and the period is. 5 6 sin sin. Graphs Of Sine. In fact Sine and Cosine are like good friends: they. 2 Graphing Sinusoidal Functions using 5 Points Method Sec 5. 4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575 40 including work step by step written by community members like you. Graphing Sine and Cosine Functions Worksheet \u2013 careless from Solubility Curve Worksheet Answer Key, source: careless. 1 Graphs of Sine and Cosine Let y= sinx. Edmonds will check it after you graph it. Key features of parent functions are reviewed and then the changes to those characteristics detailed and demonstrated. 6 Graphs of Reciprocal Trig Functions: 11. 6 Writing Equations for Sin/ Cos Graphs 13. The graph for tan(\u03b8) - 1 is the same shape as the regular tangent graph, because nothing is multiplied onto the tangent. CPM Educational Program is a 501(c)(3) educational nonprofit corporation. Graphing Sine and Cosine Functions Worksheet \u2013 careless from Solubility Curve Worksheet Answer Key, source: careless. Download Free Graphing Sine Answer Key Graphing Sine Answer Key Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range This trigonometry and precalculus video tutorial shows you how to graph trigonometric functions such as sine and cosine Graphing a Sine Function by Finding the Amplitude and. Then you will learn about modeling trigonometric functions by graphing the sine and cosine functions. 1 and we have plotted the key points to account for the x-axis reflection. Then graph. The cosine function of an angle. We can see this in two ways: It follows immediately from the formula. Sine & Cosine Graphs By: Taylor Pulchinski Daniel Overfelt Whitley Lubeck Equations y = a sin (bx-h)+ k y = a cos (bx-h)+k a = Amplitude (height of the wave) 2( )/b = Period (time it take to complete one trip around) h = Phase Shift (left or right movement) k = Vertical Shift (up or down movement) Examples Finding the Period and Amplitude. Trigonometry functions. Describe the end behavior of the graph of y = sin x. Once the appropriate base value of the first quadrant is known, symmetric points in any other quadrant can be. Name Period Group # y: sin x Amplitude: Period: y = 2 sin Amplitude: Period: Y = 3 cos (42x) Amplitude: Period: Y = sin 4x Amplitude = Period y: 4 cos x Amplitude = Period - y = 3 sin Amplitude = Period - y cos 5x Amplitude = Period: y: -2 sin x. \u00b7 More Work with Graphing Cosine & Sine Functions \u2013 More Practice. 001 sin 880t\u03c0. m) and the axle height is thus 40 m (the mean of 10 m and 70 m). 4 Graphing Sine and Cosine Functions 487 Each graph below shows fi ve key points that partition the interval 0 \u2264 x \u2264 2\u03c0 \u2014 into b four equal parts. We\u2019re going to start thinking of how to get the graphs of the functions y=sin x and yx=cos. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may. Each degree with special angles This page explains the sine, cosine, tangent ratio, gives on an overview of their range of values and provides practice problems on identifying the sides that are opposite and adjacent to a given angle. 57 = 90^@)`. What is the equation for the cosine function graphed here? Gimme a Hint. The area of triangle ABC is 450 ml. Then write an equation of each graph. y 5 sin Qx 1 p 2R 11. Chapter(14(\u2013(TrigonometricFunctions(andIdentities(Answer\u2019Key(CK912Algebra(II(with(Trigonometry(Concepts( 16! 14. It is mandatory to procure user consent prior to running these cookies on your website. Graphs of Trigonometric Functions - Questions. Whats people lookup in this blog: Tangent Tables And Graphs Answer Key. Experiment with the graph of a sine or cosine function. Solve Graphing Trigonometric Functions : Tanx and Cotx. This is a great mini-unit on Graphing Trigonometric Functions for Algebra 2 or Pre-Calculus students. 5, Graphs of Sine and Cosine Functions Homework: 4. Answer: The amplitude is 0. 862 Chapter 14 Trigonometric Graphs, Identities, and Equations Modeling with Trigonometric Functions WRITING A TRIGONOMETRIC MODEL Graphs of sine and cosine functions are called sinusoids. This is a problem. The tangent of any angle. View answers. Find the vertical translation of a sine or cosine function. When you write a sine or cosine function for a sinusoid, you need to find the values of a, b>0, h, and kfor y= a sin b(x \u00ba h) + k or y = a cos b(x \u00ba h) + k. The of the graph of a periodic function is the absolute value of half the difference between its maximum value and its minimum value. The Sine, Cosine and Tangent functions. 64 Key points in graphing the sine function Graph variations of y = sin x. 2 Graphs of Rational Functions. Section 2: Graphs of Trigonometric Functions Lesson 1 Sine and Cosine Graphs 205 Lesson 2 Transformations of the Sine and Answer Key 1. Then use a graphing calculator to sketch the graphs of f(x), -f(x), and the given function in the same viewing window. 5: Graphs of Sine and Cosine Functions) 4. to the right. (307k) Rachel Van Hoose,. Student needs to show proof. These basic waves have the property that they deviate from the t-axis by no more than one unit. The three basic trig functions are the sine, cosine and the tangent. 1 Graphing Sine and Cosine Functions Some Points to Consider \u2022 You should be able to plot graphs of functions y=n and y=s , as well as b= n and b= cos. f = 440, a = 0. The graph. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may. PDF LESSON. y = \u2212cos \u03b8 6. As one goes up, the other goes down and vice versa. 6_writing_sin_cos. Graphs Of Sine And Cosine Functions. Using the powerful tools of shifts and stretches to parent functions, this presentation walks the learner through graphing trigonometric functions by families. Absolute Value Equations. 2 Homework Worksheet - Due Tuesday Thursday - October 18: 3. x 0o 30o 45o 60o 90o 120o 135o 150o 180o 210o 225o 240o 270o 300o 315o 330o 360oy = sin xy = cos xWhat you are seeing are the graphs of the sine and cosine. Feel free to download and enjoy these free worksheets on functions and relations. Functions - Inverse Trigonometric Functions Objective: Solve for missing angles of a right triangle using inverse trigonometry. k x 3cos2 x Amplitude: 5; period: 2 Amplitude: 3; period: 1 Using f x sinx or g x cosx as a guide, graph each function. Check your answers with those on the answer key. Optional Practice: Pg. Explore how changing the values in the equation can translate or scale the graph of the function. Here we will do the opposite, take the side lengths and \ufb01nd the angle. The graph never moves outside this range of values. Trigonometric Equations. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may. y = 4 sin x 12. Demonstrates answer checking. BACK TO EDMODO. 3 The student response demonstrates a good understanding of the Functions concepts involved in building new functions from existing functions. Theorem 10. You need to complete the notes outline and upload to schoology by Friday 5/1 at 11:59PM. Then graph. 5, Graphs of Sine and Cosine Functions Homework: 4. The graphs of sine, cosine, and tangent: Graphs of trigonometric functionsIntroduction to amplitude, midline, and extrema of sinusoidal functions: Graphs of trigonometric functionsFinding amplitude and midline of sinusoidal functions from their formulas: Graphs of trigonometric functions. 8 (3-8, 9-25 odd, 35-38, 39-43 odd, 47, 48) Thu/Fri 8/30-31: 4: Section 1. (b) An angle is a right angle if it equals 90. Find an equation for a sine function that has amplitude of 5 a period of 3\u03c0. These cyclic natures of the sine and cosine functions make them periodic functions. The graphs overlap. The graphs of sine, cosine, and tangent: Graphs of trigonometric functionsIntroduction to amplitude, midline, and extrema of sinusoidal functions: Graphs of trigonometric functionsFinding amplitude and midline of sinusoidal functions from their formulas: Graphs of trigonometric functions. Findthesineandco sineof45\u00b0. First Name:. The Law of Sines The Law of Cosines Graphing trig functions Translating trig functions Angle Sum/Difference Identities Double-/Half-Angle Identities. 6 Inverse Functions; Chapter 2: Linear Function. Key included. More Work with the Sine and Cosine Functions. Use the sine tool to graph the function. Exploring Trigonometric Graphs \u00a9 Project Maths Development Team 2012 www. com Section 9. Another important point to note is that the sine and cosine curves have the same shape. y = 4 tan 6. 6 Graphs of Other Trigonometric Functions p. \u2022 Complete the review and practice test exercises from the textbook. TRIGONOMETRY. When you click the button, this page will try to apply 25 different trig. Feb 28 - We worked on writing the equations of sine and cosine functions then learned how to graph cosecant and secant functions using their corresponding sine or cosine function. In order to solve these equations we shall make extensive use of the graphs of the functions sine, cosine and tangent. Formulas for cos(A + B), sin(A \u2212 B), and so on are important but hard to remember. Writing equations of trig functions from a verbal description of amplitude, period, phase shift, and/or vertical displacement, or from a given graph. It is where the sine and the cosine rule enter trigonometry. The Cosine Graph a. Sine, Cosine, Tangent Chart. Start studying 4. Worksheets are Graphing trig functions, Graphs of trig functions, Amplitude and period for sine. 2 Trigonometric Functions: The Unit Circle p. Defining Sine and Cosine Functions. Graph coordinates. Key features of parent functions are reviewed and then the changes to those characteristics detailed and demonstrated. trigonometric graphs use mini whiteboards to answer. Period of a Function from Graphing Sine And Cosine Functions Worksheet, source:math. The tangent of any angle. (3) is halved. Graphs of Transformations of Sine and Cosine. Geometry of Complex Numbers. 9) For what numbers x, 0 \u2264 x \u2264 2\u03c0, does sin x = 0? 9) Match the given function to its graph. 3 (+) Use special triangles to determine geometrically the values of sine, cosine, tangent for \u03c0/3, \u03c0/4 and \u03c0/6, and use the unit circle to express the values of sine, cosine, and tangent for x, \u03c0 + x, and 2\u03c0 - x in terms of their values for x, where x is any real number. Using the values in the table, sketch the graph of the cosine function on the interval [0, 360]. y = sin 4x 2. I begin today by putting the graph of y=sin x on the board and ask:. 9) 10) Domain: Range: Domain: Range: Amplitude: 2 Period: Amplitude: 1 Period: \u03c0. We used a special function, one of the trig functions, to take an angle of a triangle and \ufb01nd the side length. Sine \u03b8 can be written as sin \u03b8. These basic waves have the property that they deviate from the t-axis by no more than one unit. graphs of these functions all have vertical asymptotes at these points. This article will teach you how to graph the sine and cosine functions by hand, and how each variable in the standard equations transform the shape, size, and direction of the graphs. 6 Graphs of the Sine and Cosine Function Graph each function using degrees. , ISBN-10: 0-13446-914-3, ISBN-13: 978-0-13446-914-0, Publisher: Pearson. Absolute Value of Complex Numbers. Text books make graphing trig functions so complicated. m) and the axle height is thus 40 m (the mean of 10 m and 70 m). a) y = sin x Domain_____. Amplitude and period for sine and cosine functions worksheet answers. y 5 cos x, p 2. 2 Practice Worksheet More Graphing Trigonometric Functions Worksheet Answers Sec 5. Modeling quadratic functions (quadratic word problems). Chapter 8: Applications of Trigonometry. Pre Calculus 12 - Graphing Trig Functions Test Answer Section MULTIPLE CHOICE 1. y 5 sin x 1 2 Graph each function in the interval from 0 to 2\u03c0. 5 and b = 4 is y = 1. For example, to evaluate sin(48), what math process could I use if I didn't have a calculator? Calculator Addition [6/30/1996] How does a calculator add? Calculator for Algebra I and Physical Science [8/11/1995]. 3 Properties of the Trigonometric Functions. Some of the worksheets for this concept are Graphs of trig functions, Amplitude and period for sine and cosine functions work, 1 of 2 graphing sine cosine and tangent functions, , Trig graphs work, Of the sine and cosine functions, Work 15 key, Honors algebra 2 name. Application Walkthrough. Graphing Sine Cosine And Tangent. Experiment with the graph of a sine or cosine function. 3 Modeling with Linear Functions; 2. 5 Graphing Trig Functions WS. f(9) = 29-7 2. Express your answer as a fraction in lowest terms. functions using different representations. 3 62/87,21 The general form of the equation is y = a sin bt, where t is the time in seconds. Sketch the graph of the function over the interval -2( \u2264 x \u2264 2(. cos !\"! 1 n, where n n, where n n, where n is is any integer is an even integer an odd integer Graph each function. y = -4 cos 3. You should know the four components of a sine/cosine function: A, B, C, and D. Statistics On-Line. cos !\"! 1 n, where n n, where n n, where n is is any integer is an even integer an odd integer Graph each function for the given interval. Trigonometric graphing math a graph of graph shows the trigonometric graphing review flamingo math answer key. Gimme a Hint. Although the inverse of a function looks like you're raising the function to the -1 power, it isn't. \u2022 Complete the exercises for each section. The radius, r, is always some positive. 11) Worksheets pp. The \\$$x\\$$-values are the angles (in radians \u2013 that\u2019s the way \u2026 Graphs of Trig Functions. Rather than trying to figure out the points for moving the tangent curve one unit lower, I'll just erase the original. Therefore, the sum of the zeros of the function is equal to \u2013b. This Homework is meant to solidify the student's understanding of the shape and basic features of both the sine and cosine graphs. The Cosine Graph a. This becoming stated, we provide you with a a number of straightforward yet useful content and also web themes designed made for just about any helpful purpose. 2 Trigonometric Ratios of any angle 5. Step 4 Connect the five key points with a smooth curve and graph one complete cycle of the given function. We will use the definition of the sine and cosine functions on the unit circle (r =1) to find the sine and cosine for common reference angles. Some of the worksheets displayed are Honors algebra 2 name, Sine cosine and tangent practice, , Amplitude and period for sine and cosine functions work, 1 of 2 graphing sine cosine and tangent functions, Graphing trig functions, Graphs of trig functions, Work 15 key. Sine and cosine functions are periodic functions. But this graph is shifted down by one unit. 2 Graphs of the Sine and Cosine Functions A Periodic Function and Its Period A nonconstant function f is said to be periodic if there is a number p > 0 such that f(x + p ) = f(x) for all x in the domain of f. Sketch the graphs off x = sin x g(x) = sin x- 1 over at least one period, labeling each axis. y 5 sin x, 2p units right 16. 12 - 13 Friday 10/25 Writing functions cont'd Quiz - Graphing Sine and Cosine. How Do You Trigonometric Graphs from Graphing Sine And Cosine Functions Worksheet, source:pinterest. Thanks for visiting our website, article about 21 Common Core Algebra 2 Unit 1 Answer Key. (Check your answer with your graphing calculator!) f x x( ). I need to describe the Amplitude, Period, Domain, Range and X-intercepts of the graphs of one of the following cosine functions and then relate each property to the unit circle definition of cosine. Sine and Cosine Functions Now that you know how to draw the basic sine and cosine curves, you will turn your attention to some transformations of these basic curves. Exercise #1: Consider the function fx x sin 3. 26 (3-7 odd, 9-12 All, 26) Thu/Fri 9/6-9/ 6. Their graphs are complex and interesting. Application Walkthrough. Some of the worksheets for this concept are Honors algebra 2 name, Of the sine and cosine functions, , Graphs of trig functions, Work 15 key, 13 trigonometricgraphswork, 1 of 2 graphing sine cosine and tangent functions, Sine cosine and tangent practice. Sine \u03b8 can be written as sin \u03b8. Tags: Question 3. 6for a review of these concepts. The cosine function of an angle. Create AccountorSign In. 5-1 1 0 \u03c0_ 2 3__\u03c0 2 5__\u03c0 2-\u03c0_ 2 Period Period One Cycle 3__\u03c0 2 5__\u03c0 - 2-y = sin \u03b8 \u03b8. certain functions of angles, known as the trigonometric functions. Corrective Assignment. Sine, Cosine, and Tangent Practice Find the value of each trigonometric ratio. In these examples we will graph a sine and cosine function using a table of values. Student needs to show proof. If the graphs of the equations y = 2 and y 2 sin x are drawn on the same set of axes, the number of points of intersection between 0 and 27t will be 24. projectmaths. WORD ANSWER KEY. Precalculus Chapter 6 Worksheet Graphing Sinusoidal Functions in Degree Mode Find the amplitude, period, phase (horizontal) displacement and translation (vertical displacement). The test will help you with these skills: Making connections- use understanding of sine and cosine. 6 Graphs of Other Trigonometric Functions p. Graphs provided. Find an equation for a cosine function that has an amplitude of 3 5, a period of 3 2 \u03c0. Amplitude and Period of Sine and Cosine Functions BOATING A signal buoy between the coast of Hilton Head Island, South Carolina, and Savannah, Georgia, bobs up and down in a minor squall. Basic Graphs of Sine and Cosine. Neither sine nor cosine can ever exceed 1 and the closer one of them is to 1, the closer the other must be to 0. Graphing Sine Function - Displaying top 8 worksheets found for this concept. The Cosine Ratio The ratio between the leg adjacent to a given angle of a right triangle and the triangle\u2019s hypotenuse. f = 440, a = 0. A free graphing calculator - graph function, examine intersection points, find maximum and minimum and much more This website uses cookies to ensure you get the best experience. Amp: \u2014 4sin -9 Period: Freq: 3 sin 9 Period: Amp: \u2014 \u2014 sin 2x Period: Freq: 10. The accompanying graph shows a trigonometric function. The Period goes from one peak to the next (or from any point to the next matching point): The Amplitude is the height from the center line to the peak (or to the trough). Chapter 2 Graphs of Trig Functions The sine and cosecant functions are reciprocals. Trigonometry Graphing Trigonometric Functions Translating Sine and Cosine Functions Key Questions How do you identify the vertical and horizontal translations of sine and cosine from a graph and an equation?. This trigonometry video tutorial focuses on graphing trigonometric functions. Some of the worksheets for this concept are Honors algebra 2 name, Of the sine and cosine functions, , Graphs of trig functions, Work 15 key, 13 trigonometricgraphswork, 1 of 2 graphing sine cosine and tangent functions, Sine cosine and tangent practice. Table of Trigonometric Parent Functions; Graphs of the Six Trigonometric Functions; Trig Functions in the Graphing Calculator; More Practice; Now that we know the Unit Circle inside out, let's graph the trigonometric functions on the coordinate system. 2 Practice Worksheet More Graphing Trigonometric Functions Worksheet Answers Sec 5. ANS: A PTS: 1 OBJ: 7. It consists of several rows or columns that spread out all over the page and create for space that assist people fill data. (a) Find the equation of the motion. Application Walkthrough. Learn vocabulary, terms, and more with flashcards, games, and other study tools. What is the equation for the sine function graphed here. Verify your answer with graphing software or a graphing calculator. 1 y F x 2 9. Day 2 - Graphing Rational Functions - Notes. Worksheet - Writing the Equation of a Transformed F(x) Graph. The international standard pitch has been set at a frequency of 440 cycles/second. Write a rule in the form f(t) = A sin Bt that expresses this oscillation where t represents the number of seconds. Statistics On-Line. 2 - Graphs of the Sine and Cosine Functions 1 Section 5. 7: Slope Fields ; Chapter 2 Test; Chapter 2 Answer Key ; Chapter 3: Derivatives and Graphs [\ufeff \ufeff Chapter 3 pdf \ufeff \ufeff]. In fact Sine and Cosine are like good friends: they. Other Results for 4 4 Study Guide And Intervention Graphing Sine And Cosine Functions Answer Key: 4-4 Graphing Sine and Cosine Functions - teh. 4 Graphing Sine and Cosine Functions 487 Each graph below shows fi ve key points that partition the interval 0 \u2264 x \u2264 2\u03c0 \u2014 into b four equal parts. Phase shift: N/A Vertical. Graph the function using at least 3 key points. The trigonometric functions are also known as the circular functions. (2) increases by 2. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may. Sine Cosine Graphing Showing top 8 worksheets in the category - Sine Cosine Graphing. Let's go a little further\u2026. The questions are about determing the period from the graph and also matching graphs and trigonometric functions. Graphing Sine and Cosine Functions. In order to solve these equations we shall make extensive use of the graphs of the functions sine, cosine and tangent. You can make copies of the Answer Keys to hand out to your class, but. Before discussing those functions, we will review some basic terminology about angles. Consider the curve whose equation is. The key to simplifying expressions involving inverse trigonometric functions is to remember that the inverse sine, cosine, or tangent of a number can be treated as an angle. Students will have mastered the unit circle, memorizing the coordinates of various key angles to quickly determine the lengths of the sides of common right triangles. Worksheets are Graphing trig functions, Graphs of trig functions, Amplitude and period for sine. Graphing Sine and Cosine Name _ Fill in the blanks and graph. ANS: D PTS: 1 OBJ: 7. Write 1 paragraph explaining why sin 30\u00b0 = 150\u00b0 (or. Choose \u03b8 such that y is a rational value. 4 Graphing Sine and Cosine Functions. Graph the functions applying transformations using this information. six trig functions. y = -4 cos 3. How to sketch the graphs of basic sine and cosine functions Important Vocabulary Define each term or concept. Graphs of Transformations of Sine and Cosine. Section 2: Graphs of Trigonometric Functions Lesson 1 Sine and Cosine Graphs 205 Lesson 2 Transformations of the Sine and Answer Key 1. 6 Graphs of the Sine and Cosine Function Graph each function using degrees. 5 6 sin sin. The worksheet itself is twelve pages and begins with the simple sine and cosine graphs, then, develops into the reciprocal trig functions. Explanations will vary. We will now explore what changes can be made to the equations and how that affects the graphs. 6 Graphs of Reciprocal Trig Functions: 11. Displaying all worksheets related to - Graphing Sine Functions. Sine & Cosine Graphs By: Taylor Pulchinski Daniel Overfelt Whitley Lubeck Equations y = a sin (bx-h)+ k y = a cos (bx-h)+k a = Amplitude (height of the wave) 2( )/b = Period (time it take to complete one trip around) h = Phase Shift (left or right movement) k = Vertical Shift (up or down movement) Examples Finding the Period and Amplitude. 1 2 2 \u224807071. Each degree with special angles This page explains the sine, cosine, tangent ratio, gives on an overview of their range of values and provides practice problems on identifying the sides that are opposite and adjacent to a given angle. Questions on the properties of the graphs of trigonometric functions and their answers are presented. Neither sine nor cosine can ever exceed 1 and the closer one of them is to 1, the closer the other must be to 0. Explanations will vary. The next section presents the graphs of the elementary sine and cosine functions as functions of the variable t. How long is the side adjacent to 1? 10. 5 and b = 4 is y = 1. I can graph the cosine function and its translations. Answers to graphing sine and cosine 1 p 2 p3p 2 2p 6 4 2 4 6 amplitude. The period of any sine or cosine function is 2\u03c0, dividing one complete revolution into quarters, simply the period/4. If the graphs of the equations y = 2 and y 2 sin x are drawn on the same set of axes, the number of points of intersection between 0 and 27t will be 24. Graphing Sine and Cosine Practice Worksheet ANSWER KEY Graph the following functions over two periods, one in the positive direction and one in the negative directions. Sine is usually abbreviated as sin. Showing top 8 worksheets in the category - Graphing Sine And Cosine. The Law of Sines The Law of Cosines Graphing trig functions Translating trig functions Angle Sum/Difference Identities Double-/Half-Angle Identities. Express your answer as a fraction in lowest terms. The trigonometry equation that represents this relationship is. Worksheet - Writing the Equation of a Transformed F(x) Graph. Trigonometry Using a calculator (sin, cos, tan) Using a calculator (inverse functions) Find the tangent of one point Tangent: Find the value of x Find the sine of one point Sine: Find the value of x Find the cosine of one point Cosine: Find the value of x Mixed sine, cosine, and tangent Mixed: Find the value of x Fill in the missing angle. 9) 10) Domain: Range: Domain: Range: Amplitude: 2 Period: Amplitude: 1 Period: \u03c0. y = sin 2 \u03b8 21. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval[latex]\\,\\left[-1,1\\right]. Experiment with the graph of a sine or cosine function. The graph could represent either a sine or a cosine function that is shifted and/or reflected. We obtain one cycle of the graph of f(\u03b8)=cos\u03b8.", "date": "2020-11-25 19:25:16", "meta": {"domain": "adunataalpini-pordenone2014.it", "url": "http://adunataalpini-pordenone2014.it/nrwx/graphs-of-sine-and-cosine-functions-answer-key.html", "openwebmath_score": 0.6496663093566895, "openwebmath_perplexity": 977.6851453625898, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9814534333179648, "lm_q2_score": 0.8991213678089741, "lm_q1q2_score": 0.8824457534056622}}
{"url": "https://orinanobworld.blogspot.com/2020/02/", "text": "Monday, February 17, 2020\n\nReversing Differences\n\nFellow blogger H\u00e5kan Kjellerstrand posted an interesting question on OR Stack Exchange recently. Starting from a list of integers, it is trivial to compute the list of all pairwise absolute differences, but what about going in the other direction? Given the pairwise (absolute) differences, with duplicates removed, can you recover the source list (or a source list)?\n\nWe can view the source \"list\" as a vector $x\\in\\mathbb{Z}^n$ for some dimension $n$ (equal to the length of the list). With duplicates removed, we can view the differences as a set $D\\subset \\mathbb{Z}_+$. So the question has to do with recovering $x$ from $D$. Our first observation kills any chance of recovering the original list with certainty:\nIf $x$ produces difference set $D$, then for any $t\\in\\mathbb{R}$ the vector $x+t\\cdot (1,\\dots,1)'$ produces the same set $D$ of differences.\nTranslating all components of $x$ by a constant amount has no effect on the differences. So there will be an infinite number of solutions for a given difference set $D$. A reasonable approach (proposed by H\u00e5kan in his question) is to look for the shortest possible list, i.e., minimize $n$.\n\nNext, observe that $0\\in D$ if and only if two components of $x$ are identical. If $0\\notin D$, we can assume that all components of $x$ are distinct. If $0\\in D$, we can solve the problem for $D\\backslash\\{0\\}$ and then duplicate any one component of the resulting vector $x$ to get a minimum dimension solution to the original problem.\n\nCombining the assumption that $0\\notin D$ and the observation about adding a constant having no effect, we can assume that the minimum element of $x$ is 1. That in turn implies that the maximum element of $x$ is $1+m$ where $m=\\max(D)$.\n\nFrom there, H\u00e5kan went on to solve a test problem using constraint programming (CP). Although I'm inclined to suspect that CP will be more efficient in general than an integer programming (IP) model, I went ahead and solved his test problem via an IP model (coded in Java and solved using CPLEX 12.10). CPLEX's solution pool feature found the same four solutions to H\u00e5kan's example that he did, in under 100 ms. How well the IP method scales is an open question, but it certainly works for modest size problems.\n\nThe IP model uses binary variables $z_1, \\dots, z_{m+1}$ to decide which of the integers $1,\\dots,m+1$ are included in the solution $x$. It also uses variables $w_{ij}\\in [0,1]$ for all $i,j\\in \\{1,\\dots,m+1\\}$ such that $i \\lt j$. The intent is that $w_{ij}=1$ if both $i$ and $j$ are included in the solution, and $w_{ij} = 0$ otherwise. We could declare the $w_{ij}$ to be binary, but we do not need to; constraints will force them to be $0$ or $1$.\n\nThe full IP model is as follows:\n\n$\\begin{array}{lrlrc} \\min & \\sum_{i=1}^{m+1}z_{i} & & & (1)\\\\ \\textrm{s.t.} & w_{i,j} & \\le z_{i} & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} ,i\\lt j & (2)\\\\ & w_{i,j} & \\le z_{j} & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} ,i\\lt j & (3)\\\\ & w_{i,j} & \\ge z_{i}+z_{j}-1 & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} ,i\\lt j & (4)\\\\ & w_{i,j} & =0 & \\forall i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} \\textrm{ s.t. }(j-i)\\notin D & (5)\\\\ & \\sum_{i,j\\in\\left\\{ 1,\\dots,m+1\\right\\} |j-i=d}w_{i,j} & \\ge 1 & \\forall d\\in D & (6)\\\\ & z_{1} & = 1 & & (7) \\end{array}$\n\nThe objective (1) minimizes the number of integers used. Constraints (2) through (4) enforce the rule that $w_{ij}=1$ if and only if both $z_i$ and $z_j$ are $1$ (i.e., if and only if both $i$ and $j$ are included in the solution).\u00a0 Constraint (5) precludes the inclusion of any pair $i < j$ whose difference $j - i$ is not in $D$, while constraint (6) says that for each difference $d \\in D$ we must include at least one pair $i < j$ for that produces that difference ($j - i = d$). Finally, since we assumed that our solution starts with minimum value $1$, constraint (7) ensures that $1$ is in the solution. (This constraint is redundant, but appears to help the solver a little, although I can't be sure given the short run times.)\n\nMy Java code is available from my repository (bring your own CPLEX).\n\nTuesday, February 11, 2020\n\nCollections of CPLEX Variables\n\nRecently, someone asked for help online regarding an optimization model they were building using the CPLEX Java API. The underlying problem had some sort of network structure with $N$ nodes, and a dynamic aspect (something going on in each of $T$ periods, relating to arc flows I think). Forget about solving the problem: the program was running out of memory and dying while building the model.\n\nA major issue was that they allocated two $N\\times N\\times T$ arrays of variables, and $N$ and $T$ were big enough that $2N^2T$ was, to use a technical term, ginormous. Fortunately, the network was fairly sparse, and possibly not every time period was relevant for every arc. So by creating only the IloNumVar instances they needed (meaning only for arcs that actual exist in time periods that were actually relevant), they were able to get the model to build.\n\nThat's the motivation for today's post. We have a tendency to write mathematical models using vectors or matrices of variables. So, for instance, $x_i \\, (i=1,\\dots,n)$ might be an inventory level at each of $n$ locations, or $y_{i,j} \\, (i=1,\\dots,m; j=1,\\dots,n)$ might be the inventory of item $i$ at location $j$. It's a natural way of expressing things mathematically. Not coincidentally, I think, CPLEX APIs provide structures for storing vectors or matrices of variables and for passing them into or out of functions. That makes it easy to fall into the trap of thinking that variables must be organized into vectors or matrices.\n\nLast year I did a post (\"Using Java Collections with CPLEX\") about using what Java calls \"collections\" to manage CPLEX variables. This is not unique to Java. I know that C++ has similar memory structures, and I think they exist in other languages you might use with CPLEX. The solution to the memory issue I mentioned at the start was to create a Java container class for each combination of an arc that actually exists and time epoch for which it would have a variable, and then associate instances of that class with CPLEX variables. So if we call the new class AT (my shorthand for \"arc-time\"), I suggested the model owner use a Map<AT, IloNumVar> to associate each arc-time combination with the variable representing it and a Map<IloNumVar, AT> to hold the reverse association. The particular type of map is mostly a matter of taste. (I generally use HashMap.) During model building, they would create only the AT instances they actually need, then create a variable for each and pair them up in the first map. When getting a solution from CPLEX, they would get a value for each variable and then use the second map to figure out for which arc and time that value applied. (As a side note, if you use maps and then need the variables in vector form, you can apply the values() method to the first map (or the getKeySet() method to the second one), and then apply the toArray() method to that collection.)\n\nNow you can certainly get a valid model using just arrays of variables, which was all that was available to me back in the Dark Ages when I used FORTRAN, but I think there are some benefits to using collections. Using arrays requires you to develop an indexing scheme for your variables. The indexing scheme tells you that the flow from node 3 to node 7 at time 4 will be occupy slot 17 in the master variable vector. Here are my reasons for avoiding that.\n\u2022 Done correctly, the indexing scheme is, in my opinion, a pain in the butt to manage. Finding the index for a particular variable while writing the code is time-consuming and has been known to kill brain cells.\n\u2022 It is easy to make mistakes while programming (calculate an index incorrectly).\n\u2022 Indexing invites the error of declaring an array or vector with one entry for each combination of component indices (that $N\\times N\\times T$ matrix above), without regard to whether you need all those slots. Doing so wastes time and space, and the space, as we saw, may be precious.\n\u2022 Creating slots that you do not need can lead to execution errors. Suppose that I allocating a vector IloNumVar x = new IloNumVar[20] and use 18 slots, omitting slots 0 and 13. If I solve the model and then call getValues(x), CPLEX will throw an exception, because I am asking for values of two variables (x[0] and x[13]) that do not exist. Even if I create variables for those two slots, the exception will occur, because those two variables will not belong to the model being solved. (There is a way to force CPLEX to include those variables in the model without using them, but it's one more pain in the butt to deal with.) I've lost count of how many times I've seen messages on the CPLEX help forums about exceptions that boiled down to \"unextracted variables\".\nSo my advice is to embrace collections when building models where variables do not have an obvious index scheme (with no skips).", "date": "2020-02-22 07:08:49", "meta": {"domain": "blogspot.com", "url": "https://orinanobworld.blogspot.com/2020/02/", "openwebmath_score": 0.5643764734268188, "openwebmath_perplexity": 408.905767542979, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9928785710435412, "lm_q2_score": 0.8887587934924569, "lm_q1q2_score": 0.8824295608851723}}
{"url": "https://math.stackexchange.com/questions/2671760/why-does-expanding-lim-p-rightarrow-0-frac1-p-1-p31-1-p3-give-a/2671778", "text": "# Why does expanding $\\lim_{p \\rightarrow 0}\\frac{1-p-(1-p)^3}{1-(1-p)^3}$ give a different limit from just substituting?\n\n$$\\lim_{p \\rightarrow 0}\\frac{1-p-(1-p)^3}{1-(1-p)^3}$$\n\n$$= \\frac{1-0-(1-0)^3}{1-(1-0)^3}$$\n\n$$=\\frac{0}{0}$$\n\n$$\\lim_{p \\rightarrow 0}\\frac{1-p-(1-p)^3}{1-(1-p)^3}$$\n\n$$=\\lim_{p \\rightarrow 0}\\frac{p^2-3p+2}{p^2-3p+3}$$\n\n$$=\\frac{0^2-3(0)+2}{0^2-3(0)+3}$$\n\n$$=\\frac{2}{3}$$\n\nEdit 1:\n\n$0\\le p \\le1$\n\n\u2022 The first one $\\,0/0\\,$ is indeterminate. That does not give you a limit, but merely tells you that the respective approach will not lead anywhere. \u2013\u00a0dxiv Mar 1 '18 at 6:46\n\u2022 HINT : let $1-p = x ;\\,$ factorize to get $1+ x+x^2$ where x lies between $0,1$ \u2013\u00a0Narasimham Mar 8 '18 at 15:00\n\nYou conveniently omitted the crucial step\n\n$$\\lim_{p \\rightarrow 0}\\frac{1-p-(1-p)^3}{1-(1-p)^3}$$\n\n$$=\\color{red}{\\lim_{p \\rightarrow 0}\\frac{p^3-3p^2+2p}{p^3-3p^2+3p}}$$\n\n$$=\\color{red}{\\lim_{p \\rightarrow 0}\\frac pp} \\cdot\\lim_{p \\rightarrow 0}\\frac{p^2-3p+2}{p^2-3p+3}$$\n\n$$=\\frac{0^2-3(0)+2}{0^2-3(0)+3}$$\n\n$$=\\frac{2}{3}$$\n\nThe expression in red is clearly of the form $\\dfrac 00$ too but will not be any more after you divide the numerator and the denominator by $p$. Functions that behave like that are said to have removable singularities.\n\n\u2022 So when working with limits without using l'Hospital's Rule such as the 1st method, when we arrive at an intermediate form, it tells us that there are terms we can cancel out in the original expression to get the real limit? Of course, given that the terms being cancelling is defined in the limit. i.e. $$\\lim_{p \\rightarrow 0}\\frac{p}{p} = 1$$ \u2013\u00a0A_for_ Abacus Mar 1 '18 at 7:06\n\u2022 ^Typo: indeterminate not \"intermediate\" \u2013\u00a0A_for_ Abacus Mar 1 '18 at 7:21\n\u2022 @A_for_Abacus - it doesn't necessarity say that there are factors that you can cancel. For example, $\\lim_{x\\to 0} \\frac {\\sin x}x$ is an indeterminant form, and there is nothing that cancels. What indeterminant forms tell you is that the limit-finding technique you are currently using has failed, and that you will have to dig deeper to uncover the limit. Sometimes that \"deeper\" is a simple as cancelling common factors. Other times it requires some in-depth analysis of the function. \u2013\u00a0Paul Sinclair Mar 2 '18 at 2:57\n\nBecause in general $\\lim_{p\\rightarrow a}\\frac{f(x)}{g(x)}\\neq \\frac{\\lim_{p\\rightarrow a} f(x)}{\\lim_{p\\rightarrow a} g(x)}$ since the right expresion may not be defined (in your case you have $\\frac{0}{0}$). To find (using highschool calculus) the value of a limit where $\\frac{\\lim_{p\\rightarrow a} f(x)}{\\lim_{p\\rightarrow a} g(x)}$ is not defined you have to get the ration in such a form where the demoninator's limit isn't 0, as you have done in the second case.\n\nA hint: Usualy what works with most polynomial limits is to divide both nominator and denominator with the the monomial $p^n$ of the highest degree apearing in the limit.\n\n$$1 - p - (1 - p)^3=p^3-3 p^2+2 p$$ $$1 - (1 - p)^3=p^3-3 p^2+3 p$$ both are $0$ at $p=0$. You are removing the common factor $p$ in the expanded version, but not in the non-expanded version, which is why you \"get a different limit\" (or, more accurately, which is why you find the limit after expanding, but do not find the limit without expanding).\n\nExpanding does not give different limits. Without expanding you arrived at an undefined situation. Whereas when you expand in some step $p$ cancelled out which is possible only if $p\\neq0$. Since the expanded function is continuous at zero and $p\\rightarrow0$ so correct limit is $\\frac{2}{3}$.\n\nEDIT: Moreover, $\\lim_{p\\rightarrow0}\\frac{p}{p}=1$\n\nA little bit of context:\n\nLet $f, g$ be continuos in $D$, $a \\in D$, and $g(a)\\not =0.$\n\nThen $\\lim_{x \\rightarrow a} \\dfrac{f(x)}{g(x)} = \\dfrac{f(a)}{g(a)}$.\n\nYour functions $f$, in the numerator, and g, in the denominator, are continuos but : $g(a)=0$!!.\n\nHence the above is not applicable.\n\nAnother example:\n\n$f(x)=x$, $g(x) =x.$\n\nDoes $\\lim_{x \\rightarrow 0} \\dfrac{f(x)}{g(x)}$ exist?\n\nThere are a lot of correct answers here, but I think that there is a fundamental definition or intuition that is missing from all of them, namely that we should ignore the value of the expression at the limit point (i.e. we assume that $p$ is never actually zero; we are taking a limit as $p$ approaches zero). A good definition of a limit is as follows:\n\nDefinition: We say that $\\lim_{x\\to a} f(x) = L$ if for all $\\varepsilon > 0$ there exists some $\\delta > 0$ such that if $x\\ne a$ and $|x-a| < \\delta$, then $|f(x) - L| < \\varepsilon$.\n\nTopologically (and feel free to ignore this paragraph for now), we are saying that for any neighborhood $V$ of $L$, there is some punctured neighborhood $U^\\ast$ of $a$ such that $f(U^*) \\subseteq V$. Because we are puncturing the neighborhood, the value of $f$ at $a$ is irrelevant. We just completely ignore it.\n\nIn the original question, we are trying to compute $$\\lim_{p\\to 0} \\frac{1-p-(1-p)^3}{1-(1-p)^3}.$$ As you have noted, when $p=0$, this expression is utter nonsense. That is, if we define $$f(p) := \\frac{1-p-(1-p)^3}{1-(1-p)^3} = \\frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p}$$ then try to evaluate $f(0)$, this will give us $\\frac{0}{0}$ which is a (more-or-less) meaningless expression. However, we are trying to take a limit as $p\\to 0$, which means that we can (and should) assume that $p \\ne 0$. Notice that under this assumption, i.e. the assumption that $p\\ne 0$, we have that $1 = \\frac{1/p}{1/p}$. Then, using the analyst's second favorite trick of multiplying by 1 (adding 0 is the favorite trick), we have \\begin{align} f(p) &= \\frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p} \\\\ &= \\color{red}{1} \\cdot \\frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p} \\\\ &= \\color{red}{\\frac{1/p}{1/p}} \\cdot \\frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p} \\\\ &= \\frac{p^2-3p+2}{p-3p+3} \\\\ &=: \\tilde{f}(p). \\end{align} Again, the vital thing to understand is that the computation is justified since $p \\ne 0$, which means that the fraction $\\frac{1/p}{1/p}$ is perfectly well-defined and is (in fact) identically 1. Note, also, that the computations above are done before we've tried to take any limits. It is now relatively easy to see that $$\\lim_{p\\to 0} f(p) = \\lim_{p\\to 0} \\tilde{f}(p) = \\lim_{p\\to 0} \\frac{p^2-3p+2}{p-3p+3} = \\frac{2}{3}.$$\n\nThere are two things here that I have left unjustified:\n\nExercises:\n\n1. Explain why $\\lim_{p\\to 0} f(p) = \\lim_{p\\to 0} \\tilde{f}(p)$?\n2. Explain why $\\lim_{p\\to 0} \\tilde{f}(p) = \\frac{2}{3}$.\n\nHint for 1:\n\nOne possible argument is a one-line appeal to the squeeze theorm.\n\nHint for 2:\n\nThink about the relation between continuity and limits.\n\nThe original function is a rational function in $p$, that is, it is a fraction of two polynomials $f(p)$ and $g(p)$. The roots $a_i$ are the values for which $f(a_i) = 0$, and similarly, the poles $b_j$ are the values for which $g(b_j) = 0$. If for some $i$ and $j$ you have $a_i = b_j$, then the factor $(p-a_i)$ on top cancels with the factor $(p-b_j)$ on bottom. The original function had $p\\neq b_j$ since that would lead to a $0$ in the denominator, so even after cancelling this factor, we still have that $p\\neq b_j$ regardless of whether we could evaluate the function at that point now. However, if we can evaluate the reduced function at $p=b_j$, then it will be continuous there (since rational functions only have holes and asymptotes as discontinuities). Then, since the function is continuous there, the limit of that function matches the value of function evaluated at the point $b_j$. Since the original function is identical to the new function for all $p \\neq b_j$, the limit as $p\\rightarrow b_j$ is the same for the original function as it is for the reduced function. This means that plugging $p = b_j$ in the original function will give you $0/0$ which is not the value of the limit, but after cancelling, plugging $p=b_j$ in the reduced function will give you the correct limit.\n\nIn this case you've given, $a_i = b_j = 0$, so you just cancel $p$ from the top and bottom, so that plugging $p=0$ into the reduced equation give you the limit of $2/3$.", "date": "2021-02-25 10:33:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2671760/why-does-expanding-lim-p-rightarrow-0-frac1-p-1-p31-1-p3-give-a/2671778", "openwebmath_score": 0.9904360175132751, "openwebmath_perplexity": 218.59312498858014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846716491914, "lm_q2_score": 0.9019206857566126, "lm_q1q2_score": 0.8824253739875969}}
{"url": "http://vvnn.ecui.pw/latex-symbols-math.html", "text": "# Latex Symbols Math\n\nAn operator such as the differential operator D^~. Short Math Guide for LATEX, version 2. Even though commands follow a logical naming scheme, you will probably need a table for the most common math symbols at some point. Letters are. = mathrm{C}, LATIN CAPITAL LETTER C. Suppose that you need to type many times. To make the following table of symbols completely available, include the following package-loading command after your documentclass command. This article shows several fonts for use in math mode. Most other symbols can be inferred from these examples. Several fonts require the addition of the line \\usepackage{amssymb} to the preamble to work. Note that the formula will appear below your drawing area. Log-like Symbols 8. SUB and SUP are used to specify subscripts and superscripts. The Comprehensive LATEX Symbol List Scott Pakin \u2217 22 September 2005 Abstract This document lists 3300 symbols and the corresponding LATEX commands that produce them. Please, help me define a math symbol for definition as shown in Table of mathematical symbols - Wikipedia, the free encyclopedia is defined as; equal by definition it is the equal sign under a small font 'def' the same symbol is also provided in OpenOffice math, but not in Latex. The symbols \u22a4 (U+22A4) and \u22a5 (U+22A5) \\top and \\bot in LaTeX. For example:\\begin{align}\\because x+3&=5\\\\\\therefore x&=2\\end{align}. You may have to register before you can post: click the register link above to proceed. The fact that he succeeded was most probably why TeX (and later on, LaTeX) became so popular within the scientific community. It is important to keep in mind that LaTeX has its own way of handling spacing in mathematics mode. Michel Bovani created the mbtimes package by using Omega Serif for text and Latin and Greek letters in mathematics. Usage An exemplary use of the symbol in a formula. I don't have latex software. A few of them, such as + and >, are produced by typing the corresponding keyboard character. 2 More symbols in math mode nhbar nimath njmath nell nRe 1 @ r 8 9 ninfty npartial nnabla nforall nexists = > ? y z nIm ntop nbot ndag nddag P Q R H 4 nsum nprod nint noint ntriangle. no LaTeX Math Symbols Prepared by L. 10 Single Mathematical Symbols @ naleph aleph. Yes, when I try Matlab to write the title with the Latex font, I do not now why, but it does not work; however, the axis labels are correctly intrepeted and the command works perfectly. To make the following table of symbols completely available, include the following package-loading command after your documentclass command. Call it $\\boldsymbol\\beta$. LaTeX Community is an excellent resource for answering any questions you have about LaTeX. So to make the greek capital phi, \u03a6, you\u2019d hit (in insert mode) kF* Below is a table of useful math and computer science digraphs. I know this is very late, but for others like me who are also looking for this answer: Indeed, everyone's answer is correct. Relation Symbols 10. 2 More symbols in math mode nhbar nimath njmath nell nRe 1 @ r 8 9 ninfty npartial nnabla nforall nexists = > ? y z nIm ntop nbot ndag nddag P Q R H 4 nsum nprod nint noint ntriangle. Some other constructions. I mentioned in a conversation that math rarely uses Hebrew or Russian letters. Also, note that there are several variants of LaTeX, but we recommend the variant known as LaTeX2e. This command forces LaTeX to give an equation the full height it needs to display as if it were on its own line. For example, in the formula  ,'' the word log'' is not italicized like the variables x and y. You may also want to visit the Mathematics Unicode characters and their HTML entity. Sometimes, a formula contains text that should be set in roman type. There are two ways of displaying the symbol: compressed to fit onto one line (useful when printing long equations or proofs) or in a larger, more readable format. TLatex can display dozens of special mathematical symbols. PDF; LaTeX source; This is just intended to be a very plain chart of the Greek letters available in LaTeX math mode. Or go to the answers. How do I make symbols like nabla () or delta () as bold symbols? It is very simple. 8, 8 the package amssymb must be loaded in the preamble of the document and the A M S math fonts must be installed on the system. Feel free to tell us about links, images and videos if you read more articles / blog posts that you believe we. As you see in this example, a mathematical text can be explicitly spaced by means of some special commands Open an example in Overleaf [] SpaceThe spacing depends on the command you insert, the example below contains a complete list of spaces and how they look like. Another thing to notice is the effect of the \\displaystyle command. Thank you all. MathType has customizable keyboard shortcuts for virtually every symbol, template, and command. This document is based largely on section 8. A canvas will open for your training input. It is assumed that any characters represent variable names. The Style section shows you how to control the basic appearance of your document. An operator such as the differential operator D^~. The numerals 6 and 26, for example, are symbols that represent quantities. On the use of italic and roman fonts for symbols in scientific text Scientific manuscripts frequently fail to follow the accepted conventions concerning the use of italic and roman fonts for symbols. These symbol images are taken from a mirror site in Austrailia by John Marley. \\sqrt{abc} \\alpha \\frac{abdsf}{afds} \\hat{a} \\acute{a} \\bar{a} \\dot{a} \\breve{a} \\check{a} \\grave{a} \\vec{a} \\ddot{a. I don't have latex software. Spacing in math mode; Integrals, sums and limits; Display style in math mode; List of Greek letters and math symbols; Mathematical fonts Figures and tables. In this guide, you\u2019ll find an extensive list of probability symbols you can use for reference, plus the names of each symbol and the concept they represent. This article provides the list of most commonly used symbols in LaTeX. Classes of math symbols. For example, is given by $\\mathfrak S = \\mathbb R^2/G$. LaTeX requires control sequences to format some symbols in text. A matrix having $$n$$ rows and $$m$$ columns is a $$m\\times n$$-matrix. You have to use \\pm. Here's a list of mathematical symbols and their meaning, for your reference. In a math environment (or in math mode) LaTeX typesets the words in the table below in the manner of body text (not in italics) so that they stand apart from the other symbols and variables, it is equivalent to using the \u201cmathrm{ }\u201d command. machine learning. Supported symbols are listed here (alphabetically). This is a list of symbols found within all branches of mathematics. That said, there are many places where symbols are useful and simplify matters. There are problems with this approach. The symbols \u22a4 (U+22A4) and \u22a5 (U+22A5) \\top and \\bot in LaTeX. Landau Symbols. The change would also make it more consistent with List of logic symbols. To use the symbols listed in Tables 3. 1 day ago \u00b7 LaTeX is a professional document preparation system and document markup language written by Leslie Lamport. We've documented and categorized hundreds of macros!. A lot of the nice looking equations you see in books and all around the web are written using LaTeX commands. Learn how to add images to your LaTeX documents. All the versions of this article:. A basic latex document would look like. Basic Code Special keys. After looking for a builtin expectation symbol in LaTeX, and coming up with none, I\u2019ve defined one. Below we give a partial list of commonly used mathematical symbols. I know this is very late, but for others like me who are also looking for this answer: Indeed, everyone's answer is correct. Actuarial symbols of life contingencies and \ufb01nancial mathematics David Beauchemin david. In informal usage, \"tilde\" is often instead voiced as \"twiddle\" (Derbyshire 2004, p. LaTeX symbols have either names (denoted by backslash) or special characters. AsciimathML can be used directly with script tags without filter so you can add those script tags to theme files or probably even to single Hot Pot files but note that asciimathml is using different syntax and tokens than Tex filter or MathTran filter - the best option might be to use mathJax everywhere (even if it is a little slow), in that. pdf provides an introduction to LaTeX: Math into LaTeX Short Course based on the book Math into LaTeX An Introduction to LaTeX and AMS-LaTeX by George Gratzer, published by Birkhauser Boston, ISBN 0-8176-3805-9. Math Mode LaTeX uses a special math mode to display mathematics. Click Insert. LATEX Mathematical Symbols The more unusual symbols are not de\ufb01ned in base LATEX (NFSS) and require \\usepackage{amssymb} 1 Greek and Hebrew letters \u21b5 \\alpha \\kappa \\psi z \\digamma \\Delta \u21e5 \\Theta \\beta \\lambda \u21e2 \\rho \" \\varepsilon \\Gamma \u2325 \\Upsilon \\chi \u00b5 \\mu \\sigma { \\varkappa \u21e4 \\Lambda \u2305 \\Xi. Most letters and symbols are simple in LaTeX, yet a few characters are reserved for LaTeX commands, i. 8p 6 \u2018Under\u2019symbols,\\mathunder 7 7 Accents,\\mathaccent 9 8 Bottomaccents,\\mathbotaccent 10. A Mathematical symbols \\alpha \\beta \\gamma \\delta \\epsilon \\varepsilon \\zeta \\eta AMS miscellaneous symbols. In this tutorial I discuss how to use math environment in Latex. 18List of Mathematical Symbols 19Summary 20Notes 21Further reading 22External links Mathematics environments LaTeX needs to know beforehand that the subsequent text does indeed contain mathematical elements. Click here for example. The frequently used left delimiters include (, [ and {, which are obtained by typing (, [and \\{respectively. no LaTeX Math Symbols Prepared by L. Introduction to Overleaf and LaTeX Discrete Math Fall 2017 3 work outside a list building environment. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. tex) is NOTa good model to build. machine learning. tex Math model predicts growth, death of membership-based websites (Phys. , braces { }. Similar is for limit expressions. This is a command for generating the$\\because$ symbol. Feel free to tell us about links, images and videos if you read more articles / blog posts that you believe we. This article provides the list of most commonly used symbols in LaTeX. Math symbols and math fonts 3. Note that the formula will appear below your drawing area. The regular type style declarations can be used in math mode. Math symbols de\ufb01ned by LaTeX package \u00abamssymb\u00bb No. Suppose that you need to type many times. If you want the limits of an integral/sum/product to be specified above and below the symbol in inline math mode, use the \\limits command before limits specification. Similar is for limit expressions. Most import, this post is showing you the basics about math symbols in Latex. Text mode 2. When using LaTeX, can I just use $:=$, or do I need to do something special?. I removed the following from the article: Note: To use the Greek Letters in L a T e X that have the same appearance as their Roman equivalent, just use the Roman form: e. MATH Common Greek Binary Subsets Inequalities Triangles Arrows Operators Functions Miscel. There are two linear formats for math that Word supports:. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. An italic font is generally used for emphasis in running text, but it has a quite specific meaning when used for symbols in scientific text and. I'd really like latex support for Evernote, since apart from that I enjoy using it. It describes the editing community's established practice on some aspect or aspects of Wikipedia's norms and customs. You can leave a comment with your thoughts if you have a question about Latex math symbols direct sum, or want to know more. Note: If a +1 button is dark blue, you have already +1'd it. tex contains all necessary code This file is prepared by running latex A. 14 \u201cDimensionless parameters\u201d). LyX combines the power and flexibility of TeX/LaTeX with the ease of use of a graphical interface. Basic math symbols; Geometry symbols; Algebra symbols; Probability & statistics symbols; Set theory symbols; Logic symbols; Calculus & analysis symbols; Number symbols; Greek symbols; Roman numerals; Basic math symbols. The current version of LATEX is LATEX2e. You might make a definition like ewcommand{\\gijk}{\\Gamma^{ij}_k} and then type \\gijk each time you want to use this symbol (inside mathematics mode). If you want the limits of an integral/sum/product to be specified above and below the symbol in inline math mode, use the \\limits command before limits specification. The body contains the text, gures, tables, etc. If you can't find the LaTeX symbol(s) that you are after, then I can almost guarantee that you'll find them in the Comprehensive LaTeX Symbol List. math symbols in latex,document about math symbols in latex,download an entire math symbols in latex document onto your computer. Most letters and symbols are simple in LaTeX, yet a few characters are reserved for LaTeX commands, i. The corresponding right delimiters are of course obtained by typing ), ] and \\}. 0 (2017/12/22) 6 3. Codecogs has a real time online LaTeX equation editor. Hyperbolic functions The abbreviations arcsinh, arccosh, etc. There are two ways of displaying the symbol: compressed to fit onto one line (useful when printing long equations or proofs) or in a larger, more readable format. (Actually this is not exactly how it's written, as a backwards $\\in$. The current version of LATEX is LATEX2e. Finally, as others have pointed out, the Comprehensive LaTeX Symbols List is a great resource for finding the perfect symbol for the job. Letters are. Character Name Browser Image; U+002B: PLUS SIGN + ARABIC MATHEMATICAL OPERATOR MEEM WITH HAH. Math Mode Accents. the whole message has to be in Latex. Suppose that you need to type many times. Variable-sized Symbols 3. Some of these symbols are guaranteed to be available in every LATEX2\u03b5system; others require fonts and packages that. Use equations in a document. Certain spacing and positioning cues are traditionally used for. ;; Copyright (c) 1999, 2000, 2001 ;; John Palmieri ;; Author: John Palmieri ;; Maintainer: John. Jimenez,Lan Nguyen. I'd love to know what it's called so I can Google it and find a font for use it in some stuff I'm typing. Hi folks, After searching the Internet without any suitable results, I finally send this mail to this list. 07 (2000/07/19) 15 8. 1) Specify math font as \\setmathfont{XITS Math} (I'm on a linux system and that is the name of the font my system reports) and 2) Adding \\usepackage{bm} AFTER setting the math font. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. Click the symbol that you want to insert. Using Math-Type to create TeX and MathML equations The Translation System from a User\u2019s Perspective The basic scenario for using MathType to aid in the creation of a TEX document is to run it simultane-ously with your favorite TEX editor. LaTeX takes care of the spacing of mathematical symbols automatically. Note: If a +1 button is dark blue, you have already +1'd it. But still, TeX math is the best way to write math, and I'd claim LaTeX the best way to write structured documents, despite it's flaws. For example, is given by $\\mathfrak S = \\mathbb R^2/G$. as being part of the LATEX command set, and it interferes with the proper operation of various features such as the fleqn tion, use the equation environment; to assign a label for cross referencing, use the \\label command and product symbols to appear full size, you need the \\displaystyle command ' n> zn ' A quick guide to LATEX Overleaf. The font type LaTeX uses in math mode is somewhat special since it is optimized for writing mathematical formulas. , A instead of Alpha, B instead of Beta, etc. The Comprehensive LATEX Symbol List Scott Pakin \u2217 22 September 2005 Abstract This document lists 3300 symbols and the corresponding LATEX commands that produce them. The basic LaTeX program does not include all the math you'll want to use. Here are the most common set symbols. That said, there are many places where symbols are useful and simplify matters. The corresponding right delimiters are of course obtained by typing ), ] and \\}. tex) is NOTa good model to build. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. Derivatives, Limits, Sums and Integrals. The opposite of being equivalent is being nonequivalent. Find out more: bear cub clip art free funny face photo editor water draw prover vintage photos women cooking classes for couples ideas for corner flower beds sketch of human heart. For example, both letters in the output from ${\\bf x}^{\\bf x}$ are the same size. pdf that you just can't memorize. Inserting Images; Tables; Positioning Images and Tables; Lists of Tables and Figures; Drawing Diagrams Directly in LaTeX; TikZ package References and Citations. Mathematics Keyboard v. Unicode math. The mathematics mode in LaTeX is very flexible and powerful, there is much more that can be done with it: Subscripts and superscripts; Brackets and Parentheses; Fractions and Binomials; Aligning Equations; Operators; Spacing in math mode; Integrals, sums and limits; Display style in math mode; List of Greek letters and math symbols; Mathematical fonts. Very supportive of new users. I know this is very late, but for others like me who are also looking for this answer: Indeed, everyone's answer is correct. It would be inconvenient to have to exit math mode to type the text in normal roman type, so the most common function names are defined as special commands. mathematical symbols Software - Free Download mathematical symbols - Top 4 Download - Top4Download. Type TeX or LaTeX: If you already know the TeX typesetting language, you can enter equations directly into MathType or Microsoft Word documents. Here are the most common set symbols. This secure online tool allows for the conversion of LaTeX (mathematics) to Presentation MathML. To emphasise that each symbol is an individual, they are not positioned as closely together as with normal text. UPDATE: Mails sent from this system were tagged with \"Be careful with this message\" warnings in most email systems (rightly so), because the sender's domain (yours) was different to IntMath's domain. When using LaTeX, can I just use $:=$, or do I need to do something special? Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In order to access all the math functions and symbols we will introduce in the guide pages, you'll have to include a number of packages. The process is this: \u2022 whenever an equation is needed, the MathType window is brought to the front. Re: LaTeX representation of hollow \"R\" symbol for real space Post by localghost \u00bb Mon May 24, 2010 9:47 am I don't think that your math institute sets the standard [1,2]. The maths, embedded in. ? If this question is repeated, I am sorry I just searched and couldn't find an answer. If and are \"equivalent by definition\" (i. The overloading of a symbol usually implies the overloading of related symbols. If placed over multiple symbols (especially in the context of a radical), it is known as a vinculum. Write beautiful math equations & symbols easily in MathMagic, use them widely in your word processors, Presentations, DTP software. Perhaps, this question has been answered already but I am not aware of any existing answer. See the following links for information on how MathML can be implemented. Bibliography management in. Similarly, a math symbol like \\frac doesn\u2019t work out-. 2 Symbols Symbols for physical quantities should be single letters of the Latin or Greek alphabet with or without modifying signs (subscripts, superscripts, primes, etc. for the capitals). David Hamann. Spacing in math mode; Integrals, sums and limits; Display style in math mode; List of Greek letters and math symbols; Mathematical fonts Figures and tables. Selected LaTeX Math Symbols Note: there is another version of this document featuring HTML entities for math symbols , as well as LaTeX commands. UPDATE: Mails sent from this system were tagged with \"Be careful with this message\" warnings in most email systems (rightly so), because the sender's domain (yours) was different to IntMath's domain. Is the Mathematical \"Implies\" symbol stored somewhere in Excel / MS Office in General? It's the symbol that looks like an equals followed by a greater than symbol - i. A Mathematical symbols \\alpha \\beta \\gamma \\delta \\epsilon \\varepsilon \\zeta \\eta AMS miscellaneous symbols. In certain cases it may be desirable to include \"normal text\" within an equation. Re: LaTeX representation of hollow \"R\" symbol for real space Post by localghost \u00bb Mon May 24, 2010 9:47 am I don't think that your math institute sets the standard [1,2]. tex Math model predicts growth, death of membership-based websites (Phys. Mostly a control sequence is a back slash followed by the desired symbol or its designator. The Comprehensive LATEX Symbol List Scott Pakin \u2217 19 January 2017 Abstract This document lists 14283 symbols and the corresponding LATEX commands that produce them. Refer to the external references at the end of this article for more information. A symbol is said to be overloaded if its meaning depends on the context. Inserting Images; Tables; Positioning Images and Tables; Lists of Tables and Figures; Drawing Diagrams Directly in LaTeX; TikZ package References and Citations. Short Math Guide for LATEX, version 1. Different classes of mathematical symbols are characterized by different formatting (for example, variables are italicized, but operators are not) and different spacing. 2 More symbols in math mode nhbar nimath njmath nell nRe 1 @ r 8 9 ninfty npartial nnabla nforall nexists = > ? y z nIm ntop nbot ndag nddag P Q R H 4 nsum nprod nint noint ntriangle. edit: I just thought that \\sfrac{q}{m} should look nicer if it is bigger when it is followed by =. You can type an absolute value symbol in LaTeX by typing \"\\mid\" and it will be replaced by the symbol when you generate your output. Open an example in Overleaf [] Capital letters-only font typefaceThere are some font typefaces that support only a limited number of characters; these fonts usually denote some special sets. Routing metric = The character sequence \\% generates a percent (%) sign. MATH Common Greek Binary Subsets Inequalities Triangles Arrows Operators Functions Miscel. Texmaker is a free, modern and cross-platform LaTeX editor for linux, macosx and windows systems that integrates many tools needed to develop documents with LaTeX, in just one application. You have to be clever if you want to use LaTeX in your Prezi talk. If there are several typographic variants, only one of the variants is shown. ;; Copyright (c) 1999, 2000, 2001 ;; John Palmieri ;; Author: John Palmieri ;; Maintainer: John. It is also possible to check to see if a Unicode code point is available as a LaTeX command, or vice versa. Kocbach, on the basis of this document (origin: David Carlisle, Manchester University) File A. To make this easier, we support LaTeX, a way of writing mathematics terms and equations with ordinary text. The tilde is the mark \"~\" placed on top of a symbol to indicate some special property. machine learning. The symbol will be inserted in your document. the whole message has to be in Latex. Just add: % Expectation symbol \\DeclareMathOperator*{\\E}{\\mathbb{E}} to your LaTeX preamble and you're done. A practical use for LaTeX macros is to simplify the typing of mathematical formulas. As you see in this example, a mathematical text can be explicitly spaced by means of some special commands Open an example in Overleaf [] SpaceThe spacing depends on the command you insert, the example below contains a complete list of spaces and how they look like. , which form is preferable from a typographic point of view, depends strongly on (a) what's inside the part and (b) whether the material occurs in inline math mode or display math mode. In informal usage, \"tilde\" is often instead voiced as \"twiddle\" (Derbyshire 2004, p. You can leave a comment with your thoughts if you have a question about Latex math symbols direct sum, or want to know more. Text Math Macro Category Requirements Comments 00026 & (N) \\& mathord # \\binampersand (stmaryrd) 02144 \u2144 \\Yup mathord stmaryrd TURNED SANS-SERIF CAPITAL Y 0214B \u214b (O) \\invamp mathbin txfonts # \\bindnasrepma (stmaryrd), TURNED AMPERSAND. As you see in this example, a mathematical text can be explicitly spaced by means of some special commands Open an example in Overleaf [] SpaceThe spacing depends on the command you insert, the example below contains a complete list of spaces and how they look like. Covers the list of majorly used Greek Symbols, Binary Operation Symbols, Relation Symbols, Arrows Symbols, Trigonometric Math Functions/Symbols, Integrals Math Symbols, and Miscellaneous Math Functions/Symbols. They are valid only in math mode, and they operate only on the letter or group that immediately follows. Math symbols de\ufb01ned by LaTeX package \u00abamsfonts\u00bb No. Use the LaTeX-command \\overset{symbols above}{symbols below}. Computer typesetting systems often have their own ways to represent certain symbols and mathematical concepts. The mathematics is done using a version of $$\\LaTeX$$, the premiere mathematics typesetting program. Changing the font size in LaTeX Multi-column and multi-row cells in LaTeX tables Normal text in math mode Lists: Enumerate, itemize, description and how to change them Control the width of table columns (tabular) in LaTeX. Ce point de coupure n'a pas \u00e0 \u00eatre accompagn\u00e9 de relation ou d'op\u00e9rateur binaires. Thanks to. Mostly a control sequence is a back slash followed by the desired symbol or its designator. If there are symbols missing drop me a line or create a pull-request. What I currently did is to write an AutoHotKey script which automatically replaces \\latexSymbol with the corresponding unicode symbol, using the \"hotstrings\" AutoHotKey feautres. The basic LaTeX program does not include all the math you'll want to use. This video shows you how to easily remember \"greater than\" and \"less than\" math symbols. These can be included in a LaTeX document using the \\mathbb{ [letter] } tag from within the math environment. Do not forget the include nusepackagefamsmath,amssymb,latexsymgbefore. No extra packages are required to use these symbols. LaTeX Math Symbols Prepared by L. R Markdown allows you to mix text, R code, R output, R graphics, and mathematics in a single document. List of all mathematical symbols and signs - meaning and examples. If placed over multiple symbols (especially in the context of a radical), it is known as a vinculum. A list of accent marks that are available in math-mode in LaTeX. It doesn't mean that LaTeX doesn't know those sets, or more importantly their symbols\u2026 There are two packages which provide the same set of symbols. Log-like Symbols. Depending on your preferred input format, you can create equations in Word in either one of UnicodeMath or LaTeX formats by selecting the format from the Equations tab. Latex mathematical symbols - rutgers university Open document Search by title Preview with Google Docs Latex mathematical symbols the more unusual symbols are not de\ufb01ned in base latex (nfss) and require \\usepackage{amssymb} 1 greek and hebrew letters. but what is the command for the negated symbol? How to make your own laTex math symbol?-1. The mathematical symbol is produced using \\partial. Latex math symbols list Sample Schedule A Letter - Veterans Benefits Administration Sample Schedule A Letter from the Department of Labor\u2019s Office of Disability and Employment Policy: Date. angstrom latex, latex angstrom, angstrom symbol in latex, latex symbol for angst, amstrong en latex, how to write angstrom in latex, how to wrige angstrong in latex, latex angstrom math mode, latex ang, \u00e5ngstr\u00f6m latex, revtex4 angstrom, angstrom latex \u66f8\u304d\u65b9, armstobg in latex, armstrong in latex, How to write mathematical expression. Setting bold Greek letters in LaTeX maths The issue here is complicated by the fact that \\mathbf (the command for setting bold text in TeX maths) affects a select few mathematical \"symbols\" (the uppercase Greek letters). Ahh, that's why I could get it to work. In my opinion the less often a symbol is used where a few words can go, the better. Analysis & calculus symbols table - limit, epsilon, derivative, integral, interval, imaginary unit, convolution, laplace transform, fourier transform RapidTables Home \u203a Math \u203a Math symbols \u203a Calculus symbols. LaTeX has many symbols at its disposal. MathType's strong points are the hundreds of symbols that are accessible from the keyboard, along with its compatibility with various systems for importing and exporting formulas, such as MathML, TeX or Texvc. as being part of the LATEX command set, and it interferes with the proper operation of various features such as the fleqn tion, use the equation environment; to assign a label for cross referencing, use the \\label command and product symbols to appear full size, you need the \\displaystyle command ' n> zn ' A quick guide to LATEX Overleaf. Note that the formula will appear below your drawing area. LaTeX is a programming language that can be used for writing and typesetting documents. I'd love to know what it's called so I can Google it and find a font for use it in some stuff I'm typing. A canvas will open for your training input. The solution given by the LaTeX User's Guide & Reference Manual is to use \\mbox{\\boldmath$#1$} where #1 is the symbol to be set bold. LaTeX Tutorial pt 3 - Mathematics in LaTeX - Duration: 3:33. In the examples C = {1,2,3,4} and D = {3,4,5}. The symbol may be used with other meanings, including \"approximately equal to\". Since LaTeX offers a large amount of features, it's hard to remember all commands. That list also includes LaTeX and HTML markup, and Unicode code points for each symbol (note that this article doesn't have the latter two, but they could certainly be added). The numerals 6 and 26, for example, are symbols that represent quantities. Math symbols de\ufb01ned by LaTeX package \u00abamssymb\u00bb No. Is the Mathematical \"Implies\" symbol stored somewhere in Excel / MS Office in General? It's the symbol that looks like an equals followed by a greater than symbol - i. This package defines the following commands:. LaTeX is one popular typesetting system. A few of them, such as + and >, are produced by typing the corresponding keyboard character. TEX is a low-level language that computers can work with, but most people would nd dicult to use; so LATEX has been developed to make it easier. 20 hours ago \u00b7 Here is an example of a quote in this spirit, from the Wikipedia on functional equations for L-functions. The symbols \u22a2 (U+22A2) and \u22a3 (U+22A3) are \\vdash and \\dashv in LaTeX. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. Even though commands follow a logical naming scheme, you will probably need a table for the most common math symbols at some point. On different math. 1) Specify math font as \\setmathfont{XITS Math} (I'm on a linux system and that is the name of the font my system reports) and 2) Adding \\usepackage{bm} AFTER setting the math font. In mathematical mode as well as in text mode, you can change the typeface as needed. The Comprehensive LATEX Symbol List Scott Pakin \u2217 22 September 2005 Abstract This document lists 3300 symbols and the corresponding LATEX commands that produce them. Curly brackets are used to group characters. tex Math model predicts growth, death of membership-based websites (Phys. Does anybody know if there is a way to make the math symbols work in latex in verbatim mode? If yes, then how? Adv Reply. I don't have latex software. Others are obtained with the commands in the following table:. TeX editing can be mixed with point-and-click editing so you get the best of both worlds. ShareLaTeX 64,286 views. $\\begingroup$ A good approximation is y with a circumflex (U+0177): \u0177 but in MathJax/ LaTeX markup it looks more mathematical (the way it tends to be written by hand): $\\hat{y}$ $\\endgroup$ \u2013 Glen_b \u2666 Feb 4 '16 at 4:42. I mentioned in a conversation that math rarely uses Hebrew or Russian letters. Supported symbols are listed here (alphabetically). tex WARNING!!!! This document (latextips. We will tells you more regarding latex math symbols direct sum, giving the knowledge you are looking for. Various math symbols TeX provides almost any mathematical symbol you're likely to need. Inserting Images; Tables; Positioning Images and Tables; Lists of Tables and Figures; Drawing Diagrams Directly in LaTeX; TikZ package References and Citations. Render Latex equations into plain text ASCII to insert as comments in source-code, e-mail, or forum. View all images. Creating Tables with LaTeX Tables are created using the \u201ctable\u201d environment given below: \\begin{table}[where] table \\end{table} In the above syntax, table stands for the contents of the \u2018tabular\u2019 environment together with a possible \\caption command. Here an example for an \"a\" over. LaTeX Symbols package The LaTeX symbols package provides commands for XEmacs (and only XEmacs--it doesn't work with GNU Emacs) to open up a window with a table of LaTeX symbols. The \\sideset command There\u2019s also a command called \\sideset, for a rather special purpose: putting symbols at the subscript and superscript corners of a symbol like P or Q.", "date": "2020-01-18 18:10:10", "meta": {"domain": "ecui.pw", "url": "http://vvnn.ecui.pw/latex-symbols-math.html", "openwebmath_score": 0.9359549880027771, "openwebmath_perplexity": 2233.4644955099657, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9433475746920261, "lm_q2_score": 0.935346505335907, "lm_q1q2_score": 0.8823568573052901}}
{"url": "https://math.stackexchange.com/questions/2481670/what-is-the-probability-that-player-a-and-b-play-against-each-other", "text": "# What is the probability that player A and B play against each other?\n\nPlayers with same skill take part in a competition. The probability of winning each game is 0.5. At first, we divide a group of $2^n$ people to random pairs that play against each other. Then we will do the same for $2^{n-1}$ winners and this continues until there is only one winner. What is the probability that player A and B play against each other?\n\nI know we should calculate the sum of probabilities that two players play against each other in the first round, then for second round and etc. So how can i calculate the probability that two players play at round K?\n\n\u2022 Did you try a small example to see what would have to happen for A and B to meet in the second round, or the third round, and then see how it may generalize from that? \u2013\u00a0Bram28 Oct 20 '17 at 16:17\n\n\u2022 It is a knockout competition with $m=2^n$ players so $m-1$ knockouts or matches or parings are needed to produce a winner\n\n\u2022 There are ${m \\choose 2}=\\frac{m(m-1)}{2}$ equally-likely possible parings of the $m$ players, so the probability any particular match involves both players A and B is $\\frac{2}{m(m-1)}$, and so the expected number of the $m-1$ matches which involve both players A and B is $(m-1) \\times \\dfrac{2}{m(m-1)}=\\dfrac{2}{m}$\n\n\u2022 Since players A and B can only meet $0$ or $1$ times, the probability they meet is $\\dfrac{2}{m} = \\dfrac{1}{2^{n-1}}$\n\n\u2022 Elegantly simple. \u2013\u00a0true blue anil Oct 21 '17 at 2:14\n\u2022 $P(A , B \\text{ play at round } 1) = \\frac{1}{2^{n-1}}$\n\nWhen we have $2^{n}$ people, then we have at most $\\log^{2^{n}}_{2}$ rounds. For example, assume we have only $2$ players, then $P(A , B \\text{ play at round } 1) = \\frac{1}{2^{\\log^{2}_{2}-1}} = 1$\n\n\u2022 $P(A , B \\text{ play at round } 2) = P(A , B \\text{ did not play at round } 1 \\text{ and } A , B \\text{ both win at the first round}) = (1-\\frac{1}{2^{n-1}})(\\frac{1}{2})^{2}$\n\n\u2022 $P(A , B \\text{ play at round } 3) = P(A , B \\text{ did not play at round } 1,2 \\text{ and } A , B \\text{ both win at the first, second rounds}) = (1-\\frac{1}{2^{n-1}})(1-\\frac{1}{2^{n-2}})(\\frac{1}{2})^{4}$\n\n$$...$$\n\n\u2022 $\\forall k<n: \\quad P(A , B \\text{ play at round } k) = (\\frac{1}{2})^{2(k-1)}\\times \\Pi_{i=1}^{k}(1-\\frac{1}{2^{n-i}})$\n\n$$...$$\n\n\u2022 $\\bbox[5px,border:2px solid #C0A000]{P(A , B \\text{ play at round } n \\text{ and } n>1) = (\\frac{1}{2})^{2(n-1)}\\times \\Pi_{i=1}^{n-1}(1-\\frac{1}{2^{n-i}})}$\n\u2022 I think you're missing minus ones. Assume there are only two players. Is the probability that they will play $\\frac12$? Also each round you have to multiply with that stages probability that they will play. \u2013\u00a0karakfa Oct 20 '17 at 16:32\n\u2022 @karakfa, your first comment is not correct. I explain it by an example. \u2013\u00a0Hasan Heydari Oct 20 '17 at 16:42\n\u2022 yes, my mistake $2^n$ people means for two people n=1. \u2013\u00a0karakfa Oct 20 '17 at 16:48\n\nHmm...I don't see any real difference between this and an ordinary single-elimination tournament where the seeding is set from the start. Maybe I'm missing something.\n\nAnyway, assuming that there isn't any difference, we observe that Player A has $2^n-1$ different possible opponents in the first round. Of those, $2^{n-1}$ are in the opposite half of the draw, and both they and Player A would have to win $n-1$ games to meet; this happens with probability $\\frac{1}{4^{n-1}}$. $2^{n-2}$ are in the same half, but opposite quarters, and both they and Player A would have to win $n-2$ games to meet; this happens with probability $\\frac{1}{4^{n-2}}$. And so on, until we get to the $2^{n-n} = 2^0 = 1$ single player who meets Player A in the first round. Altogether, the probability of Player A and a given Player B meeting eventually is\n\n\\begin{align} \\frac{1}{2^n-1} \\sum_{k=1}^n 2^{n-k}\\frac{1}{4^{n-k}} & = \\frac{1}{2^n-1} \\sum_{k=1}^n \\frac{1}{2^{n-k}} \\\\ & = \\frac{1}{2^n-1} \\sum_{k=0}^{n-1} \\frac{1}{2^k} \\\\ & = \\frac{1}{2^n-1} \\left(2-\\frac{1}{2^{n-1}}\\right) \\\\ & = \\frac{1}{2^n-1} \\times \\frac{2^n-1}{2^{n-1}} \\\\ & = \\frac{1}{2^{n-1}} \\end{align}\n\nI'll come back to edit this if I think (or someone points out) that there's a substantive difference between random reseeding and not reseeding.\n\nThere's also a pretty simple proof by induction: For $n = 1$ (two players), the probability is clearly $\\frac{1}{2^n-1} = 1$. For $n > 1$, the probability that they meet in the first round is $\\frac{1}{2^n-1}$. If they don't meet in the first round (with probability $\\frac{2^n-2}{2^n-1}$), then they must both win their first games (with probability $\\frac14$) to get to the next round. With reseeding, this is clearly the case of $n-1$, so with the premise that the probability of them meeting at that point is $\\frac{1}{2^{n-1-1}} = \\frac{1}{2^{n-2}}$, the overall probability for case $n$ is\n\n\\begin{align} \\frac{1}{2^n-1}+\\frac{2^n-2}{2^n-1} \\times \\frac14 \\times \\frac{1}{2^{n-2}} & = \\frac{1}{2^n-1} \\left(1+\\frac{2^n-2}{2^n}\\right) \\\\ & = \\frac{1}{2^n-1} \\times \\frac{2^{n+1}-2}{2^n} \\\\ & = \\frac{1}{2^n-1} \\times \\frac{2^n-1}{2^{n-1}} \\\\ & = \\frac{1}{2^{n-1}} \\end{align}", "date": "2019-12-14 13:59:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2481670/what-is-the-probability-that-player-a-and-b-play-against-each-other", "openwebmath_score": 0.9961243867874146, "openwebmath_perplexity": 336.79968621321535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9945307260497234, "lm_q2_score": 0.8872045862611168, "lm_q1q2_score": 0.8823522213289129}}
{"url": "http://gmatclub.com/forum/to-find-the-units-digit-of-a-large-number-155363.html", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 30 Aug 2015, 00:00\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# To find the units digit of a large number?\n\nAuthor Message\nTAGS:\nDirector\nJoined: 29 Nov 2012\nPosts: 926\nFollowers: 12\n\nKudos [?]: 528 [0], given: 543\n\nTo find the units digit of a large number?\u00a0[#permalink] \u00a003 Jul 2013, 21:58\n1\nThis post was\nBOOKMARKED\nSo if we are given a question what is the units digit of $$777^{777}$$\n\nwe find the pattern for 7 (7,9,3,1)\n\nthen we divide $$\\frac{777}{4}$$ and the remainder is 1 so the units digit is $$7^1$$ which is 7?\n\nIs this correct?\n_________________\n\nClick +1 Kudos if my post helped...\n\nAmazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/\n\nGMAT Prep software What if scenarios gmat-prep-software-analysis-and-what-if-scenarios-146146.html\n\nDirector\nJoined: 24 Aug 2009\nPosts: 507\nSchools: Harvard, Columbia, Stern, Booth, LSB,\nFollowers: 10\n\nKudos [?]: 489 [0], given: 241\n\nRe: To find the units digit of a large number?\u00a0[#permalink] \u00a003 Jul 2013, 22:08\nfozzzy wrote:\nSo if we are given a question what is the units digit of $$777^{777}$$\n\nwe find the pattern for 7 (7,9,3,1)\n\nthen we divide $$\\frac{777}{4}$$ and the remainder is 1 so the units digit is $$7^1$$ which is 7?\n\nIs this correct?\n\nAbsolutely Correct.\n_________________\n\nIf you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS.\nKudos always maximizes GMATCLUB worth\n-Game Theory\n\nIf you have any question regarding my post, kindly pm me or else I won't be able to reply\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 29110\nFollowers: 4726\n\nKudos [?]: 49758 [2] , given: 7403\n\nRe: To find the units digit of a large number?\u00a0[#permalink] \u00a003 Jul 2013, 22:10\n2\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nfozzzy wrote:\nSo if we are given a question what is the units digit of $$777^{777}$$\n\nwe find the pattern for 7 (7,9,3,1)\n\nthen we divide $$\\frac{777}{4}$$ and the remainder is 1 so the units digit is $$7^1$$ which is 7?\n\nIs this correct?\n\nYes.\n\nThe units digit of 777^777 = the units digit of 7^777.\n\n7^1 has the units digit of 7;\n7^2 has the units digit of 9;\n7^3 has the units digit of 3;\n7^4 has the units digit of 1.\n7^5 has the units digit of 7 AGAIN.\n\nThe units digit repeats in blocks of 4: {7, 9, 3, 1}...\n\nThe remainder of 777/4 is 1, thus the units digit would be the first number from the pattern, so 7.\n\nHope it's clear.\n_________________\nDirector\nJoined: 29 Nov 2012\nPosts: 926\nFollowers: 12\n\nKudos [?]: 528 [0], given: 543\n\nRe: To find the units digit of a large number?\u00a0[#permalink] \u00a003 Jul 2013, 22:23\nOne final question here is another example\n\nIf we have the find the units digit of $$344^{328}$$\n\n$$4^1$$ is 4\n$$4^2$$ is 16\n$$4^3$$ is 4\n\nso the repeating block over here {4,6}\n\nIn this case the remainder is 0 so the units digit of this expression is 6?\n\nso if there are 4 repeating blocks and the remainder is 0 we raise it to the 4th power ( some examples would be 3,7 etc) in this current example its the 2nd power?\n_________________\n\nClick +1 Kudos if my post helped...\n\nAmazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/\n\nGMAT Prep software What if scenarios gmat-prep-software-analysis-and-what-if-scenarios-146146.html\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 29110\nFollowers: 4726\n\nKudos [?]: 49758 [1] , given: 7403\n\nRe: To find the units digit of a large number?\u00a0[#permalink] \u00a003 Jul 2013, 22:26\n1\nKUDOS\nExpert's post\nfozzzy wrote:\nOne final question here is another example\n\nIf we have the find the units digit of $$344^{328}$$\n\n$$4^1$$ is 4\n$$4^2$$ is 16\n$$4^3$$ is 4\n\nso the repeating block over here {4,6}\n\nIn this case the remainder is 0 so the units digit of this expression is 6?\n\nso if there are 4 repeating blocks and the remainder is 0 we raise it to the 4th power ( some examples would be 3,7 etc) in this case its the 2nd power?\n\nYes, if the remainder is 0, then take the last digit from the block.\n_________________\nSenior Manager\nJoined: 19 Apr 2009\nPosts: 435\nLocation: San Francisco, California\nFollowers: 77\n\nKudos [?]: 284 [2] , given: 5\n\nRe: To find the units digit of a large number?\u00a0[#permalink] \u00a007 Jul 2013, 20:00\n2\nKUDOS\nfozzy,\n\nyou are correct in both cases. here is how i like to think about it:\n7^{777} example:\nIf we divide 777 by 4, the quotient is 194 and the remainder is 1.\nThis means that we will have 194 of {7, 9, 3, 1} repeating blocks, and we will have 1 more term left, and the units digit of 7^{777} will be 7, the first term in the repeating block.\n\nWhen we look at the case of 344^{328}, the units digit of powers of 4 cycle as {4, 6}, when we divide 328 by 2, the remainder is 0, this means that there will be exactly 164 blocks consisting of {4,6} without any remainder and the units digit of 344^{328} will be 6.\n\nhere are problems using the same concept(some hard):\n#1) if-n-is-a-positive-integer-what-is-the-remainder-when-82380.html\n#2) what-is-the-remainder-when-7-345-7-11-2-is-divided-by-26794.html\n#3) remainder-when-7-4n-3-6-n-104848.html\n#4) if-3-4n-1-is-divided-by-10-can-the-remainder-be-0-a-4662.html\n#5) what-s-the-remainder-of-2-x-divided-by-10-1-x-is-an-even-9651.html\n\ncheers,\ndabral\n_________________\n\nNew!2016 OFFICIAL GUIDE FOR GMAT REVIEW: Free Video Explanations.\nhttp://www.gmatquantum.com\n\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 6107\nFollowers: 340\n\nKudos [?]: 69 [0], given: 0\n\nRe: To find the units digit of a large number?\u00a0[#permalink] \u00a016 Mar 2015, 15:27\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nIntern\nJoined: 01 Apr 2015\nPosts: 10\nFollowers: 0\n\nKudos [?]: 0 [0], given: 2\n\nRe: To find the units digit of a large number?\u00a0[#permalink] \u00a001 Apr 2015, 23:35\nFinding the UNIT\u2019S DIGIT - please share some more useful tricks if its there.\n_________________\n\nFounder, aptiDude\n\"Get useful study materials at aptiStore!\"\n\nRe: To find the units digit of a large number? \u00a0 [#permalink] 01 Apr 2015, 23:35\nSimilar topics Replies Last post\nSimilar\nTopics:\n3 Shortcut to find Tens digit (Last two digit) of a number 1 17 Nov 2014, 21:27\n1 Units Digit Number Properties & Exponents 1 06 Apr 2014, 06:24\n1 How to find last digit of a number 13 03 Dec 2012, 23:28\n1 Find the sum of 4 digit numbers 3 24 Aug 2012, 01:31\nFinding prime numbers for large numbers 2 27 Mar 2012, 15:29\nDisplay posts from previous: Sort by", "date": "2015-08-30 08:00:57", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/to-find-the-units-digit-of-a-large-number-155363.html", "openwebmath_score": 0.32343626022338867, "openwebmath_perplexity": 2334.657257537312, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 0.9738443862008362, "lm_q2_score": 0.9059898216525935, "lm_q1q2_score": 0.882293101771475}}
{"url": "https://mathhelpboards.com/threads/find-all-the-integer-values-of-k-for-x-3-kx-k-11-0-has-at-least-one-positive-integer-solution.25900/", "text": "# Find all the integer values of k for x^3 - kx + (k + 11) = 0 has at least one positive integer solution\n\n#### Taran\n\n##### New member\nHi, this question was in a year 11 extension maths textbook in the enrichment section. I have the answer as k>17 and k<-11 because I graphed it on GeoGebra. The Graph can be found here: https://ggbm.at/xpegwwtq. While I know the answers I would like to know how to work it out using algebra.\n\nHere is the Question:\nConsider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x\n\nThanks, Taran\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nHi, this question was in a year 11 extension maths textbook in the enrichment section. I have the answer as k>17 and k<-11 because I graphed it on GeoGebra. The Graph can be found here: https://ggbm.at/xpegwwtq. While I know the answers I would like to know how to work it out using algebra.\n\nHere is the Question:\nConsider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x\n\nThanks, Taran\nHi Taran , and welcome to MHB.\n\nIf $x=n$ is a positive integer solution of the equation, then $n^3 - kn + k + 11 = 0$, so that $$k = \\frac{n^3+11}{n-1} = \\frac{(n-1)(n^2+n+1) + 12}{n-1} = n^2+n+1 + \\frac{12}{n-1}.$$ For that to be an integer, $n-1$ must be a factor of $12$. You can then tabulate the possible values of $n$ and $k = \\frac{n^3+11}{n-1}$, as follows: $$\\begin{array}{c|cccccc} n-1&1&2&3&4&6&12 \\\\ n&2&3&4&5&7&13 \\\\ k&19&19&25&34&59&184 \\end{array}.$$ So the only possible values for $k$ are $19,\\ 25,\\ 34,\\ 59,\\ 184$ (which all agree with your condition that $k>17$).\n\nLast edited:\n\n#### MarkFL\n\n##### Pessimist Singularitarian\nStaff member\nHi Taran , and welcome to MHB.\n\nIf $x=n$ is a positive integer solution of the equation, then $n^3 - kn + k + 11 = 0$, so that $$k = \\frac{n^3+11}{n-1} = \\frac{(n-1)(n^2+n+1) + 12}{n-1} = n^2+n+1 + \\frac{12}{n-1}.$$ For that to be an integer, $n-1$ must be a factor of $12$. You can then tabulate the possible values of $n$ and $k = \\frac{n^3+11}{n-1}$, as follows: $$\\begin{array}{c|cccccc} n-1&1&2&3&4&6&12 \\\\ n&2&3&4&5&7&13 \\\\ k&19&19&25&34&59&184 \\end{array}.$$ So the only possible values for $k$ are $19,\\ 25,\\ 34,\\ 59,\\ 184$ (which all agree with your condition that $k>17$).\nHello, Chris!\n\nThis question was posted on another site, and I found your reply so insightful, I took the liberty of posting it there, for the benefit of several there trying to solve it.\n\n#### Taran\n\n##### New member\nHi, Thank you so much!!! This question had my class stumped. That answer makes so much sense. It's been bugging me for a while and I'm very thankful for your help.\n\nThanks again, Taran", "date": "2020-06-04 04:50:00", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-all-the-integer-values-of-k-for-x-3-kx-k-11-0-has-at-least-one-positive-integer-solution.25900/", "openwebmath_score": 0.7407698035240173, "openwebmath_perplexity": 254.31747904766357, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232945094719, "lm_q2_score": 0.897695298265595, "lm_q1q2_score": 0.8822758505070551}}
{"url": "https://math.stackexchange.com/questions/955294/prove-that-int-0-pi-frac-cos-x21-cos-x-sin-x-mathrmdx-i", "text": "# Prove that $\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x =\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x$\n\nIn a related question the following integral was evaluated $$\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x =\\int_0^{\\pi} \\frac{\\mathrm{d}x/2}{1 + \\cos x \\sin x} =\\int_0^{2\\pi} \\frac{\\mathrm{d}x/2}{2 + \\sin x} \\,\\mathrm{d}x =\\int_{-\\infty}^\\infty \\frac{\\mathrm{d}x/2}{1+x+x^2}$$ I noticed something interesting, namely that \\begin{align*} \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x & = \\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x \\\\ & = \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 - \\cos x \\sin x} \\,\\mathrm{d}x = \\int_0^{\\pi} \\frac{(\\sin x)^2}{1 - \\cos x \\sin x} \\,\\mathrm{d}x \\end{align*} The same trivially holds if the upper limits are changed to $\\pi/2$ as well ($x \\mapsto \\pi/2 -u$). But I had problems proving the first equality. Does anyone have some quick hints?\n\n\u2022 consider evaluating $$\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} - \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x = \\int_0^{\\pi} \\frac{\\cos (2x)}{1 +\\frac{1}{2}\\sin(2 x)}\\,\\mathrm{d}x$$ \u2013\u00a0ganeshie8 Oct 2 '14 at 10:27\n\u2022 Are you not satisfied yet with the answers below? \u2013\u00a0Anastasiya-Romanova \u79c0 Oct 7 '14 at 12:07\n\u2022 @N3buchadnezzar Oh sorry, I didn't know that. Get well soon \u1559( ^\u2092^ c) \u2013\u00a0Anastasiya-Romanova \u79c0 Oct 8 '14 at 8:21\n\u2022 Someday ;)${}{}{}$ \u2013\u00a0N3buchadnezzar Oct 8 '14 at 10:12\n\u2022 math.stackexchange.com/questions/61605/\u2026 If you want further discussion ping me in chat. \u2013\u00a0N3buchadnezzar Oct 8 '14 at 10:32\n\nSplit the integral into two terms with limit $\\left[0,\\frac{\\pi}{2}\\right]$ and $\\left[\\frac{\\pi}{2},\\pi\\right]$ \\begin{align} \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x =\\int_0^{\\pi/2} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x +\\int_{\\pi/2}^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x \\\\ \\end{align} Using identity \\begin{align} \\int_a^{b} f(x) \\,\\mathrm{d}x = \\int_a^{b} f(a+b-x) \\,\\mathrm{d}x \\end{align} Also the facts that $\\cos\\left(\\frac{\\pi}{2}-x\\right)=\\sin x$, $\\sin\\left(\\frac{\\pi}{2}-x\\right)=\\cos x$, $\\cos\\left(\\frac{3\\pi}{2}-x\\right)=-\\sin x$, and $\\sin\\left(\\frac{3\\pi}{2}-x\\right)=-\\cos x$, we get \\begin{align} \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x &=\\int_0^{\\pi/2} \\frac{(\\sin x)^2}{1 + \\sin x \\cos x} \\,\\mathrm{d}x +\\int_{\\pi/2}^{\\pi} \\frac{(-\\sin x)^2}{1 + \\sin x \\cos x} \\,\\mathrm{d}x \\\\ &=\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\tag{1} \\end{align}\n\nLet \\begin{align} I=\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x \\end{align} Since \\begin{align} \\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x =\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x \\end{align} then \\begin{align} 2I&=\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x +\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\\\ &=\\int_0^{\\pi} \\frac{(\\cos x)^2 + (\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\\\ I&=\\frac{1}{2}\\int_0^{\\pi} \\frac{1}{1 + \\cos x \\sin x} \\,\\mathrm{d}x\\tag{2} \\end{align}\n\nUsing $(2)$, we get \\begin{align} I&=\\int_0^{\\pi} \\frac{1}{2 + 2\\cos x \\sin x} \\,\\mathrm{d}x\\\\ &=\\int_0^{\\pi} \\frac{1}{2 + \\sin (2x)} \\,\\mathrm{d}x\\qquad\\Rightarrow\\qquad x\\mapsto2x\\\\ &=\\frac{1}{2}\\int_0^{2\\pi} \\frac{1}{2 + \\sin x} \\,\\mathrm{d}x\\tag{3} \\end{align}\n\nSubtract the two integrals in question and get\n\n$$\\int_0^{\\pi} dx \\frac{\\cos{2 x}}{1+\\frac12 \\sin{2 x}} = \\frac12 \\int_0^{2 \\pi} du \\frac{\\cos{u}}{1+\\frac12 \\sin{u}}$$\n\nThis may be shown to be equal to the complex integral\n\n$$-\\frac{i}{4} \\oint_{|z|=1} \\frac{dz}{z} \\frac{z+z^{-1}}{1+\\frac{1}{4 i} (z-z^{-1})} = \\oint_{|z|=1}\\frac{dz}{z} \\frac{z^2+1}{z^2+i 4 z-1}$$\n\nThe poles of the integrand within the unit circle are at $z=0$ and $z=-(2-\\sqrt{3}) i$; their respective residues are $-1$ and $1$. By the residue theorem, therefore, the integral is zero and the two original integrals are equal.\n\nA quick hint is noticing the symmetry. A rigorous proof is that $$\\int\\limits_a^b {\\frac{{f(\\cos x)}}{{g(\\sin x\\cos x)}}dx} = \\int\\limits_a^b {\\frac{{f\\left( {\\sin \\left( {\\frac{\\pi }{2} - x} \\right)} \\right)}}{{g\\left( {\\cos \\left( {\\frac{\\pi }{2} - x} \\right)\\sin \\left( {\\frac{\\pi }{2} - x} \\right)} \\right)}}dx} = \\int\\limits_{\\frac{\\pi }{2} - b}^{\\frac{\\pi }{2} - a} {\\frac{{f\\left( {\\sin u} \\right)}}{{g\\left( {\\cos u\\sin u} \\right)}}du}$$Hope it helps ;)\n\n$$\\frac{\\cos(\\pi/2-x)^2}{1+\\cos(\\pi/2-x)\\sin(\\pi/2-x)} = \\frac{\\sin(x)^2}{1+\\sin(x)\\cos(x)}$$\n\nThe integrands are both periodic with period $\\pi$, so it suffices to verify the identity over any interval of length $\\pi$, including $\\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right]$. Then, since both integrands are even functions, it suffices to check that the identity holds when integrating over $\\left[0, \\frac{\\pi}{2}\\right]$, that is, that $$\\int_0^{\\frac{\\pi}{2}} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x =\\int_0^{\\frac{\\pi}{2}} \\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x.$$\n\nBut one can show this simply by substituting $x = \\frac{\\pi}{2} - u$ on either side.\n\n\u2022 Your answer was the first I got, andthe first one I used. However for later readers who also stumble uppon this integral I feel the answer given by Anastasiya is clearer. \u2013\u00a0N3buchadnezzar Oct 7 '14 at 17:30\n\u2022 The important thing is that you've learned something from posting here, which it seems you have. \u2013\u00a0Travis Oct 8 '14 at 3:23\n\n$$I=\\int_0^{\\pi} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x$$ $$J=\\int_0^{\\pi/2} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x$$\n\nUse $\\int_a^b f(x)dx=\\int_a^nf(a+b-x)dx$ and $\\int_0^{2a}f(x)dx=\\int_0^a f(x)dx+f(2a-x)dx$\n\n$$I=\\int_0^{\\pi/2} \\frac{(\\cos (\\pi-x))^2}{1 + \\cos (\\pi-x) \\sin (\\pi-x)} \\,\\mathrm{d}x+\\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x$$ $$I=\\int_0^{\\pi/2} \\frac{(\\cos x)^2}{1 - \\cos x \\sin x} \\,\\mathrm{d}x+\\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x$$ $$I=\\int_0^{\\pi/2} \\frac{(\\cos (\\pi/2-x))^2}{1 - \\cos (\\pi/2-x) \\sin (\\pi/2-x)} \\,\\mathrm{d}x+\\frac{(\\cos (\\pi/2-x))^2}{1 + \\cos (\\pi/2-x) \\sin (\\pi/2-x)} \\,\\mathrm{d}x$$ $$I=\\int_0^{\\pi/2} \\frac{(\\sin x)^2}{1 - \\sin x \\cos x} \\,\\mathrm{d}x+\\frac{(\\sin x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x$$ $$I=\\int_0^{\\pi/2} \\frac{(\\sin x)^2}{1 - \\sin x \\cos x} \\,\\mathrm{d}x+\\frac{(\\sin (\\pi-x))^2}{1 - \\cos (\\pi-x) \\sin(\\pi-x)} \\,\\mathrm{d}x$$ $$I=\\int_0^{\\pi} \\frac{(\\sin x)^2}{1 - \\sin x \\cos x} \\,\\mathrm{d}x$$\n\nHope you can now show that: $$J=\\int_0^{\\pi/2} \\frac{(\\cos x)^2}{1 + \\cos x \\sin x} \\,\\mathrm{d}x=\\int_0^{\\pi/2} \\frac{(\\cos (\\pi/2-x))^2}{1 + \\cos (\\pi/2-x) \\sin (\\pi/2-x)} \\,\\mathrm{d}x=\\int_0^{\\pi/2} \\frac{(\\sin x)^2}{1 + \\sin x \\cos x} \\,\\mathrm{d}x$$", "date": "2019-06-27 10:41:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/955294/prove-that-int-0-pi-frac-cos-x21-cos-x-sin-x-mathrmdx-i", "openwebmath_score": 0.9998661279678345, "openwebmath_perplexity": 1534.2748462268687, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232930001334, "lm_q2_score": 0.8976952927915968, "lm_q1q2_score": 0.8822758437721561}}
{"url": "http://stats.stackexchange.com/questions/32824/how-to-minimize-residual-sum-of-squares-of-an-exponential-fit", "text": "# How to minimize residual sum of squares of an exponential fit?\n\nI have the following data and would like to fit a negative exponential growth model to it:\n\nDays <- c( 1,5,12,16,22,27,36,43)\nEmissions <- c( 936.76, 1458.68, 1787.23, 1840.04, 1928.97, 1963.63, 1965.37, 1985.71)\nplot(Days, Emissions)\nfit <- nls(Emissions ~ a* (1-exp(-b*Days)), start = list(a = 2000, b = 0.55))\ncurve((y = 1882 * (1 - exp(-0.5108*x))), from = 0, to =45, add = T, col = \"green\", lwd = 4)\n\n\nThe code is working and a fitting line is plotted. However, the fit is visually not ideal, and the residual sum of squares seems to be quite huge (147073).\n\nHow can we improve our fit? Does the data allow a better fit at all?\n\nWe could not find a solution to this challenge on the net. Any direct help or linkage to other websites/posts is greatly appreciated.\n\n-\nIn this case, if you consider a regression model $\\text{Emissions}_i=f(\\text{Days}_i,a,b)+\\epsilon_i$, where $\\epsilon_i\\sim N(0,\\sigma)$, then you obtain similar estimators. By plotting the confidence regions, one can observe how these values are contained in the confindence regions. You cannot expect a perfect fit unless you interpolate the points or use a more flexible nonlinear model. \u2013\u00a0user10525 Jul 23 '12 at 13:12\nI changed the title because \"negative exponential model\" means something different than described in the question. \u2013\u00a0whuber Jul 23 '12 at 13:15\nThanks for making the question clearer (@whuber) and thanks for your answer (@Procrastinator). How can I calculate and plot the confidence regions. And, what would be a more flexible nonlinear model? \u2013\u00a0Strohmi Jul 23 '12 at 13:18\nYou need an additional parameter. See what happens with fit <- nls(Emissions ~ a* (1- u*exp(-b*Days)), start = list(a = 2000, b = 0.1, u=.5)); beta <- coefficients(fit); curve((y = beta[\"a\"] * (1 - beta[\"u\"] * exp(-beta[\"b\"]*x))), add = T). \u2013\u00a0whuber Jul 23 '12 at 13:20\n@whuber - maybe you should post that as an answer? \u2013\u00a0jbowman Jul 23 '12 at 13:25\n\nA (negative) exponential law takes the form $y=-\\exp(-x)$. When you allow for changes of units in the $x$ and $y$ values, though, say to $y = \\alpha y' + \\beta$ and $x = \\gamma x' + \\delta$, then the law will be expressed as\n\n$$\\alpha y' + \\beta = y = -\\exp(-x) = -\\exp(-\\gamma x' - \\delta),$$\n\nwhich algebraically is equivalent to\n\n$$y' = \\frac{-1}{\\alpha} \\exp(-\\gamma x' - \\delta) - \\beta = a\\left(1 - u\\exp(-b x')\\right)$$\n\nusing three parameters $a = -\\beta/\\alpha$, $u = 1/(\\beta\\exp(\\delta))$, and $b = \\gamma$. We can recognize $a$ as a scale parameter for $y$, $b$ as a scale parameter for $x$, and $u$ as deriving from a location parameter for $x$.\n\nAs a rule of thumb, these parameters can be identified at a glance from the plot:\n\n\u2022 The parameter $a$ is the value of the horizontal asymptote, a little less than $2000$.\n\n\u2022 The parameter $u$ is the relative amount the curve rises from the origin to its horizontal asymptote. Here, the the rise therefore is a little less than $2000 - 937$; relatively, that's about $0.55$ of the asymptote.\n\n\u2022 Because $\\exp(-3) \\approx 0.05$, when $x$ equals three times the value of $1/b$ the curve should have risen to about $1-0.05$ or $95\\%$ of its total. $95\\%$ of the rise from $937$ to almost $2000$ places us around $1950$; scanning across the plot indicates this took $20$ to $25$ days. Let's call it $24$ for simplicity, whence $b \\approx 3/24 = 0.125$. (This $95\\%$ method to eyeball an exponential scale is standard in some fields that use exponential plots a lot.)\n\nLet's see what this looks like:\n\nplot(Days, Emissions)\ncurve((y = 2000 * (1 - 0.56 * exp(-0.125*x))), add = T)\n\n\nNot bad for a start! (Even despite typing 0.56 in place of 0.55, which was a crude approximation anyway.) We can polish it with nls:\n\nfit <- nls(Emissions ~ a * (1- u * exp(-b*Days)), start=list(a=2000, b=1/8, u=0.55))\nbeta <- coefficients(fit)\nplot(Days, Emissions)\ncurve((y = beta[\"a\"] * (1 - beta[\"u\"] * exp(-beta[\"b\"]*x))), add = T, col=\"Green\", lwd=2)\n\n\nThe output of nls contains extensive information about parameter uncertainty. E.g., a simple summary provides standard errors of estimates:\n\n> summary(fit)\n\nParameters:\nEstimate Std. Error t value Pr(>|t|)\na 1.969e+03  1.317e+01  149.51 2.54e-10 ***\nb 1.603e-01  1.022e-02   15.69 1.91e-05 ***\nu 6.091e-01  1.613e-02   37.75 2.46e-07 ***\n\n\nWe can read and work with the entire covariance matrix of the estimates, which is useful for estimating simultaneous confidence intervals (at least for large datasets):\n\n> vcov(fit)\na             b             u\na 173.38613624 -8.720531e-02 -2.602935e-02\nb  -0.08720531  1.044004e-04  9.442374e-05\nu  -0.02602935  9.442374e-05  2.603217e-04\n\n\nnls supports profile plots for the parameters, giving more detailed information about their uncertainty:\n\n> plot(profile(fit))\n\n\nHere is one of the three output plots showing variation in $a$:\n\nE.g., a t-value of $2$ corresponds roughly to a 95% two-sided confidence interval; this plot places its endpoints around $1945$ and $1995$.\n\n-\nOh, I almost forgot: res <- residuals(fit); res %*% res tells us that introducing the third parameter $u$ reduces the sum of squares to $2724$ (compared to $147073$ as stated in the question). \u2013\u00a0whuber Jul 23 '12 at 14:23\nAll well and good whuber. But maybe the OP had some reason to pick the exponential model (or maybe it is just because it is well known). I think first the residuals should be looked at for the exponential model. Plot them against potential covariates to see if there is structure there and not just large random noise. Before jumping into more sophisticated models try to see if a fancier model could possibly help. \u2013\u00a0Michael Chernick Jul 23 '12 at 14:25\nWhy don't you look at the original plot, Michael? It will make it abundantly obvious why at least one additional parameter is needed. Please note, too, that in a comment to the question the OP has asked, \"what would be a more flexible nonlinear model?\" One implication of the initial analysis offered in this answer is that it should be considered out of the ordinary to fit an exponential with fewer than three parameters: there must be some inherent constraint operating in such cases (such as an intrinsically determined unit of measurement or an intrinsic location for $x$). \u2013\u00a0whuber Jul 23 '12 at 14:32\nI was not criticizing your answer! I did not see any residual plots. All I was suggesting is that plots of residuals vs potential covariates should be the first step in finding a better model. If I thought I had an answer to put up there I would have given an answer rather than raised my point as a constant. I thought you gave a great response and i was among the ones who gave you a +1. \u2013\u00a0Michael Chernick Jul 23 '12 at 14:58", "date": "2013-05-24 01:16:46", "meta": {"domain": "stackexchange.com", "url": "http://stats.stackexchange.com/questions/32824/how-to-minimize-residual-sum-of-squares-of-an-exponential-fit", "openwebmath_score": 0.789470911026001, "openwebmath_perplexity": 980.0923183511632, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.964321450147636, "lm_q2_score": 0.9149009532527357, "lm_q1q2_score": 0.8822586139821327}}
{"url": "https://scicomp.stackexchange.com/questions/37372/is-there-a-general-threshhold-for-which-a-large-condition-number-becomes-proble/37379", "text": "Is there a general threshhold for which a large condition number becomes \u201cproblematic?\u201d\n\nI think the answer is probably no in the linear algebra community, but I'd say anything above $$10^8$$ and you're starting to lose too much precision, making the situation problematic. In some statistics settings, where they use OLS, I've seen it stated that even values over $$20$$ are problematic, but that seems pretty extreme. How do you guys assess whether a condition number is \"problematic?\"\n\n\u2022 I would consider anything above $1/\\sqrt{\\epsilon_{\\text{mach}}}$ problematic, so it would be floating point system dependent. But it also depends on the size of the linear system. You can create small linear systems (3x3) with a condition number at O(1000) and the solution a computer finds could be out of acceptable tolerance. In general, I tell my students if they are using double precision the danger zone starts at 1000, and they need to be particularly careful if it is above $1/\\sqrt{\\epsilon_{\\text{mach}}}$ \u2013\u00a0Abdullah Ali Sivas May 7 at 13:30\n\u2022 @AbdullahAliSivas Is it also dependent on what method you're using to solve the linear system? If it's a direct method, I imagine a large condition number is (relatively) less problematic than an iterative method? Also, what is the logic behind $\\frac{1}{\\sqrt{\\epsilon_{mach}}}$. I know what machine epsilon is, but I think I'm having trouble seeing the direct connection. \u2013\u00a0David May 7 at 13:35\n\u2022 The condition number does not have much to do with stability of the method you are using. For example, say you are using the conjugate gradient method. The error coming from the matrix is at the mat-vec step. So the issue there would be caused by the condition number of the mat-vec operation not the condition number of the matrix. So both a direct method and CG would struggle similarly in that sense. But that doesn't have anything to do with the fact that an ill-conditioned problem is hard to solve independent of however stable the algorithm is. \u2013\u00a0Abdullah Ali Sivas May 7 at 14:08\n\u2022 $1/\\sqrt{\\epsilon_{\\text{mach}}}$ is some kind of heuristic for me; that is just where I notice things start to break. Like a small perturbation of $\\epsilon_{\\text{mach}}$ to the matrix (or the right-hand side vector) may cause upto an error of $\\sqrt{\\epsilon_{\\text{mach}}}$ even if you assume that you have a perfect algorithm which does not introduce and propagate errors and is not affected by the floating point arithmetic. And my error tolerance is usually at $10^{-8}$ level which is close to $1/\\sqrt{\\epsilon_{\\text{mach}}}$ in double precision :). \u2013\u00a0Abdullah Ali Sivas May 7 at 14:26\n\u2022 @AbdullahAliSivas Why not make all of this into an answer? :-) \u2013\u00a0Wolfgang Bangerth May 7 at 16:10\n\nThis answer is my comments compiled and edited together. I apologize for the repetition.\n\nI would consider anything above $$1/\\sqrt{\\epsilon_{\\text{mach}}}$$ problematic, so it would be floating point system dependent. But it also depends on the size of the linear system. You can create small linear systems (3x3) with a condition number at O(1000) and the solution a computer finds could be out of acceptable tolerance. In general, I tell my students if they are using double precision the danger zone starts at 1000, and they need to be particularly careful if it is above $$1/\\sqrt{\\epsilon_{\\text{mach}}}$$.\n\nI want to note that the condition number does not have much to do with stability of the method you are using. Even when it is connected, for example, if you are using the conjugate gradient method, the error coming from the matrix is at the mat-vec step. So the issue there would be caused by the condition number of the mat-vec operation not the condition number of the matrix.\n\nAs Federico Poloni pointed out in the comments, the condition number of the matrix-vector multiplication problem is bounded from above by the condition number of the matrix $$A$$, e.g. $$\\|A\\|\\|A^{-1}\\|$$. However, I think we should rather consider the slightly more general bound for the condition number of mat-vec $$Ax$$: $$\\|A\\|\\|x\\|/\\|Ax\\|$$. While we can immediately take the condition number of the problem equal to the condition number of the matrix, I usually see it written as $$\\alpha\\kappa(A)$$ with $$\\alpha=(\\|x\\|/\\|Ax\\|)/(\\|A^{-1}\\|)$$. The idea, I believe, is that mat-vec and msolve are inverses of each other, and if one of them is well-conditioned, the other one will be ill-conditioned, similar to multiplication and division (some exceptions apply: $$1\\times x$$ and $$x/1$$ are both well-conditioned). So, for matrices with high condition number msolves, it still may be the case that $$(\\|x\\|/\\|Ax\\|)\\ll \\|A^{-1}\\|$$. Hence, the condition number of the mat-vec operation may be much lower than the condition number of the matrix. As an extreme example, if $$A$$ is singular $$Ax$$ is still meaningful and may be well-conditioned, but $$\\kappa(A)=\\infty$$ and $$A^{-1}x$$ is an ill-posed problem.\n\nSo both a direct method and CG would struggle similarly if the elementary steps of the algorithm have issues related to input errors. But this is not the same thing as the conventional wisdom that an ill-conditioned problem is hard to solve independent of how stable the algorithm you employ is.\n\nNote that my arbitrary choice of $$1/\\sqrt{\\epsilon_{\\text{mach}}}$$ is basically a heuristic for me and, in my experience, it is just where things start to break. As an example, a small perturbation of the relative magnitude $$\\epsilon_{\\text{mach}}$$ to the matrix (or the right-hand side vector) may cause up to an error of $$\\sqrt{\\epsilon_{\\text{mach}}}$$, even if you assume that you have a perfect algorithm which does not introduce and propagate errors and is not affected by the floating point arithmetic. And my error tolerance is usually at $$10^{\u22128}$$ level which is close to $$\\sqrt{\\epsilon_{\\text{mach}}}$$ in double precision :)\n\n\u2022 Thanks for the expanded answer! I still disagree on some points though: (1) if we are speaking about the usual relative condition number, then scalar matrix multiplication and division have both condition number 1, for each $x$. (2) I view matrix multiplication and solving linear systems as perfectly symmetrical operations, when seen as abstract maps: multiplying by $A$ isn't any different than multiplying by $A^{-1}$. The same vector-dependent modification to the condition number can be done for solving linear systems, by multiplying by $\\alpha = ||b|| / ||A^{-1}b|| ||A||$. \u2013\u00a0Federico Poloni May 8 at 7:04\n\u2022 I was thinking more in terms of this answer: math.stackexchange.com/a/3031746 . They are definitely symmetrical when thought as maps, and probably there are matrices for which msolve is well-conditioned but not mat-vec, I just haven't heard of them. \u2013\u00a0Abdullah Ali Sivas May 8 at 7:43", "date": "2021-06-15 01:03:25", "meta": {"domain": "stackexchange.com", "url": "https://scicomp.stackexchange.com/questions/37372/is-there-a-general-threshhold-for-which-a-large-condition-number-becomes-proble/37379", "openwebmath_score": 0.793489396572113, "openwebmath_perplexity": 277.9344189141476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692325496974, "lm_q2_score": 0.903294214513915, "lm_q1q2_score": 0.8822196672558871}}
{"url": "https://themortgagestudent.com/tag/550a36-what-is-the-cube-root-of-125", "text": "# what is the cube root of 125\n\nThe cube root of -8 is written as $$\\sqrt[3]{-8} = -2$$. ... \u221b125: 5 \u221b216: 6 \u221b1,000: 10 \u221b1,000,000: 100 \u221b1,000,000,000: 1000: The calculations were performed using this cube root calculator. When did organ music become associated with baseball? 5\u00b3 = 5 * 5 * 5 = 25 * 5 = 125 So the cube root of 125 is 5. So, in this case the cube root of 125 is 5. Here is the answer to questions like: What is the cube root of 125 or what is the cube root of 125? Someone help me with this ASAP ANSWER QUICK! The nearest previous perfect cube is \u2026 Thus, each edge of the cube is 5 cm long. That number is 5. Calculating n th roots can be done using a similar method, with modifications to deal with n. While computing square roots entirely by hand is tedious. Thus, each edge of the cube is 5 cm long. The cube root of 125 is what number cubes is 125. 80% of questions are answered in under 10 minutes Answers come with explanations, so that \u2026 See next answers. So, in this case the cube root of 125 is 5. Step 1) Set up 125 in pairs of two digits from right to left and attach one set of 00 because we want one decimal: \u2026 Learn more with Brainly! What is plot of the story Sinigang by Marby Villaceran? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. cube root of 125/512 = 5/8. Estimating higher n th roots, even if using a calculator for intermediary steps, is \u2026 The length of a side (edge) of a cube is equal to the cube root of the volume. Does the calculator support fractions? What is the birthday of carmelita divinagracia? While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. The length of a side (edge) of a cube is equal to the cube root of the volume. Learn more with Brainly! 125 can be written as 125(e^0), 125(e^((2pi)i)), 125(e^((4pi)i)) where e is euler\u2019s number, i is the imaginary unit, i^2=-1, and e^((theta)i)=cos(theta)+(i)(sin(theta)) where theta is an angle measured in radians. First we will find all factors under the cube root: 125 has the cube factor of 125. All Rights Reserved. Get free help! Find the interest rate \u2026 that Sally earned from the bank. Having trouble with your homework? Guess: 5.125 27 \u00f7 5.125 = 5.268 (5.125 + 5.268)/2 = 5.197 27 \u00f7 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 \u00f7 5.196 = 5.196 Estimating an n th Root. Is evaporated milk the same thing as condensed milk? The cube root of a number answers the question \"what number can I multiply by itself twice to get this number?\". 5\u00b3 = 5 * 5 * 5 = 25 * 5 = 125 So the cube root of 125 is 5 Just right click on the above image, then choose copy link address, then past it in your HTML. The length of a side (edge) of a cube is equal to the cube root of the volume. For example, 5 is the cube root of 125 because 53 = 5\u20225\u20225 = 125, -5 is cube root of -125 because (-5)3 = (-5)\u2022(-5)\u2022(-5) = -125. Why don't libraries smell like bookstores? Yes, simply enter the fraction as a decimal floating point number and you will get the corresponding cube root. Who is the longest reigning WWE Champion of all time? Not sure about the answer? See also our cube root table from 1 to 1000. How to find the square root of 125 by long division method Here we will show you how to calculate the square root of 125 using the long division method with one decimal place accuracy. New questions in Mathematics. Therefore the cube roots of 125 are 5(e^0),5(e^((2pi)i/3)),5(e^((4pi)i/3)). 'CUBE ROOT OF 125' is a 13 letter phrase starting with C and ending with 5 Crossword clues for 'CUBE ROOT OF 125' Synonyms, crossword answers and other related words for CUBE ROOT OF 125 [five] We hope that the following list of synonyms for the word five will help you to finish your crossword today. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. How long does it take to cook a 23 pound turkey in an oven? 5(e^0)=5(1)=0 How long will the footprints on the moon last? For \u2026 As you can see the radicals are not in their simplest form. Let's check this with \u221b125*1=\u221b125. Now extract and take out the cube root \u221b125 * \u221b1. Inter state form of sales tax income tax? The cube root of -64 is written as $$\\sqrt[3]{-64} = -4$$. The cube root of 125 is 5 so therefore each edge is 5 cm which is about 2 inches A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Please link to this page! Thus, each edge of the cube is 5 cm long. All information in this site is provided \u201cas is\u201d, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. 125 is said to be a perfect cube because 5 x 5 x 5 is equal to 125. Cube of \u221b125=5 which results into 5\u221b1; All radicals are now simplified. For example, 5 is the cube root of 125 because 5 3 = 5\u20225\u20225 = 125, -5 is cube root of -125 because (-5) 3 = (-5)\u2022 (-5)\u2022 (-5) = -125. Therefore, the real cube root of 125 is 5. Copyright \u00a9 2020 Multiply Media, LLC. How will understanding of attitudes and predisposition enhance teaching? What is the conflict of the story of sinigang? So, in this case the cube root of 125 is 5. This is the lost art of how they calculated the square root of 125 by hand before modern technology was invented. That number is 5. The cube root of 125 is what number cubes is 125. Cube Root of 125. Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. Since 125 is a whole number, it is a perfect cube. Who of the proclaimers was married to a little person? ( cube root of 125 is 5 and cube root of 512 is 8) hope it helps you... i want full solution The Brain; Helper; Not sure about the answer? A cube root of a number a is a number x such that x3 = a, in other words, a number x whose cube is a. After 42 months, Sally earned $238 in simple interest. Sally deposited$850 into her bank account for 42 months. WHO WANNA JOIN MY \u2026 How long will it take to cook a 12 pound turkey? What is the contribution of candido bartolome to gymnastics? Cube roots (for integer results 1 through 10) Cube root of 1 is 1; Cube root of 8 is 2; Cube root of 27 is 3; Cube root of 64 is 4; Cube root of 125 is 5; Cube root of 216 is 6; Cube root of 343 is 7; Cube root of 512 is 8; Cube root of 729 is 9 What details make Lochinvar an attractive and romantic figure?", "date": "2021-01-19 15:22:30", "meta": {"domain": "themortgagestudent.com", "url": "https://themortgagestudent.com/tag/550a36-what-is-the-cube-root-of-125", "openwebmath_score": 0.5166188478469849, "openwebmath_perplexity": 636.7021015367188, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9843363512883316, "lm_q2_score": 0.8962513759047848, "lm_q1q2_score": 0.8822128091952628}}
{"url": "https://math.stackexchange.com/questions/2853732/prove-log-2x-log-xy-log-y8-geq-sqrt381", "text": "# Prove: $\\log_2(x)+\\log_x(y)+\\log_y(8)\\geq \\sqrt[3]{81}$\n\nProve that for every $x$,$y$ greater than $1$: $$\\log_2(x)+\\log_x(y)+\\log_y(8)\\geq \\sqrt[3]{81}$$\n\nWhat I've tried has got me to:\n\n$$\\frac{\\log_y(x)}{\\log_y(2)}+\\log_x(y)+3\\log_y(2)\\geq \\sqrt[3]{81}$$\n\nI didn't really get far.. I can't see where I can go from here, especially not what to do with $\\sqrt[3]{81}$.\n\nThis is taken out of the maths entry tests for TAU, so this shouldn't be too hard.\n\n\u2022 $\\log_xy=\\frac1{\\log_yx}$ \u2013\u00a0MalayTheDynamo Jul 16 '18 at 19:04\n\u2022 A neat thing about $\\log_a b = \\frac 1{\\log_b a}$ and $\\frac {\\log_b c}{\\log_b a} = \\log_a c$.... It means $\\log_a b* \\log_b c = \\log_a c$ and so $\\log_a b*\\log_b c* \\log_c d*....... *\\log_y z = \\log_a z$...... \u2013\u00a0fleablood Jul 16 '18 at 19:40\n\nWe have $$\\log_2x+\\log_xy+\\log_y8=\\frac{\\log x}{\\log2}+\\frac{\\log y}{\\log x}+\\frac{3\\log2}{\\log y}$$\n\nand since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.\n\nNow use AM-GM:\n\n$$\\sqrt[3]{\\frac{\\log x}{\\log2}\\cdot\\frac{\\log y}{\\log x}\\cdot\\frac{3\\log2}{\\log y}}\\le\\frac{\\frac{\\log x}{\\log2}+\\frac{\\log y}{\\log x}+\\frac{3\\log2}{\\log y}}3$$ giving $$\\frac{\\log x}{\\log2}+\\frac{\\log y}{\\log x}+\\frac{3\\log2}{\\log y}\\ge3\\sqrt[3]3=\\sqrt[3]{81}$$ as required.\n\n\u2022 While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence? \u2013\u00a0L0wRider Jul 17 '18 at 12:41\n\u2022 @Maxim AM-GM is usually the second inequality that people learn (after $x^2 \\ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it. \u2013\u00a0Ovi Jul 17 '18 at 17:37\n\u2022 @Ovi That's embarrassing. I learned about AM/GM/HM just last week :) \u2013\u00a0TheSimpliFire Jul 17 '18 at 19:22\n\u2022 @TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate. \u2013\u00a0Ovi Jul 17 '18 at 21:39\n\nLet's convert everything to $$\\log$$ base $$2$$ so we have a common something to work with:\n\n$$\\log_2 x + \\dfrac {\\log_2 y}{\\log_2 x} + \\dfrac {\\log_2 8}{\\log_2 y} \\ge \\sqrt [3]{81} = 3 \\sqrt[3]{3}$$\n\nNow let $$a = \\log_2 x$$ and $$b = \\log_2 y$$. The condition $$x, y > 1$$ implies that $$a, b > 0$$. So now we have to prove\n\n$$a + \\dfrac ba + \\dfrac {3}{b} \\ge3 \\sqrt[3]{3}$$\n\nThe cube root in there especially may remind us of AM-GM with $$3$$ terms. And indeed, the inequality $$\\text{AM}\\left(a, \\dfrac ba, \\dfrac 3b\\right) \\ge \\text{GM}\\left(a, \\dfrac ba, \\dfrac 3b\\right)$$ gives exactly what is desired.\n\nHint: Show that $a+\\frac{b}{a}+\\frac{3}{b}\\geq \\sqrt[3]{81}$ for all $a,b>0$, using AM-GM.\n\nTwo things:\n\n$\\log_a b = \\frac 1{\\log_b a}$ and $\\frac {\\log_b c}{\\log_b a} = \\log_a c$ so $\\log_a b\\log_b c = \\frac {\\log_b c}{\\log_b a} = \\log_a c$.\n\nAnd AM-GM says $\\frac {a + b+ c}3 \\ge \\sqrt[3]{abc}$.\n\nSo.....\n\n$\\frac {\\log_2 x + \\log_x y + \\log_y 8}3 \\ge \\sqrt[3]{\\log_2 x \\log_x y \\log_y 8} = \\sqrt[3]{\\log_2 8}$", "date": "2020-02-23 11:52:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2853732/prove-log-2x-log-xy-log-y8-geq-sqrt381", "openwebmath_score": 0.7880185842514038, "openwebmath_perplexity": 535.426234789077, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683484150416, "lm_q2_score": 0.8933094110250331, "lm_q1q2_score": 0.8822040996696056}}
{"url": "https://math.stackexchange.com/questions/560816/find-the-sum-of-the-series-sum-frac1nn1n2", "text": "# Find the sum of the series $\\sum \\frac{1}{n(n+1)(n+2)}$\n\nI got this question in my maths paper\n\nTest the condition for convergence of $$\\sum_{n=1}^\\infty \\frac{1}{n(n+1)(n+2)}$$ and find the sum if it exists.\n\nI managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.\n\nUsing Partial Fraction Decomposition, $$\\frac1{n(n+1)(n+2)}=\\frac An+\\frac B{n+1}+\\frac C{n+2}$$\n\n$$\\implies 1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$ $$\\implies 1=n^2(A+B+C)+n(3A+2B+C)+2A$$\n\nComparing the coefficients of the different powers (namely, $0,1,2$) of $n,$ we get $A=\\frac12,B=-1,C=\\frac12$\n\n$$\\implies\\frac1{n(n+1)(n+2)}=\\frac12\\cdot\\frac1n-\\frac1{n+1}+\\frac12\\cdot\\frac1{n+2}$$ $$=-\\frac12\\left(\\underbrace{\\frac1{n+1}-\\frac1n}\\right)+\\frac12\\left(\\underbrace{\\frac1{n+2}-\\frac1{n+1}}\\right)$$\n\nCan you recognize the two Telescoping series?\n\nHint\n\n$$\\frac{2}{n(n+2)}=\\frac{1}{n}-\\frac{1}{n+2}$$\n\nNow multiply both sides by $\\frac{1}{n+1}$.\n\n\u2022 This is a good hint. \u2013\u00a0Mark Bennet Oct 28 '17 at 17:51\n\nHint. You may write $$\\frac{1}{n(n+1)(n+2)} =\\frac{1}{2}\\left(\\frac{1}{n}-\\frac{1}{(n+1)}\\right)+\\frac{1}{2}\\left(\\frac{1}{(n+2)}-\\frac{1}{(n+1)}\\right)$$ giving two telescoping sums $$\\sum_{n=1}^N\\frac{1}{n(n+1)(n+2)} =\\frac{1}{2}-\\frac{1}{2(N+1)}+\\frac{1}{2(N+2)}-\\frac{1}{4}.$$\n\n\u2022 That is a great clean answer!! +1 Not like that show off of Jack who always has to show his \"ability\" in complicating easy things. \u2013\u00a0Von Neumann Sep 20 '16 at 20:44\n\u2022 @FourierTransform: the benefit of complicating things is that my approach also computes $$\\sum_{k\\geq 1}\\frac{1}{k(k+1)(k+2)(k+3)(k+4)(k+5)}$$ in two steps. It is, in fact, equivalent to telescoping. \u2013\u00a0Jack D'Aurizio Sep 20 '16 at 20:47\n\u2022 @DarioGutierrez Thanks. I just started with your identity:$\\frac{1}{n(n+1)(n+2)} =\\frac{1}{2}(\\frac{1}{n}+\\frac{1}{(n+2)})-\\frac{1}{(n+1)}$, but $\\frac{1}{2}(\\frac{1}{n}+\\frac{1}{(n+2)})-\\frac{1}{(n+1)}=\\frac{1}{2}\\left(\\frac{1}{n}-\\frac{1}{(n+1)}\\right)+\\frac{1}{2}\\left(\\frac{1}{(n+2)}-\\frac{1}{(n+1)}\\right)$. \u2013\u00a0Olivier Oloa Sep 20 '16 at 20:51\n\u2022 @FourierTransform It's always nice to have different points of view. That was what $\\texttt{Jack D'Aurizio}$ did it. Usually, I receive a lot of negative comments or/and down-votes in my own answers because I posted some weird overkill answer. What 'some' people don't understand is that I'm adding something different to the bag of answers. By the way, behind me I have a guy who behaves like a 'MSE-police officer' who down-vote, propose to close, review my $\\LaTeX$, change my editions, etc... a kind of insane user. Also, I enjoy your answers too which are 'full creative'. Best Regards. \u2013\u00a0Felix Marin Sep 20 '16 at 21:43\n\u2022 @FourierTransform We should have different answers and different ideas, else, what would be the point of answering in the first place? Also, different ideas lead to learning and building of intuition, which is one reason I dislike the school system to some level. I agree with Felix, we enjoy both of your answers, and we want both. Sometimes, answers aren't meant for the OP, but for the community. \u2013\u00a0Simply Beautiful Art Sep 20 '16 at 22:04\n\nThere is an alternate method and is as follows.\n\nNotice that $$\\frac{1}{n(n+1)(n+2)} = \\frac{(n-1)!}{(n+2)!} = \\frac{1}{2!} \\, B(n,3)$$ where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes \\begin{align} S &= \\sum_{n=1}^{\\infty} \\frac{1}{n \\, (n+1) \\, (n+2)} \\\\ &= \\frac{1}{2} \\, \\int_{0}^{1} \\left( \\sum_{n=1}^{\\infty} x^{n-1} \\right) \\, (1-x)^{2} \\, dx \\\\ &= \\frac{1}{2} \\, \\int_{0}^{1} \\frac{(1-x)^{2}}{1-x} \\, dx = \\frac{1}{2} \\, \\int_{0}^{1} (1-x) \\, dx \\\\ &= \\frac{1}{4} \\end{align}\n\nThis leads to the known result \\begin{align} \\sum_{n=1}^{\\infty} \\frac{1}{n \\, (n+1) \\, (n+2)} = \\frac{1}{4}. \\end{align}\n\nUsing Euler's beta function, $$S=\\sum_{k\\geq 1}\\frac{(k-1)!}{(k+2)!}=\\sum_{k\\geq 1}\\frac{\\Gamma(k)}{\\Gamma(k+3)}=\\frac{1}{2}\\sum_{k\\geq 1}B(k,3)$$ hence: $$S = \\frac{1}{2}\\int_{0}^{1}(1-x)^2\\sum_{k\\geq 1}x^{k-1}\\,dx=\\frac{1}{2}\\int_{0}^{1}(1-x)\\,dx = \\color{red}{\\frac{1}{4}}.$$ A straightforward generalization of this approach gives the identity: $$\\sum_{n\\geq 1}\\frac{1}{n(n+1)\\cdots(n+N)}=\\color{red}{\\frac{1}{N\\cdot N!}}.$$\n\nWriting\n\n$$\\frac{1}{n(n+1)(n+2)} = \\frac{1}{n+1}\\left(\\frac{1}{2n} - \\frac{1}{2(n+2)}\\right) = \\frac{1}{2n(n+1)} - \\frac{1}{2(n+1)(n+2)},$$\n\nwe see that the sum telescopes to\n\n$$\\frac{1}{2(1)(2)} = \\frac{1}{4}.$$\n\n\u2022 This is the best form to use - simple and straightforward. \u2013\u00a0Mark Bennet Oct 28 '17 at 17:51\n\nAlternatively, take $$\\frac{1}{1-x} = \\sum_{n=1}^{\\infty} x^{n-1},$$ and integrate three times with lower limit $0$, giving \\begin{align*} -\\log{(1-x)} &= \\sum_{n=1}^{\\infty} \\frac{x^n}{n} \\\\ x + (1-x)\\log{(1-x)} &= \\sum_{n=1}^{\\infty} \\frac{x^n}{n(n+1)} \\\\ \\frac{3}{4}x^2 - \\frac{1}{2}x - \\frac{1}{2} (1-x)^2 \\log{(1-x)} &= \\sum_{n=1}^{\\infty} \\frac{x^n}{n(n+1)(n+2)}, \\end{align*} and (as @Clement C reminds me) we then apply Abel's theorem to take the limit as $x \\to 1$, which gives $1/4$ as the answer.\n\n\u2022 Isn't there a bit more to the argument, then? The original power series has radius of convergence $1$, so the equality and all integration theorems only apply on $(-1,1)$. To apply the conclusion to $x=1$ at the end (outside the open disc of convergence of the original series, and where the final LHS is only defined by continuity), don't you need Abel's theorem? \u2013\u00a0Clement C. Mar 26 '15 at 14:12\n\u2022 I suppose, since you also have to take the limit to get $0\\log{0} \"=\" 0$. I'll put a note. \u2013\u00a0Chappers Mar 26 '15 at 14:17\n\nYou are on the right track. Fix $N \\geq 1$. Then \\begin{align} \\sum_{n=1}^{N} \\frac{1}{n \\cdot (n+1) \\cdot (n+2)} &= \\sum_{n=1}^N \\left(\\frac{1}{2n}-\\frac{1}{n+1}+\\frac{1}{2n+4}\\right) = \\frac{1}{2}\\sum_{n=1}^N \\frac{1}{n}-\\sum_{n=1}^{N}\\frac{1}{n+1}+\\frac{1}{2}\\sum_{n=1}^{N} \\frac{1}{n+2}\\\\ &= \\frac{1}{2}\\sum_{n=1}^N \\frac{1}{n}-\\sum_{n=2}^{N+1}\\frac{1}{n}+\\frac{1}{2}\\sum_{n=3}^{N+2} \\frac{1}{n} \\\\ &= \\frac{1}{2}\\left(1+\\frac{1}{2}\\right) + \\frac{1}{2}\\sum_{n=3}^N \\frac{1}{n}-\\left(\\frac{1}{2}+\\frac{1}{N+1}\\right)-\\sum_{n=3}^{N}\\frac{1}{n}+\\\\ &\\quad \\left(\\frac{1}{N+1}+\\frac{1}{N+2}\\right)+\\frac{1}{2}\\sum_{n=3}^{N} \\frac{1}{n} \\\\ \\end{align}\n\nCan you continue from there? (the partial sums cancel out, and you only have a few remaining terms. Taking the limit $N\\to\\infty$ will give you the limit.)\n\n\u2022 Interesting approach... Different from everything I learned.... Thanks! \u2013\u00a0Frank Mar 26 '15 at 14:09\n\nHere's another way to do it:$$\\sum_{n\\ge 1}\\frac{1}{n(n+1)(n+2)}=\\frac{1}{2}\\sum_{n\\ge 1}\\bigg(\\frac{1}{n(n+1)}-\\frac{1}{(n+1)(n+2)}\\bigg)=\\frac{1}{2}\\cdot\\frac{1}{1\\cdot 2}=\\frac{1}{4}.$$One advantage is an easy generalisation to a problem for which the partial fractions decomposition would get thorny:$$\\sum_{n\\ge 1}\\frac{1}{\\sum_{j=0}^k(n+j)}=\\frac{1}{k!\\cdot k}.$$", "date": "2019-05-20 09:03:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/560816/find-the-sum-of-the-series-sum-frac1nn1n2", "openwebmath_score": 0.9998956918716431, "openwebmath_perplexity": 804.6634061231066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683469514966, "lm_q2_score": 0.8933094067644466, "lm_q1q2_score": 0.8822040941545867}}
{"url": "http://nyvb.ilcampanileborgo.it/analytic-trigonometry-pdf.html", "text": "# Analytic Trigonometry Pdf\n\nTrigonometry is a special subject of its own, so you might like to visit: Introduction to Trigonometry; Trigonometry Index. Most of the topics that appear here have already been discussed in the Algebra book and often the text here is a verbatim copy of the text in the other book. Find the length of side AB in the figure below. There are two values of x that satisfy this condition: x=\u00c5\u00c5p\u00c5\u00c5 6 and x=p. Following a unified approach, the authors obtain estimates for these sums similar to the classical I. Algebra And Trigonometry With Analytic Geometry. Euler\u2019s Formula and Trigonometry Peter Woit Department of Mathematics, Columbia University September 10, 2019 These are some notes rst prepared for my Fall 2015 Calculus II class, to give a quick explanation of how to think about trigonometry using Euler\u2019s for-mula. Since angles are so important in the analysis of the muscu-loskeletal system, trigonometry is a very useful biomechanics tool. Clear explanations, an uncluttered and appealing layout, and examples and exercises featuring a variety of real-life applications have made this book popular among students year after year. The trigonometric functions sine and cosine have four important limit properties: You can use these properties to evaluate many limit problems involving the six basic trigonometric functions. Collaborative Project \u2014 Analytic Trigonometry 1. Resources are IEB and CAPS aligned. new and clearer proofs, expansion of some problem sets, a review of Analytic Trigonometry, a few remarkable airbrush figures, and bonuses on pages previously only partially used. 2, Using fundamental identities and verifying trigonometric identities Video 1: Solving trig identities A; AHSMESA, 8:45. The book presents the theory of multiple trigonometric sums constructed by the authors. An \"identity\" is a tautology, an equation or statement that is always true, no matter what. Check the book if it available for your country and user who already subscribe will have. 4 Analytic Trigonometry 69 Chapter 4 Analytic Trigonometry 4. We apply the novel inequality to the generalized trigonometric functions and establish several Redheffer-type inequalities for these functions. Substituting 0 for x, you find that cos x approaches 1 and sin x \u2212 3 approaches \u22123; hence, Example 2: Evaluate. If we know the length of two sides of the triangle, we are able to work out the. Trigonometric Identities Sum and Di erence Formulas sin(x+ y) = sinxcosy+ cosxsiny sin(x y) = sinxcosy cosxsiny cos(x+ y) = cosxcosy sinxsiny cos(x y) = cosxcosy+ sinxsiny tan(x+ y) = tanx+tany 1 tanxtany tan(x y) = tanx tany 1+tanxtany Half-Angle Formulas sin 2 = q 1 cos 2 cos 2 = q 1+cos 2 tan 2 = q 1+cos tan 2 = 1 cosx sinx tan 2 = sin 1+cos. CHAPTER 5: ANALYTIC TRIGONOMETRY 5. So this book is not just about mathematical content but is also about the process of learning and doing mathematics. This study guide provides practice questions for all 34 CLEP exams. 2 Verifying Trigonometric Identities 5. Unit 8 Analytic Trigonometry. Improve your math knowledge with free questions in \"Trigonometric identities I\" and thousands of other math skills. Evaluate the function at each specified value of the independent variable and simply: f(x) = x 2 + 1 (a) f(t 2) (b) f(t + 1)remember: just put x3 for x 3 (place commas between multiple answers) in order. Why you should learn it Fundamenta l trigonometric. 5 Trigonometric Form of a Complex Number Selected Applications Triangles and vectors have many real-life applications. We dare you to prove us wrong. We will use all the ideas we've been building up as we've been studying vectors to be able to solve these questions. Right Triangle Definition. Powered by Create your own unique website with customizable templates. With the knowledge of trigonometry anything from calculus to vector algebra can be solved. Yes, Analytical Trigonometry isn't particularly exciting. net : Allows you online search for PDF Books - ebooks for Free downloads In one place. Similar analyses are studied in the car industry, where numerical simulation is used in virtually every aspect of design and car production. sec x Matches (d). Chapter 5 Analytic Trigonometry Section 5. 294-296, Exercises # 1-105 (every other odd) (4) 4. 2, Using fundamental identities and verifying trigonometric identities Video 1: Solving trig identities A; AHSMESA, 8:45. Round to the nearest tenth. Some examples are. c_n = {1 \\over T}\\int\\limits_T {x(t)e^{ - jn\\omega _0 t} dt} \\quad \\quad Analysis \\cr}  The derivation is similar to that for the Fourier cosine series given above. is in Quadrant II. 3) Equation (16. Free PDF download of RD Sharma Class 11 Solutions Chapter 11 Trigonometric Equations solved by Expert Maths Teachers on NCERTBooks. 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Answers to the text problems are in the (sold-separately) Solution Key. It takes some subjects of arithmetics and geometry as any source. of Units 3 units Course Description: Math 11 is the first of a series of Math courses taken by non-science majors. About ACT math practice problems worksheet pdf. Waqas Quraish rated it it was amazing Oct 02, Stars are assigned as follows:. Unit 5 \u2013 Analytical Trigonometry \u2013 Classwork. 2 Proving Identities 11. Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles 2. Trigonometry Functions and Unit Circle TEST STUDY GUIDE Test covers: Given a right triangle, find 6 trig functions. There is a class of trigonometric functions known as inverse or \u201darc\u201d functions which will do just that: \u201dundo\u201d a regular trigonometric function so as to leave the angle by itself. For the following problems, use the given values to evaluate (if possible) the remaining trigonometric functions. co s 4 x = (cos2 x)2 = a 1 + cos 2x 2 b 2 Use cos 2 u = 1 + cos 2u 2. Trig Identities 1 Trig Identity - Proof (4) 4. The word \u201cTrigonometry\u201d comes from the Greek words: \u2018 Trigonon \u2019 meaning \u2018triangle\u2019 and \u2018 metron \u2019 meaning a \u2018measure\u2019. 2 sin2x \u2014 sin x \u20143 0, 6. Following a unified approach, the authors obtain estimates for these sums similar to the classical I. Average Value of a Function. oop&system analysis design analysis ebook pdf links; trigonometry by loney pdf. Trigonometry, 8th Edition Algebra and Trigonometry, 3rd Edition Just-in-Time Algebra and Trigonometry for Calculus (4th Edition) Algebra and Trigonometry: Structure and Method, Book 2 Algebra and Trigonometry with Analytic Geometry (with CengageNOW Printed Access Card). Trigonometry Functions and Unit Circle TEST STUDY GUIDE Test covers: Given a right triangle, find 6 trig functions. Trigonometry Trigonometry is derived from Greek words trigonon (three angles) and metron ( measure). 25sin 92 x 6. A Correlation of Sullivan Algebra and Trigonometry, \u00a92016 to the Florida Mathematics Standards for Analysis of Functions Honors (Course #1201315) 1 \u2605 = Modeling Standard SE = Student Edition TE = Teacher\u2019s Edition Florida Mathematics Standards for Analysis of Functions Honors (#1201315) Sullivan Algebra & Trigonometry 10th Edition, \u00a92016. Figure 1 To define the trigonometric functions, we may. The applications listed below represent a. c sc 1 sin 1 2 tan sin > 0. Get Algebra And Trigonometry With Analytic Geometry With Cengagenow Printed Access Card Available Titles Cengagenow PDF file for free from our online library Created Date: 20200429033911+01'00'. Trigonometry (Note-1) Pythogoras theorem : In a right- angled triangle, the square of hypotenuse is sum of the squares of the base and the perpendicular. Download PDF Calculus With Trigonometry and Analytic Geometry FREE. The book presents the theory of multiple trigonometric sums constructed by the authors. Sample text. Type your expression into the box to the right. This course is structured to not leave you behind in the dust. Description: Intuitive Guide to Fourier Analysis \u2013 Chapter 1. Essential Math 10 - Trigonometry Problems. As early as in the work of Balthazard et al. 3 sec secxx x 10. EbookNetworking. co s 4 x = (cos2 x)2 = a 1 + cos 2x 2 b 2 Use cos 2 u = 1 + cos 2u 2. UNIT 1: ANALYTICAL METHODS FOR ENGINEERS Unit code: A/601/1401 QCF Level: 4 Credit value: 15 OUTCOME 1 TUTORIAL 2 EXPONENTIAL, TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS Unit content 1 Be able to analyse and model engineering situations and solve problems using algebraic methods. (Precalculus by Larsen, Hostetler) We have numbered the videos for quick reference so it's reasonably obvious that each subsequent video presumes knowledge of the previous videos. Suppose sinx=\u00c5\u00c51\u00c5\u00c5 2. Trigonometry plays a major role in industry, where it allows manufacturers to create everything from automobiles to zigzag scissors. Arc Length, Parametric Curves 57 2. Analytic Trigonometry - Your Complete Guide 4. 4: Multiple-Angle Formulas (19) 7. Refresh and try again. Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles 2. Objectives: All students are expected to learn and demonstrate mastery of the above topics. This math plays a major role in automotive engineering, allowing car. Mathematics Chart. Analytic geometry is a great invention of Descartes and Fermat. Trigonometry 1-3 Definitions of the six trig functions, finding the trig functions when given a point on the terminal side, learning the functions of the TRIGONOMETRY This video is for physics NEET JEE student. 115 Franklin Turnpike, #203, Mahwah, NJ 07430. Surface Area and Volume Review. This page will try to simplify a trigonometric expression. Even though the subject is is easy, it is sometimes complicated for students to get their heads around basics concepts like angles, what pi is, angles in a circle and their use, right triangle using sine and cosine. Chapter 3: Graphing Linear Functions. 4 Sum and Difference Formulas 5. Suppose sinx=\u00c5\u00c51\u00c5\u00c5 2. 7 Trigonometric Equations and Inequalities. Cole offers clear explanations, an uncluttered and appealing layout, and examples and exercises featuring a variety of real-life applications. The Trigonometry problems with solutions that we provide are correct ones that you can rely on. View any or all of the videos to the right. Structural geology uses descriptive geometry, trigonometry, and analytical geometry to portray map and profile planar and linear geological features \u2022Wikipedia -Descriptive geometry is the branch of geometry which allows the representation of three-dimensional objects. 35 PLANE TRIGONOMETRY 36 In general we call a line segment positive if it generated by a point moving from left to right, as AB Figure 15. 4: Multiple-Angle Formulas (19) 7. Analyze functions. Use the hints provided. Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p 2 -2 years of high school algebra or the equivalent. Plane and Spherical Trigonometry with Tables. Created Date: 5/19/2015 1:28:14 PM. Download ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link or read online here in PDF. 1 Applications of Sinusoids 11. A keen aptitude for math improves critical thinking and promotes problem-solving abilities. Read online ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link book now. Besides other fields of mathematics, trig is used in physics, engineering, and chemistry. Post a Comment or Review: Unit 1: Unit 2: Unit 3: Unit 4: Unit 5: Unit 6: Unit 7: EOC: Unit 1-1 : Basic Compass Constructions (Doc, PDF, Key): Unit 1-2 :. 1 Law of Sines 6. com - Stu Schwartz Unit 5 - Analytical Trigonometry - Classwork A) Verifying Trig Identities: Definitions to know: Equality: a statement that is always true. Trigonometry is an important tool for evaluating measurements of height and distance. 12 KB (Last Modified Yesterday at 11:57 AM). 2 Factoring Formulas A. Linear Systems and Matrices VIII. Stewart/Clegg/Watson Calculus: Early Transcendentals, 9e, is now published. We apply the novel inequality to the generalized trigonometric functions and establish several Redheffer-type inequalities for these functions. The questions found in the worksheets are in high standard and practicing these questions will definitely make the students to reach their goal in ACT. TRIGONOMETRIC IDENTITIES Reciprocal identities sinu= 1 cscu cosu= 1 secu tanu= 1 cotu cotu= 1 tanu cscu= 1 sinu secu= 1 cosu Pythagorean Identities sin 2u+cos u= 1 1+tan2 u= sec2 u 1+cot2 u= csc2 u Quotient Identities tanu= sinu cosu cotu= cosu sinu Co-Function Identities sin(\u02c7 2 u) = cosu cos(\u02c7 2 u) = sinu tan(\u02c7 2 u) = cotu cot(\u02c7 2 u. Trigonometry cab be used in many areas such as astronomy and architecture as they aid in calculating. Latest information, websites and videos. sin 2sin 12 xx 2 3. Solid Geometry. Part IV Answer Key. 1 Inverse Functions 24 5. 1 Introduction to Identities 11. Pre-Calculus WS Trig Identities. / Thermal flexural analysis of cross-ply laminated plates using trigonometric shear deformation theory 1003 Latin American Journal of Solids and Structures 10(2013) 1001 \u2013 1023 2 THEORETICAL FORMULATION Consider a rectangular cross-ply laminated plate of length a, width b, and total thickness h com-. 4sin cos 2sinx x x 5sec 2. 429 for a lower limit of -1. Instructor: Sara Jamshidi, McAllister Bldg 419, [email\u00a0protected] An Amusing Equation: From Euler\u2019s formula with angle \u2026, it follows that the equation: ei\u2026 +1 = 0 (2) which involves \ufb01ve interesting math values in one short equation. ;Wang, Frank Y. We dare you to prove us wrong. This is a challenging course, comparable to other Advanced Placement courses; students should be able to pass the AP test or the CLEP test after completing this course. This is a 4 part worksheet: Part I Model Problems. Course Information Math 41: Trigonometry and Analytic Geometry I Course Number: 113341 I Section: 07 I MyMathLab ID: hair97316 I Class Time: MTWF 4:40 PM - 5:30 PM I Location: 105 Wartik Lab. Download PDF Calculus With Trigonometry and Analytic Geometry FREE. Download ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link or read online here in PDF. 1 Using Fundamental Identities 1. Free-eBooks. 2tan2 x csc x \u2013 tan2 x 8. Solid Geometry. This guide covers the story of trigonometry. 1 Inverse Trigonometric Functions The trigonometric functions act as an operator on the variable (angle) x, resulting in an output value y. This discipline combines many of the trigonometric, geometric, and algebraic techniques needed to prepare students for the study of calculus. 417721 0321304349 0. Examples 5 7. cot2 x csc2 x cot2 x 1 18. With the knowledge of trigonometry anything from calculus to vector algebra can be solved. Download ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link or read online here in PDF. tan x csc x sin x cos x 1 sin x 1 cos x 15. Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function. Read online ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link book now. On this page you can read or download 8 4 trigonometry worksheet glencoe geometry answers in PDF format. The text is suitable for a typical introductory Algebra & Trigonometry course, and was developed to be used flexibly. tan 162 x 5. It presented a concise account of the main results then known, but was on a scale which limited the amount of detailed discussion possible. Substituting 0 for x, you find that cos x approaches 1 and sin x \u2212 3 approaches \u22123; hence, Example 2: Evaluate. Theoretical and computational aspects of matrix inverse trigonometric and inverse hyperbolic functions are studied. 2tan2x\u2014 10. Inverse Trig Functions 21. variety over kstudied in algebraic geometry. Trigonometry is a math topic that is introduced in class 10 students. Using Trigonometry to Find Angle Measures Date_____ Period____ Find each angle measure to the nearest degree. Trigonometry. Ana-lysis of biological traces in the form of blood spots is a well-known method applicable for reconstruction and verification of the credi-bility of events. 5 j 4AXlBlf krNimgthst9s O Gr seesse nr 7v1e ed L. 2 Factoring Formulas A. Prepared for the New York State Education Department by Pearson January 2015. Part IV Answer Key. (Precalculus by Larsen, Hostetler) We have numbered the videos for quick reference so it's reasonably obvious that each subsequent video presumes knowledge of the previous videos. Trig Ratios. New Book Geometry and Trigonometry for Calculus (Wiley Self-Teaching Guides). 1) 16 28? 2) 30 24? 3) 45 13? 4) 31 37? 5) 15 37? 6) 18 34? 7) 56 33 65?. Swokowski and Jeffery A. Download ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link or read online here in PDF. The word \u201cTrigonometry\u201d comes from the Greek words: \u2018 Trigonon \u2019 meaning \u2018triangle\u2019 and \u2018 metron \u2019 meaning a \u2018measure\u2019. 3cos 4cos 42 xx 4. Powered by Create your own unique website with customizable templates. This is then applied to calculate certain integrals involving trigonometric. If we know the length of two sides of the triangle, we are able to work out the. 3 Vectors in the Plane 6. Average Value of a Function. Mathematics: Overall Analysis Grade-wise Weightage: 11th (44%), 12th (56%) Chapters with more than one questions: Application of Derivative (3) Matrices and Determinants (2) Vectors and 3D (2) Sequence and Series (2). Functions and their Graphs II. 1) x 10 27\u00b0 2) x 10 25\u00b0 3) x 7 39\u00b0 4) 8 x 46\u00b0 5) x 6 23\u00b0 6) 7 x 46\u00b0 7) x 20 72\u00b0 8) x 12 54\u00b0-1-. It takes place on the x-y plane. Swokowski and Jeffery A. Students will be held responsible for reviewing these concepts by their Math Analysis/Trig teacher. Week 2: Complex analytic functions, harmonic functions, M\u00f6bius transforms. trigonometric functions. TRIGONOMETRY Definition of the Six Trigonometric Functions Right triangle definitions, where 0 2. 4 Trigonometric Identities 10. 2 Law of Cosines. Siyepu International Journal of STEM Education Analysis of errors in derivatives of trigonometric functions Sibawu Witness Siyepu 0 0 Fundani Centre for Higher Education and Training, Cape Peninsula University of Technology , cnr Keizergracht & Tennant Str Zonnebloem 7925, Cape Town 8000 , South Africa Background: This article reports on an analysis of errors that were displayed by students. Trigonometric Equations, trigonometric identities. An Amusing Equation: From Euler\u2019s formula with angle \u2026, it follows that the equation: ei\u2026 +1 = 0 (2) which involves \ufb01ve interesting math values in one short equation. Right Triangle Trigonometry Project Due Date: Background: Trigonometry is the only way to mathematically figure out the length of a side in a right triangle given another side and an angle or to find out the measure of an angle given two sides. 2 Law of Cosines 6. Analytic Trigonometry details the fundamental concepts and underlying principle of analytic geometry. It plays an important role in surveying, navigation, engineering, astronomy and many other branches of physical science. Thanks to Andrew Solomon for this correction. The material covered includes many. Unit 5: Analytic Trig Review Solve each equation. The examination covers the subjects of Algebra, Trigonometry, and Analytic. Department of Mathematics, Purdue University 150 N. This is referred to as \u2018spectral analysis\u2019 or analysis in the \u2018frequency. 1219 83\u00b0 3) tan Y = 0. Trigonometric Ratios of certain Angles. Circular function definitions, where is any angle. Find all of the values of x in the interval @0,2 pD for which 4 sin2 x=1. All books are in clear copy here, and all files are secure so don't worry about it. Basic Trigonometric Equations An equation that contains trigonometric functions is called a trigonometric equation. Below are a few standard hints. There is no \"sum of squares\" formula, i. Analytic geometry is a great invention of Descartes and Fermat. 4 new maths courses and all free material. Given the value of one trig ratio, find the other 5 trig ratios. With the knowledge of trigonometry anything from calculus to vector algebra can be solved. Pines Use this website to download homework and review power points for upcoming tests. 2 Law of Cosines 6. tan x csc x sin x cos x 1 sin x 1 cos x 15. Remembering the de\ufb01nitions 4 6. HINT COLLECT LIKE TERMS HINT EXTRACT SQUARE ROOTS 1. This is a 4 part worksheet: Part I Model Problems. This course is structured to not leave you behind in the dust. Use the hints provided. For the most complete and up-to-date course information, contact the instructor Instructor. identities that it knows about to simplify your expression. China Phone: +86 10 8457 8802. Kouba And brought to you by : eCalculus. 12 KB (Last Modified Yesterday at 11:57 AM). Vedamurthy and Dr. 7) 55 51? 68\u00b0 8) 19 27? 45\u00b0 9) 34 55. Each worksheet contains 50 questions with answers. solu-tions of the planar Laplace equation. Answers Pre-requisite- Chapter 2. 3/9 Solving Trig Equations 3/10 Solving Trig Equations 3/13 Sum and Difference Formulas 3/14 Pi Day 3/15 Optional. CHAPTER 5: ANALYTIC TRIGONOMETRY 5. Problem 1: Students are asked to text in one thing they already know about reciprocal trig functions. Solve right triangles. Comments 1. 39,41,47,49. 2 Factoring Formulas A. Chapter 7: Trigonometric Functions and their Derivatives Chapter 8: Inverse Functions and their Derivatives Chapter 9: Numerical Differentiation, and Non-Differentiable Functions Chapter 10: Review of Differentiation Chapter 11: Application of Differentiation to Solving Equations Chapter 12: The Anti-Derivative. org Resources For The Calculus Student. The skills and concepts are in the areas of Arithmetic, Algebra, Geometry, and Data Analysis. e Chapter 2: Analytics Trigonometry. sin(39) = 5. The Calculus exam is approximately 60% limits and differential calculus and 40% integral calculus. Waqas Quraish rated it it was amazing Oct 02, Stars are assigned as follows:. students to think about where the triangle is applicable to AC circuit analysis, and not just to use it blindly. Cole and Publisher Cengage Learning. 4 Analytic Trigonometry 69 Chapter 4 Analytic Trigonometry 4. sin \u20141=0 7. Learning about analytic trigonometry can be tough. 1 Using Fundamental Identities 5. 2 Verifying Trigonometric Identities 5. 5 Multiple-Angle Formulas (double angle, power reduction) CHAPTER 6: ADDITIONAL TOPICS IN TRIGONOMETRY 6. Lesson 15: Solving Vector Problems in Two Dimensions We can now start to solve problems involving vectors in 2D. Solid Geometry. Find all solutions to the equation in the interval [0,2\ud835\udf0b) a. Save up to 80% by choosing the eTextbook option for ISBN: 9781133037644, 113303764X. Trigonometry is the branch of mathematics concerned with the measurement of the parts, sides, and angles of a triangle. net : Allows you online search for PDF Books - ebooks for Free downloads In one place. Here is the link to the video lecture on Youtube. \u2022 Each of these divisions is divided into pure or abstract, which considers magnitude or quantity abstractly, without relation to matter; and mixed/applied, which treats of magnitude. A South African based, learner centred site for those doing Maths in Grade 10 to 12. In this post we are providing you the Trigonometry question pdf that is having detailed solution. Observe that, in complex analysis, the trig functions can all be de ned in terms of the exponential function. and analysis of a teaching unit aimed to teach trigonometry to 10th grade students in a DGS environment and, at the same time, to induce students to prove the conjectures they get from their explorations. How high is the kite (from the ground)? Basic Steps: 1) Draw a picture 2) Label the parts 3) Isolate the triangle 4) Solve 5) Answer the question. Pines Use this website to download homework and review power points for upcoming tests. Trigonometry can be defined as the calculation part of geometry. Each subcategory in this sections has a collection of calculators to ease your calculation work. edu, (814) 863-9049 O ce Hours: MWF 5:30pm - 6:30pm (after class), R 4:30pm - 5:30pm, or by appointment. All books are in clear copy here, and all files are secure so don't worry about it. Round to the nearest tenth. RESEARCH FRAMEWORK The analysis of students\u2019 answers focused on their abilities to prove the conjectures raised as answer to the activities. (A triangle cannot have more than one right angle) The standard trigonometric ratios can only be used on. Plane and Spherical Trigonometry with Tables. GEOMETRY AND TRIGONOMETRY Summary Sheet \ufeff Geometry & Trigonometry summary: File Size: 467 kb: File Type: pdf: Download File. Trigonometry is used in finding out the heights of tall objects like mountains, trees, etc. The modular approach and the richness of content ensures that the book meets the needs of a variety of programs. Swokowski/Cole's Algebra and Trigonometry with Analytic Geometry, Classic Edition, 12th was bright colored and of course has pictures on the website. Students learn to determine angles and side lengths in 30-60-90 and 45-45-90 right triangles using the law of sines and the law of cosines, as well as how to identify similar triangles and determine proportions using proportionality. variety over kstudied in algebraic geometry. Analytic Trigonometry Virtual Lab File. Unit 5: Analytic Trig Review Solve each equation. This discipline combines many of the trigonometric, geometric, and algebraic techniques needed to prepare students for the study of calculus. The subtle mathematical relationships between the right triangle, the. Numerical Integration 41 1. Chapter 5 Analytic Trigonometry Section 5. One can generalize the notion of a solution of a system of equations by allowing K to be any commutative k-algebra. Derivatives of trig functions 11. For the most complete and up-to-date course information, contact the instructor Instructor. Trigonometry in basic words is the mathematics of triangles and trigonometric functions. org Resources For The Calculus Student. Trigonometric functions (\u00a77. Tutorial 16: Laws of Sines. 1 Using Fundamental Identities 5. We solve this using a specific method. Trigonometry 8 1 1 cos \u03b1 \u03b1 sin \u03b1 cot \u03b1 tan \u03b1 sec \u03b1 csc \u03b1 S1 S2 fig. 2, Using fundamental identities and verifying trigonometric identities Video 1: Solving trig identities A; AHSMESA, 8:45. trigonometric functions. S20 Math 1022 Syllabus Sections 1 - 22. Find all of the values of x in the interval @0,2 pD for which 4 sin2 x=1. no formula for. Finding trigonometric functions of angles in a 30-60-90 right triangle and 45-45-90 right triangle without using a calculator. Free-eBooks. Games, activities and quizzes to help you learn and practice trigonometry, We have games for SOHCAHTOA, Right Triangles, Trig Ratios, Unit Circle, Trig Identities, Trig Formulas, Law of Sines, Law of Cosines, Trigonometric Graphs, Inverse Trigonometry and Quizzes, examples with step by step solutions, worksheets. As early as in the work of Balthazard et al. 2 Law of Cosines 6. Collaborative Project \u2014 Analytic Trigonometry 1. pdf: Download File. On this page you can read or download 8 4 trigonometry worksheet glencoe geometry answers in PDF format. This course is structured to not leave you behind in the dust. Comments 1. Analytic Geometry Study Guide Unit 2: Right Triangle Trigonometry You should be able to: \u2022 Make connections between the angles and sides of right triangles \u2022 Select appropriate trigonometric functions to find the angles/sides of a right triangle \u2022 Use right triangle trigonometry to solve realistic problems Vocabulary and Properties. Objectives: All students are expected to learn and demonstrate mastery of the above topics. This page will try to simplify a trigonometric expression. This math plays a major role in automotive engineering, allowing car. ;Wang, Frank Y. 4 Trigonometric Functions of Real Numbers Oct. All problems have step-by-step written solutions View the lesson. I start off each section with basic definitions and processes you will need to know moving through the course. com only do ebook promotions online and we does not distribute any free download of ebook on this site. Round to the nearest tenth. Read online ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link book now. Brannan et al. Core Unit #4 \u2013 Analytic Trigonometry. Trigonometric Equations, trigonometric identities. Pines Use this website to download homework and review power points for upcoming tests. 1) 16 28? 2) 30 24? 3) 45 13? 4) 31 37? 5) 15 37? 6) 18 34? 7) 56 33 65?. c sc 1 sin 1 2 tan sin > 0. Part III Challenge Problems. The problems have been consistently praised for being at just the right level for precalculus students. Introducing the tangent ratio 2 3. The examination covers the subjects of Algebra, Trigonometry, and Analytic. Technical Report. Trigonometry is based on certain ratios, called trigonometric functions, to be defined in the next chapter. Sinusoidal function, harmonic motion, periodic functions, applications. sec x cos x 1 cos x cos x 1 17. All books are in clear copy here, and all files are secure so don't worry about it. Suggested ebook readers (i) Your COMPUTER (ii) a KINDLE or (iii) an IPAD or (iv) other ebook reader PDF files can be uploaded to an ipad by way of itunes PDF ipad apps for viewing are named--- Kindle, ibook, goodreader,etc Plus many other PDF viewers (which. Unit 5: Analytic Trig Review Solve each equation. Substituting 0 for x, you find that cos x approaches 1 and sin x \u2212 3 approaches \u22123; hence, Example 2: Evaluate. Current search Algebra Trigonometry And Analytic Geometry. no formula for. 1 - Trigonometry review Recall that the three most useful trigonometric ratios are relationships between the lengths of the sides in a right triangle as defined by the following table: Full ratio name Standard abbreviation Ratio sine(A) sin(A) A s nuse O H cosine(A) cos(A) side adjacent to A. (A triangle cannot have more than one right angle) The standard trigonometric ratios can only be used on. Older, obsolete versions are posted below: Math 221 \u2013 First Semester Calculus. no formula for. [pdf] Exploring C By Yashavant Kanetkar free Pdf download www. Trigonometric Functions V. There are three primary ones that you need to understand completely: Sine (sin) Cosine (cos) Tangent (tan) The other three are not used as often and can be derived from the three primary functions. 6820 47\u00b0 6) sin C = 0. As we know that book Trigonometry (11th Edition) has many kinds or. Basics trigonometry problems and answers pdf for grade 10. This study guide provides practice questions for all 34 CLEP exams. Trigonometry: Addition Formulas 18. 417721 0321304349 0. Right Triangle Definition. 4 Vectors and Dot Products 6. Read Online and Download PDF Ebook Algebra And Trigonometry With Analytic Geometry With Cengagenow Printed Access Card Available Titles Cengagenow. Polar Coordinates 22. Princeton Asia (Beijing) Consulting Co. All books are in clear copy here, and all files are secure so don't worry about it. ;Wang, Frank Y. Trigonometry Review with the Unit Circle: All the trig. Concept development moves from the concrete to abstract to engage the student. New Book Geometry and Trigonometry for Calculus (Wiley Self-Teaching Guides). i q QA1lJle lrhiyg1h 8tCsE srUels de3rUvgeOdH. Continuous time trigonometric representation of of CT periodic signals \u2013 Chapter 1. 2 Factoring Formulas A. pdf: File Size: 135 kb: File Type: pdf. TRIGONOMETRY - WORKSHEET WinAtSchool. Save up to 80% by choosing the eTextbook option for ISBN: 9781133037644, 113303764X. Analytical Trigonometry. The modular approach and the richness of content ensures that the book meets the needs of a variety of programs. The power triangle is an interesting application of trigonometry applied to electric circuits. Trigonometric Ratios of certain Angles. University Street, West Lafayette, IN 47907-2067 Phone: (765) 494-1901 - FAX: (765) 494-0548 Contact the Webmaster for technical and content concerns about this webpage. Finding trigonometric functions of angles in a 30-60-90 right triangle and 45-45-90 right triangle without using a calculator. 1: Verifying Trigonometric Identities (20) 7. Structural geology uses descriptive geometry, trigonometry, and analytical geometry to portray map and profile planar and linear geological features \u2022Wikipedia -Descriptive geometry is the branch of geometry which allows the representation of three-dimensional objects. Description: Intuitive Guide to Fourier Analysis \u2013 Chapter 1. Use the identity tan (x) = sin (x) / cos (x) in the left hand side of the given identity. 4 Polynomial Worksheet (pdf) Download. After completing this course, students are expected to be able to solve problems showing all appropriate details in the following topics: 1. pdf File history uploaded by Steve Sweeney 11 months, 2 weeks ago No preview is available for MCR3U - Trigonometry - Review Handout. Free math lessons and math homework help from basic math to algebra, geometry and beyond. 648 Chapter 5 Analytic Trigonometry SOLUTION Our goal is to rewrite cos4x without powers of trigonometric functions greater than 1. Analytic Trigonometry 5. Mathematics - High School. Using the discrete Fourier transform, we represent the discrete input data set as the sum of deterministic continuous trigonometric functions. The word \u201cTrigonometry\u201d comes from the Greek words: \u2018 Trigonon \u2019 meaning \u2018triangle\u2019 and \u2018 metron \u2019 meaning a \u2018measure\u2019. Sinusoidal function, harmonic motion, periodic functions, applications. The best thing to try is using trigonometric identities (see transc. Once you feel you mastered one type of problem you get stumped on the next. functions, identities and formulas, graphs: domain, range and transformations. 4cos2x + 4cosx \u20143 = 0, O < 9. 4KB 2019-10-25. sec x Matches (d). All books are in clear copy here, and all files are secure so don't worry about it. Solve Trig Problems With Double- or Half-Angles. This PDf download consists of the following\u2026 Trigonometry. Basic Trigonometry involves the ratios of the sides of right triangles. Pre-req for Math Analysis/Trigonometry Honors (For students entering Math Analysis/Trig Honors in 2019) Students entering Math Analysis/Trig Honors should make sure they are proficient with these topics before school begins on August 19, 2019. Professor Zygmund's Trigonometric Series, first published in Warsaw in 1935, established itself as a classic. Trigonometry math tests for GCSE maths. Grade 12 trigonometry problems and questions with answers and solutions are presented. 5 Solving Trigonometric Equations 41088_11_p_795-836 10/11/01 2:06 PM Page 795. and analysis of a teaching unit aimed to teach trigonometry to 10th grade students in a DGS environment and, at the same time, to induce students to prove the conjectures they get from their explorations. Learning about analytic trigonometry can be tough. Free PDF download of RD Sharma Class 11 Solutions Chapter 11 Trigonometric Equations solved by Expert Maths Teachers on NCERTBooks. Algebra and Trigonometry with Analytic Geometry 13th Edition by Earl W. 010 Summer II 2019 Trigonometry and Analytic Geometry MCS 212 @10:00 -11:45 AM M - F J Montemayor Disclaimer This syllabus is current and accurate as of its posting date, but will not be updated. simple & compound interest, loan payments, store credit. Formulas Perfect Square Factoring: Difference of Squares: Difference and Sum of Cubes: B. Cole is the author of 'Algebra and Trigonometry with Analytic Geometry (College Algebra and Trigonometry)', published 2011 under ISBN 9780840068521 and ISBN 0840068522. UNIT 1: ANALYTICAL METHODS FOR ENGINEERS Unit code: A/601/1401 QCF Level: 4 Credit value: 15 OUTCOME 1 TUTORIAL 2 EXPONENTIAL, TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS Unit content 1 Be able to analyse and model engineering situations and solve problems using algebraic methods. All books are in clear copy here, and all files are secure so don't worry about it. Page 482 #'s 5-17 odd. 3 sec secxx x 10. 2 Law of Cosines. Simply download the PDF below and start putting in the answers into MyMathLab. Euler's formula states that for any real number x: = \u2061 + \u2061,. Chapter 7: Polynomial Equations and Factoring. 1) sin 75\u00b0 2) tan 105\u00b0 Use the angle difference identity to find the exact value of each. This study guide provides practice questions for all 34 CLEP exams. 417721 0321304349 0. All cheat sheets, round-ups, quick reference cards, quick reference guides and quick reference sheets in one page. 2 Factoring Formulas A. Remembering the de\ufb01nitions 4 6. Trig Definition Math Help. There is no \"sum of squares\" formula, i. Students learn to determine angles and side lengths in 30-60-90 and 45-45-90 right triangles using the law of sines and the law of cosines, as well as how to identify similar triangles and determine proportions using proportionality. The / value at 95% was 1. Pedoe or D. Princeton Asia (Beijing) Consulting Co. Suppose this process is reversed: given a y-value, is it possible to work backward to determine the angle x that produced y? In simplest terms, we wish to solve equations such as. pdf: File Size: 135 kb: File Type: pdf. Without really climbing a tree using trigonometry. help you solve your math problems and understand the concepts behind them! \u00b7\u00b7\u00b7 or use our quick links below \u00b7\u00b7\u00b7 Fraction Calculator. Solving with Trigonometry(Doc, PDF, Key) Georgia Standards of Excellence (Click to Expand) MGSE9-12. PDF | 241. Looking at the graph of the sine function on @0,2 pD, we can find this value. Trigonometry is based on certain ratios, called trigonometric functions, to be defined in the next chapter. Equation Solver. sin 9 = -0. trigonometry and the functions of right triangles. There is a class of trigonometric functions known as inverse or \u201darc\u201d functions which will do just that: \u201dundo\u201d a regular trigonometric function so as to leave the angle by itself. After completing this course, students are expected to be able to solve problems showing all appropriate details in the following topics: 1. Recall that this means that Kis a commutative unitary ring equipped with a structure of vector space over k so that the multiplication law in Kis a bilinear map K K!K. All the exercise of Chapter 4 Inverse Trigonometric Functions RD Sharma Class 12 questions with Solutions to help you to revise complete Syllabus and Score More marks in JEE Mains, JEE Advanced, and Engineering entrance exams. students to think about where the triangle is applicable to AC circuit analysis, and not just to use it blindly. Trigonometry. Analytic geometry is a great invention of Descartes and Fermat. Use the hints provided. Use calculator to determine the value of sine, cosine or tangent. A NALYTIC TRIGONOMETRY is an extension of right triangle trigonometry. It includes 29+ Questions to help you find your strengths and weaknesses prior to taking a Trig or Math Analysis course. The applica-tions listed below represent a small sample of the applications in this chapter. Barnett Ziegler Bylean. Princeton Asia (Beijing) Consulting Co. Apply trigonometric ratios to real-world angles of elevation and depression. There are six functions that are the core of trigonometry. For this reason, it is common to focus attention on the exponential function and push the trig functions to the sidelines. Homework: pgs. , 1998), and. The examination covers the subjects of Algebra, Trigonometry, and Analytic. Then use a graphing utility to graph the function(s) at the right with the given viewing window and domain restrictions to sketch the figure described. Instructor: Sara Jamshidi, McAllister Bldg 419, [email\u00a0protected] 4, including exercises) An integral of a rational function of sinx and cosx can always be reduced to integrating a rational function. The alternate version Stewart/Clegg/Watson Calculus, 9e, will publish later this spring. Trigonometry is the study of angles. Right Triangle Trigonometry Project Due Date: Background: Trigonometry is the only way to mathematically figure out the length of a side in a right triangle given another side and an angle or to find out the measure of an angle given two sides. In the course of the work reference is made to preceding writers from whom assistance has been obtained; besides these. Download ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link or read online here in PDF. One can generalize the notion of a solution of a system of equations by allowing K to be any commutative k-algebra. Our equation becomes two seperate identities and then we solve. Evaluate the expression by identifying the values of the given trigonometric ratios. Trigonometry With Analytic Geometry 13th Printable 2019 books might be more convenient and much easier. This course is designed to review and strengthen your trigonometry knowledge to help prepare you for the scientific calculus sequence, and. 1 Inverse Trigonometric Functions The trigonometric functions act as an operator on the variable (angle) x, resulting in an output value y. Visit Cosmeo for explanations and help with your homework problems!. This is referred to as \u2018spectral analysis\u2019 or analysis in the \u2018frequency. 6820 47\u00b0 6) sin C = 0. Chapter 6 Analytic Trigonometry a. Solution: We first find the missing length of side RS. Analytic Geometry Study Guide Unit 2: Right Triangle Trigonometry You should be able to: \u2022 Make connections between the angles and sides of right triangles \u2022 Select appropriate trigonometric functions to find the angles/sides of a right triangle \u2022 Use right triangle trigonometry to solve realistic problems Vocabulary and Properties. Check this site frequently, this site is designed to help you be successful in this class. Its not exactly the same thing as the book since it is a draft and may have typos etc. Students, teachers, parents, and everyone can find solutions to their math problems instantly. The idea of this book is to give an extensive description of the classical complex analysis, here \u201dclassical\u201d means roughly \u2026 [Download] Complex Analysis PDF | Genial eBooks Download the eBook Complex Analysis in PDF or EPUB format and read it directly on your mobile phone, computer or any device. This study guide provides practice questions for all 34 CLEP exams. you\u2019ll ever need to know in Calculus Objectives: This is your review of trigonometry: angles, six trig. Trigonometric functions. A NALYTIC TRIGONOMETRY is an extension of right triangle trigonometry. These problems are on Pages 3-5 of the Discovering Trig Identities Flipchart. 2 Right Triangle Trigonometry 27 5. Some examples are. Students, teachers, parents, and everyone can find solutions to their math problems instantly. The / value at 95% was 1. pdf: File Size: 135 kb: File Type: pdf. 2tan2x\u2014 10. How high is the kite (from the ground)? Basic Steps: 1) Draw a picture 2) Label the parts 3) Isolate the triangle 4) Solve 5) Answer the question. 1 The inverse sine, cosine, and tangent functions 1. Save up to 80% by choosing the eTextbook option for ISBN: 9781133037644, 113303764X. Download ANALYTIC TRIGONOMETRY WITH APPLICATIONS 11TH EDITION book pdf free download link or read online here in PDF. In ch 1 we first defined angles \u2013 our way of measuring them was based on a circle We then narrowed our focus to angles of a triangle and explored similarity of triangles Slideshow 2169034 by dannon. All books are in clear copy here, and all files are secure so don't worry about it.\n6s7t2f2e9e 8ppg8yo0000 8raopkw6yf szm0cwd85ckdw 0pu74t5i3f16 owd5xyprqagdw q895vvsw89hv1l7 c5igzqypqefx5i8 lherq570emkff ituy879wkx9qf 60hy3jmcj6 nwtmtdyr9ei jr1sdekwd3 tp4m3woeoa6yqmf h3bt9fuajj4ro9 eje4zg4isaq 2kl587yb5r kw9vukepa9 q4tk0hlcv5qa xsasi6mflrztrc wdu1qpnd1rbdtm txpnrc9d3n qk64i8h20a17b xh62ibj59psx icf6v47rvwpgx", "date": "2020-07-06 23:09:17", "meta": {"domain": "ilcampanileborgo.it", "url": "http://nyvb.ilcampanileborgo.it/analytic-trigonometry-pdf.html", "openwebmath_score": 0.5778871178627014, "openwebmath_perplexity": 2061.6495554520557, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9908743631497862, "lm_q2_score": 0.8902942261220292, "lm_q1q2_score": 0.8821697243245974}}
{"url": "https://math.stackexchange.com/questions/2186634/how-to-draw-a-sublattice-to-exhibit-diagonalization/2186851", "text": "# How to draw a sublattice to exhibit diagonalization?\n\nGiven the matrix:\n\n$$A=\\begin{pmatrix} 3 & 1 \\\\ -1 & 2 \\\\ \\end{pmatrix}.$$\n\nLet $V =\\mathbb Z^2$ and $L = AV$. We want to find basis for $V$ and $L$ and draw the sublattice that exhibit the diagonalization.\n\nI found the diagonal form of the matrix to be:\n\n$$\\begin{pmatrix} 1 & 0 \\\\ 0 & -7 \\\\ \\end{pmatrix}$$\n\nI think we need that because the solution says the basis for $V$ is $\\{(1,2) , (0,1)\\}$ while the basis for $L$ is $\\{(1,2),(0,7)\\}$.\n\nBut it says we can conclude that from the sublattice. So, can someone briefly point out how I can sketch the sublattice?\n\nEDIT:\n\nAlso my attempt at drawing it:\n\n\u2022 Do you mean that $A$ is your name for the first matrix you display and $V$ is the module $\\mathbb Z^2$? And how did you find that purported diagonalized matrix? It doesn't have the same determinant as the original one ($-7$ versus $7$), doesn't have the same trace ($-6$ versus $5$), and neither $1$ nor $-7$ are eigenvalues as far as I can see. \u2013\u00a0Henning Makholm Mar 14 '17 at 18:29\n\u2022 @HenningMakholm I think the OP has correctly calculated the Smith normal form of the given matrix. Why should quantities such as the determinant, trace, or eigenvalues be preserved under this transformation? \u2013\u00a0Andr\u00e9 3000 Mar 14 '17 at 20:19\n\u2022 @Quasicoherent: Hmm, possibly. I was thrown off by the matrix being described as the \"diagonalization\" of the matrix. \u2013\u00a0Henning Makholm Mar 14 '17 at 20:22\n\u2022 In my textbook they call it the \"diagonal form\" sorry about that! \u2013\u00a0Lana Mar 14 '17 at 20:23\n\u2022 @HenningMakholm Ah okay, I see why that is confusing. Computing the Smith normal form corresponds to changing bases in both the domain and codomain of the linear map, so it's not like usual diagonalization where we choose the same basis in both the domain and codomain. \u2013\u00a0Andr\u00e9 3000 Mar 14 '17 at 20:25\n\nThe row and column operations you use in computing the Smith normal correspond to invertible matrices $P,Q \\in \\operatorname{GL}_2(\\mathbb{Z})$ such that $$D = PAQ$$ where $D$ is the diagonal matrix you found. (Since $-1$ is invertible in $\\mathbb{Z}$, we can actually take $D$ to be $$D = \\begin{pmatrix} 1 & 0\\\\ 0 & 7\\end{pmatrix}$$ which is what I will use for the rest of the answer.) In this case, I get $$D = \\left(\\begin{array}{rr} 1 & 0 \\\\ 0 & 7 \\end{array}\\right) \\qquad P = \\left(\\begin{array}{rr} 0 & 1 \\\\ 1 & -4 \\end{array}\\right) \\qquad Q =\\left(\\begin{array}{rr} 1 & 2 \\\\ 1 & 1 \\end{array}\\right) \\, .$$\n\nWe can interpret these as change of basis matrices for $V$. (For more on this, see this post or this post.) The change of basis matrix $P$ contains the information we seek: since $P^{-1} = \\begin{pmatrix} 4 & 1\\\\ 1 & 0\\end{pmatrix}$, then the set $$\\{v_1, v_2\\} = \\left\\{\\begin{pmatrix} 4\\\\ 1\\end{pmatrix}, \\begin{pmatrix} 1\\\\ 0\\end{pmatrix}\\right\\}$$ is a basis for $V$ such that $L = v_1 \\mathbb{Z} \\oplus 7v_2 \\mathbb{Z}$.\n\nAs for drawing the sublattice, take a look at pp. 4-5 of this set of notes by Keith Conrad. He draws a lattice and sublattice with respect to so-called unaligned and aligned bases, which I've copied below.\n\nI think this is the sort of picture you have in mind. All right, below is my attempt at drawing the lattice $V$ and its sublattice $L$. The blue parallelograms are the fundamental parallelograms of $V$ and $L$ using my basis, and the red is the same for the book's. This shows quite clearly that both answers are correct.\n\n\u2022 It's {(1,2),(0,1)} that's a basis for v, no? \u2013\u00a0Lana Mar 14 '17 at 23:03\n\u2022 That's not what I got. I'll edit to add a few more details. Could this be a result of you choosing $-7$ rather than $+7$ as the bottom-right entry of $D$? \u2013\u00a0Andr\u00e9 3000 Mar 14 '17 at 23:08\n\u2022 I just added the sample answer \u2013\u00a0Lana Mar 14 '17 at 23:10\n\u2022 ok I tried to draw according to your reference lol.. not the prettiest drawing \u2013\u00a0Lana Mar 14 '17 at 23:17\n\u2022 I have a feeling that both my answer and the book's are correct, meaning that the answer is not unique. You can see that $(4,1)^T$ and $(7,0)^T$ definitely generate $L$ since $(3,-1)^T = (7,0)^T - (4,1)^T$ and $(1,2)^T = 2 \\cdot (4,1)^T - (7,0)^T$, so my answer works. \u2013\u00a0Andr\u00e9 3000 Mar 14 '17 at 23:26", "date": "2019-06-17 19:18:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2186634/how-to-draw-a-sublattice-to-exhibit-diagonalization/2186851", "openwebmath_score": 0.8203396201133728, "openwebmath_perplexity": 195.23123817738258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446494481298, "lm_q2_score": 0.9059898184796793, "lm_q1q2_score": 0.8821121392172221}}
{"url": "https://stonedynamicsco.com/what-does-oiypld/topological-sort-problems-b4c8cf", "text": "A topological sort takes a directed acyclic graph and produces a linear ordering of all its vertices such that if the graph $$G$$ contains an edge $$(v,w)$$ then the vertex $$v$$ comes before the vertex $$w$$ in the ordering. Also try practice problems to test & improve your skill level. if the graph is DAG. The approach is based on: A DAG has at least one vertex with in-degree 0 and one vertex with out-degree 0. There are 2 vertices with the least in-degree. Topological sorting of vertices of a Directed Acyclic Graph is an ordering of the vertices v 1, v 2,... v n in such a way, that if there is an edge directed towards vertex v j from vertex v i, then v i comes before v j. In the previous post, we discussed Topological Sorting and in this post, we are going to discuss two problems based on it. Topological sorting forms the basis of linear-time algorithms for finding the critical path of the project, a sequence of milestones and tasks that controls the length of the overall project schedule. For the given graph, following 2 different topological orderings are possible-, For the given graph, following 4 different topological orderings are possible-. Remove vertex-C since it has the least in-degree. 2/24. SPOJ TOPOSORT - Topological Sorting [difficulty: easy] UVA 10305 - Ordering Tasks [difficulty: easy] UVA 124 - Following Orders [difficulty: easy] UVA 200 - Rare Order [difficulty: easy] Every day he makes a list of things which need to be done and enumerates them from 1 to n. However, some things need to be done before others. We will first create the directed Graph and perform Topological Sort to it and at last we will return the vector which stores the result of Topological Sort.NOTE if Topological Sorting is not possible, return empty vector as it is mentioned in the problem statement.Let\u2019s see the code. P and S must appear before R and Q in topological orderings as per the definition of topological sort. Topological Sort Examples. LeetCode 210 - Course Schedule II Problem : Alien Dictionary Topological Sort will be : b d a c and it will be our result.Let\u2019s see the code. Topological Sort. Excerpt from The Algorithm Design Manual: Topological sorting arises as a natural subproblem in most algorithms on directed acyclic graphs. Your email address will not be published. Topological Sorting for a graph is not possible if the graph is not a DAG. Scheduling problems \u2014 problems that seek to order a sequence of tasks while preserving an order of precedence \u2014 can be solved by performing a topological sort \u2026 then \u2018u\u2019 comes before \u2018v\u2019 in the ordering. If you're thinking Makefile or just Program dependencies, you'd be absolutely correct. Topological sort Medium Accuracy: 40.0% Submissions: 42783 Points: 4 . Which of the following statements is true? \u2022 Question: when is this problem tractable? So, here graph is Directed because [0,1] is not equal to [1,0]. Directed acyclic graphs are used in many applications to indicate the precedence of events. In the previous post, we have seen how to print topological order of a graph using Depth First Search (DFS) algorithm. Remove vertex-3 and its associated edges. Note that for every directed edge u -> v, u comes before v in the ordering. It is quite obvious that character appearing first in the words has higher preference. It is easy to notice that this is exactly the problem of finding topological \u2026 The graph does not have any topological ordering. ... ordering of V such that for any edge (u, v), u comes before v in. It is a simple Topological Sort question. The Topological Sort Problem Let G = (V;E)be a directed acyclic graph (DAG). So, let\u2019s start. Remove vertex-2 and its associated edges. Problem link\u2014 SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 \u2026 The graph is directed because \u2026 (The solution is explained in detail in the linked video lecture.). The goal of topological sortis to produce a topological order of G. COMP3506/7505, Uni of Queensland Topological Sort on a DAG. We wish to organize the tasks into a linear order that allows us to complete them one at a time without violating any prerequisites. Save my name, email, and website in this browser for the next time I comment. Each test case contains two lines. There may exist multiple different topological orderings for a given directed acyclic graph. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge u v, vertex u comes before v in the ordering. Given a sorted dictionary of an alien language having N words and k starting alphabets of standard dictionary. Find any Topological Sorting of that Graph. There's actually a type of topological sorting which is used daily (or hourly) by most developers, albeit implicitly. Scheduling or grouping problems which have dependencies between items are good examples to the problems that can be solved with using this technique. Problem: Find a linear ordering of the vertices of $$V$$ such that for each edge $$(i,j) \\in E$$, vertex $$i$$ is to the left of vertex $$j$$. For example, a \u2026 A typical Makefile looks like this: With this line we define which files depend on other files, or rather, we are defining in which topological orderthe files should be inspected to see if a rebuild \u2026 The main part of the above problem is to analyse it how approach the above problem step by step, how to reach to the point where we can understand that Topological Sort will give us the desired result.Now let\u2019s see another problem to make concept more clear. There are a total of n courses you have to take, labeled from 0 to n-1.Some courses may have prerequisites, for example, to take course 0 you have to first take course 1, which is expressed as a pair [0,1].Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.There may be multiple correct orders, you just need to return one of them. no tags Sandro is a well organised person. The main function of the solution is topological_sort, which initializes DFS variables, launches DFS and receives the answer in the vector ans. Any of the two vertices may be taken first. Then in the next line are sorted space separated values of the alien dictionary.Output:For each test case in a new line output will be 1 if the order of string returned by the function is correct else 0 denoting incorrect string returned.Your Task:You don\u2019t need to read input or print anything. Also go through detailed tutorials to improve your understanding to the topic. Remove vertex-2 since it has the least in-degree. I just finish the Course Schedule II problem and get a neat and simple answer. We learn how to find different possible topological orderings of a given graph. Topological Sort is a linear ordering of the vertices in such a way that, Topological Sorting is possible if and only if the graph is a. Topological Sort or Topological Sorting is a linear ordering of the vertices of a directed acyclic graph. Explanation: There are a total of 4 courses to take. It is a simple Topological Sort question. Review: Topological Sort Problems; Leetcode: Sort Items by Groups Respecting Dependencies Topological Sorting\u00b6 To demonstrate that computer scientists can turn just about anything into a graph problem, let\u2019s consider the difficult problem of stirring up a batch of pancakes. What does DFS Do? While the exact order of the items is unknown (i.e. Find any Topological Sorting of that Graph. From every words, we will find the mismatch and finally we will conclude the graph.Let\u2019s take the another example to make things clear. Problems; tutorial; Topological Sorting; Status; Ranking; TOPOSORT - Topological Sorting. A topological sort takes a directed acyclic graph and produces a linear ordering of all its vertices such that if the graph $$G$$ contains an edge $$(v,w)$$ then the vertex $$v$$ comes before the vertex $$w$$ in the ordering. A topological ordering is possible if and only if the graph has no directed cycles, i.e. Solve practice problems for Topological Sort to test your programming skills. if there is an edge in the DAG going from vertex \u2018u\u2019 to vertex \u2018v\u2019. As a particular case of Alexandro spaces, it can be considered the nite topological spaces. Remove vertex-D since it has the least in-degree. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: His hobbies are Topological Sort pattern is very useful for finding a linear ordering of elements that have dependencies on each other. Your email address will not be published. Explanation: There are total of 2 courses to take. Input: The first line of input takes the number of test cases then T test cases follow . | page 1 Graph : b->d->a->cWe will start Topological Sort from 1st vertex (w). Then, update the in-degree of other vertices. We have discussed two problems on Graph and hope both problem and approach to them is clear to you. Input: The first line of input takes the number of test cases then T test cases follow . An acyclic graph always has a topological sort. Topological sorting has many applications in scheduling, ordering and ranking problems, such as. If you have doubts comment below.It\u2019s Ok if you are not able to solve the problem, just learn the approach to tackle it. Topological Sort Topological sorting problem: given digraph G = (V, E) , find a linear ordering of vertices such that: for any edge (v, w) in E, v precedes w in the ordering A B C F D E A B E C D F Not a valid topological sort! See all topologicalsort problems: #topologicalsort. I think above code doesn\u2019t require any explanation. no tags Sandro is a well organised person. Abhishek is currently pursuing CSE from Heritage Institute of Technology, Kolkata. Consider the following directed acyclic graph-, For this graph, following 4 different topological orderings are possible-, Few important applications of topological sort are-, Find the number of different topological orderings possible for the given graph-, The topological orderings of the above graph are found in the following steps-, There are two vertices with the least in-degree. Note : The input prerequisites is a graph represented by a list of edges, not adjacency matrices.You may assume that there are no duplicate edges in the input prerequisites. Remove vertex-3 since it has the least in-degree. Topological Sort\u00b6 Assume that we need to schedule a series of tasks, such as classes or construction jobs, where we cannot start one task until after its prerequisites are completed. You have solved 0 / 6 problems. Topological Sorting Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge ( u v ) from \u2026 The first line of \u2026 For example, another topological \u2026 Remove vertex-D and its associated edges. Let\u2019s draw the directed graph for the above example. For above example, order should be 0 1 2 3. Given a digraph , DFS traverses all ver-tices of and constructs a forest, together with a set of source vertices; and outputs two time unit arrays, . The first line of each test case contains two integers E and V representing no of edges and the number of vertices. Remove vertex-C and its associated edges. In the above example, From first two words, it is clear that t comes before f. (t->f)From 2nd and 3rd words, it is clear that w comes before e. (w->e)From 3rd and 4th words, it is also clear that r comes before t. (r->t)Lastly, from 4th and 5th words, it is clear that e comes before r. (e->r)From above observation, order will be (w->e->r->t->f).Hope you got the Logic, it is highly recommended to think of the approach to solve this problem before moving ahead. In computer science, a topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. For example, the pictorial representation of the topological order [7, 5, 3, 1, 4, 2, 0, 6] is:. And if the graph contains cycle then it does not form a topological sort, because no node of the cycle can appear before the other nodes of the cycle in the ordering. If it is impossible to finish all courses, return an empty array. \u2022 Output: is there a topological sort that falls in L? Using the DFS for cycle detection. Now, update the in-degree of other vertices. He has a great interest in Data Structures and Algorithms, C++, Language, Competitive Coding, Android Development. Topological Sort Topological sorting problem: given digraph G = (V, E) , find a linear ordering of vertices such that: for any edge (v, w) in E, v precedes w in the ordering A B C F D E A B F C D E Any linear ordering in which all the arrows go to the right is a valid solution. Learning new skills, Content Writing, Competitive Coding, Teaching contents to Beginners. Since, we had constructed the graph, now our job is to find the ordering and for that Topological Sort will help us.We already have the Graph, we will simply apply Topological Sort on it. So the correct course order is [0,1] . a b a b b a a b a b b a... not in L! 14.4.1. I make a simple version for this Course Schedule problem. For example, a topological sorting of the following graph is \u201c5 4 \u2026 Subscribe to see which companies asked this question. Topological Sorting for a graph is not possible if the graph is not a DAG. Topological Sort | Topological Sort Examples. Watch video lectures by visiting our YouTube channel LearnVidFun. There are $n$ variables with unknown values. (d->a)From 3rd and 4th words, a comes before c. (a->c)Lastly, from 4th and 5th words, b comes before d. (b->d)From above observation order will be (b->d->a->c).Let\u2019s see the code for constructing the Graph. Lecture 8: DFS and Topological Sort CLRS 22.3, 22.4 Outline of this Lecture Recalling Depth First Search. Here\u2019s simple Program to implement Topological Sort Algorithm Example in C Programming Language. Implementation of Source Removal Algorithm. Objective : Find the order of characters in the alien language.Note: Many orders may be possible for a particular test case, thus you may return any valid order.Input:The first line of input contains an integer T denoting the no of test cases. Implementations. In a real-world scenario, topological sorting can be utilized to write proper assembly instructions for Lego toys, cars, and buildings. Practice Problems. I came across this problem in my work: We have a set of files that can be thought of as lists of items. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: Vertex-A has the least in-degree. For example, consider below graph: For some variables we know that one of them is less than the other. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- \u2026 | page 1 Hope code is clear, we constructed the graph first and then did Topological Sort of that Graph, since graph is linear, Topological Sort will be unique. | page 1 To practice previous years GATE problems on Topological Sort. Comment: Personally, I prefer DFS solution as it's more straightforward for this problem # Topological Sort | Time: O(m * n) | Space: O(m * n) # m: height of matrix # n: width of matrix # Slower than the DFS approach due to graph and indegree construction. Also since, graph is linear order will be unique.Let\u2019s see a example. graph can contain many topological sorts. Query [2, 0] means course 0 should be completed before course 2, so (0 -> 2) represents course 0 first, then course 2.Now, our objective is to return the ordering of courses one should take to finish all courses. So, following 2 cases are possible-. 2.Initialize a queue with indegree zero vertices. We will be discussing other applications of Graph in next post.That\u2019s it folks..!!! The given graph is a directed acyclic graph. | page 1 Input. So, remove vertex-B and its associated edges. ?Since, problem statement has some \u2018n\u2019 keys along with some sets which are dependent on each other, clearly it is a Graph Question.In the above problem, n represent the number of vertices and prerequisites define the edges between them.Also, [0,1] means course 1 should be completed first, then course 0. Topological Sorting for a graph is not possible if the graph is not a DAG. Problems. R. Rao, CSE 326 6 Step 1: Identify vertices that have no incoming edge \u2022The \u201c in-degree\u201d of these vertices is zero A B C F D E Also go through detailed tutorials to improve your understanding to the topic. To take course 1 you should have finished course 0. the ... \u2013 A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow.com - id: 275642-ZDc1Z Let\u2019s take an example. Running time is about 264ms. To take course 3 you should have finished both courses 1 and 2. Now, the above two cases are continued separately in the similar manner. Topological Sort Algorithms. Another correct ordering is [0,2,1,3]. A common problem in which topological sorting occurs is the following. # - Note that we could have not built graph and explore the neighbors according to matrix in topological sort proccess. So, remove vertex-1 and its associated edges. So, remove vertex-A and its associated edges. The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \\over 4$$ cup of milk. Directed acyclic graphs are used in many applications to indicate the precedence of events. right with no problems) . (b->a)From 2nd and 3rd words, d comes before a. Also go through detailed tutorials to improve your understanding to the topic. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. Detailed tutorial on Topological Sort to improve your understanding of Algorithms. The Problem Statement is very much clear and I don\u2019t think any further explanation is required.NOTE that prerequisite [0,1] means course 1 should be completed first, then course 0.How to approach the problem..? You have to check whether these constraints are contradictory, and if not, output the variables in ascending order (if several answers are possible, output any of them). A topological sort of a graph $$G$$ can be represented as a horizontal line with ordered vertices such that all edges point to the right. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. The time-stamp structure. Given a Directed Graph. An Example. We are all familiar with English Dictionary and we all know that dictionary follows alphabetical order (A-Z).Alien Dictionary also follows a particular alphabetical order, we just need to find the order with the help of given words.There are some N words given which uses k letters, we have to find the order of characters in the alien language.Hope, problem statement is clear to you, please refer to above examples for more clarity. In this task you have to find out whether Sandro can solve all his duties and if so, print the correct order. 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Contents to Beginners ( the solution is topological_sort, which initializes DFS variables, launches DFS and topological will... Of test cases then T test cases follow good examples to the.. Linear ordering of elements that have dependencies on each other Language having N words k! Directed because [ 0,1 ] is not equal to [ 1,0 ] such! The Algorithm Design Manual: topological Sorting and in this browser for above! Pair of words explained in detail in the words has higher preference 3 1 0 \u201d directed,! Print the correct order C++, Language, Competitive Coding, Android Development allows us to complete them one a! For Flood-fill Algorithm to test & improve your skill level as a natural subproblem in most Algorithms directed.", "date": "2021-07-30 08:06:16", "meta": {"domain": "stonedynamicsco.com", "url": "https://stonedynamicsco.com/what-does-oiypld/topological-sort-problems-b4c8cf", "openwebmath_score": 0.3852830231189728, "openwebmath_perplexity": 931.7901930903615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446440948805, "lm_q2_score": 0.905989822921759, "lm_q1q2_score": 0.8821121386922398}}
{"url": "http://milanoporteeserramenti.it/tec/circumference-of-a-semicircle-calculator.html", "text": "Circumference C= Diameter Area Radius \u03c0 Learn the formula for circumference: \u03c0d Learn the formula for the area of a circle: A= \u03c0r 2 Learn the value of pi correct to d. Calculate the circumference of a circle. Calculate the area of the square, and divide that by the surface area of the facing-out side of a brick. How to Use the Material Needed Calculator (Circle) This calculator is easy to use. Paper 4 (Calculator) Circle Theorems Past Paper Questions Arranged by Topic Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. If the perimeter of the semi-circle is needed, add together the lengths of one-half of the circumference + diameter. 14159) and diameter is the distance across the circle (along the edge of the half circle). In the result you will get all unknown variables presented. When you substitute the values into the formula, you obtain: A=\u03c0r 2 /2 A = (3. Set up each circle's circumference to its diameter. Thus the surface area is: SA = 2(pR 2)+ 2pRH. perimeter (circumference) of a semicircle: pi times radius. A connected section of the circumference of a circle. Suppose we want to start from 45 degrees. Online Circumference Memory Game. A Trapezium is a four-sided polygon with two non-adjacent parallel sides or one set of parallel sides. Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles Areas Related to Circles Class 10 Important Questions Very Short Answer (1 Mark) Question 1. Find A, C, r and d of a circle. Calculate the total square inches in column F by using =E2*D2. out Finds circumference of circle ----- Enter radius: 8 Circumference of Circle: 50. Paper 4 (Calculator) Circle Theorems Past Paper Questions Arranged by Topic Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Students learn about finding the circumference of a circle as a multiple of \u03c0 and as a decimal correct to 3 significant figures. A circle has a circumference of \u00eccm. (a) Calculate the area of the whole target. Then tap or click the Calculate button. A = \\frac { 1 }{ 2 } \u00d7 \u03c0 \u00d7 { r }^{ 2 } where \u03c0 = \\frac { 22 }{ 7 } (or) we can also take \u03c0 as 3. Circumference of a circle is defined as the distance around it. Then, do the following: Measure the diameter of each circle by folding in half. Once you have the radius, you can calculate the circumference with 2 x Pi x R. Yes, the area of this rectangle would be \u03c0 R 2. Instructions to Candidates. The result is A=5/4*Pi*a\u00b2. Online Circumference Memory Game. Find (a) the circumference of a circle with radius 6 centimeters and (b) the radius of a circle with circumference 31 meters. Write the formula that you will use (e. \u00a9P j260 r1w2 d 8K fukt 5a8 rS Moof qtcwxaJr1eI fLELOCD. 70 So, the circumference is about 37. It is calculated just by multiplying the diameter of the circle with \u03c0 value. When a circle is divided into two equal halves, it forms two semicircles. !Shown below is a compound shape made from a rectangle and semi-circle. An Arc is a part of the circumference of a circle. The laundry basket that Mrs. I can calculate the volume of cylinders. o Calculate the circumference of a circle. \u2022 The measure of a semicircle, or an arc that is equal to half of a circle, is 180\u00ba. A Quonset roof is typically a half-round shape of any length that can support or accommodate a cathedral or domed interior ceiling structure. 14 and \u201cr\u201d is the radius of the given semi circle. Draw a picture to represent the window. Find the circumference of a circle which has a radius of 4 cm. What is the radius of the cup? Round your answer to a suitable degree of accuracy. Where the perpendicular bisectors of the chords intersect is the expected center of the concentric circles forming the original path. G) Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle. Find the magnitude of the electric field at the point P -- the center of the semicircle. Solution : Here r = 4. The radius is: r=D/2 r=30. We'll use 3. I can calculate the volume of cylinders. Circumference of a semi-circle = = \u03c0r and the perimeter of a semi-circular shape = (\u03c0 + 2) r units. A sector of a circle) is made by drawing two lines from the centre of the circle to the circumference. Calculate the exact area of the shaded sector. Because the semicircle is only half of this divide your answer by 2. lcm 2Tt x 4cm 9. Here we are writing a simple C program that calculates the area and circumference of circle based on the radius value provided by user. How to find the area of a circle: The area of a circle can be found by multiplying pi ( \u03c0 = 3. on your calculator as an approximation for \ud835\udf0b\ud835\udf0b. TOPIC: Circumference, measurement, advanced calculations INTERACTIVITY: Participants will measure the radius of the circumference of the rainbow (semicircle). Answer: The shaded area is 314. Let me ask you a question: How do you calculate the circumference of a circle? As you know, the formula for the circumference of a circle is: C = \u03c0 \u00b7 d or C = 2 \u00b7 \u03c0 \u00b7 r The circumference is equal to \u03c0 times the length of the diameter, or it is equ. Follow these simple instructions: 1. Then tap or click the Calculate button. If you know the length of the radius, you can calculate its perimeter using the following formula:. An easy to use, free perimeter calculator you can use to calculate the perimeter of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, and sector of a circle. Circumference C= Diameter Area Radius \u03c0 Learn the formula for circumference: \u03c0d Learn the formula for the area of a circle: A= \u03c0r 2 Learn the value of pi correct to d. The radius of a circle is 1. We have step-by-step solutions for your textbooks written by Bartleby experts!. [5] A rubber gasket forms a seal between the glass and the window frame. Equation of a Circle Through Three Points Calculator. Students can play by themselves or with a partner. Online Circle Tool. And in this case, r is equal to 6. Step-by-step explanation: Here we note that the shape consists of two small circles and one larger circle. A segment is the shape formed between the chord and the arc. You can use numerous different inputs and choose to submit the radius, the diameter or the circumference of the circle. The radius of a circle is any of the line segments from its center to its perimeter. In this article, we will consider a geometric figure that does not involve line segments, but is instead curved: the circle. The total height of the window is 90cm and the total width is 60cm. Then work out 37. Give your answers to 2 d. In this equation, \"C\" represents the circumference of the circle, and \"d\" represents its diameter. The only basic figure is the semi-circle, that appears four times. Calculate the diameter of the. This circumference to diameter calculator is used to find the diameter of a circle given its circumference. Thus the surface area is: SA = 2(pR 2)+ 2pRH. TOPIC: Circumference, measurement, advanced calculations INTERACTIVITY: Participants will measure the radius of the circumference of the rainbow (semicircle). For example, if the circumference in this example equals 11 inches, multiply 11 by 0. Perimeter Of A Circle With Examples Example 1: If the perimeter of a semi-circular protractor is 66 cm, find the diameter of the protractor (Take \u03c0 = 22/7). If the handle is 28 cm long, what is the diameter of the pot? Give your answer correct to 3 significant figures. 14 or 22 \u2014 7 for \ud835\uded1. Program to calculate Area and Circumference based on user input. Smith can determine where to place it. \u2022 Write the formula: Area = \u03c0r2 \u2022 Substitute the words with the measurements you have been given. A circle has a radius of \ud835\udc5f cm and a circumference of cm. The ratio circumference. Textbook solution for Single Variable Calculus: Early Transcendentals 8th Edition James Stewart Chapter 3. The perimeter of a semicircle is half of the circumference plus the. 0 2cm 0 4cm. It is in circle geometry that the concepts of congruence and similarity. Explain in words how to determine the perimeter of a semicircle. What is 282. Perimeter of a Semicircle Formulas & Calculator. The notation for semicircle and major arc is similar to that of minor arc. A major arc is an arc that is larger than a semicircle. Therefore, all we need to do is add the measurement of each side of the shape and we will have the perimeter. The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half the circumference of a circle of radius $$r$$. Circumference (C): The distance the edge of a circle Radius and Diameter Circumference d: ltd 2Ttr Ex 1) Find the circumference of the circle above. They will also have to move the leprechaun around the rainbow throughout the activity. 14 r is the radius of the semicircle 2. We can find the perimeter of a semicircle with the help of this below formula: where, R = Radius of the semicircle Use our below online perimeter of a semicircle calculator to find the perimeter. \u00a9P j260 r1w2 d 8K fukt 5a8 rS Moof qtcwxaJr1eI fLELOCD. Next we calculate the circumference of the full circle. Circumference is equal to 2 pi r. The centroid of Semicircle $$(0, \\frac{4r}{3\\pi})$$ Circumference of Semicircle Formulas $$P = \\pi r$$ Summary of Semicircle Formula. Its unit length is a portion of the circumference. (3) diagram not to scale (b) Find AOB , giving your answer in radians. Add them to the table above, measure the circumference and diameter for each, and then compute the ratio Circumference Diameter for each object. Find three other circular objects. Our circumference calculator provides circumference of a circle by entering its radius. It depends on what characteristics of the semicircle you are given. Give your answers to 2 d. The surface of the pond is a semicircle of radius 1. Write a formula that expresses the value of in terms of \ud835\udc5f and \ud835\udf0b. The width of each small ring is 3 centimetres. Our area of a sector calculator is flexible and reliable. Give your answer to a sensible degree of accuracy. The region bounded by the circle is also often referred to as a circle (as in when we speak of the area of a circle), and the curve is referred to as the perimeter or the circumference of the circle. An arc is a part of the circumference. So it's equal to 2 pi times 6, which is going to be equal to 12pi. This is a great start. These slightly more advanced circle worksheets require students to calculate area or circumference from different measurements of a circle. The missing length is 13. Andlearning. The earlier worksheets in this section require calculating the area and the circumference given either a radius or a diameter. \u2022 Volume: The volume of a three-dimensional object is a measure of the total space it occupies, measured in cubic units. How to find the area of a circle: The area of a circle can be found by multiplying pi ( \u03c0 = 3. c) The answer of 16. You are not making allowances for the space needed for mortar, if any, and you are including bricks for the spaces at the corners of the box. We have step-by-step solutions for your textbooks written by Bartleby experts!. Find the circumference of the \ufb02 ying disc. Thus the surface area is: SA = 2(pR 2)+ 2pRH. Click here to learn about how to calculate the circumference of a circle. Calculate Areas & Properties of Annulus Rings Ring shapes are essentially formed by one circle within another. MEMORY METER. Examples of Circle and Semi-circle functions We look at a number of examples of circle and semi-circle functions, sketch their graphs, work out their domains and ranges, determine the centre and radius of a circle given its function, etc. The radius for this circle is 45. Another fact that you need to keep in mind about inscribed angles is the fact that any inscribed angle that intercepts a semicircle has to be a right angle. If d=2r where r is the radius, then C=2\u03c0r. 3 Problem 56E. C 5pd or C 52pr 26. This figure consists of a rectangle and semicircle. The length of the circumference is given by the formula: C = \u03c0d, where d is the diameter of the circle. The distance around a rectangle or a square is as you might remember called the perimeter. Assume that the radius of a circle is 21 cm. Clearly stating the units of your answers, calculate do Co(reck the circumference of his \u00a32 coin, giving your answer to an appropriate degree of accuracy. Tangent of circle: a line perpendicular to the radius that touches ONLY one point on the circle. And in this case, r is equal to 6. A chord is a straight line joining any two parts of the circumference. If the handle is 28 cm long, what is the diameter of the pot? Give your answer correct to 3 significant figures. To find the perimeter, P, of a semicircle, you need half of the circle's circumference, plus the semicircle's diameter: P = 1 2 ( 2 \u03c0 r ) + d The 1 2 and 2 cancel each other out, so you can simplify to get this perimeter of a semicircle formula. Calculate the area of one surface of the table mat. o Pi (\u03c0) Objectives. Area of a Semicircle In the case of a circle, the formula for area, A, is A = pi * r^2, where r is the circle\u2019s radius. If someone could please help with the following question? What is the formula to calculate the distance perpendicular from the section line to the nearest edge of the circumference at any point X along the section line? Thanks in advance. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. A sector of a circle) is made by drawing two lines from the centre of the circle to the circumference. 1 foot \u2248 0. Its unit length is a portion of the circumference. The area of a circle is. This is the perimeter of the semi-circle!. perimeter (circumference) of a semicircle: pi times radius. A segment is the shape formed between the chord and the arc. A major arc is an arc that is larger than a semicircle. 14(15) Replace d with 15. WIIat is the perimeter of the figure shown below (semicircle attached to rectangle)? Give the exact answer. Find the circumference of the quad-rant with radius 4. Shapes and Figures. Finally, you can find the diameter - it is simply double the radius: D = 2 * R = 2 * 14 = 28 cm. Please find teh circumference of the basket so Mrs. !Shown below is a compound shape made from a rectangle and semi-circle. When the area is fixed and the perimeter is a minimum, or when the perimeter is fixed and the area is a maximum, use Lagrange multipliers to verify that the length of the rectangle is twice its height. The arc that remains as a part of the semicircle is half as long as the circumference of the whole circle. Write a formula that expresses the value of \ud835\udc36\ud835\udc36 in terms of \ud835\udc5f\ud835\udc5f and \ud835\udf0b\ud835\udf0b. Title: Print Layout - Mathster Created Date: 20140105104657Z. A connected section of the circumference of a circle. This method yields a ceiling estimate of the number of bricks you will need. First, calculate the circumference of the whole circle. 14 (2 cm); C = 6. Find its arc length A. (Answer in units of N/C. That\u2019s because a semi circle has an arc length that is exactly 1/2 of the circumference of a complete circle. A semicircle is an arc that is \u201chalf\u201d of a circle. If we know the diameter then we can calculate the area of a circle using formula: A=\u03c0/4*D\u00b2 (D is the diameter). Draw a diagram of a circle and label all given information 2. G7-M3-Lesson 18: More Problems on Area and Circumference 1. Floor Area Area of the floor. You can also select units of measure for both input data and results. Use this calculation to calculate the area of a circle. 8 11 14 4 7 6 Now that we have all the lengths of the sides, we can simply calculate the perimeter by adding the lengths together to get 4+14+11+8+7+6=50. The perimeter of the semicircle also includes the straight line that connects the two ends of that arc. C = \u03c0d Circumference of a circle C = 3. \u2022 Circumference: The circumference of a circle is the distance around the outside of a circle and is denoted. 7 m and 9557 m2) [2+3](Curved S. out Finds circumference of circle ----- Enter radius: 8 Circumference of Circle: 50. WIIat is the perimeter of the figure shown below (semicircle attached to rectangle)? Give the exact answer. To calculate area and circumference we must know the radius of circle. 24 Circumference of a Semi-circle formula Circumference of a Semi-circle = \u03c0 * radius C Program to find the circumference of a semi-circle. DCO is a straight line. The radius of a circle is 1. What is 282. This is also known as the longest chord of the circle. ) cm\u00b2 (to. Radius: r Diameter: d Circumference: C Area: K : d = 2r C = 2 Pi r = Pi d K = Pi r 2 = Pi d 2 /4 C = 2 sqrt(Pi K) K = C 2 /4 Pi = Cr/2 To read about circles, visit The Geometry Center. We'll use 3. The calculations are done \"live\": How to Calculate the Area. Program to calculate Area and Circumference based on user input. Find the diameter. Leave your answer in terms of \u03c0 1 Area = \u220f x 11 x 11 = 121\u220f cm2 Diameter of circle = 22 cm Circumference = 22\u220f cm Circumference = 6\u03c0 m Area = 32 \u03c0 m2 = 9 \u03c0 m2 Circle = \u03c0 x 52 = 25\u03c0 Semi-circle = (\u03c0 x 22) / 2 = 4. If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, calculate the diameter of the [\u2026]. A sector is the part of a circle between two radii. Contrast radius and diameter. Show Answer. The circle has a diameter of 30 cm. If you have the radius instead of the diameter, multiply it by 2 to get the diameter. Magnetic field around a circular wire is calculated by the formula; B=2\u03c0k. [5] A rubber gasket forms a seal between the glass and the window frame. Yes, the area of this rectangle would be \u03c0 R 2. For interactive applets, worksheets, and more videos, please visit http://www. Do NOT confuse a 'sector' with a 'segment'. Input: Store the constant pI in M1: M1 = 3. asked by tom on February 7, 2014; Math. As a semicircle is half of a circle, the length of the arc of the semicircle will be \ud835\udf0b\ud835\udc51 divided by two. That is, we want -90 + a / total_semi_circles = -45, so a / total_semi_circles = 45. 7cm 5cm Ttx8 = 25. lcm 2Tt x 4cm 9. You can get your required area of the section by just putting radius value and angle. Using \u201c \u201d to represent the diameter of the circle, write an algebraic expression that will result in the perimeter of a. Find the diameter or radius of a circle using the formulas: C = \u03c0d; C = 2\u03c0r. Give your answer to a sensible degree of accuracy. Find three other circular objects. For a complete analysis of a circular arc please check out The Complete Circular Arc Calculator. Circumference of a circle 1 Work out the circumference of each circle. Use this calculation to calculate the area of a circle. A line that is drawn straight through the midpoint of a circle and that has its end points on the circle border is called the diameter (d). Ex2) 14 in. The figure below is in the shape of a semicircle. Ellipse In geometry, an ellipse is a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane that does not. ! C=2\"r) 3. How to Use the Material Needed Calculator (Circle) This calculator is easy to use. Need to subtract a semicircle? BC. Write a formula that expresses the value of in terms of \ud835\udc5f and \ud835\udf0b. Area of a Circle Calculator. If you need an anticipatory set you might want to use this online circumference memory game. In order to find the area of semi circle use the below given formula. Enter the radius: 1 The area of circle is: 3. (AQA June 2007 Intermediate Paper 1 NO Calculator) A semi-circle is cut from a circle. Define tangent. Its length is always less than half of the circumference. Note that measuring circumference requires the use of \ud835\udf0b, an irrational number customarily approximated as 3. 2) Chemistry periodic calculator. Unit 2: Two-Dimensional Geometry. The semi-circle has a diameter of 20 cm. : We know the diameter of the semi circles = the width of the rectangle, y Therefore the circumference of the semi circles = pi*y The total perimeter divide both sides by 2: total area = 2 semicircle area + rectangular potion area A = Replace x A = is the area as a function of y:. Calculate Perimeter Of a Square Calculate Perimeter Of a Rectangle Calculate Perimeter Of a Triangle. The ratio of the circumference C to diameter d of both circles simplify to the same value, 3. Calculate E at center of semicircle: A uniformly charged insulating rod of length L = 12 cm is bent into the shape of a semicircle as shown. Arcs are measured in two ways: as the measure of the central angle , or as the length of the arc itself. I can calculate the radius and diameter of a circle when I know the Circumference. Area and Circumference Worksheets. The perimeter formula is one of the easier formulas to remember in math! The perimeter of a shape is the distance around the outside of the shape. 1416\u2026 Browse more Topics Under Areas Related To Circles. In the result you will get all unknown variables presented. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. asked by tom on February 7, 2014; Math. 19 8 cm 5 cm 18 A wheel of radius 20 cm rolls without slipping on level ground. I suspect you may be visiting this page to calculate the area of an arc, or piece of a circle, because semi circle can sometimes be considered ambiguous. An arc is a segment of a 6 cm Chapter 50 cm A semicircle is of a whole circle. Circumference calculator is a free tool used to calculate the circumference of a circle when the radius is given. Circumference is the linear distance around the circle edge. Solve for a missing angle in a tangent problem. The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half the circumference of a circle of radius $$r$$. If r is the radius of a circle, then (i) Circumference = 2\u03c0r or \u03c0d, where d = 2r is the diameter of the circle. The other two sides should meet at a vertex somewhere on the circumference. Perform the following steps: 1. Solutions for the assessment Circles, Perimeters and Sectors 1) Area = 56. The radius of this circle is 8 cm. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. The diameter of a circle is known as the straight line segment which passes through the center of the circle. \u2022 Volume: The volume of a three-dimensional object is a measure of the total space it occupies, measured in cubic units. Area of a Circle Calculator. Revision exercise 3. The distance around a circle is called the circumference. A semicircle is an arc that is \u201chalf\u201d of a circle. You can also select units of measure for both input data and results. your calculator as an approximation for \ud835\udf0b. Find the perimeter of the shape. The area of a semicircle of radius r is given by A = int_0^rint_(-sqrt(r^2-x^2))^(sqrt(r^2-x^2))dxdy (1) = 2int_0^rsqrt(r^2-x^2)dx (2) = 1/2pir^2. She wants to move the basket to another place in her classroom. The circumference of the traf\ufb01 c circle is approximately 16. If you know the length of the radius, you can calculate its perimeter using the following formula:. The diameter of a circle is known as the straight line segment which passes through the center of the circle. A line that is drawn straight through the midpoint of a circle and that has its end points on the circle border is called the diameter (d). A semicircle can be used to construct the arithmetic and geometric means of two lengths using straight-edge and compass. When a circle is divided into two equal halves, it forms two semicircles. You are given a semicircle of radius 1 ( see the picture on the left ). Circumference Word Problems For each problem, you are expected to complete all of the following: 1. C 5pd or C 52pr 26. The circumference is the outside edge of the circle. 3 divide circumference in inches by the width of the wallblocks you want to use. The handle of a paint pot is half the circumference of the pot (a semi\u2013circle). Semicircle Radius using Circumference Calculation. Draw the diagram (as above on right) to understand the problem and arrive at the required answer. If you want to calculate the outer boundary of a semicircle - the circumference or perimeter of a semicircle - you need to be careful to not fall into any traps. A semicircle is exactly half a circle. your calculator as an approximation for \ud835\udf0b. The POWER function will take any number and raise it to the power of any other number. Semicircle is half of a circle. 14159, which is equal to the ratio of the circumference of any circle to its diameter. Calculate E at center of semicircle: A uniformly charged insulating rod of length L = 12 cm is bent into the shape of a semicircle as shown. Perimeter Of Sector calculator provides for the same. Find the circumference of the circles below. \u2022 The measure of a semicircle, or an arc that is equal to half of a circle, is 180\u00ba. Then r r r is a constant, and its derivative is 0 0. Work out the circumference of the circle. the areas of two circles are in a ratio 4:9 find the ratio between their circumference AB is a line segment whose midpoint is M. Need to subtract a semicircle? BC. Leave your answer in terms of \u03c0 1 Area = \u220f x 11 x 11 = 121\u220f cm2 Diameter of circle = 22 cm Circumference = 22\u220f cm Circumference = 6\u03c0 m Area = 32 \u03c0 m2 = 9 \u03c0 m2 Circle = \u03c0 x 52 = 25\u03c0 Semi-circle = (\u03c0 x 22) / 2 = 4. The figure shows a circle within a square. Circumference of a circle (C) = 2\u03c0r, where \u03c0 is pi and \u201cr\u201d is radius. 56 This is the length of just the curvy part of the semi-circle. Just enter the measurement you know. 1% Convert the percentages from a pie chart to central angle measures?. The result is A=5/4*Pi*a\u00b2. Perimeter and Area of Circle and Semicircle; Arcs of a Circle. In the result you will get all unknown variables presented. The area of a circle will be shown in the selected units. Another circle is inscribed in the semi\u2013circle so that it touches the diameter at point A and also touches the circumference of the semi\u2013 circle. The area of a semi-circle is 1/2 \u03c0r 2. The rectangle has a width of 15 meters and a height of 12 meters. Click here to learn about how to calculate the circumference of a circle. 3 divide circumference in inches by the width of the wallblocks you want to use. 14 r is the radius of the semicircle 2. Circumference = 2 \u00d7 \u03c0 \u00d7 radius = 2 \u00d7 3 \u00d7 21 [Using 3 as an estimated value for \u03c0]. The perimeter of the semicircle also includes the straight line that connects the two ends of that arc. Area and Circumference of a Circle (Grades 6-7) Given a Slope and Points, calculate the missing coordinate Determine if the Points are Collinear. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. Its unit length is a portion of the circumference. Answer by noguf (162) (pi)x(diameter)/2 + diameter. A semicircle is an arc that is \u201chalf\u201d of a circle. Circumference of a semi-circle = = \u03c0r and the perimeter of a semi-circular shape = (\u03c0 + 2) r units. More About Arc. Circumference to Diameter Calculator. It can also be thought of as a sector with an angle of 180 degrees. Step 1: say the radius of the field is r, then the total area of the field = (r 2)/2. In the article below, we provide the semicircle. Degree measure of a minor arc: Defined as the same as the measure of its corresponding central angle. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. The circumference of a circle is found using the circumference formula: Circumference = 2 * \u03c0 * radius So, all you have to do is plug in the radius into the formula, and approximate \u03c0 to be 3. Contrast radius and diameter. Perimeter Formulas and Circumference of a Circle Formula. 28 cm 2 The circumference of a bicycle wheel is 50. The ratio circumference. Determine the length of this semicircle, as measured along the curve. A compass is an instrument used to draw circles or the parts of circles called arcs. You can click on the link about the calculator to learn more about the area of semi circles and it\u2019s relationship to arc length. The circumference of a circle is an arc measuring 360 o. From geometry, you know that \"A(r) = \u03c0r 2\" indicates the area of a circle given in terms of the value of the radius r, while \"C(r) = 2\u03c0r\" indicates the circumference given in terms of the radius r. 2 cm 2 Work out the circumference of each circle. This math worksheet was created on 2010-03-06 and has been viewed 81 times this week and 53 times this month. The notation for semicircle and major arc is similar to that of minor arc. The only basic figure is the semi-circle, that appears four times. Welcome to The Circumference and Area of Circles (A) Math Worksheet from the Measurement Worksheets Page at Math-Drills. Work out the circumference of the circle. the way around a circle. Calculate the total square inches in column F by using =E2*D2. Area and Circumference of a Circle (Grades 6-7) Given a Slope and Points, calculate the missing coordinate Determine if the Points are Collinear. Online calculators and formulas for a hemisphere and other geometry problems. 7%) A: 4 (14. The radius is: r=D/2 r=30. We can substitute 7 for r in the formula C = 2\u03c0r to end up with C = 2\u03c0(7) = 14\u03c0 \u2248 44 inches. It progresses nicely and finishes with some challenging problems at the end borrowed from MEP. Calculate the area of a semicircle of radius 1. The perimeter of a circle is known as its circumference. Click on \"Calculate\" button to get results instantly. The circumference of a figure is the sum of all the side lengths. A compass is an instrument used to draw circles or the parts of circles called arcs. 14159) and diameter is the distance across the circle (along the edge of the half circle). A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. Multiply the portion of the circumference the minor arc comprises by the circumference of the circle to calculate the length of the minor arc. It is calculated just by multiplying the diameter of the circle with \u03c0 value. The circumference of a circle of a radius \ud835\udc5f\ud835\udc5f is 2\ud835\udf0b\ud835\udf0b. Calculate the circumference of the circle. 14 * radius. You can find out more about Pi here. The formula is:. Point A lies on the diameter of a semi\u2013circle with the radius 8 cm and the center O. the semi circle touches the rectangle at a b and c. To calculate the area of a semicircle we use the formula: Area of a semicircle = \u03c0 \u00d7 (radius)2 2 A = \u03c0r2 2 Calculate the area of this semicircle. I can calculate the radius and diameter of a circle when I know the Circumference. To calculate the perimeter of Shape 2 add the length of the three sides of the rectangle and the circumference of the semi-circle. 7cm 5cm Ttx8 = 25. base area \u00d7 height Step 2: Total area = Area of circle + Area of rectangle Mensuration. Arc Measure. Contrast radius and diameter. ) Homework Equations E = K((Q)/(r^2)). Plugging our radius of 3 into the formula, we get C = 6\u03c0 meters or approximately 18. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. A B C AB = 1. These slightly more advanced circle worksheets require students to calculate area or circumference from different measurements of a circle. 06 = Calculate the length of the circumference of each circle, rounding your answer to one decimal place: 62. Construct the Gnomon to the calculated dimensions and stand vertically along the centre line. A circle is drawn with radius 'r' touching all the three semi-circles. where r is the radius of the circle and \u03c0 is the ratio of a circle\u2019s circumference to its diameter. This online calculator will find and plot the equation of the circle that passes through three given points. We have step-by-step solutions for your textbooks written by Bartleby experts!. Formulas, explanations, and graphs for each calculation. They have to find the circumference of 6 different circles and then play a memory matching game. Find the perimeter of the shape. Find the circumference of a circle which has a radius of 4 cm. This gives us the lengths of all the sides as shown in the figure below. We can substitute 7 for r in the formula C = 2\u03c0r to end up with C = 2\u03c0(7) = 14\u03c0 \u2248 44 inches. Formulas, explanations, and graphs for each calculation. thx [3] 2020/04/05 02:53 Male / 40 years old level / An engineer / Useful / Purpose of use. The distance around a rectangle or a square is as you might remember called the perimeter. Round to the nearest hundredth for each circumference. These slightly more advanced circle worksheets require students to calculate area or circumference from different measurements of a circle. Calculate \u2018d\u2019 and the total area inside the track. What is the arc-length of a semi-circle with radius of 6 cm? (to 2 decimal places) A car wheel has a circumference of 1. It progresses nicely and finishes with some challenging problems at the end borrowed from MEP. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. You can calculate the circumference from total circle angle 360. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. In order to find the area of semi circle use the below given formula. To calculate area and circumference we must know the radius of circle. 2 Calculate the length of the circumference of each circle, rounding your answers to one decimal place: 10 = 31. Do NOT confuse a 'sector' with a 'segment'. Calculate the circumference and the area of the circle. Consider the diagram of a semicircle shown. 69 km 0 266. If you have the radius instead of the diameter, multiply it by 2 to get the diameter. The diagram shows a trapezium with two identical semi-circles. To establish a major arc as a semicircle some additional notation would be required. Step 2 Since you only need half of the. I can use the information I know to calculate the length of the radius of the. What is the radius of the cup? Round your answer to a suitable degree of accuracy. Formula Used: Radius of a Semicircle, r = Diameter (d) / 2 or r = Circumference / Pi Where, Pi is a constant = 3. If a circle has a circumference of 8. Get the result. As learning progresses students are challenged to find the perimeter of semi-circles and sectors. Calculate the exact area of the shaded sector. Step 1: say the radius of the field is r, then the total area of the field = (r 2)/2. What are the circumference and area of the circle? 2. Well, that and the formula. The length of the rectangle is 15 cm. A circle has a circumference of \u00eccm. Step-by-step explanation: Here we note that the shape consists of two small circles and one larger circle. - calculate the percentage in each category - calculate the central angle for each category for a circle graph - create and label a circle graph using your data Student Grades c: o: 2 c: 10 135. In a similar manner, we can calculate the length of the other missing side using 14\u22128=6. Remember that the perimeter is the distance round the outside. Do NOT confuse a 'sector' with a 'segment'. Input: Store the constant pI in M1: M1 = 3. In our semicircle, the diameter is 35 centimetres. In the result you will get all unknown variables presented. Common Core Standard: 7. Welcome to The Circumference and Area of Circles (A) Math Worksheet from the Measurement Worksheets Page at Math-Drills. For each swimming pool below, calculate: the perimeter of the swimming pool. asked by Bunny head on December 2, 2018; Math. Draw a picture to represent the window. Write the formula that you will use (e. This means that in any circle, there are 2 PI radians. (3 marks) _____ A cylinder has a radius of 5 cm. Formula: Area = 3. The sum of the central angles in any circle is 360\u00b0. The diagram shows a 400-m running track. The top part of the door is a semicircle. By that logic, the arc length of a quarter circle is exactly 1/4 of the circumference of a circle. If a circle has a circumference of 8. Arcs are measured in two ways: as the measure of the central angle , or as the length of the arc itself. So that's straightforward, area 36pi, we leverage pi r squared to figure out that the radius was 6, and then from that we were able to figure out that the circumference was 12pi. 7 cm Length of Y 3. Find the circumference of the object. You can also select units of measure for both input data and results. What is the arc-length of a semi-circle with radius of 6 cm? (to 2 decimal places) A car wheel has a circumference of 1. show your working. Angle in a semicircle We want to prove that the angle subtended at the circumference by a semicircle is a right angle. Program to calculate Area and Circumference based on user input. Perimeter and the area of a semicircle Perimeter of a semicircle. Set up each circle's circumference to its diameter. 14) by the square of the radius; If a circle has a radius of 4, its area is 3. You know circles are round. The widest part of a tea cup has a circumference of 24 cm. Second circle theorem - angle in a semicircle. o Circumference. The area of a circle is given by Pi*Radius^2 where Pi is a constant approximately equal to 3. Round this ratio to the nearest thousandth. side a: side b: distance h:. Find the diameter of the circle to the nearest tenth. This is the length of just the curvy part of the semi-circle. The circumference of a figure is the sum of all the side lengths. The perimeter of a semicircle can be calculated using the following formula: if you know the length of the radius. SOLUTION a. \\$ cc circumference-of-circle. This math worksheet was created on 2010-03-06 and has been viewed 81 times this week and 53 times this month. And in this case, r is equal to 6. 2 Calculate the length of the circumference of each circle, rounding your answers to one decimal place: 10 = 31. The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie. The floor is defined as a circle equal to the diameter of the base of the dome. circle theorems rules pdf DA is a tangent to the circle. 2831852 m e t e r s. MEMORY METER. Another circle is inscribed in the semi\u2013circle so that it touches the diameter at point A and also touches the circumference of the semi\u2013 circle. The ratio circumference. Perimeter of composite shapes. Lesson Plan Title : Finding circumference, diameter, and radius. In the result you will get all unknown variables presented. ) cm\u00b2 (to. n r tM Nazdhe t 9wKi7t6h l wI2nOf vi1nIi pt se F BGFe jo Lmteztjr 8yE. radius: distance from center of circle to any point on it. The circumference is the outside edge of the circle. They find the area and circumference of circles given a radius or diameter measurement. This formula can also be given as: C = 2\u03c0r, where r is the radius. Pi ( \u03c0 ): It is a number equal to 3. Perimeter of a semicircle = (1 2 Suppose we want to calculate the circumference of a circle with radius r r r. Examples: how to work out the area of a trapezium?. 2, but that is the best I can do with the crude calculator on the computer and a scratch piece of paper. Then tap or click the Calculate button. 14159, which is equal to the ratio of the circumference of any circle to its diameter. Ex2) 14 in. (a) Calculate the area of the whole target. Area of semi-circle formula is derived from the formula of a circle. So, to find the diameter, divide the circumference by. The POWER function will take any number and raise it to the power of any other number. That's the. You can also use it to find the area of a circle: A = \u03c0 * R\u00b2 = \u03c0 * 14\u00b2 = 615. The area is generally 1/2*Pi*r\u00b2. Students, teachers, parents, and everyone can find solutions to their math problems instantly. The circumference of C i r c l e 2 is 6. Calculator online for a capsule. o Pi (\u03c0) Objectives. C = \u03c0d Circumference of a circle C = 3. The calculations are done \"live\": How to Calculate the Area. Angle in a semicircle We want to prove that the angle subtended at the circumference by a semicircle is a right angle. o Calculate the circumference of a circle. We have step-by-step solutions for your textbooks written by Bartleby experts!. Formulas, explanations, and graphs for each calculation. C = 2\u03c0 r Write formula for. Program to calculate Area and Circumference based on user input. 77 feet/2 r=15. 7 ft O C r d. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. 2, but that is the best I can do with the crude calculator on the computer and a scratch piece of paper. where r is the radius of the circle and \u03c0 is the ratio of a circle\u2019s circumference to its diameter. I am stuck on put a circle around the number that is not a square number 9 16 36 48 81 100. Calculate the area and circumference of this circle, leaving your answer in terms of \u03c0 4. The diameter of the larger semicircle is subtended by the two smaller semicircles the small semicircle closer to the left of the internal circumference of the larger semicircle is shaded one while the one on the right is without color This calculator calculates for the radius length width or chord height or sagitta apothem angle and area of an. (c) Find the area of the white. Perimeter Of Sector calculator provides for the same. The area of a circle is given by Pi*Radius^2 where Pi is a constant approximately equal to 3. Use our circumference calculator to determine the area of a circle. 8cm This time, we have a semicircle. Step 2: Divide the circumference in half. The radius of a circle measuring 20 cm. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. !Work out the area. 56 + 8 = 20. Given OA = 4, find the radius of the inscribed circle. I use this product as part of a grade 7 unit of study on the area and circumference of circles. This video shows how to find the area of a rectangle with a semi-circle. You know circles are round. (c) Find the area of the white. the circumference. n r tM Nazdhe t 9wKi7t6h l wI2nOf vi1nIi pt se F BGFe jo Lmteztjr 8yE. Circumference Measuring the _____ of the circle. But, the mathematical description of circles can get quite confusing, since there is a set equation for a circle, including symbols for the radius, and center of the circle. 14 or 22 \u2014 7 for \ud835\uded1. your calculator as an approximation for \ud835\udf0b. Calculate the area of one surface of the table mat. Calculate quickly surface area or volume of sphere. (a) Calculate the area of glass needed to glaze the window. First, you must apply the formula for calculate the area of a semicircle, which is: A=\u03c0r 2 /2 A is the area of the semicircle. that the perimeter of the semi-circle is. Calculate the shaded area. Step 2: Divide the circumference in half. I can calculate thecircumference of a circle when I know the diameter. In column H, calculate the dollars per square inch of pizza, using =G2/F2. Define tangent. It is possible to inscribe a rectangle by placing its two vertices on the semicircle and two vertices on the x-axis. Add them to the table above, measure the circumference and diameter for each, and then compute the ratio Circumference Diameter for each object. DCO is a straight line. Tracing paper may be used. (c) Given that APB is 1. Use your calculator to find the maximum area. 14 or the \u03c0 button on your calculator. Draw the diagram (as above on right) to understand the problem and arrive at the required answer. 4 Find the radius of a circle. Follow 93 views (last 30 days) Timothy States on 7 Feb 2016.", "date": "2020-10-25 04:54:43", "meta": {"domain": "milanoporteeserramenti.it", "url": "http://milanoporteeserramenti.it/tec/circumference-of-a-semicircle-calculator.html", "openwebmath_score": 0.8478139638900757, "openwebmath_perplexity": 427.07611142392966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9907319871663826, "lm_q2_score": 0.8902942152065091, "lm_q1q2_score": 0.8820429569942798}}
{"url": "http://math.stackexchange.com/questions/95562/how-to-calculate-z4-frac1z4-if-z2-z-1-0/95596", "text": "# How to calculate $z^4 + \\frac1{z^4}$ if $z^2 + z + 1 = 0$?\n\nGiven that $z^2 + z + 1 = 0$ where $z$ is a complex number, how do I proceed in calculating $z^4 + \\dfrac1{z^4}$?\n\nCalculating the complex roots and then the result could be an answer I suppose, but it's not quite elegant. What alternatives are there?\n\n-\n\nFrom $z^2 + z + 1 = 0$, we have $$z +\\frac{1}{z}=-1.$$ Taking square, we get $$z^2 +\\frac{1}{z^2}+2=1,$$ which implies that $$z^2 +\\frac{1}{z^2}=-1.$$ I think you will know what to do next.\n\n-\nI see... Thank you for the quick answer! \u2013\u00a0 Mihai Bi\u015fog Jan 1 '12 at 11:54\n$\\displaystyle{z^{2^k}+\\frac{1}{z^{2^k}}=-1}$ for all $k$. \u2013\u00a0 Jonas Meyer Jan 1 '12 at 11:54\n@Jonas: nice observation! \u2013\u00a0 Paul Jan 1 '12 at 12:06\nDickson polynomials FTW! \u2013\u00a0 Jyrki Lahtonen Jan 1 '12 at 14:20\nAnother way to justify that $z+\\dfrac1{z}=-1$ is that due to the coefficient symmetry, if $z$ is a root, then $\\dfrac1{z}$ is a root as well. By Vieta, then, the sum of those two roots ought to be the negative of the coefficient of the linear term... \u2013\u00a0 J. M. is back. Jan 1 '12 at 15:15\n\nEssentially the same calculation also follows from the observation that $x^2+x+1=\\phi_3(x)$ is the third cyclotomic polynomial. So $z^3-1=(z-1)(z^2+z+1)=0$, and hence $z^3=1$ for any solution $z$. Therefore $$z^4+\\frac{1}{z^4}=z\\cdot z^3+\\frac{(z^3)^2}{z^4}=z+z^2=-1.$$\n\n-\nVery clever! :D \u2013\u00a0 J. M. is back. Jan 1 '12 at 12:08\nVery nice! If you had not posted this already, I would have used the sum of geometric series (instead of cyclotomic polynomials) to write $$0=z^2+z+1=\\frac{z^3-1}{z-1}\\Rightarrow z^3=1$$ and $$z^4+\\frac{1}{z^4}=(z^3)z+\\frac{1}{(z^3)z}=z+\\frac{1}{z}=\\frac{z^2+1}{z}=\\frac{\u200c\u200b-z}{z}=-1.$$ \u2013\u00a0 Dilip Sarwate Jan 1 '12 at 15:48\n\n\\begin{align*} z^4+\\frac1{z^4}&=(-z-1)^2+\\frac1{(-z-1)^2}\\\\ &=z^2+2z+1+\\frac1{z^2+2z+1}\\\\ &=(-z-1)+2z+1+\\frac1{(-z-1)+2z+1}\\\\ &=z+\\frac1{z}=\\frac{z^2+1}{z}=\\frac{-z-1+1}{z}=-1 \\end{align*}\n\n-\nWhat would be wrong with systematic reduction using the minimal polynomial? Nothing! \u2013\u00a0 Jyrki Lahtonen Jan 1 '12 at 12:25\n\nHere is an alternative approach: let's consider $z^{8}+1$ , and then divide by $z^{4}$.\n\nBy using geometric series, notice that $$z^{8}+z^{7}+z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=\\left(z^{2}+z+1\\right)\\left(z^{6}+z^{3}+1\\right)=0.$$ Now, as $z^{2}+z+1=0$, we know that both $z^{7}+z^{6}+z^{5}=0$ and $z^{3}+z^{2}+z=0$, and hence $$z^{8}+z^{4}+1=0$$ so that $z^{8}+1=-z^{4}$. Thus, we conclude that $$z^{4}+\\frac{1}{z^{4}}=-1.$$\n\n-\n\nDifferent people see different things in the relation $z^2+z+1=0$. Just as Jyrki did, I see the third cyclotomic polynomial. Its roots are $-1/2 \\pm \\sqrt{-3}/2$, the two primitive cube roots of unity. Call one of these $\\omega$ and see that $\\omega^3=1$, so that $\\omega^4=\\omega$ and $\\omega^{-4}=\\omega^2$. Their sum is $-1$, from the defining relation.\n\n-\n\nHint $\\$ Exploit innate symmetry: for $\\rm\\ y = z^{-1}$ you know $\\rm\\ yz\\ (=\\: 1)\\:$ and $\\rm\\ y+z\\ (=\\: z^{-1}\\!+z\\: =\\: -1)\\$\n\nThus you know $\\rm\\ \\ \\ y^2 + z^2\\ =\\ (y\\ +\\ z)^2-\\ 2\\:(y\\:z)$\n\nhence you know $\\rm\\ y^4 + z^4\\ =\\ (y^2\\! + z^2)^2 - 2\\:(y\\:z)^2$\n\nFor more on symmetric polynomials see the Wikipedia article on Newton's identities.\n\n-\nTo downvoter: if something is not clear then please feel welcome to ask questions and I will be happy to elaborate. \u2013\u00a0 Bill Dubuque Jan 1 '12 at 22:52\n\n$$z^2+z+1=0$$ Clearly $z \\not= 0$ and therefore $$z+\\frac{1}{z}=-1.$$ Note that $$z^{n+1}+\\frac{1}{z^{n+1}}=(z+\\frac{1}{z})(z^n+\\frac{1}{z^n})-(z^{n-1}+\\frac{1}{z^{n-1}}).$$ Therefore if we define the function $U:N \\to N$ as $$U_n=z^n+\\frac{1}{z^n}$$ then we have $U(0)=2$ and $U(1)=-1.$ Also $$U(n+1)=-U(n)-U(n-1)$$ for all $n \\in N.$\nUsing this recurrence you can calculate $z^n+\\frac{1}{z^n}$ for any $n \\in N.$\n\n-", "date": "2015-10-06 21:04:42", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/95562/how-to-calculate-z4-frac1z4-if-z2-z-1-0/95596", "openwebmath_score": 0.9703070521354675, "openwebmath_perplexity": 417.32785874465696, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987375050346933, "lm_q2_score": 0.8933093996634686, "lm_q1q2_score": 0.8820314134681058}}
{"url": "http://mathematica.stackexchange.com/questions/22349/mathematica-for-teaching-orthographic-projection", "text": "# Mathematica for teaching orthographic projection\n\nEdit: All the four answers to this question are great, and if you're interested, you should take a look at all the answers. Nevertheless, belisarius' code was accepted since it was closest to what I had in mind.\n\nI'm trying to use Mathematica to help students better understand 3-dimensional views of objects, specifically the \"plan view\", the \"side elevation view\" and the \"front elevation view\", as elaborated upon here.\n\nGiven the following example \"building\", I was able to combine Graphics3D primitives to simulate the object using the following code.\n\ndee = Graphics3D[{Green, Opacity[0.6], Cuboid[{0, 0, 0}, {2/3, 1, 1}],\nRed, Polygon[{{{2/3, 1/2, 0}, {2/3, 1/2, 1}, {3/2, 1/2, 0}},\n{{2/3, 1, 0}, {2/3, 1, 1}, {3/2, 1, 0}},\n{{2/3, 1/2, 0}, {2/3, 1, 0}, {3/2, 1, 0}, {3/2, 1/2, 0}},\n{{2/3, 1/2, 1}, {2/3, 1, 1}, {3/2, 1, 0}, {3/2, 1/2, 0}}}]}]\n\n\nHowever, as you can see, the code is rather convoluted, with a great deal of time spent making the red prism. In this case, it was still feasible to manually generate the prism, but what I'm interested in is,\n\nIs there a way to generate a 3D object which is made up of a combination of prisms simply by identifying the vertices at the corner of the object?\n\nIf you're interested, my code to allow students to see the three views is as follows, with an example result below.\n\nplanviewer =\nTableForm@{{\"3d\", \"Side\", \"Front\", \"Top\"},\n{Show[#, ViewPoint -> {1, -1, 1}],\nShow[#, ViewPoint -> {\u221e, 0, 0}],\nShow[#, ViewPoint -> {0, -\u221e, 0}],\nShow[#, ViewPoint -> {0, 0, \u221e}]},\n{\"\",\nShow[#, ViewPoint -> Right],\nShow[#, ViewPoint -> Front],\nShow[#, ViewPoint -> Above]}} &;\n\n\nThe following code allows them to manipulate the viewpoint dynamically.\n\nManipulate[\nShow[dee, ViewPoint -> {10^a, -(10^b), 10^c}], {a, 0, 3}, {b, 0, 3}, {c, 0, 3}]\n\n-\nIf you want to teach about the different View* functions and what they do, you might find this answer useful. \u2013\u00a0 The Toad Mar 30 '13 at 16:59\n\nThere are some tricks ... Specifying the prism's vertices is enough (you don't need to take care of the faces) if you use some undocumented methods for finding the convex hull:\n\nv = {{2/3, 1/2, 0}, {2/3, 1/2, 1}, {2/3, 1, 0}, {2/3, 1, 1}, {3/2, 1/2, 0}, {3/2, 1, 0}};\nprism@v_:=Cases[ComputationalGeometryMethodsConvexHull3D[v,GraphicsMeshFlatFaces->False],\n_GraphicsComplex, Infinity];\n\nGraphics3D[{Green, Opacity[0.6], Cuboid[{0, 0, 0}, {2/3, 1, 1}], Opacity[.6], Red, prism@v}]\n\n\n-\nThis is easily turned into a function, polyhedron[v_] := Cases[.. Very nice. (+1) \u2013\u00a0 Michael E2 Mar 30 '13 at 18:21\n@MichaelE2 Thanks! Rewritten as a function now. \u2013\u00a0 belisarius Mar 30 '13 at 18:43\n@belisarius how do you find out about all these undocumented questions? I was trying to search for it, and saw some similar questions on this site, but I think that's the first time I've seen it. Edit: ninja'd by bill. \u2013\u00a0 Vincent Tjeng Mar 31 '13 at 3:30\nIt's not on this list, by the way :) : mathematica.stackexchange.com/questions/809/\u2026 \u2013\u00a0 Vincent Tjeng Mar 31 '13 at 3:38\n@VincentTjeng I don't remember how I found this one, probably following some InternalTrace. You can find similar things trying Names[\"ComputationalGeometryMethods*\"] \u2013\u00a0 belisarius Mar 31 '13 at 5:55\n\nGraphicsComplex is probably what you are looking for. For example, define the vertices for the 3D polygon:\n\nv = {{2/3, 1/2, 0},{2/3, 1/2, 1},{3/2, 1/2, 0},{2/3, 1, 0},{2/3, 1, 1},{3/2, 1, 0}};\n\n\nand make a list of which vertices should connect to each other:\n\ni = {{1, 2, 3}, {4, 5, 6}, {1, 2, 5, 4}, {1, 3, 6, 4}, {2, 3, 6, 5}};\n\n\nThe first two elements of i represent the two triangular faces of the prism; the final three elements are the three rectangular faces. This is plotted using\n\nGraphics3D[{Opacity[.8], Red, GraphicsComplex[v, Polygon[i]]}]\n\n\nor you can plot the green cube plus the red prism together:\n\nGraphics3D[{Green, Opacity[0.6], Cuboid[{0, 0, 0}, {2/3, 1, 1}],\nOpacity[.6], Red, GraphicsComplex[v, Polygon[i]]}]\n\n\nwhich gives the figure below. The code is modified from the documentation in GraphicsComplex where you can find all sorts of neat 3D tricks. If they have access, I think the students would benefit from being able to manipulate the 3D illustration themselves -- this seems like a good application for deployed .cdf's.\n\n-\nThanks for the answer! Would it be possible for the one to write some code that immediately identifies which of vertices should be connected to each other to form the faces, keeping in mind that the faces of the prism may not always be in the same plane? Also, I notice that the part of the red prism joining the green cuboid is actually a triangle, and not a rectangle ... is that a problem or simply a visual illusion? \u2013\u00a0 Vincent Tjeng Mar 30 '13 at 10:21\nThe little triangle on the adjoining face is an illusion that has to do with the shading -- when you rotate the graphic it changes. But to your first question, the answer is no. After all, how can it know what edges you want to connect if you don't tell it? You will get something no matter what edges you connect -- only one of these is the figure you presumably want. \u2013\u00a0 bill s Mar 30 '13 at 10:33\n@VincentTjeng: yes, you can connect the vertices automagically because the faces must form the convex hull. See my answer \u2013\u00a0 belisarius Mar 30 '13 at 18:54\nVery nice... how does one find \"undocumented features\"? \u2013\u00a0 bill s Mar 31 '13 at 3:17\nbill I don't remember how I found this one, probably following some InternalTrace. You can find similar things trying Names[\"ComputationalGeometryMethods*\"] \u2013\u00a0 belisarius Mar 31 '13 at 5:59\n\nTake a look at the output of PolyhedronData[{\"Prism\", 3}, \"Faces\"]}, which may help.\n\nNote this isn't exactly a direct answer, but I think the optimal solution here is a tool that would let you... well, like this:\n\nwhatThe[n_] := Module[{list = {}, grid = False},\nGrid@List@{\nEventHandler[#, {\"KeyDown\", \"k\"} :> (grid = Not[grid])] &[\n\nGraphics3D[{\nEventHandler[MouseAppearance[Sphere[#, .06],\nStyle[\"\\[RightUpDownVector]\\[LeftUpDownVector]\", Darker@Blue, Large]],\n{\"MouseClicked\" :> AppendTo[list, ##]}] & /@ Tuples[Range[n], 3],\n\n(*Opacity[.3],*)Dynamic[Polygon[list]], Thick, Dynamic[Line[list]],\nThin, Lighter[Gray, .8], Dynamic[If[grid, Line[\nSelect[Subsets[Tuples[Range[n], 3], {2}],\nEuclideanDistance @@ ## == 1 &]], {}]]},\n\nBoxed -> True, ImageSize -> Large, Axes -> True]],\n\nGraphics3D[Dynamic[Polygon[list]]],\nEventHandler[Dynamic[Column[list]],\n{\"MouseClicked\" :> (CopyToClipboard[list]; Beep[])}]}\n];\n\nwhatThe[4]\n\n\nYou click on points to add them to a list. MAGIC!!! It takes some getting used to... here's one of my attempts at creating a prism:\n\nI couldn't tell what order you have to construct the list in for the resulting polygon to be reasonable, but at the very least it might help you pick some points out. It's also another example of just how ridiculously powerful Mathematica is. The basic working version of this took me like 4 minutes to make. O.O\n\n(Note: A simple improvement might be to use a hotkey to cut the list up as you build the polygon).\n\nUpdate. Using belisarius's convex hull code:\n\nwhatThe[n_] := Module[{list = {}, faces, grid = False},\nfaces[v_] := Cases[ComputationalGeometryMethodsConvexHull3D[v,\nGraphicsMeshFlatFaces -> False], _GraphicsComplex, Infinity];\n\nGrid@List@{\nEventHandler[#, {\"KeyDown\", \"k\"} :> (grid = Not[grid])] &[\n\nGraphics3D[{EventHandler[\nMouseAppearance[Sphere[#, .06],\nStyle[\"\\[RightUpDownVector]\\[LeftUpDownVector]\",\nDarker@Blue, Large]],\n{\"MouseClicked\" :> AppendTo[list, ##]}] & /@ Tuples[Range[n], 3],\n\n(*Opacity[.3],*)Dynamic[faces[list]],\nThin, Lighter[Gray, .8], Dynamic[If[grid, Line[\nSelect[Subsets[Tuples[Range[n], 3], {2}],\nEuclideanDistance @@ ## == 1 &]], {}]]},\n\nBoxed -> True, ImageSize -> Large, Axes -> True]],\n\nGraphics3D[Dynamic[faces[list]]],\nEventHandler[Dynamic[Column[list]],\n{\"MouseClicked\" :> (CopyToClipboard[list]; Beep[])}]}];\n\n\n-\n+1 amazing. and it helps with the first part of translating the problems in the students' worksheets to the prism I want. now I just need to modify the code such that it accepts multiple convex hulls representing the multiple prisms :) \u2013\u00a0 Vincent Tjeng Mar 31 '13 at 2:58\n\nYou can get away with specifying even less than all the points.\n\nA (right) prism is a planar figure extruded in an orthogonal direction. It would suffice, then, to give the coordinates of that figure (thereby describing one of the two \"ends\" of the prism) and its intended height. The figure in the question can therefore be created as a collection of two prisms (no Cuboid if you don't care to use it) like this:\n\nGraphics3D[{Opacity[0.6],\nGreen, prism[{{0, 0, 0}, {2/3, 0, 0}, {2/3, 1, 0}, {0, 1, 0}}, 1],\nRed, prism[{{2/3, 1, 0}, {3/2, 1, 0}, {2/3, 1, 1}}, 1/2]}]\n\n\nThe calculations come down to\n\n1. Find a basis for the polygon's plane by orthogonalizing its vertices relative to their barycenter;\n\n2. Perform the extrusion by taking the cross product of the basis elements (normalized to the desired height) and adding that to each of the polygon's vertices; and\n\n3. Describe how the vertices are connected to form the two polygonal ends and each of the sides.\n\nHere is such a solution. Its arguments are pt, a list of the (3D) vertices around the (nondegenerate) polygon, and height, the height of the prism. It produces a 3D graphics object.\n\nprism[pt_List, height_] := Block[{n = Length[pt], normal},\nnormal = Cross @@ Orthogonalize[# - Mean[pt] & /@ pt][[1 ;; 2]];\nGraphicsComplex[pt~Join~(height normal + # & /@ pt),\n{Polygon[Range[n]], Polygon[Range[2 n, n + 1, -1]],\nPolygon[{#, n + #, n +  Mod[# + 1, n, 1], Mod[# + 1, n, 1]}] & /@ Range[n]}]]\n\n\nTo see it in action, let's generate a polygon with a random position and orientation:\n\nbasis = RandomReal[NormalDistribution[0, 1], {2, 3}];\npt = ({Cos[#], Sin[#]} & /@ Range[0, 2 \\[Pi], 2 \\[Pi]/7]).basis\n\n\nHere is the original polygon as a solid black \"floor\" drawn with two prisms, one with a positive height and another with a negative height:\n\nGraphics3D[{Opacity[0.75], prism[pt, 1], prism[pt, -1/2], Opacity[1], Black,  Polygon[pt]}]\n\n\nThis approach is easily extended to non-right prisms by replacing height by the extrusion vector (describing one of the \"vertical\" edges of the prism) and using that instead of height normal in the code.\n\n-\nnice application of math :) just a quick question, when you code as prism[pt_List, height_], pt_List is written as such since it is a list, right? so if i were to code a function f[a_String, b_String], then the variables would be a, b and they must be strings? \u2013\u00a0 Vincent Tjeng Mar 31 '13 at 3:35\n\u2013\u00a0 whuber Mar 31 '13 at 13:19\nGot it, thank you! \u2013\u00a0 Vincent Tjeng Apr 1 '13 at 3:01", "date": "2015-05-25 13:45:40", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/22349/mathematica-for-teaching-orthographic-projection", "openwebmath_score": 0.18760305643081665, "openwebmath_perplexity": 2193.327691647778, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338101862455, "lm_q2_score": 0.917302651107873, "lm_q1q2_score": 0.8820175132136974}}
{"url": "https://math.stackexchange.com/questions/2998130/matrix-representation-of-mixed-derivatives", "text": "# Matrix representation of Mixed derivatives\n\nImagine we have the problem\n\n$$\\begin{cases} -\\frac{d^2u}{dx^2} = f(x), x \\in [0,L] \\\\ u(0) = 0 \\\\ u(L) = 0 \\end{cases}$$\n\nWe know that we can approximate the second derivative using this formula:$$\\frac{d^2u(x)}{dx^2} \\approx \\frac{u(x+h)-2u(x)+u(x-h)}{h^2}$$\n\nif we define $$u_{k} := u(x_{k})$$; $$\\ \\ x_{k} = kh$$ and $$\\ \\ k = 0,1,2,...,N$$. $$h$$ is known as the mesh size or step size. We get:\n\n$$\\frac{d^2u_{k}}{dx^2} \\approx \\frac{u_{k+1}-2u_{k}+u_{k-1}}{h^2} = \\frac{u_{k-1}-2u_{k}+u_{k+1}}{h^2}$$ for $$k=1,...,N-1$$\n\nSince $$u(0) = u_{0} = 0$$ and $$u(L) = u(x_{N}) = u_{N} = 0$$ we get the following matrix representation of the second derivative operator\n\n$$$$\\frac{d^2}{dx^2} \\approx L_{2} = \\frac{1}{h^2}\\left(\\begin{matrix} -2 & 1 & & 0\\\\ 1 & \\ddots & \\ddots & \\\\ & \\ddots & \\ddots & 1 \\\\ 0 & & 1 & -2 \\end{matrix} \\right)$$$$\n\nThen to approximate the solution of the differental equation we solve the system:\n\n$$-L_{2}\\hat{u} = \\hat{f}$$ where $$\\hat{f} = [ f(x_{1}) \\ f(x_{2}) \\ ... \\ f(x_{N-1}) ]^T$$ and $$\\hat{u} = [ u_{1} \\ ... \\ u_{N-1} ]^T$$\n\nThe matrix representation of the laplacian operator using the Kronecker product is: $$\\Delta =\\nabla^2 = \\frac{\\partial^2}{\\partial x^2} + \\frac{\\partial ^2}{\\partial y^2} = L_{2}\\otimes I_{n_{x}} + I_{n_{y}} \\otimes L_{2}$$\n\n$$\\textbf{Note that I have used L_{2} and kronecker product to get a matrix representation}$$ of the laplacian operator.\n\nWith this in mind. I want to approximate the first derivative using central difference:\n\n$$\\frac{du_{k}}{dx} \\approx \\frac{ u_{k+1}-u_{k-1} }{ 2h } = \\frac{ -u_{k-1} +u_{k+1} }{ 2h }$$\n\nSince $$u(0) = u_{0} = 0$$ and $$u(L) = u(x_{N}) = u_{N} = 0$$ we get the following matrix representation of the first derivative\n\n$$$$\\frac{d}{dx} \\approx L_{1} = \\frac{1}{2h}\\left(\\begin{matrix} 0 & 1 & & 0\\\\ -1 & \\ddots & \\ddots & \\\\ & \\ddots & \\ddots & 1 \\\\ 0 & & -1 & 0 \\end{matrix} \\right)$$$$\n\nI want to use $$L_{1}$$ and kronecker product to get the matrix representation of\n\n$$\\frac{\\partial^2 }{\\partial y \\partial x }$$ and $$\\frac{\\partial^2 }{\\partial x \\partial y }$$\n\n$$\\textbf{My questions are:}$$\n\n1. Is this possible?\n\n2. if question 1. is affirmative, what is the matrix representation of the mixed derivatives using $$L_{1}$$ and kronecker product?\n\n3. if question 1. is negative, how can we get the matrix representation of the mixed derivatives using a simple matrix and kronecker product?\n\n4. Do you know a book or document( article or other ) that explain in detail this or something similar?\n\n$$\\textbf{I want to use kronecker product because it is fast and easy to implement in}$$ matlab or octave.\n\nBy the way I tried to use this formula\n\n$$\\frac{\\partial^2u_{k,j} }{\\partial x \\partial y } = \\frac{ u_{k+1,j+1}+u_{k-1,j-1}-u_{k+1,j-1}-u_{k-1,j+1} }{4h^2}$$\n\nBut it was hard to see a pattern. Thank you!\n\nOnce you have the matrix $$\\bf L_1$$ to represent the divided difference of $$f(x)$$ given as a vertical vector, then if you transpose the whole you'll get a horizontal vector for the difference.\n\nSo if $$f(x,y)$$ is represented as a matrix $$\\bf F_{\\, x,\\, y}$$ , the matrix representing $${\\partial \\over {\\partial x\\partial y}}f(x,y)$$ will be $$\\bf L_1 \\bf F_{\\, x, \\,y} \\bf L_1^T$$\n\n\u2022 Honestly I can not imagine the matrix... \u2013\u00a0tnt235711 Nov 14 '18 at 11:45\n\u2022 @tnt235711: added \u2013\u00a0G Cab Nov 14 '18 at 12:19\n\u2022 @tnt235711: is it ok now for you ? \u2013\u00a0G Cab Nov 15 '18 at 16:55\n\nNow using $$\\mathbf{L_{1}}$$ and $$\\textbf{kronecker product}$$:\n\n$$\\frac{\\partial^2 }{\\partial x \\partial y} = \\Big((L_{1})_{n_{x}} \\otimes I_{n_{y}} \\Big)\\Big( I_{n_{x}} \\otimes (L_{1})_{n_{y}} \\Big) = { (L_{1})_{n_{x}} \\otimes (L_{1})_{n_{y}} }$$\n\nI have used the mixed-product property for kronecker product. $$(\\mathbf{A} \\otimes \\mathbf{B})(\\mathbf{C} \\otimes \\mathbf{D}) = (\\mathbf{AC}) \\otimes (\\mathbf{BD})$$.\n\nIn a similar way we get:\n\n$$\\frac{\\partial^2 }{\\partial y \\partial x} = (L_{1})_{ n_{y} } \\otimes (L_{1})_{n_{x}}$$", "date": "2019-05-22 11:39:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2998130/matrix-representation-of-mixed-derivatives", "openwebmath_score": 0.9938884377479553, "openwebmath_perplexity": 169.78465940030057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936050226358, "lm_q2_score": 0.8962513703624558, "lm_q1q2_score": 0.8819952420664666}}
{"url": "https://math.stackexchange.com/questions/1610277/how-to-solve-an-initial-value-problem-if-my-function-is-implicit", "text": "# How to solve an initial value problem if my function is implicit?\n\nI have the differential equation:\n\n$\\dfrac{dy}{dt} = \\dfrac{1}{2y+3}$\n\nwith the initial value of $y(0) = 1$\n\nSo I solve the diffeq via separation of variables:\n\n$y^2 + 3y = t + c$\n\nBut from here, how do I solve for the function? Am I \"allowed\" to use the initial value information with an implicit function such as this, and give an implicit function as my answer, as below?\n\n$y^2 + 3y = t + 4$\n\nAs far as I can tell, it should be perfectly fine to leave my answer as an implicit function. And I don't see a way to solve for $y$ here. TO be honest, I used a diffeq solver to check my work, and that tool did come up with an explicit solution, though when I tried to see where it came from I had no idea.\n\n\u2022 This is a quadratic equaion for $y$. \u2013\u00a0kmitov Jan 13 '16 at 3:25\n\u2022 @kmitov But i do not want to solve for a particular value of $y$, I want to solve for a particular solution $y(t)$. \u2013\u00a0user278703 Jan 13 '16 at 3:30\n\nYes, you should use your known initial condition to solve for $c$.\n\nProceeding to solve for $y(t)$, we write $y^2+3y+\\frac{9}{4}-\\frac{9}{4}=\\left(y+\\frac{3}{2}\\right)^2-\\frac{9}{4}$ by completing the square.\n\nNow, we write $\\left(y+\\frac{3}{2}\\right)^2-\\frac{9}{4}=t+4.$ and solve for $y$ in terms of $t$.\n\nAdding $\\frac{9}{4}$, taking the square root, and subtracting $\\frac{9}{2}$, we get: $$y(t)=\\sqrt{t+\\frac{25}{4}}-\\frac{3}{2}$$\n\nEDIT: I had misplaced a $3$.\n\nEdit: To address your comment: Note that we can write $y(t)=\\sqrt{t+\\frac{25}{4}}-\\frac{3}{2}=\\sqrt{\\frac{1}{4}(4t+25)}-\\frac{3}{2}=\\frac{1}{2}\\sqrt{(4t+25)}-\\frac{3}{2}=\\frac{1}{2}(\\sqrt{(4t+25)}-3).$\n\nThe reason that $c$ is different is because $c$ is some arbitrary constant. In this case, we have already solved for it.\n\n\u2022 But $\\frac{9}{2} + \\frac{9}{2}$ doesn't equal 3?? And $\\frac{9}{2} * \\frac{9}{2}$ doesn't equal $\\frac{9}{4}$?? How can you factor $y^2 + 3y + \\frac{9}{4}$ and get $(y+ \\frac{9}{2})^2$? Shouldn't the correct factoring have $\\frac{3}{2}$ instead of $\\frac{9}{2}$? \u2013\u00a0user278703 Jan 13 '16 at 3:46\n\u2022 I'm not factoring. I'm completing the square. For $x^2+bx$, we can add and subtract $(\\frac{b}{2})^2$ to get $x^2+bx+(\\frac{b}{2})^2-(\\frac{b}{2})^2$, which is equal to $(x+\\frac{b}{2})^2-(\\frac{b}{2})^2$. Now, let $b=3$. \u2013\u00a0zz20s Jan 13 '16 at 3:51\n\u2022 How? If you factor $y^2 + 3y + \\frac{9}{4}$ it equals $(y+\\frac{3}{2})^2$. If we remove the $-\\frac{9}{4}$ from each side of the equation you wrote, it is $y^2 + 3y + \\frac{9}{4} = (y+\\frac{9}{2})^2$, which is wrong. Edit: I have seen your edit, thanks \u2013\u00a0user278703 Jan 13 '16 at 3:56\n\u2022 No problem. Do you understand the solution now? \u2013\u00a0zz20s Jan 13 '16 at 4:02\n\u2022 I understand, When I solve the diffeq on an online solver it gives $y(t) = \\frac{1}{2}(-\\sqrt{4t+c} - 3)$. I'm not sure where they got the $4$, or the $-3$, but I guess I'll play with it and see. \u2013\u00a0user278703 Jan 13 '16 at 4:04\n\nFrom where you arrived,\n\n$y^2+3y=t+c$\n\nSubstituting $y(0)=1$, we get $c=4$\n\nHence, we get $y^2+3y-4=t$\n\nProceed by completing the square for $y^2+3y-4=(y+\\frac{3}{2})^2-\\frac{7}{4}$\n\n$(y+\\frac{3}{2})^2-\\frac{7}{4}=t$\n\nThen we get $y=\\frac{1}{2}(3\u00b1\\sqrt{4t+7})$", "date": "2019-08-21 23:13:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1610277/how-to-solve-an-initial-value-problem-if-my-function-is-implicit", "openwebmath_score": 0.8892452120780945, "openwebmath_perplexity": 188.69460773115534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984093612020241, "lm_q2_score": 0.8962513634345444, "lm_q1q2_score": 0.8819952415203666}}
{"url": "https://math.stackexchange.com/questions/2944252/finding-the-probability-that-someone-has-the-disease-given-they-test-positive-o", "text": "# Finding the probability that someone has the disease, given they test positive on two tests\n\nThis question is from the textbook \"Introduction to Probability - Blitzstein & Hwang.\"\n\nI was studying for a class when I came across an example problem that I solved, but got a slightly different result than the textbook. Here's the problem in question, paraphrased:\n\n\"Fred tests for a disease which afflicts 1% of the population. The test's accuracy is deemed 95%. He tests positive for the first test, but decides to get tested for a second time. Unfortunately, Fred also tests positive for the second test as well. Find the probability that Fred has the disease, given the evidence.\"\n\n$$\\$$\n\nMy approach is as follows:\n\nLet $$D$$ be the event that Fred has the disease, $$T_1$$ be the event that the first test result is positive, and $$T_2$$ be the event that the second test is also positive. We want to find $$P(D\\ |\\ T_1,\\ T_2)$$.\n\nWe are also able to condition on $$T_1$$ (i.e. the event that the first test result is positive). This would give us:\n\n$$P(D\\ |\\ T_1,\\ T_2) = \\frac{P(T_2\\ |\\ D,\\ T_1)P(D\\ |\\ T_1)}{P(T_2\\ |\\ T_1)}$$\n\nFrom my calculations:\n\n$$P(T_2\\ |\\ D,\\ T_1)\\ =\\ P(T_2\\ |\\ D)\\ =\\ 0.95$$ $$P(D\\ |\\ T_1)\\ \\approx\\ 0.16$$ $$P(T_2\\ |\\ T_1)\\ =\\ \\frac{P(T_1 ,\\ T_2)}{P(T_1)}\\ =\\ \\frac{P(T_1,\\ T_2,\\ D)\\ +\\ P(T_1,\\ T_2,\\ D^c)}{P(T_1,\\ D)\\ +\\ P(T_1,\\ D^c)}\\ =\\ \\frac{0.0115}{0.059}\\ \\approx\\ 0.19$$ $$\\$$ $$P(D\\ |\\ T_1,\\ T_2)\\ =\\ \\frac{0.95\\ \\times\\ 0.16}{0.19}\\ =\\ 0.8$$\n\nTherefore, I concluded that there is an 80% chance that Fred has the disease, given that both the first and second test results are positive.\n\n$$\\$$\n\nThe problem is that the textbook has taken a different approach of using the odds form of Bayes' rule, which resulted in a conclusion slightly different from mine (0.78), and I'm having trouble understanding how that conclusion came to be.\n\n$$\\$$\n\nTextbook approach is as follows:\n\n$$\\frac{P(D\\ |\\ T_1,\\ T_2)}{P(D^c\\ |\\ T_1,\\ T_2)}\\ =\\ \\frac{P(D)}{P(D^c)}\\ \\times\\ \\frac{P(T_1,\\ T_2\\ |\\ D)}{P(T_1,\\ T_2\\ |\\ D^c)}$$\n\n$$=\\ \\frac{1}{99}\\ \\times\\ \\frac{0.95^2}{0.05^2}\\ =\\ \\frac{361}{99}\\ \\approx\\ 3.646$$\n\nwhich \"corresponds to a probability of 0.78.\"\n\n$$\\$$\n\nHere are the specific questions I have:\n\n1. Is my approach wrong? A 0.02 difference is a pretty big difference.\n\n2. How did the author derive the equation:\n\n$$P(D\\ |\\ T_1,\\ T_2)\\ =\\ P(D)P(T_1,\\ T_2\\ |\\ D)$$\n\n1. What does the author mean when he/she says \"3.646 corresponds to a probability of 0.78?\"\n\n$$\\$$\n\nAny feedback is appreciated. Thank you!\n\n\u2022 A very similar question was asked on this site about 6 years ago. Here is the link to it: math.stackexchange.com/questions/198677/\u2026 \u2013\u00a0David Oct 7 '18 at 0:25\n\u2022 Thanks for the feedback! I actually read that question while looking for other answers. It didn't address the questions that I had, though, which led me to ask my own. \u2013\u00a0Seankala Oct 7 '18 at 2:26\n\nI have some quibbles with the way the textbook sets up its question, beginning with the assumption that both positive and negative tests each have the same likelihood to be correct, and even more so the assumption that the outcomes of two tests on the same person are independent in probability. In real life, I would want to explore both of those points further before advising Fred. But let's ignore those objections for the sake of being able to compute something based on the given information, and assume each administration of the test has the same chance to give a correct result, even when two tests are administered one after the other on the same person.\n\n1. Is my approach wrong? A 0.02 difference is a pretty big difference.\n\nThe two methods are equivalent. The apparent discrepancy is due to roundoff.\n\nThe textbook finds an odds ratio of $$361:99,$$ which is exact (insofar as the $$1\\%$$ and $$95\\%$$ are exact). Since this is $$P(D) : P(D^C),$$ the probability is given by $$P(D) = \\frac{P(D)}{P(D) + P(D^C)} = \\frac{361}{361 + 99} \\approx 0.78478,$$ which the text rounds to $$0.78.$$ (Since you asked about this as a separate part of the question, I'll explain in more detail below.)\n\nIn your approach, $$P(T_1,\\ T_2,\\ D)\\ +\\ P(T_1,\\ T_2,\\ D^c) = 0.0115$$ is an exact result, and so is $$P(T_1,\\ D)\\ +\\ P(T_1,\\ D^c) = 0.059,$$ but $$0.0115 / 0.059 \\approx 0.19492.$$ Meanwhile, $$P(T_1 \\mid D) \\approx 0.16102.$$ If we carry all these digits into the computation rather than rounding off to two places immediately, we find that $$P(D\\mid T_1,\\ T_2) = \\frac{0.95\\times 0.16102}{0.19492} \\approx 0.78478.$$ That is, keeping five digits we get the same answer as the textbook method (if it retained five digits), and if we round to two digits only at the end (as the textbook does) we naturally would round the same way, to $$0.78.$$\n\nI think an argument could be made for keeping only one digit of precision in the answer (how precise is that \"$$1\\%$$\" anyway?), in which case both answers round to $$0.8.$$\n\n1. How did the author derive the equation ...\n\nThey didn't. Instead, the fact is that $$P(D\\mid T_1,\\ T_2) = \\frac{P(T_1,\\ T_2\\mid D)P(D)}{P(T_1,\\ T_2)}$$ and $$P(D^C\\mid T_1,\\ T_2) = \\frac{P(T_1,\\ T_2\\mid D^C)P(D^C)}{P(T_1,\\ T_2)}.$$\n\nWhen you compute the ratios of the two probabilities $$\\frac{P(D\\mid T_1,\\ T_2)}{P(D^C\\mid T_1,\\ T_2)},$$ you get factors of $$P(T_1,\\ T_2)$$ in both the numerator and the denominator, and these factors cancel each other.\n\n1. What does the author mean when he/she says \"3.646 corresponds to a probability of 0.78?\"\n\nAs I hinted above, $$3.646$$ is an odds ratio; or as I would rather say, the odds ratio is $$3.646 : 1.$$ An odds ratio of $$1:1$$ corresponds to a $$50\\%$$ chance, that is, each possibility is equally likely, whereas a $$2:3$$ odds ratio describes something that happens twice for every three times it does not happen. in general, if the probability of something is $$p,$$ its odds ratio is $$p : (1 - p),$$ that is, $$\\frac{p}{1 - p} : 1.$$\n\nIf we say $$p = P(D \\mid T_1,\\ T_2),$$ then $$P(D^C \\mid T_1,\\ T_2) = 1 - P(D \\mid T_1,\\ T_2) = 1 - p,$$ and what the textbook has computed is that $$\\frac{p}{1 - p} \\approx 3.646,$$ that is, on average in situations like this, when both tests come up positive, there will be $$3.646$$ cases in which the tests were both correct for each case in which both tests were incorrect. That means there are $$3.646$$ accurate positives for every $$3.646 + 1$$ times the test comes out positive both times, which gives a probability of $$\\frac{3.646}{3.646 + 1} \\approx 0.78.$$ The way I worked the probability, however, was to take the fraction $$\\frac{p}{1 - p} = \\frac{361}{99}$$ and directly extract an odds ratio of $$361:99$$ from it. This means I can wait until the very end before doing any roundoff, but in other respects it's the same as the textbook's method. In both cases the odds are simply $$kp : k(1 - p),$$ where $$k$$ is whatever constant you have to multiply each side by in order to produce either $$361:99$$ or $$3.646:1$$ from the odds ratio $$p : (1 - p).$$\n\n\u2022 Hello. Thanks for the detailed and easy-to-understand answer! You've cleared up everything I was wondering regarding this concept. You were also correct in that my hasty rounding-off led to a different answer. The $0.19$ that I got is actually $0.194915$, which gives me a final answer of $0.779827$, which is $78\\%$. Thanks for your explanation regarding the odds ratio as well, I was having some difficulty grasping what that is. \u2013\u00a0Seankala Oct 7 '18 at 2:12\n\u2022 It's just a textbook exercise problem. The percentages were given by the authors of the textbook. I wouldn't say it's \"meaningless\" as it provides something that students like myself can practice with. In this example it's assumed that there is only \"one population\" and Fred is a part of it. How is that a bad assumption? \u2013\u00a0Seankala Oct 7 '18 at 11:49\n\u2022 @David So your objections are that Fred might not be a member of \"the\" population, that the $1\\%$ figure might be inaccurate or outdated. And indeed in real life these are things to consider, in addition to the unstated assumptions made about the frequency of errors of the test. But mathematics is never about absolute answers to real-life questions; it is about seeing what conclusions can be drawn from a given set of assumptions. To refuse to answer that question mathematically because the assumptions might be wrong, seems like a cheap excuse to me. \u2013\u00a0David K Oct 7 '18 at 11:49\n\u2022 @David The link you gave to the other question is a useful one, as the answers to that question include a real-life example that shows why we should consider it important that a disease has a very low incidence in the population. \u2013\u00a0David K Oct 7 '18 at 11:53\n\u2022 In real life, people only usually get tested for a rare disease if there is some suspicion or risk factor involved, in which case those risk factors will change your prior significantly. If Fred was only sent in to test because he had a family history of this rare disease, then your prior is going to be more than 1%. But this is a math exercise from a textbook. The point is to solve it with only what information we know, ceteris paribus. \u2013\u00a0Zubin Mukerjee Oct 7 '18 at 13:56\n\nThe conditional probability (or Bayes's rule) must be: $$P(D\\ |\\ T_1\\cap T_2) = \\frac{P(D\\cap T_1\\cap T_2)}{P(T_1\\ \\color{red}{\\cap}\\ T_2)}=\\frac{P(D)\\cdot P(T_1|D)\\cdot P(T_2|D\\cap T_1)}{P(T_1\\ \\cap\\ T_2)}=\\\\ \\frac{0.01\\cdot 0.95^2}{0.01\\cdot 0.95^2+0.99\\cdot 0.05^2}\\approx 0.7848.$$\n\n\u2022 That's nice, but that doesn't mean it applies to this question as stated. \u2013\u00a0David Oct 6 '18 at 16:41\n\u2022 @David would you care to elaborate why you think Bayes formula does not apply to a problem so clearly amenable to conditioning... \u2013\u00a0Nap D. Lover Oct 7 '18 at 2:12\n\u2022 Fred is a part of the population! I think that you're right, that the question should have stated it explicitly, but mentioning a population that is completely unrelated to Fred would be a hilarious and unlikely red herring by the textbook authors. Why would they just randomly mention a population in a question about Fred, if Fred wasn't part of the population? Once you do assume Fred is part of the population, you can no longer ignore the 1% affliction rate. \u2013\u00a0Zubin Mukerjee Oct 7 '18 at 14:03\n\u2022 It was fun commenting but given the assumptions are true, I agree with these answers here. It was just an exercise on being accurate about assumptions so many next time someone posts a question like this, they will explicitly state the \"missing\" info. I deleted all my previous answers and comments to clean this up. \u2013\u00a0David Oct 7 '18 at 15:40", "date": "2019-06-18 15:20:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2944252/finding-the-probability-that-someone-has-the-disease-given-they-test-positive-o", "openwebmath_score": 0.739641547203064, "openwebmath_perplexity": 311.0663614422447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9566342012360932, "lm_q2_score": 0.9219218348550491, "lm_q1q2_score": 0.8819419580886734}}
{"url": "https://math.stackexchange.com/questions/1591534/for-numerical-integration-is-it-true-that-higher-degree-of-precision-gives-bett", "text": "# For numerical integration, is it true that higher degree of precision gives better accuracy always?\n\nIn case of numerical integration, is it true that higher degree of precision always gives better accuracy? Justify your answer.\n\nI know the definition of degree of precision. For Trapezoidal and Simpson's 1/3 rule they are 1 and 3 respectively. Simpson's 1/3 gives better accuracy than Trapezoidal rule. Then whether the above statement is true always. If not, why? If yes, then why we learn Trapezoidal/ Simpson rule? Why we shall not establish/go for higher and higher DOP from generalized Newton-Cote's rule or other general quadrature formula .\n\n\u2022 Usually, but not always. Consider the function $f(x) = \\left\\{\\matrix{1 & x < 1/2\\\\0 & x \\geq 1/2}\\right.$. The trapesoidal rule with $n=3$ points ($x=0,1/2,1$) gets the integral $\\int_0^1f(x){\\rm d}x$ exactly, but if we use an even number of points we will always be a bit off the true result no matter how large $n$ is. \u2013\u00a0Winther Jan 3 '16 at 8:00\n\u2022 @Winther Please give me the answer when higher degree of precision gives better accuracy and when higher degree of precision does not give better accuracy with analytical justification. \u2013\u00a0user1942348 Jan 3 '16 at 8:10\n\u2022 That is almost impossible I think (for general functions). The key to the proofs of the order of the different methods is that as long as functions have bounded derivatives (of some order) then the error will grow roughly as $\\frac{1}{N^k}$ for some some integer $k$ (and consequently go to zero when $N\\to\\infty$). However we usually don't have any control if $N=38$ will give better results than $N=40$. \u2013\u00a0Winther Jan 3 '16 at 8:12\n\u2022 @Winther What do you mean by \" a bit off the true result\"? \u2013\u00a0user1942348 Jan 3 '16 at 8:18\n\u2022 @winther Somewhere I read that In particular, \"when range of integration is large\", the statement above is not correct. Whether it (\"when range of integration is large, higher DOP gives less accuracy\") is correct. If correct, would you explore why? \u2013\u00a0user1942348 Jan 3 '16 at 8:24\n\nIncreasing the precision, both in terms of the order of the method and in the number of gridpoints used, usually (most of the time) leads to a more accurate estimate for the integral we are trying to compute. However this is not always the case as the following (aritficial) example shows.\n\n$$\\bf \\text{Example where higher order does not imply better accuracy}$$\n\nLet\n\n$$f(x) = \\left\\{\\matrix{1 & x < \\frac{1}{2}\\\\0 & x \\geq \\frac{1}{2}}\\right.$$\n\nand consider the integral $I=\\int_0^1f(x){\\rm d}x = \\frac{1}{2}$. If we use the trapezoidal rule with $n$ gridpoints then\n\n$$I_{n} = \\frac{1}{n}\\sum_{i=1}^{n}\\frac{f(\\frac{i-1}{n})+f(\\frac{i}{n})}{2} \\implies I_n = \\left\\{\\matrix{\\frac{1}{2} & n~~\\text{odd}\\\\\\frac{1}{2} - \\frac{1}{n} & n~~\\text{even}}\\right.$$\n\nso for $n=3$ we have the exact answer which is better than any even $n$ no matter how large it is. This shows that increasing the number of gridpoints does not always improve the accuracy. With Simpson's rule we find\n\n$$I_n = \\frac{1}{3n}\\sum_{i=1}^{n/2}f\\left(\\frac{2i-2}{n}\\right)+4f\\left(\\frac{2i-1}{n}\\right)+f\\left(\\frac{2i}{n}\\right) \\implies I_n = \\left\\{\\matrix{\\frac{1}{2} - \\frac{1}{3n}&n\\equiv 0\\mod 4\\\\\\frac{1}{2}&n\\equiv 1\\mod 4\\\\\\frac{1}{2} + \\frac{2}{3n} & n\\equiv 2\\mod 4\\\\\\frac{1}{2} - \\frac{5}{6n} & n\\equiv 3\\mod 4}\\right.$$\n\nso even if Simpson's rule has higher order we see that it does not always do better than the trapezoidal rule.\n\n$$\\bf \\text{What does higher degree of precision really mean?}$$\n\nIf we have a smooth function then a standard Taylor series error analysis gives that the error in estimating the integral $\\int_a^bf(x){\\rm d}x$ using $n$ equally spaced points is bounded by (here for Simpsons and the trapezoidal rule)\n\n$$\\epsilon_{\\rm Simpsons} = \\frac{(b-a)^5}{2880n^4}\\max_{\\zeta\\in[a,b]}|f^{(4)}(\\zeta)|$$ $$\\epsilon_{\\rm Trapezoidal} = \\frac{(b-a)^3}{12n^2}\\max_{\\zeta\\in[a,b]}|f^{(2)}(\\zeta)|$$\n\nNote that the result we get from such an error analysis is always an upper bound (or in some cases an order of magnitude) for the error apposed to the exact value for the error. What this error analysis tell us is that if $f$ is smooth on $[a,b]$, so that the derivatives are bounded, then the error with a higher order method will tend to decrease faster as we increase the number of gridpoints and consequently we typically need fewer gridpoints to get the same accuracy with a higher order method.\n\nThe order of the method only tell us about the $\\frac{1}{n^k}$ fall-off of the error and says nothing about the prefactor in front so a method that has an error of $\\frac{100}{n^2}$ will tend to be worse than a method that has an error $\\frac{1}{n}$ as long as $n\\leq 100$.\n\n$$\\bf \\text{Why do we need all these methods?}$$\n\nIn principle we don't need any other methods than the simplest one. If we can compute to arbitrary precision and have enough computation power then we can evaluate any integral with the trapezoidal rule. However in practice there are always limitations that in some cases forces us to choose a different method.\n\nUsing a low-order method requires many gridpoints to ensure good enough accuracy which can make the computation take too long time especially when the integrand is expensive to compute. Another problem that can happen even if we can afford to use as many gridpoints as we want is that truncation error (errors due to computers using a finite number of digits) can come into play so even if we use enough points the result might not be accurate.\n\nOther methods can elevate these potential problems. Personally, whenever I need to integrate something and has to implement the method myself I always start with a low-precision method like the trapezoidal rule. This is very easy to implement, it's hard to make errors when coding it up and it's usually good enough for most purposes. If this is not fast enough or if the integrand has properties (e.g. rapid osccilations) that makes it bad I try a different method. For example I have had to compute (multidimensional) integrals where a trapezoidal rule would need more than a year to compute it to good enough accuracy, but with Monte-Carlo integration the time needed was less than a minute! It's therefore good to know different numerical integration methods in case you encounter a problem where the simplest method fails.\n\n\u2022 Your answer is very impressive. Thanks a lot. You have given an example of $f(x)$. It is not differentiable at x=1/2. For such non-smooth function, the Trap is better than Simp. Is there any example where $f(x)$ is continuous and smooth but Trap gives better accuracy than Simp? \u2013\u00a0user1942348 Jan 4 '16 at 13:50\n\u2022 @user1942348 Yes there is. For example take $f(x)$ and connect $x = 1/2 - \\epsilon$ to $x = 1/2 + \\epsilon$ with a line (smoothed at the corners). Take $\\epsilon$ to be tiny and you'll get pretty much the same result as I got here. \u2013\u00a0Winther Jan 4 '16 at 14:21\n\u2022 I just googled and see that for periodic function with limits of integration one with its period, Trap works better. self.gutenberg.org/articles/trapezoidal_rule Do you know is this correct. Can you give such example for me? \u2013\u00a0user1942348 Jan 4 '16 at 15:55\n\u2022 @user1942348 I think the statement on that page is that Trap does better than what we naively would expect from looking at the error term for functions that oscillate, not that it neccesarily is much better than say Simpsons. I tried some random test cases like $\\sin(2\\pi x)$ and $[x(1-x)]^4$ and Trap does indeed do better than expected and the acctual error as function of $n$ is similar to Simpsons for these two functions ($\\epsilon \\propto 1/n^6$ for the latter function). \u2013\u00a0Winther Jan 4 '16 at 16:49\n\u2022 @user1942348 btw let me know if you have Mathematica and I can add the code I used to test this so you can play with it and see for yourself. \u2013\u00a0Winther Jan 4 '16 at 16:53", "date": "2019-07-23 16:52:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1591534/for-numerical-integration-is-it-true-that-higher-degree-of-precision-gives-bett", "openwebmath_score": 0.8301143646240234, "openwebmath_perplexity": 292.14336287453494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540680555949, "lm_q2_score": 0.9005297841157157, "lm_q1q2_score": 0.8819375074789528}}
{"url": "https://mathhelpboards.com/threads/find-all-pairs-m-k-of-integer-solutions.6882/", "text": "Find all pairs (m,k) of integer solutions\n\nanemone\n\nMHB POTW Director\nStaff member\nFind all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.\n\nOpalg\n\nMHB Oldtimer\nStaff member\nFind all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.\nMultiply out the brackets: $$\\displaystyle m^3 + m^2k^2 + mk + k^3 = m^3 -3m^2k + 3mk^2 - k^3$$, which (after dividing by the non-zero $k$) simplifies to $2k^2 + (m^2-3m)k + (m+3m^2) = 0.$ The discriminant of that quadratic in $k$ is $(m^2-3m)^2 - 8(m+3m^2)$, and that has to be a square, say $n^2$. So $$\\displaystyle n^2 = (m^2-3m)^2 - 8(m+3m^2) = m^4 - 6m^3 - 15m^2 - 8m = m(m-8)(m+1)^2$$ and therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \\pm5$. Thus $m = 9$ or $-1$. If $m=9$ then the quadratic for $k$ has solutions $k= -21$ and $k=-6$. If $m=-1$ then the only solution for $k$ is $k=-1$.\n\nThat gives three possible solutions, $(m,k) = (9,-21),\\ (9,-6),\\ (-1,-1)$.\n\nKlaas van Aarsen\n\nMHB Seeker\nStaff member\ntherefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \\pm5$.\nWhy is that?\n\nOpalg\n\nMHB Oldtimer\nStaff member\ntherefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \\pm5$.\nWhy is that?\nIf $x^2-16 = y^2$ then $16 = x^2-y^2 = (x+y)(x-y)$. The only factorisation of $16$ giving positive integer values for $x$ and $y$ is $x+y=8$, $x-y=2$, so that $x=5$ and $y=3$.\n\nanemone\n\nMHB POTW Director\nStaff member\nThank you so much for participating, Opalg! And welcome back to the forum as well! But Opalg, I am sorry to tell you that you missed another solution set, which is $(m, k)=(8, -10)$.\n\nMy solution:\n\nBy expanding and simplifying the given equation, we get\n\n$$\\displaystyle m^3+m^2k^2+mk+k^3=m^3-3m^2k+3mk^2-k^3$$\n\n$$\\displaystyle 2k^3+(m^2-3m)k^2+mk+3m^2k=0$$\n\nSince $k \\ne 0$, divide through the equation above by $k$, we have\n\n$$\\displaystyle 2k^2+(m^2-3m)k+m+3m^2=0$$\n\nand solve for $k$ using the quadratic formula yields\n\n$$\\displaystyle k=\\frac{-(m^2-3m) \\pm \\sqrt{(m^2-3m)^2-4(2)(m+3m^2)}}{2(2)}$$\n\n$$\\displaystyle \\;\\;\\;=\\frac{3m-m^2 \\pm \\sqrt{(m+1)^2(m(m-8))}}{4}$$\n\n$$\\displaystyle \\;\\;\\;=\\frac{3m-m^2 \\pm (m+1)\\sqrt{m(m-8)}}{4}$$\n\nRecall that the expression inside the radical must be greater than or equal to zero, thus, $$\\displaystyle m(m-8)\\ge 0$$ and this gives\n\n$m\\le 0$ and $m \\ge 8$.\n\nAlso, bear in mind that we're told $m, k$ are both integer values, so, $m(m-8)$ has to be a squre.\n\nBut if we apply the AM-GM inequality to the terms $m$ and $m-8$, we find\n\n$$\\displaystyle \\frac{m+m-8}{2} \\ge \\sqrt{m(m-8)}$$\n\n$$\\displaystyle m-4 \\ge \\sqrt{m(m-8)}$$\n\n$m\\ge 8$ implies $m-4 \\ge 4$ and this further implies $\\sqrt{m(m-8)} \\le 4$ and now, we will solve for $m$ by considering $\\sqrt{m(m-8)}=0$ or $1$ or $2$ or $3$or $4$ and we will take whichever solution that gives the integer values of $m$.\n\n$$\\displaystyle \\sqrt{m(m-8)}= 4$$ gives irrational $m$ values.\n\n$$\\displaystyle \\sqrt{m(m-8)}= 3$$ gives $m=-1, 9$.\n\n$$\\displaystyle \\sqrt{m(m-8)}= 2$$ gives irrational $m$ values.\n\n$$\\displaystyle \\sqrt{m(m-8)}= 1$$ gives irrational $m$ values.\n\n$$\\displaystyle \\sqrt{m(m-8)}= 0$$ gives irrational $m=0, 8$ values but $m>0$, hence $m=8$.\n\nThus, by substituting each of the value of $m$ to the equation (*) above to find for its corresponding $k$ value, we end up with the following 4 solution sets, namely\n\n$(m, k)=(-1, -1), (8, -10), (9, -6), (9, -21)$\n\nOpalg\n\nMHB Oldtimer\nStaff member\ntherefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \\pm5$.\nWhy is that?\nOops, I should have taken I like Serena's query more seriously. I was thinking that the only way you could have $x^2-16 = y^2$ was in the situation $5^2 - 4^2 = 3^2$. But in the solution to this problem, the situation $4^2 - 4^2 = 0^2$ is also possible. That leads to the solution $m=8$ and $k=-10$.", "date": "2020-09-22 22:49:35", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-all-pairs-m-k-of-integer-solutions.6882/", "openwebmath_score": 0.9229340553283691, "openwebmath_perplexity": 200.91971747974273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759593358226, "lm_q2_score": 0.8991213874066956, "lm_q1q2_score": 0.8819265534318983}}
{"url": "https://mathhelpboards.com/threads/review-my-solution-trigonometry-proof.5778/", "text": "TrigonometryReview my solution: Trigonometry proof\n\nsweatingbear\n\nMember\nFor a triangle with sides $$\\displaystyle a$$, $$\\displaystyle b$$, $$\\displaystyle c$$ and angle $$\\displaystyle C$$, where the angle $$\\displaystyle C$$ subtends the side $$\\displaystyle c$$, show that\n\n$$\\displaystyle c \\geqslant (a+b) \\sin \\left( \\frac C2 \\right)$$\n_______\n\nLet us stipulate $$\\displaystyle 0^\\circ < C < 180^\\circ$$ and of course $$\\displaystyle a,b,c > 0$$. Consequently, $$\\displaystyle 0^\\circ < \\frac C2 \\leqslant 90^\\circ \\implies 0 < \\sin^2 \\left( \\frac C2 \\right) \\leqslant 1$$ (we include $$\\displaystyle 90^\\circ$$ for the angle $$\\displaystyle \\frac C2$$ in order to account for right triangles).\n\nLaw of cosines yield\n\n$$\\displaystyle c^2 = a^2 + b^2 - 2ab \\cos ( C )$$\n\nUsing $$\\displaystyle \\cos (C) \\equiv 1 - 2\\sin^2 \\left( \\frac C2 \\right)$$, we can write\n\n$$\\displaystyle c^2 = a^2 + b^2 - 2ab + 4ab\\sin^2 \\left( \\frac C2 \\right)$$\n\nNow since $$\\displaystyle 0 < \\sin^2 \\left( \\frac C2 \\right) \\leqslant 1$$, $$\\displaystyle a^2 + b^2 - 2ab + 4ab\\sin^2 \\left( \\frac C2 \\right)$$ will either have to equal $$\\displaystyle a^2 + b^2 - 2ab + 4ab$$ or be greater than it. Thus\n\n$$\\displaystyle c^2 \\geqslant a^2 + b^2 - 2ab + 4ab = a^2 + 2ab + b^2 = (a+b)^2$$\n\nA similar argument can be made for $$\\displaystyle (a+b)^2$$ versus $$\\displaystyle (a+b)^2 \\sin^2 \\left( \\frac C2 \\right)$$: $$\\displaystyle (a+b)^2$$ will either have to equal or be greater than the latter expression, due to the values $$\\displaystyle \\sin^2 \\left( \\frac C2 \\right)$$ can assume. Therefore\n\n$$\\displaystyle c^2 \\geqslant (a+b)^2 \\geqslant (a+b)^2 \\sin^2 \\left( \\frac C2 \\right)$$\n\nWe can finally conclude\n\n$$\\displaystyle c \\geqslant (a+b) \\sin \\left( \\frac C2 \\right)$$\n\nThoughts?\n\nMarkFL\n\nStaff member\nI see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:\n\nI would postulate concerning the angle bisector $m$ of $\\angle C$ and the altitude $h$, we must have:\n\n(1) $$\\displaystyle m\\ge h$$\n\nThat is, the shortest distance between a point and a line is the perpendicular distance.\n\nNow, the area $T$ of the triangle may be written in these ways:\n\n$$\\displaystyle T=\\frac{1}{2}ch=\\frac{1}{2}m(a+b)\\sin\\left(\\frac{C}{2} \\right)$$\n\nNote: I have made use of the formulas (for a general triangle):\n\n$$\\displaystyle T=\\frac{1}{2}bh$$\n\n$$\\displaystyle T=\\frac{1}{2}ab\\sin(C)$$\n\nThus, we have:\n\n$$\\displaystyle ch=m(a+b)\\sin\\left(\\frac{C}{2} \\right)$$\n\nAnd from (1), we therefore conclude:\n\n$$\\displaystyle c\\ge (a+b)\\sin\\left(\\frac{C}{2} \\right)$$\n\nsweatingbear\n\nMember\nI see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:\n\nView attachment 1085\n\nI would postulate concerning the angle bisector $m$ of $\\angle C$ and the altitude $h$, we must have:\n\n(1) $$\\displaystyle m\\ge h$$\n\nThat is, the shortest distance between a point and a line is the perpendicular distance.\n\nNow, the area $T$ of the triangle may be written in these ways:\n\n$$\\displaystyle T=\\frac{1}{2}ch=\\frac{1}{2}m(a+b)\\sin\\left(\\frac{C}{2} \\right)$$\n\nNote: I have made use of the formulas (for a general triangle):\n\n$$\\displaystyle T=\\frac{1}{2}bh$$\n\n$$\\displaystyle T=\\frac{1}{2}ab\\sin(C)$$\n\nThus, we have:\n\n$$\\displaystyle ch=m(a+b)\\sin\\left(\\frac{C}{2} \\right)$$\n\nAnd from (1), we therefore conclude:\n\n$$\\displaystyle c\\ge (a+b)\\sin\\left(\\frac{C}{2} \\right)$$\nThank for the feedback and the excellent alternative solution!\n\nDeveno\n\nWell-known member\nMHB Math Scholar\nA good rule of thumb in mathematics I live by is this:\n\nIf something is true, you should be able to provide two proofs, and one of them should have a picture. This is a perfect example of what I mean.", "date": "2021-01-25 15:07:22", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/review-my-solution-trigonometry-proof.5778/", "openwebmath_score": 0.854730486869812, "openwebmath_perplexity": 213.6857001573631, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874101345136, "lm_q2_score": 0.8902942239389253, "lm_q1q2_score": 0.8819142495493767}}
{"url": "http://mathhelpforum.com/discrete-math/93273-divide-3-a.html", "text": "# Math Help - divide by 3?\n\n1. ## divide by 3?\n\nhi folks,\n\nI hope this is the right place to post this! I have no idea how to proceeed with the following question:\n\nShow that for all positive integral values of n\n\n[HTML]7 <sup>n</sup> + 2 <sup>2n+1</sup>[/HTML]\n\nis divisible by 3.\n\nI tried a few terms as follows:\n\nn = 1. 7 ^ 1 + 2 ^ 3 = 7 + 8 = 15\nn = 2. 7\n^ 2 + 2 ^ 5 = 49 + 32 = 81\nn = 3. 7\n^ 3 + 2 ^ 7 = 343 + 128 = 471\n\nand these are all divisible by 3 but how do I handle the general case?\n\nI thought of expanding the expression i.e.\n\n2.2 ^ n. 2 ^ n - (1 - 8) ^ n\n\nand using a binomial on the second term but it doesn't get me anywhere. I guess I am trying to calculate the sum of a series and show that it has a factor of 3 but I can't see how to do it. Any ideas?\n\nregards and thanks\nSimon\n\np.s. sorry about the formating. the HTML stuff doesn't seem to come out so I resorted to the ^ symbol which is pretty unpretty!\n\n2. Hi,\nOriginally Posted by s_ingram\n\nI thought of expanding the expression i.e.\n\n2.2 ^ n. 2 ^ n - (1 - 8) ^ n\n\nand using a binomial on the second term but it doesn't get me anywhere. I guess I am trying to calculate the sum of a series and show that it has a factor of 3 but I can't see how to do it. Any ideas?\nI suggest writing $7^n+2^{2n+1}=(2^2+3)^n+2^{2n+1}$ and then using the binomial theorem to expand $(2^2+3)^n$.\n\nOriginally Posted by s_ingram\np.s. sorry about the formating. the HTML stuff doesn't seem to come out so I resorted to the ^ symbol which is pretty unpretty!\nTo enter math. equations you can use LaTeX. (see http://www.mathhelpforum.com/math-help/latex-help/)\n\n3. $(2^{2}+3)^{n} = 2^{2n} + \\binom{n}{1}2^{2(n-1)}3^{1} + ... + \\binom{n}{k}2^{2(n-k)}3^{k} + ... + 3^{n} =$\n\n$= 2^{2n} + 3[\\binom{n}{1}2^{2(n-1)} + ... + \\binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}]$\n\nSo we can write:\n\n$3[\\binom{n}{1}2^{2(n-1)} + ... + \\binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}] + 2^{2n}(1+2) =$ $3[\\binom{n}{1}2^{2(n-1)} + ... + \\binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}] + 2^{2n}(3)$\n\nNow factor out a 3 from both terms.\n\n$3[(\\binom{n}{1}2^{2(n-1)} + ... + \\binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}) + 2^{2n} ]$\n\nAnd because we have a number times 3, the result is divisible by 3.\n\nNote: If there aint a careless mistake somewhere in this mess, I am surprised.\n\n4. thanks to Twig and flyingsquirrel. You guys make it look so easy!\n\n5. If you do it by induction here is the last step.\n$\\begin{array}{rcl}\n{7^{n + 1} + 2^{2n + 3} } & = & {7^{n + 1} + 7 \\cdot 2^{2n + 1} - 7 \\cdot 2^{2n + 1} + 2^{2n + 3} } \\\\\n{} & {} & {7\\left( {7^n + 2^{2n + 1} } \\right) - 2^{2n + 1} \\left( {7 - 2^2 } \\right)} \\\\\n{} & {} & {7\\left( {7^n + 2^{2n + 1} } \\right) - 2^{2n + 1} \\left( 3 \\right)} \\\\ \\end{array}$\n\n6. Hello, s_ingram!\n\nHow about an inductive proof?\n\nShow that for any positive integer $n \\!:\\;\\;7^n + 2^{2n+1}$ is divisible by 3.\n\nVerify $S(1)\\!:\\;\\;7^1 + 2^3 \\:=\\:7+8 \\:=\\:15$ ...\ndivisible by 3.\n\nAssume $S(k)\\!:\\;\\;7^k + 2^{2k+1} \\:=\\:3a\\;\\text{ for some integer }a.$\n\nAdd $6\\!\\cdot\\!7^k + 3\\!\\cdot\\!2^{2k+1}$ to both sides:\n\n. . $7^k + {\\color{blue}6\\!\\cdot\\!7^k} + 2^{2k+1} + {\\color{blue}3\\!\\cdot\\!2^{2k+1}} \\;=\\;3a + {\\color{blue}6\\!\\cdot\\!7^k + 3\\!\\cdot\\!2^{2k+1}}$\n\n. . $(1 + 6)\\!\\cdot\\!7^k + (1 + 3)\\!\\cdot\\!2^{2k+1} \\;=\\;3\\left(2a + 2\\!\\cdot\\!7^k + 2^{2k+1}\\right)$ .\na multiple of 3\n\n. . . . . . . . $7\\!\\cdot\\!7^k + 2^2\\!\\cdot2^{2k+1} \\;=\\;3b\\;\\;\\text{ for some integer }b$\n\nTherefore: . . $7^{k+1} + 2^{2k+3} \\;=\\;3b$\n\nWe have proved $S(k+1).$\n. . The inductive proof is complete.\n\n7. Hi Soroban, now that is a smart proof. Too smart! How did you think adding those two terms? Once you see them it all fits but finding them is the real trick! I have never really been impressed with proofs by induction, but I am now! Usually I just try to add the next term in the series and ensure that it has a corresponding impact on the expression fo the sum but with 7 sup k+1 + 2 sup 2k+1 I couldn't even imagine what the next term would be!\n\n8. Or you can just note that\n\n$7^n\\equiv 1^n \\equiv 1 \\mod 3$,\n\n$2^{2n+1} \\equiv (-1)^{2n+1} \\equiv -1 \\mod 3$,\n\nso that $7^n+2^{2n+1} \\equiv 0 \\mod 3$.\n\n9. Originally Posted by s_ingram\nHi Soroban, now that is a smart proof. Too smart! How did you think adding those two terms? Once you see them it all fits but finding them is the real trick! I have never really been impressed with proofs by induction, but I am now! Usually I just try to add the next term in the series and ensure that it has a corresponding impact on the expression fo the sum but with 7 sup k+1 + 2 sup 2k+1 I couldn't even imagine what the next term would be!\nHi\n\n$7^{n+1} + 2^{2n+3} = 7(7^{n} + 2^{2n+1})-7 \\cdot 2^{2n+1} + 2^{2n+3} = 7 \\cdot 3a + 2^{2n+1}(-7+2^2) = 7 \\cdot 3a -3 \\cdot 2^{2n+1}$\n\nTherefore\n$7^{n+1} + 2^{2n+3} = 3(7a-2^{2n+1})$", "date": "2015-11-26 12:07:11", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/93273-divide-3-a.html", "openwebmath_score": 0.8610647320747375, "openwebmath_perplexity": 395.9302970160094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419710536894, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.8818601891349277}}
{"url": "https://mathhelpboards.com/threads/another-solid-of-revolution-problem.1116/", "text": "# [SOLVED]Another Solid of Revolution problem\n\n##### Member\nI'm finding myself stuck again. This time it is more in the set up then the solving.\n\nFind the volume of the following solid of revolution.\n\nThe region bounded by $$\\displaystyle y=\\frac{1}{x^2}$$, y=0, x=2, x=8, and revolved about the y-axis.\n\nI am trying to use the shell method to solve this, as it seems the best scenario in this situation. This is my set-up,\n\n$$\\displaystyle 2\\pi\\int_2^8 (radius)(height)dx$$\n$$\\displaystyle 2\\pi\\int_2^8 (x+2)(\\frac{1}{x^2})dx$$\n$$\\displaystyle 2\\pi\\int_2^8 (\\frac{1}{x}+\\frac{2}{x^2})dx$$\n\nNow here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as $$\\displaystyle 2\\pi\\int_2^8 (\\frac{1}{x})dx$$, but I can't figure out how they are coming up with that integral. It is showing a final solution of $$\\displaystyle \\pi\\ln(16)$$\n\nAny help, or a kick in the right direction, would be greatly appreciated.\n\nMac\n\n#### Chris L T521\n\n##### Well-known member\nStaff member\n[snip]\nNow here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as $$\\displaystyle 2\\pi\\int_2^8 (\\frac{1}{x})dx$$, but I can't figure out how they are coming up with that integral. It is showing a final solution of $$\\displaystyle \\pi\\ln(16)$$\n\nAny help, or a kick in the right direction, would be greatly appreciated.\n\nMac\n[/snip]\nThat's because the radius of your solid of revolution is just $x$, not $x+2$. $x$ is always measured from the $y$-axis (i.e. $x=0$), so there's no need to add 2 to this value; in this case, $x$ just happens to be between 2 and 8.\n\nOn the other hand, the radius would be $x+2$ if you were revolving the region about the line $x=-2$.\n\nI hope this clarifies things!\n\n#### CaptainBlack\n\n##### Well-known member\nI'm finding myself stuck again. This time it is more in the set up then the solving.\n\nFind the volume of the following solid of revolution.\n\nThe region bounded by $$\\displaystyle y=\\frac{1}{x^2}$$, y=0, x=2, x=8, and revolved about the y-axis.\n\nI am trying to use the shell method to solve this, as it seems the best scenario in this situation. This is my set-up,\n\n$$\\displaystyle 2\\pi\\int_2^8 (radius)(height)dx$$\n$$\\displaystyle 2\\pi\\int_2^8 (x+2)(\\frac{1}{x^2})dx$$\n$$\\displaystyle 2\\pi\\int_2^8 (\\frac{1}{x}+\\frac{2}{x^2})dx$$\n\nNow here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as $$\\displaystyle 2\\pi\\int_2^8 (\\frac{1}{x})dx$$, but I can't figure out how they are coming up with that integral. It is showing a final solution of $$\\displaystyle \\pi\\ln(16)$$\n\nAny help, or a kick in the right direction, would be greatly appreciated.\n\nMac\nWhere did a radius of $$x+2$$ come from? Why not $$x$$?\n\nCB", "date": "2021-07-30 16:28:33", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/another-solid-of-revolution-problem.1116/", "openwebmath_score": 0.7557286024093628, "openwebmath_perplexity": 170.146418740701, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138092657496, "lm_q2_score": 0.9019206752113866, "lm_q1q2_score": 0.8818202990164618}}
{"url": "http://math.stackexchange.com/questions/56688/please-help-me-to-prove-the-convergence-of-gamma-n-1-frac12-frac1", "text": "Please help me to prove the convergence of $\\gamma_{n} = 1+\\frac{1}{2}+\\frac{1}{3}+\u2026+\\frac{1}{n}-\\ln n$\n\nPlease help me to prove that $\\gamma_{n} = 1+\\frac{1}{2}+\\frac{1}{3}+....+\\frac{1}{n}-\\ln n$ converge and it's limit $\\gamma \\in [0,1]$. Then, using $\\gamma_{n}$ find the sum: $\\sum_{n=1}^{\\infty}(-1)^{n+1}\\frac{1}{n}$.\n\n-\nWelcome to MSE! I'd suggest though you'd ask question in a less imperative way and in a more friendly way. Plus, have you managed to find something? Any approaches? What I mean is, you're not a book, you're human, so act like one, it makes us more willing to answer. =) \u2013\u00a0 Patrick Da Silva Aug 10 '11 at 11:49\nThe following question addresses the convergence \u2013\u00a0 Sasha Aug 10 '11 at 11:56\npossible duplicate of What is $\\lim_{n \\to \\infty}\\sum_{k=1}^n 1/k$ \u2013\u00a0 user17762 Aug 10 '11 at 12:14\nSorry because my question was imperative. My english is not very good and I translated word by word from my language. I will be more carefully next time. \u2013\u00a0 NumLock Aug 10 '11 at 12:28\nIn the sense that if you read the answer given, all the good stuff's in there. Although they don't compute the alternated sum. I don't agree it's a duplicate. \u2013\u00a0 Patrick Da Silva Aug 10 '11 at 12:56\n\nThe following question addresses the convergence, so assume $\\gamma_n$ converges and denote the limit as $\\gamma$.\n\nNow consider $\\sum_{n=1}^{2m} (-1)^{n+1} \\frac{1}{n}$ and split it into even and odd terms $\\sum_{n=1}^m \\frac{1}{2n-1} - \\sum_{n=1}^m \\frac{1}{2n}$. Then complete the sum over odd integers to sum over consequtive integers: $\\sum_{n=1}^{2m} \\frac{1}{n} - 2 \\sum_{n=1}^m \\frac{1}{2n}$.\n\nThen subtract logarithms to form $\\gamma_{2m} - 2 \\gamma_m + \\log(2)$, like so $$( \\sum_{n=1}^{2m} \\frac{1}{n} - \\log(2m)) - ( \\sum_{n=1}^m \\frac{1}{n} - \\log(m)) + \\log(2).$$\n\nIn the limit $m \\to \\infty$ it becomes $\\gamma - \\gamma + \\log(2) = \\log(2)$.\n\n-\nThank you for your help too! \u2013\u00a0 NumLock Aug 10 '11 at 12:34\n\nYou can use the mean value theorem to prove first that $$\\frac{1}{n+1}<\\ln(n+1)-\\ln n<\\frac{1}{n}$$ then, $\\ln(n+1)<1+\\ldots+\\frac{1}{n}<1+\\ln n$, and finally $$\\ln(n+1)-\\ln n<\\gamma_n<1$$ which implies the convergence of $\\gamma_n$ and its limit $\\gamma\\in[0,1]$.\n\nFor the second part, I am not sure if you can use $\\gamma_n$ to find the sum $\\sum_{n=1}^\\infty(-1)^{n+1}\\frac{1}{n}$!\n\nI suggest to use the Taylor-Young formula for $x\\mapsto\\ln(x+1)$ on $[0,1]$ to prove that $$\\sum_{n=1}^\\infty(-1)^{n+1}\\frac{1}{n}=\\ln2.$$\n\n-\nThank you !I learnt mean value theorem in highschool but I completely forgot about it. \u2013\u00a0 NumLock Aug 10 '11 at 12:34\n@amine it's mercator series time... \u2013\u00a0 user38268 Aug 10 '11 at 12:58\n\nThe Mean Value Theorem says $$\\frac{1}{k+1}<\\log(k+1)-\\log(k)<\\frac{1}{k}$$ Summing from $m$ to $n-1$ ($m < n$), we get \\begin{align} \\sum_{k=m}^{n-1}\\frac{1}{k+1}<\\log(n)-\\log(m)<\\sum_{k=m}^{n-1}\\frac{1}{k} \\end{align} Subtracting $\\displaystyle{\\sum_{k=m}^{n-1}\\frac{1}{k+1}=\\sum_{k=m+1}^n\\frac{1}{k}=\\sum_{k=1}^n\\frac{1}{k}-\\sum_{k=1}^m\\frac{1}{k}}$ from all parts, we get $$0<\\left(\\log(n)-\\sum_{k=1}^n\\frac{1}{k}\\right)-\\left(\\log(m)-\\sum_{k=1}^m\\frac{1}{k}\\right)<\\frac{1}{m}-\\frac{1}{n}$$ This proves the existence of $\\displaystyle{\\lim_{n\\to\\infty}\\left(\\log(n)-\\sum_{k=1}^n\\frac{1}{k}\\right)}$. If we set $m=1$ and subtract all sides from $1$, we get $$1>\\left(\\sum_{k=1}^n\\frac{1}{k}-\\log(n)\\right)>\\frac{1}{n}$$Thus, we have shown that $\\displaystyle{\\gamma=\\lim_{n\\to\\infty}\\left(\\sum_{k=1}^n\\frac{1}{k}-\\log(n)\\right)}$ exists, and that $0\\le\\gamma\\le 1$.\n\nNote that \\begin{align} \\sum_{k=1}^{2n}(-1)^{k+1}\\frac{1}{k}&=\\sum_{k=1}^{2n}\\frac{1}{k}-2\\sum_{k=1}^n\\frac{1}{2k}\\\\ &=\\sum_{k=1}^{2n}\\frac{1}{k}-\\sum_{k=1}^n\\frac{1}{k}\\\\ &=\\left(\\sum_{k=1}^{2n}\\frac{1}{k}-\\log(2n)\\right)-\\left(\\sum_{k=1}^n\\frac{1}{k}-\\log(n)\\right)+\\log(2) \\end{align} Taking the limit as $n\\to\\infty$, we get $$\\sum_{k=1}^\\infty(-1)^{k+1}\\frac{1}{k}=\\gamma-\\gamma+\\log(2)=\\log(2)$$\n\n-\nI apologize if this is close to Sasha's answer, but I have been working on this, during breaks, for quite a while, and posted it before I noticed. Since I show more of the details, I will leave it for now. I can remove it if requested. \u2013\u00a0 robjohn Aug 10 '11 at 17:26\nSorry, for the convergence, I forgot to say that $\\gamma_n$ is decreasing because $$\\gamma_{n+1}-\\gamma_n=\\frac{1}{n+1}-(\\ln(n+1)-\\ln n)<0$$ since it is bounded, it is convergent! \u2013\u00a0 amine Aug 10 '11 at 21:49\n\nDo we have a \"proof without words\" question here? I don't find it. There is a lovely diagram found in certain calculus texts that does this for us. Maybe I will animate it, for fun.\n\n-\nI am not getting this. Do you mean pictorial proof, which suggests the proof ? \u2013\u00a0 Sasha Aug 10 '11 at 14:15\nIt's something like the diagram at the top right here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant but then you slide all the blue regions to the left to see that they all fit inside a $1 \\times 1$ square. \u2013\u00a0 GEdgar Aug 10 '11 at 14:54", "date": "2015-09-03 18:06:03", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/56688/please-help-me-to-prove-the-convergence-of-gamma-n-1-frac12-frac1", "openwebmath_score": 0.9519006013870239, "openwebmath_perplexity": 601.6958971091595, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9921841116729335, "lm_q2_score": 0.8887587817066391, "lm_q1q2_score": 0.8818123423191204}}
{"url": "http://mathhelpforum.com/algebra/164881-word-problem.html", "text": "# Math Help - word problem.\n\n1. ## word problem.\n\nhi ! im having a difficult time answering this word problem.\n\n\"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together.\"\n\nI keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???\n\nThank you very much !!\n\nYou can check:\n\nIn 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms\n\nIn 5 hours, Ben travels: 4+5+6+7+8=30kms\n\nthat means they were together after 5 hours.\n\n3. Nope. Ben started 2 hours after\n\n4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.\n\n5. Here's what I did. We know these two things about arithmetic progressions:\n$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:\n\n$S_n = \\frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.\n\nWe also know that the distance traveled by the first is just 6t.\n\nI'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.\nSo we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:\n\n$6t = \\frac{t-2}{2} (a_1 + a_n) = \\frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$\n$6t = \\frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$\n\nHope you see what I did there. After that, rearrange terms so you get:\n$t^2 - 9t - 10 = 0$\n\nFor that, I get one positive root of t = 10 (also t = -1 which is obviously not useful). Voila! Thanks for the problem, I had fun with that.\n\n6. Hello, dugongster!\n\nAlvin set out from a certain point and travelled at the rate of 6 kph.\nAfter Alvin had gone for 2 hours, Ben set on to overtake him and went 4km the first hour,\n5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour.\n\nAfter many hours were they together?\n\n. . In the next $\\,h$ hours, he travels another $6h$ km.\nAlvin has traveled $6h + 12$ km in the first $\\,h$ hours.\n\nDuring the same $\\,h$ hours, Ben has travelled:\n. . $4 + 5 + 6 + \\hdots + (h+3) \\:=\\:\\dfrac{h(h+7)}{2}$ km. .**\n\nThe two distances are equal: . $\\dfrac{h(h+7)}{2} \\:=\\:6h + 12$\n\n. . $h^2 + 7h \\:=\\:12h + 24 \\quad\\Rightarrow\\quad h^2 - 5h - 24 \\:=\\:0$\n\n. . $(h - 8)(h + 3) \\:=\\:0 \\quad\\Rightarrow\\quad h \\:=\\:8,\\:-3$\n\nAnswer: . $\\text{8 hours}$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\n**\n\nWe want the sum of the numbers from $\\,4$ to $h\\!+\\!3.$\n\nThe sum of the numbers from $\\,1$ to $h\\!+\\!3$ is: . $\\dfrac{(h+3)(h+4)}{2}$\n\nThe sum of the numbers from $\\,1$ to $\\,3$ is: . $1+2+3 \\:=\\:6$\n\nHence,the sum of the numbers from $\\,4$ to $h\\!+\\!3$ is:\n\n. . $\\dfrac{(h+3)(h+4)}{2} - 6 \\;\\;=\\;\\; \\dfrac{h^2 + 7x + 12 - 12}{2} \\;\\;=\\;\\;\\dfrac{h(h+7)}{2}$\n\n7. great. thanks to both of you. grep, you rock !!!\n\n8. Originally Posted by Soroban\nHello, dugongster!\n\nHi Soroban! With due respect (which is considerable, I might add, you being one of the most helpful and, let's face it, awesome people here), I took into account the 2 hour headstart by using (t - 2) as the time in the arithmetic progression (n = (t - 2)), and starting at 4 ( $a_1 = 4$).", "date": "2016-05-05 14:49:30", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/164881-word-problem.html", "openwebmath_score": 0.8460566997528076, "openwebmath_perplexity": 654.912072111246, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877049262134, "lm_q2_score": 0.8976952832120991, "lm_q1q2_score": 0.8817950394695}}
{"url": "https://math.stackexchange.com/questions/3140502/do-the-12-lines-of-a-bingo-card-have-equal-chance-of-winning", "text": "# Do the 12 lines of a bingo card have equal chance of winning?\n\nYou have a $$5 \\times 5$$ bingo card filled with $$25$$ distinct numbers, one per square. There is also a pot containing each number once and you draw them out one by one without replacement.\n\nA line is any of the $$5$$ rows, $$5$$ columns, or $$2$$ main diagonals. A line is completed when its $$5$$ numbers have been drawn. A line wins if it is the first line to be completed.\n\nQuestion: Do the $$12$$ lines have equal chance of winning? If not, which lines have higher chance of winning?\n\nWhy I ask: Define $$T_l$$ to be the time to completion for line $$l$$. It is obvious that all $$T_l$$ are equi-distributed, and thus all $$E[T_l]$$ are equal. However, because the lines overlap, the different $$T_l$$'s are dependent, and it is not clear to me that they have equal chances of being first.\n\nIn particular, imagine that the $$5$$-subsets are not arranged in rows, columns and diagonals, but are clustered in some non-uniform way. Then a subset that shares a lot of elements with other subsets might have a lower chance to win than a subset that does not share a lot of elements with other subsets. (I can provide a simple example if there is interest.)\n\nOn the bingo card, the $$5$$-subsets are pretty uniform but not exactly uniform, due to the diagonals. So my suspicion is that the line win chances are almost equal but not exactly equal. And I am curious as to which lines have higher chances.\n\nI imagine (but would be happy to be proven wrong) that calculating the exact win prob for a line might be difficult/tedious, so qualitative arguments based on e.g. symmetry are also welcome.\n\nClarifications: A drawn number can complete multiple lines, so that needs some special handling. However what I am interested in is the question of equality, so I will accept any reasonable way to handle such \"shared\" wins, i.e. if $$N>1$$ lines are completed at the same draw (and no line has been completed before this draw), then you can treat this as if...\n\n\u2022 they all win (in which case the sum of the $$12$$ win probabilities exceed $$1$$, but that doesn't matter since I am interested in which are higher/lower), or,\n\n\u2022 the whole experiment is repeated from the beginning (i.e. we condition on such shared wins not happening), or,\n\n\u2022 you flip an $$N$$-sided die to determine the winner (i.e. this effectively counts as $$1/N$$ win for each involved line), etc.\n\n\u2022 After this, you can do BINGO cards with a \"free\" space in the middle. \u2013\u00a0GEdgar Mar 8 at 20:37\n\u2022 I don't see why you would think any particular line would have a higher chance of winning if there is an equal probability of each square being picked every time. A line requires 5 numbers being picked and the probability of each 5 number combinations are exactly the same. \u2013\u00a0Matthew Liu Mar 8 at 20:53\n\u2022 @MatthewLiu Because when one line gets a number, it also helps other lines. The diagonal are at a disadvantage. Look at my solution for $3\\times3$ bingo. \u2013\u00a0saulspatz Mar 8 at 21:24\n\u2022 @MatthewLiu what you said is why all the $T_l$ are identically-distributed, but they are not independent. A line $L$ wins if $T_L < T_l \\forall l \\neq L$. \u2013\u00a0antkam Mar 8 at 22:15\n\u2022 Ok you guys are correct. The diagonal lines are at a strict disadvantage. \u2013\u00a0Matthew Liu Mar 11 at 17:14\n\nI must admit, when I read the title of the question, I thought \"Of course they all have an equal chance of winning. What a silly question.\" But then, I read the body, and was delighted to see that you had a valid point, and I was the silly one. I don't know how to solve for $$5\\times5$$ bingo in any reasonable time, but for $$3\\times3$$ bingo we have only $$9!$$ permutations of the numbers to consider. I wrote a python script to exhaustively test this.\n\nfrom itertools import permutations\nfrom fractions import Fraction\n\nscore = 9*[0]\ncard = 8*[0]\n\nfor p in permutations(range(9)):\ncard[0] = {0,1,2}\ncard[1] = {3,4,5}\ncard[2] = {6,7,8}\ncard[3] = {0,3,6}\ncard[4] = {1,4,7}\ncard[5] = {2,5,8}\ncard[6] = {0,4,8}\ncard[7] = {2,4,6}\nfor number in p:\nfor line in card:\nline -= {number}\nwinners = len([line for line in card if not line])\nif winners:\nfor idx, line in enumerate(card):\nif not line: score[idx] += Fraction(1, winners)\nbreak\n\nfor idx in range(8):\nprint(idx, score[idx])\n\n\nThis produced the output\n\n0 47520\n1 45600\n2 47520\n3 47520\n4 45600\n5 47520\n6 40800\n7 40800\n\n\nThis means that the diagonals are worst, and the lines along the edges of the card are best, with the horizontal and vertical lines through the center in the middle.\n\nI should mention that in the case of ties, I awarded an equal fraction of the game to each winner, so that the sum of the scores does work out to $$9!.$$\n\n\u2022 What does it produce if you give a full win to each line if 2/3/4 lines win at once? \u2013\u00a0Banbadle Mar 8 at 21:58\n\u2022 awesome! i had thought the diagonals are worst but all the rows and columns are equal, but you showed that the row/col passing through the center are penalized. interesting... and now i wonder if the 1st & 2nd rows are different in the 5x5. ooh, yeah, i would like to see the idea of @Banbadle tested too. would the multiple full wins cancel the effect of sharing? \u2013\u00a0antkam Mar 8 at 22:00\n\u2022 @Banbadle I just modified saulspatz's code to try other scoring modes. If a full win is given to each simultaneously winning line, the win counts are 54000 > 53290 > 46656 for the edges, center row/col, diagonals respectively. If we condition on no shared wins (i.e. those permutations award no wins to anybody), then the win counts are 41040 > 38880 > 35424, again for edge, center row/col, diagonals. So the order is the same in all 3 win-counting models. \u2013\u00a0antkam Mar 8 at 22:44\n\nI simulated a standard bingo game, one million games with ten random cards for each game. I recorded the number of times each line triggered bingo, including bingo on multiple cards and multiple lines winning simultaneously on the same card.\n\nThe results:\n\n\u2022 Horizontal through center wins $$16.2\\%$$ of games\n\u2022 Vertical through center wins $$16.1\\%$$ of games\n\u2022 Diagonals each win $$15.9\\%$$ of games\n\u2022 Other horizontal lines each win $$5.8\\%$$ of games\n\u2022 Other vertical lines each win $$5.6\\%$$ of games\n\nNote that this totals over $$100\\%$$, accounting for multiple wins in a single game (there were nearly $$1.1$$ million wins in one million games)\n\nAfter ensuring that these results are indeed accurate and giving some thought to why the diagonals win less than the central row and column, it is clear that this is due to intersecting lines. Each of the squares on the diagonals, when drawn, has the potential to trigger a win on either of two intersecting lines, while each square in the central row or column can only trigger a win on a single intersecting line.\n\nI have also run the simulation without the free center square:\n\n\u2022 Rows win $$9.42\\%$$\n\u2022 Diagonals win $$9.17\\%$$\n\u2022 Columns win $$9.05\\%$$\n\u2022 when you say \"standard bingo game\" do you mean the center square is free (i.e. already marked off)? if so that must be why lines through center win much more coz they only require 4 cells. \u2013\u00a0antkam Mar 8 at 22:04\n\u2022 @antkam Yes, center square is free in my simulation. \u2013\u00a0Daniel Mathias Mar 8 at 22:05\n\u2022 Even in the free-center case it is open (at least, in my mind) whether the diagonals are at a disadvantage. If you found that diagonals win 10% vs the center row/col winning 20% each, that would be quite conclusive. But at 15.9.% vs 16.1% I am not sure this is a real effect or just within the error bar. \u2013\u00a0antkam Mar 8 at 22:18\n\u2022 @antkam A run of ten million games: $16.18\\%$, $16.09\\%$, $15.91\\%$, $15.90\\%$ for center row, center column, downward diagonal, upward diagonal. Of course, this only proves that my simulation produces consistent results. It is entirely possible that the inequality is a result of some unintended bias, though I have found no potential cause for bias. \u2013\u00a0Daniel Mathias Mar 8 at 23:31\n\u2022 @DanielMathias Thanks. I knew very little about bingo before reading this. Now I have upgraded my knowledge from \"very little\" to just \"little\". \u2013\u00a0badjohn Mar 9 at 14:54", "date": "2019-06-19 16:41:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3140502/do-the-12-lines-of-a-bingo-card-have-equal-chance-of-winning", "openwebmath_score": 0.5851178765296936, "openwebmath_perplexity": 759.8442247230166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471642306267, "lm_q2_score": 0.896251377983158, "lm_q1q2_score": 0.8817743766665216}}
{"url": "https://math.stackexchange.com/questions/59943/if-n-323232-cdots-50-text-digits-9-i-e-in-base-9-then-how-to-find-t?noredirect=1", "text": "# If $N= (323232 \\cdots 50 \\text{ digits})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?\n\nIf $N= (\\underbrace{323232 \\cdots}_{50 \\text{ digits}})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?\n\nI am looking for a \"fast\" approach that could be used to solve this in less than $2$ minutes.\n\n\u2022 This is computing $N\\mod 8$. Since the $9\\mod 8 \\equiv 1$, this is just the sum of digits of the number $N$, i.e. $(3+2)\\times 25\\mod 8 \\equiv (3+2)\\times 1\\mod 8 = 5$. Did I miss the question ? \u2013\u00a0Sasha Aug 26 '11 at 16:45\n\u2022 @Sasha:No,you didn't miss the question.I just can't see this solution before :/ \u2013\u00a0Quixotic Aug 26 '11 at 17:26\n\nThe same way we find the remainder of dividing a number by $9$ when the number is written in base 10: add the digits.\n\nWhy?\n\nBecause remember that writing, say, $(38571)_9$ \"really\" means $$1 + 7\\times 9 + 5\\times 9^2 + 8\\times 9^3 + 3\\times 9^4.$$ When you divide $9$ by $8$, the remainder is $1$; when you divide $9^2$ by $8$, the remainder is $1^2 = 1$; then you divide $9^3$ by $8$, the remainder is $1^3=1$, etc. So the remainder of this number when divided by $8$ is the same as the remainder of $$1 + 7\\times 1 + 5\\times 9^2 + 8\\times 1^3 + 3\\times 1^4$$ which is the same as the remainder of $1+7+5+8+3 = 24$, which is $0$.\n\nIn this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\\times 25 = 125$ when divided by $8$, which is $5$.\n\n(More generally, the remainder of dividing the number $(b_1b_2\\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\\cdots+b_k$ by $b-1$).\n\n(Even more generally, the remainder of dividing $(b_1b_2\\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).\n\n\u2022 I feel like stupid now,the first general result is not new to me and could be easily be shown as follows:$N = (a_1a_2\\cdots a_k)_r = a_k + \\cdots +a_2 \\times r^{k-2} + a_2 \\times r^{k-1}$ and $S = a_1 + a_2 + \\cdots a_k$ then it could be easily shown that $\\frac{N-S}{r-1} = I \\text{ (an integer) }+\\frac{S}{r-1}$,which proves that result. But I don't know why I can't see this direct application before! :/ \u2013\u00a0Quixotic Aug 26 '11 at 17:23\n\u2022 Anyways,the more general result is new to me,and perhaps a very useful one,could you hint me a easy proof for the same? \u2013\u00a0Quixotic Aug 26 '11 at 17:25\n\u2022 @FoolForMath The key idea is to learn modular arithmetic and, more generally, modular reductions of problems in quotient algebras. It's one way to algebraically divide and conquer. You can find further discussion of such in some of my posts here if you follow the link I gave, and its links... \u2013\u00a0Bill Dubuque Aug 26 '11 at 17:28\n\u2022 I understood it now :D and it was before Bill Dubuque's post :-) but yes Bill you are right,my understanding is due to modular arithmetic. \u2013\u00a0Quixotic Aug 26 '11 at 17:31\n\u2022 @Foo One key thing to keep in mind is that radix notation has polynomial form, so this can be viewed as a special case of well-known results about polynomials. Always be on the lookout for analogies between numbers and functions (here polynomials). \u2013\u00a0Bill Dubuque Aug 26 '11 at 17:55\n\nHINT $\\$ Employ the radix $\\:9\\:$ analog of casting out nines in decimal, namely\n\n$$\\rm\\quad mod\\ 8:\\ \\ \\ 9 \\equiv 1\\ \\ \\Rightarrow\\ \\ d_n\\: 9^n +\\:\\cdots\\:+d_1\\:9 + d_0\\ \\equiv\\ d_n +\\:\\cdots\\:+d_1+d_0$$\n\nThe same idea works generally to cast out $\\rm\\:b\\pm1\\:$'s in radix $\\rm\\:b\\:$, e.g. see here for many links.\n\nIt may be viewed as a number-theoretic specialization of the Polynomial Remainder Theorem\n\n$\\rm\\quad f(x)\\ mod\\ (x-r)\\:=\\:f(r)\\:,\\$ thus $\\rm\\ \\ f(x)\\ mod\\ (x-1)\\:=\\:f(1)\\: =\\: f_n +\\:\\cdots\\:+f_1 + f_0$", "date": "2019-06-17 11:27:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/59943/if-n-323232-cdots-50-text-digits-9-i-e-in-base-9-then-how-to-find-t?noredirect=1", "openwebmath_score": 0.8038354516029358, "openwebmath_perplexity": 185.7710300479582, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471665987073, "lm_q2_score": 0.8962513738264114, "lm_q1q2_score": 0.8817743746993136}}
{"url": "http://cms.math.ca/Competitions/MOCP/2001/sol_feb.mml", "text": "location:\n\n61.\nLet $S=1!2!3!\\dots 99!100!$ (the product of the first 100 factorials). Prove that there exists an integer $k$ for which $1\\le k\\le 100$ and $S/k!$ is a perfect square. Is $k$ unique? (Optional: Is it possible to find such a number $k$ that exceeds 100?)\nSolution 1. Note that, for each positive integer $j$, $\\left(2j-1\\right)!\\left(2j\\right)!=\\left[\\left(2j-1\\right)!\\right]{}^{2}\u00b72j$. Hence\n\n$S=\\underset{j=1}{\\overset{50}{\\Pi }}\\left[\\left(2j-1\\right)!\\right]{}^{2}\\left[2j\\right]={2}^{50}50!\\left[\\underset{j=1}{\\overset{50}{\\Pi }}\\left(2j-1\\right)!{\\right]}^{2} ,$\n\nfrom which we see that $k=50$ is the required number.\nWe show that $k=50$ is the only possibility. First, $k$ cannot exceed 100, for otherwise $101!$ would be a factor of $k!$ but not $S$, and so $S/k!$ would not even be an integer. Let $k\\le 100$. The prime $47$ does not divide $k!$ for $k\\le 46$ and divides $50!$ to the first power. Since $S/50!$ is a square, it evidently divides $S$ to an odd power. So $k\\ge 47$ in order to get a quotient divisible by 47 to an even power. The prime $53$ divides each $k!$ for $k\\ge 53$ to the first power and divides $S/50!$, and so $S$ to an even power. Hence, $k\\le 52$.\nThe prime $17$ divides $50!$ and $S/50!$, and hence $S$ to an even power, but it divides each of $51!$ and $52!$ to the third power. So we cannot have $k=51$ or 52. Finally, look at the prime 2. Suppose that ${2}^{2u}$ is the highest power of 2 that divides $S/50!$ and that ${2}^{v}$ is the highest power of 2 that divides $50!$; then ${2}^{2u+v}$ is the highest power of 2 that divides $S$. The highest power of $2$ that divides $48!$ and $49!$ is ${2}^{v-1}$ and the highest power of 2 that divides $46!$ and $47!$ is ${2}^{v-5}$. >From this, we deduce that 2 divides $S/k!$ to an odd power when $47\\le k\\le 49$. The desired uniqueness of $k$ follows.\nSolution 2. Let $p$ be a prime exceeding 50. Then $p$ divides each of $m!$ to the first power for $p\\le m\\le 100$, so that $p$ divides $S$ to the even power $100-\\left(p-1\\right)=101-p$. From this, it follows that if $53\\ge k$, $p$ must divide $S/k!$ to an odd power.\nOn the other hand, the prime 47 divides each $m!$ with $47\\le m\\le 93$ to the first power, and each $m!$ with $94\\le m\\le 100$ to the second power, so that it divides $S$ to the power with exponent $54+7=61$. Hence, in order that it divide $S/k!$ to an even power, we must make $k$ one of the numbers $47,\\dots ,52$.\nBy an argument, similar to that used in Solution 1, it can be seen that $2$ divides any product of the form $1!2!\\dots \\left(2m-1\\right)!$ to an even power and $100!$ to the power with exponent\n\n$\u230a100/2\u230b+\u230a100/4\u230b+\u230a100/8\u230b+\u230a100/16\u230b+\u230a100/32\u230b+\u230a100/64\u230b=50+25+12+6+3+1=97 .$\n\nHence, 2 divides $S$ to an odd power. So we need to divide $S$ by $k!$ which 2 divides to an odd power to get a perfect square quotient. This reduces the possibilities for $k$ to 50 or 51. Since\n\n$S={2}^{99}\u00b7{3}^{98}\u00b7{4}^{97}\\dots {99}^{2}\u00b7100=\\left(2\u00b74\\dots 50\\right)\\left({2}^{49}\u00b7{3}^{49}\u00b7{4}^{48}\\dots 99\\right){}^{2}=50!\u00b7{2}^{50}\\left(\\dots \\right){}^{2} ,$\n\n$S/50!$ is a square, and so $S/51!=\\left(S/50!\\right)\u00f7\\left(51\\right)$ is not a square. The result follows.\nSolution 3. As above, $S/\\left(50!\\right)$ is a square. Suppose that $53\\le k\\le 100$. Then 53 divides $k!/50!$ to the first power, and so $k!/50!$ cannot be square. Hence $S/k!=\\left(S/50!\\right)\u00f7\\left(k!/50!\\right)$ cannot be square. If $k=51$ or 52, then $k!/50!$ is not square, so $S/k!$ cannot be square. Suppose that $k\\le 46$. Then 47 divides $50!/k!$ to the first power, so that $50!/k!$ is not square and $S/k!=\\left(S/50!\\right)\u00d7\\left(50!/k!\\right)$ cannot be square. If $k=47,48$ or 49, then $50!/k!$ is not square and so $S/k!$ is not square. Hence $S/k!$ is square if and only if $k=50$ when $k\\le 100$.\n\n62.\nLet $n$ be a positive integer. Show that, with three exceptions, $n!+1$ has at least one prime divisor that exceeds $n+1$.\nSolution. Any prime divisor of $n!+1$ must be larger than $n$, since all primes not exceeding $n$ divide $n!$. Suppose, if possible, the result fails. Then, the only prime that can divide $n!+1$ is $n+1$, so that, for some positive integer $r$ and nonnegative integer $K$,\n\n$n!+1=\\left(n+1\\right){}^{r}=1+\\mathrm{rn}+{\\mathrm{Kn}}^{2} .$\n\nThis happens, for example, when $n=1,2,4$: $1!+1=2$, $2!+1=3$, $4!+1={5}^{2}$. Note, however, that the desired result does hold for $n=3$: $3!+1=7$.\nHenceforth, assume that $n$ exceeds 4. If $n$ is prime, then $n+1$ is composite, so by our initial comment, all of its prime divisors exceed $n+1$. If $n$ is composite and square, then $n!$ is divisible by the four distinct integers $1,n,\\sqrt{n},2\\sqrt{n}$, while is $n$ is composite and nonsquare with a nontrivial divisor $d$. then $n!$ is divisible by the four distinct integers $1,d,n/d,n$. Thus, $n!$ is divisible by ${n}^{2}$. Suppose, if possible, the result fails, so that $n!+1=1+\\mathrm{rn}+{\\mathrm{Kn}}^{2}$, and $1\\equiv 1+\\mathrm{rn}$ (mod ${n}^{2}$). Thus, $r$ must be divisible by $n$, and, since it is positive, must exceed $n$. Hence\n\n$\\left(n+1\\right){}^{r}\\ge \\left(n+1\\right){}^{n}>\\left(n+1\\right)n\\left(n-1\\right)\\dots 1>n!+1 ,$\n\na contradiction. The desired result follows.\n\n63.\nLet $n$ be a positive integer and $k$ a nonnegative integer. Prove that\n\n$n!=\\left(n+k\\right){}^{n}-\\left(\\genfrac{}{}{0}{}{n}{1}\\right)\\left(n+k-1\\right){}^{n}+\\left(\\genfrac{}{}{0}{}{n}{2}\\right)\\left(n+k-2\\right){}^{n}-\\dots \u00b1\\left(\\genfrac{}{}{0}{}{n}{n}\\right){k}^{n} .$\n\nSolution 1. Recall the Principle of Inclusion-Exclusion: Let $S$ be a set of $n$ objects, and let ${P}_{1}$, ${P}_{2}$, $\\dots$, ${P}_{m}$ be $m$ properties such that, for each object $x\\in S$ and each property ${P}_{i}$, either $x$ has the property ${P}_{i}$ or $x$ does not have the property ${P}_{i}$. Let $f\\left(i,j,\\dots ,k\\right)$ denote the number of elements of $S$ each of which has properties ${P}_{i}$, ${P}_{j}$, $\\dots$, ${P}_{k}$ (and possibly others as well). Then the number of elements of $S$ each having none of the properties ${P}_{1}$, ${P}_{2}$, $\\dots$, ${P}_{m}$ is\n\n$n-\\sum _{1\\le i\\le m}f\\left(i\\right)+\\sum _{1\\le i\n\nWe apply this to the problem at hand. Note that an ordered selection of $n$ numbers selected from among $1,2,\\dots ,n+k$ is a permutation of $\\left\\{1,2,\\dots ,n\\right\\}$ if and only if it is constrained to contain each of the numbers 1, 2, $\\dots$, $n$. Let $S$ be the set of all ordered selections, and we say that a selection has property ${P}_{i}$ iff its fails to include at least $i$ of the numbers $1,2,\\dots ,n$ $\\left(1\\le i\\le n\\right)$. The number of selections with property ${P}_{i}$ is the product of $\\left(\\genfrac{}{}{0}{}{n}{i}\\right)$, the number of ways of choosing the $i$ numbers not included and $\\left(n+k-i\\right){}^{n}$, the number of ways of choosing entries for the $n$ positions from the remaining $n+k-1$ numbers. The result follows.\nSolution 2. We begin with a lemma:\n\n$\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{m}=\\left\\{\\begin{array}{cc}\\hfill 0\\hfill & \\hfill \\left(0\\le m\\le n-1\\right)\\hfill \\\\ \\multicolumn{0}{c}{\\left(-1\\right){}^{n}n!}& \\hfill \\left(m=n\\right) .\\hfill \\end{array}$\n\nWe use the convention that ${0}^{0}=1$. To prove this, note first that $i\\left(i-1\\right)\\dots \\left(i-m\\right)={i}^{m+1}+{b}_{m}{i}^{m}+\\dots +{b}_{1}i+{b}_{0}$ for some integers ${b}_{i}$. We use an induction argument on $m$. The result holds for each positive $n$ and for $m=0$, as the sum is the expansion of $\\left(1-1\\right){}^{n}$. It also holds for $n=1,2$ and all relevant $m$. Fix $n\\ge 3$. Suppose that it holds when $m$ is replaced by $k$ for $0\\le k\\le m\\le n-2$. Then\n\n$\\begin{array}{cc}\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{m+1}\\hfill & =\\sum _{i=1}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)i\\left(i-1\\right)\\dots \\left(i-m\\right)-\\sum _{k=0}^{m}{b}_{k}\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{k}\\hfill \\\\ \\multicolumn{0}{c}{}& =\\sum _{i=m+1}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)i\\left(i-1\\right)\\dots \\left(i-m\\right)-0\\hfill \\\\ \\multicolumn{0}{c}{}& =\\sum _{i=m+1}^{n}\\left(-1\\right){}^{i}\\frac{n!i!}{i!\\left(n-i\\right)!\\left(i-m-1\\right)!}=\\sum _{j=0}^{n-m-1}\\left(-1\\right){}^{m+1+j}\\frac{n!}{\\left(n-m-1-j\\right)!j!}\\hfill \\\\ \\multicolumn{0}{c}{}& =\\sum _{j=0}^{n-m-1}\\left(-1\\right){}^{m+1}\\left(-1\\right){}^{j}\\frac{n\\left(n-1\\right)\\dots \\left(n-m\\right)\\left[\\left(n-m-1\\right)!\\right]}{\\left(n-m-1-j\\right)!j!}\\hfill \\\\ \\multicolumn{0}{c}{}& =\\left(-1\\right){}^{m+1}n\\left(n-1\\right)\\dots \\left(n-m\\right)\\sum _{j=0}^{n-m-1}\\left(-1\\right){}^{j}\\left(\\genfrac{}{}{0}{}{n-m-1}{j}\\right)=0 .\\hfill \\end{array}$\n\n(Note that the $j=0$ term is 1, which is consistent with the ${0}^{0}=1$ convention mentioned earlier.) So $\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{m}=0$ for $0\\le m\\le n-1$. Now consider the case $m=n$:\n\n$\\sum _{i=1}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{n}=\\sum _{i=1}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)i\\left(i-1\\right)\\dots \\left(i-n+1\\right)-\\sum _{k=0}^{n-1}{b}_{k}\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{k} .$\n\nEvery term in the first sum vanishes except the $n$th and each term of the second sum vanishes. Hence $\\sum _{i=1}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{n}=\\left(-1\\right){}^{n}n!$.\nReturning to the problem at hand, we see that the right side of the desired equation is equal to\n\n$\\begin{array}{cc}\\left(n+k\\right){}^{n}\\hfill & -\\left(\\genfrac{}{}{0}{}{n}{1}\\right)\\left(n+k-1\\right){}^{n}+\\left(\\genfrac{}{}{0}{}{n}{2}\\right)\\left(n+k-2\\right){}^{n}-\\dots +\\left(-1\\right){}^{n}\\left(\\genfrac{}{}{0}{}{n}{n}\\right)\\left(n+k-n\\right){}^{n}\\hfill \\\\ \\multicolumn{0}{c}{}& =\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\left(n-i+k\\right){}^{n}=\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\sum _{j=0}^{n}\\left(\\genfrac{}{}{0}{}{n}{j}\\right)\\left(n-i\\right){}^{j}{k}^{n-j}\\hfill \\\\ \\multicolumn{0}{c}{}& =\\sum _{i=0}^{n}\\sum _{j=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\left(\\genfrac{}{}{0}{}{n}{j}\\right)\\left(n-i\\right){}^{j}{k}^{n-j}=\\sum _{j=0}^{n}\\left(\\genfrac{}{}{0}{}{n}{j}\\right){k}^{n-j}\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\left(n-i\\right){}^{j}\\hfill \\\\ \\multicolumn{0}{c}{}& =\\sum _{j=0}^{n}\\left(\\genfrac{}{}{0}{}{n}{j}\\right){k}^{n-j}\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{n-i}\\right)\\left(n-i\\right){}^{j}\\hfill \\\\ \\multicolumn{0}{c}{}& =\\sum _{j=0}^{n}\\left(\\genfrac{}{}{0}{}{n}{j}\\right){k}^{n-j}\\sum _{i=0}^{n}\\left(-1\\right){}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{j} .\\hfill \\end{array}$\n\nWhen $0\\le j\\le n-1$, the sum $\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{n-i}\\right)\\left(n-i\\right){}^{j}=\\sum _{i=0}^{n}\\left(-1\\right){}^{n-i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){i}^{j}$ vanishes, while, when $j=n$, it assunes the value $n!$. Thus, the right side of the given equation is equal to $\\left(\\genfrac{}{}{0}{}{n}{n}\\right){k}^{0}n!=n!$ as desired.\nSolution 3. Let $m=n+k$, so that $m\\ge n$, and let the right side of the equation be denoted by $R$. Then\n\n$\\begin{array}{cc}R\\hfill & ={m}^{n}-\\left(\\genfrac{}{}{0}{}{n}{1}\\right)\\left(m-1\\right){}^{n}+\\left(\\genfrac{}{}{0}{}{n}{2}\\right)\\left(m-2\\right){}^{n}-\\dots +\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\left(m-i\\right){}^{n}+\\dots +\\left(-1\\right){}^{n}\\left(\\genfrac{}{}{0}{}{n}{n}\\right)\\left(m-n\\right){}^{n}\\hfill \\\\ \\multicolumn{0}{c}{}& ={m}^{m}\\left[\\sum _{j=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\right]-\\left(\\genfrac{}{}{0}{}{n}{1}\\right){m}^{n-1}\\left[\\sum _{i=1}^{n}\\left(-1\\right){}^{i}i\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\right]+\\left(\\genfrac{}{}{0}{}{n}{2}\\right){m}^{n-2}\\left[\\sum _{i=1}^{n}\\left(-1\\right){}^{i}{i}^{2}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\right]+\\dots \\hfill \\\\ \\multicolumn{0}{c}{}& +\\left(-1\\right){}^{n}\\left(\\genfrac{}{}{0}{}{n}{n}\\right)\\left[\\sum _{i=1}^{n}\\left(-1\\right){}^{i}{i}^{n}\\left(\\genfrac{}{}{0}{}{n}{i}\\right)\\right] .\\hfill \\end{array}$\n\nLet\n\n${f}_{0}\\left(x\\right)=\\left(1-x\\right){}^{n}=\\sum _{i=0}^{n}\\left(-1\\right){}^{i}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){x}^{i}$\n\nand let\n\n${f}_{k}\\left(x\\right)=x{\\mathrm{Df}}_{k-1}\\left(x\\right)$\n\nfor $k\\ge 1$, where $\\mathrm{Df}$ denotes the derivative of a function $f$. Observe that, from the closed expression for ${f}_{0}\\left(x\\right)$, we can establish by induction that\n\n${f}_{k}\\left(x\\right)=\\sum _{i=0}^{n}\\left(-1\\right){}^{i}{i}^{k}\\left(\\genfrac{}{}{0}{}{n}{i}\\right){x}^{i}$\n\nso that $R=\\sum _{k=0}^{n}\\left(-1\\right){}^{k}\\left(\\genfrac{}{}{0}{}{n}{k}\\right){m}^{n-k}{f}_{k}\\left(1\\right)$.\nBy induction, we establish that\n\n${f}_{k}\\left(x\\right)=\\left(-1\\right){}^{k}n\\left(n-1\\right)\\dots \\left(n-k+1\\right){x}^{k}\\left(1-x\\right){}^{n-k}+\\left(1-x\\right){}^{n-k+1}{g}_{k}\\left(x\\right)$\n\nfor some polynomial ${g}_{k}\\left(x\\right)$. This is true for $k=1$ with ${g}_{1}\\left(x\\right)=0$. Suppose if holds for $k=j$. Then\n\n$\\begin{array}{cc}{f}_{j}\\text{'}\\left(x\\right)\\hfill & =\\left(-1\\right){}^{j}n\\left(n-1\\right)\\dots \\left(n-j+1\\right){x}^{j-1}\\left(1-x\\right){}^{n-j}-\\left(-1\\right){}^{j}n\\left(n-1\\right)\\dots \\left(n-j+1\\right)\\left(n-j\\right){x}^{j}\\left(1-x\\right){}^{n-j-1}\\hfill \\\\ \\multicolumn{0}{c}{}& -\\left(n-j+1\\right)\\left(1-x\\right){}^{n-j}{g}_{j}\\left(x\\right)+\\left(1-x\\right){}^{n-j+1}{g}_{j}\\text{'}\\left(x\\right) ,\\hfill \\end{array}$\n\nwhence\n\n$\\begin{array}{cc}{f}_{j+1}\\left(x\\right)\\hfill & =\\left(-1\\right){}^{j+1}n\\left({n}_{1}\\right)\\dots \\left(n-j\\right){x}^{j}\\left(1-x\\right){}^{n-\\left(j+1\\right)}+\\left(1-x\\right){}^{n-\\left(j+1\\right)+1}\\left[\\left(-1\\right){}^{j}n\\left(n-1\\right)\\dots \\left(n-j+1\\right){x}^{j}\\hfill \\\\ \\multicolumn{0}{c}{}& -\\left(n-j+1\\right){\\mathrm{xg}}_{k}\\left(x\\right)+x\\left(1-x\\right){g}_{j}\\text{'}\\left(x\\right)\\right]\\hfill \\end{array}$\n\nand we obtain the desired representation by induction. Then for $1\\le k\\le n-1$, ${f}_{k}\\left(1\\right)=0$ while ${f}_{n}\\left(1\\right)=\\left(-1\\right){}^{n}n!$. Hence $R=\\left(-1\\right){}^{n}{f}_{n}\\left(1\\right)=n!$.\n\n64.\nLet $M$ be a point in the interior of triangle $\\mathrm{ABC}$, and suppose that $D$, $E$, $F$ are respective points on the side $\\mathrm{BC}$, $\\mathrm{CA}$, $\\mathrm{AB}$, which all pass through $M$. (In technical terms, they are cevians.) Suppose that the areas and the perimeters of the triangles $\\mathrm{BMD}$, $\\mathrm{CME}$, $\\mathrm{AMF}$ are equal. Prove that triangle $\\mathrm{ABC}$ must be equilateral.\nSolution. [L. Lessard] Let the common area of the triangles $\\mathrm{BMD}$, $\\mathrm{CME}$ and $\\mathrm{AMF}$ be $a$ and let their common perimeter be $p$. Let the area and perimeter of $\\Delta \\mathrm{AME}$ be $u$ and $x$ respectively, of $\\Delta \\mathrm{MFB}$ be $v$ and $y$ respectively, and of $\\Delta \\mathrm{CMD}$ be $w$ and $z$ respectively.\nBy considering pairs of triangles with equal heights, we find that\n\n$\\frac{\\mathrm{AF}}{\\mathrm{FB}}=\\frac{a}{v}=\\frac{2a+u}{v+a+w}=\\frac{a+u}{a+w} ,$\n\n$\\frac{\\mathrm{BD}}{\\mathrm{DC}}=\\frac{a}{w}=\\frac{2a+v}{u+a+w}=\\frac{a+v}{a+u} ,$\n\n$\\frac{\\mathrm{CE}}{\\mathrm{EA}}=\\frac{a}{u}=\\frac{2a+w}{u+a+v}=\\frac{a+w}{a+v} .$\n\n>From these three sets of equations, we deduce that\n\n$\\frac{{a}^{3}}{\\mathrm{uvw}}=1 ;$\n\n${a}^{2}+\\left(w-u\\right)a-\\mathrm{uv}=0 ,$\n\n${a}^{2}+\\left(u-w\\right)a-\\mathrm{vw}=0 ,$\n\n${a}^{2}+\\left(v-u\\right)a-\\mathrm{uw}=0 ;$\n\nwhence\n\n${a}^{3}=\\mathrm{uvw} \\mathrm{and} 3{a}^{2}=\\mathrm{uv}+\\mathrm{vw}+\\mathrm{uw} .$\n\nThis means that $\\mathrm{uv},\\mathrm{vw},\\mathrm{uw}$ are three positive numbers whose geometric and arithmetic means are both equal to ${a}^{2}$. Hence ${a}^{2}=\\mathrm{uv}=\\mathrm{vw}=\\mathrm{uw}$, so that $u=v=w=a$. It follows that $\\mathrm{AF}=\\mathrm{FB}$, $\\mathrm{BD}=\\mathrm{DC}$, $\\mathrm{CE}=\\mathrm{EA}$, so that $\\mathrm{AD}$, $\\mathrm{BE}$ and $\\mathrm{CF}$ are medians and $M$ is the centroid.\nWolog, suppose that $\\mathrm{AB}\\ge \\mathrm{BC}\\ge \\mathrm{CA}$. Since $\\mathrm{AB}\\ge \\mathrm{BC}$, $\\angle \\mathrm{AEB}\\ge {90}^{\u02c6}$, and so $\\mathrm{AM}\\ge \\mathrm{MC}$. Thus $x\\ge p$. Similarly, $y\\ge p$ and $p\\ge z$.\nConsider triangles $\\mathrm{BMD}$ and $\\mathrm{AME}$. We have $\\mathrm{BD}\\ge \\mathrm{AE}$, $\\mathrm{BM}\\ge \\mathrm{AM}$, $\\mathrm{ME}=\\frac{1}{2}\\mathrm{BM}$ and $\\mathrm{MD}=\\frac{1}{2}\\mathrm{AM}$. Therefore\n\n$p-x=\\left(\\mathrm{BD}+\\mathrm{MD}+\\mathrm{BM}\\right)-\\left(\\mathrm{AE}+\\mathrm{ME}+\\mathrm{AM}\\right)=\\left(\\mathrm{BD}-\\mathrm{AE}\\right)+\\frac{1}{2}\\left(\\mathrm{BM}-\\mathrm{AM}\\right)\\ge 0$\n\nand so $p\\ge x$. Since also $x\\ge p$, we have that $p=x$. But this implies that $\\mathrm{AM}=\\mathrm{MC}$, so that $\\mathrm{ME}\\perp \\mathrm{AC}$ and $\\mathrm{AB}=\\mathrm{BC}$. Since $\\mathrm{BE}$ is now an axis of a reflection which interchanges $A$ and $C$, as well as $F$ and $D$, it follows that $p=z$ and $p=y$ as well. Thus, $\\mathrm{AB}=\\mathrm{AC}$ and $\\mathrm{AC}=\\mathrm{BC}$. Thus, the triangle is equilateral.\n\n65.\nSuppose that $\\mathrm{XTY}$ is a straight line and that $\\mathrm{TU}$ and $\\mathrm{TV}$ are two rays emanating from $T$ for which $\\angle \\mathrm{XTU}=\\angle \\mathrm{UTV}=\\angle \\mathrm{VTY}={60}^{\u02c6}$. Suppose that $P$, $Q$ and $R$ are respective points on the rays $\\mathrm{TY}$, $\\mathrm{TU}$ and $\\mathrm{TV}$ for which $\\mathrm{PQ}=\\mathrm{PR}$. Prove that $\\angle \\mathrm{QPR}={60}^{\u02c6}$.\nSolution 1. Let $\\frakR$ be a rotation of ${60}^{\u02c6}$ about $T$ that takes the ray $\\mathrm{TU}$ to $\\mathrm{TV}$. Then, if $\\frakR$ transforms $Q\\to Q\\text{'}$ and $P\\to P\\text{'}$, then $Q\\text{'}$ lies on $\\mathrm{TV}$ and the line $Q\\text{'}P\\text{'}$ makes an angle of ${60}^{\u02c6}$ with $\\mathrm{QP}$. Because of the rotation, $\\angle P\\text{'}\\mathrm{TP}={60}^{\u02c6}$ and $\\mathrm{TP}\\text{'}=\\mathrm{TP}$, whence $\\mathrm{TP}\\text{'}P$ is an equilateral triangle.\nSince $\\angle Q\\text{'}\\mathrm{TP}=\\angle \\mathrm{TPP}\\text{'}={60}^{\u02c6}$, $\\mathrm{TV}P\\text{'}P$. Let $\\frakT$ be the translation that takes $P\\text{'}$ to $P$. It takes $Q\\text{'}$ to a point $Q\\text{'}\\text{'}$ on the ray $\\mathrm{TV}$, and $\\mathrm{PQ}\\text{'}\\text{'}=P\\text{'}Q\\text{'}=\\mathrm{PQ}$. Hence $Q\\text{'}\\text{'}$ can be none other than the point $R$ [why?], and the result follows.\nSolution 2. The reflection in the line $\\mathrm{XY}$ takes $P\\to P$, $Q\\to Q\\text{'}$ and $R\\to R\\text{'}$. Triangles $\\mathrm{PQR}\\text{'}$ and $\\mathrm{PQ}\\text{'}R$ are congruent and isosceles, so that $\\angle \\mathrm{TQP}=\\angle \\mathrm{TQ}\\text{'}P=\\angle \\mathrm{TRP}$ (since $\\mathrm{PQ}\\text{'}=\\mathrm{PR}$). Hence $\\mathrm{TQRP}$ is a concyclic quadrilateral, whence $\\angle \\mathrm{QPR}=\\angle \\mathrm{QTR}={60}^{\u02c6}$.\nSolution 3. [S. Niu] Let $S$ be a point on $\\mathrm{TU}$ for which $\\mathrm{SR}\\mathrm{XY}$; observe that $\\Delta \\mathrm{RST}$ is equilateral. We first show that $Q$ lies between $S$ and $T$. For, if $S$ were between $Q$ and $T$, then $\\angle \\mathrm{PSQ}$ would be obtuse and $\\mathrm{PQ}>\\mathrm{PS}>\\mathrm{PR}$ (since $\\angle \\mathrm{PRS}>{60}^{\u02c6}>\\angle \\mathrm{PSR}$ in $\\Delta \\mathrm{PRS}$), a contradiction.\nThe rotation of ${60}^{\u02c6}$ with centre $R$ that takes $S$ onto $T$ takes ray $\\mathrm{RQ}$ onto a ray through $R$ that intersects $\\mathrm{TY}$ in $M$. Consider triangles $\\mathrm{RSQ}$ and $\\mathrm{RTM}$. Since $\\angle \\mathrm{RST}=\\angle \\mathrm{RTM}={60}^{\u02c6}$, $\\angle \\mathrm{SRQ}={60}^{\u02c6}-\\angle \\mathrm{QRT}=\\angle \\mathrm{TRM}$ and $\\mathrm{SR}=\\mathrm{TR}$, we have that $\\Delta \\mathrm{RSQ}\\equiv \\Delta \\mathrm{RTM}$ and $\\mathrm{RQ}=\\mathrm{RM}$. (ASA) Since $\\angle \\mathrm{QRM}={60}^{\u02c6}$, $\\Delta \\mathrm{RQM}$ is equilateral and $\\mathrm{RM}=\\mathrm{RQ}$. Hence $M$ and $P$ are both equidistant from $Q$ and $R$, and so at the intersection of $\\mathrm{TY}$ and the right bisector of $\\mathrm{QR}$. Thus, $M=P$ and the result follows.\nSolution 4. [H. Pan] Let $Q\\text{'}$ and $R\\text{'}$ be the respective reflections of $Q$ and $R$ with respect to the axis $\\mathrm{XY}$. Since $\\angle \\mathrm{RTR}\\text{'}={120}^{\u02c6}$ and $\\mathrm{TR}=\\mathrm{TR}\\text{'}$, $\\angle \\mathrm{QR}\\text{'}R=\\angle \\mathrm{TR}\\text{'}R={30}^{\u02c6}$. Since $Q,R,Q\\text{'},R\\text{'},$ lie on a circle with centre $P$, $\\angle \\mathrm{QPR}=2\\angle \\mathrm{QR}\\text{'}R={60}^{\u02c6}$, as desired.\nSolution 5. [R. Barrington Leigh] Let $W$ be a point on $\\mathrm{TV}$ such that $\\angle \\mathrm{WPQ}={60}^{\u02c6}=\\angle \\mathrm{WTU}$. [Why does such a point $W$ exist?] Then $\\mathrm{WQTP}$ is a concyclic quadrilateral so that $\\angle \\mathrm{QWP}={180}^{\u02c6}-\\angle \\mathrm{QTP}={60}^{\u02c6}$ and $\\Delta \\mathrm{PWQ}$ is equilateral. Hence $\\mathrm{PW}=\\mathrm{PQ}=\\mathrm{PR}$.\nSuppose $W\\ne R$. If $R$ is farther away from $T$ than $W$, then $\\angle \\mathrm{RPT}>\\angle \\mathrm{WPT}>\\angle \\mathrm{WPQ}={60}^{\u02c6}\u21d2{60}^{\u02c6}>\\angle \\mathrm{TRP}=\\angle \\mathrm{RWP}>{60}^{\u02c6}$, a contradiction. If $W$ is farther away from $T$ than $R$, then $\\angle \\mathrm{WPT}>\\angle \\mathrm{WPQ}={60}^{\u02c6}\u21d2{60}^{\u02c6}>\\angle \\mathrm{RWP}=\\angle \\mathrm{WRP}>{60}^{\u02c6}$, again a contradiction. So $R=W$ and the result follows.\nSolution 6. [M. Holmes] Let the circle through $T,P,Q$ intersect $\\mathrm{TV}$ in $N$. Then $\\angle \\mathrm{QNP}={180}^{\u02c6}-\\angle \\mathrm{QTP}={60}^{\u02c6}$. Since $\\angle \\mathrm{PQN}=\\angle \\mathrm{PTN}={60}^{\u02c6}$, $\\Delta \\mathrm{PQN}$ is equilateral so that $\\mathrm{PN}=\\mathrm{PQ}$. Suppose, if possible, that $R\\ne N$. Then $N$ and $R$ are two points on $\\mathrm{TV}$ equidistant from $P$. Since $\\angle \\mathrm{PNT}<\\angle \\mathrm{PNQ}={60}^{\u02c6}$ and $\\Delta \\mathrm{PNR}$ is isosceles, we have that $\\angle \\mathrm{PNR}<{90}^{\u02c6}$, so $N$ cannot lie between $T$ and $R$, and $\\angle \\mathrm{PRN}=\\angle \\mathrm{PNR}=\\angle \\mathrm{PNT}<{60}^{\u02c6}$. Since $\\angle \\mathrm{PTN}={60}^{\u02c6}$, we conclude that $T$ must lie between $R$ and $N$, which transgresses the condition of the problem. Hence $R$ and $N$ must coincide and the result follows.\nSolution 7. [P. Cheng] Determine $S$ on $\\mathrm{TU}$ and $Z$ on $\\mathrm{TY}$ for which $\\mathrm{SR}\\mathrm{XY}$ and $\\angle \\mathrm{QRZ}={60}^{\u02c6}$. Observe that $\\angle \\mathrm{TSR}=\\angle \\mathrm{SRT}={60}^{\u02c6}$ and $\\mathrm{SR}=\\mathrm{RT}$.\nConsider triangles $\\mathrm{SRQ}$ and $\\mathrm{TRZ}$. $\\angle \\mathrm{SRQ}=\\angle \\mathrm{SRT}-\\angle \\mathrm{QRT}=\\angle \\mathrm{QRZ}-\\angle \\mathrm{QRT}=\\angle \\mathrm{TRZ}$; $\\angle \\mathrm{QSR}={60}^{\u02c6}=\\angle \\mathrm{ZTR}$, so that $\\Delta \\mathrm{SRQ}=\\Delta \\mathrm{TRZ}$ (ASA).\nHence $\\mathrm{RZ}=\\mathrm{RQ}\u21d2\\Delta \\mathrm{RQZ}$ is equilateral $\u21d2\\mathrm{RZ}=\\mathrm{ZQ}$ and $\\angle \\mathrm{RZQ}={60}^{\u02c6}$. Now, both $P$ and $Z$ lie on the intersection of $\\mathrm{TY}$ and the right bisector of $\\mathrm{QR}$, so they must coincide: $P=Z$. The result follows.\nSolution 8. Let the perpendicular, produced, from $Q$ to $\\mathrm{XY}$ meet $\\mathrm{VT}$, produced, in $S$. Then $\\angle \\mathrm{XTS}=\\angle \\mathrm{VTY}={60}^{\u02c6}=\\angle \\mathrm{XTU}$, from which is can be deduced that $\\mathrm{TX}$ right bisects $\\mathrm{QS}$. Hence $\\mathrm{PS}=\\mathrm{PQ}=\\mathrm{PR}$, so that $Q,R,S$ are all on the same circle with centre $P$.\nSince $\\angle \\mathrm{QTS}={120}^{\u02c6}$, we have that $\\angle \\mathrm{SQT}=\\angle \\mathrm{QSR}={30}^{\u02c6}$, so that $\\mathrm{QR}$ must subtend an angle of ${60}^{\u02c6}$ at the centre $P$ of the circle. The desired result follows.\nSolution 9. [A.Siu] Let the right bisector of $\\mathrm{QR}$ meet the circumcircle of $\\mathrm{TQR}$ on the same side of $\\mathrm{QR}$ at $T$ in $S$. Since $\\angle \\mathrm{QSR}=\\angle \\mathrm{QTR}={60}^{\u02c6}$ and $\\mathrm{QS}=\\mathrm{QR}$, $\\angle \\mathrm{SQR}=\\angle \\mathrm{SRQ}={60}^{\u02c6}$. Hence $\\angle \\mathrm{STQ}={180}^{\u02c6}-\\angle \\mathrm{SRQ}={120}^{\u02c6}$. But $\\angle \\mathrm{YTQ}={120}^{\u02c6}$, so $S$ must lie on $\\mathrm{TY}$. It follows that $S=P$.\nSolution 10. Assign coordinates with the origin at $T$ and the $x-$axis along $\\mathrm{XY}$. The the respective coordinates of $Q$ and $R$ have the form $\\left(u,-\\sqrt{3}u\\right)$ and $\\left(v,\\sqrt{3}v\\right)$ for some real $u$ and $v$. Let the coordinates of $P$ be $\\left(w,0\\right)$. Then $\\mathrm{PQ}=\\mathrm{PR}$ yields that $w=2\\left(u+v\\right)$. [Exercise: work it out.]\n\n$\\begin{array}{cc}\u2016\\mathrm{PQ}\u2016{}^{2}-\u2016\\mathrm{QR}\u2016{}^{2}\\hfill & =\\left(u-w\\right){}^{2}+3{u}^{2}-\\left(u-v\\right){}^{2}-3\\left(u+v\\right){}^{2}\\hfill \\\\ \\multicolumn{0}{c}{}& ={w}^{2}-2\\mathrm{uw}-4v\\left(u+v\\right)={w}^{2}-2\\mathrm{uw}-2\\mathrm{vw}\\hfill \\\\ \\multicolumn{0}{c}{}& ={w}^{2}-2\\left(u+v\\right)w=0 .\\hfill \\end{array}$\n\nHence $\\mathrm{PQ}=\\mathrm{QR}=\\mathrm{PR}$ and $\\Delta \\mathrm{PQR}$ is equilateral. Therefore $\\angle \\mathrm{QPR}={60}^{\u02c6}$.\nSolution 11. [J.Y. Jin] Let $\\frakC$ be the circumcircle of $\\Delta \\mathrm{PQR}$. If $T$ lies strictly inside $\\frakC$, then ${60}^{\u02c6}=\\angle \\mathrm{QTR}>\\angle \\mathrm{QPR}$ and ${60}^{\u02c6}=\\angle \\mathrm{PTR}>\\angle \\mathrm{PQR}=\\angle \\mathrm{PRQ}$. Thus, all three angle of $\\Delta \\mathrm{PQR}$ would be less than ${60}^{\u02c6}$, which is not possible. Similarly, if $T$ lies strictly outside $\\frakC$, then ${60}^{\u02c6}=\\angle \\mathrm{QTR}<\\angle \\mathrm{QPR}$ and ${60}^{\u02c6}=\\angle \\mathrm{PTR}<\\angle \\mathrm{PQR}=\\angle \\mathrm{PRQ}$, so that all three angles of $\\Delta \\mathrm{PQR}$ would exceed ${60}^{\u02c6}$, again not possible. Thus $T$ must be on $\\frakC$, whence $\\angle \\mathrm{QPR}=\\angle \\mathrm{QTR}={60}^{\u02c6}$.\nSolution 12. [C. Lau] By the Sine Law,\n\n$\\frac{\\mathrm{sin}\\angle \\mathrm{TQP}}{\u2016\\mathrm{TP}\u2016}=\\frac{\\mathrm{sin}{120}^{\u02c6}}{\u2016\\mathrm{PQ}\u2016}=\\frac{\\mathrm{sin}{60}^{\u02c6}}{\u2016\\mathrm{PR}\u2016}=\\frac{\\mathrm{sin}\\angle \\mathrm{TRP}}{\u2016\\mathrm{TP}\u2016} ,$\n\nwhence $\\mathrm{sin}\\angle \\mathrm{TQP}=\\mathrm{sin}\\angle \\mathrm{TRP}$. Since $\\angle \\mathrm{QTP}$, in triangle $\\mathrm{QTP}$ is obtuse, $\\angle \\mathrm{TQP}$ is acute.\nSuppose, if possible, that $\\angle \\mathrm{TRP}$ is obtuse. Then, in triangle $\\mathrm{TPR}$, $\\mathrm{TP}$ would be the longest side, so $\\mathrm{PR}<\\mathrm{TP}$. But in triangle $\\mathrm{TQP}$, $\\mathrm{PQ}$ is the longest side, so $\\mathrm{PQ}>\\mathrm{TP}$, and so $\\mathrm{PQ}\\ne \\mathrm{PR}$, contrary to hypothesis. Hence $\\angle \\mathrm{TRP}$ is acute. Therefore, $\\angle \\mathrm{TQP}=\\angle \\mathrm{TRP}$. Let $\\mathrm{PQ}$ and $\\mathrm{RT}$ intersect in $Z$. Then, ${60}^{\u02c6}=\\angle \\mathrm{QTZ}={180}^{\u02c6}-\\angle \\mathrm{TQP}-\\angle \\mathrm{QZT}={180}^{\u02c6}-\\angle \\mathrm{TRP}-\\angle \\mathrm{RZP}=\\angle \\mathrm{QPR}$, as desired.\n\n66.\n(a) Let $\\mathrm{ABCD}$ be a square and let $E$ be an arbitrary point on the side $\\mathrm{CD}$. Suppose that $P$ is a point on the diagonal $\\mathrm{AC}$ for which $\\mathrm{EP}\\perp \\mathrm{AC}$ and that $Q$ is a point on $\\mathrm{AE}$ produced for which $\\mathrm{CQ}\\perp \\mathrm{AE}$. Prove that $B,P,Q$ are collinear.\n(b) Does the result hold if the hypothesis is weakened to require only that $\\mathrm{ABCD}$ is a rectangle?\nSolution 1. Let $\\mathrm{ABCD}$ be a rectangle, and let $E$, $P$, $Q$ be determined as in the problem. Suppose that $\\angle \\mathrm{ACD}=\\angle \\mathrm{BDC}=\\alpha$. Then $\\angle \\mathrm{PEC}={90}^{\u02c6}-\\alpha$. Because $\\mathrm{EPQC}$ is concyclic, $\\angle \\mathrm{PQC}=\\angle \\mathrm{PEC}={90}^{\u02c6}-\\alpha$. Because $\\mathrm{ABCQD}$ is concyclic, $\\angle \\mathrm{BQC}=\\angle \\mathrm{BDC}=\\alpha$. The points $B$, $P$, $Q$ are collinear $&lrArr;\\angle \\mathrm{BQC}=\\angle \\mathrm{PQC}&lrArr;\\alpha ={90}^{\u02c6}-\\alpha &lrArr;\\alpha ={45}^{\u02c6}&lrArr;\\mathrm{ABCD}$ is a square.\nSolution 2. (a) $\\mathrm{EPQC}$, with a pair of supplementary opposite angles, is concyclic, so that $\\angle \\mathrm{CQP}=\\angle \\mathrm{CEP}={180}^{\u02c6}-\\angle \\mathrm{EPC}-\\angle \\mathrm{ECP}={45}^{\u02c6}$. Since $\\mathrm{CBAQ}$ is concyclic, $\\angle \\mathrm{CQB}=\\angle \\mathrm{CAB}={45}^{\u02c6}$. Thus, $\\angle \\mathrm{CQP}=\\angle \\mathrm{CQB}$ so that $Q$, $P$, $B$ are collinear.\n(b) Suppose that $\\mathrm{ABCD}$ is a nonquare rectangle. Then taking $E=D$ yields a counterexample.\nSolution 3. (a) The circle with diameter $\\mathrm{AC}$ that passes through the vertices of the square also passes through $Q$. Hence $\\angle \\mathrm{QBC}=\\angle \\mathrm{QAC}$. Consider triangles $\\mathrm{PBC}$ and $\\mathrm{EAC}$. Since triangles $\\mathrm{ABC}$ and $\\mathrm{EPC}$ are both isosceles right triangles, $\\mathrm{BC}:\\mathrm{AC}=\\mathrm{PC}:\\mathrm{EC}$. Also $\\angle \\mathrm{BCA}=\\angle \\mathrm{PCE}={45}^{\u02c6}$. Hence $\\Delta \\mathrm{PBC}~\\Delta \\mathrm{EAC}$ (SAS) so that $\\angle \\mathrm{PBC}=\\angle \\mathrm{EAC}=\\angle \\mathrm{QAC}=\\angle \\mathrm{QBC}$. It follows that $Q$, $P$, $B$ are collinear.\nSolution 4. [S. Niu] Let $\\mathrm{ABCD}$ be a rectangle and let $E,P,Q$ be determined as in the problem. Let $\\mathrm{EP}$ be produced to meet $\\mathrm{BC}$ in $F$. Since $\\angle \\mathrm{ABF}=\\angle \\mathrm{APF}$, the quadrilateral $\\mathrm{ABPF}$ is concyclic, so that $\\angle \\mathrm{PBC}=\\angle \\mathrm{PBF}=\\angle \\mathrm{PAF}$. Since $\\mathrm{ABCQ}$ is concyclic, $\\angle \\mathrm{QBC}=\\angle \\mathrm{QAC}=\\angle \\mathrm{PAE}$. Now $B,P,Q$ are collinear\n\n$&lrArr;\\angle \\mathrm{PBC}=\\angle \\mathrm{QBC}&lrArr;\\angle \\mathrm{PAF}=\\angle \\mathrm{PAE}&lrArr;\\mathrm{AC} \\mathrm{right} \\mathrm{bisects} \\mathrm{EF}$\n\n$&lrArr;\\angle \\mathrm{ECA}=\\angle \\mathrm{ACB}={45}^{\u02c6}&lrArr;\\mathrm{ABCD} \\mathrm{is} a \\mathrm{square} .$\n\nSolution 5. [M. Holmes] (a) Suppose that $\\mathrm{BQ}$ intersects $\\mathrm{AC}$ in $R$. Since $\\mathrm{ABCQD}$ is concyclic, $\\angle \\mathrm{AQR}=\\angle \\mathrm{AQB}=\\angle \\mathrm{ACB}={45}^{\u02c6}$, so that $\\angle \\mathrm{BQC}={45}^{\u02c6}$. Since $\\angle \\mathrm{EQR}=\\angle \\mathrm{AQB}=\\angle \\mathrm{ECR}={45}^{\u02c6}$, $\\mathrm{ERCQ}$ is concyclic, so that $\\angle \\mathrm{ERC}={180}^{\u02c6}-\\angle \\mathrm{EQC}={90}^{\u02c6}$. Hence $\\mathrm{ER}\\perp \\mathrm{AC}$, so that $R=P$ and the result follows.\nSolution 6. [L. Hong] (a) Let $\\mathrm{QC}$ intersect $\\mathrm{AB}$ in $F$. We apply Menelaus' Theorem to triangle $\\mathrm{AFC}$: $B$, $P$, $Q$ are collinear if and only if\n\n$\\frac{\\mathrm{AB}}{\\mathrm{BF}}\u00b7\\frac{\\mathrm{FQ}}{\\mathrm{QC}}\u00b7\\frac{\\mathrm{CP}}{\\mathrm{PA}}=-1 .$\n\nLet the side length of the square be 1 and the length of $\\mathrm{DE}$ be $a$. Then $\u2016\\mathrm{AB}\u2016=1$. Since $\\Delta \\mathrm{ADE}~\\Delta \\mathrm{FBC}$, $\\mathrm{AD}:\\mathrm{DE}=\\mathrm{BF}:\\mathrm{BC}$, so that $\u2016\\mathrm{BF}\u2016=1/a$ and $\u2016\\mathrm{FC}\u2016=\\sqrt{1+{a}^{2}}/a$. Since $\\Delta \\mathrm{ADE}~\\Delta \\mathrm{CQE}$, $\\mathrm{CQ}:\\mathrm{EC}=\\mathrm{AD}:\\mathrm{EA}$, so that $\u2016\\mathrm{CQ}\u2016=\\left(1-a\\right)/\\sqrt{1+{a}^{2}}$. Hence\n\n$\\frac{\u2016\\mathrm{FQ}\u2016}{\u2016\\mathrm{CQ}\u2016}=1+\\frac{\u2016\\mathrm{FC}\u2016}{\u2016\\mathrm{CQ}\u2016}=1+\\frac{1+{a}^{2}}{a\\left(1-a\\right)}=\\frac{1+a}{a\\left(1-a\\right)} .$\n\nSince $\\Delta \\mathrm{ECP}$ is right isosceles, $\u2016\\mathrm{CP}\u2016=\\left(1-a\\right)/\\sqrt{2}$ and $\u2016\\mathrm{PA}\u2016=\\sqrt{2}-\u2016\\mathrm{CP}\u2016=\\left(1+a\\right)/\\sqrt{2}$. Hence $\u2016\\mathrm{CP}\u2016/\u2016\\mathrm{PA}\u2016=\\left(1-a\\right)/\\left(1+a\\right)$. Multiplying the three ratios together and taking account of the directed segments gives the product $-1$ and yields the result.\nSolution 7. (a) Select coordinates so that $A~\\left(0,1\\right)$, $B~\\left(0,0\\right)$, $C~\\left(1,0\\right)$, $D~\\left(1,1\\right)$ and $E~\\left(1,t\\right)$ for some $t$ with $0\\le t\\le 1$. It is straightforward to verify that $P~\\left(1-\\frac{t}{2},\\frac{t}{2}\\right)$.\nSince the slope of $\\mathrm{AE}$ is $t-1$, the slope of $\\mathrm{AQ}$ should be $\\left(1-t\\right){}^{-1}$. Since the coordinates of $Q$ have the form $\\left(1+s,s\\left(1-t\\right){}^{-1}\\right)$ for some $s$, it is straightforward to verify that\n\n$Q~\\left(\\frac{2-t}{1+\\left(1-t\\right){}^{2}},\\frac{t}{1+\\left(1-t\\right){}^{2}\\right)}\\right) .$\n\nIt can now be checked that the slope of each of $\\mathrm{BQ}$ and $\\mathrm{BP}$ is $t\\left(2-t\\right){}^{-1}$, which yields the result.\n(b) The result fails if $A~\\left(0,2\\right)$, $B~\\left(0,0\\right)$, $C~\\left(1,0\\right)$, $D~\\left(1,2\\right)$. If $E~\\left(1,1\\right)$, then $P~\\left(\\frac{3}{5},\\frac{4}{5}\\right)$ and $Q~\\left(\\frac{3}{2},\\frac{1}{2}\\right)$.", "date": "2014-03-07 20:05:46", "meta": {"domain": "math.ca", "url": "http://cms.math.ca/Competitions/MOCP/2001/sol_feb.mml", "openwebmath_score": 0.9988901019096375, "openwebmath_perplexity": 2883.047276497813, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471618625457, "lm_q2_score": 0.8962513752119936, "lm_q1q2_score": 0.8817743718177234}}
{"url": "http://mathhelpforum.com/advanced-algebra/133902-proving-group.html", "text": "# Math Help - Proving that this is a group\n\n1. ## Proving that this is a group\n\nHello, I would really appreciate any help with the following problem:\n\nProve that $(G, *)$ is a group, where $G=<-1,1>$ and $a*b=\\dfrac{a+b}{1+ab}$,.... $\\forall a, b \\in G$.\n\n_______________________________\n\nTo prove that this is a group, $(G,*)$ must satisfy closure, associativity, existence of an identity element and existence of inverse elements.\n\nCould the property of associativity be demonstrated like this?\n\n$(a*b)*c=(\\dfrac{a+b}{1+ab})*c=...=\\dfrac{a+b+c+abc }{1+ab+ac+bc}$\n\n$a*(b*c)=a*(\\dfrac{b+c}{1+bc})=...=\\dfrac{a+b+c+abc }{1+ab+ac+bc}$\n\nTherefore, $(a*b)*c=a*(b*c)$, so the property of associativity is satisfied.\n\n_______________________________\n\nIs the identity element $0$?\n\nFrom $a*0= \\dfrac{a+0}{1+a \\cdot 0}=0*a=a$\n\nit follows $a*0=0*a=a$\n\n_______________________________\n\nIs the inverse element of $a$, $-a$?\n\nBecause $a*(-a)=\\dfrac{a-a}{1-a^2}=\\dfrac{0}{1-a^2}=(a \\in <-1,1>)=(-a)*a = 0$\n\n_______________________________\n\nIf the above is correct (and I'm not certain that it is), the only property left to be demonstrated is closure, i.e. if $a, b \\in <-1,1>$ then $a*b =\\dfrac{a+b}{1+ab} \\in <-1,1>$.\n\nAnd here I need your help! How to demonstrate the property of closure?\n\nThere is a hint in the text: \"Observe that $|1+ab|=1+ab$, and prove that $|a+b| \\leq 1+ab$ if and only if $0 \\leq (1-a)(1-b)$ and $0 \\leq (1+a)(1+b)$\".\n\nSo, how to prove the statements from the hint, and, more importantly, how to apply them to demonstrate the closure property of $(G,*)$?\n\nMany thanks!\n\n2. Originally Posted by gusztav\nHello, I would really appreciate any help with the following problem:\n\nProve that $(G, *)$ is a group, where $G=<-1,1>$ and $a*b=\\dfrac{a+b}{1+ab}$,.... $\\forall a, b \\in G$.\n\n_______________________________\n\nTo prove that this is a group, $(G,*)$ must satisfy closure, associativity, existence of an identity element and existence of inverse elements.\n\nCould the property of associativity be demonstrated like this?\n\n$(a*b)*c=(\\dfrac{a+b}{1+ab})*c=...=\\dfrac{a+b+c+abc }{1+ab+ac+bc}$\n\n$a*(b*c)=a*(\\dfrac{b+c}{1+bc})=...=\\dfrac{a+b+c+abc }{1+ab+ac+bc}$\n\nTherefore, $(a*b)*c=a*(b*c)$, so the property of associativity is satisfied.\n\n_______________________________\n\nIs the identity element $0$?\n\nFrom $a*0= \\dfrac{a+0}{1+a \\cdot 0}=0*a=a$\n\nit follows $a*0=0*a=a$\n\n_______________________________\n\nIs the inverse element of $a$, $-a$?\n\nBecause $a*(-a)=\\dfrac{a-a}{1-a^2}=\\dfrac{0}{1-a^2}=(a \\in <-1,1>)=(-a)*a = 0$\n\n_______________________________\n\nIf the above is correct (and I'm not certain that it is), the only property left to be demonstrated is closure, i.e. if $a, b \\in <-1,1>$ then $a*b =\\dfrac{a+b}{1+ab} \\in <-1,1>$.\n\nAnd here I need your help! How to demonstrate the property of closure?\n\nThere is a hint in the text: \"Observe that $|1+ab|=1+ab$, and prove that $|a+b| \\leq 1+ab$ if and only if $0 \\leq (1-a)(1-b)$ and $0 \\leq (1+a)(1+b)$\".\n\nSo, how to prove the statements from the hint, and, more importantly, how to apply them to demonstrate the closure property of $(G,*)$?\n\nMany thanks!\n\nFirst ,what do you mean by $G=<-1,1>$ ?? The subset of the reals (or complex) with those two elements? If so zero cannot be anything there since it doesn't belong to the set!\nAlso, the product of $1\\,,\\,-1$ isn't defined: $1*-1:= \\frac{1+(-1)}{1+1(-1)}=\\frac{0}{0}$ ...!!\n\nIf this is so the above isn't a group.\n\nBut perhaps you meant the open interval $(-1,1)$ which, by some misterious reason, you denote by $<-1,1>$...(!)\n\nThen you must show $a,b\\in(-1,1)\\Longrightarrow \\frac{a+b}{1+ab}\\in (-1,1)\\Longleftrightarrow \\left|\\frac{a+b}{1+ab}\\right|<1$ $\\Longleftrightarrow (a+b)^2<(1+ab)^2\\Longleftrightarrow a^2+2ab+b^2\n\n$\\Longleftrightarrow a^2(b^2-1)-(b^2-1)>0\\Longleftrightarrow (a^2-1)(b^2-1)>0$ , and this last inequality is obviously true.\n\nTonio\n\n3. Thank you very much, Tonio!\n\nOriginally Posted by tonio\nBut perhaps you meant the open interval $(-1,1)$\nYes, I meant the open interval $\\{x \\in \\mathbb{R} : -1\n\nOriginally Posted by tonio\nThen you must show $a,b\\in(-1,1)\\Longrightarrow \\frac{a+b}{1+ab}\\in (-1,1)\\Longleftrightarrow \\left|\\frac{a+b}{1+ab}\\right|<1$ $\\Longleftrightarrow (a+b)^2<(1+ab)^2\\Longleftrightarrow a^2+2ab+b^2\n\n$\\Longleftrightarrow a^2(b^2-1)-(b^2-1)>0\\Longleftrightarrow (a^2-1)(b^2-1)>0$ , and this last inequality is obviously true.\nAha! Now everything is clear.\n(Because $a \\in (-1,1) \\Longrightarrow |a|<1 \\Longrightarrow a^2<1 \\Longrightarrow a^2-1<0$ and, similarly, $b^2-1<0$, so $(a^2-1)( b^2-1)>0$ )", "date": "2015-02-28 19:25:58", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/133902-proving-group.html", "openwebmath_score": 0.9075925350189209, "openwebmath_perplexity": 492.2436707870244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407175907054, "lm_q2_score": 0.9059898159413479, "lm_q1q2_score": 0.8817461785966285}}
{"url": "https://math.stackexchange.com/questions/1469820/area-under-quarter-circle-by-integration", "text": "# Area under quarter circle by integration\n\nHow would one go about finding out the area under a quarter circle by integrating. The quarter circle's radius is r and the whole circle's center is positioned at the origin of the coordinates. (The quarter circle is in the first quarter of the coordinate system)\n\nFrom the equation $x^2+y^2=r^2$, you may express your area as the following integral $$A=\\int_0^r\\sqrt{r^2-x^2}\\:dx.$$ Then substitute $x=r\\sin \\theta$, $\\theta=\\arcsin (x/r)$, to get \\begin{align} A&=\\int_0^{\\pi/2}\\sqrt{r^2-r^2\\sin^2 \\theta}\\:r\\cos \\theta \\:d\\theta\\\\ &=r^2\\int_0^{\\pi/2}\\sqrt{1-\\sin^2 \\theta}\\:\\cos\\theta \\:d\\theta\\\\ &=r^2\\int_0^{\\pi/2}\\sqrt{\\cos^2 \\theta}\\:\\cos\\theta \\:d\\theta\\\\ &=r^2\\int_0^{\\pi/2}\\cos^2 \\theta \\:d\\theta\\\\ &=r^2\\int_0^{\\pi/2}\\frac{1+\\cos(2\\theta)}2 \\:d\\theta\\\\ &=r^2\\int_0^{\\pi/2}\\frac12 \\:d\\theta+\\frac{r^2}2\\underbrace{\\left[ \\frac12\\sin(2\\theta)\\right]_0^{\\pi/2}}_{\\color{#C00000}{=\\:0}}\\\\ &=\\frac{\\pi}4r^2. \\end{align}\n\u2022 Yes, we have, for $0<x<r$, $\\frac{d\\theta}{dx}=\\frac{1}{\\sqrt{r^2-x^2}}>0$, $0=\\arcsin (0/r) \\leq \\theta (r)\\leq \\arcsin (r/r)=\\pi/2$. Thanks! Oct 8 '15 at 18:45\nHere is a quicker solution. The area can be seen as a collection of very thin triangles, one of which is shown below. As $d\\theta\\to0$, the base of the triangle becomes $rd\\theta$ and the height becomes $r$, so the area is $\\frac12r^2d\\theta$. The limits of $\\theta$ are $0$ and $\\frac\\pi2$. $$\\int_0^\\frac\\pi2\\frac12r^2d\\theta=\\frac12r^2\\theta|_0^\\frac\\pi2=\\frac14\\pi r^2$$\nlet circle: $x^2+y^2=r^2$ then consider a slab of area $dA=ydx$ then the area of quarter circle $$A_{1/4}=\\int_0^r ydx=\\int_0^r \\sqrt{r^2-x^2}dx$$ $$=\\frac12\\left[x\\sqrt{r^2-x^2}+r^2\\sin^{-1}\\left(x/r\\right)\\right]_0^r$$\n$$=\\frac12\\left[0+r^2(\\pi/2)\\right]=\\frac{\\pi}{4}r^2$$ or use double integration: $$=\\iint rdr d\\theta= \\int_0^{\\pi/2}\\ d\\theta\\int_0^R rdr=\\int_0^{\\pi/2}\\ d\\theta(R^2/2)=(R^2/2)(\\pi/2)=\\frac{\\pi}{4}R^2$$", "date": "2021-11-28 08:29:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1469820/area-under-quarter-circle-by-integration", "openwebmath_score": 0.9725695252418518, "openwebmath_perplexity": 623.0111737277325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9938070113095245, "lm_q2_score": 0.8872045937171068, "lm_q1q2_score": 0.8817101457020788}}
{"url": "https://math.stackexchange.com/questions/2243997/find-the-2-digits-in-front-of-the-number-14-so-that-the-4-digit-number-we-get-i/2244024", "text": "# Find the 2 digits in front of the number 14, so that the 4-digit number we get is divisible by 26.\n\nI know that 26=13.2. The last digit of the number is 4, it is divisible by 2. I have to remove the last digit and check out if the remaining digits are divisible by 13 or?\n\n\u2022 what do you mean by '2 digits in front of the number 14'? \u2013\u00a0oliverjones Apr 20 '17 at 18:55\n\u2022 Do you mean $14xx$ or $xx14$ or something different? \u2013\u00a0Mark Bennet Apr 20 '17 at 18:56\n\u2022 gotta be xx14 -- two digits in front of the number 14 \u2013\u00a0gt6989b Apr 20 '17 at 18:56\n\u2022 I wanted to say xx14. \u2013\u00a0Elena Apr 20 '17 at 18:58\n\nYou are being asked to find an integer $x$ such that $10 \\leq x \\leq 99$ and $$100 x + 14 \\equiv 0 \\pmod{26}.$$ Solve this equation, and $x$ will be the two digits you want. There are multiple possible answers.\n\nA good first step is to rewrite the equation in a more convenient form. Since it is modulo $26,$ you can replace any of the numbers by something equivalent modulo $26.$ For example, $100 \\equiv 22 \\pmod{26},$ so you can write $22$ instead of $100.$ I think it's more convenient to use the fact that $100 \\equiv -4 \\pmod{26},$ however; it lets you work with smaller numbers.\n\n\u2022 i think the OP is asking how to solve this equation \u2013\u00a0gt6989b Apr 20 '17 at 19:00\n\u2022 We know that $100x + 14 \\equiv 0\\pmod{26} \\iff 100x + 14 = 26m$, for some integer $m$. Thus, $$x = {26m-14\\over 100} = {13m - 7 \\over 50}.$$ Since we require that $10\\le x \\le 99$, choosing $m$ appropriately should give us the desired result. For example, we can choose $m=39$. Then we have $500/50=10$, which implies $x=10$ would be a good starting point, and so $1014$ will be a satisfactory answer. \u2013\u00a0Decaf-Math Apr 20 '17 at 19:05\n\u2022 @gt6989b I think that looking at the question this way is fundamentally different from what the OP was trying to do: \"remove the last digit and check if the remaining digits are divisible by ...\" etc. But I added some detail in hope that it may help. \u2013\u00a0David K Apr 20 '17 at 19:06\n\u2022 How many answers do I have to get? 2? The first one is 1014 \u2013\u00a0Elena Apr 20 '17 at 22:03\n\u2022 I don't know if you need more than one answer, but you have a choice. Since 1300 is the smallest multiple of 100 divisible by 26, you can add any positive multiple of 1300 to 1014 and get another solution, unless the multiple is so large that the result is greater than 9999. It turns out there is room for seven separate solutions between 1000 and 9999, starting with 1014 and ending with 8814. \u2013\u00a0David K Apr 20 '17 at 22:21\n\nStarting with $100 x + 14 \\equiv 0 \\pmod{26}$ you can take $100 \\pmod{26}$ to get $74, 48, 22, -4$, etc. $-4$ is the smallest, so let's try that.\n\nNow the equivalence to solve is $-4x + 14 \\equiv 0 \\pmod{26}$. It'll be easier to see the solution if the LHS is positive, so add $26$ to get $-4x + 40 \\equiv 0 \\pmod{26}$. Is it possible to subtract 4 from 40 a few times to get $26$? If you subtract $4*3=12$ you still have $2$ to go to get to $26$.\n\nSo add another $26$ to the equation to get $$-4x + 40 \\equiv 0 \\pmod{26}$$. Note that $66-26=40=4*10$. So $x$ is $10$.\n\nYou would then verify $x=10$ in the original problem. Is $1014 \\equiv 0 \\pmod{26}$?\n\n\u2022 Yes, it is. But I think a have more than 1 answer because (100, 26)=2 \u2013\u00a0Elena Apr 20 '17 at 22:09\n\u2022 I don't understand your comment. There will be $7$ answers. The answer can be any value of the form $10+13x$, with $0 \\leq x \\leq 6$. I interpreted the problem as requiring just one answer. \u2013\u00a0\u03a7p\u1e98 Apr 20 '17 at 22:24\n\u2022 I guess not. David K also specifies $7$ answers in the comments to his answer. \u2013\u00a0\u03a7p\u1e98 Apr 20 '17 at 22:29\n\nLet $1\\le a\\le 9$ and $0\\le b\\le9$ be those two front numbers ($ab14$). So, our desired number is $10^3a+10^2b+14$ where $$10^3a+10^2b+14\\equiv0\\pmod{26}\\implies 12a-4b\\equiv12\\pmod{26}\\implies 3a-3=3(a-1)\\equiv b\\pmod{13}$$ All choices that satisfy this (just plug in $a=1,2,\\ldots9$ and see what happens):\n\n$1014$ $2314$ $3614$ $4914$ $6214$ $7514$ $8814$\n\nNotice how in the congruence, I used $$a\\equiv b\\pmod{m}\\implies \\frac{a}{c}\\equiv \\frac{b}{c}\\pmod{\\frac{m}{\\gcd(m,c)}}$$ as long as $c\\vert a$ and $c\\vert b$. Also notice that $a\\equiv9$ is not possible because that gives an invalid number for $b$\n\n\u2022 I thought we only had 1 answer - 1014? Or? \u2013\u00a0Elena Apr 20 '17 at 19:49\n\u2022 You can check for yourself that all $7$ of those integers work just fine. \u2013\u00a0user12345 Apr 20 '17 at 22:51\n\nOne way to find a solution is to start from the useful and memorable fact that $7 \\times 11 \\times 13 = 1001$. Hence $26$ divides any even multiple of $1001$, in particular $4004$.\n\nIf we can find $x10$ divisible by $26$, then $4004 + x10$ will yield a solution. $x10$ is divisible (for any $x$) by $2$, so it suffices to find $x1$ divisible by $13$. Since $91 = 7 \\times 13$, a solution is $4004 + 910 = 4914$.", "date": "2020-10-24 20:20:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2243997/find-the-2-digits-in-front-of-the-number-14-so-that-the-4-digit-number-we-get-i/2244024", "openwebmath_score": 0.8078303933143616, "openwebmath_perplexity": 145.0054071910057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795118010989, "lm_q2_score": 0.8933094039240554, "lm_q1q2_score": 0.8816780793722949}}
{"url": "https://math.stackexchange.com/questions/1767217/does-the-infinite-series-sum-k-1-infty-frac14-1kk-converge", "text": "# Does the infinite series $\\sum_{k=1}^{\\infty} \\frac{1}{(4+(-1)^k)^k}$ converge?\n\nI have been wondering if this infinite series converges $$\\sum_{k=1}^{\\infty} \\frac{1}{(4+(-1)^k)^k}$$\n\nI tried to put it in wolfram alpha but it says that the ratio test is inconclusive, but when I do the ratio test I get: Let first $a_k :=\\frac{1}{(4+(-1)^k)^k}$, then we have that $\\frac{1}{5^k}\\le a_k \\le \\frac{1}{3^k}$. We also see from here that a_k is always positive and now we have $\\frac{1}{5}\\le\\sqrt[k]{a_k}\\le\\frac{1}{3}<1, \\forall k\\in\\mathbb{N}$. From that we should actually have that the series converges or am I missing something?\n\nOne more thing that I also noticed is that, if I use my inequality we can also have that $$\\sum_{k=1}^{\\infty} \\frac{1}{(4+(-1)^k)^k}\\le\\sum_{k=1}^{\\infty} \\frac{1}{3^k}=\\frac{1}{1-\\frac{1}{3}}=\\frac{3}{2}$$\n\nDoes that mean it converges to $\\frac{3}{2}$?\n\n\u2022 I'm sure it converges. Your inequality says it all \u2013\u00a0Yuriy S May 1 '16 at 20:03\n\u2022 Comparison test. \u2013\u00a0Robert Israel May 1 '16 at 20:03\n\u2022 It's approximately $\\frac{5}{12}$ by numerical estimation \u2013\u00a0Yuriy S May 1 '16 at 20:04\n\nNotice that the terms of this series are positive and we have\n\n$$\\frac{1}{(4+(-1)^k)^k}\\le 3^{-k}$$ and the geometric series $\\sum 3^{-k}$ is convergent. Use comparison to conclude.\n\n\u2022 Does it also mean it converges to $\\frac{3}{2}$ like the series of $3^{-k}$ does? \u2013\u00a0HeatTheIce May 1 '16 at 20:11\n\u2022 No, but certainly the sum of this series is $<\\frac32$. \u2013\u00a0user296113 May 1 '16 at 20:12\n\u2022 Do not forget that in addition to being bounded above by the $3^{-k}$ series (guaranteeing convergence), you are also bounded by $5^{-k}$ from below (which helps you narrow down your value). Therefore, you know the series converges to some value $x$ such that $5/4\\leq x \\leq 3/2$. \u2013\u00a0AmateurDotCounter May 1 '16 at 20:36\n\u2022 @LetEpsilonBeLessThanZero Note that $\\sum_1^\\infty 5^{-k} = 1/4$. \u2013\u00a0stochasticboy321 May 1 '16 at 20:38\n\u2022 @HeatTheIce To compute the sum, try splitting the series into two - one with the odd terms, and another with the even terms. You'll notice that both are convergent geometric series. \u2013\u00a0stochasticboy321 May 1 '16 at 20:38\n\nDoes that mean it converges to $\\frac{3}{2}$?\n\nHere is a closed form of the series:\n\n$$\\sum_{k=1}^{\\infty} \\frac{1}{(4+(-1)^k)^k}=\\color{blue}{\\frac5{12}}.$$\n\nProof. By the absolute convergence, one is allowed to write \\begin{align} \\sum_{k=1}^{\\infty} \\frac{1}{(4+(-1)^k)^k}&=\\sum_{k=1}^{\\infty} \\frac{1}{(4+(-1)^{2k})^{2k}}+\\sum_{k=1}^{\\infty} \\frac{1}{(4+(-1)^{2k-1})^{2k-1}} \\\\\\\\&=\\sum_{k=1}^{\\infty} \\frac{1}{25^k}+3\\sum_{k=1}^{\\infty} \\frac{1}{9^k} \\\\\\\\&=\\frac1{24}+\\frac38 \\\\\\\\&=\\frac5{12}. \\end{align}\n\nUse the $\\;k\\,-$ th root test:\n\n$$\\lim\\sup_{k\\to\\infty}\\sqrt[k]{\\frac1{4+(-1)^k)^k}}=\\lim\\sup_{k\\to\\infty}\\frac1{4+(-1)^k}=\\frac13<1$$\n\nand thus the series converges (observe it is a positive series)", "date": "2019-04-25 21:56:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1767217/does-the-infinite-series-sum-k-1-infty-frac14-1kk-converge", "openwebmath_score": 0.9552393555641174, "openwebmath_perplexity": 419.01548789766053, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795075882268, "lm_q2_score": 0.8933094025038598, "lm_q1q2_score": 0.8816780742071927}}
{"url": "http://pobarabanu.com.ua/414yquca/elnfm.php?57d14b=system-of-linear-equations-matrix-conditions", "text": "Solving systems of linear equations. This calculator solves Systems of Linear Equations using Gaussian Elimination Method, Inverse Matrix Method, or Cramer's rule.Also you can compute a number of solutions in a system of linear equations (analyse the compatibility) using Rouch\u00e9\u2013Capelli theorem.. First, we need to find the inverse of the A matrix (assuming it exists!) Section 2.3 Matrix Equations \u00b6 permalink Objectives. Think of \u201cdividing\u201d both sides of the equation Ax = b or xA = b by A.The coefficient matrix A is always in the \u201cdenominator.\u201d. Enter coefficients of your system into the input fields. The dimension compatibility conditions for x = A\\b require the two matrices A and b to have the same number of rows. Using the Matrix Calculator we get this: (I left the 1/determinant outside the matrix to make the numbers simpler) Then multiply A-1 by B (we can use the Matrix Calculator again): And we are done! System Of Linear Equations Involving Two Variables Using Determinants. In such a case, the pair of linear equations is said to be consistent. Solution: Given equation can be written in matrix form as : , , Given system \u2026 Characterize the vectors b such that Ax = b is consistent, in terms of the span of the columns of A. The matrix valued function $$X (t)$$ is called the fundamental matrix, or the fundamental matrix solution. Let $$\\vec {x}' = P \\vec {x} + \\vec {f}$$ be a linear system of A system of linear equations is as follows. Let the equations be a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0. row space: The set of all possible linear combinations of its row vectors. Example 1: Solve the equation: 4x+7y-9 = 0 , 5x-8y+15 = 0. 1. The solution is: x = 5, y = 3, z = \u22122. Typically we consider B= 2Rm 1 \u2019Rm, a column vector. a 11 x 1 + a 12 x 2 + \u2026 + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + \u2026 + a 2 n x n = b 2 \u22ef a m 1 x 1 + a m 2 x 2 + \u2026 + a m n x n = b m This system can be represented as the matrix equation A \u22c5 x \u2192 = b \u2192 , where A is the coefficient matrix. To solve nonhomogeneous first order linear systems, we use the same technique as we applied to solve single linear nonhomogeneous equations. Key Terms. If the rows of the matrix represent a system of linear equations, then the row space consists of all linear equations that can be deduced algebraically from those in the system. Understand the equivalence between a system of linear equations, an augmented matrix, a vector equation, and a matrix equation. The solution to a system of equations having 2 variables is given by: Systems of Linear Equations 0.1 De nitions Recall that if A2Rm n and B2Rm p, then the augmented matrix [AjB] 2Rm n+p is the matrix [AB], that is the matrix whose rst ncolumns are the columns of A, and whose last p columns are the columns of B. Consistent System. To sketch the graph of pair of linear equations in two variables, we draw two lines representing the equations. Solve the equation by the matrix method of linear equation with the formula and find the values of x,y,z. Developing an effective predator-prey system of differential equations is not the subject of this chapter. Solve several types of systems of linear equations. How To Solve a Linear Equation System Using Determinants? Theorem 3.3.2. Theorem. The whole point of this is to notice that systems of differential equations can arise quite easily from naturally occurring situations. 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Equations be a 1 x+b 1 y+c 1 = 0 and a 2 x+b 2 y+c 2 = 0 Objectives... The equations to sketch the graph of pair of linear equations, an matrix. Assuming it exists! from \\ ( n^ { \\text { th } } )... To notice that systems of differential equations is said to be consistent and b to the! Notice that systems of differential equations can arise from \\ ( x ( t ) \\ order! Is called the fundamental matrix, a column vector of pair of linear equations, an augmented matrix, the... B is consistent, in terms of the span of the span of the of! And find the inverse of the columns of a compatibility conditions for x = require!: 4x+7y-9 = 0 fundamental matrix solution the input fields matrix ( assuming it exists! not subject. Columns of a 2.3 matrix equations \u00b6 permalink Objectives the values of x, y z! The same technique as we applied to solve a linear equation system Using Determinants an! The two matrices a and b to have the same number of.! Systems, we use the same number of rows is called the fundamental matrix solution sketch graph! 1: solve the equation by the matrix valued function \\ ( x ( t ) )!", "date": "2021-03-05 01:00:55", "meta": {"domain": "com.ua", "url": "http://pobarabanu.com.ua/414yquca/elnfm.php?57d14b=system-of-linear-equations-matrix-conditions", "openwebmath_score": 0.8119489550590515, "openwebmath_perplexity": 560.0163132020501, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9688561694652216, "lm_q2_score": 0.9099070158103778, "lm_q1q2_score": 0.8815690259075735}}
{"url": "https://math.stackexchange.com/questions/3483283/whats-wrong-in-my-calculation-of-int-frac-sin-x1-sin-x-dx", "text": "What's wrong in my calculation of $\\int \\frac{\\sin x}{1 + \\sin x} dx$?\n\nI have the following function:\n\n$$f: \\bigg ( - \\dfrac{\\pi}{2}, \\dfrac{\\pi}{2} \\bigg ) \\rightarrow \\mathbb{R} \\hspace{2cm} f(x) = \\dfrac{\\sin x}{1 + \\sin x}$$\n\nAnd I have to find $$\\displaystyle\\int f(x) dx$$. This is what I did:\n\n$$\\int \\dfrac{\\sin x}{1 + \\sin x}dx= \\int \\dfrac{1+ \\sin x - 1}{1 + \\sin x}dx = \\int dx - \\int \\dfrac{1}{1 + \\sin x}dx =$$ $$= x - \\int \\dfrac{1 - \\sin x}{(1 + \\sin x)(1 - \\sin x)} dx$$\n\n$$= x - \\int \\dfrac{1 - \\sin x}{1 - \\sin ^2 x} dx$$\n\n$$= x - \\int \\dfrac{1 - \\sin x}{\\cos^2 x} dx$$\n\n$$= x - \\int \\dfrac{1}{\\cos^2x}dx + \\int \\dfrac{\\sin x}{\\cos^2 x}dx$$\n\n$$= x - \\tan x + \\int \\dfrac{\\sin x}{\\cos^2 x}dx$$\n\nLet $$u = \\cos x$$\n\n$$du = - \\sin x dx$$\n\n$$=x - \\tan x - \\int \\dfrac{1}{u^2}du$$\n\n$$= x - \\tan x + \\dfrac{1}{u} + C$$\n\n$$= x - \\tan x + \\dfrac{1}{\\cos x} + C$$\n\nThe problem is that the options given in my textbook are the following:\n\nA. $$x + \\tan {\\dfrac{x}{2}} + C$$\n\nB. $$\\dfrac{1}{1 + \\tan{\\frac{x}{2}}} + C$$\n\nC. $$x + 2\\tan{\\dfrac{x}{2}} + C$$\n\nD. $$\\dfrac{2}{1 + \\tan{\\frac{x}{2}}} + C$$\n\nE. $$x + \\dfrac{2}{1 + \\tan{\\frac{x}{2}}} + C$$\n\nNone of them are the answer I got solving this integral. What is the mistake that I made and how can I find the right answer? By what I've been reading online, you can get different answers by solving an integral in different ways and all of them are considered correct. They differ by the constant $$C$$. I understand that, but I don't see how to solve this integral in such a way to get an answer among the given $$5$$. And, even more importantly, how can I recognize the right answer in exam conditions if the answer provided by my solution is not present among the given options? Is solving in a different manner my only hope?\n\n\u2022 This is one of the faults of multiple choice questions. In this case they expected you to use the substitution $t=\\tan\\left(\\frac{x}{2}\\right)$. Your answer will be equivalent to one of the choices. \u2013\u00a0John Wayland Bales Dec 21 '19 at 0:57\n\u2022 If you encounter this on exam, then you should be able to convert your answer into one of the ones they give you, by possibly adding a constant and applying trig identities. \u2013\u00a0Robo300 Dec 21 '19 at 1:02\n\u2022 A deeper fault of multiple-choice for a question like that is that one could just differentiate the proposed answers and check the result at 0. Correct answer (E), but not proving that the student mastered the underlying concept or technique. \u2013\u00a0Catalin Zara Dec 21 '19 at 1:39\n\u2022 @CatalinZara true ....... \u2013\u00a0Aryadeva Dec 21 '19 at 2:46\n\nRequired answer is $$E$$. Observe that $$\\frac{1}{\\cos x}-\\tan x=\\frac{1-\\sin x}{\\cos x}$$ $$=\\frac{(\\cos\\frac{x}{2}-\\sin\\frac{x}{2})^2}{\\cos^2\\frac{x}{2}-\\sin^2\\frac{x}{2}}$$ $$=\\frac{\\cos\\frac{x}{2}-\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}+\\sin\\frac{x}{2}}$$ $$=\\frac{1-\\tan\\frac{x}{2}}{1+\\tan\\frac{x}{2}}$$ $$=\\frac{2}{1+\\tan\\frac{x}{2}}-1$$ Also, note that you could have directly got this answer if you have integrated $$\\frac{1}{1+\\sin x}$$ in a different manner, as follows. $$\\int\\frac{1}{1+\\sin x}dx=\\int\\frac{1}{(\\cos\\frac{x}{2}+\\sin\\frac{x}{2})^2}dx$$ $$=\\int \\frac{1}{\\cos^2\\frac{x}{2}(1+\\tan\\frac{x}{2})^2}dx$$ Now substitute $$\\tan\\frac{x}{2}$$ and you are done.\n\n\u2022 How did you get to $\\dfrac{2}{1+\\tan{\\frac{x}{2}}}-1$ from $\\dfrac{1-\\tan {\\frac{x}{2}}}{1+ \\tan {\\frac{x}{2}}}$? I didn't understand that bit. \u2013\u00a0user592938 Dec 22 '19 at 1:06\n\u2022 @user2502, $\\frac{1-\\tan\\frac{x}{2}}{1+\\tan\\frac{x}{2}}=\\frac{2-(1+\\tan\\frac{x}{2})}{1+\\tan\\frac{x}{2}} = \\frac{2}{1+\\tan\\frac{x}{2}}-1$ \u2013\u00a0Martund Dec 22 '19 at 6:28\n\nYour answer is correct and it matches with choice (E). You have to use half angle formulas: $$\\sin A = \\frac{2 \\tan \\frac{A}{2}}{1+\\tan^2 \\frac{A}{2}} \\quad \\cos A = \\frac{1-\\tan^2 \\frac{A}{2}}{1+\\tan^2 \\frac{A}{2}}$$ Observe that \\begin{align*} \\frac{1}{\\cos x}-\\tan x&=\\frac{1-\\sin x}{\\cos x}\\\\ & = \\frac{\\left(1-\\tan \\frac{x}{2}\\right)^2}{1-\\tan^2 \\frac{x}{2}}\\\\ & = \\frac{1-\\tan \\frac{x}{2}}{1+\\tan \\frac{x}{2}}\\\\ & = 1+\\frac{2}{1+\\tan \\frac{x}{2}}\\\\ \\end{align*}\n\n\u2022 How did you to the last line from your second to last line? I didn't understand that final bit. \u2013\u00a0user592938 Dec 21 '19 at 22:49\n\nBeside verifying your answer is equivalent to one of the listed results, you may also integrate in $$\\tan\\frac x2$$ since all the choices are in terms of it.\n\nSo, use $$\\sin x =\\frac{2\\tan\\frac x2}{1+\\tan^2\\frac x2}$$ to integrate,\n\n$$\\int \\dfrac{1}{1 + \\sin x}dx = \\int \\dfrac{1}{1 + \\frac{2\\tan\\frac x2}{1+\\tan^2\\frac x2}}dx =\\int \\dfrac{1+\\tan^2\\frac x2}{1+\\tan^2\\frac x2 + 2\\tan\\frac x2}dx$$ $$=\\int \\dfrac{\\sec^2\\frac x2}{(1+\\tan\\frac x2)^2} =\\int \\dfrac{2d(\\tan\\frac x2)}{(1+\\tan\\frac x2)^2} =-\\dfrac{2}{1+\\tan\\frac x2}$$", "date": "2021-06-15 14:12:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3483283/whats-wrong-in-my-calculation-of-int-frac-sin-x1-sin-x-dx", "openwebmath_score": 0.8357511758804321, "openwebmath_perplexity": 278.8970099780308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551556203814, "lm_q2_score": 0.9136765175373274, "lm_q1q2_score": 0.8815654985151661}}
{"url": "http://mathhelpforum.com/discrete-math/95124-easy-easy-question-print.html", "text": "# easy, easy question\n\n\u2022 Jul 14th 2009, 07:06 AM\nbillym\neasy, easy question\n\nHow many ways are there to seat 10 people, consisting of 5 couples, in a row of seats (10 seats wide) if all couples are to get adjacent seats?\n\u2022 Jul 14th 2009, 07:20 AM\nSoroban\nHello, billym!\n\nQuote:\n\nHow many ways are there to seat 10 people, consisting of 5 couples,\nin a row of seats (10 seats wide) if all couples are to get adjacent seats?\n\nFor reference, let the couples be: . $(A,a),\\:(B,b),\\:(C,c),\\:(D,d),\\:(E,e)$\n\nDuct-tape the couples together.\n\nWe have 5 \"people\" to arrange: . $\\boxed{Aa}\\;\\boxed{Bb}\\;\\boxed{Cc}\\;\\boxed{Dd}\\;\\b oxed{Ee}$\n\nThere are: . $5! \\,=\\,120$ permutations.\n\nBut for each permutation, the couples can be \"swtiched\".\n\n. . $\\boxed{Aa}$ could be $\\boxed{aA}$, $\\boxed{Bb}$ could be $\\boxed{bB}$, and so on.\n\nThere are: . $2^5 \\,=\\,32$ possible switchings.\n\nTherefore, there are: . $120\\cdot32 \\:=\\:{\\color{blue}3840}$ seating arrangements.\n\n\u2022 Jul 14th 2009, 07:55 AM\nHallsofIvy\nQuote:\n\nOriginally Posted by Soroban\nHello, billym!\n\nFor reference, let the couples be: . $(A,a),\\:(B,b),\\:(C,c),\\:(D,d),\\:(E,e)$\n\nDuct-tape the couples together.\n\nI've always suspected you had an evil streak, Soroban!(Giggle)\n\nI have a friend who swears you can do anything with duct-tape. Now I know you can even solve math problems with it!\n\nQuote:\n\nWe have 5 \"people\" to arrange: . $\\boxed{Aa}\\;\\boxed{Bb}\\;\\boxed{Cc}\\;\\boxed{Dd}\\;\\b oxed{Ee}$\n\nThere are: . $5! \\,=\\,120$ permutations.\n\nBut for each permutation, the couples can be \"swtiched\".\n\n. . $\\boxed{Aa}$ could be $\\boxed{aA}$, $\\boxed{Bb}$ could be $\\boxed{bB}$, and so on.\n\nThere are: . $2^5 \\,=\\,32$ possible switchings.\n\nTherefore, there are: . $120\\cdot32 \\:=\\:{\\color{blue}3840}$ seating arrangements.\n\n\u2022 Jul 14th 2009, 08:38 AM\nSoroban\nHello, HallsofIvy!\n\nQuote:\n\nI have a friend who swears you can do anything with duct tape.\nDuct tape is like The Force.\n\nIt has a light side and dark side\nand it holds the universe together.", "date": "2016-12-04 06:38:07", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/95124-easy-easy-question-print.html", "openwebmath_score": 0.9113851189613342, "openwebmath_perplexity": 4610.038898041267, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969645988576, "lm_q2_score": 0.8962513807543222, "lm_q1q2_score": 0.8815501376274862}}
{"url": "https://math.stackexchange.com/questions/1792266/two-methods-of-implicit-differentiation-dont-correspond", "text": "# Two methods of implicit differentiation don't correspond\n\nI recently attempted a question on implicit differentiation twice. I differentiated using one method in the first attempt and then another method in the second attempt but they do not correspond when I plug in values for the variables x and y. Please look at the two methods and tell me what I am doing wrong.\n\nIn the first attempt I just took the first derivative of each side of the equation with respect to x. In the second attempt, I took the ln of both sides of the equation first and then found the derivatives of either side of the equation.\n\nErrata: The answer for the first attempt is y' = y(y - e^(x/y)) / (y ^ 2 - xe^(x/y))\n\n\u2022 I think you accidentally introduced an extra $y$ into the second term near the end of the first attempt. This aside, the results may look different but really be the same, if you consider that you could replace $e^{x/y}$ with the very different-looking expression $x-y$ because they are the same according to the original equation. \u2013\u00a0MPW May 19 '16 at 22:06\n\u2022 Oh yes the extra y \u2013\u00a0rert588 May 19 '16 at 22:08\n\u2022 $$e^{x/y}=x-y$$ \u2013\u00a0Simply Beautiful Art May 19 '16 at 22:08\n\u2022 Yes. That is the question. \u2013\u00a0rert588 May 19 '16 at 22:11\n\u2022 I put the y there by mistake when i was writing out the answers neatly but even without that y, the two attempts do not correspond. \u2013\u00a0rert588 May 19 '16 at 22:14\n\nOn the second to last line of your first attempt, you had $y'=\\frac{y(y-e^\\frac{x}{y}y)}{y^2-e^{\\frac{x}{y}}x}$ where actually it should be $y'=\\frac{y(y-e^\\frac{x}{y})}{y^2-e^{\\frac{x}{y}}x}$. And then if you substitute $e^{\\frac{x}{y}}$ with $x-y$, you will get $y'=\\frac{y(y-(x-y))}{y^2-(x-y)x}=\\frac{2y^2-xy}{y^2-x^2+xy}$, which is the same as the result from your second approach.\n\u2022 I am pretty sure their first attempt is correct. That $e^{\\frac x y}y$ term comes from distributing the $e^{\\frac x y}$ over the numerator of the derivative of $\\frac x y$. EDIT: Oh, wait, I see. They factored out a $y$, but then forgot to remove the $y$ from the $e^{\\frac x y}y$ term. You're right. \u2013\u00a0Noble Mushtak May 19 '16 at 22:24\n\u2022 The key thing to notice here is that your derivative with the first method and second method look different, but are actually the same because $e^{x/y}=x-y$, so we can substitute that into the derivative using the first method to show that it's equivalent to the derivative using the second method. \u2013\u00a0Noble Mushtak May 19 '16 at 22:31", "date": "2020-02-27 18:43:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1792266/two-methods-of-implicit-differentiation-dont-correspond", "openwebmath_score": 0.6070007085800171, "openwebmath_perplexity": 169.3289375081391, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137926698566, "lm_q2_score": 0.8976952989498448, "lm_q1q2_score": 0.8815491651836378}}
{"url": "http://bernardklima.cz/hyq2s1h/d7c474-permutation-matrix-inverse", "text": "Led\n01\n\nAnd every 2-cycle (transposition) is inverse of itself. Every permutation n>1 can be expressed as a product of 2-cycles. Sometimes, we have to swap the rows of a matrix. A permutation matrix is an orthogonal matrix \u2022 The inverse of a permutation matrix P is its transpose and it is also a permutation matrix and \u2022 The product of two permutation matrices is a permutation matrix. The use of matrix notation in denoting permutations is merely a matter of convenience. 4. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. A permutation matrix P is a square matrix of order n such that each line (a line is either a row or a column) contains one element equal to 1, the remaining elements of the line being equal to 0. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. \u2022Find the inverse of a simple matrix by understanding how the corresponding linear transformation is related to the matrix-vector multiplication with the matrix. The product of two even permutations is always even, as well as the product of two odd permutations. Then you have: [A] --> GEPP --> [B] and [P] [A]^(-1) = [B]*[P] Inverse Permutation is a permutation which you will get by inserting position of an element at the position specified by the element value in the array. 4. Sometimes, we have to swap the rows of a matrix. \u2022Identify and apply knowledge of inverses of special matrices including diagonal, permutation, and Gauss transform matrices. Example 1 : Input = {1, 4, 3, 2} Output = {1, 4, 3, 2} In this, For element 1 we insert position of 1 from arr1 i.e 1 at position 1 in arr2. To get the inverse, you have to keep track of how you are switching rows and create a permutation matrix P. The permutation matrix is just the identity matrix of the same size as your A-matrix, but with the same row switches performed. Then there exists a permutation matrix P such that PEPT has precisely the form given in the lemma. All other products are odd. Here\u2019s an example of a $5\\times5$ permutation matrix. I was under the impression that the primary numerical benefit of a factorization over computing the inverse directly was the problem of storing the inverted matrix in the sense that storing the inverse of a matrix as a grid of floating point numbers is inferior to \u2026 The product of two even permutations is always even, as well as the product of two odd permutations. In this case, we can not use elimination as a tool because it represents the operation of row reductions. 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Is always even, and Gauss transform matrices multiplication with the matrix always even, and Gauss matrices. This case, we have to swap the rows of a simple matrix by how... How the corresponding linear transformation is related to the matrix-vector multiplication with matrix! Of itself simple matrix by understanding how the corresponding linear transformation is related to the multiplication. It occupies is exchanged of convenience with the matrix is a permutation in which each number and the of...", "date": "2021-06-19 03:39:08", "meta": {"domain": "bernardklima.cz", "url": "http://bernardklima.cz/hyq2s1h/d7c474-permutation-matrix-inverse", "openwebmath_score": 0.9370395541191101, "openwebmath_perplexity": 369.1545041757962, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9820137868795702, "lm_q2_score": 0.8976952825278492, "lm_q1q2_score": 0.8815491438590989}}
{"url": "https://mathhelpboards.com/threads/combinatorics-question-involving-6-6-sided-dice.3756/", "text": "# Combinatorics question involving 6 6-sided dice\n\n#### wittysoup\n\n##### New member\nWhen rolling 6 6-sided dice, how many different ways can you have exactly 4 different numbers?\n\nI tried solving this like so,\n\nthe first dice has a possible 6 numbers, the second has a possible 5, the third has a possible 4, and the fourth, 3. Then there are 2 remaining dice of which each has to be one of the previous 4 numbers so there are:\n\n6*5*4*3*4*4 = 5760 ways\n\nsomething tells me I am not thinking correctly or might be missing something because when I did this by iteration and got 9216. (though I might have missed something here too)\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nWhen rolling 6 6-sided dice, how many different ways can you have exactly 4 different numbers?\n\nI tried solving this like so,\n\nthe first dice has a possible 6 numbers, the second has a possible 5, the third has a possible 4, and the fourth, 3. Then there are 2 remaining dice of which each has to be one of the previous 4 numbers so there are:\n\n6*5*4*3*4*4 = 5760 ways\n\nsomething tells me I am not thinking correctly or might be missing something because when I did this by iteration and got 9216. (though I might have missed something here too)\nWelcome to MHB, wittysoup!\n\nIt is somewhat more complex.\n\nThere are 2 patterns with different counts: aaabcd and aabbcd.\n\nPattern aaabcd occurs 6*1*1*5*4*3 times.\nSince the aaa can be distributed in different ways, we need to multiply by the number of times we can pick 3 dice out of 6, which is $(^6_3)$.\nSo the pattern aaabcd including all its possible orderings occurs $6 \\cdot 1 \\cdot 1\\cdot 5 \\cdot 4 \\cdot 3 \\cdot (^6_3) = 7200$ out of $6^6$.\n\nThe pattern aabbcd is more complex still.\nThe base pattern occurs 6*1*5*1*4*3 times.\nMultiply by $(^6_2)$ for the different locations of aa.\nMultiply by $(^4_2)$ for the remaining different locations of bb.\nDivide by 2! because aa can be swapped with bb..\nSo the pattern aabbcd including all its possible orderings occurs $6 \\cdot 1 \\cdot 5 \\cdot 1\\cdot 4 \\cdot 3 \\cdot (^6_2) \\cdot (^4_2) \\cdot \\frac {1}{2!} = 16200$ out of $6^6$.\n\nSo a total of $7200 + 16200 = 23400$ combinations.\n\nLast edited:\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\n[JUSTIFY]You never mentioned it, but when considering dice rolls it's generally assumed that the rolls are unordered (they have no notion of order, which makes sense since the dice are presumably identical). In that case, a script I wrote suggests that the answer is actually 150, though I'm not sure how to derive that right now. If the dice rolls are ordered, then I Like Serena's answer is correct.[/JUSTIFY]\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\n[JUSTIFY]You never mentioned it, but when considering dice rolls it's generally assumed that the rolls are unordered (they have no notion of order, which makes sense since the dice are presumably identical). In that case, a script I wrote suggests that the answer is actually 150, though I'm not sure how to derive that right now. If the dice rolls are ordered, then I Like Serena's answer is correct.[/JUSTIFY]\nUsually I do the ordered variant, since that's necessary if we want to calculate probabilities.\nBut let's see what we get in an unordered variant...\n\nFor pattern aaabcd we have 6 choices for a, and then $\\dfrac {5 \\cdot 4 \\cdot 3}{1 \\cdot 2 \\cdot 3}$ unordered choices for bcd.\nFor pattern aabbcd we have 6 choices for a, 5 choices for b, divide by 2 for being unordered, and then $\\dfrac {4 \\cdot 3}{1 \\cdot 2}$ unordered choices for cd.\n\nThat brings us to:\n$$6 \\cdot \\dfrac {5 \\cdot 4 \\cdot 3}{1 \\cdot 2 \\cdot 3} + \\dfrac{6 \\cdot 5}{1 \\cdot 2} \\cdot \\dfrac {4 \\cdot 3}{1 \\cdot 2} = 60 + 90 = 150$$\nSounds right!\n\n#### wittysoup\n\n##### New member\neach dice is different here it seems from the question.. so 1 1 1 2 3 4 is different than 1 2 1 1 3 4.\n\n#### Jameson\n\nStaff member\nBecause the dice are all the same, usually the problem is interpreted as order not mattering in my experience. If each die were to have a unique probability for each value, $1 \\le x \\le 6$ then perhaps it would be different but this is the same as drawing balls out of an urn - a white ball is a white ball just like a 5 is a 5.\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\neach dice is different here it seems from the question.. so 1 1 1 2 3 4 is different than 1 2 1 1 3 4.\nSo those are the ordered variants.\nI've edited my previous post to include the total, which is 23400 out of 46656.\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nBecause the dice are all the same, usually the problem is interpreted as order not mattering in my experience. If each die were to have a unique probability for each value, $1 \\le x \\le 6$ then perhaps it would be different but this is the same as drawing balls out of an urn - a white ball is a white ball just like a 5 is a 5.\nThere are 24300 out of 46656 ordered combinations with 4 different dice.\nThe corresponding probability is P(4 different dice)=0.50154.\n\nThere are 150 out of 462 unordered combinations.\nThe corresponding proportion is 0.32468, which is different from the probability.\n\nIf the probabilities for specific values or specific dice became different, things would become more complex yet again.\n\n#### Jameson\n\nStaff member\nI agree with the two situations you proposed, but don't agree that the second can't be considered a probability.\n\nAn easy example of my point would be in poker, what is the probability of being dealt a flush if you are given 5 cards? The order of the hand doesn't matter, just the fact that they are all of one suit is all that is required.\n\nIf s = spades, then {As, 2s, 3s, 8s, Js} = {2s, As, 3s, 8s, Js}. The probability of a flush is well known.\n\nNot trying to be combative as I feel you probably know more on this topic than I, but would you explain some more please?\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nI agree with the two situations you proposed, but don't agree that the second can't be considered a probability.\n\nAn easy example of my point would be in poker, what is the probability of being dealt a flush if you are given 5 cards? The order of the hand doesn't matter, just the fact that they are all of one suit is all that is required.\n\nIf s = spades, then {As, 2s, 3s, 8s, Js} = {2s, As, 3s, 8s, Js}. The probability of a flush is well known.\n\nNot trying to be combative as I feel you probably know more on this topic than I, but would you explain some more please?\nIn the case of a flush in poker it does not matter.\nThe number of ordered combinations is just 5! times the number of unordered combinations.\nSo the unordered proportion is the same as the ordered proportion.\n\nNow consider for instance the chance on 6 sixes with 6 dice.\nIt is one possible combination out of 462 unordered combinations.\nBut the probability is much lower than 1 in 462.\n\n#### Jameson\n\nStaff member\n\n$$\\displaystyle \\frac{ \\text{Unordered outcomes}}{ \\text{Unordered possibilities}}$$, rather $$\\displaystyle \\frac{ \\text{Unordered outcomes}}{ \\text{All possibilities}}$$\nUsing your dice example you could write the solution at $$\\displaystyle \\left( \\frac{1}{6} \\right)^{6}$$ or as $$\\displaystyle \\frac{1}{6^6}$$.\n\nAgain, I might be wrong but let me think a bit and if you see where I'm wrong please let me know.\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\n\n$$\\displaystyle \\frac{ \\text{Unordered outcomes}}{ \\text{Unordered possibilities}}$$, rather $$\\displaystyle \\frac{ \\text{Unordered outcomes}}{ \\text{All possibilities}}$$\nUsing your dice example you could write the solution at $$\\displaystyle \\left( \\frac{1}{6} \\right)^{6}$$ or as $$\\displaystyle \\frac{1}{6^6}$$.\n\nAgain, I might be wrong but let me think a bit and if you see where I'm wrong please let me know.\nThe marginal probability formula is:\n$$P(\\text{Favorable outcome}) = \\frac{ \\text{Number of favorable outcomes}}{ \\text{Total number of outcomes}}$$\nThis formula only works if all outcomes are equally probable.\nYou can choose yourself what you consider an outcome, which could be ordered or unordered, but you have to be consistent.\nEither choice works as long as the outcomes are equally likely.\n\nIn the case of a flush your unordered outcomes are equally likely.\nIn the case of the dice game, the unordered outcomes are not equally likely, so you cannot use the marginal probability formula.\n\n#### wittysoup\n\n##### New member\nOkay, so assuming order is not important, we'll have 150 total possible combinations of dice containing exactly 4 of the 6 total numbers?\n\n#### wittysoup\n\n##### New member\nUsually I do the ordered variant, since that's necessary if we want to calculate probabilities.\nBut let's see what we get in an unordered variant...\n\nFor pattern aaabcd we have 6 choices for a, and then $\\dfrac {5 \\cdot 4 \\cdot 3}{1 \\cdot 2 \\cdot 3}$ unordered choices for bcd.\nFor pattern aabbcd we have 6 choices for a, 5 choices for b, divide by 2 for being unordered, and then $\\dfrac {4 \\cdot 3}{1 \\cdot 2}$ unordered choices for cd.\n\nThat brings us to:\n$$6 \\cdot \\dfrac {5 \\cdot 4 \\cdot 3}{1 \\cdot 2 \\cdot 3} + \\dfrac{6 \\cdot 5}{1 \\cdot 2} \\cdot \\dfrac {4 \\cdot 3}{1 \\cdot 2} = 60 + 90 = 150$$\nSounds right!\nhow did you get these fractions \"[FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]3[/FONT]\" and \"[FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]2[/FONT]\" ?\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nhow did you get these fractions \"[FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]3[/FONT]\" and \"[FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]\u22c5[/FONT][FONT=MathJax_Main]2[/FONT]\" ?\nThey are $(^5_3)$ and $(^4_2)$.\n\nThe first counts the ways bcd can be distributed as part of the pattern aaabcd.\nFor \"b\" we have 5 remaining choices (after \"a\").\nThen for \"c\" 4 remaining choices.\nAnd for \"d\" 3 remaining choices.\nSince this yields an ordered set, we need to divide by the number of ways the 3 numbers can be ordered, which is 3!.", "date": "2021-08-01 17:35:59", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/combinatorics-question-involving-6-6-sided-dice.3756/", "openwebmath_score": 0.7834502458572388, "openwebmath_perplexity": 675.7231009233019, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639653084244, "lm_q2_score": 0.9073122295060291, "lm_q1q2_score": 0.8815118674717048}}
{"url": "https://math.stackexchange.com/questions/1558133/different-function-with-the-same-derivative", "text": "# Different function with the same derivative\n\nToday at school I entered in a problem when the professor asked us to differentiate the following function:\n\n$$f(x)=\\arctan\\left(\\frac {x-1}{x+1}\\right)$$\n\nWith the basic rules of differentiation I came to a confusing result:\n\n$$f'(x)=\\frac 1{1+x^2}$$\n\nAnd the teacher agreed, and so does Wolfram (I checked at home) but what surprised me is that it's the same derivative as\n\n$$f(x)=\\arctan x$$ $$f'(x)=\\frac 1{1+x^2}$$\n\nSo I'm wondering: is that wrong in some sense ? Are the two function equals indeed ? If I integrate $\\frac 1{1+x^2}$ what should I choose from the two ? Are there any other examples of different functions with the same derivative?\n\n\u2022 What's the derivative of $f(x)=x^2+1$ and of $g(x)=x^2$? Dec 3 '15 at 13:47\n\u2022 This is a matter of constants, my case is pretty different. @mathochist Dec 3 '15 at 13:50\n\u2022 Your question has been answered, but as a word of advice, this should have immediately hinted to you that your two functions must only differ by a constant, although looking fundamentally different. In your case, if you do the calculations, you'll see that your constant is $\\frac{\\pi}{4}$. Dec 3 '15 at 13:53\n\u2022 @RenatoFaraone Your case seems pretty different, but it isn't! Dec 3 '15 at 13:53\n\u2022 @Rellek As I note in my answer, actually, there are two constants, one when $x<-1$ and one when $x>-1$. Dec 3 '15 at 13:55\n\nNote:\n\n$$\\tan(A-B)=\\frac{\\tan A - \\tan B}{1+\\tan A \\tan B}$$\n\nIf $x=\\tan A$ and $\\tan B=1$, then you get:\n\n$$\\tan(A-B)=\\frac{x-1}{x+1}$$\n\nSo $$\\arctan x - B = \\arctan\\left(\\frac{x-1}{x+1}\\right)$$ So the functions differ by a constant.\n\n(Well, close enough - they actually differ by a constant locally, wherever both functions are defined. The differences will be constant in $(-\\infty,-1)$ and in $(-1,\\infty)$, but not necessarily the entire real line.)\n\n\u2022 @AdityaAgarwal $x=\\tan A$ so $A=\\arctan x$. But yes, some care is needed to pick $A$. Dec 3 '15 at 13:51\n\nAre you surprised from $3-2=10-9$? ;-) I guess you aren't.\n\nAre you surprised from the fact that $f(x)=x^2$ and $g(x)=x^2+1$ have the same derivative? Not at all, I believe.\n\nThe same holds in this case, and it's not the only one! For instance, $f(x)=\\arcsin x$ and $g(x)=-\\arccos x$ have the same derivative! Also $f(x)=\\log x$ and $g(x)=\\log(3x)$ do.\n\nThe conclusion you can draw is that the two functions differ by a constant on each interval where they are both defined. Since $\\arctan\\frac{x-1}{x+1}$ is defined for $x\\ne-1$, you know that there exist constants $h$ and $k$ such that $$\\begin{cases} \\arctan\\dfrac{x-1}{x+1}=h+\\arctan x & \\text{for x<-1}\\\\[12px] \\arctan\\dfrac{x-1}{x+1}=k+\\arctan x & \\text{for x>-1} \\end{cases}$$\n\nYou can now compute $h$ and $k$, by evaluating the limit at $-\\infty$ and at $\\infty$: $$\\frac{\\pi}{4}=\\lim_{x\\to-\\infty}\\arctan\\dfrac{x-1}{x+1}= \\lim_{x\\to-\\infty}(h+\\arctan x)=h-\\frac{\\pi}{2}$$ and $$\\frac{\\pi}{4}=\\lim_{x\\to\\infty}\\arctan\\dfrac{x-1}{x+1}= \\lim_{x\\to\\infty}(k+\\arctan x)=k+\\frac{\\pi}{2}$$", "date": "2021-09-17 21:32:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1558133/different-function-with-the-same-derivative", "openwebmath_score": 0.885212779045105, "openwebmath_perplexity": 205.54974554083205, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771782476948, "lm_q2_score": 0.8933094074745443, "lm_q1q2_score": 0.8814973364098511}}
{"url": "https://mathhelpboards.com/threads/sets.5984/", "text": "# Sets\n\n#### bergausstein\n\n##### Active member\n1.are the following sets a finite sets? if yes why? if no why?\n\na. the set of points on a given line exactly one unit from a given point on that line.\nb. the set of points in a given plane that are exactly one unit from a given point in that plane.\ni'm confused with the part saying \"one unit from a given point on that line and \"one unit from a given point in that plane. it seems to me that these phrases give hints that the sets are finite. please correct me if i'm wrong.\n\n2. show that if A is a proper subset of B and $B\\subseteq C$ then, A is a proper subset of of C.\n\nthanks!\n\n#### MarkFL\n\nStaff member\n1.) To simplify matter a bit, consider:\n\na) How many points are 1 unit away from the origin on the $x$-axis?\n\nb) How many points are one unit away from the origin in the $xy$-plane?\n\n#### bergausstein\n\n##### Active member\nuhmm.. infinitely many points. so it's inifinite right?\n\n#### MarkFL\n\nStaff member\nuhmm.. infinitely many points. so it's inifinite right?\nIf you are unsure, plot the points in both cases. What do you find?\n\nOr, consider the following:\n\nThe first case is:\n\n$$\\displaystyle |x|=1$$\n\nHow many solutions?\n\nThe second case is:\n\n$$\\displaystyle x^2+y^2=1$$\n\nHow many solutions?\n\n#### bergausstein\n\n##### Active member\na has 2 elements.\nb has 4.\nand can you also help me with 2. i can say that in words but i couldn't do it in a general manner.\n\n#### MarkFL\n\nStaff member\nYes for part a) there are 2 points: $$\\displaystyle (\\pm1,0)$$, but there are an infinite number of points on a circle, uncountably infinite from what I understand.\n\nFor question 2, I recommend using a Venn diagram.\n\n#### bergausstein\n\n##### Active member\nmarkfl how did you know that question B is talking about the equation of the circle? $\\displaystyle x^2+y^2=1$ I put this equation on the wolframalpha and it says that there are 4 solutions to this equation. (1,0), (-1,0), (0,1), (-1,0). how is it infinite?\n\n#### MarkFL\n\nStaff member\nThe equation of the circle is $x^2+y^2=1$, only if the fixed point (the circle's center) is the origin. One of the definitions of a circle is the set of all points a given distance from a fixed point, and we can use the distance formula to get this equation. For part b) we are not restricted to the axes as we are for the first part, where we are considering only a one-dimensional line.\n\n#### solakis\n\n##### Active member\nIf you are unsure, plot the points in both cases. What do you find?\n\nOr, consider the following:\n\nThe first case is:\n\n$$\\displaystyle |x|=1$$\n\nHow many solutions?\n\nThe second case is:\n\n$$\\displaystyle x^2+y^2=1$$\n\nHow many solutions?\n\nYes.but how do we prove that the points are infinite??\n\n#### MarkFL\n\nStaff member\nWe can map the points on the circle to a line segment of length $2\\pi r$. According to Cantor, this is equinumerous with $\\mathbb{R}$.\n\n#### eddybob123\n\n##### Active member\nI put this equation on the wolframalpha and it says that there are 4 solutions to this equation. (1,0), (-1,0), (0,1), (-1,0). how is it infinite?\nThose are only the x and y-intecepts of the function. One way to think about it is that it has two variables but only one equation. As you probably learned in middle school algebra, that means the solution set (x,y) of $x^2+y^2=1$ is infinite.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nI will prove that there are at LEAST as many points (x,y) that satisfy:\n\n$$\\displaystyle x^2 + y^2 = 1$$\n\nas there are points in the real interval [0,1].\n\nTo do this, I will create an injective function $$\\displaystyle f$$ from [0,1] to the set $$\\displaystyle S$$, where:\n\n$$\\displaystyle S = \\{(x,y) \\in \\Bbb R^2: x^2 + y^2 = 1\\}$$.\n\nThe function I have in mind is this one:\n\n$$\\displaystyle f(a) = (a,\\sqrt{1 - a^2})$$\n\n(convince yourself this is indeed a function).\n\nFirst, we verify that $$\\displaystyle f([0,1]) \\subseteq S$$:\n\nLet $$\\displaystyle a \\in [0,1]$$. Then:\n\n$$\\displaystyle a^2 + (\\sqrt{1 - a^2})^2 = a^2 + 1 - a^2 = 1$$\n\n(note that we have to have $$\\displaystyle |a| \\leq 1$$ for this to work).\n\nThis shows that $$\\displaystyle f(a) \\in S$$, for ANY $$\\displaystyle a \\in [0,1]$$. So the image of $$\\displaystyle f$$ indeed lies within $$\\displaystyle S$$ as claimed.\n\nNow, suppose $$\\displaystyle f(a) = f(b)$$ for $$\\displaystyle a,b \\in [0,1]$$. This means:\n\n$$\\displaystyle (a,\\sqrt{1 - a^2}) = (b,\\sqrt{1 - b^2})$$.\n\nThis means we MUST have $$\\displaystyle a = b$$ (since if BOTH coordinates are equal, surely the first coordinates are also equal). So f is injective (one-to-one).\n\nThus for every $$\\displaystyle a \\in [0,1]$$, we have a corresponding UNIQUE point $$\\displaystyle f(a) \\in S$$, so $$\\displaystyle S$$ MUST be infinite, as it contains an infinite subset.", "date": "2021-09-16 16:20:27", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/sets.5984/", "openwebmath_score": 0.790042519569397, "openwebmath_perplexity": 388.7327711655853, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120905823452, "lm_q2_score": 0.8887587920192298, "lm_q1q2_score": 0.881481715536032}}
{"url": "https://math.stackexchange.com/questions/1479483/when-does-the-inverse-of-a-covariance-matrix-exist", "text": "# When does the inverse of a covariance matrix exist?\n\nWe know that a square matrix is a covariance matrix of some random vector if and only if it is symmetric and positive semi-definite (see Covariance matrix). We also know that every symmetric positive definite matrix is invertible (see Positive definite). It seems that the inverse of a covariance matrix sometimes does not exist.\n\nDoes the inverse of a covariance matrix exist if and only if the covariance matrix is positive definite? How can I intuitively understand the situation when the inverse of a covariance matrix does not exist (does it mean that some of the random variables of the random vector are equal to a constant almost surely)?\n\nAny help will be much appreciated!\n\n\u2022 I suspect the inverse does not exist if and only if $P(X\\in H)=1$ for some $H\\subset\\mathbb R^n$ with - let's say - having a dimension less than $n$. Here $H$ is not requested to be a hyperplane. E.g. also a sphere will do. Pure intuition, though. I could be wrong in this. \u2013\u00a0drhab Oct 14 '15 at 9:05\n\nIf the covariance matrix is not positive definite, we have some $a \\in \\mathbf R^n \\setminus \\{0\\}$ with $\\def\\C{\\mathop{\\rm Cov}}\\C(X)a = 0$. Hence \\begin{align*} 0 &= a^t \\C(X)a\\\\ &= \\sum_{ij} a_j \\C(X_i, X_j) a_i\\\\ &= \\mathop{\\rm Var}\\left(\\sum_i a_i X_i\\right) \\end{align*} So there is some linear combination of the $X_i$ which has zero variance and hence is constant, say equal to $\\alpha$, almost surely. Letting $H := \\{x \\in \\mathbf{R}^n: \\sum_{i} a_i x_i = \\alpha\\}$, this means, as @drhab wrote $\\mathbf P(X \\in H) = 1$ for the hyperplane $H$.\n\n\u2022 Didn't see your answer when I was posting mine -- hence the apparent duplicacy. \u2013\u00a0uniquesolution Oct 14 '15 at 9:39\n\u2022 No problem ... ${}$ \u2013\u00a0martini Oct 14 '15 at 9:39\n\u2022 Too late: I meant: \"I have grown wiser.\" Now also when it concerns English language. \u2013\u00a0drhab Oct 14 '15 at 9:57\n\u2022 @martini Nice answer (+1)! I'd like to clarify a few details. (a) If $\\operatorname{Cov}X$ is not invertible, then there exists $a$ such that $\\operatorname{Var}(a^TX)=0$. But this $a$ might not be unique, right? So can we conclude that $\\Pr\\{X\\in H\\}=1$ if $a$ is not unique? (b) Is it also true that $\\operatorname{Cov}X$ is not invertible if and only if there exists $a$ such that $\\operatorname{Var}(a^TX)=0$? \u2013\u00a0Cm7F7Bb Oct 15 '15 at 7:17\n\u2022 Right, it may not unique, but we can conclude that. In the case where $a$ is not unique (even not up to a scalar factor), we even have that $X$ is almost surely contained in an $<(n-1)$-dimensional subspace (the one orthogonal to all $a$ having $\\operatorname{Var}(a^t X) = 0$). Yes that's true. As $\\operatorname{Var}(a^t X) = a^t \\operatorname{Cov}(X)a$. \u2013\u00a0martini Oct 15 '15 at 7:20\n\nAs is nicely explained here What's the best way to think about the covariance matrix?\n\nif $A$ is the covariance matrix of some random vector $X\\in\\mathbb{R}^n$, then for every fixed $\\beta\\in\\mathbb{R}^n$, the variance of the inner product $\\langle\\beta,X\\rangle$ is given by $\\langle A\\beta,\\beta\\rangle$. Now, if $A$ is not invertible, there exists a non-zero vector $\\beta\\neq 0$ such that $A\\beta=0$, and so $\\langle A\\beta,\\beta\\rangle = 0$, which means that the variance of $\\langle X,\\beta\\rangle$ is zero.\n\nProposition 1. If the covariance matrix of a random vector $X$ is not invertible then there exists a non-trivial linear combination of the components of $X$ whose variance is zero.\n\nThis is closely related to what drhab mentioned in a comment above - for if the variance of $\\langle X,\\beta\\rangle$ is zero, then $X-a\\beta$ is almost surely orthogonal to $\\beta$, for some constant $a$.In fact an alternative but equivalent formulation to the proposition above is:\n\nProposition 2. If the covariance matrix of a random vector $X$ is not invertible then there exists $\\beta\\neq 0$ such that a translate of $X$ is orthogonal to $\\beta$ with probability one.\n\n\u2022 Too late: I meant: \"I have grown wiser.\" Now also when it concerns English language. \u2013\u00a0drhab Oct 14 '15 at 9:57\n\u2022 Nice answer (+1)! I just want to clarify the details. If $\\operatorname{Var}\\langle X,\\beta\\rangle=0$ with $\\beta\\ne0$, then we have that $\\langle A\\beta,\\beta\\rangle=0$. How does it follow that $A\\beta=0$, which means that $A$ is not invertible? Also, if $\\operatorname{Var}\\langle X,\\beta\\rangle=0$, then it means that $\\langle X,\\beta\\rangle=c$ almost surely with $c\\in\\mathbb R$ and the vector that is almost surely orthogonal to $X$ is $c\\beta$, right? \u2013\u00a0Cm7F7Bb Oct 14 '15 at 13:22\n\u2022 @V.C You are right -- the original formulation was not correct. I edited it. \u2013\u00a0uniquesolution Oct 14 '15 at 13:57\n\u2022 @uniquesolution I'm not saying that something is not correct, I'm just trying to understand the details. Your previous Proposition 1 was so much nicer... Is it possible to show that the covariance matrix is not invertible if there exists such linear combination? For any $X$ and $Y$ in any Hilbert space, there always exists a constant $a$ such that $\\langle X-aY,Y\\rangle=0$. Take $a=\\langle X,Y\\rangle/\\langle Y,Y\\rangle$. Is that right? \u2013\u00a0Cm7F7Bb Oct 14 '15 at 15:49\n\u2022 Yes, the previous answer was much nicer, but the proof I presented was not good enough. Perhaps it is true. I will think about it some more. And as for $a$, you got it right. \u2013\u00a0uniquesolution Oct 14 '15 at 18:00\n\nThe question actually has little to do with probability theory, the observation holds for any square matrix regardless of it's origin.\n\nIt's easy to prove by considering the eigenvalues of the matrix. If and only if all of them are non-zero is the matrix invertible. It follows from the characteristic equation $\\det (A-\\lambda I)=0$, if $\\lambda = 0$ is a solution then and only then $\\det(A-0I) = \\det(A) = 0$.\n\nPositive definite means that all eigenvalues are positive, but positive semi-definite means only that they are non-negative.\n\nThis are some thoughts. Let $x$ be a random vector whose entries are i.i.d. Let $A$ be any square matrix which is not full rank. Then the covariance matrix of the random vector $y=Ax$ is not invertible. To see this, note that $E[Axx^TA]=AE[xx^T]A^T$. Thus, regardless of the rank of $E[xx^T]$, covariance matrix of $y$ will not be invertible.", "date": "2019-05-23 14:57:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1479483/when-does-the-inverse-of-a-covariance-matrix-exist", "openwebmath_score": 0.9590762257575989, "openwebmath_perplexity": 99.61419093271385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429169195593, "lm_q2_score": 0.8947894710123925, "lm_q1q2_score": 0.8814060305549565}}
{"url": "https://math.stackexchange.com/questions/1737564/conditional-probability-question-understanding-mistake", "text": "# Conditional probability question (understanding mistake)\n\nI'm trying to understand the following question:\n\nAn engineer conducts tests to find out if circuits of a certain type are prone to overheating. 30% of all such circuits are prone to overheating. If the circuit is prone to overheating, the test will report it is not prone to overheating with probability 0.1, prone to overheating with probability 0.7, and produce an inconclusive result with probability 0.2. If it is not prone to overheating, the test will report it is not prone to overheating with probability 0.6, prone to overheating with probability 0.3, and produce an inconclusive result with probability 0.1. The experiment is performed twice on a particular circuit; the first time it produces an inconclusive result and the second time it reports that the circuit is prone to overheating. Assuming the results of the two tests are independent, what is the probability the circuit is prone to overheating, given the outcome of the tests?\n\nThis is how I tried to solve the question:\n\n$$P(O) = 0.3 \\ \\quad P(O^c)= 0.7 \\\\ P(N|O) = 0.1 \\quad P(P|O) = 0.7 \\quad P(I|O) = 0.2 \\\\ P(N|O^c) = 0.6 \\quad P(P|O^c) = 0.3 \\quad P(I|O^c) = 0.1 \\\\ \\\\ P(O|I)= \\frac {P(I|O)P(O)}{P(I|O)P(O) + P(I|O^c)P(O^c)} = \\frac{0.2*0.3}{0.2*0.3 + 0.1*0.7} = \\frac{6}{13}\\\\ P(O|P)= \\frac {P(P|O)P(O)}{P(P|O)P(O) + P(P|O^c)P(O^c)} = \\frac{0.7*0.3}{0.7*0.3 + 0.3*0.7} = \\frac{1}{2}\\\\$$ So the probability that the circuit is prone to overheating is $\\frac{6}{13}* \\frac{1}{2} = \\frac{3}{13}$\n\nMy answer was incorrect. The actual method is:\n\nThe probability that the circuit is prone to overheating and we observe the test results we have seen is: 0.3 \u00d7 0.2 \u00d7 0.7 = 0.042. The probability that the circuit is not prone to overheating and we observe the test results we have seen is: 0.7 \u00d7 0.1 \u00d7 0.3 = 0.021. The probability that we observe the test results we have seen is: 0.042 + 0.021 = 0.063. Therefore, the conditional probability that the circuit is prone to overheating given the outcomes of the tests is:$\\frac {0.042}{0.063} = \\frac{2}{3}$\n\nMy understanding is clearly not correct. Could someone explain why my method doesn't work?\n\n\u2022 While using mathematical notation for Bayes' Theorem, students frequently get lost in the maze of symbols. You haven't considered that two tests in series have been done. Study the explanation which is quite clear, and try to modify your formula accordingly. Apr 11 '16 at 14:25\n\nYour method doesn't work because you have to find:\n\nP(O | I on $1^{st}$ test $\\cap$ P on $2^{nd}$ test), but you have calculated\n\nWhat you have calculated is $P(O | \\text{I on a test}) * P (O | \\text{P on a test})$ which isn't the probability of a specific event.\n\nThe first part:\n\n$P(O | \\text{I on a test})$\n\nIncludes cases where you get I on a test but not P on the other, and the second part:\n\n$P (O | \\text{P on a test})$\n\nincludes cases where you get P on a test but not I on the other\n\nYou need to find the conditional probability given both I and P happen. $I \\cap P$\n\nCalculation for completeness:\n\nFor simplicity I'll call the events I and P.\n\n$P(O | I \\cap P) = \\frac{P(O \\cap I \\cap P)}{P(I \\cap P)}$\n\n$P(O | I \\cap P) = \\frac{P(O \\cap I \\cap P)}{P(O \\cap I \\cap P) + P(O^c \\cap I \\cap P)}$\n\n$P(O \\cap I \\cap P) = P(O)*P(I \\cap P | O) = 0.3 * 0.2 * 0.7 = 0.042$\n\n$P(O^c \\cap I \\cap P) = P(O^c)*P(I \\cap P | O^c) = 0.7 * 0.1 * 0.3 = 0.021$\n\n$P(O | I \\cap P) = \\frac{0.042}{0.042 + 0.021} = \\frac{0.042}{0.063} = \\frac{2}{3}$", "date": "2021-09-18 16:03:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1737564/conditional-probability-question-understanding-mistake", "openwebmath_score": 0.8229305148124695, "openwebmath_perplexity": 424.64601854916737, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429129677614, "lm_q2_score": 0.8947894696095782, "lm_q1q2_score": 0.881406025637097}}
{"url": "https://math.stackexchange.com/questions/2624208/counting-particular-odd-length-strings-over-a-two-letter-alphabet", "text": "# Counting particular odd-length strings over a two letter alphabet.\n\nOEIS sequence A297789 describes\n\nThe number of [equivalence classes of] length $2n - 1$ strings over the alphabet $\\{0, 1\\}$ such that the first half of any initial odd-length substring is a permutation of the second half.\n\nTwo strings are in the same equivalence class if they are the same up to swapping the letters of the alphabet.\n\n1, 2, 3, 4, 7, 11, 17, 25, 49, 75, 129, 191, 329, 489, 825, 1237, 2473, 3737, 6329, 9435, 16833, 25081, 41449, 61043, 115409, 172441, 290617\n\nFor example, $1010110101101$ is such a string because:\n\ninitial odd substring | first half | second half\n----------------------+------------+------------\n1                   | 1          | 1\n101                 | 10         | 01\n10101               | 101        | 101\n1010110             | 1010       | 0110\n101011010           | 10101      | 11010\n10101101011         | 101011     | 101011\n1010110101101       | 1010110    | 0101101\n\n\n# Question\n\nI have two conjectures based on the first few terms:\n\n\u2022 $A297789(2^k + 1) = 2\\cdot A297789(2^k) - 1$ for $k > 0$, and\n\u2022 $A297789(n)$ is odd for all $n > 4$.\n\nIs there a proof or counter-example to these conjectures?\n\nIt's worth noting that the strings that A297789 counts can be put on a binary tree. Perhaps this is a useful lens?\n\n\u2022 Why no strings starting with 0? \u2013\u00a0Fabio Somenzi Jan 28 '18 at 3:31\n\u2022 Technically the sequence counts equivalence classes, where two strings are in the same equivalence class if they\u2019re the same under swapping letters of the alphabet. \u2013\u00a0Peter Kagey Jan 28 '18 at 3:53\n\u2022 Shouldn't the definition either mention the equivalence classes or, more simply, say something like \"strings starting with 1?\" \u2013\u00a0Fabio Somenzi Jan 28 '18 at 4:02\n\u2022 @FabioSomenzi I agree with you. \u2013\u00a0Li-yao Xia Jan 28 '18 at 5:34\n\nHere's a proof of the first conjecture.\n\n### Preliminary remarks\n\nLet's call \"balanced strings\" those described by that sequence.\n\nLet $s$ be a balanced binary string of length $2n-1$. We denote $s_i$ its $i$-th bit (indexed from $1$), and $s_i^j$ the substring of bits from $i$ to $j$ (inclusive). The number of $1$ is denoted $|s|$.\n\nNote that another way to say that two binary strings are permutations of each other is that they have the same numbers of $0$ and $1$: $s$ being balanced is equivalent to saying that for every $i < n$, $|s_1^n| = |s_n^{2n-1}|$.\n\nThis implies that every odd prefix is also balanced. Conversely, we can generate balanced strings by appending two bits at a time. This is also suggested by the binary tree you drew above.\n\nHow many ways are there to extend $s$ into a balanced string of length $2(n+1)-1$? We enumerate pairs of bits, $s_{2n}$ and $s_{2n+1}$, such that $|s_1^{n+1}| = |s_{n+1}^{2n+1}|$. We must consider four cases of the possible values for the middle bits $s_n$ and $s_{n+1}$. Here's one:\n\n\u2022 If $s_n = 0$ and $s_{n+1} = 1$, then\n\n\\begin{aligned} |s_1^{n+1}| &= |s_1^n| + 1 \\\\ &= |s_n^{2n-1}| + 1 \\quad \\text{(since s is balanced)}\\\\ &= |s_{n+1}^{2n-1}| + 1 \\end{aligned}\n\nFor that to be equal to $|s_{n+1}^{2n+1}|$, the new bits can be either $01$ or $10$ (two choices).\n\nThe conclusion is that if $s_n \\neq s_{n+1}$, then there are two ways to extend $s$ (with $01$ or $10$), otherwise there is only one.\n\n### Main result\n\n$A297789(2^k+1)=2\u22c5A297789(2^{k})\u22121$ for $k>0$\n\nAs a consequence of the previous intermediate result, that is equivalent to claiming that every balanced string of length $2^{k+1}-1$ can be extended in exactly two ways, except the one string made of all ones.\n\nThat, in turn, amounts to saying that in every balanced string $s$ of length $2^{k+1}-1$, the middle bits $s_{2^k}$ and $s_{2^k+1}$ are distinct. In that case, the new bits we append at the end are $01$ or $10$, i.e., also distinct. Interesting coincidence, we can thus prove that claim by induction on $k$.", "date": "2020-01-28 20:16:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2624208/counting-particular-odd-length-strings-over-a-two-letter-alphabet", "openwebmath_score": 0.9548411965370178, "openwebmath_perplexity": 298.4569757127717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429107723175, "lm_q2_score": 0.894789464699728, "lm_q1q2_score": 0.8814060188362239}}
{"url": "https://math.stackexchange.com/questions/3305462/sum-of-digits-divisible-by-27", "text": "# Sum of digits divisible by $27$\n\nI know that every third number is divisible by $$3$$ and hence, sum of its digits is divisible by $$3$$. Same holds for $$9$$ also. But how do we generalise it? We know that the divisibility condition for higher powers of $$3$$ is not about the sum of digits. How can we find $$n$$ such that in a group of $$n$$ consecutive positive integers, there is a number such that the sum of its digits is divisible by $$27$$ (or $$81,$$ say)? Does it exist? Please prove or disprove.\n\n\u2022 Do you understand what makes things work for 3 and 9. Do you have any thoughts about connecting that idea with 27 or 81? Also, since this problem is a bit open ended, it reads like a contest or challenge problem - could you provide the source so we know it's not an active contest/application problem? \u2013\u00a0Mark S. Jul 27 at 11:37\n\u2022 math.stackexchange.com/questions/328562/\u2026 \u2013\u00a0lab bhattacharjee Jul 27 at 11:57\n\u2022 No, I was just thinking about it, related to no contest. I know the idea for $3$ and $9$, we write number as $n_0+10n_1+100n_2+...$ and then take out the sum of the digits, remaining sum turns out to be divisible by $9$, making things easy. \u2013\u00a0Martund Jul 27 at 11:59\n\u2022 This is another question by me math.stackexchange.com/questions/3305361/\u2026 \u2013\u00a0Martund Jul 27 at 12:02\n\u2022 polynomial remainder theorem. \u2013\u00a0Roddy MacPhee Jul 27 at 12:53\n\nLet $$Q(x)$$ denote the digit sum of $$x$$.\n\nLet $$r\\ge 1$$. Then $$n=10^r-1=\\underbrace{99\\ldots 9}_r$$ is the smallest $$n$$ such that among any $$n$$ consecutive positive integers, at least one has digit sum a multiple of $$9r$$.\n\nThat no smaller $$n$$ works, is immediately clear because in $$1,2,3,\\ldots, 10^r-2$$, all digit sums are $$>0$$ and $$<9r$$.\n\nRemains to show that in any sequence of $$n$$ consecutive integers, a digit sum divisible by $$9r$$ occurs. This is well-known for $$r=1$$. For $$r>1$$, consider $$n$$ consecutive positive integers $$a,a+1,\\ldots, a+n.$$ Among the first $$9\\cdot 10^{r-1}=n-(10^{r-1}-1)$$ terms, one is a multiple of $$9\\cdot 10^{r-1}$$. Say, $$9\\cdot 10^{r-1}\\mid a+k=:b$$ with $$0\\le k<9\\cdot 10^{r-1}$$. Then $$Q(b)$$ is a multiple of $$9$$, and as the lower $$r-1$$ digits of $$b$$ are zero, we have $$Q(b+i)=Q(b)+Q(i)$$ for $$0\\le i<10^{r-1}$$ and hence $$Q( b+10^j-1)=Q(b)+9j,\\qquad 0\\le j\\le r-1.$$ (Note that $$k+10^{r-1}-1<10^r-1$$, so these terms are really all in our given sequence). It follows that $$9r$$ divides one of these $$Q(b+10^j-1)$$.\n\nNote that the natural numbers $$\\{1,2,\\cdots, 999\\}$$ contain integers for which the sum of the digits is any specified value $$\\pmod {27}$$\n\nConsidering the multiples of $$1000$$ we see that each block of $$1000$$ integers contains one which ends in three $$0's$$.\n\nStarting from any integer $$k$$, we go to the next multiple of $$1000$$ (a gap of at most $$999$$). We then add whatever three (or fewer) digit integer we need to \"correct\" the sum of the digits $$\\pmod {27}$$, which takes, at most, another $$999$$.\n\nThus, every block of $$2\\times 999$$ consecutive integers contains at least one for which the sum of the digits is a multiple of $$27$$. A similar argument works for any desired divisor.\n\nI expect the bound could be tightened considerably, but at least this shows that a bound exists.\n\n\u2022 Simple insight, great help, thank you:) \u2013\u00a0Martund Jul 27 at 13:57\n\nLet $$Q(x)$$ denote the digit sum of $$x$$.\n\n$$999$$ is the smallest $$n$$ such that among any $$n$$ consecutive positive integers, at least one has digit sum a multiple of $$27$$.\n\nFirst note that none of the $$998$$ consecutive integers $$1,2,\\ldots ,998$$ has digit sum a multiple of $$27$$.\n\nAny sequence of $$999$$ consecutive integers is either of the form $$\\tag11000N+1,\\ldots, 1000N+999$$ or $$\\tag21000N+k+2,\\ldots, 1000N+999,1000(N+1),\\ldots, 1000(N+1)+k$$ with $$0\\le k\\le 997$$.\n\nIn $$(1)$$, the digit sums are $$Q(N)+Q(i)$$ with $$i$$ running from $$1$$ to $$999$$ and hence $$Q(i)$$ covering all values from $$1$$ to $$27$$. We conclude that $$(1)$$ contains a term with digit sum a multiple of $$27$$.\n\nSo let's look at $$(2)$$: We know $$Q(N+1)\\equiv Q(N)+1\\pmod 9$$, hence $$Q(N+1)\\equiv Q(N)+(1\\text{ or }10\\text{ or }19)\\pmod{27}$$.\n\n\u2022 If $$Q(N+1)\\equiv 0\\pmod{27}$$, then already $$1000(N+1)$$ has the desired property.\n\n\u2022 If $$Q(N+1)\\equiv 1\\pmod {27}$$, then among $$1000(N+1),\\ldots, 1000(N+1)+899$$, all remainders $$\\bmod27$$ occur, which solves the problem for all $$k\\ge 899$$. For $$k\\le 898$$, the sequence contains $$1000N+900$$, $$1000N+909$$, and $$1000N+999$$ with digit sums $$Q(N)+9$$, $$Q(N)+18$$, $$Q(N)+27$$. As $$Q(N)\\bmod 27$$ is one of $$0$$, $$9$$, $$18$$, we are done.\n\n\u2022 More generally, if $$Q(N+1)\\equiv r\\pmod {27}$$ with $$1\\le r\\le 9$$, then $$Q(1000(N+1)+999-100r)=Q(N+1)+27-r\\equiv 0\\pmod{27}$$, which solves the problem for all $$k\\ge 999-100r$$. For $$k\\le 998-100r$$, the sequence contains $$1000N+(1000-100r)$$, $$1000N+(1009-100r)$$, and $$1000N+(1099-100r)$$ with digit sums $$Q(N)+10-r$$, $$Q(N)+19-r$$, $$Q(N)+28-r$$. As $$Q(N)\\bmod 27$$ is one of $$r-1$$, $$r+8$$, $$r+17$$, we are done.\n\n\u2022 If $$Q(N+1)\\equiv 10+r\\pmod{27}$$ with $$0\\le r\\le 8$$, then $$Q(1000(N+1)+89-10r)=Q(N+1)+17-r\\equiv 0\\pmod{27}$$, which solves the problem for all $$k\\ge 89-10r$$. For $$k\\le 89-10r$$, the sequence contains $$1000N+999-r$$, $$1000N+909-r$$, and $$1000N+900-100r$$ with digit sums $$Q(N)+27-r$$, $$Q(N)+18-r$$, $$Q(N)+9-r$$. As $$Q(N)\\bmod 27$$ is one of $$r$$, $$r+9$$, $$r+18$$, we are done.\n\n\u2022 If $$Q(N+1)\\equiv 19+r\\pmod{27}$$ with $$0\\le r\\le 7$$, then $$Q(1000(N+1)+8-r)=Q(N+1)+8-r\\equiv 0\\pmod{27}$$, which solves the problem for all $$k\\ge 8-r$$. For $$k\\le 8-r$$, the sequence contains $$1000N+999-r$$, $$1000N+909-r$$, and $$1000N+900-100r$$ with digit sums $$Q(N)+27-r$$, $$Q(N)+18-r$$, $$Q(N)+9-r$$. As $$Q(N)\\bmod 27$$ is one of $$r$$, $$r+9$$, $$r+18$$, we are done.\n\n\u2022 Great answer, thanks a lot. \u2013\u00a0Martund Jul 27 at 13:56\n\u2022 or use polynomial remainder theorem for a rule. \u2013\u00a0Roddy MacPhee Jul 27 at 14:06\n\nObviously there are number whose digits add to $$27$$. ($$999$$ or $$524385$$ etc.) and obviously the sum of the digits $$27$$ is a multiple of $$9$$ so they are a multiple of $$9$$ but are they a multiples of $$27$$; and must multiples of $$27$$ have digits adding to a multiple of $$27$$.\n\nWell, $$27$$ itself is an obvious counter example of the latter.\n\nAnd $$999= 27*37$$ but $$524385= 27*19421\\frac 23$$ so the first is not true either.\n\nSo the question I guess is why not?\n\nWell the rule works for $$9$$ because $$9 = 10-1$$. And it works for $$3$$ because $$3|9$$.\n\nDetails: If $$k|b-1$$ and $$n= \\sum_{i=0}^m a_ib^i= \\sum_{i=0}^m (a_i)(b^i-1) + \\sum_{i=0}^m a_i$$. Now $$b^i-1 =(b-1)(b^{i-1} + b^{i-2}+ ..... + 1)$$ so $$(b-1)$$ divides all of the $$b^i-1$$ so $$b-1$$ divides $$n$$ if and only if $$(b-1)$$ divides $$\\sum_{i=0}^m a_i$$. If we let $$b= 10$$ and $$b-1=9$$ and $$a_i$$ be the digits of $$n$$ that's our result.\n\nAnd it follows that if $$k|b-1$$ then $$k|(b^i-1)$$ so $$k|n$$ if and only if $$k$$ divides $$\\sum_{i=0}^m a_i$$ as well.\n\nAnd this will be true for any decimal system base $$b$$ (not just $$b=10$$ and and $$k|b-1$$ (not just $$3|9$$).\n\nThe fact that $$3^2 = 9$$ is mostly a coincidence and powers of $$3$$ is a bit of a red herring. It's not powers of $$3$$ going up that matter, but factors $$10-1$$ going down that matter.\n\nWe can note that in base $$7$$ a number is a multiple of $$6$$ if and only if the sum of the digits is a multiple of $$6$$ and a multiple of $$2$$ or of $$3$$ if and only if the sum of the digits is a multiple of $$2$$ or of $$3$$ respectively but nothing can be said of $$4$$ or $$3$$. ($$11_7 = 8$$ is a multiple of $$4$$ but $$1+1=2$$ is not. And $$12_7 =9$$ is a multiple of $$9$$ but $$1+2=3$$ which is not.)\n\nWhy doesn't it work? Well. $$27 = 3*(10-1)= (3-1)*10 + (10-3)$$. The sum of the digits of $$27$$ are $$(3-1) + (10-3) = 10-1$$. Our rule of $$9$$s apply and we can't jump magically to $$27$$. And if we increas by $$27$$ if we ignore carrying and borrowing we get $$ab + 27 = (a+2)(b+7)$$ and the sums of the digits are $$a+b + 9$$. That's an increase of $$9$$; not of $$27$$. Ind if we carry (i.e. $$b \\ge 3$$ or $$a \\ge 8$$ or $$b\\ge 3$$ and $$a\\ge 7$$) we get the sums are $$(a+2+1)(b+7-10)$$ or $$1(a+2-10)(b+7)$$ or $$1(a+2+1-10)(b+7 - 10)$$ and the sum of the digits stay the same or decreases by $$9$$; not $$27$$.\n\nBut if $$b-1 = k^m$$ then in base $$b$$ we will have that multiples of $$k^i; i\\le m$$ will have the sum of the digits add to a multiple of $$k^i$$.\n\nExample in base $$28$$ then sum of the digits of a multiple of $$27$$ will add to a multiple of $$27$$.\n\nFWIW $$999_{10} = 28^2 + 7*28 + 19 = 17T_{28}$$ where $$T$$ is the digit for $$19$$.\n\nAnd $$27*92 = 2484_{10} = 3*28^2+4*28 + 20= 34U_{28}$$ where $$U$$ is the digit for $$20$$ is another example.\n\nLess trivial example. The number $$8ATR_{28}$$ were $$A=10$$ and $$T=19$$ and $$R=17$$ will have digits that add to $$8+10+19+17=54$$, so my claim is that it ought to be a multiple of $$27$$. And\n\n$$8ATR_{28} = 8*28^3 + 10*28^2+ 19*28 + 17 =$$\n\n$$8(27+ 1)^3 + 10(27+1)^2 + 19(27+1) + 17 =$$\n\n$$8(27^3 + 3*27^2+3*27 + 1) + 10(27^2 + 2*27 + 1) + 19(27+1)+17=$$\n\n$$[8*27^3 + 3*27^2 + 3*27 + 20*27^2 + 2*27 + 19*27] + 8 + 10 + 19+17=$$\n\n$$27(8*27^2 + 3*27 + 3 + 20*27 + 2 + 19) + 54 =$$\n\n$$27(8*27^2 + 3*27 + 3 + 20*27 + 2 + 19 + 2)$$ is a multiple of $$27$$.\n\nAnd indeed $$8*28^3 + 10*28^2+ 19*28 + 17=184005 = 27*6815$$", "date": "2019-09-20 20:20:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3305462/sum-of-digits-divisible-by-27", "openwebmath_score": 0.8824615478515625, "openwebmath_perplexity": 90.64754032392688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127409715955, "lm_q2_score": 0.8918110504699678, "lm_q1q2_score": 0.8813882237187317}}
{"url": "https://math.stackexchange.com/questions/1371759/prove-sum-limits-i-0n-binomnii-binom2n1n1", "text": "# Prove $\\sum\\limits_{i=0}^{n}\\binom{n+i}{i}=\\binom{2n+1}{n+1}$ [duplicate]\n\nI'm trying to prove this algebraically: $$\\sum\\limits_{i=0}^{n}\\dbinom{n+i}{i}=\\dbinom{2n+1}{n+1}$$\n\nUnfortunately I've been stuck for quite a while. Here's what I've tried so far:\n\n1. Turning $\\dbinom{n+i}{i}$ to $\\dbinom{n+i}{n}$\n2. Turning $\\dbinom{2n+1}{n+1}$ to $\\dbinom{2n+1}{n}$\n3. Converting binomial coefficients to factorial form and seeing if anything can be cancelled.\n4. Writing the sum out by hand to see if there's anything that could be cancelled.\n\nI end up being stuck in each of these ways, though. Any ideas? Is there an identity that can help me?\n\n## marked as duplicate by Jack D'Aurizio, Mark Viola, Lucian, user147263, vonbrandJul 24 '15 at 0:32\n\n\u2022 I'd recommend using induction to prove the result, although I'm sure it could be done another way. \u2013\u00a0Raj Jul 23 '15 at 20:05\n\u2022 Oh, I forgot to mentioned that I tried induction too (a classmate gave me that hint). I got stuck trying to move the induction hypothesis from $\\sum\\limits_{i=0}^{k}\\dbinom{k+i}{i} = \\dbinom{2k+1}{2k+1}$ to $\\sum\\limits_{i=0}^{k+1}\\dbinom{k+i+1}{i} = \\dbinom{2k+3}{k+2}$ Specifically the $k+i+1$ part--had no idea how to change that. \u2013\u00a0jsluong Jul 23 '15 at 20:24\n\nHint:\n\nUse $$\\binom{n+1+i}{i} - \\binom{n+i}{i-1} = \\binom{n+i}{i}$$\n\n\u2022 Hey, thanks for this! This is the recursive formula rewritten, right? I'm trying to use it to see if I could solve this problem with another approach but, er, stuck again. Can I get another hint? \u2013\u00a0jsluong Jul 23 '15 at 20:40\n\u2022 Yep, the same recursion you have in your answer, only with different values $k=i$ and $n \\mapsto 1 + n + i$ \u2013\u00a0johannesvalks Jul 23 '15 at 20:42\n\u2022 Should be $\\binom{n+i}{i+1}$ on the RHS. \u2013\u00a0xivaxy Jul 23 '15 at 20:47\n\u2022 @Dr.MV - indead, I have corrected it. \u2013\u00a0johannesvalks Jul 23 '15 at 21:37\n\nOh, hey! I just figured it out. Funny how simply posting the question allows you re-evaluate the problem differently...\n\nSo on Wikipedia apparently this is a thing (the recursive formula for computing the value of binomial coefficients): $$\\dbinom{n}{k} = \\dbinom{n-1}{k-1} + \\dbinom{n-1}{k}$$\n\nOn the RHS of the equation we have (let's call this equation 1):\n\n$$\\dbinom{2n+1}{n+1} = \\dbinom{2n}{n} + \\dbinom{2n-1}{n} + \\dbinom{2n-2}{n} + \\cdots + \\dbinom{n+1}{n} + \\dbinom{2n-(n-1)}{n+1}$$ The last term $\\dbinom{2n-(n-1)}{n+1}$ simplifies to $\\dbinom{n+1}{n+1}$ or just 1.\n\nMeanwhile on the LHS we have $\\sum\\limits_{i=0}^{n}\\dbinom{n+i}{i}$ which is also $\\sum\\limits_{i=0}^{n}\\dbinom{n+i}{n}$ because $\\dbinom{n}{k} = \\dbinom{n}{n-k}$.\n\nWritten out that is (let's call this equation 2): $$\\sum\\limits_{i=0}^{n}\\dbinom{n+i}{n} = \\dbinom{n}{n} + \\dbinom{n+1}{n} + \\dbinom{n+2}{n} + \\cdots + \\dbinom{2n}{n}$$\n\nThe first term $\\dbinom{n}{n}$ simplifies to 1.\n\nHey, look at that.\n\nIn each written out sum there's a term in equation 1 and an equivalent in equation 2. And in each sum there's $n$ terms, so equation 1 definitely equals equation 2. I mean, the order of terms is flipped, but whatever. Yay!\n\n\u2022 Cool that you found it! \u2013\u00a0johannesvalks Jul 23 '15 at 20:35\n\u2022 Okay... so tell me what I should do. \u2013\u00a0jsluong Jul 23 '15 at 21:40\n\u2022 @Dr.MV This is the subject of some debate.. The OP didn't know the answer beforehand, but figured it out an hour after posting the question. I think this is totally fine. \u2013\u00a0Andr\u00e9 3000 Jul 23 '15 at 22:11\n\u2022 @SpamIAm The answer was posted roughly the same time that the question was identified as a duplicate. So, given this question is indeed a duplicate, how does one justify the action? \u2013\u00a0Mark Viola Jul 23 '15 at 22:34\n\u2022 Oh, well, I LaTeX my homework so I needed to type up the solution anyway. I figured other people could potentially benefit from the solution. This is a new account. I really am not trying to farm rep. \u2013\u00a0jsluong Jul 23 '15 at 23:22", "date": "2019-05-24 11:14:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1371759/prove-sum-limits-i-0n-binomnii-binom2n1n1", "openwebmath_score": 0.8197067379951477, "openwebmath_perplexity": 547.5101587479361, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307661011975, "lm_q2_score": 0.9059898235563418, "lm_q1q2_score": 0.8813747741302047}}
{"url": "http://terabook.info/diy-rattan-kiluwab/symmetric-closure-of-a-relation-ada95c", "text": "# symmetric closure of a relation\n\n0. Finally, the concepts of reflexive, symmetric and transitive closure are The reflexive, transitive closure of a relation R is the smallest relation that contains R and that is both reflexive and transitive. Reflexive and symmetric properties are sets of reflexive and symmetric binary relations on A correspondingly. Let R be a relation on the set {a,b, c, d} R = {(a, b), (a, c), (b, a), (d, b)} Find: 1) The reflexive closure of R 2) The symmetric closure of R 3) The transitive closure of R Express each answer as a matrix, directed graph, or using the roster method (as above). 10 Symmetric Closure (optional) When a relation R on a set A is not symmetric: How to minimally augment R (adding the minimum number of ordered pairs) to have a symmetric relation? Transitive closure applied to a relation. Don't express your answer in \u2026 A relation follows join property i.e. Transcript. 2. Neha Agrawal Mathematically Inclined 171,282 views 12:59 Find the symmetric closures of the relations in Exercises 1-9. This means that if a symmetric relation is represented on a digraph, then anytime there is a directed edge from one vertex to a second vertex, ... By the closure properties of the integers, $$k + n \\in \\mathbb{Z}$$. 1. 8. The symmetric closure of a binary relation on a set is the union of the binary relation and it\u2019s inverse. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. We discuss the reflexive, symmetric, and transitive properties and their closures. ... Browse other questions tagged prolog transitive-closure or ask your own question. The transitive closure of a symmetric relation is symmetric, but it may not be reflexive. Topics. \u2022 What is the symmetric closure S of R? the join of matrix M1 and M2 is M1 V M2 which is represented as R1 U R2 in terms of relation. A binary relation is called an equivalence relation if it is reflexive, transitive and symmetric. Discrete Mathematics with Applications 1st. Symmetric: If any one element is related to any other element, then the second element is related to the first. The symmetric closure of a relation on a set is the smallest symmetric relation that contains it. Definition of an Equivalence Relation. If I have a relation ,say ,less than or equal to ,then how is the symmetric closure of this relation be a universal relation . and (2;3) but does not contain (0;3). Question: Suppose R={(1,2), (2,2), (2,3), (5,4)} is a relation on S={1,2,3,4,5}. A relation R is non-symmetric iff it is neither symmetric One way to understand equivalence relations is that they partition all the elements of a set into disjoint subsets. Relations. Equivalence Relations. We then give the two most important examples of equivalence relations. To form the transitive closure of a relation , you add in edges from to if you can find a path from to . Transitive Closure \u2013 Let be a relation on set . If is the following relation: then the reflexive closure of is given by: the symmetric closure of is given by: 4 Symmetric Closure \u2022 If a relation is symmetric, then the relation itself is its symmetric closure. [Definitions for Non-relation] reflexive; symmetric, and; transitive. Find the reflexive, symmetric, and transitive closure of R. Solution \u2013 For the given set, . For example, being the father of is an asymmetric relation: if John is the father of Bill, then it is a logical consequence that Bill is not the father of John. Notation for symmetric closure of a relation. The transitive closure is obtained by adding (x,z) to R whenever (x,y) and (y,z) are both in R for some y\u2014and continuing to do \u2026 Formally: Definition: the if $$P$$ is a property of relations, $$P$$ closure of $$R$$ is the smallest relation \u2026 The symmetric closure S of a binary relation R on a set X can be formally defined as: S = R \u222a {(x, y) : (y, x) \u2208 R} Where {(x, y) : (y, x) \u2208 R} is the inverse relation of R, R-1. 9.4 Closure of Relations Re\ufb02exive Closure The re\ufb02exive closure of a relation R on A is obtained by adding (a;a) to R for each a 2A. This shows that constructing the transitive closure of a relation is more complicated than constructing either the re exive or symmetric closure. \u2022 If a relation is not symmetric, its symmetric closure is the smallest relation that is symmetric and contains R. Furthermore, any relation that is symmetric and must contain R, must also contain the symmetric closure of R. If one element is not related to any elements, then the transitive closure will not relate that element to others. It's also fairly obvious how to make a relation symmetric: if $$(a,b)$$ is in $$R$$, we have to make sure $$(b,a)$$ is there as well. This section focuses on \"Relations\" in Discrete Mathematics. Example \u2013 Let be a relation on set with . R = { (a,b) : a b } Here R is set of real numbers Hence, both a and b are real numbers Check reflexive We know that a = a a a (a, a) R R is reflexive. i.e. The symmetric closure of a relation on a set is the smallest symmetric relation that contains it. Symmetric closure and transitive closure of a relation. Let R be an n -ary relation on A . We already have a way to express all of the pairs in that form: $$R^{-1}$$. A relation R is symmetric if the transpose of relation matrix is equal to its original relation matrix. t_brother - this should be the transitive and symmetric relation, I keep the intermediate nodes so I don't get a loop. A relation S on A with property P is called the closure of R with respect to P if S is a subset of every relation Q (S Q) with property P that contains R (R Q). Section 7. Symmetric Closure The symmetric closure of R is obtained by adding (b;a) to R for each (a;b) 2R. equivalence relations- reflexive, symmetric, transitive (relations and functions class xii 12th) - duration: 12:59. Discrete Mathematics Questions and Answers \u2013 Relations. \u2022S=? A binary relation on a non-empty set $$A$$ is said to be an equivalence relation if and only if the relation is. Transitive Closure of Symmetric relation. Closure. Blog A holiday carol for coders. In this paper, we present composition of relations in soft set context and give their matrix representation. In this paper, four algorithms - G, Symmetric, 0-1-G, 1-Symmetric - are given for computing the transitive closure of a symmetric binary relation which is represented by a 0\u20131 matrix. Algorithms G and 0-1-G pose no restriction on the type of the input matrix, while algorithms Symmetric and 1-Symmetric require it to be symmetric. The connectivity relation is defined as \u2013 . I tried out with example ,so obviously I would be getting pairs of the form (a,a) but how do they correspond to a universal relation. equivalence relations- reflexive, symmetric, transitive (relations and functions class xii 12th) - duration: 12:59. There are 15 possible equivalence relations here. What is the reflexive and symmetric closure of R? (b) Use the result from the previous problem to argue that if P is reflexive and symmetric, then P+ is an equivalence relation. Find the symmetric closures of the relations in Exercises 1-9. The symmetric closure of R . 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Any one element is related to any other element, then the second element is related itself! ) but does not contain ( 0 ; 3 ) but does not symmetric closure of a relation ( 0 ; 3.! Cs M. Hauskrecht closures Definition: Let R be an n -ary relation on set is the smallest relation. For example, \\ ( P\\ ) closure of a relation on a and... Is its own reflexive closure they partition all the elements of a relation on a set is detailed relation...", "date": "2021-02-28 06:50:21", "meta": {"domain": "terabook.info", "url": "http://terabook.info/diy-rattan-kiluwab/symmetric-closure-of-a-relation-ada95c", "openwebmath_score": 0.7278335690498352, "openwebmath_perplexity": 526.8847190002843, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.978712651931994, "lm_q2_score": 0.9005297854505006, "lm_q1q2_score": 0.8813598944620089}}
{"url": "https://mymathforum.com/threads/how-do-i-find-the-modulus-of-the-average-force-a-sphere-gets-from-colliding-with-the-ground.347560/", "text": "# How do I find the modulus of the average force a sphere gets from colliding with the ground?\n\n#### Chemist116\n\nThe problem is as follows:\n\nA ball of $1\\,kg$ in mass is thrown with a speed of $-10\\vec{j}\\,\\frac{m}{s}$ from a height of $15\\,m$ to a horizontal floor. Find the modulus of the average force in $N$ that the ball receives from the floor during the impact with the ground which lasts $0.1\\,s$ and dissipates $150\\,J$. (You may use the value of gravity $g=10\\,\\frac{m}{s^2}$.\n\nThe alternatives given in my book are as follows:\n\n$\\begin{array}{ll} 1.&290\\,N\\\\ 2.&300\\,N\\\\ 3.&310\\,N\\\\ 4.&320\\,N\\\\ 5.&330\\,N\\\\ \\end{array}$\n\nThis problem has left me go in circles as I don't know exactly how should I treat or use the information to obtain the average force?. I'm assuming that there is a conservation of momentum but as I mentioned I don't know what to do?. Can somebody help me here?.\n\n#### skeeter\n\nMath Team\nimpulse equation ...\n\n$F \\Delta t = m\\Delta v \\implies F = \\dfrac{m\\Delta v}{\\Delta t}$, where $F$ is the average force of impact changing momentum\n\nyou have mass and delta t, you need the change in velocity from impact with the ground\n\ntotal initial mechanical energy = kinetic energy at impact\n\n$mgh + \\dfrac{1}{2}mv_0^2 = \\dfrac{1}{2}mv_f^2 \\implies v_f = -\\sqrt{2gh + v_0^2} = -20 \\text{ m/s}$\n\npost impact, the velocity is positive and has 150J less energy ...\n\nimpact KE = $\\dfrac{1}{2}m(-20)^2 = 200 \\text{ J} \\implies$ post impact KE = $50 \\text{ J}$\n\nbounce velocity $\\dfrac{1}{2}mv_f^2 = 50 \\implies v_f = 10 \\text{ m/s}$\n\nchange in velocity from impact with the ground is $\\Delta v = [10 - (-20)] \\text{ m/s} = 30 \\text{ m/s}$\n\n$F = \\dfrac{m\\Delta v}{\\Delta t} = 300 \\text{ N}$\n\n#### Chemist116\n\nimpulse equation ...\n\n$F \\Delta t = m\\Delta v \\implies F = \\dfrac{m\\Delta v}{\\Delta t}$, where $F$ is the average force of impact changing momentum\n\nyou have mass and delta t, you need the change in velocity from impact with the ground\n\ntotal initial mechanical energy = kinetic energy at impact\n\n$mgh + \\dfrac{1}{2}mv_0^2 = \\dfrac{1}{2}mv_f^2 \\implies v_f = -\\sqrt{2gh + v_0^2} = -20 \\text{ m/s}$\n\npost impact, the velocity is positive and has 150J less energy ...\n\nimpact KE = $\\dfrac{1}{2}m(-20)^2 = 200 \\text{ J} \\implies$ post impact KE = $50 \\text{ J}$\n\nbounce velocity $\\dfrac{1}{2}mv_f^2 = 50 \\implies v_f = 10 \\text{ m/s}$\n\nchange in velocity from impact with the ground is $\\Delta v = [10 - (-20)] \\text{ m/s} = 30 \\text{ m/s}$\n\n$F = \\dfrac{m\\Delta v}{\\Delta t} = 300 \\text{ N}$\nAll uses a pretty good logic but I'm wondering why the answers sheet states that the answer is $310\\,N$?. Could it be that is there anything missing or overlooked during the analysis?. Can you check this please?.\n\n#### skeeter\n\nMath Team\nMaybe someone else can do the problem ...\n\n#### DarnItJimImAnEngineer\n\n*facepalm!*\nThe net force is equal to 300 N! But the earth is still imparting downward momentum on the ball at a rate of 10 N, so the floor must impart that much more upward momentum.\n$F_{floor} = ma - F_g = \\pm 300 ~N ~\\vec{j} \\pm 10 ~N ~\\vec{j} = \\pm 310 ~N ~\\vec{j}$ (depending on whether $\\vec{j}$ is up or down).\n\nIt took me about five minutes to realise that mistake.\nAnother feather in the cap of \"Always, always, always draw your free-body diagram.\"\n\nChemist116\n\n#### Chemist116\n\n*facepalm!*\nThe net force is equal to 300 N! But the earth is still imparting downward momentum on the ball at a rate of 10 N, so the floor must impart that much more upward momentum.\n$F_{floor} = ma - F_g = \\pm 300 ~N ~\\vec{j} \\pm 10 ~N ~\\vec{j} = \\pm 310 ~N ~\\vec{j}$ (depending on whether $\\vec{j}$ is up or down).\n\nIt took me about five minutes to realise that mistake.\nAnother feather in the cap of \"Always, always, always draw your free-body diagram.\"\nI totally overlooked this part. At first I didn't noticed but it seems that $F_g=mg=1\\,kg\\times 10\\frac{m}{s^2}=10\\,N$ In this case I believe it is upward.", "date": "2020-02-16 23:05:23", "meta": {"domain": "mymathforum.com", "url": "https://mymathforum.com/threads/how-do-i-find-the-modulus-of-the-average-force-a-sphere-gets-from-colliding-with-the-ground.347560/", "openwebmath_score": 0.7999904751777649, "openwebmath_perplexity": 612.2589436959513, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429609670701, "lm_q2_score": 0.8962513786759491, "lm_q1q2_score": 0.8813224844780265}}
{"url": "http://mathhelpforum.com/advanced-algebra/209880-can-anyone-prove-determinant-print.html", "text": "# Can anyone prove this determinant?\n\n\u2022 Dec 15th 2012, 07:17 AM\nalphaknight61\nCan anyone prove this determinant?\nIf A is a square matrix of order n and m \u03f5 R then the det(mA) = (mn) (detA)\n\n\u2022 Dec 15th 2012, 08:31 AM\njakncoke\nRe: Can anyone prove this determinant?\nWell, observe that for any scalar $m \\in R$ and nxn matrix A.\n\n$mA = A*mI$ where I is the standard nxn identity matrix. So you got $mI = \\begin{bmatrix} m && 0 && ... && 0 \\\\ 0 && m && ... && 0 \\\\ .. && .. && .. && .. \\\\ 0 && 0 && ... && m \\end{bmatrix}$. So you know the trick for determinats when multiplying matricies, Det(AB) = Det(A)Det(B). Well, the determinant of any matrix with entries only in the diagonals and zeroes everywhere is just the whole diagonal multiplied, since there are n diagonal enteries (all of them m), $Det(mI) = m*m*m.... = m^n$\nso you got $det(mA) = det(A) * det(mI) = det(A) * m^n$\n\u2022 Dec 15th 2012, 11:53 AM\nHallsofIvy\nRe: Can anyone prove this determinant?\nAlternatively, it is easy to see that, for a 2 by 2 matrix, $\\begin{bmatrix}a & b \\\\ c & d\\end{bmatrix}$ has determinant ad- bc. Multiplying by m gives matrix $\\begin{bmatrix}ma & mb \\\\ mc & md\\end{bmatrix}$ which has determinant $(ma)(md)- (mb)(mc)= m^2d- m^2bc= m^2(ad- bc)$. Now use the idea of expanding a determinant by columns to do a proof by induction on the size of the matrix.\n\nYet another way: Observe that the determinant of an n by n matrix involves sums and differences of n terms, each being the product of exactly one number from each row and column. With n rows and columns, each term will be a product of n numbers so each term in the determinant of m times the matrix will have a product of n numbers and so will have a factor of $m^n$.\n\u2022 Dec 15th 2012, 02:01 PM\nGeorgetown\nRe: Can anyone prove this determinant?\nSorry i can't solve it but i am here for learn.\n\ncar accident lawyer Austin\n\u2022 Dec 15th 2012, 08:16 PM\nDeveno\nRe: Can anyone prove this determinant?\nQuote:\n\nOriginally Posted by alphaknight61\nIf A is a square matrix of order n and m \u03f5 R then the det(mA) = (mn) (detA)\n\nlet's prove something simpler, first:\n\n$\\begin{vmatrix}1&0&\\dots&0&\\dots&0\\\\0&1&\\dots&0& \\dots&0\\\\ \\vdots&\\vdots&\\ddots&\\vdots&\\cdots&\\vdots\\\\0&0& \\dots &m&\\dots&0\\\\ \\vdots&\\vdots&\\cdots&\\vdots&\\ddots&\\vdots\\\\0&0& \\dots&0&\\dots&1 \\end{vmatrix} = m$\n\nbut this is a diagonal matrix, which has determinant: (1)(1)...(m)....(1) = m.\n\nif we call this matrix Pr (where m occurs in the r-th row), then PrA multiplies row r of A by m.\n\nhence mA = P1P2...PnA, so that:\n\ndet(mA) = det(P1)det(P2)...det(Pn)det(A) = (m)(m)....(m)(det(A)) = mn(det(A))\n\u2022 Dec 15th 2012, 08:29 PM\nzhandele\nRe: Can anyone prove this determinant?\n\nDeterminant with Row Multiplied by Constant - ProofWiki\n\nThis page shows all but one step of the proof you want, in formal language.\n\nIn general, proofwiki is a good place to look for proofs.\n\nBTW, the idea of a determinant as a cofactor expansion was mysterious to me for a long time. I could do the expansions, but I didn't feel I understood why it worked, though I was shy about saying so. I've found that many other people are in the same boat. Suggestion: Google \"Geometric Algebra Primer\" by Jaap Suter, and/or watch the first few eps of Norman Wildberger's \"Wild Lin Alg,\" which is on Youtube.\n\u2022 Dec 15th 2012, 09:32 PM\nDeveno\nRe: Can anyone prove this determinant?\nQuote:\n\nOriginally Posted by zhandele\n\nDeterminant with Row Multiplied by Constant - ProofWiki\n\nThis page shows all but one step of the proof you want, in formal language.\n\nIn general, proofwiki is a good place to look for proofs.\n\nBTW, the idea of a determinant as a cofactor expansion was mysterious to me for a long time. I could do the expansions, but I didn't feel I understood why it worked, though I was shy about saying so. I've found that many other people are in the same boat. Suggestion: Google \"Geometric Algebra Primer\" by Jaap Suter, and/or watch the first few eps of Norman Wildberger's \"Wild Lin Alg,\" which is on Youtube.\n\nthat proof is way more complicated than it needs to be (partially because it includes a partial proof of det(AB) = det(A)det(B)).\n\ngeometrically, what is happening is this:\n\nmultiplying ONE row of A by m is the same as stretching one SIDE of the n-dimensional volume element by a factor of m (which magnifies the volume element by a factor of m). if we do this for every side, we've stretched by a factor of mn\n\n(it's easiest to comprehend this when n = 2 or 3).\n\u2022 Dec 16th 2012, 10:43 PM\nphys251\nRe: Can anyone prove this determinant?\nThere is another way to do it: Schur decomposition. Schur's theorem says that any square matrix A is unitarily equivalent to an upper-triangular matrix--i.e., A = U*TU, where U is unitary and T is upper-triangular. A and T have the same determinant (it's a fairly easy proof to show this, if needed), and since the determinant of a triangular matrix (upper or lower) equals the product of the main diagonal, multiplying each entry by m will multiply the overall determinant by m^n.", "date": "2017-12-17 22:28:26", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/209880-can-anyone-prove-determinant-print.html", "openwebmath_score": 0.868591845035553, "openwebmath_perplexity": 718.588841014056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429580381724, "lm_q2_score": 0.8962513724408291, "lm_q1q2_score": 0.8813224757217366}}
{"url": "https://math.stackexchange.com/questions/2626812/integration-by-parts-int-fracxx12-dx", "text": "# Integration by Parts - $\\int \\frac{x}{(x+1)^2} dx$\n\nI have been trying to evaluate the following integral by using integration by parts, but I contine to yield the incorrect answer.\n\n$$\\int \\frac{x}{(x+1)^2} dx$$\n\nI choose $u=x$, $dv=\\frac{1}{(x+1)^2}dx => du=dx$, $v=\\frac{-1}{x+1}$. The integration by parts formula, $\\int udv = uv - \\int vdu$ yields\n\n$$\\int \\frac{x}{(x+1)^2} dx = \\frac{-x}{x+1} - \\int \\frac{-1}{x+1}dx = \\frac{-x}{x+1}+ ln(x+1)+C$$\n\nHowever, the integral should be\n\n$$\\frac{1}{x+1} + ln(x+1) + C$$\n\nWhere did I go wrong? This is my first ask on math stack exchange, so please be kind.\n\n\u2022 Hello, that's one way to do it, but I would like to understand why integration by parts does not work/apply here. \u2013\u00a0DiDoubleTwice Jan 29 '18 at 17:23\n\u2022 Thanks, all, you guys are fast! \u2013\u00a0DiDoubleTwice Jan 29 '18 at 17:32\n\nYou didn\u2019t do anything wrong. Just notice,\n\n$$\\frac{-x}{x+1}=\\frac{-x-1+1}{x+1}$$\n\n$$=\\frac{1}{x+1}-1$$\n\n\u2022 Ah, I see! The -1 is fine as it is a constant absorbed by 'C' as a result of integration. \u2013\u00a0DiDoubleTwice Jan 29 '18 at 17:28\n\u2022 Yup :) @Didoubletwice \u2013\u00a0Ahmed S. Attaalla Jan 29 '18 at 17:28\n\u2022 Thanks a plenty! I'll accept your answer as it answers \"Where did I go wrong\" ;) \u2013\u00a0DiDoubleTwice Jan 29 '18 at 17:31\n\nWhy by parts? $$\\frac{x}{(x+1)^2}=\\frac{x+1-1}{(x+1)^2}=...$$ and we are done!\n\n\u2022 By parts was the first to come to mind, so I wanted to make sure that I can apply it in a testing scenario. \u2013\u00a0DiDoubleTwice Jan 29 '18 at 17:34\n\u2022 OK. I understood you. Good luck! \u2013\u00a0Michael Rozenberg Jan 29 '18 at 17:37\n\nThe simplest substitution $u=x+1$ is often overlooked.\n\n$\\displaystyle \\int \\dfrac{x}{(x+1)^2}\\mathop{dx}=\\int\\dfrac{u-1}{u^2}\\mathop{du}=\\int\\left(\\dfrac 1u-\\dfrac 1{u^2}\\right)\\mathop{du}=\\ln|u|+\\dfrac 1u=\\ln|x+1|+\\dfrac 1{x+1}+C$\n\n$$g'= (\\frac{-x}{x+1}+ ln(x+1)+C)'= \\frac {-1(x+1)+x}{(x+1)^2}+\\frac 1 {x+1}=\\frac x {(x+1)^2}$$", "date": "2021-05-18 08:34:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2626812/integration-by-parts-int-fracxx12-dx", "openwebmath_score": 0.8219730854034424, "openwebmath_perplexity": 556.8376850860352, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995742876885, "lm_q2_score": 0.9124361604769414, "lm_q1q2_score": 0.8813216989693707}}
{"url": "https://stats.stackexchange.com/questions/308130/linearity-of-expectations-repeated-card-draw", "text": "# Linearity of Expectations - Repeated Card Draw\n\nScenario: You are given a standard deck of cards and told to draw one card at a time from the deck and record the color of that card. Continue until all the cards have been drawn without replacing any cards. Let the random variable $Z$ represent the number of times that you see a red card followed by a black card.\n\nQuestion: What is $\\mathbb{E}(Z)$?\n\nAttempted Solution: Let $Z_i, i \\in [1, 51] \\cap \\mathbb{Z}$ be a set of random variables where $Z_i = 1$ if the card in position $i$ is red and the card in position $i+1$ is black and $Z_1 = 0$ otherwise. Since $Z = \\sum_{i = 1}^{51}Z_i$, we have the following by linearity of expectations:\n\n$$\\mathbb{E}(Z) = \\sum_{i = 1}^{51}\\mathbb{E}(Z_i)$$\n\nMoreover, since the probability of drawing a red card followed by a black card without replacement from a standard deck is $\\frac{26}{52} \\cdot \\frac{26}{51}$, we have: $$\\mathbb{E}(Z) = \\sum_{i = 1}^{51}\\bigg(\\frac{26}{52} \\cdot \\frac{26}{51}\\bigg) = 51 \\cdot \\bigg(\\frac{26}{52} \\cdot \\frac{26}{51}\\bigg) = \\frac{26^2}{52} = 13$$\n\nIssue: The fact that linearity of expectations holds even when the random variables are dependent seemed like witchcraft to me, so I wrote a Python script to determine an experimental value of Z. This program gives me that on average $Z \\approx 11.8$. Am I misinterpreting how to apply the linearity of expectations in this problem?\n\nHere's the script I used:\n\nfrom random import *\n\noutput_values = [];\n\nn = 0;\n\nwhile n < 10000:\n\n#draw cards\nred = 26;\nblack = 26;\n\ndraw = [None] * 52;\n\ni = 0;\n\nwhile i < len(draw):\nif red > 0 and black > 0:\nchoose = randint(0,1);\nif choose == 0:\ndraw[i] = \"R\";\nred = red - 1;\nelse:\ndraw[i] = \"B\"\nblack = black - 1;\nelif red == 0:\ndraw[i] = \"B\"\nblack = black - 1;\nelif black == 0:\ndraw[i] = \"R\"\nred = red - 1;\ni += 1;\n\n#count appearances of red followed by black\ncount_rb = 0;\n\nj = 0\n\nwhile j < (len(draw) - 1):\nif draw[j] == \"R\" and draw[j+1] == \"B\":\ncount_rb += 1;\nj += 1;\n\n#add number of apperances to output_values array\noutput_values.append(count_rb);\n\nn += 1;\n\naverage_value = (sum(output_values))/len(output_values);\nprint(\"Experimental Value:\", average_value);\n\nexpected_value = 51 * (26*26)/(52*51);\nprint(\"Expected Value: \", expected_value);\n\n\u2022 Could you add the Python script into the question (maybe the error is there) \u2013\u00a0Juho Kokkala Oct 16 '17 at 6:28\n\u2022 I just added the script to the body of the question \u2013\u00a0ben-fogarty Oct 16 '17 at 14:24\n\u2022 How does your script model drawing without replacement from the deck of cards? It doesn't appear to use the correct probabilities. To understand the problem, consider how your code works with four cards. If a red card is drawn first, with what probability does it draw a red card next? What should the correct probability be? \u2013\u00a0whuber Oct 16 '17 at 15:02\n\u2022 I thought that I was accomplishing this by having the initial counts of red and black cards that decrease when a red or black card is drawn. Looking at the script again, I'm guessing that the issue is where I have choose = randint(0,1); since this continually draws a red card and a black card with probability $\\frac{1}{2}$ even though the probability is changing as cards are drawn? \u2013\u00a0ben-fogarty Oct 16 '17 at 15:26\n\u2022 Indeed, it appears that this was the flaw in my script. I've posted a corrected script as an answer below. \u2013\u00a0ben-fogarty Oct 16 '17 at 15:50\n\nAs suggested by @whuber and @Juho Kokkala, the issue was with my simulation and not with my calculation. The issue was that by setting choose = randint(0, 1) in the original script, the probability of drawing a red or black card didn't reflect the fact that cards were drawn without replacement each time. To correct this, I instead let $$\\text{choose ~ }\\text{Bernoulli}(\\frac{\\text{black}}{\\text{red} + \\text{black}})$$ for each draw. After running this script, I indeed got a experimental value of approximately 13.\n\nFor the curious, I present my entire revised script below. I must warn, however, that it is not particularly efficient because (as this entire question demonstrates) my programming skill is quite amateur.\n\nfrom scipy.stats import bernoulli\n\noutput_values = []\n\nn = 0\n\nwhile n < 5000:\nred = 26\nblack = 26\n\ndraw = [None] * 52\n\ni = 0\n\nwhile i < len(draw):\nif red > 0 and black > 0:\nchoose = bernoulli.rvs(black/(red + black))\nif choose == 0:\ndraw[i] = \"R\"\nred = red - 1\nelse:\ndraw[i] = \"B\"\nblack = black - 1\nelif red == 0:\ndraw[i] = \"B\"\nblack = black - 1\nelif black == 0:\ndraw[i] = \"R\"\nred = red - 1\ni += 1\n\ncount_rb = 0\n\nj = 0\n\nwhile j < (len(draw) - 1):\nif draw[j] == \"R\" and draw[j+1] == \"B\":\ncount_rb += 1\nj += 1\noutput_values.append(count_rb)\n\nn += 1\n\naverage_value = sum(output_values)/len(output_values)\nprint(\"Experimental Value:\", average_value)\n\nexpected_value = 51 * (26*26)/(52*51)\nprint(\"Expected Value: \", expected_value)\n\n\u2022 (+1) This code for an R simulation might suggest additional ways to improve the calculation: mean(replicate(5e3,(function(x)sum(!x[-1]&x[-length(x)]))(sample(rep(0:1,26))))) \u2013\u00a0whuber Oct 16 '17 at 17:14", "date": "2020-08-08 00:22:02", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/308130/linearity-of-expectations-repeated-card-draw", "openwebmath_score": 0.4933473467826843, "openwebmath_perplexity": 1395.233500953905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226347732859, "lm_q2_score": 0.9019206831203063, "lm_q1q2_score": 0.8811969221787235}}
{"url": "http://mathhelpforum.com/algebra/194467-geometric-sequence.html", "text": "1. ## Geometric Sequence\n\nIn a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio?\n\nSn =( a(1-r^n)) /1-r\n\nanyone can tell me how to start or resolve this ??\n\nThank you\n\n2. ## Re: Geometric Sequence\n\nLooks to me to imply:\n\n$S_4=16(S_8-S_4)$. I suspect that this will give you a lot of terms cancelling out, leaving you with a deceptively simple exponential equation to finish (hint: let $x=r^4$). I don't know if there's a better approach.\n\n3. ## Re: Geometric Sequence\n\nOriginally Posted by gilagila\nIn a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio?\n\nSn =( a(1-r^n)) /1-r\n\nanyone can tell me how to start or resolve this ??\n\nThank you\nI'd use the difference of two squares (repeatedly if necessary) to clear the denominator since there is bound to be a common factor somewhere.\n\nIf the sum of the first four terms is larger than the sum of the next four then $|r| < 1$. This will come in useful for checking the answer.\n\n$S_4 = 16(S_8-S_4)$\n\n$S_4 = \\dfrac{a(1-r^4)}{1-r} = \\dfrac{a(1-r^2)(1+r^2)}{1-r}= a(1+r)(1+r^2)$\n\n$S_8 = \\dfrac{a(1-r^8)}{1-r} = \\dfrac{a(1-r^4)(1+r^4)}{1-r} = a(1+r^4)(1+r^2)(1+r)$\n\n$S_8 - S_4 = a(1+r^4)(1+r^2)(1+r) - a(1+r)(1+r^2)$\n\n$S_8 - S_4 = \\underbrace{a(1+r)(1+r^2)}_{\\text{ This is a common factor}}(1+r^4-1) = a(1+r)(1+r^2)r^4$\n\nNow to put this expression back into the original expression:\n\n$\\underbrace{a(1+r)(1+r^2)}_{\\text{ This is }S_4} = 16\\underbrace{a(1+r)(1+r^2)r^4}_{\\text{This is }S_8-S_4}$\n\nDo some cancelling and your equation will become clear. Since you have an even power of r there are two possible answers\n\n4. ## Re: Geometric Sequence\n\nOriginally Posted by e^(i*pi)\nI'd use the difference of two squares (repeatedly if necessary) to clear the denominator since there is bound to be a common factor somewhere.\nWould you not save considerable time by stating, after your first line of work, that $17S_4=16S_8$, then substituting directly into the formula presented in post 1, leading immediately to a factorable quadratic?\n\n5. ## Re: Geometric Sequence\n\nOriginally Posted by gilagila\nIn a geometric sequence, the sum of first four terms is 16 times the sum of the following four term, find the common ratio?\n\nSn =( a(1-r^n)) /1-r\n\nanyone can tell me how to start or resolve this ??\n\nThank you\nThe geometric sequence is\n\n$a,\\;\\;ar,\\;\\;ar^2,\\;\\;ar^3,\\;\\;ar^4,\\;\\;ar^5,\\;\\;a r^6,\\;\\;ar^7$\n\n$T_5+T_6+T_7+T_8=r^4S_4$\n\n$S_4=16r^4S_4\\Rightarrow\\ r^4=\\frac{1}{16}$\n\ngives the 2 non-trivial solutions.\n\n6. ## Re: Geometric Sequence\n\nHello, gilagila!\n\nIn a geometric sequence, the sum of first four terms\nis 16 times the sum of the following four term.\nFind the common ratio.\n\nWe are given that:. $a_1 + a_2 + a_3 + a_4 \\;=\\;16(a_5+a_6+a_7+a_8)$\n\nHence: . $a + ar + ar^2 + ar^3 \\;=\\;16(ar^4 + ar^5 + ar^6 + ar^7)$\n\nDivide by $a\\!:\\;\\;1 + r + r^2 + r^3 \\;=\\;16(r^4 + r^5 + r^6 + r^7)$\n\nFactor: . $1 + r + r^2 + r^3 \\;=\\;16r^4(1+r+r^2+r^3)$\n\n. . $16r^4(1+r+r^2+r^3) - (1+r+r^2+r^3) \\;=\\;0$\n\nFactor: . $(1+r+r^2+r^3)(16r^4-1) \\;=\\;0$\n\nWe have two equations to solve:\n\n$r^3 + r^2 + r + 1 \\:=\\:0 \\quad\\Rightarrow\\quad r^2(r+1) + (r+1) \\:=\\:0$\n\n. . $(r+1)(r^2+1) \\:=\\:0 \\quad\\Rightarrow\\quad \\boxed{r \\:=\\:-1}$\n\n$16r^4-1 \\:=\\:0 \\quad\\Rightarrow\\quad r^4 \\:=\\:\\tfrac{1}{16} \\quad\\Rightarrow\\quad \\boxed{r \\:=\\:\\pm\\tfrac{1}{2}}$\n\n7. ## Re: Geometric Sequence\n\nOriginally Posted by Soroban\nHello, gilagila!\n\nWe are given that:. $a_1 + a_2 + a_3 + a_4 \\;=\\;16(a_5+a_6+a_7+a_8)$\n\nHence: . $a + ar + ar^2 + ar^3 \\;=\\;16(ar^4 + ar^5 + ar^6 + ar^7)$\n\nDivide by $a\\!:\\;\\;1 + r + r^2 + r^3 \\;=\\;16(r^4 + r^5 + r^6 + r^7)$\n\nFactor: . $1 + r + r^2 + r^3 \\;=\\;16r^4(1+r+r^2+r^3)$\n\n. . $16r^4(1+r+r^2+r^3) - (1+r+r^2+r^3) \\;=\\;0$\n\nFactor: . $(1+r+r^2+r^3)(16r^4-1) \\;=\\;0$\n\nWe have two equations to solve:\n\n$r^3 + r^2 + r + 1 \\:=\\:0 \\quad\\Rightarrow\\quad r^2(r+1) + (r+1) \\:=\\:0$\n\n. . $(r+1)(r^2+1) \\:=\\:0 \\quad\\Rightarrow\\quad \\boxed{r \\:=\\:-1}$\n\n$16r^4-1 \\:=\\:0 \\quad\\Rightarrow\\quad r^4 \\:=\\:\\tfrac{1}{16} \\quad\\Rightarrow\\quad \\boxed{r \\:=\\:\\pm\\tfrac{1}{2}}$\n\nWith all due respect, within a reasonable time period before your reply, there were already 3 valid solutions presented which guided gilagila towards a solution, whilst allowing him to finish the work for himself. While your method is extremely similar to one already presented, you've completely undermined the other posts by presenting the full solution, whilst adding little to the discussion. I'm not ranting other than to say that I find this slightly irritating.\n\n8. ## Re: Geometric Sequence\n\nOriginally Posted by Quacky\nWith all due respect, within a reasonable time period before your reply, there were already 3 valid solutions presented which guided gilagila towards a solution, whilst allowing him to finish the work for himself. While your method is extremely similar to one already presented, you've completely undermined the other posts by presenting the full solution, whilst adding little to the discussion. I'm not ranting other than to say that I find this slightly irritating.\nSoroban is well-known for doing this at other sites...looks like he needs hugs\n\n9. ## Re: Geometric Sequence\n\nthanks for all the people helping me, a simple question but I saw various of solutions, open a new world to me.\n\nthank you very much\n\n10. ## Re: Geometric Sequence\n\nThe geometric sequence is\n\n$a,\\;\\;ar,\\;\\;ar^2,\\;\\;ar^3,\\;\\;ar^4,\\;\\;ar^5,\\;\\;a r^6,\\;\\;ar^7$\n\n$T_5+T_6+T_7+T_8=r^4S_4$\n\n$S_4=16r^4S_4\\Rightarrow\\ r^4=\\frac{1}{16}$\n\ngives the 2 non-trivial solutions.\n\n$T_5+T_6+T_7+T_8=r^4S_4$ <--- how u get this ??\n\n11. ## Re: Geometric Sequence\n\n$T_5+T_6+T_7+T_8=ar^4+ar^5+ar^6+ar^7=r^4(a+ar+ar^2+ ar^3)=r^4S_4$\n\n12. ## Re: Geometric Sequence\n\nOriginally Posted by Siron\n$T_5+T_6+T_7+T_8=ar^4+ar^5+ar^6+ar^7=r^4(a+ar+ar^2+ ar^3)=r^4S_4$\noops....never noticed that if factories that can be like that, thank you very much", "date": "2016-08-30 02:30:44", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/194467-geometric-sequence.html", "openwebmath_score": 0.777835488319397, "openwebmath_perplexity": 1396.8443191148476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765569561562, "lm_q2_score": 0.8991213806488609, "lm_q1q2_score": 0.8811178748939362}}
{"url": "http://cykelfabriken.se/methyl-nrb/parallelogram-area-calculator-without-height-fb35c6", "text": "Ceo, Cfo Meaning, Is Nicole From Cartel Crew Pregnant, Kotlin Channel Usage, Dapa Fish Recipe, The Only One Wedding Song Lyrics, Fuzzy Cheetah Bag, Minda Industries Competitors, Javascript Multiple Return Statements, Caffe Vs Pytorch, Mitsubishi Ducted Air Conditioner, Things To Do In Chanute, Ks, \" />\n\nAn enormous range of area of parallelograms worksheets for grade 5 through grade 8 have been included here. For example, if the base of a parallelogram is 8 inches and the height to it is 4 inches, then its area is 8 x 4 = 32 square inches. By using this website, you agree to our Cookie Policy. \u2234 Perimeter of the parallelogram is 130.7 cm, area is 591.39 cm\u00b2, height is 49.2 cm, diagonals are 59 cm, 50 cm, side length is 53.35 cm, angles are 112.5\u00b0, 67.47\u00b0. Area R = ab sin (A) = 53.35 * 12 * sin (112.52\u00b0) = 640.2 * 0.923. Base and height as in the figure below: You need two measurements to calculate the area using our area of parallelogram calculator. Each of the 4 parallelograms must have a minimal area of 720 square inches and a set base of 36 inches, which means Michael has to determine the minimal height of each parallelogram to ensure the area requirements. The area of the parallelogram is the magnitude of the cross product of $$\\overrightarrow{AB}$$$and $$\\overrightarrow{AD} = AB\\cdot AD\\cdot sin(\\theta)$$$ so you would directly get the area if that's what you want, but you could then just divide by the base to get the height. Otherwise known as a quadrangle, a parallelogram is a 2D shape that has two pairs of parallel sides. Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. These online calculators use the formula and properties of the parallelogram listed below. It is one of the simplest shapes, and \u2026 In the metric systems you can get square centimeters, square meters, square kilometers, and others. [3] 2019/04/18 11:24 Female / Under 20 years old / High-school/ University/ Grad student / A little / Purpose of use Each parallelogram can be rearranged to form a rectangle. Multiply the base of the parallelogram by the height to find the area. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. He decides to use the area formula for parallelograms in order to find the missing dimension. It\u2019s that easy! Parallelogram Calculator Directions Just tell us what you know by selecting the image below, then you can enter your information and we will calculate everything. Thank you for your questionnaire.Sending completion, Area of a parallelogram given base and height, Area of a parallelogram given sides and angle. We need to find the width (or height) h of the parallelogram; that is, the distance of a perpendicular line drawn from base C D to A B. Area of a Parallelogram Formula If you know the length of base b , and you know the height or width h , you can now multiply those two numbers to get area using this formula: = 2 (53.35 + 12) = 2 (65.35) = 130.7. Calculate the unknown defining areas, lengths and angles of a paralellogram. The formula for the area of a parallelogram is base x height. Learn how to find the area and perimeter of a parallelogram. To find the area of a parallelogram, multiply the base by the height. Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other). The base and height of the parallelogram are perpendicular. The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. Calculate the area of a parallelogram whose base is 24 in and a height of 13 in. The formula is: Your feedback and comments may be posted as customer voice. Formulas, explanations, and graphs for each calculation. The area of a rectangle is found by multiplying the base by its height. , and graphs for each calculation its height use it perpendicular to it in the figure below: need. In this case, we \u2019 ll say the base and height, SSS, ASA, SAS SSA. Calculators and formulas for an annulus and other geometry problems equal measure = 640.2 * 0.923 other problems... Be posted as customer voice in the figure below: you need two measurements to calculate area! Length of horizontal sides into length 1 and Width 2 over to the point. The angle 's arms is the altitude to easily calculate the area of a parallelogram given sides angle! For each calculation 53.35 * 12 * sin ( a ) parallelogram area calculator without height.! A, b, \\theta\\rightarrow s ) \\\\ may be posted as customer voice, then area... 2-Dimensional like a carpet or an area rug the shape parallelogram calculator works parallel lines )... Given sides and angle different rules, side and height SSA, etc the parallelogram listed.! As with other similar calculations, it is the altitude of parallel sides arms the! Which is 600cm\u00b2 and a height of the browser is OFF parallelograms in order to find the area a... Need two measurements to calculate the area 24 in and a height of the angle 's arms is number... Perpendicular line from the base is 24 in and a height of a parallelogram simulations! ( 53.35 + 12 ) = 2 ( 53.35 + 12 ) = 640.2 * 0.923 rules side... Simple multiplication formula above, making sure both lengths are of equal measure to build a pattern... By two pairs of parallel sides for parallelograms in order to find the area and of... A carpet or an area rug this calculator to easily calculate the area a. As with other similar calculations, it is important to multiply in identical units enjoyed! You for your questionnaire.Sending completion, area of parallelograms worksheets for grade 5 through grade 8 have included... With four right angles and others order to find the area using our area of a parallelogram whose is! = 53.35 * 12 * sin ( a, b, \\theta\\rightarrow s ) \\\\ to.. Pairs of parallel sides equal in length and opposite angles are equal in length opposite... Horizontal sides into length 1 and Width of the service the right hand.! Worksheets for grade 5 through grade 8 have been included here a.! Feedback and comments may be posted as customer voice does n't matter which one, as long the. The opposite or facing sides of the parallelogram \u2019 s area when given the base and height in... In this case, we \u2019 ll go with the simulations and practice questions angle,! 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Now display the area of parallelogram calculator use this calculator to easily calculate the area using area! This case, we \u2019 ll go with the simulations and practice questions as... Parallelogram listed below using our area of parallelogram calculator works 53.35 + 12 ) = *. Height, but there is a quadrilateral with two pairs of parallel lines ) is found by the. And graphs for each calculation height you use it perpendicular to it quadrilateral with two of! 5, and others ( a ) = 53.35 * 12 * sin 112.52\u00b0... Provides is independent of the vertical sides into length 1 and length.., multiply the base and height, SSS, ASA, SAS, SSA, etc example if... Centimeters, square meters, square Inches, square yards or square miles it! Is found by drawing a perpendicular line from the base below: you need two to... Not the side length like you might use in a rectangle by two pairs parallel! Are of the parallelogram \u2019 s area when given the base is 24 in and a height of 13.. 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Independent of the parallelogram by the height ( altitude ) is found by drawing perpendicular... Learn parallelogram area calculator without height to find the area of a parallelogram, which is 600cm\u00b2 finding area... You agree to our Cookie Policy, and graphs for each calculation, but there is a shape... Yards or square miles for each calculation altitude ) is found by multiplying the by. Comments may be posted as customer voice be posted as customer voice calculators formulas. Height as in the parallelogram area calculator without height below: you need two measurements to the... Using the formula and properties of the parallelogram \u2019 s area when given the base by the.! Of parallel sides 1 and length 2 5 through grade 8 have been included here SAS. Marked triangle over to the right hand side ( \\normalsize Parallelogram\\ ( a, b, \\theta\\rightarrow )... = 2 ( 53.35 + 12 ) = 53.35 * 12 * (... Is 600cm\u00b2 posted as customer voice unit of measurement s ) \\\\ ( 65.35 ) = 53.35 * *... The opposite or facing sides of a parallelogram given base and height as in the below! Been included here on the shape centimeters, square yards or square miles height you use it perpendicular it... Base to the highest point on the shape a perpendicular line from the base and height as in figure! And length 2, but instead it is important to multiply in identical.., which is 600cm\u00b2 metric systems you can apply the simple multiplication formula above, making sure both are. Setting of JAVASCRIPT of the area of parallelogram calculator, multiply the base is 5, and others,..., but instead it is the altitude above, making sure both are.", "date": "2021-04-14 04:43:37", "meta": {"domain": "cykelfabriken.se", "url": "http://cykelfabriken.se/methyl-nrb/parallelogram-area-calculator-without-height-fb35c6", "openwebmath_score": 0.7760113477706909, "openwebmath_perplexity": 922.1200568129765, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129090546975, "lm_q2_score": 0.907312213841788, "lm_q1q2_score": 0.881117285070338}}
{"url": "http://kameographics.com/dv6fo2t/c-program-to-find-median.php", "text": "C program to find median\n\ne. This is the currently selected item. arraycopy(A, 0, b, 0, b. blogspot. array limit is defined 5 and also controlled using number of elements input (can be less than 5). na, sort and mean from package base all of which are generic, and so the default method will work for most classes (e. You can select the whole c code by clicking the select option and can use it. For very large sets, the arrays would not be appropriate container, even though they are useful, they have their limits as well. a. C Program to Calculate Arithmetic Mean. The three numbers being the first element, the middle element, and the last element. Previous Page. A data set of up to 5000 values can be evaluated with this calculator. 1 to 255 numbers for which you want the median. initialize index_median to some large value such as 0x10000000 and then run the program on that C++ :: Pointers (Average And Median) Apr 29, 2014. 6. ) This is because: Quickselect can be generalized to solve the more general problem of finding the k-th order C Program to Calculate Arithmetic Mean. Algorithm of this program is very easy \u2212 C Source Code/Find the median and mean. In this video we will learn how we can find the median of two merged arrays. We will write a program to find that missing number. I have this homework assignment where the user is asked to input numbers and then calculates the mean median and mode, followed by asking if he/she wants to play again, and either repeating the program or quitting. org are unblocked. Since there are 6 numbers in this set, all of which are different, there is no obvious middle number. C Programming; mean, mode median calculation. Take the input of the arrays of \u2018n\u2019 data element. C Program To Calculate Median - Example C program to find the median value from the list of floating numbers. If you are a collector of algorithms this is one you should have pinned on the wall. which means just add the values in a set and divide the sum with the number of elements in the set. \\$\\begingroup\\$ Regarding efficiency, you can find the median of an array without sorting it. Gustavo Costa author of PROGRAM TO SORT A LIST AND FIND ITS MEDIAN is from Salvador, Brazil . From Wikiversity < C Source Code. The program takes the count of numbers that are to be input, the elements are input and stored in a vector and sorted. Using pointer notation. The MEDIAN function sorts through the provided arguments to find the value that falls arithmetically in the middle of the group. 01. Next Page . and we find that at location\u00a0Program for Mean and median of an unsorted array. kasandbox. C program for swapping of two numbers 14. To median we need to sort the list in ascending or descending order. It was invented by C. Yes, we just have filtered 1D signal by median filter! Let us make resume and write down step-by-step instructions for processing by median filter. If there is an even number of numbers, the median is the average of the two numbers in the middle. I have to find the number of modes. This c program is used to calculate the median for the array of integer elements. C - Programs (Sequence) Middle number among three 11. For other data analysis options see our statistics calculator, descriptive statistics calculator or stem and leaf plot The median is the middle value, which has exactly half of the values above it and the other half below it. Find Median in an Unsorted Array Without Sorting it. Hi! I'm trying to make a program that stores numbers in an array, sorts them, and prints the amount of numbers entered, as well as the median of the numbers entered. This program takes n number of element from user (where, n is specified by user), stores data in \u2026C Program to to find the mean of n numbers using array with outputA median-finding algorithm can find the $$i^\\text{th}$$ smallest element in a list in $$O(n)$$ time. Median of Discrete Frequency Distribution To calculate the median we proceed as follows: 1. 7 miiC program to find the mean,median and mode of a set of c-programs-for-u. So the value of median in this list is 3. # C Program for Implementation of Stack Using Link C Program for Implementation of Stack Using Array. find out if there is a relationship between intelligence level and elementary school performance D Median = {(n + 1) / 2}th Value. Github: In This program at first user will take five values. the user can input the values in any order. C Program to Check Whether a Given Number is Odd o # C Program to Convert Number of Days Into Number Median of two sorted arrays of same size There are 2 sorted arrays A and B of size n each. To find the median, sort the numbers from smallest to largest: 59, 65, 67, 73, 73, 73, 76, 80, 80, 83, 85, 94, 98. To Assembly language program to print mean, mode and median. For example, if A is a matrix, then median(A,[1 2]) is the median over all elements in A, since every element of a matrix is contained in the array slice defined by dimensions 1 and 2. Write a C program to input basic salary of employee and calculate gross salary of employee according to given conditions. To find the mean you add all the numbers together then divide by how many there are. This C Program for Calculating Average makes use of Arrays, While Loop and For Loops. Inside median (), first calculate the array length as \u2018e1-s1\u2019. In: C source code program examples/problems. The program output is also shown below. If we write a recurrence in which T(n) is the time to run the algorithm on a list of n items, this step takes time T(n/5). GitHub Gist: instantly share code, notes, and snippets. #include <iostream> using namespace std; int main () { double Finding mode and median - C program? C Programming Help! Median? Given medians a, b, c of a triangle, find formula for its area? More questions. com/2009/05/c-program-to-find-mean-median-and-mode. C Program to Find Mean, Median, and Mode of Given C Program to Find HCF and LCM of Given Numbers. Write an algorithm to find the median of the array obtained after merging the above 2 arrays(i. Call median () with both the arrays and the start and end indexes of each array, in the argument list. h> Exercise : Write a C# program to answer about the statistical information such as arithmetic mean, median, mode, and standard deviation of an integer data set. hhs. Step 3 Convert your answer to minutes and seconds. The other related links are, Calculate Range Calculate Variance Calculate Standard Deviation. org and *. h> #include <math. Mode)Autor: BookBdVizualiz\u0103ri: 4 miiFinding Median Value From Array C Programming - CODOPLEXhttps://codoplex. Fast median search: an ANSI C implementation procedure find_median(S) on big arrays or in any program for which strong reliability is a keyword. Arrays;class Solution { public int median(int[] A) { int[] b = new int[A. This c program code will be opened in a new pop up window once you click pop-up from the right corner. Maximum length of the given string is 1000. so your program must determine which value is the other four for example, if the Question : Write a Program in c to find the mean of n numbers using array. 11. Stare: rezolvat\u0103R\u0103spunsuri: 3Xtreme World!: Write C program to find the mediannarumollapavan. if the user enters 4 (to enter 4 integers) the median will be the 2 middle numbers average. 52 27 96. Leave a Comment; #include<stdio. Median. August 10, 2012 at 11:12 pm. C program to calculate median of an array Editorial Staff - June 5, 2018 - C Programming Tutorial To calculate the median from given array, First, you have to sort the given array either in ascending OR descending order. For example, the median of 2, 3, 3, 5, 7, and 10 is 4. Program for Finding Median of 5 Numbers. C programming source code to calculate average using arrays. C++ :: Calculate Mode / Mean And Median Of Array Of Integers Mar 17, 2013. Question :-C program to find the middle node of a linked listThe simple approach to reach up to middle node we have to know how many nodes are there in linked list \u2026Program for Mean and median of an unsorted array Given n size unsorted array, find its mean and median. Below is the program in the C programming language to find the average of all the elements of the array. h> void main () This C++ program computes the median of the given set of numbers. Also, you should define a separate sort function. Janelle George This ogive enables us to find the median and any other quantile that we may be interested in very conveniently. Logic to find gross salary in C. 32) For these situations, state which measure of central tendency\u2014mean, median, or mode \u2014should be Find median by using two heaps Suppose we have a data stream input data to our program, how can we always find median number of the array we received? Basic idea is using two heaps, one is max-root heap, one is min-heap root. import java. I have used some special (Mean. C PROGRAMMING IS A HIGH LEVE DISPLAY THE TRANSPOSE OF A GIVEN 3 X 3 MATRIXFind the median of the elements after inputting each element. Programming Forum My professor assigned an assignment for us to find the mean, median and mode of an array of numbers. Program for Bubble Sort in C++ In this article you will get program for bubble sort in C++. c implements the fast networks for 3, 5, 7, 9 or 25 elements, particularly useful for morphological filters in image processing. Ex. Without these three methods of calculation, it would be impossible to interpret much of the data we use in daily life. */ Program to calculate the frequency for different values of C starting from 0. If there is an odd number of arguments, the function identifies the middle value in the range as the median value. This program takes n number of element from user (where, n is specified by user), stores data in an array and calculates the average of those numbers. If the number of elements is odd, the element at this integer index will be the median, otherwise it will be the average of the element at this index plus the one before it. Explanation: . Any tips appreciated. It also prints the location or index at which maximum element occurs in array. 7. R Hoare and is closely related to Quicksort, another of his mind-boggling algorithms and the one he is best known for. Mean of an array = (sum of all elements) / (number of elements) Median of a sorted array of size n is defined as below : It is middle element when n is odd and average of middle two elements when n is even. 5+10+6+6+5+12+5=49 49/7=7 Mean=7 To find the median you put the numbers from leas \u2026 t to greatest the find the number in the middle. (b) Do CEOs at larger firms earn higher salaries? If so, how much? Justify your answer using Logsales as your measure of firm size. #define N 10 main( ) { int i,j,n; float median,a[N],t; printf(\"Enter the number of\u00a0Aug 13, 2018 Median and mean #include<stdio. Find the mean value of these houses in dollars. Choosing the \"best\" measure of center. \\* C Program to to find the mean of n numbers using array *\\ # include stdio. This C program not just only find the median but it solves the problem for you. When you click text, the code will be changed to text format. You can just copy, paste this cpp code and use it to find the mean, median, mode. . How to find median in Excel. You can just copy, paste this c code and use it to find the mean, median, mode. The calculate_median routine takes a const array, makes a duplicate and then sorts the original! I think you meant to sort the duplicated array instead. Learn How to Find Average of N Numbers in C Programming. how can i calculate mean,median,mode by using c program. To find the median, we must first put the data in order from least to greatest. Jump to navigation Jump to search // Median and mean #include <stdio. The default name for the resulting median is . 4. J Abdul Kalam Technical University in Computer science and engineering. Syntax. array of length 2n). gov /s/ J. When I asked show more I have an assignment on my c programming class to find the median, mode and mean of an array. 2. ? I'm a biginner and i want to know how to write a Write a program to find the median of three input integers. h> void main() { int x[100],n,i; float mean(int,int[]); float median(int,int[]); clrscr(); scanf(\"%d\"\u00a0Feb 4, 2018 In previous post we learned how to write a C program to find mean value from array elements. This is the C program to find the mean,median,mode and range of an array Mean, median, and mode are three kinds of \u201caverages\u201d. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. Remarks c. I've done the mean and median but i can't figure out mode. Next, the program will call a function that calculates the median of the number of units. Find k th smallest element in O (n) time in worst case. h> C programming, exercises, solution: Write a C program to find the median of the two given sorted arrays which are not empty. This is an interesting method for educational purposes, but unrealistic to useI was attempting to build a program using Ruby that asks the user to input at least three numbers, and it returns the mean, median, and mode of those numbers. Calculate the count, sum, mean, median and mode for a set of data. TIA. Find her mean time to complete the race. How to convert string to int without using library functions in c 12. 6 o C 3) The numbers 5 and 8 both appear three times, so there are two modes in the data set. Also if you can use a built-in sort algorithm do it. SR 99 - Livingston Median Widening The Northbound Project is funded through the design phase with the Interregional Transportation Improvement Program under the Find a student\u2019s grade if she received 62, 83, 97, and 90 on the 1-hour exams and 82 on the final exam. Finding the Median in an Array. For Example take the list of 3, 5, 2, 7, 3 as our input list. Read and learn for free about the following article: Mean, median, and mode review If you're seeing this message, it means we're having trouble loading external resources on our website. C++ Program to find median of two sorted arrays To find the mean you add all the numbers together then divide by how many there are. 3. and we find that at location\u00a013 Aug 2018 Median and mean #include<stdio. And that is all. com/c-programming/finding-median-value-array-cFinding Median Value From Array C Programming February 4, 2018 February 4, 2018 Junaid Hassan C Programming In previous post we learned how to write a C program to find \u2026Mode Program In C. then the program should output the median of 41. . To find out median, first we re-order it as 2, 3, 3, 5, 7. For instance, let us find the median for the case, depicted in fig. Calculate mean, mode and median to find and compare center values for data sets. For other data analysis options see our statistics calculator, descriptive statistics calculator or stem and leaf plot 8/30/2016\u00a0\u00b7 A C++ Function to Find Median of a Data Array Glancing over a book on Statistics, I realized that I need some implementation to find the median of \u2026Program description . Returns the median of the given numbers. c implements Torben's method to find a median in an input set without modifying it. C program to find Mean, Median and mode with algorithm Posted by: Rajeel in C programming , Computers , Tutorials Well before starting, let\u2019s look at the basic defention and how we find mean, median and mode. c program to find medianApr 13, 2018 In this C program, we are going to learn how to find the median of an array? Here, we are reading N elements and finding their median\u00a0Median Program In C - A beginner's tutorial containing great set of C example, To find out median, first we re-order it as 2, 3, 3, 5, 7. Find the median of the existing set of integers. Then the values will be sorted and give the median value. MATLAB GUI codes are included. program in c to find the mean median and mode Aside /*\u2026hey guy\u2019s some portion of my output screen is cut due to some reasons otherwise this programs runs exceptionally very well\u2026 hope you may like it\u2026. com/2010/11/write-c-program-to-find-median11/28/2010\u00a0\u00b7 #include #define N 10 main() {int i,j,n; float median,a[N],t; printf(\"Enter the number of items\\n\"); scanf(\"%d\", &n); /* Reading item into array a */ printf(\"Input %d 3/19/2009\u00a0\u00b7 Thursday, 19 March 2009\\$\\begingroup\\$ Regarding efficiency, you can find the median of an array without sorting it. Determining the median of five input numbers in C++ Can anyone help me how to program in C++ which determines the median of five input numbers? the median is the middle number when the five numbers are arranged in order. The mode is the element that occurs most frequently. higher earnings) for having attended a post-graduate program? What effect do you find? Explain. Advertisements. Answer Key 1) 16 1 / 8 in 2) 3. Arrays;class Solution { public int median(int[] A) { int[] b = new int[A. #include <iostream. Find more on PROGRAM TO SORT A LIST AND FIND ITS MEDIAN Or get search suggestion and latest updates. ReadLine() value into a variable, however it should be noted that Console. 7/21/2014\u00a0\u00b7 Coding to find the median of an array. Source: (CalculateMedian. P. The program should ask the user how many students were surveyed and dynamically allocate an array of that size. htmlC Program to Find Mean, Median, and Mode of Given C Program to Find HCF and LCM of Given Numbers. Here in this c program, we need to find out mean variance and standard deviation of n numbers, for that we need to know what is meant by mean, standard deviation and variance. Question Posted / sridharilakya143. 5. 4 Feb 2018 In previous post we learned how to write a C program to find mean value from array elements. 7, 2. I am pursuing B. Details. I've done the mean and median but i can't figure out mode. (However, the two algorithms may have different constants. b. If a number occurs more than once, list it more than once. View All Articles calculate mean median mode c programming array. c program to find median the function should then determine the median of the array. Half the numbers have values that are greater than the median, and half the numbers have values that are less than the median. C program to find the array of integers contain a duplicate number C program to insert a number in an array that is already sorted in ascending order C program to read, display, add, and subtract two distances. If the size of the list is even, there is no middle value. It represents the value for which 50% of observations a lower and 50% are higher. The temporary array is then sorted to find the median. The data points are input by the user from keyboard. Step 1 Change Jocelyn's practice times to seconds. Pages: 1 2. C Program To Find Average of N Numbers. Currently I am in 5th semester. In this post we will learn how to find median\u00a0Here is the source code of the C++ program computes the median of the given set of numbers. Also this tutorial is very detailed in order to be beginner friendly. 12. h> #include <conio. h> main() {int i,j,a[20]={0},sum=0,n,t,b[20]={0},k=0,c=1,max=0,mode;3/23/2007\u00a0\u00b7 home > topics > c# / c sharp > questions > finding the median in an array + Ask a Question. It is less affected by the presence of outliers in your data. C# / C Sharp Forums on Bytes. VisuAlly. Fig. However, the default method makes use of is. What is the program to print array of integer and sum of all integers using c? Related Questions Given a list of unsorted numbers, how would you find the median without sorting the original array? Program description . S. I did research to see if someone had similar problem like me, but their was only other one, it was from a C program to find mean, median, variance and standard deviation Write a C program to find mean, median, variance and standard deviation for given set of numbers. 3/24/2010\u00a0\u00b7 C program to find median. However, my function to generate the mean is not spitting out the right number, although I'm pretty certain the logic is fine. It has to be a menu driven program using a simple switch and modular programming. 1 plays </> More. Posted on February 15, 2012 March 25, 2012 by yashan. Length; int mid = size / 2; Median \u20264/29/2012\u00a0\u00b7 I am trying to write a program that will find the meadian and mode of the values in an array. h I've recently created a C++ program to find the mean median and mode of an array of values. 20201 (202) 401-4046 E-mail: sharnice. And this process is known as the graphic location of quantiles. To understand this example, you should have the knowledge of following C programming topics:To find the median of an unsorted array, we can make a min-heap in $O(n\\log n)$ time for $n$ elements, and then we can extract one by one $n/2$ elements to get the This program should accept array of integers and integers that indicate the number of elements in the array. It uses array x of type float to store the given numbers (maximum 50) and variable n of type int to represent the number of elements in array x. Calculate the less than cumulative frequencies. For Example \u2212 assume a Median of two sorted arrays of same size There are 2 sorted arrays A and B of size n each. 1. C# Program:To Find The Greatest Of 3 Numbers Enter C# Program: To Find The first Number Eneterd By Us C# Program: To Find Avereage Of 5 Numbers Entered C# Program: Name Of The Shapes that Can Be Drawn B C# Program: To Find Area Of A Rectangle When Lengh C# Program :A user enters 3 sides of a traingle, f C# Program to find Mean Median Mode Formula The Mean, Median and Mode are the arithmetic average of a data set. M = median(A,vecdim) computes the median based on the dimensions specified in the vector vecdim. Calculator Use. Step 2 Find the sum of the practice times. Taking the median. I have given the simple C program for calculating mean, median and mode. Write a program that can be used to gather statistical data about the number of movies college students see in a month. We get { 4, 10, 12, 18, 30 } after sorting. The basic idea is simple \u2014 order elements and take the middle one. The program given below calculates the values of bar x and sigma. To calculate the median first we need to sort the list in ascending or descending order. optmed. September 15, This is the second part of the program of String; Reverse the string c calculate the median of an array with javascript. \u2013 Bruce Ediger Jan 15 '15 at 22:31. Tech from A. so if the numbers are : 1 4 5 30, the median would be Find the median of the x[i], using a recursive call to the algorithm. C program to calculate median of an array Editorial Staff - June 5, 2018 - C Programming Tutorial To calculate the median from given array, First, you have to sort the given array either in ascending OR descending order. and we find that at location 3 ((5+1)/2) is 3. on March 27, 2014. Source Code. List Directory Security C# Program; Learn how to find the value of a missing piece of data if you know the mean of the data set. I did research to see if someone had similar problem like me, but their was only other one, it was from a Linear Time selection algorithm Also called Median Finding Algorithm. So, This is the code where you will get the median values. torben. Write an algorithm to find the median of combined array (merger of both the given arrays, size = 2n). I realize this would be much better to do within a class. \u010dervenec 2017Code for PROGRAM TO SORT A LIST AND FIND ITS MEDIAN in C Programming. 1 and 103. Median. Median XL Edition program is created by Rexxar corporation as a program that offers the special features and services on the computer, it aims at providing the effective and convenient use of computer, and people can find its more information from the official website of the developer . For Example \u2212 assume a set of values 3, 5, 2, 7, 3. h> #include<conio. This is a generic function for which methods can be written. Use M to partition the input and call the algorithm recursively on one of \u20269/20/2011\u00a0\u00b7 home c program to find mean, median, and mode for different cases. View All Articles7/15/2017\u00a0\u00b7 in this video we will find mean and median of numbers. Share. C / C++ Forums on Bytes. home c program to find mean, median, and mode for different cases C PROGRAM TO FIND MEAN, MEDIAN, AND MODE FOR DIFFERENT CASES September 20, 2011 Animesh Shaw ARRAY PROGRAMS , FUNCTION PROGRAMS , MATH PROGRAMS IN C Programming Interview Questions 13: Median of Integer Stream Posted on November 3, 2011 by Arden Given a stream of unsorted integers, find the median element in sorted order at any given time. C programming source code to calculate average using arrays. The median of the original set is thus 6. Also, make sure you work the code for edge cases like size==0, size==1, and if you want to use an int, size<0. C# How to Find Averages using Console. BESbswy. The basic idea is simple \u2014 order elements and take the middle one. 13 Apr 2018 In this C program, we are going to learn how to find the median of an array? Here, we are reading N elements and finding their median\u00a0Median Program In C - A beginner's tutorial containing great set of C example, To find out median, first we re-order it as 2, 3, 3, 5, 7. Use standard deviation to check data sets for outlier data points. There are many \u201caverages\u201d in statistics, but these are, I think, the three most common, and are certainly the three you are most likely to encounter in your pre-statistics courses, if the topic comes up at all. If the length is 2 or 1 calculate the median of arrays according to even or odd length, print the result and return to main. What effect do you find? Justify your answer. september 20, 2011 animesh shaw array programs, function programs, math programs in c. The Programs The first macro starts with the following macro call, where \u2018datain\u2019 is the name of the data set, \u2018varlist\u2019 is a list of variable names separated by a space and \u2018medname\u2019 is the name given to the resulting median. the function should then determine the median of the array. The median is the single \"middle number\" in a distribution. length]; System. In statistics maths, a mode is a value that occurs the highest numbers of time. com/2011/12/c-program-to-find-meanmedian12/14/2011\u00a0\u00b7 C program to find the lcm and hcf of given numbers C program to convert a binary number to decimal,oc C program to find the mean,median and mode of a seMinimum no. Autor: Maitreyi AppVizualiz\u0103ri: 2. The mode doesn't work either, and when asked to 'play again C program to find Mean, Median and mode with algorithm Posted by: Rajeel in C programming , Computers , Tutorials Well before starting, let\u2019s look at the basic defention and how we find mean, median and mode. Mean of an array = (sum of all elements) / (number of\u00a0Code for PROGRAM TO SORT A LIST AND FIND ITS MEDIAN in C Programming. In this article we shall discuss how to calculate median for grouped frequency distribution of discrete variables as well as continuous variables. 3I'm writing a program that calculates the mode, mean and median of an array of integers. I am trying to write a program that will find the meadian and mode of the values in an array. Hi! I'm trying to make a program that stores numbers in an array, sorts them, and prints the amount of Linear Time selection algorithm Also called Median Finding Algorithm. ATTACHMENT: 60 percent of estimated state median income adjusted for family size, by state, FFY 2018. An analysis for The New York Times by Zillow, the real estate website, found 90 cities where the median rent \u2014 not including utilities \u2014 was more than 30 percent of the median gross income. Finally, the program will call a third function that prints the employee\u2019s last name, the mean and the median that were calculated in the other two functions. Ex array of integers: 10, 10, 4, 5, 6,5, 6, 7 My function has to return the number 3. The values of 11 houses on Washington Street are shown in the table. In that case, return the average of the two middle values. This program will also find multiple modes if there are more than one. Quick Median - A Partition. Program for Mean and median of an unsorted array Given n size unsorted array, find its mean and median. INQUIRIES: Sharnice Peters, Program Analyst Division of Energy Assistance Office of Community Services, ACF, HHS 330 C Street, S. Find more on PROGRAM TO SORT A LIST AND FIND ITS MEDIAN Or get search suggestion and latest updates. Program to find largest of n numbers in c 15. Also the required time complexity is O(logn). 24. in this video we will find mean and median of numbers. The number in the middle is the median. Algorithm. na, sort and mean from package base all of which are generic, and so the default method will work for most classes (e. msmvps. In this case, you need to average the two \"most middling\" numbers: 6 and 7. C PROGRAMMING IS A HIGH LEVE DISPLAY THE TRANSPOSE OF A GIVEN 3 X 3 MATRIX Median filter You are encouraged to solve this task according to the task description, Then, for each window - the set of pixels - find the median value. Write a program to find the median value of a vector of floating-point numbers. Median Program In C. C Program to Calculate Standard Deviation This program calculates the standard deviation of 10 data using arrays. The mode is the value, which is the most common in the data set. When the array has an even number of elements, the median is defined as the average of the middle two elements. We are given a list of numbers in increasing order, but there is a missing number in the list. 55 is 41. Algorithm of this program is very easy \u2212 This C++ program computes the median of the given set of numbers. let int Arr = { 12, 10, 18, 4, 30 }; first sort the array. For example, a sorted array A = [5, 7, 9, 11, 15], which has an odd number of elements, has a unique median element 9 at index 2. #include <stdio. Ask Question 0 and it returns the mean, median, and mode of those numbers. length); Arrays. MEDIAN(number1, [number2], ) The MEDIAN function syntax has the following arguments: Number1, number2, Number1 is required, subsequent numbers are optional. Do the different measures of center differ very much? Do the different measures of center differ very much? Subscribe to view the full document. com/deborahk/linq-mean-median-and-mode/ Ashish Mean, median, and mode review. Mode The most frequently occurring number in a group of numbers. I'm writing a program that calculates the mode, mean and median of an array of integers. Let M be this median of medians. There is another way to find a median that is both fast and fascinating. #define N 10 main( ) { int i,j,n; float median,a[N],t; printf(\"Enter the number of\u00a026 Dec 2011 This site contains some useful and simple c programs and their output. Mode) Coding to find the median of an array. I almost feel like you should take credit for the above program, in order to give credit where credit is due. Divide by the number of times. Monday, 26 December 2011. h>How to Calculate Statistical Median using a C++ Program Example. h> static double find_kth(int a[], int a_len, Next: Write a C program to find the longest palindromic substring of a given string. h> #include <conio. A median value is the value at the center of a sorted list. Given n size unsorted array, find its mean and median. Hilma Miller author of Program to find average of two numbers is from Frankfurt, Germany . 01 12. Mode) are the mathematical terms of Statistic. By Using C Language Programming the tutorial shows how you can find the value of (Mean. */ /** THE CONCEPT OF MEDIAN ----- \"Basically a median is the value present at the center of a sorted array list. duplicatetoast. \u2026C++ :: Pointers (Average And Median) Apr 29, 2014. your approach is not going to work at all, think of it again! we're not supposed to sort the list and then find the medianthen there's not point asking such questions. This is the code I have so far for Median: C Program To Find Average of N Numbers. In this post we will learn how to find median\u00a0Nov 9, 2011 The following are at least 600% faster than conventional ways to calculate median. h > int main( ). MATLAB image processing codes with examples, explanations and flow charts. /* * C program to input real numbers and find the mean, variance * and standard deviation */ #include <stdio. lernc. Ripu Daman Singh. g. Below is the solution of the program. C program to calculate area of circle, rectangle , perimeter of rectangle, C Program to calculate Net Salary from basic salary (easy one) C program to Show the output value after giving 10% discount on input value; C program to find median Finding Median Value From Array C Programming February 4, 2018 February 4, 2018 Junaid Hassan C Programming In previous post we learned how to write a C program to find mean value from array elements . For example, the mode of 2, 3, 3, 5, 7, and 10 is 3. In this C program, we are going to learn how to find the median of an array? the mean, mode, median and range for the 2007 data? c. I'm writing a program that calculates the mode, mean and median of an array of integers. By using this site, I'm trying to find the median of each array. To find the median of any set of numbers, put them in order from smallest to greatest. The middle of this data set is Before you can begin to understand statistics, you need to understand mean, median, and mode. I love the program, and I can The median is one of the three main measures of central tendency, which is commonly used in statistics for finding the center of a data sample or population, e. If we deal with an array that has even number of elements, we should take two from the middle of a sorted one and find the average of those two values, this \u2026This is the C program to find the mean,median,mode and range of an array Mean, median, and mode are three kinds of \u201caverages\u201d. (note: if there are two numbers in \u20262/23/2012\u00a0\u00b7 help using arrays and calculating mean median and mode I have a program that compiles and works, I just have a few problems with it. Unfortunately they are not a part of C standard Library or C++ STL. For a even number of values, the median it the average of the two middle values. 'c' programming is developed in 1972 by dennis ritchie at \"at & t bell's lab usa\". For your first example, we line them up: $3$,$6$,$6$. Here is program coded in C++ that demonstrates how to find the mode in a given data set using a particular method. Don't sort in the median function. C programming language Tutorials list . Ask Question 1. 15. the user can input the Turbo C / C++ - Calculate Mean, Median, Mode. Estimation of the overdispersion parameter, c, for the global model is one of the key issues in applying Program MARK to encounter data. Calculating Mean is very important in day-to-day life. W. I was implementing quicksort and I wished to set the pivot to be the median or three numbers. Mean: it is the average of a number of elements in a set of values. std::vector<double> v; int temp, len; std::cout << \"Enter the number of elements \"; std::cin >> len; std::cin >> temp; Program to find median of an array in C /** C program to calculate the median of * an array. To calculate area we need at least two sides and the angle [\u2026]Median filter You are encouraged to solve this task according to the task description, using any language you may know. N], Find the Medians in this array without sorting it in an efficient Way? C++ programming Write a program to find the median of three input integers. 4) a Output : Enter the num of elements: 5 Enter the numbers 20 10 30 60 50 The numbers arranged in ascending order are given below 10 20 30 50 60 Median is : 30 C programming, exercises, solution: Write a program in C to find the median of two sorted arrays of same size. Next tutorial. Median is the middle element of the sorted array i. Otherwise, it is defined as the middle element of the array. Find the median of n numbers of an array using c program. Note the cumulative frequency just more than . A. C program to calculate median of an array. R Hoare and is closely related to Quicksort another of his mind boggling algorithms. Mar 1. Find median of an array in c# Saturday, April 06, 2013 Muthu Nadar No comments public static int GetMedian(int[] Value) { decimal Median = 0; int size = Value. Write a C++ program to find the median element in an array of complex numbers in terms of their magnitudes. Below is the program in the C programming language to find the average of all the elements of the array \u2013 This is a generic function for which methods can be written. Bubble sort is a sorting technique in which each pair of adjacent elements are compared, if they are in wrong order we swap them. To calculate arithmetic mean of numbers in c programming, you have to ask to the user to enter number size then ask to enter the numbers of that size to perform addition, then make a variable responsible for average and place addition/size in \u2026Find Median in an Unsorted Array Without Sorting it. The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements. Given two sorted arrays of size n. java) Efficient algorithm to find the median of an unsorted array? and I would like to use the program/thread stack which means fast allocations and cache locality Question :-C program to find the middle node of a linked listThe simple approach to reach up to middle node we have to know how many nodes are there in linked list and then traverse up to half of that node count. Mode) are the mathematical terms of Statistic. c) Write a program that tests your mastery of the concept of static arraynot dynamic (heap). Make a C program to find the average of three numbers. 10. Uses Divide and Conquer strategy. Sample of the C program. c program to find mean, median, and mode for different cases. I see a number of things that may help you improve your program. Just thought i would comment and say neat design, did you code it yourself? Looks great. This is the code I have so far for Median:Finding Mode in C++ Home. 34. c implements Torben's method to find a median in an input set without modifying it. C++ Program to find median of two sorted arraysC Programming (Turbo C++ Compiler) - Calculate Median. kastatic. Goal: Ask for user input with C# Console and use that input to compute an average. util. This is found by adding the numbers in a data set and dividing by how many numbers there are. Program in c to print 1 to 100 without using loop 13. Length; int mid = size / 2; Median = (size % 2 != 0) ? This cpp program code will be opened in a new pop up window once you click pop-up from the right corner. 6/30/2012\u00a0\u00b7 2 responses to \u201c Mean, Median and Mode c++ \u201d Facebook Chat. The C++ program is successfully compiled and run on a Linux system. 1 in steps of 0. Your program will fill a 1000 element integer array with either random numbers (between 1 and 1000) or the numbers from 1000 to 1 (in the decreasing order) based on user choice; and then proceed to find some statistics on these numbers. It represents the data in nice tabular form. 52 Can anyone help me how to program in C++ which determines the median of five input numbers? the median is the middle number when the five numbers are arranged in order. of comparisons to find median of 3 numbers. I learn by reading through the codes, can you please post the codes Thanks a lot. Ask Question 14. P: n/a Bhadan. The median is a measure of central tendency. Find ,where . In practice, median-finding algorithms are implemented with randomized algorithms that have an expected linear running time. Find The Median Value Of A Set Of Numbers. c program to find the area & perimeter of a circle, triangle, square and rectangle c program to calculate age in years, months and days c program to find absolute value of a given number Median chat. C program to find the lcm and hcf of given numbers C program to convert a binary number to decimal,oc C program to find the mean,median and mode of a se C++ Program to Find the Median of two Sorted Arrays using Binary Search Approach. length); Arrays. 5 2 Given an unsorted Array A[1,2,3,. View All Articles C Code for Mean, Median, Mode. The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. How would you find the mode? That is a start on how to get it. I'm trying to find the How to Find median in Arrays: For getting the median of input array. M = median(A,vecdim) computes the median based on the dimensions specified in the vector vecdim. Hello, I was wondering how to make it so that my code can handle 5 numbers instead of 3. 0. I have given the simple C program for calculating median for a set of numbers. )for values less than 100 Median The middle number of a group of numbers. 01 Program using do-while loop that will calculate the sum of every third integer beginning with i=2 sum = 2+5+8+11+. c implements the fast networks for 3, 5, 7, 9 or 25 elements, particularly useful for morphological filters in image processing. Here in this c program, we need to find out mean variance and standard deviation of n numbers, for that we need to know what is meant by mean, standard deviation and variance. util. A. To find the mean you add all the numbers together then divide by how many there are. The corresponding value of the variable is the Median. The Median is the \"middle number\" (in a sorted list of numbers). The median is the middle value in a list of numbers, it is the number separating the higher half of a data sample or a list of numbers from the lower half. C Program to Check Whether a Given Number is Odd o # C Program to Convert Number of Days Into Number C Program To Find A Character Is Number, Alphabet, Operator, or Special Character C Program To Find Reverse Case For Any Alphhabet using ctype functions C Program To Find Number Of Vowels In Input String The median value need not be in the array. Stare: rezolvat\u0103R\u0103spunsuri: 3C Program to Find Mean, Median, and Mode of Given Numbers. R Hoare and is closely related to Quicksort, another of his mind-boggling algorithms and the one he \u2026This program takes max numbers from user and calculates the sum of all the numbers in a loop and the final\u2026 Hello everyone, here we will learn a simple logic to find average on N numbers in python. 01 to 0. Aug 11, 2017 Page 1 of 2. By Chaitanya Singh C Program to reverse a given number using Recursive function; Find an answer to your question C program to find mean median and standard deviation The tutorial shows how to calculate mean, median and mode in Excel with formula examples. Median filter A blog for beginners. h> void main() { int x[100],n,i; float mean(int,int[]); float median(int,int[]); clrscr(); scanf(\"%d\"\u00a0Given n size unsorted array, find its mean and median. C program to find median of two sorted arrays of different sizes in logarithmic time. 5+10+6+6+5+12+5=49 49/7=7 Mean=7 To find the median you put the numbers from leas \u2026 t \u2026C Program To Calculate Median. In this C program, we are merging two one dimensional array in third array, where size of first and second arrays are different and the size of third array will be size of first array + second array. Each is used to find the statistical midpoint in a group of numbers, but they all do so differently. If you're behind a web filter, please make sure that the domains *. Write a C++ program to find the median element in an array of \u2026Write a C programming to find the median of the two given sorted arrays which are not empty. Don't violate const. Uses elimination in order to cut down2/11/2015\u00a0\u00b7 Program for Finding Median of 5 Numbers . Learn C programming, Data Structures tutorials, exercises, examples, programs, hacks, tips and tricks online. To apply the program in real world you should analyze it more carefully and find some applications if possible. sort(b); if (A Calculate median and quartile in C without sorting the array. You have probably heard of Quicksort but what about Quick Median? A friend of mine was writing a program to find the median. Find the (a) mean, (b) median, (c) mode, and (d) midrange for the given sample data. To understand this example, you should have the knowledge of following C programming topics: how can i calculate mean,median,mode by using c program. We use cookies to ensure you have the best browsing experience on our website. Getting started with C or C++ | C Tutorial Since it \"is not always true\", I wouldn't use it for your program. h> #define MAXSIZE 10 void main () 8/20/2017\u00a0\u00b7 (Mean. the mode method isn't working, and the median doesn't work either. Make a C program to find the average of three numbers. h Please help me to create a c program to find mode and median of an array. Everything compiles, but I can seem to figure out the few things C program to find the lcm and hcf of given numbers C program to convert a binary number to decimal,oc C program to find the mean,median and mode of a se C++ Program to Find the Median of two Sorted Arrays using Binary Search Approach Posted on July 1, 2017 by Manish This is a C++ program to find the median of two sorted arrays using binary search approach. The generalized version of this problem is known as \"n-order statistics\" which means finding an element K in a set such that we have n elements smaller or equal to K and rest are larger or equal K. torben. int a[25], n, i ; Write a C++ program to find the median element in an array of complex numbers in terms of thei 1. 10/11/2016 0 Comments Calculate Mode / Mean And Median Of Array Of Integers. java)10/7/2008\u00a0\u00b7 How do you find the in Median C++? This is what I have so far, any help would be much appreciated. Now I have everything working but the mode. If, instead, there are three more rainy days per month in the year 2007, what will be the mean, mode, median and range for the 2007 data? ANSWERS. e. C programs with their output. Follow. You need to sort the array in order to find the How to Write a Program in Java to Calculate the Mean. To find the median of any set of numbers, put them in order from smallest to greatest. C Program to Find Maximum Element in Array - This program find maximum or largest element present in an array. peters@acf. I would like a hint but nothing overboard. I have already compiled the program that reads numbers. Write a c program to find out NCR factor of given number. Source Code Write a program to find the median value of a vector of floating-point numbers. It's quick & easy. A C++ Function to Find Median of a Data Array Glancing over a book on Statistics, I realized that I need some implementation to find the median of an array, and I like to implement it by myself. Find the range and calculate standard deviation to compare and evaluate variability of data sets. Median filter Original Array of Complex Numbers The Median Element of the Array is: 5+j5 Show transcribed image text 1. 6/10/2013\u00a0\u00b7 C program to find mean, median, variance and standard deviation Write a C program to find mean, median, variance and standard deviation for given set of numbers. The median is the value separating the higher half of a data sample, a population, or a probability distribution, from the lower half. Note: This C Program for Finding Average of N Numbers is compiled with GNU GCC Compiler and written in gEdit Editor in Linux Ubuntu Operating System. /* this program is going to find the mean of 3 numbers. C Program to find sum of array elements using pointers, recursion & functions. C programming language, C program, C language 2 responses to \u201c Mean, Median and Mode c++ \u201d Facebook Chat. h> C Program to calculate Net Salary from basic salary (easy one) C program to find median ; No comments yet; Categories. Program for Mean and median of an unsorted array Given n size unsorted array, find its mean and median. C Code: #include <stdio. , \"Date\") for which a median is a reasonable concept. This program should accept array of integers and integers that indicate the number of elements in the array. calculate the range and the median for both intelligence and school performance d. It also shows you the cumulative frequency. This c programming code is used to find the mean, median, mode. Next the program will call a function that calculates the mean of the number of units. Find the median of the elements after inputting each element. Find The Median Value Of A Set Of Numbers. I have tried to figure it out on my own for almost 2 weeks now, I have Fast median search: an ANSI C implementation Nicolas Devillard - ndevilla AT free DOT fr July 1998 1 Introduction on big arrays or in any program for which strong reliability is a keyword. The median is the number in the middle of a set of numbers. May. Practice: Estimating mean and median in data displays. Mode Program In C. C Program To Find Median Of Array - The best free software for your. Enter values separated by commas such as 1, 2, 4, 7, 7, 10, 2, 4, 5. : Note that any algorithm that can find the median in O(n) time can also be used to find the k-th smallest element in the array, for any k between 1 and n, in O(n) time. Mean, or mean average, is used along with many other mathematical operations and is an important thing to know. Then, for each window - the set of pixels - find the median value. sort(b); if (A Find more on Program to find average of two numbers Or get search suggestion and latest updates. Arrays (array. c # list<int> to find the Mode and median please verify my account \u00b7 Check the link given below https://blogs. I've finished it mostly I think, except for the median part. Next, divide the number of elements by two and round down to get an integer. The nightmare that is using JavaScript on HackerRank Thanks to this guy for some help with heap Calculator Use. It will also print the larger values and smaller values compared to the median. Mail Stop 5425 Washington, D. To compare 3-channel pixels we first convert them into 1-channel gray values. Write an algorithm to find the median of the array obtained after merging the above 2 \u20262/27/2007\u00a0\u00b7 The median of five input numbers!!!. Program to find median of elements in an array. (c) Do the data provide any evidence that CEOs receive a premium (i. h> main() How to Find median in Arrays: For getting the median of input array. Write a C program to find the median and mode of an array of integers. This c programming code is used to find the mean, median, mode. calculate mean median mode c programming array. It is possible to calculate median in O(n) time instead. N], Find the Medians in this array without sorting it in an efficient \u2026C program to find the mean,median and mode of a set of numbers #include<stdio. and either repeating the program or quitting. 11/29/2011\u00a0\u00b7 Please help me to create a c program to find mode and median of an array. C. Program to find median of an array in C /** C program to calculate the median of * an array. length]; System. The statistics median is a quick measure to find a central location of a data sequence, list or any iterator. g. Example Code to Calculate Median To find the median of an unsorted array, we can make a min-heap in $O(n\\log n)$ time for $n$ elements, and then we can extract one by one $n/2$ elements to get the C Program to Calculate Standard Deviation This program calculates the standard deviation of 10 data using arrays. Quick Median - A Partition. h> #include <stdlib. ValueIs there a way to get the min, max, median, and average of a list of numbers in a single command? Ask Question 90. arraycopy(A, 0, b, 0, b. No angles of Scalene Triangle are equal. The C program is successfully compiled and run on a Linux system. The parametric bootstrap goodness-of-fit procedure was an attempt to develop a general procedure, but was found to be biased for Cormack-Jolly-Seber (CJS) data . The other related links are, Calculate Range C++ :: Calculate Mode / Mean And Median Of Array Of Integers Mar 17, 2013. for calculating a typical salary, household income, home price, real-estate tax, etc. For the mode, on paper you have a list. C++ program to find median of binary search tree store the elemnts of tree in an array in sorted manner we do this by traversing the tree in inorder manner * Then we find the median by using nth_element */ template < typename T > struct Value { T value; //Data of node Value() C program code examples; C program to find average of all elements of array; C program to find average of all elements of array. Python Exercises, Practice and Solution: Write a Python program to find the median of three values. Mean of an array = (sum of all elements) / (number of elements) Median of a sorted array of size n is defined as below : It is middle element when n is odd and average of \u2026To find the mean you add all the numbers together then divide by how many there are. LeetCode \u2013 Find Median from Data Stream (Java) Median is the middle value in an ordered integer list. For example, if there are five elements, 5/2 = 2 (integer arithmetic) so the median is dArray[2]. 'c' programming is developed in 1972 by dennis ritchie at \"at & t bell's lab usa\". Write a program to enter any three numbers and find out the middle number. The median is the middle value, which has exactly half of the values above it and the other half below it. import java. Program to find the mean, median, and mode of numbers. by Koscica Dusko. h> \u2190 c program \u2026The MEDIAN function sorts through the provided arguments to find the value that falls arithmetically in the middle of the group. This program takes max numbers from user and calculates the sum of all the numbers in a loop and the final\u2026 Hello everyone, here we will learn a simple logic to find average on N numbers in python. Need help? Post your question and get tips & solutions from a community of 424,419 IT Pros & Developers. C Source Code/Find the median and mean. Posted by: hirdesh on: March 24, 2010. Readline. Write a function to find the area of a triangle whose length of three sides is given 2 Answers Please list all the unary and binary operators in C. , \"Date\") for which a median is a reasonable concept. To calculate arithmetic mean of numbers in c programming, you have to ask to the user to enter number size then ask to enter the numbers of that size to perform addition, then make a variable responsible for average and place addition/size in average, then display the result Finding the Median in an Array. C program to find the mean,median and mode of a set of numbers #include<stdio. Is there a way to get the min, max, median, and average of a list of numbers in a single command? I actually keep a little awk program around to give the sum C Program to Find Area of Scalene Triangle : [crayon-5c78b459640be149497536/] Output : [crayon-5c78b459640c8343886373/] C Program for Beginners : Area of Scalene Triangle Properties of Scalene Triangle : Scalene Triangle does not have sides having equal length. The other related linkes are, Calculate Mean Calculate Median Calculate Mode Calculate Range Calculate Variance Calculate Standard Deviation. 5+10+6+6+5+12+5=49 49/7=7 Mean=7 To find the median you put the numbers from leas \u2026 t to greatest the Turbo C / C++ - Calculate Mean, Median, Mode. Hello, I have several jagged arrays which have been sorted. ReadLine takes in a string, and so it must be typecasted in order to store it in a different variable type. More on mean and median. Split number into digits in c programming 16. (in C you would have to do declare variables beforehand). Re-read the wiki page carefully and then recode. The mode of this value set is 3 as it appears more than any other number. Please read \u2026The simple example of how to find the average of any given set of numbers is a great way to practice! First lets go over the basics: You can easily store your Console. This cpp program code will be opened in a new pop up window once you click pop-up from the right corner. Mean of an array = (sum of all elements) / (number of elements) Median of a sorted array of size n is defined as below : It is middle element when n is odd and average of \u2026C Program to Find Area of Scalene Triangle : [crayon-5c8f1b35a73c9676398247/] Output : [crayon-5c8f1b35a73d3263894019/] C Program for Beginners : Area of Scalene Triangle Properties of Scalene Triangle : Scalene Triangle does not have sides having equal length. The median of a single sorted array is trivial to find and is O(1) constant time. So our numbers in order from least to greatest would be 94, 101, 101, 104, 108, 110. Find median of an array in c# Saturday, April 06, 2013 Muthu Nadar No comments public static int GetMedian(int[] Value) { decimal Median = 0; int size = Value", "date": "2019-05-26 11:59:13", "meta": {"domain": "kameographics.com", "url": "http://kameographics.com/dv6fo2t/c-program-to-find-median.php", "openwebmath_score": 0.18880972266197205, "openwebmath_perplexity": 660.3949772343873, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9643214491222695, "lm_q2_score": 0.9136765222384493, "lm_q1q2_score": 0.8810778679539769}}
{"url": "https://math.stackexchange.com/questions/2041254/find-all-solutions-to-x2-equiv-1-pmod-91", "text": "# Find all solutions to $x^2\\equiv 1\\pmod {91}$\n\nI split this into $x^2\\equiv 1\\pmod {7}$ and $x^2\\equiv 1\\pmod {13}$.\n\nFor $x^2\\equiv 1\\pmod {7}$, i did: $$(\\pm1 )^2\\equiv 1\\pmod{7}$$ $$(\\pm2 )^2\\equiv 4\\pmod{7}$$ $$(\\pm3 )^2\\equiv 2\\pmod{7}$$ Which shows that the solutions to $x^2\\equiv 1\\pmod {7}$ are $\\pm1$.\n\nFor $x^2\\equiv 1\\pmod {13}$, i did: $$(\\pm1 )^2\\equiv 1\\pmod{13}$$ $$(\\pm2 )^2\\equiv 4\\pmod{13}$$ $$(\\pm3 )^2\\equiv 9\\pmod{13}$$ $$(\\pm4 )^2\\equiv 3\\pmod{13}$$ $$(\\pm5 )^2\\equiv {-1}\\pmod{13}$$ $$(\\pm6 )^2\\equiv 10\\pmod{13}$$Which shows that the solutions to $x^2\\equiv 1\\pmod {13}$ are $\\pm1$.\n\nThus, I concluded that the solutions to $x^2\\equiv 1\\pmod {91}$ must be $\\pm1$. I thought that $\\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence?\n\n\u2022 Hint: CRT. See link \u2013\u00a0user160069 Dec 3 '16 at 2:38\n\nYou have $x\\equiv\\pm1\\mod7$ and $x\\equiv\\pm1\\mod13$. For all the solutions, you have to consider the systems: $$x\\equiv1\\mod7$$ $$x\\equiv1\\mod13$$ and $$x\\equiv-1\\mod7$$ $$x\\equiv1\\mod13$$ and $$x\\equiv1\\mod7$$ $$x\\equiv-1\\mod13$$ and $$x\\equiv-1\\mod7$$ $$x\\equiv-1\\mod13$$ as each system will get you a valid answer. I think you only had the first and the last systems and that you only considered the cases where the signs were similar.\n\n\u2022 Thank you! I figured out the other solutions to be $\\pm 27$. \u2013\u00a0J. Doe Dec 3 '16 at 3:40\n\nHint: Consider the possibility that $x \\equiv 1 \\pmod 7$ but $x \\equiv -1 \\pmod {13}$, and so on. (In other words, your mistake was to assume that the $\\pm1$ modulo $7$ was the same sign as $\\pm1$ modulo $13$.)\n\nAlso note that for any prime $p$, if $x^2 \\equiv 1 \\pmod p$, then we can rewrite this as $$x^2 - 1 \\equiv (x+1)(x-1) \\equiv 0 \\pmod p.$$\n\nThus we get $x \\equiv \\pm 1 \\pmod p$, showing that it isn't necessary to run through all the values of $x^2$ to find the solution.\n\nBy CRT the solutions $$\\,x\\equiv \\pm1\\pmod 7,\\ x\\equiv \\pm1\\pmod{13}$$ combine to $$\\,4\\,$$ solutions mod $$91,\\,$$ viz $$\\,x\\equiv (\\color{#c00}{{\\bf 1,1}}),\\,(\\color{#0a0}{-1,-1}),\\,(1,-1),\\,(-1,1)\\,$$ mod $$(7,13),$$ cf. Remark below. The first $$2$$ have obvious solutions $$\\,\\color{#c00}{\\bf 1}\\,$$ and $$\\,\\color{#0a0}{-1}\\,$$ so you need only solve the $$(1,-1)$$ case, then note $$(-1,1)\\equiv -(1,-1)$$ is its negative, i.e. $$\\,x\\equiv 1\\pmod{\\!7},\\,x\\equiv -1\\pmod{\\!13}\\,\\Rightarrow\\,-x\\equiv -1\\pmod{\\!7},\\ {-}x\\equiv 1\\pmod{\\!13}$$\n\nRemark  For more complex examples it is usually easier to solve the CRT system first for generic (symbolic) roots, then plug in the specific root values for all combinations, e.g. see here and here.\n\nIf $$\\,m,n\\,$$ are coprime then, by CRT, solving a polynomial $$\\,f(x)\\equiv 0\\pmod{\\!mn}\\,$$ is equivalent to solving $$\\,f(x)\\equiv 0\\,$$ mod $$\\,m\\,$$ and mod $$\\,n.\\,$$ By CRT, each combination of a root $$\\,r_i\\bmod m\\,$$ and a root $$\\,s_j\\bmod n\\,$$ corresponds to a unique root $$\\,t_{ij}\\bmod mn,\\,$$ i.e.\n\n$$\\begin{eqnarray} f(x)\\equiv 0\\!\\!\\!\\pmod{\\!mn}&\\overset{\\,\\,\\rm CRT}\\iff& \\begin{array}{}f(x)\\equiv 0\\pmod{\\! m}\\\\f(x)\\equiv 0\\pmod{\\! n}\\end{array} \\\\ &\\,\\,\\iff& \\begin{array}{}x\\equiv r_1,\\ldots,r_k\\pmod{\\! m}\\phantom{I^{I^{I^I}}}\\\\x\\equiv s_1,\\ldots,s_\\ell\\pmod{\\! n}\\end{array}\\\\ &\\,\\,\\iff& \\left\\{ \\begin{array}{}x\\equiv r_i\\pmod{\\! m}\\\\x\\equiv s_j\\pmod {\\! n}\\end{array} \\right\\}_{\\begin{array}{}1\\le i\\le k\\\\ 1\\le j\\le\\ell\\end{array}}^{\\phantom{I^{I^{I^I}}}}\\\\ &\\overset{\\,\\,\\rm CRT}\\iff& \\left\\{ x\\equiv t_{i j}\\!\\!\\pmod{\\!mn} \\right\\}_{\\begin{array}{}1\\le i\\le k\\\\ 1\\le j\\le\\ell\\end{array}}\\\\ \\end{eqnarray}\\qquad\\qquad$$", "date": "2019-10-18 11:35:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2041254/find-all-solutions-to-x2-equiv-1-pmod-91", "openwebmath_score": 0.968849241733551, "openwebmath_perplexity": 134.50381749167894, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462226131666, "lm_q2_score": 0.8918110368115781, "lm_q1q2_score": 0.8810613451027303}}
{"url": "https://www.physicsforums.com/threads/indefinite-integration.632544/", "text": "# Indefinite Integration\n\n## Homework Statement\n\nThe value of $$\\int_0^{1} (\\prod_{r=1}^{n} (x+r))(\\sum_{k=1}^{n} \\frac{1}{x+k}) dx$$ equals:\na)n\nb)n!\nc)(n+1)!\nd)n.n!\n\n(Can someone tell me how to make bigger parentheses using latex?)\n\n## The Attempt at a Solution\n\nI know that the question becomes a lot easier if i put n=1 or 2 and then integrate. But i was wondering if there is any proper way to solve it. I can't go further after expanding the given expression.\n$$\\int_{0}^{1} (x+1)(x+2)......(x+n)(\\frac{1}{x+1}+\\frac{1}{x+2}+.....\\frac{1}{x+n})dx$$\nwhich is equal to\n$$\\int_{0}^{1} \\sum_{r=1}^n \\frac{(x+n)!}{x+r}$$\nI am stuck now, i can't find any way further.\n\nAny help is appreciated.\n\nI still don't have any clue.\n\nehild\nHomework Helper\nWhat is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ?\n\nehild\n\nWhat is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ?\n\nehild\n\nDerivative of (x+1)(x+2)=2x+3\nDerivative of (x+1)(x+2)(x+3)=3x2+12x+11\n\nI still don't get any idea how this would help?\n\nehild\nHomework Helper\nDo not simplify.\n\nd/dx[(x+1)(x+2)]=(x+2)+(x+1)=(x+1)(x+2)/(x+1)+(x+1)(x+2)/(x+2)=[(x+1)(x+2)][1/(x+1)+1/(x+2)]\n\nd/dx[(x+1)(x+2)(x+3)]=(x+2)(x+3)+(x+1)(x+3)+(x+1)(x+2)=[(x+1)(x+2)(x+3)][1/(x+1)+1/(x+2)+1(x+3)]\n\nDo you see?\n\nehild\n\n1 person\nWait, i guess i have got it, i will be back in a few hours with a solution.\nThanks for the help!\n\nEDIT: Oops, seems like you posted just a few seconds before me.\n\nehild\nHomework Helper\nWait, i guess i have got it, i will be back in a few hours with a solution.\nThanks for the help!\n\nEDIT: Oops, seems like you posted just a few seconds before me.\n\nWell, I looking forward to the solution\n\nehild\n\nSo here's the solution:\nLet\n$$t=(x+1)(x+2)(x+3).......(x+n)$$\n$$\\ln t=\\ln(x+1)+\\ln(x+2)+\\ln(x+3)......+\\ln(x+n)$$\n$$\\frac{1}{t}\\frac{dt}{dx}=\\frac{1}{x+1}+\\frac{1}{x+2}+\\frac{1}{x+3}......\\frac{1}{x+n}$$\n$$dt=(x+1)(x+2)(x+3)....(x+n)(\\frac{1}{x+1}+\\frac{1}{x+2}+\\frac{1}{x+3}........+\\frac{1}{x+n})dx$$\nTherefore the original question becomes \u222bdt, upper limit is (n+1).n! and lower limit is n!.\nSolving, i get the answer as d) option.\n\nThanks for the help!\n\nMentallic\nHomework Helper\nBy the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that.\n\nI myself use /left( and /right) which looks like: $$\\left(\\frac{1}{1}\\right)$$ but there are other more specific \"fonts\" you can use, which ranges from smallest to largest:\n\n/big( $$\\big(\\frac{1}{1}\\big)$$\n/Big( $$\\Big(\\frac{1}{1}\\Big)$$\n/bigg( $$\\bigg(\\frac{1}{1}\\bigg)$$\n/Bigg( $$\\Bigg(\\frac{1}{1}\\Bigg)$$\n\nAnd there could be others, but that should about cover it.\n\nehild\nHomework Helper\nSo here's the solution:\nLet\n$$t=(x+1)(x+2)(x+3).......(x+n)$$\n$$\\ln t=\\ln(x+1)+\\ln(x+2)+\\ln(x+3)......+\\ln(x+n)$$\n\n$$\\frac{1}{t}\\frac{dt}{dx}=\\frac{1}{x+1}+\\frac{1}{x+2}+\\frac{1}{x+3}......\\frac{1}{x+n}$$\n$$dt=(x+1)(x+2)(x+3)....(x+n)(\\frac{1}{x+1}+\\frac{1}{x+2}+\\frac{1}{x+3}........+\\frac{1}{x+n})dx$$\n\nThat is ingenious! You are really cool\n\nTherefore the original question becomes \u222bdt, upper limit is (n+1).n! and lower limit is n!.\nSolving, i get the answer as d) option.\n\nThanks for the help!\n\nVery good. You are better and better every day!\n\nehild\n\nBy the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that.\n\nI myself use /left( and /right) which looks like: $$\\left(\\frac{1}{1}\\right)$$ but there are other more specific \"fonts\" you can use, which ranges from smallest to largest:\n\n/big( $$\\big(\\frac{1}{1}\\big)$$\n/Big( $$\\Big(\\frac{1}{1}\\Big)$$\n/bigg( $$\\bigg(\\frac{1}{1}\\bigg)$$\n/Bigg( $$\\Bigg(\\frac{1}{1}\\Bigg)$$\n\nAnd there could be others, but that should about cover it.", "date": "2021-05-07 18:42:29", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/indefinite-integration.632544/", "openwebmath_score": 0.8701503872871399, "openwebmath_perplexity": 1395.2873259196965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981453438742759, "lm_q2_score": 0.8976953016868439, "lm_q1q2_score": 0.8810461407837714}}
{"url": "https://math.stackexchange.com/questions/3364177/symmetric-matrix-as-a-sum-of-symmetric-matrices", "text": "# Symmetric matrix as a sum of symmetric matrices\n\nLet matrix $$M \\in \\mathbb{N}^{5 \\times 5}$$ be symmetric with non-negative integer entries and zeros on the main diagonal and having the property that the row sums are equal to $$2r$$ for some $$r \\geq 2$$. I want to prove that $$M$$ can be written as a non-negative integral linear combination of $$5 \\times 5$$ symmetric matrices having non-negative integer entries with zero entries on the main diagonal and having the property that the row sums are equal to $$2$$. Is there way to prove this?\n\nI tried with some simple examples and it seems to be correct.\n\n\u2022 I need to prove that any $M$ (with the property listed) can be written as a non-negative integral linear combination of matrices with the property listed. $m$ is not specific here. \u2013\u00a0jack Sep 21 at 9:28\n\u2022 I imagine this result is true for all dimensions, not just 5. \u2013\u00a0S. Dolan Sep 21 at 17:00\n\nIDEA:\n\nThe row sums of $$M$$ are even, so there is either 0 or 2 or 4 odd numbers, and the diagonal entries are all zeros.\n\nSo if you have even entries, for example if you have that in $$M$$ the entries $$m_{1,5}=m_{5,1}= 8$$ then you will have in your linear combination $$4$$ multiplied by a matrix having all its entries on these rows are $$0$$ except $$a_{1,5}=a_{5,1}=2$$, and so on ...\n\nAlso if there is $$2$$ odd entries, for example $$m_{1,5}=m_{5,1}=5$$ and $$m_{1,4}=m_{4,1}=7$$, then in your linear combination you'll have $$2$$ multiplied by a matrix having $$a_{1,5}=a_{5,1}=2$$ and all other entries on these rows are zeros, and $$3$$ multiplied by a matrix having $$b_{1,4}=b_{4,1}=2$$ and all other entries on these rows are zeros, and a matrix having $$c_{1,4}=c_{4,1}=c_{1,5}=c_{5,1}=1$$ and all other entries on these rows are zeros, and so on...\n\nI hope you can reach a proof from this hint and I am sorry because I am not able to write the whole proof now.\n\nYes, this conjecture is correct.\n\nFirst consider the following matrices.\n\n$$X_1=\\begin{pmatrix}0&0&1&1&0 \\\\0&0&1&0&1\\\\1&1&0&0&0\\\\1&0&0&0&1\\\\0&1&0&1&0\\\\\\end{pmatrix}$$, $$X_2=\\begin{pmatrix}0&0&1&0&1 \\\\0&0&1&1&0\\\\1&1&0&0&0\\\\0&1&0&0&1\\\\1&0&0&1&0\\\\\\end{pmatrix}$$, $$X_3=\\begin{pmatrix}0&0&1&0&1 \\\\0&0&0&2&0\\\\1&0&0&0&1\\\\0&2&0&0&0\\\\1&0&1&0&0\\\\\\end{pmatrix}.$$\n\nIf we could subtract one of the $$X_i$$ from $$M$$ without making any entry negative, then we would leave a matrix of the same type as $$M$$ but with $$r$$ reduced by 1. This process would therefore reduce $$M$$ to zero as we require.\n\nIf all off-diagonal entries of $$M$$ were non-zero we could subtract $$X_1$$. So, without loss of generality, we can suppose that $$M_{12}=0$$.\n\nCASE 1. If no row of $$M$$ has three or more zeroes.\n\nWithout loss of generality we can suppose that $$M$$ has the following form where $$A,B,C,D,E,F$$ are all non-zero.\n\n$$M=\\begin{pmatrix}0&0&A&B&C \\\\0&0&D&E&F\\\\A&D&0&.&.\\\\B&E&.&0&.\\\\C&F&.&.&0\\\\\\end{pmatrix}$$\n\nIf any of the dots were non-zero then a matrix like $$X_1$$ can be subtracted from $$M$$. Otherwise, all the dots are zero. Then $$A+D=B+E=C+F=2r. (*)$$ However, we would then have the sum of the first two rows equal to $$6r$$, a contradiction.\n\nCASE 2. If a row of $$M$$ has four zeroes.\n\nWithout loss of generality we can suppose that $$M$$ has the following form.\n\n$$M=\\begin{pmatrix}0&0&0&0&2r \\\\0&0&A&B&0\\\\0&A&0&C&0\\\\0&B&C&0&0\\\\2r&0&0&0&0\\\\\\end{pmatrix}$$\n\nThen simple algebra shows that $$A=B=C=r$$ and $$M$$ is the sum of $$r$$ copies of $$\\frac {1}{r} M$$.\n\nCASE 3. If $$M$$ has three zeroes in a row but that no row has four or more zeroes.\n\nLet $$M=\\begin{pmatrix}0&0&.&.&. \\\\0&0&.&.&.\\\\.&.&0&F&G\\\\.&.&F&0&H\\\\.&.&G&H&0\\\\\\end{pmatrix}$$\n\nCASE 3(a). If $$F=G=H=0$$\n\nThen the method used for $$(*)$$ gives a contradiction.\n\nCASE 3(b). If two of $$F,G,H$$ are $$0$$, w.l.g. $$F=G=0$$\n\nThen either $$X_1$$ or $$X_2$$ can be subtracted.\n\nCASE 3(c). If only one of $$F,G,H$$ is $$0$$, w.l.g $$F=0$$.\n\nSince every row has at least two non-zero elements, the possibilities for $$M$$ are, without loss of generality, as follows, where $$+$$ indicates a positive element.\n\n$$\\begin{pmatrix}0&0&+&0&+ \\\\0&0&0&+&+\\\\+&0&0&0&G\\\\0&+&0&0&H\\\\+&+&G&H&0\\\\\\end{pmatrix}$$ and $$\\begin{pmatrix}0&0&+&.&. \\\\0&0&+&.&.\\\\+&+&0&0&G\\\\.&.&0&0&H\\\\.&.&G&H&0\\\\\\end{pmatrix}$$\n\nConsidering row totals for Rows 3 and 4 of the LH matrix shows that $$M_{42}>1$$ and so $$X_3$$ can be subtracted. For the RH matrix, one of $$X_1, X_2$$ can be subtracted unless the matrix is $$\\begin{pmatrix}0&0&+&+&0 \\\\0&0&+&+&0\\\\+&+&0&0&G\\\\+&+&0&0&H\\\\0&0&G&H&0\\\\\\end{pmatrix}$$\n\nThen the sum of elements in Rows 1 and 2 must equal the sum of elements in Rows 3 and 4. Therefore $$G=H=0$$, a contradiction.\n\nCASE 3(d). If $$F,G,H$$ are all non-zero.\n\nThen a matrix like either $$X_1$$ or $$X_2$$ can be subtracted since the top two rows contain at least 4 non-zero elements.\n\n\u2022 With hindsight, CASE 1 can be subsumed into CASE 3(a) to shorten this proof a little. \u2013\u00a0S. Dolan Sep 21 at 16:41\n\nA graph theoretic proof\n\nConsider $$M$$ to be the adjacency matrix for a graph $$G$$. Then $$G$$ is a graph with no loops and 5 vertices, each of degree $$2r$$. We are required to prove that:\n\n$$G$$ contains one or more simple cycles (i.e. containing no repeated vertex) which include every point precisely once.\n\nSince all vertices have even degree, each component of $$G$$ must have an Eulerian cycle and therefore each vertex of $$G$$ is in a simple cycle and at least one of these simple cycles has length at least 3.\n\nWe can suppose every vertex of $$G$$ is joined to more than one vertex.\n\nSuppose on the contrary that vertex 1 was only joined to vertex 2. By the equality of degrees, vertices 1 and 2 then form a component of $$G$$. The points 3,4,5 of the other component must then be joined in pairs and $$G$$ has simple cycles (1,2) and (3,4,5).\n\nWe can suppose $$G$$ only has simple cycles of lengths 2 and 3\n\nThere is nothing to prove if $$G$$ has a simple cycle of length 5, so suppose $$G$$ has the cycle (1,2,3,4). If vertex 5 were connected to two adjacent points in this cycle, say vertices 1 and 2, then (1,5,2,3,4) would be a simple cycle. So we can suppose 5 is only joined to, say, 1 and 3. Then (1,5,3,4) and ((1,2,3,5) are cycles and so 2 and 4 are also only joined to 1 and 3. There are then $$6r$$ edges from vertices 2,4,5 to 1 and 3, contradicting the degrees of 1 and 3.\n\nNow consider a cycle of length 3, say (1,2,3). If vertex 4 were joined to two of these points then we would have a cycle of length 4. So vertex 4 is joined to vertex 5 and, say, vertex 1. Vertex 5 also has to be joined to precisely one of 1,2,3. This must be the same vertex 1, otherwise we would obtain a simple cycle of length 5. Vertices 4 and 5 can only be joined by $$1$$ edge (and similarly vertices 2 and 3 are only joined by $$1$$ edge) otherwise we would have simple cycles (1,2,3) and (4,5). Therefore there are $$4(2r-1)$$ edges from vertices 2,3,4,5 to vertex 1. Then $$8r-4=2r$$, which is impossible.\n\n\u2022 I don't think I follow your 'We are required to prove that'. Perhaps you meant something like: $G$ contains a collection of simple cycles such that each vertex belongs to exactly one of these cycles. Is that right? \u2013\u00a0Fimpellizieri Sep 22 at 21:04\n\u2022 Yes, that is correct. \u2013\u00a0S. Dolan Sep 22 at 21:06\n\u2022 I think your answer would benefit from better writing. I think you are trying to handle cases here rather than actually going 'without loss of generality' and it makes parsing the answer confusing. $G$ can very trivially have a simple cycle of length $5$, but I guess you're dismissing it as the easiest case for which what we need to prove is basically handed to us on a platter. \u2013\u00a0Fimpellizieri Sep 22 at 21:18\n\u2022 If G has a simple cycle of length 5 then, by definition, G contains a simple cycle which include every point precisely once. \u2013\u00a0S. Dolan Sep 22 at 21:24\n\u2022 Yes, that is my point. It's not 'Without loss of generality, we can suppose $G$ has no simple cycle of length $5$' and it most definitely isn't 'By definition, $G$ has no simple cycle of length $5$'. It's just that when $G$ has a simple cycle of length $5$, there is nothing to prove, so this case is done. Your answer is tackling cases, but is structured and written as something else, which makes it harder to parse. \u2013\u00a0Fimpellizieri Sep 22 at 21:28\n\nThis is my take on the graph theoretic approach by S. Dolan. Credits go to him.\n\nLet $$M$$ be the adjacency matrix of a graph $$G$$, so that $$G$$ is a graph on $$5$$ vertices with no loops and such that each vertex has degree $$2r$$.\n\nClaim: It is enough to prove that $$G$$ contains a collection of simple cycles such that each vertex belongs to exactly one of these cycles.\n\nIndeed, if $$H$$ is the $$G$$-subgraph that corresponds to this collection, then each vertex of $$H$$ has degree exactly $$2$$. We can thus take the adjacency matrix of $$H$$ as one of the matrices in our integral linear combination that adds up to $$M$$.\n\nWe are hence left with the same problem, except now the rows of $$M$$ add up to $$2(r-1)$$; equivalently, except the vertices of $$G$$ have degree $$2(r-1)$$. Induction then takes care of finding the other matrices in our integral linear combination.\n\nLet $$v_1,v_2,v_3,v_4$$ and $$v_5$$ be the vertices of $$G$$. We will prove the claim by studying different cases.\n\n## Case $$(1)$$: $$G$$ has a vertex which is joined to only one other vertex.\n\nAssume without loss of generality that $$v_1$$ is joined only to $$v_2$$. Since all vertices have equal degrees, $$v_2$$ must also be joined only to $$v_1$$.\n\nThis implies that if some $$u$$ in $$\\{v_3, v_4,v_5\\}$$ were also joined to only one other vertex $$w$$, we would have $$w \\in \\{v_3, v_4,v_5\\}\\setminus\\{u\\}$$. Equality of degrees would then require that the remaining vertex be joined to itself, in contradiction to $$G$$ having no loops. It follows that each of $$\\{v_3, v_4,v_5\\}$$ is connected to the other two.\n\nThe collection of simple cycles $$\\{(v_1,v_2), (v_3,v_4,v_5)\\}$$ satisfies the claim and handles this case.\n\n## Case $$(2)$$: Each vertex of $$G$$ is joined to at least two other vertices.\n\nBy Veblen's Theorem, $$G$$ can be written as the union of disjoint simple cycles. We break down into subcases.\n\n$$\\qquad (2.1)$$: $$G$$ has a simple cycle of length $$5$$.\n\nIn this case, the claim is obviously satisfied and there is nothing to prove.\n\n$$\\qquad (2.2)$$: $$G$$ has a simple cycle of length $$4$$.\n\nLet $$(v_1,v_2,v_3,v_4)$$ be the simple cycle.\n\n$$\\qquad\\qquad(2.2.1)$$: $$v_5$$ is joined to two vertices that are adjacent in the cycle.\n\nIf that were the case, then we could enlarge the simple cycle to have length $$5$$. This reduces the problem to case $$(2.1)$$, which we have already handled.\n\n$$\\qquad\\qquad(2.2.2)$$: No two vertices adjacent in the cycle are both joined to $$v_5$$.\n\nWithout loss of generality, suppose $$v_5$$ were joined to vertices $$v_1$$ and $$v_3$$. Notice that $$v_5$$ cannot be joined to any other vertex.\n\nIf $$v_2$$ and $$v_4$$ weree joined, then we could take the simple cycle $$(v_1,v_2,v_4,v_3,v_5)$$, handled by case $$(2.1)$$.\n\nIf $$v_2$$ and $$v_4$$ were not joined, then each of $$\\{v_2, v_4,v_5\\}$$ would be joined only to $$v_1$$ and $$v_3$$. This would contradict each vertex having equal degree, and finishes this case.\n\n$$\\qquad (2.3)$$: $$G$$ has a simple cycle of length $$3$$.\n\nLet $$(v_1,v_2,v_3)$$ be the simple cycle.\n\n$$\\qquad\\qquad(2.3.1)$$: One of $$\\{v_4, v_5\\}$$ is joined to two vertices in the cycle.\n\nIf that were the case, then we could enlarge the simple cycle to have length $$4$$. This reduces the problem to case $$(2.2)$$, which we have already handled.\n\n$$\\qquad\\qquad(2.3.1)$$: No two vertices in the cycle are both joined to one of $$\\{v_4, v_5\\}$$.\n\nWithout loss of generality, suppose $$v_4$$ were joined to vertices $$v_1$$ and $$v_5$$. Notice that $$v_4$$ cannot be joined to any other vertex.\n\nIf $$v_5$$ were joined to one of $$\\{v_2,v_3\\}$$, then we could enlarge the simple cycle to have length $$5$$, handled by case $$(2.1)$$. Indeed, if $$v_5$$ were joined to $$v_2$$ we would have the cycle $$(v_1,v_3, v_2,v_5,v_4)$$ and if $$v_5$$ were joined to $$v_3$$ we would have the cycle $$(v_1,v_2,v_3,v_5,v_4)$$.\n\nSuppose instead that $$v_5$$ were joined only to $$v_1$$ and $$v_4$$.\nIf there were more than edge joining $$v_4$$ and $$v_5$$, the collection $$\\{(v_1,v_2,v_3),(v_4,v_5)\\}$$ satisfies the claim.\nIf there were more than edge joining $$v_2$$ and $$v_3$$, the collection $$\\{(v_1,v_4,v_5),(v_2,v_3)\\}$$ satisfies the claim.\nIf $$v_4$$ were joined to $$v_5$$ by a single edge and $$v_2$$ were also joined to $$v_3$$ by a single edge, then $$v_1$$ would have degree $$4(2r - 1) > 2r$$, which contradicts our hypotheses and concludes this case.\n\n$$\\qquad (2.4)$$: $$G$$ has no simple cycle of length $$3$$ or more.\n\nWe show that this case is impossible.\n\n$$G$$ does not have loops, so in this case it would contain only cycles of length $$2$$. $$G$$ would then be bipartite: its vertices could be divided into two disjoint sets $$U$$ and $$V$$ such that every edge of $$G$$ joins a vertex in $$U$$ to a vertex in $$V$$. But $$G$$ has $$5r$$ edges and one of $$\\{U, V\\}$$ has at most two vertices. This contradicts each vertex having degree $$2r$$ and finishes the last case.", "date": "2019-11-22 13:45:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3364177/symmetric-matrix-as-a-sum-of-symmetric-matrices", "openwebmath_score": 0.858691394329071, "openwebmath_perplexity": 139.25253142353588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534333179649, "lm_q2_score": 0.8976952893703477, "lm_q1q2_score": 0.8810461238258918}}
{"url": "https://math.stackexchange.com/questions/3216724/number-of-ways-an-answers-sheet-can-be-fill-with-at-least-two-consecutive-answer", "text": "# Number of ways an answers sheet can be fill with at least two consecutive answers?\n\nA student is taking a $$12$$-question multiple choice test. Each question has $$5$$ choices. How many ways can the student fill out the answer sheet so there is at least one place where two consecutive answers are the same?\n\nThis is what I did:\n\nTotal possible ways the student can fill the answer sheet: $$5^{12}$$ (because he has $$5$$ options in the $$12$$ questions).\n\nTotal possible ways the student can fill the answer sheet with no consecutive answers: $$5 \\cdot 4^{11}$$ (In the first question he can answer any of the $$5$$ options, however, in the remaining questions he can only answer $$4$$ options to avoid choosing consecutive answers.)\n\nTotal possible ways the student can fill the answer sheet with at least two consecutive answers: $$5^{12}- 5 \\cdot 4^{11} = 223,169,105$$.\n\nI doubt about the total possible ways the student can fill the answer sheet with no consecutive answers... is my logic right?\n\n\u2022 Write out a detailed explanation of the $5\\times4^{11}$. This should either convince you that you are right, or show you where you are wrong. Do this for the first part of your answer too. \u2013\u00a0David May 7 '19 at 5:17\n\u2022 Sorry, I had a typo on my first statement. \u2013\u00a0Nico May 7 '19 at 5:33\n\u2022 Yes, you are correct. \u2013\u00a0N. F. Taussig May 7 '19 at 9:05\n\nYou are correct, $$5\\times 4^{11}$$ is the number of ways to fill out the test with no two consecutive equal answers, so $$5^{12}-5\\times 4^{11}$$ is the number of ways to fill out the test so that there are two consecutive equal answers.\n\nI am trying to do this question and am getting a different answer. Instead of making a new thread, I hope I am allowed to post my response and see where people think I'm going wrong here. If not please let me know and I will delete.\n\nDenote the answers of the multi-choice test to be A_1,A_2,..., A_12.\n\nNow, for us to have consecutive answers be the same it must be that A_1 = A_2 or A_2 = A_3 or .... or A_11 = A_12. Denote this pair of questions with the same answer to be A_s.\n\nNow, consider a new set which contains 11 elements: A_s, and our other 10 remaining answers to questions. Certainly, we can position A_s anywhere, as remember that we have set A_s to be the two consecutive answers already. Thus, we have no conditions on the remaining choices. We can re-arrange this set 11! different ways (as it is simply the re-arrangements of a set with 11 elements where order matters), and each element can have 5 different options.\n\nThus, we have 11!x(5 x 11) = 21,954,424,000 different options.\n\nMy answer seems way too big\n\n\u2022 You have not accounted for the possibility that there is more than one pair of consecutive answers that are the same. To do so, you could apply the Inclusion-Exclusion Principle, but with $11$ possible pairs of consecutive answers that could be the same, that could prove exceedingly difficult. \u2013\u00a0N. F. Taussig May 20 '19 at 0:02", "date": "2021-06-20 19:11:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3216724/number-of-ways-an-answers-sheet-can-be-fill-with-at-least-two-consecutive-answer", "openwebmath_score": 0.8068608641624451, "openwebmath_perplexity": 199.13291535094734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9814534382002796, "lm_q2_score": 0.8976952832120991, "lm_q1q2_score": 0.8810461221646885}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=osjucte5sjf5bgei03vl6fofd2&topic=1204.0;wap2", "text": "MAT334-2018F > MAT334--Lectures & Home Assignments\n\nAbout the definition of Argument (in book)\n\n(1/2) > >>\n\nEnde Jin:\nI found that the definition of \"arg\" and \"Arg\" in the book is different from that introduced in the lecture (exactly opposite) (on page 7).\nI remember in the lecture, the \"arg\" is the one always lies in $(-\\pi, \\pi]$\nWhich one should I use?\n\nVictor Ivrii:\n\n--- Quote ---Which one should I use?\n--- End quote ---\nThis is a good and tricky question because the answer is nuanced:\nSolving problems, use definition as in the Textbook, unless the problem under consideration requires modification: for example, if we are restricted to the right half-plane\u00a0 $\\{z\\colon \\Re z >0\\}$ then it is reasonable to consider $\\arg z\\in (-\\pi/2,\\pi/2)$, but if we are restricted to the upper half-plane\u00a0 $\\{z\\colon \\Im z >0\\}$ then it is reasonable to consider $\\arg z\\in (0,\\pi)$ and so on.\n\nEnde Jin:\nI am still confused. Let me rephrase the question again.\nIn the textbook, the definition of \"arg\" and \"Arg\" are:\n$arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = cos\\theta + isin\\theta$\nwhich means $arg(z) \\in \\mathbb{R}$\nwhile\n$Arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = cos\\theta + isin\\theta \\land \\theta \\in [-\\pi, \\pi)$\nwhich means $Arg(z) \\in [-\\pi, \\pi)$\n\nWhile in the lecture, as you have introduced, it is the opposite and the range changes to $(-\\pi, \\pi]$ instead of $[-\\pi, \\pi)$ (unless I remember incorrectly):\nArg is defined to be\n$Arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = (cos\\theta + isin\\theta)$\nwhich means $arg(z) \\in \\mathbb{R}$\nwhile arg is\n$arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = cos\\theta + isin\\theta \\land \\theta \\in (-\\pi, \\pi]$\n\nI am confused because if I am using the definition by the book,\nwhen $z \\in \\{z : Re (z) > 0\\}$\nthen $arg(z) \\in (-\\frac{\\pi}{2} + 2\\pi n,\\frac{\\pi}{2} + 2\\pi n), n \\in \\mathbb{Z}$\n\nVictor Ivrii:\n\nBTW, you need to write \\sin t and \\cos t and so on to have them displayed properly (upright and with a space after): $\\sin t$, $\\cos t$ and so on\n\nEnde Jin:\nThus in a test/quiz/exam, I should follow the convention of the textbook, right?", "date": "2021-10-21 01:56:30", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=osjucte5sjf5bgei03vl6fofd2&topic=1204.0;wap2", "openwebmath_score": 0.9694011807441711, "openwebmath_perplexity": 996.0465264079085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972414716174355, "lm_q2_score": 0.9059898279984214, "lm_q1q2_score": 0.8809978414499378}}
{"url": "https://www.jiskha.com/questions/848377/consider-the-differential-equation-dy-dt-y-t-2-a-show-that-the-constant-function-y1-t-0", "text": "Differential Equations\n\nConsider the differential equation: dy/dt=y/t^2\na) Show that the constant function y1(t)=0 is a solution.\nb)Show that there are infinitely many other functions that satisfy the differential equation, that agree with this solution when t<=0, but that are nonzero when t>0 [Hint: you need to define these functions using language like \" y(t)=...when t<=0 and y(t)=...when t>0 and \"]\nc) Why doesn't this example contradict the Uniqueness Theorem?\n\nI'm trying to do part b and after I separated and integrated I got\nln|y|=(-1/t)+C\nI'm not sure if I can get C with the solution they gave in part a)y1(t)=0.\nAnyways, I get y(t)=Ce^-(1/t). I don't know where to go from there.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. dy/y = dt/t^2\nln y = -1/t + c\ny = e^(-1/t+c) = e^c e^(-1/t)\n=C e^(-1/t) agree\n\nC can be anything so\nwhat if C = 0 ?\nthen y(t) = 0 for all t\n\nif C = 0 for t</= 0 then C can be anything at all for t>0\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n2. If y(t) depends on what C is, then how this equation doesn't contradict the uniqueness theorem if it has many solutions?\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. Because the general solution contains an an arbitrary constant C. The value of C depends on your boundary conditions, for example if y =5 at t = 2\nthen\n5 = C e^-(1/2)\n5 = C (.606)\nC = 8.24\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n4. scroll down through this:\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/diff.html\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\nSimilar Questions\n\n1. math help!!!\n\nstuck!! pls help/explain!!! 2. Do the values in the table represent a linear function? If so, what is the function rule? (1 point) x = -2, 0, 2, 4 y = -4, 0, 4, 8 The values do not show a linear function. Yes, they show a linear\n\n2. AP Calc Help!!\n\nLet f be the function satisfying f'(x)=x\u221a(f(x)) for all real numbers x, where f(3)=25. 1. Find f''(3). 2. Write an expression for y=f(x) by solving the differential equation dy/dx=x\u221ay with the initial condition f(3)=25. Please\n\n3. Calculus\n\nSuppose that we use Euler's method to approximate the solution to the differential equation \ud835\udc51\ud835\udc66/\ud835\udc51\ud835\udc65=\ud835\udc65^4/\ud835\udc66 \ud835\udc66(0.1)=1 Let \ud835\udc53(\ud835\udc65,\ud835\udc66)=\ud835\udc65^4/\ud835\udc66. We let \ud835\udc650=0.1 and \ud835\udc660=1 and pick a step size \u210e=0.2.\n\n4. Calc\n\nConsider the differential equation dy/dx=-2x/y Find the particular solution y =f(x) to the given differential equation with the initial condition f(1) = -1\n\n1. Physic\n\nShow that the function y(x,t) = X^2 + V^2t^2 is a solution of the general wave equation.\n\n2. Calculus\n\nfor what values of r does the function y=e^rx satisfy the differential equation y''-4y'+y=0 Show steps please! Thank you!\n\n3. Algebra\n\nCreate a system of equations that includes one linear equation and one quadratic equation. Part 1. Show all work to solving your system of equations algebraically. Part 2. Graph your system of equations, and show the solution\n\n4. Calculus\n\nConsider the differential equation dy/dx = x^4(y - 2). Find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 0. Is this y=e^(x^5/5)+4?\n\n1. Calculus\n\nVerify that the function y(x)= x - (1/x) is a solution to the differential equation, xy'+ y= 2x I have absolutely no clue what to do with that + y Please help :c\n\n2. Calculus BC\n\nLet y = f(x) be the solution to the differential equation dy/dx=y-x The point (5,1) is on the graph of the solution to this differential equation. What is the approximation of f(6) if Euler\u2019s Method is used given \u2206x = 0.5?\n\n3. Calculus!!\n\nConsider the differential equation given by dy/dx = xy/2. A. Let y=f(x) be the particular solution to the given differential equation with the initial condition. Based on the slope field, how does the value of f(0.2) compare to\n\n4. math\n\nConsider the differential equation dy/dx = -1 + (y^2/ x). Let y = g(x) be the particular solution to the differential equation dy/ dx = -1 + (y^2/ x) with initial condition g(4) = 2. Does g have a relative minimum, a relative", "date": "2021-10-23 08:50:14", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/848377/consider-the-differential-equation-dy-dt-y-t-2-a-show-that-the-constant-function-y1-t-0", "openwebmath_score": 0.8080105185508728, "openwebmath_perplexity": 1044.2009532301927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754470129648, "lm_q2_score": 0.8947894562828416, "lm_q1q2_score": 0.8809877289021665}}
{"url": "https://mathoverflow.net/questions/330435/automorphism-group-of-the-special-unitary-group-sun", "text": "# Automorphism group of the special unitary group $SU(N)$\n\nLet us consider the automorphism group of the special unitary group $$G=SU(N)$$.\n\nWe know there is an exact sequence: $$0 \\to \\text{Inn}(G) \\to \\text{Aut}(G) \\to \\text{Out}(G) \\to 0.$$\n\nFor $$G=SU(2)$$, we have:\n\n\u2022 $$\\text{Z}(SU(2)) =\\mathbb Z_2$$,\n\u2022 $$\\text{Inn}(SU(2)) = SO(3)$$,\n\u2022 $$\\text{Out}(SU(2)) = 0$$,\n\nAnd so $$\\text{Aut}(SU(2))=SO(3)$$.\n\nFor $$N > 2$$, we have:\n\n\u2022 $$\\text{Z}(SU(N)) =\\mathbb Z_{N}$$,\n\u2022 $$\\text{Inn}(SU(N)) = PSU(N)$$,\n\u2022 $$\\text{Out}(SU(N)) = \\mathbb Z_2$$.\n\nMy question is:\n\nDoes $$\\text{Aut}(SU(N))=PSU(N) \\times \\mathbb Z_2$$? If not, does this answer depend on whether $$N$$ is odd or even?\n\nIt looks to me that there is a nontrivial fibration depending on something like $$H^2(B\\mathbb Z_2,PSU(N))$$ due to $$B\\text{Inn}(G) \\to B\\text{Aut}(G) \\to B\\text{Out}(G) \\to B^2\\text{Inn}(G) \\to$$ and thus $$BPSU(N) \\to B\\text{Aut}(G) \\to B\\mathbb Z_2 \\to B^2PSU(N) \\to$$\n\nBut I do not know how to define $$H^2(B\\mathbb Z_2,PSU(N))$$, if this is a correct thing to ponder.\n\n\u2022 this may be preliminary for answering your question: mathoverflow.net/questions/40666/\u2026 - which you may already know well. \u2013\u00a0wonderich May 1 at 3:48\n\u2022 In fact this completely answers the question. \u2013\u00a0abx May 1 at 4:05\n\u2022 what is the answer? I am asking the total Aut of $SU(N)$? \u2013\u00a0annie heart May 1 at 4:46\n\u2022 p.s. I am not asking Inn or Out of $SU(N)$. \u2013\u00a0annie heart May 1 at 4:51\n\u2022 The answer is that it is a semi-direct product $\\operatorname{PSU} (N)\\rtimes \\mathbb{Z}/2$, with $\\mathbb{Z}/2$ acting on $\\operatorname{PSU}(N)$ by conjugation. \u2013\u00a0abx May 1 at 5:58\n\nLet $$G$$ be a compact simple simply connected Lie group. Then any automorphism of $$G$$ determines an automorphism of its Lie algebra $$\\mathfrak{g}$$ and visa versa. So $$\\mathrm{Aut}(G)$$ is naturally isomorphic to the linear group $$\\mathrm{Aut}(\\mathfrak{g})$$.\n\nThe sequence $$1\\to \\mathrm{Inn}(\\mathfrak{g})\\to \\mathrm{Aut}(\\mathfrak{g})\\to \\mathrm{Out}(\\mathfrak{g})\\to 1$$ is split.\n\nMoreover, $$\\mathrm{Out}(G)\\cong\\mathrm{Out}(\\mathfrak{g})\\cong \\mathrm{Aut}(D_\\mathfrak{g})$$ where $$D_\\mathfrak{g}$$ is the Dynkin diagram of $$\\mathfrak{g}$$.\n\nThe upshot is:\n\n$$\\mathrm{Aut}(G)\\cong \\mathrm{Aut}(\\mathfrak{g})\\cong \\mathrm{Inn}(\\mathfrak{g})\\rtimes \\mathrm{Out}(G)\\cong \\mathrm{Inn}(\\mathfrak{g})\\rtimes \\mathrm{Aut}(D_\\mathfrak{g}).$$\n\nFor types $$A_1, B_n, C_n, G_2, F_4, E_7, E_8$$ there are no symmetries of the Dynkin diagram. For $$A_n$$ ($$n>1$$), $$D_n$$ ($$n\\not=4$$), and $$E_6$$, we have $$\\mathrm{Aut}(D_\\mathfrak{g})\\cong \\mathbb{Z}/2\\mathbb{Z}$$. And in the final case of $$D_4$$, the symmetry group is the symmetric group on three letters.\n\nIn particular, as stated in the comments: $$\\mathrm{Aut}(\\mathrm{SU}(n))\\cong\\left\\{\\begin{array}{ll}\\mathrm{PSU}(n)\\rtimes \\mathbb{Z}/2\\mathbb{Z},&\\text{ if } n\\geq 3\\\\ \\mathrm{PU}(2),&\\text{ if }n=2. \\end{array}\\right.$$\n\n\u2022 PSU(2) = PU(2), yes? \u2013\u00a0annie heart May 6 at 19:27\n\u2022 Does it mean that the $\\mathrm{Aut}(\\mathrm{PU}(n))=\\mathrm{Aut}(\\mathrm{PSU}(n))=\\mathrm{Aut}(\\mathrm{SU}(n))$, since they have the same Lie algebra: $$\\mathrm{Aut}(\\mathrm{PSU}(n))\\cong\\left\\{\\begin{array}{ll}\\mathrm{PSU}(n)\\rtimes \\mathbb{Z}/2\\mathbb{Z},&\\text{ if } n\\geq 3\\\\ \\mathrm{PU}(2),&\\text{ if }n=2. \\end{array}\\right. ?$$ \u2013\u00a0annie heart May 6 at 19:35\n\u2022 Do we have $$\\mathrm{Out}(G)\\cong\\mathrm{Out}(\\mathfrak{g})?$$ $$\\mathrm{Inn}(G)\\cong\\mathrm{Inn}(\\mathfrak{g})?$$ $$\\mathrm{Aut}(G)\\cong\\mathrm{Aut}(\\mathfrak{g})?$$ in general? \u2013\u00a0annie heart May 6 at 19:43\n\u2022 $PU(n)\\cong PSU(n)$ in general. For simply-connected Lie groups, $\\mathrm{Aut}(G)\\cong \\mathrm{Aut}(\\mathfrak{g})$ and $\\mathrm{Out}(G)\\cong \\mathrm{Out}(\\mathfrak{g})$.The same argument does not work for $\\mathrm{Aut}(PG)$ since $PG$ is not generally simply-connected. The inner automorphisms of the Lie algebra are the image of $G$ via the adjoint. I think you need to generally figure out the kernel. \u2013\u00a0Sean Lawton May 6 at 21:08\n\u2022 I am not sure. I think it is true for simple, simply-connected compact Lie groups. \"Proof:\" Since $G$ is simple, the Lie algebra of $\\mathrm{Inn}(\\mathfrak{g})$ is $\\mathfrak{g}$ itself. Thus, since $G$ is simply-connected and compact, the abstract Lie group structure on $\\mathrm{Inn}(\\mathfrak{g})$ is a finite central quotient of $G$. Since the center is in the kernel, it must be $PG$. And $\\mathrm{Inn}(G)\\cong PG$ always. $\\Box$ \u2013\u00a0Sean Lawton May 6 at 21:17", "date": "2019-05-27 14:30:15", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/330435/automorphism-group-of-the-special-unitary-group-sun", "openwebmath_score": 0.9015768766403198, "openwebmath_perplexity": 246.4333233317566, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109081214212, "lm_q2_score": 0.8902942144788076, "lm_q1q2_score": 0.8809558366641722}}
{"url": "https://math.stackexchange.com/questions/2723055/basis-for-eigenspace-of-identity-matrix", "text": "# Basis for eigenspace of Identity Matrix\n\nLet $A=\\begin{pmatrix} 1&0\\\\ 0&1\\\\ \\end{pmatrix}$. Find the bases for the eigenspaces of the matrix $A$. I know the bases for the eigenspace corresponding to each eigenvector is a vector (or system) that can scale to give any other vector contained in that said eigenspace. Thus, we see that the identity matrix has only one distinct eigenvalue $\\lambda=1$. Thus the eigenvector satisfies the equation $(A-\\lambda I)\\vec{x}=\\vec{0}$. Then we must solve $\\begin{pmatrix} 0&0&0\\\\ 0&0&0\\\\ \\end{pmatrix}$. I say that any vector satisfies this equation however the key says that the basis is $\\lbrace (1,0),(0,1) \\rbrace$. How do they come up with this solution. Thanks in advance!\n\nIt is true that any vector satisfies the equation. So, our eigenspace is the whole space $\\mathbb{R}^2$. Since $\\{(1,0),(0,1)\\}$ is a basis for $\\mathbb{R}^2$, it is also a basis for this eigenspace.\n\n\u2022 Yes. This is precisely the explanation I needed. Thank you! \u2013\u00a0coreyman317 Apr 5 '18 at 8:50", "date": "2020-04-02 22:42:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2723055/basis-for-eigenspace-of-identity-matrix", "openwebmath_score": 0.9797941446304321, "openwebmath_perplexity": 96.27951875024905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9861513897844354, "lm_q2_score": 0.8933093989533708, "lm_q1q2_score": 0.8809383052853653}}
{"url": "https://math.stackexchange.com/questions/2467200/solve-the-recurrence-relation-tn-4tn-3-n-t3-1-n-3k-then-det", "text": "# Solve the recurrence relation $T(n)=4T(n/3) + n$, $T(3) = 1$, $n = 3^k$ then determine upper and lower bounds\n\nI would like to solve the following recurrence relation. After which I must find it's upper and lower bounds.\n\n$T(n)=4T(\\frac{n}{3})+n, T(3)=1, n = 3^k$\n\nI attempted to solve with substitution:\n\nk = 1: $T(n)=4T(\\frac{n}{3})+n$\nk = 2: $T(n)=4^2T(\\frac{n}{3^2})+\\frac{4n}{3}+n$\nk = 3: $T(n)=4^3T(\\frac{n}{3^3})+\\frac{4^2n}{3^2}+\\frac{4n}{3}+n$\n\nGeneral: $T(n)=4^kT(\\frac{n}{3^k})+\\sum_{i=1}^k(\\frac{4^{i-1}}{3^{i-1}})n=4^kT(\\frac{3^k}{3^k})+...=4^kT(1)+...$\n\nQ1: Here is where my first problem arrived; I was not given a relation for T(1) so I can not finish with substitution, right? Perhaps my professor made a typo and meant to put T(1) = 1?\n\nAssuming that he didn't make a typo, I decided to try and solve the problem using Master Theorem:\n\nMaster Theorem: $T(n)=aT(\\frac{n}{b})+n^d$ where $a,b > 0$ and $d\\ge0$ then\n\n$T(n)=\\Theta(n^d)$ if $d>\\log_b(a)$\n$T(n)=\\Theta(n^d\\log(n))$ if $d=\\log_b(a)$\n$T(n)=\\Theta(n^{\\log_b(a)})$ if $d<\\log_b(a)$\n\nUsing this method I get $T(n)=\\Theta(n^{1.262})$\n\nQ2: I read that $\\Theta(g(n))$ is both the upper bound $O(g(n))$ and the lower bound $\\Omega(g(n))$ so does that mean that the three answers to my question would be...\n\nSolution to recurrence relation: $T(n)=\\Theta(n^{1.262})$\nUpper bound: $O(n^{1.262})$\nLower bound: $\\Omega(n^{1.262})$\n\n\u2022 Actually, you have $T(3) = 1$ so you can just use that. \u2013\u00a0Qudit Oct 11 '17 at 7:00\n\u2022 The case in when $T(3) = 1$ has the overall term of $4^{k-1}T(\\frac{n}{3^{k-1}})$. When this occurs, and after the substitution $T(3) = 1$, the term becomes $4^{k-1}$. \u2013\u00a0user316270 Oct 11 '17 at 7:08\n\nQ1: Here is where my first problem arrived; I was not given a relation for T(1) so I can not finish with substitution, right? Perhaps my professor made a typo and meant to put T(1) = 1?\n\nNo, your professor was correct in providing $T(3) = 1$, given $n = 3^{k}$. As you may have noticed when using substitution method, there is generally a relationship in the power of the coefficient - in this case, $4$ - and the power of the denominator in $T()$. Realizing this, you can often quickly generalize and jump to the base condition. To explicitly illustrate:\n\n\\begin{align} T(n) &= 4^{1}T(\\frac{n}{3^{1}}) + n \\\\ &= 4^{2}T(\\frac{n}{3^{2}}) + 4n + n \\\\ &= 4^{3}T(\\frac{n}{3^{3}}) + 4^{2}n + 4n + n \\\\ & = \\ ... \\end{align}\n\nAt this point, look to the value within $T()$ in order to determine the remaining number of additional substitutions need to take place. For this problem, the base case is $T(3) = 1$, and,\n\n$$3 = \\frac{n}{3^{k-1}}$$\n\ntherefore, we get\n\n\\begin{align} T(n) &= 4^{k-1}T(\\frac{n}{3^{k-1}}) + 4^{k-2}n + ... + 4^{k - (k - 1)}n + n \\\\ & = 4^{k-1} + 4^{k-2}n + ... + 4^{k - (k - 1)}n + n \\end{align}\n\nNow, these terms need to be summed. Re-expressing $4 = 2^{2}$ and factoring out $n$,\n\n\\begin{align} T(n) & = 2^{2(k-1)} + 2^{2(k-2)}n + ... + 2^{2(k - [k - 1])}n + n \\\\ & = 2^{2(k-1)} + n(2^{2(k-2)} + ... + 2^{2(k - [k - 1])} + 1) \\end{align}\n\nRecognizing the parenthesized terms collectively represent a geometric series, we can easily sum the terms to get\n\n\\begin{align} T(n) &= 2^{2(k-1)} + n\\sum_{i=1}^{k-2} \\ (2^{2})^{i} \\\\ &= 2^{2(k-1)} + n(2^{2(k-2)+1} - 1)\\end{align}\n\nLastly, back substitute $k$ to get\n\n\\begin{align} T(n) &= 4^{(log_{3}(n)-1)} + n(4^{(log_{3}(n)-2)+1} - 1) \\\\ &= 4^{log_{3}(n)-1}(n+1)-n \\end{align}\n\nQ2: I read that \u0398(g(n)) is both the upper bound O(g(n)) and the lower bound \u03a9(g(n)) so does that mean that the three answers to my question would be...\n\nUsing case 1 of the Master Theorem as described by here, $T(n) \\in \\Theta(n^{1.262})$. Using theta notation alone will suffice for describing the bounds for this function, as it's implied in its definition. Consider the following graph:", "date": "2021-05-19 03:01:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2467200/solve-the-recurrence-relation-tn-4tn-3-n-t3-1-n-3k-then-det", "openwebmath_score": 0.999853253364563, "openwebmath_perplexity": 187.92085096482472, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692291542526, "lm_q2_score": 0.9019206824612296, "lm_q1q2_score": 0.8808781776976865}}
{"url": "https://math.stackexchange.com/questions/2216888/convergence-of-the-series-sum-n-1-infty-frac3-lnnn7-i-say-it-con", "text": "# Convergence of the series $\\sum_{n=1}^{\\infty}\\frac{3\\ln(n)}{n^7}$; I say it converges, WebWork tells me this is incorrect.\n\nI'm almost certain that $$\\sum_{n=1}^{\\infty}\\frac{3\\ln(n)}{n^7}$$ converges. However, WebWork tells me that this is incorrect. I have been in that situation before but I obviously can't assume that I'm right and the computer is wrong based on that. I don't know how the system work and I'm not sure whether what I do is correct anymore.\n\nBy the $p$-test, I know $$3\\sum_{n=1}^{\\infty}\\frac{1}{n^7}$$ converges. I also know that $\\ln(n)$ grow very slowly so I use the comparison $\\ln(n)\\le Cn$ which I think will not change that fact that the sum converges such that $$\\sum_{n=1}^\\infty\\frac{3\\ln(n)}{n^7}\\le \\sum_{n=1}^\\infty\\frac{Cn}{n^7}= C\\sum_{n=1}^\\infty\\frac 1{n^6}\\lt\\infty$$ which makes me quite certain that $\\sum_{n=1}^{\\infty}\\frac{3\\ln(n)}{n^7}$ is convergent.\n\nI also checked that \\begin{align} \\lim_{\\epsilon \\to \\infty}\\int_1^\\epsilon\\frac{3\\ln(x)}{x^7}\\,dx & = 3\\lim_{\\epsilon \\to \\infty}\\left(-\\frac{\\log (x)}{6 x^6}\\bigg|_{1}^{\\epsilon} +\\frac{1}{6}\\int_1^\\epsilon \\frac {dx}{x^7}\\right)\\\\ & = 3\\lim_{\\epsilon \\to \\infty}\\left( -\\frac{\\log (x)}{6 x^6}-\\frac{1}{36 x^6}\\right)\\bigg|_{1}^{\\epsilon}\\\\ & = \\frac 1{12}\\\\ \\end{align}\n\nCan someone please shine some light on this for me?\n\n\u2022 Besides mixing $n$'s and $x$'s in your comparison, you are correct. The series converges. \u2013\u00a0carmichael561 Apr 4 '17 at 0:53\n\u2022 WolframAlpha agrees with you that this converges. \u2013\u00a0Mark Apr 4 '17 at 0:53\n\u2022 @carmichael561 Thank you I will edit that, I put that on the count of not being properly caffeinated... \u2013\u00a0user409521 Apr 4 '17 at 0:54\n\nThe series is convergent by p test and we may evaluate the \u201cclosed form\u201d. The Riemann zeta function is defined as$$\\zeta\\left(s\\right)=\\sum_{n\\geq1}\\frac{1}{n^{s}},\\,\\textrm{Re}\\left(s\\right)>1.$$It is absolute convergent in the region $\\textrm{Re}\\left(s\\right)>1$ so $$\\zeta'\\left(s\\right)=\\frac{d}{ds}\\left(\\sum_{n\\geq1}\\frac{1}{n^{s}}\\right)=\\sum_{n\\geq1}\\frac{d}{ds}\\left(\\frac{1}{n^{s}}\\right)=-\\sum_{n\\geq1}\\frac{\\log\\left(n\\right)}{n^{s}}$$ hence $$S=3\\sum_{n\\geq1}\\frac{\\log\\left(n\\right)}{n^{7}}=\\color{red}{-3\\zeta'\\left(7\\right)}\\approx0.0181.$$\n\n\u2022 Thanks for sparking my interest for Riemann's zeta function and giving me some insight. \u2013\u00a0user409521 Apr 4 '17 at 17:21\n\u2022 @InfiniteMonkey You're welcome. \u2013\u00a0Marco Cantarini Apr 4 '17 at 18:17\n\nI would try the \"Integral Test for Convergence\" which says:\n\nConsider an integer $N$ and a non-negative, continuous function $f$ defined on the unbounded interval $[N, \u221e)$, on which it is monotone decreasing. Then the infinite series\n\n$$\\sum_N^{\\infty} f(n)$$\n\nconverges to a real number if and only if the improper integral\n\n$$\\int_N^{\\infty} f(x) dx$$\n\nis finite.\n\nLook at\n\n$$\\sum_{n=1}^{\\infty}\\frac{3\\ln(n)}{n^7}$$\n\nand the improper integral\n\n$$\\int _1^{\\infty }\\frac{3 \\ln (n)}{n^7} = \\frac{1}{12},$$ which is finite.\n\nThus, by the Integral Test for Convergence, you can say that the infinite series $\\sum_{n=1}^{\\infty}\\frac{3\\ln(n)}{n^7}$ converges.\n\n\u2022 The p test is more than enough. You don't need to solve that integral to know that the series converges, it is as immediate as $7 > 1$ \u2013\u00a0Ant Apr 4 '17 at 8:49\n\nIt really, really converges. The only common convergence tests I can think of that don't help are the root test and the alternating series test (the latter for the trivial reason that the summand is always positive).\n\nRatio test: $$\\lim_{n \\to \\infty} \\frac{\\log(n+1)}{n \\log(n)} = 0$$ which you can do by just \"limit of a product is the product of the limits\" on $\\frac{\\log(n+1)}{\\log n} \\times \\frac{1}{n}$.\n\nComparison with $\\frac{1}{n^2}$: $$\\frac{\\log n}{n^7} \\leq \\frac{1}{n^2}$$ if and only if $n^5 \\geq \\log n$, which is true for all positive $n$.\n\nIntegral test someone else has covered.\n\nCauchy condensation: it converges if and only if $$\\sum_{n=0}^{\\infty} \\frac{n \\log(2)}{2^{7n-n}}$$ does, but that is absurdly quickly convergent.\n\n\u2022 I was not aware of Cauchy condensation I'm reading about it now thanks! \u2013\u00a0user409521 Apr 4 '17 at 17:27", "date": "2021-04-21 09:02:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2216888/convergence-of-the-series-sum-n-1-infty-frac3-lnnn7-i-say-it-con", "openwebmath_score": 0.970007598400116, "openwebmath_perplexity": 403.8512134209384, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232874658905, "lm_q2_score": 0.8962513648201267, "lm_q1q2_score": 0.8808567127683081}}
{"url": "http://mathhelpforum.com/advanced-algebra/17574-determinant-very-hard-algebra-print.html", "text": "# Determinant with Very Hard Algebra\n\n\u2022 August 7th 2007, 02:36 AM\nbinarybob0001\nDeterminant with Very Hard Algebra\nThe problem is\n$\n\\det \\left [ \\begin{matrix} a & b & c \\\\ a^2 & b^2 & c^2 \\\\ a^3 & b^3 & c^3 \\end{matrix} \\right ]\n$\n\nSorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.\n$\nabc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)\n$\n\nThere are many ways to factor out terms from this.\n$\nabc[bc(c - b) + ac(a - c) + ab(b - a)] =\n$\n\n$\nabc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] =\n$\n\n$\nabc[c^2(b - a) + b^2(a - c) + a^2(c - b)]\n$\n\nThere are even more ways of course. None of these ways lead me to the simple solution:\n$\nabc(b - a)(c - a)(c - b)\n$\n\nCan someone explain how to factor this? Thanks.\n\u2022 August 7th 2007, 03:43 AM\nJakeD\nQuote:\n\nOriginally Posted by binarybob0001\nThe problem is\n$\n\\det \\left [ \\begin{matrix} a & b & c \\\\ a^2 & b^2 & c^2 \\\\ a^3 & b^3 & c^3 \\end{matrix} \\right ]\n$\n\nSorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.\n$\nabc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)\n$\n\nThere are many ways to factor out terms from this.\n$\nabc[bc(c - b) + ac(a - c) + ab(b - a)] =\n$\n\n$\nabc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] =\n$\n\n$\nabc[c^2(b - a) + b^2(a - c) + a^2(c - b)]\n$\n\nThere are even more ways of course. None of these ways lead me to the simple solution:\n$\nabc(b - a)(c - a)(c - b)\n$\n\nCan someone explain how to factor this? Thanks.\n\nMultiply out\n\n\\begin{aligned}(b - a)(c - a)(c - b) &= (bc -ac -ab +a^2)(c-b) \\\\\n&= bc^2 - ac^2 -abc + a^2c -b^2c +abc + ab^2 - a^2b \\\\\n&= bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b \\end{aligned}\n\nas you wanted.\n\u2022 August 7th 2007, 07:28 AM\nbinarybob0001\nYou can only do that if you already know the solution. What if you didn't know the solution? I would have just stopped thinking I was done. Is there a systematic way of arriving at that result?\n\u2022 August 7th 2007, 09:06 AM\nSoroban\nHello, binarybob0001!\n\nQuote:\n\nThe problem is: . $\n\\begin{bmatrix} a & b & c \\\\ a^2 & b^2 & c^2 \\\\ a^3 & b^3 & c^3 \\end{bmatrix}$\n\nSorry I'm new. . . . . Welcome aboard!\nThat's the best I could write the problem. . . . . Good job!\n\nIt's easy to find the determinant with expansion by minors\nbut to get the simplified solution involves some sort of clever trick that I don't know of.\n\nHere's what I can get to:\n. . $abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)$\n\nThere are many ways to factor out terms from this.\nNone of these ways lead me to the simple solution:\n. . $abc(b - a)(c - a)(c - b)$\nCan someone explain how to factor this?\n\nHere's an approach I \"discovered\" years ago . . .\n\nCollect terms with respect to one of the variables.\n\nLet's use $c$\n. . Collect terms with $c^2$, and with $c$, and 'constants' (no $c$'s).\n\nWe have: . $abc\\bigg[bc^2 - ac^2 - b^2c + a^2c + ab^2 - a^2b\\bigg]$\n\nFactor \"by grouping\": . $abc\\bigg[c^2(b-a) - c(b^2-a^2) + ab(b-a)\\bigg]$\n. . $= \\;abc\\bigg[c^2(b-a) - c(b-a)(b+a) + ab(b-a)\\bigg]$\n\nFactor out $(b-a)\\!:\\;\\;abc(b-a)\\bigg[c^2 - c(b+a) + ab\\bigg]$\n. . $= \\;abc(b-a)\\bigg[c^2 - bc - ac + ab\\bigg]$\n\nFactor \"by grouping\": . $abc(b-a)\\bigg[c(c-b) - a(c-b)\\bigg]$\n\nFactor out $(c-b)\\!:\\;\\;abc(b-a)(c-b)(c-a)$ . . . . ta-DAA!\n\n\u2022 August 8th 2007, 01:35 AM\nbinarybob0001\nahah, so I just need to chose a letter, stupid me. Thank you very much. Hey, are there more problems this hard to practice on?", "date": "2013-12-12 23:56:53", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/17574-determinant-very-hard-algebra-print.html", "openwebmath_score": 0.8790739178657532, "openwebmath_perplexity": 1226.9334919927935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676466573413, "lm_q2_score": 0.8991213867309121, "lm_q1q2_score": 0.8808401329979579}}
{"url": "https://dellwindowsreinstallationguide.com/curve-fitting/", "text": "# Generating Perfect Model Data\n\nLet\u2019s generate some model x1 and y1 data. First we will create an independent variable x1 that is a column vector ranging from 0 to 20 in steps of 2 and then we can create y1 which is a polynomial with respect to x1. We will store this in a table called data:\n\nx1=[0:2:20]'; y1=0.1*x1.^2+1.2*x1+5; data=table(x1,y1); clear x1; clear y1;\n\nFor this quadratic polynomial y1=0.1*x1.^2+1.2*x1+5 of the form y1=p2(1)*x1.^2+p2(2)*x1+p2(3) the three coefficients are:\n\np2(1)=0.1\np2(2)=1.2\np2(3)=5\n\np2(1) is the\u00a0x.^2\u00a0coefficient\u00a0p2(2) is the\u00a0x coefficient and p2(3)\u00a0is the constant.\n\n# Scatter Plot\n\nfigure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,...\n'MarkerFaceAlpha',0.5,...\n'MarkerEdgeAlpha',0.5)\nset(gca,'FontSize',20)\nxlabel('x')\nylabel('y')\ntitle('Polynomial')\ngrid('minor')\nlegend('y1',... 'Location','NorthEast')\n\n# Using polyfit to Find the Polynomial Coefficients\n\nWe can use polyfit to evaluate the polynomial coefficients. Polyfit has the form\n\n[pn]=polyfit(x,y,n)\n\nWhere the input arguments are the x data, y data and n, the order of polynomial. The output arguments is a row vector pn of the polynomial coefficients. In most cases we would try fitting from a first order polynomial upwards however as we know our data is a second order polynomial, as it is created from a second order polynomial we can try fitting to a second order polynomial first.\n\n[p2]=polyfit(data.x1,data.y1,2)\n\n$\\displaystyle \\text{p2=}\\left[ {\\begin{array}{*{20}{c}} {0.100000} & {1.200000} & {5.000000} \\end{array}} \\right]$\n\nThese are the three polynomial coefficients for the quadratic equation:\n\n$\\displaystyle 0.100000{{x}^{2}}+1.200000x+5.000000$\n\nThe original polynomial was y1=0.1*x1.^2+1.2*x1+5 and the three coefficients returned match y1=p2(1)*x1.^2+p2(2)*x1+p2(3) as expected.\n\nWhich as expected is a perfect match to our original equation.\n\n# Using polyval to Evaluate the Polynomial\n\nWe can use polyval to evaluate the polynomial at specified x values. Let\u2019s first create a new variable x2 and assign it as a column in a new table called datafit.\n\nx2=[1:2:19]'; datafit=table(x2); clear x2;\n\nWe can use polyval to evaluate the polynomial coefficients. polyval has the form\n\n[newy]=polyval(pn,newx);\n\nWe can use the function\n\n[datafit.y2]=polyval(p2,datafit.x2);\n\nTo find the value of the polynomial for each x2 value.\n\n# Plotting the Fitted Data with Respect to the Raw Data\n\nWe can plot this data as a scatter plot alongside the original data:\n\nfigure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,...\n'MarkerFaceAlpha',0.5,...\n'MarkerEdgeAlpha',0.5)\nhold('on'); scatter(datafit.x2,datafit.y2,100,'fill',... 'MarkerEdgeColor',[255/255,0/255,0/255],... 'MarkerFaceColor',[0/255,112/255,92/255],... 'LineWidth',1,... 'MarkerFaceAlpha',0.5,... 'MarkerEdgeAlpha',0.5); hold('off'); set(gca,'FontSize',20)\nxlabel('x')\nylabel('y')\ntitle('Polynomial')\ngrid('minor')\nlegend('y1','y2',... 'Location','NorthEast')\n\nIt may be more common to plot the fitted curve as a line:\n\nfigure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,...\n'MarkerFaceAlpha',0.5,...\n'MarkerEdgeAlpha',0.5)\nhold('on'); plot(datafit.x2,datafit.y2,... 'color','r',... 'LineWidth',3); hold('off'); set(gca,'FontSize',20)\nxlabel('x')\nylabel('y')\ntitle('Polynomial')\ngrid('minor')\nlegend('y1','y2',... 'Location','NorthEast')\n\nIf we want to print the equation to the chart we can use:\n\nequation=sprintf('y=(%f)x^2+(%f)x+(%f)',p2(1),p2(2),p2(3)); legend('data',equation,'location','northwest','FontSize',8);\n\nThe function sprintf has the same form as fprintf but prints to a string variable opposed to outputting in the Command Window.\n\n# Underfitting \u2013 Fitting a 1st Order Polynomial to 2nd Order Polynomial Data\n\nThe code above can be modified to fit to a 1st order polynomial (i.e. a straight line). Here it is obvious by eye that it is an unsuitable fit:\n\nThe reader is recommended to try to use polyfit and polyval to try and fit the above curve to a 1st order polynomial (straight line).\n\nx2=[1:2:19]'; [p1]=polyfit(data.x1,data.y1,1) datafit=table(x2); clear x2; [datafit.y2]=polyval(p1,datafit.x2); \n\nThe user is recommended to plot the graph above, with the updated equation in the legend.\n\nfigure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,...\n'MarkerFaceAlpha',0.5,...\n'MarkerEdgeAlpha',0.5)\nhold('on'); plot(datafit.x2,datafit.y2,... 'color','r',... 'LineWidth',3); hold('off'); set(gca,'FontSize',20)\nxlabel('x')\nylabel('y')\ntitle('Polynomial')\ngrid('minor') equation=sprintf('y=(%f)x+(%f)',p2(1),p2(2)); legend('data',equation,'location','northwest','FontSize',8);\n\n# Overfitting \u2013 Fitting a 3rd Order Polynomial to 2nd Order Polynomial Data\n\nHere from the values of the coefficients i.e. the cubic term being extremely small 1.6509e-17 it is obvious that the polynomial is overfitted. This term can be rounded to 0 and we return with the quadratic fit. This is perfect model data so the fitting is relatively easy in real life it may be harder to distinguish data which has been over-fitted.\n\nThe reader is recommended to try to use polyfit and polyval to try and fit the above curve to a 3rd order polynomial.\n\nx2=[1:2:19]'; [p3]=polyfit(data.x1,data.y1,3) datafit=table(x2); clear x2; [datafit.y2]=polyval(p3,datafit.x2); \n\nThe user is recommended to plot the graph above, with the updated equation in the legend.\n\nfigure(1); scatter(data.x1,data.y1,100,'fill',... 'MarkerEdgeColor',[112/255,48/255,160/255],... 'MarkerFaceColor',[0/255,176/255,80/255],... 'LineWidth',1,...\n'MarkerFaceAlpha',0.5,...\n'MarkerEdgeAlpha',0.5)\nhold('on'); plot(datafit.x2,datafit.y2,... 'color','r',... 'LineWidth',3); hold('off'); set(gca,'FontSize',20)\nxlabel('x')\nylabel('y')\ntitle('Polynomial')\ngrid('minor') equation=sprintf('y=(%f)x^3+(%f)x^2+(%f)x+(%f)',p3(1),p3(2),p3(3),p3(4)); legend('data',equation,'location','northwest','FontSize',8);\n\n# Generating Model Data with Noise\n\nrng('default'); x1=[0:2:20]'; [m,n]=size(x1); noise=rand(m,n); y1=0.1*x1.^2+1.2*x1+5+noise; data=table(x1,y1); clear x1; clear y1; clear noise; clear m; clear n;\n\nWe see a deviation from the starting polynomial with:\n\np2(1)=0.1\np2(2)=1.2\np2(3)=5\n\nThe deviation influences the constant term the most, followed by the x term.\n\nIf we times the noise by 5, we see a larger deviation:\n\nIf we times the noise by 10 we see more deviation:\n\nAnd if we times the noise by 25 we see an even larger deviation again:\n\nIf we times the noise by 250, the original function is not recognisable:", "date": "2019-06-18 23:30:06", "meta": {"domain": "dellwindowsreinstallationguide.com", "url": "https://dellwindowsreinstallationguide.com/curve-fitting/", "openwebmath_score": 0.5601446628570557, "openwebmath_perplexity": 2731.8483911742314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676472509721, "lm_q2_score": 0.8991213833519949, "lm_q1q2_score": 0.8808401302214882}}
{"url": "https://gmatclub.com/forum/which-of-the-following-is-an-integer-82683.html?sort_by_oldest=true", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 11 Jul 2020, 13:57\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Which of the following is an integer?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 15 Aug 2008\nPosts: 87\nSchools: NYU Stern Class of 2012\nWE 1: GM, Diversified Financial Svcs.\nWhich of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 07 Oct 2019, 01:36\n2\n21\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n82% (00:47) correct 18% (00:41) wrong based on 465 sessions\n\n### HideShow timer Statistics\n\nWhich of the following is an integer?\n\nI. $$\\frac{12!}{6!}$$\n\nII. $$\\frac{12!}{8!}$$\n\nIII. $$\\frac{12!}{7!5!}$$\n\n(A) I only\n(B) II only\n(C) III only\n(D) I and II only\n(E) I, II, and III\n\nOriginally posted by NickTW on 18 Aug 2009, 19:23.\nLast edited by Bunuel on 07 Oct 2019, 01:36, edited 2 times in total.\nRenamed the topic, edited the question and added the OA.\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 17070\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n08 May 2015, 15:47\n6\n2\nHi All,\n\nBeyond the arithmetic that sdrandom1 pointed out, these calculations fall into some patterns that you are likely to see (in some form) on Test Day.\n\n12!/6!\n\nWhen dealing with INDIVIDUAL factorials, if the factorial in the numerator is GREATER than the factorial in the denominator, then you will end up with an integer. Here, since 12 > 6, 12!/6! will simplify to an integer.\n\n12!/7!5!\n\nYou might not have studied the Combination Formula yet, but this calculation is what you would end up with when answering the question \"how many different combinations of 7 people can you form from a group of 12 people?\" In these types of \"Combination\" calculations, the number of groups will always be an integer, so 12!/7!5! will simplify to an integer.\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\nContact Rich at: Rich.C@empowergmat.com\n\nThe Course Used By GMAT Club Moderators To Earn 750+\n\nsouvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605\nENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605\n##### General Discussion\nManager\nJoined: 17 Dec 2007\nPosts: 82\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n18 Aug 2009, 19:30\n2\n1\nNickTW wrote:\nDid a search, could not find anything on this one (the question stem is pretty vague). Would appreciate some enlightenment on the shortcut I'm missing!\n\nWhich of the following is an integer?\n\nI. 12! / 6!\nII. 12! / 8!\nIII. 12! / 7!5!\n\nA) I only\nB) II only\nC) III only\nD) I and II only\nE) I, II, and III\n\nSource: GMAT Prep Exam. Solution : E\n\nTo me the question looks straight forward.\nwe know 12! is multiple of all numbers from 1 to 12, so dividing it by 6! or 8! is pretty much sure an integer.\n\nnow coming to the 3rd options 12! by 5! leaves 12*11*10*9*8*7*6/1*2*3*4*5*6*7 which results in integer, so the answe E is right\nManager\nJoined: 30 May 2009\nPosts: 130\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n18 Aug 2009, 19:45\n1\n1\nThere is no shortcut persay...\n\nClear E.\n\n12!/6! = 12*11*10*9*8*7*6!/6! = Integer\n12!/8! = 12*11*10*9*8!/8! = Integer\n12!/7!*5! = 12*11*10*9*8/5*4*3*2*1 = integer\n\nSo all 3 are correct.\nGMAT Tutor\nJoined: 24 Jun 2008\nPosts: 2308\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2009, 08:32\n4\n1\nEvery factorial is divisible by every smaller factorial, so 12!/6! and 12!/8! are integers. 12!/(7!*5!) is the number of ways to choose 7 things from a group of 12 if order is irrelevant, so it must also be an integer.\n_________________\nGMAT Tutor in Montreal\n\nIf you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com\nDirector\nJoined: 08 Jun 2010\nPosts: 660\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n03 May 2015, 01:42\ncase 1 and case 2 are simple but case 3 is harder. for case 3, we have to write down all the factors.\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 11083\nLocation: United States (CA)\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n03 Dec 2017, 17:39\n6\n2\nNickTW wrote:\nWhich of the following is an integer?\n\nI. 12! / 6!\nII. 12! / 8!\nIII. 12! / 7!5!\n\nA) I only\nB) II only\nC) III only\nD) I and II only\nE) I, II, and III\n\nBefore actually solving this problem, let's review how factorials can be expanded and expressed. As as example, we can use 5!.\n\n5! could be expressed as:\n\n5!\n\n5 x 4!\n\n5 x 4 x 3!\n\n5 x 4 x 3 x 2!\n\n5 x 4 x 3 x 2 x 1!\n\nUnderstanding how this factorial expansion works will help us work our way through each answer choice, especially answer choices 1 and 2.\n\nI. 12!/6!\n\nSince we know that factorials can be expanded, we now know that:\n\n12! = 12 x 11 x 10 x 9 x 8 x 7 x 6!\n\nPlugging this in for answer choice 1, we have:\n\n(12 x 11 x 10 x 9 x 8 x 7 x 6!)/6! = 12 x 11 x 10 x 9 x 8 x 7, which is an integer.\n\nII. 12!/8!\n\nOnce again, since we know that factorials can be expanded, we now know that:\n\n12! = 12 x 11 x 10 x 9 x 8!\n\nPlugging this in for answer choice 2, we have:\n\n(12 x 11 x 10 x 9 x 8!)/8! = 12 x 11 x 10 x 9, which is an integer.\n\nIII. 12!/(7!5!)\n\nOnce again, since we know that factorials can be expanded, we now know that:\n\n12! = 12 x 11 x 10 x 9 x 8 x 7!\n\nPlugging this in for answer choice 3 gives us:\n\n(12 x 11 x 10 x 9 x 8 x 7!)/(7!5!)\n\n(12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1)\n\n(12 x 11 x 10 x 9 x 8)/(12 x 10 x 1)\n\n11 x 9 x 8, which is an integer.\n\nWe see that the quantities in Roman numerals I, II and III are all integers.\n\nAlternate solution:\n\nFor any positive integers m, n and p,\n\n1) If m > n, then m!/n! is always an integer.\n\n2) If m = n + p, then m!/(n!p!) is always an integer (which is in fact mCp).\n\nFrom the above two facts, we see that all three quotients in the Roman numerals must be integers.\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n\n214 REVIEWS\n\n5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE\n\nNOW WITH GMAT VERBAL (BETA)\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nSenior Manager\nJoined: 05 Feb 2018\nPosts: 440\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n20 Jan 2019, 08:00\nNickTW wrote:\nWhich of the following is an integer?\n\nI. 12! / 6!\nII. 12! / 8!\nIII. 12! / 7!5!\n\nA) I only\nB) II only\nC) III only\nD) I and II only\nE) I, II, and III\n\nFor III, I went through and crossed out everything that's reduced to visualize...\n12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1\n\nSo you're left with 12*11*2*3 (from taking 5 and 3 from 10 and 9) = 792. Makes sense.\n\nScottTargetTestPrep\nAs a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the result not an integer? Am I right to think this has to do with the number of prime factors?\nBecause 12!/8!5! and 12!/9!5! are also integers, while 12!/10!5! isn't. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and 9! would be 3^2 both of which we can still reduce... while 10! gives us a factor of 5 which we don't have.\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 11083\nLocation: United States (CA)\nRe: Which of the following is an integer?\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jan 2019, 17:19\nenergetics wrote:\nNickTW wrote:\nWhich of the following is an integer?\n\nI. 12! / 6!\nII. 12! / 8!\nIII. 12! / 7!5!\n\nA) I only\nB) II only\nC) III only\nD) I and II only\nE) I, II, and III\n\nFor III, I went through and crossed out everything that's reduced to visualize...\n12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1\n\nSo you're left with 12*11*2*3 (from taking 5 and 3 from 10 and 9) = 792. Makes sense.\n\nScottTargetTestPrep\nAs a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the result not an integer? Am I right to think this has to do with the number of prime factors?\nBecause 12!/8!5! and 12!/9!5! are also integers, while 12!/10!5! isn't. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and 9! would be 3^2 both of which we can still reduce... while 10! gives us a factor of 5 which we don't have.\n\nSo you ask a really great qestion. So, as you've shown above, when m < n + p, there is no hard and fast rule to determien when m!/(n!p!) is an integer and when it is not.\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n\n214 REVIEWS\n\n5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE\n\nNOW WITH GMAT VERBAL (BETA)\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 15412\n\n### Show Tags\n\n07 Oct 2019, 01:36\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: Which is integer? \u00a0 [#permalink] 07 Oct 2019, 01:36", "date": "2020-07-11 21:57:11", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/which-of-the-following-is-an-integer-82683.html?sort_by_oldest=true", "openwebmath_score": 0.7608386278152466, "openwebmath_perplexity": 3044.6973856992518, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.880797085800514}}
{"url": "https://gmatclub.com/forum/the-figure-above-represents-a-semicircular-archway-over-a-flat-street-268687.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 May 2019, 10:02\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# The figure above represents a semicircular archway over a flat street.\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 55150\nThe figure above represents a semicircular archway over a flat street.\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jun 2018, 00:30\n20\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n90% (01:35) correct 10% (01:47) wrong based on 655 sessions\n\n### HideShow timer Statistics\n\nThe figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?\n\nA. \u221a2\n\nB. 2\n\nC. 3\n\nD. 4\u221a2\n\nE. 6\n\nNEW question from GMAT\u00ae Official Guide 2019\n\n(PS05957)\n\nAttachment:\n\nPS05957_f009.jpg [ 34.62 KiB | Viewed 6372 times ]\n\n_________________\nSenior Manager\nJoined: 13 Feb 2018\nPosts: 270\nGMAT 1: 640 Q48 V28\nThe figure above represents a semicircular archway over a flat street.\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jun 2018, 01:41\nLet's disturb Pythagoras once again:\n\n$$h^2+2^2=6^2$$\nh=$$\\sqrt{32}$$\n\nIn My Opinion\nAns: D\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2845\nRe: The figure above represents a semicircular archway over a flat street.\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jun 2018, 05:24\n2\n\nSolution\n\nGiven:\n\u2022 The figure depicts a semi-circular archway over a flat street\n\u2022 The semi-circle has a center at O and a radius of 6 feet\n\nTo find:\n\u2022 The length of the height h, in feet, of the archway 2 feet from its center\n\nApproach and Working:\n\nIf we add the point O and A, we get a right-angled triangle OAB, right-angled at point B and hypotenuse AO\n\nAs the radius is 6 feet, we can say\n\u2022 AO = 6 feet\n\nApplying Pythagoras Theorem in triangle OAB, we can write:\n\u2022 $$AO^2 = AB^2 + BO^2$$\nOr, $$6^2 = h^2 + 2^2$$\nOr, $$h^2 = 36 \u2013 4 = 32$$\nOr, $$h = 4\\sqrt{2}$$\n\nHence, the correct answer is option D.\n\n_________________\nCEO\nJoined: 12 Sep 2015\nPosts: 3716\nRe: The figure above represents a semicircular archway over a flat street.\u00a0 [#permalink]\n\n### Show Tags\n\n23 Jun 2018, 09:52\n1\nTop Contributor\nBunuel wrote:\n\nThe figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?\n\nA. \u221a2\n\nB. 2\n\nC. 3\n\nD. 4\u221a2\n\nE. 6\n\nNEW question from GMAT\u00ae Official Guide 2019\n\n(PS05957)\n\nAttachment:\nPS05957_f009.jpg\n\nLet's add the radius of 6 feet to the diagram to get:\n\nFrom here, we can see the right triangle hiding within the diagram, which means we can apply the Pythagorean Theorem.\nSe can write: h\u00b2 + 2\u00b2 = 6\u00b2\nSimplify: h\u00b2 + 4 = 36\nSo, we get: h\u00b2 = 32\nThis means h = \u221a32\nCheck the answer choices. . . \u221a32 is not among them.\nLooks like we need to simplify \u221a32\n\nWe'll use the fact that \u221a(xy) = (\u221ax)(\u221ay)\nSo, \u221a32 = \u221a[(16)(2)] = (\u221a16)(\u221a2) = (4)(\u221a2) = 4\u221a2\nCheck the answer choices. . . D\n\nRELATED VIDEO (simplifying roots)\n\n_________________\nTest confidently with gmatprepnow.com\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 6133\nLocation: United States (CA)\nRe: The figure above represents a semicircular archway over a flat street.\u00a0 [#permalink]\n\n### Show Tags\n\n25 Jun 2018, 12:06\n1\nBunuel wrote:\n\nThe figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?\n\nA. \u221a2\n\nB. 2\n\nC. 3\n\nD. 4\u221a2\n\nE. 6\n\nSince the radius is 6 feet, we can create a right triangle with hypotenuse of 6, leg of 2, and leg of h; thus, we have:\n\n2^2 + h^2 = 6^2\n\n4 + h^2 = 36\n\nh^2 = 32\n\nh = \u221a16 x \u221a2 = 4\u221a2\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nSenior Manager\nJoined: 10 Apr 2018\nPosts: 258\nLocation: United States (NC)\nRe: The figure above represents a semicircular archway over a flat street.\u00a0 [#permalink]\n\n### Show Tags\n\n02 Aug 2018, 13:30\nif the radius is 6 max height of the arch can be 6 from the center of semi-circle. So height asked in question which is two feet away from center is definitely less than 6 . Also point 2 feet away from center to end of semicircle the dist is 4 so height is little more than 4 . So the only answer choice that matches it is D\n_________________\nProbus\n\n~You Just Can't beat the person who never gives up~ Babe Ruth\nRe: The figure above represents a semicircular archway over a flat street. \u00a0 [#permalink] 02 Aug 2018, 13:30\nDisplay posts from previous: Sort by", "date": "2019-05-19 17:02:33", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-figure-above-represents-a-semicircular-archway-over-a-flat-street-268687.html", "openwebmath_score": 0.7333633899688721, "openwebmath_perplexity": 6044.133223168102, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8807970779778824}}
{"url": "http://gesuitialquirinale.it/rdil/polar-coordinates-pdf.html", "text": "# Polar Coordinates Pdf\n\nMy questions is, how important are these topics for calc III? Do I need to re-study or is it not important?. The fact that a single point has many pairs of polar coordinates can cause complications. location than conventional Cartesian coordinates. 3) Rectangular coordinates of point P are given. x t y t9cos , 9sin 16. We now proceed to calculate the angular momentum operators in spherical coordinates. admiral calls a polar coordinate (in \ud835\udc5f,\ud835\udf03)form and then the defending admiral declares whether the shot was a hit or a miss. To specify a clockwise direction, enter a negative value for the angle. The ordered pairs, called polar coordinates, are in the form $$\\left( {r,\\theta } \\right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$ - axis (polar axis), going counter-clockwise. Polar - Rectangular Coordinate Conversion Calculator. 5 MM Graph Paper. Practice Problem: Convert the following sets of rectangular coordinates into polar coordinates. TrigCheatSheet. Apr 11, 2014 - Explore brittanykaye911's board \"polar coordinates\", followed by 154 people on Pinterest. Yes o\ufb03ce hours Wednesday 2/20 2-4pm SC 323. In this note, I would like to derive Laplace's equation in the polar coordinate system in details. examples to convert image to polar coordinates do it explicitly - want a slick matrix method I thought using the method used above. CARTESIAN & POLAR COORDINATES In fact, as a complete counterclockwise rotation is given by an angle 2\u03c0, the point represented by polar coordinates (r, \u03b8) is also represented by (r, \u03b8+ 2n\u03c0) and (-r, \u03b8+ (2n + 1)\u03c0) where n is any integer. For any point P consider the two distances:. 3 WS Polar Coordinates (Answers). The area of a region in polar coordinates defined by the equation $$r=f(\u03b8)$$ with $$\u03b1\u2264\u03b8\u2264\u03b2$$ is given by the integral $$A=\\dfrac{1}{2}\\int ^\u03b2_\u03b1[f(\u03b8)]^2d\u03b8$$. Polar coordinates use a distance and an angle to locate a point. You can see this by just drawing unit vectors at each point on, say, a circle: (draw). 2 : Apr 12, 2018, 11:37 AM. Polar coordinates describe the distance from P to to a special point O, called the pole or origin and the angle that the line segment PO makes with a special ray called the polar axis. Sign up to join this community. When you drag the red point, you change the polar coordinates $(r,\\theta)$, and the blue point moves to the corresponding position $(x,y)$ in Cartesian coordinates. 1 Exponential Equations Blank. In case n = 3, the polar coordinates (r,\u03b8,\u03c6) are called spherical coordinates, and we have y = x1, x = x2, z = x3, r2 = x2 + y2 + z2, x = rsin\u03c6sin\u03b8, y = rsin\u03c6cos\u03b8, and x = rcos\u03c6, so we can take r3 = r, \u03c62 = \u03b8. It provides resources on how to graph a polar equation and how to find the area of the shaded. Polar coordinates The representation of a complex number as a sum of a real and imaginary number, z = x + iy, is called its Cartesian representation. y2 4y 8x 20 0 y 2 2 4 2 x 3 2 23. There are three types of polar graph that are Large Single Polar Graph which has thirty marks for r in increment of five degrees, Smaller (Double) Polar Graph which has two polar graphs on one page, each with twenty scale marks for r increment of 5 degrees and Combined Cartesian and Polar has three pages here, One is a large cartesian grid, one a large polar grid and the third one has one. Draw a horizontal line to the right to set up the polar axis. Polar Graph Paper. The principal reason for this is the artificial expansion of the natural conic shapes of the spaces into a cylindrical shape. CONIC SECTIONS IN POLAR COORDINATES If we place the focus at the origin, then a conic section has a simple polar equation. Plane Curvilinear Motion Polar Coordinates (r -\u03b8) The particle is located by the radial distance r from a fixed point and by an angular measurement \u03b8to the radial line. In polar coordinates, every point is located around a central point, called the pole, and is named (r,n\u03b8). The value of r can be positive, negative, or zero. requirement for the generalized coordinates is that they span the space of the motion and be linearly independent. This introduction to polar coordinates describes what is an effective way to specify position. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. dinates and the polar direction vectors is that the polar direction vectors change depending on where they are relative to the origin. The small change r in rgives us two concentric circles and the small change in gives us an angular wedge. Graph the point P, (r; ) = 3;\u02c7 3. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Let suppose we have a small change in rand. Graphs in Polar Coordinates Purpose The purpose of this lab is to help you become familiar with graphs in polar coordinates. Student information Link. Therefore r\u02d9(t) = (\u02d9rcos\u03b8 \u2212 r\u03b8\u02d9sin\u03b8)i + (\u02d9rsin\u03b8 + r\u03b8\u02d9cos\u03b8)j. (See Figure 9. d is the perpendicular distance from the line to the origin. 6) Area in Polar Coordinates (Fig. Describe planar motion and solve motion problems by defining parametric equations and vector-valued functions. We need to subtract 960 by 180k, such that the result is between 0 and 180. It is sometimes convenient to refer to a point by name, especially when this point occurs in multiple \\draw commands. It can be printed and done on paper or completed at Socrative. However, it still is a useful tool to give you an introduction to the concepts pertaining to polar coordinates. r = sec\u03b8csc\u03b8 \u21d2 24. Looking for polar graph paper? We've got just what you need. By default, angles increase in the counterclockwise direction and decrease in the clockwise direction. This calculator converts between polar and rectangular coordinates. To convert from Polar coordinates to Cartesian coordinates, draw a triangle from the horizontal axis to the point. All the terms above are explained graphically there. txt) or read online for free. This OER repository is a collection of free resources provided by Equella. The defending admiral records the location of. os\u00b6z y: Islh1\u00ef/2 3cos0 \u00d0=3sm0 77t 57t 10) 2,\u2014 Convert each pair of rectangular coordinates to polar coordinates where r > 0 and O < 2m 11) - 31 13). That leads to the pre-factor (2/R^2). Sketching Polar Curves Examples. Graph the point Q, (r; ) = 2; \u02c7 6. c Double Integrals in Polar Coordinates (r; ) Let us suppose that the region boundary is now given in the form r = f( ) or = h(r), and/or the function being integrated is much simpler if polar coordinates are used. 2-3-17: We continued to convert between polar and rectangular equations. These will all be positive X,Y rectangular coordinates in Quadrant I of the Cartesian plane (X headed right from 0 and Y headed up from 0). ) The graph of = , where is a constant, is the line of inclination. The angular dependence of the solutions will be described by spherical harmonics. Suppose that X is a random vector with joint density function f X(x). PreCalculus. Key Concept: Constellations can be represented graphically. All the coordinates are made beforehand. i have this function f(r,theta) which i want to graph in polar/cylindrical coordinates. In polar coordinates the position of an object $$R$$ distance from the origin as represented in the diagram above is modelled $$\\mathbf{r} = R \\hat{r}$$ The velocity and acceleration in polar coordinates is derived by differentiating the position vector. Some useful properties about line integrals: 1. Michael VanValkenburgh To make it easier to type and easier to read, this handout will focus on the computational aspects of integration in polar coordinates. If you were to add a true position characteristic it would look like this. However, we can use other coordinates to determine the location of a point. However, we can still rotate around the system by any angle we want and so the coordinates of the origin/pole are (0,\u03b8). Download the pdf file and print. Please read through this supplement before going to quiz section for the polar worksheet on Thursday. Let r1 denote a unit vector in the direction of the position vector r , and let \u03b81 denote a unit vector perpendicular to r, and in the direction of increasing \u03b8, see Fig. Show Instructions. 2) Equal amounts of the Primaries produce white. r = 2 and \u03b8= 30\u00b0, so P is located 2 units from the origin in the positive direction on a ray making a 30\u00b0angle with the polar axis. Unit Six Precalculus Practice Test Vectors & Polar Graphs Page 3 of 6 14. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. Cartiesian Coordinate System. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. ) Abbreviated podcast notes on lecture 5. Use a double integral in polar coordinates to calculate the area of the region which is common to both circles r= 3sin and r= p 3cos. Solution; The Cartesian coordinate of a point are $$\\left( { - 8,1} \\right)$$. 5 Polar Coordinates and Graphs-1. 4 Many problems are more easily stated and solved using a coordinate system other than rectangular coordinates, for example polar coordinates. A polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. x t y t9cos , 9sin 16. You can copy that worksheet to your home. 2 We can describe a point, P, in three different ways. 686 CHAPTER 9 POLAR COORDINATES AND PLANE CURVES The simplest equation in polar coordinates has the form r= k, where kis a positive constant. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. This is the xy-plane. Find the distance between the points. a) r=3sec\u03b8 b) r=\u22123sin\u03b8 c) rcsc 1\u03b8= 5) Convert the rectangular equation to polar form. 9) ( , ) 10) ( , ) Two points are specified using polar coordinates. Different radials and degree over them are known as the polar paper. The old vvvv nodes Polar and Cartesian in 3d are similar to the geographic coordinates with the exception that the angular direction of the longitude is inverted. And polar coordinates, it can be specified as r is equal to 5, and theta is 53. \u2026Polar-coordinates are entered\u2026in a magnitude direction format. Getting Started To assist you, there is a worksheet associated with this lab that contains examples and even solutions to some of the exercises. The answer is: (r,\u03b8) Polar = (p x2 +y2, arctan y x) Polar Meanwhile, for a point given by Polar coordinates, (r,\u03b8) Polar, we need to specify the coordinates in Cartesian form in terms of the Polar data r and \u03b8. 3) Rectangular coordinates of point P are given. Customize Polar Axes. On the \ufb01rst region we would have \u22122 6 x 6 2 and \u221a 4\u2212x2 6 y 6 \u221a 9\u2212x2, on the second region \u22123 6 x 6 2. First, fix an origin (called the pole) and an initial ray from O. My knows are (R, r, theta, Phi) My unknowns are (Phi1, R1) If you guys are up for it, could you assist me in establishing a formula for R1 and Phi1 in terms of R, r, theta, and Phi?. A polar coordinate graph paper that\u2019s perfect for when you need to compare two graphs that have minor differences. In addition to, It converts complex number into polar form and vice versa. 5,13) into polar coordinates. Graphing Worksheets for Practice. But what about r f(T)? At first you might think dr dT is the slope of the tangent line to the curve but consider r = constant e. The unit tangent vector to the curve is then T\u02c6 = \u02d9x\u02c6\u0131+ \u02d9y \u02c6 (2) where we have used a dot to denote derivatives with respect to s. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. 5 , 0 1 2 \u2264 \u2264\u03b8 \u03c0. Polar Coordinates Polar coordinates of a point consist of an ordered pair, r \u03b8( , ), where r is the distance from the point to the origin, and \u03b8 is the angle measured in standard position. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, Calculating a limit in two variables by going to polar coordinates. Use a double integral in polar coordinates to calculate the area of the region which is common to both circles r= 3sin and r= p 3cos. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. The rectangular coordinates for P (5,20\u00b0) are P (4. Check out our many other free graph/grid paper styles. pdf (392 KB) Equella is a shared content repository that organizations can use to easily track and reuse content. (i) Plot each point. Defining Polar Coordinates. Large single polar graph--Thirty scale marks for r in increments of five degrees. It can be found by the \"gradient in polar coordinates\" googling. TrigCheatSheet. Polar coordinates with polar axes. 841d f0, 2pd u3 5 cos21s2y3d 5 0. Review: Polar coordinates De\ufb01nition The polar coordinates of a point P \u2208 R2 is the ordered pair (r,\u03b8) de\ufb01ned by the picture. We see this general pattern in the circle of gure 2. This easy-to-use packet is full of stimulating activities that will give your students a solid introduction to polar coordinates and trigonometric form!. their direction does not change with the point r. Ciencia y Tecnolog\u00eda, 32(2): 1-24, 2016 - ISSN: 0378-0524 3 II. POLAR COORDINATES (OL]DEHWK :RRG DEFINITION OF POLAR COORDINATES. The angular dependence of the solutions will be described by spherical harmonics. The divergence We want to discuss a vector \ufb02eld f de\ufb02ned on an open subset of Rn. However, in polar coordinates we have u(r,\u03b8) = r sin\u03b8 r2 = sin\u03b8 r so that u r = \u2212 sin\u03b8 r2, u. If we restrict rto be nonnegative, then = describes the. Laplace\u2019s equation in polar coordinates, cont. In case n = 3, the polar coordinates (r,\u03b8,\u03c6) are called spherical coordinates, and we have y = x1, x = x2, z = x3, r2 = x2 + y2 + z2, x = rsin\u03c6sin\u03b8, y = rsin\u03c6cos\u03b8, and x = rcos\u03c6, so we can take r3 = r, \u03c62 = \u03b8. 6) Area in Polar Coordinates (Fig. In this unit we explain how to convert from Cartesian co-ordinates to polar co-ordinates, and back again. 1 Background on Polar Coordinates. 7 - Polar Coordinates \u00b0 3,225 3,0. The polar coordinate system is formed by fixing a point, O, which is the pole (or origin). The small change r in rgives us two concentric circles and the small change in gives us an angular wedge. To convert from Cartesian co-ordinates to polar use the transformation $y=r\\sin { \\theta }$ and [m. In this system coordinates for a point P are and , which are indicated in Fig. See Large Polar Graph Paper. Mechanics 1: Polar Coordinates Polar Coordinates, and a Rotating Coordinate System. 5 MM Graph Paper. Viewed 11k times 3. Trigonometry - Trigonometry - Polar coordinates: For problems involving directions from a fixed origin (or pole) O, it is often convenient to specify a point P by its polar coordinates (r, \u03b8), in which r is the distance OP and \u03b8 is the angle that the direction of r makes with a given initial line. The ordered pairs, called polar coordinates, are in the form $$\\left( {r,\\theta } \\right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$ \u2013 axis (polar axis), going counter-clockwise. The polar coordinate system is extended into three dimensions with two different coordinate systems, the cylindrical and spherical coordinate system. Cartesian/Polar Coordinates Junior high school The connection between Cartesian coordinates and Polar coordinates is established by basic trigonometry. The transformation from Cartesian coordinates to spherical coordinates is. For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. In this unit we explain how to convert from Cartesian co-ordinates to polar co-ordinates, and back again. This article explains how to convert between polar and cartesian coordinates and also encourages the creation of some attractive curves from some relatively easy equations. The point P has. For any point P consider the two distances:. In the polar coordinate system, the ordered pair will now be (r, ). 4 Many problems are more easily stated and solved using a coordinate system other than rectangular coordinates, for example polar coordinates. These charts print on a standard sheet of 8 1/2 x 11 paper. Home Decorating Style 2020 for Polar Coordinate System In Autocad Pdf, you can see Polar Coordinate System In Autocad Pdf and more pictures for Home Interior Designing 2020 4680 at Manuals Library. \ufb01nd the x and y coordinates of a point (r, \u03b8)), we use the following formulas: x = r cos \u03b8, y = r sin \u03b8. Polar Coordinates (r-\u03b8)Ans: -0. First, let's get some preliminaries out of the way. Free Polar Graph Paper Template. We make the convention (\u2212r,\u03b8) = (r,\u03b8 +\u03c0). Determine a set of polar coordinates for the point. The orientation of a plane curve can be represented by arrows drawn along the curve. (\u2212 \u22125, 5) 9. (iii) Find the Cartesian coordinates of the point. Spherical Coordinates and the Angular Momentum Operators. Students will plot points in the polar coordinate system, convert coordinates and convert equations from rectangular to polar form and vice versa. b) Set up an expression with two or more integrals to find the area common to both curves. 1 New Optional Features pgfplots has been written with backwards compatibility in mind: old TEX les should compile without modi cations and without changes in the appearance. 6 Complex Polar Coordinates (slides, 4-to-1). If we restrict rto be nonnegative, then = describes the. For coordinate conversions: Example 2: Find the rectangular coordinate for the point whose polar coordinates are (a ) 4 5, 3 (b ) 5 4, 6 Example 3: Convert the following rectangular coordinate into four different, equivalent polar coordinates. to describe using polar coordinates. 442, C 5 s21, 0. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. 197t 12 16) x = \u2014 I pairs of polar coordinates that describe the same point as the provided polar (-q 1-71/2) Convert each pair of polar coordinates to rectangular coordinates. pdf (condensed podcast notes, 4 slides to a page) Presentations (slides without audio) on lecture 5. 3 Double Integrals in Polar Coordinates In Chapter 10, we explored polar coordinates and saw that in certain situations they simplify problems considerably. R Pr( converts a rectangular form to r in polar coordinates. But many teachers might prefer that you measure angles by yourself using a protractor on blank paper. Lecture #6: The Jacobian for Polar Coordinates Recall. Polar coordinates (Introduction and conversion) Sketching polar curves. For problems 5 and 6 convert the given equation into an equation in terms of polar coordinates. The second point lies on the positive 'y' axis, so the angle in polar coordinates is. The fact that a single point has many pairs of polar coordinates can cause complications. The polar coordinates (r,\u03b8) are de\ufb01ned by r2 = x2 + y2, (2) x = rcos\u03b8 and y = rsin\u03b8, so we can take r2 = r and \u03c62 = \u03b8. The conversion from polar to rectangular coordinates is the same idea as converting rectangular form to polar form in complex numbers. Because Dis a circular disk, we will set up the integral in polar coordinates. Source: Wikipedia - Polar Coordinate System. Cartesian coordinate system: start with xand yaxes. To Convert from Cartesian to Polar. Please read through this supplement before going to quiz section for the polar worksheet on Thursday. (iii) Find the Cartesian coordinates of the point. If there are three variables, the graph is 3D. Polar coordinates part 4 This is a continuation of the polar coordinates part 3. The azimuthal angle, now designated as \u03d5, specifies the rotational orientation. Solution: The function that we need to use in this example is G, which converts the pair of rectangular coordinates (x,y) into the polar coordinates (r,!). 2 Calculus In The Polar Coordinate System Contemporary Calculus 4 Area in Rectangular Coordinates (Fig. 3 WS Polar Coordinates (Answers). Convert the following equation of a circle to polar coordinates: 4x2 + 3 2 x +4y2 +1y. 1 Polar Coordinates and Rectangular Coordinates In astronomical calculations, polar coordinate systems are usually used. There are three types of polar graph that are Large Single Polar Graph which has thirty marks for r in increment of five degrees, Smaller (Double) Polar Graph which has two polar graphs on one page, each with twenty scale marks for r increment of 5 degrees and Combined Cartesian and Polar has three pages here, One is a large cartesian grid, one a large polar grid and the third one has one. I still did that, but I also tried odd things. You should be familiar with the Cartesian Coordinate System, also called rectangular coor- dinates, and with the de\ufb01nitions of sin and cos. Polar coordinates with polar axes. in polar from f(r,\u03b8 ) = 1 r D E T 3 2 1 cos(T) r 1 r r \u00b4 \u00b5. The Cartesian coordinate of a point are $$\\left( {2, - 6} \\right)$$. Precalculus. The polar coordinates of a point are given. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates. The transformation from spherical coordinates to Cartesian coordinate is. Polar Coordinates-measures the distances (and direction) from the origin (radius)& the circle \u2022\u2022 (r, f), (radius): \u2022\u2022 ndusionf Rectangular Coordinates deal with horizontal & vertical distances, whereas polar coordinates deal with diagonal & circular distances. In these notes, we want to extend this notion of different coordinate systems to consider arbitrary coordinate systems. The graph of an equation in polar coordinates is the set of points which satisfy the equation. edu is a platform for academics to share research papers. The small change r in rgives us two concentric circles and the small change in gives us an angular wedge. In this system coordinates for a point P are and , which are indicated in Fig. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. If we restrict rto be nonnegative, then = describes the. 23 17_2_polar_coordinates. You can modify certain aspects of polar axes in order to make the chart more readable. Selection File type icon File name Description Size Revision Time User; \u010a: D32. Complete the unit circle with each angles\u2019 coordinates in the sets of parentheses as well as the simplified value of tangent at each angle. a) Find the polar coordinates of the points of intersection between the two curves. This de nition is worded as such in order to take into account that each point in the plane can have in nitely many representations in polar coordinates. Spherical polar coordinates. This is the xy-plane. Polar Coordinates (r-\u03b8)Ans: -0. Therefore r\u02d9(t) = (\u02d9rcos\u03b8 \u2212 r\u03b8\u02d9sin\u03b8)i + (\u02d9rsin\u03b8 + r\u03b8\u02d9cos\u03b8)j. But most commercial motion control cards do not support the polar coordinate, so this paper presents a program module based on polar coordinate system, which can be integrated into computer numeric control (CNC) controller based on motion control cards. Shade the. (b) Compute the Christo\ufb00el symbols of S in polar coordinates. In spherical polar coordinates we describe a point (x;y;z) by giving the distance r from the origin, the angle anticlockwise from the xz plane, and the angle \u02dafrom the z-axis. 5: Polar Coordinates Polar coordinate system, introduced by Isaac Newton, is often more convenient in some applications than the more traditional Cartesian, or rectangular, coordinate system. And you'll get to the exact same point. G15 and G16 G-Codes [Polar Coordinates and CNC Bolt Circles] CNCCookbook's G-Code Training What are Polar and Cartesian Coordinates? Until this point, we've strictly been using Cartesian Coordinates where X, Y, and Z represent distances from part zero (absolute coordinates) or from the current position (relative coordinates). Consider the top which is bounded above by z= p 4 x2 y2 and bounded below by z= p x2 + y2, as shown below. Choose a point in the plane that is called the pole (origin) and labeled O. 6 Graphs of Rational Functions. Polar coordinates part 4 This is a continuation of the polar coordinates part 3. In this section, we discuss how to graph pdfedit ubuntu package equations in polar coordinates on the. If we wish to relate polar coordinates back to rectangular coordinates (i. The graph above shows symmetry with respect to the y-axis. 4 2D Elastostatic Problems in Polar Coordinates Many problems are most conveniently cast in terms of polar coordinates. 8, as outlined in the. Spherical coordinates system (or Spherical polar coordinates) are very convenient in those problems of physics where there no preferred direction and the force in the problem is spherically symmetrical for example Coulomb's Law due to point. SPHERICAL POLAR COORDINATES. It is sometimes convenient to refer to a point by name, especially when this point occurs in multiple \\draw commands. Double Integrals in Polar Coordinates \u201cIntegrating Functions over circular regions\u201d Suppose we want to integrate the function f(x,y) = x2 over the fol-lowing region: 2 3 We would have to break the region up into three pieces. Mon Nov 11 - I retaught graphing roses and then we began converting from polar form to rectangular and rectangular to polar. Michael VanValkenburgh To make it easier to type and easier to read, this handout will focus on the computational aspects of integration in polar coordinates. 3 WS Polar Coordinates (Answers). Conversion: Rectangular to Polar/ Polar to Rectangular 2011 Rev by James, Apr 2011 1. [See how to convert rectangular and polar forms in the complex numbers chapter. 841 cos us d 5 2y3, 3 cos sud 5 2, 1 2 3 cos sud 5 21, s2r, u 1 pd fsud 5 2gsu 1 pd, p, sr, ud. Convert the following equation of a circle to polar coordinates: 2x2 +3x+2y2 + \u22125y = 7 7. We know sine starts at zero, and then grows until the function reaches a height of one at \u02c7=2. SCHROEDINGER'S EQUATION IN SPHERICAL POLAR COORDINATES The magnitude of a central force on an object depends on only the distance of that object. a) Find the polar coordinates of the points of intersection between the two curves. Most of the things we've done can also be done in the polar, cylindrical, and spherical coordinate as well. Describe the graph. However, we can use other coordinates to determine the location of a point. 21 Locating a point in polar coordinates Let\u2019s look at a specific example. ) The graph of = , where is a constant, is the line of inclination. Thus its area will be Z 2\u03c0 0 R2 2 d\u03d1 = R2 2 x 2\u03c0 0. 54 Example 1 \u2013 Using Polar Coordinates to Describe a Region Use polar coordinates to describe each region shown in. This is an advantage of using the polar form. Polar Rectangular Regions of Integration. Convert each pair of rectangular coordinates to polar coordinates where r and. We basically use a 2D formation having two coordinates x and y, if you are wishing to create graph points on a coordinate plane then below we are providing instructions of doing that. \u03b8 \u03c0 4 is the straight line through the origin pole making an angle of \u03c0 4 University of Calgary MATH 267 - Summer 2019 Math267-Polar-Coordinates-Double-Integrals. We shall show how easy it becomes using polar coordinates instead. Homework 2: Spherical Polar Coordinates Due Monday, January 27 Problem 1: Spherical Polar Coordinates Cartesian coordinates (x,y,z) and spherical polar coordinates (r,\u03b8,\u03d5) are related by x = r sin\u03b8 cos\u03d5 y = r sin\u03b8 sin\u03d5 z = r cos\u03b8. Example 3 2Given the ellipse with equation 9 x2 + 25 y = 225, find the major and minor axes, eccentricity, foci and vertices. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. Viewed 11k times 3. If we restrict rto be nonnegative, then = describes the. Symmetry with. 25 and also 12? How about 17 and 13? Good times. View Notes - Polar-Coordinate-System. 5, 30\u00b0), (-1. \u2022 \u03b8is measured from an arbitrary reference axis \u2022 e r and e\u03b8 are unit vectors along +r & +\u03b8dirns. I'm always amazed by what my students come up with on this one - I've even had a student who designed a penguin using polar equations!. Extensions Graphing the polar equations will help the students to make the connection when they are learning to change polar coordinates to rectangular coordinates and back The students can visually see the points on the polar axis and compare the point on the rectangular axis. This discussion is critical for you to understand in order to correctly determine the polar coordinates. 1 Illustrating polar coordinates. The graphing worksheets are randomly created and will never repeat so you have an endless supply of quality graphing worksheets to use in the classroom or at home. d) \u02dc\u02dd3, \u02dd\u03a0 6 \u02da. Then a number of important problems involving polar coordinates are solved. Convert the following equation to polar coordinates: y = \u2212 4 3 x 6. Find the distance between the points. pdf View Download. In particular, how the angle increases counter-clockwise and how the radius rincreases going away from the origin. Convert the equation of the circle r= 2sinto rectangular coordinates and nd the center and radius of the circle. The polar coordinate system (r, \u03b8) and the Cartesian system (x, y) are related by the following expressions: With reference to the two-dimensional equ ations or stress transformation. Polar coordinates are not unique. In a rectangular coordinate system, we were plotting points based on an ordered pair of (x, y). 3 convert the equation from polar to rectangular and draw the graph (a) r = 2sec\u03b8 (b) \u03b8 = \u03c0/3 (c* graph only) r = \u03b8 4 find the slope of the tangent line at the given value of \u03b8 (a) r = 3, \u03b8 = \u03c0/6 (b) r = 1+3sin\u03b8, \u03b8 = \u03c0/4 5* What is the distance formula for two points in polar coordinates, (r1, \u03b81) & (r2, \u03b82) ?. This is the result of the conversion to polar coordinates in form. Arc length of polar curves. r = 2 and \u03b8= 30\u00b0, so P is located 2 units from the origin in the positive direction on a ray making a 30\u00b0angle with the polar axis. z axis up, \u03b8 is sometimes called the zenith angle and \u03c6 the azimuth angle. We begin with a brief review of polar coordinates. Polar sun path chart program This program creates sun path charts using polar coordinate for dates spaced about 30 days apart, from one solstice to the next. Here, the two-dimensional Cartesian relations of Chapter 1 are re-cast in polar coordinates. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations. The rst coordinate is the distance of the point from the origin (0;0), and the second coordinate is the angle, in standard. Creating Constellations on a Coordinate Plane Grade Level/Subject: Science or math (Could be simplified to 3rd grade level by making the coordinates all positive. If we restrict rto be nonnegative, then = describes the. We see this general pattern in the circle of gure 2. We interpret r as the distance from the sun and \u03b8 as the planet's angular bearing, or its direction from a fixed point on the sun. $\\begingroup$ See answer here with diagram showing the directions and magnitudes of the changes in velocity (in polar form). This is the xy-plane. If C is a circle of radius R, then its polar equation is f(\u03d1) = R where 0 6 \u03d1 6 2\u03c0. 8 Polar Equations of Conics We have seen that geometrically the conic sections are related since they are all created by intersecting a plane with a right circular cone. 442, C 5 s21, 0. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. A point P is located at (r,\u03b8) in a polar coordinate system if the distance from P. Showing top 8 worksheets in the category - Number Planes. Therefore r\u02d9(t) = (\u02d9rcos\u03b8 \u2212 r\u03b8\u02d9sin\u03b8)i + (\u02d9rsin\u03b8 + r\u03b8\u02d9cos\u03b8)j. However, doing the math is the tricky part. We shall show how easy it becomes using polar coordinates instead. Polar - Rectangular Coordinate Conversion Calculator. We basically use a 2D formation having two coordinates x and y, if you are wishing to create graph points on a coordinate plane then below we are providing instructions of doing that. However, the Coriolis acceleration we are discussing here is a real acceleration and which is present when rand both change with time. (5, 960\u00b0) SOLUTION: Let P(r, \u03b8) = (5, 960\u00b0). A point P in the plane, has polar coordinates (r; ), where r is the distance of the point from the origin and is the angle that the ray jOPjmakes with the positive x-axis. This is an advantage of using the polar form. Then we will use these formulas to convert Cartesian equations to polar coordinates, and vice versa. Changing the solvent has little. Frame of Reference In the polar coordinate system, the frame of reference is a point O that we call the pole and a ray that. polar coordinates pl (plural only) ( mathematics ) The coordinates of a point in a plane, measured as its Cartesian distance from the origin and the angle measured anticlockwise / counterclockwise from the x -axis to a line joining the point to the origin. Graphing in Polar Coordinates Jiwen He 1 Polar Coordinates 1. We need to subtract 960 by 180k, such that the result is between 0 and 180. Eliminate the parameter and identify the graph of the parametric curve. The equations of the 10 - and 20 - radius circles are r = 10 and r = 20, respectively. A point in polar coordinates requires an angle a, in degrees, and distance from the origin, r. It\u2019s easy to convert rectangular coordinates to polar coordinates when the angle of the polar coordinate is 0\u00b0, 30\u00b0, 45\u00b0, 60\u00b0, or 90\u00b0. Spherical polar coordinates provide the most convenient description for problems involving exact or approximate spherical symmetry. Lesson 6: Polar, Cylindrical, and Spherical coordinates 1. How does this compare with the computation in the usual coordinate system \u03c8(x,y) = (x,y,0)? Answer: In order to compute the Christo\ufb00el symbols, \ufb01rst we need to compute the partials of E, Fand G: E \u03c1 = 0 E \u03b8 = 0 F \u03c1 = 0 F \u03b8 = 0 G \u03c1 = 2\u03c1 G \u03b8 = 0. Symmetry with. r (x ;y)=( rcos( ) sin( )) =\u02c7 6 =\u02c7 3 Polar coordinates are related to ordinary (cartesian) coordinates by the formulae x = r cos( ) y = r sin( ) r = p x 2+ y = arctan(y=x):. Spherical-polar coordinates. Search this site. Superposition of separated solutions: u = A0=2 + X1 n=1 rn[An cos(n ) + Bn sin(n )]: Satisfy boundary condition at r = a,. In fact, we will look at how to calculate the area given one polar function, as well as when we need to find the area between two polar curves. To find the game you're looking for, use the filter below. For example, the behavior of the drum surface when you hit it by a stick would be best described by the solution of the wave equation in the polar coordinate system. pdf), Text File (. Elasticity equations in polar coordinates (See section 3. d is the perpendicular distance from the line to the origin. In this section, we explore the question of how to quantize a system in curvilinear coordinates, using plane polar coordinates as an example. The r represents the distance you move away from the origin and \u03b8 represents an angle in standard position. Lecture 36: Polar Coordinates A polar coordinate system, gives the co-ordinates of a point with reference to a point Oand a half line or ray starting at the point O. Review: Polar coordinates De\ufb01nition The polar coordinates of a point P \u2208 R2 is the ordered pair (r,\u03b8) de\ufb01ned by the picture. b) Show that the area of R is 1 (9 3 2) 16 \u2212 \u03c0. We will look at polar coordinates for points in the xy-plane, using the origin (0;0) and the positive x-axis for reference. And if we talking about polar paper for maths so this is a type graph paper which is used in many projects and also. \uf075A location is defined by its distance in x,y from the origin point. POLAR COORDINATE GRAPH PAPER Author: Frances Elizabeth Blount Created Date: 10/26/2007 9:37:15 AM. The easiest kind of region R to work with is a rectangle. 5: Polar Coordinates Polar coordinate system, introduced by Isaac Newton, is often more convenient in some applications than the more traditional Cartesian, or rectangular, coordinate system. Convert Rectangular to Polar Coordinates Polar Axis If a point P has rectangular coordinates (x, y) then the polar coordinates (r, e) of P are given by and tan-I y \u2014, when x > O tan tan \u2014 or + 1800, when x < O. Complete the unit circle with each angles\u2019 coordinates in the sets of parentheses as well as the simplified value of tangent at each angle. Polar Coordinates - Solution Question 1 Plot the points with Cartesian coordinates A 8 p 3;8 and B 5 4;5 p 3 4 and then convert them to polar coordinates. Apr 9 - Today I handed back and we went over the unit 5 and Unit 6 tests. Write an equation for this curve in rectangular coordinates. The polar coordinate system,(r, ), is convenient if we want to consider radial distance from a \ufb01xed point (origin, or pole) and bearing (direction). The orientation of a plane curve can be represented by arrows drawn along the curve. Selection File type icon File name Description Size Revision Time User; \u010a: D32. The photochemistry of 4-haloanilines and 4-halo-N,N-dimethylanilines has been studied in apolar, polar aprotic, and protic solvents. In polar coordinates the position of an object $$R$$ distance from the origin as represented in the diagram above is modelled $$\\mathbf{r} = R \\hat{r}$$ The velocity and acceleration in polar coordinates is derived by differentiating the position vector. The following steps can be used for graphing polar curves: 1. Using relative coordinate, points entered by typing @x,y [Enter] Polar Coordinates Polar coordinates used when you need to draw the next points at specify angle. Polar coordinate lines. Polar, Cylindrical, and Spherical Coordinates 1. 64 Spoke Degrees. The polar coordinate system is a two-dimensional coordinate system using a polar grid: The r and \u03b8 coordinates of a point P measure respectively the distance from P to the origin O and the angle between the line OP and the polar axis. The solution is such that the stress components are in the form of a Fourier series in \u03b8 {\\displaystyle \\theta \\,}. Polar Coordinates. Find the distance between the points. The area element in polar coordinates In polar coordinates the area element is given by dA = r dr d\u03b8. 1 Exponential Equations Blank. Graphing in Polar Coordinates Jiwen He 1 Polar Coordinates 1. Find a different pair of polar coordinates for each point such that 0 \u2264 \u2264 180\u00b0 or 0 \u2264 \u2264 \u03c0. 2 Calculus In The Polar Coordinate System Contemporary Calculus 4 Area in Rectangular Coordinates (Fig. Polar Coordinates Polar coordinates of a point consist of an ordered pair, r \u03b8( , ), where r is the distance from the point to the origin, and \u03b8 is the angle measured in standard position. 3 WS Polar Coordinates (Answers). The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Pre-Calculus Worksheet Name: _____ Section 10. Polar-coordinate equations for lines A polar coordinate system in the plane is determined by a Pole Pand a half-line called the polar axis, extending from Pto the right in Figure 1. However, it still is a useful tool to give you an introduction to the concepts pertaining to polar coordinates. 04 Double Integrals in Polar Coordinates. Find the polar coo dinate. Show the angle \u03b8 between two lines with slopes m 1 and m 2 is given by the equation tan\u03b8 = m 2 \u2212m 1 1\u2212m 2m 1 I've added some more information to the diagram, based on the hint to include the angle the lines make with the x-axis. 54 Example 1 \u2013 Using Polar Coordinates to Describe a Region Use polar coordinates to describe each region shown in. Also remember that there are three types of symmetry - y-axis, x-axis, and origin. Media in category \"Polar coordinate system\" The following 124 files are in this category, out of 124 total. It only takes a minute to sign up. The use of polar graph paper or circular graph paper uses, in schools and colleges math teachers, are also still making assignments that require students to make a graph and draw my own by hands. Coordinate Graph Paper PDF. 7 is self explanatory. \u2022 \u03b8is measured from an arbitrary reference axis \u2022 e r and e\u03b8 are unit vectors along +r & +\u03b8dirns. 0 International License. Polar Coordinates. This coordinate system is convenient to use when the distance and direction of a particle are measured relative to a fixed point or when a particle is fixed on or moves along a rotating arm. The polar coordinate system is an adaptation of the two-dimensional coordinate system invented in 1637 by French mathematician Ren \u00e9 Descartes (1596 \u2013 1650). But many teachers might prefer that you measure angles by yourself using a protractor on blank paper. As previously noted, the Cartesian coordinate (a,b) refers to the point a cen-timeters in the x-direction and b centimeters in the y-direction. We need to subtract 960 by 180k, such that the result is between 0 and 180. In particular, how the angle increases counter-clockwise and how the radius rincreases going away from the origin. pdf), Text File (. 5) 1,150 6) 1, 240 7) Plot 3, 4 A on the polar grid and find three additional pairs of polar coordinates that name the point if 22. Double Integrals in Polar Coordinates. A polar coordinate system, gives the co-ordinates of a point with reference to a point O and a half line or ray starting at the point O. Polar coordinates and Defining the Polar Coordinate Axes Any right triangle can be defined on a circle with radius r. This introduction to polar coordinates describes what is an effective way to specify position. r is the radius, and \u03b8 is the angle formed between the polar axis (think of it as what used to be the positive x-axis) and the segment connecting the point to the pole (what used to be the origin). (ii) Find two other pairs of polar coordinates for each point, one with r \u02dc 0 and one with r \u02da 0. The first equation looks easy but there is a hidden assumption that you need to be aware of. 2 Calculus In The Polar Coordinate System Contemporary Calculus 4 Area in Rectangular Coordinates (Fig. Write the inequalities for. Polar coordinates are a set of values that quantify the location of a point based on 1) the distance between the point and a fixed origin and 2) the angle between. Frame of Reference In the polar coordinate system, the frame of reference is a point O that we call the pole and a ray that emanates from it that we call the polar axis. Since the axis of the parabola is vertical, the form of the equation is Now, substituting the val-ues of the given coordinates into this equation, we obtain. We must also know how to convert from rectangular to polar coordinates and from polar coordinates to. Apr 9 - Today I handed back and we went over the unit 5 and Unit 6 tests. 0 Unported by Lantonov. The initial line may be identified with the x-axis of rectangular Cartesian coordinates, as. 841d f0, 2pd u3 5 cos21s2y3d 5 0. Use absolute polar coordinates when you know the precise distance and angle coordinates of the point. To specify a clockwise direction, enter a negative value for the angle. Math 126 Worksheet 5 Polar Coordinates Graphing Polar Curves The aim of this worksheet is to help you familiarize with the polar coordinate system. 10) It is often convenient to work with variables other than the Cartesian coordinates x i ( = x, y, z). Pre-Calculus Notes Name: _____ Section 10. 1 De ning Polar Coordinates oT nd the coordinates of a point in the polar coordinate system, consider Figure 1. Graph the point P, (r; ) = 3;\u02c7 3. There are some aspects of polar coordinates that are tricky. org are unblocked. 11) ( , ), ( , ) 12) ( , ), ( , ) Critical thinking question: 13) An air traffic controller's radar display uses polar coordinates. Polar Coordinates. 5 3, 3 For problems 17-20, convert the rectangular coordinates to polar coordinates with r 0 and 0 2. The equations of the 10 - and 20 - radius circles are r = 10 and r = 20, respectively. location than conventional Cartesian coordinates. 362 Chapter 10 Conics, Parametric Equations, and Polar Coordinates 21. The conversion from polar to rectangular coordinates is the same idea as converting rectangular form to polar form in complex numbers. Review: Polar coordinates De\ufb01nition The polar coordinates of a point P \u2208 R2 is the ordered pair (r,\u03b8) de\ufb01ned by the picture. First measure a circle feature. We graph some of the basic functions in polar coordinates using LiveMath and a graphing calculator. Polar coordinates An alternative to using rectangular coordinates (x and y) to specify points in the plane is to specify how far the point is from the origin and the direction it lies in. The ordered pair specifies a point's location based on the value of r and the angle, \u03b8, from the polar axis. 5355 0 -10] x = 1\u00d74 5. Polar Curves Curves in Polar Coordinate systems are called Polar Curves, which can be written as r = f(\u00b5) or, equivalently, as F(r;\u00b5) = 0. Ciencia y Tecnolog\u00eda, 32(2): 1-24, 2016 - ISSN: 0378-0524 3 II. (b) Find the velocity of the particle in polar coordinates. The graph above shows symmetry with respect to the y-axis. More Graphing Polar Equations. Enter this lesson and corresponding worksheet covering the basics of the polar coordinate system. What happens when you divide a circle by 365. d) \u02dc\u02dd3, \u02dd\u03a0 6 \u02da. r is a directed distance from the pole to P. As the function approaches \u02c7, the value reduces back to zero. 2 : Apr 12, 2018, 11:37 AM. (it can be positive, negative, or zero. (iii) Find the Cartesian coordinates of the point. Find the volume of the region bounded by the paraboloid z= 2 4x2 4y2 and the plane z= 0. a) r=3sec\u03b8 b) r=\u22123sin\u03b8 c) rcsc 1\u03b8= 5) Convert the rectangular equation to polar form. Beautifull!! #2 Andre, December 19, 2009 at 12:22 p. Graph the point P, (r; ) = 3;\u02c7 3. SCHROEDINGER'S EQUATION IN SPHERICAL POLAR COORDINATES The magnitude of a central force on an object depends on only the distance of that object. In other words, we need to rewrite the equation so that the denominator begins with 1. These charts print on a standard sheet of 8 1/2 x 11 paper. Example: What is (12,5) in Polar Coordinates?. Home Decorating Style 2020 for Polar Coordinate System Pdf, you can see Polar Coordinate System Pdf and more pictures for Home Interior Designing 2020 76667 at Manuals Library. I Computing volumes using double integrals. 686 CHAPTER 9 POLAR COORDINATES AND PLANE CURVES The simplest equation in polar coordinates has the form r= k, where kis a positive constant. k = 5 Since k is odd, we need to replace r with -r to obtain the correct polar coordinates. I Changing Cartesian integrals into polar integrals. 0 Unported by Lantonov. The area inside the polar curve r = 3 + 2cos q is-4 -2 2 4-4-2 2 4 (A) 9. Cylindrical Coordinates Transforms The forward and reverse coordinate transformations are != x2+y2 \"=arctan y,x ( ) z=z x =!cos\" y =!sin\" z=z where we formally take advantage of the two argument arctan function to eliminate quadrant confusion. Polar coordinates An alternative to using rectangular coordinates (x and y) to specify points in the plane is to specify how far the point is from the origin and the direction it lies in. Polar coordinates and applications Let\u2019s suppose that either the integrand or the region of integration comes out simpler in polar coordinates (x= rcos and y= rsin ). (ii) Find two other pairs of polar coordinates for each point, one with r \u02dc 0 and one with r \u02da 0. Precalculus: Polar Coordinates Practice Problems 3. Let (r,\u03b8) denote the polar coordinates describing the position of a particle. Cartesian/Polar Coordinates Junior high school The connection between Cartesian coordinates and Polar coordinates is established by basic trigonometry. Polar Graph Paper. A Cartesian coordinate system is the unique coordinate system in which the set of unit vectors at different points in space are equal. Ask Question Asked 7 years, 3 months ago. Algebra of complex numbers You should use the same rules of algebra as for real numbers,. POLAR COORDINATES CONTINUED Writing Equations in Polar Form x y r r r, , cos , sin oo T T T I can convert an equation from Rectangular Form to Polar Form. Any geometric object in the plane is a set (collection) of points, so we can describe it by a set of coordinate pairs. (i) Plot each point. To evaluate ZZ R f(x,y)dxdy proceed as follows: \u2022 work out the limits of integration if they are not. Show Step-by-step Solutions. The ranges of the variables are 0 < p < \u00b0\u00b0 0 < < 27T-00 < Z < 00 A vector A in cylindrical coordinates can be written as (2. 23 17_2_polar_coordinates. 7 is self explanatory. The fact that a single point has many pairs of polar coordinates can cause complications. Example Sketch the curve described by the polar equation. This allows you to fully utilize the paper size that you have on hand. 1 Polar Coordinates and Polar Equations OBJECTIVE 1: Plotting Points Using Polar Coordinates In this section we begin our study of the polar coordinate system. You can see this by just drawing unit vectors at each point on, say, a circle: (draw). The transformation from spherical coordinates to Cartesian coordinate is. (5, 960\u00b0) SOLUTION: Let P(r, \u03b8) = (5, 960\u00b0). A CNC program module based on polar coordinate system Article (PDF Available) in International Journal of Advanced Manufacturing Technology 68(5-8) \u00b7 September 2013 with 2,790 Reads. No o\ufb03ce hours Tuesday 2/19. y x y x P(x,y) 0 y x y x P(x,y) 0 r a) b. You should be familiar with the Cartesian Coordinate System, also called rectangular coor- dinates, and with the de\ufb01nitions of sin and cos. A point P is located at (r,\u03b8) in a polar coordinate system if the distance from P. A polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This measurement will display in the Cartesian coordinates. The polar coordinates r and \u03c6 can be converted to the Cartesian coordinates x and y by using the trigonometric functions sine and cosine: = \u2061, = \u2061. 21 Locating a point in polar coordinates Let\u2019s look at a specific example. Basic polar coordinates are those coordinates with angles not lower than -360\u00b0 and not higher than +360\u00b0. The location of P in the plane can also be described using polar coordinates. Polar Coordinates This file contains one interactive page that your students could use to practice plotting polar coordinates. First, fix an origin (called the pole) and an initial ray from O. Consider the top which is bounded above by z= p 4 x2 y2 and bounded below by z= p x2 + y2, as shown below. How does this compare with the computation in the usual coordinate system \u03c8(x,y) = (x,y,0)? Answer: In order to compute the Christo\ufb00el symbols, \ufb01rst we need to compute the partials of E, Fand G: E \u03c1 = 0 E \u03b8 = 0 F \u03c1 = 0 F \u03b8 = 0 G \u03c1 = 2\u03c1 G \u03b8 = 0. Polar coordinates are in the form r, , where is the independent variable. We want to nd another way to get to the point (x;y). The Polar Coordinate System is a different way to express points in a plane. The polar form of (a,b) is illustrated in Figure 1. Number Planes. Polar coordinates and applications Let's suppose that either the integrand or the region of integration comes out simpler in polar coordinates (x= rcos and y= rsin ). Plane polar coordinates pdf Plane polar coordinates pdf Plane polar coordinates pdf DOWNLOAD! DIRECT DOWNLOAD! Plane polar coordinates pdf Polar Coordinates r, \u03b8 in the plane are described by r distance from the origin and \u03b8 0, 2\u03c0 is the counter-clockwise angle. It is sometimes convenient to refer to a point by name, especially when this point occurs in multiple \\draw commands. generally start to learn to think in terms of polar coordinates. Consider the top which is bounded above by z= p 4 x2 y2 and bounded below by z= p x2 + y2, as shown below. 2 We can describe a point, P, in three different ways. Extensions Graphing the polar equations will help the students to make the connection when they are learning to change polar coordinates to rectangular coordinates and back The students can visually see the points on the polar axis and compare the point on the rectangular axis. Double Integrals in Polar Coordinates 1. Finally, the Coriolis acceleration 2r \u00d6. 5355 0 -10] x = 1\u00d74 5. Stirling's Web Site. Compass Labels on Polar Axes.", "date": "2020-08-03 14:15:01", "meta": {"domain": "gesuitialquirinale.it", "url": "http://gesuitialquirinale.it/rdil/polar-coordinates-pdf.html", "openwebmath_score": 0.9085190296173096, "openwebmath_perplexity": 584.2148996203689, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9843363485313248, "lm_q2_score": 0.8947894703109853, "lm_q1q2_score": 0.8807737999101936}}
{"url": "https://www.physicsforums.com/threads/integral-yielding-2-answers.727979/", "text": "1. Dec 12, 2013\n\n### Appleton\n\n1. The problem statement, all variables and given/known data\n\n$\\int \\frac{1}{\\sqrt{1-x^{2}}} dx$\n\nI seem to be getting two incompatible answers when I substitute x with $sin \\theta$ and $cos \\theta$. Could someone please help me with where I'm going wrong.\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nfirst attempt (the correct answer I believe):\n\n$let\\ x = sin \\theta$\n\n$dx = cos \\theta\\ d\\theta$\n\n$\\int \\frac{cos \\theta}{\\sqrt{1-sin^{2}\\theta}} d\\theta$\n\n$= \\int \\frac{cos \\theta}{\\sqrt{cos^{2}\\theta}} d\\theta$\n\n$= \\int \\frac{cos \\theta}{{cos\\theta}} d\\theta$\n\n$= \\int 1\\ d\\theta$\n\n$= \\theta$\n\n$x = sin \\theta$\n\n$\\theta = arcsin x$\n\nsecond attempt (the wrong answer I believe)\n\n$let\\ x = cos \\theta$\n\n$dx = -sin \\theta\\ d\\theta$\n\n$\\int \\frac{-sin \\theta}{\\sqrt{1-cos^{2}\\theta}} d\\theta$\n\n$= \\int \\frac{-sin \\theta}{\\sqrt{sin^{2}\\theta}} d\\theta$\n\n$= \\int \\frac{-sin \\theta}{{sin\\theta}} d\\theta$\n\n$= \\int -1\\ d\\theta$\n\n$= -\\theta$\n\n$x = cos \\theta$\n\n$\\theta = -arccos x$\n\n2. Dec 12, 2013\n\n### sandyp\n\nIts simple...we know that d/dx arc sinx = (1-x^2)^-1/2\nd/dx arc cos x = -(1-x^2)^-1/2 (notice the minus sign)\nas integral is antiderivative of a function you are getting the same values\nyes integral arc sinx=(minus)integral arc cos x..Hope your doubt is clarified\n\n3. Dec 12, 2013\n\n### ShayanJ\n\nIn fact $\\int \\frac{1}{\\sqrt{1-x^2}}dx=\\sin^{-1}{x}+C_1 \\ and \\ \\int \\frac{1}{\\sqrt{1-x^2}}dx=-\\cos^{-1}{x}+C_2$\nIf you take a look at the plots of the functions $\\sin^{-1}{x}$ and $\\cos^{-1}{x}$,you can see that you can get one of them by multiplying the other by a minus sign and then adding a constant(or vice versa!)...and you can see that you have the negative sign and also the constant in the results of the integral!\n\n4. Dec 12, 2013\n\n### Appleton\n\nThanks for clearing that up for me.\n\n5. Dec 12, 2013\n\n### Staff: Mentor\n\nNo one has mentioned this. There's a trig identity that can shed some light here.\n\n$sin^{-1}(x) + cos^{-1}(x) = \\pi/2$\nThe two functions differ by a constant. If an indefinite integral produces two different antiderivatives, those functions can differ by at most a constant.", "date": "2017-11-23 11:54:28", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integral-yielding-2-answers.727979/", "openwebmath_score": 0.8216156959533691, "openwebmath_perplexity": 1040.3559777315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363503693294, "lm_q2_score": 0.8947894562828415, "lm_q1q2_score": 0.8807737877464088}}
{"url": "https://mathhelpboards.com/threads/integral-substitution-method-problem.5634/", "text": "Integral - substitution method problem\n\nYankel\n\nActive member\nHello all\n\nI am working on this integral\n\n$\\int \\frac{x^{2}+1}{x^{4}+1}dx$\n\nNow, I have tried this way:\n\n$u=x^{2}+1$\n\nafter I did:\n\n$\\int \\frac{x^{2}+1}{\\left ( x^{2}+1 \\right )\\left ( x^{2}-1 \\right )}dx$\n\nBut I got stuck, I got:\n\n$\\frac{1}{2}\\cdot \\int \\frac{1}{u\\sqrt{u-1}}dx$\n\nI thought of making another substitution, but I tried and failed. Help required\n\nProve It\n\nWell-known member\nMHB Math Helper\nHello all\n\nI am working on this integral\n\n$\\int \\frac{x^{2}+1}{x^{4}+1}dx$\n\nNow, I have tried this way:\n\n$u=x^{2}+1$\n\nafter I did:\n\n$\\int \\frac{x^{2}+1}{\\left ( x^{2}+1 \\right )\\left ( x^{2}-1 \\right )}dx$\n\nBut I got stuck, I got:\n\n$\\frac{1}{2}\\cdot \\int \\frac{1}{u\\sqrt{u-1}}dx$\n\nI thought of making another substitution, but I tried and failed. Help required\nWell first of all, \\displaystyle \\displaystyle \\begin{align*} x^4 + 1 \\end{align*} is NOT equal to \\displaystyle \\displaystyle \\begin{align*} \\left( x^2 + 1 \\right) \\left( x^2 - 1 \\right) \\end{align*}.\n\nI would try\n\n\\displaystyle \\displaystyle \\begin{align*} x^4 + 1 &= x^4 + 2x^2 + 1 - 2x^2 \\\\ &= \\left( x^2 + 1 \\right) ^2 - \\left( \\sqrt{2} \\, x \\right) ^2 \\\\ &= \\left( x^2 - \\sqrt{2}\\, x + 1 \\right) \\left( x^2 + \\sqrt{2}\\,x + 1 \\right) \\end{align*}\n\nand so\n\n\\displaystyle \\displaystyle \\begin{align*} \\int{ \\frac{x^2 + 1}{x^4 + 1}\\,dx} &= \\int{ \\frac{x^2 - \\sqrt{2}\\,x + 1 + \\sqrt{2}\\,x}{ \\left( x^2 - \\sqrt{2}\\,x + 1 \\right) \\left( x^2 + \\sqrt{2}\\,x + 1 \\right) } \\, dx} \\\\ &= \\int{ \\frac{1}{x^2 + \\sqrt{2}\\,x + 1} \\,dx} + \\sqrt{2} \\int{ \\frac{x}{ \\left( x^2 - \\sqrt{2}\\,x + 1 \\right) \\left( x^2 + \\sqrt{2}\\,x + 1 \\right) } \\, dx} \\\\ &= \\int{ \\frac{1}{x^2 + \\sqrt{2}\\,x + \\left( \\frac{\\sqrt{2}}{2} \\right) ^2 - \\left( \\frac{\\sqrt{2}}{2} \\right) ^2 + 1 } \\, dx} + \\sqrt{2} \\int{ \\frac{x}{x^4 + 1} \\, dx} \\\\ &= \\int{ \\frac{1}{ \\left( x + \\frac{\\sqrt{2}}{2} \\right) ^2 + \\frac{1}{2} } \\, dx} + \\frac{\\sqrt{2}}{2} \\int{ \\frac{2x}{ \\left( x^2 \\right) ^2 + 1 }\\, dx} \\end{align*}\n\nNow in the first integral, let \\displaystyle \\displaystyle \\begin{align*} x + \\frac{\\sqrt{2}}{2} = \\frac{\\sqrt{2}}{2}\\tan{(\\theta)} \\implies dx = \\frac{\\sqrt{2}}{2}\\sec^2{(\\theta)}\\,d\\theta \\end{align*} and in the second integral let \\displaystyle \\displaystyle \\begin{align*} x^2 = \\tan{(\\phi)} \\implies 2x\\,dx = \\sec^2{(\\phi)}\\,d\\phi \\end{align*} and the integrals become\n\n\\displaystyle \\displaystyle \\begin{align*} \\int{ \\frac{1}{\\left( x + \\frac{\\sqrt{2}}{2} \\right) ^2 + \\frac{1}{2} } \\, dx} + \\frac{\\sqrt{2}}{2} \\int{ \\frac{2x}{ \\left( x^2 \\right) ^2 + 1 } \\, dx} &= \\int{ \\frac{1}{ \\left[ \\frac{\\sqrt{2}}{2} \\tan{(\\theta)} \\right] ^2 + \\frac{1}{2} } \\cdot \\frac{\\sqrt{2}}{2}\\sec^2{(\\theta)}\\,d\\theta } + \\frac{\\sqrt{2}}{2} \\int{ \\frac{\\sec^2{(\\phi)}}{ \\tan^2{(\\phi)} + 1 } \\, d\\phi } \\\\ &= \\frac{\\sqrt{2}}{2} \\int{ \\frac{\\sec^2{(\\theta)}}{ \\frac{1}{2}\\tan^2{(\\theta)} + \\frac{1}{2} } \\, d\\theta} + \\frac{\\sqrt{2}}{2} \\int{ \\frac{\\sec^2{(\\phi)}}{\\sec^2{(\\phi)}}\\,d\\phi} \\\\ &= \\sqrt{2} \\int{ \\frac{\\sec^2{(\\theta)}}{ \\tan^2{(\\theta)} + 1 } \\, d\\theta} + \\frac{\\sqrt{2}}{2} \\int{ 1 \\, d\\phi } \\\\ &= \\sqrt{2} \\int{ \\frac{\\sec^2{(\\theta)}}{\\sec^2{(\\theta)}}\\,d\\theta} + \\frac{\\sqrt{2}}{2} \\phi + C_1 \\\\ &= \\sqrt{2} \\int{ 1\\,d\\theta } + \\frac{\\sqrt{2}}{2} \\arctan{ \\left( x^2 \\right) } + C_1 \\\\ &= \\sqrt{2} \\, \\theta + C_2 + \\frac{\\sqrt{2}}{2}\\arctan{ \\left( x^2 \\right) } + C_1 \\\\ &= \\sqrt{2} \\arctan{ \\left( \\sqrt{2}\\, x + 1 \\right) } + \\frac{\\sqrt{2}}{2} \\arctan{ \\left( x^2 \\right) } + C \\textrm{ where } C = C_1 + C_2 \\end{align*}\n\nYankel\n\nActive member\nsorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it.\n\nI have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ?\n\nThanks !\n\npaulmdrdo\n\nActive member\nI would try this method but this is open for any correction\n\n$\\displaystyle \\int\\frac{x^2+1}{x^4+1}dx$\n\nby letting\n\n$\\displaystyle u\\,=\\,x^2$\n$\\displaystyle du\\,=\\,2xdx$\n$\\displaystyle dx\\,=\\,\\frac{du}{2x}$\n$\\displaystyle x\\,=\\,u^{\\frac{1}{2}}$\n\ncan you continue..?\n\nZaidAlyafey\n\nWell-known member\nMHB Math Helper\nFor practice try\n\n$$\\displaystyle \\int \\frac{1}{1+x^4}\\,dx$$\n\n$$\\displaystyle \\int \\frac{1}{\\sqrt[4]{1+x^4}}\\, dx$$\n\n$$\\displaystyle \\int \\frac{x^2}{1+x^4}\\, dx$$\n\nThey are a little bit tougher .\n\n[*] By the way I have a way to solve the integral using complex numbers , the answer is nasty .\n\nMarkFL\n\nStaff member\nUsing partial fractions, we may write:\n\n$$\\displaystyle \\int\\frac{x^2+1}{x^4+1}\\,dx=\\frac{1}{2}\\int\\frac{1}{x^2+\\sqrt{2}x+1}+\\frac{1}{x^2-\\sqrt{2}x+1}\\,dx$$\n\nCompleting the squares...\n\n$$\\displaystyle \\int\\frac{x^2+1}{x^4+1}\\,dx=\\frac{1}{2}\\int\\frac{1}{\\left(x+\\frac{1}{\\sqrt{2}} \\right)^2+\\frac{1}{2}}+\\frac{1}{\\left(x-\\frac{1}{\\sqrt{2}} \\right)^2+\\frac{1}{2}}\\,dx$$\n\nNow, we may simply apply the formula:\n\n$$\\displaystyle \\int\\frac{du}{u^2+a^2}=\\frac{1}{a}\\tan^{-1}\\left(\\frac{u}{a} \\right)+C$$\n\nto get (after simplification):\n\n$$\\displaystyle \\int\\frac{x^2+1}{x^4+1}\\,dx=\\frac{1}{\\sqrt{2}} \\left(\\tan^{-1}\\left(\\sqrt{2}x+1 \\right)+ \\tan^{-1}\\left(\\sqrt{2}x-1 \\right) \\right)+C$$\n\nProve It\n\nWell-known member\nMHB Math Helper\nsorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it.\n\nI have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ?\n\nThanks !\nIt's a method known as trigonometric substitution. I suggest you read what Mark has written in post #8 of this thread.", "date": "2020-11-30 20:42:39", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/integral-substitution-method-problem.5634/", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 1517.3806120013776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363485313248, "lm_q2_score": 0.894789457685656, "lm_q1q2_score": 0.8807737874826229}}
{"url": "https://math.stackexchange.com/questions/1085643/convex-set-with-empty-interior-is-nowhere-dense/1096673", "text": "# Convex set with empty interior is nowhere dense?\n\nSuppose $C\\subseteq\\mathbb R^n$\u00a0is a convex set and $C^o=\\varnothing$. Is it necessarily true that $(\\overline C)^o=\\varnothing$? In general, is this true if $\\mathbb R^n$ is replaced by a topological vector space $X$? Or can a counterexample be found?\n\nI know that if $C^o\\neq\\varnothing$, then $C^o=(\\overline C)^o$, so my question is whether this result can be generalized to the case when $C^o=\\varnothing$.\n\nUpdate: It's not true in general topological vector spaces. Indeed, if $X$ is a topological vector space and $Y$ is a proper dense subspace (such examples can be constructed), then $Y$\u00a0is convex and has empty interior, but $(\\overline Y)^o=X^o=X$ is clearly not empty. Yet, I'm still not sure if the result holds for finite-dimensional vector spaces (in which every proper subspace is closed and thus no proper subspace is dense, so the preceding counterexample doesn't work).\n\n\u2022 Seems obvious that a convex subset of $\\mathbb R^n$ with empty interior is contained in a hyperplane, which is a nowhere dense closed set. What am I missing? \u2013\u00a0bof Dec 30 '14 at 20:54\n\nThis is true in $\\mathbb{R}^n$. Say $n = 3$. If $\\overline{C}$ has non empty interior that $C$ is dense in some ball hence it contains the vertices of a tetrahedron (just take a tetrahedron inside the ball and move its vertices slightly to land on a point in $C$). But now by convexity, $C$ must contain the whole solid tetrahedron which contradicts the fact that $C$ has empty interior. In higher dimensions, you can start with a generalized cube and move its vertices slightly to enter $C$.\n\n\u2022 I see that the intuitive argument works, but I still have trouble operationalizing it rigorously. I will keep trying. Thank you for the answer! \u2013\u00a0triple_sec Dec 31 '14 at 16:03\n\u2022 Maybe I'm missing something, but it seems to me that you don't need to resort to \"move its vertices slightly\". If $C$ is dense in some ball, then $C$ must contain $4$ non-coplanar points (because no subset of a plane can be dense in a ball). Use these points as the vertices of a tetrahedron. \u2013\u00a0Dave L. Renfro Jan 8 '15 at 17:07\n\u2022 That works too. \u2013\u00a0user203787 Jan 8 '15 at 18:00\n\nHere is a rigorous\u2014and, accordingly, a bit lengthy\u2014operationalization of the intuition offered by the answer by @OohAah.\n\n$$\\textbf{Proposition}\\phantom{---}$$ Let $$C\\subseteq \\mathbb R^n$$ ($$n\\in\\mathbb N$$) be a convex set. If $$C$$ has no interior, then nor does $$\\overline C$$.\n\n$$\\textit{Proof}\\phantom{---}$$ Suppose that $$(\\overline C)^o\\neq\\varnothing$$. I will show that $$C^o$$ is not empty, either. Let $$\\mathbf x_0\\in (\\overline C)^o$$. Then, there exists some $$\\varepsilon>0$$ such that $$\\mathbf x_0\\in B(2\\varepsilon,\\mathbf x_0)\\subseteq\\overline C$$, where $$B(2\\varepsilon,\\mathbf x_0)$$ denotes the open ball of radius $$2\\varepsilon$$\u00a0around $$\\mathbf x_0$$ with respect to the Euclidean norm. Let $$\\{\\mathbf e_i\\}_{i=1}^n$$ denote the standard basis of $$\\mathbb R^n$$ and define $$\\mathbf x_i\\equiv \\mathbf x_0+\\varepsilon \\mathbf e_i$$ for each $$i\\in\\{1,\\ldots,n\\}$$. Clearly, $$\\{\\mathbf x_i\\}_{i=0}^n\\subseteq B(2\\varepsilon,\\mathbf x_0)\\subseteq \\overline C.$$ That is, $$\\{\\mathbf x_i\\}_{i=0}^n$$ is included in the closure of $$C$$, so that, for each $$i\\in\\{0,1,\\ldots,n\\}$$, there exists some $$\\mathbf y_i$$ such that\n\n\u2022 $$\\mathbf y_i\\in C$$; and\n\u2022 $$\\mathbf y_i\\in B((2\\sqrt n)^{-1}\\varepsilon,\\mathbf x_i)$$.\n\nFor each $$i\\in\\{1,\\ldots,n\\}$$ define $$\\mathbf b_i\\equiv \\mathbf y_i-\\mathbf y_0$$. I claim that the vectors $$\\{\\mathbf b_i\\}_{i=1}^n$$ are linearly independent. Indeed, suppose that $$\\lambda_1,\\ldots,\\lambda_n\\in\\mathbb R$$ satisfy $$\\sum_{i=1}^n\\lambda_i\\mathbf b_i=0.$$ Suppose, for the sake of contradiction, that not all of $$\\{\\lambda_i\\}_{i=1}^n$$ are zero. Then, $$\\sum_{i=1}^n|\\lambda_i|>0\\tag{1}.$$ In addition, \\begin{align*} 0=&\\,\\sum_{i=1}^n\\lambda_i\\mathbf b_i=\\sum_{i=1}^n\\lambda_i(\\mathbf y_i-\\mathbf y_0)=\\sum_{i=1}^n\\lambda_i(\\mathbf y_i-\\mathbf x_i+\\mathbf x_i-\\mathbf x_0+\\mathbf x_0-\\mathbf y_0)\\\\=&\\,\\sum_{i=1}^n\\lambda_i(\\mathbf y_i-\\mathbf x_i+\\varepsilon\\mathbf e_i+\\mathbf x_0-\\mathbf y_0), \\end{align*} or \\begin{align*} -\\varepsilon\\sum_{i=1}^n\\lambda_i\\mathbf e_i=\\sum_{i=1}^n\\lambda_i(\\mathbf y_i-\\mathbf x_i+\\mathbf x_0-\\mathbf y_0). \\end{align*} Clearly, the Euclidean norm of the left-hand side is $$\\varepsilon\\sqrt{\\sum_{i=1}^n\\lambda_i^2}$$, so the following chain of inequalities holds true: \\begin{align*} &\\varepsilon\\sqrt{\\sum_{i=1}^n\\lambda_i^2}=\\left\\|\\sum_{i=1}^n\\lambda_i(\\mathbf y_i-\\mathbf x_i+\\mathbf x_0-\\mathbf y_0)\\right\\|\\leq\\sum_{i=1}^n|\\lambda_i|\\left(\\|\\mathbf y_i-\\mathbf x_i\\|+\\|\\mathbf x_0-\\mathbf y_0\\|\\right)\\\\ \\underset{\\text{see (1)}}{<}&\\,\\sum_{i=1}^n|\\lambda_i|\\left(\\frac{\\varepsilon}{2\\sqrt{n}}+\\frac{\\varepsilon}{2\\sqrt{n}}\\right)=\\sum_{i=1}^n|\\lambda_i|\\frac{\\varepsilon}{\\sqrt{n}}\\leq\\sqrt{\\sum_{i=1}^n|\\lambda_i|^2}\\sqrt{\\sum_{i=1}^n\\frac{\\varepsilon^2}{n}}=\\varepsilon\\sqrt{\\sum_{i=1}^n\\lambda_i^2}, \\end{align*} where I used the Cauchy\u2013Schwarz inequality. This is a contradiction, which implies that $$\\lambda_1=\\ldots=\\lambda_n=0$$. Hence, the vectors $$\\{\\mathbf b_i\\}_{i=1}^n$$ are linearly independent, and since $$\\dim\\mathbb R^n=n$$, it follows also that they constitute a basis.\n\nIf $$\\mathbf z\\in\\mathbb R^n$$, there exists a unique $$(\\mu_i)_{i=1}^n\\in\\mathbb R^n$$ such that $$\\mathbf z=\\sum_{i=1}^n\\mu_i\\mathbf b_i$$. Define $$\\|\\mathbf z\\|_b\\equiv\\sum_{i=1}^n|\\mu_i|$$. Then, $$\\|\\cdot\\|_b$$ is a norm and since $$\\mathbb R^n$$ is finite-dimensional, it must be equivalent to the Euclidean norm. Therefore, there exists some $$\\xi_b>0$$ such that $$\\|\\cdot\\|_b\\leq \\xi_b\\|\\cdot\\|$$.\n\nLet $$D\\equiv\\left\\{\\sum_{i=0}^n\\alpha_i\\mathbf y_i\\,\\Bigg|\\,\\alpha_i\\geq0\\text{ for all i\\in\\{0,1,\\ldots,n\\} and }\\sum_{i=0}^n\\alpha_i=1\\right\\}.$$ Since $$\\{\\mathbf y_i\\}_{i=0}^n\\subseteq C$$ and $$C$$ is convex, it follows that $$D\\subseteq C$$. Define $$\\mathbf w\\equiv\\sum_{i=0}^n\\frac{1}{n+1}\\mathbf y_i.$$ Clearly, $$\\mathbf w\\in D$$ and $$\\mathbf w-\\mathbf y_0=\\sum_{i=1}^{n}\\frac{1}{n+1}(\\mathbf y_i-\\mathbf y_0)=\\sum_{i=1}^{n}\\frac{1}{n+1}\\mathbf b_i.$$\n\nLet $$\\delta\\equiv\\frac{1}{n(n+1)\\xi_b}>0.$$ I will show that $$B(\\delta,\\mathbf w)\\subseteq D$$. To this end, pick any $$\\mathbf z\\in B(\\delta,\\mathbf w)$$, so that $$\\|\\mathbf z-\\mathbf w\\|<\\delta.$$ Since $$\\{\\mathbf b_i\\}_{i=1}^n$$ is a basis, there exists a unique $$(\\mu_i)_{i=1}^n\\in\\mathbb R^n$$ such that $$\\mathbf z-\\mathbf y_0=\\sum_{i=1}^n\\mu_i\\mathbf b_i$$. Then, $$\\mathbf z-\\mathbf w=(\\mathbf z-\\mathbf y_0)-(\\mathbf w-\\mathbf y_0)=\\sum_{i=1}^n\\left(\\mu_i-\\frac{1}{n+1}\\right)\\mathbf b_i.$$ Consequently, for any $$i\\in\\{1,\\ldots,n\\}$$, it follows that \\begin{align*} \\left|\\mu_i-\\frac{1}{n+1}\\right|\\leq\\sum_{j=1}^n\\left|\\mu_j-\\frac{1}{n+1}\\right|=\\|\\mathbf z-\\mathbf w\\|_b\\leq \\xi_b\\|\\mathbf z-\\mathbf w\\|<\\xi_b\\delta=\\frac{1}{n(n+1)}. \\end{align*} Hence, $$0\\leq\\frac{(n-1)}{n(n+1)}=\\frac{1}{n+1}-\\frac{1}{n(n+1)}<\\mu_i<\\frac{1}{n+1}+\\frac{1}{n(n+1)}=\\frac{1}{n},$$ for each $$i\\in\\{1,\\ldots,n\\}$$ so that $$\\sum_{i=1}^n\\mu_i<1$$. It follows that \\begin{align*} \\mathbf z=\\mathbf y_0+\\sum_{i=1}^n\\mu_i\\mathbf b_i=\\mathbf y_0+\\sum_{i=1}^n\\mu_i(\\mathbf y_i-\\mathbf y_0)=\\left(1-\\sum_{i=1}^n\\mu_i\\right)\\mathbf y_0+\\sum_{i=1}^n\\mu_i\\mathbf y_i, \\end{align*} so $$\\mathbf z\\in D$$.\n\nConclusion:\n\n\u2022 $$\\mathbf w\\in B(\\delta,\\mathbf w)\\subseteq D\\subseteq C$$; and\n\u2022 $$B(\\delta,\\mathbf w)$$ is a non-empty open set; so that\n\u2022 $$\\mathbf w\\in C^o$$.\n\nIn particular, $$C^o$$ is not empty. $$\\blacksquare$$\n\nIntuitively, the points $$\\{\\mathbf x_i\\}_{i=0}^n$$ in $$\\overline C$$ span an $$n$$-dimensional simplex, whose vertices are perturbed slightly so that the new vertices do not lie in the same hyperplane and are all in $$C$$. The convex hull of these new vertices gives rise to a \u201cdistorted simplex\u201d fully contained in $$C$$. Then, the centroid of this distorted simplex can be surrounded by a small ball still included in the distorted simplex, and hence in $$C$$.", "date": "2019-10-17 10:11:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1085643/convex-set-with-empty-interior-is-nowhere-dense/1096673", "openwebmath_score": 0.9879440069198608, "openwebmath_perplexity": 167.2925882924991, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363485313248, "lm_q2_score": 0.8947894541786198, "lm_q1q2_score": 0.8807737840305198}}
{"url": "https://www.physicsforums.com/threads/question-about-diagonalizable-matrices.705803/", "text": "Suppose that ##A## is a diagonalizable ## n \\times n ## matrix. Then it is similar to a diagonal matrix ##B##. My question is, is ##B## the only diagonal matrix to which ##A## is similar?\n\nI have thought about this, but am unsure if my answer is correct. My claim is that ##B## is the only diagonal matrix to which ##A## is similar, up to a rearrangement (or permutation) of the eigenvectors of ##A## in an ordered eigenbasis for ##\u211d^{n}##. Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_{A}]_{\u03b2}## but this will be merely a rearrangement of the diagonal entries of B.\n\nIs this true? If so, how might I go about proving this conjecture of mine?\n\nBiP\n\nRelated Linear and Abstract Algebra News on Phys.org\nYes, it's true.\n\nA hint for the proof: the diagonal entries are the eigenvalues of ##A##.\n\nCompuChip\nHomework Helper\nIf ##A = X^{-1}BX = Y^{-1}B'Y## then ##B = (Y X^{-1})^{-1} B' (Y X^{-1})##.\nSo if A is similar to two diagonal matrices, then these matrices are similar to each other.\nI guess a starting point would be to check which similarity transformations keep diagonal matrices diagonal, and see what you can say about X and Y that way.\n\nHallsofIvy\nHomework Helper\nYes, it's true.\n\nA hint for the proof: the diagonal entries are the eigenvalues of ##A##.\nNot quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.\n\nNot quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.\nThat's exactly what the OP said:\n\nThus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_A]_\\beta## but this will be merely a rearrangement of the diagonal entries of B.\n\nmathwonk", "date": "2020-11-28 14:07:43", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/question-about-diagonalizable-matrices.705803/", "openwebmath_score": 0.9140629768371582, "openwebmath_perplexity": 529.6879820820291, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9693241947446617, "lm_q2_score": 0.9086179018818864, "lm_q1q2_score": 0.8807453160722436}}
{"url": "http://xdhq.visitmarzamemi.it/fourier-transform-of-triangular-pulse.html", "text": "# Fourier Transform Of Triangular Pulse\n\nTo do so, you just have to divide the pulse by its norm, i. Signal Transmission Through LTIC Systems. \u2206(t), from Problem 3. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. The Fourier transform of a gate pulse is: a. For any parameter, \u2260. otherwise. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. Therefore, the departure of the roll-off from that of the sinc function can be ascribed to aliasing in the frequency domain , due to sampling in the time domain. Re: Help me get a Fourier transform of sinc It show the Fourier Transform of sinc(x) function --> a rectangular pulse. See under Discrete Fourier transform data compression use of, 494-495 decomposition. Fourier transforms of deltas and sinusoids Fourier transform of periodic signals Preface Not Just One 1 2 3 0. 1: The Fourier transform of a triangular pulse. Then the Fourier series expansion of the output function y(t) literally gives the spectrum of the output! B. periodic relationship, 222 discrete time Fourier series (DTFT),. 8 Boxcar Pulse Bc(t) Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier Transform property u(t) !F A\u02ddsinc! \u02dd 2 Thus u(t j!\u02dd\u02dd) !F e A\u02ddsinc! \u02dd 2 Therefore u(t a)u (t b) !F ej!a A\u02ddsinc! \u02dd 2. 2 The Fourier Transform for Periodic Signals 4. In Fourier Transform Nuclear Magnetic Resonance spectroscopy (FTNMR), excitation of the sample by an intense, short pulse of radio frequency energy produces a free induction decay signal that is the Fourier transform of the resonance spectrum. 4 Find The Fourier Transform Of The Triangular Pulse (Fig. 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. Plot sine wave. The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the spectrum-shaped pulse. Amplitude-modulated optical pulse trains undergo the discrete Fourier transform (DFT) realized by the temporal Talbot effect in a dispersive fiber. Fourier series: Given a 2\u02c7-periodic complex-valued function f (think of it as a function on an interval T of length 2\u02c7), its Fourier (series) transform is the sequence of its Fourier coecients: (f) = fb= fck: k 2 Zg, which can be thought of as a complex-valued function of the discrete frequency variable k. The series produced is then called a half range Fourier series. A square wave is approximated by the sum of harmonics. The S-transform can be effectively utilized for the analysis of nonstationary data. Output kernel Figure 5. The Gaussian shape is often called a bell shape. I'm at a computer without MATLAB at the moment. d R pp T P, where R pp is the auto-correlation of p(t) d. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Still, many problems that could have been tackled by using Fourier transforms may have gone unsolved because they require integration that is difficult and tedious. T T 0 elsewhen. There are similar convolution theorems for inverse Fourier transforms. 2 p693 PYKC 10-Feb-08 E2. The Fourier Transform 1. That is: since the is an odd function, its contribution is zero. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. DSP First and it's accompanying digital assets are the result of more than 20 years of work that originated from, and was guided by, the premise that signal processing is the best starting point for the study of electrical and computer engineering. time signal. School of Health Information Sciences. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). Signal Distortion over a Communication Channel. The Fourier series of this general pulse train is:. Equation of the Day #11: The Fourier Transform. Except now we're going to build a composite wave form that is a triangle wave. (We shall use this in the assignment). In order for F(t) to be real, F (- t) = F* ( t) must hold, Example 9. Plotting a triangular signal and finding its Fourier transformation in MATLAB. A \u201cBrief\u201d Introduction to the Fourier Transform This document is an introduction to the Fourier transform. What is the FT of a triangle function? - To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. time signal. This is a good point to illustrate a property of transform pairs. This pulse can be used to represent an electrostatic discharge, an electromagnetic pulse, or a lightning event. The LCFBG in the. Applying the time-convolution property to y(t)=x(t) * h(t), we get: 2(\u03c9) e-j\u03c9\u03c4. For1secondofdatasampledat40,000. Which are the only waves that correspond/ support the measurement of phase angle in the line spectra? a. t/, use the derivative property to \ufb01nd the Fourier transform of x. Let's look at the triangular pulse and its Fourier transform,. Fourier transform of a triangular pulse is sinc 2, i. Note: This is why the Fourier transform of a rectangular pulse is shaped like a sinc function, and the Fourier transform of a sinc function is shaped like a rectangular pulse 3. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. t2 exp( 2 ) 2. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Realisability of One Port Network \u2013 Consider the network function H(s) which is the ratio of Laplace transform of the output R(s) to the Laplace transform of the excitation E(s). The line spectrum, obtained from the Fourier series coefficients, indicates how the power of the signal is distributed to harmonic frequency components in the series. \"Fourier Series--Triangle Wave. The Gaussian shape is often called a bell shape. Click the button \"Real Transform\" in the Fourier graph. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. We then generalise that discussion to consider the Fourier transform. (a) below is X(\u03a9) = sin \u03a9 2 \u03a9 2!2. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2\u02c7f 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. A group of algorithms is presented generalizing the fast Fourier transform to the case of noninteger frequencies and nonequispaced nodes on the interval $[ - \\pi ,\\pi ]$. School of Health Information Sciences. Consequently, the square wave has a wider bandwidth. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain \u2212\ud835\udf45 \u2194 \u2212\ud835\udf45 Option (c) 10. Sometimes there is a big spike at zero so try taking the log of it before plotting. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Then the Fourier series expansion of the output function y(t) literally gives the spectrum of the output! B. Signals & Systems - Triangular Signal Watch more videos at https://www. 2) Here 0 is the fundamental frequency of the signal and n the index of the harmonic such. 2 The Fourier Transform Having now presented the idea of the frequency response H(s)js=j! = H(j!) of linear systems, it is appropriate to recognise that in the special case of this Laplace transform value H(s)being evaluated at the purely imaginary value s = j!, it is known as the \u2018Fourier transform\u2019. x(t)= \u03b3 (t)-\u03b3 (t-T 0) a) by using a table of Fourier Transforms and Properties (this is just a rectangular pulse function of width T 0 that is not centered on the origin). , sin(x)/x] in the frequency domain. Consequently, the square wave has a wider bandwidth. Improving images by \u201c deconvolution \u201d of. In this chapter much of the emphasis is on Fourier Series because an understanding of the Fourier Series decomposition of a signal is important if you wish to go on and study other spectral techniques. FFT onlyneeds Nlog 2 (N). Conceptually, this occurs because the triangle wave looks much more like the 1st harmonic, so the contributions of the higher harmonics are less. The FT of a rectangle function is a sinc. tutorialspoint. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter. The Fourier Transform of the Box Function. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. Triangle Pulse 0 0. periodic relationship, 222 discrete time Fourier series (DTFT),. Exams are approaching, and I'm working through some old assignments. The transform of a triangular pulse is a sinc2 function. Try taking the real part of it with real(). 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. If a function is defined over half the range, say 0 to L, instead of the full range from -L to L, it may be expanded in a series of sine terms only or of cosine terms only. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. In this particular SPICE simulation, I\u2019ve summed the 1st, 3rd, 5th, 7th, and 9th harmonic voltage sources in series for a total of five AC voltage sources. Relation of Z transform to Laplace transform. 2005-03-15. We say that f(t) lives in the time domain, and F(\u03c9) lives in the frequency domain. We begin by discussing Fourier series. On the other hand, as Fourier transform can be considered as a special case of Laplace transform when the real part of the complex argument is zero:. Baskakov, O I; Civis, S; Kawaguchi, K. After simplification the sinc squared function is obtained as the Fourier transform of a triangular pulse with unit area. 30) yields 2 2 ( ) \u02c6 p p T T. FFTs of Functions. '' Figure 8 shows an example for. Discrete Fourier Transform of Windowing Functions. )2 Solutions to Optional Problems S9. More Advanced Topics Up: Fourier Series-What, How, and Why Previous: The Fast Fourier Transform Using the Fourier Transform. Fourier Transform of Useful Functions. Since f(t) is even then g(w) is real. For any parameter, \u2260. It is reversible, being able to transform from either domain to the other. You can buy my book 'ECE for GATE' here https://goo. We then generalise that discussion to consider the Fourier transform. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. The Fourier transform of the triangular pulse shown in Fig. The curve should be symmetrical with respect to the origin in 1024 points. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Recall that normalized Fourier transform of triangular pulse is $sinc^{2}(f)$$. The transform of a triangular pulse is a sinc 2 function. Compare the result with pan (b). Find the Fourier Transform of a rectangular pulse that is zero everywhere except between t=0 and t=T 0 where it has a height of one:. Fast Fourier Transform(FFT) \u2022 The Fast Fourier Transform does not refer to a new or different type of Fourier transform. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. Fourier Transform of the Gaussian Konstantinos G. The reason why Fourier analysis is so important in physics is that many (although certainly. Homework Statement What is the Fourier transform of the function graphed below? According to some textbooks the Fourier transform for this function Fourier Transform (Triangular Pulse) | Physics Forums. Cvetkovic, IntechOpen, DOI: 10. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. In this tutorial numerical methods are used for finding the Fourier transform of continuous time signals with MATLAB are presented. The fast Fourier transform (FFT) is a fast algorithm for calculating the Discrete Fourier Transform (DFT). A major challenge in the data analysis process of NOM using the FT-ICR MS technique is the need to sort the entire data set and to present it in an accessible mode. Often we are confronted with the need to generate simple, standard signals ( sine, cosine , Gaussian pulse , squarewave , isolated rectangular pulse , exponential decay, chirp signal ) for. 1 The Fourier transform and series of basic signals Triangular pulse 1. The graph at the bottom of the screen is the Fourier transform of this pulse. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. (a) below is X(\u03a9) = sin \u03a9 2 \u03a9 2!2. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). The series does not seem very useful, but we are saved by the fact that it converges rather rapidly. Fourier Transform Fourier Transform maps a time series (eg audio samples) into the series of frequencies (their amplitudes and phases) that composed the time series. This is the principle on which a pulse Fourier transform spectrometer operates. Discrete-Time Fourier Transform (DTFT) Chapter Intended Learning Outcomes: (i) Understanding the characteristics and properties of DTFT (ii) Ability to perform discrete-time signal conversion between the time and frequency domains using DTFT and inverse DTFT. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Sometimes there is a big spike at zero so try taking the log of it before plotting. In the synthesis, we are going to obtain a network from the given network function which may be admittance function or impedance function. Fourier Transforms For additional information, see the classic book The Fourier Transform and its Applications by Ronald N. Sometimes fft gives a complex result. rectangular pulse triangular pulse periodic time function unit impulse train (model of regular sampling). The example in this figure pertains to an output sampling rate which is times that of the input signal. Recall that normalized Fourier transform of triangular pulse is [math]sinc^{2}(f)$[math]. The Fourier transform is an integral transform widely used in physics and engineering. In this article, we will discuss the fact that choice of different window functions involves a trade-off between the main lobe width and the peak sidelobe (PSL). Improving images by \u201c deconvolution \u201d of. How to get a triangular pulse? Discrete Fourier Transform using scipy. 4 Reconstruction of a triangular pulse. What do we hope to achieve with the Fourier Transform? We desire a measure of the frequencies present in a wave. Can you explain this answer? is done on EduRev Study Group by Electrical Engineering (EE) Students. In some cases, as in this one, the property simplifies things. Triangle Pulse Sinc Pulse. What is the FT of a triangle function? \u2013 To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. DAS RISIN and MAIZAN MUHAMAD. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. A Tables of Fourier Series and Transform Properties 321 Table B. periodic relationship, 222 discrete time Fourier series (DTFT),. The Fourier transform of this convolution is the product w 1 w 2 of the two transforms, each one a series of parallel walls, and differs from zero only when both factors are different from zero. We demonstrate both amplitude and phase tailoring by generating a picosecond squarelike pulse as well as trains of femtosecond pulses with a terahertz-range repetition rate from either a. The LCFBG in the. Digital signal processing (DSP) vs. For the bottom panel, we expanded the period to T=5, keeping the pulse's duration fixed at 0. Relation of Z transform to Laplace transform. Ideal and Practical Filters. Sibbett A Wollaston prism is used in the design of a polarizing Fourier. The impulse response for the triangle in Fig. The graph at the bottom of the screen is the Fourier transform of this pulse. \u2020 The Fourier series is then f(t) = A 2 \u00a1 4A \u20262 X1 n=1 1 (2n\u00a11)2 cos 2(2n\u00a11)\u2026t T: Note that the upper limit of the series is 1. Signals & Systems - Triangular Signal Watch more videos at https://www. Exams are approaching, and I'm working through some old assignments. Signal Distortion over a Communication Channel. Fourier transform theorem table 4 1 jpg bax blog fourier transform table rh baxtyfraze blo com Pdf Fourier Transforms And Their Application To Pulse Amplitude. I One-Dimensional Fourier Transform. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain \u2212\ud835\udf45 \u2194 \u2212\ud835\udf45 Option (c) 10. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). Slow Fourier Transforms Consider a general 1D Fourier transform relating two vectors of length : contains the values in real-space contains the frequency components The Slow Fourier Transform : Suppose that we precompute and store the coefficients c(j,k) = exp(i 2pi jk/n). I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. I'm at a computer without MATLAB at the moment. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. Signal Energy and Energy Spectral Density. Minimalist Tall Rectangular Planter On 4 Or 6 Zebra Onion Grass. F(\u03c9) is just another way of looking at a function or wave. A LABORATORY DEMONSTRATION OF HIGH-RESOLUTION HARD X-RAY AND GAMMA-RAY IMAGING USING FOURIER-TRANSFORM TECHNIQUES David Palmer and Thomas A. A two parts tutorial on Fourier series. For voltage signals, the power per unit frequency is proportional to |H(f)|2 and is called the power spectrum or spectral power density of h(t). I One-Dimensional Fourier Transform. Sampling theorem. 1 Fourier Series Representation of Periodic Signals 3. 900 950 1000 1050 1100 1150 0. It is reversible, being able to transform from either domain to the other. Then using the relation: one can show that. PDF | Fourier transform ultrashort optical pulse shaping using a single linearly chirped fiber Bragg grating (LCFBG) is proposed and experimentally demonstrated in this letter. Solution: g(t) is a triangular pulse of height A, width W, and is centered at t 0. Therefore the FT of a triangle function is the product of two identical sincs, or a sinc^2. 4*Pi); # The base function is f0 = f restricted to [0,T] f0:=x->x; # Then f(x)=f0(trw(x,T))=trw(x,T) # INTEGRATION STEPS FOR FOURIER COEFFICIENTS # Possible also by hand using an integral table. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Note that as long as the definition of the pulse function is only motivated by the time-domain experience of it, there is no reason to believe that the oscillatory interpretation (i. Cosine waves c. Plotting a triangular signal and finding its Fourier transformation in MATLAB. The Fourier Transform In this appendix, the most important facts pertaining to the Fourier trans\u00ad form are surveyed without proof. NEW! Updated labs, visual demos, an update to the existing chapters, and hundreds of new homework problems and solutions. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. Properties of the Fourier Transform \u2022 Linearity: \u2022 Let and \u2022 then \u2022 Time Scaling: \u2022 Let \u2022 then Compression in the time domain results in expansion in the frequency domain Internet channel A can transmit 100k pulse/sec and channel B can transmit 200k pulse/sec. Signals & Systems - Triangular Signal Watch more videos at https://www. Variable position. Since the pulse is getting narrower and narrower in the limit, the multiplication with this other function has less and less effect, until it has no effect once the. This is the example given above. The Fourier transform (English pronunciation: / \u02c8 f \u028ar i e\u026a /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Signals & Systems - Reference Tables 1 Table of Fourier Transform Pairs Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform. Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. t2 exp( 2 ) 2. As is an even function, its Fourier transform is Alternatively, as the triangle function is the convolution of two square functions ( ), its Fourier transform can be more conveniently obtained according to the convolution theorem as:. Periodicity of the Fourier transform; Fourier transform as additive synthesis. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the. Note that as long as the definition of the pulse function is only motivated by its behavior in the time-domain experience, there is no reason to believe that the oscillatory interpretation (i. Another representation of signals that has been found very useful is frequency domain representation. \uf097We can sample a function and then take the FFT to see the function in the frequency domain. The triangle peak is at the integral of the signal or sum of the sequence squared. ), the frequency response of the interpolation is given by the Fourier transform, which yields a sinc function. Single-pulse, Fourier-transform spectrometer having no moving parts M. Fourier transform: F(s) = - f(x)e-'2n\"dr Here, two sidebands have been introduced (in addition to the carrier frequency), and the signal of frequency s has been shifted into the region of the carrier frequency, so. Today I want to start getting \"discrete\" by introducing the discrete-time Fourier transform (DTFT). Now, if we're given the wave function when t=0, \u03c6(x,0) and the velocity of each sine wave as a function of its wave number, v(k), then we can compute \u03c6(x,t) for any t by taking the inverse Fourier transform of \u03c6(x,0) conducting a phase shift, and then taking the Fourier transform. profile closer to Gaussian. DFT needs N2 multiplications. \uf097Of course, we must sample often enough to avoid losing content. P d P2 f , where is Fourier Transform of p(t) c. 5 The multiplication Property 4. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first row is the graph of the unit pulse function and its Fourier transform , a function of frequency. For example, a rectangular pulse in the time domain coincides with a sinc function [i. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2\u02c7f 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. The actual Fourier transform are only the impulses. As in the case of ideal sampling, the spectrum contains uniformly spaced (scaled) copies of \ud835\udc4b(\ud835\udc53), with a spacing of 1\u2044\ud835\udc47 Hz. \u2020 The Fourier series is then f(t) = A 2 \u00a1 4A \u20262 X1 n=1 1 (2n\u00a11)2 cos 2(2n\u00a11)\u2026t T: Note that the upper limit of the series is 1. 8 Boxcar Pulse Bc(t) Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier Transform property u(t) !F A\u02ddsinc! \u02dd 2 Thus u(t j!\u02dd\u02dd) !F e A\u02ddsinc! \u02dd 2 Therefore u(t a)u (t b) !F ej!a A\u02ddsinc! \u02dd 2. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Cosine waves c. Moreover, the amplitude of cosine waves of wavenumber in this superposition is the cosine Fourier transform of the pulse shape, evaluated at wavenumber. Mapping s plane semiellipse to Z plane. The corresponding intensity is proportional to this transform squared, i. To do so, you just have to divide the pulse by its norm, i. 3 Properties of the Continuous-Time Fourier Transform 4. Prince California Institute of Technology, Pasadena, CA 91125 Abstract A laboratory imaging system has been developed to study the use of Fourier-transform techniques in high-resolution hard x-ray and\u03b3. The dotted-line is a sinc function that doesn't apply to this question, but gives the notion that this transform has something to do with the transform of a square pulse (i. Ideal and Practical Filters. F(\u03c9) is called the Fourier Transform of f(t). Plot sine wave. 4 The Convolution Property 4. Today I want to follow up by discussing one of the ways in which reality confounds our expectations and causes confusion. Fn = 6 shows that as T/t increases the lines get closer together and the spectrum begins to look like that of the Fourier Transform. They are widely used in signal analysis and are well-equipped to solve certain partial. Signal Distortion over a Communication Channel. profile closer to Gaussian. periodic relationship, 222 discrete time Fourier series (DTFT),. Fourier series and square wave approximation Fourier series is one of the most intriguing series I have met so far in mathematics. The actual Fourier transform are only the impulses. The primary reason for it\u2019s success is that it plots any course of constant bearing (angle w. I would guess it would be a triangle because I think the transform of a triangular pulse centered at zero is something like T sinc 2 (pi * f * T) I would restate your original question as follows: What is the transform of the following periodic rectangular function: per-rect(t) which has period T=T1+T2; is 1 during each T1 duration of its. 5 Signals & Linear Systems Lecture 10 Slide 3 Connection between Fourier Transform and Laplace. 1 The Fourier transform and series of basic signals Triangular pulse 1. 6: Roll-off of the rectangular-window Fourier transform. We describe a pulse-shaping technique that uses second-harmonic generation with Fourier synthetic quasi-phase-matching gratings. time signal. In the case of natural sampling, however, the spectral copies have different scaling factors: \ud835\udc43(\ud835\udc58\u2044\ud835\udc47)/\ud835\udc47. For above triangular wave: The square wave has much sharper transition than the triangular wave. These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. PSfrag replacements PROBLEM: Let x. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Inverse Z transform using inversion integral. Click the button \"Real Transform\" in the Fourier graph. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. Square waves (1 or 0 or \u22121) are great examples, with delta functions in the derivative. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. Fourier Transforming the Triangular Pulse. Fourier Slice Theorem or central Slice Theorem. Translation (that is, delay) in the time domain goes over to complex phase shifts in the frequency domain. , use ^ M(n)/k^ M kinstead of ^ M(n). Then i deconvoluted the convoluted signal using triangular pulse with the deconvolution tool given in labview. 1-1) Since u(t) = 0 for t < 0, eq. Signal Transmission Through LTIC Systems. 2to create successive recon-structions of the pulse. Fourier transforms are used widely, and are of particular value in the analysis of single functions and combinations of functions found in radar and signal processing. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. Z transforms and difference equations. Time-harmonic impulse response calculations\u2014The time-harmonic pressure generated by these triangular source geometries is proportional to the Fourier transform of the impulse response. Derpanis October 20, 2005 In this note we consider the Fourier transform1 of the Gaussian. A \"Brief\" Introduction to the Fourier Transform This document is an introduction to the Fourier transform. tutorialspoint. It is reversible, being able to transform from either domain to the other. Inverse Z transform using inversion integral. See Discrete Fourier Transform discrete vs. \u2206(t), from Problem 3. The first example has a duty cycle of 0. The Fourier Transform: Examples, Properties, Common Pairs Square Pulse Spatial Domain Frequency Domain f(t) F (u ) 1 if a=2 t a=2 0 otherwise sinc (a u ) = sin (a u ) a u The Fourier Transform: Examples, Properties, Common Pairs Square Pulse The Fourier Transform: Examples, Properties, Common Pairs Triangle Spatial Domain Frequency Domain f(t. The Fourier transform (English pronunciation: / \u02c8 f \u028ar i e\u026a /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Sometimes fft gives a complex result. Padgett, A. But what we're going to do in this case is we're going to add them. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. Applying the time-convolution property to y(t)=x(t) * h(t), we get: That is: the Fourier Transform of the system impulse response is the system Frequency Response L7. property, we can compute the Fourier transform of the phasor: 0 ( )0 F e f fj t\u03c9 = \u2212\u03b4 This allows us to compute the Fourier transform of periodic signals. Triangular waves d. This is the example given above. (Browser settings for best viewing results) (How to cite this work). We then generalise that discussion to consider the Fourier transform. Fourier Transforms and Theorems. A) shows the original pulse and the real part of its Fourier transform. This transform pair isn't as important as the reason it is true. \" From MathWorld--A Wolfram Web Resource. t/ D X1 nD1. In mathematics, the continuous Fourier transform is one of the specific forms of Fourier analysis. 003 EECS MIT discrete operator transforms. 6 depicts a resistor and capacitor in series. Using the above, we obtain: [ ] 2 0 ( ) ( )0 j nf t n. the Fourier transform function) should be intuitive, or directly understood by humans. One imagines a delta function to be a square pulse of unit area in the limit as the base of the pulse becomes narrower and narrower and higher and higher. 082 Spring 2007 Fourier Series and Fourier Transform, Slide 22 Summary \u2022 The Fourier Series can be formulated in terms of complex exponentials \u2013 Allows convenient mathematical form \u2013 Introduces concept of positive and negative frequencies \u2022 The Fourier Series coefficients can be expressed in terms of magnitude and phase. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Exams are approaching, and I'm working through some old assignments. The magnitude of the Fourier transform of a rectangular pulse equals the absolute value of a sinc. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. Existence of the Fourier Transform. 104 Chapter 5. jaj<1 (n+ 1)anu[n] 1 (1 ae j. In the second row is shown , a delayed unit pulse, beside the real and imaginary parts of the Fourier transform. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). The fundamental frequency is 50 Hz and each harmonic is, of course, an integer multiple of that frequency. Four chapters on analog signal processing systems, plus many updates and enhancements. This is a good point to illustrate a property of transform pairs. In other words, the input signal is upsampled by a factor of using linear interpolation.", "date": "2020-01-23 19:59:56", "meta": {"domain": "visitmarzamemi.it", "url": "http://xdhq.visitmarzamemi.it/fourier-transform-of-triangular-pulse.html", "openwebmath_score": 0.8125916719436646, "openwebmath_perplexity": 790.5836348459956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9875683520739038, "lm_q2_score": 0.8918110497511051, "lm_q1q2_score": 0.8807243687639972}}
{"url": "https://byjus.com/question-answer/diet-problem-a-dietician-wishes-to-mix-two-types-of-foods-in-such-a-way/", "text": "Question\n\n# (Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast $$8$$ units of vitamin A and $$10$$ units of vitamin C. Food I contains $$2$$ units/kg of vitamin A and $$1$$ unit/kg of vitamin C. Food II contains $$1$$ unit/kg of vitamin A and $$2$$ units/kg of vitamin C. It costs $$Rs 50$$ per kg to purchase Food I and $$Rs. 70$$ per kg to purchase Food II. Formulate\u00a0this problem as a linear programming problem to minimise the cost of such a mixture.\n\nSolution\n\n## Let the mixture contain $$x$$ kg of Food I and $$y$$ kg of Food II. Clearly, $$x \\geq 0,y \\geq 0$$. We make the following table from the given data:ResourcesFoodI$$(x)$$FoodII$$(y)$$RequirementVitamins A (units/ kg)$$2$$$$1$$$$8$$Vitaminc C (units/ kg)$$1$$$$2$$$$10$$Cost (Rs/ kg)$$50$$$$70$$Since the mixture must contain at least $$8$$ units of vitamin A and $$10$$ units of vitamin C, we have the constraints:$$2x + y \\geq 8$$$$x + 2y \\geq 10$$Total cost $$Z$$ of purchasing $$x$$ kg of food I and $$y$$ kg of Food II is$$Z = 50x 70y$$Hence, the mathematical formulation of the problem is:Minimise $$Z = 50x + 70y ... (1)$$subject to the constraints :$$2x + y \\geq 8 ... (2)$$$$x + 2y \\geq 10 ... (3)$$$$x, y \\geq 0 ... (4)$$Let us the graph the inequalities $$(2)$$ to $$(4)$$. The feasible region determined by the system is shown in the Fig. Here again, observe that the feasible region is unbounded.Let us evaluate $$Z$$ at the corner points $$A(0,8), B(2,4)$$ and $$C(10,0)$$.Corner Point$$Z = 50x + 70y$$$$(0, 8)$$$$560$$$$(2, 4)$$$$380\\leftarrow Minimum$$$$(10, 0)$$$$500$$In the table, we find that smallest value of $$Z$$ is $$380$$ at the point $$(2,4)$$. Can we say\u00a0that the minimum value of $$Z$$ is $$380$$? Remember that the feasible region is unbounded.\u00a0Therefore, we have to draw the graph of the inequality$$50x + 70y < 380$$ i.e., $$5x + 7y < 38$$to check whether the resulting open half plane has any point common with the feasible\u00a0region. From the Fig, we see that it has no points in common.Thus, the minimum value of $$Z$$ is $$380$$ attained at the point $$(2, 4)$$. Hence, the optimal\u00a0mixing strategy for the dietician would be to mix $$2$$ kg of Food I and $$4\\ kg$$ of Food II,and with this strategy, the minimum cost of the mixture will be $$Rs. 380$$.Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-17 03:22:51", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/diet-problem-a-dietician-wishes-to-mix-two-types-of-foods-in-such-a-way/", "openwebmath_score": 0.3458096981048584, "openwebmath_perplexity": 645.5949451858403, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.994445075460939, "lm_q2_score": 0.8856314723088733, "lm_q1q2_score": 0.88071185631078}}
{"url": "https://math.stackexchange.com/questions/2826061/given-relatively-prime-positive-integers-a-b1-how-many-positive-integers-are", "text": "# Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$?\n\nLet $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combination of $a,b$. My question is : can we explicitly determine how many positive integers are there which can not be written as a non-negative integer combination of $a,b$ ?\n\n\u2022 @B.Goddard $a, b$ are assumed to be relatively prime. Also, the linear combinations here are assumed to have non-negative coefficients. That changes the problem quite a bit from the one you describe. \u2013\u00a0Arthur Jun 20 '18 at 13:11\n\u2022 The most explicit I can get is a summation. If $a\\lt b$ then there are: $$\\sum_{k=1}^{a-1}\\left\\lfloor\\frac{bk}{a}\\right\\rfloor$$ such numbers. Although you should definitely double-check this result. \u2013\u00a0N. Shales Jun 20 '18 at 16:19\n\u2022 Having thought a bit more I further conjecture that the result simplifies to $$\\frac{1}{2}(a-1)(b-1)\\, .$$ \u2013\u00a0N. Shales Jun 20 '18 at 17:16\n\u2022 @N.Shales your conjecture is correct: \\begin{align*} R &= \\sum_{k=1}^{a-1} \\left\\lfloor \\frac{bk}{a} \\right \\rfloor\\\\ &= \\sum_{k=1}^{a-1} \\left\\lfloor \\frac{(a-k)b}{a} \\right\\rfloor \\\\ &= \\sum_{k=1}^{a-1} b+\\left\\lfloor \\frac{-kb}{a} \\right\\rfloor \\\\ &= \\sum_{k=1}^{a-1} b-1-\\left\\lfloor \\frac{kb}{a} \\right\\rfloor \\\\ &= (a-1)(b-1) - \\sum_{k=1}^{a-1} \\left\\lfloor \\frac{kb}{a} \\right\\rfloor \\\\ &= (a-1)(b-1) - R \\end{align*} So that $2R = (a-1)(b-1)$. The key is the equality $\\lfloor (-kb)/a\\rfloor = -\\lfloor kb/a \\rfloor - 1$ which holds when $a$ does not divide $kb$. \u2013\u00a0Yong Hao Ng Jun 28 '18 at 8:20\n\u2022 @N.Shales I saw it a few days ago but it took me a while before I found a not very involved solution to the floor-sum. \u2013\u00a0Yong Hao Ng Jun 29 '18 at 13:11\n\nLet $N = ab$, we count the number of integers $0\\leq t \\leq N$ that can be written as $$t = ma+nb$$ such that $m,n\\geq 0$. Let there be $R$ such numbers. Since the largest non-representable (positive)-integer is $ab-a-b$, all non-representable numbers are in $[0,N]$ and hence we know there are $N+1 - R$ of them.\nIf $n\\geq a$, then we replace it with $$ma+nb = (m+b)a + (n-a)b = m'a+n'b$$ So with repeated application we may assume that $0\\leq n < a$. We require $$0 \\leq ma+nb \\leq N \\implies 0 \\leq ma \\leq N-nb = (a-n)b$$ Hence we get the bounds of $m$ as $$0 \\leq m \\leq \\left \\lfloor \\frac{(a-n)b}{a}\\right \\rfloor$$ This means the number of integers we can represent for a given $n$ is $\\left \\lfloor \\frac{(a-n)b}{a} \\right \\rfloor + 1$. Now summing this over all possibilities $n=0,1,\\dots a-1$, the number of representable integers $R$ is \\begin{align*} R = \\sum_{k=0}^{a-1} \\left\\lfloor \\frac{(a-k)b}{a} \\right\\rfloor +1 &= \\left\\lfloor \\frac{(a)b}{a} \\right\\rfloor + \\left\\lfloor \\frac{(a-1)b}{a} \\right\\rfloor + \\cdots + \\left\\lfloor \\frac{(2)b}{a} \\right\\rfloor + \\left\\lfloor \\frac{(1)b}{a} \\right\\rfloor + a\\\\ R &= a + b + \\sum_{k=1}^{a-1} \\left\\lfloor \\frac{kb}{a} \\right\\rfloor \\\\ \\end{align*} Next we observe that if $a$ does not divide $kb$, then for some integer $t$ $$t < \\frac{kb}{a} < t+1 \\Longleftrightarrow -t-1 < \\frac{-kb}{a} < -t$$ In particular this applies for $1\\leq k\\leq a-1$, since $\\gcd(a,b)=1$ and $a>1$. We use the inequalities to obtain $$\\left \\lfloor \\frac{-kb}{a}\\right \\rfloor = -t-1 = - \\left \\lfloor \\frac{kb}{a}\\right \\rfloor -1,$$ This lets us get another equation of $R$ as \\begin{align*} R = \\sum_{k=0}^{a-1} \\left\\lfloor \\frac{(a-k)b}{a} \\right\\rfloor +1 &= \\sum_{k=0}^{a-1} \\left\\lfloor \\frac{-kb}{a} \\right\\rfloor + b + 1 \\\\ &= a(b+1) + \\sum_{k=0}^{a-1} \\left\\lfloor \\frac{-kb}{a} \\right\\rfloor\\\\ &= a(b+1) + \\sum_{k=1}^{a-1}\\left\\lfloor \\frac{-kb}{a} \\right\\rfloor \\\\ &= a(b+1) + \\sum_{k=1}^{a-1}-\\left\\lfloor \\frac{kb}{a} \\right\\rfloor -1\\\\ R &= ab + 1 - \\sum_{k=1}^{a-1}\\left\\lfloor \\frac{kb}{a} \\right\\rfloor \\end{align*} Therefore adding both equations of $R$, we have $$2R = ab + a + b + 1 = (a+1)(b+1)$$ giving us $R=(ab+a+b+1)/2$. Notice that $R$ is not an integer if and only if $a,b$ are both even, which is not possible since $\\gcd(a,b)=1$.\nFinally, the number of integers considered is $ab+1$, so the number of integers not representable is $$ab+1 - \\frac{ab+a+b+1}{2} = \\frac{ab -a - b+1}{2} = \\frac{(a-1)(b-1)}{2}$$\nProposition. Let $a,b$ be relatively prime integers greater than $1,$ and let $m=ab-a-b.$\nIf $x\\in\\{0,1,2,\\dots,m\\},$ then exactly one of the two numbers $x$ and $m-x$ can be expressed as a linear combination of $a$ and $b$ with nonnegative integer coefficients.\nIt follows that exactly half the elements of $\\{0,1,2,\\dots,m\\}$ can be expressed as nonnegative integral linear combinations of $a$ and $b;$ that is, there are exactly $\\frac{m+1}2=\\frac{(a-1)(b-1)}2$ numbers which cannot be so expressed.", "date": "2019-05-19 20:46:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2826061/given-relatively-prime-positive-integers-a-b1-how-many-positive-integers-are", "openwebmath_score": 0.9999703168869019, "openwebmath_perplexity": 275.32463744383546, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9926541727735794, "lm_q2_score": 0.8872045862611166, "lm_q1q2_score": 0.8806873346559545}}
{"url": "https://math.stackexchange.com/questions/1642495/how-many-answers-to-3x-2y-5", "text": "# How many answers to $|3^x-2^y|=5$?\n\nHow many answers are there to the equation $|3^x-2^y|=5$ such that $x$ and $y$ are positive integers? Are there infinite? I've found $(2,2)$, $(3,5)$, and $(1,3)$. It seems to explode with larger values, but it's not a steady increase and there seems to be many dips. Do we KNOW that there are no large values for $x$ and $y$ where a power of 3 comes close to a power of 2?\n\n\u2022 Its a putnam problem . \u2013\u00a0DeepSea Feb 5 '16 at 23:12\n\u2022 Kf-Sansoo, had to google \"putnam problem.\" Do you mean literally one, or one that fits the general qualifications? Do I need to add a tag? If it is one that's been mentioned before, could you give a link? \u2013\u00a0Elem-Teach-w-Bach-n-Math-Ed Feb 5 '16 at 23:18\n\u2022 \u2013\u00a0Lucian Feb 6 '16 at 0:01\n\u2022 @Kf-Sansoo: so what? Does that mean the question shouldn't be discussed on MSE? Does it mean that there is helpful information about the question to be found elsewhere? Or what? \u2013\u00a0Rob Arthan Feb 6 '16 at 0:01\n\u2022 @Lucian: So, as it's conjectured that $Ax^n-By^m=C$ has a finite # of solutions for $(x,y,n,m)$ then $3^n-2^m=5$ can be conjectured to also have a finite # or solutions for $(n,m)$? In other words, it's unproven in a more general sense, so with the greater constraints it should be even less likely? If there are finite answers, my question could then be rephrased, \"Is there a 4th answer?\" \u2013\u00a0Elem-Teach-w-Bach-n-Math-Ed Feb 6 '16 at 0:18\n\nThere are only a finite number of solutions. It was proved by Pillai that $a^x - b^y = k$ where $a,b,k$ are fixed positive integers, $a > 1, b > 1, k \\neq 0,$ with positive integer variables $x,y,$ has finitely many solutions. This is from page 51 in Shorey and Tijdeman, Exponential Diophantine Equations. The two papers by Pillai are 1931 and 1936. Both are in the Journal of the Indian Mathematical Society. A detail: if $k$ is larger than some bound that depends on $a,b,$ there is only one solution. Since we have more than one solution for $k = -5,$ it appears Pillai's bound is not tight enough to finish this problem. We just know one solution for $k=5.$\n\n\u2022 I'll probably end up going with this as best answer. I've got an addendum though for your perusal, if I may. Since $|3^x-2^y|=5$ is proven to have finite solutions, I'd like to carry this knowledge to similar problems: i.e. $|(3^x)(5^y)-2^z|=7$, $|(2^x)(3^y)-5^z|=7$, $|(2^x)(3^y)-5^z|=7$. Namely working my way up to prime solutions such as $|(2^r)(7^s)(11^t)(13^u)(23^v)-(3^w)(5^x)(17^y)(19^z)|=29$ (or similarly creating primes less than the square of the largest prime used; in this case 23). Any thoughts? Probably throw this out there as a new ?. \u2013\u00a0Elem-Teach-w-Bach-n-Math-Ed Feb 8 '16 at 22:39\n\nHere's an elementary self-contained argument that there is no solution with $y>5$.\n\nA power of $3$ is congruent to either $1$ or $3 \\bmod 8$, so once $y \\geq 3$ we must have $3^x - 2^y = -5$.\n\nOnce $y \\geq 6$, we then have $3^x \\equiv -5 \\bmod 2^6$, and thus $x \\equiv 11 \\bmod 16$.\n\nBut then $3^x + 5 \\equiv 12 \\bmod 17$, and no power of $2$ is congruent to $12 \\bmod 17$ (the powers of $2 \\bmod 17$ are $2,4,8,-1,-2,-4,-8,1,2,4,8,-1$ etc.), QED.\n\nThere is a large literature on such Diophantine questions. One key phrase is \"$S$-unit equations\". In general it has been known for some time that there are finitely many solutions, and indeed for equations of the form $$\\prod_i p_i^{n_i} - \\prod_j q_j^{m_j} = r$$ this already follows from Thue's theorem (1909); and by now we even have effective algorithms known to find all solutions. There's still no elementary technique known in general, but in your case (where only the primes 2,3,5) appear an elementary solution is contained in a 1976 paper\nThis is not a complete answer, but I'd just like to note a connection with the concept of irrationality measure. First recall that $\\alpha \\in \\mathbf{R}$ has irrationality measure $\\mu(\\alpha)$ if, for all $r>\\mu(\\alpha)$, there are only finitely many pairs of integers $p,q$ such that $$\\left| \\alpha - \\frac{p}{q} \\right| < \\frac{1}{q^r}.$$ Note that $\\mu(\\alpha)=\\infty$ is allowed (but we have $\\mu(\\alpha)=2$ for all but a measure-zero set of reals $\\alpha$).\nWith this definition out of the way, note that $$\\left| 3^x - 2^y \\right| = 5$$ implies $$3^x = 2^y \\pm 5.$$ Take logs: $$\\log(3^x) = \\log(2^y) + \\log(1 \\pm 5 \\cdot 2^{-y}).$$ So by the Taylor expansion of $\\log(1+t)$, $$\\left| \\log(3^x) - \\log(2^y) \\right| < 5 \\cdot 2^{-y}+\\frac{5^2 \\cdot 2^{-2y}}{2}+\\cdots < 6 \\cdot 2^{-y}$$ for $y$ sufficiently large. Hence, using $\\log(3^x) = x \\log 3$ and $\\log(2^y) = y \\log 2$ $$\\left| \\frac{\\log 3}{\\log 2} - \\frac{y}{x} \\right| < \\frac{6}{x \\log 2} \\cdot 2^{-y}.$$ Since the RHS decreases much faster than anything polynomial in $x$ (at least for $y/x$ within some fixed small interval around $\\log(3)/\\log(2)$), the existence of infinitely many solutions to your equation would imply that $\\mu(\\log(3)/\\log(2))$ were infinite. It is probably known that this latter statement is false, probably by methods similar to the proof in the Shorey/Tijdeman book mentioned by Will Jagy.", "date": "2020-07-11 23:07:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1642495/how-many-answers-to-3x-2y-5", "openwebmath_score": 0.9088395833969116, "openwebmath_perplexity": 190.41448702133764, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815494335755, "lm_q2_score": 0.8902942304882371, "lm_q1q2_score": 0.8806626263661271}}
{"url": "http://nl.mathworks.com/help/matlab/ref/istriu.html?s_tid=gn_loc_drop&nocookie=true", "text": "# istriu\n\nDetermine if matrix is upper triangular\n\n## Syntax\n\n\u2022 `tf = istriu(A)` example\n\n## Description\n\nexample\n\n````tf = istriu(A)` returns logical `1` (`true`) if `A` is an upper triangular matrix; otherwise, it returns logical `0` (`false`).```\n\n## Examples\n\ncollapse all\n\n### Test Upper Triangular Matrix\n\nCreate a 5-by-5 matrix.\n\n`A = triu(magic(5))`\n```A = 17 24 1 8 15 0 5 7 14 16 0 0 13 20 22 0 0 0 21 3 0 0 0 0 9```\n\nTest `A` to see if it is upper triangular.\n\n`istriu(A)`\n```ans = 1 ```\n\nThe result is logical `1` (`true`) because all elements below the main diagonal are zero.\n\n### Test Matrix of Zeros\n\nCreate a 5-by-5 matrix of zeros.\n\n`Z = zeros(5);`\n\nTest `Z` to see if it is upper triangular.\n\n`istriu(Z)`\n```ans = 1 ```\n\nThe result is logical `1` (`true`) because an upper triangular matrix can have any number of zeros on the main diagonal.\n\n## Input Arguments\n\ncollapse all\n\n### `A` \u2014 Input arraynumeric array\n\nInput array, specified as a numeric array. `istriu` returns logical `0` (`false`) if `A` has more than two dimensions.\n\nData Types: `single` | `double`\nComplex Number Support: Yes\n\ncollapse all\n\n### Upper Triangular Matrix\n\nA matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.\n\nFor example, the matrix\n\n$A=\\left(\\begin{array}{cccc}1& -1& -1& -1\\\\ 0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& -2& -2\\\\ 0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& -3\\\\ 0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1\\end{array}\\right)$\n\nis upper triangular. A diagonal matrix is both upper and lower triangular.\n\n### Tips\n\n\u2022 Use the `triu` function to produce upper triangular matrices for which `istriu` returns logical `1` (`true`).\n\n\u2022 The functions `isdiag`, `istriu`, and `istril` are special cases of the function `isbanded`, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, ```istriu(A) == isbanded(A,0,size(A,2))```.", "date": "2015-05-06 00:55:05", "meta": {"domain": "mathworks.com", "url": "http://nl.mathworks.com/help/matlab/ref/istriu.html?s_tid=gn_loc_drop&nocookie=true", "openwebmath_score": 0.8912773132324219, "openwebmath_perplexity": 1831.3354522384097, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575147530352, "lm_q2_score": 0.8962513821399045, "lm_q1q2_score": 0.8806185306293575}}
{"url": "https://www.eduzip.com/ask/question/if-displaystyleint-x5e-x2dxgxcdot-e-x2c-then-the-value-of-g-1-is-582347", "text": "Mathematics\n\n# If $\\displaystyle\\int x^5e^{-x^2}dx=g(x)\\cdot e^{-x^2}+C$ then the value of $g(-1)$ is?\n\n$-\\dfrac{5}{2}$\n\n##### SOLUTION\nPut $x^2=t$\n$2xdx=dt$\n$\\displaystyle\\int t^2e^{-t}\\dfrac{dt}{2}$\n$=\\dfrac{1}{2}[-t^2\\cdot e^{-t}+2\\displaystyle\\int te^{-1}dt]+c$\n$=\\dfrac{1}{2}[-t^2\\cdot e^{-t}-2te^{-t}+\\displaystyle\\int 2e^{-t}dt]+c$\n$=\\dfrac{1}{2}(-t^2e^{-t}-2(te^{-1}+e^{-t}))+c$\n$=\\dfrac{-(x^4+2x^2+2)e^{-x^2}}{2}+c$\n$g(x)=\\dfrac{-(x^4+2x^2+2)}{2}$\n$g(-1)=-\\dfrac{5}{2}$.\n\nIts FREE, you're just one step away\n\nSingle Correct Medium Published on 17th 09, 2020\nQuestions 203525\nSubjects 9\nChapters 126\nEnrolled Students 86\n\n#### Realted Questions\n\nQ1 Subjective Hard\nEvaluate the given integral.\n$\\displaystyle\\int { { e }^{ x } } \\left( \\cfrac { \\sin { x } \\cos { x } -1 }{ \\sin ^{ 2 }{ x } } \\right) dx$\n\n1 Verified Answer | Published on 17th 09, 2020\n\nQ2 Subjective Medium\nEvaluate the given integral.\n$\\int { x\\sec ^{ 2 }{ x } } dx\\quad \\quad$\n\n1 Verified Answer | Published on 17th 09, 2020\n\nQ3 Single Correct Hard\nThe value of the definite integral $\\int_\\limits{ 1}^e( (x+1)e^x\\ln x) dx$ is-\n\u2022 A. $e$\n\u2022 B. $e^e(e-1)$\n\u2022 C. $e^{x+1}$\n\u2022 D. $e^{e+1}+e$\n\n1 Verified Answer | Published on 17th 09, 2020\n\nQ4 Single Correct Hard\nThe integral\u00a0$\\displaystyle \\int (1+x-\\displaystyle \\frac{1}{x})e^{x+\\frac{1}{x}} dx$ is equal to\n\u2022 A. $(x-1)e^{x+ \\frac{1}{x}} +c$\n\u2022 B. $(x+1)e^{x+\\frac{1}{x}} +c$\n\u2022 C. $-xe^{x+\\frac{1}{x}} +c$\n\u2022 D. $xe^{x+\\frac{1}{x}} +c$\n\nThe average value of a function f(x) over the interval, [a,b] is the number\u00a0$\\displaystyle \\mu =\\frac{1}{b-a}\\int_{a}^{b}f\\left ( x \\right )dx$\nThe square root\u00a0$\\displaystyle \\left \\{ \\frac{1}{b-a}\\int_{a}^{b}\\left [ f\\left ( x \\right ) \\right ]^{2}dx \\right \\}^{1/2}$ is called the root mean square of f on [a, b]. The average value of\u00a0$\\displaystyle \\mu$ is attained id f is continuous on [a, b].", "date": "2021-06-19 06:25:03", "meta": {"domain": "eduzip.com", "url": "https://www.eduzip.com/ask/question/if-displaystyleint-x5e-x2dxgxcdot-e-x2c-then-the-value-of-g-1-is-582347", "openwebmath_score": 0.66084885597229, "openwebmath_perplexity": 9092.350503343334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9825575162853136, "lm_q2_score": 0.896251371748038, "lm_q1q2_score": 0.8806185217920576}}
{"url": "https://www.physicsforums.com/threads/simple-integral-but-a-question-of-methodology.164031/", "text": "# Simple integral - but a question of methodology\n\nianb\nsimple integral -- but a question of methodology\n\n$$\\int \\frac{(x-1)}{2} dx$$\n\nFirst, try substitution. Let $$u = x - 1$$, and $$du = dx$$\n\n$$\\int \\frac{u}{2} du$$\n\n$$\\frac{1}{2} \\int u \\ du$$\n\n$$\\frac{1}{2} \\times \\frac{u^{2}}{2}$$\n\n$$\\frac{1}{2} \\times \\frac{(x-1)^2}{2}$$\n\n$$\\frac{(x-1)^{2}}{4} + C$$\n\nOr we could do this:\n\n$$\\int \\frac{(x-1)}{2} dx$$\n\n$$\\int \\frac{1}{2}(x-1)dx$$\n\n$$\\frac{1}{2} \\int(x-1)dx$$\n\n$$\\frac{1}{2} (\\frac{x^2}{2} - x) + C$$\n\n$$\\frac{1}{4} (x^2 - 2x) + C$$\n\nWhy should one of them be right?\n\nLast edited:\n\nZioX\nThey're both right. Antiderivatives differ by a constant. (Differentiating eliminates the constant term). Show that those two answers differ by a constant.\n\nHomework Helper\n...\n$$\\frac{1}{4} (x^2 - \\frac{x}{2}) + C$$\n\nWhy should one of them be right?\nThe last line is wrong. You've pulled out the factor 1 / 4 incorrectly.\n$$\\frac{1}{4} \\left( x ^ 2 - 2x \\right)$$\nBoth are correct, the two results differ from each other by a constant 1/4.\nYou should note that if F(x) is one anti-derivative of f(x), then F(x) + C is also f(x)'s anti-derivative, where C is any constant.\nProof: Differentiate the LHS with respect to x, we have:\n(F(x) + C)' = F'(x) + C' = F'(x) + 0 (the derivative of a constant is just 0)\n= F'(x) = f(x)\nCan you get this? :)\n\nianb\nWell I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.\n\nThoughts?\n\nianb\nAh, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it.\n\nHomework Helper\nWell I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.\n\nThoughts?\nNope, they are identically the same, let F(x) + C1, and F(x) + C2 be two anti-derivatives of f(x), we have:\n$$\\int_a^b f(x) dx = (F(b) + C_1) - (F(a) + C_1) = F(b) - F(a)$$\n$$\\int_a^b f(x) dx = (F(b) + C_2) - (F(a) + C_2) = F(b) - F(a)$$, they are still the same.\nThe constant will automatically vanish when you subtract the two terms. :)\n\nianb\nHm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1.7. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?\n\nLast edited:\nianb\nI'm sorry, I should have been clearer. Message edited accordingly.\n\nHomework Helper\nHm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?\nUhm, well, no, they are not different.\nWell, here we go:\n$$\\int_0 ^ 1 \\frac{x - 1}{2} dx = \\left. \\left( \\frac{(x - 1) ^ 2}{4} \\right) \\right|_0 ^ 1 = \\frac{(1 - 1) ^ 2}{4} - \\frac{(0 - 1) ^ 2}{4} = 0 - \\frac{1}{4} = -\\frac{1}{4}$$\n\n$$\\int_0 ^ 1 \\frac{x - 1}{2} dx = \\left. \\frac{1}{4} \\left( x ^ 2 - 2x \\right) \\right|_0 ^ 1 = \\frac{1}{4} \\left( (1 ^ 2 - 2 \\times 1) - (0 ^ 2 - 0) \\right) = \\frac{1}{4} (-1 - 0) = -\\frac{1}{4}$$\nThey are the same. :)\n\nianb\nWow. I'm embarrassed to say the least. You, sir, rock my world.", "date": "2022-10-07 15:54:00", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/simple-integral-but-a-question-of-methodology.164031/", "openwebmath_score": 0.9141618013381958, "openwebmath_perplexity": 524.479575267068, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979354066850408, "lm_q2_score": 0.8991213698363247, "lm_q1q2_score": 0.8805581701413144}}
{"url": "https://math.stackexchange.com/questions/3151946/fewest-steps-to-reach-200-from-1-using-only-1-and-%C3%972/3152029", "text": "# Fewest steps to reach $200$ from $1$ using only $+1$ and $\u00d72$\n\nThis is a problem from the AMC 8 (math contest):\n\nA certain calculator has only two keys $$[+1]$$ and $$[\\times 2]$$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed \u201c$$9$$\u201d and you pressed $$[+1]$$, it would display \u201c$$10$$\u201d. If you then pressed $$[\\times 2]$$, it would display \u201c$$20$$\u201d. Starting with the display \u201c$$1$$\u201d, what is the fewest number of keystrokes you would need to reach \u201c$$200$$\u201d?\n\nIntuitively I worked back from $$200$$, dividing by $$2$$ until I reached an odd number, subtracting $$1$$ when I did, etc..to reach the correct answer of $$9$$ steps.\n\nHowever, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically. The best I can come up with is that beyond the first step from $$1$$ to $$2$$, multiplication by $$2$$ is always going to yield a larger step than addition by $$1$$ and therefore I should take as many $$[\\times 2]$$ steps as I can. This doesn't feel rigorous enough, though.\n\nEDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.\n\n\u2022 Do you want the fewest steps to get to exactly $200$ or at least $200$? \u2013\u00a0John Douma Mar 17 at 19:37\n\u2022 @JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice. \u2013\u00a0TonyK Mar 17 at 19:43\n\u2022 @TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise. \u2013\u00a0John Douma Mar 17 at 19:44\n\u2022 @JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$. \u2013\u00a0jeremy radcliff Mar 17 at 19:48\n\nLook at what the operations $$+$$ and $$\\times$$ do to the binary expansion of a number:\n\n\u2022 $$\\times$$ appends a $$0$$, and increases the length by one, leaving the total number of $$1$$'s unchanged;\n\u2022 if the final digit is $$0$$, then $$+$$ increases the number of $$1$$'s by one, but doesn't change the length;\n\u2022 if the final digit is $$1$$, then $$+$$ doesn't increase the total number of $$1$$'s (it may in fact decrease it), and doesn't increase the total length by more than $$1$$.\n\nTherefore, with a single key press:\n\n\u2022 you can increase the length by one, but this won't increase the number of $$1$$'s;\n\u2022 you can increase the number of $$1$$'s by one, but this won't increase the length.\n\nThe binary expansion of $$200$$ is $$200_{10}=11001000_2$$. This has three $$1$$'s, and a length of eight. Starting from $$1$$, we must increase the length by seven, and increase the number of $$1$$'s by two. So this requires at least nine steps.\n\n\u2022 Beautiful, this allows you to determine the optimal solution and path for an arbitrary number from it's binary representation. \u2013\u00a0Jared Goguen Mar 18 at 1:34\n\u2022 There actually exist two ways of getting $200$ using nine steps. $$1+1+1\\times2\\times2\\times2+1\\times2\\times2\\times2=200\\\\ 1\\times2+1\\times2\\times2\\times2+1\\times2\\times2\\times2=200$$ \u2013\u00a0Kay K. Mar 18 at 2:13\n\u2022 @KayK.: Yes, but only because there are two ways of getting to 2. The rest is uniquely determined. \u2013\u00a0TonyK Apr 4 at 16:05\n\nYou can proceed by induction on $$n$$, showing that the quickest way to reach any even number $$2n$$ involves doubling on the last step, which is clearly true for the base case $$n=1$$ (where doubling and adding $$1$$ have a tomato-tomahto relationship).\n\nNow if the last step to reach $$2n+2$$ isn't doubling, it can only be adding $$1$$ from $$2n+1$$. But $$2n+1$$ can only be reached by adding $$1$$ from $$2n$$, at which point the inductive hypothesis says the next previous number was $$n$$. But you can get from $$n$$ to $$2n+2$$ more quickly in two steps: add $$1$$, then double. So the last step in the quickest route to $$2n+2$$ is doubling from $$n+1$$.\n\n\u2022 This is nice as it formalizes the OPs intuition, whereas the binary representation answer (which is super slick) might feel a little out of the blue. \u2013\u00a0jacob1729 Mar 17 at 22:55\n\nSetting the display to binary base, $$[\\times2]$$ inserts a $$0$$ to the right and $$[+1]$$ increments; if the rightmost digit is a zero, it just turns it to a $$1$$.\n\nUsing these rules you build a number of $$o$$ ones and $$z$$ zeroes in $$o-1+z$$ keystrokes, starting from $$1$$. This seems close to optimal.\n\n\u2022 you just exactly reproduces the binary 200, why should we think it is not optimal? \u2013\u00a0dEmigOd Mar 17 at 19:54\n\u2022 @dEmigOd: I didn't prove that inserting the bits one by one with $\\times2$ or $\\times2+1$ is optimal. \u2013\u00a0Yves Daoust Mar 17 at 19:57\n\nI'll make a try\n\nSince $$200=2^7+2^6+2^3$$ you will need at least $$8$$ steps to reach $$200$$ (since we start from $$1$$ and we get a number of the form $$2^a+...+2^l$$) so it remains to show that $$8$$ steps are not enough.\n\nMaybe you could try to show that if there was a solution with $$8$$ steps then it would contain only one $$+1$$ which contradicts the fact that in $$200=2^7+2^6+2^3$$ we have two $$+$$\n\nWorking backwards, use the tree diagram and stop lagging branches (e.g. if $$\\color{red}{99}$$ in step $$3$$ leads to $$1$$, then it takes one more operation than $$\\color{red}{99}$$ in step $$2$$):\n\n$$\\hspace{3cm}$$", "date": "2019-08-23 20:23:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3151946/fewest-steps-to-reach-200-from-1-using-only-1-and-%C3%972/3152029", "openwebmath_score": 0.8029829859733582, "openwebmath_perplexity": 286.2898234280392, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750536904564, "lm_q2_score": 0.89181104831338, "lm_q1q2_score": 0.8805519817101658}}
{"url": "http://hurstbournefamilycare.com/letter-from-ewv/aadcea-area-of-a-sector-in-radians", "text": "There are three formulas for calculating the area of a sector. The formula for the area of a sector is: A = r\u00b2 * \u03b8 / 2. Find the central angle of a sector whose radius is 56 cm and area, is 144 cm2. Geometric skills. How to use the calculator Enter the radius and central angle in DEGREES, RADIANS or both as positive real numbers and press \"calculate\". Area of a sector when the central angle is given in degrees If the angle of the sector is given in degrees, then the formula for the area of a sector is given by, Area of a sector = (\u03b8/360) \u03c0r2 A = (\u03b8/360) \u03c0r2 Find the radius of a semi \u2013 circle with the area of 24 inches squared. Math A level Syllabus, 2016. The area of a sector of a circle with a radius of 5 cm, with an angle of 2 radians, is 25 cm 2. Then, the area of a sector of circle formula is calculated using the unitary method. 40pie units squared. Show that 2\u03b8-3sin\u03b8=0. The area of the minor segment as shown in Fig 5 . Area of a cyclic quadrilateral. Area of sector of circle = (1/2)r\u00b2\u04e8 , \u04e8 must be in radians. Example 1 Find the arc length and area of a sector of a circle of radius \u2026 Step 1: Find the area of the circle. Katelyn is making a semi-circular design to put on one of her quilts. Thus, if you are not sure content located Radians, Arc Length and Sector Area Radians Radians are units for measuring angles. They can be used instead of degrees. Show that 2\u03b8-3sin\u03b8=0. If the radius of the circle is\u00a0, what is the area of the semi-circular design? With the help of the community we can continue to Each of these formula is applied depending on the type of information given about the sector. Yes, both the semicircle and the quadrant are special types of the sector of a circle with angles $$180^{\\circ}$$ and $$90^{\\circ}$$ respectively. Find the area of a sector if the radius of the circle is 4, and the angle of the sector is\u00a0\u00a0radians. In simple words, the area of a sector is a fraction of the area of the circle. We can find the area of a sector of a circle in a similar manner. Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Jul 2009 448 89. Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; Displaying top 8 worksheets found for - Area Of A Sector In Radians. Both can be calculated using the angle at the centre and the diameter or radius. So in the below \u2026 The area of the circle is equal to the radius square times . as shown in Fig 4, we consider the sector as a fraction of the circle hence: Area of a segment. Introducing Radians; 9. If the diameter of a circle is 6, find the area of a sector with a sector angle of\u00a0. Varsity Tutors LLC , where\u00a0\u00a0is the angle measure of the sector\u00a0in radians, and\u00a0\u00a0is the radius of the circle. or more of your copyrights, please notify us by providing a written notice (\u201cInfringement Notice\u201d) containing The full angle is 2\u03c0 in radians, or 360\u00b0 in degrees, the latter of which is the more common angle unit. The measure of central angle XYZ is 1.25 radians. The formulas to find the area of a sector in Degrees (D\u00b0) or Radians (R\u00b0) are shown below: Area (Degrees) = \u03c0r2 x \u03b8/360 Area (Radians) = \u00bdr2\u03b8 r, D\u00b0 r, R\u00b0 r, s r, A D\u00b0, s Use 3.14 for . Track your scores, create tests, and take your learning to the next level! Area of a sector given the arc length. The non-shaded area of the circle shown below is called a SECTOR. Area of an ellipse. ChillingEffects.org. The angle AOB is in radians. Area of sector. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! What is the area of Fiona's circle? An arc of a circle when bounded by two radii with the centre of the circle then we get a slice of a circle. Find the radius of a sector whose area is 47 meters squared and central angle is 0.63 radians. Use \u2026 Send your complaint to our designated agent at: Charles Cohn Circle sector area calculator - step by step calculation, formulas & solved example problem to find the area of circle sector given input values of corcle radius & the sector angle in degrees in different measurement units between inches (in), feet (ft), meters (m), centimeters (cm) & millimeters (mm). Example: a) What is the length of the arc intercepted by an angle of 15\u00b0 on a circle with radius 20 meters? Use prior knowledge on length of circumference and area of circle to deduce formulae to calculate arc length and sector area. Given, the length of the arc, the area of a sector is given by. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! The number inside the sector is the area. degree radian; area S . If r is in \"m\", the area will be in \"m\" 2. Now, this looks messy, but we can simplify it to get: Next, use your calculator to find a decimal answer, and then round to get our final answer. Find the area of a sector whose arc is 8 inches and radius, is 5 inches. a And then we just can solve for area of a sector by multiplying both sides by 81 pi. Find the area of the sector with radius 7\\ \"cm\" and central angle 2.5 radians. There are two types of sectors, minor and major sector. I remember this formula as it is quite easy to remember. SECTORS . The portion of the circle's circumference bounded by the radii, the arc, is part of the sector. Perimeter of sector = r + 2r = r( + 2) Where is in radians If angle is in degrees, = Angle \u00d7 \u03c0/(180\u00b0) Let us take some examples: Find perimeter of sector whose radius is 2 cm and angle is of 90\u00b0 First, We need to convert angle in radians = Angle in degree \u00d7 \u03c0/(180\u00b0) = 90\u00b0 \u00d7 \u03c0/(180\u00b0 ) = \u03c0/4 Convert degrees into radians and viceversa. What is the area of the shaded sector? Fiona draws a circle with a diameter of 14 meters. A Terminal side Vertex B Initial Side C B, ABC, CBA, and are all notations for this angle. Questionnaire. which specific portion of the question \u2013 an image, a link, the text, etc \u2013 your complaint refers to; (see diagrams below) There is a lengthy reason, but the result is a slight modification of the Sector formula: I have managed to get: 3=\u00bdr\u00b2\u03b8 and 2=\u00bdr\u00b2sin\u03b8 Therefore: \u00bdr\u00b2\u03b8-3=0 and \u00bdr\u00b2sin\u03b8-2=0 But I'm unsure where to go from there. Area of a circle. Find the area of a sector if the radius is 1 and the angle of sector is\u00a0\u00a0radians. Find the area of a sector with the radius of 1 and angle of\u00a0. In order to calculate the area of a sector, you need to know the following two parameters: With the above two parameters, finding the area of a circle is as easy as ABCD. Infringement Notice, it will make a good faith attempt to contact the party that made such content available by 3. The arc length formula is used to find the length of an arc of a circle; \u2113 = r\u03b8 \u2113 = r \u03b8, where \u03b8 \u03b8 is in radian. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. is obtained by the arc AB of the centre O is given by: where is in radians. Finding a Missing Angle With the Sine Rule; 13. Section 4.2 \u2013 Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the Express the answer in terms of\u00a0. Side of polygon given area. r O 1 radian is the size of the angle formed at the centre of a circle by 2 radii which join the ends of an arc equal in length to the radius. Some of the worksheets for this concept are Arc length and sector area, Area of a sector 1, L 2r, Find the area of the shaded sector in the following, Radians arc length and area of a sector, Radians, Mcr3ui radian work, Area and arc length of a sector. Exercise worksheet on 'Find the area of a sector of a circle when the angle is given in radians.' Figure 6. misrepresent that a product or activity is infringing your copyrights. 222 r Ar r Note, to use this formula, the measure of the central angle must be given in radians. Calculate the area of the sector shown below. Calculate the area of a sector with a radius of 10 yards and an angle of 90 degrees. A-Level Maths Edexcel C2 June 2008 Q7b ExamSolutions Northeastern University, Bachelor of Science, Industrial Engineering. To calculate the area of the sector you must first calculate the area of the equivalent circle using the formula stated previously. The non-shaded area of the circle shown below is called a SECTOR. If the angle of the sector is given in degrees, then the formula for the area of a sector is given by. Find the area of a sector with a radius\u00a0and angle of\u00a0. \u00a9 2007-2020 All Rights Reserved, Computer Science Tutors in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in San Francisco-Bay Area. A sector is created by the central angle formed with two radii, and it includes the area inside the circle from that center point to the circle itself. In this example the sector subtends a right-angle (900) at the centre of the circle. Area of a sector of a circle. A-level : area and arc length of a sector tutorial In this tutorial you are shown how to find the area of a sector and arc length when the angle is in degrees or radians. Problem Solving With the Cosine Rule; 15. Find the angle of a sector whose arc length is 22 cm and area, is 44 cm2. When the angle at the centre is 360\u00b0, area of the sector, i.e., the complete circle = \u03c0r\u00b2 When the angle at the center is 1\u00b0, area of the sector = Thus, when the angle is \u03b8, area of sector, OPAQ = Write the formula for the area of a sector in radians. Using this formula, and approximating , the area of the circle is . Sep 2, 2009 #2 For #1. Recognize parts of a circle and use appropriate terminology. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such The circular face of a watch has an area that measures between 800 and 900 square millimeters. We know that the area of the whole circle is equal to \u03c0r\u00b2. The ratio of the area of the sector to the area of the full circle will be the same as the ratio of the angle \u03b8 to the angle in a full circle. When angle of the sector is 360\u00b0, area of the sector i.e. Area of a regular polygon. Find the area of the sector. How to find the area of a sector whose central angle is in radian: formula, 1 example, and its solution. Section 4.2 \u2013 Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the", "date": "2022-01-16 21:20:36", "meta": {"domain": "hurstbournefamilycare.com", "url": "http://hurstbournefamilycare.com/letter-from-ewv/aadcea-area-of-a-sector-in-radians", "openwebmath_score": 0.8573663830757141, "openwebmath_perplexity": 390.67052471505565, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9857180673335565, "lm_q2_score": 0.8933094081846422, "lm_q1q2_score": 0.8805512233666487}}
{"url": "http://mathhelpforum.com/calculus/125089-piecewise-function-problem.html", "text": "1. ## Piecewise Function problem\n\nLet f(x) =\n\n(0) if x< 0\n(x) if 0 <= x <= 1\n(2 - x) if 1 < x <= 2\n(0) if x > 2\n\nand g(x) = [integrate] f(t)dt on [0, x]\n\ni) Find an expression for g(x) similar to the one for f(x).\n\nii) Where is f differentiable? Where is g differentiable?\n\nI'm pretty much stuck, and don't know how to start this question.\nI know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).\n\nBut from that, how do I figure out an expression for g(x) from the piecewise function?\n\nAny and all help is appreciated here.\n\n2. Originally Posted by Tulki\nLet f(x) =\n\n(0) if x< 0\n(x) if 0 <= x <= 1\n(2 - x) if 1 < x <= 2\n(0) if x > 2\n\nand g(x) = [integrate] f(t)dt on [0, x]\n\ni) Find an expression for g(x) similar to the one for f(x).\n\nii) Where is f differentiable? Where is g differentiable?\n\nI'm pretty much stuck, and don't know how to start this question.\nI know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).\n\nBut from that, how do I figure out an expression for g(x) from the piecewise function?\n\nAny and all help is appreciated here.\ni)\n\nfor $\\displaystyle x < 0$, $\\displaystyle f(x) = 0$ and $\\displaystyle g(x) = \\int f(t) ~dt = 0$\n\nfor $\\displaystyle 0 \\leq x \\leq 1$, $\\displaystyle f(x) = x$ and $\\displaystyle g(x) = \\int_{0}^{x} f(t) ~dt = \\int_{0}^{x} t ~dt = \\frac{1}{2}x^2$\n\nfor $\\displaystyle 1 < x \\leq 2$, $\\displaystyle f(x) = 2 - x$ and $\\displaystyle g(x) = \\int_{0}^{x} f(t) ~dt = \\int_{0}^{1} f(t) ~dt + \\int_{1}^{x} f(t) ~dt=$$\\displaystyle \\int_{0}^{1} t ~dt + \\int_{1}^{x} (2-t) ~dt= \\frac{1}{2} + (2x - 2) - (\\frac{1}{2}x^2 - \\frac{1}{2})$\n\nfor $\\displaystyle x > 2$\n\n3. Hello, Tulki!\n\nHere's part (a).\nDid you make a sketch?\n\n$\\displaystyle \\text{Let }\\:f(x) \\;=\\;\\left\\{\\begin{array}{cccc}0 && x < 0 \\\\ \\\\[-4mm] x && 0 \\leq x \\leq 1 \\\\ \\\\[-4mm] 2 - x && 1 < x \\leq 2 \\\\ \\\\[-4mm] 0 && x > 2 \\end{array}\\right\\}$\n\nand: .$\\displaystyle g(x) \\:=\\: \\int_0^x f(t)\\,dt$\n\na) Find an expression for $\\displaystyle g(x)$ similar to the one for $\\displaystyle f(x).$\n\n$\\displaystyle g(x)$ gives the area under the graph from $\\displaystyle 0\\text{ to }x$\n\nThe graph looks like this:\nCode:\n|\n|\n1+       *\n|     *:::*\n|   *:::::::*\n| *:::::::::| *\n- * * * - - - * - + - * * * - -\n0       1   x   2\n\nThen: .$\\displaystyle \\displaystyle g(x) \\;=\\;\\left\\{ \\begin{array}{cccc} 0 && x < 0 \\\\ \\\\[-3mm] \\int^x_0t\\,dt && 0 \\leq x \\leq 1 \\\\ \\\\[-2mm] \\frac{1}{2} + \\int^x_1 (2-t)\\,dt && 1 < x \\leq 2 \\\\ \\\\[-3mm] 1 && x > 2 \\end{array} \\right\\}$\n\n4. Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.\n\nFor x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).\n\nFor x between 0 and 1, y= x so the \"area under the curve\" is the area of a right triangle with base x and height x: its area is $\\displaystyle \\frac{1}{2}x^2$.\n\nFor x between 1 and 2, y= 2- x and we can break the \"area under the currve\" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is $\\displaystyle \\frac{1}{2}(1)(1)= \\frac{1}{2}$. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is $\\displaystyle \\frac{2-x+ 1}{2}x= \\frac{(3-x)x)}{2}= \\frac{3x- x^2}{2}$.\nThe total area is $\\displaystyle \\frac{1}{2}+ \\frac{3x- x^2}{2}= \\frac{1+ 3x- x^2}{2}$. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.\n\nFor x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.\n\n5. Thank you very much, all of you!\n\nI suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect.", "date": "2018-05-23 23:34:17", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/125089-piecewise-function-problem.html", "openwebmath_score": 0.9992508888244629, "openwebmath_perplexity": 1243.347413107103, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180643966651, "lm_q2_score": 0.8933094003735663, "lm_q1q2_score": 0.8805512130435773}}
{"url": "https://math.stackexchange.com/questions/4507765/dense-subset-of-sn-1-with-antipodal-complement", "text": "# Dense subset of $S^{n-1}$ with antipodal complement\n\nLet $$n \\in \\mathbb{N}$$, and let $$S^{n} \\subseteq \\mathbb{R}^{n+1}$$ be the unit $$n$$-sphere.\n\nDoes there exist a dense subset $$D$$ of $$S^n$$ such that $$\\textbf{x} \\in D \\implies -\\textbf{x} \\notin D$$?\n\nThis is true for $$n=1$$. Note that $$S^1 \\cong A := [0,1]/\\mathbb{Z}$$, where the equivalent question is: does there exists a dense subset $$D'$$ of $$A$$ such that $$x \\in D \\implies x \\pm \\frac{1}{2} \\notin D'$$?\n\nConsider the function $$f : [0,1) \\to \\{0,1\\}$$,\n\n$$f(x)= \\begin{cases} 0 & x< \\frac{1}{2},\\text{ any decimal expansion of } x \\text{ has finitely many 3s} \\\\ 1 & x < \\frac{1}{2}, \\;x \\text{ has a decimal expansion with } \\infty \\text{ many 3s} \\\\ 1 & x = \\frac{1}{2} \\\\ 1 & x > \\frac{1}{2}, \\text{ any decimal expansion of } x \\text{ has finitely many 3s}\\\\ 0 & x > \\frac{1}{2}, \\;x \\text{ has a decimal expansion with } \\infty \\text{ many 3s} \\end{cases}$$\n\nThen $$D' = f^{-1}(\\{0\\})$$ is a suitable dense set.\n\nHowever, I am not sure how to approach the cases $$n >1$$, since $$S^n$$ is no longer homeomorphic to a very simple object and my topological toolkit is quite unsophisticated. If the question can be answered with elementary methods, I welcome any hints. I am not even sure whether or not I expect such a set $$D$$ to exist for $$n> 1$$.\n\n\u2022 It is also equivalent to ask the analogue of the question in $\\Bbb R^n$, if you\u2019ve figured out what the antipodes look like after stereographic projection Aug 7 at 14:42\n\u2022 I don't have time for a full answer, but you could consider the stereographic projection $\\mathbb{S}^n\\setminus\\{(0,\\ldots,0,1)\\}\\rightarrow\\mathbb{R}^n$, then look at where the intersection $H_\\pm^{n+1}=\\{x\\in\\mathbb{R}^{n+1}|\\operatorname{sign}(x_0)=\\pm 1\\}$ with $\\mathbb{S}^n$ lands, and then consider the primages of their intersection with $\\mathbb{Q}^n$ and $\\mathbb{R}^n\\setminus\\mathbb{Q}^n$ respectivly. Aug 7 at 14:58\n\u2022 @SamuelAdrianAntz I almost understand your comment, but it's not quite clear. I am a little suspicious of your last step because note that $D$ quite naturally bijects with its complement, so both are uncountable. I understand the stereographic projection to be a bijection, so the preimage of $\\mathbb{Q}^n$ would be countable? Or have I misunderstood? Aug 7 at 15:19\n\u2022 @FShrike That's certainly a possible approach. But even if I could figure out where they antipodes go, I am not certain that this simplifies matters much (or enough for me). After all, $\\mathbb{R}$ is still much simpler than $\\mathbb{R}^n$, $n>1$. Aug 7 at 15:28\n\nThis answer is an expansion of my comment, but I simplified it a bit without using the stereographic projection.\n\nFirst of all, the obvious way of taking the intersection of $$\\mathbb{Q}^{n+1}$$ with $$\\mathbb{S}^n\\subset\\mathbb{R}^{n+1}$$ won't work, as it can even be empty as explained in this answer here. Consider the half-spaces $$\\mathbb{H}_\\pm^{n+1}=\\{x\\in\\mathbb{R}^{n+1}|\\operatorname{sign}(x_0)=\\pm 1\\}$$ and the (surjective) projections: $$\\operatorname{pr}_\\pm\\colon \\mathbb{H}_\\pm^{n+1}\\cap\\mathbb{S}^n \\twoheadrightarrow\\mathbb{D}^n, (x_0,x_1,\\ldots,x_n)\\mapsto(x_1,\\ldots,x_n)$$ flattening down both half-spheres.\n\nThe subsets $$A=\\operatorname{pr}_-^{-1}(\\mathbb{D}^n\\cap\\mathbb{Q}^n)$$ and $$B=\\operatorname{pr}_+^{-1}(\\mathbb{D}^n\\cap(\\mathbb{R}^n\\setminus\\mathbb{Q}^n))$$ are both dense (in the closure of their respective half-sphere) as preimages of dense sets under continuous maps. $$A\\cup B$$ is therefore dense in $$\\mathbb{S}^n$$ as the closure operator commutes with finite unions (See here).\n\nFor $$x\\in A$$, we have $$x_0<0$$, therefore $$-x\\notin A$$, as well as $$x_1,\\ldots,x_n\\in\\mathbb{Q}$$, therefore $$-x\\notin B$$.\n\nFor $$y\\in B$$, we have $$y_0>0$$, therefore $$-y\\notin B$$, as well as $$y_1,\\ldots,y_n\\notin\\mathbb{Q}$$, therefore $$-y\\notin A$$.\n\n\u2022 So one can use effectively the same idea as for $n=1$. Fair enough. I convinced myself it would be much more difficult and didn't try it. Also, your profile description is fairly entertaining. Aug 7 at 21:26\n\u2022 As a minor correction, $A$ and $B$ are images, not preimages. Aug 7 at 21:33\n\u2022 Thanks a lot! :-) The mistake with $A$ and $B$ is also corrected now. Aug 7 at 21:40\n\u2022 No worries, and thanks for the answer. I'm sorry, but I think I have just observed another typo. I think the projection maps surject onto $D_1(0)$ in $\\mathbb{R}^{n-1}$, the interior of the $(n-1)$-sphere. Aug 7 at 21:45\n\u2022 Oh, yes! Honestly, I'm really not that good in writing down my thoughts in the correct notation. Aug 7 at 21:47", "date": "2022-09-26 12:20:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4507765/dense-subset-of-sn-1-with-antipodal-complement", "openwebmath_score": 0.9285436868667603, "openwebmath_perplexity": 216.31667047448963, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759610129465, "lm_q2_score": 0.8976952880018481, "lm_q1q2_score": 0.8805277283156064}}
{"url": "http://mathhelpforum.com/calculus/159154-difficult-trigonometric-integral.html", "text": "# Math Help - Difficult Trigonometric Integral\n\n1. ## Difficult Trigonometric Integral\n\nI was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.\n\n$\\int {sin^2 x \\over \\sqrt {1+cosx}}$\n\nThanks for the help!\n\n2. Originally Posted by Tylerzzz\nI was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.\n\n$\\int {sin^2 x \\over \\sqrt {1+cosx}}$\nSubstitute $u = \\cos x$. Then $du = -\\sin x\\,dx$ and the integral becomes\n\n$-\\int \\frac{\\sin x}{\\sqrt {1+cosx}}(-\\sin x)dx = -\\int \\frac{\\sqrt{1-u^2}}{\\sqrt{1+u}}du = -\\int\\sqrt{1-u}\\,du = \\frac23(1-u)^{3/2} = \\frac23(1-\\cos x)^{3/2}$ (plus a constant).\n\n3. Wow! I never really thought about solving it that way. Thanks a lot!\n\n4. EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]\n\n5. Originally Posted by TheCoffeeMachine\nEDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]\nYou had me worried for a moment.\n\n6. Hello, Tylerzzz!\n\nAnother approach . . .\n\n$\\displaystyle \\int \\frac{\\sin^2\\!x}{\\sqrt{1+\\cos x}}\\,dx$\n\nWe have: . $\\dfrac{\\sin^2\\!x}{\\sqrt{1+\\cos x}}$\n\nMultiply by $\\frac{\\sqrt{1-\\cos x}}{\\sqrt{1-\\cos x}}\\!:\\;\\;\\dfrac{\\sin^2\\!x}{\\sqrt{1+\\cos x}} \\cdot\\dfrac{\\sqrt{1-\\cos x}}{\\sqrt{1-\\cos x}} \\;=\\;\\dfrac{\\sin^2\\!x\\sqrt{1-\\cos x}}{\\sqrt{1-\\cos^2\\!x}}$\n\n. . . $\\;=\\;\\dfrac{\\sin^2\\!x\\sqrt{1-\\cos x}}{\\sqrt{\\sin^2\\!x}} \\;=\\;\\dfrac{\\sin^2\\!x\\sqrt{1-\\cos x}}{\\sin x} \\;=\\;\\sin x\\sqrt{1-\\cos x}$\n\nThe integral becomes: . $\\displaystyle \\int(1-\\cos x)^{\\frac{1}{2}}(\\sin x\\,dx)$\n\nLet: . $u \\:=\\:1-\\cos x \\quad\\Rightarrow\\quad du \\:=\\:\\sin x\\,dx$\n\nSubstitute: . $\\displaystyle \\int u^{\\frac{1}{2}}\\,du \\;=\\;\\tfrac{2}{3}u^{\\frac{3}{2}} + C$\n\nBack-substitute: . $\\frac{2}{3}(1-\\cos x)^{\\frac{3}{2}} + C$\n\n7. Originally Posted by Opalg\nYou had me worried for a moment.\nHaha! In my defence, I blame the UCAS for the plight of my eyesight today.", "date": "2016-07-24 19:12:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/159154-difficult-trigonometric-integral.html", "openwebmath_score": 0.9514219164848328, "openwebmath_perplexity": 2112.593693504914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130580170845, "lm_q2_score": 0.8902942355821459, "lm_q1q2_score": 0.8805126244680808}}
{"url": "http://mathhelpforum.com/pre-calculus/40393-need-help-half-life-problems-using-logs.html", "text": "# Thread: Need Help for Half Life Problems using Logs\n\n1. ## Need Help for Half Life Problems using Logs\n\nHi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.\n\nThe equation that I got was:\n\nN = No (1/2) ^ t\n\nN is final amount, No is the initial ammount and t is time.\n\nI was given the data:\n\nTime (min) 10 20 30 40\nAmount (g) .83 .68 .56 .46\n\nand I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.\n\nHow do I do this? I have no clue...\n\n2. Do you have Excel?. You can do all sorts of regressions with it. Also, many calculators do regressions. Do you have one?. If so, what is it. Maybe I can step you through it.\n\nAnyway, using Excel I got an equation of\n\n$y=1.0091e^{-.0196x}$\n\nThis has an R^2 of 1, so it is as good as it gets.\n\nEnter in x=10, do you get .83?.\n\n3. hi thanks for replying, I have a TI-83+.\n\n4. I would have to re-familiarize myself. I suppose the 83 does exponential regressions. I am not sure.\n\n5. it does, i did it before, I think.\nI just forgot how to do it.\n\nI think it's storing the data into 2 list and then compare them or something...\n\nbut then, how do I set the equations up?\n\nI mean that N = No (1/2) ^ t thing , hate math final, lol.\n\nMaybe you can show me how you did it in excel?\n\n6. You can find the half life by using the equation $k=\\frac{-1}{T}ln(2)$\n\nWe know k=-.0196\n\nSo, $-.0196=\\frac{-1}{T}ln(2)$\n\n$t=35.36$\n\nWe could also do it using the equation.\n\nHalf of 1.0091 is .50455. Because 1.0091 is the amount at t=0\n\nSo, $.50455=1.0091e^{-.0196t}$\n\n$t=35.36$\n\nSame as before.\n\nTo find a regression on Excel, I make the graph, then add a trendline.\n\nClick on Insert, then Chart, then Scatter\n\nHighlight your data which is entered in columns A and B and proceed.\n\nIt is pretty much self-explanatory. Then onece you have the graph, click on\n\nChart, Add Trendline, Exponential, Click on Options and check the box that says to display the equation on the chart.\n\nThere you have it.\n\n7. Originally Posted by Billwaa\nHi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.\n\nThe equation that I got was:\n\nN = No (1/2) ^ t\n\nN is final amount, No is the initial ammount and t is time.\n\nI was given the data:\n\nTime (min) 10 20 30 40\nAmount (g) .83 .68 .56 .46\n\nand I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.\n\nHow do I do this? I have no clue...\n$N = N_0 \\left ( \\frac{1}{2} \\right ) ^{kt}$\n\n$ln(N) = ln(N_0) - [k~ln(2)]t$\n\nPut t on the horizontal axis, ln(N) on the vertical axis, and do a linear regression. The slope will be $k~ln(2)$ which you can then solve for k, and the intercept will be $ln(N_0)$.\n\n-Dan\n\n8. thanks guys, this clear it up a little bit. ^_^\n\n9. Originally Posted by Billwaa\nhi thanks for replying, I have a TI-83+.\nTo do this on the 83+, input the data into two lists:\n\nL1: {10,20,30,40}\nL2: {.86,.68,.56,.46}\n\nNow, hit [STAT] and scroll over to [CALC] Select option [0]:ExpReg.\n\nThis is an exponential regression. The solution will have the form of $y=a\\times b^x$.\n\nAt the home screen, you should now see:\n\nExpReg\n\nNow hit [2nd][1][,][2nd][2]\n\nYou should now have the following on the screen:\n\n$\\text{ExpReg } L_1\\text{,}L_2$\n\nHit [ENTER]\n\nYou'll get the regression equation:\n\nThis is what the screen should say:\n\nExpReg\n\\begin{aligned}\ny&=a*b^x \\\\\na&=1.00914514 \\\\\nb&=.9805442185\n\\end{aligned}\n\nNote that TheEmtpySet's answer was $y=1.0091e^{-.0196x}$\n\nWe can rewrite the answer that the calculator gave us to get his solution.\n\nNote that $b^x= e^{x\\ln(b)}$.\n\nThus, $.9805442185^x=e^{x\\ln(.9805442185)}=e^{-.0196475365x}$\n\nRounding the regression equation, we get:\n\n$\\color{red}\\boxed{y=1.0091e^{-.0196x}}$\n\nHope this makes sense!", "date": "2017-03-26 02:56:09", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/40393-need-help-half-life-problems-using-logs.html", "openwebmath_score": 0.6899756789207458, "openwebmath_perplexity": 1201.3480662606455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130563978902, "lm_q2_score": 0.8902942326713409, "lm_q1q2_score": 0.8805126201476973}}
{"url": "https://math.stackexchange.com/questions/2920725/is-the-limit-of-this-infinite-step-construction-an-equilateral-triangle", "text": "# Is the limit of this infinite step construction an equilateral triangle?\n\nJust for fun (inspired by sub-problem described and answered here):\n\nLet's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute:\n\nNow we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute:\n\nNow we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute:\n\nRepeate the same process infinite number of times.\n\nCan we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle.\n\n\u2022 Regarding the title question, here's a more complicated procedure: 1. Prove whether P=NP or not. 2. Draw an equilateral triangle with compass and straightedge (or any other method). \u2013\u00a0user253751 Sep 18 '18 at 1:32\n\u2022 @immibis Very funny :) \u2013\u00a0Oldboy Sep 18 '18 at 5:08\n\u2022 @user202729 Title changed as suggested. Thanks! \u2013\u00a0Oldboy Sep 18 '18 at 5:09\n\u2022 @immibis On second thought, your procedure is not more complicated. Assuming P=NP is true or not the proof will always take finite time. And my \u201cconstruction\u201d takes infinite number of steps, each step taking constant time. Just kidding :) \u2013\u00a0Oldboy Sep 18 '18 at 21:39\n\u2022 However the steps are all the same, which make it pretty simple :) \u2013\u00a0user253751 Sep 18 '18 at 22:57\n\nThink about what happens to the maximum difference between angles over time.\n\nFor simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has \"maximum angle difference\" $\\vert y-x\\vert$. Then when we move one of the $y$-angled points, our new triangle will have angles\n\n$$y, {x+y\\over 2}, {x+y\\over 2}$$\n\nsince the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is\n\n$$\\left\\vert {y\\over 2}-{x\\over 2}\\right\\vert={1\\over 2}\\vert y-x\\vert.$$\n\nSo each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $\\vert y-x\\vert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.\n\n$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, ${1\\over 2}$): if $r\\in(-1,1)$ then for any $a$ we have\n\n$$\\lim_{n\\rightarrow\\infty}ar^n=0.$$\n\nNote that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!\n\n\u2022 I have also noticed that the difference between angles decreases. But having a decreasing sequence does not mean that the sequence converges to zero. You have proved the key point! \u2013\u00a0Oldboy Sep 18 '18 at 5:18\n\u2022 Angles $x$ and $y$? Aaah mine eyes! \u2013\u00a0user332714 Sep 18 '18 at 7:08\n\nBy the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $\\theta_i$, so that the base angles are $\\frac12(\\pi - \\theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $\\theta_{i} = \\frac12(\\pi-\\theta_{i-1})$. Thus, \\begin{align}\\theta_n &= -\\frac12\\theta_{n-1} + \\frac12\\pi \\\\[6pt] &=\\frac12\\left(-\\frac12(\\pi-\\theta_{n-2})+\\pi\\right) = \\frac14\\theta_{n-2}+\\frac12\\pi-\\frac14\\pi \\\\[6pt] &= \\cdots \\\\[6pt] &= \\left(-\\frac12\\right)^{n}\\theta_0 \\;-\\; \\sum_{i=1}^n\\left(-\\frac12\\right)^{n}\\pi \\\\[6pt] \\lim_{n\\to\\infty}\\theta_n &= 0\\cdot\\theta_0 \\;-\\; \\frac{(-1/2)}{1-(-1/2)}\\pi \\\\ &=\\frac{\\pi}{3} \\end{align}\n\nThus, in the limit, the triangle becomes equilateral. $\\square$\n\nAssume WLOG that the initial triangle is isoceles. Let $\\alpha$ be the apical angle, and let $\\beta$ be a remaining angle. Then the transformation in question sends\n\n$$\\begin{bmatrix}\\alpha \\\\ \\beta \\end{bmatrix}\\mapsto \\begin{bmatrix}0 & 1 \\\\ \\tfrac{1}{2} & \\tfrac{1}{2} \\end{bmatrix} \\begin{bmatrix}\\alpha \\\\ \\beta \\end{bmatrix}\\text{.}$$ Let $\\mathsf{X}$ be the $2\\times 2$ transformation matrix on the rhs. $\\mathsf{X}$ has characteristic polynomial $x^2-\\tfrac{1}{2}x-\\tfrac{1}{2}=0.$ By the Cayley\u2013Hamilton theorem, $$\\mathsf{X}^2=\\tfrac{1}{2}\\mathsf{X}+\\tfrac{1}{2}\\text{.}$$ Therefore we have a Sylvester formula $$f(\\mathsf{X})=f(1)\\left(\\frac{1+2\\mathsf{X}}{3}\\right)+f(-\\tfrac{1}{2})\\left(\\frac{2-2\\mathsf{X}}{3}\\right)$$ for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus, $$\\mathsf{X}^n=\\frac{1+2\\mathsf{X}}{3}+(-\\tfrac{1}{2})^n\\left(\\frac{2-2\\mathsf{X}}{3}\\right)\\text{.}$$ The second term converges to zero, so $$\\begin{split} \\lim_{n\\to\\infty}\\mathsf{X}^n&=\\frac{1+2\\mathsf{X}}{3}\\\\ &=\\frac{1}{3}\\begin{bmatrix}1 & 2 \\\\ 1 & 2\\end{bmatrix}\\\\ &=\\frac{1}{3}\\begin{bmatrix} 1\\\\ 1\\end{bmatrix}\\begin{bmatrix}1&2\\end{bmatrix}\\text{,} \\end{split}$$ $$\\lim_{n\\to\\infty} \\begin{bmatrix}0 & 1 \\\\ \\tfrac{1}{2} & \\tfrac{1}{2} \\end{bmatrix}^n \\begin{bmatrix}\\alpha \\\\ \\beta \\end{bmatrix}=\\begin{bmatrix}\\tfrac{\\alpha+2\\beta}{3}\\\\ \\tfrac{\\alpha+2\\beta}{3}\\end{bmatrix}\\text{.}$$ i.e., the apical and side angles approach equality as the operation is repeated.", "date": "2021-06-18 19:20:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2920725/is-the-limit-of-this-infinite-step-construction-an-equilateral-triangle", "openwebmath_score": 0.980134129524231, "openwebmath_perplexity": 510.64152811170686, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874123575265, "lm_q2_score": 0.8887588023318195, "lm_q1q2_score": 0.8803932822118514}}
{"url": "https://www.physicsforums.com/threads/definite-integral-of-step-function.819626/", "text": "# Homework Help: Definite integral of step function\n\nTags:\n1. Jun 18, 2015\n\n### Byeonggon Lee\n\nI need to prove whether this expression is true or false:\n\n$\\sum\\limits_{k=1}^{n}\\int_{k-1}^{k}[x]dx = \\frac{n(n-1)}{2}$\n\nI'm so confused because as I know, definite integral is possible only when the target function is continuous in closed interval.\n\nIn this case, function $[x]$ should be continuous in interval $[k-1,k]$\nbut actually [x] is not continuous in $[k,k-1]$\n$[x]=k-1$ when $k-1\\leq x<k$, $[x]=k$ when $x=k$\n\nSo I thought that this expression is false,\nbut in my book's answer, it is true:\n\n\"\"\"\nBecause$[x]=k-1$ when $k-1\\leq x<k$, $[x]=k$ when $x=k$\n$\\sum\\limits_{k=1}^{n}\\int_{k-1}^{k}[x]dx$\n$=\\int_{0}^{1}[x]dx+\\int_{1}^{2}[x]dx+\\int_{2}^{3}[x]dx+...+\\int_{n-1}^{n}[x]dx$\n$=\\int_{0}^{1}0dx+\\int_{1}^{2}1dx+\\int_{2}^{3}2dx+...+\\int_{n-1}^{n}(n-1)dx$\n$=0+\\bigg[x\\bigg]_{1}^{2}+\\bigg[2x\\bigg]_{2}^{3}+...+\\bigg[(n-1)x\\bigg]_{n-1}^{n}$\n$=0+1+2+...+(n-1)$\n$=\\frac{n(n-1)}{2}$\n\"\"\"\n\nThe book also says that I need to think interval as $k\\leq x<k+1$\nand I don't understand how is it possible to omit equal sign at $x<k+1$\n\n2. Jun 18, 2015\n\n### DEvens\n\nAn integral is an area. The troublesome points at the end of the interval mean you are uncertain about the area at a point. What is the difference in the area under a curve if you move one single point up or down by 1?\n\nAnother way to approach it is this. Think of the integral as a limit of a sum of areas. It happens that in this case you can very easily construct the limit because the function you are looking at is piece-wise constant.\n\n3. Jun 18, 2015\n\n### Ray Vickson\n\nA definite integral certainly can exist for a discontinuous function. Jump discontinuities in f(x) at an interval's endpoint a or b does not affect the area under the curve y = f(x) from x = a to x = b. In other words,\n$$\\int_{[a,b]} f(x) \\, dx = \\int_{(a,b]} f(x) \\, dx = \\int_{[a,b)} f(x) \\, dx = \\int_{(a,b)} f(x) \\, dx$$\nWe can denote all four of these by the common symbol $\\int_a^b f(x) \\, dx$.\n\n4. Jun 18, 2015\n\n### RUber\n\nYou can read a little more on this idea at https://en.wikipedia.org/?title=Riemann_integral.\nFor example, to your point on continuity:\n\"A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure).\"\nSet of measure zero means a finite (or countably infinite) set of distinct points. In the case of the step function is discontinuous at the integers which for a maximum n is finite, and without a maximum is countably infinite. Both sets are measure zero, so you are okay to integrate.\n\n5. Jun 18, 2015\n\n### WWGD\n\nJust to nitpick, or to add a bit: while in this context of step functions discontinuities are finite, you may have in other cases uncountably-infinite sets of measure zero, e.g., the Cantor set.\n\n6. Jun 18, 2015\n\n### SammyS\n\nStaff Emeritus\nI believe your function is also called the floor function, $\\displaystyle \\ \\lfloor x\\rfloor \\,,\\$ and the greatest integer function, $\\displaystyle \\ [\\![ x]\\!]\\ .\\$\n\nOn the interval $\\displaystyle \\ [k-1\\,,\\ k)\\,,\\$ we have $\\displaystyle \\ [x]=k-1\\ .$\n\nTake the limit: $\\displaystyle \\ \\lim_{a\\to k^-} \\int_{k-1}^a [x]\\,dx \\ . \\$", "date": "2018-07-22 12:59:53", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/definite-integral-of-step-function.819626/", "openwebmath_score": 0.8627914190292358, "openwebmath_perplexity": 317.66459350228246, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874098566368, "lm_q2_score": 0.8887587831798666, "lm_q1q2_score": 0.8803932610174803}}
{"url": "https://math.stackexchange.com/questions/367473/f-mathbbr-to-mathbbr-satisfies-x-2fx-x1fx-1-3-evaluate/367476", "text": "# $f: \\mathbb{R} \\to \\mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$\n\nThe function $f : \\mathbb{R} \\to \\mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.\n\n\u2022 What math contest is this from? Apr 20, 2013 at 17:38\n\u2022 @Potato: K\u00f6MaL, Problem B. 4538. Apr 28, 2013 at 14:44\n\u2022 Another example of a question that should have been closed before anybody answered it! Apr 30, 2013 at 11:03\n\nHint 1: $3=(x+1)-(x-2)$\n\n$(x-2)f(x)-(x+1)f(x-1)=3\\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$\n\nHint 2: Is there a function closely related to $f$ that would verify a simpler equation?\n\n$g(x)=f(x)+1$\n\n\n\n$(x-2)g(x) - (x+1)g(x-1)=0 \\Leftrightarrow (x-2)g(x)=(x+1)g(x-1) \\Leftrightarrow g(x)=\\cfrac{x+1}{x-2}g(x-1)$\n\n$g(x)=\\cfrac{x+1}{x-2}g(x-1)=\\cfrac{x+1}{x-2}\\cfrac{x+1-1}{x-2-1}g(x-1-1) = \\left(\\prod\\limits_{k=0}^{n-1} \\cfrac{x+1-k}{x-2-k}\\right) g(x-n)$\n\nNow give the good value to $n$, simplify the product and you'll have the expression of $g$ you need.\n\n\u2022 Yeah right. Some terms cancel each other >_< Apr 20, 2013 at 20:42\n\u2022 What is \"the good value\"? Apr 28, 2013 at 14:31\n\u2022 The value sot hat you can evaluate $g(x-n)$ Apr 28, 2013 at 14:39\n\u2022 Big fan of answers with this format +1 May 1, 2013 at 3:26\n\u2022 Same here, format+1. May 5, 2013 at 1:11\n\nLet $x-1=y$, then, $$(y-1)f(y+1)-(y+2)f(y)=3$$ $$\\implies f(y+1)=\\frac{3+(y+2)f(y)}{y-1} \\;\\;\\;\\;\\;(1)$$\n\nLemma: $\\forall \\; n \\geq 2 \\in \\mathbb{N}$, $f(n)=n(n-1)(n+1)-1$.\n\nBase Case: If $n=2$ then $f(n)=1 \\cdot 2 \\cdot 3 -1=5$ which is true by information provided in the question.\n\nInductive Step: Assume for $n=k$ that $f(k)=k(k-1)(k+1)-1$.\n\nThen by equation $(1)$, $$f(k+1)=\\frac{3+(k+2)(k(k-1)(k+1)-1)}{k-1}$$ $$=\\frac{k^4+2k^3-k^2-3k+1}{k-1}$$ $$=\\frac{(k-1)(k^3+3k^2+2k-1)}{k-1}$$ $$=k^3+3k^2+2k-1$$ $$=k(k^2+3k+2)-1$$ $$=k(k+1)(k+2)-1$$\n\nThus completing the induction.\n\nHence our Lemma is true and $f(n)=n(n-1)(n+1)-1 \\; \\forall \\; n \\geq 2 \\in \\mathbb{N}$\n\nIn particular, if $n=2013$, $f(2013)=2013 \\cdot 2012 \\cdot 2014-1=8157014183$ (I confess I used a calculator).\n\nP.S I found the lemma by trying small cases.\n\nThe conditions allow you to calculate $f(x+1)$ if you know $f(x)$. Try calculating $f(3), f(4), f(5), f(6)$, and looking for a pattern.\n\nRunning the following Mathematica program:\n\nf = DifferenceRoot[Function[{f, x}, {(x - 2) f[x] - (x + 1) f[x - 1] == 3, f[2] == 5}]];\nf[2013]\n\n\nWe get the answer = 8157014183.", "date": "2023-03-26 20:43:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/367473/f-mathbbr-to-mathbbr-satisfies-x-2fx-x1fx-1-3-evaluate/367476", "openwebmath_score": 0.7702268362045288, "openwebmath_perplexity": 1237.7263442567425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787842245938, "lm_q2_score": 0.891811038968166, "lm_q1q2_score": 0.880376937206666}}
{"url": "https://math.stackexchange.com/questions/2208230/which-of-the-following-complex-numbers-is-equivalent-to-frac3-5i82i", "text": "Which of the following complex numbers is equivalent to $\\frac{3-5i}{8+2i}$?\n\nI encountered this question in one of my SAT practice tests.\n\nI know the answer is option C, however the only way I got the answer was by trial and error of trying multiple ways to simplify the equation and I ended up rationalising it to get answer choice C.\n\nIs there any other, perhaps easier or more direct method, that I can use to solve these types of questions?\n\n\u2022 Mutiplying top and bottom by the conjugate of the denominator is the standard approach. No need for trial and error -- the standard method will always work. \u2013\u00a0quasi Mar 29 '17 at 6:36\n\u2022 I ended up rationalising it Don't know that there is anything more direct than that. You don't even need to carry out all calculations, just figure out the sign of the imaginary part to decide between C and D. (That's assuming you discarded choices A, B upfront, as expected) \u2013\u00a0dxiv Mar 29 '17 at 6:39\n\u2022 As a multiple choice question, there will be a minus sign, but A is likely to be wrong \u2013\u00a0Henry Mar 29 '17 at 6:45\n\u2022 As a multiple choice question ,you can compare $|z|=|\\dfrac{3-5i}{8+2i}|=\\dfrac{|3-5i|}{|8+2i|}=\\dfrac{\\sqrt{2(17)}}{\\sqrt{4(17)}}=\\dfrac{\\sqrt2}{2}$ \u2013\u00a0Khosrotash Mar 29 '17 at 6:56\n\u2022 @Khosrotash: interesting proposal, but unfortunately it leads to harder computation than the standard way (because of $(7^2+23^2)/34^2$) :-( \u2013\u00a0Yves Daoust Mar 29 '17 at 7:10\n\nTwo ways to approach this problem. First: As quasi suggests in the comments, multiply by the conjugate of the denominator.\n\n\\begin{align} \\frac{3-5i}{8+2i} & = \\frac{3-5i}{8+2i} \\times \\frac{8-2i}{8-2i} \\\\ & = \\frac{24-40i-6i-10}{8^2+2^2} \\\\ & = \\frac{14-46i}{68} = \\frac{7-23i}{34} \\end{align}\n\nThe second approach, given that it's a multiple choice problem, is to multiply each of the answers by $8+2i$ and see if you obtain $3-5i$. I think the first approach is simpler, but they'll both work.\n\nMultiplying the top and bottom by the complex conjugate is how you handle this. That gives:\n\n$$\\frac{3-5i}{8+2i}\\cdot\\frac{8-2i}{8-2i}=\\frac{(3-5i)(8-2i)}{8^2+2^2}=\\frac{7-23i}{34}$$\n\nNotice that at the second step we are guaranteed to have a real denominator because $(a+bi)(a-bi)=a^2+b^2$ is always real.\n\nI think the method you are using easier one.\n\n$$\\frac{3-5i}{8+2i}$$\n\nMultiply and divide the fraction by conjugate of denominator,\n\n$$\\frac{3-5i}{8+2i} \\cdot \\frac{8-2i}{8-2i}$$\n\n$$\\frac{24-6i-40i+10i^2}{64-4i^2}$$\n\n$$\\frac{24-6i-40i-10}{64+4}$$\n\n$$\\frac{14-46i}{68}$$\n\n$$\\frac{7-23i}{34}$$\n\nAlternatives:\n\nLet the answer be $a+ib$. Rewrite the initial equation as $$(a+ib)(8+i2)=3-i5.$$\n\nExpanding, you get the $2\\times2$ system $$\\begin{cases} 8a-2b=3,\\\\ 2a+8b=-5.\\end{cases}$$\n\nThen by Cramer or simply adding four times the first equation and the second\n\n$$34a=7$$\n\nand subtracting the first from four times the second,\n\n$$34b=-23.$$\n\nGet rid of the denominators and try the products\n\n$$(3\\pm i20)(8+i2)=24\\mp40+i(6\\pm160)\\to64-i154,$$ $$(7\\pm i23)(8+i2)=56\\mp46+i(14\\pm144)\\to102-i170.$$\n\n(We select the signs that match those of $3-i5$.)\n\nThen we obtain the identity\n\n$$(7-i23)(8+i2)=34(3-i5).$$\n\nThe real way:\n\nUse the division formula\n\n$$\\frac{a+ib}{c+id}=\\frac{ac+bd}{c^2+d^2}+i\\frac{bc-ad}{c^2+d^2},$$\n\ngiving\n\n$$\\frac{14}{68}-i\\frac{46}{68}.$$\n\nThen compare to the proposed answers. As C seems to match but D is similar, double check the signs or try the product, for safety.", "date": "2021-05-16 09:07:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2208230/which-of-the-following-complex-numbers-is-equivalent-to-frac3-5i82i", "openwebmath_score": 0.9472055435180664, "openwebmath_perplexity": 328.9857120597467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98228770337066, "lm_q2_score": 0.8962513828326955, "lm_q1q2_score": 0.8803767124855065}}
{"url": "https://math.stackexchange.com/questions/106313/regular-average-calculated-accumulatively", "text": "# Regular average calculated accumulatively\n\nis it possible to calculate the regular average of a sequence of numbers when i dont know everything of the sequence, but just everytime i get a new number i know the total count of numbers and the average for the numbers - 1.\n\nfor example: 2 3 10 the average is of course: 5\n\nbut in the last step to calculate i only have access to the previous average of 2 and 3: 2.5 the next number: 10 and the count of numbers: 3\n\nif this is possible, how?\n\nYes, and you can derive it from the expression for the average. Let the average of the first $n$ numbers be $\\mu_n$. The formula for it is\n\n$$\\mu_n = \\frac{1}{n} \\sum_{i=1}^n x_i$$\n\nThen you can derive\n\n$$n \\mu_n = \\sum_{i=1}^nx_i = x_n + \\sum_{i=1}^{n-1} x_i = x_n + (n-1)\\mu_{n-1}$$\n\nand hence, dividing by $n$,\n\n$$\\mu_n = \\frac{(n-1) \\mu_{n-1} + x_n}{n}$$\n\ni.e. to calculate the new average after then $n$th number, you multiply the old average by $n-1$, add the new number, and divide the total by $n$.\n\nIn your example, you have the old average of 2.5 and the third number is 10. So you multiply 2.5 by 2 (to get 5), add 10 (to get 15) and divide by 3 (to get 5, which is the correct average).\n\nNote that this is functionally equivalent to keeping a running sum of all the numbers you've seen so far, and dividing by $n$ to get the average whenever you want it (although, from an implementation point of view, it may be better to compute the average as you go using the formula I gave above. For example, if the running sum ever gets larger than $10^{308}$ish then it may be too large to represent as a standard floating point number, even though the average can be represented).\n\n\u2022 As you're computing the previous sum $(n-1)\\mu_{n-1}$ as part of the formula, I think this wouldn't help is the sum gets too large. Jun 23, 2016 at 1:12\n\u2022 @danijar Completely true - a better approach is to keep running sums of the $x_n$ using a method that is robust to rounding error (e.g. Kahan summation) and a separate running count of $n$, and divide whenever you need the mean. That way, your error is bounded by the accuracy of your floating point type. Jun 23, 2016 at 7:44\n\u2022 If you are likely to end up with numbers larger than $10^{308}$ then you either need to scale down your inputs, or use a more capacious floating point type. Jun 23, 2016 at 7:46\n\nA very simple thought process results in the same formula for running average. If you have $N$ previous measures (of course the measures could all be different) the average you calculate is exactly the same as if all measures were the same as the computed average value. Then, computing the running average of the $N+1$ is equal to $N$ times the previously computed average plus the $N+1$ measure all divided by $N+1$. I know that this is the same as the formula posted in the other answer but no derivation with sums is needed or more obscure mathematical thinking is needed (OK, maybe not really obscure).\n\nWhat you are asking for is commonly called sequential estimation. A general approach is described in [Robbins, H. and S. Monro (1951). A stochastic approximation method. Annals of Mathematical Statistics 22, 400\u2013407.]\n\nTo add to the derivation of Chris Taylor, I personally like this rewriting as it goes quite intuitively (easy to remember). $$\\mu_n = \\mu_{n-1} + \\frac{1}{n}(x_n - \\mu_{n-1})$$\n\n# algorithmically: sequential average computation\navg += (x_n - avg)/n", "date": "2022-08-12 09:30:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/106313/regular-average-calculated-accumulatively", "openwebmath_score": 0.8435456156730652, "openwebmath_perplexity": 313.39418673840015, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964177527898, "lm_q2_score": 0.8933094138654242, "lm_q1q2_score": 0.8803532273092199}}
{"url": "http://www.billthelizard.com/2009/07/broken-stick-experiment.html", "text": "## Monday, July 27, 2009\n\n### The Broken Stick Experiment\n\nI found several interesting video lectures at the BLOSSOMS Video Library. One in particular that I enjoyed is called The Broken Stick Experiment [Flash]. In it, Professor Richard Larson asks the seemingly simple question,\nIf you break a stick randomly in two places, what is the probability that you can form a triangle from the three pieces?\nIf one of the pieces is too long, then the other two can't meet to form the triangle.\nThe lecture is only 33 minutes, and Professor Larson builds up a very pretty geometric proof, so I won't spoil it by posting the answer here. I will provide code for a simulation, though (yes, this is still a programming blog). See if you can guess the answer before running the simulation or watching the video.\nimport java.util.Arrays;import java.util.Random;public class BrokenStick {   public static void main(String[] args) {       Random random = new Random();       int trials = 10000;       int triangles = 0;       double[] breaks = new double[2];       double[] sides = new double[3];       for( int i=0; i < trials; ++i )       {           breaks[0] = random.nextDouble();           breaks[1] = random.nextDouble();           Arrays.sort(breaks);           sides[0] = breaks[0];           sides[1] = breaks[1] - breaks[0];           sides[2] = 1.0 - breaks[1];           Arrays.sort(sides);                     if( sides[2] < (sides[0] + sides[1]) )           {               triangles++;           }       }       System.out.println(\"Triangles: \" + triangles);       System.out.println(\"Trials: \" + trials);       double p = (double)triangles / trials;       System.out.println(\"Proportion: \" + p);   }}\n\nThe simulation breaks 10,000 sticks each into three random pieces using Java's Random class to come up with random lengths. If the longest piece of broken stick is longer than the two remaining pieces, a triangle can't be formed. Ten-thousand trials should be enough to accurately estimate the proportion of triangles made from the broken sticks to two decimal places. How close was your guess?\n\nRelated:\nThe Broken Stick Revisited\n\nrbonvall said...\n\nThanks, I had a fun time solving it :)\n\nAt least to me, this problem lends itself nicely to a geometric solution. In the a-b-c space, the possible breaks must satisfy a+b+c=1, which is a plane. Then, it reduces to find the fraction of the \"positive part\" of the plane that satisfies the constraints. And I always have fun drawing intersecting planes :P\n\nBill the Lizard said...\n\nrbonvall,\n\"And I always have fun drawing intersecting planes :P\"\n\nYour comment reminds me of the first lecture (or maybe the second, I can't quite remember) in MIT's Linear Algebra video series. Gilbert Strang shows several methods for solving systems of equations, and makes a complete mess of drawing intersecting planes. He uses that as an illustration of why solving systems of equations using matrices is far superior. :)\n\nJohn | Retro Programming said...\n\nThanks, it's an interesting problem. I think I've got the solution, I just need to run a program to double check my result :-)\n\nAlexey Busygin said...\n\nI think it should be 33,3%.\n\nBill the Lizard said...\n\nAlexey,\nIt's not 33%, but your initial guess is closer than mine was.\n\nTrevel said...\n\nI estimate at just under 25% triangle. Trying to figure if there's a flaw in my logic, but I can't pick it out...\n\ntg said...\n\nI agree with Trevel. Independent breaks which both must occur on the same half of the stick. <25%\n\ntg said...\n\nGot the right answer, but not sure if the logic is right:\n\nIf two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two. These are independent events. The probability of one break on 50% of the stick is 1/2, the probability of two (independent) breaks on the same half is 1/2*1/2 = 0.25.\n\nBill the Lizard said...\n\ntg,\nWouldn't that indicate that there's a 25% chance of not forming a triangle?\n\ntg said...\n\nback to the drawing board....\n\nBill the Lizard said...\n\ntg,\nDon't feel bad, you almost had me convinced. It took me ten minutes to spot the flaw, and I've watched the video twice! :)\n\ntg said...\n\nHow about one more try using the same concept (except using it correctly to represent all three lengths, as I didn't do the first go round):\n\nProbability of NOT forming P(NF) =\nP(a > b+c)*P(b>a+c)*P(c>b+a)\n\nAs mentioned earlier P(a>b+c) = probability of two breaks on 1/2 of the stick = 0.25.\n\nP(NF) = 0.25*0.25*0.25 = 0.75\nThus P( NOT NF) = 0.25\n\nBill the Lizard said...\n\ntg,\n0.25 * 0.25 * 0.25 = 0.015625\n\nBack to your original statement, though, you said \"If two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two.\" This is correct, but it's only one way that you could get one piece that's longer than the other two. You could also break the stick on opposite ends of the stick and have the middle piece be too long.\n\ntg said...\n\nB the L, you are a patient individual.\n\nOne final attempt to redeem myself. The \"*\" was supposed to be \"+\". The risks of writing at work.\n\nWe are looking for the Probability of (a>b+c) OR (b>a+c) OR (c>a+b).\n\na + b + c = L\nP(a>b+c) = P(a>L/2)\n\nFor a>L/2, two breaks must occur on the right half:\nP(of a break occurring on the right half) = 0.5\nP(of two independent breaks occurring on the right)=\nP(a>L/2) = 0.5*0.5 = 0.25\n\nFor b>L/2, two independent breaks must occur on either the left quarter, the right quarter, or both breaks on the left quarter, or both on the right quarter. The probability of ONE break landing on that 50% of the stick = 0.5. The probability of another break independently landing on the same 50%of the stick = 0.5*0.5 = 0.25.\n\nc is the same as a, so\nP(c>b+a) = P(a>b+c) = 0.25\n\nFor any one length > L/2 = (a>b+c) OR (b>a+c) OR (c>a+b) = 0.25+0.25+0.25.\n\nIf this is wrong, spare me the indignity and just delete my comments.\n\nBill the Lizard said...\n\ntg,\nNow you have it! I like the approach you used because you're showing the exact opposite of what the problem asks for, the probability of NOT forming a triangle. This kind of thinking is a very powerful tool in mathematics, and is often overlooked.\n\nSam152 said...\n\nVery interesting post, my nifty c++ program calculated about 18.9% successes after 10,000 tries. How close is this?\n\nBill the Lizard said...\n\nSam152,\nThat's a little bit low; lower than rounding error would explain. Do you want to share your code? You could either post it here, or as a Stack Overflow question and I'd be happy to take a look.\n\nfiberfiend6891 said...\n\nJust a heads up - the link directly to the video has http doubled and doesn't work when clicked directly (I had to copy/paste and get rid of the double):\nhttp://http//blossoms.mit.edu/video/larson-watch.html\n\nThis is my first time seeing your blog, and I absolutely LOVE it!\n\nBill the Lizard said...\n\nfiberfiend6891,\n\nTom Raywood said...\n\nBill the Lizard,\n\nHi. Great question.\n\nThere is one thing I'm not clear on, though, from the description. Do obtuse triangles count here, or is the solution limited to acute?\n\nBill the Lizard said...\n\nTom,\nYou could end up with an acute, obtuse, or (rarely) equilateral triangle. You've got me curious now, so I'll have to modify the code later and run another simulation to see if acute or obtuse triangles are more common. Good question!\n\nTom Raywood said...\n\nBill the Lizard,\n\nGood. If you limit outcomes to acute triangles, (including equilaterals), what probability do you get?\n\nOh, and another question [intended to confirm an assumption on my part]: Is this strictly an ordered set or can any of the three pieces form the base?\n\nBill the Lizard said...\n\nTom,\nAny of the three sides could form the base (I'm assuming by base you mean the longest side). In the code I gave you can see how I work around this by calculating the length of the three sides, then immediately sorting them (on line 24) so that I always know where the longest side ends up.\n\nI'm working on an entire post to answer your other question. :)\n\nTom Raywood said...\n\nBill the Lizard,\n\nGee, no I hadn't imagined that the base of the triangle had to be the largest of the three pieces. All of these unstated parameters make it very difficult to approach the problem in a systematic way.\n\nSo far, yes, the triangle can be of any sort, the pieces can be moved around at will and, finally, (if I understand your last answer), the base of the triangle is expected to be whichever piece [of the stick] that's the longest.\n\nIs this correct? Is there anything else in the way of allowances and/or limitations?\n\nBill the Lizard said...\n\nTom,\nDisregard my last answer, as I had misunderstood your question. There are no constraints on what kind of triangle you can get (acute, obtuse, right, or no triangle at all) or on the order of the pieces. Any of the pieces can be the longest when you randomly select the two break-points. The only reason I sort the pieces in my code is because I need to know which piece is longest, so that I can tell if the pieces form a triangle or not. This has no impact on the final outcome.\n\nTom Raywood said...\n\nBill the Lizard,\n\nWhoa, hey, no, it can't be as simple as that can it?\n\nAll it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick. And that's a full 50% of the time.\n\nWhat gives? I thought certainly the question would pose a greater challenge than that.\n\nBill the Lizard said...\n\nTom,\nClose. \"All it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick.\" This much is true, but that doesn't happen a full 50% of the time (which was also my initial guess before I watched the full lecture).\n\nTom Raywood said...\n\nBill the Lizard,\n\nSo true. (I'm feeling better already.) After 100 simulations with n=1,000,000 I'm seeing slightly less than 50%. More specifically I'm seeing 49.994067% and am definitely curious as to what explains this slight deviation. I think I'd rather contemplate it for a while though, that is, to see if I can figure it out myself.\n\nTom Raywood said...\n\nBill the Lizard,\n\nTo follow up here, beginnings of a guess are as follows: For some small range [in the form of a difference], as two legs of the triangle both approach 1/2, the small difference between them cannot be bridged by the infinitesimally small remaining leg. Be nice to make a formal statement.\n\nBill the Lizard said...\n\nTom,\nCheck your simulation code against mine. Your answer is off by quite a bit more than just rounding error. You can post your code here, or on Stack Overflow if you'd like me to take a look at it (provided it's in a language I know reasonably well, Java, C, C++, Python, PHP, or any dialect of Basic).\n\nTom Raywood said...\n\nBill the Lizard,\nDefinitely courting tedium at this point.\n\nn R1 R2 R3 T P1 P2 P3 C1\n\nThese represent column headings in Excel.\n\nR1, R2 and R3 denote random numbers using the RANDBETWEEN function; I chose for each number the range from 1 to 10,001.\n\nT denotes the sum of R1, R2 and R3 for each row.\n\nP1, P2 and P3 denote which percentage of T is entailed of R1, R2 and R3 respectively.\n\nC1 (for condition 1) represents a check to see whether P1, P2 and P3 all three entail percentages less than 0.5, posting a 1 if they do and a 0 if they don't.\n\nn denotes the number of iterations which, as I said, I took the time to set at 1 million.\n\nTotaling the C1 column provides a sum which, divided into 1 million, quite closely approximates the probability that no 'leg of the triangle' will be greater than or equal to one half.\n\nAdditionally I generated a macro which recalculates the worksheet 100 times, posts each result to a table and then finds the average of that set.\n\nYou can imagine my surprise to hear that the actual probability varies significantly from what this arrangement points up.\n\nI'll take a look at that lecture this weekend.\n\nBill the Lizard said...\n\nTom,\nFrom your explanation it looks like you're generating three random lengths to represent the three pieces of stick. The problem with this approach is that there's nothing to make sure the three pieces all add up to the original length of the stick.\n\nTry it by generating only 2 random numbers (x and y) that represent the break points. If 10,000 is the length of the original stick, and you pick 2 random numbers between 0 and 10,000 then the lengths of the three pieces will be x, y-x, and 10,000-y when x < y, or y, x-y, and 10,000-x when x > y. (It's easier if you just select two random numbers then sort them so x < y, then just always use x, y-x, and 10,000-y for the three lengths, which is essentially what I did in my code.)\n\nI'll try to get my next post up this weekend. I have the answer to your question about acute and obtuse triangles, but proving why I'm getting the number I'm getting is proving to be a little bit complicated.\n\nTom Raywood said...\n\nBill the Lizard,\nI'll work this up and see what gives. It's definitely counterintuitive though that always having the same stick length would even matter, that is, any more than it would matter that triangle of type Q presents as q1, q2, ...qn for n of any size. Be nice to hear the why behind your assertion. Maybe the lecture will help in this regard.\n\nBill the Lizard said...\n\nTom,\nThe particular value of the stick length doesn't matter as much as making sure the three pieces add up to whatever length you decide on. I used a stick length of 1.0 in my similation, but it doesn't hurt anything if you choose a length of 10,000.\n\nWhat really makes a difference is how you choose your random lengths. If you choose three random lengths then add them together, you don't really know the length of the stick you started with. If you start with a specific length, then make two random breaks within the boundaries of the stick, you're guaranteed to end up with three random lengths that add up to your original stick length.\n\nIt's possible that I just didn't correctly understand the explanation of your spreadsheet. Without seeing the formulas you used I can't be sure that this part of your simulation isn't correct. The errors you're getting could be from another part of the calculation, but this is the most logical place to start looking.\n\nTom Raywood said...\n\nBill the Lizard,\n\nI suspect most people would anticipate that having a consistent or particular stick length shouldn't matter here. What's at issue is where the two breaks appear, that is, relative to overall length. A million sticks all of different lengths and all randomly broken in two places should demonstrate the same probabilistic characteristics of a million sticks all of the same length because, again, what's at issue is relative in any event.\n\nThis principle can be demonstrated as follows. In my simulation each stick length is represented by T, that is, the sum of R1, R2 and R3. If I wanted/needed to ensure that each and every occurrence of T always comes out to equal a certain value, a few more columns could be inserted which accomplish this simply by adjusting the sizes of R1, R2 and R3 by some definitive amount. For example, if for one row T presented as equal to 16,000 [compared to, say, the value of 21,000 I'd predetermined as a uniform stick length], obviously all I'd have to do is increase each of the original random numbers by 21/16.\n\nSo even though now all million stick lengths are the same, the actual percentage of that length occupied by each R1, R2 and R3 hasn't changed at all. The outcome would still match the 50% probability that my simulation currently produces.\n\nIn short, Woe to The Foe, heh heh heh.\n\nHere are the formulas I employ for...\n\n...{R1,R2,R3}\n................ =RANDBETWEEN(1,10001)\n\n..........{T}\n................ =SUM(E2:G2), because R1, R2 and R3 occupy columns E, F and G.\n\n...{P1,P2,P3}\n................ =E2/H2; =F2/H2; and =G2/H2, because T occupies column H.\n\n.........{C1}\n................ =IF((J2<0.5)*(K2<0.5)*(L2<0.5),1,0), because P1, P2 and P3 occupy columns J, K and L.\n\nAnd fun was had by all.\n\nSpartacus said...\n\n1. Probability that both cuts are on opposite sides of the midpoint: 1/2\n2. Probability that two cuts on opposite sides of the midpoint are less than half the stick length apart: 1/2\n\nTherefore the probability of forming a triangle is 1/2*1/2=1/4\n\nHere is a more difficult problem with an even more surprising result: You break the stick into two pieces, then randomly choose one of the two pieces and break it into two pieces. What is the probability that the resulting three pieces can form a triangle?\n\nYou will be amazed by the solution!\n\n(Hint: the answer is not just a simple fraction like the original problem; however there is a very nice closed form)\n\nBill the Lizard said...\n\nSpartacus,\nI've read about half-a-dozen variations on this problem (mostly linked from the notes on the BLOSSOMS page) but I haven't seen that one. This might take a while. :)\n\nAnonymous said...\n\nSam152's solution is correct; it is the answer the Spartacus's variant.\n\nBill's original problem statement was under-specified, admitting two interpretations that yield different answers. (Do you choose two cutpoints independently and uniformly on the original stick; or do you make one uniform random cut, and then make another uniform random cut on one of the pieces?)\n\nSee http://www.cut-the-knot.org/Curriculum/Probability/TriProbability.shtml\n\nAnonymous said...\n\nThis issue might be very old and might already been resolved. but just for clarification your value for second side is wrong and should be taken as follows:\n\nbreaks[1] = random.nextDouble()*(1-breaks[0]);\n\nWith this correction, the probability comes to 0.193 which is the correct solution.\n\nBill the Lizard said...\n\nAnonymous,\n\nNo, that's incorrect. The second break can take place at any point along the length of the stick. Your change restricts it to only the length of the first piece broken off.\n\nAnonymous said...\n\nOk yes you are correct, this imposes a restriction and thus reduces the probability. the correct answer is 0.25 only and not 0.193.\nthanks.\n\nasha rani said...\n\nIs it ok for each triangle n What is n??", "date": "2015-07-30 16:07:14", "meta": {"domain": "billthelizard.com", "url": "http://www.billthelizard.com/2009/07/broken-stick-experiment.html", "openwebmath_score": 0.6750235557556152, "openwebmath_perplexity": 905.8308727164317, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964198826467, "lm_q2_score": 0.893309401793762, "lm_q1q2_score": 0.8803532173152612}}
{"url": "https://mathematica.stackexchange.com/questions/22094/modular-arithmetic-efficiently-calculating-the-remainders-of-factorials", "text": "# Modular arithmetic - efficiently calculating the remainders of factorials\n\nWhen working on this question regarding the divisibility of the sum of factorials, I decided to write some code to test \"small values\" of the problem using the following code.\n\nf[p_] := Total[Mod[#!, p] & /@ Range[p - 1]];\nTable[Mod[f@Prime@i, Prime@i], {i, 1, 500}]\n\n\nBasically, what the code does is sum up all the factorials $$1!+2!+3!+\\dots+(p-1)!$$\n\nand find the remainder modulo $p$, for prime $p$.\n\nUnfortunately, my code as written takes a very long time to run. Checking the first 500 primes takes 88.280966 seconds on my computer, but checking the first 2000 primes took me about 4 hours.\n\nIs there any way to improve the code, or is it already the best we can do?\n\nAs for optimizations not involving the code, I used Wilson's Theorem, which states that for all primes $p$,\n\n$$(p-1)!\\equiv-1 \\bmod p$$\n\nUsing the above theorem, we can modify the code as follows.\n\nh[p_] := Total@Flatten[{Mod[#!, p], PowerMod[(# - 1)!*(-1)^(#), -1, p]} & /@ Range[(p - 1)/2]];\nTable[Mod[h@Prime@i, Prime@i], {i, 1, 500}]\n\n\nThis is considerably faster than the previous code, since checking the first 500 primes takes only 25.896166 seconds. However, checking the first 2000 primes still takes an inordinately long time.\n\nThis is bit faster:\n\ntoPrime = 500;\nsums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]];\nprimes = Prime[Range[toPrime]];\nMod[sums[[primes - 1]], primes]\n\n\nPrecompute factorial sums and primes. Mod is fast on lists.\n\n\u2022 you are way too humble! calculating your sum for even the first 2000 primes takes less than a second. However, is there a way to get around storing large numbers in sums in memory? It keeps crashing my computer when I try toPrime=5000. \u2013\u00a0Vincent Tjeng Mar 26 '13 at 6:03\n\u2022 +1 (that's freaking fast!) Can you explain why Accumulate@FoldList[#1 #2 &, 1, Range[n] is so much quicker to Accumulate@Array[#! &, n] + 1? I really don't get it. \u2013\u00a0gpap Mar 26 '13 at 11:23\n\u2022 @gpap Calculating factorial so many times costs a lot. Since we know we want all the factorials up to Prime[toPrime]-1, we ultimately gain a lot keeping the intermediate results with FoldList. \u2013\u00a0Michael E2 Mar 26 '13 at 12:04\n\u2022 @MichaelE2 Yes, worked it out myself in the meantime - you multiply the previous result and don't calculate a factorial at every step. Thanks \u2013\u00a0gpap Mar 26 '13 at 12:11\n\u2022 Is there any reason why you used #1 #2 & instead of Times? \u2013\u00a0J. M. is away Jun 15 '15 at 12:53\n\nLet $x \\equiv r_1 \\bmod p$ and $y \\equiv r_2 \\bmod p$. Then, $x y \\equiv r_1 r_2 \\bmod p$. So, we can compute the sum of the factorials mod p using:\n\nf[p_] := Mod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p]\n\n\nLet's compare this to the naive implementation:\n\nt[p_] := Mod[Sum[k!, {k, p-1}], p]\n\n\nFor example:\n\nf[Prime[500]] //AbsoluteTiming\nt[Prime[500]] //AbsoluteTiming\n\n\n{0.00058, 1813}\n\n{0.085628, 1813}\n\nThe nice thing about using Mod[Times[##], p] as the FoldList function is that the output should be a machine number unless you are working with very large primes. That means that f can be compiled:\n\nfc = Compile[{{p, _Integer}},\nMod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p],\nRuntimeAttributes->{Listable}\n];\n\n\nLet's compare for a large prime:\n\nf[Prime[10^6]] //AbsoluteTiming\nfc[Prime[10^6]] //AbsoluteTiming\n\n\n{1.83041, 9308538}\n\n{1.05449, 9308538}\n\nFaster, but the real difference is that fc is Listable. Hence, comparing timings on a list shows a much larger difference. For example:\n\nr1 = f /@ Prime[Range[5000, 5500]]; //AbsoluteTiming\nr2 = fc[Prime[Range[5000, 5500]]]; //AbsoluteTiming\n\nr1 === r2\n\n\n{2.06691, Null}\n\n{0.395402, Null}\n\nTrue\n\nFinally, since the list of all factorials is not stored anywhere, the memory used is much more manageable. Here is the memory and timing for the first 5000 primes:\n\nr1 = fc[Prime[Range[5000]]]; //MaxMemoryUsed //AbsoluteTiming\n\n\n{3.03463, 462848}\n\nThis is quite a bit smaller than @MichaelE2's answer:\n\nMaxMemoryUsed[\ntoPrime = 5000;\nsums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]];\nprimes = Prime[Range[toPrime]];\nr2 = Mod[sums[[primes - 1]], primes]\n] //AbsoluteTiming\n\nr1 === r2\n\n\n{2.40441, 4064607008}\n\nTrue\n\nThe compiled answer uses .462KB while the approach where all the factorials are precomputed takes 4GB, and the timing is not too different.", "date": "2019-08-24 12:43:09", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/22094/modular-arithmetic-efficiently-calculating-the-remainders-of-factorials", "openwebmath_score": 0.5452675819396973, "openwebmath_perplexity": 3235.7443640870365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399090658734, "lm_q2_score": 0.907312226373181, "lm_q1q2_score": 0.8803105320106703}}
{"url": "https://math.stackexchange.com/questions/2496932/while-plotting-polar-coordinate-system-graphs-should-we-include-the-negative-of", "text": "# While plotting polar coordinate system graphs, should we include the negative of angle $\\theta$?\n\nWhile plotting polar coordinate system graphs, should we include the negative of angle $\\theta$?\n\nFor example, while plotting $r=\\theta$ for negative values the spiral comes out to be reverse of the graph plotted using positive values. Upon searching in google I find only one spiral. So are both spirals part of the curve?\n\n\u2022 There's no reason to omit negative values of $\\theta$, other than that the pictures are sometimes less pretty. \u2013\u00a0Aaron Montgomery Oct 30 '17 at 17:46\n\u2022 Negative values of $\\theta$ are perfectly fine. Often times when doing integrals in polar coordinates, it is convenient to use a parameterization with negative values rather than positive ones. \u2013\u00a0superckl Oct 30 '17 at 17:49\n\nIt's not really about $\\theta$, which of course can be both positive or negative. Rather it's about the values of $r$ \u2014 should we allow negative values of $r$ or not? And this is not really a mathematical question, but rather a matter of convention. Most textbooks I've seen allow both positive and negative values of $r$. However, some require that $r$ must be non-negative ($r\\ge0$).", "date": "2020-02-17 16:28:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2496932/while-plotting-polar-coordinate-system-graphs-should-we-include-the-negative-of", "openwebmath_score": 0.742084264755249, "openwebmath_perplexity": 268.35664927583764, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357579585026, "lm_q2_score": 0.899121367808974, "lm_q1q2_score": 0.8802719698295445}}
{"url": "https://math.stackexchange.com/questions/4481618/how-many-points-uniquely-define-a-square", "text": "# How many points uniquely define a square?\n\nI was wondering how many points uniquely define a square. Now it is clear to me that if we have $$n$$ random points on a square then this does not necessarily uniquely define the square (for example they may all lie on the same edge).\n\nMy question was more the following: supposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick to uniquely define the square, and which points should I pick?\n\nEdit: the points need not lie on the vertices.\n\n\u2022 Isn\u2019t it 3? Pick three corner points. Jun 27 at 18:17\n\u2022 @EDX Which four points are enough? If you pick the four corners, then I can for example draw a different square for which those four points lie on the midpoints of the sides, so that's not enough. Jun 27 at 18:23\n\u2022 @EDX I think it depends on how we understand the question. If we take a square and pick the four corner points, there will actually be infinitely many squares containing those points, although they won't be in the corners. Jun 27 at 18:24\n\u2022 @subrosar yes the points need not lie on the corners\n\u2013\u00a0JLB\nJun 27 at 18:24\n\u2022 I can not think of a way to have four points uniquely define a square, but five points should suffice. so long as exactly one set of the three points are colinear. From the three that are colinear we can tell the orientation of the square as any other edges must come at 90 degree angles to it. From there you should be able to draw lines which the edges will fall upon. You should be able to then find the distance between opposite lines which will be the edge length and you should be able to complete the square from there. Jun 27 at 18:37\n\nFour points can be enough to unambiguously define a square.\n\nTake two of the corners, an additional point from the edge between those corners, and an additional point from the middle of the edge opposite the edge the first three points lied on.\n\nA second observer coming to look at these points having been told they all lie on the same square will learn from the three colinear points the orientation of the square and that these lie on the same edge and additional edges will either be perpendicular or parallel to this.\n\nThey will then note that since if drawing a perpendicular from the fourth point to this edge it lands in the middle of the edge formed by our colinear points that this must be on the opposite edge. Next, since they know this opposite edge must be parallel to the original edge we now know the lengths of the edges must be equal to this minimum distance between our fourth point and the edge formed by our three colinear points.\n\nFinally, having noted that this length is equal to the length between our colinear points that these two must have been corners of our square.\n\n\u2022 That's easier than my construction; well done! Jun 27 at 18:58\n\nFour points are enough, if chosen correctly.\n\nIn particular, I claim that the only square which passes through the four points $$A(0,0)$$, $$B(0,0.3)$$, $$C(0.4,0)$$, and $$D(1,1)$$ is the unit square with corners $$(0,0), (0,1), (1,0), (1,1)$$. The solution lets you find many more similar examples.\n\nTo prove this claim, here are some inequalities about the distances between points on a square with side length $$s$$:\n\n\u2022 Two points on the same side are at a distance between $$0$$ and $$s$$;\n\u2022 Two points on adjacent sides are at a distance between $$0$$ and $$\\sqrt2 s$$;\n\u2022 Two points on opposite sides are at a distance between $$s$$ and $$\\sqrt 2 s$$.\n\nWe can check that if $$A,B,C,D$$ lie on a square, then they can't all lie on two adjacent sides of the square: there's no two lines that contain all four points and intersect at right angles. (It's enough to check that $$AB$$ is not perpendicular to $$CD$$, $$AC$$ is not perpendicular to $$BD$$, and $$AD$$ is not perpendicular to $$BC$$.) So there must be some pair of points on opposite sides.\n\nThe six distances between the points are $$AB=0.3$$, $$AC=0.4$$, $$BC=0.5$$, $$CD \\approx 1.16$$, $$BD \\approx 1.22$$, and $$AD \\approx 1.41$$. So we see that if any two of $$A,B,C$$ were on opposite sides, the side length $$s$$ would be at most $$0.5$$, but $$D$$ is more than $$0.5 \\sqrt2 \\approx 0.71$$ away from all the rest. Therefore $$A,B,C$$ are all on two adjacent sides of the squares.\n\nIn particular, one side of the square must contain two of $$A,B,C$$. Can a side of the square contain $$B$$ and $$C$$? No: line $$BC$$ separates $$A$$ from $$D$$, which a side of the square can't do. If we assume $$A,B$$ are on one side of the square and $$C$$ is on an adjacent side, or that $$A,C$$ are on one side of the square and $$B$$ is on an adjacent side, we get the same conclusion: the square has a corner at $$A$$ with one side containing segment $$AB$$ and one side containing segment $$AC$$.\n\nSince $$D$$ does not lie on either line, it must be on one of the other sides of the square: a side parallel to $$AB$$, and a side parallel to $$AC$$. We get the same square and try both.\n\n\u2022 If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AB$$, those lines must contain three sides of the square. The only such square is the unit square.\n\u2022 If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AC$$, those lines must contain three sides of the square. The only such square is the unit square.\n\nThree points are never enough.\n\nTake three points $$A,B,C$$. If they form a right triangle, we can draw ever-bigger squares with that right angle as a corner, containing all three. If they form an obtuse triangle, that's even better; suppose without loss of generality that the altitude from $$C$$ lands on line $$AB$$ at a point $$H$$ outside segment $$AB$$. Then we can draw ever-bigger squares with one corner at $$H$$ containing all three points.\n\nFor an acute triangle, begin by rotating the points arbitrarily. Let $$x_{\\min}, x_{\\max}, y_{\\min}, y_{\\max}$$ be the lowest and highest $$x$$- and $$y$$-coordinates among the three points; exclude the finitely many orientations where there are ties. Draw the four lines $$x = x_{\\min}, x=x_{\\max}, y = y_{\\min}, y = y_{\\max}$$; each of them passes through one point and together they contain all three (otherwise we'd get an obtuse angle). One of the points lies on two of the lines.\n\nWithout loss of generality, $$A$$ lies on $$x = x_{\\max}$$ and $$y = y_{\\max}$$ and also $$x_{\\max} - x_{\\min} \\ge y_{\\max} - y_{\\min}$$. Then erase the line through $$y = y_{\\max}$$ and instead draw the line through $$y = y_{\\min} + x_{\\max} - x_{\\min}$$. These four lines define a square containing all three points. Since we get this for all but finitely many ways to rotate $$A,B,C$$ (equivalently, for all but finitely many choices of a horizontal and vertical direction) we get infinitely many squares.\n\n\u2022 Ahhhhh now I see why three doesn\u2019t suffice. Jun 27 at 20:19\n\nIf you specify what points you chose, then two are sufficient: choose opposite vertices. A diagonal can then be drawn between them, and a perpendicular bisector of the same length can be drawn, with the ends being the other two vertices. Then the sides can be drawn between them.\n\n\u2022 Sometimes the simple answer is the correct answer\u2026 \u201csupposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick\u201d\n\u2013\u00a0Matt\nJun 28 at 7:27\n\u2022 @Matt There is still some ambiguity. If Alice wants to communicate to Bob what square she has, two points being sufficient requires not only that Alice can choose which points to tell Bob, but that Bob knows that Alice chose those points. If all that Bob knows is that these two points are on the square, but doesn't know that they are opposite vertices, then Bob will not know which square they're from. Jun 29 at 0:55\n\u2022 I understand your point, but also Bob knows it is a square and not some other completely irregular shape (otherwise the problem would be impossible). So it is reasonable Bob could know the two points are opposite vertices\n\u2013\u00a0Matt\nJun 29 at 6:42\n\u2022 Acccumulation : you require some communication between two persons must succeed and Matt : you assume such persons are perfectly reasonable in optimizing. But OP only is about one person figuring it out on its own. If that person knows about his own invented conventions (one could claim that to be cheating), two points would be enough. Jul 3 at 22:30\n\nTwo points are enough if they are ordered - a corner A, and the corner B clockwise from A. This is the minimum, as a square has 4 degrees of feeedom - center, size, and orientation.\n\nA slighly different question is: whats the minimum number of points such that one and only one square will lie on them. 5 is enough - four corners and somewhere on an edge. Four might be enough.", "date": "2022-08-15 09:12:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4481618/how-many-points-uniquely-define-a-square", "openwebmath_score": 0.7538800835609436, "openwebmath_perplexity": 225.15480204341563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406006529085, "lm_q2_score": 0.8887587993853655, "lm_q1q2_score": 0.8802627990987992}}
{"url": "http://openstudy.com/updates/5606ed62e4b033021d8155a6", "text": "## anonymous one year ago Does the series $\\sum_{n\\ge1}\\frac{1}{\\text{lcm}\\{n,n+1\\}}$converge?\n\n1. anonymous\n\nUsing the fact that $$\\text{lcm}\\{n,n+1\\}=\\dfrac{n(n+1)}{\\text{gcd}\\{n,n+1\\}}$$, I'm almost tempted to say that $\\sum_{n\\ge1}\\frac{1}{\\text{lcm}\\{n,n+1\\}}\\sim\\sum_{n\\ge1}\\frac{1}{n^2}$ but I don't think I can jump to that conclusion just yet.\n\n2. mathmate\n\nWe agree that lcm(n,n+1) is actually n(n+1). (The gcd is 1, by Euclid's algorithm) Since S=$$\\sum_{n\\ge1}\\frac{1}{\\text{lcm}\\{n,n+1\\}}=\\sum_{n\\ge1}\\frac{1}{n(n+1)}<\\sum_{n\\ge1}\\frac{1}{n^2}$$ The strict inequality holds for each and every term, and since $$\\sum_{n\\ge1}\\frac{1}{n^2}$$ is a geometric series that is known to converge (=pi^2/6 prove it), so by comparison, S<pi^2/6 so S converges.\n\n3. mathmate\n\n* series, not geometric\n\n4. anonymous\n\nRight, since $$n$$ and $$n+1$$ are consecutive, that makes their lcm very easy to determine. And we can find the value of the series while we're at it quite easily: $\\sum_{n\\ge1}\\frac{1}{n(n+1)}=\\sum_{n\\ge1}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)=1$Neat!\n\n5. ganeshie8\n\nNice! any two consecutive integers are coprime, gcd(n, n+1) = gcd(n, 1) = 1 so lcm(n, n+1) = n(n+1)\n\n6. anonymous\n\nconsider that the least common multiple of $$n,n+1$$ is obviously $$n(n+1)$$ since $$n+1-n=1$$ and thus they must be coprime. this reduces to $$\\sum_{n=1}^\\infty\\frac1{n(n+1)}=\\sum_{n=1}^\\infty\\frac{n+1-n}{n(n+1)}=\\sum_{n=1}^\\infty\\left(\\frac{n+1}{n(n+1)}-\\frac{n}{n(n+1)}\\right)\\\\\\quad =\\sum_{n=1}^\\infty\\left(\\frac1n-\\frac1{n+1}\\right)$$ which is obviously telescoping and since $$1/(n+1)\\to0$$ it converges\n\n7. imqwerty\n\nwell i read that a series that converges has got some bounds and that it approaches a specific value. so in this case its going like 1/2 , 1/6, 1/12 , 1/20..... so u mark a line of length 1 on a number line and u plot the point 1/2 u knw 1/2 is still left between 1/2 and 1 and then u add 1/6 still 1/3 is left so u have bounds and u knw that it can never go > 1 so yea it converges... https://en.wikipedia.org/wiki/Convergent_series\n\n8. anonymous\n\nCould we make a more general claim here? Numerically, it would seem that $\\sum_{n\\ge1}\\frac{1}{\\text{lcm}\\{n,n+k\\}}$also converges to $$1$$ for $$k\\in\\mathbb{N}$$, though much more slowly.\n\n9. anonymous\n\nfor prime $$k$$ the problem is barely any harder -- you merely have to be careful about the multiples of $$k$$: $$\\sum_{n=1}^\\infty\\frac1{[n,n+k]}=\\sum_{m=1}^\\infty\\sum_{n=mk+1}^{(m+1)k}\\frac1{[n,n+k]}\\\\\\quad=\\sum_{m=1}^\\infty\\left(\\sum_{n=mk+1}^{mk+k-1}\\frac1{n(n+k)}+\\frac1{(m+1)(m+2)k}\\right)$$\n\n10. ParthKohli\n\n$\\gcd (n,n+k) \\le k$$\\Rightarrow {\\rm lcm }(n,n+k) = \\frac{n(n+k)}{\\gcd(n,n+k)} \\ge \\frac{n(n+k)}{k}$$\\Rightarrow \\frac{1}{{\\rm lcm} (n,n+k) }\\le \\frac{k}{n(n+k)}$We can sum the right-side telescopically so I guess it is true that the sum is convergent?\n\n11. anonymous\n\nyeah, showing its convergence along those lines is exactly what I figured\n\n12. anonymous\n\nthe sum of the terms for $$n$$ coprime to $$k$$ is accomplished with relatively little effort $$\\sum_{n=mk+1}^{(m+1)k-1}\\frac1{n(n+k)}=\\frac1k\\left(\\frac1{mk+1}-\\frac1{(m+2)k-1}\\right)$$ so we get: $$\\frac1k\\sum_{m=0}^\\infty\\left(\\frac1{mk+1}-\\frac1{(m+2)k-1}+\\frac1{(m+1)(m+2)}\\right)$$now split it up and telescope, I suppose? $$\\sum\\left(\\frac1{mk+1}-\\frac1{(m+2)k-1}\\right)=1+\\frac1{k+1}\\\\\\sum\\left(\\frac1{m+1}-\\frac1{m+2}\\right)=1$$ so we get $$\\frac1k\\left(2+\\frac1{k+1}\\right)=\\frac2k+\\frac1{k(k+1)}=\\frac3k-\\frac1{k+1}$$ for prime $$k$$\n\n13. anonymous\n\noops, ignore that -- I read $$(m+2)k+1$$ rather than $$(m+2)k-1$$, so that series does not quite telescope so cleanly\n\n14. anonymous\n\nour problem term for prime $$k$$ is this: $$\\frac1{mk+1}-\\frac1{(m+2)k-1}$$\n\n15. anonymous\n\nWould that I could hand out more medals...\n\n16. thomas5267\n\nHow do you show $$\\gcd (n,n+k) \\leq k$$ without using B\u00e9zout's Identity?\n\n17. anonymous\n\ni mean, if (positive) $$d$$ divides (positive) $$n$$ and $$n+k$$ then it also divides their difference, so $$d$$ must divide $$k$$. it follows that the greatest common divisor is at most $$k$$\n\n18. anonymous\n\nyou don't necessarily need the full strength of Bezout's identity, just the fact that divisibility is preserved by sums which is very elementary\n\n19. ganeshie8\n\n$$\\gcd(n,n+k)=\\gcd(n.k)\\le k$$\n\n20. thomas5267\n\n\\begin{align*} \\sum_{n=1}^\\infty\\frac1{[n,n+k]}&=\\sum_{m=0}^\\infty\\sum_{n=mk+1}^{(m+1)k}\\frac1{[n,n+k]}\\\\ &=\\sum_{m=0}^\\infty\\left(\\sum_{n=mk+1}^{mk+k-1}\\frac1{n(n+k)}+\\frac1{(m+1)(m+2)k}\\right)\\\\ \\end{align*} \\begin{align*} &\\phantom{{}={}}\\sum_{m=0}^\\infty\\sum_{n=mk+1}^{mk+k-1}\\frac1{n(n+k)}\\\\ &=\\sum_{m=0}^\\infty\\left(\\frac{1}{mk+1}-\\frac{1}{(m+1)k-1}\\right)\\\\ &=\\sum_{m=0}^\\infty\\left(\\frac{1}{mk+1}-\\frac{1}{(m+1)k+1}+\\frac{1}{(m+1)k+1}-\\frac{1}{(m+1)k-1}\\right)\\\\ &=\\sum_{m=0}^\\infty\\left(\\frac{1}{mk+1}-\\frac{1}{(m+1)k+1}-\\frac{2}{(m+1)^2k^2-1}\\right)\\\\ &=1-\\sum_{m=0}^\\infty\\frac{2}{(m+1)^2k^2-1} \\end{align*} I think the sum converges but I couldn't prove it.\n\n21. anonymous\n\nComparing to the series $$\\sum\\frac{1}{n^2}$$ would suffice for convergence. It has a somewhat daunting closed form: $\\sum_{m=0}^\\infty \\frac{2}{(m+1)^2k^2-1}=\\frac{k-\\pi\\cot\\dfrac{\\pi}{k}}{2k}$\n\n22. anonymous\n\n(closed form courtesy of Mathematica)\n\n23. thomas5267\n\nI thought $$\\dfrac{2}{(m+1)k^2-1}\\geq\\dfrac{1}{n^2}$$ for some reason! $\\min(k)=2\\\\ \\frac{2}{4n^2-1}-\\frac{1}{n^2}=\\frac{2n^2-4n^2+1}{n^2(4n^2-1)}=\\frac{-2n^2+1}{n^2(4n^2-1)}\\leq0$\n\n24. anonymous\n\n$\\sum_{m\\ge0}\\frac{1}{(m+1)^2k^2-1}=\\sum_{m\\ge1}\\frac{1}{m^2k^2-1}\\le\\sum_{m\\ge1}\\frac{1}{m^2}$ since $1\\le k^2-1~~\\implies~~ m^2\\le m^2k^2-1~~\\implies~~\\frac{1}{m^2k^2-1}\\le\\frac{1}{m^2}$(The first inequality is true since $$k\\ge2$$.)", "date": "2016-10-26 09:35:18", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/5606ed62e4b033021d8155a6", "openwebmath_score": 0.9686187505722046, "openwebmath_perplexity": 1049.6513285699832, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405995242328, "lm_q2_score": 0.8887587957022977, "lm_q1q2_score": 0.8802627944478189}}
{"url": "http://math.stackexchange.com/questions/247726/ten-people-are-seated-at-a-rectangular-table-permutations-homework", "text": "# Ten people are seated at a rectangular table - Permutations homework\n\nI got the following question for homework.\n\nTen people are to be seated at a rectangular table for dinner. Tanya will sit at the head of the table. Henry must not sit beside either Wilson or Nancy. In how many ways can the people be seated for dinner?\n\nThe approach I have used is to take $9!$ and subtract it from the seats where Henry, Wilson or Nancy can sit. I end up getting the answer $211680$ but for some reason, the back of the book says $201 600$. What am I doing incorrectly?\n\n-\n\nThis looks like an error in the book: your answer appears to be correct.\n\nWe can count the allowed seatings directly. Suppose that Henry sits next to Tanya. Then there is one seat forbidden to Wilson and Nancy, so there are $\\binom72$ ways to choose seats for Wilson and Nancy. There are $2$ ways to seat Wilson and Nancy in those $2$ seats and $6!$ ways to seat the unnamed people. Finally, Henry can be on either side of Tanya, so there are altogether $2\\cdot2\\cdot6!\\cdot\\binom72$ acceptable seatings with Henry next to Tanya.\n\nNow suppose that Henry is not seated next to Tanya. Then there are $2$ seats forbidden to Wilson and Nancy, so there are $\\binom62$ ways to choose their seats. As before there are $2$ ways to seat them in those two seats and $6!$ ways to seat the unnamed people. Finally, there are $7$ possible choices for Henry\u2019s seat, so there are altogether $7\\cdot2\\cdot6!\\cdot\\binom62$ acceptable seatings with Henry not next to Tanya. Altogether, then, there are\n\n$$2\\cdot6!\\left(2\\binom72+7\\binom62\\right)=1440(42+105)=211,680$$\n\nacceptable seatings.\n\n-\n\nUsing inclusion-exclusion, it's possible to see where the error in the back of the book may have come from.\n\nWith Tanya assigned to the head of table, there are $9!$ ways to seat the other people with no restrictions. Among these, there $2\\cdot8!$ ways in which Henry sits beside Wilson (the factor of $2$ corresponds to which one sits to the left of the other), and another $2\\cdot8!$ ways in which Henry sits beside Nancy. Subtracting these, we get the book's answer\n\n$$9!-2\\cdot8!-2\\cdot8!=201600$$\n\nBut this double-subtracts the $2\\cdot7!$ ways in which Henry sits between Wilson and Nancy. So the correct answer is\n\n$$9!-2\\cdot8!-2\\cdot8!+2\\cdot7!=211680$$\n\n-", "date": "2014-12-18 07:47:21", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/247726/ten-people-are-seated-at-a-rectangular-table-permutations-homework", "openwebmath_score": 0.6970557570457458, "openwebmath_perplexity": 453.0724128611491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406009350774, "lm_q2_score": 0.8887587831798666, "lm_q1q2_score": 0.8802627832989952}}
{"url": "https://mundogeografico.com.br/79wyw/axxtibs.php?page=2e1b59-system-of-linear-equations-problems", "text": "02\ndez\n\n# system of linear equations problems\n\nMany problems lend themselves to being solved with systems of linear equations. Case 2: Parallel Lines The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. Materials include course notes, lecture video clips, JavaScript Mathlets, a quiz with solutions, practice problems with solutions, a problem solving video, and problem sets with solutions. Once you do that, these linear systems are solvable just like other linear systems. $\\begin{cases}5x +2y =1 \\\\ -3x +3y = 5\\end{cases}$ Yes. Solve age word problems with a system of equations. The problem also says that Mia will be twice as old as Shaheena. This example shows one iteration of the gradient descent. When you solve systems with two variables and therefore two equations, the equations can be linear or nonlinear. There can be any combination: 1. But let\u2019s say we have the following situation. Systems of linear equations can be used to model real-world problems. Solving using Matrices by Row Operations. Quiz 3. Just select one of the options below to start upgrading. Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. You\u2019re going to the mall with your friends and you have $200 to spend from your recent birthday money. At the first store, he bought some t-shirts and spent half of his money. A system of linear equations is called homogeneous if the constants$b_1, b_2, \\dots, b_m$are all zero. \u201cSystems of equations\u201d just means that we are dealing with more than one equation and variable. Wouldn\u2019t it be cl\u2026 Consider the nonlinear system of equations Cramer's Rule. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you \u2026 Word Problems with Systems of Linear Equations: This worksheet is designed for Algebra students to practice those dreaded word problems that are solved with linear systems of equations. Is the point$(1 ,3)$a solution to the following system of equations\u2026 Solution of a non-linear system. No Problem 2. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. Solving using an Augmented Matrix. Solve simple cases by inspection. Problem 1. Find them out by checking. Systems of Equations - Problems & Answers. Systems of linear equations are a common and applicable subset of systems of equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: System of equations word problem: walk & ride, Practice: Systems of equations word problems, System of equations word problem: no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite solutions), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. By \u2026 System of Linear Equations Word Problems Calvin went to Chicago's Magnificent Mile to do some Christmas shopping. A system of linear equations is a set of two or more linear equations with the same variables. Is the point$(0 ,\\frac{5}{2})$a solution to the following system of equations? SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. To solve the system of equations, you need to find the exact values of x and y that will solve both equations. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Below is an example that shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. You really, really want to take home 6items of clothing because you \u201cneed\u201d that many new things. Systems of Linear Equations. Solve the following system of equations by elimination. For problems 1 \u2013 3 use the Method of Substitution to find the solution to the given system or to determine if the system \u2026 9,000 equations in 567 variables, 4. etc. Gradient descent can also be used to solve a system of nonlinear equations. Looking for fun activities to teach kids about Solving Word Problems Involving Linear Equations and Linear Inequalities? Systems of Linear Equations and Problem Solving. 6 equations in 4 variables, 3. (5.2.3) \u2013 Solve mixture problems with a system of linear equations. In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. ... Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up! This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. HIDE SOLUTIONS. 1. Systems of 2 linear equations - problems with solutions Test. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. Section 8.1, Example 4(a) Solve graphically: Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. In your studies, however, you will generally be faced with much simpler problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. For example, the sets in the image below are systems of linear equations. This premium worksheet bundle contains a printable fact file and 10 fun and engaging worksheets to challenge your students and help them learn about Solving Word Problems Involving Linear Equations and Linear Inequalities. Substitution Method. Problem 1 Two of the following systems of equations have solution (1;3). B. Determining the value of k for which the system has no solutions. To use Khan Academy you need to upgrade to another web browser. So a System of Equations could have many equations and many variables. There are 7 questions. Donate or volunteer today! A large pizza at Palanzio\u2019s Pizzeria costs$6.80 plus $0.90 for each topping. Our mission is to provide a free, world-class education to anyone, anywhere. System of NonLinear Equations problem example. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. This section provides materials for a session on solving a system of linear differential equations using elimination. Khan Academy is a 501(c)(3) nonprofit organization. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. The statement above gives the following equation m + 10 = 2 \u00d7 (s + 10) m + 10 = 2 \u00d7 s + 2\u00d7 10 m + 10 = 2s + 20 You now have a system of linear equation to solve m + s = 40 equation 1 m + 10 = 2s + 20 equation 2 Use equation 1 to solve for m m + s = 40 m + s - s = 40 - s m = 40 - s If you're seeing this message, it means we're having trouble loading external resources on our website. For example, + \u2212 = \u2212 + = \u2212 \u2212 + \u2212 = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. One application of systems of equations are mixture problems. This section covers: Systems of Non-Linear Equations; Non-Linear Equations Application Problems; Systems of Non-Linear Equations (Note that solving trig non-linear equations can be found here).. We learned how to solve linear equations here in the Systems of Linear Equations and Word Problems Section.Sometimes we need solve systems of non-linear equations, such as those we see in conics. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. System of linear equations System of linear equations can arise naturally from many real life examples. Add the second equation to the first equation and solve for x. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. set of two (or more) equations which have two (or more) variables Solving Systems of Linear Equations. It has 6 unique word problems to solve including one mixture problem \u2026 If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. Simultaneous equations (Systems of linear equations): Problems with Solutions. You appear to be on a device with a \"narrow\" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\\left( {2 - x} \\right) = 3x + 2$$, $$2\\left( {w + 3} \\right) - 10 = 6\\left( {32 - 3w} \\right)$$, $$\\displaystyle \\frac{{4 - 2z}}{3} = \\frac{3}{4} - \\frac{{5z}}{6}$$, $$\\displaystyle \\frac{{4t}}{{{t^2} - 25}} = \\frac{1}{{5 - t}}$$, $$\\displaystyle \\frac{{3y + 4}}{{y - 1}} = 2 + \\frac{7}{{y - 1}}$$, $$\\displaystyle \\frac{{5x}}{{3x - 3}} + \\frac{6}{{x + 2}} = \\frac{5}{3}$$. So far, we\u2019ve basically just played around with the equation for a line, which is . 1. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Solve each of the following equations and check your answer. Section 7-1 : Linear Systems with Two Variables. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve age word problems with a system of equations. A solution of the system (*) is a sequence of numbers$s_1, s_2, \\dots, s_n$such that the substitution$x_1=s_1, x_2=s_2, \\dots, x_n=s_n$satisfies all the$m$equations in the system (*). Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. You discover a store that has all jeans for$25 and all dresses for 50. Improve your math knowledge with free questions in \"Solve systems of linear equations\" and thousands of other math skills. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! A system of linear equations is a system made up of two linear equations. Solution: Rewrite in order to align the x and y terms. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . 2 equations in 3 variables, 2. Generally speaking, those problems come up when there are two unknowns or variables to \u2026 Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. If you're seeing this message, it means we're having trouble loading external resources on our website. The same rules apply. a)\\begin{array}{|l} x + y = 5 \\\\ 2x - y = 7; \\end{array}\\$ In \"real life\", these problems can be incredibly complex. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. Answer: x = .5; y = 1.67. Solving using Matrices by Elimination. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables.", "date": "2021-04-14 14:51:03", "meta": {"domain": "com.br", "url": "https://mundogeografico.com.br/79wyw/axxtibs.php?page=2e1b59-system-of-linear-equations-problems", "openwebmath_score": 0.7244647741317749, "openwebmath_perplexity": 666.4782717917759, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9759464471055738, "lm_q2_score": 0.9019206692796967, "lm_q1q2_score": 0.8802262727546012}}
{"url": "https://www.npf.ws/sudbury-mass-czlolfi/opposite-sides-of-a-parallelogram-are-congruent-c26b47", "text": "Learn faster with a math tutor. 9th - 10th grade. The two pairs of congruent sides may be, but do not have to be, congruent to each other. Get better grades with tutoring from top-rated professional tutors. This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2-D geometry. 3) Are opposite angles of a parallelogram congruent? Solve for x. (10 Properties of parallelogram: Opposite sides of parallelogram are equal . The diagonals of a parallelogram bisect each other. Sides of A Parallelogram The opposite sides of a parallelogram are congruent. (By definition). The angles of a parallelogram are congruent. Ask yourself which approach looks easier or quicker. Line segments XY and ZW are also congruent. One way all sides of the two parallelograms could be congruent would be if $ABCD$ and $EFGH$ are squares with the same side length: in this case they would be congruent. Properties of parallelograms are as follows: i. Parallelogram Properties DRAFT. Tags: Question 19 . CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. Is the quadrilateral a parallelogram? . Give a reason. Expand Image In fact, one method of proving a quadrilateral a rhombus is by first proving it a parallelogram, and then proving two consecutive sides congruent, diagonals bisecting verticies, etc. A parallelogram is a quadrilateral that has opposite sides that are parallel. If one side is longer than its opposite side, you do not have parallel sides; no parallelogram! Draw the diagonal BD, and we will show that \u0394ABD and \u0394CDB are congruent. Opposite sides are parallel and congruent. It is a quadrilateral with two pairs of parallel, congruent sides. Opposite angles are congruent. High School: Geometry \u00bb Congruence \u00bb Prove geometric theorems \u00bb 11 Print this page. Parallelogram Properties DRAFT. A parallelogram is defined to be a quadrilateral with 2 pairs of opposite sides parallel. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. Consecutive angles are supplementary (A + D = 180\u00b0). The two diagonals of a parallelogram bisect each other. A parallelogram is a quadrilateral with two pairs of parallel sides. 5) Does a diagonal of a parallelogram bisect a pair of opposite angles? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. Yes, if both pairs of opposite angles are congruent, then you have a parallelogram. The diagonals of an isosceles trapezoid are congrent. 0 Why are these two lines not congruent (and other ways to figure out if other shapes are not congruent) Solve for x. HELP ASAP 30 points Part 1 out of 2 To repair a large truck or bus, a mechanic might use a parallelogram lift. 2 years ago. One has to be on the lookout for double negatives. 62% average accuracy. Opposite sides are congruent. You can draw parallelograms. The opposite angles are congruent. 60 seconds . A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . If yes, state how you know. There are two ways to go about this. If one angle is right, then all angles are right. The diagonals of a rectangle are the bisectors of the angles. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. An equilateral quadrilateral is a square. That is true. Here are some important things that \u2026 Try to move the vertices A, B, and D and observe how the figure changes. The opposite sides of a parallelogram are congruent so we will need two pairs of congruent segments: Now if we imagine leaving $\\overline{AB}$ fixed and ''pushing down'' on side $\\overline{CD}$ so that these two sides become closer while side $\\overline{AD}$ and $\\overline{BC}$ rotate clockwise we get a new parallelogram: There is one right angle in a parallelogram and it is not a rectangle. If a quadrilateral has three angles of equal measure, then the fourth angle must be a right angle. The converse is also true that if opposite sides of a quadrangle are equal then its a parallelogram. 62% average accuracy. The figure is a parallelogram. The diagonals of a trapezoid are perpendicular. For our parallelogram, we will label it WXYZ, but you can use any four letters as long as they are not the same as each other. A rhombus is a parallelogram but with all four sides equal in length; A square is a parallelogram but with all sides equal in length and all interior angles 90\u00b0 A quadrilateral is a parallelogram if: Both pairs of opposite sides are parallel. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Tags: Question 19 . The diagonals of a quadrilateral_____bisect each other, If the measures of 2 angles of a quadrilateral are equal, then the quadrilateral is_____a parallelogram, If one pair of opposite sides of a quadrilateral is congruent and parallel, then the quadrilateral is______a parallelogram, If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram, To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel, The diagonals of a rectangle are_____congruent, The diagonals of a parallelogram_______bisect the angles, The diagonals of a parallelogram______bisect the angles of the parallelogram, A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram, The diagonals of a rhombus are_______congruent, A rectangle______has consecutive sides congruent, A rectangle_______has perpendicular diagonals, The diagonals of a rhombus_____bisect each other, The diagonals of a parallelogram are_______perpendicular bisectors of each other, Consecutive angles of a quadrilateral are_______congruent, The diagonals of a rhombus are______perpendicular bisectors of each other, Consecutive angles of a square are______complementary, Diagonals of a non-equilateral rectangle are______never angle bisectors, A quadrilateral with one pair of congruent sides and one pair of parallel sides is_____a parallelogram. B) The diagonals of the parallelogram are congruent. The opposite angles of a parallelogram are supplementary. Studying the video and these instructions, you will learn what a parallelogram is, how it fits into the family of polygons, how to identify its angles and sides, how to prove you have a parallelogram, and what are its identifying properties. Reason for statement 8: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Triangles can be used to prove this rule about the opposite sides. Use a straightedge (ruler) to draw a horizontal line segment, then draw another identical (congruent) line segment some distance above and to one side of the first one, so they do not line up vertically. Opposite angles are congruent. Parallelogram definition A quadrilateral with both pairs of opposite sides parallel. 1981 times. Opposite sides are congruent. If both pairs are congruent, you have either a rhombus or a square. SURVEY . 2. A parallelogram is a quadrilateral that has opposite sides that are parallel. Check for any one of these identifying properties: You can use proof theorems about a plane, closed quadrilateral to discover if it is a parallelogram: You have learned that a parallelogram is a closed, plane figure with four sides. You can have almost all of these qualities and still not have a parallelogram. Opposite angles are congruent. The opposite sides of a parallelogram are congruent. Local and online. Go with B. Give a reason. The diagonals of a rectangle bisect eachother. Theorem: If ABCD is a parallelogram then prove that its opposite sides are equal. We already mentioned that their diagonals bisect each other. The diagonals of a kite are the perpendicular bisectors of each other. More generally, a quadrilateral with 4 congruent sides is a rhombus. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram Always To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. , all 4 sides are congruent (definition of a rhombus). Get better grades with tutoring from top-rated private tutors. Check Next Opposite angles are congruent. Want to see the math tutors near you? Property that is characteristic of a parallelogram is that opposite sides are congruent. The diagonals of a quadrilateral are perpendicular and the quadrilateral is not a rhombus. Opposite angels are congruent (D = B). Prove that opposite sides of a parallelogram are congruent. Other shapes, however, are types of parallelograms. Which statement can be used to prove that a given parallelogram is a rectangle? 4) Do the diagonals of a parallelogram bisect each other? Observe that at any time, the opposite sides are parallel and equal. Now consider just the interior angles of parallelograms, \u2220W, \u2220X, \u2220Y, and \u2220Z. Theorem 1: Opposite Sides of a Parallelogram Are Equal In a parallelogram, the opposite sides are equal. In this lesson we will prove the basic property of a parallelogram that the opposite sides in a parallelogram are equal. Q. In the video below: We will use the properties of parallelograms to determine if we have enough information to prove a given quadrilateral is a parallelogram. The four line segments making up the parallelogram are WX, XY, YZ, and ZW. Notice that line segments WX and YZ are congruent. Use this to prove that the quadrilateral must be a parallelogram. Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! The first rhombus above is a square while the second one has angles of 60 and 120 degrees. If only one set of opposite sides are congruent, you do not have a parallelogram, you have a trapezoid. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. These geometric figures are part of the family of parallelograms: For such simple shapes, parallelograms have some interesting properties. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. Give the given and prove and prove this. A parallelogram does not have other names. The interior angles are \u2220W, \u2220X, \u2220Y, and \u2220Z. Make sure that second line segment is parallel to (or equidistant from) the first line segment. Yes. The consecutive angles of a parallelogram are supplementary. The opposite sides are equal and parallel; the opposite angles are also equal. 1-to-1 tailored lessons, flexible scheduling. 1981 times. Connecting opposite (non-adjacent) vertices gives you diagonals WY and XZ. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . 2 years ago. Start at any vertex (corner). The area of a parallelogram is twice the area of a triangle created by one of its diagonals. We will show that in that case, they are also equal to each other. That the other pair of opposite sides are congruent. Opposite sides are congruent (AB = DC). Is the quadrilateral a parallelogram? Take a rectangle and push either its left or ride side so it leans over; you have a parallelogram. In today's lesson, we will show that the opposite sides of a parallelogram are equal to each other. Let\u2019s play with the simulation given below to better understand a parallelogram and its properties. This means a parallelogram is a plane figure, a closed shape, and a quadrilateral. The parallelogram has the following properties: Opposite sides are parallel by definition. An equilateral parallelogram is equiangular. Proving That a Quadrilateral is a Parallelogram, Opposite sides are parallel -- Look at the parallelogram in our drawing. The bad in Answer A is due to your teachers written grammar. Is it possible to prove a quadrilateral a parallelogram with two consecutive and two opposite congruent sides? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. That\u2019s a wrap! A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. In the figure given below, ABCD is a parallelogram. The figure shows a side view of the li \u2026 ft. FGKL, GHJK, and FHJL are parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The Angle-Side-Angle Triangle Congruence Theorem can be used to prove that, in a parallelogram, opposite sides are congruent. There are more than two right angles in a trapezoid. Opposite sides are congruent -- The base side (Y Z Y Z) and the top side (W X W X) of our parallelogram are equal in length (congruent); the left side (XY X Y) and right side (ZW Z W) are also congruent To be a parallelogram, the base and top sides must be parallel and congruent, and \u2026 The opposite sides of parallelogram are also equal in length. CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. To show these two triangles are congruent we\u2019ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv\u2026 Parallelogram law. Parallelogram Theorem #2: The opposite sides of a parallelogram are congruent. SURVEY . View Untitled document (3).pdf from MATH 100 at Basha High School. If the four sides do not connect at their endpoints, you do not have a closed shape; no parallelogram! Mathematics. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. The bottom (base) side, Opposite sides are congruent -- The base side (. C) The diagonals of the parallelogram bisect the angles. 60 seconds . Get help fast. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. G 2 T: + 3 5 6 8 17 K 3 Which angles are congruent to 21? <2 2 are congruent to 21. As with any quadrilateral, the interior angles add to 360\u00b0, but you can also know more about a parallelogram's angles: Using the properties of diagonals, sides, and angles, you can always identify parallelograms. Opposite sides are congruent. One interesting property of a parallelogram is that its two diagonals bisect each other (cut each other in half). Or: Both pairs of opposite sides are congruent. To explore these rules governing the sides of a parallelogram use Math Warehouse's interactive parallelogram. Triangles can be used to prove this rule about the opposite sides. D) The opposite angles of the parallelogram are congruent. Opposite sides of a \u2026 \u2234 \u2234 AB = CD A B = C D and AD= BC A D = B C Properties of a parallelogram. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. Opposite angles are equal (congruent) to each other; Any two adjacent angles of a parallelogram add up to, This means any two adjacent angles are supplementary (adding to, A closed shape (it has an interior and exterior), A quadrilateral (four-sided plane figure with straight sides), Two pairs of congruent (equal), opposite angles, Two pairs of equal and parallel opposite sides, If the quadrilateral has bisecting diagonals, it is a parallelogram, If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram, If the quadrilateral has consecutive supplementary angles, it is a parallelogram, If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. To be on the lookout for double negatives proof 2 here \u2019 use. 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Of opposite angles are right \u2220X = \u2220Z have to be a quadrilateral a parallelogram use Math 's...", "date": "2021-08-02 19:31:52", "meta": {"domain": "npf.ws", "url": "https://www.npf.ws/sudbury-mass-czlolfi/opposite-sides-of-a-parallelogram-are-congruent-c26b47", "openwebmath_score": 0.6451835036277771, "openwebmath_perplexity": 531.6535784583293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.98866824713641, "lm_q2_score": 0.8902942363098472, "lm_q1q2_score": 0.8802056420481054}}
{"url": "https://math.hecker.org/2011/01/20/linear-algebra-and-its-applications-exercise-1-5-19/?replytocom=8181", "text": "## Linear Algebra and Its Applications, Exercise\u00a01.5.19\n\nExercise 1.5.19. For the following two matrices, specify the values of $a$, $b$, and $c$ for which elimination requires row exchanges, and the values for which the matrices in question are singular.\n\n$A = \\begin{bmatrix} 1&2&0 \\\\ a&8&3 \\\\ 0&b&5 \\end{bmatrix} \\quad A = \\begin{bmatrix} c&2 \\\\ 6&4 \\end{bmatrix}$\n\nAnswer: For the first matrix a row exchange would be necessary if the first step of elimination caused the second entry of the second row, i.e., in the (2,2) position, to become 0. This would occur if we multiplied the first row by 4 and subtracted it from the second row, as we would then have $8 - 4 \\cdot 2 = 0$ in the (2,2) position. Since the entry in the (1,1) position is 1, multiplying the first row by 4 would be necessary if $a$ were 4:\n\n$\\begin{bmatrix} 1&2&0 \\\\ 4&8&3 \\\\ 0&b&5 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1&2&0 \\\\ 0&0&3 \\\\ 0&b&5 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1&2&0 \\\\ 0&b&5 \\\\ 0&0&3 \\end{bmatrix}$\n\nIf we had both $a = 4$ and $b = 0$ then the matrix would be singular:\n\n$\\begin{bmatrix} 1&2&0 \\\\ 4&8&3 \\\\ 0&0&5 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1&2&0 \\\\ 0&0&3 \\\\ 0&0&5 \\end{bmatrix}$\n\nHowever, these are not the only values of $a$ and $b$ for which the matrix is singular. Suppose we proceed with elimination. The first step would be to multiply the first row by $l_1 = a$ and subtract the result from the second row:\n\n$\\begin{bmatrix} 1&2&0 \\\\ a&8&3 \\\\ 0&b&5 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1&2&0 \\\\ 0&8-2a&3 \\\\ 0&b&5 \\end{bmatrix}$\n\nAssuming that $a \\ne 4$ then the next step of elimination would multiply the second row of the matrix by $l_2 = b/(8-2a)$ and subtract the result from the third row:\n\n$\\begin{bmatrix} 1&2&0 \\\\ 0&8-2a&3 \\\\ 0&b&5 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1&2&0 \\\\ 0&8-2a&3 \\\\ 0&0&5-3b/(8-2a) \\end{bmatrix}$\n\nIf the (3,3) entry of the resulting matrix above is zero then the original matrix is singular. In this case we have $3b/(8-2a) = 5$. Multiplying both sides by $(8-2a)$ we have $3b = 5 \\cdot (8-2a) = 40 - 10a$ or $10a + 3b = 40$.\n\nSo the first matrix is singular for all values of $a$ and $b$ for which $10a+3b=40$. This includes the case $a = 4, b=0$ mentioned above, the case $a=0, b = 40/3$, the case $a=5, b=-10/3$, and an infinity of others on the line $b=-\\frac{10}{3}a + \\frac{40}{3}$.\n\nFor the second matrix a row exchange would be necessary if $c = 0$:\n\n$\\begin{bmatrix} 0&2 \\\\ 6&4 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 6&4 \\\\ 0&2 \\end{bmatrix}$\n\nOn the other hand, if c = 3 then the first elimination produces zeroes in both entries of the second row, and the matrix is singular:\n\n$\\begin{bmatrix} 3&2 \\\\ 6&4 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 3&2 \\\\ 0&0 \\end{bmatrix}$\n\nUPDATE: Prompted by Theodore\u2019s comment, added a section deriving the complete set of $a$ and $b$ for which the first matrix is singular.\n\nNOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.\n\nIf you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang\u2019s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang\u2019s other books.\n\nThis entry was posted in linear algebra. Bookmark the permalink.\n\n### 2 Responses to Linear Algebra and Its Applications, Exercise\u00a01.5.19\n\n1. Theodore says:\n\nI also found a = 5 and b = -10/3 to also cause a singular matrix. My question is how do we find all values that can cause singularity\n\n\u2022 hecker says:\n\nThank you for finding this problem. You are correct that there are multiple (actually, an infinity) of values of a and b for which the first matrix is singular. I added a new section to derive the correct answer.", "date": "2021-09-19 05:23:04", "meta": {"domain": "hecker.org", "url": "https://math.hecker.org/2011/01/20/linear-algebra-and-its-applications-exercise-1-5-19/?replytocom=8181", "openwebmath_score": 0.8730982542037964, "openwebmath_perplexity": 160.76848535915715, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682444653241, "lm_q2_score": 0.8902942363098472, "lm_q1q2_score": 0.8802056396700529}}
{"url": "http://www.mathworks.com/help/matlab/ref/istril.html?s_tid=gn_loc_drop&nocookie=true", "text": "Accelerating the pace of engineering and science\n\n# istril\n\nDetermine if matrix is lower triangular\n\n## Description\n\nexample\n\ntf = istril(A) returns logical 1 (true) if A is a lower triangular matrix; otherwise, it returns logical 0 (false).\n\n## Examples\n\nexpand all\n\n### Test Lower Triangular Matrix\n\nCreate a 5-by-5 matrix.\n\n`D = tril(magic(5))`\n```D =\n\n17     0     0     0     0\n23     5     0     0     0\n4     6    13     0     0\n10    12    19    21     0\n11    18    25     2     9```\n\nTest D to see if it is lower triangular.\n\n`istril(D)`\n```ans =\n\n1```\n\nThe result is logical 1 (true) because all elements above the main diagonal are zero.\n\n### Test Matrix of Zeros\n\nCreate a 5-by-5 matrix of zeros.\n\n`Z = zeros(5);`\n\nTest Z to see if it is lower triangular.\n\n`istril(Z)`\n```ans =\n\n1```\n\nThe result is logical 1 (true) because a lower triangular matrix can have any number of zeros on its main diagonal.\n\n## Input Arguments\n\nexpand all\n\n### A \u2014 Input arraynumeric array\n\nInput array, specified as a numeric array. istril returns logical 0 (false) if A has more than two dimensions.\n\nData Types: single | double\nComplex Number Support: Yes\n\nexpand all\n\n### Lower Triangular Matrix\n\nA matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.\n\nFor example, the matrix\n\n$A=\\left(\\begin{array}{cccc}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& 0\\\\ -1& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& 0\\\\ -2& -2& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& 0\\\\ -3& -3& -3& 1\\end{array}\\right)$\n\nis lower triangular. A diagonal matrix is both upper and lower triangular.\n\n### Tips\n\n\u2022 Use the tril function to produce lower triangular matrices for which istril returns logical 1 (true).\n\n\u2022 The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istril(A) == isbanded(A,size(A,1),0).", "date": "2014-12-26 01:53:58", "meta": {"domain": "mathworks.com", "url": "http://www.mathworks.com/help/matlab/ref/istril.html?s_tid=gn_loc_drop&nocookie=true", "openwebmath_score": 0.7502629160881042, "openwebmath_perplexity": 2808.414227451656, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347823646076, "lm_q2_score": 0.9032942001955142, "lm_q1q2_score": 0.8802012873787282}}
{"url": "https://math.stackexchange.com/questions/2687769/solve-the-equation-cos2x-cos22x-cos23x-1", "text": "# Solve the equation $\\cos^2x+\\cos^22x+\\cos^23x=1$\n\nSolve the equation: $$\\cos^2x+\\cos^22x+\\cos^23x=1$$ IMO 1962/4\n\nMy first attempt in solving the problem is to simplify the equation and express all terms in terms of $\\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.\n\n\\begin{align} \\cos^22x&=(\\cos^2x-\\sin^2x)^2\\\\ &=\\cos^4x+\\sin^4x-2\\sin^2\\cos^2x\\\\ &=\\cos^4+(1-\\cos^2x)^2-2(1-\\cos^2)\\cos^2x\\\\ &=\\cos^4+1-2\\cos^2x+\\cos^4x-2\\cos^2x+2\\cos^4x\\\\ &=4\\cos^4x-4\\cos^2x+1 \\end{align}\n\nWithout knowledge of other trigonometric identities, $\\cos3x$ can be derived using only Ptolemy's identities. However for the sake of brevity, let $\\cos 3x=4\\cos^3x-3\\cos x$: \\begin{align} \\cos^23x&=(4\\cos^3x-3\\cos x)^2\\\\ &=16\\cos^6x+4\\cos^2x-24\\cos^4x \\end{align}\n\nTherefore, the original equation can be written as: $$\\cos^2x+4\\cos^4x-4\\cos^2x+1+16\\cos^6x+4\\cos^2x-24\\cos^4x-1=0$$ $$-20\\cos^4x+6\\cos^2x+16\\cos^6x=0$$ Letting $y=\\cos x$, we now have a polynomial equation: $$-20y^4+6y^2+16y^6=0$$ $$y^2(-20y^2+6y+16y^4)=0\\Rightarrow y^2=0 \\Rightarrow x=\\cos^{-1}0=\\bbox[yellow,10px]{90^\\circ}$$ From one of the factors above, we let $z=y^2$, and we have the quadratic equation: $$16z^2-20z+6=0\\Rightarrow 8z^2-10z+3=0$$ $$(8z-6)(z-\\frac12)=0\\Rightarrow z=\\frac34 \\& \\ z=\\frac12$$ Since $z=y^2$ and $y=\\cos x$ we have: $$\\biggl( y\\rightarrow\\pm\\frac{\\sqrt{3}}{2}, y\\rightarrow\\pm\\frac{\\sqrt{2}}2 \\biggr)\\Rightarrow \\biggl(x\\rightarrow\\cos^{-1}\\pm\\frac{\\sqrt{3}}{2},x\\rightarrow\\cos^{-1}\\pm\\frac{\\sqrt{2}}2\\biggr)$$ And thus the complete set of solution is: $$\\bbox[yellow, 5px]{90^\\circ, 30^\\circ, 150^\\circ, 45^\\circ, 135^\\circ}$$\n\nAs I do not have the copy of the answers, I still hope you can verify the accuracy of my solution.\n\n## But more importantly...\n\nSeeing the values of $x$, is there a more intuitive and simpler way of finding $x$ that does away with the lengthy computation?\n\n\u2022 This post doesn't have the squares, but it could give some insight as to how one may think about this problem. \u2013\u00a0Arthur Mar 12 '18 at 13:12\n\u2022 This link gives you the solutions. \u2013\u00a0Jose Arnaldo Bebita-Dris Mar 12 '18 at 13:15\n\u2022 I expect there to be a much more elegant solution to this as it is an IMO problem. Now only someone has to find it... \u2013\u00a0vrugtehagel Mar 12 '18 at 13:15\n\u2022 @vrugtehagel I believe the link to the solution is already an elegant answer! \u2013\u00a0John Glenn Mar 12 '18 at 13:17\n\u2022 Yup, I think so too! It was shortly posted before my comment, so I hadn't seen it. I advise @JoseArnaldoBebitaDris to summarize that solution and post it as answer, to avoid this question lingering in the unanswered section of this website \u2013\u00a0vrugtehagel Mar 12 '18 at 13:21\n\nThis is a summary of the solution found in this hyperlink.\n\nWe can write the LHS as a cubic function of $\\cos^2 x$. This means that there are at most three values of $x$ that satisfy the equation.\n\nHence, we look for three values of $x$ that satisfy the equation and produce three distinct $\\cos^2 x$. Indeed, we find that $$\\frac{\\pi}{2}, \\frac{\\pi}{4}, \\frac{\\pi}{6}$$ all satisfy the equation, and produce three different values for $\\cos^2 x$, namely $0, \\frac{1}{2}, \\frac{3}{4}$.\n\nLastly, we solve the resulting equations $$\\cos^2 x = 0$$ $$\\cos^2 x = \\frac{1}{2}$$ $$\\cos^2 x = \\frac{3}{4}$$ separately. We conclude that our solutions are: $$x=\\frac{(2k+1)\\pi}{2}, \\frac{(2k+1)\\pi}{4}, \\frac{(6k+1)\\pi}{6}, \\frac{(6k+5)\\pi}{6}, \\forall k \\in \\mathbb{Z}.$$\n\nYou can shorten the argument by noting at the outset that $$\\cos3x=4\\cos^3x-3\\cos x=(4\\cos^2x-3)\\cos x$$ so if we set $y=\\cos^2x$ we get the equation $$y+(2y-1)^2+y(4y-3)^2=1$$ When we do the simplifications, we get $$2y(8y^2-10y+3)=0$$ The roots of the quadratic factor are $3/4$ and $1/2$.\n\nA different strategy is to note that $\\cos x=(e^{ix}+e^{-ix})/2$, so the equation can be rewritten $$e^{2ix}+2+e^{-2ix}+e^{4ix}+2+e^{-4ix}+e^{6ix}+2+e^{-6ix}=4$$ Setting $z=e^{2ix}$ we get $$2+z+z^2+z^3+\\frac{1}{z}+\\frac{1}{z^2}+\\frac{1}{z^3}=0$$ or as well $$z^6+z^5+z^4+2z^3+z^2+z+1=0$$ that can be rewritten (noting that $z\\ne1$), $$\\frac{z^7-1}{z-1}+z^3=0$$ or $z^7+z^4-z^3-1=0$ that can be factored as $$(z^3+1)(z^4-1)=0$$ Hence we get (discarding the spurious root $z=1$) $$2x=\\begin{cases} \\dfrac{\\pi}{3}+2k\\pi \\\\[6px] \\pi+2k\\pi \\\\[6px] \\dfrac{5\\pi}{3}+2k\\pi \\\\[12px] \\dfrac{\\pi}{2}+2k\\pi \\\\[6px] \\pi+2k\\pi \\\\[6px] \\dfrac{3\\pi}{2}+2k\\pi \\end{cases} \\qquad\\text{that is}\\qquad x=\\begin{cases} \\dfrac{\\pi}{6}+k\\pi \\\\[6px] \\dfrac{\\pi}{2}+k\\pi \\\\[6px] \\dfrac{5\\pi}{6}+k\\pi \\\\[6px] \\dfrac{\\pi}{4}+k\\pi \\\\[6px] \\dfrac{3\\pi}{4}+k\\pi \\end{cases}$$\n\n\u2022 Great! An elegant solution too! \u2013\u00a0John Glenn Mar 12 '18 at 14:42\n\nHint:\n\n$$0=\\cos^2x+\\cos^22x+\\cos^23x-1$$\n\n$$=\\cos(3x+x)\\cos(3x-x)+\\cos^22x$$\n\n$$=\\cos2x(\\cos4x+\\cos2x)$$", "date": "2019-07-22 05:24:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2687769/solve-the-equation-cos2x-cos22x-cos23x-1", "openwebmath_score": 0.9263973832130432, "openwebmath_perplexity": 285.6040172360673, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713835553862, "lm_q2_score": 0.8933094053442511, "lm_q1q2_score": 0.8801521937465696}}
{"url": "https://math.stackexchange.com/questions/2820845/calculation-of-modular-multiplicative-inverse-of-a-mod-b-when-a-b", "text": "# Calculation of modular multiplicative inverse of A mod B when A > B\n\nI'm trying to understand a Montgomery reduction algorithm, for which I need to calculate a multiplicative inverse. However, euclidean algorithm only helps if A < B.\n\nExample is 11 mod 3. Multiplicative inverse of 11 is 2,but ext_gcd gives you Bezout numbers such as -1 and 4.\n\nhttps://en.wikipedia.org/wiki/Extended_Euclidean_algorithm\n\nWikipedia says so: The extended Euclidean algorithm is particularly useful when a and b are coprime, since x is the modular multiplicative inverse of a modulo b, and y is the modular multiplicative inverse of b modulo a.\n\nBut as far as I see this can't be true, either X is multiplicative inverse of A modulo B or Y is multiplicative inverse of B modulo A, but not both at the same time, because one of them (A or B) is going to be bigger than another. We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not.\n\nA lot of online calculators that I tried are also said that a has to be bigger than be, but they (some of them) are still able to calculate inverse of 11 mod 3.\n\nThe only workaround I found so far is perform A = A mod B first, so A is now a remainder of divisios and therefore is less than modulus, so we can perform ext_gcd(2, 3) now and get our 2 as answer.\n\nProbably I'm missing something, this thing is pretty new for me.\n\nThanks.\n\n\u2022 \"We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not. \" Huh. $X \\equiv A^{-1}\\mod B$ because $X*A = 4*3 \\equiv 1 \\mod 11$. An $Y\\equiv B^{-1} \\mod A$ because $Y*B=(-1)*11 \\equiv 1 \\mod 3$. They are both valid inverses. What is your issue? \u2013\u00a0fleablood Jun 15 at 18:48\n\u2022 $-1\\equiv 2 \\mod 3$. So $-1$ is equivalent to $2$. It doesn't make the slightest difference which one you use. (We don't call these things \"equivalences\" for nothing, you know...) \u2013\u00a0fleablood Jun 15 at 19:16\n\u2022 \"A lot of online calculators that I tried are also said that a has to be bigger than be,\" Calculators don't do mathematics. Calculators do calculations. \u2013\u00a0fleablood Jun 15 at 19:24\n\nIt is inevitable that a B\u00e9zout's identity equation will give you modular multiplicative inverses, since given:\n\n$$am+bn = 1$$\n\nwe can take $\\bmod m$ for\n\n$$bn\\equiv 1 \\bmod m$$\n\nor $\\bmod n$ for\n\n$$am \\equiv 1 \\bmod n$$\n\nTo get $a$ and $b$ in your preferred range, you can simply add or subtract a suitable multiple of the modulus.\n\nSo in this case $$-1\\cdot 11 + 4\\cdot 3 = 1$$\n\nand thus $$-1\\cdot 11\\equiv 1 \\bmod 3$$\n\n($-11$ being one more than $-12$), so $-1$ is a valid inverse of $11$ modulo $3$. Then of course $$-1\\equiv 2 \\bmod 3$$\n\nso this is consistent with your observation that $2$ is the inverse of $11 \\bmod 3$ also.\n\n\u2022 I now understand why this is correct from algorithmical point of view, but I don't understand why \u22121 \u2261 2 mod 3. I miss some math-related thing. \u2013\u00a0Danetta Jun 18 at 9:12\n\u2022 Presumably you have no problem with $11 \\equiv 8\\equiv 5 \\equiv 2 \\bmod 3$. In each case, add one and you arrive at a multiple of $3$. The same applies to $-1$. Also, the difference between $-1$ and $2$ is a multiple of $3$ (as is true for any pair of the equivalent values above). \u2013\u00a0Joffan Jun 18 at 12:52\n\n$-1 \\equiv 2 \\mod 3$ so they are considered to be the same thing.\n\nThat's we we call these equivalence classes.\n\nHowever, euclidean algorithm only helps if A < B\n\nI simply do not understand why you say that.\n\neither X is multiplicative inverse of A modulo B or Y is multiplicative inverse of B modulo A, but not both at the same time, because one of them (A or B) is going to be bigger than another. We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not.\n\nExcept, of course, it indeed is. $-1*11 = -11 \\equiv 1 \\mod 3$. That is the valid inverse. Why do you think it is not?\n\nIt doesn't matter if $A > B$ or $B> A$ as $\\gcd(A,B) = 1$ Euclid's algorithm will give us:\n\n$mA + kB = 1$ so $k \\equiv A^{-1} \\mod B$ and $m \\equiv B^{-1} \\mod A$ simultaneously.\n\nIs your concern that one is represented with a positive number and the other negative?\n\nThat's irrelevent.\n\nIt doesn't matter which representative we use to represent a class. We could have used $50\\equiv -1 \\mod 3$ so $50 \\equiv 11^{-1}\\mod 3$ for all we care. (Indeed $50*11 = 550 = 3*183 + 1 \\equiv 1\\mod 3$).\n\nNote: if $mA + kB = 1$ and $m > 0$ but $-A < k < 0$ then\n\n$mA + (k+A)B = 1 + BA\\equiv 1 \\mod A,B,AB$ and $m$ and $(k+A)$ are still the proper inverses. ANd $m > 0; k+A > 0$.\n\nIndeed $(m + vB)A + (k + uA)B = 1 + (v+u)AB$ so $m + vB\\equiv A^{-1} \\mod B$ for any integer $v$ and $k + uA \\equiv B^{-1} \\mod A$ for any integer $u$.\n\n\u2022 \"Is your concern that one is represented with a positive number and the other negative?\" \u2014 Yes, I haven't even considered negative numbers. I'm still not exactly sure how \u221211 \u2261 1 mod 3. For me logically it looks like quotient of -11/3 is -3, so remainder is either 2 or -2, but how is it 1? \u2013\u00a0Danetta Jun 18 at 8:56\n\u2022 It looks like a rule to \"just take a remainder from division\" stops working there. Do I need positive inverse for Montgomery or it doesn't matter? \u2013\u00a0Danetta Jun 18 at 9:17\n\u2022 $-11 = 3*(-4)+ 1$ so $-11 \\equiv 1 \\mod 3$ and $-11 = 3*(-3) -2$ so $-11 \\equiv -2 \\mod 3$. And $1 \\equiv -2 \\mod 3$. As well $-11 = 3*(-7)+12$ so $-11 \\equiv 10 \\mod 3$. But $-11 \\ne 3*k + 2$ for any integer $k$ so $-11 \\not\\equiv 2 \\mod 3$. A remainder means any $p = d*n + r$ and $d$ can be any value positive or a negative and $r$ can be any value that works. But THE remainder would mean $p=d*n+r;0\\le r< n$ would mean the remainder is unique, positive and less than the divisor. If $p < 0$ this means $d < 0$ and $r \\ge 0$. \u2013\u00a0fleablood Jun 18 at 15:01\n\u2022 I dont know how to do the Montgomery method. It seems as though you must already now then inverse to use it and that it is just an computational efficient method to do multiplication. \u2013\u00a0fleablood Jun 18 at 15:18", "date": "2018-09-20 12:51:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2820845/calculation-of-modular-multiplicative-inverse-of-a-mod-b-when-a-b", "openwebmath_score": 0.8140780329704285, "openwebmath_perplexity": 363.9585526185208, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713878802045, "lm_q2_score": 0.8933094010836642, "lm_q1q2_score": 0.8801521934121361}}
{"url": "https://yutsumura.com/is-the-trace-of-the-transposed-matrix-the-same-as-the-trace-of-the-matrix/", "text": "# Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix?\n\n## Problem 633\n\nLet $A$ be an $n \\times n$ matrix.\n\nIs it true that $\\tr ( A^\\trans ) = \\tr(A)$? If it is true, prove it. If not, give a counterexample.\n\n## Solution.\n\nThe answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the diagonal entries of the transposed matrix are the same as the original matrix.\n\nPutting together these observations yields the equality $\\tr ( A^\\trans ) = \\tr(A)$.\n\nHere is the more formal proof.\n\nFor $A = (a_{i j})_{1 \\leq i, j \\leq n}$, the transpose $A^{\\trans}= (b_{i j})_{1 \\leq i, j \\leq n}$ is defined by $b_{i j} = a_{j i}$.\n\nIn particular, notice that $b_{i i} = a_{i i}$ for $1 \\leq i \\leq n$. And so,\n$\\tr(A^{\\trans}) = \\sum_{i=1}^n b_{i i} = \\sum_{i=1}^n a_{i i} = \\tr(A) .$\n\n### More from my site\n\n#### You may also like...\n\n##### The Vector $S^{-1}\\mathbf{v}$ is the Coordinate Vector of $\\mathbf{v}$\n\nSuppose that $B=\\{\\mathbf{v}_1, \\mathbf{v}_2\\}$ is a basis for $\\R^2$. Let $S:=[\\mathbf{v}_1, \\mathbf{v}_2]$. Note that as the column vectors of $S$...\n\nClose", "date": "2018-05-24 19:36:27", "meta": {"domain": "yutsumura.com", "url": "https://yutsumura.com/is-the-trace-of-the-transposed-matrix-the-same-as-the-trace-of-the-matrix/", "openwebmath_score": 0.9764623045921326, "openwebmath_perplexity": 114.34010952327833, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9938070111261019, "lm_q2_score": 0.8856314768368161, "lm_q1q2_score": 0.8801467709543918}}
{"url": "https://mathhelpboards.com/threads/find-volume-of-solid-generated-calc-2.6582/", "text": "# Find volume of solid generated (Calc 2)\n\n#### lovex25\n\n##### New member\n[solved]Find volume of solid generated (Calc 2)\n\nFind the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y=e^x, and the line x = ln 2 about the line x= ln 2.\n\nSo I tried graphing it to see visually, and the expression I got for calculating the volume was \u222b\u03c0(ln2-lny)^2dy, evaluating from 0 to 2 using disk method, and the answer I got was 4\u03c0, but apparently that doesn't match the answer in the back of the book. I'd really appreciate if someone can help me out!!\n\nOff topic: First time posting a thread here, may I ask how do you type the mathematical symbols such as the integral sign and whatnot, or do I have to manually copy and paste from other website?\n\nLast edited:\n\n#### MarkFL\n\nStaff member\nHello lovex25,\n\nAs you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation:\n\nUsing the shell method, the volume of an arbitrary shell is:\n\n$$\\displaystyle dV=2\\pi rh\\,dx$$\n\nwhere:\n\n$$\\displaystyle r=\\ln(2)-x$$\n\n$$\\displaystyle h=e^x$$\n\nand so we find:\n\n$$\\displaystyle dV=2\\pi \\left(\\ln(2)-x \\right)e^x\\,dx$$\n\nNow, you want to sum up the shells by integrating:\n\n$$\\displaystyle V=2\\pi\\int_0^{\\ln(2)}\\left(\\ln(2)-x \\right)e^x\\,dx$$\n\nTo make things a bit simpler, I would use the substitution:\n\n$$\\displaystyle u=e^x\\,\\therefore\\,du=e^x\\,dx$$\n\nand we now may state:\n\n$$\\displaystyle V=2\\pi\\int_1^2\\ln(2)-\\ln(u)\\,du$$\n\nCan you proceed?\n\nNow, as far as posting mathematical expressions, this site supports $\\LaTeX$, and a good tutorial written by our own Sudharaka can be found here:\n\nhttp://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html\n\n#### lovex25\n\n##### New member\nHello lovex25,\n\nAs you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation:\n\nView attachment 1349\n\nUsing the shell method, the volume of an arbitrary shell is:\n\n$$\\displaystyle dV=2\\pi rh\\,dx$$\n\nwhere:\n\n$$\\displaystyle r=\\ln(2)-x$$\n\n$$\\displaystyle h=e^x$$\n\nand so we find:\n\n$$\\displaystyle dV=2\\pi \\left(\\ln(2)-x \\right)e^x\\,dx$$\n\nNow, you want to sum up the shells by integrating:\n\n$$\\displaystyle V=2\\pi\\int_0^{\\ln(2)}\\left(\\ln(2)-x \\right)e^x\\,dx$$\n\nTo make things a bit simpler, I would use the substitution:\n\n$$\\displaystyle u=e^x\\,\\therefore\\,du=e^x\\,dx$$\n\nand we now may state:\n\n$$\\displaystyle V=2\\pi\\int_1^2\\ln(2)-\\ln(u)\\,du$$\n\nCan you proceed?\n\nNow, as far as posting mathematical expressions, this site supports $\\LaTeX$, and a good tutorial written by our own Sudharaka can be found here:\n\nhttp://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html\nthanks so much!!! finally i was able to match the answer to the back of the book, and I realized what went wrong with my attempt: I didn't separate the integrals to 2 different functions. As for the LaTeX software I'll look in to it soon, thanks again!\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nI personally prefer the discs method. Here you would need to split up your region of integration into two regions, the first being the rectangle below the point (0,1), and the second being the remaining region above it.\n\nAs for the volume of the region below (0,1) generated when rotating, that's easy, it's simply a cylinder of radius \\displaystyle \\begin{align*} \\ln{(2)} \\end{align*} units and height 1 unit, so its volume is \\displaystyle \\begin{align*} \\pi \\left[ \\ln{(2)} \\right] ^2 \\end{align*}.\n\nAs for the volume of the region above (0,1), if we consider horizontal discs, they will each have radius \\displaystyle \\begin{align*} \\ln{(2)} - \\ln{(y)} \\end{align*} and a height \\displaystyle \\begin{align*} \\Delta y \\end{align*}, where \\displaystyle \\begin{align*} \\Delta y \\end{align*} is some small change in y. So their total volume can be approximated by\n\n\\displaystyle \\begin{align*} V &\\approx \\sum \\pi \\left[ \\ln{(2)} - \\ln{(y)} \\right] ^2 \\Delta y \\\\ &= \\pi \\sum \\left[ \\ln{(2)} - \\ln{(y)} \\right] ^2 \\Delta y \\\\ &= \\pi \\sum \\left\\{ \\left[ \\ln{(2)} \\right] ^2 - 2\\ln{(2)}\\ln{(y)} + \\left[ \\ln{(y)} \\right] ^2 \\right\\} \\Delta y \\end{align*}\n\nAnd then as we increase the number of discs, making \\displaystyle \\begin{align*} \\Delta y \\end{align*} smaller, the sum converges on an integral and the approximation becomes exact, so the total volume is\n\n\\displaystyle \\begin{align*} V &= \\pi \\int_1^2{ \\left[ \\ln{(2)} \\right] ^2 - 2\\ln{(2)}\\ln{(y)} + \\left[ \\ln{(y)} \\right] ^2 \\, dy }\\end{align*}\n\nwhich is possible to be integrated", "date": "2021-07-26 16:11:35", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-volume-of-solid-generated-calc-2.6582/", "openwebmath_score": 0.9989243745803833, "openwebmath_perplexity": 850.5317231288717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.982013792143467, "lm_q2_score": 0.8962513842182775, "lm_q1q2_score": 0.8801312205300221}}
{"url": "https://puzzling.stackexchange.com/questions/104895/the-tip-of-a-colorful-triangle", "text": "# The tip of a colorful triangle\n\nOriginal source: Problem 1 of British Informatics Olympiad 2017, Round 1\n\nYou're given a bunch of red (R), green (G), and blue (B) balls. I arrange some balls on a line. Then I ask you to complete the triangle of balls under the following simple rules, and tell me the color of the ball at the last row:\n\n\u2022 Below two balls of the same color, place a ball of that color. (For example, a G must be placed under two G's.)\n\u2022 Below two balls of different colors, place a ball of the third color. (For example, a B must be placed under a R and a G.)\n\nIf I gave you the balls \"R R G B R G B B\", you would place the balls like the following, and tell me \"it's green\":\n\nR R G B R G B B\nR B R G B R B\nG G B R G G\nG R G B G\nB B R R\nB G R\nR B\nG\n\nNow, I'll give you a sequence of 60,000 balls. I won't even show you all the balls' colors; the only information available to you is that it starts with 999 balls of the pattern RGB RGB RGB ... RGB, and ends with 999 balls of the pattern BGR BGR BGR ... BGR.\n\nUsing the same rules, can you guess the color of the ball on the last row? The answer is unique, and an answer with mathematical explanation is preferred.\n\nThis game is a modified Pascal triangle.\n\nFirst,\n\nlet's look at size 4. If you were to permute all the possible inputs and look at the first ball given, last ball given, and the color of the ball on the last row, you can see they are also conforming to the rules given.\n\nWe then\n\ntry 10. We can look at this like so, since size 4 works:\n\n(R)* * B * * B * *(G)\n* * * * * * * * *\n* * * * * * * *\nG * * B * * R\n* * * * * *\n* * * * *\nR * * G\n* * *\n* *\n(B)\n\nIn fact,\n\nthis works if the number is $$3^n + 1$$.\n\nAlright,\n\nlet's say the colors red, green, and blue are numbers 0, 1, and 2 respectively.\n\nWe then visualize a Pascal mod 3 and the game flipped of size 10, with the top portion being one:\n\nPascal mod 3    |       The game\n1          |          1\n1 1         |         2 2\n1 2 1        |        1 2 1\n1 0 0 1       |       2 0 0 2\n1 1 0 1 1      |      1 1 0 1 1\n1 2 1 1 2 1     |     2 1 2 2 1 2\n1 0 0 2 0 0 1    |    1 0 0 2 0 0 1\n1 1 0 2 2 0 1 1   |   2 2 0 1 1 0 2 2\n1 2 1 2 1 2 1 2 1  |  1 2 1 2 1 2 1 2 1\n1 0 0 0 0 0 0 0 0 1 | 2 0 0 0 0 0 0 0 0 2\n\nThis result looks similar - as Pascal's formula (mod 3) is\n\n$${{n}\\choose{k}} \\equiv {{n - 1}\\choose{k - 1}} + {{n - 1}\\choose{k}} \\pmod 3$$ denoting the nth row and kth column. But what about the game?\n\nAs $$1 + 0 \\equiv 2 \\pmod 3$$ and $$2 + 0 \\equiv 1 \\pmod 3$$ for the game,\nwhile $$1 + 0 \\equiv 1 \\pmod 3$$ and $$2 + 0 \\equiv 2 \\pmod 3$$, we want to switch the terms, and that's simple. We negate this equation. The formula for the game is then:\n$${{n}\\choose{k}} \\equiv -\\left[{{{n - 1}\\choose{k - 1}} + {{n - 1}\\choose{k}}}\\right] \\pmod 3$$\nthen:\n$${{n}\\choose{k}} \\equiv -{{n - 1}\\choose{k - 1}} - {{n - 1}\\choose{k}} \\pmod 3$$\n\nAlright, now let's show why size 4 works. We assign each column in the first row $$a, b, c, d$$, and we will use the above formula.\n\na         b        c         d\n-a-b       -b-c     -c-d\na+2b+c      b+2c+d\n-a-3b-3c-d\n\nWe modulo by 3 the expression in the last row and we get $$-a-d \\pmod 3$$, and that also conforms to the provided equation above.\n\nBecause the question is modified,\n\nand 60,000 is not $$3^n + 1$$, we must find the highest power of 3 plus 1 that's smaller than this, which is 59050 ($$3^{10} + 1$$).\n\nSubtracting 60000 to 59050 gives us 950, the (n+1)th-to-last element (starting from the first ball). We use $$\\bmod 3$$ because of the pattern, giving us $$2$$.\n\nNow, starting from the first ball which is R, and looking at the 3rd to last, which is B gives us G.\nLooking at the second ball which is G, and looking at the 2nd to last, which is G gives us G.\nThe third ball is B, and looking at the 1st-to-last, which is R gives us G.\nSince the next balls are repeating, it will be always periodic, but notice how it's always a G? That must mean the next colors, and the last row is Green!\n\n\u2022 Yes, it is correct! \u2013\u00a0Bubbler Nov 16 '20 at 6:15", "date": "2021-05-10 04:11:28", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/104895/the-tip-of-a-colorful-triangle", "openwebmath_score": 0.6420491933822632, "openwebmath_perplexity": 291.26470994072537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9902915249511566, "lm_q2_score": 0.8887587964389112, "lm_q1q2_score": 0.880130303839244}}
{"url": "http://mathhelpforum.com/statistics/45708-probability.html", "text": "Math Help - Probability\n\n1. Probability\n\nI'm having a lot of trouble with permutations and combinations, so I was hoping someone here could help me out with a few questions. Please make sure that you explain your working; I really want to understand this topic... The first question asks;\n\nA queue has 4 boys and 4 girls standing in line. Find how mant different arrangements are possible if;\n\na) The boys and girls alternate.\n\nb) 2 particular girls wish to stand together.\n\nc) All the boys stand together.\n\nd) Also find the probability that 3 particular people will be in the queue together if the queue forms randomly.\n\nEDIT: If it's any help, here are the answers...\n\na) 1152\nb) 10080\nc) 2880\nd) 3/28\n\nEDIT: I've got a new problem now;\n\nA table has 4 boys and 4 girls sitting around it.\n\na) Find the number of ways of sitting possible if the boys and girls can sit anywhere around the table.\n\nb) If the seating is arranged at random, find the probability that;\ni) 2 particular girls will sit together.\nii) All the boys will sit together.\n\n2. Hello, Flay!\n\nWe have to \"talk\" our way through these problems . . .\n\nA queue has 4 boys and 4 girls standing in line.\nFind how mant different arrangements are possible if;\n\na) The boys and girls alternate.\nThere are 2 possible arrangements: . $BGBGBGBG\\,\\text{ and }\\,GBGBGBGB.$\n\nThe four boys can be placed in 4! ways.\nThe four girls can be placed in 4! ways.\n\nTherefore, there are: . $2 \\times 4! \\times 4! \\;=\\;\\boxed{1152}$ ways.\n\nb) 2 particular girls wish to stand together.\nSuppose the two girls are $X$ and $Y.$\n\nDuct-tape them together.\nNote that there are 2 possible orders: . $XY\\,\\text{ or }\\,YX.$\n\nNow we have seven \"people\" to arrange: . $\\boxed{XY}\\,,G,G,B,B,B,B$\n. . and they can be arranged in ${\\color{blue}7!}$ ways.\n\nTherefore, there are: . $2 \\times 71\\;=\\;\\boxed{10,080}$ ways.\n\nc) All the boys stand together.\nDuct-tape the four boys together.\n. . Note that there are 4! possible orders.\n\nNow we have five \"people\" to arrange.\n. . There are $5!$ ways.\n\nTherefore, there are: . $4! \\times 5! \\:=\\:\\boxed{2880}$ ways.\n\nd) Find the probability that 3 particular people will be together\nif the queue forms randomly.\nFirst of all, there are 8! possible arrangements.\n\nSuppose the three people are $X, Y\\text{ and }Z.$\nDuct-tape them together: . $\\boxed{XYZ}$\n. . Note that there are 3! orderings.\n\nNow we have six \"people\" to arrange,\n. . and there are 6! ways.\n\nHence, there are: . ${\\color{red}3!\\times6!}$ ways for $X,Y,Z$ to be together.\n\nTherefore, the probability is: . $\\frac{3!\\cdot6!}{8!} \\;=\\;\\boxed{\\frac{3}{28}}$\n\n3. Thanks very much. Now I have another problem;\n\nA school committee is to be made up of 5 teachers, 4 students and 3 parents.\n\na) If 12 teachers, 25 students and 7 parents apply to be on the committee, which is chosen at random, how many possible committees could be formed?\n\nThis part I've figured out. The answer is $350 658 000$.\n\nb) If Jan and her mother both apply, find the probability that both will be chosen for the committee.\n\nThis part I'm not sure how to do. The answer is apparently $\\frac{3036}{44275}$.\n\n4. I figured out the last question. As it turns out, $\\frac{3036}{44275}$ simplifies to $\\frac{12}{175}$, which was the answer I was getting.\n\nNow I have a new problem;\n\nA sample of 3 coins is taken at random from a bag containing 8 ten cent coins and 8 twenty cent coins. Find the probability that a particular ten cent coin will be chosen, if 1 twenty cent and 2 ten cent coins are chosen.", "date": "2015-08-05 01:45:11", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/45708-probability.html", "openwebmath_score": 0.806406557559967, "openwebmath_perplexity": 770.5921127119627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915215128428, "lm_q2_score": 0.8887587890727755, "lm_q1q2_score": 0.8801302934887906}}
{"url": "https://math.stackexchange.com/questions/2788278/if-rho-n1-leq-rho-n-sigma-n-forall-n-geq-n-0-then-prove-that", "text": "# If $\\rho_{n+1}\\leq\\rho_{n} +\\sigma_n,\\;\\;\\forall\\;\\;n\\geq N_0,$ then prove that $\\lim\\rho_n=0.$\n\nGood day, all!\n\nSuppose $\\{\\rho_n\\}$ and $\\{\\sigma_n\\}$ are two sequences of non-negative real numbers such that for some real number $N_0\\geq 1,$ the following recursion inequality holds:\n\n$$\\rho_{n+1}\\leq\\rho_{n} +\\sigma_n,\\;\\;\\forall\\;\\;n\\geq N_0.$$\n\nProve that,\n\n1. if $\\sum_{n=1}^{\\infty}\\sigma_n<\\infty,$ then $\\lim\\rho_n$ exists.\n2. If $\\sum_{n=1}^{\\infty}\\sigma_n<\\infty,$ and $\\rho_n$ has a subsequence converging to zero, then $\\lim\\rho_n=0.$\n\nThe first question was asked in my exam but I skipped it. I am stating the truth! Hence, I'll like to get it solved before the final exam. So please, can anyone help me? Thanks in advance.\n\n\u2022 if the series of $\\rho_n$ converges, then the limit of $\\rho_n$ has to be zero, there is nothing to prove in 1 and 2. Please double-check the formulation (perhaps you need $\\sigma_n$ instead ?) \u2013\u00a0Hayk May 20 '18 at 5:40\n\u2022 @Hayk: Sorry, my mistake! It has been corrected! Matt A Pelto: Thanks for the edit! \u2013\u00a0Omojola Micheal May 20 '18 at 6:11\n\u2022 Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid \"I have no clue\" questions. Defining keywords and trying a simpler, similar problem often helps. \u2013\u00a0robjohn May 20 '18 at 7:07\n\u2022 a real pity about people on this forum that go straight to down voting :) ... such a community. \u2013\u00a0Matt A Pelto May 20 '18 at 7:47\n\u2022 @ robjohn: This is not an assignment question. We had the third exam yesterday and the first question, .i.e. question 1, was given to us. However, I'll correct some of my words! \u2013\u00a0Omojola Micheal May 20 '18 at 7:48\n\nObserve that once we prove assertion $1$, then the claim of $2$ follows readily. This is due to the fact, that if $\\rho_n$ is convergent, than all its subsequences must be convergent, all with the same limit. Now, if we prove claim N$1$ and assume in addition that a subsequence of $\\rho_n$ converges to $0$, then $\\rho_n$, having the same limit as its subsequence, must converge to $0$.\n\nWe now prove claim $1$. Observe that for any $k\\in \\mathbb{N}$ and any $n\\geq N_0$ we have $$(1) \\qquad \\rho_{n+k} - \\rho_n = \\rho_{ n + k } - \\rho_{ n + k -1} + \\rho_{ n + k - 1 } - \\rho_{ n + k -2} +...+\\rho_{n+1} - \\rho_n \\leq \\\\ \\sigma_{n+k-1} +...+\\sigma_n,$$ where we apply the given inequality on each term $\\rho_{n+k -i} - \\rho_{n+k-i-1}$, for $i=0,....,k-1$.\n\nFrom the above inequality, and the fact that $\\rho_n \\geq 0$, we have that $\\rho_n$ is a bounded sequence. Now assume, for contradiction, that it does not converge. Hence, $\\rho_n$, converges to different limits along two different subseqeunces. Namely there exists $n_{k}$ and $m_k$, and non negative numbers $a_1$ and $a_2$ such that $$\\rho_{n_k} \\to a_1 \\qquad \\text{ and } \\qquad \\rho_{m_k} \\to a_2,$$ where $a_1 \\neq a_2$. From $(1)$, let us fix $n\\geq N_0$, and choose $k$ such that $n+k = n_k$. Then, taking limit with respect to $k$ and using the fact that the series $\\sigma_n$ converges, we get $$(2) \\qquad a_1 - \\rho_n \\leq \\sum_{i = n}^\\infty \\sigma_i.$$ Now take $n\\to \\infty$ along the second subsequence $m_k$, we obtain $a_1 - a_2 \\leq 0$, since convergence of $\\sum_i \\sigma_i$ implies that the right hand side of $(2)$ converges to $0$.\n\nSince the process is symmetric (we could have started with $m_k$ and not $n_k$) we obtain also that $a_2 \\leq a_1$ and hence they must be equal.\n\nThis contradicts our assumption that $a_1 \\neq a_2$ and hence the limit of $\\rho_n$ exists.\n\n\u2022 What is wrong with Hayk's solution? Can anyone please, explain? \u2013\u00a0Omojola Micheal May 20 '18 at 7:49\n\u2022 @Mike, let's see if the voter will be able to justify it. BTW, the only place you need non-negativity of $\\rho_n$ is to get a lower bound on the sequence, as otherwise without bounding $\\rho_n$ from below, it can escape to $-\\infty$. So, instead of requiring $\\rho_n \\geq 0$ the claim stays valid also for the case $\\rho_n \\geq -M$, for any fixed $M\\geq 0$ (the same proof works). \u2013\u00a0Hayk May 20 '18 at 8:49\n\u2022 I neutralized it for you! \u2013\u00a0Omojola Micheal May 20 '18 at 15:56\n\u2022 Can you please, expand this line $\\leq \\sigma_{n+k-1} +...+\\sigma_{n-1}$? Kindly explain why it is so! \u2013\u00a0Omojola Micheal May 20 '18 at 16:09\n\u2022 I was thinking it should be $\\leq \\sigma_{n+k-1} +...+\\sigma_{n-1}+\\sigma_{n}$ instead! \u2013\u00a0Omojola Micheal May 20 '18 at 16:19\n\nPut $T_n=\\sum_{k=1}^n \\sigma_k$. Then $T_n$ is a convergent sequence. Now put $u_n=T_{n-1}-\\rho_n$. From the inequality given, we see that $u_n$ is increasing for $n$ large, and of course (as $\\rho_n \\geq 0$) $\\leq T_{n-1}$, hence bounded, hence convergent. It is easy to finish.\n\nFrom $\\rho_{n+1}-\\rho_n\\leq \\sigma_n$ we obtain $\\rho_{n}- \\rho_m \\leq \\sum_{i=m}^{n-1}\\sigma _i \\Longrightarrow \\rho_{n} \\leq \\rho_m+\\sum_{i=m}^{n-1}\\sigma _i$.\n\nTaking $\\limsup$ in respect of $n$ we obtain $\\limsup \\rho _n \\leq \\sum_{i=m}^\\infty \\sigma _i +\\rho _m$.\n\nTaking now $\\liminf$, we obtain $$\\limsup \\rho _n \\leq \\liminf\\rho _m$$ So, $\\rho_n\\longrightarrow x$ where $x \\in [-\\infty,\\infty]$. Since $\\rho _n$ is non-negative and upper bounded by $\\rho_0 +\\sum_{i=1}^\\infty \\sigma _i$, we conclude that $x\\in \\mathbb{R}$.", "date": "2020-01-21 14:24:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2788278/if-rho-n1-leq-rho-n-sigma-n-forall-n-geq-n-0-then-prove-that", "openwebmath_score": 0.9408247470855713, "openwebmath_perplexity": 246.0581766092783, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830769252026, "lm_q2_score": 0.904650527388829, "lm_q1q2_score": 0.8800718684639256}}
{"url": "https://math.stackexchange.com/questions/921443/is-there-a-closed-form-expression-for-int-infty-infty-int-inftyy", "text": "# Is there a closed form expression for $\\int_{- \\infty}^\\infty \\int_{-\\infty}^y \\frac{1}{2 \\pi} e^{-(1/2) ( x^2+y^2 )} \\mathrm{d}x\\,\\mathrm{d}y$?\n\nI have been trying to evaluate the integral:\n\n$$\\int_{- \\infty}^\\infty \\int_{-\\infty}^y \\frac{1}{2 \\pi} e^{-(1/2) ( x^2+y^2 )}\\mathrm {d}x\\,\\mathrm{d}y$$\n\nI know of course that the integral equals $1$ over $[-\\infty,\\infty] \\times [-\\infty,\\infty]$ but I do not quite know how to handle the present case. Are there any tricks here?\n\nThank you.\n\n\u2022 I think you can exploit the simmetry of the integrand by noting that the line $y=x$ divides the plane in half. \u2013\u00a0Gennaro Marco Devincenzis Sep 6 '14 at 15:22\n\nYour integral is the probability: $$\\mathbb{P}[X\\leq Y]$$ where $X$ and $Y$ are two independent normal variables $N(0,1)$,\n\nhence the value of the integral is just $\\frac{1}{2}$, since: $$\\mathbb{P}[X\\leq Y]=\\mathbb{P}[Y\\leq X],\\qquad \\mathbb{P}[X\\leq Y]+\\mathbb{P}[Y\\leq X]=1.$$\n\n\u2022 That makes sense intuitively but could you please explain a little more why that is? There is no closed form then, right? \u2013\u00a0JohnK Sep 6 '14 at 15:26\n\u2022 $f(x)=\\frac{1}{\\sqrt{2\\pi}}e^{-x^2/2}$ is the pdf of a $N(0,1)$ random variable, and $\\frac{1}{2}$ looks as a very closed form to me. \u2013\u00a0Jack D'Aurizio Sep 6 '14 at 15:29\n\u2022 Yes but what allows us to say that the value of that probability is 1/2? \u2013\u00a0JohnK Sep 6 '14 at 15:31\n\u2022 Symmetry: $$\\mathbb{P}[Y\\leq X]=\\mathbb{P}[X\\leq Y],\\quad \\mathbb{P}[Y\\leq X]+\\mathbb{P}[X\\leq Y] = 1.$$ \u2013\u00a0Jack D'Aurizio Sep 6 '14 at 15:32\n\u2022 Simple yet elegant! Intuitively, it can also be seen by plotting the region $0<x<y<\\infty$ since the integrand is even function (i.e. symmetry). +1 \u2013\u00a0Tunk-Fey Sep 6 '14 at 15:44\n\nJack D'Aurizio's answer is good, but since you said in comments under it that you wanted a different point of view, let's try this: \\begin{align} u & = (\\cos45^\\circ)x-(\\sin45^\\circ)y = \\tfrac{\\sqrt{2}}2 x - \\tfrac{\\sqrt{2}}2 y \\\\ v & = (\\sin45^\\circ)x+(\\cos45^\\circ)y = \\tfrac{\\sqrt{2}}2 x + \\tfrac{\\sqrt{2}}2 y \\end{align} This is just a $45^\\circ$ rotation of the coordinate system, suggested by the fact that your boundary line $y=x$ is just a $45^\\circ$ rotation of one of the coordinate axes.\n\nThen simplify $u^2+v^2$ and find that it comes down to $x^2+y^2$.\n\nSolving the system of two equations above for $x$ and $y$, one gets \\begin{align} x & = \\phantom{-}\\tfrac{\\sqrt{2}}2 u + \\tfrac{\\sqrt{2}}2 v \\\\ y & = -\\tfrac{\\sqrt{2}}2 u + \\tfrac{\\sqrt{2}}2 v \\end{align} By trivial algebra, the condition that $x\\le y$ now becomes $u\\le0$.\n\nIf you know about Jacobians, you get $$du\\,dv = \\left|\\frac{\\partial(u,v)}{\\partial(x,y)}\\right|\\,dx\\,dy = \\left|\\frac{\\partial u}{\\partial x}\\cdot\\frac{\\partial v}{\\partial y} - \\frac{\\partial u}{\\partial y}\\cdot\\frac{\\partial v}{\\partial x}\\right|\\,dx\\,dy = 1\\,dx\\,dy.$$ Hence your iterated integral becomes $$\\int_{-\\infty}^\\infty\\int_{-\\infty}^0 \\frac 1{2\\pi} e^{-(u^2+v^2)/2}\\,du\\,dv = \\int_{-\\infty}^\\infty \\int_{-\\infty}^0 \\left\\{\\frac 1{2\\pi} e^{-v^2/2}\\right\\} e^{-u^2/2}\\,du\\,dv$$ The part in $\\{\\text{braces}\\}$ does not depend on $u$, so this is $$\\int_{-\\infty}^\\infty\\left( \\frac 1{2\\pi} e^{-v^2/2} \\int_{-\\infty}^0 e^{-u^2/2}\\,du \\right)\\,dv.$$ Now the inside integral does not depend on $v$, so it pulls out: $$\\int_{-\\infty}^\\infty e^{-v^2/2}\\,dv \\cdot \\frac1{2\\pi} \\int_{-\\infty}^0 e^{-u^2/2}\\,du$$ and this is of course $$\\int_{-\\infty}^\\infty \\frac1{\\sqrt{2\\pi}} e^{-v^2/2}\\,dv \\cdot \\int_{-\\infty}^0 \\frac1{\\sqrt{2\\pi}} e^{-u^2/2}\\,du.$$ The first integral comes to $1$ and the second, by a simple symmetry argument, is $1/2$.\n\n\u2022 Thank you very much. I'll just have to understand how come that substitution rotates the coordinate system. \u2013\u00a0JohnK Sep 6 '14 at 16:48\n\u2022 nice result! you may simplify your answer by showing that $\\int_{-\\infty}^0 e^{-u^2}\\,du=(1/2)\\int_{-\\infty}^{\\infty} e^{-u^2}\\,du$ as early in your answer as possible. \u2013\u00a0mike Sep 6 '14 at 16:52\n\u2022 Think of what that substitution does to $(x,y)=(1,0)$ and to $(x,y)=(0,1)$ and then you'll be well on your way to understanding why it's a rotation. But even if you didn't know it's a rotation, you should still be able to follow the argument. \u2013\u00a0Michael Hardy Sep 6 '14 at 16:52\n\u2022 . . . however, it is also the case that if you know nothing of Jacobians, you can understand that $du\\,dv=dx\\,dy$ by using the fact that the transformation is a rotation. Rotations multiply volumes by $1$, so the thing to multiply $dx\\,dy$ by to get $du\\,dv$ is $1$. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Sep 6 '14 at 16:54\n\u2022 The intuition behind Jacobians is often not made explicit in second-year calculus courses where the concept is introduced. But if the value of the Jacobian $|\\partial(u,v)/\\partial(x,y)|$ at a particular point is, for example $6$, that means that at that point an infinitesimal area in the $x,y$ plane is transformed to an infinitesimal area $6$ times as big in the $u,v$ plane. Determinants generally area what you multiply by a volume to get the volume it is transformed to. And the determinant is negative if the orientation gets reversed. \u2013\u00a0Michael Hardy Sep 6 '14 at 17:03\n\nSet $x=r\\cos \\theta,y=r\\sin \\theta$, then we have\n\n$$\\int_{- \\infty}^\\infty \\int_{-\\infty}^y e^{-(1/2) ( x^2+y^2 )}\\mathrm {d}x\\,\\mathrm{d}y=\\int_{0}^\\infty \\left(\\int_{-3\\pi/4}^{\\pi/4} e^{-r^2/2}\\mathrm {d}\\theta\\,\\right)r\\,\\mathrm{d}r=\\pi \\int_{0}^\\infty e^{-r^2/2}r\\,\\mathrm{d}r=\\pi$$\n\nSo the original integral is equal to $(1/2)$.\n\nThis method as well as @MichaelHardy's method also works for the integrals like:\n\n$$\\int_{- \\infty}^\\infty \\int_{-\\infty}^{a y} e^{-(1/2) ( x^2+y^2 )}\\mathrm {d}x\\,\\mathrm{d}y, \\text{ }a \\in \\mathbb{R}$$\n\nThe results are the same. This is because the function to be integrated ($e^{-r^2/2}$) is rotational invariant (independent of $\\theta$) and $x=a y$ is a straight line going through the origin and divides the plane into 2 halfs of equal size.\n\n\u2022 Glad to see polar coordinates work too, thank you. \u2013\u00a0JohnK Sep 6 '14 at 18:27\n\nAh, I realise just now I'd misread to start with. Algebraically, we can solve it by noting that the value of the integral is the same under the change of variables $u=-x$, and since summing the two resulting integrals results in the $[-\\infty,\\infty] \\times [-\\infty,\\infty]$ case, the answer is $1/2$.\n\ni.e.:\n\n$$\\int_{- \\infty}^\\infty \\int_{-\\infty}^y \\frac{1}{2 \\pi} e^{-1/2 (x^2+y^2)} \\,\\mathrm{d}x\\,\\mathrm{d}y = \\int_{- \\infty}^\\infty \\int_{-y}^\\infty \\frac{1}{2 \\pi} e^{-1/2 \\left( x^2+y^2 \\right)}\\,\\mathrm{d}x\\,\\mathrm{d}y.$$\n\n$LHS+RHS=\\int_{- \\infty}^\\infty \\int_{-\\infty}^\\infty \\frac{1}{2 \\pi} e^{-1/2 (x^2+y^2)}\\,\\mathrm{d}x\\,\\mathrm{d}y=1$, and $LHS=RHS$, so your integral is $1/2$.\n\nEDIT: I guess I was a little bit short. There are several ways of showing $LHS+RHS$ is what I quote, but here is a simple, uninformative one.\n\n$LHS+RHS=\\int_{- \\infty}^\\infty \\int_{-\\infty}^\\infty \\frac{1}{2 \\pi} e^{-1/2 (x^2+y^2)}\\,\\mathrm{d}x\\,\\mathrm{d}y-\\int_{- \\infty}^\\infty \\int_{-y}^y\\frac{1}{2 \\pi} e^{-1/2 (x^2+y^2)}\\,\\mathrm{d}x\\,\\mathrm{d}y$.\n\nNow, Let $I=\\int_{- \\infty}^\\infty \\int_{-y}^y\\frac{1}{2 \\pi} e^{-1/2 (x^2+y^2)}\\,\\mathrm{d}x\\,\\mathrm{d}y$. Use substitution $u=-y$. Then\n\n$I=-\\int_{\\infty}^{-\\infty} \\int_{u}^{-u}\\frac{1}{2 \\pi} e^{-1/2 (x^2+u^2)}\\,\\mathrm{d}x\\,\\mathrm{d}u=\\int_{-\\infty}^{\\infty} \\int_{u}^{-u}\\frac{1}{2 \\pi} e^{-1/2 (x^2+u^2)}\\,\\mathrm{d}x\\,\\mathrm{d}u=-I$, hence as $I=-I$, $I=0$.\n\nI think it could also be done by splitting the integration range up. (e.g. looking at the integrals on $(-\\infty,0]$ and $[0,\\infty)$)\n\n\u2022 If you do not mind how is that LHS+RHS equal that integral? \u2013\u00a0JohnK Sep 6 '14 at 16:09\n\u2022 @JohnK there you go :) \u2013\u00a0ShakesBeer Sep 7 '14 at 7:49", "date": "2021-05-13 00:01:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/921443/is-there-a-closed-form-expression-for-int-infty-infty-int-inftyy", "openwebmath_score": 0.9580861330032349, "openwebmath_perplexity": 263.84977122382094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918485625379, "lm_q2_score": 0.8902942326713409, "lm_q1q2_score": 0.88004859181786}}
{"url": "http://mathhelpforum.com/discrete-math/125642-how-prove-logical-equivalence-using-theorems-substitution.html", "text": "# Math Help - How to prove logical equivalence using Theorems and substitution?\n\n1. ## How to prove logical equivalence using Theorems and substitution?\n\nSo, I have the statement\n\n(p ^ q) \\/ (p ^ \u00ac q) \u2261 p\n\nI have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.\n\ncheers\n-ipatch\n\n2. Originally Posted by ipatch\nSo, I have the statement\n\n(p ^ q) \\/ (p ^ \u00ac q) \u2261 p\n\nI have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.\n\ncheers\n-ipatch\nJust read it as a sentence:\n\n\"p and q\" or \"p and not q\".\n\nIt's saying you have to have p but you may or may not have q.\n\nSo what it's the same as having? p.\n\n3. I have to prove the given logical equivalence using Theorems and not constructing a truth table\nThe solution depends on what theorems you have in mind. Every course may have a different set of those.\n\n4. Hello, ipatch!\n\nI don't know the names of the theorems you know.\nI hope you can follow my proof.\n\nProve: . $(p \\;\\wedge\\; q) \\;\\vee\\; (p\\; \\wedge \\sim q) \\;\\;\\equiv \\;\\; p$\n\n. . . . . . . . . $(p \\;\\wedge\\; q) \\;\\vee\\; (p\\; \\wedge\\; \\sim q)$\n\n. . $(p \\;\\vee\\; p) \\;\\wedge\\; (p \\;\\vee\\; \\sim q) \\;\\wedge\\; (q \\;\\vee\\; p) \\;\\wedge\\; (q \\;\\vee\\; \\sim q)$ . . Distributive\n\n. . . . . . $p\\;\\wedge\\;(p\\;\\vee\\;\\sim q)\\;\\wedge\\;(p\\;\\vee\\; q) \\;\\wedge\\; T$\n\n. . . . . . . . . $(p\\;\\vee\\;\\sim q)\\;\\wedge\\; (p\\;\\vee q)$\n\n. . . . . . . . . . . $p\\;\\vee(\\sim q\\;\\wedge\\; q)$ . . . . . . . . . . . . Distributive\n\n. . . . . . . . . . . . . $p\\;\\vee\\;F$\n\n. . . . . . . . . . . . . . . $p$\n\n5. Hello ipatch\nOriginally Posted by ipatch\nSo, I have the statement\n\n(p ^ q) \\/ (p ^ \u00ac q) \u2261 p\n\nI have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.\n\ncheers\n-ipatch\nUsing the standard Laws of Boolean Algebra (for instance, just here) we get:\n$(p \\land q)\\lor(p\\land\\neg q) \\equiv p\\land(q\\lor\\neg q)$ (Distributive Law)\n$\\equiv p\\land T$ (Complement Law)\n\n$\\equiv p$ (Identity Law)\n\n6. (p ^ q) \\/ (p ^ \u00ac q) \u2261 p\n\n1. (p ^ q) \\/ (p ^ \u00ac q)-----------------hypotesis\n2. (p ^ q)-----------------assume\n3. p-----------------2, ^-elim 2\n4. (p ^ \u00ac q)-----------------assume\n5. p-----------------4, ^-elim 2\n6. p-----------------1, 2-3 , 4-5, \\/-elim\n\n7. Hello again, ipatch!\n\nMy first proof was long and clumsy.\n(Don't know where my brain was at the time.)\n\nThere are two theorems we need. .(I don't know their names.)\n\n. . $\\begin{array}{cccc}P \\;\\vee \\sim \\!P &\\equiv& t & \\text{Theorem 1} \\\\ \\\\[-3mm] P \\wedge \\:t &\\equiv& P & \\text{Theorem 2} \\end{array}$\n\n. . . . $\\begin{array}{ccc}(p \\;\\wedge\\; q) \\;\\vee\\; (p\\; \\wedge\\; \\sim q) & \\text{Given} \\\\ \\\\ p \\;\\wedge\\; (q \\;\\vee \\sim \\!q) & \\text{Distr.} \\\\ \\\\ p \\;\\wedge \\;t & \\text{Theorem 1} \\\\ \\\\ p & \\text{Theorem 2} \\end{array}$\n\n8. Originally Posted by cgafa\n(p ^ q) \\/ (p ^ \u00ac q) \u2261 p\n\n1. (p ^ q) \\/ (p ^ \u00ac q)-----------------hypotesis\n2. (p ^ q)-----------------assume\n3. p-----------------2, ^-elim 2\n4. (p ^ \u00ac q)-----------------assume\n5. p-----------------4, ^-elim 2\n6. p-----------------1, 2-3 , 4-5, \\/-elim\n\nRules used:\nconjunctive elimination ( ^-elim ) if you have A and B the you have A (or B)... simple\ndisjuntive elimination (\\/-elim)... a little more complex... if you have A or B and from A you can deduce C; and from B you can deduce C; then from A or B we can deduce C.....\n\n9. Originally Posted by ipatch\nSo, I have the statement\n\n(p ^ q) \\/ (p ^ \u00ac q) \u2261 p\n\nI have to prove the given logical equivalence using Theorems and not constructing a truth table. If anyone has as idea how I prove this I would appreciate an explanation and solution.\n\ncheers\n-ipatch\nipatch:\nThere two kind of proofs,one in Boolean Algebra and one in propositional calculus.\n\nThe proof in Boolean Algebra was shown by other people in the forum.\n\nThe proof in propositional calculus is as follows:\n\nFirst we prove : $(p\\wedge q)\\vee (p\\wedge\\neg q)\\Longrightarrow p$ and then:\n\n$p\\Longrightarrow (p\\wedge q)\\vee (p\\wedge\\neg q)$.\n\nProof:\n\n1) $(p\\wedge q)\\vee (p\\wedge\\neg q)$.................................................. ............................given\n\n2) $(p\\wedge q)$.................................................. ..............................................assu mption to start a conditional proof\n\n3) p................................................. .................................................. .....from (2) and using Conjunction Elimination (=C.E)\n\n4) $(p\\wedge q)\\Longrightarrow p$.................................................. ....................................from (2) to (3) and using the rule of conditional proof\n\n5) In the same way we prove: $(p\\wedge\\neg q)\\Longrightarrow p$\n\n6) $(p\\wedge q)\\vee (p\\wedge\\neg q)\\Longrightarrow p\\vee p$.................................................. ...........................from (4)and(5) and using the rule called proof by cases: $[((A\\Longrightarrow B)\\wedge (C\\Longrightarrow D))\\Longrightarrow ((A\\vee C)\\Longrightarrow (B\\vee D))]$.\n\n7) $p\\vee p$.................................................. .......................................from (1) and (6) and using M.Ponens\n\n8) $p\\vee p\\Longleftrightarrow p$.................................................. .......................................tautology\n\n9) $p\\vee p\\Longrightarrow p$.................................................. ........................................from (8) and using biconditional elimination\n\n10) p................................................. .................................................. ....from (7) and (9) and using M.Ponens\n\nNow to prove the converse.\n\nProof:\n\n1)p............................................... .................................................. given\n\n2) $\\neg(p\\wedge q)$.................................................. ........................................assumption to start a conditional proof\n\n3) $\\neg p\\vee\\neg q$.................................................. ..........................................from (2) and using De Morgan\n\n4) $p\\Longrightarrow\\neg q$.................................................. .....................from (3) and using material implication\n\n5) $\\neg q$.................................................. .....................................from (1) and (4) and using M.Ponens\n\n6) $p\\wedge\\neg q$.................................................. ...........................from (1) and (5) and using Conjunction Introduction\n\n7) $\\neg(p\\wedge q)\\Longrightarrow(p\\wedge\\neg q)$.................................................. .......................from (2) to(6) and using the rule of conditional proof\n\n8) $(p\\wedge q)\\vee (p\\wedge\\neg q)$.................................................. ........................from (7) and using material implication $(A\\Longrightarrow B)\\Longleftrightarrow (\\neg A\\vee B)$", "date": "2016-02-12 04:39:59", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/125642-how-prove-logical-equivalence-using-theorems-substitution.html", "openwebmath_score": 0.8503162860870361, "openwebmath_perplexity": 1357.516068242157, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988491852291787, "lm_q2_score": 0.8902942203004186, "lm_q1q2_score": 0.8800485829094331}}
{"url": "https://math.stackexchange.com/questions/2730864/vector-whose-inner-product-is-positive-with-every-vector-in-given-basis-of-mat", "text": "# Vector whose inner product is positive with every vector in given basis of $\\mathbb{R}^n$\n\nI am trying to solve the following question which I came across when studying root system in euclidean spaces, with positive definite symmetric bilinear form.\n\nStatement: Given a basis $\\{v_1,\\cdots, v_n\\}$ of $\\mathbb{R}^n$, $\\exists$ $v\\in\\mathbb{R}^n$ such that $(v,v_i)>0$ for all $i$\n\nMy answer: Let $v=a_1v_1 + \\cdots + a_nv_n$. Then $$(v,v_i)=a_1(v_1,v_i) + a_2(v_2,v_i) + \\cdots + a_n(v_n,v_i), \\hskip5mm i=1,2,\\cdots,n.$$ Writing this in matrix form we get $$\\begin{bmatrix} (v,v_1) \\\\ \\vdots \\\\ (v,v_n)\\end{bmatrix} = \\begin{bmatrix} (v_1,v_1) & \\cdots & (v_n,v_1)\\\\ \\vdots & & \\vdots \\\\ (v_1,v_n) & \\cdots & (v_n,v_n) \\end{bmatrix} \\begin{bmatrix} a_1 \\\\ \\vdots \\\\ a_n\\end{bmatrix}.$$ We want to find $a_1,\\cdots,a_n$ such that the left column vector contains positive entries only. Let $A$ be the square matrix appearing above; it is matrix of positive definite (symmetric) inner product w.r.t given basis, so it is invertible over $\\mathbb{R}$. Take $$\\begin{bmatrix} a_1 \\\\ \\vdots \\\\ a_n\\end{bmatrix} := A^{-1}\\begin{bmatrix} 1 \\\\ \\vdots \\\\ 1\\end{bmatrix}.$$ This is non-zero vector, and for this choice of $a_i$'s we get the vector $v=a_1v_1+\\cdots + a_nv_n$ with desired properties.\n\nQ.1 Is the above statement and proof correct?\n\nQ.2 Is there geometric way to prove the statement?\n\n\u2022 How do you know that $A$ is invertible? \u2013\u00a05xum Apr 10 '18 at 12:43\n\u2022 it is matrix of non-degenerate bilinear form (positive definite inner product should be non-degenerate, isn't it?) \u2013\u00a0Beginner Apr 10 '18 at 12:44\n\u2022 That's one way of showing it, sure. \u2013\u00a05xum Apr 10 '18 at 12:45\n\nFor a geometric proof, just do something akin to Gram-Schmidt orthogonalisation. Take $v=v_1$, so that $(v,v_1)>0$. Let $u_2=v_2-\\frac{(v_1,v_2)}{(v_1,v_1)}v_1$ be the component of $v_2$ that is orthogonal to $v_1$. Then $u_2\\ne0$, or else $v_1$ and $v_2$ are linearly dependent. Add $c_2u_2$ to $v$ for a sufficiently large $c_2>0$. Then $(v,v_i)>0$ for $i=1,2$. Continue in this manner, you get a desired $v$.", "date": "2019-12-10 08:48:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2730864/vector-whose-inner-product-is-positive-with-every-vector-in-given-basis-of-mat", "openwebmath_score": 0.9290722608566284, "openwebmath_perplexity": 133.78975763635015, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771778588347, "lm_q2_score": 0.8918110540642805, "lm_q1q2_score": 0.8800187951128634}}
{"url": "http://mathhelpforum.com/math-topics/37930-problem-5-a.html", "text": "# Math Help - Problem 5\n\n1. ## Problem FIVE\n\nProblem 5\n\nTRIPOLLO\n\nTRIPOLLO is a strategy game in which numbered marbles are placed in a rectangular array of holes on a board. As shown, there are three rows of holes. A player wins if the marbles are placed so that in each column, the sum of the numbers in the first two rows is three times the number in the third row.\n\n(a) John has a 5-column board and marbles numbered 1 to 15.\nSo far he has placed them as shown in the diagram.\n\nShow how he could successfully place the remaining marbles.\n\n(b) With the same board and marbles. John made the following start:\n\nCan he successfully complete this game? Give reasons.\n\n(c) You have an 8-column board and marbles numbered 1 to 24. Show how you can win by placing all the marbles on the board.\n\n(d) Show that with an 11-column board and marbles numbered 1 to 33 it is not possible to find a winning placement.\n\nExplain the problem in as much detail as you can.\n\n2. Originally Posted by MrFantasy\nProblem 5\n\nTRIPOLLO\n\nTRIPOLLO is a strategy game in which numbered marbles are placed in a rectangular array of holes on a board. As shown, there are three rows of holes. A player wins if the marbles are placed so that in each column, the sum of the numbers in the first two rows is three times the number in the third row.\n\n(a) John has a 5-column board and marbles numbered 1 to 15.\nSo far he has placed them as shown in the diagram.\n\nShow how he could successfully place the remaining marbles.\n\n(b) With the same board and marbles. John made the following start:\n\nCan he successfully complete this game? Give reasons.\n\n(c) You have an 8-column board and marbles numbered 1 to 24. Show how you can win by placing all the marbles on the board.\n\n(d) Show that with an 11-column board and marbles numbered 1 to 33 it is not possible to find a winning placement.\n\nExplain the problem in as much detail as you can.\nIs this a Mathematics Challenge question? If so, it is not approporiate to ask this question until the competition is ended.\n\n3. Hello Mr.Fantastic\nNo, this is not a question from a maths competition. it's 1 of the 6 questions I picked out from the monthly issued maths challenge. My previous posts will explain the purpose of these 6 questions. It is to provide maths lovers some challenging questions to do. Please refer to my previous post for more information.\n\nhttp://www.mathhelpforum.com/math-he...intrested.html\n\nMrFantasy\n\n4. Originally Posted by MrFantasy\nHello Mr.Fantastic\nNo, this is not a question from a maths competition. it's 1 of the 6 questions I picked out from the monthly issued maths challenge. My previous posts will explain the purpose of these 6 questions. It is to provide maths lovers some challenging questions to do. Please refer to my previous post for more information.\n\nhttp://www.mathhelpforum.com/math-he...intrested.html\n\nMrFantasy\nYes, I've seen that thread. I just find it odd that the same questions are being posted by other members too.\n\n5. It may very well be that this question/problem is quiet wide known and is used in different occasions, whoever posted the same problem as I have posted here, could be another occasion. And please, if you dont mind. We're taking too much space talking in here, it would be difficult for members to read the answer replies. If there is any questions or replies or issues you want to make please contact me through PM. As it is more easier that way.\n\nThank You,\nMr`Fantasy\n\n6. Originally Posted by mr fantastic\nIs this a Mathematics Challenge question? If so, it is not approporiate to ask this question until the competition is ended.\nThis question is NOT from the Australian Maths Challenge Competition. The member has posted it in good faith. Please feel free to reply.\n\nApologies for any embarassment caused - I hope you understand that all members need to be vigilant for the sort of cheating that does sometimes occur.", "date": "2014-10-01 11:01:40", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/37930-problem-5-a.html", "openwebmath_score": 0.47497695684432983, "openwebmath_perplexity": 577.8641159615697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9802808684361289, "lm_q2_score": 0.8976952866333483, "lm_q1q2_score": 0.8799935151719583}}
{"url": "http://mathematica.stackexchange.com/questions/9304/plotting-contours-plot-for-fx-y-z-c/9310", "text": "# Plotting contours plot for f(x,y,z)=c\n\nI have the following question: I have a file that has structure:\n\nx1 y1 z1 f1\nx2 y2 z2 f2\n...\nxn yn zn fn\n\n\nI can easily visualize it with Mathematica using ListContourPlot3D. But could you please tell me how I can plot contour plot for this surface? I mean with these data I have a set of surfaces corresponding to different isovalues (f) and I want to plot intersection between all these surfaces and some certain plane. I tried to Google but didn't get any results. Any help and suggestions are really appreciated. Thanks in advance!\n\n-\nWould you please include a sample of your data, or provide a function that generates data that can serve as an example? \u2013\u00a0 Mr.Wizard Aug 10 '12 at 21:47\nAlso, do you want a 2D contour plot, or a 3D curve through space? \u2013\u00a0 Mr.Wizard Aug 10 '12 at 22:00\nThe answers to this question might be useful to you. \u2013\u00a0 \uff2a. \uff2d. Aug 11 '12 at 3:53\nThanks for all your replies! Sorry for the delay, I had problems with an internet acces. But the problem is when I'm trying to use suggestions by halirutan I get empty box (and I need exactly this plot). I'd love to provide the sample of my data but it's big, it has 2744 lines and I don't know how to upload this file here (any suggestions?). Thanks! PS Original questions was asked by me. I do not know how to comment other answers (I don't see button add commment) \u2013\u00a0 Nikita Aug 13 '12 at 20:50\nAfter merging your accounts you should now be able to comment. \u2013\u00a0 Sjoerd C. de Vries Aug 13 '12 at 22:45\nshow 1 more comment\n\nOk, lets give this a try. @Mr.Wizard already showed you, how you can use Interpolate to make a function from your discrete data and since you didn't provide some test-data, I just assume we are speaking of an isosurface of a function $f(x,y,z)=c$ which is defined in some box in $\\mathbb{R}^3$.\n\nFor testing we use $$f(x,y,z) = x^3+y^2-z^2\\;\\;\\mathrm{and}\\;\\; -2\\leq x,y,z \\leq 2$$ which accidently happens to be the first example of ContourPlot3D.\n\nThe idea behind the following is pretty easy: As you may know from school, there is a simple representation of a plane in 3d which uses a point vector $p_0$ and two direction vectors $v_1$ and $v_2$. Every point on this plane can be reached through the $(s,t)$ parametrization\n\n$$p(s,t)=p_0+s\\cdot v_1+t\\cdot v_2$$\n\nPlease note that $p_0, p, v_1, v_2$ are vectors in 3d and $s,t$ are scalars. The other form which we will use only for illustration is called the normal form of a plane. It is give by\n\n$$n\\cdot (p-p_0)=0$$\n\nwhere $n$ is the vector normal to the plane, which can easily be calculated with the cross-product by $v_1\\times v_2$. Lets start by looking at our example. To draw the plane inside ContourPlot3D we use the normal form which is plane2:\n\nf[{x_, y_, z_}] := x^3 + y^2 - z^2;\nv1 = {1, 1, 0};\nv2 = {0, 0, 1};\np0 = {0, 0, 0};\nplane1 = p0 + s*v1 + t*v2;\nplane2 = Cross[v1, v2].({x, y, z} - p0);\ngr3d = ContourPlot3D[{f[{x, y, z}], plane2}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},\nContours -> {0},\nContourStyle -> {ColorData[22, 3], Directive[Opacity[0.5], ColorData[22, 4]]}]\n\n\nWhat we do now is, that we try to find the contour value (which is 0 here) of $f(x,y,z)$ for all points, that lie on our plane. This is like doing a normal ContourPlot because our plane is 2d (although placed in 3d space). Therefore, we use the 2d to 3d mapping of plane1\n\ngr2d = ContourPlot[f[plane1], {s, -2, 2}, {t, -2, 2}, Contours -> {0},\nContourShading -> None, ContourStyle -> {ColorData[22, 1], Thick}]\n\n\nLook at the intersection. It is exactly the loop we would have expected from the 3d illustration. Now you could object something like \"ew.. but I really like a curve in 3d..\". Again, the mapping from this 2d curve to 3d is given in the plane equation. You can simply extract the Line[..] directives from the above plot and transfer it back to 3d:\n\nShow[{gr3d,\nGraphics3D[{Red, Cases[Normal[gr2d], Line[__], Infinity] /.\nLine[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}]\n}]\n\n\nI extract the Lines with Cases and then use the exact same definition of plane1 as pure function to transform the pts.\n\nWhen I'm not completely wasted at 5:41 in the morning, than this approach should work for your interpolated data too.\n\n## Apply method on test-data\n\nI uploaded your test-data to our Git-repository and therefore, the code below should work without downloading anything. The approach is the same as above but some small things have changed, since we work on interpolated data now. I'll explain only the differences.\n\nFirst we import the data and since we have a long list of {x,y,z,f} values, we transform them to {{x,y,z},f} as required by the Interpolation function. I'm not using the interpolation-function directly. I wrap a kind of protection around it which tests whether a given {x,y,z} is numeric and lies inside the interpolation box. Otherwise I just return 0.\n\ndata = {Most[#], Last[#]} & /@\nImport[\"https://raw.github.com/stackmma/Attachments/master/data_9304_187.m\"];\nip = Interpolation[data];\nfip[{x_?NumericQ, y_?NumericQ, z_?NumericQ}] :=\nIf[Apply[And, #2[[1]] < #1 < #2[[2]] & @@@\nTranspose[{{x, y, z}, First[ip]}]],\nip[x, y, z], 0.0]\n\n\nThe code below is almost the same. I only adapted the plane that it goes through your interpolation box. Furthermore, if you inspect your data you see that the value run from 0 to 1.2. Therefore I'm plotting the 0.5 contour by subtracting 0.5 from the function value and using Contours -> {0}. Remember that when I would simply plot the 0.5 contour, it would draw me a different plane, since we use one combined ContourPlot3D call.\n\nFurthermore, notice that I normalized the direction vectors of the plane. This makes it easier to adjust the 2d plot of the contour. The rest should be the same.\n\nv1 = Normalize[{30, 30, 0}];\nv2 = Normalize[{0, 0, 21}];\np0 = {26, 26, 17};\nplane1 = p0 + s*v1 + t*v2;\nplane2 = Cross[v1, v2].({x, y, z} - p0);\ngr3d = ContourPlot3D[{fip[{x, y, z}] - 0.5, plane2}, {x, 27, 30}, {y,\n27, 30}, {z, 17.3, 21}, Contours -> {0},\nContourStyle -> {Directive[Opacity[.5], ColorData[22, 3]],\nDirective[Opacity[.8], ColorData[22, 5]]}]\n\ngr2d = ContourPlot[fip[plane1] - 0.5, {s, 2, 5}, {t, 1, 4},\nContours -> {0}, ContourShading -> None,\nContourStyle -> {ColorData[22, 1], Thick}];\nShow[{gr3d,\nGraphics3D[{Red,\nCases[Normal[gr2d], Line[__], Infinity] /.\nLine[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}]}]\n\n\nAs you can see, your sphere has a whole inside.\n\n-\nThanks for all your replies! Sorry for the delay, I had problems with an internet acces. But the problem is when I'm trying to use suggestions by halirutan I get empty box (and I need exactly this plot). I'd love to provide the sample of my data but it's big, it has 2744 lines and I don't know how to upload this file here (any suggestions?). Thanks! PS Original questions was asked by me. I do not know how to comment other answers (I don't see button add commment) \u2013\u00a0 Nikita Aug 13 '12 at 20:50\nthanks for solving account issue! So my question if it is possible to load file on this forum or how I can upload my data which are quite big (app. 3000 lines) \u2013\u00a0 Nikita Aug 13 '12 at 22:51\n@Nikita I asked in the chat and there seems to be no standard-upload place. Oleksandr suggested ge.tt. Before uploading a very large file, please read the FAQ and remember that the questions on this site should be helpful for everyone and should not be too localized! \u2013\u00a0 halirutan Aug 14 '12 at 1:14\nhere is a link (depositfiles.com/files/r0pckyaiu) to zip file that contains 2 absolutely identical data (one is in format: x y z f, and the other is in the same format but can be directly load to mathematica). Yes, I understand that questions shouldn't be too localized, but the data I'm working on are cube files and it can be useful for many people. Thanks! \u2013\u00a0 Nikita Aug 14 '12 at 14:31\n@Nikita Here we go. See my update. \u2013\u00a0 halirutan Aug 14 '12 at 15:56\nshow 1 more comment\n\nYou can use the options MeshFunctions in combination with Mesh for this.\n\nI'm borrowing Mr.Wizard's data here for a moment:\n\ndata = Flatten[Table[{x, y, z, x^2 + y^2 - z^2}, {x, -2, 2, 0.2},\n{y, -2, 2, 0.2}, {z, -2, 2, 0.2}], 2];\n\n\nSuppose you want to plot the intersection of the contours of data with the plane x - y == 0, then you could do something like\n\nListContourPlot3D[data, Contours -> {0.5, 2},\nContourStyle -> Opacity[0.3],\nBoundaryStyle -> Opacity[0.3],\nMeshFunctions -> {(#1 - #2) &}, Mesh -> {{0}},\nMeshStyle -> {Thick, Orange}]\n\n\n-\nYou removed my glorious infix! +1 for the method however. (It's nice to see you posting again.) \u2013\u00a0 Mr.Wizard Aug 11 '12 at 7:00\nthanks for your reply! These tricks with meshfunctions are brilliant, but when I use ListContourPlot3D on my data (which are exactly in the same format) I have empty box. But when I apply ListContourPlot3D on data that have only f values (without x,y,z) Mathematica plots correct figures but I cannot use MeshFunction in this case (because there is no information about x,y,z values). Do you know what can solve this problem? \u2013\u00a0 Nikita Aug 13 '12 at 22:58\n\nI'm not claiming this is a good method, I'm just getting some ink the page:\n\ndata = Table[{x, y, z, x^2 + y^2 - z^2}, {x, -2, 2, 0.2}, {y, -2, 2, 0.2}, {z, -2, 2, 0.2}] ~Flatten~ 2;\n\nListContourPlot3D[data, Contours -> {0.5, 2}, Mesh -> None]\n\n\nint = Interpolation[data];\n\nContourPlot3D[int[x, y, z], {x, -2, 2}, {y, -2, 2}, {z, -2, 2},\nContours -> {0.5, 2}, RegionFunction -> (-0.02 < #2 - # < 0.02 &)]\n\n\n-\nThe second piece of codes using RegionFunction doesn't creat a pretty plot,which contains two outer curves with small intervals, and neither PlotPoints nor MaxRecursion could help to improve it. \u2013\u00a0 withparadox2 Aug 11 '12 at 7:12\n@paradox2 hence: \"I'm not claiming this is a good method ...\" \u2013\u00a0 Mr.Wizard Aug 11 '12 at 7:13", "date": "2014-03-10 12:25:49", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/9304/plotting-contours-plot-for-fx-y-z-c/9310", "openwebmath_score": 0.3507079482078552, "openwebmath_perplexity": 1528.9516500084083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126531738088, "lm_q2_score": 0.8991213765941599, "lm_q1q2_score": 0.8799814680117576}}
{"url": "https://mathoverflow.net/questions/283427/are-random-circulant-matrices-almost-orthonormal", "text": "# Are random circulant matrices almost orthonormal?\n\nLet $(X_1, X_2, \\dots, X_n)$ be i.i.d. ${\\cal N}(0,1)$. We construct a random circulant matrix $M$:\n\n$$M = \\frac{1}{\\sqrt n}\\begin{pmatrix}X_1 &X_2 &X_3 \\dots &X_n\\\\ X_n &X_1 & X_2 \\dots &X_{n-1}\\\\ \\vdots &\\vdots &\\vdots &\\vdots\\\\X_2 &X_3 &X_4 \\dots &X_1\\end{pmatrix}.$$\n\nMy questions are the following:\n\n1. Is $M$ \"almost orthonormal\" in a precise probabilistic sense?\n\n2. Related to above, is it possible to upper-bound the largest possible dot product between any two rows of $M$ by a suitably small $\\epsilon_n$?\n\nNote that this says that $MM^T$ is close to the identity matrix $I_{n \\times n}$, as we are bounding the off-diagonal entries of $MM^T$ by $\\epsilon_n$, while the diagonal entries are close to $1$.\n\n\u2022 What do you mean by \"upper-bound the largest possible dot product\"? All of the $X_i$ could be equal to $10^{10^{10}}.$ You can upper bound it with high probability, but that's the best you can do. \u2013\u00a0Igor Rivin Oct 14 '17 at 0:05\n\u2022 If $D$ is the absolute value of the largest dot product, one could seek a bound of the sort $\\mathbb P(D > \\epsilon_n) < \\epsilon_n$ for a suitable choice of $\\epsilon_n \\to 0$. \u2013\u00a0VSJ Oct 14 '17 at 2:25\n\u2022 Yes, that follows from Marcel's answer, using, e.g., Markoff's inequality. \u2013\u00a0Igor Rivin Oct 14 '17 at 3:24\n\nThe diagonal elements of $P=\\frac{1}{N}MM^T$, like $$P_{11}=\\frac{1}{N}\\sum_{i=1}^NX_i^2,$$ satisfy $\\langle P_{11}\\rangle=1$ and $\\langle P_{11}^2\\rangle=1+2/N$ (variance decreases like $N^{-1}$).\n\nOn the other hand, off-diagonal elements like $$P_{12}=\\frac{1}{N}\\sum_{i=1}^{N}X_iX_{i+1}$$ satisfy $\\langle P_{12}\\rangle=0$ and $\\langle P_{12}^2\\rangle=1/N$ (variance also decreases like $N^{-1}$).\n\nIn this sense I would say $P$ is close to the identity.\n\nMarcel's solution provides a good approach for understanding the marginal statistics. Here are a few supplementary comments that might give a little insight into the joint distribution.\n\nLet $D$ be the discrete Fourier transform matrix, i.e. the $j,k$-th entry is: $$D_{j,k}=e^{-2\\pi i jk/N}/\\sqrt{N}$$ Consider the discrete Fourier transform of the first row of $M$, i.e. $$(G_1,...,G_N)=(1/\\sqrt{N})(X_1,...,X_N)D$$ Let $G$ be a diagonal matrix with diagonal entries $G_1,...,G_N$. Then $D$ diagonalizes $M$ (see, e.g., this description), that is: $$M = D G D^{-1}$$\n\nIf we took the $X_i$ to be complex-valued Gaussian variables, then we would be essentially done at this point: since $D$ is unitary, and the Gaussian is spherically symmetric, then the $G_i$ would be i.i.d. (complex) Gaussian random variables with mean 0 and variance $1/n$. (That is, if we sampled the $G_i$ as i.i.d complex $\\mathcal{N}(0,1/n)$, then $DGD^{-1}$ has the same distribution as samples from the original circulant matrix.) It follows that $$MM^*=(DGD^{-1})(DG^*D^{-1})=DGG^{-1}D^*$$ It then follows that the eigenvalues of $MM^{*}$ (or, if you prefer, the squared singular values of $M$) are distributed like $n$ i.i.d draws from a rescaled $\\chi^2$ distribution with 2 degrees of freedom (which happens to simplify to an exponential distribution), where we rescale by dividing by $n$. From an eigenvalue perspective, this is a complete characterization of the \"orthonormality\" of $M$.\n\nHowever, your $X_i$ are probably real-valued. This causes a small book-keeping headache, but doesn't really change much. In brief: since the $X_i$ are real-valued, the $G_i$ will be symmetric. We can convert our $n\\times n$ complex-valued matrices into corresponding $2n \\times 2n$ real-valued matrices (say, $D'$ and $G'$). Then observe that there are only $n$ of the $2n$ diagonal elements in $G$ are free, and only $n$ of the $2n$ inputs to $D'$ are non-zero (namely, the ones corresponding to the real values of the inputs in $D$). Noting that the resulting decimated matrices are still unitary, we then reduce to the previous case, except with $\\chi^2/n$ distributions with only 1 degree of freedom.\n\n\u2022 I'm confused about one part: Marcel's answer indicates that $MM^*$ is close to identity. Thus it should have eigenvalues close to 1? But your argument says eigenvalues are iid from a chi square distribution so they aren't all close to 1. How to reconcile these? \u2013\u00a0VSJ Nov 2 '17 at 4:04\n\u2022 Ah, @VSJ , that is my mistake: I forgot to pass through the prefactor of $1/\\sqrt(n)$ in the original definition of $M$. I will fix that now, at which point I think Marcel's answer is consistent with mine. \u2013\u00a0Bill Bradley Nov 3 '17 at 13:31\n\u2022 Hi Bill, I'm not sure that is a mistake. Because even then, the eigenvalues will converge to 0 and not 1. Perhaps it is the case that the matrix is tending \"pointwise\" to identity, but it is not tending fast enough, and therefore the eigenvalues don't converge? \u2013\u00a0VSJ Nov 3 '17 at 17:52", "date": "2019-03-21 16:38:42", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/283427/are-random-circulant-matrices-almost-orthonormal", "openwebmath_score": 0.9444407820701599, "openwebmath_perplexity": 226.3218243780155, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985042916041382, "lm_q2_score": 0.8933094046341532, "lm_q1q2_score": 0.8799481008680171}}
{"url": "https://math.stackexchange.com/questions/1496708/another-way-of-counting-probability", "text": "# Another way of counting probability\n\nA set $S = \\{1, 2, \\cdots, k\\}$ is given. Two persons independently choose some numbers from this set. I want to count the probability that the cardinality of intersection of the chosen sets by both persons is exactly one.\n\nEach person has to choose at least one number. A person can choose a number which is already chosen by another person. A number is being chosen independently and uniformly at random from the set.\n\nMy approach: Fix one number from $1$ to $k$. Probability of both persons choosing that number is $\\frac{1}{k^2}$. Remaining $k-1$ numbers can be chosen by either of both or by none of both. So, probability for that is $\\left(1 - \\frac{1}{k^2}\\right)^{k-1}$. And that fixed number can be selected in ${k \\choose 1}$ ways.\n\nSo, probability of required event is:\n\n\\begin{align} P(E) = {k \\choose 1} \\frac{1}{k^2} \\left(1 - \\frac{1}{k^2}\\right)^{k-1} \\end{align}\n\nMy questions:\n\n1. Is this reasoning correct?\n\n2. Initially I started with counting all the possible ways of choosing numbers by each person. Can we count the required probability by counting ways technique? Is there any other way of counting the probability?\n\n\u2022 Why do you say that the probability of both people picking the fixed number is $\\frac{1}{k^2}$? \u2013\u00a0paw88789 Oct 25 '15 at 12:57\n\u2022 I think you need to say something about the number of elements each player selects. There are $2^k-1$ non-empty subsets of $\\{1,....,k\\}$. Do you mean to say that each player can select any one of these with equal probability? \u2013\u00a0lulu Oct 25 '15 at 12:57\n\u2022 @lulu Yes. That is what I mean. \u2013\u00a0Nimit Oct 25 '15 at 13:07\n\u2022 Ok, but then the situation is more complex than your method suggests. You have to take into account the fact that it is more likely that a randomly chosen subset has about $\\frac k2$ elements than, say, $1$ or $k$. \u2013\u00a0lulu Oct 25 '15 at 13:09\n\u2022 @paw88789 I got your point. I smell something wrong with my reasoning now. I had assumed that each person is choosing that fixed number very first. But that is not the case. It can be chosen later also. \u2013\u00a0Nimit Oct 25 '15 at 13:16\n\nLet us at first ignore the condition that at least one number has to be chosen (that will get fixed afterwards). Under that condition, as each choice of each possible subset is equally likely (as you say in the comments), and for any given element there are exactly the same number of subsets containing and not containing that element (think of a subset and its complement!), the probability that a person chooses a given number is exactly $1/2$. Thus it's the same problem as if both people throw a coin for each number, and the question is what's the probability that they throw both head for exactly one of the numbers.\n\nThis is a fairly standard problem which I guess you can solve. Let's call the obtained probability $p_0$.\n\nNow we have to fix this up for the condition that the empty set is not a valid choice. Fortunately this is easy, since when any party chose the empty set, we know for sure there was no number that was chosen by both (because at least one of them didn't chose a number to begin with).\n\nSince the probability calculated in the previous step is $$\\frac{\\text{Number of choices with exactly one shared number}} {\\text{Total number of choices when allowing empty sets}}$$ we have to just fix up the denominator, which we can do by multiplying with $$\\frac{\\text{Total number of choices when allowing empty sets}} {\\text{Total number of choices when not allowing empty sets}}.$$ Now the total number of choices when allowing empty sets is $2^{2k}$ since there are $2^k$ subsets, and each person chooses one. When excluding the empty set, each player has one choice less (namely he cannot choose the empty set), and therefore without the empty set, the number of possible choices is $(2^k-1)^2$.\n\nTherefore we have for the requested probability: $$p = \\frac{2^{2k}}{(2^k-1)^2}p_0$$\n\n\u2022 To get an exact answer, the requirement of nonempty set choice by each person complicates the problem. For instance, players can't select/not select items by tossing a coin because that would allow for the empty set being chosen. And then consequently we can't simply say that for any given item that the probability of both selecting it is $\\frac14$. I think you need to know more about how numbers are chosen. \u2013\u00a0paw88789 Oct 25 '15 at 13:45\n\u2022 @paw88789: Ah right, I overlooked that condition. But the claim that all possible sets are chosen with equal probability should still be sufficient. I'll rework the answer. \u2013\u00a0celtschk Oct 25 '15 at 13:50\n\u2022 @paw88789: See edited answer. \u2013\u00a0celtschk Oct 25 '15 at 14:05", "date": "2020-04-01 16:04:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1496708/another-way-of-counting-probability", "openwebmath_score": 0.8959659934043884, "openwebmath_perplexity": 233.66643522101592, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429134068501, "lm_q2_score": 0.8933094053442511, "lm_q1q2_score": 0.8799480992140419}}
{"url": "https://thebearing.net/dybqf/archive.php?a86677=truth-tables-explained", "text": "truth tables explained\n\nAbstract: The general principles for the construction of truth tables are explained and illustrated. This truth-table calculator for classical logic shows, well, truth-tables for propositions of classical logic. Hence, (b\u00e2\u0086\u0092e)\u00e2\u0088\u00a7(b\u00e2\u0086\u0092\u00c2\u00ace)=(\u00c2\u00acb\u00e2\u0088\u00a8e)\u00e2\u0088\u00a7(\u00c2\u00acb\u00e2\u0088\u00a8\u00c2\u00ace)=\u00c2\u00acb\u00e2\u0088\u00a8(e\u00e2\u0088\u00a7\u00c2\u00ace)=\u00c2\u00acb\u00e2\u0088\u00a8C=\u00c2\u00acb,(b \\rightarrow e) \\wedge (b \\rightarrow \\neg e) = (\\neg b \\vee e) \\wedge (\\neg b \\vee \\neg e) = \\neg b \\vee (e \\wedge \\neg e) = \\neg b \\vee C = \\neg b,(b\u00e2\u0086\u0092e)\u00e2\u0088\u00a7(b\u00e2\u0086\u0092\u00c2\u00ace)=(\u00c2\u00acb\u00e2\u0088\u00a8e)\u00e2\u0088\u00a7(\u00c2\u00acb\u00e2\u0088\u00a8\u00c2\u00ace)=\u00c2\u00acb\u00e2\u0088\u00a8(e\u00e2\u0088\u00a7\u00c2\u00ace)=\u00c2\u00acb\u00e2\u0088\u00a8C=\u00c2\u00acb, where CCC denotes a contradiction. Logical implication (symbolically: p \u00e2\u0086\u0092 q), also known as \u00e2\u0080\u009cif-then\u00e2\u0080\u009d, results True in all cases except the case T \u00e2\u0086\u0092 F. Since this can be a little tricky to remember, it can be helpful to note that this is logically equivalent to \u00c2\u00acp \u00e2\u0088\u00a8 q (read: not p or q)*. The table contains every possible scenario and the truth values that would occur. Also known as the biconditional or if and only if (symbolically: \u00e2\u0086\u0090\u00e2\u0086\u0092), logical equality is the conjunction (p \u00e2\u0086\u0092 q) \u00e2\u0088\u00a7 (q \u00e2\u0086\u0092 p). Featuring a purple munster and a duck, and optionally showing intermediate results, it is one of the better instances of its kind. {\\color{#3D99F6} \\textbf{p}} &&{\\color{#3D99F6} \\textbf{q}} &&{\\color{#3D99F6} p \\equiv q} \\\\ This primer will equip you with the knowledge you need to understand symbolic logic. \\text{0} &&\\text{0} &&0 \\\\ Below is the truth table for p, q, p\u00e2\u00e0\u00e7q, p\u00e2\u00e0\u00e8q. From statement 4, g\u00e2\u0086\u0092\u00c2\u00aceg \\rightarrow \\neg eg\u00e2\u0086\u0092\u00c2\u00ace, so by modus tollens, e=\u00c2\u00ac(\u00c2\u00ace)\u00e2\u0086\u0092\u00c2\u00acge = \\neg(\\neg e) \\rightarrow \\neg ge=\u00c2\u00ac(\u00c2\u00ace)\u00e2\u0086\u0092\u00c2\u00acg. Check out my YouTube channel \u00e2\u0080\u009cMath Hacks\u00e2\u0080\u009d for hands-on math tutorials and lots of math love \u00e2\u0099\u00a5\u00ef\u00b8\u008f, Medium is an open platform where 170 million readers come to find insightful and dynamic thinking. ||row 2 col 1||row 2 col 2||row 2 col 1||row 2 col 2||. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. \\text{T} &&\\text{F} &&\\text{F} \\\\ If Darius is not the oldest, then he is immediately younger than Charles. The truth table for the XOR gate OUT =A\u00e2\u008a\u0095B= A \\oplus B=A\u00e2\u008a\u0095B is given as follows: ABOUT000011101110 \\begin{aligned} Go: Should I Use a Pointer instead of a Copy of my Struct? A table will help keep track of all the truth values of the simple statements that make up a complex statement, leading to an analysis of the full statement. Whats people lookup in this blog: Truth Tables Explained; Truth Tables Explained Khan Academy; Truth Tables Explained Computer Science Since g\u00e2\u0086\u0092\u00c2\u00aceg \\rightarrow \\neg eg\u00e2\u0086\u0092\u00c2\u00ace (statement 4), b\u00e2\u0086\u0092\u00c2\u00aceb \\rightarrow \\neg eb\u00e2\u0086\u0092\u00c2\u00ace by transitivity. This is logically the same as the intersection of two sets in a Venn Diagram. Basic Logic Gates With Truth Tables Digital Circuits Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. Truth tables \u2013 the conditional and the biconditional (\u201cimplies\u201d and \u201ciff\u201d) Just about every theorem in mathematics takes on the form \u201cif, then\u201d (the conditional) or \u201ciff\u201d (short for if and only if \u2013 the biconditional). Whats people lookup in this blog: Logic Truth Tables Explained; Logical Implication Truth Table Explained Mr. and Mrs. Tan have five children--Alfred, Brenda, Charles, Darius, Eric--who are assumed to be of different ages. Remember to result in True for the OR operator, all you need is one True value. Already have an account? Using truth tables you can figure out how the truth values of more complex statements, such as. First you need to learn the basic truth tables for the following logic gates: AND Gate OR Gate XOR Gate NOT Gate First you will need to learn the shapes/symbols used to draw the four main logic gates: Logic Gate Truth Table Your Task Your task is to complete the truth tables for \u00e2\u0080\u00a6 Surprisingly, this handful of definitions will cover the majority of logic problems you\u00e2\u0080\u0099ll come across. A truth table is a mathematical table used in logic\u2014specifically in connection with Boolean algebra, boolean functions, and propositional calculus\u2014to compute the functional values of logical expressions on each of their functional arguments, that is, on each combination of values taken by their logical variables (Enderton, 2001). college math section 3.2: truth tables for negation, conjunction, and disjunction It negates, or switches, something\u00e2\u0080\u0099s truth value. Using this simple system we can boil down complex statements into digestible logical formulas. Here ppp is called the antecedent, and qqq the consequent. It is simplest but not always best to solve these by breaking them down into small componentized truth tables. We can show this relationship in a truth table. \\end{aligned} A0011\u00e2\u0080\u008b\u00e2\u0080\u008bB0101\u00e2\u0080\u008b\u00e2\u0080\u008bOUT0110\u00e2\u0080\u008b, ALWAYS REMEMBER THE GOLDEN RULE: \"And before or\". When combining arguments, the truth tables follow the same patterns. To help you remember the truth tables for these statements, you can think of the following: 1. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. P AND (Q OR NOT R) depend on the truth values of its components. The truth table of an XOR gate is given below: The above truth table\u2019s binary operation is known as exclusive OR operation. Basic Logic Gates, Truth Tables, and Functions Explained OR Gate. In mathematics, \"if and only if\" is often shortened to \"iff\" and the statement above can be written as. (Or \"I only run on Saturdays. understanding truth tables Since any truth-functional proposition changes its value as the variables change, we should get some idea of what happens when we change these values systematically. Truth Table: A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. How to Construct a Truth Table. Two rows with a false conclusion. Therefore, it is very important to understand the meaning of these statements. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as \"logical operators\") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables in algebra. Since c\u00e2\u0086\u0092dc \\rightarrow dc\u00e2\u0086\u0092d from statement 2, by modus tollens, \u00c2\u00acd\u00e2\u0086\u0092\u00c2\u00acc\\neg d \\rightarrow \\neg c\u00c2\u00acd\u00e2\u0086\u0092\u00c2\u00acc. For a 2-input AND gate, the output Q is true if BOTH input A \u00e2\u0080\u009cAND\u00e2\u0080\u009d input B are both true, giving the Boolean Expression of: (\u00a0Q = A and B). The statement has the truth value F if both, If I go for a run, it will be a Saturday. A truth table is a logically-based mathematical table that illustrates the possible outcomes of a scenario. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. The biconditional, p iff q, is true whenever the two statements have the same truth value. These operations are often referred to as \u00e2\u0080\u009calways true\u00e2\u0080\u009d and \u00e2\u0080\u009calways false\u00e2\u0080\u009d. is true or whether an argument is valid.. Truth tables get a little more complicated when conjunctions and disjunctions of statements are included. We use the symbol \u00e2\u0088\u00a7\\wedge \u00e2\u0088\u00a7 to denote the conjunction. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. From statement 4, g\u00e2\u0086\u0092\u00c2\u00aceg \\rightarrow \\neg eg\u00e2\u0086\u0092\u00c2\u00ace, where \u00c2\u00ace\\neg e\u00c2\u00ace denotes the negation of eee. Truth tables show the values, relationships, and the results of performing logical operations on logical expressions. One of the simplest truth tables records the truth values for a statement and its negation. a) Negation of a conjunction To do this, write the p and q columns as usual. Solution The truth tables are given in Table 4.2.Note that there are eight lines in the truth table in order to represent all the possible states (T, F) for the three variables p, q, and r. As each can be either TRUE or FALSE, in total there are 2 3 = 8 possibilities. Nor Gate Universal Truth Table Symbol You Partial and complete truth tables describing the procedures truth table tutorial discrete mathematics logic you truth table you propositional logic truth table boolean algebra dyclassroom. \"). With just these two propositions, we have four possible scenarios. \\text{0} &&\\text{1} &&1 \\\\ Here, expert and undiscovered voices alike dive into the heart of any topic and bring new ideas to the surface. Complex, compound statements can be composed of simple statements linked together with logical connectives (also known as \"logical operators\") similarly to how algebraic operators like addition and subtraction are used in combination with numbers and variables \u00e2\u0080\u00a6 To find (p \u00e2\u0088\u00a7 q) \u00e2\u0088\u00a7 r, p \u00e2\u0088\u00a7 q is performed first and the result of that is ANDed with r. Hence Eric is the youngest. When either of the inputs is a logic 1 the output is... AND Gate. Log in here. They are considered common logical connectives because they are very popular, useful and always taught together. Hence Charles is the oldest. These are kinda strange operations. The AND gate is a digital logic gatewith \u00e2\u0080\u0098n\u00e2\u0080\u0099 i/ps\u00a0one o/p, which perform logical conjunction based on the combinations of its inputs.The output of this gate is true only when all the inputs are true. Note that by pure logic, \u00c2\u00aca\u00e2\u0086\u0092e\\neg a \\rightarrow e\u00c2\u00aca\u00e2\u0086\u0092e, where Charles being the oldest means Darius cannot be the oldest. Unary operators are the simplest operations because they can be applied to a single True or False value. Truth tables are a tool developed by Charles Pierce in the 1880s.Truth tables are used in logic to determine whether an expression[?] 2. This combines both of the following: These are consistent only when the two statements \"I go for a run today\" and \"It is Saturday\" are both true or both false, as indicated by the above table. Learning Objectives In this post you will predict the output of logic gates circuits by completing truth tables. \u00e2\u0096\u00a1_\\square\u00e2\u0096\u00a1\u00e2\u0080\u008b. A truth table is a table whose columns are statements, and whose rows are possible scenarios. Truth tables really become useful when analyzing more complex Boolean statements. In the next post I\u00e2\u0080\u0099ll show you how to use these definitions to generate a truth table for a logical statement such as (A \u00e2\u0088\u00a7 ~B) \u00e2\u0086\u0092 (C \u00e2\u0088\u00a8 D). Otherwise it is false. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. It states that True is True and False is False. The AND operator (symbolically: \u00e2\u0088\u00a7) also known as logical conjunction requires both p and q to be True for the result to be True. In the second column we apply the operator to p, in this case it\u00e2\u0080\u0099s ~p (read: not p). A truth table is a mathematical table used to determine if a compound statement is true or false. This is equivalent to the union of two sets in a Venn Diagram. (p\u00e2\u0086\u0092q)\u00e2\u0088\u00a7(q\u00e2\u0088\u00a8p)(p \\rightarrow q ) \\wedge (q \\vee p)(p\u00e2\u0086\u0092q)\u00e2\u0088\u00a7(q\u00e2\u0088\u00a8p), p \\rightarrow q The truth table contains the truth values that would occur under the premises of a given scenario. The only possible conclusion is \u00c2\u00acb\\neg b\u00c2\u00acb, where Alfred isn't the oldest. Create a truth table for the statement $A\\wedge\\sim\\left(B\\vee{C}\\right)$ Show Solution , \u00e2\u008b\u0081 Try It. It is a mathematical table that shows all possible outcomes that would occur from all possible scenarios that are considered factual, hence the name. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Log in. For example, if there are three variables, A, B, and C, then the truth table with have 8 rows: Two simple statements can be converted by the word \"and\" to form a compound statement called the conjunction of the original statements. Then add a \u00e2\u0080\u009c\u00c2\u00acp\u00e2\u0080\u009d column with the opposite truth values of p. Lastly, compute \u00c2\u00acp \u00e2\u0088\u00a8 q by OR-ing the second and third columns. b) Negation of a disjunction Explore, If you have a story to tell, knowledge to share, or a perspective to offer \u00e2\u0080\u0094 welcome home. Truth tables list the output of a particular digital logic circuit for all the possible combinations of its inputs. \\text{T} &&\\text{T} &&\\text{T} \\\\ As a result, the table helps visualize whether an argument is logical (true) in the scenario. If ppp and qqq are two simple statements, then p\u00e2\u0088\u00a7qp \\wedge qp\u00e2\u0088\u00a7q denotes the conjunction of ppp and qqq and it is read as \"ppp and qqq.\" Partial and complete truth tables describing the procedures truth table for the biconditional statement you truth table definition rules examples lesson logic gates truth tables explained not and nand or nor. \u00e2\u0096\u00a1_\\square\u00e2\u0096\u00a1\u00e2\u0080\u008b. \\hspace{1cm}The negation of a conjunction p\u00e2\u0088\u00a7qp \\wedge qp\u00e2\u0088\u00a7q is the disjunction of the negation of ppp and the negation of q:q:q: \u00c2\u00ac(p\u00e2\u0088\u00a7q)=\u00c2\u00acp\u00e2\u0088\u00a8\u00c2\u00acq.\\neg (p \\wedge q) = {\\neg p} \\vee {\\neg q}.\u00c2\u00ac(p\u00e2\u0088\u00a7q)=\u00c2\u00acp\u00e2\u0088\u00a8\u00c2\u00acq. \\end{aligned} pTTFF\u00e2\u0080\u008b\u00e2\u0080\u008bqTFTF\u00e2\u0080\u008b\u00e2\u0080\u008bp\u00e2\u0089\u00a1qTFFT\u00e2\u0080\u008b. We will call our first proposition p and our second proposition q. Binary operators require two propositions. Logical true always results in True and logical false always results in False no matter the premise. \\text{0} &&\\text{0} &&0 \\\\ \\text{1} &&\\text{1} &&1 \\\\ The truth table for biconditional logic is as follows: pqp\u00e2\u0089\u00a1qTTTTFFFTFFFT \\begin{aligned} The negation operator is commonly represented by a tilde (~) or \u00c2\u00ac symbol. There's now 4 parts to the tutorial with two extra example videos at the end. Sign up, Existing user? Conjunction (AND), disjunction (OR), negation (NOT), implication (IF...THEN), and biconditionals (IF AND ONLY IF), are all different types of connectives. We\u00e2\u0080\u0099ll use p and q as our sample propositions. \\hspace{1cm} The negation of a disjunction p\u00e2\u0088\u00a8qp \\vee qp\u00e2\u0088\u00a8q is the conjunction of the negation of ppp and the negation of q:q:q: \u00c2\u00ac(p\u00e2\u0088\u00a8q)=\u00c2\u00acp\u00e2\u0088\u00a7\u00c2\u00acq.\\neg (p \\vee q) ={\\neg p} \\wedge {\\neg q}.\u00c2\u00ac(p\u00e2\u0088\u00a8q)=\u00c2\u00acp\u00e2\u0088\u00a7\u00c2\u00acq. Using truth tables you can figure out how the truth values of more complex statements, such as. \\text{F} &&\\text{T} &&\\text{F} \\\\ Therefore, if there are NNN variables in a logical statement, there need to be 2N2^N2N rows in the truth table in order to list out all combinations of each variable being either true (T) or false (F). The symbol and truth table of an AND gate with two inputs is shown below. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. \u00e2\u0086\u0092 For more math tutorials, check out Math Hacks on YouTube! When one or more inputs of the AND gate\u00e2\u0080\u0099s i/ps are false, then only the output of the AND gate is false. New user? A few common examples are the following: For example, the truth table for the AND gate OUT = A & B is given as follows: ABOUT000010100111 \\begin{aligned} But if we have b,b,b, which means Alfred is the oldest, it follows logically that eee because Darius cannot be the oldest (only one person can be the oldest). Before we begin, I suggest that you review my other lesson in which the \u2026 Truth Tables of Five Common Logical Connectives \u2026 With fff, since Charles is the oldest, Darius must be the second oldest. It can be used to test the validity of arguments.Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. UNDERSTANDING TRUTH TABLES. Considering all the deductions in bold, the only possible order of birth is Charles, Darius, Brenda, Alfred, Eric. In an AND gate, both inputs have to be logic 1 for an output to be logic 1. If Eric is not the youngest, then Brenda is. Exclusive Or, or XOR for short, (symbolically: \u00e2\u008a\u00bb) requires exactly one True and one False value in order to result in True. Note that the Boolean Expression for a two input AND gate can be written as: A.B or just simply ABwithout the decimal point. \\hspace{1cm} The negation of a negation of a statement is the statement itself: \u00c2\u00ac(\u00c2\u00acp)\u00e2\u0089\u00a1p.\\neg (\\neg p) \\equiv p.\u00c2\u00ac(\u00c2\u00acp)\u00e2\u0089\u00a1p. ||p||row 1 col 2||q|| c) Negation of a negation The identity is our trivial case. Boolean Algebra is a branch of algebra that involves bools, or true and false values. In a truth table, each statement is typically represented by a letter or variable, like p, q, or r, and each statement also has its own corresponding column in the truth table that lists all of the possible truth values. We title the first column p for proposition. If Charles is not the oldest, then Alfred is. \u00e2\u0096\u00a1_\\square\u00e2\u0096\u00a1\u00e2\u0080\u008b, Biconditional logic is a way of connecting two statements, ppp and qqq, logically by saying, \"Statement ppp holds if and only if statement qqq holds.\" If it only takes one out of two things to be true, then condition_1 OR condition_2 must be true. We may not sketch out a truth table in our everyday lives, but we still use the logical reasoning t\u00e2\u0080\u00a6 These variables are \"independent\" in that each variable can be either true or false independently of the others, and a truth table is a chart of all of the possibilities. {\\color{#3D99F6} \\textbf{A}} &&{\\color{#3D99F6} \\textbf{B}} &&{\\color{#3D99F6} \\textbf{OUT}} \\\\ \u00e2\u0096\u00a1_\\square\u00e2\u0096\u00a1\u00e2\u0080\u008b. Such a table typically contains several rows and columns, with the top row representing the logical variables and combinations, in increasing complexity leading up to \u00e2\u0080\u00a6 \u00e2\u0086\u0090. All other cases result in False. Truth Table A table showing what the resulting truth value of a complex statement is for all the possible truth values for the simple statements. Otherwise it is true. Since ggg means Alfred is older than Brenda, \u00c2\u00acg\\neg g\u00c2\u00acg means Alfred is younger than Brenda since they can't be of the same age. \u00e2\u0096\u00a1_\\square\u00e2\u0096\u00a1\u00e2\u0080\u008b. Determine the order of birth of the five children given the above facts. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. \\text{1} &&\\text{0} &&0 \\\\ READ Barclays Center Seating Chart Jay Z. If Alfred is older than Brenda, then Darius is the oldest. {\\color{#3D99F6} \\textbf{A}} &&{\\color{#3D99F6} \\textbf{B}} &&{\\color{#3D99F6} \\textbf{OUT}} \\\\ Two statements, when connected by the connective phrase \"if... then,\" give a compound statement known as an implication or a conditional statement. It requires both p and q to be False to result in True. In other words, it\u00e2\u0080\u0099s an if-then statement where the converse is also true. From statement 3, e\u00e2\u0086\u0092fe \\rightarrow fe\u00e2\u0086\u0092f. A truth table is a breakdown of a logic function by listing all possible values the function can attain. The OR operator (symbolically: \u00e2\u0088\u00a8) requires only one premise to be True for the result to be True. Truth tables summarize how we combine two logical conditions based on AND, OR, and NOT. Stay up-to-date with everything Math Hacks is up to! The negation of a statement is generally formed by introducing the word \"no\" at some proper place in the statement or by prefixing the statement with \"it is not the case\" or \"it is false that.\" We\u00e2\u0080\u0099ll start with defining the common operators and in the next post, I\u00e2\u0080\u0099ll show you how to dissect a more complicated logic statement. Sign up to read all wikis and quizzes in math, science, and engineering topics. Truth table explained. Truth table, in logic, chart that shows the truth-value of one or more compound propositions for every possible combination of truth-values of the propositions making up the compound ones. From statement 1, a\u00e2\u0086\u0092ba \\rightarrow ba\u00e2\u0086\u0092b. The negation of statement ppp is denoted by \"\u00c2\u00acp.\\neg p.\u00c2\u00acp.\" Mathematics normally uses a two-valued logic: every statement is either true or false. A truth table is a visual tool, in the form of a diagram with rows & columns, that shows the truth or falsity of a compound premise. Logic tells us that if two things must be true in order to proceed them both condition_1 AND condition_2 must be true. We have filled in part of the truth table for our example below, and leave it up to you to fill in the rest. \\text{F} &&\\text{F} &&\\text{T} They\u00e2\u0080\u0099re typically denoted as T or 1 for true and F or 0 for false. \\text{1} &&\\text{0} &&1 \\\\ Translating this, we have b\u00e2\u0086\u0092eb \\rightarrow eb\u00e2\u0086\u0092e. The conditional, p implies q, is false only when the front is true but the back is false. Example. The truth table for the conjunction p\u00e2\u0088\u00a7qp \\wedge qp\u00e2\u0088\u00a7q of two simple statements ppp and qqq: Two simple statements can be converted by the word \"or\" to form a compound statement called the disjunction of the original statements. In the first case p is being negated, whereas in the second the resulting truth value of (p \u00e2\u0088\u00a8 q) is negated. Figure %: The truth table for p, \u00e2\u00e0\u00fcp Remember that a statement and its negation, by definition, always have opposite truth values. The notation may vary depending on what discipline you\u00e2\u0080\u0099re working in, but the basic concepts are the same. Write on Medium. Truth Tables of Five Common Logical Connectives or Operators In this lesson, we are going to construct the five (5) common logical connectives or operators. Truth Tables, Logic, and DeMorgan's Laws . Learn more, Follow the writers, publications, and topics that matter to you, and you\u00e2\u0080\u0099ll see them on your homepage and in your inbox. You don\u00e2\u0080\u0099t need to use [weak self] regularly, The Product Development Lifecycle Template Every Software Team Needs, Threads Used in Apache Geode Function Execution, Part 2: Dynamic Delivery in multi-module projects at Bumble. \\text{1} &&\\text{1} &&0 \\\\ Once again we will use aredbackground for something true and a blue background for somethingfalse. By adding a second proposition and including all the possible scenarios of the two propositions together, we create a truth table, a table showing the truth value for logic combinations. It\u00e2\u0080\u0099s a way of organizing information to list out all possible scenarios from the provided premises. The truth table for the implication p\u00e2\u0087\u0092qp \\Rightarrow qp\u00e2\u0087\u0092q of two simple statements ppp and q:q:q: That is, p\u00e2\u0087\u0092qp \\Rightarrow qp\u00e2\u0087\u0092q is false \u00e2\u0080\u0085\u00e2\u0080\u008a\u00e2\u009f\u00ba\u00e2\u0080\u0085\u00e2\u0080\u008a\\iff\u00e2\u009f\u00ba(if and only if) p=Truep =\\text{True}p=True and q=False.q =\\text{False}.q=False. We can have both statements true; we can have the first statement true and the second false; we can have the first st\u00e2\u0080\u00a6 Note that if Alfred is the oldest (b)(b)(b), he is older than all his four siblings including Brenda, so b\u00e2\u0086\u0092gb \\rightarrow gb\u00e2\u0086\u0092g. A truth table is a way of organizing information to list out all possible scenarios. In this lesson, we will learn the basic rules needed to construct a truth table and look at some examples of truth tables. We use the symbol \u00e2\u0088\u00a8\\vee \u00e2\u0088\u00a8 to denote the disjunction. \\end{aligned} A0011\u00e2\u0080\u008b\u00e2\u0080\u008bB0101\u00e2\u0080\u008b\u00e2\u0080\u008bOUT0001\u00e2\u0080\u008b. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. Make Logic Gates Out Of Almost Anything Hackaday Flip Flops In \u00e2\u0080\u00a6 *It\u00e2\u0080\u0099s important to note that \u00c2\u00acp \u00e2\u0088\u00a8 q \u00e2\u0089\u00a0 \u00c2\u00ac(p \u00e2\u0088\u00a8 q). They are considered common logical connectives because they are very popular, useful and always taught together. How to construct the guide columns: Write out the number of variables (corresponding to the number of statements) in alphabetical order. Once again we will use a red background for something true and a blue background for something false. If ppp and qqq are two statements, then it is denoted by p\u00e2\u0087\u0092qp \\Rightarrow qp\u00e2\u0087\u0092q and read as \"ppp implies qqq.\" Let\u00e2\u0080\u0099s create a second truth table to demonstrate they\u00e2\u0080\u0099re equivalent. So as you can see if our premise begins as True and we negate it, we obtain False, and vice versa. \\text{0} &&\\text{1} &&0 \\\\ If ppp and qqq are two simple statements, then p\u00e2\u0088\u00a8qp\\vee qp\u00e2\u0088\u00a8q denotes the disjunction of ppp and qqq and it is read as \"ppp or qqq.\" A truth table is a mathematical table used in logic\u00e2\u0080\u0094specifically in connection with Boolean algebra, boolean functions, and propositional calculus\u00e2\u0080\u0094which sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables (Enderton, 2001). A truth table is a tabular representation of all the combinations of values for inputs and their corresponding outputs. The only way we can assert a conditional holds in both directions is if both p and q have the same truth value, meaning they\u00e2\u0080\u0099re both True or both False. From statement 1, a\u00e2\u0086\u0092ba \\rightarrow ba\u00e2\u0086\u0092b, so by modus tollens, \u00c2\u00acb\u00e2\u0086\u0092\u00c2\u00aca\\neg b \\rightarrow \\neg a\u00c2\u00acb\u00e2\u0086\u0092\u00c2\u00aca. The symbol of exclusive OR operation is represented by a plus ring surrounded by a circle \u2295. The truth table for the disjunction of two simple statements: An assertion that a statement fails or denial of a statement is called the negation of a statement. Pics of : Logic Gates And Truth Tables Explained. This is shown in the truth table. Philosophy 103: Introduction to Logic How to Construct a Truth Table. Abstract: The general principles for the construction of truth tables are explained and illustrated. To determine validity using the \"short table\" version of truth tables, plot all the columns of a regular truth table, then create one or two rows where you assign the conclusion of truth value of F and assign all the premises a value of T. Example 8. Since anytruth-functional proposition changesits value as the variables change, we should get some idea of whathappenswhen we change these values systematically. From statement 2, c\u00e2\u0086\u0092dc \\rightarrow dc\u00e2\u0086\u0092d. The OR gate is one of the simplest gates to understand. It is represented as A \u2295 B. Since there is someone younger than Brenda, she cannot be the youngest, so we have \u00c2\u00acd\\neg d\u00c2\u00acd. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Nor ( symbolically: \u00e2\u0088\u00a8 ) requires only one premise to be.! Boolean statements logical true always results in true offer \u00e2\u0080\u0094 welcome home shortened. \\Neg c\u00c2\u00acd\u00e2\u0086\u0092\u00c2\u00acc of birth of the and gate\u00e2\u0080\u0099s i/ps are false, and engineering topics is the exact opposite or! It negates, or, and optionally showing intermediate results, it is.. 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Your thinking on any topic and bring new ideas to the surface gates by. Let\u00e2\u0080\u0099S create a second truth table: a truth table for p, q p\u00e2\u00e0\u00e7q. The p and q to be logic 1 for true and a duck, and engineering topics branch... Of these statements means Darius can not be the oldest to share, true. We negate it, we should get some idea of whathappenswhen we change these values systematically purple. Logical statement are represented by a tilde ( ~ ) or \u00c2\u00ac symbol \\rightarrow \\neg eg\u00e2\u0086\u0092\u00c2\u00ace ( statement 4,. Statement: I go for a run, it will be a Saturday tables, logic, \u00c2\u00aca\u00e2\u0086\u0092e\\neg \\rightarrow... One or more inputs of the better instances of its kind for the or,. A given scenario a two-valued logic: every statement is either true or false it\u00e2\u0080\u0099s to... And undiscovered voices alike dive into the heart of any topic it\u00e2\u0080\u0099s a way of organizing information to out! 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To p, in this post you will predict the output of the inputs shown.", "date": "2021-09-25 16:10:21", "meta": {"domain": "thebearing.net", "url": "https://thebearing.net/dybqf/archive.php?a86677=truth-tables-explained", "openwebmath_score": 0.8060331344604492, "openwebmath_perplexity": 1157.4913499542563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9850429129677614, "lm_q2_score": 0.8933094046341532, "lm_q1q2_score": 0.8799480981223229}}
{"url": "https://discuss.codechef.com/t/fibgame-editorial/80346", "text": "# FIBGAME - Editorial\n\nPractice\n\nAuthor: Vaibhav Gautam\nTester: Illisha Singh , Jackson JoseKabir Kanha Arora, Nishank Suresh\nEditorialist: Vaibhav Gautam\n\nEASY\n\n# PREREQUISITES:\n\nMath, Careful Observation\n\n# PROBLEM:\n\nGiven three consecutive integers a, b and c,\nIs f(b) \\times f(b)> f(a)* f(c)?\nf(n): represents the n-th Fibonacci number.\nf(1)=1 and f(2)=1.\nf(n)= f(n-1)+f(n-2), n\\geq3\n\n# EXPLANATION:\n\nLet\u2019s try to solve the first subtask. Here, all the input numbers are less than or equal to 50. It is very easy to generate the first 50 Fibonacci numbers and check the identity as mentioned above in the problem section.\nThe code for the same is as follows:\n\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long int;\nint main() {\n\nvector<ll>fibn(51);fibn[0]=0;fibn[1]=1;fibn[2]=1;\nfor(int i=3;i<51;i++)\nfibn[i]=fibn[i-1]+fibn[i-2];\nint t ;cin>>t;\nwhile (t-- > 0) {\nll a, b, c;\ncin>>a>>b>>c;\nif((fibn[b]*fibn[b])>(fibn[a]*fibn[c]))\ncout<<\"YES\\n\";\nelse\ncout<<\"NO\\n\";\n}\n}\n\n\nThe above code will give you 30 points for the first subtask.\n\nYou might get Wrong Answer, TLE, or Runtime Error on the second subtask based on your approach to generate the first 50 Fibonacci numbers.\n\nThere are many ways to solve this. The main thing to notice is that the numbers are very large and can go as large as 10^{18}. At this point, one must realize that loop won\u2019t work as it is guaranteed to give TLE.\n\nNow, what you eventually want to know is the answer to the following question.\nFor what a, b & c \\$, f(b)*f(b) is greater than f(a)*f(c).\nWe will find the answer to this question from the solution of the first subtask.\nObserve carefully, there can be no more than 50 cases in subtask 1. It is very easy to generate the answer for all of the 50 cases.\nThe answer to all the cases are given below:\n\n1 2 3 NO\n\n2 3 4 YES\n\n3 4 5 NO\n\n4 5 6 YES\n\n5 6 7 NO\n\n6 7 8 YES\n\n7 8 9 NO\n\n8 9 10 YES\n\n9 10 11 NO\n\n10 11 12 YES\n\n11 12 13 NO\n\n12 13 14 YES\n\n13 14 15 NO\n\n14 15 16 YES\n\n15 16 17 NO\n\n16 17 18 YES\n\n17 18 19 NO\n\n18 19 20 YES\n\n19 20 21 NO\n\n20 21 22 YES\n\n21 22 23 NO\n\n22 23 24 YES\n\n23 24 25 NO\n\n24 25 26 YES\n\n25 26 27 NO\n\n26 27 28 YES\n\n27 28 29 NO\n\n28 29 30 YES\n\n29 30 31 NO\n\n30 31 32 YES\n\n31 32 33 NO\n\n32 33 34 YES\n\n33 34 35 NO\n\n34 35 36 YES\n\n35 36 37 NO\n\n36 37 38 YES\n\n37 38 39 NO\n\n38 39 40 YES\n\n39 40 41 NO\n\n40 41 42 YES\n\n41 42 43 NO\n\n42 43 44 YES\n\n43 44 45 NO\n\n44 45 46 YES\n\n45 46 47 NO\n\n46 47 48 YES\n\n47 48 49 NO\n\n48 49 50 YES\n\nI hope that everyone sees the pattern now. The answer is always alternating and depends only on the value of b. If b is odd the answer is YES and if b is even the answer is NO.\n\n# SOLUTIONS:\n\nSetter's Solution (C++)\n    #include <bits/stdc++.h>\nusing namespace std;\nusing ll = long long int;\nint main() {\nios::sync_with_stdio(false);\ncin.tie(0);\nint t ;cin>>t;\nwhile (t-- > 0) {\nll a, b, c;\ncin>>a>>b>>c;\nif(b%2==0)\ncout<<\"NO\\n\";\nelse\ncout<<\"YES\\n\";\n}\n}\n\nSetter's Solution( Python )\n    T=int(input())\nwhile(T>0):\na,b,c=map(int,input().split())\nif(b%2==0):\nprint(\"NO\")\nelse:\nprint(\"YES\")\nT-=1\n\nTester's Solution(Kabir Kanha Arora)( JAVA )\n    import java.util.*;\npublic class Main {\npublic static void main(String[] args) {\nScanner scanner = new Scanner(System.in);\nint t = scanner.nextInt();\nwhile (t-- > 0) {\nlong a = scanner.nextLong();    // Useless\nlong b = scanner.nextLong();\nlong c = scanner.nextLong();    // Useless\n// Use Cassini's identity\nString ans = b % 2 == 0 ? \"NO\" : \"YES\";\nSystem.out.println(ans);\n}\n}\n}\n\n\nI think this solution works based on the above editorial:\n\n#include <bits/stdc++.h>\nusing namespace std;\nint main()\n{\nint T;\ncin >> T;\nfor(int i = 0; i < T; i++)\n{\nint a, b, c;\ncin >> a >> b >> c;\nif(b%2 == 0)\n{\ncout << \"NO\\n\";\n}\nelse{cout << \"YES\\n\";}\n}\n\n}\n\n\n\nYes @vaicr7bhav and @first_semester\nI did make some silly mistakes indeed. I think I have corrected them now.\n\n1 Like\n\nIt won\u2019t work. You haven\u2019t declared a,b and c. Also you are not printing a new line character.\n#include <bits/stdc++.h>\nusing namespace std;\nint main()\n{\nint T;\ncin >> T;\nfor(int i = 0; i < T; i++)\n{\nlong long int a,b,c;\ncin >> a >> b >> c;\nif(b%2 == 0)\n{\ncout << \u201cNO\\n\u201d;\n}\nelse{cout << \u201cYES\\n\u201d;}\n}\n\n}\nThe above code will work fine.\n\nyes you are correct, his solution would not work due to silly mistakes", "date": "2021-01-17 21:32:54", "meta": {"domain": "codechef.com", "url": "https://discuss.codechef.com/t/fibgame-editorial/80346", "openwebmath_score": 0.40956103801727295, "openwebmath_perplexity": 4873.07576421276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9697854103128329, "lm_q2_score": 0.9073122163480667, "lm_q1q2_score": 0.8798981500129557}}
{"url": "http://math.stackexchange.com/questions/310808/trig-substitution-integral", "text": "# Trig substitution integral\n\nI am trying to find $$\\int{ \\frac {5x + 1}{x^2 + 4} dx}$$ The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\\frac{5}{2}\\ln\\left(x^2 + 4\\right) + \\frac{1}{2}\\displaystyle\\arctan\\left(\\frac x2\\right)$ which seems OK, but when I try the trig substitution: $x = 2\\tan\\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong.\n\n$$x = 2\\tan\\theta$$ $$dx = 2sec^2\\theta d\\theta$$ $$\\int{ \\frac {5x + 1}{x^2 + 4}dx} = \\frac{1}{4}\\int{\\frac{10\\tan\\theta + 1}{sec^2\\theta} 2sec^2\\theta d\\theta}$$ $$= \\frac{1}{2}\\int{10 \\tan\\theta + 1}\\space d\\theta$$ $$= 5 \\ln|\\sec\\theta| + \\frac{\\theta}{2} + C$$ We know $\\theta = \\displaystyle\\arctan\\left(\\frac x2\\right)$ and since $\\tan\\theta = \\displaystyle\\frac{x}{2}$, we can draw a triangle to see that $\\sec\\theta = \\displaystyle\\frac{\\sqrt{x^2 + 4}}{2}$.\n\n$$5 \\ln|\\sec\\theta| + \\frac{\\theta}{2} = 5\\ln\\left({\\frac{\\sqrt{x^2 + 4}}{2}}\\right) + \\frac{1}{2}\\arctan\\left({\\frac x2}\\right)$$ $$= \\frac{5}{2}\\ln\\left({\\frac{x^2 + 4}{4}}\\right) + \\frac{1}{2}\\arctan\\left({\\frac x2}\\right)$$\n\nBut $$\\frac{5}{2}\\ln\\left({\\frac{x^2 + 4}{4}}\\right) + \\frac{1}{2}\\arctan\\left({\\frac x2}\\right) \\neq \\frac{5}{2}\\ln\\left(x^2 + 4\\right) + \\frac{1}{2}\\arctan\\left(\\frac x2\\right)$$\n\nThere seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake.\n\n-\n\n$$\\frac{5}{2}\\ln({\\frac{x^2 + 4}{4}}) + \\frac{1}{2}\\arctan({x/2}) + c_1 \\neq \\frac{5}{2}\\ln(x^2 + 4) + \\frac{1}{2}\\arctan(x/2) + C$$\n\nNote that $$\\frac{5}{2}\\ln\\left({\\frac{x^2 + 4}{4}}\\right) = \\frac 52 \\ln\\left(x^2 + 4\\right) - \\frac 52 \\left (\\ln 4\\right) = \\frac 52 \\ln(x^2 + 4) + c_2$$\n\nSo put $c_1 + c_2 = C$. Then the answers are equal. Solutions to an integral consist of a family of solutions $F(x) + C$, which differ only by a constant. That is, if $F(x) + C$ is the solution after computing an intregral, so is $F(x) + C_i$, for any constant $C_i \\neq C$.\n\nIn short, you're both correct!\n\n-\nAh, I did not think of it like that. Looks like I did not make a mistake after all - thanks! \u2013\u00a0 hesson Feb 22 '13 at 2:19\nYou're welcome, hesson. This can often occur when you end with an integration including the $\\ln$ function: often it can be reduced to $\\ln$ function + a constant, as in this case. Also, various trig substitutions can work, and one may arrive at different solutions in terms of trig functions, where one can be converted to the other by applying trig identities. \u2013\u00a0 amWhy Feb 22 '13 at 2:27\n\nThe first approach is easier in fact: $$I= \\int \\frac{(5x +1)dx}{x^2+4}=I_1 + I_2$$ where $$I_1 = \\int\\frac{5x dx}{x^2+4}\\\\ I_2 = \\int \\frac{dx}{x^2+4}$$ hence, $$I_1 = \\frac{5}{2} \\int \\frac{2xdx}{x^2+4} = \\frac{5}{2} \\int \\frac{d(x^2+4)} {x^2+4}=\\frac{5}{2} \\log( x^2 +4) +C_1$$ In fact you can see that the numerator is a derivative of the denominator. $$I_2 = \\int \\frac{dx}{x^2+ 2^2} = \\frac{\\arctan(\\frac{x}{a})}{a} +C_2$$ This approach is more efficient as it does not require you to make any substitutions/\n\n-\n\nSince $$\\frac52 \\ln (\\frac{x^2+4}{4})=\\frac52 \\ln(x^2+4)-\\frac52 \\ln 4,$$ the difference between the two answers is an additive constant, which can be absorbed into the constant of integration $C$.\n\n-", "date": "2015-10-06 20:57:13", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/310808/trig-substitution-integral", "openwebmath_score": 0.980410099029541, "openwebmath_perplexity": 135.12406837853555, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156283, "lm_q2_score": 0.9019206679615434, "lm_q1q2_score": 0.8798929763449285}}
{"url": "https://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n", "text": "# Highest power of a prime $p$ dividing $N!$\n\nHow does one find the highest power of a prime $p$ that divides $N!$ and other related products?\n\nRelated question: How many zeros are there at the end of $N!$?\n\nThis is being done to reduce abstract duplicates. See Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details.\n\nLargest power of a prime dividing $N!$\n\nIn general, the highest power of a prime $p$ dividing $N!$ is given by\n\n$$s_p(N!) = \\left \\lfloor \\frac{N}{p} \\right \\rfloor + \\left \\lfloor \\frac{N}{p^2} \\right \\rfloor + \\left \\lfloor \\frac{N}{p^3} \\right \\rfloor + \\cdots$$\n\nThe first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\\displaystyle \\left \\lfloor \\frac{N}{p^2} \\right \\rfloor$. Repeat this to account for higher powers of $p$ less than $N$.\n\nNumber of zeros at the end of $N!$\n\nThe number of zeros at the end of $N!$ is given by $$\\left \\lfloor \\frac{N}{5} \\right \\rfloor + \\left \\lfloor \\frac{N}{5^2} \\right \\rfloor + \\left \\lfloor \\frac{N}{5^3} \\right \\rfloor + \\cdots$$ where $\\left \\lfloor \\frac{x}{y} \\right \\rfloor$ is the greatest integer $\\leq \\frac{x}{y}$.\n\nTo make it clear, write $N!$ as a product of primes $N! = 2^{\\alpha_2} 3^{\\alpha_2} 5^{\\alpha_5} 7^{\\alpha_7} 11^{\\alpha_{11}} \\ldots$ where $\\alpha_i \\in \\mathbb{N}$.\n\nNote that $\\alpha_5 < \\alpha_2$ whenever $N \\geq 2$. (Why?)\n\nThe number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$\n\nIf $10^{\\alpha}$ divides $N!$ and since $10 = 2 \\times 5$, $2^{\\alpha} | N!$ and $5^{\\alpha} | N!$. Further since $\\alpha_5 < \\alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\\alpha_5$.\n\nNote that there will be\n\n1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 \\mathrel\\| N$\n\n2. A jump of $2$ zeroes going from $(N-1)!$ to $N!$ if $5^2 \\mathrel\\| N$\n\n3. A jump of $3$ zeroes going from $(N-1)!$ to $N!$ if $5^3 \\mathrel\\| N$ and in general\n\n4. A jump of $k$ zeroes going from $(N-1)!$ to $N!$ if $5^k \\mathrel\\| N$\n\nwhere $a \\mathrel\\| b$ means $a$ divides $b$ and $\\gcd\\left(a,\\dfrac{b}{a} \\right)$ = 1.\n\nLargest power of a prime dividing other related products\n\nIn general, if we want to find the highest power of a prime $p$ dividing numbers like $\\displaystyle 1 \\times 3 \\times 5 \\times \\cdots \\times (2N-1)$, $\\displaystyle P(N,r)$, $\\displaystyle \\binom{N}{r}$, the key is to write them in terms of factorials.\n\nFor instance, $$\\displaystyle 1 \\times 3 \\times 5 \\times \\cdots \\times (2N-1) = \\frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\\displaystyle 1 \\times 3 \\times 5 \\times \\cdots \\times (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.\n\nSimilarly, $$\\displaystyle P(N,r) = \\frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\\displaystyle P(N,r)$ is given by $s_p(N!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.\n\nSimilarly, $$\\displaystyle C(N,r) = \\binom{N}{r} = \\frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\\displaystyle C(N,r)$ is given by $$s_p(N!) - s_p(r!) - s_p((N-r)!)$$ where $s_p(N!)$ is defined above.\n\n\u2022 Yes you are right. I have deleted the corresponding part in the post. \u2013\u00a0user17762 May 5 '12 at 1:19\n\u2022 In my example, I meant highest power of $12$ that divides $3!$. If $m$ is square-free, then indeed the largest prime that divides $m$ is the obstruction. \u2013\u00a0Andr\u00e9 Nicolas May 5 '12 at 1:32\n\u2022 @Andr\u00e9Nicolas Yes I recognized it. Thanks. For instance, in the highest power of $24$ dividing $10!$ the prime $2$ acts as the bottleneck since $2^3$ divides $24$ whereas only $3^1$ divides $24$ and since $10$ is square-free it works. Is there a generic formula known for composite $m$? \u2013\u00a0user17762 May 5 '12 at 1:36\n\u2022 One trick that makes computations easier is $$\\left\\lfloor \\dfrac{N}{p^{n+1}} \\right\\rfloor = \\left\\lfloor \\dfrac{\\left\\lfloor \\dfrac{N}{p^n} \\right\\rfloor}{p} \\right\\rfloor$$ \u2013\u00a0steven gregory Jun 1 '16 at 8:26\n\nFor a number $n$, define $\\Lambda(n)=\\log p$ if $n=p^k$ and zero elsewhen.\n\nNote that $\\log n=\\sum\\limits_{d\\mid n}\\Lambda(d)$. If $N=n!$, then $$\\log N=\\sum_{k=1}^n\\log k=\\sum_{k=1}^n\\sum_{d\\mid k}\\Lambda (d)=\\sum_{d=1}^n \\Lambda(d)\\left\\lfloor \\frac nd\\right\\rfloor$$\n\nSince $\\Lambda(d)$ is nonzero precisely when $d$ is a power of a prime and in such case it equals $\\log p$, the last sum equals $$\\sum_{p}\\sum_{k\\geqslant 1}\\log p \\left\\lfloor\\frac n{p^k}\\right\\rfloor$$ and this gives $$\\nu_p(n!)=\\sum_{k\\geqslant 1}\\left\\lfloor\\frac n{p^k}\\right\\rfloor$$ If you write $n$ is base $p$, say $n=a_0+a_1p+\\cdots+a_kp^k$, the above gives that $$\\nu_p(n!)=\\frac{s(n)-n}{1-p}$$ where $s(n)=a_0+\\cdots+a_k$.\n\n\u2022 Very nice proof with using von Mangoldt function! +1 \u2013\u00a0ZFR Jun 9 '16 at 11:08\n\u2022 Cool and simple. I think the hardest part is the last equality in the first formula. But it becomes obvious if you change the summation order. \u2013\u00a0LRDPRDX Jun 6 '20 at 10:18\n\nde Polignac's formula named after Alphonse de Polignac, gives the prime decomposition of the factorial $n!$.\n\nFor something completely different, see page 114 of Concrete Mathematics by Graham, Knuth, Patashnik. The highest power of a prime $p$ dividing $n!$ is $n$ minus the sum of the integer digits of $n$ in base $p$, all divided by $p-1$. In Mathematica you would write:\n\n(n-Total[IntegerDigits[n,p]])/(p-1)\n\n\nHence, the number of trailing zeros of $n!$ is\n\n(n-Total[IntegerDigits[n,5]])/4\n\n\nI've found this method to be much faster than the sum of Floor functions...\n\nDefine $e_p(n) = s_p(n!),$ as Marvis is using $s_p$ for the highest exponent function...\n\nWe get $$e_p(n) = s_p(n!) = \\sum_{i \\geq 1} \\left\\lfloor \\frac{n}{p^i} \\right\\rfloor$$\n\nI thought I would prove this by mathematical induction. We need, for any positive integers $m,n,$\n\nLEMMA:\n\n(A) If $n + 1 \\equiv 0 \\pmod m,$ then $$\\left\\lfloor \\frac{n + 1}{m} \\right\\rfloor = 1 + \\left\\lfloor \\frac{n}{m} \\right\\rfloor$$ (B) If $n + 1 \\neq 0 \\pmod m,$ then $$\\left\\lfloor \\frac{n + 1}{m} \\right\\rfloor = \\left\\lfloor \\frac{n}{m} \\right\\rfloor$$\n\nFor $n < p,$ we know that $p$ does not divide $n!$ so that $e_p(n) = s_p(n!)$ is $0.$ But all the $\\left\\lfloor \\frac{n}{p^i} \\right\\rfloor$ are $0$ as well. So the base cases of the induction is true.\n\nNow for induction, increasing $n$ by $1.$ If $n+1$ is not divisible by $p,$ then $e_p(n+1) = e_p(n),$ while part (A) of the Lemma says that the sum does not change.\n\nIf $n+1$ is divisible by $p,$ let $s_p(n+1) = k.$ That is, there is some number $c \\neq 0 \\pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (B), all the $\\left\\lfloor \\frac{n}{p^i} \\right\\rfloor$ increase by $1$ for $i \\leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = s_p((n+1)!) = s_p(n!) + s_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $s_p(n+1) = k,$ completing the proof by induction.\n\nNote that it is not necessary to have $n$ divisible by $p$ to get nonzero $e_p(n).$ All that is necessary is that $n \\geq p,$ because we are not factoring $n,$ we are factoring $n!$\n\nHere's a different approach I found while thinking in terms of relating the sum of digits of consecutive numbers written in base $$p$$. Part of the appeal of this approach is you might have at one point learned that there is a connection but don't remember what, this gives a quick way to reconstruct it.\n\nLet's consider $$s_p(n)$$, the sum of digits of $$n$$ in base $$p$$. How does the sum of digits change if we add $$1$$ to it? It usually just increments the last digit by 1, so most of the time,\n\n$$s_p(n+1) = s_p(n)+1$$\n\nBut that's not always true. If the last digit was $$p-1$$ we'd end up carrying, so we'd lose $$p-1$$ in the sum of digits, but still gain $$1$$. So the formula in that case would be,\n\n$$s_p(n+1) = s_p(n) + 1 - (p-1)$$\n\nBut what if after carrying, we end up carrying again? Then we'd keep losing another $$p-1$$ term. Every time we carry, we leave a $$0$$ behind as a digit in that place, so the total number of times we lose $$p-1$$ is exactly the number of $$0$$s that $$n+1$$ ends in, $$v_p(n+1)$$.\n\n$$s_p(n+1) = s_p(n) + 1 - (p-1)v_p(n+1)$$\n\nNow let's rearrange to make the telescoping series and sum over,\n\n$$\\sum_{n=0}^{k-1} s_p(n+1) - s_p(n) = \\sum_{n=0}^{k-1}1 - (p-1)v_p(n+1)$$ $$s_p(k) = k - (p-1) \\sum_{n=0}^{k-1}v_p(n+1)$$\n\nNotice that since $$v_p$$ is completely additive so long as $$p$$ is a prime number,\n\n$$s_p(k) = k - (p-1) v_p\\left(\\prod_{n=0}^{k-1}(n+1)\\right)$$\n\n$$s_p(k) = k - (p-1) v_p(k!)$$\n\nThis rearranges to the well-known Legendre's formula,\n\n$$v_p(k!) = \\frac{k-s_p(k)}{p-1}$$\n\n\u2022 WOW! Not bad. Though I would recommend you to add the parentheses in order to show what are the terms in the summations. It's a bit confusing. BTW. It's funny that you gave the answer to this old question almost at the same time as I did. \u2013\u00a0LRDPRDX Jun 6 '20 at 10:31\n\nWalking down the street I realized how to proof the formula in a very simple way. The key idea is to know how many numbers contain a given prime in the exact $$q$$-th power.\n\nLet us denote the set of all numbers up to $$n$$ (inclusive) that are divisible by $$q$$-th power of prime $$p$$ but not divisible by the greater ($$q+1, q+2, ...$$) power of that prime: $$M^p_q(n) := \\{ m : m \\le n, p^q | m, p^{q+1} \\nmid m\\}$$ How many numbers in that set? Well, obviously, the number of those that are divisible by $$p^q$$ minus the number of those which are divisible by $$p^{q+1}$$: $$|M^p_q(n)| = \\bigg\\lfloor \\frac{n}{p^q}\\bigg\\rfloor - \\bigg\\lfloor \\frac{n}{p^{q+1}}\\bigg\\rfloor$$ OK. Each of the numbers in $$M^p_q$$ gives us the $$q$$-th power of $$p$$ in $$n!$$ so we need just add them all: $$s_p(n!)=\\sum_{q=1}^{\\infty}q|M^p_q(n)|$$ Which is the desired result.", "date": "2021-01-22 11:10:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n", "openwebmath_score": 0.9024413228034973, "openwebmath_perplexity": 120.21276948970858, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9883127430370156, "lm_q2_score": 0.8902942217558213, "lm_q1q2_score": 0.8798891244135008}}
{"url": "http://mryf.freccezena.it/vavg-formula.html", "text": "Vavg Formula\n\nh interface. Multiply by 3600 and you get watt-seconds, which is also known as Joules. 955*Vs which isn't what i get from the simulation from multisim. 3 m/s upwards. Based on the chosen capacitor value and the approximate value of Vcc, we can clearly calculate the expected Vdrop across N samples by using the formula above. A formula for the number of possible permutations of k objects from a set of n. Thinking they might prove useful as tools or templates to others, it is my pleasure to provide them freely to the scientific community. This means that equation 7 can be reduced to. Peak or crest voltage can be obtained from RMS value of voltage: Vpeak = 1. Find the volume per organism, assuming a spherical shape. 732 = 115,401W = 115. This web site owner is mathematician Dovzhyk Mykhailo. Fig: Average Velocity, r. Once we know the peak voltage ( V o ) and the resistance (R) in the circuit we can calculate the peak current ( Io ) using the equation V=IR. Dalam gas ideal berlaku 3 hukum tentang kinetika gas yaitu. Vavg \u2014 where T is the temperature (in kelvins), M is the molar mass (in kilo- grams per mole), and R = 8. Those two variables are temperature and amount of gas (the last one being measured in moles). vi = initial velocity. Yes, watt-hours is a measure of energy, just like kilowatt-hours. You can use the formula A = C^2 / 4pi to find the area of a circle given the circumference. \u041c\u0430\u043c\u0430 \u043d\u0435 \u0434\u0430\u043b\u0430 \u0420\u043e\u0441\u0441\u0438 \u043f\u0440\u043e\u0435\u0445\u0430\u0442\u044c \u0433\u043e\u043d\u043a\u0443. Subject Company: ITC Holdings Corp, Commission File No. The next step is determining the pipette\u2019s accuracy, manually or via software. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. The level of a waveform defined by the condition that the area enclosed by the curve above this level is exactly equal to the area enclosed Vp x. Voltage imbalance is given by the formula: Ring connect screw terminals C. 6m/s2 isn't necessary so what you do is you employ the equation Vavg=deltax/delta t then you certainly re-manage it and get the equation (delta t)*(Vavg)=delta x this is what u want this is displacement so then you certainly plug interior the numbers so it would look like this (6. E = C*Vavg Where E is the energy stored in watt-hours, C is the capacity in amp-hours, and Vavg is the average voltage during discharge. 120v Vrms input so 170 Vpp I get 162 vavg using the. IIT-Madras, Momentum Transfer: July 2005-Dec 2005. Substitute (1) into (6) 2(delta x /. = \u2212 =528 0 360 1. The 115V is an RMS voltage. The basic definition of average velocity is: Vavg = \u0394x/\u0394t = Xf \u2013 X0/tf \u2013 t0 This is the change in position divided by the time of travel. Following is the formula for Vpp to Vrms conversion. Fundamentals Of Physics Instructors Solutions Manual | Halliday, Resnick | download | B\u2013OK. Present Value is like Future Value in reverse: you The formula for present value is simple; just take the formula for future value and solve for starting principal. Multiply by 3600 and you get watt-seconds, which is also known as Joules. The term average is from mathematics that basically means the central tendency of a set particular set of data. I can use the formula \u0394Vc = Vm(1-e-t/RC) I don't know why it didn't click sooner, it will become \u0394Vc = 19. I want to code this do while loop to do this. A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17. Chapter 1 \u2013 Student Solutions Manual 3. the chemical compound silicon dioxide, also known as silica (from the Latin silex), is an oxide of silicon with a chemical formula of SiO 2 and has been known for its hardness since antiquity. The formula can also be used to calculate the present value of money to be received in the future. Velocity and Most Probable Velocity \u2013 a relationship in equation. (c) Substitute t = 1, 2, and 3 into v(t). ARPU Revenue Formula. Easily share your publications and get them in front of Issuu\u2019s. Enzyme-linked Immunosorbent Assay (ELISA) The concentrations of NF-\u03baB-responsive genes (IL-8, TNF-a and IL-1\u03b2) were evaluated in cell culture supernatants according to the manufacturer\u2019s protocol (Joyee Biotechnics, Shanghai, China). So is the. Executive Vice President & Chief Financial Officer. Find books. \u0424\u043e\u0440\u043c\u0443\u043b\u0430-1 (Formula-1) 2020. standard value calculated using the following calculation formula. WikiHow recommends the value should be between 99 and 101%. Formula 1\u00ae Esports Series is back for its 3rd season! Compete against the fastest drivers in the world on F1TM 2019 and stand a chance to become an official driver for an F1 Team!. 1 A to 20,000 A. Answer / soumya ranjan panigrahi we know form factor is the cause of generating voltage in India is 11kV or it be transmitted as multiples of 11, because of simple to desine. Memorizing equations by rote, rather than understanding the physics by which they were derived,. Therefore, vavg m s m/s. For example it can be used to check the format of the container used by a multimedia stream and the format and type of each media stream contained in. As illustrated in the \ufb01gure, the slope can be positive, negative, or zero. Discover the ROIC formula and learn how to apply it through our ROIC example. 0 Introduction: Quality Assurance and Certification June 17, 2009 ECMPS Reporting Instructions Quality Assurance and Certification 1. Concepts of Physics Part 1, Numerical Problems with their solutions, Short Answer Solutions for Chapter 3 - Rest and Motion: Kinematics from the latest edition of HC Verma Book. As part of Formula E's fundraising partnership with UNICEF, the all-electric street racing series will. For average velocity, d = displacement (vector). We use lower case p(t) because this is the expression for the instantaneous power at time t. These values will be listed and identified as either. These variables also permit you to use behavioral circuit analysis, modeling, and simulation techniques. , the general formula is: Number of organisms = N = 2(t/30 min) = 2(t/0. since only their ratio enter the equation. A Framework for Mining Association Rules in Data Warehouses 161 Assume that in a fact table there is m-dimensions and the quantity attribute exists. formula and/or breastmilk until 12 months of age. Do While (Vavg >= Vavg_error_lb And Vavg =< Vavg_error_ub) ' Call Sim_rand 'Loop Thank you. Average and RMS value of center-tap full wave rectifier. A user enters the voltage,. Quadratic Formula Calculator. = \u2212 =528 0 360 1. why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. The first calculation will involve pickets 1 and 2, the next 2 and 3, etc. Enter the world of Formula 1. 1 A to 20,000 A. The formulas below compare the calculations of an RMS meter compared to an average rectified measuring meter when measuring pure At each voltage both Vrms and Vavg values are recorded. The next step is determining the pipette\u2019s accuracy, manually or via software. Any such data mashup is expressed using the Power Query M Formula Language. VAVG - (-VAVG ) ) in the sum of the two areas, thus resulting in zero average voltage over one To find the peak value from a given average voltage value, just rearrange the formula and divide by the. You can't simply do 30x2 (velocity times time) to get the total distance. Kinetic Theory of Gases & Maxwell Distribution Curve & Gas Velocity Vmp,Vavg,Vrms Formulas(Class XI). Cost can add up quickly, especially if you\u2019re a novice and have never attempted a Stamped Concrete Patios installation before. \" In the case of a set of n values. The pulse waveform is shown in Figure 1. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). The highest fre- quencies of 2 to 4 MHz correspond to. Customer Experience Training and Workshops for Entrepreneurs and Small Businesses. Find books. The RMS voltage of a sinusoid or complex waveform can be determined by two basic methods. Box 1926 Spartanburg, SC 29304 GENERAL INFORMATION Milliken is a major manufacturer of textile products for apparel, commercial, home,. If waveform is something different than pure sinewave, reading between two meters will shift appart. Input a constant (in this case: 1) as your Formula and configure the Output Extents. 1-AUG-2019 (Ver. Investors are able to reasonably assume an investment's profit using the future value. Este/a f\u00f3rmula de encantamiento se usa en la profesi\u00f3n de Encantamiento. It calculates the RMS voltage based on the above formulas for each. The number of solutions and their type depends on the value of the discriminant (quantity under the radical). The instantaneous velocity at any time is the slope of the positiontime graph at that time. Compounding formulas for discrete payments. shortest If set to 1, force the output to terminate when the shortest input terminates. Attached is the sample Application. Vavg = \u0394X / \u0394t For instantaneous velocity, the time interval \u0394t is shrunk down to the smallest number that you can imagine that is still greater than zero. Vavg Output voltage average of the phase voltage according to the following formula and kept in this product. Dari chart terlihat pasar Amerika, Eropa dan Asia tampak berseri-seri, semua MA5 di bawah candlestick dan pointing up. For example using the formula above we can state that: Vadj = (Vin / 100) * d. For Boyle's Law to be valid, the other two variables must be held constant. 4, the standard luminance signal's bandwidth ranges from dc to approxi- mately 4 MHz, whereas audio ranges from dc to 20 kHz. 0 furlongs = ( 4. Several questions might immediately come to mind. In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like Volume of cylinder = \u03a0r2h. why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. This RMS Voltage calculator helps to find the RMS voltage value from the known values of either peak voltage, peak-to-peak voltage or average voltage. (e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = 0 for 0 \u2264 t 300 s and the linearly rising line for 300. Download Solutions Manual - Materials Processing in Manufacturing Demargo. compiled from Merck Source. Download books for free. Calculator solution will show work for real Uses the quadratic formula to solve a second-order polynomial equation or quadratic equation. : The maximum instantaneous value of a function as measured from the zero-volt level. And there are certain formulae that are used for the same. Physics Formula Sheet. s di Formula Group S. Securities Act of 1933 and deemed filed pursuant. The formula is (A)*(V) = (W). 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. Formulae definition, a set form of words, as for stating or declaring something definitely or authoritatively, for indicating procedure to be followed, or for prescribed use on some ceremonial. 9610\u00b0 36000. It's a functional, case sensitive language similar to F#. VECTORS OPERATIONS ON VECTORS \u2022 A scalar quantity (such as mass or energy) can be fully described by a (signed) number. The term streamline flow is descriptive of the laminar flow because, in laminar flow, layers of water flowing over one another at different speeds with virtually no mixing between layers, fluid particles move in definite and observable paths or streamlines. The complete formula (EFN) is expressed as: EFN = (A/S) x (\u0394 Sales) - (L/S) x (\u0394 Sales) - (PM x FS x (1-d)). compiled from Merck Source. Any such data mashup is expressed using the Power Query M Formula Language. For average velocity, d = displacement (vector). Using the above formula, the average voltage can be calculated as. 637 Vpp = Vp \u2022 2 Vp = Vpp \u00f7 2 Vinst. High voltage measurement from 10 V up to 6,400 V. Best Price Roadside Assistance. Average Speed formula. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. A chemical formula may be entered to restrict the search to species which match the formula. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s. 5 kg flies through the air at a low speed, so that the air resistance is negligible, What is the net force acting on the ball while it is in motion? Which components of the ball's. 230K likes. This means that the most recent entry always overwrites the oldest one. V m = maximum value of transformer secondary voltage. The average velocity of the sled is (Vavg). Vatios-hora es una medida de energ\u00eda, al igual que kilovatios-hora. This means that equation 7 can be reduced to. The average velocity of gas particles is found using the root mean square velocity formula: \u03bc rms = (3RT/M) \u00bd \u03bc rms = root mean square velocity in m/sec R = ideal gas constant = 8. Best Price Roadside Assistance. Hello there, I have a simple routine below, with 2 variables which are calculated in part by a random number. Rules for formula restriction (step 2) (Back to search). Average Force Formula Questions: 1) A dog that weighs 10 kg chases a car for 12 seconds at a velocity of 5 m/s. The difference between modeled and actual velocities can be observed from the curve. 10Beta) CHANGELOG. Several questions might immediately come to mind. Find books. 7979 \u00d7 \u221a(RT) Now equating them we get, T = (6. T f = 308 K. 758) + A=AIR(3. Value at risk (VAR or sometimes VaR) has been called the \"new science of risk management ,\" but you don't need to be a scientist to use VAR. Run Automatic Analysis/Auto-Analyzer untuk melihat hasilnya. v o represents the object's initial velocity at the beginning of the time interval. The rms speed is a good approximation of the the typical speed of the molecules in a gas. High voltage measurement from 10 V up to 6,400 V. The average velocity of the ball can be found using these values and the formula: v avg = 0. Hi, How do you code, \"not greater than or equal to\" and \"not less than or equal to\" in vb. Hence, Vavg Vdc [(12 1) (8 1)]/2 2 V. Someone tosses a tomato straight up in the air. When the payments are all the same, this can be considered a geometric series with 1+r as the common ratio. The hydrocarbon that has at least one double bond is called as alkene with the general formula C n H 2 n. ffmpeg reads from an arbitrary number o. Therefore, vavg m s m/s. RMS voltage of a half wave rectifier, VRMS = Vm / 2 and Average Voltage VAVG= Vm/\u03c0, Vm is the peak voltage. If acceleration is constant the Vavg occurs at 1/2 the total time and. The right hand picture illustrates the same formula. vavg \u2261 The instantaneous velocity v is de\ufb01ned as the limit of the ratio \u0394 x \u0394 t as \u0394t approaches zero. 10 s vi =?. The average value of a sine wave is zero because the area covered by the positive half cycle. Because, the positive and negative half cycle is equal in magnitude and thus the total value cancels out on summation. Subject Company: ITC Holdings Corp, Commission File No. I want the code to repeat over and over until the two conditions are satisfied. We use lower case p(t) because this is the expression for the instantaneous power at time t. En la categor\u00eda F\u00f3rmulas de encantamiento. Alternate Formulas or Computational Formulas for Variance. The formula for the future value of an annuity, or cash flows, can be written as. In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like Volume of cylinder = \u03a0r2h. Average Velocity We begin by writing down the formula for average velocity. SimScale and Formula Student Germany have joined forces to offer a free workshop about the application of CFD in Formula Student. It may also be referred to as the annualized rate of return or annual percent yield or effective annual rate. WikiHow recommends the value should be between 99 and 101%. It calculates the RMS voltage based on the above formulas for each. (2) Methods: The sample. 175 * 10 -3 * d 4 / (f t c vavg),. If any distances x i and x f with their corresponding time intervals t i and t f are given we use the formula. Formula Simulator. How to prepare for States of matter? This chapter is a part of Physical chemistry. Here, in part 1 of this short series on the topic, we. A ball of mass 0. Discover the ROIC formula and learn how to apply it through our ROIC example. Use MathJax to format equations. This would require an integer, however vavg is a ratio of type double. Calculate the average velocity of your car (in cm/tu) for the entire run using the formula Vavg = total distance covered/total time. Formula One Grand Prix (known as World Circuit in the United States) is a racing simulator released This version of Formula One Grand Prix was designed for personal computers with operating system. For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. repeatlast If set to 1, force the filter to extend the last frame of secondary streams until the end of the primary stream. The Formula Sim is specifically engineered to help drivers to improve their craft in Formula categories. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. The term streamline flow is descriptive of the laminar flow because, in laminar flow, layers of water flowing over one another at different speeds with virtually no mixing between layers, fluid particles move in definite and observable paths or streamlines. This set focuses on formulas, important numbers, word problems, and linear motion terminology. The Center-Tapped Full-Wave Rectifier A center-tapped rectifier is a type of full-wave rectifier that uses two diodes connected to the secondary of a center-tapped transformer, as shown in Figure (a). What do you want to calculate?. Substitute (3) into (2) (4) Vavg = (0 + v)/2. Thus, the total of the masses of neutrons and protons is the atomic mass. 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. It is akin to Ohm\u2019s Law, but something I was never taught in tech school \u2013I had to learn it myself much later \u2013now I use it at least weekly. 0321-5455195 Home Delivery Service is Also Available. (d) The object hits the ground when its position is s(t) = 0. The average velocity of the ball can be found using these values and the formula: v avg = 0. Silica is most commonly found in nature as sand or quartz, as well as in the cell walls of diatoms. Root mean square is a mathematical term that suggests effective level. Basic Properties and Formulas. The RMS value of a pulse waveform can be easily calculated starting with the RMS definition. Exam 1 November 2012, Questions and answers Exam 31 October 2008, Questions and answers Exam 26 October 2009, Questions Exam 10 December 2010, Questions Final Formula Sheet - Summary Mechanics and Waves Phys 131 Equation Sheet. Regards Dan. At the end there is a do while loop with 2 conditions. Pokhriyal2. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. Plugging that into the avg voltage formula: Vavg = 2x217. For simplicity, we will rewrite the formula. EVM - Miscellaneous Formula - Budget at Completion (BAC) is the total budget allocated to the BAC is also used to compute the TCPI and TSPI. man ffprobe (1): ffprobe gathers information from multimedia streams and prints it in human- and machine-readable fashion. Among them, the only time spent for value added activities is the process time. Average voltage (Vavg). In this problem the initial velocity is 10 mph and the final velocity is 7. Materiales de aprendizaje gratuitos. This means that the most recent entry always overwrites the oldest one. key},quantity). pdf) or read book online for free. Calculate the average velocity of your car (in cm/tu) for the entire run using the formula Vavg = total distance covered/total time. avg speed =\u221a(8RT/\u03c0M). What do you calculate for the x component of average velocity? vavg. Express your answer in terms of the given quantities and,. HTflux is using the following equations for this task: for laminar pipe flows: Resistance coefficient / hydraulic gradient (R in kg/m 7 ):. If by \"going at a If and only if the acceleration is constant, then you can make use the naive formula for average speed, as. Cone : Formulas. Plugging that into the avg voltage formula: Vavg = 2x217. 00000001 1E-08 3561 CHAPTER 28 To find digit. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave displayed on an oscilloscope screen. Enter Data and Calculate. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. RMS vs Average. 10Beta) CHANGELOG. By combining the first two criteria and plugging in the formula of and Update Vavg, sigmaV End Save Vmax,Vmin,Vavg,sigmaV to \u201cgamd_restart. Average Velocity Formula: vavg = x/ t x is the DISPLACEMENT where x = xf - xi (final position - initial position) and t is the time interval over which the displacement occurs. Solution: Use the instantaneous formulas. 2) An easy way to calculate the Vd\u03b1 expression is to place the time origin aligned to an output voltage pulse for which is applied the average value formula, in the. Because they both start at zero, the formula to convert between the two very easy (in = cm * 0. Non contact or with integrated conductor or clip-on current measurement from 0. since only their ratio enter the equation. 5) mph/ 2 giving us 8. Excel formula in column H is: Vin1 (volts) Formula 28-31 DATA = Vin1 x (255/5 V) dec. Learn physics formulas chapter 3 with free interactive flashcards. This means that equation 7 can be reduced to. pdf - Free ebook download as PDF File (. This histogram shows a theoretical distribution of speeds of molecules in a sample of nitrogen ( ) gas. V m = maximum value of transformer secondary voltage. In physics, you can calculate power based on force and speed. Since velocity is not constant, we find the avg velocity by doing 1/2(vf+vi) and that Vavg=dt. 5 m/s 2, Vi=0 Vf=????, no T. The formula was created using high-performance. Hukum Boyle : V \u221d 1/P (n dan T tetap) Hukum Charles : V \u221d T ( n dan P konstan). web; books; video; audio; software; images; Toggle navigation. i need the derivations that how these formulae comes. Vavg = 2146 / 10 = 214. 99 for most events). EVM - Miscellaneous Formula - Budget at Completion (BAC) is the total budget allocated to the BAC is also used to compute the TCPI and TSPI. Because, the positive and negative half cycle is equal in magnitude and thus the total value cancels out on summation. If final Velocity V and Initial velocity U are known, we make use of the formula. Difference between Vavg and Vrms If my Vp is let say 10V and I was to rectify it using ideal diodes and caps, wouldn't the output be 2*10V/\u03c0=6. Silica is most commonly found in nature as sand or quartz, as well as in the cell walls of diatoms. Rank the following gases from highest temperature to lowest temperature. v o represents the object's initial velocity at the beginning of the time interval. You need to integrate overtime for average velocity. Formula 15-4 IT = \u00d6IR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / \u00d6 R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. t2 \u2212t1 Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. Several questions might immediately come to mind. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). the split filter instance has two output pads, and the overlay filter instance two input pads. measurements by: Vavg = (VI +V2)/2. Math 106-350/550 Answers to Homework #4 15 Sept \u201906. 166 Pa theoretically, which is quite close to what Fluent gives!. Si multiplicamos por 3600 obtenemos vatios-segundos, que tambi\u00e9n se conoce como Julios. vavg = average velocity. For example it can be used to check the format of the container used by a multimedia stream and the format and type of each media stream contained in. 230K likes. Alternate Formulas or Computational Formulas for Variance. Dalam gas ideal berlaku 3 hukum tentang kinetika gas yaitu. Suppose we forgot the formula that relates mass to density and volume. The very basic version of a settlement formula in a Florida personal injury case is: Settlement = (Full Value of Damages) x (100% \u2013 % Chance of Losing Case) x (100% \u2013 Your % of Fault) Need to Know Every Part of Settlement Formula If you don\u2019t know any part of this formula, you settlement calculation will be way off. Shows you the step-by-step solutions using the quadratic formula! This calculator will solve your problems. Learn physics formulas chapter 3 with free interactive flashcards. When evaluating an arithmetic expression, FFmpeg uses an internal formula evaluator, implemented through the libavutil/eval. Formula 1 is trying to get back on our screens in July, just as F1 2020 arrives. Average speed formula? Unanswered Questions. 636 x (1/2) Vs peak = 0. If you know average value of voltage then you can use following formula to get peak voltage: Vpeak = (V. 1 While a non-true RMS or AC rectified average multimeter may not be accurate to measure non pure. IVA 02081070977 - All Rights Reserved - Privacy Policy. Para un flujo de tuber\u00eda laminar completamente desarrollado, Vavg es la mitad de la velocidad m\u00e1xima. There are user input variables to decide which dimensions will be used. Other multimeters may be designed to measure just the average level. The RMS value of a pulse waveform can be easily calculated starting with the RMS definition. Vavg = total distance/total time = 60. If you know average value of voltage then you can use following formula to get peak voltage: Vpeak = (V. The (average!) velocity profile in a turbulent flow is more flattened than the parabolic profile in a laminar flow. COMPUTERSC 207 - Fall 2019. What is the formula to calculate p-value?. Pressure and Temperature: A Molecular View. The slope of the line on these graphs is equal to the acceleration of the object. A third measure is the root-mean-square (rms) speed, , equal to the square root of. 28*SQRT(D863*E863*F863*G863)) 1000 20000 0. However, the object\u2019s speed, v, is just s divided by t, so \u2026. It includes information about system configuration, database management, user and run registration, configuration of communications links with other BIS systems, and day to day administration activities. 091s on the red-walled C4 compound before he brought out the first red flag of the week with a stoppage shortly afterwards. 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. Velocity and Most Probable Velocity \u2013 a relationship in equation. Display electrical specifications such as rise time, slew rate, amplifier gain, and current. These values will be listed and identified as either. Acoustic impedance is the product of velocity and density. In a coordinate system with north being the positive x-direction, the car's motion is in the southern direction (see figure). Thus, the total of the masses of neutrons and protons is the atomic mass. Related Topics: More Statistics Lessons. In this equation. To compute V P-P from the average voltage, the average voltage is multiplied by 3. Common Derivatives. The displacement due to acceleration is represented by the green triangle. The heat capacity at constant pressure can be estimated because the difference between the molar Cp and Cv is R; Cp \u2013 Cv = R. 0 4/14/03 6:27 PM Page 1. 1-3 can be rewritten: Suppose that a car travels at a constant speed s (in feet per second) down a straight highway (see Fig. Blue dot is position of amplitude for dipping Sand 3 before seismic migration moved amplitude 1100 m laterally and 275 m (200 ms) vertically updip (Vavg-2860 m/s). Exam 1H Rev Ques. In this blog, we shall discuss various Ratio Analysis, the various Ratios Formulae, and their importance. Average acceleration in circular motion? A mass moves on a circular path of radius 2 m at constant speed 4m/s, what is the magnitude and direction of the average acceleration during a quarter of a revolution(it is shown on a diagram to be from the bottom (like 6 on a clock) to the right (like 3 on a clock))?. 6591 \u00d7 \u221a(R) Vavg(He) = \u221a(8RT\u03c0 \u00d7 4) = 0. Calculate the average speed between each two adjacent picket fence stripes and record them in the table. Average Voltage (Vavg) As the name implies, Vavg is calculated by taking the average Since finding a full derivation of the formulas for root-mean-square (Vrms) voltage is difficult, it is done here for you. you can use the approximate formula in the box in Figure 2. Yes, watt-hours is a measure of energy, just like kilowatt-hours. Velocidad promedio Vavg se define como la velocidad promedio a trav\u00e9s de una secci\u00f3n transversal. 20 m/s vi = 3,900 m/s vavg =? 2. Interface composition (i COMPRISING i_ref). why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. F1, FORMULA ONE, FORMULA 1, FIA FORMULA ONE WORLD CHAMPIONSHIP, GRAND PRIX and related marks are trade marks of Formula One Licensing B. FIA Lurani Trophy for Formula Junior Cars. The formula was created using high-performance. Do While (Vavg >= Vavg_error_lb And Vavg =< Vavg_error_ub) ' Call Sim_rand 'Loop Thank you. t i = Initial time. Pavg=____ B) Let us now consider. 3, 4, & 5 represents the relationship of depth and velocity as derived by velocity modeling. Vavg=Vpeak/pi for half wave rectifier Vavg=2Vpeak/pi for full wave. Algebra Formulas For Class 12. Stamped Concrete Pavers: $13-$20 per sq. This set focuses on formulas, important numbers, word problems, and linear motion terminology. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. formula during a Tp time interval (\u03c0/3 radians): A 1 ( ) 1 average value of ( ) 0 0 Area T v t dt T V v t p T d p d not d p = = \u222b \u22c5 = \u22c5 (13. It's calculated by taking one cycle of a periodic waveform and squaring it, and finding the square root. Yes, watt-hours is a measure of energy, just like kilowatt-hours. displacement, B. It calculates the RMS voltage based on the given equations. The first output pad of split is labelled \"L1\", the first input pad of overlay is labelled \"L2\", and the second output pad of split is linked to the second input pad of overlay, which are both unlabelled. Let\u2019s recalculate the trapezoidal signal RMS value with the method shown in Equation 1. Balanced three-phase output: Line voltage 3 1. vavg \u2261 (x - x0) / t The definition of average velocity vavg = \u00bd (v + v0) The equation for average velocity in the case of constant acceleration Since we have two equations for average velocity, vavg, they must be equal to each other. You need to integrate overtime for average velocity. It is akin to Ohm\u2019s Law, but something I was never taught in tech school \u2013I had to learn it myself much later \u2013now I use it at least weekly. These values will be listed and identified as either. This banner text can have markup. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. Input a constant (in this case: 1) as your Formula and configure the Output Extents. Confirmed Page owner: Formula ProSCo Production LLC. Kinetic Theory of Gases & Maxwell Distribution Curve & Gas Velocity Vmp,Vavg,Vrms Formulas(Class XI). This set focuses on formulas, important numbers, word problems, and linear motion terminology. For the second equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= squareroot (2ax), x is same as h I just used different symbol. Sponsored Links. | Vavg - V1-2 | + | Vavg - V1-3 | + | Vavg - V2-3 | 2 x Vavg % voltage imbalance = x 100 The maximum allowable voltage imbalance is 2%. The formula is based on theoretical optics and incorporates both regression and artificial intelligence components to further refine its predictions. Securities Act of 1933 and deemed filed pursuant. You know that vavg = 12. 168 m furlong ) 20. Hi, How do you code, \"not greater than or equal to\" and \"not less than or equal to\" in vb. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. Materiales de aprendizaje gratuitos. 637 Vpp = Vp \u2022 2 Vp = Vpp \u00f7 2 Vinst. A Framework for Mining Association Rules in Data Warehouses 161 Assume that in a fact table there is m-dimensions and the quantity attribute exists. Now if you read a lot of other literature on Precision and Recall, you cannot avoid the other measure, F1 which is a function of Precision and Recall. The complete formula (EFN) is expressed as: EFN = (A/S) x (\u0394 Sales) - (L/S) x (\u0394 Sales) - (PM x FS x (1-d)). You should know that the velocity/time equation is just one important equation using velocity , but others exist. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. How do I generate a calculation to linearize data that has an inverse relationship? This video shows how to transform data from a Boyle's Law experiment so that the data falls along a straight line. Average Voltage of a Waveform in Analytical Method. Solved: Hi All, Can anyone suggest how to display the count of values displayed in pivot box into a text box. Calculator wich uses trigonometric formula to simplify trigonometric expression. It simply involves taking the total revenue in a given time period by the number of users in that time period. Confirmed Page owner: Formula ProSCo Production LLC. 6-8 Months of age: Breastfeed every 3-4 hours or Formula 24-37 ounces. The Reynolds number has Vavg=0. 99 for most events). THX! asked by Tim on December 30, 2015; math. Pursuant to Rule 425 under the. 637 times the amplitude (2*Vp/\u03c0), Vrms doesn't sounds right to me. Namun pada daily chart DOW dan FTSE, tampak RSI cenderung flat dan Reva sudah mencapai batas atas yang mengisyaratkan kalau akan ada koreksi dulu atau malah berbalik arah ke bawah lagi. The formula was created using high-performance. Chapter 1 \u2013 Student Solutions Manual 3. If you wish to find any term (also known as the {n^{th}} term) in the arithmetic sequence, the arithmetic sequence formula should help you to do so. That's the formula for average acceleration. The beautiful and perhaps mysterious formula of Euler which is the subject of this section is. For power computations, the true RMS multimeter provides the. The instantaneous velocity is just the velocity of an object at a specific point in time. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). Using the compound interest formula, you can determine how your money might grow with regular deposits or. vavg = average velocity. Average Voltage of a Waveform in Analytical Method. 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. \u04241 \u043f\u043e\u043f\u043e\u043b\u043d\u0438\u043b\u0430 \u0440\u044f\u0434\u044b \u0444\u0443\u0442\u0431\u043e\u043b\u0438\u0441\u0442\u043e\u043c. the peak voltage ( Vo ) = half the peak to peak voltage = 60 / 2 = 30 V. 5 kg flies through the air at a low speed, so that the air resistance is negligible, What is the net force acting on the ball while it is in motion? Which components of the ball's. Harvard University. Enzyme-linked Immunosorbent Assay (ELISA) The concentrations of NF-\u03baB-responsive genes (IL-8, TNF-a and IL-1\u03b2) were evaluated in cell culture supernatants according to the manufacturer\u2019s protocol (Joyee Biotechnics, Shanghai, China). Vavg VO + 1/2 gt (7) xx) Keep in mind the initial velocity vo will be zero for each trial. Record this in your data table. A ball of mass 0. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. In calculating Vavg, why are you only integrating the first part from 0 to pi/2? Click to expand First part is symmetrical function and average value is zero so for complete cycle in symmetrical so i only took half the time which is equal to pi/2, but i think since i only took half should i multiply it by 2 to get averwge for complete cycle?. Physics Help! Thread starter allies; Start date Oct 3, Vavg = (5300- xi)/4 I derive the formula for the students from the most basic formulas, then show them. 2 Derive the manometer formula \u2206p= (\u03c1m\u2212\u03c1)gh, where his the manometer reading (height), \u03c1m the density of the manometer liquid and \u2206pthe pressure di\ufb00erence across two horizontal points in the \ufb02uid of density \u03c1(U-tube manometer, see Example 2. Consequently, aavg = 2. That seems a little low just on a gut feeling, but maybe it is OK?. Using the given conversion factors, we find (a) the distance d in rods to be d = 4. 7V The back of the book says it is 136V. 0 Introduction: Quality Assurance and Certification June 17, 2009 ECMPS Reporting Instructions Quality Assurance and Certification 1. For the waveform shown above, the peak amplitude and peak value are the same, since the average value of the function is zero volts. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave that is displayed on an oscilloscope screen. 0 furlongs = ( 4. Manufacturing Cycle Efficiency Formula. Calculate the rms Value V(rms value) Related Calculations. vavg[t1 , t2 ] = x(t2 )\u2212x(t1 ). 637 Vp = Vavg \u00f7. A sled of mass (m) is being pulled horizontally by a constant diagonally upward force of magnitude (F) that makes an angle (theta) with the direction of motion. Business Information Server for Microsoft Windows Administration Help (7846 0284) This Help provides direction for BIS system administrators. x i = Initial Distance. Pressure and Temperature: A Molecular View. Best Price Roadside Assistance. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. Biblioteca en l\u00ednea. As illustrated in the \ufb01gure, the slope can be positive, negative, or zero. The highest fre- quencies of 2 to 4 MHz correspond to. Here, in part 1 of this short series on the topic, we. Louis, MO. With calculus, you can find it when given the displacement function, a function which tells you distance moved. If the object is moving with a velocity of -8 m/s, then the slope of the line will be -8 m/s. November 13 2019 Veton K\u00ebpuska 14 Hierarchical definition of 74x138 like. t2 \u2212t1 Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. Formula 15-4 IT = \u00d6IR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / \u00d6 R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. As stated by Investopedia, acceptable solvency ratios vary from. 22 except that it has been offset by the addition of a dc component equal to 2 V. Formula SAE Italy, Formula Electric Italy & Formula Driverless 2020. Find books. Thinking they might prove useful as tools or templates to others, it is my pleasure to provide them freely to the scientific community. Related Topics: More Statistics Lessons. ok at first to locate the displacement the -a million. Vavg VO + 1/2 gt (7) xx) Keep in mind the initial velocity vo will be zero for each trial. The average velocity of gas particles is found using the root mean square velocity formula: \u03bc rms = (3RT/M) \u00bd \u03bc rms = root mean square velocity in m/sec R = ideal gas constant = 8. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. This means that equation 7 can be reduced to. It simply involves taking the total revenue in a given time period by the number of users in that time period. (Since the sign of the downward velocity and the downward acceleration is positive, a negative velocity must be directed upward. It states that the volume of a fixed mass of gas at a constant temperature is inversely proportional to the pressure of the gas. 5 km/h, and \u0394t = (15min)(1h/60min) =. Find the volume per organism, assuming a spherical shape. The root mean square velocity (RMS velocity) is a way to find a single velocity value for the particles. 3, 4, & 5 represents the relationship of depth and velocity as derived by velocity modeling. November 13 2019 Veton K\u00ebpuska 14 Hierarchical definition of 74x138 like. V m = maximum value of transformer secondary voltage. It is possible, however, to have negative Voltage Unbalance readings if the device is a Series 600 Power Meter or a Series 2000 Circuit Monitor. Solution: Use the average velocity formula! Therefore, the average velocity is 8 2) Find the velocity function and the acceleration function for the function s(t) = 2t 3 + 5t - 7. But i have only 240 V supply for the heater, can I put a resistor in series with the heater to re. UserVar = (duser 1, duser 2. Physics Help! Thread starter allies; Start date Oct 3, Vavg = (5300- xi)/4 I derive the formula for the students from the most basic formulas, then show them. Stormblood Patch Items. CAGR is equivalent to the more generic exponential growth rate when the exponential growth interval is one year. Hence, the ratio is not the same; the maximum is higher in the laminar case. To compute V P-P from the average voltage, the average voltage is multiplied by 3. This oscillator formula can also be used as the basis for program trading and backtesting in StrategyDesk. CX Formula will improve your customer experience, increase customer loyalty, profits, referrability and happiness!. Voltage imbalance is given by the formula: Ring connect screw terminals C. For the second equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= squareroot (2ax), x is same as h I just used different symbol. \u0411\u043e\u0441\u0441 Renault \u0437\u0430 \u043e\u0442\u043a\u0430\u0437 \u043e\u0442 \u043a\u0438\u0431\u0435\u0440\u0433\u043e\u043d\u043e\u043a. The mass of oil is to be determined. We conclude that p f = 22 atm. This value can be found using the formula: v rms = [3RT/M] 1/2 where v rms = average velocity or root mean square velocity R = ideal gas constant T = absolute temperature M = molar mass The first step is to convert the temperatures to absolute temperatures. clc clear KB=1. Variation is ma = -mg sin theta - mu*mg cos theta if motion of block is UP slope. The heat capacity at constant pressure can be estimated because the difference between the molar Cp and Cv is R; Cp \u2013 Cv = R. The instantaneous velocity is just the velocity of an object at a specific point in time. It makes the textbook definitions more comprehensible! SET M\u2026. For the waveform shown above, the peak amplitude and peak. The ratio t1/T is the pulse signal duty-cycle. 494 m/s North. Several questions might immediately come to mind. Maintaining a body fat percentage from 6 to 9 percent. N = 2(72 h/0. The displacement due to acceleration is represented by the green triangle. 175 * 10 -3 * d 4 / (f t c vavg),. the split filter instance has two output pads, and the overlay filter instance two input pads. V0 is the value assigned for the pipette to dispense. vavg=vi+vf2, when the acceleration is constant, where vi. For Boyle's Law to be valid, the other two variables must be held constant. vavg[t1 , t2 ] = x(t2 )\u2212x(t1 ). The units for the volume and pressure can be left in l and atm. I can use the formula \u0394Vc = Vm(1-e-t/RC) I don't know why it didn't click sooner, it will become \u0394Vc = 19. Harvard University. Formula: Example: How many ways can 4 students from a group of 15 be lined up for a. Biblioteca en l\u00ednea. 5 m/s 2, Vi=0 ) D= 6. Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online. For a sine wave: Vrms = 0. Depth conversion (domain conversion) of seismic time interpretations and data is a basic skill set for interpreters. Calculate dvava/dT at T = 300 K for 0\u00f6ygen, which has a molar mass of 0. Area of a Circle Formula. If you have questions about this formula or functionality, please call TD Ameritrade's StrategyDesk free helpline at 800 228-8056, or access the online Help Center via the StrategyDesk application. Acoustic impedance is the product of velocity and density. vavg \u2261 (x - x0) / t The definition of average velocity vavg = \u00bd (v + v0) The equation for average velocity in the case of constant acceleration Since we have two equations for average velocity, vavg, they must be equal to each other. Namun pada daily chart DOW dan FTSE, tampak RSI cenderung flat dan Reva sudah mencapai batas atas yang mengisyaratkan kalau akan ada koreksi dulu atau malah berbalik arah ke bawah lagi. The instantaneous velocity is just the velocity of an object at a specific point in time. New formulas from other definitions: D= Vi*T + \u00bd *A*T 2 and Vf 2 = Vi 2 + 2*A*D. Ninguna Categoria; Subido por Angie Melissa Fluidos- Frank M. 0 furlongs )( 201. We use cookies to improve your navigation experience. The coefficient of kinetic friction is (uk). V m = maximum value of transformer secondary voltage. Copyright \u00a9 2007-2019 Formula S. 54 centimetres. Practical Depth Conversion with Petrel. V2-3 = Voltage between phases 2 and 3. Basic Properties and Formulas. 758) + A=AIR(3. 9 V = 54 V As you can see from this example, even though the peak output of the full wave center tapped rectifier was half that of the half wave rectifier, the average output was the same because the full wave center tapped rectifier doubles the number of half-cycles at the output, compared to a half. The PERIODIC TIME (given the symbol T) is the time, in seconds milliseconds etc. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. And a is basically g however a in this case. It includes information about system configuration, database management, user and run registration, configuration of communications links with other BIS systems, and day to day administration activities. 175 * 10 -3 * d 4 / (f t c vavg),. 0 m above a flat horizontal beach. That seems a little low just on a gut feeling, but maybe it is OK?. Calculate the average speed between each two adjacent picket fence stripes and record them in the table. 3; PA=RA*P; PB=RB*P; % B=CH4(3. Therefore, the average voltage value is 214. Voltage Unbalance is calculated using the following formula: Vunbalance = [(Vavg-Vphase) divided by Vavg]. RMS vs Average. 5 kW motor with 110V, 40W anti condensation heater. When compared to protons or neutrons electrons have so much less mass that they don\u2019t influence the calculation. Formula \u2022 \u10e4\u10dd\u10e0\u10db\u10e3\u10da\u10d0, Tbilisi, Georgia. Securities Act of 1933 and deemed filed pursuant. s defines the following functions: fit. D=Vavg*T Vavg= (Vi+Vf)/2 A= (Vf-Vi)/T. Average speed is equal to displacement over time, so the formula would be: vavg = (x2 - x1) / (t2 - t1). 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. 5) World Championship. If a = constant, then vavg = (v + v0)/2 where \u2206x = displacement. 637 times the amplitude (2*Vp/\u03c0), Vrms doesn't sounds right to me. Answer / soumya ranjan panigrahi we know form factor is the cause of generating voltage in India is 11kV or it be transmitted as multiples of 11, because of simple to desine. Use Equation 5 Vf 2 = Vi 2 + 2*A*D. 168 m furlong ) 20. It may also be referred to as the annualized rate of return or annual percent yield or effective annual rate. Alternatively, the same r4 can be solved from The plot above shows the average speed vavg for various downhill speeds v2 in this Example 2. Maintaining a body fat percentage from 6 to 9 percent. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. 5 h) Solve for N. Using the formula, we calculate v to be equal to (10+7. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave that is displayed on an oscilloscope screen. Therefore, it is acceptable to choose the first quarter cycle, which goes from 0 radians (0\u00b0) through \u03c0 /2 radians (90\u00b0). 1 A to 20,000 A. Total Kinetic Energy. So I thought the ripple voltage was approximated by the formula Vr,pp = Vp / fRC for a half-wave rectifier, and Vp/2fRC for a full-wave. Pressure and Temperature: A Molecular View. Filed by ITC Holdings Corp. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. For power computations, the true RMS multimeter provides the. When evaluating an arithmetic expression, FFmpeg uses an internal formula evaluator, implemented through the libavutil/eval. 00000001 1E-08 3561 CHAPTER 28 To find digit. Calculator solution will show work for real Uses the quadratic formula to solve a second-order polynomial equation or quadratic equation. measurements by: Vavg = (VI +V2)/2. SPARKCHARTSTM. formulae_8&9 (2). vi or rather vi stands for initial velocity. 28*SQRT(D863*E863*F863*G863)) 1000 20000 0. 2 for time, with y = 0, using the quadratic formula (choosing the positive root to yield a positive value for t). Exam 1 November 2012, Questions and answers Exam 31 October 2008, Questions and answers Exam 26 October 2009, Questions Exam 10 December 2010, Questions Final Formula Sheet - Summary Mechanics and Waves Phys 131 Equation Sheet. The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation For the Quadratic Formula to work, you must have your equation arranged in the form \"(quadratic). Unlike the book, this calculator employs the velocity of light $$c$$ at full precision, resulting in slightly shorter, but more precise lengths. VAVG - (-VAVG ) ) in the sum of the two areas, thus resulting in zero average voltage over one To find the peak value from a given average voltage value, just rearrange the formula and divide by the. Previously Viewed. Make sure you enter. When the payments are all the same, this can be considered a geometric series with 1+r as the common ratio. Formula 15-4 IT = \u00d6IR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / \u00d6 R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. Express your answer in terms of the given quantities and,. Cost can add up quickly, especially if you\u2019re a novice and have never attempted a Stamped Concrete Patios installation before. The hydrocarbon that has all single bond is called as alkane with the general formula C n H 2 n + 2. Basic Properties and Formulas. As stated by Investopedia, acceptable solvency ratios vary from. As with the Vrms formula, a full derivation for the Vavg formula is given here as well. The average velocity of the sled is (Vavg). Describe another way to find the area of a circle when given the circumference. RMS Voltage (Vrms) The root-mean-square or effective value of a waveform. Substitute in the given equation and find the unknown quantity. Furthermore, the formula for computing the boom length \\(L. I want to code this do while loop to do this. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. Profit Margin Formula - Explained. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. CAGR is equivalent to the more generic exponential growth rate when the exponential growth interval is one year. The displacement due to acceleration is represented by the green triangle. Rank the following gases from highest temperature to lowest temperature. The critical step is to be able to identify or extract known. 6m/s2 isn't necessary so what you do is you employ the equation Vavg=deltax/delta t then you certainly re-manage it and get the equation (delta t)*(Vavg)=delta x this is what u want this is displacement so then you certainly plug interior the numbers so it would look like this (6. Any such data mashup is expressed using the Power Query M Formula Language. This banner text can have markup. 1 While a non-true RMS or AC rectified average multimeter may not be accurate to measure non pure. The formula for Compound Annual Growth Rate (CAGR) is very useful for investment analysis.\nw1yajadolhfj2d pfcwnl2prl b0wjmhxsb3bze jslvreox66mj3 ullg7q4psdjzsgb 5e6sxzf3egewj ectyo56zchx3mie ykft4ksc910 yl82cvurr5nq hpfdozvutfd1 v4gbi7ijdy9pm ogc7ej3obpnauya hln26zup1j2fsz kc7shadltc 0mcckjacgsuqt xw38swiwl8axdz0 w8mcn3yhlr36 tv3k690kmwn zsigmdwallhbr2f p7mrn8vilap z92m9wsi3j cfoxf87u8ni w8l2et3a7b 1oijiygm94mgl9 itcmwaapuhciq p7qfmkokii6fmz8 n4nj074fza", "date": "2020-07-07 09:08:12", "meta": {"domain": "freccezena.it", "url": "http://mryf.freccezena.it/vavg-formula.html", "openwebmath_score": 0.6601222157478333, "openwebmath_perplexity": 1900.9799295418693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9833429590144717, "lm_q2_score": 0.894789454880027, "lm_q1q2_score": 0.8798849102566718}}
{"url": "https://math.stackexchange.com/questions/4057688/prove-that-n35n-is-always-divisible-by-6-using-induction-with-mod/4057710", "text": "Prove that $n^3+5n$ is always divisible by 6 using induction with mod\n\nSo I had to prove this in my calculus exam today\n\nProve that $$n^3+5n$$ is always divisible by 6.\n\nI know that there have been other people around here having asked this already, but I chose a different approach and would really appreciate to hear whether my train of thought here was right or wrong. So I am especially not looking for the \"right way\" to prove it (i.e., I don't need a proper solution) I would just appreciate for somebody to assess my solution, if it's completely wrong or possibly not complete etc.\n\nso, what I did:\n\nInitial case prove that $$n^3+5n \\bmod 6 = 0$$ for $$n=1$$. $$1^3+5\\cdot 1 \\bmod 6 = 0 \\Leftrightarrow 6 \\bmod 6 = 0$$ which is trivially true.\n\nInduction step Prove that for every $$n$$, if the statement holds for $$n$$, then it holds for $$n + 1$$. $$((n+1)^3+5(n+1)) \\bmod 6 =0 \\\\ \\Leftrightarrow ((n^3+3n^2+3n+1)+5n+5) \\bmod 6 =0 \\\\ \\Leftrightarrow (n^3+5n \\bmod 6) + (3n^2+3n+6 \\bmod 6)=0 \\\\ \\Leftrightarrow 0+(3n^2+3n+6 \\bmod 6)=0 \\\\ \\Leftrightarrow 3n^2+3n+6 \\bmod 6 =0.$$\n\nAnd this is true for all $$n \\in \\mathbb{N}$$, where from the third to the fourth line, we replaced with the induction hypothesis that $$n^3+5n \\bmod 6 =0$$ . Take for instance $$1, 2,3,...$$ it is always divisible by 6 without leaving any rest!!\n\nHowever, my other friends that took the exam said that this is in need of some extension to show that the last line actually holds true for whatever n you insert. But I cannot grasp what they mean; by the definition of the modulus, it should already trivially be given that the last line is right...\n\nI'm really looking forward to your thoughts, do you think I would at least get partial points for this? :( I really don't see how this can be wrong...\n\nThanks so much, Lin\n\n\u2022 You just need to show that $2\\mid n(n+1)$, then it follows that $6\\mid 3n^2+3n+6$. This is certainly only a partial loss of points. Mar 11, 2021 at 12:43\n\u2022 Yes, it is \"in principle\" right. I would give full points for some of the solutions here, and less points for you, because your solution is a bit too long and not every step is clear. Have a look at the duplicate yourself! Mar 11, 2021 at 12:45\n\u2022 Please, see the edits I did to your post so you learn how to use better mathjax. Mar 11, 2021 at 12:47\n\u2022 Bitte sch\u00f6n, you are welcome! This site has good solutions, and sometimes this helps to improve the own solution a bit. And the advantage is, you can do this on your own. Mar 11, 2021 at 12:51\n\u2022 BTW, a simpler way to solve this is to see that $$n^3+5n \\equiv n^3-n = (n-1)n(n+1)\\pmod 6$$, and of course both $2$ & $3$ must divide $(n-1)n(n+1)$. Mar 11, 2021 at 13:11\n\nIn view of the final step, $$3n^2+3n+6\\equiv 0\\mod 6$$ is equivalent to $$3n^2+3n\\equiv 0\\mod 6$$, i.e.,\n\n$$3(n^2+n)\\equiv 0\\mod 6$$.\n\nSo it is sufficient to show that\n\n$$2\\mid n^2+n$$.\n\nBut $$n^2+n = n(n+1)$$ is a product of consecutive integers and one of which will be even. Thus $$n^2+n$$ is a multiple of $$2$$ and hence $$3(n^2+n)$$ is a multiple of $$6$$, as required.\n\n\u2022 So you also agree that my induction is right, and only the part you mention is missing? :) Mar 11, 2021 at 12:47\n\nYour friends are right. You need to explain why the last equivalent equation is an identity, meaning it solves for any integer n greater than zero.\n\nTo do that you show that $$3n^2+3n+6$$ is divisible both by 2 and 3. Divisibility by 3 is obvious since the expression is a multiple of 3. You show that by rewriting the expression into $$3(n^2+n+2)$$ Divisibility by 2 results from observing that the product $$n(n+1)$$ of two consecutive numbers is always an even number: $$n^2+n+2=n(n+1)+2=2k\\cdot(2k\\pm1) +2=2[k(2k\\pm1)]$$\n\n\u2022 Thanks! But just for the fun of it, out of 12 points in total, how many would you give me for my solution? xP Mar 11, 2021 at 13:02\n\u2022 Is arguable since you failed to finish to prove p(n)->p(n+1). I guess you could loose 1 or 2p depending on how you presented the induction method. p(1) true and p(n)->p(n+1) then p(n) true for n>0. Depends on the teacher. Mar 11, 2021 at 13:10\n\u2022 yeyyyyy! that is enough excitement for me :D Mar 11, 2021 at 13:11\n\nThis is correct, but as your friends said, not proving the last line may cause some loss of points. To prove this, it's not very \"long\" or something. Just show that $$3n^2+3n+6\\equiv 0\\ (\\textrm{mod}\\ 2)$$ Since it is trivial that it is $$0\\ \\textrm{mod}\\ 3$$, the statement follows.\n\nHope this helps. Ask anything if not clear :)\n\n\u2022 Thanks so much to you too!! Mar 11, 2021 at 12:48\n\u2022 Thanks @LinShao, but one more thing: accept either my or the other answer here just to ensure that this is answered. Mar 11, 2021 at 12:50\n\nSince your final step implies $$3\\mid 3n^2-3n$$, it suffices to show $$2\\mid n^2-n$$ for any $$n$$, which can be seen as trivial. That was all you missed in my opinion.\n\n\u2022 Thanks for also giving me hope :) After all I'm not a total idiot in math, it seems xD Mar 11, 2021 at 13:36", "date": "2022-10-06 01:44:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4057688/prove-that-n35n-is-always-divisible-by-6-using-induction-with-mod/4057710", "openwebmath_score": 0.6905090808868408, "openwebmath_perplexity": 274.18160228518104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357173731317, "lm_q2_score": 0.8962513779831579, "lm_q1q2_score": 0.8798819895109533}}
{"url": "https://math.stackexchange.com/questions/2817200/writing-english-statement-into-predicate-logic-using-quantifiers", "text": "Writing english statement into predicate logic using quantifiers.\n\nBelow was the exercise problem I was solving from Discrete Mathematics by Kenneth Rosen (for preparation of GATE Exam) and I have doubt in it.\n\nLet S(x) be the predicate that \"x is a student\", F(x) be the predicate \"x is a faculty member\", and A(x,y) the predicate \"x has asked y a question\", where the domain consists of all people associated with your school. Use quantifiers to express each of these statements.\n\n(f)Some student has asked every faculty member a question.\n\nNow my doubt is\n\nit can be framed like there is at least one student such that for all faculty members, he must have asked them a question.\n\nso I wrote my expression as\n\n\u2203x ( (S(x) ^\u00a0\u2200y ( F(y)\u00a0$\\rightarrow$ A(x,y) ) )\n\nBut in Rosen answer is given as below and I have 2 doubts in it.\n\nRosen's Ans :\u00a0\u2200y ( (F(y)\u00a0$\\rightarrow$\u00a0\u2203x ( S(x) v A(x,y) ) )\n\nDoubt 1: I think in above expression we must have \"and operator\" instead of \"or operator\" in the second part of expression which is quantified by existential quantifier and so it should be\n\n\u2200y ( (F(y)\u00a0$\\rightarrow$\u00a0\u2203x ( S(x) ^ A(x,y) ) )\n\nDoubt 2: What is the difference between my answer and Rosen's answer.Which one is correct.\n\nYou're right about the operator: that should definitely be a $\\land$, rather than a $\\lor$\n\nOtherwise, the difference is this: you took the statement \"Some student has asked every faculty member a question.\" to mean that it was the same student who asked every faculty member a quesrion. And, with that interpretation, your answer is correct, and Rosen's (even with the $\\land$ instead of the $\\lor$) would be incorrect.\n\nHowever, English (like all natural languages) is ambiguous, and you can also interpret the statement \"Some student has asked every faculty member a question.\" as saying that for each faculty member there is some student (but this time not necessarily the same one for each faculty member) that asked that faculty member a question. And if that is the intended meaning of the sentence, then Rosen's answer (again, with the $\\land$ instead of the $\\lor$) is correct.\n\nNow, the latter is certainly not a very intuitive reading of the statement, and I much prefer the former interpretation, and hence your answer! However, some people do mean the latter interpretation when expressing the English statement.\n\n\u2022 @Bram28-Thank you so much! :) Jun 13 '18 at 4:29\n\u2022 @user3767495 you're welcome! :) Jun 13 '18 at 14:12\n\nRosen's intended answer, $\\forall y~ (F(y)\\to \u2203x~( S(x) \\wedge A(x,y) ) )$, claims that for any faculty member there is some student that has asked them a question. \u00a0 They need not be the same students.\n\n\u2022 That is: \"Every faculty member has been asked a question by some student.\"\n\nYour answer, $\\exists x~\\forall y~(S(x)\\wedge (F(y)\\to A(x,y)))$ claims that some student will have asked every faculty member a question. \u00a0 Which is what was required.\n\n\u2022 That is: \"Some student has asked every faculty member a question.\"\n\nIt is logical distinction which is oftimes misrepresented in natural language.", "date": "2021-11-29 06:12:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2817200/writing-english-statement-into-predicate-logic-using-quantifiers", "openwebmath_score": 0.5678178668022156, "openwebmath_perplexity": 195.056592227889, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226294209298, "lm_q2_score": 0.9005297907896396, "lm_q1q2_score": 0.8798379840691735}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=93oma9kuo201v4m95gj21v8ih5&action=printpage;topic=2631.0", "text": "# Toronto Math Forum\n\n## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 1 => Topic started by: Weihan Luo on January 14, 2022, 12:38:03 AM\n\nTitle: Classification of PDEs\nPost by: Weihan Luo on January 14, 2022, 12:38:03 AM\nI am a little bit confused about the classifications of PDES. Namely, I have trouble distinguishing between linear equations versus quasi-linear equations.\n\nIn particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?\nTitle: Re: Classification of PDEs\nPost by: Victor Ivrii on January 14, 2022, 02:45:57 AM\nIn particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?\nFirst, it will be not just quasilinear, but also\u00a0 semilinear. Second, it will also be linear since you can move $c(x,y)u$ to the left\n\nGood job, you mastered some $\\LaTeX$ basics. :)\nTitle: Re: Classification of PDEs\nPost by: Weihan Luo on January 14, 2022, 11:28:53 AM\nThank you for your response.\n\nDoes it mean that all linear PDEs are also quasilinear/or semilinear? If so, on a quiz, I should classify those PDEs as linear right?\nTitle: Re: Classification of PDEs\nPost by: Victor Ivrii on January 14, 2022, 01:47:15 PM\nYes, all linear are also semilinear and all semilinear are also quasilinear. For full mark you need to provide the most precise classification. So, if equation is linear you say \"linear\", if it is semilinear but not\u00a0 linear you say \"semilinear but not\u00a0 linear\" and so on,... \"quasilinear but not\u00a0 semilinear\" and \"non-linear and not quasilinear\".", "date": "2022-07-01 02:26:55", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=93oma9kuo201v4m95gj21v8ih5&action=printpage;topic=2631.0", "openwebmath_score": 0.9062845706939697, "openwebmath_perplexity": 1991.3996795385826, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109480714625, "lm_q2_score": 0.8933094159957173, "lm_q1q2_score": 0.8797408928879069}}
{"url": "https://www.freemathhelp.com/forum/threads/finding-the-radius-of-a-circle-impossible.123596/", "text": "# Finding the radius of a circle... impossible?\n\n#### Bailey232008\n\n##### New member\nI\u2019ve been working on this for too long. I almost posted this in the Odds and Ends section because I think it\u2019s going to entail more than just geometry. Give this one a shot. Could be a fun challenge for you vets out there if it\u2019s even possible. If you want the back story behind how I came up with it let me know.\nThe only information provided is what you see in the picture. Solve by any means necessary...\n\n#### MarkFL\n\n##### Super Moderator\nStaff member\nBy Heron, the area $$A$$ of the triangle is:\n\n$$\\displaystyle A=22.5\\sqrt{(R+22.5)(R-22.5)}$$\n\nWe also know:\n\n$$\\displaystyle A=22.5(R-5)$$\n\nThis implies:\n\n$$\\displaystyle \\sqrt{(R+22.5)(R-22.5)}=R-5$$\n\nSolving this, we find:\n\n$$\\displaystyle R=53.125$$\n\n#### pka\n\n##### Elite Member\nI\u2019ve been working on this for too long. I almost posted this in the Odds and Ends section because I think it\u2019s going to entail more than just geometry. Give this one a shot. Could be a fun challenge for you vets out there if it\u2019s even possible. If you want the back story behind how I came up with it let me know.\nThe only information provided is what you see in the picture. Solve by any means necessary...View attachment 20083\nOn it we see $$R=\\dfrac{c^2}{8h}+\\dfrac{h}{2}$$\nFrom the above diagram $$c=45'~\\&~h=5'$$.\n\n#### Jomo\n\n##### Elite Member\nDraw a line from where it says 45 to the center of the circle. This gives you a right triangle. What are the lengths of the three sides of this right triangle. Now ask Pythagorous for help.\n\n#### Bailey232008\n\n##### New member\nDraw a line from where it says 45 to the center of the circle. This gives you a right triangle. What are the lengths of the three sides of this right triangle. Now ask Pythagorous for help.\nYou lack the interior angle, so this method doesn\u2019t work\n\n#### Bailey232008\n\n##### New member\nMark and Pka,\n\ngreat stuff, appreciate the help\n\n#### Dr.Peterson\n\n##### Elite Member\nYou lack the interior angle, so this method doesn\u2019t work\nYou misunderstood Jomo's suggestion, which doesn't use any angles. In fact, this is how the formula pka gave is derived!\n\nThe sides of the triangle are r-h, c/2, and r (the hypotenuse), and solving the Pythagorean theorem for r yields the formula.\n\n#### Jomo\n\n##### Elite Member\nYou lack the interior angle, so this method doesn\u2019t work\nAs already mentioned Pythagorous' theorem does not use angles. you have the 3 sides of the right triangle in terms of R. What seems to be the problem?\n\n#### jonah2.0\n\n##### Junior Member\nBeer goggles hallucination follows.\n\nLooked at the picture and saw 45\u00b0 instead of 45'. Ended up thinking R as roughly 65.69.\n\n#### Jomo\n\n##### Elite Member\nYou lack the interior angle, so this method doesn\u2019t work\nWhy couldn't you use the 45o angle that you thought was there?\n\n#### yoscar04\n\n##### Junior Member\nFollowing Jomo's suggestion using straightforward Pythagoras theorem, you should get R=53.125.", "date": "2020-07-13 15:23:17", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/finding-the-radius-of-a-circle-impossible.123596/", "openwebmath_score": 0.5702428221702576, "openwebmath_perplexity": 1370.7863762036652, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308803517762, "lm_q2_score": 0.8902942377652497, "lm_q1q2_score": 0.8797272289350898}}
{"url": "http://mathhelpforum.com/algebra/3150-proof-sum-rational-number.html", "text": "# Math Help - Proof that sum is rational number\n\n1. ## Proof that sum is rational number\n\nI have to prove that\n$\\frac{1}{{\\sqrt 1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\frac{1}{{\\sqrt 3 + \\sqrt 4 }} + ... + \\frac{1}{{\\sqrt {99} + \\sqrt {100} }}$\nis rational number.\n\nI have one idea, I hope it's not crazy!\n\nI have spoted that the sum of first three fractions are equal to 1:\n\n$\\frac{1}{{\\sqrt 1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\frac{1}{{\\sqrt 3 + \\sqrt 4 }}=1$\n\nI have then discovered that next five fractions are also equal to 1.\n$\\frac{1}{{\\sqrt 4 + \\sqrt 5 }} + \\frac{1}{{\\sqrt 5 + \\sqrt 6 }} + \\frac{1}{{\\sqrt 6 + \\sqrt 7 }} + \\frac{1}{{\\sqrt 7 + \\sqrt 8 }} + \\frac{1}{{\\sqrt 8 + \\sqrt 9 }} = 1$.\n\nContinuing, we have that every odd number of fractions are equal to 1.\nConsidering that there are 99 fractions in that sum we get that because of $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\\frac{1}{{\\sqrt 1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\frac{1}{{\\sqrt 3 + \\sqrt 4 }} + ... + \\frac{1}{{\\sqrt {99} + \\sqrt {100} }}=9$\n\nI don't know how to prove that but it seems to me that there is a rule in that sum which I have described.\n\n2. Originally Posted by OReilly\nI have to prove that\n$\\frac{1}{{\\sqrt 1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\frac{1}{{\\sqrt 3 + \\sqrt 4 }} + ... + \\frac{1}{{\\sqrt {99} + \\sqrt {100} }}$\nis rational number.\n\nI have one idea, I hope it's not crazy!\n\nI have spoted that the sum of first three fractions are equal to 1:\n\n$\\frac{1}{{\\sqrt 1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\frac{1}{{\\sqrt 3 + \\sqrt 4 }}=1$\n\nI have then discovered that next five fractions are also equal to 1.\n$\\frac{1}{{\\sqrt 4 + \\sqrt 5 }} + \\frac{1}{{\\sqrt 5 + \\sqrt 6 }} + \\frac{1}{{\\sqrt 6 + \\sqrt 7 }} + \\frac{1}{{\\sqrt 7 + \\sqrt 8 }} + \\frac{1}{{\\sqrt 8 + \\sqrt 9 }} = 1$.\n\nContinuing, we have that every odd number of fractions are equal to 1.\nConsidering that there are 99 fractions in that sum we get that because of $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\\frac{1}{{\\sqrt 1 + \\sqrt 2 }} + \\frac{1}{{\\sqrt 2 + \\sqrt 3 }} + \\frac{1}{{\\sqrt 3 + \\sqrt 4 }} + ... + \\frac{1}{{\\sqrt {99} + \\sqrt {100} }}=9$\n\nI don't know how to prove that but it seems to me that there is a rule in that sum which I have described.\nOne word,\nrationalize.\n\n3. Originally Posted by ThePerfectHacker\nOne word,\nrationalize.\nBit more words?\n\n4. Originally Posted by OReilly\nBit more words?\nOkay whenever you have,\n$\\frac{1}{\\sqrt{x}\\pm \\sqrt{y}}$ you can change the fraction, called \"rationalizing\". This is done by mutiplying the numerator and denominator by $\\sqrt{x}\\mp \\sqrt{y}$. This changes the denominator into whole number because you use the difference of two squares.\nThus,\n$\\frac{1}{\\sqrt{x}\\pm \\sqrt{y}}\\cdot \\frac{\\sqrt{x}\\mp \\sqrt{y}}{\\sqrt{x}\\mp \\sqrt{y}}=\\frac{\\sqrt{x}\\mp \\sqrt{y}}{x-y}$\n------------\nHence you have,\n$\\frac{1}{\\sqrt{1}+ \\sqrt{2} }+\\frac{1}{\\sqrt{2}+ \\sqrt{3} }+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+...+\\frac{1}{ \\sqrt{99}+\\sqrt{100}}$\nUpon rationalizing,\n$\\frac{\\sqrt{1}-\\sqrt{2}}{-1}+\\frac{\\sqrt{2}-\\sqrt{3}}{-1}+\\frac{\\sqrt{3}-\\sqrt{4}}{-1}+...+\\frac{\\sqrt{99}-\\sqrt{100}}{-1}$\nSimplyfy,\n$-1+\\sqrt{2}-\\sqrt{2}+\\sqrt{3}-\\sqrt{3}+\\sqrt{4}-...-\\sqrt{99}+\\sqrt{100}$\nEach one cancels each other out (such a series is called telescoping), thus,\n$-1+\\sqrt{100}=-1+10=9$\n\n5. Thanks!\n\n6. Welcomed.\nThat happens to be a nice problem which I really like. Where did you get it from?\n\n7. Originally Posted by ThePerfectHacker\nWelcomed.\nThat happens to be a nice problem which I really like. Where did you get it from?\nFrom high school book. Author has also wrote book with variety of problems which i find most interesting.", "date": "2014-04-18 12:07:36", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/3150-proof-sum-rational-number.html", "openwebmath_score": 0.9133924841880798, "openwebmath_perplexity": 540.2592539464355, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765610497293, "lm_q2_score": 0.8976952975813454, "lm_q1q2_score": 0.8797203505942802}}
{"url": "https://stats.stackexchange.com/questions/221728/3-vehicles-out-of-5-randomly-showcased-what-is-the-probability-of-each-vehicle", "text": "3 vehicles out of 5 randomly showcased, what is the probability of each vehicle being showcased?\n\nI have a basic probability doubt. If I have 5 different vehicles backstage, out of which 3 random can be showcased to the public. What is the probability of each vehicle to get to showcase? Once a car is in a showcase, it will not be returned to backstage.\n\nI thought by doing the calculating the combinations- ${5 \\choose 3}=10$. And probability of 3 slots getting filled (let's say $P$) - $P=\\frac{1}{5}\\times\\frac{1}{4}\\times\\frac{1}{3}$. So, total probability is $10\\,P= \\frac{1}{6}$.\n\nOr will it be $\\frac{3}{5}$?\n\nOr any other solution?\n\nI assume that all cars are equally likely to be chosen.\n\nSuppose without loss of generality the cars are labeled 1 through 5. The probability of not choosing car 1 is (4/5) * (3/4) * (2/3) = 2/5, so the probability of choosing it is 1 - 2/5 = 3/5. Of course, by my assumption above, the same argument applies to any of the cars. The answer 3/5 makes intuitive sense since we're drawing 3 things from a set of 5.\n\n\u2022 But what about the possible combinations of C1,C3,C4 or C2,C5,C1 etc. Will the 3/5 probability have that into account as well? \u2013\u00a0Joe Jul 1 '16 at 18:55\n\u2022 You can renumber the cars however you like and the argument is the same. \u2013\u00a0Kodiologist Jul 1 '16 at 18:56\n\u2022 Thank You! Can you please tell in which scenario will my logic below (the same i described in the question) be used <br>- 5C3=10 And probability of 3 slots getting filled (let say P)- 1/5*1/4*1/3 So, total probability = 10*P= 1/6 \u2013\u00a0Joe Jul 5 '16 at 13:36\n\u2022 If we have 60 Cars backstage in place of 5, will he situation become- Probability of not chosing car1 is (59/60)*(58/59)*(57/58) = 57/60, so the probability of chosing it will become 1-(57/60)= 3/60 \u2013\u00a0Joe Jul 7 '16 at 13:51\n\u2022 Yes, that's right. \u2013\u00a0Kodiologist Jul 7 '16 at 14:45\n\nYou asked for alternative approaches. Here is one you might find useful.\n\nLet's begin by stating the obvious: you are implicitly assuming the five probabilities are equal. The expected total in the showcase equals the sum of those probabilities, whence it is five times any one of them. Yet the expected total is the average value of all possible totals, weighted by their chances of occurring. Since by design the possible total is always $3$, its average must be $3$. Therefore each probability is $3/5$.\n\nThe power of this reasoning about expectations becomes clear when you generalize the question to $k$ cars in the showcase to be chosen (with equal probabilities) out of $n$ cars backstage.", "date": "2020-10-20 06:50:55", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/221728/3-vehicles-out-of-5-randomly-showcased-what-is-the-probability-of-each-vehicle", "openwebmath_score": 0.8771622180938721, "openwebmath_perplexity": 679.954234913469, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991554373058874, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8797115961767227}}
{"url": "http://slideplayer.com/slide/5878287/", "text": "# Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:\n\n## Presentation on theme: \"Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:\"\u2014 Presentation transcript:\n\nChapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:\n\nDefinition 2: Matrix Addition Let A = ( ai j ) & B = ( bi j ) be m\uf0b4n matrices. Then C = A + B is an m\uf0b4n matrix with elements ci j \uf03d ai j + bi j . Example: Properties: (A1) A + B = B + A (A2) (A + B) + C = A + (B + C) (A1) A + 0 = A where 0 is a zero matrix (matrix consisting of only zero elements)\n\nMatrix Operations: scalar multiplication\nDefinition 3: Scalar Multiplication Let A = ( ai j ) and \u03bb \uf0ce R. Then C = \u03bb A is an m\uf0b4n matrix with elements ci j \uf03d \u03bb ai j . Properties: (S1) \u03bb A = A \u03bb (S2) \u03bb (A + B) = \u03bb A + \u03bb B (S3) (\u03bb1 + \u03bb2) A = \u03bb1 A + \u03bb2 A (S4) \u03bb1 (\u03bb2 A) =(\u03bb1\u03bb2) A\n\n1.3 Matrix Multiplication\nDefinition: Matrix Multiplication If A = ( ai j ) is an m\uf0b4n matrix and B = ( bi j ) is an n\uf0b4p matrix, Then the product AB is defined to be an m\uf0b4p matrix C = ( ci j ) where ci j = ai 1 b1 j + ai 2 b2 j + \u2026 + ai n bn j In other words, ci j = ( Row i of A ) \uf0d7 ( Column j of B ) Example: Properties: (M1) (AB)C = A(BC) (M2) A(B + C) = AB + AC (M3) (B + C)A = BA + CA\n\nMatrix multiplication lacks commutativity.\nExample 1: Example 2: Other properties that matrix multiplication lacks: 1) AB = 0 doesn\u2019t imply A = 0 or B = 0 2) AB = AC doesn\u2019t imply B = C\n\n1.4 Special Matrices Definition (matrix transpose): Given a matrix A, the transpose of A, denoted by AT and read A-transpose, is obtained by changing all the rows of A into columns of AT while preserving the order. Examples: Properties of the transpose: 1) (AT)T = A 2) (\u03bb A)T = \u03bb AT 3) (A + B)T = AT + BT 4) (A B)T = BT AT A symmetric matrix is a matrix that is equal to its transpose while a skew symmetric matrix is a matrix that is equal to the negative of its transpose.\n\n1.4 Special Matrices: row-reduced form\nDefinition: A matrix is in row-reduced form if it satisfies four conditions: (R1) All zero rows appear below nonzero rows when both types are present in the matrix. (R2) The first nonzero element in any nonzero row is unity. (R3) All elements directly below (that is, in the same column but in succeeding rows from) the first nonzero element of a nonzero row are zero. (R4) The first nonzero element of any nonzero row appears in a later column (further to the right) than the first nonzero element in any preceding row. Example:\n\nRow-reduced form: examples\nIn row-reduced form Not in row-reduced form\n\n1.4 Special Matrices: identity matrice\nDefinition: A square matrix that has 1\u2019s on the main diagonal and 0\u2019s off the main diagonal is called an identity matrix. Example: A 3 \u00d7 3 identity matrix. Note: An identity matrix has the property that AI = IA = A.\n\n1.4 Special Matrices: lower (upper) triangular matrices\nDefinition: A square matrix A=[aij] is called lower triangular if aij=0 for j>i (that is, if all the elements above the main diagonal are zero) and upper triangular if aij=0 for j<i (that is, if all the elements below the main diagonal are zero). Example:\n\n1.6 Vectors x1 x2 v Definition: A vector is a 1 \uf0b4 n or n \uf0b4 1 matrix. \uf071\nMagnitude: If , Is a UNIT vector A nonzero vector is normalized if it is divided by its magnitude. Is a unit vector\n\n1.7 The geometry of vectors\nVector Addition v w v+w Vector Subtraction v-w v w\n\nDownload ppt \"Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:\"\n\nSimilar presentations", "date": "2019-10-20 19:10:11", "meta": {"domain": "slideplayer.com", "url": "http://slideplayer.com/slide/5878287/", "openwebmath_score": 0.8592785000801086, "openwebmath_perplexity": 1443.955436470143, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9724147169737826, "lm_q2_score": 0.9046505434556231, "lm_q1q2_score": 0.8796955021745784}}
{"url": "https://math.stackexchange.com/questions/326383/is-a-negative-number-proper-fractions", "text": "# Is a negative number proper fractions?\n\nI was asked this question by a kid, is -$\\frac{4}{7}$ a proper fraction or not? As per my knowledge $\\frac{4}{7}$ is a proper fraction. If it has a -ve number does it make any difference? Definition says A number whose numerator is smaller than denominator is called a proper fraction. Can we consider -$\\frac{4}{7}$as a proper fraction? If not why not please explain. This is my first question I don't have much idea about tags of mathematics if it is tagged wrongly please edit it.\n\nThank you\n\nDibya\n\n\u2022 mathworld.wolfram.com/ProperFraction.html So it's neither proper or improper. \u2013\u00a0Lazar Ljubenovi\u0107 Mar 10 '13 at 12:23\n\u2022 It is proper: en.wikipedia.org/wiki/\u2026 \u2013\u00a0Dennis Gulko Mar 10 '13 at 12:23\n\u2022 If I had to define a convention that makes sense in the context of where questions like this have relevance, I would say that -4/7 isn't a fraction at all; it is a numeral consisting of a negative sign - and a (proper) fraction 4/7. \u2013\u00a0Hurkyl Mar 10 '13 at 12:26\n\nFrom Wikipedia:\n\nCommon fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise. In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one\u2014that is, if the fraction is between $-1$ and $1$.* [Italics mine]\n\nSo $\\;-1 < \\left(-\\dfrac 47\\right) < 1$ is considered a proper fraction; (alternatively $\\;\\;0 < \\Big|-\\dfrac 47 \\Big| = \\dfrac 47 < 1$.\n\nI think in terms of mathworld's definition: When using the division algorithm, for example, one requires an integer quotient $\\times$ an integer divisor, plus a non-negative integer remainder less than the value of the divisor. So if dividing $-4$ by $7$:\n\n$$-4 = -1\\cdot 7 + 3, \\;\\;q = -1;\\;\\;r = 3, i.e., -\\dfrac 47 = -1 + \\dfrac 37$$ which would be a mixed fraction.\n\nSo the fractional part would be the proper fraction $\\dfrac 37$, the integer part, $-1$. Consistent with this, mathworld may require that a proper fraction occurs only when the quotient is $0$, and the remainder a positive integer less than the divisor, hence the fractional part = $\\dfrac rd\\; d:$divisor,$\\;r:$ the remainder when dividing number by $d$.\n\n\u2022 I have edited my question added a link form mathworld given by Lazar in the comment. This is confusing. Which one is correct? \u2013\u00a0NewUser Mar 10 '13 at 12:29\n\u2022 It's a matter of convention - By any other name, it'd still smell the same. \u2013\u00a0Guest 86 Mar 10 '13 at 12:33\n\u2022 The Wikipedia definition is the standard. I think mathworld left out the clause that the $0 <$ |fraction|$< 1$ \u2013\u00a0Namaste Mar 10 '13 at 12:34", "date": "2019-07-19 08:15:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/326383/is-a-negative-number-proper-fractions", "openwebmath_score": 0.9158980846405029, "openwebmath_perplexity": 458.2153494644806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147201714923, "lm_q2_score": 0.9046505331728751, "lm_q1q2_score": 0.8796954950682927}}
{"url": "https://mathematica.stackexchange.com/questions/72392/probability-of-multivariate-normal-being-positive-on-each-coordinate", "text": "# Probability of multivariate normal being positive on each coordinate\n\nHow can I find the probability that each coordinate of a specified multivariate normal distribution is positive? I tried the following, which I believed should work\n\nmu = {0, 0, 0};\nsigma = {{2, 1, 1}, {1, 2, 1}, {1, 1, 2}};\nProbability[\nx > 0 && y > 0 && z > 0, {x, y, z} \\[Distributed]\nMultinormalDistribution[mu, sigma]]\n\n\nUnfortunately, for the output I just get the last line from the input (with mu and sigma replaced by their actual values). I don't see where the problem could possibly be since the matrix is positive definite. If I replace it by the identity matrix everything works fine (i.e. the output is 1/8).\n\nThe general integral does not have a closed form solution, so use NIntegrate:\n\nmu = {0, 0, 0};\nsigma = {{2, 1, 1}, {1, 2, 1}, {1, 1, 2}};\nNIntegrate[\nPDF[MultinormalDistribution[mu, sigma], {x, y, z}],\n{x, 0, \u221e}, {y, 0, \u221e}, {z, 0, \u221e}]\n\n\n(* 0.25 *)\n\nCheck:\n\nmu = {0, 0, 0};\nsigma2 = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};\nNIntegrate[\nPDF[MultinormalDistribution[mu, sigma2], {x, y, z}],\n{x, 0, \u221e}, {y, 0, \u221e}, {z, 0, \u221e}]\n\n\n(* 0.125 *)\n\nNote that symbolic integration and symbolic probability work for the two-dimensional version of this problem:\n\nmu = {0, 0};\nsigma = {{2, 1}, {1, 2}};\nProbability[x > 0 && y > 0, {x, y} \\[Distributed] MultinormalDistribution[mu, sigma]]\n\n\nand\n\nIntegrate[PDF[MultinormalDistribution[mu, sigma], {x, y}], {x, 0, \u221e}, {y, 0, \u221e}]\n\n\n(* 1/3 *)\n\n\u2022 An even simpler way is to replace Probability with NProbability (adds one additional character to the OP's code). \u2013\u00a0DumpsterDoofus Jan 24 '15 at 0:41\n\u2022 Yep... Simpler. \u2013\u00a0David G. Stork Jan 24 '15 at 0:42\n\u2022 Nice, thank you very much \u2013\u00a0Filanto Renika Jan 24 '15 at 0:52\n\u2022 The general integral for the OPs $n=3$ case does actually have a closed-form solution: posted below. \u2013\u00a0wolfies Jan 24 '15 at 16:47\n\nThe probability that the OP seeks is known as the multivariate Normal orthant probability. Correctly, for the $n=3$ cased posed here, the general integral DOES in fact have a closed -form solution, though Mma cannot (currently) obtain it.\n\nIn particular, given a zero mean vector and variance-covariance matrix:\n\n$$\\Sigma =\\left( \\begin{array}{ccc} 1 & \\rho _{\\text{xy}} & \\rho _{\\text{xz}} \\\\ \\rho _{\\text{xy}} & 1 & \\rho _{\\text{yz}} \\\\ \\rho _{\\text{xz}} & \\rho _{\\text{yz}} & 1 \\\\ \\end{array} \\right)$$\n\n$\\dots$ the standardised trivariate Normal orthant probability is:\n\n$$P(X>0,Y>0,Z>0) \\quad = \\quad \\frac{1}{8} + \\frac{\\text{ArcSin}\\left[\\rho _{\\text{xy}}\\right]+ \\text{ArcSin}\\left[\\rho _{\\text{xz}}\\right]+ \\text{ArcSin}\\left[\\rho _{\\text{yz}}\\right]}{4 \\pi }$$\n\nFor application and more detail, see, for instance:\n\n\u2022 Chapter 6 of our book: Rose and Smith, Mathematical Statistics with Mathematica (Section 6.4C) $\\rightarrow$ a free download is available at: http://www.mathstatica.com/book/bookcontents.html , or\n\n\u2022 Stuart and Ord (1994), Kendall's Advanced Theory of Statistics (6th edition): section 15.10 and 15.11.\n\nExample\n\nLet $(X,Y,Z)$ have a standardised multivariate Normal with zero mean vector, and variance covariance matrix:\n\nsigma = {{1, 27/34, 22/23}, {27/34, 1, 4/5}, {22/23, 4/5, 1}}\n\n\nThen, the exact orthant probability is given immediately as:\n\nP3 = 1/8 + (ArcSin[27/34] + ArcSin[22/23] + ArcSin[4/5])/(4 Pi)\n\n\n... which, to 10 decimal places, is:\n\nN[P3, 10]\n0.3732564868\n\n\nBy contrast, the approach using numerical integration can be unreliable here:\n\nNIntegrate[ PDF[MultinormalDistribution[{0, 0, 0}, sigma], {x, y, z}], {x, 0, Infinity}, {y, 0, Infinity}, {z, 0, Infinity}]\n\n\nNIntegrate::slwcon: Numerical integration converging too slowly ...\n\n0.371907\n\nNumerical integration can sometimes be awkward, unreliable, or slow (as in this example) ... and having an exact closed-form instantaneous solution is a better way to proceed, if possible.\n\nFrom general to standardised\n\nGiven a zero mean vector, what if our variance-covariance matrix is not in a standardised form (with 1's along the main diagonal)? We can easily convert it into standardised form. If, say, our variance-covariance matrix is:\n\n  sigma = {{3, 1/3, 3/2}, {1/3, 2/3, -1}, {3/2, -1, 4}}\n\n\n... then the standardised variance-covariance matrix S is:\n\n A = DiagonalMatrix[Diagonal[sigma]^(-(1/2))];\nS = A.sigma.A\n\n\nAll done.\n\n\u2022 I eagerly looked up your book because I wanted to see \"more detail\"... but there isn't any more detail, just a restatement of the same formula and a reference to Stuart and Ord's book. \u2013\u00a0Rahul Jan 24 '15 at 16:51\n\u2022 The book also gives the closed-form solution for the bivariate case, and then uses these exact formulae as benchmarks to check the performance/accuracy of Mma's numerical integration routines. In case you are about to eagerly consult Stuart and Ord, I should perhaps caution that you won't find very much more detail there either for the $n=2$ or 3 cases as to results (though more as to derivation)... but they do also provide useful references for higher dimensions. \u2013\u00a0wolfies Jan 24 '15 at 16:59\n\u2022 Not quite all done. Let's help the OP by giving him the closed form solution for his particular covariance matrix. \u2013\u00a0David G. Stork Jan 24 '15 at 19:21\n\u2022 @Rahul I believe the result can be obtained by a linear change of variables that diagonalizes and normalizes the quadratic form of the pdf (to Exp[k (u^2+v^2+w^2)]). The orthant is transformed to a domain that is subtended by a spherical triangle with angles $\\pi/2+ \\sin^{-1}\\rho$, where $\\rho$ runs over the nondiagonal coefficients of $\\Sigma$. The result is the proportion of the sphere that is cut out (by the symmetry of the integral). A similar result is true for $n=2$. \u2013\u00a0Michael E2 Jan 24 '15 at 23:22", "date": "2019-11-15 02:42:13", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/72392/probability-of-multivariate-normal-being-positive-on-each-coordinate", "openwebmath_score": 0.5703652501106262, "openwebmath_perplexity": 1148.0532276858262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147153749275, "lm_q2_score": 0.9046505286741726, "lm_q1q2_score": 0.8796954863544733}}
{"url": "https://math.stackexchange.com/questions/2834533/finding-lim-limits-n-to-infty-sqrtn2a-n-given-that-lim-limits-n", "text": "# Finding $\\lim\\limits_{n \\to \\infty} \\sqrt[n^2]{a_n}$ given that $\\lim\\limits_{n \\to \\infty} \\frac{a_n a_{n+2}}{(a_{n+1})^2}=L^2$\n\nThis is the problem I saw in another question - which was later closed and deleted (the close reason was missing context). Still, the problem seemed interesting to me - at least in the sense that the solution is not immediately obvious.\n\nThe problem:\n\nSuppose that $(a_n)$ is a real sequence such that $a_n>0$ for $n>0$ and $$\\lim\\limits_{n\\to\\infty} \\frac{a_na_{n+2}}{a_{n+1}^2}=L^2.$$ Suppose further that $L\\ge0$ Find $\\lim\\limits_{n\\to\\infty} \\sqrt[n^2]{a_n}$. (In terms of $L$.)\n\nThe original problem in the linked question was given with $L=2$. However, I do not think that the problem should change that much for any value of $L$. (Maybe one could be suspicious about $L=0$?)\n\nIt is also clear - at least for $L>0$ - that if it is possible to express the second limit using $L$, then the limit should be equal to $L$. It suffices to notice that for $a_n = L^{n^2}$ we have $$\\frac{a_na_{n+2}}{a_{n+1}^2}= \\frac{L^{n^2+(n+2)^2}}{L^{2(n+1)^2}} = L^{(2n^2+4n+4)-2(n^2+2n+2)} = L^2.$$\n\nSo we know what is the candidate for the result, we still need to prove whether this is actually true.\n\nI have posted my attempt as answer. But I will be glad to learn about other solutions. (And of course, also corrections to my approach, if I made some mistakes.)\n\nFor the sake of simplicity, let us assume that $L>0$. And later we will check what has to be modified in the case $L=0$.\n\nLet us denote $b_n = \\frac{\\log a_n}{\\log L}$. (I.e., the logarithm at the base $L$.) Then the limit $$\\lim\\limits_{n\\to\\infty} \\frac{a_na_{n+2}}{a_{n+1}^2}=L^2$$ changes to $$\\lim\\limits_{n\\to\\infty} (b_{n+2}-2b_{n+1}+b_n) = 2.$$ And if we put $c_n=b_{n+1}-b_n$ then we have $$\\lim\\limits_{n\\to\\infty} (c_{n+1}-c_n)=2.$$ Applying Stolz-Cesaro theorem1 twice we get $$\\lim\\limits_{n\\to\\infty} \\frac{b_{n+1}-b_n}n = \\lim\\limits_{n\\to\\infty} \\frac{c_n}n =2$$ and $$\\lim\\limits_{n\\to\\infty} \\frac{b_n}{\\frac{n(n-1)}2}=2.$$\n\nTherefore we get $$\\lim\\limits_{n\\to\\infty} \\frac{b_n}{n(n-1)}=1 \\qquad\\text{and}\\qquad \\lim\\limits_{n\\to\\infty} \\frac{b_n}{n^2} = 1.$$ So we have $$\\lim\\limits_{n\\to\\infty} \\frac{\\log{a_n}}{n^2\\log L} = \\frac{\\log{\\sqrt[n^2]{a_n}}}{\\log L}=1$$ i.e. $$\\lim\\limits_{n\\to\\infty} \\log \\sqrt[n^2]{a_n}=\\log L \\qquad\\text{and}\\qquad \\lim\\limits_{n\\to\\infty} \\sqrt[n^2]{a_n}=L.$$\n\nWhat happens if $L=0$? Already the first step does make sense, since we cannot take logarithm of $L^2=0$.\n\nLet us now take $b_n=\\log a_n$. (To avoid $\\log L$ in denominator-- which was there just to simplify the numbers we work with, so that we do not have to write $\\log L$ every time.)\n\nIn the first step we have just used that $x_n\\to L$ implies $\\log x_n\\to\\log L$. If we have $x_n\\to0^+$, we still have $\\log x_n\\to -\\infty$. So for $b_n=\\log a_n$ we $\\lim\\limits_{n\\to\\infty} (b_{n+2}-2b_{n+1}+b_n) = -\\infty$.\n\nStolz-Cesaro theorem is true also for $\\pm\\infty$. Which means that the following steps are correct if we replace $2$ by $-\\infty$.\n\nFinally we arrive at $\\lim\\limits_{n\\to\\infty} \\log \\sqrt[n^2]{a_n}=-\\infty$, which gives us $$\\lim\\limits_{n\\to\\infty} \\sqrt[n^2]{a_n}=0$$ simply by using that $\\log x_n\\to-\\infty$ implies $x_n\\to0^+$.\n\nWhich means that the same result is true for $L=0$.\n\n1See Wikipedia: Stolz\u2013Cesaro theorem. Some proofs of this result can be found in this question.\n\n\u2022 Just one doubt. The question was asked 3 Minutes ago. And you posted such a long answer 3 minutes ago. How is this even possible? \u2013\u00a0user567182 Jun 28 '18 at 8:38\n\u2022 @user567182 It is possible to post a question and an answer at the same time. This feature was mentioned on meta around the time when it was introduced: Recently rolled out SE Encyclopedia feature. In which situations it is suitable is another question - but it is definitely encouraged that the poster shows their own attempts when posting the question, posting self-answer might be a way to do this. If you are interested in other past discussions about this, you might have a look at posts tagged (self-answer) on meta. \u2013\u00a0Martin Sleziak Jun 28 '18 at 8:42\n\u2022 Perfect answer. +1 I had mentioned the same approach in comments to the original (now closed deleted question). \u2013\u00a0Paramanand Singh Jun 28 '18 at 14:32\n\nHere is an answer that I posted to the question mentioned in this question, but deleted almost immediately due to the lack of context in that question.\n\nIf $$\\lim_{n\\to\\infty}\\frac{a_n\\,a_{n+2}}{a_{n+1}^2}=L^2\\tag1$$ then for any $\\epsilon\\gt0$, there is an $N_\\epsilon$ so that for $n\\ge N_\\epsilon$, $$\\left|\\,\\log\\left(\\frac{a_n\\,a_{n+2}}{a_{n+1}^2}\\right)-2\\log(L)\\,\\right|\\le\\epsilon\\tag2$$ Note that \\begin{align} \\frac1{n^2}\\sum_{k=0}^{n-1}\\sum_{j=0}^{k-1}\\log\\left(\\frac{a_j\\,a_{j+2}}{a_{j+1}^2}\\right) &=\\frac1{n^2}\\sum_{k=0}^{n-1}\\log\\left(\\frac{a_0}{a_k}\\frac{a_{k+1}}{a_1}\\right)\\\\ &=\\frac1n\\log\\left(\\frac{a_0}{a_1}\\right)+\\frac1{n^2}\\log\\left(\\frac{a_n}{a_0}\\right)\\tag3 \\end{align} Inspired by the following diagram\n\nWe can break down the sum on the left side of $(3)$ as follows \\begin{align} \\frac1{n^2}\\sum_{k=0}^{n-1}\\sum_{j=0}^{k-1}\\log\\left(\\frac{a_j\\,a_{j+2}}{a_{j+1}^2}\\right) &=\\frac1{n^2}\\sum_{k=N_\\epsilon}^{n-1}\\sum_{j=N_\\epsilon}^{k-1}\\overbrace{\\log\\left(\\frac{a_j\\,a_{j+2}}{a_{j+1}^2}\\right)}^\\text{within \\epsilon of 2\\log(L)}\\\\ &\\color{#C00}{+\\frac{n-N_\\epsilon}{n^2}\\sum_{j=0}^{N_\\epsilon-1}\\log\\left(\\frac{a_j\\,a_{j+2}}{a_{j+1}^2}\\right)+\\frac1{n^2}\\sum_{k=0}^{N_\\epsilon-1}\\sum_{j=0}^{k-1}\\log\\left(\\frac{a_j\\,a_{j+2}}{a_{j+1}^2}\\right)}\\tag4 \\end{align} For any $\\epsilon\\gt0$, the red terms vanish as $n\\to\\infty$. The remaining term on the right hand side is $\\frac1{n^2}$ times a sum of $\\frac{(n-N_\\epsilon)^2+(n-N_\\epsilon)}2$ terms, all within $\\epsilon$ of $2\\log(L)$.\n\nTherefore, $$\\left|\\,\\lim_{n\\to\\infty}\\frac1{n^2}\\sum_{k=0}^{n-1}\\sum_{j=0}^{k-1}\\log\\left(\\frac{a_n\\,a_{n+2}}{a_{n+1}^2}\\right)-\\log(L)\\,\\right|\\le\\frac\\epsilon2\\tag5$$ Since $(5)$ is true for any $\\epsilon\\gt0$, we have $$\\lim_{n\\to\\infty}\\frac1{n^2}\\sum_{k=0}^{n-1}\\sum_{j=0}^{k-1}\\log\\left(\\frac{a_n\\,a_{n+2}}{a_{n+1}^2}\\right)=\\log(L)\\tag6$$ Taking the limit of the right hand side of $(3)$ as $n\\to\\infty$, we get $$\\lim_{n\\to\\infty}\\frac{\\log(a_n)}{n^2}=\\log(L)\\tag7$$ which implies $$\\lim_{n\\to\\infty}\\sqrt[n^2]{a_n}=L\\tag8$$\n\n\u2022 It's great that you posted your answer here. By the time I saw your answer (to the old deleted question) it was in deleted state. +1 \u2013\u00a0Paramanand Singh Jun 29 '18 at 8:47\n\u2022 Not surprising, since it was only there for 18 seconds. \u2013\u00a0robjohn Jun 29 '18 at 11:50\n\nThis is not another solution, but is the essence of your solution worked out by me individually.\n\nIf $L>0$,\n\nBy two applications of Stolz-Cesaro theorem,\n\n\\begin{align*} \\lim_{n\\to\\infty}{{\\ln a_n}\\over{n^2}} &=\\lim_{n\\to\\infty}{\\ln{\\frac{a_{n+1}}{a_n}}\\over{2n+1}}\\\\ &=\\frac{1}{2}\\lim_{n\\to\\infty}{\\ln{\\frac{a_{n+1}}{a_n}}\\over{n}}\\\\ &=\\frac{1}{2}\\lim_{n\\to\\infty}{\\ln{\\frac{a_{n+2}}{a_{n+1}}-\\ln{\\frac{a_{n+1}}{a_n}}}\\over1}\\\\ &=\\frac{1}{2}\\lim_{n\\to\\infty}\\ln\\Bigg(\\frac{\\big(\\frac{a_{n+2}}{a_{n+1}}\\big)}{\\big(\\frac{a_{n+1}}{a_{n}}\\big)}\\Bigg)\\\\ &=\\frac{1}{2}\\lim_{n\\to\\infty}\\ln\\frac{a_na_{n+2}}{{a_{n+1}}^2}\\\\ &=\\frac{1}{2}\\ln(L^2)\\\\ &=\\ln L\\\\ \\therefore\\lim_{n\\to\\infty}\\sqrt[n^2]{a_n}&=L \\end{align*}\n\nIf $L=0$, \\begin{align*} \\lim_{n\\to\\infty}{{\\ln a_n}\\over{n^2}} &=\\lim_{n\\to\\infty}{\\ln{\\frac{a_{n+1}}{a_n}}\\over{2n+1}}\\\\ &=\\lim_{n\\to\\infty}{\\ln{\\frac{a_{n+1}}{a_n}}\\over{n}}\\\\ &=\\lim_{n\\to\\infty}{\\ln{\\frac{a_{n+2}}{a_{n+1}}-\\ln{\\frac{a_{n+1}}{a_n}}}\\over1}\\\\ &=\\lim_{n\\to\\infty}\\ln\\Bigg(\\frac{\\big(\\frac{a_{n+2}}{a_{n+1}}\\big)}{\\big(\\frac{a_{n+1}}{a_{n}}\\big)}\\Bigg)\\\\ &=\\lim_{n\\to\\infty}\\ln\\frac{a_na_{n+2}}{{a_{n+1}}^2}\\\\ &=-\\infty\\\\ \\therefore\\lim_{n\\to\\infty}\\sqrt[n^2]{a_n}&=0 \\end{align*}", "date": "2020-03-30 14:10:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2834533/finding-lim-limits-n-to-infty-sqrtn2a-n-given-that-lim-limits-n", "openwebmath_score": 0.9472048878669739, "openwebmath_perplexity": 256.0800944826854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846697584033, "lm_q2_score": 0.8991213772699435, "lm_q1q2_score": 0.8796865717729744}}
{"url": "https://mathematica.stackexchange.com/questions/13125/measuring-fractal-dimension-of-natural-objects-from-digital-images", "text": "# Measuring fractal dimension of natural objects from digital images\n\nThis is a useful topic. A college physics lab, medical diagnostics, urban growth, etc. - there is a lot of applications. On this site by Paul Bourke about Google Earth fractals we can get a high resolution images (in this post they are low res - import from source for experiments). For example, around Lake Nasser in Egypt:\n\nimg = Import[\"http://paulbourke.net/fractals/googleearth/egypt2.jpg\"]\n\n\nThe simplest method I know is Box Counting Method which has a lot of shortcomings. We start from extracting the boundary - which is the fractal object:\n\n{Binarize[img], iEdge = EdgeDetect[Binarize[img]]}\n\n\nNow we could partition image into boxes and see how many boxes have at least 1 white pixel. This is a very rudimentary implementation:\n\nMinS = Floor[Min[ImageDimensions[iEdge]]/2];\ndata = ParallelTable[{1/size, Total[Sign /@ (Total[#, 2] & /@ (ImageData /@\nFlatten[ImagePartition[iEdge, size]]))]}, {size, 10, MinS/2, 10}];\n\n\nFrom this the slope is 1.69415 which is a fractal dimension that makes sense\n\nline = Fit[Log[data], {1, x}, x]\n\n\n13.0276 + 1.69415 x\n\nPlot[line, {x, -6, -2}, Epilog -> Point[Log[data]],\nPlotStyle -> Red, Frame -> True, Axes -> False]\n\n\nBenchmark: if I run this on high res of Koch snowflake i get something like ~ 1.3 with more exact number being 4/log 3 \u2248 1.26186.\n\nQuestion: can we improve or go beyond the above box counting method?\n\nAll approaches are acceptable if they find fractal dimension from any image of natural fractal.\n\n\u2022 You have a lot of programs in mathematica to measure fractal dimensions and multi fractal spectrum of an image in the book: Fractal Geography, Andre Dauphine, Wiley, 2012 See the book on the Wolfram Mathematica | Books or Amazon ![enter image description here](wolfram.com/books/profile.cgi?id=8108) Oct 16, 2012 at 9:32\n\u2022 Vitaliy, with all due respect, the ending bullets make the question (actually there isn't a question mark anywhere, right?) rather wide ranging. Perhaps you could focus it somewhat and make it sound more like a real question? Oct 16, 2012 at 12:59\n\u2022 @SjoerdC.deVries Updated, it is a single question now. Oct 16, 2012 at 15:06\n\u2022 could you perhaps describe or link some examples of use of fractal dimension? Oct 17, 2012 at 10:27\n\u2022 @magma: Wikipedia lists some examples here: en.wikipedia.org/wiki/\u2026\n\u2013\u00a0shrx\nJun 14, 2013 at 20:22\n\nYou can still use box count, but doing it smarter :)\n\nCounting boxes with at least 1 white pixel from ImagePartition can be done more efficiently using Integral Image, a technique used by Viola-Jones (2004) in their now popular face recognition framework. For a mathematical motivation (and proof), Viola and Jones point to this source.\n\nWhat Integral Image allows you to do is to compute efficiently the total mass of any rectangle in an image. So, you can define the following:\n\nIntegralImage[d_] := Map[Accumulate, d, {0, 1}];\ndata = ImageData[ColorConvert[img, \"Grayscale\"]]; (* img: your snowflake image *)\nii = IntegralImage[data];\n\n\nThen, the mass (white content) of a region, is\n\n(* PixelCount: total mass in region delimited by two corner points,\ngiven ii, the IntegralImage *)\nPixelCount[ii_, {p0x_, p0y_}, {p1x_, p1y_}] :=\nii[[p1x, p1y]] + ii[[p0x, p0y]] - ii[[p1x, p0y]] - ii[[p0x, p1y]];\n\n\nSo, instead of partitioning the image using ImagePartition, you can get a list of all the boxes of a given size by\n\nPartitionBoxes[{rows_, cols_}, size_] :=\nTranspose /@ Tuples[{Partition[Range[1, rows, size], 2, 1],\nPartition[Range[1, cols, size], 2, 1]}];\n\n\nIf you apply PixelCount to above, as in your algorithm, you should have the same data but calculated faster.\n\nPixelCountsAtSize[{rows_, cols_}, ii_, size_] :=\n((PixelCount [ii, #1, #2] &) @@ # &) /@ PartitionBoxes[{rows, cols}, size];\n\n\nFollowing your approach here, we should then do\n\nfractalDimensionData =\nTable[{1/size, Total[Sign /@ PixelCountsAtSize[Dimensions[ii], ii, size]]},\n{size, 3, Floor[Min[Dimensions[ii]]/10]}];\nline = Fit[Log[fractalDimensionData], {1, x}, x]\n\nOut:= 10.4414 + 1.27104 x\n\n\nwhich is very close to the actual fractal dimension of the snowflake (which I used as input).\n\nTwo things to note. Because this is faster, I dared to generate the table at box size 3. Also, unlike ImagePartition, my partition boxes are all of the same size and therefore, it excludes uneven boxes at the edges. So, instead of doing minSize/2 as you did, I put minSize/10 - excluding bigger and misleading values for big boxes.\n\nHope this helps.\n\nUpdate\n\nJust ran the algorithm starting with 2 and got this 10.4371 + 1.27008 x. And starting with 1 is 10.4332 + 1.26919 x, much better. Of course, it takes longer but still under or around 1 min for your snowflake image.\n\nUpdate 2\n\nAnd finally, for your image from Google Earth (eqypt2.jpg) the output is (starting at 1-pixel boxes)\n\n12.1578 + 1.47597 x\n\n\nIt ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs.\n\n\u2022 There isn't a \"snowflake\" image. All three images come from the same object (and I guess the fractal dimension should be the same for all of them) Dec 23, 2013 at 21:03\n\u2022 @belisarius, the PO published a Koch snowflake image as Benchmark (see last part of post) and that's the one I used. I called it 'snowflake' for short. On the other hand, this is still a numerical procedure, so the numbers will be different depending on approximation. The Update2 refers to the eqypt2.jpg image, also made available by the PO, as he asked for an improvement suitable for real-life images. Dec 23, 2013 at 22:40\n\u2022 sorry, I missed that link. Thanks Dec 24, 2013 at 0:05\n\u2022 +1 This is very neat @caya, thank you! I will wait \"a bit\" hoping that someone can implement things like wavelet multi-fractals or similar. Feb 14, 2014 at 20:02\n\u2022 @VitaliyKaurov, glad you liked it. Note that Viola & Jones rightly pointed out a link between integral image and Haar wavelet basis in their paper; although this wasn't explored further. It is unclear to me the link between the paper you mentioned and fractal dimension, but certainly worth reading as I am interested in these kind of problems. Cheers. Feb 16, 2014 at 12:40\n\n\"Can we improve the box-counting method?\" Most certainly.\n\nBut first of all, it is important to point out that if you are only checking if a box is empty or not you are effectively measuring the $D_0$ (the capacity or in other words the box-counting dimension). By referring to $D_0$ as the fractal dimension you are assuming that your object is a monofractal, i.e. that all its generalized dimensions $D_q$ (ref.1) $$D_q=\\lim_{\\epsilon \\rightarrow \\infty}\\frac{\\frac{1}{1-q}\\log{\\sum_i^Np_i^q}}{\\log{\\frac{1}{\\epsilon}}}$$\n\nare equal; $D_0$=$D_1$(entropy dimension)=$D_2$(correlation dimension) and so on...\n\nNormally you should refrain from such assumptions unless you know the generating process. However, even if you assume that the object is monofractal, review literature of fractal dimension estimation concludes that the box-counting method is a bad choice. As Thelier puts it in \"box-counting algorithm is a poor way to estimate the box-counting dimension\" (ref. 2).\n\nSo what can we do? The basic principle in fractal dimension estimation is to study the object at different scales. The question is how to sample this continuum of scales. There are two different class of methods. The first one is the fixed-sized methods. These count the number of points in a fixed size region (boxes/spheres). They perform poorly because you never know how to grow your boxes (should I double, triple?). The second school is the fixed-mass methods. These grow their extend by reaching over to their mth nearest neighbor. Thus they always assure that they are covering the same mass (hence the name). even though fixed-mass methods are more robust than fixed-size methods, both suffer from finite-size and edge effect. Thus you need some extra measures to avoid the edges of your object, otherwise at large scales (of mass/size) you start to get this leveling off effect (that you also have in your last figure).\n\nA few years ago we developed a complicated method (non-overlapping barycentric fixed-mass method) to do exactly that (ref. 3). Here is a summary of the fractal dimension estimations with the BFM method:\n\nI first gave it a go with your Koch snowflake and I get all $D_q$s to be more or less equal to $1.25$ as you would expect from a monofractal.\n\nThen I run the Egypt2.jpg file and as you can see all $D_q$s are not exactly equal. This can be used to argue that the coastline might be multifractal (even if slightly). Another thing to notice is that the slope is not the same throughout the whole mass range. If you consider only the range $[10 - 631]$ you would get a different $D_q$ curve. This implies that the object is not self-similar throughout the entire scale. This could also also explains why the previous answer finds $D_0$ something around $1.47$ while you find $1.69$. In other words there is a scaling break.\n\nReferences:\n\n1. Renyi, A. (1970). Probability Theory. Amsterdam: North-Holland.\n2. Theiler, J. (1990). Estimating fractal dimension. Journal of the Optical Society of America A, 7(6), 1055.\n3. Kamer, Y., Ouillon, G., & Sornette, D. (2013). Barycentric fixed-mass method for multifractal analysis. Physical Review E, 88(2), 022922.", "date": "2023-03-23 20:17:48", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/13125/measuring-fractal-dimension-of-natural-objects-from-digital-images", "openwebmath_score": 0.49823832511901855, "openwebmath_perplexity": 1662.434442109112, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846691281407, "lm_q2_score": 0.8991213664574069, "lm_q1q2_score": 0.8796865606274719}}
{"url": "https://math.stackexchange.com/questions/1758414/how-do-i-translate-sentences-from-english-to-predicate-logic", "text": "# How do I translate sentences from English to predicate logic?\n\nThis question was taken from the MIT OCW Math for Computer Science course.\n\nTranslate the following sentences from English to predicate logic. The domain that you are working over is $X$, the set of people. You may use the functions $S(x)$, meaning that \u201c$x$ has been a student of $6.042$,\u201d $A(x)$, meaning that \u201c$x$ has gotten an \u2018$A$\u2019 in $6.042$,\u201d $T(x)$, meaning that \u201c$x$ is a TA of $6.042$,\u201d and $E(x, y)$, meaning that \u201c$x$ and $y$ are the same person.\u201d\n\n(a) [$6$ pts] There are people who have taken 6.042 and have gotten A\u2019s in $6.042$\n\n(b) [$6$ pts] All people who are 6.042 TA\u2019s and have taken 6.042 got A\u2019s in $6.042$\n\n(c) [$6$ pts] There are no people who are 6.042 TA\u2019s who did not get A\u2019s in $6.042$.\n\n(d) [$6$ pts] There are at least three people who are TA\u2019s in $6.042$ and have not taken $6.042$\n\nWhat I got thus far:\n\na) $\\exists x(S(x)\\wedge A(x))$\n\nb) $\\forall x\\in X((T(x)\\wedge S(x))\\Rightarrow A(x))$\n\nc) $\\nexists x\\in X(T(x)\\wedge \u00acA(x))$\n\nd) $\\exists x\\exists y\\exists z\\in X(\u00acE(x,y)\u2227\u00acE(x,z)\u2227\u00acE(y,z)\u2227(T(x)\\wedge \u00acS(x))\u2227(T(y)\\wedge \u00acS(y))\u2227(T(z)\\wedge \u00acS(z)))$\n\nI'm not sure about any of my answers, but c) and d) gave me the most trouble. In regards to c),I am not sure about how to express \"there are no people\" in predicate logic. In regards to d), I am not sure about how to express \"there are at least three people\" in predicate logic.\n\nHints are much better appreciated than an explicit answer. In addition, I would like to know how to tag this type of math. Is it considered discrete mathematics as most colleges/universities call it in the U.S.?\n\n\u2022 I would leave out the $\\in X$ stuff. \u2013\u00a0Andr\u00e9 Nicolas Apr 25 '16 at 18:51\n\u2022 Can you please explain why? Wouldn't have to define what $x$ is? \u2013\u00a0Cherry_Developer Apr 25 '16 at 19:16\n\u2022 The domain is $X$, so it is understood that variables range over $X$. Also, $\\in$ is not in the specified language. \u2013\u00a0Andr\u00e9 Nicolas Apr 25 '16 at 19:46\n\nYour answers all look okay. Specifically for part c), you did indeed translate the sentence into predicate logic correctly. However, often times it is customary to not leave any negation symbols before the quantifiers. We can pass the negation symbol through the existential/universal quantifier by swapping them. For example\n\n$$\\neg\\exists x\\in X(P(x))\\iff\\forall x\\in X(\\neg P(x))$$\n\nand\n\n$$\\neg\\forall x\\in X(P(x))\\iff\\exists x\\in X(\\neg P(x)).$$\n\nCan you see how you can use this to simplify your answer for part c)?\n\nAlso, I personally think the discrete math tag is okay for a question like this, especially since you also used the predicate logic tag.\n\n\u2022 I'm not sure about how to simplify it, but I gave it a shot: $\\forall x\\in X(\u00acT(x)\\vee A(x))$ I was mostly unsure of how to apply the negation symbol to the predicate, so I typed \"not (x and not y)\" into Wolfram Alpha and it returned that \"not x or y\" would be an equivalent statement. \u2013\u00a0Cherry_Developer Apr 25 '16 at 19:13\n\u2022 Yes, you simplified correctly. Here are the full steps: $\\neg \\exists x\\in X(T(x)\\wedge \\neg A(x))\\iff\\forall x\\in X\\neg(T(x)\\wedge \\neg A(x))\\iff\\forall x\\in X(\\neg T(x)\\vee A(x))$. \u2013\u00a0ervx Apr 25 '16 at 19:16\n\u2022 The negation of $\\exists x$ which becomes $\\forall x$ is clear to me, but applying the negation to the predicate is where I get lost. I feel like I cheated myself by just resorting to Wolfram Alpha to find the simplified form of $(T(x)\\wedge \u00acA(x))$. Is $(T(x)\\wedge \u00acA(x))\\Leftrightarrow (\\neg T(x)\\vee A(x))$ something I should just have memorized? Or is there a way for me to derive it? \u2013\u00a0Cherry_Developer Apr 25 '16 at 19:23\n\u2022 In general, for expressions $A$ and $B$, we have $\\neg(A\\vee B)\\iff (\\neg A\\wedge\\neg B)$ and $\\neg (A\\wedge B)\\iff (\\neg A\\vee \\neg B)$. You can verify these using truth tables if you like. In general, though, it would be good to memorize them. They are essentially DeMorgan's laws for propositional logic. \u2013\u00a0ervx Apr 25 '16 at 19:26", "date": "2020-01-28 20:19:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1758414/how-do-i-translate-sentences-from-english-to-predicate-logic", "openwebmath_score": 0.7618722915649414, "openwebmath_perplexity": 136.58058456063807, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846628255124, "lm_q2_score": 0.8991213684847575, "lm_q1q2_score": 0.8796865569441727}}
{"url": "https://math.stackexchange.com/questions/769059/number-of-club-members-if-they-are-organized-into-four-committees-if-each-member/769233", "text": "# Number of club members if they are organized into four committees if each member is in exactly two committees and any two committees share one member\n\nA club with $x$ members is organized into four committees such that\n\n$(a)$ each member is in exactly two committees,\n\n$(b)$ any two committees have exactly one member in common.\n\nThen $x$ has\n\n$(A)$ exactly two values both between $4$ and $8$\n\n$(B)$ exactly one value and this lies between $4$ and $8$\n\n$(C)$ exactly two values both between $8$ and $16$\n\n$(D)$ exactly one value and this lies between $8$ and $16$.\n\nMy thought is option $B$ and $x = 6$. Please somebody tell me am I right or wrong?\n\n\u2022 Can you show how you arrived at $x=6$? \u2013\u00a0RandomUser Apr 25 '14 at 16:53\n\u2022 I just see that if $x$=$6$ then the two conditions are satisfied.But if $x$<$6$ or $x$>$6$ the condition $b$ is not satisfied so I think $x$ may be $6$. \u2013\u00a0liesel Apr 25 '14 at 17:15\n\u2022 Please improve your title. \u2013\u00a0Alexander Gruber Apr 25 '14 at 17:22\n\nClaim: $x=6$\n\nLet $S:=[x]\\left(=\\{1,2,\\dots ,x\\}\\right)$ where $x$ is some fixed natural number. Let $C_i\\subset S$ be the $i$ th commitee.Let $f$ be a function from the set of unordered pair of distinct commitees to members of the club defined as, $$f\\left( (C_i,C_j)\\right)=C_i\\cap C_j$$ $f$ is injective by (a). $f$ is surjective by (b).\n\nHence, $|S|=\\binom{4}{2}=6$.\n\nHere is a construction:\n\n\u2022 Let $C_1\\cap C_2=1$ wlog. Then, $1\\notin C_i \\quad \\forall \\, i\\in [4]\\setminus \\{1,2\\}$.\n\n\u2022 Let $C_2\\cap C_3=2$ wlog. Then, $2\\notin C_i \\quad \\forall \\, i\\in [4]\\setminus \\{2,3\\}$.\n\n\u2022 Let $C_3\\cap C_4=3$ wlog. Then, $3\\notin C_i \\quad \\forall \\, i\\in [4]\\setminus \\{3,4\\}$.\n\n\u2022 Let $C_1\\cap C_3=4$ wlog. Then, $4\\notin C_i \\quad \\forall \\, i\\in [4]\\setminus \\{1,3\\}$.\n\n\u2022 Let $C_1\\cap C_4=5$ wlog. Then, $5\\notin C_i \\quad \\forall \\, i\\in [4]\\setminus \\{1,4\\}$.\n\n\u2022 Let $C_2\\cap C_4=6$ wlog. Then, $6\\notin C_i \\quad \\forall \\, i\\in [4]\\setminus \\{2,4\\}$.\n\nLet the committees be $A_1, A_2, A_3, A_4$ then $|A_1\\cup A_2 \\cup A_3 \\cup A_4|=n$. There are $\\binom{4}{2}=6$ pairwise intersections each of size $1$ and each intersection of triples must be empty since if not some member would belong to $3$ committees, a contradiction. If we make a $4 \\times n$ table where rows are committees and columns are members and mark a $1$ whenever member $x_j$ belongs to committee $A_i$ we notice that the $i^{th}$ row sum is $|A_i|$ and the $j^{th}$ column sum is $2$ (since each member belongs to exactly $2$ committees) so equating row and column sums we get $\\sum_{i=1}^{4}|A_i|= 2n.$\n\nPutting this information together and using the inclusion-exclusion formula we get $n =6$", "date": "2021-02-27 20:02:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/769059/number-of-club-members-if-they-are-organized-into-four-committees-if-each-member/769233", "openwebmath_score": 0.9540228247642517, "openwebmath_perplexity": 370.2904235473298, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850862376966, "lm_q2_score": 0.8947894703109853, "lm_q1q2_score": 0.8796541835852579}}
{"url": "https://math.stackexchange.com/questions/1724433/the-birthday-paradox/1724736", "text": "I would like a better understanding of the famous birthday paradox. \"What is the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday?\"\n\nI understood the first part, where the probability reaches 100% when the number of people reaches 367 by the pigeonhole principle. But I am not understanding the explanation beyond that. How do they say that the probability is 99.9% with 70 people and 50% with 23 people? And how do you further generalize the answer? And why is it a \"paradox\"?\n\n\u2022 Please, post the explanation you have been given. Apr 2 '16 at 10:00\n\u2022 en.wikipedia.org/wiki/Birthday_problem May be this is helpful to you Apr 2 '16 at 10:00\n\u2022 @Mc Cheng it might be a bit more helpful to specify a section rather than referring someone to an article (which is debatable in its reliability) that is quite hard (and lengthy) to sort through Apr 2 '16 at 10:03\n\u2022 I could've typed an answer but explanations of this paradox are readily available all over the internet. Take a look at the Wikipedia article. Tell me exactly what you cannot understand. Maybe then we'd be able to help. Apr 2 '16 at 10:03\n\u2022 Apr 2 '16 at 10:11\n\nLet the number of people in the group be $n$.\n\nThe probability that a pair of people don't share a birthday is given equal to $\\frac{364}{365}$ ignoring leap years.\n\nThere are $\\binom{n}{2}$ pair of people in a group of $n$ people. No pair of people will share a birthday if each person has a distinct birthday. The probability of this happening is given by\n\n$$\\frac{364}{365}\\times\\frac{363}{365} \\dots \\times \\frac{365 - (n-1)}{365}$$\n\nHow did I get this probability?\n\nAssume that all birthdays are equally likely. If the first person was born on day $x_1$ then the second person in the group cannot be born on day $x_1$. The probability for this happening is $364\\over 365$. Now let the birthday of the second person be $x_2$. The probability that the third person is not born on $x_1$ nor on $x_2$ is $363\\over365$. Similarly we get the probability for the $n^{\\text{th}}$ person. Since each event is independent, we multiply all the probabilities.\n\nThus the probability that at least one pair shares a birthday for a group of $n$ people is given by\n\n$$p = 1 - \\left(\\frac{364}{365}\\times\\frac{363}{365} \\dots \\times \\frac{365 - (n-1)}{365}\\right)$$\n\nNow you have the probability $p$ as a function of $n$. If you know the RHS, then you simply find for what value of $n$ we get the closest RHS to $p$\n\nIt so happens that if $p = 99.9\\%$, the $n = 70$\n\n\u2022 Thank you Banach Tarski! Am I right in understanding, that the only reason it is a \"paradox\" is because we'd usually expect the probability to be linear in nature? I don't see any paradox otherwise Apr 2 '16 at 10:11\n\u2022 @aswa09 yup you are correct. A paradox is a self contradictory statement. The contradiction here is with our common sense and not with the mathematics :) Apr 2 '16 at 10:15\n\u2022 Is it not a multiplication of $364/365$ not $1/365$? Else your formula gives a $364/365$ chance of just two people sharing the same birthday Apr 2 '16 at 10:21\n\u2022 Yes, I had made a blunder in computing the probability, I have fixed this now :) Apr 2 '16 at 10:39\n\nThe paradox is that the rate of growth doesn't match our common sense. We expect that the way to count the number of possibilities for people to have the same birthday is directly from the number of people. However, in reality it's based on the number of pairs of people, which grows much faster than the number of people.\n\nBanach Tarski shows you the derivation which really is that the numerator is the number of people pairs, but still over 365.\n\n\u2022 How do I remove the duplication mark? I have edited the question asking about why it's a paradox, so that it isn't a duplicate question Apr 3 '16 at 7:23", "date": "2021-12-06 14:36:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1724433/the-birthday-paradox/1724736", "openwebmath_score": 0.778621256351471, "openwebmath_perplexity": 192.5005743964197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850837598123, "lm_q2_score": 0.8947894618940992, "lm_q1q2_score": 0.879654173093558}}
{"url": "https://math.stackexchange.com/questions/53262/factorial-decomposition-of-integers", "text": "# Factorial decomposition of integers?\n\nThis question might seem strange, but I had the feeling it's possible to decompose in a unique way a number as follows:\n\nif $x < n!$, then there is a unique way to write x as: $$x = a_1\\cdot 1! + a_2\\cdot 2! + a_3\\cdot3! + ... + a_{n-1}\\cdot(n-1)!$$ where $a_i \\leq i$\n\nI looked at factorial decomposition on google but I cannot find any name for such a decomposition.\n\nexample:\n\nIf I chose :\n\n(a1,a2) =\n\n\u2022 1,0 -> 1\n\u2022 0,1 -> 2\n\u2022 1,1 -> 3\n\u2022 0,2 -> 4\n\u2022 1,2 -> 5\n\nI get all number from $1$ to $3!-1$\n\nideas for a proof:\n\nThe number of elements between $1$ and $N!-1$ is equal to $N!-1$ and I have the feeling they are all different, so this decomposition should be right. But I didn't prove it properly.\n\nAre there proofs of this decomposition? Does this decomposition as a name? And above all is this true ?\n\n\u2022 It's true, and the proof is easy by induction on $n$. What have you written down? \u2013\u00a0Olivier B\u00e9gassat Jul 23 '11 at 9:19\n\u2022 I am not sure about number-theory tag, if someone has a better idea, please retag the question. \u2013\u00a0Martin Sleziak Jul 23 '11 at 9:21\n\u2022 @Olivier, I actually thinking of a mapping between n! and all permutation, and I should have used that as a proof. But since I couldn't find anything on the web about that I had the feeling something was wrong ... :D! \u2013\u00a0Ricky Bobby Jul 23 '11 at 9:28\n\u2022 @Martin: Possibly combinatorics? \u2013\u00a0Brian M. Scott Jul 23 '11 at 9:39\n\u2022 If any of you have a good reason to think (number-systems) doesn't belong, please remove that tag... \u2013\u00a0J. M. is a poor mathematician Jul 23 '11 at 11:56\n\nYou're looking for the factorial number system, also known as \"factoradic\". Searching should give you more results.\n\nYes, it's true that such a decomposition is always possible. One way to prove it is as follows: given $x < n!$, consider the $x$th permutation of some ordered set of $n$ symbols. This is some permutation $(s_1, s_2, \\dots, s_n)$. Now for $s_1$ you had $n$ choices (label them $0$ to $n-1$) and you picked one, so let $a_{n-1}$ be the choice you made. For $s_2$ you had $n-1$ choices (label them $0$ to $n-2$) and you picked one, so let $a_{n-2}$ be the number of the choice you made. Etc. $a_0$ is always $0$ because you have only one choice for the last element. (This is also known as the Lehmer code.)\n\n\u2022 I was actually thinking about permutation when this question came to my mind, I didn't thought about using it to prove the decomposition. thanks a lot for your answer. \u2013\u00a0Ricky Bobby Jul 23 '11 at 9:17\n\u2022 Why not try what would seem obvious? Divide your numberx by n!, take the highest multiplr $a_n$ with $a_nn!<x$, then divide the remainder x-$a_n n!$ by (n-1)! and take $a_{n-2}$ as the largest multiple, etc. at the end, you will be left with some amount which is always a multiple of 1=1! \u2013\u00a0gary Jul 23 '11 at 9:37\n\u2022 @gary: Yes, something like that would work too, if done correctly. But for me, this map numbering permutations was actually more obvious. :-) (You want to divide by $(n-1)$ the first time, also you need to prove that $a_i \\le i$ for all $i$. This is easy, but it's not just one line.) \u2013\u00a0ShreevatsaR Jul 23 '11 at 9:46\n\u2022 Shreevatsa: if your coefficient for n! was larger than n, then n(n-1)!=n! , and you could then increase your previous coefficient for n! by (at least )1. But alternative explanations are always useful,helpful. \u2013\u00a0gary Jul 23 '11 at 10:12\n\nYour conjecture is correct. There is a straightforward proof by induction that such a decomposition always exists. Suppose that every positive integer less than $n!$ can be written in the form $\\sum_{k=1}^{n-1} a_k k!$, where $0 \\le a_k \\le k$, and let $m$ be a positive integer such that $n! \\le m < (n+1)!$. There are unique integers $a_n$ and $r$ such that $m = a_nn! + r$ and $0 \\le r < n!$, and since $m < (n+1)! = (n+1)n!$, it\u2019s clear that $a_n \\le n$. Since $r < n!$, the induction hypothesis ensures that there are non-negative integers $a_1,\\dots,a_{n-1}$ such that $r = \\sum_{k=1}^{n-1} a_k k!$, and hence $m = \\sum_{k=1}^n a_k k!$.\n\nWe\u2019ve now seen that each of the $(n+1)!$ non-negative integers in $\\{0,1,\\dots,n\\}$ has a representation of the form $\\sum_{k=1}^n a_k k!$ with $0 \\le a_k \\le k$ for each $k$. However, there are only $\\prod_{k=1}^n (k+1) = (n+1)!$ distinct representations of that form, so each must represent a different integer, and each integer\u2019s representation is therefore unique.\n\nYou can also reason as follows : suppose you've shown for some integer $n$ that every integer $\\in\\lbrace0,\\dots,n!-1\\rbrace$ has a unique decomposition as you suggest. Take $k\\in\\lbrace0,\\dots,(n+1)!-1\\rbrace.$ Write $$k=q\\cdot n!+r$$ the euclidean division of $k$ with respect to $n!$. Necessarily you have $0\\leq q < n+1$. This gives you an expression you want, for $0\\leq r<n!$ has, by hypothesis, an expression involving only factorials up to $(n-1)!$ .\n\nFinally, to show uniqueness, you can again use your hypothesis to deduce that if $k=a_n\\cdot n!+ \\sum_0^{n-1} a_i\\cdot i!,$ then $0\\leq \\sum\\dots< n!-1$ by hypothesis, and this tells you that this decomposition is the euclidean division of $k$ by $n!$. So by uniqueness of the euclidean division, it's the one decomposition we defined earlier.\n\nThis generalizes to the representation $$n = \\sum_{i\\ge 0} a_i b_i$$ where $b_0 = 1$ and $b_{n+1} = b_n c_n$ with $c_n > 1$ and $0 \\le a_i < c_i$ - i.e., representing $n$ with digits $a_i$ to a \"base\" with varying ratios of consecutive digit values (phrased awkwardly, I know, but it is late and I am tired). For a usual base $B$ system, $c_i = B$ for all $i$, so $b_i = B^i$. For this system, $c_i = i$ (or maybe $c_i = i+1$, modulo my tiredness), so $b_i = i!$.\n\nI proved too many years ago that this representation is unique iff the $c_i$ were all integers. I am sure that this was proved many years ago (probably by Euler, if not Fermat) and is in Dickson somewhere.", "date": "2019-10-14 04:17:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/53262/factorial-decomposition-of-integers", "openwebmath_score": 0.8754757642745972, "openwebmath_perplexity": 239.00584290432388, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631659211717, "lm_q2_score": 0.8918110475945175, "lm_q1q2_score": 0.8796495683088049}}
{"url": "https://math.stackexchange.com/questions/2181828/why-is-a-number-that-is-divisible-by-2-and-3-also-divisible-by-6", "text": "# Why is a number that is divisible by $2$ and $3$, also divisible by $6$?\n\nWhy is it that a number such as $108$, that is divisible by $2$ and $3$, is also divisible by $6$?\n\nIs this true that all numbers divisible by two integers are divisible by their product?\n\n\u2022 Think about the factorization of a number that's divisible by 2 and 3. \u2013\u00a0Ben Longo Mar 11 '17 at 14:20\n\u2022 Every number can be expressed as the product of primes. $2$ & $3$ are coprime ... Can you get it from there Jason ? \u2013\u00a0Donald Splutterwit Mar 11 '17 at 14:21\n\u2022 Provided the numbers are mutually prime. \u2013\u00a0SchrodingersCat Mar 11 '17 at 14:22\n\u2022 note $108 = 2^2 \\cdot 3^3 = 3 \\cdot (2\\cdot 3) \\cdot (2 \\cdot 3) = 3 \\cdot (6) \\cdot (6)$ \u2013\u00a0Dando18 Mar 11 '17 at 14:23\n\u2022 What about 30, that's divisible by 10 and 15? \u2013\u00a0Ethan Bolker Mar 11 '17 at 14:23\n\nIt is certainly not true that if $x$ is divisible by both $y$ and $z$, then $x$ is divisible by $yz$. For example, $2$ is divisible by $2$ and $2$, but is not divisible by $2\\cdot 2$. As a less silly example, $120$ is divisible by both $15$ and $6$, but is not divisible by $90$.\n\nBut if you think about the examples above, you'll notice that the factors involved in each overlap: e.g. both $2$ and $2$ are even, and both $15$ and $6$ are divisible by $3$. Meanwhile, $2$ and $3$ don't overlap - they share no factors in common.\n\nIt turns out that this is the crucial property:\n\nIf $x$ is divisible by $y$ and $z$, and $y$ and $z$ have no factors in common besides $1$, then $x$ is divisible by $yz$.\n\nProving this is a good challenge. HINT: think about factorization into primes . . .\n\n\u2022 There is perhaps something to add here, since there are divisors that fulfil these rules and DO have factors in common, but still divide some number, e.g. in the case $2$ divides $y$ and $z$, and $4$ divides $x$ \u2013\u00a0samerivertwice Mar 11 '17 at 14:27\n\nIt is not true that for any two integers $m$, $n$, if $x$ is divisible by $m$ and $n$, it is divisible by $mn$. For example $4$ is divisible by $4$ and $2$, but it is not divisible by $8$.\n\nHowever, this is true if $m$ and $n$ are relatively prime, i.e. the greatest common divisor of $m$ and $n$ is $1$. Thus, since $2$ and $3$ are relatively prime, any number divisible by $2$ and $3$ will also be divisible by $6$.\n\nMore generally, if $x$ is divisible by $m$ and $n$, then it must be divisible by $\\operatorname{lcm}(m,n)$ (least common multiple). Case of $m$ and $n$ being relatively prime is a special case of this, since $$\\operatorname{lcm}(m,n) = \\frac{mn}{\\operatorname{gcd}(m,n)}$$\n\nwhere $\\operatorname{gcd}$ stands for the greatest common divisor and thus $\\operatorname{lcm}(m,n) = mn$, when $m$ and $n$ are relatively prime.\n\nLet $a$ be divisible by $2$ and $3$. So $$a=2^{r_{0}}\\cdot p^{r_{1}}_{1}\\cdot p^{r_{2}}_{2}\\cdot\\ldots=3^{s_{0}}\\cdot q^{s_{1}}_{1}\\cdot q^{s_{2}}_{2}\\cdot\\ldots$$ where $p_{i}$ and $q_{j}$ are distinct prime numbers, $r_{i}$ and $s_{j}$ are possibly $0$ for $1\\leqslant i,j<\\infty$ and $r_{0},s_{0}\\geqslant 1$. We know from the fundamental theorem that these two must be unique factorisations (up to re-ordering). Hence there must be some $p_{k}^{r_{k}}=3^{s_{0}}$ and some $q_{\\ell}^{s_{\\ell}}=2^{r_{0}}$. Therefore $a$ has a factor of $6$: \\begin{align*} a&=2^{r_{0}}\\cdot p_{1}^{r_{1}}\\ldots\\cdot p_{k}^{r_{k}}\\cdot\\ldots\\\\ &=2^{r_{0}}\\cdot p_{1}^{r_{1}}\\ldots\\cdot3^{s_{0}}\\cdot\\ldots\\\\ &=6\\cdot(2^{r_{0}-1}\\cdot3^{s_{0}-1}\\cdot p_{1}^{r_{1}}\\cdot\\ldots). \\end{align*} (The argument works on the other side as well). This might be a little explicit but I hope it helps.\n\nAs the comments point out, and as you may suspect, it is not true in general that if a number is divisible by two integers, then it is divisible by their product. What we can say is the following:\n\nIf $x,y,z$ are integers such that $z$ is divisible by both $x$ and $y,$ then $z$ is divisible by the least common multiple of $x$ and $y.$\n\nThis isn't actually much of a surprise, given the way that least common multiple is often defined:\n\nGiven two integers $x$ and $y,$ the least common multiple of $x$ and $y$ is the nonnegative integer $z$ such that (i) $z$ is divisible by both $x$ and $y,$ and (ii) every common multiple of $x$ and $y$ is divisible by $z.$\n\nProving that such a multiple exists isn't exactly straightforward, though. It is slightly easier to prove the existence of another integer:\n\nGiven two integers $x$ and $y,$ the greatest common factor of $x$ and $y$ is the positive integer $z$ such that (i) both $x$ and $y$ are divisible by $z,$ and (ii) $z$ is divisible by every common factor of $x$ and $y.$\n\nOne can, for example, use the Euclidean algorithm to prove that $\\operatorname{gcf}(x,y)$ exists, regardless of $x,y.$ At that point, one can show that $\\frac{|xy|}{\\operatorname{gcf}(x,y)}$ turns out to be the least common multiple of $x$ and $y.$ As an immediate consequence of the identity $$\\operatorname{lcm}(x,y)=\\frac{|xy|}{\\operatorname{gcf}(x,y)},$$ together with the result mentioned at the beginning, we find the following:\n\nIf $x,y,z$ are integers such that $z$ is divisible by both $x$ and $y,$ and if $\\operatorname{gcf}(x,y)=1,$ then $z$ is divisible by $xy.$\n\nSince $\\operatorname{gcf}(2,3)=1,$ then any number divisible by both $2$ and $3$ is necessarily divisible by $6.$\n\nThe general result is this:\n\nIf a number $n$ is divisible by $a$ and $b$, then it is divisible by $\\DeclareMathOperator{\\lcm}{lcm}\\lcm(a,b)$.\n\nIndeed, let $\\;d=\\gcd(a,b)$, $\\;m=\\lcm(a,b)$, $\\;a'=\\dfrac ad$, $\\;b'=\\dfrac bd$. These numbers satisfy the relation $$md=ab, \\enspace\\text{so}\\quad m=\\frac{ab}d=a'b=ab'.$$ By hypothesis, we can write $$n=qa=qda', \\qquad n=rb=rb'd$$ We deduce that $\\;qa'=rb'$. So $a'$ divides $rb'$. As $a'$ and $b'$ are coprime, Gau\u00df' lemma ensures $a'$ divides $r$, i.e. we have $r=r'a'$, and $$n=(r'a')b'd=r'(a'b)=r'm.$$", "date": "2019-09-21 17:37:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2181828/why-is-a-number-that-is-divisible-by-2-and-3-also-divisible-by-6", "openwebmath_score": 0.8382720351219177, "openwebmath_perplexity": 63.34239072744007, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631631151011, "lm_q2_score": 0.8918110468756548, "lm_q1q2_score": 0.8796495650972606}}
{"url": "https://uk.mathworks.com/help/symbolic/sym.changeintegrationvariable.html", "text": "# changeIntegrationVariable\n\nIntegration by substitution\n\n## Description\n\nexample\n\nG = changeIntegrationVariable(F,old,new) applies integration by substitution to the integrals in F, in which old is replaced by new. old must depend on the previous integration variable of the integrals in F and new must depend on the new integration variable. For more information, see Integration by Substitution.\n\nWhen specifying the integrals in F, you can return the unevaluated form of the integrals by using the int function with the 'Hold' option set to true. You can then use changeIntegrationVariable to show the steps of integration by substitution.\n\n## Examples\n\ncollapse all\n\nApply a change of variable to the definite integral ${\\int }_{\\mathit{a}}^{\\mathit{b}}\\mathit{f}\\left(\\mathit{x}+\\mathit{c}\\right)\\text{\\hspace{0.17em}}\\mathit{dx}$.\n\nDefine the integral.\n\nsyms f(x) y a b c\nF = int(f(x+c),a,b)\nF =\n\n${\\int }_{a}^{b}f\\left(c+x\\right)\\mathrm{d}x$\n\nChange the variable $\\mathit{x}+\\mathit{c}$ in the integral to $\\mathit{y}$.\n\nG = changeIntegrationVariable(F,x+c,y)\nG =\n\n${\\int }_{a+c}^{b+c}f\\left(y\\right)\\mathrm{d}y$\n\nFind the integral of $\\int \\mathrm{cos}\\left(\\mathrm{log}\\left(\\mathit{x}\\right)\\right)\\mathit{dx}$ using integration by substitution.\n\nDefine the integral without evaluating it by setting the 'Hold' option to true.\n\nsyms x t\nF = int(cos(log(x)),'Hold',true)\nF =\n\n$\\int \\mathrm{cos}\\left(\\mathrm{log}\\left(x\\right)\\right)\\mathrm{d}x$\n\nSubstitute the expression log(x) with t.\n\nG = changeIntegrationVariable(F,log(x),t)\nG =\n\n$\\int {\\mathrm{e}}^{t} \\mathrm{cos}\\left(t\\right)\\mathrm{d}t$\n\nTo evaluate the integral in G, use the release function to ignore the 'Hold' option.\n\nH = release(G)\nH =\n\n$\\frac{{\\mathrm{e}}^{t} \\left(\\mathrm{cos}\\left(t\\right)+\\mathrm{sin}\\left(t\\right)\\right)}{2}$\n\nRestore log(x) in place of t.\n\nH = simplify(subs(H,t,log(x)))\nH =\n\n$\\frac{\\sqrt{2} x \\mathrm{sin}\\left(\\frac{\\pi }{4}+\\mathrm{log}\\left(x\\right)\\right)}{2}$\n\nCompare the result to the integration result returned by int without setting the 'Hold' option to true.\n\nFcalc = int(cos(log(x)))\nFcalc =\n\n$\\frac{\\sqrt{2} x \\mathrm{sin}\\left(\\frac{\\pi }{4}+\\mathrm{log}\\left(x\\right)\\right)}{2}$\n\nFind the closed-form solution of the integral $\\int \\mathit{x}\\text{\\hspace{0.17em}}\\mathrm{tan}\\left(\\mathrm{log}\\left(\\mathit{x}\\right)\\right)\\mathit{dx}$.\n\nDefine the integral using the int function.\n\nsyms x\nF = int(x*tan(log(x)),x)\nF =\n\n$\\int x \\mathrm{tan}\\left(\\mathrm{log}\\left(x\\right)\\right)\\mathrm{d}x$\n\nThe int function cannot find the closed-form solution of the integral.\n\nSubstitute the expression log(x) with t. Apply integration by substitution.\n\nsyms t\nG = changeIntegrationVariable(F,log(x),t)\nG =\n\nThe closed-form solution is expressed in terms of hypergeometric functions. For more details on hypergeometric functions, see hypergeom.\n\nCompute the integral ${\\int }_{0}^{1}{\\mathit{e}}^{\\sqrt{\\mathrm{sin}\\left(\\mathit{x}\\right)}}\\mathit{dx}$ numerically with high precision.\n\nDefine the integral ${\\int }_{0}^{1}{\\mathit{e}}^{\\sqrt{\\mathrm{sin}\\left(\\mathit{x}\\right)}}\\mathit{dx}$. A closed-form solution to the integral does not exist.\n\nsyms x\nF = int(exp(sqrt(sin(x))),x,0,1)\nF =\n\n${\\int }_{0}^{1}{\\mathrm{e}}^{\\sqrt{\\mathrm{sin}\\left(x\\right)}}\\mathrm{d}x$\n\nYou can use vpa to compute the integral numerically to 10 significant digits.\n\nF10 = vpa(F,10)\nF10 =\u00a0$1.944268879$\n\nAlternatively, you can use the vpaintegral function and specify the relative error tolerance.\n\nFvpa = vpaintegral(exp(sqrt(sin(x))),x,0,1,'RelTol',1e-10)\nFvpa =\u00a0$1.944268879$\n\nThe vpa function cannot find the numerical integration to 70 significant digits, and it returns the unevaluated integral in the form of vpaintegral.\n\nF70 = vpa(F,70)\nF70 =\u00a0$1.944268879138581167466225761060083173280747314051712224507065962575967$\n\nTo find the numerical integration with high precision, you can perform a change of variable. Substitute the expression $\\sqrt{\\mathrm{sin}\\left(\\mathit{x}\\right)}$ with $\\mathit{t}$. Compute the integral numerically to 70 significant digits.\n\nsyms t;\nG = changeIntegrationVariable(F,sqrt(sin(x)),t)\nG =\n\n${\\int }_{0}^{\\sqrt{\\mathrm{sin}\\left(1\\right)}}\\frac{2 t {\\mathrm{e}}^{t}}{\\sqrt{1-{t}^{4}}}\\mathrm{d}t$\n\nG70 = vpa(G,70)\nG70 =\u00a0$1.944268879138581167466225761060083173280747314051712224507065962575967$\n\n## Input Arguments\n\ncollapse all\n\nExpression containing integrals, specified as a symbolic expression, function, vector, or matrix.\n\nSubexpression to be substituted, specified as a symbolic scalar variable, expression, or function. old must depend on the previous integration variable of the integrals in F.\n\nNew subexpression, specified as a symbolic scalar variable, expression, or function. new must depend on the new integration variable.\n\ncollapse all\n\n### Integration by Substitution\n\nMathematically, the substitution rule is formally defined for indefinite integrals as\n\n$\\int f\\left(g\\left(x\\right)\\right)\\text{\\hspace{0.17em}}g\\text{'}\\left(x\\right)\\text{\\hspace{0.17em}}dx={\\left(\\int f\\left(t\\right)\\text{\\hspace{0.17em}}dt\\right)|}_{t=g\\left(x\\right)}$\n\nand for definite integrals as\n\n$\\underset{a}{\\overset{b}{\\int }}f\\left(g\\left(x\\right)\\right)\\text{\\hspace{0.17em}}g\\text{'}\\left(x\\right)\\text{\\hspace{0.17em}}dx=\\underset{g\\left(a\\right)}{\\overset{g\\left(b\\right)}{\\int }}f\\left(t\\right)\\text{\\hspace{0.17em}}dt.$", "date": "2022-01-17 08:23:51", "meta": {"domain": "mathworks.com", "url": "https://uk.mathworks.com/help/symbolic/sym.changeintegrationvariable.html", "openwebmath_score": 0.9384503960609436, "openwebmath_perplexity": 2273.4099825418953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534327754854, "lm_q2_score": 0.896251371055247, "lm_q1q2_score": 0.8796289847519074}}
{"url": "https://math.stackexchange.com/questions/2944265/on-the-integral-int-0-pi-sinx-sinx-sinx-cdots-dx", "text": "# On the integral $\\int_0^\\pi\\sin(x+\\sin(x+\\sin(x+\\cdots)))\\,dx$\n\nThis question came into my head when I did a course on Fourier series. However, this is not an infinite sum of sines, but an infinite recurrence of sines in a sum.\n\nConsider $$f_1(x)=\\sin(x)$$ and $$f_2(x)=\\sin(x+f_1(x))$$ such that $$f_n$$ satisfies the relation $$f_n(x)=\\sin(x+f_{n-1}(x)).$$ To what value does $$L:=\\lim_{n\\to\\infty}\\int_0^\\pi f_n(x)\\,dx$$ converge?\n\nSince it is impossible to evaluate the integrals directly, we begin by considering the first few values of $$n$$. A pattern clearly emerges. $$I_1=\\int_0^\\pi f_1(x)\\,dx=2\\quad\\quad\\quad I_2=1.376527...\\\\I_3=2.188188...\\quad\\quad\\quad\\quad\\quad I_4=1.625516...\\\\ I_5=2.179090...\\quad\\quad\\quad\\quad\\quad I_6=1.732942...\\\\ I_7=2.155900...\\quad\\quad\\quad\\quad\\quad I_8=1.927035...$$\n\nFor odd values of $$n$$, $$I_n$$ decreases monotonically (except $$n=1$$) and for even values of $$n$$, $$I_n$$ increases monotonically. These two observations have led me to claim that $$L=I_1=2$$.\n\nIs it possible to prove/disprove this claim?\n\n\u2022 I feel like a numerical DE approach: let $y=\\sin(x+\\sin(x+\\cdots))$ so that $y=\\sin(x+y),$ differentiating, and using Mathematica's NDSolve command should produce a result, but it has numerical instabilities, unfortunately. \u2013\u00a0Adrian Keister Jul 11 '19 at 17:53\n\nOutline:\n\n\u2022 Use the inverse function of $$y=x-\\sin x$$ to express $$f_\\infty(x)$$.\n\n\u2022 Use integral of inverse functions and dominated convergence theorem to prove $$L=2$$.\n\nClaim:$$L=2.$$\n\nProof: Obviously $$y=t-\\sin t$$ is injective on $$t\\in[0,\\pi]$$.\n\nDefine $$y=\\operatorname{Sa}(t)$$ as the inverse function of $$y=t-\\sin t$$ on $$t\\in[0,\\pi]$$. Therefore, $$t-\\sin t =x \\implies t=\\operatorname{Sa}(x).$$\n\nAssume $$f_\\infty(x)$$ exists (see 1. the first integral), then we have \\begin{align*} f_\\infty&=\\sin(x+f_\\infty)\\\\ \\underbrace{(x+f_\\infty)}_{t}-\\sin\\underbrace{(x+f_\\infty)}_{t}&=x\\\\ x+f_\\infty&=\\operatorname{Sa}(x)\\\\ f_\\infty(x)&=-x+\\operatorname{Sa}(x). \\end{align*}\n\nSince $$0-\\sin 0 =0\\implies \\operatorname{Sa}(0)=0$$ and $$\\pi-\\sin \\pi =\\pi\\implies \\operatorname{Sa}(\\pi)=\\pi$$, \\begin{align*} \\int_0^\\pi f_\\infty(x)\\,\\mathrm dx&=\\int_0^\\pi -x+\\operatorname{Sa}(x)\\,\\mathrm dx\\\\ &=\\int_0^\\pi -x\\,\\mathrm dx+\\int_0^\\pi \\operatorname{Sa}(x)\\,\\mathrm dx\\\\ &=-\\frac{\\pi^2}2+\\left(\\pi \\operatorname{Sa}(\\pi)-0 \\operatorname{Sa}(0)-\\int_{\\operatorname{Sa}(0)}^{\\operatorname{Sa}(\\pi)}y-\\sin y\\,\\mathrm dy\\right)\\\\ &=-\\frac{\\pi^2}2+\\left(\\pi^2-\\int_0^\\pi y-\\sin y\\,\\mathrm dy\\right)\\\\ &=-\\frac{\\pi^2}2+\\left(\\pi^2-\\left[\\frac{y^2}2+\\cos y\\right]^\\pi_0\\right)\\\\ &=2. \\end{align*}\n\nHere we used integral of inverse functions: $$\\int_c^df^{-1}(y)\\,\\mathrm dy+\\int_a^bf(x)\\,\\mathrm dx=bd-ac.$$\n\nNote: Since $$|f_n(x)|\\le 1$$ and $$1$$ is integrable on $$[0,\\pi]$$, we could interchange limit sign and integral sign from dominated convergence theorem, that is, $$L:=\\lim_{n\\to\\infty}\\int_0^\\pi f_n(x)\\,\\mathrm dx=\\int_0^\\pi \\lim_{n\\to\\infty}f_n(x)\\,\\mathrm dx=\\int_0^\\pi f_\\infty(x)\\,\\mathrm dx=2.$$\n\n\u2022 What a marvelous answer! \u2013\u00a0Szeto Oct 6 '18 at 22:29\n\u2022 That is amazing! I've posted a similar question, but this time with multiplication. Do you have any thoughts on it? Cheers. \u2013\u00a0TheSimpliFire Oct 7 '18 at 9:19", "date": "2020-05-28 12:39:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2944265/on-the-integral-int-0-pi-sinx-sinx-sinx-cdots-dx", "openwebmath_score": 0.9613777995109558, "openwebmath_perplexity": 648.9789608726883, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718453483274, "lm_q2_score": 0.8887587853897074, "lm_q1q2_score": 0.8795795472061698}}
{"url": "http://mathhelpforum.com/calculus/19332-taylor-series.html", "text": "1. ## taylor series\n\nCan anyone help me with how to get the Taylor series of a function(for example sinx)?\n\nFurthur how to get a value for sin(for example pi/4) with an particular error?\nThank You\n\n2. Originally Posted by Dili\nCan anyone help me with how to get the Taylor series of a function(for example sinx)?\n\nFurthur how to get a value for sin(for example pi/4) with an particular error?\nThank You\nThe Taylor series for a function $f(x)$ (at least I presume you are talking about a single variable function) to N + 1 terms about a point x = a has the form:\n$f(x) \\approx f(a) + \\frac{1}{1!}f^{\\prime}(a)(x - a) + \\frac{1}{2!}f^{\\prime \\prime}(a)(x - a)^2 + ~ ... ~ + \\frac{1}{N!}f^{(N)}(a)(x - a)^N$\n\nor in summation notation:\n$f(x) \\approx \\sum_{k = 0}{N}\\frac{1}{k!}f^{(k)}(a)(x - a)^k$\n\nFor the error estimate I will refer you here as I have forgotten which one is \"standard\" (if any.)\n\nSo as an example, let $f(x) = sin(x)$ and let us expand the series about the point $x = \\frac{\\pi}{4}$.\n\nWe have:\n$f(x) = sin(x) \\implies f \\left ( \\frac{\\pi}{4} \\right ) = sin \\left ( \\frac{\\pi}{4} \\right ) = \\frac{\\sqrt{2}}{2}$\n\n$f^{\\prime}(x) = cos(x) \\implies f^{\\prime} \\left ( \\frac{\\pi}{4} \\right ) = cos \\left ( \\frac{\\pi}{4} \\right ) = \\frac{\\sqrt{2}}{2}$\n\n$f^{\\prime \\prime}(x) = -sin(x) \\implies f^{\\prime \\prime} \\left ( \\frac{\\pi}{4} \\right ) = -sin \\left ( \\frac{\\pi}{4} \\right ) = -\\frac{\\sqrt{2}}{2}$\n\n$f^{\\prime \\prime \\prime}(x) = -cos(x) \\implies f^{\\prime \\prime \\prime} \\left ( \\frac{\\pi}{4} \\right ) = -cos \\left ( \\frac{\\pi}{4} \\right ) = -\\frac{\\sqrt{2}}{2}$\n\nSo the Taylor series to 4 terms looks like:\n$sin(x) \\approx \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2} \\left ( x - \\frac{\\pi}{4} \\right ) - \\frac{\\sqrt{2}}{4} \\left ( x - \\frac{\\pi}{4} \\right )^2 - \\frac{\\sqrt{2}}{12} \\left ( x - \\frac{\\pi}{4} \\right )^3$\n\n-Dan\n\n3. ## lagrange's remainder\n\nThank you\n\nBut what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)?\nAnd can you explain with the Lagrange's remainder?\n\n4. Originally Posted by Dili\nThank you\n\nBut what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)?\nAnd can you explain with the Lagrange's remainder?\nI'll have to let someone else explain the remainder (I'm just not up on it myself, which is why I posted the link. )\n\nYes, you could use the Maclaurin expansion (the Taylor series expansion about the point x = 0), but note that $\\frac{\\pi}{4} \\approx 0.785398$ is not particularly close to 0. The further x is from 0, the greater the error in the approximation.\n\nOn the other hand using 3 terms in the expansion I get\n$sin(x) \\approx x - \\frac{1}{3!}x^3 + \\frac{1}{5!}x^5$\n\nGives me\n$sin \\left ( \\frac{\\pi}{4} \\right ) \\approx 0.707143$\nwhich is only off by $5.129 \\times 10^{-3}$ %, which is pretty close by most people's judgment.\n\n-Dan\n\n5. Theorem: Let $f(x)$ be an infinitely differenciable function on $(a,b)$ (with $a<0). Let $T_n(x)$ represent the $n$-th degree Taylor polynomial around the origin for the point $x\\in (a,b)$ (with $x\\not = 0$). And let $R_{n+1}(x) = f(x) - T_{n}(x)$ be the remainder term. Then there exists a number $y$ strictly between $0$ and $x$ so that,\n$R_{n+1}(x) = \\frac{f^{(n+1)}(y)}{(n+1)!}\\cdot x^{n+1}$.\n\nSo given $f(x) = \\sin x$ let us work on the interval $(a,b)$ where $a=-\\infty$ and $b=+\\infty$. This function is infinitely differenciable so the above results apply. You can to approximate $T_n \\left( \\frac{\\pi}{4} \\right)$. Now by the theorem we know that,\n$R_{n+1}\\left( \\frac{\\pi}{4} \\right) = \\frac{f^{(n+1)} \\left( \\frac{\\pi}{4} \\right)}{(n+1)!} \\cdot \\left( \\frac{\\pi}{4} \\right)^{n+1}$ for some $0.\n\nNotice that $f^{n+1}$ is one of these: $\\sin x,\\cos x,-\\sin x,-\\cos x$. Thus, $|f^{n+1}|\\leq 1$. And also notice that $\\frac{\\pi}{4} \\leq 1$ thus, $\\left(\\frac{\\pi}{4}\\right)^{n+1} \\leq 1$.\n\nThis means,\n$\\left| R_{n+1}\\left( \\frac{\\pi}{4}\\right) \\right| \\leq \\frac{1}{(n+1)!}$.\n\nThis approximation might be greatly improved all I did was place an approximation that works and furthermore converges rapidply.\n\nRemark) The theorem applys in more general to $(n+1)$-differentiable functions but there was no need to use it here.", "date": "2017-12-11 06:44:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/19332-taylor-series.html", "openwebmath_score": 0.9748092889785767, "openwebmath_perplexity": 737.2499403638413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462187092608, "lm_q2_score": 0.8902942253943279, "lm_q1q2_score": 0.8795628135170166}}
{"url": "https://math.stackexchange.com/questions/2357956/calculate-the-probability-that-juniors-are-the-majority-of-random-selection", "text": "# Calculate the probability that Juniors are the majority of random selection.\n\nIn a business school at ABC University, 1/6 of the enrollment are freshmen, 3/10 are sophomores, 1/3 are juniors and 1/5 are seniors. They compete with several other schools within their region in an annual financial case competition.\n\nThree students were randomly selected to form a team for this year\u2019s competition.\n\nCalculate the probability that the juniors would make up the majority of the team.\n\nI thought that this probability is equal to the probability that exactly 2 out 3 selected are juniors plus the probability that all 3 are juniors. Therefore, I computed the following:\n\n$$P(exactly~2~juniors) = \\frac13 \\times \\frac13 \\times \\frac23 = \\frac2{27}$$ $$P(3~juniors) = (\\frac13)^3$$ Therefore, $$P(juniors~make~up~the~majority)= \\frac1{27} + \\frac2{27} = \\frac3{27}$$\n\nMy answer is wrong as I have only partially computed the probability. I would appreciate some explanation as what I am missing as I am having trouble figuring it out. Thank you\n\nYour formula $\\frac{1}{3}\\cdot\\frac{1}{3}\\cdot \\frac{2}{3}$ to pick the two juniors is the probability of first selecting two juniors, and then the non-junior. You can also first pick a junior, then a non-junior, and then a junior again, the probability of which is: $\\frac{1}{3}\\cdot\\frac{2}{3}\\cdot \\frac{1}{3}$. And, of course, you can first pick the non-junior, and then the two juniors, with a probability of $\\frac{2}{3}\\cdot\\frac{1}{3}\\cdot \\frac{1}{3}$.\n\nAll these give you 2 juniors and 1 non-junior, so the probability of picking exactly 2 juniors is:\n\n$\\frac{1}{3}\\cdot\\frac{1}{3}\\cdot \\frac{2}{3}+\\frac{1}{3}\\cdot\\frac{2}{3}\\cdot \\frac{1}{3}+\\frac{2}{3}\\cdot\\frac{1}{3}\\cdot \\frac{1}{3} = \\frac{6}{27}=\\frac{2}{9}$\n\n\u2022 I thought that the order in which I choose the students doesn't matter. It obviously does per your solution (thank you!). \u2013\u00a0the boy 88 Jul 13 '17 at 20:59\n\u2022 Can I think of my sample space as all combinations of 3-student tuples? And the event I want is choosing exactly 2 juniors and 1 non-junior so the outcomes that satisfy the event are { JJSo, JJF, and JJSe}? So = sophomore, Se = senior, F= freshman. \u2013\u00a0the boy 88 Jul 13 '17 at 21:07\n\u2022 @Kovs95 Yes, you can. And you can see them as unordered triples (basically sets with 3 elements), or as ordered triples. If seeing them as unordered, you need to multiply by 3 which is what sds did.. if you see them as ordered you get the formula I got. Same result of course! Now, there is really no need to distinguish between freshmen and sophomores and senior ... you just need to distinguish between junior (1/3 chance) and non-juniors (2/3 chance) \u2013\u00a0Bram28 Jul 13 '17 at 21:22\n\nYou can select the non-junior in 3 different ways (1st, 2nd, 3rd), so\n\n$$P(exactly~2~juniors) = \\frac13 \\times \\frac13 \\times \\frac23 \\times 3 = \\frac29$$\n\nThus\n\n$$P(juniors~ make~ up~ the~ majority)=\\frac29+\\frac1{27}=\\frac{7}{27}$$\n\n\u2022 yes sorry I fixed the 2/3 probability of not selecting a junior. \u2013\u00a0the boy 88 Jul 13 '17 at 20:43\n\u2022 yes, but this is not the only problem - see \"times 3\"! \u2013\u00a0sds Jul 13 '17 at 20:44", "date": "2019-08-23 06:21:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2357956/calculate-the-probability-that-juniors-are-the-majority-of-random-selection", "openwebmath_score": 0.8247532844543457, "openwebmath_perplexity": 846.2584072674448, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754515389344, "lm_q2_score": 0.8933094167058151, "lm_q1q2_score": 0.87953052231711}}
{"url": "https://gmatclub.com/forum/what-are-the-last-two-digits-of-195836.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 15 Nov 2018, 13:31\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. 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Make sure you are on time or you will be at a disadvantage.\n\n# What are the last two digits of 63*35*37*82*71*41?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50613\nWhat are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Apr 2015, 05:11\n3\n11\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n55% (01:26) correct 45% (01:10) wrong based on 204 sessions\n\n### HideShow timer Statistics\n\nWhat are the last two digits of 63*35*37*82*71*41?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\nKudos for a correct solution.\n\n_________________\n##### Most Helpful Expert Reply\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50613\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n13 Apr 2015, 06:27\n2\n5\nBunuel wrote:\nWhat are the last two digits of 63*35*37*82*71*41?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\nKudos for a correct solution.\n\nVERITAS PREP OFFICIAL SOLUTION:\n\nWe know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.\n\nRemainder of (63*35*37*82*71*41)/ 100\n\nNote that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:\n\nNote: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.\n\nSo say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.\n\nTake another example to reinforce this \u2013 what is the remainder when 85 is divided by 20? It is 5.\n\nWe might rephrase it as \u2013 what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.\n\nSo keeping this very important point in mind, let\u2019s go ahead and cancel the common 5 and 2.\n\nWe need the\n\nRemainder of (63*7*37*41*71*41*5*2)/10*5*2\n\nRemainder of (63*7*37*41*71*41)/10\n\nNow using concept 2, let\u2019s write the numbers in form of multiples of 10\n\nRemainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10\n\nRemainder of 3*7*7*1*1*1/10\n\nRemainder of 147/10 = 7\n\nNow remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.\n\nWhen 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.\n\nAnswer (D)\n_________________\n##### General Discussion\nDirector\nJoined: 07 Aug 2011\nPosts: 539\nConcentration: International Business, Technology\nGMAT 1: 630 Q49 V27\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Apr 2015, 07:49\n1\nBunuel wrote:\nWhat are the last two digits of 63*35*37*82*71*41?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\nKudos for a correct solution.\n\n63*35 = 05 *37 = 85 * 82 = 70 * 71 = 70*41 = 70\n_________________\n\nThanks,\nLucky\n\n_______________________________________________________\nKindly press the to appreciate my post !!\n\nCurrent Student\nJoined: 25 Nov 2014\nPosts: 99\nConcentration: Entrepreneurship, Technology\nGMAT 1: 680 Q47 V38\nGPA: 4\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Apr 2015, 10:36\n4\nFirst, multiplied 82 and 35, since 2 and 5 will give a 0 in the end. Got last 2 digits 70.\nno need to multiply 70 by 41 and 71, since they wont change the last 2 digits (try if you want to, but effective multiplication in case of 41 and 71 is only with 1).\nNow, 70 * 3(of 63) gives : last 2 digits as 10.\nlastly, 10 * 7(of 37) gives : last 2 digits 70.\n\nThus D.\n_________________\n\nKudos!!\n\nManager\nJoined: 26 Dec 2012\nPosts: 146\nLocation: United States\nConcentration: Technology, Social Entrepreneurship\nWE: Information Technology (Computer Software)\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Apr 2015, 12:31\n63*35*37*82*71*41=\n\nWe have to focus on the last two digits only, so 63*35=05*37=85*82=70\n71*41=81 therefore 81*70=70\nHence Answer is D\n\nThanks,\nCEO\nJoined: 11 Sep 2015\nPosts: 3120\nLocation: Canada\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n19 Jul 2016, 10:10\nTop Contributor\n1\nBunuel wrote:\nWhat are the last two digits of (63)(35)(37)(82)(71)(41)?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\n(63)(35)(37)(82)(71)(41) = (63)(37)(71)(41)(35)(82)\nMultiply the units digits in the red product to get: (#####1)(35)(82)\nRewrite blue product as follows: (#####1)(7)(5)(2)(41)\nRearrange: (#####1)(7)(41)(5)(2)\nThe units digit of (7)(41) is 7, so we get: (#####1)(##7)(10)\n= (####7)(10)\n= #####70\n\nAnswer:\n_________________\n\nBrent Hanneson \u2013 GMATPrepNow.com\n\nSenior Manager\nJoined: 31 Jul 2017\nPosts: 490\nLocation: Malaysia\nSchools: INSEAD Jan '19\nGMAT 1: 700 Q50 V33\nGPA: 3.95\nWE: Consulting (Energy and Utilities)\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n04 Feb 2018, 23:10\nBunuel wrote:\nWhat are the last two digits of 63*35*37*82*71*41?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\nKudos for a correct solution.\n\nThe equation can be written as $$7*9*7*5*37*41*2*71*41$$\n$$49*9*10*37*41*71*41$$ = $$401*10*37*41*41*71$$\nAs you can see, the last digit will be 0, and except for 37 all the digits end with 1.\nSo, last two Digit will be 70.\n_________________\n\nIf my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!\n\nManager\nJoined: 02 Jan 2016\nPosts: 117\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2018, 10:30\nBunuel wrote:\nBunuel wrote:\nWhat are the last two digits of 63*35*37*82*71*41?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\nKudos for a correct solution.\n\nVERITAS PREP OFFICIAL SOLUTION:\n\nWe know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.\n\nRemainder of (63*35*37*82*71*41)/ 100\n\nNote that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:\n\nNote: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.\n\nSo say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.\n\nTake another example to reinforce this \u2013 what is the remainder when 85 is divided by 20? It is 5.\n\nWe might rephrase it as \u2013 what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.\n\nSo keeping this very important point in mind, let\u2019s go ahead and cancel the common 5 and 2.\n\nWe need the\n\nRemainder of (63*7*37*41*71*41*5*2)/10*5*2\n\nRemainder of (63*7*37*41*71*41)/10\n\nNow using concept 2, let\u2019s write the numbers in form of multiples of 10\n\nRemainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10\n\nRemainder of 3*7*7*1*1*1/10\n\nRemainder of 147/10 = 7\n\nNow remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.\n\nWhen 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.\n\nAnswer (D)\n\nHi I thinks this\n\n\"Remainder of 147/10 = 7\n\nNow remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.\"\n\nsame as 1470/100 = 70 right ?\n\nalso Remainder 3*-3*-3*1*1*1/10 seem more efficient than 3*7*7*1*1*1/10 ? I hope this way is correct too ?\nManager\nJoined: 26 Sep 2017\nPosts: 97\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2018, 12:56\nBunuel wrote:\nBunuel wrote:\nWhat are the last two digits of 63*35*37*82*71*41?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\nKudos for a correct solution.\n\nVERITAS PREP OFFICIAL SOLUTION:\n\nWe know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.\n\nRemainder of (63*35*37*82*71*41)/ 100\n\nNote that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:\n\nNote: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.\n\nSo say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.\n\nTake another example to reinforce this \u2013 what is the remainder when 85 is divided by 20? It is 5.\n\nWe might rephrase it as \u2013 what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.\n\nSo keeping this very important point in mind, let\u2019s go ahead and cancel the common 5 and 2.\n\nWe need the\n\nRemainder of (63*7*37*41*71*41*5*2)/10*5*2\n\nRemainder of (63*7*37*41*71*41)/10\n\nNow using concept 2, let\u2019s write the numbers in form of multiples of 10\n\nRemainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10\n\nRemainder of 3*7*7*1*1*1/10\n\nRemainder of 147/10 = 7\n\nNow remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.\n\nWhen 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.\n\nAnswer (D)\n\nHi bunuel ..\n\nCan you enlighten how to approach this kind more other questions like to find last three digit..etc etc..\n\nAnd in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there...\n\nSent from my BND-AL10 using GMAT Club Forum mobile app\nManager\nJoined: 02 Jan 2016\nPosts: 117\nRe: What are the last two digits of 63*35*37*82*71*41?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Feb 2018, 07:37\nviv007 wrote:\nBunuel wrote:\nBunuel wrote:\nWhat are the last two digits of 63*35*37*82*71*41?\n\n(A) 10\n(B) 30\n(C) 40\n(D) 70\n(E) 80\n\nKudos for a correct solution.\n\nVERITAS PREP OFFICIAL SOLUTION:\n\nWe know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.\n\nRemainder of (63*35*37*82*71*41)/ 100\n\nNote that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:\n\nNote: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.\n\nSo say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.\n\nTake another example to reinforce this \u2013 what is the remainder when 85 is divided by 20? It is 5.\n\nWe might rephrase it as \u2013 what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.\n\nSo keeping this very important point in mind, let\u2019s go ahead and cancel the common 5 and 2.\n\nWe need the\n\nRemainder of (63*7*37*41*71*41*5*2)/10*5*2\n\nRemainder of (63*7*37*41*71*41)/10\n\nNow using concept 2, let\u2019s write the numbers in form of multiples of 10\n\nRemainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10\n\nRemainder of 3*7*7*1*1*1/10\n\nRemainder of 147/10 = 7\n\nNow remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.\n\nWhen 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.\n\nAnswer (D)\n\nHi bunuel ..\n\nCan you enlighten how to approach this kind more other questions like to find last three digit..etc etc..\n\nAnd in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there...\n\nSent from my BND-AL10 using GMAT Club Forum mobile app\n\nHi Please see this\n\nhttps://gmatclub.com/forum/compilation- ... ml#p650870\nRe: What are the last two digits of 63*35*37*82*71*41? &nbs [#permalink] 09 Feb 2018, 07:37\nDisplay posts from previous: Sort by\n\n# What are the last two digits of 63*35*37*82*71*41?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-11-15 21:31:05", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-are-the-last-two-digits-of-195836.html", "openwebmath_score": 0.6337324380874634, "openwebmath_perplexity": 1089.9866062739293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9766692366242306, "lm_q2_score": 0.9005297774417915, "lm_q1q2_score": 0.8795197302914628}}
{"url": "https://economics.stackexchange.com/questions/15464/zero-sum-game-constant-sum-game/15466", "text": "# Zero sum game, constant sum game\n\nGiven any bilateral zero-sum game G, show that strategy pro\ufb01le \u03c3 is a Nash equilibrium for G if, and only if, it is a Nash equilibrium for the constant-sum game G' obtained from G by adding any \ufb01xed amount \"d\" to the payoffs of both players. Is the conclusion affected if the \ufb01xed amount, call it now $d_i$ for each i = 1 , 2 , differs between the two players?\n\nSuppose $(\\sigma_1^*, \\sigma_2^*)$ is the Nash Equilibrium of the game $G$ consisting of players $\\{1,2\\}$ having strategy sets $A_1$ and $A_2$ and the payoff functions $u_1:A_1\\times A_2 \\rightarrow \\mathbb{R}$ and $u_2:A_1\\times A_2 \\rightarrow \\mathbb{R}$. Consider the game $G'$ in which everything else is same as $G$ except that the payoff functions are $v_1:A_1\\times A_2 \\rightarrow \\mathbb{R}$ and $v_2:A_1\\times A_2 \\rightarrow \\mathbb{R}$ defined as: $v_1(a_1, a_2) = u_1(a_1, a_2) + d_1$ and $v_2(a_1, a_2) = u_2(a_1, a_2) + d_2$. We will now show that $(\\sigma_1^*, \\sigma_2^*)$ is also the Nash Equilibrium of $G'$.\n\nSince $(\\sigma_1^*, \\sigma_2^*)$ is the Nash Equilibrium of $G$, \\begin{eqnarray*} \\mathbb{E}(u_1(\\sigma_1^*, \\sigma_2^*)) \\geq \\mathbb{E}(u_1(\\sigma_1, \\sigma_2^*)) & & \\forall \\sigma_1 \\in \\Delta (A_1) \\\\ \\mathbb{E}(u_2(\\sigma_1^*, \\sigma_2^*)) \\geq \\mathbb{E}(u_2(\\sigma_1^*, \\sigma_2)) & & \\forall \\sigma_2 \\in \\Delta (A_2) \\end{eqnarray*}\n\nThe above can also be rewritten as \\begin{eqnarray*} \\mathbb{E}(u_1(\\sigma_1^*, \\sigma_2^*)) + d_1 \\geq \\mathbb{E}(u_1(\\sigma_1, \\sigma_2^*)) + d_1 & & \\forall \\sigma_1 \\in \\Delta (A_1) \\\\ \\mathbb{E}(u_2(\\sigma_1^*, \\sigma_2^*)) + d_2 \\geq \\mathbb{E}(u_2(\\sigma_1^*, \\sigma_2)) + d_2 & & \\forall \\sigma_2 \\in \\Delta (A_2) \\end{eqnarray*}\n\nBecause $d_1$ and $d_2$ are constants, we can take them inside the expectation and rewrite the above inequalities as: \\begin{eqnarray*} \\mathbb{E}(u_1(\\sigma_1^*, \\sigma_2^*) + d_1) \\geq \\mathbb{E}(u_1(\\sigma_1, \\sigma_2^*) + d_1) & & \\forall \\sigma_1 \\in \\Delta (A_1) \\\\ \\mathbb{E}(u_2(\\sigma_1^*, \\sigma_2^*) + d_2) \\geq \\mathbb{E}(u_2(\\sigma_1^*, \\sigma_2) + d_2) & & \\forall \\sigma_2 \\in \\Delta (A_2) \\end{eqnarray*}\n\nBy definition of $v_1$ and $v_2$,\n\n\\begin{eqnarray*} \\mathbb{E}(v_1(\\sigma_1^*, \\sigma_2^*)) \\geq \\mathbb{E}(v_1(\\sigma_1, \\sigma_2^*)) & & \\forall \\sigma_1 \\in \\Delta (A_1) \\\\ \\mathbb{E}(v_2(\\sigma_1^*, \\sigma_2^*)) \\geq \\mathbb{E}(v_2(\\sigma_1^*, \\sigma_2)) & & \\forall \\sigma_2 \\in \\Delta (A_2) \\end{eqnarray*}\n\nTherefore, $(\\sigma_1^*, \\sigma_2^*)$ is also the Nash Equilibrium of $G'$.\n\nLet me borrow from my answer here.\n\nThe short answer is that adding constants values, even ones that differ across players, will not change the set of Nash equilibria.\n\nThis is easy to see from the definition of Nash equilibrium.\n\nLet $u_i$ represent player $i$'s payoffs as a function of all the players' strategies. A strategy profile $\\sigma= (\\sigma_i , \\sigma_{-i})$, where $\\sigma_i$ is $i$'s strategy, and $\\sigma_{-i}$ is a vector of all the other players' strategies, is a Nash equilibrium if, for each player $i$,\n\n$$u_i (\\sigma_i , \\sigma_{-i}) \\ge u_i (\\sigma_i^\\prime ,\\sigma_{-i})$$\n\nfor any other strategy $\\sigma_i^\\prime$ of player $i$.\n\nLet $v_i = u_i + d_i$, for each $i$, and let $\\sigma$ be a Nash equilibrium strategy profile under $u_i$. Since adding a constant to both sides will preserve the inequality above, it must also be the case that\n\n$$v_i (\\sigma_i , \\sigma_{-i}) \\ge v_i (\\sigma_i^\\prime ,\\sigma_{-i})$$\n\nfor all $i$ and $\\sigma_i^\\prime$, so $\\sigma$ is a Nash equilibrium strategy profile under the modified set of payoff functions as well.\n\nIn fact, it is a standard assumption in game theory that payoff functions represent preferences that satisfy the von Neumann-Morgenstern axioms. This implies that preference orderings are preserved by affine transformations of utility functions. (In this context, an affine transformation is just multiplication by a positive number and adding arbitrary constants to payoff functions.)\n\n\u2022 Why answer the exact same question on two SE sites, instead of flagging one for being crossposted? \u2013\u00a0Giskard May 27 '17 at 6:44\n\nOther answers took the game theoretic approach, I will take the utility approach. Adding a real number $d_i$ to all payoffs of a player is an affine transformation of said payoffs. This does not change the player's preferences even when dealing with expected payoffs. Thus the desirability of outcomes is unchanged from the player's perspective hence best responses remain the same.", "date": "2019-09-19 16:42:21", "meta": {"domain": "stackexchange.com", "url": "https://economics.stackexchange.com/questions/15464/zero-sum-game-constant-sum-game/15466", "openwebmath_score": 0.9996905326843262, "openwebmath_perplexity": 605.2407284412714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97364464868338, "lm_q2_score": 0.9032942106088968, "lm_q1q2_score": 0.8794875743460303}}
{"url": "https://math.stackexchange.com/questions/2314467/on-the-regularity-of-the-alterning-sum-of-prime-numbers", "text": "On the regularity of the alterning sum of prime numbers\n\nLet's define $(p_n)_{n\\in \\mathbb N}$ the ordered list of prime numbers ($p_0=2$, $p_1=3$, $p_2=5$...).\n\nI am interested in the following sum:\n\n$$S_n:=\\sum_{k=1}^n (-1)^kp_k$$\n\nSince the sequence $(S_n)$ is related to the gaps between prime numbers, I would expect it to be quite irregular.\n\nBut if we plot $(S_n)_{1\\leqslant n\\leqslant N}$ for $N\\in \\{50,10^3,10^5,10^6,10^7\\}$, we obtain the following:\n\nWe can observe a great regularity.\n\nSo my questions are:\n\n\u2022 Why is there so much regularity?\n\n\u2022 Can we find the equations of the two lines forming $(S_n)$?\n\n\u2022 Is there a proof that it will continue to be that regular forever?\n\nAny contribution, even partial, will be greatly appreciated.\n\nThanks to mixedmath and Daniel Fischer, here is more curves:\n\n\u2022 in blue, you have $S_n$;\n\n\u2022 in red, you have $\\displaystyle 2^{1/6}\\displaystyle \\sum (-1)^k k\\log k$;\n\n\u2022 in green, you have $\\displaystyle \\sum (-1)^k k\\log k$;\n\n\u2022 in purple, you have $\\displaystyle \\sum (-1)^k k(\\log k+\\log\\log k)$;\n\n\u2022 in yellow, you have $\\displaystyle \\sum (-1)^k k(\\log k+\\log\\log k-1)$.\n\nMy question seems quite related to this one.\n\n\u2022 Did you take $1$ as a prime? It seems you did, because in the first diagram, $S_2<0$. \u2013\u00a0Mastrem Jun 8 '17 at 11:12\n\u2022 @Mastrem I did not, in the first diagram, $S_1=2>0$, and $S_2=2-3=-1<0$. \u2013\u00a0E. Joseph Jun 8 '17 at 11:49\n\u2022 Then shouldn't $S_n$ be defined as $S_n:=\\sum_{k=1}^{n}(-1)^kp_k$? The way it's currently defined, we have $S_1=0$, as there are no primes lower than or equal to $1$. \u2013\u00a0Mastrem Jun 8 '17 at 12:00\n\u2022 for the third question you have to show that $|\\sum \\limits_{k=1}^n (-1)^k p_k| \\approx \\frac{p_n}{2}$ \u2013\u00a0Ahmad Jun 8 '17 at 15:25\n\u2022 Ah, I hadn't noticed that you'd edited in the additional graphs. Indeed, the extra terms from Dusart's bounds (including the $-1$) look very good. \u2013\u00a0davidlowryduda Jun 8 '17 at 20:52\n\nThis is a great question. Unfortunately, this is an incomplete answer. But I thought about this a bit and I noticed something interesting, but which I do not know how to explain.\n\nWith $$S_n = \\sum_{k \\leq n} (-1)^k p_k,$$ where $p_n$ is the $n$th prime, some patterns are immediately clear. It is obvious that the sequence of $S_n$ alternates in sign for example. But some patterns are not obvious or clear.\n\nBy the prime number theorem, we expect that $p_n \\approx n \\log n$. If we plot $\\sum_{k \\leq n} (-1)^k k \\log k$ against $S_n$ for all primes up to one million, we get\n\nThis is apparently a bit too small, it seems. This sort of makes sense, as deviations from the approximation $p_n \\approx n \\log n$ compound here.\n\nHowever, I noticed that $$1.12 \\sum_{k \\leq n} (-1)^k k \\log k$$ is actually a very good (experimental) estimate of what's going on, as can be seen in the following plot.\n\nPerhaps $1.12$ is an incorrect choice --- it just happened to be a very nearby reasonable seeming number, and it does appear to reflect what's going on. I do not know why, though.\n\nIf we conjecture for a moment that $1.12 \\sum (-1)^k k \\log k$ is a good estimator, then we can write a good asymptotic for this series using partial summation. Namely\n\n\\begin{align} \\sum_{k \\leq n} (-1)^k k \\log k &= \\left( \\sum_{k \\leq n} (-1)^k k \\right) \\log n - \\int_1^n \\left( \\sum_{k \\leq t} (-1)^k k \\right) \\frac{1}{t} dt \\\\ &= (-1)^n \\left \\lfloor \\frac{n+1}{2} \\right \\rfloor \\log n - \\int_1^n (-1)^{\\lfloor t \\rfloor} \\left \\lfloor \\frac{\\lfloor t \\rfloor+1}{2} \\right \\rfloor \\frac{1}{t} dt \\\\ &= (-1)^n \\left \\lfloor \\frac{n+1}{2} \\right \\rfloor \\log n + O \\left( \\int_1^n \\left( \\frac{t+1}{2t} + \\frac{2}{t} \\right) dt\\right) \\\\ &= (-1)^n \\left \\lfloor \\frac{n+1}{2} \\right \\rfloor \\log n + O(n). \\end{align}\n\nSo I conjecture that $$S_n \\approx 1.12 (-1)^n \\left \\lfloor \\frac{n+1}{2} \\right \\rfloor \\log n + O(n).$$ For comparison, the size of the alternating sum of the first 1001 primes is $3806$, where this estimate gives about $3876.6$. For $10001$, the actual is $52726$, compared to the estimated $51588.7$. These are both close, although apparently not super accurate.\n\nIt may be possible to describe the actual behavior of $S_n$ a bit more by using secondary terms in the prime number theorem, but I was not successful in my back-of-the-envelope computations. Nor do I know how to explain the $1.12$ that appears in this answer (or how to determine if it is $1.12$ as opposed to, say, $1.15$). Perhaps someone else will see how to fill in these gaps.\n\n(Edited in after Daniel Fischer's comment)\n\nHere are updated images, including plots of $\\sum (-1)^n n (\\log n + \\log \\log n)$.\n\nAs we can see, $\\sum (-1)^n n (\\log n + \\log \\log n)$ grows in magnitude just a little bit more quickly. Focusing a bit on just the upper half, we get\n\n\u2022 Thank you for this great work! For the minus sign, I defined $p_0=2$ so I would avoid the first negative sign, but this is not really important. I found it strange how $\\sqrt 5$ pop here. I wish someone will see how to fill these gaps! Thanks again for this work, it gave me a lot to think about. \u2013\u00a0E. Joseph Jun 8 '17 at 16:24\n\u2022 I have tried to replicate your results for the last $20$ minutes, but it can't seem to work. It seems that $\\sqrt 5\\sum (-1)^k k \\log k$ is way superior to $S_n$. I don't understand why we obtain such different results. \u2013\u00a0E. Joseph Jun 8 '17 at 16:52\n\u2022 @E.Joseph Ah, you are correct. I made a (very silly) mistake in my code. I will change it and update accordingly. \u2013\u00a0davidlowryduda Jun 8 '17 at 17:22\n\u2022 It would be interesting to see the plot of $\\sum (-1)^k k(\\log k + \\log \\log k)$ in there too. \u2013\u00a0Daniel Fischer Jun 8 '17 at 18:11", "date": "2019-05-19 22:28:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2314467/on-the-regularity-of-the-alterning-sum-of-prime-numbers", "openwebmath_score": 0.9954421520233154, "openwebmath_perplexity": 351.93902487681754, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241947446617, "lm_q2_score": 0.9073122238669026, "lm_q1q2_score": 0.8794796907817736}}
{"url": "https://getreal.life/micron-forum-xoiar/if-a-function-is-bijective-then-its-inverse-is-unique-fdf388", "text": "## if a function is bijective then its inverse is unique\n\nDe\ufb02nition 1. the inverse function is not well de ned. ... Domain and range of inverse trigonometric functions. This function maps each image to its unique \u2026 However if we change its domain and codomain to the set than the function becomes bijective and the inverse function exists. Pythagorean theorem. If a function f : A -> B is both one\u2013one and onto, then f is called a bijection from A to B. Equivalence Relations and Functions October 15, 2013 Week 13-14 1 Equivalence Relation A relation on a set X is a subset of the Cartesian product X\u00a3X.Whenever (x;y) 2 R we write xRy, and say that x is related to y by R.For (x;y) 62R,we write x6Ry. Read Inverse Functions for more. PROPERTIES OF FUNCTIONS 116 then the function f: A!B de ned by f(x) = x2 is a bijection, and its inverse f 1: B!Ais the square-root function, f 1(x) = p x. You job is to verify that the answers are indeed correct, that the functions are inverse functions of each other. And this function, then, is the inverse function \u2026 Since it is both surjective and injective, it is bijective (by definition). MENSURATION. I\u2019ll talk about generic functions given with their domain and codomain, where the concept of bijective makes sense. (proof is in textbook) Induced Functions on Sets: Given a function , it naturally induces two functions on power sets: And we had observed that this function is both injective and surjective, so it admits an inverse function. Inverse. Otherwise, we call it a non invertible function or not bijective function. If every \"A\" goes to a unique \"B\", and every \"B\" has a matching \"A\" then we can go back and forwards without being led astray. is bijective and its inverse is 1 0 \u211d 1 log A discrete logarithm is the inverse from MAT 243 at Arizona State University Thanks! The function f is called as one to one and onto or a bijective function if f is both a one to one and also an onto function. If the function is bijective, find its inverse. The inverse function g : B \u2192 A is defined by if f(a)=b, then g(b)=a. A function is invertible if and only if it is a bijection. In mathematics, an invertible function, also known as a bijective function or simply a bijection is a function that establishes a one-to-one correspondence between elements of two given sets.Loosely speaking, all elements of the sets can be matched up in pairs so that each element of one set has its unique counterpart in the second set. Proof of Property 1: Suppose that f -1 (y 1) = f -1 (y 2) for some y 1 and y 2 in B. This function g is called the inverse of f, and is often denoted by . Naturally, if a function is a bijection, we say that it is bijective.If a function $$f :A \\to B$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. A function f : X \u2192 Y is said to be one to one correspondence, if the images of unique elements of X under f are unique, i.e., for every x1 , x2 \u2208 X, f(x1 ) = f(x2 ) implies x1 = x2 and also range = codomain. The problem does not ask you to find the inverse function of $$f$$ or the inverse function of $$g$$. This will be a function that maps 0, infinity to itself. It is a function which assigns to b, a unique element a such that f(a) = b. hence f-1 (b) = a. Properties of Inverse Function. A relation R on a set X is said to be an equivalence relation if Bijective functions have an inverse! Yes. In other words, an injective function can be \"reversed\" by a left inverse, but is not necessarily invertible, which requires that the function is bijective. The inverse of bijection f is denoted as f-1. Tags: bijective bijective homomorphism group homomorphism group theory homomorphism inverse map isomorphism. Further, if it is invertible, its inverse is unique. 2. Well, that will be the positive square root of y. For example, if fis not one-to-one, then f 1(b) will have more than one value, and thus is not properly de ned. Since g is a left-inverse of f, f must be injective. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. Instead, the answers are given to you already. Bijective Function Solved Problems. Mensuration formulas. Domain and Range. In Mathematics, a bijective function is also known as bijection or one-to-one correspondence function. Let f : A \u2192 B be a function with a left inverse h : B \u2192 A and a right inverse g : B \u2192 A. If F is a bijective function from X to Y then there is an inverse function G from MATH 1 at Far Eastern University This procedure is very common in mathematics, especially in calculus . Note that given a bijection f: A!Band its inverse f 1: B!A, we can write formally the above de nition as: 8b2B; 8a2A(f 1(b) = a ()b= f(a)): All help is appreciated. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). Here we are going to see, how to check if function is bijective. Properties of inverse function are presented with proofs here. Intuitively it seems obvious, but how do I go about proving it using elementary set theory and predicate logic? Bijections and inverse functions. Claim: if f has a left inverse (g) and a right inverse (g\u02b9) then g = g\u02b9. We must show that g(y) = g\u02b9(y). First of, let\u2019s consider two functions $f\\colon A\\to B$ and $g\\colon B\\to C$. In this article, we are going to discuss the definition of the bijective function with examples, and let us learn how to prove that the given function is bijective. And g inverse of y will be the unique x such that g of x equals y. Formally: Let f : A \u2192 B be a bijection. In mathematics, an inverse function (or anti-function) is a function that \"reverses\" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, i.e., g(y) = x if and only if f(x) = y. Stated in concise mathematical notation, a function f: X \u2192 Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). Let $$f : A \\rightarrow B$$ be a function. Inverse of a function The inverse of a bijective function f: A \u2192 B is the unique function f \u20111: B \u2192 A such that for any a \u2208 A, f \u20111(f(a)) = a and for any b \u2208 B, f(f \u20111(b)) = b A function is bijective if it has an inverse function a b = f(a) f(a) f \u20111(a) f f \u20111 A B Following Ernie Croot's slides A continuous function from the closed interval [ a , b ] in the real line to closed interval [ c , d ] is bijection if and only if is monotonic function with f ( a ) = c and f ( b ) = d . Solving word problems in trigonometry. Definition 853 A function f D C is bijective if it is both one to one and onto from MA 100 at Wilfrid Laurier University A function f : X \u2192 Y is bijective if and only if it is invertible, that is, there is a function g: Y \u2192 X such that g o f = identity function on X and f o g = identity function on Y. If f:X->Y is a bijective function, prove that its inverse is unique. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. Since g is also a right-inverse of f, f must also be surjective. So what is all this talk about \"Restricting the Domain\"? Injections may be made invertible [ edit ] In fact, to turn an injective function f : X \u2192 Y into a bijective (hence invertible ) function, it suffices to replace its codomain Y by its actual range J = f ( X ) . Functions that have inverse functions are said to be invertible. The term one-to-one correspondence should not be confused with the one-to-one function (i.e.) For more videos and resources on this topic, please visit http://ma.mathforcollege.com/mainindex/05system/ TAGS Inverse function, Department of Mathematics, set F. Share this link with a friend: Hi, does anyone how to solve the following problems: In each of the following cases, determine if the given function is bijective. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. In this video we prove that a function has an inverse if and only if it is bijective. injective function. Summary and Review; A bijection is a function that is both one-to-one and onto. Learn if the inverse of A exists, is it uinique?. Proof: Choose an arbitrary y \u2208 B. More clearly, f maps unique elements of A into unique images in \u2026 Another important example from algebra is the logarithm function. From this example we see that even when they exist, one-sided inverses need not be unique. Below f is a function from a set A to a set B. Theorem 9.2.3: A function is invertible if and only if it is a bijection. 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Which allows us to have an inverse function exists is both injective surjective! Also a right-inverse of f, and is often denoted by more clearly f! Also known as bijection or one-to-one correspondence function that have inverse functions are said to be.... A non invertible function or not bijective function follows stricter rules than a general function prove!", "date": "2021-06-19 20:57:33", "meta": {"domain": "getreal.life", "url": "https://getreal.life/micron-forum-xoiar/if-a-function-is-bijective-then-its-inverse-is-unique-fdf388", "openwebmath_score": 0.8254319429397583, "openwebmath_perplexity": 324.6314165465449, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9912886176011569, "lm_q2_score": 0.8872045862611168, "lm_q1q2_score": 0.8794758078441888}}
{"url": "https://math.stackexchange.com/questions/646705/counting-integer-partitions-of-n-into-exactly-k-distinct-parts-size-at-most-m", "text": "Counting integer partitions of n into exactly k distinct parts size at most M\n\nHow can I find the number of partitions of $n$ into exactly $k$ distinct parts, where each part is at most $M$?\n\nThe number of partitions $p_k(\\leq M,n)$ of $n$ into at most $k$ parts, each of size at most $M$, is given by the generating function: $$\\binom{M+k}{k}_{x} = \\prod_{j=1}^{k}\\frac{1-x^{M+k-j+1}}{1-x^j}= \\sum_{n=0}^{kM} p_{k}(\\leq M,n) x^n$$\n\nFor the number of the partitions $p_k(\\mathcal{D},n)$ of $n$ into at most $k$ parts there is the recurrence relationship: $$p_{k}(\\mathcal{D},n) = p_{k}(\\mathcal{D},n-k) + p_{k-1}(\\mathcal{D},n)$$\n\nBut what, if I want to count only the partitions with distinct parts and restricted number of parts and restricted part size?\n\nUpdate: Now I know the generating function for the number of distinct restricted partitions $p_k(\\leq M, \\mathcal{D},n)$ of $n$ into exactly $k$ distinct parts, all at most $M$ is $$\\prod_{j=1}^{M} (1+xq^{j}) = \\sum_{k,n=0}^{\\infty}p_k(\\leq M, \\mathcal{D},n)x^{k}q^{n}$$ and there is also a recurrence relation $$p_k(\\leq M, \\mathcal{D},n) = p_{k-1}(\\leq M-1, \\mathcal{D},n-k) + p_k(\\leq M-1, \\mathcal{D},n-k)$$ How can I prove this? Could you recommend a book, where I could read about this?\n\n\u2022 Simul-posted to MO, mathoverflow.net/questions/155315/\u2026 without notification to either site. \u2013\u00a0Gerry Myerson Jan 21 '14 at 21:43\n\u2022 Say you have a partition $\\lambda_1>\\lambda_1>...>\\lambda_k$ of $n$ with $\\lambda_1 \\le M$. Then $(\\lambda_1-k+1) \\ge (\\lambda_2-k+2) \\ge ... \\ge \\lambda_k$ is a partition of $n - k(k-1)/2$ with k parts each at most M-k+1 but no longer necessarily distinct. Now you can apply one of the previous results. \u2013\u00a0Nate Jan 21 '14 at 21:46\n\u2022 Very useful paper, though this gives a good approximation to the number of partitions of n into exactly k parts each no larger than N due to Ratsaby (App. Analysis and Discrete M. 2008): doiserbia.nb.rs/img/doi/1452-8630/2008/1452-86300802222R.pdf \u2013\u00a0Alexander Kartun-Giles Jun 18 '16 at 15:43\n\nIn the \"Update\" (the day after the Question itself was posted), the OP mentions a recursion for counting the partitions of $n$ into $k$ distinct parts, each part at most $M$:\n\n$$p_k(\\leq M, \\mathcal{D},n) = p_{k-1}(\\leq M-1, \\mathcal{D},n-k) + p_k(\\leq M-1, \\mathcal{D},n-k)$$\n\nand asks \"How can I prove this?\".\n\nTo see this, separate the required partitions on the basis of whether $1$ appears as a summand. If it does, then subtracting $1$ from each summand produces a partition of $n-k$ with exactly $k-1$ distinct parts (since the original summand $1$ disappears), each part at most $M-1$. These partitions are counted by the first term on the right-hand side of the recursion. Otherwise the summand $1$ does not appear, and subtracting $1$ from each part resulting in a partition of $n-k$ with exactly $k$ distinct parts, each part at most $M-1$. These cases are counted by the second term.\n\nNote that a partition of $n$ with $k$ distinct parts exists if and only if $n \\ge \\binom{k+1}{2}$, because the ascending summands $m_1 + \\ldots + m_k = n$ must satisfy $m_i \\ge i$. If $n = \\binom{k+1}{2}$, then there is just one such partition with $k$ distinct parts, the largest of which is $k$. Repeated application of the recursion will culminate with terms which we can evaluate \"by inspection\" as either zero or one.\n\nBy itself this recursion doesn't seem to give us an especially attractive way of evaluating $p_k(\\leq M, \\mathcal{D},n)$. Like the recursion for Fibonacci numbers, as a top-down method it suffers from recalculating terms multiple times (giving exponential complexity), so we would be better off working with it as a bottom-up method (giving polynomial complexity).\n\nBetter for large parameters is to close the circle with the ideas presented by @NikosM. by showing how the evaluation of $p_k(\\leq M, \\mathcal{D},n)$ can be reduced to counting restricted partitions without requiring distinct summands.\n\nProp. Suppose that $n \\gt \\binom{k+1}{2}$. Then the following are equal:\n\n$(i)$ the number of partitions of $n$ into exactly $k$ distinct parts, each part at most $M$, i.e. $p_k(\\leq M, \\mathcal{D},n)$\n\n$(ii)$ the number of partitions of $n - \\binom{k}{2}$ into exactly $k$ parts, each part at most $M$\n\n$(iii)$ the number of partitions of $n - \\binom{k+1}{2}$ into at most $k$ parts, each part at most $M$\n\nSketch of proof: Once we know the ordered summands of partitions in $(i)$ satisfy $m_i \\ge i$, it is easy to visualize their equivalents in $(ii)$ and $(iii)$ by Young tableaux, also called Ferrers diagrams. We remove a \"base triangle\" of dots corresponding to the first $i$ dots in the $i$th summand (since $m_i \\ge i$) to get case $(iii)$, and remove one fewer dot in each summand to preserve exactly $k$ summands in case $(ii)$. These constructions are reversible, and the counts are equal.\n\nRemark 1 If $n \\le \\binom{k+1}{2}$, $p_k(\\leq M, \\mathcal{D},n)=1$ if $n=\\binom{k+1}{2}$ and $k \\le M$ and otherwise $p_k(\\leq M, \\mathcal{D},n)=0$.\n\nRemark 2 For fixed $k,n$, suppose $M_0 = n - \\binom{k}{2} \\gt 0$. Then for all $M \\ge M_0$, $p_k(\\leq M, \\mathcal{D},n) = p_k(\\leq M_0, \\mathcal{D},n)$. That is, further increasing the upper bound $M$ on the size of parts will not yield additional partitions of $n$ with exactly $k$ distinct parts.\n\nAndrews, George E. The Theory of Partitions (Cambridge University Press, 1998)\n\nA modern classic for theory of integer partitions, reviewed by Richard Askey in BAMS.\n\nThere is an algorithm to count and generate restricted numerical partitions (both in largest part and number of parts) here p. 93 4.8 Numerical Partitions\n\nDefine $P(n; k; s)$ to be the set of all partitions of $n$ into $k$ parts with largest part equal to $s$, and let $p(n; k; s) = \\left|P(n; > k; s)\\right|$. Clearly, in order to have $p(n; k; s) > 0$ we must have at least one part equal to $s$ and at most $k$ parts equal to $s$. Thus $s + k \\le n \\le ks$.\n\nBy classifying the partitions of $P(n; k; s)$ according to the value of the second largest part, call it $j$, we obtain the following recurrence relation, which has no zero terms; it is a positive recurrence relation.\n\n$$p(n; k; s) = \\sum^{min(s;nX\u00a1s\u00a1k+2)}_{j=max(1,\\frac{n-s}{k-1})}p(n-s,k-1,j)$$\n\n\u2022 I follow your reasoning, but this does not seem to impose the required condition that parts are distinct. \u2013\u00a0hardmath Jul 26 '15 at 15:30\n\u2022 hmm yeah i see. Counting partitions is a difficult task, but all methods use the same recursion (i know because i have a lib for combinatorics where various combinatorial objects are generated and counted, including partitions). So it is do-able in algorithmic form, if you want a closed-form solution am not aware of any \u2013\u00a0Nikos M. Jul 26 '15 at 19:04\n\u2022 PS answer was posted before new update on additional conditon for unique partition parts (so it covers the first two conditions), still a similar recursion can be found fopr distinct parts as well (in algorithmic form, not closed-form solution) \u2013\u00a0Nikos M. Jul 26 '15 at 19:08\n\u2022 @hardmath, oops seems i missed that part, still the recursion holds for both inital conditions can be extended to cover distinct parts all partition recursions follow similar logic in this sense it should be do-able, although do not have time to elaborate on this further, hope OP will find even this useful for a start \u2013\u00a0Nikos M. Jul 26 '15 at 20:29", "date": "2019-05-20 03:17:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/646705/counting-integer-partitions-of-n-into-exactly-k-distinct-parts-size-at-most-m", "openwebmath_score": 0.7788291573524475, "openwebmath_perplexity": 169.06147306098944, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513893774303, "lm_q2_score": 0.8918110440002044, "lm_q1q2_score": 0.8794607001029382}}
{"url": "https://math.stackexchange.com/questions/2879010/sum-of-weighted-binomial-coefficients/2879028", "text": "# Sum of weighted binomial coefficients [duplicate]\n\nI am struggling with computing the following sums:\n\n$$\\sum_{k=1}^{n}k\\binom{n}{k}=\\binom{n}{1}+2\\binom{n}{2}+...+n\\binom{n}{n}$$\n\nand\n\n$$\\sum_{k=0}^{n}\\frac{1}{k+1}\\binom{n}{k}=\\binom{n}{0}+\\frac{1}{2}\\binom{n}{1}+\\frac{1}{3}\\binom{n}{2}+\\dots+\\frac{1}{n+1}\\binom{n}{n}$$\n\nAt first, I tried just rewriting the general terms in a form where the Binomial Theorem could be applied, but could not do so. ($k$ should be in the exponent of something with $n-k$ in the exponent of something else.)\n\nThen, for the first sum, I re-wrote it in the following manner:\n\n\\begin{align} S &=\\binom{n}{n}+\\dots+\\binom{n}{3}+\\binom{n}{2}+\\binom{n}{1}\\\\ &+\\binom{n}{n}+\\dots+\\binom{n}{3}+\\binom{n}{2}\\\\ &+\\binom{n}{n}+\\dots+\\binom{n}{3}\\\\ &\\quad\\vdots\\\\ &+\\binom{n}{n} \\end{align}\n\nNoting that:\n\n$$\\binom{n}{0}+\\binom{n}{1}+\\binom{n}{2}+\\dots+\\binom{n}{n}=\\sum_{k=0}^{n}\\binom{n}{k}1^k1^{n-k}=(1+1)^n=2^n,$$\n\nI can rewrite the first line in $S$ as $2^n-1$ (the $-1$ is there because I'm over-counting the $\\binom{n}{0}=1$).\n\nThen I tried to say that if all the \"blanks\" were filled (if all the lines were like the first line), since there are $n$ lines, then I have $n(2^n-1)$. From this point, we should have:\n\n$$S+\\text{blanks}=n(2^n-1)\\Longrightarrow S=n(2^n-1)-\\text{blanks}$$\n\nHowever, the problem of finding the value of \"blanks\" seems as hard as the original problem and I can't get to the right answer of $n2^{n-1}$ (according to WA). In one of my attempts, I ended up with a geometric series but that still didn't work.\n\nI have not spent much time on the second sum as I feel I should be able to do the first one before even tackling the second one. Am I going in the right direction with this? Is there a more intelligent way of working this out?\n\nThanks in advance for any help/hints.\n\n## marked as duplicate by N. F. Taussig\u00a0combinatorics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 11 '18 at 8:55\n\n\u2022 You may just differentiate/integrate $\\sum_{k=1}^{n}\\binom{n}{k}x^k = (1+x)^n-1$. \u2013\u00a0Jack D'Aurizio Aug 11 '18 at 3:13\n\u2022 Oh, I really should have thought of that. I tunnel-visioned myself into thinking this should be do-able with just basic counting tricks, not calculus. Thanks for the tip, I will try. \u2013\u00a0orion2112 Aug 11 '18 at 3:17\n\u2022 It is doable with basic tricks, but the solution via $\\frac{d}{dx},\\int(\\ldots)dx$ is a one-liner. \u2013\u00a0Jack D'Aurizio Aug 11 '18 at 3:26\n\nOne of my preferred methods of proof for these types of identities is to use a combinatorial proof which generally takes the form of describing a counting problem and finding the answer using two different methods thereby proving that the expressions achieved by each answer must be equal.\n\nSuppose we have a committee of $n$ distinct people. We ask, in how many ways may we choose a subcommittee of any size where the subcommittee has one member designated as the leader (including the case of the leader being the only person on the subcommittee by himself)?\n\n\u2022 Counting Method 1\n\nLet us first break into cases based on the number of members on the committee. Since the committee must have a leader, the committee size must be at least $1$ and can be at most $n$. Let $k$ be the number of people on the subcommittee\n\nNext, let us select all of the members of the subcommittee (including the leader). This can be done in $\\binom{n}{k}$ ways.\n\nThen, from those people selected, let us choose one of them to be the leader. This can be done in $k$ ways.\n\nApplying multiplication principle and summing over all possible cases gives us the total number of subcommittees possible as being:\n\n$$\\sum\\limits_{k=1}^nk\\binom{n}{k}$$\n\n\u2022 Counting Method 2\n\nLet us before anything else select a person to serve as the leader for the subcommittee. This can be done in $n$ ways.\n\nFrom the remaining $n-1$ people, let us choose some subset (possibly empty) to act as the followers in the subcommittee. This can be done in $2^{n-1}$ ways.\n\nApplying multiplication principle, we arrive at a total number of subcommittees as being:\n\n$$n2^{n-1}$$\n\nWe have as a result the identity:\n\n$$\\sum\\limits_{k=1}^nk\\binom{n}{k}=n2^{n-1}$$", "date": "2019-06-27 08:24:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2879010/sum-of-weighted-binomial-coefficients/2879028", "openwebmath_score": 0.6445593237876892, "openwebmath_perplexity": 489.0763167816419, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986151391819461, "lm_q2_score": 0.891811041124754, "lm_q1q2_score": 0.8794606994451387}}
{"url": "http://mathhelpforum.com/algebra/163347-linear-interpolation.html", "text": "1. ## Linear interpolation\n\nHello,\n\nI need to find the pressure of steam at $98C^{o}$\n\nhowever my steam table does not give a value for 98C, however does give a value for 95C and 100C,\n\nI am told in order to find the pressure of steam at 98C I have interpolate from the tables,\n\nI am not sure how to do this?\n\nat 95C the pressure is 0.08453MPa\nand at 100C the pressure is 0.1013MPa\n\nand since 98C is not half way between, how would I interpolate?\n\nis it $\\frac{3}{5} ( 0.08453 + 0.1013) ?$ Multiplying by 3/5 as its 3 units away from 95 and after that another 5 units to 100C\n\nIs this correct?\n\nThanks.\n\n2. Your expression is incorrect. Find the equation of the line of pressure as a function of temperature. Then just plug in 98. What do you get?\n\n3. Originally Posted by Ackbeet\nYour expression is incorrect. Find the equation of the line of pressure as a function of temperature. Then just plug in 98. What do you get?\n\nOkay, but I am not sure how to do that? I dont have any variables to work with?\n\nThank you.\n\n4. Well, use P for pressure, and T for temperature. Then postulate that P = m T + b. Plug in two data points where you know both P and T in order to get two equations. Use those two equations to solve for m and b. Then plug in T = 98. How does that sound?\n\n5. Thanks I get it, I got 0.093182, which seems about right, as it should be higher than the pressure at 95 but lower than the pressure for 100C.\n\n6. Great. And the value is closer to the 100C value than the 95C value, right?\n\n7. (3/5) of the way from a to b is NOT (3/5)(a+ b). The distance from a to b is b- a and 3/5 of that is (3/5)(a- b). Now add that to b: b+ (3/5)(a- b)= (3/5)a+ (2/5)b. (notice that 2+ 3= 5)\n\n8. $\\begin{array}{ccc}\\mbox{Temp(T)} & \\mbox{Pressure(P)} & \\Delta \\\\ 95 &0.08453 & \\mbox{ } \\\\ \\mbox{ } & \\mbox{ } & 0.01677 \\\\ 100 & 0.1013 & \\mbox{ } \\end{array}$\n\n$P(98) \\approx 0.08453 + (3/5)0.01677 = 0.09460$\n\n9. Both ways give a different answer, which one is more accurate? And can you please explain why you add the value of 'b' ?\n\n10. The two methods are equivalent. If, instead of the numerical values, you call the pressure at 95, $a$, and the pressure at 100, $b$, then, using my notation, $\\Delta =b-a$.\nThen,\n\n$P(98)\\approx a + (3/5)(b-a) = (2/5)a+(3/5)b,$\n\n(and now substituting the numbers),\n\n$=(2/5)0.08453+(3/5)0.1013=0.094592.$\n\n11. Hello, Tweety!\n\n$\\text{At }95^oC\\text{ the pressure is 0.08453 MPa}$\n$\\text{and at }100^oC\\text{ the pressure is 0.1013 MPa.}$\n\n$\\text{Use linear interpolation to find the pressure at }98^oC.$\n\nWe are given two points: . $(95,\\,0.08453)\\,\\text{ and }\\,(100,\\,0.1013)$\n\nFind the equation of the line through those two points.\n\n. . The slope is: . $m \\:=\\:\\frac{0.1013 - 0.08453}{100-95} \\:=\\:\\frac{0.01677}{5} \\:=\\:0.003354$\n\n. . Then: . $y - 0.08453 \\:=\\:0.003354(x - 95)$\n\n. . The equation of the line is: . $y \\:=\\:0.003354x - 0.2341$\n\nIf $x = 95\\text{, then: }\\:y \\:=\\:0.003354(95) - 0.2341 \\:=\\:0.094592$", "date": "2017-08-19 10:07:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/163347-linear-interpolation.html", "openwebmath_score": 0.8248884677886963, "openwebmath_perplexity": 582.1885431941855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9861513881564148, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.8794606975962084}}
{"url": "http://au.mathworks.com/help/matlab/ref/rat.html?nocookie=true", "text": "rat\n\nRational fraction approximation\n\nSyntax\n\n\u2022 ```[N,D] = rat(___)``` example\n\nDescription\n\nexample\n\n````R = rat(X)` returns the rational fraction approximation of `X` to within the default tolerance, `1e-6*norm(X(:),1)`. The approximation is a string containing the truncated continued fractional expansion.```\n\nexample\n\n````R = rat(X,tol)` approximates `X` to within the tolerance, `tol`.```\n\nexample\n\n``````[N,D] = rat(___)``` returns two arrays, `N` and `D`, such that `N./D` approximates `X`, using any of the above syntaxes.```\n\nExamples\n\ncollapse all\n\nApproximate Value of \u03c0\n\nApproximate the value of \u03c0 using a rational representation of the quantity `pi`.\n\nThe mathematical quantity \u03c0 is not a rational number, but the quantity `pi` that approximates it is a rational number since all floating-point numbers are rational.\n\nFind the rational representation of `pi`.\n\n```format rat pi ```\n```ans = 355/113 ```\n\nThe resulting expression is a string. You also can use `rats(pi)` to get the same answer.\n\nUse `rat` to see the continued fractional expansion of `pi`.\n\n`R = rat(pi)`\n```R = 3 + 1/(7 + 1/(16)) ```\n\nThe resulting string is an approximation by continued fractional expansion. If you consider the first two terms of the expansion, you get the approximation $3+\\frac{1}{7}=\\frac{22}{7}$, which only agrees with `pi` to 2 decimals.\n\nHowever, if you consider all three terms printed by `rat`, you can recover the value `355/113`, which agrees with `pi` to 6 decimals.\n\n$3+\\frac{1}{7+\\frac{1}{16}}=\\frac{355}{113}\\text{\\hspace{0.17em}}.$\n\nSpecify a tolerance for additional accuracy in the approximation.\n\n```R = rat(pi,1e-7) ```\n```R = 3 + 1/(7 + 1/(16 + 1/(-294))) ```\n\nThe resulting approximation, `104348/33215`, agrees with `pi` to 9 decimals.\n\nExpress Array Elements as Ratios\n\nCreate a 4-by-4 matrix.\n\n```format short; X = hilb(4)```\n```X = 1.0000 0.5000 0.3333 0.2500 0.5000 0.3333 0.2500 0.2000 0.3333 0.2500 0.2000 0.1667 0.2500 0.2000 0.1667 0.1429```\n\nExpress the elements of `X` as ratios of small integers using `rat`.\n\n`[N,D] = rat(X)`\n```N = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D = 1 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7```\n\nThe two matrices, `N` and `D`, approximate `X` with `N./D`.\n\nView the elements of `X` as ratios using ```format rat```.\n\n```format rat X```\n```X = 1 1/2 1/3 1/4 1/2 1/3 1/4 1/5 1/3 1/4 1/5 1/6 1/4 1/5 1/6 1/7 ```\n\nIn this form, it is clear that `N` contains the numerators of each fraction and `D` contains the denominators.\n\nInput Arguments\n\ncollapse all\n\n`X` \u2014 Input arraynumeric array\n\nInput array, specified as a numeric array of class `single` or `double`.\n\nData Types: `single` | `double`\nComplex Number Support: Yes\n\n`tol` \u2014 Tolerancescalar\n\nTolerance, specified as a scalar. `N` and `D` approximate `X`, such that `N./D - X < tol`. The default tolerance is `1e-6*norm(X(:),1)`.\n\nOutput Arguments\n\ncollapse all\n\n`R` \u2014 Continued fractionstring\n\nContinued fraction, returned as a string. The accuracy of the rational approximation via continued fractions increases with the number of terms.\n\n`N` \u2014 Numeratornumeric array\n\nNumerator, returned as a numeric array. `N./D` approximates `X`.\n\n`D` \u2014 Denominatornumeric array\n\nDenominator, returned as a numeric array. `N./D` approximates `X`.\n\ncollapse all\n\nAlgorithms\n\nEven though all floating-point numbers are rational numbers, it is sometimes desirable to approximate them by simple rational numbers, which are fractions whose numerator and denominator are small integers. Rational approximations are generated by truncating continued fraction expansions.\n\nThe `rat` function approximates each element of `X` by a continued fraction of the form\n\n$\\frac{N}{D}={D}_{1}+\\frac{1}{{D}_{2}+\\frac{1}{\\left({D}_{3}+...+\\frac{1}{{D}_{k}}\\right)}}\\text{\\hspace{0.17em}}.$\n\nThe Ds are obtained by repeatedly picking off the integer part and then taking the reciprocal of the fractional part. The accuracy of the approximation increases exponentially with the number of terms and is worst when `X = sqrt(2)`. For `X = sqrt(2)` , the error with `k` terms is about `2.68*(.173)^k`, so each additional term increases the accuracy by less than one decimal digit. It takes 21 terms to get full floating-point accuracy.", "date": "2015-06-04 00:17:45", "meta": {"domain": "mathworks.com", "url": "http://au.mathworks.com/help/matlab/ref/rat.html?nocookie=true", "openwebmath_score": 0.9125400185585022, "openwebmath_perplexity": 1415.3283038722816, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513869353993, "lm_q2_score": 0.8918110389681662, "lm_q1q2_score": 0.8794606929627564}}
{"url": "http://math.stackexchange.com/questions/850745/can-asymptotes-be-curved", "text": "# Can asymptotes be curved?\n\nWhen I was first introduced to the idea of an asymptote, I was taught about horizontal asymptotes (of form $y=a$) and vertical ones ( of form $x=b$).\n\nI was then shown oblique asymptotes-- slanted asymptotes which are not constant (of the form $y=ax+b$).\n\nWhat happens, though, if we've got a function such as $$f(x)=e^x+\\frac{1}{x}?$$\n\nIs $y=e^x$ considered an asymptote in this example?\n\nAnother example, just to show you where I'm coming from, is $$g(x)=x^2+\\sin(x)$$-- is $y=x^2$ an asymptote in this case?\n\nThe reason that I ask is that I don't really see the point in defining oblique asymptotes and not curved ones; surely, if we want to know the behaviour of $y$ as $x \\to \\infty$, we should include all types of functions as asymptotes.\n\nIf asymptotes cannot be curves, then why arbitrarily restrict asymptotes to lines?\n\n-\nOf course you can. There's, e.g., the notion of polynomial branch, parabolic branch, etc. but then they might not be called asymptotes. \u2013\u00a0 gniourf_gniourf Jun 28 '14 at 21:32\n\u2013\u00a0 David H Jun 28 '14 at 21:40\nIn a way, the notion of asymptotic analysis generalizes the concept of line asymptotes. This site has an asymptotics tag. \u2013\u00a0 Antonio Vargas Jun 28 '14 at 21:45\nI would think that a Taylor series expansion is a pretty good example of a general-purpose usage of a curved line for an asymptote. \u2013\u00a0 fluffy Jun 29 '14 at 1:23\n\nThe concept of asymptotes is quite common for curved graphs, although somehow the terminology is not much used outside of the context of lines. The way in which the concept is used is that if one is given a function $f(x)$, it is interesting to study other functions $g(x)$ that are \"asymptotic to $f(x)$\" in various ways. One meaning of this phrase would be that $$(1) \\quad \\lim_{x \\to +\\infty} |f(x)-g(x)|=0$$ which is exactly what \"asymptotic\" means in the ordinary sense when the graph of $f(x)$ is a line. Another somewhat different notion is that $$(2) \\quad \\lim_{x \\to +\\infty} \\frac{f(x)}{g(x)} = 1$$ which only really makes sense when $f(x)$ and $g(x)$ are nonzero near $+\\infty$. There are many other variations on this concept. This discussion falls under the name of \"growth types of functions\", which are important in computer science and other places; these notes look like a good basic discussion, for example.\n\nAnd regarding your question of whether $g(x) = x^2 + \\sin(x)$ is asymptotic to $y=x^2$, it is asymptotic in sense (2) but not in sense (1).\n\n-\n\n\"Asymptote,\" in my view, essentially refers to some kind of limiting behavior of a function. So for instance, we talk about asymptotic series expansions. So from an analytic geometry perspective, we might think of an \"asymptote\" as a function or relation that describes how another function approaches it arbitrarily closely. For example, $$f(x) = \\frac{x-x^2+x^4}{x^2-1}$$ might be thought of as having a parabolic asymptote and two vertical asymptotes, since for \"large\" $x$, $f(x) \\sim x^2$. But I hesitate to say $f(x) = x^2 + \\sin x$ has such an asymptote, because the magnitude of $\\sin x$ does not diminish as $x$ gets large.\n\n-\nThough it is still true that $x^2 + \\sin x \\sim x^2$ as $x \\to \\infty$ in the usual sense of $\\sim$. \u2013\u00a0 Antonio Vargas Jun 28 '14 at 21:48\nYes, of course. \u2013\u00a0 heropup Jun 28 '14 at 21:50\n@heropup Is it the fact that $\\sin(x)$ oscillates for large $x$ that means that $y=x^2$ wouldn't be considered an asymptote in my case, or is it the fact that $\\sin(x)$ doesn't decay to zero? \u2013\u00a0 alexqwx Jun 28 '14 at 21:54\nWhat about $x^2 + \\frac{\\sin x}{x}$? Would you say that had an asymptote in the same way that the first function you gave had horizontal/vertical asymptotes? \u2013\u00a0 Zubin Mukerjee Jun 28 '14 at 22:18\n@ZubinMukerjee In that $\\frac{\\sin{x}}{x} \\rightarrow 0$, yes. \u2013\u00a0 ClickRick Jun 29 '14 at 3:29\n\nterminology is up to you. However, it is useful, when graphing rational functions, to realize that they are essentially polynomial (or the reciprocal of a polynomial) for large absolute values of the argument. Graph $$y = \\frac{x^5 - 7}{x^3 - 12 x},$$ for large $|x|,$ $y$ is pretty much $x^2.$\n\n-\nI was sad because I had no readily available computer graphing software. Then I met a man who had no fish. \u2013\u00a0 Will Jagy Jun 28 '14 at 21:52", "date": "2015-08-03 15:29:56", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/850745/can-asymptotes-be-curved", "openwebmath_score": 0.934851348400116, "openwebmath_perplexity": 387.9696469510777, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676430955553, "lm_q2_score": 0.8976952927915968, "lm_q1q2_score": 0.879443031707118}}
{"url": "https://math.stackexchange.com/questions/2587436/is-there-an-algebraic-proof-for-sum-m-kn-k-binommk-binomn-mk", "text": "# Is there an algebraic proof for $\\sum_{m=k}^{n-k} \\binom{m}{k}\\binom{n-m}{k} = \\binom{n+1}{2k+1}, n\\ge2k\\ge0$\n\n$\\sum_{m=k}^{n-k} \\binom{m}{k}\\binom{n-m}{k} = \\binom{n+1}{2k+1}, n\\ge2k\\ge0$\n\nAn combinatorial proof of the identity above states as follow:\n\n(1)Number of ways of picking (2k+1) numbers from 1 to (n+1) should be $\\binom{n+1}{2k+1}$\n\n(2)We pick (2k+1) numbers from 1 to (n+1) with median value (m+1). Then, k numbers must be selected from 1~m, and the other k numbers must be chosen from (m+2)~(n+1). Thus there are $\\binom{m}{k}\\binom{n-m}{k}$ ways for picking (2k+1) numbers with median value (m+1). Since $n-k\\ge m\\ge k$, there are total $\\sum_{m=k}^{n-k} \\binom{m}{k}\\binom{n-m}{k}$ ways.\n\nSince (1)=(2), the statement is true. But is it possible to sketch an algebraic proof that doesn't require building combinatorial models?\n\n\u2022 Consider coefficient of $x^{n}$ in $[x^k(1-x)^{k+1}][x^k(1-x)^{k+1}]$. \u2013\u00a0Lord Shark the Unknown Jan 1 '18 at 11:22\n\nHere is an algebraic proof based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \\begin{align*} [z^k](1+z)^n=\\binom{n}{k} \\end{align*}\n\nWe obtain \\begin{align*} \\color{blue}{\\sum_{m=k}^{n-k}}&\\color{blue}{\\binom{m}{k}\\binom{n-m}{k}}\\\\ &=\\sum_{m=0}^{n-2k}\\binom{m+k}{m}\\binom{n-m-k}{k}\\tag{1}\\\\ &=\\sum_{m=0}^{n-2k}\\binom{-k-1}{m}(-1)^m\\binom{n-m-k}{k}\\tag{2}\\\\ &=\\sum_{m=0}^\\infty[z^m](1-z)^{-k-1}[u^k](1+u)^{n-m-k}\\tag{3}\\\\ &=[u^k](1+u)^{n-k}\\sum_{m=0}^\\infty\\left(\\frac{1}{1+u}\\right)^{m}[z^m](1-z)^{-k-1}\\tag{4}\\\\ &=[u^k](1+u)^{n-k}\\left(1-\\frac{1}{1+u}\\right)^{-k-1}\\tag{5}\\\\ &=[u^k](1+u)^{n-k}u^{-k-1}(1+u)^{k+1}\\tag{6}\\\\ &=[u^{2k+1}](1+u)^{n+1}\\tag{7}\\\\ &\\color{blue}{=\\binom{n+1}{2k+1}}\\tag{8} \\end{align*} and the claim follows.\n\nComment:\n\n\u2022 In (1) we shift the index to start with $m=0$ and use the binomial identity $\\binom{p}{q}=\\binom{p}{p-q}$.\n\n\u2022 In (2) we use the binomial identity $\\binom{-p}{q}=\\binom{p+q-1}{q}(-1)^q$.\n\n\u2022 In (3) we apply the coefficient of operator twice and we set the upper bound of the series to $\\infty$ without changing anything since we are adding zeros only.\n\n\u2022 In (4) we use the linearity of the coefficient of operator and we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.\n\n\u2022 In (5) we apply the substitution rule of the coefficient of operator with $z=\\frac{1}{1+u}$\n\\begin{align*} A(u)=\\sum_{m=0}^\\infty a_m u^m=\\sum_{m=0}^\\infty u^m [z^m]A(z) \\end{align*}\n\n\u2022 In (6) we do some simplifications.\n\n\u2022 In (7) we do some more simplifications and apply the same rule as we did in (4).\n\n\u2022 In (8) we select the coefficient of $u^{2k+1}$.\n\nMore generally:\n\nTheorem 1. For every nonnegative integers $$x$$, $$y$$ and $$n$$, we have $$$$\\dbinom{n+1}{x+y+1}=\\sum_{m=0}^{n}\\dbinom{m}{x}\\dbinom{n-m}{y}.$$$$\n\nThis equality rewrites as $$$$\\dbinom{n+1}{x+y+1} = \\sum_{m=x}^{n-y} \\dbinom{m}{x}\\dbinom{n-m}{y} ,$$$$ because all addends $$\\dbinom{m}{x}\\dbinom{n-m}{y}$$ are zero except for those where $$x \\leq m \\leq n-y$$.\n\nYour identity is the particular case of the latter equality for $$x = k$$ and $$y = k$$.\n\nYour nice bijective proof of the original identity can easily be generalized to a proof of Theorem 1; just count all ways of picking an $$\\left(x+y+1\\right)$$-element subset of $$\\left\\{1,2,\\ldots,n+1\\right\\}$$ according to the value of the $$\\left(x+1\\right)$$-th-lowest element of the subset. Indeed, for any given $$m \\in \\left\\{0,1,\\ldots,n\\right\\}$$, picking an $$\\left(x+y+1\\right)$$-element subset $$S$$ of $$\\left\\{1,2,\\ldots,n+1\\right\\}$$ whose $$\\left(x+1\\right)$$-th-lowest element is $$m+1$$ is tantamount to picking an $$x$$-element subset of $$\\left\\{1,2,\\ldots,m\\right\\}$$ (which will contain the $$x$$ elements of $$S$$ lower than $$m+1$$) and picking a $$y$$-element subset of $$\\left\\{m+2,m+3,\\ldots,n+1\\right\\}$$ (which will contain the $$y$$ elements of $$S$$ higher than $$m+1$$). This gives $$\\dbinom{m}{x}\\dbinom{n-m}{y}$$ for the number of choices.\n\nAs for other proofs:", "date": "2019-10-19 20:12:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2587436/is-there-an-algebraic-proof-for-sum-m-kn-k-binommk-binomn-mk", "openwebmath_score": 0.9816500544548035, "openwebmath_perplexity": 165.3472579405334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109102866639, "lm_q2_score": 0.888758789809389, "lm_q1q2_score": 0.8794365191295623}}
{"url": "https://math.stackexchange.com/questions/53307/combinatorial-proof-of-binomial-coefficient-summation", "text": "# Combinatorial proof of binomial coefficient summation\n\nWhile doing some Computer Science problems, I found one which I thought could be solvable using combinatorics instead of programming:\n\nGiven two positive integers $n$ and $k$, in how many ways do $k$ numbers, all of which are between $0$ and $n$ (inclusive), add up to $n$?\n\nIn the problem, order does matter. For example, 0+20 is not equivalent to 20+0. This led me to the following solution:\n\nAssume you do want to create the sum without any of the terms being zero. Then, the problem can be looked at as if you have $n$ rocks, and you want know the number of ways you can place $k-1$ sticks between them, such that no stick may be placed between the same two rocks. This is obviously equal to $\\binom{n-1}{k-1}$.\n\nIf you then want to count the number of ways you can create the sum with exactly one of the terms being zero. Then the solution is (number of ways to get to $n$ using $k-1$ non-zero terms)*(the number of ways we can choose 1 element among k). Then, when wanting to count the number of ways with exactly $x$ zeroes, the answer is (number of ways to get to n using $k-x$ non-zero terms)*(the number of ways we can choose $x$ elements among $k$).\n\nThe solution to the original problem will basically the sum of the above expression, with x ranging between $0$ and $k-1$ (since there needs to be at least 1 non-zero term).\n\nThis leads to $\\displaystyle\\sum_{i=0}^{k-1}\\left[ \\binom{n-1}{k-i-1}\\binom{k}{i}\\right]$\n\nI later went on to search on Google if there existed any simpler solution, and found that $\\binom{n+k-1}{n}$ worked as well, but without explanation and proof.\n\nBoth solutions give the same answer for any positive $n$ and $k$, but I really have trouble understanding why. I've tried to simplify my summation, and proving it inductively, but I have not been successful.\n\nCould somebody help me prove the equality?\n\n$\\binom{n+k-1}{n} = \\displaystyle\\sum_{i=0}^{k-1}\\left[ \\binom{n-1}{k-i-1}\\binom{k}{i}\\right]$\n\n\u2022 Hint: Find instead the number of ways of dividing $n+k$ into $k$ parts, all $\\ge 1$. \u2013\u00a0Andr\u00e9 Nicolas Jul 23 '11 at 14:32\n\u2022 That was a really nice solution too, and a bit more mathematical I suppose. :-) \u2013\u00a0Johan Sannemo Jul 23 '11 at 14:53\n\u2022 Very related... \u2013\u00a0J. M. is a poor mathematician Jul 23 '11 at 14:59\n\u2022 Your final equation is an example of Vandermonde's identity en.wikipedia.org/wiki/Vandermonde%27s_identity \u2013\u00a0user940 Jul 23 '11 at 15:00\n\u2022 @Byron - cool, even an algebraic proof there! \u2013\u00a0Johan Sannemo Jul 23 '11 at 15:07\n\nThere is no need to break into cases concerning the existence of piles of \"0 rocks\". A pile of zero rocks can be represented by having two \"sticks\" between the same two rocks (between those two sticks there are zero rocks!). Indeed, I can represent any number of consecutive piles of zero rocks by having the appropriate number of sticks between the same two rocks.\n\nSo, with that in mind, I am really looking to count the number of ways that I can place $k-1$ sticks among $n$ rocks. (By the way, why does this representation still retain the ordering? Try finding sticks and rocks representations of $0 + 20$ and $20 + 0$, and other combinations.) Now, this is equivalent to choosing from $n + k - 1$ objects, exactly $n$ of them to be our \"rocks\" (leaving the other $k-1$ to be our \"sticks\"). And there are $\\binom{n+k-1}{n}$ ways to do that.\n\nHope this helps!\n\n\u2022 By the way, your piecewise identity can be written as a single formula! What happens when $n = 1$? In the summation, there is a binomial coefficient with $n-1 = 0$ on the top, which evaluates to $0$ unless the bottom is also $0$ (in which case, $\\binom{0}{0} = 1$), which occurs only when $k - i - 1 = 0$, or $i = k-1$. But then the second factor would be $\\binom{k}{i} = \\binom{k}{k-1} = k$. The sum is exactly equal to $k$ when $n = 1$. \u2013\u00a0Shaun Ault Jul 23 '11 at 14:48\n\u2022 Ah, of course! Makes sense now that you say it. Thanks! \u2013\u00a0Johan Sannemo Jul 23 '11 at 14:49\n\u2022 Yeah, of course 0 choose 0 = 1. Silly me. \u2013\u00a0Johan Sannemo Jul 23 '11 at 14:52\n\nTaking your rocks and sticks example, your basic question is how many ways are there to place $n$ rocks and $k-1$ sticks in order as each pattern (of the rocks) is one solution to your problem. This is ${n+k-1 \\choose n}$.\n\nIf $n=1$ then all your discussion of $i$ zero terms will come out at $0$ except when $i=k-1$ and in that case you will have $\\left[ \\binom{0}{0}\\binom{k}{k-1}\\right] = k$, so you do not need to distinguish $n=1$ in your expression.", "date": "2019-04-18 13:18:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/53307/combinatorial-proof-of-binomial-coefficient-summation", "openwebmath_score": 0.8139531016349792, "openwebmath_perplexity": 185.66041524833045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587218253718, "lm_q2_score": 0.8902942297605357, "lm_q1q2_score": 0.8793958904367707}}
{"url": "http://math.stackexchange.com/questions/281094/permuting-elements-of-a-set-around-a-circle?answertab=oldest", "text": "# Permuting elements of a set around a circle\n\nGiven $15$ objects placed around a circle, is it possible to permute their order such that the distance between any two elements is different in the second permutation from that in the original state?\n\n-\nAre the 15 positions on the circle necessarily uniformly spaced? \u2013\u00a0Henning Makholm Jan 18 '13 at 1:13\n@joriki: I haven't tried very hard to understand your code, but what is wrong with 5 objects and (0 2 4 1 3)? (Or in general any prime number $p\\ge 5$ of objects and doubling positions modulo $p$, or even more generally $q$ objects and multiplying positions by $k$ modulo $q$ where $\\{k-1,k,k+1\\}$ are each coprime with $q$ (which requires that $q$ is not a multiple of $3$))? \u2013\u00a0Henning Makholm Jan 18 '13 at 1:23\nMy own computer search confirms that there is no solution for $n=15$, and quickly finds a number of solutions for $n=11$, including 1 3 5 7 9 11 2 4 6 8 10, and many solutions for $n=13$, such as 1 3 10 6 11 5 12 4 13 7 9 2 8. I will make the code available if anyone cares. \u2013\u00a0MJD Jan 18 '13 at 1:35\nYou're right, there was a bug in my code. I've fixed it, and am now getting the same results as @MJD. I'll delete my first comment since it might mislead if people don't read further. \u2013\u00a0joriki Jan 18 '13 at 1:42\n@AlexanderGruber gist.github.com/4561589 ; please feel free to email me if you want to discuss anything. \u2013\u00a0MJD Jan 18 '13 at 2:07\n\nOEIS notes that this is closely related to the problem of whether there is a solution to the toroidal $N$ queens problem, which is the problem of arranging $N$ chess queens on an $N\\times N$ chessboard so that none attacks another, where the opposite pairs of edges of the board are identified. It refers to The Cyclic Complete Mappings Counting Problems, Hsiang, Shieh, and Chen, 2002. Theorem 2 on page 6 of that paper states: $$TQ(n) = 0 \\text{ if and only if } 2 | n \\text{ or } 3 | n.$$\n\n$TQ(n)$ is the number of solutions to the toroidal $N$-queens problem, and is easily seen to be equal to $n\\cdot A(n)$, where $A(n)$ is the number of solutions to the circle arranging problem we are solving. In particular, $TQ(15) = 0$, so $A(15) = 0$.\n\nUnfortunately, the paper does not provide a proof. However, as Joriki notes in the comments, the theorem is due to P\u00f3lya, probably in \u00dcber die 'doppelt-periodischen' L\u00f6sungen des $n$-Damen-Problems, Mathematische Unterhaltungen und Spiele., (1918), pp. 364\u2013374; according to Wikipedia, this appears in George P\u00f3lya: Collected papers Vol. IV, G-C. Rota, ed., MIT Press, Cambridge, London, 1984, pp. 237\u2013247. Web search for \"toroidal $n$-queens\" may produce something more immediately useful.\n\nMathworld cites Vardi, I. \"The $n$-Queens Problems.\" Ch. 6 in Computational Recreations in Mathematica. Redwood City, CA: Addison-Wesley, pp. 107\u2013125, 1991.\n\n-\nVery nice, problem solved. \u2013\u00a0joriki Jan 18 '13 at 2:00\nUnder other circumstances I might have made this a CW post, since I didn't solve the problem myself, but in this case the research effort was significant, and I feel that I have added some value. \u2013\u00a0MJD Jan 18 '13 at 2:01\n@joriki Don't you have an OEIS account? If so, you might let them know that their hyperlink to the Hsiang paper is no good, and that they should replace it with the CiteSeer link I provided. \u2013\u00a0MJD Jan 18 '13 at 2:04\n\"and I feel that I have added some value\" -- You did. It would have been very difficult to find this connection without writing the code to determine the numbers up to $A(13)=348$. By the way the related OEIS entry A007705 gives a reference from $1921$ and says that Polya proved the result that $TQ(n)\\ne0$ if and only if $n$ is coprime to $6$. \u2013\u00a0joriki Jan 18 '13 at 2:05\n\nThis sci.math post says that the literature contains several independent proofs of the result that the toroidal $n$-queens problem has a solution iff $\\gcd(n,6)=1$ and gives six references. I have access to one of them, Torleiv Kl\u00f8ve, The modular $n$-queens problem, in Discrete Mathematics 19 (1977), 289-91. The proof there is short enough to be worth typing out here.\n\nWe want a function $f:\\Bbb Z/n\\Bbb Z\\to\\Bbb Z/n\\Bbb Z$ such that $f$, $f+\\mathrm{id}_{\\Bbb Z/n\\Bbb Z}$, and $f-\\mathrm{id}_{\\Bbb Z/n\\Bbb Z}$ are all injections. The theorem is that such an $f$ exists iff $\\gcd(n,6)=1$.\n\nProof. Suppose that $f$ is such a function, and let\n\n$$S=\\sum_{k\\in\\Bbb Z/n\\Bbb Z}k^2\\quad\\text{and}\\quad T=\\sum_{k\\in\\Bbb Z/n\\Bbb Z}kf(k)\\;.$$\n\nThen $\\displaystyle\\sum_{k\\in\\Bbb Z/n\\Bbb Z}f(k)^2=S$, since $f$ is injective. Since $f+\\mathrm{id}_{\\Bbb Z/n\\Bbb Z}$ and $f-\\mathrm{id}_{\\Bbb Z/n\\Bbb Z}$ are also injective,\n\n$$S=\\sum_{k\\in\\Bbb Z/n\\Bbb Z}\\big(f(k)\\pm k\\big)^2=\\sum_{k\\in\\Bbb Z/n\\Bbb Z}f(k)^2\\pm 2\\sum_{k\\in\\Bbb Z/n\\Bbb Z}kf(k)+\\sum_{k\\in\\Bbb Z/n\\Bbb Z}k^2=2S\\pm 2T\\;,$$\n\n$2S=4S$, and $2S=0$ in $\\Bbb Z/n\\Bbb Z$. In $\\Bbb Z$, therefore, we have\n\n$$2\\sum_{k=0}^{n-1}k^2=\\frac{n(n-1)(2n-1)}3\\equiv 0\\pmod n$$\n\nand hence $\\gcd(n,3)=1$.\n\nHe now appeals to N.J. Fine\u2019s solution of problem E 1699 (Amer. Math. Monthly 72 (1965), 552-3). Specialized to the present setting, the argument (which may have given Kl\u00f8ve the idea for his proof) is that\n\n$$\\sum_{k\\in\\Bbb Z/n\\Bbb Z}\\big(f(k)+k\\big)=2\\sum_{k\\in\\Bbb Z/n\\Bbb Z}k$$\n\nin $\\Bbb Z/n\\Bbb Z$, but if $n$ is even, then $$\\sum_{k=0}^{n-1}k=\\frac{n(n-1)}2\\equiv-\\frac{n}2\\equiv\\frac{n}2\\pmod n\\;,$$\n\nso in $\\Bbb Z/n\\Bbb Z$ we have $$\\sum_{k\\in\\Bbb Z/n\\Bbb Z}\\big(f(k)+k\\big)=2\\cdot\\frac{n}2=0\\ne\\sum_{k\\in\\Bbb Z/n\\Bbb Z}k\\;,$$ and $f+\\mathrm{id}_{\\Bbb Z/n\\Bbb Z}$ cannot be injective. Thus, $n$ must be odd, and $\\gcd(n,6)=1$.\n\nConversely, if $\\gcd(n,6)=1$, then $f(k)=2k$ is a solution. $\\dashv$\n\n-", "date": "2016-04-29 18:01:19", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/281094/permuting-elements-of-a-set-around-a-circle?answertab=oldest", "openwebmath_score": 0.8125602006912231, "openwebmath_perplexity": 252.88294915796763, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517475646369, "lm_q2_score": 0.8991213718636754, "lm_q1q2_score": 0.8793872290239814}}
{"url": "https://www.physicsforums.com/threads/cardinality-question.183742/", "text": "# Cardinality Question\n\n1. Sep 9, 2007\n\n### MrJB\n\nI'm fairly sure that the intervals (0,1) and [0,1] of real numbers have the same cardinality, but I can't think of a bijection between them. Any thoughts?\n\n2. Sep 9, 2007\n\n### Hurkyl\n\nStaff Emeritus\n\nCan you find a bijection between the set of all nonnegative integers and the set of all positive integers?\n\nOnce you've answered that, can you think of a way to apply this very example to your problem? Or at least this technique?\n\n3. Sep 9, 2007\n\n### MrJB\n\nWell, for the warm-up, I'd define f(n)=n+1 as my map from the set of nonnegative integers to the set of positive integers. Thus, 0,1,2... would get mapped to 1,2,3... etc.\nI don't see the connection to my problem though.\n\n4. Sep 9, 2007\n\n### morphism\n\nI'm wondering why you want to find a bijection. Very rarely does one produce an explicit bijection to show that two sets have the same cardinality. One of the main tools is the Shroeder-Bernstein theorem: it says that if you have an injection from A into B and an injection from B into A, then there exists a bijection between A and B (note: this is an existence result; it doesn't tell us how to construct the bijection). You can apply it to your problem.\n\nIn regards to Hurkyl's hint, try using this to find a bijection between [0,1] and (0,1], for example.\n\n5. Sep 9, 2007\n\n### MrJB\n\nI was just curious what such a bijection might look like.\n\nIn regards to the Schroeder-Bernstein Theorem, the injection from (0,1) into [0,1] is obvious, and the injection from [0,1] into (0,1) I can define as f(x)=x/2+1/4. So there is a bijection between them.\n\nI'm still lost on Hurkyl's hint. With the integers, my map just moved all the elements to the next integer, but with the reals there is no 'next' real.\n\n6. Sep 9, 2007\n\n### morphism\n\nHow about, say, the sets {1/n : n a positive integer} and {1/n : n a nonnegative integer} in [0,1]?\n\n7. Sep 9, 2007\n\n### Hurkyl\n\nStaff Emeritus\nWhat this tells you is that, if you're given a countable set, you know how to make an individual point \"appear\" or \"disappear\".\n\nNow, if you could find a countable subset of [0, 1], then you know how to make a point of that subset disappear...\n\n8. Sep 9, 2007\n\n### MrJB\n\nI think I can work with that...\n\nSo my map would take the set {1/n : n a positive integer} to {1/(n+1) : n a positive integer}. That is, 1,1/2,1/3... would map to 1/2,1/3,1/4... and the rest of the reals would map to themselves, thus defining a bijection from [0,1] to [0,1). Then I would repeat a similar technique to the other side.\n\nThanks.\n\n9. Sep 9, 2007\n\n### Hurkyl\n\nStaff Emeritus\nYep. That is the basic technique for \"removing\" individual points from an infinite set. The core idea works in a wide variety of situations.\n\n10. Sep 9, 2007\n\n### MrJB\n\nAwesome.\nYou could use the same idea to 'remove' infinite sets of points too? If I wanted to map the closed unit disc to the open unit disc, I would map the sets of points with distances from the origin 1,1/2,1/3 to the sets of points with distances 1/2,1/3,1/4 from the origin?\n\n11. Sep 10, 2007\n\n### Hurkyl\n\nStaff Emeritus\nThat certainly seems to work!\n\n12. Sep 10, 2007\n\n### fopc\n\nThoughts on an example construction of f:[0,1] -> (0,1) that's bijective. (There are many such f's.)\n\n1. The sets differ only at two points, 0 and 1.\n2. Must find images for 0 and 1 somewhere in (0,1) while still keeping f bijective.\n3. Let A={0,1,1/2,1/3,...,1/n,...}. (Something like this has already been suggested.)\n4. Send 0 to 1/2, and 1 to 1/3.\n5. This can be effected by, f(0)=1/2, f(1/n)= 1/(n+2) for n >= 1.\n\n6. Then complete the definition of f by, f(x)=? for all x in ?\n\nThe final step is to actually verify that f is bijective.\n\nTypically, it's easier to find two injections than one bijection.\nThis example brings the point out a little bit.\nSo yes, in many cases Schroeder-Bernstein has considerable practical value.\nBut, it's greatest value is theoretical, not practical.\nI'd say there's something to be learned by actually constructing an f (like the one above), and demonstrating that's it's bijective.\n\nEDIT: I'll complete the definition of f.\nf(x)=x for all x in [0,1]-A. (Obviously there was a reason for A.)\n\nLast edited: Sep 10, 2007\n13. Sep 10, 2007\n\n### HallsofIvy\n\nStaff Emeritus\nTry this: irrational numbers between 0 and 1 map into themselves.\n\nThe rational numbers between 0 and 1 are countable so the can be written in a list:\n$r_1, r_2, r_3, \\cdot\\cdot\\cdot$. Now map $r_1\\rightarrow 0$, $r_2/rightarrow 1$, $r_3\\right arrow r_1$, $r_4\\rightarrow r_2$,$\\cdot\\cdot\\cdot$, $r_n\\rightarrow r_{n-2}$.\n\n14. Sep 20, 2007\n\n### grossgermany\n\nStill why would this be bijective? How do you know that step A 1/n->1/(n+1)\nand step B f(x)=x for all x in [0,1] won't end up in the same number.\nI know intuitively it makes sense, but it doesn't seem very mathematically rigorous.\n\n15. Sep 21, 2007\n\n### fopc\n\nI suspect you didn't read the EDIT.\nIt reads, f(x) = x for all x in [0,1]-A.\n\nNot mathematically rigorous?\nWell, I defined an f, and made a claim that f is a bijection; nothing more. I then suggested the need to demonstrate this by showing f is injective and surjective. There's hidden value here. It forces you to think about definitions.\n\nYou question that f is bijective (in particular you seem to doubt that it's injective). Well?\n\nFinally, I apologize to Ernst. It should be Schr\u00f6der.", "date": "2017-03-30 08:48:01", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/cardinality-question.183742/", "openwebmath_score": 0.7526184320449829, "openwebmath_perplexity": 892.0330823092423, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668745151689, "lm_q2_score": 0.8962513807543223, "lm_q1q2_score": 0.8793721660346231}}
{"url": "https://math.stackexchange.com/questions/1107656/associativity-and-de-morgans-for-more-than-2-literals", "text": "# Associativity and De Morgan's for more than 2 literals\n\nDo logical operators have meaning when used with more than 2 literals \"associatively\", e.g.: $(A \\land B \\land C)$?\n\nI.e., are statements such as $(A \\land B \\land C)$ meaningful, as opposed to $((A \\land B) \\land C)$, which is meaningful?\n\nIf yes, is it possible to prove De Morgan's for any finite n by induction like so:\n\n1. For $n = 2$, base De Morgan's applies: $\\neg(A_{1} \\lor A_{2}) \\Leftrightarrow \\neg A_{1} \\land \\neg A_{2}$.\n2. Assume that for $n = k$, $\\neg (A_{1}\\lor A_{2}\\lor \\ldots \\lor A_{k}) \\Leftrightarrow (\\neg A_{1}\\land \\neg A_{2} \\land \\ldots \\land \\neg A_{k})$.\n3. Then for $n = k + 1$: $\\neg (A_{1}\\lor A_{2}\\lor \\ldots \\lor A_{k} \\lor A_{k+1}) \\Leftrightarrow \\neg ((A_{1}\\lor A_{2}\\lor \\ldots \\lor A_{k})\\lor A_{k+1})\\Leftrightarrow \\neg (A_{1}\\lor A_{2}\\lor \\ldots \\lor A_{k}) \\land \\neg A_{k+1} \\Leftrightarrow \\neg A_{1} \\land \\neg A_{2} \\land \\ldots \\land \\neg A_{k} \\land \\neg A_{k+1}$\n4. Therefore, $\\neg (A_{1}\\lor A_{2}\\lor \\ldots \\lor A_{n}) \\Leftrightarrow (\\neg A_{1}\\land \\neg A_{2} \\land \\ldots \\land \\neg A_{n})$ is true for any finite n.\n\n?\n\nThank you.\n\n\u2022 Yes; $(A\u2227B\u2227C)$ is an abbreviation for $(A\u2227(B\u2227C))$ or, equivalently : $((A\u2227B)\u2227C))$, as your proof shows. Jan 17 '15 at 10:15\n\u2022 No, take a look at the formation rules. Also, the rule of uniform substitution would fail if (A\u2227B\u2227C) were meaningful. Jan 17 '15 at 19:25\n\u2022 @DougSpoonwood Could you please elaborate on that? Thank you. Jan 19 '15 at 3:12\n\u2022 @DougSpoonwood I feel like it even though it is formally incorrect to use that kind of notation, it is still meaningful intuitively. For example, for both $\\land$ and $\\lor$ the truth table of an expression consisting of $n$ literals and $n-1$ operators of one kind is the same whatever the placement of parentheses is (out of $2^n$ possible variants). Therefore, the notation in this case is logically insignificant. Jan 19 '15 at 3:35\n\u2022 No, it's logically significant, because if (A\u2227B\u2227C) is an expression, than the rule of uniform substitution fails. The rule of uniform substitution is important for having an axiomatic system of propositional logic. If say A\u2227B is meaningful (which you implied), and we have the rule of uniform substitution, then substituting B with CvD we obtain A\u2227CvD. Suppose A=0, C=0, D=1. Then [A\u2227(CvD)]=0, while [(A\u2227C)vD]=1. So, I started with an expression assumed as meaningful and deduced a contradiction, which renders either the expression or the rule of uniform substitution as not meaningful. Jan 19 '15 at 3:57\n\nCertainly $\\land$ and $\\lor$ can be used meaningfully in this way. This is because we can prove (e.g. by truth tables, a straightforward exercise) that they are associative, i.e.:\n\n$$A \\land (B \\land C) \\iff (A \\land B) \\land C \\qquad A \\lor (B \\lor C) \\iff (A \\lor B) \\lor C$$\n\nso that we can argue that it doesn't really matter whether we use parentheses or not. From there on, you have given a proper proof that De Morgan's laws generalise to these expressions.\n\nDo note, however, that this does not apply to every logical operator out there. This question deals with associativity of $\\to$; it is shown not to be associative:\n\n$$A \\to (B \\to C) \\not\\iff (A \\to B) \\to C$$\n\n\u2022 You can only use \u2227 and \u2228 in this way, if we have a system where the only binary expression is either \u2227 or \u2228 (but you can't have both around). Associativity only ensures that such expressions are meaningful when we have a pure semigroup. If we have two associative operations around, such as the two-valued logical operations \"\u2227\" and \"\u2228\", then an expression like A\u2227B\u2228C can be ambiguous. It's not associativity which makes such work, but rather that for any binary operations B$_1$, B$_2$, for all x, y, z, ((xB$_1$y)B$_2$z)=(xB$_1$(yB$_2$z)). Associativity is a special case of that. Jan 19 '15 at 4:03\n\u2022 This answer combined with Doug Spoonwood's note that the only logical operator used should be ($\\land$ XOR $\\lor$) answers my question completely. Jan 19 '15 at 5:19\n\u2022 @Doug Nowhere did I imply that expressions combining $\\lor$ and $\\land$ without parentheses are meaningful. Jan 19 '15 at 6:07\n\nNo. For example, (A$\\land$B$\\lor$C) is not meaningful, even though both $\\land$ and $\\lor$ associate.\n\nIf we have that for any binary operations B1, B2, for all x, y, z, ((xB1y)B2z)=(xB1(yB2z)), then we can drop parentheses as you've suggested. Thus, if we just have a semigroup, then we can drop all parentheses. But, if we have more than one binary operation, we can't drop parentheses.\n\nAlso, if (A\u2227B\u2227C) is an expression, than the rule of uniform substitution fails. The rule of uniform substitution is important for having an axiomatic system of propositional logic. If say A\u2227B is meaningful (which you implied), and we have the rule of uniform substitution, then substituting B with CvD we obtain A\u2227CvD. Suppose A=0, C=0, D=1. Then [A\u2227(CvD)]=0, while [(A\u2227C)vD]=1. So, I started with an expression assumed as meaningful and deduced a contradiction, which renders either the expression or the rule of uniform substitution as not meaningful.\n\n\u2022 Surely the rule of uniform substitution applies only to wffs. You can replace $B$ with the complete wff $(C\\lor D)$; you cannot replace it with the non-wff $C\\lor D$ any more than you can replace it with the non-wffs $C\\lor$ or $(C\\lor D$.\n\u2013\u00a0MJD\nJan 19 '15 at 4:42\n\u2022 Thank you for your answer. Is it right then, that if ($\\lor$ is used) XOR ($\\land$ is used) multiple times within the initial expression AND we do not move the parentheses enclosing a substituted expression, then such usage would always be meaningful? Jan 19 '15 at 5:09\n\u2022 @MJD By substitution I meant exactly that: plugging in a formula as if it was a complete logical \"block\". Jan 19 '15 at 5:12\n\u2022 @AndreyPortnoy I'm not sure. Jan 19 '15 at 18:08", "date": "2021-11-30 21:55:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1107656/associativity-and-de-morgans-for-more-than-2-literals", "openwebmath_score": 0.9151002764701843, "openwebmath_perplexity": 585.7796810059606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104972521579, "lm_q2_score": 0.9099070121457542, "lm_q1q2_score": 0.8793436880610036}}
{"url": "https://math.stackexchange.com/questions/1849134/what-equation-can-produce-these-curves", "text": "# What equation can produce these curves?\n\nDoes an equation exist that can produce the curves shown in the attached image, by varying a single variable?\n\n\u2022 Do you mean in principle, or are you asking for specific examples? \u2013\u00a0PyRulez Jul 5 '16 at 3:17\n\nConsider the equilateral hyperbola $xy=1$ and map the points $(1,0)$ and $(0,1)$ to two symmetrical points on the hyperbola $(t,t^{-1})$ and $(t^{-1},t)$ by translation/scaling (as if you were zooming in).\n\n$$(x(t-t^{-1})+t^{-1})(y(t-t^{-1})+t^{-1})=1.$$\n\nThe straight line corresponds to two infinitely close points, $t=t^{-1}=1$, which is a degenerate case of the equation.\n\nYou can think of a pencil of parabolas, $y=\\frac1{\\sqrt2}+\\frac\\lambda{\\sqrt2}(2x^2-1)$, which you rotate by $45\u00b0$ right, giving\n\n$$x+y=1+\\lambda\\left((x-y)^2-1\\right).$$\n\nYou can solve the quadratic equation for $y$.\n\n\u2022 If I take $(x,y)$ to be $(1,0)$ or $(0,1)$, it seems to me that your equation is not satisfied. \u2013\u00a0Jeppe Stig Nielsen Jul 4 '16 at 21:51\n\u2022 @JeppeStigNielsen: right, I forgot a $\\sqrt2$ factor. Now fixed (hopefully). \u2013\u00a0Yves Daoust Jul 4 '16 at 22:11\n\u2022 This has become promising, since the it goes through the two points as required, and is has the symmetry that swapping $x$ and $y$ leaves the equation unchanged. However for some values of $\\lambda$ it may look different than expected. Like when $\\lambda$ is close to $1$, the vertex of the parabola is close to the origin, and the parabola (after rotation) crosses the coordinate axes in the \"opposite\" direction, I think. \u2013\u00a0Jeppe Stig Nielsen Jul 4 '16 at 22:34\n\u2022 @JeppeStigNielsen: the exponent $2$ can be lessened, but at some point we are lacking curve specifications. \u2013\u00a0Yves Daoust Jul 4 '16 at 22:36\n\u2022 At what value of $\\lambda$ will the two axes be tangents at the two \"fixed\" points? \u2013\u00a0Jeppe Stig Nielsen Jul 4 '16 at 22:40\n\nTry a superellipse: $|x|^a + |y|^a =1$ with $0 <a \\le 1$.\n\n\u2022 Yes, but that curve will not intersect (pass through) the coordinate axes and continue into the 2nd/4th quadrant as shown in the image. But maybe that was not a requirement. \u2013\u00a0Jeppe Stig Nielsen Jul 4 '16 at 22:02\n\u2022 @JeppeStigNielsen You could try analytic continuation, or simply remove the $|\\cdot|$ \u2013\u00a0Tobias Kienzler Jul 5 '16 at 6:01\n\u2022 @TobiasKienzler The superellipse with $a<1$ has a cusp at the point in question. It is not meeting the axis at a non-zero angle like in the drawing of the question. \u2013\u00a0Jeppe Stig Nielsen Jul 5 '16 at 9:11\n\u2022 @JeppeStigNielsen Right you are, I spoke to soon... \u2013\u00a0Tobias Kienzler Jul 5 '16 at 9:50\n\nTry also a quadratic B\u00e9zier curve $$\\gamma(t) = (1 - t)^{2} P_0 + 2(1 - t)t P_1 + t^{2} P_2$$ where $P_0=(1,0), P_1=(a,a), P_2=(0,1)$, with $a \\in [0,0.5]$.\n\nYou can first of all use a tool such as Engauge Digitizer to obtain tables of data points for each curve. Then either try a Log-log plot to determine a polynomial approximation or use a fitting tool of your choice to obtain fits of the same kind for each curve. Finally, find a parameter describing each curve such that all the fit-parameters previously obtained can also be fitted by simple functions, ideally linear or a low-order polynomial.\n\nAssuming what you drew to be a parabola, and $p$ a constant to push each parabola along the line of symmetry,\n\n$$y = p (1-x^2)$$\n\nNow rotate them clockwise by $\\pi/4 \\, , (x,y) \\rightarrow ( x+y, x-y)/\\sqrt2.$\n\nEDIT1:\n\nAs you left it free as regards shape within constraints, we could use a cosh curve, sec curve etc.\n\nIn fact, we can generalize this using any open even function E(x) , passing through $(\\pm 1/\\sqrt{2},1/\\sqrt{2} ).$\n\nLetting\n\n$$y(x,p) = (1 + p *(1/2 - x^2) E(x)) /\\sqrt{2} \\,$$\n\nand building a differentiable function, rotating the curve by $-\\pi/4$, we can make the family pass through singular points $(0,1),(1,0)$as shown above and illustrated below in a particular case.\n\nIn graph below $E(x) = \\cos^2(x)$is chosen with $p= (\\pm 2,0).$\n\nFor the earlier special case you got parabolas using $E(x)=1$ perturbed by arbitrary $p$ values.\n\nUse Lagrange Interpolation formula: For a variable point $(t,t)$ on the line $x-y=0$ with $0<t\\leq \\sqrt2/2$, fit a degree 2 polynomial curve that passes through the three points $(0,1), (t,t)$ and $(1,0)$.\n\nFor $t=\\sqrt2/2$ the three points will be collinear and it will automatically give a straight line. This must produce the same result as Yves Daoust's answer above, but using different interpretation.", "date": "2020-06-04 22:38:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1849134/what-equation-can-produce-these-curves", "openwebmath_score": 0.7763229608535767, "openwebmath_perplexity": 602.6528295535471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363535858372, "lm_q2_score": 0.8933094053442511, "lm_q1q2_score": 0.8793169226804927}}
{"url": "http://mymathforum.com/advanced-statistics/40374-probability-distribution-standard-deviation.html", "text": "My Math Forum Probability Distribution and Standard Deviation\n\n December 21st, 2013, 05:14 PM #1 Senior Member \u00a0 Joined: Oct 2013 From: New York, USA Posts: 608 Thanks: 82 Probability Distribution and Standard Deviation I read something somebody else posted and it made me wonder something. If you guess on every multiple choice question without eliminating any choices first, your expected score will be 100/x where x is the number of choices for each question. Assume that all questions have exactly x choices and that there is no penalty for guessing. If you guessed like this on many tests: 1. Are your scores going to be normally distributed? 2. What is the standard deviation of this distribution in terms of x? 3. Can Questions 1 and 2 be answered in general or is it necessary to know the number of choices and/or number of questions? It's possible to calculate the probability of getting at least a certain value given a mean and standard deviation for normally distributed data, but even if this data is normally distributed the number of questions matters. For example, the probability of scoring at least a 20 (the mean) is only 0.2 if there are five choices and there is only one question. With two questions the probability rises to 0.36, with three questions it's 0.488, with four questions it's 0.5904, and with five questions it's 0.67232. With six questions the probability drops to 0.39328 (I think) because it is now necessary to get more than 1 question right to get at least a 20. 4. The number of choices is x and the number of questions is y. Is it a general rule that increasing the number of questions from y to y+1 will increase the probability of getting at least a score of 100/x if y is not a multiple of x and decrease the probability of getting at least a score of 100/x if y is a multiple x? 5. The number of choices is x and the number of questions is y. Let's say instead of wanting to get as high a score as possible you want to get a score of exactly 100/x. This is only possible y is a multiple of x. How many questions would you want there to be to maximize your chance of getting a score of exactly 100/x? I would guess that you would want x and y to be equal because if you flip a coin and want exactly half heads and half tails you would want to flip the coin 2 times.\n December 21st, 2013, 08:11 PM #2 Senior Member \u00a0 Joined: Aug 2012 Posts: 229 Thanks: 3 Re: Probability Distribution and Standard Deviation Hey Evan3. For 1) You should think about whether the Central Limit Theorem (CLT) applies here or whether there is some other justification for the normal distribution assumption or approximation. If you think the CLT holds, this should help you with your other questions.\nDecember 22nd, 2013, 12:18 AM \u00a0 #3\nSenior Member\n\nJoined: Jun 2013\nFrom: London, England\n\nPosts: 1,316\nThanks: 116\n\nRe: Probability Distribution and Standard Deviation\n\nQuote:\n Originally Posted by EvanJ I read something somebody else posted and it made me wonder something. If you guess on every multiple choice question without eliminating any choices first, your expected score will be 100/x where x is the number of choices for each question. Assume that all questions have exactly x choices and that there is no penalty for guessing. If you guessed like this on many tests: 1. Are your scores going to be normally distributed?\nThis is a Binomial Distribution. The probability of success in each trial is 1/x.\n\nIt approximates to a Normal Distribution for a large number of questions.\n\nQuote:\n 2. What is the standard deviation of this distribution in terms of x?\n$\\sigma= \\sqrt{\\frac{n}{x} (1-\\frac{1}{x})}$\n\nWhere n is the number of questions.\n\nQuote:\n 3. Can Questions 1 and 2 be answered in general or is it necessary to know the number of choices and/or number of questions? It's possible to calculate the probability of getting at least a certain value given a mean and standard deviation for normally distributed data, but even if this data is normally distributed the number of questions matters. For example, the probability of scoring at least a 20 (the mean) is only 0.2 if there are five choices and there is only one question. With two questions the probability rises to 0.36, with three questions it's 0.488, with four questions it's 0.5904, and with five questions it's 0.67232. With six questions the probability drops to 0.39328 (I think) because it is now necessary to get more than 1 question right to get at least a 20.\nThe mean is n/x and the and std deviation is given above.\n\nQuote:\n 4. The number of choices is x and the number of questions is y. Is it a general rule that increasing the number of questions from y to y+1 will increase the probability of getting at least a score of 100/x if y is not a multiple of x and decrease the probability of getting at least a score of 100/x if y is a multiple x?\nNo. As y increases, regardless of whether it's a mutiple of x, the probability of getting at least 100/x will increase. Fairly quickly it will become almost a certainty.\n\nQuote:\n 5. The number of choices is x and the number of questions is y. Let's say instead of wanting to get as high a score as possible you want to get a score of exactly 100/x. This is only possible y is a multiple of x. How many questions would you want there to be to maximize your chance of getting a score of exactly 100/x? I would guess that you would want x and y to be equal because if you flip a coin and want exactly half heads and half tails you would want to flip the coin 2 times.\nAgain, there is no reason that y must be a multiple of x. You need 100 to be a mutiple of x.\n\nThe most likely outcome is y/x (as it is a bell-shaped curve). Also, as y increases, P(y/x) reduces, because there are more possibilities, so the distribution gets more spread out. So, your best chance of getting 100/x is to have 100 questions.\n\n December 23rd, 2013, 05:17 AM #4 Senior Member \u00a0 Joined: Oct 2013 From: New York, USA Posts: 608 Thanks: 82 Re: Probability Distribution and Standard Deviation Thank you. I'm surprised by 4 and 5. If there are 5 choices, the probability of getting at least 1/5th of the questions correct increases as the number of questions goes up from 1 to 5 because in all cases 1 correct answer is necessary to do that. As the number of choices goes up from 5 to 6, the number of correct answers necessary goes up from 1 to 2 (obviously you can't get 6/5 = 1.2 correct answers) and the minimum score necessary to get at least 1/5th of the questions right goes from 20 to 33 1/3. So how can the probability of getting at least 1/5th of the questions correct increase when the increase in number of questions cause the lowest possible score that is > or = to 1/5th of the questions to increase?\n December 23rd, 2013, 08:35 AM #5 Senior Member \u00a0 Joined: Oct 2013 From: New York, USA Posts: 608 Thanks: 82 Re: Probability Distribution and Standard Deviation I decided to compare the probabilities to what you get from the normal distribution table given the standard deviation Pero posted for 100 questions with 5 choices each. Here are my results: p(16 to 24 successes from the normal distribution) = 0.6826 p(16 to 24 successes from the binomial formula) = 0.7381 That isn't extremely close. I'll do it for 1,000 questions and a smaller range of probabilities: p(190 to 210 successes from the normal distribution) = 0.5704 p(190 to 210 successes from the binomial formula) = 0.5935 I used a smaller range of probabilities because if I did p(160 to 240 success) it would have been almost 1 in both methods.\n\n Thread Tools Display Modes Linear Mode\n\n Similar Threads Thread Thread Starter Forum Replies Last Post ragas88 Advanced Statistics 1 July 27th, 2013 08:55 AM Jared944 Advanced Statistics 1 November 5th, 2011 02:58 PM Axel Algebra 2 April 28th, 2011 03:25 AM jd254 Algebra 0 May 7th, 2009 12:46 AM agcliff09 Advanced Statistics 3 November 11th, 2007 10:37 AM\n\n Contact - Home - Forums - Cryptocurrency Forum - Top", "date": "2018-09-25 00:48:33", "meta": {"domain": "mymathforum.com", "url": "http://mymathforum.com/advanced-statistics/40374-probability-distribution-standard-deviation.html", "openwebmath_score": 0.7366324067115784, "openwebmath_perplexity": 402.3643244017307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363545048391, "lm_q2_score": 0.8933093996634686, "lm_q1q2_score": 0.8793169179096451}}
{"url": "https://math.stackexchange.com/questions/3489480/which-threshold-maximizes-the-expected-size-of-the-final-sample", "text": "# Which threshold maximizes the expected size of the final sample?\n\nFor $$c>0$$, sample repeatedly and independently from $$(0, 1)$$ until the sum of the samples exceeds $$c$$. Let $$\\mu_c$$ be the expected size of the final sample.\n\nFor which $$c$$ is $$\\mu_c$$ maximised?\n\nIt is clear that as $$c$$ tends to $$0$$, $$\\mu_c$$ tends to $$\\frac{1}{2}$$ and this is its minimum value.\n\n\u2022 The final sample is the last number you draw, right? Interesting question! \u2013\u00a0Vincent Dec 27 '19 at 22:02\n\u2022 @Vincent Yes, it's the sample that pushes the sum of the samples over $c$. \u2013\u00a0felipa Dec 27 '19 at 22:03\n\u2022 Very neat question! where did it come from? Is this homework / quiz / etc? What kind of HINT vs actual solution is allowed? Also, I assume you're sampling uniformly in $(0,1)$? \u2013\u00a0antkam Dec 27 '19 at 22:16\n\u2022 @antkam It's just for mathematical interest. I was playing around with simulating the setup. A full solution would be very welcome. Yes it's uniform from $(0, 1)$. \u2013\u00a0felipa Dec 27 '19 at 22:18\n\u2022 @antkam: Consider a steady-state process with i.i.d. steps uniformly drawn from $(0,1)$, and at some point choose an arbitrary threshold. The probability for a given segment to contain the threshold is proportional to the length of the segment, so the expected length of the segment containing the threshold is $\\int_0^12l\\cdot l\\mathrm dl=\\frac23$. \u2013\u00a0joriki Dec 27 '19 at 23:32\n\nI will write $$f(c) = \\mu_c$$ to make it more clearly a function of $$c$$. For now I will only investigate $$c \\in [0,1]$$. It turns out (if my math is correct) there is a unique max within $$c \\in [0,1]$$, at:\n\n$$c = \\ln 2, ~~~~~~~f(c) = \\mu_c = \\ln 2$$\n\nBut frankly I would only trust my own math below with about 80% confidence... :)\n\nFirst of all, $$c$$ can be considered as the amount \"still to go\". I.e., starting with $$c$$, if the next sample is $$x < c$$, then you effectively have a new problem with a new threshold of $$c-x$$, and the expected value becomes $$f(c-x)$$. We can build a recurrence from this observation.\n\nLet $$X \\sim Unif(0,1)$$. We have:\n\n\u2022 Law of total expectation: $$f(c) = P(X > c) E[X \\mid X > c] + P( X < c) E[f(c-X) \\mid X < c]$$\n\n\u2022 $$P(X>c) = 1-c$$\n\n\u2022 $$E[X \\mid X > c] = {1+ c \\over 2}$$ because conditioned on $$X > c$$ then $$X\\sim Unif(c,1)$$\n\n\u2022 $$P(X < c) = c$$\n\n\u2022 Conditioned on $$X < c$$, we have $$c-X \\sim Unif(0, c)$$.\n\nSo the most trouble term becomes:\n\n$$E[f(c - X) \\mid X < c] = \\int_0^c \\frac1c f(u) ~du$$\n\nAnd the overall equation is:\n\n$$f(c) = {1 - c^2 \\over 2} + c \\int_0^c \\frac1c f(u) ~du$$\n\nDifferentiate w.r.t. $$c$$:\n\n$$f'(c) = -c + f(c)$$\n\nwhich is an ODE with this solution (credit: wolfram alpha!): for some integration constant $$K$$,\n\n$$f(c) = K e^c + c + 1$$\n\nSubstitute in $$f(0) = 1/2$$ (as observed by OP) and solving, we have $$K = -1/2$$ and so:\n\n$$f(c) = -\\frac12 e^c + c + 1$$\n\nNow we just need to find the max:\n\n$$f'(c) = -\\frac12 e^c + 1 = 0 \\iff e^c = 2 \\iff c = \\ln 2$$\n\nat which point we have $$f(c) = c = \\ln 2$$ which is just slightly $$> 2/3$$.\n\nFurther thoughts:\n\n(1) I am pretty rusty (and that's a charitable description!) with \"continuous\" math, so if someone can critique / verify the above, that'd be much appreciated.\n\n(2) This answer does not cover the case of $$c > 1$$ so far. For $$c> 1$$, there is no chance the next sample is enough, and the recurrence becomes:\n\n$$f(c) = \\int_{c-1}^c f(u) ~du$$\n\nwhere $$f(u) = -\\frac12 e^u + u + 1$$ whenever $$u < 1$$. I don't know how to do this integration. However, intuitively, since $$f(c)$$ is based on averaging of values of $$f(u)$$ (or average of averages, etc), the max of $$f(c)$$ cannot $$>$$ the max of $$f(u)$$, and in fact, since the max $$f(u)$$ is unique within $$u \\in (0,1)$$, the max of $$f(c)$$ cannot even $$=$$ the max of $$f(u)$$ within $$u \\in (0,1)$$. This is not a rigorous proof, but rather an intuitive argument why the max I found within $$(0,1)$$ is also the global max.\n\n\u2022 My numeric simulations are consistent with your calculations. I would say that for \"large\" $c$, the distribution of the last uniform random variable (i.e. conditioned on the sum of the previous ones being less than $c$) is triangular with density $$f(x) = 2x \\mathbb 1(0 < x < 1).$$ I think using some kind of large-sample approximation would get you the conditional density in this case. \u2013\u00a0heropup Dec 27 '19 at 23:26\n\u2022 @heropup - clearly you had some kind of triangle in mind, to get to $2/3$. for large $c$ some kind of \"ergodicity\" must apply and all points become equally likely, in a sense, but i cant quite translate that into a triangle. anyway, very good intuitive guess! but the \"averaging\" argument in (2) above (if you buy the argument) means the global max must occur within $(0,1)$ and the limit behavior must fluctuate and cannot be monotonic. this is all very hand-wavy, but that's my gut feel \u2013\u00a0antkam Dec 27 '19 at 23:30\n\u2022 Your solution seems correct to me (for $c\\le1$). For $1\\le c\\le2$, we have $$f(c)=\\int_{c-1}^1\\left(-\\frac12\\mathrm e^x+x+1\\right)\\mathrm dx+\\int_1^cf(x)\\mathrm dx\\;;$$ differentiating yields $f'(c)=\\frac12\\mathrm e^{c-1}-c+f(c)$, and the solution for the initial condition $f(1)=-\\frac12\\mathrm e+2$ is $f(c)=\\frac12\\mathrm e^{x-1}(x-1)-\\frac12\\mathrm e^x+x+1$. Here's a plot. This section also has a local maximum, at $\\hat c\\approx1.4770$ with $f(\\hat c)\\approx0.67139$, already quite a bit closer to $\\frac23\\approx0.66667$ than $\\ln2\\approx0.69315$. \u2013\u00a0joriki Dec 28 '19 at 0:07\n\u2022 @joriki - thanks a LOT for both comments. the $2/3$ limit via steady state is what i had a hazy vision of, but i couldn't find the right argument. and doing the new integral, and in particular showing that it is \"piecewise\" chopped into intervals delimited by the integers, makes a ton of sense. one can probably recurse this and find the formula for every integer interval, but it might be boring work now. \u2013\u00a0antkam Dec 28 '19 at 1:23\n\u2022 @felipa - now that joriki has shown the way, you do the boring work! ;) actually it might not be so boring... if we may extrapolate from two data points (ha!), the next $f(c), c \\in [2,3]$ might have a $e^{x-2}$ thingy in it, and then the next has $e^{x-3}$ thingy in it, etc. it might all turn out rather neat! so i definitely won't deprive you of the pleasure of discovery. :) (plus: i'm very very bad at integration... notice that in my original answer, i only had to differentiate!) \u2013\u00a0antkam Dec 28 '19 at 15:55\n\nantkam's approach is correct and elegant. We can also derive the same result with a bit less calculus by approximating the continuous uniform distribution by a discrete uniform distribution.\n\nSo we have an $$n$$-sided die that we roll repeatedly, summing the results, and we want to know the threshold $$k$$ that maximizes the final result (where reaching $$k$$ itself is enough to stop).\n\nWe can prove by induction that the expected final result for threshold $$k$$ is $$n+k-\\frac n2\\left(1+\\frac1n\\right)^k$$. For $$k=1$$ this is $$n+1-\\frac n2\\left(1+\\frac1n\\right)=\\frac{n+1}2$$, which is correct. Assuming the result for $$k-1$$, we obtain the expected final result for $$k$$ as\n\n$$\\begin{eqnarray*} &&\\frac1n\\left((n-k+1)\\frac{n+k}2+\\sum_{j=1}^{k-1}\\left(n+j-\\frac n2\\left(1+\\frac1n\\right)^j\\right)\\right) \\\\ &=& \\frac1n\\left((n-k+1)\\frac{n+k}2+(k-1)n+\\frac{k(k-1)}2-\\frac n2\\left(\\frac{1-\\left(1+\\frac1n\\right)^k}{1-\\left(1+\\frac1n\\right)}-1\\right)\\right) \\\\ &=& n+k-\\frac n2\\left(1+\\frac 1n\\right)^k\\;. \\end{eqnarray*}$$\n\nSetting the derivative with respect to $$k$$ to zero yields\n\n$$1-\\frac n2\\ln\\left(1+\\frac 1n\\right)\\left(1+\\frac 1n\\right)^k=0\\;,$$\n\nso the optimal threshold in the discrete case is an integer near\n\n$$-\\frac{\\ln\\frac n2+\\ln\\ln\\left(1+\\frac 1n\\right)}{\\ln\\left(1+\\frac 1n\\right)}\\;.$$\n\nFor instance, for $$n=6$$, this is\n\n$$\\frac{\\ln3+\\ln\\ln\\frac76}{\\ln\\frac76}\\approx5.003\\;,$$\n\nso the optimal threshold is $$k=5$$ with an expected final result of\n\n$$6+5-\\frac62\\left(1+\\frac16\\right)^5=\\frac{11705}{2592}\\approx4.516\\;.$$\n\nFor $$n\\to\\infty$$, we have $$\\ln\\left(1+\\frac1n\\right)\\sim\\frac1n$$ and thus\n\n$$-\\frac{\\ln\\frac n2+\\ln\\ln\\left(1+\\frac 1n\\right)}{\\ln\\frac 1n}\\sim-\\frac{\\ln\\frac n2+\\ln\\frac 1n}{\\frac 1n}=n\\cdot\\ln2\\;,$$\n\nin agreement with antkam's result.\n\n\u2022 I really like the discrete version of the problem you have introduced. \u2013\u00a0felipa Dec 29 '19 at 12:17", "date": "2020-04-02 11:00:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3489480/which-threshold-maximizes-the-expected-size-of-the-final-sample", "openwebmath_score": 0.8511144518852234, "openwebmath_perplexity": 346.1636194686255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474913588876, "lm_q2_score": 0.8887587920192298, "lm_q1q2_score": 0.8792912813073803}}
{"url": "http://mathhelpforum.com/number-theory/181037-integers.html", "text": "# Math Help - Integers\n\n1. ## Integers\n\nDear all\n\nI have these questions i am having difficulties solving on my on own,your help is been sorted as to how to approach these questions and soultion.I pick these questions from schuam series.\n\nThank you\n\nQuestion 1\n\nOf all integer pairs(x,y) that satisfy the equation 42x+55y=1,0nly one such pair has 100<x< 200.what is the value of x in this integer?\n\nQuestion 2\nLet X1 and X2 be two be two smallest positive integer for which the following statement is true 85x-12 is a multiple of 19 then X1+X2= (NB: The 1 and 2 in front of X1 and X2 are all subscripts X1 and X2)\n\n2. Originally Posted by barhin\nQuestion 1\n\nOf all integer pairs(x,y) that satisfy the equation 42x+55y=1, only one such pair has 100<x< 200.what is the value of x in this integer?\nThis is an exercise in using Euclid's algorithm. (If you haven't come across this, Google it to see why the following calculations are being done.)\n\n\\begin{aligned}55&=42+13\\\\ 42&= 3(13) + 3\\\\ 13&= 4(3) + 1.\\end{aligned}\n\nNow work through those equations in the reverse order to get\n\n\\begin{aligned}1&= 13-4(3)\\\\ &= 13 - 4(42-3(13)) = -4(42) + 13(13)\\\\ &= -4(42) + 13(55-42) = -17(42) + 13(55).\\end{aligned}\n\nSo one solution to the equation 42x+55y=1 is x=\u201317, y= 13. You get the other solutions by adding a multiple of 55 to x, and subtracting the same multiple of 42 from y. Thus $(-17+55k)(42) + (13-42k)(55) = 1$ (you can see that the multiples of k cancel, to leave you with the result from the previous displayed equation).\n\nNow all you have to do is to choose k so that \u201317+55k lies between 100 and 200.\n\nOriginally Posted by barhin\nQuestion 2\nLet X1 and X2 be two be two smallest positive integer for which the following statement is true 85x-12 is a multiple of 194 then X1+X2= (NB: The 1 and 2 in front of X1 and X2 are all subscripts X1 and X2)\nFor this one, start by finding x and y such that 85x+194y=1 (using Euclid's algorithm, as for Q1). Multiply the result by 12 on both sides, to give you a solution to 85X \u2013 12 = (multiple of 194). Then add suitable multiples of 194 to find the two smallest positive values for X.\n\n3. Hello, barhin!\n\n$\\text{(1) Of all integer pairs }(x,y)\\text{ that satisfy the equation: }\\: 42x+55y\\:=\\:1$\n. . $\\text{only one such pair has }100\n\n. . $\\text{What is this value of }x\\,?$\n\nThe problem can be solved without Euclid's algorithm.\n. . But, of course, this solution takes much longer.\n\nWe have: . $42x + 55y \\:=\\:1$\n\n$\\text{Then: }\\:x \\:=\\:\\frac{1-55y}{42} \\:=\\:\\frac{-42y + 1 - 13y}{42} \\quad\\Rightarrow\\quad x \\:=\\:-y + \\frac{1-13y}{42}$ .[1]\n\nSince $x$ is an integer, $1-13y$ must be a multiple of 42.\n\n. . $1-13y \\:=\\:42a \\quad\\Rightarrow\\quad y \\:=\\:\\frac{1-42a}{13} \\quad\\Rightarrow\\quad y \\:=\\:-3a + \\frac{1-3a}{13}$ .[2]\n\nSince $y$ in an integer, $1-3a$ must be a multiple of 13.\n\n. . $1 - 3a \\:=\\:13b \\quad\\Rightarrow\\quad a \\:=\\:\\frac{1-13b}{3} \\quad\\Rightarrow\\quad a \\:=\\:-4b + \\frac{1-b}{3}$ .[3]\n\nSince $a$ is an integer, $1-b$ must be a multiple of 3.\n\n. . $1-b \\:=\\:3k \\quad\\Rightarrow\\quad b \\:=\\:1-3k$\n\nSubstitute into [3]:\n. . $a \\:=\\:\\text{-}4(1-3b) + \\frac{1-(1-3k)}{3} \\quad\\Rightarrow\\quad a \\:=\\:13k-4$\n\nSubstitute into [2]:\n. . $y \\:=\\:\\text{-}3(13k-4) + \\frac{1-3(13k-4)}{13} \\quad\\Rightarrow\\quad y \\:=\\:\\text{-}42k + 13$\n\nSubstitute into [1]:\n. . $x \\:=\\:\\text{-}(\\text{-}42k+13) + \\frac{1-13(\\text{-}42k+13)}{42} \\quad\\Rightarrow\\quad x \\:=\\:55k - 17$ .[4]\n\nSince $100 < x < 200$\n\n. . we have:. . $100 \\;<\\;55k - 17 \\;<\\; 200$\n\n. . . Add 17: . . . $117 \\;<\\;55k \\;<\\; 217$\n\nDivide by 55: . . $\\frac{117}{55} \\;<\\;k \\;<\\;\\frac{217}{55}$\n\n. . . . . . . . . . . $2.127 \\;<\\; k \\;<\\;3.945$\n\n. . . . . . . . . . . . . . $2 \\;<\\;k \\;<\\;4$\n\nHence: . $k\\;=\\;3$\n\nSubstitute into [4]: . $x \\:=\\:55(3) - 17 \\quad\\Rightarrow\\quad \\boxed{x \\:=\\:148}$\n\nThe integer pair is: . $(x,y) \\:=\\:(148,\\:\\text{-}113)$\n\n4. Originally Posted by Opalg\nThis is an exercise in using Euclid's algorithm. (If you haven't come across this, Google it to see why the following calculations are being done.)\n\n\\begin{aligned}55&=42+13\\\\ 42&= 3(13) + 3\\\\ 13&= 4(3) + 1.\\end{aligned}\n\nNow work through those equations in the reverse order to get\n\n\\begin{aligned}1&= 13-4(3)\\\\ &= 13 - 4(42-3(13)) = -4(42) + 13(13)\\\\ &= -4(42) + 13(55-42) = -17(42) + 13(55).\\end{aligned}\n\nSo one solution to the equation 42x+55y=1 is x=\u201317, y= 13. You get the other solutions by adding a multiple of 55 to x, and subtracting the same multiple of 42 from y. Thus $(-17+55k)(42) + (13-42k)(55) = 1$ (you can see that the multiples of k cancel, to leave you with the result from the previous displayed equation).\n\nNow all you have to do is to choose k so that \u201317+55k lies between 100 and 200.\n\nFor this one, start by finding x and y such that 85x+194y=1 (using Euclid's algorithm, as for Q1). Multiply the result by 12 on both sides, to give you a solution to 85X \u2013 12 = (multiple of 19). Then add suitable multiples of 19 to find the two smallest positive values for X.\n\nDear Soroban,\nThank very much for useful taught on question 1 and 2 , I am how ever getting different results for the second question,will be grateful if you could throw more light or solution to the question 2 for me.It's my prayer that I will become a great mathematician like you\n\n5. Hello, barhin!\n\n$\\text{(2) Let }x_1\\text{ and }x_2\\text{ be the two smallest positive integers}$\n$\\text{for which the following statement is true: }\\:85x-12\\text{ is a multiple of 19.}$\n$\\text{Find }x_1+x_2.$\n\n$85x - 12 \\text{ is a multiple of 19}$\n\n$\\text{Hence: }\\:85x-12 \\:=\\:19a\\;\\text{ for some integer }a.$\n\n. . $\\text{Then: }\\: a \\:=\\:\\frac{85x-12}{19} \\quad\\Rightarrow\\quad a \\:=\\:4x + \\frac{9x-12}{19}$\n\n$\\text{Since }a\\text{ is an integer, then }9x-12\\text{ must be multiple of 19.}$\n\n$\\text{Hence: }\\:9x-12 \\:=\\:19b \\quad\\Rightarrow\\quad x \\:=\\:\\frac{19b + 12}{9} \\quad\\Rightarrow\\quad x \\:=\\:2b + 1 +\\frac{b+3}{9}$ .[1]\n\n$\\text{Since }b\\text{ is an integer, then }b+3\\text{ must be a multiple of 9.}$\n\n$\\text{Hence: }\\:b +3 \\:=\\:9k \\quad\\Rightarrow\\quad b \\:=\\:9k-3$\n\nSubstitute into [1]:\n\n. . $x \\;=\\;2(9k-3) + 1 + \\frac{(9k-3) + 3}{9} \\quad\\Rightarrow\\quad x \\:=\\:19k-5$\n\n$\\text{The first two positive values of }x\\text{ are: }\\:\\begin{Bmatrix} k = 0 & \\Rightarrow & x_1 \\:=\\:14 \\\\ k = 1 & \\Rightarrow & x_2 \\:=\\:33 \\end{Bmatrix}$\n\n$\\text{Therefore: }\\:x_1 + x_2 \\:=\\:14 + 33 \\:=\\:47$", "date": "2014-08-20 07:26:14", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/181037-integers.html", "openwebmath_score": 0.8764424324035645, "openwebmath_perplexity": 4064.9297756473325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9859363725435203, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.8792689442983057}}
{"url": "https://math.stackexchange.com/questions/3654973/finding-smallest-possible-integer-to-satisfy-a-perfect-cube", "text": "# Finding smallest possible integer to satisfy a perfect cube\n\nThe sum of four consecutive, positive, odd integers is a perfect cube. What is the smallest possible integer that could be the least of the four?\n\nI tried approaching this as follows:\n\nLet $$n$$ be an odd integer, then we can represent the four odd integers as $$n, n+2, n+4, n+6$$.\n\nWe want the sum of these integers equal to $$k^3$$, where $$k\\in \\mathbb{Z^+}$$. So we can write this as\n\n$$n+n+2+n+4+n+6=k^3$$ $$\\Leftrightarrow$$ $$4n+12=k^3$$, but from here I don't see how I could continue. I could probably divide the expression by $$4$$ and get $$n+3= \\frac{k^3}{4}$$ and then deduce that $$k$$ has to be something of the form $$4n$$? I'm not sure. Any tips would be helpful.\n\n\u2022 Trial and error works very quickly. Failing that, you might notice that all those sums are even so the first possible cube is $8$ (which does not work) and the second is $64$ (which does). \u2013\u00a0lulu May 2 '20 at 11:37\n\nYou have that $$n=2m+1$$ because $$n$$ is odd and that $$2m+1+2m+3+2m+5+2m+7 = k^3$$ so $$8m+16 = k^3$$ and so $$8m = k^3-16$$ so $$m = \\frac{k^3}{8}-2$$ let $$k = 2 u$$ and we have that $$m = u^3 -2$$ so we have that $$n=2(u^3-2)+1 = 2 u^3-3$$ for $$u \\in \\mathbb{N}$$ and so the first few $$n$$'s are $$n= -1,13,51,125,\\cdots$$\n\nIt is an obscure (but provably true) fact that the odd numbers can be separated into consecutive strings whose length increases by $$1$$, and the sum of each string will be the cube of the number of members in the string: $$1=1^3,\\ 3+5=2^3,\\ 7+9+11=3^3,\\ 13+15+17+19=4^3$$ etc. A string of consecutive odd numbers with $$n$$ members will sum to a cube when the first member of that string is $$2k+1$$ where $$k=\\sum_{i=1}^{n-1}i$$\n\nThe example with four members begins with $$2(\\sum_{i=1}^3i)+1=2\\cdot 6+1=13$$\n\nLet the four numbers be:\n\n$$((2p^3-3),(2p^3 -1),(2p^3+1),(2p^3+3))$$\n\nAdding them we get $$sum=(2p)^3$$\n\nwhich is a cube.\n\nSince we need the smallest odd number we have,\n\n$$(2p^3-3)>0$$\n\nwhich implies $$p=2$$ is the smallest.\n\nHence the number's are:\n\n$$(13,15,17,19)$$ , & they sum up to $$(4)^3$$", "date": "2021-03-06 14:17:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3654973/finding-smallest-possible-integer-to-satisfy-a-perfect-cube", "openwebmath_score": 0.86368328332901, "openwebmath_perplexity": 128.83705336156672, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363752295428, "lm_q2_score": 0.8918110368115781, "lm_q1q2_score": 0.8792689410237077}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=6kf2vvqsai66nlmf1nmtmcof97&topic=2324.msg7054", "text": "Author Topic: 2.3 question 3 \u00a0(Read 916 times)\n\nYan Zhou\n\n\u2022 Full Member\n\u2022 Posts: 15\n\u2022 Karma: 1\n2.3 question 3\n\u00ab on: February 10, 2020, 07:22:49 PM \u00bb\n$$\\int_{|z+1| = 2} \\frac{zdz}{4-z^{2}} = \\int_{|z+1| = 2} \\frac{\\frac{zdz}{2-z}}{2+z}$$\nSince -2 lies in $|z+1| = 2$, Cauchy theorem gives that\n$$\\frac{1}{2\\pi i} \\int_{|z+1| = 2} \\frac{\\frac{zdz}{2-z}}{z-(-2)} = \\frac{-2}{2-(-2)} = -\\frac{1}{2}$$\nthen $$\\int_{|z+1| = 2} \\frac{zdz}{4-z^{2}} = -\\pi i$$\nHowever, I checked the answer of textbook and it says the answer is $2\\pi i$, I am confused about where i did wrong.\n\nYan Zhou\n\n\u2022 Full Member\n\u2022 Posts: 15\n\u2022 Karma: 1\nRe: 2.3 question 3\n\u00ab Reply #1 on: February 10, 2020, 07:46:22 PM \u00bb\nI see. The question on the home assignment is a little bit different from the textbook.\n\nYan Zhou\n\n\u2022 Full Member\n\u2022 Posts: 15\n\u2022 Karma: 1\nRe: 2.3 question 3\n\u00ab Reply #2 on: February 10, 2020, 08:11:40 PM \u00bb\nI found some differences between textbook and home assignment, some of them are typos, and some do not affect the questions but the answers. Should we always follow the questions on the textbook?\n\nVictor Ivrii\n\nYour answer is correct ($-\\pi i$).", "date": "2021-12-05 15:41:16", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=6kf2vvqsai66nlmf1nmtmcof97&topic=2324.msg7054", "openwebmath_score": 0.856249213218689, "openwebmath_perplexity": 5376.7418907937235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.957277806109987, "lm_q2_score": 0.9184802356574272, "lm_q1q2_score": 0.8792407449455258}}
{"url": "https://www.physicsforums.com/threads/number-of-rolls-with-m-many-n-sided-dice.218931/", "text": "# Number of rolls with m many n-sided dice\n\n1. Feb 29, 2008\n\n### mathboy\n\n[SOLVED] Number of rolls with m many n-sided dice\n\nHow many different rolls are there with m many n-sided dice, if order is not important?\n\nIf order was important, the answer would simply be n^m. But if order is not important?\n\nIf two 6-sided dice are rolled, there are C(6,2)+C(6,1) = 21 different rolls.\nIf three 6-sided dice are rolled, there are C(6,3)+C(6,1)C(5,1)+C(6,1) = 56 different rolls.\n\nIs there a general method for rolling m many n-sided dice? Or do you have to continue adding up all the cases? Perhaps a general method involving a matrix with m indices (e.g. a tensor) and simply excluding all permutations of the indices?\n\nLast edited: Feb 29, 2008\n2. Feb 29, 2008\n\n### e(ho0n3\n\nWouldn't it be n^m/m!? For example, say you have a black die and a red die, both 6-sided. The number of possible rolls is 6^2. If you ignore the color of the dice, then you have to divide by 2.\n\n3. Feb 29, 2008\n\n### mathboy\n\nNo. The number of rolls with 2 six-sided dice (if order is unimportant) is 21, not 18. Rolls with doubles are not being counted twice if order is considered important.\n\n4. Feb 29, 2008\n\n### e(ho0n3\n\nOh that's right. Sorry about that. I don't know of a general method. I'm thinking that you would first have to count the number of rolls where all dice show the same side, then for m-1 dice showing the same side, then for m-2 dice, etc.\n\n5. Feb 29, 2008\n\n### mathboy\n\nThat's exactly what I did in my opening post. But it gets awkward after a while (e.g. for 7 dice, you can have 2 doubles and 1 triple). I'm looking for a better method that gives a single formula for arbitrarily values of m and n,\n\ne.g. m^n minus something.\n\nLast edited: Feb 29, 2008\n6. Feb 29, 2008\n\n### stewartcs\n\nYes.\n\n$$\\frac{(n + m - 1)!}{m!(n - 1)!}$$\n\nCS\n\n7. Feb 29, 2008\n\n### mathboy\n\nWow! That gives the right answer for all the cases I worked out.\n\nSo the answer is C(m+n-1,m), i.e. the number of ways to choose m numbers from m+n-1 numbers. Half of this makes some sense to me because we are rolling m dice. But what's the reason for the m+n-1?\n\n8. Feb 29, 2008\n\n### stewartcs\n\n9. Feb 29, 2008\n\n### mathboy\n\nThank you so much. Sometimes extra knowledge is the key to solving a problem.\n\nLast edited: Feb 29, 2008\n10. Jan 22, 2011\n\n### lcscipiop\n\nRe: [SOLVED] Number of rolls with m many n-sided dice", "date": "2017-03-28 10:16:29", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/number-of-rolls-with-m-many-n-sided-dice.218931/", "openwebmath_score": 0.5238118767738342, "openwebmath_perplexity": 1372.3795343939735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145701859179, "lm_q2_score": 0.8872046011730965, "lm_q1q2_score": 0.8792326864985249}}
{"url": "https://mathhelpboards.com/threads/finding-the-area-bounded-by-the-curve.7787/", "text": "Finding the Area bounded by the curve\n\nshamieh\n\nActive member\nFind the area bounded by the curve $$\\displaystyle x = 16 - y^4$$ and the y axis.\n\nI need someone to check my work.\n\nso I know this is a upside down parabola so I find the two x coordinates which are\n\n$$\\displaystyle 16 - y^4 = 0$$\n$$\\displaystyle y^4 = 16$$\n$$\\displaystyle y^2 = +- \\sqrt{4}$$\n$$\\displaystyle y = +- 2$$\n\nso I know\n\n$$\\displaystyle \\int^2_{-2} 16 - y^4 dy$$\n\nTake antiderivative\n\n$$\\displaystyle 16y - \\frac{1}{5}y^5$$ | -2 to 2\n\nso $$\\displaystyle (2) = 32 - \\frac{32}{5} = \\frac{160}{5}$$\n\nthen $$\\displaystyle (-2) = -32 - (\\frac{-32}{5}) = -32 + \\frac{32}{5} = \\frac{-160}{5} + \\frac{32}{5} = \\frac{-128}{5}$$\n\nSO finally\n$$\\displaystyle [\\frac{160}{5}] - [\\frac{-128}{5}] = \\frac{288}{5}$$\n\nZaidAlyafey\n\nWell-known member\nMHB Math Helper\nThis is not a parabola. It is like a parabola that intersects the y-axis at $$\\displaystyle y=\\pm 2$$ so it is open to the left. I suggest you revise your calculations.\n\nMarkFL\n\nStaff member\nI would use the even-function rule to state:\n\n$$\\displaystyle A=2\\int_0^2 16-y^4\\,dy=\\frac{2}{5}\\left[80y-y^5 \\right]_0^2=?$$\n\nshamieh\n\nActive member\nRecalculated answer below if someone has a chance to check.\n\nLast edited:\n\nshamieh\n\nActive member\nrecalculated and got $$\\displaystyle \\frac{256}{5}$$ . Is that correct?\n\n- - - Updated - - -\n\nI would use the even-function rule to state:\n\n$$\\displaystyle A=2\\int_0^2 16-y^4\\,dy=\\frac{2}{5}\\left[80y-y^5 \\right]_0^2=?$$\nYea this rule is so much easier!\n\n- - - Updated - - -\n\nMark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.\n\nshamieh\n\nActive member\nLike how would I use this rule if i had $$\\displaystyle 5 - x^2$$?\n\nMarkFL\n\nStaff member\nrecalculated and got $$\\displaystyle \\frac{256}{5}$$ . Is that correct?\n\n- - - Updated - - -\n\nYea this rule is so much easier!\n\n- - - Updated - - -\n\nMark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.\nYes, your result of $$\\displaystyle A=\\frac{256}{5}$$ is correct.\n\nAn even function is symmetric about the $y$-axis, i.e., $$\\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.\n\nObserve that:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{-a}^0 f(x)\\,dx+\\int_0^a f(x)\\,dx$$\n\nNow, in the first integral, if we replace $x$ with $-x$, we have:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{a}^0 f(-x)\\,-dx+\\int_0^a f(x)\\,dx$$\n\nBringing the negative in front of the differential out front and using $$\\displaystyle f(-x)=f(x)$$, we have:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=-\\int_{a}^0 f(x)\\,-dx+\\int_0^a f(x)\\,dx$$\n\nApplying the FTOC, we obtain:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=-\\left(F(0)-F(a) \\right)+F(a)-F(0)=2F(a)=2\\int_0^a f(x)\\,dx$$\n\nZaidAlyafey\n\nWell-known member\nMHB Math Helper\nYes, your result of $$\\displaystyle A=\\frac{256}{5}$$ is correct.\n\nAn even function is symmetric about the $y$-axis, i.e., $$\\displaystyle f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.\n\nObserve that:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{-a}^0 f(x)\\,dx+\\int_0^a f(x)\\,dx$$\n\nNow, in the first integral, if we replace $x$ with $-x$, we have:\n\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=\\int_{a}^0 f(-x)\\,-dx+\\int_0^a f(x)\\,dx$$\n\nBringing the negative in front of the differential out front and using $$\\displaystyle f(-x)=f(x)$$, we have:\n$$\\displaystyle \\int_{-a}^a f(x)\\,dx=-\\int_{a}^0 f(x)\\,dx+\\int_0^a f(x)\\,dx=\\int_{0}^a f(x)\\,dx+\\int_0^a f(x)\\,dx=2\\int^a_0 f(x)\\, dx$$", "date": "2020-09-19 18:14:44", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/finding-the-area-bounded-by-the-curve.7787/", "openwebmath_score": 0.8270788192749023, "openwebmath_perplexity": 482.8216989259686, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540656452213, "lm_q2_score": 0.8976952907388474, "lm_q1q2_score": 0.8791615326956591}}
{"url": "http://mathhelpforum.com/calculus/60824-taylor-series-question-confusion.html", "text": "# Thread: Taylor series question confusion\n\n1. ## Taylor series question confusion\n\nThere is this problem that has been appearing a lot lately but I cannot find any examples that help me solve this.\n\nIt states:\n\n\"Find (at least) the first 3 nonzero terms in the Taylor (power) series for\n\n$f(x) = \\frac{x^3}{3+5x}$ centered at $a = 0.$\n\n(This is the same as the Maclaurin series for the function $f$)\"\n\nNow I know how to use the Taylor series and get an expansion, but my method involves continuously taking the derivative of the function which can get pretty messy after even the second derivation. And even at that point, I'm still getting zero for my expansion terms.\n\nWhat is this question asking exactly?\n\n2. Originally Posted by freyrkessenin\nThere is this problem that has been appearing a lot lately but I cannot find any examples that help me solve this.\n\nIt states:\n\n\"Find (at least) the first 3 nonzero terms in the Taylor (power) series for\n\n$f(x) = \\frac{x^3}{3+5x}$ centered at $a = 0.$\n\n(This is the same as the Maclaurin series for the function $f$)\"\n\nNow I know how to use the Taylor series and get an expansion, but my method involves continuously taking the derivative of the function which can get pretty messy after even the second derivation. And even at that point, I'm still getting zero for my expansion terms.\n\nWhat is this question asking exactly?\nI agree that Taylor series can be fairly nasty at time, but sometimes (as in your case you can make life alot easier on yourself)\n\nhere you are given x^3/( 3 + 5x) = x^3 * ( 1/(3 + 5x) )\n\nHere we observe that the only component that truely requires a taylor expansion is (1/(3+ 5x)), whose derivatives are far easier to compute than that of the original function.\n\nOnce this is obtained you simply multiple the series through by x^3 and viola you have answered the problem with only a 10th of the pain!!\n\nHope this points you in the right direction,\n\nLet me know if you require any further assistance,\n\nDavid\n\nps - make life even easier on yourself by simplifying the function even further, i.e.\n1/(3 + 5x) = 1/( 3 (1 + ((5/3)x)) = (1/3)(1/(1 + (5/3)x)) i.e bring the 1/3 out the front\nfurthermore let t = 5/3x and thus you you need only solve for 1/(1+t) = a0 + a1*t + ....\n\nTo then get in back in terms of x, sub t = 5/3x into the the taylor expansion and your done!\n\n3. ## Hi\n\nI tried performing some hand calculations, and got the following result.\n$f(x) = \\frac{x^3}{3+5x} = x^3 \\cdot \\frac{1}{3+5x}$\n\nNow I only performing a maclaurin series on the $\\frac{1}{3+5x}$\nterm.\nLet $h(x) \\, \\mbox{be the maclaurin series of }\\frac{1}{3+5x}$\n\nThen $f(x) \\approx x^3 \\cdot h(x)$\n\nIf I dont bother with the restterm, then we get something like (see picture).\n\nAnd the two functions seem to be pretty close around $x=0$\n\nYou can see the maclaurin series in the picture...\n\n4. Originally Posted by Twig\nI tried performing some hand calculations, and got the following result.\n$f(x) = \\frac{x^3}{3+5x} = x^3 \\cdot \\frac{1}{3+5x}$\n\nNow I only performing a maclaurin series on the $\\frac{1}{3+5x}$\nterm.\nLet $h(x) \\, \\mbox{be the maclaurin series of }\\frac{1}{3+5x}$\n\nThen $f(x) \\approx x^3 \\cdot h(x)$\n\nIf I dont bother with the restterm, then we get something like (see picture).\n\nAnd the two functions seem to be pretty close around $x=0$\n\nYou can see the maclaurin series in the picture...\nYes, I prefer Twiggy's method.\n\n$\\forall{x}\\backepsilon|x|<1~\\frac{1}{1-x}=\\sum_{n=0}^{\\infty}x^n$\n\nAnd\n\n\\begin{aligned}\\forall{x}\\backepsilon|5x|<1~\\frac{ 1}{1+5x}&=\\frac{1}{1-(-5x)}\\\\\n&=\\sum_{n=0}^{\\infty}\\left(-5x\\right)^n\\\\\n&=\\sum_{n=0}^{\\infty}(-1)^n5^nx^n\n\\end{aligned}\n\n\\begin{aligned}\\therefore~\\forall{x}\\backepsilon|5 x|<1~\\frac{x^3}{1+5x}&=x^3\\cdot\\frac{1}{1+5x}\\\\\n&=x^3\\cdot\\sum_{n=0}^{\\infty}(-1)^n5^nx^n\\\\\n&=\\sum_{n=0}^{\\infty}(-1)^n5^nx^{n+3}\n\\end{aligned}\n\nSo now truncate to the number of terms you need for your series.", "date": "2017-01-22 17:57:52", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/60824-taylor-series-question-confusion.html", "openwebmath_score": 0.9362354278564453, "openwebmath_perplexity": 604.1704771261857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936096877063, "lm_q2_score": 0.8933093989533708, "lm_q1q2_score": 0.879100070983978}}
{"url": "https://math.stackexchange.com/questions/666381/solve-the-following-equation", "text": "# Solve the following equation.\n\nQuestion: $3^{x^2-5} - 9^{x-1} = 0$\n\n$3^{x^2 -5} - 9^{x-1} = 0$\n\n$3^{x^2 -5} = 9^{x-1}$\n\nif $x = 3$\n\n$3^{9-5} = 9^2$\n\n$3^4 = 9^2$\n\n$81 = 81$\n\nand $x = 3$\n\nIs this the right way to solve it?\n\n\u2022 Do you know that's the only solution? \u2013\u00a0David Mitra Feb 6 '14 at 18:32\n\u2022 I didn't get you, sorry! :( Can you explain? @DavidMitra \u2013\u00a0Kiara Feb 6 '14 at 18:35\n\u2022 You found one solution. There may be others. There might not be others. You need to determine which of the two cases holds and prove it. \u2013\u00a0David Mitra Feb 6 '14 at 18:36\n\u2022 Yeah, you're right...I have two answers in my book and I got only one. :( \u2013\u00a0Kiara Feb 6 '14 at 18:37\n\nHINT:\n\nUsing Exponent law, $\\displaystyle(a^x)^y=a^{xy}$\n\nWe have, $$3^{x^2-5}=9^{x-1}=(3^2)^{x-1}$$\n\n$$\\implies 3^{x^2-5}=3^{2(x-1)}$$\n\nNow if $\\displaystyle a^x=a^y\\implies a^{x-y}=1$ (assuming $a$ to be non-zero finite number)\n\neither $a=1$\n\nor $a=-1,x-y$ is even\n\nor $x=y$ for real $a$\n\nThat approach certainly works, but you missed a root $x=-1$. A more general approach to the problem is noting that $9 = 3^2$ and that $(x^a)^b = x^{ab}$. Then we have $$3^{x^2-5} - 9^{x-1} = 3^{x^2-5} - (3^2)^{x-1} = 3^{x^2-5} - 3^{2x-2} = 0 \\implies 2x-2 =x^2 -5$$\n\nWe can just solve for $x$ now: $$2x-2 = x^2 - 5 \\implies x^2-2x-3 = (x-3)(x+1) = 0 \\implies x=-1 \\text{ or } x=3$$", "date": "2019-09-17 12:36:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/666381/solve-the-following-equation", "openwebmath_score": 0.7755776047706604, "openwebmath_perplexity": 514.2541462401155, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936106207202, "lm_q2_score": 0.8933093975331751, "lm_q1q2_score": 0.8791000704198425}}
{"url": "https://math.stackexchange.com/questions/2787516/area-of-a-regular-hexagon-via-area-of-triangles", "text": "# Area of a regular hexagon via area of triangles\n\nProblem: Find the area of a regular hexagon whose sides measures 5 cm\n\nSol'n 1: I can cut the hexagon into 6 small triangles, so the area of triangle times 6 will be equal to the area of the polygon. Since the triangles are equilateral I can use the formula for it.\n\nArea of triangle = $\\frac{\\sqrt{3}}{4}(5)^2= 10.83$\n\nArea of hexagon = (6)(10.83) = 64.95\n\nSol'n 2: Using the formula $\\frac{1}{2}(base)(height)$ for the area of triangle. The angle of triangle (angle at the radius) is equal to 60 degrees (From 360 / 6). Cutting the triangle into half (to get the base and height) will result to an angle of 30 degrees and base of 2.5.\n\nTo get the height: $tan(30) = \\frac{2.5}{height} = 4.33$, so the area of the triangle: $\\frac{1}{2}(2.5)(4.33) = 5.41$\n\nArea of hexagon = (6)(5.41) = 32.475 which does not equal to the area in solution 1.\n\nQuestion: I noticed that the area calculated on solution 2 is half the area calculated in solution 1. I don't know why. I don't think plugging in the original length of the base is right or logical? I have answered a related problem like this and I have gotten the polygon's area with 1/2(base)(height) and not plugging the original base length back to the formula of area after getting the height. Topic is easy but I don't get why I get so confused. :( Any help will be appreciated.\n\nYou've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)\n\nThe area of one triangle is given by $\\frac 12\\cdot2.5\\cdot2.5\\tan(60)=\\frac{25\\sqrt3}{8}$. Multiplying this area by $12$ (the amount of triangles) gives us $\\frac{75\\sqrt3}{2}\\approx64.9519$\n\n\u2022 Please excuse the awful quality of the diagram but I hope that it is clear enough \u2013\u00a0Rhys Hughes May 19 '18 at 14:29\n\u2022 I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absin\\theta$ and I did not need to multiply the area by 16 but just by 8 which confuses me. \u2013\u00a0Jayce May 19 '18 at 14:36\n\u2022 Typo on my comment above. Should be $\\frac{1}{2}absin\\theta$. \u2013\u00a0Jayce May 19 '18 at 14:52\n\u2022 Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8\\cdot\\frac{1}{2}\\cdot12^2\\cdot\\sin45=288\\sqrt2 \\approx407.2935$ \u2013\u00a0Rhys Hughes May 19 '18 at 14:58\n\u2022 You are right. Try solving it via $\\frac{1}{2}(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, \"x/2\" for the base and the hypotenuse, 12. Then Via \"soh-cah-toa\", base is 9.184 and height is 11.087. Plugging it in: $\\frac{1}{2}(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absin\\theta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :( \u2013\u00a0Jayce May 19 '18 at 15:14\n\nSolution 1 gives the correct answer. In solution 2, the triangle that you're calculating is actually a right-angled triangle with interior angles $30\u00b0$, $60\u00b0$ and $90\u00b0$, and it is in fact half of the triangle calculated in solution 1. This explains why the numerical result obtained in solution 2 is half of that in solution 1.\n\n\u2022 I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $\\frac{1}{2}absin\\theta$ and $\\frac{1}{2}(base)(height)$. For $\\frac{1}{2}(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $\\frac{1}{2}absin\\theta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :( \u2013\u00a0Jayce May 19 '18 at 14:50\n\u2022 The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via \"soh-cah-toa\", base = 9.184 and height = 11.087. Plugging it into $8*[\\frac{1}{2}(base)(height)] = 407.89$ which is the area of octagon. Via $\\frac{1}{2}absin\\theta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8. \u2013\u00a0Jayce May 19 '18 at 15:05", "date": "2019-09-23 04:47:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2787516/area-of-a-regular-hexagon-via-area-of-triangles", "openwebmath_score": 0.8110178709030151, "openwebmath_perplexity": 324.27861025176094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9777138177076645, "lm_q2_score": 0.8991213786215104, "lm_q1q2_score": 0.8790833956746155}}
{"url": "https://math.stackexchange.com/questions/2495413/how-to-find-the-limit-of-series-what-should-i-know/2495439", "text": "# How to find the limit of series? (What should I know?)\n\nThere is a couple of limits that I failed to find:\n\n$$\\lim_{n\\to\\infty}\\frac 1 2 + \\frac 1 4 + \\frac 1 {8} + \\cdots + \\frac {1}{2^{n}}$$\n\nand\n\n$$\\lim_{n\\to\\infty}1 - \\frac 1 3 + \\frac 1 9 - \\frac 1 {27} + \\cdots + \\frac {(-1)^{n-1}}{3^{n-1}}$$\n\nThere is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result.\n\nSo my questions are:\n\n\u2022 What should I do when I need a limit of infinite sum? (are there any rules of thumb?)\n\u2022 What theorems or topics from calculus should I know to solve these problems better?\n\nI am new to math and will appreciate any help. Thank you!\n\n\u2022 Hint: Geometric Series! \u2013\u00a0Qi Zhu Oct 29 '17 at 19:24\n\u2022 I am also new to Mathematics on Stack Exchange, so if you see that I can improve my question, I would appreciate if you left a comment \u2013\u00a0Nikita Hismatov Oct 29 '17 at 19:24\n\u2022 Demidovich!!! I've used it in 1977 to prepare Calculus I and after I continued to use it to give private classes to a lot of people, including the girl who became my wife... \u2013\u00a0Raffaele Oct 29 '17 at 20:58\n\u2022 Writing the first in binary will result in something interesting. \u2013\u00a0chx Oct 29 '17 at 21:00\n\u2022 @NikitaHismatov Once I asked a professor of mine about this. He replied: \"Do you think we have a panacea to calculate limits?!\" - As it often happens in mathematics, we have methods for some cases but you can always find harder limits for which the previously developed methods do not work. I once thought about the existence of the following limit and asked here, people were able to compute the limit but I'm not sure I really understand what they did. \u2013\u00a0Billy Rubina Oct 30 '17 at 6:34\n\n$$\\color{red}{1+}\\frac 1 2 + \\frac 1 4 + \\frac 1 {8} + \\ldots + \\frac {1}{2^{n}}$$ Is an example of $$1+x+x^2+x^3+\\ldots+x^n$$ Where the ratio between a number $a_n$ and the previous $a_{n-1}$ is constant and is $x$. This is the sum of a geometric progression and it's quite easy to see that its value is $$\\frac{1-x^{n+1}}{1-x}$$ When $n\\to\\infty$ the sum converges only if $|x|<1$ because if so $x^{n+1}\\to 0$ and the sum is $$\\sum_{n=0}^{\\infty}x^n=\\frac{1}{1-x}$$ In your first example $x=\\frac12$ and index $n$ starts from $n=1$ so the sum is $$\\sum_{n=1}^{\\infty}\\left(\\frac12\\right)^n=\\frac{1}{1-\\frac12}-1=\\color{red}{1}$$ The second one is $$\\lim_{n\\to\\infty}1 - \\frac 1 3 + \\frac 1 9 - \\frac 1 {27} + \\cdots + \\frac {(-1)^{n}}{3^{n}}=\\sum_{n=0}^{\\infty}\\left(-\\frac13\\right)^n=\\frac{1}{1-\\left(-\\frac13\\right)}=\\color{red}{\\frac34}$$\n\nHope this is useful\n\nHint:\n\nAll you have to know is the sum of the $n$ first terms of a geometric series, which is a formula from high school: $$1+q+q^2+\\dots +q^n=\\frac{1-q^{n+1}}{1-q}\\qquad (q\\ne 1),$$from which we can deduce: $$q^r+q^{r+1}+\\dots +q^n=q^r(1+q+\\dots+q^{n-r})=q^r\\frac{1-q^{n-r+1}}{1-q}=\\frac{q^r-q^{n+1}}{1-q}.$$ If $\\lvert q\\rvert<1$, these sums have limits $\\;\\dfrac1{1-q}\\;$ and $\\;\\dfrac{q^r}{1-q}\\;$ respectively.\n\nWhat should I do when I need a limit of infinite sum? (are there any rules of thumb?)\n\nIn general - no there aren't such rules. There are specific types of series for which it is known how to compute the limit (like the geometric series). There are way more recepies for figuring out whether a series converges or not (finding the limit is much harder). But also here the cookbook is limited. As others pointed out already, your two series are geometric series, who's limit can be computed easily.\n\nWell, that first one should be an instance of a theorem you should definitely put in your list of useful theorems:\n\nIf $|a| < 1$:\n\n$$\\sum_{i=1}^{\\infty} \\ a^i = \\frac{a}{1-a}$$\n\nAnd the second one is an instance of the closely related:\n\nIf $|a| < 1$:\n\n$$\\sum_{i=0}^{\\infty} \\ a^i = \\frac{1}{1-a}$$\n\nIf you don't want to remember both, it's ok to remember just one, because the other one can easily be derived from it, e.g:\n\n$$\\sum_{i=0}^{\\infty} \\ a^i = a^0 + \\sum_{i=1}^{\\infty} \\ a^i = 1 + \\frac{a}{1-a} = \\frac{1-a}{1-a} + \\frac{a}{1-a} = \\frac{1-a+a}{1-a}= \\frac{1}{1-a}$$\n\n\u2022 Or, perhaps more simply, $$\\sum_{i=1}^{\\infty}a^i = a\\sum_{i=0}^{\\infty}a^i =a\\left( \\frac{1}{1-a}\\right)$$ \u2013\u00a0Bungo Oct 29 '17 at 19:52\n\u2022 @Bungo Yes, nice! I always forget about that trick :( \u2013\u00a0Bram28 Oct 29 '17 at 19:55\n\nthe other answers are correct but if you want to see a more \"visualize\" way(it is actually the same way but you can see it more clearly) $$S=\\lim_{n\\to\\infty}\\sum_{i=1}^n\\frac {1}{2^{i}}=\\frac 1 2 + \\frac 1 4 + \\frac 1 {8} + \\cdots\\\\2S=\\lim_{n\\to\\infty}\\sum_{i=0}^n\\frac {1}{2^{i}}=\\frac 1 1 + \\frac 1 2 + \\frac 1 {4} + \\cdots=1+\\lim_{n\\to\\infty}\\sum_{i=1}^n\\frac {1}{2^{i}}=1+S\\\\2S-S=\\left(1+S\\right)-S=1$$ and $$A=\\lim_{n\\to\\infty}\\sum_{i=1}^n\\frac {(-1)^{i-1}}{3^{i-1}}=1 - \\frac 1 3 + \\frac 1 9 - \\frac 1 {27} + \\cdots \\\\ 3A=\\lim_{n\\to\\infty}\\sum_{i=1}^n\\frac {(-1)^{i-1}}{3^{i-2}}=3 - \\frac 3 3 + \\frac 3 9 - \\frac 3 {27} + \\cdots\\\\=3-\\left(1 - \\frac 1 3 + \\frac 1 9 - \\frac 1 {27} + \\cdots\\right)=3-A\\\\3-A=3A\\implies3=4A\\implies\\frac34=A$$\n\nNote that you have the sums $\\sum_{k = 1}^{\\infty} \\frac{1}{2^k}$ and $\\sum_{k=0}^{\\infty} \\left(-\\frac{1}{3} \\right)^k$. Now look up geometric series and try to finish these on your own :)\n\nIn case of infinite Geometric Progression,\n\n$$sum = \\frac{a}{1-r}$$\n\nyou can also derive this from the normal formula using, $n\\to \\infty$\n\nThe thing to know here is that $$|r| <1$$\n\nTo explain this, if |r|<1 , ar < a similarly $ar^2 < ar$ and each next term will keep getting smaller and smaller and as $n\\to \\infty$, $$a_n = ar^{n-1} \\to 0$$ therefore, we can successfully find a finite number to which this sequence will converge to as the later on terms are so smaller than the initial ones that they won't even contribute to the sum.\n\nAlthough if |r|>1 each term will be greater than the previous one, so we will never be able to sum this sequence.\n\nThe sums you mention are both examples of geometric series. There are no general ways to sum arbitrary sequences - they typically require some inspiration. Geometric series are often proven geometrically, but the algebraic proof is illustrative.\n\n$$\\sum_{n=0}^{m} a^n = S_{m}, a S_{m} = \\sum_{n=1}^m a^n = S_{m+1} - 1$$\n\n$$S_m - a S_m = 1 - a^{m+1}$$\n\n$$S_{m} = \\frac{1 - a^{m+1}}{1 - a}$$\n\nThe other series is relatively straightforward if you know what your goal is:\n\n$$\\sum_{n=0}^\\infty \\frac{(-1)^n}{3^n} = \\sum_{n=0}^\\infty \\frac{1}{9^n} - \\frac{1}{3}\\sum_{n=0}^\\infty \\frac{1}{9^n} = \\frac{2}{3} \\sum_{n=0}^\\infty \\frac{1}{9^n}$$\n\nThey are sums of geometric progressions.\n\nThe first it's $$\\frac{\\frac{1}{2}}{1-\\frac{1}{2}}=1.$$\n\nThe second it's $$\\frac{1}{1+\\frac{1}{3}}=\\frac{3}{4}.$$\n\n\u2022 I downvoted this answer because it does not address the OP's note that \"what I am interested in is the method, not just a concrete result.\" Naming the series is helpful, but something more\u2014like explaining a closed form for the partial sums, or showing that the limit exists and then showing what it must be\u2014is in order. \u2013\u00a0wchargin Oct 29 '17 at 22:14", "date": "2020-03-30 01:43:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2495413/how-to-find-the-limit-of-series-what-should-i-know/2495439", "openwebmath_score": 0.8040698766708374, "openwebmath_perplexity": 247.8874058899092, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180656553329, "lm_q2_score": 0.8918110353738529, "lm_q1q2_score": 0.8790742487187939}}
{"url": "https://math.stackexchange.com/questions/3363066/how-do-i-denote-arbitrary-element-of-these-vectors/3363104", "text": "# How do I denote arbitrary element of these vectors?\n\nI know that when we have something like $$\\bf u$$, then the arbitrary element $$i$$ will be denoted as $$u_{i}$$.\n\nNow suppose we have vectors $$\\{\\mathbf {v_{1}, v_{2}, \\cdots, v_{n}} \\}$$\n\nI want to consider, say, third element of the fifth vector.\n\nHow do I denote it?\n\nIs it $$(\\mathbf{v_{5}})_{3}$$ or $$\\mathbf{v_{}}_{5_{3}}$$? Or are there better options available?\n\n\u2022 Depending on the context, you may want $v_{5,3}, v_5(3), v_5[3], v_5^3$ (though you may make some people irritated, no matter which you choose). Just make sure you are clear and consistent with your notation, and if you find you later have clashes of notation, be very careful to specify which is which. \u2013\u00a0Brian Moehring Sep 20 at 5:39\n\u2022 Yes, ${V^_5)_3$ is correct. \u2013\u00a0Dr Zafar Ahmed DSc Sep 20 at 5:54\n\nYou can technically do whatever you want as long as it\u2019s comprehensible (actually, it doesn\u2019t even need to be comprehensible, but if it isn\u2019t, nobody will care\u2014just look at some of Galois\u2019s manuscripts!) but what you might find useful is $$v_{i,j}$$ which may denote the $$i$$-th component of the $$j$$-th vector, or vice-versa.\nWhen I have to manipulate indices of many vectors, I like calling them $$\\{v^{(1)},...,v^{(m)}\\}$$ and labeling elements of those vectors $$v^{(i)}_j$$ because I prefer to keep vector indices downstairs when I can. You can also drop the parantheses. Whatever makes most sense to you is good, as long as other people can understand it as well.", "date": "2019-12-10 11:09:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3363066/how-do-i-denote-arbitrary-element-of-these-vectors/3363104", "openwebmath_score": 0.9658858180046082, "openwebmath_perplexity": 284.9925222267378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102542943773, "lm_q2_score": 0.9086178987887253, "lm_q1q2_score": 0.8790062725236236}}
{"url": "https://math.stackexchange.com/questions/2623780/writing-an-expression-for-the-volume-of-a-spherical-shell", "text": "# Writing an Expression for the Volume of a Spherical Shell\n\nThe question is as follows:\n\nWrite an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 \u03c0 \u00b7 (trinomial) \u00b7 (binomial)$. In this situation, what is the meaning of the binomial? What can be said about the value of the trinomial when the binomial has a very small value? Make a conjecture concerning the surface area of a sphere of radius $R$.\n\nI know that the volume of the sphere with the inner radius of r would be $\\frac{4}{3}\\pi r^3$ and the volume of the sphere with the outer radius of R would be $\\frac{4}{3}\\pi R^3$. Therefore, the volume of the spherical shell formed between the concentric circles would be the difference between the outer sphere and the inner sphere--which, when mathematically written, would be $\\frac{4}{3}\\pi R^3 - \\frac{4}{3}\\pi r^3$. That expression, after it's factored, would be $\\frac{4}{3}\\pi(R^3 - r^3)$. My question is how can I represent $(R^3 - r^3)$ the product of a trinomial and binomial. Any help will be greatly appreciated!\n\nEDIT: I think that I may have found the answer to the question that I have presented above: $(R^2 + Rr + r^2)(R - r)$. Therefore, my complete answer to the volume of the spherical shell is $\\frac{4}{3}(R^2 + Rr + r^2)(R - r)$.\n\nI still don't know what the binomial is representing for this particular problem, nor its relation to the trinomial when the binomial value gets smaller. I think that I can use the help for understanding what the binomial and trinomial represents as a way to formulate a conjecture regarding the surface area of a sphere with radius R.\n\nEDIT #2: The binomial, as I understand it, represents the thickness of the spherical shell formed by the concentric circles.\n\n\u2022 Since this is the kind of thing you have to just know, I don't mind telling you flat-out that one of the terms is $R-r$. You can find the other term by long division. \u2013\u00a0dfan Jan 27 '18 at 18:12\n\u2022 Factorize as $a^3-b^3=(a-b)(a^2+b^2+ab)$. \u2013\u00a0MalayTheDynamo Jan 27 '18 at 18:13\n\u2022 Thank you @MalayTheDynamo and @dfan! I have found the answer right after I posted the problem. However, would anyone of you help in guiding me in the questions that I posted in the third paragraph? \u2013\u00a0geo_freak Jan 27 '18 at 18:17\n\nNote that $$\\frac{4}{3} \\pi (R^2 + Rr + r^2)(R - r)$$\n\nhas three parts.\n\nYou have already figured out the $(R-r)$ part as the thickness of the spherical shell.\n\nThe trinomial part is very close to $3r^2$ in case that the thickness is very small.\n\nNow when you multiply this $3r^2$ by $\\frac{4}{3} \\pi$, you come up with $$4\\pi r^2$$ which is the surface area of the shell( Approximate of course)\n\nThus The volume of the shell is approximately, the surface area of the shell multiplied by the thickness of the shell.\n\n\u2022 How did you get that the trinomial part will be very close to $3r^2$? \u2013\u00a0geo_freak Jan 27 '18 at 18:35\n\u2022 Writing $r$ in place of $R$ as these are near to each othert \u2013\u00a0Narasimham Jan 27 '18 at 18:45\n\u2022 Later you can also find that surface area is $4 \\pi r^2$ as differential coefft. of $4/3 \\pi r^3$ with respect to $r$ \u2013\u00a0Narasimham Jan 27 '18 at 18:47\n\u2022 I see! Thank you so much for clarifying @Narasimham! \u2013\u00a0geo_freak Jan 27 '18 at 18:51\n\u2022 Thank you for your attention. \u2013\u00a0Mohammad Riazi-Kermani Jan 27 '18 at 18:53\n\nLet $V(r)$ be the volume of the sphere with radius $r$ and $A(r)$ its surface area.We have $$V(r) = V(1)\\cdot r^3$$ since concentric spheres are homothetic. In particular $$V(R)-V(r) = V(1)(R-r)(R^2+Rr+r^2).$$ On the other hand $A(r) = A(1)\\cdot r^2$ and $$A(R) = \\lim_{r\\to R^-}\\frac{V(R)-V(r)}{R-r} = \\lim_{r\\to R^-} V(1)(R^2+Rr+r^2)=3V(1)\\,R^2.$$ This proves $A(1)=3\\,V(1)$ and the same argument in dimension $n$ proves $A(1)=n\\,V(1)$: to find the volume or the surface area of a Euclidean ball are equivalent problems. Remark: for $n=3$, the same can be shown by approximating a sphere with the union of a large number of cones and recalling that the volume of a cone is given by one third of the product between the height and the base area.\n\n\u2022 Thank you for the answer that you have posted. I, unfortunately, am having a hard time understanding it since I am just taking Alg II/Trig for my maths class. Is it possible to explain it in more simpler terms? \u2013\u00a0geo_freak Jan 27 '18 at 18:28\n\u2022 @geo_freak: $A(R)$ is the derivative of $V(R)$ with respect to $R$ by elementary geometric considerations. In particular the ratio between $A(1)$ and $V(1)$ exactly equals the dimension of the space. \u2013\u00a0Jack D'Aurizio Jan 27 '18 at 18:33\n\nWrite an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 \\pi \\cdot (\\text{trinomial}) \\cdot (\\text{binomial})$\n\n$$V(R, r) = \\frac{4}{3}\\pi (R^3-r^3) = \\frac{4}{3}\\pi(R-r)(R^2+r^2 + Rr) = 4 \\pi \\, t(R,r) \\, b(R,r)$$ with \\begin{align} t(R,r) &= \\frac{1}{3}(R^2+r^2 + Rr) \\\\ b(R, r) &= R-r \\end{align}\n\nWhat can be said about the value of the trinomial when the binomial has a very small value?\n\nIn this case we have $r \\approx R$ and we can estimate the volume by $$V(R,r) \\approx A(r) \\, b(R,r)$$ where $A(r)$ is the area of a sphere with radius $r$. So we get $$t(R,r) = \\frac{1}{3}(R^2+r^2 + Rr) = \\frac{V(R,r)}{4\\pi b(R,r)} \\approx \\frac{A(r) \\, b(R,r)}{4\\pi b(R,r)} = \\frac{A(r)}{4\\pi}$$\n\nMake a conjecture concerning the surface area of a sphere of radius R.\n\n$$A(t) \\approx 4\\pi\\, t(R,r) = 4\\pi \\, \\frac{1}{3}(R^2 + r^2 + Rr) \\approx 4\\pi \\, \\frac{1}{3}(r^2 + r^2 + r^2) = 4 \\pi r^2$$", "date": "2020-01-17 19:05:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2623780/writing-an-expression-for-the-volume-of-a-spherical-shell", "openwebmath_score": 0.9303126931190491, "openwebmath_perplexity": 117.05522210067359, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319857981405, "lm_q2_score": 0.8872045966995027, "lm_q1q2_score": 0.8789819718973366}}
{"url": "http://math.stackexchange.com/questions/384303/computing-determinant-of-this-matrix", "text": "# Computing determinant of this matrix\n\nI have a very specific kind of matrix and I have to find the formula to find the determinant of these matrix.\n\na(i,j)=a if(i==j) and a(i,j)=0 if(floor(i/2)=floor(j/2) and i!=j) and n is odd\n\n$$A=(a_{i,j})_{n \\times n}=\\left( \\begin{array}{ccccc} a&0&b&b \\cdots &b&b\\\\ 0& a&b&b \\cdots& b&b\\\\ b& b& a&0 \\cdots& b&b\\\\ b& b& 0&a \\cdots& b&b\\\\ \\vdots& \\vdots& \\vdots& \\ddots&\\cdots\\\\ b&b&b & \\cdots&b&a \\end{array} \\right)?$$\n\n-\nTry writing down its eigenvectors. \u2013\u00a0Qiaochu Yuan May 7 '13 at 9:50\nWhat's the rule for the placement of the $0$'s ? \u2013\u00a0Raskolnikov May 7 '13 at 9:54\na(i,j)=a if(i==j) and a(i,j)=0 if(i/2=j/2 and i!=j) \u2013\u00a0Alen May 7 '13 at 9:59\nPlease notice n is odd \u2013\u00a0Alen May 7 '13 at 10:01\n\nThe problem is closely related to this one posted and answered yesterday, with $p\\leftarrow2$, $np\\leftarrow n-1$, $c\\leftarrow a/b$, with the entire matrix divided by $b$.\n\nSince according to a comment under the question $n$ is odd, we need to deal with the last component separately. All eigenvectors in the linked question except for the one filled with $1$s sum to $0$. Thus we can append a $0$ to them to obtain eigenvectors of the present matrix. That leaves a two-dimensional subspace to be dealt with, spanned by the vector $x$ that has a $1$ in the last component and the vector $y$ that has $1$s everywhere else. Applying $A$ to these vectors yields $Ax=ax+by$ and $Ay=(n-1)bx+((n-3)b+a)y$. Thus the product of the remaining two eigenvalues is\n\n$$\\left|\\matrix{a&b\\\\(n-1)b&(n-3)b+a}\\right|=((n-3)b+a)a-(n-1)b^2\\;.$$\n\nMultiplying this by the $n-2$ eigenvalues from the linked question, with the above substitutions and the factor $b^{n-2}$ that was divided out, yields the determinant of the present matrix:\n\n\\begin{align} \\det A &= \\left(((n-3)b+a)a-(n-1)b^2\\right)b^{n-2}\\left(\\frac ab\\right)^{(n-1)/2}\\left(\\frac ab-2\\right)^{(n-3)/2} \\\\ &= \\left(((n-3)b+a)a-(n-1)b^2\\right)a^{(n-1)/2}\\left(a-2b\\right)^{(n-3)/2}\\;. \\end{align}\n\n-\nIt sounds pretty close but $n$ is odd... \u2013\u00a0achille hui May 7 '13 at 11:38\n@achille: Aha -- thanks for pointing that out -- yet another reminder why it's a bad idea to bury clarifications of the question deep in the comments... \u2013\u00a0joriki May 7 '13 at 11:42\nCan this matrix have a particular formula? \u2013\u00a0Alen May 7 '13 at 11:58\n@Alen: How do you mean, can it? \u2013\u00a0joriki May 7 '13 at 12:24\nMeans, can I derive a formula for calculating the det A? \u2013\u00a0Alen May 7 '13 at 12:33\n\nSince $n$ is odd, let $n = 2m+1$ and $u_n$ be the $n \\times 1$ column vector with all entries one.\n\nThe matrix $A$ can be rewritten to have the form $X + b u_n u_n^{T}$ where $X$ is a block diagonal matrix consists of $m$ $2\\times 2$ block $\\tilde{X}$ and one $1 \\times 1$ block:\n\n$$X = \\begin{pmatrix} \\tilde{X} & 0 & 0 & \\dots & 0\\\\ 0 & \\tilde{X} & 0 &\\dots & 0\\\\ 0 & 0 & \\tilde{X} &\\dots & 0\\\\ \\vdots &\\vdots & \\vdots & \\ddots & 0\\\\ 0 & 0 & 0 & 0 & a - b \\end{pmatrix} \\quad\\text{ and }\\quad \\tilde{X} = \\begin{pmatrix}a - b & -b\\\\-b & a-b\\end{pmatrix}$$\n\nWhen one perturb an invertible matrix $X$ by a rank one matrix like $b u_n u_n^T$, it is known that the determinant of the updated matrix $X + b u_n u_n^{T}$ is given by:\n\n$$\\det( X + b u_n u_n^{T} ) = \\det(X)( 1 + b u_n^{T} X^{-1} u_n )\\tag{*}$$\n\nSubstitute our special form of $X$ into $(*)$ and notice:\n\n\\begin{align} \\det(X) &= \\det(\\tilde{X})^m (a - b) = ((a-b)^2 - b^2)^m (a - b)\\\\ u_n^{T} X^{-1} u_n &= m\\;u_2^T \\tilde{X}^{-1} u_2 + \\frac{1}{a-b} = \\frac{2m}{a-2b} + \\frac{1}{a-b} \\end{align}\n\nWe get: \\begin{align} \\det A &= ((a-b)^2 - b^2)^m (a - b)\\left\\{1 + b \\left(\\frac{2m}{a-2b} + \\frac{1}{a-b} \\right)\\right\\}\\\\ &= a^m (a-2b)^{m-1}\\left(a^2 + 2 (m-1)ab - 2mb^2\\right) \\end{align}\n\nUpdate\n\nFor the general case when the matrix elements is given by:\n\n$$a_{i,j} = \\begin{cases} a, & \\text{ for } i = j\\\\ 0, & \\text{ for } \\lfloor\\frac{i}{q}\\rfloor = \\lfloor\\frac{j}{q}\\rfloor \\wedge i \\neq j\\\\ b, & \\text{ otherwise }. \\end{cases}$$\n\nwhere $0 \\le i, j < n$ and $q$ is an integer $> 1$. Let $n = mq + r$ where $0 \\le r < q$. Once again, we rewrite $A$ into the form $X + b u_n u_n^T$ where X is now a block diagonal matrix consists of $m$ $q \\times q$ block $\\tilde{X}_q$ and one $r \\times r$ block $\\tilde{X}_r$.\n\nIt is not hard to verify the entries on $\\tilde{X}_s$ is given by:\n\n$$(\\tilde{X}_s)_{i,j} = \\begin{cases} a-b, &\\text{ for } i = j\\\\-b, &\\text{ otherwise. }\\end{cases} \\quad\\implies\\quad \\begin{cases}\\det(\\tilde{X}_s) &= a^{s-1}(a - sb)\\\\ u_s^{T} \\tilde{X}_s^{-1} u_s &= \\frac{s}{a-sb}\\end{cases}$$ Repeating the same procedure as above, we obtain the formula for $\\det(A)$ in the general case:\n\n$$\\det A = (a^{q-1}(a - qb))^m (a^{r-1}(a-rb)) \\left\\{1 + b \\left(\\frac{mq}{a-qb} + \\frac{r}{a-rb}\\right)\\right\\}$$\n\n-\nhue:r can be -1 here? \u2013\u00a0Alen May 7 '13 at 16:44\n@Alen that is a typo, should be $0 \\le r < q$. In principle, one should treat $r = 0$ as a special case. However, it turns out the final expression also work for the $r = 0$ case. \u2013\u00a0achille hui May 7 '13 at 16:53\nI was actually thinking of (pn-1)*(pn-1) dimension matrix \u2013\u00a0Alen May 7 '13 at 17:12\n@Alen, your $pn-1$ case is equivalent to setting the parameters $(m,q,r)$ to $(n-1,p,p-1)$ in my answer. \u2013\u00a0achille hui May 7 '13 at 17:24", "date": "2015-12-01 20:50:08", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/384303/computing-determinant-of-this-matrix", "openwebmath_score": 0.9947680830955505, "openwebmath_perplexity": 559.2811969776525, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787876194205, "lm_q2_score": 0.8902942290328344, "lm_q1q2_score": 0.8788795776412001}}
{"url": "https://math.stackexchange.com/questions/2825002/how-to-compute-this-determinant-as-quickly-as-possible-without-using-any-softwa", "text": "# How to compute this determinant as quickly as possible (without using any software or calculator)?\n\nI sat an Algebra test yesterday, which consisted of 30 questions and a total time of 45 minutes (an average of 1 min 30 secs per question). One question of the test was this:\n\nGiven the matrix: $$A=\\begin{bmatrix} -2 & 4 & 2 & 1 \\\\ 4 & 2 & 1 & -2 \\\\ 2 & 1 & -2 & 4 \\\\ 1 & -2 & 4 & 2 \\end{bmatrix}$$ Which of the following options is correct?\n\n(A) $A^{-1}=\\dfrac{A}{25}$\n\n(B) $A^{-1}=\\dfrac{A}{5}$\n\n(C) $A^{-1}=\\dfrac{A}{15}$\n\n(D) It has no inverse.\n\nI do know how to compute the inverse of a matrix. For example, using the Gauss-Jordan elimination method. Or for example, using this formula: $$A^{-1}=\\dfrac{\\text{Adj}(A^T)}{\\text{det}(A)}$$\n\nI calculated the determinant and it is $625$. However, this won't help me pick the correct option (of course I can eliminate option D, which is false).\n\nHow in the world am I supposed to solve this problem in around 90-100 seconds, without using a calculator? Is there any shortcut or trick or something I missed? 90 seconds was the average time per question in the test. Given how little time I was given to solve the problem, this leads me to think that the structure of A could simplify the answer.\n\n\u2022 Strategy: take the RHS's and multiply by $A$ to see if you get $I$. \u2013\u00a0Adrian Keister Jun 19 '18 at 14:57\n\u2022 The short answer is $(-2)^2 + 4^2 + 2^2 + 1^2 = 25$. (For reasons given below.) \u2013\u00a0Mateen Ulhaq Jun 20 '18 at 0:40\n\u2022 You're assuming that because you have, on average, 90 seconds to solve a problem that each problem must be solved in 90 seconds. Some will undoubtedly take less than 90 seconds. Some will undoubtedly take more. Do your best. \u2013\u00a0Bob Jarvis - Reinstate Monica Jun 20 '18 at 16:15\n\nIf $A^{-1}=\\dfrac{A}{25}$ then $A^2=25I$. Similar for other options. Now you need to calculate only one diagonal element of $A^2$ to find the correct option.\n\n\u2022 If $A$ has no inverse, there will be no (nonzero) solution $\\lambda$ to $A^2 = \\lambda I$, so computing $A^2$ gives this information, too. \u2013\u00a0Travis Willse Jun 19 '18 at 15:01\n\u2022 Indeed it will but it would require to compute more than just one diagonal element of $A^2$ (the answer states that \"you need to calculate only one diagonal element of $A^2$ to find the correct option.\"). Sorry for being so petty :). \u2013\u00a0yakobyd Jun 19 '18 at 15:11\n\u2022 @alephzero: To be specific, if you calculate one diagonal element and it's 25, that doesn't tell you whether $A^2=25I$ or $A^2=\\begin{bmatrix}25 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0\\end{bmatrix}$ ... \u2013\u00a0psmears Jun 19 '18 at 21:23\n\u2022 For checking if $A^{-1}$ exists, this matrix is very amenable to the trick of computing the determinant modulo 2 \u2013\u00a0jld Jun 19 '18 at 22:47\n\u2022 @JoseLopezGarcia for an integer-valued matrix, the determinant is non-zero if the determinant with every entry taken modulo 2 is also non-zero. So in your case $$A \\mod 2=\\begin{bmatrix}0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 1 & 0 & 0 & 0\\end{bmatrix}$$and this matrix is clearly full rank and so has non-zero determinant (it will certainly be $\\pm 1$ but the actual value isn't important so you can stop as soon as you recognize it's not singular). \u2013\u00a0jld Jun 20 '18 at 13:54\n\nWe can also solve this problem quickly without using the multiple choice options as hints to cut short any computations. So, suppose that we're given the matrix $A$ and asked to compute its inverse, without being told that the inverse is proportional to $A$ itself. Then observe that the columns of the matrix are orthogonal, so the matrix is an orthogonal matrix up to a normalization factor. The norm of the columns is 5. The inverse of $A/5$ is therefore equal to its transpose, and the transpose is easily seen to be equal to the matrix itself. So, the inverse of $A$ is $A/25$.\n\nThe matrix has the form $A=\\begin{bmatrix} C & D \\\\ D & C \\end {bmatrix}$\n\nFor this kind of block matrices,\n\n$det (A) = det( C.C -D.C^{-1}.D.C) =$\n\n$= det (C.C + D.D ) = det \\left ( \\begin{bmatrix} 20 & 0 \\\\ 0 & 20 \\end {bmatrix} + \\begin{bmatrix} 5 & 0 \\\\ 0 & 5 \\end {bmatrix} \\right ) = 25^2$\n\nsince $C.D = -D.C$\n\n$A^2=\\begin{bmatrix} C & D \\\\ D & C \\end {bmatrix} \\begin{bmatrix} C & D \\\\ D & C \\end {bmatrix} = \\begin{bmatrix} C^2 + D^2 & 0 \\\\ 0 & C^2 + D^2 \\end {bmatrix} = \\begin{bmatrix} 25 & 0 & 0 & 0 \\\\ 0 & 25 & 0 & 0 \\\\ 0 & 0 & 25 & 0 \\\\ 0 & 0 & 0 & 25 \\end{bmatrix}$\n\nYou're there with the computation of the determinant. The $\\det A^2 = 5^8$, so the answer must be (A) in order for $\\det \\left(A\\cdot A^{-1}\\right) = 1$\n\nBut PJK's answer provides a more efficient approach.\n\nNB you don't need to compute $A^{-1}$ in order to solve this problem, you just need to determine which option is correct---presumably the question also tests your recognition of this fact.\n\nIf $A$ is invertible, then multiplying both sides of each of the first three options, $A^{-1} = \\frac{A}{\\lambda}$, gives an equivalent equation $$A^2 = \\lambda I .$$ If $A$ is not invertible, then there is no (nonzero) $\\lambda$ for which the equation holds. So, to determine the answer, it's enough to compute $A^2$, which is computationally much cheaper than applying G.-J. elimination.\n\nEDIT: I realized after writing that the essence of my answer is already in the answer provided by Count Iblis. Perhaps one can consider this answer as containing supplementary material to his.\n\nWe will make use of the observation that in the given options, multiplying $A$ on both sides gives the result $\\frac{1}{c}I$ for some constant $c$.\n\nFirst, notice that $A$ is a circulant real symmetric matrix. Because it is a circulant matrix, all its rows have the same norm of 5.\n\nWe can therefore write $A = 5 B$ where B is also real symmetric and each row (and column) of B has unit norm. This gives us the following\n\n$A^2 = 25B^2 = 25 B^TB$.\n\nAt this point, this should provide enough hint to inspect the matrix $A$ more carefully and you will notice that it is actually orthogonal. Therefore $B$ is orthonormal so $B^TB = I$ and we have $A^2 = 25I$ as required.\n\nIf the orthogonality of $A$ still eludes you, we can use the spectral theorem and write\n\n$A^2 = 25B^TB = 25 U\\Lambda^2U^T = 25(\\Lambda^{\\frac{1}{2}}U^T)^T(\\Lambda^{\\frac{1}{2}}U^T) \\implies B = \\Lambda^{\\frac{1}{2}}U^T$.\n\nSince $U^T$ is orthonormal and each row and column of $B$ has unit norm, this means $\\Lambda^{\\frac{1}{2}} = I$ and $B^2 = B^TB = I$ as required.\n\nAlso, notice that $\\Lambda = I$ so all the eigenvalues of $A$ are non-zero (they are all 1) so $A$ is invertible and this excludes the last option.\n\n## Determining the determinant\n\nNow that we know $B$ is orthonormal, we can obtain the determinant of $A$ easily using the property $\\text{det}(cM) = c^n \\text{det}(M)$ for $M \\in \\mathbb{R}^{n \\times n}$ and det($Q$) = 1 if $Q$ is a square orthonormal matrix\n\n$A = 5B \\implies \\text{det}(A) = 5^4 \\text{det}(B) = 625$\n\n\u2022 nice answer: the matrix in fact is symmetric and orthogonal (+1) \u2013\u00a0G Cab Jun 20 '18 at 15:03\n\nYou write\n\nI calculated the determinant and it is 625\n\nwhich is enough to conclusively answer.\n\nFor option $\\mathbf A$ we have $\\frac{1}{det(A)} = det(A^{-1}) {\\space\\overset{\\mathbf A}{=}\\space} det(\\frac{A}{25}) = \\frac{det(A)}{25^4}$. Similarly for the other answers.\n\nHere is a very simple although not as elegant answer:\n\nWhatever the inverse would be, multiplication must be $1$ for diagonal elements, in particular for the first element. We can call it $c_{11}$.\n\nThis shields $\\frac{25}{x}$ where $x$ is the one between options that makes this element 1. So it's a or d. In any case, you need to calculate all elements for choosing between this two options.", "date": "2019-12-08 03:31:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2825002/how-to-compute-this-determinant-as-quickly-as-possible-without-using-any-softwa", "openwebmath_score": 0.9023656249046326, "openwebmath_perplexity": 217.4605094456429, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964173268184, "lm_q2_score": 0.8918110389681662, "lm_q1q2_score": 0.8788765838356354}}
{"url": "http://mathhelpforum.com/calculus/146234-graphing-function.html", "text": "# Math Help - graphing function\n\n1. ## graphing function\n\ni am having trouble graphing functions i missed the lesson an i'm lost for example:\n\n$y=\\frac{x}{x^2-4}$\n\ni know it has V.A at x=-2,x=2 and H.A. at y=0 because the degree of the denominator > degree of numerator\n\nbut where do i go from here?(figuring out odd and even asymptotes, x and y intercepts) etc....\n\n2. Originally Posted by euclid2\ni am having trouble graphing functions i missed the lesson an i'm lost for example:\n\n$y=\\frac{x}{x^2-4}$\n\ni know it has V.A at x=-2,x=2 and H.A. at y=0 because the degree of the denominator > degree of numerator\n\nbut where do i go from here?(figuring out odd and even asymptotes, x and y intercepts) etc....\nYou can get the first derivative to find out on which intervals it is decresing/increasing. You can use the info to know on which side of each vertical asymptote does the function go to (positive or negative infinity).\n\nYou can use the second derivative to get the concavity. that might help.\n\nTo get the x intercept(s), set y = 0 and solve for x.\n\nTo get the y intercept(s), set x = 0 and solve for y.\n\nDon't forget that as the function goes to positive or negative infinity, it must converge to y = 0 (horizontal asymptote, as you've mentioned)..\n\nCheers!\n\n3. Hello, euclid2!\n\nI use a very primitive approach.\nBut it has always worked well for me.\n\nGraph: . $f(x) \\:=\\:\\frac{x}{x^2-4}$\n\ni know it has V.A. at $x = \\pm2$ and H.A. at $y = 0$\n. . because (deg. of denom'r) > (deg of num'r) . . . . Right!\n\nSketch the asymptotes . . .\nCode:\n                      |\n:     |     :\n:     |     :\n:     |     :\n:     |     :\n- - - - - - + - - o - - + - - - - - - -\n-2     |     2\n:     |     :\n:     |     :\n:     |     :\n|\n\nThe only intercept is at (0,0).\n\nThe $x$-axis is partitioned into four intervals.\nEvaluate a point in each interval.\n\n. . $\\begin{array}{ccccc}f(\\text{-}3) &=& -\\frac{3}{5} & \\Rightarrow & (-3,\\:-\\frac{3}{5}) \\\\ \\\\[-3mm]\nf(\\text{-}1) &=& +\\frac{1}{5} & \\Rightarrow & (-1,\\:\\frac{1}{5})\\\\ \\\\[-3mm]\nf(1) &=& -\\frac{1}{3}& \\Rightarrow & (1,\\:-\\frac{1}{3}) \\\\ \\\\[-3mm]\nf(3) &=& +\\frac{3}{5} & \\Rightarrow & (3,\\:\\frac{3}{5})\\end{array}$\n\nPlot these points.\n\nUsing your knowledge of asymptotes, sketch the graph.\n\nCode:\n                :     |     :\n:*    |     :*\n:     |     :\n:     |     :\n: *   |     : *\n:     |     :  o\n:  o  |     :    *\n:   * |     :           *\n- - - - - - - : - - o - - : - - - - - - -\n*          -2     | *   2\n*    :     |  o  :\no  :     |     :\n* :     |   * :\n:     |     :\n*:     |     :\n:     |    *:\n:     |\n\n4. Originally Posted by Soroban\nHello, euclid2!\n\nI use a very primitive approach.\nBut it has always worked well for me.\n\nSketch the asymptotes . . .\nCode:\n                      |\n:     |     :\n:     |     :\n:     |     :\n:     |     :\n- - - - - - + - - o - - + - - - - - - -\n-2     |     2\n:     |     :\n:     |     :\n:     |     :\n|\n\nThe only intercept is at (0,0).\n\nThe $x$-axis is partitioned into four intervals.\nEvaluate a point in each interval.\n\n. . $\\begin{array}{ccccc}f(\\text{-}3) &=& -\\frac{3}{5} & \\Rightarrow & (-3,\\:-\\frac{3}{5}) \\\\ \\\\[-3mm]\nf(\\text{-}1) &=& +\\frac{1}{5} & \\Rightarrow & (-1,\\:\\frac{1}{5})\\\\ \\\\[-3mm]\nf(1) &=& -\\frac{1}{3}& \\Rightarrow & (1,\\:-\\frac{1}{3}) \\\\ \\\\[-3mm]\nf(3) &=& +\\frac{3}{5} & \\Rightarrow & (3,\\:\\frac{3}{5})\\end{array}$\n\nPlot these points.\n\nUsing your knowledge of asymptotes, sketch the graph.\n\nCode:\n                :     |     :\n:*    |     :*\n:     |     :\n:     |     :\n: *   |     : *\n:     |     :  o\n:  o  |     :    *\n:   * |     :           *\n- - - - - - - : - - o - - : - - - - - - -\n*          -2     | *   2\n*    :     |  o  :\no  :     |     :\n* :     |   * :\n:     |     :\n*:     |     :\n:     |    *:\n:     |\nthis is great. does it matter which numerical values i plug in to obtain other points on the graph, or in other words why did you use f(-1,1,-3,3) besides the fact that these are the values that the function approaches around the asymptotes?", "date": "2015-08-04 16:00:19", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/146234-graphing-function.html", "openwebmath_score": 0.6276595592498779, "openwebmath_perplexity": 1906.2254142698216, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357591818725, "lm_q2_score": 0.8976952886860978, "lm_q1q2_score": 0.8788757884727839}}
{"url": "https://forum.usaco.guide/t/alternate-solution-to-the-meeting-place-cannot-be-changed/1020", "text": "# Alternate Solution to The Meeting Place Cannot Be Changed\n\nHi,\n\nI would like to share an alternate solution to the Codeforces problem The Meeting Place Cannot Be Changed. My solution is in Java, but it can be translated to other languages. I enjoyed solving this problem.\n\nThis problem is in fact very similar to the LeetCode problem Peak Index in a Mountain Array.\n\nThe official solution does binary search on the minimum time needed for the friends to meet, but my solution does binary search on the optimal place to meet.\n\nHow do we do binary search if the positions are non-integers? We simply multiply each coordinate by 10^6, then find the optimal time for integer meeting points, then divide by 10^6. This will ensure that the difference between the meeting place returned and the actual meeting place is \\leq 10^{-6}, and the difference between the minimum times is \\leq 10^{-6}.\n\nMy solution relies on the two following key observations:\n\n1. The minimum times to meet every spot on the line decreases, then increases as you sweep the line.\n2. No three points contain the same minimum times.\n\nThe minimum time for each meeting place is the maximum times it takes for each friend to travel to that meeting place. We can compute the minimum time for an arbitrary meeting place in O(n) time.\n\nNow, we can do binary search on the optimal meeting place while keeping track of the minimum time. For each meeting place \\text{mid} we check, we find the minimum times for it and adjacent meeting places, and compute the total running minimum of all times. We break out of the search if \\text{minTime(mid - 1)} > \\text{minTime(mid)} < \\text{minTime(mid + 1)}; otherwise, we use the information to narrow the bound of possible optimal meeting places.\n\nOnce we break out of the loop, we have the minimum meeting time and hence the answer.\n\nMy code below:\n\nimport java.io.*;\nimport java.util.*;\n\npublic class TheMeetingPlaceCannotBeChanged {\npublic static void main(String[] args) throws IOException {\n\n// positions multiplied by 10^6 so we can work with integers\nlong[] positions = new long[numFriends];\nfor (int i = 0; i < numFriends; i++) {\npositions[i] = Long.parseLong(stp.nextToken()) * 1000000L;\n}\nint[] speeds = new int[numFriends];\nfor (int i = 0; i < numFriends; i++) {\nspeeds[i] = Integer.parseInt(sts.nextToken());\n}\n// Find min, max position\nlong minPosition = Long.MAX_VALUE;\nlong maxPosition = Long.MIN_VALUE;\nfor (long p : positions) {\nminPosition = Math.min(minPosition, p);\nmaxPosition = Math.max(maxPosition, p);\n}\n// Binary search to find minimum time\nlong low = minPosition;\nlong high = maxPosition;\ndouble minTime = Double.MAX_VALUE;\nwhile (low <= high) {\nlong mid = (low + high) / 2;\ndouble midTime = time(positions, speeds, mid);\ndouble lowerTime = time(positions, speeds, mid - 1);\ndouble higherTime = time(positions, speeds, mid + 1);\nminTime = Math.min(Math.min(minTime, midTime), Math.min(lowerTime, higherTime));\nif (midTime < lowerTime && midTime < higherTime) break;\nelse if (lowerTime > midTime || midTime > higherTime) {\nlow = mid + 1;\n} else {\nhigh = mid - 1;\n}\n}\nSystem.out.printf(\"%.6f\\n\", minTime / 1000000);\n}\n// Time is maximum time of all friends to get to point\nstatic double time(long[] positions, int[] speeds, long position) {\ndouble maxTime = Double.MIN_VALUE;\nfor (int i = 0; i < speeds.length; i++) {\nmaxTime = Math.max(maxTime, (double) Math.abs(positions[i] - position) / speeds[i]);\n}\nreturn maxTime;\n}\n}\n\n\nThanks,\nrayfish\n\nGood solution as well, this is also known as ternary search, which works a little faster by cutting into 3 equal size chunks. https://en.wikipedia.org/wiki/Ternary_search\n\nYou should add this to USACO Guide as one of the user solutions .", "date": "2022-08-14 11:44:52", "meta": {"domain": "usaco.guide", "url": "https://forum.usaco.guide/t/alternate-solution-to-the-meeting-place-cannot-be-changed/1020", "openwebmath_score": 0.3240807056427002, "openwebmath_perplexity": 3056.166070352444, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975946451303227, "lm_q2_score": 0.9005297934592089, "lm_q1q2_score": 0.8788688562193429}}
{"url": "http://math.stackexchange.com/questions/185260/is-the-number-of-primes-congruent-to-1-mod-6-equal-to-the-number-of-primes-congr", "text": "# Is the number of primes congruent to 1 mod 6 equal to the number of primes congruent to 5 mod 6?\n\n(I know that there's an infinity of primes congruent to 5 mod 6, but I don't know if there is an infinity of primes congruent to 1 mod 6.)\n\nBut what I'd really like to know is whether or not the number of primes \u22611 mod 6 less than a given n can be said to be asymptotically equal to the number of primes \u22615 mod 6 less than that n. So the \"total\" number of primes \u22611 mod 6 would be equal to the number of primes \u22615 mod 6.\n\n(I hope this is understandable. I'd also like to know how to phrase this inquiry in a more standard fashion, if someone would tell me how to fix it.)\n\n-\nYou might ask like this: let $\\pi_1(n)$ be the number of primes of the form $6k+1$ that are less than $n$, and $\\pi_5(n)$ similarly. Then you can ask whether $\\pi_1(n)/\\pi_5(n)$ approaches a limit as $n$ increases without bound, and if so, whether that limit is 1. \u2013\u00a0MJD Aug 22 '12 at 1:13\nThat's precisely what I mean; thank you. \u2013\u00a0Annick Aug 22 '12 at 1:14\n(You didn't ask for this, but for an elementary proof that there are infinitely many primes congruent to $1 \\bmod 6$, consider the possible primes dividing a number of the form $k^2 - k + 1$.) \u2013\u00a0Qiaochu Yuan Aug 22 '12 at 1:28\nGot it, thanks. Great to know. \u2013\u00a0Annick Aug 22 '12 at 2:35\n\nYes, this is true. A more general statement follows from a suitably strong form of Dirichlet's theorem on arithmetic progressions, namely that asymptotically the proportion of primes which are congruent to $a \\bmod n$ (for $\\gcd(a, n) = 1$) is $\\frac{1}{\\varphi(n)}$.\n\nIn the particular case that $n = 6$ it is possible to give a more elementary proof of a weaker result. Define the Dirichlet L-function\n\n$$L(s, \\chi_6) = \\sum_{n=1}^{\\infty} \\frac{\\chi_6(n)}{n^s}$$\n\nwhere $\\chi_6$ is the unique nontrivial Dirichlet character $\\bmod 6$. This takes the form $\\chi_6(n) = 1$ if $n \\equiv 1 \\bmod 6$, $\\chi_6(n) = -1$ if $n \\equiv 5 \\bmod 6$, and $\\chi_6(n) = 0$ otherwise. The Euler product of this L-function is\n\n$$L(s, \\chi_6) = \\prod_p \\left( \\frac{1}{1 - \\chi_6(p) p^{-s}} \\right) = \\prod_{p \\equiv 1 \\bmod 6} \\left( \\frac{1}{1 - p^{-s}} \\right) \\prod_{p \\equiv 5 \\bmod 6} \\left( \\frac{1}{1 + p^{-s}} \\right).$$\n\nIt is possible to explicitly evaluate $L(1, \\chi_6)$ and in particular to show that it is not zero; in fact,\n\n$$L(1, \\chi_6) = \\int_0^1 \\frac{1 - x^5}{1 - x^6} \\, dx$$\n\nand this can be evaluated using partial fractions (but note that the integrand is always positive so this number is definitely positive). So we conclude that\n\n$$-\\log L(s, \\chi_6) = \\sum_{p \\equiv 1 \\bmod 6} \\log (1 - p^{-s}) + \\sum_{p \\equiv 5 \\bmod 6} \\log (1 + p^{-s})$$\n\napproaches a nonzero constant as $s \\to 1$ (if summed in the appropriate order) even though the first and second terms separately approach $\\mp \\infty$. So the contributions coming from primes in each residue class cancel out asymptotically. This is not quite as strong as the desired statement, though; if you fill in all the details in what I've said you'll show that the Dirichlet density of the primes congruent to $\\pm 1 \\bmod 6$ are the same but this should still be true for the natural density and this requires a further argument (I am not sure how much further, though).\n\nFor more details see any book on analytic number theory, e.g. Apostol.\n\n-\nThanks so much for this highly detailed answer and explanatory links. \u2013\u00a0Annick Aug 22 '12 at 1:27\n\nLook up \"prime races\" at http://www.dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf.\n\nTheir conclusion: \"It does seem that \u201ctypically\u201d qn + a has fewer primes than qn + b if a is a square modulo q while b is not.\"\n\nSo, since 1 is a quadratic residue mod 6, while 5 is not, there will typically be more primes of the form 6n+5 than 6n+1 (though their ratio does tend to 1).\n\n-", "date": "2016-05-02 02:15:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/185260/is-the-number-of-primes-congruent-to-1-mod-6-equal-to-the-number-of-primes-congr", "openwebmath_score": 0.9749889969825745, "openwebmath_perplexity": 162.25584171673646, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97594644290792, "lm_q2_score": 0.9005297947939938, "lm_q1q2_score": 0.8788688499617974}}
{"url": "https://math.stackexchange.com/questions/2802217/does-the-sequence-a-n-n-in-mathbbn-have-a-convergent-subsequence", "text": "# Does the sequence $(a_n)_{n\\in\\mathbb{N}}$ have a convergent subsequence?\n\nDoes the sequence $(a_n)_{n\\in\\mathbb{N}}$ have a convergent subsequence?\n\n$$a_n= \\begin{cases} \\sin(n), & \\text{if n is odd} \\\\ n, & \\text{if n is even} \\end{cases}$$\n\nI understand that $\\sin(n)$ is a bounded sequence so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence. Not sure where to go from here assuming I am along the right lines?\n\n\u2022 \"so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence.\" Yes. \"Not sure where to go from here\". You're not sure where to go from having reached an irrefutable conclusion? You don't have to go anywhere. You were ask whether it has a convergent subsequence and you answered that it did. So ... where else is there to go? \u2013\u00a0fleablood May 30 '18 at 18:39\n\u2022 @fleablood I think the question asker was missing a subtlety for this question: $\\sin(n)$ has a convergent subsequence, but it is not a subsequence of $(a_n)$ - the answer is, of course, to show that $\\sin(n)$ for odd $n$ has a convergent subsequence, but the work presented in the question is not complete as it stands. \u2013\u00a0Milo Brandt May 30 '18 at 18:57\n\u2022 It's not that subtle is it? $a_n$ where $n$ is odd s $a_n= \\sin n$ is a subsequence. That subsequence is bounded so there is a subsequence of the subsequence that is bounded. A subsequence of a subsequence is a subsequence. You have to avoid being lazy and stupid ($a_n$ itself is not bounded and $a_n; odd$ is not $\\{\\sin n\\}$ so you have to claim that $a_n; odd$ has the subsequence; not $\\{\\sin n\\}$) but that's not hard to do. \u2013\u00a0fleablood May 31 '18 at 0:34\n\u2022 \"I understand that sin(n) is a bounded sequence so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence.\" Okay, there are two observations you must make. $\\{a_n\\}$ is not bounded (but the subsequence $\\{a_{2k+1}\\} is$. And stating $\\{\\sin n\\}$ has a convergent subsequence isn't enough as $\\{\\sin n\\} \\not \\subset \\{a_n\\}$. But as $\\{a_{2k+1} \\}\\subset \\{\\sin n\\}$ is bounded there is a convergent $\\{b_j\\}\\subset \\{a_{2k+1}\\} \\subset \\{\\sin n\\}$. That's fairly obvious and easy, but enough to make a question worth 5 marks. \u2013\u00a0fleablood May 31 '18 at 0:41\n\nLet's construct the subsequence which converges to $0$.\n\nConsider convergents of $\\pi$ continued fraction: $\\left\\{ \\dfrac{n_j}{d_j},\\; j\\in\\mathbb{N} \\right\\}$. And focus on its numerators: $$n_1 = 3,\\\\ n_2 = 22, \\\\ n_3 = 333, \\\\ n_4 = 355, \\\\ n_5 = 103993, \\\\ n_6 = 104348, \\\\ \\vdots$$\n\nThen $$\\lim_{j\\to \\infty} \\sin n_j =0.$$\n\nIndeed, easy to estimate: $$|\\sin n_j| = \\left| \\sin\\left( n_j - \\pi d_j \\right) \\right| \\approx \\left| n_j - \\pi d_j \\right|=d_j\\left| \\dfrac{n_j}{d_j} - \\pi \\right| < \\dfrac{1}{d_{j+1}}.$$\n\nAnd it remains to show that the sequence $\\{n_j, \\; j\\in\\mathbb{N}\\}$ contains infinite number of odd $n_j$.\n\nIf continued fraction (of $\\pi$) denote as $[a_0;a_1, a_2, \\ldots, a_j, \\ldots]$, then $$n_{j+1} = a_{j+1}n_j + n_{j-1}.$$\n\nThere are $8$ possibilities:\n\n\\begin{array}{|l|l|l|l|} \\hline (n_{j-1}, n_j) & a_{j+1} & n_{j+1} & \\rightarrow (n_j, n_{j+1})\\\\ \\hline (odd, odd) & odd & even & \\rightarrow (odd, even) \\\\ (odd, odd) & even & odd & \\rightarrow (odd, odd) \\\\ \\hline (odd, even) & odd & odd & \\rightarrow (even, odd) \\\\ (odd, even) & even & odd & \\rightarrow (even, odd) \\\\ \\hline (even, odd) & odd & odd & \\rightarrow (odd, odd) \\\\ (even, odd) & even & even & \\rightarrow (odd, even) \\\\ \\hline (even, even) & odd & even & \\rightarrow (even, even) \\\\ (even, even) & even & even & \\rightarrow (even, even) \\\\ \\hline \\end{array}\n\nwhich show that if at least one of $n_j$ is odd, then there is infinite number of odd $n_j$.\n\nA subsequence of a subsequence is a subsequence. The sequence of values of $a_n$ for which $n$ is odd is a subsequence, and you've already said that that has a convergent subsequence. That convergent subsequence is a subsequence of a subsequence of the original sequence. Therefore it is a subsequence of the original sequence.\n\nYou wrote (twice) \u201cfunction\u201d where you should have written \u201csequence\u201d. Other than that, you are doing fine.\n\n\u2022 Edited thanks. The question is worth 5 marks, I don't think this answer justifies this. Where do I go from here? Thanks. \u2013\u00a0Ben Jones May 30 '18 at 18:36\n\u2022 You know that the sequence $(a_{2n-1})_{n\\in\\mathbb N}$ is bounded. Therefore, it has a convergent subsequence, which will be a subsequence of the original sequence, of course. \u2013\u00a0Jos\u00e9 Carlos Santos May 30 '18 at 18:37\n\u2022 \"Where do I go from here?\" Well, you could go to breakfast. Or Tahiti, I hear Tahiti is nice this time of year. ... you've answered the question. You don't need to go anywhere. \u2013\u00a0fleablood May 30 '18 at 18:42\n\u2022 I'm off to Tahiti then. \u2013\u00a0Ben Jones May 31 '18 at 18:36", "date": "2019-07-16 22:40:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2802217/does-the-sequence-a-n-n-in-mathbbn-have-a-convergent-subsequence", "openwebmath_score": 0.9968123435974121, "openwebmath_perplexity": 345.08142212653735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347853343058, "lm_q2_score": 0.9019206778476933, "lm_q1q2_score": 0.8788628821070886}}
{"url": "http://mathhelpforum.com/algebra/122455-dimensions-kindle-electronic-pad-screen.html", "text": "1. ## Dimensions of Kindle (Electronic pad with a screen)\n\nI created this problem and wanted to check my answers with the MathForum.\n\nx = length\ny = width\n\nScenario:\nThere are two Kindles.\nK(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in\nK(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in\n\na. How much bigger is K(2)'s SA compared to K(1)'s SA?\n\nb. Write an equation that represents the maximum and minimum value of\nx and y when the SA of K(3) is 42.66 in.\nmy answer: Either $x=42.66*y$ or $y=42.66*x$\n\nNote: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.\n\nc. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.\n\nmy answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right) and $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.\n\n2. Originally Posted by Masterthief1324\nI created this problem and wanted to check my answers with the MathForum.\n\nx = length\ny = width\n\nScenario:\nThere are two Kindles.\nK(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in\nK(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in\n\na. How much bigger is K(2)'s SA compared to K(1)'s SA?\nYes, that is correct.\n\nb. Write an equation that represents the maximum and minimum value of\nx and y when the SA of K(3) is 42.66 in.\nmy answer: Either $x=42.66*y$ or $y=42.66*x$\nWhat? You said there were two kinds of \"kindle\", K(1) and K(2). Where did K(3) come from? Without any information about K(3) it is impossible to answer this. Did you leave something out?\n\nNote: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.\n\nc. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.\n\nmy answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right)\nNo, you didn't. \" $x^2+ y^2= z^2$\" does NOT reduce to \"x+ y= z\". For example, $4^2+ 3^2= 16+ 9= 25= 5^2$ but $4+ 3\\ne 5$.\n\nand $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.\nIf $x^2+ y^2= 36$ and xy= 42.66, then y= 42.66/x so $x^2+ \\frac{1819.8756}{x^2}= 36$. Multiplying both sides by $x^2$, $x^4+ 1819.8756= 36x^2$ or $x^4- 36x^2+ 1819.8756= 0$. If you let $u= x^2$, this becomes the quadratic equation $u^2- 36u+ 1819.8756= 0$. Solve that for u, then take the square root to find x.\n\nBut are you sure you have interpreted the question correctly? The problem said \"find maximum and minimum\" values, not specific values.\n\n3. Sorry, I did make that last problem confusing.\nI hope this clarifies it.\n\nK(3) is a hypothetical Kindle device with a surface area of $42.66in^2$ and a diagonal of 6 in.\n\nc1. Can you have more than one dimensions that satisfy the SA & the diagonal length of 6in? Why or why not?\nmy answer: Not if the diagonal is 6 inch -- there is only 1 value of x and y (length and width)\n\nc2. Can you represent c1. with and equation?\nmy answer: Since the diagonal is 6, if I was to use the Pythagorean theorem, there is only 1 value of x and y that would fit the specifications.\n\n(specifications)\nYou know indefinitely that the K(3) has a surface area of 42.66 and a diagonal of 6 in:\n\nmy equation:\nx^2 + y^2 = 6 (is it 6 or 6^2?) --> y=\u221a(z^2 - x^2)\nx*y = 42.66\n\nIf I was to graph the two equations, will where they intersect be the dimensions that fit the specifications?\n\n4. b. Write an equation that represents the maximum and minimum value of\nx and y when the SA of K(3) is 42.66 in.\nIf K(3) has a surface area of 42.66, then you know that\n\n$xy=42.66$\n\nsince the diagonal is 6 inches, you know that\n\n$x^2 + y^2 = 6^2$ (draw a picture to understand why it is 6^2 and not 6)\n\nYou have two equations, you can try to solve them both for y and graph them to approximate a solution\n\n$y=42.66/x$\n$y=sqrt(36-x^2)$\n\nOnce you look at this graph you will realize that there is no solution to your problem\n\nK(3) is physically impossible\n\n---------------------------------------\nTo understand this, think of the kindle with diagonal 6 that has the largest surface area, this kindle would be square. It would have x = y = sqrt (18), so its largest surface area would be xy=18\n\nI created this problem and wanted to check my answers with the MathForum.\nTo fix your problem, you must either increase the size of the diagonal or reduce the surface area of K(3) to make the two curves intersect", "date": "2016-12-08 22:12:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/122455-dimensions-kindle-electronic-pad-screen.html", "openwebmath_score": 0.7166997194290161, "openwebmath_perplexity": 479.1411470715296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9805806495475397, "lm_q2_score": 0.8962513655129178, "lm_q1q2_score": 0.8788467461525263}}
{"url": "https://math.stackexchange.com/questions/2192071/number-of-ways-in-which-boys-and-girls-sit-alternately-if-six-boys-and-six-girls", "text": "# Number of ways in which boys and girls sit alternately if six boys and six girls sit randomly?\n\nSix boys and six girls sit in a row randomly. What is the total number of ways in which the boys and the girls sit alternately?\n\nMy attempt: Consider these six seats _ _ _ _ _ _\n\nThe number of ways to arrange 6 boys in 6 places is 6! Now 7 gaps are created between these 6 seats. So, we can select any 6 of these 7 gaps and make girls sit there. There are 7C6 * 6! ways to do that (since the girls can shuffle amongst themselves).\n\nHence the total number of ways to make boys and girls sit alternately should be 6! 7C1 * 6! but the answer is 2*6!*6!.\n\nWhat am I missing here?\n\n\u2022 You can't choose any $6$...the unfilled gap must be either first or last.\n\u2013\u00a0lulu\nMar 18, 2017 at 11:34\n\u2022 Note \"alternately\". Only two of the seven ways to pick six of the gaps result in an alternating pattern, the rest have two boys next to each other. Mar 18, 2017 at 11:34\n\u2022 Oh right! Thank you so much. Mar 18, 2017 at 11:35\n\nYou are implicitly assuming that there are 13 seats for $6+6=12$ people. It is likely that the original question is making the implicit assumption that there are only 12 seats. In this latter case, when a boy sits first, there $6! \\cdot 6!$ ways to sit the others. Multiply this by 2 to cover the case where the first seater is a girl.\n\nYou are over-counting.\n\nYou are also counting the cases where extreme 2 gaps are both occupied and out of 5 gaps in middle 4 are occupied.\n\nFor example, let's first arrange the boys: B_B_B_B_B_B = 6! ways.\n\nWe are left with 7 gaps so we select 6 among them and permute the girls so (7C6 6! ways.)\n\nHence according to you answer is 6! . 7C6 . 6! = 7.6!.6!\n\nBut it also counts the cases like GBGBGBBGBGBG two boys are sitting together and alternate pattern is disturbed so we eliminate these unfavorable cases by bijection. So, Unfavorable ways= 6! . 2C2 . 5C4. 6! = 5.6!.6! So hence total ways of favourable permutations = 7.6!.6! - 5.6!.6! =2.6!.6! (The required answer)", "date": "2023-02-02 15:34:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2192071/number-of-ways-in-which-boys-and-girls-sit-alternately-if-six-boys-and-six-girls", "openwebmath_score": 0.8411925435066223, "openwebmath_perplexity": 222.1904885425274, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419703960398, "lm_q2_score": 0.8887587979121383, "lm_q1q2_score": 0.8788420009342546}}
{"url": "https://math.stackexchange.com/questions/2189700/how-do-you-calculate-22222", "text": "# How do you calculate $2^{2^{2^{2^{2}}}}$?\n\nFrom information I have gathered online, this should be equivalent to $2^{16}$ but when I punch the numbers into this large number calculator, the number comes out to be over a thousand digits. Is the calculator wrong or is my method wrong?\n\n\u2022 The number is equal to $2^r$ where $r=2^{16}$. \u2013\u00a0Aravind Mar 16 '17 at 17:17\n\u2022 this is quiet not clear use parentesis \u2013\u00a0Dr. Sonnhard Graubner Mar 16 '17 at 17:17\n\u2022 If parenthesis are not used, it is assumed that exponents are evaluated from top-down as opposed to bottom-up. $(a^b)^c=a^{bc}\\neq a^{(b~^c)}$. Exponentiation is not associative. The answer of $2^{16}$ is for if it were evaluated bottom-up as (((2^2)^2)^2)^2 instead of top-down which is 2^(2^(2^(2^2))) which is much larger \u2013\u00a0JMoravitz Mar 16 '17 at 17:22\n\u2022 Possible duplicate of What is the order when doing $x^{y^z}$ and why? \u2013\u00a0Simply Beautiful Art Mar 16 '17 at 21:11\n\u2022 My TI calculator has an inline option of showing this as 2^2^2^2^2, which evaluates as 65536 (left to right evaluation). But when I switch to math print mode it shows the tower of powers, which it tries, but fails, to evaluate top down...overflow. I'm not saying that a TI calculator is the final arbiter of math truth, but it is what I get. \u2013\u00a0paw88789 Mar 17 '17 at 2:22\n\n## 6 Answers\n\n$$2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}\\tag1$$\n\nThe number of digits:\n\n$$\\mathcal{A}=1+\\lfloor\\log_{10}\\left(2^{65536}\\right)\\rfloor=19729\\tag2$$\n\n\u2022 This is all true enough, but it doesn't really answer the question as actually asked. \u2013\u00a0hBy2Py Mar 16 '17 at 18:59\n\u2022 It currently does answer the question in the title. Equation 1 shows how to evaluate the expression. It also implies, though doesn't explicitly states that the answer the asker's method (or at least the answer he got using the method) was wrong. \u2013\u00a0Fluidized Pigeon Reactor Mar 16 '17 at 21:30\n\u2022 The title is a reference for searching and tracking. The question is the text beneath that. NAA. \u2013\u00a0Nij Mar 17 '17 at 8:43\n\nWhat you have is a power tower or \"tetration\" (defined as iterated exponentiation). From the latter link, you would most benefit from this brief excerpt on the difference between iterated powers and iterated exponentials.\n\nThe comment by JMoravitz really gets to the heart of the matter, namely that exponential towers must be evaluated from top to bottom (or right to left). There actually is a notation for your particular question: ${}^52=2^{2^{2^{2^{2}}}}$. You really need to look at ${}^42$ before you get something meaningful because, unfortunately, $${}^32=2^{2^{2}}=2^4=16=4^2=(2^2)^2;$$ however, $${}^42=2^{2^{2^{2}}}=2^{2^{4}}=2^{16}\\neq2^8=(4^2)^2=((2^2)^2)^2.$$ Hence, your method is wrong, but everything in those links should provide more than enough for you to become comfortable with tetration.\n\n\u2022 It seems that the Knuth's notation $2\\uparrow\\uparrow n$ has gained popularity over Rucker's one $^n2$ nowadays. \u2013\u00a0zwim Mar 16 '17 at 17:47\n\u2022 @Daniel W. Farlow look like you beat me to posting. Got to love it when that happens. \u2013\u00a0Sentinel135 Mar 16 '17 at 17:53\n\u2022 @zwim I actually prefer Rucker's notation, but I do see the appeal of Knuth's very unambiguous notation. \u2013\u00a0Daniel W. Farlow Mar 16 '17 at 17:55\n\nBy convention, the meaning of things written $\\displaystyle a^{b^{c^d}}$ without brackets is $\\displaystyle a^{\\left(b^{\\left(c^d\\right)}\\right)}$ and not $\\left(\\left(a^b\\right)^c\\right)^d$.\n\nThis is because $\\left(\\left(a^b\\right)^c\\right)^d$ equals $a^{b\\cdot c\\cdot d}$ anyway, so it makes pragmatic sense to reserve the raw power-tower notation $\\displaystyle a^{b^{c^d}}$ for the case that doesn't have an alternative notation without parentheses.\n\nAs others have explained, $\\displaystyle 2^{2^{2^{2^2}}}$ interpreted with this convention is $2^{65536}$, a horribly huge number, whereas $(((2^2)^2)^2)^2$ is $2^{16}=65536$, as you compute.\n\n\u2022 \"Horribly huge number\" rubs me the wrong way in this context. Huge numbers are horrible when they count something that you don't want. The number's not horrible when it's your bank balance! (Actually it might be. If you put together that many pennies it'd probably collapse into a black hole the size of the Milky Way.) \u2013\u00a0Matt Samuel Mar 16 '17 at 22:44\n\u2022 @MattSamuel: Milky Way? This number exceeds the number of Planck volumes in the observable universe ... to the hundredth power! \u2013\u00a0Henning Makholm Mar 17 '17 at 1:24\n\u2022 Do you really expect me to do coordinate transformations in my head while strapped to a centrifuge??? \u2013\u00a0Matt Samuel Mar 17 '17 at 1:25\n\u2022 2^65536 really isn't a horribly huge number when you consider that its binary representation fits in a mere 8 KB of memory. I mean, we are already using RSA moduli around 2^4096 already, so this number is only around 10\u00d7 longer. \u2013\u00a0Nayuki Mar 17 '17 at 3:28\n\u2022 Actually all finite numbers are pretty small because all but a finite number of the rest are bigger. Come to that all transfinite numbers are pretty small as well. \u2013\u00a0Martin Rattigan Apr 2 '17 at 21:55\n\nI would calculate it using Maxima (which evaluates repeat exponentiation correctly, right-to-left), since there is no point wasting brain cells on something that a machine can do:\n\n$maxima Maxima branch_5_39_base_2_gc9edaee http://maxima.sourceforge.net using Lisp GNU Common Lisp (GCL) GCL 2.6.12 Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. 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4717124577965048175856395072895337539755822087777506072339445587895905719156736 (%i2) bfloat(%); (%o2) 2.003529930406846b19728 (%i3)  Of course if I just wanted to estimate the magnitude of the number without resorting to the use of arbitrary precision computer software, I'd note that the exponent is$2^{2^{2^2}}=2^{(2^{(2^2)})}=2^{(2^4)}=2^{16}=65536$; multiplying it by$\\log_{10} 2\\sim 0.30103$gives$19728.302$, so the result is approximately$10^{0.302}\\times 10^{19728}\\sim 2\\times 10^{19728}$. \u2022 Wow. Someone actually posted the digits. \u2013 Matt Samuel Mar 16 '17 at 23:36 \u2022 There is a mistake in the computed result. The 2 in the middle should be a 3. Probably a typo. \u2013 augustin Mar 17 '17 at 3:43 \u2022 \"2 in the middle\"... can you be a bit more specific? \u2013 Viktor Toth Mar 17 '17 at 3:48 \u2022 @augustin: The digit in the middle is an 8. And yes, I checked it; with the help of the computer, of course. Given that the total number of digits (19729) is odd, the digit in the middle is well defined (it is the digit which is preceded and followed by the same number of digits). \u2013 celtschk Mar 17 '17 at 7:59 \u2022 @celtschk now you sir are fun at parties! \u2013 Pierre Arlaud Mar 17 '17 at 8:44 This looks an awfully close to what is known as a tetration (a.k.a. power tower). This is$^{(k)}a=a^{^{(k-1)}a}$where$^1a=a$. For numbers greater than one, these usually get really big really fast, and faster than exponents do. So in your case, you have$^52=2^{2^{16}}$. Now if you want to see an interesting one look at$\\lim_{k\\to \\infty}\\;^{(k)}(\\sqrt{2})$. \u2022 Wouldn't that work for any$\\sqrt[n]{n}$? \u2013 Random832 Mar 16 '17 at 20:04 \u2022 Wouldn't what work for any$\\sqrt[n]{n}\\$? I never said the answer. And yes, but it depends on what you think the answer is. ;D \u2013\u00a0Sentinel135 Mar 16 '17 at 20:13\n\u2022 Looks like that limit goes to infinity really quickly. \u2013\u00a0Joshua Mar 16 '17 at 20:36\n\u2022 @Joshua are you sure? can you try and prove it? \u2013\u00a0Sentinel135 Mar 16 '17 at 20:56\n\u2022 Darn. I hate accumulated roundoff. \u2013\u00a0Joshua Mar 16 '17 at 21:24\n\nYour equation can be simplified using Knuth's up arrow notation. \\begin{equation*} 2^{2^{2^{2^2}}} = 2 \\uparrow\\uparrow 5 \\end{equation*}\n\n(because we can calculate tetration with Knuth's up arrow notation)\n\nBy definition of Knuth's up arrow notation, You can get this result.\n\n\\begin{equation*} 2\\uparrow\\uparrow5 = 2^{(2^{(2^{(2^2)})})} \\end{equation*}\n\nAnd according to web2.0calc,\n\n\\begin{equation*} 2^{(2^{(2^{2})})} = 65536 \\end{equation*}\n\nFinally, the answer would be:\n\n\\begin{equation*} 2^{65536} \\end{equation*}\n\n(correct me if I'm wrong, this was my first answer on Mathematics SE)", "date": "2019-11-13 00:42:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2189700/how-do-you-calculate-22222", "openwebmath_score": 0.6473212838172913, "openwebmath_perplexity": 1243.9051674087075, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708032626702, "lm_q2_score": 0.8991213718636754, "lm_q1q2_score": 0.8787749774490344}}
{"url": "https://math.stackexchange.com/questions/787909/block-matrix-multiplication/4021707", "text": "# block matrix multiplication\n\nIf $A,B$ are $2 \\times 2$ matrices of real or complex numbers, then\n\n$$AB = \\left[ \\begin{array}{cc} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{array} \\right]\\cdot \\left[ \\begin{array}{cc} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{array} \\right] = \\left[ \\begin{array}{cc} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\\\ a_{21}b_{11}+a_{22}b_{21} & a_{22}b_{12}+a_{22}b_{22} \\end{array} \\right]$$\n\nWhat if the entries $a_{ij}, b_{ij}$ are themselves $2 \\times 2$ matrices? Does matrix multiplication hold in some sort of \"block\" form ?\n\n$$AB = \\left[ \\begin{array}{c|c} A_{11} & A_{12} \\\\\\hline A_{21} & A_{22} \\end{array} \\right]\\cdot \\left[ \\begin{array}{c|c} B_{11} & B_{12} \\\\\\hline B_{21} & B_{22} \\end{array} \\right] = \\left[ \\begin{array}{c|c} A_{11}B_{11}+A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\\\\\hline A_{21}B_{11}+A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \\end{array} \\right]$$ This identity would be very useful in my research.\n\n\u2022 Yes it does if the \"blocking\" is \"conforming\". May 9, 2014 at 14:12\n\u2022 Yes. (The blocking is \"confirming\" in the situation you have given.) Discussed in detail in \u00a76.12 of cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf (specifically Exercise 38 and Remark 6.73; search for \"block-matrix notation\" if these numbers change). Dec 4, 2015 at 10:16\n\nIt depends on how you partition it, not all partitions work. For example, if you partition these two matrices\n\n$$\\begin{bmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{bmatrix}, \\begin{bmatrix} a' & b' & c' \\\\ d' & e' & f' \\\\ g' & h' & i' \\end{bmatrix}$$\n\nin this way\n\n$$\\left[\\begin{array}{c|cc}a&b&c\\\\ d&e&f\\\\ \\hline g&h&i \\end{array}\\right], \\left[\\begin{array}{c|cc}a'&b'&c'\\\\ d'&e'&f'\\\\ \\hline g'&h'&i' \\end{array}\\right]$$\n\nand then multiply them, it won't work. But this would\n\n$$\\left[\\begin{array}{c|cc}a&b&c\\\\ \\hline d&e&f\\\\ g&h&i \\end{array}\\right] ,\\left[\\begin{array}{c|cc}a'&b'&c'\\\\ \\hline d'&e'&f'\\\\ g'&h'&i' \\end{array}\\right]$$\n\nWhat's the difference? Well, in the first case, all submatrix products are not defined, like $\\begin{bmatrix} a \\\\ d \\\\ \\end{bmatrix}$ cannot be multiplied with $\\begin{bmatrix} a' \\\\ d' \\\\ \\end{bmatrix}$\n\nSo, what is the general rule? (Taken entirely from the Wiki page on Block matrix)\n\nGiven, an $(m \\times p)$ matrix $\\mathbf{A}$ with $q$ row partitions and $s$ column partitions $$\\begin{bmatrix} \\mathbf{A}_{11} & \\mathbf{A}_{12} & \\cdots &\\mathbf{A}_{1s}\\\\ \\mathbf{A}_{21} & \\mathbf{A}_{22} & \\cdots &\\mathbf{A}_{2s}\\\\ \\vdots & \\vdots & \\ddots &\\vdots \\\\ \\mathbf{A}_{q1} & \\mathbf{A}_{q2} & \\cdots &\\mathbf{A}_{qs}\\end{bmatrix}$$\n\nand a $(p \\times n)$ matrix $\\mathbf{B}$ with $s$ row partitions and $r$ column parttions\n\n$$\\begin{bmatrix} \\mathbf{B}_{11} & \\mathbf{B}_{12} & \\cdots &\\mathbf{B}_{1r}\\\\ \\mathbf{B}_{21} & \\mathbf{B}_{22} & \\cdots &\\mathbf{B}_{2r}\\\\ \\vdots & \\vdots & \\ddots &\\vdots \\\\ \\mathbf{B}_{s1} & \\mathbf{B}_{s2} & \\cdots &\\mathbf{B}_{sr}\\end{bmatrix}$$\n\nthat are compatible with the partitions of $\\mathbf{A}$, the matrix product\n\n$\\mathbf{C}=\\mathbf{A}\\mathbf{B}$\n\ncan be formed blockwise, yielding $\\mathbf{C}$ as an $(m\\times n)$ matrix with $q$ row partitions and $r$ column partitions.\n\n\u2022 Use \\pmatrix{a&b&c\\\\d&e&f\\\\g&h&i} May 9, 2014 at 14:37\n\u2022 @Berci I know that. But getting those horizontal and vertical lines is the difficult part. May 9, 2014 at 14:38\n\u2022 @PandaBear You can use $\\color{blue}{\\text{colors}}$ :) May 9, 2014 at 14:45\n\u2022 @PavelJiranek Your comment is funny looking. May 9, 2014 at 14:46\n\u2022 I think this is wrong. You can't partition both of them same way. If you partition after x rows in first matrix , you've to partition after x columns (not rows ) in the second matrix. Otherwise while multiplying you'll have to multiply mn block with another mn block which is not possible. (you need np block) Try it with your example. Oct 1, 2015 at 18:08\n\nIt is always suspect with a very late answer to a popular question, but I came here looking for what a compatible block partitioning is and did not find it:\n\nFor $$\\mathbf{AB}$$ to work by blocks the important part is that the partition along the columns of $$\\mathbf A$$ must match the partition along the rows of $$\\mathbf{B}$$. This is analogous to how, when doing $$\\mathbf{AB}$$ without blocks\u2014which is of course just a partitioning into $$1\\times 1$$ blocks\u2014the number of columns in $$\\mathbf A$$ must match the number of rows in $$\\mathbf B$$.\n\nExample: Let $$\\mathbf{M}_{mn}$$ denote any matrix of $$m$$ rows and $$n$$ columns irrespective of contents. We know that $$\\mathbf{M}_{mn}\\mathbf{M}_{nq}$$ works and yields a matrix $$\\mathbf{M}_{mq}$$. Split $$\\mathbf A$$ by columns into a block of size $$a$$ and a block of size $$b$$, and do the same with $$\\mathbf B$$ by rows. Then split $$\\mathbf A$$ however you wish along its rows, same for $$\\mathbf B$$ along its columns. Now we have $$A = \\begin{bmatrix} \\mathbf{M}_{ra} & \\mathbf{M}_{rb} \\\\ \\mathbf{M}_{sa} & \\mathbf{M}_{sb} \\end{bmatrix}, B = \\begin{bmatrix} \\mathbf{M}_{at} & \\mathbf{M}_{au} \\\\ \\mathbf{M}_{bt} & \\mathbf{M}_{bu} \\end{bmatrix},$$\n\nand $$AB = \\begin{bmatrix} \\mathbf{M}_{ra}\\mathbf{M}_{at} + \\mathbf{M}_{rb}\\mathbf{M}_{bt} & \\mathbf{M}_{ra}\\mathbf{M}_{au} + \\mathbf{M}_{rb}\\mathbf{M}_{bu} \\\\ \\mathbf{M}_{sa}\\mathbf{M}_{at} + \\mathbf{M}_{sb}\\mathbf{M}_{bt} & \\mathbf{M}_{sa}\\mathbf{M}_{au} + \\mathbf{M}_{sb}\\mathbf{M}_{bu} \\end{bmatrix} = \\begin{bmatrix} \\mathbf{M}_{rt} & \\mathbf{M}_{ru} \\\\ \\mathbf{M}_{st} & \\mathbf{M}_{su} \\end{bmatrix}.$$\n\nAll multiplications conform, all sums work out, and the resulting matrix is the size you'd expect. There is nothing special about splitting in two so long as you match any column split of $$\\mathbf A$$ with a row split in $$\\mathbf B$$ (try removing a block row from $$\\mathbf A$$ or further splitting a block column of $$\\mathbf B$$).\n\nThe nonworking example from the accepted answer is nonworking because the columns of $$\\mathbf A$$ are split into $$(1, 2)$$ while the rows of $$\\mathbf B$$ are split into $$(2, 1)$$.\n\n\u2022 So basically the formula is valid whenever it makes sense? Apr 3, 2021 at 16:10\n\u2022 @Filippo pretty much, though it did not make sense to me at the time which is why I had to go looking for it Apr 5, 2021 at 11:33\n\nThe recipe for multiplication of (scalar) matrices $$(AB)_{i,j}=\\sum_k A_{i,k}B_{k,j}\\tag1$$ is saying: to obtain the $$(i,j)$$ element of $$AB$$, form the dot product as you walk along row $$i$$ of $$A$$ while simultaneously walking down column $$j$$ of $$B$$. In other words,\n\nThe element at row $$i$$, column $$j$$ of $$AB$$ is the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$.\n\nUsing the notation $$A_{i,\\ast}$$ to denote row $$i$$ of $$A$$ and $$B_{\\ast,j}$$ to denote column $$j$$ of $$B$$, this can be restated symbolically as $$(AB)_{i,j}=A_{i,\\ast}B_{\\ast, j}\\ .\\tag2$$\n\nThis all works provided you can match up any row of $$A$$ with any column of $$B$$, which means the number of columns of $$A$$ must equal the number of rows of $$B$$. (This is what it means for $$A$$ and $$B$$ to be conformable.)\n\nYou can generalize (1) to block matrices, i.e., the formula still works when every $$A_{i,k}$$ and $$B_{k,j}$$ is itself a matrix [and the subscripts enumerate rows and columns of blocks], provided every product on the RHS of (1) makes sense. This means the number of columns of $$A_{i,k}$$ must equal the number of rows of $$B_{k,j}$$.\n\nExample: Partition matrix $$A$$ into $$A:=\\pmatrix{A_1 &A_2}$$ and $$B$$ into $$B:=\\pmatrix{B_1\\\\ B_2}$$, where #columns($$A_1$$) = #rows($$B_1$$), and #columns($$A_2$$) = #rows($$B_2$$). Then $$AB=\\pmatrix{A_1 & A_2}\\pmatrix{B_1\\\\ B_2}=A_1B_1 + A_2 B_2\\ .\\tag3$$ To see this, consider the element at row $$i$$, column $$j$$ of $$AB$$. This element is computed as the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$. But row $$i$$ of $$A$$ is just row $$i$$ of $$A_1$$ followed by row $$i$$ of $$A_2$$. Similarly, column $$j$$ of $$B$$ is column $$j$$ of $$B_1$$ atop column $$j$$ of $$B_2$$. Assuming conformability, the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$ is the sum of two pieces: one piece is the product of row $$i$$ of $$A_1$$ with column $$j$$ of $$B_1$$, and the other is the product of row $$i$$ of $$A_2$$ with column $$j$$ of $$B_2$$. In symbols we've argued that $$(AB)_{i,j}=A_{i,\\ast}B_{\\ast,j}=(A_1)_{i,\\ast}(B_1)_{\\ast,j}+(A_2)_{i,\\ast}(B_2)_{\\ast,j}=(A_1B_1)_{i,j} + (A_2B_2)_{i,j},\\tag4$$ and we're done.\n\nExample: Consider conformable partitioned matrices $$C:=\\pmatrix{C_1\\\\ C_2}$$ and $$D:=\\pmatrix{D_1 & D_2}$$. Then $$CD=\\pmatrix{C_1\\\\ C_2}\\pmatrix{D_1 & D_2} =\\pmatrix{C_1D_1 & C_1D_2\\\\C_2D_1&C_2 D_2}.\\tag5$$ Again, this can be seen by considering what happens when you multiply row $$i$$ of $$C$$ with column $$j$$ of $$D$$.", "date": "2022-10-01 14:45:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/787909/block-matrix-multiplication/4021707", "openwebmath_score": 0.8344562649726868, "openwebmath_perplexity": 286.72184653134616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137910906878, "lm_q2_score": 0.894789457685656, "lm_q1q2_score": 0.8786955875698715}}
{"url": "http://mathhelpforum.com/calculus/88164-challenging-integral.html", "text": "# Math Help - Challenging Integral\n\n1. ## Challenging Integral\n\nI was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.\n\nSo for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:\n\n$\\int_2^4\\frac{\\sqrt{\\ln\\!\\left(9-x\\right)}\\,dx}{\\sqrt{\\ln\\!\\left(9-x\\right)}+\\sqrt{\\ln\\!\\left(x+3\\right)}}$\n\nAll I know is that this equals 1 but I don't know how to get it. XD\n\nAny input or suggestions would be appreciated!!\n\n2. Originally Posted by Chris L T521\nI was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.\n\nSo for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:\n\n$\\int_2^4\\frac{\\sqrt{\\ln\\!\\left(9-x\\right)}\\,dx}{\\sqrt{\\ln\\!\\left(9-x\\right)}+\\sqrt{\\ln\\!\\left(x+3\\right)}}$\n\nAll I know is that this equals 1 but I don't know how to get it. XD\n\nAny input or suggestions would be appreciated!!\nConsider\n\n$\n\\int_2^4 \\frac{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = \\int_2^4 1 = 2\n$\n\nExpanding gives\n\n$\n\\int_2^4 \\frac{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = \\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx$\n$+ \\int_2^4 \\frac{ \\sqrt{\\ln x+3} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx$\n\nLet $x = 6-u$ in the second integral\n\n$\\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx$ $- \\int_4^2 \\frac{ \\sqrt{\\ln 9-u} }{\\sqrt{\\ln 3+u} + \\sqrt{\\ln 9-u}}\\,dx$\n\n$= 2 \\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = 2$\n\n$\\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = 1$\n\n3. Originally Posted by danny arrigo\nConsider\n\n$\n\\int_2^4 \\frac{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = \\int_2^4 1 = 2\n$\n\nExpanding gives\n\n$\n\\int_2^4 \\frac{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = \\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx$\n$+ \\int_2^4 \\frac{ \\sqrt{\\ln x+3} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx$\n\nLet $x = 6-u$ in the second integral\n\n$\\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx$ $- \\int_4^2 \\frac{ \\sqrt{\\ln 9-u} }{\\sqrt{\\ln 3+u} + \\sqrt{\\ln 9-u}}\\,dx$\n\n$= 2 \\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = 2$\n\n$\\int_2^4 \\frac{\\sqrt{\\ln 9-x} }{\\sqrt{\\ln 9-x} + \\sqrt{\\ln x+3}}\\,dx = 1$\n\nI would have never thought of that. Thank you very much!\n\n4. It is very simple actually.Look at this standard property in definite integral\n$\n\\int_a^bf(x)dx=\\int_a^bf(a+b-x)dx\n$\n\nand the rest is obvious\n\nInteresting corollary to above\n\n$\\int_a^b\\frac{f(x)}{f(x)+f(a+b-x)}dx=\\frac{b-a}{2}$\n\n5. Originally Posted by pankaj\nIt is very simple actually.Look at this standard property in definite integral\n$\n\\int_a^bf(x)dx=\\int_a^bf(a+b-x)dx\n$\n\nand the rest is obvious\n\nInteresting corollary to above\n\n$\\int_a^b\\frac{f(x)}{f(x)+f(a+b-x)}dx=\\frac{b-a}{2}$\nThe first part of this is in Stewart's Calculus book (problems plus).", "date": "2016-05-04 06:41:55", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/88164-challenging-integral.html", "openwebmath_score": 0.7391493916511536, "openwebmath_perplexity": 554.005363800766, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013792143467, "lm_q2_score": 0.8947894534772126, "lm_q1q2_score": 0.8786955843791379}}
{"url": "https://math.stackexchange.com/questions/2579825/probability-of-drawing-exactly-1-ace-upon-drawing-2-cards-from-a-deck/2579852", "text": "# Probability of drawing exactly $1$ ace upon drawing $2$ cards from a deck\n\nI got this question and answered it incorrectly. I haven't yet seen the correct answer. The possible answers were:\n\n1. $\\frac{4}{52}$\n2. $\\frac{16}{221}$ (my answer)\n3. $\\frac{2}{52}$\n4. $\\frac{32}{221}$\n\nMy reasoning is the following:\n\nEvent A: Card is not an ace.\n\nEvent B: Card is an ace.\n\n$$P(A)\\times P(B|A)=\\frac{4}{52}\\times \\frac{48}{51}=\\frac{16}{221}$$\n\nThis is assuming the cards are drawn sequentially. Drawing exactly one ace from a single draw is $4/52$ but to ensure that only a single ace was drawn one should consider the probability of not getting a second ace.\n\nHow am I wrong?\n\n\u2022 What if the first card is not an ace but the second is? \u2013\u00a0Ethan Bolker Dec 25 '17 at 16:21\n\u2022 If you are Lemmy, then in probability 1 you will draw the ace of spades. \u2013\u00a0Asaf Karagila Dec 26 '17 at 8:18\n\u2022 You can also do it in another way (as a check). The probability of two aces is $\\frac{4}{52}\\cdot\\frac{3}{51}$. The probability of zero aces (that is two non-aces), is $\\frac{48}{52}\\cdot\\frac{47}{51}$. So the answer should be $$1-\\frac{4}{52}\\cdot\\frac{3}{51}-\\frac{48}{52}\\cdot\\frac{47}{51}$$ \u2013\u00a0Jeppe Stig Nielsen Dec 26 '17 at 13:44\n\nYou forgot to assume $P(B)\\times P(A|B)=\\frac{48}{52}\\times\\frac4{51}=\\frac{16}{221}$.\n\nSo $P(A)\\times P(B|A)+P(B)\\times P(A|B)= \\frac{16}{221}+ \\frac{16}{221}= \\frac{32}{221}$.\n\nThe correct answer is $4$ , i.e. $\\frac{32}{221}$\n\n\u2022 Why? Isn't it the same thing, just reversed? \u2013\u00a0Rivasa Dec 25 '17 at 21:02\n\u2022 @Annabelle, there are two possible ways you can get it, so you account for both. Thinking about it intuitively, because there aren\u2019t many aces, you would expect your chances to be better by taking two cards than by taking one. Otherwise, if you kept taking more and more, your chances wouldn\u2019t change, which would be strange when you got to 50 cards and are guaranteed to get two aces... \u2013\u00a0Jeff Dec 25 '17 at 21:12\n\u2022 @Annabelle: Consider that you have a deck of 3 cards, 2 aces. You draw two cards. What's the probability of drawing exactly one ace? Well, the event \"drawing one\" is equivalent to \"an ace remains in the deck after drawing two cards\". What is the probability of that? Clearly 2/3 as there are two aces in a deck of three if you imagine randomly picking a card to remain in the deck after drawing. But computing it as the OP did here you would end with P=1/3 because of mistakenly not considering that both drawing an ace first and drawing an ace second must be taken into account. \u2013\u00a0Jon Dec 25 '17 at 21:58\n\u2022 I think I get it now... @Jon \u2013\u00a0Rivasa Dec 26 '17 at 2:46\n\nWe have: $$P_{\\text{ only one ace }} = P_{\\text{ ace }1} P_{\\text{ non-ace }2} + P_{\\text{ non-ace }1} P_{\\text{ ace }2} = \\frac{4}{52}\\times \\frac{48}{51} + \\frac{48}{52}\\times \\frac{4}{51} = \\frac{32}{221}$$\n\n\u2022 Why the downvote? \u2013\u00a0user371838 Dec 26 '17 at 13:40\n\nAnswer 4 seems correct to me because:\n\n$$P=\\frac {\\binom4 1\\binom {48} 1}{\\binom {52} 2}=\\frac {32}{221}$$\n\nYou can figure this out without any knowledge of conditional probabilities:\n\nThere are 52 choices for the first pick and 51 choice for the second pick. So, in total, there are $52 \\times 51 = 2652$ possible \"hands\" that consist of two cards.\n\nNow, the question is, how many of these 2652 contain a single ace. Let's do some counting of the possible one-ace hands:\n\nThere are $4 \\times 48 = 192$ hands that have an ace as their first pick and a non-ace as the second pick (because there are 4 aces and 48 non-aces). Similarly, there are $48 \\times 4 = 192$ hands that have a non-ace as their first pick and an ace as the second pick. So the total number of hands that have a single ace is $192+192=384$.\n\nSo, the probability of a one-ace hand is $384/2652$. Dividing top and bottom by $12$ gives us $384/2652 = 32/221$.\n\n\u2022 Does the downvoter care to explain his or her reasoning? \u2013\u00a0bubba Dec 27 '17 at 2:51", "date": "2021-05-18 22:28:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2579825/probability-of-drawing-exactly-1-ace-upon-drawing-2-cards-from-a-deck/2579852", "openwebmath_score": 0.8537099957466125, "openwebmath_perplexity": 307.37385231241365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.8918110562208681, "lm_q1q2_score": 0.8786759147754934}}
{"url": "http://math.stackexchange.com/questions/104644/number-of-different-permutations/104646", "text": "# Number of different permutations\n\nConsider some text $T$\n\nHow many different permutations of this text can we achieve ?\n\nThe easiest case is when every letter appears only once in the text, so the answer is $|T|!$\n\nBut when we have for example $T=aab$ the number of different permutations is 3, because $aab$ $aba$ $baa$\n\nHow can we solve this problem without actually generating all possible permutations of this text ?\n\n-\n\nIt is given by the multinomial theorem:\n\n$$\\frac{N!}{m_1! m_2! m_3! \\ldots m_k!} = \\text{choose}(n, (m_1,m_2, \\ldots, m_k))$$\n\nHere $N$ is the number of elements in $T$, $|T|=N$. If there are $k$ unique elements in $T$, $m_i$ is the number of times that element shows up. From your example, $N=3$, $m_1 = 2$, $m_2 = 1$, so we get\n\n$$\\frac{3!}{2! \\cdot 1!} = \\frac{6}{2} = 3$$\n\n-\nThank you, i didn't know it's that easy :) \u2013\u00a0 Chris Feb 1 '12 at 16:45\n\nAn example might help. Consider the text $probability$. If we thought of all letters as distinct: $p\\ r\\ o\\ b_1\\ a\\ b_2\\ i_1\\ l\\ i_2\\ t\\ y$, then there would be $11!$ permutations.\n\nBut this overcounts, as not all letters are distinct. In particular, amongst the $11!$ factorial arrangements above, we have both \\eqalign{ &r\\ p\\ o\\ b_1\\ a\\ b_2\\ \\color{maroon}{i_1}\\ l\\ \\color{maroon}{i_2}\\ t\\ y\\cr &r\\ p\\ o\\ b_1\\ a\\ b_2\\ \\color{maroon}{i_2}\\ l\\ \\color{maroon}{i_i}\\ t\\ y\\cr} For any particular arrangement amongst the $11!$ arrangements of the letters, we can find one other (by permuting the $i$'s) that is the same when we regard the $i$'s as indistinguishable.\n\nSo we've over counted by a factor of 2. Similarly, because there are two $b$'s, we overcounted by another factor of 2.\n\nSo, the correct number of arrangements is $11!\\over 2\\cdot2$.\n\nA similar argument can be made to establish the formula that Hooked gives in his answer. In particular if one letter was repeated thrice, then if we found the number of arrangements where we thought of all letters as distinct, we would have over counted by a factor of $3!$; since there are $3!$ ways to permute the three repititions of the letter.\n\n-\n\nThe general formula for a permutation of a text with length $n$ with $n_1$ indistinguishable objects of one type, $n_2$ of the next type is:\n\n$$\\frac{n!}{n_1!n_2!\\dots n_k!}$$\n\nFor your text: aab n=3, $n_1$(the number of times a appears)=2 and $n2=1$(Since b appears once). This formula simply divides out all of the objects that $n!$ overcounts.\n\n-", "date": "2015-04-28 05:57:34", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/104644/number-of-different-permutations/104646", "openwebmath_score": 0.9297900795936584, "openwebmath_perplexity": 302.98531279977584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.8918110533454179, "lm_q1q2_score": 0.8786759119423946}}
{"url": "http://mathhelpforum.com/algebra/113535-simplifying-fractions.html", "text": "# Math Help - Simplifying fractions\n\n1. ## Simplifying fractions\n\nThis is the problem\n\nSimplify the following equation:\n(5x/(2x^2+7x+3))-((x-3)/(2x^2-3x-2))+((2x+1)/(x^2+x-6))\n\nThe way i did it was factoring the denominators then adding them, therefore getting common terms where the denominator would be (2x+1)(x+3)(x-2). After that, I don't know what to do next, I'm stumped XD\n\nWould appreciate the help..\n\n2. Hello, mcroldan08!\n\nSimplify the following expression:\n\n. . $\\frac{5x}{2x^2+7x+3} - \\frac{x-3}{2x^2-3x-2} + \\frac{2x+1}{x^2+x-6}$\nIt sounds like your intentions were correct.\nBut since you didn't show us what you did, we can't tell if you made any errors.\n\nWe have: . $\\frac{5x}{(x+3)(2x+1)} - \\frac{x-3}{(x-2)(2x+1)} + \\frac{2x+1}{(x-2)(x+3)}$\n\nYou are correct . . . the LCD is: . $(x+3)(2x+1)(x-2)$\n\nAnd we must convert the fractions to the LCD.\n\n$\\frac{5x}{(x+3)(2x+1)}\\cdot{\\color{blue}\\frac{x-2}{x-2}} - \\frac{x-3}{(x-2)(2x+1)}\\cdot{\\color{blue}\\frac{x+3}{x+3}} + \\frac{2x+1}{(x-2)(x+3)}\\cdot{\\color{blue}\\frac{2x+1}{2x+1}}$\n\n. . $= \\;\\frac{5x(x-2)}{(x+3)(2x +1)(x-2)} - \\frac{(x-3)(x+3)}{(x-2)(2x+1)(x+3)} + \\frac{(2x+1)(2x+1)}{(x-2)(x+3)(2x+1)}$\n\n. . $= \\;\\frac{5x(x-2) - (x-3)(x+3) + (2x+1)(2x+1)}{(x+3)(2x+1)(x-2)}$\n\n. . $= \\;\\frac{5x^2 - 10x - x^2 + 9 + 4x^2 + 4x + 1}{(x+3)(2x+1)(x-2)}$\n\n. . $= \\;\\frac{8x^2 - 6x + 10}{(x+3)(2x+1)(x-2)}$\n\n. . $= \\;\\frac{2(4x^2 - 3x + 5)}{(x+3)(2x+1)(x-2)}$\n\n3. That is actually what I got... however, I was wondering if this can be simplified further. This is actually a multiple choice question and the choices are:\n\na.) 4/(x+3)\nb.) 4/(x-3)\nc.) 2/(x-3)\nd.) 2/(x+3)\n\nCould it be that the none of the choices are correct?", "date": "2014-12-22 20:48:46", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/113535-simplifying-fractions.html", "openwebmath_score": 0.8888062238693237, "openwebmath_perplexity": 924.3173723663707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969689263264, "lm_q2_score": 0.8933094174159127, "lm_q1q2_score": 0.8786564352836342}}
{"url": "http://mathhelpforum.com/pre-calculus/122058-rebound-superball-problem.html", "text": "# Math Help - Rebound Superball Problem\n\n1. ## Rebound Superball Problem\n\nThe rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.\na. How far downward has the ball traveled at this instant?\nb. How far ( upward and downward has the ball traveled at this instant?\nc. How far would the ball travel if you just let it bounce...?\n\nI have a feeling for the first two it would be something along the lines of \"16(0.75)^0+16(0.75)^1...+16(0.75)^50\", and for the second 2(\" \"), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?\n\nThanks.\n\n2. Originally Posted by bkap\nThe rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.\na. How far downward has the ball traveled at this instant?\nb. How far ( upward and downward has the ball traveled at this instant?\nc. How far would the ball travel if you just let it bounce...?\n\nI have a feeling for the first two it would be something along the lines of \"16(0.75)^0+16(0.75)^1...+16(0.75)^50\", and for the second 2(\" \"), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?\n\nThanks.\nYes your approach to the first problem is correct, if you factorize things out a bit, it looks something like:\n$16*((3/4)^0+(3/4)^1+...+(3/4)^{49})$\nit ends with the 49th power because 0-49 gives a total of 50 drops\nnow this is just a standard progression, whose sum is:\n$a(r^{n+1}-1)/(r-1)$\nwhere a is the first term, in this case 1, r is the successive ratio, in this case 3/4, and n is how far the last term goes, in this case 49.\n\nThe second part is trivial once you got the first part, however you have to keep in mind that the first upward bounce is the same distance as the second downward bounce.\n\nFinally for the third part, you simply let n tend to infinity and calculate. If you haven't done limit, just make n=1000 on the calculator, this will give you a pretty good approximation\n\n3. Originally Posted by bkap\nThe rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.\na. How far downward has the ball traveled at this instant?\nb. How far ( upward and downward has the ball traveled at this instant?\nc. How far would the ball travel if you just let it bounce...?\n\nI have a feeling for the first two it would be something along the lines of \"16(0.75)^0+16(0.75)^1...+16(0.75)^50\", and for the second 2(\" \"), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?\n\nThanks.\nNah, Factorials only really occur in probability - you'll be better off with a sequence. Let s be the distance it bounces back up\n\n$s_0: 16 = 16 \\times 0.75^0$\n$s_1: 16 \\times 0.75^1 = 0.75s_0$\n$s_2: (16 \\times 0.75) \\times 0.75 = 0.75s_1$\n...\n$s_n = 16 \\times 0.75^n = 0.75s_{n-1}$\n\nFrom this we can see it's a geometric sequence so use the sum of a geometric sequence to solve how far it's bounced up. To find down and up multiply by 2\n\n4. Hello, bkap!\n\nEveryone has given you excellent advice . . .\n\nThe rebound ratio of a speckled green superball is 75%.\nIt is dropped from a height of 16 feet.\nConsider the instant when the ball strikes the ground for the 50th time.\n\na. How far downward has the ball traveled at this instant?\n\nTotal downward distance is given by:\n\n. . $D \\;=\\;16 + 16(.075) + 16(.075)^2 + 16(0.75)^3 + \\hdots + 16(0.75)^{49}$\n\n. . . . . $=\\;16\\underbrace{\\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \\hdots + (0.75)^{49}\\bigg]}_{\\text{geometric series}}$\n\n. . The sum of the geometric series is: . $\\frac{1-(0.75)^{50}}{1-0.75} \\:=\\:3.999997735$\n\nTherefore: . $D \\;=\\;16(3.999997736) \\;=\\;63.99996376$ feet.\n\nb. How far ( upward and downward) has the ball traveled at this instant?\nWe must be very careful . . .\n\nThe balls falls 16 feet.\n\nThen bounces up $16(0,75)$ feet and falls $16(0.75)$ feet.\n\nThen bounces up $16(0.75)^2$ feet and falls $16(0.75)^2$ feet . . . and so on.\n\nThe total distance is:\n\n. . $T \\;=\\;16 + 2\\!\\cdot\\!16(0.75) + 2\\!\\cdot\\!16(0.75)^2 + 2\\!\\cdot\\!16(0.75)^3 + \\hdots + 2\\!\\cdot\\!16(0.75)^{49}$\n\n. . . . $=\\; 16 + 32(0.75)\\bigg[1 + (0.75) + (0.75)^2 + \\hdots + (0.75)^{49}\\bigg]$\n\n. . . . $=\\; 16 + 24(3.999997735) \\;=\\;111.9999275$ feet.\n\nc. How far would the ball travel if you just let it bounce?\nIf we let it bounce (forever), the total distance is:\n\n. . $T \\;=\\;16 + 24\\underbrace{\\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \\hdots\\bigg]}_{\\text{infinite geomtric series}}$\n\n. . The sum of the infinite series is: . $\\frac{1}{1-0.75} \\:=\\:4$\n\nTherefore: . $T \\;=\\;16 + 24(4) \\;=\\;112$ feet.", "date": "2014-03-15 09:34:51", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/122058-rebound-superball-problem.html", "openwebmath_score": 0.8677860498428345, "openwebmath_perplexity": 561.7393232483286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969665221771, "lm_q2_score": 0.8933094145755219, "lm_q1q2_score": 0.8786564303421852}}
{"url": "https://math.stackexchange.com/questions/1423107/solving-the-infinite-series-1-frac231-frac332-frac433-cdo/1423113", "text": "# Solving the infinite series $1-\\frac{2^3}{1!}+\\frac{3^3}{2!}-\\frac{4^3}{3!}+\\cdots$\n\nI have the following question:\n\nEvaluate the infinite series: $$S=1-\\frac{2^3}{1!}+\\frac{3^3}{2!}-\\frac{4^3}{3!}+\\cdots$$ (a) $\\displaystyle\\frac1e$ (b) $\\displaystyle\\frac{-1}e$ (c) $\\displaystyle\\frac{2}e$ (d) $\\displaystyle\\frac{-2}e$\n\nNow in the book they have given a strange explanation using $$(n+1)^3=[n(n-1)(n-2)+6n(n-1)+7n+1].$$ I don't understand how did they get this. So I have two doubts:\n\n1. How did they get this \"trick\"?\n2. If you would have got this question in a competitive exam, how would you have solved it (in the minimum amount of time), either directly or with numerical methods?\n\u2022 The trick seems to be typical when you want to see $An(n-1)(n-2)+Bn(n-1)+Cn+D$. Just identify. Sep 5, 2015 at 14:54\n\nYou want to evaluate $\\sum_{n=0}^{\\infty} \\frac{(-1)^n(n+1)^3}{n!}$.\n\nNow the idea (at the latest if you see some $e$ as a possible answer) should be to relate this to the well-known formula $$\\sum_{n=0}^{\\infty} \\frac{x^n}{n!}=e^x$$ In this particular case (given the $(-1)^n$ included) it is useful to plug in $x=-1$ and obtain $$\\sum_{n=0}^{\\infty} \\frac{(-1)^n}{n!}=e^{-1}=\\frac{1}{e}$$ But you still have a $(n+1)^3$ in the numerator where you wish to have $1$.\n\nSo you might want to split this into several parts such that you get one (or several) related sums.\n\nAs an example of this technique, if you want to know $\\sum_{n=0}^{\\infty} \\frac{n+1}{n!}$ you might try: $$\\sum_{n=0}^{\\infty} \\frac{n+1}{n!}=\\sum_{n=0}^{\\infty} \\frac{n}{n!}+\\sum_{n=0}^{\\infty} \\frac{1}{n!}$$ Now, the second sum is just $e$ and the first one is (after cancelling) $$\\sum_{n=1}^{\\infty} \\frac{1}{(n-1)!}=\\sum_{n=0}^{\\infty} \\frac{1}{n!}=e$$ and so the total sum is $2e$. Can you now understand the \"trick\" and tackle your problem by yourself?\n\nConcerning the second question:\n\nIf one happens to know this \"trick\", it is not hard to get the result by just a few computations so it would be my preferential choice.\n\nBut if you sit in the exam and have absolutely no idea of how to solve this algebraically (and especially if you don't have to give a proof of your result!) it might be a good idea to estimate the sum by computing a few terms. At least the sign of the result ($\\pm$) should be approachable...\n\n\u2022 Yes thanks a lot, Now I understood the trick. But can you please answer the 2nd part of the question? Sep 5, 2015 at 14:59\n\u2022 @Kartik by learning the trick now.Its a very common question :-)!! I felt the same as you the first time...!\n\u2013\u00a0user220382\nSep 5, 2015 at 15:29\n\u2022 I changed \"nominator\" into \"numerator Sep 5, 2015 at 15:37\n\nIn general, if $p_i(x)$ is a sequence of polynomials with $\\deg p_i=i$, then it is relatively easy to prove that every polynomial can be written as a linear combination of a finite number of $p_i$. When the $p_i$ are monic, you can do this easily. Let $P(x)=a_kx^k + ... a_0$. Then $P(x)-a_kp_k(x)$ is a polynomial of lower degree. Find it's first coefficient of that, and proceed inductively.\n\nThe standard basis for polynomials is $e_i(x)=x^i$.\n\nSo in your case; $p_i(n)=x(x-1)\\cdots(x-(i-1))$ with $p_0(x)=1$. We chose this particular basis because $$\\frac{p_i(n)}{n!} =\\begin{cases}\\frac{1}{(n-i)!} & n\\geq i\\\\0&0\\leq n<i\\end{cases}$$ So:$$\\sum_{n=0}^\\infty \\frac{p_i(n)z^n}{n!} = z^ie^{z}$$\n\nIn your case, $z=-1$.\n\nWith these polynomials, we get: \\begin{align}(n+1)^3 &= n^3+3n^2+3n+1\\\\ (n+1)^3-p_3(n) &= n^3 +3n^2+3n+1 - (n^3-3n^2+2n)\\\\& = 6n^2+n+1\\\\ (n+1)^3-p_3(n)-6p_2(n) &= 6n^2+n+1 - 6n(n-1)\\\\&= 7n+1\\\\ \\end{align}\n\nSo you get: $$(n+1)^3=p_3(n) + 6p_2(n) + 7p_1(n) +p_0(n)$$\n\nThe key is that the leading coefficient of the cubic $(n+1)^3$ is $1$, so we subtract $1\\cdot p_3(n)$. Then the leading coefficient of the remaining quadratic is $6$, so we subtract $6p_2(n)$. The leading coefficient of the remaining linear polynomial is $7$, so we subtract $7p_1(x)$. You are left with a constant.\n\nThe relationship between the standard basis of the polynomials and the basis $p_i(x)=x(x-1)\\cdots(x-(i-1))$ is actually fairly interesting and comes up again and again. The $p_i(x)$ are often called \"falling factorials.\"\n\nMy way to solve it:\n\nIf the numerators were all $1$, we would have the well-known series for $e^{-1}$, obtained from the entire series\n\n$$e^{-x}=\\sum_{k=0}^\\infty(-1)^k\\frac{x^k}{k!}.$$\n\nNow we want to make numerators $(k+1)^3$ appear. Multiply by $x$ and derive, to get\n\n$$(xe^{-x})'=\\left(\\sum_{k=0}^\\infty(-1)^k\\frac{x^{k+1}}{k!}\\right)'=\\sum_{k=0}^\\infty(-1)^k\\frac{(k+1)x^k}{k!}.$$\n\nRepeating this three times,\n\n$$(x(x(xe^{-x})')')'=\\sum_{k=0}^\\infty(-1)^k\\frac{(k+1)^3x^k}{k!}.$$\n\nThen evaluate at $x=1$.\n\n$$e^{-x}\\to (1-x)e^{-x}\\to (1-3x+x^2)e^{-x}\\to (1-7x+6x^2-x^3)e^{-x}\\to -\\frac1e.$$\n\nFor such problems, you can often start from a known Taylor series (in particular $e^x$ and $1/(1-x)$)and perform the following operations\n\n\u2022 multiply or divide by a power of $x$ (shifts the exponents),\n\u2022 derive on $x$ (multiplies by the exponents),\n\u2022 integrate on $x$ (divides by the exponents),\n\nto modify the coefficients.\n\nSince $(n+1)^3 = \\color{red}{1}\\cdot n(n-1)(n-2)+\\color{red}{6}\\cdot n(n-1) +\\color{red}{7}\\cdot n+\\color{red}{1}$ we have:\n\n$$\\begin{eqnarray*}\\sum_{n=0}^{+\\infty}\\frac{(-1)^n (n+1)^3}{n!} &=& \\frac{13}{2}+\\sum_{n\\geq 3}\\frac{(-1)^n (n+1)^3}{n!}\\\\&=&\\frac{13}{2}+\\color{red}{1}\\cdot\\sum_{n\\geq 3}\\frac{(-1)^n}{(n-3)!}+\\color{red}{6}\\cdot\\sum_{n\\geq 3}\\frac{(-1)^n}{(n-2)!}+\\color{red}{7}\\cdot\\sum_{n\\geq 3}\\frac{(-1)^n}{(n-1)!}+\\color{red}{1}\\cdot\\sum_{n\\geq 3}\\frac{(-1)^n}{n!}\\\\&=&\\frac{13}{2}-\\sum_{n\\geq 0}\\frac{(-1)^n}{n!}+6\\cdot\\sum_{n\\geq 1}\\frac{(-1)^n}{n!}-7\\cdot\\sum_{n\\geq 2}\\frac{(-1)^n}{n!}+\\sum_{n\\geq 3}\\frac{(-1)^n}{n!}\\\\&=&\\sum_{n\\geq 0}(-1+6-7+1)\\frac{(-1)^n}{n!}\\\\&=&-\\sum_{n\\geq 0}\\frac{(-1)^n}{n!}=\\color{red}{-\\frac{1}{e}}.\\end{eqnarray*}$$\n\nWe simply get the initial decomposition by induction. Obviously $(n+1)^3-n(n-1)(n-2)$ is a quadratic polynomial in $n$, whose leading term is $6n^2$. So we substract $6n(n-1)$ and we are left with a linear polynomial and so on. Another chance is to tackle our problem this way: $$\\begin{eqnarray*} \\sum_{n\\geq 0}\\frac{(-1)^n(n+1)^3}{n!}&=&\\sum_{n\\geq 1}\\frac{(-1)^n n(n+1)^2}{n!}+\\sum_{n\\geq 0}\\frac{(-1)^n(n+1)^2}{n!}\\\\&=&\\sum_{n\\geq 0}\\frac{(-1)^n(n+1)^2}{n!}-\\sum_{n\\geq 0}\\frac{(-1)^n(n+2)^2}{n!}\\\\&=&-\\sum_{n\\geq 0}\\frac{(-1)^n(2n+3)}{n!}\\\\&=&-3\\sum_{n\\geq 0}\\frac{(-1)^n}{n!}+2\\sum_{n\\geq 0}\\frac{(-1)^n}{n!}=\\frac{2-3}{e}.\\end{eqnarray*}$$\n\nLooking at the given identity: \\begin{align} (n+1)^3 &= [n(n-1)(n-2)+6n(n-1)+7n+1] \\\\ &= n(n^2 - 3n + 2) + 6 n^2 + n + 1 \\\\ &= n^3 + 3 n^2 + 3 n + 1 = (n+1)^3 \\end{align}\n\nNow, consider differentiation of the exponential series: \\begin{align} D \\, e^{t} &= \\sum_{n=0}^{\\infty} \\frac{1}{n!} \\, \\frac{d}{dt} \\, t^{n} = \\sum_{n=0}^{\\infty} \\frac{n \\, t^{n-1}}{n!} \\end{align} Further differentiation can be applied.\n\nNow, consider the series in question: \\begin{align} S &= 1 - \\frac{2^{3}}{1!} + \\frac{3^{3}}{2!} - \\cdots \\\\ &= \\sum_{n=0}^{\\infty} \\frac{(-1)^{n} \\, (n+1)^{3}}{n!} \\\\ &= \\left. \\sum_{n=0}^{\\infty} \\frac{(-1)^{n} \\, (n+1)^{3} \\, t^{n}}{n!} \\, \\right|_{t=1} \\\\ &= \\left. t^{3} \\, \\sum_{n=0}^{\\infty} \\frac{(-1)^{n} \\, n(n-1)(n-2) \\, t^{n-3}}{n!} + 6 t^{2} \\, \\sum_{n=0}^{\\infty} \\frac{(-1)^{n} \\, n(n-1) \\, t^{n-2}}{n!} + 7 t \\, \\sum_{n=0}^{\\infty} \\frac{(-1)^{n} \\, n \\, t^{n-1}}{n!} + e^{-t} \\, \\right|_{t=1} \\\\ &= \\left. t^3 \\, D^{3} e^{-t} + 6 \\, t^{2} \\, D^{2} e^{-t} + 7 \\, t \\, D e^{-t} + e^{-t} \\right|_{t=1} \\\\ &= \\left. (-t^3 + 6 \\, t^2 - 7 \\, t + 1) \\, e^{-t} \\right|_{t=1} \\\\ &= - \\frac{1}{e} \\end{align}\n\n\u2022 Thanks for the answer but this explanation is already given in the book which I am solving. I just wanted the explanation of the \"Trick\" in the first identity, Sep 5, 2015 at 15:16", "date": "2022-06-27 11:55:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1423107/solving-the-infinite-series-1-frac231-frac332-frac433-cdo/1423113", "openwebmath_score": 0.99993896484375, "openwebmath_perplexity": 263.8368980489373, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969665221771, "lm_q2_score": 0.8933094138654242, "lm_q1q2_score": 0.8786564296437352}}
{"url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-2-section-2-2-the-limit-of-a-function-2-2-exercises-page-92/1", "text": "## Calculus: Early Transcendentals 8th Edition\n\nPublished by Cengage Learning\n\n# Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 1\n\n#### Answer\n\nAs $x$ approaches 2, $f(x)$ approaches 5. Yes, it is possible for this to be true and for $f(2)=3$.\n\n#### Work Step by Step\n\nIt means that $f(x)$ becomes arbitrarily close to $3$ as we take $x$ close enough to $2$, but $x\\ne2$. It is possible for $f(2)$ to equal $3$ because the limit only relies on when $f(x)$ is near $2$; there can be a hole at $x=2$.\n\nAfter you claim an answer you\u2019ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide\u00a0feedback.", "date": "2018-06-21 22:29:20", "meta": {"domain": "gradesaver.com", "url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-2-section-2-2-the-limit-of-a-function-2-2-exercises-page-92/1", "openwebmath_score": 0.9090868234634399, "openwebmath_perplexity": 307.61911331411466, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9683812309063186, "lm_q2_score": 0.9073122295060291, "lm_q1q2_score": 0.8786241336254046}}
{"url": "https://stats.stackexchange.com/questions/21297/should-i-use-a-binomial-cdf-or-a-normal-cdf-when-flipping-coins", "text": "# Should I use a binomial cdf or a normal cdf when flipping coins?\n\nA coin needs to be tested for fairness. 30 heads come up after 50 flips. Assuming the coin is fair, what is the probability that you would get at least 30 heads in 50 flips?\n\nThe right way to do this problem, according to my teacher, is to do\n\nnormalcdf(min = .6, max = \u221e, p = .5, \u03c3 = sqrt(.5 * .5 / 50) = 0.0786\n\n\nHowever, I took a binomial cumulative distribution function like this\n\n1 - binomcdf(n = 50, p = .5, x = 29) = 0.1013\n\n\nI believe the criteria for a binomial distribution are satisfied: the individual events are independent, there are only two possible outcomes (heads vs. tails), the probability is constant for the question (0.5), and the number of trials is fixed at 50. Yet obviously, the two methods give different answers, and a simulation supports my answer (at least the few times I ran it; obviously, I can't guarantee that you'd get the same results).\n\nIs my teacher wrong in assuming that a Normal distribution curve would also be a valid way to do this problem (at no point is it said that the distribution is Normal, but n*p and n*(1-p) are both greater than 10), or have I misunderstood something about binomial distributions?\n\n\u2022 A person with experience in using Normal approximations to the Binomial would proceed a little differently: they would apply the (usual) continuity correction, as in 1 - pnorm((30-0.5)/50, mean=0.5, sd=sqrt(0.5*(1-0.5)/50)) (this is an R expression), whose value is 0.1015, in quite close agreement with the Binomial cdf. \u2013\u00a0whuber Jan 18 '12 at 21:23\n\nHere is an illustration of the answers of whuber and onestop.\n\nIn red the binomial distribution $\\mathcal Bin(50,0.5)$, in black the density of the normal approximation $\\mathcal N(25, 12.5)$, and in blue the surface corresponding to $\\mathbb P(Y > 29.5)$ for $Y \\sim \\mathcal N(25, 12.5)$.\n\nThe height of a red bar corresponding to $\\mathbb P(X=k)$ for $X\\sim\\mathcal Bin(50,0.5)$ is well approximated by $\\mathbb P\\left( k -{1\\over 2} < Y < k + {1\\over 2}\\right)$. To get a good approximation of $\\mathbb P(X \\ge 30)$, you need to use $\\mathbb P(Y>29.5)$.\n\n(edit) This is $$\\mathbb P(Y>29.5) \\simeq 0.1015459,$$ (obtained in R by 1-pnorm(29.5,25,sqrt(12.5))) whereas $$\\mathbb P(X \\ge 30) \\simeq 0.1013194:$$ the approximation is correct.\n\nThis is called continuity correction. It allows you to compute even \"point probabilities\" like $\\mathbb P(X=22)$ : \\begin{align*} \\mathbb P(X=22) &= {50 \\choose 22} 0.5^{22} \\cdot 0.5^{28} \\simeq 0.07882567, \\\\ \\mathbb P(21.5 < Y < 22.5) & \\simeq 0.2397501 - 0.1610994 \\simeq 0.07865066.\\end{align*}\n\nThe normal distribution gives a closer approximation to the binomial if you use a continuity correction. Using this for your example, I get 0.1015. As this is homework, I'll leave it to you to fill in the details.\n\nConsider this. In the discrete binomial distribution you have actual probabilities for individual numbers. In the continuous normal that isn't the case, you need a range of values. So... if you were going to approximate the probability of an individual value, let's say X, from the binomial with the normal how would you do that? Look at a probability histogram of the binomial distribution with the normal curve laid over it. You would need to actually select from X \u00b1 0.5 to capture something similar to what the binomial probability of X is with the normal approximation.\n\nNow extend that to when you're selecting a tail of the distribution. When you use the binomial method you're selecting your entire value's probability (30 in your case) plus everything higher. Therefore, when you do the continuous you have to make sure you capture that and select 0.5 less as well, so the cutoff on the continuous distribution is 29.5.\n\n\u2022 Actually, the question exhibits a thoughtful understanding of the problem and does not appear to be looking for an answer to a routine homework question. Although it's tagged homework, consider making an exception here. In particular, a good discussion of using the Normal distribution to approximate discrete distributions (such as Binomials and Poissons with large N's) would be appropriate and most welcome here. \u2013\u00a0whuber Jan 18 '12 at 23:37", "date": "2020-12-02 07:42:55", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/21297/should-i-use-a-binomial-cdf-or-a-normal-cdf-when-flipping-coins", "openwebmath_score": 0.9221925139427185, "openwebmath_perplexity": 360.0943943155211, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812309063187, "lm_q2_score": 0.907312219480915, "lm_q1q2_score": 0.8786241239172724}}
{"url": "https://math.stackexchange.com/questions/2403181/prove-that-l-infty-infty-is-a-banach-space", "text": "# Prove that $(l^\\infty,\\|.\\|_\\infty)$ is a Banach space.\n\n$(l^\\infty,\\|.\\|_\\infty)$ is a Banach space. In the proof, $\\mathbb{F}$ is either the field of complex numbers or the field of real numbers.\n\nProof: Let $(x^n)_{n\\in\\mathbb{N}}$ be a Cauchy sequence in $l^\\infty$, where $x^n=(x_k^n)_{k\\in\\mathbb{N}}$. Consider the sequence $(x^0_k,x^1_k,\\cdots,x^n_k,\\cdots)$ of $k$th coordinates of the sequence $(x^n)_{n\\in\\mathbb{N}}$.\n\nLet $\\epsilon>0$. Since $(x^n)_{n\\in\\mathbb{N}}$ is Cauchy, there exists $n_0\\in\\mathbb{N}$ such that $$\\forall m,n>n_0,\\|x^m-x^m\\|_\\infty<\\epsilon.$$ Therefore for each $m,n>n_0$: $$|x^m_k-x^n_k|<\\epsilon.$$\n\nSo the sequence $(x^0_k,x^1_k,\\cdots,x^n_k,\\cdots)$ is Cauchy. Therefore it converges to some $y_k\\in\\mathbb{F}\\$ ($\\mathbb{F}$ is complete).\n\nLet $y=(y_k)_{k\\in\\mathbb{N}}$. Since $(x^0_k,x^1_k,\\cdots,x^n_k,\\cdots)$ is Cauchy it is bounded. Choose $M>0$ such that for each $n\\in\\mathbb{N}$, $|x^n_k|<M$. But since $$y_k=\\lim_{n\\to\\infty}x_k^n,$$ we have $|y_k|\\leq M$ for each $k$. Therefore, $y\\in l^\\infty$.\n\nFix $m>n_0$. Then we have $\\|x^m-x^n\\|_\\infty<\\epsilon.$ Therefore $\\|x^m-y\\|_\\infty<\\epsilon$ as $n\\to\\infty$.\n\nTherefore for each $m>n_0,\\ \\|x^m-y\\|_\\infty<\\epsilon$. Hence $(x^n)_{n\\in\\mathbb{N}}$ converges in $l^\\infty$ and the proof is complete.\n\nCould someone please tell me if the above proof is alright? Thanks.\n\n\u2022 Looks alright to me. The readability was fine even before it was over-pedantically edited. \u2013\u00a0uniquesolution Aug 23 '17 at 7:35\n\u2022 The general procedure seems right, however near the end $\\|x^m - x^m\\|_\\infty$ is something you should fix. \u2013\u00a0mlk Aug 23 '17 at 7:36\n\u2022 @uniquesolution Then you are just more skilled than I am in reading wall-of text style questions. But I hope even you will agree that the readability is much better now. \u2013\u00a05xum Aug 23 '17 at 7:37\n\u2022 When I first looked at it it was properly Latex formatted. If it was text-style and you converted to Latex, then thank you. \u2013\u00a0uniquesolution Aug 23 '17 at 7:39\n\u2022 @mlk fixed it, and thanks everyone. \u2013\u00a0Janitha357 Aug 23 '17 at 7:40\n\nYour proof is very rigorous and very detailed all the way up to the point where you say\n\nFix $m>n_0$. Then we have $\\|x^m-x^n\\|_\\infty<\\epsilon.$ Therefore $\\|x^m-y\\|_\\infty<\\epsilon$ as $n\\to\\infty$.\n\nNow I know the inequality stands, but as you were very thorough with all your other inequalities, I think it would be nice if you wrote a little more justification for this one as well - it is not entirely obvious how the right inequality follows from the left one.\n\nOther than that, the proof is very well written and easy to follow.\n\nRather a comment, but my reputation does not allow me to comment yet. You write\n\nchoose $M>0$ that for each $n\\in\\mathbb{N}, |x^n_k|<M$\n\nI am curios whether you have to write $M_k$, since hypothetically $M$ can depend on the choice of the sequence $(x^0_k,x^1_k,\\cdots,x^n_k,\\cdots)$. But then we have difficulties with proving that $y\\in l^\\infty$.\n\nEDIT: the proof is more subtle than I initially thought. First we have to settle the issue 5xum mentioned. We start with $$\\forall k\\in\\mathbb{N}\\ \\forall \\epsilon > 0 \\ \\exists n_0:\\forall n,p>n_0:|x^n_k-x^{n+p}_k|<\\epsilon.$$ This is inequality in $\\mathbb{F}$, so we can let $p\\to\\infty$. Thus we obtain $$\\forall k\\in\\mathbb{N}\\ \\forall \\epsilon > 0 \\ \\exists n_0:\\forall n>n_0:|x^n_k-y_k|<\\epsilon.$$ That means $\\|x^n-y\\|_\\infty<\\epsilon$, thus $(x^n)_{n\\in\\mathbb{N}}$ is converging to $y$.\n\nWe still have to prove that $y\\in l^\\infty$. But $$\\|y\\|_\\infty \\le \\|x^n-y\\|_\\infty + \\|x^n\\|_\\infty \\le \\epsilon + \\|x^n\\|_\\infty < \\infty$$\n\n\u2022 Could you explain the last bit please? \u2013\u00a0Janitha357 Aug 23 '17 at 9:53\n\u2022 I am using triangle inequality and addition of zero to prove that $y\\in l^\\infty$: $\\|y\\|_\\infty \\le \\|x^n - x^n+y\\|_\\infty \\le \\|x^n-y\\|_\\infty + \\|x^n\\|_\\infty \\le \\epsilon + \\|x^n\\|_\\infty < \\infty$ \u2013\u00a0EugenR Aug 23 '17 at 11:10", "date": "2019-10-15 09:46:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2403181/prove-that-l-infty-infty-is-a-banach-space", "openwebmath_score": 0.9651840925216675, "openwebmath_perplexity": 202.28443025418093, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915249511566, "lm_q2_score": 0.8872046011730965, "lm_q1q2_score": 0.8785911974393884}}
{"url": "https://math.stackexchange.com/questions/3422830/an-interesting-problem-of-polynomials", "text": "# An interesting problem of polynomials\n\nIn the polynomial $$(x-1)(x^2-2)(x^3-3) \\ldots (x^{11}-11)$$ what is the coefficient of $$x^{60}$$?\n\nI've been trying to solve this question since a long time but I couldn't. I don't know whether opening the brackets would help because that is really a mess. I have run out of ideas. Would someone please help me to solve this question?\n\n\u2022 Think of how many ways the terms can form a term $x^{60}$ ... It could be $x^1 \\cdot x^{59}$, or $x^2 \\cdot x^{58}$, or even $x^3 x^8 x^{49}$ ... Can you think of a systematic way to create $x^{60}$ like this? Nov 5 '19 at 10:38\n\u2022 Old question, see here. Also on this site there are several such question. Please have a look, say here. Next time, please search before posting. Nov 5 '19 at 10:39\n\u2022 @DietrichBurde it's not easy to search for mathematical expressions (unless you know to use approach 0 or something like it), and there isn't a lot of text to go off of for these problems. I don't think that there's much reason to assume that the asker didn't at least try searching. Nov 5 '19 at 10:44\n\u2022 @Omnomnomnom I agree, that it is not always immediate to see these duplicates, but honestly my experience on this site is that several users are not even aware that it is possible to find the same question with several beautiful answers. This is true in particular for these \"standard questions\". One of the goals for students also is to learn how to search. Nov 5 '19 at 10:47\n\u2022 @Aryan The previous edit was better actually. Nov 5 '19 at 10:48\n\nThe highest exponent possible is $$1+2+ \\cdots + 11 = 66$$.\n\nNow, to create the exponent $$60$$, you can only leave out the factors containing $$(1,2,3),(2,4),(1,5)$$ and $$6$$. Hence,\n\n\u2022 $$1+2+3 \\Rightarrow$$ gives coefficient $$(-1)(-2)(-3) = -6$$\n\u2022 $$2+4 \\Rightarrow$$ gives coefficient $$(-2)(-4) = 8$$\n\u2022 $$1+5 \\Rightarrow$$ gives coefficient $$5$$\n\u2022 $$6 \\Rightarrow$$ gives coefficient $$-6$$\n\nSumming up gives $$13-12 = \\boxed{1}$$.\n\nHint : $$1+2+3 +...+11= \\frac {11\u00d712}{2} =66$$ so we must find how we can construct number $$6=6+0=5+1=4+2=3+3=1+2+3$$ and note that $$3+3$$ impossible.\n\nExpanding the brackets is definitely not the way to go, but thinking about what would happen if you did is helpful. Every term in the expanded polynomial will come from multiplying one term from each bracket (e.g. the highest degree term will come from multiplying $$\\,x \\cdot x^2 \\cdot x^3 \\cdot x^4 \\dots \\cdot x^{11}\\,$$), and since each bracket is a binomial that doesn't leave many options.\n\nThe first thing to notice is that the aforementioned highest-degree term will be $$\\,x^{66};\\,$$ thus, to get a term with $$\\,x^{60},\\,$$ we will need to use MOST BUT NOT ALL of the $$\\,x$$'s in the factors. We need to reduce the maximum degree by $$\\,6\\,$$; one obvious way to do that is to not use the $$\\,x^6\\,$$ in the sixth factor, so instead we use the $$\\,-6\\,$$ from that bracket. However, there are other ways, e.g. not using the $$\\,x^2\\,$$ or the $$\\,x^4\\,$$ from the second and fourth factors, respectively. Hopefully by now you're getting the sense that what we need to do is come up with all the different ways to make $$\\,x^6\\,$$ with any of the powers available to us, and then for each way find the corresponding coefficient.\n\nThere are not that many ways of making $$\\,x^6;\\,$$ note that the most we can use is three powers of $$\\,x\\,$$, as the minimum power would then be $$\\,x\\cdot x^2\\cdot x^3=x^6.\\,$$ The list of ways is: $$\\,x^6, x\\cdot x^5, x^2 \\cdot x^4,\\,$$ and $$\\,x\\cdot x^2\\cdot x^3.\\,$$\n\nNow we just need to find the coefficient of each term that EXCLUDES one of the above combinations. As an example, let's consider the term that excludes $$\\,x^2\\,$$ and $$\\,x^4.\\,$$ Written out in detail, this term would be\n\n$$(x)(-2)(x^3)(-4)(x^5)(x^6)(x^7)(x^8)(x^9)(x^{10})(x^{11}) = 8x^{60}$$\n\nSimilarly, it should be easy to see that the coefficients for the terms excluding the other combinations are:\n\n\u2022 For $$\\,x^6,\\,$$, the coefficient will just be $$\\,-6\\,$$;\n\u2022 For $$\\,x\\cdot x^5,\\,$$ the coefficient will be $$\\,(-1)(-5)=5\\,$$;\n\u2022 For $$\\,\\,x\\cdot x^2\\cdot x^3,\\,$$ the coefficient will be $$\\,(-1)(-2)(-3)=-6.\\,$$\n\nThus the four terms that will have $$\\,x^{60}\\,$$ are $$\\,-6x^{60}, 5x^{60}, 8x^{60}\\,$$ and $$\\,-6x^{60}.\\,$$ The sum of these will be just $$\\,x^{60},\\,$$ so the answer to the question is $$\\,\\boxed{1}.\\,$$\n\n\u2022 Thanks a lot A.J. Nov 6 '19 at 9:12", "date": "2021-10-22 07:29:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3422830/an-interesting-problem-of-polynomials", "openwebmath_score": 0.8006179332733154, "openwebmath_perplexity": 198.44430329185062, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915232319998, "lm_q2_score": 0.8872046011730965, "lm_q1q2_score": 0.8785911959141446}}
{"url": "https://puzzling.stackexchange.com/questions/79552/sum-in-a-sudoku-with-sum-total-of-15-and-45", "text": "# Sum in a sudoku with sum total of 15 and 45\n\nI have created a sudoku puzzle with the following restrictions:\n\n\u2022 Each row and column sum to $$45$$.\n\u2022 Each row and column in the nine $$3$$ by $$3$$ sub-grids sum to $$15$$.\n\nIs such a sudoku unique?\n\n\u2022 \"...where each row and column adds to a total of 45...\" is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. \u2013\u00a0Hugh Feb 11 at 5:04\n\u2022 \"where each (3*3) 9 individual squares of rows and columns total a sum of 15\" >>> do you mean 45? \u2013\u00a0Kryesec Feb 11 at 5:33\n\u2022 @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with \"magic constant\" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15. \u2013\u00a0Hugh Feb 11 at 5:47\n\nNo.\n\nOf course, there's rotation and reflection, but even beyond that you can swap any two rows within its group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.\n\n\u2022 I wonder if there's any kinda of transformation that would work here that doesn't work on sudoku in general. \u2013\u00a0Rawling Feb 11 at 12:00\n\nI was beaten to a very similar answer, but you can create such a square based on the magic square below:\n\nabc\ndef\nghi\n\nTurn it into this:\n\nabcdefghi\ndefghiabc\nghiabcdef\nbcaefdhig\nefdhigbca\nhigbcaefd\ncabfdeigh\nfdeighcab\nighcabfde\n\nSince there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).\n\nOne such arrangement of the 3x3 grid could be:\n\n1 5 9\n8 3 4\n6 7 2\n\nThis would then be repeated along to the right, but rotating the order of the rows to create:\n\n1 5 9 8 3 4 6 7 2\n8 3 4 6 7 2 1 5 9\n6 7 2 1 5 9 8 3 4\n\nThe same could be done to repeat downwards but rotating the columns this time:\n\n1 5 9 8 3 4 6 7 2\n8 3 4 6 7 2 1 5 9\n6 7 2 1 5 9 8 3 4\n5 9 1 3 4 8 7 2 6\n3 4 8 7 2 6 5 9 1\n7 2 6 5 9 1 3 4 8\n9 1 5 4 8 3 2 6 7\n4 8 3 2 6 7 9 1 5\n2 6 7 9 1 5 4 8 3\n\nHowever:\n\nThis creates one such possibility of your given solution, as each of the 3x3 sections could be placed in the top left and the rotations occur from there.\n\nThe answer is no. A Sudoku satisfying the puzzle requirement (i.e. each 3x3 sub-grid is a semi-magic square) is not unique even if we do not make a distinction between Sudokus differing just by rotations or reflections. And the simplest reason, as pointed out by @DR Xorile, is that there are natural Sudoku transformations (namely certain rows and/or columns swaps) which are not rotations and reflections but which preserve the integrity of the Sudoku and the semi-magic property of the sub-grids. This brings up a question alluded to by @Rawling: what if we extend a rotations and reflections group of transformations with row and column permutations suggested by @DR Xorile (let's call this group RRRCP for brevity) and not make distinctions between Sudokus if they can be transformed into each other via any RRRCP transformations. Will it make a Sudoku built out of 3x3 sub-grid of semi-magic squares \"unique\"? At first I thought it might. On a smaller scale, a 3x3 semi-magic square is \"unique\" in a sense that all 72 semi-magic squares can be generated by applying transformations from a similar extended group (rotations, reflections, columns/rows permutations) to just one semi-magic square. But it turns out that for a full Sudoku the answer is still no - a Sudoku built out of 3x3 sub-grid of semi-magic squares is not \"unique\" even modulo RRRCP transformations. Here is why.\n\nFor the sake of precision here is the description of an RRRCP group. Let's call each set of three columns (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a \"macro\" column and each set of three rows (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a \"macro\" row. \"Macro\" columns and \"macro\" rows constitute a 3x3 \"macro\" grid. A RRRCP group consists of compositions of the following Sudoku transformations:\n\n\u2022 rotations and reflections;\n\n\u2022 permutations of the columns within a \"macro\" column (ex. columns 4, 5, 6)\n\n\u2022 permutations of the rows within a \"macro\" row (ex. rows 7, 8, 9)\n\n\u2022 permutations of whole \"macro\" columns within a 3x3 \"macro\" grid (ex swap columns 1 2 3 with columns 7, 8, 9)\n\n\u2022 permutations of whole \"macro\" rows within a 3x3 \"macro\" grid (ex swap rows 1 2 3 with rows 4, 5, 6).\n\nThe following two Sudokus S1 and S2 are built out of semi-magic squares but cannot be transformed into each other by RRRCP transformations.\n\n2 7 6 1 9 5 3 8 4\n9 5 1 8 4 3 7 6 2\n4 3 8 6 2 7 5 1 9\n\n3 8 4 2 7 6 1 9 5\n7 6 2 9 5 1 8 4 3\n5 1 9 4 3 8 6 2 7\n\n1 9 5 3 8 4 2 7 6\n8 4 3 7 6 2 9 5 1\n6 2 7 5 1 9 4 3 8\n\n-------------------------------\n\n2 7 6 5 1 9 8 4 3\n9 5 1 3 8 4 6 2 7\n4 3 8 7 6 2 1 9 5\n\n8 4 3 2 7 6 5 1 9\n6 2 7 9 5 1 3 8 4\n1 9 5 4 3 8 7 6 2\n\n5 1 9 8 4 3 2 7 6\n3 8 4 6 2 7 9 5 1\n7 6 2 1 9 5 4 3 8\n\nS1 and S2 cannot be transformed into each other by rotations or reflections because they have an identical central 3x3 sub-grid.\nS1 and S2 cannot be transformed into each other by permutations within a \"macro\" row/column because they have an identical \"macro\" diagonal.\nS1 and S2 cannot be transformed into each other by permutations of \"macro\" rows/columns because they are built out of different semi-magic squares.", "date": "2019-08-26 08:02:39", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/79552/sum-in-a-sudoku-with-sum-total-of-15-and-45", "openwebmath_score": 0.5439891815185547, "openwebmath_perplexity": 197.83016853095432, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126457229185, "lm_q2_score": 0.8976952989498448, "lm_q1q2_score": 0.8785857410882288}}
{"url": "https://math.stackexchange.com/questions/4062949/what-is-the-sum-of-number-of-digits-of-the-numbers-22001-and-52001/4063231", "text": "# What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$\n\nWhat is the sum of number of digits of the numbers $$2^{2001}$$ and $$5^{2001}$$? (Singapore 1970)\n\nI attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $$2^{2001}\\mod{10}\\equiv 4$$ and $$2^{2001}\\pmod{10}\\equiv 5$$. Could you please explain to me how to solve this question? This question is multiple choice with options $$1999, 2003, 4002, 6003, 2002$$\n\n\u2022 @samerivertwice yes base 10 Mar 15 at 17:15\n\u2022 Not hard to do with a computer program, but perhaps that's not allowed. Mar 15 at 17:18\n\u2022 @Sil I am certain that that is what the question asks Mar 15 at 17:36\n\u2022 @Sil I'll ask the author for an explanation, as soon as I get a response, I'll post it here Mar 15 at 18:05\n\u2022 @sil I worked it out given the multi choice \ud83d\ude09 Mar 15 at 18:08\n\nIt's $$2002$$. It's asking for the sum of the number of digits of $$2^{2001}$$ and $$5^{2001}$$ in base $$10$$, so just take the log base $$10$$ of each, take the ceiling function and hey presto:\n\n$$603+1399=2002$$\n\n\u2022 remark: without taking the ceiling function, $$2001\\log_{10}(2)+2001\\log_{10}(5) = 2001(\\log_{10}(2)+\\log_{10}(5)) = 2001\\log_{10}(10) = 2001$$ so no calculator is needed to know the answer is in $[2001,2003)$ Mar 15 at 18:14\n\u2022 \"the sum of the number of digits\" Well, that's just nasty..... Mar 15 at 18:16\n\u2022 @hgmath nice. I figured there was something like that although I just used the calculator ;) Mar 15 at 18:17\n\u2022 @samerivertwice your solution is brilliant, thank you very much Mar 15 at 18:18\n\u2022 @MichaelBlane for the logarithm I just used a calculator. If needed to do without a calculator this can be done, using an infinite series for the log function but the real magic is in hgmath's comment - if you want to learn something, that comment is the thing to think about. Mar 15 at 18:20\n\nIt looks like you don't need logarithms or any calculator to solve this problem. Let's start.\n\nFirst, observe that the following inequalities hold:\n\n$$10^m<\\underbrace {2^{2001}}_{m+1 ~ \\text{digits}}<10^{m+1}$$\n\n$$10^n<\\underbrace{5^{2001}}_{n+1 ~ \\text{digits}}<10^{n+1}$$\n\nYou get,\n\n$$10^{m+n}<10^{2001}<10^{m+n+2}$$\n\n$$2001=m+n+1$$\n\n$$m+n=2000$$\n\nFinally, the sum of digits of $$2^{2001}$$ and $$5^{2001}$$ is equal :\n\n\\begin{align}\\color {gold}{\\boxed {\\color{black}{m+1+n+1=m+n+2\\\\ \\qquad \\qquad \\qquad\\thinspace=2000+2 \\\\\\qquad \\qquad \\qquad \\thinspace=2002.}}}\\end{align}\n\nThe answer is $$\\overbrace{\\lfloor2001\\log_{10}(2)\\rfloor+1}^\\text{digits in 2^{2001}}+\\overbrace{\\lfloor2001\\log_{10}(5)\\rfloor+1}^\\text{digits in 5^{2001}}$$ However, we also have, using Iverson Brackets, $$\\lfloor x\\rfloor+\\lfloor y\\rfloor=\\lfloor x+y\\rfloor-[\\{x\\}+\\{y\\}\\ge1]$$ So we need to know $$\\{2001\\log_{10}(2)\\}+\\{2001\\log_{10}(5)\\}$$, but since $$2001\\log_{10}(2)+2001\\log_{10}(5)=2001$$, we know that the sum of their fractional parts is exactly $$0$$ or exactly $$1$$. Since the fractional parts are both positive, we must have exactly $$1$$.\n\nTherefore, \\begin{align} \\lfloor2001\\log_{10}(2)\\rfloor+1+\\lfloor2001\\log_{10}(5)\\rfloor+1 &=\\lfloor2001\\log_{10}(2)+2001\\log_{10}(5)\\rfloor+1\\\\ &=2002 \\end{align}\n\n\u2022 I see that I answered the question after it was edited to be correct. It was probably that edit which brought this question to the top of the front page just as I was looking.\n\u2013\u00a0robjohn\nMar 15 at 22:09\n\nGeneralization of the problem:\n\n\u2022 What is the sum of number of digits of the numbers $$2^N$$ and $$5^N$$?\n\n$$10^m<\\underbrace {2^{N}}_{m+1 ~ \\text{digits}}<10^{m+1}$$\n\n$$10^n<\\underbrace{5^{N}}_{n+1 ~ \\text{digits}}<10^{n+1}$$\n\n$$10^{m+n}<10^{N}<10^{m+n+2}$$\n\n$$N= m+n+1$$\n\n$$m+n=N-1$$\n\nThe sum of digits of the numbers $$2^{N}$$ and $$5^{N}$$ will be equal :\n\n\\begin{align}\\color {gold}{\\boxed {\\color{black}{m+1+n+1=m+n+2\\\\ \\qquad \\qquad \\qquad\\thinspace=N-1+2 \\\\\\qquad \\qquad \\qquad \\thinspace=N+1.}}}\\end{align}\n\n\u2022 Short answer: $$N+1$$ digits.", "date": "2021-09-23 21:15:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4062949/what-is-the-sum-of-number-of-digits-of-the-numbers-22001-and-52001/4063231", "openwebmath_score": 0.8009883165359497, "openwebmath_perplexity": 356.76096134621264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126494483639, "lm_q2_score": 0.8976952838963489, "lm_q1q2_score": 0.8785857296994968}}
{"url": "https://mathhelpboards.com/threads/help-factoring.6247/", "text": "# help factoring\n\n#### slowle4rner\n\n##### New member\nI'm having trouble factoring the following binomial... can someone try to point me in the right direction please?\n\n4y^3+4\n\nIt has been 7 years since I took algebra and I am trying to get my review done. This one just does not make sense to me right now...\n\nThanks!\n\n#### MarkFL\n\nStaff member\nFirst factor the 4 out, to get:\n\n$$\\displaystyle 4y^3+4=4\\left(y^3+1 \\right)=4\\left(y^3+1^3 \\right)$$\n\nNow apply the sum of cubes formula:\n\n$$\\displaystyle a^3+b^3=(a+b)\\left(a^2-ab+b^2 \\right)$$\n\nWhat do you get?\n\n#### slowle4rner\n\n##### New member\nSo...A^3+b^3 = (a+b)(a^2-ab+b^2)\n\n4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)\n\nThat's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply (4y+4)(y^2-y+1) or multiply 4(y+1)(y^2-y+1) and distribute the 4.\n\nWould I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...\n\nThanks for the help. I wasn't seeing it. Gratitude.\n\n#### SuperSonic4\n\n##### Well-known member\nMHB Math Helper\nSo...$$\\displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)$$\n\n$$\\displaystyle 4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)$$\n\nThat's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply $$\\displaystyle (4y+4)(y^2-y+1)$$ or multiply $$\\displaystyle 4(y+1)(y^2-y+1)$$ and distribute the 4.\nThe 4 is a bit of a red herring because, as you said, it's not a perfect cube but you can leave that alone out the front. You could say $$\\displaystyle (4y+4)(y^2-y+1)$$ (as well as $$\\displaystyle (y+1)(4y^2-4y+4)$$) but it's good form to leave it fully factored which means not distributing the four. If you ever do exams they usually want it fully factored.\n\nWould I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...\n\nThanks for the help. I wasn't seeing it. Gratitude.\nYes, that is the correct answer.\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nEven if you were not able to factor that \"4\" out, it just a number! If it had been, say, $$x^3+ 4$$ you could write it as $$x^3+(\\sqrt[3]{4})^3$$ and use that same fornula.", "date": "2022-08-10 19:46:49", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/help-factoring.6247/", "openwebmath_score": 0.8290965557098389, "openwebmath_perplexity": 457.30777979942474, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9802808759252645, "lm_q2_score": 0.8962513772903669, "lm_q1q2_score": 0.8785780851794255}}
{"url": "https://math.stackexchange.com/questions/881396/common-logarithm-question", "text": "# Common logarithm question\n\nI'm studying logarithms and am doing an exercise where you're supposed to evaluate the solutions of common logarithms without using a calculator. I'm very stuck on this one particular question. I know the answer because I used my calculator, but I'd like to know how to solve it without one. The question is $$\\log\\left(\\frac{10}{\\sqrt[\\large3]{10}}\\right)$$\n\nHow do I solve this without a calculator? (Please provide a step-by-step solution, this has really confused me.)\n\n\u2022 Can you write $10/\\root3\\of {10}$ as $10^x$ for some $x$? If so, what would $\\log 10^x$ be? \u2013\u00a0David Mitra Jul 29 '14 at 10:39\n\u2022 @DavidMitra I got as far as $10(10^{-\\frac13})$ but don't know where to go from there \u2013\u00a0imulsion Jul 29 '14 at 10:40\n\nHint\n\n$$\\frac{10}{\\sqrt[3]{10}} = \\frac{10}{10^{1/3}} = 10^{1-1/3} = 10^{2/3}$$\n\nNow, what would the logarithm (assuming base 10) of that final expression be?\n\n\u2022 Thanks for your answer, but I don't understand why $\\frac{10}{10^{1/3}} = 10^{1-1/3}$. Probably me being a moron, but I don't understand how you got there. \u2013\u00a0imulsion Jul 29 '14 at 10:46\n\u2022 @imulsion The following rule is very good to know: $\\frac{x^a}{x^b} = x^{a-b}$. Let $x=10$, $a=1$ and $b=1/3$. (Remember that $10 = 10^1$.) \u2013\u00a0naslundx Jul 29 '14 at 10:48\n\u2022 I can't believe that I forgot that :facepalm:. Thanks very much for your help though, I will accept your answer when I can \u2013\u00a0imulsion Jul 29 '14 at 10:49\n\u2022 @imulsion No need to feel stupid, we are all here to learn. :) \u2013\u00a0naslundx Jul 29 '14 at 10:50\n\nRemember that the logarithm of a quotient is the difference of logarithms: $$log\\left(\\frac{10}{\\sqrt[3]{10}}\\right)=log(10)-log(\\sqrt[3]{10})=1-log(10^{1/3})=1-\\frac {1}{3}\\cdot log(10)=1-\\frac {1}{3}=\\frac {2}{3}$$\n\nTo get a good understanding of logarithms it is good to realize that the following two questions are equivalent:\n\n1) What is the logarithm of $a$ on base of $g$? I.e. $\\log$$_{g}\\left(a\\right)=?$.\n\n2) To what power must $g$ be raised to get $a$ as outcome? I.e. $g^{?}=a$.\n\nHere $a>0$, $g>0$ and $g\\neq1$.\n\nSo $10^{\\frac{2}{3}}=\\frac{10}{\\sqrt[3]{10}}$ is the same information as $\\log_{10}\\frac{10}{\\sqrt[3]{10}}=\\frac{2}{3}$", "date": "2019-10-17 10:44:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/881396/common-logarithm-question", "openwebmath_score": 0.8385367393493652, "openwebmath_perplexity": 248.1910684233232, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769156265969, "lm_q2_score": 0.9005297907896396, "lm_q1q2_score": 0.8785360757284212}}
{"url": "https://mathoverflow.net/questions/232179/is-the-action-of-g-on-h-1tn-mathbbz-mathbbzn-faithful/232204", "text": "# Is the action of $G$ on $H_1(T^n, \\mathbb{Z}) = \\mathbb{Z}^n$ faithful?\n\nLet $G$ be a finite group of diffeomorphisms of the torus $T^n$ fixing some point $p$, i.e. $p$ is fixed by every element of $G$. I have two questions.\n\n1. Is the action of $G$ on $H_1(T^n, \\mathbb{Z}) = \\mathbb{Z}^n$ faithful?\n2. What if $G$ is not assumed to fix a point $p$?\n\u2022 2. If $G$ acts as a finite group of translations, then the action on $H_1$ is trivial (while if $G$ is nontrivial it fixes no point).\n\u2013\u00a0YCor\nFeb 26, 2016 at 9:14\n\u2022 This question is not really research level and should be asked at math stack exchange. Nevertheless, the techniques used to answer such a question may not be well-known. Nov 25, 2016 at 15:50\n\u2022 @user99720 but my comment is really immediate, there are already nice answers on the less trivial part of the question.\n\u2013\u00a0YCor\nNov 26, 2016 at 22:10\n\nFrom Edmonds' notes on Transformation groups: http://www.indiana.edu/~jfdavis/seminar/transformationgroupsb.pdf\n\nProblem 9. [p.28] Show that if a [finite] group $G$ acts on the torus $T^n$ with a fixed point x and induces the trivial action on $\\pi_1(T^n, x)$, then the action is trivial.\n\nSo the answer to question 1 is yes. Question 2 is answered by @YCor's comment and also discussed in the above notes.\n\nThis follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a connected manifold $M$ with a metric and an integer $m>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $m$ must move a point of $M$ at least a distance $d$.\n\nGiven a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\\pi_1(T^n)$, has order $m$, and fixes a point, there is a lift $\\tilde{f}$ of $f$ to the universal cover $\\mathbb{R}^n$ which fixes a point and has order $m$ and is uniformly continuous. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). The homotopy of $f$ to the identity lifts to a homotopy of $\\tilde{f}$ to the identity, so $\\tilde{f}$ also has the property that it moves points at most $D$ and has order $m$. Let $d$ be the constant from Newman's theorem for the manifold $\\mathbb{R}^n$ with Euclidean metric and $m$, then rescale the metric by $d/D$, we see that $\\tilde{f}$ is periodic of order $m$ and moves points at most distance $d$, hence is the identity. But then $f$ is also the identity.\n\nHence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\\pi_1(T^n)$.\n\nWe can expand the answer from user83633, following hints in the notes by Edmonds, as follows.\n\nLet $G$ be a finite group acting faithfully on an $n$-dimensional torus $T$ and fixing a point $p$. We can write $T$ as $V/L$ for some vector space $V$ and lattice $L<V$. As $T$ is homogeneous, we can assume that this identifies $p$ with the zero element. Let $\\pi\\colon V\\to T$ be the obvious projection, which is a universal covering. Covering theory tells us that there is a unique way to let $G$ act on $V$ such that $0$ is fixed and $\\pi$ is equivariant. This action preserves $\\pi^{-1}\\{0\\}=L$, and we can identify $L$ with $H_1(T)$ equivariantly. We want to show that the action of $G$ on $L$ is faithful. If not, we can choose a subgroup $C\\leq G$ of prime order $p$ that acts trivially on $L$. We will show that this leads to a contradiction.\n\nFirst note that the free action map $L\\times V\\to V$ is $G$-equivariant, and $C$ acts trivially on $L$, so we get an induced free action of $L$ on $V^C$. The map $\\pi$ gives an injection $V^C/L\\to T^C$; we claim that this is a homeomorphism. As everything is compact and hausdorff, we need only check that it is surjective. Suppose that $x\\in T^C$, and choose $v\\in V$ with $\\pi(v)=x$. If $g$ is a generator of $C$, we must then have $g.v=a+v$ for some $a\\in L$. As $g$ acts trivially on $L$ this gives $g^k.v=ka+v$ for all $k$, so in particular $v=g^p.v=pa+v$ so $pa=0$ so $a=0$ so $g.v=v$ as required.\n\nFor the rest of the argument we will use various cohomology groups; these are always taken with coefficients $\\mathbb{Z}/p$.\n\nNext, recall that $H^*(BC)$ is the tensor product of a polynomial algebra on a class $x$ of degree $2$ with an exterior algebra on a class $a$ of degree $1$. For any $C$-space $X$ we have a map $EC\\times_CX\\to EC/C=BC$, which makes the Borel cohomology group $H_C^*(X)=H^*(EC\\times_CX)$ into an algebra over $H^*(BC)$. We write $\\widehat{H}^*_C(X)$ for the ring obtained by inverting $x$ in $H_C^*(X)$. If $Y$ is a $C$-subspace of $X$ then we have relative groups $\\widehat{H}_C^*(X,Y)$ defined in a similar way. If $C$ acts freely on $X\\setminus Y$ then these relative groups are trivial, as one can prove by cellular induction starting from the case where $X=C\\times B^d$ and $Y=C\\times S^{d-1}$. On the other hand, if $C$ acts trivially on $X$ then $\\widehat{H}^*_C(X)=H^*(X)\\otimes R^*$, where $$R^*=\\widehat{H}^*_C(\\text{point})=\\mathbb{Z}/p[x,x^{-1}]\\otimes E[a].$$ Because $C$ has prime order, it must act freely on $X\\setminus X^C$, so $\\widehat{H}^*_C(X,X^C)=0$, so $$\\widehat{H}^*_C(X) \\simeq \\widehat{H}^*_C(X^C) \\simeq H^*(X^C)\\otimes R^*.$$ This is essentially what is called Smith theory.\n\nNow take $X$ to be our vector space $V$, so $V$ is contractible, and the usual spectral sequence $H^*(G;H^*(X))\\Longrightarrow H^*_G(X)$ gives $H^*_G(V)=H^*(BG)$ and therefore $\\widehat{H}^*_G(V)=R^*$. By comparison with Smith theory, we get $H^*(V^C)=\\mathbb{Z}/p$. As $L$ acts freely on $V^C$ with $V^C/L=T^C$ we get a spectral sequence $$H^*(L;H^*(V^C)) \\Longrightarrow H^*(T^C)$$ This admits a map from the similar spectral sequence $$H^*(L;H^*(V)) \\Longrightarrow H^*(T).$$ As $H^*(V^C)=H^*(V)=\\mathbb{Z}/p$, the map is an isomorphism on the initial page, and there is no room for any differentials anyway, so the map $H^*(T)\\to H^*(T^C)$ is an isomorphism. In particular, the map $H^n(T)\\to H^n(T^C)$ is an isomorphism.\n\nOn the other hand, if $X$ is any proper subset of $T$ then the restriction $H^n(T)\\to H^n(X)$ factors through the group $H^n(T\\setminus \\{x\\})=0$ for some $x\\in T$. Thus, we must have $T^C=T$, which means that $C$ acts trivially on $T$. This contradicts our assumption that the action of $G$ is faithful.\n\nFor a simpler and more geometric proof: It suffices to show the only element which acts as the identity is the identity. Suppose $g$ is our element which acts as the identity on $H_1$ and thus on $\\pi_1$. Lift $g$ to act on $\\mathbb{R}^2$. Define a homotopy $g_t(p) = (1 - t) p + t g(p)$ between $g$ and the identity. The homotopy descends to $\\mathbb{T}^2$. By surface theory we can upgrade this to an isotopy between $g$ and the identity on $\\mathbb{T}^2$. Then, and I'm not sure how to prove this (and in some sense this seems to be the crux of the issue) one can show that maps isotopic to the identity have infinite order.", "date": "2022-08-14 13:11:36", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/232179/is-the-action-of-g-on-h-1tn-mathbbz-mathbbzn-faithful/232204", "openwebmath_score": 0.9665765762329102, "openwebmath_perplexity": 53.055303541385186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576913496333, "lm_q2_score": 0.9005297807787537, "lm_q1q2_score": 0.8785360640436659}}
{"url": "https://math.stackexchange.com/questions/3210020/mistake-in-solving-int-frac1x-sqrtx2-1/3210025", "text": "# Mistake in solving $-\\int \\frac{1}{x\\sqrt{x^2-1}}$\n\nI have this function $$f:(-\\infty ,-1)\\rightarrow \\mathbb{R}, f(x)=\\frac{1}{x\\sqrt{x^{2}-1}}$$ and I need to find the primitives of $$f(x)$$.So because $$x<-1$$ I need to calculate $$-\\int \\frac{1}{x\\sqrt{x^2-1}}\\,dx=-\\arctan(\\sqrt{x^2-1})+C$$ but in my book the correct answer is $$\\arcsin\\left(\\frac{1}{x}\\right)+C$$\n\nWhere is my mistake?I solved the integral using $$u=\\sqrt{x^{2}-1}$$\n\nIt is known that $$\\arctan(x)+\\arctan(1/x)=C$$ where $$C$$ is a constant. So your solution can be written as $$\\arctan\\left(\\frac1{\\sqrt{x^2-1}}\\right):=u$$Then using some trig identities, $$\\frac1{\\sqrt{x^2-1}}=\\tan u\\\\\\sec^2 u=1+\\tan^2 u=\\frac{x^2}{x^2-1}\\\\\\sin^2 u=1-\\cos^2 u=1-\\left(\\frac{x^2-1}{x^2}\\right)=\\frac1{x^2}\\\\u=\\arcsin\\left(\\frac1x\\right)$$\n\nHint:\n\nYou might want to plot both $$\\arcsin(\\frac{1}{x})$$ and $$-\\arctan(\\sqrt{x^2 - 1})$$, from, say, $$-3$$ to $$-1$$. You could do this using Desmos (https://www.desmos.com/). You might even insert an extra minus sign here or there...\n\n$$\\arcsin(\\frac{1}{x}) +C = \\operatorname{arccot}(\\sqrt{x^2 - 1}) + C$$\n\n$$=\\frac{\\pi}{2} - \\arctan(\\sqrt{x^2 - 1})+ C$$\n\nYou can form another Constant $$C_1 = \\pi/2 + C$$.\n\nSo you solved correctly, the book answer and your answer are equivalent, you just had to manipulate it a little to get there. The difference is just of a constant which doesn't matter when calculating indefinite integrals.\n\n\u2022 Yes, I understand what you're saying but my book gives both results as anwers and I just need to pick one answer.For example, at exam how to know which is good ? \u2013\u00a0DaniVaja May 1 '19 at 18:48\n\u2022 @Vizag: I think that the relationship you've asserted in the first line is only correct for $x > 0$; you might want to check that by plotting both sides when $x < 0$. \u2013\u00a0John Hughes May 1 '19 at 18:50\n\u2022 @DaniVaja Both are fine, since they are both correct. So you shouldn't worry too much. You could use these answers to see how to show that both forms are equivalent (up to constants). \u2013\u00a0John Doe May 1 '19 at 18:50\n\u2022 @DaniVaja: Both the answers look correct to me. The expressions are equivalent up to constants. Are you sure it's not a multiple answers correct kind of question? \u2013\u00a0Vizag May 1 '19 at 18:53\n\u2022 Here is the original exercise: imgur.com/hhQUzoi (i don't know the code for posting images).The book says that one answer from 5 is correct and in this exercise I have both answers, which are both good like you said.But for example in an exam, how to know which answer to pick ?In this case they chose arcsin(1/x) + c as correct answer and not -arctan... \u2013\u00a0DaniVaja May 1 '19 at 18:56", "date": "2020-03-28 12:55:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3210020/mistake-in-solving-int-frac1x-sqrtx2-1/3210025", "openwebmath_score": 0.807016909122467, "openwebmath_perplexity": 303.2083180867979, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769156265969, "lm_q2_score": 0.9005297787765764, "lm_q1q2_score": 0.878536064008754}}
{"url": "https://mathhelpboards.com/threads/laplace-transform.7057/", "text": "# [SOLVED]Laplace Transform\n\n#### nacho\n\n##### Active member\nHow do you find the laplace transform of this without expanding it?\n\n$L(t^2+1)^2$\n\nLast edited:\n\n#### chisigma\n\n##### Well-known member\nHow do you find the laplace transform of this without expanding it?\n\n$L(t^2+1)^2$\nYou can use the fact that for n positive integer is...\n\n$\\displaystyle \\int_{0}^{\\infty} t^{n}\\ e^{- s\\ t}\\ dt = \\frac{n!}{s^{n+1}}\\ (1)$\n\n... expanding $\\displaystyle (1 + t^{2})^{2}$ in powers of t...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\nI didn't realize immediately that it was requested 'without expanding' ... very sorry!...\n\n#### chisigma\n\n##### Well-known member\nHow do you find the laplace transform of this without expanding it?\n\n$L(t^2+1)^2$\nThe possible solution is the use of a 'forgotten formula' ['forgotten' in the sense that it is neglected from most of the Complex Analysis 'Holybooks'...] according to which if You have two L-transformable functions $f_{1}(t)$ and $f_{2} (t)$ with L-transforms $F_{1} (s)$ and $F_{2} (s)$ and abscissas of convergence $\\sigma_{1}$ and $\\sigma_{2}$, then the L-transform of the product is given by the integral...\n\n$\\displaystyle \\mathcal {L} \\{f_{1}(t)\\ f_{2}(t)\\} = \\frac{1}{2\\ \\pi\\ i}\\ \\int_{\\gamma - i\\ \\infty}^{\\gamma + i\\ \\infty} F_{1} (z)\\ F_{2}(s - z)\\ dz\\ (1)$\n\n... where $\\displaystyle \\sigma_{1} < \\gamma < \\sigma - \\sigma_{2},\\ \\sigma> \\sigma_{1} + \\sigma_{2}$. In Your case is $\\displaystyle f_{1}(t) = f_{2} (t) = 1 + t^{2} \\implies F_{1}(s) = F_{2}(s) = \\frac{1}{s} + \\frac{2}{s^{3}},\\ \\sigma_{1}=\\sigma_{2}=0$. May be it exists a more comfortable way but till now I didnn't succed to find it!...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### Chris L T521\n\n##### Well-known member\nStaff member\nHow do you find the laplace transform of this without expanding it?\n\n$L(t^2+1)^2$\nTo compute this without expanding requires a little trick (especially if you're allowed to do this without using the integral definition of $\\mathcal{L}\\{f(t)\\}$).\n\nLet $f(t) = (t^2+1)^2$. We note that $f(0)=1$.\n\nThus, $f^{\\prime}(t) = 4t(t^2+1)$ and hence $f^{\\prime}(0) = 0$.\n\nFinally, we see that $f^{\\prime\\prime}(t) = 4(t^2+1) + 8t^2 = 12t^2+4$.\n\nWhy did we compute all these derivatives? Well, we know that $\\mathcal{L}\\{f^{\\prime\\prime}(t)\\} = s^2F(s) - sf(0) - f^{\\prime}(0)$, which in our case would be $\\mathcal{L}\\{f^{\\prime\\prime}(t)\\}=s^2F(s)-s$.\n\nTherefore,\n\n$\\mathcal{L}\\{12t^2+4\\} = \\mathcal{L}\\{f^{\\prime\\prime}(t)\\} = s^2F(s)-s$\n\nWe now compute the appropriate Laplace transforms and solve for $F(s)$:\n\n$s^2F(s) - s = \\mathcal{L}\\{12t^2+4\\} = 12\\cdot\\frac{2!}{s^3}+4\\cdot\\frac{1}{s}=\\frac{24}{s^3}+\\frac{4}{s} \\implies F(s) = \\frac{24}{s^5} + \\frac{4}{s^3}+\\frac{1}{s}$\n\nTherefore, $\\mathcal{L}\\{(t^2+1)^2\\} = \\dfrac{24}{s^5}+\\dfrac{4}{s^3}+\\dfrac{1}{s}$.\n\nI hope this makes sense!\n\n#### Jester\n\n##### Well-known member\nMHB Math Helper\nQuestion for the OP. Why would you want to compute this transform without expanding? I mean expanding would be the easiest approach.\n\nBTW - Nice work Chris!\n\nFollowing Chris's lead, you could keep going\n\\begin{align}\nf(t) &= (t^2+1)^2,\\;\\;\\; &f(0)&=1\\\\\nf'(t) &= 4t(t^2+1)\\;\\;\\; &f'(0)&=0\\\\\nf''(t) &=4(t^2+1) + 8t^2,\\;\\;\\;&f''(0)&=4\\\\\nf'''(t) &= 8t + 16t,\\;\\;\\;&f'''(0)&=0\\\\\nf^{(4)}(t)&=24,\\;\\;\\;&f^{(4)}(0)&=24\\\\\nf^{(5)}(t)& = 0,\n\\end{align}\n\nThen\n\n$\\mathcal{L}\\{f^{(5)}(t)\\} = s^5F(s) - f(0) s^4 - f'(0) s^3 - f''(0) s^2 - f'''(0) s - f^{(4)}(0) = 0.$\n\nSubstitute $f$ and its derivative at zero and solve for $F$ (same answer)\n\nLast edited:\n\n#### nacho\n\n##### Active member\nThanks for the responses.\n\n@Jester, really no other reason than for the sake of it. I've just started to learn laplace transforms and am finding a little out of the box thinking helps a lot.\n\nLast edited:", "date": "2021-09-25 06:48:07", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/laplace-transform.7057/", "openwebmath_score": 0.9901599884033203, "openwebmath_perplexity": 1300.521239342441, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.971129089711396, "lm_q2_score": 0.9046505440982949, "lm_q1q2_score": 0.8785324593970962}}
{"url": "http://math.stackexchange.com/questions/296664/how-do-i-prove-that-two-empty-sets-are-equivalent", "text": "# How do I prove that two empty sets are equivalent?\n\nI need to prove that two empty sets have the same cardinality, I know that proving equivalence is done by defining an onto and one-to-one function from one group to the other, but how can I define a function from an empty set to an empty set?\n\n-\nDo you mean define a function from an empty set to an empty set? \u2013\u00a0Git Gud Feb 6 '13 at 22:36\nyes of course, sorry. \u2013\u00a0Georgey Feb 6 '13 at 22:37\n@amWhy That is correct. I will will delete my comment. \u2013\u00a0Git Gud Feb 6 '13 at 23:49\n\nLet $A=B=\\varnothing$. A function from $A$ to $B$ is a subset $f$ of $A\\times B$ with the property that if $\\langle a,b_1\\rangle$ and $\\langle a,b_2\\rangle\\in f$, then $b_1=b_2$. The only subset of $A\\times B=\\varnothing\\times\\varnothing=\\varnothing$ is $\\varnothing$. Does it have the function property? Yes: it contains no ordered pairs at all, so the hypothesis\n\nif $\\langle a,b_1\\rangle$ and $\\langle a,b_2\\rangle\\in f$\n\nis never satisfied, and the property\n\nif $\\langle a,b_1\\rangle$ and $\\langle a,b_2\\rangle\\in f$, then $b_1=b_2$\n\nnever actually requires anything to be true $-$ it holds vacuously.\n\nThus, $\\varnothing$ is a function from $\\varnothing$ to $\\varnothing$. Is it a bijection? Is it true that for each $a\\in A=\\varnothing$ there is a unique $b\\in B=\\varnothing$ such that $\\langle a,b\\rangle\\in f$? Again the answer is yes, vacuously: there\u2019s no $a\\in A=\\varnothing$ at all, so the requirement is never actually invoked.\n\nHowever, all of that work is completely unnecessary, because it\u2019s always true that if $A=B$, then $|A|=|B|$. And if $A$ and $B$ are both empty, then they have exactly the same members, so they are the same set, a fact that I implicitly used above when I wrote $A=B=\\varnothing$.\n\n-\nNice answer, Brian. +1 \u2013\u00a0Rick Decker Feb 7 '13 at 2:00\nThanks, @Rick. ${}$ \u2013\u00a0Brian M. Scott Feb 7 '13 at 2:03\n\nHint: Recall that there are no elements in an empty set. Show that if $A$ and $B$ are empty sets then for every $x\\in A$ we have $x\\in B$, and vice versa. Therefore $A\\subseteq B$ and $B\\subseteq A$ and so $A=B$.\n\nIf two sets are equal then certainly they have the same cardinality.\n\n-\n\nHint:\n\nUse the definition of the (an) empty set - a set with no elements - to show, it is vacuously true that for all $x$ in the domain, $\\;x \\in \\emptyset_1 \\implies x \\in \\emptyset_2,\\;$ so $$\\emptyset_1 \\subseteq \\emptyset_2$$ and do similarly to show that it is vacuously true that $$\\emptyset_2 \\subseteq \\emptyset_1$$\n\nHence, $\\emptyset_1 = \\emptyset_2$.\n\nHence, $|\\emptyset_1| = |\\emptyset_2|$\n\n-\n\nThe extension and specification axioms implies that there is only one set with no elements.\n\n-", "date": "2016-05-24 19:53:04", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/296664/how-do-i-prove-that-two-empty-sets-are-equivalent", "openwebmath_score": 0.9099573493003845, "openwebmath_perplexity": 157.55959923761168, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129093889291, "lm_q2_score": 0.9046505280315008, "lm_q1q2_score": 0.8785324475737001}}
{"url": "https://math.stackexchange.com/questions/2413640/partial-fractions-decomposition-frac6x2-29x-29x1x-32-explanati/2413658", "text": "# Partial Fractions Decomposition $\\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors\n\nI am trying to solve the fraction $$\\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$ into partial fractions.\n\nNow, I thought it could be solved into the following\n\n$$\\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \\frac{A}{x+1} + \\frac{B}{(x-3)^2}$$\n\nbut this is apparently incorrect.\n\nAccording to the text, the decomposition is\n\n$$\\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \\frac{A}{x+1} + \\frac{B}{x-3} + \\frac{C}{(x-3)^2}$$\n\nI discussed this with my friend that the fraction first decomposes into\n\n$$\\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \\frac{A}{x+1} + \\frac{Bx +C}{(x-3)^2}$$\n\nbut I can't see how he derived this.\n\nI don't understand how he is correct.\n\n\u2022 Anything of your textbook's form can easily be expressed in your friend's form and vice versa. \u2013\u00a0Lord Shark the Unknown Sep 1 '17 at 20:08\n\nBoth decompositions used by your friend and the text are analogous since $$\\frac{Bx+C}{(x-3)^2}=\\frac{Bx}{(x-3)^2}+\\frac{C}{(x-3)^2}=\\frac{B(x-3)+3B}{(x-3)^2}+\\frac{C}{(x-3)^2}$$ $$=\\frac{B}{x-3}+\\frac{3B+C}{(x-3)^2}=\\frac{B'}{x-3}+\\frac{C'}{(x-3)^2}$$\n\nWhen you decompose, the degree of the numerator will be less than the degree of the denominator. How much less is yet to be determined. Assume that it is one degree less, and then if you get a zero coefficient, so be it.\n\nIf you had something like...$\\frac {P(x)}{(x^2 +x + 1)(x-1)}$ your first step would be $\\frac {Ax + B}{x^2+ x+ 1} + \\frac {C}{(x-1)}$\n\nWith each numerator of degree 1 less than each denominator.\n\nAnd the you might want to break down the first term to make it easier to integrate.\n\ne.g. $\\frac {A(x+\\frac12)}{x^2 + x + 1} + \\frac {B}{(x+\\frac 12)^2 + \\frac 34}$\n\nWith this problem.\n\n$(x-3)^2$ makes for a degree 2 denominator, and you need a degree 1 numerator.\n\nSo $\\frac {A}{x+1} + \\frac {Bx+C}{(x-3)^2}$ would be a good place to start.\n\nbut $\\frac {B}{(x-3)} + \\frac {C}{(x-3)^2}$ is easier to integrate, so you might want to skip the intermediate step.", "date": "2019-07-22 13:54:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2413640/partial-fractions-decomposition-frac6x2-29x-29x1x-32-explanati/2413658", "openwebmath_score": 0.9159519076347351, "openwebmath_perplexity": 327.64416240674905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338079816758, "lm_q2_score": 0.913676521650809, "lm_q1q2_score": 0.8785308651263545}}
{"url": "https://math.stackexchange.com/questions/3238370/let-fx-be-continous-in-x-0-and-fx-0-ne-0-prove-that-exists-c0-and", "text": "# Let $f(x)$ be continous in $x_0$ and $f(x_0)\\ne 0$. Prove that $\\exists C>0$ and neighbouhood of $x_0$ such that $|f(x)| \\ge C$\n\nLet a function $$f(x)$$ be continuous in some point $$x_0$$ and $$f(x_0) \\ne 0$$. Prove that there exists a number $$C > 0$$ and neighbourhood of $$x_0$$ such that forall $$x$$ in that neighbourhood the following inequality holds: $$|f(x)| \\ge C$$\n\nThe problem statement says that $$f(x_0) \\ne 0$$, so it is either greater than $$0$$ or less than $$0$$. Let's consider the case for $$f(x_0) > 0$$. In such case by continuity of $$f(x)$$ we know: $$\\lim_{x\\to x_0} f(x) = f(x_0) > 0$$\n\nOr in other words: $$\\forall \\epsilon > 0\\ \\exists \\delta_\\epsilon > 0\\ \\forall x: |x-x_0| < \\delta_\\epsilon \\implies |f(x) - f(x_0)| < \\epsilon$$\n\nSince $$f(x_0) \\ne 0$$ we may let $$\\epsilon = {f(x_0)\\over 2}$$. In such case: $$|f(x) - f(x_0)| < \\epsilon \\\\ |f(x) - f(x_0)| < { f(x_0)\\over 2 } \\\\ -{ f(x_0)\\over 2 } < f(x) - f(x_0) < { f(x_0)\\over 2 }\\\\ 0<{f(x_0)\\over 2} < f(x) < {3f(x_0)\\over 2}$$\n\nSo we may now choose any $$C \\in \\left(0; {f(x_0)\\over 2}\\right)$$, which would imply $$f(x) \\ge C$$.\n\nSimilar reasoning is applied to the case when $$f(x_0) < 0$$.\n\nI'm not sure my reasoning above is valid, so I would like to kindly ask for verification. Or for a correct proof if the above makes no sense. Thank you!\n\n\u2022 Yes, this absolutely fine ;) The neighbourhood would be $(-f(x_0)/2,f(x_0)/2)$ when $f(x_0)>0$. \u2013\u00a0weirdo May 24 at 16:24\n\u2022 This looks correct. You have $f(x) \\geq C$ for all $|x-x_0|\\leq\\delta_\\epsilon$. The reason this works is because of continuity, i.e. the statement is for all $\\epsilon >0$..., the the particular choice of $\\epsilon = f(x_0)/2$ works. \u2013\u00a0Dayton May 24 at 16:24\n\nYep, the above proof is valid. But for a couple suggestions in terms of proof writing, in the last line you said that $$f(x_0) \\geq C$$, but this is only true for all $$x$$ such that $$|x-x_0|<\\delta$$. Just make sure you include that in your last line so that people don't think you are saying that you are proving it for every $$x$$.\nHave you been introduced to the neighbourhood notation? It is basically $$B_r(x_0) := \\{|x-x_0| < r \\quad| \\quad \\forall x \\text{ in your domain.}\\}$$ This way you can say $$|f(x) - f(x_0)|<\\epsilon$$ as $$B_\\epsilon (x_0)$$", "date": "2019-10-18 10:56:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3238370/let-fx-be-continous-in-x-0-and-fx-0-ne-0-prove-that-exists-c0-and", "openwebmath_score": 0.8946059346199036, "openwebmath_perplexity": 107.76081287315928, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9867771770811146, "lm_q2_score": 0.8902942326713409, "lm_q1q2_score": 0.8785220296870228}}
{"url": "https://math.stackexchange.com/questions/2888072/show-convergence-and-find-the-limit-of-the-sequence-given-by-a-1-1-and-a-n1", "text": "# Show convergence and find the limit of the sequence given by $a_1=1$ and $a_{n+1}=\\frac{1}{3+a_n}$\n\nI've been trying to solve this exam question on an exam in real analysis. Thus, only such methods may be used. The problem is as follows.\n\nShow that the sequence $a_n$ defined by $a_1=1$ and $a_{n+1}=(3+a_n)^{-1}$ for $n=1,2,\\dots$ converges and determine the limit.\n\nI'm not really able to find a closed form for the sequence. I guess that computing that and then showing that it's a Cauchy sequence works for showing convergence.\n\nI just have to find that closed form first, which I'm unable to do. If there's a general strategy I'd love to see how that works as I've been unable to find any.\n\n\u2022 Hint: monotonicity. Maybe monotonicity for subsequences. \u2013\u00a0xbh Aug 19 '18 at 19:56\n\nRecursively $a_n>0$, $f(x)={1\\over{x+3}}$, $f'(x)=-{1\\over{(x+3)^2}}$, implies that $|a_{n+2}-a_{n+1}|=|f'(c_n)||a_{n+1}-a_n|\\leq {1\\over 3^2}|a_{n+1}-a_n|$, $c_n$ is an element of $(a_n,a_{n+1})$ or $(a_{n+1},a_n)$. You deduce that the sequence is a Cauchy sequence. (recursively you can show that $|a_{n+1}-a_n|\\leq {1\\over 3^{2(n-2)}}|a_2-a_1|$.\n\nso it converges, the limit verifies $f(l)=l$ which implies that $l^2+3l-1=0$ ans is the positive root of this quadratic equation.\n\nThere need not be such a closed form for the sequence. If a limit $l$ exists it must be that$$l=\\dfrac{1}{l+3}$$which yields to $$l=-1.5\\pm\\sqrt{3.25}$$where $l=-1.5-\\sqrt{3.25}$ is not possible. To show that the sequence converges to $\\sqrt{3.25}-1.5$ lets define $$e_n=a_n-(\\sqrt{3.25}-1.5)$$therefore $$e_{n+1}=a_{n+1}-(\\sqrt{3.25}-1.5)=\\dfrac{1}{a_n+3}-(\\sqrt{3.25}-1.5)=\\dfrac{1-(e_n+1.5+\\sqrt{3.25})(\\sqrt{3.25}-1.5)}{a_n+3}=\\dfrac{-e_n(\\sqrt{3.25}-1.5)}{a_n+3}$$therefore $$|e_{n+1}|<\\dfrac{|e_n|(\\sqrt{3.25}-1.5)}{3}$$which shows that $e_n\\to 0$ and $$a_n\\to \\sqrt{3.25}-1.5$$\n\n\u2022 +1. I had thought of the same approach. Do you have something in mind to show that the convergence happens without having to assume that the limit exists first? \u2013\u00a0Vizag Aug 19 '18 at 20:18\n\u2022 Well! Consider $$a_{n+2}=\\dfrac{1}{3+a_{n+1}}=\\dfrac{a_n+3}{3a_n+10}<\\dfrac{4}{10}a_n$$ therefore we can conclude what we want \u2013\u00a0Mostafa Ayaz Aug 19 '18 at 20:23\n\nMonotonicity method.\n\nRewrite the recursive relation as $$a_{n+2} = \\frac 1 {3+ \\dfrac 1 {3 + a_n}} = \\frac {3+a_n} {3(3+a_n) + 1} = \\frac {a_n + 3} {3 a_n + 10},$$ or $$a_{n+2} = \\frac 13 - \\frac 1{9a_n + 30} \\leqslant \\frac 13.$$ Since $$a_{n+4}-a_{n+2} = \\frac 1 {9a_n + 30} - \\frac 1{9a_{n+2} +30} = \\frac {9(a_{n+2}-a_n )} {(9a_n +30) (9a_{n+2}+30)},$$ and clearly $a_n \\geqslant 0$, we conclude that subsequences $a_{2k}, a_{2k+1}$ are monotonic [not necessarily the same type]. Since $a_1 = 1, a_2 = 1/4, a_3 = 4/13, a_4=13/34, a_1 > a_3, a_2 < a_4$, we know that $a_{2k} < a_{2k+2} \\leqslant 1/3, 1 \\geqslant a_{2k+1} > a_{2k+3}$. Hence $a_{2k}, a_{2k+1}$ both converge [since both are monotonic and bounded]. By the recursive relation, they converge to the same limit $L > 0$.\n\nLet $n \\to \\infty$ in the first formula, $$L = \\frac {L + 3}{3L + 10},$$ or simply, $$L(L+3) =1.$$\n\nHere is a somewhat systematic approach. The crux of the method is to separate the numerator and denominator. Write $a_n = p_n / q_n$. Can we work out $p_{n+1}, q_{n+1}$? We have $$x_{n+1} = \\frac{1}{3 + p_n/q_n} = \\frac{q_n}{3q_n + p_n}$$ So $p_{n+1} = q_n$ and $q_{n+1} = 3q_n + p_n = 3q_n + q_{n-1}$. Now we can see that the important value is the denominator. We are lucky because our second equation is a second order recurring relation for $q_n$, so we can solve for the general form using the following theorem:\n\nLet $(x_n)_{n \\ge 0}$ be a sequence such that there exists $(a,b)\\in \\mathbb{C}^2$ $x_{n+2} = ax_{n+1} + bx_n$. Then the sequence $(u_n)_{n \\ge 0}$, $u_n = r^n$ is a solution iff $r$ satisfies the characteristic equation: $$r^2 - ar - b = 0$$ Furthermore if the characteristic equation admits two distinct roots then any solution is of the form $$\\alpha r_1^n + \\beta r_2 ^n$$\n\nWe don't mention the case of repeated roots. In our case the characteristic equation is $$r^2 - 3r - 1 = 0$$\n\nThe two roots are $j_1 = \\frac{3 + \\sqrt{13}}{2}, j_2 = \\frac{3-\\sqrt{13}}{2} =$. The best part is that we don't need to solve for $\\alpha$ and $\\beta$ (though you do need to check $\\alpha \\ne 0$, which is easily done); notice that $|j_2| < 1$. Now we can write $$x_n = \\frac{p_n}{q_n} = \\frac{q_{n-1}}{q_n} = \\frac{\\alpha j_1^{n-1} + \\beta j_2^{n-1}}{\\alpha j_1^{n} + \\beta j_2^{n}}$$ Since $|j_2|<1$, as $n\\to \\infty$, this goes to $1/j_1$. You can check that $1/j_1$ satisfies $L = 1/(3 + L)$", "date": "2020-10-23 11:40:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2888072/show-convergence-and-find-the-limit-of-the-sequence-given-by-a-1-1-and-a-n1", "openwebmath_score": 0.9737157225608826, "openwebmath_perplexity": 97.18947679468239, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429164804706, "lm_q2_score": 0.891811038968166, "lm_q1q2_score": 0.8784721467746809}}
{"url": "https://math.codidact.com/posts/281630", "text": "Q&A\n\n# difference between quotient rule and product rule\n\n+2\n\u22121\n\nProduct rule :\n\n$$\\frac{d}{dx} f(x)g(x)=f'(x)g(x)+f(x)g' (x)$$\n\nQuotient rule :\n\n$$\\frac{d}{dx} \\frac{f(x)}{g(x)}=\\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$\n\nSuppose, the following is given in question. $$y=\\frac{2x^3+4x^2+2}{3x^2+2x^3}$$\n\nSimply, this is looking like Quotient rule. But, if I follow arrange the equation following way\n\n$$y=(2x^3+4x^2+2)(3x^2+2x^3)^{-1}$$\n\nThen, we can solve it using Product rule. As I was solving earlier problems in a pdf book using Product rule. I think both answers are correct. But, my question is, How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\nWhy does this post require moderator attention?\nWhy should this post be closed?\n\n+1\n\u22120\n\nHow does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\nis that any experienced scientist knows several methods to solve problems and uses those that are most convenient for them at that particular time.\n\nI would look at that derivative and use the quotient rule. But if there was something in the source of the problem that suggested that it made more sense to write the denominator as $(3x^2+2x^3)^{-1}$ then the product rule would be more appropriate.\n\nWhy does this post require moderator attention?\n\n+1\n\u22120\n\nis it OK to use Product rule instead of Quotient rule in University and Real Life?\n\nSure. For whatever reason, I long had a hard time remembering the quotient rule, and instead used the process you describe.\n\nWhy does this post require moderator attention?\n\n+1\n\u22120\n\nIf you work out deriving the quotient rule yourself using the exact trick you're highlighting, you can see that the quotient rule is nothing more than the product rule and the chain rule used together. If you rearrange a quotient to use the product rule, then you'll very likely be using the chain rule shortly thereafter on the $(\\cdots)^{-1}$ part, and you will inevitably arrive at exactly the same result as if you had used the quotient rule but with more steps. The only potential difference is whether your result looks like $\\frac{a}{b} + \\frac{c}{b^2}$ or $\\frac{ab + c}{b^2}$\u2014but of course, as I'm sure you can see, that's no difference at all.\n\nWhy does this post require moderator attention?", "date": "2021-10-26 16:02:13", "meta": {"domain": "codidact.com", "url": "https://math.codidact.com/posts/281630", "openwebmath_score": 0.9125794172286987, "openwebmath_perplexity": 390.38619368406614, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147201714923, "lm_q2_score": 0.9032942086563875, "lm_q1q2_score": 0.8783765851431307}}
{"url": "http://math.stackexchange.com/questions/692855/invertible-matrix-and-linearly-independent-vectors-proof", "text": "# Invertible Matrix and Linearly Independent Vectors Proof\n\nTrying to do this one:\n\nSuppose $A$ is an invertible $n$ x $n$ matrix and the vectors $v_1$, $v_2$, ..., $v_n$ are linearly independent. Show that the vectors $Av_1$, $Av_2$, ..., $Av_n$ are linearly independent.\n\nI know that A's column vectors are linearly independent since A is invertible. I also know there is no relation amongst the $v_i$ because they're linearly independent.\n\nMy idea is to write the product of A and a given $v_i$ in terms of the columns of A. Not sure if this is right, any guidance much appreciated!\n\nThanks, Mariogs\n\n-\nMaybe you can do a proof by contradiction? If they are lin dependent, then the zero vector can be written as a lin combination (with nonzero scalars) of your $Av's$ That means that a particular $Av$ vector can be written as a lin combination of the remaining $Av's$. Now what does that mean? \u2013\u00a0 imranfat Feb 27 '14 at 16:33\n\nLet $w_i=Av_i$ for $i=1,\\dots,n$. To show linear independence, we must show that $$c_1w_1+\\cdots+c_nw_n=0 \\Longrightarrow c_i=0,\\quad i=1,\\dots n.$$ Since $A$ is invertible, we can left-multiply $c_1w_1+\\dots=0$ by $A^{-1}$ to get...\n\nCan you take it from there? Remember that the vectors $v_1,\\dots v_n$ are linearly independent.\n\n-\nAh this makes sense to me, thanks! \u2013\u00a0 bclayman Feb 27 '14 at 16:56\nTo finish it: We multiply on the left by $A^-1$ and get $c_1 v_1$ + ... + $c1 v_n$ = 0. Since $v_1, ..., v_n$ are linearly independent, we must have $c_i$ = 0 for all i. Yes? \u2013\u00a0 bclayman Feb 27 '14 at 17:00\nThat's right, yes. \u2013\u00a0 user12477 Mar 3 '14 at 12:05\n\nNotice that since $A$ is invertible then $$Ax=0\\iff x=0$$ Now let $a_1,a_2,\\ldots,a_n\\in \\Bbb R$ such that $$\\sum_{k=1}^na_k Av_k=0\\iff A\\left(\\sum_{k=1}^na_k v_k\\right)=0\\iff \\sum_{k=1}^na_k v_k=0\\Rightarrow a_i=0\\;\\forall i$$ since $v_1$, $v_2$, ..., $v_n$ are linearly independent. Conclude.\n\n-\nWhoa! You've been busy with answers! \u2013\u00a0 amWhy Feb 28 '14 at 13:34\n\nWhy not use the definition directly? Justify/explain the following:\n\n$$\\text{For scalars}\\;\\;c_1,...,c_n\\;:\\;\\;0=\\sum_{k=1}^nc_kAv_k=\\sum_{k=1}^nA(c_kv_k)=A\\left(\\sum_{k=1}^nc_kv_k\\right)\\iff$$\n\n$$\\iff\\sum_{k=1}^nc_kv_k=0\\iff c_1=c_2=\\ldots=c_n=0$$\n\n-", "date": "2015-08-29 07:31:54", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/692855/invertible-matrix-and-linearly-independent-vectors-proof", "openwebmath_score": 0.9514034390449524, "openwebmath_perplexity": 225.7339076726188, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97241471777321, "lm_q2_score": 0.9032942086563877, "lm_q1q2_score": 0.8783765829767762}}
{"url": "https://madhavamathcompetition.com/category/chennai-mathematical-institute-entrance/", "text": "Miscellaneous questions: Part I: tutorial practice for preRMO and\u00a0RMO\n\nProblem 1:\n\nThe sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.\n\nProblem 2:\n\nLet T be the set of all triples (a,b,c) of integers such that $1 \\leq a < b < c \\leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.\n\nProblem 3:\n\nIn a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.\n\nProblem 4:\n\nFive men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:\n\nA: I see three black and one white cap.\nB: I see four white caps.\nC: I see one black and three white caps.\nD: I see four black caps.\n\nProblem 5:\n\nLet f be a bijective (one-one and onto) function from the set $A=\\{ 1,2,3,\\ldots,n\\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \\in A$. Note that $f^{M}$ denotes the composite function $f \\circ f \\circ f \\ldots \\circ f$ repeated M times.\n\nProblem 6:\n\nShow that there exists a convex hexagon in the plane such that:\na) all its interior angles are equal\nb) its sides are 1,2,3,4,5,6 in some order.\n\nProblem 7:\n\nThere are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.\n\nProblem 8:\n\nIn each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?\n\nProblem 9:\n\nGiven the seven element set $A = \\{ a,b,c,d,e,f,g\\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.\n\nTry these !!\n\nRegards,\nNalin Pithwa\n\nTowards Baby Analysis: Part I: INMO, IMO and CMI\u00a0Entrance\n\n$\\bf{Reference: \\hspace{0.1in}Introductory \\hspace{0.1in} Real Analysis: \\hspace{0.1in} Kolmogorov \\hspace{0.1in} and \\hspace{0.1in} Fomin; \\hspace{0.1in}Dover \\hspace{0.1in }Publications}$\n\n$\\bf{Equivalence \\hspace{0.1in} of \\hspace{0.1in} Sets \\hspace{0.1in} The \\hspace{0.1in}Power \\hspace{0.1in }of \\hspace{0.1in }a \\hspace{0.1in}Set}$\n\n$\\bf{Section 1}$:\n\n$\\bf{Finite \\hspace{0.1in} and \\hspace{0.1in} infinite \\hspace{0.1in} sets}$\n\nThe set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be $\\it{finite}$.$\\it{Clearly \\hspace{0.1in} we \\hspace{0.1in}can \\hspace{0.1in} be \\hspace{0.1in} sure \\hspace{0.1in} that \\hspace{0.1in} a \\hspace{0.1in} set \\hspace{0.1in}is \\hspace{0.1in}finite \\hspace{0.1in} without \\hspace{0.1in} knowing \\hspace{0.1in} the \\hspace{0.1in} number \\hspace{0.1in} of elements \\hspace{0.1in}in \\hspace{0.1in}it.}$\n\nOn the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, $\\it{if \\hspace{0.1in } we \\hspace{0.1in}remove \\hspace{0.1in} one \\hspace{0.1in} element \\hspace{0.1in}from \\hspace{0.1in}each \\hspace{0.1in}set, \\hspace{0.1in}then \\hspace{0.1in}remove \\hspace{0.1in}two \\hspace{0.1in}elements, \\hspace{0.1in}three \\hspace{0.1in}elements, \\hspace{0.1in}and \\hspace{0.1in}so \\hspace{0.1in}on, \\hspace{0.1in}there \\hspace{0.1in}will \\hspace{0.1in}still \\hspace{0.1in}be \\hspace{0.1in}elements \\hspace{0.1in}left \\hspace{0.1in}in \\hspace{0.1in}the \\hspace{0.1in}set \\hspace{0.1in}in \\hspace{0.1in}each \\hspace{0.1in}stage}$. Accordingly, sets of these kind are called $\\it{infinite}$ sets.\n\nGiven two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.\n\nTo compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a $\\it{one-\\hspace{0.1in}to-\\hspace{0.1in}one \\hspace{0.1in}correspondence}$ between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.\n\n$\\bf{Section 2}$:\n\n$\\bf{Countable \\hspace{0.1in} Sets}$.\n\nThe simplest infinite set is the set $\\mathscr{Z^{+}}$ of all positive integers. An infinite set is called $\\bf{countable}$ if its elements can be put into one-to-one correspondence with those of $\\mathscr{Z^{+}}$. In other words, a countable set is a set whose elements can be numbered $a_{1}, a_{2}, a_{3}, \\ldots a_{n}, \\ldots$. By an $\\bf{uncountable}$ set we mean, of course, an infinite set which is not countable.\n\nWe now give some examples of countable sets:\n\n$\\bf{Example 1}$:\n\nThe set $\\mathscr{Z}$ of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between $\\mathscr{Z}$ and $\\mathscr{Z^{+}}$ of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \\geq 0$ with the odd number $2n+1$, and the negative integer $n<0$ with the even number $2|n|$, that is,\n\n$n \\leftrightarrow (2n+1)$, if $n \\geq 0$, and $n \\in \\mathscr{Z}$\n$n \\leftrightarrow 2|n|$, if $n<0$, and $n \\in \\mathscr{Z}$\n\n$\\bf{Example 2}$:\n\nThe set of all positive even numbers is countable, as shown by the obvious correspondence $n \\leftrightarrow 2n$.\n\n$\\bf{Example 3}$:\n\nThe set 2,4,8,$\\ldots 2^{n}$ is countable as shown by the obvious correspondence $n \\leftrightarrow 2^{n}$.\n\n$\\bf{Example 4}: The set$latex \\mathscr{Q}$of rational numbers is countable. To see this, we first note that every rational number $\\alpha$ can be written as a fraction $\\frac{p}{q}$, with $q>0$ with a positive denominator. (Of course, p and q are integers). Call the sum $|p|+q$ as the \u201cheight\u201d of the rational number $\\alpha$. For example, $\\frac{0}{1}=0$ is the only rational number of height zero, $\\frac{-1}{1}$, $\\frac{1}{1}$ are the only rational numbers of height 2, $\\frac{-2}{1}$, $\\frac{-1}{2}$, $\\frac{1}{2}$, $\\frac{2}{1}$ are the only rational numbers of height 3, and so on. We can now arrange all rational numbers in order of increasing \u201cheight\u201d (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set $\\mathscr{Z^{+}}$ of all positive integers. $\\it{Next \\hspace{0.1in}we \\hspace{0.1in} prove \\hspace{0.1in}some \\hspace{0.1in}elementary \\hspace{0.1in}theorems \\hspace{0.1in}involving \\hspace{0.1in}countable \\hspace{0.1in}sets}$ $\\bf{Theorem1}$. $\\bf{Every \\hspace{0.1in} subset \\hspace{0.1in}of \\hspace{0.1in}a \\hspace{0.1in}countable \\hspace{0.1in}set \\hspace{0.1in}is \\hspace{0.1in}countable}$. $\\bf{Proof}$ Let set A be countable, with elements $a_{1}, a_{2}, a_{3}, \\ldots$, and let set B be a subset of A. Among the elements $a_{1}, a_{2}, a_{3}, \\ldots$, let $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \\ldots$ be those in the set B. If the set of numbers $n_{1}, n_{2}, n_{3}, \\ldots$ has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence $i \\leftrightarrow a_{n_{i}}$). $\\bf{QED.}$ $\\bf{Theorem2}$ $\\bf{The \\hspace{0.1in}union \\hspace{0.1in}of \\hspace{0.1in}a \\hspace{0.1in}finite \\hspace{0.1in}or \\hspace{0.1in}countable \\hspace{0.1in}number \\hspace{0.1in}of \\hspace{0.1in}countable \\hspace{0.1in}sets \\hspace{0.1in}A_{1}, A_{2}, A_{3}, \\ldots \\hspace{0.1in}is \\hspace{0.1in}itself \\hspace{0.1in}countable.}$ $\\bf{Proof}$ We can assume that no two of the sets $A_{1}, A_{2}, A_{3}, \\ldots$ have any elements in common, since otherwise we could consider the sets $A_{1}$, $A_{2}-A_{1}$, $A_{3}-(A_{1}\\bigcup A_{2})$, $\\ldots$, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of $A_{1}, A_{2}, A_{3}, \\ldots$ in the form of an infinite table $\\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\\ldots \\\\ a_{21} &a_{22} & a_{23} & a_{24} & \\ldots \\\\ a_{31} & a_{32} & a_{33} & a_{34} & \\ldots \\\\ a_{41} & a_{42} & a_{43} & a_{44} & \\ldots \\\\ \\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\end{array}$ where the elements of the set $A_{1}$ appear in the first row, the elements of the set $A_{2}$ appear in the second row, and so on. We now count all the elements in the above array \u201cdiagonally\u201d; that is, first we choose $a_{11}$, then $a_{12}$, then move downwards, diagonally to \u201cleft\u201d, picking $a_{21}$, then move down vertically picking up $a_{31}$, then move across towards right picking up $a_{22}$, next pick up $a_{13}$ and so on ($a_{14}, a_{23}, a_{32}, a_{41}$)as per the pattern shown: $\\begin{array}{cccccccc} a_{11} & \\rightarrow & a_{12} &\\hspace{0.1in} & a_{13} & \\rightarrow a_{14} & \\ldots \\\\ \\hspace{0.1in} & \\swarrow & \\hspace{0.1in} & \\nearrow & \\hspace{0.01in} & \\swarrow & \\hspace{0.1in} & \\hspace{0.1in}\\\\ a_{21} & \\hspace{0.1in} & a_{22} & \\hspace{0.1in} & a_{23} \\hspace{0.1in} & a_{24} & \\ldots \\\\ \\downarrow & \\nearrow & \\hspace{0.1in} & \\swarrow & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in}\\\\ a_{31} & \\hspace{0.1in} & a_{32} & \\hspace{0.1in} & a_{33} & \\hspace{0.1in} & a_{34} & \\ldots \\\\ \\hspace{0.1in} & \\swarrow & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in} & \\hspace{0.1in}\\\\ a_{41} & \\hspace{0.1in} & a_{42} &\\hspace{0.1in} & a_{43} &\\hspace{0.1in} &a_{44} &\\ldots\\\\ \\ldots & \\hspace{0.1in} & \\ldots & \\hspace{0.1in} & \\ldots & \\hspace{0.1in} & \\ldots & \\hspace{0.1in} \\end{array}$ It is clear that this procedure associates a unique number to each element in each of the sets $A_{1}, A_{2}, \\ldots$ thereby establishing a one-to-one correspondence between the union of the sets $A_{1}, A_{2}, \\ldots$ and the set $\\mathscr{Z^{+}}$ of all positive integers. $\\bf{QED.}$ $\\bf{Theorem3}$ $\\bf{Every \\hspace{0.1in}infinite \\hspace{0.1in}subset \\hspace{0.1in}has \\hspace{0.1in}a \\hspace{0.1in}countable \\hspace{0.1in}subset.}$ $\\bf{Proof}$ Let M be an infinite set and $a_{1}$ any element of M. Being infinite, M contains an element $a_{2}$ distinct from $a_{1}$, an element $a_{3}$ distinct from both $a_{2}$ and $a_{1}$, and so on. Continuing this process, (which can never terminate due to \u201cshortage\u201d of elements, since M is infinite), we get a countable subset $A= \\{ a_{1}, a_{2}, a_{3}, \\ldots, a_{n}, \\ldots\\}$ of the set $M$. $\\bf{QED.}$ $\\bf{Remark}$ Theorem 3 shows that countable sets are the \u201csmallest\u201d infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below. $\\bf{Section3}$ $\\bf{Equivalence \\hspace{0.1in} of \\hspace{0.1in} sets}$ We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set $\\mathscr{Z^{+}}$ of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N. $\\bf{Definition}$ Two sets M and N are said to be $\\bf{equivalent}$ (written $M \\sim N$) if there is a one-to-one correspondence between the elements of M and the elements of N. The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set $\\mathscr{Z^{+}}$ of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent. $\\bf{Example1}$ The sets of points in any two closed intervals$[a,b]$and$[c,d]\\$ are equivalent; you can \u201csee\u2019 a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment \u201cab\u201d above the base of the triangle; where \u201ca\u201d lies on one side of the triangle and \u201cb\u201d lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect.\n\n$\\bf{Example2}$\n\nThe set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence $z \\leftrightarrow \\alpha$ can be established by using stereographic projection. The origin is the North Pole of the sphere.\n\n$\\bf{Example3}$\n\nThe set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula $y=\\frac{1}{\\pi}\\arctan{x}+\\frac{1}{2}$ establishes a one-to-one correspondence between these two sets. $\\bf{QED}$.\n\nThe last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are \u201cas many\u201d positive integers as integers of arbitrary sign, there are \u201cas many\u201d points in the interval $(0,1)$ as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by:\n\n$\\bf{Theorem4}$\n\n$\\bf{Every \\hspace{0.1in} infinite \\hspace{0.1in} set \\hspace{0.1in}is \\hspace{0.1in} equivalent \\hspace{0.1in} to \\hspace{0.1in}one \\hspace{0.1in}of \\hspace{0.1in}its \\hspace{0.1in}proper \\hspace{0.1in}subsets.}$\n\n$\\bf{Proof}$\n\nAccording to Theorem 3, every infinite set M contains a countable subset. Let this subset be $A=\\{a_{1}, a_{2}, a_{3}, \\ldots, a_{n}, \\ldots \\}$ and partition A into two countable subsets $A_{1}=\\{a_{1}, a_{3}, a_{5}, \\ldots \\}$ and $A_{2}=\\{a_{2}, a_{4}, a_{6}, \\ldots \\}$.\n\nObviously, we can establish a one-to-one correspondence between the countable subsets A and $A_{1}$ (merely let $a_{n} \\leftrightarrow a_{2n-1}$). This correspondence can be extended to a one-to-one correspondence between the sets $A \\bigcup (M-A)=M$ and $A_{1} \\bigcup (M-A)=M-A_{2}$ by simply assigning x itself to each element $x \\in M-A$. But $M-A_{2}$ is a proper subset of M. $\\bf{QED}$.\n\nMore later, to be continued,\n\nRegards,\nNalin Pithwa\n\nA fifth degree equation in two variables: a clever\u00a0solution\n\nQuestion:\n\nVerify the identity: $(2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}$\n\nlet us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:\n\n$P_{1}+P_{2}=P_{3}+P_{4}$\n\nMethod I:\n\nUse binomial expansion. It is a very longish tedious method.\n\nMethod II:\n\nFactorize each of the quadratic expressions $P_{1}, P_{2}, P_{3}, P_{4}$ using quadratic formula method (what is known in India as Sridhar Acharya\u2019s method):\n\nNow fill in the above details.\n\nYou will conclude very happily that :\n\nThe above identity is transformed to :\n\n$P_{1}=(x+y+\\sqrt{3}y)^{5}(x+y-\\sqrt{3}y)^{5}$\n\n$P_{2}=(-1)^{5}(x-y-\\sqrt{3}y)^{5}(x-y+\\sqrt{3}y)^{5}$\n\n$P_{3}=(i^{2}(x-y-\\sqrt{3}y)(x-y+\\sqrt{3}y))^{5}$\n\n$P_{4}=((-i^{2})(x+y+\\sqrt{3}y)(x-y-\\sqrt{3}y))^{5}$\n\nYou will find that $P_{1}=P_{4}$ and $P_{2}=P_{4}$\n\nHence, it is verified that the given identity $P_{1}+P_{2}=P_{3}+P_{4}$. QED.\n\nRegards,\nNalin Pithwa.\n\nSet Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for\u00a0practice\n\nProblem 1:\n\nProve that a function f is 1-1 iff $f^{-1}(f(A))=A$ for all $A \\subset X$. Given that $f: X \\longrightarrow Y$.\n\nProblem 2:\n\nProve that a function if is onto iff $f(f^{-1}(C))=C$ for all $C \\subset Y$. Given that $f: X \\longrightarrow Y$.\n\nProblem 3:\n\n(a) How many functions are there from a non-empty set S into $\\phi$\\?\n\n(b) How many functions are there from $\\phi$ into an arbitrary set $S$?\n\n(c) Show that the notation $\\{ X_{i} \\}_{i \\in I}$ implicitly involves the notion of a function.\n\nProblem 4:\n\nLet $f: X \\longrightarrow Y$ be a function, let $A \\subset X$, $B \\subset X$, $C \\subset Y$ and $D \\subset Y$. Prove that\n\ni) $f(A \\bigcap B) \\subset f(A) \\bigcap f(B)$\n\nii) $f^{-1}(f(A)) \\supset A$\n\niii) $f(f^{-1}(C)) \\subset C$\n\nProblem 5:\n\nLet I be a non-empty set and for each $i \\in I$, let $X_{i}$ be a set. Prove that\n\n(a) for any set B, we have $B \\bigcap \\bigcup_{i \\in I}X_{i}=\\bigcup_{i \\in I}(B \\bigcap X_{i})$\n\n(b) if each $X_{i}$ is a subset of a given set S, then $(\\bigcup_{i \\in I}X_{i})^{'}=\\bigcap_{i \\in I}(X_{i})^{'}$ where the prime indicates complement.\n\nProblem 6:\n\nLet A, B, C be subsets of a set S. Prove the following statements:\n\n(i) $A- (B-C)=(A-B)\\bigcup(A \\bigcap B \\bigcap C)$\n\n(ii) $(A-B) \\times C=(A \\times C)-(B \\times C)$\n\n\ud83d\ude42 \ud83d\ude42 \ud83d\ude42\n\nNalin Pithwa\n\nSet Theory, Relations, Functions: Preliminaries: Part IX: (tutorial\u00a0problems)\n\nReference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.\n\nProblem 1:\n\nProve that if $A \\bigcup B=A$ and $A \\bigcap B=A$, then $A=B$.\n\nProblem 2:\n\nShow that in general $(A-B)\\bigcup B \\neq A$.\n\nProblem 3:\n\nLet $A = \\{ 2,4, \\ldots, 2n, \\ldots\\}$ and $B= \\{ 3,6,\\ldots, 3n, \\ldots\\}$. Find $A \\bigcap B$ and $A - B$.\n\nProblem 4:\n\nProve that (a) $(A-B)\\bigcap (C)=(A \\bigcap C)-(B \\bigcap C)$\n\nProve that (b) $A \\Delta B = (A \\bigcup B)-(A \\bigcap B)$\n\nProblem 5:\n\nProve that $\\bigcup_{a}A_{\\alpha}-\\bigcup_{a}B_{\\alpha}=\\bigcup_{\\alpha}(A_{\\alpha}-B_{\\alpha})$\n\nProblem 6:\n\nLet $A_{n}$ be the set of all positive integers divisible by $n$. Find the sets (i) $\\bigcup_{n=2}^{\\infty}A_{n}$ (ii) $\\bigcap_{n=2}^{\\infty}A_{n}$.\n\nProblem 7:\n\nFind (i) $\\bigcup_{n=1}^{\\infty}[n+\\frac{1}{n}, n - \\frac{1}{n}]$ (ii) $\\bigcap_{n=1}^{\\infty}(a-\\frac{1}{n},b+\\frac{1}{n})$\n\nProblem 8:\n\nLet $A_{\\alpha}$ be the set of points lying on the curve $y=\\frac{1}{x^{\\alpha}}$ where $(0. What is $\\bigcap_{\\alpha \\geq 1}A_{\\alpha}$?\n\nProblem 9:\n\nLet $y=f(x) = $ for all real x, where $$ is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval $\\frac{1}{4} \\leq y \\leq \\frac{3}{4}$? Partition the real line into classes of points with the same image.\n\nProblem 10:\n\nGiven a set M, let $\\mathscr{R}$ be the set of all ordered pairs on the form $(a,a)$ with $a \\in M$, and let $aRb$ if and only if $(a,b) \\in \\mathscr{R}$. Interpret the relation R.\n\nProblem 11:\n\nGive an example of a binary relation which is:\n\n\u2022 Reflexive and symmetric, but not transitive.\n\u2022 Reflexive, but neither symmetric nor transitive.\n\u2022 Symmetric, but neither reflexive nor transitive.\n\u2022 Transitive, but neither reflexive nor symmetric.\n\nWe will continue later, \ud83d\ude42 \ud83d\ude42 \ud83d\ude42\n\nPS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.\n\nNalin Pithwa\n\nSet Theory, Relations, Functions: Preliminaries: part\u00a0VIIIA\n\n(We continue from part VII of the same blog article series with same reference text).\n\nTheorem 4:\n\nA set M can be partitioned into classes by a relation R (acting as a criterion for assigning two elements to the same class) if and only R is an equivalence relation on M.\n\nProof of Theorem 4:\n\nEvery partition of M determines a binary relation on M, where $aRb$ means that \u201ca belongs to the same class as b.\u201d It is then obvious that R must be reflexive, symmetric and transitive, that is, R is an equivalence relation on M.\n\nConversely, let R be an equivalence relation on M, and let $K_{a}$ be the set of all elements $x \\in M$ such that $xRa$ (clearly, $a \\in K_{a}$, since R is reflexive). Then, two classes $K_{a}$ and $K_{b}$ are either identical or disjoint. In fact, suppose that an element c belongs to both $K_{a}$ and $K_{b}$, so that $cRa$ and $cRb$. But by symmetry of R, being an equivalence relation, we can infer that $aRc$ also and, further by transitivity, we say that $aRb$. If now, $x \\in K_{a}$ then we have $xRa$ and hence, $xRb$ (since we already have $aRb$ and using transitivity).\n\nSimilarly, we can prove that $x \\in K_{b}$ implies that $x \\in K_{a}$.\n\nTherefore, $K_{a}=K_{b}$ if $K_{a}$ and $K_{b}$ have an element in common. Therefore, the distinct sets $K_{a}$ form a partition of M into classes.\n\nQED.\n\nRemark:\n\nBecause of theorem 4, one often talks about the decomposition of a set M into equivalence classes.\n\nThere is an intimate connection between mappings and partitions into classes, as illustrated by the following examples:\n\nExample 1:\n\nLet f be a mapping of a set A into a set B and partition A into sets, each consisting of all elements with the same image $b=f(a) \\in B$. This gives a partition of A into classes. For example, suppose f projects the xy-plane onto the x-axis by mapping the point $(x,y)$ into the point $(x,0)$. Then, the preimages of the points of the x-axis are vertical lines, and the representation of the plane as the union of these lines is the decomposition into classes corresponding to f.\n\nExample 2:\n\nGiven any partition of a set A into classes, let B be the set of these classes and associate each element $a \\in A$ with the class (that is, element of B) to which it belongs. This gives a mapping of A into B. For example, suppose we partition three-dimensional space into classes by assigning to the same class all points which are equidistant from the origin of coordinates. Then, every class is a sphere of a certain radius. The set of all these classes can be identified with the set of points on the half-line $[0, \\infty)$ each point corresponding to a possible value of the radius. In this sense, the decomposition of 3-dimensional space into concentric spheres corresponds to the mapping of space into the half-line $[0,\\infty)$.\n\nExample 3:\n\nSuppose that we assign all real numbers with the same fractional part to the same class. Then, the mapping corresponding to this partition has the effect of \u201cwinding\u201d the real line onto a circle of unit circumference. (Note: The largest integer $\\leq x$ is called the integral part of x, denoted by [x], and the quantity $x -[x]$ is called the fractional part of x).\n\nIn the next blog article, let us consider a tutorial problem set based on last two blogs of this series.\n\n\ud83d\ude42 \ud83d\ude42 \ud83d\ude42\n\nNalin Pithwa\n\nA quadratic and trigonometry combo question: RMO and IITJEE maths\u00a0coaching\n\nQuestion:\n\nGiven that $\\tan {A}$ and $\\tan {B}$ are the roots of the quadratic equation $x^{2}+px+q=0$, find the value of\n\n$\\sin^{2}{(A+B)}+ p \\sin{(A+B)}\\cos{(A+B)} + q\\cos^{2}{(A+B)}$\n\nSolution:\n\nLet $\\alpha=\\tan{A}$ and $\\beta=\\tan{B}$ be the two roots of the given quadratic equation: $x^{2}+px+q=0$\n\nBy Viete\u2019s relations between roots and coefficients:\n\n$\\alpha+\\beta=\\tan{A}+\\tan{B}=-p$ and $\\alpha \\beta = \\tan{A}\\tan{B}=q$ but we also know that $\\tan{(A+B)}=\\frac{\\tan{A}+\\tan{B}}{1-\\tan{A}\\tan{B}}=\\frac{-p}{1-q}=\\frac{p}{q-1}$\n\nNow, let us call $E=\\sin^{2}{(A+B)}+p\\sin{(A+B)\\cos{(A+B)}}+\\cos^{2}{(A+B)}$ which in turn is same as\n\n$\\cos^{2}{(A+B)}(\\tan^{2}{(A+B)}+p\\tan{(A+B)}+q)$\n\nWe have already determined $\\tan{(A+B)}$ in terms of p and q above.\n\nNow, again note that $\\sin^{2}{\\theta}+\\cos^{2}{\\theta}=1$ which in turn gives us that $\\tan^{2}{\\theta}+1=\\sec^{2}{\\theta}$ so we get:\n\n$\\sec^{2}{(A+B)}=1+\\tan^{2}{(A+B)}=1+\\frac{p^{2}}{(q-1)^{2}}=\\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}}$ so that\n\n$\\cos^{2}{(A+B)}=\\frac{1}{\\sec^{2}{(A+B)}}=\\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}$\n\nHence, the given expression E becomes:\n\n$(\\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\\frac{p^{2}}{(q-1)^{2}}+\\frac{p^{2}}{q-1}+q)$, which is the desired solution.\n\n\ud83d\ude42 \ud83d\ude42 \ud83d\ude42\n\nNalin Pithwa.", "date": "2019-11-12 08:50:17", "meta": {"domain": "madhavamathcompetition.com", "url": "https://madhavamathcompetition.com/category/chennai-mathematical-institute-entrance/", "openwebmath_score": 0.9011602997779846, "openwebmath_perplexity": 294.34835768822796, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127433812521, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8783716472247783}}
{"url": "http://math.stackexchange.com/questions/909680/how-would-i-find-a-point-on-the-y-axis-equidistant-from-two-other-points", "text": "# How would I find a point on the y-axis equidistant from two other points?\n\nThe points are$$(5,-5) and (1,1)$$\n\nI tried doing this visually and came up with (0,-5). This wasn't correct once I applied the distance formula to check the distance between that point and the two others.\n\n-\nLet the point be $(0,y)$. Then $5^2+(y+5)^2=1^2+(y-1)^2$. Expand and solve. (There are other ways.) \u2013\u00a0Andr\u00e9 Nicolas Aug 26 '14 at 11:49\nAs for what you did wrong, I'm not sure, but $(-5,0)$ is not on the $y$-axis... \u2013\u00a0fixedp Aug 26 '14 at 11:51\nthat was a typo \u2013\u00a0Cherry_Developer Aug 26 '14 at 11:52\n\nA point on the $y-$axis is of the form $(0,y)$.\n\nThe distance between $(0,y)$ and $(5,-5)$ is: $$\\sqrt{(5-0)^2+(-5-y)^2}=\\sqrt{25+(5+y)^2}$$\n\nThe distance between $(0,y)$ and $(1,1)$ is: $$\\sqrt{(1-0)^2+(1-y)^2}=\\sqrt{1+(1-y)^2}$$\n\nThe two distances are equal, so also their squares:\n\n$$25+(5+y)^2=1+(1-y)^2$$\n\nNow you have to solve for $y$.\n\n-\nIt's probably more convenient to equate the squares of the distances, rather than the distances themselves. \u2013\u00a0fixedp Aug 26 '14 at 11:50\nI have doubts but I think it's (0,-4) \u2013\u00a0Cherry_Developer Aug 26 '14 at 11:59\nIt is correct!! \u2013\u00a0Mary Star Aug 26 '14 at 12:04\n\nYou want the point where the perpendicular bisector of the two points cuts the $y$-axis.\n\nThe slope of the line between $(5,-5)$ and $(1,1)$ is $-\\frac{3}{2}$, so the slope of the normal is $\\frac23$. Hence the normal through the mid-point $(3,-2)$ cuts the $y$-axis at $(0,-4)$.\n\n-\n+1 because on seeing the question I thought to myself, \"well personally I'd just solve the equation, but what would that teach you about the set of points equidistant between A and B?\" :-) \u2013\u00a0Steve Jessop Aug 26 '14 at 13:19\n@Steve: Yes! I was disturbed by all those quadratic terms in the other answers, which always cancelled each other out. \u2013\u00a0TonyK Aug 26 '14 at 13:22\nThat's an insight in itself, mind you. They cancel out so that the set of equidistant points can be a straight line, as opposed to something curved if there were quadratic terms left in there! \u2013\u00a0Steve Jessop Aug 26 '14 at 13:23\nThis also illustrates why it's always possible to find such a point, unless the given points lie on a horizontal line (which would make the perpendicular bisector parallel to the $y$-axis. \u2013\u00a0cjm Aug 26 '14 at 14:31", "date": "2016-06-25 05:38:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/909680/how-would-i-find-a-point-on-the-y-axis-equidistant-from-two-other-points", "openwebmath_score": 0.8329616785049438, "openwebmath_perplexity": 425.24283634401104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864314449021, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8783205200293828}}
{"url": "https://math.stackexchange.com/questions/2738419/intuitively-or-precisely-which-functions-can-be-approximated-by-straight-line", "text": "Intuitively (or precisely), which functions can be approximated by straight lines?\n\nI am trying to build up some intuition of what the various notions of analysis mean. By intuition of continuity is that it means the graph of the function is 'connected', but not necessarily in any nice way. This lead me to ask the question 'Given a function $f: \\mathbb{R} \\to \\mathbb{R}$ and a point $x_0 \\in \\mathbb{R}$, what assumptions must we impose on $f$ so that we can say $f$ is 'approximately a line' in some neighbourhood of $x_0$?'\n\nTo apply Taylor's theorem, we need that $f$ is twice continuously differentiable at $x_0$. However I am only interested in some tiny neighbourhood of $x_0$, so can we weaken these assumptions?\n\nWe certainly can't weaken them all the way down to continuity at $x_0$; for example the function $g: \\mathbb{R} \\to \\mathbb{R}$ defined by $$g(x) = \\begin{cases} x \\sin(\\frac{1}{x}), \\ x \\neq 0 \\\\ 0, \\qquad \\quad \\ x=0 \\end{cases}$$ is continuous at $x_0 = 0$, but certainly cannot be approximated by a line there. Differentiability at $x_0$ also won't do; consider $h: \\mathbb{R} \\to \\mathbb{R}$ given by $$h(x) = \\begin{cases} x^2 \\sin(\\frac{1}{x}), \\ x \\neq 0 \\\\ 0, \\qquad \\quad \\ \\ \\ x=0. \\end{cases}$$ This is differentiable and hence continuous at $x_0 = 0$. Indeed it is differentiable everywhere, but it is not continuously differentiable at $0$; it's derivative is $h': \\mathbb{R} \\to \\mathbb{R}$ given by $$h'(x) = \\begin{cases} 2x\\sin(\\frac{1}{x}) - \\cos(\\frac{1}{x}), \\ x \\neq 0 \\\\ 0, \\qquad \\qquad \\qquad \\qquad x=0 \\end{cases}$$ whose limit at $0$ doesn't exist. Going one step further, we consider $k: \\mathbb{R} \\to \\mathbb{R}$ given by $$k(x) = \\begin{cases} x^3 \\sin(\\frac{1}{x}), \\ x \\neq 0 \\\\ 0, \\qquad \\quad \\ \\ \\ x=0. \\end{cases}$$ Again this is differentiable everywhere, but now it is continuously differentiable at $0$ (and so everywhere); it's derivative is $k': \\mathbb{R} \\to \\mathbb{R}$ given by $$k'(x) = \\begin{cases} 3x^2\\sin(\\frac{1}{x}) - x\\cos(\\frac{1}{x}), \\ x \\neq 0 \\\\ 0, \\qquad \\qquad \\qquad \\qquad \\quad \\ x=0 \\end{cases}$$ which is continuous everywhere. However this function obviously still can't be approximated by a straight line near $0$ (since it takes on positive, negative, and zero values arbitrarily near the origin, and a straight line through the origin can only take on positive, negative, $\\textit{or}$ zero values near the origin).\n\nAt first I thought the problem was that the value of the derivatives of $g, h$ and $k$ around $x_0 = 0$ were unbounded, i.e. their 'slopes' grow without bound, however this is only the case for $g$; the derivative of $h$ near $0$ is approximately between $-1$ and $1$ and the derivative of $k$ near $0$ is approximately $0$. Then I thought the problem was that that the derivatives of $g, h,$ and $k$ had arbitrarily many sign changes near $x_0 = 0$, however this also can't be right.\n\nSo now I'm just confused. Do we in fact need all the assumptions warranted by Taylor's theorem, i.e. twice continuously differentiable at $x_0$, to guarantee that a function is 'like a line' at a point? Do we actually need more hypotheses than Taylor's theorem in that we have to assume information about $f$ not at $x_0$, i.e. in some neighbourhood?\n\ntl;dr: what restrictions must we place on a function $f: \\mathbb{R} \\to \\mathbb{R}$ so that when we 'zoom in close enough', it looks like a straight line?\n\nThere are several related concepts at work here. The standard term in Calculus/Analysis is the linearization $L(x)=f^{\\prime}(x_{0}) (x-x_{0}) + f(x_{0})$ of a function $f(x)$ at a point $x_{0}$, also sometime referred to as the linear approximation to $f(x)$ at $x_{0}$. This can obviously be done whenever $f(x)$ is differentiable at $x_{0}$, but to actually consider it an \"approximation\" to $f(x)$ in any meaningful sense, we want the behavior of $f(x)$ and $L(x)$ to match at least under some assumptions. This is where we consider the second derivative: if the second derivative $f^{\\prime\\prime}= 0$ in an interval, then $f^{\\prime}$ will be constant, and so $f(x) = L(x)$. If $f^{\\prime\\prime}$ is not constant but bounded, then we can make an estimate about how far away $f(x)$ and $L(x)$ can get. If $f^{\\prime\\prime}$ is unbounded or undefined at some points in that interval, then $f(x)$ can diverge wildly from $L(x)$ and our approximation won't be any good. We still can form the \"linear approximation\", it just might not be very good to actually approximate our function.\nIf you only need to consider a tiny neighborhood around $x_{0}$, the proper way to weaken your conditions is that you only need $f^{\\prime\\prime}$ to be defined in that tiny neighborhood. You will unfortunately probably have lots of problems if $f^{\\prime\\prime}(x_{0})$ is itself undefined.\nI personally wouldn't ever say $f$ is \"approximately a line\", just that we can approximate $f(x)$ by a linear function in that neighborhood. The closest exact definition for saying to functions are \"approximately the same\" has to do with the concept of the functions being equal \"almost everywhere\", so saying $f$ is \"approximately a line\" might be confused to mean that you have a linear function $L(x)$ where $f(x) = L(x)$ for almost all values of $x$ with a few exceptions. Saying that one function approximates another function basically says that they are close, essentially saying that $|f(x)-L(x)| < A$ for some $A$. Then we can say it is a \"good\" approximation if $A$ is \"small\" in some sense.\nAbout $k(x)$: \"this function [...] can't be approximated by a straight line near 0 (since it takes on positive, negative, and zero values arbitrarily near the origin, and a straight line through the origin can only take on positive, negative, or or zero values\". We judge an approximation by the amount of error, or the distance between $f(x)$ and $L(x)$ (where $L(x)$ is the linear approximation). If this difference is small, we don't care if one is positive and the other is negative.\n\u2022 Thanks, I hadn't considered the more statistical approach to approximating a function by a line (involving errors). Although I'm not sure how '$f$ is approximately a line on a neighbourhood' and '$f$ can be approximated by a line on a neighbourhood$' are any different semantically, i.e. 'is approximately = can be approximated by' (at least to me). Would you mind elaborating the distinction for me? \u2013 Joel Brennan Apr 15 '18 at 17:37 \u2022 @JoelBrennan I've updated my answer to explain what I think the difference is between these two phrasings. \u2013 Morgan Rodgers Apr 15 '18 at 21:01 Let us first define what it means that a curve$c\\colon I\\to \\mathbb R^2$, defined on an interval$I$of$\\mathbb R$, possesses a tangent in$\\tau\\in I$: There exists a one-dimensional subspace$W$of$\\mathbb R^2$and an$\\epsilon>0$such that for all$s,t\\in I\\cap(\\tau-\\epsilon,\\tau+\\epsilon)$it holds that $$c(s)\\neq c(t)\\text{ and } \\lim_{s,t\\to\\tau}\\mathbb R\\cdot(c(t)-c(s))=W, \\text{ if s\\leq\\tau\\leq t and s\\neq t}.$$ For those$\\tau$where$c$doesn\u2019t have a tangent, but the left- resp. rightsided tangents exist we will say that$c$has a tip resp. an edge in$\\tau$if the left- and rightsided tangents are equal resp. not equal. And now strange things may happen. Consider one slightly modified of your examples, namely$c(t)=(t,5+t^2\\cdot\\sin(\\pi/2t))$for$t\\neq0$and$c(0)=(0,0)$. It is immersive in$\\tau=0$, has a tangent in$\\tau=0$but it is a limit point of$t$\u2019s, for which$c$has a tip. (Consider the sequences$s_n=1/(2n+2)$and$t_n=1/(2n+1)$.). This doesn\u2019t fit with the notion of \u201csmoothness\u201d. Therefore let us define$c$is smooth in$\\tau$if there\u2019s a neighborhood$U$of$\\tau$such that$c$possesses a tangent$c(t)+W_t$for all$t\\in U$and$\\lim_{t\\to\\tau}W_t=W_{\\tau}$, i.e., in a neighborhood the tangents exist an they converge to the tangent in$c(\\tau)$. PS. Even more weirdness reveals the following beast. Define$\\phi(t)=e^{-1/t}$for positive$t$and$0$else. Define the injective path $$c(t)=(\\phi(t)+2\\phi(-t))\\cdot(\\cos(1/|t|),\\sin(1/|t|))$$ for$t\\neq0$and$c(0)=(0,0)$. Now$c$has neither a tangent nor a tip nor an edge in$t=0$. Please feel free to plot this curve. \u2022 Thanks for your reply. The maths here is a tiny bit over my head but I will do my best to understand it. I presume it's differential geometry? What does the$\\cdot$which appears next to the limit mean? Also can you verify that$W$is a one-dimensional subspace of$\\mathbb{R}$and not of$\\mathbb{R}^2$; the only 1D subspaces of$\\mathbb{R}$I can think of (in the linear algebra sense) are$\\mathbb{R}$itself as well as incomplete sets such as$\\mathbb{Q}, n\\mathbb{Z}$, etc, but the$\\lim$seems to suggest you want a complete set. \u2013 Joel Brennan Apr 15 '18 at 21:45 \u2022 Sorry,$W$is supposed to be a line, that is a one-dimensional subspace of$\\mathbb R^2$, I\u2018ve corrected. The dot means the multiplication sign, hence$\\mathbb R\\cdot(c(t)-c(s))$describes the line through$c(t)-c(s)\\$ \u2013\u00a0Michael Hoppe Apr 16 '18 at 10:14", "date": "2019-07-20 18:28:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2738419/intuitively-or-precisely-which-functions-can-be-approximated-by-straight-line", "openwebmath_score": 0.935771107673645, "openwebmath_perplexity": 168.20632088211147, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864302631449, "lm_q2_score": 0.8872045966995027, "lm_q1q2_score": 0.8783205115995938}}
{"url": "https://in.mathworks.com/help/matlab/ref/spline.html", "text": "# spline\n\nCubic spline data interpolation\n\n## Description\n\nexample\n\ns = spline(x,y,xq) returns a vector of interpolated values s corresponding to the query points in xq. The values of s are determined by cubic spline interpolation of x and y.\n\nexample\n\npp = spline(x,y) returns a piecewise polynomial structure for use by ppval and the spline utility unmkpp.\n\n## Examples\n\ncollapse all\n\nUse spline to interpolate a sine curve over unevenly-spaced sample points.\n\nx = [0 1 2.5 3.6 5 7 8.1 10];\ny = sin(x);\nxx = 0:.25:10;\nyy = spline(x,y,xx);\nplot(x,y,'o',xx,yy)\n\nUse clamped or complete spline interpolation when endpoint slopes are known. To do this, you can specify the values vector $\\mathit{y}$ with two extra elements, one at the beginning and one at the end, to define the endpoint slopes.\n\nCreate a vector of data $\\mathit{y}$ and another vector with the $\\mathit{x}$-coordinates of the data.\n\nx = -4:4;\ny = [0 .15 1.12 2.36 2.36 1.46 .49 .06 0];\n\nInterpolate the data using spline and plot the results. Specify the second input with two extra values [0 y 0] to signify that the endpoint slopes are both zero. Use ppval to evaluate the spline fit over 101 points in the interpolation interval.\n\ncs = spline(x,[0 y 0]);\nxx = linspace(-4,4,101);\nplot(x,y,'o',xx,ppval(cs,xx),'-');\n\nExtrapolate a data set to predict population growth.\n\nCreate two vectors to represent the census years from 1900 to 1990 (t) and the corresponding United States population in millions of people (p).\n\nt = 1900:10:1990;\np = [ 75.995  91.972  105.711  123.203  131.669 ...\n150.697 179.323  203.212  226.505  249.633 ];\n\nExtrapolate and predict the population in the year 2000 using a cubic spline.\n\nspline(t,p,2000)\nans = 270.6060\n\nGenerate the plot of a circle, with the five data points y(:,2),...,y(:,6) marked with o's. The matrix y contains two more columns than does x. Therefore, spline uses y(:,1) and y(:,end) as the endslopes. The circle starts and ends at the point (1,0), so that point is plotted twice.\n\nx = pi*[0:.5:2];\ny = [0  1  0 -1  0  1  0;\n1  0  1  0 -1  0  1];\npp = spline(x,y);\nyy = ppval(pp, linspace(0,2*pi,101));\nplot(yy(1,:),yy(2,:),'-b',y(1,2:5),y(2,2:5),'or')\naxis equal\n\nUse spline to sample a function over a finer mesh.\n\nGenerate sine and cosine curves for a few values between 0 and 1. Use spline interpolation to sample the functions over a finer mesh.\n\nx = 0:.25:1;\nY = [sin(x); cos(x)];\nxx = 0:.1:1;\nYY = spline(x,Y,xx);\nplot(x,Y(1,:),'o',xx,YY(1,:),'-')\nhold on\nplot(x,Y(2,:),'o',xx,YY(2,:),':')\nhold off\n\nCompare the interpolation results produced by spline, pchip, and makima for two different data sets. These functions all perform different forms of piecewise cubic Hermite interpolation. Each function differs in how it computes the slopes of the interpolant, leading to different behaviors when the underlying data has flat areas or undulations.\n\nCompare the interpolation results on sample data that connects flat regions. Create vectors of x values, function values at those points y, and query points xq. Compute interpolations at the query points using spline, pchip, and makima. Plot the interpolated function values at the query points for comparison.\n\nx = -3:3;\ny = [-1 -1 -1 0 1 1 1];\nxq1 = -3:.01:3;\np = pchip(x,y,xq1);\ns = spline(x,y,xq1);\nm = makima(x,y,xq1);\nplot(x,y,'o',xq1,p,'-',xq1,s,'-.',xq1,m,'--')\nlegend('Sample Points','pchip','spline','makima','Location','SouthEast')\n\nIn this case, pchip and makima have similar behavior in that they avoid overshoots and can accurately connect the flat regions.\n\nPerform a second comparison using an oscillatory sample function.\n\nx = 0:15;\ny = besselj(1,x);\nxq2 = 0:0.01:15;\np = pchip(x,y,xq2);\ns = spline(x,y,xq2);\nm = makima(x,y,xq2);\nplot(x,y,'o',xq2,p,'-',xq2,s,'-.',xq2,m,'--')\nlegend('Sample Points','pchip','spline','makima')\n\nWhen the underlying function is oscillatory, spline and makima capture the movement between points better than pchip, which is aggressively flattened near local extrema.\n\n## Input Arguments\n\ncollapse all\n\nx-coordinates, specified as a vector. The vector x specifies the points at which the data y is given. The elements of x must be unique.\n\nData Types: single | double\n\nFunction values at x-coordinates, specified as a numeric vector, matrix, or array. x and y typically have the same length, but y also can have exactly two more elements than x to specify endslopes.\n\nIf y is a matrix or array, then the values in the last dimension, y(:,...,:,j), are taken as the values to match with x. In that case, the last dimension of y must be the same length as x or have exactly two more elements.\n\nThe endslopes of the cubic spline follow these rules:\n\n\u2022 If x and y are vectors of the same size, then the not-a-knot end conditions are used.\n\n\u2022 If x or y is a scalar, then it is expanded to have the same length as the other and the not-a-knot end conditions are used.\n\n\u2022 If y is a vector that contains two more values than x has entries, then spline uses the first and last values in y as the endslopes for the cubic spline. For example, if y is a vector, then:\n\n\u2022 y(2:end-1) gives the function values at each point in x\n\n\u2022 y(1) gives the slope at the beginning of the interval located at min(x)\n\n\u2022 y(end) gives the slope at the end of the interval located at max(x)\n\n\u2022 Similarly, if y is a matrix or an N-dimensional array with size(y,N) equal to length(x)+2, then:\n\n\u2022 y(:,...,:,j+1) gives the function values at each point in x for j = 1:length(x)\n\n\u2022 y(:,:,...:,1) gives the slopes at the beginning of the intervals located at min(x)\n\n\u2022 y(:,:,...:,end) gives the slopes at the end of the intervals located at max(x)\n\nData Types: single | double\n\nQuery points, specified as a scalar, vector, matrix, or array. The points specified in xq are the x-coordinates for the interpolated function values yq computed by spline.\n\nData Types: single | double\n\n## Output Arguments\n\ncollapse all\n\nInterpolated values at query points, returned as a scalar, vector, matrix, or array.\n\nThe size of s is related to the sizes of y and xq:\n\n\u2022 If y is a vector, then s has the same size as xq.\n\n\u2022 If y is an array of size Ny = size(y), then these conditions apply:\n\n\u2022 If xq is a scalar or vector, then size(s) returns [Ny(1:end-1) length(xq)].\n\n\u2022 If xq is an array, then size(s) returns [Ny(1:end-1) size(xq)].\n\nPiecewise polynomial, returned as a structure. Use this structure with the ppval function to evaluate the piecewise polynomial at one or more query points. The structure has these fields.\n\nFieldDescription\nform\n\n'pp' for piecewise polynomial\n\nbreaks\n\nVector of length\u00a0L+1 \u00a0with strictly increasing elements that represent the start and end of each of\u00a0L\u00a0intervals\n\ncoefs\n\nL-by-k \u00a0matrix with each row\u00a0coefs(i,:)\u00a0containing the local coefficients of an order\u00a0k\u00a0polynomial on the\u00a0ith interval,\u00a0[breaks(i),breaks(i+1)]\n\npieces\n\nNumber of pieces, L\n\norder\n\nOrder of the polynomials\n\ndim\n\nDimensionality of target\n\nSince the polynomial coefficients in coefs are local coefficients for each interval, you must subtract the lower endpoint of the corresponding knot interval to use the coefficients in a conventional polynomial equation. In other words, for the coefficients [a,b,c,d] on the interval [x1,x2], the corresponding polynomial is\n\n$f\\left(x\\right)=a{\\left(x-{x}_{1}\\right)}^{3}+b{\\left(x-{x}_{1}\\right)}^{2}+c\\left(x-{x}_{1}\\right)+d\\text{\\hspace{0.17em}}.$\n\n## Tips\n\n\u2022 You also can perform spline interpolation using the interp1 function with the command interp1(x,y,xq,'spline'). While spline performs interpolation on rows of an input matrix, interp1 performs interpolation on columns of an input matrix.\n\n## Algorithms\n\nA tridiagonal linear system (possibly with several right-hand sides) is solved for the information needed to describe the coefficients of the various cubic polynomials that make up the interpolating spline. spline uses the functions ppval, mkpp, and unmkpp. These routines form a small suite of functions for working with piecewise polynomials. For access to more advanced features, see interp1 or the Curve Fitting Toolbox\u2122 spline functions.\n\n## References\n\n[1] de Boor, Carl. A Practical Guide to Splines. Springer-Verlag, New York: 1978.\n\n## Version History\n\nIntroduced before R2006a", "date": "2023-01-29 23:23:35", "meta": {"domain": "mathworks.com", "url": "https://in.mathworks.com/help/matlab/ref/spline.html", "openwebmath_score": 0.6367332935333252, "openwebmath_perplexity": 1685.007936668553, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765598801371, "lm_q2_score": 0.8962513759047847, "lm_q1q2_score": 0.8783053401470104}}
{"url": "http://moshtarakin.com/g1nj4q/719034-diagonalizable-matrix-example", "text": "Example # 3: Diagonalize the matrix, \"A\". In other words, when is diagonalizable, then there exists an invertible matrix such that where is a diagonal matrix, that is, a matrix whose non-diagonal entries are zero. 5.3 Diagonalization The goal here is to develop a useful factorization A PDP 1, when A is n n. We can use this to compute Ak quickly for large k. The matrix D is a diagonal matrix (i.e. has three different eigenvalues. Let A be an n n matrix. Compute D2 and D3. We also showed that A is diagonalizable. A square matrix in which every element except the principal diagonal elements is zero is called a Diagonal Matrix. Free Matrix Diagonalization calculator - diagonalize matrices step-by-step This website uses cookies to ensure you get the best experience. There are many types of matrices like the Identity matrix.. Properties of Diagonal Matrix A square matrix D = [d ij] n x n will be called a diagonal matrix if d ij = 0, whenever i is not equal to j. In fact, there is a general result along these lines. Diagonalization Example Example If Ais the matrix A= 1 1 3 5 ; then the vector v = (1;3) is an eigenvector for Abecause Av = 1 1 3 5 1 3 = 4 12 = 4v: The corresponding eigenvalue is = 4. Dk is trivial to compute as the following example illustrates. Remark Note that if Av = v and cis any scalar, then A(cv) = cAv = c( v) = (cv): Consequently, if v is an eigenvector of A, then so is cv for any nonzero scalar c. Select the incorrectstatement: A)Matrix !is diagonalizable B)The matrix !has only one eigenvalue with multiplicity 2 C)Matrix !has only one linearly independent eigenvector D)Matrix !is not singular Let A be a square matrix of order n. Assume that A has n distinct eigenvalues. (1)(a) Give an example of a matrix that is invertible but not diagonalizable. Theorem: An n x n matrix, \"A\", is diagonalizable if and only if \"A\" has \"n\" linearly independent eigenvectors. entries off the main diagonal are all zeros). (2)Given a matrix A, we call a matrix B a s Moreover, if P is the matrix with the columns C 1, C 2, ..., and C n the n eigenvectors of A, then the matrix P-1 AP is a diagonal matrix. Then A is diagonalizable. -Compute across the 2nd row = -2 - 1 - 2 + 0 = -5 0 => { 1, 2, 3} linearly independent. By using this website, you agree to our Cookie Policy. Theorem: An $n \\times n$ matrix with $n$ distinct eigenvalues is diagonalizable. if and only if the columns of \"P\" are \"n\" linearly independent eigenvectors of \"A\". Since A is not invertible, zero is an eigenvalue by the invertible matrix theorem , so one of the diagonal entries of D is necessarily zero. The diagonal entries of \"D\" are eigenvalues of \"A\" that correspond, respectively to the eigenvectors in \"P\". (1)(b): Give an example of a matrix that is diagonalizable but not invertible. Theorem. Remark: It is not necessary for an $n \\times n$ matrix to have $n$ distinct eigenvalues in order to be diagonalizable. The above theorem provides a sufficient condition for a matrix to be diagonalizable. (i) A is diagonalizable (ii) c A(x) = (x 1)m 1(x 2)m 2 (x r)m r and for each i, A has m i basic vectors. The following conditions are equivalent. Example The eigenvalues of the matrix:!= 3 \u00e2\u0088\u009218 2 \u00e2\u0088\u00929 are \u00e2\u0080\u0099.=\u00e2\u0080\u0099 /=\u00e2\u0088\u00923. EXAMPLE: Let D 50 04. Example (A diagonalizable 2 \u00d7 2 matrix with a zero eigenvector) In the above example, the (non-invertible) matrix A = 1 3 A 2 \u00e2\u0088\u0092 4 \u00e2\u0088\u0092 24 B is similar to the diagonal matrix D = A 00 02 B . Theorem. Example Define the matrix and The inverse of is The similarity transformation gives the diagonal matrix as a result. 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To be diagonalizable Properties of diagonal matrix has three different eigenvalues element except the principal diagonal is... Order n. Assume that a has diagonalizable matrix example distinct eigenvalues this website, you agree our... In fact, there is a general result along these lines as a result a has n eigenvalues... 1 ) ( a ) Give an example of a matrix that is diagonalizable but not diagonalizable the theorem! Order n. Assume that a has n distinct eigenvalues the similarity transformation gives the matrix! That correspond, respectively to the eigenvectors in  P '' are  n '' linearly eigenvectors! A has n distinct eigenvalues is diagonalizable but not invertible, respectively to the eigenvectors in P... Of a matrix to be diagonalizable example Define the matrix:! = 3 \u00e2\u0088\u009218 2 \u00e2\u0088\u00929 are \u00e2\u0080\u0099.=\u00e2\u0080\u0099.... P '' elements is zero is called a diagonal matrix has three different eigenvalues a... Matrix of order n. Assume that a has n distinct eigenvalues provides a sufficient condition for matrix! D '' are  n '' linearly independent eigenvectors of  a '' Define the matrix and the of... Diagonal entries of  a '' that correspond, respectively to the eigenvectors . Different eigenvalues matrix to be diagonalizable matrix has three different eigenvalues result along these.. If the columns of  a '' if and only if the columns of  ''... N distinct eigenvalues matrix as a result '' that correspond, respectively to the eigenvectors in  ''. Matrix in which every element except the principal diagonal elements is zero is called a matrix! Except the principal diagonal elements is zero is called a diagonal matrix if columns! Assume that a has n distinct eigenvalues Define the matrix,  a '' linearly independent eigenvectors ... Are all zeros ) ) Give an example of a matrix that is but... The inverse of is the similarity transformation gives the diagonal matrix a square matrix of order n. that! D '' are  n '' linearly independent eigenvectors of  a '' that,... Sufficient condition for a matrix that is diagonalizable but not invertible of matrices like the Identity matrix.. of! Diagonalizable but not diagonalizable  n '' linearly independent eigenvectors of  D '' eigenvalues! A ) Give an example of a matrix to be diagonalizable every except. Website, you agree to our Cookie Policy a has n distinct eigenvalues these lines lines... Is called a diagonal matrix the columns of  D '' are eigenvalues of the matrix the! General result along these lines as the following example illustrates the main diagonal are all ). ) ( b ): Give an example of a matrix to be diagonalizable example. Is zero is called a diagonal matrix are eigenvalues of  D '' are eigenvalues of the matrix: =! Matrix has three different eigenvalues is a general result along these lines:! = \u00e2\u0088\u009218. Diagonal matrix has three different eigenvalues eigenvectors in  P '' Diagonalize the matrix:! = 3 2. An example of a matrix that is invertible but not diagonalizable distinct eigenvalues to the eigenvectors in P... All zeros ) 2 \u00e2\u0088\u00929 are \u00e2\u0080\u0099.=\u00e2\u0080\u0099 /=\u00e2\u0088\u00923 a matrix that is but... Website, you agree to our Cookie Policy a '' that correspond, respectively to the eigenvectors in P. 1 ) ( b ): Give an example of a matrix is! Is called a diagonal matrix using this website, you agree to our Cookie Policy and the inverse is. Eigenvectors in  P '' are  n '' linearly independent eigenvectors of  a '' types of matrices the... The main diagonal are all zeros )  n '' linearly independent eigenvectors of  a '' correspond, to... Example illustrates sufficient condition for a matrix to be diagonalizable a general result along lines. ( b ): Give an example of a matrix to be.!", "date": "2021-05-06 21:17:32", "meta": {"domain": "moshtarakin.com", "url": "http://moshtarakin.com/g1nj4q/719034-diagonalizable-matrix-example", "openwebmath_score": 0.9041623473167419, "openwebmath_perplexity": 589.5670543626902, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765616345253, "lm_q2_score": 0.8962513738264114, "lm_q1q2_score": 0.8783053396826263}}
{"url": "https://www.physicsforums.com/threads/integration-by-substituition-question.59811/", "text": "# Integration by substituition question.\n\n1. Jan 14, 2005\n\n### misogynisticfeminist\n\nI'm able to do this question but my answer is different from that in the book.\n\n$$\\int \\frac {x^3}{\\sqrt{ x^2 -1}} dx$$\n\nwhat I did was to take the substituition $$u= (x^2-1)^1^/^2$$\n\nso, $$x^2 -1 = u^2$$\n\n$$x^2 = u^2+1$$\n$$x= (u^2+1)^1^/^2$$\n$$x^3= (u^2+1)^3^/^2$$\n$$dx= \\frac {1}{2}(u^2+1)^-^1^/^2 (2u)du$$\n$$dx= u(u^2+1)^-^1^/^2 du$$\n\n$$\\int \\frac {x^3}{\\sqrt{ x^2 -1}} dx= \\int \\frac {(u^2+1)^3^/^2}{u} . \\frac {u}{(u^2+1)^1^/^2} du$$\n\nwhich simplifies to,\n\n$$\\int u^2+1 du$$\n$$\\frac {1}{3} u^3 +u+C$$\n$$\\frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2$$\n\nthat's my final answer but the book gave,\n\n$$\\frac {1}{3} (x^2+2)\\sqrt{x^2-1}+C$$\n\nwhere does my mistake lie?\n\n2. Jan 14, 2005\n\n### Hurkyl\n\nStaff Emeritus\n\n3. Jan 14, 2005\n\n### dextercioby\n\nYap,it's the same \"animal\".It's just the fur is a little shady...\n\nDaniel.\n\n4. Jan 14, 2005\n\n### MathStudent\n\n5. Jan 14, 2005\n\n### vincentchan\n\n$$\\frac {1}{3} (x^2-1)^3^/^2 + (x^2-1)^1^/^2$$\n$$= (x^2-1)^1^/^2 (\\frac {1}{3}(x^2-1) +1)$$\n$$= (x^2-1)^1^/^2 (\\frac {1}{3}x^2-\\frac {1}{3}+1)$$\n$$= (x^2-1)^1^/^2 (\\frac {1}{3}x^2+\\frac {2}{3})$$\n$$\\frac {1}{3} (x^2+2)\\sqrt{x^2-1}$$ :surprised:\n\n6. Jan 14, 2005\n\n### Curious3141\n\nIsn't y^(3/2) = (y)(y^(1/2)) ? Rearrange and simplify what you've got and it should come out the same.\n\nEDIT : NM, Vincent has shown the working\n\n7. Jan 15, 2005\n\n### misogynisticfeminist\n\nhey thanks, that's a very good tip !\n\nI am wondering if the answer which I gave is in a different form from the one in the answer script during an exam, will I be penalised?\n\n8. Jan 15, 2005\n\n### dextercioby\n\nUnless your teacher is a narrow minded s.o.b.,i don't see why.If i were u,on this integral i would have gone for another substitution,using hyperbolic sine and cosine.\n\nDaniel.", "date": "2017-02-27 03:10:21", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integration-by-substituition-question.59811/", "openwebmath_score": 0.7655555605888367, "openwebmath_perplexity": 4938.873807318634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409524, "lm_q2_score": 0.8962513710552469, "lm_q1q2_score": 0.8783053332980796}}
{"url": "http://fkug.realizzazionifaidate.it/write-an-algorithm-to-compute-pi.html", "text": "Mathematicians have found several different mathematical series that, if carried out infinitely, will accurately calculate pi to a great number of decimal places. 1 Principle of the algorithm. Pi is an irrational number -- a number with an unending string of non-repeating digits after the decimal point. Write a program to calculate pow(x,n) write a function to compute x n. The other answers in this page give valid methods for computing PI. Algorithm and flowchart 1. Using R code I have to write a pseudo code and real code to answer this question. By BBP algorithm, one is able to calculate any digits of pi without calculating any prior digits. Yes, \u03c0 can be an angle. #define PI 3. Algorithm. Shor\u2019s algorithm is famous for factoring integers in polynomial time. 4*M pi = --- N Although the Monte Carlo Method is often useful for solving problems in physics and mathematics which cannot be solved by analytical means, it is a rather slow method of calculating pi. of the numbers to calculate its L. No Input Output 1 30 4410 2 10 490 19. Looking at Pi and Pi[PDF], there are a lot of formulas. It uses the random numbers to calculate the area of a circle inscribed within a square and from that, one can calculate pi. One of the Borwein iterative algorithms is \"quartically convergent\", meaning that if you start an iteration with D correct digits of pi, it gives you 4D correct digits at the end. Before writing an algorithm for a problem, one should find out what is/are the inputs to the algorithm and what is/are expected output after running the algorithm. Write a program using your preferred IDE (Integrated Development Environment) like Eclipse, CodeBlock, Visual C++, Dev C++, etc. Software Development in the UNIX Environment Sample C Program. A value for pi is needed. Create a Python project to get the value of Pi to n number of decimal places. That\u2019s about 664 bits, and so 108 terms of the series should be enough. Calculate area=PI*R*R 5. The number of active processes in the system and the length of the queue to the disk-buffer cache are considered in the algorithm design. I found 3 methods to check, they are: 1. No Input Output 1 30 4410 2 10 490 19. If you want to calculate pi, first measure the circumference of a circle by wrapping a piece of string around the edge of it and then measuring the length of the string. More precisely, their method permits to obtain the n-th bit of p in time O(nlog 3 n) and space O(log(n)). The 18th digit is a check digit which is used to validate the rest of that code and check that it is correct. Bearing Between Two Points Date: 12/19/2001 at 20:32:39 From: Doug Subject: Latitude/Longitude calculations I have found numerous solutions for finding the distance between two Lat/Long points on the earth (including the Haversine Formula), but I can't seem to find a reference that shows how to also calculate the direction between those points. The number of iterations could exceed 250,000. On a quantum computer, to factor an integer , Shor's algorithm runs in polynomial time (the time taken is polynomial in \u2061, the size of the integer. Dynamic Programming Algorithm to compute the n-th Tribonacci number. If I have to give you one general advice on the matter is do not try to write your own SVD algorithms unless you have successfully taken a couple of classes on Numerical Linear Algebra already and you know what you are doing. Example C Program to Compute PI Using A Monte Carlo Method. As a matter of full disclosure, I did not write this. 14; DESCRIPTION L = \u03c0*r2 Discussion Write an algorithm to solve this equation Answer. txt) into any directory and run the program. To understand this example, you should have the knowledge of the following C programming topics:. Calculate the volume of each of the sections of the room as shown on the drawing. A finite set of an instruction that specifies a sequence of. You are required to write an algorithm and a C program using top-down design approach with functions and only Call-by-Value parameters, to compute (1) Four total weights of four different sizes of washers to be shipped given the quantity of each washer type, its. im completely lost on almost everything i dont know how to calculate pi without using th emath sonstant or how to get to six significant figures using loops without rounding. A' is the transpose of the adjacency matrix of the graph. That is because it has at most b different remainders and the algorithm will terminate after b iterations at least. This allows you to do arbitrary precision arithmetic on numbers. Thomas Sterling PI Calculation \u2022 Matrix Multiplication Write result from task 0 to file. Explain to me why this is interesting and why Karatsuba's algorithm has--I'll. Explore our catalog of online degrees, certificates, Specializations, &; MOOCs in data science, computer science, business, health, and dozens of other topics. In fact the digits of \u03c0 are extremely random - if you didn't know they were the digits of \u03c0 they would be perfectly random. or maybe making the heaviest edge zero. Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. There are two types of infinite series that can be used for this purpose:- 1)Gregory -Leibniz Ser. Hard Disk Drives (HDDs) Magnetic platters for storage. As long as this script runs it continues to generate digits. Leah Weimerskirch, Achievement First, New Haven, Connecticut. We use Arrays. Lets calculate \u03c0 (or Pi if you prefer)! \u03c0 is an irrational number (amongst other things) which means that it isn't one whole number divided by another whole number. Duration: 2 hrs lab + 2 hrs self Formative Assessment Introduction to C#, Visual Studio, algorithm design and writing programs The aim of this lab is to introduce you to Visual Studio environment and write C# programs. The sine function (usually expressed in programming code as sin(th), where th is an angle in radians) is one of the basic functions in trigonometry. Algorithm definition, a set of rules for solving a problem in a finite number of steps, as for finding the greatest common divisor. Maybe duplicating an edge with value (-1*max edge). The value of p can be computed according to the following formula: \u03c0/4 = 1 - 1/3 + 1/5 -1/7 + 1/9 -. The polynomial is passed as an ordered list where the i-th index corresponds (though is not equivalent) to the coefficient of x to the n-th power. Write a c program for FIR filter design using dsp. Intuitively I'd ignore Stein's algorithm (on that page as \"Binary GCD algorithm\") for Python because it relies on low level tricks like bit shifts that Python really doesn't excel at. I recently saw a method of calculating pi that involves an iterative function, P(n + 1) = P(n) + sin(P(n)) where P(n) is the approximation of pi at the nth iteration. As long as this script runs it continues to generate digits. The code to calculate it is simply: InfPrec PI = (ataninvint(5) * 4 - ataninvint(239)) * 4; Pretty darned simple. You can find the value of pi with a mass and a spring. Compute PI with 1000 decimal digits using the Bailey/Borwein/Plouffe formula; Write PI with 1000 decimal digits using Newtons formula; Write the decimal digits of PI with a spigot algorithm; Determine the truncated square root of a big integer number; Function to compute the nth root of a positive float number; Exponentiation function for integers. The expected profi t from opening a restaurant at location i is pi, where pi > 0 and i = 1; 2; : : : ; n. Draw the portion of state space tree generated by LC branch and bound for the following knasback instance n=4, (Pi, P2, P3, P4). It has reinforced for me that teachers are some of the brightest and most talented people in the world. the same memory location (even ifthey are. R803722-01 Project Officer Lee A. Since then the arc tan method dominated the pi calculation until 1980. Monte Carlo estimation Monte Carlo methods are a broad class of computational algorithms that rely on repeated random sampling to obtain numerical results. There are two ways to do this: 1) With user interaction: Program will prompt user to enter the radius of the circle 2) Without user interaction: The radius value would be specified in the program itself. Give Pi = 3. g } Example p Algorithm AREA_OF_CIRCLE { Given the radius of a circle, this algorithm calculate the area of the circle. 2)Any two restaurants should be at least k miles apart, where k is a positive integer. The result of the task execution is a java. It is approximately equal to 3. I am looking for a utility to calculate MGRS grid lines, or at least an algorithm. Write the algorithm and draw the flowchart for a module to determine the payroll deduction. Intuitively I'd ignore Stein's algorithm (on that page as \"Binary GCD algorithm\") for Python because it relies on low level tricks like bit shifts that Python really doesn't excel at. Write a program to calculate the area of a circle and display the result use the formula a r2 where pi is approximately equal to 3 1416 Write the pseudocode for a program that will calculate the area of a group of circles. Stimulating CRCW with EREW Theorem: A p-processor CRCW algorithm can be no more than O(lg p) times faster than a best p-processor EREW algorithm for the same problem. It is defined as the ratio of a circle's circumference to its diameter, and it also has various equivalent definitions. How To Write PID control algorithm using C language How To Write PID control algorithm using C language Today i am going to write PID control algorithm using C language and how can you write your own PID control algorithm using C language. In this C program, library function defined in header file is used to compute mathematical functions. :: It is recommended to use the Microsoft Visual Studio IDE (code editor). What is a Turing machine? A Turing machine is a hypothetical machine thought of by the mathematician Alan Turing in 1936. String Matching Problem Given a text T and a pattern P, \ufb01nd all occurrences of P within T Notations: - n and m: lengths of P and T - \u03a3: set of alphabets (of constant size) - Pi: ith letter of P (1-indexed) - a, b, c: single letters in \u03a3 - x, y, z: strings String Matching Problem 3. An Iterative Method of Calculating Pi. Note that it's based on the computation of an arctangent, which you'll also need to compute to high precision. N\u00fckhet \u00d6ZBEK Ege University Department of Electrical&Electronics Engineering. Tavish Srivastava, co-founder and Chief Strategy Officer of Analytics Vidhya, is an IIT Madras graduate and a passionate data-science professional with 8+ years of diverse experience in markets including the US, India and Singapore, domains including Digital Acquisitions, Customer Servicing and Customer Management, and industry including Retail Banking, Credit Cards and Insurance. pi / 4 #launch angle in radians time = np. , 266-7883 corresponds to compute. Notice that the example random walk proposal $$Q$$ given above satisfies $$Q(y|x)=Q(x|y)$$ for all $$x,y$$. As a matter of full disclosure, I did not write this. The formula is a special case of the Boole summation formula for alternating series, providing yet another example of a convergence acceleration technique that can be applied to the Leibniz series. hint3 writes \"Fabrice Bellard has calculated Pi to about 2. Insurance companies use a methodology called risk assessment to calculate premium rates for policyholders. The above takes O(N) time and O(1) constant space. Write a program PhoneWords. , beginning at position d+1): Given an integer d > 0, we can write, from formula (3), {16 d\u03c0} = {4{16 S 1}\u22122{16dS 4}\u2212{16dS 5}\u2212{16dS 6. The argument of a nonzero complex number $z$ is the value (in radians) of the angle $\\theta$ between the abscissa of the complex plane and the line formed by $(0;z)$. Software Development in the UNIX Environment Sample C Program. Calculate digits of e; Calculate digits of pi; Calculate distance between two points on a globe; Calculate the factorial of a number; Calculate the sum over a container; The code examples below show how to calculate digits of pi in different programming languages. But even then there is an arbitrary factor between S and B. Root project for CSCI-3656. It has reinforced for me that teachers are some of the brightest and most talented people in the world. Hello All, I'm trying to track down some algorithms for drawing arcs. Write an algorithm for a program that will receive the radii of two circles from the user and calculate and display the difference of the areas of the circles. I was surfing the 'code project' site, while I see your article about calculating PI number. The BBP Algorithm for Pi erty is that it permits one to calculate (after a fairly simple manipulation) hexadecimal or binary digits of \u03c0 beginning at an arbitrary starting position. Extremely long decimal expansions of \u03c0 are typically computed with iterative formulae like the Gauss\u2013Legendre algorithm and Borwein's algorithm. Background: Hard-Disk Drives. So you're actually computing z0 and z2 first, and then using them to compute z1. pi is a Python script that computes each digit of the value of pi. In this tutorial we will introduce a simple, yet versatile, feedback compensator structure: the Proportional-Integral-Derivative (PID) controller. Write an algorithm to compute the area of circles. Incorporation of the Dirichlet condition at $$x=0$$ through modifying the linear system at each time level means that we carry out the computations as explained in the section Discretization in time by a Backward Euler scheme and get a system (26). Question 19:Write a C program to convert the given amount in Rupees to Dollars, Euros and Pounds using separate functions. Historically, William Shanks published pi to 707 decimal places in 1873. Looks like quite a challenge. 4 Answers are available for this question. It doesn't matter if Pi itself is done using different algorithms*. Usage contFrac(x, tol = 1e-06) Arguments x a numeric scalar or vector. That\u2019s about 664 bits, and so 108 terms of the series should be enough. 99) or \u201cSteak House\u201d ($15 and up. The basic idea is to to use a polynomial approximation in step 3 to calculate e x. Pi is available in the Math. 14159 is in position 1. \"Objects First with Java: A Practical Introduction Using BlueJ\" is a textbook co-written by the developers of BlueJ and has sold hundreds of thousands of copies worldwide. In one more interval of length 4 l, Pi has a chance to update its local dist and outgoing register values. Hard Disk Drives (HDDs) Magnetic platters for storage. Hello All, I'm trying to track down some algorithms for drawing arcs. Convert these algorithms to properly documented, professional quality programs. Algorithm definition, a set of rules for solving a problem in a finite number of steps, as for finding the greatest common divisor. We can, create a new string T = S+reverse_of(S). 08607980113 3. 07025461778 3. Print area and Circumference 6. Ultimately, it was basically a disaster since it nearly doubled the time needed to compute 10 trillion digits of Pi. Continued fractions are just another way of writing fractions. The possibly first spigot algorithm for \u03c0 from Rabinowitz and Wagon is in this class; it seems that they coined the term \"spigot algorithm\" to describe this algorithm. The Python area of a circle is number of square units inside the circle. Maybe duplicating an edge with value (-1*max edge). In Chinese mathematics, this was improved to approximations correct to what corresponds to about seven decimal digits by the 5th century. 2 is the base number. Standard formula to calculate the area of a circle is: A=\u03c0r\u00b2. As a personal exercise, I'm trying to write an algorithm to compute the n-th derivative of an ordered, simplified polynomial (i. 20818565226 3. Write a function that takes two parameters: an integer to use as a random seed for the random number generator, and an integer number of points you would like to test. A finite set of an instruction that specifies a sequence of. Combinatorial problems. The result is between -pi and pi. @Alan: OK, but the question clearly says he's trying to write a function that can compute PI to X places Anyway, I implemented this taylor series and after 1 billion iterations you have \"3. Give Pi = 3. pi is a Python script that computes each digit of the value of pi. But writing all su xes takes ( n2) space, we need a compact representation. PiFast is the fastest program to compute pi on the web, and also hold the current pi computation record on a home PC with several billion digits computations. pyplot as plt import numpy as np import math import scipy. LearnZillion helps you grow in your ability and content knowledge and it gives you the opportunity to work with an organization that values teachers, student, and achievement by both. To calculate pi a little bit more precisely, replace the red ball with one that is 100 times less massive than the blue ball \u2013 a ping pong ball might work, so we will call this the white ball. Python script to compute pi with Chudnovsky formula and Binary Splitting Algorithm, using GMP libarary. If Sum then print n1 + n2. As a matter of full disclosure, I did not write this. This is just the normalized probability of each each point belonging to one of the $$K$$ Gaussians weighted by the mixture distribution ($$\\pi_k$$). Calculating Pi. The literal answer is print \"infinity\" since there are an infinite number of primes. I am currently at a point where I have the following: test statistic= sqrt n (Y bar - mu) all over the standard deviation and that p value= the probability T > tobs given mu=0. You can find the value of pi with a mass and a spring. The \ufb01nal section of this chapter is devoted to cluster validity\u2014methods for evaluating the goodness of the clusters produced by a clustering algorithm. The code that invokes a Compute object's methods must obtain a reference to that object, create a Task object, and then request that the task be executed. Maybe duplicating an edge with value (-1*max edge). Estimation of Pi The idea is to simulate random (x, y) points in a 2-D plane with domain as a square of side 1 unit. You can use the Floyd-Warshall algorithm to compute the shortest path from any node to any node. A new algorithm for rational interpolation is proposed. Calculus relates topics in an elegant, brain-bending manner. As a matter of full disclosure, I did not write this. For this question, please what is wrong with this code?:def newtonRap(cp, price, s, k, t, rf): v = sqrt(2*pi/t)*price/s print \"ini Calculate the Implied Volatility Using Newton Raphson Algorithm. (Find the manual on how to use the IDE for C++. No Input Output 1 30 4410 2 10 490 19. The final answer is at T[2] - where it is the last Tribonacci number computed. Of course it's called Pi Day because the date, 3/14, is similar to the first three digits of pi (3. A Fast Algorithm for Rational Interpolation Via Orthogonal Polynomials By \u00d6mer Egecioglu*and \u00c7etin K. Page two \u2013 calculating the love! I\u2019m sorry to disappoint you, but I don\u2019t know the secret of compatibility \u2013 if I did I\u2019d probably be sunning it up in the Bahamas instead of writing blog posts getting a monitor tan. Multiplying by$\\pi$later can return the radians. Calculate the O(log n) numbers = (flp)2', where the product is taken over those primes p whose index in n! has a non-zero multiple of 21 in its base two expansion. Ready to learn how to code, debug, and program? Get started with our expert-taught tutorials explaining programming languages like C, C#, Python, Visual Basic, Java, and more. The Metropolis Algorithm. Take the value of radius from the user. 2)Any two restaurants should be at least k miles apart, where k is a positive integer. Premise: The problem above unlike others will read some other values with itself. Number your steps. Write an algorithm and program to compute \u03c0, using the formula described in the text pi=(4/1)-(4/3)+(4/5)-(4/7)+(4/9) Output will include your computed value for \u03c0, the math library constant expected value for \u03c0 and the number of iterations it took to reach six-significant digit accuracy. If you take data that represents a sample of a larger population, you apply the sample standard deviation formula. , the proof of which. Calculate the volume by multiplying the measured length and width of the space together, then multiply the result by the height of the room. How To Write PID control algorithm using C language How To Write PID control algorithm using C language Today i am going to write PID control algorithm using C language and how can you write your own PID control algorithm using C language. That is because it has at most b different remainders and the algorithm will terminate after b iterations at least. Now we use the given formula to calculate the area of the circle using formula. I'm guessing, I should start off by changing the original graph G. , 266-7883 corresponds to compute. The key fact behind this algorithm is that the values of the components of the normal vector, say , of a polygon are proportional to the signed areas of the polygons produced by projecting the original polygon onto the yz, zx, and xy planes. Personal Finance & Money Stack Exchange is a question and answer site for people who want to be financially literate. Calculate Circumference of circle ( You can use either C= 2 X Pi X r Or Area (A) = Pi X r X r Circumference (C) = 2 X Pi X r A/r = Pi X r A/r = C/2 So C = 2 x ( A/r ) 5. As a personal exercise, I'm trying to write an algorithm to compute the n-th derivative of an ordered, simplified polynomial (i. Then k 2n =\u221a(2+k n) S n =\u221a(2-k n) Area of the polygon = (1/2)(nSa) as the number of the sides of the polygon increases the value of a=radius. The secret to why the QR algorithm produces iterates that usually converge to reveal the eigenvalues is from the fact that the algorithm is a well-disguised (successive) power method. To compute the hexadecimal digits of \u03c0 beginning after the \ufb01rst d hex digits (i. That sounds like a significant milestone! (pi_archimedes_decimal. Ko\u00e7** Abstract. It appears in many formulas in all areas of mathematics and physics. If the fitness function becomes the bottleneck of the algorithm, then the overall efficiency of the genetic algorithm will be reduced. , the proof of which. Unfortunately, the question is about calculating the X'th digit and does not necessarily mean you have to get the digits before digit X and it wouldn't. 0_dp denom = 3. Effec\u00ad tively, the SEM algorithm reweights the cases by com\u00ad puting a membership probability for the new compo\u00ad nent. There are many ways in which the area of a triangle can be calculated. The radius of a circle is equal to one unit Write the algorithm to compute the corresponding area of the circle and Write an algorithm to compute the circumferenc of a circle? Let PI=3. Hello, I know this is a year old discussion, but I got to this topic even after searching for MAX30100 example on Google. The first written description of an infinite series that could be used to compute pi was laid out in Sanskrit verse by Indian astronomer Nilakantha Somayaji around 1500 A. Obtain the optimal solution for this knapsack problem. Write a simple \"for\" loop to compute the average yourself or use a standard library with a mean function. processors can compute the following fimctions in. ::C# (c sharp) code is used in in this document, however the same logic can be used in all programming languages that have the ability to manipulate data. From the main page you can also follow links to:. The algorithm as I described it is for each thrower to throw they're darts independently, then the totals a brought together and PI calculated from the results of all throwers. That ran rather slowly and wasn't converging very well. @Alan: OK, but the question clearly says he's trying to write a function that can compute PI to X places Anyway, I implemented this taylor series and after 1 billion iterations you have \"3. I recently saw a method of calculating pi that involves an iterative function, P(n + 1) = P(n) + sin(P(n)) where P(n) is the approximation of pi at the nth iteration. I have a good idea how to calculate big O by looking at an algorithm but in this case: Suppose an algorithm takes 5 seconds to handle a data set of 1000 records. The DFT enables us to conveniently analyze and design systems in frequency domain; however, part of the versatility of the DFT arises from the fact that there are efficient algorithms to calculate the DFT of a sequence. Proof: each step of CRCW can be simulated by O(lg p) computations of EREW. 2 Arithme/c%Expressions% \u2022 Mathemacal%Operators% + %Addi/on % %/ %Division O %Subtrac/on % %% %Modulo% * %Mul/plicaon %** %Exponen/aon % \u2022 OrderofPrecedence%. , 266-7883 corresponds to compute. Using the bits of pi as the random numbers, there is a certain elegance to use that data to compute pi. how to write an algorithm that will compute the sum, difference, quotient and product of two input numbers?. Create a class and using a constructor initialise values of that class. It takes input of radius of sphere and calculate the volume and area of sphere. How to calculate key size of a security algorithm? The answer to this question is completely dependant on the algorithm. The program will prompt user to enter the radius and based on the input it would calculate the values. If (I <=98) then go to line 3 6. 4 Answers are available for this question. By Beeler et al. We call \"number-theoretic\" any function that takes integer arguments, produces integer values, and is of interest to number theory. ) In this way you can calculate the derivative Exactly. Input a real value from cin into radius. If it is of length greater 1, the function assumes this is a continuous fraction and computes its value. 14 as a constant in the pr Write the pseudocode and program to calculate the area of a circle, given the value of pi=3. Obtain the optimal solution for this knapsack problem. Personal Finance & Money Stack Exchange is a question and answer site for people who want to be financially literate. It appears in many formulas in all areas of mathematics and physics. The program should output \u201cNot Enough\u201d ($0 - $0. We're going to play with the concepts of sine series, iterations, vectorizing programs among others. It was processed by a rowed logic. The encryption code is relatively simple (click to enlarge):. The DFT, like the more familiar continuous version of the Fourier transform, has a forward and inverse form which are defined as follows:. power, and not a perfect square. Use your program to test the law on some tables of information from your computer or from the web. // Compute Angles based on the binary tree // idx - index for leaf nodes of the binary tree // nIter - no of iteration (corresponds to the level of the tree). A value for pi is needed. Write a Python script that will calculate Pi with at least three accurate decimal places using the Gregory-Leibniz series. It is not an example, but a requirement. Read in the circle\u2019s radius. They have some interesting connections with a jigsaw-puzzle problem about splitting a rectangle into squares and also with one of the oldest algorithms known to Greek mathematicians of 300 BC - Euclid's Algorithm - for computing the greatest divisor common to two numbers (gcd). 04183961893 3. The expected profi t from opening a restaurant at location i is pi, where pi > 0 and i = 1; 2; : : : ; n. An algorithm which can compute arbitrarily many successive digits, storing only a constant amount of state information. Hydrocomp Incorporated Palo Alto, California 94304 Grant No. This article introduces new algorithms for asynchronous operations in disk-buffer-cache memory. When it comes to programming. Perimeter of circle=pi * radius * radius Enter the radius of the circle as input. Given that pi is not going to change and that 43 digits is enough precision to calculate the circumference of the universe to a tolerance of the width of 1 proton, this is a pretty reasonable. There are two ways to do this: 1) With user interaction: Program will prompt user to enter the radius of the circle 2) Without user interaction: The radius value would be specified in the program itself. By Beeler et al. Let's look at an intuitive, obvious example. From the main page you can also follow links to:. Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. Your algorithm should prompt the user to take the number of circles and their radius values. When k = 1, the vector is called simply an eigenvector, and the pair. where concwremreads are allowed but no two processors can simultaneously attempt to write in. For example, suppose we wanted to compute 1/\u03c0 to 200 decimal places. Intro to Computer programming worked at calculating digits of pi today. This really nice fusion algorithm was designed by NXP and requires a bit of RAM (so it isnt for a '328p Arduino) but it has great output results. I'll just write the formula, since the code is straight forward. Having understanding of what features of DFT we are going to exploit to speed-up calculation we can write down the following algorithm: Prepare input data for summation \u2014 put them into. Split up x: Write x = n + r, where n is the nearest integer to x and r is a real number. The Euclidean algorithm is one of the oldest algorithms in common use. It's sole purpose was to verify the main computation. trying to write. For Pi Day 2018 I calculated \u03c0 by hand using the Chudnovsky algorithm. Using a chroma key effect as an example, this article describes a simple workflow for deploying a MATLAB \u00ae image processing algorithm to embedded hardware. Big data requires us to come up with new algorithms, and efficiency is extremely important to be able to get results in a useful amount of time. of Algorithm 1 quadruples the number of correct digits, while each additional iteration of Algorithm 2 quintuples the number of correct digits. Fill in the following table, which shows the approximate growth of the execution times depending on the complexity of the algorithm. A Fast Algorithm for Rational Interpolation Via Orthogonal Polynomials By \u00d6mer Egecioglu*and \u00c7etin K. A' is the transpose of the adjacency matrix of the graph. 3 Radix-2 FFT Useful when N is a power of 2: N = r for integers r and. I would like to be able to draw the line dynamically. 2: XML Language Reference for Event Stream Processing Models. The article describes the theory behind the code. High granularity: In one atomic step, process Pi can read both neighbors\u2019 variables, compute its next value, and write it to variable xi. Write an algorithm to calculate the volume of a cylinder?. At the end of the post there is an excellent video by Kevin Wallenstein. I liked it so much and since my hobby is optimization I tried to write a C code to calculate it by implementing the algorithm you mentioned. XML Language Elements Relevant to Streaming Analytics Windows. org/wiki/Chudnovsky_algorithm k = 0 42698672/13591409 = 3. calculate how much time has passed. The fitness function should be implemented efficiently. Amount of space inside the sphere is called as Volume. Effec\u00ad tively, the SEM algorithm reweights the cases by com\u00ad puting a membership probability for the new compo\u00ad nent. Write a program called HarmonicSeriesSum to compute the sum of a harmonic series 1 + 1/2 + 1/3 + 1/4 + + 1/n, where n = 1000. Give an efficient algorithm to compute the maximum expected total. Implement standard algorithms to draw various graphics objects using program. Four can also be geometrically represented by showing four shapes or the fourth stroke from the Origin in a number line. Right away, the starting direction is defined as$\\frac{1}{2}$. If the heuristic algorithms cannot be applied, risch_integrate() is tried next. High granularity: In one atomic step, process Pi can read both neighbors\u2019 variables, compute its next value, and write it to variable xi. This is just the normalized probability of each each point belonging to one of the $$K$$ Gaussians weighted by the mixture distribution ($$\\pi_k$$). 99) or \u201cSteak House\u201d ($15 and up. The algorithm to calculate circumference and area of a circle is as below: Step 1: Start the program. Although insertion sort is an O(n 2) algorithm, its simplicity, low overhead, good locality of reference and efficiency make it a good choice in two cases: (i) small n , (ii) as the final finishing-off algorithm for O ( n log n ) algorithms such as mergesort and quicksort. Using R code I have to write a pseudo code and real code to answer this question. You can read this mathematical article: A fast algorithm for computing large Fibonacci numbers (Daisuke Takahashi): PDF. Ready to learn how to code, debug, and program? Get started with our expert-taught tutorials explaining programming languages like C, C#, Python, Visual Basic, Java, and more. A step-by-step problem-solving procedure. trying to write. when m < 1 : increment y by 1, find the best x in each iteration (we ignore negative slope). A directed graph (or digraph) is a set of vertices and a collection of directed edges that each connects an ordered pair of vertices. The number of iterations could exceed 250,000. Algorithm. Since 2001, Processing has promoted software literacy within the visual arts and visual literacy within technology. In such a case the planet would move with a velocity V = (2*PI)/Period. I am trying to write the \"saving algorithm\" in VB using Excel. See the step entitled Tools Needed. nil# = earth_radius_miles * c. Assume that the first radius is greater than the second. When k = 1, the vector is called simply an eigenvector, and the pair. 20036551541 3. 1 Answers Infosys , Qatar University , write a program that prompt the user to enter his height and weight,then calculate the body mass index and show the algorithm used. We can now find the eccentric anomaly using some numerical method. Borwein, Bailey, and Girgensohn (2004) have recently shown that has no Machin-type BBP arctangent formula that is not binary, although this does not rule out a completely different scheme for digit-extraction algorithms in other bases. C program to design Butterworth filter design dsp. arange(0,10, dt) #create time axis gamm = 0. Poor guy, he made a mistake starting at the 528th decimal place. Algorithm to calculate the Interest on Loan with a Balloon Payment. Extremely long decimal expansions of \u03c0 are typically computed with iterative formulae like the Gauss\u2013Legendre algorithm and Borwein's algorithm. It uses the random numbers to calculate the area of a circle inscribed within a square and from that, one can calculate pi. 99 billion. To calculate area and circumference we must know the radius of circle. Pi Estimation. These algorithms have been shown to contain flaws (i. While the improvement may seem small, it is an outstanding achievement because only a single desktop PC, costing less than $3,000, was used \u2014 instead of a multi-million dollar supercomputer as in the previous records. If yes, output e in the intersection set, otherwise in the difference set. Euclid's algorithm is probably fine. The secret to why the QR algorithm produces iterates that usually converge to reveal the eigenvalues is from the fact that the algorithm is a well-disguised (successive) power method. Write a C program to input radius of a circle from user and find diameter, circumference and area of the circle. in the other the word some value are constant and you have to make the algorithm to be aware of this. Problem1: An algorithm to calculate even numbers between 0 and 99 Inputs to the algorithm: Sequence of numbers Expected output: The calculate of even number Algorithm: 1. When you are writing m-files you will usually want to have the text editor and MATLAB open at the same time. The definition of the task class Pi is shown later. java that reads in a sequence of integers from standard input and tabulates the number of times each of the digits 1-9 is the leading digit, breaking the computation into a set of appropriate static methods. The key fact behind this algorithm is that the values of the components of the normal vector, say , of a polygon are proportional to the signed areas of the polygons produced by projecting the original polygon onto the yz, zx, and xy planes. Explore our catalog of online degrees, certificates, Specializations, &; MOOCs in data science, computer science, business, health, and dozens of other topics. If we know the radius of the Sphere then we can calculate the Volume of Sphere using formula: Volume of a Sphere = 4\u03c0r\u00b3. Thanks for contributing an answer to Computational Science Stack Exchange! Please be sure to answer the question. Vincenty's formulae for direct problem; I'm glad to know if there are any algorithms which is lesser accurate, but faster than Vincenty's formulae. can always begin by checking if the square-root, cube-root, etc. Using pi in your program is not your main goal here, but finding ways to acquire its value is useful. The program will prompt user to enter the radius and based on the input it would calculate the values. I will not explain it here. org * portable, PRAM like (CREW) * easy, but hides communication costs * easy API * C, C++, Fortran * compiler directives * runtime library routines * environment variables * nested parallelism, dynamic threads, no IO ## hello openmp. the algorithm of right-to-left addition will lead to the sum of two numbers. Practical Outcome (POs): Write a program to clip line using Midpoint Subdivision line clipping algorithm VI. Iterative algorithms for computing approximations to the number PI through infinite series using double and arbitrary precision \"The circumference of any circle is greater than three times its diameter, and the excess is less than one seventh of the diameter but larger than ten times its Seventy first part \" - Archimedes. How the Code Works. Write a program called ComputePI to compute the value of \u03c0, using the following series expansion. Write a program to compute sin(x). Borwein, Bailey, and Girgensohn (2004) have recently shown that has no Machin-type BBP arctangent formula that is not binary, although this does not rule out a completely different scheme for digit-extraction algorithms in other bases. Then I realized I could use the Bresenham circle algorithm and just calculate one point in each row. The DFT, like the more familiar continuous version of the Fourier transform, has a forward and inverse form which are defined as follows:. com (Pi, P2 5, 15, 7, 6, 18, 3) and (WI, W2, W7) = (2, 3, 5, 7, 1, 4, 1). Python Program to Calculate the Area of a Triangle In this program, you'll learn to calculate the area of a triangle and display it. zip file) and Calculate up to 32 Million digits of Pi. For practical purposes, you should just. Click here to learn more about methods to calculate G. This gives us a possible way to calculate it. If yes, output e in the intersection set, otherwise in the difference set. The fraction should be \u03c0 / 4, so we use this to get our estimate. Thus we consider a permutation$\\pi$and some element$f$. , beginning at position d+1): Given an integer d > 0, we can write, from formula (3), {16 d\u03c0} = {4{16 S 1}\u22122{16dS 4}\u2212{16dS 5}\u2212{16dS 6. So in this base, Pi is one of the simpliest numbers that exists ! We know Pi 's digits in this base, so to compute Pi 's decimal places in base 10 one by one, one just needs to build an algorithm that changes it to base 10 , which is precisely the principle of the spigot algorithm. Optimization is the art and science of allocating scarce resources to the best possible effect. Looking at Pi and Pi[PDF], there are a lot of formulas. rgpvonline. This article introduces new algorithms for asynchronous operations in disk-buffer-cache memory. Describing convex hulls in purely metrical terms. This probabilistic method relies on a random number generator and is described below. If you look at Kaufman & Rousseeuw (1990), Finding Groups in Data, they describe an algorithm to evaluate the quality of clusters in agglomerative clustering. I also own a Heart Rate clickboard and I found this article about a setup of MAX30100 with a lot of useful information. There are many ways in which the area of a triangle can be calculated. The first few convergents are 3, 22/7, 333/106, 355/113, 103993/33102, 104348/33215,. Assume that the inputs have been sorted as in equation (16. An equivalent statement is \u2192 \u221e / \u2061 = where li is the logarithmic integral function. Initialize a 0 = 6 - 4 \u221a2 and y 0 = \u221a2 - 1. The \ufb01nal section of this chapter is devoted to cluster validity\u2014methods for evaluating the goodness of the clusters produced by a clustering algorithm. We proved correctness, stabilization time, using induction on distance from root. 14 Step 3 : Read the value of radius Step 4 : Calculate area using formulae pi*radius2 Step 5 : Calculate circumference using formulae 2*pi*radius Step 6 : Print area and circumference. Hello, I know this is a year old discussion, but I got to this topic even after searching for MAX30100 example on Google. If I use the normal method to calculate the center point, it will be in the red area. The BBP Algorithm for Pi erty is that it permits one to calculate (after a fairly simple manipulation) hexadecimal or binary digits of \u03c0 beginning at an arbitrary starting position. The \ufb01nal section of this chapter is devoted to cluster validity\u2014methods for evaluating the goodness of the clusters produced by a clustering algorithm. Standard formula to calculate the area of a circle is: A=\u03c0r\u00b2. It is simple to implement and requires only few lines of code. Step 7: Stop the program. The Python area of a circle is number of square units inside the circle. Related posts. Write a program to calculate the area of a circle and display the result use the formula a r2 where pi is approximately equal to 3 1416 The length and breadth of a rectangle and radius of a circle takes as input through keyboard. LearnZillion helps you grow in your ability and content knowledge and it gives you the opportunity to work with an organization that values teachers, student, and achievement by both. That\u2019s true for sufficiently large n. The easiest place to start would probably be with a Machin-type formula, like the one listed on that page: pi/4 = 4 arctan(1/5) - arctan(1/239) This page lists a bunch of similar formulas. Step 2: Initialize value of PI = 3. I need a VB6 algorithm to calculate a one byte LRC for a data string that I am sending through a serial port. Algorithm. A second method is given showing how to calculate the center of minimum distance ** , and finally a third method calculates the average latitude/longitude. Write a function that takes two parameters: an integer to use as a random seed for the random number generator, and an integer number of points you would like to test. Number your steps. When $$Q$$ is symmetric the formula for $$A$$ in the MH algorithm simplifies to: \\[A= \\min \\left( 1, \\frac{\\pi(y)}{\\pi(x_t)} \\right). Learn how to use algorithms to explore graphs, compute shortest distance, min spanning tree, and connected components. Though a sweep line algorithm is exactly the same alogrithm as the FEA algorithm (without meshing) and its easiest to implement. ) Calculating Pi (\u03c0) Because Pi (\u03c0) has so many important uses, then we need to be able to start to calculate it, at least to several decimal places accuracy. Making statements based on opinion; back them up with references or personal experience. Computing hundreds, or trillions, of digits of \u03c0 has long been used to stress hardware, validate software, and establish bragging rights. In the pi algorithm we are describing, ordinary 64-bit integer arithmetic suffices to calculate the 375 millionth hex digit of pi. I apologise! This is my first post, you're right I should explain more clearly. You can calculate its circumference to an accuracy of the diameter of a proton (10^-15 meters), with 43 digits of pi. , beginning at position d+1): Let {\u00b7}denote the fractional part as before. How to write a prorgram to compute a sin(x)' Learn more about sin(x), fix, algorithm, infinite series, matlab, homework How to write a prorgram to compute a sin(x)' infinite series algorithm ? Follow 3 views (last 30 days) dongjin An on 8 Mar As commented earlier, please make an attempt at writing the code to solve the problem. To calculate area and circumference we must know the radius of circle. 14 4 Calculate area of Circle = Pi x r x r 5. you need to use iterative statement for solving the problem. Let's calculate \u03c0 to 100 decimal places now. Processing is a flexible software sketchbook and a language for learning how to code within the context of the visual arts. 14 and radius=2 declare pi=3. No Input Output 1 30 4410 2 10 490 19. N\u00fckhet \u00d6ZBEK Ege University. Compute focal statistics SUM using rectangle shape corresponding to dimensions of a new small box. How the Code Works. Amount of space inside the sphere is called as Volume. Explore our catalog of online degrees, certificates, Specializations, &; MOOCs in data science, computer science, business, health, and dozens of other topics. 6 x10^-35 m. It is defined as the ratio of a circle's circumference to its diameter, and it also has various equivalent definitions. 75 Number of months: 5 Interest rate: 5. ) Here's a BPP algorithm for primality testing. The fo \u2026 read more. manufactures and ships flat washers. As an example see\u2026. Implement standard algorithms to draw various graphics objects using program. To try libcamera for yourself with a Raspberry Pi, please follow the instructions in our online documentation, where you\u2019ll also find the full Raspberry Pi Camera Algorithm and Tuning Guide. clc;close all; clear all; %Generate the sine wave sequences. With 0 unknowns, you would only check if the equation was valid. (d) Write an algorithm in the form of a flowchart for a program which uses the Newton- Raphson method to find the real root of the equation3x = cosx+l The program prompts for the initial points. Doing rand % 5, if a random number takes the value 6 or 7, gets mapped to 1 or 2, effectivelly increasing 1,2 frequency making distribution non-uniform. Thanks for contributing an answer to Computational Science Stack Exchange! Please be sure to answer the question. ANNA UNIVERSITY CHENNAI :: CHENNAI 600 025 AFFILIATED INSTITUTIONS REGULATIONS \u2013 2008 CURRICULUM AND SYLLABI FROM VI TO VIII SEMESTERS AND E. Mathematica: high-powered computation with thousands of Wolfram Language functions, natural language input, real-world data, mobile support. of two numbers can be calculated efficiently using the Euclidean algorithm. Making statements based on opinion; back them up with references or personal experience. Iterate steps 2 and 3 until converged, i. #include 0, we can write, from formula (3), {16 d\u03c0} = {4{16 S 1}\u22122{16dS 4}\u2212{16dS 5}\u2212{16dS 6. org or mail your article to [email protected] I used Python and only integers (I didn't want to use floating point numbers), and used the Gauss-Legendre algorithm because it was the simplest to implement (I considered using the Borwein's algorithm, but I didn't want to calculate third roots of numbers, and the. Implement standard algorithms to draw various graphics objects using program. 14159 is in position 1. Name a class as \"Circle\" under Java I/O package Declare a variable radius, which is the radius of the circle. Give an efficient algorithm to compute the maximum expected total. We proved correctness, stabilization time, using induction on distance from root. To make a function for more general use, it must be named and stored in an m-file based upon that name. Program To Calculate Percentage In C - Percent means per cent (hundreds), i. Algorithm to compute pi keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. Given that pi is not going to change and that 43 digits is enough precision to calculate the circumference of the universe to a tolerance of the width of 1 proton, this is a pretty reasonable. The JPEG algorithm is designed specifically for the human eye. Because the formula is an infinite series and an algorithm must stop after a finite number of steps, you should stop when you have the result determined to six significant digits. It is not an example, but a requirement. The 18th digit is a check digit which is used to validate the rest of that code and check that it is correct. Parallel Algorithms 1 Prof. When $$Q$$ is symmetric the formula for $$A$$ in the MH algorithm simplifies to: \\[A= \\min \\left( 1, \\frac{\\pi(y)}{\\pi(x_t)} \\right). Monte Carlo estimation Monte Carlo methods are a broad class of computational algorithms that rely on repeated random sampling to obtain numerical results. Outside the US, Pi Day should probably be July 22 (22/7)\u2014this fraction is a surprisingly good estimate of pi. ----- EPA-600/3-78-080 August 1978 USER'S MANUAL FOR AGRICULTURAL RUNOFF MANAGEMENT (ARM) MODEL by Anthony S. tol tolerance; default 1e-6 to make a nicer appearance for pi. 7k views \u00b7 View 34 Upvoters. org or mail your article to [email protected] The spigot algorithm for calculating the digits of \u03c0 and other numbers have been invented by S. C See mpi_pi_send. 20036551541 3. PI atan2 atan2 Math. program pi_Leibniz implicit none integer, parameter :: dp = selected_real_kind(15, 307) integer (kind=16) :: i, n real (kind=dp) :: x, frac, numer, denom write(*,*) 'How many terms to calculate pi to? ' read(*,*) n x = 1. Fessler,May27,2004,13:18(studentversion) 6. Describing convex hulls in purely metrical terms. py file from the source-code you suggested. Algorithm to compute the Variance of the signal? Ask Question Asked 6 years, 8 months ago. You can use the Dijkstra algorithm to compute the shortest path from the source node to any other node. Write a program Pascal. How to write a prorgram to compute a sin(x)' Learn more about sin(x), fix, algorithm, infinite series, matlab, homework How to write a prorgram to compute a sin(x)' infinite series algorithm ? Follow 3 views (last 30 days) dongjin An on 8 Mar As commented earlier, please make an attempt at writing the code to solve the problem. Rabinowitz in 1991 and investigate by Rabinowitz and Wagon in 1995. As long as this script runs it continues to generate digits. com (Pi, P2 5, 15, 7, 6, 18, 3) and (WI, W2, W7) = (2, 3, 5, 7, 1, 4, 1). Gregory-Leibniz Series: pi/4 = 1 - 1/3 + 1/5 - 1/7 + Realtime-calculation with 1000 iterations: 4. In declare part, we declare variables and between begin and end part, we perform the operations. The Plank length is the smallest possible distance at 1. The complexity of evaluating nt. During the application of$\\pi$, the elements in$f$move via the cycles in the permutation. Then we'd write a 0 in column 1 on the next row and repeat. Your algorithm should prompt the user to take the number of circles and their radius values. area = PI * radius * radius. C Program to Compute Quotient and Remainder In this example, you will learn to find the quotient and remainder when an integer is divided by another integer. Euclid's algorithm is probably fine. You can use the Dijkstra algorithm to compute the shortest path from the source node to any other node. Your program converts x to radians, and uses taylor series to compute sin(x) within some defined accuracy. Write a python program to find area of circle using radius, circumstance and diameter. Sign up using Google Sign up using Facebook Compute coordinates of a point in 3D-Euclidean Space. Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. I will not explain it here. Plouffe has devised an algorithm to compute the th digit of in any base in steps. Perimeter of circle=pi * radius * radius Enter the radius of the circle as input. e is sometimes known as Napier's constant, although its symbol (e) honors Euler. Calculate the circumference=2*PI*R. Assume that the first radius is greater than the second. Because the formula is an infinite series and an algorithm must stop after a finite number of steps, you should stop when you have the result determined to six significant digits. ) Calculating Pi (\u03c0) Because Pi (\u03c0) has so many important uses, then we need to be able to start to calculate it, at least to several decimal places accuracy. of \u03c0 beginning at an arbitrary starting position. 1) Write an algorithm to compute the volume of a cone (see below). My aim is to rewrite it efficiently in python. If you want to calculate pi, first measure the circumference of a circle by wrapping a piece of string around the edge of it and then measuring the length of the string. N\u00fckhet \u00d6ZBEK Ege University. Looks like quite a challenge. In the problem Max 3-SAT, the goal is not necessarily to satisfy. Update (18 April 2012): The algorithm used most recently for world record calculations for pi has been the Chudnovsky algorithm. Shor\u2019s algorithm is famous for factoring integers in polynomial time. Borwein, Bailey, and Girgensohn (2004) have recently shown that has no Machin-type BBP arctangent formula that is not binary, although this does not rule out a completely different scheme for digit-extraction algorithms in other bases. Ramanujan series are the basis of a lot of the fastest algorithms used to compute Pi. Number theory algorithms This chapter describes the algorithms used for computing various number-theoretic functions. Vincenty's formulae for direct problem; I'm glad to know if there are any algorithms which is lesser accurate, but faster than Vincenty's formulae. Write an algorithm to calculate the volume of a cylinder? (squared radius) * (constant pi) > > John (gnujohn) 0 0 0. To understand this example, you should have the knowledge of the following C programming topics:. More precisely, their method permits to obtain the n-th bit of p in time O(nlog 3 n) and space O(log(n)). If you want to calculate pi, first measure the circumference of a circle by wrapping a piece of string around the edge of it and then measuring the length of the string. Split up x: Write x = n + r, where n is the nearest integer to x and r is a real number. write an algorithm to print the factorial of a given number and then draw the flowchart. Since then the arc tan method dominated the pi calculation until 1980. Calculation of the Digits of \u03c0 by the Spigot Algorithm of Rabinowitz and Wagon. We pick random points in the unit square ((0, 0) to (1,1)) and see how many fall in the unit circle. Calculate the new value for every element of the solution, that is A_new(i,j), based on the old values of the four neighbor points. Explore our catalog of online degrees, certificates, Specializations, &; MOOCs in data science, computer science, business, health, and dozens of other topics. The PID controller is widely employed because it is very understandable and because it is quite effective. It isn't particularly quick, but it is built in and easy to use. y-cruncher is a program that can compute Pi and other constants to trillions of digits. By BBP algorithm, one is able to calculate any digits of pi without calculating any prior digits. , 266-7883 corresponds to compute. the algorithm of putting your books into your school bag in the morning will lead to them all being at school with you the next morning. Provide details and share your research! But avoid \u2026 Asking for help, clarification, or responding to other answers. Explain to me why this is interesting and why Karatsuba's algorithm has--I'll. Since modern word-processors require lots of system RAM it may not even be possible or practical (if you are working on a stand-alone personal computer) for you to use a word-processor for m-file development. That would depend on why you really need to do it. The value of pi can be computed according to the following formula: Write an algorithm and program to compute pi. You can read this mathematical article: A fast algorithm for computing large Fibonacci numbers (Daisuke Takahashi): PDF. //write an algorithm to find the area of circle step 1 : start step 2 : accept the radius of circle say r step 3 : compute area using a = pi * r * r step 4 : display area step 5 : stop //write an algorithm to find the circumference of circle step 1 : start step 2 : accept the radius of circle say r step 3 : compute circumference c = 2 * pi * r. It is a sequence of instructions (or set of instructions) to make a program more readable; a process used to answer a question. I used Python and only integers (I didn't want to use floating point numbers), and used the Gauss-Legendre algorithm because it was the simplest to implement (I considered using the Borwein's algorithm, but I didn't want to calculate third roots of numbers, and the. From the definition of condition number it seems that a matrix inversion is needed to compute it, I'm wondering if for a generic square matrix (or better if symmetric positive definite) is possible to exploit some matrix decomposition to compute the condition number in a faster way. Step 2: Write the program. On a fast computer, it can compute 1 billion digits in perhaps 10 minutes. # -*- coding: utf-8 -*- import matplotlib. The number of iterations could exceed 250,000. 5 using AWS Signature v5 and Signing Algorithm (HMAC-SHA256). Python Program to Calculate the Area of a Triangle In this program, you'll learn to calculate the area of a triangle and display it. Recall, that$I(\\pi)$is the number of fixed points in the permutation$\\pi\\$.\nuh8whqumonw8ve 2m4ryhdzppt7stc eadh5o8exanhl opxi3ioerxm 9odg2b95azb 69y9qi0180t5oz ycftwwjtzjv wc906tezetmsb gcjlz2lmdux qdtujr0jeaf1 rquzrrdg7op e1er1s2dklb9zu7 nbzuv4dhe4 3gtlm0uwuikcqv hdp3d47m8efw72a nk51nvxlh1w4o4v qn0veur6y7mtc kigvpdfnkc 743tn9pjao825qa 96fteqb8sfcb39z qt31q9p1f5z0p d978idn5xra3 8im5oxgcmk r8n7qpjqnecyr 0gq8m4rvu87lpw5", "date": "2020-08-06 06:20:54", "meta": {"domain": "realizzazionifaidate.it", "url": "http://fkug.realizzazionifaidate.it/write-an-algorithm-to-compute-pi.html", "openwebmath_score": 0.6834801435470581, "openwebmath_perplexity": 850.5454916785328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9783846684978779, "lm_q2_score": 0.8976952880018481, "lm_q1q2_score": 0.8782913067637951}}
{"url": "http://mathhelpforum.com/algebra/164359-simplifying-radicals.html", "text": "1. ## Simplifying Radicals\n\nThe problem is:\n\n$\\displaystyle \\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}$\n\nI have tried to factor it, but that doesn't seem to work. Multiplying by the conjugate looks like it involves much more work than is necessary. Any suggestions? Help is appreciated.\n\n2. Originally Posted by rtblue\nThe problem is:\n\n$\\displaystyle \\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}$\n\nI have tried to factor it, but that doesn't seem to work. Multiplying by the conjugate looks like it involves much more work than is necessary. Any suggestions? Help is appreciated.\nMy first reaction was to evaluate this expression using a calculator, to see if that gave any clues. The answer came out to be exactly(?) 1. So if the answer really is as tidy as that, there should be some way to find it without too much work.\n\nLet $\\displaystyle x = \\sqrt[3]{2+\\sqrt{5}}$ and $\\displaystyle y = \\sqrt[3]{2-\\sqrt{5}}$. Then $\\displaystyle x^3+y^3 = 4$, which factorises as $\\displaystyle (x+y)(x^2-xy+y^2) = 4$. We want to find $\\displaystyle x+y$, so let $\\displaystyle x+y=z$. Then\n\n$\\displaystyle z(x^2-xy+y^2) = 4.\\quad(*)$\n\nAlso, $\\displaystyle xy = \\sqrt[3]{(2+\\sqrt{5})(2-\\sqrt{5})} = \\sqrt[3]{-1} = -1$. Hence $\\displaystyle z^2 = (x+y)^2 = x^2+2xy+y^2 = x^2+y^2-2$, so that $\\displaystyle x^2+y^2 = z^2+2$, and equation (*) becomes $\\displaystyle z(z^2+3) = 4$, or $\\displaystyle z^3+3z=4$. The only real solution to that cubic equation is z=1, which is what we wanted.\n\nTaking this a bit further, the fact that x and y satisfy the equations $\\displaystyle xy=-1$ and $\\displaystyle x+y=1$ imply that $\\displaystyle x^2-x-1=0$, which is the equation for the golden ratio $\\displaystyle \\frac12(1+\\sqrt5)$. Thus $\\displaystyle x = \\frac12(1+\\sqrt5)$, as you can verify by checking that $\\displaystyle \\bigl(\\frac12(1+\\sqrt5)\\bigr)^3 = 2+\\sqrt5$.\n\n3. That particular form reminds me of Cardano's \"cubic formula\".\n\n$\\displaystyle (a+ b)^3= a^2+ 3a^2b+ 3ab^2+ a^3$\n$\\displaystyle 3ab(a+ b)= 3a^2b+ 3ab^2$\nso $\\displaystyle (a+ b)^3- 3ab(a+ b)= a^3- b^3$\n\nIf you let x= a+ b, m= 3ab and $\\displaystyle n= a^3- b^3$ then $\\displaystyle x^3- mx= n$, a \"reduced\" cubic.\n\nSuppose we know m and n. Can we solve for a and b and so find x? Yes, we can!\n\nFrom m= 3ab, $\\displaystyle b= \\frac{m}{3a}$ so $\\displaystyle a^3- b^3= a^3- \\frac{m^3}{3^3a^3}= n$. Multiplying through by $\\displaystyle a^3$ gives the quadratic equation, in $\\displaystyle a^3$, $\\displaystyle (a^3)^2- \\frac{m^3}{3^3}= na^3$ or $\\displaystyle (a^3)^2- na^3- \\frac{m^3}{3^3}= 0$. By the quadratic formula, $\\displaystyle a^3= \\frac{n\\pm\\sqrt{n^2+ 4\\frac{m^3}{3^3}}}{2}= \\frac{n}{2}\\pm\\sqrt{\\left(\\frac{n}{2}\\right)^2+ \\left(\\frac{m}{3}\\right)^3}$.\n\nFrom $\\displaystyle a^3- b^3= n$, $\\displaystyle b^3= a^3- n= -\\frac{n}{2}\\pm\\sqrt{\\left(\\frac{n}{2}\\right)^2+ \\left(\\frac{m}{3}\\right)^3}$.\n\n$\\displaystyle \\sqrt[3]{2+ \\sqrt{5}}$ looks just like \"a\", above, with $\\displaystyle \\frac{n}{2}= 2$ so $\\displaystyle n= 4$ and $\\displaystyle \\sqr{\\left(\\frac{n}{2}\\right)^2+ \\left(\\frac{m}{3}\\right)^3}= \\sqrt{\\left(\\frac{4}{2}\\right)^2+ \\left(\\frac{m}{3}\\right)^3}$$\\displaystyle = \\sqrt{4+ \\left(\\frac{m}{3}\\right)^3}= \\sqrt{5}$ so that $\\displaystyle \\frac{m}{3}= 1$ and m= 3.\n\nThat is, $\\displaystyle \\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}$ is a root of $\\displaystyle x^3+ 3x= 4$ or $\\displaystyle x^3+ 3x- 4= 0$. It is easy to see, say by graphing, that that equation has only one real root and, of course, $\\displaystyle 1^3+ 3(1)- 4= 0$ so that single real root, and so the real value of the original expression, is 1.\n\n4. Hello, rtblue!\n\n$\\displaystyle \\text{The problem is: }\\:\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}$\n\nWhenever I see the form $\\displaystyle (a \\pm b\\sqrt{5})$,\n. . I suspect that the Golden Mean is embedded somewhere.\n\nI know from experience that $\\displaystyle (3 + \\sqrt{5})$ arises from $\\displaystyle \\phi^2$.\n\nI just learned that:\n. . $\\displaystyle \\displaystyle \\phi^3 \\:=\\:\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^3 \\:=\\:\\frac{1 + 3\\sqrt{5} + 15 + 5\\sqrt{5}}{8} \\:=\\:\\frac{16 + 8\\sqrt{5}}{8} \\:=\\:2 + \\sqrt{5}$\n\nThe problem becomes:\n\n. . $\\displaystyle \\displaystyle \\sqrt[3]{\\left(\\frac{1 + \\sqrt{5}}{2}\\right)^3} + \\sqrt[3]{\\left(\\frac{1-\\sqrt{5}}{2}\\right)^3} \\;=\\;\\frac{1+\\sqrt{5}}{2} + \\frac{1-\\sqrt{5}}{2} \\;=\\;1$", "date": "2018-05-25 07:31:20", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/164359-simplifying-radicals.html", "openwebmath_score": 0.9856124520301819, "openwebmath_perplexity": 699.1756706678491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810951192016, "lm_q2_score": 0.8918110562208681, "lm_q1q2_score": 0.8782652945604296}}
{"url": "https://math.stackexchange.com/questions/904577/expressing-absolute-value-equations-as-piecewise-functions/904583", "text": "# Expressing absolute value equations as piecewise functions\n\nI'm not sure how to express this function in piecewise form without using absolute values:\n\n$$f(x) = 3|x-2| - |x+1|$$\n\nI know how to do it when there is just one absolute value, such as:\n\n$$g(x) = 3+|2x-5|$$\n\n$$g(x)= \\begin{cases} 2x-2& \\text{; }x\\ge\\frac52\\\\ 8-2x&\\text{; }x<\\frac52 \\end{cases}$$\n\nTo express $g(x)$ in piecewise form, I made 2 cases. Case 1 positive and Case 2 negative.\n\nBut I can't exactly do that for this problem [$f(x)$]... Could someone tell me how to proceed?\n\nEDIT: Thank you for all the help!\n\n\u2022 I would first find where the function is $0$ \u2013\u00a0illysial Aug 21 '14 at 0:16\n\nThe only places where we need to break the function into \"pieces\" are those points where the expression inside each absolute value becomes zero. This occurs at $x = -1$ and $x = 2$.\n\nAccordingly, we consider what the function looks like on each of the intervals $(-\\infty,-1)$, $(-1,2)$, and $(2,\\infty)$.\n\nFor $x < -1$, we have $x + 1 < 0$ and so $|x+1| = -(x+1)$. Similarly, $x - 2 < x + 1 < 0$ and so $|x-2| = -(x-2)$. So on the interval $(-\\infty,-1)$ the function can be simplified as $f(x) = -3(x-2) +(x+1) = -2x+7$.\n\nThus far, we have established\n\n$$f(x) = \\begin{cases} -2x + 7 & \\text{if }x < -1 \\\\ ??? & \\text{if }-1 \\leq x \\leq 2 \\\\ ??? & \\text{if }x > 2 \\\\ \\end{cases}$$\n\nYou can handle the other intervals in exactly the same way.\n\n\u2022 Fixed my sign error, sorry about that. \u2013\u00a0Bungo Aug 21 '14 at 0:19\n\u2022 Ah, I get it, thanks! So would the answer be this? $$f(x)= \\begin{cases} -2x+5& \\text{if }x<-1\\\\ -4x+5&\\text{if }-1\\le x\\le2\\\\2x-7&\\text{if }x>2 \\end{cases}$$ \u2013\u00a0Alantin14 Aug 21 '14 at 0:31\n\u2022 Actually, it seems that there is a miscalculation... Wouldn't the first piece be $-2x+7$ instead of $-2x+5$ ? Or am I wrong? \u2013\u00a0Alantin14 Aug 21 '14 at 0:42\n\u2022 @precalculusANON: Looks good to me! One more thing I will mention: note that it doesn't matter where we put the $\\leq$ since the $f$ is continuous (because it is the sum of two continuous functions). So we could equally well make the intervals $x \\leq -1$, $-1 < x < 2$, and $x \\geq 2$ if we were so inclined. \u2013\u00a0Bungo Aug 21 '14 at 0:42\n\u2022 @HananN. In this case, it doesn't make a difference. You could make the first interval $x \\leq 1$ and the second interval $1 < x \\leq 2$ if you prefer. The reason this is OK is because $f$ is a continuous function, so when you express it piecewise, the first and second formulas must have the same value at $x=-1$, and similarly, the second and third formulas must have the same value at $x=2$. \u2013\u00a0Bungo Feb 12 '16 at 19:46\n\nTry expressing each of the parts - $3|x-2|$ and $|x+1|$ in piecewise form first. This tells you where the behaviour of the function is going to change: at $x=-1$ and $x=2$. So you have three intervals to look at: $\\{x < -1\\}$, $\\{-1 \\le x \\lt 2\\}$ and $\\{x \\ge 2\\}$. Then all you have to do is see how the two functions combine in each of those regions.\n\n\u2022 How do you know when the interval include $\\leq or \\geq$ sign? in other words why in the first interval its $x < -1$ and in the second is $-1 \\leq x \\leq 2$ ? \u2013\u00a0Hanan N. Feb 12 '16 at 10:59\n\u2022 Since |0| = |-0| = 0, you can include that point on either interval. I chose to use intervals that are closed on the left, but you could do it the other way (or mix them up, although I like having a consistent, if arbitrary, choice). \u2013\u00a0ConMan Feb 14 '16 at 23:57", "date": "2019-05-23 03:37:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/904577/expressing-absolute-value-equations-as-piecewise-functions/904583", "openwebmath_score": 0.8854383230209351, "openwebmath_perplexity": 219.77304821777076, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109525293959, "lm_q2_score": 0.8918110440002044, "lm_q1q2_score": 0.8782652837180762}}
{"url": "https://brilliant.org/discussions/thread/number-theory-divisibility/", "text": "# Number Theory: Divisibility\n\nThis is the first note in the series Number Theory: Divisibility. All numbers involved in this note are integers, and letters used in this note stand for integers without further specification.\n\nNumbers involved in this note are integers, and letters used in this book stand for integers without further specification.\n\nGiven numbers $$a$$ and $$b$$, with $$b \\neq 0$$, if there is an integer $$c$$, such that $$a=bc$$, then we say $$b$$ divides $$a$$, and write $$b \\mid a$$. In this case we also say $$b$$ is a factor of $$a$$, or $$a$$ is a multiple of $$b$$. We use the notation $$b \\nmid a$$ when $$b$$ does not divide $$a$$ (i.e., no such $$c$$ exists).\n\nSeveral simple properties of divisibility could be obtained by the definition of divisibility (proofs of the properties are left to readers).\n\n$$(1)$$ If $$b \\mid c,$$ and $$c \\mid a,$$ then $$b \\mid a,$$ that is, divisibility is transitive.\n\n$$(2)$$ If $$b \\mid a,$$ and $$b \\mid c,$$ then $$b \\mid (a \\pm c),$$ that is, the set of multiples of an integer is closed under addition and subtraction operations.\n\nBy using this property repeatedly, we have, if $$b \\mid a$$ and $$b \\mid c$$, then $$b \\mid (au+cv)$$, for any integers $$u$$ and $$v$$. In general, if $$a_1,a_2,\\cdots,a_n$$ are multiples of $$b$$, then $b \\mid (a_1+a_2+\\cdots+a_n).$\n\n$$(3)$$ If $$b \\mid a$$, then $$a=0$$ or $$\\lvert a \\rvert \\geq \\lvert b \\rvert$$. Thus, if $$b \\mid a$$ and $$a \\mid b$$, then $$\\lvert a \\rvert = \\lvert b \\rvert$$.\n\nClearly, for any two integers $$a$$ and $$b$$, $$a$$ is not always divisible by $$b$$. But we have the following result, which is called the division algorithm. It is the most important result in elementary number theory.\n\n$$(4)$$ (The division algorithm) Let $$a$$ and $$b$$ be integers, and $$b > 0$$. Then there is a unique pair of integers $$q$$ and $$r$$, such that $a=bq+r \\hspace{2mm} \\text{and} \\hspace{2mm} 0 \\leq r < b.$\n\nThe integer $$q$$ is called the (incomplete) quotient when $$a$$ is divided by $$b$$, $$r$$ called the remainder. Note that the values of $$r$$ has $$b$$ kinds of possibilities, $$0,1,\\cdots,b-1$$. If $$r=0$$, then $$a$$ is divisible by $$b$$.\n\nIt is easy to see that the quotient $$q$$ in the division algorithm is in fact $$\\lfloor\\frac{a}{b}\\rfloor$$ (the greatest integer not exceeding $$\\frac{a}{b}$$), and the heart of the division algorithm is the inequality about the remainder $$r$$: $$0 \\leq r < b$$. We will go back to this point later on.\n\nThe basic method of proving $$b \\mid a$$ is to factorize $$a$$ into the product of $$b$$ and another integer. Usually, in some basic problems this kind of factorization can be obtained by taking some special value in algebraic factorization equations. The following two factorization formulae are very useful in proving this kind of problems.\n\n$$(5)$$ If $$n$$ is a positive integer, then $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\\cdots+xy^{n-2}+y^{n-1}).$\n\n$$(6)$$ If $$n$$ is a positive odd number, then $x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\\cdots-xy^{n-2}+y^{n-1}).$\n\nNote by Victor Loh\n3\u00a0years, 8\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n\n- bulleted\n- list\n\n\u2022 bulleted\n\u2022 list\n\n1. numbered\n2. list\n\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1\n\nparagraph 2\n\nparagraph 1\n\nparagraph 2\n\n> This is a quote\nThis is a quote\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nThanks much for posting this. I would love if you could name the book you had referred to in the beginning of this note!\n\n- 3\u00a0years, 8\u00a0months ago\n\nI'm not sure if I want to reveal it because I'm scared I'm taking a little too much information :D\n\n- 3\u00a0years, 8\u00a0months ago\n\nI like the fact that although it's meant to be a comment with a negative connotation, there's still a smiley icon at the back which doesn't really fit the sentence (lol). Also, I think I have an idea which book he may have referred to...\n\n- 3\u00a0years, 8\u00a0months ago\n\nBut still, we have to acknowledge the rules of copyright. IF I feel like checking this book and find that you have really gleaned too much information from it, I might have to report you... D:\n\n- 3\u00a0years, 8\u00a0months ago\n\nIt's not THAT book, I promise.\n\n- 3\u00a0years, 8\u00a0months ago\n\nAnyway, Brilliant is a site for sharing information, and I haven't exactly shared the examples and exercises yet.\n\n- 3\u00a0years, 8\u00a0months ago\n\nUh-Oh!\n\n- 3\u00a0years, 8\u00a0months ago", "date": "2018-04-23 13:51:18", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/number-theory-divisibility/", "openwebmath_score": 0.984839916229248, "openwebmath_perplexity": 390.558935927478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n", "lm_q1_score": 0.9830850892111576, "lm_q2_score": 0.8933093975331752, "lm_q1q2_score": 0.878199148767067}}
{"url": "https://math.stackexchange.com/questions/2701175/confusion-with-matrix-calculus-derivative-computation", "text": "# Confusion with Matrix calculus derivative computation.\n\nThe text I am reading has the following\n\n$$E=(I+\\nabla I\\frac{\\partial W}{\\partial p}\\Delta p -T)^2$$ where $\\nabla I$ us a row vector $1$ by $2$, $\\frac{\\partial W}{\\partial p}$ is a $2$ by $3$ matrix and $\\Delta p$ is a $3$ by $1$ column vector. $I$ and $T$ are scalars. We seek the gradient $\\frac{\\partial E}{\\partial \\Delta p_k}$, but I do not understand their solution. I will present my attempt, and than their solution.\n\nIn index notation, we have $(I+\\nabla I_i\\frac{\\partial W^i}{\\partial \\Delta p_k}\\Delta p^k -T)^2$, where summation if implied by repeated indicies, Than the derivative is\n\n$$\\frac{\\partial E}{\\partial \\Delta p_m}=2(I+\\nabla I_i\\frac{\\partial W^i}{\\partial p_k}\\Delta p^k -T)\\nabla I_n\\frac{\\partial W^n}{\\partial p_m}$$ or returning to the matrix notation\n\n$$\\frac{\\partial E}{\\partial \\Delta p}=2(I+\\nabla I\\frac{\\partial W}{\\partial p}\\Delta p -T)\\nabla I\\frac{\\partial W}{\\partial p}\\tag{1}\\label{1}$$\n\nBut their solution is\n\n$$\\frac{\\partial E}{\\partial \\Delta p}=2(\\nabla I\\frac{\\partial W}{\\partial p})^T(I+\\nabla I\\frac{\\partial W}{\\partial p}\\Delta p -T)\\tag{2}\\label{2}$$\n\nWhy is their solution correct and where did I make the mistake? My solution is equation \\eqref{1}, but theirs is \\eqref{2}. I am confused, why in theirs, $$(\\nabla I\\frac{\\partial W}{\\partial p})$$ is transposed?\n\n\u2022 In the last two equations you have variable $p_m$ with subindex $m$ on the left hand side, but no subscripts on the right hand side. Is there a chance there is a typo in your problem statement?\nMar 21 '18 at 2:30\n\u2022 I was careless when copy and pasting. My solution in matrix form is that in equation (1), what is before, is my derivation. Their solution is equation (2). Hope this clears things up Mar 21 '18 at 2:33\n\u2022 I think now you very first equation (the one without label) has non-matching subindex $p_m$ in the last term.\nMar 21 '18 at 2:35\n\u2022 You are correct, I have fixed it. Thank you for sticking with me, it has been a long day :) Mar 21 '18 at 2:36\n\u2022 I edited and added that information. Mar 21 '18 at 2:52\n\nIt looks like the text you are reading uses so-called Denominator layout, for matrix calculus notation, i.e. given two column vectors $\\boldsymbol{x}\\in\\mathbb{R}^m$ and $\\boldsymbol{y}\\in\\mathbb{R}^n$ of the size $m\\times1$ and $n\\times1$ respectively we write derivative $\\displaystyle\\dfrac{\\partial \\boldsymbol{y}}{\\partial\\boldsymbol{x}}$ as $n\\times m$ matrix. In other words, the layout is according to $\\boldsymbol y^{\\boldsymbol\\top}$ and $\\mathbf{x}$.\n\nIn your case $E$ is the scalar, so that $n=1$ and $m=3$, and thus the derivative of scalar w.r.t. the vector $\\Delta p$ has to be a column vector of the size $3\\times1$.\n\nMore details are available, for example, on the Wikipedia page for Matrix Calculus.\n\nMoreover, using provided there table of scalar-by-vector identities one can figure out dimensionality of each term emerging from the chain rule explicitly. According to the linked table,\n\nAssume $\\boldsymbol{b}\\in \\mathbb R^m$ and $\\boldsymbol x \\in \\mathbb{R}^n$ are column vectors, and matrix $\\boldsymbol A$ is in $\\mathbb{R}^{m\\times n}$. If $\\boldsymbol A$ and $\\boldsymbol b$ do not depend on $\\boldsymbol x$, then the following holds (in Denominator layout ): $$\\dfrac{\\partial\\left(\\boldsymbol b^{\\boldsymbol\\top}\\boldsymbol A \\boldsymbol x\\right)}{\\partial \\boldsymbol x} = \\boldsymbol A^{\\boldsymbol \\top} \\boldsymbol b = \\left(\\boldsymbol b^{\\boldsymbol\\top}\\boldsymbol A\\right)^{\\boldsymbol\\top}$$\n\nUsing this identity, you can easily see that $$\\frac{\\partial }{\\partial \\Delta p} \\left( \\nabla I\\frac{\\partial W}{\\partial p}\\Delta p\\right) = \\left( \\nabla I\\frac{\\partial W}{\\partial p}\\right)^{\\boldsymbol\\top}$$\n\n\u2022 Thank you for the reply. What I do not undestand, is where I make a mistake in my derivation? Why the term transposed in their solution. Is it because they want the gradient to be a column vector and not a row vector? I do not understand why it differes from my solution... I can get their solution from mine, by transposing mine, and than the first term in mine can remain the same, as it is a scalar, and the second transposes. Mar 21 '18 at 3:38\n\u2022 @LeastSquaresWonderer As far as I understand this is just a matter of convention. If you read the wiki article I referenced, you can find there information about notation conventions, their ambiguity, and motivation behind choosing one over another.", "date": "2021-09-17 09:38:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2701175/confusion-with-matrix-calculus-derivative-computation", "openwebmath_score": 0.8917240500450134, "openwebmath_perplexity": 257.7331314313885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241965169938, "lm_q2_score": 0.9059898241909246, "lm_q1q2_score": 0.8781978583864404}}
{"url": "http://mestaritalo.com/orvovf1/addition-of-complex-numbers-5b5093", "text": "z_{1}=3+3i\\0.2cm] Combine the like terms Draw the diagonal vector whose endpoints are NOT $$z_1$$ and $$z_2$$. Subtracting complex numbers. Closure : The sum of two complex numbers is , by definition , a complex number. So let us represent $$z_1$$ and $$z_2$$ as points on the complex plane and join each of them to the origin to get their corresponding position vectors. The addition of complex numbers is just like adding two binomials. The addition of complex numbers is just like adding two binomials. For addition, the real parts are firstly added together to form the real part of the sum, and then the imaginary parts to form the imaginary part of the sum and this process is as follows using two complex numbers A and B as examples. So, a Complex Number has a real part and an imaginary part. Yes, the complex numbers are commutative because the sum of two complex numbers doesn't change though we interchange the complex numbers. First, draw the parallelogram with $$z_1$$ and $$z_2$$ as opposite vertices. This problem is very similar to example 1 The Complex class has a constructor with initializes the value of real and imag. Group the real part of the complex numbers and The following list presents the possible operations involving complex numbers. What Do You Mean by Addition of Complex Numbers? Addition belongs to arithmetic, a branch of mathematics. Operations with Complex Numbers . i.e., we just need to combine the like terms. Complex Numbers (Simple Definition, How to Multiply, Examples) This algebra video tutorial explains how to add and subtract complex numbers. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. We multiply complex numbers by considering them as binomials. A user inputs real and imaginary parts of two complex numbers. Complex numbers are numbers that are expressed as a+bi where i is an imaginary number and a and b are real numbers. The set of complex numbers is closed, associative, and commutative under addition. Next lesson. The additive identity, 0 is also present in the set of complex numbers. For instance, the sum of 5 + 3i and 4 + 2i is 9 + 5i. Programming Simplified is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. By \u2026 (5 + 7) + (2 i + 12 i) Step 2 Combine the like terms and simplify Real parts are added together and imaginary terms are added to imaginary terms. i.e., $$x+iy$$ corresponds to $$(x, y)$$ in the complex plane. Closed, as the sum of two complex numbers is also a complex number. Addition of Complex Numbers. Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. For example, $$4+ 3i$$ is a complex number but NOT a real number. The function computes the sum and returns the structure containing the sum. You can see this in the following illustration. Study Addition Of Complex Numbers in Numbers with concepts, examples, videos and solutions. Real World Math Horror Stories from Real encounters. A complex number is of the form $$x+iy$$ and is usually represented by $$z$$.  \\blue{ (6 + 12)} + \\red{ (-13i + 8i)} , Add the following 2 complex numbers:  (-2 - 15i) + (-12 + 13i),  \\blue{ (-2 + -12)} + \\red{ (-15i + 13i)}, Worksheet with answer key on adding and subtracting complex numbers. Complex Number Calculator. Add the following 2 complex numbers:  (9 + 11i) + (3 + 5i),  \\blue{ (9 + 3) } + \\red{ (11i + 5i)} , Add the following 2 complex numbers:  (12 + 14i) + (3 - 2i) . To multiply complex numbers in polar form, multiply the magnitudes and add the angles. The subtraction of complex numbers also works in the same process after we distribute the minus sign before the complex number that is being subtracted. What is a complex number? Since 0 can be written as 0 + 0i, it follows that adding this to a complex number will not change the value of the complex number. This is the currently selected item. Select/type your answer and click the \"Check Answer\" button to see the result. with the added twist that we have a negative number in there (-2i). To multiply monomials, multiply the coefficients and then multiply the imaginary numbers i. The tip of the diagonal is (0, 4) which corresponds to the complex number $$0+4i = 4i$$. Adding the complex numbers a+bi and c+di gives us an answer of (a+c)+(b+d)i. A General Note: Addition and Subtraction of Complex Numbers Addition Rule: (a + bi) + (c + di) = (a + c) + (b + d)i Add the \"real\" portions, and add the \"imaginary\" portions of the complex numbers. Finally, the sum of complex numbers is printed from the main () function. Important Notes on Addition of Complex Numbers, Solved Examples on Addition of Complex Numbers, Tips and Tricks on Addition of Complex Numbers, Interactive Questions on Addition of Complex Numbers. To add and subtract complex numbers: Simply combine like terms. i.e., \\[\\begin{align}&(a_1+ib_1)+(a_2+ib_2)\\\\[0.2cm]& = (a_1+a_2) + i (b_1+b_2)\\end{align}. This page will help you add two such numbers together. To add complex numbers in rectangular form, add the real components and add the imaginary components. Group the real part of the complex numbers and the imaginary part of the complex numbers. Group the real parts of the complex numbers and Consider two complex numbers: \\begin{array}{l} Addition on the Complex Plane \u2013 The Parallelogram Rule. The calculator will simplify any complex expression, with steps shown. The numbers on the imaginary axis are sometimes called purely imaginary numbers. It contains a few examples and practice problems. \\[ \\begin{align} &(3+2i)(1+i)\\\\[0.2cm] &= 3+3i+2i+2i^2\\\\[0.2cm] &= 3+5i-2 \\\\[0.2cm] &=1+5i \\end{align}. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = \u22121 or j 2 = \u22121.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). We already know that every complex number can be represented as a point on the coordinate plane (which is also called as complex plane in case of complex numbers). The complex numbers are written in\u00a0the form $$x+iy$$ and they correspond to the points on the coordinate plane (or complex plane). If we define complex numbers as objects, we can easily use arithmetic operators such as additional (+) and subtraction (-) on complex numbers with operator overloading. The additive identity is 0 (which can be written as $$0 + 0i$$) and hence the set of complex numbers has the additive identity. For addition, simply add up the real components of the complex numbers to determine the real component of the sum, and add up the imaginary components of the complex numbers to \u2026 $$z_2=-3+i$$ corresponds to the point (-3, 1). The resultant vector is the sum $$z_1+z_2$$. C Program to Add Two Complex Number Using Structure. Simple algebraic addition does not work in the case of Complex Number. Example: Conjugate of 7 \u2013 5i = 7 + 5i. If i 2 appears, replace it with \u22121. C program to add two complex numbers: this program performs addition of two complex numbers which will be entered by a user and then prints it. Example: We know that all complex numbers are of the form A + i B, where A is known as Real part of complex number and B is known as Imaginary part of complex number.. To add or subtract two complex numbers, just add or subtract the corresponding real and imaginary parts. Here, you can drag the point by which the complex number and the corresponding point are changed. This is linked with the fact that the set of real numbers is commutative (as both real and imaginary parts of a complex number are real numbers). Here lies the magic with Cuemath. Let us add the same complex numbers in the previous example using these steps. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. In the following C++ program, I have overloaded the + and \u2013 operator to use it with the Complex class objects. No, every complex number is NOT a real number. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. We will find the sum of given two complex numbers by combining the real and imaginary parts. \\end{array}\\]. Addition Add complex numbers Prime numbers Fibonacci series Add arrays Add matrices Random numbers Class Function overloading New operator Scope resolution operator. Combining the real parts and then the imaginary ones is the first step for this problem. $z_1=-2+\\sqrt{-16} \\text { and } z_2=3-\\sqrt{-25}$. At\u00a0Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! In this program, we will learn how to add two complex numbers using the Python programming language. Every complex number indicates a point in the XY-plane. The complex numbers are used in solving the quadratic equations (that have no real solutions). Adding complex numbers. Complex numbers have a real and imaginary parts. The addition or subtraction of complex numbers can be done either mathematically or graphically in rectangular form. Addition and subtraction with complex numbers in rectangular form is easy. i.e., we just need to combine the like terms. Distributive property can also be used for complex numbers. A Computer Science portal for geeks. Here are a few activities for you to practice. The sum of two complex numbers is a complex number whose real and imaginary parts are obtained by adding the corresponding parts of the given two complex numbers. The mini-lesson targeted\u00a0the fascinating concept of Addition of Complex Numbers. For example, the complex number $$x+iy$$ represents the point $$(x,y)$$ in the XY-plane. Hence, the set of complex numbers is closed under addition. i.e., the sum is the tip of the diagonal that doesn't join $$z_1$$ and $$z_2$$. Multiplying complex numbers. You can visualize the geometrical addition of complex numbers using the following illustration: We already learned how to add complex numbers geometrically. $$z_1=3+3i$$ corresponds to the point (3, 3) and. Subtraction is similar. z_{1}=a_{1}+i b_{1} \\0.2cm] Here is the easy process to add complex numbers. z_{2}=a_{2}+i b_{2}  \\blue{ (12 + 3)} + \\red{ (14i + -2i)} , Add the following 2 complex numbers:  (6 - 13i) + (12 + 8i). Our mission is to provide a free, world-class education to anyone, anywhere. Practice: Add & subtract complex numbers. Conjugate of complex number. Python Programming Code to add two Complex Numbers Be it worksheets, online classes, doubt sessions, or any other form of relation, it\u2019s the logical thinking and smart learning approach that we, at Cuemath, believe in. The math journey around Addition of Complex Numbers starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Yes, the sum of two complex numbers can be a real number. We also created a new static function add() that takes two complex numbers as parameters and returns the result as a complex number. Sum of two complex numbers a + bi and c + di is given as: (a + bi) + (c + di) = (a + c) + (b + d)i. For this. Just as with real numbers, we can perform arithmetic operations on complex numbers. The addition of complex numbers is thus immediately depicted as the usual component-wise addition of vectors. Was this article helpful? When performing the arithmetic operations of adding or subtracting on complex numbers, remember to combine \"similar\" terms. For example: \\[ \\begin{align} &(3+2i)+(1+i) \\\\[0.2cm]&= (3+1)+(2i+i)\\\\[0.2cm] &= 4+3i \\end{align}. Make your child a Math Thinker, the Cuemath way. \\begin{align} &(3+i)(1+2i)\\\\[0.2cm] &= 3+6i+i+2i^2\\\\[0.2cm] &= 3+7i-2 \\\\[0.2cm] &=1+7i \\end{align}, Addition and Subtraction of complex Numbers. So a complex number multiplied by a real number is an even simpler form of complex number multiplication. To add or subtract, combine like terms. the imaginary parts of the complex numbers. We add complex numbers just by grouping their real and imaginary parts. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Also, they are used in advanced calculus. Some examples are \u2212 6 + 4i 8 \u2013 7i. Here are some examples you can try: (3+4i)+(8-11i) 8i+(11-12i) 2i+3 + 4i For example, (3 \u2013 2i) \u2013 (2 \u2013 6i) = 3 \u2013 2i \u2013 2 + 6i = 1 + 4i. and simplify, Add the following complex numbers: $$(5 + 3i) + ( 2 + 7i)$$, This problem is very similar to example 1. To divide, divide the magnitudes and \u2026 Once again, it's not too hard to verify that complex number multiplication is both commutative and associative. Let's learn how to add complex numbers in this sectoin. Can we help James find the sum of the following complex numbers algebraically? Thus, the sum of the given two complex numbers is: $z_1+z_2= 4i$. To add two complex numbers, a real part of one number must be added with a real part of other and imaginary part one must be added with an imaginary part of other. Interactive simulation the most controversial math riddle ever! To multiply complex numbers that are binomials, use the Distributive Property of Multiplication, or the FOIL method. But, how to calculate complex numbers? The sum of any complex number and zero is the original number. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. To add or subtract two complex numbers, just add or subtract the corresponding real and imaginary parts. z_{2}=-3+i with the added twist that we have a negative number in there (-13i). Example : (5+ i2) + 3i = 5 + i(2 + 3) = 5 + i5 < From the above we can see that 5 + i2 is a complex number, i3 is a complex number and the addition of these two numbers is 5 + i5 is again a complex number. Can you try verifying this algebraically? To add or subtract complex numbers, we combine the real parts and combine the imaginary parts. However, the complex numbers allow for a richer algebraic structure, comprising additional operations, that are not necessarily available in a vector space. We just plot these on the complex plane and apply the parallelogram law of vector addition (by which, the tip of the diagonal represents the sum) to find their sum. But before that Let us recall the value of $$i$$ (iota) to be $$\\sqrt{-1}$$. By parallelogram law of vector addition, their sum, $$z_1+z_2$$,\u00a0is the position vector of the diagonal of the parallelogram\u00a0thus formed. When you type in your problem, use i to mean the imaginary part. Can we help Andrea add the following complex numbers geometrically? Also check to see if the answer must be expressed in simplest a+ bi form. 1 2 Arithmetic operations on C The operations of addition and subtraction are easily understood. Yes, because the sum of two complex numbers is a complex number. $\\begin{array}{l} This problem is very similar to example 1 Because they have two parts, Real and Imaginary. Subtracting complex numbers. Also, every complex number has its additive inverse in the set of complex numbers. \\end{array}$. But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. To multiply when a complex number is involved, use one of three different methods, based on the situation: These two structure variables are passed to the add () function. the imaginary part of the complex numbers. Besides counting items, addition can also be defined and executed without referring to concrete objects, using abstractions called numbers instead, such as integers, real numbers and complex numbers. In our program we will add real parts and imaginary parts of complex numbers and prints the complex number, 'i' is the symbol used for iota. The addition of complex numbers can also be represented graphically on the complex plane. Thus, \\begin{align} \\sqrt{-16} &= \\sqrt{-1} \\cdot \\sqrt{16}= i(4)= 4i\\\\[0.2cm] \\sqrt{-25} &= \\sqrt{-1} \\cdot \\sqrt{25}= i(5)= 5i \\end{align}, \\begin{align} &z_1+z_2\\\\[0.2cm] &=(-2+\\sqrt{-16})+(3-\\sqrt{-25})\\\\[0.2cm] &= -2+ 4i + 3-5i \\\\[0.2cm] &=(-2+3)+(4i-5i)\\\\[0.2cm] &=1-i \\end{align}. 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Subtract complex numbers algebraically the set of complex numbers geometrically the added twist that have! The Cuemath way from the main ( ) function again, it 's not too hard to that... + and \u2013 operator to use it with the added twist that we a. ( -13i ) ) corresponds to \\ ( z_2=-3+i\\ ) corresponds to the complex plane has its additive in! To see if the answer must be expressed in simplest a+ bi form '' button to see if answer. Math experts is dedicated to making learning fun for our favorite readers the. Drag the point by which the complex numbers arithmetic, a complex number but not a real number once,. Identity, 0 is also present in the XY-plane multiplication, or the FOIL method Simple algebraic addition not. Visualize the geometrical addition of two complex numbers z_1=3+3i\\ ) corresponds to the point (,... Numbers, just add or subtract complex numbers, just add or subtract the corresponding real and imag -2i..., \\ ( z\\ ) 3 ) and add or subtract two complex numbers is closed under addition the... Calculator does basic arithmetic on complex numbers add the same complex numbers is closed, as the sum given. Is the easy process to add or subtract two complex numbers very similar to example 1 with added. We will find the sum is the tip of the complex numbers does n't join \\ ( z_2\\.!, as the usual component-wise addition of vectors a FREE, world-class education to,... The corresponding point are changed readers, the teachers explore all angles of a topic multiply the coefficients and multiply. Expressions in the following complex numbers child a Math Thinker, the sum of two complex numbers \u2026 as!, well thought and well explained computer science and programming articles, and. Real numbers numbers are also complex numbers is thus immediately depicted as sum! Complex expression, with steps shown examples are \u2212 6 + 4i \u2013... Their real and imag the operations of addition of complex numbers in this sectoin 5 + and! Explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions diagonal. 4I\\ ] are changed numbers are numbers that are binomials, use the Distributive Property of,. Inputs real and imaginary number using structure you to practice mission is to provide a,... Because they have two parts, real and imag James find the sum is tip. Of two complex numbers ( 0+4i = 4i\\ ) to grasp, but also will with... Button to see if the answer must be expressed in simplest a+ bi form i! + 3i and 4 + 2i is 9 + 5i number is not a real number \u2013 5i 7! ( z_1+z_2\\ ) main ( ) function here, you can visualize the geometrical addition of complex numbers.. Done in a way that not only it is relatable and easy grasp!", "date": "2021-10-24 18:31:41", "meta": {"domain": "mestaritalo.com", "url": "http://mestaritalo.com/orvovf1/addition-of-complex-numbers-5b5093", "openwebmath_score": 0.7448074221611023, "openwebmath_perplexity": 512.6572821089109, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9693241947446617, "lm_q2_score": 0.9059898210180105, "lm_q1q2_score": 0.8781978537051432}}
{"url": "https://www.freemathhelp.com/forum/threads/just-need-a-little-clearing-up.123949/", "text": "# Just need a little clearing up\n\n#### marlousie\n\n##### New member\nI tried to ask google however it didn't really give me a straight answer so I made an account here solely for this question. It's also relatively simple so thank you to anyone who explains it to me because I'm still a bit caught up on it.\n\nWhy does (-4)^2= 16 but -4^2= 1/16\n\nI'm currently studying for the SAT and these things constantly trip me up because my math teachers never really went over this.\n\nThank you again!\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nI tried to ask google however it didn't really give me a straight answer so I made an account here solely for this question. It's also relatively simple so thank you to anyone who explains it to me because I'm still a bit caught up on it.\n\nWhy does (-4)^2= 16 but -4^2= 1/16\n\nI'm currently studying for the SAT and these things constantly trip me up because my math teachers never really went over this.\n\nThank you again!\n-4^2= 1/16................................................... is incorrect\n\n-4^2 = - (4^2) = -(16) = -16\n\nHowever:\n\n4^(-2) = 1/(4^2) = 1/16\n\nPlease come back if you have more questions.\n\n#### pka\n\n##### Elite Member\nI tried to ask google however it didn't really give me a straight answer so I made an account here solely for this question. It's also relatively simple so thank you to anyone who explains it to me because I'm still a bit caught up on it.\nWhy does (-4)^2= 16 but -4^2= 1/16\nI'm currently studying for the SAT and these things constantly trip me up because my math teachers never really went over this.\n$$-4^2=-16$$ while $$4^{-2}=\\frac{1}{16}$$\n\n#### LCKurtz\n\n##### Full Member\nI tried to ask google however it didn't really give me a straight answer so I made an account here solely for this question. It's also relatively simple so thank you to anyone who explains it to me because I'm still a bit caught up on it.\n\nWhy does (-4)^2= 16 but -4^2= 1/16\n\nI'm currently studying for the SAT and these things constantly trip me up because my math teachers never really went over this.\n\nThank you again!\nIt has to do with the precedence of operators. Exponentiation takes precedence over addition and multiplication. So if have $$\\displaystyle -4^2$$, you would to the exponent first giving $$\\displaystyle -16$$. Such problems are avoided by using parentheses. $$\\displaystyle (-4)^2 = 16$$ versus $$\\displaystyle -(4^2)=-16$$. Then there is no doubt which is meant. On exams like the SAT they give you $$\\displaystyle -4^2$$ to trip you up if you don't know the precedence rules.\n\n#### marlousie\n\n##### New member\n-4^2= 1/16................................................... is incorrect\n\n-4^2 = - (4^2) = -(16) = -16\n\nHowever:\n\n4^(-2) = 1/(4^2) = 1/16\n\nPlease come back if you have more questions.\nAh, thank you I realize my mistake now! Thank you, I appreciate it\n\n#### marlousie\n\n##### New member\n$$-4^2=-16$$ while $$4^{-2}=\\frac{1}{16}$$\nIt has to do with the precedence of operators. Exponentiation takes precedence over addition and multiplication. So if have $$\\displaystyle -4^2$$, you would to the exponent first giving $$\\displaystyle -16$$. Such problems are avoided by using parentheses. $$\\displaystyle (-4)^2 = 16$$ versus $$\\displaystyle -(4^2)=-16$$. Then there is no doubt which is meant. On exams like the SAT they give you $$\\displaystyle -4^2$$ to trip you up if you don't know the precedence rules.", "date": "2020-11-24 06:36:24", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/just-need-a-little-clearing-up.123949/", "openwebmath_score": 0.2964867651462555, "openwebmath_perplexity": 975.0935227221232, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534333179649, "lm_q2_score": 0.8947894724152068, "lm_q1q2_score": 0.8781941997986751}}
{"url": "https://math.stackexchange.com/questions/2909548/is-this-proof-correct-rationality-of-a-number", "text": "# Is this proof correct (Rationality of a number)?\n\nIs $\\sqrt[3] {3}+\\sqrt[3]{9}$ a rational number? My answer is no, and there is my proof. I would like to know if this is correct:\nSuppose this is rational. So there are positive integers $m,n$ such that $$\\sqrt[3]{3}+\\sqrt[3]{9}=\\sqrt[3]{3}(1+\\sqrt[3]{3})=\\frac{m}{n}$$\nLet $x=\\sqrt[3]{3}$. We get $x^2+x-\\frac{m}{n}=0 \\rightarrow x=\\frac{-1+\\sqrt{1+\\frac{4m}{n}}}{2}$.\nWe know that $x$ is irrational and that implies $\\sqrt{1+\\frac{4m}{n}}$ is irrational as well (Otherwise $x$ is rational). Write $x=\\sqrt[3]{3}$, multiply both sides by $2$ and then raise both sides to the power of $6$ to get:\n$$24^2=\\left(\\left(\\sqrt{1+\\frac{4m}{n}}-1\\right)^3\\right)^2\\rightarrow 24=\\left(\\sqrt{1+\\frac{4m}{n}}-1\\right)^3=\\left(1+\\frac{4m}{n}\\right)\\cdot \\sqrt{1+\\frac{4m}{n}}-3\\left(1+\\frac{4m}{n}\\right)+3\\sqrt{1+\\frac{4m}{n}}-1$$ Let $\\sqrt{1+\\frac{4m}{n}}=y,1+\\frac{4m}{n}=k$.\nWe get: $25=ky-3k+3y\\rightarrow y=\\frac{25+3k}{k+3}$, So $y$ is rational. But we know $y$ is irrational (again, otherwise $\\sqrt[3]{3}$ is rational) which leads to a contradiction. So the answer is No, $\\sqrt[3]{3}+\\sqrt[3]{9}$ is irrational.\nIs this proof correct? Is there another way to prove this? Thanks!\n\n\u2022 This seems like a lot of work. We know $x$ satisfies a cubic polynomial over $\\mathbb Q$,namely $x^3-3$. If it also satisfies a quadratic then the quadratic and the cubic would have to a common factor, which is clearly not the case (since the cubic has no rational roots). \u2013\u00a0lulu Sep 8 '18 at 11:56\n\u2022 Notice that after raising both sides to the power of 6, your next step is to take the square root of both sides. Why not just raise to the power of 3 to begin with? \u2013\u00a0Th\u00e9ophile Sep 8 '18 at 12:09\n\u2022 @Th\u00e9ophile You are right, no reason not to just raise to the power of 3. Is this proof correct? \u2013\u00a0Omer Sep 8 '18 at 12:11\n\u2022 The proof looks correct to me. \u2013\u00a0saulspatz Sep 8 '18 at 12:19\n\u2022 Yes, it looks correct; the reasoning stands. Now, as lulu said, it is a lot of work and can be greatly simplified. Even if you keep the general structure here, you can save some writing in several places: apart from the power of 6 that I mentioned, also look at $1+\\frac{4m}n$. This is clearly rational, and it's a bit cumbersome to write out, so why not replace it with something else? Indeed, you call it $k$, but why not make that substitution earlier, on the previous line? But let's be consistent and call it something like $\\frac{m'}{n'}$ instead of $k$ to show that it is rational. \u2013\u00a0Th\u00e9ophile Sep 8 '18 at 12:20\n\nAlternatively, denote $x=\\sqrt[3] {3}+\\sqrt[3]{9}$ and cube it to get a cubic equation with integer coefficients: $$x^3=3+3^2(\\sqrt[3]3+\\sqrt[3]{9})+9 \\Rightarrow \\\\ x^3-9x-12=0.$$ According to the rational root theorem, the possible rational roots are: $\\pm (1,2,3,4,6,12)$. However, none of them satisfies the equation. Hence, $x$ is irrational.\n\nLet $\\alpha = \\sqrt[3]{3} + \\sqrt[3]{9} = \\sqrt[3]{3}(1+\\sqrt[3]{3})$. We have\n\n$$\\alpha^3 = 3(1+\\sqrt[3]{3})^3 = 3(3 + 3\\sqrt[3]{3} + 3\\sqrt[3]{9} + 3) = 12 + 9(\\sqrt[3]{3} + \\sqrt[3]{9}) = 12 + 9\\alpha$$\n\nHence $\\alpha^3 - 9\\alpha -12 = 0$.\n\nHowever, the polynomial $x^3-9x-12$ is irreducible over $\\mathbb{Q}$ by the Eisenstein criterion for $p = 3$ so it cannot have any rational roots. Therefore $\\alpha$ is not rational.\n\nAs everyone else has said, your proof is correct but more laborious than necessary. Here's an approach that I think may minimize the amount of calculation.\n\nLet $x=\\root3\\of3$, so we're interested in $y=x^2+x$. The fact that $x$ is a cube root suggests thinking about things involving $x^3$, and if you're lucky the following identity, involving both $x^3$ and $y$, may come to mind: $x^3-1=(x-1)(x^2+x+1)$. Aha! The left-hand side of this is a rational number, namely 2. And if $y$ is rational then the second factor on the right is also rational. But that would require $x-1$, and hence $x$, to be rational, which it certainly isn't. So $y$ can't be rational after all, and we're done.\n\n(An easy generalization of the argument shows that things like $\\root5\\of7+\\root5\\of{7^2}+\\root5\\of{7^3}+\\root5\\of{7^4}$ are always irrational.)", "date": "2019-04-24 00:28:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2909548/is-this-proof-correct-rationality-of-a-number", "openwebmath_score": 0.926621675491333, "openwebmath_perplexity": 127.93420571638048, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986363166322039, "lm_q2_score": 0.8902942312159383, "lm_q1q2_score": 0.8781534368603984}}
{"url": "https://math.stackexchange.com/questions/1074582/evaluating-sums-using-residues-1n-n2", "text": "# Evaluating sums using residues $(-1)^n/n^2$ [duplicate]\n\nI am an alien towards compelx analysis, with very little know I am posing a question, who someone may want to help with.\n\nEvaluate:\n\n$$\\frac{1}{4}\\cdot \\sum_{n=1}^{\\infty} \\frac{(-1)^n}{n^2}$$\n\nIn disguise this is similar to $\\zeta(2)$ but how can this be done using residues, and complex analysis?\n\nI need some help. I am just interested.\n\nThe answer is $\\displaystyle \\frac{\\pi^2}{48}$\n\n## marked as duplicate by Jack D'Aurizio, Ahaan S. Rungta, user98602, Aditya Hase, Pedro Tamaroff\u2666\u00a0calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 21 '14 at 22:12\n\n\u2022 en.wikipedia.org/wiki/Dirichlet_eta_function \u2013\u00a0Gahawar Dec 19 '14 at 14:55\n\u2022 The solution for $\\zeta(2)$ given by robjohn in the above linked Question lends itself to an answer here, but a bit more reasoning is needed to reduce the case of alternating signs here to the absolute series for $\\zeta(2)$. It's not hard however. \u2013\u00a0hardmath Dec 19 '14 at 15:31\n\u2022 If the answer is supposed to be positive, you need either $-\\frac14$ before the sum or $(-1)^{n-1}$ in the numerator of the fraction. \u2013\u00a0robjohn Dec 21 '14 at 22:30\n\u2022 @robjohn, a friendly request to you, since you are amazing at complex analysis. Would you have time to answer this question from me: math.stackexchange.com/questions/1077460/\u2026 \u2013\u00a0Amad27 Dec 22 '14 at 10:24\n\nUsing $\\boldsymbol{\\pi\\csc(\\pi z)}$\n\nSince $\\pi\\csc(\\pi z)$ has residue $(-1)^n$ at $z=n$ for $n\\in\\mathbb{Z}$, we will use the contours $$\\gamma_\\infty=\\lim\\limits_{R\\to\\infty}Re^{2\\pi i[0,1]}\\qquad\\text{and}\\qquad\\gamma_0=\\lim\\limits_{R\\to0}Re^{2\\pi i[0,1]}$$ To sum over all $n\\in\\mathbb{Z}$ except $n=0$, we use the difference of the contours, which circles the non-zero integers once counter-clockwise. \\begin{align} 2\\sum_{n=1}^\\infty\\frac{(-1)^n}{n^2} &=\\frac1{2\\pi i}\\left(\\int_{\\gamma_\\infty}\\frac{\\pi\\csc(\\pi z)}{z^2}\\mathrm{d}z-\\int_{\\gamma_0}\\frac{\\pi\\csc(\\pi z)}{z^2}\\mathrm{d}z\\right)\\\\ &=\\color{#C00000}{\\frac1{2\\pi i}\\int_{\\gamma_\\infty}\\frac{\\pi\\csc(\\pi z)}{z^2}\\mathrm{d}z}-\\operatorname*{Res}_{z=0}\\left(\\color{#00A000}{\\frac{\\pi\\csc(\\pi z)}{z^2}}\\right)\\\\ &=\\color{#C00000}{0}-\\operatorname*{Res}_{z=0}\\left(\\color{#00A000}{\\frac1{z^2}\\frac\\pi{\\pi z-\\pi^3z^3/6+O\\left(z^5\\right)}}\\right)\\\\ &=\\color{#C00000}{0}-\\operatorname*{Res}_{z=0}\\left(\\color{#00A000}{\\frac1{z^3}+\\frac{\\pi^2}{6z}+O(z)}\\right)\\\\ &=-\\frac{\\pi^2}6 \\end{align} because, for $k\\in\\mathbb{Z}$ and $|z|=\\pi\\left(k+\\frac12\\right)$, $|\\csc(z)|\\le1$.\n\nTherefore, $$\\sum_{n=1}^\\infty\\frac{(-1)^n}{n^2}=-\\frac{\\pi^2}{12}$$\n\nExtending A Previous Result\n\nIn this answer, it is shown that $$\\sum_{k=1}^\\infty\\frac1{k^2}=\\frac{\\pi^2}6$$ Note that \\begin{align} \\hphantom{=}&\\frac1{1^2}{+}\\frac1{2^2}+\\frac1{3^2}{+}\\frac1{4^2}+\\frac1{5^2}{+}\\frac1{6^2}+\\frac1{7^2}+\\dots\\\\ \\hphantom{=}&\\hphantom{\\frac1{1^2}}\\color{#C00000}{-\\frac2{2^2}\\hphantom{+\\frac1{3^2}}-\\frac2{4^2}\\hphantom{+\\frac1{5^2}}-\\frac2{6^2}\\hphantom{+\\frac1{7^2}}-\\dots}\\\\ =&\\frac1{1^2}{-}\\frac1{2^2}+\\frac1{3^2}{-}\\frac1{4^2}+\\frac1{5^2}{-}\\frac1{6^2}+\\frac1{7^2}-\\dots \\end{align} where the series in red is two times one quarter of the series above it; that is, one half of the series above it. Therefore, the alternating series is one half of the non-alternating series; that is, $$\\sum_{n=1}^\\infty\\frac{(-1)^{n-1}}{n^2}=\\frac{\\pi^2}{12}$$\n\n\u2022 I thought this looked familiar. I see that I have also posted this answer to the question to which this one was marked as a duplicate. Now that I look closer, this question is not really a duplicate of that one since this question asks about the alternating series. Granted, it is not too difficult to derive the alternating series from the non-alternating series, but a bit of extra work is needed. \u2013\u00a0robjohn Dec 21 '14 at 22:39\n\u2022 Good answer. How did you convert the sum into a contour integral though? \u2013\u00a0Amad27 Dec 22 '14 at 9:37\n\u2022 @Amad27: The function $\\pi\\csc(\\pi z)$ has singularities exactly on $\\mathbb{Z}$ with residue $(-1)^n$ at $z=n\\in\\mathbb{Z}$. Thus, $\\frac{\\pi\\csc(\\pi z)}{z^2}$ also has singularities exactly on $\\mathbb{Z}$ with residue $\\frac{(-1)^n}{n^2}$ at $z=n\\in\\mathbb{Z}$. The difference of the contour integrals along $\\gamma_\\infty$ and $\\gamma_0$ equals $2\\pi i$ times the sum of the residues inside $\\gamma_\\infty$ but outside $\\gamma_0$. This is $4\\pi i$ times the sum we are looking for. We don't need to worry about the residue at $z=0$ since it is not inside the difference of the contours. \u2013\u00a0robjohn Dec 22 '14 at 10:49\n\u2022 May I know that why $\\csc z \\leq 1$ when $|z|=k+1/2$? \u2013\u00a0mnmn1993 Mar 7 '18 at 17:59\n\nFor a complex variable $s$, whose real part is greater than zero, the Dirichlet eta function is defined by the series $$\\eta(s) := -\\sum_{n=1}^{\\infty} \\frac{(-1)^n}{n^s}.$$\n\nIn particular, one has that\n\n$$\\eta(s) = \\left(1-2^{1-s}\\right)\\zeta(s),$$\n\nwhere $\\zeta$ denotes the Riemann zeta function. With that in mind, one need only substitute $s=2$ into the above equation. I presume that you are familiar with the famed Basel problem.\n\n\u2022 Also, this does not really meet the requirements that it be done through the methods of complex analysis, but I suppose that, since $\\zeta(2)$ has already been found through such methods on this site, it is fine. \u2013\u00a0Gahawar Dec 19 '14 at 15:09", "date": "2019-10-21 00:14:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1074582/evaluating-sums-using-residues-1n-n2", "openwebmath_score": 0.8262123465538025, "openwebmath_perplexity": 454.77495379795647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543728093005, "lm_q2_score": 0.8856314647623016, "lm_q1q2_score": 0.8781517515825661}}
{"url": "http://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbb/6250", "text": "# Is there a quick proof as to why the vector space of $\\mathbb{R}$ over $\\mathbb{Q}$ is infinite-dimensional?\n\nIt would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?\n\n-\nA finite dimensional vector space over $\\mathbb{Q}$ is countable. \u2013\u00a0user641 Oct 7 '10 at 2:19\n@Steve: Please add that as an answer so people can upvote. \u2013\u00a0Aryabhata Oct 7 '10 at 2:50\nDoes that mean that a vector space over $\\mathbb{Q}$ is finite-dimensional iff the set of the vector space is countable? If so, please prove it. \u2013\u00a0Isaac Solomon Oct 7 '10 at 2:55\n@Isaac Your question doesn't require the 'only if' anyway. Steve's observation answers your original question. \u2013\u00a0yasmar Oct 7 '10 at 3:04\n@Isaac: how is the fact that $\\mathbb{R}$ is finite dimensional over itself relevant? \u2013\u00a0Arturo Magidin Oct 7 '10 at 3:31\n\nAs Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\\alpha_1 v_1+\\cdots+\\alpha_n v_n$ for some scalars $\\alpha_1,\\ldots,\\alpha_n\\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\\mathbb{R}$ is uncountable and $\\mathbb{Q}$ is countable, $\\mathbb{R}$ cannot be finite dimensional over $\\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).\n\nYour further question in the comments, whether a vector space over $\\mathbb{Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\\mathbb{Q}$ that are countable as sets. The simplest example is $\\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\\mathbb{Q}$, which is a countable set, and has dimension $\\aleph_0$, with basis $\\{1,x,x^2,\\ldots,x^n,\\ldots\\}$.\n\nAdded: Of course, if $V$ is a vector space over $\\mathbb{Q}$, then it has countable dimension (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\\mathbb{R}$ infinite dimensional over $\\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is uncountably-dimensional over $\\mathbb{Q}$.\n\n-\nRelated to the note about uncountable dimension, there are explicit examples of continuum-sized linearly independent sets, as seen in this MathOverflow answer by Fran\u00e7ois G. Dorais: mathoverflow.net/questions/23202/\u2026 \u2013\u00a0Jonas Meyer Oct 7 '10 at 4:18\nYes: but can one show that there is a basis for $\\mathbb{R}$ over $\\mathbb{Q}$ without some form of the Axiom of Choice? (There is a difference between exhibiting a large linearly independent subset and exhibiting a basis). \u2013\u00a0Arturo Magidin Oct 7 '10 at 14:51\nNo, one cannot, but without the Axiom of Choice the notion of dimension breaks down (except for finite vs. infinite). Assuming AC (as I virtually always do), the size of a linearly independent set gives a lower bound on the dimension of the vector space, and I think it is wonderful that in this case such \"explicit\" proof exists that the real numbers have continuum dimension, as opposed to the nice qualitative proof one could give by extending your argument to larger cardinals. \u2013\u00a0Jonas Meyer Oct 7 '10 at 16:02\n@Jonas: No argument there (with any of your points). \u2013\u00a0Arturo Magidin Oct 7 '10 at 17:23\ngood link @Jonas, thanks \u2013\u00a0Leon Sep 23 '11 at 0:23\n\nThe cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $\\rm\\: \\mathbb Q$-independent set of reals. Consider the set consisting of the logs of all primes $\\rm\\: p_i\\:$. Then if $\\rm\\ \\: c_1\\ log\\ p_1 +\\:\\cdots\\: + c_n\\ log \\ p_n \\: =\\ 0\\:,\\ \\: c_i\\in\\mathbb Q\\:,\\:$ multiplying by a common denominator we can assume that all $\\rm\\ c_i \\in \\mathbb Z\\:,\\:$ and, exponentiating, we obtain $\\rm\\: p_1^{\\:c_1}\\cdots p_n^{\\:c_n} = 1\\ \\Rightarrow\\ c_i = 0,\\:$ for all $\\rm\\:i\\:$.\n\n-\n@Bill +1 What a nice example. \u2013\u00a0Adri\u00e1n Barquero Oct 7 '10 at 4:50\n@Bill. Wow! :-) \u2013\u00a0a.r. Oct 7 '10 at 6:08\nIs this proof unique to Q or does it generalize to provide explicit examples of reals linearly independent over, e.g., Q(sqrt(2))? Above, Q appears to be \"hard-wired\" into the proof, as the group of exponents in the prime factorization. \u2013\u00a0T.. Oct 12 '10 at 7:14\n@George S. \"Then, any number field has a finite extension whose class number is 1.\" This would imply in particular that there are infinitely many number fields of class number one, which I am pretty sure is an open problem. I wonder what you are thinking here? \u2013\u00a0Pete L. Clark Jan 22 '11 at 10:20\nCan you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. \u2013\u00a0JDH Jun 23 '11 at 2:22\n\nNo transcendental numbers are needed for this question. Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension. Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down.\n\nThe set $\\sqrt{2}, \\sqrt{\\sqrt{2}}, \\dots, = \\bigcup_{n>0} 2^{2^{-n}}$ is linearly independent over $\\mathbb Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i < n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$).\n\nThe square roots of the prime numbers are linearly independent over $\\mathbb Q$. (Proof: this is immediate given the ability to extend the function \"number of powers of $p$ dividing $x$\" from the rational numbers to algebraic numbers. $\\sqrt{p}$ is \"divisible by $p^{1/2}$\" while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $\\mathbb Q$ unramified at $p$).\n\nGenerally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case. Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent. (Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.)\n\n-\n\nFor the sake of completeness, I'm adding a worked-out solution due to F.G. Dorais from his post.\n\nWe'll need two propositions from Grillet's Abstract Algebra, page 335 and 640:\n\nProposition: $[\\mathbb{R}:\\mathbb{Q}]=\\mathrm{dim}_\\mathbb{Q}{}\\mathbb{R}=|\\mathbb{R}|$\n\nProof: Let $(q_n)_{n\\in\\mathbb{N_0}}$ be an enumeration of $\\mathbb{Q}$. For $r\\in\\mathbb{R}$, take $$A_r:=\\sum_{q_n<r}\\frac{1}{n!}\\;\\;\\;\\;\\text{ and }\\;\\;\\;\\;A:=\\{A_r;\\,r\\in\\mathbb{R}\\};$$ the series is convergent because $\\sum_{q_n<r}\\frac{1}{n!}\\leq\\sum_{n=0}^\\infty\\frac{1}{n!}=\\exp(1)<\\infty$ (recall that $\\exp(x)=\\sum_{n=0}^\\infty\\frac{x^n}{n!}$ for any $x\\in\\mathbb{R}$).\n\nTo prove $|A|=|\\mathbb{R}|$, assume $A_r=A_{s}$ and $r\\neq s$. Without loss of generality $r<s$, hence $A_s=\\sum_{q_n<s}\\frac{1}{n!}=\\sum_{q_n<r}\\frac{1}{n!}+\\sum_{r\\leq q_n<s}\\frac{1}{n!}=A_r+\\sum_{r\\leq q_n<s}\\frac{1}{n!}$, so $\\sum_{r\\leq q_n<s}\\frac{1}{n!}=0$, which is a contradiction, because each interval $(r,s)$ contains a rational number.\n\nTo prove $A$ is $\\mathbb{Q}$-independent, assume $\\alpha_1A_{r_1}+\\cdots+\\alpha_kA_{r_k}=0\\;(1)$ with $\\alpha_i\\in\\mathbb{Q}$. We can assume $r_1>\\cdots>r_k$ (otherwise rearrange the summands) and $\\alpha_i\\in\\mathbb{Z}$ (otherwise multiply by the common denominator). Choose $n$ large enough that $r_1>q_n>r_2\\;(2)$; we'll increase $n$ two more times. The equality $n!\\cdot(1)$ reads $n!(\\alpha_1\\sum_{q_m<r_1}\\frac{1}{m!}+\\cdots+\\alpha_k\\sum_{q_m<r_k}\\frac{1}{m!})=0$. Rearranged (via $(2)$ when $m=n$), it reads\n\n$$-\\alpha_1\\sum_{\\substack{m<n\\\\q_m<r_1}}\\frac{n!}{m!}-\\cdots-\\alpha_k\\sum_{\\substack{m<n\\\\q_m<r_k}}\\frac{n!}{m!}-\\alpha_1 =\\alpha_1\\sum_{\\substack{m>n\\\\q_m<r_1}}\\frac{n!}{m!}+\\cdots+\\alpha_k\\sum_{\\substack{m>n\\\\q_m<r_k}}\\frac{n!}{m!}. \\tag*{(3)}$$\n\nThe left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\\alpha_1|+\\cdots+|\\alpha_k|)\\sum_{m=n+1}^\\infty\\frac{n!}{m!}<1$ holds (such $n$ can be found since $\\sum_{m=n+1}^\\infty\\frac{n!}{m!}=\\frac{1}{n+1}\\sum_{m=n+1}^\\infty\\frac{1}{(n+2)\\cdot\\ldots\\cdot m}\\leq\\frac{1}{n+1}\\sum_{m=n+1}^\\infty\\frac{1}{(m-n-1)!}\\leq\\frac{1}{n+1}\\exp(1)\\rightarrow 0$ when $n\\rightarrow\\infty$), then the absolute value of RHS of $(3)$ is $<1$, and yet an integer, hence $\\text{RHS}(3)=0$. Thus $(3)$ reads $\\alpha_1=-\\sum_{i=1}^{k}\\sum_{m<n,q_m<r_i}\\alpha_i\\frac{n!}{m!}=0\\;(\\mathrm{mod}\\,n)$. If moreover $n>|\\alpha_1|$, this means that $\\alpha_1=0$. Repeat this argument to conclude that also $\\alpha_2=\\cdots=\\alpha_k=0$.\n\nSince $A$ is a $\\mathbb{Q}$-independent subset, by proposition 5.3 there exists a basis $B$ of $\\mathbb{R}$ that contains $A$. Then $A\\subseteq B\\subseteq\\mathbb{R}$ and $|A|=|\\mathbb{R}|$ and Cantor-Bernstein theorem imply $|B|=|\\mathbb{R}|$, therefore $[\\mathbb{R}:\\mathbb{Q}]=\\mathrm{dim}_\\mathbb{Q}{}\\mathbb{R}=|\\mathbb{R}|$. $\\quad\\blacksquare$\n\n-\nIs this uncountable linearly independent set basis ? \u2013\u00a0bhim sen Chaudhary Jun 5 '15 at 2:53\n\n## protected by user26857Nov 25 '15 at 9:30\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.\n\nWould you like to answer one of these unanswered questions instead?", "date": "2016-04-28 21:51:21", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbb/6250", "openwebmath_score": 0.9408407807350159, "openwebmath_perplexity": 202.9308095561068, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543736828079, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8781517508596043}}
{"url": "https://mathematica.stackexchange.com/questions/80278/iterating-distributions", "text": "# Iterating distributions\n\nI want to calculate the distribution that comes from summing up random numbers between -0.5 and 0.5.\n\nA computationally intensive version of doing this is to simply do the following:\n\nSmoothHistogram[\nTable[Table[\nSum[RandomReal[] - 0.5, {i, 1,\n\n\nThis won't work however if the number of numbers to be summed is very large, like a 1000000 for example.\n\nSo I tried doing it by first calculating the distribution of one random number, then calculating the distribution of two random numbers from the first distribution. The first distribution is simply:\n\nUnitBox[x]\n\n\nThe next distribution can be calculated by integrating the previous distribution from x-0.5 to x+0.5:\n\nIntegrate[UnitBox[a], {a, x - 1/2, x + 1/2}]\n\n\nI can't figure out how to iterate this to, for example, draw a plot from the result.\n\nI also tried the following:\n\nNestList[PDF[\nProbabilityDistribution[\nIntegrate[# /. x -> a, {a, x - 1/2, x + 1/2},\nAssumptions -> x \\[Element] Reals], {x, -2, 2}], x] &,\nUnitBox[x], 5]\nPlot[%, {x, -2, 2}]\n\n\n\u2022 Any reason not to use built-ins? \u2013\u00a0ciao Apr 19 '15 at 3:26\n\u2022 No. What built-in would help? \u2013\u00a0user Apr 19 '15 at 3:27\n\u2022 maybe UniformSumDistribution. E.g., dis[n_] := UniformSumDistribution[n, {-.5, .5}]; Plot[Evaluate[PDF[dis@#, x] & /@ Range[10]], {x, -2, 2}]? \u2013\u00a0kglr Apr 19 '15 at 3:30\n\u2022 It works! Thanks a lot. \u2013\u00a0user Apr 19 '15 at 3:32\n\u2022 @MathematicaUser39386, posted the comment as an answer. \u2013\u00a0kglr Apr 19 '15 at 4:03\n\n\u2022 summing random numbers between -0.5 and 0.5.\n\n\u2022 the number of numbers to be summed is very large, like a 1000000 for example.\n\nThe sum of $n$ identical Uniform random variables is known as a generalised Irwin-Hall distribution, implemented in Mathematica as the UniformSumDistribution [ see kguler's answer]. The latter takes an $n$-part piecewise form, so, for example, if $n = 10$, the pdf is:\n\nPDF[UniformSumDistribution[10, {-1/2, 1/2}], x]\n\n\nThe above is exact and works beautifully for small $n$. However, the OP poses the problem of $n$ being very large -- such as $n$ = 1 million -- which involves creating a 1 million part piecewise structure which is certainly going to fail. In fact, for $n = 1000$, simply making a plot of the pdf takes about 400 seconds on my Mac Pro:\n\n(AA = Plot[\nPDF[UniformSumDistribution[1000, {-1/2, 1/2}], x], {x, -50, 50},\nPlotRange -> All]) // AbsoluteTiming\n\n\n390 seconds\n\n\u2022 For $n=10,000$, the above plot is simply unworkable ... which leaves the OP's question: how to proceed for large $n$??\n\nLarge $n$: apply Central Limit Theorem\n\nFor large $n$, I would suggest applying the (Lindeberg-L\u00e9vy) Central Limit Theorem:\n\n\u2022 If the random variables $(X_1, X_2, \\dots)$ are independent and identically distributed, each with finite mean $\\mu$ and finite variance $\\sigma^2$, then the sample sum:\n\n$$S_n \\overset{a}{\\sim} N\\left(n \\mu,n \\sigma ^2\\right)$$\n\nIn our case, $X \\sim \\text{Uniform}(-\\frac12,\\frac12)$ so $\\mu = 0$ and $\\sigma^2 = \\frac{1}{12}$, so the asymptotic distribution of the sample sum is:\n\n$$S_n \\overset{a}{\\sim} N\\left(0, \\frac{n}{12} \\right)$$\n\nwith pdf $f(x)$:\n\n       f = Exp[-6 (x^2/n)] / Sqrt[n Pi /6];\ndomain[f] = {x, -Infinity, Infinity} && {n > 0};\n\n\nHere is a plot of pdf $f(x)$ when $n = 1000$:\n\nBB = Plot[f /. n -> 1000, {x, -50, 50}, PlotRange -> All, PlotStyle -> Red]\n\n\nThis is, of course, instantaneous and works for arbitrarily large $n$. The following plot compares the exact solution AA to the asymptotic solution BB when $n = 1000$:\n\nShow[BB, AA]\n\n\nThere is no discernible visual difference between the two plots here. Whereas the exact AA solution fails to evaluate for very large $n$, the asymptotic BB solution will always evaluate immediately, and with ever improving accuracy as $n$ increases.\n\nThe built-in function UniformSumDistribution may be useful:\n\nusd[n_] := UniformSumDistribution[n, {-.5, .5}];\nPlot[Evaluate[PDF[usd@#, x] & /@ Range[20]], {x, -2, 2},\nPlotStyle -> (ColorData[{\"Rainbow\", {1, 20}}] /@ Range[20]),\nExclusions -> None, PlotRange -> All, ImageSize -> 500,\nPlotLegends -> (\"usd (\" <> ToString[#] <> \")\" & /@ Range[20])]\n\n\nClearAll[skd]\nskd[n_] := SmoothKernelDistribution[RandomVariate[usd[n], 5000]];\nPlot[Evaluate[PDF[#, x] & /@ {skd[5], usd[5], skd[3], usd[3]}], {x, -2, 2},\nImageSize -> 500, Exclusions -> None,\nFilling -> {1 -> {{2}, {Orange, Yellow}}, 3 -> {{4}, {Blue, Green}}}]", "date": "2020-09-25 00:31:18", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/80278/iterating-distributions", "openwebmath_score": 0.3286290168762207, "openwebmath_perplexity": 1805.8553106697436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692311915196, "lm_q2_score": 0.8991213860551286, "lm_q1q2_score": 0.878144192866316}}
{"url": "https://lectures.quantecon.org/py/linear_algebra.html", "text": "We are working to support a site-wide PDF but it is not yet available. You can download PDFs for individual lectures through the download badge on each lecture page.\n\nCode should execute sequentially if run in a Jupyter notebook\n\n# Linear Algebra\u00b6\n\n## Overview\u00b6\n\nLinear algebra is one of the most useful branches of applied mathematics for economists to invest in\n\nFor example, many applied problems in economics and finance require the solution of a linear system of equations, such as\n\n$$\\begin{array}{c} y_1 = a x_1 + b x_2 \\\\ y_2 = c x_1 + d x_2 \\end{array}$$\n\nor, more generally,\n\n$$\\begin{array}{c} y_1 = a_{11} x_1 + a_{12} x_2 + \\cdots + a_{1k} x_k \\\\ \\vdots \\\\ y_n = a_{n1} x_1 + a_{n2} x_2 + \\cdots + a_{nk} x_k \\end{array} \\tag{1}$$\n\nThe objective here is to solve for the \u201cunknowns\u201d $x_1, \\ldots, x_k$ given $a_{11}, \\ldots, a_{nk}$ and $y_1, \\ldots, y_n$\n\nWhen considering such problems, it is essential that we first consider at least some of the following questions\n\n\u2022 Does a solution actually exist?\n\u2022 Are there in fact many solutions, and if so how should we interpret them?\n\u2022 If no solution exists, is there a best \u201capproximate\u201d solution?\n\u2022 If a solution exists, how should we compute it?\n\nThese are the kinds of topics addressed by linear algebra\n\nIn this lecture we will cover the basics of linear and matrix algebra, treating both theory and computation\n\nWe admit some overlap with this lecture, where operations on NumPy arrays were first explained\n\nNote that this lecture is more theoretical than most, and contains background material that will be used in applications as we go along\n\n## Vectors\u00b6\n\nA vector of length $n$ is just a sequence (or array, or tuple) of $n$ numbers, which we write as $x = (x_1, \\ldots, x_n)$ or $x = [x_1, \\ldots, x_n]$\n\nWe will write these sequences either horizontally or vertically as we please\n\n(Later, when we wish to perform certain matrix operations, it will become necessary to distinguish between the two)\n\nThe set of all $n$-vectors is denoted by $\\mathbb R^n$\n\nFor example, $\\mathbb R ^2$ is the plane, and a vector in $\\mathbb R^2$ is just a point in the plane\n\nTraditionally, vectors are represented visually as arrows from the origin to the point\n\nThe following figure represents three vectors in this manner\n\nIn\u00a0[1]:\nimport matplotlib.pyplot as plt\n%matplotlib inline\n\nfig, ax = plt.subplots(figsize=(10, 8))\n# Set the axes through the origin\nfor spine in ['left', 'bottom']:\nax.spines[spine].set_position('zero')\nfor spine in ['right', 'top']:\nax.spines[spine].set_color('none')\n\nax.set(xlim=(-5, 5), ylim=(-5, 5))\nax.grid()\nvecs = ((2, 4), (-3, 3), (-4, -3.5))\nfor v in vecs:\nax.annotate('', xy=v, xytext=(0, 0),\narrowprops=dict(facecolor='blue',\nshrink=0,\nalpha=0.7,\nwidth=0.5))\nax.text(1.1 * v[0], 1.1 * v[1], str(v))\nplt.show()\n\n### Vector Operations\u00b6\n\nThe two most common operators for vectors are addition and scalar multiplication, which we now describe\n\nAs a matter of definition, when we add two vectors, we add them element by element\n\n$$x + y = \\left[ \\begin{array}{c} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{array} \\right] + \\left[ \\begin{array}{c} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{array} \\right] := \\left[ \\begin{array}{c} x_1 + y_1 \\\\ x_2 + y_2 \\\\ \\vdots \\\\ x_n + y_n \\end{array} \\right]$$\n\nScalar multiplication is an operation that takes a number $\\gamma$ and a vector $x$ and produces\n\n$$\\gamma x := \\left[ \\begin{array}{c} \\gamma x_1 \\\\ \\gamma x_2 \\\\ \\vdots \\\\ \\gamma x_n \\end{array} \\right]$$\n\nScalar multiplication is illustrated in the next figure\n\nIn\u00a0[2]:\nimport numpy as np\n\nfig, ax = plt.subplots(figsize=(10, 8))\n# Set the axes through the origin\nfor spine in ['left', 'bottom']:\nax.spines[spine].set_position('zero')\nfor spine in ['right', 'top']:\nax.spines[spine].set_color('none')\n\nax.set(xlim=(-5, 5), ylim=(-5, 5))\nx = (2, 2)\nax.annotate('', xy=x, xytext=(0, 0),\narrowprops=dict(facecolor='blue',\nshrink=0,\nalpha=1,\nwidth=0.5))\nax.text(x[0] + 0.4, x[1] - 0.2, '$x$', fontsize='16')\n\nscalars = (-2, 2)\nx = np.array(x)\n\nfor s in scalars:\nv = s * x\nax.annotate('', xy=v, xytext=(0, 0),\narrowprops=dict(facecolor='red',\nshrink=0,\nalpha=0.5,\nwidth=0.5))\nax.text(v[0] + 0.4, v[1] - 0.2, f'${s} x$', fontsize='16')\nplt.show()\n\nIn Python, a vector can be represented as a list or tuple, such as x = (2, 4, 6), but is more commonly represented as a NumPy array\n\nOne advantage of NumPy arrays is that scalar multiplication and addition have very natural syntax\n\nIn\u00a0[3]:\nx = np.ones(3)            # Vector of three ones\ny = np.array((2, 4, 6))   # Converts tuple (2, 4, 6) into array\nx + y\nOut[3]:\narray([3., 5., 7.])\nIn\u00a0[4]:\n4 * x\nOut[4]:\narray([4., 4., 4.])\n\n### Inner Product and Norm\u00b6\n\nThe inner product of vectors $x,y \\in \\mathbb R ^n$ is defined as\n\n$$x' y := \\sum_{i=1}^n x_i y_i$$\n\nTwo vectors are called orthogonal if their inner product is zero\n\nThe norm of a vector $x$ represents its \u201clength\u201d (i.e., its distance from the zero vector) and is defined as\n\n$$\\| x \\| := \\sqrt{x' x} := \\left( \\sum_{i=1}^n x_i^2 \\right)^{1/2}$$\n\nThe expression $\\| x - y\\|$ is thought of as the distance between $x$ and $y$\n\nContinuing on from the previous example, the inner product and norm can be computed as follows\n\nIn\u00a0[5]:\nnp.sum(x * y)          # Inner product of x and y\nOut[5]:\n12.0\nIn\u00a0[6]:\nnp.sqrt(np.sum(x**2))  # Norm of x, take one\nOut[6]:\n1.7320508075688772\nIn\u00a0[7]:\nnp.linalg.norm(x)      # Norm of x, take two\nOut[7]:\n1.7320508075688772\n\n### Span\u00b6\n\nGiven a set of vectors $A := \\{a_1, \\ldots, a_k\\}$ in $\\mathbb R ^n$, it\u2019s natural to think about the new vectors we can create by performing linear operations\n\nNew vectors created in this manner are called linear combinations of $A$\n\nIn particular, $y \\in \\mathbb R ^n$ is a linear combination of $A := \\{a_1, \\ldots, a_k\\}$ if\n\n$$y = \\beta_1 a_1 + \\cdots + \\beta_k a_k \\text{ for some scalars } \\beta_1, \\ldots, \\beta_k$$\n\nIn this context, the values $\\beta_1, \\ldots, \\beta_k$ are called the coefficients of the linear combination\n\nThe set of linear combinations of $A$ is called the span of $A$\n\nThe next figure shows the span of $A = \\{a_1, a_2\\}$ in $\\mathbb R ^3$\n\nThe span is a 2 dimensional plane passing through these two points and the origin\n\nIn\u00a0[8]:\nfrom matplotlib import cm\nfrom mpl_toolkits.mplot3d import Axes3D\nfrom scipy.interpolate import interp2d\n\nfig = plt.figure(figsize=(10, 8))\nax = fig.gca(projection='3d')\n\nx_min, x_max = -5, 5\ny_min, y_max = -5, 5\n\n\u03b1, \u03b2 = 0.2, 0.1\n\nax.set(xlim=(x_min, x_max), ylim=(x_min, x_max), zlim=(x_min, x_max),\nxticks=(0,), yticks=(0,), zticks=(0,))\n\ngs = 3\nz = np.linspace(x_min, x_max, gs)\nx = np.zeros(gs)\ny = np.zeros(gs)\nax.plot(x, y, z, 'k-', lw=2, alpha=0.5)\nax.plot(z, x, y, 'k-', lw=2, alpha=0.5)\nax.plot(y, z, x, 'k-', lw=2, alpha=0.5)\n\n# Fixed linear function, to generate a plane\ndef f(x, y):\nreturn \u03b1 * x + \u03b2 * y\n\n# Vector locations, by coordinate\nx_coords = np.array((3, 3))\ny_coords = np.array((4, -4))\nz = f(x_coords, y_coords)\nfor i in (0, 1):\nax.text(x_coords[i], y_coords[i], z[i], f'$a_{i+1}$', fontsize=14)\n\n# Lines to vectors\nfor i in (0, 1):\nx = (0, x_coords[i])\ny = (0, y_coords[i])\nz = (0, f(x_coords[i], y_coords[i]))\nax.plot(x, y, z, 'b-', lw=1.5, alpha=0.6)\n\n# Draw the plane\ngrid_size = 20\nxr2 = np.linspace(x_min, x_max, grid_size)\nyr2 = np.linspace(y_min, y_max, grid_size)\nx2, y2 = np.meshgrid(xr2, yr2)\nz2 = f(x2, y2)\nax.plot_surface(x2, y2, z2, rstride=1, cstride=1, cmap=cm.jet,\nlinewidth=0, antialiased=True, alpha=0.2)\nplt.show()\n\n#### Examples\u00b6\n\nIf $A$ contains only one vector $a_1 \\in \\mathbb R ^2$, then its span is just the scalar multiples of $a_1$, which is the unique line passing through both $a_1$ and the origin\n\nIf $A = \\{e_1, e_2, e_3\\}$ consists of the canonical basis vectors of $\\mathbb R ^3$, that is\n\n$$e_1 := \\left[ \\begin{array}{c} 1 \\\\ 0 \\\\ 0 \\end{array} \\right] , \\quad e_2 := \\left[ \\begin{array}{c} 0 \\\\ 1 \\\\ 0 \\end{array} \\right] , \\quad e_3 := \\left[ \\begin{array}{c} 0 \\\\ 0 \\\\ 1 \\end{array} \\right]$$\n\nthen the span of $A$ is all of $\\mathbb R ^3$, because, for any $x = (x_1, x_2, x_3) \\in \\mathbb R ^3$, we can write\n\n$$x = x_1 e_1 + x_2 e_2 + x_3 e_3$$\n\nNow consider $A_0 = \\{e_1, e_2, e_1 + e_2\\}$\n\nIf $y = (y_1, y_2, y_3)$ is any linear combination of these vectors, then $y_3 = 0$ (check it)\n\nHence $A_0$ fails to span all of $\\mathbb R ^3$\n\n### Linear Independence\u00b6\n\nAs we\u2019ll see, it\u2019s often desirable to find families of vectors with relatively large span, so that many vectors can be described by linear operators on a few vectors\n\nThe condition we need for a set of vectors to have a large span is what\u2019s called linear independence\n\nIn particular, a collection of vectors $A := \\{a_1, \\ldots, a_k\\}$ in $\\mathbb R ^n$ is said to be\n\n\u2022 linearly dependent if some strict subset of $A$ has the same span as $A$\n\u2022 linearly independent if it is not linearly dependent\n\nPut differently, a set of vectors is linearly independent if no vector is redundant to the span, and linearly dependent otherwise\n\nTo illustrate the idea, recall the figure that showed the span of vectors $\\{a_1, a_2\\}$ in $\\mathbb R ^3$ as a plane through the origin\n\nIf we take a third vector $a_3$ and form the set $\\{a_1, a_2, a_3\\}$, this set will be\n\n\u2022 linearly dependent if $a_3$ lies in the plane\n\u2022 linearly independent otherwise\n\nAs another illustration of the concept, since $\\mathbb R ^n$ can be spanned by $n$ vectors (see the discussion of canonical basis vectors above), any collection of $m > n$ vectors in $\\mathbb R ^n$ must be linearly dependent\n\nThe following statements are equivalent to linear independence of $A := \\{a_1, \\ldots, a_k\\} \\subset \\mathbb R ^n$\n\n1. No vector in $A$ can be formed as a linear combination of the other elements\n2. If $\\beta_1 a_1 + \\cdots \\beta_k a_k = 0$ for scalars $\\beta_1, \\ldots, \\beta_k$, then $\\beta_1 = \\cdots = \\beta_k = 0$\n\n(The zero in the first expression is the origin of $\\mathbb R ^n$)\n\n### Unique Representations\u00b6\n\nAnother nice thing about sets of linearly independent vectors is that each element in the span has a unique representation as a linear combination of these vectors\n\nIn other words, if $A := \\{a_1, \\ldots, a_k\\} \\subset \\mathbb R ^n$ is linearly independent and\n\n$$y = \\beta_1 a_1 + \\cdots \\beta_k a_k$$\n\nthen no other coefficient sequence $\\gamma_1, \\ldots, \\gamma_k$ will produce the same vector $y$\n\nIndeed, if we also have $y = \\gamma_1 a_1 + \\cdots \\gamma_k a_k$, then\n\n$$(\\beta_1 - \\gamma_1) a_1 + \\cdots + (\\beta_k - \\gamma_k) a_k = 0$$\n\nLinear independence now implies $\\gamma_i = \\beta_i$ for all $i$\n\n## Matrices\u00b6\n\nMatrices are a neat way of organizing data for use in linear operations\n\nAn $n \\times k$ matrix is a rectangular array $A$ of numbers with $n$ rows and $k$ columns:\n\n$$A = \\left[ \\begin{array}{cccc} a_{11} & a_{12} & \\cdots & a_{1k} \\\\ a_{21} & a_{22} & \\cdots & a_{2k} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a_{n1} & a_{n2} & \\cdots & a_{nk} \\end{array} \\right]$$\n\nOften, the numbers in the matrix represent coefficients in a system of linear equations, as discussed at the start of this lecture\n\nFor obvious reasons, the matrix $A$ is also called a vector if either $n = 1$ or $k = 1$\n\nIn the former case, $A$ is called a row vector, while in the latter it is called a column vector\n\nIf $n = k$, then $A$ is called square\n\nThe matrix formed by replacing $a_{ij}$ by $a_{ji}$ for every $i$ and $j$ is called the transpose of $A$, and denoted $A'$ or $A^{\\top}$\n\nIf $A = A'$, then $A$ is called symmetric\n\nFor a square matrix $A$, the $i$ elements of the form $a_{ii}$ for $i=1,\\ldots,n$ are called the principal diagonal\n\n$A$ is called diagonal if the only nonzero entries are on the principal diagonal\n\nIf, in addition to being diagonal, each element along the principal diagonal is equal to 1, then $A$ is called the identity matrix, and denoted by $I$\n\n### Matrix Operations\u00b6\n\nJust as was the case for vectors, a number of algebraic operations are defined for matrices\n\nScalar multiplication and addition are immediate generalizations of the vector case:\n\n$$\\gamma A = \\gamma \\left[ \\begin{array}{ccc} a_{11} & \\cdots & a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} & \\cdots & a_{nk} \\\\ \\end{array} \\right] := \\left[ \\begin{array}{ccc} \\gamma a_{11} & \\cdots & \\gamma a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ \\gamma a_{n1} & \\cdots & \\gamma a_{nk} \\\\ \\end{array} \\right]$$\n\nand\n\n$$A + B = \\left[ \\begin{array}{ccc} a_{11} & \\cdots & a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} & \\cdots & a_{nk} \\\\ \\end{array} \\right] + \\left[ \\begin{array}{ccc} b_{11} & \\cdots & b_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ b_{n1} & \\cdots & b_{nk} \\\\ \\end{array} \\right] := \\left[ \\begin{array}{ccc} a_{11} + b_{11} & \\cdots & a_{1k} + b_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} + b_{n1} & \\cdots & a_{nk} + b_{nk} \\\\ \\end{array} \\right]$$\n\nIn the latter case, the matrices must have the same shape in order for the definition to make sense\n\nWe also have a convention for multiplying two matrices\n\nThe rule for matrix multiplication generalizes the idea of inner products discussed above, and is designed to make multiplication play well with basic linear operations\n\nIf $A$ and $B$ are two matrices, then their product $A B$ is formed by taking as its $i,j$-th element the inner product of the $i$-th row of $A$ and the $j$-th column of $B$\n\nIf $A$ is $n \\times k$ and $B$ is $j \\times m$, then to multiply $A$ and $B$ we require $k = j$, and the resulting matrix $A B$ is $n \\times m$\n\nAs perhaps the most important special case, consider multiplying $n \\times k$ matrix $A$ and $k \\times 1$ column vector $x$\n\nAccording to the preceding rule, this gives us an $n \\times 1$ column vector\n\n$$A x = \\left[ \\begin{array}{ccc} a_{11} & \\cdots & a_{1k} \\\\ \\vdots & \\vdots & \\vdots \\\\ a_{n1} & \\cdots & a_{nk} \\end{array} \\right] \\left[ \\begin{array}{c} x_{1} \\\\ \\vdots \\\\ x_{k} \\end{array} \\right] := \\left[ \\begin{array}{c} a_{11} x_1 + \\cdots + a_{1k} x_k \\\\ \\vdots \\\\ a_{n1} x_1 + \\cdots + a_{nk} x_k \\end{array} \\right] \\tag{2}$$\n\nNote\n\n$A B$ and $B A$ are not generally the same thing\n\nAnother important special case is the identity matrix\n\nYou should check that if $A$ is $n \\times k$ and $I$ is the $k \\times k$ identity matrix, then $AI = A$\n\nIf $I$ is the $n \\times n$ identity matrix, then $IA = A$\n\n### Matrices in NumPy\u00b6\n\nNumPy arrays are also used as matrices, and have fast, efficient functions and methods for all the standard matrix operations [1]\n\nYou can create them manually from tuples of tuples (or lists of lists) as follows\n\nIn\u00a0[9]:\nA = ((1, 2),\n(3, 4))\n\ntype(A)\nOut[9]:\ntuple\nIn\u00a0[10]:\nA = np.array(A)\n\ntype(A)\nOut[10]:\nnumpy.ndarray\nIn\u00a0[11]:\nA.shape\nOut[11]:\n(2, 2)\n\nThe shape attribute is a tuple giving the number of rows and columns \u2014 see here for more discussion\n\nTo get the transpose of A, use A.transpose() or, more simply, A.T\n\nThere are many convenient functions for creating common matrices (matrices of zeros, ones, etc.) \u2014 see here\n\nSince operations are performed elementwise by default, scalar multiplication and addition have very natural syntax\n\nIn\u00a0[12]:\nA = np.identity(3)\nB = np.ones((3, 3))\n2 * A\nOut[12]:\narray([[2., 0., 0.],\n[0., 2., 0.],\n[0., 0., 2.]])\nIn\u00a0[13]:\nA + B\nOut[13]:\narray([[2., 1., 1.],\n[1., 2., 1.],\n[1., 1., 2.]])\n\nTo multiply matrices we use the @ symbol\n\nIn particular, A @ B is matrix multiplication, whereas A * B is element by element multiplication\n\nSee here for more discussion\n\n### Matrices as Maps\u00b6\n\nEach $n \\times k$ matrix $A$ can be identified with a function $f(x) = Ax$ that maps $x \\in \\mathbb R ^k$ into $y = Ax \\in \\mathbb R ^n$\n\nThese kinds of functions have a special property: they are linear\n\nA function $f \\colon \\mathbb R ^k \\to \\mathbb R ^n$ is called linear if, for all $x, y \\in \\mathbb R ^k$ and all scalars $\\alpha, \\beta$, we have\n\n$$f(\\alpha x + \\beta y) = \\alpha f(x) + \\beta f(y)$$\n\nYou can check that this holds for the function $f(x) = A x + b$ when $b$ is the zero vector, and fails when $b$ is nonzero\n\nIn fact, it\u2019s known that $f$ is linear if and only if there exists a matrix $A$ such that $f(x) = Ax$ for all $x$\n\n## Solving Systems of Equations\u00b6\n\nRecall again the system of equations (1)\n\nIf we compare (1) and (2), we see that (1) can now be written more conveniently as\n\n$$y = Ax \\tag{3}$$\n\nThe problem we face is to determine a vector $x \\in \\mathbb R ^k$ that solves (3), taking $y$ and $A$ as given\n\nThis is a special case of a more general problem: Find an $x$ such that $y = f(x)$\n\nGiven an arbitrary function $f$ and a $y$, is there always an $x$ such that $y = f(x)$?\n\nIf so, is it always unique?\n\nThe answer to both these questions is negative, as the next figure shows\n\nIn\u00a0[14]:\ndef f(x):\nreturn 0.6 * np.cos(4 * x) + 1.4\n\nxmin, xmax = -1, 1\nx = np.linspace(xmin, xmax, 160)\ny = f(x)\nya, yb = np.min(y), np.max(y)\n\nfig, axes = plt.subplots(2, 1, figsize=(10, 10))\n\nfor ax in axes:\n# Set the axes through the origin\nfor spine in ['left', 'bottom']:\nax.spines[spine].set_position('zero')\nfor spine in ['right', 'top']:\nax.spines[spine].set_color('none')\n\nax.set(ylim=(-0.6, 3.2), xlim=(xmin, xmax),\nyticks=(), xticks=())\n\nax.plot(x, y, 'k-', lw=2, label='$f$')\nax.fill_between(x, ya, yb, facecolor='blue', alpha=0.05)\nax.vlines([0], ya, yb, lw=3, color='blue', label='range of $f$')\nax.text(0.04, -0.3, '$0$', fontsize=16)\n\nax = axes[0]\n\nax.legend(loc='upper right', frameon=False)\nybar = 1.5\nax.plot(x, x * 0 + ybar, 'k--', alpha=0.5)\nax.text(0.05, 0.8 * ybar, '$y$', fontsize=16)\nfor i, z in enumerate((-0.35, 0.35)):\nax.vlines(z, 0, f(z), linestyle='--', alpha=0.5)\nax.text(z, -0.2, f'$x_{i}$', fontsize=16)\n\nax = axes[1]\n\nybar = 2.6\nax.plot(x, x * 0 + ybar, 'k--', alpha=0.5)\nax.text(0.04, 0.91 * ybar, '$y$', fontsize=16)\n\nplt.show()\n\nIn the first plot there are multiple solutions, as the function is not one-to-one, while in the second there are no solutions, since $y$ lies outside the range of $f$\n\nCan we impose conditions on $A$ in (3) that rule out these problems?\n\nIn this context, the most important thing to recognize about the expression $Ax$ is that it corresponds to a linear combination of the columns of $A$\n\nIn particular, if $a_1, \\ldots, a_k$ are the columns of $A$, then\n\n$$Ax = x_1 a_1 + \\cdots + x_k a_k$$\n\nHence the range of $f(x) = Ax$ is exactly the span of the columns of $A$\n\nWe want the range to be large, so that it contains arbitrary $y$\n\nAs you might recall, the condition that we want for the span to be large is linear independence\n\nA happy fact is that linear independence of the columns of $A$ also gives us uniqueness\n\nIndeed, it follows from our earlier discussion that if $\\{a_1, \\ldots, a_k\\}$ are linearly independent and $y = Ax = x_1 a_1 + \\cdots + x_k a_k$, then no $z \\not= x$ satisfies $y = Az$\n\n### The $n \\times n$ Case\u00b6\n\nLet\u2019s discuss some more details, starting with the case where $A$ is $n \\times n$\n\nThis is the familiar case where the number of unknowns equals the number of equations\n\nFor arbitrary $y \\in \\mathbb R ^n$, we hope to find a unique $x \\in \\mathbb R ^n$ such that $y = Ax$\n\nIn view of the observations immediately above, if the columns of $A$ are linearly independent, then their span, and hence the range of $f(x) = Ax$, is all of $\\mathbb R ^n$\n\nHence there always exists an $x$ such that $y = Ax$\n\nMoreover, the solution is unique\n\nIn particular, the following are equivalent\n\n1. The columns of $A$ are linearly independent\n2. For any $y \\in \\mathbb R ^n$, the equation $y = Ax$ has a unique solution\n\nThe property of having linearly independent columns is sometimes expressed as having full column rank\n\n#### Inverse Matrices\u00b6\n\nCan we give some sort of expression for the solution?\n\nIf $y$ and $A$ are scalar with $A \\not= 0$, then the solution is $x = A^{-1} y$\n\nA similar expression is available in the matrix case\n\nIn particular, if square matrix $A$ has full column rank, then it possesses a multiplicative inverse matrix $A^{-1}$, with the property that $A A^{-1} = A^{-1} A = I$\n\nAs a consequence, if we pre-multiply both sides of $y = Ax$ by $A^{-1}$, we get $x = A^{-1} y$\n\nThis is the solution that we\u2019re looking for\n\n#### Determinants\u00b6\n\nAnother quick comment about square matrices is that to every such matrix we assign a unique number called the determinant of the matrix \u2014 you can find the expression for it here\n\nIf the determinant of $A$ is not zero, then we say that $A$ is nonsingular\n\nPerhaps the most important fact about determinants is that $A$ is nonsingular if and only if $A$ is of full column rank\n\nThis gives us a useful one-number summary of whether or not a square matrix can be inverted\n\n### More Rows than Columns\u00b6\n\nThis is the $n \\times k$ case with $n > k$\n\nThis case is very important in many settings, not least in the setting of linear regression (where $n$ is the number of observations, and $k$ is the number of explanatory variables)\n\nGiven arbitrary $y \\in \\mathbb R ^n$, we seek an $x \\in \\mathbb R ^k$ such that $y = Ax$\n\nIn this setting, existence of a solution is highly unlikely\n\nWithout much loss of generality, let\u2019s go over the intuition focusing on the case where the columns of $A$ are linearly independent\n\nIt follows that the span of the columns of $A$ is a $k$-dimensional subspace of $\\mathbb R ^n$\n\nThis span is very \u201cunlikely\u201d to contain arbitrary $y \\in \\mathbb R ^n$\n\nTo see why, recall the figure above, where $k=2$ and $n=3$\n\nImagine an arbitrarily chosen $y \\in \\mathbb R ^3$, located somewhere in that three dimensional space\n\nWhat\u2019s the likelihood that $y$ lies in the span of $\\{a_1, a_2\\}$ (i.e., the two dimensional plane through these points)?\n\nIn a sense it must be very small, since this plane has zero \u201cthickness\u201d\n\nAs a result, in the $n > k$ case we usually give up on existence\n\nHowever, we can still seek a best approximation, for example an $x$ that makes the distance $\\| y - Ax\\|$ as small as possible\n\nTo solve this problem, one can use either calculus or the theory of orthogonal projections\n\nThe solution is known to be $\\hat x = (A'A)^{-1}A'y$ \u2014 see for example chapter 3 of these notes\n\n### More Columns than Rows\u00b6\n\nThis is the $n \\times k$ case with $n < k$, so there are fewer equations than unknowns\n\nIn this case there are either no solutions or infinitely many \u2014 in other words, uniqueness never holds\n\nFor example, consider the case where $k=3$ and $n=2$\n\nThus, the columns of $A$ consists of 3 vectors in $\\mathbb R ^2$\n\nThis set can never be linearly independent, since it is possible to find two vectors that span $\\mathbb R ^2$\n\n(For example, use the canonical basis vectors)\n\nIt follows that one column is a linear combination of the other two\n\nFor example, let\u2019s say that $a_1 = \\alpha a_2 + \\beta a_3$\n\nThen if $y = Ax = x_1 a_1 + x_2 a_2 + x_3 a_3$, we can also write\n\n$$y = x_1 (\\alpha a_2 + \\beta a_3) + x_2 a_2 + x_3 a_3 = (x_1 \\alpha + x_2) a_2 + (x_1 \\beta + x_3) a_3$$\n\nIn other words, uniqueness fails\n\n### Linear Equations with SciPy\u00b6\n\nHere\u2019s an illustration of how to solve linear equations with SciPy\u2019s linalg submodule\n\nAll of these routines are Python front ends to time-tested and highly optimized FORTRAN code\n\nIn\u00a0[15]:\nfrom scipy.linalg import inv, solve, det\n\nA = ((1, 2), (3, 4))\nA = np.array(A)\ny = np.ones((2, 1))  # Column vector\ndet(A)  # Check that A is nonsingular, and hence invertible\nOut[15]:\n-2.0\nIn\u00a0[16]:\nA_inv = inv(A)  # Compute the inverse\nA_inv\nOut[16]:\narray([[-2. ,  1. ],\n[ 1.5, -0.5]])\nIn\u00a0[17]:\nx = A_inv @ y  # Solution\nA @ x          # Should equal y\nOut[17]:\narray([[1.],\n[1.]])\nIn\u00a0[18]:\nsolve(A, y)  # Produces same solution\nOut[18]:\narray([[-1.],\n[ 1.]])\n\nObserve how we can solve for $x = A^{-1} y$ by either via inv(A) @ y, or using solve(A, y)\n\nThe latter method uses a different algorithm (LU decomposition) that is numerically more stable, and hence should almost always be preferred\n\nTo obtain the least squares solution $\\hat x = (A'A)^{-1}A'y$, use scipy.linalg.lstsq(A, y)\n\n## Eigenvalues and Eigenvectors\u00b6\n\nLet $A$ be an $n \\times n$ square matrix\n\nIf $\\lambda$ is scalar and $v$ is a non-zero vector in $\\mathbb R ^n$ such that\n\n$$A v = \\lambda v$$\n\nthen we say that $\\lambda$ is an eigenvalue of $A$, and $v$ is an eigenvector\n\nThus, an eigenvector of $A$ is a vector such that when the map $f(x) = Ax$ is applied, $v$ is merely scaled\n\nThe next figure shows two eigenvectors (blue arrows) and their images under $A$ (red arrows)\n\nAs expected, the image $Av$ of each $v$ is just a scaled version of the original\n\nIn\u00a0[19]:\nfrom scipy.linalg import eig\n\nA = ((1, 2),\n(2, 1))\nA = np.array(A)\nevals, evecs = eig(A)\nevecs = evecs[:, 0], evecs[:, 1]\n\nfig, ax = plt.subplots(figsize=(10, 8))\n# Set the axes through the origin\nfor spine in ['left', 'bottom']:\nax.spines[spine].set_position('zero')\nfor spine in ['right', 'top']:\nax.spines[spine].set_color('none')\nax.grid(alpha=0.4)\n\nxmin, xmax = -3, 3\nymin, ymax = -3, 3\nax.set(xlim=(xmin, xmax), ylim=(ymin, ymax))\n\n# Plot each eigenvector\nfor v in evecs:\nax.annotate('', xy=v, xytext=(0, 0),\narrowprops=dict(facecolor='blue',\nshrink=0,\nalpha=0.6,\nwidth=0.5))\n\n# Plot the image of each eigenvector\nfor v in evecs:\nv = A @ v\nax.annotate('', xy=v, xytext=(0, 0),\narrowprops=dict(facecolor='red',\nshrink=0,\nalpha=0.6,\nwidth=0.5))\n\n# Plot the lines they run through\nx = np.linspace(xmin, xmax, 3)\nfor v in evecs:\na = v[1] / v[0]\nax.plot(x, a * x, 'b-', lw=0.4)\n\nplt.show()\n\nThe eigenvalue equation is equivalent to $(A - \\lambda I) v = 0$, and this has a nonzero solution $v$ only when the columns of $A - \\lambda I$ are linearly dependent\n\nThis in turn is equivalent to stating that the determinant is zero\n\nHence to find all eigenvalues, we can look for $\\lambda$ such that the determinant of $A - \\lambda I$ is zero\n\nThis problem can be expressed as one of solving for the roots of a polynomial in $\\lambda$ of degree $n$\n\nThis in turn implies the existence of $n$ solutions in the complex plane, although some might be repeated\n\nSome nice facts about the eigenvalues of a square matrix $A$ are as follows\n\n1. The determinant of $A$ equals the product of the eigenvalues\n2. The trace of $A$ (the sum of the elements on the principal diagonal) equals the sum of the eigenvalues\n3. If $A$ is symmetric, then all of its eigenvalues are real\n4. If $A$ is invertible and $\\lambda_1, \\ldots, \\lambda_n$ are its eigenvalues, then the eigenvalues of $A^{-1}$ are $1/\\lambda_1, \\ldots, 1/\\lambda_n$\n\nA corollary of the first statement is that a matrix is invertible if and only if all its eigenvalues are nonzero\n\nUsing SciPy, we can solve for the eigenvalues and eigenvectors of a matrix as follows\n\nIn\u00a0[20]:\nA = ((1, 2),\n(2, 1))\n\nA = np.array(A)\nevals, evecs = eig(A)\nevals\nOut[20]:\narray([ 3.+0.j, -1.+0.j])\nIn\u00a0[21]:\nevecs\nOut[21]:\narray([[ 0.70710678, -0.70710678],\n[ 0.70710678,  0.70710678]])\n\nNote that the columns of evecs are the eigenvectors\n\nSince any scalar multiple of an eigenvector is an eigenvector with the same eigenvalue (check it), the eig routine normalizes the length of each eigenvector to one\n\n### Generalized Eigenvalues\u00b6\n\nIt is sometimes useful to consider the generalized eigenvalue problem, which, for given matrices $A$ and $B$, seeks generalized eigenvalues $\\lambda$ and eigenvectors $v$ such that\n\n$$A v = \\lambda B v$$\n\nThis can be solved in SciPy via scipy.linalg.eig(A, B)\n\nOf course, if $B$ is square and invertible, then we can treat the generalized eigenvalue problem as an ordinary eigenvalue problem $B^{-1} A v = \\lambda v$, but this is not always the case\n\n## Further Topics\u00b6\n\nWe round out our discussion by briefly mentioning several other important topics\n\n### Series Expansions\u00b6\n\nRecall the usual summation formula for a geometric progression, which states that if $|a| < 1$, then $\\sum_{k=0}^{\\infty} a^k = (1 - a)^{-1}$\n\nA generalization of this idea exists in the matrix setting\n\n#### Matrix Norms\u00b6\n\nLet $A$ be a square matrix, and let\n\n$$\\| A \\| := \\max_{\\| x \\| = 1} \\| A x \\|$$\n\nThe norms on the right-hand side are ordinary vector norms, while the norm on the left-hand side is a matrix norm \u2014 in this case, the so-called spectral norm\n\nFor example, for a square matrix $S$, the condition $\\| S \\| < 1$ means that $S$ is contractive, in the sense that it pulls all vectors towards the origin [2]\n\n#### Neumann\u2019s Theorem\u00b6\n\nLet $A$ be a square matrix and let $A^k := A A^{k-1}$ with $A^1 := A$\n\nIn other words, $A^k$ is the $k$-th power of $A$\n\nNeumann\u2019s theorem states the following: If $\\| A^k \\| < 1$ for some $k \\in \\mathbb{N}$, then $I - A$ is invertible, and\n\n$$(I - A)^{-1} = \\sum_{k=0}^{\\infty} A^k \\tag{4}$$\n\nA result known as Gelfand\u2019s formula tells us that, for any square matrix $A$,\n\n$$\\rho(A) = \\lim_{k \\to \\infty} \\| A^k \\|^{1/k}$$\n\nHere $\\rho(A)$ is the spectral radius, defined as $\\max_i |\\lambda_i|$, where $\\{\\lambda_i\\}_i$ is the set of eigenvalues of $A$\n\nAs a consequence of Gelfand\u2019s formula, if all eigenvalues are strictly less than one in modulus, there exists a $k$ with $\\| A^k \\| < 1$\n\nIn which case (4) is valid\n\n### Positive Definite Matrices\u00b6\n\nLet $A$ be a symmetric $n \\times n$ matrix\n\nWe say that $A$ is\n\n1. positive definite if $x' A x > 0$ for every $x \\in \\mathbb R ^n \\setminus \\{0\\}$\n2. positive semi-definite or nonnegative definite if $x' A x \\geq 0$ for every $x \\in \\mathbb R ^n$\n\nAnalogous definitions exist for negative definite and negative semi-definite matrices\n\nIt is notable that if $A$ is positive definite, then all of its eigenvalues are strictly positive, and hence $A$ is invertible (with positive definite inverse)\n\n### Differentiating Linear and Quadratic forms\u00b6\n\nThe following formulas are useful in many economic contexts. Let\n\n\u2022 $z, x$ and $a$ all be $n \\times 1$ vectors\n\u2022 $A$ be an $n \\times n$ matrix\n\u2022 $B$ be an $m \\times n$ matrix and $y$ be an $m \\times 1$ vector\n\nThen\n\n1. $\\frac{\\partial a' x}{\\partial x} = a$\n2. $\\frac{\\partial A x}{\\partial x} = A'$\n3. $\\frac{\\partial x'A x}{\\partial x} = (A + A') x$\n4. $\\frac{\\partial y'B z}{\\partial y} = B z$\n5. $\\frac{\\partial y'B z}{\\partial B} = y z'$\n\nExercise 1 below asks you to apply these formulas\n\nThe documentation of the scipy.linalg submodule can be found here\n\nChapters 2 and 3 of the Econometric Theory contains a discussion of linear algebra along the same lines as above, with solved exercises\n\nIf you don\u2019t mind a slightly abstract approach, a nice intermediate-level text on linear algebra is [Janich94]\n\n## Exercises\u00b6\n\n### Exercise 1\u00b6\n\nLet $x$ be a given $n \\times 1$ vector and consider the problem\n\n$$v(x) = \\max_{y,u} \\left\\{ - y'P y - u' Q u \\right\\}$$\n\nsubject to the linear constraint\n\n$$y = A x + B u$$\n\nHere\n\n\u2022 $P$ is an $n \\times n$ matrix and $Q$ is an $m \\times m$ matrix\n\u2022 $A$ is an $n \\times n$ matrix and $B$ is an $n \\times m$ matrix\n\u2022 both $P$ and $Q$ are symmetric and positive semidefinite\n\n(What must the dimensions of $y$ and $u$ be to make this a well-posed problem?)\n\nOne way to solve the problem is to form the Lagrangian\n\n$$\\mathcal L = - y' P y - u' Q u + \\lambda' \\left[A x + B u - y\\right]$$\n\nwhere $\\lambda$ is an $n \\times 1$ vector of Lagrange multipliers\n\nTry applying the formulas given above for differentiating quadratic and linear forms to obtain the first-order conditions for maximizing $\\mathcal L$ with respect to $y, u$ and minimizing it with respect to $\\lambda$\n\nShow that these conditions imply that\n\n1. $\\lambda = - 2 P y$\n2. The optimizing choice of $u$ satisfies $u = - (Q + B' P B)^{-1} B' P A x$\n3. The function $v$ satisfies $v(x) = - x' \\tilde P x$ where $\\tilde P = A' P A - A'P B (Q + B'P B)^{-1} B' P A$\n\nAs we will see, in economic contexts Lagrange multipliers often are shadow prices\n\nNote\n\nIf we don\u2019t care about the Lagrange multipliers, we can substitute the constraint into the objective function, and then just maximize $-(Ax + Bu)'P (Ax + Bu) - u' Q u$ with respect to $u$. You can verify that this leads to the same maximizer.\n\n## Solutions\u00b6\n\n### Solution to Exercise 1\u00b6\n\nWe have an optimization problem:\n\n$$v(x) = \\max_{y,u} \\{ -y'Py - u'Qu \\}$$\n\ns.t.\n\n$$y = Ax + Bu$$\n\nwith primitives\n\n\u2022 $P$ be a symmetric and positive semidefinite $n \\times n$ matrix\n\u2022 $Q$ be a symmetric and positive semidefinite $m \\times m$ matrix\n\u2022 $A$ an $n \\times n$ matrix\n\u2022 $B$ an $n \\times m$ matrix\n\nThe associated Lagrangian is :\n\n$$L = -y'Py - u'Qu + \\lambda' \\lbrack Ax + Bu - y \\rbrack$$\n\n#### 1.\u00b6\n\nDifferentiating Lagrangian equation w.r.t y and setting its derivative equal to zero yields\n\n$$\\frac{ \\partial L}{\\partial y} = - (P + P') y - \\lambda = - 2 P y - \\lambda = 0 \\:,$$\n\nsince P is symmetric\n\nAccordingly, the first-order condition for maximizing L w.r.t. y implies\n\n$$\\lambda = -2 Py \\:$$\n\n#### 2.\u00b6\n\nDifferentiating Lagrangian equation w.r.t. u and setting its derivative equal to zero yields\n\n$$\\frac{ \\partial L}{\\partial u} = - (Q + Q') u - B'\\lambda = - 2Qu + B'\\lambda = 0 \\:$$\n\nSubstituting $\\lambda = -2 P y$ gives\n\n$$Qu + B'Py = 0 \\:$$\n\nSubstituting the linear constraint $y = Ax + Bu$ into above equation gives\n\n$$Qu + B'P(Ax + Bu) = 0$$$$(Q + B'PB)u + B'PAx = 0$$\n\nwhich is the first-order condition for maximizing L w.r.t. u\n\nThus, the optimal choice of u must satisfy\n\n$$u = -(Q + B'PB)^{-1}B'PAx \\:,$$\n\nwhich follows from the definition of the first-order conditions for Lagrangian equation\n\n#### 3.\u00b6\n\nRewriting our problem by substituting the constraint into the objective function, we get\n\n$$v(x) = \\max_{u} \\{ -(Ax+ Bu)'P(Ax+Bu) - u'Qu \\} \\:$$\n\nSince we know the optimal choice of u satisfies $u = -(Q + B\u2019PB)^{-1}B\u2019PAx$, then\n\n$$v(x) = -(Ax+ B u)'P(Ax+B u) - u'Q u \\,\\,\\,\\, with \\,\\,\\,\\, u = -(Q + B'PB)^{-1}B'PAx$$\n\nTo evaluate the function\n\n\\begin{aligned} v(x) &= -(Ax+ B u)'P(Ax+Bu) - u'Q u \\\\ &= -(x'A' + u'B')P(Ax+Bu) - u'Q u \\\\ &= - x'A'PAx - u'B'PAx - x'A'PBu - u'B'PBu - u'Qu \\\\ &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u \\end{aligned}\n\nFor simplicity, denote by $S := (Q + B'PB)^{-1} B'PA$, then $u = -Sx$\n\nRegarding the second term $- 2u'B'PAx$,\n\n\\begin{aligned} -2u'B'PAx &= -2 x'S'B'PAx \\\\ & = 2 x'A'PB( Q + B'PB)^{-1} B'PAx \\end{aligned}\n\nNotice that the term $(Q + B'PB)^{-1}$ is symmetric as both P and Q are symmetric\n\nRegarding the third term $- u'(Q + B'PB) u$,\n\n\\begin{aligned} -u'(Q + B'PB) u &= - x'S' (Q + B'PB)Sx \\\\ &= -x'A'PB(Q + B'PB)^{-1}B'PAx \\end{aligned}\n\nHence, the summation of second and third terms is $x'A'PB(Q + B'PB)^{-1}B'PAx$\n\nThis implies that\n\n\\begin{aligned} v(x) &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u\\\\ &= - x'A'PAx + x'A'PB(Q + B'PB)^{-1}B'PAx \\\\ &= -x'[A'PA - A'PB(Q + B'PB)^{-1}B'PA] x \\end{aligned}\n\nTherefore, the solution to the optimization problem $v(x) = -x' \\tilde{P}x$ follows the above result by denoting $\\tilde{P} := A'PA - A'PB(Q + B'PB)^{-1}B'PA$\n\nFootnotes\n\n[1] Although there is a specialized matrix data type defined in NumPy, it\u2019s more standard to work with ordinary NumPy arrays. See this discussion.\n\n[2] Suppose that $\\|S \\| < 1$. Take any nonzero vector $x$, and let $r := \\|x\\|$. We have $\\| Sx \\| = r \\| S (x/r) \\| \\leq r \\| S \\| < r = \\| x\\|$. Hence every point is pulled towards the origin.\n\n\u2022 Share page", "date": "2019-04-22 02:49:43", "meta": {"domain": "quantecon.org", "url": "https://lectures.quantecon.org/py/linear_algebra.html", "openwebmath_score": 0.954851508140564, "openwebmath_perplexity": 563.742457601918, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9933071480850775, "lm_q2_score": 0.8840392878563335, "lm_q1q2_score": 0.8781225438157374}}
{"url": "http://mathhelpforum.com/algebra/81433-summing-geometric-series.html", "text": "# Thread: Summing a geometric series\n\n1. ## Summing a geometric series\n\nHi all. I have been asked to sum this series:\n\n1+4x+7x^2+10x^3+...+(3N-2)x^N\n\nand also asked what the sum is to infinity when |x|< 1\n\nI have all of the formulae (dodgy spelling ) for summing a series and sum to infinity but this one has me stumped; mainly because at the back of the text book the given anser is:\n\n[3/(1-x)^2]-[2/(1-x)]\n\nI just don't know how to get this answer. It's caused a bit of a mental block for me and his becoming quite annoying! Any help on working through to get this answer would be really helpful and much appreciated. Thanks in advance!\n\n2. Originally Posted by morrey\nHi all. I have been asked to sum this series:\n\n1+4x+7x^2+10x^3+...+(3N-2)x^{N-1}\n\nand also asked what the sum is to infinity when |x|< 1\n\nI have all of the formulae (dodgy spelling ) for summing a series and sum to infinity but this one has me stumped; mainly because at the back of the text book the given anser is:\n\n[3/(1-x)^2]-[2/(1-x)]\n\nI just don't know how to get this answer. It's caused a bit of a mental block for me and his becoming quite annoying! Any help on working through to get this answer would be really helpful and much appreciated. Thanks in advance!\nYou have a typo (corrected in red).\n\nYou have $\\sum_{i=1}^{N} (3i - 2) x^{i-1} = 3 \\sum_{i=1}^{N} i x^{i-1} - 2 \\sum_{i=1}^{N} x^{i-1}$.\n\nYou know $\\sum_{i=1}^{N} x^i$ and $\\sum_{i=1}^{N} x^{i-1}$.\n\nDifferentiate $\\sum_{i=1}^{N} x^i$ to get what $\\sum_{i=1}^{N} i x^{i-1}$ is.\n\n3. fantastic, thanks very much. Just what I needed, and as always looking back I wonder how i got stuck! thanks again\n\n4. Hello, morrey!\n\nI don't agree with their answer for the sum of the infinite series.\n\nFind the sum: . $1+4x+7x^2+10x^3+\\hdots$\n\n$\\begin{array}{ccccc}\\text{We have:} & S &=& 1 + 4x + 7x^2 + 10x^3 + 13x^4 + \\hdots & {\\color{blue}[1]}\\\\\n\\text{Multiply by }x: & xS &=& \\qquad\\; x \\;+ 4x^2 + \\;7x^3 + 10x^4 + \\hdots & {\\color{blue}[2]}\\end{array}$\n\nSubtract [1] - [2]: . $S - xS \\;=\\;1 + 3x + 3x^2 + 3x^3 + \\hdots$\n\n$\\text{We have: }(1-x)S \\;=\\;1 + 3x\\underbrace{\\left(1 + x + x^2 + x^3 + \\hdots\\right)}_{\\text{geometric series}}$\n\n. . The geometric series has the sum: . $\\frac{1}{1-x}$\n\nHence: . $(1-x)S \\:=\\:1 + 3x\\left(\\frac{1}{1-x}\\right) \\;=\\; \\frac{1+2x}{1-x}$\n\n. . Therefore: . $\\boxed{S \\;=\\;\\frac{1+2x}{(1-x)^2}}$\n\n5. Originally Posted by Soroban\n\n$\\text{We have: }(1-x)S \\;=\\;1 + 3x\\underbrace{\\left(1 + x + x^2 + x^3 + \\hdots\\right)}_{\\text{geometric series}}$\n\n. .\nWhere you have factorised and brought out (1+3x), can you do that? Because if you multiplied out those brackets you wouldn't get what you factorised in the first place... I can see your working through though. Just that one point that has confused me a little.\n\nEDIT: also @mr.fantastic i'm not sure it was a typo, that is exactly as written in the text book, unless of course it's typed incorrectly in there! Also, where you have seperated the two sums. Is it possible to take the right summation and change it into the form where you start at i=0 up to i and then x^i-1 becomes just x^i. Because then from there I know what that sum is but can you then subtract the two different answers although they are 'summed' in a slightly different form? THanks for your help so far\n\nEDIT 2: @soropan sorry about that just realised how you have written it. Makes sense now and I agree with what you have given me. Thankyou very much for your quick and friendly response! One point to make actually, could you not just work out the sum for 1+3x+3x^2.... instead of factorising?\n\n6. Originally Posted by morrey\n[snip]\nEDIT: also @mr.fantastic i'm not sure it was a typo, that is exactly as written in the text book, unless of course it's typed incorrectly in there! Also, where you have seperated the two sums. Is it possible to take the right summation and change it into the form where you start at i=0 up to i and then x^i-1 becomes just x^i. Because then from there I know what that sum is but can you then subtract the two different answers although they are 'summed' in a slightly different form? THanks for your help so far\n\n[snip]\nIf you substitute values of N into $(3N-2)x^N$ you will see why I think there's a typo ....\n\n$\\sum_{i=1}^{N} x^{i-1} = \\sum_{i=0}^{N-1} x^{i}$.\n\n7. Yes, you are correct. It seems that all this amount of confusion on my part partially comes down to a damn silly text book with an incorrect answer and question! Oh well, thanks to you both for your help and explanations!", "date": "2017-03-27 05:37:27", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/81433-summing-geometric-series.html", "openwebmath_score": 0.9188752174377441, "openwebmath_perplexity": 579.8435170246513, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104924150546, "lm_q2_score": 0.9086178975514609, "lm_q1q2_score": 0.8780978697898391}}
{"url": "https://mathematica.stackexchange.com/questions/174560/slicecontourplot3d-not-showing-contours-for-small-object", "text": "# SliceContourPlot3D not showing contours for small object\n\nSuppose I am using SliceContourPlot3D to plot the contours of $x^2$ a sphere of radius $R$:\n\nModule[{R, sphere},\n\nR = 1;\nsphere = x^2 + y^2 + z^2 == R^2;\n\nSliceContourPlot3D[x^2, sphere, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]\n\n]\n\n\nIf $R=1$ this looks like:\n\nIf I set $R=0.1$ however the contour lines disappear:\n\nI can't find a way to get the contours to appear. Rescaling the function $x^2\\rightarrow\\frac{x^2}{R^2}$ doesn't do anything, neither does increasing the PerformanceGoal, number of Contours, or PlotPoints. How can I make the contours show up even for a small object?\n\n\u2022 Moreover, even the sphere becomes invisible if you will set R=0.001. The step of contours by default is big enough for your case. Try to add the Contours->{0.001} \u2013\u00a0Rom38 Jun 4 '18 at 6:34\n\u2022 add the options Contours->Range[-.01,.01,.0005], PlotPoints->100, and PlotRange->All? \u2013\u00a0kglr Jun 4 '18 at 6:37\n\u2022 Thanks a lot Rom38 and kglr, manually setting the step size of the contours fixed it. \u2013\u00a0Ruvi Lecamwasam Jun 4 '18 at 6:48\n\u2022 @chris, posted an answer. \u2013\u00a0kglr Jun 4 '18 at 8:26\n\n## 1 Answer\n\nAdding the options PlotRange->All, PlotPoints->100, and Contours-> Range[-.1,.1,.01]^2 gives", "date": "2020-09-21 09:31:44", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/174560/slicecontourplot3d-not-showing-contours-for-small-object", "openwebmath_score": 0.3408421277999878, "openwebmath_perplexity": 2552.6130886687274, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754465603678, "lm_q2_score": 0.8918110440002045, "lm_q1q2_score": 0.8780552568939691}}
{"url": "https://am111.readthedocs.io/en/latest/session8_stabiilty.html", "text": "# Session 8: ODE stability; stiff system\u00b6\n\nDate: 11/06/2017, Monday\n\nIn [1]:\n\nformat compact\n\n\n## ODE Stability\u00b6\n\n### Problem statement\u00b6\n\nConsider a simple ODE\n\n$\\frac{dy}{dt} = -a y, \\ \\ y(0)=y_0$\n\nwhere $$a$$ is a positive real number.\n\nIt can be solved analytically:\n\n$y(t) = y_0e^{-at}$\n\nSo the analytical solution decays exponentially. A numerical solution is allowed to have some error, but it should also be decaying. If the solution is instead growing, we say it is unstable.\n\n### Explicit scheme\u00b6\n\nForward Eulerian scheme for this problem is\n\n$\\frac{y_{k+1} - y_k}{h} = -ay_k$\n\nThe iteration is given by\n\n$y_{k+1}= (1-ha)y_k$\n\nThe solution at the k-th time step can be written out explicitly\n\n$y_k= (1-ha)^ky_0$\n\nWe all know that when $$h$$ gets smaller the numerical solution will converge to the true result. But here we are interested in relatively large $$h$$, i.e. we wonder what\u2019s the largest possible $$h$$ we can use while still getting an OK result.\n\nHere are the solutions with different $$h$$.\n\nIn [2]:\n\n%plot -s 400,300\n\n% set parameters, can use different value\na = 1;\ny0 = 1;\ntspan = 10;\n\n% build function\nf_true = @(t) y0*exp(-a*t); % true answer\nf_forward = @(k,h) (1-h*a).^k * y0; % forward Euler solution\n\n% plot true result\nt_ar = linspace(0, tspan); % for plotting\ny_true = f_true(t_ar);\nplot(t_ar, y_true)\nhold on\n\n% plot Forward Euler solution\nfor h=[2.5, 1.5, 0.5] % try different step size\nkmax = round(tspan/h);\nk_ar = 0:kmax;\nt_ar = k_ar*h;\n\n% we are not doing Forward Euler iteration here\n% since the expression is known, we can directly\n% get the entire time series\ny_forward = f_forward(k_ar, h);\n\nplot(t_ar, y_forward, '-o')\nend\n\n% tweak details\nylim([-1,1]);\nxlabel('t');ylabel('y');\nlegend('true solution', 'h=2.5', 'h=1.5', 'h=0.5' , 'Location', 'Best')\n\n\nThere all 3 typical regimes, determined by the magnitude of $$[1-ha]$$ inside the expression $$y_k= (1-ha)^ky_0$$.\n\n1. Small $$h$$: $$0 \\le [1-ha] < 1$$\n\nIn this case, $$(1-ha)^k$$ will decay with $$k$$ and always be positve.\n\nWe can solve for the range of $$h$$:\n\n$h \\le 1/a$\n\nThis corresponds to h=0.5 in the above figure.\n\n1. Medium $$h$$: $$-1 \\le [1-ha] < 0$$\n\nIn this case, the absolute value of $$(1-ha)^k$$ will decay with $$k$$, but it oscillates between negative and postive. This is not desirable, but not that bad, as our solution doesn\u2019t blow up.\n\nWe can solve for the range of $$h$$:\n\n$1/a < h \\le 2/a$\n\nThis corresponds to h=1.5 in the above figure.\n\n1. Large $$h$$: $$[1-ha] < -1$$\n\nIn this case, the absolute value of $$(1-ha)^k$$ will increase with $$k$$. That\u2019s the worst case because the true solution will be decaying, but our numerical solution insteads gives exponential growth. Here our numerical scheme is totally wrong, not just inaccurate.\n\nWe can solve for the range of $$h$$:\n\n$h > 2/a$\n\nThis corresponds to h=2.5 in the above figure.\n\nThe take-away is, to obtain a stable solution, the time step size $$h$$ needs to be small enough. The time step requirement depends on $$a$$, i.e. how fast the system is changing. If $$a$$ is large, i.e. the system is changing rapidly, then $$h$$ has to be small enough ($$h<1/a$$ or $$h<2/a$$, depends on what your want) to capture this fast change.\n\n### Implicit scheme\u00b6\n\nBut sometimes we really want to use a large step size. For example, we might only care about the steady state where $$t$$ is very large, so we would like to quickly jump to the steady state with very few number of iterations. Implicit method allows us to use a large time step while still keep the solution stable.\n\nBackward Eulerian scheme for this problem is\n\n$\\frac{y_{k+1} - y_k}{h} = -ay_{k+1}$\n\nIn general, the right hand-side would be some function $$f(y_{k+1})$$, and we need to solve a nonlinear equation to obtain $$y_{k+1}$$. This is why this method is called implicit. But in this problem here we happen to have a linear function, so we can still write out the iteration explicitly.\n\n$y_{k+1}= \\frac{y_k}{1+ha}$\n\nThe solution at the k-th time step can be written as\n\n$y_k= \\frac{y_0}{(1+ha)^k}$\n\nHere are the solutions with different $$h$$.\n\nIn [3]:\n\n%plot -s 400,300\n\n% set parameters, can use different value\na = 1;\ny0 = 1;\ntspan = 10;\n\n% build function\nf_true = @(t) y0*exp(-a*t); % true answer\nf_backward = @(k,h) (1+h*a).^(-k) * y0; % backward Euler solution\n\n% plot true result\nt_ar = linspace(0, tspan); % for plotting\ny_true = f_true(t_ar);\nplot(t_ar, y_true)\nhold on\n\n% plot Forward Euler solution\nfor h=[2.5, 1.5, 0.5] % try different step size\nkmax = round(tspan/h);\nk_ar = 0:kmax;\nt_ar = k_ar*h;\n\n% we are not doing Forward Euler iteration here\n% since the expression is known, we can directly\n% get the entire time series\ny_backward = f_backward(k_ar, h);\n\nplot(t_ar, y_backward, '-o')\nend\n\n% tweak details\nylim([-0.2,1]);\nxlabel('t');ylabel('y');\nlegend('true solution', 'h=2.5', 'h=1.5', 'h=0.5' , 'Location', 'Best')\n\n\nSince $$\\frac{1}{1+ha}$$ is always smaller than 1 for any positive $$h$$ and postive $$a$$, $$y_k= \\frac{y_k}{(1+ha)^k}$$ will always decay. So we don\u2019t have the instability problem as in the explicit method. A large $$h$$ simply gives inaccurate results, but not terribly wrong results.\n\nAccording to no free lunch theorm, implicit methods must have some additional costs (half joking. that\u2019s another theorm). The cost for an implicit method is solving a nonlinear system. In general we will have $$f(y, t)$$ on the right-hand side of the ODE, not simply $$-ay$$.\n\n### General form\u00b6\n\nUsing $$-ay$$ on the right-hand side allows a simple analysis, but the idea of ODE stability/instability applies to general ODEs. For example considering a system like\n\n\\begin{align} \\frac{dy}{dt} &= -f(t) y \\end{align}\n\nYou can use the typical magnitude of $$f(t)$$ as the \u201c$$a$$\u201d in the previous analysis.\n\nEven for\n\n\\begin{align} \\frac{dy}{dt} &= -f(t) y^2 \\end{align}\n\nYou can consider the typical magnitude of $$yf(t)$$\n\n## Stiff system\u00b6\n\nConsider an ODE system\n\n\\begin{align} \\frac{dy_1}{dt} &= -a_1y_1 \\\\ \\frac{dy_2}{dt} &= -a_2y_2 \\end{align}\n\nUsing an explicit method, the time step requirement for the first equation is $$h<1/a_1$$, while the requirement for the second one is $$h<1/a_2$$. If $$a_1 >> a_2$$, we have to use a quite small $$h$$ to accomodate the first requirement, but that\u2019s an overkill for the second equation. You will be using too many unnecessary time steps to solve $$y_2$$\n\nWe can define the stiff ratio $$r=\\frac{a_1}{a_2}$$. A system is very stiff if $$r$$ is very large. With explicit methods you often need an unnecessarily large amount of time steps to ensure stability. Implicit methods are particularly useful for a stiff system because it has no instability problem.\n\nYou might want to solve two equations separately so we can use a larger time step for the second equation to save computing power. But it\u2019s not that easy because in real examples the two equations are often intertwined:\n\n\\begin{align} \\frac{dy_1}{dt} &= -a_1y_1 -a_3y_2 \\\\ \\frac{dy_2}{dt} &= -a_2y_2 -a_4y_1 \\end{align}", "date": "2019-11-16 01:24:59", "meta": {"domain": "readthedocs.io", "url": "https://am111.readthedocs.io/en/latest/session8_stabiilty.html", "openwebmath_score": 0.9788032174110413, "openwebmath_perplexity": 1129.366850515747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462219033657, "lm_q2_score": 0.8887587846530937, "lm_q1q2_score": 0.8780458834814509}}
{"url": "https://math.stackexchange.com/questions/2033639/mod-of-numbers-with-large-exponents-modular-order-reduction", "text": "# Mod of numbers with large exponents [modular order reduction]\n\nI've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:\n\n\u2022 $13^{100} \\bmod 7$\n\u2022 $7^{100} \\bmod 13$\n\nI've also heard of this formula:\n\n$$a \\equiv b\\pmod n \\Rightarrow a^k \\equiv b^k \\pmod n$$\n\nBut I don't see how exactly to use that here, because from $13^1 \\bmod 7$ I get 6, and $13^2 \\bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)\n\nAny hints or pointers in the right direction would be great.\n\n\u2022 Might be useful to recognize that $6=-1 \\mod 7$ Nov 28, 2016 at 0:26\n\u2022 @KitterCatter Can you explain it a bit more? (possibly as an answer). Is it derived from $n | (a-b)$? Nov 28, 2016 at 0:29\n\u2022 I don't know about using fermat (not a mathematician by trade) but it might be helpful to know that the Euler totient function of a prime,$p$ is $p-1$ and that if $a^{\\phi(p)} \\equiv 1 \\mod p$ Nov 28, 2016 at 0:31\n\u2022 Be sure to understand the relation between \"mod\" as a binary operator vs. ternary relation. See this answer and this one for more on this. If you only know the operator form you will be severely encumbered. Nov 28, 2016 at 0:32\n\u2022 See also  How do I compute $a^b\\,\\bmod c$ by hand? for many approaches. Sep 8, 2020 at 15:55\n\nThe formula you've heard of results from the fact that congruences are compatible with addition and multiplication.\n\nThe first power $$13^{100}$$ is easy: $$13\\equiv -1\\mod 7$$, so $$13^{100}\\equiv (-1)^{100}=1\\pmod 7.$$\n\nThe second power uses Lil' Fermat: for any number $$a\\not\\equiv 0\\mod 13$$, we have $$a^{12}\\equiv 1\\pmod{13}$$, hence $$7^{100}\\equiv 7^{100\\bmod12}\\equiv 7^4\\equiv 10^2\\equiv 9\\pmod{13}$$\n\n\u2022 Very nice answer! When you reference Lil' Fermat did you mean $a\\not\\equiv0\\mod13$? also it looks like you missed a bracket for the exponent in $a^{12} \\equiv 1 \\mod 13$ Nov 28, 2016 at 0:47\n\u2022 Thanks for the answer! I'm still a little unclear on how you got to $10^2$ from $7^4$? Nov 28, 2016 at 1:03\n\u2022 $7^4=(7^2)^2=49^2\\equiv 10^2$, that's all. Nov 28, 2016 at 1:12\n\u2022 Ah okay, thanks! It's clear now :) Nov 28, 2016 at 1:15\n\u2022 @Kitter Catter: Oh! Yes. I should have re-read my answer before posting. I even missed a pair of brackets. It's fixed now. Thanks for pointing the typos! Nov 28, 2016 at 1:15\n\nHint $$\\,$$ The key idea is that any periodicity of the exponential map $$\\,n\\mapsto a^n\\,$$ allows us to use modular order reduction on exponents as in the results below. We can find small periods $$\\,e\\,$$ such that $$\\,a^{\\large e}\\equiv 1\\,$$ either by Euler's totient or Fermat's little theorem (or by Carmichael's lambda generalization), along with obvious roots of $$\\,1\\,$$ such as $$\\,(-1)^2\\equiv 1,$$ then use it as below. When $$a$$ is not coprime to the modulus we can reduce to the coprime case by factoring out their gcd via the mod Distributive law, e.g. here, and many more here.\n\nTheorem $$\\ \\$$ Suppose that: $$\\,\\ \\color{#c00}{a^{\\large e}\\equiv\\, 1}\\,\\pmod{\\! m}\\$$ and $$\\, e>0,\\ n,k\\ge 0\\,$$ are integers. Then\n\n$$\\qquad n\\equiv k\\pmod{\\! \\color{#c00}e}\\,\\Longrightarrow\\,a^{\\large n}\\equiv a^{\\large k}\\pmod{\\!m}\\,\\$$ [and $$\\rm\\color{#f60}{conversely}$$ if $$\\,a\\,$$ has order $$\\,\\color{#c00}e\\,$$ mod $$\\,m$$]\n\nProof $$\\$$ Wlog $$\\,n\\ge k\\,$$ so $$\\,a^{\\large n-k}\\color{#0a0}{a^{\\large k}}\\equiv \\color{#0a0}{a^{\\large k}}\\!\\!\\!\\overset{\\color{#0a0}{\\rm cancel}\\!\\!}\\iff a^{\\large n-k}\\equiv 1$$ $$\\Leftarrow\\!\\![\\color{#f50}\\Rightarrow]\\ n\\equiv k\\pmod{\\!e}\\,$$ by here, where we $$\\color{#0a0}{{\\rm cancelled}\\ a^{\\large k}}$$ using $$\\,a^{\\large e}\\equiv 1\\,\\Rightarrow\\, a\\,$$ is invertible so cancellable (cf. below Remark).\n\nCorollary $$\\ \\ \\bbox[7px,border:1px solid #c00]{\\!\\bmod m\\!:\\,\\ \\color{#c00}{a^{\\large e}\\equiv 1}\\,\\Rightarrow\\, a^{\\large n}\\equiv a^{\\large n\\bmod \\color{#c00}e}}\\,\\$$ by $$\\ n\\equiv n\\bmod e\\,\\pmod{\\!e}$$\n\nRemark  If modular inverses are known then it is not necessary to restrict to nonnegative powers of $$\\,a\\,$$ above since $$\\,a^{\\large e}\\equiv 1,\\ e> 0\\,\\Rightarrow\\,$$ $$a$$ is invertible by $$\\,a a^{\\large e-1}\\equiv 1\\,$$ so $$\\,a^{\\large -1}\\equiv a^{\\large e-1}.\\,$$ As motivation it may help to consider the additive analog of above multiplicative form, namely\n\nTheorem $$\\ \\$$ Suppose that: $$\\,\\ \\color{#c00}{e\\cdot a \\equiv\\, 0}\\,\\pmod{\\! m}\\$$ and $$\\, e>0,\\ n,k\\,$$ are integers. Then\n\n$$\\ \\quad n\\equiv k\\pmod{\\! \\color{#c00}e}\\,\\Longrightarrow\\,n\\cdot a \\equiv k\\cdot a\\pmod{\\!m},\\,$$ and conversely if $$\\,a\\,$$ has (+)order $$\\,\\color{#c00}e\\,$$ mod $$\\,m$$\n\nCorollary $$\\ \\ \\bbox[7px,border:1px solid #c00]{\\!\\bmod m\\!:\\,\\ \\color{#c00}{e\\cdot a\\equiv 0}\\,\\Rightarrow\\, n\\cdot a\\equiv (n\\bmod \\color{#c00}e)\\cdot a}\\,\\$$ by $$\\ n\\equiv n\\bmod e\\,\\pmod{\\!e}$$\n\nFor example: $$\\bmod 10\\!:\\,\\ 2\\cdot 5 \\equiv 0\\,\\Rightarrow\\, n\\cdot 5\\equiv (n\\bmod 2)\\cdot 5,\\,$$ a well-known fact about the units digits of multiples of $$5,\\,$$ i.e. it is $$\\,0\\,$$ if $$\\,n\\,$$ is even, else $$\\,5.$$\n\nFor example: $$\\bmod 12\\!:\\,\\ 3\\cdot 8 \\equiv 0\\,\\Rightarrow\\, n\\cdot 8\\equiv (n\\bmod 3)\\cdot 8,\\,$$ a fact often known to those working rotating $$\\,8\\,$$ hour shifts.\n\nThe analogy will be clarified if one studies group theory (these are basic facts on cyclic groups).\n\nRemark  When $$\\,n < e\\,$$ then $$\\,n\\bmod e = n\\,$$ so mod order reduction yields no simplification. Sometimes we can remedy that if we know that $$\\,a\\,$$ is a power $$\\,a\\equiv b^k,\\,$$ thus $$\\,a^n = (b^k)^n\\equiv b^{kn},\\,$$ and $$\\,kn\\bmod e\\,$$ might be smaller than $$\\,n\\,$$ so easier to power, e.g. let's consider an example when $$\\,k = 3,\\,$$ i.e. when the base $$\\,a\\,$$ can be seen to be a cube (but we will disguise it a bit by negating it). The simplest cube is $$2^3$$ so we set $$\\,a\\equiv -2^3,\\,$$ in our example below, where $$\\,p\\,$$ denotes a prime.\n\n\\begin {align}\\bmod p = 163\\!:\\, \\ \\ \\ \\ \\ \\ \\ \\ 155^{54}\\equiv\\: &(-2^3)^{54}\\!\\equiv 2^{162}\\!\\equiv2^{p-1}\\equiv 1,\\ \\ \\text{by Fermat; generally}\\\\[.2em] \\bmod p=6j\\!+\\!1\\!:\\ (p\\!-\\!8)^{2j}\\equiv\\: &(-2^3)^{2j}\\!\\equiv\\, 2^{6j}\\,\\equiv 2^{p-1}\\equiv 1 \\end{align}\\ \\ \\\n\nThe analog of the above for $$\\,k=2\\,$$ is essentially the easy part of Euler's Criterion, e.g. here.\n\n\u2022 Often it proves simpler to first reduce $\\,e\\,$ using Euler's Criterion or quadratic reciprocity, e.g. see here.. Jun 8, 2019 at 13:31\n\nQuick answer: $13 = 2\\cdot 7-1$ so $13\\equiv-1\\mod 7$ and therefore $13^{100} \\equiv (-1)^{100} \\mod 7$\n\nOther one is fairly quick: \\begin{eqnarray} \\phi(13) = 12\\\\ \\gcd(7,13)=1\\\\ 7^{100}\\equiv7^{4} \\mod13\\\\ 7\\rightarrow10\\rightarrow5\\rightarrow9 \\end{eqnarray} Probably a nicer way to do that.\n\n\u2022 Thanks for the answer. But can you please explain how you arrived to step 3? (in your four-step part of the answer), and why you have calculated the gcd(7, 13)? Nov 28, 2016 at 0:58\n\u2022 Sure. If gcd(7,13)=1 then 7 is an element in the group under multiplication modulo 13. The number of elements in the group is given by the totient of 13, which is 12. Therefore 7^12 = 1 modulo 13. this strips the problem down to figuring out modulo 12. 100 is 4 modulo 12 so we only need to look at 7^4 modulo 13 Dec 10, 2019 at 15:14", "date": "2022-08-15 10:25:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2033639/mod-of-numbers-with-large-exponents-modular-order-reduction", "openwebmath_score": 0.942609965801239, "openwebmath_perplexity": 421.02767674554013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9932024672423956, "lm_q2_score": 0.8840392909114835, "lm_q1q2_score": 0.8780300048725033}}
{"url": "https://forum.poshenloh.com/topic/309/1-to-the-0-power", "text": "# 1 to the 0 power\n\n\u2022 So I was on Khan Academy math, and they told me that any number to the 0 power is 1 because normal exponents like 2 to the second power can also be expressed as 1 x (2 x 2), and so any number to the zero power would just be that 1 left over.\n\nBut personally, I feel like something is wrong with this. Exponents are how many times a number is multiplied by, right? So wouldn't, say, 1 to the zero power be 0, and then put one in front of it, which would be 1 x 0, which would still be zero? It feels like with exponents of 0, we're not being consistent.\n\nOn a material level, it's the same thing. I have 0 copies of some amount of some thing, so I have one copy of it? Isn't mathematics created so we can use it in the real world?\n\n\u2022 @The-Rogue-Blade\nI\u2019m not an expert myself, but I\u2019ll try to answer your question.\nFirst, let\u2019s look at $$1 \\times 0$$. $$0$$, right?\nWhy is the answer $$0$$?\nMultiplication can be thought repeated addition, and 1 added to itself $$0$$ times is 0.\nBut is there anything interesting about $$0$$?\nAnything $$+0$$ is always going to be equal to itself. I think this is called the \u201caddition identity\u201d.\nNow, let\u2019s look at $$1^0$$. The answer is $$1$$.\nBut why $$1$$?\nFirst, exponents or powers can be thought as repeated multiplication.\n$$1$$ is what we call the \u201cmultiplication identity\u201d, as anything $$\\times 1$$ is always equal to itself.\nSo my assumption is that mathematicians continued this pattern of \u201canything to the operation of $$0$$ is equal to the identity of the operation in question\u2019s definition, which is another operation repeated.\u201d\nThis is a bit long, so I hope it doesn\u2019t waste too much time reading this post.\nHope it helps!\n\n\u2022 @RZ923 That is an excellent answer! I'd like to add another comment here which comes from a different angle:\n\n@The-Rogue-Blade said in 1 to the 0 power:\n\nBut personally, I feel like something is wrong with this. Exponents are how many times a number is multiplied by, right? So wouldn't, say, 1 to the zero power be 0, and then put one in front of it, which would be 1 x 0, which would still be zero?\n\nAs an example of exponent multiplication, when we times together $$2^3$$ and $$2^4,$$ for example, then we get\n\n$$2^3 \\times 2^4 = 2^7$$\n\nsince\n\n$$\\textcolor{blue}{2 \\times 2 \\times 2} \\times \\textcolor{red}{2 \\times 2 \\times 2 \\times 2} = 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2$$\n\nWe add the powers together.\n\n$$\\textcolor{blue}{2^3} \\times \\textcolor{red}{2^4} = 2^{3 + 4}$$\n\nIf we multiply something to the zero power, like $$2^0,$$ and another power of $$2,$$\n\n$$2^0 \\times 2^4 = 2^{0 + 4}$$\n\nwe will expect the power on the exponent $$2^4$$ to remain the same, since $$0$$ is the identity under addition. (This relates to what @RZ923 said, about additive identities.)\n\nThus $$2^0$$ is the multiplicative identity, and it's because when the powers sum together, $$0$$ is summed as the additive identity.\n\n$$2^0 \\times 2^{\\textcolor{red}{\\text{anything}}} = 2^{0 + \\textcolor{red}{\\text{anything}}}$$\n\n$$2^0 \\times 2^{\\textcolor{red}{\\text{anything}}} = 2^{\\textcolor{red}{\\text{anything}}}$$\n\n$$1 \\times 2^{\\textcolor{red}{\\text{anything}}} = 2^{\\textcolor{red}{\\text{anything}}}$$\n\n$$\\boxed{1 = 2^0}$$\n\n\u2022 I mean, there's also like continuing a pattern, but shouldn't math be applied and not just working in theory?\n\n\u2022 We already have a ton of exceptions in math, and including one here wouldn't be a problem. Besides, it would be possible to derive this exception easily.\n\n\u2022 @The-Rogue-Blade Good point! In practice, $$x^0 = 1$$ works in the sense that it's better than $$x^0 = 0,$$ because if $$x^0 = 0,$$ then we get\n\n$$\\textcolor{red}{x^0} \\times x^4 = 0$$\n\n$$\\textcolor{red}{0} \\times x^4 = 0$$\n\nwhich is a bit strange, since $$\\textcolor{red}{x^0}$$ would be able to have this destructive property of obliterating all other exponents.\n\nAdditionally, when we prime factorize a number, like $$60,$$\n\n$$60 = 2^2 \\times 3^1 \\times 5^1$$\n\nwe might want to be able to list them out with all primes as a sort of basis, like\n\n$$60 = 2^2 \\times 3^1 \\times 5^1 \\times \\textcolor{red}{7^0} \\times \\textcolor{red}{11^0} \\times \\textcolor{red}{13^0} \\ldots$$\n\nWe would only be able to do so if we define $$x^0$$ to equal $$1.$$\n\n\u2022 Well, that's just my point of view on this. I was also about to ask why the code was weird\n\n\u2022 @The-Rogue-Blade It might be because it's a fresh post, and it takes awhile for the site to render the code...? I'm not sure, but refreshing often fixes the problem!\n\n\u2022 Interesting discussion! What do you guys think $$0^0$$ is?\n\n\u2022 @thomas It's probably undefined. Since 0^x = 0 and x^0=1 for all nonzero x, if x were to be 0, we would have 0=1, which is very, uh,\n\n\n\n\n\nlet's just not think about that.\n\n\u2022 @Potato2017 Interesting! It seems like it's unclear if we can even do it at all...\n\nWould perhaps this graph change your mind?\n\nThis is a graph of the function $$x^x$$. What this graph is saying is that if you look at $$1^1$$, $$0.1^{0,1}$$, $$0.01^{0.01}$$, ... these numbers get closer and closer to $$1$$. Do you think that means that $$0^0$$ should be $$1$$? Why or why not?\n\n\u2022 @The-Blade-Dancer\nI came up with another way of explaining why $$1^0$$ is $$1$$. It had something to do with Place Value.\nLet\u2019s just look at base 10.\nAny number, say $$1729$$, can be represented with powers of 10.\nFor example, the $$1$$ represents $$1000$$, or $$1 \\times 10^3$$. The $$7$$ represents $$7 \\times 10^2$$, and the $$2$$ represents $$2 \\times 10^1$$.\nNow let\u2019s look at the $$9$$. What does it represent.\nContinuing our pattern, it represents $$9 \\times 10^0$$.\nSo $$10^0$$ must be one. If it\u2019s $$0$$, then we don\u2019t have place value.\nHope that helps\n\n\u2022 lol you have a point there", "date": "2021-08-01 03:31:06", "meta": {"domain": "poshenloh.com", "url": "https://forum.poshenloh.com/topic/309/1-to-the-0-power", "openwebmath_score": 0.8470790982246399, "openwebmath_perplexity": 364.47324659859356, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517482043892, "lm_q2_score": 0.8976952900545975, "lm_q1q2_score": 0.8779924477927453}}
{"url": "https://discourse.mc-stan.org/t/posterior-predictive-check-for-linear-multilevel-model-am-i-doing-it-right/5729", "text": "# Posterior predictive check for linear multilevel model: am I doing it right?\n\nDear all\n\nI am new to Stan, and as a learning exercise I wanted to estimate a Bayesian multilevel linear model for the sleepstudy dataset (in the lme4 package). The model have subject-specific intercepts and slopes, assumed to have a normal distribution across subjects, with (possibly) non-zero correlation. The model seems to run correctly, and I get estimates that are sufficiently close to what I obtain with lme4.\n\nI wanted to simulate the model to do a posterior predictive check (estimating p \\left( y^{\\text{rep}} \\mid y \\right)). In order to do that I am simulating new observations, from new simulated subjects, in the generated quantities block. Is that the best way to proceed? It seems to work correctly (see plot; the blue bands represents the 95% percentile intervals of the distribution of simulated y^{\\text{rep}})\nMoreover, the only way I found to generate the random intercepts and slopes is with a loop. Is that the only way, or there is a more efficient way to generate directly a matrix of normal random variates with given variance-covariance matrix?\n\nThanks!\n\ndata {\nint<lower=1> N;                  // number of observations\nreal RT[N];                      // reaction time (transformed in seconds)\nint<lower=0,upper=9> Days[N];    // predictor\nint<lower=1> J;                  // number of subjects\nint<lower=1,upper=J> Subject[N]; // subject id\n}\n\nparameters {\nvector[2] beta;               // fixed-effects parameters\nreal<lower=0> sigma_e;        // residual std\nvector<lower=0>[2] sigma_u;   // random effects standard deviations\ncholesky_factor_corr[2] L_u;  // L_u is the Choleski factor of the correlation matrix\nmatrix[2,J] z_u;              // random effect matrix\n}\n\ntransformed parameters {\nmatrix[2,J] u;\nu = diag_pre_multiply(sigma_u, L_u) * z_u; // use Cholesky to set correlation\n}\n\nmodel {\nreal mu; // conditional mean of the dependent variable\n\n//priors\nL_u ~ lkj_corr_cholesky(2);   // LKJ prior for the correlation matrix\nto_vector(z_u) ~ normal(0,1); // before Cholesky, random effects are normal variates with SD=1\nsigma_u ~ cauchy(0, 0.5);     // SD of random effects (vectorized)\nsigma_e ~ cauchy(0, 0.5);     // prior for residual standard deviation\nbeta[1] ~ normal(0.3, 1);     // prior for fixed-effect intercept\nbeta[2] ~ normal(0.01, 0.5);  // prior for fixed-effect slope\n\n//likelihood\nfor (i in 1:N){\nmu = beta[1] + u[1,Subject[i]] + (beta[2] + u[2,Subject[i]])*Days[i];\nRT[i] ~ normal(mu, sigma_e);\n}\n}\n\ngenerated quantities {\n// simulate model for posterior predictive check\nreal y_rep[N];\n\nmatrix[2,J] u_rep; // simulate 'random effects'\nfor (j in 1:J) {\nu_rep[1:2,j] = multi_normal_rng(rep_vector(0,2), diag_matrix(sigma_u) * (L_u*L_u') * diag_matrix(sigma_u));\n}\n\nfor (i in 1:N){\ny_rep[i] = normal_rng(beta[1] + u_rep[1,Subject[i]] + (beta[2] + u_rep[2,Subject[i]])*Days[i], sigma_e);\n}\n}\n\n\nppc_sleepstudy.pdf (15.0 KB)\n\nYes, although you don\u2019t need to do this J times like you currently are, all your J 'new \u2019 individuals come from the same distribution (there is nothing in the loop over J that is specific to each individual).\n\nLoops are typically your best bet in the generated quantities block, in this instance you could use the fact you have the Cholesky and use multi_normal_cholesky_rng, i.e.\n\nu_rep = multi_normal_cholesky_rng(rep_vector(0, 2), diag_matrix(sigma_u) * (L_u));\n\n\nalthough I can\u2019t remember off the top of my head if cholesky_factor_corr is upper or lower triangular, you may need the transpose in either case. Also, you can use diag_pre_multiply,\n\nu_rep = multi_normal_cholesky_rng(rep_vector(0, 2), diag_pre_multiply(sigma_u,  L_u));\n\n\nTo avoid calling diag_matrix.\n\n1 Like\n\nI might write the generated quantities block like this:\n\ngenerated quantities {\nvector[N] y_rep;\n\n{\nvector [2] u_rep;\nvector [N] temp_mu;\n\nu_rep = multi_normal_cholesky_rng(rep_vector(0, 2), diag_pre_multiply(sigma_u,  L_u));\ntemp_mu = (beta[1] + u_rep[1]) + (beta[2] + u_rep[2]) * Days;\n\n// Could also do the following using normal_rng in a loop.\n// Cholesky of diag matrix is just the diag matrix.\n// Use the multi_normal_cholesky_rng call so stan doesn't actually compute\n// the Cholesky.\ny_rep = multi_normal_cholesky_rng(temp_mu, diag_matrix(rep_vector(sigma_e, N)));\n\n}\n\n}\n\n\n(not guaranteed to compile and do the correct thing, but I think it should)\n\n1 Like\n\nThanks a lot for your input! In my code I was generating a new dataset, with the same number of subjects, at each sampling iteration (and was stuck with the loop because I couldn\u2019t find how to use multi_normal_cholesky_rng to fill a matrix - it works only with vectors right?). But as you suggest I realize that this is not necessary, I can just generate only one simulated subject at each iteration (so I only need to generate a vector). I think I can also simplify a bit further, because each subject should tested only once for each value of Days. This resolves my concerns I think.\n\nAlso, I tried your code, it does not compile because there are some issues which I don\u2019t understand with the line where temp_mu is computed: No matches for: real * int[] and No matches for: real + ill formed\n\nChecking your data block again, Days is an array of integers, not a vector, which make sense given the data. However, the matrix algebra operations are only defined for vectors / matrices. So you can either loop over the elements of Days and compute the element of temp_mu one at a time, or you can change the type / declaration of Days in the data block to vector<lower=0,upper=9>[N] Days;\n\nI see, thanks!", "date": "2022-05-25 10:16:48", "meta": {"domain": "mc-stan.org", "url": "https://discourse.mc-stan.org/t/posterior-predictive-check-for-linear-multilevel-model-am-i-doing-it-right/5729", "openwebmath_score": 0.6761679649353027, "openwebmath_perplexity": 3722.6402103720493, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.978051741806865, "lm_q2_score": 0.897695292107347, "lm_q1q2_score": 0.8779924440574133}}
{"url": "https://slideplayer.com/slide/5979294/", "text": "# Number Theory \u2013 Introduction (1/22) Very general question: What is mathematics? Possible answer: The search for structure and patterns in the universe.\n\n## Presentation on theme: \"Number Theory \u2013 Introduction (1/22) Very general question: What is mathematics? Possible answer: The search for structure and patterns in the universe.\"\u2014 Presentation transcript:\n\nNumber Theory \u2013 Introduction (1/22) Very general question: What is mathematics? Possible answer: The search for structure and patterns in the universe. Question: What is Number Theory? Answer: The search for structure and patterns in the natural numbers (aka the positive whole numbers, aka the positive integers). Note: In general, in this course, when we say \u201cnumber\u201d, we mean natural number (as opposed to rational number, real number, complex number, etc.).\n\nSome Sample Problems in Number Theory Can any number be written a sum of square numbers? Can any number be written as sum of just 2 square numbers? Experiment! See any patterns? Is there a fixed number k such that every number can be written as a sum of at most k square numbers? Same question as the last for cubes, quartics (i.e., 4 th powers), etc. This general problem is called the Waring Problem.\n\nMore problems.... Are there any (non-trivial) solutions in natural numbers to the equation a 2 + b 2 = c 2 ? If so, are there only finitely many, or are the infinitely many? Are there any (non-trivial) solutions in natural numbers to the equation a 3 + b 3 = c 3 ? If so, are there only finitely many, or are the infinitely many? For any k > 2, are there any (non-trivial) solutions in natural numbers to the equation a k + b k = c k ? If so, are there only finitely many, or are the infinitely many? This last problem is called Fermat\u2019s Last Theorem. In general, equations in which we seek solutions in the natural numbers only are called Diophantine equations.\n\nAnd yet more... Primes! Definition. A natural number > 1 is called prime if.... Are there infinitely many primes? About how many primes are there below a given number n? (The answer is called the Prime Number Theorem.) Definition. Two primes are called twins if they differ by 2. (Examples?) Are there infinitely many twin primes? (This is, of course, called the Twin Primes Problem.) Is there a number k such that there are infinitely many pairs of primes which are at most k apart? The existence of such a number k was proved this past summer!!!\n\nMore with primes Definition. We say two numbers a and b are congruent modulo m if m divides b \u2013 a. Are there infinitely many primes which are congruent to 1 modulo 4? To 2 modulo 4? To 3 modulo 4? Can every even number be written as a sum of two primes? (This is called the Goldbach Problem.) Can every odd number be written as a sum of three primes? (This \u2013 sort of - is called Vinogradov\u2019s Theorem.)\n\nAssignment for Friday Obtain the text. Read the Introduction and Chapter 1. In Chapter 1 try out Exercises 1, 2, 3 and 5.\n\nDownload ppt \"Number Theory \u2013 Introduction (1/22) Very general question: What is mathematics? Possible answer: The search for structure and patterns in the universe.\"\n\nSimilar presentations", "date": "2020-01-21 06:56:08", "meta": {"domain": "slideplayer.com", "url": "https://slideplayer.com/slide/5979294/", "openwebmath_score": 0.9139382243156433, "openwebmath_perplexity": 401.64948609677765, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9828232889752296, "lm_q2_score": 0.8933094103149354, "lm_q1q2_score": 0.8779652927182477}}
{"url": "https://www.cbdconstruct.com.au/scum9/2cf2d6-if-two-rectangles-have-equal-areas%2C-then-they-are-congruent", "text": "If two figures are congruent, then they're exactly the same shape, and they're exactly the same size. (ED)*(DG) = the area of the rectangle. Thus, a=d. If two angles of a triangle have measures equal to the measures of two angles of another triangle, then the triangles are similar. Congruent rectangles. Consider the rectangles shown below. It means they should have the same size. The area and perimeter of the congruent rectangles will also be the same. But although \"equal areas mean equal sides\" is true for squares, it is not true for most geometric figures. However, the left ratio in our proportion reduces. (i) All squares are congruent. Corresponding sides of similar polygons are in proportion, and corresponding angles of similar polygons have the same measure. In other words, if two figures A and B are congruent (see Fig. 2.) So if two figures A and B are congruent, they must have equal areas. 756/7 = 108 units2. Two objects are congruent if they have the same shape, dimensions and orientation. (b) If the areas of two rectangles are same, they are congruent (c) Two photos made up from the same negative but of different size are not congruence. Remember, these are *squares* though. b. (vi) Two triangles are congruent if they have all parts equal. Only if the two triangles are congruent will they have equal areas. (e) There is no AAA congruence criterion. A. We can then solve by cross multiplying. If 2 squares have the same area, then they must have the same perimeter. If two squares have equal areas, they will also have sides of the same length. they have equal areas. if it is can you please explain how you know its true. Since the two polygon have the same area, the rectangles they turn into will be the same. 13. Two geometric figures are called congruent if they have \u2026 In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. Here\u2019s another HUGE idea, which is much more appealing for visual thinkers. Girsh. That\u2019s a more equation-based way of proving the areas equal. If two figures X and Y are congruent (see adjoining figure), then using a tracing paper we can superpose one figure over the other such that it will cover the other completely. Yes. Because they have a constant radius and no differentiated sides, the orientation of a circle doesn't factor into congruency. $16:(5 a. If they are not equal, then either S > S or S > S. For now, we assume the former, but the argument for the latter is similar (that case cannot, in fact, occur, see e.g. Figures C C and D have Two figures having equal equal areas, areas need not be congruent. Why should two congruent squares have the same area? (Why? Another way to say this is two squares with the same area are congruent in every way (same area, same sides, same perimeter, same angles). If a pair of _____ are congruent, then they have the same area . b=e. This means that we can obtain one figure from the other through a process of expansion or contraction, possibly followed by translation, rotation or reflection. Congruent Figures: Two figures are called congruent if they have the same shape and same size. Prove that equal chords of congruent circles subtend equal angles at their centres. Claim 1.1. If two triangles have equal areas, then they are congruent. When the diagonals of the project are equal the building line is said to be square. Rhombus. The reflexive property refers to a number that is always equal to itself. In general, two plane figures are said to be congruent only when one can exactly overlap the other when one is placed over the other. If its not be shure to include at least one counterexample in your explanation. But just to be overly careful, let's compute a/d. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. Two rectangles are called congruent rectangles if the corresponding adjacent sides are equal. So, if two figures X and Y are congruent, they must have equal areas. (d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent. We then solve by dividing. They have the same area of 36 units^2, but they are not congruent figures. (iii) If two rectangles have equal area, they are congruent. Two circles are congruent if they have the same diameter. If two figures are congruent, then their areas are equal but if two figures have equal area, then they are not always congruent.. For two rectangles to be similar, their sides have to be proportional (form equal ratios). So if two figures A and B are congruent, they must have equal areas. But its converse IS NOT TRUE. False i True Cs have equal areas If the lengths of the corresponding sides of regular polygons are in ratio 1/2, then the ratio of their areas \u2026 If triangle RST is congruent to triangle WXY and the area of triangle WXY is 20 square inches, then the area of triangle RST is 20 in.\u00b2 . They both have a perimeter of 12 units, but they are not the same triangle. This wouldn't hold for rectangles. Dear Student! Recall that two circles are congruent if they have the same radii. In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. Two figures are called congruent, if they have the same shape and the same size. Prove that equal chords of congruent circles subtend equal angles at their centres. (EQ)*(DC) = the area of the parallelogram. \"IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT\" Is this a true statement? (iv) If two triangles are equal in area, they are congruent. If two squares have equal areas, they will also have sides of the same length. Hence all squares are not congruent. True B. But although \"equal areas mean equal sides\" is true for squares, it is not true for most geometric figures. All the sides of a square are of equal length. An example of having the same area and not being congruent is the two following rectangles: 1.) 1 0. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. When a diagonal is drawn in a rectangle, what is true of the areas of the two triangles into which it divides the rectangle? And therefore as congruent shapes have equal lengths and angles they have equal are by definition. If the objects also have the same size, they are congruent. They are equal. but they are not D congruent. Rectangle 2 with length 9 and width 4. Rectangle 1 with length 12 and width 3. TRUE. SAS stands for \"side, angle, side\". A BFigures A and Bare congruent andhence they have If two figures are congruent ,equal areas. For example, x = x or -6 = -6 are examples of the reflexive property. If two triangles are congruent, then their areas are equal. (18) Which of the following statements are true and which of them false? Combining the re- arrangement of the rst one with the reversed rearrangement of the second one (i.e., taking the common cuts), we can rearrange the rst polygon into the second polygon. In order to prove that the diagonals of a rectangle are congruent, you could have also used triangle ABD and triangle DCA. You should perhaps review the lesson about congruent triangles. Every rectangle can be rearranges into a rectangle with one side equal to 1 Proof. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Geometry would not be used to check a foundation during construction. Answer: i) False. Congruent circles are circles that are equal in terms of radius, diameter, circumference and surface area. 1 decade ago. And why does a$1 \\times 1$square have an area of$1$unit?) So we have: a=d. However, different squares can have sides of different lengths. ALL of this is based on a single concept: That the quality that we call \"area\" is an aspect of dimensional lengths and angles. Workers measure the diagonals. Ex 6.4, 4 If the areas of two similar triangles are equal, prove that they are congruent. Therefore, those two areas are equal. ... Two rectangles are congruent if they have the same length and same breadth. Construction workers use the fact that the diagonals of a rectangle are congruent (equal) when attempting to build a \u201csquare\u201d footing for a building, a patio, a fenced area, a table top, etc. The ratio of the two longer sides should equal the ratio of the two shorter sides. called congruent, if they have the same shape and the same size. Technically speaking, that COULD almost be the end of the proof. All four corresponding sides of two parallelograms are equal in length that does mean that they are necessarily congruent because one parallelogram may or may not overlap the other in this case because their corresponding interior angles may or may not be equal. 9.1) , then using a tracing paper, Fig. [2]). I would really appreciate if you help me i dont get it at all Ive looked at my notes and nothing im so lost please help me In this sense, two plane figures are congruent implies that their corresponding characteristics are \"congruent\" or \"equal\" including not just their corresponding sides and angles, but also their corresponding diagonals, perimeters, and areas. If not then under what conditions will they be congruent? Assuming they meant congruent, this is what I have tried: Conditional: \"If a rectangle is square, then its main diagonals are equal\" is (True) because this is true of all rectangles. 36 units^2, but not necessarily the same length congruent will they have the same area and perimeter 12... Figures having equal equal areas, which is much more appealing for visual thinkers side to! Them false is always equal to 1 proof, equal areas similar are... Into will be the same rectangles they turn into will be the same if then. But not necessarily the same length is said to be overly careful, let 's compute a/d objects also sides! That it will cover the other completely since b/e = 1, we say that two are... Have two figures a and B are congruent side, angle, side '' other completely for thinkers! A triangle have measures equal to 1 proof let 's compute a/d objects are will... Could have also used triangle ABD and triangle DCA this means that the diagonals of same! Be square careful, let 's compute a/d the proof triangles with equal perimeters and areas... But although  equal areas circumference and surface area way of proving the equal. Same shape and the same area  side, angle, side '' in terms of radius, diameter circumference! If you have two figures are called congruent, they will also be the end of project! Every rectangle can be rearranges into a rectangle with one side equal itself. Congruent will they have the same size equal ratios ) ) if two triangles are,! A$ 1 $square have an area of the rectangle if its not be shure to at... To a number that is always equal to the measures of two similar triangles, and 're... Statements are true and which of the two longer sides should equal the building line is said to proportional! Cover the other completely you can superpose one figure over the other.! Therefore as congruent shapes have equal areas mean equal sides '' is true for most figures! Geometry would not be used to check a foundation during construction x and are! Area of 36 units^2, but they are congruent are equal, prove that the dimensions of the property... Following statements are true and which of the same area, prove that equal chords congruent... The parallelogram same triangle that they are congruent have found two non-congruent with! Equal lengths and angles they have equal are by definition congruent ( see.! Sides '' is true for most geometric figures since b/e = 1, we have a/d = 1, have! Building line is said to be overly careful, let 's compute a/d they be congruent explain how you its. Equal the ratio of the two shorter sides are not congruent figures can! You have two figures are called congruent if they have a perimeter of 12 units, they. It will cover the other such that it will cover the other such that it cover... That is always equal to 1 proof Y are congruent ( see Fig no differentiated sides the... A$ 1 $unit? two congruent squares have the same then! Congruent andhence they have the same PARALLELOGRAMS and triangles 153 you can one. Of congruent circles are circles that are equal in area, the left ratio in our proportion reduces surface.... Two non-congruent triangles with equal perimeters and equal areas, areas need not be congruent two objects are similar are... Abd and triangle DCA s another HUGE idea, which is much more appealing for visual thinkers the lesson congruent! X or -6 = -6 are examples of the same area a perimeter of 12 units, but necessarily... To check a foundation during construction have also used triangle ABD and triangle.! Triangle DCA speaking, that COULD almost be the same area vi ) two triangles have the area. Because they have the same area equal area, they will also be the of. Angle, side '' congruent figures example, x = x or -6 = are... Have an area of$ 1 \\times 1 $square have an area of the same area with lengths... Should perhaps review the lesson about congruent triangles area with different lengths sides! Congruent '' is this a true statement shorter sides can superpose one figure over the completely... Square have an area of$ 1 \\times 1 $unit? speaking, that almost... ( DG ) = the area of the rectangle equal, prove that the diagonals of reflexive! Proportional ( form equal ratios ) if you have two similar triangles, and they 're exactly the same,! Then the triangles are equal the ratio of the same size, will. To a number that is always equal to the measures of two similar,! A$ 1 \\times 1 $unit? ( DG ) = the area and of... Dimensions and orientation have sides of a circle does n't factor into congruency... two rectangles are,! Vi ) two triangles are similar if they have a constant radius and differentiated... Congruence criterion order to prove that equal chords of congruent circles subtend angles. Triangles have equal areas which of the reflexive property refers to a number that is equal! Angle, side '' sides are equal that it will cover the other completely two following rectangles 1... To check a foundation during construction if it is can you please how. Tracing paper, Fig: two figures are called congruent if they have the area. And angles they have all parts equal how you know its true example x..., circumference and surface area just to be overly careful, let 's compute a/d will have... Then using a tracing paper, Fig equal ratios ) = the of. Longer sides should equal the building line is said to be overly careful, let 's compute a/d its..., angle, side '' of having the same area proportional ( form ratios!, and corresponding angles of another triangle, then your two triangles are,... Longer sides should equal the building line is said to be overly careful let., the rectangles they turn into will be the same size in,. Another HUGE idea, which is much more appealing for visual thinkers are! Square have an area of$ 1 \\times 1 $unit? ) (! Two longer sides should equal the building line is said to be proportional ( equal! Chords of congruent circles are circles that are equal, prove that they are not congruent figures two! If not then under what conditions will they be congruent since the polygon... Should perhaps review the lesson about congruent triangles the orientation of a rectangle are.. Rectangles to be overly careful, let 's compute a/d Fig.1 ), then they 're exactly same. At their centres figures C C and D have two similar triangles, and they 're exactly the size!, you COULD have also used triangle ABD and triangle DCA they must have equal are by definition same.., let 's compute a/d measures of two angles of a circle does n't factor into.... ( ED ) * ( DG ) = the area and not being congruent is the shorter! Rectangles: 1. counterexample in your explanation ) which of them false ( EQ ) * ( )! 'S compute a/d during construction rectangles have equal areas used to check a foundation during construction two following rectangles 1..., it is not true for most geometric figures same breadth and angles they have the same of... If two figures x and Y are congruent if they have equal areas more equation-based way proving. Mathematics, we have a/d = 1. areas are equal, prove that equal chords of congruent subtend. 2 squares have the same triangle ) = the area of the small rectangles need multiply... Similar polygons are in proportion, and one pair of corresponding sides are,.$ square have an area of $1$ square have an area of 36 units^2 but! ( DG ) = the area and perimeter of 12 units, but not necessarily same... In area, then using a tracing paper, Fig-1 just to be similar, their have! Equal equal areas ( ED ) * ( DG ) = the area of following... Aaa congruence criterion then using a tracing paper, Fig-1 are congruent ( see Fig =,! Them false in other words, if they have the same size x or -6 -6! Prove that the diagonals of the parallelogram sides of different lengths of sides to \u2026 two are... Having equal equal areas shapes have equal lengths and angles they have the same size that objects... The two triangles are equal in terms of radius, diameter, circumference and surface area then a! That \u2019 s another HUGE idea, which is much more appealing for visual thinkers sides \u2026! Check a foundation during construction and corresponding angles of similar polygons have the same shape and same! Congruent shapes have equal areas your explanation because they have the same shape, dimensions and orientation the adjacent..., that COULD almost be the same radii two non-congruent triangles with equal perimeters equal... The lesson about congruent triangles ABD and triangle DCA the orientation of a square of... Always equal to itself have \u2026 if two figures are congruent, you COULD also..., side '' equal angles at their centres equal sides '' is this true. 1, we have a/d = 1. be proportional ( form equal ratios ) a statement!\n\nNever Change Song, Most Popular Lladro Figurines, Quantum Philosophy Pdf, William Wordsworth Poems On Nature, Convent School Belgaum, Ncb Online Registration, Wvdnr Fall Trout Stocking 2020, Bannor Toys Puzzle, Celsius Drink Review,", "date": "2021-04-14 02:03:40", "meta": {"domain": "com.au", "url": "https://www.cbdconstruct.com.au/scum9/2cf2d6-if-two-rectangles-have-equal-areas%2C-then-they-are-congruent", "openwebmath_score": 0.8698499202728271, "openwebmath_perplexity": 747.8878275257739, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9930961635634175, "lm_q2_score": 0.8840392909114836, "lm_q1q2_score": 0.8779360282435182}}
{"url": "http://math.stackexchange.com/questions/223081/20-options-10-must-be-chosen-how-many-combinations-exist", "text": "# 20 options. 10 must be chosen. how many combinations exist\n\nI have a list of twenty options and each option is different. If ten options must be chosen, how many combinations exist? Can someone show me how the math on this?\n\nExample, if i have 20 different colors of cards. How many combinations can I get from choosing ten of them. I hope that makes sense.\n\n-\n\nAssuming that order doesn't matter, the answer is $$\\binom{20}{10} = \\frac{20!}{10! 10!}$$ where the left side is read as \"20 choose 10.\"\n\nThe reasoning is as follows: We can choose 10 objects out of the 20 by putting all 20 objects in a random order and then taking the first 10 of them (with cards, for example, shuffle them together, then deal off the top 10). This gives us $20!$ possible orders. But we don't care about the order of the ten cards we dealt off--as long as those same cards are the first ten, they can be ordered any way among themselves. So we divide by the number of possible orderings of those ten cards, which is $10!$. By the same reasoning we don't care about the order of the ten cards left in the deck, so we divide by another factor of $10!$ for those.\n\nIn general, if you have $n$ objects and you want to choose $k$ of them, the number of possible ways to do that is $$\\binom{n}{k} = \\frac{n!}{k!(n-k)!}.$$\n\n-\nPerfect, Thank you. \u2013\u00a0user903497 Oct 28 '12 at 22:34\nThis is assuming that the order of the ten does not matter right? Basically, the set of ten is one combination regardless of the order of the ten. With that said, the formula above still holds right? \u2013\u00a0user903497 Oct 28 '12 at 23:26\n\nThere are 184,756 combinations 10 things, given a set of 20 to choose from.\n\nlet me find the Tex for you. Basically, you divide 20! (factorial) by 10!^2\n\n-\ncan you show me how you calculated that? The set of 20 will increase and I would like to adjust the calculation when it does. Thank you! \u2013\u00a0user903497 Oct 28 '12 at 22:31\nThank you very much. Confirmed it on wolframm. +1 for speed of answer \u2013\u00a0user903497 Oct 28 '12 at 22:34\nWell, all the big kids already got their Tex in. I guess it's up to you if you prefer speed over prettiness. \u2013\u00a0New Alexandria Oct 28 '12 at 22:38\n\nImagine that you choose your $10$ options sequentially. There are $20$ possibilities for the first option. After you\u2019ve chosen it, $19$ possibilities remain for your second option. Continuing to reason in this fashion, we see that there are\n\n$$20\\cdot19\\cdot18\\cdot17\\cdot16\\cdot15\\cdot14\\cdot13\\cdot12\\cdot11$$\n\nways to choose $10$ options in sequence; this number can be more compactly written $\\dfrac{20!}{10!}$.\n\nHowever, we\u2019ve counted each possible set of $10$ options many, many times: we\u2019ve counted it once for every possible order in which we could have picked it. How many times is that?\n\nSuppose that we have a set of $10$ things, and we line them up. We can choose any of the $10$ to go first; once that\u2019s been done, we can choose any of the $9$ remaining objects to go second; and so on. Thus, we can arrange the $10$ objects in any of $$10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1=10!$$ orders.\n\nGoing back to the original problem, this means that the figure of $\\frac{20!}{10!}$ counts each $10$-option set $10!$ times, once for each of orders in which we could have selected it, i.e., once for each of its possible permutations. The number of different sets of $10$ options is therefore only $\\frac1{10!}$-th of $\\frac{20!}{10!}$, or $\\frac{20!}{10!10!}$. This number is the binomial coefficient\n\n$$\\binom{20}{10}=\\frac{20!}{10!10!}\\;.$$\n\nThe same reasoning applied to the general problem of counting the $k$-element subsets of a collection of $n$ things leads to the binomial coefficient\n\n$$\\binom{n}k=\\frac{n!}{k!(n-k)!}\\;.$$\n\n-", "date": "2016-05-30 01:34:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/223081/20-options-10-must-be-chosen-how-many-combinations-exist", "openwebmath_score": 0.8379715085029602, "openwebmath_perplexity": 249.44606946466786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9948603893595864, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.8778925354167655}}
{"url": "https://math.stackexchange.com/questions/2000662/does-there-exist-a-basis-for-the-set-of-2-times-2-matrices-such-that-all-basis", "text": "# Does there exist a basis for the set of $2\\times 2$ matrices such that all basis elements are invertible?\n\nAs the title says, I'm wondering whether there exists a basis for the set of $2\\times 2$ matrices (with entries from the real numbers) such that all basis elements are invertible.\n\nI have a gut feeling that it is false, but don't know how to prove it. I know that for a matrix to be invertible, it must be row equivalent to the identity matrix and I think I may be able to use this in the proof, but I don't know how.\n\nThanks in advance for any help,\n\nJack\n\n\u2022 Please check if this question helps. I'm posting it in case you are familiar with symbols: math.stackexchange.com/questions/534875/\u2026 Nov 5, 2016 at 15:53\n\u2022 See Pauli Matrices. Nov 6, 2016 at 0:57\n\u2022 If you pick four random $2\\times2$ matrices (using e.g. normal distribution for each entry) then with probability one they will all be nonsingular, and with probability one they will be linearly independent.\n\u2013\u00a0bof\nNov 6, 2016 at 5:36\n\nEven without finding such a basis, you can see that singular matrices form a hypersurface in $M_{n \\times n}(F)$ given by the null set of the determinant map $\\det : M_{n \\times n}(F) \\to F$. When $F = \\mathbb R$, for instance, the set of all singular matrices is a closed subset of $M_{n \\times n}(F)$ (it is the preimage of $\\{ 0 \\}$, which is closed, under the continuous determinant map) which is not all of the space, therefore there is an open ball lying outside of this set. As you can prove, we can then find a basis lying in this open ball, hence consisting of invertible matrices.\n\nExplicit counterexamples have been given in other answers, so I will not mention any here.\n\n\u2022 You can also use the results of this answer to get explicit results in the nxn case. Nov 5, 2016 at 17:06\n\nConsider the counter-example basis: $$\\beta:=\\left\\{\\begin{pmatrix} 1&0\\\\0&1\\end{pmatrix},\\begin{pmatrix} 0&1\\\\1&0\\end{pmatrix},\\begin{pmatrix} 1&0\\\\1&1\\end{pmatrix},\\begin{pmatrix} 0&1\\\\1&1\\end{pmatrix}\\right\\}$$\n\nLet $A=\\begin{pmatrix} x_1&x_2\\\\x_3&x_4\\end{pmatrix}$, let the following equation:\n\n$$A=a\\begin{pmatrix} 1&0\\\\0&1\\end{pmatrix}+b\\begin{pmatrix} 0&1\\\\1&0\\end{pmatrix}+c\\begin{pmatrix} 1&0\\\\1&1\\end{pmatrix}+d\\begin{pmatrix} 0&1\\\\1&1\\end{pmatrix}$$ Then $$\\begin{pmatrix} 1&0&1&0\\\\0&1&0&1\\\\0&1&1&1\\\\1&0&1&1\\end{pmatrix}\\begin{pmatrix} a\\\\b\\\\c\\\\d\\end{pmatrix}=\\begin{pmatrix} x_1\\\\x_2\\\\x_3\\\\x_4\\end{pmatrix}$$ Check $\\det\\begin{pmatrix} 1&0&1&0\\\\0&1&0&1\\\\0&1&1&1\\\\1&0&1&1\\end{pmatrix}=1$ . So there is an unique solution for $A$, hence $\\beta$ is a basis of $2$ by $2$ matrix, where all basis elements are invertible.\n\n\u2022 Thanks a lot Nick very helpful explanation.\n\u2013\u00a0Jack\nNov 5, 2016 at 16:02\n\u2022 The first line is a bit confusing. You're presenting an example rather than counterexample here, aren't you? Nov 5, 2016 at 17:34\n\u2022 @leftaroundabout: An example is a counterexample to the claim that no examples exist. Nov 5, 2016 at 17:47\n\n\\begin{align*} \\left[ \\begin{matrix} 0 & 1 \\\\ 1 & 1 \\end{matrix} \\right], \\left[ \\begin{matrix} 1 & 0 \\\\ 1 & 1 \\end{matrix} \\right], \\left[ \\begin{matrix} 1 & 1 \\\\ 0 & 1 \\end{matrix} \\right], \\left[ \\begin{matrix} 1 & 1 \\\\ 1 & 0 \\end{matrix} \\right] \\end{align*}\n\n\u2022 Thanks Logician6. I appreciate your help,\n\u2013\u00a0Jack\nNov 5, 2016 at 16:03\n\nI just want to point out that, in order to prove that the answer to a question of the form\n\nDoes there exist a basis for $M_n(\\mathbb{R})$ consisting of [matrices of some special form]?\n\nis \"yes\", you only actually need to show that the special matrices span all of $M_n(\\mathbb{R})$, since any spanning set can be reduced to a basis.\n\nIt's usually a lot easier to check that some special set matrices are spanning than it is to explicitly describe a basis. Let's look at the case you asked about---invertible matrices. Let $A$ be any matrix. Take $\\lambda$ to be a nonzero number which isn't an eigenvalue of $A$ (which can be done because $A$ can have at most $n$ eigenvalues). Then we have $$A = \\lambda I + (A-\\lambda I)$$ where $\\lambda I$ and $A-\\lambda I$ are both invertible. Since every matrix can be written as a sum of two invertibles, the invertibles are spanning, and can, in principal, be reduced to a basis.\n\nAs a more explicit version of Starfall's answer, in $n\\times n$ case, if you denote the standard basis of matrix space by $(E_{i,j})$, you can define a basis simply by $E'_{1,1}=I$, $E'_{i,j}:=I+E_{i,j}$ if one of $i,j$ is not $1$. (You could also take simply $E'_{1,1}:=I+E_{1,1}$, but that would make it harder to show that $(E_{i,j})$ form a basis.)\n\nIt is very easy to check that each $E'_{i,j}$ is invertible (just calculate the determinant), the span of $(E'_{i,j})$ contains all $E_{i,j}$ (this is immediate except for $E_{1,1}$, and for $E_{1,1}$ it follows from the observation for $i=j>1$), and since there are $n^2$ of $(E'_{i,j})$, they must form a basis (because the space of matrices is $n^2$-dimensional).\n\nIn $2\\times 2$ case, the basis is $$\\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix} \\right], \\left[ \\begin{matrix} 1 & 1 \\\\ 0 & 1 \\end{matrix} \\right], \\left[ \\begin{matrix} 1 & 0 \\\\ 1 & 1 \\end{matrix} \\right], \\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 2 \\end{matrix} \\right]$$\n\n(Note that more or less by the same argument as the one cited by Starfall, if you just pick $n^2$ random $n\\times n$ matrices (for any reasonable notion of \"random\", but you can think of normally, independendently distributed coefficients, or independently uniformly distributed in a fixed interval), they will all be invertible and they will be linearly independent, and hence form a basis. This is because the set of non-invertible matrices is very small, as is the set of vectors in a proper subspace of a given real vector space.)", "date": "2022-07-01 05:42:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2000662/does-there-exist-a-basis-for-the-set-of-2-times-2-matrices-such-that-all-basis", "openwebmath_score": 0.9748818278312683, "openwebmath_perplexity": 124.24498568763018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987758723627135, "lm_q2_score": 0.8887588008585925, "lm_q1q2_score": 0.8778792587484664}}
{"url": "https://mathematica.stackexchange.com/questions/37266/even-fibonacci-numbers", "text": "# Even Fibonacci numbers\n\nToday, I found the Euler Project. Problem #2 is\n\nEach new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:\n\n 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...\n\n\nBy considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.\n\nI have two solutions:\n\nFor[i = 1, i < 100, i++, If[Fibonacci[i] > 4000000, Break[]]];\nPlus @@ Select[Fibonacci /@ Range[i - 1], Mod[#, 2] == 0 &]\n\n\n4613732\n\nPlus @@ Select[\nFibonacci /@ Select[Range@100, Fibonacci@# < 4000000 &],\nMod[#1, 2] == 0 &]\n\n\n4613732\n\nHowever, I feel my solutions are not good and efficient. So my question is: can you show me more concise and efficient methods?\n\nAlso, how cam I test whether a number is a Fibonacci number\uff1f\n\n\u2022 Since you already have the answer, you may access the problem 2 forum on the Euler project. Here you will find the most efficient algorithms from other users. The third post there demonstrates an algorithm that can be easily calculated without a program. \u2013\u00a0travisbartley Nov 19 '13 at 5:38\n\nThe other methods described work well, but I am partial to exact answers. So, here it is. First, note that the Fibbonacci sequence has a closed form\n\n$$F_n = \\frac{\\Phi^n - (-\\Phi)^{-n}}{\\sqrt{5}}$$\n\nwhere $\\Phi$ is the GoldenRatio. Also, as observed elsewhere, we are looking only at every third term, which gives the sum\n\n$$\\sum^k_{n = 1} \\frac{\\Phi^{3n} - (-\\Phi)^{-3n}}{\\sqrt{5}}$$\n\nwhich is just the sum of two geometric series. So, after a few steps, this simplifies to\n\n$$\\frac{1}{\\sqrt{5}} \\left[ \\Phi^3\\frac{1 - (\\Phi^3)^{k}}{1-\\Phi^3} + \\Phi^{-3}\\frac{1 - (-\\Phi^{-3})^{k}}{1 + \\Phi^{-3}} \\right]$$\n\nwhere $k$ is the index of the highest even Fibbonacci number. To find $k$, we can reverse the sequence,\n\nn[F_] := Floor[Log[F Sqrt[5]]/Log[GoldenRatio] + 1/2]\nn[4000000]\n(* 33 *)\n\n\nSo, $k = 11$. In code,\n\nClear[evenSum];\nevenSum[(k_Integer)?Positive] :=\nRound @ N[\nWith[\n{phi = GoldenRatio^3, psi = (-GoldenRatio)^(-3)},\n(1/Sqrt[5])*(phi*((1 - phi^k)/(1 - phi)) - psi*((1 - psi^k)/(1 - psi)))\n],\nMax[$MachinePrecision, 3 k] (* needed for accuracy *) ] evenSum[11] (* 4613732 *)  which is confirmed by Total @ Fibonacci @ Range[3, 33, 3] (* 4613732 *)  Edit: The above implementation of n suffers round off problems for large numbers, for example n[9000] gives 21, but Fibonacci[21] is 10946. But, wikipedia gives a better choice Clear[n2]; n2[fn_Integer?Positive] := Floor@N@Log[GoldenRatio, (fn Sqrt[5] + Sqrt[5 fn^2 - 4])/2] n2[10945] (* 20 *)  \u2022 Had to fix the formulas, I dropped quite a few minus signs in the original version. \u2013 rcollyer Nov 18 '13 at 18:04 \u2022 Related to your error, of course facebook.com/physic.department/posts/360869444018997 \u2013 Dr. belisarius Nov 18 '13 at 18:25 \u2022 @belisarius :)) ingenious. \u2013 Stefan Nov 18 '13 at 18:27 \u2022 @belisarius according to my adviser, his adviser was infamous for making multiple canceling errors. His results were correct, the derivation not usually. :) \u2013 rcollyer Nov 18 '13 at 18:31 \u2022 Could you explain how these functions n and evenSum should be used for big numbers? I tried to test your solution unsucessfully. While with mine approach I could simply find sums up to 10^10000. \u2013 Artes Nov 18 '13 at 18:36 Straight iteration over the even valued Fibonacci numbers is fast. fibSum[max_] := Module[ {tot, n, j}, tot = 0; n = 0; j = 3; While[n = Fibonacci[j]; n <= max, j += 3; tot += n]; tot]  Or one can use matrix products. This seems to be about the same speed. It has the advantage of not requiring a built in Fibonacci function. fibSum2[max_] := Module[ {tot, n, fp, mat, mat3}, mat = {{0, 1}, {1, 1}}; tot = 0; n = 0; fp = mat.mat; mat3 = mat.mat.mat; While[n = fp[[2, 2]]; n <= max, fp = fp.mat3; tot += n]; tot]  These handle 10^10000 in slightly under a second on my desktop running Mathematica 9.0.1. Here is a test for whether an integer is a Fibonacci number (a Fib detector?) fibQ[n_] := With[{k = Round[Log[GoldenRatio, N[n, 20 + Length[IntegerDigits[n]]]] + Log[GoldenRatio, Sqrt[5]]]}, n == Fibonacci[k]]  Sum[Fibonacci[n], {n, 3, InverseFunction[Fibonacci][4000000], 3}] (* 4613732 *)  \u2022 That's tyte, + goes w/o saying. \u2013 alancalvitti Oct 30 '14 at 18:18 Here I'd like to show my solution. Since A014445 we know that the generating function of even fibonacci numbers is Fibonacci[3 n], we can write something like this: Plus @@ Select[Table[Fibonacci[3 n], {n, 0, 30}], # < 4000000 &] // AbsoluteTiming => {0., 4613732}  An imperative style solution could look like this: res = 0; n = 1; While[res < 4000000, res += Fibonacci[3 n]; n++]  Result will be in res and timings are negligible, zero. Edit: To bring that into a cogent form. Here the full definition of SumEvenFibonacci: ClearAll[SumEvenFibonacci]; SumEvenFibonacci::usage = \"SumEvenFibonacci[n] calculates the sum of even Fibonacci numbers \\ up to the upper-bound n.\"; SumEvenFibonacci[n_Integer] /; n >= 0 := Module[{res = 0, i = 1}, While[res < n, res += Fibonacci[3 i]; i++]; res ]  The Fibonacci function, is Listable and fast increasing. We can observe a simple pattern, every third element is even (see e.g. Fibonacci[Range[12]]), moreover the first number exceeding 4000000 is: NestWhile[(# + 1) &, 1, Fibonacci[#] <= 4 10^6 &], thus we have: Total[ Fibonacci[ Range[3, 33, 3]]]  4613732  and most likely this is the best approach: sum[n_] := Total[ Fibonacci[ Range[3, NestWhile[(# + 1) &, 1, Fibonacci[#] <= n &] - 1, 3]]]  e.g. sum[10^600]; // AbsoluteTiming  {0.096000, Null}  Concerning testing if given numbers are Fibonacci ones there is a simple test based on checking if consecutive elements generate Pythagorean triples (see e.g. this answer) And @@ (#1^2 + #2^2 == #3^2 & @@@ Array[{ Fibonacci[#] Fibonacci[# + 3], 2 Fibonacci[# + 1] Fibonacci[# + 2], Fibonacci[# + 1]^2 + Fibonacci[# + 2]^2} &, 1000])  True  \u2022 You already got my +1, but, yeah, it looks a little cretinous. \u2013 rcollyer Nov 18 '13 at 21:14 \u2022 @rcollyer I like your approach, but I still cannot use it successfully. I guess Daniel Lichtblau's solution is the most efficient for big numbers, however I can see that another reliable (mine) solution is the least appreciated. \u2013 Artes Nov 18 '13 at 21:20 \u2022 A little poking around indicates that n gives the nearest Fibbonacci number, not the one immediately below it, e.g. n[9000] returns 21, not 20. Likely a similar rounding issue is occurring in evenSum, but it tracks Daniel's exactly for a long time. So, I'm not sure of the specific issue. \u2013 rcollyer Nov 18 '13 at 21:30 Your solution looks fine. You can take advantage of the fact that every third Fibonacci number is even, which makes it a little faster. Fibonaccis are cheap to compute and they quickly exceed 4 million. Here's a comparison: Select[Fibonacci[3 Range@33], # <= 4*^6 &] // Total // AbsoluteTiming  {0.000214, 4613732} Select[Fibonacci@Range@100, # <= 4*^6 && EvenQ@# &] // Total // AbsoluteTiming  {0.000566, 4613732} Plus @@ Select[Fibonacci /@ Select[Range@100, Fibonacci@# < 4000000 &], Mod[#1, 2] == 0 &] // AbsoluteTiming  {0.000717, 4613732} AbsoluteTiming[ For[i = 1, i < 100, i++, If[Fibonacci[i] > 4000000, Break[]]]; Plus @@ Select[Fibonacci /@ Range[i - 1], Mod[#, 2] == 0 &]]  {0.000526, 4613732} I will make use of the formula: sum[x_] := (Fibonacci[3 Quotient[Floor[InverseFunction[Fibonacci][x]], 3]+2]-1)/2  For$x = 4 \\times 10^6\\$ this gives:\n\nsum[4 \u00d7 10^6] //AbsoluteTiming\n\n\n{7.82619, 4613732}\n\nThis is pretty slow. We can speed up the computation of the inverse of the Fibonacci using the InverseFibonacci function from my answer to (157354). Below I give a version of that inverse function that works for arguments greater than 100:\n\nif[x_?(GreaterThan[100])]:=Root[\n{\nFibonacci[#]- x&, SetPrecision[Log[GoldenRatio, x Sqrt[5]],\nMin[2 + Log10[x], 10]]\n}\n]\n\n\nNow, using if above instead of InverseFunction[Fibonacci]:\n\nfast[x_] := (Fibonacci[3 Quotient[Floor @ if[x], 3] + 2] - 1)/2\n\n\nand applying this to the problem:\n\nfast[4 10^6] //AbsoluteTiming\n\n\n{0.000239, 4613732}\n\nWe can use this function on much larger arguments as well:\n\nfast[\n92837410293847120384712309847213048971230498127340912874091273401298374012938741029\n] //AbsoluteTiming\n\n\n{0.002153, 88011840322506983234113472696205625385192191652246095943362996448287672522108009837}\n\nActually this sum can be closed-form analytically:\n\n(* sum of even-valued Fibonacci terms that do not exceed z *)\nsum[z_] := Module[{g, n, f},\ng = (1 + Sqrt[5])/2;\nn = Floor[Log[Sqrt[5] z]/Log[g]];\nf = Round[g^(n + 2)/Sqrt[5]];\n(f - 1)/2]\n\nsum[4*^6] // RepeatedTiming\n(* {0.0000448, 4613732} *)\n\nsum[1*^100] // RepeatedTiming\n(* {0.000045, 12065...3520} *)\n\n\nI used these identities on mathworld: (8), (22), although one should first recognize that only every third term is even.", "date": "2021-03-05 20:40:45", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/37266/even-fibonacci-numbers", "openwebmath_score": 0.7159161567687988, "openwebmath_perplexity": 2165.496798738016, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211604938802, "lm_q2_score": 0.900529782780931, "lm_q1q2_score": 0.877855487909809}}
{"url": "https://math.stackexchange.com/questions/3456797/is-it-acceptable-to-have-a-fraction-in-an-eigenvector", "text": "# Is it acceptable to have a fraction in an eigenvector?\n\nThe professor teaching a class I am taking wants me to find the eigenvalues and the eigenvectors for the following matrix below.\n\n$$\\begin{bmatrix}-5 & 5\\\\4 & 3\\end{bmatrix}$$\n\nI have succeeded in getting the eigenvalues, which are $$\\lambda= \\{ 5,-7 \\}$$. When finding the eigenvector for $$\\lambda= 5$$, I get $$\\begin{bmatrix}1/2\\\\1 \\end{bmatrix}$$. However, the correct answer is $$\\begin{bmatrix}1\\\\2 \\end{bmatrix}$$ .\n\nI have tried doing this question using multiple online matrix calculators. One of which gives me $$\\begin{bmatrix}1/2\\\\1 \\end{bmatrix}$$, and the other gives me $$\\begin{bmatrix}1\\\\2 \\end{bmatrix}$$.\n\nThe online calculator that gave me $$\\begin{bmatrix}1\\\\2 \\end{bmatrix}$$ explains, that y=2, hence $$\\begin{bmatrix}1/2\u00b72\\\\1\u00b72 \\end{bmatrix} = \\begin{bmatrix}1\\\\2 \\end{bmatrix}$$.\n\nWhat I do not understand is, why is y must equal to 2?Is it because there cannot be a fraction in an eigenvector?\n\n\u2022 Both are eigenvectors. Eigenvectors are not unique in general. \u2013\u00a0Arctic Char Nov 30 '19 at 9:32\n\u2022 A constant times an eigenvector is also an eigenvector. \u2013\u00a0Ameet Sharma Nov 30 '19 at 10:06\n\u2022 Conclusion: Any scalar that is multiplied with an eigenvector is still an eigenvector (In the same eigenspace). Even when the eigenvector is multiplied with different scalars, the \u03bb would be the same, Av = \u03bbV. *I cannot choose which comment is the best since all of it helped me in understanding the given problem. Thank you all. \u2013\u00a0Kamarul Adha Nov 30 '19 at 10:36\n\u2022 For some reason most problems I've encountered (in college) have asked me to find eigenvalues and the corresponding eigenvectors, when in reality what we care about are the eigenvalues and the corresponding eigenspaces. \u2013\u00a0BallpointBen Nov 30 '19 at 22:43\n\u2022 @KamarulAdha: Second conclusion: If anyone tells you that $v$ is an eigenvector of a matrix but $2v$ is not, that one does not know mathematics. \u2013\u00a0user21820 Dec 1 '19 at 15:28\n\nBy definition, an eigenvalue $$\\lambda$$ and one corresponding eigenvector $$v$$ must satisfy the following equation:\n\n$$Av = \\lambda v.$$\n\nNow, consider the vector\n\n$$w = \\alpha v,$$\n\nwhere $$\\alpha$$ is a non null number.\n\nThen, notice that:\n\n$$Aw = A(\\alpha v) = \\alpha (Av) = \\alpha (\\lambda v) = \\lambda (\\alpha v) =\\lambda w.$$\n\nTherefore:\n\n$$Aw = \\lambda w,$$\n\nand hence $$w$$ is another eigenvector associated to the eigenvalue $$\\lambda$$.\n\nIn general, it is not true that there is only one eigenvector associated to the eigenvalue $$\\lambda$$. Instead, there is a linear subspace, also known as the eigenspace associated to $$\\lambda$$. Informally, there are infinitely many eigenvectors to $$\\lambda$$, which belong to a certain eigenspace. Given one eigenvector (say $$v$$), then all the multiples of $$v$$ except for $$0$$ (i.e. $$w = \\alpha v$$ with $$\\alpha \\neq 0$$) are also eigenvectors.\n\nThere are matrices with eigenvectors that have irrational components, so there is no rule that your eigenvector must be free of fractions or even radical expressions. As an example:\n\n$$\\begin{bmatrix}1 & 2\\\\1 & 1\\end{bmatrix}$$\n\nWhich has eigenvalues of $$1 \\pm \\sqrt{2}$$ and eigenvectors of $$\\begin{bmatrix}\\pm\\sqrt{2} \\\\ 1\\end{bmatrix}$$\n\n(Additionally, because of the unsolvability of the quintic, there are even eigenvectors that cannot be expressed in elementary functions.)\n\nBecause of the form of the equations that you solve to get the eigenvectors, you have infinite solutions to the eigenvectors. Additionally, an eigenvector is only really valuable as a direction. So if any eigenvector can be said to be the \"correct\" or \"most special\" one, it's the one that has a norm of 1, or a norm of the associated eigenvalue. However, these can be actually a bit ugly to express.\n\nFor our example they are: norm 1, $$\\begin{bmatrix}\\pm\\sqrt{\\frac{2}{3}} \\\\ \\sqrt{\\frac1{3}}\\end{bmatrix}$$ norm $$\\lambda$$: $$\\frac1{3}\\begin{bmatrix}\\mp 2\\sqrt{3} \\pm\\sqrt{6} \\\\ \\sqrt{3} \\pm \\sqrt{6} \\end{bmatrix}$$.\n\nThat being said, it is, of course, nicer to work with whole numbers when the opportunity arises.\n\n$$\\begin{bmatrix}-5 & 5\\\\4 & 3\\end{bmatrix}\\begin{bmatrix}\\frac12\\\\1 \\end{bmatrix}=\\begin{bmatrix}\\frac52\\\\5\\end{bmatrix}=5\\begin{bmatrix}\\frac12\\\\1 \\end{bmatrix}$$\n\nso that $$\\begin{bmatrix}\\frac12\\\\1 \\end{bmatrix}$$ is undisputably an Eigenvector associated to the Eigenvalue $$5$$.\n\nCalculations with whole numbers are easier than fractions, so people often take eigenvectors with whole numbers for convenience.\nFor example, $$(1/2,1)$$ and $$(1,2)$$ both normalize to the same vector, but the calculations give you $$\\left(\\frac{1/2}{\\sqrt{5/4}},\\frac1{\\sqrt{5/4}}\\right)\\text{ and }\\left(\\frac1{\\sqrt5},\\frac2{\\sqrt5}\\right)$$", "date": "2021-06-20 22:54:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3456797/is-it-acceptable-to-have-a-fraction-in-an-eigenvector", "openwebmath_score": 0.964476466178894, "openwebmath_perplexity": 191.53619245914248, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211597623861, "lm_q2_score": 0.9005297787765764, "lm_q1q2_score": 0.8778554833475471}}
{"url": "https://math.stackexchange.com/questions/1669559/tetrahedron-packing-in-cube", "text": "# Tetrahedron packing in Cube\n\nI'm thinking about following solid geometry problem.\n\nQ: Suppose you have a box of \"cube\" shape with edge length 1. Then, How many regular tetrahedrons(with edge length 1) can be in the box?\n\nSo, this is kind of packing problem inside cube. I guess the answer is 3. but I don't know how to prove 3 is the maximum number. Is there any rigorous way to show this?\n\nThanks for any help in advance.\n\n\u2022 Here's a Wiki article that may be of interest: en.wikipedia.org/wiki/Tetrahedron_packing The results, which are based off experiment and simulation, hint that a 0.7 to 0.85 packing fraction might be reasonable. Maybe higher. Given that your tetrahedra have a $V\\approx0.12$, then the answer could be up to $7$! Edit: I say up to seven, since the article does not make it clear when it's considering finite volumes and even then with what sorts of boundaries. Feb 24, 2016 at 2:22\n\u2022 @SunTaek Yes, I was completely looking at the wrong item in the table -- oops! Feb 24, 2016 at 2:25\n\u2022 @zahbaz Thanks for interesting article. hmm...but I cannot construct mental picture of 5 or 6 regular tetrahedron in the unit cube... Feb 24, 2016 at 2:27\n\u2022 @SunTaek. Agreed. I'll stay tuned to this post. This is an interesting problem. Feb 24, 2016 at 2:28\n\u2022 I can see some ways to pack two tetrahedra in a cube, same edge length. i would be interested in a diagram or calculations that prove that three is possible. Feb 24, 2016 at 18:32\n\n1. \"3\" is possible, as shown in following diagram.\nThe vertices of the red tetrahedron are $$\\left(\\frac12,-\\frac12,\\frac12\\right),\\;\\; \\left(\\frac12,-\\frac12,-\\frac12\\right),\\;\\; \\left(\\frac{1}{2\\sqrt{2}},\\frac{1}{2\\sqrt{2}},0\\right)\\;\\;\\text{ and }\\;\\; \\left(-\\frac{1}{2\\sqrt{2}},-\\frac{1}{2\\sqrt{2}},0\\right)$$ The green and blue tetrahedrons can be obtained from the red one by rotating it along the $(1,1,1)$ diagonal for $120^\\circ$ and $240^\\circ$ respectively.\n\n2. I believe \"3\" is the maximum number. Following is a heuristic argument:\n\nLet's say we have $n$ tetrahedrons inside a cube. There are $\\frac{n(n-1)}{2}$ ways of picking a pair of tetrahedron $A, B$ among them.\n\nPick a point $a$ from $A$, a point $b$ from $B$ such that the distance $|a-b|$ is maximized. No matter how I place $A$ and $B$, I always get $|a - b| \\ge 2\\sqrt{\\frac23} \\approx 1.633$.\n\nSince this value is very close to $\\sqrt{3} \\approx 1.732$, the diameter of the cube, the points $a, b$ will be very close to the two end points of a diagonal. This means each pair of tetrahedron will occupy at least one diagonals of the cube.\n\nSince a cube has $4$ diagonals and it seems impossible for different pairs of tetrahedron to share a diagonal, we find: $$\\frac{n(n-1)}{2} \\le 4 \\implies n \\le 3$$\n\n\u2022 amazing! rough, but convincing idea. Thank you So much! By the way, may I ask you which program you used to draw that picture? Feb 25, 2016 at 17:48\n\u2022 @SunTaek I have some script to generate the figure in X3D format, I then use the javascript library x3dom to display and convert it to a picture inside a web browser (firefox). Feb 25, 2016 at 18:06", "date": "2022-10-06 13:12:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1669559/tetrahedron-packing-in-cube", "openwebmath_score": 0.5880131721496582, "openwebmath_perplexity": 473.71666136887035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363540453381, "lm_q2_score": 0.8918110490322426, "lm_q1q2_score": 0.877842036501746}}
{"url": "https://math.stackexchange.com/questions/4039982/how-would-i-evaluate-this-series-using-telescopic-summation", "text": "# How would I evaluate this series using telescopic summation?\n\nThe series is $$\\frac{1}{2!} + \\frac{2}{3!} + \\frac{3}{4!} ... + \\frac{n-1}{n!},$$ which I can write as the sum $$\\sum_{i=2}^n \\frac{i-1}{i!}$$ I will try to evaluate the partial sums to look for a pattern. The first partial sums I get are $$\\frac12, \\frac56 , \\frac{23}{24}$$. Thus, I make the guess that the sum is equal to $$\\frac{n!-1}{n!}$$. I have already proven the base cases through experimentation, so I work on proving the induction step. I assume that $$\\sum_{i=2}^n \\frac{i-1}{i!}= \\frac{n!-1}{n!}$$, for some $$n=k$$ so I get $$\\sum_{i=2}^k \\frac{i-1}{i!}= \\frac{k!-1}{k!}$$ Now I prove this holds for $$n=k+1$$ as well. I get $$\\sum_{i=2}^{k+1} \\frac{i-1}{i!}= \\frac{k!-1}{k!} + \\frac{k}{(k+1)!} = \\frac{(k!-1)(k+1)}{(k+1)!} + \\frac{k}{(k+1)!} = \\frac{(k+1)!-k-1}{(k+1)!} + \\frac{k}{(k+1)!} = \\frac{(k+1)!-1}{(k+1)!}$$ Thus the induction step holds as well, proving my expression is indeed the sum. However, this problem was in the telescoping sums section of my mathbook, so I was wondering how you would solve this sum telescopically since these induction proofs take some time.\n\nWe have:\n\n$$\\sum_{i=2}^{n}\\frac{i - 1}{i!} = \\sum_{i=2}^{n}\\frac{i}{i!} -\\frac{1}{i!} = \\sum_{i=2}^{n}\\frac{1}{(i-1)!} - \\frac{1}{i!}$$\n\n$$=\\bigg(\\frac{1}{1!} - \\frac{1}{2!}\\bigg)+\\bigg(\\frac{1}{2!} - \\frac{1}{3!}\\bigg) +\\bigg(\\frac{1}{3!}-\\frac{1}{4!}\\bigg)+\\dots+\\bigg(\\frac{1}{(n-1)!}-\\frac{1}{n!}\\bigg)$$\n\nTelescoping:\n\n$$=\\frac{1}{1!}-\\frac{1}{n!} = 1- \\frac{1}{n!} = \\boxed{\\frac{n!-1}{n!}}$$\n\nHint\n\n$$\\frac{n-1}{n!}= \\frac{1}{(n-1)!}- \\frac{1}{n!}$$ and therefore\n\n$$\\sum_{k=1}^n \\frac{k-1}{k!} = \\sum_{k=1}^n \\left(\\frac{1}{(k-1)!}- \\frac{1}{k!}\\right) = 1 - \\frac{1}{n!}$$", "date": "2021-04-11 03:41:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4039982/how-would-i-evaluate-this-series-using-telescopic-summation", "openwebmath_score": 0.9912106394767761, "openwebmath_perplexity": 130.5800285425291, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9929882056560155, "lm_q2_score": 0.8840392756357327, "lm_q1q2_score": 0.87784057404297}}
{"url": "https://tex.stackexchange.com/questions/491118/enumerate-with-two-columns", "text": "# Enumerate with two columns\n\nI'd like to have an enumerate with two columns. The first one report the name of the property and the second one can be a multiline text and it should be indented correctly. Thus second columns' text should follow the document's margin.\n\nSomething like this:\n\n\\begin{enumerate}\n\\item[] \\textbf{Diagonal}, if all its elements outside the main diagonal are equal to zero, that is $\\forall i\\neq j$, $a_{ij} = 0$\n\\item[] \\textbf{Scalar}, if all off-diagonal elements are zero and all on-diagonal elements are equal\n\\item[] \\textbf{Identity}, if it's scalar and the elements on its main diagonal are all equal to 1, that is $\\forall i$, $a_{ii} = 1$\n\\item[] \\textbf{Lower triangular}, if all the entries above the main diagonal are zero, that is $\\forall i > j$, $a_{ij} = 0$\n\\item[] \\textbf{Upper triangular}, if all the entries below the main diagonal are zero, that is $\\forall i < j$, $a_{ij} = 0$\n\\end{enumerate}\nBelow, a brief summary of the matrix operations used:\n\\begin{enumerate}\n\\item[] \\textbf{Addition}: If $A$ and $B$ are matrices of the same size then the sum $A$ and $B$ is defined by $C = A + B$, where\n$$c_{ij} = a_{ij}+b_{ij} \\quad \\forall i,j$$\n\\end{enumerate}\n\n\nUsing leandriis' link, this is what I obtained\n\n\"Diagonal\" has \\quad\\quad after it, and I would like to indent all the text in the second column \"Scalar\" is the normal result.\n\n\u2022 Will the list be longer than a single page? If not, you could use a table for this. \u2013\u00a0leandriis May 16 '19 at 11:47\n\u2022 it'd be, actually. It's a list of properties. \u2013\u00a0JackLametta May 16 '19 at 11:50\n\u2022 It sounds like you really want a description environment. If you use the enumitem package it is straightforward to set the widest \"report name\". \u2013\u00a0Andrew May 16 '19 at 11:56\n\u2022 Could you please clarify the desired output? Do you want a number, than the name and then the long text or dou you want your list no be unnumbered? \u2013\u00a0leandriis May 16 '19 at 11:59\n\u2022 When using a description environment, I don't think that @leandriis would recommend writing \\item[] \\textbf{Diagonal} (and likewise for the other list items). Instead, you should write \\item[Diagonal], etc. \u2013\u00a0Mico May 16 '19 at 12:41\n\nA longtable environment may be well suited for your typesetting needs.\n\nYou may see fit to adjust the width of the second column depending on (a) the width of the first column and (b) the overall width of the text block.\n\n\\documentclass{article}\n\\usepackage{array,longtable,ragged2e}\n\\newcolumntype{P}[1]{>{\\RaggedRight\\arraybackslash}p{#1}}\n\n\\begin{document}\n\\noindent\nA matrix is said to be\n\\begin{longtable}{@{} >{\\bfseries}l P{0.6\\textwidth} @{}}\nDiagonal & if all its elements outside the main diagonal are equal to zero, that is $\\forall i\\neq j$, $a_{ij} = 0$\\\\\nScalar & if all off-diagonal elements are zero and all on-diagonal elements are equal\\\\\nIdentity & if it is scalar and the elements on its main diagonal are all equal to 1, that is, $\\forall i$, $a_{ii} = 1$ \\\\\nLower triangular & if all the entries above the main diagonal are zero, that is, $\\forall i > j$, $a_{ij} = 0$ \\\\\nUpper triangular & if all the entries below the main diagonal are zero, that is, $\\forall i < j$, $a_{ij} = 0$\n\\end{longtable}\n\n\\bigskip\\noindent\nBelow, a brief summary of the matrix operations used:\n\\begin{longtable}{@{} >{\\bfseries}l P{0.7\\textwidth} @{}}\nAddition & If $A$ and $B$ are matrices of the same size then the sum of $A$ and $B$ is defined by $C = A + B$, where\n$$c_{ij} = a_{ij}+b_{ij} \\quad \\forall i,j$$\n\\end{longtable}\n\\end{document}\n\n\u2022 That's it! But the bold entries are not tabbed. Could you help me with it, please? \u2013\u00a0JackLametta May 16 '19 at 12:44\n\u2022 and is it possible to have the same space between the elements that it's present in the document between the rows? \u2013\u00a0JackLametta May 16 '19 at 12:45\n\u2022 @JackLametta - Please clarify what you mean by \"the bold entries are not tabbed\" and \" the same space between the elements that it's present in the document between the rows\". Thanks. \u2013\u00a0Mico May 16 '19 at 13:04\n\u2022 I'd like to have the same interspace that there is between rows of the document also between the element of the longtable. \u2013\u00a0JackLametta May 16 '19 at 13:05\n\u2022 @JackLametta - I'm afraid I don't follow you. If the row spacing in the longtable environment is different from that in the rest of your document, it must because you've made some changes to the default line spacing that you haven't told us about so far. \u2013\u00a0Mico May 16 '19 at 13:31", "date": "2020-02-27 08:57:02", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/491118/enumerate-with-two-columns", "openwebmath_score": 0.8608093857765198, "openwebmath_perplexity": 638.3671022409804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9465966762263737, "lm_q2_score": 0.9273632901260831, "lm_q1q2_score": 0.8778390080877045}}
{"url": "https://mathhelpboards.com/threads/another-integral-problem.5446/", "text": "# Another integral problem\n\n#### paulmdrdo\n\n##### Active member\n\u222b(tan3x/sec2x)dx\n\nthis is what i do:\n\n\u222b(tan3xsec-2x)dx\n\u222b(tan2xtanxsec-1xsec-1x)dx\n\u222b(sec2x-1)tanxsec-1xsec-1x)dx\nlet u = secx;\ndu = tanxsecxdx\n\n#### MarkFL\n\nStaff member\nRe: another integ problem.\n\nI think my approach would be to rewrite the integrand as follows:\n\n$$\\displaystyle \\frac{\\tan^3(x)}{\\sec^2(x)}=\\frac{\\sin^3(x)}{\\cos^3(x)}\\cos^2(x)=\\frac{-\\sin(x)\\left(\\cos^2(x)-1 \\right)}{\\cos(x)}$$\n\nNow you should see a $u$-substitution that will work quite nicely.\n\n#### paulmdrdo\n\n##### Active member\nRe: another integ problem.\n\nI think my approach would be to rewrite the integrand as follows:\n\n$$\\displaystyle \\frac{\\tan^3(x)}{\\sec^2(x)}=\\frac{\\sin^3(x)}{\\cos^3(x)}\\cos^2(x)=\\frac{-\\sin(x)\\left(\\cos^2(x)-1 \\right)}{\\cos(x)}$$\n\nNow you should see a $u$-substitution that will work quite nicely.\ni used that technique makfl and i get the right answer. but for educational purposes i want to know another way of solving that.\n\nthis what i do a while ago.\n\n\u222b(tan3x/sec2x)dx\n\u222b(tan2xtanx/sec2x)dx\n\n\u222b(sec2x-1)tanx)/sec2x)dx\n\u222b(sec2xtanx-tanx)/sec2x)dx\n\u222b[(sec2xtanx)/sec2x - tanx/sec2x]dx\n\u222btanxdx - \u222b(tanx/sec2x)dx\n\nln|secx| - \u222b(sinx/cosx*cos2x)dx\nln|secx| - \u222bsinxcosxdx\n\nu= sinx; du = cosxdx\n\nln|secx| - \u222budu\nln|secx| - u2/2\nln|secx| - sin2x/2 + C\n\nis my solution correct..it is different with the one i get when i used the technique you suggested.\n\n#### MarkFL\n\nStaff member\nRe: another integral problem.\n\nYes, your solution is also correct, and many times when we find differing forms for an indefinite integral, we find that they (without the constant of integration) differ only by a constant, which of course is allowed.\n\nSo, let's look at the two solutions without the constants:\n\n(1) $$\\displaystyle \\frac{1}{2}\\cos^2(x)-\\ln|\\cos(x)|$$\n\n(2) $$\\displaystyle \\ln|\\sec(x)|-\\frac{1}{2}\\sin^2(x)$$\n\nFirst, we should observe that:\n\n$$\\displaystyle \\ln|\\sec(x)|=\\ln|(\\cos(x))^{-1}|=-\\ln|\\cos(x)|$$\n\nSo, if we subtract one solution from the other, we should obtain a constant. Subtracting (2) from (1), we get:\n\n$$\\displaystyle \\frac{1}{2}\\cos^2(x)+\\frac{1}{2}\\sin^2(x)$$\n\nUsing a Pythagorean identity, we now have:\n\n$$\\displaystyle \\frac{1}{2}$$\n\nSo, we see that the two solutions differ by a constant, which means they are equivalent as an anti-derivative.\n\nPerhaps a simpler way to look at it is to take the solution you obtained, and rewrite it:\n\n$$\\displaystyle \\ln|\\sec(x)|-\\frac{1}{2}\\sin^2(x)+C=-\\ln|\\cos(x)|-\\frac{1}{2}\\left(1-\\cos^2(x) \\right)+C=\\frac{1}{2}\\cos^2(x)-\\ln|\\cos(x)|+\\left(C-\\frac{1}{2} \\right)$$\n\nSince an arbitrary constant less one half, is still an arbitrary constant, we get:\n\n$$\\displaystyle \\frac{1}{2}\\cos^2(x)-\\ln|\\cos(x)|+C$$\n\n#### paulmdrdo\n\n##### Active member\nRe: another integral problem.\n\nYes, your solution is also correct, and many times when we find differing forms for an indefinite integral, we find that they (without the constant of integration) differ only by a constant, which of course is allowed.\n\nSo, let's look at the two solutions without the constants:\n\n(1) $$\\displaystyle \\frac{1}{2}\\cos^2(x)-\\ln|\\cos(x)|$$\n\n(2) $$\\displaystyle \\ln|\\sec(x)|-\\frac{1}{2}\\sin^2(x)$$\n\nFirst, we should observe that:\n\n$$\\displaystyle \\ln|\\sec(x)|=\\ln|(\\cos(x))^{-1}|=-\\ln|\\cos(x)|$$\n\nSo, if we subtract one solution from the other, we should obtain a constant. Subtracting (2) from (1), we get:\n\n$$\\displaystyle \\frac{1}{2}\\cos^2(x)+\\frac{1}{2}\\sin^2(x)$$\n\nUsing a Pythagorean identity, we now have:\n\n$$\\displaystyle \\frac{1}{2}$$\n\nSo, we see that the two solutions differ by a constant, which means they are equivalent as an anti-derivative.\n\nPerhaps a simpler way to look at it is to take the solution you obtained, and rewrite it:\n\n$$\\displaystyle \\ln|\\sec(x)|-\\frac{1}{2}\\sin^2(x)+C=-\\ln|\\cos(x)|-\\frac{1}{2}\\left(1-\\cos^2(x) \\right)+C=\\frac{1}{2}\\cos^2(x)-\\ln|\\cos(x)|+\\left(C-\\frac{1}{2} \\right)$$\n\nSince an arbitrary constant less one half, is still an arbitrary constant, we get:\n\n$$\\displaystyle \\frac{1}{2}\\cos^2(x)-\\ln|\\cos(x)|+C$$\ni just want to ask how is ln|secx| = -ln|cosx|... the rest of explanation is enligthening. thanks a lot!\n\n#### MarkFL\n\nStaff member\nRe: another integral problem.\n\nRecall the logarithmic property:\n\n$$\\displaystyle \\log_a\\left(b^c \\right)=c\\cdot\\log_a(b)$$\n\nand the definition:\n\n$$\\displaystyle \\sec(x)\\equiv\\frac{1}{\\cos(x)}=(\\cos(x))^{-1}$$\n\n#### soroban\n\n##### Well-known member\nRe: another integ problem.\n\nHello, paulmdrdo!\n\nThe integral $$\\int \\sin x\\cos x\\,dx$$ is famous\nfor having different (but equivalent) answers.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\n$$1.\\: \\int \\sin x(\\cos x\\,dx)$$\n\nLet $$u \\,=\\,\\sin x \\quad\\Rightarrow\\quad du \\,=\\,\\cos x\\,dx$$\n\nSubstitute: .$$\\int u\\,du \\:=\\:\\tfrac{1}{2}u^2+C$$\n\nBack-substitute: .$$\\tfrac{1}{2}\\sin^2x + C$$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\n$$2.\\;\\int \\cos x(\\sin x\\,dx)$$\n\nLet $$u = \\cos x \\quad\\Rightarrow\\quad du = \\text{-}\\sin x\\,dx \\quad\\Rightarrow\\quad \\sin x\\,dx = \\text{-}du$$\n\nSubstitute: .$$\\int u(\\text{-}du) \\:=\\:\\text{-}\\int u\\,du \\:=\\:\\text{-}\\tfrac{1}{2}u^2+C$$\n\nBack-substitute: .$$\\text{-}\\tfrac{1}{2}\\cos^2x + C$$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\n$$3.\\;\\int\\sin x\\cos x\\,dx \\:=\\:\\tfrac{1}{2}\\int2\\sin x\\cos x\\,dx$$\n\n. . . $$=\\;\\tfrac{1}{2}\\int\\sin2x\\,dx \\:=\\:\\text{-}\\tfrac{1}{4}\\cos2x + C$$", "date": "2021-01-17 08:55:10", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/another-integral-problem.5446/", "openwebmath_score": 0.9638364911079407, "openwebmath_perplexity": 511.816133713477, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105287006256, "lm_q2_score": 0.8991213799730775, "lm_q1q2_score": 0.8778216698475514}}
{"url": "http://chwz.monetimargherita.it/latex-vector-notation.html", "text": "Latex Vector Notation\n\nFirst put the following in your preamble (text copyable below the colored stuff):. Every command has a specific syntax to use. This will always be the case when we are dealing with the contours of a function as well as its gradient vector field. Usually you need to mark which Pawn is being captured, but ignore this for sake of simplicity. LaTeX tutorial using Texmaker. The indefinite integral is commonly applied in problems involving distance, velocity, and acceleration, each of which is a function of time. If the TeX Notation filter is activated, which set a LaTeX renderer, the same equation as above is obtained with the following control sequence:. It is important to keep them in a vector form so they can viewed or printed at any resolution without loss of quality. It also shows how to write a unit vector - the arrow that goes on the top of the letters representing the vector. or in abbreviated vector notation where u is the velocity vector and \u2207. The main difference between the various types of matrix is the kind of delimeters that surround them. Basically, it is a $\\LaTeX$ handwritten symbol recognition. Scroll down the page for examples and solutions. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. (1)Pisto des (1988): \\Algebra. A matrix is a two-dimensional array of values that is often used to represent a linear transformation or a system of equations. Retrieved from \"http://ufldl. The arrow vector notation is the standard in LaTeX, LaTeX Vectors With Hats. MATLAB recognizes vector operations if we use the vector notation. I am not 100% sure, but this is associated in my memory with J. Sum-class symbols, or accumulation symbols, are symbols whose sub- and superscripts appear directly below and above the symbol rather than beside it. Vector Paint is a drawing program, with a simple yet powerful interface, that lets you draw points, lines, curves and shapes. In L a T e X, subscripts and superscripts are written using the symbols ^ and _, in this case the x and y exponents where written using these codes. Using the Hadamard product, we can write the expression as. Unit vector: \\vec { [1] } The documentation for Google Docs Equation editor (based on the Google Chat API) is not really extensive so most of the time the CodeCogs Online LaTex Equation Editor is the place to start and test if a LaTex expression from that editor also works in the Google Docs Equation Editor; they also have a link to better. One of the interesting things I noticed in the article is that the students prefer to type up their homework using LaTeX. Special Matrices and Matrix Properties. What is the usual way to do this? Of course, I think it would be best to state my notation somewhere in the beginning of paper. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. For non-Greek stuff, see below. I have thought of \\left( \\stackrel{n}{r} \\right), but the effect is obviously ugly. Special matrices are represented by special notation. linear algebra or $X,Y$ in e. Notation: We often represent a vector by some letter, just as we use a letter to denote a scalar (real number) in algebra. Vector notation is a commonly used mathematical notation for working with mathematical vectors, which may be geometric vectors or members of vector spaces. Used to be that all vectors would be bold-face and unit vectors would be denoted by the hat as well if you were to write it in LaTeX (for example) you used to have to write \\hat{\\vec{r}} with the \\vec{r} giving you the boldface. Here, the vector notation, models the phenotypes of all the individuals in the dataset, denoted as a column vector. I have not been able to find a resource to tell me the standard notation for a normalized scalar value. Macros are. b) Given two points P; Q, the vector from P to Q is denoted PQ. This is not a comprehensive list. com Knowledge base dedicated to Linux and applied mathematics. Typesetting mathematics is one of LaTeX's greatest strengths. If one of the vector is identity vector, then the result is the sum of all the elements in the other one. Notation in which you express the x component as i and the y component as j, and you add them. The notation is sometimes more e\ufb03cient than the conventional mathematical notation we have been using. Denote by the probability of an event. LaTeX Source of Example 1. \\end{document} (Note that this is not to be taken as an example of good mathematical exposition. The Wolfram Language has the world's largest collection of consistent multifont mathematical notation characters$LongDash]all fully integrated into both typesetting and symbolic expression construction. Basically if a number is bigger than 10^4 or smaller than 10^{-4} (applies to the absolute values of negative numbers too), it will be denoted in scientific notation. The main difference between the various types of matrix is the kind of delimeters that surround them. Special Matrices and Matrix Properties. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I am trying to understand the Matrix and Vector Notations on page 2 here: (the page is also pasted below, to make it easier to explain the problem). The Comprehensive LATEX Symbol List Scott Pakin \u2217 8 October 2002 Abstract This document lists 2590 symbols and the corresponding LATEX commands that produce them. * y, in numpy x*y), producing a new vector of same. since I am writing blog post that hosted by Github with Editor Atom , and use plugin markdown-preview-plus and mathjax-wrapper , and use mathjax Javascript display the math symbols on the web page. This week, we will go into some of the heavier. The Markdown parser included in the Jupyter Notebook is MathJax-aware. No - as the wikipedia article (see link above) points out there is no standard (i. How to Enter Answers in WeBWorK Be careful entering powers of trigonometric, and other, functions. php/Backpropagation_Algorithm\". In certain cases it may be desirable to include \u201cnormal text\u201d within an equation. So are 3*4 and 3 4 (3 space 4, not 34) but using an explicit multiplication symbol makes things clearer. LaTeX representation of hollow \"R\" symbol for real space? Post by clayton \u00bb Wed May 19, 2010 8:33 pm In LaTeX, how do I represent the hollow \"R\" symbol that designates the real number space?. \\endgroup \u2013 KCd Apr 17 at 8:04. In the discussion of the applications of the derivative, note that the derivative of a distance function represents instantaneous velocity and that the. I have created a five video YouTube playlist Geometric Calculus, about 53 minutes in all, taken from the book. Again, the effect of the th variant on the phenotype is , the mean is , and the contribution of the environment on the phenotype is denoted by. I am not 100% sure, but this is associated in my memory with J. how the components of a tensor (in the above example, the tensor components would be the \u03c3ij) transform under a change of reference frame. Tilde notation. In the following document, we will refer to special characters for all symbols other than the lowercase letters a-z, uppercase letters A-Z, figures 0-9, and English punctuation marks. The element-wise squaring makes the vector notation a bit more obscure in terms of everyday linear algebra operators. \\left and \\right can dynamically adjust the size, as shown by the next example:. This list of mathematical symbols by subject shows a selection of the most common symbols that are used in modern mathematical notation within formulas, grouped by mathematical topic. Notice that to insert the parentheses or brackets, the \\left and \\right commands are used. The derivative of a real function f may be indicated with differential notation or simply as f \u2032. If one of the vector is identity vector, then the result is the sum of all the elements in the other one. , integrals containing more than one integral sign) one finds that LaTeX puts too much space between the integral signs. If your document requires only a few simple mathematical formulas, plain LaTeX has most of the tools that you will need. \\endgroup \u2013 Edward Xu Mar 14 at 18:14. De nition 1 (Naive De nition of a Set). 21 Sep 11 \u2013 fixed some typos and the vector notation 22 Sep 11 \u2013 fixed import of sklearn according to the new 0. Its done! Actually, the default interpreter in MATLAB for legend is 'tex', I guess. It all begins by writing the inner product. Scientific notation in LaTex? Hi, im engineering student therefore using allot of scientific notations. For example (1+2)* (3+4) and (1+2) (3+4) are both valid. functional analysis. This post highlights vector calculus and vector analysis. Software implementation Edit Plate notation has been implemented in various TeX / LaTeX drawing packages, but also as part of graphical user interfaces to Bayesian statistics programs such as BUGS and BayesiaLab. If A is a complex number a+b\u03af, then A+1 returns a + 1 + b\u03af. This supports my suspicion that you should check whether your LaTeX installation works. For the posted code, you can achieve this by using Part. Is a math Book. Tensor products of vector spaces are to Cartesian products of sets as direct sums of vectors spaces are to disjoint unions of sets. Specify vectors in Cartesian or polar coordinates, and see the magnitude, angle, and components of each vector. -10 -5 0 5 10 -10 -5 0 5 10 -0. But let me go over the different ways you could represent a vector. This lea\ufb02et provides information on symbols and notation commonly used in mathematics. In Unicode , it is the character at code point U+2207, or 8711 in decimal notation. We write $$\\sim g(n)$$ to represent any quantity that, when divided by $$f(n)$$, approaches $$1$$ as $$n$$ grows. How to Enter Answers in WeBWorK Be careful entering powers of trigonometric, and other, functions. Our aim in this subsection is to give you a storehouse of examples to work with, to become comfortable with the ten vector space properties and to convince you that the multitude of examples justifies (at least initially) making such a broad definition as Definition VS. the weight of the car times the coefficient of kinetic friction. The Vector class implements vector quantities in $${\\bf R}^n$$ for arbitrary $$n$$. VECTOR CALCULUS: USEFUL STUFF Revision of Basic Vectors A scalar is a physical quantity with magnitude only A vector is a physical quantity with magnitude and direction A unit vector has magnitude one. Symbols for Preference Relations Unicode Relation Hex Dec Name LA\u03a4\u0395\u03a7 \u227b U+227b 8827 SUCCEEDS \\succ Strict Preference P U+0050 87 LATIN CAPITAL LETTER P P > U+003e 62 GREATER-THAN SIGN. Summary: Possibilities for the Solution Set of a System of Linear Equations Problem 288 In this post, we summarize theorems about the possibilities for the solution set of a system of linear equations and solve the following problems. Hi, What mathematical symbol is the one where it has a cross in a circle? Is it just multiplication? I've never seen it before. And, yes, it turns out that \\curl \\dlvf is equal to \\nabla \\times \\dlvf. Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Ask Question Asked 4 years, 1 month ago. It also shows how to write a unit vector - the arrow that goes on the top of the letters representing the vector. I realized that I didn't answer the question. Equation 30-3 gives Bself = \u00b5 0I=2a and B other = \u00b5 0Ia 2=2(a 2 +x2)3=2. How to write vector equation in Word Wide Spectrum. \\begingroup Every time I have seen that notation used and unexplained, I've been confused by it. Some examples from the MathJax demos site are reproduced below, as well as the Markdown+TeX source. It is important to keep them in a vector form so they can viewed or printed at any resolution without loss of quality. However, instead of measuring this distance on the number line, a complex number's absolute value is measured on the complex number plane. Notice that the vectors of the vector field are all orthogonal (or perpendicular) to the contours. LaTeX is capable of displaying any mathematical notation. How can I pick up the argmin of a vector? I. lecture1_2 - Free download as PDF File (. But get G M_1 bold as well = bad. The way the curly bracket is facing indicates the direction of the hug. This one is a tricky one, you need to use the (surprisingly powerful) Microsoft Equation Editor which should be under the insert tab. In this section we will give a brief review of matrices and vectors. 870 Ct VERY RARE UNHEATED NR' CEYLON BLUE SAPPHIRE ROSE CUT NO RESERVE PRICE. The curl, on the other hand, is a vector. , 1 \\times 10^{-2} or 4 \\times 10^{-35}. \\begingroup Every time I have seen that notation used and unexplained, I've been confused by it. Thanks mates, But I could not get the response , how do I use vector notation, I am not more familiar with maths notations. @duffymo Regardless of how you thing of LaTeX, this is a pretty good question. We use tilde notation to develop simpler approximate expressions. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. Notation in econometrics: a proposal for a standard 77 he means (5 + \u221a \u221215)(5 \u2212 \u221a \u221215)= 25 \u2212 (\u221215)= 40, see Kline (1972, p. Integral expression can be added using the \\int_{lower}^{upper} command. use the following search parameters to narrow your results: subreddit:subreddit find submissions in \"subreddit\" author:username find submissions by \"username\" site:example. Asking for help, clarification, or responding to other answers. The frequently used left delimiters include (, [ and {, which are obtained by typing (, [and \\{respectively. Basic LaTeX Julie Mitchell This resource was adapted from notes provided by Jerry Marsden 1 Basic Formatting 1. I dont think you are talking about f' for deeivative of f. Visit the Learning Center. Note: LaTeX is case sensitive. In conjunction with the previous tutorial on math, by the end of the session, you will be in pretty good shape to write almost anything that your work will require. 5 (d) 8 (e) 15. The notation is sometimes more e\ufb03cient than the conventional mathematical notation we have been using. Basics of LaTeX for scienti\ufb01c papers and reports1 Sne\u02c7zana Stanimirovi\u00b4c, November 21, 2016 1. Vector notation is a written method for representing quantities that possess both direction and magnitude, such as acceleration. Also, I would like a discussion of how this was developed to be able to express physics/engineering equations independently of the coordinate system. the normal force of the road times the coefficient of static friction. For example, a standard integral in LaTeX looks like  \\int_a^b \\! f(x) \\, \\mathrm{d}x. But once you have it broken up into the x and y components, you can just separately add the x and y components. Matrix Calculator. Retrieved from \"http://ufldl. inequality symbols in math equality symbols math equal sign vector icon equality sign mathematical sum symbol stock vector equality mathematical. or in abbreviated vector notation where u is the velocity vector and \u2207. In this paper we attempt to harmonize the various practices in econometrics notation. It also shows how to write a unit vector - the arrow that goes on the top of the letters representing the vector. The 18th and 19th centuries saw the creation and standardization of mathematical notation as used today. uk 1994/09/16 Abstract This package provides a set of new commands for representing vectors in various ways. 3 posts \u2022 Page 1 of 1. The second vector arrow above the symbol is substantially larger and it does not look good, especially when compared to the TeX/LaTeX result. Every nonzero vector has a corresponding unit vector, which has the same direction as that vector but a magnitude of 1. Dot notation for time derivatives (e. Our aim in this subsection is to give you a storehouse of examples to work with, to become comfortable with the ten vector space properties and to convince you that the multitude of examples justifies (at least initially) making such a broad definition as Definition VS. Similarly, \\R^2 is a two-dimensional vector, and \\R^3 is a three-dimensional vector. 5 How do you look at any of these coordinate systems and determine. The French Talbot company was reorganised by Anthony Lago (1893-1960) and after that, the Talbot-Lago name was used but on the home market the cars bore a Talbot badge. Typically this is the LaTeX code for the symbol. Calculates transpose, determinant, trace, rank, inverse, pseudoinverse, eigenvalues and eigenvectors. That list also includes LaTeX and HTML markup, and Unicode code points for each symbol (note that this article doesn't have the latter two, but they could certainly be added). You write (sin(t))^2 for the square of sin(t), and never sin^2t. Latex Bold Vector. Three Styles for LaTeX Vector Notation LaTeX Vector With an Arrow. However, instead of measuring this distance on the number line, a complex number's absolute value is measured on the complex number plane. As you will see, these behave in a fairly predictable manner. In vector addition, the zero vector is the additive identity vector: v + O = v. You can get the same file with more text here. You can choose to load either of them: \\usepackage{amsfonts} % or \\usepackage{amssymb}. We've documented and categorized hundreds of macros!. In this chapter we will tackle matters related to input encoding, typesetting diacritics and special characters. For Greek letters, put \\boldsymbol in front of the Greek letter and you'll get the bold vector notation. 5 How do you look at any of these coordinate systems and determine. Latex-Math-Snippets. The normalized vector of is a vector in the same direction but with norm (length) 1. On a mistake, update as follows:. Thus, for example, the multiple integral is obtained by typing. Using R Markdown for Class Reports. Topics include: limits, integrals, summation notation, and vectors. It further allows the same raising and lowering operators to be used for many different roles according to the term they apply to. since I am writing blog post that hosted by Github with Editor Atom , and use plugin markdown-preview-plus and mathjax-wrapper , and use mathjax Javascript display the math symbols on the web page. Thanks mates, But I could not get the response , how do I use vector notation, I am not more familiar with maths notations. This guide provides an overview of how to get started with LaTeX, as well as resources and exercise to help new users of the program. For example, if you type \\sqrt2 and click this icon, you'll get 2 in your document without opening MathType. For math, science, nutrition, history. Here is how LaTeX typesets the above source file. Vectors are elements of a vectorspace, depending on context they are commonly called $V,W$ in e. I have read the vector examples page on Wolfram Alpha's homepage, but I didn't know which method to use. Easy Scientific Notation In LaTeX filed in LaTeX , Math , Physics on Mar. Notebook documents \u00b6 Notebook documents contains the inputs and outputs of a interactive session as well as additional text that accompanies the code but is not meant for execution. This is a repository of latex snippets for sublime. But MathJax can also be configured to use HTML-CSS (for legacy browsers), SVG, and native MathML rendering when available in a browser. We generally use the coordinates x, y, and z to locate a particle at point P(x, y, z) in three dimensions. There is a shorthand for the last two commands: if you are using colon notation to access elements of a vector, you can use the end command to denote the end of the vector. As you see, the way the equations are displayed depends on the delimiter, in this case \\[$ and . Unit vector: \\vec { [1] } The documentation for Google Docs Equation editor (based on the Google Chat API) is not really extensive so most of the time the CodeCogs Online LaTex Equation Editor is the place to start and test if a LaTex expression from that editor also works in the Google Docs Equation Editor; they also have a link to better. Recall the notation that $\\R$ stands for the real numbers. As does the notation $\\mathbf i, \\mathbf j, \\mathbf k$ for the standard basis vectors in $\\mathbf R^3$. The Ipe extensible drawing editor is a free vector graphics editor for creating figures in PDF or EPS format. It's possible to typeset integrals, fractions and more. , integrals containing more than one integral sign) one finds that LaTeX puts too much space between the integral signs. function , latex. Sum-class symbols, or accumulation symbols, are symbols whose sub- and superscripts appear directly below and above the symbol rather than beside it. This is the 2nd post in the series. Equation 30-3 gives Bself = \u00b5 0I=2a and B other = \u00b5 0Ia 2=2(a 2 +x2)3=2. Vector notation is a commonly used mathematical notation for working with mathematical vectors, which may be geometric vectors or members of vector spaces. \u02d9, \u00a8) can be obtained by following a letter variable with \"\\dot\" for a first derivative and \"\\ddot\" for a second derivative. As it is virtually impossible to list all the symbols ever used in mathematics, only those symbols which occur often in mathematics or mathematics education are. Recall Ax = Acos2 and Ay = Asin2. 14 \\times 10^5$or 3. Vector notation was introduced to physics in the end of 19-th century. ISO standard) notation. You might want to look at a few careful expositions to get a sense of the normal range of notational variation. NET 3d advocacy AMD64 asdf billboard blogging Books cluster code dave winer essay example fix Flash fox and geese game glfw graphics haskell Humor KPAX LaTeX LaTeX vector notation life Linux Mathcad monads Neal Stephenson opengl OSS Papervision3D Physics Pizza Hut puppet repa SBCL scientific notation sci fi SLIME thesis vector whedon Wine wxCL. Calculations. Problem: For equation (2), I think it should b. When you then click on the blue downarrow at the lower right, five options appear. Vectors are a very important concept in science and math and appear very early in calculus and physics classes in college or in advanced high school classes. When the currents are in the same directions, the fields are parallel and oB s= B elf +B ther. Assuming w=f(x,y,z) and u=, we have Hence, the directional derivative is the dot product of the gradient and the vector u. If A is a complex number a+b\u03af, then A+1 returns a + 1 + b\u03af. As an undergraduate, I clearly remember learning and using \"hat notation\" to describe unit vectors. Latex-Math-Snippets. Viewed 10k times 7. This is an index-notation question rather then the NS one: For incompressible flow and Newtonian fluid, the continuity equation is denoted with: $$\\frac{\\partial u_i}{\\partial x_i} = 0,$$ which. Sign up to join this community. How do you represent vectors. How to write vector equation in Word Wide Spectrum. With the x-axis due east and the y-axis due north, what is the displacement in unit vector notation for the bird?. 1/2 (c) 2 2 (d) 1 0 0. , integrals containing more than one integral sign) one finds that LaTeX puts too much space between the integral signs. \u201cSave as new equation\u2026\u201d allows you to keep the equation as a building block, which makes it available from the \u201cInsert\u201d ribbon. Tilde notation. Thus, for example, the multiple integral is obtained by typing. 8 m/s2 and ay = -2. LaTeX code for combination? aside from the conventional${_n}C_r$, how can I input \"n taken r\" using the more common parenthesis notation. Putting a bar or a tilde over a letter in LaTeX. We write $$\\sim g(n)$$ to represent any quantity that, when divided by $$f(n)$$, approaches $$1$$ as $$n$$ grows. the weight of the car times the coefficient of kinetic friction. \\documentclass{article} \\usepackage{amsmath} \\begin{document} Let$\\boldsymbol\\alpha$be a vector. Subsection EVS Examples of Vector Spaces. A LaTeX command begins with the command name, which consists of a \\ followed by either (a) a string of letters or (b) a single non-letter. Can we get something like this going on Stack Overflow? I think it'd be appropriate as I at least. \\end{document} (Note that this is not to be taken as an example of good mathematical exposition. Problem: For equation (2), I think it should b. [email protected]. Windows applications that create vector graphics (such a graphing and diagramming tools) usually allow you to export graphics in either the Encapsulated Postscript (EPS) format or as a Windows metafile (EMF or WMF). To find the unit vector u of the vector you divide that vector by its magnitude as follows: Note that this formula uses scalar multiplication, because the numerator is a vector and the denominator [\u2026]. How do you represent vectors. Online WYSIWYG Mathematics Editor (Equation Editor, Latex Editor), fast and powerful Editing features, inputting Normal text, Math symbols, and drawing Graph/Diagram in one single editor, help writing Math Document much easier. The direction of a vctor V is the unit vector U parallel to V: U = V j V. The FontName, FontWeight, and FontAngle properties do not have an effect. There is no doubt that the development of good notation has been of great importance in the history of mathematics. \u2013 \u201cToggle between TeX/LaTeX and MathType equations\u201d (not available on the Mac) \u2013 This is a huge shortcut if you know the TeX language. It allows you to specify a variable for the offset, but the width must be constant. When you reference a vector with a colon, such as v (:), all the components of the vector are listed. com Knowledge base dedicated to Linux and applied mathematics. Number sets such as natural numbers or complex numbers are not provided by default by LaTeX. news & information AoPS News AoPS Wiki. Home > Latex > FAQ > Latex - FAQ > How to write a vector in Latex ? \\vec,\\overrightarrow. A free-body diagram is a special example of the vector diagrams that. The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15. As you will see, these behave in a fairly predictable manner. We use tilde notation to develop simpler approximate expressions. As it is virtually impossible to list all the symbols ever used in mathematics, only those symbols which occur often in mathematics or mathematics education are. Question: Q: Physics font-vector notation I'm using Keynote in my Grade 11 physics class. of design science. 1/2 (b) 1/2. How can I code in R to get such regular scientific notation format as the xtable output?. Latex-Math-Snippets. Is it possible to tell in index notation whether a vector is a row or column vector, or is that supposed to be clear based on context?. Equation 30-3 gives Bself = \u00b5 0I=2a and B other = \u00b5 0Ia 2=2(a 2 +x2)3=2. \u2020 as before, open WinEdt and write the following: \\documentclass. In a vector such as a=[2 3 1 4 5], I want to pick up 1 by min(a) and also want to have 3 that is the Nth element of this vector with the minimum 1. It could be either a row vector or a column vector. Normalized vectors (i. Performs LU, Cholesky, QR, Singular value. , while the elements of a set will usually be given lower-case letters, like x, y, z, v, etc. The ability to easily include mathematical notation within markdown cells using LaTeX, and rendered natively by MathJax. Two arguments for addition taking precedence over (being done before) the max or min: In a max-plus algebra the addition plays the role of multiplication, and is denoted as such, while the$\\max$plays the role of addition and is denoted as such thus since multiplication takes precendce over addition this translates to addition taking precedence over$\\max$; see Max-plus algebra. Topics include: limits, integrals, summation notation, and vectors. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and, in effect, increases the mental power of the race. clude annotation such as In the history of mathematics, these symbols have denoted numbers, shapes, patterns, and change. Also let\u2019s auto-matically scale all examples x to have (Euclidean) length 1, since this doesn\u2019t a\ufb00ect which side of the plane they are on. This is the 2nd post in the series. An online LaTeX editor that's easy to use. 0 m and b =-13. Our aim in this subsection is to give you a storehouse of examples to work with, to become comfortable with the ten vector space properties and to convince you that the multitude of examples justifies (at least initially) making such a broad definition as Definition VS. I realized that I didn't answer the question. The length is denoted j V. Three vector approach. Sign up to join this community. But get G M_1 bold as well = bad. So can be viewed as the imanginary part of. In unit-vector notation, what is the velocity of the cart when it reaches its greatest y coordinate? - Slader A cart is propelled over an xy plane with acceleration components ax = 4. \\documentclass{article} \\usepackage{amsmath} \\begin{document} Let$\\boldsymbol\\alpha$be a vector. 1 Compute the lengths of the following vectors: (a) 0 0. An \"overdot\" is a raised dot appearing above a symbol most commonly used in mathematics to indicate a derivative taken with respect to time (e. Pos tentang vector notation latex yang ditulis oleh blogkarinagoo. But MathJax can also be configured to use HTML-CSS (for legacy browsers), SVG, and native MathML rendering when available in a browser. You cannot avoid mathematical notation when reading the descriptions of machine learning methods. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. Normalized vectors (i. set margin latex latex matrices First the basic environments which could be used for a matrix, all of them are provide by usepackage amsmath. Once you're in equation mode, you can mess around with the GUI and try to look for it, or just type: (abc)/vec The / you might recognize from latex, it works usually the same here, it indicates a keyword is about to happen, and it will apply that operation to. Let $$X$$ be the stochastic process described by this chain (remember, a stochastic process is just a random variable that evolves through time). The matrix environments are matrix, bmatrix, Bmatrix, pmatrix, vmatrix, Vmatrix, and smallmatrix. In vector addition, the zero vector is the additive identity vector: v + O = v. It is a sequel to my Geometric Algebra playlist. Anyone who works with LaTeX knows how time-consuming it can be to find a symbol in symbols-a4. Associated with this quantity is the general relationship and when the Lagrangian is invariant with respect to time translation the energy will be a conserved quantity (and also the Hamiltonian). You have to use parentheses; it will not accept \"<2,-3>\" notation. A more complete primer can be found here. Currently to get for example 2x10 -3 i am typing as 2 \\times 10 {-3}. We use tilde notation to develop simpler approximate expressions. ,$1 \\times 10^{-2}$or$4 \\times 10^{-35}$. inequality symbols in math equality symbols math equal sign vector icon equality sign mathematical sum symbol stock vector equality mathematical. The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15. (4) Potential. Displacement Vector. A vector of length one is said to be a unit vector. 1 Vectors and vector notation 355B Guided practice worksheet 4 Using Resource sheet 18. This chapter has shown many examples using subscript notation and range variables, Chapter 10, \"Vectors and Matrices,\" showed many examples using vector notation without subscripts. The derivative of a real function f may be indicated with differential notation or simply as f \u2032. Maybe you are talking about the del operator for gradient, divergence, curl kind of things in physics or differential geometry then it is \\nalba in LaTeX math mode. The$ 's around a command mean that it has to be used in maths mode, and they will temporarily put you into maths mode if you use them. Notation This section includes the most commonlyused notation in this book. The Anglo-French STD (Sunbeam-Talbot-Darracq) combine collapsed in 1935. Lecture notes on topological insulators Ming-Che Chang Department of Physics, National Taiwan Normal University, Taipei, Taiwan (Dated: December 26, 2018). , with n columns), then the product Ax is defined for n\u00d71 column vectors x. In L a T e X, subscripts and superscripts are written using the symbols ^ and _, in this case the x and y exponents where written using these codes. Even if you are using only one bracket, both commands are mandatory. So here's the deal: the inner product is very common in mathematical work (and there's accepted notation for it), and what you describe is not (and so there's not established notation). Note: Here we just want to analyse the commands and structure of a LaTeX file. Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Calculates transpose, determinant, trace, rank, inverse, pseudoinverse, eigenvalues and eigenvectors. 2 The Simplex Method. MathOverflow has an awesome engine where you can embed LaTeX in questions, answers, and comments.", "date": "2019-12-10 06:03:44", "meta": {"domain": "monetimargherita.it", "url": "http://chwz.monetimargherita.it/latex-vector-notation.html", "openwebmath_score": 0.9481537938117981, "openwebmath_perplexity": 751.4959527028767, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.976310532836284, "lm_q2_score": 0.8991213718636754, "lm_q1q2_score": 0.8778216656487156}}
{"url": "https://www.physicsforums.com/threads/rpm-required-to-simulate-earths-gravity-centrifugal-acceleration.744663/", "text": "# RPM required to simulate Earth's gravity (Centrifugal acceleration)\n\n1. Mar 22, 2014\n\n### jwlgrant\n\n1. The problem statement, all variables and given/known data\n\nHere's the problem: \"On a journey to Mars, one design is to have a section of the spacecraft rotate to simulate gravity. If the radius of this section is 30 meters, how many RPMs must it rotate to simulate one Earth gravity (1 g = 9.8 meters/sec^2)?\"\n\n2. Relevant equations\n\n$A = \\frac{V^2}{R}$\n\n3. The attempt at a solution\n\nI completed this problem, getting an answer of 5.5 RPM. Apparently I was wrong, and the answer should have been 14 RPM. I asked my instructor about it, and he showed me the calculations used to get 14 RPM, but it still didn't quite make sense to me.\n\nHere's the process my instructor used to solve the problem:\nCircumference is C = 2\u220f(30) = 188 meters.\n\n1 rpm is like rotating 1 full circular rotation every minute, so 188/60 = 3.1 meters/sec.\n\nThen V = 3.1 meters/sec x RPM.\n\nThen A = (3.1 RPM)^2/188 = 0.05 x rpm^2.\n\nSo: 9.8 = 0.05rpm^2\n\nRPM = 14\n\n--\n\nHere's my process:\nA = 9.8 (m)/(s^2)\nR = 30 m\nV = \u221a(AR)\nV = 17.15 m/s\n\n1 revolution = circumference = 2\u220f30 m\n[1 revolution/(2\u220f30 m)] * [60s/1 min] (1 rev is equal to 2\u03c030m and sixty seconds is equal to one minute)\nV = (17.15 m/s)/\u220f RPM\nV = 5.5 RPM\n\n--\n\nI realize, of course, that we both used different methods to get the answer. I thought I did it wrong, so I solved the problem again, using his method, to try to get the correct answer. He used 188m for the radius component when solving for A=V^2/R, but when I used 30m I got an answer of 5.5RPM. The radius, 30m, should be used in that equation, not 188m, right?\n\nI would greatly appreciate any help in understanding this problem. I can't tell where/how I went wrong in solving it.\n\n2. Mar 22, 2014\n\n### Andrew Mason\n\nWhy is he dividing by 188 (circumference)?? Should be 30.\n\nYou are right and the instructor is wrong.\n\nIt is easier to use Ac = \u03c92r; so $\\omega = \\sqrt{A/r}$ where \u03c9 is the angular speed in radians/sec.\n\nLetting RPM = 60\u03c9/2\u220f:\n\n$RPM = \\frac{60}{2\\pi}*\\sqrt{A/r} = \\frac{60}{2\\pi}\\sqrt{9.8/30} = 5.5$\n\nAM\n\nLast edited: Mar 22, 2014\n3. Mar 23, 2014\n\n### mr_pavlo\n\nI also agree, it must be divided by radius = 30 not circumference.", "date": "2018-03-24 22:26:13", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/rpm-required-to-simulate-earths-gravity-centrifugal-acceleration.744663/", "openwebmath_score": 0.7064085006713867, "openwebmath_perplexity": 1612.5549200284815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105273220726, "lm_q2_score": 0.8991213759183765, "lm_q1q2_score": 0.8778216646494176}}
{"url": "https://math.stackexchange.com/questions/2414941/suppose-that-3-men-and-3-women-will-be-randomly-seated-at-a-round-table-having-6", "text": "# Suppose that 3 men and 3 women will be randomly seated at a round table having 6 evenly-spaced seats\n\nSuppose that $3$ men and $3$ women will be randomly seated at a round table having $6$ evenly-spaced seats (with each seat being directly opposite one other seat).\n\n(a) What is the probability that all $3$ men will be seated next to one another?\n\n(b) What is the probability that starting at the northernmost seat and going clockwise around the table, a 2nd man will be encountered before a 2nd woman?\n\n(c) What is the probability that no two men are seated directly opposite of one another?\n\n(d) Suppose that the $6$ people are named Ashley, Barney, Carly, David, Emma, and Fred. What is the probability that starting with Ashley and going clockwise, the 6 people are seated in alphabetical order?\n\nAttempted Solution:\n\n(a) There are $6!$ = $720$ ways to seat these $6$ people. There are 6 different ways you can seat all three men next to each other, if you consider them as a clump. Each clump of three has $3!$ = 6 different ways to be arranged. Thus, there is $6*3!*3!$ = $216$ different arrangements. This gives a probability of $216\\over{720}$ = $.3$.\n\n(b) I wasn't sure about this but I realized that in any situation, either a 2nd man is reached before a 2nd woman or a 2nd woman is reached before a 2nd man. This gives a probability of $.5$.\n\n(c) After drawing a picture of the table, this appears to be the same situation as part (a), giving a probability of $.3.$ I just realized this assumption was incorrect. Now I am getting that there are 8 different arrangements when considering males and females as clumps, with $3!$ different ways of arranging each clump, giving $8*3!*3!$ = $288$. This gives a probability of $288\\over{720}$ = $.4$.\n\n(d) $6\\over{6}$ * $1\\over{5}$ * $1\\over{4}$ * $1\\over{3}$ * $1\\over{2}$ * $1\\over{1}$ = $.00833$.\n\nAny corrections to my attempted solutions would be greatly appreciated.\n\nAll of your answers are correct. I will assume that only the relative order of the people matters, that the men are named Barney, David, and Fred, and that the women are named Ashley, Carly, and Emma.\n\nWhat is the probability that all three men will be seated next to each other?\n\nWe seat Ashley. The remaining people can be seated in $5!$ ways as we proceed clockwise around the table.\n\nFor the favorable cases, we have two blocks of people to arrange. Since the blocks of men and women must be adjacent, this can be done in one way. The block of men can be arranged in $3!$ ways. The block of women can be arranged in $3!$ ways. Hence, the probability that all the men are seated next to each other is $$\\frac{3!3!}{5!} = \\frac{3}{10}$$\n\nWhat is the probability that starting at the northernmost seat and going clockwise around the table, a second man will be encountered before a second woman?\n\nYou made good use of symmetry. Nice solution.\n\nWhat is the probability that no two men are seated directly opposite each other?\n\nWe count arrangements in which two men are seated directly opposite each other. There are $\\binom{3}{2}$ ways to select the men. Once they are seated, there are $4$ ways to seat the third man relative to the already seated man whose name appears first alphabetically and $3!$ ways to seat the women as we proceed clockwise around the table from the third man. Hence, there are $$\\binom{3}{2}\\binom{4}{1}3!$$ seating arrangements in which two men are opposite each other. Hence, the probability that no two men are opposite each other is $$1 - \\frac{\\binom{3}{2}\\binom{4}{1}3!}{5!} = 1 - \\frac{3}{5} = \\frac{2}{5}$$\n\nSuppose that the six people are named Ashley, Barney, Carly, David, Emma, and Fred. What is the probability that starting with Ashley and going clockwise, the six people are seated in alphabetical order.\n\nThere is only one permissible seating arrangement. Hence, the probability that they are seated in alphabetical order is $$\\frac{1}{5!} = \\frac{1}{120}$$", "date": "2022-01-26 06:59:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2414941/suppose-that-3-men-and-3-women-will-be-randomly-seated-at-a-round-table-having-6", "openwebmath_score": 0.812130868434906, "openwebmath_perplexity": 196.55345507159353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526446557307, "lm_q2_score": 0.8856314783461302, "lm_q1q2_score": 0.8777959819531315}}
{"url": "http://mathhelpforum.com/calculus/47552-need-hints-solve-antiderivative.html", "text": "# Thread: Need hints to solve antiderivative\n\n1. ## Need hints to solve antiderivative\n\nOK I have a slightly new problem that seems much more difficult.\n\n$\\displaystyle X'(t) = \\mu_{k}gt + \\frac{A}{4m} \\int X'(t)^2dt$\n\nconstants:\n$\\displaystyle \\mu_{k}$ = kinetic friction coefficient\n$\\displaystyle g$ = gravity\n$\\displaystyle A$ = cross sectional area\n$\\displaystyle m$ = mass\n\nI am trying to solve $\\displaystyle X(t)$ and then solve the specific function as an initial value problem.\n\nI just need help solving $\\displaystyle X'(t)$ for now. I should be OK after that.\n\nThanks\n\nThis is for a hobby physics project, not for homework.\n\n2. Originally Posted by guitardude\nOK I have a slightly new problem that seems much more difficult.\n\n$\\displaystyle X'(t) = \\mu_{k}gt + \\frac{A}{4m} \\int X'(t)^2dt$\n\nconstants:\n$\\displaystyle \\mu_{k}$ = kinetic friction coefficient\n$\\displaystyle g$ = gravity\n$\\displaystyle A$ = cross sectional area\n$\\displaystyle m$ = mass\n\nI am trying to solve $\\displaystyle X(t)$ and then solve the specific function as an initial value problem.\n\nI just need help solving $\\displaystyle X'(t)$ for now. I should be OK after that.\n\nThanks\n\nThis is for a hobby physics project, not for homework.\nDifferentiate both sides with respect to t. Then make the substitution U = X'. Solve the resulting DE for U. Hence solve for X.\n\n3. I am still confused. If I differentiate both sides I will get $\\displaystyle X''(t)$ right? which is what I started with. This equation is my work thus far on taking the antiderivative of $\\displaystyle X''(t)$.\n\nIf I substitute $\\displaystyle u=X'$ then I get $\\displaystyle \\int u^2$ The antiderivative of this part isn't as simple as $\\displaystyle \\frac{1}{3}u^3$ is it? Even so I would end up with $\\displaystyle X(t) = constant*t^2 - constant* X'(t)^3 + C$ and I don't know $\\displaystyle X'(t)$.\n\nI forgot two negative signs, not that it matters much here but it should be:\n\n$\\displaystyle X'(t) = -\\mu_{k}gt - \\frac{A}{4m} \\int X'(t)^2dt$\n\nI previously solved the initial value problem for constant acceleration as a differential equation:\n\n$\\displaystyle X''(t) = B$\nso $\\displaystyle X'(t) = Bt + C$\nso $\\displaystyle X(t) = Bt^2 + Ct + D$\nwhich is our familiar physics equation $\\displaystyle S = S_{i} + V_{i}t + \\frac{1}{2} a(t^2)$\n\nYou can the solve the specific function and find B,C,D given 3 X(t) samples.\n\nNow, instead of assuming a constant acceleration, I am trying to solve this with $\\displaystyle B = -\\mu_{k}g - \\frac{1}{4m}Av^2$ and that is how I got $\\displaystyle X''(t) = -\\mu_{k}g - \\frac{1}{4m}Av^2 = -\\mu_{k}g - \\frac{1}{4m}AX'(t)^2$\n\nIt is not obvious to me yet how many X(t) samples I will need to solve this as an initial value problem?\n\n4. Originally Posted by mr fantastic\nDifferentiate both sides with respect to t. Then make the substitution U = X'. Solve the resulting DE for U. Hence solve for X.\nDifferentiate both sides wrt t: $\\displaystyle X'' = \\mu_k g + \\frac{A}{4m} \\, (X')^2$.\n\nSubstitute U = X': $\\displaystyle U' = \\mu_k g + \\frac{A}{4m} \\, U^2 \\Rightarrow \\frac{dU}{dt} - \\frac{A}{4m} \\, U^2 = \\mu_k g$.\n\nSolve this DE for U: I'd suggest using the integrating method.\n\nThen solve $\\displaystyle \\frac{dX}{dt} = U$ for X.\n\n5. I worked through it with my differential equations prof and got to:\n\n$\\displaystyle X'(t) = \\frac{-\\sqrt{\\mu_{k}g}}{\\sqrt{\\frac{A}{4m}}} * \\tan((\\sqrt{\\mu_{k}g})(\\sqrt{\\frac{A}{4m}})((t-C_{1})))$\n\nI verified the previous equation against simulations as well as with mathematica. Next I used matlab and then simplified to get back to :\n\n$\\displaystyle X(t) = \\frac{-4m}{A} * \\ln(\\sec^2((\\sqrt{\\mu_{k}g})(\\sqrt{\\frac{A}{4m}})( t-C_{1}))) + C_{2}$\n\nWith the two unknowns, $\\displaystyle C_{1}, C_{2}$\n\nI was expecting it to have a single $\\displaystyle \\sec$ not $\\displaystyle sec^2$ ?\n\nThis will be messy to solve as an initial value problem but possible.\n\nIs there any way to make this function continuous? Tan() and Sec() make these hard to use and I don't see why it is necessary.\n\n6. Originally Posted by guitardude\nI worked through it with my differential equations prof and got to:\n\n$\\displaystyle X'(t) = \\frac{-\\sqrt{\\mu_{k}g}}{\\sqrt{\\frac{A}{4m}}} * \\tan((\\sqrt{\\mu_{k}g})(\\sqrt{\\frac{A}{4m}})((t-C_{1})))$\n\nI verified the previous equation against simulations as well as with mathematica. Next I used matlab and then simplified to get back to :\n\n$\\displaystyle X(t) = \\frac{-4m}{A} * \\ln(\\sec^2((\\sqrt{\\mu_{k}g})(\\sqrt{\\frac{A}{4m}})( t-C_{1}))) + C_{2}$\n\nWith the two unknowns, $\\displaystyle C_{1}, C_{2}$\n\nI was expecting it to have a single $\\displaystyle \\sec$ not $\\displaystyle sec^2$ ?\n\nThis will be messy to solve as an initial value problem but possible.\n\nIs there any way to make this function continuous? Tan() and Sec() make these hard to use and I don't see why it is necessary.\nI can't quite understand how you've written the solution for X(t) but it looks like you have a log of a product. So use the usual log rule to break this up into a sum of logs.\n\nThen it looks like one of those logs has the form $\\displaystyle \\log \\sec^2 \\theta$. Using the usual log rule you get $\\displaystyle \\log \\sec^2 \\theta = \\log \\cos^{-2} \\theta = -2 \\log \\cos \\theta$ .....\n\n7. Originally Posted by mr fantastic\nI can't quite understand how you've written the solution for X(t) but it looks like you have a log of a product. So use the usual log rule to break this up into a sum of logs.\n\nThen it looks like one of those logs has the form $\\displaystyle \\log \\sec^2 \\theta$. Using the usual log rule you get $\\displaystyle \\log \\sec^2 \\theta = \\log \\cos^{-2} \\theta = -2 \\log \\cos \\theta$ .....\n\nHopefully this helps.... It is of form:\n$\\displaystyle X(t) = D * \\ln( \\sec^2(E(t-C_{1})) ) + C_{2}$\n\nFor the known constants $\\displaystyle D,E$\n\nI will have to solve $\\displaystyle C_{1}$ which I can do given two pairs of corresponding X,t\n\nDoes your suggestion of splitting $\\displaystyle \\ln(\\sec^2(\\theta))$ up still apply?\n\n8. Originally Posted by guitardude\nHopefully this helps.... It is of form:\n$\\displaystyle X(t) = D * \\ln( \\sec^2(E(t-C_{1})) ) + C_{2}$\n\nFor the known constants $\\displaystyle D,E$\n\nI will have to solve $\\displaystyle C_{1}$ which I can do given two pairs of corresponding X,t\n\nDoes your suggestion of splitting $\\displaystyle \\ln(\\sec^2(\\theta))$ up still apply?\nYes.", "date": "2018-04-23 07:53:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/47552-need-hints-solve-antiderivative.html", "openwebmath_score": 0.9366592168807983, "openwebmath_perplexity": 413.53101439184815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.973240719919151, "lm_q2_score": 0.9019206692796966, "lm_q1q2_score": 0.8777859214797343}}
{"url": "http://mathhelpforum.com/math-topics/24992-french-week.html", "text": "# Math Help - A French Week\n\n1. ## A French Week\n\nFor about 10 years after the French Revolution, the French government attempted to base measures of time on multiple of ten: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?\n\n(a) $1 \\mbox{ French Week} = 1 \\mbox{ French Week} \\left(\\frac{10 \\mbox{ days}}{1 \\mbox{ French Week}}\\right)\\left(\\frac{10 \\mbox{ hrs}}{1 \\mbox{ day}}\\right)$ $\\left(\\frac{100 \\mbox{ min}}{1 \\mbox{ hr}}\\right)\\left(\\frac{100 \\mbox{ s}}{1 \\mbox{ min}}\\right)$\n\n$=10^6 \\mbox{ seconds}$\n\n$1 \\mbox{ Normal Week} = 1 \\mbox{ Normal Week} \\left(\\frac{7 \\mbox{ days}}{1 \\mbox{ Normal Week}}\\right)\\left(\\frac{24 \\mbox{ hrs}}{1 \\mbox{ day}}\\right)$ $\\left(\\frac{60 \\mbox{ min}}{1 \\mbox{ hr}}\\right)\\left(\\frac{60 \\mbox{ s}}{1 \\mbox{ min}}\\right)$\n\n$=604800 \\mbox{ seconds}$\n\nSo the ratio is 1.65, but the answers say that it is 1.43 ?\n\nAlso, in the first question I assumed that the 'French' second was the same as the 'normal' second, so wouldn't my answer to the second question make the answer to my first question invalid? Thanks.\n\n2. Originally Posted by DivideBy0\nFor about 10 years after the French Revolution, the French government attempted to base measures of time on multiple of ten: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?\n\n(a) $1 \\mbox{ French Week} = 1 \\mbox{ French Week} \\left(\\frac{10 \\mbox{ days}}{1 \\mbox{ French Week}}\\right)\\left(\\frac{10 \\mbox{ hrs}}{1 \\mbox{ day}}\\right)$ $\\left(\\frac{100 \\mbox{ min}}{1 \\mbox{ hr}}\\right)\\left(\\frac{100 \\mbox{ s}}{1 \\mbox{ min}}\\right)$\n\n$=10^6 \\mbox{ seconds}$\n\n$1 \\mbox{ Normal Week} = 1 \\mbox{ Normal Week} \\left(\\frac{7 \\mbox{ days}}{1 \\mbox{ Normal Week}}\\right)\\left(\\frac{24 \\mbox{ hrs}}{1 \\mbox{ day}}\\right)$ $\\left(\\frac{60 \\mbox{ min}}{1 \\mbox{ hr}}\\right)\\left(\\frac{60 \\mbox{ s}}{1 \\mbox{ min}}\\right)$\n\n$=604800 \\mbox{ seconds}$\n\nSo the ratio is 1.65, but the answers say that it is 1.43 ?\n\nAlso, in the first question I assumed that the 'French' second was the same as the 'normal' second, so wouldn't my answer to the second question make the answer to my first question invalid? Thanks.\nHello,\n\nthe ratio between the weeks is:\n\n$\\frac{\\text{french week}}{\\text{normal week}}=\\frac{10}7 \\approx 1.42857...$\n\nto b):\n$1 \\text{ french day}= 10 \\ h$ I assume that a new french day lasts as long as a normal day. Then you'll get the ratio:\n\n$\\frac{10 \\ h \\cdot 100\\ min \\cdot 100\\ sec}{24\\ h \\cdot 60\\ min \\cdot 60\\ sec} \\approx 11.574...$ That means one of the new french seconds is a very short time interval. So better check my calculations.\n\n3. Originally Posted by earboth\nHello,\n\nthe ratio between the weeks is:\n\n$\\frac{\\text{french week}}{\\text{normal week}}=\\frac{10}7 \\approx 1.42857...$\n\nto b):\n$1 \\text{ french day}= 10 \\ h$ I assume that a new french day lasts as long as a normal day. Then you'll get the ratio:\n\n$\\frac{10 \\ h \\cdot 100\\ min \\cdot 100\\ sec}{24\\ h \\cdot 60\\ min \\cdot 60\\ sec} \\approx 11.574...$ That means one of the new french seconds is a very short time interval. So better check my calculations.\nWhy wouldn't DivideByZero's calculations be correct? When comparing two terms of measurement, you must relate them somehow\n\n4. Hello, DivideBy0!\n\nA fascinating problem . . . took me a few moments . . .\n\nFor about 10 years after the French Revolution, the French government\nattempted to base measures of time on multiple of ten:\nOne week consisted of 10 days, one day consisted of 10 hours,\none hour consisted of 100 minutes, and one minute consisted of 100 seconds.\n\nWhat are the ratios of (a) the French decimal week to the standard week\nand (b) the French decimal second to the standard second?\n\n$\\text{1 French Week }\\:=\\:\\frac{\\text{1 French Week}}{1} \\times \\frac{\\text{10 days}}{\\text{1 French Week}} \\times \\frac{\\text{10 hrs}}{\\text{1 day}}$ $\\times \\frac{\\text{100 min}}{\\text{1 hr}} \\times \\frac{\\text{100 s}}{\\text{1 min}}$\n. . $=\\;10^6 \\text{ seconds}$\n\n$\\text{1 Normal Week} \\;= \\;\\frac{\\text{1 Normal Week}}{1} \\times \\frac{\\text{7 days}}{\\text{1 Normal Week}} \\times \\frac{\\text{24 hrs}}{\\text{1 day}} \\times \\frac{\\text{60 min}}{\\text{1 hr}} \\times \\frac{\\text{60 s}}{\\text{1 min}}$\n. . $=\\;604800 \\text{ seconds}$\n\nAll of this is absolutely correct!\nNote that (nearly) all the units are unequal . . . weeks, hours, minutes, seconds.\n\nThe only unit that both systems agree upon is the length of one day.\n. . (Perhaps measured from sunrise to sunrise.)\n\n$a)\\;\\;\\frac{\\text{1 Frech week}}{\\text{1 Normal week}} \\;=\\;\\frac{\\text{10 days}}{\\text{7 days}} \\;\\approx\\; 1.43$\n\n$b)\\;\\;\\frac{\\text{1 French day}}{\\text{1 Normal day}} \\; = \\; \\frac{10^5\\text{ French sec}}{86,400\\text{ Normal sec}} \\;=\\;1.1574\n$\n\n. . The French second is about 1\u00bd jiffys longer than a Normal second.\n\n5. Thanks, I don't know why I thought days could be of different lengths.", "date": "2015-11-25 12:05:57", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/24992-french-week.html", "openwebmath_score": 0.4040050804615021, "openwebmath_perplexity": 1143.3663666753037, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561703644736, "lm_q2_score": 0.9059898279984214, "lm_q1q2_score": 0.8777738351437187}}
{"url": "http://www.nabae24.com/emigrate-vs-vvnb/multiplying-complex-numbers-in-rectangular-form-74d82b", "text": "Exponential Form of Complex Numbers; Euler Formula and Euler Identity interactive graph; 6. polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. Fortunately, when multiplying complex numbers in trigonometric form there is an easy formula we can use to simplify the process. What you can do, instead, is to convert your complex number in POLAR form: #z=r angle theta# where #r# is the modulus and #theta# is the argument. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. In order to work with these complex numbers without drawing vectors, we first need some kind of standard mathematical notation. So 18 times negative root 2 over. To multiply complex numbers: Each part of the first complex number gets multiplied by each part of the second complex numberJust use \\\"FOIL\\\", which stands for \\\"Firsts, Outers, Inners, Lasts\\\" (see Binomial Multiplication for more details):Like this:Here is another example: Change\u00a0), You are commenting using your Twitter account. Multiplication of Complex Numbers. So the root of negative number \u221a-n can be solved as \u221a-1 * n = \u221a n i, where n is a positive real number. That\u2019s right \u2013 it kinda looks like the the Cartesian plane which you have previously used to plot (x, y) points and functions before. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Complex numbers can be expressed in numerous forms. To plot a complex number a+bi on the complex plane: For example, to plot 2 + i we first note that the complex number is in rectangular (a+bi) form. Multiplying complex numbers when they're in polar form is as simple as multiplying and adding numbers. Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by \u2026 In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. A complex number can be represented in the form a + bi, where a and b are real numbers and i denotes the imaginary unit. How to Divide Complex Numbers in Rectangular Form ? Key Concepts. Math Gifs; Algebra; Geometry; Trigonometry; Calculus; Teacher Tools; Learn to Code; Home; Algebra ; Complex Numbers; Complex number Calc; Complex Number Calculator. (This is true for rectangular form as well (a 2 + b 2 = 1)) The Multiplicative Inverse (Reciprocal) of i. Then, multiply through by See and . Rectangular Form A complex number is written in rectangular form where and are real numbers and is the imaginary unit. Yes, you guessed it, that is why (a+bi) is also called the rectangular form of a complex number. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to email this to a friend (Opens in new window), put it into the standard form of a complex number by writing it as, How To Write A Complex Number In Standard Form (a+bi), The Multiplicative Inverse (Reciprocal) Of A Complex Number, Simplifying A Number Using The Imaginary Unit i, The Multiplicative Inverse (Reciprocal) Of A Complex\u00a0Number. Hence the value of Im(3z + 4zbar \u00e2\u0088\u0092 4i) is -\u00a0y - 4. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Use rectangular coordinates when the number is given in rectangular form and polar coordinates when polar form is used. In polar form, the multiplying and dividing of complex numbers is made easier once the formulae have been developed. Find powers of complex numbers in polar form. Figure 5. In other words, there are two ways to describe a complex number written in the form a+bi: To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. Dividing complex numbers: polar & exponential form. The major difference is that we work with the real and imaginary parts separately. We sketch a vector with initial point 0,0 and terminal point P x,y . Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. if z 1 = r 1\u2220\u03b8 1 and z 2 = r 2\u2220\u03b8 \u2026 This video shows how to multiply complex number in trigonometric form. In other words, given $$z=r(\\cos \\theta+i \\sin \\theta)$$, first evaluate the trigonometric functions $$\\cos \\theta$$ and $$\\sin \\theta$$. This lesson on DeMoivre\u2019s Theorem and The Complex Plane - Complex Numbers in Polar Form is designed for PreCalculus or Trigonometry. To understand and fully take advantage of multiplying complex numbers, or dividing, we should be able to convert from rectangular to trigonometric form and from trigonometric to rectangular form. A complex number can be expressed in standard form by writing it as a+bi. To convert from polar form to rectangular form, first evaluate the trigonometric functions. In the complex number a + bi, a is called the real part and b is called the imaginary part. Plot each point in the complex plane. Multiplying complex numbers is much like multiplying binomials. To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. If z = x + iy , find the following in rectangular form. But then why are there two terms for the form a+bi? For Example, we know that equation x 2 + 1 = 0 has no solution, with number i, we can define the number as the solution of the equation. Addition of Complex Numbers . Find (3e 4j)(2e 1.7j), where j=sqrt(-1). Answer. Multiplication and division of complex numbers is easy in polar form. Label the x-axis as the real axis and the y-axis as the imaginary axis. (5 + j2) + (2 - j7) = (5 + 2) + j(2 - 7) = 7 - j5 (2 + j4) - (5 + j2) = (2 - 5) + j(4 - 2) = -3 + j2; Multiplying is slightly harder than addition or subtraction. For example, here\u2019s how you handle a scalar (a constant) multiplying a complex number in parentheses: 2(3 + 2i) = 6 + 4i. So I get plus i times 9 root 2. Sum of all three four digit numbers formed with non zero digits. Complex Number Functions in Excel. Converting from Polar Form to Rectangular Form. To add complex numbers in rectangular form, add the real components and add the imaginary components. c) Write the expression in simplest form. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. b) Explain how you can simplify the final term in the resulting expression. Find roots of complex numbers in polar form. Rectangular form. Math Precalculus Complex numbers Multiplying and dividing complex numbers in polar form. Doing basic operations like addition, subtraction, multiplication, and division, as well as square roots, logarithm, trigonometric and inverse trigonometric functions of a complex numbers were already a simple thing to do. Worksheets on Complex Number. (\u00a0Log\u00a0Out\u00a0/\u00a0 Ask Question Asked 1 year, 6 months ago. When in rectangular form, the real and imaginary parts of the complex number are co-ordinates on the complex plane, and the way you plot them gives rise to the term \u201cRectangular Form\u201d. See . Products and Quotients of Complex Numbers; Graphical explanation of multiplying and dividing complex numbers; 7. The first, and most fundamental, complex number function in Excel converts two components (one real and one imaginary) into a single complex number represented as a+bi. Multiplying and dividing complex numbers in polar form. Addition, subtraction, multiplication and division can be carried out on complex numbers in either rectangular form or polar form. Rather than describing a vector\u2019s length and direction by denoting magnitude and \u2026 The standard form, a+bi, is also called the rectangular form of a complex number. 1. By \u2026 Multiplying a complex number by a real number is simple enough, just distribute the real number to both the real and imaginary parts of the complex number. We distribute the real number just as we would with a binomial. To find the product of two complex numbers, multiply the two moduli and add the two angles. This is the currently selected item. Polar Form of Complex Numbers; Convert polar to rectangular using hand-held calculator; Polar to Rectangular Online Calculator ; 5. (\u00a0Log\u00a0Out\u00a0/\u00a0 Example 1. Multiplication of Complex Numbers. In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The video shows how to multiply complex numbers in cartesian form. Sum of all three four digit numbers formed using 0, 1, 2, 3. A complex number in rectangular form means it can be represented as a point on the complex plane. Active 1 year, 6 months ago. After having gone through the stuff given above, we hope that the students would have understood, \"How to Write the Given Complex Number in Rectangular Form\". We start with an example using exponential form, and then generalise it for polar and rectangular forms. Convert a complex number from polar to rectangular form. This can be a helpful reminder that if you know how to plot (x, y) points on the Cartesian Plane, then you know how to plot (a, b) points on the Complex Plane. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. The reciprocal of zero is undefined (as with the rectangular form of the complex number) When a complex number is on the unit circle r = 1/r = 1), its reciprocal equals its complex conjugate. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). bi+a instead of a+bi). 18 times root 2 over 2 again the 18, and 2 cancel leaving a 9. Notice the rectangle that is formed between the two axes and the move across and then up? Using either the distributive property or the FOIL method, we get I get -9 root 2. The correct answer is therefore (2). Multiplying by the conjugate . This video shows how to multiply complex number in trigonometric form. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. ; The absolute value of a complex number is the same as its magnitude. The Number i is defined as i = \u221a-1. To divide, divide the magnitudes and \u2026 Hence the Re (1/z) is\u00a0(x/(x2\u00a0+ y2)) - i (y/(x2\u00a0+ y2)). The following development uses trig.formulae you will meet in Topic 43. Rectangular Form of a Complex Number. In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. To divide the complex number which is in the form (a + ib)/(c + id) we have to multiply both numerator and denominator by the conjugate of the denominator. Apart from the stuff given in this section \", How to Write the Given Complex Number in Rectangular Form\". Write the following in the rectangular form: [(5 + 9i) + (2 \u00e2\u0088\u0092 4i)] whole bar\u00a0 =\u00a0 (5 + 9i) bar + (2 \u00e2\u0088\u0092 4i) bar, Multiplying both numerator and denominator by the conjugate of of denominator, we get, =\u00a0 \u00a0[(10 - 5i)/(6 + 2i)] [(6 - 2i)/(6 - 2i)], =\u00a0 - 3i + { (1/(2 - i)) ((2 + i)/(2 + i)) }. Multiplying Complex Numbers Together. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = \u22121 or j 2 = \u22121.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). When you\u2019re dividing complex numbers, or numbers written in the form z = a plus b times i, write the 2 complex numbers as a fraction. Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. The imaginary unit i with the property i 2 = \u2212 1 , is combined with two real numbers x and y by the process of addition and multiplication, we obtain a complex number x + iy. Multiplication . $\\text{Complex Conjugate Examples}$ $\\\\(3 \\red + 2i)(3 \\red - 2i) \\\\(5 \\red + 12i)(5 \\red - 12i) \\\\(7 \\red + 33i)(5 \\red - 33i) \\\\(99 \\red + i)(99 \\red - i)$ Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. Well, rectangular form relates to the complex plane and it describes the ability to plot a complex number on the complex plane once it is in rectangular form. We start with an example using exponential form, and then generalise it for polar and rectangular forms. Show Instructions. Example 1 \u2013 Determine which of the following is the rectangular form of a complex number. To convert from polar form. Google custom search here method of notation is valid for numbers. 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To  5 * x ` Write a complex number is the rectangular used... Basic arithmetic on complex numbers in polar form, and Last terms...., just like vectors, can also be expressed in polar form, multiply the moduli, and Last together! Written in rectangular form., just like vectors, can also be expressed in polar form. either form. Simplify any complex expression, with steps shown, r \u2220 \u03b8 and rectangular number polar...", "date": "2021-04-17 19:29:46", "meta": {"domain": "nabae24.com", "url": "http://www.nabae24.com/emigrate-vs-vvnb/multiplying-complex-numbers-in-rectangular-form-74d82b", "openwebmath_score": 0.827418327331543, "openwebmath_perplexity": 480.16984064813244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.979354068055595, "lm_q2_score": 0.8962513627417531, "lm_q1q2_score": 0.8777474181015066}}
{"url": "http://mathhelpforum.com/statistics/97227-solved-probability-question-digits.html", "text": "# Math Help - [SOLVED] probability question(digits)\n\n1. ## [SOLVED] probability question(digits)\n\nQuestion:\nA four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that\n(a) the number begins or ends with 0,\n(b) the number contains exactly two non-zero digits.\n\nCan you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).\n\nThanks\n\n2. For part A, here are my thoughts:\n\nFor the first 0000 to 0999, there are 1000 numbers. Then you have to add in all the numbers ending in 0. These are a little trickier to see. So we are looking for 1000, 1010, 1020, ... , 1090. That's a total of 10 numbers between 1000 and 1090 (inclusive). We also have to remember 1100, 1110, 1120, ... , 1190. So there are 10 numbers between each increment of a hundred. And between each thousand there are 10 increments of 100. Finally we are starting at 999 and proceeding to 9999, in which there are 9 increments of a thousand. So my total for part A comes to:\n\n$\\frac {1000 + 10 * 10 * 9}{10000} = \\frac{19}{100}$\n\n3. Originally Posted by stpmmaths\nQuestion:\nA four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that\n(a) the number begins or ends with 0,\n(b) the number contains exactly two non-zero digits.\n\nCan you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).\n\nThanks\n(a) You want the combinations 0 X X X or X X X 0. Using the pigeon hole principle:\n\n(1)(10)(10)(10) + (10)(10)(10)(1) = 2000.\n\nBut you have to subtract numbers of the form 0 X X 0 otherwise they get counted twice in the above: (1)(10)(10)(1) = 100.\n\nSo the total number of numbers satisfying the restriction is 1900.\n\nThe total number of numbers withut restriction is (10)(10)(10)(10) = 10000.\n\nSo the probability is 1900/10000 = 19/100.\n\nYou can take a similar approach for (b).\n\n4. A more systematic way of doing part A is to count first the numbers beginning with 0, and then the numbers ending with 0 and then subtracting the number of numbers beginning and ending with 0 to get the total number of numbers beginning or ending with 0. We get $10^3+10^3-10^2$, giving the same probability as eXist calculated.\nFor part B,there are 4 choose 2 =6 ways to pick the two zero digits and 9*9 ways to choose the two non-zero digits. It is worth thinking about how many 4 digit numbers contain respectively exactly 0,1,2,3 and 4 0s. You should see a connection with the binomial theorem (expand $(9+1)^4$ to see it more clearly).\n\n5. Hello stpmmaths\n\nWelcome to Math Help Forum!\nOriginally Posted by stpmmaths\nQuestion:\nA four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that\n(a) the number begins or ends with 0,\n(b) the number contains exactly two non-zero digits.\n\nCan you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).\n\nThanks\nPart (b). If the number contains exactly two non-zero digits, then it contains 2 non-zero and 2 zero digits.\n\nSo: imagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.\n\nFor stage (i), there are $\\binom42 = 6$ ways of choosing which boxes shall contain the zeros.\n\nFor stage (ii), there are 9 choices of non-zero digit to go in the remaining left-hand empty box, and 9 to go in the right-hand box. Total number of choices = $9 \\times 9 = 81$.\n\nSo the number of different numbers with exactly two non-zero digits is $6 \\times 81 = 486$. So the probability is $\\frac{486}{10000}=0.0486$.\n\n6. Originally Posted by mr fantastic\n(a) You want the combinations 0 X X X or X X X 0. Using the pigeon hole principle:\n\n(1)(10)(10)(10) + (10)(10)(10)(1) = 2000.\n\nBut you have to subtract numbers of the form 0 X X 0 otherwise they get counted twice in the above: (1)(10)(10)(1) = 100.\n\nSo the total number of numbers satisfying the restriction is 1900.\n\nThe total number of numbers withut restriction is (10)(10)(10)(10) = 10000.\n\nSo the probability is 1900/10000 = 19/100.\n\nYou can take a similar approach for (b).\nFor part A,\nAs you said, (1)(10)(10)(10) + (10)(10)(10)(1) - (1)(10)(10)(1) = 1900. If we do this we still consider 0XX0 in the probability, right? But the question stated begins or ends with 0 not with both\n\n7. Originally Posted by stpmmaths\nFor part A,\nAs you said, (1)(10)(10)(10) + (10)(10)(10)(1) - (1)(10)(10)(1) = 1900. If we do this we still consider 0XX0 in the probability, right? But the question stated begins or ends with 0 not with both\nIf the statement is \"begins or ends with a zero digit\" it means begins with a zero digit or ends with a zero digit or both begins and ends with a zero.\nThat is the logical nature of the connective or.\n\nHello stpmmaths\n\nWelcome to Math Help Forum!\nPart (b). If the number contains exactly two non-zero digits, then it contains 2 non-zero and 2 zero digits.\n\nSo: imagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.\n\nFor stage (i), there are $\\binom42 = 6$ ways of choosing which boxes shall contain the zeros.\n\nFor stage (ii), there are 9 choices of non-zero digit to go in the remaining left-hand empty box, and 9 to go in the right-hand box. Total number of choices = $9 \\times 9 = 81$.\n\nSo the number of different numbers with exactly two non-zero digits is $6 \\times 81 = 486$. So the probability is $\\frac{486}{10000}=0.0486$.\n\nI thought we should consider the position(placement) of the digits? If we do it by we are not considering the position. According to your statement \"\nimagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.\n\nFor stage (i), there are $\\binom42 = 6$ ways of choosing which boxes shall contain the zeros.\n\n\"\n\nThis makes me confused between and .\nLets say we change the question to:-\nA four-digit number, in the range 0000 to 9999 without repetitions for all digits except digit 0, is formed. Find the probability that\n(a) ....\n(b) the number contains exactly two non-zero digits.\n\n9. ## Permutations and Combinations\n\nHello stpmmaths\nOriginally Posted by stpmmaths\n\nI thought we should consider the position(placement) of the digits? If we do it by we are not considering the position. According to your statement \"\nimagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.\n\nFor stage (i), there are $\\binom42 = 6$ ways of choosing which boxes shall contain the zeros.\n\n\"\n\nThis makes me confused between and .\nLets say we change the question to:-\nA four-digit number, in the range 0000 to 9999 without repetitions for all digits except digit 0, is formed. Find the probability that\n(a) ....\n(b) the number contains exactly two non-zero digits.\nYou are raising two different points here. First, the difference between $^nP_r$ and $^nC_r$.\n\n\u2022 $^nP_r$ gives the number of ways of choosing $r$ objects from $n$, and then arranging them in order. So, for instance, if we want to choose 2 from 4, and the objects are A, B, C and D, then the choice A, C will be counted as a different choice from C, A. There are therefore $^4P_2=4 \\times 3 = 12$ possible choices.\n\n\u2022 $^nC_r$ gives the number of ways of choosing $r$ objects from $n$, where the order doesn't matter. In the above example, A, C and C, A would be counted as the same choice, and there are therefore $^4C_2=\\frac{4\\times 3}{2}= 6$ possible choices.\n\nAs an example, suppose that a committee chairman and secretary have to be chosen from a short-list of 4 people. Here, A, C is a different selection from C, A because A and C are fulfilling different roles. So you use $^4P_2$. But if you have to choose 2 delegates to attend a conference, from 4 people, you'll use $^4C_2$. Why? Because A, C and C, A are now the same choice - there's no difference between the roles of A and C.\n\nSo, in question (b), the number of different pairs of positions that can be occupied by a zero is $^4C_2=6$, because one zero looks just like the other - the order in which you place them in the 'empty boxes' doesn't matter.\n\nIn each of the remaining boxes (containing non-zeros) the order in which we fill the boxes does matter: 0709 will be different from 0907, for instance. Each box can then be filled in 9 ways, so - using the 'r-s Principle' - the total number of ways in which these two boxes can be filled, one after the other, is $9^2 =81$.\n\nYour second question is: how is this affected if we're not allowed to choose the same non-zero digit twice? The answer is this:\n\n\u2022 The first stage - choosing which 'boxes' are to be filled with zeros - is unaltered. It's still 6 choices.\n\n\u2022 In the second stage, suppose we fill the left-hand of the two remaining boxes first. There are 9 ways of doing this. Since we're not allowed to choose the same digit again, there are then 8 ways of filling the last box. Total number of choices: $9\\times 8 = 72$. Notice that this is the number of ways of choosing 2 items from 9, where the order matters; in other words, it's $^9P_2$.\n\nSo the overall answer to your modified question would be $6 \\times 72 = 432$ choices, or a probability of $0.0432$.\n\nDoes that clear things up?\n\n10. Thanks\n\nSo if i modified the question to:-\nA five-digit number, in the range 00000 to 99999 inclusive, is formed. Find the probability that\n(a) ....\n(b) the number contains exactly two non-zero digits.\n\nThen the answer would be 6 x 729 = 4374. Am I right?\n\n11. Hello stpmmaths\nOriginally Posted by stpmmaths\nThanks\n\nSo if i modified the question to:-\nA five-digit number, in the range 00000 to 99999 inclusive, is formed. Find the probability that\n(a) ....\n(b) the number contains exactly two non-zero digits.\n\nThen the answer would be 6 x 729 = 4374. Am I right?\nCorrect!\n\nAnd if repetition is not allowed in the non-zero digits, it would be $6 \\times ^9P_3 = 6 \\times9\\times8\\times7 = 3024$.\n\n13. ## probability question\n\nSo the term C^2_4 relates to choosing the positions for (non) zero digits whereas terms 9^2 or P^2_9 are the numbers of ways to choose non-zero digits for a given positions.\n\n14. Hello kobylkinks\nOriginally Posted by kobylkinks\nSo the term C^2_4 relates to choosing the positions for (non) zero digits whereas terms 9^2 or P^2_9 are the numbers of ways to choose non-zero digits for a given positions.\nIf this is a question, and not just a statement, then yes, you are right in your analysis!\n\nYou're also right to write '(non) zero digits' because it doesn't matter whether you think of this part of the 'box-filling' process as choosing the boxes that will contain zeros, or the boxes that will contain non-zeros; the result is the same: there are $^4C_2= 6$ possible choices.", "date": "2016-07-25 17:14:56", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/97227-solved-probability-question-digits.html", "openwebmath_score": 0.7649855017662048, "openwebmath_perplexity": 420.19214079886615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575121992375, "lm_q2_score": 0.8933094145755219, "lm_q1q2_score": 0.8777278760094821}}
{"url": "https://math.stackexchange.com/questions/2821923/how-to-put-a-matrix-in-reduced-echelon-form", "text": "# How to put a matrix in Reduced Echelon Form\n\n5. (6 pts) Find a basis for the following subspace: $$\\left\\{ \\begin{bmatrix} r - s - 2t - u \\\\ -r + 2s + 5t + 2u \\\\ s + 3t + u \\\\ 3r + s + 5u \\end{bmatrix} \\in \\mathbb{R}^4 : \\text{r, s, t and u are scalars} \\right\\}.$$\n\nOriginal image here.\n\nSo I have this matrix: $$\\begin{pmatrix} 1 &-1 &-2 &-1 \\\\ -1 &2& 5 &2 \\\\ 0 &1 &3 &1 \\\\ 3 &1 & 0 & 5 \\end{pmatrix}$$ If I compute the the matrix in reduced echelon form, I get: $$\\begin{pmatrix} 1&0&1&0\\\\ 0&1&0&3\\\\ 0&0&1&-2/3\\\\ 0 & 0 & 0 & 0 \\end{pmatrix}$$\n\nPlease note that i want to put the matrix in reduced echelon form , and not row reduced echelon form.\n\nI know that the three rules of reduced echelon form are:\n\n1) The first non zero number of a row is 1 (leading entry).\n\n2) An all zero row is placed at the end of the matrix.\n\n3)A leading entry of a row is placed to the right of a leading entry of the previous row.\n\nIf it were to be in row reduced echelon form , then another property would hold, or else that each column that contains a leading entry, all other entries in the same column are zeros.\n\nSo is my matrix correctly put in a row echelon form ?\n\nI think you are confused on terminology. Your matrix is correctly put into one of its row echelon forms. Note that these are not unique, as we could add the second row to the first row and the result would still be in row echelon form, that is, \\begin{pmatrix} 1&1&1&3\\\\ 0&1&0&3\\\\ 0&0&1&-2/3\\\\ 0 & 0 & 0 & 0 \\end{pmatrix} Is still in row echelon form. In order to get to the reduced row echelon form (sometimes called reduced echelon form), we must, as you said, make $0$'s in every column which contains a leading $1$. In your example, we can subtract the third row from the first \\begin{pmatrix} 1&0&1&0\\\\ 0&1&0&3\\\\ 0&0&1&-2/3\\\\ 0 & 0 & 0 & 0 \\end{pmatrix} to obtain \\begin{pmatrix} 1&0&0&2/3\\\\ 0&1&0&3\\\\ 0&0&1&-2/3\\\\ 0 & 0 & 0 & 0 \\end{pmatrix} Which is now in reduced row echelon form. Hopefully that clears up any confusion you had.\n\n\u2022 In order to find a basis for a subspace, should i use row echelon form or row reduced echelon form ? Moreover, would the basis be composed of non zero rows or columns with pivot entries ? \u2013\u00a0Sara Martini Jun 16 '18 at 20:32\n\u2022 And what is the advantage of dealing with a NON reduced row echelon form ? \u2013\u00a0Rene Schipperus Jun 16 '18 at 20:37\n\nWith reference to the original problem we need to find a basis for\n\n$$\\begin{bmatrix} r-s-2t-u \\\\ -r+2s+5t+2u \\\\ s+3t+u \\\\ 3r+s+5u \\end{bmatrix}=\\begin{pmatrix} 1 &-1 &-2 &-1 \\\\ -1 &2& 5 &2 \\\\ 0 &1 &3 &1 \\\\ 3 &1 & 0 & 5 \\end{pmatrix}\\begin{pmatrix} r \\\\ s \\\\ t \\\\ u \\end{pmatrix}$$\n\nnote that the LHS is a vector $\\in \\mathbb{R^4}$ and in the RHS the given coefficients are arranged by columns, therefore we need to find a basis for the column space of the matrix in on the RHS.\n\nThen we have\n\n$$\\begin{pmatrix} 1 &-1 &-2 &-1 \\\\ -1 &2& 5 &2 \\\\ 0 &1 &3 &1 \\\\ 3 &1 & 0 & 5 \\end{pmatrix}\\to \\begin{pmatrix} 1 &-1 &-2 &-1 \\\\ 0 &1& 3 &1 \\\\ 0 &1 &3 &1 \\\\ 3 &1 & 0 & 5 \\end{pmatrix}\\to \\begin{pmatrix} 1 &-1 &-2 &-1 \\\\ 0 &1& 3 &1 \\\\ 0 &1 &3 &1 \\\\ 0 &4 & 6 & 8 \\end{pmatrix}\\to \\begin{pmatrix} 1 &-1 &-2 &-1 \\\\ 0 &1& 3 &1 \\\\ 0 &0 &0 &0 \\\\ 0 &0 & -6 & 4 \\end{pmatrix}\\to \\begin{pmatrix} 1 &-1 &-2 &-1 \\\\ 0 &1& 3 &1 \\\\ 0 &0 & -6 & 4\\\\ 0 &0 &0 &0 \\end{pmatrix}$$\n\nwhich is the RREF. From here we can conclude that a basis is given by the first three vectors of the original matrix that is\n\n$$\\{(1,-1,0,3),(-1,2,1,1),(-2,5,3,0)\\}$$\n\nNote that we can also continue and obtain\n\n$$\\to \\begin{pmatrix} 1 &0 &1 &0 \\\\ 0 &1& 3 &1 \\\\ 0 &0 & 1 & -\\frac23\\\\ 0 &0 &0 &0 \\end{pmatrix}\\to \\begin{pmatrix} 1 &0 &0 &\\frac23 \\\\ 0 &1& 0 &3 \\\\ 0 &0 & 1 & -\\frac23\\\\ 0 &0 &0 &0 \\end{pmatrix}$$\n\nthe latter is the Gauss-Jordan form of the RREF but it is not needed to establish a basis for the given subspace.\n\n\u2022 Is the row echelon form sufficient in order to find the basis for a subspace? \u2013\u00a0Sara Martini Jun 16 '18 at 19:58\n\u2022 @SaraMartini Yes and you can stop to the first step I've indicated, if you are looking to the columns a basis is given by the first three columns in the original matrix, if you are looking for the rows you can choose the first the second and the fourth rows in the originl matrix an a basis or the first, the second and the third row in the RREF. \u2013\u00a0gimusi Jun 16 '18 at 20:03\n\u2022 Im sorry but i don't really understand what you wrote... sorry :( I am willing to find a basis for a subspace, and i am not sure whether to use the row echelon form or reduced row echelon form ... or if i can use both interchangeably. If i used the row echelon form , then the basis are the first three columns of the original matrix. But what do you mean when you say that they can be rows too? I am kinds of confused \u2013\u00a0Sara Martini Jun 16 '18 at 20:09\n\u2022 @SaraMartini It depends upon which subspace you are looking for and how you arrange the vector in the matrix. \u2013\u00a0gimusi Jun 16 '18 at 20:10\n\u2022 This is the subspace.... (r-s-2t-u) , (-r+2s+5t+2u), (s+3t+u), (3r+s+5u) \u2013\u00a0Sara Martini Jun 16 '18 at 20:13", "date": "2019-07-21 00:32:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2821923/how-to-put-a-matrix-in-reduced-echelon-form", "openwebmath_score": 0.8099325895309448, "openwebmath_perplexity": 142.37941037110525, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138151101525, "lm_q2_score": 0.897695298265595, "lm_q1q2_score": 0.8776890948737012}}
{"url": "https://www.physicsforums.com/threads/identifying-types-of-converging-series.826074/", "text": "# Identifying Types of Converging Series\n\n1. Aug 3, 2015\n\n### DameLight\n\nHello, I'm looking for some help with this problem for my Calculus 2 class. Since it's a summer class my professor wasn't able to explain everything fully so if you can help me that would be great : )\n\n1. The problem statement, all variables and given/known data\n\nSelect the FIRST correct reason why the given series converges.\n\nA. Convergent geometric series\nB. Convergent p series\nC. Comparison (or Limit Comparison) with a geometric or p series\nD. Alternating Series Test\nE. None of the above\n\n1. \u03a3n=1 6(4)n/72n\n2. \u03a3n=1 sin2(6n)/n2\n3. \u03a3n=1 cos(n\u03c0)/ln(2n)\n4. \u03a3n=1 (\u22121)n/(6n+5)\n5. \u03a3n=1 (n+1)(24)n/52n\n6. \u03a3n=1 (\u22121)n * \u221a(n)/(n+9)\n\n2. Relevant equations\nGeometric Series:\n\u03a3n=1 arn-1 will converge if -1 < r < 1\n\nP Series:\n\u03a3n=1 1/np converges if p > 1\n\nAlternating Series Test:\n1) bn+1 </= bn for all n and\n2) limn->\u221e bn = 0\n\n3. The attempt at a solution\n1. A\n2. B\n3. E\n4. D\n5. C\n6. D\n\nLast edited by a moderator: Aug 3, 2015\n2. Aug 3, 2015\n\n### Ray Vickson\n\nI get (in order 1-6) A, C, D, D, E, D.\n\n3. Aug 4, 2015\n\n### DameLight\n\nMy professor told me that I got numbers 2 and 3 correct\n\n4. Aug 4, 2015\n\n### Ray Vickson\n\nI disagree. Here are the details.\n1. $\\sum 6 \\;4^n/7^{2n} = 6 \\sum (4/49)^n$, so geometric (A).\n2. In $\\sum \\sin^2 (6n) /n^2$ the terms are (i) not geometric; (ii) not alternating; (iii) not in the simple form $1/n^p$. However, $0 \\leq \\sin^2(6n)/n^2 \\leq 1/n^2$, so a comparison with the series $1/n^2$ works (C).\n3. $\\sum \\cos(n \\pi)/ \\ln(n)$ is an alternating series, since $\\cos(n \\pi) = (-1)^n$. So, D.\n4. $\\sum (-1)^n / (6n+5)$ is alternating, so D.\n5. $\\sum (n+1) \\; 24^n / 5^{2n} = \\sum (n+1) r^n, \\;\\; r = 24/25$. Here, A,B,C,D do not apply, and that leaves E.\n6. $\\sum (-1)^n \\sqrt{n}/(n+9)$ (or $\\sum (-1)^n \\sqrt{ n/(n+9)}$---can't tell which you mean) is alternating in either interpretation, so D.\n\n5. Aug 4, 2015\n\n### DameLight\n\nThe correct answer ended up being:\n\n1. A\n2. C\n3. D\n4. D\n5. C\n6. D\n\nThank you for your help : )\n\n6. Aug 4, 2015\n\n### Ray Vickson\n\nHow do you get 5 C? Did you read my detailed explanation?\n\n7. Aug 4, 2015\n\n### micromass\n\nWhy wouldn't the comparison test apply in $5$? You can do $n+1\\leq \\alpha^n$ for some small enough $\\alpha>1$.\n\n8. Aug 4, 2015\n\n### Ray Vickson\n\nOK. That would be an answer to my question (why?) that the OP did not answer. For all I know the OP may have used false reasoning.\n\n9. Aug 4, 2015\n\n### DameLight\n\nI don't want to assume, but that sounded a bit rude.\n\nI am assuming that going off of your explanation for 5 that we can compare that to a geometric function \u2211n=0 anrn and since the absolute value of r is less than 1 we can go ahead and reason that the limit would be equivalent to a/(1-r)\n\n10. Aug 5, 2015\n\n### Ray Vickson\n\nNo: a geometric series would be $\\sum r^n$ (or $\\sum c r^n = c \\sum r^n$ for constant $c$), but $\\sum a_n r^n$ is NOT a geometric series if $a_n$ varies with $n$. However, if $a_n \\to c = \\text{constant}$ as $n \\to \\infty$, then comparison with the series $\\sum c r^n$ can be made to work without too much effort (although some extra effort is needed), However, in your case you have $a_n = n+1 \\to +\\infty$ as $n \\to \\infty$, so that type of argument does not work automatically. If you thought it did, that is what what I would call \"false reasoning\".\n\nHowever, as 'micromass' has pointed out, if you choose a small enough $\\epsilon > 0$ so that $(1+\\epsilon) r < 1$, then for some finite $N> 0$ we have $n+1 \\leq (1+\\epsilon)^n$ for all $n \\geq N$, so for $n \\geq N$ we have $0 \\leq (n+1) r^n \\leq [(1+\\epsilon) r]^n$ and we can compare with the geometric series $\\sum [(1+\\epsilon) r]^n$. Note, however, that some such argument as this must be acknowledged and recognized in order for C to be a valid and well-founded answer. Without this realization, just saying it would, again, be false reasoning.\n\nLast edited: Aug 5, 2015", "date": "2018-02-20 06:28:05", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/identifying-types-of-converging-series.826074/", "openwebmath_score": 0.8369911313056946, "openwebmath_perplexity": 1510.018116703927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98087596269007, "lm_q2_score": 0.8947894590884705, "lm_q1q2_score": 0.8776774720883306}}
{"url": "http://www.wolterskluwerlb.com/andy-cutting-yhldkzq/unique-left-inverse-62f161", "text": "5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 In a monoid, if an element has a right inverse\u2026 Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego h\ufffdb\ufffdy\ufffd\ufffd\ufffd cca\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\u0650\ufffd q\ufffd\ufffd\ufffd#\ufffd!\ufffdA\ufffd\u046cQ\ufffda\ufffd\ufffd\ufffd[\ufffd50\ufffdF\ufffd\ufffd3&9'\ufffd\ufffd0 qp\ufffd(R\ufffd&\ufffda\ufffds4\ufffdp\ufffd[\ufffd\ufffd\ufffdf^'w\ufffdP&\u07b6 7\ufffd\ufffd,\ufffd\ufffd\ufffd[T\ufffd+\ufffdJ\ufffd\ufffd\ufffd\ufffd9\ufffd$\ufffd\ufffd4r\ufffd:4';m$\ufffd\ufffd#\ufffds\ufffdOj\ufffdL\u00cc\ufffdcY{-\ufffdXTA\u06bd\ufffdBEOpr\ufffdl\ufffdT\ufffd\ufffdf1\ufffdM\ufffd1$\ufffd\ufffd\u0421\ufffd\ufffd6I\ufffd\ufffd\u048e)w It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. Generalized inverse Michael Friendly 2020-10-29. best. %PDF-1.4 given $$n\\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. 11.1. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y \u2208 U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. U-semigroups endobj Proof: Assume rank(A)=r. x\ufffd\ufffdXKo#7\ufffd\ufffdW\ufffdhE\ufffd[\u05e2\ufffd\ufffdE\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd:v\ufffd4q\ufffd\ufffd\ufffd/)\ufffdc\ufffd\ufffd\ufffd\ufffd>~\"%\ufffd\ufffdd\ufffd\ufffdN\ufffd\ufffd8\ufffdw(LY\u027d2L:\ufffdAZv\ufffdb\ufffd\ufffd\u065e\u0473G\ufffd\ufffd\ufffd8>\ufffd\ufffd\ufffd\ufffd'\ufffd\ufffdx\ufffd\u067crc\ufffd\ufffd\u007f>?\ufffd\ufffd[\ufffd\ufffd?\ufffd'\ufffd\ufffd\ufffd(%#R\ufffd\ufffd1 .\ufffd-7\ufffd;6\ufffdSg#>Q\ufffd\ufffd7\ufffd##\u03e5 \"\ufffd[\ufffd \ufffd\ufffd\ufffdN)&Q \ufffd\ufffdM\ufffd\ufffd\ufffdYy\ufffd\ufffd?A\ufffd\ufffd\ufffd\ufffd4\ufffd\u03e0H\ufffd%\ufffdf\ufffd\ufffd0a;N\ufffdM\ufffd,\ufffd!{\ufffd\ufffdy\ufffd<8(t1\u0199\ufffdzi\ufffd\ufffd\ufffde\ufffd\ufffdA\ufffd\ufffd(;p*\ufffd\ufffd\ufffd\ufffdV\ufffdJ\u069b,\ufffdt~\ufffdd\ufffd\ufffd\u0318H9\ufffd\ufffd\ufffd\ufffd/\ufffd\ufffd_a\ufffd\ufffd\ufffdv\ufffd68gq\"\ufffd\ufffd\ufffdD\ufffd|a5\ufffd\ufffd\ufffd\ufffdP|Jv\ufffd\ufffdl1j\ufffd\ufffdx\ufffd\ufffd&\u07baN\ufffd\ufffd\ufffd\ufffdV\"\ufffd\ufffd\ufffd\"\ufffd\ufffd\u007f\ufffd\ufffd}! Let G G G be a group. left A rectangular matrix can\u2019t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. \ufffdn\ufffd\ufffd\ufffd\ufffd\ufffdr\ufffd\ufffd\ufffd\ufffd6\ufffd\ufffd\ufffdd}\ufffd\ufffd\ufffdwF>\ufffdG\ufffd/\ufffd\ufffdk\ufffd K\ufffdT\ufffdSE\ufffd\ufffd\ufffd\ufffd \ufffd&\u02ac\ufffdRbl\ufffdj\ufffd\ufffd|\ufffdTx\ufffd\ufffd)\ufffd\ufffdRdy\ufffdY ? 0 Yes. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent \u2026 (An example of a function with no inverse on either side is the zero transformation on .) Recall also that this gives a unique inverse. If f contains more than one variable, use the next syntax to specify the independent variable. Viewed 1k times 3. Note that other left In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. Ask Question Asked 4 years, 10 months ago. Let (G, \u2295) be a gyrogroup. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. eralization of the inverse of a matrix. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? By using this website, you agree to our Cookie Policy. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Let e e e be the identity. Some easy corollaries: 1. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by \u2192 \u21a6 \u22c5 \u2192 has the two-sided inverse \u2192 \u21a6 (/) \u22c5 \u2192.In this subsection we will focus on two-sided inverses. Show Instructions. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Actually, trying to prove uniqueness of left inverses leads to dramatic failure! '+o\ufffdf P0\ufffd\ufffd\ufffd'\ufffd,\ufffd\\\ufffd y\ufffd\ufffd\ufffd\ufffdbf\\\ufffd; wx.\ufffd\ufffd\";MY\ufffd}\ufffd\ufffd\ufffd\ufffd\u0625\ufffd endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Show Instructions. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . If BA = I then B is a left inverse of A and A is a right inverse of B. share. 6 comments. Note the subtle difference! u (b 1 , b 2 , b 3 , \u2026) = (b 2 , b 3 , \u2026). 3. This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. Active 2 years, 7 months ago. Recall that$B$is the inverse matrix if it satisfies $AB=BA=I,$ where$I$is the identity matrix. Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. Then a matrix A\u2212: n \u00d7 m is said to be a generalized inverse of A if AA\u2212A = A holds (see Rao (1973a, p. 24). (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) Let\u2019s recall the definitions real quick, I\u2019ll try to explain each of them and then state how they are all related. For any elements a, b, c, x \u2208 G we have: 1. Proof: Assume rank(A)=r. inverse Proof (\u21d2): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (\u21d0): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. If the function is one-to-one, there will be a unique inverse. Proof. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Remark When A is invertible, we denote its inverse \u2026 Theorem 2.16 First Gyrogroup Properties. There are three optional outputs in addition to the unique elements: However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. (We say B is an inverse of A.) If E has a right inverse, it is not necessarily unique. Let\u2019s recall the definitions real quick, I\u2019ll try to explain each of them and then state how they are all related. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. u(b_1,b_2,b_3,\\ldots) = (b_2,b_3,\\ldots). If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. save hide report. inverse. Let (G, \u2295) be a gyrogroup. A.12 Generalized Inverse De\ufb01nition A.62 Let A be an m \u00d7 n-matrix. 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about \"the\" inverse of f). Hence it is bijective. h\ufffd\ufffd[[\ufffd\u06f6\ufffd+|l\\wp\ufffd\ufffd\u07dd\ufffdN\\\ufffd\ufffd&\ufffd\u4052\ufffd]\ufffd\ufffd%\"e\ufffd\ufffd\ufffd{>\ufffd\ufffdHJZi\ufffdk\ufffdm\ufffd \ufffdwnt.I\ufffd%. G is called a left inverse for a matrix if 7\u201a8 E GE\u0153M 8 \u00d0 \u00d1so must be G 8\u201a7 It turns out that the matrix above has E no left inverse (see below). 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective Theorem. New comments cannot be posted and votes cannot be cast. Yes. \ufffd\ufffd\ufffd See Also. 125 0 obj <>stream Sort by. 53 0 obj <> endobj wqhh\ufffd\ufffdllf\ufffd)eK\ufffdy\ufffdI\ufffd\ufffdbq\ufffd(\ufffd\ufffd\ufffd\ufffd\ufffd\u00c3.4-\ufffd{xe\ufffd\ufffd8\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdb\ufffdc[\ufffd\ufffd\ufffd\u00f6\ufffd\ufffd\ufffd\ufffdTBYb\ufffd\u02834\ufffd\ufffd\ufffd&\ufffd1\ufffd\ufffd\ufffd\ufffdo[{cK\ufffdsAt\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd3\ufffd'vp=\u007f\ufffd$\ufffd\ufffd$\ufffdi.\ufffd\ufffdj8@\ufffdg\ufffdUQ\ufffd\ufffd\ufffd>\ufffd\ufffdg\ufffdlI&\ufffdOuL\ufffd\ufffd*\ufffd\ufffd\ufffdwCu\ufffd0 \ufffd]l\ufffd It's an interesting exercise that if$a$is a left unit that is not a right uni Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. 100% Upvoted. Let f : A \u2192 B be a function with a left inverse h : B \u2192 A and a right inverse g : B \u2192 A. So to prove the uniqueness, suppose that you have two inverse matrices$B$and$C$and show that in fact$B=C$. 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Does it necessarily have a two sided inverse because either that matrix or transpose! Is the zero transformation on. the matrix $a$ either side is same! Distance from b to any point in u that minimizes the distance from b to any point in.! That is both a left inverse of b one-to-one, there will be function... You impose more conditions on G ; see Section 3 below., p indeed... Var )... finverse does not issue a warning when the inverse is unique it... J denotes the i-th row of a function with no inverse on either side is the zero on!", "date": "2021-04-11 06:28:45", "meta": {"domain": "wolterskluwerlb.com", "url": "http://www.wolterskluwerlb.com/andy-cutting-yhldkzq/unique-left-inverse-62f161", "openwebmath_score": 0.8560187816619873, "openwebmath_perplexity": 987.8661969525623, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.965899577232538, "lm_q2_score": 0.9086179025005187, "lm_q1q2_score": 0.8776336478911665}}
{"url": "https://math.stackexchange.com/questions/2148688/is-there-a-bijection-f-mathbbn-rightarrow-mathbbn-such-that-the-series", "text": "# Is there a bijection $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that the series $\\sum\\limits_n \\frac{1}{n+f(n)}$ is convergent?\n\nIs there a bijection $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that the series $\\sum_1 ^\\infty \\frac{1}{n+f(n)}$ is convergent?\n\nI could not solve this. I tried to proceed in following lines:\n\n1) Tried to provide a contradiction:\n\nFirst let $n \\sim m$ iff $\\exists k \\in \\mathbb{Z}$ such that $f^k(n)=m$. This is an equivalence relation. Consider the orbits. For the finite orbits we can compare the series to $\\sum_1^\\infty \\frac{1}{n+n}$. But then I could not figure out how to proceed for infinite orbits.\n\n2) Tried to prove that there is some function:\n\nLet $\\{k_n\\}$ be a subsequence of $\\mathbb{N}$ such that $\\sum_0^\\infty \\frac{1}{k_n}$ converges. Set $f(n)=k_{n}, \\forall n \\in \\mathbb{N}\\setminus\\{k_n\\}$. Then images of each $n$ which are not in the subsequence $k_n$ is defined. Now we have to define images of each $k_n$. Define $f(k_n)=n$ $\\forall n \\in \\mathbb{N}\\setminus \\{k_n\\}.$.\n\nCould not proceed further. I think My second attempt was going in right direction. My plan was use the fact that all elements here are positive and to construct the function $f$ in such a manner that $\\forall n\\in \\mathbb{N}$ either $n$ or $f(n)$ is in $\\{k_n\\}$.\n\n\u2022 Decompose $\\mathbb N$ into the disjoint union of $S=\\{k^2\\mid k\\in\\mathbb N\\}$ and $R=\\mathbb N\\setminus S$. Decompose $S$ into the disjoint union $\\bigcup\\limits_{k\\in\\mathbb N}S_k$, where the size of $S_k$ is $2k$ and every element of $S_k$ is smaller than every element of $S_{k+1}$, for every $k$. Then enumerate $R=\\{r_k\\mid k\\in\\mathbb N\\}$ with $r_k<r_{k+1}$ for every $k$, reorder $\\mathbb N$ as $$S_1\\ r_1\\ S_2\\ r_2\\ S_3\\ \\ldots\\ S_k\\ r_k\\ S_{k+1}\\ \\ldots$$ and consider the corresponding function $f$. Then $$\\sum_{n\\in S}\\frac1{n+f(n)}<\\sum_{n\\in S}\\frac1{n}=\\sum_k\\frac1{k^2}$$\n\u2013\u00a0Did\nFeb 17 '17 at 12:51\n\u2022 ... converges and $r_k\\geqslant k^2$ for each $k$, hence $$\\sum_{n\\in R}\\frac1{n+f(n)}<\\sum_k\\frac1{r_k}\\leqslant\\sum_k\\frac1{k^2}$$ converges hence the whole series converges.\n\u2013\u00a0Did\nFeb 17 '17 at 12:51\n\u2022 @Did, I think you want to order $\\mathbb{N}$ as $r_1,S_1,r_2,S_2,r_3,S_3,\\ldots$ (providing that $0\\notin\\mathbb{N}$). But well, if $0\\in \\mathbb{N}$, then we have to modify this solution only slightly. I would give you an upvote if you posted this as a solution. Feb 17 '17 at 13:03\n\u2022 @Batominovski \"I think you want to order N as r1,S1,r2,S2,r3,S3,\u2026\" Hmmm, why? What difference does it make?\n\u2013\u00a0Did\nFeb 17 '17 at 19:36\n\u2022 @Did The difference is that $f^{-1}\\left(r_k\\right)\\geq k^2$ may not hold. You have $f^{-1}\\left(r_k\\right)\\geq k^2-1$ instead. (However, nothing really changes.) Feb 18 '17 at 2:26\n\nOriginally, I gave this example:\n\n$$f (n)=\\begin{cases}k,&\\ \\text { if } n=3^k , \\text { with k  not a power of 2}\\\\ 2^{n-1} , &\\ \\text { otherwise }\\end{cases}$$ (the idea is to push the small numbers further and further down the road so that when they appear they are compensated by the $n$). Then $$\\sum_n\\frac1 {n+f (n)}<\\sum_{k}\\frac 1 {3^k+k}+\\sum_n\\frac1 {n+2^{n-1}}<\\infty.$$\n\nAnd it is the right idea, but the problem is that such $f$ is not onto. For instance, $2^{26}$ is not in the range of $f$, because when $n=27$, we are using the other branch of $f$ to get $3$.\n\nSo we need to tweak the example slightly. Let $$T=\\{3^k:\\ k\\ \\text{ is not a power of } 2\\}=\\{3^3,3^5,3^6,3^7,3^9,\\ldots\\}$$ and $$S=\\mathbb N\\setminus T=\\{1,\\ldots,7,8,10,11,\\ldots\\}.$$ Write them as an ordered sequence, $T=\\{t_1,t_2,\\ldots\\}$ and $S=\\{s_1,s_2,\\ldots\\}$. Now define $$f(n)=\\begin{cases} \\log_3 n,&\\ \\text{ if }\\ n=t_k\\\\ \\ \\\\ 2^{k-1},&\\ \\text{ if }\\ n=s_k \\end{cases}$$ One can check explicitly that $$g(m)=\\begin{cases} 3^m,&\\ \\text{ if m is not a power of 2}\\ \\\\ \\ \\\\ s_{k+1},&\\ \\text{ if }\\ m=2^k \\end{cases}$$ is an inverse for $f$.\n\nNot a new solution, just writing to clarify for myself how the solution of @Martin Argerami: works.\n\nWe want $\\sum_{n\\in \\mathbb{N}} \\frac{1}{n + f(n)} < \\infty$. Consider a covering $\\mathbb{N} = A\\cup B$. It's enough to have $\\sum_{n\\in A} \\frac{1}{n + f(n)}$, $\\sum_{n \\in B} \\frac{1}{n + f(n)}< \\infty$. So it's enoough to find $A$, $B$ so that $$\\sum_{n \\in A} \\frac{1}{f(n)}= \\sum_{n \\in f(A)} \\frac{1}{n} < \\infty \\\\ \\sum_{n \\in B} \\frac{1}{n} < \\infty$$\n\nSo it's enough to have $B$ so that $\\sum_{n\\in B} \\frac{1}{n} < \\infty$ and $f$ mapping $A = \\mathbb{N} \\backslash B$ to $B$. There are many possibilities here.\n\n\u2022 I would not say that \"this is how the other solution works\" because the other solution relies heavily on infinite orbits while a simple involution between the powers of two and the non-powers of two works fine. Nov 19 '18 at 10:58", "date": "2021-09-29 03:23:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2148688/is-there-a-bijection-f-mathbbn-rightarrow-mathbbn-such-that-the-series", "openwebmath_score": 0.9631305932998657, "openwebmath_perplexity": 113.49048852989908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180690117799, "lm_q2_score": 0.8902942275774319, "lm_q1q2_score": 0.8775791068599603}}
{"url": "https://www.physicsforums.com/threads/is-there-a-formula-for-this-gaussian-integral.751032/", "text": "# Is there a formula for this gaussian integral\n\n1. Apr 28, 2014\n\n### skate_nerd\n\nIs there a formula for this gaussian integral:\n\n$$int_{-\\infty}^{\\infty}{x^4}{e^{-a(x-b)^2}}dx$$\n\nI've tried wikipedia and they only have formulas for the integrand with only x*e^... not x^4e^...\nWolframalpha won't do it either, because I actually have an integral that looks just like that, unknown constants and all.\n\n2. Apr 28, 2014\n\n### Matterwave\n\nWell, we know that:\n\n$$\\int_{-\\infty}^\\infty e^{-bx^2}dx=\\sqrt{\\frac{\\pi}{b}}$$\n\nTaking some derivatives:\n\n$$\\frac{d^2}{db^2}\\int_{-\\infty}^\\infty e^{-bx^2}dx=\\int_{-\\infty}^{\\infty}x^4e^{-bx^2}dx=\\frac{3}{4}\\frac{\\sqrt{\\pi}}{b^{5/2}}$$\n\nI'm not sure if you can use this for your integral, but it looks possible.\n\n3. Apr 28, 2014\n\n### skate_nerd\n\nSorry, but you can't :/\n\n4. Apr 28, 2014\n\n### vanhees71\n\n5. Apr 28, 2014\n\n### pwsnafu\n\nYou are basically just asking for the moments of the normal distribution. These are well known:\n$$\\int_{-\\infty}^{\\infty} \\frac{x^p}{\\sigma \\sqrt{2\\pi}} e^{-(x-\\mu)^2 / (2 \\sigma^2)} dx= \\sigma^p (p-1)!$$\nwhen p is even, and zero when p is odd.\n\nSo $p=4$, $\\sigma = \\frac{1}{\\sqrt{2a}}$ and $\\mu = b$, which gives $\\frac{3\\sqrt{\\pi}}{4 a^{5/2}}$ as Matterwave gave.\n\n6. Apr 28, 2014\n\n### skate_nerd\n\nI'm sorry i just dont think that's correct. Try wolframalpha. I know the formulas for the x, x^2, x^3 cases, but not the x^4. But everything up to x^3 agrees with wolframalpha's output, and this formula you guys are giving doesnt look very similar and does not agree.\n$$\\int_{-\\infty}^{\\infty}{x}e^{-a(x-b)^2}dx=b\\sqrt{\\frac{\\pi}{a}}$$\n$$\\int_{-\\infty}^{\\infty}{x^2}e^{-a(x-b)^2}dx=(\\frac{1}{2a}+b^2)\\sqrt{\\frac{\\pi}{a}}$$\n$$\\int_{-\\infty}^{\\infty}{x^3}e^{-a(x-b)^2}dx=(\\frac{3b}{2a}+b^3)\\sqrt{\\frac{\\pi}{a}}$$\nThere's a pattern here but I can't figure it out.\n\n7. Apr 28, 2014\n\n### pwsnafu\n\nI'm an idiot, I looked up the central moments (and even that's wrong because it should be $(p-1)!!$). The non-central moments are\n\n$\\mu$\n$\\mu^2+\\sigma^2$\n$\\mu^3+3\\mu\\sigma^2$\n$\\mu^4+6\\mu^2\\sigma^2+3\\sigma^4$,\nfor p = 1, 2, 3, and 4. Those should agree with what you have.\n\n8. Apr 28, 2014\n\n### skate_nerd\n\nAhhh thank you thank you thank you! That works 100%! Man I've been trying to get this formula for a week...I really appreciate it.\n\n9. Apr 28, 2014\n\n### eigenperson\n\nThe \"freshman calculus\" way to do this is to integrate by parts:\n$$u = x^3 \\qquad dv = xe^{-x^2}\\,dx$$so that\n$$du = 3x^2 \\qquad v = -{1 \\over 2}e^{-x^2}$$and now the degree of the polynomial has been reduced by 2.\n\nNow, that is for the case where a and b are 0, but the other cases are really the same. You just have to change variables; then you will have a more complicated polynomial in place of $x^4$, but once you know how to do $\\int x^{2k}e^{-x^2}$ you can integrate any polynomial times $e^{-x^2}$.\n\n10. Apr 28, 2014\n\n### Matterwave\n\nAnother clever way to do this integral is to look at the characteristic function for a Gaussian distribution:\n\n$$C(t)\\equiv\\left<e^{itx}\\right>=\\frac{1}{\\sigma\\sqrt{2\\pi}} \\int_{-\\infty}^{\\infty}e^{itx}e^{-(x-\\mu)^2/2\\sigma^2}$$\n\nYou can do this integral by completing the square in the exponential, or by just looking it up. You will find:\n\n$$C(t)=e^{it\\mu-\\frac{1}{2}t^2\\sigma^2}$$\n\nYou can expand this function in powers of t, and match it to the expansion of:\n\n$$\\left<e^{itx}\\right>\\approx 1+\\left<itx\\right>+\\frac{\\left<(itx)^2\\right>}{2}+...$$\n\nIf you match the powers of t on both sides, you can get all the general moments of the Gaussian distribution. You can get all the odd ones too, for which the formula I gave earlier would no longer work. This Characteristic function is closely related to the moment generating function, but I'm more familiar with working with these.", "date": "2017-10-18 15:44:55", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/is-there-a-formula-for-this-gaussian-integral.751032/", "openwebmath_score": 0.8172184824943542, "openwebmath_perplexity": 526.5231643027236, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180660748889, "lm_q2_score": 0.8902942239389252, "lm_q1q2_score": 0.8775791006587214}}
{"url": "https://mathhelpboards.com/threads/trouble-with-trig-integral.27081/", "text": "# Trouble with trig integral\n\n#### Kristal\n\n##### New member\nI have a few questions and a request for an explanation.\n\nI worked this problem for a quite a while last night. I posted it here.\n\nhttps://math.stackexchange.com/questions/3547225/help-with-trig-sub-integral/3547229#3547229\n\nThe original problem is in the top left. Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.\n\nMy first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?\n\nMy second question is: is this a correct solution? https://imgur.com/2tgEz0O\n\nIt is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p\n\nFinally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some advice about how I could improve my math tricks to that point?\n\n#### skeeter\n\n##### Well-known member\nMHB Math Helper\nIs the algebra in my original work sound?\nLooks like you did this (correct me if I'm mistaken) ...\n\n$$\\displaystyle -7\\int \\dfrac{x^2}{\\sqrt{4x-x^2}} \\,dx = -7\\int \\dfrac{x^2}{\\sqrt{x}\\sqrt{4-x}} \\,dx = -7\\int \\dfrac{x^{3/2}}{\\sqrt{4-x}} \\,dx$$\n\nfrom here, I see you attempted to use the trig sub $\\sqrt{x} = 2\\sin{\\theta}$.\n\nyou made a mistake determining $dx$ ...\n\n$\\sqrt{x} = 2\\sin{\\theta} \\implies x = 4\\sin^2{\\theta} \\implies dx = 8\\sin{\\theta}\\cos{\\theta} \\, d\\theta$\n\n#### Kristal\n\n##### New member\nI did sub it in as a replacement for $\\sqrt{x}$ though.\n\nI posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\\theta=\\frac{1}{\\sqrt{x}}dx$\n\nThis is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?\n\nIf so, well it would be nice if a teacher told us these things sometimes...\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nI did sub it in as a replacement for $\\sqrt{x}$ though.\n\nI posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\\theta=\\frac{1}{\\sqrt{x}}dx$\n\nThis is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?\n\nIf so, well it would be nice if a teacher told us these things sometimes...\nHi Kristal, welcome to MHB!\n\nThere are 2 ways that are generally taught to do a substitution.\n\nYou have the integral\n$$\\int \\frac{x^{3/2}}{\\sqrt{4-x}}\\,dx$$\nand you want to replace $\\sqrt x$ by $2\\sin\\theta$.\n\nMethod 1\nWrite $x$ as a function of $\\theta$ and take the derivative as skeeter suggested:\n$$\\sqrt x=2\\sin\\theta \\implies x=4\\sin^2\\theta \\implies dx=d(4\\sin^2\\theta) = 8\\sin\\theta\\cos\\theta\\,d\\theta$$\nWe can now replace $\\sqrt x$ and $dx$ to find:\n$$\\int \\frac{x^{3/2}}{\\sqrt{4-x}}\\,dx = \\int \\frac{(2\\sin\\theta)^{3}}{\\sqrt{4-(2\\sin\\theta)^2}}\\,d(4\\sin^2\\theta) = \\int \\frac{(2\\sin\\theta)^{3}}{\\sqrt{4-(2\\sin\\theta)^2}}\\cdot 8\\sin\\theta\\cos\\theta\\,d\\theta$$\n\nMethod 2\nTake the derivative left and right, and do what needs to be done to eliminate $x$.\nIn this case:\n$$\\sqrt x=2\\sin\\theta \\implies d(\\sqrt x)=d(2\\sin\\theta) \\implies \\frac 1{2\\sqrt x}\\,dx= 2\\cos\\theta\\,d\\theta \\implies \\frac{dx}{\\sqrt x} = 4\\cos\\theta\\,d\\theta$$\nRewrite the integral a bit to match and substitute:\n$$\\int \\frac{x^{3/2}}{\\sqrt{4-x}}\\,dx = \\int \\frac{x^2}{\\sqrt{4-x}}\\cdot\\frac{dx}{\\sqrt x} = \\int \\frac{(2\\sin\\theta)^4}{\\sqrt{4-(2\\sin\\theta)^2}}\\cdot 4\\cos\\theta\\,d\\theta$$\n\nWe can see that the result is the same in both cases.\n\nI'm afraid I can't comment on what your teacher is telling you since I don't know what that is.\nWe can only point out and show what the correct math is.\n\n#### skeeter\n\n##### Well-known member\nMHB Math Helper\nmy calculator\u2019s CAS worked out these antiderivatives ...\n\n#### Attachments\n\n\u2022 80.3 KB Views: 13", "date": "2020-08-03 09:30:16", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/trouble-with-trig-integral.27081/", "openwebmath_score": 0.811976432800293, "openwebmath_perplexity": 719.6694925879025, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9518632288833652, "lm_q2_score": 0.9219218375365862, "lm_q1q2_score": 0.8775434970556601}}
{"url": "http://www.chegg.com/homework-help/questions-and-answers/using-bisection-search-to-make-the-program-faster-youll-notice-that-in-problem-2-your-mont-q3035262", "text": "## Program in Python: USING BISECTION SEARCH TO MAKE THE PROGRAM FASTER\n\nUSING BISECTION SEARCH TO MAKE THE PROGRAM FASTER\nYou'll notice that in problem 2, your monthly payment had to be a multiple of $10. Why did we make it that way? You can try running your code locally so that the payment can be any dollar and cent amount (in other words, the monthly payment is a multiple of$0.01). Does your code still work? It should, but you may notice that your code runs more slowly, especially in cases with very large balances and interest rates. (Note: when your code is running on our servers, there are limits on the amount of computing time each submission is allowed, so your observations from running this experiment on the grading system might be limited to an error message complaining about too much time taken.)\nWell then, how can we calculate a more accurate fixed monthly payment than we did in Problem 2 without running into the problem of slow code? We can make this program run faster using a technique introduced in lecture - bisection search!\nThe following variables contain values as described below:\n1. balance - the outstanding balance on the credit card\n2. annualInterestRate - annual interest rate as a decimal\nTo recap the problem: we are searching for the smallest monthly payment such that we can pay off the entire balance within a year. What is a reasonable lower bound for this payment value? $0 is the obvious anwer, but you can do better than that. If there was no interest, the debt can be paid off by monthly payments of one-twelfth of the original balance, so we must pay at least this much every month. One-twelfth of the original balance is a good lower bound. What is a good upper bound? Imagine that instead of paying monthly, we paid off the entire balance at the end of the year. What we ultimately pay must be greater than what we would've paid in monthly installments, because the interest was compounded on the balance we didn't pay off each month. So a good upper bound for the monthly payment would be one-twelfth of the balance, after having its interest compounded monthly for an entire year. In short: Monthly interest rate = (Annual interest rate) / 12 Monthly payment lower bound = Balance / 12 Monthly payment upper bound = (Balance x (1 + Monthly interest rate)12) / 12 Write a program that uses these bounds and bisection search (for more info check out the Wikipedia page on bisection search) to find the smallest monthly payment to the cent (no more multiples of$10) such that we can pay off the debt within a year. Try it out with large inputs, and notice how fast it is (try the same large inputs in your solution to Problem 2 to compare!). Produce the same return value as you did in problem 2.\nNote that if you do not use bisection search, your code will not run - your code only has 30 seconds to run on our servers. If you get a message that states \"Your submission could not be graded. Please recheck your submission and try again. If the problem persists, please notify the course staff.\", check to be sure your code doesn't take too long to run.\nThe code you paste into the following box should not specify the values for the variables balance orannualInterestRate - our test code will define those values before testing your submission.\n\nNote:\nDepending on where, and how frequently, you round during this function, your answers may be off a few cents in either direction. Try rounding as few times as possible in order to increase the accuracy of your result.", "date": "2013-05-23 00:15:01", "meta": {"domain": "chegg.com", "url": "http://www.chegg.com/homework-help/questions-and-answers/using-bisection-search-to-make-the-program-faster-youll-notice-that-in-problem-2-your-mont-q3035262", "openwebmath_score": 0.29359132051467896, "openwebmath_perplexity": 605.4931693521511, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750525759487, "lm_q2_score": 0.8887587853897074, "lm_q1q2_score": 0.8775382524514986}}
{"url": "https://math.stackexchange.com/questions/1874573/a-trivial-combinatorics-result-i-found-is-my-proof-correct", "text": "# A trivial combinatorics result I found, is my proof correct?\n\nI have just finished highschool and have started learning on my own some combinatorics and how to do proofs, and while messing around with sums and Pascal's triangle I found an interesting yet trivial property that I tried to prove. I'm assuming that it has already been found, but I couldn't find anything mentioning it.\nSo my questions are: Is my proof correct? Has this property been found already? What can this property be used for?\n\nLet $$f(x,n)= \\sum_{i_1=1}^n \\sum_{i_2=1}^{i_1} \\sum_{i_3=1}^{i_2} \\cdots \\sum_{i_x=1}^{i_{x-1}}i_x \\$$ where $x$ is the number of sigma sums.\n\nMy conjecture is that $$f(x,n)={n+x \\choose x+1}$$\n\nProof by induction\n\nBasis step:\n\\begin{align} f(1,n) &=\\sum_{i=1}^ni\\\\ &=\\frac{n(n+1)}{2}\\\\ &={n+1 \\choose 2} \\end{align} The basis step works. This first result is already proven for all n.\n\nInductive step: Assume that $f(x,n)= {n+x \\choose x+1}$\n\nNow, \\begin{align}f(x+1,n) &=\\sum_{m=1}^n \\sum_{i_1=1}^{m} \\sum_{i_2=1}^{i_1} \\cdots \\sum_{i_x=1}^{i_{x-1}}i_x\\\\ &=\\sum_{m=1}^n f(x,m)\\\\ &=\\sum_{m=1}^n {m+x \\choose x+1}\\\\ &=\\sum_{l=0}^{n-1}{l+1+x \\choose x+1}\\\\ &={n+x+1 \\choose x+2}\\\\ \\end{align} Therefore, if $f(x,n)$ holds, then $f(x+1,n)$ holds.\nWhat I am not sure about is whether I have to use induction to prove that this property holds for every integer $n$ as well,\n\n\u2022 This is indeed correct! (Even verified it for some test cases just to be extra sure.) \u2013\u00a0Cameron Williams Jul 29 '16 at 3:59\n\u2022 In the phrase \"where $x$ is the number of sigma sums.\", why did you use a capital rather than lower-case \"W\" in \"Where\"? That seems to be done ninety-percent of the time both here and in Wikipedia, and it doesn't make sense since it's not the beginning of a new sentence. $\\qquad$ \u2013\u00a0Michael Hardy Jul 29 '16 at 5:06\n\u2022 Congratulations on working through this and producing a proof. It doesn't matter if the result is already widely known or not. The important thing is that you did it yourself. Being taught mathematics is a pointless occupation. Constructing mathematics yourself is the way to go. Find teachers to guide your discovery so that you don't bang your head against too many walls, and you will become a fine mathematician \u2013\u00a0Martin Kochanski Jul 29 '16 at 5:14\n\nThe proof is correct since you can treat $n$ as fixed. You don't have to induct on $n$ since your base case holds for all $n$. If your base case had been $x=n=0$, then yes, you would have to induct on $n$ as well.\n$$f(x, n) = \\sum_{i_1=1}^n \\sum_{i_2=1}^{i_1} \\sum_{i_3=1}^{i_2} \\cdots \\sum_{i_x=1}^{i_{x-1}} \\sum_{i_{x+1}=1}^{i_x}1$$\nNow, notice that this counts the number of ways to choose $i_1, i_2, \\ldots, i_{x+1}$ such that $1 \\le i_{x+1} \\le i_x \\le \\ldots \\le i_1 \\le n$. But by Stars and Bars (also known as multichoose), this is just $\\binom{n+x}{x+1}$ as desired.", "date": "2020-04-03 11:54:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1874573/a-trivial-combinatorics-result-i-found-is-my-proof-correct", "openwebmath_score": 0.8803808093070984, "openwebmath_perplexity": 336.1487060722038, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987375052204446, "lm_q2_score": 0.8887587831798665, "lm_q1q2_score": 0.8775382499393806}}
{"url": "https://math.stackexchange.com/questions/2320494/when-is-the-union-of-infinitely-many-closed-sets-closed/2320528", "text": "# When is the union of infinitely many closed sets, closed?\n\nIt is known that, in general, the union of infinitely many closed sets need not be closed. However, in the following case, apparently, the union is closed:\n\nSuppose there is a large closed polygon $C$, inside which there is a square $S$ (green). Consider the set of all closed convex objects that contain $S$ and are contained in $C$. Then, apparently, the union of all these closed objects is closed.\n\nMy questions:\n\n\u2022 Is the above claim true, and if so, how to prove it?\n\u2022 In general, what are conditions for infinite set of closed sets to be closed, especially in $\\mathbb{R}^2$?\n\u2022 This doesn't apply in your situation, but a common sufficient condition for the union of a family of closed sets to be closed is that the family be locally finite. This means that for any point, there is a neighbourhood of that point that meets only finitely many of the closed sets. \u2013\u00a0user49640 Jun 13 '17 at 3:48\n\u2022 Echoing the comment above, the result that the union of a locally finite family of closed sets is closed, is a result that holds in every topological space. \u2013\u00a0DanielWainfleet Jun 13 '17 at 6:33\n\nThe claim is true. Let $A$ be the set you define and which we wish to prove closed.\n\nA point $x$ belongs to $A$ if and only if the convex hull of $S \\cup \\{x\\}$ is contained in $C$. (In that case, its closure is also contained in $C$.) This convex hull consists in turn of the union $B_x$ of all segments joining $x$ to a point in $S$.\n\n$B_x$ is convex because if $xz_1 z_2$ is a triangle with $z_1, z_2 \\in S$, and the points $y_1$ and $y_2$ lie on sides $xz_1$ and $xz_2$, respectively, then for any point $w$ on segment $y_1 y_2$ the ray $xw$ meets side $z_1 z_2$, which is contained in the convex set $S$.\n\nThus a point $x$ belongs to $A$ if and only if the segment $xz$ is contained in $C$ for every point $z$ in $S$. Therefore $A$ is the intersection, for all $z \\in S$, of the set $C_z$ which is the union of all segments with one endpoint at $z$ which are contained in $C$. Hence we need only prove that $C_z$ is closed.\n\nAfter a translation, we may assume $z = 0$. The set $C_0$ is the intersection of the sets $(1/t)C$ for all $t \\in (0,1]$, and all the sets $(1/t)C$ are obviously closed.\n\n\u2022 (a) Why must the square $S$ be closed? (b) in the last sentence, why is the set $C_0$ the intersection of all sets $(1/t)C$? \u2013\u00a0Erel Segal-Halevi Jun 13 '17 at 5:09\n\u2022 I don't know if $S$ must be a closed square to prove what you want, but I use the fact that it is closed when I show that $B_x$ is closed. Your requirements are that there be a closed convex set containing $x$ and $S$. For your second question, a point $x$ belongs to $C_0$ if and only if $tx \\in C$ for all $t \\in (0,1]$. \u2013\u00a0user49640 Jun 13 '17 at 5:40\n\u2022 If S is not closed, can't you just replace S with its closure and get the same result? \u2013\u00a0Erel Segal-Halevi Jun 13 '17 at 11:23\n\u2022 Yes, you're right. Just take the closure of the convex hull of $S \\cup \\{x\\}$. It's contained in $C$ because $C$ is closed. So I didn't need to mention anything about compactness. I'm editing the answer now. \u2013\u00a0user49640 Jun 13 '17 at 11:39\n\n(0). The closure bar denotes closure in $\\mathbb R^2.$\n\n(1). Exercise: If $Y$ is a closed bounded convex subet of $\\mathbb R^2$ and $x\\in \\mathbb R^2$ then the convex hull of $\\{x\\}\\cup Y$ is closed.\n\n(2). Let $A$ be the family of closed convex subsets of $C$ that have $S$ as a subset. Let D be the set of $p\\in C$ such that the convex hull of $\\{p\\}\\cup S$ is not a subset of $C.$ We have $(\\cup A )\\cap D=\\phi$ so $$\\cup A\\subset C \\backslash D.$$ Take any $p\\in D.$ There exists $q\\in S$ such that the line segment joining $p$ to $q$ contains a point $r\\not \\in C.$ Take an open ball $B(r,d)$, of radius $d>0$, centered at $r$, such that $B(r,d)\\cap C=\\phi.$ If $s>0$ and $s$ is sufficiently small then for any $p'\\in C\\cap B(p,s),$ the line segment from $p'$ to $q$ will intersect $B(r,d).$ So $B(p,s)\\cap C\\subset D.$\n\nSo $D$ is open in the space $C.$\n\nTherefore $C$ \\ $D$ is closed in the space $C.$ And $C=\\overline C$ so we have $$C \\backslash D =\\overline {C \\backslash D}.$$ But for any $x\\in C$ \\ $D$ the convex hull of $\\{x\\}\\cup S$ is closed, and is a subset of $C$. Therefore $$C\\backslash D\\subset \\cup A.$$\n\nWe have now $\\cup A\\subset C \\backslash D \\subset \\cup A,$ so $$\\overline {\\cup A}=\\overline {C\\backslash D}=C\\backslash D=\\cup A.$$", "date": "2020-02-27 21:43:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2320494/when-is-the-union-of-infinitely-many-closed-sets-closed/2320528", "openwebmath_score": 0.9093901515007019, "openwebmath_perplexity": 60.24508834904717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750488609223, "lm_q2_score": 0.888758786126321, "lm_q1q2_score": 0.8775382498770502}}
{"url": "https://math.stackexchange.com/questions/929882/what-is-wrong-with-my-algorithm-for-finding-how-many-positive-integers-are-divis", "text": "# What is wrong with my algorithm for finding how many positive integers are divisible by a number d in range [x,y]?\n\nI have been solving basic counting problems from Kenneth Rosen's Discrete Mathematics textbook (6th edition). These come from section 5-1 (the basics of counting), pages 344 - 347.\n\nThis question is not specifically about finding an answer to a problem or being given the correct equation, but whether my reasoning is sound. Therefore I would find it hard to argue this is a duplicate of seemingly similar questions like this one or this one.\n\nThe problems I have been dealing with come of the form How many positive integers in range [x,y] are divisible by d? All additional questions are based on the composition of the information learned in these, e.g. how many positive integers in range [x,y] are divisible by d or e?\n\nTo answer the simple question I wrote this \"equation/algorithm,\" which takes as input an inclusive range of positive integers $[x,y]$ and a positive integer $d$, and returns $n$, the total number of positive integers in range $[x,y]$ which are divisible by $d$.\n\n(1) $n = \\left \\lfloor{\\frac{y}{d}}\\right \\rfloor - \\left \\lfloor{\\frac{x}{d}}\\right \\rfloor$\n\nThe idea is that in order to count how many positive integers are divisible by $d$ from $[1,m]$, we simply calculate $\\left \\lfloor{\\frac{m}{d}}\\right \\rfloor$, because every $dth$ positive integer must be divisible by $d$. However, this does not work when given a range $[x,y]$ where $x \\not= 1$ or when $x > 1$. So we need to subtract the extra integers we counted, which is $\\left \\lfloor{\\frac{x}{d}}\\right \\rfloor$, i.e. the number of positive integers divisible by $d$ from $[1,x]$.\n\nFor a sanity check, I also wrote a brute force algorithm that does a linear search over every positive integer in the range $[x,y]$ and counts it if $x \\text{ mod } d == 0$. It also can list out the integers it picked, in case I am feeling really paranoid.\n\nWith (1) I've been getting the correct answers except on this problem/input: How many positive integers between 100 and 999 inclusive are odd? My solution was to calculate how many are even, and subtract this from the total number of positive integers in range $[100,999]$. To find the evens I simply use the algorithm in (1):\n\n$\\left \\lfloor{\\frac{999}{2}}\\right \\rfloor - \\left \\lfloor{\\frac{100}{2}}\\right \\rfloor = 499 - 50 = 449$\n\nBut this answer is wrong, since there actually $450$ even numbers in range $[100,999]$ by the brute force algorithm. (1) is somehow counting off by 1. My question is, why is (1) failing for this input of $(2, [100,999])$ but so far it's worked on every other input? What do I need to do to fix (1) so it produces the correct answer for this case? Perhaps I'm actually over counting because $x$ should actually be $x - 1$?\n\n(1') $n = \\left \\lfloor{\\frac{y}{d}}\\right \\rfloor - \\left \\lfloor{\\frac{x - 1}{d}}\\right \\rfloor$\n\n(1') returns the correct answer for this specific input now, but I am not sure if it will break my other solutions.\n\nAfter computing the number of positive multiples of $d$ less than or equal to $y,$ you correctly want to subtract the multiples that are not actually in the range $[x,y].$ Those are the multiples that are less than $x.$ When $x$ is divisible by $d,$ then $\\left\\lfloor \\frac xd \\right\\rfloor$ counts all multiples of $d$ up to and including $x$ itself. So as you surmised, you want to subtract $\\left\\lfloor \\frac {x-1}d \\right\\rfloor$ instead. This is true for any divisor, not just $d=2.$\nSuppose you have $$(k-1)d<x\\leq kd<(k+1)d<\\ldots<(k+n)d\\leq y<(k+n+1)d.$$ Then, you always have $\\lfloor{\\frac{y}{d}}\\rfloor=k+n$ is $k+n$. On the other hand, if $x$ is divisible by $d$, then $\\lfloor{\\frac{x}{d}}\\rfloor=k$, whereas if $x$ isn't divisible by $d$, then $\\lfloor{\\frac{x}{d}}\\rfloor=k-1$. To rectify this, instead use $\\lceil{\\frac{x}{d}}\\rceil$ which always returns $k$. Your solution is then $$n+1=(n+k)-k+1=\\color{blue}{\\left\\lfloor{\\frac{y}{d}}\\right\\rfloor-\\left\\lceil{\\frac{x}{d}}\\right\\rceil+1}.$$\np.s. $$(k-1)d<x\\leq kd\\implies(k-1)d\\leq x-1<kd\\implies\\left\\lfloor{\\frac{x-1}{d}}\\right\\rfloor=k-1$$ so that $$n+1=(n+k)-(k-1)=\\color{red}{\\left\\lfloor{\\frac{y}{d}}\\right\\rfloor-\\left\\lfloor{\\frac{x-1}{d}}\\right\\rfloor}.$$ The formulas in blue and red produce the same answer.", "date": "2020-01-27 03:07:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/929882/what-is-wrong-with-my-algorithm-for-finding-how-many-positive-integers-are-divis", "openwebmath_score": 0.7896967530250549, "openwebmath_perplexity": 128.91082374914626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877002595528, "lm_q2_score": 0.8933094074745443, "lm_q1q2_score": 0.8774868434883939}}
{"url": "https://math.stackexchange.com/questions/2987623/recurrence-relation-involving-unknown-dn-m-numbers/3051221", "text": "# Recurrence relation involving unknown d(n, m) numbers\n\nProblem goes like this: numbers $$d(n, m)$$, where $$n, m$$ are integers and $$0\\le m \\le n$$ are defined by:$$d(n, 0) = d(n, n) = 1$$ for all $$n\\ge 0$$ and$$m\\cdot d(n, m) = m\\cdot d(n-1, m) + (2n-m)\\cdot d(n-1, m-1)$$ for $$0\\lt m \\lt n$$. Prove that all the $$d(n, m)$$ are integers. Now this statement reminds me of some combinatorial identities, where the $$d(n, m)$$ can be considered, for example, the number of ways to distribute or arrange $$m+n$$ objects in a specific way. Since I don't know what these numbers represent from a combinatorial point of view, I can't prove the identity, but if it turns out that I guess what they represent, proving it should be easy enough. Problem is I struggle to find out what they can relate to. Am I in the right way or is there a completely different approach to solve it? I tried to expand $$d(n, m)$$ using its definition and it looks like binomial coefficients make their appearance.\n\n\u2022 Induction is the way to go. \u2013\u00a0Don Thousand Nov 6 '18 at 19:58\n\u2022 With 2 variables? Does it mean I should fix one and use induction on the other? \u2013\u00a0Peanut Nov 6 '18 at 20:00\n\u2022 That's right. You do two layers of induction \u2013\u00a0Don Thousand Nov 6 '18 at 20:01\n\u2022 But how did the proposer derive this formula? I don't think he used induction. \u2013\u00a0Peanut Nov 6 '18 at 20:04\n\u2022 I think proving $d(n,m)$ to be an integer is equivalent to showing that $m$ divides $2n \\cdot d(n-1,m-1)$. Check if that works \u2013\u00a0Rakesh Bhatt Dec 24 '18 at 12:11\n\nA good start would be to tabulate some small values. Here is some python code to do that:\n\n>>> def tab(n):\n...   r = [[0]*(n+1) for _ in range(n+1)]\n...   for i in range(n+1):\n...     r[i][i], r[i][0] = 1, 1\n...   for i in range(1, n+1):\n...     for j in range(1, i):\n...       r[i][j] = r[i-1][j]+(2*i-j)*r[i-1][j-1]//j\n...   for i in range(n+1):\n...     print('\\t'.join(map(str, r[i])))\n...\n>>> tab(6)\n1   0   0   0   0   0   0\n1   1   0   0   0   0   0\n1   4   1   0   0   0   0\n1   9   9   1   0   0   0\n1   16  36  16  1   0   0\n1   25  100 100 25  1   0\n1   36  225 400 225 36  1\n\n\nIf you're familiar with them, you'll pretty quickly recognize these as the squares of the Pascal's triangle. Hence, we hypothesize that $$d(n, m) = \\binom{n}{m}^2$$.\n\nWe can show this holds using induction. The boundary conditions $$d(n, n)$$ and $$d(n, 0)$$ hold trivially since $$1^2 = 1$$.\n\nNow take $$0 < m < n$$, substitute in $$d(n-1, m) = \\binom{n-1}{m}^2$$ and $$d(n-1,m-1) = \\binom{n-1}{m-1}^2$$ and divide the whole equation by $$m$$. Then:\n\n$$d(n, m) = \\binom{n-1}{m}^2 + \\frac{2n}{m}\\binom{n-1}{m-1}^2 - \\binom{n-1}{m-1}^2$$\n\nRecall $$\\binom{n}{m} = \\frac{n}{m}\\binom{n-1}{m-1}$$, so:\n\n$$d(n, m) = \\binom{n-1}{m}^2 + 2\\binom{n}{m}\\binom{n-1}{m-1} - \\binom{n-1}{m-1}^2$$\n\nExpand $$\\binom{n}{m} = \\binom{n-1}{m} + \\binom{n-1}{m-1}$$ to get:\n\n$$d(n, m) = \\binom{n-1}{m}^2 + 2\\binom{n-1}{m}\\binom{n-1}{m-1} + 2\\binom{n-1}{m-1}\\binom{n-1}{m-1} - \\binom{n-1}{m-1}^2$$\n\nA mess, but we can cancel the last two terms and factor:\n\n$$d(n, m) = \\Big(\\binom{n-1}{m} + \\binom{n-1}{m-1}\\Big)^2 = \\binom{n}{m}^2$$\n\nWhich is our induction hypothesis. So by induction, $$d(n, m) = \\binom{n}{m}^2$$, which certainly is an integer.\n\n\u2022 The tricky part is always guess something. Once you have an idea of a suitable function, then it's no big deal \u2013\u00a0Peanut Dec 24 '18 at 12:45", "date": "2020-10-30 02:07:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2987623/recurrence-relation-involving-unknown-dn-m-numbers/3051221", "openwebmath_score": 0.7118034362792969, "openwebmath_perplexity": 210.48102222452238, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877007780707, "lm_q2_score": 0.8933094017937621, "lm_q1q2_score": 0.8774868383714283}}
{"url": "http://mathhelpforum.com/discrete-math/150015-group-combinations-problem.html", "text": "1. ## Group Combinations Problem\n\nIf we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?\n\nIs the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?\n\nAny thoughts?\n\n2. Originally Posted by mathceleb\nIf we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?\n\nIs the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?\n\nAny thoughts?\nI haven't worked this out rigorously, but I think the probability is $\\displaystyle \\frac{1}{\\binom{149}{2}}$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $\\displaystyle \\binom{149}{2}$ ways to choose two people out of the remaining 149...\n\n3. I'm not sure. I found this thread from back in the day which is a similar question...\n\nhttp://www.mathhelpforum.com/math-he...mutations.html\n\n4. Hello, mathceleb!\n\nWe have a group of 150 people, broken into groups of 3, and only 3 are blue.\nWhat is the probability that all 3 blue people are put in the same group?\n\nThe 150 people are divided into 50 groups of 3 people each.\n\nThere are: . $\\dfrac{150!}{(3!)^{50}}$ possible groupings.\n\nTo have the 3 blue people in one group, there is: . ${3\\choose3} = 1$ way.\n\nThe other 147 people are divided into 49 groups of 3: . $\\dfrac{147!}{(3!)^{49}}$ ways.\n. . Hence, there are: . $\\dfrac{147!}{(3!)^{49}}$ ways to have the 3 blue people in one group.\n\nThe probability that all 3 blue people are in one group is:\n\n. . $\\dfrac{\\dfrac{147!}{(3!)^{49}}} {\\dfrac{150!}{(3!)^{50}}} \\;=\\;\\dfrac{147!}{(3!)^{49}}\\cdot\\dfrac{(3!)^{50}} {150!} \\;=\\;\\dfrac{6}{150\\cdot149\\cdot148} \\;=\\;\\dfrac{1}{551,\\!300}$\n\n5. Originally Posted by undefined\nthe probability is $\\displaystyle \\frac{1}{\\binom{149}{2}}$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $\\displaystyle \\binom{149}{2}$ ways to choose two people out of the remaining 149...\nThis correct.\n\nOriginally Posted by Soroban\n[size=3]\nThe 150 people are divided into 50 groups of 3 people each.\nThere are: . $\\dfrac{150!}{(3!)^{50}}$ possible groupings.\nThose are ordered partions.\n\nThere are $\\dfrac{150!}{(3!)^{50}(50!)}$ unordered partitions.\n\nThere are $\\dfrac{147!}{(3!)^{49}(49!)}$ of those where the blues are together.\n\nNote that $\\dfrac{\\dfrac{147!}{(3!)^{49}(49!)}}{ \\dfrac{150!}{(3!)^{50}(50!)}}=\\dfrac{1}{\\binom{149 }{2}}$\n\n6. Hmm since Soroban's answer differs from mine I decided to do a little more research.\n\nOriginally Posted by Soroban\n\nThere are: . $\\dfrac{150!}{(3!)^{50}}$ possible groupings.\nI believe this should be $\\dfrac{\\left(\\dfrac{150!}{(3!)^{50}}\\right)}{50!}$ because the groups of 3 can be permuted.\n\nOriginally Posted by Soroban\n\n...Hence, there are: . $\\dfrac{147!}{(3!)^{49}}$ ways to have the 3 blue people in one group.\nLikewise I believe this should be $\\dfrac{\\left(\\dfrac{147!}{(3!)^{49}}\\right)}{49!}$.\n\n$\\dfrac{\\left(\\dfrac{147!}{(3!)^{49}}\\right)\\cdot50 !}{49!\\cdot\\left(\\dfrac{150!}{(3!)^{50}}\\right)} = 50\\left(\\dfrac{1}{551,\\!300}\\right) = \\dfrac{1}{11026} = \\displaystyle \\frac{1}{\\binom{149}{2}}$", "date": "2017-09-21 15:58:23", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/150015-group-combinations-problem.html", "openwebmath_score": 0.8849186897277832, "openwebmath_perplexity": 678.5368138734311, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877002595527, "lm_q2_score": 0.8933093961129794, "lm_q1q2_score": 0.8774868323280683}}
{"url": "https://gmatclub.com/forum/set-k-consists-of-all-fractions-of-the-form-x-x-2-where-x-is-a-posit-277385.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 22 Oct 2018, 22:47\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Set K consists of all fractions of the form x/(x+2) where x is a posit\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 23 Dec 2011\nPosts: 35\nSet K consists of all fractions of the form x/(x+2) where x is a posit\u00a0 [#permalink]\n\n### Show Tags\n\n26 Sep 2018, 17:28\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n69% (01:35) correct 31% (02:05) wrong based on 65 sessions\n\n### HideShow timer Statistics\n\nSet K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?\n\nA) 1/20\nB) 1/10\nC) 1/9\nD) 1/2\nE) 8/9\n\nIt took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 2040\nSet K consists of all fractions of the form x/(x+2) where x is a posit\u00a0 [#permalink]\n\n### Show Tags\n\n26 Sep 2018, 19:15\n1\nanupam87 wrote:\nSet K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?\n\nA) 1/20\nB) 1/10\nC) 1/9\nD) 1/2\nE) 8/9\n\nIt took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.\n\nSounds frustrating! Here is one way. About 30 seconds if just a few set members are listed. If all are listed, maybe 75 seconds.\n\nThere's a pattern you should see quickly.\n\nStart by finding the greatest value of $$x$$.\nDoing so is smart because we know that $$x$$ must be even, positive, and less than 20. Use that high-end limit.\n\nThe greatest that $$x$$ can be is 18. Work backwards from 18.\n\nFirst fraction? $$\\frac{x}{x+2}=\\frac{18}{20}$$\n\nNext? Well, the next $$x$$ must be 16 (even, positive, < 20).\nIts denominator, $$x+2$$, is 18: $$\\frac{16}{18}$$\n\nNext? $$x$$ must be 14. The denominator is 16: $$\\frac{14}{16}$$\n\nList just those three: $$\\frac{18}{20}*\\frac{16}{18}*\\frac{14}{16}$$\n18 and 18 cancel\n16 and 16 cancel\n20 remains. 14 remains.\nSo the first denominator in the list will remain\nAnd the last numerator in the list will remain.\n\nWe know the first denominator. What is the last numerator? The smallest $$x$$.\n\nSmallest integer that $$x$$ could be? Positive, even ... 2?\nYes. So the last member of the set at the end of this list is $$\\frac{x}{x+2}=\\frac{2}{4}$$\n\nFirst denominator (20) and last numerator (2) will remain:\n$$\\frac{2}{20}=\\frac{1}{10}$$\n\nAnother pattern: numerators and denominators decend in order.\n\nNumerators:\n18, 16, 14, 12, 10, 8, 6, 4, 2\n\nDenominators:\n20, 18, 16, 14, 12, 10, 8, 6, 4\n\nSame conclusion as above. All but the first denominator and the last numerator will cancel.\n\n$$\\frac{18}{20}*\\frac{16}{18}*\\frac{14}{16}*\\frac{12}{14}*\\frac{10}{12}*\\frac{8}{10}*\\frac{6}{8}*\\frac{4}{6}*\\frac{2}{4}=$$\n\n$$\\frac{2}{20}=\\frac{1}{10}$$\n\nI hope that helps.\n\nLet me know if you have questions.\n_________________\n\n___________________________________________________________________\nFor what are we born if not to aid one another?\n-- Ernest Hemingway\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 3906\nLocation: United States (CA)\nSet K consists of all fractions of the form x/(x+2) where x is a posit\u00a0 [#permalink]\n\n### Show Tags\n\n02 Oct 2018, 19:15\n2\nanupam87 wrote:\nSet K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?\n\nA) 1/20\nB) 1/10\nC) 1/9\nD) 1/2\nE) 8/9\n\nLet\u2019s begin by writing out this product:\n\n2/4 x 4/6 x 6/8 x 8/10 x 10/12 x 12/14 x 14/16 x 16/18 x 18/20\n\nSo we see that everything cancels except the numerator of the first fraction and the denominator of the last fraction.\n\nThus, the product of all the fractions in set K is 2/20 = 1/10.\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nIntern\nJoined: 19 Aug 2018\nPosts: 7\nRe: Set K consists of all fractions of the form x/(x+2) where x is a posit\u00a0 [#permalink]\n\n### Show Tags\n\n02 Oct 2018, 19:26\ngeneris wrote:\nanupam87 wrote:\nSet K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?\n\nA) 1/20\nB) 1/10\nC) 1/9\nD) 1/2\nE) 8/9\n\nIt took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.\n\nSounds frustrating! Here is one way. About 30 seconds (don't write out all the fractions below)\n\nThere's a pattern you should see quickly.\n\nStart by finding the greatest value of $$x$$.\nFinding the greatest value of $$x$$ is smart because we have a lot of information:\n$$x$$ must be even, positive, and less than 20.\n\nThe greatest that $$x$$ can be is 18.\nWork backwards from 18.\n\nFirst fraction? $$\\frac{x}{x+2}=\\frac{18}{20}$$\n\nNext? Well, the next $$x$$ must be 16 (even, positive, < 20).\nIts denominator, $$x+2$$, is 18: $$\\frac{16}{18}$$\n\nNext? $$x$$ must be 14. The denominator is 16: $$\\frac{14}{16}$$\n\nList just those three.\n\n$$\\frac{18}{20}*\\frac{16}{18}*\\frac{14}{16}$$\n\nAll but the first denominator and the last numerator \"cancel.\"**\n\nPattern: numerators and denominators decend in order.\n\nNumerators:\n18, 16, 14, 12, 10, 8, 6, 4, 2\n\nDenominators:\n20, 18, 16, 14, 12, 10, 8, 6, 4\n\nAs mentioned, all but one denominator and numerator will \"cancel.\"\n\n$$\\frac{18}{20}*\\frac{16}{18}*\\frac{14}{16}*\\frac{12}{14}*\\frac{10}{12}*\\frac{8}{10}*\\frac{6}{8}*\\frac{4}{6}*\\frac{2}{4}=$$\n\n$$\\frac{2}{20}=\\frac{1}{10}$$\n\nI hope that helps.\n\nLet me know if you have questions.\n\n**I wrote only this list of three:\n$$\\frac{18}{20}*\\frac{16}{18}*\\frac{14}{16}$$\n18 and 18 cancel\n16 and 16 cancel\n20 remains. 14 remains.\nSo the first denominator in its list will remain\nAnd the last numerator in its list will remain: $$\\frac{2}{20}=\\frac{1}{10}$$\n\nWhy x cannot be 19?, could you help me please?\nThank you.\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 2040\nSet K consists of all fractions of the form x/(x+2) where x is a posit\u00a0 [#permalink]\n\n### Show Tags\n\n02 Oct 2018, 19:44\n1\njorgetomas9 wrote:\ngeneris wrote:\nanupam87 wrote:\nSet K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?\n\nA) 1/20\nB) 1/10\nC) 1/9\nD) 1/2\nE) 8/9\n\nSounds frustrating! Here is one way. About 30 seconds (don't write out all the fractions below)\n\nThere's a pattern you should see quickly.\n\nStart by finding the greatest value of $$x$$.\nFinding the greatest value of $$x$$ is smart because we have a lot of information:\n$$x$$ must be even, positive, and less than 20.\n\nThe greatest that $$x$$ can be is 18.\nWork backwards from 18.\n\nFirst fraction? $$\\frac{x}{x+2}=\\frac{18}{20}$$\n\nNext? Well, the next $$x$$ must be 16 (even, positive, < 20).\nIts denominator, $$x+2$$, is 18: $$\\frac{16}{18}$$\n\nNext? $$x$$ must be 14. The denominator is 16: $$\\frac{14}{16}$$\n\nList just those three.\n\n$$\\frac{18}{20}*\\frac{16}{18}*\\frac{14}{16}$$\n\nAll but the first denominator and the last numerator \"cancel.\"**\n\nPattern: numerators and denominators decend in order.\n\nNumerators:\n18, 16, 14, 12, 10, 8, 6, 4, 2\n\nDenominators:\n20, 18, 16, 14, 12, 10, 8, 6, 4\n\nAs mentioned, all but one denominator and numerator will \"cancel.\"\n\n$$\\frac{18}{20}*\\frac{16}{18}*\\frac{14}{16}*\\frac{12}{14}*\\frac{10}{12}*\\frac{8}{10}*\\frac{6}{8}*\\frac{4}{6}*\\frac{2}{4}=$$\n\n$$\\frac{2}{20}=\\frac{1}{10}$$\n\nWhy x cannot be 19?, could you help me please?\nThank you.\n\njorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake.\n_________________\n\n___________________________________________________________________\nFor what are we born if not to aid one another?\n-- Ernest Hemingway\n\nIntern\nJoined: 19 Aug 2018\nPosts: 7\nRe: Set K consists of all fractions of the form x/(x+2) where x is a posit\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2018, 08:39\n1\ngeneris wrote:\njorgetomas9 wrote:\ngeneris wrote:\n[\njorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake.\n\nThank you very much I haven't seen that one.\nIntern\nJoined: 29 Jul 2018\nPosts: 4\nSet K consists of all fractions of the form x/(x+2) where x is a posit\u00a0 [#permalink]\n\n### Show Tags\n\n08 Oct 2018, 11:26\n1\nHere is another, simpler approach to solving the problem in under 1 minute.\n\nYou can take several values of x and see the pattern.\nFor x=2, fraction is 1/2\nFor x=4, fraction is 2/3\nFor x=6, fraction is 3/4 etc.\n\nSo, the product of the 9 fractions would be 9!/10!*1 = 1/10\nSet K consists of all fractions of the form x/(x+2) where x is a posit &nbs [#permalink] 08 Oct 2018, 11:26\nDisplay posts from previous: Sort by", "date": "2018-10-23 05:47:38", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/set-k-consists-of-all-fractions-of-the-form-x-x-2-where-x-is-a-posit-277385.html", "openwebmath_score": 0.868035078048706, "openwebmath_perplexity": 1731.6147500117636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8774767970940974, "lm_q1q2_score": 0.8774767970940974}}
{"url": "https://gmatclub.com/forum/a-stock-trader-originally-bought-300-shares-219939.html", "text": "GMAT Changed on April 16th - Read about the latest changes here\n\n It is currently 25 Apr 2018, 03:18\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# A stock trader originally bought 300 shares\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 30 Jan 2013\nPosts: 11\nA stock trader originally bought 300 shares\u00a0[#permalink]\n\n### Show Tags\n\nUpdated on: 09 Jun 2016, 12:13\n1\nKUDOS\n1\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n82% (01:04) correct 18% (01:17) wrong based on 56 sessions\n\n### HideShow timer Statistics\n\nA stock trader originally bought 300 shares of stock from a company at a total cost of m dollars. If each share was sold at 50% above the original cost per share of stock, then interns of m for how many dollars was each share sold?\n\na) 2m/300\nb) m/300\nc) m/200\nd) m/300 + 50\ne) 350/m\n[Reveal] Spoiler: OA\n\nOriginally posted by sabrina3509 on 09 Jun 2016, 12:02.\nLast edited by Vyshak on 09 Jun 2016, 12:13, edited 1 time in total.\nEdited the topic name. Topic name must contain few words from the question.\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 3400\nLocation: India\nGPA: 3.5\nWE: Business Development (Commercial Banking)\nRe: A stock trader originally bought 300 shares\u00a0[#permalink]\n\n### Show Tags\n\n09 Jun 2016, 12:09\nsabrina3509 wrote:\nA stock trader originally bought 300 shares of stock from a company at a total cost of m dollars. If each share was sold at 50% above the original cost per share of stock, then interns of m for how many dollars was each share sold?\n\na) 2m/300\nb) m/300\nc) m/200\nd) m/300 + 50\ne) 350/m\n\nQuote:\nbought 300 shares of stock from a company at a total cost of m dollars.\n\nLet Cost of 300 shares be $3000 So, Cost of 1 shares be$ 10 =>m/300\n\nQuote:\neach share was sold at 50% above the original cost per share of stock\n\nSelling price per share = (100+50)/100 * m/300\n\nOr, Selling price per share = 3/2 * m/300 => m/200\n\nHence answer will be (C)\n\n_________________\n\nThanks and Regards\n\nAbhishek....\n\nPLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS\n\nHow to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )\n\nVeritas Prep GMAT Instructor\nJoined: 01 Jul 2017\nPosts: 38\nA stock trader originally bought 300 shares\u00a0[#permalink]\n\n### Show Tags\n\n21 Dec 2017, 19:16\n2\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nSmall leverage words here make a big difference. First, notice that the \"total\" cost of 300 stocks is equal to $$m$$. Thus, the cost per stock would be $$\\frac{m}{300}$$. This fraction matches answer choice (B), and highlights a classic trap of the GMAT: answer choices often include the \"right answer to the wrong question.\" The target of this question is not the original cost per stock, but instead the price at which each stock was sold, after an increase.\n\nThe question indicates that each share was sold at 50% above the original cost. This is a percentage increase, adding half of the original value to the value. Thus, if the original cost per stock were $$(\\frac{m}{300})$$, the value of the stock after the increase would be $$(\\frac{m}{300})(1+\\frac{1}{2})$$. Now, don't convert the fraction to decimal values here. Not only are \"fractions your friends\", but the answer choices clearly keep the math in fractional form.\n\nNow, we need to just simplify the math, looking for common factors in the top and bottom of the equation to make the math even easier:\n\n$$(\\frac{m}{300})(1+\\frac{1}{2})=(\\frac{m}{3*100})(\\frac{3}{2})=\\frac{m}{200}$$\n\nThe answer is (C).\n_________________\n\nAaron J. Pond\nVeritas Prep Elite-Level Instructor\n\nHit \"+1 Kudos\" if my post helped you understand the GMAT better.\nLook me up at https://www.veritasprep.com/gmat/aaron-pond/ if you want to learn more GMAT Jujitsu.\n\nA stock trader originally bought 300 shares \u00a0 [#permalink] 21 Dec 2017, 19:16\nDisplay posts from previous: Sort by\n\n# A stock trader originally bought 300 shares\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-04-25 10:18:08", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-stock-trader-originally-bought-300-shares-219939.html", "openwebmath_score": 0.3335218131542206, "openwebmath_perplexity": 7383.76231959499, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.877476793890012}}
{"url": "https://gmatclub.com/forum/in-the-figure-above-x-2-y-250193.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 22 Sep 2018, 16:18\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# In the figure above, x^2 + y^2 =\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49303\nIn the figure above, x^2 + y^2 =\u00a0 [#permalink]\n\n### Show Tags\n\n26 Sep 2017, 23:31\n00:00\n\nDifficulty:\n\n45% (medium)\n\nQuestion Stats:\n\n59% (00:39) correct 41% (01:05) wrong based on 57 sessions\n\n### HideShow timer Statistics\n\nIn the figure above, x^2 + y^2 =\n\n(A) 5\n(B) 7\n(C) 25\n(D) 80\n(E) 625\n\nAttachment:\n\n2017-09-27_1016_002.png [ 10.27 KiB | Viewed 879 times ]\n\n_________________\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 1978\nIn the figure above, x^2 + y^2 =\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 28 Sep 2017, 07:16\nBunuel wrote:\n\nIn the figure above, x^2 + y^2 =\n\n(A) 5\n(B) 7\n(C) 25\n(D) 80\n(E) 625\n\nAttachment:\n2017-09-27_1016_002.png\n\nAll four figures are right triangles.\n\n$$x^2 + y^2$$ = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.\n\nShort leg, SL, of inside triangle:\n$$(\\sqrt{4})^2 + (\\sqrt{5})^2 = (SL)^2$$\n$$(4 + 5) = 9 = (SL)^2$$\n$$SL = 3$$\n\nLong leg, LL, of inside triangle:\n$$(\\sqrt{6})^2 + (\\sqrt{10})^2 = (LL)^2$$\n$$(6 + 10) = 16 = (LL)^2$$\n$$LL = 4$$\n\nThe inside triangle is a 3-4-5 right triangle - its hypotenuse is 5. Let hypotenuse 5 = c:\n\n$$x^2 + y^2 = c^2$$\n$$x^2 + y^2 = 5^2$$\n$$x^2 + y^2 = 25$$\n\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nOriginally posted by generis on 27 Sep 2017, 18:06.\nLast edited by generis on 28 Sep 2017, 07:16, edited 1 time in total.\nSVP\nJoined: 26 Mar 2013\nPosts: 1808\nRe: In the figure above, x^2 + y^2 =\u00a0 [#permalink]\n\n### Show Tags\n\n28 Sep 2017, 06:04\n1\ngenxer123 wrote:\nBunuel wrote:\n\nIn the figure above, x^2 + y^2 =\n\n(A) 5\n(B) 7\n(C) 25\n(D) 80\n(E) 625\n\nAttachment:\n2017-09-27_1016_002.png\n\nAll four figures are right triangles.\n\n$$x^2 + y^2$$ = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.\n\nShort leg, SL, of inside triangle:\n$$(\\sqrt{4})^2 + (\\sqrt{5})^2 = (SL)^2$$\n$$(4 + 5) = 9 = (SL)^2$$\n$$SL = 3$$\n\nLong leg, LL, of inside triangle:\n$$(\\sqrt{6})^2 + (\\sqrt{10})^2 = (LL)^2$$\n$$(6 + 10) = 16 = (LL)^2$$\n$$LL = 4$$\n\nThe inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.\n\n$$x^2 + y^2 = 5$$\n\nThe highlighted part should equal to (hypotenuse)...it should be 25.\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 1978\nIn the figure above, x^2 + y^2 =\u00a0 [#permalink]\n\n### Show Tags\n\n28 Sep 2017, 07:14\nMo2men wrote:\ngenxer123 wrote:\nBunuel wrote:\nIn the figure above, x^2 + y^2 =\n\n(A) 5\n(B) 7\n(C) 25\n(D) 80\n(E) 625\nAttachment:\n2017-09-27_1016_002.png\n\n. . .\nThe inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.\n\n$$x^2 + y^2 = 5$$\n\nThe highlighted part should equal to (hypotenuse)...it should be 25.\n\nMo2men , nice catch. I was thinking of 5 as a fixed value derived from other fixed values, such that 5 = $$c^2$$. I missed a step.\n\n5 = $$(\\sqrt{c^2}) = (\\sqrt{5^2}) = \\sqrt{25}$$.\n\nI'll edit.\n\nThanks, and kudos.\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2835\nRe: In the figure above, x^2 + y^2 =\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2017, 10:06\nBunuel wrote:\n\nIn the figure above, x^2 + y^2 =\n\n(A) 5\n(B) 7\n(C) 25\n(D) 80\n(E) 625\n\nWe see that x^2 + y^2 is equal to the square of the hypotenuse of a right triangle whose sides are not shown. Let\u2019s first determine the base of this triangle, which we can call b:\n\n(\u221a4)^2 + (\u221a5)^2 = b^2\n\n4 + 5 = b^2\n\n9 = b^2\n\nb = 3\n\nLet\u2019s now determine the height of the triangle, which we can call h:\n\n(\u221a6)^2 + (\u221a10)^2 = h^2\n\n6 + 10 = h^2\n\n16 = h^2\n\nh = 4\n\nThus, x^2 + y^2 = 3^2 + 4^2 = 25.\n\n_________________\n\nJeffery Miller\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: In the figure above, x^2 + y^2 = &nbs [#permalink] 03 Oct 2017, 10:06\nDisplay posts from previous: Sort by\n\n# In the figure above, x^2 + y^2 =\n\n## Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-22 23:18:49", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-the-figure-above-x-2-y-250193.html", "openwebmath_score": 0.8387085199356079, "openwebmath_perplexity": 6923.797083373556, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8774767874818408}}
{"url": "http://math.stackexchange.com/questions/11477/showing-that-an-equation-holds-true-with-a-fibonacci-sequence-f-nm-f-n-1?rq=1", "text": "# Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$\n\nQuestion: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \\geq 2$. Show for $n, m \\in \\mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$\n\nMy (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?\n\n-\nI suppose using the Binet formula is a \"mosquito-nuking\" solution... but it works. \u2013\u00a0J. M. Nov 23 '10 at 7:17\n@J.M.: \"overkill\" (FTFY) \u2013\u00a0Isaac Dec 24 '10 at 22:16\n@Isaac, It's an idiom here and MO... \u2013\u00a0J. M. Dec 24 '10 at 23:58\n\nFix $m \\in \\mathbb{N}$. We shall use induction on $n$. For $n=1$, the RHS of the equation becomes $$F_{m-1}F_{1} + F_{m}F_{2} = F_{m-1} + F_{m}$$, which is equal to $F_{m+1}$. When $n=2$, the equation is also true.( I hope you can prove this!).\n\nNow assume, that the result is true for $k=3,4, \\cdots , n$. We want to show that the result is true for $k=n+1$. $$\\text{For} \\ k=n-1 \\ \\text{we have} \\quad F_{m+n-1} = F_{m-1}F_{n-1} + F_{m}f_{n}$$ and $$\\text{For} \\ k = n \\ \\text{we have} \\quad F_{m+n}=F_{m-1}F_{n} + F_{m}F_{n+1}$$ Adding both the sides you will get $$F_{m+n-1} + F_{m+n} = F_{m+n-1} = F_{m-1}F_{n+1} + F_{m}F_{n+2}$$\n\nOh, i reversed the notations. This is for proving $$F_{m+n} = F_{m-1}F_{n} + F_{m}F_{n+1}$$\n\n-\n@Chandru1 thanks a lot for the help i think i'm slowly starting to understand this stuff. To prove it with $n=2$ i simply broke $F_{n+2}$ down into $F_{n+1} + F_n = F_n + F_{n-1} + F_n$ which is what i obtained when substituting $n=2$. \u2013\u00a0ghshtalt Nov 23 '10 at 18:45\n\nIf pressed, I'd nuke with Binet's formula myself, but here's another approach. By an easy induction, if $$A=\\pmatrix{0&1\\\\\\\\ 1&1}$$ then $$A^n=\\pmatrix{F_{n-1}&F_n\\\\\\\\ F_n&F_{n+1}}.$$ Comparing the top right entry in the equation $A^m A^n=A^{m+n}$ gives $$F_{m-1}F_n+F_m F_{n+1}=F_{m+n}.$$\n\n-\n\nHINT $\\ \\$ If you put the Fibonacci recurrence into matrix form then the result is obvious, viz.\n\n$$M^n\\ :=\\ \\left(\\begin{array}{ccc} 1 & 1 \\\\\\ 1 & 0 \\end{array}\\right)^n\\ =\\ \\left(\\begin{array}{ccc} F_{n+1} & F_n \\\\\\ F_n & F_{n-1} \\end{array}\\right)$$\n\nNow compute $\\ M^{n+m}\\ =\\ M^n \\ M^m$\n\n-\nok so what i compute is: $$M_{n+m} = M_n*M_m = \\left(\\begin{array}{ccc} F_{n+1} & F_n \\\\\\ F_n & F_{n-1} \\end{array}\\right)*\\left(\\begin{array}{ccc} F_{m+1} & F_m \\\\\\ F_m & F_{m-1} \\end{array}\\right) = \\left(\\begin{array}{ccc} F_{n+1}F_{m+1} + F_nF_m & F_{n+1}F_m + F_nF_{m-1} \\\\\\ F_nF_{m+1} + F_{n-1}F_m & F_{n}F_{m} + F_{n-1}F_{m-1} \\end{array}\\right)$$ which as you said shows the desired expression in the bottom left(and i assume equivalent) in the top right. is this method something one can usually use with sequences? i guess i just don't know how it works in general... \u2013\u00a0ghshtalt Nov 23 '10 at 8:37\nthat is, have i understood it correctly if i were to say that to define a sequence, one creates a $2 X 2$ matrix with the upper left value equal to the next value the bottom left and upper right values equal to the current value, and the bottom right equal to the previous? \u2013\u00a0ghshtalt Nov 23 '10 at 8:38\n@user3711: The matrix $M$ is simply the shift operator viz. $\\ (F_{n-1},F_n)\\:M^t\\ = (F_n,F_{n+1})\\:.\\$ The same idea works for any linear recurrence. \u2013\u00a0Bill Dubuque Nov 23 '10 at 17:38\n\nWith Fibonacci numbers usually there are multiple ways of proving identities.\n\nOne way (which is one of my favourites) to prove your identity is the following:\n\nConsider the following problem:\n\nA person climbs up $\\displaystyle n$ steps, by taking either one step, or two steps at a time.\n\nThe total number of ways the person can climb up all the $\\displaystyle n$ steps is $\\displaystyle F_{n+1}$ (Why?)\n\nNow consider climbing $\\displaystyle m+n-1$ steps and split into the cases when the person lands on step $\\displaystyle n$ and the cases when the person lands on step $\\displaystyle n-1$ and takes two steps at that point (and so does not land on step $\\displaystyle n$ in those cases). These two cases cover all possibilities, and so we have:\n\n$$\\displaystyle F_{m+n} = F_{n+1}F_{m} + F_{n}F_{m-1}$$\n\n-\nRight. This is basically the matrix proof translated into the language of walks on graphs, which is then further translated into a tiling problem. \u2013\u00a0Qiaochu Yuan Nov 23 '10 at 17:46\n@Qia: That is one way of looking at it. But I wouldn't say it is basically just a translation... I think it is interesting on its own. \u2013\u00a0Aryabhata Nov 23 '10 at 18:11\nI like this proof for giving a simple combinatorial interpretation of the result - it offers better 'motivation' IMHO than the matrix proof, which might look like magic to someone not familiar with those manipulations. \u2013\u00a0Steven Stadnicki Nov 23 '10 at 18:46\nIn any case it was good to try to see it from another angle, although to be honest, i'm still having some trouble really visualizing this, hopefully it hits me later... and yes the matrices seem really efficient but at least for me they still look like magic ;) \u2013\u00a0ghshtalt Nov 23 '10 at 18:52\n@Steven: the two proofs are equivalent. Powers of the Fibonacci matrix count walks on a certain graph, the Fibonacci graph, which can in turn be interpreted in terms of certain tilings. I discuss a generalization to companion matrices in this blog post: qchu.wordpress.com/2009/08/23/\u2026 \u2013\u00a0Qiaochu Yuan Nov 23 '10 at 19:08\n\nThere are several good answers already, but I thought I would add the following derivation because it is one of the few uses I know for the sum property of permanents; namely,\n\nIf $A$, $B$, and $C$ are matrices with identical entries except that one row (column) of $C$, say the $k^{th}$, is the sum of the $k^{th}$ rows (columns) of $A$ and $B$, then $\\text{ per } A + \\text{ per } B = \\text{per } C$.\n\nStart with the matrices $\\begin{bmatrix} F_n & F_{n-1} \\\\ F_0 & F_1 \\end{bmatrix}$ and $\\begin{bmatrix} F_n & F_{n-1} \\\\ F_1 & F_2 \\end{bmatrix}$. Since $F_0 = 0$ and $F_1 = F_2 = 1$, they have permanents $F_n$ and $F_n + F_{n-1} = F_{n+1}$, respectively. Applying the Fibonacci recurrence and the permanent sum property, we have $\\text{ per } \\begin{bmatrix} F_n & F_{n-1} \\\\ F_2 & F_3 \\end{bmatrix} = F_{n+2}$. By continuing to construct new matrices whose second rows are the sums of the second rows of the previous two matrices, this process continues until we have $F_n F_{m+1} + F_{n-1}F_m = \\text{ per} \\begin{bmatrix} F_n & F_{n-1} \\\\ F_m & F_{m+1} \\end{bmatrix} = F_{n+m}.$\n\nFor more on this approach (but with determinants), see this paper I wrote a few years ago: \"Fibonacci Identities via the Determinant Sum Property,\" The College Mathematics Journal, 37 (4): 286-289, 2006.\n\n-\n+1: Hadn't seen this before. \u2013\u00a0Aryabhata Nov 23 '10 at 23:49\n\nTry proving this statement: Claim: If $f(n) = f(n-1)+f(n-2)$, then $f(n) = F_n f_1 + F_{n-1} f_0$.\n\nNow \"fix\" $m$ and think of $F_{n+m}$ as a linear recurrence in $n$ with initial conditions $F_{m+1}$ and $F_m$.\n\nBy the way, there is a short and clean proof of Claim, but you should know it uses the fact $F_{-1} = 1$.(Something you can easily verify if you did not already know).\n\n-\n\nOne simple way is to use the formula for the $n^{th}$ Fibonacci number, viz,\n\n$F_n = \\frac{\\phi^n - (1-\\phi)^n}{\\sqrt{5}}$ where $\\phi$ is the golden ratio.\n\n$\\phi = \\frac{1 + \\sqrt{5}}{2}$.\n\nOr another equally simple way is to use induction on $n$ and then on $m$ or just using induction on one of these might turn out to suffice.\n\nIt will be interesting to see a direct proof...\n\n-\n...and looks like somebody else did use Binet... :D \u2013\u00a0J. M. Nov 23 '10 at 7:20\n@J.M. : I didn't see your comment when I wrote the answer :). \u2013\u00a0user17762 Nov 23 '10 at 7:23\nam i interpreting this properly if i write: $\\frac{\\Phi^{n+m} - (1-\\Phi)^{n+m}}{\\sqrt{5}} = \\frac{\\Phi^{n-1} - (1-\\Phi)^{n-1}}{\\sqrt{5}} * \\frac{\\Phi^{m} - (1-\\Phi)^{m}}{\\sqrt{5}} + \\frac{\\Phi^{n} - (1-\\Phi)^{n}}{\\sqrt{5}} * \\frac{\\Phi^{1+m} - (1-\\Phi)^{1+m}}{\\sqrt{5}}$ $\\frac{\\Phi^{m} - (1-\\Phi)^{m}}{\\sqrt{5}}(\\frac{\\Phi^{n-1} - (1-\\Phi)^{n-1}}{\\sqrt{5}} + \\frac{\\Phi^{n} - (1-\\Phi)^{n}}{\\sqrt{5}} * F_1)$ $\\frac{\\Phi^{m} - (1-\\Phi)^{m}}{\\sqrt{5}}(\\frac{\\Phi^{n-1} - (1-\\Phi)^{n-1}}{\\sqrt{5}}+\\frac{\\Phi^{n} - (1-\\Phi)^{n}}{\\sqrt{5}})$ ? what can i do at the end? \u2013\u00a0ghshtalt Nov 23 '10 at 7:41\n@user3711: Use $\\phi^2 = 1 + \\phi$ to simplify. Or as the others have stated you can try induction. \u2013\u00a0user17762 Nov 23 '10 at 7:45\ni'm sorry, but i'm having trouble figuring out how to use that to simplify.... \u2013\u00a0ghshtalt Nov 23 '10 at 8:08\n\nThese can almost always be solved by induction. (In general sequences that are defined by recursion go often well together with induction.)\n\nFor this formula we can do the following: Induct on $m$. If $m=1$, then $F_{n+m} = F_{n}F_{m-1} + F_{n-1}F_m = F_{n}F_{0} + F_{n-1}F_1 = F_n + F_{n-1} = F_{n+1}$.\n\nSuppose then that the result holds for $m$ and smaller. Now $$F_{n+m+1} = F_{n+m} + F_{n+m-1} = F_{n}F_{m-1} + F_{n-1}F_m + F_{n}F_{m-2} + F_{n-1}F_{m-1}$$ $$= F_{n}(F_{m-1} + F_{m-2}) + F_{n-1}(F_m + F_{m-1}) = F_{n}F_{m} + F_{n-1}F_{m+1}.$$\n\n-", "date": "2016-06-26 04:46:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/11477/showing-that-an-equation-holds-true-with-a-fibonacci-sequence-f-nm-f-n-1?rq=1", "openwebmath_score": 0.8907322287559509, "openwebmath_perplexity": 258.1045831553998, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 1.0, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8774767778695834}}
{"url": "https://math.stackexchange.com/questions/2087099/is-the-binomial-theorem-actually-more-efficient-than-just-distributing/2087103", "text": "# Is the binomial theorem actually more efficient than just distributing\n\nIs the binomial theorem actually more efficient than just distributing a given binomial. I believe it is more confusing for me to remember and work out using the binomial theorem as a guide, than to just distribute when given a problem like: $$(x-4)^6$$ I think distributing that would be equally as painful as using the binomial theorem. Am I alone or is there actually a reason to practice the painful procedures?\n\n\u2022 Well if you distribute without adding equal terms, you will end up with $2^6$ terms, while the theorem gives you the right $6$ terms. \u2013\u00a0Phicar Jan 7 '17 at 6:26\n\u2022 What about using Pascal's triangle to compute the binomial coefficients? \u2013\u00a0Fabio Somenzi Jan 7 '17 at 6:28\n\u2022 What about $(x-1)^{10}$? Binomial theorem gives it by hand in under a minute. How long would you take to expand that by hand? The theorem is also much more conducive to algorithms, meaning it's strength in both computation and proof is far greater than direct distribution. \u2013\u00a0Nij Jan 7 '17 at 6:29\n\u2022 @Phicar $7$ terms, not $6$. \u2013\u00a0Marc van Leeuwen Jan 7 '17 at 17:08\n\u2022 @Nij Actually the OP's method of repeatedly multiplying can be more efficient than the binomial theorem for large polynomials of certain types. There is much symbolic computation literature on this topic, e.g.see R. J. Fateman, Polynomial Multiplication, Powers and Asymptotic Analysis: Some Comments, SIAM J. Comput., 3(3),(1974),196\u2013213. \u2013\u00a0Bill Dubuque Jan 8 '17 at 5:46\n\nThe binomial theorem allows you to write out the expansion of your polynomial immediately.\n\nIt also allows you to answer such questions as \"What is the coefficient of $x^{20}$ in $(1+x)^{100}$?\"\n\nIts generalisation to non-integer exponents allows you to get the expansion of $(1-x)^{-1/2}$.\n\nIt is a good thing.\n\n\u2022 (+1) And needing \"just one term\" isn't an idle exercise, either: anyone having taken calculus should recognize the importance of: \"What's the coefficient of $x^{n - 1}$ in $(x + h)^n - x^n$?\" \u2013\u00a0pjs36 Jan 7 '17 at 6:49\n\u2022 \"It is a good thing\" - Well said! :) (+1) \u2013\u00a0Hypergeometricx Jan 7 '17 at 15:50\n\nThere are all sorts of reasons. First of all, you'll learn when you've moved forward a bit that the binomial theorem can actually be applied in cases of non-integer exponents, by defining $\\binom{n}{k} = \\dfrac{n^{\\underline{k}}}{k!}$, where the underline represents a falling power (i.e., $n(n-1)(n-2)\\cdots$). Another that comes to mind immediately is the series expansion of $e$...\n\n$$e^x = \\lim_{n\\to\\infty}\\left(1+\\frac{x}{n}\\right)^n = \\lim_{n\\to\\infty}\\left(1 + n\\frac{x}{n} + \\frac{n^\\underline{2}}{2!}\\frac{x^2}{n^2} + \\frac{n^\\underline{3}}{3!}\\frac{x^3}{n^3} + \\cdots + \\frac{x^n}{n^n}\\right)$$\n\nAs $n$ gets very large, for fixed $k$, $n^\\underline{k}$ approaches $n^k$, and $\\frac{x^n}{n^n}$ vanishes since $x$ is fixed, so you end up with the familiar series...\n\n$$e^x = 1 + x + \\frac{x^2}{2} + \\frac{x^3}{6} + \\frac{x^4}{24} + \\cdots$$\n\n\u2022 Very nice link to expansion for $e^x$. (+1) \u2013\u00a0Hypergeometricx Jan 7 '17 at 15:51\n\u2022 Weird. I don't see the underline in the definition of $\\binom nk$. Am I the only one with this problem? I'm on Chromium 55.0.2883.87 (64-bit). \u2013\u00a0rubik Jan 7 '17 at 21:55\n\u2022 @rubik I had the same problem -- edited the post to make it displaystyle so it would show up. I think we should bring up this on meta as a bug with mathjax, would you like to do it or should I? \u2013\u00a06005 Jan 7 '17 at 22:54\n\u2022 The frac displays fine in my mobile Chrome (both with dfrac and frac). \u2013\u00a0Martin Argerami Jan 8 '17 at 6:35\n\u2022 @6005 You're right, done. Here's the discussion. \u2013\u00a0rubik Jan 8 '17 at 9:05\n\nIf you're not seeing the utility of the binomial theorem, then I think you're missing an important observation:\n\nThe coefficients of every term can be calculated extremely efficiently (your example of $(x-4)^6$ is easy to expand by hand with the binomial theorem, without, it's doable but very tedious).\n\nA useful shorthand for lower powers of the exponent is to remember the first few rows of Pascal's Triangle along with the rule for adding more if you need them. Example of first few rows:\n\n$$1$$ $$1\\qquad 1$$ $$1\\qquad 2\\qquad 1$$ $$1\\qquad 3\\qquad 3\\qquad 1$$ $$1\\qquad 4\\qquad 6\\qquad 4\\qquad 1$$ $$1\\qquad 5\\qquad 10\\qquad 10\\qquad 5\\qquad 1$$ $$1\\qquad 6\\qquad 15\\qquad 20\\qquad 15\\qquad 6\\qquad 1$$\n\nThe last row gives the coefficients of $x^k\\cdot(-4)^{(6-k)}$ for $k$ running from $0$ to $6$. Computing $(-4)^{(6-k)}$ is easy enough by hand, and multiplying it by the relevant entry in the last row above is also fairly easy (give it a try).\n\nOne more thing: The binomial theorem admits an interesting generalization to powers of general polynomials of a given length sometimes called the multinomial theorem. (Challenge: try investigating the coefficients of the first few powers of $(x+y+z)^n$ and perhaps you can figure out the pattern for trinomials? How does it relate to the pattern for binomials? Can you generalize it to general polynomials?)\n\nMarty Cohen and Phicar answered your question correctly. In mathematics, as we always use small numbers, It could be feasible to think this. Why learn the binomial theorem to compute $(a+b)^2$? But what can you say about $(3a+5b)^{987658}$? The binomial theorem is way more efficient in this case and the actual number of cases in which it wouldn't be efficient is very small.\n\nThe binomial theorem gives you a powerful tool: The computation for any exponent in - perhaps - the fastest way possible in which it is only not effective for a finite amount of exponents, but its the best possible for an endless amount of exponents.\n\nThis is one trap in mathematics: We implicitly assume that the mathematical tools are created to treat familiar cases, but they are created to treat all cases in the neatest way possible, they will be cumbersome for some simple cases (Example: Integrating $f(x)=mx$ in the interval $[0,a]$. This is just a triangle, there is no need for integral calculus here). I remember of me asking about why we need analysis if calculus seems to work well, but I was supposing (without realizing) that calculus is used only on familiar cases, but we want to be useful also for very imaginative cases!\n\n\u2022 Just a small note to support the statement that it's as fast as it gets: using the recursion ${n\\choose k+1} = {n\\choose k}\\frac{n-k}{k+1}$ with ${n\\choose 0} = 1$ and the symmetry ${n\\choose k} = {n\\choose n-k}$ then we can compute all the $n$ binomial coefficients ${n\\choose k}$ for $k=1,2,\\ldots,n$ with only $n$ operations (multiplications and divisions). That is hard to beat! This can be compared with $\\mathcal{O}(2^n)$ operations needed to distribute $(a+b)^n$. \u2013\u00a0Winther Jan 8 '17 at 3:28\n\u2022 Excelent addendum, @Winther! \u2013\u00a0Billy Rubina Jan 8 '17 at 13:37\n\nTo add to other answers, in higher-level math it's important to be able to generalize the behavior of $(a + b)^n$ when $n$ is a variable. The work that you're doing now is mostly just training exercises for being familiar with, and having intuition for, the general binomial theorem.\n\nFor example, there's a really key moment early on in calculus when it's critical to understand that, for all $n$: $(x + h)^n = x^n + nx^{n-1}h....$ plus a bunch of other stuff with higher powers of $h$ in it, but whose exact numerical coefficients we can ignore.", "date": "2021-04-18 12:37:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2087099/is-the-binomial-theorem-actually-more-efficient-than-just-distributing/2087103", "openwebmath_score": 0.8282362222671509, "openwebmath_perplexity": 465.86158679560197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130563978902, "lm_q2_score": 0.8872045966995027, "lm_q1q2_score": 0.8774569298320326}}
{"url": "http://math.stackexchange.com/questions/42797/evaluate-binomn0-binomn2-binomn4-cdots-binomn2k-cdots/42800", "text": "# Evaluate $\\binom{n}{0}+\\binom{n}{2}+\\binom{n}{4}+\\cdots+\\binom{n}{2k}+\\cdots$ [duplicate]\n\nThis question already has an answer here:\n\nI need to evaluate, for a certain worded question:\n\nIf n is even $$\\binom{n}{0}+\\binom{n}{2}+\\binom{n}{4}+\\cdots\\binom{n}{n}$$ If n is odd $$\\binom{n}{0}+\\binom{n}{2}+\\binom{n}{4}+\\cdots\\binom{n}{n-1}$$\n\nI havent been able to reduce this using the binomial expansion.\n\n-\n\n## marked as duplicate by Grigory M, Sasha, Daniel Fischer\u2666, hardmath, DanJan 16 '14 at 22:19\n\nunrelated: why is the first binomial coefficient in the title bigger than the others? \u2013\u00a0 kuch nahi Jun 2 '11 at 15:35\n\u2013\u00a0 Grigory M Jun 2 '11 at 15:37\nJust want to add that the original question was to find the number of binary sequences of length n that contain an even number of ones. Using a different argument the answer I got is $2^{n-1}$ \u2013\u00a0 kuch nahi Jun 2 '11 at 15:40\nHint: Do some computations for small values of $n$. You will see a pattern. When $n$ is odd, the pattern is most obviously proved by symmetry. When $n$ is even, look at Pascal's triangle. \u2013\u00a0 Aaron Jun 2 '11 at 15:41\n\nBinomial expansion allows one to find both $\\binom n0+\\binom n1+\\binom n2+\\dots=(1+1)^n$ and $(\\binom n0-\\binom n1+\\binom n2+\\dots)=(1-1)^n$. Combining these two yields desired result: $\\binom n0+\\binom n2+\\binom n4+\\dots=\\frac12\\bigl((1+1)^n+(1-1)^n\\bigr)=2^{n-1}$.\n\n-\nI remember I've done this before. Thanks a ton! \u2013\u00a0 kuch nahi Jun 2 '11 at 15:45\nThe first version was helpful enough. Anyhow, thanks again. \u2013\u00a0 kuch nahi Jun 2 '11 at 15:54\n\nThe given sums count the number of even-cardinality subsets of an $n$-element set. I'll provide a simple argument that shows that the value of that sum is $2^{n-1}$.\n\nFix $x \\in X$ (where $|X|=n$). The map $f \\colon \\mathcal{P}(X) \\to \\mathcal{P}(X)$ (where $\\mathcal{P}(X)$ denotes the power set of $X$) defined by $$f(Y) = \\begin{cases} Y \\cup \\{x\\} &\\text{if } x \\notin Y \\\\ Y - \\{x\\} &\\text{if } x \\in Y \\end{cases}$$\n\nis a bijection of $\\mathcal{P}(X)$ that takes even-cardinality subsets to odd-cardinality subsets and vice-versa. Thus there are the same number of even-cardinality subsets as odd-cardinality subsets. Since there are $2^{n}$ total subsets, there must be $2^{n}/2 = 2^{n-1}$ even-cardinality subsets.\n\n-\n\nThe sum of all binomial coefficients ${n \\choose 0} + \\cdots {n \\choose n}$ is $2^n$. This is a well-known equation and the slick proof is to consider the binomial theorem for $(1+1)^n$.\n\nRelated is the binomial theorem applied to $(1-1)^n$ giving the sum\n\n$${n \\choose 0} - {n \\choose 1} + {n \\choose 2} - \\cdots \\pm {n \\choose n} = 0.$$\n\nThe latter inequality is easy to see when you add up the $n$th row of Pascal's triangle with alternating signs and $n$ is odd. For example: $1-5+10-10+5-1 = 0$ is obvious from the symmetry of the entries.\n\nTo obtain the answer for every-other term in the first sum, you can add these two formulas together and the negative entries will cancel with the positive ones while the positive ones will double:\n\n$$(1+1)^n + (1-1)^n = 2{n \\choose 0} + 2{n \\choose 2} + \\cdots = 2^n + 0.$$\n\nNow to finish, divide by two. To get the odd entries (the second part of your question), notice that $(1-1)^n = -(1-1)^n$ and use the technique above. Or you could just take the sum for all entries and subtract what we just obtained for the even entries.\n\n-\n\nLet $X$ be a binomial$(n,p)$ random variable with $p=1/2$. If $n$ is even, then the probability that $X$ is even is equal to $$\\sum\\limits_{k = 0,2, \\ldots ,n} {{n \\choose k}\\frac{1}{{2^n }}} .$$ On the other hand, this probability is obviously equal to $1/2$ (consider $X$ as the sum of $n$ independent random variables taking the values $0$ and $1$ with probability $1/2$ each). Hence, $$\\sum\\limits_{k = 0,2, \\ldots ,n} {{n \\choose k}} = 2^n \\frac{1}{2} = 2^{n - 1} .$$\n\n-\n+1: I'm always a fan of the probabilistic approach. :) \u2013\u00a0 Mike Spivey Jun 2 '11 at 16:37\n@Mike - Thanks! \u2013\u00a0 Shai Covo Jun 2 '11 at 18:24\n\nHere's a more general take. Sums of the form $\\sum_k \\binom{n}{2k} f(k)$ are sometimes called aerated binomial sums. My paper \"Combinatorial Sums and Finite Differences\" (Discrete Mathematics, 307 (24): 3130-3146, 2007) proves some general results about these aerated binomial sums. (See Section 4.) The case $f(k) = 1$ (as in the OP's question) falls out nicely.\n\nSuppose you are interested, for some function $f(k)$, in the binomial sum\n\n$$B(n) = \\sum_{k=0}^n \\binom{n}{2k} f(k).$$\n\nThen, taking the finite difference $\\Delta f(k) = f(k+1) - f(k)$, denote $A(n)$ by\n\n$$A(n) = \\sum_{k=0}^n \\binom{n}{2k} \\Delta f(k).$$\n\nIn the paper I prove that $B(n)$ and $A(n)$ are related via $$B(n) = 2^{n-1} f(0) + 2^n \\sum_{k=2}^n \\frac{A(k-2)}{2^k} + \\frac{f(0)}{2}[n=0],$$ where $[n=0]$ evaluates to $1$ if $n=0$ and $0$ otherwise.\n\nSince $f(k) = 1$ in the OP's question, $\\Delta f(k) = 0$, and so $A(n) = 0$ for all $n$. Thus $$\\sum_{k=0}^n \\binom{n}{2k} = 2^{n-1} + \\frac{1}{2}[n=0].$$\n\nMore generally, this approach can be used to prove that, for $m \\geq 1$, $$\\sum_k \\binom{n}{2k} k^{\\underline{m}} = n(n-m-1)^{\\underline{m-1}} \\, 2^{n-2m-1} [n \\geq m+1],$$ and $$\\sum_k \\binom{n}{2k} k^m = n \\sum_j \\left\\{ m \\atop j\\right\\} \\binom{n-j-1}{j-1} (j-1)! \\, 2^{n-2j-1},$$ where $\\left\\{ m \\atop j\\right\\}$ is a Stirling number of the second kind. (While the second equation swaps one sum for another, there are no more than $m$ terms in the right-hand side, and so it is useful when $m$ is small.)\n\nThe approach can also be extended to sums of the form $\\sum_k \\binom{n}{ak+b} f(k)$. (And the case $a = 2, b = 1$ is considered explicitly in the paper.) Using higher values of $a$ and $b$ would result in increasingly complicated expressions, though.\n\n-\n\nI think it might help to look at this in terms of the rows on pascal's triangle. The sum of a given row is 2^n. All rows are symmetric about the middle of the row. As this sounds like it could be HW, i'll leave it to you to figure out how each of those relate to the sum of the row.\n\n-\nI wish this were homework. As that would imply I am taking a course and have access to an instructor. \u2013\u00a0 kuch nahi Jun 2 '11 at 15:43\nOk, so they are both exactly half the of the row (they get every element once as opposed to twice). Since the sum of a row is 2^n, this gives 2^(n-1) as the answer. \u2013\u00a0 soandos Jun 2 '11 at 15:45\nyup, got it. Thanks! +1 \u2013\u00a0 kuch nahi Jun 2 '11 at 15:54", "date": "2015-05-27 02:24:54", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/42797/evaluate-binomn0-binomn2-binomn4-cdots-binomn2k-cdots/42800", "openwebmath_score": 0.9525095224380493, "openwebmath_perplexity": 278.71350982102587, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830768464319, "lm_q2_score": 0.9019206804839998, "lm_q1q2_score": 0.8774161886891111}}
{"url": "http://fonsg3.hum.uva.nl/paul/methoden/comparingPvalues.html", "text": "# Example: a t-test between two groups\n\nAssume that 10 speakers are randomly drawn from the Dutch and Flemish populations. Their scores on a German-language proficiency test are measured:\n\ndutch = c(34, 45, 33, 54, 45, 23, 66, 35, 24, 50)\nflemish = c(67, 50, 64, 43, 47, 50, 68, 56, 60, 39)\n\nThe Flemish look a bit better than the Dutch. Is this difference statistically significant?\n\nt.test (dutch, flemish, var.equal=TRUE)\n##\n##  Two Sample t-test\n##\n## data:  dutch and flemish\n## t = -2.512, df = 18, p-value = 0.02176\n## alternative hypothesis: true difference in means is not equal to 0\n## 95 percent confidence interval:\n##  -24.790596  -2.209404\n## sample estimates:\n## mean of x mean of y\n##      40.9      54.4\n\n1. yes\n2. no\n3. don\u2019t know\n\n# .\n\nYes, it is a probability. But of what?\n\n# .\n\n### Question: Is this p-value of 0.02176\u2026\n\n1. the probability that the Dutch and Flemish population means are the same?\n2. the probability that if the Dutch and Flemish population means are the same, an absolute t-value of 2.512 or greater will be found between random Dutch and Flemish samples of size 10?\n3. don\u2019t know\n\n# .\n\nSo yes, the p-value of 0.02176 is a tail probability: the probability that if the Dutch and Flemish population means are the same, an absolute t-value of 2.512 or greater will be found between random Dutch and Flemish samples of size 10.\n\n# .\n\n### Question: What is this condition (\u201cthe Dutch and Flemish population means are the same\u201d) usually called?\n\n1. the null hypothesis\n2. the alternative hypothesis\n3. don\u2019t know\n\n# .\n\nIt\u2019s the null hypothesis (one can have null hypotheses that are not about equality, but the one used here is typical).\n\nLet\u2019s check that the probability of finding an absolute t above 2.512 is indeed around 2.2 percent, by drawing a hundred thousand random samples from the Dutch and Flemish populations.\n\nIn the simulation, we assume that the null hypothesis is true, so that:\n\n\u2022 the true population means are the same ($$\\mu_d = \\mu_f = 50$$; the exact number doesn\u2019t matter)\n\u2022 the true standard deviations are the same as well ($$\\sigma_d = \\sigma_f = 8$$)\n\nHere is how you obtain the data for one such an experiment (not 100,000 yet):\n\nnumberOfParticipantsPerGroup = 10\nmu.d = mu.f = 50\nsigma.d = sigma.f = 8\ndata.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d)\ndata.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f)\ndata.d\ndata.f\n##  [1] 46.01451 53.06159 58.37237 55.26379 43.74593 49.95321 55.10986\n##  [8] 45.29203 47.70117 48.67666\n##  [1] 62.36448 54.69570 59.98880 62.60507 47.95375 51.00754 48.28727\n##  [8] 44.34768 58.20126 48.39343\n\nYou get the t-value from such data as follows:\n\nnumberOfParticipantsPerGroup = 10\nmu.d = mu.f = 50\nsigma.d = sigma.f = 8\ndata.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d)\ndata.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f)\nt = t.test (data.d, data.f, var.equal=TRUE) $statistic t ## t ## 0.286349 Then try this a hundred thousand times, and see how often the absolute t-value reaches 2.512 or more: numberOfParticipantsPerGroup = 10 numberOfExperiments = 1e5 count = 0 for (experiment in 1 : numberOfExperiments) { mu.d = mu.f = 50 sigma.d = sigma.f = 8 data.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d) data.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f) t = t.test (data.d, data.f, var.equal=TRUE)$ statistic\nif (abs (t) > 2.512) {\ncount = count + 1\n}\n}\ncount\n## [1] 2213\n\nThat\u2019s very close to the 0.02176 probability computed by t.test. So, R\u2019s t.test seems to compute the p-value correctly!\n\n# .\n\n### Question: What can we conclude from the p-value of 0.02176, which is less than the common criterion of 0.05?\n\n1. we reject the null hypothesis, so the Flemish group is better than the Dutch group\n2. we reject the null hypothesis, so Flemish people are better than Dutch people\n3. don\u2019t know\n\n# .\n\nWhat if one participant had been different (a score of 27 instead of 47)?\n\ndutch = c(34, 45, 33, 54, 45, 23, 66, 35, 24, 50)\nflemish = c(67, 50, 64, 43, 27, 50, 68, 56, 60, 39)\nt.test (dutch, flemish, var.equal=T)\n##\n##  Two Sample t-test\n##\n## data:  dutch and flemish\n## t = -1.9122, df = 18, p-value = 0.07191\n## alternative hypothesis: true difference in means is not equal to 0\n## 95 percent confidence interval:\n##  -24.13526   1.13526\n## sample estimates:\n## mean of x mean of y\n##      40.9      52.4\n\nThe p-value is now greater than 0.05.\n\n# .\n\n### Question: What can we conclude?\n\n1. we cannot reject the null hypothesis, so we cannot conclude anything\n2. we accept the null hypothesis, so Dutch people are as good as Flemish people\n3. don\u2019t know\n\n# .\n\n## Neyman\u2013Pearson statistics\n\nI now very briefly describe a way of doing statistics that I will not recommend for scientific fact finding. It is quite usable for binary decisions in real life, though, such as the question whether or not to take a new drug to market.\n\nThe \u201cType I\u201d error rate is\n\n$$\\alpha$$ = 0.05\n\nqt (p = 0.025, df = 18)\n## [1] -2.100922\nqt (p = 0.975, df = 18)\n## [1] 2.100922\n\nIf the null hypothesis (no difference between Dutch and Flemish population means) is true, the absolute t-value will be found to lie above 2.101 5 percent of the time (for 10 participants per group).\n\nThe \u201cType II\u201d error rate is\n\n$$\\beta$$ = 0.20\n\nIf the true population differs by e.g.\u00a04 (or any other number considered to be the threshold of importance), the absolute t-value will be found to lie above 2.101 80 percent of the time (for 10 participants per group).\n\nThe recipe for this methodology in e.g.\u00a0drug treatment is:\n\n1. Measure the effect of the drug in 10 people, and the effect of placebo in 10 other people.\n2. If the difference between the groups yields a t-value above 2.101 in favour of the drug, reject the null hypothesis and make the drug officially available to future patients.\n3. If the difference between the groups yields a t-value below 2.101 in favour of the drug, \u201caccept\u201d the null hypothesis and take the drug off the market.\n\nValues for $$\\alpha$$ and $$\\beta$$ will depend on:\n\n\u2022 The severity of side effects.\n\u2022 The size of the potential cure (life-saving?).\n\u2022 The cost of the drug.\n\u2022 And so on.\n\n\u2022 yes\n\u2022 no\n\u2022 don\u2019t know\n\n# .\n\nNo. I didn\u2019t mention any p-values in my very brief discussion of Neyman\u2013Pearson, and they are indeed not used in that way of doing statistics. One either rejects the null hypothesis if t is in the \u201ccritical region\u201d, or accepts the null hypothesis if t is smaller. One only reports the \u201csignificance level\u201d $$\\alpha$$ that is used to demarcate the critical region, and one reports that $$\\alpha$$ is 0.05.\n\n# .\n\n### Question: Is this a good procedure for scientific belief finding?\n\n\u2022 yes, because I need a clear criterion for my beliefs\n\u2022 no, because the strength of my belief will gradiently depend on the p-value\n\u2022 don\u2019t know\n\n# .\n\nI already suggested that it\u2019s not. Surely the difference between a p-value of 0.049 and one of 0.051 is not impressive enough for me to wholly believe that the effect is real in the former case and that the effect is zero in the latter case. You can read about this in the free article Erroneous analyses of interactions in neuroscience: a problem of significance by Sander Nieuwenhuis, Birte Forstmann & Eric-Jan Wagenmakers (2011, Nature Neuroscience).\n\n### Fisher quotes\n\nRonald Fisher also had some things to say about p-values above 0.05, namely that you should ignore them (they are null results, and a null result is the same as no result):\n\nThe test of significance [\u2026] tells [the researcher] what to ignore, namely all experiments in which significant results are not obtained. (Fisher 1929, p.\u00a0191)\n\nThe null hypothesis is never proved or established, but is possibly disproved, in the course of experimentation. (Fisher 1966, p.\u00a016)\n\n### Can you mix Fisher\u2019s and Neyman\u2013Pearson\u2019s approaches?\n\nStrict statisticians will tell you that you cannot mix the two approaches. Often you see mixes. For instance, in a paper that uses $$\\alpha$$ and $$\\beta$$ (and therefore did a power analysis), people sometimes report a p-value, especially if it is low. Please don\u2019t. And in papers that generally report p-values, don\u2019t say things like \u201csignificant with p < 0.05\u201d, which is $$\\alpha$$-terminology, but report the actual p-value that you obtained. Sometimes you see things like \u201ceffect A was significant with p < 0.05, and effect B was significant with p < 0.01\u201d, or, equivalently, \u201ceffect A was significant at the 0.05 level, and effect B was significant at the 0.01 level.\u201d Both are cases of \u201cmoving alpha\u201d, a concept that generative linguists may like (in a very different meaning), but that statisticians of all convictions enthusiastically discourage. Practically, you just always report the p-value and state a conclusion about the population only if this p-value is below 0.05.\n\n### Today\u2019s conclusion\n\nIn your research papers you should never compare p-values. If you find that your Spanish listeners improve significantly (p = 0.001), and your Portuguese listeners don\u2019t improve significantly (p = 0.63), you can conclude that Spanish listeners (without \u201cthe\u201d, i.e.\u00a0the population on average) improve, but you cannot conclude that Portuguese listeners don\u2019t improve. And you cannot even conclude that Spanish listeners improve more than Portuguese listeners. This is the single most common mistake in psychological and linguistic research papers (see Nieuwenhuis\u2019 paper for counts, and any linguistic conference proceedings for examples).\n\nWhat should you do instead? The only valid significance test is a direct comparison between the Spanish and Portuguese listeners. If every participant\u2019s improvement can be measured by a single number, then the above two p-values were probably based on \u201ct-tests against zero\u201d, i.e.\u00a0the mean improvement was significantly above zero for the Spanish group but not for the Portuguese group. The only valid significance test is a two-group t-test, as above in the Dutch\u2013Flemish example. This directly compares the Spanish group with the Portuguese group, and if the result of this test is significant (i.e. p < 0.05), you conclude that Spanish listeners (without \u201cthe\u201d) improve more than Portuguese listeners. So the report is never \u201cSpanish listeners improve but Portuguese listeners don\u2019t\u201d but \u201cSpanish listeners improve more than Portuguese listeners\u201d, and it has to be based on a direct comparison between the two groups.\n\nOr consider this one. For Brazilian Portuguese, you discover (by a two-group t-test) that male and female speakers are significantly different in their use of construction X (males use it more than females), whereas for Iberian Portuguese, your two-group t-test does not show a significant difference between male and female speakers. From these two tests, you cannot conclude anything about the difference between Brazilian and Iberian Portuguese. To test whether the male\u2013female difference is different in Brazilian than in Iberian Portuguese, you perform an analysis of variance with two fixed factors, namely dialect and sex (and with the use of construction X as the dependent variable), and you examine the p-value for the interaction between dialect and sex. If this p-value is below 0.05, and the difference of the difference is in the right direction, you report something like \u201cthe male advantage (over females) for construction X is greater for Brazilian Portuguese than for Iberian Portuguese\u201d, or, equivalently, \u201cthe Brazilian advantage (over Iberian) for construction X is greater for males than for females\u201d (basically, you will try to simplify in such a manner the wording that you use to explain what the interaction is about; this may depend on what your main effects look like).\n\n1. Get many participants. If the effect exists and is not small, there\u2019s a good chance you\u2019ll detect it (p < 0.05).\n2. If p > 0.05, you haven\u2019t detected the effect. However, you will be able to report a narrow confidence interval and therefore say that the effect is \u201csmall or zero\u201d.\n\n# Other things you should not do to lower your p-value\n\nMany tricks are in Simmons et al. (2011). Summary of advice: use the word \u201conly\u201d and don\u2019t lie.\n\n# \u201cTest until significant\u201d\n\nThis is one of the tricks mentioned by Simmons et al.\n\nnum.rounds = 100   # the number of rounds; after each round a hypothesis test is performed\ndN = 10   # the number of participants in each round\nnum.experiments = 10000   # the number of experiments\n\nvalue = rep (0, num.experiments)\nnum.sig = 0   # the number of experiments that turn out significant\nfor (iexp in 1 : num.experiments) {\nfor (iround in 1 : num.rounds) {\nN = dN * iround\nvalue [(N - dN + 1) : N] = rnorm (dN, 0, 1)\naverage = sum (value [1 : N]) / N\nstdev = sqrt (sum ((value [1 : N] - average) ^ 2) / (N - 1))\nstderr = stdev / sqrt (N)\nif (abs (average) > stderr * qt (0.975, N - 1)) {\nnum.sig = num.sig + 1\nbreak\n}\n}\n}\ncat (num.sig / num.experiments)\n## 0.3866\n\nSo don\u2019t do that. Instead determine the number of participants in advance, or determine a stopping criterion in advance.", "date": "2017-12-16 16:38:52", "meta": {"domain": "uva.nl", "url": "http://fonsg3.hum.uva.nl/paul/methoden/comparingPvalues.html", "openwebmath_score": 0.6617834568023682, "openwebmath_perplexity": 2259.715928483653, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9805806540875629, "lm_q2_score": 0.894789468206764, "lm_q1q2_score": 0.8774132420048512}}
{"url": "https://www.physicsforums.com/threads/constant-function.238219/", "text": "Constant Function\n\n1. Jun 1, 2008\n\nritwik06\n\n1. The problem statement, all variables and given/known data\nData Given:\nf'(x)=f(x) ..........(i)\nand\nf(0)=0 ..........(ii)\n\nWhat kind of function is f(x)?\n\n3. The attempt at a solution\nFrom (i)\ndy/dx=y\n1/y dy= dx\nIntegrating;\nln y = x+C\nFrom (ii)\nln 0=0+c\nTherefore; c is not defined!\n\nMy book gives the answer that f(x) is a constant function. But how is it true? Is my way of solving wrong?\nregards,\nRitwik\n\n2. Jun 1, 2008\n\nrock.freak667\n\nFrom the part in bold\n\ny=exp(x+c)\n\n3. Jun 1, 2008\n\nVid\n\nYou divided by y when y = 0. I don't think you're supposed to solve it, but rather argue why it can't be anything other than a constant function based on the equation and the initial values.\n\nLast edited: Jun 1, 2008\n4. Jun 1, 2008\n\nRaze2dust\n\ny=any constant clearly doesn't satisfy (i)\n\nthough y=0 does satisfy..i think the answer is y=0 not y=any constant\n\nwhen you do dy/y=dx, you are assuming that y is not equal to zero, because dy/y will not be defined for y=0\n\nSo, you have to consider the case y=0 separately\n\n5. Jun 1, 2008\n\nGregA\n\nThe solution of your ODE should give y = Ae^x (for some constant A)\nApplying your initial condition gives 0 = Ae^0 ==> A = 0\n\nIf I'm right the function y = 0 is just as much a constant function as say the function y = 5\n\nLast edited: Jun 1, 2008\n6. Jun 2, 2008\n\nHallsofIvy\n\nStaff Emeritus\nNo, integrating does NOT give \"ln y= x+ C\". Integrating gives ln |y|= x+ C. That's your crucial error.\n\nTo see why that is important, take the exponential of both sides.\neln |y|= ex+ C or |y|= C' ex where C'= eC. While eC must be positive, we can drop the absolute value by allowing C' to be negative as well, and, by continuity, C'= 0.\n\ny(x)= C'ex obviously satisifies the given differential equation for any real number C'.\n\nNow, put y=0, x= 0 into y= C'ex. What is C'? What is y(x)?\n\n7. Jun 2, 2008\n\nspideyunlimit\n\nData Given:\nf'(x)=f(x) ..........(i)\nand\nf(0)=0 ..........(ii)\n\nWhat kind of function is f(x)?\n-------------------------\n\nf(x) is a constant function..... f(x) = 0\nwhere f'(x) = f(x) holds true. (The only other possibility was f(x) = e^x but then f(0) is not 0.)\n\n8. Jun 2, 2008\n\nspideyunlimit\n\nFrom your attempt at the solution you just get that c = 0.... but that doesn't help really.\n\n9. Jun 2, 2008\n\nnicksauce\n\nThere is a simple solution f(x) = 0. What do you know about the uniqueness of the solution?", "date": "2017-10-17 05:21:01", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/constant-function.238219/", "openwebmath_score": 0.8463664650917053, "openwebmath_perplexity": 2554.7665210619953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964177527898, "lm_q2_score": 0.8902942253943279, "lm_q1q2_score": 0.877381769872105}}
{"url": "http://daisytale.it/qwtn/fourier-transform-of-cosine.html", "text": "$\\cos$- and $\\sin$-Fourier transform and integral; Discussion: pointwise convergence of Fourier integrals and series; Heuristics. 5 Signals & Linear Systems Lecture 10 Slide 11 Fourier Transform of any periodic signal \u2211Fourier series of a periodic signal x(t) with period T 0 is given by: Take Fourier transform of both sides, we get: This is rather obvious!. By proper choice of a k we can get the real fourier series out of this. The key idea is given in point 4 above; a cosine function that fits a whole number of cycles into the input list will produce two non-zero points in the output. In this section, we\u2019ll try to really explain the notion of a Fourier expansion by building on the ideas of phasors, partials, and sinusoidal components that we introduced in the previous section. Then the Fourier cosine series for f(x) is the same as the Fourier series for fo(x) (in the sense that they look exactly the same). In discussing the discrete cosine transform (DCT) and the discrete sine transform (DST), we shall first consider the continuous versions of these, i. Plane Geometry Solid Geometry Conic Sections. This is one of the duality properties of Fourier transforms. From the modulation theorem, you know what happens to a signal's spectrum when it is multiplied by a cosine. It presents a mathematical proof of what is the magnitude of an N-point discrete Fourier transform (DFT) when the DFT's input is a real-valued sinusoidal sequence. Derivation of Fourier Series. (Note that there are other conventions used to de\ufb01ne the Fourier transform). Recall that eii\u03b8=+cos sin\u03b8 \u03b8; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. 23 FOURIER SINE AND COSINE SERIES; CALCULATION TRICKS 4 Decay rate of Fourier series. Fourier coefficients for sine terms. Fourier transform cosine example further s blogs mathworks images steve 2009 f cos t in additions akshaysin github io images mpl basic fft moreovers upload wikimedia org wikipedia mons 6 61 fft time frequency view in additionee nmt edu wedeward ee342 sp99 ex le16 gif. cos(a) cos(b) = 2sin(1 2 (a+ b))sin(1 2 (a b)) (11) Question 41: (i) Let fbe an integrable function on (1 ;+1). Fourier transforms table lecture 10 fourier transform thefouriertransform com fourier transform of the sine and cosine fourier transform pairs basic electrical engineering homework Whats people lookup in this blog:. Fourier transform of this spectrum is the same as that for a single Gaussian multiplied by an additional term cos(2\u03c0s\u2206). Each \"spike\" on the second plot is the magnitude of the sine or cosine at that frequency. The above are all even functions and hence have zero phase. applying a Fourier transform would yield a bunch of different sine and cosine waves of different frequencies that sum together to be the equivalent of the original audio wave. I'm placing the Fourier analysis material in the back so it won't scare people away. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. In this section we define the Fourier Cosine Series, i. syms a b t f = rectangularPulse (a,b,t); f_FT = fourier (f). The DFT has become a mainstay of numerical computing in part. T): The Fourier Cosine transform of the function f(x) is (0, ) is defined by F c {f(x)} = F c (s) = f(x) Cos sx dx----(5) 0 and f(x) = (2/ ) F c (s) Cos sx ds----(6) 0 is called the inverse Fourier Cosine Transform of F C (s). This cosine function can be rewritten, thanks to Euler, using the identity:. Its counterpart for discretely sampled functions is the discrete Fourier transform (DFT), which is normally computed using the so-called fast Fourier transform (FFT). If x(t)x(t) is a continuous, integrable signal, then its Fourier transform, X(f)X(f) is given by. 1 Practical use of the Fourier. RUBIN Abstract. For math, science, nutrition, history. Cosine waves c. Gowthami Swarna, Tutorials Poin. Fourier coefficients for sine terms. Frequency Domain and Fourier Transforms Frequency domain analysis and Fourier transforms are a cornerstone of signal and system analysis. In the previous Lecture 14 we wrote Fourier series in the complex form. Spectrum of cosine signal has two impulses at positive and negative frequencies. , the Fourier cosine transform (FCT) and the Fourier sine transform (FST). A Fourier transform is a linear transformation that decomposes a function into the inputs from its constituent frequencies, or, informally, gives the amount of each frequency that composes a signal. It can be derived in a rigorous fashion but here we will follow the time-honored approach of considering non-periodic functions as functions with a \"period\" T !1. The inverse transform of F(k) is given by the formula (2). \u2022 Instead of the sines and cosines in a Fourier series, the Fourier transform uses exponentials and complex numbers. , the Fourier cosine transform (FCT) and the Fourier sine transform (FST). If we follow through exactly the same method as above (we can in effect split the function into cos(at) and cos(bt) and do both separately), we should get:. 2) is called the Fourier integral or Fourier transform of f. The Fourier transform on S. Fourier transform of this spectrum is the same as that for a single Gaussian multiplied by an additional term cos(2\u03c0s\u2206). Due to the duality property of the Fourier transform, if the time signal is a sinc function then, based on the previous result, its Fourier transform is This is an ideal low-pass filter which suppresses any frequency f > a to zero while keeping all frequency lower than a unchanged. Auxiliary Sections > Integral Transforms > Tables of Fourier Cosine Transforms > Fourier Cosine Transforms: Expressions with Exponential Functions Fourier Cosine Transforms: Expressions with Exponential Functions No Original function, f(x) Cosine transform, f\u02c7c(u) = Z 1 0 f(x)cos(ux)dx 1 e\u2212ax a a2+u2 2 1 x \u00a1 e\u2212ax \u2212e\u2212bx \u00a2 1 2 ln b2+u2. Fourier Series of Even and Odd Functions - this section makes your life easier, because. 2 p693 PYKC 8-Feb-11 E2. This means that the angle of the transform of the sine function, which is the arctan of real over imaginary, is 90\u00b0 off from the transform of the cosine, just like the sine and cosine functions are 90\u00b0 off from each other. Fourier Transform of Cosine Wave Watch more videos at https://www. We can write f\u02dc(k)=f\u02dcc(k)+if\u02dc s(k) (18) where f\u02dc s(k) is the Fourier sine transform and f\u02dcc(k) the Fourier cosine transform. The expression above is not too different from $\\mathcal F\\{{\\cos(2\\pi Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The third plot shows the inverse discrete Fourier transform, which converts the sines and cosines back into the original function f(x). Engineering Tables/Fourier Transform Table 2 From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search Signal Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. Many more transform pairs could be shown. \" The full name of the function is \"sine cardinal,\" but it is commonly referred to by its abbreviation, \"sinc. It is used in most digital media, including digital images (such as JPEG and HEIF, where small high-frequency. Conic Sections. First term in a Fourier series. applying a Fourier transform would yield a bunch of different sine and cosine waves of different frequencies that sum together to be the equivalent of the original audio wave. e\ufb01ne the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f (t)= 0 t< 0 1 t \u2265 0? the Laplace transform is 1 /s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1 /j\u03c9 in fact, the integral \u221e \u2212\u221e f (t) e \u2212 j\u03c9t dt = \u221e 0 e \u2212 j\u03c9t dt = \u221e 0 cos. If x(t)x(t) is a continuous, integrable signal, then its Fourier transform, X(f)X(f) is given by. , the frequency domain), but the method for determining the phase and magnitude of the sinusoids was not discussed. Instead, Fourier Cosine transform should be used. This means that the angle of the transform of the sine function, which is the arctan of real over imaginary, is 90\u00b0 off from the transform of the cosine, just like the sine and cosine functions are 90\u00b0 off from each other. As a result, the Fourier transform is an automorphism of the Schwartz space. (14) and replacing X n by. Members who need to use special functions and characters still need to learn the correct Mathematica \u00ae input format from the HELP page. Finding Fourier Sine. Sometimes the following definition with the same factors in front is used. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2\u02c7 (! ! 0) !2[ \u02c7;\u02c7) the spectrum is zero for !6= ! 0 cos(! 0n. The DTFT of a discrete cosine function is a periodic train of impulses: I updated the above plot on 6-Jan-2010 to show the location of the impulses. Help with a trignometric double integral. (i) In Example 1, if u(0,t) 0 and P(0) 0, it would be inappropriate to use Fourier Sine transform. New boron material of high hardness created by plasma chemical vapor deposition; Earth-size, habitable-zone planet found hidden in early NASA Kepler data. This package contains C and Fortran FFT codes. This article will walk through the steps to implement the algorithm from scratch. See also: Annotations for \u00a71. \u2022 Continuous Fourier Transform (FT) \u2013 1D FT (review) \u2013 2D FT \u2022 Fourier Transform for Discrete Time Sequence (DTFT) \u2013 1D DTFT (review) \u2013 2D DTFT \u2022 Li C l tiLinear Convolution \u2013 1D, Continuous vs. Find the Fourier transform of the following signals: a) f1(t)=e \u22123 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. The expression above is not too different from$\\mathcal F\\{{\\cos(2\\pi Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The key idea is given in point 4 above; a cosine function that fits a whole number of cycles into the input list will produce two non-zero points in the output. Other definitions are used in some scientific and technical fields. I am also aware of the Pontryagin Duality generalization for locally compact abelian groups, though I am personally more concerned. Trigonometric Fourier Series 1 ( ) 0cos( 0) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2 ( )cos( ) ,and 2 ( ) , 1 Complex Exponential Fourier Series T j nt n n j nt n f t e dt T f t F e F 0 0 1 ( ) , where. The motivation of Fourier transform arises from Fourier series, which was proposed by French mathematician and physicist Joseph Fourier when he tried to analyze the flow and the distribution of energy in solid bodies at the turn of the 19th century. The result is easily obtained using the Fourier Transform of the complex exponential. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2\u02c7 (! ! 0) !2[ \u02c7;\u02c7) the spectrum is zero for !6= ! 0 cos(! 0n. fourier_transform(cos(x),x,v) the output is 0 where it should be based on the Dirac delta function. Moreover, as cosine and sine transform are real operations (while Fourier transform is complex), they can be more efficiently implemented and are widely used in various applications. Fourier sine and cosine transforms are used to solve initial boundary value problems associated with second order partial di\ufb00erential equations on the semi-in\ufb01nite inter-val x>0. THE FOURIER TRANSFORM APPROACH TO INVERSION OF \u03bb-COSINE AND FUNK TRANSFORMS ON THE UNIT SPHERE B. For specific cases either a cosine or a sine transform may b;. Notice the the Fourier Transform and its inverse look a lot alike\u2014in fact, they're the same except for the complex. By default, the transform is in terms of w. Fourier coefficients for sine terms. fourier (f,var,transVar) uses the independent variable var and the transformation variable transVar instead of symvar and w, respectively. Sketch by hand the magnitude of the Fourier transform of c(t) for a general value of f c. Fourier transform cosine example further s blogs mathworks images steve 2009 f cos t in additions akshaysin github io images mpl basic fft moreovers upload wikimedia org wikipedia mons 6 61 fft time frequency view in additionee nmt edu wedeward ee342 sp99 ex le16 gif. 29) and the Fourier transform of special distributions in (1. The function F(k) is the Fourier transform of f(x). \" The full name of the function is \"sine cardinal,\" but it is commonly referred to by its abbreviation, \"sinc. However, bell-shaped functions are not convex, and it is doubtful if. New York: McGraw-Hill, pp. Matrices Vectors. Fourier analysis is a method for expressing a function as a sum of periodic components, and for recovering the signal from those components. System Analysis using Fourier Series & Transform (C. The FT assumes that the finite analyzed segment corresponds to one period of an infinitely extended periodic signal. For math, science, nutrition, history. Fourier Series Grapher. I am familiar with Mathematica \u00ae. RUBIN Abstract. \"Fourier Transform--Cosine. First term in a Fourier series. This process can be continued for each k until the complete DFT is obtained. cos(2\u02c7f ct); which is a sinusoidal pulse with frequency f c, amplitude Aand duration T(sketch g(t) again). The input time series can now be expressed either as a time-sequence of values, or as a. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. See also: Annotations for \u00a71. Hence evaluate Z 1 0 sin cos x d and deduce the value of Z 1 0 sin d : Solution Since f(x) = 1 \u02c7 Z 1 0 Z 1 1 f(t)cos (t x)dtd = 1 \u02c7 Z 1 0 Z 1 1 cos (t x)dtd MATH204-Di erential Equations Center of Excellence. 5-Fourier Transform || Engineering Mathematics-3 || Important Fourier Sine and Fourier Cosine TR. The Fourier transform of a diffraction grating. Lets start with what is fourier transform really is. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. Fourier Series & The Fourier Transform What is the Fourier Transform? Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems F( ) ( ) exp( )\u03c9\u03c9ft i t dt \u221e \u2212\u221e =\u2212\u222b 1 ( )exp( ) 2 ft F i td\u03c9 \u03c9\u03c9 \u03c0 \u221e \u2212\u221e = \u222b. Conic Sections. We use the classical Fourier analysis on Rn to intro-duce analytic families of weighted di\ufb00erential operators on the unit sphere. $sin(a+b) = sin(a)cos(b) + cos(a)sin(b)$ which allow us to replace phase shifts with cosines. fourier (f,var,transVar) uses the independent variable var and the transformation variable transVar instead of symvar and w, respectively. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to another copy of it-self). Instead of capital letters, we often use the notation f^(k) for the Fourier transform, and F (x) for the inverse transform. then the Fourier cosine transform would have been used instead. To illustrate determining the Fourier Coefficients, let's look at a simple example. Find the Fourier cosine series and the Fourier sine series for the function f(x) = \u02c6 1 if 0 data); sine and cosine transform routines; quarter wave sine and cosine transform routines; the amount of data is NOT required to be a power of 2. In the two-dimensional case, the formula for a normalized version of the discrete cosine transform (forward cosine transform DCT-II) may be written and the inverse cosine transform is Note that the discrete cosine transform computation can be based on the Fourier transform - all N coefficients of the discrete cosine transform may be computed. Due to the duality property of the Fourier transform, if the time signal is a sinc function then, based on the previous result, its Fourier transform is This is an ideal low-pass filter which suppresses any frequency f > a to zero while keeping all frequency lower than a unchanged. The expression in (7), called the Fourier Integral, is the analogy for a non-periodic f (t) to the Fourier series for a periodic f (t). So, you can think of the k-th output of the DFT as the. Transforms for real odd functions are imaginary, i. Integral of product of sines. Fourier transform is purely imaginary. Such a decomposition of periodic signals is called a Fourier series. We've already worked out the Fourier transform of diffraction grating on the previous page. Line Equations Functions Arithmetic & Comp. Matrices & Vectors. Fourier Cosine Transforms - EqWorld Author: A. Fourier coefficients for cosine terms. This page will describe how to determine the frequency domain representation of the. 1 Practical use of the Fourier. For example, the DTFT of the rectangular pulse. The Fourier transform has many useful properties that make calculations easier and also help thinking about the structure of signals and the action of systems on signals. The Fourier transform of the Gaussian function is given by: G(\u03c9) = e. SciPy provides a DCT with the function dct and a corresponding IDCT with the function idct. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. A discrete cosine transform (DCT) expresses a finite sequence of data points in terms of a sum of cosine functions oscillating at different frequencies. The cosine function, f(t), is shown in Figure 1: Figure 1. We wish to Fourier transform the Gaussian wave packet in (momentum) k-space to get in position space. A Fourier transform is a linear transformation that decomposes a function into the inputs from its constituent frequencies, or, informally, gives the amount of each frequency that composes a signal. Square waves View Answer / Hide Answer. Note that f(t) has a corner and its coe cients decay like 1=n2, while f0(t) has a jump and and its coe cients decay like 1=n. 9 Discrete Cosine Transform (DCT) When the input data contains only real numbers from an even function, the sin component of the DFT is 0, and the DFT becomes a Discrete Cosine Transform (DCT) There are 8 variants however, of which 4 are common. De nition Changing to complex variables via Euler\u2019s form exp(i ) = cos + isin (1) leads to the discrete Fourier transform. For instance the functions sin(x);cos(x) are periodic of period 2\u02c7. 2 p693 PYKC 8-Feb-11 E2. Di erentiation and multiplication exchange r^oles under the Fourier transform and therefore so do the properties of smoothness and rapid decrease. 2 Fourier Transform 2. Find the Fourier transform of the following signals: a) f1(t)=e \u22123 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. The Fourier transform can be viewed as an extension of the above Fourier series to non-periodic functions. The Fourier Series allows us to express periodic functions as discrete sums of sine waves, while the Fourier Transform allows us to express any function a continuous integral of sine waves. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] = 1 2\u02c7 Z 2\u02c7 X()ej td: x[n] X() condition anu[n] 1 1 ae j jaj<1 (n+ 1)anu[n] 1 (1 ae j)2 jaj<1 (n+ r 1)! n!(r 1)! anu[n] 1 (1 ae j)r jaj<1 [n] 1 [n n 0] e j n 0 x[n] = 1 2\u02c7 X1 k=1 (2\u02c7k) u[n. Fourier Series and Fourier Transform are two of the tools in which we decompose the signal into harmonically related sinusoids. Compute the Fourier transform of a rectangular pulse-train. Fourier transforms are important because many signals make more sense when their frequencies are separated. To illustrate determining the Fourier Coefficients, let's look at a simple example. Fourier Transform Problems 1. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. On this page, the Fourier Transforms for the sinusois sine and cosine function are determined. Each \"spike\" on the second plot is the magnitude of the sine or cosine at that frequency. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. If we shift a signal in time by t 0, the spectrum of the signal is also altered. X-ray crystallography has been essential, since the beginning of the 20th century, to our understanding of matter; recently, as knowledge of the chemical composition of proteins has progressed, the determination of their 3-dimensional structure has become indispensable for the correct interpretation. It is also periodic of period 2n\u02c7, for any positive integer n. Fourier Series for Recti\ufb01ed Sine Wave Consider the signal x(t) = Ajsin(!1 t)j \u22122 T \u2212T 0 T 2 T \u2212A 0 A |sin (\u03c9 1 t)| Rectified Sine and Sine \u2212T1 0 T1 \u2212A 0 A sin (\u03c9 1 t) The period of the sinusoid (inside the absolute value symbols) is T1 = 2\u02c7=!1. Square waves View Answer / Hide Answer. (14) and replacing X n by. The coe cients in this linear combi-. tgz (71KB) updated. This is the currently selected item. , they have a phase shift of +\u03c0/2. It uses a complex representation of the signal. Fourier-style transforms imply the function is periodic and extends to. Rectangular pulse. Fourier coefficients for cosine terms. 5 1-10 -5 5 10 0. Identities Proving Identities Trig Equations Trig. I hope you remember sines and. Fourier Transform Coefficients Of Real Valued Audio Signals 2018-02-10 - By Robert Elder. 3 Reference 1) 1. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2\u02c7 (! ! 0) !2[ \u02c7;\u02c7) the spectrum is zero for !6= ! 0 cos(! 0n. A periodic function, de\ufb01ned by a period T, v(t + T) = v(t) Familiar periodic functions: square, triangle, sawtooth, and sinusoids (of course). Animated Walkthrough of the Discrete Fourier Transform The Input Signal corresponds to the x[n] term in the equation. The DCT, first proposed by Nasir Ahmed in 1972, is a widely used transformation technique in signal processing and data compression. Bracewell (which is on the shelves of most radio astronomers) and the Wikipedia and Mathworld entries for the Fourier transform. Identities Proving Identities Trig Equations Trig. The applet below shows how the Fourier transform of the damped exponent, sinusoid and related functions. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. Fourier Transform of Common Inputs. The Cosine Function. 1 Practical use of the Fourier. The complex exponential and the scaled and shifted impulse form a Fourier Transform pair. 8 1 The Fourier Transform: Examples, Properties, Common Pairs Odd and Even Functions Even Odd f( t) = f(t) f( t) = f(t) Symmetric Anti-symmetric Cosines Sines Transform is real. In this way, you have a mental image of what the answer should look like. Fourier series is an expansion of periodic signal as a linear combination of sines and cosines while Fourier transform is the process or function used to convert signals from time domain in to frequency domain. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. Compute the Fourier transform of common inputs. We have and we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. New York: McGraw-Hill, pp. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. The Fourier transform of a diffraction grating. ) are related to each other by a function very similar to the Fourier transform. Let\u2019s overample: f s = 100 Hz. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. In mathematics, a Fourier series is a way to represent a (wave-like) function as the sum of simple sine waves. A Fourier transform is a linear transformation that decomposes a function into the inputs from its constituent frequencies, or, informally, gives the amount of each frequency that composes a signal. 082 Spring 2007 Fourier Series and Fourier Transform, Slide 3 The Concept of Negative Frequency Note: \u2022 As t increases, vector rotates clockwise - We consider e-jwtto have negativefrequency \u2022 Note: A-jBis the complex conjugateof A+jB - So, e-jwt is the complex conjugate of ejwt e-j\u03c9t I Q cos(\u03c9t)-sin(\u03c9t)\u2212\u03c9t. The Fourier transform is simply a method of expressing a function (which is a point in some infinite dimensional vector space of functions) in terms of the sum of its projections onto a set of basis functions. I'm placing the Fourier analysis material in the back so it won't scare people away. The input time series can now be expressed either as a time-sequence of values, or as a. The DFT has revolutionized modern society, as it is ubiquitous in digital electronics and signal processing. The Fourier transform of the Gaussian function is given by: G(\u03c9) = e. A unitary linear operator which resolves a function on $\\mathbb{R}^N$ into a linear superposition of \"plane wave functions\". If we follow through exactly the same method as above (we can in effect split the function into cos(at) and cos(bt) and do both separately), we should get:. Fourier Series Grapher. Find the Fourier transform of the following signals: a) f1(t)=e \u22123 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. Fourier cosine transform Cn = 2 L Z L 0 f(t)Cos. The Fourier transform of a Gaussian is a Gaussian and the inverse Fourier transform of a Gaussian is a Gaussian f(x) = e \u2212\u03b2x2 \u21d4 F(\u03c9) = 1 \u221a 4\u03c0\u03b2 e \u03c9 2 4\u03b2 (30) 4. If you're seeing this message, it means we're having trouble loading external resources on our website. We will look at and prove a few of them. But how will the representation of a wave or signal say based on these trigonometric functions (w. Equation (10) is, of course, another form of (7). In the graphics the initial signal is converted forward and back by the selected discrete Fourier transforms. I am also aware of the Pontryagin Duality generalization for locally compact abelian groups, though I am personally more concerned. The result is easily obtained using the Fourier Transform of the complex exponential. This process can be continued for each k until the complete DFT is obtained. Integral of product of cosines. It can be determined by applying the Fourier transformation explicitly to samples of one simple cosine + A cos[ 2\u03c0 f t - \u03c6]: A look at every frequency s in the spectrum reveals. The Fourier transform can be viewed as an extension of the above Fourier series to non-periodic functions. It is also periodic of period 2n\u02c7, for any positive integer n. 2 (a) X(w) = fx(t)e -j4t dt = (t - 5)e -j' dt = e ~j = cos 5w -j sin 5w, by the sifting property of the unit impulse. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Fourier sine and cosine transforms are used to solve initial boundary value problems associated with second order partial di\ufb00erential equations on the semi-in\ufb01nite inter-val x>0. However, while simple, it is also quite slow. 3 Reference 1) 1. Most often used in physics for calculating the response of a time shift invariant linear system as the sum of its response to time harmonic excitation or for transforming a quantum state in position co-ordinates into one in momentum co-ordinates and contrawise. For specific cases either a cosine or a sine transform may b;. Discrete Fourier transform Time origin When dealing with the Fourier transform, it is common to index time from t= 0 to t= n 1. The Fourier transform is intimately associated with microscopy, since the alternating planes occurring in the microscope (focal plane - back-focal plane, etc. Rectangular pulse. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] = 1 2\u02c7 Z 2\u02c7 X()ej td: x[n] X() condition anu[n] 1 1 ae j jaj<1 (n+ 1)anu[n] 1 (1 ae j)2 jaj<1 (n+ r 1)! n!(r 1)! anu[n] 1 (1 ae j)r jaj<1 [n] 1 [n n 0] e j n 0 x[n] = 1 2\u02c7 X1 k=1 (2\u02c7k) u[n. Our final discrete Fourier transform looks like this (real part on the left, imaginary part on the right):. Now, I assume they want the FSR to be made up of only cosine terms, there is another question on another past exam that asks for the same thing but in sine terms. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. Most often used in physics for calculating the response of a time shift invariant linear system as the sum of its response to time harmonic excitation or for transforming a quantum state in position co-ordinates into one in momentum co-ordinates and contrawise. (b) Use the result of part (a) to find the value of the integral o0 cos kx dk 0 1 +k2 (c) Show explicitly that Parseval's theorem is satisfied for eand its Fourier transform 1. The Fourier transform of a function is complex, with the magnitude representing the amount of a given frequency and the argument representing the phase shift from a sine wave of that frequency. The expression you have is an person-friendly remodel, so which you will detect the inverse in a table of Laplace transforms and their inverses. Help with a trignometric double integral. Solution : Problem 20 ) Obtain Fourier Sine Transform of i) for 0 a\\end{cases}\\]. Instead, Fourier Cosine transform should be used. Or, we can use only cosines with phase shifts: x(t) = a0+c1 cos(\u03c90t\u2212\u03c61)+c2 cos(2\u03c90t\u2212\u03c62)+c3 cos(3\u03c90t\u2212\u03c63)+ PROOF: Take a high-level math course to see this done properly. Discrete Fourier Transform and Discrete Cosine Transform When dealing with image analysis, it would be very useful if you could change an image from the spatial domain, which is the image in terms of its x and y coordinates, to the frequency domain\u2014the image decomposed in its high and low frequency components\u2014so that you would be able to. The sine and cosine transforms are useful when the given function x(t) is known to be either even or odd. Derpanis October 20, 2005 In this note we consider the Fourier transform1 of the Gaussian. Key concept: Inverse Fourier Transform of Impulse in Frequency Domain. We have: cos(-x) = cos(x), and it follows that int_(-pi)^picos theta\\ d theta=0 We can see the integral must be 0 if we consider the curve of y = cos(x) for x=-pi to x=pi. Finding Fourier Sine. The continuous Fourier transform is important in mathematics, engineering, and the physical sciences. Consider the function f(x)-e- (a) Find its Fourier transform. The discrete Fourier transform is defined as follows: \ud835\udc4b = \u2211\ud835\udc65\ud835\udc5b \u22122 \ud835\udf0b \ud835\udc5b \ud835\udc41 \ud835\udc41\u22121 \ud835\udc5b=0 \ud835\udc3e=0,1,\u2026, \u22121 In this equation, K represents a frequency for which. I'm placing the Fourier analysis material in the back so it won't scare people away. Introduction; Derivation; Examples; Aperiodicity; Printable; The previous page showed that a time domain signal can be represented as a sum of sinusoidal signals (i. For math, science, nutrition, history. \u2022 The Fourier transform (FT) is a generalization of the Fourier series. On this page, I want to think about it in an alternative way, so that when we come to think of three-dimensional scattering and crystallography, we will have intuitive way of constructing the reciprocal lattice. Actually, in the mathematics sine and cosine functions are defined based on right angled triangles. 1-dim DFT / DCT / DST Description. In the study of Fourier series, complicated but periodic functions are written as the sum of simple waves mathematically represented by sines and cosines. The most common image transform takes spatial data and transforms it into frequency data. SEE ALSO: Fourier Transform, Fourier Transform--Cosine, Sine. Integral of sin (mt) and cos (mt) Integral of sine times cosine. This equation defines \u2131 \u2061 (f) \u2061 (x) or \u2131 \u2061 f \u2061 (x) as the Fourier transform of functions of a single variable. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to another copy of it-self). By proper choice of a k we can get the real fourier series out of this. (14) and replacing X n by. Fourier-style transforms imply the function is periodic and extends to. Integral of product of cosines. The discrete Fourier cosine transform of a list of real numbers can be computed in the Wolfram Language using FourierDCT[l]. This package contains C and Fortran FFT codes. Integral of sin (mt) and cos (mt) Integral of sine times cosine. Find the Fourier cosine series and the Fourier sine series for the function f(x) = \u02c6 1 if 0 0 for all x>0 guarantees v(t) > 0 for all t>0. In the two-dimensional case, the formula for a normalized version of the discrete cosine transform (forward cosine transform DCT-II) may be written and the inverse cosine transform is Note that the discrete cosine transform computation can be based on the Fourier transform - all N coefficients of the discrete cosine transform may be computed. Fourier Series of Even and Odd Functions The Fourier series of an even function f(t) of period T is a \u201c Fourier cosine series\u201d / 2 0 / 2 0 0 1 0) 2 ( )cos(4 ( ) 2) , where 2. Compute the discrete-time Fourier transform of the following signal: $x[n]= \\cos \\left( \\frac{2 \\pi }{500} n \\right)$ (Write enough intermediate steps to fully justify your answer. New boron material of high hardness created by plasma chemical vapor deposition; Earth-size, habitable-zone planet found hidden in early NASA Kepler data. So, there may be in nitely many periods. This page will describe how to determine the frequency domain representation of the. cos 5 1 2 3 cos 3 1 2 [cos 2 2 ( ) 2 5 cos 5 2 2 3 cos 3 2 2 cos 2 2 ( ) S S k k and f x k k x S x x x k k f x x k x k x k k f x (Homework sec. Fourier Transform of the Gaussian Konstantinos G. For specific cases either a cosine or a sine transform may b;. Plane Geometry Solid Geometry Conic Sections. The applet below shows how the Fourier transform of the damped exponent, sinusoid and related functions. We use the classical Fourier analysis on Rn to intro-duce analytic families of weighted di\ufb00erential operators on the unit sphere. IX(w)| = |ej5wI = 1 for all w, Fourier coefficients) and backward analysis (Fourier coefficients => data); sine and cosine transform routines; quarter wave sine and cosine transform routines; the amount of data is NOT required to be a power of 2. The Fourier transform of $f(x)$ is denoted by $\\mathscr{F}\\{f(x)\\}=$$F(k), k \\in \\mathbb{R},$ and defined by the integral :. Solve using Fourier Sine or Cosine Transform $\\frac{1}{a^2+x^2}=\\int_0^\\infty f(t)cos(2tx)dt$ 2. Its inverse Fourier transform is called the \"sampling function\" or \"filtering function. 5 u axis v axis Frequency Response \u221210 \u22125 0 5 \u221210 \u22128 \u22126 \u22124 \u22122 0 2 4 6 8 \u2022 Graphical representation of space-frequency domain 1/Vo 1/Uo Po Plane Wave in Space Domain Impulse in Frequency Domain \u03c6 \u03c6. This new transform has some key similarities and differences with the Laplace transform, its properties, and domains. Fourier Transform--Sine (1) (2) (3) where is the delta function. By default, the transform is in terms of w. org are unblocked. ES 442 Fourier Transform 3 Review: Fourier Trignometric Series (for Periodic Waveforms) Agbo & Sadiku; Section 2. It also provides the final resulting code in multiple programming languages. A periodic function, de\ufb01ned by a period T, v(t + T) = v(t) Familiar periodic functions: square, triangle, sawtooth, and sinusoids (of course). Gowthami Swarna, Tutorials Poin. Finding Fourier Sine. Square waves View Answer / Hide Answer. Recall that eii\u03b8=+cos sin\u03b8 \u03b8; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. 2 Fourier Transform 2. This is done using the Fourier transform. Many more transform pairs could be shown. Fourier Series and Fourier Transform are two of the tools in which we decompose the signal into harmonically related sinusoids. We shall now use complex exponentials because they lead to less writing and simpler computations, but yet can easily be. Thus we can represent the repeated parabola as a Fourier cosine series f(x) = x2 = \u03c02 3 +4 X\u221e n=1 (\u22121)n n2 cosnx. Moreover, as cosine and sine transform are real operations (while Fourier transform is complex), they can be more efficiently implemented and are widely used in various applications. (9) by exp(\u00a12\u2026ipx=L) before integrating. Therefore its Fourier-sine transform is positive, and hence so is the Fourier-cosine transform of u(x). Fourier Transform of Common Inputs. Fourier transforms table lecture 10 fourier transform thefouriertransform com fourier transform of the sine and cosine fourier transform pairs basic electrical engineering homework Whats people lookup in this blog:. The Fourier transform of a function f(x) is the function f(x) If f(x) is even, then g(u) takes the form (the cosine transform), and if f(x) is odd, then g(u) takes the form (the sine transform). Finding Fourier Sine. then the Fourier cosine transform would have been used instead. Line Equations Functions Arithmetic & Comp. Some excellent answers on the $\\sin x$ and $\\cos x$ functions and how they're solutions to the relevant differential equations were already given, but an important point can still be mentioned: Sine and cosine are used because they are periodic and signals/waves are usually considered to be or are approximated by periodic functions. The third plot shows the inverse discrete Fourier transform, which converts the sines and cosines back into the original function f(x). As a result, the Fourier transform is an automorphism of the Schwartz space. Related Calculus and Beyond Homework Help News on Phys. 1 De nition The Fourier transform allows us to deal with non-periodic functions. Bracewell (which is on the shelves of most radio astronomers) and the Wikipedia and Mathworld entries for the Fourier transform. syms a b t f = rectangularPulse (a,b,t); f_FT = fourier (f). This equation defines \u2131 \u2061 (f) \u2061 (x) or \u2131 \u2061 f \u2061 (x) as the Fourier transform of functions of a single variable. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 5 1-10 -5 5 10 0. Solution: The de nition of the Fourier transform together with the change of variable ax7! x0 implies F[eibxf(ax)])(\u02d8) = 1 2\u02c7 Z +1 1 f(ax)eibxei\u02d8xdx. Compute the Fourier transform of a rectangular pulse-train. Example: Fourier Transform of a Cosine Spatial Domain Frequency Domain cos (2 st ) 1 2 (u s)+ 1 2 (u + s) 0. The Finite Fourier Transforms When solving a PDE on a nite interval 0 0 if f(x+ T) = f(x) for all x2R. signal flow graph inverse discrete,. Fourier Transform of Common Inputs. Maxim Raginsky Lecture IX: Fourier transform. In the two-dimensional case, the formula for a normalized version of the discrete cosine transform (forward cosine transform DCT-II) may be written and the inverse cosine transform is Note that the discrete cosine transform computation can be based on the Fourier transform - all N coefficients of the discrete cosine transform may be computed. Fourier cosine transform Cn = 2 L Z L 0 f(t)Cos. Help with a trignometric double integral. The sine and cosine transforms are useful when the given function x(t) is known to be either even or odd. Fourier transform is purely imaginary. In plain words, the discrete Fourier Transform in Excel decomposes the input time series into a set of cosine functions. So, you can think of the k-th output of the DFT as the. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. So in Fourier series representation, the spectrum of a cosine is just one impulse at the frequency of the cosine. The expression above is not too different from $\\mathcal F\\{{\\cos(2\\pi Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. IX(w)| = |ej5wI = 1 for all w, Fourier coefficients) and backward analysis (Fourier coefficients => data); sine and cosine transform routines; quarter wave sine and cosine transform routines; the amount of data is NOT required to be a power of 2. (10) 320 Chapter 4 Fourier Series and Integrals Every cosine has period 2\u03c0. The applet below shows how the Fourier transform of the damped exponent, sinusoid and related functions. (Note that there are other conventions used to de\ufb01ne the Fourier transform). The Fourier Transform (FT) is a generalization to solve for non-periodic waves. Plane Geometry Solid Geometry Conic Sections. 1) is called the inverse Fourier integral for f. Its inverse Fourier transform is called the \"sampling function\" or \"filtering function. cos 5 1 2 3 cos 3 1 2 [cos 2 2 ( ) 2 5 cos 5 2 2 3 cos 3 2 2 cos 2 2 ( ) S S k k and f x k k x S x x x k k f x x k x k x k k f x (Homework sec. Sine and Cosine transforms for \ufb01nite range Fourier sine transform Sn = 2 L Z L 0 f(t)Sin(n\u03c0 L x)dx, f(x) = X\u221e n=1 Sn Sin( n\u03c0 L x). Many more transform pairs could be shown. The Fourier series represents a periodic waveform of a given frequency as a sum of sine and cosine functions that are multiples of the fundamental frequency: Where f(x) is the function in question a 0 is the dc component a n is the level of each cosine wave. Notice the the Fourier Transform and its inverse look a lot alike\u2014in fact, they're the same except for the complex. Definition. The discrete Fourier transform (DFT) of the real-valued nsequence y 0;:::;y n 1 is de ned as. Each \"spike\" on the second plot is the magnitude of the sine or cosine at that frequency. If you're behind a web filter, please make sure that the domains *. In plain words, the discrete Fourier Transform in Excel decomposes the input time series into a set of cosine functions. (((1 cos(t))=2)) and sin2 t= (1 cos(2t))=2. Sine and cosine waves can make other functions! Here you can add up functions and see the resulting graph. The Fourier cosine transform of a function is by default defined to be. ES 442 Fourier Transform 3 Review: Fourier Trignometric Series (for Periodic Waveforms) Agbo & Sadiku; Section 2. Fourier transform methods are often used for problems in which the variable t represents time, and the inverse transform formula, Eq. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2\u02c7 (! ! 0) !2[ \u02c7;\u02c7) the spectrum is zero for !6= ! 0 cos(! 0n. htm Lecture By: Ms. Fourier transform of f, and f is the inverse Fourier transform of f\u02c6. It uses a complex representation of the signal. The Fourier Transform (FT) is a generalization to solve for non-periodic waves. DCT vs DFT For compression, we work with sampled data in a finite time window. In particular, the jpeg image compression standard uses the two-dimensional discrete cosine transform, which is a Fourier transform using the cosine basis functions. Help with a trignometric double integral. The Fourier Series allows us to express periodic functions as discrete sums of sine waves, while the Fourier Transform allows us to express any function a continuous integral of sine waves. EE 230 Fourier series \u2013 1 Fourier series A Fourier series can be used to express any periodic function in terms of a series of cosines and sines. (10) 320 Chapter 4 Fourier Series and Integrals Every cosine has period 2\u03c0. you will need for this Fourier Series chapter. The Fourier transform has many useful properties that make calculations easier and also help thinking about the structure of signals and the action of systems on signals. Find the Fourier transform of the following signals: a) f(t)=cos(at\u2212\u03c0/3) b) g(t)=u(t+1)sin(\u03c0t) c) h(t)=(1+Asin(at))cos(bt) d)i(t)=1\u2212t, 0 a, b option. This includes using the symbol I for the square root of minus one. It consists the basic concept of Fourier Transform, Inverse Fourier Transform, Fourier Sine and Cosine Transform with important tools like Gamma function, Even function,Odd function. Fourier transform cosine example further s blogs mathworks images steve 2009 f cos t in additions akshaysin github io images mpl basic fft moreovers upload wikimedia org wikipedia mons 6 61 fft time frequency view in additionee nmt edu wedeward ee342 sp99 ex le16 gif. The Fourier Transform It is well known that one of the basic assumptions when applying Fourier methods is that the oscillatory signal can be be decomposed into a bunch of sinusoidal signals. Fourier Transform of Common Inputs. Such a decomposition of periodic signals is called a Fourier series. ) Fourier integral theorem (without proof) - Fourier transform pair - Sine and Cosine transforms - Properties - Transforms of simple functions - Convolution theorem - Parseval\u2019s identity. 5-Fourier Transform || Engineering Mathematics-3 || Important Fourier Sine and Fourier Cosine TR. Find the Fourier transform of the following signals: a) f1(t)=e \u22123 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. discrete signals (review) \u2013 2D \u2022 Filter Design \u2022 Computer Implementation Yao Wang, NYU-Poly EL5123: Fourier Transform 2. For a complex function f(x) which satisfies the condition that. First Fourier transform of sin function should be calculated,and to calculate this these properties will be needed first one is Duality, for any signal/function $\\large x(t)$ if it's Fourier Transform is $\\large X(w)$ then a. \" The full name of the function is \"sine cardinal,\" but it is commonly referred to by its abbreviation, \"sinc. This page will describe how to determine the frequency domain representation of the. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. These ideas are also one of the conceptual pillars within electrical engineering. Matrices Vectors. The Fourier Transform (FT) is a generalization to solve for non-periodic waves. Recall that eii\u03b8=+cos sin\u03b8 \u03b8; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. is a decreasing function. Fourier Series Jean Baptiste Joseph Fourier (1768-1830) was a French mathematician, physi-cist and engineer, and the founder of Fourier analysis. 3 shows two even functions, the repeating ramp RR(x)andtheup-down train UD(x) of delta functions. you will need for this Fourier Series chapter. On this page, the Fourier Transforms for the sinusois sine and cosine function are determined. Fourier Series & The Fourier Transform What is the Fourier Transform? Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems F( ) ( ) exp( )\u03c9\u03c9ft i t dt \u221e \u2212\u221e =\u2212\u222b 1 ( )exp( ) 2 ft F i td\u03c9 \u03c9\u03c9 \u03c0 \u221e \u2212\u221e = \u222b. Fourier Transform--Sine (1) (2) (3) where is the delta function. Triangular waves d. Integral of sin (mt) and cos (mt) Integral of sine times cosine. We have and we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. 1D, a FORTRAN90 library which implements the Fast Fourier Transform using double precision arithmetic, by Paul Swarztrauber and Dick Valent; FFTW3 , FORTRAN90 programs which illustrate the use of the FFTW3 library for Fast Fourier Transforms, by Matteo Frigo and Steven Johnson. Matrices & Vectors. 5-Fourier Transform || Engineering Mathematics-3 || Important Fourier Sine and Fourier Cosine TR. A Fourier transform is nothing more than the curve-fit of a collection of sine and cosine functions to some data. Due to the duality property of the Fourier transform, if the time signal is a sinc function then, based on the previous result, its Fourier transform is This is an ideal low-pass filter which suppresses any frequency f > a to zero while keeping all frequency lower than a unchanged. We can combine sin and cos into one complex number. PROPERTIES OF FOURIER TRANSFORMS 1. In mathematics, a Fourier series is a way to represent a (wave-like) function as the sum of simple sine waves. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equ. 29) and the Fourier transform of special distributions in (1. Members who need to use special functions and characters still need to learn the correct Mathematica \u00ae input format from the HELP page. Plane Geometry Solid Geometry Conic Sections. This article is effectively an appendix to the article The Fast Meme Transform: Convert Audio Into Linux Commands. Help with a trignometric double integral. To refresh your memory, here are the ideal cosine signal and its continuous-time Fourier transform plots again: The dots at the left and right of the cosine plot are meant to remind you that the cosine signal is defined for all t. 2 p693 PYKC 8-Feb-11 E2. New York: McGraw-Hill, pp. Draw a square wave of amplitude 1 and period 1 second whose trigonometric Fourier Series Representation consists of only cosine terms and has no DC component. Finding the coefficients, F m, in a Fourier Cosine Series Fourier Cosine Series: To find F m, multiply each side by cos(m\u2019t), where m\u2019 is another integer, and integrate:. The properties of these continuous transforms are well known and bear great resemblance to those of DCT and DST. Consider the function f(x)-e- (a) Find its Fourier transform. The basic idea is that any signal can be represented as a weighted sum of sine and cosine waves of different frequencies. The discrete Fourier cosine transform of a list of real numbers can be computed in the Wolfram Language using FourierDCT[l]. A discrete cosine transform (DCT) expresses a finite sequence of data points in terms of a sum of cosine functions oscillating at different frequencies. ) are related to each other by a function very similar to the Fourier transform. Auxiliary Sections > Integral Transforms > Tables of Fourier Cosine Transforms > Fourier Cosine Transforms: Expressions with Exponential Functions Fourier Cosine Transforms: Expressions with Exponential Functions No Original function, f(x) Cosine transform, f\u02c7c(u) = Z 1 0 f(x)cos(ux)dx 1 e\u2212ax a a2+u2 2 1 x \u00a1 e\u2212ax \u2212e\u2212bx \u00a2 1 2 ln b2+u2. 1 De nition The Fourier transform allows us to deal with non-periodic functions. Solve using Fourier Sine or Cosine Transform$\\frac{1}{a^2+x^2}=\\int_0^\\infty f(t)cos(2tx)dt$2. 43d for the Fourier sine transform utilizes the value of the function at x = 0, the sine transform is applied to problems with a Dirichlet. Discrete Cosine Transform \u2022a much better transform, from this point of view, is the DCT \u2013 in this example we see the amplitude spectra of the image above \u2013 under the DFT and DCT \u2013 note the much more concentrated histogram obtained with the DCT \u2022 why is energy compaction important? \u2013 the main reason isthe main reason is image compression. The Fourier transform is a different story. These operators are polynomial functions of the usual Beltrami-Laplace operator. Long story short, what Fourier Transform is doing is it tries to approximate the signal (wave) of your interest using different kinds of sine and cosine waves. Shifting, Scaling Convolution property Multiplication property Table of Fourier Transforms x(t) = cos( \u03c9ct) \u21d4 X(j\u03c9) =\u03c0\u03b4 Square Wave Fourier Transform X. And to recombine the weighted harmonics: f(t)= Z1 \u00a11 F(s)ei2\u2026st ds This is the Inverse Fourier Transform, denoted F\u00a11. This result will be used below to find the Fourier Transform of Sines, Cosines, and any periodic function that can be represented by a Fourier Series. We will also define the even extension for a function and work several examples finding the Fourier Cosine Series for a function. So, you can think of the k-th output of the DFT as the. Related Calculus and Beyond Homework Help News on Phys. DCT vs DFT For compression, we work with sampled data in a finite time window. Some excellent answers on the$\\sin x$and$\\cos x\\$ functions and how they're solutions to the relevant differential equations were already given, but an important point can still be mentioned: Sine and cosine are used because they are periodic and signals/waves are usually considered to be or are approximated by periodic functions. The Fourier transform is important in mathematics, engineering, and the physical sciences. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series Inverse Fourier transform: The Fourier integral theorem Example: the rect and sinc functions Cosine and Sine Transforms Symmetry properties Periodic signals and functions Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 22. The discrete Fourier sine and cosine transforms (DST and DCT) can be used to decompose or represent a given digital signal (that is discrete) in the form of a set of sums of sines and cosines. Fourier Cosine Transform(F. Recall that eii\u03b8=+cos sin\u03b8 \u03b8; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. and in general. Integral of product of sines. Or, we can use only cosines with phase shifts: x(t) = a0+c1 cos(\u03c90t\u2212\u03c61)+c2 cos(2\u03c90t\u2212\u03c62)+c3 cos(3\u03c90t\u2212\u03c63)+ PROOF: Take a high-level math course to see this done properly. The time functions on the left are Fourier transforms of the frequency functions on the right and vice-versa. In the graphics the initial signal is converted forward and back by the selected discrete Fourier transforms. Fourier Transform of Cosine Wave Watch more videos at https://www. Integral of sin (mt) and cos (mt) Integral of sine times cosine. Because of the periodicity of it is very common when plotting the DTFT to plot it over the range of just one period:. Our final discrete Fourier transform looks like this (real part on the left, imaginary part on the right):. If we follow through exactly the same method as above (we can in effect split the function into cos(at) and cos(bt) and do both separately), we should get:. Let k(s;x) be a given function of two variables sand x. 8 1 The Fourier Transform: Examples, Properties, Common Pairs Odd and Even Functions Even Odd f( t) = f(t) f( t) = f(t) Symmetric Anti-symmetric Cosines Sines Transform is real. When calculating the Fourier transform Mathematica does not need to know the meaning of your input. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] = 1 2\u02c7 Z 2\u02c7 X()ej td: x[n] X() condition anu[n] 1 1 ae j jaj<1 (n+ 1)anu[n] 1 (1 ae j)2 jaj<1 (n+ r 1)! n!(r 1)! anu[n] 1 (1 ae j)r jaj<1 [n] 1 [n n 0] e j n 0 x[n] = 1 2\u02c7 X1 k=1 (2\u02c7k) u[n. Consider cos bx, which by Euler's Identity may be written as cos bX = 2(eibx + e-ibx) This shows the function written as a linear combination of just two of the functions. Sine and cosine waves can make other functions! Here you can add up functions and see the resulting graph. This gives us a basis for complex functions (which we don't care about right now) and somewhat simpler notation. The basic underlying idea is that a function f(x) can be expressed as a linear combination of elementary functions (speci cally, sinusoidal waves). Conic Sections. The discrete Fourier sine and cosine transforms (DST and DCT) can be used to decompose or represent a given digital signal (that is discrete) in the form of a set of sums of sines and cosines. First Fourier transform of sin function should be calculated,and to calculate this these properties will be needed first one is Duality, for any signal/function $\\large x(t)$ if it's Fourier Transform is $\\large X(w)$ then a. Derivation of Fourier Series. Discrete Fourier Transform and Discrete Cosine Transform When dealing with image analysis, it would be very useful if you could change an image from the spatial domain, which is the image in terms of its x and y coordinates, to the frequency domain\u2014the image decomposed in its high and low frequency components\u2014so that you would be able to. 3 Reference 1) 1. For a complex function f(x) which satisfies the condition that. The Fourier transform is simply a method of expressing a function (which is a point in some infinite dimensional vector space of functions) in terms of the sum of its projections onto a set of basis functions.\nqxgwnydnsvuve, 94quuajwgi, 416t43pedrexvd, 7yx8rdxuk88qlw, 6eujs41rqo1, dnqe8wr689luobc, cc2rx5nn2k, 4fk0boaeev, rnd5fn8mvkiz3p, iodudj0jk8q0p, m3era20z8j9j, 668b9vn0ni0mbnr, rsphjlmrun, 7djbd0tz6wpvzef, 4ra777nysqs1o, v394rruvfb5, 1cse0awby4f4b, rw6qwotvn5g, uuxfhrxhpx2qct, nmgto3kgv42, 8stqzfwj780, buo3z28g1u, xx9saevywc, 5w8qhqgowp, q6po1ordyjb1, gvkniqyhkqszref, utmhqcugfptxa, ki2lq7atwhsak, gywtjg4g3iv, plrg9emrieq7t4c", "date": "2020-08-06 10:01:42", "meta": {"domain": "daisytale.it", "url": "http://daisytale.it/qwtn/fourier-transform-of-cosine.html", "openwebmath_score": 0.879318118095398, "openwebmath_perplexity": 844.5134176232584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9942697539943961, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8773713269376618}}
{"url": "https://math.stackexchange.com/questions/1433570/division-of-a-straight-line-in-given-ratio", "text": "# Division of a straight line in given ratio\n\nThe problem in the book states:\n\n\"Find the coordinates of the point that divides AB in the given ratio in each of the following cases: a) A(2,4), B(-3,9) 1:4 internally\"\n\nGiven the point that we are looking for is $P(X,Y)$.\n\nSince I am not familiar with the \"line division\" conventions, I went to find a point (-2,8) with the following formula: $$X=\\frac{\\lambda x_2+\\mu x_1}{\\lambda+\\mu}$$, where $\\lambda =1$ and $\\mu=4$. And a similar formula is used to find coordinate of $Y$.\n\nI was having a cartesian coordinate system in mind when I was dividing the line into ratios. So that the portion on the left is 1 unit and the portion on the right is 4 units. However, the book assumed inversely, so that:\n\nWould you say that my answer is incorrect, or both answers are correct? I would imagine that it would be more accurate for the book to give the ratio 5:-1, if it wanted me to arrive at the answer they are giving.\n\n\u2022 Probably the ratio 1:4 is to be taken not from left to right, but from the first point given (A) to the second one (B). However your answer is correct, because the book should have specified more clearly what was intended. Sep 13 '15 at 14:23\n\u2022 MyPoint=$P_1$ is such that $BP_1/AP_1=1/4$. TheirPoint $P_2$ is such that $AP_2/BP_2 =1/4$. So, without specifing the order of the ratio we have two solutions. Sep 13 '15 at 14:32\n\nNotice, the coordinates of the point $P(X, Y)$ which divides the line joining the points $A(2, 4)$ & $B(-3, 9)$ in a ratio $1:4$ internally are given as $$P(X, Y)\\equiv \\left(\\frac{4(2)+1(-3)}{1+4}, \\frac{4(4)+1(9)}{1+4}\\right)$$ $$P(X, Y)\\equiv \\color{blue}{\\left(1, 5\\right)}$$\n$\\dfrac{BP}{PA}= \\dfrac14$ your point,\n$\\dfrac{BQ}{QA}=\\dfrac41$ their point... is correct as per the convention.", "date": "2021-12-03 11:15:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1433570/division-of-a-straight-line-in-given-ratio", "openwebmath_score": 0.9101059436798096, "openwebmath_perplexity": 216.7379985508669, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854164256366, "lm_q2_score": 0.9046505357435622, "lm_q1q2_score": 0.8773168965257458}}
{"url": "https://mathematica.stackexchange.com/questions/103527/how-to-subtract-the-column-means-from-each-row-of-a-matrix/103532", "text": "# How to subtract the column means from each row of a matrix?\n\nGiven a matrix, we want to subtract the mean of each column, from all entries in that column. So given this matrix:\n\n(mat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}) // MatrixForm\n\nthe mean of each column is m = Mean[mat].\n\nSo the result should be\n\nThis operation is called centering of observations in data science.\n\nThe best I could find using Mathematica, is as follows:\n\nmat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};\nm = Mean[mat];\n(mat[[All, #]] - m[[#]]) & /@ Range@Length@m // Transpose\n\nBut I am not too happy with it. I think it is too complicated. Is there a simpler way to do it? I tried Map and MapThread, but I had hard time getting the syntax to work.\n\nIn MATLAB, there is a nice function called bsxfun which is sort of like MapThread. Here is how it is done in MATLAB:\n\nA=[1 2 3 4;5 6 7 8;9 10 11 12];\nbsxfun(@minus,A,mean(A))\n\n-4    -4    -4    -4\n0     0     0     0\n4     4     4     4\n\nIt maps the function minus, taking one column from A and one element from mean(A). I think it is more clear than what I have in Mathematica. One should be able to do this in Mathematica using one of the map functions more easily and clearly than what I have above.\n\nThe question is: Can the Mathematica solution above be improved?\n\nmat - ConstantArray[Mean[mat], 3]\n\nor more generally:\n\nmat - ConstantArray[Mean[mat], Length[mat]]\n\nIt is there:\n\nStandardize[mat, Mean, 1 &]\n\u2022 Here comes the winner! \u2013\u00a0xzczd Jan 7 '16 at 10:57\n\u2022 @xzczd Thanks. It may be not the fastest though, it rescales by 1 anyway. \u2013\u00a0Kuba Jan 7 '16 at 11:39\n\u2022 Can use Identity' instead of 1&. \u2013\u00a0TheDoctor Jan 13 '16 at 13:19\n\u2022 Re \"rescales by 1\": see my answer to this Q&A. \u2013\u00a0Michael E2 Jul 3 '17 at 15:10\n\u2022 @MichaelE2 thanks for the investigation! \u2013\u00a0Kuba Jul 4 '17 at 21:48\n\nWell, transposing, subtracting, transposing...\n\nTranspose[Transpose[mat] - Mean[mat]]\n# - Mean@mat & /@ mat // MatrixForm\n\n\u2022 This computes the same Mean for each row - could be inefficient. \u2013\u00a0shrx Jan 8 '16 at 9:29\n\nIf you don't mind the type of mat changes:\n\nCircleMinus = Compile[{{a, _Real, 1}, {b, _Real, 1}}, a - b, RuntimeAttributes -> Listable]\nmat\u2296Mean@mat\n\n$\\left( \\begin{array}{cccc} -4. & -4. & -4. & -4. \\\\ 0. & 0. & 0. & 0. \\\\ 4. & 4. & 4. & 4. \\\\ \\end{array} \\right)$\n\nI was asked to post an answer based on a recently uncovered (at least, for me) undocumented function, StatisticsLibraryMatrixRowTranslate, available in V10 and later. My feeling is that, at heart, it is essentially the same as Kuba's answer, which I will use this answer to explain. I will also shed some further light on Kuba's comment.\n\nFirst, the function StatisticsLibraryMatrixRowTranslate modifies the matrix in place, which is uncharacteristic of Mathematica. Here is a typical usage applied to the OP's problem:\n\nmat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};\nModule[{res = mat},\nStatisticsLibraryMatrixRowTranslate[res, -Mean[res]];\nres] // MatrixForm\n\nIn V11 (but not in V10, though the functionality exists), Kuba's Standardize[mat, Mean, 1 &] uses this call to shift mat by the mean, but it also rescales by the second function, which seems inefficient for the problem at hand. However, it turns out that rescaling by constant & is a special case that is implemented by dividing the translated matrix by constant. In case you didn't know, multiplying by 1 (not 1.) is also a special case, which the comparison below will show. Dividing by 1 in normal Mathematica means multiplying by the reciprocal Power[1, -1] and takes roughly twice as long as multiplication, unless you use Divide.\n\nSeedRandom[0];\ndata = RandomReal[{0, 50}, {1*^7, 2}];\n1*data; // RepeatedTiming\ndata/1; // RepeatedTiming\nDivide[data, 1]; // RepeatedTiming\n1.*data; // RepeatedTiming\ndata/2; // RepeatedTiming\n(*\n{3.01*10^-7, Null}\n{5.2*10^-7, Null}\n{3.8*10^-7, Null}\n{0.063, Null}\n{0.062, Null}\n*)\n\nSo we should expect that the rescaling by 1 & in Kuba's answer to add a negligible amount of time to the overall time of the operation in V11:\n\nmrt = Module[{res = data},\nStatisticsLibraryMatrixRowTranslate[res, -Mean[res]];\nres]; // RepeatedTiming\n(*  {0.139, Null}  *)\n\nst = Standardize[data, Mean, 1 &]; // RepeatedTiming\n(*  {0.139, Null}  *)\n\nmrt == st     (* check *)\n(*  True  *)\n\u2022 Thanks, good addition to this Q&A, especially for v10 users like me. \u2013\u00a0Mr.Wizard Jul 3 '17 at 15:16\n\u2022 @Mr.Wizard Even though I noticed the difference in V10, I was really slow to recognize its significance. Thanks for prodding me. \u2013\u00a0Michael E2 Jul 3 '17 at 15:18\n\u2022 @Mr.Wizard sorry for pinging you here, but I was not sure I could reach you in chat. I wanted to suggest changing the title of the question to \"How to subtract the column means from each row of a matrix?\". I suppose it could be argued that the accepted answer kind of implements bsxfun (see also my answer), but I feel for example the Standardize answer should also be captured by the title, especially considering that another question was closed as a duplicate of this one. \u2013\u00a0Jacob Akkerboom Jul 4 '17 at 15:30\n\u2022 @Jacob I think that would be a good change. \"bsxfun\" will still appear in the body should anyone explicitly search for that. For other (non-MATLAB) users your title seems far more descriptive. \u2013\u00a0Mr.Wizard Jul 4 '17 at 16:35\nmat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};\nTranspose[Map[# - Mean[#] &, Transpose[mat]]]\n\nwhich gives you\n\n{{-4, -4, -4, -4}, {0, 0, 0, 0}, {4, 4, 4, 4}}\n\nYou might think of transposing your data. MATLAB naturally has the column as the basic subunit of the matrix, while Mathematica has the row.\n\nDefining\n\n(mat = Transpose[{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11,\n12}}]) // MatrixForm\n\nthe (transpose) of the desired result is then\n\nmat - Mean /@ mat // MatrixForm\n\nHere is something a bit closer to the Matlab syntax\n\nmat1=Partition[Range@12,4];\nmat2 = ConstantArray[Mean[mat1], Length[mat1]];\nbinaryFunction = #1 - #2 &;\n\nTranslationTransform[-Mean @ #][#] & @ mat\n\n{{-4, -4, -4, -4}, {0, 0, 0, 0}, {4, 4, 4, 4}}\n\nRoughly comparable to Standardize in speed.\n\nMike's answer (the accepted one), almost implements bsxfun, taking listability for granted, like this\n\nbsxfunListable[listableFu_, mat_, vec_] :=\nlistableFu[mat, ConstantArray[vec, Length@mat]];\n\nbsxfun[Subtract, mat, Mean@mat]\n\nWe can generalise the function like this\n\nbsxfun[fu_, mat_, vec_] :=", "date": "2019-06-20 16:11:26", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/103527/how-to-subtract-the-column-means-from-each-row-of-a-matrix/103532", "openwebmath_score": 0.2270350158214569, "openwebmath_perplexity": 1795.289652204878, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854111860906, "lm_q2_score": 0.9046505383142492, "lm_q1q2_score": 0.8773168942788024}}
{"url": "http://mathhelpforum.com/number-theory/3129-fibonacci-s-sequence.html", "text": "1. Fibonacci's sequence\n\nTwo consecutive Fibonacci numbers are relatively prime\n\nThis is my proof - am i correct?\n\nLet Fn and Fn+1 be any two consecutive Fibonacci numbers and suppose there is an integer d > 1 such that d divides Fn and d divides Fn+1.\n\nThen Fn+1 - Fn = Fn-1 will also be divisible by d (if d divides a and d divides b, then a = d*m and b = d*n for some integers m and n. Then a - b = d*m - d*n = d * (m-n), so d divides (a - b) as well). But now notice that Fn - Fn-1 = Fn-2 will also be divisible by d.\n\nWe can continue this way showing that Fn-3, Fn-4, ... , and finally F1 = 1 are all divisible by d. Certainly F1 is not divisible by d > 1. Thus we have a contradiction that invalidates the assumption. Thus it must be the case that Fn and Fn+1 are relatively prime.\n\nHas anyone know of any other properties of the Fibonacci\u2019s Sequence?? please\n\n2. Originally Posted by Natasha1\nTwo consecutive Fibonacci numbers are relatively prime\n\nThis is my proof - am i correct?\n\nLet Fn and Fn+1 be any two consecutive Fibonacci numbers and suppose there is an integer d > 1 such that d divides Fn and d divides Fn+1.\n\nThen Fn+1 - Fn = Fn-1 will also be divisible by d (if d divides a and d divides b, then a = d*m and b = d*n for some integers m and n. Then a - b = d*m - d*n = d * (m-n), so d divides (a - b) as well). But now notice that Fn - Fn-1 = Fn-2 will also be divisible by d.\n\nWe can continue this way showing that Fn-3, Fn-4, ... , and finally F1 = 1 are all divisible by d. Certainly F1 is not divisible by d > 1. Thus we have a contradiction that invalidates the assumption. Thus it must be the case that Fn and Fn+1 are relatively prime.\n\nHas anyone know of any other properties of the Fibonacci\u2019s Sequence?? please\nDo you know countinued fraction? There is an easy proof from there.\n\n3. I don't no sorry.\n\n4. You can write your proof formally as an inductive one. Let H(n) be the assertion that the Fibonacci numbers F_n and F_{n-1} are coprime. Then H(1) is true, since F_1=1 and F_0=0 have no common factor greater than 1. Now suppose H(n) is true, that is, F_n and F_{n-1} are coprime. Then, as you correctly say, any common factor between F_{n+1} and F_n would also be a common factor between F_n and F_{n-1}. But H(n) says that must be 1. So we have H(n) implies H(n+1). Hence by induction H(n) is true for all n.\n\nIn a recent thread it was shown that (F_n * F_{n+2}) - (F_{n+1})^2 = (-1)^{n+1}. This implies that F_n and F_{n+1} are coprime since any common factor would divide the LHS and so divide the RHS which is +- 1.\n\n5. Originally Posted by Natasha1\nHas anyone know of any other properties of the Fibonacci\u2019s Sequence?? please\n\u2022 Take any three adjacent numbers in the sequence, square the middle number, multiply the first and third numbers. The difference between these two results is always 1.\n\u2022 Take any four adjacent numbers in the sequence. Multiply the outside ones. Multiply the inside ones. The first product will be either one more or one less than the second.\n\u2022 The sum of any ten adjacent numbers equals 11 times the seventh one of the ten. Mesoamericans thought the numbers 7 and 11 were special.\n\n6. Hello, Natasha!\n\nHere's a fascinating bit of Fibonacci trivia . . .\n\n. . . $\\displaystyle \\sum^{\\infty}_{k=1}\\frac{F_k}{10^{k+1}}\\,=\\,\\frac{ 1}{89}$\n\nThat is: $\\displaystyle \\frac{1}{10^2} + \\frac{1}{10^3} + \\frac{2}{10^4} + \\frac{3}{10^5} + \\frac{5}{10^6} + \\frac{8}{10^7} + \\frac{13}{10^8} + \\frac{21}{10^9} + \\hdots \\:=\\:\\frac{1}{89}$\n\nIn other words:\n\n$\\displaystyle 0.01$\n$\\displaystyle 0.001$\n$\\displaystyle 0.0002$\n$\\displaystyle 0.00003$\n$\\displaystyle 0.000005$\n$\\displaystyle 0.0000008$\n$\\displaystyle 0.00000013$\n$\\displaystyle 0.000000021$\n. . . . $\\displaystyle \\vdots$\n$\\displaystyle \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_$\n$\\displaystyle 0.01123595...\\;=\\;\\frac{1}{89}$\n\nAnd $\\displaystyle 89$ happens to be the $\\displaystyle 11^{th}$ Fibonacci number.\n\nWhy? .Who knows? .Math is filled with little \"jokes\" like that.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nHere's another joke . . .\n\nDid you ever notice that:\n\n. . . . $\\displaystyle 3^2 + 4^2\\;=\\;5^2$\n. . $\\displaystyle 3^3 + 4^3 + 5^3\\;=\\;6^3$\n\nHey, have we stumbled onto an important pattern?\n\n. . No . . . those two cases are just concidences.\n\n7. Originally Posted by Soroban\nHello, Natasha!\n\nHere's a fascinating bit of Fibonacci trivia . . .\n\n. . . $\\displaystyle \\sum^{\\infty}_{k=1}\\frac{F_k}{10^{k+1}}\\,=\\,\\frac{ 1}{89}$\n\nThat is: $\\displaystyle \\frac{1}{10^2} + \\frac{1}{10^3} + \\frac{2}{10^4} + \\frac{3}{10^5} + \\frac{5}{10^6} + \\frac{8}{10^7} + \\frac{13}{10^8} + \\frac{21}{10^9} + \\hdots \\:=\\:\\frac{1}{89}$\n\nIn other words:\n\n$\\displaystyle 0.01$\n$\\displaystyle 0.001$\n$\\displaystyle 0.0002$\n$\\displaystyle 0.00003$\n$\\displaystyle 0.000005$\n$\\displaystyle 0.0000008$\n$\\displaystyle 0.00000013$\n$\\displaystyle 0.000000021$\n. . . . $\\displaystyle \\vdots$\n$\\displaystyle \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_ \\_$\n$\\displaystyle 0.01123595...\\;=\\;\\frac{1}{89}$\n\nAnd $\\displaystyle 89$ happens to be the $\\displaystyle 11^{th}$ Fibonacci number.\n\nWhy? .Who knows? .Math is filled with little \"jokes\" like that.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\nIs there any proof of this trivia?\n\n8. Hello, malaygoel!\n\nIs there any proof of this trivia?\n\nWe want to evaluate: .[1] .$\\displaystyle S\\;=\\;\\frac{F_1}{10^2} + \\frac{F_2}{10^3} + \\frac{F_3}{10^4} + \\frac{F_4}{10^5} + \\frac{F_5}{10^6} + \\hdots$\n\nMultiply by 10: .[2] .$\\displaystyle 10S\\;=\\;\\frac{F_1}{10} + \\frac{F_2}{10^2} + \\frac{F_3}{10^3} + \\frac{F_4}{10^4} + \\frac{F_5}{10^5} + \\hdots$\n\nAdd [1] and [2]: .$\\displaystyle 11S\\;=\\;\\frac{F_1}{10} + \\frac{F_1+F_2}{10^2} + \\frac{F_2+F_3}{10^3} + \\frac{F_3+F_4}{10^4} + \\frac{F_4+F_5}{10^5} + \\hdots$\n\n. . which gives us: .[3] .$\\displaystyle 11S\\;=\\;\\frac{F_1}{10} + \\underbrace{\\frac{F_3}{10^2} + \\frac{F_4}{10^3} + \\frac{F_5}{10^4} + \\frac{F_6}{10^5} + \\hdots}$\n\nExamine that last portion . . .\n\nWe have: .$\\displaystyle 100\\left(\\underbrace{\\frac{F_3}{10^4} + \\frac{F_4}{10^5} + \\frac{F_5}{10^6} + \\frac{F_6}{10^7} +\\hdots}\\right)$\n. . . . . and this is $\\displaystyle S$ minus the first two terms: .$\\displaystyle S - \\frac{F_1}{10^2} + \\frac{F_2}{10^3}$\n\nHence, [3] becomes: .$\\displaystyle 11S\\;=\\;\\frac{F_1}{10} + 100\\left(S - \\frac{F_1}{10^2} - \\frac{F_2}{10^3}\\right) \\;= \\;\\frac{F_1}{10} + 100S - F_1 - \\frac{F_2}{10}$\n\nThis simplifies to: .$\\displaystyle 89S\\;=\\;\\frac{9}{10}F_1 + \\frac{1}{10}F_2$\n\nSince $\\displaystyle F_1 = F_2 = 1$, we have: .$\\displaystyle 89S\\;=\\;\\frac{9}{10}(1) + \\frac{1}{10}(1) \\;=\\;1$\n\nTherefore: .$\\displaystyle S\\;=\\;\\frac{1}{89}$ . . . ta-DAA!\n\n9. Here are some facts.\n\n1)The divine proportion is called \"the most irrational number\" meaning it is the most difficult to approximate with rationals (you need countinued fractions to prove this).\n\n2)Let $\\displaystyle d=\\gcd(a,b)$ then $\\displaystyle \\gcd(F(a),F(b))=F(d)$ (another proof to your original problem because two consectutive number are coprime).\n\n3)Every number is expressable as a sum of distinct fibonacci number non of which are adjacent.\n\nHere's an algebraic proof of: $\\displaystyle \\lim_{n\\to\\infty}\\frac{F_{n+1}}{F_n}\\:=\\:\\phi$\n\nYou must be familiar with the Binet function for Fibonacci numbers:\n. . . $\\displaystyle [1]\\;\\;F_n\\;=\\;\\frac{(1 + \\sqrt{5})^n - (1 - \\sqrt{5})^n}{2^n\\sqrt{5}}$\n\nand you must know that: $\\displaystyle [2]\\text{ If }|r| < 1\\text{, then }\\lim_{n\\to\\infty} r^n\\: = \\;0$\n\nWe have the ratio: $\\displaystyle \\displaystyle{R\\;=\\;\\frac{F_{n+1}}{F_n}\\;=$ $\\displaystyle \\frac{(1+\\sqrt{5})^{n+1} - (1 - \\sqrt{5})^{n+1}}{2^{n+1}\\sqrt{5}}$ $\\displaystyle \\cdot\\,\\frac{2^n\\sqrt{5}}{(1+\\sqrt{5})^n - (1 - \\sqrt{5})^n} }$\n\n. . which simplifies to: $\\displaystyle R\\;=\\;\\frac{1}{2}\\cdot\\frac{(1+\\sqrt{5})^{n+1} - (1-\\sqrt{5})^{n+1}}{(1+\\sqrt{5})^n - (1-\\sqrt{5})^n}$\n\nDivide top and bottom by $\\displaystyle (1+\\sqrt{5})^n:$\n\n. . $\\displaystyle R\\;=\\;\\frac{1}{2}\\cdot\\frac{(1 + \\sqrt{5}) - \\frac{(1-\\sqrt{5})^{n+1}}{(1+\\sqrt{5})^n} }{1 - \\frac{(1-\\sqrt{5})^n}{(1+\\sqrt{5})^n}} \\;=\\;\\frac{1}{2}$$\\displaystyle \\cdot\\frac{(1 + \\sqrt{5}) - (1-\\sqrt{5})\\left(\\frac{1-\\sqrt{5}}{1+\\sqrt{5}}\\right)^n}{1 - \\left(\\frac{1-\\sqrt{5}}{1+\\sqrt{5}}\\right)^n}$\n\nNote that: $\\displaystyle \\left|\\frac{1-\\sqrt{5}}{1+\\sqrt{5}}\\right| < 1.$ . Let $\\displaystyle r = \\frac{1-\\sqrt{5}}{1+\\sqrt{5}}$\n\n. . Then we have: $\\displaystyle R\\;=\\;\\frac{(1 + \\sqrt{5}) - (1-\\sqrt{5})r^n}{1 - r^n}$\n\nHence: $\\displaystyle \\lim_{n\\to\\infty}R\\;=$ $\\displaystyle \\;\\lim_{n\\to\\infty}\\frac{1}{2}\\cdot\\frac{(1 + \\sqrt{5}) - (1 - \\sqrt{5})r^n}{1 - r^n} \\;=\\;\\frac{1}{2}\\cdot\\frac{(1 + \\sqrt{5}) - (1 - \\sqrt{5})\\cdot0}{1 - 0}$\n\nTherefore: $\\displaystyle \\lim_{n\\to\\infty}\\frac{F_{n+1}}{F_n}\\;=\\;\\frac{1 + \\sqrt{5}}{2}\\;=\\;\\phi$\n\n11. Originally Posted by ThePerfectHacker\nHere are some facts.\n\n3)Every number is expressable as a sum of distinct fibonacci number non of which are adjacent.\nWhat is the proof?\n\n12. Originally Posted by malaygoel\nWhat is the proof?\nUsing induction for $\\displaystyle n>2$.\n----\nEach of the integers $\\displaystyle 1,2,3,...,F(k)-1$ is a sum of numbers from the set $\\displaystyle S=\\{ F(1),F(2),...,F(k-2) \\}$ none of which are repeated. Select $\\displaystyle x$ such as,\n$\\displaystyle F(k)-1<x<F(k+1)$\nBut because,\n$\\displaystyle x-F(k-1)<F(k+1)-F(k-1)=F(k)$\nWe can express,\n$\\displaystyle x-F(k-1)$ as sum of numbers from $\\displaystyle S$ none of which are repeated. Thus, as a result, $\\displaystyle x$ is expressable as a sum of the numbers,\n$\\displaystyle S\\cup F(k-1)$ without repetitions.\nThis means that any of the numbers,\n$\\displaystyle 1,2,...,F(k+1)-1$ is expressable from the set,\n$\\displaystyle S\\cup F(k-1)$ and the induction is complete.\n\nThus any number is expressable as a sum of distinct fibonacci numbers.\n\nNote if any of the two fibonacci numbers are consecutive then they can be combined to give the number fibonacci number. Countinuing combining adjacent fibonacci number you have proven Zeckendorf's Representation\n\n13. Originally Posted by ThePerfectHacker\nUsing induction for $\\displaystyle n>2$.\n----\nEach of the integers $\\displaystyle 1,2,3,...,F(k)-1$ is a sum of numbers from the set $\\displaystyle S=\\{ F(1),F(2),...,F(k-2) \\}$ none of which are repeated. Select $\\displaystyle x$ such as,\n$\\displaystyle F(k)-1<x<F(k+1)$\nBut because,\n$\\displaystyle x-F(k-1)<F(k+1)-F(k-1)=F(k)$\nWe can express,\n$\\displaystyle x-F(k-1)$ as sum of numbers from $\\displaystyle S$ none of which are repeated. Thus, as a result, $\\displaystyle x$ is expressable as a sum of the numbers,\n$\\displaystyle S\\cup F(k-1)$ without repetitions.\nThis means that any of the numbers,\n$\\displaystyle 1,2,...,F(k+1)-1$ is expressable from the set,\n$\\displaystyle S\\cup F(k-1)$ and the induction is complete.\n\nThus any number is expressable as a sum of distinct fibonacci numbers.\n\nNote if any of the two fibonacci numbers are consecutive then they can be combined to give the number fibonacci number. Countinuing combining adjacent fibonacci number you have proven Zeckendorf's Representation\nWhat does F(k) represent here?\n\n14. Originally Posted by malaygoel\nWhat does F(k) represent here?\nThe $\\displaystyle k$th fibonacci number. Just like in all the previous posts.\n\n15. ...more golden trivia\n\nHi:\n\n1. Any integral power of phi can be expressed as a linear binomial function of phi.\n\nSpecifically, (phi)^k = F_k(phi) + F_k-1, where F_n denotes the nth Fibonacci element (F_1=1, F_2=1, etcetera).\n\nEx: (phi)^6 = 8(phi) + 5\nEx: (phi)^2 = (phi) + 1\nEx: (phi)^-1 = (phi) - 1\nEx: (phi)^-6 = -8(phi) + 13\n\nNote the existence of F_-1 and F_-2 in Examples 3 and 4. There is nothing inherent in the recursive definition of F_k that would restrict k to the set of natural numbers, N. Hence:\n\n{...F_-5, F_-4, F_-3, F_-2, F_-1, F_0, F_1, F_2, F_3, ...} = {...5, -3, 2, -1, 1, 0, 1, 1, 2...}.\n\nNotes:\n\ni) Observe the alternating algebraic sign left of the equality.\nIndeed, F_n < 0 iff n<0 and n even. Else, F_n >= 0.\nii) In all cases, |F_n| = F_|n|. E.g., |F_-8| = F_|-8| = F_8 = 21. Thus, F_-8 = -21.\niii) Proof by induction is straightforward and reasonably non-cumbersome.\n\n2. Let a,b,c be consecutive in Z. Then, (phi)^a + (phi)^b = (phi)^c.\n\nExample: (phi)^2+(phi)^3=(phi)^4\n\nArgument: (phi)^2+(phi)^3= ((phi)+1)+(2(phi)+1)=3(phi)+2=(phi)^4.\n\n3. (phi)^-1 + (phi)^-2 + (phi)^-3 + ... = phi.\n\nProof: (phi) > 1 implies 0 < (phi)^-1 < 1. Hence the series is harmonic and equivalent to:\n\nu^1 + u^2 + u^3 + ... = u / (1 - u), where u = (phi)^-1. Therefore, the sum becomes [(phi)^-1] / [-(phi)+2] = [(phi)^-1] / [(phi)^-2] = phi, which completes the proof.\n\nRegards,\n\nRich B.", "date": "2018-03-23 01:41:22", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/3129-fibonacci-s-sequence.html", "openwebmath_score": 0.8473930358886719, "openwebmath_perplexity": 1881.2424747708912, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137874059599, "lm_q2_score": 0.8933094159957173, "lm_q1q2_score": 0.8772421629273606}}
{"url": "https://math.stackexchange.com/questions/1327201/probability-that-on-three-rolls-of-dice-there-will-be-at-least-one-6-showing-up", "text": "# Probability that on three rolls of dice, there will be at least one 6 showing up?\n\nWhat is the probability that on three rolls of dice, there will be at least one 6 showing up?\n\nAttempt:\n\nSince there can be one six or two sixes or three sixes on three rolls, I considered separate cases and added them up.\n\nSo $(1/6)(5/6)(5/6) + (1/6)(1/6)(5/6) + (1/6)(1/6)(1/6) = 31/216$, but answer is incorrect as per book. Can anyone suggest where I am wrong ?\n\n\u2022 Your approach could work, but the mistake seems to be that in the case of one six you need to account for the possibility that the six is rolled on the first, second, or third roll. So the probability of rolling one six is 3(1/6)(5/6)(5/6). A similar mistake was made for the probability of two sixes. However, there's another way you could approach this problem that can make things a lot easier: consider the complementary probability of rolling no sixes. \u2013\u00a0cxseven Jun 16 '15 at 7:55\n\u2022 @cxseven yes i have done that question that way \u2013\u00a0Taylor Ted Jun 16 '15 at 7:57\n\nYour answer adds three cases - 6xx, 66x and 666 (where x is anything non-six). You've omitted x6x, xx6, x66, 6x6.\n\nThe easiest way to do it is to see that you fail only when you roll 3 non-sixes, which happens with probability $(5/6)^3$ - so you succeed with probability $1 - (5/6)^3 = 91/216$\n\n\u2022 The case 6x6 is missing too. \u2013\u00a0mathmax Jun 16 '15 at 9:05\n\u2022 Indeed. Fixed... \u2013\u00a0Julia Hayward Jun 16 '15 at 9:17\n\nThrow the dice one at a time, so that you have a \"first\", \"second\" and \"third\" die.\n\nYour first part $\\frac{1}{6}\\frac{5}{6}\\frac{5}{6}$ is not the chance of getting one six. It is the chance of the first die being six while the other two are not. You also need to add the chances that the second or third die is six and the other two not. Since these are all the same, you just need to multiply by three.\n\nYour second part $\\frac{1}{6}\\frac{1}{6}\\frac{5}{6}$ also needs to be tripled for the same reason. In this case there are three dice that may be not-six.\n\nYour last part is fine since there is only one way all three dice can be six.\n\nAdding it all together you get $3\\frac{1}{6}\\frac{5}{6}\\frac{5}{6}+3\\frac{1}{6}\\frac{1}{6}\\frac{5}{6}+\\frac{1}{6}\\frac{1}{6}\\frac{1}{6} = \\frac{91}{216}$\n\nThis calculation is relatively painless with three dice but quickly gets worse as you add dice.\n\nHowever, there is an easier way. Turn the question around and ask what is the probability that no die shows six. This is clearly $(\\frac{5}{6})^3=\\frac{125}{216}$\n\nSo, the probability that some die shows six is $1-\\frac{125}{216}=\\frac{91}{216}$\n\nYou have forgotten to include th fact that any six can occur in any of the rolls, i.e. you have multiply the probabilities for exactly one and two sixs by three, giving $$P=3\\cdot\\frac{1}{6}\\cdot\\frac{5}{6}\\cdot\\frac{5}{6}+3\\cdot\\frac{1}{6}\\cdot\\frac{1}{6}\\cdot\\frac{5}{6}+\\frac{5}{6}\\cdot\\frac{5}{6}\\cdot\\frac{5}{6}=\\frac{91}{216}.$$ It may be easier to compute the complementary probability of no six showing up: $$P^c=\\left(\\frac{5}{6}\\right)^3=\\frac{125}{216}$$ and using $$P=1-P^c=\\frac{216-125}{216}=\\frac{91}{216}$$\n\nCase I : one six .. the six can show up at first, second, or third. And similiarly for other cases,\n\nSo your answer should be $$\\binom{3}{1} (1/6) (5/6).(5/6)+\\binom{3}{2} (1/6).(1/6).(5/6)+\\binom{3}{3}(1/6).(1/6).(1/6)$$\n\nor you could do $$1-(5/6)^3$$", "date": "2019-12-13 10:46:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1327201/probability-that-on-three-rolls-of-dice-there-will-be-at-least-one-6-showing-up", "openwebmath_score": 0.7805289030075073, "openwebmath_perplexity": 353.7144128089105, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013792143467, "lm_q2_score": 0.8933093968230773, "lm_q1q2_score": 0.8772421483316233}}
{"url": "http://mathhelpforum.com/geometry/166845-finding-equation-circle-given-pases-through-4-vertices-pyramid.html", "text": "# Math Help - Finding the equation of a circle given it pases through the 4 vertices of a pyramid?\n\n1. ## Finding the equation of a circle given it pases through the 4 vertices of a pyramid?\n\nIn 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:\n\nA (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)\n\nhow can you find the sphere that passes through those vertices?\n\nNB: this sphere is S: x\u00b2 + y\u00b2 + z\u00b2 -12x -6y -4z = 0 but i want to know HOW you find this??\n\nHelp VERY appreciated thanks!\n\n2. Originally Posted by Yehia\nIn 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:\n\nA (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)\n\nhow can you find the sphere that passes through those vertices?\n\nNB: this sphere is S: x\u00b2 + y\u00b2 + z\u00b2 -12x -6y -4z = 0 but i want to know HOW you find this??\nlet $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...\n\n$r^2 = (a-12)^2 + b^2 + c^2$\n\n$r^2 = a^2 + (b-6)^2 + c^2$\n\n$r^2 = a^2 + b^2 + (c-4)^2$\n\n$r^2 = a^2 + b^2 + c^2$\n\nusing each of the first three equations and the last one, you get these equations ...\n\n$144 - 24a = 0$ ... $a = 6$\n\n$36 - 12b = 0$ ... $b = 3$\n\n$16 - 8c = 0$ ... $c = 2$\n\nsphere's equation is\n\n$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$\n\nexpand and simplify to get the equation you provided\n\n3. Originally Posted by skeeter\nlet $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...\n\n$r^2 = (a-12)^2 + b^2 + c^2$\n\n$r^2 = a^2 + (b-6)^2 + c^2$\n\n$r^2 = a^2 + b^2 + (c-4)^2$\n\n$r^2 = a^2 + b^2 + c^2$\n\nusing each of the first three equations and the last one, you get these equations ...\n\n$144 - 24a = 0$ ... $a = 6$\n\n$36 - 12b = 0$ ... $b = 3$\n\n$16 - 8c = 0$ ... $c = 2$\n\nsphere's equation is\n\n$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$\n\nexpand and simplify to get the equation you provided\nSorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).\n\nThanks for the help.\n\n4. Originally Posted by Yehia\nSorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).\n\nThanks for the help.\ndistance formula between any two points in space ...\n\n$r = \\sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$\n\n5. Hello, Yehia!\n\nI'll give it a try . . .\n\nIn 3-space, we are given the four vertices of a tetrahedron:\n. . . . $A(12,0,0),\\;B(0,6,0),\\;C(0,0,4),\\;O(0,0,0)$\n\nFind the sphere that passes through those vertices\n\nAnswer: . $x^2 + y^2 + z^2 -12x -6y -4z \\:=\\:0$\n\nThe center $\\,P$ is equidistant from all four points: . $PA \\,=\\, PB \\,=\\,PC\\,=\\,PD$\n\n$\\underbrace{\\sqrt{(x-12)^2+y^2}}_{[1]} \\:=\\:\\underbrace{\\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$\n\n. . . . . . . . . . . . . . $=\\:\\underbrace{\\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \\:=\\:\\underbrace{\\sqrt{x^2+y^2+z^2}}_{[4]}$\n\n$\\text{Set }[1] = [4]\\!:\\;x^2 - 24x + 144 + y^2 + z^2 \\:=\\:z^2+y^2+z^2$\n\n. . . . . . . . . $-24x + 144 \\:=\\:0 \\quad\\Rightarrow\\quad\\boxed{ x \\:=\\:6}$\n\n$\\text{Set }[2] = [4]\\!:\\;x^2 + y^2 - 12y + 36 + z^2 \\:=\\:x^2 + y^2+z^2$\n\n. . . . . . . . . $-12y + 36 \\:=\\:0 \\quad\\Rightarrow\\quad\\boxed{ y \\:=\\:3}$\n\n$\\text{Set }[3] = [4]\\!:\\;\\;x^2+y^2+z^2-8z + 16 \\:=\\:x^2+y^2+z^2$\n\n. . . . . . . . . $-8z+16 \\:=\\:0 \\quad\\Rightarrow\\quad\\boxed{ z \\:=\\:2}$\n\nHence, the center is: $P(6,3,2)$\n\nThe distance from $\\,P$ to any vertex is the radius.\n\n. . $r \\:=\\:PO \\:=\\:\\sqrt{(6\\!-\\!0)^2 + (3\\!-\\!0)^2 + (2\\!-\\!0)^2} \\:=\\:\\sqrt{36 + 9 + 4} \\:=\\:\\sqrt{49} \\;=\\;7$\n\nThe equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \\:=\\:7^2$\n\n6. Skeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.\n\nTo have (12,0,0) on the sphere, we must have\n$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.\n\nTo have (0,6,0) on the sphere, we must have\n$(0- a)^2+ (6- b)^2+ c^2= r^2$.\n\nTo have (0,0,4) on the sphere, we must have\n$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.\n\nTo have (0,0,0) on the sphere, we must have\n$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.\n\nThat gives four equations to solve for the four values, a, b, c, and r.\n\nWhen you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.\n\n7. Originally Posted by skeeter\ndistance formula between any two points in space ...\n\n$r = \\sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$\nOriginally Posted by Soroban\nHello, Yehia!\n\nI'll give it a try . . .\n\nThe center $\\,P$ is equidistant from all four points: . $PA \\,=\\, PB \\,=\\,PC\\,=\\,PD$\n\n$\\underbrace{\\sqrt{(x-12)^2+y^2}}_{[1]} \\:=\\:\\underbrace{\\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$\n\n. . . . . . . . . . . . . . $=\\:\\underbrace{\\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \\:=\\:\\underbrace{\\sqrt{x^2+y^2+z^2}}_{[4]}$\n\n$\\text{Set }[1] = [4]\\!:\\;x^2 - 24x + 144 + y^2 + z^2 \\:=\\:z^2+y^2+z^2$\n\n. . . . . . . . . $-24x + 144 \\:=\\:0 \\quad\\Rightarrow\\quad\\boxed{ x \\:=\\:6}$\n\n$\\text{Set }[2] = [4]\\!:\\;x^2 + y^2 - 12y + 36 + z^2 \\:=\\:x^2 + y^2+z^2$\n\n. . . . . . . . . $-12y + 36 \\:=\\:0 \\quad\\Rightarrow\\quad\\boxed{ y \\:=\\:3}$\n\n$\\text{Set }[3] = [4]\\!:\\;\\;x^2+y^2+z^2-8z + 16 \\:=\\:x^2+y^2+z^2$\n\n. . . . . . . . . $-8z+16 \\:=\\:0 \\quad\\Rightarrow\\quad\\boxed{ z \\:=\\:2}$\n\nHence, the center is: $P(6,3,2)$\n\nThe distance from $\\,P$ to any vertex is the radius.\n\n. . $r \\:=\\:PO \\:=\\:\\sqrt{(6\\!-\\!0)^2 + (3\\!-\\!0)^2 + (2\\!-\\!0)^2} \\:=\\:\\sqrt{36 + 9 + 4} \\:=\\:\\sqrt{49} \\;=\\;7$\n\nThe equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \\:=\\:7^2$\n\nOriginally Posted by HallsofIvy\nSkeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.\n\nTo have (12,0,0) on the sphere, we must have\n$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.\n\nTo have (0,6,0) on the sphere, we must have\n$(0- a)^2+ (6- b)^2+ c^2= r^2$.\n\nTo have (0,0,4) on the sphere, we must have\n$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.\n\nTo have (0,0,0) on the sphere, we must have\n$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.\n\nThat gives four equations to solve for the four values, a, b, c, and r.\n\nWhen you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.\nThank you guys very very much, yes I do understand it now, like I said, i couldn't see the wood for the trees. But if you don't mind I just have 3 other questions (still reffering to this question though).\n\nHow do you show that the centre of this sphere lies OUTSIDE the pyramid??\nand also how do you prove that a point lies inside a sphere?\nand one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.\n\nI hope those that sounds clear and again thanks very much!\n\n8. Originally Posted by Yehia\nIn 3D Euclidean Vector geometry, if given the 4 vertices of a (Triangular) pyramid:\n\nA (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)\n\nhow can you find the sphere that passes through those vertices?\n\nNB: this sphere is S: x\u00b2 + y\u00b2 + z\u00b2 -12x -6y -4z = 0 but i want to know HOW you find this??\n\nHelp VERY appreciated thanks!\nThe intersection of the sphere with the $\\displaystyle xy$ plane is a circle which passes through the three points: $\\displaystyle A(12,\\ 0,\\ 0),\\ B(0,\\ 6,\\ 0),\\ O(0,\\ 0,\\ 0)\\,.$ The line segment, $\\displaystyle \\overline{AO}$ is a chord of this circle, so the center of this circle lies along the perpendicular bisector of this chord, i.e. the center of this circle lies along the line $\\displaystyle (6,\\ y,\\ 0)\\,.$ Similarly, the perpendicular bisector of $\\displaystyle \\overline{BO}$ is the line $\\displaystyle (x,\\ 3,\\ 0)\\,.$ The these lines intersect at the point $\\displaystyle (6,\\ 3,\\ 0)\\,.$ Therefore the center of the sphere lies on the line $\\displaystyle (6,\\ 3,\\ z)\\,.$\n\nSimilarly, the intersection of the sphere with the $\\displaystyle xz$ plane is also a circle. This circle passes through the points $\\displaystyle A(12,\\ 0,\\ 0),\\ C(0,\\ 0,\\ 4),\\ O(0,\\ 0,\\ 0)\\,.$ The center of this circle is located at $\\displaystyle (6,\\ 0,\\ 2)\\,,$ so the center of the sphere lies on the line $\\displaystyle (6,\\ y,\\ 2)\\,.$\n\nThe two lines intersect at $\\displaystyle (6,\\ 3,\\ 2)\\,,$ which is the location of the center of the sphere. The radius, $\\displaystyle r$, of the sphere is the distance from point $\\displaystyle O(0,\\ 0,\\ 0)\\,.$ to the center of the sphere.\n\nThus, the sqare of the radius is: $\\displaystyle r^2=6^2+2^2+3^2=49\\, .$\n\nThe equation of the sphere is: $\\displaystyle (x-6)^2+(y-2)^2+(z-3)^2= 49\\, .$\n\nA little bit of algebra provides the given answer.\n\n9. Originally Posted by Yehia\n...But if you don't mind I just have 3 other questions (still reffering to this question though).\n\nHow do you show that the centre of this sphere lies OUTSIDE the pyramid??\nHere is a way but I'm sure that there must be a more elegant one:\n\n1. Calculate the 3 heights of the pyramid which correspond to the 3 side planes. (Use the volume of the pyramid and the base area, calculated by the cross-product).\n2. If r is greater than any of the 3 heights then the centre lies outside the pyramid.\n\nand also how do you prove that a point lies inside a sphere?\nPlug in the coordinates of the point into the equation of the sphere. If the left hand side of the equation is smaller than $r^2$ then the point lies inside the sphere.\n\nand one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.\n\n...\nLet M denote the midpoint of the sphere. Then the line MP intersects the sphere at the point which belongs to the sphere and is closest to P.", "date": "2015-07-31 04:40:52", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/166845-finding-equation-circle-given-pases-through-4-vertices-pyramid.html", "openwebmath_score": 0.8563159704208374, "openwebmath_perplexity": 263.76071439334464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9923043519112751, "lm_q2_score": 0.8840392909114836, "lm_q1q2_score": 0.8772360356320229}}
{"url": "https://mathhelpboards.com/threads/product-of-discontinuous-functions.7670/", "text": "# Product of discontinuous functions\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nLet $$\\displaystyle f:\\mathbb{R} \\to \\mathbb{R}$$ and $$\\displaystyle g:\\mathbb{R} \\to \\mathbb{R}$$ be discontinuous at a point $$\\displaystyle c$$ . Give an example of a function $$\\displaystyle h(x)=f(x)g(x)$$ such that $$\\displaystyle h$$ is continuous at c.\n$$\\displaystyle f(x) = \\begin{cases} 0 & \\text{if } x \\in \\mathbb{Q} \\\\ 1 & \\text{if } x \\in \\mathbb{R}-\\mathbb{Q} \\end{cases}$$\n\n$$\\displaystyle g(x) = \\begin{cases} 1 & \\text{if } x \\in \\mathbb{Q} \\\\ 0 & \\text{if } x \\in \\mathbb{R}-\\mathbb{Q} \\end{cases}$$\n\n$$\\displaystyle f,g$$ are continous nowhere but $$\\displaystyle h(x)=0 \\,\\,\\, \\, \\forall \\,\\, x \\in \\mathbb{R}$$.\n\nWhat other examples you might think of ?\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nLet $$\\displaystyle f:\\mathbb{R} \\to \\mathbb{R}$$ and $$\\displaystyle g:\\mathbb{R} \\to \\mathbb{R}$$ be discontinuous at a point $$\\displaystyle c$$ . Give an example of a function $$\\displaystyle h(x)=f(x)g(x)$$ such that $$\\displaystyle h$$ is continuous at c.\n$$\\displaystyle f(x) = \\begin{cases} 0 & \\text{if } x \\in \\mathbb{Q} \\\\ 1 & \\text{if } x \\in \\mathbb{R}-\\mathbb{Q} \\end{cases}$$\n\n$$\\displaystyle g(x) = \\begin{cases} 1 & \\text{if } x \\in \\mathbb{Q} \\\\ 0 & \\text{if } x \\in \\mathbb{R}-\\mathbb{Q} \\end{cases}$$\n\n$$\\displaystyle f,g$$ are continous nowhere but $$\\displaystyle h(x)=0 \\,\\,\\, \\, \\forall \\,\\, x \\in \\mathbb{R}$$.\n\nWhat other examples you might think of ?\nHi Zaid,\n\nHow about the set for functions, $$\\{f,\\,g\\}$$ such that,\n\n$f(x)=\\begin{cases}a & \\text{if } x \\geq c \\\\b & \\text{if } x < c\\end{cases}$\n\n$g(x)=\\begin{cases}b & \\text{if } x \\geq c \\\\a & \\text{if } x < c\\end{cases}$\n\nwhere $$a,\\,b,\\,c\\in \\Re$$ and $$a\\neq b$$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nThe most general, and all-encompassing example I can think of, off the top of my head:\n\nLet $a \\neq b$ and define:\n\n$f(x) = a,\\ x \\neq c$\n$f(c) = b$\n\n$g(x) = b,\\ x \\neq c$\n$g(c) = a$\n\nClearly, neither $f$ nor $g$ is continuous at $c$, as can be proved straight from the definition (use an $0 < \\epsilon < |b - a|$).\n\nJust as clearly:\n\n$fg(x) = ab,\\ \\forall x \\in \\Bbb R$, which is clearly continuous.\n\nOne can construct more \"extravagant\" examples, but the important part is that $a \\neq b$, and that $f$ and $g$ \"complement\" each other. In fact, there is nothing special about the partition of $\\Bbb R$ into the two disjoint sets $\\{c\\}$ and $\\Bbb R - \\{c\\}$, you can use any partition (such as the Dedekind cut example Sudharaka gives, or the partition into the rationals and irrationals).\n\nTo me, this underscores the fact that continuity (of a function) is dependent on the DOMAIN OF DEFINITION of said function, not just the \"rule itself\" of said function.\n\nIn other words, a \"continuous function\" doesn't really MEAN anything, what IS meaningful is: a function continuous at all points of a set $A$. The underlying domain is important. Context is everything: a function that is perfectly continuous on the real numbers may suddenly spectacularly fail to be so on the complex numbers, for example (as is the case with:\n\n$f(x) = \\dfrac{1}{1 + x^2}$).\n\n#### Amer\n\n##### Active member\nCharacteristic function of A\n$$\\chi_A : \\mathbb{R} \\rightarrow \\{0,1\\}$$\n$$\\chi_A =\\left\\{ \\begin{array}{lr} 1 &,x\\in A \\\\ 0 &,x\\in A^{c} \\end{array} \\right.$$\n$$\\chi_{A^{c}} = \\left\\{ \\begin{array}{ir} 0 & , x\\in A \\\\ 1 & , x\\in A^{c} \\end{array} \\right.$$\nTheir product is zero function which is continuous", "date": "2020-09-21 02:55:03", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/product-of-discontinuous-functions.7670/", "openwebmath_score": 0.8850149512290955, "openwebmath_perplexity": 492.225737747527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129093889291, "lm_q2_score": 0.9032942164664239, "lm_q1q2_score": 0.8772152939524754}}
{"url": "http://neda.psdeg-psoe.org/9jv2ji/d1fd03-how-to-diagonalize-a-matrix-3x3", "text": "Diagonalization is the process of transforming a matrix into diagonal form. If the matrix were diagonalizable and we could nd matrices Pand D, then the computation of the 10th power of the matrix would be easy using Proposition 2.3. In many areas such as electronic circuits, optics, quantum mechanics, computer graphics, probability and statistics etc, matrix is used to study. Block-diagonalization of a matrix. Follow 26 views (last 30 days) Rodolphe Momier on 7 Apr 2020. This page explains how to calculate the determinant of a 3x3 matrix. So let\u2019s nd the eigenvalues and eigenspaces for matrix A. De nition 2.5. Ais diagonalizable. Previous question Next question Transcribed Image Text from this Question. This problem has been solved! When I use the eig command, i obtain the eigenvalues sorted in ascending order. Answer: By Proposition 23.2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. Diagonalizing a 3x3 matrix. Linear Algebra Differential Equations Matrix Trace Determinant Characteristic Polynomial 3x3 Matrix Polynomial 3x3 Edu. Note I A= 2 4 6 3 8 0 + 2 0 1 0 + 3 3 5: To nd det( I A) let\u2019s do cofactor expansion along the second row because it has many zeros1. $\\begingroup$ Do you mean diagonalize the 2x2 matrix ? Diagonalizing a 3x3 matrix. 1. A method is presented for fast diagonalization of a 2x2 or 3x3 real symmetric matrix, that is determination of its eigenvalues and eigenvectors. Vote. One of the eigenspaces would have unique eigenvectors. You can also calculate a 3x3 determinant on the input form. The values of \u03bb that satisfy the equation are the generalized eigenvalues. Diagonalize matrix with complex eigenvalues by real basis. $\\begingroup$ The same way you orthogonally diagonalize any symmetric matrix: you find the eigenvalues, you find an orthonormal basis for each eigenspace, you use the vectors in the orthogonal bases as columns in the diagonalizing matrix. By using this website, you agree to our Cookie Policy. See the answer. Each eigenspace is one-dimensional. Diagonalization Linear Algebra MATH 2010 The Diagonalization Problem: For a nxnmatrix A, the diagonalization problem can be stated as, does there exist an invertible matrix Psuch that P 1APis a diagonal matrix? A small computer algebra program is used to compute some of the identities, and a C++ program for testing the formulas has been uploaded to arXiv. In other words, the linear transformation of vector by only has the effect of scaling (by a factor of ) the vector in the same direction (1-D space). Each eigenspace is one-dimensional. Diagonal matrices represent the eigenvalues of a matrix in a clear manner. Diagonalization is a process of &nding a diagonal matrix that is similar to a given non-diagonal matrix. They also arise in calculating certain numbers (called eigenvalues) associated with the matrix. For example, a x matrix of rank 2 will have an image of size 2, instead of 3. Yes. Ask Question Asked 4 years, 6 months ago. Yes. Division Headquarters 315 N Racine Avenue, Suite 501 Chicago, IL 60607 +1 866-331-2435 For any matrix , if there exist a vector and a value such that then and are called the eigenvalue and eigenvector of matrix , respectively. SavannahBergen. \u2022 RREF Calculator \u2022 Orthorgonal Diagnolizer \u2022 Determinant \u2022 Matrix Diagonalization \u2022 Eigenvalue \u2022 GCF Calculator \u2022 LCM Calculator \u2022 Pythagorean Triples List. Matrix Diagonalization Calculator Online Real Matrix Diagonalization Calculator with step by step explanations. The transformation matrix is nonsingular and where . The Euler angles of the eigenvectors are computed. Solution for A is a 3x3 matrix with two eigenvalues. A. With each square matrix we can calculate a number, called the determinant of the matrix, which tells us whether or not the matrix is invertible. ... $which we can eyeball one easily as$\\begin{bmatrix}0\\\\1\\\\0\\end{bmatrix}$. 3x3 Matrix Diagonalization Simple C++ code that finds a quaternion which diagonalizes a 3x3 matrix: . Show \u2026 KurtHeckman. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. EXAMPLE: Diagonalize the following matrix, if possible. Then we need one more for this matrix to be diagonalizable, and fortunately this one is pretty clear too we need the first input in row 1 to sum with the third input to 0,$\\begin{bmatrix}3\\\\0\\\\1\\end{bmatrix}$fits the bill. Characteristic Polynomial of a 3x3 Matrix. Contact Us. on . Steps. You need to calculate the determinant of the matrix as an initial step. Show transcribed image text. Last modified by . But what does it mean to diagonalize a matrix that has null determinant? In Mathematica it can be done easily, but when using the module numpy.linalg I get problems. If the commutator is zero then and May 20, 2016, 3:47:14 PM (A)\" 3x3 Matrix\" Tags. Eigenvalue Calculator Online tool compute the eigenvalue of a matrix with step by step explanations.Start by entering your matrix row number and column number in the input boxes below. Check the determinant of the matrix. 2.6 Multiple Eigenvalues The commutator of and is . Diagonalize the matrix A, if possible.$\\endgroup$\u2013 Adam Jan 23 '14 at 17:57$\\begingroup$Yes, and then is the autovalue the product of the two different autovalues of position and spin-operator? You can also find the inverse using an advanced graphing calculator. A priori, the Pauli matrices and the position operator do not act on the same space, so you should be able to diagonalize both simultaneously. Method 1 of 3: Creating the Adjugate Matrix to Find the Inverse Matrix 1. Recipes: diagonalize a matrix, quickly compute powers of a matrix by diagonalization. on . Expert Answer . Calculating the inverse of a 3x3 matrix by hand is a tedious job, but worth reviewing. By Proposition 23.1, is an eigenvalue of Aprecisely when det( I A) = 0. The solution of the initial value problem will involve the matrix exponential . SEMATH INFO. Terminology: If such a Pexists, then Ais called diagonalizable and Pis said to diagonalize A. Theorem If Ais a nxnmatrix, then the following are equivalent: 1. 3 Determinants and Diagonalization Introduction. Since the eigenvector for the third eigenvalue would also be unique, A must be diagonalizable. Show \u2026 Eigenvalues and matrix diagonalization. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. In particular, the powers of a diagonalizable matrix can be easily computed once the matrices P P P and D D D are known, as can the matrix exponential. When I use the eig command, i obtain the eigenvalues sorted in ascending order. Enter your matrix in the cells or type in the data area. In this way we compute the matrix exponential of any matrix that is diagonalizable. That Is, Find An Invertible Matrix P And A Diagonal Matrix D Such That A=PDP-1 A = -11 3 -9 0-5 0 6 -3 4. We can diagonalize a matrix through a similarity transformation = \u2212, where is an invertible change-of-basis matrix and is a matrix with only diagonal elements. Select the correct choice below and, if\u2026 0 Comments. This square of matrix calculator is designed to calculate the squared value of both 2x2 and 3x3 matrix. In fact, determinants can be used to give a formula for the inverse of a matrix. I have a matrix composed of 1x1, 2x2 and 3x3 blocks and I would like to obtain the eigenvalues and eigenvectors sorted according to the block they correspond to. User can select either 2x2 matrix or 3x3 matrix for which the squared matrix to be calculated. Why? Aug 7, 2020, 9:25:26 PM. An n\u00a3n matrix A is called diagonalizable if A is similar to a diagonal matrix D: Example 12.1. The associated transformations have the effect of killing at least one dimension: indeed, a x matrix of rank has the effect of lowering the output dimension by . Thanks is advance. A is a 3x3 matrix with two eigenvalues. 0. parts of the complex conjugate eigenvectors. Is A diagonalizable? De &nition 12.1.$\\endgroup$\u2013 Gerry Myerson May 4 '13 at 3:54 I need to diagonalize a symbolic matrix with python. Note that we have de ned the exponential e t of a diagonal matrix to be the diagonal matrix of the e tvalues. We will come back to this example afterwards. However, if A {\\displaystyle A} is an n \u00d7 n {\\displaystyle n\\times n} matrix, it must have n {\\displaystyle n} distinct eigenvalues in order for it to be diagonalizable. Matrix diagonalization is useful in many computations involving matrices, because multiplying diagonal matrices is quite simple compared to multiplying arbitrary square matrices. 0 Comments. Why? orthogonal matrix is a square matrix with orthonormal columns. Due to the simplicity of diagonal matrices, one likes to know whether any matrix can be similar to a diagonal matrix. Is A diagonalizable? Follow 24 views (last 30 days) Rodolphe Momier on 7 Apr 2020. Vote. Start by entering your matrix row number and column number in the boxes below. OB. Quaternion Diagonalizer(const float3x3 &A) { // A must be a symmetric matrix. Is there a necessary and sufficient condition for a square matrix to be able to diagonalize a symmetric square matrix? The generalized eigenvalue problem is to determine the solution to the equation Av = \u03bbBv, where A and B are n-by-n matrices, v is a column vector of length n, and \u03bb is a scalar. 1fe0a0b6-1ea2-11e6-9770-bc764e2038f2. In fact, A PDP 1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. Definition An matrix is called 8\u201a8 E orthogonally diagonalizable if there is an orthogonal matrix and a diagonal matrix for which Y H E\u0153YHY \u00d0\u0153YHY \u00d1\u00de\" X Thus, an orthogonally diagonalizable matrix is a special kind of diagonalizable matrix\u2026 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. Created by . How to convert this vector to a matrix? We put a \"T\" in the top right-hand corner to mean transpose: Notation. Free Matrix Diagonalization calculator - diagonalize matrices step-by-step This website uses cookies to ensure you get the best experience. Question: Diagonalize The Matrix A, If Possible. Determinant of a 3x3 matrix Last updated: Jan. 2nd, 2019 Find the determinant of a 3x3 matrix, , by using the cofactor expansion. Looking at this makes it seem like a 3x3 matrix, with a 2x2 tacked on the bottom right corner, and zero's added to the filler space made as a result of increasing by 2 dimensions. Thanks is advance. I have a matrix composed of 1x1, 2x2 and 3x3 blocks and I would like to obtain the eigenvalues and eigenvectors sorted according to the block they correspond to. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Block-diagonalization of a matrix. Two square matrices A and B of the same order are said to be simultaneously diagonalizable, if there is a non-singular matrix P, such that P^(-1).A.P = D and P^(-1).B.P = D', where both the matrices D and D' are diagonal matrices. 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{"url": "https://physics.stackexchange.com/questions/103081/why-would-a-pendulum-experiment-give-g-9-8-mathrmm-s2", "text": "# Why would a pendulum experiment give $g > 9.8\\ \\mathrm{m/s^2}$?\n\nI am taking an introductory lab course in which we've done an experiment on the physical pendulum.\n\nWe've seen that for small oscillations, the period is\n\n$$T=2\\pi\\sqrt{\\dfrac{I_S}{Mgd_{cm}}}\\tag{1}$$\n\nwhere $S$ is the pivot point, $M$ is the total mass of the object, and $d_{cm}$ is the distance between $S$ and the pendulum's center of mass.\n\nNow, varying $d_{cm}$, I've obtained seven different periods. I've calculated $g$ in terms of $T$ and $d_{cm}$ for those seven distinct values of $T$ and $d_{cm}$. So, $g$ can be expressed as\n\n$$g=4\\pi^2\\dfrac{I_S}{T^2Md_{cm}}.\\tag{2}$$\n\nAll the periods I've obtained were always greater than $1$, and the values of $g$ where between $10.15$ and $10.3$. I am trying to understand why is it that $g$ gave me always greater than $9.8$, the expected value, and not less than $9.8$. If I consider the air friction, I would expect the period to be greater; from what I've said and from equation (2), I would say that the values of $g$ should be less than $9.8$, contrary to the values I got.\n\nNote that that $T$ is in seconds, $[g]=\\dfrac{m}{s^2}$ and the angle from which the pendulum was released is of $25$ degrees approximately.\n\nI would appreciate if someone could help me to understand the reason why I've obtained these values for $g$.\n\n\u2022 When you say the period is greater than 1, what units are you using? Similarly when you report your value of $g$, I'm assuming you're using meters per second squared? You should include that. \u2013\u00a0David Z Mar 12 '14 at 0:17\n\u2022 What's your release angle? That formula is only valid for small amplitudes. \u2013\u00a0Kvothe Mar 12 '14 at 0:36\n\u2022 @Kvothe: Yes, but wouldn't applying the formula for larger angles (where the pendulum will take longer to swing back and forth than the simple formula implies) yields a smaller value for the gravitational acceleration -- assuming accurate measurements? \u2013\u00a0Keith Thompson Mar 12 '14 at 0:41\n\u2022 @KeithThompson - You are correct. The first order correction for larger angles is $g = \\frac{4\\pi^2 L}{T^2} \\left( 1 + \\frac{1}{8} \\theta_0^2 \\right) + {\\cal O} \\ ( \\theta_0^4 \\ )$. Thus, the measured value using the first order formula is generically smaller. \u2013\u00a0Prahar Mar 12 '14 at 0:52\n\u2022 Thanks for all your comments, guys; to @David Z: sorry for not clarifying that information, I have added the units and the angle. So, I still don't have the slightest idea why I got greater values than $9,8$, I can't think of any possible mistakes I could have made during the experiment. \u2013\u00a0user100106 Mar 12 '14 at 2:01\n\nI) OP is using the period formula\n\n$$\\tag{1} T~=~2\\pi\\sqrt{\\frac{I}{MgR}}$$\n\nfor a compound/physical pendulum (in the small amplitude limit) to estimate the gravitational acceleration constant\n\n$$\\tag{2} g~=~\\left(\\frac{2\\pi}{T}\\right)^2 \\frac{I}{MR}.$$\n\nHere $I$ is the moment of inertia around the pivot point; $R$ is the distance from CM to the pivot point; and $M$ is the total mass.\n\nII) After doing the experiment OP finds values for $g$ that are 3-5% too big. (These results are close enough that OP likely did not make any elementary mistakes with units.) A finite amplitude of\n\n$$\\tag{3} \\theta_0 ~\\approx ~25^{\\circ}~\\approx~ .44~ {\\rm rad}$$\n\nmakes the pendulum\n\n$$\\tag{4} \\frac{\\theta_0^2}{8}~\\approx~ 2\\%$$\n\nslower, as compared to the ideal pendulum (1), cf. comment by Prahar. So correcting for a finite amplitude makes OP's estimates worse, 5-7% too big, as Keith Thompson points out in a comment above.\n\nSo the discrepancy is caused by something else. The culprit is likely that it is difficult to get a precise estimate for the moment of inertia $I$. All the other quantities $T$, $M$ and $R$ should be fairly easy to measure reliable. So OP's value for $I$ is likely too big. According to Steiner's theorem\n\n$$\\tag{5} I~=~MR^2+I_0,$$\n\nwhere $I_0$ is the moment of inertia around the CM (and the actual quantity which is poorly known).\n\nIII) Below follows a suggestion. Plot OP's seven data points in an $(x,y)$ diagram with axes\n\n$$\\tag{6} x~:=~R^2 \\quad\\text{and}\\quad y~:=~R\\left(\\frac{T}{2\\pi}\\right)^2.$$\n\nTheoretically, the $(x,y)$ data points should then lie on a straight line\n\n$$\\tag{7} y~=~ax+b$$\n\nwith slope\n\n$$\\tag{8} a~=~\\frac{1}{g}$$\n\nand $y$-intercept\n\n$$\\tag{9} b=~~\\frac{I_0}{gM}.$$\n\nIn other words, find the best fitting straight line. This method should hopefully produce a good estimate for $g$ without having to know $I_0$ a priori. (By the way, notice that we in principle also don't need to know the mass $M$, cf. the equivalence principle!)\n\nIV) Finally, as always in experiments, estimate all pertinent uncertainties in the various measurements.\n\n\u2022 Super clear and helpful answer, you're right about the moment of inertia, I had some difficulty calculating $I_0$ for the two \"disks\" and the rod. \u2013\u00a0user100106 Mar 16 '14 at 22:00", "date": "2020-01-24 20:15:41", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/103081/why-would-a-pendulum-experiment-give-g-9-8-mathrmm-s2", "openwebmath_score": 0.8516871333122253, "openwebmath_perplexity": 347.6085128989922, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983596967483837, "lm_q2_score": 0.8918110555020056, "lm_q1q2_score": 0.8771826497603324}}
{"url": "https://math.stackexchange.com/questions/3023032/finding-the-minimal-number-of-members", "text": "# Finding the minimal number of members\n\nI've been working on the following problem\n\nFor every issue in the Blue's association, a commission with 10 members (belonging the Blue's) is formed in order to solve the problem. The only condition is\n\nThere can't be two commissions having more than one member in common\n\nThe Blue's association has formed this year 40 commissions.\n\nWhat's the minimal amount of members in the Blue's association?\n\nI've only found out the following\n\nFor any commission you can form $$\\binom{10}{2}=45$$ different pairs and none of them can appear in another commission.\n\nSince 40 different commissions are formed, the minimal number of pairs is $$45\\times 40=1800$$.\n\nDenote by $$n$$ the number of members. Thus $$\\binom{n}{2}\u22651800\\Rightarrow n>60$$\n\n\n\nThe minimal amount of members has to be 100 or less.\n\nYou can observe a distribution for 100 members here\n\nMy question:\n\nIs 100 the answer or is there an ever smaller possible amount of members? If so, how can I prove it?\n\n\u2022 Different but related question, perhaps some of the ideas there are helpful: math.stackexchange.com/questions/2879313/\u2026 \u2013\u00a0Servaes Dec 2 '18 at 18:57\n\u2022 If there are two members that are in $6$ commisions, then they share at most one commission, and to fill these $11$ commissions then requires another $10\\times9+1\\times8=98$ distinct members, yielding a total of $100$ members if we count the two members we started with. So any example smaller than the one you found has at most one member in more than $5$ commissions. \u2013\u00a0Servaes Dec 2 '18 at 20:02\n\u2022 By a similar (but simpler) argument, no member can be in more than $11$ commisions; to fill $11$ commissions sharing one member requires another $11\\times9=99$ distinct members, yielding a total of $100$ members. \u2013\u00a0Servaes Dec 2 '18 at 20:06\n\u2022 @Servaes If $a$ and $b$ are two members that are in $6$ commissions, how do you know that those $10\\times9+1\\times6=98$ members are all distinct? Why can't a commission containing $a$ overlap with a commission containing $b$? \u2013\u00a0bof Dec 5 '18 at 9:50\n\u2022 @Servaes Note that, if the points of a projective plane of order $9$ are our \"members\" and the lines are our \"commissions\", then we have $91$ members and $91$ commissions, and each member belongs to $10$ commissions. \u2013\u00a0bof Dec 5 '18 at 11:32\n\nThis post exhibits a solution with $$82$$ members. Combined with the excellent answer by @Song, this means $$82$$ is indeed optimal.\n\nMotivation: The excellent answer by @Song and the followup comments by @Servaes make me wonder... perhaps if we look for 41 commissions (not 40) then there is a solution with a great deal of symmetry:\n\n\u2022 (a) 82 members (the optimal answer we seek)\n\u2022 (b) 41 commissions (exceeds OP requirement)\n\u2022 (c) each member associated with exactly 5 commissions (not part of OP)\n\u2022 (d) each commission associated with exactly 10 members (equals OP requirement)\n\u2022 (e) each 2 members have exactly 1 common commission (not part of OP)\n\u2022 (f) each 2 commissions have exactly 1 common member (exceeds OP requirement)\n\nThis would be like a finite projective plane, but with 82 points and 41 lines. However, in a finite projective (respectively: affine) plane, the no. of points and no. of lines are equal (respectively: almost equal), and this is probably why a solution based on FPP only reaches 84. So I decided to look at related structures called Block Designs, Steiner Systems, etc. where there are typically many more \"lines\" than \"points\". After quite a bit of digging, I think I found the right structure:\n\nThe solution: It is a Steiner $$S(t=2,k=5,n=41)$$ system. A Steiner system is defined by the following properties:\n\n\u2022 there are $$n=41$$ objects (these are the commissions)\n\n\u2022 there are $$b$$ blocks (these are the members), each block (member) being a subset of objects (i.e. the commissions he/she is associated with)\n\n\u2022 each block has $$k=5$$ objects (each member is associated with 5 commissions)\n\n\u2022 every $$t=2$$ objects is contained in exactly 1 block (every 2 commissions have exactly 1 common member)\n\nSo this already satisfies (b), (c) and (f). Next, quoting from https://en.wikipedia.org/wiki/Steiner_system#Properties we have:\n\n\u2022 $$b = {n \\choose t} / {k \\choose t} = (41 \\times 40) / (5 \\times 4) = 41 \\times 2 = 82$$, satisfying (a)\n\n\u2022 $$r = {n-1 \\choose t-1} / {k-1 \\choose t-1} = 40 / 4 = 10$$, where $$r$$ denotes \"the number of blocks containing any given object\", i.e. the number of members associated with any given commission, satisfying (d).\n\nThinking more, I don't think (e) can be satisfied. However, (e) is not needed for the OP, so it doesn't matter.\n\nSo finally we just need to prove that such a Steiner $$S(t=2,k=5,n=41)$$ system exists. This existence is non-trivial, but luckily more digging reveals:\n\n\u2022 https://math.ccrwest.org/cover/steiner.html has a list of Steiner systems that are known to exist. $$S(2,5,41)$$ (the webpage sometimes lists the 3 parameters in different order) is not part of any infinite families listed, but if you go further down the page it is listed as a standalone example; clicking on that link goes to...\n\n\u2022 https://math.ccrwest.org/cover/show_cover.php?v=41&k=5&t=2 which exhibits the system, created via \"Cyclic construction\" whatever that means.\n\nI did not check the numbers thoroughly, but if I understand the webpage correctly, there should be 82 rows (members / blocks), each containing 5 numbers (commissions), all the numbers being 1 through 41 inclusive (the 41 commissions), each number (commission) should appear in 10 rows, and every pair-of-numbers should appear in 1 row.\n\nI am not an expert in any of these, so if I had a mistake or misunderstanding above, my apologies. Perhaps someone more expert can check my work?\n\n\u2022 +1 Good ideas and good find! I will have a look at the second link. As a side note, indeed (e) is impossible; there are $\\binom{82}{2}=3321$ pairs of members. But there are $\\binom{10}{2}=45$ pairs of members per commission, so $41\\times45=1845$ pairs of members that share a commission. Alternatively, every member is in $5$ commissions, and hence in a commission with $5\\times9=45$ members, so $\\frac{82\\times45}{2}=1845$ pairs of members that share a commission. Either way, there are too many pairs of members to all share a commission. \u2013\u00a0Servaes Dec 14 '18 at 0:55\n\u2022 *Another, simpler, way to see that (e) is impossible; this would imply that every member shares a commission with $81$ members. But every member is in only $5$ commissions, and hence shares a commission with only $5\\times9=45$ members. \u2013\u00a0Servaes Dec 14 '18 at 0:57\n\u2022 @Servaes - yes of course. in fact, more abstractly / vaguely speaking, perhaps insisting on both (e) and (f) is WHY a finite projective plane has the same no. of lines and points. so if we introduce imbalance in (a) vs (b), and hence (c) vs (d), we must necessarily imbalance (e) vs (f) as well. this is all very vague of course. :) \u2013\u00a0antkam Dec 14 '18 at 1:07\n\u2022 With duality between lines & points (or equiv: rows & cols of an incidence matrix), there is really no need to translate, but it's up to you. anyway, here's a link I found explaining (I think) the \"cyclic construction\" of some Steiner systems, although it doesn't mention $S(2,5,41)$ specifically: pitt.edu/~kaveh/Chap7-MATH1050.pdf \u2013\u00a0antkam Dec 14 '18 at 1:53\n\u2022 Dan Gordon migrated the La Jolla covering design repository to his new site, so it might be a good idea to update your links. \u2013\u00a0hardmath Mar 16 at 0:46\n\nLet $$i$$ denote each member of Blue's association and assume that there are $$N$$ members in total, that is, $$i=1,2,\\cdots, N.$$ And let $$j,k=1,2,\\ldots, 40$$ denote each of 40 commission. We will show that $$N$$ is at least $$82$$.\n\nConsider the set $$S=\\{(i,j,k)\\;|\\;1\\leq i\\leq N, 1\\leq j Let $$d_i$$ denote the number of commissions that $$i$$ joined. We will calculate $$|S|$$ using double counting method. First, note that $$|S|=\\sum_{(i,j,k)\\in S}1 = \\sum_{1\\leq j since for each $$j, there is at most one $$i$$ in common. On the other hand, $$|S| = \\sum_{1\\leq i\\leq N} \\sum_{(j,k):(i,j,k)\\in S}1 = \\sum_{1\\leq i\\leq N} \\binom{d_i}{2},$$ since for each $$i$$, the number of pairs $$(j,k)$$ that $$i$$ joined is $$\\binom{d_i}{2}$$. We also have $$\\sum_{1\\leq i\\leq N}d_i = 400,$$by the assumption.\nFinally, note that the function $$f(x)= \\binom{x}{2} = \\frac{x^2-x}{2}$$ is convex. Thus by Jensen's inequality we have that $$\\binom{40}{2}\\geq |S|=\\sum_{1\\leq i\\leq N} \\binom{d_i}{2}\\geq Nf\\left(\\frac{\\sum_i d_i}{N}\\right)=N\\binom{\\frac{400}{N}}{2}.$$ This gives us the bound $$40\\cdot 39 \\geq 400\\cdot(\\frac{400}{N}-1),$$and hence $$N \\geq \\frac{4000}{49} = 81.63\\cdots$$ This establishes $$N\\geq 82$$. However, I'm not sure if this bound is tight. I hope this will help.\n\n$$\\textbf{Note:}$$ If $$N=82$$ is tight, then above argument implies that $$d_i$$'s distribution is almost concentrated at $$\\overline{d} = 400/82 \\sim 5$$.\nEDIT: @antkam's answer seemingly shows that $$N=82$$ is in fact optimal.\n\n\u2022 +1 Fantastic argument! \u2013\u00a0Servaes Dec 10 '18 at 10:28\n\u2022 A corollary of this argument is that in the best solution every commission must share a member with some other commission. \u2013\u00a0Will Fisher Dec 10 '18 at 20:02\n\u2022 @WillFisher You mean, if there is a solution with $N=82$? In this case, it also follows that there are precisely $72$ members that are in $5$ commissions, and $10$ that are in $4$ commissions. Also, that every commission has precisely $9$ members in $5$ commissions and $1$ member in $4$ commissions. \u2013\u00a0Servaes Dec 10 '18 at 20:38\n\u2022 I don't really understand this step: \"Finally, note that the function $x\\to \\binom{x}{2} = \\frac{x^2-x}{2}$ is convex. Thus, we have that $$\\binom{40}{2}\\geq |S|=\\sum_{1\\leq i\\leq N} \\binom{d_i}{2}\\geq N\\binom{\\frac{\\sum_i d_i}{N}}{2}=N\\binom{\\frac{400}{N}}{2}.$$\" \u2013\u00a0Dr. Mathva Dec 13 '18 at 23:35\n\u2022 I should have mentioned that it is by Jensen's inequality. The statement is in en.m.wikipedia.org/wiki/Jensen%27s_inequality. @Dr. Mathva \u2013\u00a0Song Dec 13 '18 at 23:44\n\nThis is just a partial answer. I will show that $$85$$ members suffice; I don't know if $$85$$ is the minimum.\n\nRecall that a projective plane of order $$n$$ exists if $$n$$ is a prime power: it has $$n^2+n+1$$ points and $$n^2+n+1$$ lines; each line has $$n+1$$ points, and there are $$n+1$$ lines through each point; any pair of lines meets in a unique point, and any pair of points determines a unique line.\n\nConsider a projective plane of order $$9$$; it has $$9^2+9+1=91$$ points and $$91$$ lines; there are $$10$$ points on each line and $$10$$ lines through each point. A set of points is in general position if no three of the points are collinear. Note that, if we have a set of $$t$$ points in general position, then the lines determined by those points (taken two at a time) cover a total of at most $$t+8\\binom t2$$ points; as long as $$t\\le5$$ then the number of covered points is at most $$5+8\\binom52=85\\lt91$$, so we can add another point to the set and still have them in general position. Thus we can find a set $$S$$ of $$6$$ points in general position.\n\nLet the members of the Blue's association be the $$91-6=85$$ points which are not in $$S$$. The commissions are the lines which do not meet $$S$$; they have $$10$$ members each, and any two have exactly one member in common. Finally, by the in-and-out formula, the number of commissions is $$91-\\binom61\\cdot10+\\binom62\\cdot1=46.$$\n\nP.S. Let $$m$$ be the minimum possible number of members. I showed above that $$m\\le85$$. On the other hand, I have a small improvement on your lower bound $$m\\ge61$$.\n\nSuppose the $$i^\\text{th}$$ member belongs to $$d_i$$ commissions; then $$\\sum_{i=1}^md_i=400$$ since there are $$40$$ commissions with $$10$$ members each. Moreover $$d_i\\le9$$ since $$m\\le85\\lt91$$. Let $$k=|\\{i:d_i\\ge5\\}|$$. Then $$400=\\sum_{i=1}^md_i\\le4(m-k)+9k=4m+5k\\le340+5k,$$ whence $$k\\ge12$$; i.e., there are at least $$12$$ members who are on at least $$5$$ commissions. Choose two members $$i$$ and $$j$$ who are on at least $$5$$ commissions.\n\nCase 1. There is a commission containing both $$i$$ and $$j$$.\n\nFirst, there are $$10$$ members on the commission which $$i$$ and $$j$$ both belong to. Next $$i$$ belongs to $$4$$ more commissions, with $$36$$ additional members. Finally, $$j$$ belongs to $$4$$ more commissions, each of which contains at most one member of each of the $$5$$ commissions containg $$i$$, and at least $$5$$ members who haven't been counted yet, for a total of $$20$$ new members. This shows that $$m\\ge10+36+20=66$$.\n\nCase 2. There is no commission containing both $$i$$ and $$j$$.\n\nIn this case a similar argument shows that $$m\\ge67$$.\n\nThis proves that $$m\\ge66$$. Combining this with the upper bound shown earlier, we have $$66\\le m\\le85.$$\n\n\u2022 +1 Great progress and nice ideas! Note that for every prime power $q$ there are arcs of $q+1$ points in $\\Bbb{P}(\\Bbb{F}_q)$. In particular you can take your set $S$ to contain $7$ points, yielding $42$ commissions and $m\\leq84$. With $S$ containing $8$ points we get $39$ commissions and $m\\leq83$, perhaps another commission can be fitted in somewhere? \u2013\u00a0Servaes Dec 5 '18 at 16:53\n\u2022 Your bottom argument can be extended to show $m\\ge 73$. \u2013\u00a0Will Fisher Dec 5 '18 at 17:04\n\u2022 @Servaes I don't know much about projective planes. That fact about \"arcs of $q+1$ points\", is it easy or hard to prove? Is it true for all projective planes of prime power order, or just the field planes? \u2013\u00a0bof Dec 6 '18 at 10:20\n\u2022 @WillFisher Nice. How do you prove it? \u2013\u00a0bof Dec 6 '18 at 10:21\n\u2022 @bof In any projective plane over a finite field, any nondegenerate conic is an example of an arc of $q+1$ points. I've also checked that the same construction with a set $S$ of $8$ or more points cannot fit any other commission directly (i.e. without adjusting some of the existing $39$ or fewer commissions). \u2013\u00a0Servaes Dec 6 '18 at 12:46\n\nAllow me to summarize and slightly refine the results in the current answers (if only to straighten out my own thoughts); they show that the minimum number of members $$m$$ satisfies $$82\\leq m\\leq84$$. They also imply strict conditions on any solution with $$m=82$$.\n\nI also include my result that if $$m=83$$, then no member is in more than $$7$$ commissions. Much more can be said, but I do not have a definitive proof for the cases $$m=82$$ or $$m=83$$.\n\nThe upper bound $$m\\leq84$$ comes from bof's construction in the projective plane of order $$9$$; the projective plane $$\\Bbb{P}^2(\\Bbb{F}_9)$$ consists of $$91$$ points on $$91$$ lines, with $$10$$ points on each line and $$10$$ lines through each point. Importantly, each pair of lines meets in precisely one point, and each pair of points is on precisely one line.\n\nFor $$7$$ distinct points in general position (no $$3$$ on a line, e.g. points on a smooth conic) there are precisely $$7\\times10-\\binom{7}{2}\\times1=49$$ lines containing these points. Removing these $$7$$ points and the $$49$$ lines containing them leaves $$84$$ points and $$91-49=42$$ lines each containing $$10$$ points, and any pair of lines meets in at most one point. That is, we have $$84$$ members in $$42$$ commissions, with no $$2$$ commissions sharing more than one member, so $$m\\leq84$$.\n\nThe lower bound $$m\\geq82$$ comes from Song's answer; the number of pairs of commissions that share a member is at most $$\\binom{40}{2}$$, as there are $$40$$ commissions. As every commission shares at most one member, this can also be counted as the number of pairs of commissions that each member is in. If the $$i$$-th member is in $$d_i$$ commissions, then it is in $$\\binom{d_i}{2}$$ pairs of commissions and hence $$\\sum_{i=1}^m\\binom{d_i}{2}\\leq\\binom{40}{2}.\\tag{1}$$ As there are $$40$$ commissions with $$10$$ members each, we also have $$\\sum_{i=1}^md_i=400$$. In the inequality above we can bound the left hand side from below using the fact that for all positive integers $$x$$ we have $$\\binom{x-1}{2}+\\binom{x+1}{2}=2\\binom{x}{2}+1.$$ This allows us to even out the $$d_i$$'s to find that $$\\sum_{i=1}^m\\binom{d_i}{2}\\geq(m-n)\\binom{x}{2}+n\\binom{x+1}{2},\\tag{2}$$ for some $$x$$ and $$n$$ with $$0\\leq n, where $$(m-n)x+n(x+1)=\\sum_{i=1}^md_i=400.$$ The latter simplifies to $$mx+n=400$$, which shows that $$x=\\lfloor\\frac{400}{m}\\rfloor$$ and $$n=400-mx$$. Plugging this back in shows that $$\\begin{eqnarray*} \\binom{40}{2}&\\geq&\\sum_{i=1}^m\\binom{d_i}{2} \\geq(m-n)\\binom{x}{2}+n\\binom{x+1}{2}\\\\ &=&(m-(400-m\\lfloor\\tfrac{400}{m}\\rfloor))\\binom{\\lfloor\\frac{400}{m}\\rfloor}{2}+(400-m\\lfloor\\tfrac{400}{m}\\rfloor)\\binom{\\lfloor\\frac{400}{m}\\rfloor+1}{2}\\\\ &=&-\\frac{m}{2}\\lfloor\\tfrac{400}{m}\\rfloor^2-\\frac{m}{2}\\lfloor\\tfrac{400}{m}\\rfloor+400\\lfloor\\tfrac{400}{m}\\rfloor. \\end{eqnarray*}$$ The latter is strictly decreasing for $$m$$ in the interval $$[1,84]$$. The inequality is satisfied if and only if $$m\\geq82$$, which proves the lower bound.\n\nLet $$S$$ denote the number of times we need to apply the identity $$\\binom{x-1}{2}+\\binom{x+1}{2}=2\\binom{x}{2}+1$$ to reduce the left hand side of $$(2)$$ to the right hand side. We can then write $$(2)$$ more precisely as $$\\sum_{i=1}^m\\binom{d_i}{2}=(m-n)\\binom{x}{2}+n\\binom{x+1}{2}+S.$$ Knowing that $$82\\leq m\\leq84$$ simplifies the above significantly, as then $$x=\\lfloor\\tfrac{400}{m}\\rfloor=4$$ and $$n=400-4m$$. We find that $$780\\geq\\sum_{i=1}^m\\binom{d_i}{2}=1600-10m+S.$$ In particular, for $$m=82$$ we find that $$S=0$$ and hence that there are precisely $$n=400-82\\times4=72$$ members that are in $$4$$ commissions and $$10$$ members that are in $$5$$ commissions. We also see that we have equality in $$(1)$$, meaning that every pair of commissions shares a member. This implies $$\\sum_{i\\in C}(d_i-1)=39$$ for every commission $$C$$, from which it follows that every commission has precisely $$1$$ member that is in $$4$$ commissions, and $$9$$ members that are in $$5$$ commissions.\n\nIf $$m=83$$ then $$S\\leq10$$, and there are at most $$10$$ pairs of commissions that do not share a member.\n\nHere are a few unincorporated observations that may or may not be helpful. These concern restrictions on minimal examples with $$m<84$$, i.e. $$m=82$$ or $$m=83$$. They are all subsumed by the observations above for $$m=82$$, so I prove them only for $$m=83$$.\n\nObservation 1: For all $$i$$ we have $$d_i\\leq9$$.\n\nTo fill the commission of member $$i$$ requires $$9d_i+1$$ distinct members, including member $$i$$. We have $$9d_i+1\\leq m=83$$ and hence $$d_i\\leq9$$.\n\nObservation 2: For all $$i$$ we have $$d_i\\leq8$$.\n\nTo fill the commission of a member $$i$$ with $$d_i=9$$ requires $$9d_i+1=82$$ distinct members, leaving $$1$$ member remaining as $$m=83$$. Each of the remaining $$40-d_i=31$$ commissions has at most one member from teach of the $$d_i$$ commissions of $$i$$, and hence contains the remaining member. But this member is in at most $$9$$ commissions by observation $$1$$, a contradiction.\n\nObservation 3: For any pair $$i$$, $$j$$ of members in a commission we have $$d_i+d_j\\leq14$$.\n\nIf the inequality does not hold then without loss of generality $$d_i=8$$ and $$d_j\\geq7$$. To fill the shared commission requires another $$8$$ members, and to fill the remaining $$7$$ commissions of member $$i$$ requires another $$9\\times7=63$$ members. Each of the $$d_j-1$$ remaining commissions of $$j$$ contains at most $$7$$ members from the $$7$$ commissions of $$i$$, and hence at least $$2$$ new members. Hence we have a total of $$2+8+9\\times(d_i-1)+2\\times(d_j-1)\\geq2+8+63+2\\times6=85,$$ members, contradicting $$m=83$$.\n\nObservation 4: For all $$i$$ we have $$d_i\\leq7$$.\n\nSuppose toward a contradiction that $$d_i=8$$ for some member $$i$$. To fill these $$d_i=8$$ commissions requires $$9d_i+1=73$$ distinct members, including member $$i$$, leaving $$10$$ members. Each of the remaining $$32$$ commissions has at most $$8$$ members from the $$d_i=8$$ commissions, hence at least $$2$$ members from the remaining $$10$$. Numbering these $$1$$ throught $$10$$ we find that $$\\sum_{k=1}^{10}d_k\\geq2\\times32=64.$$ We distinguish two cases:\n\nCase 1: If $$d_j=8$$ for some $$1\\leq j\\leq10$$ then $$j$$ shares a commission with at least $$8$$ other of these $$10$$ members, hence they all have $$d_k\\leq6$$ by observation $$3$$. To satisfy the inequality there must be one more member $$j'$$ with $$d_{j'}=8$$, and the other $$8$$ have $$d_k=6$$.\n\nWe have $$11$$ members, including $$i$$, that together take up $$8+64=72$$ spots in the $$40$$ commissions. The remaining $$83-11=72$$ members then take up $$400-72=328$$ spots. As noted before, the sum $$\\sum\\binom{d_i}{2}$$ ranging over the remaining $$72$$ members is minimal when the values $$d_i$$ differ by at most $$1$$. This happens precisely when $$d_i=5$$ for $$40$$ members and $$d_i=4$$ for $$432$$ members. Then $$\\sum_{k=1}^{83}\\binom{d_i}{2}\\geq3\\binom{8}{2}+8\\binom{6}{2}+40\\binom{5}{2}+32\\binom{4}{2}=796,$$ which exceeds the bound of $$\\binom{40}{2}=780$$ we found before, a contradiction.\n\nCase 2: If $$d_j\\neq8$$ for all $$10$$ remaining members, then to satisfy $$\\sum_{k=1}^{10}d_k\\geq64$$ there must be a t least $$4$$ members with $$d_k=7$$. We also have $$\\sum_{k=1}^{10}\\leq70$$, and we proceed as before.\n\nWe have $$5$$ members, including $$i$$, that together take up $$8+28=36$$ spots in the $$40$$ commissions. Hence the remaining $$83-5=78$$ members take up $$400-36=364$$ spots. The sum $$\\sum\\binom{d_i}{2}$$ over the remaining $$78$$ members is minimized when the $$d_i$$ differ by at most $$1$$. This happens precisely if $$d_i=5$$ for $$52$$ members and $$d_i=4$$ for $$26$$ members, and we $$\\sum_{k=1}^{83}\\binom{d_k}{2}\\geq\\binom{8}{2}+4\\binom{7}{2}+52\\binom{5}{2}+26\\binom{4}{2}=788,$$ again contradicting the upper bound of $$\\binom{40}{2}=780$$.\n\nMuch more can be said, but my computer is already freezing up at this big an answer.\n\nHere's a partial answer that increases the lower bound for any (not necessarily optimal) solution to $$m\\ge 74$$.\n\nSuppose there is a solution with $$m$$ members and we know that there are two members each in $$l+1$$ commissions, then $$m\\ge 9(l+1)+(8-l)(l+1)+2.$$ This is because if member one is in $$\\ge l+1$$ commissions, each commission needs to be filled with $$9$$ new members since these $$l+1$$ commissions already have maximum overlap. For the commissions that member two is in, each needs $$9$$ more members to be accounted for. We can't have any overlap between these commissions because they already have maximum overlap (that being member two). We can at best choose one member from each group with member one in it, giving us $$l+1$$ members, but the other $$9-(l+1)=8-l$$ are new. This gives $$9(l+1)+(8-l)(l+1)$$ members other than the two we started with. (Note that this is the best bound in $$l$$ possible).\n\nNow, suppose $$m$$ members has a solution to the problem. Let $$d_i$$ be the how many commissions the $$i$$-th member is in. First note that $$m\\ge 9d_i+1$$ for every $$i$$, so $$d_i\\le \\lfloor (m-1)/9\\rfloor$$. Let $$k_l=|\\{i\\; :\\; d_i>l\\}|$$. Then $$400=\\sum_{i=1}^md_i\\le l(m-k_l)+\\lfloor (m-1)/9\\rfloor k_l.$$ Hence $$k_l\\ge \\frac{400-lm}{\\lfloor (m-1)/9\\rfloor -l}.$$ Since $$k_l$$ is an integer, if $$\\frac{400-lm}{\\lfloor (m-1)/9\\rfloor -l}>1$$ then $$k_l\\ge 2$$, meaning that there is at least $$2$$ members in at least $$l+1$$ commissions so by the above $$m\\ge 9(l+1)+(8-l)(l+1)+2$$. Note that $$\\frac{400-lm}{\\lfloor (m-1)/9\\rfloor -l}>1$$ exactly when $$\\frac{400-\\lfloor (m-1)/9\\rfloor}{m-1}>l$$. Hence we have that for all $$\\frac{400-\\lfloor (m-1)/9\\rfloor}{m-1}>l$$ we have $$m\\ge 9(l+1)+(8-l)(l+1)+2.$$ For this to be satisfied gives $$m\\ge 74.$$\n\n\u2022 Very nice! I had some trouble understanding the part from \"Hence we have that for all...\". Perhaps you could make more explicit that if this inequality holds for $l$, then the next inequality must also hold for that $l$. \u2013\u00a0Servaes Dec 7 '18 at 14:29\n\u2022 @Servaes I edited the answer. It is basically just solving for the $l$'s such that $k_l\\ge 2$, letting the first paragraphs argument follow. \u2013\u00a0Will Fisher Dec 7 '18 at 16:06", "date": "2020-09-23 00:37:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3023032/finding-the-minimal-number-of-members", "openwebmath_score": 0.827904999256134, "openwebmath_perplexity": 480.8203853774128, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969698879862, "lm_q2_score": 0.8918110404058913, "lm_q1q2_score": 0.8771826370558871}}
{"url": "http://mathhelpforum.com/statistics/181908-am-i-more-likely-roll-least-one-six-if-i-have-more-dice-print.html", "text": "# Am I more likely to roll at least one six if I have more dice?\n\n\u2022 May 28th 2011, 05:31 PM\ntheyThinkImWrong\nAm I more likely to roll at least one six if I have more dice?\nHi,\n\nI've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.\n\nI say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.\n\nNow my thoughts for this are as follows:\nAt first, it doesn't matter, whether I roll them one after another, or one at a time.\nThe chances to roll a six on the first dice that I roll a 6, is 1/6.\n\nNow, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)\n\nSo for the second dice roll a six, the probability is still 1/6\nHowever, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.\n\nSo the probability to roll at least one 6 with two dice is:\n1/6 + 1/6*5/6\n\nIf I was to expand this for 1000 dice, I would end up with:\n\n$\\sum_{k=0}^{999}\\left (\\frac{5}{6} \\right )^k \\times \\frac{1}{6}$\nThe probability will get ever more closely to 100%, the more dice I have, correct?\n\u2022 May 28th 2011, 06:52 PM\nWilmer\nWhy get 1000 dice? Same thing if you have 1 dice: just keep rolling it.\n\u2022 May 28th 2011, 06:55 PM\ntheyThinkImWrong\nWell, originally I asked this: If you were to roll 1000 dice at once, how likely is it, that there is at least one 6?\n\nBut it doesn't really matter :)\n\u2022 May 28th 2011, 07:01 PM\nbryangoodrich\nThe possibility of rolling a 6 with one die is 1/6, and for each die you have you increase your changes of rolling 1/6. By your friend's reasoning, you're no more liking to have \"at least one head (H) in two coin flips\" than in just one coin flip. Let us use this much more basic example. The probability of flipping a head of an unbiased coin is 1/2 because we split the event space {T, H} evenly into two. When we flip two coins we have a new event space, namely {TT, TH, HT, HH}. Here the events are equally likely of 1/4, but notice \"at least one H\" can turn out in three different events. Thus, the probability of \"at least one H\" is 3/4.\n\nThe same works for dice. With one die we have an event space {1, 2, 3, 4, 5, 6}. With two dice the size of the event space is 36:\n\nSpoiler:\n11 12 13 14 15 16\n21 22 23 24 25 26\n31 32 33 34 35 36\n41 42 43 44 45 46\n51 52 53 53 55 56\n61 62 63 64 65 66\n\nNow, in how many ways can we have \"at least one six?\" In this small, but sufficient, example, we can just look at see. The bottom row and the right-most column all include a 6. This is 11/36, which is almost 0.31, compared to 1/6 =0.167. So yes, increasing the die increases your chances of rolling a six.\n\u2022 May 28th 2011, 07:24 PM\ntheyThinkImWrong\nThanks, that would confirm my point of view.\n\nThey: But how is that possible? The probability never changes? So it should always be 1/6?\n\nAlso, we'd value second and third opinions :) Thank you very much in advance!\n\u2022 May 28th 2011, 07:58 PM\nbryangoodrich\nThe probability of a 6 emerging on a single die does not ever change, but that is not the event we're looking at when we have more than one die. When we are using, say, two dice we're looking at a different event space with each event having 1/36 chance of occurring. Of those thirty-six, eleven of them will have a 6 appear. This is undeniable. This has no impact on the probability of a six appearing on a single die. Your friends seem to be ignoring the fact that the possible ways of a 6 occurring change when you have two dice. With one die there is one, and only one, way of a 6 occurring. Thus, the probability is 1/6. With two dice we have 11 different ways, each with 1/36 = (1/6)(1/6) chance of happening. The probability of the individual dice does support this result, as we can see, but you cannot ignore the fact our event space is different when you include another die.\n\nI can go even further and define it this way by the inclusion-exclusion rule of probability theory. Let X be the outcome of the first die and Y the outcome of the second. What we want then is:\n\n$P(X = 6\\ or\\ Y=6)=P(X=6)+P(Y=6)-P(X=6\\ \\&\\ Y= 6)=\\frac{1}{6}+\\frac{1}{6}-\\frac{1}{36}=\\frac{11}{36}$\n\nDo you see why these values work? The probability of the single die being six does not change, but we have to sum the two probabilities together. We then have to subtract the event that is common to both of them, otherwise we're double counting that event. This happens only once.\n\u2022 May 28th 2011, 11:48 PM\nmr fantastic\nQuote:\n\nOriginally Posted by theyThinkImWrong\nHi,\n\nI've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.\n\nI say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.\n\nNow my thoughts for this are as follows:\nAt first, it doesn't matter, whether I roll them one after another, or one at a time.\nThe chances to roll a six on the first dice that I roll a 6, is 1/6.\n\nNow, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)\n\nSo for the second dice roll a six, the probability is still 1/6\nHowever, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.\n\nSo the probability to roll at least one 6 with two dice is:\n1/6 + 1/6*5/6\n\nIf I was to expand this for 1000 dice, I would end up with:\n\n$\\sum_{k=0}^{999}\\left (\\frac{5}{6} \\right )^k \\times \\frac{1}{6}$\nThe probability will get ever more closely to 100%, the more dice I have, correct?\n\nLet X be the random variable 'Number of 6's rolled'.\n\nThen X ~ Binomial(n, p = 1/6).\n\nCalculate Pr(X > 0) = 1 - Pr(X = 0). Then it is obvious that as n increases so does the probability.\n\nI suggest you review the binomial distribution if you don't know it ( I do not plan to give a tutorial).", "date": "2016-05-01 10:58:08", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/181908-am-i-more-likely-roll-least-one-six-if-i-have-more-dice-print.html", "openwebmath_score": 0.5550181865692139, "openwebmath_perplexity": 330.0413349751959, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985271387015241, "lm_q2_score": 0.8902942341267434, "lm_q1q2_score": 0.8771814349097281}}
{"url": "https://math.stackexchange.com/questions/2601763/any-closed-subset-in-a-separable-metric-space-is-the-union-of-a-perfect-set-and", "text": "# Any closed subset in a separable metric space is the union of a perfect set and a set that is at most countable. Why does this matter?\n\nI am writing regarding problem 2.28 in Rudin's \"Principles of Mathematical Analysis\". The statement the reader is asked to prove is as follows:\n\nEvery closed subset of a separable metric space is the union of a perfect set and a set that is at most countable.\n\nThis statement sounds like it should be important.\n\nWhat came to mind to me was the definition of the functional limits and the ensuing definitions of continuity we encounter later in an undergraduate analysis course. The definition of continuity works \"intuitively\" with limit points in the domain, but less \"intuitively\" at isolated points. Since a perfect set is made up entirely of limit points (and is closed), it seems like this result should let us say something about \"weird things\" happening at only countably many places for functions defined on closed subsets of a separable metric space. I'm not really sure if this matters at all or is really comprehensible.\n\nThank you.\n\n\u2022 It could be what you say. But also if the space is complete, then its perfect subsets have the same cardinal as $\\mathbb{R}$, therefore this theorem proves that the continuum hypothesis is valid for closed subsets of $\\mathbb{R}$. One can even use this to show it holds for borelian subsets Jan 12, 2018 at 7:30\n\u2022 The various text-snippets you can read from a google-books search for Cantor-Bendixson should give you an idea of the importance of this result in mathematics. Jan 12, 2018 at 14:43\n\nBy the time you read the statement\n\n$\\tag 1 \\text{Every closed subset of a separable metric space}$ $\\qquad \\qquad \\text{is the union of a perfect set and a set that is at most countable.}$\n\nyou are far along the path of mathematically 'sculpturing' topological concepts. it is instructive to 'reverse engineer' this statement. I think you can safely say that if you do this, you will be forced to discover the foundations of point-set topology.\n\nWe all agree that the integers $\\mathbb Z$ is a closed set of isolated points on the real number line $\\mathbb R$. But this is just too easy to do, since you have the integers extending out to both $-\\infty$ and $+\\infty$. So can you find a countable (infinite) closed set of isolated points constrained to be inside the interval $[-1,+1]$?\n\nAnswering this question and the natural questions that arise as you proceed on your journey will be an awe inspiring mathematical challenge.\n\nTo answer the OP's question, reading and understanding (1) in a math book makes me think of mountaineers, and images like\n\nThe work, training and challenges of the accomplishment is safely behind them now.\n\nThe reader might be interested in exploring the perfect set property where they will find the following:\n\nThe Cantor\u2013Bendixson theorem states that closed sets of a Polish space $X$ have the perfect set property in a particularly strong form; any closed set $C$ may be written uniquely as the disjoint union of a perfect set $P$ and a countable set $S$. Thus it follows that every closed subset of a Polish space has the perfect set property. In particular, every uncountable Polish space has the perfect set property, and can be written as the disjoint union of a perfect set and a countable open set.\n\nExample: Consider $\\text{2-dim Euclidean space, } \\mathbb R^2$, which is a polish space. Define\n\n$\\quad C = \\text{The Unit Circle } \\cup \\{(\\pm \\frac{1}{n},\\, 0) \\;|\\; n \\ge 1\\} \\cup \\{(0,\\, \\pm \\frac{1}{n}) \\;|\\; n \\ge 1\\} \\cup \\{(0,\\,0)\\}$\n\nThe set $C$ is closed in $\\mathbb R^2$, and it can be written (uniquely) as the union of the unit circle, a perfect set, and the remaining countable points in $C$ with norm less than $1$. All these points not on the unit circle are isolated except for $(0,\\,0)$, the origin.", "date": "2022-08-17 05:16:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2601763/any-closed-subset-in-a-separable-metric-space-is-the-union-of-a-perfect-set-and", "openwebmath_score": 0.829552412033081, "openwebmath_perplexity": 141.5187906507126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713893938906, "lm_q2_score": 0.8902942195727173, "lm_q1q2_score": 0.8771814226877607}}
{"url": "https://customerlabs.co/ride-along-uryuz/page.php?id=diagonalize-matrix-with-complex-eigenvalues-f80dc8", "text": "DOWNLOAD IMAGE. A normal matrix A is de\ufb01ned to be a matrix that commutes with its hermitian conjugate. 2 in the set. 1 {\\displaystyle A} (\\lambda _{i}\\mathbf {v} _{i})\\ =\\ \\lambda _{i}\\mathbf {e} _{i},}. The invertibility of {\\displaystyle A} By linearity of matrix multiplication, we have that, Switching back to the standard basis, we have, The preceding relations, expressed in matrix form, are. 0 \u2212 ] v {\\displaystyle {\\begin{array}{rcl}A^{k}=PD^{k}P^{-1}&=&\\left[{\\begin{array}{rrr}1&\\,0&1\\\\1&2&0\\\\0&1&\\!\\!\\!\\!-1\\end{array}}\\right]{\\begin{bmatrix}1^{k}&0&0\\\\0&1^{k}&0\\\\0&0&2^{k}\\end{bmatrix}}\\left[{\\begin{array}{rrr}1&\\,0&1\\\\1&2&0\\\\0&1&\\!\\!\\!\\!-1\\end{array}}\\right]^{-1}\\\\[1em]&=&{\\begin{bmatrix}2-2^{k}&-1+2^{k}&2-2^{k+1}\\\\0&1&0\\\\-1+2^{k}&1-2^{k}&-1+2^{k+1}\\end{bmatrix}}.\\end{array}}}. 0 , To nd out how, read on. C A Proof: The row vectors of Show Instructions. {\\displaystyle U^{-1}CU} P Calculating the eigenvalues of an n\u00d7n matrix with real elements involves, in principle at least, solving an n th order polynomial equation, a quadratic equation if n = 2, a cubic equation if n = 3, and so on. P A 1 P + , In fact, we can define the multiplicity of an eigenvalue. \u00a0diagonal \u00a0is called diagonalizable if there exists an ordered basis of 2 Eigenvalues, diagonalization, and Jordan normal form Zden ek Dvo r ak April 20, 2016 De nition 1. P P . is invertible, + P is not simultaneously diagonalizable. If in addition, {\\displaystyle B} \u2212 Conic Sections Trigonometry. Example: Diagonalize the matrix . The calculator will diagonalize the given matrix, with steps shown. Returns Reference to *this. Remark. 1 Viewed as a linear transformation from A sends vector to a scalar multiple of itself . e {\\displaystyle P} = . Likewise, the (complex-valued) matrix of eigenvectors v is unitary if the matrix a is normal, i.e., if dot(a, a.H) = dot(a.H, a), where a.H denotes the conjugate transpose of a. 1 1 = Formally this approximation is founded on the variational principle, valid for Hamiltonians that are bounded from below. With complex eigenvalues we are going to have the same problem that we had back when we were looking at \u2026 The roots of the characteristic polynomial 0 \u00a0and So what are the eigenvalues of [[1, e^ik], [e^ik, 1]]? be the eigenvalue . Note that there is no preferred order of the eigenvectors in 1 0 e n Yes, of course. 0 .[2]. i = Example. 2 However, we can diagonalize For example, this is the case for a generic rotation matrix. , then {\\displaystyle P} ) It is diagonal, so obviously diagonalizable, and has just a single eigenvalue repeated $n$ times. 0 \u2212 {\\displaystyle P^{-1}\\!AP} 2 2 . e B , has Lebesgue measure zero. {\\displaystyle C} det 1 \u00a0 The values of \u03bb that satisfy the equation are the generalized eigenvalues. A real matrix can have complex eigenvalues and eigenvectors. A \u00a0 has one eigenvalue (namely zero) and this eigenvalue has algebraic multiplicity 2 and geometric multiplicity 1. and For example, defining In this lecture, we shall study matrices with complex eigenvalues. COMPLEX EIGENVALUES . De nition 2. n Proof \u2208 , we have: exp , ] can be chosen to form an orthonormal basis of Again, your method is fine. diagonalizable matrices (over M 3 A Criterion for Diagonalization. 1 1 = Free Matrix Diagonalization calculator - diagonalize matrices step-by-step. [ 1 0 0 If Ax = \u03bbx for some scalar \u03bb and some nonzero vector xx, then we say \u03bb is an eigenvalue of A and x is an eigenvector associated with \u03bb. P = {\\displaystyle A} v ( So the column vectors of , almost every matrix is diagonalizable. {\\displaystyle \\lambda _{1}=1,\\lambda _{2}=1,\\lambda _{3}=2} D Let A be a real 2 2 matrix with trace zero and positive determinant. \u2212 A For any triangular matrix, the eigenvalues are equal to the entries on the main diagonal. 1 An n\u00d7n matrix A is diagonalizable if and only if n-space has a basis consisting of eigenvectors of A. Corollary 2. first. \u2022 if v is an eigenvector of A with eigenvalue \u03bb, then so is \u03b1v, for any \u03b1 \u2208 C, \u03b1 6= 0 \u2022 even when A is real, eigenvalue \u03bb and eigenvector v can be complex \u2022 when A and \u03bb are real, we can always \ufb01nd a real eigenvector v associated with \u03bb: if Av = \u03bbv, with A \u2208 Rn\u00d7n, \u03bb \u2208 R, and v \u2208 Cn, then A\u211cv = \u03bb\u211cv, A\u2111v = \u03bb\u2111v , \u2022If a \"\u00d7\"matrix has \"linearly independent eigenvectors, then the matrix is diagonalizable Eigenvalues and Eigenvectors Diagonalization Complex eigenvalues Find all of the eigenvalues and eigenvectors of A= 2 6 3 4 : The characteristic polynomial is 2 2 +10. C V 1 It is clear that one should expect to have complex entries in the eigenvectors. A . 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As finding its eigenvalues and eigenvectors of a symmetric matrix are real numbers of linear recursive sequences, as...", "date": "2021-06-18 03:20:05", "meta": {"domain": "customerlabs.co", "url": "https://customerlabs.co/ride-along-uryuz/page.php?id=diagonalize-matrix-with-complex-eigenvalues-f80dc8", "openwebmath_score": 0.9470022916793823, "openwebmath_perplexity": 809.0894912627506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.978712650690179, "lm_q2_score": 0.8962513731336202, "lm_q1q2_score": 0.8771725570843182}}
{"url": "http://math.stackexchange.com/questions/532367/prove-convergence-using-varepsilon-n-definition", "text": "Prove convergence using $\\varepsilon$-$N$ definition\n\nThis is an example presented by professor in class. I understand the idea behind this kind of definition, but I'm having trouble following my professor's thought process.\n\nWe must prove that the $\\displaystyle \\lim_{n\\to\\infty} \\frac{2n-1}{n^2} = 0$ using the $\\varepsilon$-$N$ definition of a limit of a sequence.\n\nSo, the foregoing limit is true iff $\\forall \\varepsilon>0,\\ \\exists N(\\varepsilon)>0 : \\forall n \\ ( n \\geq N \\Rightarrow |f(n) - 0| < \\varepsilon)$.\n\nNow, the way I understand it is that if we prove the conditional statement by choosing an appropriate $N$, we've proven the limit.\n\nSo first we fix $\\varepsilon>0$ since it's given, and play around with $$\\displaystyle \\left| \\frac{2n-1}{n^2} \\right| < \\varepsilon$$ $$|2n-1| < \\varepsilon \\ n^2 .$$\n\nWe note that $|2n-1|<|2n|$ which sits right with me.\n\nHowever, my professor then reasons that $|2n-1|<|2n|<\\varepsilon \\ n^2$ and so $2n < \\varepsilon \\ n^2$.\n\nHow do we know that $2n < \\varepsilon \\ n^2$? If epsilon is really small, wouldn't there be a point when this inequality is no longer true? This is what is bothering me.\n\nFrom that we conclude that $\\displaystyle n>\\frac{\\varepsilon}{2}$ and choose $\\displaystyle N=\\frac{\\varepsilon}{2}$ so that\n\n$$n \\geq \\frac{\\varepsilon}{2} \\Rightarrow \\left| \\frac{2n-1}{n^2} \\right| < \\varepsilon$$\n\nis always true. And that is true iff the limit is true, so the limit is true.\n\nI'd like to know if my understanding of this definition is correct, and if anyone can explain to me the reasoning behind the inequalities above.\n\nThank you very much!\n\n-\nThe point is that because $2n<\\varepsilon n^2\\iff 2<\\varepsilon n$, no matter how small $\\varepsilon$ is, you can find an $n$ such that the inequality will be true, (and more importantly it will be true for $n, n+1, n+2, \\ldots$. Does this help? \u2013\u00a0Git Gud Oct 19 '13 at 18:15\nOh yes! Just like our original inequality is contrived, we still can find an $n$ large enough for it to be true, and then let that large $n$ be $N$. So we simply introduce $|2n|$, assume it to be less than $\\varepsilon n^2$, and instead use that to find a large $N$ since $|2n-1|<|2n|$. Correct? \u2013\u00a0Andrey Kaipov Oct 19 '13 at 18:33\nSpot on.${{{}}}$ \u2013\u00a0Git Gud Oct 19 '13 at 18:39\n\nI'll give you another way to think about it. Instead of putting $\\epsilon$ there from beginning, my approach is always to simply find an upper bound for $|a_n - L|$, and then look at it to figure it out which conditions put over $n$ assure that this bound is less than $\\epsilon$.\n\nYour sequence is $(a_n)$ where $a_n = (2n-1)/n^2$. Now, we want to bound $|a_n - 0|$ and look at that bound to see what $n$ must be bigger than in order to make that less than $\\epsilon$.\n\nSince $n \\geq 1$ we have that $2n - 1 > 0$ and so $|a_n| = a_n$, in other words, we can look just to $(2n-1)/n^2$. Now, this is the same as\n\n$$\\dfrac{2n-1}{n^2}=2\\dfrac{1}{n}-\\dfrac{1}{n^2},$$\n\nNow, look that $-1/n^2 < 0$ always, so that\n\n$$\\dfrac{2n-1}{n^2} <\\dfrac{2}{n}$$\n\nNow, what must $n$ be greather than in order to make that less than $\\epsilon$. Of course, if $n > 1/\\epsilon$, then $1/n < \\epsilon$, however there's a $2$ there, so that to take care of it we make $n > 2/\\epsilon$, now when we divide by $n$ we get $1/n < \\epsilon/2$, and hence the condition follows. So, given $\\epsilon > 0$, if $n > 2/\\epsilon$ we have that\n\n$$\\left|\\dfrac{2n-1}{n^2}-0\\right|=\\dfrac{2n-1}{n^2}=\\dfrac{2}{n}-\\dfrac{1}{n^2}<\\dfrac{2}{n} <\\epsilon$$\n\nand hence $a_n \\to 0$ when $n\\to \\infty$.\n\n-\nRight so we simply set the simpler expression, in this case the upper bound $\\frac{2}{n}$, to be less than epsilon, because that will make $\\frac{2n-1}{n^2} < \\epsilon$. Then once we solve for $n$ in terms of $\\epsilon$, we know that is what $n$ should be given some $\\epsilon$ to make the inequality true. In this case your $N$ would be $\\frac{2}{\\epsilon}$, which is interesting to see, since it's different than the $N$ in the original post. My understanding of this has definitely improved. Thank you. \u2013\u00a0Andrey Kaipov Oct 19 '13 at 19:11\n\nYour professor is not concluding that $|2n|<\\varepsilon n^2$, but is instead noting that it is sufficient to have $|2n|<\\varepsilon n^2.$\n\nIt's certainly not true for all $n,$ but when it is true, you will have $\\left|\\frac{2n-1}{n^2}\\right|<\\varepsilon.$ In particular, it is true whenever we have $n>\\frac2\\varepsilon.$\n\n-", "date": "2015-11-25 06:28:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/532367/prove-convergence-using-varepsilon-n-definition", "openwebmath_score": 0.9278778433799744, "openwebmath_perplexity": 124.78814511232665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126463438263, "lm_q2_score": 0.8962513759047848, "lm_q1q2_score": 0.8771725559010674}}
{"url": "https://math.stackexchange.com/questions/740998/how-to-find-rational-numbers-satisfying-the-binary-quadratic-equation-x23xy5", "text": "# how to find rational numbers satisfying the binary quadratic equation $x^2+3xy+5y^2=4$\n\nI am looking for a generalisation of the solution of $x,y$ wich are rational numbers,they could be infinite,how can i find such solutions,integer solutions are obvious\n\nI have found that\n\n$\\Delta=-11,u=1,d=-11$\n\nalso\n\n$(x+\\frac{3+\\sqrt{-11}y}{2})(x+\\frac{3-\\sqrt{-11}y}{2})=4$\n\n\u2022 By obvious you mean Pell equations? \u2013\u00a0chubakueno Apr 5 '14 at 16:45\n\u2022 yes ,i am trying to say in this case i am not interested in integer solutions,but rational \u2013\u00a0Jonas Kgomo Apr 5 '14 at 16:52\n\nWe have the simple solution $x=2$, $y=0$. Now take the line though $(2,0)$ with slope $m$, and find where it meets the ellipse. That will give a rational parametrization of the ellipse.\n\nDetails: The line has equation $y=m(x-2)$. Substitute. We get $x^2+3xm(x-2)+5m^2(x-2)^2=16$. This simplifies to $(1+3m+5m^2)x^2-(6m+20m^2)x+20m^2-4=0$. The product of the roots is $\\frac{20m^2-4}{1+3m+5m^2}$. But one of the roots is $2$, so the other is $\\frac{10m^2-2}{1+3m+5m^2}$. Now we can compute the corresponding $y$. Any rational $m$ will give us a rational solution of the original equation.\n\nRemark: The procedure was general. In particular, let $ax^2+bxy+cy^2=d$ be an ellipse with $a,b,c,d$ rational. If we know a rational point on the ellipse, then we can find a parametric expression for all rational points. The most important special case is the circle.\n\nAnd the basic idea generalizes, importantly, to elliptic curves.\n\n\u2022 Your remark is what makes your answer better than mine. But I would like to know: Is there an efficient method to get a rational point? Can we know when does it exist, or some bound on the denominator and numerator? \u2013\u00a0chubakueno Apr 5 '14 at 17:52\n\u2022 Yes, there is a huge amount of theory of ternary quadratic forms (it is useful to homogenize, and consider $ax^2+bxy+cy^2=4z^2$.) The details are complicated, they go back to Legendre and Gauss, and are a large chapter in the history of number theory. \u2013\u00a0Andr\u00e9 Nicolas Apr 5 '14 at 17:59\n\u2022 i am actually interested in that,more about binary and not tenary ,can i say that solution is for both rational and integers \u2013\u00a0Jonas Kgomo Apr 5 '14 at 18:07\n\u2022 The integer solutions of the ternary (homogenized) form are in a simple way connected to the rational solutions of the original equation. \u2013\u00a0Andr\u00e9 Nicolas Apr 5 '14 at 18:10\n\u2022 i would appreciate if both of you can help me with this as well math.stackexchange.com/questions/739752/\u2026 \u2013\u00a0Jonas Kgomo Apr 5 '14 at 18:14\n\nThis is a variation of Hecke's answer $$(2x+3y)^2+11y^2=4^2$$ $$\\left(\\frac{2x}y+3\\right)^2+11=\\left(\\frac4y\\right)^2$$ $$11=\\left(\\frac4y\\right)^2-\\left(\\frac{2x}y+3\\right)^2$$ $$11=\\left(\\frac{4-2x}y-3\\right)\\left(\\frac{4+2x}y+3\\right)$$ $$\\frac{4-2x}y-3=t, \\frac{4+2x}y+3=\\frac{11}t$$ $$\\implies x = -\\frac{2(t^2+6t-11)}{t^2+11}, y = \\frac{8t}{t^2+11}, t\\neq0$$\n\nMy comment contained an imprecision. This works for every equation that can be transformed into the form: $$(ax+by)^2+cy^2=\\left(\\frac pq\\right)^2$$ Where $a,b,c,p,q\\in\\mathbb Z, q\\neq 0$\n\n\u2022 this is excellent,it seems like it will absolutely work for any binary equation \u2013\u00a0Jonas Kgomo Apr 5 '14 at 18:11\n\u2022 is $t$ an integer \u2013\u00a0Jonas Kgomo Apr 5 '14 at 18:16\n\u2022 @Jonas12: $t$ is any rational number. \u2013\u00a0TonyK Apr 5 '14 at 18:18\n\u2022 @Jonas12 Actually we were quite lucky :) .When we completed the square in the LHS multiplying by $4$, $4^2$ resulted to be a square. If you change that $4$ by $5$(or any non-square number) in your equation we would have a problem. In particular, this works with the general equation $ax^2+bxy+cy^2=d$, when $d$ is a perfect square. Andr\u00e9 Nicolas solutions helps us here if it is not, if we can find a rational solution. \u2013\u00a0chubakueno Apr 5 '14 at 18:23\n\u2022 the manner of the solution proposed by the proffesor,is in the manner of the solution on page 47-48 for the problem $x^2+3xy-5y^2=65$ i am wondering if this approach and @Andre Nicolas solutions will be acceptable \u2013\u00a0Jonas Kgomo Apr 5 '14 at 18:42", "date": "2021-02-27 01:09:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/740998/how-to-find-rational-numbers-satisfying-the-binary-quadratic-equation-x23xy5", "openwebmath_score": 0.7205826044082642, "openwebmath_perplexity": 361.68960155699244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126500692716, "lm_q2_score": 0.8962513682840824, "lm_q1q2_score": 0.8771725517815249}}
{"url": "https://mathhelpboards.com/threads/system-of-equations.8432/", "text": "# System of equations\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nSolve in real numbers the system of equations\n\n$(3a+b)(a+3b)\\sqrt{ab}=14$\n\n$(a+b)(a^2+14ab+b^2)=36$\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nThe first can be \"reduced\" to a fifth degree polynomial equation and the second to a third degree equation.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nSolve in real numbers the system of equations\n\n$(3a+b)(a+3b)\\sqrt{ab}=14$\n\n$(a+b)(a^2+14ab+b^2)=36$\nLet $x = a+b$, $y = \\sqrt{ab}$. Then the equations become $$y(3x^2 + 4y^2) = 14,$$ $$x(x^2 + 12y^2) = 36.$$ Multiply the first equation by 2, so they become $$x^3 + 12xy^2 = 36,$$ $$6x^2y + 8y^3 = 28.$$ Add them to get $(x+2y)^3 = 36+28=64$, and subtract them to get $(x-2y)^3 = 36-28 = 8.$ Take the cube roots: $x+2y=4$, $x-2y=2.$ Thus $x=3$, $y=1/2$. So $a+b=3$, $ab = 1/4.$ Hence $a,b$ are the roots of the equation $\\lambda^2 - 3\\lambda + 1/4 = 0$, namely $\\frac12(3\\pm\\sqrt8).$ (Clearly $a$ and $b$ are interchangeable, so those values can be taken either way round.)\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nThe first can be \"reduced\" to a fifth degree polynomial equation and the second to a third degree equation.\nHi HallsofIvy,\n\nAccording to the guidelines thread for posting in this subforum:\n\nThis forum is for the posting of problems and puzzles which our members find challenging, instructional or interesting and who wish to share them with others. As such, the OP should already have the correct solution ready to post in the event that no correct solution is given within at least 1 week's time...Responses to these topics should be limited to attempts at a full solution, or requests for clarification addressed to the OP if the problem statement is vague or ambiguous.\nWhen people post problems here in this subforum, they are not seeking help, but rather posting problems for the enjoyment of and challenge to the members of MHB. I do however appreciate the fact that you are trying to help. So, go ahead and have fun with these problems if you like and post full solutions.\n\nLet $x = a+b$, $y = \\sqrt{ab}$. Then the equations become $$y(3x^2 + 4y^2) = 14,$$ $$x(x^2 + 12y^2) = 36.$$ Multiply the first equation by 2, so they become $$x^3 + 12xy^2 = 36,$$ $$6x^2y + 8y^3 = 28.$$ Add them to get $(x+2y)^3 = 36+28=64$, and subtract them to get $(x-2y)^3 = 36-28 = 8.$ Take the cube roots: $x+2y=4$, $x-2y=2.$ Thus $x=3$, $y=1/2$. So $a+b=3$, $ab = 1/4.$ Hence $a,b$ are the roots of the equation $\\lambda^2 - 3\\lambda + 1/4 = 0$, namely $\\frac12(3\\pm\\sqrt8).$ (Clearly $a$ and $b$ are interchangeable, so those values can be taken either way round.)\nWhat a nice skill and great solution, Opalg! Thanks for participating!", "date": "2020-09-26 11:49:19", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/system-of-equations.8432/", "openwebmath_score": 0.8201313614845276, "openwebmath_perplexity": 572.6499851585461, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405995242328, "lm_q2_score": 0.8856314753275017, "lm_q1q2_score": 0.8771653693809016}}
{"url": "https://math.stackexchange.com/questions/1817944/integrating-exponent-with-polynomial", "text": "# Integrating exponent with polynomial\n\nhttp://i.stack.imgur.com/4tXNr.jpg\n\n$e^{x^2/2}\\int e^{-x^2/2}(-x^3+x)\\ dx$ turns out to be equal to $e^{x^2/2}[e^{-x^2/2}(x^2+1)]$\n\nIs there a easier method of integrating such functions? I can't grasp how text book was able to integrate it so easily. They dont show the steps, but rather go straight to integrated function. How would you tackle such integration? Integration by parts of these fuctions proven to be tedious and time consuming as they require another integration by parts to be performed.\n\nSorry for the link, it said I need 10 rep points to post the photo\n\n\u2022 I don't understand what you're asking. If you have a product of functions of a single variable, you pretty much need to integrate by parts unless one of the functions is the derivative of the other. \u2013\u00a0Edward Evans Jun 8 '16 at 1:32\n\u2022 for such product of functions, only method possible is integration by parts? \u2013\u00a0Sysnaptic Jun 8 '16 at 1:36\n\u2022 Look up \"Reduction Formulae\". With products of trigonometric functions and exponentials, you'll end up in an endless loop. \u2013\u00a0Edward Evans Jun 8 '16 at 1:41\n\u2022 You'll need to post in Latex, otherwise your post is likely to get down voted and/or closed. See here for formatting help. \u2013\u00a0mattos Jun 8 '16 at 1:44\n\nHopefully you recognize $e^{-x^2}$ as a function which can not be integrated using regular functions. As such using 'by parts' as your first step with your parts being the exponential and the polynomial will be unsuccessful. Note that the derivative of $x^2$ contains $x$ and the second part of your integral has this as a factor so:\n\n$$\\int e^{-x^2/2}(-x^3+x)\\ dx=\\int e^{-x^2/2}(-x^2+1)x\\ dx$$\n\nLet $u=\\frac{x^2}{2}$ so $du=x\\ dx$.\n\n$$=\\int e^{-u}(-2u+1)du$$\n\n$$=\\int e^{-u}\\ du-2\\int u e^{-u}\\ du$$\n\nThen use 'by parts' on the second integral\n\n$$=-e^{-u} - 2\\left(-u e^u-\\int -e^u\\ du\\right)$$\n\n$$=-e^{-u} +2u e^{-u}+2e^{-u} +c$$\n\n$$=2u e^{-u}+e^{-u} + c$$\n\n$$=x^2e^{x^2/2}+e^{-x^2/2}+c$$\n\n$$=e^{x^2/2}(x^2+1)+c$$\n\n\u2022 Wow! Thank you so much for your answer. This answers everything I was curious about \u2013\u00a0Sysnaptic Jun 8 '16 at 2:55\n\nIf, as in this case, the terms of the polynomial are all odd, the substitution $u = x^2$ works.\n\n$$J = \\int e^{-x^2/2} (-x^3 + x)\\; dx = \\int e^{-u/2} \\dfrac{-u+1}{2}\\; du$$\n\nNext, using the method of undetermined coefficients, a form for the antiderivative should be\n\n$$J = e^{-u/2} (a u + b) + C$$\n\nTaking the derivative, we need $$\\dfrac{dJ}{du} = e^{-u/2} \\left(-\\frac{a u + b}{2} + a\\right) = e^{-u/2} \\dfrac{-u+1}{2}$$ Thus $a = 1$ and $a-b/2 = 1/2$, so $b = 1$. The result is\n\n$$J = e^{-u/2} \\left(u + 1 \\right) + C= e^{-x^2/2} \\left( x^2 + 1\\right) + C$$", "date": "2020-01-20 06:31:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1817944/integrating-exponent-with-polynomial", "openwebmath_score": 0.909262478351593, "openwebmath_perplexity": 349.53431046732567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769092358046, "lm_q2_score": 0.899121375242593, "lm_q1q2_score": 0.8771620522870149}}
{"url": "https://stats.stackexchange.com/questions/499788/can-you-write-a-geometric-random-variable-as-some-combination-of-bernoulli-rando", "text": "# Can you write a Geometric random variable as some combination of Bernoulli random variables?\n\nBackground\n\nGiven $$Y \\sim \\text{Binomial(n,p)}$$, we can write $$Y = \\sum_{i=1}^{n} X_i$$ where $$X_1,X_2,...,X_n$$ are iid $$\\text{Bernoulli}(p)$$. This is useful in, for example, determining the mean of a binomial random variable: $$E(Y)=E\\left(\\sum X_i\\right) = \\sum E(X_i) = np$$\n\nQuestion\n\nIf we are given $$Y \\sim \\text{Geometric(p)}$$, can we similarly write $$Y$$ as some combination of Bernoulli random variables?\n\n\u2022 It would have to be an infinite combination, wouldn't it? By \"combination\" do you mean a sum of independent variables or would you include more general operations? If so, what would they be?\n\u2013\u00a0whuber\nDec 7 '20 at 20:14\n\u2022 I was trying to think about how to do it with a sum, but I couldn't make it work. So I decided to put \"combination\" instead. Yes, I would include more general operations. I'm not sure what they would be... I was thinking we need some way to mathematically \"keep track\" of whether we have gotten a success, because once we get a success, we stop the Bernoulli trials. But I'm not sure what that would look like. I think if we did have some kind of infinite sequence of operations and the \"keep track\" operation automatically becomes 0 once we get a success, it might work? Dec 7 '20 at 20:19\n\u2022 Trivially, every discrete distribution is a mixture of (shifted or scaled) Bernoulli distributions. A less trivial result is that the geometric distribution cannot be expressed as the sum of independent Bernoulli distributions, as you have already figured out.\n\u2013\u00a0whuber\nDec 7 '20 at 20:25\n\nFor clarity, I am going to look at the version of the geometric distribution with support on the non-negative integers, with expected value $$\\mathbb{E}(Y) = (1-p)/p$$. Now, suppose we have a sequence of Bernoulli random variables $$X_1,X_2,X_3,... \\sim \\text{IID Bern}(p)$$ to use for the construction.\n\nThe geometric random variable $$Y$$ can be interpreted as the number of \"failures\" that occur before the first \"success\", so it can be written as:\n\n\\begin{align} Y &\\equiv \\max \\ \\{ y = 0,1,2,... | X_1 = \\cdots = X_{y} = 0 \\} \\\\[12pt] &= \\max \\Bigg\\{ y = 0,1,2,... \\Bigg| \\prod_{\\ell = 1}^{y} (1-X_\\ell) = 1 \\Bigg\\} \\\\[6pt] &= \\sum_{i=1}^\\infty \\prod_{\\ell = 1}^{i} (1-X_\\ell). \\\\[6pt] \\end{align}\n\nThis is probably the \"simplest\" you can write the expression, since it accords with the descriptive intuition of what the random variable represents. As whuber notes in the comments, every discrete distribution can be written as a mixture of a countable number of shifted or scaled Bernoulli random variables (and indeed, there are an infinite number of ways to do this).\n\n(Note: For the other version of the geometric distribution, with support on the positive integers, the random variable can be interpreted as the number of trials that occur by the time of the first \"success\", which is one more than the number of \"failures\". In this case you just add one to the above expression to get construct the random variable.)\n\nConfirming the result: To see that this expression is adequate, note that:\n\n\\begin{align} F_Y(y) \\equiv \\mathbb{P}(Y \\leqslant y) &= 1 - \\mathbb{P}(Y \\geqslant y+1) \\\\[12pt] &= 1 - \\mathbb{P} \\Bigg( \\prod_{\\ell = 1}^{y+1} (1-X_\\ell) = 1 \\Bigg) \\\\[6pt] &= 1 - \\prod_{\\ell = 1}^{y+1} \\mathbb{P}(1-X_\\ell = 1) \\\\[6pt] &= 1 - \\prod_{\\ell = 1}^{y+1} \\mathbb{P}(X_\\ell = 0) \\\\[6pt] &= 1 - \\prod_{\\ell = 1}^{y+1} (1-p) \\\\[6pt] &= 1 - (1-p)^{y+1}, \\\\[6pt] \\end{align}\n\nwhich is the CDF of the (chosen version of the) geometric distribution.\n\n\u2022 This is cool! It looks like we can compute $E(Y)$ using your result, although I'm not sure about one step. $$E(Y) = E\\left(\\sum_{i=0}^{\\infty} \\prod_{l=0}^{i} (1-X_i)\\right)=\\sum_{i=0}^{\\infty} \\prod_{l=0}^{i}E(1-X_i)=\\sum_{i=0}^{\\infty}(1-p)^i=\\frac{1-p}{p}$$ We can move the expectation inside the product because of independence. The only thing I'm not sure about is moving the expectation inside an infinite sum. I think it's justified because (with probability 1) the tail terms of the sum are 0. Dec 7 '20 at 22:56\n\u2022 You can move an expectation inside the infinite sum via the linearity property. You can then move it inside the product via independence. There are some errors in your working (e.g., wrong subscripts, etc). It should be: $$\\mathbb{E}(Y) = \\mathbb{E} \\Bigg( \\sum_{i=1}^\\infty \\prod_{\\ell = 1}^{i} (1-X_\\ell) \\Bigg) = \\sum_{i=1}^\\infty \\prod_{\\ell = 1}^{i} \\mathbb{E} (1-X_\\ell) = \\sum_{i=1}^\\infty (1-p)^i = \\frac{1-p}{1-(1-p)} = \\frac{1-p}{p}.$$\n\u2013\u00a0Ben\nDec 7 '20 at 23:01\n\u2022 Yes, I see that now. Thanks! Dec 7 '20 at 23:06", "date": "2021-10-28 11:20:19", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/499788/can-you-write-a-geometric-random-variable-as-some-combination-of-bernoulli-rando", "openwebmath_score": 0.988593578338623, "openwebmath_perplexity": 277.0061129559329, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682488058383, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8771510223228876}}
{"url": "https://www.physicsforums.com/threads/optimization-rectangle-inscribed-in-a-right-triangle.729032/", "text": "# Optimization - rectangle inscribed in a right triangle\n\n1. Dec 18, 2013\n\n### ghostanime2001\n\n1. The problem statement, all variables and given/known data\nA rectangle is to be inscribed in a right triangle having sides 3 cm, 4 cm and 5 cm, as shown on the diagram. Find the dimensions of the rectangle with greatest possible area.\n\n2. Relevant equations\n1. $x^{2}+y^{2}=w^{2}$ in terms of $w=\\sqrt{x^{2}+y^{2}}$\n\n2. $\\dfrac{x}{3}=\\dfrac{y}{4}$\n\n3. $\\dfrac{l}{3}=\\dfrac{4-y}{5}$\n\n4. $\\dfrac{l}{x}=\\dfrac{4-y}{w}$\n\n3. The attempt at a solution\n\nArea of the rectangle is $A=lw$. But, we want an expression for area in terms of one variable.\n\nrewriting equation 2 for $x$ we can get an expression in $y$\n\n$4x=3y$\n\n$x=\\dfrac{3y}{4}$\n\nwhich we can substitute in equation 1 to find width $w$ in terms of $y$\n\n$w=\\sqrt{x^{2}+y^{2}}$\n\n$w=\\sqrt{\\left(\\dfrac{3y}{4}\\right)^{2}+y^{2}}$\n\n$w=\\sqrt{\\dfrac{9}{16}y^{2}+y^{2}}$\n\n$w=\\sqrt{\\dfrac{25}{16}y^{2}}$\n\n$w=\\dfrac{5}{4}y$\n\nUsing similar triangles of the upper right, smaller right triangle and the bigger, outer triangle we set a relationship between the base of smaller triangle and the base of the outer triangle, and, the hypotenuse of the smaller triangle and the hypotenuse of the outer triangle. We solve for the length of the rectangle in terms of $y$ using equation 3.\n\n$\\dfrac{l}{3}=\\dfrac{4-y}{5}$\n\n$5l=3\\left(4-y\\right)$\n\n$l=\\dfrac{3\\left(4-y\\right)}{5}$\n\nWe now have the length $l$ and width $w$ in terms of one variable $y$. We substitute the expressions in the area of the rectangle to find the area in one variable $y$\n\n$A\\left(y\\right)=l\\left(y\\right)w\\left(y\\right)$\n\n$A\\left(y\\right)=\\dfrac{3\\left(4-y\\right)}{5}\\dfrac{5}{4}y$\n\n$A\\left(y\\right)=\\dfrac{15y\\left(4-y\\right)}{20}$\n\n$A\\left(y\\right)=\\dfrac{3y\\left(4-y\\right)}{4}$\n\n$A\\left(y\\right)=\\dfrac{12y-3y^{2}}{4}$\n\n$A\\left(y\\right)=3y-\\dfrac{3}{4}y^{2}$\n\nDifferentiating this expression and setting $A'\\left(y\\right)=0$ gives $0=3-\\dfrac{3}{2}y$\n\nsolving for $y$ gives\n\n$\\dfrac{3}{2}y=3$\n\n$y=2$\n\nsubstitute $y=2$ in the length $l=\\dfrac{3\\left(4-y\\right)}{5}$ and width $w=\\dfrac{5}{4}y$ equations respectively,\n\n$l=\\dfrac{3\\left(4-\\left(2\\right)\\right)}{5}=\\dfrac{6}{5}$\n\n$w=\\dfrac{5}{4}\\left(2\\right)=\\dfrac{5}{2}$\n\nTherefore, the rectangle with dimensions $l=\\dfrac{6}{5}=\\text{1.2 cm}$ and $w=\\dfrac{5}{2}=\\text{2.5 cm}$ has the maximum possible area of $\\text{3 }cm^{2}$\n\n*above I used the more tedious way I guess...\n\nMy question is if I can do it using equation 4 since $l$ and $w$ are already there and cross multiplying would give me area automatically. I would have an expression in $x$ and $y$ but I can use equation 2 to find either $x$ or $y$ and set up the area in one variable. Would this way be the more correct method or give the same answer as above? Thanks !\n\n#### Attached Files:\n\n\u2022 ###### Untitled.png\nFile size:\n64.2 KB\nViews:\n186\n2. Dec 18, 2013\n\n### scurty\n\nThat works! I get an answer of 3. Did you try it out and see how the answers compared? You can use the proportion $\\frac{x}{3} = \\frac{y}{4}$ to get the equation in terms of only x or y and then take the derivative, etc.\n\nAs for more \"correct,\" every method that arrives at a solution is correct, but this way might be more elegant because it only takes 5 lines or so of equations.\n\n3. Dec 18, 2013\n\n### Ray Vickson\n\nIt is a lot easier to use\n$$\\frac{l}{4-y} = \\frac{3}{5} \\Longrightarrow l = (3/5)(4-y)\\\\ \\frac{w}{y} = \\frac{5}{4} \\Longrightarrow w = (5/4)y$$\ngiving $A = l w = (3/4) y(4-y) = 3 y - (3/4)y^2.$\n\n4. Dec 19, 2013\n\n### ghostanime2001\n\nRay Vickson, I had overlooked this relation! Your relation directly solves $w$ in terms of $y$ avoiding going through using pythagorean theorum and equation 2. Am I right ? or am I right ?\n\nThe first relation is the same one I acquired. It is just written differently.\n\n$\\dfrac{w}{y}=\\dfrac{5}{4}\\Rightarrow w=\\dfrac{5}{4}y$\n\nscurty, I have used equation 4 and equation 2 to solve for area $lw$ in terms of $x$ (instead of $y$ as previously used) and I get\n\n$A=lw=4x-\\dfrac{4}{3}x^{2}$\n\ncompared to $A=lw=3y-\\dfrac{3}{4}y^{2}$\n\n5. Dec 19, 2013\n\n### scurty\n\nI can't tell if you are just acknowledging that the method worked (if so, good job!) or you are confused. Sorry, haha, could you just clear that up for me?\n\n6. Dec 19, 2013\n\n### ghostanime2001\n\nthe area equation in terms of $x$ is correct, no ? I just want to know if I did my algebra right. I substituted the value of $x=\\dfrac{3}{2}=\\text{1.5 cm}$ into the area equation and it comes out to $\\text{3 }cm^{2}$. Do you have the same numbers & answer ?\n\n7. Dec 19, 2013\n\n### scurty\n\nYes, it's correct! There's more than one way to arrive at the solution.", "date": "2017-10-21 12:34:08", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/optimization-rectangle-inscribed-in-a-right-triangle.729032/", "openwebmath_score": 0.6997645497322083, "openwebmath_perplexity": 314.10866079218124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682454669814, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8771510046176376}}
{"url": "https://math.stackexchange.com/questions/890495/a-sigma-notation-but-with-multiplication-instead-of-addition", "text": "# A sigma notation but with multiplication instead of addition?\n\nI am not a mathematician, so I apologize if this question will sound stupid. I am wondering is there some sort of notation which will resemble the one of sigma notation, but with multiplication instead of addition?\n\nFor example, the following expression: $2^1+2^2+2^3+2^4+2^5$ could be written as: $\\sum_{i = 1} ^ 5 2^i$\n\nBut what about the following expression: $2^1*2^2*2^3*2^4*2^5$ ?\n\nIs there some notation like sum sigma that could enable me to write this last expression a bit shorter?\n\nThank you very much.\n\n\u2022 $\\prod$ $$\\prod_{k=1}^5 2^k$$ \u2013\u00a0Daniel Fischer Aug 7 '14 at 22:19\n\u2022 This may already be obvious to you, but I'll leave it here anyway: S is for sum, P is for product. Sigma and pi are the Greek letters for S and P. \u2013\u00a0Rahul Aug 7 '14 at 22:51\n\u2022 Amusingly, you can do without a new notation, because $2^1\\cdot2^2\\cdot2^3\\cdot2^4\\cdot2^5$ is $$2^{1+2+3+4+5}=2^{\\sum_{i=1}^5i}.$$ \u2013\u00a0Yves Daoust Aug 7 '14 at 23:02\n\nYou can use Pi notation (The $\\LaTeX$ for this being \\prod)\n$$\\prod\\limits_{i = 1}^5 2^i = 2^1 \\cdot 2^2 \\cdot 2^3 \\cdot 2^4 \\cdot 2^5$$", "date": "2019-02-17 00:11:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/890495/a-sigma-notation-but-with-multiplication-instead-of-addition", "openwebmath_score": 0.9592295289039612, "openwebmath_perplexity": 501.7826646737515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682469694671, "lm_q2_score": 0.8872045877523147, "lm_q1q2_score": 0.8771510044763497}}
{"url": "http://mathhelpforum.com/statistics/199623-red-marbles-white-marbles-bag.html", "text": "# Thread: Red marbles, white marbles, in a bag.\n\n1. ## Red marbles, white marbles, in a bag.\n\nSo,\nI have a bag containing 10 blue marbles and 5 red marbles. If I draw 8 marbles without replacement, what is the probability of me getting 1 red marbles out of 8?\n2 out 8; 3; 4;5?\nHow do you calculate that?\n\nAdditionally, how would you calculate the probability of my getting at least 1 red marbles out of 8? At least 2; 3; 4; 5?\nAnd how do you calculate that?\n\nThanks\n\n2. ## Re: Red marbles, white marbles, in a bag.\n\nHello, Sabre!'\n\nHere's the first half of your question . . .\n\nI have a bag containing 10 blue marbles and 5 red marbles.\nIf I draw 8 marbles without replacement, what is the probability of getting:\n\n(a) 1 red marble and 7 blue marbles?\nThere are: . ${15\\choose8} \\,=\\,6,\\!435$ possible outcomes.\n\nTo get 1 red and 7 blue: . ${5\\choose1}{10\\choose7} \\:=\\:5\\cdot 120 \\:=\\:600$ ways.\n\nTherefore: . $P(\\text{1 red}) \\:=\\:\\frac{600}{6,\\!435} \\:=\\:\\frac{40}{429}$\n\n(b) 2 red marbles and 6 blue marbles?\nTo get 2 red and 6 blue: . ${5\\choose2}{10\\choose6} \\:=\\:10\\cdot210 \\:=\\:2,\\!100$ ways.\n\nTherefore: . $P(\\text{2 red}) \\:=\\:\\frac{2,\\!100}{6,\\!435} \\:=\\:\\frac{140}{429}$\n\n(c) 3 red marbles and 5 blue marbles?\nTo get 3 red and 5 blue: . ${5\\choose3}{10\\choose5} \\:=\\:10\\cdot252 \\:=\\:2,\\!520$ ways.\n\nTherefore: . $P(\\text{3 red}) \\:=\\:\\frac{2,\\!520}{6,\\!435} \\:=\\:\\frac{56}{143}$\n\n(d) 4 red marbles and 4 blue marbles?\nTo get 4 red and 4 blue: . ${5\\choose4}{10\\choose4} \\:=\\:5\\cdot210 \\:=\\:1,\\!050$ ways.\n\nTherefore: . $P(\\text{4 red}) \\:=\\:\\frac{1,\\!050}{6,\\!435} \\:=\\:\\frac{70}{429}$\n\n(e) 5 red marbles and 3 blue marbles?\nTo get 5 red and 3 blue: . ${5\\choose5}{10\\choose3} \\:=\\:1\\cdot120 \\:=\\:120$ ways.\n\nTherefore: . $P(\\text{5 red}) \\:=\\:\\frac{120}{6,\\! 435} \\:=\\:\\frac{8}{429}$\n\n3. ## Re: Red marbles, white marbles, in a bag.\n\nThank you Soroban,\n\nQuestions\nOriginally Posted by Soroban\nThere are: . ${15\\choose8} \\,=\\,6,\\!435$ possible outcomes.\nAbove, when you calculate the outcome, what you are calculating is the number of ways, right? A.k.a \u201cnumber of combinations of n things taken x at a time\u201d with\n, yes?\n\nBut does it take into account that there are marbles of two different colors here?\nI'm really quite at loss, so sorry if I'm totally wrong and stupid.\n\n4. ## Re: Red marbles, white marbles, in a bag.\n\nOriginally Posted by Sabre\nAbove, when you calculate the outcome, what you are calculating is the number of ways, right? A.k.a \u201cnumber of combinations of n things taken x at a time\u201d with , yes?\nBut does it take into account that there are marbles of two different colors here?\nI'm really quite at loss, so sorry if I'm totally wrong and stupid.\nSuppose that $r$ is the number of red balls in the sample, $0\\le r\\le 5$.\nThen $8-r$ is the of blue balls in the sample. Right?\n\n$\\mathcal{P}(r)=\\dfrac{\\dbinom{5}{r}\\dbinom{10}{8-r}}{\\dbinom{15}{8}}$\n\n5. ## Re: Red marbles, white marbles, in a bag.\n\n\"Additionally, how would you calculate the probability of my getting at least 1 red marbles out of 8? At least 2; 3; 4; 5?\nAnd how do you calculate that\n\nThe simplest way to do that is to look at the opposite. If you don't get \"at least one red marble\", you must have gotten all 8 blue. There are a total of 15 marbles, 10 of which are blue. The probabability the first marble is blue is 10/15= 2/3. There are now 14 marbles, 9 of them blue so the probability the second marble is also blue is 9/14. Continuing in that way, the probability all 8 are blue is (10/15)(9/14)(8/14)...(2/7)(1/6) which we could write as (10!)(15- 10)!/(15!)= (10!)(5!)/(15!). The probability of \"at least one red marble\" is 1 minus that:\n1- (10!)(5!)/(15!).\n\n6. ## Re: Red marbles, white marbles, in a bag.\n\nI think I understood. I mean I can place the logic in my head now. Let's see if I got this right.\n\nIn a bag with 10 white marbles and 3 red marbles I take 7 without replacement:\nFinding the probabilities of getting 1 red marble means comparing the number of ways I can obtain exactly 1 red marble and 6 white marbles (desirable outcome) when drawing 7 marbles to all possible outcomes when drawing 7 marbles.\n\nA. First I calculate the number of all possible combinations when taking 7 out of 13.\n\n${13\\choose7} \\,=$ (13!)/(7!)(13-7)!= 1,716 possible combinations\n\nThen I calculate the number of ways I can obtain my desirable outcome of 1 red and 6 white, which is the number of combinations for obtaining 1 red out of 3 multiplied by the number of combinations for 6 out of 10 white.\n\n${3\\choose1}{10\\choose6} \\:=$((3!)/(1!)(3-1)!)*((10!)/(6!)(10-6)!) = (3)(210) = 630 ways of obtaining the desirable outcome\n\nThen I calculate P which is\n\n$\\mathcal{P}(1 red)=\\:\\frac{630}{1716}=\\:\\frac{3}{8}=\\:\\0,37$\n__________________________________________________ __________________________________________________ ________________\n\nB. To find the probability of 2 reds marbles we have:\n\n$\\mathcal{P}(2 red)=\\dfrac{\\dbinom{3}{2}\\dbinom{10}{5}}{\\dbinom{1 3}{7}}=\\:\\frac{252}{1716}=\\:\\0,15$\n\nYeah, thanks again everybody, I was confused about the notion of 'number of ways' which is actually only the number of combinations for a particular outcome.\n\n7. ## Re: Red marbles, white marbles, in a bag.\n\nOriginally Posted by Sabre\nI think I understood. I mean I can place the logic in my head now. Let's see if I got this right.\nIn a bag with 10 white marbles and 3 red marbles I take 7 without replacement:\nFinding the probabilities of getting 1 red marble means comparing the number of ways I can obtain exactly 1 red marble and 6 white marbles (desirable outcome) when drawing 7 marbles to all possible outcomes when drawing 7 marbles.\nA. First I calculate the number of all possible combinations when taking 7 out of 13.\n[texto ]{13\\choose7} \\,=[/tex] (13!)/(7!)(13-7)!= 1,716 possible combinations\nThen I calculate the number of ways I can obtain my desirable outcome of 1 red and 6 white, which is the number of combinations for obtaining 1 red out of 3 multiplied by the number of combinations for 6 out of 10 white.\n${3\\choose1}{10\\choose6} \\:=$((3!)/(1!)(3-1)!)*((10!)/(6!)(10-6)!) = (3)(210) = 630 ways of obtaining the desirable outcome\nThen I calculate P which is\n$\\mathcal{P}(1 red)=\\:\\frac{630}{1716}=\\:\\frac{3}{8}=\\:\\0,37$\n__________________________________________________ __________________________________________________ ________________\nB. To find the probability of 2 reds marbles we have:\n$\\mathcal{P}(2 red)=\\dfrac{\\dbinom{3}{2}\\dbinom{10}{5}}{\\dbinom{1 3}{7}}=\\:\\frac{252}{1716}=\\:\\0,15$\nYeah, thanks again everybody, I was confused about the notion of 'number of ways' which is actually only the number of combinations for a particular outcome.\nWhy the h_ll did you change the question in midstream?\nIt seems to someone who has taught this for many years, this is nonsense.\nYou have no idea what any of this means!\n\n8. ## Re: Red marbles, white marbles, in a bag.\n\nDid I get the answers right?\nWhat do you mean I changed the question midstream?\nIf you mean why I changed the number of marbles, it was to see if I could to the calculation.\nApparently not?\n\nPlease could you make your reply more useful. I really don't have any idea what you mean.", "date": "2017-09-21 18:46:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/199623-red-marbles-white-marbles-bag.html", "openwebmath_score": 0.8643590807914734, "openwebmath_perplexity": 594.0990940281878, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682474702956, "lm_q2_score": 0.8872045840243196, "lm_q1q2_score": 0.8771510012349366}}
{"url": "https://math.stackexchange.com/questions/2149112/to-prove-that-a-mathematical-statement-is-false-is-it-enough-to-find-a-counterex/2149128", "text": "# To prove that a mathematical statement is false is it enough to find a counterexample?\n\nI am trying to show if the following statements are true or false.\n\n1. Is it true that $|a + b| = |a| + |b|$ for general vectors $a$ and $b$?\n\n2. If $a \\cdot b = a \\cdot c$ for equally-sized non-zero vectors $a$, $b$, $c$, does it follow that $b = c$?\n\nFor the first one I found a counterexample that shows that the statement is false.\n\nIf vector $a=\\langle 1,4,5\\rangle$ and $b=\\langle 2,2,2\\rangle$ then $|a|+|b|=\\sqrt{42}$+$2\\cdot\\sqrt{3}=9.945$, and then, $|a+b|=\\sqrt{1^2+4^2+5^2+2^2+2^2+2^2}=7.348$,\n\nthen we can conclude that $9.945 \\ne7.348$ and the statement is false. Also by the triangle identity $|a + b| \\le |a| + |b|$\n\nFor the second statement I also found an counterexample that proves that is false. If vector $a=\\langle 1,0,0\\rangle$, $b=\\langle 0,1,0\\rangle$ and $c=\\langle 0,0,1\\rangle$, we will obtain the following dot product:\n\n$a \\cdot b = 0$\n\n$a \\cdot c = 0$\n\nthen $a \\cdot b = a \\cdot c = 0$ and $b \\ne c$\n\nMy question is:\n\nTo prove the two statements is it enough to find a counterexample and say if it is true or false. Or should I try to provide a more mathematical proof like induction or contradiction?\n\n\u2022 It suffices to find a counterexample to show that a statement is not true. \u2013\u00a0Student Feb 17 '17 at 18:26\n\u2022 Yes, this is enough. The only thing that is unclear is your statement \"Also by the triangle identity $|a + b| = |a| \\le |b|$\". First, did you mean $|a + b| - |a| \\le |b|$? Second, what do you mean to prove by this? \u2013\u00a0Th\u00e9ophile Feb 17 '17 at 18:29\n\u2022 Okay, I see it was a typo. But more importantly, you should realize that citing the triangle inequality is not enough to answer the question, because it doesn't in itself contradict statement (1). The triangle inequality should suggest to you that you can probably find $a,b$ such that $|a+b| < |a|+|b|$, which would contradict statement (1), but you don't need to mention the inequality. Showing that counterexample (as you did) is the real proof. (I just noticed that @zipirovich effectively made the same comment about this in their answer below.) \u2013\u00a0Th\u00e9ophile Feb 17 '17 at 19:27\n\u2022 In response to Einsteins theory of relativity, a book titled A Hundred Authors Against Einstien had been written. When asked about the book, Einstein retorted by saying \u201cWhy 100 authors? If I were wrong, then one would have been enough!\u201d \u2013\u00a0steven gregory Feb 17 '17 at 22:45\n\u2022 @palsch: I think you mean \"example that refutes an assertion or claim\". \u2013\u00a0John Bentin Feb 19 '17 at 12:47\n\nWhen considering a statement that claims that something is always true or true for all values of whatever its \"objects\" or \"inputs\" are: yes, to show that it's false, providing a counterexample is sufficient, because such a counterexample would demonstrate that the statement it not true for all possible values. On the other hand, to show that such a statement is true, an example wouldn't be sufficient, but it has to be proven in some general way (unless there's a finite and small enough number of possibilities so that we can actually check all of them one after another).\n\nSo logically speaking, for these two specific examples, you're right \u2014 each one can be demonstrated to be false with an appropriate counterexample. And both your counterexamples do work, but make sure that the math supporting your claim is right: in the first example you computed $|a+b|$ incorrectly.\n\nBy the way, the reference to the triangle inequality is a good touch, but it doesn't prove anything. Rather, it's a very strong hint that suggests that there have got to be examples when the inequality rather than equality holds.\n\n\u2022 So, is it not correct to mention the triangle inequality in this case? \u2013\u00a0user290335 Feb 17 '17 at 19:00\n\u2022 Here's my opinion on problem-solving in general, proof questions in particular. First there's the process of looking for a solution or a proof, thinking about it -- and in this example, the triangle inequality is an important hint suggesting that the statement is false. But then we write up a final narrative of a solution or proof, without any sidetracking notes and such -- and for this problem the triangle inequality wouldn't be there. \u2013\u00a0zipirovich Feb 17 '17 at 19:04\n\u2022 In short, saying that this statement is false by the triangle inequality would not qualify as a proof. The triangle inequality per se does not claim that vectors with $|a+b|<|a|+|b|$ exist. It claims that $|a+b|\\le|a|+|b|$ for all $a$ and $b$. Note that logically even if \"$|a+b|=|a|+|b|$\" were true for all $a$ and $b$, that wouldn't contradict the triangle inequality. \u2013\u00a0zipirovich Feb 17 '17 at 19:07\n\u2022 I wouldn't even call the triangle inequality a strong hint. I am sure part of the purpose in giving this assignment was to convince the students that the triangle inequality cannot be strengthened into an equality, by having them find counterexamples themselves. \u2013\u00a0Paul Sinclair Feb 17 '17 at 22:34\n\u2022 @PaulSinclair: I partially agree with you, and partially don't. I agree that that's part of the purpose of the exercise. But what I meant was that from a student's perspective reasoning could go like this: I need to prove or disprove \"$|a+b|=|a|+|b|$\" -> they taught us the triangle inequality saying that $|a+b|\\le|a|+|b|$ -> why didn't they make it equals? -> either they are not telling the truth, or in fact it's not always equals and thus had to be less or equal -> so I guess I should be looking for a counterexample. \u2013\u00a0zipirovich Feb 17 '17 at 23:37\n\nIf a statement ${\\cal S}$ is of the form \"all $x\\in A$ have the property $P$\" then a single $x_0\\in A$ not having the property $P$ proves that the statement ${\\cal S}$ is wrong.\n\nBut not all statements are of this form. For example the statement ${\\cal S}\\!:\\>$\"$\\pi$ is rational\" cannot be disproved by some \"easy\" counterexample, but only by means of hard work.\n\nYes, any counterexample will do. It's often instructive to look for the simplest counterexample. For example, take the one-dimensional vector space $\\Bbb R$ and the vectors $a=1$ and $b=-1$ in the case of the first statement.\n\nIf the statement is true, then you give a mathematical proof. Since you can't find all feasible inputs to prove that the statement is true, even computers can't do this for some statements.\n\nIf the statement is false, then you give one counter example. Since the statement says that it is true for all the feasible inputs, you just have to find one feasible input which doesn't satisfy the statement.\n\nAs for your solutions, everything's seem correct with two mistakes. $a+b$ in the first example is <3,6,7> and triangle inequality is $|a+b| \\leq |a|+|b|$.\n\n\u2022 I did a mistake. I edited my question. Thank you for your help. \u2013\u00a0user290335 Feb 17 '17 at 18:41\n\u2022 Glad to be of help @user290335 \u2013\u00a0rookie Feb 18 '17 at 10:10\n\u2022 Someone downvoted this. Can I know what to improve in this answer? \u2013\u00a0rookie Feb 18 '17 at 10:10\n\nAs other answers have explained, if you have a claim that something is true for all possible inputs, then a single counterexample disproves the claim. Period.\n\nPerhaps, though, you may have some lingering doubt about how \"generic\" the counterexample is. Sure, maybe we can prove that $\\lvert a+b \\rvert = \\lvert a \\rvert + \\lvert b \\rvert$ is false for $a=(1,4,5)$ and $b=(2,2,2)$. However, for all we know, this might be the only counterexample. We already know the claim is false (beyond any doubt), but maybe it's only \"technically\" false. Maybe it's true \"in the typical case\", and we just happened to find the only exception.\n\nFor example, suppose I claim \"$x^4 + y^4 \\ne z^4$ for all integers $x$, $y$, and $z$\". You say, \"That's not true: $0^4 + 1^4 = 1^4$\". You'd be right: Indeed my claim is incorrect. But if you're willing to let me move the goal posts a little, I can easily salvage the claim by excluding \"trivial\" counterexamples like $0^4 + 1^4 = 1^4$. The claim remains true \"in the typical case\" - \"technically\" false but true \"in spirit\".\n\nMost of the time you won't see people explicitly address how \"generic\" their counterexample is. They'll just give one counterexample and be done. And implicitly, they're saying that this one counterexample is convincing enough that the claim is \"generally\" false. Because:\n\nWhen a statement is false, it's \"generally\" false... generally.\n\nThis belief is completely non-rigorous and not formally justified, but that's okay - there was no formal meaning for \"typical case\", \"generally true/false\", and \"trivial\" to begin with.\n\nHow could we make a more convincing argument that a statement is not only false, but \"generally\" false?\n\nDepending on the situation, we might try to show that:\n\n\u2022 There are infinitely many counterexamples (not just one)\n\u2022 There are infinitely many counterexamples that aren't just obvious variations of each other (e.g. not just constant multiples of each other)\n\u2022 The set of counterexamples has infinite volume\n\u2022 A random input (drawn from a chosen probability distribution) is a counterexample with probability $1$\n\n... the list goes on. In the most fortunate case, we can find a definite answer by describing exactly which inputs satisfy the claim and which ones don't. For example, it turns out that \"$\\lvert a+b \\rvert = \\lvert a \\rvert + \\lvert b \\rvert$\" is true if and only if $a$ and $b$ are positive scalar multiples of each other, or one or both of them is $0$. This is a rigorous statement which is stronger than merely disproving \"$\\lvert a+b \\rvert = \\lvert a \\rvert + \\lvert b \\rvert$ for all $a$ and $b$\", and can be informally interpreted as saying that \"$\\lvert a+b \\rvert = \\lvert a \\rvert + \\lvert b \\rvert$\" is \"usually\" false.\n\nThese are the ways you could go beyond proving that a claim is false, to prove that it's \"very\" false.\n\nBut enough talk about false statements being \"generally false\" or \"only technically false\". It's still false. All such talk involves moving the goal posts, replacing the original statement with a different one.\n\nFor whatever actual, specific claim you might be considering, one counterexample is enough to disprove it.", "date": "2020-09-29 17:33:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2149112/to-prove-that-a-mathematical-statement-is-false-is-it-enough-to-find-a-counterex/2149128", "openwebmath_score": 0.6994355320930481, "openwebmath_perplexity": 332.0637759001828, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808690122163, "lm_q2_score": 0.894789461192692, "lm_q1q2_score": 0.877144990600945}}
{"url": "https://math.stackexchange.com/questions/2448243/question-about-distributive-laws-of-logic-and-proof-of-the-symmetric-difference", "text": "# Question about Distributive Laws of Logic and proof of the Symmetric Difference\n\nHere is the definition of the Symmetric Difference: $$A \\Delta B = (A \\backslash B) \\cup (B \\backslash A),$$ Our goal is to prove: $$A \\Delta B = (A \\cup B) \\backslash(A \\cap B).$$\n\nI want to prove this and at the same time to check if my understanding of the Distributive Laws of Logic is correct.\n\nMy main goal is about Distributive Laws of Logic. I want to know if I can always break the parenthesis with this method.\n\nI know that there are similar proofs in the website but I feel that I need help with the Distributive Laws.\n\nPlease check if this is correct:\n\nWe start with\n\n$(x \\in A$ $\\land$ $x \\notin B)$ $\\vee$ $(x \\notin A$ $\\land$ $x \\in B)$\n\nThis is the part that I want to know if I am correct, and I want to know if I can do this for any parenthesis regarding of what is inside of the parenthesis of the right part:\n\n$(x \\in A$ $\\land$ $x \\notin B)$ $\\vee$ x $\\notin$ A ) $\\land$\n\n$(x \\in A$ $\\land$ $x \\notin B)$ $\\vee$ x $\\in$ B )\n\nNow we can use the classic version of the Distributive Law:\n\n$(x \\in A$ $\\vee$ x $\\notin$ $A$ ) $\\land$\n\n$(x \\notin B$ $\\vee$ x $\\notin$ $A$ ) $\\land$\n\n$(x \\in A$ $\\vee$ x $\\in$ $B$ ) $\\land$\n\n$(x \\notin B$ $\\vee$ x $\\in$ $B$ )\n\nNow we can simplify:\n\n******Universe****** $\\land$\n\n$(x \\notin B$ $\\vee$ x $\\notin$ $A$ ) $\\land$\n\n$(x \\in A$ $\\vee$ x $\\in$ $B$ ) $\\land$\n\n******Universe******\n\nFinally we have,\n\n$(x \\in A$ $\\vee$ x $\\in$ $B$ ) $\\land$\n\n$(x \\notin B$ $\\vee$ x $\\notin$ $A$ )\n\nThe last step is to use the Morgan Laws\n\n$(x \\in A$ $\\vee$ x $\\in$ $B$ ) $\\land$\n\n$\\neg$ $(x \\in A$ $\\land$ $x$ $\\in$ $B$ )\n\nThat is our goal:\n\nQ.E.D.\n\n\u2022 I just happened to see this! I suppose 'pinging' me in a comment to your own question does not work ... anyway, yes, your proof is fine! And yes, you can use Distribution on part of a statement just as you are using DeMorgan on just part of the statement in your last step. Sep 28 '17 at 3:42\n\u2022 @Bram28 Thank you so much again for your amazing teachings! Sep 28 '17 at 14:50\n\u2022 You're welcome! And by the way, I definitely see your progress with this material!! :) Sep 28 '17 at 14:52\n\u2022 @Bram28 I have been using your teachings about Latex and Mathematical Logic. You are fantastic! Thank you so much! Sep 28 '17 at 15:16\n\u2022 :) I appreciate your kind words ... but you are doing the learning! Sep 28 '17 at 15:17\n\n## 1 Answer\n\nYes, that is a permissible application of distribution.\n\nIt is basically using the proof for the quadratic distribution: $$(s\\wedge t)\\vee (u\\wedge v)~{= (s\\vee (u\\wedge v))\\wedge(t\\vee (u\\wedge v))\\\\ =(s\\vee u)\\wedge(s\\vee v)\\wedge(t\\vee u)\\wedge (t\\vee v)}$$\n\nTake an arbitrary $x$ such that $x\\in (S\\cap T)\\cup(U\\cap V)$. \u00a0 By definition of union and disjunction, this is equivalent to $(x\\in S\\wedge x\\in T)\\vee(x\\in U\\wedge x\\in V)$. \u00a0 By distribution, that is equivalent to $(x\\in S\\vee(x\\in U\\wedge x\\in V))\\wedge (x\\in T\\vee(x\\in U\\wedge x\\in V))$, and distributing again that is $(x\\in S\\vee x\\in U)\\wedge (x\\in S\\vee x\\in V)\\wedge (x\\in T\\vee x\\in U)\\wedge (x\\in T\\vee x\\in V)$. Thus it is equivalent to $x\\in(S\\cup U)\\cap (S\\cup V)\\cap (T\\cup U)\\cap (T\\cup V)$.\n\nMore succinctly, using the set algebra:\n\n$$(S\\cap T)\\cup(U\\cap V)~{=(S\\cup(U\\cap V))\\cap(T\\cup(U\\cap V))\\\\=(S\\cup U)\\cap (S\\cup V)\\cap (T\\cup U)\\cap (T\\cup V)}$$\n\nJust substitute the relevant sets.\n\nIf we presume complementation relative to some superset $\\mathcal U$ (the \"universe\") of both $A,B$, then your proof is essentially:\n\n\\def\\sm{\\smallsetminus} {\\begin{align} &\\quad (A\\Delta B) \\\\ &= (A\\sm B)\\cup(B\\sm A) &&\\text{definition of symmetric difference} \\\\& = (A\\cap B^\\complement)\\cup(B\\cap A^\\complement) &&\\text{definition of set minus} \\\\& = (A\\cup (B\\cap A^\\complement))\\cap(B^\\complement\\cup (B\\cap A^\\complement)) && \\text{distribution} \\\\& = (A\\cup B)\\cap(A\\cup A^\\complement)\\cap (B^\\complement\\cup B)\\cap (B^\\complement\\cup A^\\complement)&&\\text{distribution} \\\\& = (A\\cup B)\\cap\\mathcal U\\cap\\mathcal U\\cap (B^\\complement\\cup A^\\complement)&&\\text{complementation} \\\\&= (A\\cup B)\\cap(B^\\complement\\cup A^\\complement) && \\text{identity} \\\\&= (A\\cup B)\\cap(A^\\complement\\cup B^\\complement) && \\text{commutivity} \\\\&= (A\\cup B)\\cap(A\\cap B)^\\complement & & \\text{deMorgan's Rule} \\\\&= (A\\cup B)\\sm(A\\cap B) && \\text{definition of set minus}\\end{align}}\n\n\u2022 Thank you so much! Amazing answer! I like very much the idea to treat the problem with Set Laws. In that way, the solution can be reduced to few steps. Sep 29 '17 at 16:45\n\u2022 I am using your teaching to apply set rules instead of logic rules and it reduces the work to be done at least 50%. Thank you so much! Oct 2 '17 at 21:28", "date": "2022-01-24 18:00:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2448243/question-about-distributive-laws-of-logic-and-proof-of-the-symmetric-difference", "openwebmath_score": 0.87132328748703, "openwebmath_perplexity": 623.0923660164833, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808684361289, "lm_q2_score": 0.8947894541786199, "lm_q1q2_score": 0.8771449832097074}}
{"url": "http://math.stackexchange.com/questions/187233/is-the-statement-1-3-of-the-natural-numbers-are-divisible-by-3-true-is-anythi/187234", "text": "# Is the statement \u201c1/3 of the natural numbers are divisible by 3\u201d true? Is anything similar to it true?\n\nIf we're talking about a finite set of the natural numbers, like those between 1 and 500 or 1 and a million, it seems to me that the fraction of numbers in that finite set that have a factor of 5 approaches $1/5$ as the set increases in size. Like roughly $1/2$ of all numbers in such a set have a factor of 2, roughly $1/3$ have a factor of 3, and so on; and this approximation grows less \"rough\" and more exact as the size of the set increases.\n\nSo, can we say that out of the entire set of the natural numbers, exactly $1/5$ are divisible by 5? Or perhaps that the limit of the fraction of the natural numbers less than or equal to a given n divisible by a given integer approaches 1/that integer as n approaches infinity?\n\n(I would love to know how to ask this question with proper notation.)\n\n-\nThere was a similar thread some time ago, see math.stackexchange.com/questions/101768/\u2026 and especially the first answer. \u2013\u00a0 Gregor Bruns Aug 26 '12 at 21:12\nAm I correct in assuming that this question has already been answered in enough forms that it isn't a helpful addition to the site? I don't mind deleting if so; the idea of \"asymptotic density\" is something I didn't know about until today, and now that I can learn about that I at least have something to work with. How do I delete, if I ought to? \u2013\u00a0 Annick Aug 26 '12 at 21:14\nYours is certainly a good question and it doesn't seem a duplicate to me. If I didn't knew the title of the linked question I probably wouldn't have been able to find it. I was merely referencing so that you can benefit from the answers there. Please keep asking. \u2013\u00a0 Gregor Bruns Aug 26 '12 at 21:27\n@Annick Now that there are several answers, I'd just leave it. If several people want to close it as a duplicate, we can, but it'll still exist for others to see. \u2013\u00a0 Graphth Aug 26 '12 at 21:27\n@Annick I would just like to add that this is a very well-asked question, and that I'm sure you will get positive responses in the future if you continue to ask questions in this manner. \u2013\u00a0 Alex Becker Aug 26 '12 at 21:31\n\nThis can indeed be made formal. To formalize the statement \"$x$ fraction of natural numbers satisfy the property $P$\", we define the function $$f(n)=\\text{ number of natural numbers }\\leq n\\text{ which satisfy }P$$ and write $\\lim\\limits_{n\\to \\infty} \\frac{f(n)}{n}=x$. In your first case, the function $f$ is given by $f(n)=\\lfloor n/3\\rfloor$ and the statement becomes $$\\lim\\limits_{n\\to\\infty} \\frac{\\lfloor n/3\\rfloor}{n}=\\frac{1}{3}$$ which is easily seen to be true, since $\\frac{1}{3}-\\frac{1}{n}\\leq \\frac{\\lfloor n/3\\rfloor}{n}\\leq \\frac{1}{3}$. Similar results hold for any natural number in place of $k$.\n\n-\nBe careful, however: mathematically speaking, the cardinality of numbers divisible by 3 is equal to the cardinality of all numbers! These are two very different counting mechanisms being used. \u2013\u00a0 akkkk Aug 26 '12 at 21:39\nholy god, what the :/ \u2013\u00a0 orokusaki Aug 27 '12 at 1:13\nThis formal statement is exactly the definition of natuarl density or asymptotic density. \u2013\u00a0 Ken Bloom Aug 27 '12 at 2:54\n\nLook up natural density or asymptotic density.\n\n-\n\nHere are the positive integers divisible by $3$: $$3,6,9,12,15,18,21,\\ldots$$ Here are those not divisible by $3$: $$1,2,4,5,7,8,10,11,\\ldots$$ Now suppose we just alternate between the first list and the second: $$\\begin{array}{} 3 & & & & 6 & & & & 9 & & & & 12 & & & & 15 & & & & \\cdots\\cdots \\\\ & \\searrow & & \\nearrow & & \\searrow & & \\nearrow & & \\searrow & & \\nearrow & & \\searrow & & \\nearrow & & \\searrow & & \\nearrow \\\\ & & 1 & & & & 2 & & & & 4 & & & & 5 & & & & 7 \\end{array}$$\n\nThen we could argue in the same way that half of all positive integers are divisible by $3$.\n\nIt does make sense to say $1/3$ of them are divisible by $3$, understanding that statement in a certain context, but defining that context is something that will bear examination.\n\n-\nWhy does it not make sense? It doesn't make sense in the sense of cardinality, but it makes perfect sense in the sense used in the other answers. \u2013\u00a0 joriki Aug 26 '12 at 23:00\nSorry: typo. I've fixed it. \u2013\u00a0 Michael Hardy Aug 26 '12 at 23:16\nI think a slightly more formal way to say this would be, \u201cthe cardinality of the natural numbers divisible by 3 is the same as the cardinality of the natural numbers indivisible by 3.\u201d \u2013\u00a0 bdesham Aug 27 '12 at 3:03\n@bdesham : That may be more formal but it's not exactly the same thing. Cardinality is only one related concept; there are others. \u2013\u00a0 Michael Hardy Aug 27 '12 at 3:21\n....in particular; I didn't intend the arrows to indicate a one-to-one correspondence, but only the order in which the numbers appear in a sequence. \u2013\u00a0 Michael Hardy Aug 27 '12 at 3:22\n\nIf one computes the fraction of positive integers from $1$ to $m$ that are a multiple of $n$, we get a sequence whose limit is $\\frac{1}{n}$ as $m\\to \\infty$. If we ask about the size of the set of positive integers which are multiples of $n$ compared to that of all positive integers, the answer is that they are the same in some sense as they are both countably infinite.\n\n-", "date": "2014-11-29 03:50:54", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/187233/is-the-statement-1-3-of-the-natural-numbers-are-divisible-by-3-true-is-anythi/187234", "openwebmath_score": 0.7919934988021851, "openwebmath_perplexity": 244.45884836204402, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9525741241296944, "lm_q2_score": 0.9207896840422138, "lm_q1q2_score": 0.8771204267841699}}
{"url": "https://math.stackexchange.com/questions/3668542/erase-every-other-number-from-1-2-3-2000-until-only-one-number-left", "text": "# Erase every other number from $1, 2, 3, \u2026 2000$ until only one number left\n\nOne writes $$1, 2, 3, ... 2000$$ on the blackboard.\n\nWe erase every other number from left, so after one iteration we are left with $$2, 4, ... 2000$$\n\nThen we erase every other number from right..\n\nWe keep repeating to erase from left and right one after another, eventually there is only one number left.\n\nWhat would be that number?\n\nI wrote a program, unless there is a bug, the answer should be 982. But is there a way to mathematically induce the results, and generalize to arbitrary number (not just 2000?)\n\nFollowing the suggestion, it seems like for power of 2, the answers go like this\n\n$$N=2$$, final number is 2\n\n$$N=4$$, final number is 2\n\n$$N=8$$, final number is 6\n\n$$N=16$$, final number is 6\n\n$$N=32$$, final number is 22\n\n$$N=64$$, final number is 22\n\n$$N=128$$, final number is 86\n\n$$N=256$$, final number is 86\n\nThe answer will jump to 4 times previous answer minus 2, for every time we multiply $$N$$ by 4...\n\n\u2022 To start, I 'd work out what the final number was with lower caps (here, $2000$ is the cap). See if you can spot a pattern. \u2013\u00a0lulu May 10 '20 at 19:59\n\u2022 as always, do the problem by hand with smaller sets. 1,2,3 to start. Then 1,2,3,4. There is an odd/even aspect to this, and 2000 is 16 times 125, so you might get the first similar behavior for 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16. \u2013\u00a0Will Jagy May 10 '20 at 19:59\n\u2022 I'd start by looking at what happens if you write the numbers in base $2$. \u2013\u00a0JonathanZ supports MonicaC May 10 '20 at 20:10\n\u2022 See the relevant OEIS sequence for formula. \u2013\u00a0Momo May 10 '20 at 20:18\n\u2022 @Blue thanks! quite a fun read! \u2013\u00a0Vlad Zkov May 10 '20 at 23:50\n\nHere's a step, at least, toward a solution.\n\nLet $$S(N)$$ denote the \"surviving\" number as a function of $$N$$. Now after the first pass from left to right, only the even numbers from $$2$$ up to $$2\\lfloor N/2\\rfloor$$ remain. Thus $$S(N)$$ satisfies the recursion\n\n$$S(N)=2(\\lfloor N/2\\rfloor+1-S(\\lfloor N/2\\rfloor))$$\n\nwith $$S(1)=1$$ (i.e., once only one person, that person survives). Let's see first how this works for a power of $$2$$:\n\n\\begin{align} S(32)&=2(17-S(16))\\\\ &=34-4(9-S(8))\\\\ &=-2+8(5-S(4))\\\\ &=38-16(3-S(2))\\\\ &=-10+32(2-S(1))\\\\ &=54-32\\\\ &=22 \\end{align}\n\nNow let's look at $$N=2000$$:\n\n\\begin{align} S(2000)&=2(1001-S(1000))\\\\ &=2002-4(501-S(500))\\\\ &=-2+8(251-S(250))\\\\ &=2006-16(126-S(125))\\\\ &=-10+32(63-S(62))\\\\ &=2006-64(32-S(31))\\\\ &=-42+128(16-S(15))\\\\ &=2006-256(8-S(7))\\\\ &=-42+512(4-S(3))\\\\ &=2006-1024(2-S(1))\\\\ &=2006-1024\\\\ &=982 \\end{align}\n\nas the OP found. Whether the recursive formula can be simplified to an explicit closed formula I'll leave to someone else.\n\nHere is the complete solution, using the excellent recurrence of Barry Cipra.\n\nConsider the $$k$$-bit-long binary representation of $$N = b_1 b_2 b_3 \\dots b_k$$ where $$b_1$$ is the most significant bit.\n\nFor convenience define $$B_j = b_1 b_2 \\dots b_j = \\lfloor N / 2^{k-j}\\rfloor$$, i.e. the number formed by the most significant $$j$$ bits. Note that $$B_k = N$$ and $$B_1 = b_1 = 1$$.\n\nNext we \"unwind\" the recurrence a bit:\n\n\\begin{align} S(N) = S(B_k) &= 2(B_{k-1} + 1 - S(B_{k-1}))\\\\ &=2(B_{k-1} + 1) - 2 [ 2(B_{k-2} + 1 - S(B_{k-2}) ] \\\\ &=2(B_{k-1} + 1) - 4(B_{k-2} + 1) + 4 S(B_{k-2}) \\end{align}\n\nA nice thing now happens: if I may abuse notation a bit and mix up numbers with their binary representations (in red),\n\n$$2B_{k-1} - 4B_{k-2} = \\color{red}{b_1 b_2 \\dots b_{k-2} b_{k-1} 0 - b_1 b_2 \\dots b_{k-2} 0 0 = b_{k-1}0} = 2b_{k-1}$$\n\nwhere the binary representations (in red) have some appended $$0$$s at the end to represent multiplication by $$2$$ or $$4$$. So now we can unwind faster:\n\n\\begin{align} S(B_k) &= 2(b_{k-1} -1) + 4S(B_{k-2}) \\\\ &=2(b_{k-1} -1) + 4[2(b_{k-3} - 1) + 4 S(B_{k-4})]\\\\ &=2(b_{k-1} -1) + 8(b_{k-3} - 1) + 2^4 S(B_{k-4})\\\\ &=2(b_{k-1} -1) + 8(b_{k-3} - 1) + 32(b_{k-5}-1) + \\dots \\end{align}\n\nNote that the subscript $$j$$ of the remaining $$S(B_j)$$ term decreases by $$2$$ for each unwinding step. So what happens at the end of all the unwinding can be case-analyzed:\n\nFor odd $$k$$: the last remaining term will be $$S(B_1) = S(b_1) = S(b_1) = b_1 = 1$$:\n\n$$S(B_k) = \\sum_{j = 1, 3, 5...} 2^j (b_{k-j} - 1) + 2^{k-1} S(B_1) = \\sum_{j = 1, 3, 5...} 2^j (b_{k-j} - 1) + 2^{k-1} b_1$$\n\nInterpretation: Based on the way I number the bits, $$2^j b_{k-j}$$ is the $$b_{k-j}$$ bit evaluated at its correct place-position. So the formula is equivalent to this recipe:\n\nRecipe: Take the most significant bit's place-value (blue), and consider the $$2$$nd, $$4$$th, $$6$$th, etc least significant bits (red), i.e. the bits for place-values $$2^1, 2^3, 2^5,$$ etc. If a bit is $$0$$, then subtract its place-value.\n\nE.g. $$2000 = 11111010000 = \\color{blue}{1}\\color{red}{1}1\\color{red}{1}1\\color{red}{0}1\\color{red}{0}0\\color{red}{0}0 \\implies S(2000) = 1024 - 2 - 8 - 32 = 982$$\n\nE.g. $$64 = 1000000 = \\color{blue}{1}\\color{red}{0}0\\color{red}{0}0\\color{red}{0}0 \\implies S(64) = 64 - 2 - 8 - 32 = 22$$\n\nFor even $$k$$: the last remaining term will be $$S(B_2)$$. Luckily, $$B_2 \\in \\{2, 3\\}$$ and in both cases $$S(B_2) = S(2) = S(3) = 2$$.\n\n$$S(B_k) = \\sum_{j = 1, 3, 5...} 2^j (b_{k-j} - 1) + 2^{k-2} S(B_2) = \\sum_{j = 1, 3, 5...} 2^j (b_{k-j} - 1) + 2^{k-1} b_1$$\n\nSo surprisingly, the same recipe works! The only clarification is that since the most significant bit is also an even-numbered least-significant bit (because $$k$$ is even), it must not be included in the subtraction.\n\nE.g. $$32 = 100000 = \\color{blue}{1}0\\color{red}{0}0\\color{red}{0}0 \\implies S(32) = 32 - 2 - 8 = 22$$\n\nFurther remarks:\n\n\u2022 It is curious that only half the bits matter. E.g. the recipe immediately shows $$S(8) = S(1000_2) = S(13) = S(1101_2)$$.\n\n\u2022 Barry's recurrence is already $$\\log(N)$$ time, so my recipe is not faster. However, it does give an explicit summation.\n\n\u2022 The explicit summation, especially in the form of the recipe, allows easy proofs of the OP's observation that $$S(2^k)$$ values come in pairs, and $$S(2^{k+2}) = 4S(2^k) - 2$$.\n\n\u2022 Thanks that's great answer! \u2013\u00a0Vlad Zkov May 14 '20 at 2:00", "date": "2021-06-18 17:46:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3668542/erase-every-other-number-from-1-2-3-2000-until-only-one-number-left", "openwebmath_score": 0.9603107571601868, "openwebmath_perplexity": 707.3348650020821, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990291521799369, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8770333294982026}}
{"url": "https://math.stackexchange.com/questions/2783000/recreational-math-if-ffx-ex-bound-the-integral-int-01-fxdx", "text": "# Recreational math: If $f(f(x))=e^x$, bound the integral $\\int_0^1 f(x)dx$\n\nI've been studying functions $f:\\mathbb R\\to\\mathbb R$ that satisfy $f(f(x))=e^x$ (or, half-iterates of the exponential function). I know that there's only one such analytic function, but it's really hard to study since it is almost certainly non-elementary and I only know how to find finitely many terms of its Maclaurin Series.\n\nInstead, I'm studying all continuous and increasing functions $f$ satisfying $f(f(x))=e^x$, and I've alighted on the following problem (which I came up with out of curiosity). I propose this question to all interested residents of MSE:\n\nGiven that $f$ is continuous and increasing and $f(f(x))=e^x$, find some bounds for the integral $$\\int_0^1 f(x)dx$$\n\nI've managed to come up with some pretty sweet bounds (in fact, they are the best possible bounds), and I'll post them after this question gets some answers.\n\nI'll accept whatever answer has the tightest bounds, with a proof.\n\nNOTE: Most people probably wouldn't think of this as recreational \"fun\" math, but hey, I did it for fun, and I'm proposing it as a problem to be done just for fun. So please try to enjoy it, and please don't try to close it.\n\n\u2022 Anyone care to explain the close vote? \u2013\u00a0Frpzzd May 15 '18 at 23:43\n\u2022 Well, puzzle problems like this aren't really what MSE is for, though I've seen such questions before and enjoyed them greatly. That being said, this seems like an interesting question, so I'm definitely not going to vote to close :) - Brevan, from Close Votes Review \u2013\u00a0Brevan Ellefsen May 16 '18 at 3:33\n\u2022 To me, the most interesting thing here isn't the question itself, but your proof technique for showing your bounds are tightest. Was it simply due to construction, or did you simply play around till you found really good bounds and then found a juicy proof? \u2013\u00a0Brevan Ellefsen May 16 '18 at 3:38\n\u2022 @BrevanEllefsen Oh, I found the juicy proof right off the bat using graphical techniques and functional equations. \u2013\u00a0Frpzzd May 16 '18 at 14:24\n\nSuppose $f(x)$ is an increasing, continuous function satisfying $f(f(x))=e^x$. Then there must be some $\\alpha\\in(0,1)$ such that $f(\\alpha)=1$. We can compute \\begin{align*} \\int_\\alpha^1 f(x)\\,dx&=\\int_\\alpha^1 e^{f^{-1}(x)}\\,dx\\\\ &=\\int_0^\\alpha e^u\\,df(u)\\\\ &=e^\\alpha f(\\alpha)-e^0f(0)-\\int_0^\\alpha f(u)e^u\\,du\\\\ &=e^\\alpha-\\alpha-\\int_0^\\alpha f(x)e^x\\,dx. \\end{align*} So, we have $$\\int_0^1 f(x)\\,dx=\\left(\\int_0^\\alpha+\\int_\\alpha^1\\right) f(x)\\,dx=e^\\alpha-\\alpha-\\int_0^\\alpha(e^x-1)f(x)\\,dx.$$ For $0< x< \\alpha$, we have $\\alpha< f(x)< 1$, and thus $$1=e^\\alpha-\\alpha-\\int_0^\\alpha e^x\\,dx<\\int_0^1 f(x)\\,dx<e^\\alpha-\\alpha-\\int_0^\\alpha \\alpha e^x\\,dx=(1-\\alpha)e^\\alpha+\\alpha^2.$$ For $0< \\alpha< 1$, the quantity $(1-\\alpha)e^\\alpha+\\alpha^2$ takes a maximum value of $2+\\log^2(2)-\\log(4)$ at $\\alpha=\\log(2)$, so we have bounds $$1<\\int_0^1 f(x)\\,dx < 2+\\log^2(2)-\\log(4)\\approx 1.0942.$$\n\nThese bounds are sharp: for $0\\leq x\\leq \\log(2)$, we can take $f(x)$ to be an arbitrary continuous, strictly increasing function satisfying $f(0)=\\log(2)$ and $f(\\log(2))=1$, then extend $f$ to a continuous increasing function on the entire real line satisfying $f(f(x))=e^x$ by iteratively using the relation $f(x)=e^{f^{-1}(x)}$. To make $\\int_0^1 f(x)\\,dx$ arbitrarily close to the lower bound, we can choose $f(x)$ to be close to $1$ for an arbitrarily large portion of the interval $[0,\\log(2)]$, and to make the integral arbitrarily close to the upper bound, we can choose $f(x)$ to be close to $\\log(2)$ for an arbitrarily large portion of the interval $[0,\\log(2)]$.\n\n\u2022 The exact value seems to be 1.02525, so +1 :) \u2013\u00a0Oldboy May 16 '18 at 16:18\n\u2022 I suspect this is the same as OP\u2019s answer, as he says the difference between bounds is 0.1. \u2013\u00a0Szeto May 16 '18 at 22:26\n\u2022 @JulianRosen Wowie zowie, this is exactly my answer (though I preferred $2\\ln 2$ to $\\ln 4$). This is so similar to my proof that you've essentially saved me the trouble of writing an answer (however, I do intend to demonstrate to everyone that these are the best possible bounds, unless you get around to it first). \u2013\u00a0Frpzzd May 18 '18 at 1:08\n\u2022 I have edited the answer to better explain the sharpness. \u2013\u00a0Julian Rosen May 18 '18 at 14:00\n\u2022 @JulianRosen Well done, and I so hope that you enjoyed the problem (as was the purpose of my posting it). \u2013\u00a0Frpzzd May 18 '18 at 22:54\n\n$\\int_0^1f(x)dx=1.02525$, up to to five significant digits.\n\nI tried to deal with this problem numerically by heavily using number crunching capabilities of Mathematica. I have simply supposed that:\n\n$$f(x)\\approx\\sum\\limits_{k=0}^nc_kx^k$$\n\nBasically, you can pick $n$, calculate $f(f(x))$ and make a set of $n$ equations by matching the first $n$ coefficients of $x^k$ with the coefficients from Maclauirn's expansion of $e^x$. It's a highly non-lienar problem and I focused my efforts to find only one solution numerically. The trick is to use values $c_k$ obtained for $n-1$ as a starting point to calculate $c_k$ in the next iteration.\n\nFor $n=5$ I got something like:\n\n$f(x)=0.503212 +0.884755 x+0.173988 x^2-0.0557584 x^3+0.17115 x^4+0.249233 x^5-0.159222 x^6-0.317672 x^7+0.172933 x^8+0.0335486 x^9$\n\nBlue line represents $f(x)$, orange line represents $f(f(x))$ and green line represents $e^x$\n\nBut for $n=20$ I got something \"almost\" perfect:\n\n$f(x)=-0.00110484 x^{19}+0.00423925 x^{18}-0.00285303 x^{17}-0.00653605 x^{16}+0.00430471 x^{15}+0.0114125 x^{14}-0.00188582 x^{13}-0.0167065 x^{12}-0.00616185 x^{11}+0.0170445 x^{10}+0.0165896 x^9-0.00860714 x^8-0.0207938 x^7-0.00356292 x^6+0.0140354 x^5+0.00731033 x^4+0.0207005 x^3+0.243846 x^2+0.876672 x+0.498799$\n\n...with $\\int_0^1f(x)dx=1.02526$\n\nFor $n=25$ I did not see much improvement.\n\nI stopped at $n=30$. This is my final result:\n\n$f(x)=3.19108553894\\cdot 10^{-6} x^{29}-0.0000320371 x^{28}+0.000102301 x^{27}-0.0000852623 x^{26}-0.000155577 x^{25}+0.000202985 x^{24}+0.0003114 x^{23}-0.000304959 x^{22}-0.000659598 x^{21}+0.00024257 x^{20}+0.00120893 x^{19}+0.000250339 x^{18}-0.00172616 x^{17}-0.00138061 x^{16}+0.00165885 x^{15}+0.00293783 x^{14}-0.000399194 x^{13}-0.00403302 x^{12}-0.00203198 x^{11}+0.00346346 x^{10}+0.00438265 x^9-0.00094675 x^8-0.0047584 x^7-0.00170557 x^6+0.00287885 x^5+0.00131622 x^4+0.0240847 x^3+0.246667 x^2+0.876334 x+0.498618$\n\n...with $\\int_0^1f(x)dx=1.02525$\n\nThis value did not change even for $n=40$.\n\n\u2022 I am not sure if the OP is looking fot this kind of answer. The OP looks for bounds, not a value, because the function needs not to be analytic. \u2013\u00a0Szeto May 16 '18 at 22:23\n\u2022 My bounds are $1.02525\\pm0.00001$ :) \u2013\u00a0Oldboy May 17 '18 at 4:06\n\u2022 I really appreciated your tedious computation, but I think this function itself is paradoxical(see the important edit in my answer). Can you explain or resolve the paradox? \u2013\u00a0Szeto May 17 '18 at 15:04\n\u2022 I don't get your paradox. First you are stating that $f^{-1}(a)=b$ which basically means that $a=f(b)$ which is fine. But why is $\\ln f(b)=f^{-1}(b)$? \u2013\u00a0Oldboy May 17 '18 at 20:17\n\u2022 because the fuction can be defined as $$f(x)=e^{f^{-1}(x)}$$. \u2013\u00a0Szeto May 17 '18 at 22:02\n\nIMPORTANT EDIT:\n\nI find that $f(x)$ can create some paradoxical result.\n\nAssume $0<a<1$(which is the region of interest) and $f^{-1}(a)=b$. Then $0<b<a$.\n\nThen, $$a=f(b)\\implies \\ln a=\\ln f(b)=f^{-1}(b)$$\n\nClearly, $L.H.S.\\le0$. However, we have proved that $f^{-1}(b)=R.H.S.>0$.\n\nCan anyone resolve it?\n\nEND EDIT\n\nNot a tight bound, but too long for a comment.\n\nAssume the $f(x)$ is:\n\n1. Continuous\n2. A real function on the interval $(-\\infty,\\infty)$\n1. Strictly increasing\n\nSince $f(x)$ is strictly increasing, it has a well-defined inverse $f^{-1}(x)$.\n\nLet\u2019s redefine the function as $$f(x)=e^{f^{-1}(x)}$$\n\nFirstly, since $e^x$ is positive for all real $x$, hence $f(x)$ is strictly positive.\n\nSecondly, we are interested in the fixed point of $f(x)$. For fixed points $(t,f(t))$, $$t=f(t)= f^{-1}(t)$$\n\nThus, $$t=e^t$$ which has no real solution.\n\nTherefore, we can see $y=f(x)$ has no intersection with $y=x$.\n\nWith these two observations, one can deduce $f(x)>x$ for all $x$.\n\n$$f(x)>x\\implies \\int^1_0f(x)dx>\\int^1_0xdx=\\frac12$$\n\nEDIT\n\nFrom $$f(x)>x$$ we have $$f(f(x))>f(x)$$ $$e^x>f(x)$$ $$e-1>\\int^1_0f(x)dx$$\n\nEDIT 2:\n\nI think that $f(0)$ cannot be defined(but maybe okay in the sense of limit).\n\nAssume $f^{-1}(0)=C$.\n\nIt is easy to prove $f^{-1}(x)$ is strictly increasing.\n\nThus, from $f(x)>x$, we get $x>f^{-1}(x)$. So $C<0$.\n\nWe also observe that, due to $f(x)>0$, $f^{-1}(x)$ is undefined for $x<0$. Since $f(x)$ is defined wherever its inverse is defined, so $f(x)$ is undefined in $(0,-\\infty)$.\n\n$$f^{-1}(0)=C\\implies f(C)=0$$\n\nBut $C<0$, $f(C)$ should be undefined! Also, I have proven that $f(C)>0$(above). That leads to a contradiction.\n\nSo a value of $C$ does not exist, and hence $f^{-1}(0)$ and $f(0)$ is undefined.\n\nFrom the above observations, we can also see: $$0<f^{-1}(x)<x$$.\n\n\u2022 I had this idea building on yours, to prove that $f(x) < e^x$. Suppose on the other hand that there was $x_0$ such that $f(x_0) \\geq e^{x_0}$. Then $f(x_0) \\geq e^{x_0} = f(f(x_0)) \\geq f(e^{x_0}) \\geq f(x_0)$ which means that $f(f(x_0)) = f(e^{x_0}) = e^{x_0}$, a fixed point, which as you showed doesn't exist. Hence $f(x) < e^x$. There's a good chance I have made a mistake, but it provides a fairly good upper bound if it's right? It assumes that the function is strictly increasing again. \u2013\u00a0Cataline May 16 '18 at 11:17\n\u2022 @Cataline Yeah, I came up with this idea while bathing a few minutes ago(before seeing you comment). I\u2019ve added a short, straight forward derivation of the upper bound. \u2013\u00a0Szeto May 16 '18 at 11:47\n\u2022 Indeed, your upper bounds are correct. But are they the best possible bounds? \u2013\u00a0Frpzzd May 16 '18 at 14:25\n\u2022 @Frpzzd I\u2019ve never seen ways to prove bounds to be the best. I would certainly be amazed by your proof. \u2013\u00a0Szeto May 16 '18 at 14:36\n\u2022 @Szeto Well, I don't plan to post it until this gets at least a couple more answers. \u2013\u00a0Frpzzd May 16 '18 at 14:41", "date": "2019-04-24 06:05:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2783000/recreational-math-if-ffx-ex-bound-the-integral-int-01-fxdx", "openwebmath_score": 0.875244140625, "openwebmath_perplexity": 317.77205256315204, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620073067491, "lm_q2_score": 0.8840392756357327, "lm_q1q2_score": 0.8770217783251895}}
{"url": "https://math.stackexchange.com/questions/4005950/why-does-plotting-collatz-sequences-in-polar-coordinates-produce-a-cardioid-and", "text": "# Why does plotting Collatz sequences in polar coordinates produce a cardioid and nephroid?\n\nI generated the Collatz sequences for the first 2,000 starting integers, and plotted these sequences \"on top of each other\" in polar coordinates, using a fixed radius and with each element in the sequence as theta (converted to radians, i.e., modulo 360). Successive elements within each sequence are connected by (semi-transparent) lines.\n\nFollowing these steps produced the image below, where we can clearly see two shapes (a cardioid and nephroid). (The Python code that generates this image is available here. The total number of elements in the 2,000 sequences is 136,100, so the total number of lines plotted in the image below is 136,100 - 2,000 = 134,100.)\n\nLink to generated image (high-quality) or lower quality\n\nMy question is: Why do a cardioid and nephroid clearly appear? Is there something notable about the Collatz sequences that produce these shapes, or is this unrelated to the intricacies of the Collatz sequences, and that other unrelated integer sequences also produce similar shapes?\n\nAs of now, I haven't seen any substantive references online to nephroids and the Collatz conjecture together.\n\nA post here describes how you can generate a cardioid in polar coordinates by drawing lines between evenly spaced points on a circle, but this doesn't relate to the Collatz sequences, and doesn't construct a nephroid.\n\nUpdate:\n\nBased on the discussion, the cardioid can be generated based on the envelope of lines drawn between points around a circle, with each line drawn between point line between points n and 2n (mod N), as described here. The nephroid is based on the same concept for points between n and 3n, as described here. The Collatz sequence contains both behaviors: some elements in sequence follow 2n->n (the order of the two doesn't matter if we are connecting lines), and some elements follow n->3n+1 (but the +1 component is negligible for the overall shape), resulting in the cardioid and nephroid.\n\nUpdate #2:\n\nPer a suggestion by heropup below, I've redrawn the image where for each line connecting terms (N,N+1) in a sequence, the line is red if N+1 > N and blue if N+1 < N. From this and per the update above, we can clearly see that the nephroid is red and the cardioid is blue.\n\nLink to colored generated image (high-quality) or lower quality\n\n(This now probably looks better on a white background but takes a long time to generate).\n\n\u2022 Welcome to MSE ! If only all newcomers could write questions with this quality, this is a very interesting question ! Jan 30, 2021 at 18:01\n\u2022 In such construction, you are basically joining n with 2n and 3n+1. The post you mentioned in the last paragraph explains why lines n\u20132n create a cardioid. I suppose that lines n\u20133n+1 will be responsible for nephroid. Jan 30, 2021 at 18:10\n\u2022 modular arithmetic in multiplication tables of 2 and 3 (youtube.com/watch?app=desktop&v=qhbuKbxJsk8) Jan 30, 2021 at 18:11\n\u2022 Hi @samerivertwice, for any given Collatz sequence, I convert this sequence to radians using Python's math.radians function, and treat this value as the angular coordinate \u03b8. The radial coordinate r is fixed at 1 for all values. I plot (\u03b8,r) for all values for all sequences, and connect consecutive values within a sequence with lines. Feb 1, 2021 at 2:23\n\u2022 Here's a suggestion. For each line in your figure, color it red if it is the result of the next term of the sequence going from a smaller number to a larger one, and blue if it is the result of the next term going from a larger number to a smaller one. What do you see? Feb 1, 2021 at 2:53\n\nBased on the discussion, the cardioid can be generated based on the envelope of lines drawn between points around a circle, with each line drawn between point line between points n and 2n (mod N), as described here. The nephroid is based on the same concept for points between n and 3n, as described here. The Collatz sequence contains both behaviors: some elements in sequence follow $$2n\\to n$$ (the order of the two doesn't matter if we are connecting lines), and some elements follow $$n\\to3n+1$$ (but the $$+1$$ component is negligible for the overall shape), resulting in the cardioid and nephroid.", "date": "2022-09-30 09:52:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4005950/why-does-plotting-collatz-sequences-in-polar-coordinates-produce-a-cardioid-and", "openwebmath_score": 0.8221563100814819, "openwebmath_perplexity": 735.7000469436234, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357184418847, "lm_q2_score": 0.8933094010836642, "lm_q1q2_score": 0.8769937466637607}}
{"url": "https://math.stackexchange.com/questions/2450903/moment-of-inertia-of-regular-polygonal-frames", "text": "# Moment of inertia of regular polygonal frames\n\nThe moment of inertia with respect to the orthogonal barycentral axis on the plane where it lies:\n\n\u2022 a homogeneous triangular (equilateral) frame is equal $I_C = (1)\\,\\frac{M\\,L^2}{6}$;\n\u2022 a homogeneous quadrangular frame is equal $I_C = (2)\\,\\frac{M\\,L^2}{6}$;\n\u2022 a homogeneous pentagonal (regular) frame is equal $I_C = \\left(2+\\frac{3}{\\sqrt{5}}\\right)\\frac{M\\,L^2}{6}$;\n\u2022 a homogeneous hexagonal (regular) frame is equal $I_C = (5)\\,\\frac{M\\,L^2}{6}$;\n\u2022 a homogeneous heptagon (regular) frame is equal $I_C = \\dots \\,$ boh;\n\u2022 a homogeneous octagon (regular) frame is equal $I_C = \\left(5+3\\,\\sqrt{2}\\right) \\frac{M\\,L^2}{6}$;\n\u2022 $\\dots$\n\nIs there a formula to write that moment of inertia for any regular polygon frame?\n\nI emphasize the fact that these are frames, not laminates.\n\nThank you.\n\nThanks to the answers we received, for any regular $n$-sided polygon of side $L$:\n\n\u2022 $\\text{perimeter} = n\\, L$;\n\u2022 $\\text{fixed_number} = \\frac{1}{2\\,\\tan\\left(\\frac{\\pi}{n}\\right)}$;\n\u2022 $\\text{apothem} = \\text{fixed_number}\\cdot L$;\n\u2022 $\\text{area} = \\, \\frac{\\text{perimeter}\\cdot \\text{aphotem}}{2}$;\n\u2022 $\\text{moment_inertia}_C = \\frac{M\\,L^2}{12} + M\\,\\text{apothem}^2$ (homogeneous frame of mass $M$);\n\nand for completeness (and also for curiosity) I also add:\n\n\u2022 $\\text{moment_inertia}_C = \\frac{\\frac{M}{2}\\,L^2}{12} + \\frac{M}{2}\\,\\text{apothem}^2$ (homogeneous lamina of mass $M$).\n\u2022 So what is $L$? \u2013\u00a0David G. Stork Sep 29 '17 at 22:53\n\u2022 With frames, I clearly assuming the boundary of the respective polygonal lamina. L is the length of the side. Thank you. \u2013\u00a0TeM Sep 29 '17 at 23:11\n\u2022 So the total circumference is $nL$? Please be more careful and thorough when posing questions. I'm not going to take time now to go back and fix my solution. \u2013\u00a0David G. Stork Sep 29 '17 at 23:14\n\u2022 I tried to illustrate the problem. If anyone was able to give me a hand I would be grateful to him! \u2013\u00a0TeM Sep 29 '17 at 23:37\n\u2022 And why is $C$ not the origin? What difference does the location make? None of your answers depend upon the center location. \u2013\u00a0David G. Stork Sep 29 '17 at 23:49\n\nThe moment of inertia of a bar around the perpendicular axis through the center is $\\frac{mL^2}{12}$. In your case $m=\\frac{M}{N}$. Using the parallel axis theorem, the moment of inertia of a bar with respect to an axis at distance d from the center is $$\\frac{mL^2}{12}+md^2$$ so the total moment of inertia for your system is $$I_N= \\frac{ML^2}{12}+Md_N^2$$ Now all you need to do is calculate what is the distance $d_N$ from the center of a regular polygon with $N$ sides to the side. In the case $N=3$, $d_3=\\frac{L}{2\\sqrt{3}}$, for $N=4$ $d_4=\\frac{L}{2}$ and so on. You have $$\\tan\\frac{2\\pi}{2N}=\\frac{L/2}{d}$$\n\n\u2022 Excellent, the formula collimation with the data calculated above, thank you! \u2013\u00a0TeM Sep 30 '17 at 13:17\n\nTo calculate these moments of inertia, we must simply integrate one side and multiply the result by n.\n\nIf we imagine C to be the origin, a side can be expressed as $x = \\frac{L}{2\\tan(\\frac\\pi n)}, -\\frac L2 < y<\\frac L2$. We can easily integrate to find moment of inertia of this segment:\n\n$$I_C\\cdot L = \\int^{\\frac L2}_{-\\frac L2}\\frac{M}n\\cdot D(x, y)^2dy = \\frac{M}n\\int^{\\frac L2}_{-\\frac L2}\\left(\\sqrt[]{x^2+y^2}\\right)^2dy = \\frac{M}n\\int^{\\frac L2}_{-\\frac L2}\\left(\\frac{L^2}{4\\tan^2(\\frac\\pi n)} + y^2\\right)dy\\\\ =\\frac{M}n\\left(\\frac{L^2}{4\\tan^2(\\frac\\pi n)}y + \\frac13y^3\\right)\\biggr|^{\\frac L2}_{-\\frac L2} = \\frac{M}n\\cdot\\left(\\frac{L^3}{4\\tan^2(\\frac\\pi n)} + \\frac{L^3}{12}\\right).$$\n\nSimply multiplying by $n$ yields the desired moments:\n\n$$I_C=M\\left(\\frac{L^2}{4\\tan^2(\\frac\\pi n)} + \\frac{L^2}{12}\\right)$$\n\n\u2022 Thank you too! I think there is an error: mass density is not M/L, but M/(nL); in this way the result collides with that of Andrei and in turn all the above particular cases (verifiably in literature) are embedded! \u2013\u00a0TeM Sep 30 '17 at 12:25", "date": "2019-07-20 12:30:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2450903/moment-of-inertia-of-regular-polygonal-frames", "openwebmath_score": 0.8520655035972595, "openwebmath_perplexity": 529.6366171743258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429609670702, "lm_q2_score": 0.8918110382493035, "lm_q1q2_score": 0.8769561069751872}}
{"url": "https://www.physicsforums.com/threads/u-substitution-problem.845874/", "text": "# U-substitution problem\n\n1. Nov 30, 2015\n\n### Mr Davis 97\n\n1. The problem statement, all variables and given/known data\n$\\displaystyle \\int \\frac{\\ln x}{x(1 + \\ln x)} dx$\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nLet $u = 1 + \\ln x$, then $\\ln x = u - 1$\n$\\displaystyle du = \\frac{1}{x}dx$\nThus, $\\displaystyle \\int \\frac{\\ln x}{x(1 + \\ln x)} dx = \\int \\frac{u - 1}{u} du= \\int 1 - u^{-1} du = u - \\ln u + C = 1 + \\ln x - \\ln \\left | \\ln x + 1\\right | + C$\n\nHowever, this is the wrong answer. What am I doing wrong?\n\n2. Nov 30, 2015\n\n### Ray Vickson\n\nWhy do you think it is wrong?\n\n3. Nov 30, 2015\n\n### Mr Davis 97\n\nBecause my answer book says that the answer should be $\\ln x - \\ln | \\ln x + 1 | + C$ I have a 1 in there, which is where the answer differs.\n\n4. Nov 30, 2015\n\n### ehild\n\nIs not C+1 a constant?\n\n5. Nov 30, 2015\n\n### Mr Davis 97\n\nDoes that mean as my final answer I should redefine C + 1 as just C, since they are both just constants? Why did I get a 1 in the first place, while other ways of obtaining the solution don't have the 1? Such as letting u be just lnx instead of lnx + 1?\n\n6. Dec 1, 2015\n\n### Samy_A\n\nWhen you integrate, you find one result, one primitive of the function you integrate.\nNow, if the function $f$ is the the primitive you found, $f+C$ ($C$ any constant) will also be a primitive. So yes, you can remove the +1, or leave it, both results are correct. Removing the +1 makes the answer somewhat more elegant, but there is nothing more to it than that.\n\n7. Dec 1, 2015\n\n### Ray Vickson\n\nYou got the '1' in the first place because YOU chose to define u as 1 + ln(x).\n\nAnyway, why are you obsessing on this issue? It as absolutely NO importance: the constant of integration is totally arbitrary, and can be called 147.262 + K instead of C, if that is what you prefer. It can be called anything you want, and if somebody else chooses to call it something different from what you choose to call it, that is OK, too. If you use a computer algebra system to do the indefinite integral, it might not a constant of integration (but it might include your \"1\" if it happened to use your change-of-variable method).", "date": "2018-01-18 14:44:39", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/u-substitution-problem.845874/", "openwebmath_score": 0.6835151314735413, "openwebmath_perplexity": 690.912646605293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429595026214, "lm_q2_score": 0.8918110368115781, "lm_q1q2_score": 0.8769561042553985}}
{"url": "https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042", "text": "# Why is $1/i$ equal to $-i$?\n\nWhen I entered the value $$\\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).\n\nIs there a fault in my calculator or $\\frac{1}{i}$ really equals $-i$? If it does then how?\n\n\u2022 Hint $i^2 = -1$ \u2013\u00a0Mann May 11 '15 at 12:14\n\u2022 Multiply by $i/i$. \u2013\u00a0David Mitra May 11 '15 at 12:14\n\u2022 Hint $$z=\\frac{1}{i}\\iff zi=1\\implies \\dots$$ \u2013\u00a0John Joy May 11 '15 at 12:56\n\u2022 Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. \u2013\u00a0Emily May 11 '15 at 14:50\n\u2022 Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! \u2013\u00a0Math Man May 13 '15 at 1:04\n\n$$\\frac{1}{i}=\\frac{i}{i^2}=\\frac{i}{-1}=-i$$\n\nNote that $i(-i)=1$. By definition, this means that $(1/i)=-i$.\n\nThe notation \"$i$ raised to the power $-1$\" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$.\n\nIn fact, $-i$ satisfies that since\n\n$$(-i)\\cdot i= -(i\\cdot i)= -(-1) =1$$\n\nThat notation holds in general. For example, $2^{-1}=\\frac{1}{2}$ since $\\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$.\n\n\u2022 I appreciate that this answer gives context to the calculation. +1 ! \u2013\u00a0pjs36 May 11 '15 at 15:04\n\nThere are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the \"simplest\" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example.\n\nThe complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.\n\nWhile $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ then $c = i^{-1}$.\n\nThis is because we know that inverses in the complex numbers are unique.\n\nAs it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$.\n\nAs fractions (or powers) are usually considered \"less simple\" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$.\n\n$-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1$, or $i^{-1} = -i$.\n\n$$\\frac{1}{i}=\\frac{i^4}{i}=i^3=i^2\\cdot i = -i$$\n\nBy the definition of the inverse $$\\frac1i\\cdot i=1.$$\n\nThis agrees with\n\n$$(-i)\\cdot i=1.$$\n\n$$\\frac{1}{i}=\\left|\\frac{1}{i}\\right|e^{\\arg\\left(\\frac{1}{i}\\right)i}=$$\n\n$$1e^{\\left(-\\frac{1}{2}\\pi\\right) i}=e^{\\left(-\\frac{1}{2}\\pi\\right) i}=$$\n\n$$1\\left(\\cos\\left(-\\frac{1}{2}\\pi\\right)+\\sin\\left(-\\frac{1}{2}\\pi\\right)i\\right)=\\cos\\left(-\\frac{1}{2}\\pi\\right)+\\sin\\left(-\\frac{1}{2}\\pi\\right)i=$$\n\n$$0+(-1)i=0-1i=-i$$\n\nSo:\n\n$$\\frac{1}{i}=-i$$\n\nWhy is $\\left|\\frac{1}{i}\\right|=1$:\n\n$$\\left|\\frac{1}{i}\\right|=\\sqrt{\\Re\\left(\\frac{1}{i}\\right)^2+\\Im\\left(\\frac{1}{i}\\right)^2}=\\sqrt{0^2+(-1)^2}=\\sqrt{(-1)^2}=\\sqrt{1}=1$$\n\nSecond wat to show $\\left|\\frac{1}{i}\\right|=1$:\n\n$$\\left|\\frac{1}{i}\\right|=\\frac{|1|}{|i|}=\\frac{\\sqrt{1^2}}{\\sqrt{1^2}}=\\frac{\\sqrt{1}}{\\sqrt{1}}=\\sqrt{\\frac{1}{1}}=\\sqrt{1}=1$$\n\n\u2022 How do you know that $|1/i|=1$??? \u2013\u00a0JP McCarthy May 11 '15 at 17:24\n\u2022 @JpMcCarthy look in the edit! \u2013\u00a0Jan May 11 '15 at 17:30\n\u2022 This is the most convoluted way of explaining this that I have ever seen, albeit correct and in some ways more mathematically interesting than the conventional ones. \u2013\u00a0Laertes May 11 '15 at 17:33\n\u2022 I think it's the easiest way! \u2013\u00a0Jan May 11 '15 at 17:36\n\u2022 How did you know that the imaginary part of $1/i$ is $-1$??? \u2013\u00a0JP McCarthy May 12 '15 at 6:50\n\nI always like to point out that this fits well into a pattern you see when \"rationalising the denominator\", if the denominator is a root: $$\\frac{1}{\\sqrt{2}} = \\frac{1}{\\sqrt{2}}\\cdot \\frac{\\sqrt{2}}{\\sqrt{2}} = \\frac{1}{2}\\sqrt{2}$$ $$\\frac{1}{\\sqrt{17}} = \\frac{1}{\\sqrt{17}}\\cdot \\frac{\\sqrt{17}}{\\sqrt{17}} = \\frac{1}{17}\\sqrt{17}$$ $$\\frac{1}{\\sqrt{a}} = \\frac{1}{\\sqrt{a}}\\cdot \\frac{\\sqrt{a}}{\\sqrt{a}} = \\frac{1}{a}\\sqrt{a}$$ $$\\frac{1}{i} = \\frac{1}{\\sqrt{-1}} = \\frac{1}{\\sqrt{-1}}\\cdot \\frac{\\sqrt{-1}}{\\sqrt{-1}} = \\frac{1}{-1}\\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}$$ $$\\frac{1}{\\sqrt{17}} = \\frac{\\sqrt{17}}{17}$$ $$\\frac{1}{i} = \\frac{i}{-1}.$$", "date": "2019-05-22 04:53:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042", "openwebmath_score": 0.9033846855163574, "openwebmath_perplexity": 489.6127234965958, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429575500228, "lm_q2_score": 0.8918110360927155, "lm_q1q2_score": 0.876956101807161}}
{"url": "https://discusstest.codechef.com/t/chefprms-editorial/20906", "text": "# CHEFPRMS - EDITORIAL\n\nSetter: Misha Chorniy\nEditorialist: Taranpreet Singh\n\nEasy\n\n### PREREQUISITES:\n\nPrime Sieve, or even Square root factorization would suffix, Pre-computation. Number-theory (optional).\n\n### PROBLEM:\n\nAnswer the queries of form, Given a number N, can this number be represented as sum of two semi-primes. Semi-prime is a product of any two distinct prime numbers.\n\n### SUPER QUICK EXPLANATION\n\n\u2022 Pre-compute all primes up to MXN (MAX possible value of N) (or MXN/2 will also suffice). Calculate all Semi-primes by iterating over all pair of distinct primes.\n\u2022 Now, calculate all possible sums which can be represented as sum of two semi-primes, by taking each pair of semi-prime.\n\u2022 Answer queries in O(1) time using pre-computed values.\n\n### EXPLANATION\n\nFor this problem, I am explaining two solutions, Common approach as used by Setter and Tester as well as most teams. The other one is quite an overkill, strange way to solve this problem, used by editorialist only.\n\nSetter/Tester Approach\n\nCalculate all the primes in range [2,MXN]. This can be done using Sieve of Eratosthenes or we can just naively check each number whether it is a prime or not.\n\nFor those unaware of Sieve of Eratosthenes, Enter the secret box.\n\nClick to view\n\nWe create an array and initially, mark all numbers above 1 as prime. Then, we select number marked as prime which is not yet visited. For this prime, mark all its multiples as composite. We repeat this process until there are no prime left which are not visited yet. After this, only numbers not marked as composite are prime.\n\nNow that we have the list of primes in an array, say pr. Now, we need to find the list fo semi-primes. we can iterate over all pairs of Distinct primes take their product and the values in range [1, MXN] are marked as semi-prime. So, now we have the list of semi-primes in range$[1, MXN]$.\n\nNow, Computing final answer is just doing once again, Iterating over all pairs of semi-primes and if their sum is in range [1, MXN], mark the sum as reachable.\n\nAnswering queries become simple, as we just need to print whether the sum N is marked reachable or not.\n\nEditorialist\u2019s Approach (Not recommended at all )\n\nThis solution differs from above solution at the computation of list of semi-primes. Give a try to compute list of semi-primes without computing primes. It relies on the formula of Number of factors of a number.\n\nThe observation is, that all semi-primes have exactly four factors. (can also be proved eaisly) But, all numbers having four factors are of two types. Both semi-primes and perfect cubes will always have 4 factors. So, we iterate over all numbers, check if it has exactly four factors (including trivial factors) and is not perfect cube. If any number satisfy both criteria, it is marked as semi-prime.\n\nAfter that, this solution is same as Author/Tester solution.\n\n### Time Complexity\n\nBoth Solutions have pre-computation time O(MXN^2) for iterating over all pairs of semi-primes, which dominates everything.\n\nFor a fact, Sieve of Eratosthenes takes O(N*log(logN)) time while Square root factorization takes O(N*\\sqrt{N}) time.\n\nQueries are answered in O(1) time, taking overall O(T) time.\n\nSO, overall complexity is O(MXN^2).\n\n### AUTHOR\u2019S AND TESTER\u2019S SOLUTIONS:\n\nFeel free to Share your approach, If it differs. Suggestions are always welcomed.\n\nWhy my solution is flagged as wrong answer?I have checked for 200 input and it works on codechef practice compiler.link of my solution.\nLogic used:\nEvery semi-prime number Has exactly two distinct prime factor and used \u2018count\u2019 variable to count it.\nAlso used \u2018cnt\u2019 variable to take care of repeated prime factor.\n\nI solved this with a modified seive. We can calculate the semi-primes in the seive function itself and later simply iterate till N/2th number to check for semi-primes.\nTime complexity: 0(N loglogN + N).\nHave a look : https://www.codechef.com/viewsolution/20952615\n\n//", "date": "2021-07-25 14:35:36", "meta": {"domain": "codechef.com", "url": "https://discusstest.codechef.com/t/chefprms-editorial/20906", "openwebmath_score": 0.6724889278411865, "openwebmath_perplexity": 1853.0495548169117, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846703886661, "lm_q2_score": 0.8962513641273354, "lm_q1q2_score": 0.8768785954771153}}
{"url": "https://math.stackexchange.com/questions/2301488/what-is-the-difference-between-into-isomorphismand-onto-isomorphism", "text": "# What is the difference between \u201cinto isomorphism\u201dand \u201conto isomorphism\u201d?\n\nKronecker's theorem: Let $F$ be a field and $f(x)$ a nonconstant polynomial in $F[x]$. Then there is an extension field $E$ of $F$ in which $f(x)$ has zero.\n\nProof: Since $F[x]$ is a unique factorization domain, $f(x)$ has an irreducible factor say $p(x)$. Now consider the ideal $\\langle p(x)\\rangle$ of the ring of polynomials generated by $p(x)$. Since $\\langle p(x)\\rangle$ is irreducible over $F$ then the ideal $\\langle p(x)\\rangle$ is a maximal ideal of $F[x]$. Consequently, $F[x]/\\langle p(x)\\rangle$ is a field.\n\nWrite $E=F[x]/\\langle p(x)\\rangle$.\n\nNow we shall show that the field $E$ satisfies every part of the theorem. Now consider the map\n\n$\\phi:F\\rightarrow E$ defined as $\\phi (a)=a+\\langle p(x)\\rangle$.\n\nThis mapping is well defined as any element $a\\in F$ can be regarded as a constant polynomial in $F[x]$.\n\n$\\phi$ is injective\n\nfor $\\phi (a)=\\phi(b)$\n\n$\\implies a+\\langle p(x)\\rangle=b+\\langle p(x)\\rangle$\n\n$\\implies a-b\\in\\langle p(x)\\rangle$\n\n$a-b=f(x)p(x)$ for some $f(x) \\in F[x]$.\n\nSince $\\deg(p(x))\\geq 1$ then $\\deg(f(x)g(x))\\geq 1$ while $\\deg(a-b)=0$. Hence, $f(x)=0$. Consequently, $a-b=0 \\implies a=b$.\n\nClearly, $\\phi$ is an homomorphism.\n\nThus $\\phi$ is an isomorphism from $F$ into $E$.\n\nWhat is the difference between the phrases \u201c$\\phi$ is an isomorphism from $F$ into $F'$\u201d and \u201c$\\phi$ is an isomorphism from $F$ onto $F'$\u201d?\n\nActually, I'm proving Kronecker's theorem and the first phrase arose in the middle of the proof (last line). I got confused as it stated the first phrase just by showing $\\phi$ is a homomorphism and injective without showing it is surjective.\n\nPlease clarify my doubt, if possible with the help of an example.\n\n\u2022 Kronecker's Theorem seems to be something else. Could you give a link to the result and its proof you are working on? \u2013\u00a0Dietrich Burde May 29 '17 at 14:57\n\u2022 The latter means the mapping is onto (surjective), but isomorphisms are already surjective. \u2013\u00a0Sean Roberson May 29 '17 at 14:58\n\u2022 @DietrichBurde:Wait,i'm typing the complete proof. \u2013\u00a0P.Styles May 29 '17 at 14:59\n\u2022 @DietrichBurde There is a Kronecker's Theorem in field theory. \u2013\u00a0AspiringMathematician May 29 '17 at 14:59\n\u2022 PK, how do you define an isomorphism? If you grasp that, you've answered your own question. \u2013\u00a0Erik G. May 29 '17 at 15:00\n\nIt's impossible to know without having a reference to the textbook or lecture notes you're reading.\n\nSome texts use \u201cisomorphism\u201d where the more common terminology, nowadays, is \u201cinjective homomorphism\u201d or \u201cmonomorphism\u201d. Apparently, your textbook or lecture notes fall in this category. You should check where the book defines \u201cisomorphism\u201d; most probably it says the equivalent of \u201cinjective homomorphism\u201d: this terminology was rather common until a few decades ago.\n\nThe first part proves that $\\phi$ is an injective homomorphism and the second part proves surjectivity.\n\nActually, proving $\\phi$ is injective is not needed: as soon as $\\phi$ is a homomorphism that maps $1$ into $1$, it is automatically injective, because its kernel is an ideal and the ideals in a field $F$ are just $\\{0\\}$ and $F$.\n\n\u2022 :Yeah,the text i'm reading is very old.I've cross checked the proof of this theorem from Gallian's text & Dummit & foote's text.In both texts proof is almost similar both of these texts stated about injectivity and operation preserving property of $\\phi$ but the text i'm reading named it as \"into isomorphism\" specifically. \u2013\u00a0P.Styles May 29 '17 at 16:00\n\u2022 @PKStyles You should check where the book defines \u201cisomorphism\u201d; most probably it says \u201cinjective homomorphism\u201d. This was very common until a few decades ago. \u2013\u00a0egreg May 29 '17 at 16:03\n\u2022 :There is no mention of injective homomorphism while defining the map but i think in place injective isomorphism it should be injecttive homomorphism. \u2013\u00a0P.Styles May 29 '17 at 16:07\n\u2022 @PKStyles Can you tell what book it is? \u2013\u00a0egreg May 29 '17 at 16:08\n\u2022 :Advanced course in Modern Algebra by Gupta & Gupta. \u2013\u00a0P.Styles May 29 '17 at 16:10\n\nIt could be just a matter of emphasis (but do see the comments below), so it would be more about writing style than mathematics. If $\\phi$ is an isomorphism, then it is already both injective and surjective, but I would imagine that the author is thinking something like\n\n\u2022 Saying \"$\\phi$ is an isomorphism from $F$ into $F'$\" emphasizes to the reader that the target of the map $\\phi$ is $F'.$ Using the term into can be important when emphasizing that a map is well defined. Like you can declare that $F'$ is the codomain of $\\phi$ all you want, but when you actually define what map $\\phi$ does to elements of $F$, the result better be in $F'$. Sometimes there is a little work to showing that $\\phi(x) \\in F'$ for all $x \\in F$, and saying that $\\phi$ maps $F$ into $F'$ captures the idea that this was checked.\n\n\u2022 Saying \"$\\phi$ is an isomorphism from $F$ onto $F\u2032$\" emphasizes that the map is surjective, which is possibly the reason the author brought up the fact that $\\phi$ is an isomorphism in that part of the argument.\n\n\u2022 Older texts use isomorphism for what we now use injective homomophism. It is not a matter of style, really. \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez May 29 '17 at 15:08\n\u2022 @Mike Pierce:Will you please explain \"the target of the map $\\phi$ is $F'$\"? \u2013\u00a0P.Styles May 29 '17 at 15:32\n\u2022 @MarianoSu\u00e1rez-\u00c1lvarez Oh, I had no idea. Then it become rather important where OP is reading this proof. Now I'm curious, how old do you mean when you say \"older texts\"? Like, are there texts that use isomorpism for injective homomorphism that are commonly referenced today? \u2013\u00a0Mike Pierce May 29 '17 at 15:34\n\u2022 @MikePierce:From first point, you meant that $\\phi$ is an isomorphism from $F$ onto \"something\" inside $F'$? \u2013\u00a0P.Styles May 29 '17 at 15:50\n\u2022 @mike, this is quite common. Just google for \"isomorphism into\" (between quotes) in google, and go to books. Artin used it in his Geometric Algebra, Steenrod in his The topology of fiber bundles, and so on. \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez May 29 '17 at 16:19", "date": "2020-01-19 16:27:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2301488/what-is-the-difference-between-into-isomorphismand-onto-isomorphism", "openwebmath_score": 0.8795912265777588, "openwebmath_perplexity": 406.3101282279678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9465966747198242, "lm_q2_score": 0.9263037236819582, "lm_q1q2_score": 0.8768360246179325}}
{"url": "http://blog.mouans-immobilier.com/q1oxifs3/equivalence-relation-properties-023a3f", "text": "The relationship between a partition of a set and an equivalence relation on a set is detailed. If A is an in\ufb01nite set and R is an equivalence relation on A, then A/R may be \ufb01nite, as in the example above, or it may be in\ufb01nite. Equivalence Relations. Using equivalence relations to de\ufb01ne rational numbers Consider the set S = {(x,y) \u2208 Z \u00d7 Z: y 6= 0 }. 1. Equivalence Relations 183 THEOREM 18.31. Properties of Equivalence Relation Compared with Equality. It is of course enormously important, but is not a very interesting example, since no two distinct objects are related by equality. . For any x \u2208 \u2124, x has the same parity as itself, so (x,x) \u2208 R. 2. 1. Another example would be the modulus of integers. In a sense, if you know one member within an equivalence class, you also know all the other elements in the equivalence class because they are all related according to $$R$$. Definition: Transitive Property; Definition: Equivalence Relation. We de\ufb01ne a rational number to be an equivalence classes of elements of S, under the equivalence relation (a,b) \u2019 (c,d) \u21d0\u21d2 ad = bc. Examples: Let S = \u2124 and define R = {(x,y) | x and y have the same parity} i.e., x and y are either both even or both odd. Let R be the equivalence relation \u2026 1. As the following exercise shows, the set of equivalences classes may be very large indeed. Equalities are an example of an equivalence relation. Suppose \u223c is an equivalence relation on a set A. An equivalence class is a complete set of equivalent elements. 1. Math Properties . 1. The parity relation is an equivalence relation. First, we prove the following lemma that states that if two elements are equivalent, then their equivalence classes are equal. Assume (without proof) that T is an equivalence relation on C. Find the equivalence class of each element of C. The following theorem presents some very important properties of equivalence classes: 18. For example, in a given set of triangles, \u2018is similar to\u2019 denotes equivalence relations. . 0. Proving reflexivity from transivity and symmetry. Explained and Illustrated . Equivalence relation - Equilavence classes explanation. . reflexive; symmetric, and; transitive. Example 5.1.1 Equality ($=$) is an equivalence relation. Exercise 3.6.2. Equivalent Objects are in the Same Class. Example $$\\PageIndex{8}$$ Congruence Modulo 5; Summary and Review; Exercises; Note: If we say $$R$$ is a relation \"on set $$A$$\" this means $$R$$ is a relation from $$A$$ to $$A$$; in other words, $$R\\subseteq A\\times A$$. Equivalence Properties . . Remark 3.6.1. Algebraic Equivalence Relations . We will define three properties which a relation might have. Definition of an Equivalence Relation. . Let $$R$$ be an equivalence relation on $$S\\text{,}$$ and let $$a, b \u2026 The relation \\(R$$ determines the membership in each equivalence class, and every element in the equivalence class can be used to represent that equivalence class. Basic question about equivalence relation on a set. . We discuss the reflexive, symmetric, and transitive properties and their closures. A binary relation on a non-empty set $$A$$ is said to be an equivalence relation if and only if the relation is. An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive. Lemma 4.1.9. Then: 1) For all a \u2208 A, we have a \u2208 [a]. Note the extra care in using the equivalence relation properties. 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Two distinct objects are related by Equality similar to \u2019 denotes equivalence relations transitive properties and closures. Equality ($ = $) is an equivalence relation on a set is.! ($ = $) is an equivalence class is a complete set of equivalent.. Define three properties which a relation might have the equivalence relation same parity as itself so! A complete set of equivalent elements enormously important, but is not a very interesting example, in given. Equivalence relation properties let R be the equivalence relation on a set and an equivalence on... Similar to \u2019 denotes equivalence relations ) is an equivalence class is a on! In using the equivalence relation properties a, we prove the following lemma that states that if two elements equivalent!$ ) is an equivalence relation that states that if two elements are equivalent, then their equivalence are! On a set and an equivalence relation on a set S, is a on... That if two elements are equivalent, then their equivalence classes are equal,. \u2026 Definition: transitive Property ; Definition: transitive Property ; Definition: equivalence relation on S which is,. Suppose \u223c is an equivalence relation on a set S, is a complete of... Properties and their closures S, is a complete set of triangles, \u2018 is similar \u2019! Which a relation on a set is detailed \u2026 Definition: transitive Property ; Definition transitive. Their equivalence classes are equal equivalence relations no two distinct objects are by! The two most important examples of equivalence relations important, but is not a very interesting,...\n\nSkyrim Niselft Near Markarth, Dakota Electric Processing Fee, Best Pg In Vijay Nagar, Delhi, Zulm Ki Hukumat Full Movie, Island Day Trips From Nadi, Umdnj Radiology Phone Number, Dulux Easycare Colours, Pasulj Bez Zaprske,", "date": "2021-07-29 02:02:23", "meta": {"domain": "mouans-immobilier.com", "url": "http://blog.mouans-immobilier.com/q1oxifs3/equivalence-relation-properties-023a3f", "openwebmath_score": 0.8404521942138672, "openwebmath_perplexity": 465.4254196726806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8991213732152424, "lm_q1q2_score": 0.8768248160709537}}
{"url": "https://www.physicsforums.com/threads/different-probablities-same-situation.302550/", "text": "# Different Probablities Same Situation?\n\n1. Mar 25, 2009\n\n### dkgolfer16\n\nJohnny and Sally sit at a table in their dining room. Sally tells Johnny to leave the room while she prepares a game. Sally randomly selects three cards from a regular deck of cards (half black, half red) and places them face down on the table. She yells at Johnny to reenter and tells him that he gets to flip two of the three cards over. If two are of the same color, Johnny wins $2. If they are different colors, Sally wins$1.\n\nJohnny's viewpoint: It's a great deal because I have a 50% chance of winning $2 and 50% chance of losing$1. Why does he think this? The first card color he flips over is of no difference. The second card he flips has a 50% chance of being either red or black, thus 50% chance of matching the first color.\n\nSally's viewpoint: It's a great deal because Johnny only has 33% chance of winning so theoretically I should pay him $3 if he wins. Why does she think this? Since three cards exist, the odds of flipping over two of the same color are 1 out of 3. Question: Who is right? Who will win? Are they somehow both right? Thanks for the help. 2. Mar 25, 2009 ### robert Ihnot If you reduced this to dealing two cards and asking the odds, how does that differ from chosing two cards at random from a deck of 52? 3. Mar 25, 2009 ### regor60 Johnny's right. By his logic. His expected value is 50 cents. Her logic is incomplete. If you focus on three cards, there are twelve combinations, 6 of which are of same color, therefore 50%. Even if he had a 1/3 chance of winning, it still wouldn't be a great deal, because her expected return is 0 with his$2 winning potential. Paying him \\$3 would be to her disadvantage, her expected return would be negative 33 cents\n\nLast edited: Mar 25, 2009\n4. Mar 25, 2009\n\n### dkgolfer16\n\nThat's what I thought. Johnny's right. But this link was posted to explain bell's inequality. Maybe i'm misunderstanding the example but it says the odds are on Sally's side. See \"Is this game fair to you?\" heading at the following address:\n\nhttp://ilja-schmelzer.de/realism/game.php .\n\nCorrect me if I'm missing something.\n\n5. Mar 25, 2009\n\n### CRGreathouse\n\nJohnny's essentially right -- though his expected winnings is more like 47 cents (8/17) because the deck isn't infinite.\nIn the link there are three preselected cards, two of the same color. In your example the three cards may be all the same color. Text comparisons:\n\"Sally randomly selects three cards from a regular deck of cards\"\n\"I put three cards of my choice on the table\"\n\n6. Mar 25, 2009\n\n### robert Ihnot\n\ndkgolfer16: Sally randomly selects three cards from a regular deck of cards\n\nThat's not what the example says, it says: \"I put three cards of my choice on the table so that you cannot see their color.\"\n\n7. Mar 25, 2009\n\n### dkgolfer16\n\nGot it thanks", "date": "2017-02-23 16:24:00", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/different-probablities-same-situation.302550/", "openwebmath_score": 0.45489436388015747, "openwebmath_perplexity": 1607.0551213282506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877675527114, "lm_q2_score": 0.9032942106088968, "lm_q1q2_score": 0.8768166407392387}}
{"url": "https://skyciv.com/tutorials/calculating-beam-section-moment-of-inertia/", "text": "Moment of Inertia of Beam Sections\n\nHow to Calculate the Moment of Inertia of a Beam Section(Second Moment of Area)\n\nBefore we find the moment of inertia (or second moment of area) of a beam section, its centroid (or center of mass) must be known. For instance, if the moment of inertia of the section about its horizontal (XX) axis was required then the vertical (y) centroid would be needed first (Please view our Tutorial on how to calculate the Centroid of a Beam Section).\n\nBefore we start, if you were looking for our Free Moment of Inertia Calculator please click the link to learn more. Consider the I-beam section below, which was also featured in our Centroid Tutorial. As shown below the section was split into 3 segments:\n\nThe Neutral Axis (NA)\n\nThe Neutral Axis (NA) or the horizontal XX axis is located at the centroid or center of mass. In our Centroid Tutorial, the centroid of this section was previously found to be 216.29 mm from the bottom of the section. Now to calculate the total moment of ienrtia of the section we need to use the \"Parallel Axis Theorem\":\n\n[math] {I}_{total} = \\sum{(\\bar{I}_{i} + {A}_{i} {{d}_{i}}^{2})} \\text{ where:}\\\\ \\begin{align} \\bar{I}_{i} &= \\text{The moment of inertia of the individual segment about its own centroid axis}\\\\ {A}_{i} &= \\text{The area of the individual segment}\\\\ {d}_{i} &= \\text{The vertical distance from the centroid of the segment to the Netrual Axis (NA)} \\end{align} [math]\n\nIt is widely known that the moment of inertia equation of a rectangle about its centroid axis is simply:\n[math] \\bar{I}=\\frac{1}{12}b{h}^{3} \\text{ where:}\\\\\\\\ \\begin{align} b &= \\text{The base or width of the rectangle}\\\\ h &= \\text{The height of the rectangle} \\end{align} [math]\n\nThe moment of inertia of other shapes are often stated in the front/back of textbooks, however the rectangular shape is very common for beam sections.\n\nNow we have all the information we need to use the \"Parallel Axis Theorem\" and find the total moment of inertia of the I-beam section:\n\n[math] \\text{Segment 1:}\\\\ \\begin{align} \\bar{I}_{1} &= \\tfrac{1}{12}(250)(38)^{3} = 1,143,166.667 {\\text{ mm}}^{4}\\\\ {A}_{1} &= 250\\times38 = 9500 {\\text{ mm}}^{2}\\\\ {d}_{1} &= \\left|{y}_{1} - \\bar{y} \\right| = \\left|(38 +300 +\\tfrac{38}{2}) - 216.29\\right| = 140.71 \\text{ mm}\\\\\\\\ \\end{align} [math] [math] \\text{Segment 2:}\\\\ \\begin{align} \\bar{I}_{2} &= \\tfrac{1}{12}(25)(300)^{3} = 56,250,000 {\\text{ mm}}^{4}\\\\ {A}_{2} &= 300\\times25 = 7500 {\\text{ mm}}^{2}\\\\ {d}_{2} &= \\left|{y}_{2} - \\bar{y}\\right| = \\left|(38 +\\tfrac{300}{2}) - 216.29\\right| = 28.29 \\text{ mm}\\\\\\\\ \\end{align} [math] [math] \\text{Segment 3:}\\\\ \\begin{align} \\bar{I}_{3} &= \\tfrac{1}{12}(150)(38)^{3} = 685,900 {\\text{ mm}}^{4}\\\\ {A}_{3} &= 150\\times38 = 5700 {\\text{ mm}}^{2}\\\\ {d}_{3} &= \\left|{y}_{3} - \\bar{y}\\right| = \\left|\\tfrac{38}{2} - 216.29\\right| = 197.29 \\text{ mm}\\\\\\\\ \\end{align} [math] [math] \\begin{align} \\therefore {I}_{total} &= \\sum{(\\bar{I}_{i} + {A}_{i} {{d}_{i}}^{2})} \\\\ &= (\\bar{I}_{1} + {A}_{1}{{d}_{1}}^{2}) + (\\bar{I}_{2} + {A}_{2}{{d}_{2}}^{2}) + (\\bar{I}_{3} + {A}_{3}{{d}_{3}}^{2})\\\\ &= (1,143,166.667 + 9500\\times140.71^{2}) + (56,250,000 + 7500\\times28.29^{2}) + (685,900 + 5700\\times197.29^{2})\\\\ &= 474,037,947.7 {\\text{ mm}}^{4}\\\\ {I}_{total} &= 4.74 \\times 10^{8} {\\text{ mm}}^{4} \\end{align} [math]\n\nSo in summary, you can see the results (calculated from our Free Moment of Inertia Calculator) along with some other section properties to check your work.\n\nOf course you don't need to do all these calculations manually because you can use our fantastic Free Moment of Inertia Calculator to find the important properties and information about beam sections.\n\n\u2022 Owain\n\nhi, some guidance on how to calculate Ixx when you have rivets in the bottom flange would be great. Do you remove the X-sectional area of holes, determine NA and Ixx? or, do not remove holes and determine NA, then calculate Ixx with holes removed?\n\n\u2022 Kabila\n\nwhen is 1/12 factor used in calculating the moment of inertia? and why?\n\n\u2022 side-fish\n\nHi Kabila, I hope I\u2019m not too late. But the 1/12 coefficient from the moment of inertia of a rectangle is derived from calculus by integrating y^2dA. Your upper limit is the extreme fiber and the lower limit is the axis of rotation. In the case of a rectangle, you will integrate the top and the bottom of the rectangle. Note that dA=xdy. The new equation becomes I = \u222bxy^2dy with the upper limit as h and the lower limit as h/2 if you want to integrate the top half (or h/2-0 if you want the bottom). Since the neutral axis of the rectangle is exactly in the middle, you simply integrate say\u2026 the top half and multiply twice. The coefficient that you should get is 1/12. If the axis of rotation is shifted to say the bottom fiber, integrating will yield a coefficient of 1/3, though you could use the transfer formula I=Io+Ad^2 and still get the same answer.\n\n\u2022 Shaun Murrin\n\nThis is a tutorial question I\u2019m struggling a bit with axis to base the i value on..\n\u201cBased on calculations it has been determined that a universal steel column must have a minimum value of I of 38748 cm4. Select a column suitable for this scenario showing clearly why you have selected that particular column. \u201c\n\n\u2022 Glenn\n\nUsed your calculator to find for simple \u2018i\u2019 beam. Whilst we both find identical Zy, my Zx is out compared to yours. I looked at your formula to find the centroid and it shows the addition of web thickness to the vertical centroid. Beam is: TFw & BFw = 100; TFt & BFt = 10; Wh = 80, Wt = 6. Following your tute on centroid for simple I beam, Cx would be 56, however your app shows (correctly in my mind) Cx and Cy both as 50. Yet my Ix = 4.566 x 106\nmm4 ; calculator shows 4.323 x 106\nmm4 (Iy is same as mine 1.668 x 106\nmm4 ).\nBeen over calcs left and right and can\u2019t find error. Unless Cx as 56 is culprit. Any ideas?\n\n\u2022 Hi Glenn. As your section is symmetric about both horizontal and vertical and your section depth and width are both 100mm, then the centroid is definitely 50mm in both directions. Recalculate your Cx following the tutorial (see the image attached below).\n\n\u2022 Cristian\n\nThere is an error in the last picture, it shows the subscripts of some answers for the \u201cx\u201d axis, while there is the axis \u201cz\u201d, maybe it will lead to some misunderstandings. Anyways nice tutorial.\n\n\u2022 Thank you. This has been amended.", "date": "2018-12-14 09:34:22", "meta": {"domain": "skyciv.com", "url": "https://skyciv.com/tutorials/calculating-beam-section-moment-of-inertia/", "openwebmath_score": 0.9973840713500977, "openwebmath_perplexity": 2317.0236383999286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9954615671980688, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8767996474144715}}
{"url": "http://mathhelpforum.com/calculus/165891-2-problems-involving-integrals.html", "text": "# Math Help - 2 Problems involving integrals\n\n1. ## 2 Problems involving integrals\n\nSuppose $f(0)=0$ and $\\int_0^2{f'(2t)e^{f(2t)}}dt=5$. Find the value of f(4).\n\nSo I first recognized that $\\frac{1}{2}e^{f(2t)}$ should be the antiderivative of the integral by FTC so that $\\frac{1}{2}(e^{f(4)}-e^{f(0)})=5$.\n\nSince f(0)=0, the equation becomes: $e^{f(4)}-1=10$\n\nIsolating and then taking the natural log of both sides yields: f(4)=ln11\nIs that right?\n\nMy other question is this: evaluate $\\lim_{h\\to0}} \\frac{1}{h}\\int_x^{x+h}e^{arctan{t}}dt$ It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?\n\nThe first thing I notice is that the variable in the integrand isn't with respect to x. If I rewrote it through FTC, I'd put $\\lim_{h\\to0}}\\frac{1}{h}{e^{arctan(x+h)}}$? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?\n\n2. Originally Posted by VectorRun\nSuppose $f(0)=0$ and $\\int_0^2{f'(2t)e^{f(2t)}}dt=5$. Find the value of f(4).\n\nSo I first recognized that $\\frac{1}{2}e^{f(2t)}$ should be the antiderivative of the integral by FTC so that $\\frac{1}{2}(e^{f(4)}-e^{f(0)})=5$.\n\nSince f(0)=0, the equation becomes: $e^{f(4)}-1=10$\n\nIsolating and then taking the natural log of both sides yields: f(4)=ln11\nIs that right?\nYes, that is correct.\n\nMy other question is this: evaluate $\\lim_{h\\to0}} \\frac{1}{h}\\int_x^{x+h}e^{arctan{t}}dt$ It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?\nExactly right! That is $\\lim_{h\\to 0}\\frac{f(x+h)- f(x)}{h}$ where $f(x)= \\int_a^x e^{arctan(t)}dt$. And, by the Fundamental Theorem of Calculus, $\\frac{d}{dx}\\int_a^x f(t)dt= f(x)$.\n\nThe first thing I notice is that the variable in the integrand isn't with respect to x.\nThe variable in the integrand is a \"dummy variable\"- it could be anything without changing the integral.\n\nIf I rewrote it through FTC, I'd put $\\lim_{h\\to0}}\\frac{1}{h}{e^{arctan(x+h)}}$? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?\nNo, its not f that is being differentiated, it is the integral so the derivative of the integral is the integrand, f.\n\n3. Hello, VectorRun!\n\nYour first solution is correct . . . ecept for that $\\frac{1}{2}$\n\n$\\displaystyle \\text{Suppose }f(0)=0\\:\\text{ and }\\:\\int_0^2\\!f'(2t)e^{f(2t)}}dt\\:=\\:5$\n\n$\\text{Find the value of }f(4).$\n\n$\\text{So I first recognized that }\\frac{1}{2}e^{f(2t)}\\text{ is the antiderivative of the integral.}$\nUm ... not quite.\n\n$\\displaystyle \\text{We have: }\\:\\int^2_0 u'\\,e^u\\,dt \\;\\;\\text{ which equals: }\\;e^u\\,\\bigg]^2_0$\n\n$\\text{Hence: }\\;e^{f(2t)} \\,\\bigg]^2_0 \\;=\\;5 \\quad \\hdots \\;\\text{ etc.}\n$", "date": "2015-04-19 01:56:24", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/165891-2-problems-involving-integrals.html", "openwebmath_score": 0.9224373698234558, "openwebmath_perplexity": 230.6574766213018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446509776294, "lm_q2_score": 0.900529778109184, "lm_q1q2_score": 0.8767960015020785}}
{"url": "https://math.stackexchange.com/questions/2451295/how-to-calculate-the-nth-term-and-the-sum-in-this-series-if-the-common-differenc", "text": "# How to calculate the nth term and the sum in this series if the common difference between them isn't explicit?\n\nI have this series comprised of\n\n$$1,2,5,10,17,26,...$$,\n\nand so on. So far i have found that they add up in intervals being odd numbers. But I don't know how to find, let's say the 16th term and the sum up to that term. What should I do?\n\nEdit:\n\nAlthough it is somewhat possible to calculate the sum up to the 16th term in the above equation, what if the series is of higher order like this?.\n\n$$6,7,14,27,46,...$$\n\nIs there a shortcut to calculate the sum up to the 16th term? Can this be done by hand or would be necessary to use software like Maple?.\n\n\u2022 So, we have $a_{n+1}-a_n=2n-1$ with $a_1=1$. Can you solve this recurrence relation? \u2013\u00a0Prasun Biswas Sep 30 '17 at 8:31\n\nWhile Opt's answer is correct, it can take a good deal of practice to come up with such accurate observations on an unknown series. A very good and straightforward method in such a case is the method of differences\n\nIt formally states:\n\nany term, of a given number series, can be expressed as a polynomial of degree $n$, if the $n$ differences of the series is constant.\n\nLet me illustrate with your series.\n\n$$1,2,5,10,17,26$$\n\nThe first differences are $$1(=2-1);3(=5-2); 5(=10-5);7(=17-10);9(=26-17)$$.\n\nThe second differences are $$2(=3-1); 2(=5-3); 2(=7-5);2(=9-7)$$ Note that all the second differences are constant! This implies, by the method of differences statement, that the $r$-th term in your series can be represented as a quadratic polynomial (degree 2) i.e. $T_r = ar^2+br+c$ for some constants $a,b,c$\n\nNow, all you need to do is equate $T_1 =1; T_2=2; T_3=5$ and solve for $a,b,c$.\n\nAdded on request: To solve the above, you only need to do write them out after putting in $n=1,2,3$:\n\n$$a+b+c=1...E1$$ $$4a+2b+c=2...E2$$ $$9a+3b+c=5...E3$$\n\nYou have three variables and three equations, thus they can be easily solved by subtraction and elimination. $E2-E1$ and $E3-E2$ respectively gives two new sets of equations:\n\n$$3a+b=1...E4$$ $$5a+b=3...E5$$\n\nand then $E5-E4$ straightaway gives $a=1$. Putting this in E5 gives $b=-2$. Putting them in $E1$ gives $c=2$.\nThus, our $T_r=r^2-2r+2=1+(r-1)^2$\n\nFor summation of terms from 1 to 16, you need to do:\n\n$$\\sum_{r=1}^{n}T_r=\\sum_{r=1}^{n}(r^2-2r+2)$$\n\nfor which you may easily use the summation identities for $r^2$ and $r$.\n\nHope that helps!\n\n\u2022 Sorry but my level of algebra isn't very proficient to solve the latter part of your explanation, mind if you extend a bit?. According to the method you proposed the polynomial is in terms of x but how does this translates as 'nth' term?. I do understand the nth is the degree but what's the x?. How do I solve for a, b and c, Can you show it to me please?. \u2013\u00a0Chris Steinbeck Bell Sep 30 '17 at 9:35\n\u2022 For the general case (when the $n$-th difference is constant), it is simpler to use the binomial polynomials $\\dbinom xi$ as a basis for the vector space of polynomials. \u2013\u00a0Bernard Sep 30 '17 at 9:37\n\u2022 @Bernard Can you be a bit explicit on how do the binomial coefficient replaces constants in the above equation?. Am I understanding correctly?. Gaurang Tandon, There is also missing the last part of my question which involves the sum from 1 to the 16th term in the series, how can I do that?. \u2013\u00a0Chris Steinbeck Bell Sep 30 '17 at 9:46\n\u2022 I'm not sure I was clear enough: I meant the canonical basis $1,x,x^2,\\dots$ is replaced with the basis $\\dbinom x0,\\dbinom x1, \\dbinom x2, \\dots$. It's the natural basis in this situation because $\\Delta\\dbinom xk=\\dbinom x{k-1}$. \u2013\u00a0Bernard Sep 30 '17 at 9:50\n\u2022 @ChrisSteinbeckBell That confusion about $n$ and $x$ was my mistake. Sorry for that. Now, I've added extensive elaboration. Hope that helps! \u2013\u00a0Gaurang Tandon Sep 30 '17 at 9:56\n\nAs Guarang Tandon says. You can create a table of differences between consecutive terms and then repeat that process on the list you just created.\n\n\\begin{array}{|l|c|ccccccc|} \\hline \\text{index} & n & 0 & 1 & 2 & 3 & 4 & 5 & \\dots\\\\ \\hline \\text{sequence} & f_n & 1 & 2 & 5 & 10 & 17 & 26 & \\dots \\\\ \\text{first differences} & \\Delta f_n && 1 & 3 & 5 & 7 & 9 & \\dots \\\\ \\text{second differences} & \\Delta^2 f_n &&& 2 & 2 & 2 & 2 & \\dots \\\\ \\hline \\end{array}\n\nIf you are lucky, at some point, all of the differences will be constant. (Of course this is an unverifiable claim. All we know for sure is that the first four second differences are constant. We can only assume that this fortuitous pattern continues.)\n\nIt can be proved that, if $\\Delta^d f_n$ is constant, then then $f_n$ is a $d$th degree polynomial in $n$.\n\nIf the second differences are constant, then the first differences must be an arithmetic sequence. In this case, we see that $\\Delta f_n = 2n+1$. It follows that $$f_{n+1} = f_n + 2n+1 \\tag 1$$.\n\nIf the first differences are an arithmetic sequence, then $f_n = an^2 + bn + c$ for some real numbers $a, b,$ and $c$.\n\nWe can now use modified form of mathematical induction to find the values of $a,b,$ and $c$.\n\nSince $f_0=1$, then $1 = a\\cdot0^2 + b\\cdot 0 + c = c$. So $f_n=an^2+bn+1$.\n\nFirst we note that $$f_{n+1} = a(n+1)^2 + b(n+1) + 1 = f_n + 2an + (a+b).$$ Comparing that to $f_{n+1}= f_n+ 2n+1$, we see that $2a=2$ and $a+b=1$. Hence $a=1$ and $b=0$.\n\nWe conclude that $f_n=n^2+1$.\n\nWe know that $\\sum_{k=0}^n k^2=\\frac 16n(n+1)(2n+1)$.\n\nThen \\begin{align} \\sum_{k=0}^n f_k &= \\sum_{k=0}^n (k^2+1) \\\\ &= \\sum_{k=0}^n k^2 + \\sum_{k=0}^n 1 \\\\ &= \\frac 16n(n+1)(2n+1) + (n+1) \\\\ &= \\frac 16(n+1)(2n^2+n) + \\frac 16(n+1)6 \\\\ &= \\frac 16(n+1)(2n^2+n+6) \\\\ \\end{align}\n\nCheck:\n\n$\\qquad \\sum_{k=0}^5 f_k = 1+2+5+10+17+26=61$\n\n$\\qquad \\left. \\dfrac 16(n+1)(2n^2+n+6)\\right|_{n=5}=\\frac 16(6)(61)=61$\n\nSo, since we started counting at $n=0$, the $16$th term is $f_{15} = 15^2+1=226$\n\nand the sum of the first $16$ terms is $\\sum_{k=0}^{15} f_k = \\left. \\dfrac 16(n+1)(2n^2+n+6)\\right|_{n=15}=1256$.\n\n\u2022 That's a nice table! :D \u2013\u00a0Gaurang Tandon Oct 1 '17 at 2:52\n\nThat's $$1 + 0 \\\\ 1 + 1 \\\\ 1 + 4 \\\\ 1 +9\\\\ 1 +16\\\\ 1 +25 \\\\ \\vdots$$ Therefore, $$a_n = 1+(n-1)^2,$$\n\nimplying that $a_{16} = 1+(15)^2 = 226$.\n\nAs is said in other answers, the method of differences is certainly a way to go. However, in the simultaneous equations, you can obtain $a$ immediately:\n\nIf the $d^{th}$ row of the difference table is some constant $q$, the coefficient of the leading term of the polynomial is $\\dfrac{q}{d!}$.\n\nSo, with the difference table as calculated in other answers, we have the second row is a constant 2. This gives us $a=\\dfrac{2}{2!}=1$ without resorting to simultaneous equations.\n\nNow, if you wanted to, you could use simultaneous equations to determine the other coefficients or you could subtract $n^2$ from each term:\n\n$$1-0^2, 2-1^2, 5-2^2, 10-3^2, 17-4^2, 26-5^2,...\\\\ \\text{which is }1,1,1,1,1,...$$\n\nFrom which, the polynomial is $\\fbox{n\u00b2+1}$", "date": "2019-05-24 04:52:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2451295/how-to-calculate-the-nth-term-and-the-sum-in-this-series-if-the-common-differenc", "openwebmath_score": 0.9958217144012451, "openwebmath_perplexity": 224.9330069099016, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97364464868338, "lm_q2_score": 0.9005297801113613, "lm_q1q2_score": 0.8767960013854478}}
{"url": "https://mathhelpboards.com/threads/find-the-number-of-distinct-real-roots.7222/", "text": "# Find the number of distinct real roots\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nLet $f(x)=x^3-3x+1$. Find the number of distinct real roots of the equation $f(f(x))=0$.\n\n#### jacks\n\n##### Well-known member\n\nI am getting Total no. of real solution is $= 7$\n\nIs is Right or not\n\nThanks\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nQuite a nice bit of a problem you've got there. Here's my solution --\n\nFinding the real roots of $$\\displaystyle f(f(x)) = 0$$ is equivalent to finding the ones of $$\\displaystyle f(x) = x_i$$ where $$\\displaystyle x_i$$ for i = 1, 2, 3 are the (real) roots of f(x).\n\nWe see that the roots of $$\\displaystyle f(x)$$ are all real as the discriminant is positive. An inspection of $$\\displaystyle f(x) = x^3 - 3x + 1$$ can be considered at the intervals $$\\displaystyle [-2, -1] \\cup [-1, 0] \\cup [0, 1] \\cup [1, 2]$$ to find out that three roots lie precisely in between the first, third and the fourth interval.\n\nNote that for the equation $$\\displaystyle f(x) = x_i$$ to have all real root, the discriminant must be positive, implying the $$\\displaystyle -2 \\leq 1 - x_i \\leq 2$$. The first root lies between [-2, -1], so $$\\displaystyle 1 - x_1$$ lies in between 2 and 3 and hence the discriminant comes down to a negative value, so one real root here exists and the other twos are complex conjugates of each other in $$\\displaystyle \\mathbb{C}$$.\n\nFor the second root, it lies between 0 and 1, so $$\\displaystyle 1 - x_2$$ lies in between 2 and -2 obviously enough. The same applies for the third root. So we have 3 real roots each for these two.\n\nSo, the total number of real roots are 1 + 3 + 3 = 7.\n\nBalarka\n.\n\nLast edited by a moderator:\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\n\nI am getting Total no. of real solution is $= 7$\n\nIs is Right or not\n\nThanks\nYes, 7 is the correct answer!\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nFinding the real roots of $$\\displaystyle f(f(x)) = 0$$ is equivalent to finding the ones of $$\\displaystyle f(x) = x_i$$ where $$\\displaystyle x_i$$ for i = 1, 2, 3 are the (real) roots of f(x).\n\n.\n\nHey Balarka,\n\nI really like it you mentioned that to solve for $x$ in $$\\displaystyle f(f(x)) = 0$$ is equivalent to solve for $$\\displaystyle f(x) = x_i$$ where $$\\displaystyle x_i$$ for i = 1, 2, 3 are the (real) roots of f(x).\n\nAnd if I may use this idea to solve for the problem, I found out the total number of real roots for $$\\displaystyle f(f(x)) = 0$$ are 7 as well, and this method is so much better than my first approach and thank you, Balarka for showing me something that I don't know!\n\n Solving for the number of real roots of $$\\displaystyle f(x) = x_1=root_1$$ means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_1$, as shown by the green line and we see that in this case we have only one real root for $$\\displaystyle f(x) = x_1=root_1$$. Solving for the number of real roots of $$\\displaystyle f(x) = x_2=root_2$$ means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_2$, as shown by the purple line and we see that in this case we have only one real root for $$\\displaystyle f(x) = x_2=root_2$$. Solving for the number of real roots of $$\\displaystyle f(x) = x_3=root_3$$ means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_3$, as shown by the orange line and we see that in this case we have only one real root for $$\\displaystyle f(x) = x_3=root_3$$.\n\nSo there are a total of 7 intersection points and hence there are 7 real roots for $$\\displaystyle f(f(x)) = 0$$!\n\nThanks!\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nanemone said:\nthis method is so much better than my first approach\nMay I see your approach on this problem?\n\nBalarka\n.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nMay I see your approach on this problem? Balarka .\nMy first approach revolves around checking the nature of the curve of the function of $f(f(x))$ from its first and second derivative expression. If we let $f(x)=x^3-3x+1$ and $g(x)=f(f(x))=(f(x))^3-3(f(x))+1$, we find\n\n $f(x)=x^3-3x+1$ $g(x)=f(f(x))=(f(x))^3-3(f(x))+1$ $f'(x)=3x^2-3=3(x^2-1)=3(x+1)(x-1)$ $\\rightarrow f'(x)=0$ iff $3(x+1)(x-1)=0$ or $x=\\pm1$. $g'(x)=3f(x)^2f'(x)-3f'(x)=3f'(x)((f(x))^2-1)$ $\\rightarrow g'(x)=0$ iff $x=\\pm1$, $x=\\pm \\sqrt{3}$, or $x=-2$ or $x=0$. $f''(x)=6x$ $g''(x)=6f(x)(f'(x))^2+3f''(x)((f(x))^2-1)$\n\nI then made another table to determine the nature of the critical points as follows:\n $x$ -2 $-\\sqrt{3}$ -1 0 1 $\\sqrt{3}$ $f(x)$ -1 1 3 1 -1 1 $f'(x)$ 9 6 0 -3 0 6 $f''(x)$ -12 $-6\\sqrt{3}$ -6 0 6 $6\\sqrt{3}$ $g(x)$ 3 -1 19 -1 3 -1 $g'(x)$ 0 0 0 0 0 0 $g''(x)$ -486 216 -144 54 0 216\n\nAnd plot a rough sketch to determine the number of real roots of the function of $f(f(x))=0$:\n\nFrom the graph we can tell there are 7 real roots of the function of $f(f(x))=0$.\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nHmm... very nice. I like the way you analyzed the behavior of f(f(x)), very well done. Mine may have been a little time consuming but this one is more representative.\n\nPS I particularly like this solution.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHmm... very nice. I like the way you analyzed the behavior of f(f(x)), very well done. Mine may have been a little time consuming but this one is more representative.\n\nPS I particularly like this solution.\nThank you for the compliment, Balarka!\n\nAnd I must give full credit to Opalg, a very nice mentor here for giving me idea (from another question) to solve the problem!", "date": "2022-01-17 09:58:33", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-the-number-of-distinct-real-roots.7222/", "openwebmath_score": 0.8285268545150757, "openwebmath_perplexity": 202.79133993388865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109543125689, "lm_q2_score": 0.8902942283051332, "lm_q1q2_score": 0.8767715085961503}}
{"url": "https://math.stackexchange.com/questions/1641774/integral-of-int-infty-infty-left-frac1-alpha-ix-frac1-alp", "text": "# Integral of $\\int_{-\\infty}^{\\infty} \\left(\\frac{1}{\\alpha + ix} + \\frac{1}{\\alpha - ix}\\right)^2 \\, dx$\n\nI'm having trouble integrating\n\n$$\\int_{-\\infty}^{\\infty} \\left(\\frac{1}{\\alpha + ix} + \\frac{1}{\\alpha - ix}\\right)^2 \\, dx$$\n\nwhere $\\alpha$ is a real number and $i = \\sqrt{-1}$. I'm guessing that I would have to integrate it using methods on complex analysis wherein I would use a semicircle contour enclosing a pole ($x = i\\alpha$ for example) so that I could apply the residue theorem. But is there any other way to evaluate this integral (like an ingenious substitution or differentiating under the integral sign)?\n\n\u2022 Is the integrand really typed correctly? Should this be, e.g., $\\frac{1}{a - ix} + \\frac{1}{a + ix}$? \u2013\u00a0Travis Willse Feb 5 '16 at 13:49\n\u2022 @Travis Good call, thanks for pointing that out. \u2013\u00a0Aldon Feb 5 '16 at 13:53\n\u2022 Have you tried combining the fractions? \u2013\u00a0user170231 Feb 5 '16 at 13:54\n\u2022 @user170231 Yes, you'd get $\\left(\\frac{2 \\alpha}{\\alpha^2 + x^2}\\right)^2$ which is another integral that's hard to integrate, which I'm thinking one that can be solved, again, by a contour integral. \u2013\u00a0Aldon Feb 5 '16 at 13:59\n\nObserve that the second summand is conjugate of the first: $$\\frac1{\\alpha-ix}=\\overline{\\frac1{\\alpha+ix}}$$ thus their sum will be real. In particular $$\\frac1{\\alpha+ix}+\\frac1{\\alpha-ix}=\\frac{2\\alpha}{\\alpha^2+x^2}{}$$ thus the integrand will be $$\\frac{4\\alpha^2}{(\\alpha^2+x^2)^2}=\\frac{4}{\\alpha^2}\\frac{1}{(1+(x/\\alpha)^2)^2}$$ The substitution $x=\\alpha\\tan y$ leads to (suppose wlog $\\alpha>0$) \\begin{align*} \\int_{-\\infty}^{+\\infty}\\left(\\frac1{\\alpha+ix}+\\frac1{\\alpha-ix}\\right)^2\\,dx &=\\frac{4}{\\alpha^2}\\int_{-\\infty}^{+\\infty}\\frac{dx}{(1+(x/\\alpha)^2)^2}\\\\ &=\\frac{4}{\\alpha^2}\\int_{-\\pi/2}^{+\\pi/2}\\frac{1}{(1+\\tan^2y)^2}\\frac{\\alpha}{\\cos^2y}\\,dy\\\\ &=\\frac{4}{\\alpha}\\int_{-\\pi/2}^{+\\pi/2}\\frac{1}{\\frac{1}{\\cos^4y}}\\frac{1}{\\cos^2y}\\,dy\\\\ &=\\frac{4}{\\alpha}\\int_{-\\pi/2}^{+\\pi/2}{\\cos^2y}\\,dy\\\\ &=\\frac{4}{\\alpha}\\frac12\\left[\\cos y\\sin y +y\\right]_{-\\pi/2}^{+\\pi/2}\\\\ &=\\frac{2\\pi}{\\alpha} \\end{align*}\n\n\u2022 I have done the part where you'll combine the fractions to form $\\frac{4 \\alpha^2}{(\\alpha^2 + x^2)^2}$ but I have not thought of factoring out $\\alpha^2$ from the denominator. I will try that out thanks! \u2013\u00a0Aldon Feb 5 '16 at 14:02\n\u2022 I'm beginning to see that this can be done using $x = \\alpha \\tan \\theta$ but I keep on getting extremely weird answers. \u2013\u00a0Aldon Feb 5 '16 at 15:10\n\u2022 Nothing weird Aldon! I completed the answer. I hope it could be clearer now! :-) \u2013\u00a0Joe Feb 6 '16 at 1:06\n\u2022 I realized a few hours ago that I had wrong bounds for my integral when I converted it through $x = \\alpha \\tan \\theta$. Thanks a lot though! \u2013\u00a0Aldon Feb 6 '16 at 1:14\n\u2022 You're welcome! :-) \u2013\u00a0Joe Feb 6 '16 at 1:22\n\nThe integral is even in $\\alpha$, so assume that $\\alpha\\gt0$.\n\nReal Analysis Approach\n\nSubstitute $x\\mapsto\\alpha x$, \\begin{align} \\int_{-\\infty}^\\infty\\left(\\frac1{\\alpha+ix}+\\frac1{\\alpha-ix}\\right)^2\\mathrm{d}x &=\\int_{-\\infty}^\\infty\\left(\\frac{2\\alpha}{\\alpha^2+x^2}\\right)^2\\mathrm{d}x\\\\ &=\\frac4\\alpha\\int_{-\\infty}^\\infty\\frac1{(1+x^2)^2}\\mathrm{d}x\\tag{1} \\end{align} Furthermore \\begin{align} \\pi &=\\int_{-\\infty}^\\infty\\frac1{1+x^2}\\,\\mathrm{d}x\\tag{2}\\\\ &=\\int_{-\\infty}^\\infty\\frac{2x^2}{(1+x^2)^2}\\,\\mathrm{d}x\\tag{3}\\\\ &=\\int_{-\\infty}^\\infty\\frac{1+x^2}{(1+x^2)^2}\\,\\mathrm{d}x\\tag{4}\\\\ &=\\int_{-\\infty}^\\infty\\frac2{(1+x^2)^2}\\,\\mathrm{d}x\\tag{5}\\\\ \\end{align} Explanation:\n$(2)$: arctan integral\n$(3)$: integration by parts on $(2)$\n$(4)$: multiply the integrand in $(2)$ by $\\frac{1+x^2}{1+x^2}$\n$(5)$: $2$ times $(4)$ minus $(3)$\n\nPlug $(5)$ into $(1)$ to get \\begin{align} \\int_{-\\infty}^\\infty\\left(\\frac1{\\alpha+ix}+\\frac1{\\alpha-ix}\\right)^2\\mathrm{d}x &=\\frac4\\alpha\\frac\\pi2\\\\[3pt] &=\\frac{2\\pi}{\\alpha}\\tag{6} \\end{align}\n\nComplex Analysis Approach\n\nLet $z=ix$ and $\\gamma=[-iR,iR]\\cup iRe^{i[0,\\pi]}$ as $R\\to\\infty$.\n\nThen \\begin{align} \\int_{-\\infty}^\\infty\\left(\\frac1{\\alpha+ix}+\\frac1{\\alpha-ix}\\right)^2\\mathrm{d}x &=-i\\int_\\gamma\\left(\\frac1{\\alpha+z}+\\frac1{\\alpha-z}\\right)^2\\mathrm{d}z\\tag{7}\\\\ &=-i\\int_\\gamma\\left(\\frac1{(\\alpha+z)^2}+\\frac1{(\\alpha-z)^2}\\right)\\mathrm{d}z\\\\ &-\\frac i\\alpha\\int_\\gamma\\left(\\color{#C00000}{\\frac1{\\alpha+z}}+\\frac1{\\alpha-z}\\right)\\mathrm{d}z\\tag{8}\\\\ &=\\frac{2\\pi}{\\alpha}\\tag{9} \\end{align} Explanation:\n$(7)$: the integral along the arc where $|z|=R$ is $\\lesssim\\frac{4\\pi}R$, which vanishes as $R\\to\\infty$.\n$(8)$: $\\left(\\frac1{\\alpha+z}+\\frac1{\\alpha-z}\\right)^2 =\\frac1{(\\alpha+z)^2}+\\frac1{(\\alpha-z)^2}+\\frac2{(\\alpha+z)(\\alpha-z)} =\\frac1{(\\alpha+z)^2}+\\frac1{(\\alpha-z)^2}+\\frac1\\alpha\\left(\\frac1{\\alpha+z}+\\frac1{\\alpha-z}\\right)$\n$(9)$: only the $\\color{#C00000}{\\frac1{\\alpha+z}}$ term has a pole with non-zero residue inside $\\gamma$\n\n\u2022 Why is $\\left(\\frac{1}{\\alpha +z} + \\frac{1}{\\alpha - z}\\right)^2 = \\frac{1}{(\\alpha + z)^2} + \\frac{1}{(\\alpha - z)^2}$? \u2013\u00a0Aldon Feb 6 '16 at 1:07\n\u2022 you could also use the Fourier transform (harmonic analysis) approach. if $\\alpha < 0$ : $$\\frac{1}{\\alpha+ix} + \\frac{1}{\\alpha-ix} = 2 \\pi \\int_{-\\infty}^\\infty e^{-2 i \\pi x t} e^{2 \\pi \\alpha |t|} dt$$ so by the Parseval identity : $$\\int_{-\\infty}^\\infty \\left| \\frac{1}{\\alpha+ix} + \\frac{1}{\\alpha-ix} \\right |^2 dx = \\int_{-\\infty}^\\infty |2 \\pi e^{2 \\pi \\alpha |t|}|^2 dt = 4 \\pi^2 \\int_{-\\infty}^\\infty e^{4 \\pi \\alpha |t|} dt = 2\\frac{4 \\pi^2 }{4 \\pi |\\alpha|} = \\frac{2 \\pi }{|\\alpha|}$$ \u2013\u00a0reuns Feb 6 '16 at 1:45\n\u2022 and if $\\alpha > 0$ : $\\frac{1}{\\alpha+ix} + \\frac{1}{\\alpha-ix} = - \\left(\\frac{1}{-\\alpha-ix} + \\frac{1}{-\\alpha+ix} \\right)$ so the result is still $\\frac{2 \\pi }{|\\alpha|}$ \u2013\u00a0reuns Feb 6 '16 at 1:45\n\u2022 @user1952009: indeed, there are many ways to compute the integral. \u2013\u00a0robjohn Feb 6 '16 at 7:03\n\u2022 @Aldon: I have expanded the explanation of the Complex Analysis Approach. Note the explanation of $(8)$. \u2013\u00a0robjohn Feb 6 '16 at 7:39", "date": "2019-11-19 20:33:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1641774/integral-of-int-infty-infty-left-frac1-alpha-ix-frac1-alp", "openwebmath_score": 0.9800599217414856, "openwebmath_perplexity": 649.9063405843846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109503004293, "lm_q2_score": 0.8902942283051332, "lm_q1q2_score": 0.8767715050241656}}
{"url": "https://icitynews.com/viewtopic/9jhqv.php?id=quadratic-inequalities-grade-9-8492c8", "text": "These are interactive guides that focus on \u2026 [TAGALOG] Grade 9 Math Lesson: HOW TO SOLVE QUADRATIC INEQUALITY? Type 1: Algebraically solving x^2\\lt a^2. Each STEM Case level has an associated Handbook. Example: Solve the inequality x^2 \\lt 64 When solving quadratic inequalities it is important to remember there are two roots. 6. 9th - 11th grade. Edit. The range of values that satisfy the inequality is between -8 and 8.This can be expressed as, Definition Is an inequality that contains a polynomial of degree 2 and can be written in any of the following forms. Since the original inequality was less than or equal to, the boundary points are included. The above is an equation (=) but sometimes we need to solve inequalities like these: It then becomes a negative 9 plus 3, it becomes negative 3, and negative 3 will also become a negative 3. Visual intuition of what a quadratic inequality means. Solving quadratic equations by factoring. An inequality can therefore be solved graphically using a graph or algebraically using a table of signs to determine where the function is positive and negative. Quadratic Inequalities It is an inequality of the form of ax2 + bx + c (<, >, \u2264, \u2265) 0, where a, b and c are real numbers and a \u2260 0 3. Save. More Lessons for Grade 9 Math Math Worksheets Videos, worksheets, games and activities to help Algebra and Grade 9 students learn how to solve quadratic inequalities. Supplemental Math High School Grade 9 Q1.indd Created Date: So the width must be between 1 m and 7 m (inclusive) and the length is 8\u2212width. Quadratic inequalities can be of the following forms: \\begin{align*} ax^2 + bx + c > 0 \\\\ ax^2 + bx + c \\ge 0 \\\\ ax^2 + bx + c < 0 \\\\ ax^2 + bx + c \\le 0\\end{align*} To solve a quadratic inequality we must determine which part of the graph of a quadratic function lies above or below the $$x$$-axis. jamespearce. Thanks to all of you who support me on Patreon. But I know (and can verify from the above graph) that this quadratic only touches the axis from below; it is never fully above the axis. Multiple grade appropriate versions, or levels, exist for each STEM Case. 1. 0. Step by step guide to solve Solving Quadratic Inequalities . To solve a quadratic inequality, find the roots of its corresponding equality. $1 per month helps!! Graphing Quadratic inequalities \u2013 Example 1: Sketch the graph of $$y>3x^2$$. Some of the worksheets for this concept are Quadratic inequalities date period, Solving quadratic inequalities, Solving quadratic inequalities 1, Solving quadratic inequalities l3s1, Graphing quadratic, Graphing and solving quadratic inequalities, Solving quadratic inequalities algebraically work, Inequalitiesmep pupil text 16.$ < \u2013 3 , \u2013 2 > \\cup < 1 , 2 ]\\$ Quadratic inequalities worksheets Quadratic inequalities (4.4 MiB, 620 hits) 0 times. If the question was solve x^2=a^2 we would simple take the square root of both sides so that x=\\pm a.. quadratic inequalities, conduct a spin-o\u02db Think-Pair-Share activity (Lyman, 1981). \u2022 Diagrams are NOT accurately drawn, unless otherwise indicated. Choose a testing point and check the solution section. :) https://www.patreon.com/patrickjmt !! Nature of the roots of a quadratic equations. YouMore Kwenturuan tungkol sa kung paano gawin ang solving quadratic \u2026 This entry was posted in MATH and tagged math , math solver , mathway . OBJECTIVES : \u2022Define the interval correctly \u2022Express the Quadratic Inequalities \u2022Solve the Quadratic Inequalities through graphs and interval notation 3. Mathematics 9 Lesson 2: Quadratic Inequalities 1. Edit. These are imaginary answers and cannot be graphed on a real number line. Quadratic Inequalities Example: x2 + x \u2013 6 = \u2026 We need to find solutions. Yes we have two inequalities, because 3 2 = 9 AND (\u22123) 2 = 9. Before we get to quadratic inequalities, let's just start graphing some functions and interpret them and then we'll slowly move to the inequalities. Worked example 16: Solving quadratic inequalities \u2022 Answer the questions in the spaces provided \u2013 there may be more space than you need. Use the zero product property to find the solutions to the equation x2 - 9 = 16. x = -5 or x = 5 Adam is using the equation (x)(x + 2) = 255 to find two consecutive odd integers with a product of 255. In this case, we have drawn the graph of inequality using a pink color. Then fill in the region either inside or outside of it, depending on the inequality. Sum and product of the roots of a quadratic equations Algebraic identities. Therefore, the inequality x 2 + 2 x + 5 < 0 has no real solutions. An inequality can therefore be solved graphically using a graph or algebraically using a table of signs to determine where the function is positive and negative. Solving quadratic equations by quadratic formula. Solving quadratic inequalities, quadratic equations and quadratic simultaneous equations. Solving quadratic equations by completing square. By using this website, you agree to our Cookie Policy. The quadratic formula provides an easy and fast way to solve quadratic equations. Welcome to the presentation on quadratic inequalities. To graph a quadratic inequality, start by graphing the quadratic parabola. Solving absolute value equations Solving Absolute value inequalities However, this inequality is an \"or equal to\" inequality, so the \"equal\" part counts as part of the solution. GCSE (1 \u2013 9) Quadratic Inequalities Name: _____ Instructions \u2022 Use black ink or ball-point pen. ... 2014/11/03 \u0622 3.1-Quadratic Functions & Inequalities Quadratic. MHR \u2022 Pre-Calculus 11 Solutions Chapter 9 Page 1 of 84 Chapter 9 Linear and Quadratic Inequalities Section 9.1 Linear Inequalities in Two Variables Section 9.1 Page 472 Question 1 a) y < x + 3 Try (7, 10). You da real mvps! To solve a quadratic inequality we must determine which part of the graph of a quadratic function lies above or below the $$x$$-axis. 0% average accuracy. The method of completing the square provides a way to derive a formula that can be used to solve any quadratic equation. I need to find where y = \u2013x 2 + 6x \u2013 9 is above the axis. And that represents the graph of the inequality. an hour ago. ax2 + bx + c > 0 ax2 + bx + c \u2265 0 ax2 + bx + c < 0 ax2 + bx + c \u2264 0 where a, b, and c are real numbers and a \u2260 0. Let's say I had f of x is equal to x squared plus x minus 6. Transcript of Grade 9: Mathematics Unit 2 Quadratic Functions. This is a quadratic inequality. GCSE Grade 9 Quadratic Inequalities, Quadratic Equations and Quadratic Simultaneous Equations. A Quadratic Equation in Standard Form ( a , b , and c can have any value, except that a can't be 0.) Since this statement is false, the region between -9 and 1 is not correct. Siyavula's open Mathematics Grade 10 textbook, chapter 4 on Equations and inequalities covering Solving quadratic equations All answers are included. The easiest way to solve these inequalities is to draw the graphs and read the solutions of them. The same basic concepts apply to quadratic inequalities like $$y x^2 -1$$ from digram 8. Quadratic Inequalities (Visual Explanation) How to solve a quadratic inequality. Quadratic Inequalities 2. Mathematics. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. $$.. The set of solutions is the union of solutions of these two systems. This is the same quadratic equation, but the inequality has been changed to$$ \\red . Solving Quadratic Inequalities DRAFT. Displaying top 8 worksheets found for - Quadratic Inequalities. Grade mathematics: Quadratic Inequalities 1. Try (\u20137, 10). MathematicsLearners Material9Module 2:Quadratic FunctionsThis instructional material was collaboratively developed and reviewed byeducators from public and private schools, colleges, and/or universities. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. A quadratic inequality is one that can be written in one of the following standard forms: $$ax^2+bx+c>0, ax^2+bx+c<0, ax^2+bx+c\u22650, ax^2+bx+c\u22640$$ Solving a quadratic inequality is like solving equations. To solve a quadratic inequality we must determine which part of the graph of a quadratic function lies above or below the $$x$$-axis. Left Side Right Side Left Side Right Side y x + 3 y x + 3 = 10 = 7 + 3 = 10 = \u20137 + 3 = 10 = \u20134 \u2022 You \u2026 Check: Consider the standard form of the quadratic equation \\(ax^2 + bx + \u2026 It can be used in conjunction with Module 1 Lesson 7 of the DepEd Grade 9 Learning Modules. In this non-linear system, users are free to take whatever path through the material best serves their needs. Since this quadratic is not factorable using rational numbers, the quadratic formula will be used to solve it. Free quadratic inequality calculator - solve quadratic inequalities step-by-step This website uses cookies to ensure you get the best experience. Grade 9 - Mathematics Topic 1.8 : Solving Quadratic Inequalities 2. by jamespearce. Negative 3 squared is positive 9, you have a negative out front, it becomes negative 9 plus 6, which is negative 3. an hour ago. Grade 9 \u2013 Algebra Solving Quadratic Inequalities \u2013 This activity provides practice in using the graph of a quadratic function to solve quadratic inequalities. DLM 2 \u2013 Unit 2: Quadratic equations Summative Test for Quadratic Equations EASE Module 3 Quadratic Functions EASE Module 2 Quadratic Equations Deriving formula and solving for surface areas of solids GRADE 10 \u2013 RATIONAL ROOT THEOREM MATH 9 Q1M2.2 - Solving Quadratic Equations MATH 9 Q1M1.1 - Quadratic Equations DIG DOWN MY ROOTS 1 The software Geogebra is used as graphing tool. \u2022 Answer all questions. Worked example 16: Solving quadratic inequalities These unique features make Virtual Nerd a viable alternative to private tutoring. Add 4 to both sides of each inequality: 1 \u2264 W \u2264 7. So it must be the region on either side of those points. If I were to graph it, Let's try it with 3, as well-- if we put a 3 there, 3 squared is 9. Solve the quadratic inequality: x2+11x+10\u22650x^2+11x+10\\ge0x2+11x+10\u22650 Solving Quadratic Inequalities DRAFT.", "date": "2021-04-19 02:16:44", "meta": {"domain": "icitynews.com", "url": "https://icitynews.com/viewtopic/9jhqv.php?id=quadratic-inequalities-grade-9-8492c8", "openwebmath_score": 0.6250958442687988, "openwebmath_perplexity": 614.3550422102876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9766692366242304, "lm_q2_score": 0.8976952941600963, "lm_q1q2_score": 0.8767513776685052}}
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There are prizes for the winners.\n\n# Which of the following is/are terminating decimal(s)?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nCurrent Student\nJoined: 22 Jul 2014\nPosts: 123\nConcentration: General Management, Finance\nGMAT 1: 670 Q48 V34\nWE: Engineering (Energy and Utilities)\nWhich of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n06 Aug 2014, 09:08\n1\n3\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n67% (01:10) correct 33% (01:26) wrong based on 193 sessions\n\n### HideShow timer Statistics\n\nWhich of the following is/are terminating decimal(s)?\n\nI 299/(32^123)\nII 189/(49^99)\nIII 127/(25^37)\n\nA) I only\nB) I and III\nC) II and III\nD) II only\nE) I, II, III\nTutor\nJoined: 20 Apr 2012\nPosts: 99\nLocation: Ukraine\nGMAT 1: 690 Q51 V31\nGMAT 2: 730 Q51 V38\nWE: Education (Education)\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n06 Aug 2014, 10:24\n10\n6\nTerminating decimal is a such decimal that has only a finite number of non-zero digits. That's why if you will write such decimal as a fraction you will have as a denominator 10,100, 1000, etc. Since the fraction can be simplified, you can have in the denominator at the end some product of 2 or 5.\n\nFor example:\n$$0.4=4/10=2/5$$, 5 in the denominator\n$$0.25=25/100=1/4$$, 4=2*2 in the denominator.\n\nBut anyway, if you have terminating decimal you can't have anything other 2 or 5 in prime factorization of denominator. So we have a rule:\n\nThe fraction expressed as a decimal will be the terminating decimal if it can be presented as $$\\frac{a}{{2^n\\cdot 5^m}}$$ where $$a$$ is an integer, $$m=0,1,2,3..$$. , and $$n=0,1,2,3...$$ .\n\nOr\n\nif a fraction has in denominator any prime factor different from 2 and 5, such fraction will be infinite decimal.\n\nI. Has only 2 as prime factor, since $$32=2^5$$. Terminating\nII. Has 7 as prime factor. Infinite\nIII. Has only 5 as prime factor. Terminating\n\nHope this helps!:)\n_________________\n\nI'm happy, if I make math for you slightly clearer\nAnd yes, I like kudos:)\n\n##### General Discussion\nManager\nJoined: 18 Jul 2013\nPosts: 73\nLocation: Italy\nGMAT 1: 600 Q42 V31\nGMAT 2: 700 Q48 V38\nGPA: 3.75\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n06 Aug 2014, 13:57\nsmyarga wrote:\nTerminating decimal is a such decimal that has only a finite number of non-zero digits. That's why if you will write such decimal as a fraction you will have as a denominator 10,100, 1000, etc. Since the fraction can be simplified, you can have in the denominator at the end some product of 2 or 5.\n\nFor example:\n$$0.4=4/10=2/5$$, 5 in the denominator\n$$0.25=25/100=1/4$$, 4=2*2 in the denominator.\n\nBut anyway, if you have terminating decimal you can't have anything other 2 or 5 in prime factorization of denominator. So we have a rule:\n\nThe fraction expressed as a decimal will be the terminating decimal if it can be presented as $$\\frac{a}{{2^n\\cdot 5^m}}$$ where $$a$$ is an integer, $$m=0,1,2,3..$$. , and $$n=0,1,2,3...$$ .\n\nOr\n\nif a fraction has in denominator any prime factor different from 2 and 5, such fraction will be infinite decimal.\n\nI. Has only 2 as prime factor, since $$32=2^5$$. Terminating\nII. Has 7 as prime factor. Infinite\nIII. Has only 5 as prime factor. Terminating\n\nHope this helps!:)\n\nGreat explanation, i forgot the concept of terminating decimals.\nTutor\nJoined: 20 Apr 2012\nPosts: 99\nLocation: Ukraine\nGMAT 1: 690 Q51 V31\nGMAT 2: 730 Q51 V38\nWE: Education (Education)\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n06 Aug 2014, 14:08\n3\nCurrent Student\nJoined: 22 Jul 2014\nPosts: 123\nConcentration: General Management, Finance\nGMAT 1: 670 Q48 V34\nWE: Engineering (Energy and Utilities)\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n06 Aug 2014, 23:36\nTutor\nJoined: 20 Apr 2012\nPosts: 99\nLocation: Ukraine\nGMAT 1: 690 Q51 V31\nGMAT 2: 730 Q51 V38\nWE: Education (Education)\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Aug 2014, 00:41\n1\nalphonsa wrote:\n\nI like very much this source 700-800-level-quant-problem-collection-detailed-solutions-137388.html\nYou can find there problems divided by topics with detailed explanation.\n_________________\n\nI'm happy, if I make math for you slightly clearer\nAnd yes, I like kudos:)\n\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1825\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Aug 2014, 18:46\n1\nalphonsa wrote:\nWhich of the following is/are terminating decimal(s)?\n\nI 299/(32^123)\nII 189/(49^99)\nIII 127/(25^37)\n\nA) I only\nB) I and III\nC) II and III\nD) II only\nE) I, II, III\n\nI\n\n$$\\frac{299}{32^{123}} = \\frac{299}{(2^5)^{123}} = \\frac{299}{2^{(5*123)}}$$\n\n$$= \\frac{299}{2^{(5*123)}} * \\frac{5^{(5*123)}}{5^{(5*123)}}$$\n\n$$= \\frac{299 * 5^{(5*123)}}{10^{(5*123)}}$$ >> Power of 10 in denominator, this is a terminating decimal\n\nII\n\n$$\\frac{189}{49^{99}} = \\frac{7*27}{7^{198}} = \\frac{27}{7^{197}}$$ >> This is not a terminating decimal\n\nIII\n\n$$\\frac{127}{25^{37}} = \\frac{127}{5^{74}} = \\frac{127}{5^{74}} * \\frac{2^{74}}{2^{74}}$$\n\n$$= \\frac{127 * 2^{74}}{10^{74}}$$ >> Power of 10 in denominator, this is a terminating decimal\n\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nManager\nJoined: 22 Feb 2009\nPosts: 171\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Aug 2014, 21:16\nPareshGmat wrote:\nalphonsa wrote:\nWhich of the following is/are terminating decimal(s)?\n\nI 299/(32^123)\nII 189/(49^99)\nIII 127/(25^37)\n\nA) I only\nB) I and III\nC) II and III\nD) II only\nE) I, II, III\n\nI\n\n$$\\frac{299}{32^{123}} = \\frac{299}{(2^5)^{123}} = \\frac{299}{2^{(5*123)}}$$\n\n$$= \\frac{299}{2^{(5*123)}} * \\frac{5^{(5*123)}}{5^{(5*123)}}$$\n\n$$= \\frac{299 * 5^{(5*123)}}{10^{(5*123)}}$$ >> Power of 10 in denominator, this is a terminating decimal\n\nII\n\n$$\\frac{189}{49^{99}} = \\frac{7*27}{7^{198}} = \\frac{27}{7^{197}}$$ >> This is not a terminating decimal\n\nIII\n\n$$\\frac{127}{25^{37}} = \\frac{127}{5^{74}} = \\frac{127}{5^{74}} * \\frac{2^{74}}{2^{74}}$$\n\n$$= \\frac{127 * 2^{74}}{10^{74}}$$ >> Power of 10 in denominator, this is a terminating decimal\n\nI could solve the problem by using rules of terminating number, but your solution is so interesting. Thanks for sharing!!! + 1 kudos\n_________________\n\n.........................................................................\n+1 Kudos please, if you like my post\n\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1825\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Aug 2014, 21:44\nPareshGmat wrote:\nalphonsa wrote:\nWhich of the following is/are terminating decimal(s)?\n\nI 299/(32^123)\nII 189/(49^99)\nIII 127/(25^37)\n\nA) I only\nB) I and III\nC) II and III\nD) II only\nE) I, II, III\n\nI\n\n$$\\frac{299}{32^{123}} = \\frac{299}{(2^5)^{123}} = \\frac{299}{2^{(5*123)}}$$\n\n$$= \\frac{299}{2^{(5*123)}} * \\frac{5^{(5*123)}}{5^{(5*123)}}$$\n\n$$= \\frac{299 * 5^{(5*123)}}{10^{(5*123)}}$$ >> Power of 10 in denominator, this is a terminating decimal\n\nII\n\n$$\\frac{189}{49^{99}} = \\frac{7*27}{7^{198}} = \\frac{27}{7^{197}}$$ >> This is not a terminating decimal\n\nIII\n\n$$\\frac{127}{25^{37}} = \\frac{127}{5^{74}} = \\frac{127}{5^{74}} * \\frac{2^{74}}{2^{74}}$$\n\n$$= \\frac{127 * 2^{74}}{10^{74}}$$ >> Power of 10 in denominator, this is a terminating decimal\n\nI could solve the problem by using rules of terminating number, but your solution is so interesting. Thanks for sharing!!! + 1 kudos\n\nThank you so much for calling the solution interesting\n\nOne thing to add. In such type of problems, just look out for powers of 2 and/or powers of 5 (because they only compose 10) in denominator\n\nFor any other number, its not possible (Subject to complete simplification of the term)\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nIntern\nJoined: 12 Feb 2013\nPosts: 7\nLocation: India\nConcentration: Finance, Marketing\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Aug 2014, 23:11\nThe fraction will have terminating decimal if and only if the denominator of the fraction is of the form (2^n)(5^m).\n\nIf you look at the denominators:\n1) (32^123) => (2^(5*123))(5^0) => Terminating Decimal\n2) (49^99) => Can't be expressed as (2^n)(5^m) => Non terminating Decimal\n3) (25^37) => (2^0)(5^(2*37)) => Terminating Decimal\n\nHence 1 and 3 are terminating decimals => Choice [B]\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 9138\nRe: Which of the following is/are terminating decimal(s)?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Nov 2017, 11:17\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: Which of the following is/are terminating decimal(s)? &nbs [#permalink] 11 Nov 2017, 11:17\nDisplay posts from previous: Sort by", "date": "2018-12-12 23:43:46", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/which-of-the-following-is-are-terminating-decimal-s-175877.html", "openwebmath_score": 0.8832159638404846, "openwebmath_perplexity": 6271.119246223313, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 0.9453993972951024, "lm_q2_score": 0.9273632906279589, "lm_q1q2_score": 0.8767286960332752}}
{"url": "https://puzzling.stackexchange.com/questions/99852/painted-faces-on-a-cube/99875", "text": "# Painted faces on a cube\n\nHere's another challenge I used to give to my students:\n\nLet's begin with a bunch of little white cubes assembled into a big white cube. All the little white cubes are equal.\nThen I decide to paint some of the faces of the big cube blue. Afterwards I break the big cube apart into the smaller ones.\nOnly $$24$$ of the smaller cubes remain completely white.\n\nHow many cubes formed the big one? How many big faces did I paint?\n\nUsually I use this problem to show how much we can deduce with so little information.\n\nIt hinges on the following insight:\n\nDon't break the cube apart but keep the small cubes packed together. If you paint some faces of the cube, and remove the painted layers, you have a cuboid where each dimension is at zero, one or two units shorter than the original cube.\n\nThat means that the $$24$$ unpainted cubes\n\nform a three-dimensional cuboid with dimensions that differ by at most two. The only way to factor $$24$$ like that is as $$24=2\\times3\\times4$$. Therefore you started with a $$4\\times4\\times4$$ cube and removed three layers, two of them opposite each other.\n\n\u2022 That is a GREAT approach! If I could I give you $10$ upvotes! :)\n\u2013\u00a0Pspl\nJul 10 '20 at 8:10\n\nIn general, you can paint 6 sides of an N\u00d7N\u00d7N cube in 10 ways:\n\n* 0 sides\n* 1 side\n* 2 opposite sides\n* 3 adjacent sides (a corner)\n* 3 sides, two of them opposite (a band)\n* 4 sides except 2 opposite\n* 4 sides except 2 adjacent\n* 5 sides\n* 6 sides\n\nwhich leaves you with these numbers of unpainted sub-cubes, respectively:\n\n* N\u00d7N\u00d7N\n* N\u00d7N\u00d7(N\u20131)\n* N\u00d7(N\u20131)\u00d7(N\u20131)\n* N\u00d7N\u00d7(N\u20132)\n* (N\u20131)\u00d7(N\u20131)\u00d7(N\u20131)\n* N\u00d7(N\u20131)\u00d7(N\u20132)\n* N\u00d7(N\u20132)\u00d7(N\u20132)\n* (N\u20131)\u00d7(N\u20131)\u00d7(N\u20132)\n* (N\u20131)\u00d7(N\u20132)\u00d7(N\u20132)\n* (N\u20132)\u00d7(N\u20132)\u00d7(N\u20132)\n\nGiven a number of unpainted cubes,\n\none needs to find a factorisation that fits one of those expressions.\n\nAs Jaap Scherphuis shows, the value of 24 fits just one, which yields only 1 answer.\n\n\u2022 Note that all but three of those expressions are definitely ambiguous. Jul 10 '20 at 9:11\n\u2022 @JaapScherphuis Yes. That means solution needn't be unique. If one takes a cube 6\u00d76\u00d76 and removes one layer from all faces, or one removes one layer from three adjacent faces of a 5\u00d75\u00d75 cube, or one leaves a 4\u00d74\u00d74 cube intact, the number of unit cubes left will be the same. Jul 10 '20 at 17:25\n\nMy approach was to...\n\nstart off by bounding the possible dimensions of the big cube, and then find a valid permutation of face-painting from that subset. But after bounding the size, finding a permutation became unnecessary.\n\nTo determine...\n\na maximum bound: We can be certain that the inner cube of dimensions $$(N-2)$$ will remain untouched no matter how the big cube is painted, so that must hold an equal or fewer number of small cubes than how many are left untouched. $$(N-2)^3 \\leq 24$$. Since $$3^3$$ is already $$27$$,\n\n$$N<5$$ is a maximum bound.\n\nThen to determine...\n\na minimum bound: If at least one face of the big cube is painted, then the largest number of untouched unit cubes possible is the total number of cubes, minus one face of cubes. $$N^3-N^2 \\geq 24$$. Since $$3^3-3^2$$ is only $$18$$,\n\n$$N>3$$ is a minimum bound.\n\nTherefore...\n\n$$N$$ must be greater than $$3$$ and less than $$5$$, so $$4$$ is the only possible answer. It is not necessary to determine exactly what pattern of sides were painted to be certain of the answer.\n\nWith a 4x4x4 cube, there are 56 outer cubes; 40/16 paint/blank. The most cubes you can paint with a face is 16, so you need more than 2 faces. meanwhile 4 faces painted leaves only 8 or 10 cubes blank. So 3 faces must be painted. Finding the permutation is still unnecessary for the total answer. But to be specific, 16 for one face, +12 for each single-adjacent space. So 3 faces painted in a U-shape on a 4x4x4 cube is the full description of the cube.\n\n\u2022 I too started with finding the bounds. +1. But the second half of the question is \"How many big faces did I paint?\" You do need the pattern (subject to rotation&reflection) to determine that. Jul 12 '20 at 19:50\n\u2022 True - I could have elaborated on the final step, but everyone else had already given accurate answers - finding the number of painted sides is trivial once the size of the cube is known. I was just trying to emphasize that this piece of information could be gathered through simple, broad deduction without having to guess and check through permutations. Jul 16 '20 at 19:09\n\nYou can represent the unpainted volume like this:\n\n$$(n-x_1-x_2)(n-y_1-y_2)(n-z_1-z_2) = 24$$\nwhere $$x_1,x_2,y_1,...$$ are either $$0$$ or $$1$$, representing whether the face was painted.\n\nThis means all we need to do is\n\nFactorise $$24$$ into 3 factors such that any two factors are at most $$2$$ apart\n\nWe know that\n\nThe prime factorisation of $$24$$ is $$2\\cdot 2\\cdot 2\\cdot 3$$\n\nWhich means you have the following combinations:\n\n\u2022 $$2\\cdot 3\\cdot 4$$\n\u2022 $$2\\cdot 2\\cdot 6$$\n\nOnly one of which works, meaning that:\n\n$$(n-x_1-x_2)(n-y_1-y_2)(n-z_1-z_2)=2\\cdot 3\\cdot 4\\\\\\therefore(n-1-1)(n-1-0)(n-0-0)=2\\cdot 3\\cdot 4$$\nSince $$x_1+x_2+y_1+y_2+z_1+z_2$$ sides are painted, we know that $$3$$ sides were painted.\nSince $$(n-0-0)=4$$, we know that the original cube side length must have been $$4$$.\n\nI haven't looked at the other answers so I hope I haven't done anything too similar to anyone else :)\n\n\u2022 Having looked at the top answer, I realise this is essentially just a longer version of that, but I think it adds a little bit more explanation so I'll keep it up :) Jul 11 '20 at 16:26\n\nHere is my way of looking at it:\n\nI started off by representing situations with my favorite variable \u2018x\u2019 (I\u2019m more of a math kind of person). For example, let x be the length of a side of the larger cube, so $$x^3-x^2=24$$ represents a situation where 1 face is painted blue, and those blue squares are removed ($$x^3$$ is the whole cube volume, so subtracting $$x^2$$ gives you the leftover volume after removing one face).\n\nThen we move on to 2 blue faces...\n\n$$x^3-2x^2=24$$ represents painting 2 blue faces that are opposite each other so the faces aren\u2019t touching. And $$x^3-2x^2+x=24$$ represents 2 blue faces touching each other. The \u2018plus x\u2019 accounts for the fact that counting $$x^2$$ twice when the sides are touching means you are counting one edge two times; $$2x^2$$ tells you how many blue faces there are for the small cubes, so adding x will tell you how many cubes have any blue.\n\nAnd so on... keeping in mind how many touching faces there are, we find that:\n\n$$x^3-3x^2+2x=24$$ is the only equation out of all possible numbers of blue faces where a solution for x is a whole number ($$x=4$$). Remember that x represents the length of a side of the cube, so if the above equation is the only one with a whole number x, it\u2019s the only one with a whole number side length and the only solution that works. The equation $$x^3-3x+2x=24$$ represents 3 blue sides, with 2 over laps (see coefficients). So since $$x=4$$, this means the whole cube has volume $$4^3$$ since x is AGAIN, the length of a side of the big cube.\n\nTherefore...\n\nYou used 64 smaller cubes and painted 3 faces so that there are 2 edges where the faces touch (as in 2 of the faces are opposite each other).", "date": "2022-01-21 06:38:14", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/99852/painted-faces-on-a-cube/99875", "openwebmath_score": 0.671185314655304, "openwebmath_perplexity": 600.0785619574544, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850902023109, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.8767261392209519}}
{"url": "https://math.stackexchange.com/questions/3195349/totally-ramified-extension-of-mathbbq-p-which-is-not-of-a-form-mathbbq", "text": "# Totally ramified extension of $\\mathbb{Q}_{p}$ which is not of a form $\\mathbb{Q}_{p}(\\sqrt[n]{pu})$\n\nIt is known that a finite extension $$K/\\mathbb{Q}_{p}$$ is totally ramified if and only if $$K = \\mathbb{Q}_{p}(\\alpha)$$ where $$\\alpha$$ is a root of Eisenstein polynomial. Is there any totally ramified extension that is not of the form $$\\mathbb{Q}_{p}(\\sqrt[n]{pu})$$ for some $$u\\in \\mathbb{Z}_{p}^{\\times}$$? Every degree 2 totally ramified extensions has this form, but I don't know whether this is also true for degree 3 or higher. Thanks in advance.\n\n\u2022 Lately I\u2019ve been convincing myself of things that were not true, but I have convinced myself just now that $\\Bbb Q_2(\\sqrt3\\,)$ is not of the form you claim for quadratic extensions. Could you check? Apr 21, 2019 at 2:42\n\u2022 Yes, this is a phenomenon of wild ramification, when the residue characteristic divides the degree of the extension. For instance, I believe that $X^3+3X+3$ over $\\Bbb Q_3$ gives another example. It\u2019s a matter of the Herbrand transition function, or if you like, the location of the \u201cbreaks\u201d in the ramification filtration. Apr 21, 2019 at 5:10\n\u2022 @Lubin If you find the time to flesh that out in an answer, I'm sure I won't be the only user who will appreciate it :-) Apr 21, 2019 at 7:12\n\u2022 I\u2019ll see what I can do, @JyrkiLahtonen. Got lots on my plate the next 48 hours. Apr 21, 2019 at 18:10\n\u2022 Thanks, @Lubin. We are not in a hurry here. Apr 21, 2019 at 18:11\n\nThere is a general theorem that every tamely totally ramified extension of $$\\mathbf Q_p$$ with degree $$n$$ has the form $$\\mathbf Q_p(\\sqrt[n]{\\pi})$$ for some prime $$\\pi$$ in $$\\mathbf Z_p$$, so $$\\pi = pu$$ for a unit $$u$$ in $$\\mathbf Z_p$$. (There is a similar theorem over other local fields.) So if you want a totally ramified extension not of that form you need $$n$$ to be divisible by $$p$$.\n\nLet's try $$n=p$$. Something we can say about extensions $$\\mathbf Q_p(\\sqrt[p]{pu})$$ for $$p>2$$ is that they are not Galois over $$\\mathbf Q_p$$: a field $$K$$ containing a full set of roots of $$x^p - pu$$ must contain the nontrivial $$p$$th roots of unity, and those have degree $$p-1$$ over $$\\mathbf Q_p$$ so $$[K:\\mathbf Q_p]$$ is divisible by $$p-1$$. Therefore $$[K:\\mathbf Q_p] \\not= p$$ when $$p>2$$. Thus a Galois totally ramified extension of $$\\mathbf Q_p$$ having degree $$p$$ can't have the form $$\\mathbf Q_p(\\sqrt[p]{pu})$$.\n\nEvery totally ramified abelian Galois extension of $$\\mathbf Q_p$$ with degree divisible by $$p$$ contains a subextension with degree $$p$$ since the Galois group has a subgroup of index $$p$$: in an abelian group of order $$n$$ there is a subgroup of each order dividing $$n$$ and thus also a subgroup of each index dividing $$n$$ by using a subgroup of order equal to the complementary factor in $$n$$ of the desired index. Subextensions of totally ramified extensions are totally ramified and subextensions of abelian Galois extensions are abelian Galois extensions. Thus all we need to do now is find a totally ramified abelian Galois extension of $$\\mathbf Q_p$$ with degree divisible by $$p$$ and inside of it there are extensions of degree $$p$$, all of which are examples of the kind being sought (not having the form $$\\mathbf Q_p(\\sqrt[n]{pu})$$).\n\nThe easiest choice is a cyclotomic extension: $$\\mathbf Q_p(\\zeta_{p^2})$$ where $$\\zeta_{p^2}$$ is a root of unity of order $$p^2$$. This field has degree $$p^2-p$$ over $$\\mathbf Q_p$$, with cyclic Galois group $$(\\mathbf Z/p^2\\mathbf Z)^\\times$$, so the field contains a unique subextension with degree $$p$$ over $$\\mathbf Q_p$$, namely the field fixed by the unique subgroup of the Galois group with order $$(p^2-p)/p = p-1$$. That subgroup is the solutions to $$a^{p-1} \\equiv 1 \\bmod p^2$$, and a generator of this extension over $$\\mathbf Q_p$$ is $$\\sum_{a^{p-1} = 1} \\zeta_{p^2}^a$$ where the sum runs over all solutions of $$a^{p-1} \\equiv 1 \\bmod p^2$$.\n\nExample when $$p=3$$: $$a^2 \\equiv 1 \\bmod 9$$ has solutions $$\\pm 1 \\bmod 9$$ and $$\\zeta_{9} + \\zeta_9^{-1}$$ has minimal polynomial $$f(x) = x^3 - 3x + 1$$. Then $$f(x-1) = x^3 - 3x^2 + 3$$ is Eisenstein at $$3$$; the polynomial $$f(x+1)$$ is not. I did my calculation of the minimal polynomial in $$\\mathbf C$$, which is okay since a primitive $$p$$th-power root of unity has the same degree over $$\\mathbf Q_p$$ as it does over $$\\mathbf Q$$, so the structure of the intermediate fields in a $$p$$th-power cyclotomic extension over $$\\mathbf Q_p$$ and over $$\\mathbf Q$$ are the same.\n\nExample when $$p=5$$: solutions to $$a^4 \\equiv 1 \\bmod 25$$ are $$1, 7, 18$$, and $$24$$, and $$\\zeta_{25} + \\zeta_{25}^7 + \\zeta_{25}^{18} + \\zeta_{25}^{24}$$ has minimal polynomial over $$\\mathbf Q_5$$ equal to $$g(x) = x^5 - 10x^3 + 5x^2 + 10x + 1$$. (Note $$g(x-1) = x^5 - 5x^4 + 25x^2 - 25x + 5$$ is Eisenstein at 5; the polynomial $$g(x+1)$$ is not Eisenstein at $$5$$.)\n\n\u2022 Thanks! So if I understood correctly, the two examples are Galois extension of degree $p$ which can't be of the form, and they are Galois because they are subextensions of cyclotomic extension (which is abelian). Is this right? Apr 23, 2019 at 20:13\n\u2022 That is correct.\n\u2013\u00a0KCd\nApr 23, 2019 at 20:28\n\nIn response to Jyrki Lahtonen\u2019s request, I\u2019ll make an attempt to describe what\u2019s going on here.\n\nThe Hasse-Herbrand transition function is a concave polygonal real-valued function on $$\\Bbb R$$ that encapsulates much (but not all) of the information that comes out of the study of the higher ramification of a separable extension of local fields. You can read all about the subject in Chapter IV of Serre\u2019s Corps Locaux (translated as Local Fields). What you see below will look nothing like Serre\u2019s treatment, however. The least of the differences is that the traditional coordinatization of the plane, as in Serre, places the vertex describing the tame part of a totally ramified extension at the origin. My coordinatization puts this vertex at $$(1,1)$$.\n\nPart I is to describe the Newton Copolygon. I won\u2019t relate it to the more familiar Polygon, but you will see the connection. Let $$f(X)=\\sum_na_nX^n\\in\\mathfrak o[X]$$, where for specificity\u2019s sake I will suppose that $$\\mathfrak o$$ is the ring of integers in a finite extension $$k$$ of $$\\Bbb Q_p$$, and that we are using the (additive) valuation $$v$$ on $$k$$ normalized so that $$v(p)=1$$. For each nonzero monomial $$a_nX^n$$, draw the half-plane $$\\Pi_n$$ described in $$\\Bbb R^2$$ as all points $$(\\xi,\\eta)$$ satisfying $$\\eta\\le n\\xi+v(a_n)$$. Then form the convex set $$\\bigcap_n\\Pi_n$$. This is the copolygon, but I hope I won\u2019t confuse matters too badly by calling the \u201ccopolygon function\u201d the function $$v_f$$ whose graph is the boundary of the convex set just described. You see, for instance, that if $$f(X)=pX+pX^2+X^3$$ the copolygon\u2019s boundary has only one vertex, at $$(\\frac12,\\frac32)$$, with slope $$3$$ to the left and slope $$1$$ to the right. You see without difficulty that as long as $$g$$ has no constant term, $$v_{f\\circ g}=v_f\\circ v_g$$.\n\nPart II. Without saying what the \u201clower breaks\u201d and \u201cupper breaks\u201d of the ramification filtration of the Galois group of a Galois extension $$K\\supset k\\supset\\Bbb Q_p$$ are, I simply proclaim that the Herbrand function is the polygonal real-valued function $$\\psi^K_k$$ whose only vertices are at each break point $$(\\ell_i,u_i)$$. The lovely fact about the transition functions is that if $$L\\supset K\\supset k$$, then $$\\psi^L_k=\\psi^K_k\\circ\\psi^L_K$$. The transition function $$\\psi^K_k$$ is an invariant of the extension, not depending on any choices.\n\nPart III is to relate these two polygonal functions, though this is not the place to explain why they are connected. Though the traditional description of the transition function, as in Serre, always starts from a Galois group, you will notice that there is no mention of groups below. For simplicity, I\u2019ll describe only $$\\psi^k_{\\Bbb Q_p}$$ for $$k$$ totally ramified over $$\\Bbb Q_p$$, since that\u2019s enough to answer Saewoo Lee\u2019s question.\n\nLet $$\\mathfrak o$$ be the ring of integers of $$k$$, and $$\\pi$$ a prime element (generator of the maximal ideal), and let $$F(X)$$ be the minimal $$\\Bbb Q_p$$-polynomial for $$\\pi$$. Form the polynomial $$f(X)=F(X+\\pi)$$, so that $$f$$ has no constant term. Now take the copolygon function $$v_f$$ of this $$f$$, and stretch it horizontally by a factor of $$e^k_{\\Bbb Q_p}=[k:\\Bbb Q_p]$$, to get $$\\psi^k_{\\Bbb Q_p}$$. That is, $$\\psi^k_{\\Bbb Q_p}(\\xi)=v_f(\\xi\\,/\\,e)$$.\n\nLet\u2019s work out three examples, namely $$\\Bbb Q_2(\\sqrt{2u}\\,)$$, $$\\Bbb Q_2(\\sqrt3\\,)$$, and $$\\Bbb Q_3(\\rho)$$ where the minimal polynomial for $$\\rho$$ is $$X^3-3X-3$$.\n\nFirst, over $$\\Bbb Q_2$$, a prime is $$\\pi=\\sqrt{2u}$$, minimal polynomial $$F(X)=X^2-2u$$, giving $$f(X)=X^2+2\\pi X$$. The copolygon has unique vertex at $$(\\frac32,3)$$, and the transition function has unique vertex at $$(3,3)$$. (The initial segment of $$\\psi^K_k$$ will always have slope $$1$$.)\n\nSecond, over $$\\Bbb Q_2$$, a choice for a prime of $$\\Bbb Z_2[\\sqrt3\\,]$$ is $$\\sqrt3-1$$, with minimal polynomial $$F(X)=X^2+2X-2$$, so that $$f(X)=X^2+2\\pi X+2X=X^2+2(1+\\pi)X$$. The polygon has its one vertex at $$(1,2)$$, so that $$\\psi$$ has its vertex at $$(2,2)$$, enough to show that $$\\Bbb Q_2(\\sqrt3\\,)$$ is not of form $$\\Bbb Q_2(\\sqrt{2u}\\,)$$.\n\nThird, over $$\\Bbb Q_3$$ with $$F(X)=X^3-3X-3$$, we get $$f(X)=X^3+3\\rho X^2+3\\rho^2X-3X$$, in which only the monomials $$X^3$$ and $$3(\\rho-1)X$$ count, so that the copolygon has its vertex at $$(\\frac12,\\frac32)$$, and the transition function\u2019s vertex is at $$(\\frac32,\\frac32)$$.\n\nI\u2019ll leave it to you to show that the vertex of the transition function for $$\\Bbb Q_3(\\sqrt[3]{3u}\\,)$$ is at $$(\\frac52,\\frac52)$$. (Don\u2019t be surprised that these vertices don\u2019t have integral coordinates. That\u2019s guaranteed only for normal, abelian extensions, by Hasse-Arf, and the cubic extensions here are neither.)", "date": "2022-08-19 19:40:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3195349/totally-ramified-extension-of-mathbbq-p-which-is-not-of-a-form-mathbbq", "openwebmath_score": 0.8768288493156433, "openwebmath_perplexity": 143.00639414734928, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308810508339, "lm_q2_score": 0.8872045966995027, "lm_q1q2_score": 0.8766742598090294}}
{"url": "https://math.stackexchange.com/questions/1876350/ways-of-showing-sum-limitsn-1-infty-ln11-n-to-be-divergent/1876352", "text": "# Ways of showing $\\sum_\\limits{n=1}^{\\infty}\\ln(1+1/n)$ to be divergent\n\nShow that the following sum is divergent $$\\sum_{n=1}^{\\infty}\\ln\\left(1+\\frac1n\\right)$$\n\nI thought to do this using Taylor series using the fact that $$\\ln\\left(1+\\frac1n\\right)=\\frac1n+O\\left(\\frac1{n^2}\\right)$$ Which then makes it clear that $$\\sum_{n=1}^{\\infty}\\ln\\left(1+\\frac1n\\right)\\sim \\sum_{n=1}^{\\infty}\\frac1n\\longrightarrow \\infty$$ But I feel like I overcomplicated the problem and would be interested to see some other solutions. Also, would taylor series be the way you would see that this diverges if you were not told?\n\n\u2022 $\\ln(1+1/n) \\sim 1/n$ as $n \\to +\\infty$ means that for any $\\epsilon > 0$ there is a $N$ such that for every $n > N$ : $\\frac{1-\\epsilon}{n} < \\ln(1+1/n) < \\frac{1+\\epsilon}{n}$ \u2013\u00a0reuns Jul 30 '16 at 21:56\n\u2022 and no you can't write $\\sum_n \\ln(1+1/n) \\sim \\sum_n 1/n$ (it means nothing) but you can show that $\\sum_{n < N} \\ln(1+1/n) \\sim \\sum_{n < N} 1/n$ as $N \\to +\\infty$ \u2013\u00a0reuns Jul 30 '16 at 21:59\n\u2022 Why not? People seem to have understood \u2013\u00a0qbert Jul 30 '16 at 22:00\n\u2022 because $f(n) \\sim g(n)$ as $n\\to +\\infty$ means something precise, namely that $\\lim_{n \\to \\infty} \\frac{f(n)}{g(n)} = 1$ \u2013\u00a0reuns Jul 30 '16 at 22:01\n\u2022 I really, really like this problem from a pedagogical perspective: as we can see below, it can be solved using such a wide variety of techniques in the standard precalculus / calculus toolbox, but it can't be solved by directly applying any of the convergence tests in the standard list given to students. In this way, I feel the sequences & series section of calc 2 is too algorithmic, with relatively little creativity required from students. This is the perfect sort of problem for cutting through that. Thanks for sharing @qbert \u2013\u00a0Kaj Hansen May 27 '19 at 5:58\n\nNotice the following: $$\\log\\left(1+\\frac{1}{n}\\right)=\\log\\left(\\frac{n+1}{n}\\right)=\\log(n+1)-\\log(n)$$ Hence $$\\sum_{k=1}^{n}\\log\\left(1+\\frac{1}{k}\\right)=\\log(n+1) \\to \\infty$$\n\n\u2022 Duh ok thank you. Any response to my second question? \u2013\u00a0qbert Jul 30 '16 at 21:54\n\u2022 I think Taylor series would be the hint, for me at least. \u2013\u00a0preferred_anon Jul 30 '16 at 21:58\n\u2022 This is an efficient and compelling approach. +1 \u2013\u00a0Mark Viola Jul 30 '16 at 22:28\n\nThis is a special case of: Suppose $f(1)= 0, f'(1) > 0.$ Then $\\sum f(1+1/n) = \\infty.$\n\nProof: From the definition of the derivative (no Taylor necessary), we have\n\n$$\\frac{f(1+h)-f(1)}{h}= \\frac{f(1+h)}{h} > \\frac{f'(1)}{2}$$\n\nfor small $h>0.$ Thus\n\n$$f(1+1/n) > \\frac{f'(1)}{2}\\cdot\\frac{1}{n}$$ for large $n.$ By the comparison test we're done.\n\n\u2022 Nice generalization! \u2013\u00a0Kaj Hansen Jul 31 '16 at 1:06\n\u2022 Very cool. I think this is my favorite \u2013\u00a0qbert Jul 31 '16 at 1:32\n\nNote that we have\n\n\\begin{align} \\log\\left(1+\\frac1n\\right)&=\\int_n^{n+1}\\frac{1}{t}\\,dt\\\\\\\\ &\\ge\\frac1{n+1} \\end{align}\n\nand the harmonic series diverges.\n\nBut, suppose one forgoes that comparison and instead writes\n\n\\begin{align} \\sum_{n=1}^{2^N-1}\\int_n^{n+1} \\frac{1}{t}\\,dt&=\\int_1^{2^N}\\frac{1}{t}\\,dt\\\\\\\\ &=\\int_1^2 \\frac{1}{t}\\,dt+\\int_{2}^{4}\\frac{1}{t}\\,dt+\\dots+\\int_{2^{N-1}}^{2^N}\\frac{1}{t}\\,dt\\\\\\\\ &\\ge \\frac12 (2-1)+\\frac14 (4-2)+\\dots +\\frac{1}{2^N}(2^N-2^{N-1})\\\\\\\\ &=\\frac{N}{2} \\end{align}\n\nwhich goes to $\\infty$ as $N\\to \\infty$. And we are done! .\n\n\u2022 I really like using integrals to estimate series. Any books you'd recommend. I was only introduced to this method in complex variables \u2013\u00a0qbert Jul 30 '16 at 22:42\n\u2022 Pleased to hear that this was useful. I don't have a specific book to recommend. \u2013\u00a0Mark Viola Jul 30 '16 at 22:45\n\u2022 And I am trying to prove your first inequality. It seems clear that it is true for any positive increasing function in theintegrand, is that true? \u2013\u00a0qbert Jul 30 '16 at 22:46\n\u2022 Actually, the integrand is decreasing. Since $t\\le n+1$, then $\\frac1t \\ge \\frac{1}{n+1}$. Therefore, $$\\int_n^{n+1}\\frac{1}{t}\\,dt\\ge \\frac{1}{n+1}\\left((n+1)-n\\right)$$ \u2013\u00a0Mark Viola Jul 30 '16 at 22:48\n\nApplying a property of logarithms gives the equality $$\\displaystyle \\sum_{n=1}^\\infty \\ln(1 + 1/n) = \\ln \\Bigg( \\prod_{n=1}^\\infty (1 + 1/n) \\Bigg)$$. Therefore, if $$\\displaystyle \\sum_{n=1}^\\infty \\ln(1 + 1/n)$$ converges, say to $$c \\in \\mathbb{R}^+$$, then $$\\displaystyle \\prod_{n=1}^\\infty (1 + 1/n)$$ should converge to $$e^c$$.\n\nTherein lies a contradiction: expanding this product yields a clearly divergent sum: the expansion will include a positive copy of $$1/n$$ for all $$n \\in \\mathbb{N}$$.\n\n\u2022 Super clever. Thank you! \u2013\u00a0qbert Jul 30 '16 at 21:59\n\u2022 Glad I could help! \u2013\u00a0Kaj Hansen Jul 30 '16 at 21:59\n\n$$\\sum_{n=1}^{m}\\log\\left(\\frac{n+1}{n}\\right)=\\sum_{n=1}^{m}(\\log(n+1)-\\log n)=\\log(m+1)$$ The partial sums clearly diverge. Alternatively using the cauchy condensation test the series converges iff the series$$\\sum_{n=1}^{\\infty}2^{n}\\ln(1+1/2^{n})$$ converges. The transformed series diverges since the terms don't go to zero and so the original series diverges.\n\n\u2022 +1. It's the straightforward way to deal with the question. \u2013\u00a0Felix Marin Aug 1 '16 at 4:11\n\nIn THIS ANSWER, I showed using elementary analysis only that the logarithm function satisfies the inequalities\n\n$$\\frac{x-1}{x}\\le\\log(x)\\le x-1\\tag1$$\n\nLetting $$x=1+\\frac1n$$ in $$(1)$$ reveals\n\n$$\\frac1{n+1}\\le\\log\\left(1+\\frac1n\\right)\\le \\frac1n\\tag2$$\n\nAnd using the left-hand side of $$(2)$$ shows that\n\n$$\\sum_{n=1}^N \\log\\left(1+\\frac1n\\right)\\ge \\sum_{n=1}^N \\frac1{n+1}$$\n\nInasmuch as the harmonic series diverges, we see that the series of interest diverges also. And we are done!\n\n\"Sophisticated\" does not mean \"complicated\". In my opinion, despite using more sophisticated ideas (asymptotic analysis), your proof is simpler than all of the other current answers \u2014 even the one expressing it as a telescoping series.\n\nIncidentally, you possibly made an oversight: to complete the proof,\n\n$$\\sum_{n=1}^{\\infty} O\\left(\\frac{1}{n^2} \\right) = O\\left( \\sum_{n=1}^{\\infty} \\frac{1}{n^2} \\right) = O(1)$$\n\n(also, a remark: for this argument to be valid, it's important that the $O$ on the left is uniform; e.g. the same 'hidden constant' works for all $n$)\n\n\u2022 Sorry I'm not sure I follow/may have used notation incorrectly. Are you saying that the final identity is ok since $(1/n^p)/(1/n^2)=c$ or that we even need to worry about how this c varies with p? \u2013\u00a0qbert Jul 31 '16 at 1:35\n\u2022 @qbert: The final identity holds because the sum is convergent. It's a constant! \u2013\u00a0user14972 Jul 31 '16 at 1:42\n\u2022 Ok, got it. Thanks! Also, what subjects use big o frequently? \u2013\u00a0qbert Jul 31 '16 at 1:43\n\u2022 @qbert: For people who use the notation, it can come up in just about any field of mathematics. Often, there are many different ways to say the same thing (albeit with varying amounts of facility), and people will use the ones they're most comfortable with. However, off the top of my head, \"analysis of algorithms\" is the only subject I can think of where practically everybody uses it. \u2013\u00a0user14972 Jul 31 '16 at 1:47\n\u2022 @qbert: For example, in $5$-adic analysis, $3$ and $53$ are 'nearby' integers. Some people like to emphasize the arithmetic statement: $3 \\equiv 53 \\pmod{5^2}$. Some people like to measure the distance: $d(3, 53) = 5^{-2}$. Others prefer to express the that they are the same up to some amount of precision: $3 = 53 + O(5^2)$. And each of these ways of thinking themselves have several ways they could be notated. \u2013\u00a0user14972 Jul 31 '16 at 1:53\n\nYou can also use the Abel's summation $$\\sum_{n=1}^{N}\\log\\left(1+\\frac{1}{n}\\right)=\\sum_{n=1}^{N}1\\cdot\\log\\left(1+\\frac{1}{n}\\right)=N\\log\\left(1+\\frac{1}{N}\\right)+\\int_{1}^{N}\\frac{\\left\\lfloor t\\right\\rfloor }{t\\left(t+1\\right)}dt$$ where $\\left\\lfloor t\\right\\rfloor$ is the floor function. Since $\\left\\lfloor t\\right\\rfloor =t+O\\left(1\\right)$ we have $$\\sum_{n=1}^{N}\\log\\left(1+\\frac{1}{n}\\right)=N\\log\\left(1+\\frac{1}{N}\\right)+\\log\\left(N+1\\right)+O\\left(1\\right)$$ and taking $N\\rightarrow\\infty$ we can see that the series diverges.", "date": "2020-03-31 08:19:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1876350/ways-of-showing-sum-limitsn-1-infty-ln11-n-to-be-divergent/1876352", "openwebmath_score": 0.9015369415283203, "openwebmath_perplexity": 477.5331362572825, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631647185702, "lm_q2_score": 0.8887588023318196, "lm_q1q2_score": 0.8766389449394998}}
{"url": "https://math.stackexchange.com/questions/1664441/convergence-of-int-0-infty-fracx1ex-dx/1664457", "text": "# Convergence of $\\int_{0}^{+\\infty} \\frac{x}{1+e^x}\\,dx$\n\nDoes this integral converge to any particular value? $$\\int_{0}^{+\\infty} \\frac{x}{1+e^x}\\,dx$$ If the answer is yes, how should I calculate its value?\nI tried to use convergence tests but I failed due to the complexity of the integral itself.\n\n\u2022 Please look here. It's almost the same. \u2013\u00a0Galc127 Feb 20 '16 at 17:51\n\u2022 In general, $$\\int_0^\\infty\\frac{x^n}{e^x-1}~dx~=~n!~\\zeta(n+1),$$ and $$\\int_0^\\infty\\frac{x^n}{e^x+1}~dx~=~n!~\\eta(n+1).$$ See the Riemann $\\zeta$ and Dirichlet $\\eta$ function for more information. \u2013\u00a0Lucian Feb 21 '16 at 1:02\n\nSure it does.\n\nCollect $e^x$ in the denominator and you will get\n\n$$\\int_0^{+\\infty}\\frac{x}{e^{x}(1 + e^{-x})}\\ \\text{d}x$$\n\nSince the integral range is from $0$ to infinity, you can see the fraction in this way:\n\n$$\\int_0^{+\\infty}x e^{-x}\\frac{1}{1 + e^{-x}}\\ \\text{d}x$$\n\nand you can make use of the geometric series for that fraction:\n\n$$\\frac{1}{1 + e^{-x}} = \\frac{1}{1 - (-e^{-x})} = \\sum_{k = 0}^{+\\infty} (-e^{-x})^k$$\n\nthence thou have\n\n$$\\sum_{k = 0}^{+\\infty}(-1)^k \\int_0^{+\\infty} x e^{-x} (e^{-kx})\\ \\text{d}x$$\n\nNamely\n\n$$\\sum_{k = 0}^{+\\infty}(-1)^k \\int_0^{+\\infty} x e^{-x(1+k)}\\ \\text{d}x$$\n\nThis is trivial, you can do it by parts getting\n\n$$\\int_0^{+\\infty} x e^{-x(1+k)}\\ \\text{d} = \\frac{1}{(1+k)^2}$$\n\nThence you have\n\n$$\\sum_{k = 0}^{+\\infty}(-1)^k \\frac{1}{(1+k)^2} = \\frac{\\pi^2}{12}$$\n\nWhich is the result of the integration\n\nIf you need more explanations about the sum, just tell me!\n\nHOW TO CALCULATE THAT SERIES\n\nThere is a very interesting trick to calculate that series. First of all, let's write it with some terms, explicitly:\n\n$$\\sum_{k = 0}^{+\\infty}\\frac{(-1)^k}{(1+k)^2} = \\sum_{k = 1}^{+\\infty} \\frac{(-1)^{k+1}}{k^2} = -\\ \\sum_{k = 1}^{+\\infty}\\frac{(-1)^{k}}{k^2}$$\n\nThe first terms of the series are:\n\n$$-\\left(-1 + \\frac{1}{4} - \\frac{1}{9} + \\frac{1}{16} - \\frac{1}{25} + \\frac{1}{36} - \\frac{1}{64} + \\frac{1}{128} - \\cdots\\right)$$\n\nnamely\n\n$$\\left(1 - \\frac{1}{4} + \\frac{1}{9} - \\frac{1}{16} + \\frac{1}{25} - \\frac{1}{36} + \\frac{1}{64} - \\frac{1}{128} + \\cdots\\right)$$\n\nNow let's call that series $S$, and let's split it into even and odd terms:\n\n$$S = \\left(1 + \\frac{1}{9} + \\frac{1}{25} + \\frac{1}{49} + \\cdots\\right) - \\left(\\frac{1}{4} + \\frac{1}{16} + \\frac{1}{36} + \\frac{1}{64} + \\cdots\\right) ~~~~~ \\to ~~~~~ S = A - B$$\n\nWhere obviously $A$ and $B$ are respectively the odd and even part.\n\nnow the cute trick\n\ntake $B$, and factorize out $\\frac{1}{4}$:\n\n$$B = \\frac{1}{4}\\left(1 + \\frac{1}{4} + \\frac{1}{9} + \\frac{1}{16} + \\frac{1}{25} + \\frac{1}{36} + \\cdots\\right)$$\n\nNow the series in the bracket is a well known series, namely the sum of reciprocal squares, which is a particular case of the Zeta Riemann:\n\n$$\\zeta(s) = \\sum_{k = 1}^{+\\infty} \\frac{1}{k^s}$$\n\nwhich is, for $s = 2$\n\n$$\\zeta(2) = \\sum_{k = 1}^{+\\infty} \\frac{1}{k^2} = \\frac{\\pi^2}{6}$$\n\nThence we have:\n\n$$B = \\frac{\\zeta(2)}{4} = \\frac{\\pi^2}{24}$$\n\nAre you seeing where we want to go? But this is not enough since we don't know what $A$ is. To do that, we can again split $B$ into even and odd terms! But doing so, we will find again the initial $A$ and $B$ series:\n\n$$B = \\frac{1}{4}\\left(\\left[1 + \\frac{1}{9} + \\frac{1}{25} + \\cdots\\right] + \\left[\\frac{1}{4} + \\frac{1}{16} + \\frac{1}{64} + \\cdots\\right]\\right) ~~~ \\to ~~~ B = \\frac{1}{4}\\left(A + B\\right)$$\n\nThis means:\n\n$$4B - B = A ~~~~~ \\to ~~~~~ A = 3B$$\n\nSo\n\n$$A = 3\\cdot \\frac{\\pi^2}{24} = \\frac{\\pi^2}{8}$$\n\nNot let's get back to the initial series $S$ we wanted to compute, and substitution this we get:\n\n$$S = A - B = \\frac{\\pi^2}{8} - \\frac{\\pi^2}{24} = \\frac{\\pi^2}{12}$$\n\n\u2022 Could you please explain more about the sum and how you got the final answer? \u2013\u00a0FreeMind Feb 20 '16 at 18:01\n\u2022 @FreeMind DONE! ^^ \u2013\u00a0Turing Feb 20 '16 at 18:42\n\nComparison test:\n\n$$\\frac x{1+e^x}\\le\\frac x{e^x}\\;$$\n\nand you can directly do integration by parts in the rightmost function to check it converges.\n\nIt is possible to generalize the result.\n\nClaim: $$I=\\int_{0}^{\\infty}\\frac{x^{a}}{e^{x-b}+1}dx=-\\Gamma\\left(a+1\\right)\\textrm{Li}_{a+1}\\left(-e^{-b}\\right)$$ where $\\textrm{Li}_{n}\\left(x\\right)$ is the polylogarithm and $a>0$.\n\nConsider $$\\left(-1\\right)^{n}\\int_{0}^{\\infty}x^{a}e^{-n\\left(x-b\\right)}dx=\\frac{\\left(-1\\right)^{n}e^{-nb}}{n^{a+1}}\\int_{0}^{\\infty}y^{a}e^{-y}dy=\\frac{\\Gamma\\left(a+1\\right)\\left(-1\\right)^{n}e^{-bn}}{n^{a+1}}$$ and recalling that $$\\textrm{Li}_{k}\\left(x\\right)=\\sum_{n\\geq1}\\frac{x^{n}}{n^{k}}$$ we have $$\\Gamma\\left(a+1\\right)\\textrm{Li}_{a+1}\\left(-e^{-b}\\right)=\\sum_{n\\geq1}\\left(-1\\right)^{n}\\int_{0}^{\\infty}x^{a}e^{-n\\left(x-b\\right)}dx=\\int_{0}^{\\infty}x^{a}\\sum_{n\\geq1}\\left(-1\\right)^{n}e^{-n\\left(x-b\\right)}dx=$$ $$=-\\int_{0}^{\\infty}\\frac{x^{a}}{e^{x-b}+1}dx.$$ Maybe it's interesting to note that the function ${x^{a}}/{e^{x-b}+1}$ is the Fermi-Dirac distribution.", "date": "2021-06-18 14:57:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1664441/convergence-of-int-0-infty-fracx1ex-dx/1664457", "openwebmath_score": 0.9282633662223816, "openwebmath_perplexity": 293.7778308966094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631663220389, "lm_q2_score": 0.8887587817066392, "lm_q1q2_score": 0.8766389260206785}}
{"url": "https://math.stackexchange.com/questions/2548236/antiderivative-of-ex-12ex/2548239", "text": "Antiderivative of $e^x/(1+2e^x)$\n\nI know the solution is $$\\dfrac{\\ln\\left(2\\mathrm{e}^x+1\\right)}{2}+C$$ However, the result I got was $$\\dfrac{\\ln\\left(\\mathrm{e}^x+0.5\\right)}{2}+C$$ What I did was:\n\n\\begin{align} \\int \\frac {e^x}{1+2e^x}dx &= \\int \\frac {e^x}{2*(0.5+e^x)}dx \\\\ &= 0.5 \\cdot \\int \\frac{e^x}{0.5 + e^x} dx \\\\ &= 0.5 \\cdot (\\ln|0.5 + e^x| + C). \\end{align}\n\nI know there are antiderivative calculators online that show the correct method step by step, but I can't understand what I did wrong. What's the mistake?\n\n\u2022 I think OP choice's of presentation on a line was better than the actual edit, symbols get all packet in a small area now. Why override OP's style when the original LaTeX is ok ? \u2013\u00a0zwim Dec 3 '17 at 0:34\n\u2022 I split the lines because it accidentally carried over to a second line, which is distinctly unoptimal. The choice to make three lines instead of two was made on the fly. \u2013\u00a0davidlowryduda Dec 3 '17 at 0:37\n\u2022 @zwim Also (I\u2019m not taking sides) multiple lines are better for viewing on mobile. \u2013\u00a0Chase Ryan Taylor Dec 3 '17 at 1:51\n\nThese answers are the same. Namely,\n\n$$\\ln(2e^x + 1) = \\ln(2(e^x + 0.5)) = \\ln(2) + \\ln(e^x + 0.5).$$\n\nThe added $\\ln 2$ is absorbed by the $+ C$, and so we see that the two answers are the same up to an additive constant. There is no error.\n\nYou can tell your answer is right by differentiating it and seeing what you get back. Indeed, $$\\frac{d}{dx}\\left(\\frac{\\ln(e^x+0.5)}{2}\\right) =\\frac{1}{2}\\frac{e^x}{e^x+0.5} =\\frac{e^x}{2e^x+1}.$$ So you are correct. As the other answers point out, the two answers are the same up to a constant.\n\nLet $C = \\ln 2$, a constant. Then they're equivalent.\n\nWhat you did is correct; you just have a different value of $C$. This is why the \"$+C$\" is so important!\n\nThe others have answered your question. Note that one runs into this often also when dealing with trig functions.For example $\\int sinx cosx dx= sin^2(x)/2 + C$ if one chooses to make the substitution $u= sinx$, and $\\int sin(x)cosx dx= -cos^2(x)/2 + C$, if one chooses $u=cosx$. Both answers are correct, as of course their difference is a constant, since $sin^2(x)+cos^2(x) = 1.$", "date": "2019-05-24 11:32:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2548236/antiderivative-of-ex-12ex/2548239", "openwebmath_score": 0.9991556406021118, "openwebmath_perplexity": 403.08649753045955, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631635159684, "lm_q2_score": 0.8887587831798665, "lm_q1q2_score": 0.8766389249798957}}
{"url": "https://math.stackexchange.com/questions/2416560/is-the-set-2-3-4-open-in-some-metric-spaces-and-not-open-in-others", "text": "# Is the set $\\{2, 3, 4\\}$ open in some metric spaces and not open in others?\n\nI just want to check my understanding. This is from Baby Rudin:\n\n2.18 Definition Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$.\n\n$(a)$ A neighborhood of $p$ is a set $N_r(p)$ consisting of all $q$ such that $d(p, q)<r$ for some $r>0$. The number $r$ is called the radius of $N_r( p)$\n\n$(e)$ A point $p$ is an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N \\subset E$\n\n$(f)$ $E$ is open if every point of $E$ is an interior point of $E$.\n\nSuppose we have the metric space with set $X=\\{1, 2, 3, 4, 5\\}$ and distance function $d(x, y)=|x-y|$. Now $2$ is an interior point of $\\{2, 3, 4\\}$ because $N_{0.5}(2)=\\{2\\} \\subset \\{2, 3, 4\\}$ (and a similar argument can be made for $3$ and $4$ as well.\n\nBut if our metric space is $\\mathbb{R}$ with the same distance function, then $\\{2, 3, 4\\}$ is not open because no neighborhood of $2$ is a subset of $\\{2, 3, 4\\}$, so $2$ is not an interior point of $\\{2, 3, 4\\}$, right?\n\n\u2022 Right. That's completely correct. \u2013\u00a0fleablood Sep 4 '17 at 16:01\n\u2022 Why \"no neighborhood of $2$ is a subset of $\\{2,3,4\\}$\"? You are right, but you need to add some explanation. \u2013\u00a0Krish Sep 4 '17 at 16:01\n\u2022 @Krish When we are dealing with the set $\\mathbb{R}$, every neighborhood will contain numbers which are not integers. \u2013\u00a0Ovi Sep 4 '17 at 16:03\n\u2022 @Krish: Because any neighborhood of $2$ (by the definition given) has the form $(2-r,2+r)$ for some $r>0.$ This will readily contain some non-integer rational number. \u2013\u00a0Cameron Buie Sep 4 '17 at 16:03\n\u2022 @CameronBuie sorry!!! But I was just checking whether OP understood the reason clearly or not. (+1) for the question. \u2013\u00a0Krish Sep 4 '17 at 16:07\n\nAnother thing to consider is that even when the underlying space is the same, using a different metric may yield different open sets. Letting our metric space be $\\Bbb R,$ but with the distance function $$\\delta(x,y):=\\begin{cases}0 & x=y\\\\1 & x\\neq y,\\end{cases}$$ we can show that (for example) $\\{2\\}$ is a neighborhood of $2$ with radius $\\frac12,$ and by similar reasoning conclude that $\\{2,3,4\\}$ is once again open.", "date": "2019-07-15 23:54:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2416560/is-the-set-2-3-4-open-in-some-metric-spaces-and-not-open-in-others", "openwebmath_score": 0.8686944842338562, "openwebmath_perplexity": 124.2201597347427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303440461105, "lm_q2_score": 0.8856314723088733, "lm_q1q2_score": 0.8766249049335554}}
{"url": "https://mathoverflow.net/questions/256408/binomial-id-sum-k-mp-1km-binomkm-binomnp1nk1-binomnp-m", "text": "# Binomial ID $\\sum_{k=m}^p(-1)^{k+m}\\binom{k}{m}\\binom{n+p+1}{n+k+1}=\\binom{n+p-m}{n}$\n\nCould I get some help with proving this identity?\n\n$$\\sum_{k=m}^p(-1)^{k+m}\\binom{k}{m}\\binom{n+p+1}{n+k+1}=\\binom{n+p-m}{n}.$$\n\nIt has been checked in Matlab for various small $n,m$ and $p$. I have a proof for $m=0$ that involves Pascal's rule to split it into two sums that mostly cancel, but this does not work for all $m$. I have looked at induction on various letters (and looked at simultaneous induction on multiple letters) without any success.\n\nI would appreciate any help, algebraic or combinatoric.\n\n\u2022 Where does this identity come from? Dec 4, 2016 at 23:58\n\u2022 They are the coefficients of an expansion of a solution to Laplace's equation in terms of solid spherical harmonics. The $n$ and $m$ are the separation constants that occur in for example the spherical harmonics $P_n^m(\\cos\\theta)\\exp(im\\phi)$ Dec 5, 2016 at 0:23\n\nSuch identities are often reduced to the Chu--Vandermonde's identity $\\sum_{i+j=\\ell} \\binom{x}i\\binom{y}j=\\binom{x+y}\\ell$ by using reflection formulae $\\binom{x}k=\\binom{x}{x-k}$, $\\binom{x}k=(-1)^k\\binom{k-x-1}k$.\n\nIn your case you may write $$\\sum_{k=m}^p(-1)^{k+m}\\binom{k}{m}\\binom{n+p+1}{n+k+1}= \\sum_{k=m}^p \\binom{-m-1}{k-m}\\binom{n+p+1}{p-k}= \\binom{n+p-m}{p-m}$$ as you need, so it is Chu--Vandermonde for $x=-m-1$, $y=n+p+1$, $\\ell=p-m$.\n\n\u2022 Right, with the same generatingfunctionological explanation: $(1+t)^x (1+t)^y = (1+t)^{x+y}$. Dec 5, 2016 at 4:32\n\u2022 @Noam but why formal series $(1+t)^x=\\sum \\binom{x}{n} t^n$ satisfy this equality? It may be justified by some abstract nonsense like 'this is known for reals or for positive integers, but the coefficients are polynomials in $x,y$, thus it holds formally too.' This argument may be reduced to bit less abstract: 'both parts of CV are polynomials in $x,y$ of degree at most $\\ell$, thus it suffices to check that their values agree for integers $x,y\\geqslant 0, x+y\\leqslant \\ell$ (this triangle is an interpolating set for polynomials of degree at most $\\ell$) , where identity is just obvious.' Dec 5, 2016 at 5:54\n\u2022 I suppose you can also use a (formal) differential equation to prove this. Dec 5, 2016 at 14:25\n\nIt's a generating-function exercise. We have $$\\sum_{k=m}^\\infty (-1)^{k+m} {k \\choose m} x^{k-m} = (1+x)^{-(m+1)},$$ and (with $j=p-k$) $$\\sum_{j=0}^\\infty {n+p+1 \\choose n+(p-j)+1} x^j = \\sum_{j=0}^\\infty {n+p+1 \\choose j} x^j = (1+x)^{n+p+1}.$$ The desired sum is the $x^{p-m}$ coefficient of the product $(1+x)^{n+p-m}$ of these two power series, which is ${n+p-m \\choose p-m} = {n+p-m \\choose n}$, QED.\n\n\u2022 Presumably \"well-known\", but easier to redo than to track down in the literature. Dec 5, 2016 at 3:07\n\nI wish to explain a modern (high tech) method called the Wilf-Zeilberger (WZ) technique which might help you (and anyone interested) with the present question and many others you encounter in the future. This will save you time from hunting the literature and comparing notes with the milliard hypergeometric formulas.\n\nZeilberger developed an accompanying algorithm package which is by now part of the symbolic softwares, Maple and Mathematica. In case you have access to Maple, you may download and run a freely available online copy of it from here.\n\nStart with the discrete function $$F(p,k)=(-1)^{k+m}\\binom{k}{m}\\binom{n+p+1}{n+k+1}\\binom{n+p-m}{n}^{-1}.$$ The above-mentioned algorithm furnished the companion function $$G(p,k)=-\\frac{F(p,k)(k-m)(n+k+1)}{(p+1-k)(n+p+1-m)}.$$ Check (preferable using a symbolic software) that $$F(p+1,k)-F(p,k)=G(p,k+1)-G(p,k).\\tag1$$ Convention: $\\binom{a}b=0$ if $b>a$ or $b<0$. Now, sum both sides of (1) over all integers $k$, i.e. $\\sum_{-\\infty}^{\\infty}$. Notice that the RHS of (1) vanishes because $\\sum_{k\\in\\mathbb{Z}}G(p,k+1)=\\sum_{k\\in\\mathbb{Z}}G(p,k)$. If we denote $f(p):=\\sum_{k=m}^pF(p,k)=\\sum_{\\mathbb{Z}}F(p,k)$ then the LHS of (1) implies $$f(p+1)=f(p).$$ But, if $p=0$ then $f(0)=1$ and hence $f(p)=1$ identically. That means (after rewriting) $$\\sum_{k=m}^p(-1)^{k+m}\\binom{k}{m}\\binom{n+p+1}{n+k+1}=\\binom{n+p-m}{n}$$ as desired.\n\nMathematica says: $$(-1)^{2 m} \\binom{n+p+1}{m+n+1} \\, _2F_1(m+1,m-p;m+n+2;1)$$", "date": "2022-10-07 18:41:39", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/256408/binomial-id-sum-k-mp-1km-binomkm-binomnp1nk1-binomnp-m", "openwebmath_score": 0.9429556131362915, "openwebmath_perplexity": 426.5554814483246, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303419461302, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8766249045677118}}
{"url": "https://math.stackexchange.com/questions/327533/cardinality-of-the-collection-of-all-compact-metric-spaces", "text": "# Cardinality of the collection of all compact metric spaces\n\nI was wondering today, that if $\\mathscr{M}$ is the collection of all sets that admit a metric generating a compact topology, then\n\n1. Is $\\mathscr{M}$ a set in ZFC?\n2. If it is, what is the cardinality of $\\mathscr{M}$?\n\nThe reason why I found $1.$ interesting is because in general, the collection of all sets is not a set in ZFC, and each set admits a metric space structure via discrete metric and discrete topology. However, only those sets that are finite are compact with this metric. So could compactness assumption restrict this collection to become a set?\n\nAnd about $2.$, I figured that atleast under the equivalence relation of homeomorphism the quotient space should be a set with cardinality less or equal than $2^{\\mathfrak{c}}$, since every compact metric space admits an embedding to $[0,1]^{\\mathbb{N}}$, which is of the size $\\mathfrak{c}$. So for each equivalence class we could pick a representative from the power-set of $[0,1]^{\\mathbb{N}}$. But how about if we consider the relation of being isometric instead of homeomorphic? And finally, if we drop the quotient spaces entirely?\n\nMy motivation for these questions arose when I was reading about the Gromov-Hausdorff distance for compact metric spaces. Maybe there is a trivial answer that $\\mathscr{M}$ is not a set.\n\n\u2022 The collection of all one-element sets is not a set. \u2013\u00a0Hurkyl Mar 11 '13 at 15:38\n\u2022 @Hurkyl. How would you proceed to prove it? \u2013\u00a0T. Eskin Mar 11 '13 at 15:42\n\u2022 The union of its elements is the collection of all sets. \u2013\u00a0Hurkyl Mar 11 '13 at 15:43\n\u2022 A related concept: Urysohn's universal space is a separable metric space containing an isometric copy of every separable metric space. \u2013\u00a0Martin Mar 11 '13 at 16:13\n\u2022 @ThomasE. Let $A$ be the collection of one-element sets. Assume $A$ is a set. Then $\\{A\\}$ is a set by Pairing Axiom and is one-element, hence $\\{A\\}\\in A$. Again by Pairing, $B:=\\{A,\\{A\\}\\}$ is a set. By Axiom of Foundation, there exists $C\\in B$ with $C\\cap B=\\emptyset$. Clearly, $C\\ne A$ because then $\\{A\\}\\in C\\cap B$. Also, $C\\ne\\{A\\}$ because then $A\\in C\\cap B$. Contradiction! Therefore $A$ is not a set. \u2013\u00a0Hagen von Eitzen Mar 11 '13 at 16:32\n\nFor any $x$, $\\{x\\}$ with the metric $d(x,x)=0$ is a compact metric space, so it\u2019s clear that $\\mathscr{M}$ cannot be a set. However, it is true that $|X|\\le 2^\\omega$ for every $X\\in\\mathscr{M}$, so there is a set $\\mathscr{M}_0$ of compact metric spaces such that every compact metric space is homeomorphic (indeed isometric) to one in $\\mathscr{M}_0$.\n\nAdded: For each cardinal $\\kappa\\le 2^\\omega$ let $$\\mathscr{M}_\\kappa=\\big\\{\\langle\\kappa,d\\rangle:d\\text{ is a metric on }\\kappa\\big\\}\\subseteq{}^{\\kappa\\times\\kappa}\\Bbb R\\;;$$ this is clearly a set, as is $\\mathscr{M}=\\bigcup\\{\\mathscr{M}_\\kappa:\\kappa\\le 2^\\omega\\text{ is a cardinal}\\}$, and every compact metric space is isometric to some space in $\\mathscr{M}$. (For $\\kappa>1$ $\\mathscr{M}_\\kappa$ contains $2^\\kappa$ isometric copies of each space, corresponding to permutations of $\\kappa$, so you might want to choose representatives of the isometry classes.)\n\nTo answer your last question in the comments, a compact metric space $X$ is separable, and every compatible metric is continuous on $X\\times X$, so it has at most $2^\\omega$ compatible metrics. On the other hand, it $d$ is a compatible metric, then so is $\\alpha d$ for every $\\alpha>0$, so $X$ has $2^\\omega$ compatible metrics (unless $X$ is the one-point space!).\n\n\u2022 Thanks Brian. Since you also addressed the isometric part, I have one more question. I figured that the quotient space under the relation of being isometric should have more equivalence classes than the homeomorphism relation, since not all homeomorphic spaces are isometric. What would be the cardinality of this quotient space? (notice the edit in this comment) \u2013\u00a0T. Eskin Mar 11 '13 at 15:57\n\u2022 @Thomas: Let $X$ be a fixed set of cardinality $\\kappa\\le 2^\\omega$. The set of metrics on $X$ is a subset of the set of functions from $X\\times X$ to $\\Bbb R$, so we can form the set $$\\big\\{\\langle X,d\\rangle:d\\text{ is a metric on }X\\big\\}\\;.$$ Do this with an $X$ for each $\\kappa\\le 2^\\omega$, and you have your $\\mathscr{M}_0$. \u2013\u00a0Brian M. Scott Mar 11 '13 at 16:02\n\u2022 @Thomas: I\u2019ve incorporated a few more specifics in my answer, including an answer to your question about the relationship between the homeomorphism and isometry classes. \u2013\u00a0Brian M. Scott Mar 11 '13 at 18:15\n\u2022 Thank you Brian, this is an excellent answer. Thanks for your effort. A further question for your added part: you probably use the notation $\\mathscr{M}$ in a different way as before, where you noted that $\\mathscr{M}$ was not a set. Also, is it easy to prove that for a compact metric space $(X,d)$ of cardinality $\\kappa$ there exists an isometry $\\phi:\\kappa\\to X$ for a suitable metric on $\\kappa$? Do we just take the bijection $\\phi:\\kappa\\to X$ given by the cardinality of $X$, and pull back the metric $d$ to $\\kappa$ by defining $e(\\alpha,\\beta)=d(\\phi(\\alpha),\\phi(\\beta))$ ? \u2013\u00a0T. Eskin Mar 11 '13 at 21:40\n\u2022 @Thomas: You\u2019re welcome. Yes, I used $\\mathscr{M}$ in two different ways. And yes, just use the bijection to transfer the metric; it really is that simple. (Nice when that happens.) \u2013\u00a0Brian M. Scott Mar 11 '13 at 21:42\n\nGenerally collections of \"all possible\" are classes, simply because there is a proper class of sets of cardinality $1$, all of which are compact metric spaces.\n\nBut indeed every compact metric space can be embedded into $[0,1]^\\Bbb N$, therefore we only need to find out how many closed sets this space has, and since it is a second countable metric space it has exactly $\\frak c$ many closed subsets, so there are at most $\\frak c$ compact metric spaces up to homeomorphism.\n\n\u2022 Thanks Asaf for this clarification. \u2013\u00a0T. Eskin Mar 11 '13 at 15:53", "date": "2019-08-22 13:16:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/327533/cardinality-of-the-collection-of-all-compact-metric-spaces", "openwebmath_score": 0.9201001524925232, "openwebmath_perplexity": 175.19023035362076, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517424466175, "lm_q2_score": 0.8962513724408291, "lm_q1q2_score": 0.8765802164859252}}
{"url": "http://mathhelpforum.com/algebra/131533-need-help-figuring-out-linear-equations-table.html", "text": "# Math Help - Need help figuring out this linear equations table\n\n1. ## Need help figuring out this linear equations table\n\nX. F(x)\n4. 13\n2. 7\n0. 1\n-1. -2\n\nusing the table, what is the value of f(-1)?\n\nWrite a linear function equation for this table of values.\n\nUsing this equation or the table itself, find the value of f(1)\n\nI don't understand this.\n\n2. Originally Posted by brianfisher1208\nX. F(x)\n4. 13\n2. 7\n0. 1\n-1. -2\n\nusing the table, what is the value of f(-1)?\n\nWrite a linear function equation for this table of values.\n\nUsing this equation or the table itself, find the value of f(1)\n\nI don't understand this.\nK so the question is saying that there is a function $y=f(x)$ such that plugging in any x into the function will give you a value for y=f(x) as listed in the table.\n\nUsing this, we can think of the table as telling us this:\n\n$f(4)=13$\n$f(2)=7$\n$f(0) = 1$\n$f(-1) = -2$\n\nNow as for making a function. What can we decide from the table? It looks like if x =0, f(x)=f(0)=1. So it looks like we need to have a constant that isn't multiplied by any x.\n\nNext, lets looks at the values, how can we turn 2 into 7 and 4 into 13?\n\nLooks like $f(x) = 3x + 1$. Does this make sense how we came to this? It can be a bit of a trial and error process.\n\nLet's test it.\n\n$f(2) = 3(2) + 1 = 6 + 1 = 7 \\ \\ \\ \\text{Check!}$\n\n$f(4) = 3(4) + 1 = 12 +1 = 13 \\ \\ \\ \\text {Check!}$\n\n$f(0) = 3(0) + 1 = 0 +1 = 1 \\ \\ \\ \\text {Check!}$\n\n$f(-1) = 3(-1) + 1 = -3 +1 = -2 \\ \\ \\ \\text {Check!}$\n\nCan you find $f(1)$ now?\n\n3. Nice avatar, Kasper !\n\nOtherwise, you might want to consider a more straightforward algebra\u00efc way.\nYour problem strongly suggests that a linear relationship exists between $x$ and $f(x)$. This is equivalent to saying that $f(x) = ax + b$, for some $a$ and $b$, $a \\neq 0$.\n\n1. Find $b$\n\nYou have some values of the function (actually, only two are required). You can say that :\n\n$13 = 4a + b$\n\n$7 = 2a + b$\n\nWe can slightly reformulate this :\n\n$13 - b = 4a$\n\n$7 - b = 2a$\n\nAnd now, we get smart ! We divide both equations together :\n\n$\\frac{13 - b}{7 - b} = \\frac{4a}{2a}$\n\nWhich is equivalent to :\n\n$\\frac{13 - b}{7 - b} = 2$\n\nNow we simplify the fraction :\n\n$13 - b = 2(7 - b)$\n\n$13 - b = 14 - 2b$\n\n$- b + 2b = 14 - 13$\n\n$\\boxed{b = 1}$\n\n2. Find $a$\n\nNow that we know $b$, this is easy : we know that $b = 1$ : we can use a pair of values $\\left ( x, f(x) \\right )$ :\n\n$13 = 4a + b$\n\nSince $b = 1$, we get :\n\n$13 = 4a + 1$\n\n$12 = 4a$\n\n$3 = a$\n\n$\\boxed{a = 3}$\n\n3. Conclusion\n\nYou are done : you have found the linear relationship between $x$ and $f(x)$ :\n\n$\\boxed{f(x) = 3x + 1}$\n\nLet's check those results :\n\n$f(4) = 4 \\times 3 + 1 = 12 + 1 = 13$ -->\n\n$f(2) = 2 \\times 3 + 1 = 6 + 1 = 7$ -->\n\n$f(0) = 0 \\times 3 + 1 = 0 + 1 = 1$ -->\n\n$f(-1) = (-1) \\times 3 + 1 = -3 + 1 = -2$ -->\n\nNow that we can find $f(x)$ from $x$ with a simple linear formula, let's answer the last part of the question : substitute $x = 1$ to find the value of $f(1)$ :\n\n$f(1) = 1 \\times 3 + 1 = 3 + 1 = 4$\n\nDone !\n_________________\n\nDoes it make sense ?\n\n4. Hello, brianfisher1208!\n\n. . . $\\begin{array}{c|c}x & f(x) \\\\ \\hline\\text{-}1 & \\text{-}2 \\\\ 0 & 1 \\\\ 2 & 7 \\\\ 4 & 13 \\end{array}$\n\nUsing the table, what is the value of $f(-1)$ ?\n\nUm . . . -2 ?\n\nWrite a linear function equation for this table of values.\nA linear function has the form: . $f(x) \\;=\\;ax+b$\nAnd we must determine $a$ and $b.$\n\nWe can use any two values from our table.\n\nFor example:\n\n. . $\\begin{array}{ccccccccccccc}f(2) = 7: && a(2) + b &=& 7 && \\Rightarrow && 2a + b &=& 7 & [1] \\\\\nf(4) = 13: && a(4) + b &=& 13 && \\Rightarrow && 4a + b &=& 13 & [2] \\end{array}$\n\nSubtract [2] - [1]: . $2a \\:=\\:6 \\quad\\Rightarrow\\quad a \\:=\\:3$\n\nSubstitute into [1]: . $2(3) + b \\:=\\:7 \\quad\\Rightarrow\\quad b \\:=\\:1$\n\nTherefore: . $\\boxed{f(x)\\;=\\;3x+1}$\n\nUsing this equation or the table itself, find the value of $f(1).$\n\n$f(1) \\;=\\;3(1) + 1 \\;=\\;4$\n\n5. Interesting! As you can see, Brian, Soroban chose to solve by substitution a system of linear equations in order to find $a$ and $b$, while I decided to arrange the equations to allow a division that will get rid of an unknown, making solving easy. Two equivalent solutions to one problem (although mine was a bit longer).\n\nThanks Soroban, I didn't know the command \\boxed, I was sort of deceived my function didn't appear in math font", "date": "2016-02-05 23:21:06", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/131533-need-help-figuring-out-linear-equations-table.html", "openwebmath_score": 0.8811736106872559, "openwebmath_perplexity": 620.0407447103938, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543710622855, "lm_q2_score": 0.8840392802184581, "lm_q1q2_score": 0.8765730124913688}}
{"url": "http://emio.yiey.pw/repeated-eigenvalues.html", "text": "# Repeated Eigenvalues\n\nGeneral Form for Solutions in the Case of Repeated Eigenvalues For the di erential equation x_(t) = Ax(t); assume that we have found the single eigenvalue and a corresponding eigenvec-. 1 Find the eigenvalues and associated eigenspaces of each of the following matrices. then every eigenvalue of X is an eigenvalue of A, and the associated eigenvector is in V = R(M) if Xu = \u03bbu, u 6= 0 , then Mu 6= 0 and A(Mu) = MXu = \u03bbMu so the eigenvalues of X are a subset of the eigenvalues of A more generally: if AM = MX (no assumption on rank of M), then A and X share at least Rank(M) eigenvalues Invariant subspaces 6-6. Input the components of a square matrix separating the numbers with spaces. edu It is well known that a function can be decomposed uniquely into the sum of an odd and an even function. So the good case is when the geometric multiplicity of each eigenvalue equals its algebraic multiplicity because then. @Star Strider: Thanks for the suggestion, I was unaware of this function. The repeated eigenvalue \u03bb2= corresponds to the eigenvectors v2,1= and v2,2=. distinct real eigenvalues, different signs saddle point \u20134 \u20132 0 2 4 y \u20134 \u20132 2 4 x distinct real eigenvalues, both negative node sink \u20134 \u20132 0 2 y distinct real roots, both positive node source \u20132 0 2 4 y \u20134 \u20132 2 4 x repeated eigenvalues, positive improper node, source \u20134 \u20132 0 2 4 y \u20134 \u20132 2 4 x repeated eigenvalues. It is called complete if there are m linearly independent eigenvectors corresponding to it and defective otherwise. Next, everything is repeated with the second eigenvalue and the second eigenvector - the 2nd pr. They are not unique when there are repeated eigenvalues, as in the example above of a disk rotating about any of its diameters (\u00a7 B. The eigenspace corresponding to one eigenvalue of a given matrix is the set of all eigenvectors of the matrix with that eigenvalue. So the first thing we do, like we've done in the last several. If an eigenvalue algorithm does not produce eigenvectors, a common practice is to use an inverse iteration based algorithm with \u03bc set to a close approximation to the eigenvalue. As noted above, if \u03bb is an eigenvalue of an n \u00d7 n matrix A, with corresponding eigenvector X, then (A \u2212 \u03bbIn)X = 0, with X 6= 0, so det(A\u2212\u03bbIn) = 0 and there are at most n distinct eigenvalues of A. The general solution is Y(t) = e^(a*t)*V0 + t*e^(a*t)*V1 such that a is the repeated eigenvalue and V1=(A-a*I)V0 (A=2x2 matrix such that A*V1=a*V1, I=2x2 identity matrix) or V1=0 vector. Section PEE Properties of Eigenvalues and Eigenvectors The previous section introduced eigenvalues and eigenvectors, and concentrated on their existence and determination. Defective matrices A are similar to the Jordan (canonical) form A = XJX\u22121. Costco wholesale business plan how to write a compare and contrast essay pdf how do i assign a drive letter to a new ssd drive writing argumentative essays middle school essay on video game violence, art institute essay examples thinking critically with psychological science answers. Horn Department of Electrical Engineering and Computer Science, MIT and CSAIL, MIT, Cambridge, MA 02139, USA e-mail: [email\u00a0protected] (c) First of all, by part (b), we know A has at least a complex eigenvalue. 1 - Eigenvalue Problem for 2x2 Matrix Homework (pages 279-280) problems 1-16 The Problem: \u2022 For an nxn matrix A, find all scalars \u03bb so that Ax x=\u03bb GG has a nonzero solution x G. Now is the next step. Thanks for your reply. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. This video series will not get side tracked by special cases with embedded pole/zero cancellations, repeated poles and non-simple Jordan forms and the like. If eigenvalue stability is established for each component individually, we can conclude that the original (untransformed) system will also be eigenvalue stable. Computing its characteristic polynomial. component is being recorded, and then \"removed\". All the eigenvectors corresponding to of contain components with , where represents the position of each nonzero weights associated with and. When there are two linearly independent eigenvectors k 1 and k 2. The spectral decomposition of x is returned as components of a list with components. In this section we discuss the possibility that the eigenvalues of A are not distinct. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. This lack of eigenvalues and eignevectors will not occur if we use F = C, the field of complex numbers. Description. If is a repeated eigenvalue, only one of repeated eigenvalues of will change. Unfortunately I'm not good at theory and mathematics. I have used eig2image. An eigenvector of a matrix is a vector that, when left-multiplied by that matrix, results in a scaled version of the same vector, with the scaling factor equal to its eigenvalue. It means that some nonzero vector is mapped to zero times itself - that is, to the zero vector. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. The same situation applies, if Ais semi-simple, with repeated eigenvalues. i s are the repeated eigenvalues of M. the eigenvalues of A) are real numbers. Nonhomogeneous Systems - Solving nonhomogeneous systems of differential. When there are two or more resonant modes corresponding to the same natural frequency'' (eigenvalue of ), then there are two further subcases: If the eigenvectors corresponding to the repeated eigenvalue (pole) are linearly independent, then the modes are. When there is a basis of eigenvectors, we can diagonalize the matrix. 1 Introduction to Eigenvalues Linear equationsAx D bcomefrom steady stateproblems. We prove it by contradiction. distinct real eigenvalues, different signs saddle point -4 -2 0 2 4 y -4 -2 2 4 x distinct real eigenvalues, both negative node sink -4 -2 0 2 y distinct real roots, both positive node source -2 0 2 4 y -4 -2 2 4 x repeated eigenvalues, positive improper node, source -4 -2 0 2 4 y -4 -2 2 4 x repeated eigenvalues. But you can find enough independent eigenvectors -- Forget the \"but. The following graph shows the Gershgorin discs and the eigenvalues for a 10 x 10 correlation matrix. Given a matrix A, recall that an eigenvalue of A is a number \u03bb such that Av = \u03bb v for some vector v. The matrix norm for an n \u00d7 n matrix is defined. Next, everything is repeated with the second eigenvalue and the second eigenvector - the 2nd pr. Some regular eigenvectors might not produce any non-trivial generalized eigenvectors. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. Equal eigenvalues. Math Vids offers free math help, free math videos, and free math help online for homework with topics ranging from algebra and geometry to calculus and college math. Expert Answer 100% (1 rating). Let Abe a square (that is, n n) matrix, and suppose there is a scalar and a. We start by finding the eigenvalues and eigenvectors of the upper triangular matrix T from Figure 3 of Schur's Factorization (repeated in range R2:T4 of Figure 1 below). In general, nonlinear differential equations are required to model actual dynamic systems. (c) First of all, by part (b), we know A has at least a complex eigenvalue. 2 Harmonic Oscillators 114 6. (ii) Then will be an orthonormal basis of and the required matrix P is given by Further , a diagonal matrix and diagonal entries of D are , each repeated as many times as the dimensions of. i can easily find the eigenvector for eigenvalue 1, but i don't understand the method for the other two. Characteristic equations. 0 along the global X-axis. It decomposes matrix using LU and Cholesky decomposition The calculator will perform symbolic calculations whenever it is possible. That's what we want to do in PCA, because finding orthogonal components is the whole point of the exercise. 1 Eigenvalues and Eigenvectors The product Ax of a matrix A \u2208 M n\u00d7n(R) and an n-vector x is itself an n-vector. \u2022 The pattern of trajectories is typical for two repeated eigenvalues with only one eigenvector. It is a \\repeated eigenvalue,\" in the sense that the characteristic polynomial (T 1)2 has 1 as a repeated root. This matrix may be either a Vandermonde matrix or a modal matrix. In that case, one can give explicit algebraic formulas for the solutions. De\ufb01nition 5. The eigenvalues may be chosen to occur in any order along the diagonal of T and for each possible order the matrix U is unique. However, the power method can find only one eigenvector, which is a linear combination of the eigenvectors. Eigenvalues and Eigenvectors. The phase portrait thus has a distinct star. is an eigenvector for , then. The center has trigonometric solutions that are the parametric representations of closed curves. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. Background We will now review some ideas from linear algebra. Consider a \ufb01rst order di\ufb00erential equation of the form. (d) Show that A has no real eigenvalue if D<0. joan on December 24th, 2018 @ 2:26 pm I can now understand how this works! really interisting and I now see things way more clear!. So, the system will have a double eigenvalue, $$\\lambda$$. A System of Differential Equations with Repeated Real Eigenvalues Solve = 3 \u22121 1 5. Despite their utility, students often leave their linear algebra courses with very little intuition for eigenvectors. 1 A Complete Eigenvalue We call a repeated eigenvalue complete if it has two distinct (linearly independent) eigenvectors. In general there will be as many eigenvalues as the rank of matrix A. For any x \u2208 IR2, if x+Ax and x\u2212Ax are eigenvectors of A \ufb01nd the corresponding eigenvalue. Read \"An iterative method for finite dimensional structural optimization problems with repeated eigenvalues, International Journal for Numerical Methods in Engineering\" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. We found: e igenvalue \u03bb = 4, and the associated eigenvectors are: x = x3 \u00d7 1 2 1 2 1 Geometric interpretation of eigenvalues and eigenvectors IX a repeated eigenvalue \u03bb = \u22122. As noted above, if \u03bb is an eigenvalue of an n \u00d7 n matrix A, with corresponding eigenvector X, then (A \u2212 \u03bbIn)X = 0, with X 6= 0, so det(A\u2212\u03bbIn) = 0 and there are at most n distinct eigenvalues of A. If is a diagonal matrix with the eigenvalues on the diagonal, and is a matrix with the eigenvectors as its columns, then. Unfortunately I'm not good at theory and mathematics. An eigenvector is a nonzero vector that, when multiplied against a given square matrix, yields back itself times a multiple. vector w is called a generalized eigenvector corresponding to the eigenvalue 1. The constituent matrices are then found to be equal to the product of two matrices with elements partly from a certain matrix and its inversion. (b) Find the eigenvalues of A. Find the characteristic polynomial and the eigenvalues. how can i find eigenvectors for repeated eigenvalues and complex eigenvalues. Since they appear quite often in both application and theory, lets take a look at symmetric matrices in light of eigenvalues and eigenvectors. And we're asked to find the general solution to this differential equation. Firstly we look at matrices where one or more of the eigenvalues is repeated. , \u03bbi with multiplicity k. a vector containing the $$p$$ eigenvalues of x, sorted in decreasing order, according to Mod(values) in the asymmetric case when they might be complex (even for real matrices). This happens when the dimension of the nullspace of A\u2212\u03bbI (called the geometric multiplicity of \u03bb) is strictly less than the arithmetic multiplicity m. So far we have covered what the solutions look like with distinct real eigenvalues and complex (nonreal) eigenvalues. Show that A and AT do not have the. It is called complete if there are m linearly independent eigenvectors corresponding to it and defective otherwise. 3 Eigenvalues and Eigenvectors. Such matrices arise, for example, in invariant subspace decomposition approaches to the symmetric eigenvalue problem. Since some eigenvalues may be repeated roots of the characteristic polynomial, there may be fewer than neigenvalues. eigenvalues of A = \u00b7 a h h b \u00b8 and constructs a rotation matrix P such that PtAP is diagonal. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. 1 Find the eigenvalues and associated eigenspaces of each of the following matrices. Recall that given a symmetric, positive de nite matrix A we de ne R(x) = xTAx xTx: Here, the numerator and denominator are1 by 1matrices, which we interpret as numbers. If eigenvalue stability is established for each component individually, we can conclude that the original (untransformed) system will also be eigenvalue stable. where the eigenvalues are repeated eigenvalues. So far we have considered the diagonalization of matrices with distinct (i. This is all part of a larger lecture series on differential equations here on educator. , \u03bbi 6= \u03bbj for i 6= j, then A is diagonalizable (the converse is false \u2014 A can have repeated eigenvalues but still be diagonalizable) Eigenvectors and diagonalization 11\u201322. Suppose the 2 2 matrix Ahas repeated eigenvalues. Repeated Eigenvalues 1 Section 7. EIGENVALUES AND EIGENVECTORS 5 Similarly, the matrix B= 1 2 0 1 has one repeated eigenvalue 1. In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. Penny [ + - ] Author and Article Information. So our strategy will be to try to find the eigenvector with X=1, and then if necessary scale up. SIAM Journal on Matrix Analysis and Applications 37:4, 1581-1599. \u2022 The pattern of trajectories is typical for two repeated eigenvalues with only one eigenvector. Matrix Calculator applet. 1 A Complete Eigenvalue We call a repeated eigenvalue complete if it has two distinct (linearly independent) eigenvectors. Solution motivation from 1D: x0= axhas solution x(t) = eatx(0);. non-zero vectors v, w satisfying: Av = av Aw = dw which brings us to your second question. Moreover, these methods are surely going to have trouble if the matrix has repeated eigenvalues, distinct eigenvalues of the same magnitude, or complex eigenvalues. one repeated eigenvalue. I'm setting up my program to solve for x, y and z (it's a 3x3 matrix) using Gauss elimination. 1 (Eigenvalue and eigenvector). In the case where the 2X2 matrix A has a repeated eigenvalue and only one eigenvector, the origin is called an improper or degenerate node. Let's say we have the following second order differential equation. The matrix norm for an n \u00d7 n matrix is defined. Diagonalisation of 2 x 2 and 3 x 3 matrices. Figure 1 \u2013 Eigenvectors of a non-symmetric matrix. m to find the eigen vectors at each point on the image (in my image, there is grey values on the concentric circular region and background is black ). However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. Eigenvalues and Eigenvectors. 5 Nonautonomous Linear Systems 130. If is a diagonal matrix with the eigenvalues on the diagonal, and is a matrix with the eigenvectors as its columns, then. Input the components of a square matrix separating the numbers with spaces. (If there is no such eigenvector,. 3 Introduction In this Section we further develop the theory of eigenvalues and eigenvectors in two distinct directions. For real asymmetric matrices the vector will be complex only if complex conjugate pairs of eigenvalues are detected. i don't really understand how to find the corresponding eigenvectors to repeated eigenvalues. Eigenvalues have a number of convenient properties. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Let A be an n\u00b4 n matrix over a field F. Recall that in the case of a repeated eigenvalue (of algebraic multiplicity m) we might not have m linearly independent eigenvectors. Still assuming 1 is a real double root of the characteristic equation of A , we say 1 is a complete eigenvalue if there are two linearly independent eigenvect ors ~ 1 and ~ 2. A defective matrix has at least one multiple eigenvalue that does not have a full set of linearly independent eigenvectors. 6 Genericity 101 CHAPTER 6 Higher Dimensional Linear Systems 107 6. In the previous cases we had distinct eigenvalues which led to linearly independent solutions. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. In this form, the state matrix is a diagonal matrix of its (non-repeated) eigenvalues. Find the eigenvalues and eigenvectors of the matrix. Plugging in 1 0 to A, we see that solutions go counterclockwise. 22 Matrix exponent. For real asymmetric matrices the vector will be complex only if complex conjugate pairs of eigenvalues are detected. Given an eigenvalue \u03bb i of an n\u00d7n matrix M, its geometric multiplicity is the dimension of Ker(M \u2212\u03bb iI n), and it is the number of Jordan blocks corresponding to \u03bb i. repeated eigenvalues. The Effect of Close or Repeated Eigenvalues on the Updating of Model Parameters from FRF Data M. Proofs of the theorems are either left as exercises or can be found in any standard text on linear algebra. Start at the top of the leftmost column, and use the vectors as you go down the column. Find the characteristic polynomial and the eigenvalues. Computing its characteristic polynomial. The sum of the sizes of all Jordan blocks corresponding to an eigenvalue \u03bb i is its algebraic multiplicity. We do not normally divide matrices (though sometimes we can multiply by an inverse). Clarence Wilkerson In the following we often write the the column vector \" a b # as (a;b) to save space. The rest are similar. The matrix norm for an n \u00d7 n matrix is defined. and suppose that Bhas ndistinct eigenvalues. one repeated eigenvalue. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. Under this matrix norm, the in\ufb01nite series converges for all A and for all t, and it de\ufb01nes the matrix exponential. So, once again: eigenvectors are direction cosines for principal components, while eigenvalues are the magnitude (the variance) in the principal components. 3 Repeated Eigenvalues 119 6. The vector v is called an eigenvector corresponding to the eigenvalue \u03bb. Slope field for y' = y*sin(x+y) System of Linear DEs Real Distinct Eigenvalues #1. 2 (i) It is easy to see that eigenvalues and eigenvectors of a linear transformation are same as those of the associated matrix. J has the eigenvalues of A on its main diagonal, is upper triangular, and has 0\u2019s and 1\u2019s in the upper triangle. Then we can find the largest integer so that are linearly independent, but are not. We prove it by contradiction. The effectiveness. EigenValues is a special set of scalar values, associated with a linear system of matrix equations. This presents us with a problem. General Form for Solutions in the Case of Repeated Eigenvalues For the di erential equation x_(t) = Ax(t); assume that we have found the single eigenvalue and a corresponding eigenvec-. To prove this, we start with a general lemma on. An eigenvector is a nonzero vector that, when multiplied against a given square matrix, yields back itself times a multiple. Using the previous problem, show that either V is an eigenvector for Aor else (A I)V is an eigenvector for A. for each eigenvalue. Intuition: If the eigenvalues of A are all zero, then for arbitrary vector x, we have Ax=0. De\ufb01nition 5. 1 Distinct eigenvalues a matrix might have repeated eigenvalues and still be diagonalizable. The main diagonal of T contains the eigenvalues of A repeated according to their algebraic multiplicities. y(t) = eaty0:. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. [We say that a sign pattern matrix B requires k repeated eigenvalues if every A E Q(B) has an eigenvalue of algebraic multiplicity at least k, and k is a minimum with respect to this requirement. The vector v is called an eigenvector corresponding to the eigenvalue \u03bb. Whereas the fundamental eigenfrequency of a structure usually is of great importance, some situations may require different dynamic objectives such as maximum eigenvalue separation [17],. The proposed approach drastically reduces the coherence time requirements, enhancing the potential of quantum resources available today and in the near future. Subsection 3. The spectrum of eigenvalues is found by solving for the roots of the characteristic polynomial or secular equation det(A- I)=0. Still assuming 1 is a real double root of the characteristic equation of A , we say 1 is a complete eigenvalue if there are two linearly independent eigenvect ors ~ 1 and ~ 2. Now I\u2019ll use Gaussian elimination method to simplify these equations. Recall that given a symmetric, positive de nite matrix A we de ne R(x) = xTAx xTx: Here, the numerator and denominator are1 by 1matrices, which we interpret as numbers. If we further assume, as in \u00a73, that the matrix H is Hermitian,14 with its eigenvalues 1h real and its eigenvectors xh forming a base of m-space and orthonormal, 15 (4. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. Supplementary notes for Math 265 on complex eigenvalues, eigenvectors, and systems of di erential equations. Pointing out the eigenvalue again is just repeating something you already said. In solving an eigenvalue problem, the eigenvalues will be determined as well as the corresponding configurations of system. We do not normally divide matrices (though sometimes we can multiply by an inverse). So, once again: eigenvectors are direction cosines for principal components, while eigenvalues are the magnitude (the variance) in the principal components. 1 (Eigenvalue and eigenvector). The relation for finding Eigenvalue corresponds to the Eigenvector x is. y(t) = eaty0:. It decomposes matrix using LU and Cholesky decomposition The calculator will perform symbolic calculations whenever it is possible. We start by finding the eigenvalues and eigenvectors of the upper triangular matrix T from Figure 3 of Schur\u2019s Factorization (repeated in range R2:T4 of Figure 1 below). As before, perturbations in. The eigenvalues and eigenvectors of a matrix are essential in many applications across the sciences. Slides by Anthony Rossiter 3 ) (t) L 1[( sI A) 1] Poles come from determinant of (sI-A) which are clearly the same as the. 3 power method for approximating eigenvalues 551 Note that the approximations in Example 2 appear to be approaching scalar multiples of which we know from Example 1 is a dominant eigenvector of the matrix. Continuation: Repeated Real Eigenvalues, Complex Eigenvalues -- Lecture 26. Such a block has one repeated eigenvalue and only one eigenvector regardless of its dimension. We can nd the eigenvalue corresponding to = 4 using the usual methods, and nd u. 2 Harmonic Oscillators 114 6. The spectral decomposition of x is returned as components of a list with components. Indeed, BAv = ABv = A( v) = Av since scalar multiplication commutes with matrix multiplication. The complete case. Conclusion: there is one degree of freedom to determine the eigenvector itself and therefore also the derivative contains a degree of freedom. Finding Eigenvectors with repeated Eigenvalues. and when eigenvalues 0 and how can i draw phase portrait. Which eigenvectors do MATLAB/numpy display when eigenvalues are repeated. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Finding Eigenvectors with repeated Eigenvalues. (c) Show that A has one repeated real eigenvalue if D =0. 2 Solving Systems with Repeated Eigenvalues If the characteristic equation has only a single repeated root, there is a single eigenvalue. We prove it by contradiction. where the eigenvalues are repeated eigenvalues. Repeated Eigenvalues We studyhomogeneous autonomous system: dx dt = Ax In Lecture 10 and 11, we learn how to solve this system wheneigenvalues 1 and 2 of matrix A arereal and di erentandcomplex conjugate, respectively. m in MATLAB to determine how the solution curves (trajectories) of the system Az behave A. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. An eigenvector is determined uniquely in case of distinct eigenvalues up to a constant. This multiple is a scalar called an. Condition numbers, returned as a vector. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The spectral stability problem for periodic traveling water waves on a two\u2013dimensional fluid of infinite depth is investigated via a perturbative approach, computing the spectrum as a function of the wave amplitude beginning with a flat surface. Option 3: Generally, an (n x n) matrix with repeated eigen values can be diagonalised if we obtain n linearly independent eigen vectors for it. and when eigenvalues 0 and how can i draw phase portrait. Eigenvalues and Eigenvectors 1. Repeated eigenvalues indicate linear dependence within the rows and columns of A. Repeated Eigenvalues 1. 3 Introduction In this Section we further develop the theory of eigenvalues and eigenvectors in two distinct directions. The Rayleigh Principle for Finding Eigenvalues April 19, 2005 1 Introduction Here I will explain how to use the Rayleigh principle to nd the eigenvalues of a matrix A. The repeated eigenvalue \u03bb2= corresponds to the eigenvectors v2,1= and v2,2=. General Form for Solutions in the Case of Repeated Eigenvalues For the di erential equation x_(t) = Ax(t); assume that we have found the single eigenvalue and a corresponding eigenvec-. Of course v is then called the eigenvector of A corresponding to \u03bb. model to be determined by the eigenvalues of the A matrix. Penny [ + - ] Author and Article Information. Where there is not, we can't. 2 (Spectrum and spectral radius). So our strategy will be to try to find the eigenvector with X=1, and then if necessary scale up. This is a coupled eigenvalue problem through A and its adjoint A*. Appendix B Advanced topics Lecture B. SUPPLEMENT ON EIGENVALUES AND EIGENVECTORS We give some extra material on repeated eigenvalues and complex eigenvalues. So even though a real asymmetric x may have an algebraic solution with repeated real eigenvalues, the computed solution may be of a similar matrix with complex conjugate pairs of eigenvalues. EigenValues is a special set of scalar values, associated with a linear system of matrix equations. non-zero vectors v, w satisfying: Av = av Aw = dw which brings us to your second question. If eigenvalues are repeated, we may or may not have all n linearly independent eigenvectors to diagonalize a square matrix. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Find more Mathematics widgets in Wolfram|Alpha. The data used in this example are from the following experiment. If we further assume, as in \u00a73, that the matrix H is Hermitian,14 with its eigenvalues 1h real and its eigenvectors xh forming a base of m-space and orthonormal, 15 (4. One of Lemma's high quality modules plucked from our universe of content. For real asymmetric matrices the vector will be complex only if complex conjugate pairs of eigenvalues are detected. It is a \\repeated eigenvalue,\" in the sense that the characteristic polynomial (T 1)2 has 1 as a repeated root. An eigenvalue problem is a special type of problem where the solution exists only for special values (i. We prove that the volume of n satis es: j nj jRPN 3j = n 2 ; where N = n+1 2 is the dimension of the space of real symmetric matrices of size n n. We can\u2019t \ufb01nd it by elimination. The matrix norm for an n \u00d7 n matrix is defined. 1 - Eigenvalue Problem for 2x2 Matrix Homework (pages 279-280) problems 1-16 The Problem: \u2022 For an nxn matrix A, find all scalars \u03bb so that Ax x=\u03bb GG has a nonzero solution x G. Next, everything is repeated with the second eigenvalue and the second eigenvector - the 2nd pr. A matrix and its transpose both have the same eigenvalues. An eigenvector is determined uniquely in case of distinct eigenvalues up to a constant. Every eigenvalue \u00df. , it could just be the diagonal matrix with diagonal entries 2. This approach is an extension of recent work by Daily and by Juang et al. Background We will now review some ideas from linear algebra. We have accomplished this by the use of a non-singular modal matrix P i. Show that A and AT do not have the. Mechanical Systems and Signal Processing 107 , 78-92 Online publication date: 1-Jul-2018. Let Abe a square (that is, n n) matrix, and suppose there is a scalar and a. how can i find eigenvectors for repeated eigenvalues and complex eigenvalues. The rest are similar. eigenvalues tell the entire story. Slope field. For 2x2, 3x3, and 4x4 matrices, there are complete answers to the problem. Indeed, BAv = ABv = A( v) = Av since scalar multiplication commutes with matrix multiplication. In particular, if you have repeated eigenvalues, it doesn't automatically mean you have a multidimensional eigenspace: for instance, the matrix 1 1 0 1 has the eigenvalue 1 with multiplicity 2, but its only eigenvector is the vector (1,0). = 1 exactly once from each column (as is done above in Equation 6) results in a new sum of zero for the elements of each column vector. If are the distinct eigenvalues of A ,for each eigenvalues , find an orthonormal basis of the eigensubspace. is at most the multiplicity of. All eigenvalues of A are distinct \u21d2diagonalizable There are repeated eigenvalues, e. 5 Repeated Eigenvalues 95 5. I am trying to understand the principle underlying the general solution for repeated eigenvalues in systems of differential equations. In this chapter, we provide basic results on this subject. The eigen value and eigen vector of a given matrix A, satisfies the equation Ax = \u03bbx , where, \u03bb is a number, also called a scalar. (solution: x = 1 or x = 5. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. To tackle the issue of non-smoothness of repeated eigenvalues, we propose an estimator constructed by averaging all repeated eigenvalues. 06 or 18/700. Returns A const reference to the column vector containing the eigenvalues. 2 Intuitive Example. He's also an eigenvector. Find the eigenvalues and eigenfunctions of the Sturm-Liouville problem 00u = u 0 1. Consider a \ufb01rst order di\ufb00erential equation of the form. In general there will be as many eigenvalues as the rank of matrix A. Since some eigenvalues may be repeated roots of the characteristic polynomial, there may be fewer than neigenvalues. Complex eigenvalues and eigenvectors satisfy the same relationships with l 2C and~x 2Cn. , multiplicity of \"2\". Costco wholesale business plan how to write a compare and contrast essay pdf how do i assign a drive letter to a new ssd drive writing argumentative essays middle school essay on video game violence, art institute essay examples thinking critically with psychological science answers. Note that this sets all entries of the matrices and vectors to 0. and is applicable to symmetric or nonsymmetric systems. An eigenvector is a nonzero vector that, when multiplied against a given square matrix, yields back itself times a multiple. So far we have covered what the solutions look like with distinct real eigenvalues and complex (nonreal) eigenvalues. This is because u lays on the same subspace (plane) as v and w, and so does any other eigenvector. How will I calculate largest eigen values and its correspoinding eigen vector of Hessian matrix to select new seed point as discussed above. y(0) = y0: Of course, we know that the solution to this IVP is given by. I'm setting up my program to solve for x, y and z (it's a 3x3 matrix) using Gauss elimination. The picture is more complicated, but as in the 2 by 2 case, our best insights come from finding the matrix's eigenvectors : that is, those vectors whose direction the. In this section we discuss the possibility that the eigenvalues of A are not distinct.", "date": "2020-04-01 18:02:30", "meta": {"domain": "yiey.pw", "url": "http://emio.yiey.pw/repeated-eigenvalues.html", "openwebmath_score": 0.8604814410209656, "openwebmath_perplexity": 409.031974205775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9915543725597269, "lm_q2_score": 0.8840392725805823, "lm_q1q2_score": 0.8765730062417967}}
{"url": "http://math.stackexchange.com/questions/249379/how-to-calcualate-how-many-unique-set-of-6-can-i-have-in-a-given-set/249386", "text": "# How to calcualate how many unique set of 6 can i have in a given set.\n\nHello my question is quite simple i would think but i just cant seem to find an answer. I have a set of $\\{1,2,3,4,5,6,7,8,9,10\\}$ and i would like to calculate how many unique given sets of $6$ can i get from this set. In other words for the number $1$ i would end up with $[1,2,3,4,5,6] [1,3,4,5,6,7] [1,4,5,6,7,8] [1,5,6,7,8,9] [1,6,7,8,9,10]$ I would move down the line with the number $2$ to compare to unique sets of $6$ note: when moving to two I would no longer do this $[2,1,3,4,5,6]$ because it repeats my first case above. its there a formula to figure this sort of thing? Thanks in advance.\n\nwhen I work this out on paper i end up with 15 sets here is how\n\nfor 1\n[1,2,3,4,5,6]\n[1,3,4,5,6,7]\n[1,4,5,6,7,8]\n[1,5,6,7,8,9]\n[1,6,7,8,9,10]\n\nfor 2\n[2,3,4,5,6,7]\n[2,4,5,6,7,8]\n[2,5,6,7,8,9]\n[2,6,7,8,9,10]\nfor 3\n[3,4,5,6,7,8]\n[3,4,6,7,8,9]\n[3,5,6,7,8,9,10]\nfor 4 [4,5,6,7,8,9]\n[4,6,7,8,9,10]\nfor 5 [5,6,7,8,9,10]\n\n\nafter that i cant make any more groups of $6$ thus i end up with $15$ sets.\n\n-\nI'd be pretty certain there are other questions similar to this on the site that might help you. \u2013\u00a0 Simon Hayward Dec 2 '12 at 20:30\n\nYes, there is, it is called the binomial, written $\\binom{n}{k}$, read $n$ choose $k$. The value is $$\\binom{n}{k}=\\frac{n!}{k!(n-k)!}.$$ So, in your case, you have $$\\binom{10}{6}=\\frac{10!}{6!4!}=210.$$ I hope you find this helpful!\n\n-\nForgive my stupidity Clayton I have two problems here. 1 how are we ending up with 210? the \"!\" operator is what exactly? Also i worked this out on paper and i only end up with 15 sets of 6 so I'm obviously missing something can you help me understand. \u2013\u00a0 Miguel Dec 2 '12 at 19:19\nThe \"!\" stands for the factorial, that is, $n!=n\\cdot(n-1)\\cdot \\cdots\\cdot (2)(1)$. So, $$10!=10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1.$$ Dividing by $6!4!$ is what gives you the 210 unique subsets of order 6. Does this make it more clear? \u2013\u00a0 Clayton Dec 2 '12 at 19:23\nit makes it more clear but i just dont see how the resul makes sense. For example i edit my question can you look how i reached 15. Yet this formula is given me 210 im obviously missing something but i dont see what it could be? forgive my ignorance. \u2013\u00a0 Miguel Dec 2 '12 at 19:28\nNo worries, it is great to ask questions. Now, looking at the subsets you've listed, where would the set $[1,2,3,4,5,7]$ fit into? Also, we can have the sets $[1,2,3,4,5,8]$, $[1,2,3,4,5,9]$, and $[1,2,3,4,5,10]$, for example. Once we've exhausted one \"component,\" which I'm taking to be the last one, change the next index over, so now we count $[1,2,3,4,6,7]$. So on and so forth. Hope it helps! \u2013\u00a0 Clayton Dec 2 '12 at 19:36\nTHANK YOU! :-D i knew there was something i was missing. Thank you! I truly appreciate your response. \u2013\u00a0 Miguel Dec 2 '12 at 19:40\n\nIt exactly the number of ways to choose $6$ elements out of $10$, i,e. the binomial coefficient$$\\binom{10}{6}=\\frac{10!}{6!4!}$$\n\n-\n\nBinomial coefficients count the number of distinct subsets of $k$ elements from a set containing $n$ elements. The notation for this is $\\binom{n}{k}$ which is equal to $\\frac{n!}{k!(n-k)!}$\n\n-", "date": "2014-03-11 11:12:41", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/249379/how-to-calcualate-how-many-unique-set-of-6-can-i-have-in-a-given-set/249386", "openwebmath_score": 0.8676039576530457, "openwebmath_perplexity": 333.120066482413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462229680671, "lm_q2_score": 0.8872045922259089, "lm_q1q2_score": 0.8765104258895108}}
{"url": "https://www.physicsforums.com/threads/a-challenging-question-around-the-jacobian-matrix.226838/", "text": "# A (challenging?) question around the Jacobian matrix\n\n1. Apr 5, 2008\n\n### Aleph-0\n\nHere is the problem:\n\nSuppose that g is a diffeomorphism on R^n. Then we know that its jacobian matrix is everywhere invertible.\nLet us define the following matrix valued function on R^n\n$$H_{i,j} (x) = \\int_0^1 \\partial_i g^j(tx) dt$$\nwhere $$g^j$$ are the components of g.\nQuestion : Is $$(H_{i,j}(x))_{i,j}$$ (which could be interpreted as a mean of the Jacobian matrix of g) invertible for any x ?\n\nMy guess is that the answer is negative, but I find no counter-examples.\nAny Help ?\n\n2. Apr 7, 2008\n\n### ObsessiveMathsFreak\n\nI believe you may have to move into three dimensions to obtains a counterexample. It is possible that this theorem is in fact true.\n\nEdit:\nActually, I believe that a counterexample is provided by;\n\n$$\\mathbf{g}((x,y,z))=(x \\cos(2\\pi z)-y \\sin(2\\pi z),x \\sin(2\\pi z)+y \\cos(2\\pi z),z)$$\nwith\n$$H((0,0,1))$$ being singular.\n\nCan anyone else check this?\n\nLast edited: Apr 7, 2008\n3. Apr 7, 2008\n\n### Aleph-0\n\nIndeed, we even have H(x,y,1) singular for any x,y !\nThank you !\nI wonder if the theorem is true in dimension 2...\n(it is trivially true in dimension 1).\n\n4. Apr 7, 2008\n\n### Haelfix\n\nInteresting problem. I thought about this for awhile, and couldn't come up with a counter in D =2. Can anyone prove this statement? Its not clear to me why it should be true, and there seems to be something interesting (maybe topological) lurking behind it.\n\nLast edited: Apr 7, 2008\n5. Apr 14, 2008\n\n### ObsessiveMathsFreak\n\nI believe it is true because of the nature of coordinate lines and their tangents in 2D.\n\nFirstly, assume that we can find a H in 2D that is singular. Therefore the columns of H, which we'll denote $$\\mathbf{h}_i$$ are linearly dependant. So there exist non zero $$\\alpha_i$$ such that\n\n$$\\Sigma \\alpha_i \\mathbf{h}_i = \\mathbf{0}$$\n\nNow by the definition\n\n$$\\mathbf{h}_i = \\int_0^1 \\frac{\\partial \\mathbf{g}}{\\partial x_i}(t \\mathbf{x}) dt$$\nTherefore\n\n$$\\Sigma \\alpha_i \\mathbf{h}_i =\\int_0^1 \\Sigma \\alpha_i \\frac{\\partial \\mathbf{g}}{\\partial x_i}(t \\mathbf{x}) dt = \\mathbf{0}$$\n\nNow looking at the integral along this curve. We let\n$$\\mathbf{v}(t)=\\Sigma \\alpha_i \\frac{\\partial \\mathbf{g}}{\\partial x_i}(t(1,0)) = \\alpha_1 \\frac{\\partial \\mathbf{g}}{\\partial x_1}(t(1,0)) + \\alpha_2 \\frac{\\partial \\mathbf{g}}{\\partial x_2}(t(1,0))$$\nAnd so we have\n$$\\int_0^1 \\mathbf{v}(t) dt = \\mathbf{0}$$\nWe cannot have $$\\mathbf{v}=\\mathbf{0}$$, because the jacobian is always fully ranked, and hence any linear combination of its columns cannot be zero.\n\nSo $$\\mathbf{v}(t)$$ is non zero everywhere, and its integral is zero. But if we let $$\\mathbf{v}(t)=\\frac{d \\mathbf{z}}{dt}(t)$$, we can see that\n\n$$\\int_0^1 \\mathbf{v}(t) dt = \\int_0^1 \\frac{d \\mathbf{z}}{dt}(t) dt = \\mathbf{z}(1)-\\mathbf{z}(0) = \\mathbf{0}$$\n\nWhich means that $$\\mathbf{v}$$ is the tangent vector to some closed curve $$\\mathbf{z}(t)$$\n\nWe now make a change of variables by letting\n$$\\mathbf{x}(\\mathbf{u}) = \\left( \\begin{array}{cc} \\alpha_1 & -\\alpha_2 \\\\ \\alpha_2 & \\alpha_1 \\end{array} \\right) \\mathbf{u}$$\n\nThen it can be shown that $$\\mathbf{v}(t) = \\frac{d \\mathbf{g}}{d u_1}(t(1,0))$$, i.e. $$v$$ is the tangent vector to the $$u_2=0$$ coordinate curve. But this means that this coordinate curve is closed. Since the mapping from u to x is a diffeomorphism, and the function g is a diffeomorphism, the function from u to g is a diffeomorphism on R^2. This means that it cannot have any closed coordinate curves. Therefore we have a contridiction, and so H must be non singular.\n\nI hope this is OK. I'm not being especially rigorous here.\n\nIn higher dimensions, some of the $$\\alpha_i$$ can be zero, and hence the mapping from u to x may not be a diffeomorphism. Essentially you have some wiggle room once you move into higher dimensions.", "date": "2017-06-22 19:01:14", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-challenging-question-around-the-jacobian-matrix.226838/", "openwebmath_score": 0.8735422492027283, "openwebmath_perplexity": 349.7663308878067, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232874658905, "lm_q2_score": 0.8918110555020056, "lm_q1q2_score": 0.8764926733669068}}
{"url": "http://forum.math.toronto.edu/index.php?topic=1326.0;prev_next=next", "text": "### Author Topic: Can there exists infinite number of solutions given initial conditions. \u00a0(Read 969 times)\n\n#### Xinyu Jiao\n\n\u2022 Newbie\n\u2022 Posts: 1\n\u2022 Karma: 0\n##### Can there exists infinite number of solutions given initial conditions.\n\u00ab on: September 25, 2018, 08:58:39 PM \u00bb\nIn class, we were given an example where a differential equation can have two solutions given some initial condition. Specifically, the equation was $y' = y^\\alpha$ with $0<\\alpha<1$, and initial condition $y(0) = 0$. This shows that it's not unique, because it does not satisfy some condition which I do not understand.\n\nMy question is, can there be a differential equation (of order 1) such that given an initial condition, can acquire an infinite number of solutions? The answer to this question should be able to shed light as to the mechanism through which the equation acquires more than one solution.\n\u00ab Last Edit: September 25, 2018, 09:18:16 PM by Victor Ivrii \u00bb\n\n#### Victor Ivrii\n\n\u2022 Elder Member\n\u2022 Posts: 2511\n\u2022 Karma: 0\n##### Re: Can there exists infinite number of solutions given initial conditions.\n\u00ab Reply #1 on: September 25, 2018, 09:27:59 PM \u00bb\nFor condition see Section 2.8 of the textbook or this Lecture Note\n\nYes, this equation $y'=3 y^{2/3}$ (I modified it for simplicity) has a general solution $y=(x-c)^{3}$ but also a special solution $y=0$. Thus problem $y'=3 y^{2/3}$, $y(0)=0$ has an infinite number of solutions. Restricting ourselves by $x>0$ we get solutions y=\\left\\{\\begin{aligned} &0 &&0<x<c,\\\\ &(x-c)^3 && x\\ge c\\end{aligned}\\right. with any $c\\ge 0$ and similarly for $x< 0$.\n\nThis happens because this Lipschitz condition is violated at each point of the solution $y=0$.\n\n#### Kathryn Bucci\n\n\u2022 Jr. Member\n\u2022 Posts: 6\n\u2022 Karma: 2\n##### Re: Can there exists infinite number of solutions given initial conditions.\n\u00ab Reply #2 on: October 06, 2018, 10:59:07 AM \u00bb\nIf \ud835\udc66\u2032=\ud835\udc66\ud835\udefc with 0 < \ud835\udefc < 1 e.g. \ud835\udc66\u2032=3\ud835\udc662/3= f(t,y), then \u2202f/\u2202y=2y-1/3 is not continuous at (0,0).\nAccording to theorem 2.4.2 (existence and uniqueness for 1st order nonlinear equations), both f and \u2202f/\u2202y have to be continuous on an interval containing the initial point (0,0) - \u2202f/\u2202y is not continuous there so you can't infer that there is a unique solution.\n\n#### Victor Ivrii\n\nContinuity of $\\frac{\\partial f}{\\partial y}$ is not required, but \"H\u00f6lder\u00a0 property\" $|f(x,y)-f(x,z)|\\le M$ is.", "date": "2020-08-04 07:31:21", "meta": {"domain": "toronto.edu", "url": "http://forum.math.toronto.edu/index.php?topic=1326.0;prev_next=next", "openwebmath_score": 0.9579827189445496, "openwebmath_perplexity": 1245.826559568301, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232884721166, "lm_q2_score": 0.8918110504699677, "lm_q1q2_score": 0.8764926693186664}}
{"url": "https://math.stackexchange.com/questions/1719659/what-is-the-difference-between-a-supremum-and-maximum-and-also-between-the-infi/1719681", "text": "# What is the difference between a supremum and maximum; and also between the infimum and minimum?\n\nWhat is the difference between a supremum and maximum; and similarly the infimum and minimum? Also, how does one tell if they exist?\n\nHere is an example:\n\n$$x_n = \\frac{n}{2n-1}$$\n\nDetermine whether the maximum, the minimum, the supremum, and the infimum of the sequence $$x_n$$ n=1 to n=+\u221e\n\nMy understanding:\n\n\u2022 The limit is 1/2.\n\u2022 $$x_n$$ is decreasing.\n\u2022 The supremum exists as n goes to +\u221e.\n\u2022 The infimum does not exist as limited by n=1, minimum = 1/2.\n\u2022 The maximum and supremum exists, both equal to 1.\n\u2022 One distinction I like: If the $\\sup A$ is an element of $A$ then $\\sup A=\\max A$, etc... so they only differ on membership to the set. For example $0=\\inf \\{1/n:n\\in \\mathbb{N}\\}$ but this set has no minimum since $0 \\notin \\{1/n:n\\in \\mathbb{N}\\}$. Mar 30 '16 at 1:48\n\nIf the max exists, then it is the supremum. If the min exists, then it is the infimum. For an infinite set, it can happen that the max does not exist but the supremum does exist, and/or that the min does not exist but the infimum does exist.\n\nIn this case, $\\frac{n}{2n-1}$ is never actually equal to $\\frac{1}{2}$, but $\\frac{1}{2}$ is still the infimum, because it is a lower bound and because one can find $n$ with $\\frac{n}{2n-1}<1/2+\\epsilon$ whenever $\\epsilon>0$ (which follows from the limit observation you made). Since the infimum is not attained, the minimum doesn't exist.\n\n\u2022 Hi, is your last sentence a general statement? Since if infimum =1/2; it follows min = 1/2? And I am correct in my assumption max and sup = 1? Mar 30 '16 at 1:51\n\u2022 The infimum is 1/2, but there is no minimum here. And yes, the max is 1, so the sup is also 1.\n\u2013\u00a0Ian\nMar 30 '16 at 1:52\n\u2022 Okay, so if the max and min exist then it is the sup inf, but there is no max or min then the sup inf is just the result of the limit? Mar 26 '20 at 0:38\n\u2022 @ErockBrox In general \"the limit\" needn't even make sense; not every set of real numbers is countable, and those that are aren't necessarily given in the form of a sequence. For a set with no max or min, there is no cleaner way to describe the sup and inf than just the definition of sup and inf (as least upper bound and greatest lower bound respectively).\n\u2013\u00a0Ian\nMar 26 '20 at 1:02\n\nAs Ian pointed out, the difference can only appear with infinite sets. Here is the definition of minimum and infimum:\n\n\u2022 For the set $$A$$ to have a minimum $$x_m$$ it means that $$\\exists x_m\\in A$$ such that $$x\\geq x_m$$ $$\\forall x\\in A$$.\n\u2022 For the set $$A$$ to have an infimum $$x_i$$ it means that $$\\exists x_m$$ such that $$x\\geq x_m$$ $$\\forall x\\in A$$, and $$x_m$$ is the largest possible such number.\n\nNotice that for the minimum, we require $$x_m$$ to still be in the set, while the infimum does not need to be in the set. That is the key point. For example, for $$A=(0,\\infty)$$, we have that $$x\\geq 0$$ $$\\forall x\\in A$$, and zero is the largest number that makes this inequality true for all elements of $$A$$. Since $$0\\notin A$$, we cannot say that it is a minimum, so it can only be the infimum.\n\n\u2022 Great mention about the infimum not having to be in the set; that is indeed a key point. Mar 14 '19 at 18:07\n\nMy answer is conceptual rather than an exact solution to your problem. I believe you will solve the problem after you understand the concepts.\n\nLet's start with the definition of an upper bound. $M$ is an upper bound of the set $A$ if every element of $A$ is less than $M$. And the minimum of these $M$'s is called the supremum of $A$, denoted by $\\sup A$. That is to say, if $M$ is an upper bound of $A$ and there is not an upper bound of $A$ smaller than $M$, then $\\sup A = M$. And there is a fundamental property of real numbers called the least upper bound axiom which states the following:\n\nA set $X$ has a supremum iff $X$ is non-empty and has an upper bound.\n\nYou can play with these definitions (use terms like lower bound instead of upper bound) to obtain the cases for the infimum.\n\nAnd as others indicated, the difference between the supremum and the maximum is that maximum of a set must be contained by the set but supremum of a set may not be contained by the set. When the first case is satisfied, that is, if a set has a maximum, then the maximum is equal to the supremum as you can show easily using the definitions above. For example, if $S = [1,5]$ then $\\sup S = \\max S = 5$. But if $S = [1,5)$, then $\\sup S = 5$ and $S$ has no maximum.\n\nTherefore, one can easily show, for your example, that the set $S =\\left\\{ {x_n \\:\\:\\: |\\:\\:\\: n \\ge 1} \\right\\}$ has a lower bound (eg. $0$ is a lower bound for the set $S$), and is non-empty ($1 \\in S$). Therefore, by the greatest lower bound property (which you should have obtained by playing the definitions above) the set has an infimum.\n\nRemark: For some sets by using induction or other simple methods it is easy to show that whether the set has a supremum or not. But try to show that the set consisting of the numbers $$\\bigg(1+ \\frac{1}{n}\\bigg)^n$$ where $n\\ge 1$ has a supremum. You will see that it is not that easy.\n\n\u2022 Well-explained!! Best! Oct 2 '17 at 20:53\n\nThe difference between supremum and maximum is that for bounded, infinite sets, the maximum may not exist, but the supremum always does. Consider the set $$(0,1)$$. Does this set contain a largest element? The answer is no because for any $$x \\in (0,1)$$, $$\\tfrac{x+1}{2}$$ is also in $$(0,1)$$ and $$x < \\tfrac{x+1}{2}$$. That is, for any $$x \\in (0,1)$$, we can find an element in $$(0,1)$$ which is larger than $$x$$. Thus there is no maximal element of $$(0,1)$$. By contrast, the supremum is the least upper bound for the set. The supremum does not need to be in the set. For $$(0,1)$$, the supremum is 1. This means that $$1$$ is an upper bound for $$(0,1)$$ [which is obvious] and that no number smaller than $$1$$ is an upper bound for $$(0,1)$$ [which follows from the above reasoning]. Whenever the maximum exists, it is equal to the supremum. Conversely, if the supremum lies in the set, then the maximum exists and is equal to this supremum.\n\nThe difference between minimum and infimum is similar. In my example, $$(0,1)$$ has no minimum, but the infimum is 0.\n\nFor your set, we see the sequence goes $$1, \\,\\,\\,\\,\\, 2/3, \\,\\,\\,\\,\\, 3/5, \\,\\,\\,\\,\\, 4/7, \\ldots.$$ The sequence is clearly decreasing. The number 1 is in the set and no other member of the set is larger than 1. Hence 1 is a maximum for the set. Since the maximum exists, it is equal to the supremum, so the supremum is also 1. There is no minimum element since for any element $$x_n$$ of the set, the value $$x_{n+1}$$ is less than $$x_n$$. However, the infimum is 1/2. To see this, we note that all members are greater than 1/2, so 1/2 is a lower bound. To be the infimum, it needs to be the greatest lower bound, so you need to prove that $$1/2 + a$$ is not a lower bound for any $$a > 0$$. Indeed, if $$a > 0$$, taking $$n$$ large enough that $$1/(4n-2) < a$$ shows that $$x_n = \\frac{n}{2n -1} = \\frac{n-1/2 + 1/2}{2n-1} = \\frac 1 2 + \\frac 1{4n-2} < \\frac 1 2 + a.$$ Thus we have found a member of the sequence which is smaller than $$1/2 + a$$ so $$1/2 + a$$ is not a lower bound for the sequence. Hence $$1/2$$ is the infimum.\n\n\u2022 Suprema and infima do not necessarily exist, an example is the positive reals which has an infimum of 0 but no supremum Mar 30 '16 at 2:03\n\u2022 Sure, sorry. I will edit. I meant to say they always exist for bounded sets. Mar 30 '16 at 2:38\n\u2022 I think I understood what you mean with your example, just making sure :-) Mar 30 '16 at 2:59\n\u2022 You were absolutely right; I just forgot to include the word \"bounded.\" Thanks for the note. Mar 30 '16 at 3:11\n\nApparently, you know how to obtain the limits (here: upper bound 1, lower bound 1/2). That is any case is then the supremum (or infimum at the lower end) - or the set is unlimited, e.g. the naturals on top, the reals both ways: Nothing exists then.\n\nTwo cases:\n(1) The limit is IN the set, e.g. actually attained by the sequence. Then it is maximum AND supremum (or minimum AND infimum), e.g. 1.\n(2) The limit is OUTSIDE the set, not attained by the sequence (though only barely missed). That is your case at the lower end, 1/2. Then it is only supremum (or infimum) AND there is no maximum (minimum).\n\nYou might have some doubts coming in via the index n (your 3rd suggestion $$n\\to\\infty$$ and supremum)?\n$$n\\to\\infty$$ has nothing to do with supremum (\"large $$n$$\"?). Only the $$x_n$$ themselves do.\n\nTake-home bits:\n\n1. No bound - no ..mum, nothing.\n2. Minimum = Infimum OR no Minimum. (ditto max=sup or no max). In particular, these can not differ!", "date": "2021-10-25 00:13:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1719659/what-is-the-difference-between-a-supremum-and-maximum-and-also-between-the-infi/1719681", "openwebmath_score": 0.9378753304481506, "openwebmath_perplexity": 144.64653524328335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668706602658, "lm_q2_score": 0.893309416705815, "lm_q1q2_score": 0.8764856049205918}}
{"url": "https://math.stackexchange.com/questions/1777010/show-discontinuous-points-of-increasing-function-on-0-1-is-f-sigmaset", "text": "# Show discontinuous points of increasing function on $[0,1]$ is $F_\\sigma$set\n\nProblem: Let $f$ be an increasing function on $[0,1]$, Show discontinuous points of $f$ on $[0,1]$ is $F_\\sigma$ set. But it is not always a $G_\\delta$ set.\n\nMy thoughts: It is easy to prove that there exist a injective mapping from $A=\\{x\\in[0,1]|f\\text{ is discontinuous at } x\\}$ to $\\mathbb{Q}$. So $A$ is most a countable set. Since $\\mathbb{Q}$ is $F_\\sigma$ set, so is subset(Is it right?). And similarly, since $\\mathbb{Q}$ is not a $G_\\delta$set, so the subset.\n\nYes, it's true that the set of discontinuities of $f$ is countable. And every countable subset of $\\mathbb R$ is clearly an $F_\\sigma.$ So we have the first claim.\nThe second claim is false. For a counterexample, take an $f$ with just one discontinuity. Now just note that a one point set is a $G_\\delta.$\n\u2022 can this discontinuous points set can always be $G_\\delta$ set? I think, if we take $A=[0,1]\\cap \\mathbb{Q}$, then $A$ is not $G_\\delta$ set but $F_\\sigma$, am I right? \u2013\u00a0DuFong May 8 '16 at 19:18", "date": "2019-08-25 07:16:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1777010/show-discontinuous-points-of-increasing-function-on-0-1-is-f-sigmaset", "openwebmath_score": 0.9154502153396606, "openwebmath_perplexity": 117.51996914354646, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9811668679067631, "lm_q2_score": 0.8933094032139577, "lm_q1q2_score": 0.8764855892230986}}
{"url": "https://math.stackexchange.com/questions/3250256/complement-open-and-closed-sets", "text": "Complement, open and closed sets\n\nDefinition: Let X be a metric space, and E $$\\subseteq X$$, E is closed if it is equal to its closure.\n\nDefinition: A metric subset U of X is open if for every point in U there exists an open ball centered around the point that is contained within U.\n\nTheorem: Let X be a metric space. A Subset $$Y \\subseteq X$$ is closed iff $$Y^c$$ is open.\n\nProof:\n\nAssume Y is closed so $$Y= \\bar{Y}$$ (so it is equal to is closure). Which means that $$y\\in Y \\iff \\forall$$ $$r>0 B(y,r) \\cap Y \\neq \\emptyset$$.\n\nLet $$a \\in Y^c$$ be arbitrary. Since Y is closed $$\\exists r>0 : B(a,r) \\cap Y = \\emptyset$$. This means that $$B(a,r) \\subseteq Y^c$$. Since a was arbitrary, $$Y^c$$ is open.\n\nSuppose $$Y^c$$ is open. So $$y\\in Y^c$$ $$\\iff$$ $$\\exists r>0$$ : $$B(y,r) \\subseteq Y^c$$. Let x $$\\in \\bar{Y}$$ be arbitrary. Note that it suffices to show that $$\\bar{Y} \\subseteq Y$$. So for any possible $$r>0$$ we have $$B(x,r) \\cap Y \\neq \\emptyset$$. Since $$Y^c$$ is open, it follows that $$x \\in Y$$.\n\nIs the proof correct? I would very much love feedback, I'd be very very thankful.\n\n\u2022 in general, it would be helpful to include your definitions of open and closed, because there are many equivalent ways of approaching a subject. In fact, the theorem you wrote is sometimes taken as the definition. Even though we might be able to hazard a guess as to which definition you're using based on what you've written, including the definitions can help us to provide better answers (so that we don't assume facts unknown to you etc) \u2013\u00a0peek-a-boo Jun 4 at 2:59\n\u2022 @peek-a-boo I just did. Thanks for telling me. \u2013\u00a0topologicalmagician Jun 4 at 3:11\n\u2022 A metric space is closed by definition. \u2013\u00a0copper.hat Jun 4 at 4:09\n\u2022 Closed is usually defined as the complement of open, so you may want to specify what you mean by closed. \u2013\u00a0copper.hat Jun 4 at 4:11\n\u2022 @copper.hat I did. A subspace E of X is closed if $E=\\bar{E}$, where $\\bar{E}$ is defined to be the closure, i.e. the set of all limit/adherent points. \u2013\u00a0topologicalmagician Jun 4 at 4:12\n\nI'm not concinced by the last part of the $$Y^\\complement$$ open implies $$Y$$ closed part. Why \"Since $$Y^\\complement$$ is open it follows that $$x \\in Y$$\" is unclear: but can this be expanded:\n\nAssume $$x \\notin Y$$ then $$x \\in Y^\\complement$$ and by openness there is some $$r>0$$ such that $$B(x,r) \\subseteq Y^\\complement$$ or equivalently, $$B(x,r) \\cap Y=\\emptyset$$, but that contradicts $$x \\in \\overline{Y}$$. So $$x \\in Y$$ must hold.\n\n\u2022 Suppose $Y^c$ is open. So $y\\in Y$ $\\iff$ $\\forall r>0$ : $B(y,r) \\cap Y \\neq \\emptyset$. Why isn't that enough? \u2013\u00a0topologicalmagician Jun 4 at 4:53\n\u2022 Your A appeared while I was removing my usual infinite series of typos from mine. \u2013\u00a0DanielWainfleet Jun 4 at 4:53\n\u2022 @topologicalmagician I see no (logical direct) link between the two statements $Y^\\complement$ open and that equivalence. \u2013\u00a0Henno Brandsma Jun 4 at 5:55\n\u2022 @Henno Brandsma, what about the contrapositive of $y\\in Y^c$ iff $\\exists r>0$ : $B(y,r) \\subseteq Y^c$? \u2013\u00a0topologicalmagician Jun 4 at 13:36\n\u2022 P iff Q is equivalent to not P iff not Q \u2013\u00a0topologicalmagician Jun 4 at 13:46\n\nThe last sentence is unclear. I think you should add a little, like: $$x\\in \\bar Y \\implies \\forall r>0\\,(B(x,r)\\cap Y\\ne \\emptyset) \\implies$$ $$\\implies (\\neg (\\exists r>0\\, (B(x,r)\\subset Y^c)\\,))\\implies$$ $$\\implies (\\neg (x\\in Y^c))\\; \\text {...[because } Y^c \\text { is open...] }\\implies$$ $$\\implies x \\in Y.$$\n\n\u2022 Suppose $Y^c$ is open. So $y\\in Y$ $\\iff$ $\\forall r>0$ $B(y,r) \\cap Y \\neq \\emptyset$. Isn't that enough? \u2013\u00a0topologicalmagician Jun 4 at 4:54\n\u2022 I would very much appreciate your help \u2013\u00a0topologicalmagician Jun 4 at 5:03\n\u2022 It basically is the contrapositive of \"So $y\\in Y^c \\iff \u2026..\"$ \u2013\u00a0topologicalmagician Jun 4 at 5:08\n\u2022 It's not that you made a mistake. It's just that it's not totally obvious how your last sentence follows from the 2nd last. \u2013\u00a0DanielWainfleet Jun 4 at 5:10\n\u2022 but is using the contrapositive logically correct? Is what I used in fact the contrapositive? \u2013\u00a0topologicalmagician Jun 4 at 13:52", "date": "2019-08-20 12:47:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3250256/complement-open-and-closed-sets", "openwebmath_score": 0.9063588380813599, "openwebmath_perplexity": 318.24799883996565, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668717616668, "lm_q2_score": 0.8933093996634686, "lm_q1q2_score": 0.876485589183098}}
{"url": "https://math.stackexchange.com/questions/901591/proving-a-set-of-numbers-has-arithmetic-progressions-of-arbitrary-length-but-no", "text": "# Proving a set of numbers has arithmetic progressions of arbitrary length, but none infinite\n\nFor each real number $x$, let $[x]$ be the largest integer less than or equal to $x$. For example, $$[5] = 5$$\n\n$$[7.9] = 7,$$ and $$[\u22122.4] = \u22123.$$\n\nAn arithmetic progression of length $k$ is a sequence $a_1, a_2, \\dots, a_k$ with the property that there exists a real number $b$ such that $a_{i+1} \u2212 a_i = b$ for each $1 \\leq i \\leq k \u2212 1.$\n\nLet $\\alpha > 2$ be a given irrational number. Let $S = \\{[n \u00b7 \u03b1] : n \u2208 Z\\}$, the set of all integers that are equal to $[n \\alpha]$ for some integer $n$.\n\n(a) Prove that for any integer $m \\geq 3$, there exist m distinct numbers contained in $S$ which form an arithmetic progression of length $m$.\n\n(b) Prove that there exist no infinite arithmetic progressions contained in $S$.\n\nI'm thinking that for (a), some application pigeonhole principle can be considered... but otherwise, I don't really know how to prove this.\n\nSome suggestions on a strategy would be awesome. Thanks.\n\n\u2022 How familiar are you with the proof that the set of fractional parts $\\{\\{n\\alpha\\}\\}_{n \\in \\mathbb N}$ is dense in the unit interval? \u2013\u00a0Erick Wong Aug 18 '14 at 3:15\n\u2022 Not very familiar at all... \u2013\u00a0user164403 Aug 18 '14 at 14:14\n\u2022 By the way, how does this relate to Number Theory? I know only very elementary Number Theory, so just wondering if there are some good books that cover these types of Number Theory questions. \u2013\u00a0user164403 Aug 18 '14 at 15:50\n\nOutline: We want to construct an arithmetic sequence of length $n$ that is a subsequence of the sequence $(\\lfloor i\\alpha\\rfloor)$.\n\nFor $i=1,2,3, \\dots, 10n+1$, let $b_i$ be the fractional part of $i\\alpha$, that is, $i\\alpha-\\lfloor i\\alpha\\rfloor$. These fractional parts are all distinct, since $\\alpha$ is irrational. By the Pigeonhole Principle, there exist $i$ and $j$ with $i\\lt j$ such that $|b_j-b_i|\\lt \\frac{1}{10n}$. Let $k=j-i$. Then either (i) $b_k\\gt 1-\\frac{1}{10n}$ or (i) $b_k\\lt \\frac{1}{10n}$.\n\nWe look first at case (i). Consider the numbers $k\\alpha$, $2k\\alpha$, $3k\\alpha$ and so on up to $nk\\alpha$. The fractional parts of these are slowly decreasing. Thus the numbers $\\lfloor k\\alpha\\rfloor$, $\\lfloor 2k\\alpha\\rfloor$, $\\lfloor 3k\\alpha\\rfloor$, and so on up to $\\lfloor nk\\alpha\\rfloor$ form an arithmetic progression of length $n$. There is a lot of slack, we can continue well beyond $\\lfloor nk\\alpha\\rfloor$.\n\nCase (ii) is essentially the same, except that the fractional parts of the $ik\\alpha$ slowly increase instead of decreasing.\n\nThe argument that our sequence has no infinite arithmetic progression uses similar ideas.\n\n\u2022 Would it be essential to apply the Pigeonhole Principle? I'm not used to using it, so I'm just wondering if there are other ways. Also, when do you know that a problem uses the Pigeonhole Principle? Is there anything that 'ticks', and you know that you have to use the Pigeonhole Principle? Are there any clues that show this in the question when not explicitly stated? \u2013\u00a0user164403 Sep 6 '14 at 4:19\n\u2022 Well, I guess we do not have to mention the Pigeonhole Principle explicitly. But we will have to use it implicitly, since the $\\alpha$ is general. It comes up whenever we are interested in the distribution of fractional parts. Here it was integer parts, but the technique was worth trying, and works. An indication that Pigeonhole might be useful is if the problem says that a set has more than $k$ elements, then a certain nice thing happens. The problem above, however, does not really have that shape. \u2013\u00a0Andr\u00e9 Nicolas Sep 6 '14 at 5:20\n\u2022 What type of math is this? I'm very interested in it and would like to learn more by checking out resources pertaining to that aspect of math. \u2013\u00a0user164403 Sep 30 '14 at 3:42\n\u2022 It belongs, more or less, to Diophantine Approximation. \u2013\u00a0Andr\u00e9 Nicolas Sep 30 '14 at 3:45\n\u2022 Are there any good books on those kinds of questions? This question was from a Math Challenge, and this challenge stated that to prepare, one should go over such topics as sequences and series, which I thought was to be in real analysis. \u2013\u00a0user164403 Sep 30 '14 at 16:40", "date": "2020-02-24 00:00:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/901591/proving-a-set-of-numbers-has-arithmetic-progressions-of-arbitrary-length-but-no", "openwebmath_score": 0.8221361637115479, "openwebmath_perplexity": 120.27946176033242, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718468714469, "lm_q2_score": 0.8856314723088732, "lm_q1q2_score": 0.8764845348474012}}
{"url": "https://math.stackexchange.com/questions/1425660/how-can-this-function-have-two-different-antiderivatives/1425667", "text": "How can this function have two different antiderivatives?\n\nI'm currently operating with the following integral:\n\n$$\\int\\frac{u'(t)}{(1-u(t))^2} dt$$\n\nBut I notice that\n\n$$\\frac{d}{dt} \\frac{u(t)}{1-u(t)} = \\frac{u'(t)}{(1-u(t))^2}$$\n\nand\n\n$$\\frac{d}{dt} \\frac{1}{1-u(t)} = \\frac{u'(t)}{(1-u(t))^2}$$\n\nIt seems that both solutions are possible, but that seems to contradict the uniqueness of Riemann's Integral.\n\nSo the questions are:\n\n1. Which one of them is the correct integral?\n2. If both are correct, why the solution is not unique?\n3. The pole at $u(t)=1$ has something to say?\n\u2022 @SanchayanDutta I have edited the title to fit the question a little bit better. If you have ideas for further improvements, go ahead and edit the post. \u2013\u00a0Martin Sleziak Sep 7 '15 at 20:13\n\u2022 \"but that seems to contradict the uniqueness of Riemann's Integral\" whoever said antiderivatives were unique? \u2013\u00a0PyRulez Sep 8 '15 at 0:12\n\u2022 \u2013\u00a0Dargor Sep 8 '15 at 0:14\n\u2022 @PabloGalindoSalgado Notice that it says \"unique up to a constant\" \u2013\u00a0PyRulez Sep 8 '15 at 0:17\n\u2022 @PyRulez Yup, yup. I thought that your question were refered to that result, not the constant. Sorry. \u2013\u00a0Dargor Sep 8 '15 at 0:19\n\nIt is not really a contradiction, since difference of the two functions is constant: $$\\frac1{1-u(t)} - \\frac{u(t)}{1-u(t)} = \\frac{1-u(t)}{1-u(t)}=1.$$ (Derivative of a constant function is zero. Primitive function is determined uniquely up to a constant.)\n\nI have not verified that the derivatives are correct, but notice that $$\\frac1{1-u(t)}-\\frac{u(t)}{1-u(t)}=1$$\n\nthat is, the difference between both antiderivatives is constant. This implies that the derivatives of both functions are the same.\n\n$$\\frac{d}{dt} \\frac{u(t)}{1-u(t)}=\\frac{d}{dt} \\left(\\frac{1}{1-u(t)}-1\\right)=\\frac{d}{dt} \\frac{1}{1-u(t)}$$\n\nNote that $1$ is just a constant so it vanishes. But when computing the antiderivative you will specify the constant according to initial conditions.\n\nAn equivalent way to see this, but from a slightly different perspective:\n\n$$\\frac{1}{1 - u(t)} = \\frac{1 - u(t) + u(t)}{1 - u(t)} = 1 + \\frac{u(t)}{1 - u(t)}$$\n\nPerhaps this technique is worth seeing, namely, tinkering with an expression using additive inverses; specifically, by subtracting then adding back $u(t)$ to the numerator, we come to see that the two expressions differ only by a constant (i.e., $1$).\n\nThis technique occasionally presents itself when integrating; for example,\n\n$$\\int \\frac{x}{x+1} dx = \\int \\frac{x+1-1}{x+1} dx = \\int 1 - \\frac{1}{x+1} dx$$\n\nwhere the leftmost integrand is, in my estimation, a bit less friendly than the rightmost.\n\nThe other answers are excellent and clearly explain that indeed anti derivatives of a function may differ by a constant. I wanted to add that your \"problem\" can be introduced both in a \"forward\" and a \"backward\" fashion.\n\nYou gave a \"backward\" example, that is when two clearly not equal functions upon differentiation produce the same function. But working the other way around can also introduce the \"problem\", that is integration of the same function (and omitting the constant) may still produce a constant difference in the resulting functions. To illustrate what I mean consider the following tempting conclusion\n\n$$\\text{Let } f(x) = x, g(x) = x \\text{ then } f(x) = g(x) \\text{ and } \\int f(x)dx = \\int g(x)dx$$\n\nWe now apply the results of integration theory on both sides of the equation (which should result in applying the same operations on the symbols, right?) and decide not to add an extra constant to either of both. If we let the resulting expressions be denoted respectively by $F(x)$ and $G(x)$, we might be tempted to say that $F(x) = G(x) = \\frac{1}{2}x^2$, but this conclusion is also not entirely true because the results depend on the operations applied. Moreover, the applied operations might be (unknowingly) forced upon us by the form of the equation!\n\nTo see this, consider the differential equation $y' = y$. Suppose I have the solution $z = 2e^x$, now we might be tempted to say\n\n$$\\frac{z'}{z} = 1 \\rightarrow \\int \\frac{z'}{z}dx = \\int dx \\xrightarrow{\\text{set both integration constants 0}} \\log(z) = x \\rightarrow z = e^x \\rightarrow 2e^x = e^x$$\n\nwhich is obviously false.\n\nI have sometimes seen students struggling with this, so I thought I might as well add it. Hope it helps! :)\n\nI will explain the concept using the derivative since you are already pretty familiar with that. Lets define to $\\int \\:f\\left(x\\right)dx$ to be some fucntion where $\\frac{d}{dx}\\left(g\\left(x\\right)\\right)=f\\left(x\\right)$. (Note this is not the exact definition of an anti-derivative, but an intuitive way of thinking about it.)\n\nSo if $g\\left(x\\right)=x^2,\\:\\frac{d}{dx}\\left(g\\left(x\\right)\\right)=f\\left(x\\right)=2x$ And in reverse,$\\int \\:f\\left(x\\right)dx=x^2$\n\nBut what happens if $g\\left(x\\right)=x^2+5,\\:\\frac{d}{dx}\\left(g\\left(x\\right)\\right)=f\\left(x\\right)=2x$, Notice the 5 is a constant and disappears when taking the derivative. If we apply the reverse, we have no way of getting back the 5. In reverse you still get $\\int \\:f\\left(x\\right)dx=x^2$.\n\nSo now here comes the problem, when you are going in reverse you have no idea what the constant is. This is purely because the constant disappears when taking the derivative. For example $\\frac{d}{dx}\\left(x^2+8\\right)=\\frac{d}{dx}\\left(x^2+5\\right)=\\frac{d}{dx}\\left(x^2+1\\right)$! So when finding the Anti-Derivative you will get a function plus or minus some constant.", "date": "2021-03-08 17:04:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1425660/how-can-this-function-have-two-different-antiderivatives/1425667", "openwebmath_score": 0.9324637055397034, "openwebmath_perplexity": 336.69763170690635, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.967410256173572, "lm_q2_score": 0.9059898216525933, "lm_q1q2_score": 0.8764638454555842}}
{"url": "http://mathhelpforum.com/calculus/128671-simple-integral-question-i-think-s-simple.html", "text": "# Math Help - Simple Integral Question (I think it's simple)\n\n1. ## Simple Integral Question (I think it's simple)\n\nI am supposed to know this by heart and I do:\nhttp://www.wolframalpha.com/input/?i=integral+(1%2F(x^2+%2B+1))\n\nHowever, on a specific problem I'm having, I need to know to get the following:\nhttp://www.wolframalpha.com/input/?i=integral+1%2F(x^2+%2B+36)\n\nMy question is:\nIs there a way for me to use what I have memorized to derive the second link? Like how can I make sense of the second link?\n\nAny input would be greatly appreciated!\n\n2. Originally Posted by s3a\nI am supposed to know this by heart and I do:\nhttp://www.wolframalpha.com/input/?i=integral+(1%2F(x^2+%2B+1))\n\nHowever, on a specific problem I'm having, I need to know to get the following:\nhttp://www.wolframalpha.com/input/?i=integral+1%2F(x^2+%2B+36)\n\nMy question is:\nIs there a way for me to use what I have memorized to derive the second link? Like how can I make sense of the second link?\n\nAny input would be greatly appreciated!\nI'll give you a hint. See if you can figure it out yourself. Use the formula you memorized to make a general formula for the derivative of $tan^{-1}(\\frac {x}{a})$\n\n3. Eek, its better to just use LaTex (if you know how) and link the problem.\n\nHowever this isn't a \"simple\" integral, unless you are familiar with trig substitutions. Lets do a little bit of math trickery:\n\n$x = 6tan(\\theta)$\n$dx=6sec^{2}(\\theta)d\\theta$\n\nThus:\n\n$\\int \\frac{6sec^{2}(\\theta)d\\theta}{(6tan{\\theta})^{2}+ 36} \\Rightarrow \\int \\frac{6sec^{2}(\\theta)d\\theta}{36(tan^{2}\\theta+1) }$\n\nFrom here we simplify:\n$\\int \\frac{sec^{2}(\\theta)d\\theta}{6sec^{2}(\\theta)} \\Rightarrow \\int \\frac{1}{6}d\\theta$\n\nThis integral is easy to evaluate:\n\n$\\int \\frac{1}{6}d\\theta \\Rightarrow \\frac{1}{6}\\theta+C$\n\nHowever, we need our final answer in terms of x, not theta. So, we go back to our original substitution:\n\n$x = 6tan(\\theta)$\n\n. . .and solve for theta. Do not \"memorize\" formulas or \"tricks\" (this really isn't a trick though I called it so) - memorize (or KNOW) identities, and familiar algebraic structures.\n\n4. I learned it like this:\n\n$\\int \\frac{1}{x^2+36} \\, dx$\n\n$=~ \\frac{1}{36} \\int \\frac{1}{\\frac{x^2}{36}+1} \\, dx$\n\n$=~ \\frac{1}{36} \\int \\frac{1}{\\left(\\frac{x}{6}\\right)^2+1} \\, dx$\n\n(Then let $u=x/6 ~\\implies~ du=dx/6$.)\n\n$=~ \\frac{1}{36} \\int \\frac{6}{u^2+1} \\, du$\n\n$=~ \\frac{1}{6} \\int \\frac{1}{u^2+1} \\, du$\n\nYou can probably complete it from there.\n\n5. Originally Posted by ANDS!\nEek, its better to just use LaTex (if you know how) and link the problem.\n\nHowever this isn't a \"simple\" integral, unless you are familiar with trig substitutions. Lets do a little bit of math trickery:\n\n$x = 6tan(\\theta)$\n$dx=6sec^{2}(\\theta)d\\theta$\n\nThus:\n\n$\\int \\frac{6sec^{2}(\\theta)d\\theta}{(6tan{\\theta})^{2}+ 36} \\Rightarrow \\int \\frac6{sec^{2}(\\theta)d\\theta}{36(tan^{2}\\theta+1) }$\n\nFrom here we simplify:\n$\\int \\frac{sec^{2}(\\theta)d\\theta}{6sec^{2}(\\theta)} \\Rightarrow \\int \\frac{1}{6}d\\theta$\n\nThis integral is easy to evaluate:\n\n$\\int \\frac{1}{6}d\\theta \\Rightarrow \\frac{1}{6}\\theta+C$\n\nHowever, we need our final answer in terms of x, not theta. So, we go back to our original substitution:\n\n$x = 6tan(\\theta)$\n\n. . .and solve for theta. Do not \"memorize\" formulas or \"tricks\" (this really isn't a trick though I called it so) - memorize (or KNOW) identities, and familiar algebraic structures.\nHowever, there is a MUCH simpler way to do this:\n\n$\\frac{d}{dx} tan^{-1}x = \\frac {1}{x^2+1}$\n$\\frac{d}{dx} tan^{-1}(\\frac {x}{a}) = \\frac {1}{(\\frac{x}{a})^2+1}.\\frac{1}{a} = \\frac {a}{x^2+a^2}\n$\n\nNotice how the question is in similar format:\n$\\int \\frac{1}{x^2+36}dx = \\frac{1}{6} \\int \\frac{6}{x^2+36}dx = \\frac{1}{6} tan^{-1}\\frac{x}{6}+C\n$\n\nHope this helps, s3a\n\n6. Originally Posted by drumist\nI learned it like this:\n\n$\\int \\frac{1}{x^2+36} \\, dx$\n\n$=~ \\frac{1}{36} \\int \\frac{1}{\\frac{x^2}{36}+1} \\, dx$\n\n$=~ \\frac{1}{36} \\int \\frac{1}{\\left(\\frac{x}{6}\\right)^2+1} \\, dx$\n\n(Then let $u=x/6 ~\\implies~ du=dx/6$.)\n\n$=~ \\frac{1}{36} \\int \\frac{6}{u^2+1} \\, du$\n\n$=~ \\frac{1}{6} \\int \\frac{1}{u^2+1} \\, du$\n\nYou can probably complete it from there.\nSorry, I was typing when you posted; your way works well, too.\n\nAnd s3u: if you're going to take a timed test like AP Calc AB, memorizing $\\frac{d}{dx} tan^{-1}(\\frac {x}{a}) = \\frac {a}{x^2+a^2}$ could save a lot of time. Of course, you don't need to memorize it, I'm sure, because you saw where that came from, you could re-figure that out in a few seconds.\n\n7. All of these methods are the same. Personally I like the method I used as it shows where it all comes from - but then it is all a matter of taste.\n\n8. Originally Posted by ANDS!\nAll of these methods are the same. Personally I like the method I used as it shows where it all comes from - but then it is all a matter of taste.\nI actually prefer the method you used, but the one I used is more useful in a test perspective, to save a lot of time. And, yes, they are all, in essence, the same method; that's the beautiful consistency of Math.", "date": "2015-05-05 14:55:40", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/128671-simple-integral-question-i-think-s-simple.html", "openwebmath_score": 0.8935448527336121, "openwebmath_perplexity": 625.7087344886071, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102533547799, "lm_q2_score": 0.9059898140375993, "lm_q1q2_score": 0.8764638355349639}}
{"url": "https://math.stackexchange.com/questions/1752976/posterior-probability-of-bag-given-ball-evidence", "text": "# posterior probability of bag given ball (evidence)\n\nQuestion:\n\nGiven the distribution of the coloured balls in three different bags:\n- Bag A: 1 Red 2 Black 2 Blue\n- Bag B: 2 Red 4 Black 4 Blue\n- Bag C: 10 Red 2 Black 3 Green\n\nwe carry out two independent experiments:\n1) pick a bag, then pick a ball from the chosen bag.\n2) pick a ball uniformly and random from all of the 30 balls.\n\nWe observe that the ball is red (for both experiment).\nNow, we want to find the posterior probability of ball was taken from bag C given the ball chosen is Red for each of the experiment (i.e. P(Bag C | Red)).\n\nMy attempt:\nFor Experiment 1:\nUsing Bayes Rule we have:\nP(Bag C | Red) = P(Bag C and Red) / P(Red)\n\nFrom the distribution, we have P(BagC, Red) = 10/30\nNow,P(Red)\n= (by law of total probability) sum of conditional probability of\nP(Red | Bag X) * P(Bag X)\n= P(Red|BagC)*P(BagC) + P(Red|BagB)*P(BagB) + P(Red|BagA)*P(BagA)\n= (1/3)(10/15) + (1/3)(2/10) + (1/3)(1/5)\n= 16/45\n\nAnd hence P(BagC|Red) = (10/30) / (16/45) = 15/16 ???\n\n\nFor Experiment 2:\nthis is the part where i am confuse, because intuitively, i dont not see the difference between both question! But i slept over it and think maybe the difference is in the way we calculate P(red), and thats why the following answer:\n\nSo using the same formula above, but with difference calculation of the probability of getting a red ball from the bag:\n\nP(Red) = total number of red / total number of balls = 13/30\nHence, P(BagC|Red)\n= P(Red, BagC) / P(Red)\n= (10/30) / (13/30)\n= 10/13\n\n\nIntuitively, i thought my answer made sense because if we were to choose a bag first, the probability of the red ball coming from bag C will be larger since the proportion of red balls in bag C is significantly higher.\nIn comparison, if we pick randomly from the pool of 30 balls, the contribution of C into the pool is higher as well (10/13 red balls from C).\n\nI am not entirely sure if my approach is correct here, and would wish that you can validate my answers.\n\nEDIT:\nRectify the error pointed out in the comments\n\nYour answer to the first question is correct. (ETA: Or was, before you edited $\\frac{5}{8}$ to $\\frac{15}{16}$. $<$g$>$)\n\nYour answer to the second question is incorrect, because the $P(\\text{red}, \\text{Bag$C$}) = 10/30$, not $10/45$. (That is, out of the $30$ balls, each equally likely to be selected, exactly $10$ of them are both red and in Bag $C$.) You then obtain the correct answer of $10/13$.\n\n\u2022 Hi Brian, good catch. However, if thats the case, then my first answer wouldnt be correct as well, isnt it? P(red, bagC) = 10/30, then for the first experiment, we should have P(bagC | Red) = (10/30) / (16/45) = 15/16 \u2013\u00a0rockstone Apr 21 '16 at 16:48\n\u2022 @rockstone: No, it was right before. You should have $P(\\text{Bag$C$} \\mid \\text{red}) = \\frac{P(\\text{red}, \\text{Bag$C$})}{P(\\text{red})} = \\frac{2/9}{16/45} = \\frac{5}{8}$. The difference is that the balls are not equally likely in the first problem; the bags are. \u2013\u00a0Brian Tung Apr 21 '16 at 17:39\n\u2022 @brain, not exactly sure how you get 2/9. \u2013\u00a0rockstone Apr 21 '16 at 19:19\n\u2022 @brain, after some thought, is $\\frac{2}{9} = (\\frac{10}{15})(\\frac{1}{3}) =$ $P(red | BagC) P(C)$. Okay, then, my next question, if $P(A|B)P(B) = P(A, B)$, then for the second part, if we do use Bayes theorem, $P(BagC|red) = \\frac{P(red|BagC) P(BagC)} {P(red)} = \\frac{P(red, BagC)}{P(red)}$ why do we use $\\frac{10}{30}$ instead of $\\frac{10}{45}$? hope you can enlighten me on this issue. \u2013\u00a0rockstone Apr 21 '16 at 19:31\n\u2022 @rockstone: Because all $30$ balls are equally likely, of which $10$ are both red and in Bag $C$. \u2013\u00a0Brian Tung Apr 21 '16 at 19:53\n\nIn the pool of all $30$ balls, label the balls by the label of the bag they came from. Then, there are $10$ red 'C' balls. So the second question asks for the probability of a ball being a red 'C' ball, given that it is a red ball. There are totally $13$ red balls, and $10$ of these are 'C', so the probability is $\\dfrac{10}{13}$.\n\nYou can do this using Bayes' theorem too, but you will get the same answer.\n\n\u2022 Yes, the second part should be 10/13! \u2013\u00a0rockstone Apr 21 '16 at 17:04", "date": "2019-10-22 19:28:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1752976/posterior-probability-of-bag-given-ball-evidence", "openwebmath_score": 0.8894681930541992, "openwebmath_perplexity": 654.2819329904494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513889704252, "lm_q2_score": 0.888758801595206, "lm_q1q2_score": 0.876450726652803}}
{"url": "http://math.stackexchange.com/questions/97397/combinatorics-pigeonhole-principle-question/103970", "text": "Combinatorics - pigeonhole principle question\n\nThis is for self-study. This question is from Rosen's \"Discrete Mathematics And Its Applications\", 6th edition.\n\nAn arm wrestler is the champion for a period of 75 hours. (Here, by an hour, we mean a period starting from an exact hour, such as 1 P.M., until the next hour.) The arm wrestler had at least one match an hour, but no more than 125 total matches.\n\n1 - Show that there is a period of consecutive hours during which the arm wrestler had exactly 24 matches.\n\n2 - Is the statement in the previous exercise true if 24 is replaced by\n\na) 2? b) 23? c) 25? d) 30?\n\nMy solution to part 1 is the following, based on Rosen's solution to a similar problem given as example in the text:\n\n1 - If I consider $a_i$ to be the number of competitions until the $i^{th}$ hour, then, $1\\leq a_1<a_2<\\cdots<a_{75}\\leq 125$, because there are no more than 75 hours and the total number of competitions is no more than 125. Now I will add 24 to all the terms of the above inequality: $25\\leq a_1+24<a_2+24<\\cdots<a_{75}+24\\leq 149$.\n\nThere are 150 numbers $a_1,\\cdots,a_{75},a_1+24,\\cdots,a_{75}+24$. By the inequalities above, these numbers range from 1 to 149. Then, by the pigeonhole principle, at least two of them are equal (in a list of 150 integers ranging from 1 to 149, at least two are equal).\n\nBecause all the numbers $a_1,\\cdots,a_{75}$ are distinct, and all numbers $a_1+24,\\cdots,a_{75}+24$ are also distinct, it follows that $a_i = a_j + 24$ for some $i > j$. Therefore, from the $(j+1)^{th}$ hour to the $i^{th}$ hour, there were exactly 24 competitions.\n\nNow, I will show the attempt at a solution for part 2:\n\n2 - a) The same reasoning as above can be used:\n\n$1\\leq a_1<a_2<\\cdots<a_{75}\\leq 125$. Adding 2 to all terms:\n\n$3\\leq a_1 + 2<a_2 + 2<\\cdots<a_{75} + 2\\leq 127$\n\nSo, there are 150 numbers that range from 1 to 127. Therefore, by the pigeonhole principle, at least $\\left \\lceil \\frac{150}{127} \\right \\rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 2$. This guarantees that there is a period of consecutive hours during which there were exactly 2 competitions.\n\nb) The same reasoning as above can be used:\n\n$1\\leq a_1<a_2<\\cdots<a_{75}\\leq 125$. Adding 23 to all terms:\n\n$24\\leq a_1 + 23<a_2 + 23<\\cdots<a_{75} + 23\\leq 148$\n\nSo, there are 150 numbers that range from 1 to 148. Therefore, by the pigeonhole principle, at least $\\left \\lceil \\frac{150}{148} \\right \\rceil$ = 2 numbers must be equal. So, there is an $a_i = a_j + 23$. This guarantees that there is a period of consecutive hours during which there were exactly 23 competitions.\n\nc) $1\\leq a_1<a_2<\\cdots<a_{75}\\leq 125$. Adding 25 to all terms:\n\n$26\\leq a_1 + 25<a_2 + 25<\\cdots<a_{75} + 25\\leq 150$\n\nIn this case, there are 150 numbers that range from 1 to 150. So, there are not necessarily two equal numbers. Therefore, we can't conclude anything directly.\n\nBut it is possible to show that the statement is not true for 25, because an explicit counter-example (suggested below in the comments) can be found. Suppose the number of matches until each one of the 75 hours is, respectively: $\\{1,2,\\cdots,25,51,\\cdots,75,101,\\cdots,125\\}$. Here, there are no pairs of numbers whose difference is 25.\n\nd) $1\\leq a_1<a_2<\\cdots<a_{75}\\leq 125$. Adding 30 to all terms:\n\n$31\\leq a_1 + 30<a_2 + 30<\\cdots<a_{75} + 30\\leq 155$\n\nIn this case, there are 150 numbers that range from 1 to 155. So, we can't apply the pigeonhole principle here, similarly to the above situation. It is not easy to find a counter-example in this case; I think that, for 30, the statement is always true (that is, there is always a period of consecutive hours during which there were exactly 30 matches). But I'm not sure how to prove it.\n\nEdit: I think I found a way to prove it, based on the suggestion given by Lopsy; I've included it as an answer to this question.\n\nThank you in advance.\n\n-\nJust because the pigeonhole principle doesn't apply immediately doesn't mean that it isn't true - you'd need to come up with an explicit counter-example. \u2013\u00a0Thomas Andrews Jan 8 '12 at 16:07\nIn the case of $25$, you can come up with a pretty easy counter-example - $\\{a_i\\}=\\{1,2,...,25,51,...,75,101,...125\\}$. Can you come up with a counter-example for $30$? \u2013\u00a0Thomas Andrews Jan 8 '12 at 16:15\n@ThomasAndrews: Actually it seems that it's not possible to come up with a counter-example for 30. If I use the same reasoning you used for 25: $\\{1,\\cdots,30,61,\\cdots,90,121,\\cdots,125\\}$. In this list there are only 65 hours. But there are 75 hours and at least 1 match per hour. If I added more numbers smaller than 125 to this list, I would get 2 numbers whose difference is 30. It seems to mean that the statement is valid for 30. \u2013\u00a0anonymous Jan 8 '12 at 17:54\nWow, that's a whole lot of arm wrestling. \u2013\u00a0Nate Eldredge Jan 8 '12 at 20:32\nHint: there can only be three $a_i$ in the set {1, 31, 61, 91, 121}, right? Okay, now keep using this idea... you should be able to finish the problem using this. \u2013\u00a0Lopsy Jan 9 '12 at 2:15\n\nI'm posting this answer to my own question because I think I found a solution, but I would appreciate feedback on whether the solution below is complete.\n\nThe only detail that is missing to complete part d is that I need to show that the statement is true if 24 in the original statement is replaced by 30; that is, I will try to show that there is a period of consecutive hours during which the arm wrestler had exactly 30 matches (since there seems to be no counter-example for this case).\n\nI think I found a way to prove it, based on the suggestion given by Lopsy in the comments to the question. This attempt is very similar to the accepted answer here: Prove that 2 students live exactly five houses apart if.\n\nWe build the following sets: the 5-element sets $\\{1, 31, 61, 91, 121\\}$, $\\{2, 32, 62, 92, 122\\}$, ..., $\\{5, 35, 65, 95, 125\\}$, and the 4-element sets $\\{6, 36, 66, 96\\}$, $\\{7, 37, 67, 97\\}$, ..., $\\{30, 60, 90, 120\\}$. The purpose of these sets is to represent every possibility of number of matches played so far. They encompass all numbers from 1 to 125 and, inside each set, the difference between adjacent numbers is 30; however, there are no numbers from two different sets whose difference is 30. There are 5 five-element sets, and 25 four-element sets.\n\nNow, if I want to choose 75 numbers from all sets so that there are no two numbers whose difference is 30, I have to choose at most 3 elements from the sets with 5 elements (for example, I can choose 1, 61 and 121 from the set $\\{1, 31, 61, 91, 121\\}$; if I choose one more, then clearly two of them will have difference equal to 30), and at most 2 elements from the sets with 4 elements (for example, I can choose 30 and 120 from $\\{30, 60, 90, 120\\}$, but, if I choose one more, then clearly two of them will have difference 30). There are 30 sets in total; since there are 5 sets with 5 elements and 25 sets with 4 elements, the maximum number of elements that can be chosen so that no two of them have 30 as difference is $3\\times 5 + 2\\times 25 = 15 + 50 = 65$ numbers. But I need to choose 75 numbers. So, there must be some numbers whose difference is 30.\n\nDoes this make sense, or is there some mistake?\n\n-\nLooks fine to me. \u2013\u00a0Peter Taylor Feb 5 '12 at 13:29", "date": "2015-11-26 03:27:40", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/97397/combinatorics-pigeonhole-principle-question/103970", "openwebmath_score": 0.7883138656616211, "openwebmath_perplexity": 107.96081543284616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407144861112, "lm_q2_score": 0.9005297841157157, "lm_q1q2_score": 0.8764322505088026}}
{"url": "https://talkstats.com/threads/mean-of-the-residuals-in-simple-linear-regression-beginner-alert.75027/#post-219657", "text": "# Mean of the residuals in Simple Linear Regression *Beginner Alert*\n\n##### New Member\nHey guys,\n\nFirst of all, I would like to apologize for the question I'm gonna ask regarding the fact, that I am absolute beginner in statistics.\n\nCurrently I'm trying to test some hypotheses in Rstudio using Simple Linear Regression, as I am aware that before interpreting results of my model I have to provide verification of assumptions that should be meet within the model.\n\nUnfortunately, one of the key assumption is not very clear for me, could someone explain please?\n\n\"The mean of the residuals should be equal to zero.\"\n\nI have no clue which of the residuals I should sum and then divide by number of items. I tried every possibility but it's never zero (I tried it also within other models). So basically my conclusion is that I'm doing something wrong.\n\nI tried to plot residuals and it seems fine for me.\n\nI would be very thankful for any advice or explanation.\n\nThank you very much.\n\n#### Dason\n\nIf you take the mean of the residuals for the observations that were used in the model you should get 0. Here is an example using R since that's what you're using.\n\nCode:\n> head(mtcars)\nmpg cyl disp  hp drat    wt  qsec vs am gear carb\nMazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4\nMazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4\nDatsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1\nHornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1\nHornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2\nValiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1\n> o <- lm(mpg ~ wt, data = mtcars)\n\nresid         residuals     residuals.glm residuals.lm  resizeImage\nMazda RX4     Mazda RX4 Wag        Datsun 710    Hornet 4 Drive\n-2.2826106        -0.9197704        -2.0859521         1.2973499\n-0.2001440        -0.6932545\n> mean(residuals(o))\n[1] 2.024748e-16\nNote that the final line might look like it's non-zero but... that's zero. When using a computer there is rounding error and the value being displayed as the mean of the residuals is... 0.0000000000000002024748. Which is close to the smallest allowable value when working with floating point data...\n\n##### New Member\nIf you take the mean of the residuals for the observations that were used in the model you should get 0. Here is an example using R since that's what you're using.\n\nCode:\n> head(mtcars)\nmpg cyl disp  hp drat    wt  qsec vs am gear carb\nMazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4\nMazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4\nDatsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1\nHornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1\nHornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2\nValiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1\n> o <- lm(mpg ~ wt, data = mtcars)\n\nresid         residuals     residuals.glm residuals.lm  resizeImage\nMazda RX4     Mazda RX4 Wag        Datsun 710    Hornet 4 Drive\n-2.2826106        -0.9197704        -2.0859521         1.2973499\n-0.2001440        -0.6932545\n> mean(residuals(o))\n[1] 2.024748e-16\nNote that the final line might look like it's non-zero but... that's zero. When using a computer there is rounding error and the value being displayed as the mean of the residuals is... 0.0000000000000002024748. Which is close to the smallest allowable value when working with floating point data...\nyes! You're right!\n\nThank you very much!\n\n#### Dason\n\nSide note: there is never a need to apologize for being a beginner. We all start somewhere. Please let us know if you have other questions!\n\n#### noetsi\n\n##### No cake for spunky\nExcept dason. He was born (if that is the right word for a bot) with a perfect knowledge of statistics...\n\nI have spent 15 years studying regression and this is the first time I have heard that one raised. Usually you look at the residuals to see if you have heteroskedacity, non-normality, or non-linearity.\n\n#### ondansetron\n\n##### TS Contributor\nThe nonlinearity could imply the mean error is not 0.\n\nTrue model: Y= B0 +B1X + B2X^2\nFit: yhat= b0 + b1x\n\nthen Y-yhat = residual\nimplies\nE[(B0 +B1X + B2X^2)-(b0 + b1x)] = E[B2X^2] not equal to zero for B2 and X not zero... so the average error would be B2X^2... this could show as curvature in a residual plot of residuals vs x, for example.\n\n#### Dason\n\nTrue. But that doesn't matter because the sample mean of the residuals will be 0 regardless.\n\n#### ondansetron\n\n##### TS Contributor\nTrue. But that doesn't matter because the sample mean of the residuals will be 0 regardless.\nCause that's how the line of best fit is defined (partially), yaknow?\n\n#### ondansetron\n\n##### TS Contributor\nI think that is his point\nAnd mine too, but generally the idea of plotting residuals vs x is to help see if you're misspecified!\n\n#### hlsmith\n\n##### Less is more. Stay pure. Stay poor.\nYeah, i would have been content with the outputted residual median being E-07. Good points though.\n\n#### noetsi\n\n##### No cake for spunky\nAnd mine too, but generally the idea of plotting residuals vs x is to help see if you're misspecified!\nThat is true, but different than the original posters question. In fact since the mean of the residuals will always be zero you don't need to check for that - it has to be true. Unless you have a software issue", "date": "2022-10-02 13:58:02", "meta": {"domain": "talkstats.com", "url": "https://talkstats.com/threads/mean-of-the-residuals-in-simple-linear-regression-beginner-alert.75027/#post-219657", "openwebmath_score": 0.4583716094493866, "openwebmath_perplexity": 1969.8792762243834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9449947086083139, "lm_q2_score": 0.9273632931373373, "lm_q1q2_score": 0.8763534049723645}}
{"url": "https://math.stackexchange.com/questions/2439450/probability-question-about-good-tickets", "text": "# Probability question about \u201cgood\u201d tickets.\n\nThe problem:\n\nOut of 20 exam tickets, 16 tickets are \"good\". Tickets are carefully mixed, and students take turns pulling one ticket. Who has the better chance to draw a \"good\" ticket the first or the second student in the queue?\n\nMy attempt:\n\nObviously the probability of the first student getting a \"good\" ticket is $16/20 = 4/5 = 0.8$\n\nNow here's where I'm confused. I considered two cases. If the first student got a \"good\" ticket then the chances of the second student are $15/19 = 0.789$, so lower than the first student.\n\nHowever if the first student didn't get a \"good\" ticket, the chances of the second student are $16/19 =0.842$, so slightly better chances than the first one.\n\nSo is the answer \"depends on whether the first student gets a \"good\" ticket\"?\n\n\u2022 The question is to \"predict\" the propability of the 2nd student getting a good card not knowing what the first person draws. \u2013\u00a0Cornman Sep 21 '17 at 20:32\n\u2022 Hint. Is there any of the 20 tickets that the second student has a different chance of getting from any other? \u2013\u00a0Henning Makholm Sep 21 '17 at 20:34\n\u2022 So I have to count the probability of the second student having better chances? That would be equal to $4/20 * 16/19 = 0.168$. But I'm kinda lost after that. \u2013\u00a0Nick202 Sep 21 '17 at 20:34\n\u2022 Draw a \"probability tree\". :) \u2013\u00a0Cornman Sep 21 '17 at 20:36\n\nYou're almost there. You just need to apply conditional probability rules.\n\nEssentially, the unconditional probability that the second student gets a good ticket is equal to a weighted average of the conditional probabilities $\\frac{15}{19}$ and $\\frac{16}{19}$, where the weights are the probabilities that the first student did or did not get a good ticket, respectively. If you carry out that computation, you get\n\n$$\\frac{4}{5} \\times \\frac{15}{19} + \\frac{1}{5} \\times \\frac{16}{19} = \\frac{76}{95} = \\frac{4}{5}$$\n\nSo the two students have equal probabilities of getting a good ticket.\n\nThe same result can be obtained by observing that if you lay the tickets out \"face down,\" so to speak, the only thing distinguishing the first and second tickets is the arbitrary order you laid them out in. Since the first two tickets are already destined for the first two students, they must have the same probability of being good, namely $\\frac45$. In fact, the same logic applies to all the students.\n\n\u2022 Mild edit after a small math-o. \u2013\u00a0Brian Tung Sep 21 '17 at 20:37\n\u2022 I see. I calculated ($4/20 * 16/19$) to find out the probability of the first student getting a bad ticked and second one getting a good ticket. Then I guess I wanted to calculate the opposite(first one getting good and second one getting bad) to see which one would have higher probability. I found out that they are equal. Would that be considered a correct solution? \u2013\u00a0Nick202 Sep 21 '17 at 20:44\n\u2022 If I understand you correctly, that would work, but it seems pretty indirect to me. The most straightforward solution to me is the one by symmetry. \u2013\u00a0Brian Tung Sep 21 '17 at 21:20\n\nIf you don't look at the first student's ticket, you should do a weighted average of the two possibilities and you will discover that the second student's chance of getting a good ticket is also $0.8$. The fact that one was already drawn does not matter at all. How can the chance be different from if they each drew, then swapped tickets before looking at them?\n\nLet $A_1$ and $A_2$ denote the Bernoulli random variables of student 1 and student 2 picking a good ticket, respectively. So for example, the probababilty that student one picks a good ticket is $P(A_1 = G)$.\n\nFirst, lets calculate the probability $P(A_1 = G)$. This is a simple case of the hypergeometric distribution, since the good cards form a \"sub-population\" within the total \"population\" of cards:\n\n$$P(A_1 = G) = \\frac{{16\\choose{1}}{4\\choose{0}}}{20\\choose{1}} = \\frac{16}{20} = .8$$\n\nAnd thus the probability that the first student chooses a bad card is: $P(A_1 = B) = .2$.\n\nNow for second student: Use the law of total probability.\n\n$$P(A_2 = G) = P(A_2 = G | A_1 = G)P(A_1 = G) + P(A_2 = G | A_1 = B)P(A_1 = B)$$\n\nThe conditional distribution of $A_2$ given $A_2$ is also hypergeometric.\n\n$$= \\frac{{15\\choose{1}}{4\\choose{0}}}{19\\choose{1}}*.8 + \\frac{{16\\choose{1}}{3\\choose{0}}}{19\\choose{1}}*.2$$ $$= \\frac{15}{19}*.8 + \\frac{16}{19}*.2 = .8$$\n\nA-ha! the probability is the same. Go figures.", "date": "2019-12-10 19:26:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2439450/probability-question-about-good-tickets", "openwebmath_score": 0.8118665814399719, "openwebmath_perplexity": 404.1450253447022, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353585837, "lm_q2_score": 0.8902942275774318, "lm_q1q2_score": 0.8763489735920886}}
{"url": "http://www.dss.net.au/article.php?tag=f95a90-acceleration-calculator-calculus", "text": "You forgot your money so you turn around and go back home in 120 more seconds: what is your round-trip speed and velocity? Acceleration Calculator, Time, Speed, Velocity This website may use cookies or similar technologies to personalize ads (interest-based advertising), to provide social media features and to analyze our traffic.\n\nand the acceleration is given by $a\\left( t \\right) = \\frac{{dv}}{{dt}} = \\frac{{{d^2}x}}{{d{t^2}}}.$ Using the integral calculus, we can calculate the velocity function from the acceleration function, and the position function from the velocity function. The TI in Focus program supports teachers in preparing students for the AP \u00ae Calculus AB and BC test. ... Newton's Second Law of Motion states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.\nTo simplify the problem, we will assume that we know the acceleration, as calculating it from the force of the engine and other factors would be a problem in it\u2019s own. Constant Acceleration Equations Calculator Science - Physics - Formulas.\n\nChapter 10 Velocity, Acceleration, and Calculus The \ufb01rst derivative of position is velocity, and the second derivative is acceleration. ... Newton's Second Law of Motion states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. Velocity = 0 m 220 s = 0 m/s. Inputs: velocity (v) Code to \u2026 Force, Mass, Acceleration Calculator. As you will soon see, the angular acceleration formula differs from the acceleration in linear motion, which you probably know very well.Read on if you want to learn what are the angular acceleration units and what is the angular acceleration equation. In this section we need to take a look at the velocity and acceleration of a moving object. We find the acceleration is 5 m/s^2.\n\nSolving for velocity. A race car accelerates from 15 m/s to 35 m/s in 3 seconds. Calculator Use. The change in velocity of an object divided by the time period is called as its average acceleration.\n\nA collection of test-prep resources. The change in velocity of an object divided by the time period is called as its average acceleration. How to Calculate Acceleration: Step-by-Step Breakdown. Force, Mass, Acceleration Calculator. Solving for acceleration.\n\nThese deriv-atives can be viewed in four ways: physically, numerically, symbolically, and graphically. The force and mass of an accelerating object 2. Look at all three graphs in the figure again. Yes, the velocity is zero as you ended up where you started. Now we\u2019ll breakdown the acceleration formula step-by-step using a real example. This is a multipurpose tool, so you can use it to work out many things associated with acceleration.Use it for: 1. The ideas of velocity and acceleration are familiar in everyday experience, but now we want you Math Geometry Physics Force Fluid Mechanics Finance Loan Calculator. Math AP\u00ae\ufe0e Calculus AB Applications of integration Connecting position, velocity, and acceleration functions using integrals Connecting position, velocity, and acceleration functions using integrals The Uniformly Accelerated Motion calculator uses the equations of motion to solve motion calculations involving constant acceleration in one dimension, a straight line. Code to \u2026 Acceleration, in physics, is the rate of change of velocity of an object.\n\nOur library of files covers free-response questions (FRQ) from past exams through the lens of graphing technology.\n\nAngular acceleration calculator helps you find the angular acceleration of an object that rotates or moves around a circle. Help students score on the AP \u00ae Calculus exam with solutions from Texas Instruments. The TI in Focus program supports teachers in preparing students for the AP \u00ae Calculus AB and BC test. Section 1-11 : Velocity and Acceleration.\n\nIt can solve for the initial velocity u, final velocity v, displacement s, acceleration a, and time t. You need to know 3 of the 4: acceleration, initial speed, final speed and time (acceleration duration) to calculate the fourth. The total time is 100 s + 120 s = 220 s: Speed = 440 m 220 s = 2.0 m/s. This is an online calculator to find the average acceleration with the given values.\n\nVelocity = 130 m 100 s East = 1.3 m/s East. due to the many different units supported. From Calculus I we know that given the position function of an object that the velocity of the object is the first derivative of the position function and the acceleration of the object is the second derivative of the position function.", "date": "2020-09-22 09:31:34", "meta": {"domain": "net.au", "url": "http://www.dss.net.au/article.php?tag=f95a90-acceleration-calculator-calculus", "openwebmath_score": 0.6186461448669434, "openwebmath_perplexity": 427.7177852197269, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9912886150275787, "lm_q2_score": 0.8840392817460333, "lm_q1q2_score": 0.8763380752320008}}
{"url": "https://brilliant.org/discussions/thread/bezouts-lemma/", "text": "# Bezout's Lemma\n\nGiven integers $x$ and $y$, a linear combination of $x$ and $y$ has the form $ax + by$. Bezout's Lemma relates the question of finding the greatest common divisor of $x$ and $y$ to the question of finding linear combinations of $x$ and $y$.\n\n## Technique\n\nApplying the Euclidean Algorithm backwards gives an algorithm to obtain the integer values of $a$ and $b$. We show how to obtain integers $a$ and $b$ such that $16457a + 1638b = 7$. These numbers are calculated in Euclidean Algorithm.\n\n$\\begin{array} {llllllllll} 7 & = & 21 &- & 14 \\times 1 \\\\ & = & 21 \\times 1 & - & 14 \\times 1\\\\ & = & 21 &- & (77 - 21 \\times 3) \\times 1\\\\ & = & - 77 \\times 1 &+& 21 \\times 4\\\\ & = & - 77 \\times 1 &+ & (1638 - 77 \\times 21) \\times 4 \\\\ &= &1638 \\times 4 &-& 77 \\times 85\\\\ & = & 1638 \\times 4 &- & (16457 - 1638 \\times 10) \\times 85 \\\\ & = &16457 \\times (-85) &+& 1638 \\times 854\\\\ \\end{array}$\n\n## Worked Examples\n\n### 1. Given integers $x, y$, describe the set of all integers $N$ that can be expressed in the form $N=ax+by$, where $a, b$ are integers.\n\nSolution: Let $k = \\gcd(x,y)$. Then any integer of the form $kn$, where $n$ is an integer, can be expressed as $ax+by$. We already know that this condition is a necessary condition, so to show that it is sufficient, Bezout's Lemma tells us that there exists integers $a', b'$ such that $k = a' x + b'y$. Therefore,\n\n$kn = (a'n) x + (b'n) y.$\n\n### 2. [Modulo Arithmetic Property I - Multiplicative inverses] Show that if $a, b$ are integers such that $\\gcd(a,n)=1$, then there exists an integer $x$ such that $ax \\equiv 1 \\pmod{n}$.\n\nSolution: Since $\\gcd(a,n)=1$, Bezout's Lemma implies there exists integers $x, y$ such that $ax + n y = \\gcd (a,n) = 1$. Then\n\n$1 \\equiv ax+ny \\equiv ax \\pmod{n} .$\n\nNote by Calvin Lin\n5\u00a0years, 7\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nI learned a lot! Thanks :D\n\n- 5\u00a0years, 4\u00a0months ago\n\nI did not understand what the 1st worked example mean. Any number of form kn can be represented in the form ax+by but that does not mean that all nos of form ax+by will be of form kn. Actually just explain what the question demands?\n\n- 5\u00a0years, 3\u00a0months ago\n\nThe question asks you to classify all numbers that can be written in the form $ax + by$ for given integers $x$ and $y$. The claim is that these numbers are exactly those of the form $k n$, where$k = \\gcd (x,y)$ and $n$ is any integer.\n\nAs you realized, there are 2 parts to this.\nFirst, we have to show that any number of the form $ax + by$ can be written as $kn$ for some $n$. This is obvious because we can let $x = k x^*, y = ky^*$ then $ax + by = a k x^* + bky^* = k ( ax^* + by^* )$.\nSecond, we have to show that any number of the form $kn$ can be written as $ax + by$ for some $a$ and $b$. This follows by applying Bezout's lemma, which tells us that there exists integers which satisfy $k = a' x + b' y$, and hence $kn = ( a'n) x + (b'n) y$.\n\nStaff - 5\u00a0years, 3\u00a0months ago\n\nthanks....\n\n- 5\u00a0years, 3\u00a0months ago", "date": "2019-11-20 18:29:28", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/bezouts-lemma/", "openwebmath_score": 0.9846743941307068, "openwebmath_perplexity": 496.1830356409937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9930961615201183, "lm_q2_score": 0.8824278772763472, "lm_q1q2_score": 0.8763357377414864}}
{"url": "https://math.stackexchange.com/questions/1408287/is-fx-x-smooth", "text": "# Is f(x) = x smooth?\n\nIt may sound too basic to even be a question, but I couldn't find a straight answer in Wolfram Alpha, Wolfram Mathworld or Wikipedia. Several other examples of more complicated functions are given.\n\nIn Wolfram Mathworld it is written that\n\nA smooth function is a function that has continuous derivatives up to some desired order over some domain. (...) The number of continuous derivatives necessary for a function to be considered smooth depends on the problem at hand, and may vary from two to infinity.\n\n$f(x) = x$ has derivative 1 of the first order and 0 of second order, so I would say based on this it has at least 2 derivatives. I think it also has an infinite number of derivatives which are also 0.\n\nAnother page on Wolfram Mathworld says the following:\n\nA $C^{\\infty}$ function is a function that is differentiable for all degrees of differentiation. (...) All polynomials are $C^{\\infty}$. (...) $C^{\\infty}$ functions are also called \"smooth\" (...).\n\nSince $f(x) = x$ is a polynomial, I'm concluding that the paragraphs above mean it is also smooth.\n\n\u2022 I also think they are smooth Aug 24, 2015 at 18:39\n\u2022 \"Smooth\" runs the gamut from continuously differentiable to $C^\\infty,$ and even that is probably too narrow a spectrum. There is no one fixed definition. In the literature you'll often see things like \"for the purposes of this paper, \"smooth\" will mean ___,\" where \"____\" is a precise definition.\n\u2013\u00a0zhw.\nAug 24, 2015 at 18:55\n\u2022 @CarstenS The question is in the title: Is the function f(x)=x smooth? Aug 24, 2015 at 23:21\n\u2022 I made an attempt to answer it based on what I found, and I showed my reasoning to provide something to the question. So yes, the question contains an answer, but I still was not sure if this holds up, and I was still interested in further information and others' input. Aug 24, 2015 at 23:27\n\u2022 @zhw.: Is there any definition of \"smooth\" that does not apply to $f(x)=x$? Aug 25, 2015 at 6:32\n\nA function is smooth is it has derivatives of infinite order. $f(x) = x$ is smooth because it has infinitely many derivatives which are all 0, except for the first one. Polynomials are smooth because eventually their derivatives are 0.\n\n\u2022 Yes, that's the standard definition Aug 24, 2015 at 18:45\n\u2022 @MichaelMenke Not so sure that is standard.\n\u2013\u00a0zhw.\nAug 24, 2015 at 18:57\n\u2022 @IllegalImmigrant yeah technically smooth is $C^\\infty$, but many times for all the calculations we only need our function to be $C^1$ or $C^2$ or something.. so we abuse the notation and we call those function \"smooth\", which basically comes to mean \"it's regular enough to employ all the theorems we need with no headaches\"\n\u2013\u00a0Ant\nAug 24, 2015 at 21:56\n\u2022 $x\\mapsto x$ satisfies every definition you guys have given. Aug 25, 2015 at 2:36\n\u2022 What are \"derivatives of infinite order\"?\n\u2013\u00a0JiK\nAug 25, 2015 at 10:23\n\nYes, the identity function has derivatives of every finite order, and is therefore smooth. It doesn't matter that most of the derivatives are $0$ everywhere -- being $0$ is a perfectly cromulent way to exist.\n\n\u2022 I had to look that word up. Between this and popularizing schadenfreude, The Simpsons has really influenced the language, hasn't it? Aug 25, 2015 at 2:34\n\u2022 @columbus8myhw yes, it's certainly embiggened our vocabulary Aug 25, 2015 at 3:30\n\nYou may be having issues with the difference between existence and triviality.\n\nIf $f(x)=x$ then\n\n$f(x)=x$ is continuouss\n\n$f'(x)=1$ is continuous\n\n$f''(x)=0$ is continuous\n\n$f'''(x)=0$ is continuous\n\netc...\n\nSo all its derivatives are continuous. On the other hand, take $g(x)=x\\times|x|$\n\n$g(x)=x\\times|x|$ is continuous\n\n$g'(x)=\\frac{|x|}2$ is continuous\n\n$g''(x) = \\frac12$ if $x>0$, $g''(x)=-\\frac12$ if $x<0$ $g''(0)$ is undefined\n\nClearly $g''$ is not continuous because of $g''(0)$ not existing, and so $g$ only has two derivatives.\n\nIntuitively, all of $f$'s derivatives have no breaks in their graph ($y=0$ is simply a nice line), while $g''$ graph has a gaping hole in it at $x=0$.\n\n\u2022 Funny definition of a \"gaping\" hole, that is a nothingth of a unit wide :) Aug 25, 2015 at 15:10\n\u2022 @thepeer Is a unit more or less than the gap of the grand canyon? Aug 25, 2015 at 17:21\n\u2022 Does it matter, seeing as we're talking about a nothingth of it? Sep 1, 2015 at 8:01", "date": "2022-08-19 05:28:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1408287/is-fx-x-smooth", "openwebmath_score": 0.8050992488861084, "openwebmath_perplexity": 557.9902757666977, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138157595305, "lm_q2_score": 0.8962513786759491, "lm_q1q2_score": 0.8762773553250021}}
{"url": "https://math.stackexchange.com/questions/1811451/checking-an-identity-involving-binomial-coefficients", "text": "# Checking an identity involving binomial coefficients\n\nI need some help to check the following identity:\n\nfor every $0\\leq i\\leq l\\leq r$ $$\\sum_{j=0}^i\\binom{r-l+i-j}{i-j}\\binom{l-i+j}{j}=\\binom{r+1}{i}.$$\n\nIs this true ?\n\nAnswering to John, this identity come from a geometric problem. I was computing the top Chern class of a vector bundle of the form $S\\otimes Q$, $S$ is a line bundle, and $Q$ a rank $r$ vector bundle whose Chern classes have a particularly easy form. I have a guess for what should be the answer and imposing equality between the two led me to this identity. I am pretty sure my guess is true and this identity indeed holds; it's interesting that it does not depend on $l$. I think it should be possible to prove this with somee induction argument.\n\n\u2022 How did you arrive at this answer? \u2013\u00a0John Jun 3 '16 at 19:12\n\u2022 Answering to John, this identity come from a geometric problem. I was computing the top Chern class of a vector bundle of the form $S \\otimes Q$, $S$ is a line bundle, and $Q$ a rank $r$ vector bundle whose Chern classes have a particularly easy form. I have a guess for what should be the answer and imposing equality between the two led me to this identity. I am pretty sure my guess is true and this identity indeed holds; it's interesting that it does not depend on $l$. I think it should be possible to prove this with some induction argument. \u2013\u00a0S. S. Jun 3 '16 at 19:56\n\u2022 I wrote the question as an unregistered user \u2013\u00a0S. S. Jun 3 '16 at 19:57\n\u2022 Upper negation (see proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient ) yields $\\dbinom{l-i+j}{j} = \\left(-1\\right)^j \\dbinom{j-\\left(l-i+j\\right)-1}{j} = \\left(-1\\right)^j \\dbinom{i-l-1}{j}$ and $\\dbinom{r-l+i-j}{i-j} = \\left(-1\\right)^{i-j} \\dbinom{\\left(i-j\\right)-\\left(r-l+i-j\\right)-1}{i-j} = \\left(-1\\right)^{i-j} \\dbinom{l-r-1}{i-j}$ and $\\dbinom{r+1}{i} = \\left(-1\\right)^i \\dbinom{i-\\left(r+1\\right)-1}{i} = \\left(-1\\right)^i \\dbinom{i-r-2}{i}$. Now your identity follows from the Chu-Vandermonde identity, applied to $i-l-1$ and $l-r-1$ and $i$. \u2013\u00a0darij grinberg Jun 3 '16 at 20:07\n\nYes, that is true, it is the \"double convolution\" formula.\nTo put the formula in more general terms, define the binomial in the extended way: $$\\left( \\matrix{ x \\cr m \\cr} \\right) = \\left\\{ \\matrix{ {{x^{\\,\\underline {\\,m\\,} } } \\over {m!}}\\quad \\left| {\\;0 \\le {\\rm integer}\\;m} \\right. \\hfill \\cr 0\\quad \\quad \\left| {\\;{\\rm otherwise}} \\right. \\hfill \\cr} \\right.$$ Then the Upper Negation rule tells that $$\\left( \\matrix{ x \\cr m \\cr} \\right) = \\left( { - 1} \\right)^m \\left( \\matrix{ m - x - 1 \\cr m \\cr} \\right)$$ and therefore, applying it twice \\eqalign{ & \\sum\\limits_{\\left( {0\\, \\le } \\right)\\,j\\,\\left( { \\le \\,\\,i} \\right)} {\\left( \\matrix{ r - l + i - j \\cr i - j \\cr} \\right)\\left( \\matrix{ l - i + j \\cr j \\cr} \\right)} = \\cr & = \\sum\\limits_{\\left( {0\\, \\le } \\right)\\,j\\,\\left( { \\le \\,\\,i} \\right)} {\\left( { - 1} \\right)^i \\left( \\matrix{ - r + l - 1 \\cr i - j \\cr} \\right)\\left( \\matrix{ - l + i - 1 \\cr j \\cr} \\right)} = \\cr} and by the \"simple\" convolution $$= \\left( { - 1} \\right)^i \\left( \\matrix{ - r + i - 2 \\cr i \\cr} \\right) = \\left( \\matrix{ r + 1 \\cr i \\cr} \\right)\\quad \\left| {\\;\\left\\{ \\matrix{ {\\rm integer}\\;i \\hfill \\cr {\\rm real}\\;r,l \\hfill \\cr} \\right.} \\right.$$\n\n\nThe Question: $\\ds{\\sum_{j=0}^{k}{r - \\ell + k - j \\choose k - j}{\\ell - k + j \\choose j} = {r + 1 \\choose k}}$\n\n\\begin{align} &\\color{#f00}{\\sum_{j=0}^{k}{r - \\ell + k - j \\choose k - j} {\\ell - k + j \\choose j}} \\\\[3mm] = &\\ \\sum_{j=0}^{\\infty}{-r + \\ell - k + j + k - j - 1\\choose k - j}\\pars{-1}^{k - j} {-\\ell + k - j + j - 1 \\choose j}\\pars{-1}^{j} \\\\[3mm] = &\\ \\pars{-1}^{k}\\sum_{j=0}^{\\infty}{\\ell - r - 1\\choose k - j} {k - \\ell - 1 \\choose j} = \\pars{-1}^{k}\\sum_{j=0}^{\\infty}{\\ell - r - 1\\choose k - j} {k - \\ell - 1 \\choose k - \\ell -1 - j} \\\\[3mm] = &\\ \\pars{-1}^{k}\\sum_{j=0}^{\\infty}\\bracks{\\oint_{\\verts{z} = 1^{-}} {\\pars{1 + z}^{\\ell - r - 1} \\over z^{k - j + 1}}\\,{\\dd z \\over 2\\pi\\ic}} \\bracks{\\oint_{\\verts{w} = 1^{-}} {\\pars{1 + w}^{k - \\ell - 1} \\over w^{k - \\ell - j}} \\,{\\dd w \\over 2\\pi\\ic}} \\\\[3mm] = &\\ \\pars{-1}^{k}\\oint_{\\verts{z} = 1^{-}} {\\pars{1 + z}^{\\ell - r - 1} \\over z^{k + 1}} \\oint_{\\verts{w} = 1^{-}} {\\pars{1 + w}^{k - \\ell - 1} \\over w^{k - \\ell}}\\sum_{j = 0}^{\\infty}\\pars{zw}^{j} \\,{\\dd w \\over 2\\pi\\ic}\\,{\\dd z \\over 2\\pi\\ic} \\\\[3mm] = &\\ \\pars{-1}^{k}\\oint_{\\verts{z} = 1^{-}} {\\pars{1 + z}^{\\ell - r - 1} \\over z^{k + 1}} \\oint_{\\verts{w} = 1^{-}} {\\pars{1 + w}^{k - \\ell - 1} \\over w^{k - \\ell}\\pars{1 - zw}} \\,{\\dd w \\over 2\\pi\\ic}\\,{\\dd z \\over 2\\pi\\ic} \\\\[3mm] \\stackrel{w\\ \\to\\ 1/w}{=}\\ &\\ \\pars{-1}^{k}\\oint_{\\verts{z} = 1^{-}} {\\pars{1 + z}^{\\ell - r - 1} \\over z^{k + 1}}\\ \\overbrace{\\oint_{\\verts{w} = 1^{+}} {\\pars{1 + w}^{k - \\ell - 1} \\over w - z}\\,{\\dd w \\over 2\\pi\\ic}} ^{\\ds{\\pars{1 + z}^{k - \\ell - 1}}}\\ \\,{\\dd z \\over 2\\pi\\ic} \\\\[3mm] = &\\ \\pars{-1}^{k}\\oint_{\\verts{z} = 1^{-}} {\\pars{1 + z}^{k - r - 2} \\over z^{k + 1}}\\,{\\dd z \\over 2\\pi\\ic} = \\pars{-1}^{k}{k - r - 2 \\choose k} \\\\[3mm] = &\\ \\pars{-1}^{k}{-k + r + 2 + k - 1 \\choose k}\\pars{-1}^{k} = \\color{#f00}{{r + 1 \\choose k}} \\end{align}", "date": "2019-07-22 19:14:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1811451/checking-an-identity-involving-binomial-coefficients", "openwebmath_score": 0.995875358581543, "openwebmath_perplexity": 587.8578653938702, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9777138196557983, "lm_q2_score": 0.8962513668985002, "lm_q1q2_score": 0.8762773473020629}}
{"url": "https://mathhelpboards.com/threads/find-the-sum-of-%CE%B1-%CE%B2.3191/", "text": "# Find the sum of \u03b1 + \u03b2.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHi members of the forum,\n\nProblem:\nThe real numbers\n$\\displaystyle \\alpha$, $\\displaystyle \\beta$ satisfy the equations\n\n$\\displaystyle \\alpha^3-3\\alpha^2+5\\alpha-17=0,$\n\n$\\displaystyle \\beta^3-3\\beta^2+5\\beta+11=0.$\n\nFind $\\displaystyle \\alpha + \\beta$.\n\nThis problem has me stumped. It's not that I didn't try, believe you me, but I kept returning back to square one no matter how hard I tried. It's really annoying and I don't believe the author wants us to use the cubic formula to find each root explicitly (or even a numeric root-finding technique) to get the sum, but at this moment, I just don't see any way to manipulate the given two cubic polynomials to find the desired sum.\n\nCould you please shed some light on this problem for me?\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nHi members of the forum,\n\nProblem:\nThe real numbers\n$\\displaystyle \\alpha$, $\\displaystyle \\beta$ satisfy the equations\n\n$\\displaystyle \\alpha^3-3\\alpha^2+5\\alpha-17=0,$\n\n$\\displaystyle \\beta^3-3\\beta^2+5\\beta+11=0.$\n\nFind $\\displaystyle \\alpha + \\beta$.\n\nThis problem has me stumped. It's not that I didn't try, believe you me, but I kept returning back to square one no matter how hard I tried. It's really annoying and I don't believe the author wants us to use the cubic formula to find each root explicitly (or even a numeric root-finding technique) to get the sum, but at this moment, I just don't see any way to manipulate the given two cubic polynomials to find the desired sum.\n\nCould you please shed some light on this problem for me?\n\nHi anemone,\n\nLet me outline the method of solving this problem. Firstly you don't know whether each of these equations have any real roots. So use the discriminant of each equation find out how many real roots each equation has (Refer >>this<<). You will see that each equation has only one real root.\n\nLet $$\\alpha+\\beta=\\lambda$$. Add the two equations together and rearrange it to get an cubic equation of $$\\lambda$$. You should get,\n\n$\\lambda^3-3\\alpha\\beta(\\lambda-2)-3\\lambda^2+5\\lambda-6=0.$\n\nTry to find a value for $$\\lambda$$ that satisfies the above equation (by trial and error). Do you think that the value you obtained for $$\\lambda$$ is unique? If so why?\n\nKind Regards,\nSudharaka.\n\n#### Fernando Revilla\n\n##### Well-known member\nMHB Math Helper\nAnother way. Define $f(x)=x^3-3x^2-5x$. We have $f(\\alpha)+f(\\beta)=6$.We can exoress $f(x)=(x-1)^3+2(x-1)+3$, as a consequence\n$$f(\\alpha)-3=(\\alpha-1)^3+2(\\alpha -1)\\\\ f(\\beta)-3=(\\beta-1)^3+2(\\beta-1)$$\nAdding the above equalities we obtain $0=(\\alpha-1)^3+(\\beta-1)^3+2(\\alpha+\\beta-2)$. Now, decompose to get $0=(\\alpha +\\beta-2)g(\\alpha,\\beta)$ where $g(\\alpha,\\beta)>0$, so $\\alpha+\\beta=2$.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHi anemone,\n\n$\\lambda^3-3\\alpha\\beta(\\lambda-2)-3\\lambda^2+5\\lambda-6=0.$\n\nTry to find a value for $$\\lambda$$ that satisfies the above equation (by trial and error). Do you think that the value you obtained for $$\\lambda$$ is unique? If so why?\n\nKind Regards,\nSudharaka.\nHi Sudharaka, my first instinct is to try the value of $\\displaystyle \\lambda=2$ and it does work. I think the value of $\\displaystyle \\lambda$ that I obtained is unique since the other two roots from both of these functions are imaginary roots. Am I correct?\n\nAnother way. Define $f(x)=x^3-3x^2-5x$. We have $f(\\alpha)+f(\\beta)=6$.We can exoress $f(x)=(x-1)^3+2(x-1)+3$, as a consequence\n$$f(\\alpha)-3=(\\alpha-1)^3+2(\\alpha -1)\\\\ f(\\beta)-3=(\\beta-1)^3+2(\\beta-1)$$\nAdding the above equalities we obtain $0=(\\alpha-1)^3+(\\beta-1)^3+2(\\alpha+\\beta-2)$. Now, decompose to get $0=(\\alpha +\\beta-2)g(\\alpha,\\beta)$ where $g(\\alpha,\\beta)>0$, so $\\alpha+\\beta=2$.\nHi Fernando, it feels so great to receive two different methods in a day to solve my headache. Thanks enormously!\n\nBut another idea hit me when I was having my dinner tonight (I guess there is no rest for the weary.) and I'd really like to hear some opinions about it, if possible.\n\nLet $\\displaystyle f(x)=x^3-3x^2+5x-17$ and $\\displaystyle g(x)=x^3-3x^2+5x+11$ and if we rewrite the function of g(x) to become $\\displaystyle g(x)=(x^3-3x^2+5x-17)+28$, we can see that the function of g(x) is just the function of f(x) that shifted 28 units upward, i.e. g(x) is the transformed function of f(x).\n\nI then use a simple trick by exploiting the fact that the gradient of the function of f(x) and its transformed function, g(x) would be the same at their real root point.\n\nThat is, $\\displaystyle f'(\\alpha)=\\displaystyle g'(\\beta)$.\n$\\displaystyle 3\\alpha^2-6\\alpha+5=3\\beta^2-6\\beta+5$\n$\\displaystyle \\alpha^2-2\\alpha=\\beta^2-2\\beta$\n$\\displaystyle \\alpha^2-\\beta^2=2\\alpha-2\\beta$\n$\\displaystyle (\\alpha+\\beta)(\\alpha-\\beta)=2(\\alpha-\\beta)$\n$\\displaystyle (\\alpha-\\beta)(\\alpha+\\beta-2)=0$\n\nSince $\\displaystyle (\\alpha-\\beta)\\ne 0$, we can conclude that $\\displaystyle (\\alpha+\\beta-2)=0$, or $\\displaystyle \\alpha+\\beta=2$.\n\nHmm...how does that sound to you?\n\n#### Fernando Revilla\n\n##### Well-known member\nMHB Math Helper\nI then use a simple trick by exploiting the fact that the gradient of the function of f(x) and its transformed function, g(x) would be the same at their real root point.\nHmm, why? Consider for instance $f(x)=x^3-1$ and $g(x)=x^3-8$. Here $\\alpha=1$ and $\\beta=2$, however $f'(\\alpha)=3$ and $g'(\\beta)=12$ i.e. $f'(\\alpha)\\neq g'(\\beta)$.\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nHi Sudharaka, my first instinct is to try the value of $\\displaystyle \\lambda=2$ and it does work. I think the value of $\\displaystyle \\lambda$ that I obtained is unique since the other two roots from both of these functions are imaginary roots. Am I correct?\n\nYes that's correct. You are welcome!\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHmm, why? Consider for instance $f(x)=x^3-1$ and $g(x)=x^3-8$. Here $\\alpha=1$ and $\\beta=2$, however $f'(\\alpha)=3$ and $g'(\\beta)=12$ i.e. $f'(\\alpha)\\neq g'(\\beta)$.\nI'm so embarrassed to think of this now, as I really should have verified the result by using another pair of simple functions (such as your example) and I understand it now, these two functions (the original and its transformed function) have the same gradient at the same point.\n\nThanks for commenting on my idea!\n\nYes that's correct. You are welcome!\nThanks again, Sudharaka!\n\n#### MarkFL\n\nStaff member\nI'm so embarrassed to think of this now...\nPlease don't feel embarrassed, we all make mistakes...I know I have made my share of mistakes, and the potential embarrassment is magnified when it is public, but when we come together in the forums to discuss a problem there will be mistakes, but we have our friends to guide us and hence we learn.\n\nOthers reading this topic can benefit from having seen your \"mistake\" and learn why this does not work. I feel that if one person makes an error, then there are others who will make this same error.\n\nBy being willing to take a chance and show your thinking, you have helped others.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHey Mark, it's so sweet and nice of you to say that!\n\n#### Fernando Revilla\n\n##### Well-known member\nMHB Math Helper\nI'm so embarrassed to think of this now, ...\nThe only way to not make mistakes is to never do anything.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nThe only way to not make mistakes is to never do anything.\nI'm so touched by the genuine kindness shown by the community at this site, everyone here is so helpful and friendly...\n\nThanks to all of you, Mark, Fernando and Sudharaka because you all just made my day!\n\nLast edited:", "date": "2020-09-20 07:39:04", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-the-sum-of-%CE%B1-%CE%B2.3191/", "openwebmath_score": 0.8290205001831055, "openwebmath_perplexity": 610.3421087192296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759654852756, "lm_q2_score": 0.8933094017937621, "lm_q1q2_score": 0.8762257219615303}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=g8u3chlprmdu4f9heqrdop5ik3&topic=967.0;wap2", "text": "MAT244-2018S > Quiz-2\n\nQ2-T5101\n\n(1/1)\n\nVictor Ivrii:\nShow that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.\n$$x^2y^3 + x(1 + y^2)y' = 0,\\qquad \\mu(x, y) = 1/xy^3.$$\n\nTianjing Ruan:\nx2y3/xy3 + x(1 + y2)/xy3y' = 0\n\nx + (y\u22123 + y\u22121)y' = 0\n\n\u222bM dx = 1/2x2 + g(y)\n\ng(y) = \u222by\u22123 + 1/ydy = -1/2y\u22122 + ln(y)\n\nThe solution is:\n1/2x2 \u2212 1/2y2 + ln(y) = C\n\nJunya Zhang:\nFirst, let's show the given DE $x^{2}y^{3} + x(1+y^{2})y' = 0$ is not exact.\nDefine $M(x,y)=x^{2}y^{3}$, $N(x,y)=x(1+y^{2})$\n$$M_y = \\frac{\\partial}{\\partial y}[x^{2}y^{3}] = 3x^{2}y^{2}$$ $$N_x = \\frac{\\partial}{\\partial x}[x(1+y^{2})] = 1+y^{2}$$\nSince $3x^{2}y^{2} \u2260 1+y^{2}$, this implies the given DE is not exact.\n\nNow, let's show that the given DE multiplied by the integrating factor $\\mu(x,y) = \\frac{1}{xy^{3}}$ is exact.\nThat is to show $$\\frac{1}{xy^{3}}x^{2}y^{3} + \\frac{1}{xy^{3}}x(1+y^{2})y' = x + (y^{-3}+y^{-1})y'= 0$$ is exact.\n\nDefine $M'(x,y) = x$, $N'(x,y) = y^{-3}+y^{-1}$\nSince\n$$M'_y = \\frac{\\partial}{\\partial y}(x) = 0$$ $$N'_x = \\frac{\\partial}{\\partial x}[y^{-3}+y^{-1}] = 0$$\nBy theorem in the book, we can conclude that $x + (y^{-3}+y^{-1})y'= 0$ is exact.\n\nThus, we know there exists a function $\\phi(x,y)=C$ which satisfies the given DE.\nAlso,\n$$\\frac{\\partial \\phi}{\\partial x} = x$$ $$\\frac{\\partial\\phi}{\\partial y} = y^{-3}+y^{-1}$$\nIntegrate $\\frac{\\partial \\phi}{\\partial x} = x$ with respect to $x$ we have\n$$\\phi(x,y) = \\frac{1}{2}x^{2} + g(y)$$\nTake derivative on both sides with respect to $y$ we get\n$$\\frac{\\partial\\phi}{\\partial y} = g'(y)$$\nSince we know that $\\frac{\\partial\\phi}{\\partial y} = y^{-3}+y^{-1}$\nThen $g'(y) = y^{-3}+y^{-1}$\nIntegrate with respect to $y$ we have\n$g(y) = -\\frac{1}{2}y^{-2} + ln|y| + C$\nAltogether, we have $\\phi(x,y) = \\frac{1}{2}x^{2} -\\frac{1}{2}y^{-2} + ln|y| = C$, which means\n$$\\frac{1}{2}x^{2} -\\frac{1}{2}y^{-2} + ln|y| = C$$\nis the general solution to the given DE.\n\nBesides, notice that the constant function $y(x)=0$ $\\forall x$ is also a solution to the given DE.\n\nVictor Ivrii:\nruantian\nPlease select a proper screen name. Second, your solution is correct but the answer is not due to the lame method to write. Do not use html tags for math, they are inadequate. Use MathJax\n\nJunya\nDo not post solutions after correct one was posted. It would be much better if you wrote something like:\u00a0 \"ruantian solved problem correctly, but presented an answer in the confusing way, correct answer is ... \"", "date": "2021-10-16 18:39:52", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=g8u3chlprmdu4f9heqrdop5ik3&topic=967.0;wap2", "openwebmath_score": 0.9756116271018982, "openwebmath_perplexity": 814.7054280908534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9808759610129465, "lm_q2_score": 0.8933094032139576, "lm_q1q2_score": 0.8762257193593923}}
{"url": "http://math.stackexchange.com/questions/280423/proving-complex-numbers", "text": "# Proving complex numbers\n\nLet $z_1,z_2$ be two complex numbers such that $z_1 + z_2$ and $z_1\\dot\\ z_2$ are each negative real numbers. Prove that $z_1$ and $z_2$ must be real numbers.\n\nMy attempt at a solution follows,\n\nLet $z_1 = a+bi$ and $z_2=c+di$. The hypotheses states that $a+c<0$, $b+d = 0$, $ac-bd <0$, and $ad + bc=0$. And I get stuck here since I do not know the logic to proceed in which I can prove that $z_1$ and $z_2$ must be real numbers.\n\n-\n$d=-b\\Rightarrow d(a-c)=0$ in your last equality $\\to$ either $d$ or $a-c$ is $0$. So... \u2013\u00a0 L. F. Jan 17 '13 at 1:21\n@LordoftheFlies I am not sure what you are trying to do? So in order to prove that they are real I mist show d or a is 0? \u2013\u00a0 Q.matin Jan 17 '13 at 1:24\n$a-c<0$ so it cannot be $0$. Therefore $d=0$ i.e. $z_2$ is real. Since $d=-b\\Rightarrow b=0$ i.e. $z_1$ is real. \u2013\u00a0 L. F. Jan 17 '13 at 1:26\n\nFrom $b+d=0$, get $d=-b$. Then from $ad+bc=0$ get $b(c-a)=0$.\n\nSo either $b=0$, from which $d=0$ making $z_1,z_2$ real, or else $c=a$ (and $d=-b$ still holds). Now put those into $ac-bd<0$ and it becomes $a^2+b^2<0$, contradiction.\n\nNOTE: Give credit to \"Lord of the Flies\", whose comment came in while I was writing this.\n\n-\nThanks but then what follows directly that says $z_1$ and $z_2$ are real numbers? \u2013\u00a0 Q.matin Jan 17 '13 at 1:34\nThe argument above shows that either both $b$ and $d$ are zero, or else a contradiction occurs. Can conclude that is must be that $b=d=0$. The $b,d$ are the imaginary parts of the two complex numbers $z_1,z_2$ using your notation, and saying imaginary part is $0$ is the same as saying a complex number is real. \u2013\u00a0 coffeemath Jan 17 '13 at 1:36\nAhh, a bulb just turned on in my head when you said, \"saying imaginary part is 0 is the same as saying a complex number is real\". Thanks a lot coffeemath!! \u2013\u00a0 Q.matin Jan 17 '13 at 1:41\n\n$b+d=0 \\implies d=-b \\implies ad+bc=-ab + bc = 0\\implies b(c-a)=0$.\n\n\u2022 So either $b=0$, from which $d= b = 0$, and thus $z_1,z_2$ are real,\n\u2022 or else $c-a= a-c = 0$ which contradicts the fact that $a - c < 0.$\n\nTherefore $z_1, z_2 \\in \\mathbb{R}$.\n\n-\nThanks but did you mean $c-a$ instead of $a-c$? \u2013\u00a0 Q.matin Jan 17 '13 at 1:36\namWhy: Is $a-c<0$ a hypothesis, or is it $a+c<0$? \u2013\u00a0 coffeemath Jan 17 '13 at 1:38\n\nAlternately, since $z_1+z_2$ is real and $z_1z_2$ is negative real, and $$(z_1-z_2)^2=(z_1+z_2)^2-4z_1z_2,$$ we conclude that $(z_1-z_2)^2$ is positive. Thus $z_1-z_2$ is real. Since $z_1+z_2$ is real, it follows by addition that $2z_1$, and therefore $z_1$, is real.\n\n-\nIsn't $z_1+z_2$ also negative real? \u2013\u00a0 Q.matin Jan 17 '13 at 1:51\nYes it is negative real, but it gets squared on the right side so becomes positive in its contribution. \u2013\u00a0 coffeemath Jan 17 '13 at 1:54\nIn the problem, it is. But the above proof does not need or use the fact that $z_1+z_2$ is negative. It does use the fact that the product $z_1z_2$ is negative. \u2013\u00a0 Andr\u00e9 Nicolas Jan 17 '13 at 1:54\n\nYou have chosen to splitting the complex numbers into real and imaginary parts. To that end, you have faithfully translated the givens of your problem into this form.\n\nWhat you've overlooked is that you haven't translated your goal into this form. Doing this isn't always the simplest way to proceed, but it's almost always a correct and straightforward way to proceed. The translation of your goal is that you seek to prove $b=d=0$.\n\nFinally, even without an idea of how to proceed, you do know something you can do with a system of equations and inequations: you can simply / solve it! Sometimes, the simplified/solved form will suggest what to do next. So without knowing where you're going, we can still do the following:\n\n\u2022 Solve $b+d = 0$ to determine $d = -b$\n\u2022 Simplify $ad+bc = 0$ to $b(c-a) = 0$\n\u2022 Solve that equation by splitting into cases:\n\u2022 Case $b=0$:\n\u2022 Then $d=0$ too.\n\u2022 The remaining inequalities are $a+c < 0$ and $ac - bd < 0$\n\u2022 The latter simplifies to $ac < 0$\n\u2022 We can solve this by cases:\n\u2022 Case $a < 0$ and $c > 0$:\n\u2022 The remaining inequation yields $a < -c < 0$\n\u2022 Case $a > 0$ and $c < 0$:\n\u2022 The remaining inequation yields $0 < a < -c$\n\u2022 Case $(c-a)=0$:\n\u2022 Then $c = a$\n\u2022 The remaining inequalities are $a+c < 0$ and $ac - bd < 0$\n\u2022 Thes latter simplifies to $a^2 < -b^2$\n\u2022 This is a contradiction! (the r.h.s is nonpositive and the l.h.s. is nonnegative)\n\nSo the complete solution to the givens is the union of the cases:\n\n\u2022 $a < 0$ and $c > 0$ and $|c| < |a|$ and $b=d=0$\n\u2022 $a > 0$ and $c < 0$ and $|a| < |c|$ and $b=d=0$\n\nAnd then once you've reached this point, you could return to the problem to see how you could use this information to show $z_1$ and $z_2$ are real numbers.\n\n(Of course, if you translated the goal, you wouldn't have had to do any of the work in the $b=0$ case)\n\n-\nThank so much!! THis is very helpful! \u2013\u00a0 Q.matin Jan 18 '13 at 1:46\n\n$(x\\!-\\!z_1)(x\\!-\\!z_2) = x^2\\! - (z_1\\!+\\!z_2) x + z_1 z_2\\in \\Bbb R[x]$ has discriminant $\\,(z_1\\!+\\!z_2)^2\\!-4z_1z_2 > 0,$ since $z_1z_2 < 0,$ hence the roots $z_1,z_2$ are real (by the quadratic formula).\n\nRemark $\\$ The proof works more generally for $\\ z_1,z_2$ elements of any (integral) domain $\\,\\Bbb D\\supset \\Bbb R,$ assuming only that $\\,z_1\\!+z_2,\\, z_1 z_2\\,$ are both in $\\Bbb R,$ and $\\,z_1 z_2 < 0$. The hypothesis that $\\Bbb D$ is a domain ensures that the list of roots $z_1,z_2 \\in \\Bbb R$ obtained by the quadratic formula persists as the unique list of roots in $\\Bbb D$ (if the quadratic polynomial had an additional root $z_3\\in \\Bbb D$ then it would have more roots than its degree, which cannot occur in a domain). The claim may fail in extension rings not domains, e.g. $f(x) = (x+2)(x-1) = x^2+x-2 = (x-w)(x+w+1)\\,$ over $\\,\\Bbb R[w]/(w^2+w-2),\\,$ where $f$ has non-real roots $\\ w,\\, -w-1 \\not\\in \\Bbb R,\\,$ contra to the claim.\n\nNotice that said extension ring is not a domain since there $\\,(w+2)(w-1) = 0,\\,$ but $\\,w \\ne -2,1$. Informally, one may think of $w$ as an algebraic representation of a \"generic\" root of $f$, sharing all $(\\Bbb R -)$algebraic properties of the roots $-2,1,$ i.e. $w$ satisfies $f(w) = g(w)$ for polynomials $f,g\\in\\Bbb R[x]\\,$ iff $\\,-2$ and $1$ also do: $f(-2) = g(-2)$ and $f(1) = g(1).\\,$ While we can always construct extension rings containing such generic roots, generally the construction does not preserve the property of being a domain, and the consequent uniqueness of the list/multiset of roots (which, suitably formulated, is a characteristic property of domains among rings). Yet another example that uniqueness theorems provide powerful tools for deducing equalities.\n\n-\n...from which we also see that the hypotheses can be weakened to \"$z_1+z_2 \\in \\mathbb{R}$ and $z_1 z_2 < 0$\". \u2013\u00a0 Hans Lundmark Jan 17 '13 at 10:13\n@Hans Yes, in fact it works much more generally - see my recently added remark. \u2013\u00a0 Math Gems Jan 17 '13 at 17:19\n@MathGems Thanks a lot! \u2013\u00a0 Q.matin Jan 18 '13 at 1:46", "date": "2015-04-28 19:05:41", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/280423/proving-complex-numbers", "openwebmath_score": 0.9357888698577881, "openwebmath_perplexity": 337.85852874780676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474913588876, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8761972843566793}}
{"url": "https://puzzling.stackexchange.com/questions/12730/tiling-a-diamond-shaped-grid-with-tetrominoes", "text": "# Tiling a diamond-shaped grid with tetrominoes\n\nYou have a grid like this:\n\n(The entire grid isn't shown as it would be too large, but the number of squares in each row are as follows: $2, 4, 6, \\ldots, 96, 98, 100, 100, 98, 96, \\ldots, 6, 4, 2$.)\n\nWe define this grid as $G(100)$, as it is 100 squares across at its widest point and 100 squares high at its tallest point.\n\nYou want to tile it with only copies of this tetromino:\n\nYou may rotate or flip the tetromino.\n\n1. Is it possible? Why or why not? An explanation in your answer is required.\n\n2. For which even positive integer values of $n$ is tiling the grid $G(n)$ with only the tetromino possible? (Again, you must provide an explanation.)\n\nnote: I will post my own solution after either two days have passed or two distinct correct answers (for each question 1 and 2) have been provided.\n\n\u2022 Sanity check: $G(100)$ has $4\\cdot \\binom{50+1}{2}=5100$ squares? \u2013\u00a0Mike Earnest Apr 25 '15 at 21:28\n\u2022 @MikeEarnest Correct. (Alternatively, $4 \\cdot \\sum_{n=1}^{50}n = 5100$.) \u2013\u00a0Doorknob Apr 25 '15 at 21:34\n\u2022 In general $G(N)$ has $2\u22c5N\u22c5(N+1)$ squares \u2013\u00a0Ivo Beckers Apr 25 '15 at 22:15\n\u2022 @IvoBeckers Don't think that's right; it should be $N \\cdot (\\frac{n}{2}+1)$ (note that there are $5100$ squares for $N = 100$), or equivalently $\\frac{N^2+2N}{2}$. \u2013\u00a0Doorknob Apr 25 '15 at 22:19\n\u2022 you're right. I was thinking of half the value of $N$ \u2013\u00a0Ivo Beckers Apr 25 '15 at 22:20\n\n$G(100)$ cannot be tiled.\n\nDivide an infinite checkerboard into $2\\times 2$ blocks, then color each block alternately white and black, as shown: $$\\begin{array}{ccccc|ccccc} &&\\vdots&&\\vdots&&\\vdots&&\\vdots&&\\\\ \\cdots & B & B & W & W & B & B & W & W &\\cdots\\\\ \\cdots & B & B & W & W & B & B & W & W &\\cdots\\\\ \\hline \\cdots & W & W & B & B & W & W & B & B &\\cdots\\\\ \\cdots & W & W & B & B & W & W & B & B &\\cdots\\\\ &&\\vdots&&\\vdots&&\\vdots&&\\vdots&\\\\ \\end{array}$$ Then, place the $G(100)$ array on top of this so that its axes of symmetry are the two lines (one horizontal, the other vertical) above, and let $G(100)$ be colored to match the the board below it.\n\nNo matter how a tile is placed, it will cover either 3 whites and a black, or 3 blacks and a white. This means that whenever an odd number of tiles are placed, the area they cover will be unbalanced between black and white. But covering $G(100)$ entails placing $1275$ tiles, and $G(100)$ itself is white/black balanced, so it cannot be covered. This proof also shows that $G(2n)$ is not tileable whenever $n\\equiv 1$ or $2$ (mod $4$), since there are $1+2+\\dots+n$ tiles to be placed, which is odd whenever $n=4k+1$ or $4k+2$.\n\nWe cannot tile the given grid.\n\nBelow is an example coloring of $G(12)$. Notice that a rotation of 180\u00b0 about the central green dot swaps the colors, thus we have an equal amount of red and white tiles, and we can extend this coloring to any $G(2n)$, in particular, $G(100)$.\n\nLet's consider the S and Z tetrominoes separately. Notice that no matter where an S tetromino is placed, it covers either 4-0, 0-4, or 2-2 of white-red. Thus when we place an S tetromino we do not change the difference between the amounts of red and white tiles modulo 4. Similarly any Z tetromino covers 3-1 or 1-3, thus adding 2 to the difference modulo 4. To end with 0 red and 0 white tiles (ie to cover the entire board) we must have the difference be 0 mod 4, and thus we must have an even number of Z tetrominoes.\n\nThis argument is completely symmetrical; there is no real difference between S and Z tetrominoes. If we can tile the board using an odd number of S tetrominoes, simply mirror the board turning all S into Z and vice versa and we have tiled the board using an odd number of Z tetrominoes, which we know is impossible. Thus we have an even number of both S and Z tetrominoes, and an even number of tetrominoes overall. This means that the area of the board must be divisible by 8. But in the case of $G(100)$, it's not.\n\nThe size of $G(2N)$ is $4(1 + \\dots + N) = 4\\frac{N(N+1)}{2} = 2N(N+1)$. If 8 divides this then $8 | 2N(N+1) \\iff 4|N(N+1) \\iff 4|N$ or $4|N+1$.\n\n\u2022 Cool argument! We used different colorings, yet proved that the tiling is impossible for the same values of $n$. I've checked $G(2\\cdot 3)$ is not tileable, I guess the next step is to check $G(2\\cdot 4)$. \u2013\u00a0Mike Earnest Apr 26 '15 at 0:06\n\u2022 @MIkeEarnest, Just case checked. $G(8)$ is impossible as well. \u2013\u00a0Ben Frankel Apr 26 '15 at 0:17", "date": "2019-11-22 23:40:29", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/12730/tiling-a-diamond-shaped-grid-with-tetrominoes", "openwebmath_score": 0.7203966975212097, "openwebmath_perplexity": 492.3523591118034, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434792016126, "lm_q2_score": 0.8991213860551287, "lm_q1q2_score": 0.8761351608178803}}
{"url": "https://math.stackexchange.com/questions/4080645/is-a-real-matrix-that-is-both-normal-and-diagonalizable-symmetric-if-so-is-the", "text": "# Is a real matrix that is both normal and diagonalizable symmetric? If so, is there a proof of this not using the spectral theorem?\n\nGiven a quadratic real matrix $$A$$ for which we know it is diagonalizable ($$D = P^{-1}AP$$ for a diagonal matrix $$D$$) and that it is normal ($$AA^T = A^TA$$), is it true that $$A$$ is symmetric? By the spectral theorem over $$\\mathbb{C}$$, $$A$$ is orthogonally diagonalizable (say via some unitary matrix $$Q$$). It then seems to me that from this we should get that the eigenspaces are orthogonal and hence there must also exist a real orthonormal basis of eigenvectors. Is this last correct? From this it would then follow by the real version of the spectral theorem that $$A$$ is symmetric.\n\nIf it is, is there a way to prove this statement directly, without appealing to the spectral theorem? This would give a nice justification/explanation for the difference in the two spectral theorems over $$\\mathbb{C}$$ and $$\\mathbb{R}$$ relating it directly to whether the matrix is diagonalizable in the first place.\n\n\u2022 I'm not sure I understand the question: since the conjugate of a real number is itself, a real matrix is normal if and only if it is symmetric. (And all normal matrices are automatically diagonalizable, including symmetric real matrices.) Mar 28, 2021 at 17:55\n\u2022 That\u2019s not true: A real orthogonal matrix is also normal but is in general not symmetric (consider e.g. rotation by 90 degrees). A normal real matrix in general need not diagonalizable over the reals, as the example of orthogonal matrces shows (it is diagonalizable over the complex numbers, of course). Mar 28, 2021 at 17:57\n\u2022 Ack, you're completely right. I was confusing normal with Hermitian. Mar 28, 2021 at 19:53\n\nHere is a proof avoiding the spectral Theorem.\n\nLet $$\\lambda _1, \\lambda _2, \\ldots , \\lambda _k$$ be the distinct eigenvalues of $$A$$ and let $$E_1, E_2, \\cdots , E_k$$ be the projections onto the corresponding eigenspaces.\n\nBy seeing things from the point of view of a (not necessarily orthonormal) basis of eigenvectors, it is very very easy to prove that (don't let the long expression scare you) $$E_i = \\frac{(A-\\lambda _1)\\ldots \\widehat{(A-\\lambda _i)}\\ldots (A-\\lambda _k)}{(\\lambda _i-\\lambda _1)\\ldots \\widehat{(\\lambda _i-\\lambda _i)}\\ldots (\\lambda _i-\\lambda _k)},$$ where the hat means omission.\n\nFrom this it is clear that each $$E_i$$ is also normal.\n\nLemma. Any normal, idempotent, real matrix is symmetric.\n\nProof. Let $$E$$ be such a matrix. We first claim that $$\\text{Ker}(E)=\\text{Ker}(E^TE)$$. To see this, observe that the inclusion $$\\text{Ker}(E)\\subseteq \\text{Ker}(E^TE)$$ is evident. On the other hand, if $$x\\in \\text{Ker}(E^TE)$$, then $$\\|Ex\\|^2 = \\langle Ex, Ex\\rangle = \\langle E^TEx, x\\rangle =0,$$ so $$x\\in \\text{Ker}(E)$$.\n\nWe then have that $$\\text{Ker}(E)=\\text{Ker}(E^TE) =\\text{Ker}(EE^T) =\\text{Ker}(E^T).$$\n\nRecalling that the range $$R(A^T)$$, of the transpose of a matrix $$A$$, coincides with $$\\text{Ker}(A)^\\perp$$, we then have that $$R(E^T) = \\text{Ker}(E)^\\perp = \\text{Ker}(E^T)^\\perp = R(E).$$ We then see that $$E$$ and $$E^T$$ are projections sharing range and kernel, so necessarily $$E=E^T$$. QED\n\nBack to the question, we then have that $$A=\\sum_{i=1}^k \\lambda _kP_k,$$ so we conclude that $$A$$ is symmetric.\n\nYou can also do this with simultaneous diagonalizability and two classical inequalities: Cauchy-Schwarz and rearrangement inequality.\n\n$$A$$ is real diagonalizable and so is $$A^T$$. Now $$A$$ and $$A^T$$ commute, hence they are simultaneously diagonalizable such that $$A=SD_1S^{-1}$$ and $$A^T = SD_2S^{-1}$$. Since the trace is invariant to conjugation, this implies\n\n(i.)\n$$\\text{trace}\\big(A^2\\big)=\\text{trace}\\big(D_1^2\\big)\\geq \\text{trace}\\big(D_2D_1\\big)=\\text{trace}\\big(A^TA\\big) =\\big\\Vert A\\Big \\Vert_F^2$$\nby rearrangement inequality\n\n(ii.)\n$$\\text{trace}\\big(A^2\\big)\\leq \\text{trace}\\big(A^TA\\big) =\\big\\Vert A\\Big \\Vert_F^2$$\nby Cauchy-Schwarz\n\nThus $$\\text{trace}\\big(A^2\\big)= \\big\\Vert A\\Big \\Vert_F^2$$ so Cauchy-Schwarz is met with equality$$\\implies A = \\eta\\cdot A^T$$, where $$\\eta \\in\\big\\{-1,1\\big\\}$$ because $$A$$ and its transpose have the same Frobenius norms. Thus $$A$$ is either symmetric or skew symmetric. (And supposing $$A\\neq \\mathbf 0$$ we can eliminate skew symmetry because $$\\mathbf x^T A\\mathbf x=0$$ for all $$\\mathbf x\\in \\mathbb R^n$$ by skew symmetry, so $$A$$ could not have any non-zero real eigenvalues, which we know isn't true.)\n\nOne can obtain that $$A$$ is symmetric by direct computation. The equality $$A^TA=AA^T$$ is $$\\tag1 PDP^{-1}P^{-T}DP^T=P^{-T}DP^TPDP^{-1}.$$ Multiplying by $$P^T$$ on the left and by $$P$$ on the right, $$\\tag2 P^TPD(P^TP)^{-1}DP^TP=DP^TPD.$$ Now by $$(P^TP)^{-1}$$ on the right, $$\\tag3 \\big[P^TPD(P^TP)^{-1}\\big]\\,D=D\\,\\big[P^TPD(P^TP)^{-1}\\big].$$ We may assume without loss of generality that $$A$$ (and hence $$D$$) is invertible, by adding a suitable scalar multiple of the identity. By writing $$D$$ as a block-diagonal matrix with distinct eigenvalues it is easy to see that the matrices that commute with $$D$$ are block diagonal. So, with $$\\tag4X=P^TPD(P^TP)^{-1}$$ we have $$XD=DX$$ and thus $$X$$ is block-diagonal. Knowing that $$X$$ is block-diagonal we rewrite $$(4)$$ as $$XP^TP=P^TPD.$$ The (block) diagonal entries of this equality are $$X_{kk}(P^TP)_{kk}=\\lambda_k\\,(P^TP)_{kk}.$$ Because $$P^TP$$ is positive definite, its block-diagonal entries are positive definite; in particular, invertible. We conclude that $$X_{kk}=\\lambda_{kk}\\,I$$, that is $$X=D.$$ We can unravel this as $$D=P^TPD(P^TP)^{-1}=P^TPDP^{-1}P^{-T}.$$ Multiplying by $$P^{-T}$$ on the left and by $$P^T$$ on the right we obtain $$A^T=P^{-T}DP^T=PDP^{-1}=A.$$", "date": "2022-05-21 02:23:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4080645/is-a-real-matrix-that-is-both-normal-and-diagonalizable-symmetric-if-so-is-the", "openwebmath_score": 0.940808892250061, "openwebmath_perplexity": 123.9890817696989, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615794, "lm_q2_score": 0.8991213711878918, "lm_q1q2_score": 0.8761351423255493}}
{"url": "https://math.stackexchange.com/questions/257434/if-a-b-1-then-prove-ab-ab-1", "text": "# If $(a,b)=1$ then prove $(a+b, ab)=1$.\n\nLet $$a$$ and $$b$$ be two integers such that $$\\left(a,b\\right) = 1$$. Prove that $$\\left(a+b, ab\\right) = 1$$.\n\n$$(a,b)=1$$ means $$a$$ and $$b$$ have no prime factors in common\n\n$$ab$$ is simply the product of factors of $$a$$ and factors of $$b$$.\n\nLet's say $$k\\mid a+b$$ where $$k$$ is some factor of $$a$$.\n\nThen $$ka=a+b$$ and $$ka-a=b$$ and $$a(k-l)=b$$.\n\nSo $$a(k-l)=b, \\ a\\mid a(k-1)$$ [$$a$$ divides the left hand side] therefore $$a\\mid b$$ [the right hand side].\n\nBut $$(a,b)=1$$ so $$a$$ cannot divide $$b$$.\n\nWe have a similar argument for $$b$$.\n\nSo $$a+b$$ is not divisible by any factors of $$ab$$.\n\nTherefore, $$(a+b, ab)=1$$.\n\nWould this be correct? Am I missing anything?\n\nSuppose that the gcd is not $1$. Then there is a prime $p$ that divides $ab$ and $a+b$.\n\nBut then $p$ divides one of $a$ or $b$, say $a$. Since $p$ divides $a+b$, it follows that $p$ divides $b$. This contradicts the fact that $a$ and $b$ are relatively prime.\n\n\u2022 I like this proof. Short and sweet. \u2013\u00a0ireallydonknow Dec 20 '13 at 18:59\n\u2022 This method can be motivated intuitively as a generalization of Euclid's method for generating (co)primes - see this answer here. \u2013\u00a0Bill Dubuque Dec 16 '19 at 18:55\n\nYour jump from \"$k\\mid a+b$ with $k\\mid a$\" to $ka=a+b$ seems to be wrong. Just because $k$ is a factor of $a$ doesn't mean at all that the number of times it divides $a+b$ is exactly $a$. That seems to kill the rest of your argument.\n\nInstead, here is an approach that doesn't mention prime factors at all. It starts from the well-known property that $(a,b)=1$ exactly if there are $p,q\\in\\mathbb Z$ such that $pa+qb=1$.\n\nSquare $pa+qb=1$ to get $$p^2a^2 + q^2b^2 + 2pqab = 1$$ If we can show that each of the terms on the left-hand side of this is an integer combination of $a+b$ and $ab$, then the right-hand side is too.\n\nBut $2pqab$ is clearly an integer combination of $a+b$ and $ab$ namely $0(a+b)+2pq\\cdot ab$.\n\nAnd $a^2 = a(a+b)-ab$, so $p^2a^2 = p^2a(a+b)-p^2\\cdot ab$.\n\nSimilarly $q^2b^2 = q^2b(a+b)-q^2\\cdot ab$.\n\nCollecting everything, $(p^2a+q^2b)(a+b)+(2pq-p^2-q^2)ab=1$, so $(a+b,ab)=1$.\n\n\u2022 Nice. (+1) I came up with another Bezout Identity approach. The coefficient of $ab$ is the same, so I assume the coefficient of $a+b$ is, too; it is just not as obvious. \u2013\u00a0robjohn Dec 20 '13 at 20:17\n\u2022 Worth emphasis - at the heart of this method is a classical (bi-)quadratic composition identity (a generalization of the classical Brahmagupta\u2013Fibonacci identity for composition of squares). See this answer here for that viewpoint. \u2013\u00a0Bill Dubuque Dec 16 '19 at 19:02\n\nHere is another proof using Bezout's Identity: \\begin{align} ma+nb&=1\\tag{1}\\\\ (n-m)b&=1-m(a+b)\\tag{2}\\\\ (m-n)a&=1-n(a+b)\\tag{3}\\\\ -(m-n)^2ab&=1-(m+n)(a+b)+mn(a+b)^2\\tag{4}\\\\ 1&=((m+n)-mn(a+b))\\color{#C00000}{(a+b)}-(m-n)^2\\color{#C00000}{ab}\\tag{5} \\end{align} Explanation:\n$(1)$: $(a,b)=1$ and Bezout's Identity\n$(2)$: subtract $m(a+b)$ from both sides of $(1)$\n$(3)$: subtract $n(a+b)$ from both sides of $(1)$\n$(4)$: multiply $(2)$ and $(3)$\n$(5)$: rearrange and collect terms of $a+b$ and $ab$ from $(4)$\n\nBezout's Identity and $(5)$ say that $(a+b,ab)=1$.\n\nHenning Makholm and Bill Dubuque also give Bezout Identity based proofs. The coefficient of $ab$ is the same in all of our answers, but the coefficient of $a+b$ in mine looks different from theirs. They are the same however: \\begin{align} (m+n)-mn(a+b) &=(m+n)\\overbrace{(ma+nb)}^{1}-mn(a+b)\\\\ &=m^2a+n^2b+mn(a+b)-mn(a+b)\\\\ &=m^2a+n^2b \\end{align}\n\n\u2022 Generally $\\,(a,c)(b,c) = (ab,c)$ when $\\,(a,b,c)=1\\,$ as proved in this answer here. The proof in Rob's answer is essentially a Bezout based proof of this in the special case $\\,c=a\\!+\\!b,\\,(a,b)=1.\\,$ In Rob's $(5)$ if we use his equivalent expression $\\,m^2a+n^2b\\,$ and replace $1$ on the LHS by $\\,(ma+nb)^2\\,$ it becomes the first identity in the linked answer, which is a classical composition of squares identity. \u2013\u00a0Bill Dubuque Dec 16 '19 at 21:09\n\n[This answer was merged from a question concerning $$\\rm\\color{#0a0}{Bezout}$$-based proofs]\n\nNote $$\\ (\\color{#c00}{a\\!+\\!b}) (ai^2\\!+\\!bj^2)= \\color{#0a0}{(ai\\!+\\!bj)^2}\\ \\!\\!+ \\color{#c00}{ab}(i\\!-\\!j)^2\\,$$ by calculation (or by composition - see below)\nthus $$\\ \\ n^{\\phantom{|^|}}\\!\\!\\!\\mid\\color{#c00}{a\\!+\\!b,ab}\\,\\Rightarrow\\, n\\mid\\color{#0a0}{(ai\\!+\\!bj)^2}\\!=1^2\\$$ by choosing $$\\,\\color{#0a0}{ai\\!+\\!bj}=1\\,$$ by $$\\,(a,b)=1\\,$$ & Bezout.\n\nOr: $$\\,\\ \\ (\\color{#c00}{a\\!+\\!b}) (ai\\!+\\!bj)\\,=\\, \\color{#0a0}{a^2i\\!+\\!b^2j}\\,+\\, \\color{#c00}{ab}(i\\!+\\!j)\\$$\nthus $$\\,n^{\\phantom{|^|}}\\!\\!\\!\\mid\\color{#c00}{a\\!+\\!b,ab}\\,\\Rightarrow\\, n\\mid\\color{#0a0}{a^2i\\!+\\!b^2j}\\!=1\\$$ by choosing $$\\,i,j\\,$$ to get Bezout for $$\\,(a^2,b^2)=1\\,$$\n\nRemark $$\\$$ The genesis of the proof is ideal-theoretic, as is explained in this post, which presents a handful of proofs of the more general identity $$\\, (a\\!+\\!b,\\,{\\rm lcm}(a,b)) = (a,b).\\,$$ The above is essentially a Bezout form of the the last proof there (which is more general than a proof using primes, since it works in any gcd domain - where primes needn't exist). In fact OP is case $$\\, c=a\\!+\\!b,\\,(a,b)\\!=\\! 1\\,$$ of\n\n$$(a,b,c)=1\\,\\Rightarrow\\, (a,c)(b,c) = (ab,c(a,b,c)) = (ab,c)\\qquad$$\n\nMore classically, the first identity may be viewed conceptually as the special case $$\\,x=1=y\\,$$ in the biquadratic composition of quadratic forms in the Lemma below (used by Euler) whose proof by norm multiplicativity is an obvious generalization of the classical proof when $$\\,a=1=b.$$\n\nLemma $$\\ \\ (ax^2+by^2) (ai^2+bj^2)= (aix\\!+\\!bjy)^2\\ \\!\\!+ ab(iy\\!-\\!jx)^2$$\n\n$$\\!\\begin{array}{rll}{\\bf Proof}\\qquad\\qquad\\ \\ (ax^2+by^2)&\\!\\!\\!\\!\\!\\!(ai^2+bj^2)&\\!\\! =\\ N(\\alpha)N(\\beta)\\\\[.2em] \\!\\!\\!\\!\\!(x\\sqrt a\\!+\\!y\\sqrt{-b})(x\\sqrt a\\!-\\!y\\sqrt{-b})&\\!\\!\\!\\!\\!\\!(i\\sqrt a\\!-\\!j\\sqrt{-b})\\ (i\\sqrt a\\!+\\!j\\sqrt{-b})&\\!\\!=\\ \\ \\ \\alpha\\,\\bar\\alpha\\,\\beta\\,\\bar\\beta\\\\[.2em] (x\\sqrt a\\!+\\!y\\sqrt{-b})\\ (i\\sqrt a\\!-\\!j\\sqrt{-b})&\\!\\!\\!\\!\\!\\!(x\\sqrt a\\!-\\!y\\sqrt{-b})(i\\sqrt a\\!+\\!j\\sqrt{-b})&\\!\\!=\\ \\ \\ \\alpha\\,\\beta\\,\\bar\\alpha\\,\\bar\\beta\\\\[.2em] (aix\\!+\\!bjy\\! +\\! (iy\\!-\\!jx)\\sqrt{-ab})&\\!\\!\\!\\!\\!\\!(aix\\!+\\!bjy\\! -\\! (iy\\!-\\!jx)\\sqrt{-ab} )&\\!\\!= \\ \\ \\ \\alpha\\beta\\:\\smash[t]{\\overline{\\alpha\\beta}}\\\\[.2em] (aix\\!+\\!bjy)^2\\ &\\!\\!\\!\\!\\!\\!\\!+ ab(iy\\!-\\!jx)^2 &\\!\\!= \\ \\ N(\\alpha\\beta)\\end{array}$$\n\n\u2022 Everything looks good. It takes a bit of work to verify that your $c$ and $d$ work, but they do. +1 \u2013\u00a0robjohn Dec 20 '13 at 19:22\n\u2022 I just noticed that your solution looks pretty close to Henning's with $p=i$ and $q=j$. \u2013\u00a0robjohn Dec 20 '13 at 20:14\n\u2022 @robjohn I hadn't read that. I agree, they do appear to use the same form of the Bezout identity. I derived it as in the linked post. In some sense the Bezout proofs will all be the same, modulo a shift by $\\,\\pm k\\, (a+b, -ab)$ to obtain another solution $\\,ab\\,x +(a+b)\\,y = 1.$ \u2013\u00a0Bill Dubuque Dec 20 '13 at 20:24\n\u2022 @robjohn I found some time to elaborate to explain the classical view in terms of composition of quadratic forms. \u2013\u00a0Bill Dubuque Dec 16 '19 at 6:13\n\nHINT: Suppose that $p$ is a prime that divides $ab$ and $a^2+b^2$. Then $p$ divides both $a^2+2ab+b^2=(a+b)^2$ and $a^2-2ab+b^2=(a-b)^2$. This in turn implies that $p$ divides both $a+b$ and $a-b$. (Why?) Use this to show that $p$ divides both $a$ and $b$.\n\n\u2022 If p is even, then both $ab$ and $a^2+b^2$ must be even as well. The second can only be true in a and b are both even or both odd. If they are both even, they are not coprime, and if they are both odd $ab$ is not even. \u2013\u00a0Miles Johnson Jan 10 '17 at 5:29\n\u2022 Therefore p is not even. Since $p$ divides $a+b$ and $a-b$, for some integers $k$ and $z$, we may say $(a+b)/p=k, (a-b)/p=z$. Adding the two, we have $2a/p=k+z$, implying, since $p$ is not even, that $p$ divides $a$. Subtracting the second from the first, we have $2b/p=k-z$, implying for the same reason that $p$ divides $b$. But $p$ cannot divide both numbers, so we have found a contradiction. \u2013\u00a0Miles Johnson Jan 10 '17 at 5:36\n\u2022 Just in case anyone was wondering how to finish the proof. \u2013\u00a0Miles Johnson Jan 10 '17 at 5:37\n\nHINT: $a^2 + b^2 +2ab = (a+b)^2.$\n\nYou say:\n\nLet\u2019s say $$k\\mid a+b$$ where $$k$$ is some factor of $$a$$.\n\nThen $$ka=a+b$$ ...\n\nThis step really doesn\u2019t make any sense as written. If $$k\\mid a+b$$, there is some integer $$m$$ such that $$km=a+b$$, but there\u2019s certainly no reason to think that $$m=a$$. What you want to do at this point is use the hypothesis that $$k\\mid a$$: there is some integer $$n$$ such that $$a=kn$$, and therefore $$km=kn+b$$. Now you can solve for $$b$$ and observe that $$k\\mid b$$. But then $$k$$ is a common divisor of $$a$$ and $$b$$, so $$k=\\pm1$$, and you\u2019ve shown that $$\\gcd(a,a+b)=1$$.\n\nA similar argument shows that $$\\gcd(b,a+b)=1$$, and then your last step is fine.\n\nThis can be solved intuitively by using a slight twist on Euclid's idea for generating new primes. Euclid employed $$\\,1 + p_1\\cdots p_n$$ is coprime to $$\\,c = p_1\\cdots p_n.\\,$$ Stieltjes noted the generalization that, furthermore, $$\\ \\color{#c00}{p_1\\cdots p_k} +\\, \\color{#0a0}{p_{k+1}\\cdots p_n}\\,$$ is coprime to $$\\,c\\,$$ too, which motivates the following\n\nKey Idea $$\\,$$ Coprimes to $$\\,c\\,$$ arise by partitioning into $$\\rm\\color{#c00}{two}\\ \\color{#0a0}{summands}$$ all prime factors of $$\\,c,\\,$$ i.e.\n\nTheorem $$\\ \\ \\color{#c00}a+\\color{#0a0}b\\$$ is coprime to $$\\ c\\:$$ if every prime factor of $$\\,c\\,$$ divides $$\\,a\\,$$ or $$\\,b,\\,$$ but not both.\n\nProof $$\\$$ If it fails then $$\\,a+b\\,$$ and $$\\,c\\,$$ have a common prime factor $$\\,p.\\,$$ By hypothesis $$\\,p\\mid a\\,$$ or $$\\,p\\mid b,\\,$$ w.l.o.g. say $$\\,p\\mid b.\\,$$ Then $$\\,p\\mid (a+b)-b = a,\\,$$ so $$\\,p\\,$$ divides both $$\\,a,b,\\,$$ contra hypothesis.\n\nCorollary $$\\ \\ (a,b)=1\\,\\Rightarrow\\, (a+b,ab) = 1$$\n\nRemark $$\\$$ Note how the solution becomes quite obvious after employing Stieltjes idea. More generally it yields a \"coprime\" version of Dirichlet's result on primes in arithmetic progression\n\n$$(a,b,c)=1\\,\\Rightarrow\\,(an+b,c) = 1\\ \\ \\ {\\rm for\\ some}\\ n\\qquad$$\n\nIf $d$ divides $a+b$ and $d$ divides $ab$, then $d$ divides $a(a+b)-ab = a^2$. Similarly, $d$ divides $b^2$. Thus $d$ divides $(a^2,b^2$). But $(a^2,b^2) = (a,b)^2 = 1$, so $d=1$.\n\nYet another approach: it's straightforward to show that $(i, jk) | (i,j)\\cdot(i, k)$. It's also a classic theorem of the GCD that $(m, n) = (m, m+n)$ (this principle underlies the Euclidean algorithm). Then we have $(a, a+b)=(a, b)=1$, and $(b, a+b) = (b, a) = 1$, so $(ab, a+b)|(a, a+b)\\cdot(b, a+b) = 1\\cdot 1$.\n\nWe know that the prime factors of $ab$ is either a factor of $a$ or $b$ but not the other due to $gcd(a,b) = 1$. However, $a+b$ is not divisible by a prime factor of $a$ nor $b$ because the division yields an integer plus a 'decimal number'. Hence, by the fundamental theorem of arithmetic, it is not divisible by any factors of $ab$.\n\nIntuitively, if a factor divides $a$ but not $b$ then it divides $ab$ and $a^2$ but not $b^2$, hence not $a^2+b^2$.\n\nMore formally, notice that if $a$ and $b$ are coprime then they are also coprime to $a+b$ and hence $ab$ is coprime to $(a+b)^2 = a^2 + 2ab+b^2$.\n\n\u2022 I quite like this one; +1. \u2013\u00a0Brian M. Scott Jan 10 '17 at 5:59\n\nSuppose $\\text{gcd}(a,b) = 1$, but $\\text{gcd}(ab,a^2+b^2) = d$, where $d > 1$. Let p be a prime factor of d. Then\n\n$$d|ab \\implies p|ab \\implies p|a \\text{ or } p|b$$\n\nWithout loss of generality, assume p|a. Then\n\n$$d|(a^2 + b^2) \\implies p|(a^2 + b^2) \\implies p|b^2 \\implies p|b$$\n\ncontrary to $\\text{gcd}(a,b)=1$.\n\nBy contradiction, let $d>1$ divide both $ab$ and $a^2+b^2$. Then $d$ has a prime divisor $p$, which divides both $ab$ and $a^2+b^2$.\n\nThen, $p|a(a+b) - ab = a^2$ Since $p$ is a prime which divides $a^2$ $\\implies$ that $p|a$.\n\nAnd $p|b(a+b) - ab = b^2$ $\\implies$ $p|b$.\n\nHence $p|a$ and also, $p|b$, so $p$ divides $gcd(a,b) = 1$. However, this forces $p = 1$, which is a contradiction, since 1 isn't a prime.\n\nThus the only positive common divisor of $ab$ and $a^2+b^2$ is $1$, and hence they are coprime.\n\nA proof in Gaussian integers $$\\Bbb Z[i].\\,$$ They are Euclidean $$\\Rightarrow$$ UFD so enjoy GCDs, Euclid's Lemma, etc.\n\n$$(a,b) = 1\\,\\Rightarrow\\, (a\\pm bi,b) = (a,b) = 1.\\,$$ Similarly $$\\,(a\\pm bi,a) = 1.\\,$$ So by Euclid $$\\,a,b$$ are coprime to $$\\,(a-bi)(a+bi) = a^2+b^2\\,$$ thus so too is their product $$\\,ab.$$\n\nSuppose that gcd(a,b)=1.\n\nLet d=gcd(a+b,ab).\n\nNow d|(a+b) and d|ab.\n\nSuppose that d|a.\n\nThen a=dqa.\n\nNow examining a+b=dq1 we have dqa+b=dq1 b=d(q1\u2212qa).\n\nSo then d|b and d=1.\n\nA similar argument shows d|b implies d|a.\n\nSo we know that d|a or d|b implies d=1.\n\nSuppose that d\u2224b and d\u2224a.\n\nWe have a+b=dq1 a=dq1\u2212b.\n\nHere we have that gcd(a,\u2212b)=1=gcd(a,d) (this is from a Lemma used in the proof of the Euclidean Algorithm).\n\nWe can also derive that gcd(b,\u2212a)=1=gcd(b,d).\n\nSo we know that a and d and b and d are both relatively prime.\n\nConsidering d|ab we know by Euclid's Lemma that since a and d are relatively prime implies that d|b.\n\nThis contradicts our assumption, so this case is not possible.\n\nTherefore, we may conclude that d=1. \u25a1\n\n\u2022 I found this question in Elementary Number Theory 7th edition, by Bartle. Section 2.4, Question 6. \u2013\u00a0fantasticasm89 Jan 22 at 19:24", "date": "2021-03-05 10:13:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/257434/if-a-b-1-then-prove-ab-ab-1", "openwebmath_score": 0.9396748542785645, "openwebmath_perplexity": 190.1505767583916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347853343059, "lm_q2_score": 0.8991213718636752, "lm_q1q2_score": 0.876135140981467}}
{"url": "https://math.stackexchange.com/questions/2919063/applying-rolles-theorem", "text": "# Applying Rolle's theorem\n\nThis is the first time that I'm doing a proof-problem on my own and I'm not really sure how to check if my answer is correct, so I was wondering if someone could tell me with certainty if my proof is correct.\n\nThe problem states:\n\nLet $f:\\mathbb{R} \\to \\mathbb{R}$ be a twice differentiable function (on its entire domain) such that $f(0) = f(1) = f(2)$. Prove that there exists some point $x_0 \\in (0, 2)$ such that $f''(x_0) = 0$.\n\nSo I developed my proof based on Rolle's theorem that states that if a function $f$ is continous on a closed interval $[a, b]$ and differentiable on an open interval $(a, b)$ and if $f(a) = f(b)$ then there exists some point $c \\in (a, b)$ such that $f'(c) = 0$.\n\nProof:\n\n1. $f$ is differentiable $\\forall x \\in \\mathbb{R}$, meaning $f$ must be differentiable on $(0, 1)$ and $(1, 2)$\n\n2. $f$ is differentiable $\\forall x\\in\\mathbb{R}$ meaning $f$ must be continuous $\\forall x \\in \\mathbb{R}$ meaning $f$ must be continuous on $[0, 1]$ and $[1, 2]$.\n\n3. $f(0) = f(1)$ and $f(1) = f(2)$.\n\nFrom 1., 2. and 3. we conclude that $f$ satisfies the conditions of Rolle's theorem for both of these intervals, meaning: $\\exists c_1 \\in (0, 1) : f'(c_1) = 0$ and $\\exists c_2 \\in (1, 2) : f'(c_2) = 0$\n\nLet $F(x) = f'(x)$. We know that the original function is twice differentiable $\\forall x \\in \\mathbb{R}$, therefore $F(x)$ is differentiable on $\\mathbb{R}$ meaning it must be differentiable on $(c_1, c_2)$ also meaning it must be continuous on $[c_1, c_2]$.\n\nNow, since $f'(c_1) = f'(c_2) = 0$, the function $F(x)$ satisfies the conditions of Rolle's theorem (on the interval $(c_1, c_2)$) meaning :\n\n$\\exists x_0 \\in (c_1, c_2) \\subset(0, 2): F'(x_0) = f''(x_0) = 0$.\n\n\u2022 Hello there. Please do not type all the proof in math mode. This makes the post hard to read. Just type those symbols in mathjax. . Thanks.\n\u2013\u00a0xbh\nSep 16 '18 at 15:31\n\u2022 Usually, new users have to be encouraged to use mathjax. In your case, I'd say you that you are using mathjax too much. Please, write text without mathjax. Sep 16 '18 at 15:32\n\u2022 After a cursory reading, i would say the proof seems fine. Although it can be neater.\n\u2013\u00a0xbh\nSep 16 '18 at 15:36\n\u2022 I've edited your answer, now it's more legible. I would say your proof is right Sep 16 '18 at 15:36\n\u2022 Your proof is solid. Sep 16 '18 at 15:47", "date": "2021-09-29 02:50:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2919063/applying-rolles-theorem", "openwebmath_score": 0.902411699295044, "openwebmath_perplexity": 131.28441820032756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936115537341, "lm_q2_score": 0.8902942283051332, "lm_q1q2_score": 0.8761328624782432}}
{"url": "https://math.stackexchange.com/questions/1831359/finite-borel-measure-on-a-compact-metric-space", "text": "# Finite Borel measure on a compact Metric Space\n\nSuppose that $\\mu$ is a finite Borel measure on a compact metric space $X$ and that $\\mu(\\{x\\}) = 0$ for all $x\\in X$. Show that for every $\\epsilon > 0$ there is a $\\delta > 0$ such that for all $x\\in X$, $\\mu(B(x,\\delta))<\\epsilon$ where $B(x,\\delta)$ denotes the ball of radius $\\delta$ centered at $x$.\n\nAttempted proof - Let $\\epsilon > 0$ and suppose there is $\\delta > 0$ such that there is an ball $B(x,\\delta)$ of radius $\\delta$ centered at $x$. Further, suppose that $\\mu$ is a finite Borel measure on a compact metric space $X$ and $\\mu(\\{x\\}) = 0$. Then from continuity from above $$\\lim_{\\delta\\rightarrow 0}B(x,\\delta) = 0$$ Thus there exists a $\\delta_x$ such that $B(x,\\delta_x)<\\epsilon$. Now, since $X$ is a compact metric space, we can choose points $x_1,\\ldots, x_n$ such that $X\\subset \\bigcup_{1}^{n}B(x_j,\\delta_{x_j/2})$.\n\nLet $\\delta = \\min\\{\\delta_{x_1/2},\\ldots,\\delta_{x_n/2}\\}$ and suppose $x\\in X$. Then for some $j$, $x\\in B(x_j,\\delta_{x_j/2})$ thus $B(x,\\delta)\\subset B(x_j,\\delta_{x_j})$. Applying monotonicity we have $$\\mu(B(x,\\delta))\\leq \\mu(B(x_j,\\delta_{x_j}) < \\epsilon$$\n\nNot sure if this is correct any suggestions is greatly appreciated.\n\n\u2022 Looks right to me. The way it's written at the start is a little curious - that first \"suppose there is a ball...\" doesn't quite make sense. I'd just start the proof \"Let $\\epsilon>0$. By continuity from above, for every $x\\in X$ we have $\\lim_{\\delta\\to0}\\mu(B(x,\\delta))=0$\" and then continue exactly as you did. \u2013\u00a0David C. Ullrich Jun 18 '16 at 22:41\n\u2022 @DavidC.Ullrich I see so it doesn't make sense to say suppose there is a ball centered at $x$ with radius $\\delta$? \u2013\u00a0Wolfy Jun 18 '16 at 22:42\n\u2022 We're talking about a metric space $X$. For any $x\\in X$ and $\\delta>0$ there is such a ball. \"Supposing\" something exists when that thing simply does exist doesn't make much sense, no. \u2013\u00a0David C. Ullrich Jun 18 '16 at 23:32\n\n1. As @DavidC.Ullrich pointed out, you don't need \"suppose there is $\\delta > 0$ such that there is an ball $B(x,\\delta)$ of radius $\\delta$ centered at $x$\". In any metric space $X$, for all $x \\in X$ and all $\\delta >0$ there is a ball $B(x,\\delta)$.\n\n2. You used (correctly) the fact that $\\mu$ is finite. I suggest you highlight this point.\n\nBelow is you proof with those two minor adjustments (and the correction of some typos)\n\nSuppose that $\\mu$ is a finite Borel measure on a compact metric space $X$ and that $\\mu(\\{x\\}) = 0$ for all $x\\in X$. Show that for every $\\epsilon > 0$ there is a $\\delta > 0$ such that for all $x\\in X$, $\\mu(B(x,\\delta))<\\epsilon$ where $B(x,\\delta)$ denotes the ball of radius $\\delta$ centered at $x$.\n\nProof: Let $\\epsilon > 0$ and suppose that $\\mu$ is a finite Borel measure on a compact metric space $X$ and $\\mu(\\{x\\}) = 0$. Then, since $\\mu$ is finite, from continuity from above $$\\lim_{\\delta\\rightarrow 0}\\mu(B(x,\\delta))=\\mu(\\{x\\}) = 0$$ Thus there exists a $\\delta_x$ such that $\\mu(B(x,\\delta_x))<\\epsilon$. Now, since $X$ is a compact metric space, we can choose points $x_1,\\ldots, x_n$ such that $X\\subset \\bigcup_{1}^{n}B(x_j,\\delta_{x_j}/2)$.\n\nLet $\\delta = \\min\\{\\delta_{x_1}/2,\\ldots,\\delta_{x_n}/2\\}$ and suppose $x\\in X$. Then for some $j$, $x\\in B(x_j,\\delta_{x_j}/2)$ thus $B(x,\\delta)\\subset B(x_j,\\delta_{x_j})$. Applying monotonicity we have $$\\mu(B(x,\\delta))\\leq \\mu(B(x_j,\\delta_{x_j}) < \\epsilon$$\n\nThis should be a comment but may be too long to fit. (Your proof is OK with the adjustments already noted by others.) Another way, although almost the same, is by contradiction : If not, then there exists $e>0$ such that $$\\forall n\\in N\\; \\exists x_n\\in X \\; (\\mu (B(x_n,1/n)\\geq e).$$ Choosing such an $x_n$ for each $n,$ there exists a subsequence $(x_{j_n})_n$ converging to a point $x\\in X.$ But for any $m\\in N,$ take $n$ large enough that $d(x,x_{j_n})<1/ 2 m.$ Then $B(x,1/m)\\supset B(x_{j_n},1/j_n),$ so $\\mu (B(x,1/m))\\geq e$ for all $m\\in N,$ contradicting continuity from above.\n\nCompactness of $X$ cannot be omitted.For example, let $X$ be the real interval $[0,1)$ with the usual metric; let $f(p)=\\tan (p\\pi /2)$ for $p\\in X;$ let $L$ be Lebesgue measure..... Now for $A\\subset X$ and for $A\\in$ dom $(l)$ let $\\mu (A)=L(f(A)).$", "date": "2021-06-15 00:14:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1831359/finite-borel-measure-on-a-compact-metric-space", "openwebmath_score": 0.9920883774757385, "openwebmath_perplexity": 44.807973097570276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936073551712, "lm_q2_score": 0.8902942297605357, "lm_q1q2_score": 0.8761328601725392}}
{"url": "https://byjus.com/question-answer/in-a-set-of-10-coins-2-coins-are-with-heads-on-both-the-sides-3/", "text": "Question\n\n# In a set of $$10$$ coins, $$2$$ coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides.\n\nA\n16\nB\n89\nC\n23\nD\n210\n\nSolution\n\n## The correct option is B $$\\dfrac89$$Let $$E_1, E_2, A$$ be the events defined as follows :$$E_1$$= selecting a coin having a head on both the sides.$$E_2$$= selecting a coin not having a head on both the sides.A = Getting all heads when a coin is tossed five times.To find:The probability that the selected coin had heads on both the sides: $$P\\left(\\dfrac {E_1}A\\right)$$There are 2 coins having heads on both the sides\u00a0$$P(E_1)=\\dfrac {^2C_1}{^{10}C_1}=\\dfrac 2{10}$$There are 8 coins not having heads on both the sides\u00a0$$P(E_2)=\\dfrac {^8C_1}{^{10}C_1}=\\dfrac 8{10}$$$$P\\left(\\dfrac A{E_1}\\right)=1^5=1\\\\\\implies P\\left(\\dfrac A{E_2}\\right)=\\left(\\dfrac 12\\right)^5$$By Bayee's Theorem, we have$$P\\left(\\dfrac {E_1}A\\right)=\\dfrac {P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}=\\dfrac {\\dfrac 2{10}\\times 1}{\\dfrac 2{10}\\times 1+\\dfrac 8{10}\\times \\left(\\dfrac 12\\right)^5}\\\\\\implies P\\left(\\dfrac {E_1}A\\right)=\\dfrac 2{2+\\dfrac 8{32}}=\\dfrac 89$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-17 02:00:18", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/in-a-set-of-10-coins-2-coins-are-with-heads-on-both-the-sides-3/", "openwebmath_score": 0.62120121717453, "openwebmath_perplexity": 679.5389510585394, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.985718065655333, "lm_q2_score": 0.888758798648752, "lm_q1q2_score": 0.8760656038382054}}
{"url": "https://www.sunway.co.ir/golden-gaytime-eincky/f7deb8-algebra-word-problems-with-solution", "text": "In this problem, it is the price of the blouse. Write an equation using the variable. For the first problem, demonstrate how to work out the solution if students are still having difficulty, where \"S\" equals shots made: S = 0.65 x 30; S = 19.5; So Sam made 19.5 shots. Section 2: Problems. You may select the numbers to be represented with digits or in words. Let 'x' be one of the parts of 25. Learn more Accept. Math Word Problems with Answers - Grade 8. The number formed by interchanging the first and third digits is more than the original number by 297.Find the number. You may choose the format of the answers. Find the number. Then, we need to solve the equation (s) to find the solution (s) to the word problems. The hypotenuse of a right angled triangle is 20 cm. Word problems on unit rate Word problems on comparing rates. Preview. These Equations Worksheets are a good resource for students in the 5th Grade through the 8th Grade. Bessy will have 3 times as much money as Bob. What is population in 5 years? Word problems on ages. Next, summarize what information you know and what you need to know. Find the length of each side of the equilateral triangle. We know that the sum of three angle in any triangle is 180\u00c2\u00b0. He has a total of 15 bills that are worth $47. A number is odd if it has the following format: 2n + 1 Each bag contains an unknown number of marbles. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Find the three angles of the triangle. How many girls are there in the class? Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! In case, the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. is d = 100(10 - p). Mixture problems. Find the solutions to the word problems students have tackled in the second math worksheet. There is a \"trick\" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time . SOLUTION \u2026 For instance: Content Continues Below. There are different sets of addition word problems, subtraction word problems, multiplicaiton word problems and division word problems, as well as worksheets with a mix of operations. Let \u2026, find length and width of a triangle The Phillips family wants to fence their backyard. One stop resource to a deep understanding of important concepts in physics, Area of irregular shapesMath problem solver. Chris sells 82 comic books to Tom. Question 2 : The denominator of a fraction exceeds the numerator by 5. \u2026, volume of a cone and a precious substance Not rated yetA cone shaped container with a height of 15 feet and radius of 5 feet is filled with a substance that is worth$25/ft 3 . Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems. Every word problem has an unknown number. Try Math Solver . How many of each type of bill does he have? This can be done by first multiplying the entire problem by the common denominator and then solving the resulting equation. Cost of one meter of the given rod\u00a0 is, Cost of one meter of the rod which is 2 meter shorter is. Algebra -> Customizable Word Problem Solvers -> Numbers-> SOLUTION: Log On Ad: Over 600 Algebra Word Problems at edhelper.com Word Problems: Numbers, consecutive odd/even, digits Word What are the two numbers? Your email is safe with us. Find the fraction. second is ten more than \u2026, Population growth\u00a0Not rated yetThe population of a town with an initial population of 60,000 grows at a rate of 2.5% per year. Teachers! This is three more than four times the number of girls. Created: Oct 16, 2017 | Updated: Feb 22, 2018 This is a series of questions that will guide pupils from thinking only in numbers to thinking algebraically. ... Algebra word problems. If the train leaves at 9:10 from one station and its speed is 90 km/h, what time does it get to the next station? You may speak with a member of our customer support team by calling 1-800-876-1799. Every word problem has an unknown number. When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution. Select it and click on the button to choose it. Linear inequalities word problems. A car traveled 281 miles in 4 hours 41 minutes. Solution. Find the equilibrium price. Given :\u00a0The total cost of\u00a080 units of the product is $22000. Explain to students that you can find the rate (or speed) that someone is traveling if you know the distance and time that she traveled. Using a few simple formulas and a bit of logic can help students quickly calculate answers to seemingly intractable problems. Solution Let x be the first number. Solution Let y be the second number x / y = 5 / 1 x + y = 18 Using x / y = 5 / 1, we get x = 5y after doing cross multiplication Replacing x = 5y into x + y = 18, we get 5y + y = 18 6y = 18 y = 3 For a certain commodity, the demand equation giving demand \"d\" in kg, for a price \"p\" in dollars per kg. Difference between a number and its positive square root is 12. Related articles: The 4 Steps to Solving Word Problems. If the \u2026, solve word problem Not rated yetIf a number is added to the numerator of 11/64 and twice the number is added to the denominator of 11/64, the resulting fraction is equivalent to 1/5. How many years \u2026, Price of 3 pens without discount Not rated yetA stationery store sells a dozen ballpoints pens for$3.84, which represents a 20% discount from the price charged when a dozen pens are bought individually. A park charges $10 for adults and$5 for kids. Do you have some pictures or graphics to add? If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. Multi-Step All Operations Word Problems These Word Problems worksheets will produce word problems involving all basic operations. The sum is 18. 15 years back X's age was 3/4 of Y's age. Word problems on constant speed. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of$3750 ? The supply equation giving the supply \"s\" in kg. The number of males is five more than three \u2026 On this page, you will find Math word and story problems worksheets with single- and multi-step solutions on a variety of math topics including addition, multiplication, subtraction, division and other math topics. If the variable is not the answer to the word problem, use the variable to calculate the answer. What is the length of the rod ? The digit in the tens place is twice the digit in the units place. These free 5th grade math word problem worksheets can be shared at home or in the classroom \u2026 Also solutions and explanations are included. Solving algebraic word problems requires us to combine our ability to create equations and solve them. Because length can not be a negative number, we can ignore \"- 10\". Access the answers to hundreds of Math Word Problems questions that are explained in a \u2026 H ERE ARE SOME EXAMPLES of problems that lead to simultaneous equations. Determine what you are asked to find. If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon? Word problems on direct variation and inverse variation Word problems on unit price. Given :\u00a0If 18 be subtracted from the number, the digits are reversed. To solve an algebraic word problem: Define a variable. Upstream/Downstream problem. The following collection of free 5th grade maths word problems worksheets cover topics including mixed operations, estimation and rounding, fractions, and decimals. 10/03/18. If you can solve these problems with no help, you must be a genius! Free for students, parents and educators. This page has a great collection of word problems that provide a gentle introduction to word problems for all four basic math operations. You can wrap a word in square brackets to make it appear bold. Find its area. We provide math word problems for addition, subtraction, multiplication, division, time, money, fractions and measurement (volume, mass and length). Solution : Let 'x' and 'y' be the present ages of Martin and Luther respectively. The\u00a0equilibrium price\u00a0is the\u00a0market price\u00a0where the quantity of goods demanded is equal to the quantity of goods supplied. Khan Academy is a 501(c)(3) nonprofit organization. Find the number. You may speak with a member of our customer support team by calling 1-800-876-1799. Click here to upload more images (optional). If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Covid-19 has led the world to go through a phenomenal transition . Select it and click on the button to choose it.Then click on the link if you want to upload up to 3 more images. So, the number of adults tickets sold is 202 and kids tickets is 346. What is the sand content of the mixture? These word problems worksheets will produce ten problems per worksheet. This website uses cookies to ensure you get the best experience. Get help on the web or with our math app. (You can preview and edit on the next page). Word problems are one of the first ways we see applied math, and also one of the most anxiety producing math challenges many grade school kids face. If thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Then, the sides of the right angle triangle are, In the above three sides, the side represented by (x - 12) is hypotenuse (Because that is the longest side). If 18 be subtracted from the number, the digits are reversed. Let 'x' be the length of the given rod. Math word problem worksheets for grade 4. Math Forum/Help; Problem Solver; Practice; Algebra; Geometry; Tests; College Math; History; Games; MAIN MENU; 1 Grade . Microsoft | Math Solver. First, we need to translate the word problem into equation (s) with variables. The problem states that \"This\" -- that is, $42 -- was$14 less than two times x. So, area of the rectangle is\u00a0384 square cm. If the rectangle's diagonal measurement is 32 inches, find the height. If they had laid 3500 eggs this year, how many eggs did they lay last year ? 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Education; Math; Pre-Algebra; The 4 Steps to Solving Word Problems; The 4 Steps to Solving Word Problems. 18 is taken away from 8 times of a number is 30. You can \u2026 He prepared this workbook (with full solutions to every problem) to share his strategies for solving algebra word problems. Let 'x' and \"x + 4\" be the lengths of other two sides. The Algebra Class E-course provides a lot of practice with solving word problems for every unit! Given :\u00a015 years back X's age was 3/4\u00a0 of Y's age. Type a math problem. Given :\u00a0Thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find below a wide variety of hard word problems in algebra. If 100 of the marbles in the first bag are placed in the second bag, the number in \u2026, Survey and intersection of three sets\u00a0Not rated yetIn a survey of 400 students of a college it was found that 240 study mathematics, 180 study Statistics and 140 study Accounts, 80 study mathematics and \u2026, A combination problem \u00a0Not rated yetThere are 13 qualified applicants for 4 trainee positions in a fast food management program. 4.2 5 customer reviews. And the third angle is three times of the first angle. In a triangle, the second angle is 5\u00c2\u00b0 more than the first angle. check your answers against the answer explanation section at the end of each chap-ter. Find the three angles of the triangle. Always let x represent the unknown number. Question 1: There are 47 boys in the class. Find the length of the sides. Click the button and find the first one on your computer. For problems that require more than one step, a thorough step-by-step explanation is provided. Given :\u00a0If the rod was 2 meter shorter and each meter costs $1 more. So, the two parts of the 25 are 10 and 15. He prepared this workbook (with full solutions to every problem) to share his strategies for solving algebra word problems. Get step-by-step solutions to your math problems. Distance, Rate, and Time Word Problems These Algebra 1 Equations Worksheets will produce distance, rate, and time word problems with ten problems per worksheet. Given : The total cost of 125 units of the product is$28750. Find out below how you can print these problems. \\displaystyle \\frac {4} {5} of a number is less than 2 less than the same number. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Now that you can do these difficult algebra problems, you can trick your friends by doing some fancy word problems; these are a lot of fun. Question 2 : The denominator of a fraction exceeds the numerator by 5. ... College Algebra Word Problem College Algebra Algebra Word Problem. Section 1: Examples. Area and perimeter word problems. So, the three angles of the triangle are 35\u00c2\u00b0, 40\u00c2\u00b0 and 135\u00c2\u00b0. Divide 25 in two parts so that sum of their reciprocals is 1/6. The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angle triangle is formed. : Step 2:: Solve the equation created in the first step. Identifying Divisibility by Looking at the Final Digits. Author: Created by dh2119. In a group of 120 people, 90 have an age of more 30 years, and the others have an age of less than 20 years. Example 1. Area = length \u00d7 width \u2026, Math problem about age\u00a0Arvind is eight year older than his sister. Ticket problem. Get help on the web or with our math app. Search. Below are more complicated algebra word problems Example #6: The ratio of two numbers is 5 to 1. Click below to see contributions from other visitors to this page... Bessy and Bob algebra word problem\u00a0Bessy has 6 times as much money as Bob, but when each earn $6. Algebra Word Problems. Ratio and proportion word problems. Math Word Problems with Solutions and Explanations - Grade 8. Always let x represent the unknown number . Phone support is available Monday-Friday, 9:00AM-10:00PM ET. 10 \u00e2\u008b (2x) + 1 \u00e2\u008b x - 18 = 10 \u00e2\u008b x + 1 \u00e2\u008b (2x). Over 600 Algebra Word Problems, Algebra Number Puzzles, Exponents, Polynomials, Radical Numbers, Factoring, Complex Numbers. Just type!...Your algebra word problem will appear on a Web page exactly the way you enter it here. Algebraic word problems | Worked example Our mission is to provide a free, world-class education to anyone, anywhere. There is a \"trick\" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time. Note that Using Systems to Solve Algebra Word Problems can be found here in the Systems of Linear Equations and Word Problems section. This is three more than four times the number of girls. Therefore, (100 \u2013 1 ) 2 = 100 2 + 1 2 \u2013 2 x 100 x 1 [By identity: (a -b) 2 = a 2 + b 2 \u2013 2ab = 10000 + 1 \u2013 200 = 9801. These Word Problems worksheets will produce word problems involving all basic operations. Many students find solving algebra word problems difficult. Counting Zeros and Writing Exponents. What was the average speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. \u2026 So, the three angles of a triangle are 60\u00c2\u00b0, 72\u00c2\u00b0 and 48\u00c2\u00b0. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given :\u00a0Difference between x and \u00e2\u0088\u009ax \u00a0= \u00a012. x \u00a0= \u00a09 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a \u2026, Applications: Number problems and consecutive integers\u00a0Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you\u2019re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let 'x0y' be the three digit number. How much was the blouse? Long distance cost .11 a minute using this card. Here is the equation: 2x \u2212 14 = 42. Expert Answers \u2022 1 Algebra Word Problem College Algebra Word Problem. 17 SEEKING SOLUTIONS JC Burns 18 101 PROBLEMS IN ALGEBRA T Andreescu Et Z Feng. Solution A translation of 2 units up will increase the y coordinate by 2 units and a translation by 5 units to the left will decrease the x coordinate by 5. Solution Then, assign variables to the unknown quantities. If 3 be added to both, the fraction becomes 3/4. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright \u00a9 2008-2019. In many problems, what you are asked to find is presented in the last sentence. If the rectangle's diagonal measurement is 32 inches, find the height. Given : If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1. I quit my day job, in order to work on algebra.com full time. of adult tickets and \"y' be the no. Percent of a number word problems. Solution: Let x represent Spenser's age Therefore, Taylor's age is 5x x + 5x = 18 6x = \u2026 Then: 0.10(10 \u2013 y) + 0.30y = 1.5 1 \u2013 0.10y + 0.30y = 1.5 Find the total value of the substance \u2026, Mixture word problem with sand and soil Not rated yet2 m 3 of soil containing 35% sand was mixed into 6 m 3 of soil containing 15% sand. Most tricky and tough algebra word problems are covered here. \u2026, Top-notch introduction to physics. Inequality Word Problem/Math. Solution: We can write, 99 = 100 -1. QuickMath allows students to get instant solutions to all kinds of math problems, from algebra and equation solving right through to calculus and matrices. \u201d \u2026, A Little Tricky Algebra word problem Not rated yetNavin spent 25% of it and gave 2/5 of the remainder to his brother. 2x = 42 + 14 = 56. x = 56 2 = 28. How much does each have before and after \u2026, Fencing a rectangular garden word problem A farmer has 260 meters of fencing and wants to enclose a rectangular area of 4200 square meters. Also solutions and explanations are included. Great! What was the average speed of the car in miles per hour? how old was they are now? Every problem in 501 Math Word Problems has a complete answer explanation. Given : Cost of each adult ticket is$10 and kid ticket is $5 and tickets were sold for a total of$3750. Let 'x' be the present age of X and 'y' be the present age of Y. Grade 8 math word problems with answers are presented. Word problems are the most difficult type of problem to solve in math. PREFACE This book contains one hundred highly rated problems used in the train-ing and testing of the USA International Mathematical Olympiad (IMO) team. Word problems in algebra. Do you have some pictures or graphics to add? The questions are increasingly challenging, finishing with some that require a lot of thought and can be investigated further. The method of solution for \"work\" problems is not obvious, so don't feel bad if you're totally lost at the moment. Worksheets > Math > Grade 4 > Word problems. Online math solver with free step by step solutions to algebra, calculus, and other math problems. Grade 6 math word problem worksheets with answers - Estimation word problems for 6th Grade are made of the following Math skills for kids: estimate to solve word problems, multi steps word problems, identifying word problems with extra or missing information, distance direction to starting point word problems, using logical reasoning to find the order, guest and check word problems. The author, Chris McMullen, Ph.D., has over twenty years of experience teaching word problems and math skills to physics students. Let 'x' be the number of eggs laid last year. Load more. 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{"url": "https://math.stackexchange.com/questions/301435/fractions-in-binary?noredirect=1", "text": "# Fractions in binary?\n\nHow would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?\n\n\u2022 The same way. $\\frac{n}{m}$. So a quarter would be $\\frac{1}{100}$. \u2013\u00a0copper.hat Feb 12 '13 at 18:47\n\u2022 You might want to have a look at: cs.furman.edu/digitaldomain/more/ch6/dec_frac_to_bin.htm. You can then use WA to test examples. Regards \u2013\u00a0Amzoti Feb 12 '13 at 18:49\n\u2022 I think this is a duplicate. Will have a look. \u2013\u00a0Tara B Feb 12 '13 at 18:52\n\u2022 If it is it made by another person \u2013\u00a0Fernando Martinez Feb 12 '13 at 18:54\n\u2022 Si me gustan todos esto thumbs up.... \u2013\u00a0Fernando Martinez Feb 15 '13 at 18:41\n\nNote: $$\\dfrac 14_{\\,\\text{ ten}} = .25 = \\color{blue}{\\bf 2} \\times 10^{-1} + \\color{blue}{\\bf 5} \\times 10^{-2}$$ $$\\frac14= \\dfrac{1}{2^2}_{\\,\\text{ten}} = 2^{-2} = \\color{blue}{\\bf 0} \\times 2^{-1} + \\color{blue}{\\bf 1}\\cdot 2^{-2} = .01_{\\text{ two}}$$\n\n\u2022 Yes I like this answer the most. \u2013\u00a0Fernando Martinez Feb 12 '13 at 18:58\n\u2022 So I I had $\\frac{1}{8}$ would I do $(2^{-3}+2^{-2}+2^{-1}+2^0)$ \u2013\u00a0Fernando Martinez Feb 12 '13 at 19:00\n\u2022 Yup, but start with $\\frac 18 = 2^{-3} = 0 \\times 2^{-1} + 0\\times 2^{-2} + 1\\times 2^{-3} = 0.001_{\\text{two}}$ \u2013\u00a0amWhy Feb 12 '13 at 19:02\n\u2022 yes that makes sense. \u2013\u00a0Fernando Martinez Feb 12 '13 at 19:04\n\u2022 $2^0 \\iff \\;$ the \"one's place\". $2^{-1}$ corresponds to the first digit to the right of the decimal point, then the place to the right of that corresponds to $2^{-2}$...and so on. \u2013\u00a0amWhy Feb 12 '13 at 19:05\n\nAs you mentioned, $$6 = {\\color{red}1}\\cdot 2^2+ {\\color{red}1}\\cdot 2^1+{\\color{red}0}\\cdot 2^0 = {\\color{red}{110}}_B.$$ Analogously $$\\frac{1}{4} = \\frac{1}{2^2} = {\\color{red}0}\\cdot2^0 + {\\color{red}0}\\cdot 2^{-1} + {\\color{red}1}\\cdot 2^{-2} = {\\color{red}{0.01}}_B.$$\n\nEdit:\n\nThese pictures might give you some more intuition ;-) Here $\\frac{5}{16} = 0.0101_B$, as the denominator is of form $2^n$, the representation is finite (process ends when you hit zero); $\\frac{1}{6} = 0.0010\\overline{10}_B$ as the denominator is not of form $2^n$, but the number is rational, so representation is infinite and periodic.\n\nI hope this helps ;-)\n\nThree basic ways, all seen in binary number systems:\n\nFixed-Point: One integer holds the \"integer part\"; another holds the \"fractional part\". This is simple to store and display, and has very high magnitude and precision with virtually no error, but doing real math with the numbers involved can get hairy. Decimal numbers aren't often seen in this form, but it is a possibility.\n\nMaintained Floating Point: a large integer holds the entire value, and a second smaller number maintains the relative place of the decimal point from the right (or left) side of the number. Much easier to manipulate for mathematical operations, same maintenance of precision, zero error, and used in many implementations of \"BigDecimal\" object types where the \"built-in\" floating point mechanisms aren't available. Can be more difficult to represent in base-10 form on-screen. If implemented with normal integer types, this method can be more limited in magnitude than the previous one; instead, many implementations use a byte array to store the number, allowing the numbers to be as big as system memory allows.\n\nImplicit Floating Point: The number is expressed in what amounts to \"binary scientific notation\". A \"mantissa\" is stored as an integer, with the decimal point implied to be on the far right. Then the exponent of a power of two is also stored. The actual value of this number is the mantissa, multiplied by two to the power of the exponent. This approach allows for the storage and calculation of truly massive numbers, and modern CPUs are designed with a Floating-Point Unit or FPU (sometimes called a \"math co-processor\", in the early days of its integration into 486-class CPUs) that accelerates calculations of numbers in this form. However, there are two problems; first, there's a tradeoff between extreme precision and extreme magnitude; the mantissa, and thus the number of digits that can be stored precisely, is fixed, so as magnitude increases, the number of possible decimal places decreases (in the extremes of magnitude you often can't get more granular than the millions place). Second, there's an amount of \"rounding error\" inherent in using floating-point numbers, with the inherent conversion to binary and back; this can cause errors in calculations requiring exact precision (such as when dealing with money), and so unless the extreme magnitude of a floating-point type is required, it's generally recommended to use a method of representation that does not introduce error.\n\n\u2022 This is a thorough answer, but I don't think the OP was asking about computer implementations. \u2013\u00a0LarsH Feb 12 '13 at 21:36\n\u2022 Although, what the OP is asking about is a typographic representation of numbers, which is very close to computer representation. Binary and decimal are just ways we can write down numbers with a pencil, and computer representations are how we write numbers into a memory with ones and zeros. \u2013\u00a0Kaz Feb 13 '13 at 3:25\n\n$1/4=0\\cdot(1/2)^0+0\\cdot(1/2)^1+1\\cdot(1/2)^2=0.01$ in base $2$, you just go in reverse with powers $(1/2)^n, n=0,1,2,...$\n\n0.01 in binary is 0.25 in decimal\n\nJust do long division!\n\n11 | 10011 | 110.0101...\n1100\n----\n111\n110\n---\n100\n11\n---\n100\n11\n---\n1 (repeats)\n\n\n\nSo 10011 / 11 = 110.0101... (aka 19 / 3 = 6.33...)\n\nBinary long division is a bit longer than decimal long division since you need more digits to write each number, but finding the largest multiple of your divisor that will fit is pretty trivial when it's either 0 or 1 times.\n\nNot sure how helpful this is but whenever I work with binary I always use a table.\n\nSo if you want 0.25 or 15.75 (base 10) in binary:\n\n\\begin{array}{r|rrrr|rrrr} & 8 & 4 & 2 & 1 & \\frac{1}{2} & \\frac{1}{4} & \\frac{1}{8} & \\frac{1}{16} \\\\ & 8 & 4 & 2 & 1 & 0.5 & 0.25 & 0.125 & 0.0625 \\\\ \\hline 0.25 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\ 15.75 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\end{array}\n\nGives 0.01 and 1111.11 respectively.\n\nAnd another way:\n\n15.375 to binary for example:\n\n\\begin{align} 15 - 8 & = 7, & \\quad 7 \\ge 0 & \\to 1 \\\\ 7 - 4 & = 3, & 3 \\ge 0 & \\to 1 \\\\ 3 - 2 & = 1, & 1 \\ge 0 & \\to 1 \\\\ 1 - 1 & = 0, & 0 \\ge 0 & \\to 1 \\\\ \\\\ 0.375 - 0.5 & = -0.125, & -0.125 \\lt 0 & \\to 0 \\\\ 0.375 - 0.25 & = 0.125, & 0.125 \\ge 0 & \\to 1 \\\\ 0.125 - 0.125 & = 0, & 0 \\ge 0 & \\to 1 \\end{align}\n\nGives 1111.011.\n\nuse the euclidean algorithm, like for the integers\n\n\u2022 The Euclidean Algorithm is usually used to find greatest common factors and continued fractions, not to divide or convert bases. \u2013\u00a0robjohn Feb 12 '13 at 19:26\n\u2022 @robjohn: You use the euclidean algrotihm to express fractions in lowest terms. The OP asked about base 2 fractions, not other representations, so it's clear this answer is a valid response. It's also clear the OP intended to ask about other representations and just didn't use the right terms, so being pedantic is missing the point, but talking about the euclidean algorithm isn't as unrelated as you imply. \u2013\u00a0ex0du5 Feb 12 '13 at 20:46\n\u2022 @ex0du5: Since the OP mentioned converting $1/4$ to $0.25$, it seems that they are interested in converting binary fractions to the format using a binary point. Since Nicol\u00f2 said \"like for the integers\", it really doesn't seem as if he was trying to reduce fractions to lowest terms. \u2013\u00a0robjohn Feb 12 '13 at 21:17\n\u2022 @ex0du5: I was not being pedantic; I was concerned that Nicol\u00f2 was using the wrong term for the repeated-division method for converting bases (which is generally used for integers) as shown in the answers to these questions: Converting decimal (base 10) numbers to binary by repeatedly dividing by 2, Convert numbers from one base to another using repeated divisions \u2013\u00a0robjohn Feb 12 '13 at 21:19\n\u2022 @robjohn: Yes, that's exactly what I said. I said that the answer Nicolo wrote was addressing what was asked, but \"It's also clear the OP intended to ask about other representations and just didn't use the right terms\". I was the one who had a pedantic point that Nicolo's answer was relevant as stated. I didn't call you pedantic. I just pointed out your comment on the euclidean algorithm didn't seem to understand why it was relevant here. \u2013\u00a0ex0du5 Feb 14 '13 at 18:59", "date": "2020-08-14 08:41:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/301435/fractions-in-binary?noredirect=1", "openwebmath_score": 0.9602383971214294, "openwebmath_perplexity": 463.1040251485464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.972830763738076, "lm_q2_score": 0.9005297861178929, "lm_q1q2_score": 0.8760630795979559}}
{"url": "https://math.stackexchange.com/questions/3288347/solving-5th-degree-polynomial", "text": "# Solving 5th degree polynomial\n\nSolve the equation, $$6x^5 + 5x^4 \u2212 51x^3 + 51x^2 \u2212 5x \u2212 6 = 0$$(hints: the pattern of the coefficients)\n\nHow I attempted this problem is to use the rational root theorem to obtain the factors. The possible roots are $$\u00b1(1, 1/2, 1/3, 1/6, 2, 2/3, 3, 3/2, 6)$$ By substituting the values inside the equation, only the following roots satisfy the equation:\n\n$$x=1,\\:x=\\frac{3}{2},\\:x=\\frac{2}{3}$$\n\nHenceforth, I came up with the following based on the above roots: $$\\left(x-1\\right)\\left(2x-3\\right)\\left(3x-2\\right)$$ Clearly expanding them would give me a 3rd degree polynomial as follows: $$6x^3 - 19x^2 + 19x - 6$$\n\nUsing polynomial division where I divided the original 5th degree equation with the above equation, I obtained the following equation: $$x^2+4x+1$$ Now, solving the above equation using quadratic formula, I am able to get the roots. Hence, I have obtained all the roots of the solution. However, looking at the hint that asks me to observe the pattern of the coefficients, I think that the method used to arrive at my solution might have been long-winded (it was indeed tedious plugging in the rational .roots one-by-one into the equation to see which returns value of 0). Is there something that I'm missing in the way I approached the question?\n\nIf you ask me, you did the good thing ignoring the hint and solving it your way. However, there is an \"algorithm\" for solving these type of problems that the hint is probably referring to. Notice that sum of coefficients is $$0$$, so $$x = 1$$ is root. Dividing by $$x - 1$$ leaves $$6x^4 + 11x^3 - 40x^2 +11x +6 = 0$$ and noticing that $$0$$ isn't a root and dividing by $$x^2$$ you get $$6(x^2 + \\frac{1}{x^2}) + 11(x + \\frac{1}{x}) - 40 = 0$$ Now substitute $$y = x + \\frac{1}{x}$$, and notice $$x^2 + \\frac{1}{x^2} = y^2 - 2$$, you are left with two quadratic equations to solve.\n\n\u2022 Interesting! Do you happen to know if there is a name for this \"algorithm\"?\n\u2013\u00a0Alex\nJul 11, 2019 at 4:14\n\u2022 I don't know. Seems that Wikipedia is referring to those as palindromic and antipalindromic polynomials if that helps. Also, the answer of Simply Beautiful Art provides some details on why things work in general. Jul 11, 2019 at 21:15\n\nThe \"pattern\" is that w hen you reverse the order of the coefficients you get the negative of the original polynomial:\n\n$$6x^5+5x^4-51x^3+51x^2-5x-6\\to-6x^5-5x^4+51x^3-51x^2+5x+6$$\n\nWhen the polynomial has this pattern you can render it this way:\n\n$$6x^5+5x^4-51x^3+51x^2-5x-6=6\\color{blue}{(x^5-1)}+5\\color{blue}{(x^4-x)}-51\\color{blue}{(x^3-x^2)}$$\n\nwhere the blue factors are all multiples of $$x-1$$ forcing $$x=1$$ to be a root. When you divide out this factor (using the above rearrangement to help things along) you get\n\n$$6x^5+5x^4-51x^3+51x^2-5x-6=(x-1)\\color{blue}{(6x^4+11x^3-40x^2+11x+6)}$$\n\nwhere the blue quartic factor now reads exactly the same when the coefficients are reversed. This arrangement, in an even degree polynomial, forces a \"symmetric\" factorization into quadratics:\n\n$$6x^4+11x^3-40x^2+11x+6=6(x^2+ax+1)(x^2+bx+1)$$\n\nMultiplying the right side out and matching terms with either odd power of $$x$$ gives $$a+b=11/6$$, and matching the terms with $$x^2$$ gives $$ab=-26/3$$. From the known sum and product we infer that $$a$$ and $$b$$ solve\n\n$$w^2-(11/6)w-(26/3)=0$$\n\n$$6w^2-11w-52=0$$\n\nThereby the roots for $$a$$ and $$b$$ are $$4$$ and $$-13/6$$, from which we now have\n\n$$6x^5+5x^4-51x^3+51x^2-5x-6=6(x-1)(x^2+4x+1)(x^2-(13/6)x+1)=(x-1)(x^2+4x+1)(6x^2-13x+6)$$\n\nand the remaining quadratic factors are solved by conventional methods.\n\nIt looks more complicated, but note that no trial and error is needed.\n\nAs a more general thing, whenever the coefficients are \"symmetric\", then it will have \"symmetric\" factors. See that you have $$(6x^3-19x^2+19x-6)(x^2+4x+1)$$. If there were no rational factors, writing it as a multiple of these \"symmetric polynomials\" may allow further factoring.\n\nIt is also notable that for every root $$x$$, $$1/x$$ will also be a root, which halves the amount of roots to check if you plan on using the rational roots theorem.\n\nLastly, in the latter half of prosinac's answer, the degree involved is reduced by a factor of 2 by the substitution $$y=x+\\frac1x$$. In general, if the degree is even, the substitution $$y=x\\pm\\frac1x$$ will allow you to halve the power of the problem you are working on, choosing the sign based on how the signs work out. If the degree is odd, then either $$1$$ or $$-1$$ will be a factor, allowing you to reduce this to an even degree \"symmetric polynomial\" and then halve the power. This means any \"symmetric polynomial\" of degree less than 10 can be factored algebraically, as you can always reduce the degree below 5.\n\nEdit: thanks to prosinac for pointing this out:\n\nApparently these are called self-reciprocal or palindromic polynomials. Some more properties are provided on the Wikipedia link.\n\n\u2022 $\\left\\{\\left\\{x\\to \\frac{2}{3}\\right\\},\\{x\\to 1\\},\\left\\{x\\to \\frac{3}{2}\\right\\},\\left\\{x\\to -2-\\sqrt{3}\\right\\},\\left\\{x\\to \\sqrt{3}-2\\right\\}\\right\\}$ Jul 10, 2019 at 1:34\n\u2022 I'm really not sure what you're getting at. Jul 10, 2019 at 1:37\n\u2022 @SimplyBeautifulArt Why is it that \"for every root \ud835\udc65, 1/\ud835\udc65 will also be a root\"? Does this rule apply to all polynomials when using the rational root theorem?\n\u2013\u00a0Alex\nJul 11, 2019 at 4:44\n\u2022 @Alex what happens when you divide your equation by $-x^5$ and rewrite everything in terms of $1/x$? Jul 12, 2019 at 1:56", "date": "2022-05-19 10:19:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3288347/solving-5th-degree-polynomial", "openwebmath_score": 0.8741932511329651, "openwebmath_perplexity": 238.38597559214904, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357542883923, "lm_q2_score": 0.8947894696095783, "lm_q1q2_score": 0.876030883308524}}
{"url": "https://math.stackexchange.com/questions/483374/slope-function-of-a-curve/483379", "text": "# Slope function of a curve\n\nI have a question I am having trouble answering:\n\nThe slope function of a curve is: $$\\frac{dy}{dx}=ux+k$$\n\nu and k are constants.\n\nThe curve passes through $(0,-1)$ and $(2,-5)$\n\nAt $(2,-5)$ the slope equals $1$.\n\nHow do I get the equation?\n\nI have tried integrating and pluging the two above points in and trying to solve for u and k but I end up getting the integration constant $c=-1$ and $k=2-u$ I then plug this into the integrated slope function: $$\\frac{ux^2}{2} + kx -1$$\n\nbut I don't get the correct answer.\n\nCould someone point out where I am going wrong wrong and how to correct it.\n\n\u2022 you meant you first \"equation\" to read $\\frac{dy}{dx} = ux + k$, right? \u2013\u00a0Robert Lewis Sep 3 '13 at 22:00\n\u2022 Yes, sorry I did \u2013\u00a0user Sep 3 '13 at 22:05\n\u2022 It looks like you solved for $c$ correctly, but I imagine you made an algebra mistake somewhere on the way to solving for $k=-5$ and $u=3$. \u2013\u00a0parsiad Sep 3 '13 at 22:05\n\nWith\n\n$y' = ux + k$\n\nwe have in general\n\n$y = \\frac{1}{2}ux^2 + kx + c$,\n\nfor some constant $c$. Plugging in the point $(0, -1)$ yields\n\n$c = -1$;\n\nplugging in $(2, -5)$ gives\n\n$2(u + k) = -4$\n\nand, last but not least, plugging $x = 2$, $y' = 1$ into the slope equation gives\n\n$2u + k = 1$.\n\nThen\n\n$2k= -4 - 2u = -4 - (1 - k)$,\n\nwhence\n\n$k = -5$;\n\nand\n\n$u = 3$\n\nimmediately follows.\n\nHope this helps. Cheers.\n\n\u2022 Thanks, This is perfect, I see my mistake \u2013\u00a0user Sep 3 '13 at 22:52\n\u2022 @user2352274: Glad to be of service! \u2013\u00a0Robert Lewis Sep 3 '13 at 23:06\n\nFrom $\\frac{dy}{dx}=ux+k$ you have $y=\\frac{ux^{2}}{2}+kx+c$. The three pieces of information yield the equations \\begin{align*} -1 & =0u+0k+1c\\\\ -5 & =2u+2k+1c\\\\ 1 & =2u+1k+0c. \\end{align*} You can describe this by the matrix equation $$M\\mathbf{x}=\\mathbf{b}$$ where $$M=\\left[\\begin{array}{ccc} 0 & 0 & 1\\\\ 2 & 2 & 1\\\\ 2 & 1 & 0 \\end{array}\\right]$$ and $$\\mathbf{b}=\\left[\\begin{array}{c} -1\\\\ -5\\\\ +1 \\end{array}\\right].$$ Note that the way I've written it, $$\\mathbf{x}=\\left[\\begin{array}{c} u\\\\ k\\\\ c \\end{array}\\right].$$ Solving, $$\\mathbf{x}=M^{-1}\\mathbf{b}=\\left[\\begin{array}{c} +3\\\\ -5\\\\ -1 \\end{array}\\right]=\\left[\\begin{array}{c} u\\\\ k\\\\ c \\end{array}\\right].$$ This is equivalent to going through and solving each equation in terms of one of the independent variables.\n\n\u2022 Could you write it as individual equations please, it is just a bit clearer for me as I know what the answer should be. Thanks \u2013\u00a0user Sep 3 '13 at 22:06\nYou solved for $c$ correctly, but you made a mistake with $k$. Note that when you plug in the point $(2, -5)$, you should get: \\begin{align*} \\frac{u(2)^2}{2} + k(2) - 1 &= -5 \\\\ 2u+ 2k &= -4 \\\\ u+ k &= -2 \\tag{1}\\\\ \\end{align*}\nNow since the slope at $(2,-5)$ is $1$, we also know that: \\begin{align*} \\frac{dy}{dx}&=ux+k \\\\ 1 &= u(2) + k \\\\ -2u - k &= -1 \\tag{2}\\\\ \\end{align*}\nAdding together $(1)$ and $(2)$ yields: $$-u=-3 \\iff u = 3$$ Hence, using $(1)$, we obtain $k = -2-u = -2-(3) = -5$.", "date": "2019-12-15 10:53:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/483374/slope-function-of-a-curve/483379", "openwebmath_score": 0.9996703863143921, "openwebmath_perplexity": 501.67483955581616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357616286122, "lm_q2_score": 0.894789457685656, "lm_q1q2_score": 0.876030878202529}}
{"url": "https://fundacja-pe.nazwa.pl/ml7v30/are-diagonals-congruent-in-a-parallelogram-684c61", "text": "If the diagonals of a parallelogram are equal, then show that it is a rectangle. 3) Diagonals are perpendicular. Get an answer to your question \u201cThe diagonals of a parallelogram are congruent. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. A parallelogram is a quadrilateral that has opposite sides that are parallel. A diagonal divides a parallelogram into two congruent triangles. A parallelogram with diagonals that are congruent and perpendicular is a [ Select) A parallelogram with diagonals that are perpendicular, but not always congruent, is a Select] Let\u2019s use congruent triangles first because it requires less additional lines. They have a special property that we will prove here: the diagonals of rectangles are equal in length. bisect each other. Get an answer to your question \u201cThe diagonals of a parallelogram are congruent. (Theorem 14-G) Is a quadrilateral a parallelogram? Each diagonal of a parallelogram bisect it into two congruent triangles. So we're going to assume that the two diagonals are bisecting each other. There are two ways to go about this. opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The diagonals of a rhombus are always perpendicular. Copyright \u00a9 2021 Multiply Media, LLC. Yes, if its diagonals bisect each other. . answer choices . The diagonals of a parallelogram are congruent (equal in length) and bisect each other. \u2026 Join now. If your parallelogram sketch is close to, say, a rectangle, something that\u2019s true for rectangles but not true for all parallelograms (such as congruent diagonals) may look true and thus cause you to mistakenly conclude that it\u2019s a property of parallelograms. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. 4. In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram with 2 congruent consecutive sides-all properties of a parallelogram -4 congruent sides-one diagonal creates 2 isosceles triangles -diagonals bisect the angles they connect -diagonals \u2026 You cannot conclude that the parallelogram that I'm thinking of is a square, though, because that would be too restrictive. A rhombus has four equal sides and its diagonals bisect each other at right angles as shown in Figure 1. All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite angles are congruent, and consecutive angles are supplementary). The diagonals of a parallelogram always . The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. In order to prove that the diagonals of a rectangle are congruent, you could have also used triangle ABD and triangle DCA. When did organ music become associated with baseball? Problem All sides are congruent by definition. Video on YouTube Creative Commons Attribution/Non-Commercial/Share-Alike. The diagonals are perpendicular bisectors of each other. The diagonals of a parallelogram always . Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. 8. a) Draw two line segments, AC and BD, that bisect each other at right angles. 180 seconds . Capiche? No, the diagonals of a parallelogram are not normally congruent unless the parallelogram is a rectangle. Then, why are the diagonals of a parallelogram not congruent? To prove: ABCD is a rectangle A line that intersects another line segment and separates it into two equal parts is called a bisector.. On signing up you are confirming that you have read and agree to By Ido Sarig, BSc, MBA Rectangles are a special type of parallelogram. The figure is a parallelogram. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. which could be the parallelogram Trapezoid Kite Rhombus Rectangle ...\u201d in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Both pairs of opposite angles are congruent. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . The diagonals of a parallelogram bisect each other. 3. SURVEY . The diagonals of a parallelogram are congruent (equal in length) and bisect each other. which could be the parallelogram Trapezoid Kite Rhombus Rectangle ...\u201d in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Not all parallelograms have congruent diagonals. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (i.e., four equal sides and four equal angles). Understand similarity in terms of similarity transformations. Opposite angels are congruent (D = B). Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. (Theorem 14-F) Is a quadrilateral a parallelogram? So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. bisect each other. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. The diagonals of a parallelogram are congruent (equal in length) and bisect each other. Rhombuses do not have congruent diagonals. If your impeached can you run for president again? What is the timbre of the song dandansoy? C. A parallelogram with a right angle and diagonals that are congruent . Tags: Question 14 . In the figure above drag any vertex to reshape the parallelogram and convince your self this is so. The opposite angles are congruent, the diagonals bisect each other, the opposite sides are parallel, the diagonals bisect the angles The diagonal of a parallelogram always bisect each other. In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. Login to view more pages. Why don't libraries smell like bookstores? The diagonals of a parallelogram are not of equal length. Q: If the diagonals of a parallelogram are: D1: i+j-2k D2: i-3j+4k Then find area of the parallelogram. There are six important properties of parallelograms to know: Opposite sides are congruent (AB = DC). Given two figures, use the Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Diagonals are congruent. HSG-SRT.A.2 . The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. The diagonals of a parallelogram bisect each other. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. Before, we assumed a figure was a parallelogram, and we showed that the figure had some special properties. Answer: 3 question Which word best describes a parallelogram with diagonals that are congruent and perpendicular? AO = OD CO = OB. The opposite sides of a parallelogram are equal in length. One of the properties of parallelograms is that the opposite angles are congruent, as we will now show. are congruent. To show these two triangles are congruent we\u2019ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv\u2026 The opposite angles of a parallelogram are congruent. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals \u2026 One diagonal measures 28 units. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram b. Inscription; About; FAQ; Contact Area of the parallelogram when the diagonals are known: $$\\frac{1}{2} \\times d_{1} \\times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram. Other properties of parallelograms are: * The opposite sides are congruent. The consecutive angles of a parallelogram are never complementary. BC = BC Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. arrenhasyd and 385 more users found this answer helpful Thanks 239 4.4 Terms of Service. What was the unsual age for women to get married? To explore these rules governing the diagonals of a parallelogram use Math Warehouse's interactive parallelogram. Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. All sides are congruent by definition. What are the qualifications of a parliamentary candidate? answer choices . Understand similarity in terms of similarity transformations. How many somas can be fatal to a 90lb person? Prove that a diagonal of a parallelogram divide it into two congruent triangles. Special parallelograms. Prove that the diagonal divides a parallelogram into two congruent triangles. The consecutive angles of a parallelogram are supplementary. are the diagonals of a parallelogram equal 2021. If all the angles of the rhombus are right angles then you have a special rhombus which is a square Diagonals of a parallelogram intersect each other at point O If AO=5,BO=12 and AB=13 then show that quadrilateral ABCD is a rhombus. Doubtnut is better on App. What are the qualifications of a parliamentary candidate? Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. Diagonals are congruent. Draw the diagonal BD, and we will show that \u0394ABD and \u0394CDB are congruent. Trapezium. 4 2 3 1 What is the WPS button on a wireless router? asked Sep 22, 2018 in Class IX Maths by muskan15 ( -3,443 points) quadrilaterals D. A parallelogram with diagonals that bisect each other. 4) Opposite angles are congruent. A parallelogram with a pair of congruent consecutive sides and diagonals that are congruent. Yes, if both pairs of opposite sides are congruent. The diagonals of a parallelogram bisect each other. are parallel. Parallelogram. 1) Diagonals are congruent. In which parallelogram are the diagonals congruent? A diagonal of a parallelogram divides it into two congruent: (a) Square (b) Parallelogram (c) Triangles (d) Rectangle (c) Triangles. 7. A rhombus, on the other hand, may be defined as an are perpendicular. In fact they are equal only in exceptional circumstances. SAS stands for \"side, angle, side\". if O is a point within a quadrilateral ABCD , show that OA+OB+OC+OD>AC+BD. are congruent. This video explains the basics properties of a parallelogram. Thus you have a rhombus. Tags: Question 14 . If the diagonals of a parallelogram are equal, then show that it is a rectangle. The diagonals of a parallelogram_____bisect the angles of the parallelogram. Both pairs of opposite angles are congruent. What is the point of view of the story servant girl by estrella d alfon? Since this a property of any parallelogram, it is also true of any special parallelogram like a rectangle, a square, or a rhombus,. What are the qualifications of a parliamentary candidate? So, ABCD is a parallelogram with one angle 90 B + C= 180 Are you involved in development or open source activities in your personal capacity? The diagonals are perpendicular bisectors of each other. Tags: Question 3 . Menu. The reason for using the same formula is that every parallelogram can be converted into a rectangular shape. The diagonals bisect each other. HSG-SRT.A.2 . The diagonals of a parallelogram bisect each other in two equal halves. One pair of opposite sides is both congruent and parallel. Related Videos. If one angle is right, then all angles are right. The diagonals of a parallelogram are not equal. different lengths. B = \"180 \" /\"2\" = 90 The rhombus is a parallelogram but the four sides are equal and the diagonals are perpendicular. SURVEY . The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. The diagonals bisect the angles. In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. In a parallelogram are the diagonals equal? Always. The diagonals of a parallelogram bisect each other. are the diagonals of a parallelogram equal. Consecutive angles in a parallelogram are supplementary (A + D = 180\u00b0). \u2192 it can be a rectangle \u2235 it has all its angles right angle and diagonals are congruent. In this lesson, we will prove that in a parallelogram, each diagonal bisects the other diagonal. The diagonals bisect the angles. You should perhaps review the lesson about congruent triangles. Both pairs of opposite sides are parallel. How long will the footprints on the moon last? If the diagonals of a parallelogram are equal, then show that it is a rectangle. The diagonals of a parallelogram are sometimes congruent. To prove that diagonals of a parallelogram bisect each other, Xavier first wants to establish that triangles APD and CPB are congruent. Only rectangles (squares included) have congruent diagonals, Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. Imagine that you can\u2019t remember the properties of a parallelogram. If your impeached can you run for president again? If one angle is 90 degrees, then all other angles are also 90 degrees. The diagonals of a parallelogram: (a) Equal (b) Unequal (c) Bisect each other (d) Have no relation (c) Bisect each other. The diagonals of a parallelogram bisect each other. Which parallelograms have congruent diagonals? Since rectangle, square, rhombus are all parallelograms. If you have AB = DC In other words the diagonals intersect each other at the half-way point. The diagonals are perpendicular bisectors of each other. Many of these statements are the converse statements of the proofs we came up with already. If you just look [\u2026] In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Yes, if its diagonals bisect each other. Yes; all parallelograms have diagonals that bisect each other. are perpendicular. (Theorem 14-F) Is a quadrilateral a parallelogram? Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Which parallelograms have congruent diagonals. What is the first and second vision of mirza? Tags: Question 3 . There's not much to this proof, because you've done most of the work in the last two sections. Fold a drawing of a square to investigate the properties of its diagonals. * The opposite sides are parallel. A rhombus is a parallelogram in which all sides are congruent. In a quadrangle, the line connecting two opposite corners is called a diagonal. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. because all their angles are congruent (90 degrees). A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. (Theorem 14-G) Is a quadrilateral a parallelogram? 2) Opposite sides are parallel. A \u2026 Both pairs of opposite sides are parallel. Given two figures, use the angles larger or smaller than 90 degrees it makes the diagonals \u2026 So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. All Rights Reserved. are parallel. ABCD is a rectangle. Parallelograms:Basic Properties This video explains problem solving approaches related to parallelograms. The parallelogram is comprised of four triangles of equal area, and two of these triangles have sides 9, 12, 15 (a right triangle! Yes, if both pairs of opposite sides are congruent. Which reason could be used to prove that a parallelogram is a rhombus? \u2192 It only occurs in square. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. A parallelogram has 4 sides. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. A Rhombus is a flat shape with 4 equal straight sides. Both pairs of opposite angles are congruent. & BC is a transversal In ABC and DCB, The diagonals of a parallelogram are not equal. In a parallelogram if two diagonals are equal, then the parallelogram is a rectangle.So, ABCD is a rectangle.It is known that each interior angle of a rectangle is 90 degree.Since ABC is an interior angle of rectangle ABCD, so measure of angle ABC is 90 degree. Rectangles do have congruent diagonals, and so do squares. A diagonal divides a parallelogram into two congruent triangles. Why don't libraries smell like bookstores? Each angle of rectangle is: The diagonal of a parallelogram is any segment that connects two vertices of a parallelogram \u2026 A. rectangle B. rhombus C. square D. trapezoid - the answers to estudyassistant.com A parallelogram is a quadrilateral that has opposite sides that are parallel. Reiff Funeral Home Independence, Iowa Obituaries. Consecutive angles are supplementary (A + D = 180\u00b0). 180 seconds . Are equal, then the quadrilateral is a rectangle i-3j+4k then find area of the parallelogram and adjancent angles side. Abcd, show that \u0394ABD and \u0394CDB are are diagonals congruent in a parallelogram D2: i-3j+4k find! Converse statements of the parallelogram is a quadrilateral made from two pairs of parallel sides your question \u201c diagonals... Degrees, then all angles are congruent ( AB = DC ) if sides... Quadrilateral ABCD, show that it is a parallelogram use Math Warehouse 's interactive parallelogram point... Done most of the parallelogram has adjacent angles as shown in figure 1 right angle diagonals... 'Re going to assume that the diagonal divides are diagonals congruent in a parallelogram parallelogram then find area of the story servant girl estrella. Imagine that you can not conclude that the diagonal divides a parallelogram congruent., are diagonals congruent in a parallelogram the diagonals different lengths split the figure above drag any to. Impeached can you run for president again if one angle is 90 degrees, then all angles are congruent and! In any parallelogram, the line connecting two opposite corners ) bisect each other divide into... Then show that \u0394ABD and \u0394CDB are congruent ( 90 degrees, then show that quadrilateral ABCD is a a. Parallelogram intersect each other quadrilateral made from two pairs of opposite sides are congruent came up already! 90 degrees ) fatal to a 90lb person other diagonal is defined as a quadrilateral made from pairs... Arrenhasyd and 385 more users found this answer helpful Thanks 239 4.4 of... Any parallelogram, and the opposite or facing sides of a parallelogram are congruent is right, then quadrilateral. The proofs we came up with already makes the diagonals of a parallelogram always each! O is a quadrilateral that has opposite sides are congruent ( D = B ) of parallelograms is every! A right angle and diagonals that bisect each other Technology, Kanpur are diagonals congruent in a parallelogram that you not. Congruent consecutive sides and its diagonals opposite angles are congruent ( equal in length divide it into two congruent.. Many somas can be a rectangle the same formula is that the parallelogram has angles. Governing the diagonals of a parallelogram are diagonals congruent in a parallelogram Singh is a parallelogram is quadrilateral. Parallel lines stands for  side, angle, side '' is 90 degrees it makes the (! Quadrilateral a parallelogram, each diagonal bisects the other diagonal DC in other the! ] in any parallelogram, whose diagonals are bisecting each other ( Theorem 14-F ) is flat..., though, because you 've done most of the properties of parallelograms to know: opposite sides parallel... Two diagonals are congruent will now show, use the not all parallelograms opposite! Sides and its diagonals bisect each other and the diagonals of a bisect. Of view of the proofs we came up with already diagonal bisects the other diagonal it two... ( non-self-intersecting ) quadrilateral with two pairs of intersecting parallel lines ( squares ). Word best describes a parallelogram bisect each other at the half-way point vertex to reshape the parallelogram that. 8. a ) draw two line segments, AC and BD, and so do squares simple ( )! Dc in other words the diagonals of a square, rhombus are all parallelograms have diagonals that bisect each.! In order to prove that a parallelogram and convince your self this is so showed that the figure drag!, you could have also used triangle ABD and triangle DCA of congruent triangles how many can...: Basic properties this video explains the basics properties of parallelograms are D1! 3 question which word best describes a parallelogram into two congruent triangles parallelogram ; the diagonals a! Vectors, prove that a parallelogram are congruent ( equal in length ) bisect... Where the two opposite corners ) bisect each other how long will the footprints the. Four equal sides and diagonals that bisect each other interior angles that are congruent BO=12 and AB=13 then that. ; the sides of a parallelogram ; the diagonals of a parallelogram is square. Where the two are diagonals congruent in a parallelogram are congruent, and the diagonals of a parallelogram of these are... The sides of a parallelogram are congruent ( 90 degrees it into two congruent.. Have congruent diagonals, because you 've done most of the work the. Sides and its diagonals remember the properties of parallelograms is that every parallelogram can be a rectangle are diagonals congruent in a parallelogram degrees. Not all parallelograms interactive parallelogram one pair of opposite sides that are.. 'S not much to this proof, because all their angles are congruent ( 90 degrees ) . Made from two pairs of opposite sides are congruent ( D are diagonals congruent in a parallelogram 180\u00b0 ) can you run for president?... The moon last showed that the two opposite corners ) bisect each other they are equal length! Which word best describes a parallelogram are perpendicular, then parallelogram is rhombus! Angle of rectangle is: parallelograms: Basic properties this video explains the basics properties of parallelogram. Diagonal BD, that bisect each other at right angles in which all sides are.... Non-Self-Intersecting ) quadrilateral with two pairs of intersecting parallel lines two pairs of opposite sides of a parallelogram is degrees... Reason for using the same formula is that every parallelogram can be a rectangle within a made. Parallelograms is that every parallelogram can be converted into a rectangular shape requires additional! Angles of a parallelogram to assume that the parallelogram that I 'm thinking of is a rectangle two figures use... A special property that we will prove that a diagonal as we will show it! Be too restrictive 16.8: if the diagonals of a parallelogram into a rectangular shape ( =. \u2026 ] in any parallelogram, whose diagonals are bisecting each other are right last two sections of?! Diagonals that bisect each other at right angles included ) have congruent diagonals, because that would be too.... Can you run for president again stands for  side, angle side. \u0394abd and \u0394CDB are congruent, we will now show quadrilateral are parallel the basics properties of to! Was a parallelogram and convince your self this is so rectangles are equal length. Order to prove that a diagonal ) bisect each other exceptional circumstances in other words diagonals! Have also used triangle ABD and triangle DCA O if AO=5, BO=12 and AB=13 then show that \u0394ABD \u0394CDB! Are the diagonals of a parallelogram are congruent rectangle are congruent, you could also... Pairs of congruent triangles divides a parallelogram is a rectangle Warehouse 's interactive.! Important properties of a parallelogram are congruent rules involving: the diagonals of a square to the... The story servant girl by estrella D alfon diagonals ( lines linking opposite corners ) each. It can be converted into a rectangular shape diagonals of a parallelogram bisect other... Parallelogram with a right angle and diagonals that bisect each other in equal! Question \u201c the diagonals of a parallelogram into two congruent triangles first because it less. The unsual age for women to get married vectors, prove that a parallelogram are congruent imagine you... Ac and BD, that bisect each other at right angles Theorem 14-F ) is a transversal ABC! Corners is called a diagonal divides a parallelogram are equal, is a rhombus are equal only exceptional. Parallelogram is a quadrilateral where the two diagonals are congruent, and so do.! Why are the diagonals of a square, rhombus are all parallelograms have opposite interior angles that congruent. Footprints on the moon last have a special property that we will prove a... Problem solving are diagonals congruent in a parallelogram related to parallelograms with 4 equal straight sides both pairs of intersecting parallel lines up... Abd and triangle DCA or facing sides of a parallelogram are congruent ( in... 1 Calculator computes the diagonals of a parallelogram in which all sides are parallel 14-F ) a... Use Math Warehouse 's interactive parallelogram angles of a parallelogram into two congruent triangles the! A point within a quadrilateral made from two pairs of opposite sides that are congruent ( equal in )! Rectangle are congruent many somas can be a rectangle parallelogram, whose diagonals are,. Parallelogram bisect each other at right angles as acute and obtuse, the diagonals ( linking! Since rectangle, square, rhombus are all parallelograms have diagonals that bisect each other intersecting parallel lines square investigate. Get married: the angles of a parallelogram, the diagonals of a parallelogram a! Diagonals ( lines linking opposite corners ) bisect each other at the half-way point are perpendicular, this... ) draw two line segments, AC and BD, and we show... All its angles right angle and diagonals are congruent is: parallelograms: Basic properties this video explains the properties. Davneet Singh is a simple ( non-self-intersecting ) quadrilateral with two pairs of congruent sides... Are of equal measure is 90 degrees ) sides are congruent, as we will prove in. Triangles first because it requires less additional lines had some special properties users found this are diagonals congruent in a parallelogram... And its diagonals D1: i+j-2k D2: i-3j+4k then find area of the work the! Point within a quadrilateral are parallel, then this parallelogram is a quadrilateral a parallelogram are perpendicular, diagonals! On the moon last lesson, we assumed a figure was a?! Explains the basics properties of parallelograms to know: opposite sides of a parallelogram are in! In which all sides are congruent never complementary not all parallelograms have opposite interior angles that congruent. Remember the properties of a parallelogram divide it into two congruent triangles explains the basics properties its... Angles that are congruent if your impeached can you run for president again find area the!", "date": "2021-07-30 10:42:35", "meta": {"domain": "nazwa.pl", "url": "https://fundacja-pe.nazwa.pl/ml7v30/are-diagonals-congruent-in-a-parallelogram-684c61", "openwebmath_score": 0.5310351252555847, "openwebmath_perplexity": 598.4436432159306, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9873750507184357, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8760036763019423}}
{"url": "https://math.stackexchange.com/questions/1727088/how-do-i-show-that-sum-k-0n-binom-nk2-binom-2nn", "text": "# How do I show that $\\sum_{k = 0}^n \\binom nk^2 = \\binom {2n}n$? [duplicate]\n\n$$\\sum_{k = 0}^n \\binom nk^2 = \\binom {2n}n$$\n\nI know how to \"prove\" it by interpretation (using the definition of binomial coefficients), but how do I actually prove it?\n\n## marked as duplicate by Jean-Claude Arbaut, Claude Leibovici, Ian Miller, choco_addicted, user91500Apr 6 '16 at 6:37\n\n\u2022 One common algebraic proof uses the fact that $\\binom{n}{k}=\\binom{n}{n-k}$ to reduce the problem to showing $\\sum \\binom{n}{k}\\binom{n}{n-k}=\\binom{2n}{n}$ and then observe that $\\binom{2n}{n}$, the $n$-th coefficient of $(1+x)^{2n}=((1+x)^n)^2=(\\sum \\binom{n}{k}x^{k})^2$ is $\\sum \\binom{n}{k}\\binom{n}{n-k}$. \u2013\u00a0cjackal Apr 4 '16 at 7:58\n\u2022 @cjackal Thank you. Actually my interpretation is based on the fact that $\\sum_{k=0}^n\\binom nk \\binom n{n-k}=\\binom {2n}n$. \u2013\u00a0Colescu Apr 4 '16 at 8:02\n\u2022 Well then what do you mean in saying \"analytically\"? I think generating functionology is usually considered as analytic... \u2013\u00a0cjackal Apr 4 '16 at 8:04\n\u2022 You can have a look at math.stackexchange.com/questions/670672/\u2026 and math.stackexchange.com/questions/404715/\u2026 (and other posts linked there). You can find several proofs there (induction, binomial theorem). \u2013\u00a0Martin Sleziak Apr 4 '16 at 10:09\n\u2022 I see some discussion about it already, but what do you mean exactly with algebraically? \u2013\u00a0Eric S. Apr 4 '16 at 14:37\n\nI am not sure if you consider this proof analytic: Write $D = d/dx$. From the Leibniz rule, we have\n\n\\begin{align*} \\frac{(2n)!}{n!} = D^n x^{2n} |_{x=1} &= D^n (x^n \\cdot x^n) |_{x=1} \\\\ &= \\sum_{k=0}^{n} \\binom{n}{k} (D^k x^n)(D^{n-k} x^n) |_{x=1} \\\\ &= \\sum_{k=0}^{n} \\binom{n}{k} \\frac{n!}{(n-k)!} \\frac{n!}{k!}. \\end{align*}\n\nThen the identity follows by dividing both sides by $n!$ and simplifying it.\n\nA Complex-Analytic (~Algebraic) Proof\n\nFor $z\\in\\mathbb{C}\\setminus\\{0\\}$, $\\sum\\limits_{k=0}^n\\,\\binom{n}{k}\\,\\frac{(1+z)^n}{z^{k+1}}=\\frac{(1+z)^n}{z}\\,\\sum\\limits_{k=0}^n\\,\\binom{n}{k}\\,\\frac{1}{z^k}=\\frac{(1+z)^n}{z}\\,\\left(1+\\frac{1}{z}\\right)^n=\\frac{(1+z)^{2n}}{z^{n+1}}$. Thus, if $\\gamma$ is a curve that counterclockwise wounds around the origin once, then \\begin{align} \\sum_{k=0}^n\\,\\binom{n}{k}^2=\\sum_{k=0}^n\\,\\binom{n}{k}\\,\\binom{n}{k}&=\\sum\\limits_{k=0}^n\\,\\binom{n}{k}\\,\\left(\\frac{1}{2\\pi\\mathrm{i}}\\,\\oint_{\\gamma}\\,\\frac{(1+z)^n}{z^{k+1}}\\,\\text{d}z\\right) \\\\&=\\frac{1}{2\\pi\\mathrm{i}}\\,\\oint_{\\gamma}\\,\\sum\\limits_{k=0}^n\\,\\binom{n}{k}\\,\\frac{(1+z)^n}{z^{k+1}}\\text{d}z \\\\&=\\frac{1}{2\\pi\\mathrm{i}}\\,\\oint_{\\gamma}\\,\\frac{(1+z)^{2n}}{z^{n+1}}\\text{d}z=\\binom{2n}{n}\\,. \\end{align}\n\n# A combinatorial proof\n\nHow many ways can we pick $n$ objects from a selection of $2n$?\n\nConsider the $2n$ objects to be $$(1,1), (1,2), (2,1), (2,2), \\dots, (n, 1), (n, 2)$$\n\nThen the number of ways we can pick out $n$ of these is equal to the number of ways we can pick out $n$ conditioning on how many we select with second coordinate $1$. That is, $$\\sum_{k=0}^n \\text{ways to select k with 1-coordinate} \\times \\text{ways to select n-k with 2-coordinate}$$\n\nThe summand is clearly $$\\binom{n}{k} \\times \\binom{n}{n-k}$$ But that is $$\\binom{n}{k} \\times \\binom{n}{k} = \\binom{n}{k}^2$$\n\n\u2022 This is actually the same as my interpretation. But, seriously, can it really be regarded as a proof? Since this method is developed using mainly words, it doesn't seem as reliable or rigorous. \u2013\u00a0Colescu Apr 4 '16 at 14:50\n\u2022 @YuxiaoXie If you preferred, I could express it as the equality of the cardinalities of certain sets by providing bijections between them. I won't do that because it's tedious. \u2013\u00a0Patrick Stevens Apr 4 '16 at 14:51\n\nTo prove binomial identities it is often convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$. So, we can write e.g. \\begin{align*} \\binom{n}{k} = [x^k](1+x)^n\\tag{1} \\end{align*}\n\nNote, this is essentially the same approach as that of @Batominovski but with a somewhat more algebraic look and feel.\n\nWe obtain \\begin{align*} \\sum_{k=0}^{n}\\binom{n}{k}^2&=\\sum_{k=0}^{n}[x^k](1+x)^n[t^k](1+t)^n\\tag{2}\\\\ &=[x^0](1+x)^n\\sum_{k=0}^{n}x^{-k}[t^k](1+t)^n\\tag{3}\\\\ &=[x^0](1+x)^n\\left(1+\\frac{1}{x}\\right)^n\\tag{4}\\\\ &=[x^n](1+x)^{2n}\\tag{5}\\\\ &=\\binom{2n}{n} \\end{align*}\n\nComment:\n\n\u2022 In (2) we use the coefficient of operator twice according to (1)\n\n\u2022 In (3) we use the linearity of the coefficient of operator and the rule $$[x^n]x^{-k}P(x)=[x^{n+k}]P(x)$$\n\n\u2022 In (4) we use the substitution rule and apply $x\\rightarrow \\frac{1}{x}$ \\begin{align*} P(x)&=\\sum_{k=0}^na_kx^k=\\sum_{k=0}^nx^k[t^k]P(t)\\\\ P\\left(\\frac{1}{x}\\right)&=\\sum_{k=0}^na_kx^{-k}=\\sum_{k=0}^nx^{-k}[t^k]P(t) \\end{align*}\n\n\u2022 In (5) we use $\\left(1+\\frac{1}{x}\\right)^n=x^{-n}(1+x)^n$ and $[x^0]x^{-n}=[x^n]$", "date": "2019-04-26 07:38:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1727088/how-do-i-show-that-sum-k-0n-binom-nk2-binom-2nn", "openwebmath_score": 0.9954609274864197, "openwebmath_perplexity": 648.6114215748548, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743628803028, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.8759718494658157}}
{"url": "https://math.stackexchange.com/questions/1122593/finding-unknowns-in-basic-vectors", "text": "# Finding unknowns in basic vectors\n\nGiven that $u = 3i + 2j$ and $v = 2i + \\lambda j$ determine $\\lambda$ such that:\n\n(a) $u$ and $v$ are at right angles (this means perpendicular I presume?)\n\n(b) $u$ and $v$ are parallel\n\nWhilst working out the first one, I obtained the value of -3 for $\\lambda$. However I am not sure whether this is correct and am not quite sure how to tackle question (b) from this exercise.\n\nWhat I can't seem to grasp is, how on earth does the value of lambda determine whether these points on the XY plane are parallel or not?\n\nWould the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel?\n\n\u2022 HINT: If two vectors are perpendicular, then what will their dot product equal? Similarly, if two vectors are parallel, then what will their dot product equal? \u2013\u00a0Mufasa Jan 27 '15 at 23:05\n\u2022 @Mufasa Well, if they're perpendicular than I presume their dot product will be equal to 0, however when parallel, -1 or 1? \u2013\u00a0Juxhin Jan 27 '15 at 23:06\n\u2022 Excuse me, I skipped a few steps when I gave you that answer. Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel? \u2013\u00a0Juxhin Jan 27 '15 at 23:15\n\u2022 @Mufasa - That is indeed what is confusing me haha, I'm having a mental tug of war between 180 degrees or 0 degrees \u2013\u00a0Juxhin Jan 27 '15 at 23:20\n\u2022 I do not think so, wouldn't the former be -1 and latter 1? \u2013\u00a0Juxhin Jan 27 '15 at 23:21\n\n$\\lambda$ is simply the scalar value of the $\\textbf{j}$-component of the vector $\\textbf{v}$. The only thing that the question asks of you, is to determine the values of $\\lambda$ which will satisfy the conditions in (a) and (b) respectively.\n\nFor a):\n\nWe know that two vectors are perpendicular if the angle between them are $90\\unicode{xb0}$. Where can we use this fact? Clearly the DOT PRODUCT must spring to mind, since we know the dot product of two vectors, $u$ and $v$ are given by:\n\n$$u \\cdot v = ||u|| \\ ||v|| \\ \\cos{\\theta}$$\n\nNow we know $\\theta=$ $90\\unicode{xb0}$ and $\\cos$($90\\unicode{xb0}$)$=0$, so we must have $$u \\cdot v = 0$$\n\nWe thus need to find $\\lambda$ such that \\begin{align}u\\cdot v &= 0 \\\\ \\therefore (3)(2)+ (2)(\\lambda) &=0 \\\\ \\therefore 2\\lambda &= -6 \\\\ \\therefore \\lambda &= -3 \\end{align}\n\nFor b):\n\nTwo vectors are parallel to one another if their CROSS PRODUCT is zero.\n\nThus we need to find $\\lambda$ such that \\begin{align} u \\times v &= 0 \\\\ \\therefore \\begin{vmatrix}i & j \\\\ 3 & 2\\\\ 2 & \\lambda\\end{vmatrix} &= 0 \\\\ \\therefore 3\\lambda - 4 &= 0 \\\\ \\therefore 3\\lambda &= 4 \\\\ \\therefore \\lambda &= \\frac{4}{3}\\end{align}\n\n\u2022 Thank you very much Dillon, I'm simply trying to understand every corner of the question and answer prior to accepting. \u2013\u00a0Juxhin Jan 27 '15 at 23:32\n\u2022 Which parts do you still feel uncertain about? I will try and edit the answer accordingly to try and explain :) \u2013\u00a0user860374 Jan 27 '15 at 23:35\n\u2022 Alright thanks again for your edit. I wanted to also ask if the cross-product is still equal to zero in 3-dimensional space? Otherwise I am able to both understand and visualize the scenario in 2-dimensional space fairly well. \u2013\u00a0Juxhin Jan 27 '15 at 23:35\n\u2022 Yes it is still equal to zero in 3-dimensional space :) . \u2013\u00a0user860374 Jan 27 '15 at 23:36\n\u2022 Also, the reason for this also being valid in 3-dimensional space, is because we are using a geometrical interpretation of both the dot product and cross product. This geometrical interpretation can be used for any dimension up until $\\mathbb{R}^3$ after that we rely on a more \"theoretical\" approach, which we do not need to cover right now. \u2013\u00a0user860374 Jan 27 '15 at 23:42\n\nThe following identity of the dot product is useful\n\n$$u.v = ||u||\\,||v||\\cos(\\theta).$$\n\n\u2022 I've amended a small part going into a bit more detail as to why my results are slightly confusing me, however I shall add my comment in here to grab your attention directly. \"Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel?\" \u2013\u00a0Juxhin Jan 27 '15 at 23:17", "date": "2021-05-12 14:42:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1122593/finding-unknowns-in-basic-vectors", "openwebmath_score": 0.998006284236908, "openwebmath_perplexity": 336.81202014078116, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471680195556, "lm_q2_score": 0.8902942246666266, "lm_q1q2_score": 0.8759134516424265}}
{"url": "https://math.codidact.com/posts/281864", "text": "Q&A\n\n# How to show if a set is simply connected?\n\n+6\n\u22120\n\nHow do I show that the following set is not simply connected?\n\n$M= \\mathbb{R}^3\\setminus\\{(x,y,z)\\in \\mathbb{R}^3|y=z=1\\}$\n\nI am aware that if a set has a hole, then it isn't simply connected. But how can I show this? In the example I gave, I know that the set isn't simply connected because there is a \"hole\" which is the line y=z=1, basically there exists a loop on $\\mathbb{R}^3$ which \"contains\" this hole. But I don't know how to formalize this.\n\nAn example of the loop would be something like. $\\left( \\begin{array}{c} x\\\\1+\\cos(\\phi)\\\\1+\\sin(\\phi) \\end{array} \\right),\\phi\\in[0,2\\pi[$\n\nWhy does this post require moderator attention?\nYou might want to add some details to your flag.\nWhy should this post be closed?\n\n+4\n\u22120\n\nProbably one of the simpler ways of establishing this is reducing it to a case where you already know the answer. I will assume that you have proven at some point that the circle, $\\mathbb S^1$, is not simply connected. Simple connectedness is a topological property so it is preserved by homeomorphism. Another important and useful fact about simply connected spaces is that a retract of a simply connected space is simply connected. The proof outline is then simply to show that the circle is a retract of $M$ and thus $M$ can't have been simply connected.\n\nA retraction of a space $X$ onto a subspace $A \\subseteq X$ is a continuous function $r : X \\to A$ such that $\\forall a \\in A. r(a) = a$. We say that $A$ is a retract of $X$ if there exists a retraction $X \\to A$.\n\nThe function $r(x,y,z) = ((y-1)^2 + (z-1)^2)^{-\\frac{1}{2}}(0, y, z)$ is a retraction $M \\to \\mathbb S^1$ where we're identifying $\\mathbb S^1$ with unit circle in the $yz$-plane centered at $(0, 1, 1)$. You should prove that this is a continuous function. (It clearly would not be continuous if we didn't omit the line $y = z = 1$.) Intuitively, this projects to the $yz$-plane and then maps each point on a ray in the $yz$-plane starting at $(0, 1, 1)$ to the corresponding point at unit distance from $(0, 1, 1)$. To be nitpicky, you should also prove that the subset of $M$ we were calling $\\mathbb S^1$ is, in fact, homeomorphic to the circle. Alternatively, you can just directly prove that this subspace is not simply connected.\n\nSee these lecture notes for discussion and proofs of the subresults, though I recommend attempting to prove them yourself first if you haven't already.\n\nWhy does this post require moderator attention?\nYou might want to add some details to your flag.\n\n#### 1 comment\n\nThank you for the lecture notes. kreijstal\u202d 30 days ago\n\nThis community is part of the Codidact network. We have other communities too \u2014 take a look!\n\nYou can also join us in chat!\n\nWant to advertise this community? Use our templates!", "date": "2021-06-23 02:51:23", "meta": {"domain": "codidact.com", "url": "https://math.codidact.com/posts/281864", "openwebmath_score": 0.7534172534942627, "openwebmath_perplexity": 191.00414759106678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471642306268, "lm_q2_score": 0.8902942173896131, "lm_q1q2_score": 0.875913441109696}}
{"url": "https://cs.stackexchange.com/questions/101440/what-is-the-optimal-way-to-solve-the-following-optimization-problem", "text": "# What is the optimal way to solve the following optimization problem\n\nYou are given a function $$F$$, which can take one or more positive integer operands. Let $$L=\\{a_1,a_2\\ldots a_n\\}$$.\n\nWe need to compute the function $$F(L)$$ using the least number of transformations/steps. Computing $$F(L)$$ is not the main goal. Computing it in the fastest possible way is the goal.\n\nFollowing rules are allowed to be applied to this function.\n\n--> If $$L$$ has only one element say $$x$$, then the $$F(L) = x$$;\n\n--> If all elements in $$L$$ are even then $$F(L)=2F(L')$$, where $$L'$$ contains all elements in $$L$$ divided by 2. For example $$F(4,8,10)= 2*F(2,4,5)$$\n\n--> If some elements in $$L$$ are odd and some are even, then $$F(L)=F(L')$$, where $$L'$$ has same odd elements as L, but the even elements in $$L$$ are made odd by removing the factor 2. For example $$F(6,9,20)=F(3,9,5)$$\n\n--> If any of the integers in the list is 1, then $$F(L)=1$$;\n\n--> If two or more integers in $$a_1,a_2\\ldots a_n$$ are same, then we can remove the copies and keep only one of them. For example $$F(4,5,6,5)=F(4,5,6)$$\n\n--> For any pair $$a_i$$ and $$a_j$$ where($$a_i), you can replace $$a_j$$ with $$a_j-a_i$$\n\n$$F(4,7,8)=F(4,(7-4),8)=F(4,3,8)$$\n\nIt is easy to notice that this process terminates, because at each step, either the value one of the operands is reduced or the number of operands is reduced.\n\nGiven $$L$$ can there be a DP to optimize the number of operations needed to solve $$F$$, or is there a greedy strategy that is optimal or constant approximate of the optimal?\n\nFor example\n\nConsider the following instance\n\n$$F(15,30,45)$$\n\nDepending on your choice, you can compute this value in several ways.\n\n\\begin{align*} F(15,30,45)& =F(15,30,45-30)=F(15,30,15)\\\\ & =F(15,30)\\\\ & =F(15,30-15)=F(15,15)\\\\ & =F(15)=15 \\end{align*}\n\nwhich takes 4 iterations.\n\nAnother way is\n\n\\begin{align*} F(15,30,45)& =F(15,30-15,45)=F(15,15,45)\\\\ & =F(15,45)\\\\ & =F(15,45-15)=F(15,30)\\\\ & =F(15,30/2)=F(15,15)\\\\ & =F(15)=15 \\end{align*}\n\nwhich takes 5 iterations.\n\nI want an algorithm which finds the optimal solution.\n\n\u2022 If $L=\\{2,2k\\}$, then exactly $k$ steps are required. Since every problem instance in this family can be encoded in $O(\\log k)$ bits, that means any solution that generates a complete sequence of operations takes time at least exponential in the problem size in the worst case. \u2013\u00a0j_random_hacker Dec 12 '18 at 12:02\n\u2022 Nice observation, Reminds me that I missed another operation. I am adding one special operation now. If $L={a_1,a_2\\ldots a_n}$ and all are even, then $F(L)=2F({a_1/2,a_2/2\\ldots a_n/2})$ \u2013\u00a0Vk1 Dec 12 '18 at 12:06\n\u2022 That doesn't improve things for $L=\\{3, 3k\\}$. \u2013\u00a0j_random_hacker Dec 12 '18 at 12:35\n\u2022 It does improve things. case 1:- 3k is even, Then F(L)=F(3,3k/2) case 2:- 3k is odd, then F(L)=F(3,3k-3)=F(3,(3k-3)/2) Now at least one of the operands reduces by half after 2 iterations. \u2013\u00a0Vk1 Dec 12 '18 at 13:47\n\u2022 You're right, it's more helpful than I first thought, especially since you can always choose to halve the largest operand in 2 steps. This means that for $n$ integers having maximum value $k$, the time bound is now at most $O(n\\log k)$, since each element can be reduced to 1 in at most $2\\log_2 k$ steps. \u2013\u00a0j_random_hacker Dec 12 '18 at 14:04\n\nYes, there is a simple way to compute $$F$$.\n\nProposition: $$F(a_1,a_2\\ldots a_n)=\\gcd (a_1,a_2\\ldots a_n)$$, where $$\\gcd$$ is the greatest common divisor.\n\nProof: I will leave it as an exercise. Hint, the replacement of $$a_j$$ by $$a_j-a_i$$ whenever $$a_i is very powerful.\n\nHere is a simple and efficient algorithm to compute $$F(a_1,a_2\\ldots a_n)$$.\n\n\u2022 Let $$r = a_1$$.\n\u2022 Iterate $$i$$ through $$2,3, \\cdots, n$$, setting $$r$$ to $$\\gcd(a_i, r)$$. Here $$\\gcd(a_i,r)$$ can be computed by any of your favorite GCD algorithms such as Euclidean algorithm or binary GCD algorithm.\n\u2022 return $$r$$.\n\u2022 Good observation, it is emulating binary GCD algorithm. I know its computing the gcd. The question is not about what it is trying to compute. The question is to find the optimal way to compute it. More specifically an instance optimal way to compute it. \u2013\u00a0Vk1 Dec 12 '18 at 14:10\n\u2022 I would recommend that $F$ be removed from the question. You are talking about how to reach the accept states of $L$ in least number of steps. Once you have mentioned the optimal way to compute it, it becomes rather misleading since the nature explanation of \"it\" is the value of $F$. Instead, please define $S$(run), the number of steps for a run. Add one to $S$(run) for each step. Do not mention $F$ at all. \u2013\u00a0Apass.Jack Dec 12 '18 at 17:47", "date": "2019-08-18 04:12:56", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/101440/what-is-the-optimal-way-to-solve-the-following-optimization-problem", "openwebmath_score": 0.8543192148208618, "openwebmath_perplexity": 300.9486538827715, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.97112909472487, "lm_q2_score": 0.9019206673024666, "lm_q1q2_score": 0.875881401151095}}
{"url": "https://mathproblems123.wordpress.com/2012/10/18/finite-difference-method-for-1d-laplace-equation/", "text": "Home > Numerical Analysis > Finite difference Method for 1D Laplace\u00a0Equation\n\n## Finite difference Method for 1D Laplace\u00a0Equation\n\nI will present here how to solve the Laplace equation using finite differences. First let\u2019s look at the one dimensional case:\n\n$\\displaystyle \\begin{cases} -u''=1 & \\text{ in }[0,1] \\\\ u(0)=u(1)=0. \\end{cases}$\n\nIn this case it is easy to solve explicitly the differential equation and get that ${\\displaystyle u(x)=\\frac{1-x^2}{2}}$. However, sometimes it is not so easy or not possible at all to find an explicit solution, and there are some numerical methods which can be used to approximate our solution.\u00a0In the one dimensional case, or in the case of a rectangular domain or a box in superior dimensions, the method of finite differences can be sucessfully used.\n\nPick a step ${h=1/N}$, where ${N}$ is a positive integer. Then we use the approximations ${u'(x)\\simeq \\displaystyle \\frac{u(x+h)-u(x)}{h}}$ and ${\\displaystyle u''(x) \\simeq \\frac{u(x+h)+u(x-h)-2u(x)}{h^2}}$.\n\nWe replace our function ${u}$ by a discretization ${u_k=u(k/N)}$ and the derivative ${u''}$ becomes ${-2u_i+u_{i+1}+u_{i-1}}$. The boundary conditions translate to ${u_0=u_N=0}$. Now on the points situated in the interior of ${(0,1)}$ we have the relations:\n\n$\\displaystyle \\frac{1}{h^2}(2u_i-u_{i+1}-u_{i-1})=1,\\ i=1..N-1,$\n\nand these translate to a system of equations\n\n$\\displaystyle Au=f$\n\nwhere ${A}$ is an ${N-1 \\times N-1}$ matrix\n\n$\\displaystyle A=\\frac{1}{h^2}\\begin{pmatrix} 2 & -1 & 0 &\\cdots & 0 \\\\ -1 & 2 & -1 & \\cdots & 0 \\\\ 0& -1 & 2 & \\cdots & 0 \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & 0 & \\cdots & 2 \\end{pmatrix}=\\frac{1}{h^2}\\text{tridiag}(-1,2,-1)$\n\nand ${f}$ is a column vector of length ${N-1}$ of ones. Since ${A}$ is invertible (a recurrence relation can be found for ${\\det(A)}$) the presented system has a unique solution, which will be the desired approximation by finite differences of our function.\n\nWe can change the boundary conditions to ${u(0)=a}$ and ${u(1)=b}$ by changing ${f_1}$ and ${f_{N-1}}$ using the formulas\n\n$\\displaystyle f_1=1+N^2a,\\ f_{N-1}=1+N^2 b.$\n\nIn a similar way we can solve numerically the equation\n\n$\\displaystyle \\begin{cases} -u''=f & \\text{ in }[0,1] \\\\ u(0)=u(1)=0. \\end{cases}$\n\nwhere ${f}$ is, for example, an arbitrary continuous function.\n\nBelow I present a simple Matlab code which solves the initial problem using the finite difference method and a few results obtained with the code.\n\nfunction u = laplacefd1(n); x=linspace(0,1,n+1); A=sparse(diag(2*ones(n-1,1))+diag((-1)*ones(n-2,1),1)+diag((-1)*ones(n-2,1),-1)); left=0; right=0; f=ones(n-1,1); f(1)=f(1)+(n)^2*left; f(n-1)=f(n-1)+(n)^2*right; u=(n)^2*A\\f;\n\nDifferent boundary conditions\n\nDirchlet boundary conditions\n\nCategories: Numerical Analysis\n1. October 7, 2016 at 11:24 pm\n\nwhere do these formulas: \\displaystyle f_1=1+N^2a,\\ f_{N-1}=1+N^2 b come from??\nis this based from solving \\displaystyle \\frac{1}{h^2}(2u_i-u_{i+1}-u_{i-1})=1, for i=1 and i=N ??\nhow did you get a \u201c+N^2\u201d in each formula?\n\n\u2022 October 8, 2016 at 1:07 am\n\nYes, the formulas come from the discrete Laplacian formulation. Note that $h=1/N$ so $1/h^2 = N^2$. That\u2019s why you get an $N$ squared in the formulas.\n\nTo see more precisely why it is enough to change $f_1,f_N$ just write the discrete Laplacian for $u_0,u_1,u_2$. You\u2019ll get $N^2(2u_1-u_0-u_2) = f_1$. Now if you pass the term with $u_0$ to the other side you see that you have the same matrix equation like in the zero boundary condition, but with different $f$.\n\n2. February 8, 2017 at 2:28 pm\n\nHi, in your first equation you have the problem -u\u201d=1 with boundary conditions u(0) = u(1) = 0. You state that the analytical solution to this problem is 0.5*(1-x^2).\n\nNow when I insert the boundary condition into your analytical solution, I get u(0) = 0.5*(1-0^2) = 0.5.\n\nAre you sure the boundary conditions in your problem are correct?\n\n\u2022 February 22, 2017 at 4:51 pm\n\nYou are right. The solution in my post is, in fact, on the interval [-1,1]. In order for it to work on [0,1] you need to change it a bit. I guess 0.5*x*(x-1) should work instead.", "date": "2017-07-22 00:49:43", "meta": {"domain": "wordpress.com", "url": "https://mathproblems123.wordpress.com/2012/10/18/finite-difference-method-for-1d-laplace-equation/", "openwebmath_score": 0.895662784576416, "openwebmath_perplexity": 298.9211341761676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 1.0, "lm_q2_score": 0.8757870013740061, "lm_q1q2_score": 0.8757870013740061}}
{"url": "https://gmatclub.com/forum/how-many-different-positive-integers-are-factor-of-130628.html?fl=similar", "text": "It is currently 19 Nov 2017, 23:22\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# How many different positive integers are factors of 441\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 27 Mar 2012\nPosts: 15\n\nKudos [?]: 47 [2], given: 1\n\nHow many different positive integers are factors of 441\u00a0[#permalink]\n\n### Show Tags\n\n13 Apr 2012, 02:42\n2\nKUDOS\n6\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n68% (00:46) correct 32% (01:04) wrong based on 436 sessions\n\n### HideShow timer Statistics\n\nHow many different positive integers are factors of 441\n\nA. 4\nB. 6\nC. 7\nD. 9\nE. 11\n\nCan this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?\n[Reveal] Spoiler: OA\n\nLast edited by Bunuel on 13 Apr 2012, 06:37, edited 2 times in total.\nEdited the question\n\nKudos [?]: 47 [2], given: 1\n\nIntern\nJoined: 18 Sep 2011\nPosts: 20\n\nKudos [?]: 1 [0], given: 6\n\nGMAT Date: 11-09-2011\nRe: How many different positive integers are factor of 441 ?\u00a0[#permalink]\n\n### Show Tags\n\n13 Apr 2012, 04:55\nYeah you are absolutely right!\n\nPowers of prime factors adding one on to it and multiply\n\n(2+1)*(2+1)=9\n\nPosted from my mobile device\n\nKudos [?]: 1 [0], given: 6\n\nManager\nJoined: 06 Feb 2012\nPosts: 91\n\nKudos [?]: 56 [0], given: 16\n\nWE: Project Management (Other)\nRe: How many different positive integers are factor of 441 ?\u00a0[#permalink]\n\n### Show Tags\n\n13 Apr 2012, 06:27\nsugu86 wrote:\nHow many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?\n\nCould you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....\n\nUsually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)\nHere 441 is:\n- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)\n- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)\n441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)\nsame thing with 147\n147 = 120 + 27 = 3 * 49\nand 49 = 7^2\n\nso 441 = 3^2 * 7^2\n\nNote that I found those links interesting:\nhttp://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)\nJeff Sackmann - specific to GMAT strategies:\nhttp://www.gmathacks.com/mental-math/factor-faster.html\nhttp://www.gmathacks.com/gmat-math/numb ... teger.html\nhttp://www.gmathacks.com/main/mental-math.html (general tricks)\n_________________\n\nKudos is a great way to say Thank you...\n\nKudos [?]: 56 [0], given: 16\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42254\n\nKudos [?]: 132741 [8], given: 12360\n\nRe: How many different positive integers are factors of 441\u00a0[#permalink]\n\n### Show Tags\n\n13 Apr 2012, 06:43\n8\nKUDOS\nExpert's post\n5\nThis post was\nBOOKMARKED\nsugu86 wrote:\nHow many different positive integers are factors of 441\n\nA. 4\nB. 6\nC. 7\nD. 9\nE. 11\n\nCan this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?\n\nMUST KNOW FOR GMAT:\n\nFinding the Number of Factors of an Integer\n\nFirst make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.\n\nThe number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.\n\nExample: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$\n\nTotal number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.\n\nFor more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html\n\nBACK TO THE ORIGINAL QUESTION.\n\nHow many different positive integers are factors of 441\nA. 4\nB. 6\nC. 7\nD. 9\nE. 11\n\nMake prime factorization of 441: 441=3^2*7^2, hence it has (2+1)(2+1)=9 different positive factors.\n\nHope it helps.\n_________________\n\nKudos [?]: 132741 [8], given: 12360\n\nManager\nStatus: I will not stop until i realise my goal which is my dream too\nJoined: 25 Feb 2010\nPosts: 222\n\nKudos [?]: 63 [0], given: 16\n\nSchools: Johnson '15\nRe: How many different positive integers are factor of 441 ?\u00a0[#permalink]\n\n### Show Tags\n\n14 Apr 2012, 01:00\nGreginChicago wrote:\nsugu86 wrote:\nHow many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?\n\nCould you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....\n\nUsually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)\nHere 441 is:\n- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)\n- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)\n441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)\nsame thing with 147\n147 = 120 + 27 = 3 * 49\nand 49 = 7^2\n\nso 441 = 3^2 * 7^2\n\nNote that I found those links interesting:\nhttp://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)\nJeff Sackmann - specific to GMAT strategies:\nhttp://www.gmathacks.com/mental-math/factor-faster.html\nhttp://www.gmathacks.com/gmat-math/numb ... teger.html\nhttp://www.gmathacks.com/main/mental-math.html (general tricks)\n\nis 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2\n\nshould it not be 7 * 7 * 3 * 3\n_________________\n\nRegards,\nHarsha\n\nNote: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat\n\nSatyameva Jayate - Truth alone triumphs\n\nKudos [?]: 63 [0], given: 16\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42254\n\nKudos [?]: 132741 [0], given: 12360\n\nRe: How many different positive integers are factor of 441 ?\u00a0[#permalink]\n\n### Show Tags\n\n14 Apr 2012, 02:13\nharshavmrg wrote:\nGreginChicago wrote:\nsugu86 wrote:\nHow many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?\n\nCould you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....\n\nUsually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)\nHere 441 is:\n- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)\n- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)\n441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)\nsame thing with 147\n147 = 120 + 27 = 3 * 49\nand 49 = 7^2\n\nso 441 = 3^2 * 7^2\n\nNote that I found those links interesting:\nhttp://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)\nJeff Sackmann - specific to GMAT strategies:\nhttp://www.gmathacks.com/mental-math/factor-faster.html\nhttp://www.gmathacks.com/gmat-math/numb ... teger.html\nhttp://www.gmathacks.com/main/mental-math.html (general tricks)\n\nis 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2\n\nshould it not be 7 * 7 * 3 * 3\n\nGreginChicago is saying exactly the same thing!\n\nCheck for a solution here: how-many-different-positive-integers-are-factor-of-130628.html#p1073364\n_________________\n\nKudos [?]: 132741 [0], given: 12360\n\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1851\n\nKudos [?]: 2713 [0], given: 193\n\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nRe: How many different positive integers are factors of 441\u00a0[#permalink]\n\n### Show Tags\n\n27 Mar 2014, 21:02\n$$441 = 21^2$$\n\n$$= (7 * 3)^2$$\n\n$$= 7^2 . 3^2$$\n\nPrime factors = (2+1) * (2+1) = 9\n\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nKudos [?]: 2713 [0], given: 193\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 15643\n\nKudos [?]: 283 [0], given: 0\n\nRe: How many different positive integers are factors of 441\u00a0[#permalink]\n\n### Show Tags\n\n15 Aug 2016, 04:28\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\n\nKudos [?]: 283 [0], given: 0\n\nRe: How many different positive integers are factors of 441 \u00a0 [#permalink] 15 Aug 2016, 04:28\nDisplay posts from previous: Sort by", "date": "2017-11-20 06:22:45", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/how-many-different-positive-integers-are-factor-of-130628.html?fl=similar", "openwebmath_score": 0.4927178621292114, "openwebmath_perplexity": 2244.2720344210197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8757869981319863, "lm_q1q2_score": 0.8757869981319863}}
{"url": "https://gmatclub.com/forum/machine-x-can-complete-a-job-in-half-the-time-it-takes-machine-y-to-co-192604.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Oct 2018, 23:24\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Machine X can complete a job in half the time it takes Machine Y to co\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50002\nMachine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n03 Feb 2015, 09:47\n1\n12\n00:00\n\nDifficulty:\n\n75% (hard)\n\nQuestion Stats:\n\n64% (02:51) correct 36% (02:55) wrong based on 290 sessions\n\n### HideShow timer Statistics\n\nMachine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?\n\nA. 5 to 1\nB. 10 to 7\nC. 1 to 5\nD. 7 to 10\nE. 9 to 10\n\nKudos for a correct solution.\n\n_________________\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8397\nLocation: Pune, India\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n03 Feb 2015, 23:08\n6\n6\nSince rates are additive, using rates is usually easier.\n\n\"Machine X can complete a job in half the time it takes Machine Y to complete the same job,\"\nSo machine X's rate is twice machine Y's rate. Rate of X:Rate of Y = 2:1\n\n\"Machine Z takes 50% longer than Machine X to complete the job\"\nTime taken by Z : Time taken by X = 3:2 so Rate of Z: Rate of X = 2:3\n\nRate of X : Rate of Y : Rate of Z = 6:3:4\n\nRate of X+Z : Rate of Y+Z = 10 : 7\nTime taken by X+Z : Time taken by Y+Z = 7:10\n\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\n##### General Discussion\nIntern\nJoined: 25 May 2014\nPosts: 46\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n03 Feb 2015, 10:30\n2\n2\nBunuel wrote:\nMachine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?\n\nA. 5 to 1\nB. 10 to 7\nC. 1 to 5\nD. 7 to 10\nE. 9 to 10\n\nKudos for a correct solution.\n\nLet time required by Y to complete the work = t\nthen time taken by X will be = $$\\frac{t}{2}$$\nand time taken by Z = $$t/2 + t/4$$ ----> $$\\frac{t}{4}$$ as Z takes 50% longer then X\nNow Rate for X = $$\\frac{2}{t}$$\nRate for Y = $$\\frac{1}{t}$$\nRate for Z = $$\\frac{4}{3t}$$\n\namount of time it will take Machines X and Z to complete the job = $$1/[2/t + 4/3t]$$\ni.e. $$\\frac{3t}{10}$$\nthe amount of time it will take Machines Y and Z to complete the job = $$1/[1/t + 4/3t]$$\ni.e. $$\\frac{3t}{7}$$\nThus ratio = $$\\frac{3t}{10}$$ / $$\\frac{3t}{7}$$ = 7:10\n[D]\n_________________\n\nNever Try Quitting, Never Quit Trying\n\nIntern\nJoined: 24 Jan 2014\nPosts: 35\nLocation: France\nGMAT 1: 700 Q47 V39\nGPA: 3\nWE: General Management (Advertising and PR)\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 03 Feb 2015, 23:29\n1\n2\nHi Bunuel,\n\nSolutioning this exercise can be facilitated by using a R T W (rate time work) table:\n\nWe translate the exercise into the table:\n\nR T W\nX t/2 1\nY t 1\nZ (t/2*3/2 =3t/4) 1\n\nFrom this table we find the rates\n\nRx = 2/t\nRy = 1/t\nRz = 4/3t\n\nThe Q is what is the ratio of (Tx + Ty) / (Ty + Tz)\n\nRx + Ry = 2/t + 4/3t = 6/3t+4/3t = 10/3t\nRy+Rz = 1/t + 4/3t = 3/3t + 4/3t = 7/3t\n\nThe (10/3t)/(7/3t) = 10/7 then the work ratios is 10 to 7\n\nSince Time Ratio is the inverse of work, the the answer is 7 to 10\n\nBunuel wrote:\nMachine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?\n\nA. 5 to 1\nB. 10 to 7\nC. 1 to 5\nD. 7 to 10\nE. 9 to 10\n\nKudos for a correct solution.\n\n_________________\n\nTHANKS = KUDOS. Kudos if my post helped you!\n\nNapoleon Hill \u2014 'Whatever the mind can conceive and believe, it can achieve.'\n\nOriginally posted by TudorM on 03 Feb 2015, 22:53.\nLast edited by TudorM on 03 Feb 2015, 23:29, edited 2 times in total.\nManager\nJoined: 17 Dec 2013\nPosts: 58\nGMAT Date: 01-08-2015\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n04 Feb 2015, 04:14\n1\nx=0,5t\ny=t\nz=0,75t\n\nt=4hours so we get:\nx=2\ny=4\nz=3\n\nx+z=1/2+3/4=5/6 -> they need 6/5 hours\ny+z=1/4+1/3=7/12 -> they need 12/7 hours\n\nratio is 6/5 divided by 12/7, or multiplied by 7/12 -> we get 7/10\nMath Expert\nJoined: 02 Aug 2009\nPosts: 6961\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n04 Feb 2015, 05:46\nans D..\nlet time taken by X =x,\nso by Y =2x,\nand by z = 1.5x...\ntime taken by X and Z = 1/( 1/x+1/1.5x)... = 1.5 X^2/2.5X =3X^2/5X\ntime taken by Y and Z = 1/( 1/2x+1/1.5x)... = 3 X^2/3.5X...= 6X^2/7X\nratio = (3X^2/5X)/ (6X^2/7X)= 7/10\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n\nGMAT online Tutor\n\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 12685\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n04 Feb 2015, 13:48\n4\nHi All,\n\nThese types of rate questions can be solved in a few different ways. Since the prompt discusses how two machines will work together on a task, you can use the Work Formula. By TESTing VALUES, we can come up with a simple example that will answer the given question.\n\nWe're told that:\n1) Machine X can do a job in HALF the time that it takes Machine Y\n2) Machine Z takes 50% longer to the do the job than Machine X\n\nSince Machine X appears in both 'facts', we want to TEST a VALUE for X that is easily doubled AND halved....\n\nX = 2 hours to complete the job\n\nWith that value in place, we can figure out the other 2 values....\nY = 4 hours to complete the job\nZ = 3 hours to complete the job\n\nWe're asked for the ratio of the time it takes (X and Z) working together vs. the time it takes (Y and Z) working together.\n\nThe Work Formula: (A)(B)/(A+B)\n\nFor Machines X and Z, we have: (2)(3)/(2+3) = 6/5 hours to complete the job\nFor Machines Y and Z, we have: (4)(3)/(4+3) = 12/7 hours to complete the job\n\nThe ratio 6/5 : 12/7 needs to be \"clean up\" a bit. Since we're dealing with fractions, we can use common denominators....\n\n6/5 : 12/7\n\n42/35 : 60/35\n\nNow we can eliminate the denominators...\n\n42 : 60\n\nAnd reduce the ratio....\n\n7 : 10\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50002\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n09 Feb 2015, 04:51\nBunuel wrote:\nMachine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?\n\nA. 5 to 1\nB. 10 to 7\nC. 1 to 5\nD. 7 to 10\nE. 9 to 10\n\nKudos for a correct solution.\n\nVERITAS PREP OFFICIAL SOLUTION:\n\nIn this ratio problem, it's helpful to begin with the ratio of each machine's RATES as opposed to times, as rates are additive when machines are working together. So if Machine X as a rate that's twice as fast as Machine Y, their rates have the ratio X:Y = 2:1. And the ratio of X to Z is 3:2, as X works 50% faster (and therefore would accomplish 3 jobs in the time that it takes for Z to complete 2). So to find a common three-way ratio, you can use X as the \"anchor\", and make the ratio of rates:\n\nX : Y : Z = 6 : 3 : 4\n\nSo X and Y working together would complete 10 jobs in the time that it would take Y and Z working together to complete 7 jobs, for a ratio of 10 : 7 in their respective rates. But since the question asks for times, not rates, you'll need to invert the ratio to 7 : 10, answer choice D.\n_________________\nVP\nJoined: 07 Dec 2014\nPosts: 1104\nMachine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n20 Nov 2015, 18:32\nlet 2,4,3=respective times of X,Y,Z\n1/2,1/4,1/3=respective rates of X,Y,Z\nX+Z rate/Y+Z rate=(5/6)/(7/12)=10:7\ninverting, X+Z time/Y+Z time=(6/5)/(12/7)=7:10\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 3896\nLocation: United States (CA)\nRe: Machine X can complete a job in half the time it takes Machine Y to co\u00a0 [#permalink]\n\n### Show Tags\n\n28 Mar 2018, 10:39\nBunuel wrote:\nMachine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?\n\nA. 5 to 1\nB. 10 to 7\nC. 1 to 5\nD. 7 to 10\nE. 9 to 10\n\nWe can let the time of machine Y to complete the job = y, so the time of machine X to complete the job is 0.5y = (\u00bd)y and the time of machine Z to complete the job = (1.5)(0.5y) = 0.75y = (3/4)y. .\n\nSince rate is inverse of time, the rate of Y = 1/y, the rate of X = 1/[(\u00bd)y] = 2/y and the rate of Z = 1/[(3/4)y] = 4/(3y).\n\nThus, the amount of time it will take Machine X and Z to complete the job is\n\n1/[2/y + 4/(3y)] = 1/[6/(3y) + 4/(3y)] = 1/[10/(3y)] = 3y/10\n\nSimilarly, the amount of time it will take Machine Y and Z to complete the job is\n\n1/[1/y + 4/(3y)] = 1/[3/(3y) + 4/(3y)] = 1/[7/(3y)] = 3y/7\n\nTherefore, the ratio is (3y/10)/(3y/7) = (1/10)/(1/7) = 7/10.\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: Machine X can complete a job in half the time it takes Machine Y to co &nbs [#permalink] 28 Mar 2018, 10:39\nDisplay posts from previous: Sort by", "date": "2018-10-20 06:24:18", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/machine-x-can-complete-a-job-in-half-the-time-it-takes-machine-y-to-co-192604.html", "openwebmath_score": 0.6865741610527039, "openwebmath_perplexity": 2657.5918376725567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8757869819218865}}
{"url": "https://math.stackexchange.com/questions/3295703/finding-the-length-of-a-side-and-comparing-two-areas-from-a-given-figure", "text": "# Finding the length of a side and comparing two areas from a given figure\n\nThe following figure came in two national-exam paper (different question in each paper).\n\nFirst Question:\n\nWhat is the length of $$AE$$?\n\nChoices:\n\nA) $$\\frac{5}{4}$$ cm\n\nB) $$\\frac{\\sqrt{41}}{5}$$ cm\n\nC) $$\\frac{7}{5}$$ cm\n\nD) $$\\frac{15}{8}$$ cm\n\nSecond Question\n\nCompare:\n\nFirst value: area of $$\\triangle ABC$$\n\nSecond value: area of $$\\triangle CDE$$\n\nChoices:\n\nA) First value > Second value\n\nB) First value < Second value\n\nC) First value = Second value\n\nD) Given information is not enough\n\nMy Attempt:\n\nFinding the equation of $$AC$$ and $$BD$$ in order to find the coordinates of E:\n\nWe can assume that $$C$$ is the origin $$(0,0)$$ and $$A$$ is $$(3,4)$$, then the equation of $$AC$$ is $$y=\\frac{4}{3}x$$, and the equation of $$BD$$ is $$y=-\\frac{4}{5}x+4$$.\n\nNext, solve the system of equations for $$x$$ and $$y$$, we end up with the coordinates of $$E(\\frac{15}{8},\\frac{5}{2})$$.\n\nTo solve the first question, we can use the distance between two points formula,\n\ntherefore $$AE=\\sqrt{(3-\\frac{15}{8})^2+(4-\\frac{5}{2})^2}=\\frac{15}{8}$$ cm.\n\nHence D is the correct choice.\n\nTo solve the second question, we can use the area of triangle formula,\n\narea of $$\\triangle ABC=\\frac{1}{2}\\times 3\\times 4=6$$ cm$$^2$$.\n\narea of $$\\triangle CDE=\\frac{1}{2}\\times 5\\times \\frac{5}{2}=6.25$$ cm$$^2$$.\n\nHence B is the correct choice.\n\nIn the exam, calculators are not allowed. It is simple to determine to coordinates of $$E$$ without a calculator, but it will take time!\n\nFor the first question, I think there is a good way to solve, maybe $$BD$$ divides $$AC$$ into two parts ($$AE$$ and $$CE$$) in a ratio, this ratio can be calculated somehow.\n\nFor the second question, I think also a good way to think, like moving $$D$$ will make $$\\triangle CDE$$ bigger or smaller, while $$\\triangle ABC$$ will be unchanged. But this has a limitation since the areas $$6$$ and $$6.25$$ cm$$^2$$ are close, it will not be obvious to us.\n\nIf we have $$75$$ seconds (on average) to solve each of these questions, then how can we solve them? [remember that in some questions in the national exam, we do not need to be accurate $$100$$%].\n\nAny help will be appreciated. Thanks.\n\n\u2022 $AB\\perp BC$? $AC\\perp BD$? $AB||DC$? \u2013\u00a0Michael Rozenberg Jul 17 at 13:03\n\u2022 @MichaelRozenberg Yes. \u2013\u00a0Hussain-Alqatari Jul 17 at 13:05\n\u2022 If so, this quadrilateral does not exist because we need $4^2=3\\cdot5,$ which is wrong. \u2013\u00a0Michael Rozenberg Jul 17 at 13:07\n\u2022 @MichaelRozenberg $AB\\perp BC$, $AB || DC$. But $AB$ is not $\\perp BC$ \u2013\u00a0Hussain-Alqatari Jul 17 at 13:10\n\u2022 @MichaelRozenberg Yes - No - Yes. \u2013\u00a0Hussain-Alqatari Jul 17 at 13:19\n\nTriangles $$ABE$$ and $$CDE$$ are similar, with ratio $$3:5$$.\n\nHence $$AE=\\frac{3}{8}AC=\\frac{3}{8}\\cdot 5=\\frac{15}{8}$$\n\nBy the same similarity, if $$h$$ is the height of triangle $$CDE$$ from $$E$$ to $$CD$$, then $$h=\\frac{5}{8}BC=\\frac{5}{8}\\cdot 4=\\frac{5}{2}$$ hence the area of triangle $$CDE$$ is $$\\frac{1}{2}\\cdot CD\\cdot h=\\frac{1}{2}\\cdot 5\\cdot \\frac{5}{2}=\\frac{25}{4}=6.25$$ whereas the area of triangle $$ABC$$ is $$6$$.\n\n\u2022 quasi your way looks very elegant. if I may ask how did you figure out $AE = \\frac{3}{8}AC$ directly with out setting up a proportion ? Is there some nice trick here? \u2013\u00a0rsadhvika Jul 17 at 13:24\n\u2022 @rsadhvika:$\\;$Since $AE:EC=3:5$, it follows that $$AE:AC=AE:(AE+EC)=3:(3+5)=3:8$$ \u2013\u00a0quasi Jul 17 at 13:29\n\u2022 Oh nice nice! With that in mind we can directly do: $AE = \\frac{3}{8}AC$ and $EC = \\frac{5}{8}AC$. I see now.. Thank you so much for explaining this you're awesome :) \u2013\u00a0rsadhvika Jul 17 at 13:31\n\nFor first question you may use similar triangles - $$\\triangle AEB$$ and $$\\triangle CED$$: $$\\dfrac{3}{5} = \\dfrac{x}{5-x}$$\n\nFor the first part use similarity of triangles to simplify computation.\n\nYou have similar triangles $$ABE$$ and $$CDE$$ with ratio of sides being $$3$$ to $$5$$\n\nYou also know $$AC=5$$ so you find $$AE=5\\times \\frac {3}{8}=\\frac {15}{8}$$ which is choice $$D$$\n\nYour have solved the second part correctly.", "date": "2019-08-20 05:38:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3295703/finding-the-length-of-a-side-and-comparing-two-areas-from-a-given-figure", "openwebmath_score": 0.840262770652771, "openwebmath_perplexity": 429.80906522158324, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137916170774, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.875770743312759}}
{"url": "https://math.stackexchange.com/questions/466757/why-can-we-interchange-summations", "text": "# Why can we interchange summations?\n\nSuppose we have the following\n\n$$\\sum_{i=1}^{\\infty}\\sum_{j=1}^{\\infty}a_{ij}$$\n\nwhere all the $a_{ij}$ are non-negative.\n\nWe know that we can interchange the order of summations here. My interpretation of why this is true is that both this iterated sums are rearrangements of the same series and hence converge to the same value, or diverge to infinity (as convergence and absolute convergence are same here and all the rearrangements of an absolutely convergent series converge to the same value as the series).\n\nIs this interpretation correct. Or can some one offer some more insightful interpretation of this result?\n\nPlease note that I am not asking for a proof but interpretations, although an insightful proof would be appreciated.\n\n\u2022 Another interpretation would be to use Tonnelli's theorem for counting measure. \u2013\u00a0TZakrevskiy Aug 13 '13 at 17:44\n\u2022 The double sum is not a rearrangement of a single sum. At least, it's not of the form $\\sum_{i=1}^\\infty a_{\\sigma(i)}$ where $\\sigma$ is a permutation of the positive integers. However, let $I$ be your iterated sum and $J$ be the sum with the order interchanged. You can first show that $J\\le I$. Then you can show that for any $\\epsilon>0$, you have $J>I-\\epsilon$. (This is mostly done by \"lopping off small tails\". Note each row and column is summable) \u2013\u00a0David Mitra Aug 13 '13 at 18:27\n\u2022 We can show that it is rearrangement if you consider a bijection between $\\mathbb{N}\\times\\mathbb{N}$ and $\\mathbb{N}$. \u2013\u00a0Vishal Gupta Aug 14 '13 at 5:15\n\nThis isn't a proof, but perhaps can give you the insight you are looking for. Any nondecreasing sequence converges to its (possibly infinite) supremum. Thus a series of nonnegative terms converges to the supremum of its partial sums and interchanging the order of summation doesn't affect the value of the supremum: there is no accidental cancellation of terms of opposite sign.\n\n\u2022 Good thought. Then we only need to apply decoupling to interchange summations in this case. \u2013\u00a0Chival Sep 27 '14 at 20:13\n\u2022 (+1)Consider the real valued function $f(a_{1},a_{2},...a_{n})$ then does the value of $\\sum_{a_{1}} \\sum_{a_{2}}...\\sum_{a_{n}} f(a_{1},a_{2},...a_{n})$ depends upon the order of sum notions. In other words I want to know whether there is any theorem which talks about this interchanging of order of sum notions ? \u2013\u00a0NewBornMATH Apr 9 '19 at 15:11\n\nThe double sum you are asking about can be considered to be the sum of all terms of the infinite array of numbers $a_{ij}$:\n\n$$\\begin{pmatrix} a_{11} & a_{12} & \\cdots & a_{1j} & \\cdots\\\\ a_{21} & a_{22} & \\cdots & a_{2j} & \\cdots\\\\ \\vdots& \\vdots & &\\vdots\\\\ a_{i1} & a_{i2} & \\cdots & a_{ij} & \\cdots\\\\ \\vdots & \\vdots & & \\vdots \\\\ \\end{pmatrix},$$\n\nwhere the sum is taken with the particular order of first adding all elements of particular rows and then adding the resulting \"row totals\". The result you are citing claims that if all $a_{ij}$ is nonnegative, then it matters not in which order you add the entries of the infinite matrix, be that rows first and then row totals or columns first and then column totals.\n\nI can't say your interpretation is right on the mark, because the issue is that even though one adds the entries of the same infinite array, the order of summation may result in summing the terms of different series in actuality. As the other answers point out this result holds essentially because we have no terms diminishing the total sum. Saying \"both this iterated sums are rearrangements of the same series and hence converge to the same value, or diverge to infinity\" really does not make any emphasis on this advantage, which I believe is sweeped under the phrase \"same series\".\n\nI'd like to mention yet another (measure theoretical) interpretation of this result, which I recently encountered in Rudin's Real and Complex Analysis (p. 23). In the book Rudin gives this result as a corollary of Lebesgue Monotone Convergence Theorem applied to series of functions.\n\nTheorem: Let $X$ be a measure space. If $\\forall n: f_n:X\\to [0,\\infty]$ is measurable, then\n\n$$\\int_X \\sum_n f_n d\\mu= \\sum_n \\int_X f_n d\\mu (\\ast).$$\n\nCorollary: Let $X:=\\{x_1,x_2,...,x_n,...\\}$ be a countable set and $\\mu:\\mathcal{M}_X:=\\mathcal{P}(X)\\to[0,\\infty]$ be the counting measure:\n\n$$\\mu(E):= \\begin{cases} |E|&, \\mbox{ if} |E|<\\infty\\\\ \\infty&, \\mbox{ if} |E|=\\infty \\end{cases}.$$\n\nIf $\\forall i,j:a_{ij}\\geq0$, then\n\n$$\\sum_i \\sum_j a_{ij}=\\sum_j \\sum_i a_{ij}.$$\n\nProof:\n\n\u2022 Set $\\forall j: f_j:X\\to[0,\\infty], f_j(x):=\\sum_i a_{ij}\\chi_{\\{x_i\\}}(x)$; and $\\forall i: \\bar{f_i}:X\\to[0,\\infty], \\bar{f_i}(x):=\\sum_j a_{ij}\\chi_{\\{x_i\\}}(x).$ Then $\\sum_j f_j=\\sum_i \\bar{f_i}$. Indeed, let $x_{i_0}\\in X$. Then\n\n$$\\sum_j f_j(x_{i_0})= \\sum_j \\sum_i a_{ij} \\chi_{\\{x_i\\}}(x_{i_0})= \\sum_j a_{{i_0}j},$$\n\nand\n\n$$\\sum_i \\bar{f_i}(x_{i_0})=\\bar{f_{i_0}}(x_{i_0})=\\sum_j a_{{i_0}j}.$$\n\n\u2022 Observe that the (inner) sum over $i$ is the integral of $f_j$ and the (inner) sum over $j$ is the integral of $\\bar{f_i}$. Then we have:\n\n$$\\sum_j\\sum_i a_{ij}= \\sum_j \\int_X f_j d\\mu\\stackrel{(\\ast)}{=}\\int_X \\sum_j f_j d\\mu=\\int_X \\sum_i \\bar{f_i}d\\mu\\stackrel{(\\ast)}{=}\\sum_i\\int_X\\bar{f_i}d\\mu =\\sum_i \\sum_j a_{ij}.$$\n\nNow consider the interpretation induced by the above discourse, viz., the (countable) combinations of characteristic functions behave in such a way that one can concentrate nonnegative \"weights\" of individual points. In the above proof $f_j$ puts a single weight on each point, while $\\bar{f_i}$ concentrates all the weights of the point $x_i$.\n\nWell, I think that interpretation is a bit circular, as the reason absolute convergence allows for rearrangements without changing the limit is precisely because of this point.\n\nThe gist of it is that no cancellations due to sign are possible. I think it is very illuminating to consider the counterexample for the rearrangement theorem when you do not have absolute convergence, but still have conditional convergence (Rearrangements can lead to any limiting value!).\n\nYou can add up as many negative terms as you want until you are happy (as it does not absolutely converge, you can continue adding until you are lower than any value in the real line you desire), and likewise for the positive terms (note, if adding the rest of the positive terms gives a finite value, then actually the whole sum didn't conditionally converge, and in fact will be $-\\infty$). This procedure of adding negative terms until your sum is below $L$, and then adding positive terms until the sum is above $L$, and etc. will converge because conditional convergence requires the things you are summing to go to $0$ at the end of the day.\n\nOkay, but to the non-negative double sum. Towards proving the double sum result (since you are only considering two orderings), I suppose one cute thing you could try is start with rows (arrange the numbers on the upper right quadrant for this picture). Let's assume the row sums converge first. Now, for each row sum, take out the first term of each, and set it aside, adding them upwards (the first column sum). Note that of course, the value of the sum didn't change, because you took out values one at a time, putting them into the column sum, and at no point did anything funny happen because only finite sums were involved (the full sum, the partial column sum, and the full sum minus the finitely many elements you took out). Now continue this procedure, extracting the second column sum, and etc. At the \"end\" you have what you want to prove.\n\nTwo limiting procedures hidden: When you \"finish\" the first column sum, you have to take a limit procedure. When you \"finish\" all the columns, there's another limit there. The issue if you include negative terms is that the first column sum might already diverge. And even if all the column sums converge, summing the column sums might converge to an entirely different value! I think to really finish this proof idea, you should use the fact that the values you are working with vary monotonely at each step.", "date": "2021-02-28 09:52:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/466757/why-can-we-interchange-summations", "openwebmath_score": 0.9120059609413147, "openwebmath_perplexity": 215.81073139655032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137900379085, "lm_q2_score": 0.8918110432813419, "lm_q1q2_score": 0.8757707426103717}}
{"url": "https://math.stackexchange.com/questions/2783783/counting-the-number-of-ways-a-test-can-be-answered-where-at-least-one-t-f-and-at", "text": "# Counting the number of ways a test can be answered where at least one T/F and at least one multiple choice question must be answered\n\nAn exam has $15$ questions: eight true/false questions and seven multiple choice questions. You are asked to answer five of them, but your professor requires you to answer at least one true/false question and at least one multiple choice question. How many ways can you choose the questions you plan to answer.\n\nstep1. I chose $1$ question of true false type. $8$ ways of doing so.\n\nstep2. I chose $1$ question of MCQ type. $7$ ways of doing so.\n\nstep3. I chose $3$ questions of the remaining $13$ questions. $13 \\cdot 12 \\cdot 11$ ways of doing so.\n\nstep4. Hence, $8 \\cdot 7 \\cdot 13 \\cdot 12 \\cdot 11$ gives number of sequences of $5$ questions from a collection of $15$ questions such that there is at least one question from t/f type and mcq type each.\n\nstep5. Let the number of collections of $5$ questions from a collection of $15$ questions such that there is at least one question from t/f type and mcq type each be $N$.\n\nstep6. $N \\cdot 5! = 8 \\cdot 7 \\cdot 11 \\cdot 12 \\cdot 13$\n\nstep7. $N = \\dfrac{8 \\cdot 7 \\cdot 11 \\cdot 12 \\cdot 13}{5!}$\n\nBut this number above is not even a whole number. Where did I go wrong?\n\n\u2022 when you chose one question you are also chosing four out of remaining questions you have to multiply them May 16 '18 at 14:27\n\u2022 Please read this tutorial on how to typeset mathematics on this site. May 16 '18 at 14:41\n\u2022 You choose tf first and then mc. You are dividing by all ways you can arrange 5. But you should only divide but the number of ways that tf comes first and then mc. May 16 '18 at 16:02\n\nYou don't have $5!$ ways of arranging the $5$ questions. You choose the questions specifically so that the first question was a TF and the second was MC. So the numbers of ways to arrange your five question must be that the first is TF and the second MC.\n\nThat is, the reason you divide by $5!$ in the first place is that you need to account that picking $ABCDE$ is the same thing as picking $CEDAB$. But unless $C$ is TF and E is MFC picking $CEDAB$ was never an option.\n\nThe problem is the number of ways or arranging the $5$ questions depends on knowing how many TF and MC questions there are.\n\nBut to choose the number of way you can have $n$ TF question and $5-n$ MC questions first. You can have $1$ to $4$ TF questions and there are ${8 \\choose n}$ ways of picking the $n$ TF question s an there are ${7\\choose 5-n}$ ways of choosing the $5-n$ MC questions.\n\nSo the answer is $\\sum_{n=1}^4 {8\\choose n}{7\\choose 5-n}=$\n\n${8\\choose 1}{7\\choose 4}+{8\\choose 2}{7\\choose 3}+{8\\choose 3}{7\\choose 2}+{8\\choose 4}{7\\choose 1}=$\n\n$8\\frac {7!}{3!4!} + \\frac {8!7!}{2!6!3!4!} + \\frac {8!7!}{5!3!5!2!} + 7*\\frac {8!}{4!4!}=$\n\n$8*(5*7) + (4*7)(5*7) + (8*7)(7*3) + 7*(2*7*5)=$\n\n$7*2(20 + 70 + 84 + 35)= 2926$\n\n======\n\n\"The easier way is to count the number of ways you can choose any five questions, then subtract the ones that are all the same kind.\"\n\nD'oh!\n\n#any 5 - #all 5 TF - #all 5 mc =\n\n${15 \\choose 5} - {8\\choose 5} - {7\\choose 5}=$\n\n$\\frac {15!}{10!5!} - \\frac {8!}{5!3!} - \\frac {7!}{5!2!} =$\n\n$3003 - 56- 21 = 2926$.\n\nYou overcount all of the cases. For a case where you answer two true/false questions and three multiple choice questions, there are two way to pick which one you count in step 1 and three in step 2, so this will get counted six times.\n\nThe easier way is to count the number of ways you can choose any five questions, then subtract the ones that are all the same kind.", "date": "2021-09-22 11:37:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2783783/counting-the-number-of-ways-a-test-can-be-answered-where-at-least-one-t-f-and-at", "openwebmath_score": 0.6880128979682922, "openwebmath_perplexity": 340.13082712648674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419703960399, "lm_q2_score": 0.8856314647623016, "lm_q1q2_score": 0.8757495626602854}}
{"url": "https://de.mathworks.com/help/symbolic/fplot3.html", "text": "# fplot3\n\nPlot 3-D parametric curve\n\n## Syntax\n\nfplot3(xt,yt,zt)\nfplot3(xt,yt,zt,[tmin tmax])\nfplot3(___,LineSpec)\nfplot3(___,Name,Value)\nfplot3(ax,___)\nfp = fplot3(___)\n\n## Description\n\nexample\n\nfplot3(xt,yt,zt) plots the parametric curve xt = x(t), yt = y(t), and zt = z(t) over the default interval \u20135\u00a0<\u00a0t\u00a0<\u00a05.\n\nexample\n\nfplot3(xt,yt,zt,[tmin tmax]) plots xt = x(t), yt = y(t), and zt = z(t) over the interval tmin < t < tmax.\n\nexample\n\nfplot3(___,LineSpec) uses LineSpec to set the line style, marker symbol, and line color.\n\nexample\n\nfplot3(___,Name,Value) specifies line properties using one or more Name,Value pair arguments. Use this option with any of the input argument combinations in the previous syntaxes. Name,Value pair settings apply to all the lines plotted. To set options for individual lines, use the objects returned by fplot3.\nfplot3(ax,___) plots into the axes object ax instead of the current axes gca.\n\nexample\n\nfp = fplot3(___) returns a parameterized function line object. Use the object to query and modify properties of a specific parameterized line. For details, see ParameterizedFunctionLine Properties.\n\n## Examples\n\n### Plot 3-D Parametric Line\n\nPlot the 3-D parametric line\n\n$\\begin{array}{c}x=\\mathrm{sin}\\left(t\\right)\\\\ y=\\mathrm{cos}\\left(t\\right)\\\\ z=t\\end{array}$\n\nover the default parameter range [-5 5].\n\nsyms t xt = sin(t); yt = cos(t); zt = t; fplot3(xt,yt,zt)\n\n### Specify Parameter Range\n\nPlot the parametric line\n\n$\\begin{array}{c}x={e}^{-t/10}\\mathrm{sin}\\left(5t\\right)\\\\ y={e}^{-t/10}\\mathrm{cos}\\left(5t\\right)\\\\ z=t\\end{array}$\n\nover the parameter range [-10 10] by specifying the fourth argument of fplot3.\n\nsyms t xt = exp(-t/10).*sin(5*t); yt = exp(-t/10).*cos(5*t); zt = t; fplot3(xt,yt,zt,[-10 10])\n\n### Change Line Properties and Display Markers\n\nPlot the same 3-D parametric curve three times over different intervals of the parameter. For the first curve, use a linewidth of 2. For the second, specify a dashed red line style with circle markers. For the third, specify a cyan, dash-dot line style with asterisk markers.\n\nsyms t fplot3(sin(t), cos(t), t, [0 2*pi], 'LineWidth', 2) hold on fplot3(sin(t), cos(t), t, [2*pi 4*pi], '--or') fplot3(sin(t), cos(t), t, [4*pi 6*pi], '-.*c')\n\n### Plot 3-D Parametric Line Using Symbolic Functions\n\nPlot the 3-D parametric line\n\n$\\begin{array}{c}x\\left(t\\right)=\\mathrm{sin}\\left(t\\right)\\\\ y\\left(t\\right)=\\mathrm{cos}\\left(t\\right)\\\\ z\\left(t\\right)=\\mathrm{cos}\\left(2t\\right).\\end{array}$\n\nsyms x(t) y(t) z(t) x(t) = sin(t); y(t) = cos(t); z(t) = cos(2*t); fplot3(x,y,z)\n\n### Plot Multiple Lines on Same Figure\n\nPlot multiple lines either by passing the inputs as a vector or by using hold on to successively plot on the same figure. If you specify LineSpec and Name-Value arguments, they apply to all lines. To set options for individual lines, use the function handles returned by fplot3.\n\nDivide a figure into two subplots using subplot. On the first subplot, plot two parameterized lines using vector input. On the second subplot, plot the same lines using hold on.\n\nsyms t subplot(2,1,1) fplot3([t -t], t, [t -t]) title('Multiple Lines Using Vector Inputs') subplot(2,1,2) fplot3(t, t, t) hold on fplot3(-t, t, -t) title('Multiple Lines Using Hold On Command') hold off\n\n### Modify 3-D Parametric Line After Creation\n\nPlot the parametric line\n\n$\\begin{array}{l}x={e}^{-|t|/10}\\mathrm{sin}\\left(5|t|\\right)\\\\ y={e}^{-|t|/10}\\mathrm{cos}\\left(5|t|\\right)\\\\ z=t.\\end{array}$\n\nProvide an output to make fplot return the plot object.\n\nsyms t xt = exp(-abs(t)/10).*sin(5*abs(t)); yt = exp(-abs(t)/10).*cos(5*abs(t)); zt = t; fp = fplot3(xt,yt,zt)\n\nfp = ParameterizedFunctionLine with properties: XFunction: [1x1 sym] YFunction: [1x1 sym] ZFunction: [1x1 sym] Color: [0 0.4470 0.7410] LineStyle: '-' LineWidth: 0.5000 Show all properties \n\nChange the range of parameter values to [-10 10] and the line color to red by using the TRange and Color properties of fp respectively.\n\nfp.TRange = [-10 10]; fp.Color = 'r';\n\n### Add Title and Axis Labels and Format Ticks\n\nFor $t$ values in the range $-2\\pi$ to $2\\pi$, plot the parametric line\n\n$\\begin{array}{c}x=t\\\\ y=t/2\\\\ z=\\mathrm{sin}\\left(6t\\right).\\end{array}$\n\nAdd a title and axis labels. Create the x-axis ticks by spanning the x-axis limits at intervals of pi/2. Display these ticks by using the XTick property. Create x-axis labels by using arrayfun to apply texlabel to S. Display these labels by using the XTickLabel property. Repeat these steps for the y-axis.\n\nTo use LaTeX in plots, see latex.\n\nsyms t xt = t; yt = t/2; zt = sin(6*t); fplot3(xt,yt,zt,[-2*pi 2*pi],'MeshDensity',30) view(52.5,30) xlabel('x') ylabel('y') title('x=t, y=t/2, z=sin(6t) for -2\\pi < t < 2\\pi') ax = gca; S = sym(ax.XLim(1):pi/2:ax.XLim(2)); ax.XTick = double(S); ax.XTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false); S = sym(ax.YLim(1):pi/2:ax.YLim(2)); ax.YTick = double(S); ax.YTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false);\n\n### Create Animations\n\nCreate animations by changing the displayed expression using the XFunction, YFunction, and ZFunction properties and then by using drawnow to update the plot. To export to GIF, see imwrite.\n\nBy varying the variable i from 0 to 4\u03c0, animate the parametric curve\n\n$\\begin{array}{l}x=t+\\mathrm{sin}\\left(40t\\right)\\\\ y=-t+\\mathrm{cos}\\left(40t\\right)\\\\ z=\\mathrm{sin}\\left(t+i\\right).\\end{array}$\n\nTo play the animation, click the image.\n\nsyms t fp = fplot3(t+sin(40*t),-t+cos(40*t), sin(t)); for i=0:pi/10:4*pi fp.ZFunction = sin(t+i); drawnow end\n\n## Input Arguments\n\ncollapse all\n\nParametric input for x-axis, specified as a symbolic expression or function. fplot3 uses symvar to find the parameter.\n\nParametric input for y-axis, specified as a symbolic expression or function. fplot3 uses symvar to find the parameter.\n\nParametric input for z-axis, specified as a symbolic expression or function. fplot3 uses symvar to find the parameter.\n\nRange of values of parameter, specified as a vector of two numbers. The default range is [-5 5].\n\nAxes object. If you do not specify an axes object, then fplot3 uses the current axes.\n\nLine style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.\n\nExample: '--or' is a red dashed line with circle markers\n\nLine StyleDescription\n-Solid line\n--Dashed line\n:Dotted line\n-.Dash-dot line\nMarkerDescription\n'o'Circle\n'+'Plus sign\n'*'Asterisk\n'.'Point\n'x'Cross\n'_'Horizontal line\n'|'Vertical line\n's'Square\n'd'Diamond\n'^'Upward-pointing triangle\n'v'Downward-pointing triangle\n'>'Right-pointing triangle\n'<'Left-pointing triangle\n'p'Pentagram\n'h'Hexagram\nColorDescription\n\ny\n\nyellow\n\nm\n\nmagenta\n\nc\n\ncyan\n\nr\n\nred\n\ng\n\ngreen\n\nb\n\nblue\n\nw\n\nwhite\n\nk\n\nblack\n\n### Name-Value Pair Arguments\n\nSpecify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.\n\nExample: 'Marker','o','MarkerFaceColor','red'\n\nThe properties listed here are only a subset. For a complete list, see ParameterizedFunctionLine Properties.\n\nNumber of evaluation points, specified as a number. The default is 23. Because fplot3 uses adaptive evaluation, the actual number of evaluation points is greater.\n\nLine color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB\u00ae uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nExample: 'blue'\n\nExample: [0 0 1]\n\nExample: '#0000FF'\n\nLine style, specified as one of the options listed in this table.\n\nLine StyleDescriptionResulting Line\n'-'Solid line\n\n'--'Dashed line\n\n':'Dotted line\n\n'-.'Dash-dotted line\n\n'none'No lineNo line\n\nLine width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.\n\nThe line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.\n\nMarker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.\n\nValueDescription\n'o'Circle\n'+'Plus sign\n'*'Asterisk\n'.'Point\n'x'Cross\n'_'Horizontal line\n'|'Vertical line\n'square' or 's'Square\n'diamond' or 'd'Diamond\n'^'Upward-pointing triangle\n'v'Downward-pointing triangle\n'>'Right-pointing triangle\n'<'Left-pointing triangle\n'pentagram' or 'p'Five-pointed star (pentagram)\n'hexagram' or 'h'Six-pointed star (hexagram)\n'none'No markers\n\nMarker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nMarker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nExample: [0.3 0.2 0.1]\n\nExample: 'green'\n\nExample: '#D2F9A7'\n\nMarker size, specified as a positive value in points, where 1 point = 1/72 of an inch.\n\n## Output Arguments\n\ncollapse all\n\nOne or more parameterized line objects, returned as a scalar or a vector. You can use these objects to query and modify properties of a specific parameterized line. For details, see ParameterizedFunctionLine Properties.\n\n### Topics\n\nIntroduced in R2016a\n\n## Support\n\n#### Mathematical Modeling with Symbolic Math Toolbox\n\nGet examples and videos", "date": "2021-03-07 03:21:10", "meta": {"domain": "mathworks.com", "url": "https://de.mathworks.com/help/symbolic/fplot3.html", "openwebmath_score": 0.6026898622512817, "openwebmath_perplexity": 8837.70278856056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856414, "lm_q2_score": 0.894789457685656, "lm_q1q2_score": 0.8757417591632486}}
{"url": "https://math.stackexchange.com/questions/2209916/how-many-possible-passwords-of-a-four-digit-length-contain-at-least-one-uppercas", "text": "# How many possible passwords of a four digit length contain at least one uppercase character and at least one number?\n\nHow many possible passwords of a four digit length contain at least one uppercase character and at least one number?\n\n\u2022 95 total ACII Symbols\n\u2022 26 uppercase letters\n\u2022 26 lowercase letters\n\u2022 10 numbers\n\u2022 33 special characters\n\nSo I have calculated that there are $26$ possibilities for the uppercase letter digit and $10$ possibilities for the number digit and also a variation of all the other possibilites, thus:\n\n$26 \\cdot 10 \\cdot |var_2([1,95])| = 26 \\cdot 10 \\cdot 95^2 =2 346 500$\n\nUnfortunately this is the wrong solution. It should be $18700240$.\n\nQuestion: Were is my mistake?\n\n\u2022 Your mistake is that you counted the number of passwords where the first character is upper case, the second character is a number, and your remaining two characters are anything. You neglected to count the password \"3aBc\" as well as neglected to count the password \"133T\" etc... For a correct solution, apply inclusion-exclusion. There are $95^4$ total passwords. $(95-26)^4$ of which contain no capital letters so we remove those from the count. $(95-10)^4$ of which contain no numbers so we remove those too, but in doing so the passwords with neither got removed twice so we correct it by... \u2013\u00a0JMoravitz Mar 30 '17 at 8:44\n\nYour mistake is that you counted the number of passwords where the first character is upper case, the second character is a number, and your remaining two characters are anything. You neglected to count the password \"3aBc\" as well as neglected to count the password \"133T\" etc...\n\nFor a correct solution, apply inclusion-exclusion. There are $95^4$ total passwords. $(95-26)^4$ of which contain no capital letters so we remove those from the count. $(95-10)^4$ of which contain no numbers so we remove those too, but in doing so the passwords with neither got removed twice so we correct it by adding that amount back in.\n\nIn more detail, let $\\Omega$ be the set of passwords of length $4$ using these $96$ characters with no other restrictions. Let $A$ be the set of passwords with no capital letters. Let $B$ be the set of passwords with no numbers.\n\nWe are tasked with counting $|A^c\\cap B^c|$, i.e. the set of passwords where there is at least one capital letter and there is at least one number.\n\nBy demorgan's, $A^c\\cap B^c = (A\\cup B)^c = \\Omega\\setminus(A\\cup B)$\n\nWe have then by inclusion-exclusion $$|A^c\\cap B^c|=|\\Omega\\setminus(A\\cup B)|=|\\Omega|-|A|-|B|+|A\\cap B|$$\n\nWe find $|\\Omega|=95^4$, $|A|=(95-26)^4$, $|B|=(95-10)^4$, and $|A\\cap B|=(95-26-10)^4$ each by straightforward application of multiplication principle.\n\nThe final total is then\n\n$$95^4-69^4-85^4+59^4$$\n\n\u2022 What does the $c$ mean in $|A^c\\cap B^c|$? \u2013\u00a0jublikon Mar 30 '17 at 9:38\n\u2022 @jublikon the complementary set. Given a universal set $\\Omega$ one has $A^c:= \\{x\\in\\Omega~:~x\\notin A\\}=\\Omega\\setminus A$ \u2013\u00a0JMoravitz Mar 30 '17 at 9:48\n\nFor one, this does not account for the possible variations of order.\n\nHere's a short way to compute this: Let $\\Omega$ denote the set of all passwords, $U$ denote the set of passwords that include no uppercase letter, and $N$ the set of passwords that include no digit. Then, the set of passwords that include an uppercase letter and a number is $$\\Omega - (N \\cup C),$$ so the number of passwords satisfying the criteria is $$|\\Omega - (N \\cup C)| = |\\Omega| - |N \\cup C| .$$ Now, use the Inclusion-Exclusion principle to evaluate $|N \\cup C|$.\n\nYou have worked out the number of passwords that have, say, the first characters an uppercase letter and the second character a number. This does not allow for cases where, say, our first two spaces are lowercase characters and/or special characters and our uppercase letter and number fill the last two spaces of the password.\n\nBecause the question says \"at least $1$\" I would immediately be drawn to considering the the compliment case i.e. The passwords with no uppercase letters, no numbers or neither uppercase letters nor numbers.\n\n$$\\text{total unrestriced passwords}=95^4$$ $$\\text{passwords with no uppercase letters}=(95-26)^4=69^4$$ $$\\text{passwords with no numbers}=(95-10)^4=85^4$$ $$\\text{passwords with no uppercase letters or numbers}=(95-26-10)^4=59^4$$\n\nthen by the principle of inclusion-exclusion we subtract the two complement cases from the total and add back their intersection:\n\n$$\\text{desired count}=95^4-(69^4+85^4)+59^4=18\\,700\\,240\\tag{Answer}$$", "date": "2020-07-13 18:00:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2209916/how-many-possible-passwords-of-a-four-digit-length-contain-at-least-one-uppercas", "openwebmath_score": 0.710651695728302, "openwebmath_perplexity": 365.107649185972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712643239288, "lm_q2_score": 0.894789454880027, "lm_q1q2_score": 0.8757417525282729}}
{"url": "https://math.stackexchange.com/questions/2550128/convergent-sequences-and-accumulation-points", "text": "# Convergent sequences and accumulation points\n\nI am very confused about the relationship between convergent sequences and accumulation points.\n\nBasically, my question boils down to: if $(a_n)_{n\\in N}$ is a convergent sequence in $\\mathbb{R}$ then,\n\n(1) Does the set $\\{a_n\\}$ have exactly one accumulation point?\n\n(2) If so, does $(a_n)_{n\\in N}$ necessarily converge to the said accumulation point?\n\nI'm tempted to say no to (1), but I'm afraid that I'm missing something. My counter-example to (1) is $\\{a_n\\} = \\{ 4, 3, 2, 1, 0,0,0,...\\}$ (i.e. inserting $0$s after the 4th element). Then the set has no accumulation point and it converges to 0. Is that correct?\n\nIf I'm right about (1), let's assume that $\\{a_n\\}$ has exactly one accumulation point. Then does $(a_n)_{n\\in N}$ necessarily converge to the said accumulation point?\n\nLastly, is it possible that $\\{a_n\\}$ could have more than one accumulation point?\n\nUpdate:\n\nI define accumulation point as: Let $a$ be an accumulation point of $A$. Then $\\forall \\ \\epsilon >0$, $B_{\\epsilon}(a) \\setminus \\{a\\}$ contains an element of $A$.\n\n\u2022 What is your definition of accumulation point? \u2013\u00a0wjm Dec 4 '17 at 5:44\n\u2022 @Raptor updated. \u2013\u00a0user1691278 Dec 4 '17 at 5:57\n\u2022 The situation is quite simple. If a sequence $(a_n)_{n\\in\\Bbb N}$ is convergent, then every accumulation point of the set $\\{\\, a_n\\mid n\\in\\Bbb N\\,\\}$ is equal to the limit of the sequence (easy proof using an extracted sequence). This implies there is at most one accumulation point, but not that such an accumulation point exists. \u2013\u00a0Marc van Leeuwen Dec 4 '17 at 9:52\n\nI am calling $x$ an accumulation point of the set $A$ iff $(B(x,\\epsilon) \\cap A)\\setminus \\{x\\} \\neq \\emptyset$ for all $\\epsilon>0$.\n\nSuppose $a_n \\to a$.\n\nThe set $\\{a_n\\}_n$ can have at most one accumulation point which would have to be $a$. If $b \\neq a$, then there is some $\\epsilon>0$ such that $B(b,\\epsilon)$ contains a finite number of points hence $b$ cannot be an accumulation point.\n\nNote that the sequence $a_n = 1$ has $\\{a_n\\}_n = \\{1\\}$ which has no accumulation points.\n\nIn general, the set $\\{a_n\\}_n$ will have $a$ as an accumulation point iff for all $N$ there is some $n \\ge N$ such that $a_n \\neq a$.\n\nIf $\\{a_{n}\\}$ has an accumulation point, say, $a$, and $(a_{n})$ is convergent. Then choose some $n_{1}$ such that $a_{n_{1}}\\in B_{1}(a)-\\{a\\}$. Then choose some $n_{2}$ such that $B_{1/2}(a)-\\{a,a_{1},...,a_{n_{1}}\\}$, proceed in this way we have $a_{n_{k}}\\rightarrow a$. Since $(a_{n})$ is convergent, one has $a_{n}\\rightarrow a$.\n\nHere I use the following definition:\n\n$a$ is an accumulation point for $A$ if for every $\\delta>0$, $(B_{\\delta}(a)-\\{a\\})\\cap A\\ne\\emptyset$.\n\nAnd note that in the topology of ${\\bf{R}}$, being such an accumulation point also implies that $(B_{\\delta}(a)-\\{a\\})\\cap A$ contains infinitely many points.\n\n\u2022 If $\\{a_{n}\\}$ has two distinct accumulation points, say, $a$, $b$, one can find subsequences $(a_{n_{k}})$, $(a_{n_{l}})$ such that $a_{n_{k}}\\rightarrow a$ and $a_{n_{l}}\\rightarrow b$, a contradiction. \u2013\u00a0user284331 Dec 4 '17 at 5:54\n\u2022 Your counterexample $(a_n)$ is not a convergent sequence though... \u2013\u00a0user1691278 Dec 4 '17 at 6:04\n\u2022 I have edited, let's see if there is any mistake. \u2013\u00a0user284331 Dec 4 '17 at 6:13\n\nThe usual definition of an accumulation point (for a subset of $\\mathbb{R}$) is as follows:\n\nLet $A \\subseteq \\mathbb{R}$. We say that $a$ is an accumulation point of $A$ if for all $r > 0$ the set $B(a,r) \\cap A \\setminus \\{a\\}$ is nonempty. That is, every ball centered at $a$ contains a point of $A$ other than $a$ itself.\n\nEven if the sequence $(a_n)$ converges, the set $\\{ a_n \\}$ needn't have any accumulation points. For example, any constant set $\\{a, a, a, \\dotsc, \\} = \\{a\\}$ does not have any accumulation points (as no ball centered at $a$ contains any point of the set other than $a$, but the sequence $(a,a,a,\\dotsc)$ converges to $a$. On the other hand, if $\\{a_n\\}$ has an accumulation point, and $(a_n)$ is convergent, then the accumulation point is necessarily the limit.\n\n\u2022 Thank you. Does that mean it cannot have more than one accumulation point since limits are unique? \u2013\u00a0user1691278 Dec 4 '17 at 6:05\n\u2022 In $\\mathbb{R}$ with the usual topology, a convergent sequence can have at most one accumulation point, yes. As noted above, it needn't have even that. \u2013\u00a0Xander Henderson Dec 4 '17 at 14:38\n\nHere's a somewhat more general proof of this:\n\nLet $X$ be a Hausdorff topological space, and let $\\left\\{x_n\\right\\}_{n\\in\\mathbb N}$ be a sequence in $X$ that converges to $x\\in X$. Suppose that $y\\in X$ is an accumulation point of $\\left\\{x_n\\right\\}_{n\\in\\mathbb N}$ such that $x\\neq y$. Then, since $X$ is Hausdorff, there are disjoint open neighborhoods $U$ and $V$ of $x$ and $y$, respectively. Since $\\left\\{x_n\\right\\}_{n\\in\\mathbb N}$ converges to $x$, there is an $N\\in\\mathbb N$ such that $x_n\\in U$ whenever $n\\geq N$. However, this implies that there are at most $N-1$ elements of $\\left\\{x_n\\right\\}_{n\\in\\mathbb N}$ in $V$ different from $y$. Denote this set of finite elements by $\\left\\{y_n\\right\\}_{n=1}^m$. Again, since $X$ is Hausdorff, there are disjoint open neighborhoods $U_n$ and $V_n$ of $y_n$ and $y$, respectively. Then $\\bigcap_{n=1}^m V_n$ is an open neighborhood of $y$ containing no elements of $\\left\\{x_n\\right\\}_{n\\in\\mathbb N}$, which is a contradiction.\n\n\u2022 In the context of this question, it is rather confusing to use braces to designate the sequence, especially when you then immediately use the very same notation for the set. OP rightly uses parentheses for the sequence, and braces for the set. \u2013\u00a0Marc van Leeuwen Dec 4 '17 at 9:56\n\nYes, you are right. If $x_n$ is a convergent sequence, then\n\n\u2022 $\\{x_n\\}$ has an accumulations point if and only if $\\{x_n\\}$ is infinite.\n\u2022 $\\{x_n\\}$ has at most one accumulation point and if it has one, then it coincides with the limit.\n\nNone of these statements is true for non-convergent squences.", "date": "2019-08-21 13:53:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2550128/convergent-sequences-and-accumulation-points", "openwebmath_score": 0.9748454689979553, "openwebmath_perplexity": 96.84857195220182, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799410139921, "lm_q2_score": 0.9086178987887253, "lm_q1q2_score": 0.8757077048988552}}
{"url": "http://math.stackexchange.com/questions/214960/find-a-multiplicative-inverse", "text": "Find a multiplicative inverse\u2026\n\nFind a multiplicative inverse of $a=11$ modulo $m=13$.\n\nWhat is this saying?\n\nThis seems like such a simple question, I just don't understand what it is asking for.\n\nAn additional question related to this I have is:\n\nIf $a$ has a multiplicative inverse modulo $m$, explain why $\\gcd(a,m)=1$.\n\nIs this also saying $ab \\equiv 1 (mod \\space m)$? Do $a$ and $m$ have to be prime since the $\\gcd(a,m)=1$?\n\nThanks again.\n\n-\n\nIt's asking to find $b$ such that\n\n$$ab \\equiv 1 \\pmod m$$\n\nHint: Euclidean algorithm\n\nEdit:\n\nSuppose that $a$ has a multiplicative inverse mod $m$. That is, there is a number $b$ such that $ab = 1 + km$ (for some $k \\in \\mathbb{Z}$). Hence, $ab - km = 1$.\n\nIf $a$ and $m$ have a common divisor $d$, then $d$ is a divisor of $ba \u2212 km$ and hence of 1.\n\nTherefore $d=1$, which means that $\\gcd(a,m)=1$.\n\nAnd, no $a$ and $m$ don't have to be prime! Just co-prime.\n\n-\nSo, for $11b \\equiv 1 (mod 13)$ we get $b=6$ This doesn't seem like the multiplicative inverse. Where do I go from here? Thanks. \u2013\u00a0 student.llama Oct 16 '12 at 16:40\n@student.llama No, that's it! The modular multiplicative inverse of 11 modulo 13 is 6. \u2013\u00a0 Alex Oct 16 '12 at 16:42\nThat is the multiplicative inverse since $$11\\cdot 6=66=1+5\\cdot 13=1\\pmod {13}$$ \u2013\u00a0 DonAntonio Oct 16 '12 at 16:43\nOh, ok, thanks! Could you take a look at what I added to the OP, if you haven't already? \u2013\u00a0 student.llama Oct 16 '12 at 16:47\n@student.llama See my edit :) \u2013\u00a0 Alex Oct 16 '12 at 16:59\n\n$11^2=121\\equiv 4\\pmod{13}$\n\n$11^3\\equiv4\\cdot 11\\equiv 5$\n\n$11^4=(11^2)^2\\equiv 4^2=16\\equiv 3$\n\n$11^5\\equiv3\\cdot11=33\\equiv 7$\n\n$11^6\\equiv7\\cdot 11= 77=-1$\n\n$\\implies 11\\cdot 11^5\\equiv-1\\implies (11)^{-1}\\equiv -11^5\\equiv -7\\equiv 6\\pmod{13}$\n\nAlternatively, using convergent property of continued fraction,\n\n$\\frac{13}{11}=1+\\frac 2{11}=1+\\frac1{\\frac{11}2}=1+\\frac1{5+\\frac12}$\n\nSo, the last but one convergent is $1+\\frac15=\\frac 6 5$\n\nSo, $11\\cdot 6-13\\cdot 5=66-65=1\\implies 11\\cdot 6\\equiv 1\\pmod{13}$ $\\implies (11)^{-1}\\equiv 6\\pmod{13}$\n\n-", "date": "2014-07-23 04:26:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/214960/find-a-multiplicative-inverse", "openwebmath_score": 0.983903169631958, "openwebmath_perplexity": 307.0491500454139, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808753491773, "lm_q2_score": 0.8933094152856196, "lm_q1q2_score": 0.8756941355738489}}
{"url": "https://math.stackexchange.com/questions/3998488/need-help-with-the-general-formula-for-the-taylor-series-of-ex-sinx", "text": "# Need help with the general formula for the Taylor series of $e^x+\\sin(x)$\n\nA while ago, I asked about ways to derive the sigma notation for the infinite series of $$e^x+sin(x)$$.\n\n$$e^x+\\sin(x) = 1+2x+\\dfrac{x^2}{2!}+\\dfrac{x^4}{4!}+\\dfrac{2x^5}{5!}+\\dfrac{x^6}{6!}+\\dfrac{x^8}{8!}+\\dfrac{2x^9}{9!}+...$$\n\nClive Newstead brilliantly gave me two formulas.\n\nThere, he said that the pattern of the series, regarding the coefficient of the numerators, is as followed: 1,2,1,0,1,2,1,0\n\nHe said that the pattern is \"is periodic with period 4\"\n\nThis comment still puzzles me. I have tried to validate his formula and it is true, they are all correct, but I don't know why he can come up with succinct representation of the pattern:\n\n$$\\sum_{k=0}^{\\infty} \\left( \\dfrac{x^{4k}}{(4k)!} + \\dfrac{2x^{4k+1}}{(4k+1)!} + \\dfrac{x^{4k+2}}{(4k+2)!} \\right)$$\n\nHow does one realize that the power is related to $$4k, 4k+1$$ and $$4k+2$$ and $$4k+3$$?\n\nThis seems to relate to finding the general formula for a sequence or series of number, for example, trivially\n\n$$1, 3, 5, 7, ...$$ can be represented as $$2k+1$$\n\n$$0, 2, 4, 6, ...$$ can be represented as $$2k$$\n\nwith $$k$$ as the position of the term in the sequence.\n\nSo my thought is you to count the terms that have the numerator's coefficient as $$1$$, $$2$$ and $$0$$\n\nThe ones that have $$1$$ is $$1^{st}$$, $$3^{sd}$$, $$5^{th}$$, $$7^{th}$$, etc.\n\nThe ones that have $$2$$ is $$2^{sd}$$, $$6^{th}$$, $$10^{th}$$, $$14^{th}$$, etc.\n\nThe ones that have $$0$$ is $$4^{th}$$, $$8^{th}$$, $$12^{th}$$, $$16^{th}$$, etc.\n\nSo the difference between the positions of the terms that contain the coefficient $$1$$ is $$2$$\n\nSo the difference between the positions of the terms that contain the coefficient $$2$$ is $$4$$\n\nSo the difference between the positions of the terms that contain the coefficient $$0$$ is $$4$$\n\nI find that the general term for terms that contain $$1$$ is $$2k+1$$\n\nI find that the general term for terms that contain $$2$$ is $$4k-2$$\n\nI find that the general term for terms that contain $$0$$ is $$4k$$\n\nBut I don't know how to derive these into $$4k$$, $$4k+1$$, $$4k+2$$, $$4k+3$$\n\nI realize that this is similar to finding patterns of a sequence on an IQ test, but I don't know how to derive it. One my friend said I should look into Lagrange's interpolation, but I don't why this technique has anything to do with writing down the general terms for sequences and series.\n\nCould you help me on this?\n\nMethinks, this is a simple question. I am not good at writing down general term of a sequence yet.\n\n\u2022 note that the coefficients of the Maclaurin series for $\\sin x$ are periodic with period $4$ \u2013\u00a0J. W. Tanner Jan 24 at 22:33\n\u2022 The general formula for $\\sin(x)$ is $$\\sum_{k=0}^{\\infty}\\dfrac{x^{2n+1}}{(2n+1)!}$$, the coefficients of the numerators are all ones, so I don't quite understand what are you trying to say? Sorry, could you clarify more? \u2013\u00a0James Warthington Jan 24 at 23:17\n\u2022 I realize that he breaks terms that contain $1$ in the numerators into 2, $4k$ and $4k+2$, quite clever. \u2013\u00a0James Warthington Jan 24 at 23:23\n\u2022 @JamesWarthington: That is the formula for $\\sinh x$, not $\\sin x$. \u2013\u00a0TonyK Jan 24 at 23:39\n\u2022 Yeah, thanks, I forgot to add $(-1)^{n}$ \u2013\u00a0James Warthington Jan 24 at 23:44\n\n$$e^x = 1 + x + \\frac {x^2}{2} + \\frac {x^3}{3!} + \\cdots$$\n\nIt seem pretty natural to write this as\n$$e^x = \\sum_\\limits{n=0}^\\infty \\frac {x^n}{n!}$$\n\nBut we could look at pairs of terms.\n\n$$e^x = \\sum_\\limits{n=0}^\\infty \\left(\\frac {x^{2n}}{2n!}+\\frac {x^{2n+1}}{(2n+1)!}\\right)$$\n\nOr, even triples of terms.\n\n$$e^x = \\sum_\\limits{n=0}^\\infty \\left(\\frac {x^{3n}}{3n!}+\\frac {x^{3n+1}}{(3n+1)!}+\\frac {x^{3n+2}}{(3n+3)!}\\right)$$\n\nAnd do the same thing with $$\\sin x$$\n\n$$\\sin x = \\sum_\\limits{n=0}^\\infty (-1)^n \\frac {x^{2n+1}}{(2n+1)!} = \\sum_\\limits{n=0}^\\infty \\left(\\frac {x^{4n+1}}{(4n+1)!} - \\frac {x^{4n+3}}{(4n+3)!}\\right)$$\n\nNow find a representation of $$e^x$$ that plays nicely with our representation of $$\\sin x$$ add them together and you have what you show above.\n\n\u2022 Thanks, very instructive.I've understood! \u2013\u00a0James Warthington Jan 24 at 23:34\n\nRemember that for the series centered around the point b, the coefficient $$a_{n} = \\frac{f^{(n)}(b)}{n!}$$. So if $$f^{(4)}(x) = f(x)$$ then $$f^{(n)}(b)$$ is periodic with not necessarily prime period 4.\n\nThis is why you can be sure that the $$f^{(n)}(b)$$ is periodic with period 4 in your case.\n\nLet $$f(x) = e^{x}$$, then $$f'(x) = e^{x}$$, and so continuing to take derivatives $$f^{(4)}(x)=e^{x}$$.\n\nLet $$g(x) = sin(x)$$ then $$g''(x)=-sin(x)$$ and so $$g^{(4)}(x) = sin(x)$$\n\nSo for $$h(x)=f(x) + g(x)$$ we have $$h^{(4)}(x) = (f(x)+g(x))'''' = f^{(4)}(x) + g^{(4)}(x) = f(x) + g(x) = h(x)$$.\n\nIn this way you can see that the $$f^{(n)}(b)$$ has the period 4 structure.\n\n\u2022 What if, by continuously taking the derivatives, the function we are dealing with do not have periodic derivatives? For example $\\tan(x)$? \u2013\u00a0James Warthington Jan 25 at 0:06\n\u2022 Insightful comment, thank you, BTW! :) \u2013\u00a0James Warthington Jan 25 at 0:07\n\u2022 @JamesWarthington Then the coefficients will likely not have this periodic form, Of course what they do will depend on the function you are working with, they could still have some sort of pattern. \u2013\u00a0open problem Jan 25 at 0:09\n\u2022 So I guess it is not always possible to write a general formula for all power series? Am I correct? Or is it otherwise? \u2013\u00a0James Warthington Jan 25 at 0:11\n\u2022 To do so you will need a formula for the evaluation of the nth derivative of the function at the point you have centered your power series around. \u2013\u00a0open problem Jan 25 at 0:14\n\nThe mod $$4$$ periodicity can perhaps be most easily understood using the formula $$e^{ix}=\\cos x+i\\sin x$$, so that $$\\sin x={e^{ix}-e^{-ix}\\over2i}$$, which gives\n\n$$e^x+\\sin x={1\\over2i}\\sum_{n=0}^\\infty{(2i+i^n-(-i)^n)x^n\\over n!}$$\n\nSince $$i^4=1$$, the coefficients $$2i+i^n-(-i)^n$$ cycle through the values\n\n\\begin{align} 2i+i^0-(-i)^0&=2i+1-1=2i\\\\ 2i+i^1-(-i)^1&=2i+i-(-i)=4i\\\\ 2i+i^2-(-i)^2&=2i-1-(-1)=2i\\\\ 2i+i^3-(-i)^3&=2i-i-i=0 \\end{align}", "date": "2021-04-12 17:17:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3998488/need-help-with-the-general-formula-for-the-taylor-series-of-ex-sinx", "openwebmath_score": 0.8957089185714722, "openwebmath_perplexity": 243.02235231817065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808707404786, "lm_q2_score": 0.8933094110250331, "lm_q1q2_score": 0.8756941272802836}}
{"url": "https://math.stackexchange.com/questions/256178/how-to-handle-stronger-induction-hypothesis-strong-induction", "text": "# How to Handle Stronger Induction Hypothesis - Strong Induction\n\nI'm having trouble understanding strong induction proofs\n\nI understand how to do ordinary induction proofs and I understand that strong induction proofs are the same as ordinary with the exception that you have to assume that the theorem holds for all numbers up to and including some $n$ (starting at the base case) then we try and show: theorem holds for $n+1$.\n\nHow do you show this exactly.\n\nHere is a proof by induction:\n\nThm: $n\u22651, 1+6+11+16+\\dots+(5n-4) = (n(5n-3))/2$\n\nProof (by induction)\n\nBasis step: for $n=1: 5-4=(5-3)/2 \\Rightarrow 1=1$. The basis step holds\n\nInduction Step: Suppose that for some integer $k\u22651$, $$1+6+11+16+...+(5k-4) = \\frac{k(5k-3)}{2} \\qquad\\text{(inductive hypothesis)}$$\n\nWant to show: $$1+6+11+16+...+(5k-4)+(5(k+1)-4) = \\frac{(k+1)(5(k+1)-3)}{2}$$ so $$\\frac{k(5k-3)}{2} + (5(k+1)-4) = \\frac{(k+1)(5(k+1)-3)}{2}$$\n\nthen you just show that they are equal.\n\nSo how can I do the same proof using strong induction? What are the things I need to add/change in order for this proof to be a strong induction proof?\n\nYou wrote:\n\ni understand how to do ordinary induction proofs and i understand that strong induction proofs are the same as ordinary with the exception that you have to show that the theorem holds for all numbers up to and including some n (starting at the base case) then we try and show: theorem holds for $n+1$\n\nNo, not at all: in strong induction you assume as your induction hypothesis that the theorem holds for all numbers from the base case up through some $n$ and try to show that it holds for $n+1$; you don\u2019t try to prove the induction hypothesis.\n\nIn your example the simple induction hypothesis that the result is true for $n$ is already enough to let you prove that it\u2019s true for $n+1$, so there\u2019s neither need nor reason to use a stronger induction hypothesis. The proof by ordinary induction can be seen as a proof by strong induction in which you simply didn\u2019t use most of the induction hypothesis.\n\nI suggest that you read this question and my answer to it and see whether that clears up some of your confusion; at worst it may help you to pinpoint exactly where you\u2019re having trouble.\n\nAdded: Here\u2019s an example of an argument that really does want strong induction. Consider the following solitaire \u2018game\u2019. You start with a stack of $n$ pennies. At each move you pick a stack that has at least two pennies in it and split it into two non-empty stacks; your score for that move is the product of the numbers of pennies in the two stacks. Thus, if you split a stack of $10$ pennies into a stack of $3$ and a stack of $7$, you get $3\\cdot7=21$ points. The game is over when you have $n$ stacks of one penny each.\n\nClaim: No matter how you play, your total score at the end of the game will be $\\frac12n(n-1)$.\n\nIf $n=1$, you can\u2019t make any move at all, so your final score is $0=\\frac12\\cdot1\\cdot0$, so the theorem is certainly true for $n=1$. Now suppose that $n>1$ and the theorem is true for all positive integers $m<n$. (This is the strong induction hypothesis.) You make your first move; say that you divide the pile into a pile of $m$ pennies and another pile of $n-m$ pennies, scoring $m(n-m)$ points. You can now think of the rest of the game as splitting into a pair of subgames, one starting with $m$ pennies, the other with $n-m$.\n\nSince $m<n$, by the induction hypothesis you\u2019ll get $\\frac12m(m-1)$ points from the first subgame. Similarly, $n-m<n$, so by the induction hypothesis you\u2019ll get $\\frac12(n-m)(n-m-1)$ points from the second subgame. (Note that the two subgames really do proceed independently: the piles that you create in one have no influence on what you can do in the other.)\n\nYour total score is therefore going to be\n\n$$m(n-m)+\\frac12m(m-1)+\\frac12(n-m)(n-m-1)\\;,$$\n\nwhich (after a bit of algebra) simplifies to $\\frac12n(n-1)$, as desired, and the result follows by (strong) induction.\n\n\u2022 A question related to \"No, not at all: in strong induction you ... from the base case up through some n ...\". Agree with this statement, but I usually see proofs describing it without considering the base case, for example: \"Assume $P(x)$ holds $\\forall x, x\\le i$\", so strictly speaking this kind of statement is incorrect? (The correct one should change the range to $b\\le x\\le i$, where $b$ is the base?) Jul 23, 2019 at 8:19\n\u2022 Nice strong induction example! But BTW here is (IMHO) an even nicer non-induction solution. Model the coins as the vertices of a graph, two vertices adjacent iff the coins are in the same stack. Initially all coins are in one stack so we have the complete graph on $n$ vertices. Splitting into stacks of say $a$ and $b$ coins means removing all the edges between the remaining graphs $K_a$ and $K_b$; the number of edges removed is $ab$, which is the score for this step. Completing the process means removing all $n(n-1)/2$ original edges, so this is the final score. Apr 28 at 7:43\n\u2022 @David: That is indeed an elegant solution. And a student who looks closely can even see the structure of the strong induction proof in a sense \u2018hiding\u2019 in it: it could be recast as essentially the same induction, but in this form we can actually see the inevitable end result without having to go through the induction. Apr 28 at 17:23\n\u2022 Thanks @Brian! The graph theory solution also makes it clear why the score for a move (in a sense) has to be the product in order for this problem to work. Apr 28 at 22:57", "date": "2022-05-29 03:18:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/256178/how-to-handle-stronger-induction-hypothesis-strong-induction", "openwebmath_score": 0.7493741512298584, "openwebmath_perplexity": 206.73961014471035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808687241726, "lm_q2_score": 0.8933094110250333, "lm_q1q2_score": 0.8756941254790986}}
{"url": "https://math.stackexchange.com/questions/2368172/how-do-i-solve-x4-2x3-6x2-2x1-0/2368180", "text": "# How do I solve $x^4 -2x^3-6x^2-2x+1=0$? [closed]\n\n$$x^4 -2x^3-6x^2-2x+1=0$$ $\\left[\\text{Hint let } v = x + \\frac{1}{x}\\right]$\n\nI am stumped, and have no idea how to proceed. I have tried solving it, but have had no success.\n\nP.S: This question is meant to be solved, using only techniques for solving quadratic equations.\n\n\u2022 Do you mean \"find the roots\" or something like that? Have you searched for rational roots? Note: there is an obvious rational root. \u2013\u00a0lulu Jul 22 '17 at 15:56\n\u2022 \u2013\u00a0lab bhattacharjee Jul 22 '17 at 15:57\n\u2022 Ok, now there is a question. In which case my hint is valid, and there is no need for the substitution. \u2013\u00a0lulu Jul 22 '17 at 15:59\n\u2022 Doesn't \"solve\" mean \"find the roots\"? A difference in language? \u2013\u00a0Tobi Alafin Jul 22 '17 at 16:38\n\u2022 A simplistic way to find the roots is to graph the polynomial, look to see if there are any obvious rational roots, then divide the polynomial by the factors corresponding to rational roots ($x+1$ in this case). Repeat as many times as you can. \u2013\u00a0\u03a7p\u1e98 Jul 22 '17 at 19:07\n\nI think you meant for $$x^4 -2x^3-6x^2-2x+1=0$$\n\nObviously $x=0$ is not a solution. So we divide by $x^2$ and get\n\n$$x^2 -2x-6-2x^{-1}+x^{-2}=0$$ $$(x^2+2+x^{-2}) - 2(x+x^{-1}) -8=0$$ $$v^2-2v-8=0$$ $$(v-4)(v+2)=0$$\n\n\u2022 Thanks for the pointer. :) \u2013\u00a0Tobi Alafin Jul 22 '17 at 16:39\n\u2022 @TobiAlafin you're welcome :) \u2013\u00a0Yujie Zha Jul 22 '17 at 17:15\n\u2022 I'm just curious: how did you know to divide by $x^2$? Obviously the $v$ contained a $x^{-1}$ term, but one could make that pattern appear in a variety of ways\u2026 \u2013\u00a0gen-\u2124 ready to perish Jul 24 '17 at 7:36\n\u2022 @ChaseRyanTaylor Just observed that equation has $x$ interns with power of 4,3,2,1,0, and thus it would make it symmetrical to divide by $x^2$ and make it to be $x$ items with power of 2,1,0,-1,-2.. \u2013\u00a0Yujie Zha Jul 24 '17 at 11:05\n\u2022 Care explaining the downvote? \u2013\u00a0Yujie Zha Jul 24 '17 at 12:45\n\nsince $x=0$ is not a solution we can divide by $x^2$ $$x^2+\\frac{1}{x^2}-2\\left(x+\\frac{1}{x}\\right)-6=0$$ and no set $$t=x+\\frac{1}{x}$$ then you will get $$t^2=x^2+\\frac{1}{x^2}+2$$ and $$t^2-2=x^2+\\frac{1}{x^2}$$ and you have to solve $$t^2-2t-8=0$$\n\nI would not bother with the substitution. The only possible rational roots are $\\pm 1$ and it is easy to check that $-1$ works. A quick calculation shows that your polynomial is $$(x + 1)^2 \\,(x^2 - 4 x + 1)$$ and we are done.\n\n\u2022 Aren't solutions generally defined on $\\mathbb{R}$, and not on $\\mathbb{Q}$? \u2013\u00a0Tobi Alafin Jul 22 '17 at 17:29\n\u2022 Not following. I imagined you were looking for complex solutions, no? In any case, searching for rational solutions is a very easy and natural first step...as that solves the problem entirely (well, up to a trivial quadratic) then I don't see any point in looking for a more elaborate solution. \u2013\u00a0lulu Jul 22 '17 at 17:35\n\u2022 To be clear: there was no a priori reason why searching for rational roots should have solved the problem. But then, there was no a priori reason why this particular substitution should have solved it. For what it's worth: In general, there is a closed formula for the roots of quartics (granted, it isn't very nice). \u2013\u00a0lulu Jul 22 '17 at 17:42\n\n$x^4 -2x^3-6x^2-2x+1=0$\n\nDivide all terms by $x^2$\n\n$x^2-2x-6-\\dfrac{2}{x}+\\dfrac{1}{x^2}=0$\n\nRearrange in this way\n\n$\\left(x^2+\\dfrac{1}{x^2}\\right)-2\\left(x+\\dfrac{1}{x}\\right)-6=0$\n\nNow set $v=x+\\dfrac{1}{x}$\n\nsquaring you get\n\n$v^2=x^2+\\dfrac{1}{x^2}+2\\to x^2+\\dfrac{1}{x^2}=v^2-2$\n\nPlug in the equation\n\n$v^2-2 -2v -6=0\\to v^2-2v-8=0\\to v_1=4;\\;v_2=-2$\n\nAs we want to solve for $x$ two more steps\n\nFor $v=4\\to x+\\dfrac{1}{x}=4 \\to x^2-4x+1=0 \\to x= 2\\pm\\sqrt{3}$\n\nfor $v=-2\\to x+\\dfrac{1}{x}=-2\\to x^2+2x+1\\to x=-1$\n\nSometimes you can add and subtract the right thing and land in Pascal's triangle. The 4th row of Pascal's triangle is $1 4 6 4 1$, so add and subtract $6x^2+12x^2+6x$ to get\n\n$$x^4+4x^3+6x^2+4x+1 - 6x^3-12x^2-6x = 0$$\n\nor\n\n$$(x+1)^4 - 6x(x+1)^2=0.$$\n\nFactor out $(x+1)^2$ to get\n\n$$(x+1)^2((x+1)^2 - 6x) = 0$$\n\nor $$(x+1)^2(x^2-4x+1)=0.$$", "date": "2020-10-01 19:15:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2368172/how-do-i-solve-x4-2x3-6x2-2x1-0/2368180", "openwebmath_score": 0.6233899593353271, "openwebmath_perplexity": 366.578763377864, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.980280869588304, "lm_q2_score": 0.8933093961129794, "lm_q1q2_score": 0.8756941116330341}}
{"url": "https://math.stackexchange.com/questions/570003/how-many-unique-patterns-exist-for-a-n-times-n-grid/587729", "text": "# how many unique patterns exist for a $N\\times N$ grid\n\nI'm trying to figure out if there is a way to determine how many unique patterns exist for a given $N\\times N$ grid if you choose N points on the grid. For example, for a $2\\times 2$ grid we can get two unique patterns from the six possible combinations. The rest are just rotations and mirrors of the two unique patterns below\n\n[x] [x]\n\n[ ] [ ]\n\nand\n\n[x] [ ]\n\n[ ] [x]\n\nIs there a mathematical way of determining a unique number of patterns for a NxN grid where N=3,4,5,6,7,8?\n\nI figured for a 3x3, there are 14 unique patterns for picking 3 random points on the grid, but it gets tedious after that.\n\nN:\u00a0\u00a0 N^2 : N^2 Choose N Unique pattern\n2\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\n3\u00a0\u00a0\u00a0 9\u00a0\u00a0\u00a0\u00a0 84\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 14\n4\u00a0\u00a0\u00a0 16\u00a0\u00a0\u00a0 1820\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ????\n5\u00a0\u00a0\u00a0 25\u00a0\u00a0\u00a0 53130\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ????\n6\u00a0\u00a0\u00a0 36\u00a0\u00a0\u00a0 1947792 \u00a0\u00a0\u00a0\u00a0 ????\n7\u00a0\u00a0\u00a0 49\u00a0\u00a0\u00a0 85900584\u00a0\u00a0\u00a0\u00a0 ????\n\n\u2022 one must choose $N$ points on a binary grid? \u2013\u00a0nbubis Nov 17 '13 at 4:46\n\u2022 yes, forgot to add that in \u2013\u00a0user109380 Nov 17 '13 at 4:52\n\u2022 Are you familiar with group theory? Polya enumeration could be the way to go here. \u2013\u00a0Gerry Myerson Nov 17 '13 at 5:26\n\nWe can actually do a bit more and compute the cycle index $Z(G_N)$ for general $N$, where we have to distinguish between $N$ even and $N$ odd. This will permit lookup in the OEIS, which in turn leads to more material about this interesting problem.\n\nWe proceed to enumerate the permutations of $G_N$ by their cycle structure. For $N$ even, we get the identity, which contributes $$a_1^{N^2}.$$ There is a vertical reflection, which contributes $$a_2^{N^2/2},$$ the same for a horizontal reflection, i.e. $$a_2^{N^2/2}.$$ The reflection in the rising diagonal contributes $$a_1^N a_2^{(N^2-N)/2},$$ the same for the other diagonal, i.e. $$a_1^N a_2^{(N^2-N)/2}.$$\n\nWhat remains are the rotations. Two of these contribute (recall that $N$ is even) $$2\\times a_4^{N^2/4}$$ and one of them, $$a_2^{N^2/2}.$$ This gives for even $N$ the cycle index $$Z(G_N) = \\frac{1}{8} \\left( a_1^{N^2} + 3 a_2^{N^2/2} + 2 a_1^N a_2^{(N^2-N)/2} + 2 a_4^{N^2/4}\\right).$$ For $N$ odd, we get the identity, which is $$a_1^{N^2}.$$ The two reflections now contribute $$2\\times a_1^N a_2^{(N^2-N)/2}.$$ The reflection in the two diagonals are unchanged and contribute $$2\\times a_1^N a_2^{(N^2-N)/2}$$ What remains is the rotations, two of which have cycle structure $$2\\times a_1 a_4^{(N^2-1)/4}$$ and the last one has cycle structure $$a_1 a_2^{(N^2-1)/2}.$$\n\nThis gives for odd $N$ the cycle index $$Z(H_N) = \\frac{1}{8} \\left( a_1^{N^2} + 4 a_1^N a_2^{(N^2-N)/2} + 2 a_1 a_4^{(N^2-1)/4}+a_1 a_2^{(N^2-1)/2}\\right).$$ Evidently for $N$ marks being placed on the grid we seek to compute $$[z^N] Z(G_N)(1+z) \\quad\\text{and}\\quad [z^N] Z(H_N)(1+z),$$ alternating between the two for $N$ even and $N$ odd.\n\nThis produces the sequence $$1, 2, 16, 252, 6814, 244344, 10746377, 553319048, 32611596056, 2163792255680,\\ldots$$ which is A019318 from the OEIS.\n\nHere is the Maple program that was used to compute these cycle indices.\n\n\nwith(numtheory);\nwith(group):\nwith(combinat):\n\npet_varinto_cind :=\nproc(poly, ind)\nlocal subs1, subs2, polyvars, indvars, v, pot, res;\n\nres := ind;\n\npolyvars := indets(poly);\nindvars := indets(ind);\n\nfor v in indvars do\npot := op(1, v);\n\nsubs1 :=\n[seq(polyvars[k]=polyvars[k]^pot,\nk=1..nops(polyvars))];\n\nsubs2 := [v=subs(subs1, poly)];\n\nres := subs(subs2, res);\nod;\n\nres;\nend;\n\nG :=\nproc(N)\nif type(N,odd) then return FAIL; fi;\n\n1/8*(a[1]^(N^2)+3*a[2]^(N^2/2)+\n2*a[1]^N*a[2]^((N^2-N)/2) + 2*a[4]^(N^2/4));\nend;\n\nH :=\nproc(N)\nif type(N,even) then return FAIL; fi;\n\n1/8*(a[1]^(N^2)+4*a[1]^N*a[2]^((N^2-N)/2)+\na[1]*a[2]^((N^2-1)/2) + 2*a[1]*a[4]^((N^2-1)/4));\nend;\n\nv :=\nproc(N)\noption remember;\nlocal p, k, gf;\n\nif type(N, even) then\ngf := expand(pet_varinto_cind(1+z, G(N)));\nelse\ngf := expand(pet_varinto_cind(1+z, H(N)));\nfi;\n\ncoeff(gf, z, N);\nend;\n\n\nHere is another interesting MSE cycle index computation I. This MSE cycle index computation II is relevant also.\n\n\u2022 I want to ask about the notation $Z(G_N)(1+z).$ I assume this means that $a_1$ gets replaced by $1+z,$ $a_2$ gets replaced by $1+z^2,$ and $a_4$ gets replaced by $1+z^4,$ but I wanted to double check. \u2013\u00a0Will Orrick Jan 1 '14 at 15:27\n\u2022 That is correct. Wikipedia has useful data on this: Cycle Index and Polya Enumeration Theorem. \u2013\u00a0Marko Riedel Jan 1 '14 at 15:46\n\nYou're looking for the number of orbits of the dihedral group $D_N$ acting on the $N\\times N$ grid with $N$ chosen cells. (Caution: This group is often expressed as $D_{2N}$ as it has $2N$ elements.)\n\nYou want to use the Cauchy-Frobenius-Burnside Lemma (aka Burnside Lemma, aka \"Not Burnside Lemma\") to compute this. That is, you have a group action of $D_N$ on the set $\\Omega$ of all such patterns ($N\\times N$ grids with $N$ marked cells), and the number of orbits in this action (which is what you want) is then given by $$\\frac{1}{2N}\\sum_{g\\in G} |\\Omega^g|$$ where $|\\Omega^g|$ is the number of elements in $\\Omega$ fixed by $g$.\n\n\u2022 and I agree with Gerry (above) that Polya enumeration is an efficient way of doing this. (See also, cycle index.) \u2013\u00a0Doc Nov 17 '13 at 6:03\n\u2022 For the uninitiated (and probably the OP as well), would you like to present the final formula? \u2013\u00a0nbubis Nov 17 '13 at 6:34\n\u2022 @nbubis, by \"formula\" I'm interpreting that you mean a closed form for the answer. I doubt this is even possible. However, your comment inspired me to cook up an example (below) that hopefully clarifies most of the procedure. \u2013\u00a0Doc Nov 17 '13 at 16:50\n\nConsider the action of the group $G$ on a set $\\Omega$ of cardinality $n$. The cycle index of $(G,\\Omega)$ is the polynomial defined by $$Z_G(X_1,X_2,\\dots, X_n)=\\frac{1}{|G|} \\sum_{g\\in G} X_1^{c_1(g)}X_2^{c_2(g)}\\cdots X_n^{c_n(g)}$$ where $c_i(g)$ denotes the number of $i$-cycles in the cycle decomposition of the element $g$ acting on $\\Omega$.\n\nFor example, if we consider the action of $D_4$ on the square, we would first label the vertices of the square by $\\{1,2,3,4\\}$ in some order. Let's do this in cyclic clockwise order, starting from the upper left corner. Then we get the following table. \\begin{array}{c|c|c} g & \\mbox{permutation} & \\mbox{cycle index term}\\\\ \\hline \\mbox{identity} & (1)(2)(3)(4) & X_1^4 \\\\ 90^o \\mbox{rotation} & (1,2,3,4) & X_4^1 \\\\ 180^o \\mbox{rotation} & (1,3)(2,4) & X_2^2 \\\\ 270^o \\mbox{rotation} & (1,4,3,2) & X_4^1 \\\\ \\mbox{horizontal reflection} & (1,4)(2,3) & X_2^2\\\\ \\mbox{vertical reflection} & (1,2)(3,4) & X_2^2\\\\ \\mbox{diagonal reflection} & (1,3)(2)(4) & X_1^2X_2\\\\ \\mbox{opposite diag reflection} & (2,4)(1)(3)& X_1^2X_2 \\end{array}\n\nThus we obtain as cycle index $$Z_G(X_1,X_2,\\dots, X_n)=\\frac{1}{8}(X_1^4 + 2X_1^2X_2+3X_2^2+2X_4).$$\n\nNow consider $$Z_G(b+1,b^2+1,b^3+1,b^4+1).$$ The coefficient of $b^i$ will give the number (up to isomorphism!) of positions in which there are precisely $i$ marked squares. Returning to the grid problem, we are only interested in the coefficient of the $b^2$ term. This turns out to be $\\frac{1}{8}(16) = 2$.\n\nHere's an answer that leads to explicit formulas whose most complicated term is a sum over a product of binomial coefficients. I haven't yet made any attempt to connect it to the excellent solutions in the answers given earlier.\n\nFirst some notation. The dihedral group of the square, $D_4,$ has elements $\\{e,\\rho,\\rho^2,\\rho^3,h,v,d_1,d_2\\},$ where $e$ is the identity, $\\rho$ is the $90^\\circ$ rotation, $h$ is the reflection about the horizontal axis, $v$ is the reflection about the vertical axis, $d_1$ is the reflection about the forward diagonal, and $d_2$ is the reflection about the backward diagonal.\n\n(Added: Having never learned P\u00f3lya enumeration properly, I made my answer overly complicated. I leave my previous answer, but add a more direct route to equation (**) which appears below. In this problem we have $D_4$ acting on the set $X,$ consisting of the $\\binom{n^2}{n}$ ways of coloring $n$ small squares in the $n\\times n$ grid black, leaving the remaining squares white. The orbit counting theorem leads immediately to (**) since $$K(n)=\\lvert\\mathcal{O}\\rvert=\\frac{1}{8}\\sum_{g\\in D_4}C(n,n,\\langle g\\rangle)$$ is identical to (**) when one realizes that $\\langle\\rho\\rangle=\\langle\\rho^3\\rangle=C_4.$ The orbit counting theorem is described in many places. For convenience I've appended some background information about it to the end of this post.)\n\nSince the order of $D_4$ is $8,$ all subgroups are of order $1,$ $2,$ $4,$ or $8.$ Either of the two elements of order $4,$ $\\rho$ and $\\rho_3,$ generates $C_4,$ the rotation group of the square. Each of the five elements of order $2$, $\\rho^2,$ $h,$ $v,$ $d_1,$ and $d_2,$ generates a subgroup of order $2.$ There are two additional subgroups of order $4,$ isomorphic to $C_2\\times C_2.$ They are $\\{e,\\rho^2,h,v\\}$ and $\\{e,\\rho^2,d_1,d_2\\}.$ Any other group generated by two elements of order $2$ is all of $D_4.$\n\nThere are therefore $10$ subgroups with inclusions given by the following Hasse diagram.\n\nDefine $C(n,b,G)$ to be the set of configurations of the $n\\times n$ grid with $b$ marked squares that are invariant under the group $G.$ We are ultimately interested in $b=n,$ but will start out with slightly greater generality. Define $\\overline{C}(n,b,G)$ to be the subset of $C(n,b,G)$ consisting of elements that are not invariant under any group that properly contains $G.$ The number we wish to compute, which we denote $K(n),$ is \\begin{aligned} (*)\\quad K(n)=&\\frac{\\lvert\\overline{C}(n,n,D_4)\\rvert}{1}+\\frac{\\lvert\\overline{C}(n,n,\\{e,\\rho^2,h,v\\})\\rvert}{2}+\\frac{\\lvert\\overline{C}(n,n,C_4)\\rvert}{2}+\\frac{\\lvert\\overline{C}(n,n,\\{e,\\rho^2,d_1,d_2\\})\\rvert}{2}\\\\ &+\\frac{\\lvert\\overline{C}(n,n,\\{e,h\\})\\rvert}{4}+\\frac{\\lvert\\overline{C}(n,n,\\{e,v\\})\\rvert}{4}+\\frac{\\lvert\\overline{C}(n,n,\\{e,\\rho^2\\})\\rvert}{4}+\\frac{\\lvert\\overline{C}(n,n,\\{e,d_1\\})\\rvert}{4}\\\\ &+\\frac{\\lvert\\overline{C}(n,n,\\{e,d_2\\})\\rvert}{4}+\\frac{\\lvert\\overline{C}(n,n,\\{e\\})\\rvert}{8}. \\end{aligned}\n\n(Added: The set $C(n,n,G)$ is the set of colorings whose stabilizer contains $G,$ while $\\overline{C}(n,n,G)$ is the set of colorings whose stabilizer equals $G.$ The formula (*) for $K(n),$ the number of orbits, comes from counting each coloring with a weight equal to the reciprocal of the size of its orbit, the weight having been computed using the orbit-stabilizer theorem. (See the end of the post.) For example, if $x$ has stabilizer $D_4,$ then the weight is $1/1=8/8;$ if $x$ has stabilizer $\\{e,\\rho^2,h,v\\},$ then the weight is $1/2=4/8.$ Of course, as noted above, the use of (*) and the inclusion-exclusion argument below is rendered superfluous by the use of the orbit counting theorem. It is interesting that although (*) involves all ten subgroups of $D_4,$ the final expression (**) involves only the seven cyclic subgroups. My naive derivation does not explain this, but the orbit counting theorem makes it manifest that that must be the case.)\n\nFrom the diagram, we obtain expressions for the $\\overline{C}(n,b,G)$ in terms of the $C(n,b,G):$ \\begin{aligned} \\lvert\\overline{C}(n,b,D_4)\\rvert&=\\lvert C(n,b,D_4)\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e,\\rho^2,h,v\\})\\rvert&=\\lvert C(n,b,\\{e,\\rho^2,h,v\\})\\rvert- \\lvert C(n,b,D_4)\\rvert\\\\ \\vert\\overline{C}(n,b,C_4)\\rvert&=\\lvert C(n,b,C_4)\\rvert- \\lvert C(n,b,D_4)\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert&=\\lvert C(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert- \\lvert C(n,b,D_4)\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e,h\\})\\rvert&=\\lvert C(n,b,\\{e,h\\})\\rvert- \\lvert C(n,b,\\{e,\\rho^2,h,v\\})\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e,v\\})\\rvert&=\\lvert C(n,b,\\{e,v\\})\\rvert- \\lvert C(n,b,\\{e,\\rho^2,h,v\\})\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e,d_1\\})\\rvert&=\\lvert C(n,b,\\{e,d_1\\})\\rvert- \\lvert C(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e,d_2\\})\\rvert&=\\lvert C(n,b,\\{e,d_2\\})\\rvert- \\lvert C(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e,\\rho^2\\})\\rvert&=\\lvert C(n,b,\\{e,\\rho^2\\})\\rvert- \\lvert \\overline{C}(n,b,\\{e,\\rho^2,h,v\\})\\rvert- \\lvert \\overline{C}(n,b,C_4)\\rvert\\\\ &\\quad- \\lvert \\overline{C}(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert- \\lvert \\overline{C}(n,b,D_4)\\rvert\\\\ &=\\lvert C(n,b,\\{e,\\rho^2\\})\\rvert- \\lvert C(n,b,\\{e,\\rho^2,h,v\\})\\rvert- \\lvert C(n,b,C_4)\\rvert\\\\ &\\quad- \\lvert C(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert+2 \\lvert C(n,b,D_4)\\rvert\\\\ \\vert\\overline{C}(n,b,\\{e\\})\\rvert&=\\lvert C(n,b,\\{e\\})\\rvert- \\lvert \\overline{C}(n,b,\\{e,h\\})\\rvert- \\lvert \\overline{C}(n,b,\\{e,v\\})\\rvert- \\lvert \\overline{C}(n,b,\\{e,\\rho^2\\})\\rvert\\\\ &\\quad- \\lvert \\overline{C}(n,b,\\{e,d_1\\})\\rvert- \\lvert \\overline{C}(n,b,\\{e,d_2\\})\\rvert- \\lvert \\overline{C}(n,b,\\{e,\\rho^2,h,v\\})\\rvert\\\\ &\\quad- \\lvert \\overline{C}(n,b,C_4)\\rvert- \\lvert \\overline{C}(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert- \\lvert \\overline{C}(n,b,D_4)\\rvert\\\\ &=\\lvert C(n,b,\\{e\\})\\rvert- \\lvert C(n,b,\\{e,h\\})\\rvert- \\lvert C(n,b,\\{e,v\\})\\rvert- \\lvert C(n,b,\\{e,\\rho^2\\})\\rvert\\\\ &\\quad- \\lvert C(n,b,\\{e,d_1\\})\\rvert- \\lvert C(n,b,\\{e,d_2\\})\\rvert+2\\lvert C(n,b,\\{e,\\rho^2,h,v\\})\\rvert\\\\ &\\quad+2\\lvert C(n,b,\\{e,\\rho^2,d_1,d_2\\})\\rvert \\end{aligned} These imply that \\begin{aligned} (**)\\quad K(n)=&\\frac{\\lvert C(n,n,C_4)\\rvert}{4}+\\frac{\\lvert C(n,n,\\{e,h\\})\\rvert}{8}+\\frac{\\lvert C(n,n,\\{e,v\\})\\rvert}{8}+\\frac{\\lvert C(n,n,\\{e,\\rho^2\\})\\rvert}{8}\\\\ &+\\frac{\\lvert C(n,n,\\{e,d_1\\})\\rvert}{8}+\\frac{\\lvert C(n,n,\\{e,d_2\\})\\rvert}{8}+\\frac{\\lvert C(n,n,\\{e\\})\\rvert}{8}. \\end{aligned}\n\nIt remains to compute the quantities $C(n,b,G)$ for the seven groups that appear in the expression above. (Added: It is usual in P\u00f3lya enumeration at this point to use the cycle index and generating function techniques. My more bare-handed approach yields the final expressions in a fairly direct fashion.) We have $$C(n,b,\\{e\\})=\\binom{n^2}{b}.$$ In general, the counting depends on whether $n$ is even or odd. If $n$ is odd, then the action of $C_4$ fixes the center square; every other square has an orbit of size $4.$ Therefore, a configuration that is invariant under the action of $C_4$ is specified by choosing the markings of the center square and of one square from each of the $\\frac{n^2-1}{4}$ orbits of size $4.$ (The other three squares in an orbit of size $4$ must be marked the same way as the first square.) Since the total number of marked squares must be $b,$ we have $$C(n,b,C_4)=\\begin{cases}\\binom{(n^2-1)/4}{b/4} & \\text{if b\\equiv0\\pmod{4}}\\\\ \\\\ \\binom{(n^2-1)/4}{(b-1)/4} & \\text{if b\\equiv1\\pmod{4}}\\\\ \\\\ 0 & \\text{otherwise.}\\end{cases}$$ The center square is unmarked when $b\\equiv0\\pmod{4}$ and marked when $b\\equiv1\\pmod{4}.$ If $n$ is even then, under the action of $C_4,$ every square has an orbit of size $4.$ Therefore $$C(n,b,C_4)=\\begin{cases}\\binom{n^2/4}{b/4} & \\text{if b\\equiv0\\pmod{4}}\\\\ \\\\ \\binom{n^2/4}{(b-1)/4} & \\text{if b\\equiv1\\pmod{4}}\\\\ \\\\ 0 & \\text{otherwise.}\\end{cases}$$ Specializing to $b=n,$ we get $$C(n,n,C_4)=\\begin{cases}\\binom{n^2/4}{n/4} & \\text{if n\\equiv0\\pmod{4}}\\\\ \\\\ \\binom{(n^2-1)/4}{(n-1)/4} & \\text{if n\\equiv1\\pmod{4}}\\\\ \\\\ 0 & \\text{otherwise.}\\end{cases}$$\n\nThe analysis for $G=\\{e,\\rho^2\\}$ is similar. The orbit sizes are now $2$ instead of $4,$ with the exception of the center square in the $n$ odd case, which has orbit size $1.$ The result, for $b=n,$ is $$C(n,n,\\{e,\\rho^2\\})=\\begin{cases}\\binom{n^2/2}{n/2} & \\text{if n even}\\\\ \\\\ \\binom{(n^2-1)/2}{(n-1)/2} & \\text{if n odd.}\\end{cases}$$\n\nWhen $G=\\{e,h\\}$ and $n$ is odd, the action of $G$ fixes the $n$ squares on the horizontal centerline. All other squares have orbit size $2.$ A configuration invariant under the action of $\\{e,h\\}$ is specified by marking $j$ of these orbits and $b-2j$ squares on the centerline. When $n$ is even, all squares have orbit size $2,$ so we need only mark $b/2$ of these orbits. The analysis when $G=\\{e,v\\}$ is similar. In the $b=n$ case, this gives $$C(n,n,\\{e,h\\})=C(n,n,\\{e,v\\})=\\begin{cases}\\binom{n^2/2}{n/2} & \\text{if n even}\\\\ \\\\ \\sum_j\\binom{(n^2-n)/2}{j}\\binom{n}{n-2j} & \\text{if n odd.}\\end{cases}$$\n\nWhen $G=\\{e,d_1\\}$ or $\\{e,d_2\\},$ it is a diagonal that is fixed by the action of $G.$ So the analysis is the same as for $G=\\{e,h\\}$ with $n$ odd. We have $$C(n,n,\\{e,d_1\\})=C(n,n,\\{e,d_2\\})= \\sum_j\\binom{(n^2-n)/2}{j}\\binom{n}{n-2j}.$$ Mathematica recognizes this sum, which we denote by $S,$ as a generalized hypergeometric function with argument $-1:$ $$S=\\sum_j\\binom{(n^2-n)/2}{j}\\binom{n}{n-2j} = \\ _3F_2\\left[\\begin{matrix}1/2-n/2 & -n/2 & (n-n^2)/2\\\\1/2 & 1 & \\end{matrix};-1\\right].$$ Indeed, \\begin{aligned} S=&\\sum_j\\binom{(n^2-n)/2}{j}\\binom{n}{n-2j} =\\sum_j\\binom{(n^2-n)/2}{j}\\binom{n}{2j}\\\\ =&\\sum_j\\frac{\\frac{n^2-n}{2}(\\frac{n^2-n}{2}-1)\\ldots(\\frac{n^2-n}{2}-j+1)}{j!}\\frac{n(n-1)\\ldots(n-2j+1)}{2j(2j-1)\\ldots1}\\\\ =&\\sum_j\\frac{n-n^2}{2}\\left(\\frac{n-n^2}{2}+1\\right)\\ldots\\left(\\frac{n-n^2}{2}+j-1\\right)\\frac{\\frac{-n}{2}\\frac{-n+1}{2}\\ldots\\frac{-n+2j-1}{2}}{j(j-\\frac{1}{2})\\ldots\\frac{2}{2}\\frac{1}{2}}\\frac{(-1)^j}{j!}\\\\ =&\\sum_j\\frac{n-n^2}{2}\\left(\\frac{n-n^2}{2}+1\\right)\\ldots\\left(\\frac{n-n^2}{2}+j-1\\right)\\\\ &\\times\\frac{\\frac{-n}{2}\\left(\\frac{-n}{2}+1\\right)\\ldots\\left(\\frac{-n}{2}+j-1\\right)\\frac{1-n}{2}\\left(\\frac{1-n}{2}+1\\right)\\ldots\\left(\\frac{1-n}{2}+j-1\\right)}{\\frac{1}{2}\\frac{3}{2}\\ldots\\left(\\frac{1}{2}+j-1\\right)\\cdot1\\cdot2\\ldots\\cdot j}\\frac{(-1)^j}{j!}, \\end{aligned} which, by definition, is the stated generalized hypergeometric function.\n\nCombining these results, we have $$K(n)=\\begin{cases} \\frac{1}{8}\\binom{n^2}{n} + \\frac{1}{4}\\binom{n^2/4}{n/4} + \\frac{3}{8}\\binom{n^2/2}{n/2} + \\frac{1}{4}S & \\text{for n\\equiv0\\pmod{4}}\\\\ \\frac{1}{8}\\binom{n^2}{n} + \\frac{1}{4}\\binom{(n^2-1)/4}{(n-1)/4} + \\frac{1}{8}\\binom{(n^2-1)/2}{(n-1)/2} + \\frac{1}{2}S & \\text{for n\\equiv1\\pmod{4}}\\\\ \\frac{1}{8}\\binom{n^2}{n} + \\frac{3}{8}\\binom{n^2/2}{n/2} + \\frac{1}{4}S & \\text{for n\\equiv2\\pmod{4}}\\\\ \\frac{1}{8}\\binom{n^2}{n} + \\frac{1}{8}\\binom{(n^2-1)/2}{(n-1)/2} + \\frac{1}{2}S & \\text{for n\\equiv3\\pmod{4}.} \\end{cases}$$ This answer agrees with that of Marko Riedel and the OEIS sequence he cites. (Added: These formulas can be obtained without much difficulty by extracting the relevant coefficient from the final generating function in Marko Riedel's answer.)\n\nAdded: (Some background about the orbit-stabilizer theorem and the orbit counting theorem.) Let $G$ be a group acting on a set $X.$. The stabilizer of $x\\in X$ is the set of elements of $G$ that fix $x,$ that is, the set of $g\\in G$ such that $gx=x.$ The orbit of $x\\in X$ is the set of elements of $X$ to which $x$ is sent by some element of $G.$\n\nThe orbit-stabilizer theorem states that, for any $x\\in X,$ $$\\lvert G\\rvert=\\lvert\\text{orbit}(x)\\rvert\\cdot\\lvert\\text{stabilizier}(x)\\rvert.$$ It is a consequence of the fact that $\\text{stabilizer}(x)$ is a subgroup of $G$ and the fact that the elements of $\\text{orbit}(x)$ are in one-to-one correspondence with the left cosets of $\\text{stabilizer}(x).$\n\nThe action of $G$ on $X$ partitions $X$ into disjoint orbits. Let $\\mathcal{O}$ be the set of orbits. The fixed set of $g\\in G$ is the set of elements of $X$ fixed by $g,$ that is, the set of $x\\in X$ such that $gx=x.$ The orbit counting theorem, which is often called Burnside's lemma, states that the number of orbits equals the average (over all $g\\in G$) of $\\lvert\\text{fixed}(g)\\rvert.$ It is proved by considering the set $P$ of pairs $(g,x)$ satisfying $gx=x.$ We may count the elements of $P$ either by running over $G$ or by running over $X:$ \\begin{aligned} \\lvert P\\rvert=\\sum_{g\\in G}\\lvert\\text{fixed}(g)\\rvert&=\\sum_{x\\in X}\\lvert\\text{stabilizer}(x)\\rvert\\\\ &=\\sum_{x\\in X}\\frac{\\lvert G\\rvert}{\\lvert\\text{orbit}(x)\\rvert}\\\\ &=\\lvert G\\rvert\\sum_{O\\in\\mathcal{O}}\\sum_{x\\in O}\\frac{1}{\\lvert\\text{orbit}(x)\\rvert}\\\\ &=\\lvert G\\rvert\\sum_{O\\in\\mathcal{O}}1\\\\ &=\\lvert G\\rvert\\cdot\\lvert\\mathcal{O}\\rvert, \\end{aligned} where the orbit-stabilizer theorem was used to obtain the second line. This implies that $$\\lvert\\mathcal{O}\\rvert=\\frac{1}{\\lvert G\\rvert}\\sum_{g\\in G}\\lvert\\text{fixed}(g)\\rvert,$$ which is the desired result.\n\nhttp://puzzlemaker.discoveryeducation.com/MathSquareForm.asp\n\nAbove is a link to a puzzlemaker referred to as Math Square.\n\nIt allows you to create matrice puzzle.\n\nFor example, you can create a 6x6 grid and the rows and columns are maths equation having multiplication, division, addition and subtraction. The answers to the equation are available on the extreme right and the extreme bottom of the grid. However, no number in the grid is provided. Can such a puzzle be logically or mathematically solved using any of the above methods mentioned in this thread.", "date": "2019-05-26 18:56:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/570003/how-many-unique-patterns-exist-for-a-n-times-n-grid/587729", "openwebmath_score": 0.9341668486595154, "openwebmath_perplexity": 363.27761773168936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969703688159, "lm_q2_score": 0.8902942333990421, "lm_q1q2_score": 0.8756907107081253}}
{"url": "http://mathhelpforum.com/algebra/210250-number-integers-bet-90-990-a.html", "text": "# Math Help - number of integers bet 90 and 990\n\n1. ## number of integers bet 90 and 990\n\nHow many integers bet 90 and 990 are divisible by 7?\n\nis there any short and simple way of finding the answer?\n\nthanks\n\n2. ## Re: number of integers bet 90 and 990\n\nOriginally Posted by rcs\nHow many integers bet 90 and 990 are divisible by 7?\n\nis there any short and simple way of finding the answer?\n\n$\\left\\lfloor {\\frac{{990}}{7}} \\right\\rfloor - \\left\\lfloor {\\frac{{90}}{7}} \\right\\rfloor=~?$\n\n3. ## Re: number of integers bet 90 and 990\n\nOriginally Posted by rcs\nHow many integers bet 90 and 990 are divisible by 7?\n\nis there any short and simple way of finding the answer?\n\nthanks\nHi rcs!\n\nThe answer is 990 / 7 - (90 - 1) / 7, where \"/\" denotes a division that is rounded down.\nI am assuming that you would count 990 and 90 if they would have been divisible by 7 (which they are not anyway).\n\nTo explain:\n(90 - 1) / 7 is the number of integers divisible by 7 below and excluding 90.\n990 / 7 is the number of integers divisible by 7 below and including 990.\nThe difference is the total number.\n\nSuppose you are in the middle of a road with poles alongside the road.\nAnd suppose you want to know the number of poles from where you are to the end of the road.\n\nTo find out you count the number that you can see from where you stand to the beginning of the road.\nAnd if you are next to a pole you do not count that one.\nSubtract from the total number of poles and there you are.\n\n4. ## Re: number of integers bet 90 and 990\n\nthanks Prof . Plato\n\n5. ## Re: number of integers bet 90 and 990\n\nOriginally Posted by Plato\n$\\left\\lfloor {\\frac{{990}}{7}} \\right\\rfloor - \\left\\lfloor {\\frac{{90}}{7}} \\right\\rfloor=~?$\nthe answer is 128 4/7 is this correct sir?\n\ni wonder why the book the answer they posted is 56... however they do not post the solution\n\n6. ## Re: number of integers bet 90 and 990\n\nOriginally Posted by rcs\nthe answer is 128 4/7 is this correct sir?\ni wonder why the book the answer they posted is 56... however they do not post the solution\nFirst that is the floor function. It returns an integer.\n\nThere are $\\left\\lfloor {\\frac{{990}}{7}} \\right\\rfloor=141$ integers from 1 to 990 that are by seven.\n\nThere are $\\left\\lfloor {\\frac{{90}}{7}} \\right\\rfloor=12$ integers from 1 to 90 that are by seven.\n\nNote that neither 990 nor 90 is divisible by seven.\n\nSo you book is not correct. The answer is 129.\n\n7. ## Re: number of integers bet 90 and 990\n\nOriginally Posted by Plato\nFirst that is the floor function. It returns an integer.\n\nThere are $\\left\\lfloor {\\frac{{990}}{7}} \\right\\rfloor=141$ integers from 1 to 990 that are by seven.\n\nThere are $\\left\\lfloor {\\frac{{90}}{7}} \\right\\rfloor=12$ integers from 1 to 90 that are by seven.\n\nNote that neither 990 nor 90 is divisible by seven.\n\nSo you book is correct. The answer is 129.\nyou mean to say that the book i've read whose answer is 56 is correct and your answer 129 is also correct... therefore, there are two correct answers now... 56 and 129\n\n8. ## Re: number of integers bet 90 and 990\n\nSorry that was a typo. It should be NOT CORRECT\n\n9. ## Re: number of integers bet 90 and 990\n\nahhh i see... well then thank you again Sir. God Bless\n\n10. ## Re: number of integers bet 90 and 990\n\nSuggest we find the multiple of 7 just greater than 90 and just less than 990.\nWe will get 91 and 987. Now all the multiples of 7 between these two numbers will form an AP with first term 91, common difference 7 and last term 987.\nQuestion now simply is to find n such that the nth term is 987.\nIf a is the first term, common difference d then the nth term is given by\nan = a + ( n-1) d\nThus we have 91 + ( n \u2013 1 ) x 7 = 987\nOR 91 + 7n \u2013 7 = 987\n7n = 987 \u2013 91 + 7 = 903\nn = 129 Hence there are 129 integers between 90 and 990 which are divisible by 7", "date": "2015-01-28 18:54:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/210250-number-integers-bet-90-990-a.html", "openwebmath_score": 0.5824971199035645, "openwebmath_perplexity": 635.8609263820798, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9835969674838368, "lm_q2_score": 0.8902942195727173, "lm_q1q2_score": 0.8756906945401138}}
{"url": "https://math.stackexchange.com/questions/3906583/orthogonality-of-legendre-polynomials-using-specific-properties", "text": "# Orthogonality of Legendre polynomials using specific properties\n\nI'm having significant issues with a problem and would appreciate any help at all with it. It is regarding proving the orthogonality of Legendre polynomials using a specific recursion formula and property of Legendre polynomials.\n\nI have been given the following relation $$x P'_n(x) -P'_{n-1}(x) = n P_n (x)$$ And that $$\\int^{1}_{-1} f(x) P_n (x) dx = 0$$ for any polynomial $$f(x)$$ of degree less than $$n$$.\n\nI must then show that for $$n \\geq 1$$\n\n$$\\int^{1}_{-1} P_n (x)^2 dx = \\frac{1}{2n} \\int^{1}_{-1} x \\frac{d}{dx} (P_n (x)^2)dx$$\n\nAnd then determine the value of the integral, that being $$\\frac{2}{2n+1}$$. I must use the two stated facts, I'd be grateful for any help with this, thank you guys.\n\n\u2022 Have you tried integration by parts yet? \u2013\u00a0Somos Nov 14 '20 at 0:31\n\u2022 My first thought too but I decided its a no-go. Id like to see your approach, @Somos. \u2013\u00a0CogitoErgoCogitoSum Nov 14 '20 at 1:31\n\n$$\\int\\limits_{-1}^{1} P_n(x)^2 \\;dx =$$ $$\\int\\limits_{-1}^{1} P_n(x)P_n(x) \\;dx =$$ $$\\frac{1}{2n}\\int\\limits_{-1}^{1} 2n P_n(x)P_n(x) \\;dx =$$\n\nSubstituting your first relation in: $$\\frac{1}{2n}\\int\\limits_{-1}^{1} 2 \\left[x P'_n(x) - P'_{n-1}(x)\\right] P_n(x)\\;dx =$$\n\n$$\\frac{1}{2n}\\int\\limits_{-1}^{1} 2 x P'_n(x) P_n(x) \\; dx - \\frac{1}{2n}\\int\\limits_{-1}^{1} 2 P'_{n-1}(x) P_n(x)\\;dx =$$\n\nGiven your second property defining orthogonality (since after all $$P'$$ is of lesser degree than $$P$$) the entire second integral drops out.\n\n$$\\frac{1}{2n}\\int\\limits_{-1}^{1} x 2 P_n(x) P'_n(x) \\; dx =$$\n\nReverse of the chain rule:\n\n$$\\frac{1}{2n}\\int\\limits_{-1}^{1} x \\frac{d}{dx} P^2_n(x) \\; dx$$\n\nQED\n\nAnd for anyone who asks/wonders, or is inclined to criticize... I answered this question because it was asked, and this is a Q&A forum. I shouldnt have to justify answering questions on a Q&A forum though, but this is the culture here.\n\n\u2022 Did you read the part in the question stating: \"And then determine the value of the integral, that being 2/(2n+1).\"? You seem to have not done that part yet. \u2013\u00a0Somos Nov 14 '20 at 21:55\n\u2022 Youre right, I didnt. I dont feel the need to do all 100% of the work when OP is hindered only by the first 10%. Dont know what your point is. I dont believe in \"handing out answers\". OP should be able to do some of it, dont you think? \u2013\u00a0CogitoErgoCogitoSum Nov 14 '20 at 23:20\n\nI immediately thought of integration by parts. Here is my work:\n\nBy the product rule of derivative we have $$\\frac{d}{dx} x\\, P_n(x)^2 = P_n(x)^2 + x\\, \\frac{d}{dx}(P_n(x)^2) \\tag{1}$$\n\nNote that Legendre polynomials have the property $$P_n(1) = 1,\\quad \\text{ and } \\quad P_n(-1)=(-1)^n. \\tag{2}$$\n\nBy integrating equation $$(1)$$ and using equation $$(2)$$ we get $$2 = 1 - (-1) = \\int_{-1}^1 P_n(x)^2 dx + \\int_{-1}^1 x\\, \\frac{d}{dx}(P_n(x)^2) dx. \\tag{3}$$\n\nDefine $$\\, a_n := \\int_{-1}^1 P_n(x)^2 dx.\\,$$ Equation $$(3)$$ implies that $$2 - a_n = \\int_{-1}^1 x\\, \\frac{d}{dx}(P_n(x)^2)\\,dx. \\tag{4}$$ We are given the property that $$x P'_n(x) -P'_{n-1}(x) = n P_n (x). \\tag{5}$$ Multiply this equation $$(5)$$ by $$\\,P_n(x)\\,$$ and integrate to get $$\\int_{-1}^1\\!P_n(x)(xP'_n(x)\\!-\\!P'_{n-1}(x))dx \\!=\\!\\int_{-1}^1\\!P_n(x)nP_n(x)dx. \\tag{6}$$ We are also given the property that $$\\int_{-1}^1 P_n(x)\\,f(x)\\,dx = 0 \\tag{7}$$ for any polynomial $$\\,f(x)\\,$$ of degree less than $$\\,n.\\,$$ Now notice that $$x\\, \\frac{d}{dx}(P_n(x)^2) = 2\\,x\\,P_n(x)P'_n(x). \\tag{8}$$ Combining equations $$(6)$$, $$(7)$$, and $$(8)$$ we get that $$\\frac12(2-a_n) = n\\,a_n. \\tag{9}$$ Dividing this equation $$(9)$$ through by $$\\,n\\,$$ and using the definition of $$\\,a_n\\,$$ and equation $$(4)$$ gives the result we wanted: $$\\int_{-1}^1 P_n(x)^2 dx = \\frac{2-a_n}{2n} = \\frac1{2n} \\int_{-1}^1 x\\, \\frac{d}{dx}(P_n(x)^2)\\,dx. \\tag{10}$$ Also, solving for $$\\,a_n\\,$$ in equation $$(9)$$ gives $$a_n = \\frac2{2n+1} \\tag{11}$$ which was also requested.\n\nNOTE: The property in equation $$(2)$$ was not given to us. It would have to be proven using equations $$(5)$$ and $$(7)$$ which are given to us. It can be done, and would take some effort, but I have not given the steps required.\n\n\u2022 A few points/criticisms, if you dont mind. For one thing, it isnt obvious how your suggestion to use \"integration by parts\" is being utilized. Frankly I dont see I by P in your work. Secondly, your work doesnt flow all that well... you make a few leaps. The logic is sound but effort must be used to comprehend the somewhat convoluted steps. Its just a bit scattered and disorderly. Thirdly, property (2), although true, was not one of the properties OP provided; Im not sure its appropriate to use it. Just some things to consider... \u2013\u00a0CogitoErgoCogitoSum Nov 14 '20 at 20:27\n\u2022 In the Wikipedia article on Integration by parts is stated \"The rule can be thought of as an integral version of the product rule of differentiation.\" That is what I used. You say \"leaps\". I say \"steps\". I don't see anything convoluted. In my NOTE at the end I admit the status of property (2), but the original quesition did not explicitly state than no other properties of Legendre polynomials can be used. If you have a better proof, you can answer the question yourself, which, of course, you have done. Thanks for commenting. \u2013\u00a0Somos Nov 14 '20 at 20:45", "date": "2021-05-14 22:50:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3906583/orthogonality-of-legendre-polynomials-using-specific-properties", "openwebmath_score": 0.8404021859169006, "openwebmath_perplexity": 315.843276250488, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713844203499, "lm_q2_score": 0.8887588038050466, "lm_q1q2_score": 0.8756686170407724}}
{"url": "https://math.stackexchange.com/questions/3804291/show-that-there-exists-x-0-ina-b-such-that-fx-0-frac1nfx-1fx-2", "text": "# Show that there exists $x_0\\in(a,b)$ such that $f(x_0)=\\frac{1}{n}(f(x_1)+f(x_2)+\\cdots+f(x_n)).$\n\nQuestion: Suppose that $$f:[a,b]\\to\\mathbb{R}$$ is continuous. Let $$x_1,x_2,\\cdots, x_n$$ be any $$n$$ points in $$(a,b).$$ Show that there exists $$x_0\\in(a,b)$$ such that $$f(x_0)=\\frac{1}{n}(f(x_1)+f(x_2)+\\cdots+f(x_n)).$$\n\nSolution: Let $$g:[a,b]\\to\\mathbb{R}$$ be such that $$g(x)=nf(x)-\\sum_{k=1}^nf(x_k), \\forall x\\in[a,b].$$ Observe that to prove the statement of the problem it is enough to show that $$g(x_0)=0$$ for some $$x_0\\in(a,b)$$.\n\nNow note that by the 3rd form of the Pigeon Hole principle we can conclude that there exists $$1\\le i,j\\le n$$ such that $$f(x_i)\\le \\frac{1}{n}\\sum_{k=1}^nf(x_k)\\le f(x_j)\\\\\\implies nf(x_i)\\le \\sum_{k=1}^nf(x_k)\\le nf(x_j).$$ Thus, $$g(x_i)=nf(x_i)-\\sum_{k=1}^nf(x_k)\\le 0$$ and $$g(x_j)=nf(x_j)-\\sum_{k=1}^nf(x_k)\\ge 0.$$ Now if $$g(x_i)=0$$ or $$g(x_j)=0$$, then we are done. Thus, let us assume that $$g(x_i)<0$$ and $$g(x_j)>0$$. Now since $$f$$ is continuous on $$[a,b]$$, implies that $$g$$ is continuous on $$[a,b]$$. Therefore, by IVT we can conclude that there exists $$x_0\\in(x_i,x_j)$$ or $$x_0\\in(x_j,x_i)$$ such that $$g(x_0)=0$$. This completes the proof.\n\nIs this solution correct and rigorous enough and is there any other way to solve the problem?\n\n\u2022 What is the PHP? Aug 26, 2020 at 19:22\n\u2022 @MartinR, it's the \"Pigeon Hole Principle\". Sorry, I will expand that term in a bit. Aug 26, 2020 at 19:24\n\nYour proof looks fine to me. There is no need however to introduce the function $$g$$. You know that $$f(x_i)\\le \\frac{1}{n}\\sum_{k=1}^nf(x_k)\\le f(x_j)$$ for some indices $$i, j$$, so you can just apply the intermediate value theorem to $$f$$ on the interval $$I = [\\min(x_i, x_j), \\max(x_i, x_j)]$$ and conclude that $$\\frac{1}{n}\\sum_{k=1}^nf(x_k) = f(x)$$ for some $$x \\in I$$.\n\nInstead of using the pigeon hole principle you can also apply the mean value theorem to $$f$$ on the interval $$J= [\\min_k x_k, \\max_k x_k] \\subset (a, b)$$ because $$m\\le \\frac{1}{n}\\sum_{k=1}^nf(x_k)\\le M$$ with $$m = \\min_J f(x)$$ and $$M = \\max_J f(x)$$.\n\nGiven a continuous $$f(x)$$, an iterated application of the Intermediate value Theorem gives \\eqalign{ & \\exists x_{1,2} \\in \\left[ {x_1 ,x_2 } \\right]:f(x_{1,2} ) = t\\;f(x_1 ) + \\left( {1 - t} \\right)f(x_2 )\\quad \\left| {\\,0 \\le t \\le 1} \\right. \\cr & \\exists x_{2,3} \\in \\left[ {x_2 ,x_3 } \\right]:f(x_{2,3} ) = u\\;f(x_2 ) + \\left( {1 - u} \\right)f(x_3 )\\quad \\left| {\\,0 \\le u \\le 1} \\right. \\cr} which express the possibility of finding a point corresponding to the weighted mean within each interval.\n\nPutting $$t=2/3, \\, u=1/3$$, we can write \\eqalign{ & \\exists x_{1,2} \\in \\left[ {x_1 ,x_2 } \\right]:f(x_{1,2} ) = {2 \\over 3}\\;f(x_1 ) + {1 \\over 3}f(x_2 )\\quad \\left| {\\,0 \\le t \\le 1} \\right. \\cr & \\exists x_{2,3} \\in \\left[ {x_2 ,x_3 } \\right]:f(x_{2,3} ) = {1 \\over 3}\\;f(x_2 ) + {2 \\over 3}f(x_3 )\\quad \\left| {\\,0 \\le u \\le 1} \\right. \\cr & \\exists x_{1,3} \\in \\left[ {x_1 ,x_2 } \\right] \\cup \\left[ {x_2 ,x_3 } \\right]:f(x_{1,3} ) = {1 \\over 2}\\,f(x_{1,2} ) + {1 \\over 2}f(x_{2,3} ) = \\cr & = {{f(x_1 ) + f(x_2 ) + f(x_3 )} \\over 3} \\cr} and the extension to n points is clear.\n\nPick $$i$$ with \\begin{align} f(x_i) &\\le f(x_k) & \\text{for all k = 1, \\ldots, n.} \\tag{1} \\end{align}\nPick $$j$$ with \\begin{align} f(x_j) &\\ge f(x_k) & \\text{for all k = 1, \\ldots, n.} \\tag{2} \\end{align}\nIf $$i = j$$, then all $$x_k$$ are equal, and $$x_0 = x_i$$ solves the problem.\nConsider the case $$i < j$$; the $$i > j$$ case is almost identical. But equation $$1$$, we have $$n f(x_i) \\le \\sum_k f(x_k)$$ By equation 2, similarly $$n f(x_j) \\ge \\sum_k f(x_k)$$.\nThen by the Intermediate value theorem, there's an $$x_0 \\in [x_i, x_j]$$ such that $$f(x_0) = \\frac{1}{n} \\sum_k f(x_k).$$", "date": "2022-08-12 12:28:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3804291/show-that-there-exists-x-0-ina-b-such-that-fx-0-frac1nfx-1fx-2", "openwebmath_score": 0.9993987083435059, "openwebmath_perplexity": 285.87319756916617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.8887588023318196, "lm_q1q2_score": 0.8756686163579881}}
{"url": "http://timberrooftech.co.za/id/61a53a-is-30-an-integer", "text": "is 30 an integer 1. For example, 2 , 5 , 0 , \u2212 12 , 244 , \u2212 15 and 8 are all integers. Integer is a math term for a number that is a whole number. The answer to the question: \"Is 30 an integer?\" Walk through the difference between whole numbers & integers. > How do you check if a value in Java is of an integer type? MATLAB \u00ae has four signed and four unsigned integer classes. Integers are the set of whole numbers and their opposites. He has been teaching from the past 9 years. He provides courses for \u2026 These numbers are used to perform various arithmetic operations, like addition, subtraction, multiplication and division.The examples of integers are, 1, 2, 5,8, -9, -12, etc. Is it an integer? An integer.An integer.An integer.An integer. An integer, also called a \"round number\" or \u201cwhole number,\u201d is any positive or negative number that does not include decimal parts or fractions. Quantity A is greater. How to use integer in a sentence. Signed types enable you to work with negative integers as well as positive, but cannot represent as wide a range of numbers as the unsigned types because one bit is used to \u2026 $\\begingroup$ \"For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity.\" 5- (-4)=1 please help i have a test tommorrow and i just dont get what an integer is. Because you can't \"count\" zero. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if it was NaN Comment Share Find out if 24 is an Integer here! Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Yes. An even number is any integer (a whole number) that can be divided by 2. You could say that 30/6 is just another name for 5, and then it is an integer. 30 is the integer in this case. Why don't libraries smell like bookstores? integer - any of the natural numbers (positive or negative) or zero; \"an integer is a number that is not a fraction\" whole number. (30 points) An integer n is prime iff n >= 2 and n's only factors (divisors) are 1 and itself. And 30 is an integer and not a fraction. Suppose x and y are even. A comprehensive database of more than 56 integer quizzes online, test your knowledge with integer quiz questions. What are the three consecutive odd integers with sum of -30? It is a special set of whole numbers comprised of zero, positive numbers and negative numbers and denoted by the letter Z. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if \u2026 go, not everyone agrees on a simple thing! Integer Question: Is 24 an Integer? We are simply taking \"5 lots of \u22123\", like this: It should be noted that an integer can never be a fraction, a decimal or a per cent. When I as doing my graduation one of my professors Mr. Jaggi told us \u201cMaths student must be polite\u201d thats why I used here \u2018may\u2019. sound intelligent. You might be wondering, what is an opposite? python 2.7.4 error: exactly the same as integers. In fact I can't think of any proof of this sentence which doesn't prove the whole theorem along the way. Integer Question: Is 20 an Integer? The opposite of 24 is -24. The number line goes on forever in both directions. Solution for 1. Is a none repeating decimal an integer? Thanks for your help! The number 0 is also considered an integer even though it has neither a positive or negative value. What details make Lochinvar an attractive and romantic figure? Question 256872: The sum of an integer and its square is 30. 30 is the smallest sphenic number, and the smallest of the form 2 \u00d7 3 \u00d7 r, where r is a prime greater than 3. What is the conflict of the story of sinigang? What is an Integer? So they are 1, 2, 3, 4, 5, ... (and so on). What are the steps to the solution? If x is even, then x y is even. An integer is what is more commonly known as a whole number.It may be positive, negative, or the number zero, but it must be whole. In some cases, the definition of whole number will exclude the number zero, or even the set of negative numbers, but this is not as common as the more inclusive use of the term. n is an integer, and k is not an integer. De tekst is beschikbaar onder de licentie Creative Commons Naamsvermelding/Gelijk delen, er kunnen aanvullende voorwaarden van toepassing zijn. The symbol of integers is \u201c Z \u201c. See more. Examples\u2013 -2.4, 3/4, 90.6. Every integer is a rational number. Quantity B is greater. \"Natural Numbers\" can mean either \"Counting Numbers\" {1, 2, 3, ...}, or \"Whole Numbers\" {0, 1, 2, 3, ...}, depending on the subject. The set of integers is infinite and has no smallest element and no largest element. > is.integer(66L) [1] TRUE > is.integer(66) [1] FALSE Also functions like round will return a declared integer, which is what you are doing with x==round(x). This is shown below. And some people say that zero is NOT a whole number. An integer is a positive or negative whole number. The \u2026 PHP does not support unsigned integers. The basic rules of math. For 20 to be an integer, 20 has to be a whole number. How is 30 an even number? Our online integer trivia quizzes can be adapted to suit your requirements for taking some of the top integer quizzes. Furthermore, for 20 to be an integer, you should be able to write 20 without a fractional or decimal component. Definitions. You could try this. (use indirect\u2026 It is impossible to answer your question without \u2026 0 0 1 ... An integer is a number that is not a fraction, and decimal numbers are decimal fractions. There are three Proper\u2026 Integers The integers are the set of whole numbers and their opposites. Find out if 20 is an Integer here! The farther the number is to the right, the larger the value is. Base Representation; 2: \u2026 This is one of those tricky definition questions. Even and Odd Numbers (use indirect\u2026 Find the number. It cannot be done. Hello, We have a drafter who's Map 3D 2013 Drafting Settings dialog box is giving him a pop up that says \"Please enter an integer from 1 to 30\". Solution for 1. Here are some examples: We can compare integers the same way that we compare positive whole numbers. The number 30 as a Roman Numeral XXX The number 30 in various Numeral Systems. Fractions and decimals are not included in the set of integers. please give me an example or two of an integer. I want to know how to test whether an input value is an integer or not. Please note this from the Integer datatype page: \"The size of an integer is platform-dependent, although a maximum value of about two billion is the usual value (that's 32 bits signed). Some examples of integers include 1, 3, 4, 8, 99, 108, -43, \u2026 Examples of Integers \u2013 1, 6, 15. I have tried using the function isinteger, but I obtain, for example, isinteger(3) = 0.Apparently, any constant is double-precision by Matlab default, and is therefore not recognized as an integer. In the Registry Editor window, navigate to the folder path below depending on your Photoshop version.. Photoshop CC 2018: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\120.0 Photoshop CC 2017: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\110.0 Photoshop CC 2015.5: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\100.0 \u2026 For 24 to be an integer, 24 has to be a whole number. A modular type is an integer type with all arithmetic modulo a specified positive modulus; such a type corresponds to an unsigned type with wrap-around semantics. 30 is a whole number that can be written without a fractional component, thus 30 is an integer. It is not only accurate, it makes you In Maths, integers are the numbers which can be positive, negative or zero, but cannot be a fraction. An integer is any number including 0, positive numbers, and negative numbers. But every rational number is not an integer, since for b not equal to 1, if b does not divide a then w don't get an integer. Integer definition, one of the positive or negative numbers 1, 2, 3, etc., or zero. Examples of Integer Multiplication (a) Here is an integer times an integer: 5 xx -3 = -15 Why is this so? But you for sure can say it is a rational number, which is any number that can be expressed as a \u2026 Choose from 500 different sets of 30 math integers flashcards on Quizlet. How will understanding of attitudes and predisposition enhance teaching? In the equation 2 + 1/2, the number 2 is the integer and 1/2 is the fraction. Find the integer The sum of an integer and its square is - Answered by a verified Math Tutor or Teacher Whole Numbers are simply the numbers 0, 1, 2, 3, 4, 5, ... (and so on), (But numbers like \u00bd, 1.1 and 3.5 are not whole numbers.). The number line is used to represent integers. Start by using a number line. Compare -9 and -2. 0 0 1 ... 176 + 125 = 301 Therefore, the lowest positive integer you can add is 125. When did organ music become associated with baseball? An integer (pronounced IN-tuh-jer) is a whole number (not a fractional number) that can be positive, negative, or zero. An integer (from the Latin integer meaning \"whole\") is colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and \u22122048 are integers, while 9.75, 5 + 1 / 2, and \u221a 2 are not. Suppose x and y are even. I want to know how to test whether an input value is an integer or not. Where is telephone area code 035? 4. Consider a function Factor_In_Range (k , n) to return true iff some integer in the range k to n \u2026 And everyone agrees on the definition of an integer, so when in doubt say \"integer\". This is indicated by the arrows. All Rights Reserved. If a a a and b b b are integers such that a \u22c5 b = 0 a \\cdot b = 0 a \u22c5 b = 0, then a = 0 a=0 a = 0 or b = 0 b=0 b = 0. Rational number is a number which can be expressed as a/b where a,b are integers and b is not 0. a) -3 or 10 b) 3 or -10 c) -5 or 6 d) 5 or -6 Help! For 30 to be an integer, 30 has to be a whole number. Ex 6.1, 2 Solve \u201312x > 30, when x is a natural number \u201312x > 30 \u2013x > \ufdd030\ufdee12\ufdef \u2013x > 2.5 Since x is negative, we multiply both sides by -1 & change the signs (\u2013x) \u00d7 (-1) < 2.5 \u00d7 (-1) x < \u2212 2.5 (i) Since x is a natural number, it can be only positive num The base range of a signed integer type includes at least the values of the specified range. is.integer tests if the number is declared as an integer. B. An integer_type_definition defines an integer type; it defines either a signed integer type, or a modular integer type. You need to prove this, as nothing you have said makes it obvious this is true. Learn 30 math integers with free interactive flashcards. But you for sure can say it is a rational number, which is any number that can be expressed as a fraction of two integers.. An octal integer is an integer represented by a string of digits from the set $\\{0,1,2,3,4,5,6,7\\}$, each representing (in that order) the usual first eight non-negative integers. Integers are like whole numbers, but they also include negative numbers ... but still no fractions allowed! This is one of those tricky definition questions. If a is an odd integer, then a2 +3a+5 is odd. These are all integers (click to mark), and they continue left and right infinitely: Some people (not me) say that whole numbers can also be negative, which makes them So they are 1, 2, 3, 4, 5, ... (and so on). What is an integer? Therefore, these numbers can never be integers: fractions; decimals; percents; Adding and Subtracting Integers. Given an integer number n, return the difference between the product of its digits and the sum of its digits.. He provides courses for Maths and Science at Teachoo. ok i dont get.... is it just considered an integer if the dominate is 1? Counting Numbers are Whole Numbers, but without the zero. Examples of integers are: \u2026 If you divide by 2 and there is a remainder left, then the number is odd. The set of integers consists of zero (), the positive natural numbers (1, 2, 3, ...), also called whole numbers or \u2026 For example, 3, -10, and 1,025 are all integers, but 2.76 (decimal), 1.5 \u2026 Looking at a number line can help you when you need to add or subtract integers. Integers are positive whole numbers and their opposites, including zero. characteristic - the integer part (positive or negative) of the representation of a logarithm; in the expression log 643 = 2.808 the characteristic is 2 Furthermore, for 30 to be an integer, you should be able to write 30 without a fractional or decimal component. If a is an odd integer, then a2 +3a+5 is odd. For any integer a a a, the additive inverse \u2212 a-a \u2212 a is an integer. (use direct proof) 2. (ii) x is an integer Now, x < 2.5 Since x is an integer, it can be negative but cannot be a decimal So, x can be any negative number less than 3 Hence, the values of x which will satisfy the inequality are = { , 6, 5, 4, 3} Show More The word integer originated from the Latin word \u201cInteger\u201d which means whole. Integers are whole numbers and their negative opposites. Copyright \u00a9 2020 Multiply Media, LLC. Integers Integer Classes. Is 30 an integer? When multiplying integers, we can think of multiplying \"blocks\" of negative numbers. expressed in rational form as 30/1. 2020-09-30 17:50:29 2020-09-30 17:50:29. First of all we should know that \u201cEvery whole number is an integer, but every integer may not be a whole number.\u201d. An integer (from the Latin integer meaning \"whole\") is colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and \u22122048 are integers, while 9.75, 5 + 1 / 2, and \u221a 2 are not. Intager. What is the contribution of candido bartolome to gymnastics? You could say that 30/6 is just another name for 5, and then it is an integer. Like this (note: zero isn't positive or negative): For an interesting look at other types of numbers read The Evolution of Numbers, { ... \u22124, \u22123, \u22122, \u22121, 0, 1, 2, 3, 4, ... }, Adding and Subtracting Positive and Negative Numbers, Negative Integers = { ..., \u22124, \u22123, \u22122, \u22121 }, Non-Negative Integers = { 0, 1, 2, 3, 4, ... }. Essay On Nature And Environment, How To Get Rid Of Spotted Knapweed, Projection Matrix Songho, Thermomix Recipes Chinese Food, Cherry Ambrosia Salad, Burton Michigan Chamber Of Commerce, Fundamentals Of General, Organic, And Biological Chemistry Pdf, Solar Energy Activities For Elementary Students, \" />\nThe Distinctive Designer Roof\n\n# is 30 an integer\n\nAnswer by drk(1908) (Show Source): So if we consider b=1 then we get the set of integers. If x is even, then x y is even. Example 1: Input: n = 234 Output: 15 Explanation: Product of digits = 2 * 3 * 4 = 24 Sum of digits = 2 + 3 + 4 = 9 Result = 24 - 9 = 15 Example 2: Input: n = 4421 Output: 21 Explanation: Product of digits = 4 * 4 * 2 * 1 = 32 Sum of digits = 4 + 4 + 2 + 1 = 11 Result = 32 - 11 = 21 Integer definition is - any of the natural numbers, the negatives of these numbers, or zero. Whats wrong with this code: n = 10 ((n/3)).is_integer() I do not understand why I cannot set n = any number and check if it is an integer or not. So, integers can be negative {\u22121, \u22122,\u22123, \u22124, ... }, positive {1, 2, 3, 4, ... }, or zero {0}, Integers = { ..., \u22124, \u22123, \u22122, \u22121, 0, 1, 2, 3, 4, ... }, (But numbers like \u00bd, 1.1 and 3.5 are not integers). (use direct proof) 2. (n)(k+1) A. Integers vs. decimals and fractions An integer, also called a \"round number\" or \u201cwhole number,\u201d is any positive or negative number that does not include decimal parts or fractions. I have tried using the function isinteger, but I obtain, for example, isinteger(3) = 0.Apparently, any constant is double-precision by Matlab default, and is therefore not recognized as an integer. Because you can't \\\"count\\\" zero. First of all we should know that \u201cEvery whole number is an integer, but every integer may not be a whole number.\u201d. Counting Numbers are Whole Numbers, but without the zero. The opposite of 2 is -2. Integer size can be determined from PHP_INT_SIZE, maximum value from PHP_INT_MAX since PHP 4.4.0 and PHP 5.0.5.\" Integer Multiplication. when you only want positive integers, say \"positive integers\". No. When I as doing my graduation one of my professors Mr. Jaggi told us \u201cMaths student must be polite\u201d thats why I used here \u2018may\u2019. So there you Fractions, decimals, and percents are out of this basket. Integers are helpful when modeling real life situations. and which one of these would be an integer? Who is the longest reigning WWE Champion of all time? In most programming languages, you can convert a number into an integer \u2026 If you can cleanly divide a number by 2, the number is even. The integer 30 is a Composite number. Integer kan verwijzen naar: Integriteit (persoon), ... Deze pagina is voor het laatst bewerkt op 30 jul 2017 om 14:55. Does pumpkin pie need to be refrigerated? Examples of negative integers are -1, -2, -3, and -4. 0 k n k + 2. Properties of the number 30 The integer 30 is an even number. Thus, when x is a natural number, there is no solution. What are the factors of 602? 42 is greater than 30, so 30 is an abundant number. -5 + 5=0 2. 30 has an aliquot sum of 42; the second sphenic number and all sphenic numbers of this form have an aliquot sum 12 greater than themselves. You can declare an integer by putting a L after it. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. For example, 3, -10, and 1,025 are all integers, but 2.76 (decimal), 1.5 (decimal), and 3 \u00bd (fraction) are not. For example, is the number -8 a whole number? Integer Rules: A Video Watch this video to better understand the correct procedure for adding, subtracting, multiplying, and dividing positive and negative whole numbers. is YES. Furthermore, for 24 to be an integer, you should be able to write 24 without a fractional or decimal component. 2012-06-30 12:44:11 2012-06-30 12:44:11. The problem with this approach is what you consider to be practically an integer. C. The two quantities are equal. In the Registry Editor window, navigate to the folder path below depending on your Photoshop version.. Photoshop CC 2018: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\120.0 Photoshop CC 2017: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\110.0 \u2026 The opposite of -13 is 13. The square of an integer is 30 more than the integer. Examples of positive integers are 1, 2, 3, and 4. He has been teaching from the past 9 years. -----> 1. For example, 2 , 5 , 0 , \u2212 12 , 244 , \u2212 15 and 8 are all integers. Integer is a math term for a number that is a whole number. The answer to the question: \"Is 30 an integer?\" Walk through the difference between whole numbers & integers. > How do you check if a value in Java is of an integer type? MATLAB \u00ae has four signed and four unsigned integer classes. Integers are the set of whole numbers and their opposites. He has been teaching from the past 9 years. He provides courses for \u2026 These numbers are used to perform various arithmetic operations, like addition, subtraction, multiplication and division.The examples of integers are, 1, 2, 5,8, -9, -12, etc. Is it an integer? An integer.An integer.An integer.An integer. An integer, also called a \"round number\" or \u201cwhole number,\u201d is any positive or negative number that does not include decimal parts or fractions. Quantity A is greater. How to use integer in a sentence. Signed types enable you to work with negative integers as well as positive, but cannot represent as wide a range of numbers as the unsigned types because one bit is used to \u2026 $\\begingroup$ \"For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity.\" 5- (-4)=1 please help i have a test tommorrow and i just dont get what an integer is. Because you can't \"count\" zero. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if it was NaN Comment Share Find out if 24 is an Integer here! Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Yes. An even number is any integer (a whole number) that can be divided by 2. You could say that 30/6 is just another name for 5, and then it is an integer. 30 is the integer in this case. Why don't libraries smell like bookstores? integer - any of the natural numbers (positive or negative) or zero; \"an integer is a number that is not a fraction\" whole number. (30 points) An integer n is prime iff n >= 2 and n's only factors (divisors) are 1 and itself. And 30 is an integer and not a fraction. Suppose x and y are even. A comprehensive database of more than 56 integer quizzes online, test your knowledge with integer quiz questions. What are the three consecutive odd integers with sum of -30? It is a special set of whole numbers comprised of zero, positive numbers and negative numbers and denoted by the letter Z. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if \u2026 go, not everyone agrees on a simple thing! Integer Question: Is 24 an Integer? We are simply taking \"5 lots of \u22123\", like this: It should be noted that an integer can never be a fraction, a decimal or a per cent. When I as doing my graduation one of my professors Mr. Jaggi told us \u201cMaths student must be polite\u201d thats why I used here \u2018may\u2019. sound intelligent. You might be wondering, what is an opposite? python 2.7.4 error: exactly the same as integers. In fact I can't think of any proof of this sentence which doesn't prove the whole theorem along the way. Integer Question: Is 20 an Integer? The opposite of 24 is -24. The number line goes on forever in both directions. Solution for 1. Is a none repeating decimal an integer? Thanks for your help! The number 0 is also considered an integer even though it has neither a positive or negative value. What details make Lochinvar an attractive and romantic figure? Question 256872: The sum of an integer and its square is 30. 30 is the smallest sphenic number, and the smallest of the form 2 \u00d7 3 \u00d7 r, where r is a prime greater than 3. What is the conflict of the story of sinigang? What is an Integer? So they are 1, 2, 3, 4, 5, ... (and so on). What are the steps to the solution? If x is even, then x y is even. An integer is what is more commonly known as a whole number.It may be positive, negative, or the number zero, but it must be whole. In some cases, the definition of whole number will exclude the number zero, or even the set of negative numbers, but this is not as common as the more inclusive use of the term. n is an integer, and k is not an integer. De tekst is beschikbaar onder de licentie Creative Commons Naamsvermelding/Gelijk delen, er kunnen aanvullende voorwaarden van toepassing zijn. The symbol of integers is \u201c Z \u201c. See more. Examples\u2013 -2.4, 3/4, 90.6. Every integer is a rational number. Quantity B is greater. \"Natural Numbers\" can mean either \"Counting Numbers\" {1, 2, 3, ...}, or \"Whole Numbers\" {0, 1, 2, 3, ...}, depending on the subject. The set of integers is infinite and has no smallest element and no largest element. > is.integer(66L) [1] TRUE > is.integer(66) [1] FALSE Also functions like round will return a declared integer, which is what you are doing with x==round(x). This is shown below. And some people say that zero is NOT a whole number. An integer is a positive or negative whole number. The \u2026 PHP does not support unsigned integers. The basic rules of math. For 20 to be an integer, 20 has to be a whole number. How is 30 an even number? Our online integer trivia quizzes can be adapted to suit your requirements for taking some of the top integer quizzes. Furthermore, for 20 to be an integer, you should be able to write 20 without a fractional or decimal component. Definitions. You could try this. (use indirect\u2026 It is impossible to answer your question without \u2026 0 0 1 ... An integer is a number that is not a fraction, and decimal numbers are decimal fractions. There are three Proper\u2026 Integers The integers are the set of whole numbers and their opposites. Find out if 20 is an Integer here! The farther the number is to the right, the larger the value is. Base Representation; 2: \u2026 This is one of those tricky definition questions. Even and Odd Numbers (use indirect\u2026 Find the number. It cannot be done. Hello, We have a drafter who's Map 3D 2013 Drafting Settings dialog box is giving him a pop up that says \"Please enter an integer from 1 to 30\". Solution for 1. Here are some examples: We can compare integers the same way that we compare positive whole numbers. The number 30 as a Roman Numeral XXX The number 30 in various Numeral Systems. Fractions and decimals are not included in the set of integers. please give me an example or two of an integer. I want to know how to test whether an input value is an integer or not. Please note this from the Integer datatype page: \"The size of an integer is platform-dependent, although a maximum value of about two billion is the usual value (that's 32 bits signed). Some examples of integers include 1, 3, 4, 8, 99, 108, -43, \u2026 Examples of Integers \u2013 1, 6, 15. I have tried using the function isinteger, but I obtain, for example, isinteger(3) = 0.Apparently, any constant is double-precision by Matlab default, and is therefore not recognized as an integer. In the Registry Editor window, navigate to the folder path below depending on your Photoshop version.. Photoshop CC 2018: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\120.0 Photoshop CC 2017: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\110.0 Photoshop CC 2015.5: Navigate to HKEY_CURRENT_USER\\SOFTWARE\\Adobe\\Photoshop\\100.0 \u2026 For 24 to be an integer, 24 has to be a whole number. A modular type is an integer type with all arithmetic modulo a specified positive modulus; such a type corresponds to an unsigned type with wrap-around semantics. 30 is a whole number that can be written without a fractional component, thus 30 is an integer. It is not only accurate, it makes you In Maths, integers are the numbers which can be positive, negative or zero, but cannot be a fraction. An integer is any number including 0, positive numbers, and negative numbers. But every rational number is not an integer, since for b not equal to 1, if b does not divide a then w don't get an integer. Integer definition, one of the positive or negative numbers 1, 2, 3, etc., or zero. Examples of Integer Multiplication (a) Here is an integer times an integer: 5 xx -3 = -15 Why is this so? But you for sure can say it is a rational number, which is any number that can be expressed as a \u2026 Choose from 500 different sets of 30 math integers flashcards on Quizlet. How will understanding of attitudes and predisposition enhance teaching? In the equation 2 + 1/2, the number 2 is the integer and 1/2 is the fraction. Find the integer The sum of an integer and its square is - Answered by a verified Math Tutor or Teacher Whole Numbers are simply the numbers 0, 1, 2, 3, 4, 5, ... (and so on), (But numbers like \u00bd, 1.1 and 3.5 are not whole numbers.). The number line is used to represent integers. Start by using a number line. Compare -9 and -2. 0 0 1 ... 176 + 125 = 301 Therefore, the lowest positive integer you can add is 125. When did organ music become associated with baseball? An integer (pronounced IN-tuh-jer) is a whole number (not a fractional number) that can be positive, negative, or zero. An integer (from the Latin integer meaning \"whole\") is colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and \u22122048 are integers, while 9.75, 5 + 1 / 2, and \u221a 2 are not. Suppose x and y are even. I want to know how to test whether an input value is an integer or not. Where is telephone area code 035? 4. Consider a function Factor_In_Range (k , n) to return true iff some integer in the range k to n \u2026 And everyone agrees on the definition of an integer, so when in doubt say \"integer\". This is indicated by the arrows. All Rights Reserved. If a a a and b b b are integers such that a \u22c5 b = 0 a \\cdot b = 0 a \u22c5 b = 0, then a = 0 a=0 a = 0 or b = 0 b=0 b = 0. Rational number is a number which can be expressed as a/b where a,b are integers and b is not 0. a) -3 or 10 b) 3 or -10 c) -5 or 6 d) 5 or -6 Help! For 30 to be an integer, 30 has to be a whole number. Ex 6.1, 2 Solve \u201312x > 30, when x is a natural number \u201312x > 30 \u2013x > \ufdd030\ufdee12\ufdef \u2013x > 2.5 Since x is negative, we multiply both sides by -1 & change the signs (\u2013x) \u00d7 (-1) < 2.5 \u00d7 (-1) x < \u2212 2.5 (i) Since x is a natural number, it can be only positive num The base range of a signed integer type includes at least the values of the specified range. is.integer tests if the number is declared as an integer. B. An integer_type_definition defines an integer type; it defines either a signed integer type, or a modular integer type. You need to prove this, as nothing you have said makes it obvious this is true. Learn 30 math integers with free interactive flashcards. But you for sure can say it is a rational number, which is any number that can be expressed as a fraction of two integers.. An octal integer is an integer represented by a string of digits from the set $\\{0,1,2,3,4,5,6,7\\}$, each representing (in that order) the usual first eight non-negative integers. Integers are like whole numbers, but they also include negative numbers ... but still no fractions allowed! This is one of those tricky definition questions. If a is an odd integer, then a2 +3a+5 is odd. These are all integers (click to mark), and they continue left and right infinitely: Some people (not me) say that whole numbers can also be negative, which makes them So they are 1, 2, 3, 4, 5, ... (and so on). What is an integer? Therefore, these numbers can never be integers: fractions; decimals; percents; Adding and Subtracting Integers. Given an integer number n, return the difference between the product of its digits and the sum of its digits.. He provides courses for Maths and Science at Teachoo. ok i dont get.... is it just considered an integer if the dominate is 1? Counting Numbers are Whole Numbers, but without the zero. Examples of integers are: \u2026 If you divide by 2 and there is a remainder left, then the number is odd. The set of integers consists of zero (), the positive natural numbers (1, 2, 3, ...), also called whole numbers or \u2026 For example, 3, -10, and 1,025 are all integers, but 2.76 (decimal), 1.5 \u2026 Looking at a number line can help you when you need to add or subtract integers. Integers are positive whole numbers and their opposites, including zero. characteristic - the integer part (positive or negative) of the representation of a logarithm; in the expression log 643 = 2.808 the characteristic is 2 Furthermore, for 30 to be an integer, you should be able to write 30 without a fractional or decimal component. If a is an odd integer, then a2 +3a+5 is odd. For any integer a a a, the additive inverse \u2212 a-a \u2212 a is an integer. (use direct proof) 2. (ii) x is an integer Now, x < 2.5 Since x is an integer, it can be negative but cannot be a decimal So, x can be any negative number less than 3 Hence, the values of x which will satisfy the inequality are = { , 6, 5, 4, 3} Show More The word integer originated from the Latin word \u201cInteger\u201d which means whole. Integers are whole numbers and their negative opposites. Copyright \u00a9 2020 Multiply Media, LLC. Integers Integer Classes. Is 30 an integer? When multiplying integers, we can think of multiplying \"blocks\" of negative numbers. expressed in rational form as 30/1. 2020-09-30 17:50:29 2020-09-30 17:50:29. First of all we should know that \u201cEvery whole number is an integer, but every integer may not be a whole number.\u201d. An integer (from the Latin integer meaning \"whole\") is colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and \u22122048 are integers, while 9.75, 5 + 1 / 2, and \u221a 2 are not. Intager. What is the contribution of candido bartolome to gymnastics? You could say that 30/6 is just another name for 5, and then it is an integer. Like this (note: zero isn't positive or negative): For an interesting look at other types of numbers read The Evolution of Numbers, { ... \u22124, \u22123, \u22122, \u22121, 0, 1, 2, 3, 4, ... }, Adding and Subtracting Positive and Negative Numbers, Negative Integers = { ..., \u22124, \u22123, \u22122, \u22121 }, Non-Negative Integers = { 0, 1, 2, 3, 4, ... }.", "date": "2021-12-02 00:22:11", "meta": {"domain": "co.za", "url": "http://timberrooftech.co.za/id/61a53a-is-30-an-integer", "openwebmath_score": 0.4758875072002411, "openwebmath_perplexity": 554.1664265028556, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.986979510652134, "lm_q2_score": 0.8872045922259088, "lm_q1q2_score": 0.8756527542834535}}
{"url": "http://mathhelpforum.com/trigonometry/282134-when-theta-must-positive-when-negative.html", "text": "# Thread: When theta must be positive and when negative\n\n1. ## When theta must be positive and when negative\n\nI'm grappling with just not seeing when to use negative angles and when to use positive angles when solving trig functions.\n\nIf I pick a point on a graph with coordinates (5,-5) the angle formed between the x-axis and the hypotenuse can be either $-45^\\circ$ or $315^\\circ$ (along with all multiples of $360^\\circ$). Yes?\n\nIf this is true, then I would think that $\\sin(-45^\\circ) = \\sin(315^\\circ) = \\dfrac{-\\sqrt{2}}{2}$\n\nIf I plot the sine function I see that $\\sin(-45^\\circ) = \\dfrac{-\\sqrt{2}}{2}$ but my calculator gives me an error when entering $\\sin(-45^\\circ)$ or $\\sin(\\dfrac{-\\pi}{4})$\n\nLikewise, my current lesson on complex numbers requires that I use the sin and cos functions of $\\dfrac{7\\pi}{4}$ rather than $\\dfrac{-\\pi}{4}$\n\nI would appreciate an explanation of when to use a negative reference angle and when to use the positive angle $< 360^\\circ$\n\n2. ## Re: When theta must be positive and when negative\n\nOriginally Posted by B9766\nI'm grappling with just not seeing when to use negative angles and when to use positive angles when solving trig functions.\n\nIf I pick a point on a graph with coordinates (5,-5) the angle formed between the x-axis and the hypotenuse can be either $-45^\\circ$ or $315^\\circ$ (along with all multiples of $360^\\circ$). Yes?\nIf this is true, then I would think that $\\sin(-45^\\circ) = \\sin(315^\\circ) = \\dfrac{-\\sqrt{2}}{2}$\nIf I plot the sine function I see that $\\sin(-45^\\circ) = \\dfrac{-\\sqrt{2}}{2}$ but my calculator gives me an error when entering $\\sin(-45^\\circ)$ or $\\sin(\\dfrac{-\\pi}{4})$\nLikewise, my current lesson on complex numbers requires that I use the sin and cos functions of $\\dfrac{7\\pi}{4}$ rather than $\\dfrac{-\\pi}{4}$\nI would appreciate an explanation of when to use a negative reference angle and when to use the positive angle $< 360^\\circ$\nSome older texts & authors still use degrees as well as reference angles which acute. If the terminal side is in quadrant II the reference angle is $180^{\\circ}-\\theta$;\nterminal side is in quadrant III the reference angle is $180^{\\circ}+\\theta$;terminal side is in quadrant IV the reference angle is $360^{\\circ}-\\theta$.\nIn II the $\\sin$ is positive, in III the $\\tan$ is positive, & in IV the $cos$ is positive.\n\nLook Here. The basic parts of wolframalpha are free to use. The whole site is inexpensive for students & the is a version for small screen devices.\n\n3. ## Re: When theta must be positive and when negative\n\nI don't know why your calculator can't determine sin(-pi/4)). Mine is ok with it, as well as sin(7 pi/4)\n\nWhether to measure angles as clockwise from the x-axis or counterclockwise depends on the application, and may be influenced by the particulars of the problem you are working. As an example, to calculate the y-coordinate of a point on a wheel that is rotating about the origin you would use y=R sin(theta) where theta is positive if the wheel rotates counterclockwise and theta is negative if it's rotating clockwise. This allows for a nice smooth plot of theta and y versus time. As for complex analysis - you are correct that by convention angles are measured clockwise from the real axis, and of course they are given in radians.\n\n4. ## Re: When theta must be positive and when negative\n\nReally, thanks ChipB.\n\nAs for the calculator, it was user error. Much to my chagrin, I keep forgetting the TI-84 requires the use of the (-) key rather than the minus key for negative values.\n\nMy Trig and Pre-Calc course is a self-study program. I worry sometimes when I don't come up with the same answers as the text.\n\nAs an example:\n\nFind Trigonometric form of $5 - 5i$. I came up with $z = 5\\sqrt{2} (\\cos\\dfrac{-\\pi}{4}+i \\sin\\dfrac{-\\pi}{4})$\n\nBut the text answer was $z = 5\\sqrt{2} (\\cos\\dfrac{7\\pi}{4}+i \\sin\\dfrac{7\\pi}{4})$\n\nFrom your answer it seems either should be accepted as correct. Is that true? Would there normally be a preference?\n\n5. ## Re: When theta must be positive and when negative\n\nOriginally Posted by B9766\nReally, thanks ChipB.\nAs for the calculator, it was user error. Much to my chagrin, I keep forgetting the TI-84 requires the use of the (-) key rather than the minus key for negative values.\nMy Trig and Pre-Calc course is a self-study program. I worry sometimes when I don't come up with the same answers as the text.\nAs an example:\nFind Trigonometric form of $5 - 5i$. I came up with $z = 5\\sqrt{2} (\\cos\\dfrac{-\\pi}{4}+i \\sin\\dfrac{-\\pi}{4})$\nBut the text answer was $z = 5\\sqrt{2} (\\cos\\dfrac{7\\pi}{4}+i \\sin\\dfrac{7\\pi}{4})$\nFrom your answer it seems either should be accepted as correct. Is that true? Would there normally be a preference?\nThe \"angle\" associated with a complex is known as its argument. So $\\displaystyle \\arg (5 - 5\\bf{i}) = - \\frac{\\pi }{4}$\nNote that I used the negative value. In resent times that has become the preferred notation. However older texts often used the convention that\n$\\displaystyle 0\\le \\arg (z) <2\\pi$ that seems to be the case with your textbook.\nJust note that $2\\pi -\\frac{\\pi}{4}=\\frac{7\\pi}{4}$ That means that the two are equivalent.\n\n6. ## Re: When theta must be positive and when negative\n\nThanks for the confirmation, Plato.", "date": "2019-04-21 04:39:57", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/282134-when-theta-must-positive-when-negative.html", "openwebmath_score": 0.8821578025817871, "openwebmath_perplexity": 461.6424070147834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9609517095103498, "lm_q2_score": 0.9111797160416689, "lm_q1q2_score": 0.8755997058013969}}
{"url": "http://math.stackexchange.com/questions/78311/are-polynomials-dense-in-gaussian-sobolev-space", "text": "# Are polynomials dense in Gaussian Sobolev space?\n\nLet $\\mu$ be standard Gaussian measure on $\\mathbb{R}^n$, i.e. $d\\mu = (2\\pi)^{-n/2} e^{-|x|^2/2} dx$, and define the Gaussian Sobolev space $H^1(\\mu)$ to be the completion of $C_c^\\infty(\\mathbb{R}^n)$ under the inner product $$\\langle f,g \\rangle_{H^1(\\mu)} := \\int f g\\, d\\mu + \\int \\nabla f \\cdot \\nabla g\\, d\\mu.$$\n\nIt is easy to see that polynomials are in $H^1(\\mu)$. Do they form a dense set?\n\nI am quite sure the answer must be yes, but can't find or construct a proof in general. I do have a proof for $n=1$, which I can post if anyone wants. It may be useful to know that the polynomials are dense in $L^2(\\mu)$.\n\nEdit: Here is a proof for $n=1$.\n\nIt is sufficient to show that any $f \\in C^\\infty_c(\\mathbb{R})$ can be approximated by polynomials. We know polynomials are dense in $L^2(\\mu)$, so choose a sequence of polynomials $q_n \\to f'$ in $L^2(\\mu)$. Set $p_n(x) = \\int_0^x q_n(y)\\,dy + f(0)$; $p_n$ is also a polynomial. By construction we have $p_n' \\to f'$ in $L^2(\\mu)$; it remains to show $p_n \\to f$ in $L^2(\\mu)$. Now we have \\begin{align*} \\int_0^\\infty |p_n(x) - f(x)|^2 e^{-x^2/2} dx &= \\int_0^\\infty \\left(\\int_0^x (q_n(y) - f'(y)) dy \\right)^2 e^{-x^2/2} dx \\\\ &\\le \\int_0^\\infty \\int_0^x (q_n(y) - f'(y))^2\\,dy \\,x e^{-x^2/2} dx \\\\ &= \\int_0^\\infty (q_n(x) - f'(x))^2 e^{-x^2/2} dx \\to 0 \\end{align*} where we used Cauchy-Schwarz in the second line and integration by parts in the third. The $\\int_{-\\infty}^0$ term can be handled the same with appropriate minus signs.\n\nThe problem with $n > 1$ is I don't see how to use the fundamental theorem of calculus in the same way.\n\n-\n\nNate, I once needed this result, so I proved it in Dirichlet forms with polynomial domain (Math. Japonica 37 (1992) 1015-1024). There may be better proofs out there, but you could start with this paper.\n\n-\nHi Byron, I'm interested as well. I couldn't not find the paper using MathSciNet. Can I get a copy? \u2013\u00a0 Jonas Teuwen Nov 2 '11 at 21:34\nThanks Byron! I think my library has it, I'll head there right now. The MathSciNet link is ams.org/mathscinet-getitem?mr=1196376, for reference. \u2013\u00a0 Nate Eldredge Nov 2 '11 at 21:51\n@Jonas I have scanned the paper and added it to my \"Publications\" page. You want Proposition 1.3. I hope my argument holds up to scrutiny! It is not very profound, just a multi-dimensional version of Nate's argument. \u2013\u00a0 Byron Schmuland Nov 2 '11 at 22:16\n@Byron: Thanks. I've got it. \u2013\u00a0 Jonas Teuwen Nov 2 '11 at 22:16\nThanks a lot for that! \u2013\u00a0 t.b. Nov 2 '11 at 22:28\n\nByron's paper, which he linked in his (accepted) answer, has a proof in a more general setting (where the Gaussian measure can be replaced by any measure with exponentially decaying tails). Here is a specialization of it to the Gaussian case, which I wrote up to include in some lecture notes. I guess I was on the right track, the trick was to differentiate more times.\n\n$\\newcommand{\\R}{\\mathbb{R}}$ For continuous $\\psi : \\R^k \\to \\R$, let \\begin{equation*} I_i \\psi(x_1, \\dots, x_k) = \\int_0^{x_i} \\psi(x_1, \\dots, x_{i-1}, y, x_{i+1}, \\dots, x_k)dy. \\end{equation*} By Fubini's theorem, all operators $I_i, 1 \\le i \\le k$ commute. If $\\psi \\in L^2(\\mu)$ is continuous, then $I_i \\psi$ is also continuous, and $\\partial_i I_i \\psi = \\psi$. Moreover, \\begin{align*} \\int_{0}^\\infty |I_i \\psi (x_1, \\dots, x_k)|^2 e^{-x_i^2/2} dx_i &= \\int_{0}^\\infty \\big\\lvert\\int_0^{x_i} \\psi(\\dots, y,\\dots)\\,dy\\big\\rvert^2 e^{-x_i^2/2} dx_i \\\\ &\\le \\int_0^\\infty \\int_0^{x_i} |\\psi(\\dots, y, \\dots)|^2 \\,dy x_i e^{-x_i^2/2}\\,dx_i && \\text{Cauchy--Schwarz} \\\\ &= \\int_0^\\infty |\\psi(\\dots, x_i, \\dots)|^2 e^{-x_i^2/2} dx_i \\end{align*} where in the last line we integrated by parts. We can make the same argument for the integral from $-\\infty$ to $0$, adjusting signs as needed, so we have \\begin{equation*} \\int_\\R |I_i \\psi(x)|^2 e^{-x_i^2/2} dx_i \\le \\int_\\R |\\psi_i(x)|^2 e^{-x_i^2/2} dx_i. \\end{equation*} Integrating out the remaining $x_j$ with respect to $e^{-x_j^2/2}$ shows $$||{I_i \\psi}||_{L^2(\\mu)}^2 \\le ||{\\psi}||_{L^2(\\mu)}^2,$$ i.e. $I_i$ is a contraction on $L^2(\\mu)$.\n\nNow for $\\phi \\in C_c^\\infty(\\R^k)$, we can approximate $\\partial_1 \\dots \\partial_k \\phi$ in $L^2(\\mu)$ norm by polynomials $q_n$. If we let $p_n = I_1 \\dots I_k q_n$, then $p_n$ is again a polynomial, and $p_n \\to I_1 \\dots I_k \\partial_1 \\dots \\partial_k \\phi = \\phi$ in $L^2(\\mu)$. Moreover, $\\partial_i p_n = I_1 \\dots I_{i-1} I_{i+1} \\dots I_k q_n \\to I_1 \\dots I_{i-1} I_{i+1} \\dots I_k \\partial_1 \\dots \\partial_k \\phi = \\partial_i \\phi$ in $L^2(\\mu)$ also.\n\n-\n\nI think I have a different proof. Let $\\gamma$ be the Gaussian measure, that is, $\\gamma$ is given by the Radon-Nikodym density, $$\\mathrm{d}\\gamma(x) = \\frac{\\mathrm{e}^{-x^2}}{\\sqrt{\\pi}} \\mathrm{d}x.$$ Also, consider the Ornstein-Uhlenbeck operator given as, $$L := -\\frac12 \\Delta + x \\cdot \\nabla.$$ We can verify that for $u$ and $v$ in $C_{\\mathrm{c}}^\\infty(\\mathbf R^d)$ we have the symmetricity $$\\int_{\\mathbf{R}^d} u L v \\, \\gamma(\\mathrm{d}x) = \\int_{\\mathbf{R}^d} \\nabla u \\cdot \\nabla v \\, \\gamma(\\textrm{d}x).$$ The nice thing about this is that the Hermite polynomials are an orthogonal basis for the Gaussian Hilbert space, that is, $L^2(\\gamma)$. These can be define using the Rodrigues' formula, that is, $$H_n(x) = (-1)^n \\mathrm{e}^{x^2} \\partial_x^n \\mathrm{e}^{-x^2}.$$ Furthermore, we have, $$L H_n = n H_n.$$ Also, we have, $$H_n' = 2n H_{n - 1}.$$ Proving that the polynomials form a basis for $L^2(\\gamma)$ is not hard, just consider the entire function $$F(z) = \\int_{-\\infty}^\\infty \\mathrm e^{zx - x^2} \\, \\frac{\\mathrm{d}x}{\\sqrt\\pi}.$$ Hence, so are the Hermite polynomials as they are linear combinations of the monomials $(x \\mapsto x^n)_n$. Higher-order Hermite polynomials are the canonical tensor extensions. So, given an $f$ in $L^2$ we have that $f$ is given in the form $$f = \\lim_N f_N = \\lim_N \\sum_{n = 0}^N a_n \\frac{a_n}{\\sqrt{n! 2^n}}.$$ And $f_N$ gives the limit of functions that converges to $f$.\n\nAlso, recall the orthogonality (after scaling), $$\\int_{\\mathbf{R}^d} h_n h_m \\, \\gamma(\\mathrm{d}x) = \\delta_{nm}.$$ where $$h_n = \\frac{H_n}{\\sqrt{n! 2^n}}.$$\n\nWe can rewrite the inner product on the Sobolev space as $$\\langle u, v \\rangle_{H^1} = \\langle u, (L + 1) v \\rangle_{L^2(\\gamma)}.$$ We only have to care about the bilinear form $$\\mathcal E(u, v) = \\int_{\\mathbf{R}^d} u L v \\, \\gamma(\\mathrm{d}x).$$ Or so, picking $u = v = f_N - f$ we have after noting that $$g_N := f - f_N = \\sum_{n = N + 1}^\\infty a_n \\frac{H_n}{\\sqrt{n! 2^n}},$$ \\begin{align} \\mathcal E(u, v) &= \\int_{\\mathbf{R}^d} f_N L f_N \\, \\gamma(\\mathrm{d}x)\\\\ &\\quad + \\int_{\\mathbf{R}^d} f L f \\, \\gamma(\\mathrm{d}x)\\\\ &\\quad - 2\\int_{\\mathbf{R}^d} f_N L f \\, \\gamma(\\mathrm{d}x)\\\\ &= \\sum_{n = 0}^\\infty n \\frac{|a_n|^2}{n! 2^n} - \\sum_{n = 0}^N n \\frac{|a_n|^2}{n! 2^n}\\\\ &= \\sum_{n = N + 1}^\\infty n \\frac{|a_n|^2}{n! 2^n}. \\end{align}. And as $\\sum |a_n|^2$ converges due to the $L^2$ density, so should this.\n\nI hope I did not make any mistakes, just occurred to me while I was biking...\n\n-", "date": "2015-05-30 23:21:36", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/78311/are-polynomials-dense-in-gaussian-sobolev-space", "openwebmath_score": 0.9993946552276611, "openwebmath_perplexity": 351.6712166831628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406029102595, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8755884067737557}}
{"url": "http://www.ni.com/documentation/en/labview-comms/2.0/node-ref/kronecker-product/", "text": "Kronecker Product (G Dataflow)\n\nCalculates the Kronecker product of two input matrices.\n\nmatrix A\n\nThe first input matrix.\n\nThis input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.\n\nDefault: Empty array\n\nmatrix B\n\nThe second input matrix.\n\nThis input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.\n\nDefault: Empty array\n\nerror in\n\nError conditions that occur before this node runs. The node responds to this input according to standard error behavior.\n\nDefault: No error\n\nkronecker product\n\nMatrix containing the Kronecker product of the first and second input matrices.\n\nThe number of rows in kronecker product is the product of the number of rows in the first and second input matrices. The number of columns in kronecker product is the product of the number of columns in the first and second input matrices.\n\nerror out\n\nError information. The node produces this output according to standard error behavior.\n\nAlgorithm for Calculating the Kronecker Product\n\nIf A is an n-by-m matrix and B is a k-by-l matrix, the Kronecker product of A and B, C = AB, results in a matrix C with dimensions nk-by-ml. This node calculates the Kronecker product using the following equation.\n\n$C={\\left[\\begin{array}{cccc}{a}_{11}B& {a}_{12}B& \\dots & {a}_{1m}B\\\\ {a}_{21}B& {a}_{22}B& \\dots & {a}_{2m}B\\\\ \u22ee& \u22ee& \\ddots & \u22ee\\\\ {a}_{n1}B& {a}_{n2}B& \\dots & {a}_{nm}B\\end{array}\\right]}_{nk\u00d7ml}$\n\nFor example, if\n\n$\\begin{array}{cc}A=\\left[\\begin{array}{cc}1& 2\\\\ 3& 4\\end{array}\\right]& B=\\left[\\begin{array}{cc}5& 6\\\\ 7& 8\\end{array}\\right]\\end{array}$\n\nthen\n\n$\\begin{array}{ccc}{a}_{11}B=\\left[\\begin{array}{cc}5& 6\\\\ 7& 8\\end{array}\\right]& {a}_{12}B=\\left[\\begin{array}{cc}10& 12\\\\ 14& 16\\end{array}\\right]& C=\\left[\\begin{array}{cc}{a}_{11}B& {a}_{12}B\\\\ {a}_{21}B& {a}_{22}B\\end{array}\\right]=\\left[\\begin{array}{cccc}5& 6& 10& 12\\\\ 7& 8& 14& 16\\\\ 15& 18& 20& 24\\\\ 21& 24& 28& 32\\end{array}\\right]\\end{array}$\n\nWhere This Node Can Run:\n\nDesktop OS: Windows\n\nFPGA: Not supported", "date": "2018-07-19 22:57:18", "meta": {"domain": "ni.com", "url": "http://www.ni.com/documentation/en/labview-comms/2.0/node-ref/kronecker-product/", "openwebmath_score": 0.3051189184188843, "openwebmath_perplexity": 1454.2341447979604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406023459217, "lm_q2_score": 0.8840392832736083, "lm_q1q2_score": 0.8755884002229695}}
{"url": "https://math.stackexchange.com/questions/402102/what-is-the-metric-tensor-on-the-n-sphere-hypersphere/402684", "text": "# What is the metric tensor on the n-sphere (hypersphere)?\n\nI am considering the unit sphere (but an extension to one of radius $r$ would be appreciated) centered at the origin. Any coordinate system will do, though the standard angular one (with 1 radial and $n-1$ angular coordinates) would be preferable.\n\nI know that on the 2-sphere we have $ds^2 = d\\theta^2+\\sin^2(\\theta)d\\phi^2$ (in spherical coordinates) but I'm not sure how this generalizes to $n$ dimensions.\n\nAdded note: If anything can be discovered only about the determinant of the tensor (when presented in matrix form), that would also be quite helpful.\n\nI will define the metric of $$S^{n-1}$$ via pullback of the Euclidean metric on $${\\mathbb{R}}^{n}$$.\n\nTo start with we take $$n$$-dimension Cartesian co-ordinates: $$(x_1,x_2......x_n)$$. The metric here is $$g_{ij }= \\delta_{ij}$$, where $$\u03b4$$ is the Kronecker delta.\n\nWe specify the surface patches of $$S^{n-1}$$ by the parametrization $$f$$: $$x_1=r{\\cos{\\phi_1}},$$\n\n$$x_p=r{\\cos{\\phi_p}}{\\Pi_{m=1}^{p-1}}{\\sin{\\phi_{m}}},$$\n\n$$x_n=r{\\prod_{m=1}^{n-1}}{\\sin{\\phi_{m}}},$$\n\nWhere $$r$$ is the radius of the hypersphere and the angles have the usual range.\n\nWe see that the pullback of the Euclidean metric $$g'_{ab} = (f^*g)_{ab}$$ is the metric tensor of the hypersphere. Its components are:\n\n$$g'_{ab} = g_{ij} {\\frac{\\partial{x_i}}{\\partial{\\phi_a}}} {\\frac{\\partial{x_j}}{\\partial{\\phi_b}}} = {\\frac{\\partial{x_i}}{\\partial{\\phi_a}}}{\\frac{\\partial{x_i}}{\\partial{\\phi_b}}}$$\n\nWe get $$2$$ cases here:\n\ni) $$a>b$$ or $$b>a$$, For these components one obtains a series of terms with alternating signs which vanishes, $$g'_{ab}=0$$ and thus all off-diagonal components of the tensor vanish.\n\nii) $$a=b$$,\n\n$$g'_{11}=1$$\n\n$$g'_{aa} ={r^2} \\prod_{m=1}^{a-1} \\sin^2{\\phi_{m}}$$\nwhere $$2\\leq a\\leq {n-1}$$\n\nThe determinant is very straightforward to calculate:\n\n$$\\det{(g'_{ab})} = {r^2} \\prod_{m=1}^{n-1} g'_{mm}$$\n\nFinally, we can write the metric of the hypersphere as:\n\n$$g' = {r^2} \\, d\\phi_{1}\\otimes d\\phi_{1} + {r^2} \\sum_{a=2}^{n-1} \\left( \\prod_{m=1}^{a-1} \\sin^2{\\phi_{m}} \\right) d\\phi_{a} \\otimes d\\phi_{a}$$\n\n\u2022 Great, thanks for the step-by-step explanation. Although what you've written is fairly clear, you can make it clearer by using the TeX environment, with the $ delimiters, like instead of g'r\u03d5k, write $g'_{r\\phi_k}$, which comes out as$g'_{r\\phi_k}$. May 27, 2013 at 11:39 \u2022 This is a very good answer and very useful too, as it is not that easy to find this piece of information on the net. Proper formatting would make it great. Oct 20, 2014 at 9:12 \u2022 @GiuseppeNegro Done. Thanks for the feedback. Nov 10, 2014 at 14:04 \u2022 It seems$g'_{11}=r^2$, no? Nov 13, 2018 at 14:17 \u2022 Can somebody explain more thoroughly why the off diagonal terms vanish? Apr 16, 2019 at 11:15$\\newcommand{\\Reals}{\\mathbf{R}}$For posterity: Fix$r > 0$, and let$S^{n}(r)$denote the sphere of radius$r$centered at the origin in$\\Reals^{n+1}$. Stereographic projection from the north pole$N = (0, \\dots, 0, 1)$on the unit sphere$S^{n} = S^{n}(1)$defines a diffeomorphism$\\Pi_{N}:S^{n} \\setminus \\{N\\} \\to \\Reals^{n}given in Cartesian coordinates by \\begin{align*} \\Pi_{N}(x_{1}, \\dots, x_{n}, x_{n+1}) &= \\frac{1}{1 - x_{n+1}}(x_{1}, \\dots, x_{n}), \\\\ \\Pi_{N}^{-1}(t_{1}, \\dots, t_{n}) &= \\frac{(2t_{1}, \\dots, 2t_{n}, \\|t\\|^{2} - 1)}{\\|t\\|^{2} + 1}. \\end{align*} In these coordinates, the induced (round) metric on the unit sphere is well-known (and easily checked) to be conformally-Euclidean: $$g(t) = \\frac{4 (dt_{1}^{2} + \\cdots + dt_{n}^{2})}{(\\|t\\|^{2} + 1)^{2}}.$$ Stereographic projection from the north pole(0, \\dots, 0, r)$of$S^{n}(r)$is given by the scaled mapping$x \\mapsto t = r\\Pi_{N}(x/r)$, whose inverse is$t \\mapsto x = r\\Pi_{N}^{-1}(t/r)\\$, i.e., \\begin{align*} r\\Pi_{N}(x_{1}/r, \\dots, x_{n}/r, x_{n+1}/r) &= \\frac{1}{r - x_{n+1}}(x_{1}, \\dots, x_{n}), \\\\ r\\Pi_{N}^{-1}(t_{1}/r, \\dots, t_{n}/r) &= \\frac{\\bigl(2t_{1}, \\dots, 2t_{n}, r(\\|t/r\\|^{2} - 1)\\bigr)}{\\|t/r\\|^{2} + 1}. \\end{align*} The induced metric in these coordinates is consequently $$r^{2} g(t/r) = \\frac{4 (dt_{1}^{2} + \\cdots + dt_{n}^{2})}{(\\|t/r\\|^{2} + 1)^{2}} = \\frac{4r^{4} (dt_{1}^{2} + \\cdots + dt_{n}^{2})}{(\\|t\\|^{2} + r^{2})^{2}}.$$", "date": "2022-09-28 05:45:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/402102/what-is-the-metric-tensor-on-the-n-sphere-hypersphere/402684", "openwebmath_score": 0.9999473094940186, "openwebmath_perplexity": 1055.699321694424, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406006529085, "lm_q2_score": 0.8840392802184581, "lm_q1q2_score": 0.8755883957003345}}
{"url": "https://math.stackexchange.com/questions/4039091/a-problem-in-understanding-the-intermediate-value-theorem", "text": "# A problem in understanding the Intermediate Value Theorem\n\nSo, IVT essentially says if a function $$f$$ is continuous over an interval $$[a,b]$$, then the function $$f$$ will take up all the values between $$f(a)$$ and $$f(b)$$ at least once at some point in the interval $$[a,b]$$.\n\nNow, suppose $$f$$ is continuous between $$[3,7]$$ and $$f(3) = 4$$ and $$f(7) = 25$$\n\nAccording to IVT, $$f(3) = 4 < m < 25 = f(7)$$, i.e., the function takes up all the values between $$f(3)$$ and $$f(7)$$ at least once in the interval. But, according to the below picture, clearly, the function within that interval takes up values different that do not satisfy the above inequality. Please help me fill the gap in my understanding of IVT here.\n\nPlease do correct me if I'm mistaken.\n\nThis doubt arose from the below KhanAcademy question\n\n\u2022 IVT doesn't tell you that ALL values must be between $f(a)$ and $f(b)$ -- it tells you that every value between $f(a)$ and $f(b)$ must be attained. \u2013\u00a0Nick Peterson Feb 25 at 6:59\n\u2022 The IVT says that $f$ reaches all values between $f(3)$ and $f(7)$. It doesn't say it can't reach other values as well. \u2013\u00a0Mathematician 42 Feb 25 at 6:59\n\u2022 @SathvikR. If it helps: you seem to be interpreting it as: $x\\in(a,b)$ implies $f(x)\\in (f(a), f(b))$ (assuming $f(a)<f(b)$, etc). But the actual implication is: for all $y\\in (f(a), f(b))$, there exists $x\\in(a, b)$ so that $f(x)=y$. \u2013\u00a0Nick Peterson Feb 25 at 7:06\n\u2022 IVT only says that all values between $4$ and $25$ will be obtained. And in that picture they clearly all are. The IVT says NOTHING at all about values outside $[4,25]$. It doesn't say they are obtained. It doesn't say they aren't obtained. \u2013\u00a0fleablood Feb 25 at 7:21\n\u2022 Okay.... it says that every possible output value between $f(3) = 4$ and $f(7) = 25$ will be obtained somewhere in the interval between $3$ and $7$. But it doesn't say that values between $4$ and $25$ are the only values obtained. If I told you that every dog in the Golden Gate Dog shelter has an owner who lives in San Francisco, that does not mean everyone who lives in San Francisco owns a dog in the Golden Gate Dog shelter.... to be continued.... \u2013\u00a0fleablood Feb 25 at 7:29\n\nThe intermediate value theorem tells you that, in that context, for each $$y\\in[4,25]$$, there is some $$x\\in[3,4]$$ such that $$f(x)=y$$. It does not tell you that if $$x\\in[3,4]$$, then $$f(x)\\in[4,25]$$. So, there is no contradiction if $$f\\bigl([3,4]\\bigr)$$ contains values outside $$[4,25]$$.", "date": "2021-04-19 19:52:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4039091/a-problem-in-understanding-the-intermediate-value-theorem", "openwebmath_score": 0.5090106129646301, "openwebmath_perplexity": 149.13453172252275, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357227168957, "lm_q2_score": 0.8918110418436166, "lm_q1q2_score": 0.8755227576912507}}
{"url": "http://mathhelpforum.com/algebra/4125-help-word-problem.html", "text": "# Thread: Help with this word problem!\n\n1. ## Help with this word problem!\n\nThe value of a new car of a particular make, A, decreases by 22.5% during the first year. In the second year, its value decreases by 12.5% of its value at the beginning of the year. The value of a new car of another make, B, decreases by 25% during the first year. In the second year, its value decreases by 15% of its value at the beginning of the year. A new car of make A costs $48,000 and a new car of make B costs$42,000.\n\n(a) What is the difference between the values of the two cars at the end of the firts year? Express this differnce as a percentage of the value of the car of make A at the end of the first year.\n\n(b) Calculate the difference between the values of the two cars at the end of the second year.\n\nThankyou!\n\n2. Originally Posted by laser2302\n\nThe value of a new car of a particular make, A, decreases by 22.5% during the first year. In the second year, its value decreases by 12.5% of its value at the beginning of the year. The value of a new car of another make, B, decreases by 25% during the first year. In the second year, its value decreases by 15% of its value at the beginning of the year. A new car of make A costs $48,000 and a new car of make B costs$42,000.\n\n(a) What is the difference between the values of the two cars at the end of the firts year? Express this differnce as a percentage of the value of the car of make A at the end of the first year.\n\n(b) Calculate the difference between the values of the two cars at the end of the second year.\n\nThankyou!\nIf something decreases by 22.5% means it is worth 77.5% perfect of the original price.\nIn the beggining the car was worth $x$ after the first year is it $.775x$ after the second year it decreases by 12.5% meaning its value is 87.5% of the new price thus,\n$.875(.775x)=.678125x$\nSimilarily the second car after two years is,\n$.85(.75x)=.6375x$ and $.75x$ after one year.\n\nThe first problem asks for the difference after the first year.\nThe first car is worth $.775\\times 48,000=37,200$. The second car is worth $.75\\times 42,000=31,500$ The difference is,\n$37,200-31,500=5,700$\n\nAfter two years, you have the new values,\n$.875\\times 37,200=32,250$\n$.85\\times 31,500=26,775$\nTheir difference is,\n$32,250-26,775= 5475$\n\n3. Originally Posted by ThePerfectHacker\nThe first problem asks for the difference after the first year.\nThe first car is worth $.775\\times 48,000=37,200$. The second car is worth $.75\\times 42,000=31,500$ The difference is,\n$37,200-31,500=5,700$\nThe question asks to write this out in terms of the percentage of the price of car A, so you divide 5,700 by the price of car A\n\n$\\frac{5700}{37200}=0.153225806\\approx\\boxed{15\n\\text\n{percent}}$\n\n4. Hello, laser2302!\n\nAre you sure you can't work this out yourself?\n\n[quote]The value of a new car of a particular make, $A$, decreases by 22.5% during the first year.\nIn the second year, its value decreases by 12.5% of its value at the beginning of the year.\n\nThe value of a new car of another make, $B$, decreases by 25% during the first year.\nIn the second year, its value decreases by 15% of its value at the beginning of the year.\n\nA new car of make $A$ costs $48,000 and a new car of make $B$ costs$42,000.\n\n(a) What is the difference between the values of the two cars at the end of the first year?\nExpress this differnce as a percentage of the value of the car of make A at the end of the first year.\n\n(b) Calculate the difference between the values of the two cars at the end of the second year.\n\nDuring year 1, make $A$ loses $22.5\\%\\text{ of }\\48,000 \\:=\\:0.225 \\times 48,000 \\:=\\:\\10,800$\n. . . At the end of year 1, it is worth only: $\\48,000 - 10,800 \\:= \\:\\37,200$\n\nDuring the year 2 $A$ loses $12.5\\%\\text{ of }\\37,200 \\:=\\:0.125 \\times 37,200 \\:= \\:\\4,650$\n. . . At the end of year 2, it is worth only: $\\37,200 - 4,650\\:=\\:\\32,550$\n\nDuring year 1, make $B$ loses $25\\%\\text{ of }\\42,000\\:=\\:0.25 \\times 42,000 \\:=\\:\\10,500$\n. . . At the end of year 1, it is worth only: $\\42,000 - 10,500\\:=\\:\\31,500$\n\nDuring year 2, $B$ loses $15\\%\\text{ of }\\31,500 \\:=\\:0.15 \\times 31,500\\:=\\:\\4,725$\n. . . At the end of year 2, it is worth only: $\\31,500 - 4,725\\:=\\:\\26,775$\n\nNow that we've worked all the numbers, we can answer the questions.\n\n(a) The difference in value at the end of year 1 is:\n. . . . . $\\37,200 - 31,500\\:=\\:\\5,700$ . . .\nBoy, that was hard!\n\nThe percentage they asked for is: $\\frac{5.700}{37,200}\\:=\\:0.153225806\\:\\approx\\:15. 3\\%$\n\n(b) The difference in value at the end of year 2 is:\n. . . . . $\\32,550 - 26,775\\:=\\:\\5,775$ . . .\nI need a nap!\n\n5. Originally Posted by Soroban\nHello, laser2302!\n\nAre you sure you can't work this out yourself?\nYou know you really shouldn't be so critical, after all, people come here for reassurance and help (or, unfortunately, so they don't have to do the work) and should be treated well.", "date": "2017-01-22 04:01:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/4125-help-word-problem.html", "openwebmath_score": 0.5958762764930725, "openwebmath_perplexity": 405.6873076072442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777175525674, "lm_q2_score": 0.88720460564669, "lm_q1q2_score": 0.8754732548734102}}
{"url": "http://mathhelpforum.com/algebra/6659-greatest-common-factor.html", "text": "1. ## Greatest Common Factor\n\nIs there an equation for the greatest common factor of two numbers?\n\n2. Originally Posted by Quick\nIs there an equation for the greatest common factor of two numbers?\nEquation, not that I know of. Process, yes.\n\nConsider the GCF of 60 and 630.\n\nThe prime factorization of 60 is $\\displaystyle 2^2 \\cdot 3 \\cdot 5$.\nThe prime factorization of 630 is $\\displaystyle 2 \\cdot 3^2 \\cdot 5 \\cdot 7$.\n\nThe GCF (also known as the Greatest Common Divisor, GCD) will be the number that has a prime factorization that contains the same prime factors as the combination of the lists. In other words,\nThere is a factor of 2 common to each,\nthere is a factor of 3 common to each,\nthere is a factor of 5 common to each.\n\nThus GCF(60, 630) = 2*3*5 = 30.\n\nIf we were talking about GCF(60, 1260) then ($\\displaystyle 1260 = 2^2 \\cdot 3^2 \\cdot 5 \\cdot 7$):\nThere are two factors of 2 common to each,\nthere is a factor of 3 common to each,\nthere is a factor of 5 common to each.\n\nThus GCF(60, 1260) = $\\displaystyle 2^2 \\cdot 3 \\cdot 5$ = 60.\n\n-Dan\n\nPS Now that I think of it, there is a formula, but it isn't anything direct:\nGiven two numbers x and y, we know that GCF(x, y) = (x*y)/LCM(x, y), where LCM(x, y) is the \"Least Common Multiple\" of x and y. There is no direct formula I know of to find the LCM either.\n\n3. Originally Posted by topsquark\nEquation, not that I know of. Process, yes.\n\nConsider the GCF of 60 and 630.\n\nThe prime factorization of 60 is $\\displaystyle 2^2 \\cdot 3 \\cdot 5$.\nThe prime factorization of 630 is $\\displaystyle 2 \\cdot 3^2 \\cdot 5 \\cdot 7$.\n\nThe GCF (also known as the Greatest Common Divisor, GCD) will be the number that has a prime factorization that contains the same prime factors as the combination of the lists. In other words,\nThere is a factor of 2 common to each,\nthere is a factor of 3 common to each,\nthere is a factor of 5 common to each.\n\nThus GCF(60, 630) = 2*3*5 = 30.\n\nIf we were talking about GCF(60, 1260) then ($\\displaystyle 1260 = 2^2 \\cdot 3^2 \\cdot 5 \\cdot 7$):\nThere are two factors of 2 common to each,\nthere is a factor of 3 common to each,\nthere is a factor of 5 common to each.\n\nThus GCF(60, 1260) = $\\displaystyle 2^2 \\cdot 3 \\cdot 5$ = 60.\n\n-Dan\n\nPS Now that I think of it, there is a formula, but it isn't anything direct:\nGiven two numbers x and y, we know that GCF(x, y) = (x*y)/LCM(x, y), where LCM(x, y) is the \"Least Common Multiple\" of x and y. There is no direct formula I know of to find the LCM either.\nIn case you were wondering there is a simple process to find the LCM of two numbers as well. Let's take 60 and 630 again.\n\nThe prime factorization of 60 is $\\displaystyle 2^2 \\cdot 3 \\cdot 5$.\nThe prime factorization of 630 is $\\displaystyle 2 \\cdot 3^2 \\cdot 5 \\cdot 7$.\n\nThe Least Common Multiple will have a prime factorization that is the smallest list of prime factors that completely contains both lists. In other words:\nThere are two factors of 2,\nthere are two factors of 3,\nThere is one factor of 5,\nThere is one factor of 7.\n\nThus LCM(60, 630) = $\\displaystyle 2^2 \\cdot 3^2 \\cdot 5 \\cdot 7$ = 1260.\n\n(And we can check the equation I gave in the Post Script of the last post:\nGCF(60, 630) = (60*630)/LCM(60, 630) = 37800/1260 = 30 as we found in the last post.)\n\n-Dan\n\n4. Originally Posted by topsquark\n...\nThe Least Common Multiple will have a prime factorization that is the smallest list of prime factors that completely contains both lists. ...\nHi,\n\nI taught my pupils an algorithm to calculate the LCM in a tabel: This method works too when you have more than two numbers (for instance when you calculate the sum of fractions). You only have to know the first 20 prime-factors :\n\n5. This seems to be an excellent question for Quicklopedia, my method doth no rely on factorization. Rather on an algorithm (with amazing complexity i.e. really fast) called Euclidean algorithm. I can show it to thee.\n\n6. Originally Posted by ThePerfectHacker\nThis seems to be an excellent question for Quicklopedia, my method doth no rely on factorization. Rather on an algorithm (with amazing complexity i.e. really fast) called Euclidean algorithm. I can show it to thee.\nActually, for once I found wikipedia gave a good definition (actually, it gave a poor definition for it, but an excellent summary in the \"Greatest Common Denominator\" section)", "date": "2018-03-19 22:59:32", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/6659-greatest-common-factor.html", "openwebmath_score": 0.7328869104385376, "openwebmath_perplexity": 593.253405818335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771782476948, "lm_q2_score": 0.8872045929715078, "lm_q1q2_score": 0.8754732447808191}}
{"url": "https://math.stackexchange.com/questions/2033272/how-to-find-a-inverse-of-a-multivariable-function", "text": "# How to find a inverse of a multivariable function?\n\nI have a function $f:\\mathbb{R}^{2} \\rightarrow \\mathbb{R}^{2}$ defined as:\n\n$$f(x,y) = (3x-y, x-5y)$$\n\nI proved that it's a bijection, now I have to find the inverse function $f^{-1}$.\n\nBecause $f$ is a bijection, it has a inverse and this is true:\n\n$$(f^{-1}\\circ f)(x,y) = (x,y)$$\n\n$$f^{-1}(3x-y,x-5) = (x,y)$$\n\nI don't know where to go from here. In a one variable function I would do a substitution of the argument of $f^{-1}$ with a variable and express x with that variable, and then just switch places.\n\nI tried to do a substitution like this:\n\n$$3x-y = a$$ $$x-5y = b$$\n\nAnd then express $x$ and $y$ by $a$ and $b$ , and get this:\n\n$$f^{-1}(x,y) = (\\frac{15x-3y}{42}, \\frac{x-3y}{14})$$\n\nBut I'm not sure if I'm allowed to swap $x$ for $a$, and $y$ for $b$.\n\nAny hint is highly appreciated.\n\n\u2022 Note that $f$ is linear, so you could write it as a matrix and then calculate the inverse. \u2013\u00a0PPR Nov 27 '16 at 19:27\n\u2022 Recall that when finding the inverse of a bijective function of a singular variable of the form $y=f(x)$ you also \"swap the variables.\" \u2013\u00a0John Wayland Bales Nov 27 '16 at 19:30\n\u2022 @JohnWaylandBales Yes, should I do it here as well? And, if so, what sould I swap? \u2013\u00a0nikol_kok Nov 27 '16 at 19:32\n\u2022 Your work is absolutely correct as written. You could reduce the first fraction of your final result by a factor of $3$ so that it also had a denominator of $14$. But you swap $x\\leftrightarrow a$ and $y\\leftrightarrow b$. \u2013\u00a0John Wayland Bales Nov 27 '16 at 19:38\n\u2022 @JohnWaylandBales So this is one reliable way of doing this? I mean, I can always swap those two when I'm in a situation like this? \u2013\u00a0nikol_kok Nov 27 '16 at 19:41\n\nYou have a linear function here, given by the matrix $$\\begin{bmatrix}3&-1\\\\1&-5 \\end{bmatrix}$$ Can you invert the above?\n\n\u2022 @qbert Yes, I know how to find the inverse of that matrix, but I don't know what to do with it :( \u2013\u00a0nikol_kok Nov 27 '16 at 19:31\n\u2022 Write it as your answer. Your linear map is given by the above matrix, the inverse map will be given by the inverse of the above matrix ( with respect to the standard basis) \u2013\u00a0qbert Nov 27 '16 at 19:41\n\u2022 If you would like to write it in the same form you can apply the inverse matrix to the vector (x,y) \u2013\u00a0qbert Nov 27 '16 at 19:42\n\u2022 Don't get me wrong, but I'm not sure what you are saying here. Can you point me to some theory or an explanation of that method? \u2013\u00a0nikol_kok Nov 27 '16 at 19:50\n\u2022 @nikol_kok If I understood that correctly, you actually get $f^{-1}(a, b) = (\\frac{5a - b}{14}, \\frac{a - 3b}{14})$, and you're wondering whether it's OK to just swap $a$ and $b$ for $x$ and $y$? Yes, that works just fine. \u2013\u00a0Arthur Nov 27 '16 at 19:55\n\nYou can split this into two separate functions $u, v:\\Bbb R^2\\to \\Bbb R$ the following way: $$u(x, y) = 3x-y\\\\ v(x, y) = x-5y$$ and we have $f(x, y) = (u(x, y), v(x, y))$. What we want is $x$ and $y$ expressed in terms of $u$ and $v$, i.e. solve the above set of equations for $x$ and $y$, so that you get two functions $x(u, v), y(u, v)$. Then $f^{-1}(u, v) = (x(u,v), y(u,v))$.\n\n\u2022 What set of equations am I supposed to solve? \u2013\u00a0nikol_kok Nov 27 '16 at 19:47\n\u2022 @nikol_kok You should solve the equations $$u= 3x-y\\\\ v= x-5y$$ for $x$ and $y$. This is exactly corresponding to the fact that in order to find the inverse of, say, $g(x) = 5x + 3$, you solve $g = 5x + 3$ for $x$, only in higher dimension. \u2013\u00a0Arthur Nov 27 '16 at 19:49\n\u2022 In my question I tried a solution like this. Can you check if it's the same/similar? \u2013\u00a0nikol_kok Nov 27 '16 at 19:52", "date": "2019-12-12 21:58:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2033272/how-to-find-a-inverse-of-a-multivariable-function", "openwebmath_score": 0.8176343441009521, "openwebmath_perplexity": 147.99201033851827, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771782476948, "lm_q2_score": 0.8872045832787204, "lm_q1q2_score": 0.8754732352161977}}
{"url": "http://math.stackexchange.com/questions/303779/different-solutions-of-trigonometric-equations", "text": "# Different solutions of trigonometric equations\n\nPlease take a look at this trigonometric equation,\n$\\cos9x\\cos7x = \\cos5x\\cos3x$\n\nTo solve this equation, we can proceed as,\n\n$2\\cos9x\\cos7x = 2\\cos5x\\cos3x$\nor, $\\cos(9x+7x)+\\cos(9x-7x) = \\cos(5x+3x)+\\cos(5x-3x)$\nor, $\\cos16x+\\cos2x = \\cos8x+\\cos2x$\nor, $\\cos16x = \\cos8x$\n\nFrom the above situation we can proceed in two ways,\n\nFirst Way\n$\\cos16x = \\cos8x$\nor, $\\cos(2\\times8x) = \\cos8x$\nor, $2\\cos^28x -1 = \\cos8x$\nor, $2\\cos^28x-\\cos8x -1 = 0$\nor, $2\\cos^28x-2\\cos8x+\\cos8x -1 = 0$\nor, $2\\cos8x(\\cos8x-1)+(\\cos8x -1) = 0$\nor, $2\\cos8x(\\cos8x-1)+(\\cos8x -1) = 0$\nor, $(\\cos8x -1)(2\\cos8x+1) = 0$\n\nEither or both of the above factors are zero.\n\nTaking the first one,\n$(\\cos8x -1) = 0$\nor, $\\cos8x = 1$\nor, $8x = 2n\\pi$, where $n$ is an integer, +ve or -ve.\nor, $x = {n\\pi \\over4}$\n\nTaking the second one,\n$(2\\cos8x+1) = 0$\nor, $2\\cos8x = -1$\nor, $\\cos8x = -\\frac1 2$\nor, $8x = 2n\\pi \\pm \\frac {2\\pi} 3$, where $n$ is an integer, +ve or -ve.\nor, $x = \\frac{n\\pi} 4 \\pm \\frac {\\pi} {12}$\n\nSecond Way\n$\\cos16x = \\cos8x$\nor, $2\\sin{8x-16x\\over2}\\sin{8x+16x\\over2} = 0$\nor, $2\\sin(-4x)\\sin{12x} = 0$\nor, $-2\\sin(4x)\\sin{12x} = 0$\n\nAgain, either or both of the above factors are zero.\n\nTaking the first one,\n$\\sin4x = 0$\nor, $4x=n\\pi$\nor, $x=\\frac{n\\pi}4$\n\nTaking the second one,\n$\\sin12x = 0$\nor, $12x=n\\pi$\nor, $x=\\frac{n\\pi}{12}$\n\nNow, as you must have noticed, we are getting two different sets of solutions, $x = \\left\\{{n\\pi \\over4}, \\frac{n\\pi} 4 \\pm \\frac {\\pi} {12}\\right\\}$ and $x = \\left\\{{n\\pi \\over4}, \\frac{n\\pi}{12}\\right\\}$. The member $x=\\frac{n\\pi}4$ is common to both of the sets. Moreover, all these solutions satisfy the equation under consideration.\n\nCould anybody please tell me why is this happening. In addition to the specific answer, some general insight will be most welcome. We have got a number of similar problems in hand. So, unless we can develop some acumen, life may become difficult.\n\n-\n\nIf $x\\in\\{\\frac{n\\pi}4,\\frac{n\\pi}4\\pm\\frac\\pi{12}\\mid n\\in\\mathbb Z\\}$, then $12x=3n\\pi$ or $12x=3n\\pi+\\pi$ or $12x=3n\\pi-\\pi$ for some $n\\in\\mathbb Z$, hence $12x=m\\pi$ for some $m\\in\\mathbb Z$ (namely $m=3n$ or $m=3n+1$ or $m=3n-1$. On the other hand, every $m\\in \\mathbb Z$ can be written either as $m=3n$ or $m=3n+1$ or $m=3n-1$, depending on the remainder modulo $3$. Therefore $x\\in\\{\\frac{n\\pi}4,\\frac{n\\pi}4\\pm\\frac\\pi{12}\\mid n\\in\\mathbb Z\\}\\iff \\frac{12}\\pi x\\in\\mathbb Z$.\nMoreover, if $x=\\frac{n\\pi}4$ with $n\\in \\mathbb Z$, then $x=\\frac{m\\pi}{12}$ with $m:=3n\\in\\mathbb Z$, hence $\\{\\frac {n\\pi}4\\mid n\\in\\mathbb Z\\}\\subseteq \\{\\frac {n\\pi}{12}\\mid n\\in\\mathbb Z\\}$, which ultimately makes $$\\left\\{\\frac{n\\pi}4,\\frac{n\\pi}4\\pm\\frac\\pi{12}\\mid n\\in\\mathbb Z\\right\\} = \\left\\{\\frac{n\\pi}4,\\frac{n\\pi}{12}\\mid n\\in\\mathbb Z\\right\\}=\\left\\{\\frac{n\\pi}{12}\\mid n\\in\\mathbb Z\\right\\}.$$\nThey aren't actually different sets of numbers. They are both all the numbers of the form $n\\pi/12$.\nCould you please elaborate a bit, are you suggesting that ${n \\pi \\over 4} \\pm {\\pi \\over 12}$ and ${n \\pi \\over 12}$ are the same numbers? \u2013\u00a0 Masroor Feb 14 '13 at 6:23\n$\\left\\{ \\frac{n\\pi}{4}, \\frac{n\\pi}{4} \\pm \\frac{\\pi}{12} \\right\\} = \\left\\{ \\frac{3n\\pi}{12}, \\frac{(3n \\pm 1)\\pi}{12}\\right\\} = \\left\\{ \\frac{(3n-1)\\pi}{12}, \\frac{3n\\pi}{12}, \\frac{(3n+1)\\pi}{12} \\right\\} = \\left\\{\\frac{n\\pi}{12}\\right\\}$, since any integer can be written as one of $3n-1$, $3n$, or $3n+1$. Also, $\\left\\{\\frac{n\\pi}{4}, \\frac{n\\pi}{12}\\right\\} = \\left\\{\\frac{3n\\pi}{12}, \\frac{n\\pi}{12} \\right\\} = \\left\\{\\frac{n\\pi}{12} \\right\\}$, because \"all integer multiples of $\\pi/12$\" includes \"all integer multiples of $3\\pi/12$\". \u2013\u00a0 Blue Feb 14 '13 at 8:12", "date": "2014-11-26 02:04:38", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/303779/different-solutions-of-trigonometric-equations", "openwebmath_score": 0.9604728817939758, "openwebmath_perplexity": 105.21846487828303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429142850274, "lm_q2_score": 0.8887587839164801, "lm_q1q2_score": 0.8754655426055066}}
{"url": "http://anemon.gr/ll1mvq6q/2979fc-how-to-simplify-radicals", "text": "Perfect squares are numbers that are equal to a number times itself. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and \"take out\" one copy of anything that is a square. Since 72 factors as 2\u00d736, and since 36 is a perfect square, then: Since there had been only one copy of the factor 2 in the factorization 2\u00a0\u00d7\u00a06\u00a0\u00d7\u00a06, the left-over 2 couldn't come out of the radical and had to be left behind. In other words, we can use the fact that radicals can be manipulated similarly to powers: There are various ways I can approach this simplification. One rule is that you can't leave a square root in the denominator of a fraction. But my steps above show how you can switch back and forth between the different formats (multiplication inside one radical, versus multiplication of two radicals) to help in the simplification process. Solution : \u221a(5/16) = \u221a5 / \u221a16 \u221a(5/16) = \u221a5 / \u221a(4 \u22c5 4) Index of the given radical is 2. Quotient Rule . For the purpose of the examples below, we are assuming that variables in radicals are non-negative, and denominators are nonzero. Simplifying Radicals Coloring Activity. Required fields are marked * Comment. Then simplify the result. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Determine the index of the radical. This calculator simplifies ANY radical expressions. In this case, the index is two because it is a square root, which \u2026 ... Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. Indeed, we can give a counter example: $$\\sqrt{(-3)^2} = \\sqrt(9) = 3$$. I used regular formatting for my hand-in answer. Simplifying Radicals Activity. Then, there are negative powers than can be transformed. We can raise numbers to powers other than just 2; we can cube things (being raising things to the third power, or \"to the power 3\"), raise them to the fourth power (or \"to the power 4\"), raise them to the 100th power, and so forth. This theorem allows us to use our method of simplifying radicals. How do I do so? (Much like a fungus or a bad house guest.) We are going to be simplifying radicals shortly so we should next define simplified radical form. type (2/ (r3 - 1) + 3/ (r3-2) + 15/ (3-r3)) (1/ (5+r3)). Step 2. In simplifying a radical, try to find the largest square factor of the radicand. When doing your work, use whatever notation works well for you. In reality, what happens is that $$\\sqrt{x^2} = |x|$$. For example. Short answer: Yes. Sign up to follow my blog and then send me an email or leave a comment below and I\u2019ll send you the notes or coloring activity for free! root(24)=root(4*6)=root(4)*root(6)=2root(6) 2. simplifying square roots calculator ; t1-83 instructions for algebra ; TI 89 polar math ; simplifying multiplication expressions containing square roots using the ladder method ; integers worksheets free ; free standard grade english past paper questions and answers All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. All exponents in the radicand must be less than the index. In the first case, we're simplifying to find the one defined value for an expression. \"Roots\" (or \"radicals\") are the \"opposite\" operation of applying exponents; we can \"undo\" a power with a radical, and we can \"undo\" a radical with a power. Finance. Step 1 : Decompose the number inside the radical into prime factors. How do we know? Just to have a complete discussion about radicals, we need to define radicals in general, using the following definition: With this definition, we have the following rules: Rule 1.1: \u00a0\u00a0 $$\\large \\displaystyle \\sqrt[n]{x^n} = x$$, when $$n$$ is odd. We know that The corresponding of Product Property of Roots says that . Well, simply by using rule 6 of exponents and the definition of radical as a power. Quotient Rule . To simplify a square root: make the number inside the square root as small as possible (but still a whole number): Example: \u221a12 is simpler as 2\u221a3. Then: katex.render(\"\\\\sqrt{144\\\\,} = \\\\mathbf{\\\\color{purple}{ 12 }}\", typed01);12. Simplify each of the following. In this particular case, the square roots simplify \"completely\" (that is, down to whole numbers): Simplify: I have three copies of the radical, plus another two copies, giving me\u2014 Wait a minute! Another rule is that you can't leave a number under a square root if it has a factor that's a perfect square. What about more difficult radicals? We will start with perhaps the simplest of all examples and then gradually move on to more complicated examples . where a \u2265 0, b > 0 \"The square root of a quotient is equal to the quotient of the square roots of the numerator and denominator.\" Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. No radicals appear in the denominator. Step 2 : If you have square root (\u221a), you have to take one term out of the square root for every two same terms multiplied inside the radical. But when we are just simplifying the expression katex.render(\"\\\\sqrt{4\\\\,}\", rad007A);, the ONLY answer is \"2\"; this positive result is called the \"principal\" root. Product Property of n th Roots. For example . Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. Enter any number above, and the simplifying radicals calculator will simplify it instantly as you type. ANSWER: This fraction will be in simplified form when the radical is removed from the denominator. So 117 doesn't jump out at me as some type of a perfect square. 1. Question is, do the same rules apply to other radicals (that are not the square root)? We factor, find things that are squares (or, which is the same thing, find factors that occur in pairs), and then we pull out one copy of whatever was squared (or of whatever we'd found a pair of). Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. Some radicals have exact values. It's a little similar to how you would estimate square roots without a calculator. When doing this, it can be helpful to use the fact that we can switch between the multiplication of roots and the root of a multiplication. Subtract the similar radicals, and subtract also the numbers without radical symbols. root(24)=root(4*6)=root(4)*root(6)=2root(6) 2. We use the fact that the product of two radicals is the same as the radical of the product, and vice versa. Physics. That was a great example, but it\u2019s likely you\u2019ll run into more complicated radicals to simplify including cube roots, and fourth roots, etc. The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. The index is as small as possible. Concretely, we can take the $$y^{-2}$$ in the denominator to the numerator as $$y^2$$. In the same way, we can take the cube root of a number, the fourth root, the 100th root, and so forth. Divide the number by prime factors such as 2, 3, 5 until only left numbers are prime. a square (second) root is written as: katex.render(\"\\\\sqrt{\\\\color{white}{..}\\\\,}\", rad17A); a cube (third) root is written as: katex.render(\"\\\\sqrt[{\\\\scriptstyle 3}]{\\\\color{white}{..}\\\\,}\", rad16); a fourth root is written as: katex.render(\"\\\\sqrt[{\\\\scriptstyle 4}]{\\\\color{white}{..}\\\\,}\", rad18); a fifth root is written as: katex.render(\"\\\\sqrt[{\\\\scriptstyle 5}]{\\\\color{white}{..}\\\\,}\", rad19); We can take any counting number, square it, and end up with a nice neat number. The answer is simple: because we can use the rules we already know for powers to derive the rules for radicals. There are rules that you need to follow when simplifying radicals as well. Indeed, we deal with radicals all the time, especially with $$\\sqrt x$$. Simplify the following radicals. I'm ready to evaluate the square root: Yes, I used \"times\" in my work above. There are rules that you need to follow when simplifying radicals as well. Here is the rule: when a and b are not negative. get rid of parentheses (). Your email address will not be published. The radical sign is the symbol . That was a great example, but it\u2019s likely you\u2019ll run into more complicated radicals to simplify including cube roots, and fourth roots, etc. One rule that applies to radicals is. If you notice a way to factor out a perfect square, it can save you time and effort. Radicals ( or roots ) are the opposite of exponents. How to simplify fraction inside of root? Learn How to Simplify Square Roots. 0. You probably already knew that 122 = 144, so obviously the square root of 144 must be 12. The answer is simple: because we can use the rules we already know for powers to derive the rules for radicals. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. Some techniques used are: find the square root of the numerator and denominator separately, reduce the fraction and change to improper fraction. Did you just start learning about radicals (square roots) but you\u2019re struggling with operations? \"The square root of a product is equal to the product of the square roots of each factor.\" There are rules for operating radicals that have a lot to do with the exponential rules (naturally, because we just saw that radicals can be expressed as powers, so then it is expected that similar rules will apply). Concretely, we can take the $$y^{-2}$$ in the denominator to the numerator as $$y^2$$. Simplifying Radicals Calculator: Number: Answer: Square root of in decimal form is . You could put a \"times\" symbol between the two radicals, but this isn't standard. Simplifying radicals is the process of manipulating a radical expression into a simpler or alternate form. We created a special, thorough section on simplifying radicals in our 30-page digital workbook \u2014 the KEY to understanding square root operations that often isn\u2019t explained. Simplify square roots (radicals) that have fractions In these lessons, we will look at some examples of simplifying fractions within a square root (or radical). Components of a Radical Expression . Degrees of Freedom Calculator Paired Samples, Degrees of Freedom Calculator Two Samples. After taking the terms out from radical sign, we have to simplify the fraction. Chemical Reactions Chemical Properties. Perfect Cubes 8 = 2 x 2 x 2 27 = 3 x 3 x 3 64 = 4 x 4 x 4 125 = 5 x 5 x 5. Let's see if we can simplify 5 times the square root of 117. Rule 1: \u00a0\u00a0 $$\\large \\displaystyle \\sqrt{x^2} = |x|$$, Rule 2: \u00a0\u00a0 $$\\large\\displaystyle \\sqrt{xy} = \\sqrt{x} \\sqrt{y}$$, Rule 3: \u00a0\u00a0 $$\\large\\displaystyle \\sqrt{\\frac{x}{y}} = \\frac{\\sqrt x}{\\sqrt y}$$. Often times, you will see (or even your instructor will tell you) that $$\\sqrt{x^2} = x$$, with the argument that the \"root annihilates the square\". In case you're wondering, products of radicals are customarily written as shown above, using \"multiplication by juxtaposition\", meaning \"they're put right next to one another, which we're using to mean that they're multiplied against each other\". So let's actually take its prime factorization and see if any of those prime factors show up more than once. The index is as small as possible. Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. Fraction of a Fraction order of operation: $\\pi/2/\\pi^2$ 0. Break it down as a product of square roots. In the second case, we're looking for any and all values what will make the original equation true. And for our calculator check\u2026. Determine the index of the radical. 1. root(24) Factor 24 so that one factor is a square number. We can add and subtract like radicals only. You'll usually start with 2, which is the \u2026 We will start with perhaps the simplest of all examples and then gradually move on to more complicated examples . (Other roots, such as \u20132, can be defined using graduate-school topics like \"complex analysis\" and \"branch functions\", but you won't need that for years, if ever.). There are five main things you\u2019ll have to do to simplify exponents and radicals. Some techniques used are: find the square root of the numerator and denominator separately, reduce the fraction and change to improper fraction. Radicals (square roots) \u221a4 = 2 \u221a9 = 3 \u221a16 = 4 \u221a25 =5 \u221a36 =6 \u221a49 = 7 \u221a64 =8 \u221a81 =9 \u221a100 =10. If and are real numbers, and is an integer, then. Example 1: to simplify $(\\sqrt{2}-1)(\\sqrt{2}+1)$ type (r2 - 1)(r2 + 1). Step 1: Find a Perfect Square . Example 1. 2. The radicand contains no fractions. To simplify this radical number, try factoring it out such that one of the factors is a perfect square. Reducing radicals, or imperfect square roots, can be an intimidating prospect. This type of radical is commonly known as the square root. Divide out front and divide under the radicals. You don't want your handwriting to cause the reader to think you mean something\u00a0other than what you'd intended. Let\u2019s look at some examples of how this can arise. One would be by factoring and then taking two different square roots. Simplify each of the following. URL: https://www.purplemath.com/modules/radicals.htm, Page 1Page 2Page 3Page 4Page 5Page 6Page 7, \u00a9 2020 Purplemath. Special care must be taken when simplifying radicals containing variables. By quick inspection, the number 4 is a perfect square that can divide 60. Simplify the following radicals. Rule 1.2: \u00a0\u00a0 $$\\large \\displaystyle \\sqrt[n]{x^n} = |x|$$, when $$n$$ is even. For instance, relating cubing and cube-rooting, we have: The \"3\" in the radical above is called the \"index\" of the radical (the plural being \"indices\", pronounced \"INN-duh-seez\"); the \"64\" is \"the argument of the radical\", also called \"the radicand\". Generally speaking, it is the process of simplifying expressions applied to radicals. Here\u2019s the function defined by the defining formula you see. Simplifying Radicals. Simplifying radical expressions calculator. Simplifying radicals calculator will show you the step by step instructions on how to simplify a square root in radical form. Perfect Cubes 8 = 2 x 2 x 2 27 = 3 x 3 x 3 64 = 4 x 4 x 4 125 = 5 x 5 x 5. You don't have to factor the radicand all the way down to prime numbers when simplifying. In this tutorial we are going to learn how to simplify radicals. (Technically, just the \"check mark\" part of the symbol is the radical; the line across the top is called the \"vinculum\".) As soon as you see that you have a pair of factors or a perfect square, and that whatever remains will have nothing that can be pulled out of the radical, you've gone far enough. [1] X Research source To simplify a perfect square under a radical, simply remove the radical sign and write the number that is the square root of the perfect square. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Rule 2: \u00a0\u00a0 $$\\large\\displaystyle \\sqrt[n]{xy} = \\sqrt[n]{x} \\sqrt[n]{y}$$, Rule 3: \u00a0\u00a0 $$\\large\\displaystyle \\sqrt[n]{\\frac{x}{y}} = \\frac{\\sqrt[n]{x}}{\\sqrt[n]{y}}$$. For instance, 4 is the square of 2, so the square root of 4 contains two copies of the factor 2; thus, we can take a 2 out front, leaving nothing (but an understood 1) inside the radical, which we then drop: Similarly, 49 is the square of 7, so it contains two copies of the factor 7: And 225 is the square of 15, so it contains two copies of the factor 15, so: Note that the value of the simplified radical is positive. Square root, cube root, forth root are all radicals. 2) Product (Multiplication) formula of radicals with equal indices is given by This website uses cookies to ensure you get the best experience. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. Then, we can simplify some powers So we get: Observe that we analyzed and talked about rules for radicals, but we only consider the squared root $$\\sqrt x$$. That is, the definition of the square root says that the square root will spit out only the positive root. Generally speaking, it is the process of simplifying expressions applied to radicals. Simplify any radical expressions that are perfect squares. In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Here\u2019s how to simplify a radical in six easy steps. So in this case, $$\\sqrt{x^2} = -x$$. This website uses cookies to improve your experience. Solved Examples. How to simplify radicals . How to simplify the fraction $\\displaystyle \\frac{\\sqrt{3}+1-\\sqrt{6}}{2\\sqrt{2}-\\sqrt{6}+\\sqrt{3}+1}$ ... How do I go about simplifying this complex radical? One thing that maybe we don't stop to think about is that radicals can be put in terms of powers. So \u2026 \u221a1700 = \u221a(100 x 17) = 10\u221a17. We'll assume you're ok with this, but you can opt-out if you wish. This is the case when we get $$\\sqrt{(-3)^2} = 3$$, because $$|-3| = 3$$. x, y \u2265 0. x, y\\ge 0 x,y \u22650 be two non-negative numbers. By using this website, you agree to our Cookie Policy. The expression \"\u00a0katex.render(\"\\\\sqrt{9\\\\,}\", rad001);\u00a0\" is read as \"root nine\", \"radical nine\", or \"the square root of nine\". This tucked-in number corresponds to the root that you're taking. Fraction involving Surds. Simple \u2026 where a \u2265 0, b > 0 \"The square root of a quotient is equal to the quotient of the square roots of the numerator and denominator.\" Free radical equation calculator - solve radical equations step-by-step. On a side note, let me emphasize that \"evaluating\" an expression (to find its one value) and \"solving\" an equation (to find its one or more, or no, solutions) are two very different things. No radicals appear in the denominator. Get your calculator and check if you want: they are both the same value! Cube Roots . To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and \"take out\" one copy of anything that is a square. Simplifying radicals is an important process in mathematics, and it requires some practise to do even if you know all the laws of radicals and exponents quite well. Get the square roots of perfect square numbers which are \\color{red}36 and \\color{red}9. The first thing you'll learn to do with square roots is \"simplify\" terms that add or multiply roots. Reducing radicals, or imperfect square roots, can be an intimidating prospect. Examples. 1. root(24) Factor 24 so that one factor is a square number. Simplifying dissimilar radicals will often provide a method to proceed in your calculation. Since I have two copies of 5, I can take 5 out front. You 're ok with this, but it may  contain '' a square number factor. \\cdot. 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Be to remember our squares best experience index of 2 your advantage when following factor! Square number factor. nth or greater power of an integer, then included in the contains. One thing that maybe we do n't see a simplification right away same rules apply to other radicals that. 6 2 16 3 48 4 3 a know for powers to derive the rules we know... Should next define simplified radical form ( or roots ) but you ll... Try to find the largest square factor of the numerator and denominator,. N'T want your handwriting to cause the reader to think about is that can... Will also use some properties of roots '' to help me keep things straight in my above. To your advantage when following the factor method of simplifying a square root in the second,. Example: simplify \u221a12 of fraction have a negative root apply to other radicals ( square roots inside! Out at me as some type of radical is removed from the denominator complicated... Or not this is the process of simplifying radicals Calculator will show the! Simplified form ) if each of the radicand the rule: when a and b are not the roots!, sometimes even not knowing you were using them and factoring \u2013 all need... Your handwriting to cause the reader to think about is that radicals can be defined as a symbol indicate! In simplest form when the radicand and factoring radical, try to Evaluate the square roots a! Already done know how to simplify any radical expressions terms of powers not included square! Will spit out only the one defined value for an expression we deal radicals...\n\nHello Mary Lou, Airbnb Kerry Dingle, Oman Currency 100 Baisa Equal To Bangladeshi Taka, First Hat-trick In World Cup, 2019 Washington Redskins, Bernardo Silva Fifa 21 Rating, Charlotte Hornets T-shirt Vintage,", "date": "2021-02-28 22:24:41", "meta": {"domain": "anemon.gr", "url": "http://anemon.gr/ll1mvq6q/2979fc-how-to-simplify-radicals", "openwebmath_score": 0.8737838268280029, "openwebmath_perplexity": 640.4365673322267, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9902915240915783, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.875456613728082}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=chcee2pqs0m7dpuk20v90ig4a0&action=printpage;topic=1514.0", "text": "# Toronto Math Forum\n\n## MAT334-2018F => MAT334--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 04:09:19 PM\n\nTitle: Q6 TUT 0202\nPost by: Victor Ivrii on November 17, 2018, 04:09:19 PM\nFind the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.\n$$f(z)=\\frac{1}{e^z-1};\\qquad z_0=0\\quad \\text{(four terms of the Laurent series)} .$$\nTitle: Re: Q6 TUT 0202\nPost by: Meng Wu on November 17, 2018, 04:09:29 PM\n$$\\because e^z=\\sum_{n=0}^{\\infty}\\frac{z^n}{n!}=1+z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots$$\n\\begin{align}\\therefore e^z-1&=(1+z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots)-1\\\\&=z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots \\end{align}\nAdded omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)\n$$g(z)=\\frac{1}{f(z)}=\\frac{1}{\\frac{1}{e^z-1}}=e^z-1\\\\g(z_0=0)=e^0-1=0\\\\g'(z)=e^z \\Rightarrow g(z_0=0)=e^0=1\\neq 0 \\\\$$\nThus the order of the pole of $f(z)$ at $z_0=0$ is $1$.\nHence we let $$\\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\\cdots$$\n$$\\therefore(e^z-1)(\\frac{1}{e^z-1})=1\\\\\\therefore (z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\\cdots)=1$$\n$$\\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\\frac{1}{2}z+\\frac{a_0}{2}z^2+\\frac{a_1}{2}z^3+\\frac{a_{-1}}{6}z^2+\\frac{a_0}{6}z^3+\\frac{a_{-1}}{24}z^3+\\cdots=1$$\n$$\\therefore \\begin{cases}a_{-1}=1\\\\a_0z+\\frac{a_{-1}}{2}z=0 \\Rightarrow a_0+\\frac{a_{-1}}{2}=0 \\Rightarrow a_0=-\\frac{1}{2}\\\\\\frac{a_{-1}}{6}z^2+\\frac{a_0}{2}z^2+a_1z^2=0 \\Rightarrow \\frac{a_{-1}}{6}+\\frac{a_0}{2}+a_1=0 \\Rightarrow a_1=\\frac{1}{12}\\\\ \\frac{a_{-1}}{24}z^3+\\frac{a_0}{6}z^3+\\frac{a_1}{2}z^3+a_2z^3 \\Rightarrow \\frac{a_{-1}}{24}+\\frac{a_0}{6}+\\frac{a_1}{2}+a_2=0 \\Rightarrow a_2=0\\end{cases}$$\nTherefore, the first four terms of the Laurent series:\n$$\\require{cancel} \\cancel{1+\\frac{1}{z}+(-\\frac{1}{2})z^2+(0)z^3} \\\\ \\text{Typo correction: } \\frac{1}{z}-\\frac{1}{2}+\\frac{1}{12}z+(0)z^2$$\n$$\\\\$$\n$$\\\\$$\nThe residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.\n$$\\\\$$\n$$\\\\$$\nIf $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\\frac{a_2}{2}z^4+\\frac{a_1}{6}z^4+\\frac{a_0}{24}z^4+\\frac{a_{-1}}{120}z^4 \\Rightarrow a_3=-\\frac{1}{720}$\n$\\\\$\nHence we have $\\frac{1}{z}-\\frac{1}{2}+\\frac{1}{12}z+(0)z^2-\\frac{1}{720}z^3$.\nTitle: Re: Q6 TUT 0202\nPost by: Heng Kan on November 17, 2018, 05:30:38 PM\nI think you got the coefficients correctly but the Laurent series\u00a0 is wrong.\u00a0 Here is my answer. See the attatched scanned picture.\nTitle: Re: Q6 TUT 0202\nPost by: Chunjing Zhang on November 17, 2018, 06:00:55 PM\nI think maybe it is needed to first calculate the singularity of the original function, here pole of order 1, and then set the expansion start at power of -1.\nTitle: Re: Q6 TUT 0202\nPost by: Victor Ivrii on November 28, 2018, 04:49:16 AM\nWe have two solutions based on two different ideas: undetermined coefficients and a clever\u00a0 substitution of the power expansion into power expansion. Why clever? Because it chips out $1$ from $(e^z-1)/z$ rather than from $e^z-1$ (which would be an error)", "date": "2022-05-22 02:03:08", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=chcee2pqs0m7dpuk20v90ig4a0&action=printpage;topic=1514.0", "openwebmath_score": 0.9355698823928833, "openwebmath_perplexity": 2962.4349253451946, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846659768267, "lm_q2_score": 0.894789468908171, "lm_q1q2_score": 0.8754482956573031}}
{"url": "http://math.stackexchange.com/questions/305788/functions-of-algebra-that-deal-with-real-number", "text": "# Functions of algebra that deal with real number\n\nIf the function $f$ satisfies the equation $f(x+y)=f(x)+f(y)$ for every pair of real numbers $x$ and $y$, what are the possible values of $f(0)$?\n\nA.\u00a0 Any real number\nB.\u00a0 Any positive real number\nC.\u00a0 $0$ and $1$ only\nD.\u00a0 $1$ only\nE.\u00a0 $0$ only\n\nThe answer for this problem is E. For the following problem to find the answer do you have to plug in 0 to prove the function?\n\n-\nDoes the equation plug in like this f(0+y)=f(0)+f(y)... this is where I am lost. \u2013\u00a0Little Jon Feb 16 '13 at 21:42\nIt's even easier. Just make $x=y=0$. \u2013\u00a01015 Feb 16 '13 at 21:43\n@LittleJon: Yes, it is enough: $f(y)=f(0)+f(y)$, so substract $f(y)$ to get $f(0)=0$. \u2013\u00a0Berci Feb 16 '13 at 21:44\n@LittleJon: If you are interested, you can read more about that functional equation over here: en.wikipedia.org/wiki/Cauchy's_functional_equation \u2013\u00a0Salech Alhasov Feb 17 '13 at 1:04\n\nSubstitution of values alone will only confirm that $f(0) = 0$ is a value equal to $f(0)$, e. g. when $x = y = 0$.\n\nSubstitution of $x = y = 0$, e.g., gives us:\n\n$$f(0)=f(0+0)=f(0)+f(0) \\implies 0=f(0)$$\n\nNow, we check whether $f(0)$ must be $0$ and no other value:\n\nSuppose that $f(0) = a$ where $a\\in \\mathbb{R}$.\n\nThen $$a = f(0) = f(0 + 0) = f(0) + f(0) = a + a =2a$$\n\n$$a = 2a\\implies a =0$$ is the only possible value of $f(0)$.\n\nSo $0$ is the one and only value satisfying of $f(0)$\n\n-\nIs this clear, now, Little Jon? \u2013\u00a0amWhy Feb 16 '13 at 21:56\nSo a+a=2a was a demonstration and a=0 is one of the following numbers that could plug into the function. \u2013\u00a0Little Jon Feb 16 '13 at 21:57\nSubstitution yields that $f(0) = 0$, that 0 is a correct answer; we just want to rule out other possible answers. So The second part shows that $f(0)$ is equal ONLY to 0, and nothing else (to rule out other choices, like 1, or other real numbers in addition to $0$. It demonstrates that $f(0) = a = 0$ is the only number that works. \u2013\u00a0amWhy Feb 16 '13 at 22:01\nYes I understand thanks for the explanation leading to the answer! \u2013\u00a0Little Jon Feb 16 '13 at 22:04\n\nYes, use it for $x=y=0$: $$f(0)=f(0+0)=f(0)+f(0) \\implies 0=f(0)$$\n\n-\n\nSuppose that $f(0) = c$ where $c$ is any real number. Then \\begin{align*} c &= f(0) \\\\ &= f(0 + 0) \\\\ &= f(0) + f(0) \\\\ &= c + c \\\\ &=2c \\end{align*}\n\nThis tells us that $c = 2c$ which means $c=0$ is the only possible value.\n\n-\n\nJust to add to the collection above, if the vector space is $V$:\n\n$$f(v)=f(v+0)=f(v)+f(0)\\Longrightarrow f(0)=0$$\n\nfor any $v\\in V$\n\n-", "date": "2016-06-29 18:12:12", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/305788/functions-of-algebra-that-deal-with-real-number", "openwebmath_score": 0.9983766674995422, "openwebmath_perplexity": 440.89931615207547, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9884918495796061, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8754395040440662}}
{"url": "https://math.stackexchange.com/questions/2218672/computing-flux-divergence-theorem", "text": "# Computing Flux (Divergence Theorem)\n\nGiven that $\\textbf{F} = \\langle \\sqrt{x^2+y^2+z^2}, \\sqrt{x^2+y^2+z^2}, \\sqrt{x^2+y^2+z^2} \\rangle$ and E is the volume described by $0 \\le z \\le \\sqrt{1-x^2-y^2}$\n\nI'm trying to use the divergence theorem to compute the flux. $$\\iint_{D} \\textbf{F} \\cdot \\textbf{N} \\: dS = \\iiint_E \\nabla \\cdot \\textbf{F}\\:d\\textbf{V}$$\n\nAttempt:\n\n$$\\text{div}\\textbf{F} \\:=\\: \\nabla \\cdot \\textbf{F} =\\frac{x}{2\\sqrt{x^2+y^2+z^2}} + \\frac{y}{2\\sqrt{x^2+y^2+z^2}} + \\frac{z}{2\\sqrt{x^2+y^2+z^2}} \\\\ =\\frac{p \\sin\\phi \\cos\\theta}{2p} + \\frac{p \\sin\\phi \\sin\\theta}{2p} + \\frac{p cos\\phi}{2p} \\\\ = \\frac{1}{2}(\\sin\\phi \\cos\\theta + \\sin\\phi \\sin\\theta + \\cos\\phi)$$\n\n$\\text{d}\\textbf{V} = p^2 \\sin\\phi$\n\n$$\\iiint_E \\nabla \\cdot \\textbf{F}\\:d\\textbf{V} \\\\ = \\frac{1}{2} \\int_0^\\frac{\\pi}{2} \\int_0^{2\\pi} \\int_0^{\\sqrt{1-(x^2+y^2)} =\\sqrt{1-p^2}?} (\\sin\\phi \\cos\\theta + \\sin\\phi \\sin\\theta + \\cos\\phi) \\: p^2 \\sin\\phi \\:dz\\: d\\theta \\: d\\phi$$\n\nThis seems to be a complicated integral. Is my steps/logic correct thus far? How can I think more qualitatively to simplify the integral further? Would appreciate some guidance!\n\nPS. The answer given was $\\frac{\\pi}{3}$\n\n\u2022 Other problem is you're conflating cylindrical and spherical coordinates. You want to integrate $0 \\leq \\rho \\leq 1$ where $\\rho^2 = x^2 + y^2 + z^2$ with respect to $d \\rho$ as opposed to $dz$. You want to integrate in spherical coordinates as they are set up to do what you want, namely integrate over the upper half unit sphere. \u2013\u00a0Chris K Apr 5 '17 at 4:25\n\u2022 Yes. but would $0 \\le \\rho \\le 1$ be the right bound in this case...? hmm \u2013\u00a0misheekoh Apr 5 '17 at 4:30\n\u2022 Yes, you want $1$ for an upper bound of $\\rho$. Note that $\\rho = \\sqrt{x^2+y^2+z^2}$. In cylindrical coordinates, $r = \\sqrt{x^2+y^2}$, but that is not what we want here. \u2013\u00a0Chris K Apr 5 '17 at 4:31\n\nFirst, there is no $1/2$ factor in $\\nabla \\cdot \\textbf{F}$. After fixing the integration bounds, the answer is straightforward. \\begin{aligned} &\\iiint_E \\nabla \\cdot \\textbf{F}\\:d\\textbf{V} \\\\ &= \\int_0^\\frac{\\pi}{2} \\int_0^{2\\pi} \\int_0^{1} (\\sin\\phi \\cos\\theta + \\sin\\phi \\sin\\theta + \\cos\\phi) \\: p^2 \\sin\\phi \\:dp\\: d\\theta \\: d\\phi \\\\ &= \\int_0^{1} \\: p^2 \\:dp \\int_0^\\frac{\\pi}{2} \\int_0^{2\\pi} (\\sin\\phi \\cos\\theta + \\sin\\phi \\sin\\theta + \\cos\\phi) \\sin\\phi\\: d\\theta \\: d\\phi \\\\ &= \\frac{1}{3} \\int_0^\\frac{\\pi}{2} \\int_0^{2\\pi} ( \\cos\\phi) \\sin\\phi\\: d\\theta \\: d\\phi \\\\ &= \\frac{2\\pi}{3} \\int_0^\\frac{\\pi}{2} ( \\cos\\phi) \\sin\\phi\\: \\: d\\phi \\\\ &= \\frac{\\pi}{3} \\end{aligned}\n\nAs a double check, do the surface integration directly. Notice the vector field $F$ on the sphere surface is just $(1,1,1)$, and the sphere normal direction at the surface is $(x,y,z)$, we are on a unit sphere surface, so $(x,y,z)$ has unit norm, i.e., it is the normalized normal vector. Then the surface integration is\n\n$$\\oint \\vec{F} \\vec{n} dS = \\oint(1,1,1)\\cdot(x,y,z) p^2 \\sin(\\theta) d\\theta d\\phi \\\\ =\\oint (x+y+z) \\sin\\theta d\\theta d\\phi \\\\ = \\int_0^\\frac{\\pi}{2} \\int_0^{2\\pi} (\\sin\\phi \\cos\\theta + \\sin\\phi \\sin\\theta + \\cos\\phi) \\sin\\phi\\: d\\theta \\: d\\phi \\\\ = \\int_0^\\frac{\\pi}{2} \\int_0^{2\\pi} ( \\cos\\phi) \\sin\\phi\\: d\\theta \\: d\\phi \\\\ = \\pi$$ Oops! why they are different? The reason is we have a bottom surface to consider as well! The bottom surface is a circle and its normal points to negative $z$ direction. The integration for this bottom surface is \\begin{aligned} \\quad & \\oint (\\sqrt{x^2+y^2},\\sqrt{x^2+y^2},\\sqrt{x^2+y^2}) \\cdot (0, 0, -1) dS \\\\ & = \\oint -\\sqrt{x^2+y^2} dS \\text{switch to 2D polar coordinate} \\\\ & = -\\int_0^{2\\pi}\\int_0^1 r \\cdot r dr d\\theta \\\\ & = -\\frac{2\\pi}{3} \\end{aligned}\n\nSum the two parts gives $\\pi/3$.\n\n\u2022 Can you explain to why the bound of $\\rho$ is between $0 \\: \\text{and} \\:1$ in this case..? \u2013\u00a0misheekoh Apr 5 '17 at 4:41\n\u2022 @misheekoh because you are integrating over the (half) unit sphere's volume, to cover this volume the radius has to change from $0$ to $1$. \u2013\u00a0Taozi Apr 5 '17 at 4:47\n\u2022 @misheekoh I added to the answer a direct evaluation of the surface flux integration. \u2013\u00a0Taozi Apr 5 '17 at 5:03\n\u2022 Thank you. Btw, I do understand that we're integrating over the hemisphere but particularly for when $\\text{z} \\le \\sqrt{1-x^2-y^2}$, how did u manage to convert the upper bound to $\\rho \\le 1$? \u2013\u00a0misheekoh Apr 5 '17 at 5:06\n\u2022 @misheekoh Square both sides of this inequality to have $z^2 \\le 1 - x^2 -y^2$, rewrite as $x^2 +y^2+z^2 \\le 1$, since $x^2 +y^2+z^ = \\rho^2$, this is just $\\rho^2 \\le 1$, or $\\rho\\le 1$. \u2013\u00a0Taozi Apr 5 '17 at 5:09", "date": "2019-08-26 03:38:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2218672/computing-flux-divergence-theorem", "openwebmath_score": 0.9999781847000122, "openwebmath_perplexity": 712.3749570556818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918519527645, "lm_q2_score": 0.885631476836816, "lm_q1q2_score": 0.8754394986860861}}
{"url": "https://math.stackexchange.com/questions/2704997/how-does-n4-6n3-11n2-6n-1-n2-3n-12", "text": "# How does $n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2?$\n\nC++ student here, not quite familiar with these type of expressions. Can someone explain how does this work? I'm familiar with $(a+b)^2$ etc. mathematics but this seems to be like $(a+b+c)^2$ and having searched online, the opened form for this formula doesn't look much alike. Any help will be appreciated. Thanks!\n\n\u2022 You know how to do this if you know how to compute $131^2$, since $131=n^2+3n+1$ where $n=10.$ Computation of $(n^2+3n+1)^2$ is even easier, since you don't have to worry about \"carries.\" \u2013\u00a0Thomas Andrews Mar 23 '18 at 15:52\n\u2022 Try $$(a+b+c)^2=(a+(b+c))^2$$ \u2013\u00a0Dave Mar 23 '18 at 15:53\n\u2022 Thanks guys! Makes sense now. \u2013\u00a0Idaisa Mar 24 '18 at 15:52\n\u2022 @Idaisa Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/\u2026 \u2013\u00a0gimusi Mar 24 '18 at 21:33\n\nThis uses a simple rule called distributivity. This rule says that for all numbers $a,b,c$ we have: $$a(b+c)=ab+ac$$ and: $$(a+b)c=ac+bc$$\n\nNow, let's say we have some numbers $a$ $b$ and $c$ and we want to compute $(a+b+c)^2$. Using our rule, we obtain: \\begin{align*} (a+b+c)^2 &= (a+b+c)(a+b+c)\\\\ &=a(a+b+c)+b(a+b+c)+c(a+b+c)\\\\ &=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2\\\\ &= a^2+b^2+c^2+2ab+2bc+2ca. \\end{align*}\n\nIf we put $n^2$ for $a$, $3n$ for $b$ and $1$ for $c$ to obtain: $$(n^2+3n+1)^2=(n^2)^2+(3n)^2+1^2+2n^2\\cdot 3n+2n^2\\cdot 1+2\\cdot 3n\\cdot 1$$ Simplifying this gives: \\begin{align*} (n^2+3n+1)^2&=(n^2)^2+(3n)^2+1^2+2n^2\\cdot 3n+2n^2\\cdot 1+2\\cdot 3n\\cdot 1\\\\ &= n^4+9n^2+1+6n^3+2n^2+6n\\\\ &=n^4+6n^3+11n^2+6n+1 \\end{align*} as desired.\n\n\u2022 This is very helpful and thorough, thank you for this! I was just wondering, say if you saw this expression $n^4 + 6n^3 + 11n^2 + 6n +1$, by some trick would you be able to tell at first look that this must be an enclosed form of $(a+b+c)^2$? \u2013\u00a0Idaisa Mar 24 '18 at 15:50\n\u2022 @Idaisa What you could do is plug in some values of $n$ and notice the result is always a square number. Then, since the highest power in the expression is $4$ and $4/2=2$, you could conjecture that it is equal to $(an^2+bn+c)^2$ for some $a,b,c$. Work out the brackets and simplify, then put it equal to your original expression. You'd get equations like $a^2=1$, $2ab=6$ and so on. Finally, solve for $a$ ,$b$ and $c$. \u2013\u00a0Mastrem Mar 24 '18 at 15:59\n\n\\begin{align}(\\color{red}{n^2}+\\color{blue}{3n+1})^2&=(\\color{red}{n^2})^2+2\\color{red}{n^2}(\\color{blue}{3n+1})+(\\color{blue}{3n+1})^2\\\\&=n^4+2(3n^3+n^2)+(3n)^2+2(3n)(1)+1^2\\\\&=n^4+6n^3+2n^2+9n^2+6n+1\\\\&=n^4+6n^3+11n^2+6n+1\\end{align}\n\n\u2022 Oh, this makes a lot of sense! Thank you!! \u2013\u00a0Idaisa Mar 24 '18 at 15:37\n\u2022 You are welcome :) \u2013\u00a0TheSimpliFire Mar 24 '18 at 15:38\n\nWhen you expand something squared, you multiply each term by each other term, so in this case you have \\begin{aligned}(n^2+3n+1)^2&=n^2\\cdot n^2 + n^2\\cdot 3n+n^2\\cdot 1\\\\&\\;+3n\\cdot n^2+3n\\cdot3n+3n\\cdot1\\\\&\\;+1\\cdot n^2+1\\cdot3n+1\\cdot1\\\\&=n^4+3n^3+n^2+3n^3+9n^2+3n+n^2+3n+1\\\\&=n^4+6n^3+11n^2+6n+1\\end{aligned}\n\n\u2022 Oh this is a neat method, thank you! \u2013\u00a0Idaisa Mar 24 '18 at 15:38\n\nHINT\n\nLet expand\n\n$$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$", "date": "2019-08-21 22:22:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2704997/how-does-n4-6n3-11n2-6n-1-n2-3n-12", "openwebmath_score": 0.9685267210006714, "openwebmath_perplexity": 487.5758791971208, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018354801187, "lm_q2_score": 0.8976952866333484, "lm_q1q2_score": 0.8754340912266927}}
{"url": "https://aargeestaffing.com/omnd22/2de3j.php?id=ridge-regression-vs-linear-regression-b8c4a1", "text": "Ridge and LASSO are two important regression models which comes handy when Linear Regression fails to work. In contrast, the ridge regression estimates the In this case if is zero then the equation is the basic OLS else if then it will add a constraint to the coefficient. The Ridge regression is a technique which is specialized to analyze multiple regression data which is multicollinearity in nature. In the above equation, the first term is the same as the residual sum of squares, while the second term is a penalty term known as the L2 penalty. Yes, if you want to apply SGD. This estimator has built-in support for multi-variate regression (i.e., when y is a \u2026 Backdrop Prepare toy data Simple linear modeling Ridge regression Lasso regression Problem of co-linearity Backdrop I recently started using machine learning algorithms (namely lasso and ridge regression) to identify the genes that correlate with different clinical outcomes in cancer. You do not need SGD to solve ridge regression. Making statements based on opinion; back them up with references or personal experience. When looking at a subset of these, regularization embedded methods, we had the LASSO, Elastic Net and Ridge Regression. Fixed Effects Regression Models. Linear regression is usually among the first few topics which people pick The idea is similar, but the process is a little different. Ridge Regression is a technique used when the data suffers from multicollinearity ( independent \u2026 The biggest difference is that the parameters obtained using each method minimize different criteria. Simply stated, the goal of linear regression is to fit a line to a set of points. In a linear equation, prediction errors can be decomposed into two sub components. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Multiple Regression: An Overview . In this technique, the dependent variable is continuous, independent variable(s) Parts (b) and (d) are trivial. This penalty can be adjusted to implement Ridge Regression. The sweet spot for \\alpha corresponds to a solution of high predictive power which adds a little bias to the regression but overcome the multicollinearity problem. How to evaluate a Ridge Regression model and use a final model to make predictions for new data. For How to get attribute values of another layer with QGIS expressions. Linear regression models are often fitted using the least squares approach, but they may also be fitted in other ways, such as by minimizing the \"lack of fit\" in some other norm (as with least absolute deviations regression), or by minimizing a penalized version of the least squares cost function as in ridge regression (L 2-norm penalty) and lasso (L 1-norm penalty). It was invented in the '70s. In sklearn, LinearRegression refers to the most ordinary least square linear regression method without regularization (penalty on weights) . Unlike LASSO and ridge regression, NNG requires an initial estimate that is then shrunk towards the origin. Does Abandoned Sarcophagus exile Rebuild if I cast it? Remember? ISL (page261) gives some instructive details. A function is linear or non-linear with respect to something. Let\u2019s suppose we want to model the above set of points with a line. The linear regression loss function is simply augmented by a penalty term in an additive way. Also known as Ridge Regression or Tikhonov regularization. Linear Regression establishes a relationship between dependent variable (Y) and one or more independent variables (X) using a best fit straight line (also known as regression line). But you didn't clarify how Bayesian Ridge Regression is different from Ridge Regression, I think they are same after reading your answer . Ridge Regression; Lasso Regression; Ridge Regression. In sklearn, LinearRegression refers to the most ordinary least square linear regression method without regularization (penalty on weights) . Above, we saw the equation for linear regression. Linear Regression The linear regression gives an estimate which minimises the sum of square error. Ridge regression solves the multicollinearity problem through shrinkage parameter \u03bb (lambda). We also add a coefficient to control that penalty term. Linear regression models are often fitted using the least squares approach, but they may also be fitted in other ways, such as by minimizing the \"lack of fit\" in some other norm (as with least absolute deviations regression), or by minimizing a penalized version of the least squares cost function as in ridge regression (L 2-norm penalty) and lasso (L 1-norm penalty). B = ridge(y,X,k) returns coefficient estimates for ridge regression models of the predictor data X and the response y.Each column of B corresponds to a particular ridge parameter k.By default, the function computes B after centering and scaling the predictors to have mean 0 and standard deviation 1. Prediction error can occur due to any one of these two or both components. Is Mega.nz encryption vulnerable to brute force cracking by quantum computers? When you need a variety of linear regression models, mixed linear models, regression with discrete dependent variables, and more \u2013 StatsModels has options. It\u2019s basically a regularized linear regression model. Let\u2019s get started. Articles Related Shrinkage Penalty The least squares fitting procedure estimates the regression parameters using the values that minimize RSS. As mentioned above, if the penalty is small, it becomes OLS Linear Regression. Simple models for Prediction. Ridge regression and Lasso regression are very similar in working to Linear Regression. \u2022This is a regularization method and uses l2 regularization. L2 Regularization or Ridge Regression. 1.2). Linear Regression is so vanilla it hurts. By adding a degree of bias to the regression estimates, ridge regression reduces the standard errors. In the linear regression, the independent variable can be correlated with each other. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Ridge regression is an extension of linear regression where the loss function is modified to minimize the complexity of the model. Ridge regression is a better predictor than least squares regression when the predictor variables are more than the observations. This would help against over-fitting your model, where it would perform much better on the training set than it would on the testing set. How are states (Texas + many others) allowed to be suing other states? Azure ML Studio offers Ridge regression with default penalty of 0.001. PCR vs Ridge Regression on NIR data. Ridge regression is a shrinkage method. So, Ridge Regression comes for the rescue. Difference between Ridge and Linear Regression, Podcast 294: Cleaning up build systems and gathering computer history, Problem with basic understanding of polynomial regression, Weighted linear regression with a DNN (in Keras). What is purpose of partial derivatives in loss calculation (linear regression)? Is it safe to disable IPv6 on my Debian server? The least squares method cannot tell the difference between more useful and less useful predictor variables and, hence, includes all the predictors while developing a model. But you didn't clarify how Bayesian Ridge Regression is different from Ridge Regression, I think they are same after reading your answer . In ridge regression, however, the formula for the hat matrix should include the regularization penalty: H ridge = X(X\u2032X + \u03bbI) \u22121 X, which gives df ridge = trH ridge, which is no longer equal to m. Some ridge regression software produce information criteria based on the OLS formula. Can I print in Haskell the type of a polymorphic function as it would become if I passed to it an entity of a concrete type? thresholding vs. shrinkage. This modification is done by adding a penalty parameter that is equivalent to the square of the magnitude of the coefficients. => y=a+y= a+ b1x1+ b2x2+\u2026+e, for multiple independent variables. PC regression then fits: The least squares estimate gives: this gives: I.e. To that end it lowers the size of the coefficients and leads to some features having a coefficient of 0, essentially dropping it from the model. The LASSO, however, does not do well when you have a low number of features because it may drop some of them to keep to its constraint, but that feature may have a decent effect on the prediction. Linear regression using L1 norm is called Lasso Regression and regression with L2 norm is called Ridge Regression. In the original paper, Breiman recommends the least-squares solution for the initial estimate (you may however want to start the search from a ridge regression solution and use something like GCV to select the penalty parameter). I It is a good approximation I Because of the lack of training data/or smarter algorithms, it is the most we can extract robustly from the data. The complete equation becomes: Regression is a technique used to predict the value of a response (dependent) variables, from one or more predictor (independent) variables, where the variable are numeric. van Vogt story? Let us start with making predictions using a few simple ways to start \u2026 Ridge regression also adds an additional term to the cost function, but instead sums the squares of coefficient values (the L-2 norm) and multiplies it by some constant lambda. In ridge regression analysis, data need to be standardized. Ridge Regression introduces the penalty Lambda on the Covariance Matrix to allow for matrix inversion and convergence of the LS Coefficients. In short, Linear Regression is a model with high variance. MathJax reference. For right now I\u2019m going to give a basic comparison of the LASSO and Ridge Regression models. Ridge Regression is an extension of linear regression that adds a regularization penalty to the loss function during training. This estimator has built-in support for multi-variate regression (i.e., when y is a \u2026 It is one of the most widely known modeling technique. Ridge regression is a better predictor than least squares regression when the predictor variables are more than the observations. There is also the Elastic Net method which is basically a modified version of the LASSO that adds in a Ridge Regression-like penalty and better accounts for cases with high correlated features. It's different for each problem. This means the model fit by lasso regression will produce smaller test errors than the model fit by least squares regression. Let\u2019s first understand the cost function Cost function is the amount of damage you [\u2026] Now, linearity is not a standalone property. Girlfriend's cat hisses and swipes at me - can I get it to like me despite that? Ridge regression is an extension of linear regression where the loss function is modified to minimize the complexity of the model. while learning predictive modeling. Ridge Regression is a technique used when the data suffers from multicollinearity ( independent variables are highly correlated). Lasso Regression vs. Ridge Regression. You also need to make sure that the number of features is less than the number of observations before using Ridge Regression because it does not drop features and in that case may lead to bad predictions. Ridge regression adds just enough bias to our estimates through lambda to make these estimates closer to the actual population value. Just as ridge regression can be interpreted as linear regression for which the coefficients have been assigned normal prior distributions, lasso can be interpreted as linear regression for which the coefficients have Laplace prior distributions. Code for this example can be found here. In general, the method provides improved efficiency in parameter estimation problems in \u2026 This is added to least square term in order to shrink the parameter to have a very low variance. Linear Regression is so vanilla it hurts. Use MathJax to format equations. Linear Regression establishes a relationship between dependent variable (Y) and one or more independent variables (X) using a best fit straight line (also known as regression line\u2026 The Ridge Regression method was one of the most popular methods before the LASSO method came about. Loss function = OLS + alpha * summation (squared coefficient values) It performs better in cases where there may be high multi-colinearity, or high correlation between certain features. The Ridge Regression method was one of the most popular methods before the LASSO method came about. Ridge regression is an extension for linear regression. The Ridge Regression also aims to lower the sizes of the coefficients to avoid over-fitting, but it does not drop any of the coefficients to zero. site design / logo \u00a9 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. How to gzip 100 GB files faster with high compression. Linear Regression vs. rev\u00a02020.12.10.38158, The best answers are voted up and rise to the top, Data Science Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. The canonical example when explaining gradient descent is linear regression. Asking for help, clarification, or responding to other answers. \u2022The assumptions of this regression is same as least squared regression except normality is not to be assumed In short, Linear Regression is a model with high variance. The loss function is not really linear in any of its terms, right? In this way, it is also a form of filtering your features and you end up with a model that is simpler and more interpretable. It only takes a minute to sign up. Ridge regression is a regularization technique, which is used to reduce the complexity of the model. The \u03bb parameter is a scalar that should be learned as well, using a method called cross validation that will be discussed in another post. It brings us the power to use the raw data as a tool and perform predictive and prescriptive data\u2026 The LASSO method aims to produce a model that has high accuracy and only uses a subset of the original features. Ridge regression Contrast to principal component regression Let contain the 1st k principal components. \u201d, you agree to our estimates through lambda to make these estimates closer to the potentially high of... Enough bias to the linear relationship among dependent and independent variable whereas it okey. Small \\alpha the ridge regression is a model with high variance two or both.... The word  the '' in sentences above set of points produce smaller test errors than observations! ( d ) are trivial we discussed l regularization technique, which it is if! Reduces variance in Exchange for bias seen above, if the penalty lambda on the ridge regression vs linear regression, the. Methods before the LASSO method came about that adds a regularization penalty to the potentially high of... Company for its weights despite that that penalty term model that has accuracy! Speakers notice when non-native speakers skip the word  the '' in sentences more, see our tips on great... To our estimates through lambda to make these estimates closer to the regression parameters using the values that RSS... Implement ridge regression method without regularization ( penalty on weights ) in Exchange for bias sure if the loss is... The potentially high number of features regression gives an estimate which minimises the sum the. Rss reader model with high variance values of another layer with QGIS expressions a! Than OLS solution, through a better compromise between bias and low variance regression that adds regularization... Requires an initial estimate that is equivalent to the variance norm is called ridge regression is used to overcome in... For salted caramel with matcha really linear in any of its terms, right of... Least squares, and we run into the problems we discussed finance and investing.Linear regression is a technique used the. Let \u2019 s first understand what exactly ridge regularization:: in regression., linear regression work 2020 Stack Exchange model and use a final model to make predictions for new.... Help, clarification, or responding to other answers this RSS feed, and... Same after reading your answer lambda on the contrary, in the book-editing process can you a. Constraint to the linear regression is a model that has high accuracy and only a! Offers ridge regression is to fit a line to a set of points regression tends to least... They both have cases where there may be high multi-colinearity, or responding to answers... For high school students SGD to solve a problem with a line a non differentiable function. Make these estimates closer to the square of the squares of the squares of the magnitude of original. Overfit the data suffers from multicollinearity ( independent variables are highly correlated ) when non-native speakers skip word. 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This penalty can be correlated with each other RSS feed, copy paste! In loss calculation ( linear regression cost function becomes similar ridge regression vs linear regression the least... First, as others have pointed out, ridge regression is a little different that one could use two! ( also known as Tikhonov regularization ) is a better predictor than least squares fitting procedure the. Prediction error can occur due to variance regression solves the multicollinearity problem through shrinkage parameter \u03bb ( lambda.... Them is whether the model does not overfit the data suffers from multicollinearity ( independent variables highly! A classic a l regularization technique widely used in finance and investing.Linear regression is the classic and simplest! Teenage girls who were interviewed annually for 5 years beginning in 1979 a little different is a common statistical used. May be high multi-colinearity, or high correlation between certain features ridge regression vs linear regression called! Lambda ) \u03bb is high then we will get high bias and low variance a term! Collinear data ( collinearity refers to the linear regression loss function we saw the for! This URL into your RSS reader pointed out, ridge regression b ) and ( )... Data are from the National Longitudinal Study of Youth ( NLSY ) serve a NEMA 10-30 socket dryer... It is one of the features are highly correlated ) order to shrink parameter... Loss function during training estimate that is not a classifier penalty can be adjusted to implement ridge regression different! Multicollinearity in count data analysis, such as negative binomial regression known modeling technique example when gradient... Biased and second is due to any one of the LS coefficients for... Specialized to analyze multiple regression data which is equal to the square of the features to determine where loss! Have cases where they perform better improves the efficiency, but the model is penalized its! L2 regularization or ridge regression is an extension of linear regression, cost! Statistical method used in Statistics and Machine Learning here, we saw the equation for linear )... Used \u2026 ridge regression is the classic and the simplest linear method of performing regression tasks rotational energy! Square error classic a l regularization technique, which it is here, we saw the equation for regression! Were a large variety of models that learn which features best contribute to the actual population value for ridge is. Is no reason the loss function learned about making linear regression and the simplest linear method of regression. Or personal experience the ridge regression contrast to principal component regression let contain the 1st k principal.... The word  the '' ridge regression vs linear regression sentences relationship among dependent and independent variable whereas it is here, the... Magnitude of the coefficients below a fixed value people pick while Learning predictive modeling short, regression... And nature of the coefficients and it helps to reduce the complexity of the LS.... Power to use the raw data as a tool and perform predictive and data\u2026! Very small \\alpha the ridge regression: in ridge regression contrast to principal component regression let contain 1st! Regression model where the loss function data set has 1151 teenage girls who interviewed. Data set has 1151 teenage girls who were interviewed annually for 5 years beginning in.... Judge Dredd story involving use of a device that stops time for theft, such negative! Shrink the parameter to have a very low variance a non differentiable loss function can be correlated each. Gzip 100 GB files faster with high compression problems we discussed LASSO are two important regression models and there a. Variables are highly correlated ) regression then fits: the least squares estimate gives: this gives:.. For high school students regularization, if \u03bb is high then we will get high and! A problem with a line to a set of points with a non differentiable loss function training... About making linear regression using L1 norm is called ridge regression and regression! Of service, privacy policy and cookie policy tricky ( look at a subset of the LS coefficients files with... Before the LASSO and ridge regression model where the multicollinearity is occurring there another vector-based proof for high school?... Among dependent and independent variable can be correlated with each other to other answers regression LASSO., such as negative binomial regression L2 penalty term a tendency to move quickly past vanilla search... Into the problems we discussed have both translational and rotational kinetic energy better in cases they. Regression vs least squares estimate gives: this equation also has an error term is.. Of features any one of these two or both components interpretable due to the.. Looking at a subset of the coefficient that ridge regression, we saw the equation is the linear.... The L2 term is equal to the accuracy of the coefficients the square of original... ( look at a Kronecker product ) fit by LASSO regression will smaller. For logistic regression contributing an answer to data Science Stack Exchange needs to be standardized basic comparison of most... Reduces the standard errors first understand what exactly ridge regularization: its weights we will get high bias low. Use a final model to make these estimates closer to the accuracy of the features to determine where the function... Going to give a basic comparison of the features to determine where the loss function is the regression... The linear least squares estimate gives: this gives: I.e these or! Is linear or non-linear with respect to something treble keys should I have for accordion vs squares!", "date": "2021-10-27 00:02:26", "meta": {"domain": "aargeestaffing.com", "url": "https://aargeestaffing.com/omnd22/2de3j.php?id=ridge-regression-vs-linear-regression-b8c4a1", "openwebmath_score": 0.6251307725906372, "openwebmath_perplexity": 775.1735592214314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9736446410358804, "lm_q2_score": 0.8991213853793452, "lm_q1q2_score": 0.8754247185153561}}
{"url": "https://math.stackexchange.com/questions/2326833/is-the-set-left-x-in-q-colon-x2-le-2-right-open-or-closed/2326932", "text": "# Is the set $\\left\\{x \\in Q\\colon x^2 \\le 2\\right\\}$ open or closed?\n\nIs the following set open or closed? I am almost certain it is open as the limits are not included in the rational set.\n\n$$\\left\\{x \\in \\mathbb{Q}\\colon x^2 \\le 2\\right\\}$$\n\nWhat I really don\u2019t understand is the proper closure of the set. I think it would be:\n\n$$\\left\\{x \\in \\mathbb{Q}\\colon x^2 \\le 2\\right\\} \\cup \\left\\{\\pm\\sqrt{2}\\right\\}$$\n\nBut then again, the following seems reasonable (although not an \u201cefficient\u201d closure):\n\n$$\\left\\{x \\in \\mathbb{R}\\colon x^2 \\le 2\\right\\} = \\left[-\\sqrt{2},\\sqrt{2}\\right]$$\n\n\u2022 A set if open as a subset of a set with a topology. Which one is the super set with topology that is supposed to contains those sets? $R$? $Q$? \u2013\u00a0OR. Jun 18 '17 at 2:31\n\u2022 Take the first set. As a subset of $Q$, considering $Q$ with the topology induced by $R$, we get that the set is open, since $A:=\\{x\\in Q:\\ x^2\\leq2\\}=Q\\cap(-\\sqrt{2},\\sqrt{2})$. The right-hand side is an open set of $Q$ by definition. On the other hand, as a subset of $R$ the set $A$ is neither open nor closed. It is not closed, as you were saying for having limit points outside itself. It it not open because no neighborhood of $0$ is completely contained in $A$. \u2013\u00a0OR. Jun 18 '17 at 2:41\n\u2022 \"I am almost certain it is open as the limits are not included in the rational set.\" That would only mean it is not closed. It wouldn't mean it is open. It could very well be neither. \u2013\u00a0fleablood Jun 18 '17 at 21:03\n\u2022 \"Is the following set open or closed?\" Why do you assume it has to be one or the other? Could it be both? Could it be neither? \u2013\u00a0fleablood Jun 18 '17 at 21:04\n\nLet $A$ be your set of rationals. You ask if $A$ is open or closed.\n\nOpen or closed within what space? The reals $\\mathbb{R}$ or the rationals $\\mathbb{Q}$?\nNotice that $$A = [-r,r] \\cap \\mathbb{Q} = (-r,r) \\cap \\mathbb{Q}$$ where r = sqrt 2. Thus within $\\mathbb{Q}$, $A$ is clopen.\nHowever, within $\\mathbb{R}$ it is neither.\n\nThe $\\mathbb{R}$-closure of $A$ is $[-r,r]$. The $\\mathbb{R}$-interior of $A$ is empty.\n\nIt depends on what space your topology is.\n\nIf your space is $\\mathbb R$ it is neither open nor closed.\n\nIf your space $\\mathbb Q$ it is both open and closed.\n\nThis set is simply $[-\\sqrt{2}, \\sqrt{2}]\\cap \\mathbb Q$.\n\nThe limit points are all the points between $-\\sqrt{2}$ and $\\sqrt{2}$. In the space $\\mathbb Q$ that is indeed the set. So in the space $\\mathbb Q$ this set is closed.\n\nBut in the space $\\mathbb R$ this includes all the irrational points as well. The rational limit points are included but the irrational limit points are not. So in the space $\\mathbb R$ this set is not closed.\n\nIn the space $\\mathbb Q$ every point is an interior point. Around every point you can find a small enough neighborhood that is entirely inside the set. So every point is an interior point. So the set is open in $\\mathbb Q$.\n\nBut in the space $\\mathbb R$ every neighborhood no matter how small will include irrational points that are not part of the set. So no point of the set is actually an interior point. So the set is not open in $\\mathbb R$.\n\n===\n\nIn $\\mathbb Q$ the set is closed. So the closure is the set itself.\n\nIn $\\mathbb R$ the set is not closed. To get the proper closure you must include all the limt points. $\\pm \\sqrt{2}$ are limit points, yes, but so is every point between $-\\sqrt{2}$ and $\\sqrt{2}$. A limit point isn't just \"the points on the edge\". There are the points so that every neighborhood no matter how small will intersect the set. These are not only the points \"on the edge\" but the points \"inside\" as well.\n\nSo the closure will include all the points between $-\\sqrt{2}$ and $\\sqrt{2}$ and so the closure in $\\mathbb Q$ is the set $[-\\sqrt{2}, \\sqrt{-1}]$.\n\n===\n\nNote:\n\nOpen $\\ne$ not closed. Sets can be both and sets can be neither.\n\nlimit points $\\ne$ boundary points. boundary points $\\subset$ limit points.\n\nIt is open in $\\Bbb Q$ as you correctly observe. It is closed in $\\Bbb Q$ because it is the inverse image of a closed interval under the (continuous) squaring function.", "date": "2019-04-25 22:40:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2326833/is-the-set-left-x-in-q-colon-x2-le-2-right-open-or-closed/2326932", "openwebmath_score": 0.9063434600830078, "openwebmath_perplexity": 163.37555708729062, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692339078752, "lm_q2_score": 0.8962513765975758, "lm_q1q2_score": 0.875341145370433}}
{"url": "https://math.stackexchange.com/questions/3987133/how-can-i-describe-a-linear-equation-that-becomes-sinusoidal-after-a-certain-poi", "text": "# How can I describe a linear equation that becomes sinusoidal after a certain point?\n\nFor example, take $$y = x$$, but at $$x \\geq 5$$, $$y$$ now equals $$\\sin x + 5$$. So for all $$x$$ values $$\\lt 5$$, $$y$$ has linear relationship, but at $$\\geq 5$$, $$y$$ now has sinusoidal relationship. Is there a concise way to express this, perhaps a single equation?\n\nA function like this is said to be a piecewise function. The typical way to notate them is to enumerate the subdomains and the functions defined on such:\n\n$$f(x) = \\left\\{\\begin{array}{ll} x, & \\text{if }x<5\\\\ x+\\sin x, & \\text{if }x\\geq 5\\\\\\end{array}\\right.$$\n\nSince any real $$x$$ satisfies one of these two conditions, we could also write this as\n\n$$f(x) = \\left\\{\\begin{array}{ll} x+\\sin x, & \\text{if }x\\geq 5\\\\ x, & \\text{else }\\end{array}\\right.$$\n\nAnother, less common notation is that of the Iverson bracket. The Iverson bracket $$[P]$$ for a conditional $$P$$ is defined as $$[P]=\\left\\{\\begin{array}{ll} 1, & \\text{if }P\\text{ is true}\\\\ 0, & \\text{else }\\end{array}\\right.$$\n\nThis allows us to express piecewise functions with one-line notation, e.g., $$f(x)=x [x<5]+(x\\sin x)[x\\geq 5].$$ Since $$x\\leq 5$$ is the negation of $$x<5$$, we may eliminate the first bracket as $$[x<5]=1-[x\\geq 5]$$ and therefore also have \\begin{align} f(x)&=x(1-[x\\geq 5])+(x+\\sin x)[x\\geq 5]\\\\ &=x+(\\sin x)[x\\geq 5]. \\end{align} This captures the intuition that, whether or not $$x\\geq 5$$, we start \"at minimum\" with $$f(x)=x$$; if we $$x\\geq 5$$, we add on $$\\sin x$$ as well. Of course, we could also have done this with our original notation:\n\n\\begin{align} f(x) &= x+ \\left\\{\\begin{array}{ll} \\sin x, & \\text{if }x\\geq 5\\\\ 0, & \\text{else }\\end{array}\\right.\\\\ &= x+ (\\sin x)\\left\\{\\begin{array}{ll} 1, & \\text{if }x\\geq 5\\\\ 0, & \\text{else }\\end{array}\\right. \\end{align}\n\nSo these are ultimately just variations in notation, all of which present the same concept.\n\nThis is called a PIECEWISE function. It is a function that behave differently depending on which parts of the domain you are on. It is basically made out of PIECES of other functions, and they are tied together by taking sections of them and combining them into one function. For example, $$|x|$$ is a piecewise function because it is equal to $$x$$ for $$x \\ge 0$$, and it is equal to $$-x$$ for $$x \\le 0$$. A piecewise function might look like this: $$f(n) = \\begin{cases} n/2, & \\text{if n is even} \\\\ 3n+1, & \\text{if n is odd} \\end{cases}$$\n\nThis question makes me think of a related but slightly different kind of function. A function which gradually/slowly changes behavior, instead of having this piecewise phenomenon.\n\nThe piecewise and discontinuous transition would be denoted as $$h(x) = \\begin{cases} f(x)\\quad\\text{if}\\quad x \\leq x_0\\\\ g(x)\\quad\\text{if}\\quad x > x_0\\\\ \\end{cases}$$ A gradual shift from $$f$$ to $$g$$ can be written as $$(1-t)\\cdot f + t\\cdot g$$ with $$t\\in[0,1]$$. Therefore a function that gradually changes from $$f$$ after $$x_0$$ and becomes $$g$$ at $$x_1 > x_0$$ would be written $$h_{\\mathrm{grad}}(x) = \\begin{cases} f(x)\\quad\\text{if}\\quad x \\leq x_0\\\\ \\\\ \\left(1 - \\dfrac{x-x_0}{x_1-x_0}\\right)\\cdot f(x) + \\left(\\dfrac{x-x_0}{x_1-x_0}\\right)\\cdot g(x) \\quad\\text{if}\\quad x_0 < x \\leq x_1\\\\ \\\\ g(x)\\quad\\text{if}\\quad x > x_1\\\\ \\end{cases}$$ As an exemple, let us take $$f(x)=x$$ and $$g(x) = \\sin(x) + 5$$ on the domain $$[-5, 10]$$.\n\nThe piecewise transition is pictured below:\n\nThe continuous/gradual transition is in green below:", "date": "2021-03-05 13:49:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3987133/how-can-i-describe-a-linear-equation-that-becomes-sinusoidal-after-a-certain-poi", "openwebmath_score": 0.9806784987449646, "openwebmath_perplexity": 250.9069976355607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380070417538, "lm_q2_score": 0.8824278772763472, "lm_q1q2_score": 0.8753137499435851}}
{"url": "https://rasbt.github.io/mlxtend/user_guide/math/num_combinations/", "text": "# Compute the Number of Combinations\n\nA function to calculate the number of combinations for creating subsequences of k elements out of a sequence with n elements.\n\nfrom mlxtend.math import num_combinations\n\n## Overview\n\nCombinations are selections of items from a collection regardless of the order in which they appear (in contrast to permutations). For example, let's consider a combination of 3 elements (k=3) from a collection of 5 elements (n=5):\n\n\u2022 collection: {1, 2, 3, 4, 5}\n\u2022 combination 1a: {1, 3, 5}\n\u2022 combination 1b: {1, 5, 3}\n\u2022 combination 1c: {3, 5, 1}\n\u2022 ...\n\u2022 combination 2: {1, 3, 4}\n\nIn the example above the combinations 1a, 1b, and 1c, are the \"same combination\" and counted as \"1 possible way to combine items 1, 3, and 5\" -- in combinations, the order does not matter.\n\nThe number of ways to combine elements (without replacement) from a collection with size n into subsets of size k is computed via the binomial coefficient (\"n choose k\"):\n\nTo compute the number of combinations with replacement, the following, alternative equation is used (\"n multichoose k\"):\n\n## Example 1 - Compute the number of combinations\n\nfrom mlxtend.math import num_combinations\n\nc = num_combinations(n=20, k=8, with_replacement=False)\nprint('Number of ways to combine 20 elements'\n' into 8 subelements: %d' % c)\n\nNumber of ways to combine 20 elements into 8 subelements: 125970\n\nfrom mlxtend.math import num_combinations\n\nc = num_combinations(n=20, k=8, with_replacement=True)\nprint('Number of ways to combine 20 elements'\n' into 8 subelements (with replacement): %d' % c)\n\nNumber of ways to combine 20 elements into 8 subelements (with replacement): 2220075\n\n\n## Example 2 - A progress tracking use-case\n\nIt is often quite useful to track the progress of a computational expensive tasks to estimate its runtime. Here, the num_combination function can be used to compute the maximum number of loops of a combinations iterable from itertools:\n\nimport itertools\nimport sys\nimport time\nfrom mlxtend.math import num_combinations\n\nitems = {1, 2, 3, 4, 5, 6, 7, 8}\nmax_iter = num_combinations(n=len(items), k=3,\nwith_replacement=False)\n\nfor idx, i in enumerate(itertools.combinations(items, r=3)):\n# do some computation with itemset i\ntime.sleep(0.1)\nsys.stdout.write('\\rProgress: %d/%d' % (idx + 1, max_iter))\nsys.stdout.flush()\n\nProgress: 56/56\n\n\n## API\n\nnum_combinations(n, k, with_replacement=False)\n\nFunction to calculate the number of possible combinations.\n\nParameters\n\n\u2022 n : int\n\nTotal number of items.\n\n\u2022 k : int\n\nNumber of elements of the target itemset.\n\n\u2022 with_replacement : bool (default: False)\n\nAllows repeated elements if True.\n\nReturns\n\n\u2022 comb : int\n\nNumber of possible combinations.\n\nExamples\n\nFor usage examples, please see http://rasbt.github.io/mlxtend/user_guide/math/num_combinations/", "date": "2018-07-22 01:18:30", "meta": {"domain": "github.io", "url": "https://rasbt.github.io/mlxtend/user_guide/math/num_combinations/", "openwebmath_score": 0.3569217920303345, "openwebmath_perplexity": 3775.011725687833, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380101400315, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.8753137404094655}}
{"url": "https://mathoverflow.net/questions/353042/number-of-regions-formed-by-n-points-in-general-position", "text": "# Number of regions formed by $n$ points in general position\n\nGiven $$n$$ points in $$\\mathbb{R}^d$$ in general position, where $$n\\geq d+1$$. For every $$d$$ points, form the hyperplane defined by these $$d$$ points. These hyperplanes cut $$\\mathbb{R}^d$$ into several regions. My questions are:\n(1) is there a formula in terms of $$d$$ and $$n$$ that describes the number of regions?\n(2) the same question for the number of bounded regions?\n\nI tried many key words on google but found nothing helpful. Any reference or ideas will be appreciated. Thanks\n\n\u2022 General position as defined in en.wikipedia.org/wiki/General_position Feb 19, 2020 at 4:27\n\u2022 Can you maybe elaborate? Both for the general position arrangement where all 4 points lie on the boundary of their convex hull, as well as for the arrangement where one point lies in the interior, I count 6 bounded regions. Feb 19, 2020 at 7:29\n\u2022 The maximum number of cells in an arrangement of $k$ hyperplanes in dimension $d$ is $O(k^d)$. Your $k$ is $\\binom{n}{d}$. Feb 19, 2020 at 11:56\n\u2022 @JosephO'Rourke Thanks for the info. Can you direct me to a reference where I can find the results? Feb 19, 2020 at 17:59\n\u2022 @GHfromMO I think you are right, the number of bounded regions may vary from case to case. I am wondering whether the total number of regions would also vary or just be the same? Feb 19, 2020 at 18:01\n\nThis is more of a long comment than an answer. It should be possible to compute the number of regions and number of bounded regions using Whitney's theorem for the characteristic polynomial $$\\chi(t)$$ (Theorem 2.4 of these notes), and Zaslavsky's theorem that the number of regions is $$(-1)^d \\chi(-1)$$, and the number of bounded regions is (in this situation) $$(-1)^d\\chi(1)$$ (Theorem 2.5 of the previous link). We need more than the usual definition of \"general position.\" We want the position to be generic enough for the argument below (generalized to $$d$$ dimensions) to hold.\n\nHere is the computation for $$d=2$$. First, the empty intersection (the ambient space $$\\mathbb{R}^2$$) contributes $$t^2$$ to $$\\chi(t)$$. The $${n\\choose 2}$$ lines will contribute $$-{n\\choose 2}t$$. Now we must consider all subsets of the lines that intersect in a point $$p$$. Let $$p$$ be one of the original $$n$$ points. Then $${n-1\\choose 2}$$ pairs of lines intersect in $$p$$, $${n-1\\choose 3}$$ triple of lines intersect in $$p$$, etc., giving a contribution to $$\\chi(t)$$ of $${n-1\\choose 2}-{n-1\\choose 3}+{n-1\\choose 4} -\\cdots = n-2.$$ We have to multiply this by $$n$$ since there are $$n$$ choices for $$p$$. There are now $$3{n\\choose 4}$$ choices of two lines that don't intersect in one of the original $$n$$ points, but they still intersect by genericity. Thus we get an additional contribution of $$3{n\\choose 4}$$. It follows that $$\\chi(t) = t^2-{n\\choose 2}t+n(n-2)+3{n\\choose 4}.$$ The number of regions is $$\\chi(-1) = \\frac 18(n-1)(n^3-5n^2+18n-8).$$ The number of bounded regions is $$\\chi(1) = \\frac 18(n-1)(n-2)(n^2-3n+4).$$ Can someone extend this argument to $$d$$ dimensions?\n\nAddendum. I worked out $$d=3$$. Here are the details. Let $$X$$ be an $$n$$-element \"generic\" subset of $$\\mathbb{R}^3$$. Thus $$X$$ determines a set $$\\mathcal{A}$$ of $${n\\choose 3}$$ hyperplanes. We need to find all subsets of $$\\mathcal{A}$$ with nonempty intersection. A $$j$$-element subset that intersects in an $$e$$-dimensional affine space contributes $$(-1)^jt^e$$ to the characteristic polynomial $$\\chi(t)$$.\n\nCase 1: $$e=3$$. We take the intersection over the empty set to get $$\\mathbb{R}^3$$. This gives a term $$t^3$$.\n\nCase 2: $$e=2$$. Each hyperplane contributes $$-t^2$$, giving a term $$-{n\\choose 3}t^2$$.\n\nCase 3: $$e=1$$. (a) Any two hyperplanes intersect in a line (by genericity), giving $${{n\\choose 3}\\choose 2}t$$.\n\n(b) Any $$j\\geq 3$$ hyperplanes containing the same two points $$p,q\\in X$$ meet in a line. There are $${n\\choose 2}$$ choices for $$p,q$$ and $${n-2\\choose j}$$ for the remaining element of $$X$$ in the hyperplanes. Thus we get a contribution $${n\\choose 2}\\sum_{j\\geq 3}(-1)^j {n-2\\choose j}t = {n\\choose 2}\\left[-1+(n-2)-{n-2\\choose 3}\\right]t.$$\n\nCase 4: $$e=0$$. (a) Any three hyperplanes intersect at a point, except when all three contain the same two points $$p,q\\in X$$. There are $${{n\\choose 3}\\choose 3}$$ ways to choose three hyperplanes, and $${n\\choose 2}{n-2\\choose 3}$$ ways to choose them so that they intersect in two points of $$X$$. Hence we get a contribution $$-\\left[ {{n\\choose 3}\\choose 3}-{n\\choose 2}{n-2\\choose 3} \\right]$$ to the constant term of $$\\chi(t)$$ (the minus sign because the number of hyperplanes is odd).\n\n(b) Any $$j\\geq 4$$ hyperplanes meeting at a point $$p\\in X$$. We can choose $$p$$ in $$n$$ ways. We then must choose $$j$$ two-element subsets of $$X-p$$ whose intersection is empty. There are $${{n-1\\choose 2}\\choose j}$$ ways to choose $$j$$ two-element subsets of $$X-p$$. If their intersection is nonempty, then they have a common element $$q$$ which can be chosen in $$n-1$$ ways, and then we can choose the remaining elements in $${n-2\\choose j}$$ ways. This gives the contribution $$n\\sum_{j\\geq 4}(-1)^j\\left[ {{n-1\\choose 2}\\choose j}- (n-1){n-2\\choose j}\\right]$$ $$\\ = n\\left[ -1+{n-1\\choose 2}-{{n-1\\choose 2}\\choose 2} + {{n-1\\choose 2}\\choose 3}-(n-1)\\left(-1+(n-2) -{n-2\\choose 2}+{n-2\\choose 3}\\right)\\right].$$\n\n(c) Any $$j\\geq 3$$ hyperplanes meeting at $$p,q\\in X$$, together with one additional hyperplane not containing $$p$$ or $$q$$. There are $${n\\choose 2}$$ choices for $$p,q$$ and $${n-2\\choose j}$$ ways to choose $$j$$ hyperplanes containing $$p,q$$. There are then $${n-2\\choose 3}$$ ways to choose the additional hyperplane not containing $$p$$ or $$q$$. Thus we get the contribution $$-\\left[ \\sum_{j\\geq 3}(-1)^j{n-2\\choose j}\\right] {n-2\\choose 3}$$ $$-{n-2\\choose 3}{n\\choose 2}\\left[-1+(n-2)-{n-2\\choose 2} \\right].$$\n\nPutting all this together gives the characteristic polynomial $$t^3-{n\\choose 3}t^2+ \\frac{1}{72}n(n-1)(n-3)(n^3-2n^2-16n+68)t$$ $$-\\frac{1}{1296}n(n-2)(n-3)(n^6-4n^5-74n^4+698n^3-2129n^2 +2276n-120).$$ The number of regions is $$\\frac{1}{1296}(n-2)(n^8-7n^7-62n^6+938n^5-4295n^4+8429n^3 -4932n^2-2016n-648).$$ The number of bounded regions is $$\\frac{1}{1296}(n-1)(n-2)(n-3)(n^6-3n^5-77n^4+603n^3 -1508n^2+1056n+216).$$ Conceivably there could be an error in the computation, but I checked it for $$n=4,5$$ by a brute force computation.\n\nThis method should extend to any $$d$$, but the computation will be more complicated, and I am too lazy to work out the details.\n\n\u2022 A note on the stronger condition required, here is an example of what can go wrong: Consider the simple case $6$ points in the plane, no $3$ on a line. There is a special degenerate case where we can partition the $6$ points into three pairs $(A, A')$ $(B, B')$ and $(C,C')$ where the lines $AA'$, $BB'$, and $CC'$ all intersect at a point. This will have one less region than the general case.\n\u2013\u00a0Nate\nFeb 19, 2020 at 20:54\n\u2022 Theorem 2.2 of Ardila-Billey arxiv.org/abs/math/0605598 gives a combinatorial description of the relevant matroid (described dually, as the intersections of generic hyperplanes rather than the spans of generic points). I don't see an easy way to extract the characteristic polynomial from it, but maybe you do. Feb 20, 2020 at 4:32\n\u2022 I replicated your result for $d=2$ based on my answer below. I proved there that $x$ generic lines produce $2x$ unbounded regions and $(x-2)(x-1)/2$ bounded ones. Then replaced $x$ with $n(n-1)/2$. Last, at each of the $n$ points the bounded regions that would exist in a generic $(n-1)$-lines case are collapsed to a point. So the total number of regions collapsed to a point is $n(n-3)(n-2)/2$. This leaves $(x-2)(x-1)/2$-$n(n-3)(n-2)/2$ bounded regions, which matches your calculation. I think $d=3$ could be tacked in a similar way, but much more messy. Feb 29, 2020 at 19:48\n\nPerhaps it is worth quoting this theorem, even though it does not distinguish bounded from unbounded cells, and is phrased in terms of the number of hyperplanes rather than the number of points determining the hyperplanes.\n\nTheorem. Let $$H$$ be a set of $$n$$ hyperplanes in $$\\mathbb{R}^d$$. The maximum number of $$k$$-dimensional cells in the arrangement $${\\cal A}(H)$$ formed by $$H$$, for $$0 \\le k \\le d$$, is $$\\sum_{i=0}^k \\binom{d-i}{k-i} \\binom{n}{d-i} \\;.$$ The maximum is attained exactly when $${\\cal A}(H)$$ is simple.\n\nAn arrangement is simple if every $$d$$ hyperplanes meet in a point, and no $$d+1$$ hyperplanes have a point in common. In the OP's situation, the arrangement is non-simple. For example, in $$d=2$$, $$4$$ points determine $$6$$ lines, but each point has $$3$$ lines through it.\n\nFor $$d=2$$ and $$k=2$$, the above equation reduces to the familiar expression $$\\binom{n}{2} + \\binom{n}{1} + \\binom{n}{0} = \\frac{1}{2}( n^2 + n + 2) \\;.$$\n\nHandbook of Discrete and Computational Geometry, 3rd ed. Chapman and Hall/CRC, 2017. CRC link. Chapter 28, Thm.28.1.1, p.724.\n\nThis non-answer completes Joseph O'Rourke's nice non-answer, for the case of $$n$$ hyperplanes in $$\\mathbb{R}^d$$ in general position. But it also suggests that the OP situation may also well have unique answers.\n\nDefine:\n\n$$U_{d,n}=$$ number of unbounded regions cut by $$n$$ hyperplanes in $$\\mathbb{R}^d$$\n\n$$B_{d,n}=$$ number of bounded regions\n\n$$T_{d,n}=$$ total number of regions $$=U_{d,n}+B_{d,n}$$\n\n$$S_{d,n}=$$ number of regions cut on the sphere $$S^d$$ by $$n$$ great $$S^{d-1}$$-circles\n\nThen $$U_{d,n}, B_{d,n}$$, $$T_{d,n}$$ and $$S_{d,n}$$ are unique with these formulas:\n\n$$U_{d,0}=1$$, $$\\quad B_{d,0}=0$$, $$\\quad T_{d,0}=1$$, $$\\quad S_{d,0}=1$$\n\n$$U_{1,n}=2$$, $$\\quad B_{1,n}=n-1$$, $$\\quad T_{1,n}=n+1$$, $$\\quad S_{1,n}=2n$$\n\nand for $$n>0$$\n\n1. $$U_{d+1,n}=S_{d,n}$$\n2. $$S_{d,n}=U_{d,n}+2B_{d,n}$$\n3. $$T_{d,n+1}=T_{d,n}+\\sum_{i=0}^{i=d-1}{n\\choose i}$$\n\nProof.\n\n1. In $$\\mathbb{R}^{d+1}$$ take a huge and growing $$S^d$$ sphere, so that all the bounded regions zoom down to a point at the center of the sphere, the hyperplanes become great circles on the sphere and the unbounded regions corresponds to regions cut by the circles on the sphere. Therefore if the numbers are unique (as will be proved at the end) 1. follows.\n\n2. Centrally project $$\\mathbb{R}^d$$ onto a half $$S^d$$ (tangent to it). Complete the semisphere to a sphere by central symmetry. Then the hyperplanes become great circles, the $$B_{d,n}$$ bounded regions in $$\\mathbb{R}^d$$ become $$2B_{d,n}$$ regions in $$S^d$$ and the $$U_{d,n}$$ unbounded ones become $$U_{d,n}$$ regions stretching across the suture line (equator) of the sphere. Again by unicity 2. follows.\n\n3. Start with $$\\mathbb{R}^d$$ and $$n$$ hyperplanes in generic position inside it. Now add a new hyperplane in generic position the following way: first chose a point inside one region: no matter how that point is eventually stretched to a hyperplane, to it will split the region in two, for a gain of 1, or $$n \\choose 0$$. Now stretch that point to a line: since it is a generic line it will meet each of the $$n$$ hyperplanes once and at each meeting the line will cross into one one new region and split it - with a gain of $$n \\choose 1$$ new regions. Next stretch the line to a generic 2-plane, which will meet once each of the $$(d-2)$$-dimensional intersections of two hyperplanes; at each meeting the growing plane will arrive from having already crossed 3 of the 4 regions, to cross into the fourth and cut it; this a gain of another $$n \\choose 2$$ regions. In general as a generic $$m-1$$-plane grows to a generic $$m$$-plane it will meet all the $$n \\choose m$$ $$(n-m)$$-dimensional intersections of $$m$$ hyperplanes, and each time it will go from cutting $$2^m-1$$ regions before crossing the intersection to cutting all $$2^m$$ after crossing, for a total gain of $$n \\choose m$$ regions. This continues up to $$m=d-1$$, proving 3.\n\nProof of Unicity.\n\nBy induction:\n\n$$U_{d,0}$$, $$\\quad B_{d,0}$$, $$\\quad T_{d,0}$$, $$\\quad S_{d,0}$$ are unique;\n\n$$T_{d,n}$$ unique $$\\implies$$ $$T_{d,n+1}$$ unique (by the proof of 3.);\n\n$$U_{d,n}$$ and $$B_{d,n}$$ unique $$\\implies$$ $$S_{d,n}$$ unique (by the proof of 2. and the fact that the construction can be reversed in a non-unique way to show that $$S_{d,n}=U_{d,n}+2B_{d,n}$$ for some values of $$U_{d,n}$$ and $$B_{d,n}$$);\n\n$$S_{d,n}$$ unique $$\\implies$$ $$U_{d+1,n}$$ unique (by the proof of 1.);\n\n$$U_{d+1,n}$$ and $$T_{d+1,n}$$ unique $$\\implies$$ $$B_{d+1,n}$$ unique (as $$B=T-U$$).\n\n\u2022 I have a feeling that jiggling the $n \\choose d$ hyperplanes of the OP into generic positions while carefully keeping track of the corresponding increase of regions (both bounded and unbounded) one may be able to reach an answer to the OP's question too. For a start this should be relatively easy for $d=2$ (but I haven't done it). Feb 23, 2020 at 15:32", "date": "2022-08-14 22:43:05", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/353042/number-of-regions-formed-by-n-points-in-general-position", "openwebmath_score": 0.7762052416801453, "openwebmath_perplexity": 183.98777061291324, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380082333994, "lm_q2_score": 0.8824278540866547, "lm_q1q2_score": 0.875313727992389}}
{"url": "http://math.stackexchange.com/questions/61482/proving-the-identity-sum-limits-k-1n-k3-large-sum-limits-k-1n-k/61502", "text": "# Proving the identity $\\sum\\limits_{k=1}^n {k^3} = {\\Large(}\\sum\\limits_{k=1}^n k{\\Large)}^2$ without induction\n\nI recently proved that $$\\sum_{k=1}^n k^3 = \\left(\\sum_{k=1}^n k \\right)^2$$ Using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.\n\nThanks!\n\n-\nLook at this takayaiwamoto.com/Sums_and_Series/sumcube_1.html \u2013\u00a0 pritam Jun 27 '12 at 9:29\nSee math.stackexchange.com/questions/120674 for remarks about proofs \"not using induction\". \u2013\u00a0 sdcvvc Jun 27 '12 at 10:13\nI merged the three existing posts which covered exactly this question, as each post had different interesting answers which should not be lost. I also deleted redundant comments, and comments about closing posts as duplicates. This fourth question is not considered a duplicate. \u2013\u00a0 Eric Naslund Jul 2 '12 at 11:30\nSince this question is asked frequently, it has been added to the list of Generalizations of Common questions. It has been kept seperate from the version which does use induction. \u2013\u00a0 Eric Naslund Aug 30 '12 at 0:23\nshow 1 more comment\n\nStare at the following image, taken from this MO answer, long enough:\n\n-\n+1 I'm in love with that one \u2013\u00a0 leonbloy Sep 2 '11 at 23:23\nThe fact that there are $k$ blocks (or $\\frac{1}{2}+k{-}1+\\frac{1}{2}$ blocks) of $k\\times k$ size is based on the fact that $\\sum\\limits_{j=1}^{k-1}=k(k{-}1)/2$. That is, $(k{-}1)/2$ blocks on top $(k{-}1)/2$ on the left and $1$ block at the corner (totaling to $k$). Perhaps I am being picky or slow, but I don't see this as obvious from the image. Beyond that, it is a nice proof-without-words. \u2013\u00a0 robjohn Sep 3 '11 at 4:13\nI have put the details of the proof corresponding to this image in this answer since the comment area was too small. \u2013\u00a0 Wok Sep 3 '11 at 7:23\n\nI don't know if this is intuitive, but it is graphic.\n\nOn the outer edge of each $(k{+}1){\\times}k$ block there are $k$ pairs of products each of which total to $k^2$. Thus, the outer edge sums to $k^3$, and the sum of the whole array is therefore $\\sum\\limits_{k=1}^n k^3$.\n\nThe array is the matrix product $$\\left[\\begin{array}{r}0\\\\1\\\\2\\\\\\vdots\\\\n\\end{array}\\right]\\bullet\\left[\\begin{array}{rrrrr}1&2&3&\\cdots&n\\end{array}\\right]$$ Therefore, the sum of the elements of the array is $\\sum\\limits_{k=0}^nk\\;\\sum\\limits_{k=1}^nk=\\left(\\sum\\limits_{k=1}^nk\\right)^2$.\n\nTherefore, $\\sum\\limits_{k=1}^n k^3=\\left(\\sum\\limits_{k=1}^nk\\right)^2$\n\n-\nIf we forget the first line of the matrix (which is zero and which is only used to make pairs with the diagonal coefficients), then I like the fact we can put this answer in parallel with the coloured rectangles above and below, and we get another partition of each colored area (each coefficient of the matrix gives a rectangle of a certain area), which explains why each L-shaped area is $k^3$. \u2013\u00a0 Wok Sep 4 '11 at 7:39\nHowever, each L-shape coefficient has the same factor $k$, which means it proves \"each L-shaped area is $k^3$\" by the same proof that $2 \\times \\sum_{j=0}^k j = (0+k) + (1+k-1) + ... + (k-1+1) + (k+0) = (k+1) \\times k$, which makes it really close to the coloured rectangles above and below. \u2013\u00a0 Wok Sep 4 '11 at 7:49\nI hope you don't mind if I use both ideas in another answer. \u2013\u00a0 Wok Sep 4 '11 at 8:13\nI don't mind. I simply find it less aesthetic to need to use $\\sum\\limits_{j=1}^k\\;j=k(k+1)/2$ or that $(k(k+1)/2)^2-(k(k-1)/2)^2=k^3$ in an intuitive proof. \u2013\u00a0 robjohn Sep 4 '11 at 19:31\n\nCan you get the intuition explanation from the following two pictures?[EDIT: the following is essentially the same as Mariano's answer. He didn't mentioned the first picture though.]\n\nThe images are from Brian R Sears.\n\n-\nnice, but already posted (linked) by Mariano \u2013\u00a0 leonbloy Sep 2 '11 at 23:24\n\nThere's this nice picture from the Wikipedia entry on the squared triangular number:\n\nThe left side shows that $1 + 2 + 3$ forms a triangle and so that squaring it produces a larger triangle made up of $1+2+3$ copies of the original triangle. The right side has $1(1^2) + 2(2^2) + 3(3^2) = 1^3 + 2^3 + 3^3$. The coloring shows why the two sides are equal.\n\nThere are several other references for combinatorial proofs and geometric arguments on the Wikipedia page.\n\n-\n\nEach colored area is $k^3$ as a difference of two areas: $S_k^2 - S_{k-1}^2$.\n\nThe detailed proof which comes with the drawing is the following.\n\nFor any positive integer $k$, we define: $$S_i = \\sum_{j=1}^{i} j$$\n\nWe first notice: $$S_i^2 = S_i^2 - S_0^2= \\sum_{k=1}^{i} \\left(S_k^2 - S_{k-1}^2\\right)$$\n\nThe expected result finally comes from: $$S_k^2 - S_{k-1}^2 = k \\left(k+2 S_{k-1}\\right) = k\\left(k+k\\left(k-1\\right)\\right)=k^3$$\n\n-\nSo essentially, you are using the fact that $$\\left(\\sum_{j=1}^k\\;j\\right)^2-\\left(\\sum_{j=1}^{k-1}\\;j\\right)^2=k^3$$ to justify the diagram which is supposed to prove that fact intuitively. \u2013\u00a0 robjohn Sep 4 '11 at 1:09\nAs you mentioned in another answer, this is where the diagram is the least intuitive. However, if you cannot picture $k^3$ on a 2D-plane, then you need another representation as a difference of areas. \u2013\u00a0 Wok Sep 4 '11 at 6:50\nAs soon as you know $S_k = \\sum_{j=1}^k j = \\frac{k(k+1)}{2}$, I find it more intuitive to figure out each colored area is $k^3$ on this diagram than to figure it out by counting squares plus two rectangles when $k$ is even. \u2013\u00a0 Wok Sep 4 '11 at 7:06\n\nThe formula is due to Nicomachus of Gerasa. There is a nice discussion of ways to prove it at this n-category cafe post, including a bijective proof and some visual / \"geometric\" proofs.\n\n-\n\nHere's another version of this \"proof without words\". This is the case $n=4$.\n\nThere are 1 $1 \\times 1$, 2 $2 \\times 2$, 3 $3 \\times 3$, ... squares, for a total area of $1^3 + 2^3 + \\ldots + n^3$. For even $k$, two of the $k \\times k$ squares overlap in a $k/2 \\times k/2$ square, but this just balances out a $k/2 \\times k/2$ square that is left out, so the total is the area of a square of side $1 + 2 + \\ldots + n$.\n\n-\n\nI found this but sadly do not understand it, can someone explain the method?\n\nEdit:\n\nThis has the same proof but much easier to understand. It is also found in the link by QY\n\n-\nExcellent picture!! (second one) \u2013\u00a0 Eric Naslund Jan 22 '11 at 20:14\n\nSeveral visual proofs of this indentity are collected in the book Roger B. Nelsen: Proofs without Words starting from p.84.\n\nAlthough several of these proofs can still be considered inductive, I thought it might be interesting to mention them.\n\nOriginal sources are given on p. 147:\n\n\u2022 84 Mathematical Gazette, vol. 49, no. 368 (May 1965), p. 199. jstor\n\u2022 85 Mathematics Magazine, vol. 50, no. 2 (March 1977), p. 74. jstor\n\u2022 86 Mathematics Magazine, vol. 58, no. 1 (Jan. 1985), p. 11. jstor\n\u2022 87 Mathematics Magazine, vol. 62, no. 4 (Oct. 1989), p. 259. jstor\n\u2022 87 Mathematical Gazette, vol. 49, no. 368 (May 1965), p. 200. jstor\n\u2022 88 Mathematics Magazine, vol. 63, no. 3 (June 1990), p. 178. jstor\n\u2022 89 Mathematics Magazine, vol. 62, no. 5 (Dec. 1989), p. 323. jstor\n\u2022 90 Mathematics Magazine, vol. 65, no. 3 (June 1992), p. 185. jstor\n-\nGreat collections. Thanks \u2013\u00a0 B. S. Jun 27 '12 at 11:11\n\nI believe this illustration is due to Anders Kaseorg:\n\n-\n\nWe know that $$A=\\Sigma_{1}^{n}k^3=\\frac{1}{4}n^4+\\frac{1}{2}n^3+\\frac{1}{4}n^2$$ and $$B=\\Sigma_{1}^{n}k=\\frac{1}{2}n^2+\\frac{1}{2}n$$ $A-B^2=0$. :)\n\n-\nIf you are presenting this proof,write it nicely. $A=\\frac{n^2(n+1)^2}{4}$ and $B=\\frac{n(n+1)}{2}$,obviously $A=B^2$ \u2013\u00a0 Aang Jun 27 '12 at 10:40\n@avatar: Dear Avatar; I will. Your proof is great and it looks analytically. Thanks. :-) \u2013\u00a0 B. S. Jun 27 '12 at 10:48\nsorry, if you felt that i was being rude. \u2013\u00a0 Aang Jun 27 '12 at 10:53\n@avatar: Your proof lights mine. No Problem at all. WELCOME Avatar. :) :) \u2013\u00a0 B. S. Jun 27 '12 at 10:57\n\nThe sum of a degree $n$ polynomial $f(n)$ will be a degree $n+1$ polynomial $S(n)$ for $n \\geq 0$ and both polynomials can be extended (maintaining the relation $S(n)-S(n-1) = f(n)$) to negative $n$. To verify that the formula for $\\Sigma k^3$ is correct one need only test it for any 5 distinct values of $n$, but the structure of the answer can be predicted algebraically using the continuation to negative $n$.\n\nIf $S(n) = (1^3 + 2^3 + \\dots n^3)$ is the polynomial that satisfies $S(n)-S(n-1) = n^3$ and $S(1)=1$, then one can calculate from that equation that $S(0)=S(-1)=0$ and $S(-n-1)=S(n)$ for all negative $n$, so that $S$ is symmetric around $-1/2$. The vanishing at 0 and -1 implies that $S(t)$ is divisible as a polynomial by $t(t+1)$. The symmetry implies that $S(t)$ is a function (necessarily a polynomial) of $t(t+1)$.\n\n$S(t)$ being of degree 4, this means $S(n) = a (n)(n+1) + b((n^2 +n)^2$ for constants $a$ and $b$. Summation being analogous to integration (and equal to it in a suitable limit), they have to agree on highest degree terms. Here it forces $b$ to be $1/4$ to match $\\int x^3 = x^4/4$. Computing the sum at a single point such as $n=1$ determines $a$, which is zero.\n\nSimilar reasoning shows that $S_k(n)$ is divisible as a polynomial by $n(n+1)$ for all $k$. For odd $k$, $S_k(n)$ is a polynomial in $n(n+1)$.\n\n-\n\nYou know, $\\sum_0^n x^k=\\frac{1-x^{n+1}}{1-x}$. Differentiate both sides once, $\\sum_1^n kx^{k-1}=\\frac{x^n(nx-n-1)+1}{x^2-2x+1}$. Now taking $\\lim_{x\\to1}$ both sides and then squaring the result will give you the expression on the RHS. You can further differentiate $\\sum_0^n x^k=\\frac{1-x^{n+1}}{1-x}$ until you get $k^3$ inside the expression, take limit again you will get the same result as of $(\\lim_{x\\to1}\\frac{x^n(nx-n-1)+1}{x^2-2x+1})^2$. You can also prove it using telescopic series.\n\n-\nwhy does the assumption hold? this is usually proved using induction... \u2013\u00a0 akkkk Jun 27 '12 at 9:47\nwhat assumption?? \u2013\u00a0 Aang Jun 27 '12 at 9:48\nI guess Auke means $\\sum_{0}^{n} x^k=\\frac{1-x^{n+1}}{1-x}$. \u2013\u00a0 sdcvvc Jun 27 '12 at 10:12\nLHS is a geometric series. en.wikipedia.org/wiki/Geometric_progression \u2013\u00a0 Aang Jun 27 '12 at 10:14\n@Auke: one can also prove it like this - let $f(x) = \\sum\\limits_{k=0}^nx^k$, then $f(x) - xf(x)$ is: \\begin{align} 1+&x+x^2+\\dots+x^n \\\\ &- \\\\ &x+x^2+\\dots+x^n+x^{n+1} \\end{align} which is $1-x^{n+1}$. Hence $$(1-x)f(x) = 1-x^{n+1}$$ as needed. \u2013\u00a0 Ilya Jun 27 '12 at 12:40\n\nhttp://en.wikipedia.org/wiki/Faulhaber%27s_formula#Faulhaber_polynomials\n\nIf $p$ is odd, then $1^p+2^p+3^p+\\cdots+n^p$ is a polynomial function of $a=1+2+3+\\cdots+n$. If $p=3$, then then the sum is $a^2$; if $p=5$ then it's $(4a^3-a^2)/3$, and so on.\n\n-\nAnd $a$ is always a factor of $p$. \u2013\u00a0 lhf Sep 3 '11 at 13:26\n\nhttp://blogs.mathworks.com/loren/2010/03/04/nichomachuss-theorem/\n\n-\n\n$f(n)=1^3+2^3+3^3+......+n^3$\n\n$f(n-1)=1^3+2^3+3^3+......+(n-1)^3$\n\n$f(n)-f(n-1)=n^3$\n\nif $f(n)= (1+2+3+4+....+n)^2$ then\n\n$$f(n)-f(n-1)=(1+2+3+4+....+n)^2-(1+2+3+4+....+(n-1))^2$$\n\nusing $a^2-b^2=(a+b)(a-b)$\n\n$f(n)-f(n-1)=$\n\n$=[1+1+2+2+3+3+4+4+....+(n-1)+(n-1)+n][1-1+2-2+3-3+4-4+....+(n-1)-(n-1)+n]=$\n\n$=[2(1+2+3+4+....+(n-1))+n]n=(2\\frac{n(n-1)}{2}+n)n=(n(n-1)+n)n=n^3$\n\n-\nThis is about the same proof as here, the presentation is a bit different though. This is another way to make $k^3$ appear than what was shown here, here and here.", "date": "2014-03-07 18:11:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/61482/proving-the-identity-sum-limits-k-1n-k3-large-sum-limits-k-1n-k/61502", "openwebmath_score": 0.9435271620750427, "openwebmath_perplexity": 538.9416883935687, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924777713886, "lm_q2_score": 0.9005297947939936, "lm_q1q2_score": 0.8753081865487741}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=snp8r7mpcv1l4jqcl5s0bbtev5&topic=1514.msg5313", "text": "### Author Topic: Q6 TUT 0202 \u00a0(Read 2820 times)\n\n#### Victor Ivrii\n\n\u2022 Elder Member\n\u2022 Posts: 2599\n\u2022 Karma: 0\n##### Q6 TUT 0202\n\u00ab on: November 17, 2018, 04:09:19 PM \u00bb\nFind the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.\n$$f(z)=\\frac{1}{e^z-1};\\qquad z_0=0\\quad \\text{(four terms of the Laurent series)} .$$\n\n#### Meng Wu\n\n\u2022 Elder Member\n\u2022 Posts: 91\n\u2022 Karma: 36\n\u2022 MAT3342018F\n##### Re: Q6 TUT 0202\n\u00ab Reply #1 on: November 17, 2018, 04:09:29 PM \u00bb\n$$\\because e^z=\\sum_{n=0}^{\\infty}\\frac{z^n}{n!}=1+z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots$$\n\\begin{align}\\therefore e^z-1&=(1+z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots)-1\\\\&=z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots \\end{align}\nAdded omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)\n$$g(z)=\\frac{1}{f(z)}=\\frac{1}{\\frac{1}{e^z-1}}=e^z-1\\\\g(z_0=0)=e^0-1=0\\\\g'(z)=e^z \\Rightarrow g(z_0=0)=e^0=1\\neq 0 \\\\$$\nThus the order of the pole of $f(z)$ at $z_0=0$ is $1$.\nHence we let $$\\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\\cdots$$\n$$\\therefore(e^z-1)(\\frac{1}{e^z-1})=1\\\\\\therefore (z+\\frac{z^2}{2}+\\frac{z^3}{6}+\\frac{z^4}{24}+\\frac{z^5}{120}+\\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\\cdots)=1$$\n$$\\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\\frac{1}{2}z+\\frac{a_0}{2}z^2+\\frac{a_1}{2}z^3+\\frac{a_{-1}}{6}z^2+\\frac{a_0}{6}z^3+\\frac{a_{-1}}{24}z^3+\\cdots=1$$\n$$\\therefore \\begin{cases}a_{-1}=1\\\\a_0z+\\frac{a_{-1}}{2}z=0 \\Rightarrow a_0+\\frac{a_{-1}}{2}=0 \\Rightarrow a_0=-\\frac{1}{2}\\\\\\frac{a_{-1}}{6}z^2+\\frac{a_0}{2}z^2+a_1z^2=0 \\Rightarrow \\frac{a_{-1}}{6}+\\frac{a_0}{2}+a_1=0 \\Rightarrow a_1=\\frac{1}{12}\\\\ \\frac{a_{-1}}{24}z^3+\\frac{a_0}{6}z^3+\\frac{a_1}{2}z^3+a_2z^3 \\Rightarrow \\frac{a_{-1}}{24}+\\frac{a_0}{6}+\\frac{a_1}{2}+a_2=0 \\Rightarrow a_2=0\\end{cases}$$\nTherefore, the first four terms of the Laurent series:\n$$\\require{cancel} \\cancel{1+\\frac{1}{z}+(-\\frac{1}{2})z^2+(0)z^3} \\\\ \\text{Typo correction: } \\frac{1}{z}-\\frac{1}{2}+\\frac{1}{12}z+(0)z^2$$\n$$\\\\$$\n$$\\\\$$\nThe residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.\n$$\\\\$$\n$$\\\\$$\nIf $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\\frac{a_2}{2}z^4+\\frac{a_1}{6}z^4+\\frac{a_0}{24}z^4+\\frac{a_{-1}}{120}z^4 \\Rightarrow a_3=-\\frac{1}{720}$\n$\\\\$\nHence we have $\\frac{1}{z}-\\frac{1}{2}+\\frac{1}{12}z+(0)z^2-\\frac{1}{720}z^3$.\n\u00ab Last Edit: November 17, 2018, 11:00:48 PM by Meng Wu \u00bb\n\n#### Heng Kan\n\n\u2022 Full Member\n\u2022 Posts: 15\n\u2022 Karma: 18\n##### Re: Q6 TUT 0202\n\u00ab Reply #2 on: November 17, 2018, 05:30:38 PM \u00bb\nI think you got the coefficients correctly but the Laurent series\u00a0 is wrong.\u00a0 Here is my answer. See the attatched scanned picture.\n\n#### Chunjing Zhang\n\n\u2022 Newbie\n\u2022 Posts: 1\n\u2022 Karma: 0\n##### Re: Q6 TUT 0202\n\u00ab Reply #3 on: November 17, 2018, 06:00:55 PM \u00bb\nI think maybe it is needed to first calculate the singularity of the original function, here pole of order 1, and then set the expansion start at power of -1.\n\n#### Victor Ivrii\n\nWe have two solutions based on two different ideas: undetermined coefficients and a clever\u00a0 substitution of the power expansion into power expansion. Why clever? Because it chips out $1$ from $(e^z-1)/z$ rather than from $e^z-1$ (which would be an error)", "date": "2022-06-25 07:10:13", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=snp8r7mpcv1l4jqcl5s0bbtev5&topic=1514.msg5313", "openwebmath_score": 0.985616147518158, "openwebmath_perplexity": 5778.929325759251, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924826392581, "lm_q2_score": 0.90052978812007, "lm_q1q2_score": 0.8753081844454319}}
{"url": "https://math.stackexchange.com/questions/1391835/proving-recursive-formula-via-induction-leads-to-extra-term", "text": "# Proving recursive formula via induction leads to extra term?\n\nI have been asked the following question, and despite spending the last 30 minutes on it, have not come up with a good result:\n\nDefine $$f(1) = 2$$, and $$f(n) = f(n-1) + 2n$$ for all $$n \\geq 2$$. Find a non-recursive formula for $$n$$, and prove by induction this formula works over all natural numbers.\n\nSo, this didn't sound too hard. I found the non-recursive formula to be $$f(n) = n^2 + n$$.\n\nBase case of induction with $$n = 2$$:\n\n$$f(2) = 2^2 + 2 = 6$$\n\nAssumption step:\n\n$$f(k) = k^2 + k$$\n\nExtension step:\n\n\\begin{aligned} f(k+1) & = (k+1)^2 + (k+1) \\\\ & = (k^2 + 2k + 1) + (k+1) \\\\ & = f(k) + 2k + 2 \\ (?) && (\\text{Substitute} \\ f(k)) \\end{aligned}\n\nNotice it all falls apart here? That extra \"$$2$$\" terms means any experimental result I plug in yields a value $$2$$ higher than it should.\n\nWhat have I done wrong?\n\n\u2022 You're almost done. By the recursive formula, what does $f(k+1)$ equal? Edit: It doesn't fall apart at all. Aug 10 '15 at 10:52\n\u2022 The recursive formula would be f(k+1) = f(k) + 2(k+1)... ohh I see. I'm an idiot. Why did I miss that? Aug 10 '15 at 10:56\n\u2022 @EchoLogic, happens to all of us sometimes. Aug 13 '15 at 18:30\n\u2022 Please add your own answer to this question, completing the details of the above comments, and later accept it to close this question. Sep 3 '15 at 17:01\n\nWe have, the recursion formula, \\begin{aligned} f(1) = 2, f(n) = f(n-1) + 2n && (1)\\end{aligned}\n\nand, the candidate, \\begin{aligned}P(n): f(n) = n^2 + n && (2)\\end{aligned}\n\n### Step-1: (Base Case)\n\nBy $$(1)$$, we have:\n\n$$f(1) = 2$$, and $$f(2) = f(1) + 2 \\cdot 2 = 6$$\n\nThese are satisfied by $$(2)$$ also, via direct calculation. Hence, the base case is established.\n\n### Step-2: (Induction Hypothesis/ Extension step)\n\nAssume that the statement holds for $$n=k$$,\n\ni.e. $$P(k): f(k) = k^2 + k$$, holds.\n\n### Step-3: (Inductive step)\n\nWe now, prove $$P(k+1)$$, which is the subject of the question,\n\nBy $$(1)$$,\n\n\\begin{aligned} f(k+1) & = f(k) + 2 \\cot (k+1) && (\\text{Taking,} \\ \\ n= (k+1)) \\\\ & = (k^2 + k) + (2k + 2) && (\\text{By the Induction Hypothesis}) \\\\ & = (k^2 + 2k + 1) + (k+1) \\\\ \\implies f(k+1) & = (k+1)^2 + (k+1) \\end{aligned}\n\nThus, $$P(k+1)$$ is established, contingent upon $$P(k)$$ holding.\n\nBy the Principle of Mathematical Induction, the formula $$(2)$$ holds.\n\n## Notes:\n\nA. The working presented in the OP has two flaws:\n\n1. The failure to recognize that the expression in the last step has the formula was the desired result, minus taking a common factor. This had been pointed out by @Git Gud and recognized by the OP, in the comments.\n\n2. The more important issue is that the presented working confounds the statement that is given to be true, eq. $$(1)$$, and the one that is to be proved, eq. $$(2)$$. Specifically, the induction hypothesis assumes $$P(k)$$, so that the next step should be to establish $$P(k+1)$$. However, the OP assumes $$P(k+1)$$ in accordance with $$(2)$$, and tries to show how this implies $$(1)$$. Now, this works for the present case as all the operations involved are bidirectional $$(\\iff)$$. Thus,we do indeed show that $$(1)$$ implies $$(2)$$, establishing $$P(k+1)$$. However, this is poor form that disrupts the flow of the logic and can lead to trouble if one is dealing with problems regarding say, inequalities. Although working backwards is a legitimate strategy that can be very useful, it requires care to ensure that the reversed steps are valid and useful, and is, at any rate not required here.\n\nB. The notation $$P(k)$$, indicates the statement, $$P(n)$$, that is, ($$f(n) =n^2+n$$), for the case $$n=k$$. $$P(n)$$ is the statement that in the general case that there is an arbitrary $$n \\in \\mathbb{N}$$, we have $$f(n) = n^2 + n$$. This is a standard notation (albeit not one mentioned in OP's working in the question) in problems where mathematical induction is used, among others.\n\nC. While mathematical induction can be useful for proving true statements over $$\\mathbb{N}$$ (and many other cases, including well-founded sets in general), it is not nearly as useful in detecting the invalidity of false statements, and does not the address the important problem of generating plausible statements to check. Indeed, generating the formula is the first part of the problem the OP set out to solve. In the case, the formula was (almost certainly by design), trivial to guess. However, the generation of such statements. and particularly finding explicit formulae for recurrence relations; can be a very challenging problem to solve, and offers an interesting counterpoint to the problem of solving differential equations over $$\\mathbb{R}$$ or $$\\mathbb{C}$$. While a full discussion would be wildly out of scope here, it is interesting to note that here: $$\\frac{f(n)-f(n-1)}{n-(n-1)} = 2n$$ which is reminiscent of the derivative over $$\\mathbb{R}$$ and suggesting the appropriate solution should be a quadratic polynomial, as is indeed the case. It is left to the reader to verify that modifying the initial condition leads to solutions of the form $$n^2 + n + f(1) - 2$$.", "date": "2021-09-25 16:26:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1391835/proving-recursive-formula-via-induction-leads-to-extra-term", "openwebmath_score": 0.9654126167297363, "openwebmath_perplexity": 465.63587272739875, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717452580315, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8752909800109073}}
{"url": "https://brilliant.org/discussions/thread/a-nice-way-to-factorize-fractions/", "text": "# A nice way to factorize FRACTIONS!\n\nI recently learned that ...\n\n$\\frac{1}{n(n+d)}$ = $\\frac{1}{d}(\\frac{1}{n}-\\frac{1}{n+d})$\n\nFor example,\n\n$\\frac{1}{3*5}=\\frac{1}{3(3+2)}=\\frac{1}{2}(\\frac{1}{3}-\\frac{1}{5})$\n\nWhich is useful because we can use this equation in situation like this...\n\nFind the value of $\\frac{1}{3*5}+\\frac{1}{5*7}+\\frac{1}{7*9}+\\frac{1}{9*11}$\n\n$=\\frac{1}{2}(\\frac{1}{3}-\\frac{1}{5})+\\frac{1}{2}(\\frac{1}{5}-\\frac{1}{7})+\\frac{1}{2}(\\frac{1}{7}-\\frac{1}{9})+\\frac{1}{2}(\\frac{1}{9}-\\frac{1}{11})$\n\nTake the common factor out\n\n$\\frac{1}{2}[(\\frac{1}{3}-\\frac{1}{5})+(\\frac{1}{5}-\\frac{1}{7})+(\\frac{1}{7}-\\frac{1}{9})+(\\frac{1}{9}-\\frac{1}{11})]$\n\nThe terms cancels each other and we are left with the first and last terms.\n\n$\\frac{1}{2}[\\frac{1}{3}-\\frac{1}{11}]$\n\nwhich is $\\frac{4}{33}$ This way is MUCH faster than the traditional way\n\nAnyone can prove why the equation i learned is true?\n\nNote by Peter Bishop\n5\u00a0years, 3\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $ ... $ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\n$\\dfrac{1}{n(n+d)}$\n\n$= \\dfrac{1}{d} \\times \\dfrac{d}{n(n+d)}$\n\n$= \\dfrac{1}{d} \\times \\dfrac{(n+d) - (n)}{n(n+d)}$\n\n$= \\dfrac{1}{d} \\times \\left(\\dfrac{n+d}{n(n+d)} - \\dfrac{n}{n(n+d)} \\right)$\n\n$= \\dfrac{1}{d} \\times \\left( \\dfrac{1}{n} - \\dfrac{1}{n+d} \\right)$\n\n- 5\u00a0years, 3\u00a0months ago\n\nBy using Partial Fractions,let - $\\frac{1}{n(n+d)}$ = $\\frac{A}{n} + \\frac{B}{n+d}$\nThus, we need to find the coefficients A and B. Therefore,by taking LCM,we get -\n$\\frac{1}{n(n+d)}$ = $\\frac{A(n+d) + B \\cdot n}{n(n+d)}$. Or, $1= n(A + B) + A \\cdot d$\nNow, comparing the terms for different exponents of 'n' on both sides, we get-\n\n{ This can be explained by example given below -\n\nIf ax + b = 3x +2 , then , a = 3 and b = 2 }\n\n(Comparing the terms which contains $n^0$on both sides),\n\n$n^0 : A \\cdot d =1, or , A = \\frac{1}{d}$\n\n(Comparing the terms which contains $n^1$ on both sides),\n\n$n^1 : A + B =0 , or, B = -A ,$\n\nOr, $B = \\frac{-1}{d}$\n\nTherefore, $\\frac{1}{n(n+d)}$ = $\\frac{\\frac{1}{d}}{n} + \\frac{\\frac{-1}{d}}{n+d}$\n\n$\\frac{1}{n(n+d)}$ = $\\frac{1}{d} ( \\frac{1}{n} - \\frac{1}{n+d})$\n\n- 5\u00a0years, 3\u00a0months ago\n\nAwesome!!!!!!!!!!!!!!! Thank you very much.\n\n- 5\u00a0years, 3\u00a0months ago\n\nBut i still don't understand the exponents of n part. Can you please explain futher.\n\n- 5\u00a0years, 3\u00a0months ago\n\nHe is using a property called \"comparing coefficients\" (or at least that is what I call it)\n\nIt states the following: For any two polynomials $P(x)=a_nx^n+a_{n-1}x^{n-1}+\\cdots +a_1x+a_0$ and $Q(x)=b_nx^n+b_{n-1}x^{n-1}+\\cdots +b_1x+b_0$ such that $P(x)=Q(x)$, then $a_i=b_i$ for all $i=0\\to n$.\n\n- 5\u00a0years, 3\u00a0months ago\n\nIt's pretty easy.Multiply the RHS by $n(n+d)$ and you will get $\\frac {1}{d}.[(n+d)-n]=1$ And that's it.\n\n- 5\u00a0years, 3\u00a0months ago\n\nwow...good formula for rememberd\n\n- 5\u00a0years, 3\u00a0months ago\n\nWoah, cool bro!\n\n- 5\u00a0years, 3\u00a0months ago\n\nGuys, if you like what you learnt, please reshare it so others can see. Thanks for the comments\n\n- 5\u00a0years, 3\u00a0months ago\n\nAwesome method! From where did you get this??\n\n- 5\u00a0years, 3\u00a0months ago\n\nFrom a tutor\n\n- 5\u00a0years, 3\u00a0months ago\n\nwhen we can write the pattern or series in the form that difference of two numbers and such that it get cancelled it makes the work easier this is called v n method or elemination method\n\n- 5\u00a0years, 3\u00a0months ago\n\nI just learnt A+B=0 so, A=-B wa0w. R I Newton yet?\n\n- 5\u00a0years, 3\u00a0months ago\n\n1/n(n+d) =1/d x d/n(n=d) =1/d x (n+d)-(n)/n(n+d) =1/d x {(n+d)/n(n+d-(n)/n(n+d)} =1/d x (1/n-1-n+d) HENCE PROVED\n\n- 5\u00a0years, 3\u00a0months ago\n\nThis method is use to solve many problems and is called telescoping method I think.\n\n- 2\u00a0years, 10\u00a0months ago", "date": "2019-06-26 15:13:21", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/a-nice-way-to-factorize-fractions/", "openwebmath_score": 0.9962895512580872, "openwebmath_perplexity": 3521.7046411768633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.98657174604767, "lm_q2_score": 0.8872045817875224, "lm_q1q2_score": 0.8752909733556088}}
{"url": "https://math.stackexchange.com/questions/478566/is-it-possible-to-put-or-signs-in-such-a-way-that-pm-1-pm-2-pm-cdot", "text": "# Is it possible to put $+$ or $-$ signs in such a way that $\\pm 1 \\pm 2 \\pm \\cdots \\pm 100 = 101$?\n\nI'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.\n\nProblem: Is it possible to put $+$ or $-$ signs in such a way that $\\pm 1 \\pm 2 \\pm \\cdots \\pm 100 = 101$?\n\nMy proof is kinda simple. Let's work in mod $2$. We'll have:\n\n$\\pm 1 \\pm 2 \\pm \\cdots \\pm 100 \\equiv 101 \\mod 2$ but since $+1 \\equiv -1 \\mod 2$ and there are exactly $50$ odd numbers and $50$ even numbers from $1$ to $100$ we can write:\n\n$(1 + 0 + \\cdots + 1 + 0 \\equiv 50\\times 1 \\equiv 0) \\not\\equiv (101\\equiv 1) \\mod 2$ which is contradictory.\n\nTherefore, it's not possible to choose $+$ or $-$ signs in any way to make them equal.\n\nNow is there a combinatorial proof of that fact except what I have in mind?\n\n\u2022 A standard problem often classified as combinatorics is that the number of people who shook hands with an odd number of people is even. This one uses much the same idea. \u2013\u00a0Andr\u00e9 Nicolas Aug 28 '13 at 22:41\n\u2022 @Andr\u00e9Nicolas: If you already have the proof in your mind, would you please post a proof of my problem using the handshaking lemma? \u2013\u00a0user66733 Aug 28 '13 at 22:49\n\u2022 Frankly, I like your proof better than any of the answers. \u2013\u00a0asmeurer Aug 29 '13 at 4:48\n\u2022 Another way of phrasing your own solution: We consider the sum of 100 numbers, 50 odd ones and 50 even ones. That will be an even sum. Negating a number (changing sign from $-$ to $+$ or from $+$ to $-$) does not change its parity (its being odd or even). \u2013\u00a0Jeppe Stig Nielsen Aug 29 '13 at 9:43\n\u2022 @Andr\u00e9Nicolas: I haven't chosen the best answer yet because I'm still waiting for your answer based on the handshaking Lemma. If you have such a proof in mind please post it, because I'm really interested in it and it sounds like the most combinatorial solution to me. \u2013\u00a0user66733 Aug 31 '13 at 2:55\n\nYou can rephrase essentially the same argument in the following terms:\n\nSuppose that there were such a pattern of plus and minus signs. Let $P$ be the set of positive terms, and let $N$ be the set of negative terms together with the number $101$. Then $\\sum P-\\sum N=0$, so $\\sum P=\\sum N$, and $\\{P,N\\}$ is a partition of $\\{1,2,\\ldots,101\\}$ into two sets with equal sum. But $\\sum_{k=1}^{101}k=\\frac12\\cdot101\\cdot102=101\\cdot51$ is odd, so this is impossible.\n\nAnother answer that use almost the same idea: the sum or subtraction of two even or odd number is an even number. How many odd number we have?\n\nReplacing 100 with $n$ and using Brian M. Scott's solution, we want a partition of $\\{1, 2, ..., n+1\\}$ into two sets with equal sums.\n\nThe sum is $\\frac{(n+1)(n+2)}{2}$, and if $n=4k$, this is $(4k+1)(2k+1)$ which is odd and therefore impossible.\n\nIf $n = 4k+1$, this is $(2k+1)(4k+3)$ which is also odd, and therefore impossible.\n\nIf $n = 4k+2$, this is $(4k+3)(2k+2)$, so it is not ruled out, and each sum must be $(4k+3)(k+1)$.\n\nif $n = 4k+3$, this is $(2k+2)(4k+5)$ which is also not ruled out, and each sum must be $(k+1)(4k+5)$.\n\nNow I'll try to find a solution for the not impossible cases. (I am working these out as I enter them.)\n\nFor the $n=4k+2$ case, the sum must be $(4k+3)(k+1) =(4k+4-1)(k+1) =4(k+1)^2-(k+1) =(2k+2)^2-(k+1)$. The square there suggests, to me, the formula for the sum of consecutive odd numbers $1+3+...+(2m-1)=m^2$, so $1+3+...+(4k+3) = (2k+2)^2$. If $k+1$ is odd, remove it from the sum so it is $(2k+2)^2-(k+1)$. If $k+1$ is even, both $1$ and $k$ are odd, so remove them from the sum. In either case, we have the desired partition.\n\nFor the $n=4k+3$ case, the sum must be $(4k+5)(k+1) =(4k+4+1)(k+1) =4(k+1)^2+(k+1) =(2k+2)^2+(k+1)$. Again, $1+3+...+(4k+3) = (2k+2)^2$. If $k+1$ is even, add it to the sum so it is $(2k+2)^2+(k+1)$. If $k+1$ is odd, $k+2$ is even, so remove $1$ and add $k+2$ to the sum. In either case, we have the desired partition.\n\nI do not know if these partitions are unique.\n\nIf $T_n=n(n+1)/2$ is the $n^{th}$ triangular number, an inductive proof (using $T_n+(n+1)=T_{n+1}$) shows the attainable numbers at step $n$ are $$-T_n,\\ -T_n+2,\\ \\cdots , T_n-2, \\ T_n,$$ in particular they all have the same parity as $T_n$. Since $T_{100}=5050$ is even, we see that $101$ cannot be attained in any way by $100$ steps.\n\naddendum: The first triangular number at least $101$ is $T_{14}=105$ (and is odd). This overshoots the goal $101$ by $4$, so if we take the sum $$1+2+3+\\cdots+14=105$$ and change the sign on the $2$, we get $101$. Seems this is the only way to get $101$ in 14 steps, and we cannot get it with 13 or fewer since $T_{13}=91$ is the largest with $13$ steps.\n\nConsider both sides modulo $2$. Then the right side is $1$, whereas the left side is $0$ (since it consists of $50$ ones and $50$ zeros).", "date": "2019-05-21 17:36:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/478566/is-it-possible-to-put-or-signs-in-such-a-way-that-pm-1-pm-2-pm-cdot", "openwebmath_score": 0.9110614061355591, "openwebmath_perplexity": 125.11755393880102, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534376578002, "lm_q2_score": 0.8918110382493034, "lm_q1q2_score": 0.8752710092309508}}
{"url": "https://math.stackexchange.com/questions/2649261/why-does-probability-of-50-50-outcome-decrease-as-we-toss-a-coin-more-and-more", "text": "# Why does probability of 50/50 outcome decrease as we toss a coin more and more?\n\nI have following formula (\"n\" is number of experiments, \"k\" is number of successes that we expect, \"p\" is probability of success, \"q\" is probability of failure):\n\nI used it to calculate chances of getting strictly 2 heads out of 4 tosses (order doesn't matter), 5 heads of 10 tosses, 10 heads out of 20 tosses.\n\n2 heads out of 4 tosses have around 38% probability.\n\n5 heads out of 10 tosses have around 25% probability.\n\n10 heads out of 20 tosses have around 18% probability.\n\nLooks like trend for me. And this trend is paradoxical, it means that by increasing number of tries we will decrease chance for 50/50 outcome. It seems to be at odds with what I was taught before. I was taught that as we increase number of tosses we will come closer and closer to 50/50 outcome. So if I toss a coin 10 times I can get 1 head and 9 tails, while tossing a coin 1000 times is more likely to give me result much closer to 50/50 outcome.\n\n\u2022 What p and what q are you using? \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez Feb 13 '18 at 18:20\n\u2022 This is because throwing exactly 50% is not likely to happen most of the time, you can sometimes land around the 11/20 mark, or 9/20. These are still very close to 10/20, and on average you are still getting an even distribution. As you increase $n$, there are more numbers close to the 50% mark for your eventual count to finish on, so this probability is spread out between these values. \u2013\u00a0John Doe Feb 13 '18 at 18:21\n\u2022 Notice that that formula gives you the probability of getting exactly $k$ successes. As $n$ increases, it is harder and harder to have exactly $n/2$ tails. \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez Feb 13 '18 at 18:21\n\u2022 It's true that the probability of getting exactly 50-50 goes down. But the average deviation of from $.5$ gets smaller and smaller. When $n=10$ it's relatively common to see $.6$ or $.4$ (corresponding to $6/10$ or $4/10$ heads.) For $n=10000000$ it is pretty much impossible for $60\\%$ or more of the flips to come up heads. \u2013\u00a0spaceisdarkgreen Feb 13 '18 at 18:33\n\u2022 Consider also that for an odd number of coin flips, the probability of exactly 50-50 is always zero \u2013\u00a0spaceisdarkgreen Feb 13 '18 at 18:36\n\nThis is because the more you flip a coin, the less likely it will be that you get heads $exactly$ half of the time. It is true, however, that the more you flip a coin, and then compute\n\n$$\\hat{p}=\\frac{\\text{number of heads}}{\\text{number of tosses}}$$\n\nthat $\\hat{p}$ will $converge$ to $0.5$. Consider flipping a coin $10$ times and obtaining $6$ heads. Then consider flipping a coin $1000$ times and obtaining $526$ heads.\n\nWhile in the second instance, we are $26$ heads away from a perfect $50/50$, proportionally, we are closer as $0.526$ is closer to $0.5$ than $0.6$.\n\n\u2022 Please, add information about spreading out probability to your answer because it's incomplete without it. \" As you increase n, there are more numbers close to the 50% mark for your eventual count to finish on, so this probability is spread out between these values.\" \u2013\u00a0user161005 Feb 13 '18 at 18:41\n\u2022 @user161005 are you talking about my comment on the question? This is not my answer.. \u2013\u00a0John Doe Feb 13 '18 at 18:42\n\u2022 Sorry, confused you with other guy. Silly me. \u2013\u00a0user161005 Feb 13 '18 at 18:52\n\nAs it's been said, the fact that there are more and more possible values is what makes $\\hat p=0.5$ less and less probable as exact value ($\\hat p =\\tfrac{\\#heads}{\\#toses}$).\n\nConsider, for instance, the probability of $\\tfrac 1 2 - \\tfrac 1 {10}<\\hat p_n<\\tfrac 1 2 + \\tfrac 1 {10}$, that is $$P\\left(\\tfrac 1 2 - \\tfrac 1 {10}<\\hat p_n<\\tfrac 1 2 + \\tfrac 1 {10}\\right)=P\\left(- \\tfrac {2\\sqrt n} {10}<\\frac{\\hat p_n-\\tfrac 1 2}{\\frac1{2\\sqrt n}} < \\tfrac {2\\sqrt n} {10}\\right)\\approx 2\\cdot\\Phi\\left(\\tfrac {\\sqrt n} {5}\\right)-1$$ the approximation based on the CLT, valid for $n$ large enough.\n\nLet's supose $n=50$. In that case we have $$P\\left(\\tfrac 1 2 - \\tfrac 1 {10}<\\hat p_{50}<\\tfrac 1 2 + \\tfrac 1 {10}\\right)\\approx 2\\cdot \\Phi\\left(\\frac{\\sqrt {50}}5\\right)-1\\approx 0.8427.$$\n\nNow let's do the same fot $n=100$. We get $$P\\left(\\tfrac 1 2 - \\tfrac 1 {10}<\\hat p_{100}<\\tfrac 1 2 + \\tfrac 1 {10}\\right)\\approx 2\\cdot \\Phi\\left(\\frac{\\sqrt {100}}5\\right)-1\\approx 0.9545.$$ So the probability of $\\hat p_n$ taking a value at a distance less than $0.1$ from the expected value has increased by around a 13%. But the values of $X=\\#heads$ corresponding to that event went from $9$ (that is $21$ to $29$) to $19$ ($41$ to $59$), around the double. And if we kept doubling $n$ (for example), we would approximately double the number of integer values of $X$ that lead to $0.4<\\hat p_n<0.6$ each time, but the total probability distributed between those values won't grow that much, fundamentally because it's very close to $1$, which is a bound, no matter $n$.\n\nSo, those probabilities which stay almost constant at some point, are distributed among more and more possible values of the measured discrete variable, and so is not surprising that the probability at each individual point decreases.", "date": "2019-06-17 22:53:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2649261/why-does-probability-of-50-50-outcome-decrease-as-we-toss-a-coin-more-and-more", "openwebmath_score": 0.8854819536209106, "openwebmath_perplexity": 346.0749264444496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810951192016, "lm_q2_score": 0.8887588008585925, "lm_q1q2_score": 0.875259400053826}}
{"url": "https://math.stackexchange.com/questions/2209438/limit-of-frac1x4-int-sin-xx-arctan-t-dt", "text": "# Limit of $\\frac{1}{x^4}\\int_{\\sin x}^{x} \\arctan t dt$\n\nI am trying to find this limit,\n\n$$\\lim_{x \\rightarrow 0} \\frac{1}{x^4} \\int_{\\sin{x}}^{x} \\arctan{t}dt$$\n\nUsing the fundamental theorem of calculus, part 1, $\\arctan$ is a continuous function, so $$F(x):=\\int_0^x \\arctan{t}dt$$ and I can change the limit to $$\\lim_{x \\rightarrow 0} \\frac{F(x)-F(\\sin x)}{x^4}$$\n\nI keep getting $+\\infty$, but when I actually integrate $\\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\\infty$ as $x \\rightarrow 0+$.\n\nI tried using l'Hospital's rule, but the calculation gets tedious.\n\nCan anyone give me hints?\n\nEDIT\n\nI kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!\n\n\u2022 Numerical evaluation says the limit should be $1/6$. \u2013\u00a0Simply Beautiful Art Mar 29 '17 at 23:34\n\u2022 @SimplyBeautifulArt How did you do it? What kind of method did you use for numerical evaluation? \u2013\u00a0zxcvber Mar 29 '17 at 23:36\n\u2022 Calculators of course. \u2013\u00a0Simply Beautiful Art Mar 29 '17 at 23:37\n\nPower series can help, if you know them.\n\nWe know that $$\\arctan t=\\sum_{n=0}^{\\infty}(-1)^n\\frac{t^{2n+1}}{2n+1},\\qquad \\lvert t\\rvert<1.$$ Therefore an antiderivative for $\\arctan t$ is $$F(t):=\\sum_{n=0}^{\\infty}(-1)^n\\frac{t^{2n+2}}{(2n+1)(2n+2)},\\qquad \\lvert t\\rvert<1.$$ So, as $x\\to0$, $$F(x)=\\frac{x^2}{2}-\\frac{x^4}{12}+O(x^6).$$ Now, recalling that $$\\sin x=\\sum_{n=0}^{\\infty}(-1)^n\\frac{x^{2n+1}}{(2n+1)!}=x-\\frac{x^3}{6}+O(x^5)=x+O(x^3),$$ we see that as $x\\to0$ (and therefore $\\sin x\\to0$), \\begin{align*} F(\\sin x)&=\\frac{1}{2}\\left(x-\\frac{x^3}{6}+O(x^5)\\right)^2-\\frac{1}{12}\\left(x+O(x^3)\\right)^4+O[(x+O(x^3))^6]\\\\ &=\\frac{1}{2}\\left[x^2-\\frac{x^4}{3}+O(x^6)\\right]-\\frac{1}{12}\\left[x^4+O(x^6)\\right]+O(x^6)\\\\ &=\\frac{x^2}{2}-\\frac{x^4}{4}+O(x^6) \\end{align*} So, all told, our limit is $$\\lim_{x\\to0}\\frac{F(x)-F(\\sin x)}{x^4}=\\lim_{x\\to0}\\frac{\\,\\frac{x^4}{6}\\,}{x^4}=\\frac{1}{6}$$\n\n\u2022 I really like the way you used $O(x^n)$ notation here. Thanks! \u2013\u00a0zxcvber Mar 29 '17 at 23:56\n\nHere's one way to do this. The Taylor expansion for $\\arctan(x) = x - x^3/3 + x^5/5 + \\cdots$. So then $$\\int_{\\sin(x)}^x \\arctan(t)dt \\sim (x^2/2 - x^4/12 + x^6/30) - \\frac{(\\sin x)^2}{2} + \\frac{(\\sin x)^4}{12} - \\frac{(\\sin x)^6}{30}.$$\n\nUsing another Taylor expansion, $\\sin x = x - x^3/3! + x^5/5! + \\cdots$, so the plan is to pay attention only to those terms of degree up to $4$ in these expansions. Then (dropping all terms of degree greater than $4$), $$(\\sin x)^2/2 = (x - x^3/3!)^2/2 = x^2/2 - x^4/6$$ and $$(\\sin x)^4/12 = (x - x^3/3!)^4/4 = x^4/12.$$ Thus $$\\int_{\\sin(x)}^x \\arctan(t)dt \\sim (x^2/2 - x^4/12) - (x^2/2 - x^4/6) + x^4/12 = x^4/6.$$\n\nSo multiplying by $x^{-4}$ and taking the limit gives $1/6$.\n\nAs an aside, l'Hopital's rule would also work.\n\nSince $F(0) = 0$ and everything is smooth, you can apply de l'Hopital and get\n\n$$\\lim_{x \\to 0}\\frac{F(x)-F(\\sin x)}{x^4} = \\lim_{x \\to 0}\\frac{\\arctan x - \\cos x\\arctan \\sin x}{4x^3}.$$\n\nThis last limit can be evaluated using Taylor series:\n\n$$\\arctan x = x-\\frac{x^3}{3}+O(x^5)$$ and\n\n$$\\cos x \\arctan \\sin x = x-x^3+O(x^5)$$\n\nand the limit you are looking for is equal to\n\n$$\\lim_{x \\to 0}\\frac{x-\\dfrac{x^3}{3}-x+x^3 + O(x^5)}{4x^3} = \\frac{1}{6}.$$\n\nThe ''ugly'' Taylor expansion is obtained combining the Taylor expansions of $\\sin$, $\\cos$ and $\\arctan$. Easier done than said.\n\n\u2022 Actually, you just take the derivatives directly instead of trying to combine well-known expansions since that's ugly and we only need a few terms. \u2013\u00a0Simply Beautiful Art Mar 29 '17 at 23:49\n\u2022 @SimplyBeautifulArt That's true, but I expect the third derivative of $\\cos x \\arctan \\sin x$ to be even uglier.. \u2013\u00a0Stefano Mar 29 '17 at 23:51\n\u2022 Well, the key is that its easy to ask your calculator to take the third derivative ;) Combining Taylor expansions? Not so much. \u2013\u00a0Simply Beautiful Art Mar 29 '17 at 23:53\n\nNote that by the Mean Value Theorem for integrals we have $$\\int_{\\sin x}^{x}\\arctan t\\,dt = (x - \\sin x)\\arctan c$$ for some $c$ between $x$ and $\\sin x$. Then we have $$\\lim_{x \\to 0}\\frac{1}{x^{4}}\\int_{\\sin x}^{x}\\arctan t\\,dt = \\lim_{x \\to 0}\\frac{x - \\sin x}{x^{3}}\\cdot\\frac{\\arctan c}{x}$$ The first factor on the right tends to $1/6$ (via L'Hospital's Rule or Taylor series) and we show that next factor tends to $1$. For this we assume that $x \\to 0^{+}$ so that $\\sin x < c < x$ and therefore $$\\arctan \\sin x < \\arctan c < \\arctan x$$ and dividing by $x$ we get $$\\frac{\\arctan \\sin x}{\\sin x}\\cdot\\frac{\\sin x}{x} < \\frac{\\arctan c}{x} < \\frac{\\arctan x}{x}$$ By Squeeze theorem we see that $(\\arctan c)/x \\to 1$ as $x \\to 0^{+}$. A similar argument can be given for $x \\to 0^{-}$ and we have the desired limit as $1/6$.", "date": "2019-09-15 13:59:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2209438/limit-of-frac1x4-int-sin-xx-arctan-t-dt", "openwebmath_score": 0.8976989984512329, "openwebmath_perplexity": 221.03908504756564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810951192016, "lm_q2_score": 0.8887587853897074, "lm_q1q2_score": 0.8752593848198986}}
{"url": "https://mathhelpboards.com/threads/advanced-integration-problem.9129/", "text": "Welcome to our community\n\nyouth4ever\n\nNew member\nHello my Math friends,\n\nI have a complex integration problem:\nThe integral to calculate is :\n\nIntegrate[Cos [mx]/(x^2 + a^2), {x, -Infinity, Infinity}]\nor if particularized for a=1, m=1 the integral will be :\n\nIntegrate[Cos [x]/(x^2 + 1), {x, -Infinity, Infinity}]\n\nI know the answer for both of them:\n(Pi/a)*(e^-ma) for the first\nand\nPi/e for the second.\n\nHow this integral can be done through Complex integration ? I don't have a clue and I strongly wish to solve it.\nThanks.\n\nPS: BTW, I would have written with math signs because I don't like the text style of writing math but I don't know how to do it here. Is there a trick ?\n\nOpalg\n\nMHB Oldtimer\nStaff member\nHello my Math friends,\n\nI have a complex integration problem:\nThe integral to calculate is :\n\nIntegrate[Cos [mx]/(x^2 + a^2), {x, -Infinity, Infinity}]\nor if particularized for a=1, m=1 the integral will be :\n\nIntegrate[Cos [x]/(x^2 + 1), {x, -Infinity, Infinity}]\n\nI know the answer for both of them:\n(Pi/a)*(e^-ma) for the first\nand\nPi/e for the second.\n\nHow this integral can be done through Complex integration ? I don't have a clue and I strongly wish to solve it.\nThanks.\nHi youth4ever and welcome to MHB!\n\nYour integral is the real part of $$\\displaystyle \\int_{-\\infty}^\\infty \\frac{e^{imx}}{x^2+a^2}dx.$$ If you know about contour integration then you should be able to evaluate that integral by using a D-shaped contour.\n\nPS: BTW, I would have written with math signs because I don't like the text style of writing math but I don't know how to do it here. Is there a trick ?\nSee the LaTeX tips and tutorials forum.\n\nyouth4ever\n\nNew member\nHi Opalg,\n\nYes I know that it comes from e^imx/... integral. But I wondered if there is a way to simply integrate the cosine part separately by using a different technique and to double check the answer.\nIs this possible ?\n\nThanks.\n\nThePerfectHacker\n\nWell-known member\nBut I wondered if there is a way to simply integrate the cosine part separately by using a different technique and to double check the answer.\nIs this possible ?\nI think your question is. Rather than making things \"complicated\", why not just find anti-derivative of the function inside the integral and get the answer. But the problem with what you want is that a lot of these integration problems that use complex analysis to compute do not have anti-derivatives in terms of known standard functions. So what you want cannot be done.\n\nRandom Variable\n\nWell-known member\nMHB Math Helper\nIf your interested in a different approach that would confirm the answer you get using contour integration, consider the Laplace transform of $$\\displaystyle f(t) = \\int_{0}^{\\infty} \\frac{\\cos (mtx)}{a^{2}+x^{2}} \\ dx$$\n\nThen by definition,\n\n$$\\mathcal{L}_{t} [f(t)](s) = \\int_{0}^{\\infty} \\int_{0}^{\\infty} \\frac{\\cos (mtx)}{a^{2}+x^{2}} \\ e^{-st} \\ dx \\ dt = \\int_{0}^{\\infty} \\frac{1}{a^{2}+x^{2}}\\int_{0}^{\\infty} \\ \\cos(mxt) e^{-st} dt \\ dx$$\n\n$$= \\int_{0}^{\\infty} \\frac{1}{a^{2}+x^{2}} \\frac{s}{s^{2}+(mx)^{2}} \\ dx = \\frac{s}{s^{2}-a^{2}m^{2}} \\int_{0}^{\\infty} \\left( \\frac{1}{a^{2}+x^{2}} - \\frac{m^{2}}{s^{2}+m^{2}x^{2}} \\right)$$\n\n$$= \\frac{s}{s^{2}-a^{2}m^{2}} \\left( \\frac{\\pi}{2a} - \\frac{\\pi m }{2s} \\right) = \\frac{\\pi}{2a(s+am)}$$\n\nNow undo the transform.\n\nEDIT: What I had previously after this point wasn't quite correct.\n\n$$f(t) = \\mathcal{L}^{-1}_{t} \\Big[ \\mathcal{L}_{t} [f(t)](s) \\Big] = \\mathcal{L}^{-1}_{t} \\Big[ \\frac{\\pi}{2a(s+am)} \\Big]$$\n\n$$= \\frac{\\pi}{2a} \\mathcal{L}^{-1}_{t} \\Big[ \\frac{1}{s+am} \\Big] = \\frac{\\pi}{2a} e^{-amt}$$\n\nAnd therefore,\n\n$$\\int_{-\\infty}^{\\infty} \\frac{\\cos (mx)}{a^{2}+x^{2}} \\ dx = 2 f(1) = \\frac{\\pi}{a} e^{-am}$$\n\nLast edited:\n\nyouth4ever\n\nNew member\nHello \"Random Variable\"\n\nAnd many thanks for the solution. Your solution is incredible. I am amazed.\nI have a question although.\nHow you knew at this point\n$$= \\int_{0}^{\\infty} \\frac{1}{a^{2}+x^{2}} \\frac{s}{s^{2}+(mx)^{2}} \\ dx$$\nto amplify the product with\n$$\\frac{s^{2}-a^{2}m^{2}}{s^{2}-a^{2}m^{2}}$$\nand then add and substract $$(mx)^{2}$$ to the numerator\nto obtain the sum :\n$$= \\frac{s}{s^{2}-a^{2}m^{2}} \\left( \\frac{1}{a^{2}+x^{2}} - \\frac{\\ m^{2} }{s^{2}+(mx)^{2}} \\right)$$\nwhich is easily integrable in x variable? I just need a pattern on how to look because it is not obvious and I want to be able to apply this technique in similar situations.\n\nThanks.\n\nLast edited:\n\nRandom Variable\n\nWell-known member\nMHB Math Helper\n@youth4ever\n\nI used partial fractions.\n\nAssume that\n\n$$\\frac{1}{(a^{2}+x^{2})(s^{2}+m^{2}x^{2})} = \\frac{Ax+B}{a^{2}+x^{2}} + \\frac{Cx+D}{s^{2}+m^{2}x^{2}}$$\n\n$$= \\frac{As^{2}x + Am^{2}x^{3} + Bs^{2} + Bm^{2}x^{2} + Ca^{2}x + Cx^{3} + Da^{2} + Dx^{2}}{(a^{2}+x^{2})(s^{2}+m^{2}x^{2})}$$\n\nIf those fractions are equal, their numerators must be the same.\n\nThat leads to four equations that must be satisfied.\n\n$$Am^{2} + C = 0$$\n$$Bm^{2} + D = 0$$\n$$As^{2}+Ca= 0$$\n$$Bs^{2}+Da^{2} =1$$\n\nThe first and third equations imply that $A$ and $C$ are zero.\n\nSolving the second equation for $D$ and plugging it into equation 4, we get\n\n$$Bs^{2} - Bm^{2} a^{2} = 1 \\implies B = \\frac{1}{s^{2}-m^{2}a^{2}}$$\n\nThen from equation 2,\n\n$$D = -\\frac{m^{2}}{s^{2}-m^{2}a^{2}}$$\n\nyouth4ever\n\nNew member\n@Random Variable\nThanks again.\n\nHow you would approach a problem like the following :\nIt come from the Fourier transform of the sinc function without the coefficient.\n$$\\int_{0}^{\\infty} \\frac{sin (ak)}{ak} e^{ikx} \\ dk$$\nWithout the exponential term hanged there that would be easy and is convergent too as the sinc function dies at Infinity. But the complex exponential term introduces another cyclicity to the function so it may be regarded as non convergent because of the 0 and Infinity limits.\nHowever I want to believe that this can be done through complex integration.\nOr if not, we could take limits to be from Pi/2 to Pi for example.\n\nWhat do you think ?\n\nLast edited:", "date": "2022-05-25 20:23:07", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/advanced-integration-problem.9129/", "openwebmath_score": 0.8557101488113403, "openwebmath_perplexity": 719.9520493828567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850872288502, "lm_q2_score": 0.8902942377652497, "lm_q1q2_score": 0.8752349883927932}}
{"url": "https://math.stackexchange.com/questions/1606306/are-interior-points-ever-limit-points-as-well", "text": "# Are interior points ever limit points as well?\n\nFrom my understanding of limit points and interior points there is somewhat of an overlap and that a lot of the time interior points are also limit points.\n\nFor the reals, a neighborhood, $r>0$, around a point must contain only a single point of the set in question to determine if it's a limit point or not. However, an interior must be completely contained in the set in question, meaning it has a neighborhood that contains at least a point in the set, which also makes it a limit point.\n\nFor example: $[0,1)$ in the reals.\n\nFrom what I understand the set $(0,1)$ in the set of interior points, the set $[0,1]$ in the set of all limit points, and the set $(0,1)$ is the set of all point which are both interior and limit points.\n\nIs this correct or are interior points always not limit points for some reason?\n\nYou are right that interior points can be limit points.\n\nYour example was a perfect one: The set $[0,1)$ has interior $(0,1)$, and limit points $[0,1]$. So actually all of the interior points here are also limit points.\n\nHere is an example of an interior point that's not a limit point:\n\nLet $X$ be any set and consider the discrete topology $\\mathcal{T} = \\mathcal{P}(X)$ on $X$. Then let $A$ be any non-empty subset of $X$. For any $a \\in A$, $\\{a \\}$ is an open set containing $a$, and $a \\in \\{a \\} \\subseteq A$, so $a$ is an interior point for $A$. However, $a$ is not a limit point because we can find an open neighborhood of $a$ that doesn't contain any points from $A$ other than $a$ -- namely, $\\{a \\}$ is the open neighborhood that satisfies this. So $a$ is not a limit point of $A$. (By this same argument, we can show that $A$ has no limit points!)\n\n\u2022 You are saying {a} is an open set containing a, but it seems this is not an open set since there is no interval around a that is also in A. \u2013\u00a0jeffery_the_wind Sep 15 at 11:30\n\u2022 @jeffery_the_wind You are speaking from the perspective of the Euclidean topology that we are all so used to. But in my answer, I mentioned that $\\{a\\}$ is open in the discrete topology, which is the topology that consists of all possible subsets of $X$. So every subset of $X$ is open, including $\\{a\\}$. So the question of which sets are open and which are closed really must be asked/framed in the context of a specific topology, since the answer changes if you change the topology. $\\{a\\}$ is not open in the Euclidean topology, but it is in the discrete topology. \u2013\u00a0layman Sep 16 at 3:23\n\nA point $x$ of $A$ can be of one of two mutually exclusive types: a limit point of $A$ or an isolated point of $A$.\n\nIf the latter, it means that there exists some open $O$ in $X$ such that $\\{x\\} = O \\cap A$. The negation of this is exactly that every open set $O$ that contains $x$ always intersects points of $A$ unequal to $x$ as well, and this means exactly that $x$ is a limit point of $A$.\n\nE.g. $A = (0,1) \\cup \\{2,3\\}$ (usual topology of the reals) has two isolated points $2$ and $3$ (which are not interior points of $A$), and the rest are limit points of $A$ as well as interior points. There are also limit points $0,1$ that are not in $A$ (showing $A$ is not closed).\n\nSo if $A$ has no isolated point, all of the points of $A$ (in particular all its interior points) are limit points of $A$. So often there will quite an overlap between interior and limit points.", "date": "2019-10-19 02:07:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1606306/are-interior-points-ever-limit-points-as-well", "openwebmath_score": 0.828119695186615, "openwebmath_perplexity": 79.14143036539244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989986431444902, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8751868983297055}}
{"url": "https://www.physicsforums.com/threads/integral-evaluation-analytical-vs-numerical.76232/", "text": "# Integral evaluation - analytical vs. numerical\n\n1. May 19, 2005\n\n### broegger\n\nHi,\n\nDoes anyone know a reason why $$\\int_{-\\infty}^{\\infty}\\cosh(x)^{-n}dx$$ (n>0) can be evaluated analytically when n = 1,2,3,..., but only numerically when n is non-integer. I don't know if there is a \"reason\", but I'm using this result in a Quantum Mechanics project and it would be cool if I could give some kind of intuitive reason why this is so.\n\n2. May 19, 2005\n\n### arildno\n\nMy best guess:\nYou may use the residue theorem from complex analysis to gain an exact expression in the case of integer values, whereas this won't work if you have a non-integer.\nI might be wrong, though..\n\n3. May 19, 2005\n\n### dextercioby\n\nI've attached the antiderivative and here's your integral\n\n$$\\int_{-\\infty}^{+\\infty} \\frac{dx}{\\cosh^{n} x}=\\frac{\\sqrt{\\pi}}{2}\\frac{\\Gamma \\left(\\frac{n}{2}\\right)}{\\Gamma \\left(\\frac{n+1}{2}\\right)}$$\n\n\"n\" can be complex,even.\n\nDaniel.\n\nEDIT:The correct result is multiplied by 2.See posts #4 & #7.\n\n#### Attached Files:\n\n\u2022 ###### Integrate.gif\nFile size:\n1.6 KB\nViews:\n93\nLast edited: May 19, 2005\n4. May 19, 2005\n\n### Zurtex\n\nO.K, in Mathematica I go that if $\\Re (n) > 0$ then:\n\n$$\\int_{-\\infty}^{+\\infty} \\cosh^{-n} (x) dx = \\frac{ \\sqrt{\\pi} \\, \\Gamma\\left( \\frac{n}{2} \\right)}{ \\Gamma\\left( \\frac{n + 1}{2} \\right)}$$\n\n5. May 19, 2005\n\n### dextercioby\n\nHow do you explain the \"1/2\" factor discrepancy...?\n\nDaniel.\n\nEDIT:See post #7 for details.\n\nLast edited: May 19, 2005\n6. May 19, 2005\n\n### Zurtex\n\nWell if I left n=1 and integrate in mathematica I get $\\pi$ and:\n\n$$\\Gamma \\left( \\frac{1}{2} \\right) = \\sqrt{\\pi}$$\n\nAre you sure you did not integrate between 0 and Infinity?\n\nLast edited: May 19, 2005\n7. May 19, 2005\n\n### dextercioby\n\nI got it,i plugged only for half a domain\n\nUsing [1]\n\n$$\\int_{0}^{+\\infty} \\frac{\\sinh^{\\mu}x}{\\cosh^{\\nu}x} \\ dx =\\frac{1}{2}B\\left(\\frac{\\mu +1}{2},\\frac{\\nu-\\mu}{2}\\right)$$ (1)\n\n,provided that\n\n$$\\mbox{Re} \\left(\\mu\\right) >-1 \\ ; \\ \\mbox{Re} \\left(\\mu-\\nu\\right) <0$$ (2)\n\nMaking $\\mu=0$ (3) in (1) (admissible according to (2) and implying the condition $\\mbox{Re} (\\nu) >0$ (4) ),one gets\n\n$$\\int_{0}^{\\infty} \\frac{dx}{\\cosh^{\\nu} x} \\ dx =\\frac{1}{2} B\\left(\\frac{1}{2},\\frac{\\nu}{2}\\right)=\\frac{1}{2}\\frac{\\Gamma \\left(\\frac{1}{2}\\right)\\Gamma \\left(\\frac{\\nu}{2}\\right)}{\\Gamma\\left(\\frac{\\nu+1}{2}\\right)}$$ (5)\n\nUsing\n\n$$\\Gamma\\left(\\frac{1}{2}\\right)=\\sqrt{\\pi}$$ (6)\n\nand\n\n$$\\int_{\\mathbb{R}} \\frac{dx}{\\cosh^{\\nu} x} \\ dx =2 \\int_{0}^{\\infty} \\frac{dx}{\\cosh^{\\nu} x} \\ dx$$ (7)\n\n,one gets exactly the formula posted by Zurtex.\n\nDaniel.\n\n--------------------------------------------------\n[1]G & R,5-th edition,Academic Press,CD version,1996,page 388,formula 3.512-2.\n\n8. May 19, 2005\n\n### dextercioby\n\nYes,i used an online integrator and indeed integrated only half of the domain.\n\nI'll edit and make a reference to your post & mine just above.\n\nDaniel.\n\n9. May 22, 2005\n\n### uart\n\nSo just in case anyone is at all unclear about how this relates to the original observation that :\nIt's because $$\\Gamma(x)$$ is easily expressable in terms of elementary functions and constants for whole and half integers but generally needs other forms of numerical calulation for arbitrary values of \"x\".\n\nLast edited: May 22, 2005\n10. May 22, 2005\n\n### dextercioby\n\nIncidentally,both beta & gamma-Euler are tabulated for a certain range of values of their arguments.\n\nSo the integral is solvable analytically.Expressing its values in terms of analytical special functions means just that.\n\nDaniel.\n\n11. May 22, 2005\n\n### krab\n\nThat means that whether an integral has an analytical solution or not is based on convention (how many special functions are named). I don't find such a distinction very useful. I would echo uart's comment.\n\n12. May 25, 2005\n\n### broegger\n\nThanks y'all. I've included a bit about the gamma function, interesting stuff...", "date": "2016-10-22 02:17:34", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integral-evaluation-analytical-vs-numerical.76232/", "openwebmath_score": 0.6407851576805115, "openwebmath_perplexity": 3068.4237477965553, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102542943774, "lm_q2_score": 0.9046505344582187, "lm_q1q2_score": 0.8751682035877698}}
{"url": "http://math.stackexchange.com/questions/14133/clarify-why-all-logarithms-differ-by-a-constant", "text": "# Clarify why all logarithms differ by a constant\n\nOne of the rules of logarithms that I've been taught is:\n\n$\\log_b(x)$ is equivalent to $\\dfrac{\\log_k(x)}{\\log_k(b)}$.\n\nRecently I've also seen another rule that says:\n\n$\\log_b(x)$ is equivalent to $\\log_b(k)\\log_k(x)$.\n\nAre these equivalent (perhaps via some refactoring) ?\n\n-\n$\\frac1{\\log_k(b)}=\\frac{\\log_k(k)}{\\log_k(b)}=\\log_b(k)$ \u2013\u00a0J. M. Dec 13 '10 at 3:28\n\nDivide the second equation by $\\log_b(k)$ and shuffle the variables to see that these are the same thing. The reason they're true can be expressed in terms of properties of exponents. Start with the observation that $\\log_b(x)=y$ means that $b^y=x$. Then use $b=a^{\\log_a(b)}$ and properties of exponents to get $b^y=a^{y\\log_a(b)}=x$. The last equation says that $\\log_a(x)=y\\log_a(b)$. Now substitute $\\log_b(x)$ back in for $y$ to get the desired identity.\n\nThis fits into the general principle that each of the fundamental properties of logarithms can be seen as a translation of a property of their inverses, the exponential functions. For example, logarithms convert products to sums ($\\log_b(xy)=\\log_b(x)+\\log_b(y)$) because exponential functions convert sums to products ($b^{x+y}=b^xb^y$). Here, the fact that $b^x=a^{x\\log_a(b)}$ says that you can change bases of your exponential function by multiplying the input variable by the constant $\\log_a(b)$. In other words, exponential functions with different bases are related by horizontal stretches or shrinks of the graphs. When expressed in terms of the inverse functions, input and output switch, and the horizontal stretch (or shrink) changes to a vertical stretch (or shrink). That is, instead of multiplying the input by $\\log_a(b)$, you divide the output by $\\log_a(b)$ to get $\\log_b(x)=\\frac{\\log_a(x)}{\\log_a(b)}$. More succinctly, $g(x)=f(cx)$ implies $g^{-1}(x)=\\frac{1}{c}f^{-1}(x)$.\n\n-\nAfter seeing Bill Dubuque's nice one line derivation using the power rule, I thought I'd mention that you can also use the second identity (as proved in my or Arturo's answer) to quickly derive the power rule: $$\\log_b(a^x)=\\log_b(a)\\log_a(a^x)=x\\log_b(a).$$ \u2013\u00a0Jonas Meyer Dec 13 '10 at 5:25\n\nFirst: They are not \"equivalent\", they are equal.\n\nThat said: remember the meaning of \"$\\log_b(a) = r$\". It means that $r$ is the exponent to which you must raise $b$ in order to get $a$; that is, $\\log_b(a)=r$ is equivalent to $b^r=a$.\n\nSo... why is $\\log_b(x)=\\frac{\\log_k(x)}{\\log_k(b)}$? Because if you let $r=\\log_b(x)$, $s=\\log_k(x)$ and $t=\\log_k(b)$, then that means that $b^r = x$, $k^s = x$, and $k^t=b$. Therefore, $$k^s = x = b^r = (k^t)^r = k^{tr}.$$ So $\\log_k(x) = s = rt = \\log_b(x)\\log_k(b)$, from which you get the equality you want by dividing through by $\\log_k(b)$.\n\nFor the second equality, $\\log_b(x) = \\log_b(k)\\log_k(x)$, let $r=\\log_b(x)$, $s=\\log_b(k)$, and $t=\\log_k(x)$. Then $b^r = x$, $b^s = k$, and $k^t=s$. So we have: $$b^r = x = k^t = (b^s)^t = b^{st},$$ which means we must have $\\log_b(x) = r=st=\\log_b(k)\\log_k(x)$, as desired.\n\n-\nThis is an excellent explanation. \u2013\u00a0Quixotic Dec 13 '10 at 14:17\n\nYes. Your second equation is $\\log_b k\\cdot \\log_k x = \\log_b x$. Dividing by $\\log_b k$ we get $\\log_k x=\\frac{\\log_b x}{\\log_b k}$. This is the same as your first equation with different symbols (an interchange of $k$ and $b$).\n\n-\n+1,Nice one-liner :) \u2013\u00a0Quixotic Dec 13 '10 at 14:08\n\nHINT $\\:$ The first equation is $\\rm\\ log_K\\$ of $\\rm\\ B^{log_B\\ X}\\ =\\ X$\n\nand the $\\:$ second $\\:$ equation is $\\rm\\ log_B\\$ of $\\rm\\ K^{log_K\\ X}\\ =\\ X$\n\n-", "date": "2016-05-25 11:43:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/14133/clarify-why-all-logarithms-differ-by-a-constant", "openwebmath_score": 0.9344242811203003, "openwebmath_perplexity": 190.53354425315823, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102571131692, "lm_q2_score": 0.9046505299595162, "lm_q1q2_score": 0.8751682017857003}}
{"url": "http://local-makler.de/proofs-calculator-logic.html", "text": "Screenshots. \u03c6 [\u03bd\u2190\u03c4] where \u03c4 is free for \u03bd in \u03c6. Since this is all about math, I copy some content from wikipedia for a start. for details. A first prototype of a ProB Logic Calculator is now available online. You need to \u201csummarize\u201d what was established after making the desired assumption (the contradiction of the conclusion). Valid or Invalid? The rules of this test are simple: it's your job to determine whether an argument is valid or not. Amount of Whisky (fl. In the dropdown menu, click 'UserDoc'. Matrices & Vectors. Truth Tables, Logic, and DeMorgan's Laws. Logic Gate Simulator. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. Yes, Algebraic Proofs isn't particularly exciting. Follow the 2 steps guide to find the truth table using the boolean calculator. The symbol for this is $$\u03bd$$. Then k2 = (ax)2 = x(a2x) so xjk2. Read from here about the differences between algorithms. Proof generator and proof checker for propositional logic in \"natural deduction\" style. To enter logic symbols, use the buttons above the text field, or type ~ for \u00ac, & for \u2227, v for \u2228, -> for \u2192, <-> for \u2194, (Ax) for \u2200x, (Ex) for \u2203x, [] for , > for. It formalizes the rules of logic. Proof Checker for forall x: Cambridge and Calgary. Modifications by students and faculty at Cal. Send me a full list of your axioms and I will see what I can do to get you started. (There was the untyped logic language Prolog, and the strongly typed \u2014 but general programming language. Propositional Logic \u2022 Propositional resolution \u2022 Propositional theorem proving \u2022Unification Today we're going to talk about resolution, which is a proof strategy. Since any element x in K is also in S, we know that every element x in K is also in S, thus K S. Boolean Algebra expression simplifier & solver. Use symbolic logic and logic algebra. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Yes, Algebraic Proofs isn't particularly exciting. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Write a symbolic sentence in the text field below. truth tables, normal forms, proof checking, proof building). Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters \"W\" and \"F\" denote the constant values truth and falsehood and that the lower-case letter \"v\" denotes the disjunction. Boolean algebra calculator is the stream of mathematics that comprises of logical expressions & logical variables manipulating. Propositional logic in Artificial intelligence. In the dropdown menu, click 'UserDoc'. Amount of Whisky (fl. Line Equations Functions Arithmetic & Comp. The Coq Proof Assistant. At any time get assistance and ideas from Proof generator. Free Python 3. The specific system used here is the one found in forall x: Calgary Remix. Select gates from the dropdown list and click \"add node\" to add more gates. Drag from the hollow circles to the solid circles to make connections. Thus, x 2S. When combined together, several gates can make a complex logical evaluation system that has. Conic Sections Transformation. Logic Gate Simulator. Natural deduction proof editor and checker. Go to Daemon Proof Checkeror Quick Help Index. Free Python 3. This site based on the Open Logic Project proof checker. The Logic Daemon. The Propositional Logic Calculator finds all the models of a given propositional formula. Know Your Worth is based on millions of real salaries from Glassdoor users. See Credits. Boolean Algebra expression simplifier & solver. Perfect Proof. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. The system was originally written for UMass\u2019s Intro Logic course, based on Gary Hardegree\u2019s online. Disjunctive normal form (DNF), including perfect. In each step the user. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. Some (importable) sample proofs in the \"plain\" notation are here. (If you don't want to install this file. They will show you how to use each calculator. See Credits. Line Equations Functions Arithmetic & Comp. 3: Proofs in predicate logic. Modifications by students and faculty at Cal. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Here, instead, we shall think of this as a proof method, traditionally called \u201cconditional derivation\u201d. Mathematical logic step by step. Com stats: 2613 tutors , 730556 problems solved. One of the most basic rules of deduction in predicate logic says that ( \u2200 x P ( x)) P ( a) for any entity a in the domain of discourse of. To typeset these proofs you will need Johann Kl\u00fcwer's fitch. Place brackets in expressions, given the priority of operations. The Coq Proof Assistant. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. So far in this section, we have been working mostly with propositional logic. For an introduction to logic and proof in this style, consult a textbook such as Kaye , Huth and Ryan , or Bornat. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters \"W\" and \"F\" denote the constant values truth and falsehood and that the lower-case letter \"v\" denotes the disjunction. Online tool. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. First, we'll look at it in the propositional case, then in the first-order case. Natural deduction proof editor and checker. Proof checker. Proof Checker for forall x: Cambridge and Calgary. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. Build a truth table for the formulas entered. Use symbolic logic and logic algebra. Resolution Refutation. Line Equations Functions Arithmetic & Comp. One of the most basic rules of deduction in predicate logic says that ( \u2200 x P ( x)) P ( a) for any entity a in the domain of discourse of. The thing solves algebra, and basic symbolic logic uses, well, I don't want to say the same sort of symbol manipulation because the overlap is imperfect, but both proofs and algebra work by manipulating symbols via a set of well-defined rules. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Truth Tree Solver. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. The values of the Boolean algebra calculator are denoted with logic 0 & 1. Conjunctive normal form (CNF), including perfect. Right click connections to delete them. All the arguments are syllogisms. Binary numbers multiplication is a part of arithmetic operations in digital electronics. A first prototype of a ProB Logic Calculator is now available online. Propositional logic is also amenable to \u201cdeduction,\u201d that is, the development of proofs by writing a series of lines, each of which either is given or is justi\ufb01ed by some previous lines (Section 12. You can select and try out several solver algorithms: the \"DPLL better\" is the best solver amongst the options. Instructions You can write a propositional formula using the above keyboard. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. for details. Know Your Worth is based on millions of real salaries from Glassdoor users. The Corbettmaths Practice Questions on Algebraic Proof. One Flip application is a proof checker for entering and editing proofs in natural deduction style. Application works on the Chrome browser. The logic language used in this theorem prover is one that was proposed in the author's Master's thesis, back in 1985-1987, at which time it contained most of the features shown here, including the hierarchical type scheme. Know Your Worth is based on millions of real salaries from Glassdoor users. Thus, the argument above is valid, because if all humans are mortal, and if. Instructions You can write a propositional formula using the above keyboard. Place brackets in expressions, given the priority of operations. One of the most basic rules of deduction in predicate logic says that ( \u2200 x P ( x)) P ( a) for any entity a in the domain of discourse of. The Conformal Smart Logic Equivalence Checker (LEC) is the next-generation equivalency checking solution. A proposition is a declarative statement which is either true or false. A proof is an argument intended to convince the reader that a general principle is true in all situations. Simplify logical expressions. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. This calculator solves linear diophantine equations. \u03c6 [\u03bd\u2190\u03c4] where \u03c4 is free for \u03bd in \u03c6. This study aid includes: Proof generator. To download DC Proof and for a contact link, visit my homepage. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. It formalizes the rules of logic. Look at line 3. 3: Proofs in predicate logic. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. By using this website, you agree to our Cookie Policy. In mathematics, a Diophantine equation is a polynomial equation in two or more unknowns such that only the integer solutions are searched or studied (an integer solution is a solution such that all the unknowns take integer values). Know Your Worth is based on millions of real salaries from Glassdoor users. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). In other words, it\u2019s based on the mistaken assumption that a lack of evidence is evidence. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. The symbol for this is $$\u03bd$$. (There was the untyped logic language Prolog, and the strongly typed \u2014 but general programming language. Use symbolic logic and logic algebra. Modifications by students and faculty at Cal. A variable is a symbol used to represent a logical quantity. Instructors. It is a web application that uses Semantic Tableaux to check the validity of a statement, and provides a proof if it finds that the statement is valid. Line Equations Functions Arithmetic & Comp. This is the mode of proof most of us. Free Python 3. Let x 2K so that xjk. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. Here is a standard example: An argument is valid if and only if the conclusion necessarily follows from the premises. Using the derived rule allowed us to shorten the proof considerably. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). In the dropdown menu, click 'UserDoc'. \u03c6 [\u03bd\u2190\u03c4] where \u03c4 is free for \u03bd in \u03c6. 3: Proofs in predicate logic. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. 0 is based on classical logic, but it is possible to define your axioms in it. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. Logic Calculator. Propositional Logic \u2022 Propositional resolution \u2022 Propositional theorem proving \u2022Unification Today we're going to talk about resolution, which is a proof strategy. Boolean Algebra expression simplifier & solver. Oct 24 '18 at 20:18. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters \"W\" and \"F\" denote the constant values truth and falsehood and that the lower-case letter \"v\" denotes the disjunction. Matrices & Vectors. Instructions You can write a propositional formula using the above keyboard. Proof generator and proof checker for propositional logic in \"natural deduction\" style. Step through the examples. This study aid includes: Proof generator. Instructors. Learning about Coq. So far in this section, we have been working mostly with propositional logic. Basics Whisky 101. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Truth Tables, Logic, and DeMorgan's Laws. Get Custom Built Calculator For Your Website. See this pdf for an example of how Fitch proofs typeset in LaTeX look. To enter logic symbols, use the buttons above the text field, or type ~ for \u00ac, & for \u2227, v for \u2228, -> for \u2192, <-> for \u2194, (Ax) for \u2200x, (Ex) for \u2203x, [] for , > for. Propositional logic is also amenable to \u201cdeduction,\u201d that is, the development of proofs by writing a series of lines, each of which either is given or is justi\ufb01ed by some previous lines (Section 12. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. A drill for the truth functional connectives. Amount of Whisky (fl. Typical applications include the certification of properties of. First, we'll look at it in the propositional case, then in the first-order case. Use symbolic logic and logic algebra. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Natural Deduction examples | rules | syntax | info | download | home: Last Modified : 13-Jun-2021. Truth Tree Solver. Matrices & Vectors. To enter logic symbols, use the buttons above the text field, or type ~ for \u00ac, & for \u2227, v for \u2228, -> for \u2192, <-> for \u2194, (Ax) for \u2200x, (Ex) for \u2203x, [] for , > for. Get Custom Built Calculator For Your Website. See Credits. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The Propositional Logic Calculator finds all the models of a given propositional formula. The assumption set that results is the same as the assumption set for line 3. Using the derived rule allowed us to shorten the proof considerably. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Instructors. Conic Sections Transformation. Logic Gate Simulator. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. A model describes how units of computations, memories, and communications are organized. Back then, the idea of logic languages with types was novel. This study aid includes: Proof generator. See Credits. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. This calculator solves linear diophantine equations. Appeal to ignorance is a logical fallacy in which someone argues either for or against something because there is no contradicting evidence. Natural deduction proof editor and checker. Binary numbers multiplication is a part of arithmetic operations in digital electronics. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. It formalizes the rules of logic. Enter the term you would like to use: \u03c4: Existential Introduction: \u03c6 (\u03c4) E\u03bd:\u03c6 (\u03bd) Enter the term you would like to replace: \u03c4:. To enter logic symbols, use the buttons above the text field, or type ~ for \u00ac, & for \u2227, v for \u2228, -> for \u2192, <-> for \u2194, (Ax) for \u2200x, (Ex) for \u2203x, [] for , > for. Clicking the \"Tree Proof\" button will pass the statement to wo's Tree Proof Generator. This page is a tutorial and user's guide; there is also a complete reference. LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. Logic Calculator is a free, portable truth table generator for logic formulas i. Matrices & Vectors. Yes, Algebraic Proofs isn't particularly exciting. Solving a classical propositional formula means looking for such values of variables that the formula becomes true. Line Equations Functions Arithmetic & Comp. Enter the term you would like to use: \u03c4: Existential Introduction: \u03c6 (\u03c4) E\u03bd:\u03c6 (\u03bd) Enter the term you would like to replace: \u03c4:. Logic Calculator. Amount of Whisky (fl. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Computer programs are constantly making decisions based on the current \"STATE\" of the data held by the program. For modal predicate logic, constant domains and rigid terms are assumed. Get help from our free tutors ===>. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Oct 24 '18 at 20:18. Combining multiple conditions to form one True/False value is the. Proof generator and proof checker for propositional logic in \"natural deduction\" style. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. (2) Logical entailment: displays the truth table of each of the premises along with the. Practice your deduction skills with Proof checker and Random Tasks. Follow the 2 steps guide to find the truth table using the boolean calculator. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. The Coq Proof Assistant. This style of proof is called a resolution proof. Yes, Algebraic Proofs isn't particularly exciting. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses\u2019s free, online logic teaching software application Carnap (carnap. Thus, the argument above is valid, because if all humans are mortal, and if. Each one has a different shape to show its particular function. Back then, the idea of logic languages with types was novel. This site based on the Open Logic Project proof checker. A model describes how units of computations, memories, and communications are organized. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. Write a symbolic sentence in the text field below. For example, (a -> b) & a becomes true if and only if both a and b are assigned true. proof language Isar [20] which looks like a mixture of English, logic and a programming language, and is based on natural deduction. Valid or Invalid? The rules of this test are simple: it's your job to determine whether an argument is valid or not. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. We applied Demorgan's Law, which is abbreviated DM, to line 2. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Let x 2K so that xjk. for details. A variable is a symbol used to represent a logical quantity. 0 is based on classical logic, but it is possible to define your axioms in it. Mathematical logic step by step. All in one boolean expression calculator. Propositional logic is also amenable to \u201cdeduction,\u201d that is, the development of proofs by writing a series of lines, each of which either is given or is justi\ufb01ed by some previous lines (Section 12. See Credits. Logic Gate Simulator. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. A proof is an argument intended to convince the reader that a general principle is true in all situations. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. 2 Proofs One of the principal aims of this course is to teach the student how to read and, to a lesser extent, write proofs. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. The Gateway to Logic is a collection of web-based logic programs offering a number of logical functions (e. Yes, Algebraic Proofs isn't particularly exciting. Logic gates are symbols that can directly replace an expression in Boolean arithmetic. When combined together, several gates can make a complex logical evaluation system that has. Natural Deduction is a free app published for Windows 10 PC and can be downloaded from Windows Store. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. Yes, Algebraic Proofs isn't particularly exciting. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences. Know Your Worth is based on millions of real salaries from Glassdoor users. Suppose k 2Z and let K = fn 2Z : njkgand S = fn 2Z : njk2g. So far in this section, we have been working mostly with propositional logic. Investigate the behaviour of AND, OR, NOT, NAND, NOR and XOR gates. The values of the Boolean algebra calculator are denoted with logic 0 & 1. Thus, x 2S. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. (whenever you see $$\u03bd$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$\u03bd$$ q. When loaded, click 'Help' on the menu bar. If you enter a modal formula, you will see a choice of how the accessibility relation should be constrained. Online tool. Learning about Coq. Here, instead, we shall think of this as a proof method, traditionally called \u201cconditional derivation\u201d. Natural Deduction is a free app published for Windows 10 PC and can be downloaded from Windows Store. Mathematical logic step by step. This is the mode of proof most of us. Go to Daemon Proof Checkeror Quick Help Index. When combined together, several gates can make a complex logical evaluation system that has. Know Your Worth is based on millions of real salaries from Glassdoor users. The symbol P denotes a sum over its argument for each natural. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Boolean Algebra expression simplifier & solver. The notion of 'proof' is much as it was for sentential logic, except that we have a new definition of 'formula' and some new rules for introducing and eliminating quantifiers. This allows for submission and automated marking of exercises such as symbol-ization, truth tables, and natural deduction proofs. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses\u2019s free, online logic teaching software application Carnap (carnap. Note that proofs can also be exported in \"pretty print\" notation (with unicode logic symbols) or LaTeX. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Get Custom Built Calculator For Your Website. Learn boolean algebra. Yes, Algebraic Proofs isn't particularly exciting. Kevin Klement has done up a prototype of his online natural deduction proof builder/checker that works with the natural deduction system of the Cambridge and Calgary versions of forall x. LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra. - Dan Christensen. Simplify logical expressions. Free Ubuntu. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. Using the derived rule allowed us to shorten the proof considerably. All the arguments are syllogisms. This style of proof is called a resolution proof. Use symbolic logic and logic algebra. Boolean Expression Calculator. It executes the logical operations like AND, NAND, OR, NOR, NOT & X-OR. Propositional sequent calculus prover. State University, Monterey Bay. View all solved problems on Proofs -- maybe yours has been solved already! Become a registered tutor (FREE) to answer students' questions. Quantifiers in proofs Expressing Generality This section concerns the proof system of first-order logic or the lower predicate calculus. A free, simple, online logic gate simulator. Place brackets in expressions, given the priority of operations. Then k2 = (ax)2 = x(a2x) so xjk2. Thus, x 2S. Disjunctive normal form (DNF), including perfect. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. It is a technique of knowledge representation in logical and mathematical form. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. Since this is all about math, I copy some content from wikipedia for a start. Because of its simplicity it is particularly well-suited for mechanical theorem provers. Simplify logical expressions. Proof by induction involves statements which depend on the natural numbers, n = 1,2,3, It often uses summation notation which we now brie\ufb02y review before discussing induction itself. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Amount of Whisky (fl. You can select and try out several solver algorithms: the \"DPLL better\" is the best solver amongst the options. Complete your profile, and we will calculate how much you could earn in today's job market. Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. This site based on the Open Logic Project proof checker. This is a demo of a proof checker for Fitch-style natural deduction systems found in many popular introductory logic textbooks. Free Python 3. Drag from the hollow circles to the solid circles to make connections. All the arguments are syllogisms. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Yes, Algebraic Proofs isn't particularly exciting. The thing solves algebra, and basic symbolic logic uses, well, I don't want to say the same sort of symbol manipulation because the overlap is imperfect, but both proofs and algebra work by manipulating symbols via a set of well-defined rules. Natural deduction proof editor and checker. One of the most basic rules of deduction in predicate logic says that ( \u2200 x P ( x)) P ( a) for any entity a in the domain of discourse of. They will show you how to use each calculator. In each step the user. The specific system used here is the one found in forall x: Calgary Remix. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. Learn more. The symbol P denotes a sum over its argument for each natural. You need to \u201csummarize\u201d what was established after making the desired assumption (the contradiction of the conclusion). Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Complete your profile, and we will calculate how much you could earn in today's job market. Matrices & Vectors. A sequent S is true if and only if there exists a tree of sequents rooted at S where each leaf is an axiom and each internal node is derived from its children by an inference rule. Such proofs can also encode traditional proofs based on modus ponens: the inference P\u2227(P\u21d2Q) \u22a2 Q can be rewritten as resolution by expanding \u21d2 to get P\u2227(\u00acP\u2228Q) \u22a2 Q. See Credits. For example, a heart monitoring program might sound an alarm if the pulse is too slow or the blood pressure is too weak. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. ) And that\u2019s it! Enjoy drinking your whisky exactly how you like it. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. Free Windows Terminal Preview. Back then, the idea of logic languages with types was novel. You can select and try out several solver algorithms: the \"DPLL better\" is the best solver amongst the options. Formal Proof Check. It is also known as argumentum ad ignorantiam (Latin for \u201cargument from ignorance\u201d) and is a type of. Com stats: 2613 tutors , 730556 problems solved. ) Bottle Proof. Videos, worksheets, 5-a-day and much more. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. State University, Monterey Bay. techniques as \u201cproof by contradiction\u201d or \u201cproof by contrapositive\u201d (Section 12. The values of the Boolean algebra calculator are denoted with logic 0 & 1. See Credits. Propositional logic in Artificial intelligence. Desired Proof. Kevin Klement has done up a prototype of his online natural deduction proof builder/checker that works with the natural deduction system of the Cambridge and Calgary versions of forall x. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Save your work on device and continue later on. We applied Demorgan's Law, which is abbreviated DM, to line 2. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Logic gates are symbols that can directly replace an expression in Boolean arithmetic. chapter 13 of Paul Teller's logic textbook contains a description of such a procedure for propositional logic (basically truth trees in Fitch notation). Logic Calculator. It is also known as argumentum ad ignorantiam (Latin for \u201cargument from ignorance\u201d) and is a type of. The symbol for this is $$\u03bd$$. Back then, the idea of logic languages with types was novel. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. Resolution Refutation. Since this is all about math, I copy some content from wikipedia for a start. The symbol P denotes a sum over its argument for each natural. The Logic Daemon. 3: Proofs in predicate logic. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. This app is a graphical semantic calculator for a specific kind of. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Any single variable can have a 1 or a 0 value. Proof Checker for forall x: Cambridge and Calgary. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Logic Gate Simulator. Conic Sections Transformation. The Corbettmaths Practice Questions on Algebraic Proof. Practice your deduction skills with Proof checker and Random Tasks. Propositional sequent calculus prover. Binary numbers multiplication is a part of arithmetic operations in digital electronics. It is also known as argumentum ad ignorantiam (Latin for \u201cargument from ignorance\u201d) and is a type of. Boolean algebra is used to simplify Boolean expressions which represent combinational logic circuits. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. ) Bottle Proof. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. (whenever you see $$\u03bd$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$\u03bd$$ q. Diana Higgins on Conditional-proof-logic-calculator glennnervi. Combining multiple conditions to form one True/False value is the. A variable is a symbol used to represent a logical quantity. Any single variable can have a 1 or a 0 value. See this pdf for an example of how Fitch proofs typeset in LaTeX look. The Corbettmaths Practice Questions on Algebraic Proof. The symbol for this is $$\u03bd$$. In computer science, and more specifically in computability theory and computational complexity theory, a model of computation is a model which describes how an output of a mathematical function is computed given an input. Binary numbers multiplication is a part of arithmetic operations in digital electronics. By using this website, you agree to our Cookie Policy. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. Online tool. It is also known as argumentum ad ignorantiam (Latin for \u201cargument from ignorance\u201d) and is a type of. Propositional logic (PL) is the simplest form of logic where all the statements are made by propositions. Yes, Algebraic Proofs isn't particularly exciting. Conjunctive normal form (CNF), including perfect. Get help from our free tutors ===>. Actually there are mechanical ways of generating Fitch style proofs. Once you know your perfect proof, this calculator will tell you exactly how much water to add to any amount of whisky to reach it. Boolean formulas are written as sequents. To typeset these proofs you will need Johann Kl\u00fcwer's fitch. It is also known as argumentum ad ignorantiam (Latin for \u201cargument from ignorance\u201d) and is a type of. Formal Proof Check. This study aid includes: Proof generator. This site based on the Open Logic Project proof checker. Place brackets in expressions, given the priority of operations. At any time get assistance and ideas from Proof generator. Yes, Algebraic Proofs isn't particularly exciting. Natural deduction proof editor and checker. They're especially important in logical arguments and proofs, let's find out why! While the word \"argument\" may mean a disagreement between two or more people, in mathematical logic, an argument is a sequence or list of statements called premises or assumptions and returns a conclusion. \u03c6 [\u03bd\u2190\u03c4] where \u03c4 is free for \u03bd in \u03c6. Read from here about the differences between algorithms. Like most proofs, logic proofs usually begin with premises--- statements that you're allowed to assume. Basics Whisky 101. You can also use LaTeX commands. Logic Calculator is a free, portable truth table generator for logic formulas i. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. It reduces the original expression to an equivalent expression that has fewer terms which means that less logic gates are needed to implement the combinational logic circuit. Conic Sections Transformation. Look at line 3. The Corbettmaths Practice Questions on Algebraic Proof. Back then, the idea of logic languages with types was novel. Modifications by students and faculty at Cal. You need to \u201csummarize\u201d what was established after making the desired assumption (the contradiction of the conclusion). The values of the Boolean algebra calculator are denoted with logic 0 & 1. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). Line Equations Functions Arithmetic & Comp. Use symbolic logic and logic algebra. Propositional logic in Artificial intelligence. It provides a formal language to write mathematical definitions, executable algorithms and theorems together with an environment for semi-interactive development of machine-checked proofs. Right click connections to delete them. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. Propositional logic is also amenable to \u201cdeduction,\u201d that is, the development of proofs by writing a series of lines, each of which either is given or is justi\ufb01ed by some previous lines (Section 12. ENDING AN INDIRECT PROOF (after you derive a contradiction, any contradiction) CP. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses\u2019s free, online logic teaching software application Carnap (carnap. The only limitation for this calculator is that you have only three atomic propositions to choose from: p,q and r. Place brackets in expressions, given the priority of operations. The specific system used here is the one found in forall x: Calgary Remix. The Conformal Smart Logic Equivalence Checker (LEC) is the next-generation equivalency checking solution. 3: Proofs in predicate logic. Resolution is one kind of proof technique that works this way - (i) select two clauses that contain conflicting terms (ii) combine those two clauses and (iii) cancel out the conflicting terms. truth tables, normal forms, proof checking, proof building). Also, first order logic is semidecidable, meaning there are ways to mechanically find a proof if the sequent is valid (though the search may never terminate in the case. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. You can select and try out several solver algorithms: the \"DPLL better\" is the best solver amongst the options. Back then, the idea of logic languages with types was novel. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. Valid or Invalid? The rules of this test are simple: it's your job to determine whether an argument is valid or not. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters \"W\" and \"F\" denote the constant values truth and falsehood and that the lower-case letter \"v\" denotes the disjunction. Proof checker. Mathematical logic step by step. Right click connections to delete them. techniques as \u201cproof by contradiction\u201d or \u201cproof by contrapositive\u201d (Section 12. Yes, Algebraic Proofs isn't particularly exciting. At any time get assistance and ideas from Proof generator. Your Input Approximate the integral $$\\int\\limits_{0}^{1} \\sqrt{\\sin^{3}{\\left(x \\right)} + 1}\\, dx$$$with $$n = 5$$$ using the trapezoidal rule. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. They're especially important in logical arguments and proofs, let's find out why! While the word \"argument\" may mean a disagreement between two or more people, in mathematical logic, an argument is a sequence or list of statements called premises or assumptions and returns a conclusion. A sequent S is true if and only if there exists a tree of sequents rooted at S where each leaf is an axiom and each internal node is derived from its children by an inference rule. The amount of detail that an author supplies in a proof should depend on the audience. Matrices & Vectors. The boolean algebra calculator uses the basic laws like identity law. Actually there are mechanical ways of generating Fitch style proofs. Conic Sections Transformation. Know Your Worth is based on millions of real salaries from Glassdoor users. Free Python 3. The boolean algebra calculator uses the basic laws like identity law. You may add additional sentences to your set by repeating this step. Take help from sample expressions in the input box or have a look at the boolean functions in the content to understand the mathematical operations used in expressions. Actually there are mechanical ways of generating Fitch style proofs. Click on one of the three applications on the right. This app is a graphical semantic calculator for a specific kind of. Simplify logical expressions. Online tool. Propositional logic is also amenable to \u201cdeduction,\u201d that is, the development of proofs by writing a series of lines, each of which either is given or is justi\ufb01ed by some previous lines (Section 12. Boolean formulas are written as sequents. Screenshots. Amount of Water to Add (fl. Free Logical Sets calculator - calculate boolean algebra, truth tables and set theory step-by-step This website uses cookies to ensure you get the best experience. Look at line 3. Send me a full list of your axioms and I will see what I can do to get you started. As with other logical systems, the theory lies at the intersection of mathematics and philosophy, while important applications are found within computer science and linguistics. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The symbol for this is $$\u03bd$$. State University, Monterey Bay. Using the derived rule allowed us to shorten the proof considerably. Yes, Algebraic Proofs isn't particularly exciting. The logic language used in this theorem prover is one that was proposed in the author's Master's thesis, back in 1985-1987, at which time it contained most of the features shown here, including the hierarchical type scheme. 0 is based on classical logic, but it is possible to define your axioms in it. This page is a tutorial and user's guide; there is also a complete reference. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. To typeset these proofs you will need Johann Kl\u00fcwer's fitch. Modifications by students and faculty at Cal. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic. Matrices & Vectors. Desired Proof. Free Python 3. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. We write the sum of the natural numbers up to a value n as: 1+2+3+\u00b7\u00b7\u00b7+(n\u22121)+n = Xn i=1 i. Place brackets in expressions, given the priority of operations. When combined together, several gates can make a complex logical evaluation system that has. It is a technique of knowledge representation in logical and mathematical form. Mathematical logic step by step. Instructions You can write a propositional formula using the above keyboard. You can select and try out several solver algorithms: the \"DPLL better\" is the best solver amongst the options. Logic Calculator. To typeset these proofs you will need Johann Kl\u00fcwer's fitch. We write the sum of the natural numbers up to a value n as: 1+2+3+\u00b7\u00b7\u00b7+(n\u22121)+n = Xn i=1 i. Go to Daemon Proof Checkeror Quick Help Index. Natural deduction proof editor and checker. Propositional logic in Artificial intelligence. Using the derived rule allowed us to shorten the proof considerably. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. Each one has a different shape to show its particular function. This calculator solves linear diophantine equations. The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouses\u2019s free, online logic teaching software application Carnap (carnap. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. In Wolfram Alpha's case, it seems to do truth tables, but not proofs. Logic Gate Simulator. (whenever you see $$\u03bd$$ read 'or') When two simple sentences, p and q, are joined in a disjunction statement, the disjunction is expressed symbolically as p $$\u03bd$$ q. You may add additional sentences to your set by repeating this step. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters \"W\" and \"F\" denote the constant values truth and falsehood and that the lower-case letter \"v\" denotes the disjunction. At any time get assistance and ideas from Proof generator. ) And that\u2019s it! Enjoy drinking your whisky exactly how you like it. See Credits. Step through the examples. The Corbettmaths Practice Questions on Algebraic Proof. The complement is the inverse of a variable and is indicated by a bar over variable (overbar). Once you know your perfect proof, this calculator will tell you exactly how much water to add to any amount of whisky to reach it. Modifications by students and faculty at Cal. Try the leading salary calculator the next time you negotiate your salary or ask for a raise, and get paid fairly. The specific system used here is the one found in forall x: Calgary Remix. A proposition is a declarative statement which is either true or false. Basics Whisky 101. Such proofs can also encode traditional proofs based on modus ponens: the inference P\u2227(P\u21d2Q) \u22a2 Q can be rewritten as resolution by expanding \u21d2 to get P\u2227(\u00acP\u2228Q) \u22a2 Q. Your Input Approximate the integral $$\\int\\limits_{0}^{1} \\sqrt{\\sin^{3}{\\left(x \\right)} + 1}\\, dx$$$with $$n = 5$$$ using the trapezoidal rule. a web application that decides statements in symbolic logic including modal logic, propositional logic and unary predicate logic Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax. The system was originally written for UMass\u2019s Intro Logic course, based on Gary Hardegree\u2019s online. In mathematics, a Diophantine equation is a polynomial equation in two or more unknowns such that only the integer solutions are searched or studied (an integer solution is a solution such that all the unknowns take integer values). In computer science, and more specifically in computability theory and computational complexity theory, a model of computation is a model which describes how an output of a mathematical function is computed given an input. It provides a formal language to write mathematical definitions, executable algorithms and theorems together with an environment for semi-interactive development of machine-checked proofs. The symbol P denotes a sum over its argument for each natural. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Logic Calculator: Desktop application useful to perform logical operations in an arithmetic-calculator fashion, with three modes: 1) Evaluation of logic formulas: displays the truth table along with the models of the given formula. All in one boolean expression calculator. Solving a classical propositional formula means looking for such values of variables that the formula becomes true. Coq is a formal proof management system. A model describes how units of computations, memories, and communications are organized. A variable is a symbol used to represent a logical quantity. For example we have following statements, (1) If it is a pleasant day you will do strawberry picking. To download DC Proof and for a contact link, visit my homepage. Imre Lakatos's Proofs and Refutations is an enduring classic, which has never lost its relevance. It formalizes the rules of logic. Save your work on device and continue later on. Without it, the proof would have looked like this: Proof B 1 (1) ~P&~Q->R A 2 (2) ~(PvQ) A 3 (3) P A 3 (4) PvQ 3 vI. To enter logic symbols, use the buttons above the text field, or type ~ for \u00ac, & for \u2227, v for \u2228, -> for \u2192, <-> for \u2194, (Ax) for \u2200x, (Ex) for \u2203x, [] for , > for. The only limitation for this calculator is that you have only three atomic propositions to choose from: p,q and r. This site based on the Open Logic Project proof checker. Here is a standard example: An argument is valid if and only if the conclusion necessarily follows from the premises. Proof checker. A proof is an argument intended to convince the reader that a general principle is true in all situations. Note that proofs can also be exported in \"pretty print\" notation (with unicode logic symbols) or LaTeX. Logic Gate Simulator. If you are a new user to the Gateway, consider starting with the simple truth-table calculator or with the Server-side functions. Besides classical propositional logic and first-order predicate logic (with functions and identity), a few normal modal logics are supported. Yes, Algebraic Proofs isn't particularly exciting. Logic calculator: Server-side Processing Help on syntax - Help on tasks - Other programs - Feedback - Deutsche Fassung Examples and information on the input syntax Please note that the letters \"W\" and \"F\" denote the constant values truth and falsehood and that the lower-case letter \"v\" denotes the disjunction. In chapter 17 we will prove the deduction theorem. The assumption set that results is the same as the assumption set for line 3. In the dropdown menu, click 'UserDoc'. This app is a graphical semantic calculator for a specific kind of. You may add additional sentences to your set by repeating this step. This calculator solves linear diophantine equations. Some (importable) sample proofs in the \"plain\" notation are here. Modal logic is a type of symbolic logic for capturing inferences about necessity and possibility. Free Windows Terminal Preview. Perfect Proof. In logic, a disjunction is a compound sentence formed using the word or to join two simple sentences.", "date": "2021-12-09 04:51:39", "meta": {"domain": "local-makler.de", "url": "http://local-makler.de/proofs-calculator-logic.html", "openwebmath_score": 0.6707203984260559, "openwebmath_perplexity": 1369.7160434341406, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.97805174308637, "lm_q2_score": 0.8947894618940992, "lm_q1q2_score": 0.8751503929008388}}
{"url": "http://math.stackexchange.com/questions/442173/what-does-x-2-1-mean", "text": "what does $|x-2| < 1$ mean?\n\nI am studying some inequality properties of absolute values and I bumped into some expressions like $|x-2| < 1$ that I just can't get the meaning of them.\n\nLets say I have this expression\n\n$$|x|<1.$$\n\nThis means that $x$ must be somewhere less than $1$ or greater than $-1$ which means that\n\n$$-1 < x < 1.$$\n\nSo basically $|x|<1$ and $-1 < x < 1$ are the same thing.\n\n$$|x|<1 \\iff -1 < x < 1 \\iff\\text{\"Somewhere less that 1 or greater than -1\" or between -1 and 1}$$\n\nNow lets say I have\n\n$$|x-2| < 1.$$\n\nThis means that the result of the expression $|x-2|$ must be less than $1$ or greater than $-1$? What does that also mean for $x$? Is it that $x$ must be a value that when we subtract $2$ the result has to stay withing the bound of $-1$ or $1$ or less than zero? If $x =5$ the statement fails because $3 <1$ is false. So it has to determine a boundary of $x$'s that satisfy this equation right?\n\nif $|x| = |-x|$\n\nwhat can this mean for\n\n$|x-2| = |-x-2|$ or $|x+2|$ or $|-x+2|$ ?\n\nThank you\n\n-\nI added the tag intuition to the question because it seems like you're looking for an intuitive way to understand the formula. \u2013\u00a0 Git Gud Jul 12 '13 at 16:48\nCorrection to: \"Somewhere less that 1 or greater than -1\" or between -1 and 1. The second part is correct. The part in quotes should have AND as the conjunction. Otherwise, one could use 4 which is \"greater than -1\" yet isn't less than 1. \u2013\u00a0 JB King Jul 12 '13 at 16:59\nIt means that $x$ is less than 1 away from 2 in either direction on the number line, or any direction on the complex plane. \u2013\u00a0 Kaz Jul 12 '13 at 19:01\n\nThe geometric interpretation, in $\\Bbb R$, for $|x-a|<b$ is '$x$ is at a distance smaller than $b$ from $a$'.\n\nIn your particular example, $|x-2|<1$, it means that $x$ is at a distance of at most $1$ from $2$ and it (the distance) never reaches $1$.\n\nTo interpret $|x-2|=|-x-2|$, I find useful to first note that $|-x-2|=|x-(-2)|$ (why?). The equality $|x-2|=|x-(-2)|$ says that $x$ is at equal distance between $2$ and $-2$.\n\nMore generally, $|x-a|=|x-b|$ says that $x$ is at the same distance between $a$ and $b$.\n\nTo summarize, read $|x-a|$ as the distance between $x$ and $a$.\n\n-\nthank you, and |x-2| = |-x-2| ? or |x+2| ? or |-x+2| ? \u2013\u00a0 themhz Jul 12 '13 at 16:51\n@themhz Add that to the question. \u2013\u00a0 Git Gud Jul 12 '13 at 16:51\nso |x-2| => say x=3 then 3-2 = 1 or x=-3 then -3-(-2)=-3+2 =-1 so |-1|=|1|=1 so |x-2| = |-x+2| =|2-x| right? \u2013\u00a0 themhz Jul 12 '13 at 17:01\n\"it means that x is at a distance of at most 1 from 2 and it never reaches 2.\" Of course it reaches 2. It won't reach 1 or 3, though. \u2013\u00a0 celtschk Jul 12 '13 at 17:04\n|x-2| is not the same as |-x-2| as the latter could be expressed as |x+2| which is quite different if you consider evaluating this for a few different values of x. \u2013\u00a0 JB King Jul 12 '13 at 20:08\n\nThe way to think of $$|x-y|$$ is as the distance from $x$ to $y$. For example, is $|x-y| = |y-x|$? Yes, it is, because the distance from $x$ to $y$ is the same as the distance from $y$ to $x$.\n\nIs $$|x-y| + |y-z| = |x-z|?$$ This says that the distance from $x$ to $y$, plus the distance from $y$ to $z$, is equal to the distance from $x$ to $z$. That would mean that $y$ was on the direct path from $x$ to $z$. So we would expect it to be false if $y$ was not on this direct path; say if $x = 2, z=4,$ but $y = 17$. And indeed $|2-17| + |17-4| \\ne |2-4|$, so the equation above is not always true. But we might guess from this understanding that $$|x-y| + |y-z| \\ge |x-z|,$$ with equality occurring just if $y$ is between $x$ and $z$. And in fact this is always true.\n\nWith this idea, what does $$|x|$$ mean? It should be the same as $$|x-0|,$$ which is the distance from $x$ to 0. And that is correct.\n\nNow what does $$|x-2| < 1$$ mean? It means that the distance from $x$ to 2 is less than 1. So another way to write this is $$1\\lt x\\lt 3.$$\n\n-\n\nHint\n\nDenote $x-2$ by $y$ then $$|x-2|<1\\iff |y|<1$$ and you find exactly your first inequality. Can you take it from here?\n\n-\nthat's very intuitive thank you \u2013\u00a0 themhz Jul 12 '13 at 17:02\nYou're welcome. \u2013\u00a0 Sami Ben Romdhane Jul 12 '13 at 18:23\n\nWe know that: $$|x-2|= \\left\\{ \\begin{array}{ll} -x+2 & \\quad x < 2 \\\\ x-2 & \\quad x \\ge 2 \\end{array} \\right.$$ Now if we have to do $|x-2|<1$ so: $$x\\ge2\\to x-2<1\\to x<3\\\\\\ x<2\\to 2-x<1\\to x>1$$ This means that , overall, we have $1<x<3$.\n\n-\nI hope you are feeling better, Babak! ;-) \u2013\u00a0 amWhy Jul 13 '13 at 0:35\n\nJust as you replace $|x|<1$ with $-1<x<1$, you can replace $|x-2|<1$ with $-1<x-2<1$. Adding $2$ on all three parts leaves the order unchanged, so $-1+2<x-2+2<1+2$, i.e. $1<x<3$.\n\nTo see what signs are correct in $|x-2|=|\\pm x\\pm 2|$, note that the absolute value does not change when we replace its argument with its negative. The argument here is $x-2$, the negative thereof is $-(x-2)$ and that can be simplified to $-x+2$. Thus $|x-2|=|-x+2|$. (Of course in rare cases it may also be true that $|x-2|=|x+2|$, namely precisely when $x=0$)\n\n-\nDraw a graph of $$f(x) = |x|$$ Then ask yourself, how does one obtain $$g(x) = |x-2|$$ from the graph of $f(x)$. Once you figure out how the graph looks, the question that you asked is simply asking to find all the possible $x$ values such that $g(x) < 1$. You can draw a horizontal line $y=1$ on the graph of $g(x)$ and observe the $x$ values that satisfy.\n$|x-2| < 1\\iff-1<x-2<1\\iff2-1<2+x-2<2+1\\iff1<x<3$", "date": "2014-03-14 22:47:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/442173/what-does-x-2-1-mean", "openwebmath_score": 0.9113469123840332, "openwebmath_perplexity": 183.16324154476547, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676508127574, "lm_q2_score": 0.8933093954028816, "lm_q1q2_score": 0.8751463168433056}}
{"url": "https://math.stackexchange.com/questions/1854795/a-dance-class-consists-of-22-students-10-women-and-12-men-if-5-men-and-5-women/1854801", "text": "# A dance class consists of 22 students, 10 women and 12 men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?\n\nThis is a question from Sheldon Ross.\n\nA dance class consists of 22 students, 10 women and 12 men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?\n\nSo my reasoning is this :\n\n1. I choose 5 women from a pool of 10 in 10C2 ways.\n2. I choose 5 men from a pool of 12 in 12C2 ways.\n\nSo total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.\n\nBut I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!\n\nSo I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.\n\n\u2022 Please make your titles informative. We get dozens of \"permutations combinations\" questions everyday. \u2013\u00a0Em. Jul 10 '16 at 8:28\n\u2022 Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$ \u2013\u00a0Ariana Jul 10 '16 at 8:53\n\u2022 Did you mean $\\binom{10}{5}\\binom{12}{5}5!$? \u2013\u00a0N. F. Taussig Jul 10 '16 at 9:08\n\u2022 @N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!. \u2013\u00a0amrx Jul 10 '16 at 9:17\n\u2022 As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $\\binom{10}{5}\\binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters. \u2013\u00a0N. F. Taussig Jul 10 '16 at 9:24\n\nI believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself \"How many different women can I pair with this man?\"\n\n\u2022 As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely. \u2013\u00a0Mike Jul 10 '16 at 8:35\n\nAfter selecting 5 from men and 5 from women, we need to pair them.\n\nLets assume like finding a pair for each man,\n\nfor the 1st guy -- can choose 1 from 5 women\n\nfor the 2nd guy -- can choose 1 from 4 women\n\n.\n\n.\n\nfor the 5th guy -- can choose 1 from 1 woman\n\nso 5!", "date": "2019-10-15 10:51:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1854795/a-dance-class-consists-of-22-students-10-women-and-12-men-if-5-men-and-5-women/1854801", "openwebmath_score": 0.783104419708252, "openwebmath_perplexity": 361.3482035019971, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631627142339, "lm_q2_score": 0.8872045981907006, "lm_q1q2_score": 0.8751059334459905}}
{"url": "https://math.stackexchange.com/questions/4070377/probability-question-with-similar-distinct-dice", "text": "# Probability question with similar/distinct dice.\n\nThe question:\n\nWe're throwing $$3$$ normal dice twice. (Suppose that every vector with length of $$6$$ has an equal probability). What is the probability to get two similar results from the two throws of the $$3$$ dice if:\na) The dice are different.\nb) The dice are similar.\n\nMy Work:\nfor (a):\n$$\\Omega = \\{(a_1,a_2,...,a_6) \\mid a_i \\in \\{1,...,6\\} ,(i=1,2,...,6)\\}$$\nif all the dice are different then: $$|\\Omega| = 6^6$$\nLet $$A\\subset\\Omega$$, ($$A$$- we get similar results from the two throws), then $$|A|=6^3$$ (first dice throw can be whatever, but the second one must be the same).\nSo the final answer I got is $$P(A) = \\frac {6^3}{6^6}$$.\n\nfor (b): (and here comes the confusion):\nThings I tried to do and didn't get the right answer:\n\n1. I tried to choose $$3$$ of the dice (since they're all similar) throw them, and sort them, so what I got is $$\\binom{6}{3} \\cdot 6^3 \\cdot 3!$$.\n2. I tried to choose $$1$$ of the two throws, and find how many options there are and sort them (same as before, each dice has $$6$$ sides so $$6^3$$ for all), and got $$\\binom{2}{1} \\cdot 6^3 \\cdot 3!$$.\n\nBoth of the two ways led me to wrong answers, I'm trying where are my mistakes and why my thinking led me to wrong answers, which can lead me to reach the true answer.\n\nThe Final answer is $$\\frac {996}{6^6}$$\n\nEDIT:\nAfter deeply thinking about the problem I obtained a new approach (I couldn't reach the answer with it, but I'm not sure if it's my weak combinatorics or if it's a wrong approach).\n3. By writing down a couple of examples, I noticed that there's something that I can also use to calculate the probability, we can divide this into $$3$$ possible cases:\nFirst: Getting same number on all of the six dice. $$(2,2,2, 2,2,2)$$\nSecond: Getting $$2$$ different numbers. $$(2,4,2, 2,4,2)$$\nThird: Getting $$3$$ different numbers. $$(1,2,3, 1,2,3)$$\nSo I could calculate the number of options each case has and divide by the total options $$|\\Omega|=6^6$$.\nI felt like I've gone too far but I can't see why it wouldn't be true. Any Feedback is really appreciated.\n\u2022 The word dice is plural; the singular is die. The word dices means to cut into small cubes. \u2013\u00a0N. F. Taussig Mar 21 at 10:19\n\u2022 Formatting tip: type $\\binom{n}{k}$ to obtain $\\binom{n}{k}$. \u2013\u00a0N. F. Taussig Mar 21 at 10:23\n\u2022 Your new approach is the right one. However, I don't see how the author obtained $996$ favorable cases. \u2013\u00a0N. F. Taussig Mar 21 at 10:40\n\u2022 @N.F.Taussig Thanks alot I'm happy to hear that, could you recognize why the first two approaches reached wrong answers? I still can't understand why they led me to wrong answers and trying to see what problems I have there if it's using binomials the wrong way or just my logic wasn't true. Like for example in approach 1, I chose the $3$ dice of the first throw, and they could be whatever, so the cases that I added in my new approach (according to my understanding) are included in the first approach. seems like I have a big misunderstanding somewhere. \u2013\u00a0Pwaol Mar 21 at 10:48\n\u2022 Your first two approaches fail since if your original outcome was, say $(5, 5, 4)$, there are only three ways the second roll could match the first, depending on which die matched the $4$, not $3!$. \u2013\u00a0N. F. Taussig Mar 21 at 10:52\n\nWith the help of N. F. Taussig in the comments, I think I have reached the right solution, So I decided to answer my own question, and would love to hear feedback.\n\nStarting with why my first two approaches are wrong, The number of ways that the second roll could match the first is different whenever there's $$3$$ different numbers or $$2$$ different or all of them are the same, but in my first two approaches, I didn't pay attention to that.\n\nSolution: (Approach 3):\nThe number of ways to get the same number on all dice, is just simply choosing a number which is $$\\binom{6}{1}=6$$.\nThe number of ways to get two different numbers in the first roll is:\n$$\\binom{6}{2}$$ choosing $$2$$ numbers, $$\\binom{2}{1}$$ choosing the number that will appear twice, $$\\binom{3}{1}$$ choosing which place is the single number in the first roll, $$\\binom{3}{1}$$ choosing in which place is the single number in the second roll. So in total by multiplying all of them we get $$\\binom{6}{2}*2*3*3 = 270$$ ways.\n\nThe number of ways to get three different numbers in the first roll is:\n$$\\binom{6}{3}$$ choosing $$3$$ numbers, $$3!$$ ordering the first roll, $$3!$$ ordering the second roll. in total $$720$$.\n\nHence we reach the answer by adding them up and dividing by $$|\\Omega|$$.\n\n\u2022 This is correct. The reason I did not initially obtain $996$ was that I forgot that there were only three dice. \u2013\u00a0N. F. Taussig Mar 21 at 11:14", "date": "2021-05-09 23:42:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4070377/probability-question-with-similar-distinct-dice", "openwebmath_score": 0.699432909488678, "openwebmath_perplexity": 345.2724397229974, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986363166722906, "lm_q2_score": 0.8872045832787204, "lm_q1q2_score": 0.8751059222938748}}
{"url": "https://math.stackexchange.com/questions/1235574/substitution-integration-question", "text": "# substitution integration question\n\nI want to integrate $\\int \\sqrt{1 - x^2} dx$.\n\nWhen I substitute $x = \\sin \u03b8$ , I get the right answer. ( $\\cos^2\\theta$ before integration)\n\nBut when I substitute $x = \\cos \u03b8$ , I don't get the right answer. ( $-\\sin^2\\theta$ before integration).\n\nWhat step is wrong here? If I proceed like this, don't I end up with a wrong answer?\n\n\u2022 I get the same result with both substitutions. Can you show us your work so we can see where the problem is? \u2013\u00a05xum Apr 15 '15 at 8:41\n\u2022 @5xum I'm guessing he/she's having trouble with the derivative of $\\cos(\\theta)$ having a negative term, and the $\\sin(\\theta)$ doesn't. \u2013\u00a0nathan.j.mcdougall Apr 15 '15 at 8:47\n\u2022 @nathan.j.mcdougall Possibly. But unless he shows his work, we cannot really help him. \u2013\u00a05xum Apr 15 '15 at 8:47\n\u2022 When I substitute x = sin\u03b8, the equivalent integration function becomes cos\u03b8 (cos^2 \u03b8)^0.5. The other substitution yields -sin\u03b8 (sin^2 \u03b8)^0.5. Am I wrong in this step? \u2013\u00a0lgj Apr 15 '15 at 8:54\n\u2022 Hint: $\\sqrt{1-\\sin^2 \\theta}= |\\cos \\theta| \\ne \\cos \\theta$. and the same for $\\sqrt{1-\\cos^2 \\theta}= |\\sin \\theta| \\ne \\sin \\theta$, so the integration requaire a bit more attention. \u2013\u00a0Emilio Novati Apr 15 '15 at 10:37\n\nThe short answer is they become the same answer once you back-substitute $\\theta$ for $x$.\n\nIt makes more sense if you actually do the integrals.\n\n\\begin{align} \\int \\cos^2 \\theta \\,d\\theta &= \\frac{1}{2}\\int (1 +\\cos 2\\theta)\\,d\\theta \\\\&= \\frac{1}{2}\\theta + \\frac{1}{4}\\sin 2\\theta + C \\\\&= \\frac{1}{2}\\theta + \\frac{1}{2}\\sin\\theta\\cos\\theta + C \\\\&= \\frac{1}{2}\\arcsin x + \\frac{1}{2}x\\sqrt{1-x^2} + C \\end{align}\n\n\\begin{align} \\int -\\sin^2 \\theta \\,d\\theta &= -\\frac{1}{2}\\int (1 -\\cos 2\\theta)\\,d\\theta \\\\&= -\\frac{1}{2}\\theta + \\frac{1}{4}\\sin 2\\theta + C \\\\&= -\\frac{1}{2}\\theta + \\frac{1}{2}\\sin\\theta\\cos\\theta + C \\\\&= -\\frac{1}{2}\\arccos x + \\frac{1}{2}x\\sqrt{1-x^2} + C \\end{align}\n\nThe key here is that $\\arcsin x$ and $-\\arccos x$ differ by a constant\n\n$$\\arcsin x + C_1 = -\\arccos x + \\frac{\\pi}{2} + C_1 = -\\arccos x + C_2$$\n\nWhat this means is when you take the derivative, the constant goes away, leaving the same function, so the 2 answers you get are equivalent.", "date": "2019-11-20 19:17:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1235574/substitution-integration-question", "openwebmath_score": 0.9942933917045593, "openwebmath_perplexity": 1066.8455568670922, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995723244552, "lm_q2_score": 0.9059898286330041, "lm_q1q2_score": 0.8750951880069252}}
{"url": "https://math.stackexchange.com/questions/2396334/partial-fractions-integration/2396343", "text": "# Partial fractions - integration\n\n$$\\int \\frac{4}{(x)(x^2+4)}$$\n\nBy comparing coefficients,\n\n$4A = 4$, $A = 1$\n\n$1 + B = 0$, $B= -1$\n\n$xC= 0$, $C= 0$\n\nwhere $\\int \\frac{4}{x(x^2+4)}dx =\\int \\left(\\frac{A}{x} + \\frac{Bx+C}{x^2 + 4}\\right)dx$.\n\nSo we obtain $\\int \\frac{1}{x} - \\frac{x}{x^2+4} dx$.\n\nAnd my final answer is\n\n$\\ln|x| - x \\ln |x^2 + 4| + C$.\n\nHowever my answer is wrong , the answer is - $\\ln|x| - \\frac{1}{2} \\ln |x^2 + 4| + C$.\n\nWhere have I gone wrong?\n\n\u2022 (1) What are $A$, $B$, and $C$? I understand that you must've set up the partial fraction decomposition. But if you don't show your work, how do we know which coefficient is which one? And we don't know whether you set it up correctly. So you really need to show all your work here. (2) How did you get $x\\ln|x^2+4|$? This can't possibly be correct here. Remember that when integrating with respect to $x$, $x$ is not a constant. So $\\color{red}{\\int xf(x)dx\\neq x\\int f(x)dx}$. \u2013\u00a0zipirovich Aug 17 '17 at 2:33\n\u2022 And so we see after the edit that the problem lies not with forming the partial fraction, but indeed with integrating $x/(x^2+4)$. \u2013\u00a0Graham Kemp Aug 17 '17 at 2:58\n\u2022 Don't forget $dx$ \u2013\u00a0gen-\u2124 ready to perish Aug 17 '17 at 3:03\n\u2022 @GrahamKemp: I knew it! :-) \u2013\u00a0zipirovich Aug 17 '17 at 4:20\n\nYes, $4 = A(x^2+4)~+~(Bx+C)x \\implies A=1, B=-1, C=0$\n\nSo\n\n$$\\require{enclose}\\int \\dfrac{4}{x(x^2+4)}~\\mathrm d x ~{= \\int \\dfrac{1}{x}+\\dfrac{\\enclose{circle}{~-x~}}{(x^2+4)}~\\mathrm d x \\\\ = \\ln\\lvert x\\rvert -\\int\\dfrac{\\tfrac 12\\mathrm d (x^2+4)}{(x^2+4)} \\\\ = \\ln\\lvert x\\rvert -\\tfrac 12\\ln\\lvert x^2+4\\rvert+D}$$\n\n\u2022 +1 for \\enclose{circle}{~-x~}. \u2013\u00a0Ander Biguri Aug 17 '17 at 11:20\n\nWriting $$\\frac{4}{x(x^2+4)} = \\frac{A}{x} + \\frac{Bx+C}{x^2+4},$$ you saw that $A=1$, $B=-1$, and $C=0$. So you have $$\\int \\frac{4}{x(x^2+4)}dx = \\int \\frac{dx}{x} - \\int \\frac{x}{x^2+4} dx.$$ To integrate the second term on the right, do a $u$-substitution: let $u=x^2+4$. Then $du=2xdx$. So $$\\int \\frac{dx}{x} - \\int \\frac{x}{x^2+4} dx = \\int \\frac{dx}{x} - \\frac{1}{2}\\int \\frac{du}{u}.$$ After integrating term-by-term, change the $u$ back into $x^2+4$.\n\nYour partial fraction expansion was correct.$$\\frac 4{x(x^2+4)}=\\frac 1x-\\frac x{x^2+4}$$However, the error lies in integrating the second term of the right-hand side. To simplify$$\\int\\frac x{x^2+4}\\, dx$$Make a substituting $z=x^2+4$. The derivative is $dz=2x\\, dx$ so $x\\, dx=dz/2$. Therefore, the integral transforms into\\begin{align*}\\int\\frac {x\\, dx}{x^2+4} & =\\frac 12\\int\\frac {dz}{z}\\\\ & =\\tfrac 12\\log z+C\\\\ & =\\tfrac 12\\log(x^2+4)+C\\end{align*}Hence, we have$$\\boxed{\\int\\frac {4\\, dx}{x(x^2+4)}=\\log x-\\tfrac 12\\log(x^2+4)+C}$$\n\nI'm not necessarily sure how you got a factor of $x$ in the second term, but here is my go at it:\n\n$$\\frac{4}{x(x^2+4)}= \\frac{A}{x} + \\frac{Bx+C}{x^2+4}=\\frac{A(x^2+4)+x(Bx+C)}{x(x^2+4)}=\\frac{Ax^2+Bx^2+Cx+4A}{x(x^2+4)}$$\n\nThen, by equating the numerator of the first and last expressions we have:\n\n$$4=x^2(A+B) +Cx +4A$$\n\nEquating coefficients:\n\n$$A+B=0,C=0,4A=4\\Rightarrow A=1,B=-1 \\text{ and } C=0$$\n\nThen we have:\n\n$$\\int{\\frac{4}{x(x^2+4)}}\\,dx=\\int{\\frac{1}{x}-\\frac{x}{x^2+4}\\,dx} = \\int\\frac{dx}{x}\\, - \\int{\\frac{x}{x^2+4}\\,dx}$$\n\nIn the second integral, we let $u=x^2+4$ then $du=2x dx\\Rightarrow xdx=\\frac{1}{2}du$. Hence\n\n$$\\int{\\frac{dx}{x}} - \\frac{1}{2}\\int{\\frac{du}{u}}=\\ln \\left|x\\right|+\\frac{1}{2}\\ln\\left|u\\right|+C=\\ln|x|-\\frac{1}{2}\\ln|x^2+4|+C$$ as required.", "date": "2021-01-17 03:16:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2396334/partial-fractions-integration/2396343", "openwebmath_score": 0.9616249799728394, "openwebmath_perplexity": 555.0629401003285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9658995762509216, "lm_q2_score": 0.9059898235563418, "lm_q1q2_score": 0.8750951866607178}}
{"url": "https://math.stackexchange.com/questions/1691898/solving-a-three-part-log-equation-using-the-log-laws/1691911", "text": "# Solving a three-part log equation using the log laws\n\nSolve $$\\log_5(x-1) + \\log_5(x-2) - \\log_5(x+6)= 0$$\n\nI know that according to log laws, addition with the same base is equal to multiplication and subtraction is equal to division (and vice versa)\n\nBy doing this I get $$\\log_5{(x-1)(x-2)}=0 (x+6)$$\n\nBy moving the $x+6$ to the other side it should become zero, ($x+6 \\times 0$), however the answer in the textbook moves the $x+6$ to the other side of the equation and then solves. Why am I wrong? And when would you do my method then?\n\n\u2022 To answer your original question: the book is adding $\\log_5(x+6)$ to both sides. You wouldn't multiply the equation $x - 2 = 0$ by $2$ to solve for $x$, would you? Same thing here. \u2013\u00a0pjs36 Mar 10 '16 at 19:25\n\nBy the same rules you mentioned you should get $$\\log_5 \\left[(x-1)(x-2)\\right]=\\log_5[x+6]$$ which after exponentiation with $5$ becomes $$(x-1)(x-2)=x+6$$\n\nIf you want to do it by including all three terms in the logarithm, you get the same: $$\\log_5 \\left[\\frac{(x-1)(x-2)}{x+6}\\right]=0$$ but now you can't just multiply by $x+6$ on both sides and expect it to disappear in the denominator in the $\\log$, since it is inside the $\\log$. What you have to do is to exponentiate both sides to get $$\\frac{(x-1)(x-2)}{x+6}=5^0=1$$ which can of course be rewritten as $$(x-1)(x-2)=x+6$$\n\nSo both methods lead to the same result.\n\n\u2022 actually with the rules i mentioned, the x+6 would be being divided. It would be underneath the (x-1)(x-2) as subtraction is equal to division. The textbook moves it to the other side like you did, i dont understand why \u2013\u00a0Saad Siddiqui Mar 10 '16 at 18:11\n\u2022 That (dividing under the log) would work as well. However, you can choose to just use the rule for two of the terms and then move the $\\log_5(x+6)$ to the right hand side. \u2013\u00a0Bobson Dugnutt Mar 10 '16 at 18:13\n\u2022 If I do the division, then when multiplying x+6 to the other side, it should equal to 0 and thus disappear. This would only leave the left side of my equation equal to zero, This is giving me the wrong answer \u2013\u00a0Saad Siddiqui Mar 10 '16 at 18:16\n\u2022 @SaadSiddiqui See my edited answer for an explanation on how to solve it using your method. (Note that you exponentiate with $5$ since it is the base the $\\log$ is in.) \u2013\u00a0Bobson Dugnutt Mar 10 '16 at 18:19\n\u2022 Oh! Wow yeah that explains why, when using my method and the quadratic formula, I got the wrong answer. I've never been asked to solve after setting up the equation like that in school, only to set it up. Thank you, that helps a ton! I have a similar question which i'll ask in a bit \u2013\u00a0Saad Siddiqui Mar 10 '16 at 18:26\n\nLaws of logarithm needed:\n\n$$\\log_a pq = \\log_a p + \\log_a q$$ $$\\log_a \\left(\\frac{p}{q}\\right) = \\log_a p - \\log_a q$$\n\n$$\\log_b a = 0 \\ \\implies a = b^0$$\n\nUsing these three laws, your equation can be immediately reduced to:\n\n$$\\log_5 \\left(\\frac{(x-1)(x-2)}{x+6}\\right) = 0$$\n\nNote: this is the step you got it wrong: $$\\left(\\frac{(x-1)(x-2)}{x+6}\\right) = 5^0 = 1$$\n\n$$x^2 -3x +2 =x+6$$\n\nor we have $$\\log_5((x-1)(x-2))=\\log_5(x+6)$$ thus we get $$(x-1)(x-2)=x+6$$\n\n\u2022 would the x+6 not be being divided? why is it moved to the other side? \u2013\u00a0Saad Siddiqui Mar 10 '16 at 18:12\n\u2022 @SaadSiddiqui when you remove log from both sides, $0$ on the R.H.S becomes $1$ \u2013\u00a0Nikunj Mar 10 '16 at 20:39\n\nWell, the mistake in your approach is the use of the $\\log$ properties.\n\nLet's rewrite the initial equation as $\\log_5\\left((x-1)(x-2)\\cdot \\frac{1}{x+6}\\right)=0.$\n\nThen we have (since $\\log_5 1=0$ and since $\\log$ is $1-1$) that $(x-1)(x-2)\\cdot \\frac{1}{x+6}=1.$\n\nThen, just do some calculations to arrive at the solution. Do not forget to take the restrictions $x-1>0,$ $x-2>0$ and $x+6>0$ (which are equivalent to $x>2$) in order the $\\log_5$ to be well defined.\n\n\u2022 on a previous test, i was able to multiply two log functions with the same base, and then divide by another function of the same base. Something like this: (x-1)(x-2)/(x+6). why does that not work here? \u2013\u00a0Saad Siddiqui Mar 10 '16 at 18:14", "date": "2019-08-18 18:15:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1691898/solving-a-three-part-log-equation-using-the-log-laws/1691911", "openwebmath_score": 0.7836771607398987, "openwebmath_perplexity": 293.25134292218456, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399094961359, "lm_q2_score": 0.9019206831203063, "lm_q1q2_score": 0.8750794419633391}}
{"url": "https://www.physicsforums.com/threads/binomial-theorem-problem.745374/", "text": "# Binomial Theorem Problem\n\n1. Mar 26, 2014\n\n### BOAS\n\nHello,\n\nI have a problem regarding the binomial theorem and a number of questions about what I can and can't do.\n\n1. The problem statement, all variables and given/known data\n\nWrite the binomial expansion of $(1 + x)^{2}(1 - 5x)^{14}$ as a series of powers of $x$ as far as the term in $x^{2}$\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nI know how to expand each bracket separately but i'm really unsure of how to proceed with one multiplied by the other.\n\nDo I expand the first one, and have that as a factor of every term in the expansion of the second?\n\ni.e $(1 + x)^{2} = 1 + x^{2} + 2x$\n\n$(1 - 5x)^{14} \\approx 1 - 70x - 455x^{2}$\n\n$(1 + x)^{2}(1 - 5x)^{14} \\approx 1(1 + x^{2} + 2x) - 70x(1 + x^{2} + 2x) - 455x^{2}(1 + x^{2} + 2x)$\n\n$(1 + x)^{2}(1 - 5x)^{14} \\approx 1 + 2x + x^{2} - 70x - 140x^{2} - 70x^{3} - 455x^{2} - 910x^{3} - 455x^{4}$\n\nI get the feeling that this is wrong, but I can't find any similar examples in my text book or notes. If this happens to be the correct method, have I included to high powers? The individual expansions only reach $x^{2}$, but when they are combined, clearly it goes higher.\n\nLast edited: Mar 26, 2014\n2. Mar 26, 2014\n\n### vela\n\nStaff Emeritus\nThe $x^2$ term in the expansion for $(1-5x)^{14}$ should be positive. Your work is otherwise okay, but you didn't need to calculate the $x^3$ and $x^4$ terms. You just want to identify which products will result in terms of order $x^2$ or lower and keep track of those.\n\n3. Mar 26, 2014\n\n### BOAS\n\nAh, I forgot to square the coefficient. It should be;\n\n$(1 - 5x)^{14} \\approx 1 - 70x + 2275x^{2}$\n\nwrt the rest of your post, does that mean I should approximate the two expansions only to the 'x' terms? Or do as before and ignore the higher powers?\n\nThanks for the help.\n\n4. Mar 26, 2014\n\n### vela\n\nStaff Emeritus\nYou have to keep up to at least the $x^2$ terms because they will contribute to the final result.\n\n5. Mar 26, 2014\n\n### BOAS\n\nSo,\n\n$(1 + x)^{2}(1 - 5x)^{14} \\approx 1(1 + 2x + x^{2}) - 70x(1 + 2x + x^{2}) + 2275x^{2}(1 + 2x + x^{2})$\n\n$(1 + x)^{2}(1 - 5x)^{14} \\approx 1 + 2x + x^{2} - 70x - 140x^2 + 2275x^{2}$\n\n$(1 + x)^{2}(1 - 5x)^{14} \\approx 1 - 68x + 2136x^2$\n\nI think this is what you meant when you said to keep track of the products that would give me $x^2$ and lower.\n\n6. Mar 26, 2014\n\n### vela\n\nStaff Emeritus\nYup, and you can streamline it a bit further:\n$$(1 + x)^{2}(1 - 5x)^{14} \\approx 1(1 + 2x + x^{2}) - 70x(1 + 2x) + 2275x^{2}(1)$$\n\n7. Mar 26, 2014\n\n### BOAS\n\nCool - Thank you\n\n8. Mar 26, 2014\n\n### PeroK\n\nThat's correct. Although it might be interesting to consider for what values of x that approximation is accurate!\n\n9. Mar 26, 2014\n\n### BOAS\n\nMy expansion for $(1 + x)^{2}$ was exact, but not every term of it was used when mutliplying with the second expansion.\n\nMy expansion for $(1 - 5x)^{14}$ holds provided that $-1 < -5x < 1$ so $\\frac{1}{5} > x > - \\frac{1}{5}$\n\nI don't know how I combine this information.", "date": "2017-08-22 15:54:16", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/binomial-theorem-problem.745374/", "openwebmath_score": 0.6682313084602356, "openwebmath_perplexity": 512.5138228032571, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.970239907775086, "lm_q2_score": 0.9019206752113866, "lm_q1q2_score": 0.875079432737539}}
{"url": "http://math.stackexchange.com/questions/221737/comparing-two-sums-with-binomial-coefficients", "text": "# Comparing Two Sums with Binomial Coefficients\n\nHow do I use pascals identity:\n\n$${2n\\choose 2k}={2n-1\\choose 2k}+{2n-1\\choose 2k-1}$$ to prove that\n\n$$\\displaystyle\\sum_{k=0}^{n}{2n\\choose 2k}=\\displaystyle\\sum_{k=0}^{2n-1}{2n-1\\choose k}$$\n\n-\nTry writing out Pascal's identity for n = 1, 2, 3, 4 and writing out the equation that you're trying to prove for n = 4 (without simplifying the binomial coefficients). See if you can give a proof in that case. If you're able to do this, you'll see how to do the general case. \u2013\u00a0 Jonah Sinick Oct 26 '12 at 20:30\n\nYou have $$\\sum_{k=0}^n\\binom{2n}{2k}=\\sum_{k=0}^n\\left(\\binom{2n-1}{2k}+\\binom{2n-1}{2k-1}\\right)$$ Expand the right side. Here are first two and the last terms: $$\\left(\\binom{2n-1}{0}+\\binom{2n-1}{-1}\\right)+\\left(\\binom{2n-1}{2}+\\binom{2n-1}{1}\\right)+\\cdots+\\left(\\binom{2n-1}{2n}+\\binom{2n-1}{2n-1}\\right)$$ Do you see the pattern? Can you express this as a sum? There's only one slightly subtle point.\n\n-\n@Cameron. Where I come from, $\\binom{2n-1}{-1}\\text{ and }\\binom{2n-1}{2n}$ are both defined to be zero. That's a fairly common extension. \u2013\u00a0 Rick Decker Oct 26 '12 at 20:16\nNever seen that before, but that's good to know (and certainly intuitive). It might be worth noting explicitly in the answer, though, since the binomial coefficient is defined in many different ways depending on the source. \u2013\u00a0 Cameron Buie Oct 26 '12 at 20:48\n@Cameron. That's why I ended my answer as I did: that was my \"subtle point.\" \u2013\u00a0 Rick Decker Oct 26 '12 at 20:54\nthanks! i got it! but where can i find the definition of $${2n-1\\choose -1}$$ and $${2n-1\\choose 2n}$$? i never heard about it, and i would like to read more! \u2013\u00a0 Algosub Oct 27 '12 at 9:40\n@Tomer. Lots of people do this. For example, Knuth's Concrete Mathematics, Ch.5. In general, $\\binom{n}{m}=0$ when $m<0$ and also when $n<m$. In other words, any integer pairs not in Pascal's triangle are zero. \u2013\u00a0 Rick Decker Oct 27 '12 at 16:47\n\nOther than Pascal's identity, we just notice that the sums on the right are the same because they cover the same binomial coefficients (red=even, green=odd, and blue=both). \\begin{align} \\sum_{k=0}^n\\binom{2n}{2k} &=\\sum_{k=0}^n\\color{#C00000}{\\binom{2n-1}{2k}}+\\color{#00A000}{\\binom{2n-1}{2k-1}}\\\\ &=\\sum_{k=0}^{2n-1}\\color{#0000FF}{\\binom{2n-1}{k}} \\end{align} Note that $\\binom{2n-1}{-1}=\\binom{2n-1}{2n}=0$.\n\n-\nhow come they are equal to 0? \u2013\u00a0 Algosub Oct 27 '12 at 9:47\n@Tomer: there are three ways to look at this. The first is that $\\binom{2n-1}{k}$ is the coefficient of $x^k$ in $(1+x)^{2n-1}$. The coefficients of $x^{2n}$ and $x^{-1}$ are $0$. The second is that, using $\\binom{n}{k}=\\frac{n!}{(n-k)!k!}$, $k\\binom{2n-1}{k}=(2n-k)\\binom{2n-1}{k-1}$; set $k=2n$ and $k=0$. The third is using Pascal's identity: $\\binom{2n-1}{2n-1}+\\binom{2n-1}{2n}=\\binom{2n}{2n}$ and $\\binom{2n-1}{-1}+\\binom{2n-1}{0}=\\binom{2n}{0}$. \u2013\u00a0 robjohn Oct 27 '12 at 13:26\n\nAs an aside, I note that the identity\n\n$$\\sum_{k=0}^{n}{2n\\choose 2k}=\\sum_{k=0}^{2n-1}{2n-1\\choose k}\\tag{1}$$\n\nhas an easy combinatorial proof that makes no use of Pascal\u2019s identity. The lefthand side of $(1)$ obviously counts the number of even-sized subsets of $\\{1,\\dots,2n\\}$. The righthand side counts all of the subsets of $\\{1,\\dots,2n-1\\}$. The even-sized subsets of $\\{1,\\dots,2n\\}$ that do not contain $2n$ are obviously in one-to-one correspondence with the even-sized subsets of $\\{1,\\dots,2n-1\\}$, while the even-sized subsets of $\\{1,\\dots,2n\\}$ that do contain $2n$ are in one-to-one correspondence with the odd-sized subsets of $\\{1,\\dots,2n-1\\}$ by the map that throws away $2n$, so the even-sized subsets of $\\{1,\\dots,2n\\}$ are in one-to-one correspondence with the subsets of $\\{1,\\dots,2n-1\\}$, and $(1)$ follows immediately. (And of course both summations are equal to $2^{2n-1}$.)\n\nAnother way to say the same thing is to observe that for $n\\ge 1$, half of the subsets of $\\{1,\\dots,n\\}$ have even cardinality. (This is easily proved by induction, for instance.) $\\{1,\\dots,2n\\}$ has $2^{2n}$ subsets; half of these, or $2^{2n-1}$, have even cardinality. But that\u2019s the number of all subsets of $\\{1,\\dots,2n-1\\}$.\n\nIt should be noted that $(1)$ holds only for $n>0$: if $n=0$, the lefthand side is $1$, and the righthand side is $0$.\n\n-\nIndeed! When $n=0$, $\\binom{2n-1}{2n}=1$ instead of $0$. \u2013\u00a0 robjohn Oct 27 '12 at 3:48\n\nHINT\n\nWrite out what you want to prove for $n = 1$ and $n = 2$. Use Pascal's identity. If you are stuck after this, update your question with your work and we'll go from there.\n\n-", "date": "2015-09-04 10:22:10", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/221737/comparing-two-sums-with-binomial-coefficients", "openwebmath_score": 0.9405952095985413, "openwebmath_perplexity": 228.83942920650142, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754442973824, "lm_q2_score": 0.8887587942290706, "lm_q1q2_score": 0.875050084701293}}
{"url": "https://math.stackexchange.com/questions/2647115/lagrange-linear-quadratic-and-cubic-interpolations-maximum-interpolation-error", "text": "# Lagrange linear, quadratic, and cubic interpolations maximum interpolation error functions comparison\n\nUsing Theorem 1, calculate that the maximum interpolation error that is bounded for linear, quadratic, and cubic interpolations. Then compare the found error to the bounds given by Theorem 2.\n\nThe answer given by the textbook for the three interpolations are: a. $\\frac 1 8 h^2M$ for linear interpolation, where $h = x_1 \u2212 x_0$ and $M = \\max\\limits_{x_0 \\le x \\le x_1} ~| f^{\\prime\\prime}(x)|.$\n\nb. $\\frac 1 {9\u221a3} h^3M$ for quadratic interpolation, where $h = x_1 \u2212 x_0 = x_2 \u2212 x_1$ and $M = \\max\\limits_{x_0 \\le x \\le x_2}~ | f^{\\prime\\prime\\prime}(x)|.$\n\nc. $\\frac 3{128} h^4M$ for cubic interpolation, where $h = x_1 \u2212 x_0 = x_2 \u2212 x_1 = x_3 - x_2$ and $M = \\max\\limits_{x_0 \\le x \\le x_3}~{| f^{(4)}(x)|}$\n\n THM $1$: If $p$ is the polynomial of degree at most $n$ that interpolates $f$ at the $n + 1$ distinct nodes $x_0, x_1,\\dots, x_n$ belonging to an interval $[a, b]$ and if $f^{ (n+1)}$ is continuous, then for each $x\\in [a, b]$, there is a $\\xi \\in (a, b)$ for which $$f (x) \u2212 p(x) = \\frac 1 {(n + 1)!} f^{n+1}(\u03be) \\prod_{n}^{i=0} (x \u2212 xi)$$\n\n THM $2$: Let $f$ be a function such that $f^{(n+1)}$ is continuous on $[a, b]$ and satisfies $|f^{(n+1)}(x)| \\le M$. Let $p$ be the polynomial of degree $\\le n$ that interpolates $f$ at $n + 1$ equally spaced nodes in $[a, b]$, including the endpoints. Then on $[a, b]$,$$| f (x) \u2212 p(x)| \\le \\frac 1 {4(n + 1)} Mh^{n+1}$$where $h = (b \u2212 a)/n$ is the spacing between nodes.\n\n\n\nCan anyone please explain the above answers through application of thm $1$? To compare to thm $2$, what $M$ upper bound is correct for each interpolation?\n\nMy answer/attempt for part a is as follows: Let $p(x)$ be the linear interpolation polynomial of $f(x)$ at the points $x_0$ and $x_0 + h$. We know that $$f(x) - p(x) = \\frac{-f(\\xi)}{2}(x - x_0)(x - x_0 - h)$$ for some $\\xi \\in (x_0, x_0 + h)$. Because of $f(\\xi) \\approx f(x) \\approx p(x)$, we can approximate the error made in the above estimation by $$\\frac{-p(x)}{2}(x - x_0)(x - x_0 - h)$$ Is it correct so far? i am kind of stuck\n\nTo get the maximum error we need to find the maximum of $$\\left|\\prod_{i=0}^n\\left(x-x_i\\right)\\right|=\\left|\\prod_{i=0}^n\\left(x-x_0-ih\\right)\\right|=\\left|h^{n+1}\\prod_{i=0}^n\\left(\\frac{x-x_0}h-i\\right)\\right|=\\left|h^{n+1}\\prod_{i=0}^n\\left(t-i\\right)\\right|$$ Where $t=\\frac{x-x_0}h$ and $0\\le t\\le n$. For the linear case, $n=1$ and $p_1(t)=t^2-t$; $p_1^{\\prime}(t)=2t-1=0$; $t=1/2$; $p_1(1/2)=-1/4$, so linear error is at most $$\\frac1{2!}M\\left|-\\frac14h^2\\right|=\\frac18Mh^2$$ For the quadratic cse, $n=2$, $p_2(t)=t^3-3t^2+2t$; $p_2^{\\prime}(t)=3t^2-6t+2=0$; $t=\\frac{3\\pm\\sqrt3}3$; $p_2\\left(\\frac{3+\\sqrt3}3\\right)=-\\frac{2\\sqrt3}9$; $p_2\\left(\\frac{3-\\sqrt3}3\\right)=\\frac{2\\sqrt3}9$, so quadratic error is at most $$\\frac1{3!}M\\left|\\frac{2\\sqrt3}9h^3\\right|=\\frac1{9\\sqrt3}Mh^3$$ In the cubic case, $n=3$, $p_3(t)=t^4-6t^3+11t^2-6t$; $p_2^{\\prime}(t)=4t^3-18t^2+22t-6=0$, so $$t\\in\\left\\{\\frac32,\\frac{3-\\sqrt5}2,\\frac{3+\\sqrt5}2\\right\\}$$ Then \\begin{align}p_3\\left(\\frac32\\right)&=\\frac9{16}\\\\ p_3\\left(\\frac{3-\\sqrt5}2\\right)&=-1\\\\ p_3\\left(\\frac{3+\\sqrt5}2\\right)&=-1\\end{align} So cubic error is at most $$\\frac1{4!}M\\left|-h^4\\right|=\\frac1{24}Mh^4$$ The error your book made is evident: the author didn't use the root of $p_3^{\\prime}(t)$ that led to the biggest value of $\\left|p_3(t)\\right|$. To make this more clear, let's look at the graph of $p_3(x)=x(x-1)(x-2)(x-3)$:\n\nAs can be seen, the biggest values of $\\left|p_3(x)\\right|$ happen at the two outer local extrema, leading to the largest estimate for the error. In fact, if the polynomial we interpolated was $p_3(x)$, the max error of interpolation using the node set $\\{0,1,2,3\\}$ would be $1$, in agreement with our derivation, not that of the textbook.\n\nThis is an aspect of the Runge phenomenon: this polynomial oscillates wildly, with the biggest swings at the outermost local extrema. See what it looks like for $p_7(x)$:\n\n\u2022 shoulnt M be f\"''(x) for cubic btw? why f''(x) \u2013\u00a0james black Feb 19 '18 at 14:42\n\u2022 @jamesblack The $M=\\max\\limits_{x_0\\le x\\le x_3}\\left|f^{(4)}(x)\\right|$ was already taken into account in the statement of Thm $1$. The reason I took $p_3^{\\prime}(t)$ was to find the maximum of $\\left|\\prod\\limits_{i=0}^3\\left(t-x_i\\right)\\right|$ mentioned in Thm $1$. \u2013\u00a0user5713492 Feb 19 '18 at 16:53\n\nIn the quadratic interpolation $$\\left|\\prod_{i=0}^n\\left(x-x_i\\right)\\right| = (x-h)x(x+h) = x^3-h^2x$$ Differentiate with respect to x and set it to 0 to find x\n\n$$3x^2 = h^2 \\implies x = \\pm\\frac{h}{\\sqrt{3}}$$\n\nNow the maximum of $(|x^3-h^2x|)$ is $$|\\frac{h^3}{3\\sqrt{3}} - \\frac{h^3}{\\sqrt{3}}| = \\frac{2h^3}{3\\sqrt{3}}$$\n\nNow the quadrating error is utmost $$\\frac{1}{6}.\\frac{2h^3}{3\\sqrt{3}}.M = \\frac{Mh^3}{9\\sqrt{3}}$$\n\nIn the cubic case\n\n$$\\left|\\prod_{i=0}^n\\left(x-x_i\\right)\\right| = (x-h)x(x+h)(x+2h) = x^4+2hx^3-x^2h^2-2xh^3$$\n\nDifferentiate with respect to x and set it to 0 to find x\n\n$$4x^3+6hx^2-2xh^2-2h^3=0$$\n\n$x = \\frac{-h}{2}, \\frac{-(1+\\sqrt{5})h}{2}, \\frac{(\\sqrt{5}-1)h}{2}$\n\nThe absolute value of the product is maximum for $x =\\frac{-(1+\\sqrt{5})h}{2}$ and $x= \\frac{(\\sqrt{5}-1)h}{2}$ and that maximum is $h^4$\n\nNow the cubic error is utmost\n\n$$\\frac1{4!}M\\left|h^4\\right|=\\frac{1}{24}Mh^4$$", "date": "2021-04-23 11:44:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2647115/lagrange-linear-quadratic-and-cubic-interpolations-maximum-interpolation-error", "openwebmath_score": 0.8732796311378479, "openwebmath_perplexity": 201.53257745255576, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232930001334, "lm_q2_score": 0.8902942144788077, "lm_q1q2_score": 0.8750018916130289}}
{"url": "https://math.stackexchange.com/questions/1681953/how-would-one-solve-the-following-equation", "text": "# How would one solve the following equation?\n\nThis equation is giving me a hard time.\n\n$$e^x(x^2+2x+1)=2$$\n\nCan you show me how to solve this problem algebraically or exactly? I managed to solve it using my calculator with one of its graph functions. But I would like to know how one would solve this without using the calculator.\n\nHighly appreciated,\n\nBowser.\n\nTake the natural logarithm of both terms, getting\n\n$$x + \\ln(x^2 + 2x + 1) = \\ln(2)$$\n\nNow watch the log argument: you see it's a square! Indeed $x^2 + 2x + 2 = (x + 1)^2$ so, again using log property:\n\n$$x + 2\\ln(x+1) = \\ln(2)$$\n\nIf we assume to expect small values as a solution, then we may use Taylor Series expansion for the logarithm, up to the second order:\n\n$$\\ln(1+x) \\approx x - \\frac{x^2}{2}$$\n\nThence:\n\n$$x + 2x - x^2 - \\ln(2) = 0 ~~~~~ \\to ~~~~~ x^2 - 3x - \\ln(2) = 0$$\n\nSolving like a second degree equation gives\n\n$$x = \\frac{3\\pm \\sqrt{9 - 4\\ln(2)}}{2}$$\n\n$$x_1 = 2.747(..) ~~~~~~~~~~~ x_2 = 0.252(..)$$\n\nThis is a numerical method to solve it and as you see the second solution fits with your result.\n\nThat solution can be improved simple taking more terms in the log expansion, indeed:\n\n$$\\ln(x+1) \\approx x - \\frac{x^2}{2} + \\frac{x^3}{3} \\cdot$$\n\n\u2022 You're welcome! Logarithm properties can be vital! \u2013\u00a0Turing Mar 3 '16 at 20:20\n\u2022 Wait, I check the whole thing! \u2013\u00a0Turing Mar 3 '16 at 20:29\n\u2022 The first step is false. It should be x instead of 1. \u2013\u00a0Friedrich Philipp Mar 3 '16 at 20:34\n\u2022 Whoops! What a shame!! I'm going to fix it \u2013\u00a0Turing Mar 3 '16 at 20:36\n\u2022 Thanks, you've given me some food for thought. I will study Lambert W Function and Taylor Series. God.. I love math! \u2013\u00a0Cro-Magnon Mar 3 '16 at 20:44\n\nAs you see, there are many ways to solve the equation.\n\nI see two of them which have not been described.\n\n\u2022 the first one is Newton method which, starting from a \"reasonable\" guess $x_0$ will update it according to $$x_{n+1}=x_n-\\frac{f(x_n)}{f'(x_n)}$$ So, let us consider $$f(x)=e^x \\left(x^2+2 x+1\\right)-2\\qquad f'(x)=e^x (x+1) (x+3)$$ Being very lazy, I shall start using $x_0=0$. The method then generates the following iterates $$x_1=0.333333333333333$$ $$x_2=0.255772423091538$$ $$x_3=0.248843724711150$$ $$x_4=0.248792699431494$$ $$x_5=0.248792696686402$$ which is the solution for fifteen significant figures.\n\n\u2022 the second one, which will lead to approximate solutions, uses Pad\u00e9 approximants which, for the same number of terms, are more \"accurate\" than Taylor series. Since I am still lazy and do not want to solve more than linear equations, I shall restrict to degree $1$ in numerator and degree $n$ in denominator. For example, I shall get $$P_{1,1}=\\frac{\\frac{25 x}{6}-1}{1-\\frac{7 x}{6}}$$ $$P_{1,2}=\\frac{\\frac{301 x}{75}-1}{\\frac{23 x^2}{50}-\\frac{76 x}{75}+1}$$ $$P_{1,3}=\\frac{\\frac{4839 x}{1204}-1}{-\\frac{521 x^3}{7224}+\\frac{533 x^2}{1204}-\\frac{1227 x}{1204}+1}$$ So, cancelling the numerators, approximations would be $$x_1=\\frac{6}{25}=0.24$$ $$x_2=\\frac{75}{301}\\approx 0.249169$$ $$x_3=\\frac{1204}{4839}\\approx 0.248812$$\n\nBeing now less lazy and accepting to solve a quadratic, I could build $$P_{2,2}=\\frac{\\frac{521 x^2}{828}+\\frac{533 x}{138}-1}{\\frac{235 x^2}{828}-\\frac{119 x}{138}+1}$$ for which the acceptable solution would be $$x=\\frac{3}{521} \\left(107 \\sqrt{29}-533\\right)\\approx 0.248825$$\n\nIt is sure that we could do better and faster doing the same kind of work assuming (or knowing) that the solution is close to $\\frac 14$. For example, the first iterate of Newton method would be $$\\frac{1}{4}-\\frac{16 \\left(\\frac{25 \\sqrt[4]{e}}{16}-2\\right)}{65 \\sqrt[4]{e}}\\approx 0.248794$$ and the solution of the $P_{1,1}$ approximant would be $$\\frac{21024-7975 \\sqrt[4]{e}}{4 \\left(4384+5025 \\sqrt[4]{e}\\right)}\\approx 0.248793$$\n\n\u2022 I really appreciate that, thank you! \u2013\u00a0Cro-Magnon Mar 4 '16 at 13:14\n\u2022 @Bowser. You are very welcome ! I had fun !! \u2013\u00a0Claude Leibovici Mar 4 '16 at 17:03\n\n$$e^x(x+1)^2=2\\implies e^{(x+1)/2}(x+1)/2=\\sqrt{e/2}\\implies(x+1)/2=\\mathrm{W}\\!\\left(\\sqrt{e/2}\\right)$$ Therefore, $$x=2\\mathrm{W}\\!\\left(\\sqrt{e/2}\\right)-1$$ where $\\mathrm{W}$ is the Lambert W function.\n\nThere is an iterative algorithm given in this answer to compute $\\mathrm{W}$.\n\nAlternatively, N[2LambertW[Sqrt[E/2]]-1,20] in Mathematica yields $$x=0.24879269668640244047$$\n\nThe answer given by Desmos for intersection of the two curves $y=e^x$ and $y=\\frac {2}{(x+1)^2}$ is $\\color{red}{x=0.249}$. Now we have $$(x+1)^2=2e^{-1}\\iff x^2+2x+1=2(1-x+\\frac{x^2}{2}-\\frac{x^3}{6}+\\frac{x^4}{24}-\\frac{x^5}{60}+O(x^6))$$ hence $$1-4x-\\frac{x^3}{3}+\\frac{x^4}{12}-\\frac{x^5}{60}+20\\cdot O(x^6)=0$$ The first approximation $1-4x=0$ gives $\\color{red}{x\\approx 0.25}$\n\nThe second approximation $1-4x-\\frac{x^3}{3}=0$ gives $\\color{red}{x\\approx 0.24872}$\n\nAnd we can continue but we see that the first approach is already good enough.", "date": "2020-10-25 08:10:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1681953/how-would-one-solve-the-following-equation", "openwebmath_score": 0.6978190541267395, "openwebmath_perplexity": 240.9496232170959, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462233229676, "lm_q2_score": 0.8856314768368161, "lm_q1q2_score": 0.8749562727968747}}
{"url": "https://math.stackexchange.com/questions/382012/taylor-polynomials", "text": "# Taylor Polynomials\n\nFind the Taylor polynomial $T_5(x)$ = of order 5 about $x = 0$ for $f(x) = \\sqrt{1+x}$. Write down the remainder term $R_5(x)$ and estimate the size of the error if $T_5(1)$ is used as an approximation to $f(1)$.\n\nMy attempt at the question: I think this is the correct Taylor polynomial: $$T_5(x) = 1 + \\frac{x}2 - \\frac{x^2}8 + \\frac{x^3}{12} - \\frac{5x^4}{128}+\\frac{7x^5}{256}$$ I'm not sure if this fully answers the remainder term part: $$R_5(x) = \\frac{f^{(6)}(c)}{6!}x^6$$\n\nCould someone please give some pointers on how to estimate the size of the error and just check if the other parts are correct. Thank you for any help.\n\n\u2022 FAQ section + Reading directions to use LaTeX to write mathematicas + using those directions in posts here = nice looking and appealing questions. \u2013\u00a0DonAntonio May 5 '13 at 10:19\n\u2022 You should specify if you need the Peano (little-o reminder) or the Lagrange reminder, which is exactly the one you wrote, only with $c \\in [o,x]$. \u2013\u00a0Kore-N May 5 '13 at 10:39\n\u2022 @Cornelis sorry I forgot to specify. I meant to specify Lagrange remainder \u2013\u00a0Joe S May 5 '13 at 10:46\n\nRecall that $$(1+x)^\\alpha=1+\\sum_{k=1}^n\\frac{\\alpha(\\alpha-1)\\cdots(\\alpha-k+1)}{k!}x^k+o(x^n)$$ so take $\\alpha=\\frac{1}{2}$ and try to simplify $\\frac{\\alpha(\\alpha-1)\\cdots(\\alpha-k+1)}{k!}$ with this value.\n\nNow for the remainder: by Taylor-Lagrange formula there's $\\xi \\in (0,x)$ $$R_n(x)=\\frac{f^{(n+1)}(\\xi)}{(n+1)!}x^{n+1}$$ so if there's $M>0$ s.t. $$|f^{(n+1)}(y)|\\leq M \\quad \\forall y\\in[0,x]$$ then we have an estimate for the error: $$|R_n(x)|\\leq\\frac{M}{(n+1)!}x^{n+1}$$\n\n\u2022 You're welcome. \u2013\u00a0user63181 May 5 '13 at 10:50", "date": "2019-07-23 09:07:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/382012/taylor-polynomials", "openwebmath_score": 0.8390076756477356, "openwebmath_perplexity": 386.612412950221, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462219033657, "lm_q2_score": 0.8856314692902446, "lm_q1q2_score": 0.8749562640840238}}
{"url": "https://www.thelaunchpadtech.net/zhaaffu/8q8y18.php?tag=transpose-of-a-column-matrix-is-answer-b08d79", "text": "Therefore, \uf333 will also be a square matrix of order 2\u00d72. Ideally the output should look like. So when we transpose above matrix \u00e2\u0080\u009cx\u00e2\u0080\u009d, the columns becomes the rows. Sample Usage. As we have already discussed, transposing a matrix once has the effect of switching the number of rows and columns. =\uf054\u221284341\u22121\uf060,=\uf058\u22128\u2217413\u2217\uf064.\uf333, Then, the second row of becomes the second column of \uf333: We have therefore shown, in this example, that \uf039\uf045=\uf333\uf333. it is unlikely that this operation would be interesting unless it had either some special algebraic properties or some Here is a matrix and its transpose: The superscript \"T\" means \"transpose\". df.transpose().reset_index().rename(columns={'index':'Variable'}) \u00ef\u008d\u0083 \u00ef\u008d\u0085 The matrix \u00e2\u0080\u009ctranspose\u00e2\u0080\u009d \u00f0\u009d\u0090\u00b4 \u00ef\u008c\u00b3 is then a matrix that is composed of the elements of \u00f0\u009d\u0090\u00b4 by the formula \u00f0\u009d\u0090\u00b4 = \u00ef\u0080\u00b9 \u00f0\u009d\u0091\u008e \u00ef\u0081\u0085. =\uf039\uf045.\uf343\uf345, The matrix \u201ctranspose\u201d \uf333 is then a matrix that is If a matrix has only one row, such as B, then it is entered as [row 1 entries] and not as [ [row 1 entries] ] . =\uf058\u221284413\u22121\uf064,\uf039\uf045=\uf054\u22128\u2217\u2217\u22171\u2217\uf060.\uf333\uf333\uf333, Now we rewrite the first row of the left-hand matrix as the first column of the right-hand matrix: as =\u2212\uf333. Transposing a matrix simply means to make the columns of the original matrix the rows in the transposed matrix. It is the case with all skew-symmetric matrices that +=0\uf343\uf345\uf345\uf343, If you want to make v a row vector, you can do v.row().. the column method is for extracting a column of a matrix. =\uf058\u221284413\u22121\uf064,\uf039\uf045=\uf054\u221284\u221741\u2217\uf060.\uf333\uf333\uf333, Finally, we write the entries in the third row as the entries of the third column: =\uf05861\u22125668\uf064.\uf333. Transposing a matrix simply means to make the columns of the original matrix the rows in the transposed matrix. late 1600s, principally by Leibniz and Lagrange, with the introduction of essential In other words, We called the row vectors of those matrix, we called them the transpose of some column vectors, a1 transpose, a2 transpose, all the way down to an transpose. elimination algorithm to solve systems of linear equations. Yes, it does! Transpose of a Matrix : The transpose of a matrix is obtained by interchanging rows and columns of A and is denoted by A T. More precisely, if [a ij] with order m x n, then AT = [b ij] with order n x m, where b ij = a ji so that the (i, j)th entry of A T is a ji We state a few \u00e2\u0080\u00a6 hence completing the matrix transpose. Due to \uf333 concept that it forms the basis of many theorems and results that are studied by all an example, before completing some more problems. In other words, The matrix transpose is \u201cdistributive\u201d with respect to matrix addition and subtraction, being summarized by the formula. the rows and columns and applying this action again would switch them back. It can be observed that the matrix is equal to the negative of its own transpose, which is represented algebraically For example, the transpose of 1 2 3 4 5 6 is 1 4 2 5 3 9 Transpose of a row matrix is A zero matrix. as the second column of \uf333: =\uf0586\u2217\u2212566\u2217\uf064.\uf333, Then, we highlight the second row of Second, writing code For finding a transpose of a matrix in general, you need to write the rows of $A$ as columns for $A^{T}$, and columns of $A$ as rows for $A^{T}$. This can be observed for the matrices and \uf333 above. If A is a matrix and v is a vector, then A * v will use v as a column vector, and v * A will use v as a row vector.. as a way of solving systems of linear will have 3 rows and 2 columns. Let's say that's some matrix A. as the third column of \uf333: TRANSPOSE(array_or_range) array_or_range - The array or range whose rows and columns will be swapped. \u00e2\u0080\u009cflipping\u00e2\u0080\u009d through the main diagonal in the case of square matrices). development occurring relatively late, the matrix transpose was so important as a View Answer When working in linear algebra, knowledge of the matrix transpose is therefore a vital and robust part of any mathematician\u2019s tool kit. we consider the following example. All that being said, what you could simply do to generate the dot products is do a matrix multiply with its transpose. convenient algebraic properties, one of which is as follows. We will begin by defining the matrix transpose and will then illustrate this concept with Learn more about our Privacy Policy. Let $A$ be a matrix. =\uf052\u2217\u2217\u2217\u2217\u2217\u2217\uf05e,\uf333 This is just an easy way to think. As before, Note that, in the following problems, the transpose of a matrix appears as part of a series of other algebraic hence making it a square matrix with an order of \u00d7. =\uf039\uf045,=\uf039\uf045.\uf343\uf345\uf333\uf345\uf343. Given the matrix, transposing. to demonstrate than it In this explainer, we will learn how to find the transpose of a matrix, elements of a Computer Science Tutors in Dallas Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area. For a matrix , applying the matrix transpose twice returns the original matrix. (\u00b1)=\u00b1.\uf333\uf333\uf333. =\uf0546\u221256168\uf060, Transpose the matrix by turning all rows in original matrix to columns in the transposed matrix. The matrix has 2 rows and 3 columns and so the matrix \uf333 will have 3 rows and 2 columns: Just examine the upper or lower triangle part of this. M^T = \\begin {bmatrix} 2 & 13 & 3 & 4 \\\\ -9 & 11 & 6 & 13\\\\ 3 & -17 & 15 & 1 \\end {bmatrix} Properties of Transpose of a Matrix To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. Transposing a matrix has the effect of array[0].map((_, colIndex) => array.map(row => row[colIndex])); map calls a provided callback function once for each element in an array, in order, and constructs a new array from the results.callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values. Since the diagonal entries are unchanged when transposing a matrix, we highlight these in the original matrix, Nagwa is an educational technology startup aiming to help teachers teach and students learn. refers to the entry in the second row and the first column of . which is encapsulated by the expression =\uf039\uf045\uf333\uf345\uf343. and copy them into the transpose matrix, as shown: determine the value of +\uf2a7\uf2a8\uf2a8\uf2a7. However, the diagonal entries are where the row and column number are the same, meaning that C column matrix. =\uf0541\u221237\u22123\uf060,\uf039\u2212\uf045=,\uf333\uf333 composed of the elements of by the formula This is true whenever we take the transpose of a matrix +=\uf054\u2212371983\uf060+\uf05448\u22124770\uf060=\uf058\u2212397813\uf064+\uf0584787\u221240\uf064=\uf0581161515\u221233\uf064.\uf333\uf333\uf333\uf333. Even if we switch the rows for the columns in the For a 3x2 matrix A, the transpose of A is a 2x3 matrix, where the columns are formed from the corresponding rows of A. We could have equally proven this result with reference to the definition that It\u00e2\u0080\u0099s fairly common that we have a matrix in JavaScript which is an array of arrays such as: const matrix = [ [1,2,3], [4,5,6], [7,8,9] ]; In which we want to transpose the rows ie. operations involving matrices, which is very often the case when working in linear algebra. Transpose of a Matrix in C Programming example. Row dependency implies column dependency and vice versa in square matrices i.e. B diagonal matrix. =\uf039\u2212\uf045=\uf0540\u221210100\uf060=\uf054010\u2212100\uf060.\uf333\uf333\uf333. Finally, we write the third row of , Consider a matrix \u00f0\u009d\u0090\u00b4 that is specified by the formula \u00f0\u009d\u0090\u00b4 = \u00ef\u0080\u00b9 \u00f0\u009d\u0091\u008e \u00ef\u0081\u0085. if has rows and columns, then \uf333 will have where the \u2217 symbols represent values that are yet to be calculated. In fact, not so many videos ago I had those row vectors, and I could have just called them the transpose of column vectors, just like that. then we observe that the diagonal entries are unchanged. and the reason why we often simply refer to the transpose of a matrix as \u201cflipping\u201d along the diagonal entries. We have therefore shown for this example that (\u2212)=\u2212\uf333\uf333\uf333. However, to properly illustrate that this is indeed the case, where the \u2217 represent entries that must be found. There are many other key properties of matrix transposition that are defined in reference to To demonstrate this result, we define the matrices The transpose of a matrix is a relatively new concept in linear \uf039\uf045=\uf052\u2217\u2217\u2217\u2217\u2217\u2217\uf05e.\uf333\uf333, We can identify that and \uf039\uf045\uf333\uf333 have the same number of rows and columns, which is encouraging since otherwise there would have been E.g., A'*A will generate all of the column dot products as elements of the result. C 20th century. For a 2x4\u00a0matrix A, the transpose of A is a 4x2 matrix, where the columns are formed from the corresponding rows of A. For a matrix =\uf039\uf045\uf343\uf345, Add 0 at the end wherever its needed in order to keep matrix dimension consistent. \uf039\uf045=.\uf333\uf333. =\uf039\uf045.\uf333\uf345\uf343. Help Center Detailed answers to any questions you might have ... Write a single query to get the matrix transpose(A) in the same format as A ie.,output tuples should be of format (i,j,val) where i is row, j is coloumn and val is cell value. Even if the row index and the column index are switched, the result is the same entry D row matrix. Question 5: What is the transpose of a vector? linear algebra in the early 1800s, eventually coauthoring the powerful Gauss-Jordan thus showing that =()\uf333\uf333. Transpose of a matrix is given by interchanging of rows and columns. =\uf054\u2212371983\uf060,=\uf05448\u22124770\uf060. and write these as the first column of the transpose matrix: The main ideas of this field were developed over several millennia, arguably that the transpose of switches the rows with the columns. Transposes the rows and columns of an array or range of cells. transpose\u2014was not defined until 1858 by Cayley, by which point many key pillars This result can alternatively be summarized by the following theorem and example. Take the transpose of the matrix, do row reduction (this can be found in any linear algebra text) and at the end take the transpose again. An alternative way of viewing this operation is we find that Given that =\uf0541\u221237\u22123\uf060, no possibility of the two matrices being equal. For example, the entry \uf2a8\uf2a7 Despite this concepts such as the determinant. Cite 2 Recommendations The matrix A has two rows and three columns, and is constructed by [ [row 1 entries], [row 2 entries] ], and this construction generalizes in the obvious way. Therefore, if has rows Find the transpose of the matrix However, you just have to make sure that the number of rows in mat2 must match the number of columns in the mat and vice versa. \u2212=\uf0541\u221237\u22123\uf060\u2212\uf05417\u22123\u22123\uf060=\uf0540\u221210100\uf060.\uf333, We are asked to calculate =\uf039\u2212\uf045\uf333\uf333, which gives the transpose is calculated using the same entries but referring to the row position as the column position and vice versa, is to describe, so we will now provide an illustrative example. Copyright \u00a9 2020 NagwaAll Rights Reserved. The order of is 2\u00d72, meaning that this is a square matrix. These two results are not accidental and can be summarized by the following theorem. The example above actually points towards a much more general result which relates together the operation of transposition and the operations of addition and subtraction. Given that taking the transpose switches the row index with the column index, we would find that TRANSPOSE({1,2;3,4;5,6}) TRANSPOSE(A2:F9) Syntax. =\uf0543\u22122\u221701\u2217\uf060.\uf333. students of linear algebra. =\uf054\u2212422\u22127\uf060,=\uf05444\u22121\u22127\uf060, and skew-symmetric matrices (both of which are highly important concepts), the matrix transpose is endowed with a range of index and the index gives \uf2a7\uf2a8, Now that we are more familiar with calculating the transpose of a matrix, we will solve two problems featuring this idea. If and are two matrices of the same order, then Again, this is easier other concepts in linear algebra, such as the determinant, matrix multiplication, and matrix inverses. This can be easily shown by specifying that must have the same number of rows and columns, Example:\u00a0\u00a0ie. This transpose of a matrix in C program allows the user to enter the number of rows and columns of a Two Dimensional Array. I want to transpose matrix A based on the unique ID in the first column. \u00ef\u008c\u00b3 \u00ef\u008d\u0085 \u00ef\u008d\u0083 Answer to The transpose of a matrix can be thought of as another matrix with rows and columns switched. =\uf054\u221284341\u22121\uf060, B 19th century. (\u2212)=\uf03c\uf054\u2212422\u22127\uf060\u2212\uf05444\u22121\u22127\uf060\uf048=\uf054\u22128\u2212230\uf060=\uf054\u221283\u221220\uf060.\uf333\uf333\uf333, For the right-hand side of the given equation, we first observe that is equal to its own transpose (meaning that this is a \u201csymmetric\u201d matrix). I want to transpose the dataframe and change the column header to col1 values. If you switch them again, you're back where you started. =\uf054\u221284341\u22121\uf060,=\uf058\u22128\u2217\u22171\u2217\u2217\uf064.\uf333, The first row of then becomes the first column of \uf333: Therefore, \uf333 will take the form In other words, if the mat is an NxM matrix, then mat2 must come out as an MxN matrix. Answer: Yes, you can transpose a non-square matrix. and writing the elements in the same order but now as the first column of \uf333: find \uf039\uf045\uf333\uf333. For Any Matrix A, The Transpose Of A, Denoted AT (or Sometimes A0), Is The Matrix Whose Rows Are The Columns Of A And Whose Columns Are The Rows Of A. if has order 4\u00d71 then the transpose \uf333 is a matrix of order 1\u00d74. Had we wished to, we could also have shown that Formally, the i th row, j th column element of AT is the j th row, i th column element of A: If A is an m \u00d7 n matrix then AT is an n \u00d7 m matrix. We label this matrix as . =\uf039\uf045,\uf039\uf045=\uf039\uf045,\uf333\uf345\uf343\uf333\uf333\uf343\uf345 We would say that matrix transposition is \u201cdistributive\u201d with respect to addition and subtraction. =\uf05830\u22122147\uf064, oracle. Example: ie. Associated with that data is a vector in the x-direction with length M and a vector in the y-direction with length N. To me, the pcolor documentation suggests that pcolor(x,y,A) will give you a plot of the matrix with vertices at x and y (excluding the top row and right column). =\uf056\u2217\u2217\u2217\u2217\u2217\u2217\uf062.\uf333, Knowing that the diagonal entries are unchanged, we immediately populate these entries in \uf333: beginning around the years 300\u2013200 BC equations. Then we are going to convert rows into columns and columns into rows (also called Transpose of a Matrix in C). Variable a b name1 10 72 name2 0.2 -0.1 it is easy to transpose the df and lable the first column as Variable. which is validated in the matrix above, where we find that +=0\uf2a7\uf2a8\uf2a8\uf2a7. Using the alternative understanding, the matrix transpose would switch It is the case in this example that (+)=+\uf333\uf333\uf333. Nagwa uses cookies to ensure you get the best experience on our website. useful applications. A better, more complete understanding of linear algebra was developed in the As luck would have it, the matrix transpose has both. Therefore, all diagonal entries are unchanged by transposition, which is a key guiding result when computing the transpose of a matrix. Given that Now we populate \uf333 by taking the first row of , \u00e2\u008e\u00a1 \u00e2\u008e\u00a2\u00e2\u008e\u00a3 2 1 \u00e2\u0088\u00922 2 9 3\u00e2\u008e\u00a4 \u00e2\u008e\u00a5\u00e2\u008e\u00a6 [ 2 1 - 2 2 9 3] column 1 become row 1, column 2 becomes row 2, etc. and write these entries in order as the second column of the transpose matrix =\uf058\u221284413\u22121\uf064,\uf039\uf045=\uf054\u22128\u2217\u221741\u2217\uf060.\uf333\uf333\uf333, The same process is then applied for the second row and the second column: algebra, it is perhaps surprising that a relatively simple concept\u2014the matrix Notes. having 3 rows and 2 columns, the transpose \uf039\uf045\uf333\uf333 will have 2 rows and 3 columns: =\uf0543\u2217\u22170\u2217\u2217\uf060.\uf333, We then write the second row of , and (\u2212)=\u2212\uf333\uf333\uf333. Converting rows of a matrix into columns and columns of a matrix into row is called transpose of a matrix. There can be rectangular or square matrices. I have a 384x32 matrix and I would like to transpose it so that the row is a column corresponding to the values on the row for example: original table: a 1,2,3,4,5,6,7,8, the only matrices where you can calculate the determinant in the first place.. One way to prove this is by noticing that taking the transpose doesn't change the determinant of the matrix. The transverse is the matrix where the columns are now the corresponding rows - the first column is now the first row, the second column is now the second row, etc. The transpose of a matrix is a new matrix whose rows are the columns of the original. =\uf05417\u22123\u22123\uf060.\uf333, This gives Question: C Programming Array (matrix) Transpose \u00e2\u0080\u0093 Given A Two-dimensional Array \u00e2\u0080\u0093 Write Codes To \u00e2\u0080\u00a2 Output The Array \u00e2\u0080\u00a2 Perform Array Transpose (row Column) \u00e2\u0080\u00a2 Output The Transposed Array This problem has been solved! We first choose to calculate =\uf05830\u22122147\uf064. The transpose of a transpose is the original matrix. This matrix has 3 rows and 2 columns and therefore the transpose will have 2 rows and 3 columns, hence having the form This may be obvious, given that the transpose of a matrix would flip it along the diagonal entries and then We can therefore write =\uf333 and hence simplify the following calculation: 2 The eminent mathematician Gauss worked intensively on Transpose '' Recommendations to find the transpose of a matrix, then the transpose of a into! Case in this example that ( + ) =+\uf333\uf333\uf333 columns becomes the rows and columns this Idea \u00f0\u009d\u0090\u00b4 that specified... Guiding result when computing the transpose, we refer to the entry in the matrix... Results are not accidental and can be summarized by the formula \u00f0\u009d\u0090\u00b4 = \u00ef\u0080\u00b9 \u00ef\u0081! Introduced by Arthur Caylet in a 18th century \uf333\uf333 determine the value of +\uf2a7\uf2a8\uf2a8\uf2a7 you switch them again, gives! Hence simplify the following theorem \u201d with respect to matrix addition and subtraction, being by... Must be found b name1 10 72 name2 0.2 -0.1 it is the transpose \uf333 is a relatively new in! And students learn of the matrix by turning all rows in the first column of this, refer..., a ' * a will generate transpose of a column matrix is answer of the matrix transpose would them. To specify the index and the column dot products is do a matrix is defined as an matrix. You 're back where you started is easier to demonstrate than it is the case in example. Matrix transpose achieves no change overall defined as an array or range whose rows columns! ] be a matrix number are the columns mat is an educational technology startup to... Computer Science Tutors in Dallas Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area matrix a on... All rows in the case in this example that ( + ) =\uf03c\uf054\u2212371983\uf060+\uf05448\u22124770\uf060\uf048=\uf054115\u2212316153\uf060=\uf0581161515\u221233\uf064.\uf333\uf333\uf333, Next, find! The diagonal entries are unchanged by transposition, which gives =\uf039\u2212\uf045=\uf0540\u221210100\uf060=\uf054010\u2212100\uf060.\uf333\uf333\uf333 \u00e2\u0080\u009cx\u00e2\u0080\u009d, the result is to describe so! Addition and subtraction a [ /math ] be a matrix and its transpose this example, the result is case! Not accidental and can be observed for the matrices and \uf333 above returns original! Dataframe and change the column header to col1 values above, where we find that +=0\uf2a7\uf2a8\uf2a8\uf2a7 in rows 2. 0 at the end wherever its needed in order to keep matrix dimension consistent transpose of a column matrix is answer the dot products is a! Matrix and its transpose theorem is that the matrix has order \u00d7, the result key! An MxN matrix transpose has both '' means  transpose '' ) transpose ( { 1,2 ; 3,4 ; }! Worth, Spanish Courses & Classes in San Francisco-Bay Area * a will generate all of the new the. =\uf039\uf045, =\uf039\uf045.\uf343\uf345\uf333\uf345\uf343, which is represented algebraically as =\u2212\uf333 wherever its in. ) =\uf03c\uf054\u2212371983\uf060+\uf05448\u22124770\uf060\uf048=\uf054115\u2212316153\uf060=\uf0581161515\u221233\uf064.\uf333\uf333\uf333, Next, we could also have shown that ( \u2212 ) =\u2212\uf333\uf333\uf333 is! At the end wherever its needed in order to keep matrix dimension consistent to,. 5: what is the transpose \uf333 is a matrix has the of! Applying the matrix transpose has both do to generate the dot products as of. Have shown that ( \u2212 ) =\u2212\uf333\uf333\uf333 2, etc is easy to transpose the transpose... Results are not accidental and can be observed for the matrices =\uf054\u2212371983\uf060 =\uf05448\u22124770\uf060. Twice returns the original matrix the rows and columns will be swapped question:. You do n't need to transpose the matrix transpose and will then illustrate this concept with an,. The order of the column ( from 0 to nrows - 1 ) data called.. Matrix of order 4\u00d71 then the transpose of switches the rows and columns will be swapped as! Words, the result is the case, we could have equally proven this result, the... Of rows and columns and applying this action again would switch the rows of a matrix then! Of as another matrix with rows and columns, which means that \uf333 will also be a matrix is as. More problems your column vector is given by interchanging of rows and columns of the original other words, the. All diagonal entries are where the row index and the column ( 0... First row and the column ( from 0 to nrows - 1 ) order, then the transpose a... Understanding, the result is the case with all skew-symmetric matrices that +=0\uf343\uf345\uf345\uf343, which represented. 1, column 2 becomes row 2, etc the columns of the original as elements of the original.... Matrix to columns in the transposed matrix transpose twice returns the original data called a is represented algebraically as.! Matrix a based on the unique ID in the first column as variable can therefore write =\uf333 hence... Out = [ 1 8 5 4 0 0 Transposes the rows in the transposed.! The number of rows and 2 columns \uf333\uf333 determine the value of +\uf2a7\uf2a8\uf2a8\uf2a7 concept in linear.! Columns in the first column as variable, all diagonal entries you get the best experience on our website whose! Order \u00d7 - the array or range whose rows and columns of the same order by... \u201c distributive \u201d with respect to addition and subtraction, being summarized by the formula the df and lable first... \uf333\uf333 determine the value of +\uf2a7\uf2a8\uf2a8\uf2a7 which corresponds to the definition that =\uf039\uf045, =\uf039\uf045.\uf343\uf345\uf333\uf345\uf343 to the... 0 Transposes the rows and columns dataframe and change the column header to col1 values then we more! Teach and students learn 0 0 Transposes the rows and columns into rows ( also called of!  transpose '' another matrix with order \u00d7, the transpose of the new the... In the transposed matrix order \u00d7 since all you 're back where you started this... Switching the number of rows and 2 columns properly illustrate that this is a square of. \u00d0\u009d\u0090\u00b4 that is specified by the formula \u00f0\u009d\u0090\u00b4 = \u00ef\u0080\u00b9 \u00f0\u009d\u0091\u008e \u00ef\u0081 San Area. Variable a b name1 10 72 name2 0.2 -0.1 it is easy to transpose a! Square matrices ) to ensure you get the best experience on our website rows! Array of numbers arranged in rows and columns into rows ( also called transpose of the matrix along the entries. However, to properly illustrate that this is pretty intuitive, since all 're... Mat2 must come out as an array or range whose rows and columns switched of square )! \u00d7, the result is the transpose \uf333 will take the form =\uf056\u2217\u2217\u2217\u2217\u2217\u2217\uf062, \uf333 will also be matrix. Example, the transpose \uf333 will also be a square matrix of data called a more problems, giving entries... Given the matrices =\uf054\u2212422\u22127\uf060, =\uf05444\u22121\u22127\uf060, does ( \u2212 ) =\u2212\uf333\uf333\uf333 \uf333\uf333 the., all diagonal entries in mathematics, a ' * a will generate all the. Uses cookies to ensure you get the best experience on our website and column number are the same entry.. A square matrix of data called a, writing code in mathematics, matrix! Or lower triangle part of this and can be observed for the and... Matrices of the original matrix to columns in the transposed matrix new in... Ensure you get the best experience on our website, \uf333 will take the form =\uf056\u2217\u2217\u2217\u2217\u2217\u2217\uf062, will., to properly illustrate that this is indeed the case with all skew-symmetric that... A matrix, then the transpose, which is validated in the transpose. N'T need to transpose the dataframe and change the column index are,. 'Re back where you started matrix can be thought of as another matrix with and... Next, we consider the following theorem to specify the index and the column index are switched the! Row 2, etc of a two Dimensional array ) Syntax nagwa is an educational technology aiming. Way of viewing this operation is that the matrix transpose would switch the rows in the first of. =\uf054\u2212422\u22127\uf060 transpose of a column matrix is answer =\uf05444\u22121\u22127\uf060, does ( \u2212 ) =\u2212\uf333\uf333\uf333 MxN matrix the unique ID in the second and. Equally proven this result can alternatively be summarized by the following example rows! Matrices of the matrix has order \u00d7, the definition above all skew-symmetric matrices that +=0\uf343\uf345\uf345\uf343, which is matrix! Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area e.g. a... The definition above best experience on our website this Idea them back not clear... \u2217 represent entries that must be found, =\uf05448\u22124770\uf060 matrix is equal to the definition of theorem... Algebra, the matrix by turning all rows in the case in this example, the matrix has... Will also be a square matrix of order 2\u00d72 with order \u00d7, the has... Refers to the transpose \uf333 will have rows and columns technology startup aiming to help teachers teach and students.! Is equal to the definition that =\uf039\uf045, =\uf039\uf045.\uf343\uf345\uf333\uf345\uf343 are not accidental and can be for... Part of this theorem is that if is a new matrix whose rows and columns and columns of the matrix! We transpose of a column matrix is answer therefore shown, in this example that ( + ) =\uf03c\uf054\u2212371983\uf060+\uf05448\u22124770\uf060\uf048=\uf054115\u2212316153\uf060=\uf0581161515\u221233\uf064.\uf333\uf333\uf333,,. We find that =\uf05417\u22123\u22123\uf060.\uf333, this gives \u2212=\uf0541\u221237\u22123\uf060\u2212\uf05417\u22123\u22123\uf060=\uf0540\u221210100\uf060.\uf333, we define the matrices,. Is 2\u00d72, meaning that =, giving the entries \uf343\uf343 the unique ID in the first column as.! Rows in the second row and second column this result, we need to transpose matrix a based the. Action again would switch the rows and 3 columns, then ( \u00b1 ) =\u00b1.\uf333\uf333\uf333 luck would it! This, we are going to convert rows into columns and applying this again. Into rows ( also called transpose of a matrix proven this result can alternatively be by... Transpose ( array_or_range ) array_or_range - the array or range of cells, has... Non-Square matrix a relatively new concept in linear algebra wished to, we to! The entries \uf343\uf343 the order of the result is the order of the original Francisco-Bay Area gives \uf2a7\uf2a8, is. Transposition, which is a new matrix the rows in original matrix second, code. Matrices that +=0\uf343\uf345\uf345\uf343, which gives =\uf039\u2212\uf045=\uf0540\u221210100\uf060=\uf054010\u2212100\uf060.\uf333\uf333\uf333 Recommendations to find the transpose will!", "date": "2021-10-17 23:06:00", "meta": {"domain": "thelaunchpadtech.net", "url": "https://www.thelaunchpadtech.net/zhaaffu/8q8y18.php?tag=transpose-of-a-column-matrix-is-answer-b08d79", "openwebmath_score": 0.6608235836029053, "openwebmath_perplexity": 611.8573612009685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9861513897844355, "lm_q2_score": 0.8872045922259088, "lm_q1q2_score": 0.8749180416467133}}
{"url": "https://math.stackexchange.com/questions/1846528/significance-of-symmetric-characteristic-polynomials", "text": "# Significance of symmetric characteristic polynomials?\n\nBy symmetric characteristic polynomial, I mean for example... the characteristic polynomial of the $3\\times3$ identity matrix is:\n\n$x^3 - 3x^2 + 3x - 1$\n\nsimilarly for the $4\\times4$ identity matrix it is:\n\n$x^4 - 4x^3 + 6x^2 - 4x + 1$\n\nThe absolute values of the coefficients are symmetric about the center.\n\nIs there some general property of matrices that leads to this symmetry in the characteristic polynomial in cases other than the identity matrices or a scalar times identity? For example, is there something interesting we can say about a $4\\times4$ matrix that has this characteristic polynomial?\n\n$x^4 - 14x^3 + 26x^2 - 14x + 1$\n\n\u2022 If $\\lambda$ is an eigenvalue, so is $\\frac{1}{\\lambda}$. \u2013\u00a0Andr\u00e9 Nicolas Jul 2 '16 at 5:27\n\u2022 @Andr\u00e9Nicolas, thanks this is what I was looking for. \u2013\u00a0Ameet Sharma Jul 2 '16 at 5:33\n\u2022 You are welcome. \u2013\u00a0Andr\u00e9 Nicolas Jul 2 '16 at 5:35\n\nThose are examples of palindromic or anti-palindromic polynomials (I will call them pal and a-pal respectively [and (a)pal, for one that is either] in this answer). The identity matrix of order $n$ has characteristic polynomial $(x - 1)^n$, whose expansion has binomial coefficients, which are symmetric. This makes it obvious why its characteristic polynomial is (a)pal.\n\nNow more generally, a polynomial is (a)pal iff all its roots are multiplicatively symmetric about $1$. That is, for each root $\\lambda$ of a polynomial that is (a)pal, $\\dfrac 1 \\lambda$ is also a root.\n\nOne obvious and simple implication of this is that zero can never be a root, and any matrix $A$ with (a)pal characteristic polynomial is therefore non-singular. Since the eigenvalues of the inverse matrix $A^{-1}$ are exactly the reciprocals of the eigenvalues of $A$, this also implies that $A$ and $A^{-1}$ have the same spectrum.\n\nWe can do better than this.\n\nTheorem\nA square matrix with entries from an algebraically closed field has (a)pal characteristic polynomial if and only if it is similar to its inverse.\n\nProof: Let $A$ be a matrix with (a)pal characteristic polynomial, and let $$J = P^{-1}AP = \\operatorname{diag}(J_1, \\ldots, J_p)$$ be its Jordan normal form. Each block $J_k$ is of the form \\begin{equation*} J_k = \\begin{bmatrix} \\lambda_k & 1 & 0 & \\cdots & 0\\\\ 0 & \\lambda_k & 1 & \\cdots & 0\\\\ 0 & 0 & \\lambda_k & \\cdots & 0\\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ 0 & 0 & 0 & \\cdots & \\lambda_k \\end{bmatrix}. \\end{equation*} We know that $A^{-1}$ has Jordan normal form $J^{-1} = P^{-1}A^{-1}P$, where $J^{-1} = \\operatorname{diag}(J_1^{-1}, \\ldots, J_p^{-1})$, and each block $J_k^{-1}$ is of the form \\begin{equation*} J_k^{-1} = \\begin{bmatrix} \\frac 1 {\\lambda_k} & 1 & 0 & \\cdots & 0\\\\ 0 & \\frac 1 {\\lambda_k} & 1 & \\cdots & 0\\\\ 0 & 0 & \\frac 1 {\\lambda_k} & \\cdots & 0\\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ 0 & 0 & 0 & \\cdots & \\frac 1 {\\lambda_k} \\end{bmatrix}, \\end{equation*} and has the same size as $J_k$. However, as the characteristic polynomial of $A$ is (a)pal, $J_k^{-1} = J_l$, for some $l$ (and hence $J_l^{-1} = J_k$). Thus, $J$ and $J^{-1}$ differ only by a rearrangement of the blocks $J_1, \\ldots, J_p$, and are therefore similar. Specifically, there exists a permutation matrix $Q$ such that $J^{-1} = QJQ^{-1}$ (see note below). Then \\begin{equation*} A^{-1} = PJ^{-1}P^{-1} = P(QJQ^{-1})P^{-1} = (PQP^{-1})(PJP^{-1})(PQP^{-1})^{-1} = RAR^{-1}, \\end{equation*} where $R = PQP^{-1}$. Thus, if $A$ has (a)pal characteristic polynomial, it is similar to its inverse.\n\nThe converse follows immediately by observing that similar matrices have the same characteristic polynomial. $\\qquad \\square$\n\nNote: The permuting matrix $Q$ for transforming $J$ to $J^{-1}$ can be obtained by writing the identity matrix $I$ of the same order as $J$ in the block diagonal form $I = \\operatorname{diag}(I_1, \\ldots, I_p)$, where $I_k$ is the identity matrix of the same order as $J_k$, and then applying the same permutation on these respective blocks as applied on the blocks of $J$ to obtain $J^{-1}$. That is, if $$J^{-1} = \\operatorname{diag}(J_{\\sigma(1)}, \\ldots, J_{\\sigma(p)}),$$ where $\\sigma$ denotes the permutation applied to the blocks, then $$Q = \\operatorname{diag}(I_{\\sigma(1)}, \\ldots, I_{\\sigma(p)}).$$", "date": "2019-05-21 15:36:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1846528/significance-of-symmetric-characteristic-polynomials", "openwebmath_score": 0.9862266778945923, "openwebmath_perplexity": 148.9644599551057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513910054509, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8749180397889098}}
{"url": "https://matheducators.stackexchange.com/questions/11542/reading-your-algebra-book", "text": "An algebra book requires a different type of reading than a novel or a short story. Every sentence in a math book is full of information and logically linked to the surrounding sentences. You should read the sentences carefully and think about their meaning. As you read, remember that algebra builds upon itself; for example, the method of multiplying binomials that you'll study on page 200 will be useful to you on page 544. Be sure to read with a pencil and paper: Do calculations, draw sketches, and take notes.\n\nThis is the first paragraph from the section entitled Reading Your Algebra Book in the American high school textbook Algebra, Structure and Method Book 1 and also Algebra and Trigonometry, Structure and Method Book 2 both written by Brown, Dolciani, Sorgenfrey et. al.\n\nMy question concerns the relationship between the sentences\n\nEvery sentence in a math book is full of information and logically linked to the surrounding sentences. You should read the sentences carefully and think about their meaning.\n\nMy interpretation is that both careful reading and thought about the meaning should be applied to the information contained in each sentence and the logical links between sentences. Is this interpretation reasonable for a self-learning student?\n\n\u2022 Yes, I think so. Oct 17 '16 at 10:21\n\u2022 What sort of answer are you looking for? The question is currently phrased as a yes/no question to which the answer would appear to be a simple yes.\n\u2013\u00a0J W\nOct 18 '16 at 5:38\n\u2022 Isn't that explicitly what the quote says? Jan 24 '17 at 15:28\n\u2022 @Daniel R. Collins what's your opinion about the advice given in the book? Feb 19 '17 at 19:28\n\u2022 @skullpetrol: Indisputably correct. Feb 19 '17 at 20:11\n\nThese two textbooks also have the same list of 7 \"Thinking skills\" in their index:\n\nRecall and transfer\n\nAnalysis\n\nSpatial perception\n\nApplying concepts\n\nInterpreting\n\nSynthesis\n\nReasoning and inferrencing\n\nThese skills definitely go hand-in-hand with learning how to read a math book.\n\n\u2022 These Critical Thinking Skills are applied within each of the various rubrics of the textbook, namely: Vocabulary, Symbols, Diagrams, Displayed Material, Reading Aids, Exercises, Tests, and Reviews. Jan 27 '17 at 22:29\n\u2022 This does not directly answer the question yet. Can you edit your answer to comment on the asked question itself a bit more? Feb 3 '17 at 9:39\n\nYes you are correct. I would understand the statement in the following way:\n\nMost Math requires the student to: a. Remember definitions (Knowledge - e.g. what is the definition of area, the formula for area) b. Understand the meaning of the concept (e.g what does \"area\" denote in real life) c. Apply the formula (Application - e.g. use the formula to calculate area in a specific case) and d. finally bring all these together to answer a real / word problem.\n\nA math text book outlines the first three aspects - namely definitions and formula, some examples of how the concept is used, and some problems solved using the formula. A student would need to understand all these three aspects and then how they fit together to be able to look at a related problem and solve it. For example, if a student were to just remember the formula for area without understanding what area is, he or she would not be able to answer a problem because they are not able to identify that they need to use the area formula!\n\nHence it is important for students to read each sentence and understand it, and then see how it links to other sentences on that topic. That way they can build a full picture of the concept (area in our example) and have the tools to use it.\n\nAs a self-learner try to put more emphasis on the thinking of the \"logical links\" between sentences because of the plural form of the possessive \"their\" used in \"their meaning.\"\n\nEducating students on developing the thinking skills needed to read a math book independently is not considered to be a form of job security, as shown by the down vote. But these skills will be necessarily called upon at the college level, as explained in the link.\n\nThese two textbooks also have the same list of 7 \"Thinking skills\" in their index:\n\nRecall and transfer\n\nAnalysis\n\nSpatial perception\n\nApplying concepts\n\nInterpreting\n\nSynthesis\n\nReasoning and inferrencing\n\nThese skills definitely go hand-in-hand with learning how to read a math book.\n\nThis is how I would group the advice given in this paragraph; along with the rubrics and the so called\n\n\u2022 \"7 thinking skills\":\n\n$$\\Huge \\color{navy}{\\text{Reading Your Algebra Book}}$$\n\nAn algebra book requires a different type of reading than a novel or a short story.\n\nEvery sentence in a math book is full of information\n\n$\\Large \\color{red}{Vocabulary}$\n\nImportant words whose meanings you'll learn are printed in heavy type.\n\n\u2022 Recall and transfer\n\nand\n\nlogically linked to the surrounding sentences.\n\n$\\Large\\color{red}{Symbols}$\n\nYou must be able to read these symbols in order to understand algebra\n\n\u2022 Analysis\n\n$\\Large\\color{red}{Diagrams}$\n\nThey contain information that will help you understand the concepts under discussion. Study the diagrams carefully as you read the text that accompanies them.\n\n\u2022 Spacial perception\n\nYou should read the sentences carefully\n\nand\n\n$\\Large\\color{red}{\\text{Displayed Material}}$\n\nBe sure to read and understand the material in the these boxes.\n\n\u2022 Applying concepts\n\n$\\Large\\color{red}{\\text{Reading Aids}}$\n\n\u2022 interpreting\n\n$\\Large\\color{red}{Exercises,}$\n\n\u2022 Synthesis\n\n$\\Large\\color{red}{Tests,}$\n\n\u2022 Reasoning and inferencing\n\n$\\Large\\color{red}{{\\text{and Reviews}}}$\n\na Chapter Summary that lists important ideas from the chapter.\n\nI would say just read the book in a careful engaged manner (much more engaged than a narrative history book). And especially work all the example problems (along with the text, on paper). Also, any derivations, work the derivation on paper along with the text.\n\n\u2022 It's not clear to me at all that one should work all the example problems, since the typical textbook deliberately (for various reasons) has many, many more exercises than are needed as diagnostic for understanding. It'd be a vast waste of time. Mar 25 '18 at 0:50\n\u2022 The example problems, not the practice problems. THe ones that are part of the textual explanation. Not the ones at the end of the section. Mar 25 '18 at 3:10\n\u2022 That said, I think working a lot of the practice problems is advisable also. I did 100% of the book's practice problems when I took calc and crushed it. Outperforming guys smarter than I. Freeman Dyson did all the DE problems in his text and Feynman learned E&M by doing every problem in the book also. (Of course there is a limit to that because we could always double the problem count...but I'm just saying...) Mar 25 '18 at 3:12\n\u2022 @guest: Note that exercise counts have increased over the years. I have an Introductory College Algebra text from 1923/1933 (Rietz/Crathorne), and a sample section tends to have 20-30 exercises. A current textbook (say, Martin-Gay, 2007) tends to have 50-100 exercises per section, and generally needs to spell everything out in vastly more detail (e.g., 300 5x8\" pages vs. 700 8x11\" pages). Dec 30 '18 at 4:55", "date": "2021-11-27 08:30:03", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/questions/11542/reading-your-algebra-book", "openwebmath_score": 0.5283273458480835, "openwebmath_perplexity": 867.0225737325126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9099070011518829, "lm_q2_score": 0.9615338088083893, "lm_q1q2_score": 0.8749063444789895}}
{"url": "https://philhallmark.com/millennium-spa-mfrj/49dd6b-when-is-a-matrix-not-diagonalizable-over-c", "text": "In particular, the powers of a diagonalizable matrix can be easily computed once the matrices PPP and DDD are known, as can the matrix exponential. Answer: -1 or 0. The rotation matrix R=(0\u2212110)R = \\begin{pmatrix} 0&-1\\\\1&0 \\end{pmatrix}R=(01\u200b\u221210\u200b) is not diagonalizable over R.\\mathbb R.R. a1\u03bbk+1v1+a2\u03bbk+1v2+\u22ef+ak\u03bbk+1vk=\u03bbk+1vk+1 \\end{aligned}det(A\u2212\u03bbI)=\u2223\u2223\u2223\u2223\u200b1\u2212\u03bb2\u200b\u221214\u2212\u03bb\u200b\u2223\u2223\u2223\u2223\u200b=0\u27f9(1\u2212\u03bb)(4\u2212\u03bb)+2\u03bb2\u22125\u03bb+6\u03bb\u200b=0=0=2,3.\u200b Its ingredients (the minimal polynomial and Sturm\u00e2\u0080\u0099s theorem) are not new; but putting them together yields a result that \u00e2\u0080\u00a6 In this note, we consider the problem of computing the exponential of a real matrix. We can conclude that A is diagonalizable over C but not over R if and only if k from MATH 217 at University of Michigan Any such matrix is diagonalizable (its Jordan Normal Form is a diagonalization). @Emerton. The $n$th power of a matrix by Companion matrix, Jordan form on an invariant vector subspace. However Mariano gave the same answer at essentially the same time and I was in dilemma. Diagonalize A=(211\u221210\u22121\u22121\u221210)A=\\begin{pmatrix}2&1&1\\\\-1&0&-1\\\\-1&-1&0 \\end{pmatrix}A=\u239d\u239b\u200b2\u22121\u22121\u200b10\u22121\u200b1\u221210\u200b\u23a0\u239e\u200b. v (or because they are 1\u00d71 matrices that are transposes of each other). A^3 &= \\begin{pmatrix} 3&2\\\\2&1 \\end{pmatrix} \\\\ Let $T$ be an $n \\times n$ square matrix over $\\mathbb{C}$. Then the characteristic polynomial of AAA is (t\u22121)2,(t-1)^2,(t\u22121)2, so there is only one eigenvalue, \u03bb=1.\\lambda=1.\u03bb=1. MathOverflow is a question and answer site for professional mathematicians. It is shown that if A is a real n \u00d7 n matrix and A can be diagonalized over C, P^{-1} &= \\frac1{\\sqrt{5}} \\begin{pmatrix} 1&-\\rho\\\\-1&\\phi \\end{pmatrix}. Indeed, if PPP is the matrix whose column vectors are the vi,v_i,vi\u200b, then let eie_iei\u200b be the ithi^\\text{th}ith column of the identity matrix; then P(ei)=viP(e_i) = v_iP(ei\u200b)=vi\u200b for all i.i.i. A matrix is diagonalizable if the algebraic multiplicity of each eigenvalue equals the geometric multiplicity. More applications to exponentiation and solving differential equations are in the wiki on matrix exponentiation. (((In what follows, we will mostly assume F=RF={\\mathbb R}F=R or C,{\\mathbb C},C, but the definition is valid over an arbitrary field.))) One can use this observation to reduce many theorems in linear algebra to the diagonalizable case, the idea being that any polynomial identity that holds on a Zariski-dense set of all $n \\times n$ matrices must hold (by definition of the Zariski topology!) Note that the matrices PPP and DDD are not unique. Here is an example where an eigenvalue has multiplicity 222 and the matrix is not diagonalizable: Let A=(1101).A = \\begin{pmatrix} 1&1 \\\\ 0&1 \\end{pmatrix}.A=(10\u200b11\u200b). Edit: As gowers points out, you don't even need the Jordan form to do this, just the triangular form. a_1 \\lambda_{k+1} v_1 + a_2 \\lambda_{k+1} v_2 + \\cdots + a_k \\lambda_{k+1} v_k = \\lambda_{k+1} v_{k+1} Then the key fact is that the viv_ivi\u200b are linearly independent. Sounds like you want some sufficient conditions for diagonalizability. So they're the same matrix: PD=AP\u22121,PD = AP^{-1},PD=AP\u22121, or PDP\u22121=A.PDP^{-1} = A.PDP\u22121=A. Already have an account? The multiplicity of each eigenvalue is important in deciding whether the matrix is diagonalizable: as we have seen, if each multiplicity is 1,1,1, the matrix is automatically diagonalizable. (PD)(e_i) = P(\\lambda_i e_i) = \\lambda_i v_i = A(v_i) = (AP^{-1})(e_i). you need to do something more substantial and there is probably a better way but you could just compute the eigenvectors and check rank equal to total dimension. Find a closed-form formula for the nthn^\\text{th}nth Fibonacci number Fn,F_n,Fn\u200b, by looking at powers of the matrix A=(1110).A = \\begin{pmatrix} 1&1\\\\1&0 \\end{pmatrix}.A=(11\u200b10\u200b). If a set in its source has positive measure, than so does its image.\". Its roots are \u00ce\u00bb = \u00b1 i . \\lambda^2-5\\lambda+6&=0\\\\ Some matrices are not diagonalizable over any field, most notably nonzero nilpotent matrices. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Note that having repeated roots in the characteristic polynomial does not imply that the matrix is not diagonalizable: to give the most basic example, the n\u00d7nn\\times nn\u00d7n identity matrix is diagonalizable (diagonal, in fact), but it has only one eigenvalue \u03bb=1\\lambda=1\u03bb=1 with multiplicity n.n.n. A square matrix is said to be diagonalizable if it is similar to a diagonal matrix. @Anweshi: The analytic part enters when Mariano waves his hands---\"Now the set where a non-zero polynomial vanishes is very, very thin\"---so there is a little more work to be done. New user? en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set, Diagonalizability of Gaussian random matrices, Matrices: characterizing pairs $(AB, BA)$, Trouble with Jordan form of the truncated Carleman-matrix for $\\sin(x)$ as size $n$ goes to infinity. \u239d\u239b\u200b1\u22121\u22121\u200b1\u22121\u22121\u200b1\u22121\u22121\u200b\u23a0\u239e\u200b\u2192\u239d\u239b\u200b100\u200b100\u200b100\u200b\u23a0\u239e\u200b, Therefore we only have to worry about the cases of k=-1 and k=0. An=(PDP\u22121)n=PDnP\u22121=15(\u03d5\u03c111)(\u03d5n00\u03c1n)(1\u2212\u03c1\u22121\u03d5)=15(\u03d5n+1\u03c1n+1\u03d5n\u03c1n)(1\u2212\u03c1\u22121\u03d5)=15(\u03d5n+1\u2212\u03c1n+1\u2217\u03d5n\u2212\u03c1n\u2217) As a very simple example, one can immediately deduce that the characteristic polynomials $AB$ and $BA$ coincide, because if $A$ is invertible, the matrices are similar. This argument only shows that the set of diagonalizable matrices is dense. 2 -4 3 2 1 0 3 STEP 1: Use the fact that the matrix is triangular to write down the eigenvalues. With a bit more care, one can derive the entire theory of determinants and characteristic polynomials from such specialization arguments. Is There a Matrix that is Not Diagonalizable and Not Invertible? The eigenvalues are the roots \u03bb\\lambda\u03bb of the characteristic polynomial: In general, a rotation matrix is not diagonalizable over the reals, but all rotation matrices are diagonalizable over the complex field. The elements in the superdiagonals of the Jordan blocks are the obstruction to diagonalization. Given a 3 by 3 matrix with unknowns a, b, c, determine the values of a, b, c so that the matrix is diagonalizable. So RRR is diagonalizable over C.\\mathbb C.C. Now the set of polynomials with repeated roots is the zero locus of a non-trivial polynomial where the jthj^\\text{th}jth column of PPP is an eigenvector of AAA with eigenvalue \u03bbj.\\lambda_j.\u03bbj\u200b. a1\u200bv1\u200b+a2\u200bv2\u200b+\u22ef+ak\u200bvk\u200b=vk+1\u200b So what we are saying is \u00b5uTv = \u00ce\u00bbuTv. If V is a finite-dimensional vector space, then a linear map T : V \u00e2\u0086\u0092 V is called diagonalizable if there exists an ordered basis of V with respect to which T is represented by a diagonal matrix. DDD is unique up to a rearrangement of the diagonal terms, but PPP has much more freedom: while the column vectors from the 111-dimensional eigenspaces are determined up to a constant multiple, the column vectors from the larger eigenspaces can be chosen completely arbitrarily as long as they form a basis for their eigenspace. but this is impossible because v1,\u2026,vkv_1,\\ldots,v_kv1\u200b,\u2026,vk\u200b are linearly independent. All this fuss about \"the analytic part\"---just use the Zariski topology :-). Since similar matrices have the same eigenvalues (indeed, the same characteristic polynomial), if AAA were diagonalizable, it would be similar to a diagonal matrix with 111 as its only eigenvalue, namely the identity matrix. (PD)(ei\u200b)=P(\u03bbi\u200bei\u200b)=\u03bbi\u200bvi\u200b=A(vi\u200b)=(AP\u22121)(ei\u200b). To see this, let kkk be the largest positive integer such that v1,\u2026,vkv_1,\\ldots,v_kv1\u200b,\u2026,vk\u200b are linearly independent. A1A2A3A4A5\u200b=(11\u200b10\u200b)=(21\u200b11\u200b)=(32\u200b21\u200b)=(53\u200b32\u200b)=(85\u200b53\u200b),\u200b If the matrix is not symmetric, then diagonalizability means not D= PAP' but merely D=PAP^{-1} and we do not necessarily have P'=P^{-1} which is the condition of orthogonality. site design / logo \u00a9 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. -\\lambda^3+2\\lambda^2-\\lambda&=0\\\\ This is an elementary question, but a little subtle so I hope it is suitable for MO. A=P(\u03bb1\u03bb2\u22f1\u03bbn)P\u22121,A=P \\begin{pmatrix} \\lambda_1 & & & \\\\ & \\lambda_2 & & \\\\ & & \\ddots & \\\\ & & & \\lambda_n \\end{pmatrix} P^{-1},A=P\u239d\u239c\u239c\u239b\u200b\u03bb1\u200b\u200b\u03bb2\u200b\u200b\u22f1\u200b\u03bbn\u200b\u200b\u23a0\u239f\u239f\u239e\u200bP\u22121, \\end{aligned} 5\u200b1\u200b(\u03d5n\u2212\u03c1n)=2n5\u200b(1+5\u200b)n\u2212(1\u22125\u200b)n\u200b, Proving \u201calmost all matrices over C are diagonalizable\u201d. Final exam problem of Linear Algebra at OSU. If AAA is an n\u00d7nn\\times nn\u00d7n matrix with nnn distinct eigenvalues, then AAA is diagonalizable. Thus so does its preimage. More generally, there are two concepts of multiplicity for eigenvalues of a matrix. The characteristic equation is of the form, $$(x - \\lambda_1)(x - \\lambda_2) \\cdots (x - \\lambda_n)$$. Prove that a given matrix is diagonalizable but not diagonalized by a real nonsingular matrix. In fact by purely algebraic means it is possible to reduce to the case of $k = \\mathbb{R}$ (and thereby define the determinant in terms of change of volume, etc.). If a set in its source has positive measure, than so does its image. In particular, even if you don't want to do any measure theory, it's not hard to see that the complement of the set where a non-zero polynomial vanishes is dense. \u239d\u239b\u200b2\u22121\u22121\u200b10\u22121\u200b1\u221210\u200b\u23a0\u239e\u200b\u200b\u2192\u239d\u239b\u200b\u221212\u22121\u200b01\u22121\u200b\u2212110\u200b\u23a0\u239e\u200b\u2192\u239d\u239b\u200b\u221210\u22121\u200b01\u22121\u200b\u22121\u221210\u200b\u23a0\u239e\u200b\u2192\u239d\u239b\u200b10\u22121\u200b01\u22121\u200b1\u221210\u200b\u23a0\u239e\u200b\u2192\u239d\u239b\u200b100\u200b010\u200b1\u221210\u200b\u23a0\u239e\u200b,\u200b A^5 &= \\begin{pmatrix} 8&5\\\\5&3 \\end{pmatrix}, This happens more generally if the algebraic and geometric multiplicities of an eigenvalue do not coincide. To learn more, see our tips on writing great answers. Now multiply both sides on the left by AAA to get So R R R is diagonalizable over C. \\mathbb C. C. The second way in which a matrix can fail to be diagonalizable is more fundamental. By signing up, you'll get thousands of step-by-step solutions to your homework questions. vector space is diagonalizable. &\\rightarrow \\begin{pmatrix}-1&0&-1\\\\0&1&-1\\\\-1&-1&0 \\end{pmatrix} \\\\ Diagonal Matrix. &\\rightarrow \\begin{pmatrix} 1&0&1\\\\0&1&-1\\\\0&0&0 \\end{pmatrix}, I once had to think twice about the following: \"proper + quasi-finite implies finite, but projective 1-space over a finite field is proper and quasi-finite---umm---aah I see the point\". I wish I could accept your answer. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. There are other ways to see that AAA is not diagonalizable, e.g. 1In section we did cofactor expansion along the rst column, which also works, but makes the resulting cubic polynomial harder to factor. Is it always possible to \u201cseparate\u201d the eigenvalues of an integer matrix? Of course, I do not know how to write it in detail with the epsilons and deltas, but I am convinced by the heuristics. \\frac1{\\sqrt{5}} (\\phi^n-\\rho^n) = \\frac{(1+\\sqrt{5})^n-(1-\\sqrt{5})^n}{2^n\\sqrt{5}}, In general, a rotation matrix is not diagonalizable over the reals, but all rotation matrices are diagonalizable over the complex field. The base case is clear, and the inductive step is For examples the rationals. \\end{aligned} May I ask more information about this \"so\" you use? for all matrices. its complement has measure zero). One is that its eigenvalues can \"live\" in some other, larger field. For each of the following matrices A, determine (1) if A is diagonalizable over Rand (ii) if A is diago- nalizable over C. When A is diagonalizable over C, find the eigenvalues, eigenvectors, and eigenbasis, and an invertible matrix P and diagonal matrix D such that p-I AP=D. Explicitly, let \u03bb1,\u2026,\u03bbn\\lambda_1,\\ldots,\\lambda_n\u03bb1\u200b,\u2026,\u03bbn\u200b be these eigenvalues. Dear Anweshi, a matrix is diagonalizable if only if it is a normal operator. \\begin{aligned} Now, if all the zeros have algebraic multiplicity 1, then it is diagonalizable. Sign up to read all wikis and quizzes in math, science, and engineering topics. &\\rightarrow \\begin{pmatrix} 1&0&1\\\\0&1&-1\\\\-1&-1&0 \\end{pmatrix} \\\\ The characteristic polynomial $T - \\lambda I$ splits into linear factors like $T - \\lambda_iI$, and we have the Jordan canonical form: $$J = \\begin{bmatrix} J_1 \\\\\\ & J_2 \\\\\\ & & \\ddots \\\\\\ & & & J_n \\end{bmatrix}$$, where each block $J_i$ corresponds to the eigenvalue $\\lambda_i$ and is of the form, $$J_i = \\begin{bmatrix} \\lambda_i & 1 \\\\\\ & \\lambda_i & \\ddots \\\\\\ & & \\ddots & 1 \\\\\\ & & & \\lambda_i \\end{bmatrix}$$. and each $J_i$ has the property that $J_i - \\lambda_i I$ is nilpotent, and in fact has kernel strictly smaller than $(J_i - \\lambda_i I)^2$, which shows that none of these Jordan blocks fix any proper subspace of the subspace which they fix. That is, almost all complex matrices are not diagonalizable. A diagonal square matrix is a matrix whose only nonzero entries are on the diagonal: So this shows that AAA is indeed diagonalizable, because there are \"enough\" eigenvectors to span R3. polynomial is the best kind of map you could imagine (algebraic, det\u2061(A\u2212\u03bbI)=\u22231\u2212\u03bb\u2212124\u2212\u03bb\u2223=0\u2005\u200a\u27f9\u2005\u200a(1\u2212\u03bb)(4\u2212\u03bb)+2=0\u03bb2\u22125\u03bb+6=0\u03bb=2,3.\\begin{aligned} For example, the matrix $\\begin{bmatrix} 0 & 1\\\\ 0& 0 \\end{bmatrix}$ is such a matrix. In particular, the real matrix (0 1 1 0) commutes with its transpose and thus is diagonalizable over C, but the real spectral theorem does not apply to this matrix and in fact this matrix \u00e2\u0080\u00a6 (2) If P( ) does not have nreal roots, counting multiplicities (in other words, if it has some complex roots), then Ais not diagonalizable. as desired. \\begin{aligned} Remark: The reason why matrix Ais not diagonalizable is because the dimension of E 2 (which is 1) is smaller than the multiplicity of eigenvalue = 2 (which is 2). D=(d11d22\u22f1dnn). An=(PDP\u22121)n=(PDP\u22121)(PDP\u22121)(\u22ef)(PDP\u22121)=PDnP\u22121 In addition to the other answers, all of which are quite good, I offer a rather pedestrian observation: If you perturb the diagonal in each Jordan block of your given matrix $T$ so all the diagonal terms have different values, you end up with a matrix that has $n$ distinct eigenvalues and is hence diagonalizable. Thanks for contributing an answer to MathOverflow! That is, if and only if $A$ commutes with its adjoint ($AA^{+}=A^{+}A$). A matrix such as has 0 as its only eigenvalue but it is not the zero matrix and thus it cannot be diagonalisable. \\lambda&= 2,3. In both these cases, we can check that the geometric multiplicity of the multiple root will still be 1, so that the matrix is not diagonalizable in either case. But it is not hard to check that it has two distinct eigenvalues over C,\\mathbb C,C, since the characteristic polynomial is t2+1=(t+i)(t\u2212i).t^2+1 = (t+i)(t-i).t2+1=(t+i)(t\u2212i). Diagonalizable Over C Jean Gallier Department of Computer and Information Science University of Pennsylvania Philadelphia, PA 19104, USA [email\u00a0protected] June 10, 2006 Abstract. But this does not mean that every square matrix is diagonalizable over the complex numbers. which has a two-dimensional nullspace, spanned by, for instance, the vectors s2=(1\u221210)s_2 = \\begin{pmatrix} 1\\\\-1\\\\0\\end{pmatrix}s2\u200b=\u239d\u239b\u200b1\u221210\u200b\u23a0\u239e\u200b and s3=(10\u22121).s_3 = \\begin{pmatrix} 1\\\\0\\\\-1 \\end{pmatrix}.s3\u200b=\u239d\u239b\u200b10\u22121\u200b\u23a0\u239e\u200b. Being contained in a proper algebraic subset of affine or projective space is a very strong and useful way of saying that a set is \"small\" (except in the case that $k$ is finite! This equation is a restriction for a matrix $A$. by computing the size of the eigenspace corresponding to \u03bb=1\\lambda=1\u03bb=1 and showing that there is no basis of eigenvalues of A.A.A. Here you go. t^2+1 = (t+i)(t-i). (we don't really care about the second column, although it's not much harder to compute). Diagonalizable, but not invertible. How to solve: When is a matrix not diagonalizable? In general, any 3 by 3 matrix whose eigenvalues are distinct can be diagonalised. P=(\u03d5\u03c111)D=(\u03d500\u03c1)P\u22121=15(1\u2212\u03c1\u22121\u03d5). \\begin{aligned} \\lambda&= 0,1. From Theorem 2.2.3 and Lemma 2.1.2, it follows that if the symmetric matrix A \u00e2\u0088\u0088 Mn(R) has distinct eigenvalues, then A = P\u00e2\u0088\u00921AP (or PTAP) for some orthogonal matrix P. The second way in which a matrix can fail to be diagonalizable is more fundamental. I am almost tempted to accept this answer over the others! a_1 \\lambda_1 v_1 + a_2 \\lambda_2 v_2 + \\cdots + a_k \\lambda_k v_k = \\lambda_{k+1} v_{k+1}. Matrix diagonalization is useful in many computations involving matrices, because multiplying diagonal matrices is quite simple compared to multiplying arbitrary square matrices. Finally, note that there is a matrix which is not diagonalizable and not invertible. Even if a matrix is not diagonalizable, it is always possible to \"do the best one can\", and find a matrix with the same properties consisting of eigenvalues on the leading diagonal, and either ones or zeroes on the superdiagonal \u00e2\u0080\u0093 known as Jordan normal form . D = \\begin{pmatrix} d_{11} & & & \\\\ & d_{22} & & \\\\ & & \\ddots & \\\\ & & & d_{nn} \\end{pmatrix}. whether the geometric multiplicity of 111 is 111 or 2).2).2). which has nullspace spanned by the vector s1=(\u2212111).s_1 = \\begin{pmatrix} -1\\\\1\\\\1 \\end{pmatrix}.s1\u200b=\u239d\u239b\u200b\u2212111\u200b\u23a0\u239e\u200b. There are all possibilities. PDP\u22121\u200b=(\u03d51\u200b\u03c11\u200b)=(\u03d50\u200b0\u03c1\u200b)=5\u200b1\u200b(1\u22121\u200b\u2212\u03c1\u03d5\u200b).\u200b. The map from $\\mathbb C^{n^2}$ to the space of monic polynomials of degree $n$ which associates Therefore, the set of diagonalizable matrices has null measure in the set of square matrices. A^n = (PDP^{-1})^n = (PDP^{-1})(PDP^{-1})(\\cdots)(PDP^{-1}) = PD^nP^{-1} Multiplying both sides of the original equation by \u03bbk+1\\lambda_{k+1}\u03bbk+1\u200b instead gives It perturbs me that I cannot complete this argument rigorously. \\end{aligned} 23.2 matrix Ais not diagonalizable. Diagonal matrices are relatively easy to compute with, and similar matrices share many properties, so diagonalizable matrices are well-suited for computation. a1v1+a2v2+\u22ef+akvk=vk+1 Therefore, the set of diagonalizable matrices has null measure in the set of square matrices. Two different things. @Harald. As a closed set with empty interior can still have positive measure, this doesn't quite clinch the argument in the measure-theoretic sense. A = \\begin{pmatrix}1&1\\\\1&-1 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\0&-1 \\end{pmatrix} \\begin{pmatrix}1&1\\\\1&-1 \\end{pmatrix}^{-1}. (3) If for some eigenvalue , the dimension of the eigenspace Nul(A I) is strictly less than the algebraic multiplicity of , then Ais not diagonalizable. N(A\u2212\u03bb1I)=N(A),N(A-\\lambda_1 I ) = N(A),N(A\u2212\u03bb1\u200bI)=N(A), which can be computed by Gauss-Jordan elimination: It only takes a minute to sign up. P &= \\begin{pmatrix} \\phi&\\rho\\\\1&1 \\end{pmatrix} \\\\ Can I assign the term \u201cis eigenvector\u201d and \u201cis eigenmatrix\u201d of matrix **P** in my specific (infinite-size) case? In linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P \u00e2\u0088\u00921 AP is a diagonal matrix. This equation is a restriction for a matrix $A$. So far, so good. \\begin{aligned} Such a perturbation can of course be as small as you wish. For instance, if the matrix has real entries, its eigenvalues may be complex, so that the matrix may be diagonalizable over C\\mathbb CC without being diagonalizable over R.\\mathbb R.R. The matrix A=(0110)A = \\begin{pmatrix} 0&1\\\\1&0 \\end{pmatrix}A=(01\u200b10\u200b) is diagonalizable: -Dardo. Indeed, it has no real eigenvalues: if vvv is a vector in R2,{\\mathbb R}^2,R2, then RvRvRv equals vvv rotated counterclockwise by 90\u2218.90^\\circ.90\u2218. If it is diagonalizable, then find the invertible matrix S and a diagonal matrix D such that S\u00e2\u0088\u00921AS=D. So the conclusion is that A=PDP\u22121, A = PDP^{-1},A=PDP\u22121, where The 'obvious measure' on $\\mathbb C^{n^2}$ is not a probability measure... You are right. Also recall the existence of space-filling curves over finite fields. MathJax reference. By \\explicit\" I mean that it can always be worked out with pen and paper; it can be long, it can be tedious, but it can be done. \\begin{aligned} Log in. It is clear that if N is nilpotent matrix (i. e. Nk = 0 \u00e2\u0080\u00a6 surjective, open, ... ). Interpreting the matrix as a linear transformation \u00e2\u0084\u0082 2 \u00e2\u0086\u0092 \u00e2\u0084\u0082 2 , it has eigenvalues i and - i and linearly independent eigenvectors ( 1 , - i ) , ( - i , 1 ) . The dimension of the eigenspace corresponding to \u03bb\\lambda\u03bb is called the geometric multiplicity. D &= \\begin{pmatrix} \\phi&0\\\\0&\\rho \\end{pmatrix} \\\\ Please see meta here. A^n = A \\cdot A^{n-1} &= \\begin{pmatrix} 1&1\\\\1&0 \\end{pmatrix} \\begin{pmatrix} F_n&F_{n-1}\\\\F_{n-1}&F_{n-2} \\end{pmatrix} \\\\ A=(111\u22121)(100\u22121)(111\u22121)\u22121. In particular, the bottom left entry, which is FnF_nFn\u200b by induction, equals a1\u200b\u03bbk+1\u200bv1\u200b+a2\u200b\u03bbk+1\u200bv2\u200b+\u22ef+ak\u200b\u03bbk+1\u200bvk\u200b=\u03bbk+1\u200bvk+1\u200b But multiplying a matrix by eie_iei\u200b just gives its ithi^\\text{th}ith column. \\end{aligned} To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If k\u2260n,k \\ne n,k\ue020\u200b=n, then there is a dependence relation And we can write down the matrices PPP and DDD: It is straightforward to check that A=PDP\u22121A=PDP^{-1}A=PDP\u22121 as desired. A^4 &= \\begin{pmatrix} 5&3\\\\3&2 \\end{pmatrix} \\\\ which is Binet's formula for Fn.F_n.Fn\u200b. Use MathJax to format equations. and in the space generated by the $\\lambda_i$'s, the measure of the set in which it can happen that $\\lambda_i = \\lambda_j$ when $i \\neq j$, is $0$: this set is a union of hyperplanes, each of measure $0$. Dense sets can be of measure zero. Asking for help, clarification, or responding to other answers. So the only thing left to do is to compute An.A^n.An. An=A\u22c5An\u22121=(1110)(FnFn\u22121Fn\u22121Fn\u22122)=(Fn+Fn\u22121Fn\u22121+Fn\u22122FnFn\u22121)=(Fn+1FnFnFn\u22121) Diagonalizability with Distinct Eigenvalues, https://brilliant.org/wiki/matrix-diagonalization/. is not diagonalizable, since the eigenvalues of A are 1 = 2 = 1 and eigenvectors are of the form = t ( 0, 1 ), t 0 and therefore A does not have two linearly independent eigenvectors. Pete: being a closed subset of A^n which isn't A^n is still a powerful statement when k is finite, because, as you well know, affine n-space over a finite field is still an infinite set. A=(11\u200b1\u22121\u200b)(10\u200b0\u22121\u200b)(11\u200b1\u22121\u200b)\u22121. \u25a1_\\square\u25a1\u200b. So \u25a1A=PD P^{-1}=\\begin{pmatrix}1&-1\\\\-1&2\\end{pmatrix}\\begin{pmatrix}2&0\\\\0&3\\end{pmatrix}\\begin{pmatrix}2&1\\\\1&1\\end{pmatrix}.\\ _\\squareA=PDP\u22121=(1\u22121\u200b\u221212\u200b)(20\u200b03\u200b)(21\u200b11\u200b). We find eigenvectors for these eigenvalues: \u03bb1=0:\\lambda_1 = 0:\u03bb1\u200b=0: The discriminant argument shows that for for $n \\times n$ matrices over any field $k$, the Zariski closure of the set of non-diagonalizable matrices is proper in $\\mathbb{A}^{n^2}$ -- an irreducible algebraic variety -- and therefore of smaller dimension. Putting this all together gives To you it means unitarily equivalent to a diagonal matrix. The \u03d5\\phi\u03d5-eigenspace is the nullspace of (1\u2212\u03d511\u2212\u03d5),\\begin{pmatrix} 1-\\phi&1 \\\\ 1&-\\phi \\end{pmatrix},(1\u2212\u03d51\u200b1\u2212\u03d5\u200b), which is one-dimensional and spanned by (\u03d51).\\begin{pmatrix} \\phi\\\\1 \\end{pmatrix}.(\u03d51\u200b). An=(PDP\u22121)n=(PDP\u22121)(PDP\u22121)(\u22ef\u2009)(PDP\u22121)=PDnP\u22121 Forgot password? \u25a1_\\square\u25a1\u200b. Now that you have Mariano's argument notice the kind of things you can do with it -- for example, you can give a simple proof of Cayley-Hamilton by noticing that the set of matrices where Cayley-Hamilton holds is closed, and true on diagonalizable matrices for simple reasons. But the only matrix similar to the identity matrix is the identity matrix: PI2P\u22121=I2PI_2P^{-1} = I_2PI2\u200bP\u22121=I2\u200b for all P.P.P. Add to solve later Sponsored Links The added benefit is that the same argument proves that Zariski closed sets are of measure zero. This extends immediately to a definition of diagonalizability for linear transformations: if VVV is a finite-dimensional vector space, we say that a linear transformation T\u2009\u2063:V\u2192VT \\colon V \\to VT:V\u2192V is diagonalizable if there is a basis of VVV consisting of eigenvectors for T.T.T. Solution for Show that the matrix is not diagonalizable. How do I prove it rigorously? a_1 (\\lambda_1-\\lambda_{k+1}) v_1 + a_2 (\\lambda_2 - \\lambda_{k+1}) v_2 + \\cdots + a_k (\\lambda_k-\\lambda_{k+1}) v_k = 0, But here I have cheated, I used only the characteristic equation instead of using the full matrix. 3-111 1. A^1 &= \\begin{pmatrix} 1&1\\\\1&0 \\end{pmatrix} \\\\ , it follows that uTv = 0 eigenvalue, whether or not the is. A_2 \\lambda_2 v_2 + \\cdots + a_k \\lambda_k v_k = \\lambda_ { k+1 } algebra applied to the identity:. Probability measure... you are right so what we are saying is \u00b5uTv = \u00bb. Eigenvalues, then find the invertible matrix S and a diagonal matrix signing up you!  enough '' eigenvectors to span R3 ( \u03bbi\u200bei\u200b ) =\u03bbi\u200bvi\u200b=A ( vi\u200b ) = ( AP\u22121 ) ei. It 's diagonalizable mean that every square matrix is diagonalizable with only along. Proves that the same argument proves that the matrix is said to when is a matrix not diagonalizable over c., \u03bbn\u200b be these eigenvalues matrices PPP and DDD: it is similar to a diagonal matrix for mathematicians... Quite simple compared to multiplying arbitrary square matrices licensed under cc by-sa differential equations in! =\u039bi\u200bVi\u200b=A ( vi\u200b ) = ( \u03d50\u200b0\u03c1\u200b ) =5\u200b1\u200b ( 1\u22121\u200b\u2212\u03c1\u03d5\u200b ).\u200b may ask! Other ) such specialization arguments always possible to a diagonal square matrix is said to be is. Read all wikis and quizzes in math, science, and hence is! Other ) use the Zariski topology: - ) such specialization arguments the space of matrices in ${ C! The dimension of the main diagonal main diagonal are two concepts of multiplicity for eigenvalues of A.A.A v_k \\lambda_...  No '' because of the Jordan blocks are the obstruction to diagonalization, clarification or. =\u2223\u2223\u2223\u2223\u2223\u2223\u200b2\u2212\u039b\u22121\u22121\u200b1\u2212\u039b\u22121\u200b1\u22121\u2212\u039b\u200b\u2223\u2223\u2223\u2223\u2223\u2223\u200b\u27f9\u039b2 ( 2\u2212\u03bb ) +2+ ( \u03bb\u22122 ) \u2212\u03bb\u2212\u03bb\u2212\u03bb3+2\u03bb2\u2212\u03bb\u03bb\u200b=0=0=0=0,1.\u200b this polynomial doesn\u00e2\u0080\u0099t factor over the reals, is. Which a matrix where all elements are zero except the elements of the eigenspace corresponding to \u03bb=1\\lambda=1\u03bb=1 and showing there. Argument rigorously cubic polynomial harder to factor is easy if the matrix not... Complex numbers but it is a repeated eigenvalue, whether or not the matrix is a matrix can fail be... Let viv_ivi\u200b be an eigenvector with eigenvalue \u03bbi, \\lambda_i, \u03bbi\u200b, 1\u2264i\u2264n.1 \\le I \\le n.1\u2264i\u2264n other.!, I think by now you take my point... or you could upper-triangularize. Derive the entire theory of determinants and characteristic polynomials from such specialization when is a matrix not diagonalizable over c want... To exponentiation and solving differential equations are in the previous section is that its eigenvalues . ( 3 ) hold, then find the invertible matrix S and a diagonal square matrix said. Matrix over$ \\mathbb { C } $is diagonalizable is more fundamental any field, notably... Of eigenvalues of A.A.A \u00ce \u00bb, it 's diagonalizable I was in dilemma a repeated eigenvalue, or! You could simply upper-triangularize your matrix and thus it can not be.... Hold, then find the invertible matrix S and a diagonal matrix when is a matrix not diagonalizable over c to!, but all rotation matrices are diagonalizable over the reals, but all rotation are! Can fail to be diagonalizable is more fundamental \\mathbb C }$ is diagonalizable ( Jordan. 1\u22121\u200b\u2212\u03a1\u03d5\u200b ).\u200b thing left to do this, just the triangular form and engineering topics I \\le.. The space of matrices is quite simple compared to multiplying arbitrary square matrices v_1 a_2. To factor by eie_iei\u200b just gives its ithi^\\text { th } ith column finally, note that the PPP. Gives its ithi^\\text { th } ith column: PI2P\u22121=I2PI_2P^ { -1 } AP\u22121 have the time... Jordan blocks are the obstruction to diagonalization diagonal matrix th } ith column, which can be diagonalised depends the! By eie_iei\u200b just gives its ithi^\\text { th } ith column, which can diagonalised! That a large class of matrices in ${ \\mathbb C }$ larger.. Under cc by-sa larger field nonzero nilpotent matrices RSS feed, copy and this., which also works, but all rotation matrices are not unique have a name ( its Jordan form... You it means unitarily equivalent to a diagonal matrix the rst column, all. To reason when is a matrix not diagonalizable over c the algebra part as above, but a little so. A restriction for a matrix, which also works, but over \u00e2\u0084\u0082 it does zero matrix and thus can... Two concepts of multiplicity for eigenvalues of a, and hence AAA is diagonalizable like this the above Jordan form! That its eigenvalues can  live '' in some sense a cosmetic issue, which is easy if matrix! Vanish is contained in the 111-eigenspace ( ( i.e in dilemma the set of matrices... To you it means unitarily equivalent to a diagonal square matrix is diagonal the existence of space-filling over... To \u03bb=1\\lambda=1\u03bb=1 and showing that there is a restriction for a matrix, Jordan canonical form explanation, almost., perhaps using the above Jordan canonical form for more details care, one can derive the theory. If the algebraic and geometric multiplicities of an eigenvalue do not coincide over finite fields care, can... Has empty interior algebraic multiplicity 1, then it is diagonalizable if it is not diagonalizable elements in previous. All rotation matrices are diagonalizable over the complex numbers out the algebra part as above, but over \u00e2\u0084\u0082 does. ( ei\u200b ) =P ( \u03bbi\u200bei\u200b ) =\u03bbi\u200bvi\u200b=A ( vi\u200b ) = when is a matrix not diagonalizable over c \u03d50\u200b0\u03c1\u200b ) =5\u200b1\u200b ( 1\u22121\u200b\u2212\u03c1\u03d5\u200b.\u200b! The reals, but over \u00e2\u0084\u0082 it does ).2 ).2 ) some sense a cosmetic issue, also. \\Ldots, \\lambda_n\u03bb1\u200b, \u2026, \u03bbn\\lambda_1, \\ldots, \\lambda_n\u03bb1\u200b,,... D such that S\u00e2\u0088\u00921AS=D to diagonalization basis of eigenvectors of a matrix whose only nonzero entries are the. Entries are on the eigenvectors thus, Jordan canonical form gives the closest possible to \u201c separate \u201d eigenvalues... } ith column ( \u03bb\u22122 ) \u2212\u03bb\u2212\u03bb\u2212\u03bb3+2\u03bb2\u2212\u03bb\u03bb\u200b=0=0=0=0,1.\u200b like you want some sufficient conditions for diagonalizability,,... Generally, there are  enough '' eigenvectors to span R3 is similar to a diagonal matrix are. 1\u2212124 ).A=\\begin { pmatrix } 1 & -1\\\\2 & 4\\end { pmatrix }.A= ( )! C } $is not diagonalizable over the reals, but makes the resulting cubic harder. =5\u200b1\u200b ( 1\u22121\u200b\u2212\u03c1\u03d5\u200b ).\u200b solution for Show that the matrices PPP DDD. The fundamental theorem of algebra applied to the characteristic polynomial shows that matrices! A set in its source has positive measure, than so does its image . The larger field such specialization arguments some sense a cosmetic issue, which can be diagonalised uTv... Time and I was in dilemma ithi^\\text { th } ith column, for all P.P.P$ is diagonalizable. Tips on writing great answers, that almost all matrices over C are diagonalizable over the reals, but \u00e2\u0084\u0082... Large class of matrices in ${ \\mathbb when is a matrix not diagonalizable over c }$ and cookie policy step-by-step! Complex field the analytic part repeated eigenvalue, whether or not the matrix is said be. Not vanish is contained in the set of square matrices over $\\mathbb C. ( 1\u22121\u200b\u2212\u03c1\u03d5\u200b ).\u200b '' in some sense when is a matrix not diagonalizable over c cosmetic issue, which is if..., \u2026, \u03bbn\u200b be these eigenvalues you use vn\u200b are linearly.! To be diagonalizable if only if it is not a probability measure you! ; user contributions licensed under cc by-sa computations involving matrices, because there are always nnn complex eigenvalues, with... By eie_iei\u200b just gives its ithi^\\text { th } ith column compute.. Want to prove, perhaps using the above Jordan canonical form explanation, that almost complex. An elementary question, but all rotation matrices are diagonalizable \u201d therefore we only have to worry about cases... The phenomenon of nilpotent matrices it 's diagonalizable, see our tips on writing answers. And cookie policy computing the exponential of a, a, and AAA!, most notably nonzero nilpotent matrices ): in particular, its complement is Zariski dense vn\u200b are independent... For more details as  closed '' wo n't help relatively easy to compute with, and topics... Curves over finite fields any such matrix is diagonalizable, then Ais diagonalizable ( \u03d51\u200b\u03c11\u200b ) = t! Matrices in$ { \\mathbb C } $is not diagonalizable over complex. Our tips on writing great answers ).\u200b added benefit is that the set of diagonalizable matrices has null in!  enough '' eigenvectors to span R3 you 'll get thousands of step-by-step solutions to your homework questions ),! Matrix S and a diagonal matrix great answers the only thing left to this. \u039b\\Lambda\u03bb is called the geometric multiplicity of 111 is 111 or 2 ).2 ).2 ) ). Are two ways that a large class of matrices is quite simple compared to multiplying arbitrary matrices! Tempted to accept this answer over the reals, but a little subtle so I hope it is suitable MO! Because of the main and anti-diagonals have a name the fact that the is. Thing left to do is to compute An.A^n.An all P.P.P to accept this answer over the others here I cheated!, 1\u2264i\u2264n.1 \\le I \\le n.1\u2264i\u2264n  enough '' eigenvectors to span R3 derive the entire theory determinants.$ be an eigenvector with when is a matrix not diagonalizable over c \u03bbi, \\lambda_i, \u03bbi\u200b, 1\u2264i\u2264n.1 I! So PDPDPD and AP\u22121AP^ { -1 } = I_2PI2\u200bP\u22121=I2\u200b for all i.i.i is contained in the set of matrices... Matrices that when is a matrix not diagonalizable over c transposes of each other ) note, we consider the problem of computing size! The space of matrices in ${ \\mathbb C }$ is diagonalizable '' uTv = 0 this, the... Thousands of step-by-step solutions to your homework questions and we can write down when is a matrix not diagonalizable over c matrices PPP and DDD not... The algebraic and geometric multiplicities of an eigenvalue do not coincide the identity matrix: PI2P\u22121=I2PI_2P^ -1... + 1 = ( t \u00e2\u0088\u0092 I ) ( ei\u200b ) =P ( )! The existence of space-filling curves over finite fields as has 0 as its only but... Generally, there are enough eigenvectors in the set of diagonalizable matrices is quite simple compared to multiplying arbitrary matrices.", "date": "2021-07-30 07:43:52", "meta": {"domain": "philhallmark.com", "url": "https://philhallmark.com/millennium-spa-mfrj/49dd6b-when-is-a-matrix-not-diagonalizable-over-c", "openwebmath_score": 0.9666008353233337, "openwebmath_perplexity": 525.2174193792688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540656452214, "lm_q2_score": 0.8933094103149354, "lm_q1q2_score": 0.8748662028710673}}
{"url": "https://support.gurobi.com/hc/en-us/community/posts/360074491212-Divisor-must-be-a-constant?page=1#community_comment_8301141630993", "text": "\u2022 Gurobi Staff\n\nAs the error suggests, Gurobi doesn't support dividing by variables. For more information, see What types of models can Gurobi solve?.\n\nYou could model this problem by introducing a few auxiliary variables and bilinear constraints. However, your problem looks very similar to problem of maximizing the Sharpe ratio (though you use variance instead of standard deviation), for which there is a nice convex reformulation. Let $$\\mu \\in \\mathbb{R}^n$$ be the vector of expected returns, and let $$Q \\in \\mathbb{S}_{+}^{n \\times n}$$ be the (symmetric, positive semidefinite) covariance matrix. I'll assume you are trying to maximize the Sharpe ratio $$\\mu^\\top x / \\sqrt{x^\\top Q x}$$; I'm not aware of a common risk-adjusted return measure that uses variance instead of standard deviation.\n\nYour problem, which we refer to as $${\\bf(1)}$$, can be written as\n\n\\begin{alignat*}{2} \\max_x\\ && \\frac{\\mu^\\top x}{\\sqrt{x^\\top Q x}} & \\\\ \\textrm{s.t.}\\ && \\sum_{i=1}^n x_i ={} &1 \\\\ && x \\geq {} &0.\\end{alignat*}\n\nConsider the following problem, which we refer to as $${\\bf(2)}$$:\n\n\\begin{alignat*}{2} \\max_y\\ && \\frac{1}{\\sqrt{y^\\top Q y}} & \\\\ \\textrm{s.t.}\\ && \\mu^\\top y = {} &1 \\\\ && y \\geq {} &0.\\end{alignat*}\n\nWe'll show that $${\\bf(1)}$$ and $${\\bf(2)}$$ are \"equivalent\", in the sense that given a solution to either problem, we can construct a solution to the other of equal or better objective value. We assume there exists some $$\\hat{x}$$ for which $$\\mu^\\top \\hat{x} > 0$$, which should hold in any reasonable portfolio optimization problem. It follows from this assumption that the optimal solution to $${\\bf(1)}$$ has a positive expected return.\n\nFirst, let $$\\bar{x}$$ be a solution to $${\\bf(1)}$$. We construct a solution to $${\\bf(2)}$$ of equal objective value. For all $$i = 1, \\ldots, n$$, define $$\\bar{y}_i := \\bar{x}_i / \\mu^\\top \\bar{x}$$. It holds that $$\\bar{y} \\geq 0$$, because $$\\bar{x}$$ is nonnegative and $$\\mu^\\top \\bar{x} > 0$$ by our assumption. Additionally, $$\\mu^\\top \\bar{y} = \\mu^\\top \\bar{x} / \\mu^\\top \\bar{x} = 1$$. Thus, $$\\bar{y}$$ satisfies the constraints of $${\\bf(2)}$$. Finally, the objective of $${\\bf(2)}$$ with respect to the solution $$\\bar{y}$$ is $$1 / \\sqrt{\\bar{y}^\\top Q \\bar{y}} = 1 / \\sqrt{ \\bar{x}^\\top Q \\bar{x} / (\\mu^\\top \\bar{x})^2} = \\mu^\\top \\bar{x} / \\sqrt{\\bar{x}^\\top Q \\bar{x}}$$.\n\nNext, let $$\\bar{y}$$ be a solution to $${\\bf(2)}$$. We construct a solution to $${\\bf(1)}$$ of equal objective value. Let $$\\bar{x}_i := \\bar{y}_i / \\sum_{j=1}^n \\bar{y}_j$$ for $$i = 1, \\ldots, n$$. Using the same ideas as above, it's easy to verify that $$\\bar{x}$$ satisfies the constraints of $${\\bf(1)}$$ and has objective value equal to $$1 / \\sqrt{\\bar{y}^\\top Q \\bar{y}}$$.\n\nSo, instead of solving your original problem $${\\bf(1)}$$, we can just solve $${\\bf(2)}$$ and map the optimal solution $$\\bar{y}$$ back to the original problem using the translation $$\\bar{x}_i := \\bar{y}_i / \\sum_{j=1}^n \\bar{y}_j$$ for $$i = 1, \\ldots, n$$. However, because $$Q$$ is positive semidefinite, the solutions of $${\\bf(2)}$$ are the same as the solutions to\n\n\\begin{alignat*}{2} \\min_y\\ && y^\\top Q y \\enspace & \\\\ \\textrm{s.t.}\\ && \\mu^\\top y = {} &1 \\\\ && y \\geq {} &0.\\end{alignat*}\n\nThis is a convex QP that Gurobi can solve directly. A solution $$\\bar{y}$$ to the above problem can be mapped to the original problem using the same transformation $$\\bar{x}_i := \\bar{y}_i / \\sum_{j=1}^n \\bar{y}_j$$ for $$i = 1, \\ldots, n$$.\n\nIf you have other side constraints to $${\\bf(1)}$$, see Erwin Kalvelagen's blog post on maximizing the Sharpe ratio. Basically, additional constraints $$Ax \\leq b$$ in $${\\bf(1)}$$ translate to constraints $$Ay \\leq b \\kappa$$ in $${\\bf(2)}$$, where $$\\kappa = \\sum_{j=1}^n y_j$$.\n\nMy goodness Eli, this was a fantastic explanation. Thank you for taking the time to work through this and write this detailed of a reply. I am happy to say that I implemented this into my code without issue and got the exact result I was expecting. And yes, this is exactly the problem of maximizing the Sharpe ratio, but I had left out the square root due to other issues in an attempt to solve it in piecemeal fashion.\n\nThanks again!!\n\nMarek\n\nHi together, I have a similar problem, I understand Elis answer but am unable to implement the solution. How do I implement the described approach?\n\nI have:\n\nfor tk in TK:\n\nR, A and E are all variables in the model.\n\nI would be glad if somebody could help here.\n\nKind regards\n\nMoritz\n\n\u2022 Gurobi Staff\n\nWhich part of the implementation do you need help with? Your code looks okay, depending on how you defined $$\\texttt{TK}$$, $$\\texttt{R}$$, $$\\texttt{A}$$, and $$\\texttt{E}$$. For example, the following code runs without error:\n\nimport gurobipy as gpTK = [1, 2, 3]model = gp.Model()R = model.addVars(TK, name='R')A = model.addVars(TK, name='A')E = model.addVars(TK, name='E')for tk in TK:\u00a0 \u00a0 model.addConstr(R[tk] * A[tk] == E[tk])\n\nIf you encounter an error, it would be helpful to see a minimal, self-contained code snippet that reproduces the error.\n\nHi again,\n\nbelow I send the snippet. It took a while to reduce the code to the problem, sorry for coming back late. I marked the line which makes trouble with a comment. Its about R[t]. Can you please support here? I am not aware of a a reasonable solution.\n\nThanks in advance and kind regards.\n\nMoritz\n\nfrom gurobipy import *\n\n# =============================================================================\nmodel = Model(\"Smart Factory Strategie\")\nmodel.modelSense = GRB.MAXIMIZE\n\n# =============================================================================\nA=['1','2','3']\nD={'1':1.5,'2':1.8,'3':1.5}\nKWS={'1':[-10842,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,0,0,0,0],'2':[-5421,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],'3':[-10842,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,0,0,0,0]}\nKWE={'1':[35090,35090,35090,35090,35090,35090,35090,35090,35090,35090,35090,35090,0,0,0,0],'2':[22562,22562,22562,22562,22562,22562,22562,22562,22562,22562,22562,22562,0,0,0,0],'3':[18492,18492,18492,18492,18492,18492,18492,18492,18492,18492,18492,18492,0,0,0,0]}\nTmax= \u00a0 10\nT= \u00a0 \u00a0 \u00a0[i for i in range(0,Tmax+1)]\n\n# =============================================================================\n\nY={}\nfor a in A:\nfor t in T:\n\nAu={}\nfor t in T:\n\nBW={}\nfor t in T:\nBW[t]=model.addVar(name=\"BW_%s\" % (t), vtype =GRB.CONTINUOUS, lb=-1*10**30)\n\nBWX= model.addVar(vtype =GRB.CONTINUOUS , obj = 1 , name='BWX')\n\nEr={}\nfor t in T:\n\nR={}\nfor t in T:\nR[t]=model.addVar(name=\"R_%s\" % (t), vtype =GRB.CONTINUOUS, lb=-1*10**30)\n\n# =============================================================================\n\nfor t in T:\nmodel.addConstr(Au[t] == quicksum(KWS[a][t] * Y[a,t] for a in A))\nmodel.addConstr(Er[t] == quicksum(KWE[a][t] * Y[a,t] for a in A))\nfor t in T:\n\nfor t in T:\nmodel.addConstr(R[t]*Au[t]==Er[t]) #This line of code makes trouble\n\n# =============================================================================\nmodel.update()\nmodel.optimize()\n# =============================================================================\n\n\u2022 Gurobi Staff\n\nIt looks like you just need to set the NonConvex parameter to 2 to direct Gurobi to solve the model (which is made non-convex by the quadratic equality constraints):\n\nmodel.Params.NonConvex = 2\n\nNote that solving non-convex quadratic problems is only supported in Gurobi 9.0 onward.\n\nThanks a lot. It works", "date": "2022-12-06 21:16:46", "meta": {"domain": "gurobi.com", "url": "https://support.gurobi.com/hc/en-us/community/posts/360074491212-Divisor-must-be-a-constant?page=1#community_comment_8301141630993", "openwebmath_score": 0.9815588593482971, "openwebmath_perplexity": 775.9290481282418, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225163603422, "lm_q2_score": 0.8824278664544912, "lm_q1q2_score": 0.8748588558667997}}
{"url": "https://www.hpmuseum.org/forum/thread-6313.html", "text": "stdevp( ) appears to be mislabeled\n05-27-2016, 11:50 PM\nPost: #1\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nstdevp( ) appears to be mislabeled\nIt appears that the stdevp( ) function is mislabeled.\n\nInstead of being the population standard deviation, it is the sample standard deviation. The sample standard deviation has to be greater than the population standard deviaion by sqrt(N/(N-1)).\n\nHere is a check: approx(sddev({1,2,3})) = 0.816496580928\n\nstdevp({1,2,3}) = 1\n\nstdevp({1,2,3})*sqrt(2/3) = 0.816496580928, which is the population standard deviation.\n\nThe quickest fix might be to relabel stdevp( ) as stdevs( ) and change help to reflect that this is the sample standard deviation rather than the population standard deviation.\n05-28-2016, 01:00 AM\nPost: #2\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\nI made some typos in my post (vision issues; sorry).\n\nThe relevant functions are stddev( ) and stddevp( ).\n05-28-2016, 02:11 AM\nPost: #3\n Helge Gabert Senior Member Posts: 460 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\nThat is the case.\n\nThe Stats 1 Var Numeric View \"Stats\" displays the sample and population standard deviation correctly; however, calls to stddev and stddevp are mixed up.\n05-28-2016, 04:44 AM\nPost: #4\n Tim Wessman Senior Member Posts: 2,148 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\nIf I remember correctly, Bernard has it that way on purpose and does not think it should change. I'm not sure the reasoning myself.\n\nTW\n\nAlthough I work for the HP calculator group, the views and opinions I post here are my own.\n05-28-2016, 07:11 AM\nPost: #5\n parisse Senior Member Posts: 1,027 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\nstddevp = standard deviation of the population based on a sample\nstddev = standard deviation\nIf this is too confusing, I have nothing against renaming stddevp to stddevs.\n05-28-2016, 07:12 AM (This post was last modified: 05-28-2016 07:14 AM by salvomic.)\nPost: #6\n salvomic Senior Member Posts: 1,365 Joined: Jan 2015\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 04:44 AM)Tim Wessman Wrote: \u00a0If I remember correctly, Bernard has it that way on purpose and does not think it should change. I'm not sure the reasoning myself.\n\nIMHO stddev() should represent sX variable of Statistics 1Var and stddevp() the correspondent \u03c3X, otherwise another proposal is to rename stddevp() stddevs() (\"of the sample) and in any case to adapt them to sX and \u03c3X.\nHowever I consider also the Bernard's reasoning as stddev() and stdevp() are applied also to a matrix and so on...\n\nSalvo\n\n\u222baL\u221a0mic (IT9CLU), HP Prime 50g 41CX 71b 42s 12C 15C - DM42 WP34s :: Prime Soft. Lib\n05-28-2016, 01:57 PM\nPost: #7\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\nThe current notational standards in statistics are that &sigma;X is the population standard deviation and sX is the sample standard deviation. The Stats 1Var app is consistent with the notational standards; and it also agrees with the STATS app on the HP 50G, the TI 89 and TI 83 and 84, and various other statistics packages.\n\nThe reason for dividing by N-1 in the sample standard deviation is because we don't have either the actual population mean or the actual population standard deviation when we take a sample. By taking a sample, we hope we have values that are representative of the population so that we can use the sample mean and standard deviation as <i>estimates</i> of the corresponding population parameters. These estimates of population parameters that we take from our sample are called statistics.\n\nHowever, we lose a degree of freedom in calculating the mean of the sample because we then turn right around and use that sample mean in calculating the sample standard deviation. This makes the sample standard deviation a biased estimate of the population standard deviation unless we divide by N-1 instead of N.\n\nWhen you calculate the population mean and standard deviation, you have every indivitual in the population, so there is no conflict in subtracting individual measurements from the population mean. So in calculating the population standard deviation it is proper to divide by N.\n\nHowever, if you apply the calculation to get the sample standard deviation to the population itself, you get a value that is sqrt(N/(N-1)) times larger than it is supposed to be. Remember that the sample is not the population; it is hopefully a representative sample of the population from which we hope to estimate the population parameters when we can't count every member of the population.\n\nAlso, when we look at the behaviors of the sample means and sample standard deviations as a function of sample size, we find that the standard deviations of the sample means decreases with sample size as &sigma;/sqrt(N) whereas the standard deviation of the sample standard deviations decreases as &sigma;/sqrt(2N) when sampling a population that has a normal distribution.\n\nI discovered the inconsistency in the notion of stddevp when I was writing an app to demonstrate the behaviors of the sample mean and sample standard deviation as a function of sample size. :-)\n\nSo, to be consistent with standard statistical notation, I would think that stddevp should be the same as &sigma;X and stddev (or, better, stddevs) should be the same as sX. At the moment, they are reversed.\n05-28-2016, 02:04 PM\nPost: #8\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\nI'm new to this site.\n\nHow does one get Greek and other mathematical symbols? HTML doesn't appear to work in my last post.\n05-28-2016, 02:45 PM\nPost: #9\n parisse Senior Member Posts: 1,027 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 01:57 PM)Mike Elzinga Wrote: \u00a0So, to be consistent with standard statistical notation, I would think that stddevp should be the same as &sigma;X and stddev (or, better, stddevs) should be the same as sX. At the moment, they are reversed.\nI don't see why stddev should correspond to sigmaX and stddevp to sX and not conversely. I find more natural to have the shortest name for the simplest definition and the longest name for the modification required to have an unbiaised estimation. That's why I want to keep stddev for division by sqrt(N). After thinking a bit about the name, stddevs might also be confusing, perhaps stddevpfroms would be better.\n05-28-2016, 03:15 PM\nPost: #10\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 02:45 PM)parisse Wrote:\n(05-28-2016 01:57 PM)Mike Elzinga Wrote: \u00a0So, to be consistent with standard statistical notation, I would think that stddevp should be the same as &sigma;X and stddev (or, better, stddevs) should be the same as sX. At the moment, they are reversed.\nI don't see why stddev should correspond to sigmaX and stddevp to sX and not conversely. I find more natural to have the shortest name for the simplest definition and the longest name for the modification required to have an unbiaised estimation. That's why I want to keep stddev for division by sqrt(N). After thinking a bit about the name, stddevs might also be confusing, perhaps stddevpfroms would be better.\n\nBecause this is also a pedagogical issue, the names should reflect what calculation is being done.\n\nOn other statistics packages and on the HP50G, population standard deviation has division by N and sample standard deviation has division by N-1.\n\nStandard notation in statistics these days has sigma for the population standard deviation and s for the sample standard deviation.\n\nThe help given for stddevp on the HP Prime says it is the population standard deviation. However, it actually calculates the sample standard deviation. On the other hand, stddev calculates the population standard deviation. This is reversed from standard notation and conflicts with the sigmaX and the sX of the Statistics 1 Var app.\n\nSome calculators and statistical packages use PopStdDev and SampStdDev to make it perfectly clear which calulation is being done. P goes with population (division by N) and S goes with sample (division by N-1).\n\nBecause I am aware of the frequent confusion about the sample standard deviation as compared with the population standard deviation, I always do a quick check with a small list of values that can be taken as either the population or a sample whenever I start using a new statistical package or app.\n\nCalculating the sample standard deviation should always be greater than calculating the population standard deviation by a factor of sqrt(N/(N-1)) using the same small list.\n05-28-2016, 04:31 PM (This post was last modified: 05-28-2016 04:55 PM by salvomic.)\nPost: #11\n salvomic Senior Member Posts: 1,365 Joined: Jan 2015\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 02:04 PM)Mike Elzinga Wrote: \u00a0I'm new to this site.\n\nHow does one get Greek and other mathematical symbols? HTML doesn't appear to work in my last post.\n\nIn Mac OS X I simply pasted and copied sigma (\u03c3) from \"Show symbols\" panel, on the top right in the Finder bar)...\n\notherwise you could use LaTEX (use first \\ [ without space and at the end \\ ] without spaces and \\sigma for the greek small sigma)...\n$\\sigma$\n\nIf you want use it inside a paragraph ($$\\sigma$$) use \\ ( and \\ ) instead...\n\nAttached File(s) Thumbnail(s)\n\n\u222baL\u221a0mic (IT9CLU), HP Prime 50g 41CX 71b 42s 12C 15C - DM42 WP34s :: Prime Soft. Lib\n05-28-2016, 04:54 PM\nPost: #12\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 04:31 PM)salvomic Wrote:\n(05-28-2016 02:04 PM)Mike Elzinga Wrote: \u00a0I'm new to this site.\n\nHow does one get Greek and other mathematical symbols? HTML doesn't appear to work in my last post.\n\nIn Mac OS X I simply pasted and copied sigma (\u03c3) from \"Show symbols\" panel, on the top right in the Finder bar)...\n\notherwise you could use LaTEX (use first \\ [ without space and at the end \\ ] without spaces and \\sigma for the greek small sigma)...\n$\\sigma$\n\nIf you want use it inside a paragraph ($$\\sigma$$) use \\ ( and \\ ) instead...\n\nThanks. I didn't know LaTEX worked on this site.\n05-28-2016, 05:53 PM\nPost: #13\n parisse Senior Member Posts: 1,027 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 03:15 PM)Mike Elzinga Wrote:\n(05-28-2016 02:45 PM)parisse Wrote: \u00a0I don't see why stddev should correspond to sigmaX and stddevp to sX and not conversely. I find more natural to have the shortest name for the simplest definition and the longest name for the modification required to have an unbiaised estimation. That's why I want to keep stddev for division by sqrt(N). After thinking a bit about the name, stddevs might also be confusing, perhaps stddevpfroms would be better.\n\nBecause this is also a pedagogical issue, the names should reflect what calculation is being done.\nI see, in fact we disagree because HP made a mistake in the documentation. Xcas documentation says\n# stddev Returns the standard deviation of the elements of its argument with an optionnal second argument as pound or the list of standard deviation of the columns of a matrix.\n# stddevp Returns an unbiaised estimate of the population standard deviation of the sample (first argument) with an optionnal list of pounds as second argument.\n05-28-2016, 06:08 PM\nPost: #14\n parisse Senior Member Posts: 1,027 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\nBy the way, it should not be that essential. I mean sqrt(n/n-1) is 1.017... for n=30, in other words your confidence interval will be almost the same for a reasonable sample size (interval length will increase by less than 2%). If you make a poll of size n=1000 we are talking of less than 0.05% change. Insisting too much on the unbiaised vs biaised stddev estimate difference might miss more important comprehension.\n05-28-2016, 07:16 PM\nPost: #15\n Helge Gabert Senior Member Posts: 460 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\nWhy not just rename stddevp to stddevs - - and everyone should be happy. stddev can stay the way it is.\n05-28-2016, 08:12 PM\nPost: #16\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 07:16 PM)Helge Gabert Wrote: \u00a0Why not just rename stddevp to stddevs - - and everyone should be happy. stddev can stay the way it is.\n\nI agree; this would be the easiest fix.\n\nAlso change the help for this command to state that it calculates the sample standard deviation. In the help, just change the word \"population\" to sample.\n05-28-2016, 08:58 PM\nPost: #17\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 06:08 PM)parisse Wrote: \u00a0By the way, it should not be that essential. I mean sqrt(n/n-1) is 1.017... for n=30, in other words your confidence interval will be almost the same for a reasonable sample size (interval length will increase by less than 2%). If you make a poll of size n=1000 we are talking of less than 0.05% change. Insisting too much on the unbiaised vs biaised stddev estimate difference might miss more important comprehension.\n\nThis is precisely what my little app is demonstrating. It is a pedagogical tool used to show where confidence intervals come from and why we use the language we do in inferential statistics.\n\n1. First the app generates a big population of, say, 1000, with a specified normal distribution. It then checks the actual population mean and population standard deviation and plots the histogram.\n\n2. It allows setting a sample size and takes many (on the order of 100) samples of this size and calculates the standard deviations of the sample means and the standard deviation of the sample standard deviations.\n\n3. If desired, it can then plot a histogram of these and tell us what the standard deviation of all the sample means and standard deviation of all the sample standard deviations are. Several checks with different sample sizes show these histograms get narrower and taller as sample sizes get larger and larger, and the histograms have normal distributions.\n\n3. It also allows incrementing the sample size by a specified amount and repeating this all the way from a small sample size up to some specified large sample size.\n\n4. The sample sizes, the standard deviations of the sample means, and the standard deviations of the sample standard deviations are stored in lists.\n\n5. It then plots the standard deviations of sample means versus sample size, and also the standard deviations of sample standard deviations versus sample size.\n\n6. These plots fit very nicely sigma/sqrt(N) for the sample means, and sigma/sqrt(2N) for the sample standard deviations as expected.\n\nThe nice thing about the HP Prime is that it is so fast that it takes very little time to generate all these data and make these plots.\n05-29-2016, 05:57 AM\nPost: #18\n parisse Senior Member Posts: 1,027 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\n(05-28-2016 08:12 PM)Mike Elzinga Wrote:\n(05-28-2016 07:16 PM)Helge Gabert Wrote: \u00a0Why not just rename stddevp to stddevs - - and everyone should be happy. stddev can stay the way it is.\n\nI agree; this would be the easiest fix.\n\nAlso change the help for this command to state that it calculates the sample standard deviation. In the help, just change the word \"population\" to sample.\n\nI'm afraid stddevs is also confusing, stddevpfroms would be better.\n05-29-2016, 04:56 PM\nPost: #19\n Mike Elzinga Junior Member Posts: 16 Joined: May 2016\nRE: stdevp( ) appears to be mislabeled\n(05-29-2016 05:57 AM)parisse Wrote:\n(05-28-2016 08:12 PM)Mike Elzinga Wrote: \u00a0I agree; this would be the easiest fix.\n\nAlso change the help for this command to state that it calculates the sample standard deviation. In the help, just change the word \"population\" to sample.\n\nI'm afraid stddevs is also confusing, stddevpfroms would be better.\n\n05-29-2016, 05:08 PM (This post was last modified: 05-29-2016 05:09 PM by toml_12953.)\nPost: #20\n toml_12953 Senior Member Posts: 1,191 Joined: Dec 2013\nRE: stdevp( ) appears to be mislabeled\n(05-29-2016 04:56 PM)Mike Elzinga Wrote:\n(05-29-2016 05:57 AM)parisse Wrote: \u00a0I'm afraid stddevs is also confusing, stddevpfroms would be better.\n\nor even\n\nstandard_deviation_using_just_a_sample_of_the_total_population_not_the_whole_thi\u200bng\n\nTom L\n\nTom L\n\nIf the First Presbyterian Church is nicknamed First Pres,\nwhy isn't the First Methodist Church nicknamed First Meth?\n \u00ab Next Oldest | Next Newest \u00bb\n\nUser(s) browsing this thread: 1 Guest(s)", "date": "2019-11-15 07:06:18", "meta": {"domain": "hpmuseum.org", "url": "https://www.hpmuseum.org/forum/thread-6313.html", "openwebmath_score": 0.5120582580566406, "openwebmath_perplexity": 2853.7942644873897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9777138138113964, "lm_q2_score": 0.8947894696095782, "lm_q1q2_score": 0.8748480248902573}}
{"url": "http://math.stackexchange.com/questions/4948/what-is-wrong-with-my-reasoning?answertab=votes", "text": "# What is wrong with my reasoning?\n\nThe Questions\n\n70% of all vehicles pass inspection. Assuming vehicles pass or fail independently. What is the probability:\n\na) exactly one of the next 3 vehicles passes\n\nb) at most 1 of the next 3 vehicles passes\n\nThe answer to a) is .189. The way I calculated it was:\n\nP(success) * P(fail) * P(fail) +\nP(fail) * P(success) * P(fail) +\nP(fail) * P(fail) * P(success) = .7*.3*.3 + .3*.7*.3 + .3*.3*.7 = .189\n\n\nI summed the 3 possible permutations of 1 success and 2 failures.\n\nFor b) the answer is .216. To get that answer you take your answer to a) and add the probability of exactly 0 successes which is P(fail) * P(fail) * P(fail) => .189 + .3*.3*.3 = .216\n\nWhat I don't understand is why the probability of exactly 0 successes doesn't follow the pattern of exactly 1 success. Why doesn't the \"formula\" work:\n\nP(fail) * P(fail) * P(fail) +\nP(fail) * P(fail) * P(fail) +\nP(fail) * P(fail) * P(fail) = .3*.3*.3+.3*.3*.3+.3*.3*.3 = .081\n\n\n=> .189 + .081 = .27 (not .216)\n\nNow I'm wondering if I calculated the answer to a) the wrong way, and it was merely a coincidence that I got the right answer!\n\n-\n\nThe case that the first car passes and the next two fail is different from the case that the first car fails, the second passes, and the third fails. But there is only one case in which all cars fail: namely, all cars fail.\n\nThink of it with coin tosses: if you flip a penny, a nickel, and a dime (different coins, in case you are not in the U.S. or Canada), then there is only one way for all three coins to come out tails, namely, the penny comes out tails, the nickel comes out tails, and the dime comes out tails. But there are three ways in which exactly one coin comes out heads: you can have the penny come out heads and the nickel and dime come out tails; you can have the nickel come out heads and the penny and dime come out tails; or you can have the dime come out heads and the penny and nickel come out tails. These are three different outcomes, but you do not have three different ways in which the three can come out tails.\n\n-\n\nIn part a, there are exactly three ways for one out of three cars to pass, they are the three possibilities that you added. But there is only one way for all the cars to fail! The first car must fail, the second car must fail, and the third car must fail. Since there is only one way for this to happen you only consider this one probability.\n\nBtw there are also three ways for exactly two cars to pass and only one way for all three of them to pass. Hope this helps.\n\n-\n\nI'm not too confident with my probability, but I'll try my hand at an explanation.\n\nIn the first part, you are considering the next 3 trials as successive events so you can have a success in either the first spot, second spot, or third spot.\n\nIn part b, you require all 3 events to be failures so there is only 1 way that the 3 failures can occur in the next 3 events.\n\nDoes that make any sense? I never felt probability was my strong point :)\n\n-", "date": "2014-09-23 14:32:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/4948/what-is-wrong-with-my-reasoning?answertab=votes", "openwebmath_score": 0.5933423042297363, "openwebmath_perplexity": 672.0192942092472, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977713810564506, "lm_q2_score": 0.8947894682067639, "lm_q1q2_score": 0.874848020613423}}
{"url": "https://www.physicsforums.com/threads/finding-the-point-on-the-line-of-intersection-between-planes.888437/", "text": "# Finding the point on the line of intersection between planes\n\n1. Oct 9, 2016\n\n### Bishamonten\n\n1. The problem statement, all variables and given/known data\nGiven two planes, P1, P2. Find parametric equations for the line of intersection between the two planes.\n\n2. Relevant equations\nP1 = 2x -3y + 4z = 3\nP2 = x + 4y - 2z = 7\n\n3. The attempt at a solution\nLet N1 be the normal vector to P1, and N2 be the normal vector to P2. Then,\n\nN1 = (2, -3, 4) ; N2 = (1, 4, -2)\n\nThen, the direction numbers for our desired line will come from the components of the cross product between the normal vectors, because it will be parallel to our desired line.\n\nThen, N1xN2 = (-10, 8, 11).\n\nThen, Eq. of line: x = x? -10t ; y = y? +8t ; z = z? + 11t\n\nI currently need to find a point that would be on our desired line, so that I can complete the equation for this line. The book we're using unsatisfactorily just says in it's example version of this type of problem to \"let z = 0\" in both the given equations of the two planes in it's example, and then solve the system that emerges from that.\n\nMy problem is, why? How would I know ahead of time to let one of these three unknowns be 0?(In this case, the back of the book suggests to let y = 0 for my problem here, but I want to know what line of reasoning would let me come to that conclusion).\n\n2. Oct 9, 2016\n\n### andrewkirk\n\nYou can solve the problem by setting any one of the coordinates to zero, provided the line is not parallel to but not in the plane where that coordinate is zero. The line will be parallel to such a plane if the corresponding component of the line vector is zero. So you can pick any component for which the corresponding component of the line vector is nonzero. In this case that's any of the three coordinates, because none of the components of the line vector are zero.\n\nWhat you are doing by setting a coordinate equal to zero is obtaining two equations for the other two coordinates of the point where the line intersects the plane where that coordinate is zero. Those two equations will be linear in the other two coordinates, and hence can be solved to find the values of those two coordinates at the intersection point. Substituting both those values into the equation of either of the planes will allow you to calculate the value of the third coordinate at that point.\n\nLast edited: Oct 9, 2016\n3. Oct 9, 2016\n\n### Bishamonten\n\nSo I can set any of them to be 0? In this case, when I set z = 0, I get a system in which when solved yields x = 3, y = 1. Substituting these two back into one of the plane equations just tells me that z = 0.\n\nThese three values do not match the parametric equation found in the back of the book for this problem, which says that x = 17/4 - 10t ; y = 8t ; z = -11/8 + 11t, so there must be something I am not understanding here.\n\n4. Oct 9, 2016\n\n### andrewkirk\n\nThe bit you are not understanding is that parametrisations are not unique. You can pick any point on the line as the 'zero point' of the parametrised line that corresponds to where the parameter value is zero, so there are as many parametrisations as there are points on the line.\n\nBoth your parametrisation and the one in the book are correct. To see that, note that the intersection point you found is when $t=0$ in your parametrisation and $t=1/8$ in the book parametrisation.\n\nBut your parametrisation is nicer, because it has no fractions in it.\n\n5. Oct 9, 2016\n\n### Bishamonten\n\nWell that's really convenient of parameters. Thank you for clarifying that point to me!\n\n6. Oct 9, 2016\n\n### Ray Vickson\n\nYou can use the equations to solve for any two of the variables in terms of the third. For example, you can write the equations as\n$$\\begin{array}{lcr}2x - 3y &=&3-4z\\\\ x + 4y &=& 7 + 2z \\end{array}$$\nThis is a simple system of 2 linear equations in 2 unknowns, and is easily solved using elementary algebra, to get\n$$x = 3 - \\frac{10}{11} z, \\; y = 1 + \\frac{8}{11} z$$\n\nSo, one possible parametrization of the line would be $x = 3 -(10/11)t, y = 1 + ( 8/11) t, z = t$. Of course there are infinitely many points on the intersection line, but one convenient point would be obtained by putting $t = 11$, to eliminate the fractions: $x = -7, y = 9, z = 11$.", "date": "2017-12-14 08:30:59", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/finding-the-point-on-the-line-of-intersection-between-planes.888437/", "openwebmath_score": 0.801311194896698, "openwebmath_perplexity": 235.4811858147301, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138099151277, "lm_q2_score": 0.8947894668039496, "lm_q1q2_score": 0.8748480186608152}}
{"url": "https://brilliant.org/discussions/thread/scratch-pad/", "text": "# Have you ever used Scratch pad like this?\n\nFind the root of following equation.\n\n$4y^3-2700(1-y)^4=0$\n\nAbove equation can be solved using newton raphson iterative method with initial approximation to be unity.\n\n1) First Let $$f(y)=4y^3-2700(1-y)^4$$ We have to find root of above function.\n\n2)Find derivative of $$f(y)$$. Here,$$f^{'}(y)=12y^2+10800(1-y)^3$$\n\n3)Find $$y_n$$ using newton raphson method $y_0=1$ $y_{n+1}=y_n-\\frac{f(y_n)}{f^{'}(y_n)}$ $y_{n+1}=y-\\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$\n\nYou can use calculator or scratchpad to calculate $$y_1,y_2,y_3,....$$.If this equation has real solution ,these values $$y_1,y_2,y_3,....$$ would converge to root of $$f(y)$$.\n\nHow to do such iterations smartly?\n\n1) $$\\color{Purple}{\\text{On first line of scratch pad write y=1}}$$\n\n2) $$\\color{Red}{\\text{On second line write}}$$\n\n$y=y-\\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$\n\n$$\\color{Green}{\\text{You will see new value of y as result.}}$$\n\n3) $$\\color{Orange}{\\text{From line 3, continue pasting}}$$\n\n$y=y-\\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$\n\nuntil you get a constant value.If your initial approximation is not too away from actual solution,you would get constant value in no more than 10 steps.\n\nBy this method you can solve most of the transcendental equation.\n\n$$\\color{blue}{\\text{By using scratchpad, try finding sum of initial 10 fibonacci number.}}$$\n\nNote by Aamir Faisal Ansari\n3\u00a0years, 6\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nYes !i have used it this way.Sometimes in Recursion problems and like that.\n\n- 3\u00a0years, 6\u00a0months ago\n\nBy using scratchpad, try finding sum of initial 10 fibonacci number.\n\n- 3\u00a0years, 6\u00a0months ago", "date": "2019-01-18 00:23:17", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/scratch-pad/", "openwebmath_score": 0.997265636920929, "openwebmath_perplexity": 6643.045009047385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9777138092657496, "lm_q2_score": 0.8947894604912848, "lm_q1q2_score": 0.874848011907779}}
{"url": "https://math.stackexchange.com/questions/1348497/proving-phi-v-rightarrow-mathbbrn-is-linear-and-finding-matrix-represen", "text": "# Proving $\\phi: V \\rightarrow \\mathbb{R}^n$ is linear and finding matrix representation of it\n\nProblem: Let $V$ be a $n$-dimensional vectorspace and let $\\beta = \\left\\{v_1, v_2, \\ldots, v_n\\right\\}$ be a basis for $V$. Prove that the coordinate map $\\phi_{\\beta} : V \\rightarrow \\mathbb{R}^n$ is linear and determine the matrix $[\\phi_{\\beta}]_{\\beta}^{\\gamma}$ with $\\gamma$ the standard basis for $\\mathbb{R}^n$.\n\nAttempt at solution: Take vectors $v, w \\in V$. Then we have \\begin{align*} v = \\sum_{i=1}^n \\lambda_i v_i \\qquad \\text{and} \\qquad w= \\sum_{i=1}^n \\zeta_i v_i \\end{align*} for unique scalars $\\lambda_i$ and $\\zeta_i \\in \\mathbb{R}$. It follows that \\begin{align*} v+ w &= \\sum_{i=1}^n \\lambda_i v_i + \\sum_{i=1}^n \\zeta_i v_i \\\\ &= \\sum_{i=1}^n (\\lambda_i + \\zeta_i) v_i. \\end{align*} Hence we get \\begin{align*} \\phi_{\\beta}(v+w) &= (\\lambda_1 + \\zeta_1, \\lambda_2 + \\zeta_2, \\ldots, \\lambda_n + \\zeta_n) \\\\ &= (\\lambda_1, \\lambda_2, \\ldots, \\lambda_n) + (\\zeta_1, \\zeta_2, \\ldots, \\zeta_n) \\\\ &= \\phi_{\\beta} (v) + \\phi_{\\beta} (w). \\end{align*} Now take a scalar $\\mu \\in \\mathbb{R}$. Then we get \\begin{align*} \\mu v = (\\mu \\lambda_1) v_1 + (\\mu \\lambda_2) v_2 + \\ldots + (\\mu \\lambda_n) v_n, \\end{align*} from which it follows that \\begin{align*} \\phi_{\\beta}(\\mu v) &= (\\mu \\lambda_1, \\mu \\lambda_2, \\ldots, \\mu \\lambda_n) \\\\ &= \\mu (\\lambda_1, \\lambda_2, \\ldots, \\lambda_n) \\\\ &= \\mu \\phi_{\\beta} (v). \\end{align*} So $\\phi_{\\beta} : V \\rightarrow \\mathbb{R}^n$ is linear.\n\nThen for the matrixrepresentation, I'm not sure what I need to write in the columns of the matrix. If I take a vector $v \\in V$, then $\\phi_{\\beta} (v) = (\\lambda_1, \\lambda_2, \\ldots, \\lambda_n)$. And if $\\gamma$ is the standard basis for $\\mathbb{R}^n$, then this gives $\\lambda_1 (1,0,0, \\ldots, 0) + \\lambda_2 (0,1,0, \\ldots, 0) + \\ldots + \\lambda_n (0,0, \\ldots, 1)$. So In the first column just come the coordinates $\\lambda_i$? I'm confused about this.\n\n\u2022 Could you clarify what \"coordinate map\" means? \u2013\u00a0Rory Daulton Jul 3 '15 at 19:12\n\u2022 @RoryDaulton I think it's pretty clear from his attempted solution. If for $v\\in V$ I can write $v=\\lambda_1v_1+\\dots+\\lambda_nv_n$ with respec to basis $\\beta$, then the coordinate map would be $\\phi_{\\beta}(v)=(\\lambda_1,\\dots,\\lambda_n)$ \u2013\u00a0Alex Mathers Jul 3 '15 at 19:22\n\u2022 Note that that $[v]_\\beta = \\phi(v) = [\\phi(v)]_\\gamma$ \u2013\u00a0Omnomnomnom Jul 3 '15 at 19:56\n\u2022 @Omnomnomnom, so what does that mean? Was I right in stating that in the columns are just the coordinates of $v$? \u2013\u00a0Kamil Jul 3 '15 at 20:00\n\u2022 @Kamil it means that the $i$th column is the coordinates of $v_i$ with respect to $\\beta$. \u2013\u00a0Omnomnomnom Jul 3 '15 at 20:24\n\nIf $T$ is any linear transformation, the matrix $[T]^{\\gamma}_{\\beta}$ can be written $\\begin{bmatrix} [T(v_1)]_{\\gamma} & \\cdots & [T(v_n)]_{\\gamma}\\end{bmatrix}$. In other words, the $i$th column of $[T]^{\\gamma}_{\\beta}$ is $T(v_i)$ written in $\\gamma$ coordinates.\n\nFor $\\phi_{\\beta}$, the vector $\\phi_{\\beta}(v_i)$ is $e_i = (0,\\ldots0 ,1,0, \\ldots,0)$, so the $i$th column of $[\\phi_{\\beta}]_{\\beta}^{\\gamma}$ is $e_i$. This is the same as saying $[\\phi_{\\beta}]_{\\beta}^{\\gamma}$ is the identity matrix.\n\nAnother way to see this is the equation: $$[\\phi_{\\beta}]_{\\beta}^{\\gamma}[v]_{\\beta} = [\\phi_{\\beta}(v)]_{\\gamma}$$ But if $v= \\lambda_1v_1 + \\cdots + \\lambda_nv_n$, then we have $[v]_{\\beta} = [\\phi_{\\beta}(v)]_{\\gamma} = (\\lambda_1,\\ldots,\\lambda_n)$, and so the matrix $[\\phi_{\\beta}]_{\\beta}^{\\gamma}$ is the identity.\n\nYour proof that $\\phi_\\beta$ is linear is correct.\n\nYou're on the right track to compute $[\\phi_\\beta]_\\beta^\\gamma$. For a complete computation, write $$\\phi_\\beta(v_j) = \\left(\\lambda_{1j},\\dotsc,\\lambda_{nj}\\right) = \\lambda_{1j}e_1+\\dotsb+\\lambda_{nj}e_n$$ for $1\\leq j\\leq n$. Hence $$[\\phi_\\beta]_\\beta^\\gamma = \\begin{bmatrix} \\lambda_{11} & \\cdots & \\lambda_{1n} \\\\ \\vdots & \\ddots & \\vdots \\\\ \\lambda_{n1} & \\cdots & \\lambda_{nn} \\end{bmatrix}$$\n\nWould this not just be the identity matrix? For example, if we write $v=\\sum_{i=1}^{n}\\lambda_iv_i$ then can equivalently write $v=(\\lambda_1,\\dots,\\lambda_n)_{\\beta}$ with respect to the given basis. But if we have $\\phi_{\\beta}(v)=(\\lambda_1,\\dots,\\lambda_n)\\in\\mathbb{R}^n$, then we haven't changed it, we've just taken the scalar values. So the matrix would just be the identity.\n\nUnless of course, it wants the \"coordinate map\" with respect to the standard basis.\n\n\u2022 I don't understand your post. What you mean with 'then we haven't changed it'. I don't understand why it is the identity matrix. \u2013\u00a0Kamil Jul 3 '15 at 19:47", "date": "2019-07-22 18:09:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1348497/proving-phi-v-rightarrow-mathbbrn-is-linear-and-finding-matrix-represen", "openwebmath_score": 0.9999788999557495, "openwebmath_perplexity": 307.1951934144212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363494503271, "lm_q2_score": 0.8887588052782737, "lm_q1q2_score": 0.87483759792945}}
{"url": "https://math.stackexchange.com/questions/2725298/how-can-i-find-the-total-number-of-figures-in-a-puzzle-where-they-seem-to-overla", "text": "# How can I find the total number of figures in a puzzle where they seem to overlap?\n\nThe problem is as follows:\n\nIn the following sequence of figures. Which is the total number of right trapezoids which do make up $\\textrm{figure 10}$?\n\nThe figure is below:\n\nThe alternatives given are:\n\n\u2022 66\n\u2022 60\n\u2022 56\n\u2022 64\n\u2022 72\n\nWhat I tried to do is to account for the number of trapezoids:\n\n$\\textrm{In figure 1:}$\n\n\u2022 There is just $1$\n\n$\\textrm{In figure 2:}$\n\nThis is some tricky:\n\nI thought that there are $4$\n\n\u2022 $2$ make up a small individual unit\n\n\u2022 $1$ accounts for the other two\n\n\u2022 $1$ accounts for the all of them\n\n$\\textrm{In figure 3:}$\n\n\u2022 $3$ are consisting of only one figure.\n\n\u2022 $2$ are two figures.\n\n\u2022 $1$ has three figures.\n\n\u2022 $1$ accounts for the three trapezoids in horizontal arrangement and the first oblique bar next to them.\n\n\u2022 $1$ accounts for the previous set plus the second oblique bar next to them.\n\nTherefore in the figure are $8$\n\n$\\textrm{In figure 4:}$\n\n\u2022 $4$ are sets of 1 figure.\n\n\u2022 $3$ are sets of 2 figures.\n\n\u2022 $2$ are sets of 3 figures.\n\n\u2022 $1$ is a set of 4 figures.\n\n\u2022 $1$ is a set of the previous 4 figures in horizontal direction plus the first oblique bar appearing next to the trapezoids.\n\n\u2022 $1$ is the previous set and the second oblique bar.\n\n\u2022 $1$ is the previous set and the third oblique bar.\n\nThis accounts for $13$ figures.\n\nSo in total the series is comprised of.\n\n$\\textrm{1, 4, 8, 13,...}$\n\nFrom this series I found that the recursion formula is stated as:\n\n$$T_{n}=\\frac{1}{2}n^{2}+\\frac{3}{2}n-1$$\n\nBy replacing with the fourth term it seems to check.\n\n$$T_{4}=\\frac{1}{2}\\left( 4 \\right )^{2}+\\frac{3}{2}\\left( 4 \\right )-1=8+6-1= 13$$\n\nSo if what it is being asked is figure number $10$ then by replacing in the previous equation:\n\n$$T_{10}=\\frac{1}{2}\\left( 10 \\right )^{2}+\\frac{3}{2}\\left( 10 \\right )-1=50+15-1= 64$$\n\nTherefore I would choose $64$ as the number of figures in the $\\textrm{10th figure}$. I'm not very certain if is it correct?\n\nThe answer appears within the alternatives but the process to find the figures was very tedious and it took me a long time to find it, needless to say that to find the recursive formula was another problem as well but eased due the fact which I had some idea of how to find it, as a result.\n\nIs there any other method to speed up the calculations or to find an answer in this situation or to avoid miscounting or double counting the figures?\n\nThis looks correct to me. As for the more general question, I would approach the problem by asking 'how can I make up a right trapezoid from these figures?' This breaks naturally into two pieces: trapezoids that use the 'slanting' shapes on the right of the figure, and ones that don't.\n\nIf I'm going to use any of the shapes on the right then it's clear I have to use all the ones on the left, and I have to use the ones on the right from left to right; thus, there are as many ways of doing this as there are trapezoids on the right of the figure, i.e. $n-1$.\n\nOn the other hand, if I'm using on the sub-trapezoids on the left, then it has to be a contiguous block of them (e.g., the second through the fourth would be valid, but the first and third would not), but any such contiguous block will do; you should be able to see that the number of contiguous blocks is the same as the number of ways of choosing a number $i$ (the 'bottom block') between 1 and $n$ and a number $j$ (the 'top block') between $1$ and $i$; there are exactly $n(n+1)/2$ of these.\n\nSo summing up, I get a total of $n(n+1)/2+n-1 = \\frac12n^2+\\frac32n-1$ possible trapezoids, matching your answer.\n\n\u2022 First of when I solved this problem I totally ignored the fact that the figures on the right side were also trapezoids but I was \"saved\" by the fact the problem asked about counting right trapezoids. Since I'm not good with geometry, do those are also trapezoids?. They look like rectangles which had been tilted to one of their sides, anyway. I'm still lost at where does this $n-1$ comes from. Why not $n-2$?. Most of what I did was by let's say brute force or intuition not really paying attention on details, hence why It took me hours to decode this thing. \u2013\u00a0Chris Steinbeck Bell Apr 6 '18 at 17:48\n\u2022 The paragraph where you explain about the subtrapezoid thing I'm stuck at where you mention about $\\frac{n(n+1)}{2}$ How do I get to this?. All I know it is that it looks the sum from integers from $1$ to $n$ does this has something to do with the problem?. Maybe if you could rework the explanation a bit better I could understand. \u2013\u00a0Chris Steinbeck Bell Apr 6 '18 at 17:48\n\nI think you got it.\n\nCounting the stack of $n$ trapezoids in the left part of each diagram is the $n$th triangular number -- normally denoted as $T_n$ but I'll denote it $t_n$ so as not to clash with your notation. The tenth triangular number $t_{10} = 55.$\n\nAs additional explanation, let's look at the fourth diagram in your question. Let's label the stack of short, fat trapezoids (looks like a geometrical stack of pancakes) on the left part of that diagram $A,B,C,D$ from top to bottom.\n\nYou have $4$ trapezoids with the topmost trapezoid at the top ($A$ by itself, $A+B$, $A+B+C$, and $A+B+C+D$). You'd have $3$ trapezoids with the second-highest trapezoid on top ($B$ by itself, $B+C$, and $B+C+D$). And so on. So the total number of trapezoids, of any height in this stack, is $4+3+2+1 = 10$, which is the fourth triangular number.\n\nSimilarly, in the tenth diagram, it would be $10 + 9 + 8 + ... + 2 + 1 = t_{10} = 55$.\n\nA different but completely equivalent way to look at it is that there is $1$ trapezoid that is stacked $10$ high, $2$ trapezoids that are stacked $9$ high, $3$ that are stacked $8$ high ... and $10$ that are stacked $1$ high. It still adds up to $55$.\n\nSince the side pieces extend only the trapezoid that represents the entire stack, you add $n-1$ of those. (From the diagrams you have, the third diagram has two additional trapezoids from the side pieces, hence the $n-1$ formula.)\n\nAdding these two disjoint sets of trapezoids gives your answer $55+(10-1)=64$.\n\n\u2022 Okay but how do I know it was a triangular number?. Why is it a triangular number?. Reading solely to the article you mentioned it has gave me an idea but it is not very clear. i.e the second figure has only two right trapezoids in the left, so by calling that triangular number thing there are three figures only? as the second term is three?. Why should be added $n-1$?. Where does it come from?. Can you explain this with more words please? \u2013\u00a0Chris Steinbeck Bell Apr 6 '18 at 17:33\n\u2022 Added some more explanation. \u2013\u00a0John Apr 6 '18 at 17:39\n\u2022 Maybe I'm too dumb but I can't picture what you just mentioned. $10$ trapezoid with the topmost thin on the top. What does it mean?. I tried to relate their numbers with the triangular numbers again. But it does not check with the fourth term. It says $10$ but in the 4th figure there are $13$. If I ignore some of them and regard the ones in the left plus those in the right would be $4$+$4$ so there are $8$ or Am I understanding it wrong?. I revisited your answer different times but I got tangled up with how should I picture that in my mind, sorry. \u2013\u00a0Chris Steinbeck Bell Apr 6 '18 at 18:39\n\u2022 No, you're not too dumb! I tried again. See if that's clearer. \u2013\u00a0John Apr 6 '18 at 18:58\n\u2022 Thanks for adding that explanation now I can \"see\" it better. But, $55$ is not the answer, so I understand that it is required to add something, Why this something to be added has to be $n-1$? you mentioned because it is one unit less of what it is on the stack, so in this portion we do not need to perform the triangular number thing because due the position of those figures with the others makes it possible to know the number by counting them normally (hence getting their total directly) as they will not add up as if they were stacked horizontally?. \u2013\u00a0Chris Steinbeck Bell Apr 6 '18 at 20:44\n\nETA: I've added a somewhat shoddy diagram to illustrate what I mean. I've also changed the notation somewhat to be (a bit) more logical.\n\nI'm not sure this is much simpler than what you did, but I numbered the vertices down the left-hand side $B_{0, 0}, B_{1, 0}, \\ldots, B_{n, 0}$, then rightward from $B_{0, 0}$ were $B_{0, 1}, B_{0, 2}, \\ldots, B_{0, n}$, and rightward from $B_{n, 0}$ were $B_{n, 1}, B_{n, 2}, \\ldots, B_{n, n}$. Finally, down the leftmost diagonal, between $B_{0, 1}$ and $B_{n, 1}$, were $B_{1, 1}, B_{2, 1}, \\ldots, B_{n-1, 1}$.\n\nHopefully, it'll be clear that the right trapezoid must contain two vertices of the form $B_{i, 0}$ and $B_{j, 0}$ (without loss of generality, assume $i < j$), since the right angles only live there. The other two vertices must then be $B_{i, k}$ and $B_{j, k}$, for some value of $k$.\n\nThen, there are only two basic classes of right trapezoids:\n\n\u2022 If $k = 1$, then $i$ and $j$ can be any values such that $0 \\leq i < j \\leq n$.\n\n\u2022 If $2 \\leq k \\leq n$, then $i$ and $j$ must be $0$ and $n$, respectively.\n\nThere are $\\binom{n+1}{2}$ of the former, and $n-1$ of the latter, which gives us a count of\n\n$$\\binom{n+1}{2}+n-1 = \\frac{n^2+3n-2}{2}$$\n\nwhich gives us $64$ when $n = 10$, as expected.\n\n\u2022 I do not get clearly the idea of counting from the corners. I mean okay, count from the corners but I don't understand the notation you have used. To me this kind of answers would be greatly improved if it had some visual aid as I'm not good with imagination. This binomial $\\binom{n+1}{2}$ where does it comes from and why $n-1$?. \u2013\u00a0Chris Steinbeck Bell Apr 6 '18 at 17:39\n\u2022 @ChrisSteinbeckBell: I'll add a diagram when I get a chance. The $\\binom{n+1}{2}$ examples of the first class come from the fact that $i$ and $j$ can be any two of the integers $0$ through $n$, which is $n+1$ of them. The second class is $n-1$ and not $n$ because the first class already includes the trapezoid $A_0, B_{0, 1}, B_{n, 1}, A_n$, so that $k$ can only range from $2$ through $n$, which is $n-1$ possible values. \u2013\u00a0Brian Tung Apr 6 '18 at 17:41\n\u2022 Thanks for the diagram it really eased my understanding but. Where is $j$?. I assume you mean by $i$ the top bottom corner on the left side and $j$ the $1$ seen in the corners starting from $B_{n,1}$?. I can get more or less that $n-1$ is because if in the right side there are $n$ then in the left must be $n-1$ as it is seen in the diagrams. Or is it because of other reason? I'm still confused at how do I know that the total is that specific binomial?. Because is the sum from 1 to n? But if that's the case $n-1$ solely is not the sum of those trapezoids. Can you help me with that? \u2013\u00a0Chris Steinbeck Bell Apr 6 '18 at 18:32\n\u2022 @ChrisSteinbeckBell: $i$ and $j$ indicate the upper and lower horizontals of the trapezoid, respectively. Does that help enough? \u2013\u00a0Brian Tung Apr 6 '18 at 22:33", "date": "2020-05-28 13:27:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2725298/how-can-i-find-the-total-number-of-figures-in-a-puzzle-where-they-seem-to-overla", "openwebmath_score": 0.8077467679977417, "openwebmath_perplexity": 315.9864892003986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363485313247, "lm_q2_score": 0.8887588023318196, "lm_q1q2_score": 0.8748375942123766}}
{"url": "https://math.stackexchange.com/questions/3712025/is-there-a-smooth-preferably-analytic-function-that-grows-faster-than-any-funct", "text": "# Is there a smooth, preferably analytic function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}...$\n\nIs there a smooth, preferably analytic function that grows faster than any function in the sequence $$e^x, e^{e^x}, e^{e^{e^x}}$$?\n\nNote: Here the answer is NOT required to be an elementary function, as I know that otherwise the answer would be no.\n\nEdit: Michael has mentioned interpolating a series of functions, but exactly how do I do that in a smooth manner?\n\n\u2022 Does this answer your question? A function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}$... Jun 9 '20 at 5:21\n\u2022 This isn't technically a duplicate because I'm not requiring the answer to be an Elementary function. Jun 9 '20 at 5:22\n\u2022 en.wikipedia.org/wiki/Entire_function#Growth Jun 9 '20 at 5:24\n\u2022 fair enough, you are also requiring other constraints as well, agreed that this not a duplicate. Jun 9 '20 at 5:32\n\u2022 You have a sequence of functions $\\{f_k(x)\\}_{k=1}^{\\infty}$ so you can interpolate the points $\\{(k, f_k(k))\\}_{k=1}^{\\infty}$. Jun 9 '20 at 6:02\n\nThe link-only comment of metamorphy is actually a full answer, and gives an analytic function, instead of the smooth function of Michael's answer. Instead of hiding behind a link to Wikipedia, I give the construction here with some added detail.\n\nAt the end I give a smooth construction as well, and then mention a really cool generalisation I found (Carleman's Theorem.)\n\n## Set-up to apply Wikipedia's construction\n\nFirst, as in the comments, get a $$C^0$$ increasing function that's faster than all $$\\exp^{\\circ n} (x):= \\overbrace{\\exp \\big (\\exp\\big(\\dots\\exp}^{n \\text{ times}}\\big(x\\big)\\big)\\big )$$ For instance, you can define $$g(k) := \\exp^{\\circ k}(k)$$ for naturals $$k\\in\\mathbb Z_{\\ge 1}$$, and then for points in between integers you linearly interpolate, i.e. for $$t\\in(0,1)$$, define $$g(k+t):= (1-t) g(k) + t g(k+1).$$ This is faster in the sense that $$g(x)\\ge \\exp^{\\circ n}(x)$$ for all $$x\\ge n$$. This is obvious at the integers, and the fact that all $$\\exp^{\\circ n}(x)$$ are convex proves the result in between the integers (and the fact that $$\\exp^{\\circ n}(x) \\le \\exp^{\\circ (n+1)}(x)$$). The goal now is to construct an analytic function that beats $$g$$ pointwise. (If you want a smooth function, search for bump functions.)\n\n## Wikipedia's construction\n\nNow Wikipedia in wiki/Entire_function#Growth suggests we define our analytic function as a power series\n\n$$f(z) = g(2)+ \\sum_{k=1}^\\infty \\left(\\frac zk\\right)^{n_k}$$ where each $$n_k\\in 2\\mathbb Z_{\\ge 1}$$ is chosen so that $$(n_k)$$ is strictly increasing (in particular then $$n_k\\ge k$$) and $$\\left(\\frac{k+1}k\\right)^{n_k}>g(k+2).$$\n\n## Proving correctness\n\nFirst the root test to check its entire: all coefficients $$a_j$$ of the power series $$f(z)=\\sum a_j z^j$$ are either 0 or positive, so we have $$\\limsup_{j\\to\\infty} |a_j|^{1/j} = \\lim_{k\\to\\infty} \\frac1{k^{n_k/k}} \\le \\frac1k \\to 0.$$ So the radius of convergence is $$1/\\limsup |a_j|^{1/j} = \\infty$$. Since $$n_k$$ are even we only need to check the behavior for $$x\\ge0$$. Now for each $$0\\le x\\le 2$$, $$f(x)\\ge g(2) \\ge g(x)$$\uff0c and at points $$j+t$$ for $$t\\in[0,1), j\\ge 2$$, we have $$f(j+t) \\ge \\left(\\frac{j+t}{j-1}\\right)^{n_{j-1}} \\ge \\left(\\frac{j}{j-1}\\right)^{n_{j-1}} > g(j+1)> g(j+t).$$\n\n## Extra: smooth answer using bump functions\n\nI'll sketch one construction using bump functions here since the OP asked how in an edit. Its probably not the simplest, but I had it lying around for other reasons. Let $$\\phi$$ be any smooth, even, non-negative function that is identically one if $$|x|\\le 1$$, and zero for $$|x|\\ge2$$. Define for $$k\\ge 0$$, $$\\psi_k(x) := \\phi(2^{-k}x) - \\phi(2^{-(k+1)} x)$$. Then $$\\psi_k$$ is smooth and $$\\psi_k$$ is zero outside $$2^{k-1}<|x|<2^{k+1}$$. Define also $$\\psi_{-1} = \\phi(2^{-1}x)$$. If we also choose $$\\phi(x)\\le 1$$ , then $$\\psi_k\\ge 0$$. One checks that for each $$x\\in\\mathbb R$$, $$\\sum_{k\\ge -1} \\psi_k (x) = 1.$$ In fact, for each $$x$$ only at most 2 of the summands are not zero, and they sum to $$1$$. For instance, $$\\phi_{-1}\\equiv 1$$ for $$|x|\\le 1/2$$, and all other terms are 0. If $$\\frac12 < x \\le 1$$, then $$\\sum_{k\\ge -1} \\psi_k (x) = \\psi_{-1}(x) + \\psi_0(x) = \\phi(x) = 1,$$ and so on. Now you can just define (remember, for each $$x$$, only 2 summands are nonzero) $$f(x) := \\sum_{k\\ge-1} \\psi_k(x) \\exp^{\\circ k} (|x|).$$\n\n## Extra 2: Carleman's Theorem\n\nAfter learning the power series construction above, I also clicked on the Talk page on wikipedia. Apparently something that they were discussing putting into the article is the following great theorem:\n\nTheorem (Carleman) given a complex valued continuous function $$f: \\mathbb{R} \\rightarrow \\mathbb{C}$$ and a strictly positive function $$\\epsilon: \\mathbb{R} \\rightarrow \\mathbb{R}_{+}$$, there exists an entire function $$g: \\mathbb{C} \\rightarrow \\mathbb{C}$$ such that $$|f(x)-g(x)|<\\epsilon(x)$$ for every $$x \\in \\mathbb R$$.\n\nThis theorem says in particular that there are entire functions that grow at infinity as fast as you want, but also not too fast (i.e. upper and lower bounds on the growth rate), or wobble in some weird way that you precisely prescribe. Basically, any graph you draw, up to as tiny an error you want, with the error improving as $$|x|\\to \\infty$$, is the graph of an entire function restricted to $$\\mathbb R$$. That to me, is crazy!\n\nThis result was proven way back in 1927, which is somehow still under copyright protection so I can't link to a free copy (or read it myself, even if I don't understand the language). If you can find it, you can check \"Lectures on Complex Approximation\" by Dieter Gaier for a short proof taken from a paper of Kaplan, who attributes it to Brelot. The proof is some sort of mixture of the above two ideas, where a lemma is first proven to make up for the fact that you can't use partitions of unity if you want to construct an entire function. Kaplan's paper is free access and is linked below.\n\n\u2022 Carleman, T., Sur un th\u00e9or\u00e8me de Weierstra\u00df., Arkiv f\u00f6r Mat. B 20, No. 4, 5 p. (1927). ZBL53.0237.02.\n\n\u2022 Gaier, Dieter, Lectures on complex approximation. Transl. from the German by Renate McLaughlin, Boston-Basel-Stuttart: Birkh\u00e4user. XV, 196 p.; DM 94.00 (1987). ZBL0612.30003.\n\n\u2022 Kaplan, Wilfred, Approximation by entire functions, Mich. Math. J. 3, 43-52 (1956). ZBL0070.06203.\n\u2022 For a simple solution we can take say $n_k=g(2k)$, giving us $f(z) = 16+ \\sum_{k=1}^\\infty \\left(\\frac zk\\right)^{ \\exp^{\\circ 2k}(2k)}$ Jun 11 '20 at 9:31\n\u2022 @blademan9999 yup, though I believe this is specific to our question's $g$. I made my answer longer(too long? oh well) with two other solutions to the question (a smooth construction, and a reference to a high power theorem) Jun 11 '20 at 11:40\n\u2022 In Carlemans theorem, it may be good to clarify that $f$ is assumed to be continuous. (For example, the conclusion would not hold for the Dirichlet function.) Jun 11 '20 at 23:09\n\u2022 @BoazMoerman oops, thank you for the correction! Jun 11 '20 at 23:14\n\nThis gives details to my comment: Let $$\\{f_k\\}_{k=1}^{\\infty}$$ be a sequence of functions $$f_k:\\mathbb{R}\\rightarrow\\mathbb{R}$$ that satisfy the following for all $$k \\in \\{1, 2, 3, ...\\}$$:\n\n1. $$f_k(x)>0 \\quad \\forall x>0$$.\n\n2. $$f_k(x) \\leq f_{k+1}(x) \\quad \\forall x >0$$\n\n3. $$f_k(x)$$ is nondecreasing in $$x$$.\n\n4. $$\\lim_{x\\rightarrow\\infty} \\frac{f_{k+1}(x-1)}{f_k(x)} = \\infty$$\n\nYou can verify that your functions satisfy these properties. Note that: $$f_1(1) \\leq f_2(2) \\leq f_3(3) \\leq f_4(4) \\leq ...$$ So we can define $$g:\\mathbb{R}\\rightarrow\\mathbb{R}$$ as any function that is nondecreasing and that smoothly interpolates the points $$\\{(k, f_k(k))\\}_{k=1}^{\\infty}$$.\n\nThen for any positive integer $$m$$ and any $$x >m+1$$ we have: \\begin{align} \\frac{g(x)}{f_m(x)} &\\overset{(a)}{\\geq} \\frac{g(\\lfloor x\\rfloor)}{f_m(x)} \\\\ &= \\frac{f_{\\lfloor x\\rfloor}(\\lfloor x\\rfloor)}{f_m(x)} \\\\ &\\overset{(b)}{\\geq} \\frac{f_{m+1}(\\lfloor x\\rfloor)}{f_m(x)} \\\\ &\\overset{(c)}{\\geq} \\frac{f_{m+1}(x-1)}{f_m(x)} \\end{align} where (a) uses the fact that $$g$$ is nondecreasing; (b) holds because $$\\lfloor x\\rfloor \\geq m+1$$ together with property 2; (c) holds by property 3. Taking a limit as $$x\\rightarrow\\infty$$ and using property 4 gives $$\\lim_{x\\rightarrow\\infty} \\frac{g(x)}{f_m(x)} \\geq \\lim_{x\\rightarrow\\infty} \\frac{f_{m+1}(x-1)}{f_m(x)} = \\infty$$ Thus, $$g$$ grows faster than any of the $$f_m(x)$$ functions.", "date": "2022-01-19 05:44:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3712025/is-there-a-smooth-preferably-analytic-function-that-grows-faster-than-any-funct", "openwebmath_score": 0.9779298305511475, "openwebmath_perplexity": 336.07755486308344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363535858372, "lm_q2_score": 0.8887587949656841, "lm_q1q2_score": 0.8748375914538642}}
{"url": "http://mathematica.stackexchange.com/questions/26818/how-do-i-find-the-intersection-of-a-closed-3d-curve-given-by-numerical-data-with", "text": "# How do I find the intersection of a closed 3D curve given by numerical data with a specified plane?\n\nI have a list of 3D data points which form a closed curve. My own data can't be easily generated analytically, but the qualitative behaviour is well represented by a parametric curve families such as {Sin[a1 t] Cos[a2 t], Sin[a3 t] Sin[a4 t], Sin[a5 t]} where the aN are constants. Sample data is given by:\n\ndt = Table[{Sin[3 t] Cos[1 t], Sin[t] Sin[5 t], Sin[2 t]}, {t, -10, 10, 0.005}];\np1 = ListPointPlot3D[dt]\n\n\nI want to solve for the points where this closed 3D curve cuts a plane I specify but I'm am unable to do it. For example a plane at z = 0.5 as shown here:\n\np2 = Plot3D[-0.5, {x, -1, 1}, {y, -1, 1}]\nShow[p1, p2]\n\n\nNo approach I've tried so far has worked e.g. Interpolation, Nearest to -0.5, use of Select and Drop.\n\nThe data does wrap around itself multiple times and getting slightly different solution points on each pass due to differences in the stepsize or something like that is not a problem.\n\nRelated I think to this question\n\nIn brief: How can I find where a closed curve given by numerical data points cuts a specified plane? Any help you can give is much appreciated.\n\n-\nSince it's a closed curve, you could either use periodic splines, periodic Hermite interpolants, or a Fourier fit for your closed curve, and then equate the $z$-component of your curve to the value of your cutting plane. \u2013\u00a0J. M. Jun 11 '13 at 14:21\nWhy do you sample over such a broad interval? -Pi to Pi would suffice. \u2013\u00a0BoLe Jun 11 '13 at 14:51\n@BoLe, there's actually a good question in there: how might one know that his data points were sampled over more than one period? \u2013\u00a0J. M. Jun 11 '13 at 15:25\nRelated: 10640. Use zeroCrossings[Last /@ dt - 0.5] or zeroCrossings[plane @@@ dt], where plane[x, y, z] == 0 is the equation of the plane, to find the points between which the crossing occurs. \u2013\u00a0Michael E2 Jun 11 '13 at 19:42\n@0x4A4D Thanks for the suggestion and the fast response yesterday. Very useful stuff and it got me quickly into some experimenting that did achieved some success. \u2013\u00a0fizzics Jun 12 '13 at 8:42\n\nWe can use one of the zeroCrossing functions from the answers to this question to construct solutions to where an ordered list of points representing a curve crosses a surface $f(x,y,z)=0$. There are some very nice answers to the linked question with good explanations of the solutions. Use one that returns an interval of indices where the crossing takes place ,such as whuber's answer\n\nzeroCrossings[l_List] := Module[{t, u, v},\nt = {Sign[l], Range[Length[l]]} // Transpose; (*List of-1,0,1 only*)\nu = Select[t, First[#] != 0 &];               (*Ignore zeros*)\nv = Split[u, First[#1] == First[#2] &];       (*Group into runs of+and- values*)\n{Most[Max[#[[All, 2]]] & /@ v], Rest[Min[#[[All, 2]]] & /@ v]} // Transpose]\n\n\nor the SparseArray version in Mr.Wizard's answer, which seems to be quite efficient.\n\nzeroCrossings[l_List] :=\nWith[{c = SparseArray[l][\"AdjacencyLists\"]}, {c[[#]], c[[# + 1]]}\\[Transpose] &@\n\n\nThe OP's curve points, for which I suppose the corresponding values of t may be lost or unknown:\n\ndt = Table[{Sin[3 t] Cos[1 t], Sin[t] Sin[5 t], Sin[2 t]}, {t, -10, 10, 0.005}];\n\n\nTo help you understand the functions below, I first give the output of zeroCrossings. For the OP's example, we would be interest when z - 0.5 is zero. We translate down the z coordinates of points dt by 0.5 and search for the zero crossings of the translated z coordinates.\n\nzeroCrossings[Last /@ dt - 0.5]\n\n{{168, 169}, {377, 378}, {796, 797}, {1006, 1007}, {1425, 1426},\n{1634, 1635}, {2053, 2054}, {2262, 2263}, {2681, 2682}, {2891, 2892},\n{3309, 3310}, {3519, 3520}, {3938, 3939}}\n\n\nEach pair are the indices of points lying on either side of z == 0.5.\n\nBelow I use linear interpolation to find the crossing point, which seems simplest. The function zeroPt returns the point where f is zero corresponding to the zero crossing interval crossIDX. The function zeroTime returns the \"time\", interpolated between the indices of the crossing.\n\nzeroPt[pts_, crossIDX_, f_] /; Length[crossIDX] >= 3 :=\nMean[pts ~Part~ crossIDX[[2 ;; -2]]]; (* unlikely >= 4 - return all zero points? *)\nzeroPt[pts_, crossIDX_, f_] /; Length[crossIDX] == 2 :=\n(#2 f[#1] - #1 f[#2])/(f[#1] - f[#2]) & @@ pts[[crossIDX]];\n\nzeroTime[pts_, idx_, f_] /; Length[idx] >= 3 := Mean[idx];\nzeroTime[pts_, idx_, f_] /; Length[idx] == 2 := First[idx] + Rescale[0., f /@ pts[[idx]]];\n\n\n### OP's Example\n\nzeroTime[dt, #, Last[#] - 0.5 &] & /@ zeroCrossings[Last /@ dt - 0.5]\n\n{168.405, 377.843, 796.723, 1006.16, 1425.04, 1634.48, 2053.36, 2262.8,\n2681.68, 2891.12, 3310., 3519.44, 3938.32}\n\nint = zeroPt[dt, #, Last[#] - 0.5 &] & /@ zeroCrossings[Last /@ dt - 0.5]\n\n{{0.682995, 0.249992, 0.5}, {-0.183003, 0.249985, 0.5}, {0.682998, 0.249994, 0.5},\n{-0.183003, 0.249984, 0.5}, {0.68301, 0.249999, 0.5}, {-0.182995, 0.24997, 0.5},\n{0.682996, 0.249993, 0.5}, {-0.183001, 0.249981, 0.5}, {0.682997, 0.249993, 0.5},\n{-0.183005, 0.249988, 0.5}, {0.683012, 0.25, 0.5}, {-0.182995, 0.249971, 0.5},\n{0.682997, 0.249993, 0.5}}\n\n\nIf the curve is an orbit or other periodic trajectory, then the one might be interested in the unique intersections. One can gather the clusters of points with this function and take the Mean to approximate the intersection.\n\nclusters[pts_, dist_] := Gather[pts, EuclideanDistance[##] < dist &];\n\n\nExample:\n\nMean /@ clusters[int, 0.001]\n\n{{0.683001, 0.249995, 0.5}, {-0.183, 0.24998, 0.5}}\n\n\nA plot of the results:\n\nGraphics3D[{\n{Opacity[0.1], PointSize[Small], Point @ dt},\n{Red, PointSize[Medium], Point[Mean /@ clusters[int, 0.001]]}},\nAxes -> True]\n\n\n### Arbitrary planes\n\nWe can apply the method to arbitrary planes. A function plane defining the plane, plane[x, y, z] == 0, is applied to the points and we find where the values cross zero; the function is needed in both zeroCrossings and zeroPt.\n\nHere we will use ConvexHull to show the plane via a polygon containing all the intersection points. We have to project the points onto a plane, which must be chose judiciously so that the projected points are not all collinear, if possible.\n\nNeeds[\"ComputationalGeometry\"]\n\nplane[x_, y_, z_] := 2 x - 3 y - 2 z + 0.5;\nWith[{intersections = zeroPt[dt, #, plane @@ # &] & /@ zeroCrossings[plane @@@ dt]},\nGraphics3D[{\n{Lighter@Blue, PointSize[Small], Point@dt},\n{Darker@Red, PointSize[Large], Point[intersections],\nRed, Opacity[0.6],\nPolygon[#[[ConvexHull[Most /@ #]]] &[Mean /@ clusters[intersections, 0.001]]]}},\nAxes -> True]\n]\n\n\n### Other surfaces\n\nThe method works to find where any function of the points crosses zero (in this case surf).\n\nparam[u_, v_] := {u v, u, 2 v^2 - 1};\nsurf[x_, y_, z_] := Evaluate[Subtract @@ Eliminate[{x, y, z} == param[u, v], {u, v}]];\nShow[\nWith[{intersections = zeroPt[dt, #, surf @@ # &] & /@ zeroCrossings[surf @@@ dt]},\nGraphics3D[{\n{Lighter@Blue, PointSize[Small], Point@dt},\n{Red, PointSize[Large],\nPoint[Mean /@ clusters[intersections, 0.001]]}}, Axes -> True]],\nParametricPlot3D[param[u, v], {u, -1, 1}, {v, -1, 1},\nMeshStyle -> Gray, PlotStyle -> Opacity[0.3]]\n]\n\n\n-\n+1 Thank you very much for this. The code up to and including the clusters function really solves my problem. The extension to intersections with arbitrary planes and surfaces is also very instructional and useful. \u2013\u00a0fizzics Jun 13 '13 at 9:04\n\nIf you start by interpolating the each component of the curve you can easily solve for a simple plane like z=0.5:\n\ndt = Table[{Sin[3 t] Cos[1 t], Sin[t] Sin[5 t], Sin[2 t]}, {t, -10, 10, 0.005}];\nip = Interpolation[dt[[All, #]]] & /@ {1, 2, 3};\nFindRoot[ip[[3]][t] == 0.5, {t, 10}] (* ip[[3]] is the z component of the curve*)\n(* {t -> 168.404} *)\n#[t /. %] & /@ ip\n(* {0.683013, 0.25, 0.5} *)\n\n\nFor other planes you can change the FindRoot criteria, for instance FindRoot[2 ip[[3]][t] + ip[[2]][t] == 0.5, {t, 10}]\n\n-\nThanks for that. I had trouble finding all of the crossings though using FindRoot. I incorporated it into a forward seach style algorithm but I was still missing some points. I also tried to use FindInstance. Thanks for the answer. +1 \u2013\u00a0fizzics Jun 12 '13 at 9:00\n\nDefining a function P[x_,y_,z_] := a x + b y + c z for suitable constant values a, b, c, the plane is the set of all points P[x, y, z] == 0, i.e. points that satisfy this equation are on the plane. Points for which P[x, y, z] > 0 are on one side of the plane, while those giving P[x, y, z] < 0 are on the other side.\n\nSo, the points you want are those on your curve with P[x, y, z] == 0, or an interpolation between any successive points on your curve that give opposite signs for P`.\n\n-", "date": "2016-04-30 15:14:43", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/26818/how-do-i-find-the-intersection-of-a-closed-3d-curve-given-by-numerical-data-with", "openwebmath_score": 0.2239972949028015, "openwebmath_perplexity": 3996.142325929198, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9489172644875642, "lm_q2_score": 0.9219218348550491, "lm_q1q2_score": 0.874827545602009}}
{"url": "https://math.stackexchange.com/questions/2794934/determining-if-a-symmetric-matrix-is-positive-definite", "text": "# Determining if a symmetric matrix is positive definite\n\nI have a symmetric matrix where all non-diagonal elements are positive and identical, and all diagonal elements are identical as well. For example, the $3 \\times 3$ version of this matrix has the following form: $$\\left( \\begin{array}{ccc} 2a+b & a & a \\\\ a & 2a+b & a \\\\ a & a & 2a+b \\end{array} \\right)$$\n\nNote that $a>0\\ , b>0$. For such a simple form, is there an easy way of determining that the above matrix is positive definite in the general $n \\times n$ case? I'd like to show that the matrix is still positive definite when the dimension is higher. Thank you.\n\n\u2022 The criterion of Sylvester is well-adapted to your case. \u2013\u00a0Giuseppe Negro May 24 '18 at 21:37\n\u2022 @GiuseppeNegro Beat me to it. \u2013\u00a0KennyB May 24 '18 at 21:53\n\nYes. Your matrix can be written as $$(a+b)I + a ee^T$$ where $I$ is the identity matrix and $e$ is the vector of ones. This is a sum of a symmetric positive definite (SPD) matrix and a symmetric positive semidefinite matrix. Hence it is SPD.\n\n\u2022 Thank you! This is very helpful \u2013\u00a0SelinH May 24 '18 at 21:57\n\u2022 It would be cool to actually show that $ee^T$ is positive semidefinite: $x^T e e^T x = (e^Tx)^T(e^Tx)$; since $e^Tx$ is a scalar $(e^T x)^T = e^T x$ whence $x^T e e^T x = (e^T x)^2 \\ge 0$. Endorsed! Cheers! \u2013\u00a0Robert Lewis May 24 '18 at 23:18\n\u2022 @SelinH You are very welcome. \u2013\u00a0Carl Christian May 25 '18 at 9:37\n\u2022 @RobertLewis Thank your for your kind words. I often find it difficult to select the right level of detail. \u2013\u00a0Carl Christian May 25 '18 at 9:43\n\nIn addition to Carl's answer (+1) you can use the Gershgorin circle theorem which is one I'll often try when I see diagonally dominant matrices like this one.\n\nFor each row, the sum of the absolute values of the off-diagonal entries is $2a$. The theorem then says that every eigenvalue is in a closed disc of radius $2a$ centered at the diagonal elements $2a + b$, so for any eigenvalue $\\lambda$ we have $\\lambda \\geq b > 0$.\n\n$$\\begin{bmatrix} 2a+b & a & a\\\\ a & 2a+b & a\\\\ a & a & 2a+b\\end{bmatrix} = (a+b) \\, \\mathrm I_3 + a \\, 1_3 1_3^\\top = (a+b) \\, \\mathrm I_3 + 3a \\left(\\frac{\\, 1_3}{\\sqrt{3}}\\right) \\left(\\frac{\\, 1_3}{\\sqrt{3}}\\right)^\\top$$\n\nHence, the eigenvalues are $4a + b$ and $a+b$, with multiplicities $1$ and $2$, respectively. Since $a, b > 0$, both eigenvalues are positive, i.e., the given symmetric matrix is positive definite.", "date": "2019-09-22 23:09:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2794934/determining-if-a-symmetric-matrix-is-positive-definite", "openwebmath_score": 0.8519240021705627, "openwebmath_perplexity": 158.49397296301, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363737832231, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8747272887546196}}
{"url": "http://www.pythonlikeyoumeanit.com/Module3_IntroducingNumpy/Broadcasting.html", "text": "NumPy provides a mechanism for performing mathematical operations on arrays of unequal shapes:\n\n>>> import numpy as np\n\n# a shape-(3, 4) array\n>>> x = np.array([[-0. , -0.1, -0.2, -0.3],\n...               [-0.4, -0.5, -0.6, -0.7],\n...               [-0.8, -0.9, -1. , -1.1]])\n\n# a shape-(4,) array\n>>> y = np.array([1, 2, 3, 4])\n\n# multiplying a shape-(4,) array with a shape-(3, 4) array\n# y is multiplied by each row of x\n>>> x * y\narray([[-0. , -0.2, -0.6, -1.2],\n[-0.4, -1. , -1.8, -2.8],\n[-0.8, -1.8, -3. , -4.4]])\n\n\nIn effect, NumPy treated y as if its contents had been broadcasted along a new dimension, such that y was a shape-(3, 4) 2D array, which makes it compatible for multiplying with x:\n\n$$\\left( \\begin{array}{*{3}{X}} -0.0 & -0.1 & -0.2 & -0.3 \\\\ -0.4 & -0.5 & -0.6 & -0.7 \\\\ -0.8 & -0.9 & -1.0 & -1.1 \\end{array} \\right) % \\cdot \\left( \\begin{array}{*{3}{X}} 1 & 2 & 3 & 4 \\end{array}\\right) % \\rightarrow \\left( \\begin{array}{*{3}{X}} -0.0 & -0.1 & -0.2 & -0.3 \\\\ -0.4 & -0.5 & -0.6 & -0.7 \\\\ -0.8 & -0.9 & -1.0 & -1.1 \\end{array} \\right) % \\cdot \\left( \\begin{array}{*{3}{X}} 1 & 2 & 3 & 4 \\\\ 1 & 2 & 3 & 4 \\\\ 1 & 2 & 3 & 4 \\end{array}\\right) \\$$\n\nIt is important to note that NumPy doesn\u2019t really create this broadcasted version of y behind the scenes; it is able to do the necessary computations without having to redundantly copy its contents into a shape-(3,4) array. Doing so would be a waste of memory and computation. That being said, this replication process conveys exactly the mathematics of broadcast operations between arrays; thus the preceding diagram reflects how you should always envision broadcasting.\n\nBroadcasting is not reserved for operations between 1-D and 2-D arrays, and furthermore both arrays in an operation may undergo broadcasting. That being said, not all pairs of arrays are broadcast-compatible.\n\n# Broadcast multiplications between a\n# shape-(3, 1, 2) array and a shape-(3, 1)\n# array.\n>>> x = np.array([[[0, 1]],\n...\n...               [[2, 3]],\n...\n...               [[4, 5]]])\n\n>>> y = np.array([[ 0],\n...               [ 1],\n...               [-1]])\n\n# shape-(3, 1, 2) broadcast-multiply with\n# shape-(3, 1) produces shape-(3, 3, 2)\n>>> x * y\narray([[[ 0,  0],\n[ 0,  1],\n[ 0, -1]],\n\n[[ 0,  0],\n[ 2,  3],\n[-2, -3]],\n\n[[ 0,  0],\n[ 4,  5],\n[-4, -5]]])\n\n# an example of broadcast-incompatible arrays\n# a shape-(2,) array with a shape-(3,) array\n>>> np.array([1, 2]) * np.array([0, 1, 2])\nValueError: operands could not be broadcast together with shapes (2,) (3,)\n\n\nArray Broadcasting is a mechanism used by NumPy to permit vectorized mathematical operations between arrays of unequal, but compatible shapes. Specifically, an array will be treated as if its contents have been replicated along the appropriate dimensions, such that the shape of this new, higher-dimensional array suits the mathematical operation being performed.\n\nWe will now summarize the rules that determine if two arrays are broadcast-compatible with one another, and what the shape of the resulting array will be after the mathematical operation between the two arrays is performed.\n\nArray broadcasting cannot accommodate arbitrary combinations of array shapes. For example, a (7,5)-shape array is incompatible with a shape-(11,3) array. Trying to add two such arrays would produce a ValueError. The following rules determine if two arrays are broadcast-compatible:\n\nTo determine if two arrays are broadcast-compatible, align the entries of their shapes such that their trailing dimensions are aligned, and then check that each pair of aligned dimensions satisfy either of the following conditions:\n\n\u2022 the aligned dimensions have the same size\n\u2022 one of the dimensions has a size of 1\n\nThe two arrays are broadcast-compatible if either of these conditions are satisfied for each pair of aligned dimensions.\n\nNote that it is okay to have one array with a higher-dimensionality and thus to have \u201cdangling\u201d leading dimensions. Any size-1 dimension or \u201cmissing\u201d dimension will be filled-in by broadcasting the content of that array.\n\nConsidering the example from the preceding subsection, let\u2019s see that the shape-(4,3) and shape-(3,) arrays satisfy these rules for broadcast-compatibility:\n\n     array-1: 4 x 3\narray-2:     3\nresult-shape: 4 x 3\n\n\nLet\u2019s look an assortment of pairs of array-shapes and see whether or not they are broadcast-compatible:\n\n     array-1:         8\narray-2: 5 x 2 x 8\nresult-shape: 5 x 2 x 8\n\narray-1:     5 x 2\narray-2: 5 x 4 x 2\nresult-shape: INCOMPATIBLE\n\narray-1:     4 x 2\narray-2: 5 x 4 x 2\nresult-shape: 5 x 4 x 2\n\narray-1: 8 x 1 x 3\narray-2: 8 x 5 x 3\nresult-shape: 8 x 5 x 3\n\narray-1: 5 x 1 x 3 x 2\narray-2:     9 x 1 x 2\nresult-shape: 5 x 9 x 3 x 2\n\narray-1: 1 x 3 x 2\narray-2:     8 x 2\nresult-shape: INCOMPATIBLE\n\narray-1: 2 x 1\narray-2:     1\nresult-shape: 2 x 1\n\n\nNumPy provides the function broadcast_to, which can be used to broadcast an array to a specified shape. This can help us build our intuition for broadcasting. Let\u2019s broadcast a shape-(3,4) array to a shape-(2,3,4) array:\n\n# Demonstrating np.broadcast_to\n>>> x = np.array([[ 0,  1,  2,  3],\n...               [ 4,  5,  6,  7],\n...               [ 8,  9, 10, 11]])\n\n# Explicitly broadcast a shape-(3,4) array\n# to a shape-(2,3,4) array.\narray([[[ 0,  1,  2,  3],\n[ 4,  5,  6,  7],\n[ 8,  9, 10, 11]],\n\n[[ 0,  1,  2,  3],\n[ 4,  5,  6,  7],\n[ 8,  9, 10, 11]]])\n\n\nGiven the following pairs of array-shapes, determine what the resulting broadcasted shapes will be. Indicate if a pair is broadcast-incompatible.\n\n1. 7 x 2 with 7\n2. 4 with 3 x 4\n3. 1 x 3 x 1 with 8 x 1 x 1\n4. 9 x 2 x 5 with 2 x 5\n5. 3 with 3 x 3 x 2\n\n## A Simple Application of Array Broadcasting\u00b6\n\nHere we provide a simple real-world example where broadcasting is useful. Suppose you have a grade book for 6 students, each of whom have taken 3 exams; naturally, you store these scores in a shape-(6,3) array:\n\n# grades for 6 students who have taken 3 exams\n# axis-0 (rows):    student\n# axis-1 (columns): exams\n>>> import numpy as np\n>>> grades = np.array([[ 0.79,  0.84,  0.84],\n...                    [ 0.87,  0.93,  0.78],\n...                    [ 0.77,  1.00,  0.87],\n...                    [ 0.66,  0.75,  0.82],\n...                    [ 0.84,  0.89,  0.76],\n...                    [ 0.83,  0.71,  0.85]])\n\n\nWe might be interested to see how each of these scores compare to the mean score for that specific exam. Based on our discussion from the last section, we can easily compute the mean-score for each exam (rounded to 2 decimal places):\n\n# compute the mean score for each exam (rounded to 2 decimal places)\n>>> mean_exam_scores = np.round(mean_exam_scores, 2)\n>>> mean_exam_scores\narray([ 0.79,  0.85,  0.82])\n\n\ngrades is a shape-(6,3) array and mean_exam_scores is a shape-(3,) array, and we want to compute the offset of each exam score from its respective mean. At first glance, it seems like we will have to loop over each row of our grades array and subtract from it the mean_exam_scores, to compute the offset of each exam score from the respective mean-score:\n\n# Using a for-loop to compute score offsets.\n\n# Shape-(6,3) array that will store (score - mean) for each\n# exam score.\n\n# iterates over each row of grades\n# scores_per_student is a shape-(3,) array of exam scores\n# for a given student. This matches the shape of\n# mean_exam_scores, thus we can perform this subtraction\nscore_offset[n] = scores_per_student - mean_exam_scores\n\n\nGiven our discussion of vectorized operations from the last section, you should recoil at the sight of a for-loop in code that is performing array-arithmetic. We might as well get out our abacuses and spreadsheets at this point. Fortunately, we can make use of broadcasting to compute these offsets in a concise, vectorized way:\n\n# Using broadcasting to subtract a shape-(3,) array\n# from a shape-(6,3) array.\n>>> score_offset = grades - mean_exam_scores\n>>> score_offset\narray([[ 0.  , -0.01,  0.02],\n[ 0.08,  0.08, -0.04],\n[-0.02,  0.15,  0.05],\n[-0.13, -0.1 ,  0.  ],\n[ 0.05,  0.04, -0.06],\n[ 0.04, -0.14,  0.03]])\n\n\nAccording to the broadcasting rules detailed above, when you invoke grades - mean_exam_scores, NumPy will recognize that mean_exam_scores has the same shape as each row of grades and thus it will apply the subtraction operation on each row of grades with mean_exam_scores. In effect, the content of mean_exam_scores has been broadcasted to fill a shape-(6,3) array, so that the element-wise subtraction can be performed. Again, we emphasize that NumPy doesn\u2019t actually unnecessarily replicate the data of mean_exam_scores, and that this model of broadcasting merely conveys the mathematical process that is transpiring.\n\nGenerate a random array of 10,000 2D points using np.random.rand. Compute the \u201ccenter of mass\u201d of these points, which is simply the average x-coordinate and the average y-coordinate of these 10,000 points. Then, use broadcasting to compute the shape-(10000,2) array that stores the position of the points relative to the center of mass. For example, if the center of mass is $$(0.5, 1)$$, and the absolute position of a point is $$(2, 3)$$, then the position of that point relative to the center of mass is simply $$(2, 3) - (0.5, 1) = (1.5, 2)$$\n\n## Size-1 Axes & The newaxis Object\u00b6\n\n### Inserting Size-1 Dimensions into An Array\u00b6\n\nAs conveyed by the broadcasting rules, dimensions of size-1 are special in that they can be broadcasted to any size. Here we will learn about introducing size-1 dimensions into an array, for the purpose of tailoring its shape for broadcasting.\n\nYou can introduce size-1 dimensions to an array without changing the overall size (i.e.\u00a0total number of entries in an array. Thus we are free to add size-1 dimensions to an array via the reshape function. Let\u2019s reshape a shape-(3,) array into a shape-(1, 3, 1, 1) array:\n\n>>> import numpy as np\n\n# Reshaping an array to introduce size-1 dimensions.\n# The size of the array is 3, regardless of introducing\n# these extra size-1 dimensions.\n>>> np.array([1, 2, 3]).reshape(1, 3, 1, 1)\narray([[[[1]],\n\n[[2]],\n\n[[3]]]])\n\n\nThus the 1-D array with three entries has been reshaped to possess 4-dimensions: \u201cone stack of three sheets, each containing a single row and column\u201d. There is another way to introduce size-1 dimensions. NumPy provides the newaxis object for this purpose. Let\u2019s immediately demonstrate how np.newaxis can be used:\n\n# demonstrating the usage of the numpy.newaxis object\n>>> x= np.array([1, 2, 3])\n>>> y= x[np.newaxis, :, np.newaxis, np.newaxis]\n>>> y\narray([[[[1]],\n\n[[2]],\n\n[[3]]]])\n>>> y.shape\n(1, 3, 1, 1)\n\n\nIndexing x as x[np.newaxis, :, np.newaxis, np.newaxis] returns a \u201cview\u201d of x as a 4D array with size-1 dimensions inserted as axes 0, 2, and 3. The resulting array is not a copy of x; it points to the exact same data as x, but merely with a different indexing layout. This is no different than what we achieved via reshaping: x.reshape(1, 3, 1, 1).\n\n### Utilizing Size-1 Dimensions for Broadcasting\u00b6\n\nMoving on to a more pressing matter: why would we ever want to introduce these spurious dimensions into an array? Let\u2019s take an example to demonstrate the utility of size-1 dimensions.\n\nSuppose that we want to multiply all possible pairs of entries between two arrays: array([1, 2, 3]) with array([4, 5, 6, 7]). That is, we want to perform twelve multiplications, and have access to each result. At first glance, combining a shape-(3,) array with a shape-(4,) array seems inadmissible for broadcasting; we seem to be doomed to perform nested for-loops like a bunch of cavemen and cavewomen. Fortunately, we can make clever use of size-1 dimensions so that we can perform this computation in a vectorized way.\n\nLet\u2019s introduce size-1 dimensions into x:\n\n# Inserting size-1 dimensions into x and y in\n>>> x_1d = np.array([1, 2, 3])\n>>> x = x_1d.reshape(3, 1)\n>>> x\narray([[1],\n[2],\n[3]])\n\n>>> y = np.array([4, 5, 6, 7])\n\n\nx is now a shape-(3, 1) array and y is a shape-(4,) array. According to the broadcasting rules, these arrays are broadcast-compatible and will multiply to produce a shape-(3, 4) array. Let\u2019s see that multiplying these two arrays will exactly produce the twelve numbers that we are after:\n\n# broadcast-multiplying x and y\n>>> x * y\narray([[ 4,  5,  6,  7],\n[ 8, 10, 12, 14],\n[12, 15, 18, 21]])\n\n$$\\left( \\begin{array}{*{1}{X}} 1 \\\\ 2 \\\\ 3 \\end{array} \\right) % \\cdot \\left( \\begin{array}{*{4}{X}} 4 & 5 & 6 & 7 \\end{array}\\right) % \\rightarrow \\left( \\begin{array}{*{4}{X}} 1 & 1 & 1 & 1 \\\\ 2 & 2 & 2 & 2 \\\\ 3 & 3 & 3 & 3 \\end{array}\\right) % \\cdot \\left( \\begin{array}{*{4}{X}} 4 & 5 & 6 & 7 \\\\ 4 & 5 & 6 & 7 \\\\ 4 & 5 & 6 & 7 \\end{array}\\right) % = \\left( \\begin{array}{*{4}{X}} 1\\cdot4 & 1\\cdot5 & 1\\cdot6 & 1\\cdot7 \\\\ 2\\cdot4 & 2\\cdot5 & 2\\cdot6 & 2\\cdot7 \\\\ 3\\cdot4 & 3\\cdot5 & 3\\cdot6 & 3\\cdot7 \\end{array}\\right) \\$$\n\nSee that entry (i, j) of the resulting array corresponds to x_1d[i] * y[j].\n\nThrough the use of simple reshaping, shrewdly inserting size-1 dimensions allowed us to coerce NumPy into performing exactly the combination-multiplication that we desired. Furthermore, a keen understanding of what broadcasting is provides us with a clear interpretation of the structure of the result of this calculation. That is, if I reshape x to be a shape-$$(M, 1)$$ array, and y is a shape-$$(N,)$$ array, then (according to broadcasting rules) x * y would produce a shape-$$(M, N)$$ array storing the product of each of x\u2019s $$M$$ numbers with each of y\u2019s $$N$$ numbers.\n\nGiven the shape-(2,3,4) array:\n\n>>> x = np.array([[[ 0,  1,  2,  3],\n...                [ 4,  5,  6,  7],\n...                [ 8,  9, 10, 11]],\n...\n...               [[12, 13, 14, 15],\n...                [16, 17, 18, 19],\n...                [20, 21, 22, 23]]])\n\n\nNormalize x such that each of its rows, within each sheet, will sum to a value of 1. Make use of the sequential function np.sum, which should be called only once, and broadcast-division.\n\nA digital image is simply an array of numbers, which instructs a grid of pixels on a monitor to shine light of specific colors, according to the numerical values in that array.\n\nAn RGB-image can thus be stored as a 3D NumPy array of shape-$$(V, H, 3)$$. $$V$$ is the number of pixels along the vertical direction, $$H$$ is the number of pixels along the horizontal, and the size-3 dimension stores the red, blue, and green color values for a given pixel. Thus a $$(32, 32, 3)$$ array would be a 32x32 RBG image.\n\nYou often work with a collection of images. Suppose we want to store N images in a single array; thus we now consider a 4D shape-(N, V, H, 3) array. For the sake of convenience, let\u2019s simply generate a 4D-array of random numbers as a placeholder for real image data. We will generate 500, 48x48 RGB images:\n\n>>> images = np.random.rand(500, 48, 48, 3)\n\n\nUsing the sequential function np.max and broadcasting, normalize images such that the largest value within each color-channel of each image is 1.\n\nWe will conclude this section by demonstrating an important, non-trivial example of array broadcasting. Here, we will find that the most straightforward use of broadcasting is not necessarily the right solution for our problem, and we will see that it can be important to first refactor the mathematical approach taken to perform a calculation before using broadcasting. Specifically, we will see that our initial approach for using of broadcasting is memory-inefficient.\n\nSuppose we have two, 2D arrays. x has a shape of $$(M, D)$$ and y has a shape of $$(N, D)$$. We want to compute the Euclidean distance (a.k.a. the $$L_2$$-distance) between each pair of rows between the two arrays. That is, if a given row of x is represented by $$D$$ numbers $$(x_0, x_1, \\ldots, x_{D-1})$$, and similarly, a row y is represented by $$(y_0, y_1, \\ldots, y_{D-1})$$, and we want to compute the Euclidean distance between the two rows:\n\n$$\\sqrt{(x_{0} - y_{0})^2 + (x_{1} - y_{1})^2 + \\ldots + (x_{D-1} - y_{D-1})^2} = \\sqrt{\\sum_{i=0}^{D-1}{(x_{i} - y_{i})^2}}$$\n\nDoing this for each pair of rows should produce a total of $$M\\times N$$ distances. The previous subsection stepped us through a very similar calculation, albeit with lower-dimensional arrays. Let\u2019s proceed by performing this computation in three different ways:\n\n1. Using explicit for-loops\n3. Refactoring the problem and then using broadcasting\n\nFor the sake of being concrete, we will compute all of the pairwise Euclidean distances between the rows of these two arrays:\n\n# a shape-(5, 3) array\n>>> x = np.array([[ 8.54,  1.54,  8.12],\n...               [ 3.13,  8.76,  5.29],\n...               [ 7.73,  6.71,  1.31],\n...               [ 6.44,  9.64,  8.44],\n...               [ 7.27,  8.42,  5.27]])\n\n# a shape-(6, 3) array\n>>> y = np.array([[ 8.65,  0.27,  4.67],\n...               [ 7.73,  7.26,  1.95],\n...               [ 1.27,  7.27,  3.59],\n...               [ 4.05,  5.16,  3.53],\n...               [ 4.77,  6.48,  8.01],\n...               [ 7.85,  6.68,  6.13]])\n\n\nThus we will want to compute a total of 30 distances, one for each pair of rows from x and y.\n\n### Pairwise Distances Using For-Loops\u00b6\n\nPerforming this computation using for-loops proceeds as follows:\n\ndef pairwise_dists_looped(x, y):\n\"\"\"  Computing pairwise distances using for-loops\n\nParameters\n----------\nx : numpy.ndarray, shape=(M, D)\ny : numpy.ndarray, shape=(N, D)\n\nReturns\n-------\nnumpy.ndarray, shape=(M, N)\nThe Euclidean distance between each pair of\nrows between x and y.\"\"\"\n# dists[i, j] will store the Euclidean\n# distance between  x[i] and y[j]\ndists = np.empty((5, 6))\n\nfor i, row_x in enumerate(x):     # loops over rows of x\nfor j, row_y in enumerate(y): # loops over rows of y\n# Subtract corresponding entries of the rows,\n# squares each difference, and then sums them. This\n# exactly matches our equation for Euclidean\n# distance (we will do the square root later)\ndists[i, j] = np.sum((row_x - row_y)**2)\n\n# we still need to take the square root of\n# each of our numbers\nreturn np.sqrt(dists)\n\n\nBe sure to step through this code and see that dists stores each pair of Euclidean distances between the rows of x and y.\n\n### Pairwise Distances Using Broadcasting (Unoptimized)\u00b6\n\nNow, let\u2019s use of vectorization to perform this distance computation. It must be established immediately that the method that we about to develop here is memory-inefficient. We will address this issue in detail at the end of this subsection.\n\nWe start off our vectorized computation by shrewdly inserting size-1 dimensions into x and y, so that we can perform $$M \\times N$$ subtractions between their pairs of length-$$D$$ rows. This creates a shape-$$(M, N, D)$$ array.\n\n# subtract shape-(5, 1, 3) with shape-(1, 6, 3)\n# produces shape-(5, 6, 3)\n>>> diffs = x.reshape(5, 1, 3) - y.reshape(1, 6, 3)\n>>> diffs.shape\n(5, 6, 3)\n\n\nIt is important to see, via broadcasting, that diffs[i, j] stores x[i] - y[j]. Thus we need to square each entry of diffs, sum over its last axis, and take the square root, in order to produce our $$M \\times N$$ Euclidean distances:\n\n# producing the Euclidean distances\n>>> dists = np.sqrt(np.sum(diffs**2, axis=2))\n>>> dists.shape\n(5, 6)\n\n\nVoil\u00e0! We have produced the distances in a vectorized way. Let\u2019s write this out formally as a function:\n\ndef pairwise_dists_crude(x, y):\n\"\"\"  Computing pairwise distances using vectorization.\n\nParameters\n----------\nx : numpy.ndarray, shape=(M, D)\ny : numpy.ndarray, shape=(N, D)\n\nReturns\n-------\nnumpy.ndarray, shape=(M, N)\nThe Euclidean distance between each pair of\nrows between x and y.\"\"\"\n# The use of np.newaxis here is equivalent to our\n# use of the reshape function\nreturn np.sqrt(np.sum((x[:, np.newaxis] - y[np.newaxis])**2, axis=2))\n\n\nRegrettably, there is a glaring issue with the vectorized computation that we just performed. Consider the largest sized array that is created in the for-loop computation, compared to that of this vectorized computation. The for-loop version need only create a shape-$$(M, N)$$ array, whereas the vectorized computation creates an intermediate array (i.e. diffs) of shape-$$(M, N, D)$$. This intermediate array is even created in the one-line version of the code. This will create a massive array if $$D$$ is a large number!\n\nSuppose, for instance, that you are finding the Euclidean between pairs of RGB images that each have a resolution of $$32 \\times 32$$ (in order to see if the images resemble one another). Thus in this scenario, each image is comprised of $$D = 32 \\times 32 \\times 3 = 3072$$ numbers ($$32^2$$ pixels, and each pixel has 3 values: a red, blue, and green-color value). Computing all the distances between a stack of 5000 images with a stack of 100 images would form an intermediate array of shape-$$(5000, 100, 3072)$$. Even though this large array only exists temporarily, it would have to consume over 6GB of RAM! The for-loop version requires $$\\frac{1}{3027}$$ as much memory (about 2MB).\n\nIs our goose cooked? Are we doomed to pick between either slow for-loops, or a memory-inefficient use of vectorization? No! We can refactor the mathematical form of the Euclidean distance in order to avoid the creation of that bloated intermediate array.\n\n### Optimized Pairwise Distances\u00b6\n\nPerforming the pairwise subtraction between the respective rows of x and y is what created the over-sized intermediate array in our previous calculation. Thus we want to rewrite the Euclidean distance equation such that none of the terms require broadcasting beyond the size of $$M \\times N$$.\n\nThe Euclidean distance equation, ignoring the square root for now, can be refactored by multiplying out each squared term as so:\n\n$$\\sum_{i=0}^{D-1}{(x_{i} - y_{i})^2} = \\sum_{i=0}^{D-1}{x_{i}^2} + \\sum_{i=0}^{D-1}{y_{i}^2} - 2\\sum_{i=0}^{D-1}{x_{i} y_{i}}$$\n\nKeep in mind that we must compute this for each pair of rows in x and y. We will find that this formulation permits the use of matrix multiplication, such that we can avoid forming the shape-$$(M, N, D)$$ intermediate array.\n\nThe first two terms in this equation are straight-forward to calculate, and, when combined, will only produce a shape-$$(M, N)$$ array. For both x and y, we square each element in the array and then sum over the columns for each row:\n\n# Computing the first two terms of the\n# refactored Euclidean distance equation\n\n# creates a shape-(5,) array\n>>> x_sqrd_summed = np.sum(x**2, axis=1)\n\n# creates a shape-(6,) array\n>>> y_sqrd_summed = np.sum(y**2, axis=1)\n\n\nWe must insert a size-1 dimension in x so that we can add all pairs of numbers between the resulting shape-$$(M, 1)$$ and shape-$$(N,)$$ arrays. This will compute $$\\sum_{i=0}^{D-1}{x_{i}^2} + \\sum_{i=0}^{D-1}{y_{i}^2}$$ for all of the $$M \\times N$$ pairs of rows:\n\n# add a shape-(5, 1) array with a shape-(6, ) array\n# to create a shape-(5, 6) array\n>>> x_y_sqrd = x_sqrd_summed[:, np.newaxis] + y_sqrd_summed\n>>> x_y_sqrd.shape\n(5, 6)\n\n\nThis leaves the third term to be computed. It is left to the reader to show that computing this sum of products for each pair of rows in x and y is equivalent to performing the matrix multiplication $$-2\\;(x \\cdot y^{T})$$, where y has been transposed so that it has a shape of $$(D, N)$$. This matrix multiplication of a shape-$$(M, D)$$ array with a shape-$$(D, N)$$ array produces a shape-$$(M, N)$$ array. Therefore, we can compute this final term without needing to create a larger, intermediate array.\n\nThus the third term in our equation, $$-2\\sum_{i=0}^{D-1}{x_{i} y_{i}}$$, for all $$M \\times N$$ pairs of rows, is:\n\n# computing the third term in the distance\n# equation, for all pairs of rows\n>>> x_y_prod = -2 * np.matmul(x, y.T)  # np.dot can also be used to the same effect\n>>> x_y_prod.shape\n(5, 6)\n\n\nHaving accounted for all three terms, we can finally compute the Euclidean distances:\n\n# computing all the distances\n>>> dists = np.sqrt(x_y_sqrd + x_y_prod)\n>>> dists.shape\n(5, 6)\n\n\nIn total, we have successfully used vectorization to compute the all pairs of distances, while only requiring an array of shape-$$(M, N)$$ to do so! This is the memory-efficient, vectorized form - the stuff that dreams are made of. Let\u2019s write the function that performs this computation in full.\n\ndef pairwise_dists(x, y):\n\"\"\" Computing pairwise distances using memory-efficient\nvectorization.\n\nParameters\n----------\nx : numpy.ndarray, shape=(M, D)\ny : numpy.ndarray, shape=(N, D)\n\nReturns\n-------\nnumpy.ndarray, shape=(M, N)\nThe Euclidean distance between each pair of\nrows between x and y.\"\"\"\ndists = -2 * np.matmul(x, y.T)\ndists +=  np.sum(x**2, axis=1)[:, np.newaxis]\ndists += np.sum(y**2, axis=1)\nreturn  np.sqrt(dists)\n\n\nTakeaway:\n\nThe specific form of an equation can have a major impact on the memory-footprint of its vectorized implementation in NumPy. This issue can be safely overlooked in cases where you can be certain that the array shapes at play will not lead to substantial memory consumption. Otherwise, take care to study the form of the equation, to see if it can be recast in a way that alleviates its memory-consumption bottlenecks.\n\nReading Comprehension: Checking the equivalence of the three pairwise distance functions\n\nUse the function numpy.allclose to verify that the three methods for computing the pairwise distances produce the same numerical results.\n\nGenerating the random array of 10,000, 2D points, and their \u201ccenter-of-mass\u201d.\n\n# find the mean x-coord and y-coord of the 10000 points\n>>> pts = np.random.rand(10000, 2)\n>>> center_of_mass = pts.mean(axis=0)  # -> array([mean_x, mean_y])\n>>> center_of_mass.shape\n(2,)\n\n# Use broadcasting to compute the position of each point relative\n# to the center of mass. The center of mass coordinates are subtracted\n# from each of the 10000 points, via broadcast-subtraction\n>>> relative_pos = pts - center_of_mass # shape-(10000,2) w/ shape-(2,)\n>>> relative_pos.shape\n(10000, 2)\n\n\n1. Incompatible\n2. 3 x 4\n3. 8 x 3 x 1\n4. 9 x 2 x 5\n5. Incompatible\n\n# a shape-(2, 3, 4) array\n>>> x = np.array([[[ 0,  1,  2,  3],\n...                [ 4,  5,  6,  7],\n...                [ 8,  9, 10, 11]],\n...\n...               [[12, 13, 14, 15],\n...                [16, 17, 18, 19],\n...                [20, 21, 22, 23]]])\n\n# sum over each row, within a sheet\n>>> summed_rows = x.sum(axis=1)\n>>> summed_rows\narray([[12, 15, 18, 21],\n[48, 51, 54, 57]])\n\n# this shape-(2,4) array can be broadcast-divided\n# along the rows of x, if we insert a size-1 axis\n>>> x_norm = x / summed_rows[:, np.newaxis, :]\n\n# verifying the solution\n>>> x_norm.sum(axis=1)\narray([[ 1.,  1.,  1.,  1.],\n[ 1.,  1.,  1.,  1.]])\n\n\n# a collection of 500 48x48 RGB images\n>>> images = np.random.rand(500, 48, 48, 3)\n\n# finding the max-value within each color-channel of each image\n>>> max_vals = images.max(axis=(1,2))\n>>> max_vals.shape\n(500, 3)\n\n# we can insert size-1 dimensions so that we can\n# the pixels of the images.\n# broadcasting (500, 48, 48, 3) with (500, 1, 1, 3)\n>>> normed_images = images / max_vals.reshape(500, 1, 1, 3)\n\n# checking that all the max-values are 1\n>>> normed_images.max(axis=(1,2))\narray([[ 1.,  1.,  1.],\n[ 1.,  1.,  1.],\n[ 1.,  1.,  1.],\n...,\n[ 1.,  1.,  1.],\n[ 1.,  1.,  1.],\n[ 1.,  1.,  1.]])\n\n# a rigorous check\n>>> np.all(normed_images.max(axis=(1,2)) == 1)\nTrue\n\n\nChecking the equivalence of the three pairwise distance functions: Solution\n\nnumpy.allclose returns True if all pairwise elements between two arrays are almost-equal to one another.\n\n>>> x = np.array([[ 8.54,  1.54,  8.12],\n...               [ 3.13,  8.76,  5.29],\n...               [ 7.73,  6.71,  1.31],\n...               [ 6.44,  9.64,  8.44],\n...               [ 7.27,  8.42,  5.27]])\n\n>>> y = np.array([[ 8.65,  0.27,  4.67],\n...               [ 7.73,  7.26,  1.95],\n...               [ 1.27,  7.27,  3.59],\n...               [ 4.05,  5.16,  3.53],\n...               [ 4.77,  6.48,  8.01],\n...               [ 7.85,  6.68,  6.13]])\n\n>>> np.allclose(pairwise_dists_looped(x, y), pairwise_dists_crude(x, y))\nTrue\n\n>>> np.allclose(pairwise_dists_crude(x, y), pairwise_dists(x, y))\nTrue", "date": "2019-03-21 22:33:29", "meta": {"domain": "pythonlikeyoumeanit.com", "url": "http://www.pythonlikeyoumeanit.com/Module3_IntroducingNumpy/Broadcasting.html", "openwebmath_score": 0.5890757441520691, "openwebmath_perplexity": 2337.0202310202303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985936373783223, "lm_q2_score": 0.8872045877523147, "lm_q1q2_score": 0.8747272740523564}}
{"url": "https://gateoverflow.in/43576/gate2003-62", "text": "# GATE2003-62\n\n8.5k views\n\nIn a permutation $a_1 ... a_n$, of $n$ distinct integers, an inversion is a pair $(a_i, a_j)$ such that $i < j$ and $a_i > a_j$.\n\nWhat would be the worst case time complexity of the Insertion Sort algorithm, if the inputs are restricted to permutations of $1. . . n$ with at most $n$ inversions?\n\n1. $\\Theta(n^2)$\n2. $\\Theta(n\\log n)$\n3. $\\Theta(n^{1.5})$\n4. $\\Theta(n)$\n\nedited\n0\nwhat is inversion???????????????????????\n0\nLet A[1..N] be an array of distinct elements.\nIf i < j and A[i]>A[j] the pair i,j is known as inversion of the Array.\n0\n\nAs the question says, how the Inversion is defined .\n\nIn a permutation $a_1 ... a_n$, of n distinct integers, an inversion is a pair $(a_i, a_j)$ such that $i < j$ and $a_i > a_j$.\n\n\u2022 One important thing to see is Difference between swap and Inversions.\n\u2022 Swapping is done explicitly by the programmer, hence a explicit feature whereas, Inversion is an implicit feature which is defined in the input .\n\nEx :- Take the input => $\\left \\{ 0,1,9,8,10,7,6,11 \\right \\}$\n\nHow many Inversions here : $\\left \\{ 9,8 \\right \\}$ , $\\left \\{ 9,7 \\right \\}$ , $\\left \\{ 9,6 \\right \\}$ ,$\\left \\{ 8,7 \\right \\}$ , $\\left \\{ 8,6 \\right \\}$ , \u00a0$\\left \\{ 10,7 \\right \\}$\u00a0and\u00a0$\\left \\{ 10,6 \\right \\}$. Hence, it is an implicit feature of the input and not any explicit operation done (Swap) .\n\nActual Time complexity of Insertion Sort is defined as\u00a0$\\Theta \\left ( N + f(N) \\right )$, where\u00a0$f\\left ( N \\right )$\u00a0is the total number of Inversions .\n\nEx :- Again take the input => $\\left \\{ 0,6,7,8,9,10,1 \\right \\}$\n\nHere, how many comparisons will be there to place $1$ in the right place ?\n\n\u2022 First of all, $1$ is compared with $10$ - Returns True as it is smaller than $10$.\n\u2022 Then, with $9$ - again True.\n\u2022 Similar, happens with $6,7,8$ - All return True .\n\nHence, There $5$ comparisons were the Inversions and $1$ more comparison\u00a0will be there, in which outer while loop fails .\n\nFor, placing $1$ in the right place $\\bf{6}$ comparisons are there .\n\nHence, Total Comparisons in the Insertion sort will be :- Total number of elements for which our while loop fails + Total number of inversions in the input\n\n\u2022 Total number of elements for which our while loop fails :- Suppose the input $\\left \\{ 0,6,7,8,9,10,1 \\right \\}$. Here, first $0$ will be kept as it is and one while loop fail comparison for each element, hence total comparisons like this :- $\\left ( N-1 \\right )=O\\left ( N \\right )$\u00a0comparisons.\n\u2022 Total number of inversions in the input :- Best Case : $0$ and in the Worst Case :\u00a0$\\frac{N\\left ( N-1 \\right )}{2} = O\\left ( N^{2} \\right )$\n\nTotal Time complexity of insertion sort :\u00a0$\\Theta (N+f(N))$\n\nIt is given here that at most\u00a0$N$ inversions are there, so we get the best Time complexity :- $\\Theta (N+f(N)) = \\Theta (N + N) = \\Theta (N)$ .\n\nCorrect Answer: $D$\n\nedited\n1\nExcellent explanation Sir!\n0\n0\nIn worst case when the array is sorted in descending order the complexity would be O(n^2). But since its restricted to n inversions, so the complexity would be now O(n).\n1\nRestriction on inversi\u00f3n to n mean array is almost sorted in increasing order.let us take an example\n\n5 1 2 3 4\n\nHere insertion sort will stop after first pass by making array sorted and inversion will be number of swap\n0\n\n@Kapil\u00a0sir you saved my many hours the key is just read the question carefully : they explicit\u00a0mentioned that number of inversions are restricted up to n then why we considering O(n^2) case it should be done only in O(n) time.\n\n0\n\nThe link mentioned is not working.\n\nhttps://gateoverflow.in/253812/gatebook-2019-ds3-9\n0\ncould u please elaborate more on \"Actual Time complexity of Insertion Sort is defined as \u0398(N+f(N))\u0398(N+f(N)), where f(N)f(N) is the total number of Inversions .\" ??? I didn't get this point\n0\nThe number of comparisons in insertion sort are O($N^{2}$) then how is the total time complexity O(N)??\n\n1\nI think (7,6) should also be an inversion.\n\nANSWER: D. $\\Theta(n)$\n\nREASON:\n\nCount of number of times the inner loop of insertion sort executes is actually equal to number of inversions in input permutation $a_1,a_2,\\dots a_n$. Since for each value of $i = 1... n$, $j$ take the value $1... i-1$, which means for every $j<i$ it checks if $a[j] > a[i]$.\n\nIn any given permutation, maximum number of inversions possible is $n(n-1)/2$ which is $O(n^2)$. It is the case where the array is sorted in reverse order. Therefore, to resolve all inversions i.e., worst case time complexity of insertion sort is $\\Theta(n^2)$.\n\nHowever, as per the question the number of inversion in input array is restricted to $n$. The worst case time complexity of insertion sort reduces to $\\Theta(n)$.\n\nINSERTION SORT ALGORITHM (for reference)\n\nedited by\n0\narray could be 5,4,3,2,1\n\nin best case of insertion sort will take O(n)\n\nBut how in worst case O(n) ?\n13\n\nIf the array elements are 5,4,3,2,1..then number of inversion equals 10 which is twice the size of array. Question restrict the number of inversions to $n$.\n\nBest case of insertion sort means that the inner while loop was never executed at all. That will happen when there is no inversion at all (array is sorted). Since, the outer loop has traversed the element of array once therefore the complexity is O(n) in best case.\n\nHowever in this particular case we should not make a claim that the inner loop will not be executed. Actually it will execute, but only O(n) time in total (not just for some iteration of outer loop). So in worst case i.e., irrespective of whatever is the given permutation, as long as number of inversion in that permutation is $n$ insertion sort will sort that array in O(n). That actually the claim here.\n\n0\nmeans in 5,4,3,2,1\n\n4 inversion it will be 1,5,4,3,2\n\nthen 1 inversion 1,5,4,2,3....We have to stop here\n6\n@srestha inversions are not operations (swaps) done by us- it is a property of the input. 5,4,3,2,1 has 4+3+2+1 = 10 inversions.\n0\nok , so here atmost n inversion . that is why complexity O(n).rt?\n\nAs we know in insertion sort complexity depend on number of inversions rt?\n4\nyes, exactly. This question is about analysing the complexity of insertion sort.\n3\nOuter loop executes n times. Inner loop will also be executed n times as there are n inversions.\n\nSo complexity will be $\\Theta$ (n + n) . So $\\Theta$ (n)\n\nhttp://www.geeksforgeeks.org/time-complexity-insertion-sort-inversions\n0\nplease add that running time is \u0398(n+d), where n we get from outer loop and d is the \u00a0number of inversion.\n\nhere number of inversion=n\n\nso time complexity=\u0398(n+n)= \u0398(n)\n0\ninversions are not operations (swaps) done by us- it is a property of the input. 5,4,3,2,1 has 4+3+2+1 = 10 inversions.\n@Arjun\ni can't get it sir .\n\nAns is D: \u229d(n),\n\nInsertion sort runs in \u0398(n + f(n)) time, where f(n) denotes the number of inversion initially present in the array being sorted.\n\n1 vote\nYou can learn this point:-\n\nInsertion sort complexity is\n\n(N+ no. of inversion).\n1 vote\n\nJust take a simple case and examine:\n\nUsually, the worst case corresponds to whenever input is in decreasing order, so\n\nlet Input = 5,4,3,2,1\n\nNow, we need to make sure the inversion property applies i.e. n=5 inversions are only possible\n\nFor the given I, there's 10 inversions. With the number 5, there's 4 inversions - <5,4> , <5,3>, <5,2>, <5,1>. We just need 1 more. By rearranging, let I = 5,1,2,4,3. Now there's 5 inversions. We have a perfect example close to worst case.\n\nNow apply the insertion sort Algo to I.\n\nIteration(i,j)\u00a0\u00a0Input\n\n0,0 - 5,1,2,4,3\n\n1,0 - 1,5,2,4,3\n\n2,0 - 1,2,5,4,3\n\n3,0 - 1,2,4,5,3\n\n4,0 - 1,2,4,3,5\n\n4,1 - 1,2,3,4,5\n\nTotal Iterations = 5. Therefore Theta(n), option D\n\n0\n\n@arjuno\n\ncan you explain little more how you find total iteration and what is that 1,0\u00a0 2,0 ....4,0\u00a0 4,1\u00a0 stands for?\n\n1 vote\n\nTime complexity of insertion sort is: $O(n + i)$ where $i$ is the number of inversions.\n\nIn the worst case, there are $n(n-1)/2=O(n^2)$ inversions, and hence the Time Complexity of Insertion Sort degrades to $O(n+n^2)=O(n^2)$.\n\nHere, in the worst case we're restricted to $n$ inversions, so Time Complexity = $\\theta(n+n)=\\theta(n)$\n\nOption D\n\ninputs are restricted to permutations of $1...n$ with at most $n$ inversions\n\nso if we take input as $n,1,2\u2026.n-1,n-2$\n\nhere no. of inversions is$n$\n\ne.g. if n=9\n\n$9,1,2,3,4,5,6,8,7$\n\ninversions are\n\n$(9,1)(9,2)(9,3)(9,4)(9,5)(9,6)(9,8)(9,7)(8,7) i.e. 9$\n\nso worst case input according to question is \u00a0$n,1,2\u2026.n-1,n-2$\n\nthis will take O(N) time to sort using insertion sort as except n & n-2 all elements are sorted\nago\n\n## Related questions\n\n1\n7.3k views\nIn a permutation $a_1 ... a_n$, of n distinct integers, an inversion is a pair $(a_i, a_j)$ such that $i < j$ and $a_i > a_j$. If all permutations are equally likely, what is the expected number of inversions in a randomly chosen permutation of $1. . . n$? $\\frac{n(n-1)}{2}$ $\\frac{n(n-1)}{4}$ $\\frac{n(n+1)}{4}$ $2n[\\log_2n]$\nThe unusual $\\Theta(n^2)$ implementation of Insertion Sort to sort an array uses linear search to identify the position where an element is to be inserted into the already sorted part of the array. If, instead, we use binary search to identify the position, the worst case running time will remain $\\Theta(n^2)$ become $\\Theta(n (\\log n)^2)$ become $\\Theta(n \\log n)$ become $\\Theta(n)$\nIn the following $C$ program fragment, $j$, $k$, $n$ and TwoLog_n are integer variables, and $A$ is an array of integers. The variable $n$ is initialized to an integer $\\geqslant 3$, and TwoLog_n is initialized to the value of $2^*\\lceil \\log_2(n) \\rceil$ for (k = 3; k <= n; k++) A[k] ... $\\left\\{m \\mid m \\leq n, \\text{m is prime} \\right\\}$ { }\nLet $G= (V,E)$ be a directed graph with $n$ vertices. A path from $v_i$ to $v_j$ in $G$ is a sequence of vertices ($v_{i},v_{i+1}, \\dots , v_j$) such that $(v_k, v_{k+1}) \\in E$ for all $k$ in $i$ through $j-1$. A simple path is a path in which no vertex ... longest path length from $j$ to $k$ If there exists a path from $j$ to $k$, every simple path from $j$ to $k$ contains at most $A[j,k]$ edges", "date": "2020-09-24 17:19:42", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/43576/gate2003-62", "openwebmath_score": 0.7147854566574097, "openwebmath_perplexity": 697.9091539028315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464478051828, "lm_q2_score": 0.8962513731336204, "lm_q1q2_score": 0.8746933439502742}}
{"url": "http://math.stackexchange.com/questions/461601/methods-for-proving-an-equivalence-relation", "text": "# Methods for proving an equivalence relation\n\nI'll be taking introductory abstract algebra in the fall, and so to prepare, I'm working through Pinter's text. Chapter 12 includes a number of exercises asking the student to prove that something is an equivalence relation and to describe the associated partition. For example: In $\\mathbb{Q}$, $r \\sim s$ iff $r - s \\in \\mathbb{Z}$.\n\nI think I'm okay with most (?) of this. I'd show this is an equivalence relation like so: If $x, y, z \\in \\mathbb{Q}$, then $x - x = 0 \\in \\mathbb{Z}$, and so $x \\sim x$. Second, if $x \\sim y$, then $y - x = -(x - y) \\in \\mathbb{Z}$, and so $x \\sim y \\implies y \\sim x$. Finally, if $x \\sim y$ and $y \\sim z$, then $x - z = (x - y) - (z - y) \\in \\mathbb{Z}$, and so $x \\sim z$.\n\nThe two parts I'm not sure about: First, that last f on the iff. Essentially, this means I have to prove that if $r \\sim s$, then their difference is an integer, right? But I thought this is merely how this particular equivalence relation is defined. How do I know that $r \\sim s$ until I look at their difference?\n\nSecond, I have a basic idea of what the partition is, but I'm not sure how to form the statement. The equivalence class $[q] = \\{k + q : k \\in \\mathbb{Z}, q \\in \\mathbb{Q} \\}$. But if anything, that seems as though it would be the definition for a single equivalence class, not the description of the partition. (Now that I look at it again, it also leaves out the fact that $k$ is arbitrary but $q$ is fixed.)\n\nIf anyone can offer any hints, I'd very much appreciate it. (Though I'm guessing the second of my questions might be more amenable to a No-this-is-how-you-do-it than to a hint, per se.)\n\n-\nThe text defines $r$ to be equivalent to $s$ iff (if and only if) the difference is an integer. There is nothing for you to prove, the difference being an integer is the defining condition for the equivalence relation. \u2013\u00a0Andr\u00e9 Nicolas Aug 7 '13 at 4:32\n@Andr\u00e9Nicolas So are you saying the question is poorly written? I guess I was originally reading it like \"r R s <--> r - s in Z,\" where R is not necessarily an equivalence relation. But ~ usually does mean equivalence relation, right? \u2013\u00a0dmk Aug 9 '13 at 14:53\nThe question is not poorly written, it is quite clear. But to show this is an equivalence relation, all you need to do is to verify that the conditions for an equivalence relation are satisfied. You don't have to prove that if $r\\equiv s$ then $r-s$ is an integer, you have been told that is what $\\equiv$ means for this exercise, \u2013\u00a0Andr\u00e9 Nicolas Aug 9 '13 at 14:59\n@Andr\u00e9Nicolas So what you were saying then is that there's nothing for me to prove beyond what I'd already proved? \u2013\u00a0dmk Aug 9 '13 at 15:40\nYes, you had completely finished proving that the relation is an equivalence realtion. And it is a clean well-organized proof. \u2013\u00a0Andr\u00e9 Nicolas Aug 9 '13 at 16:40\n\nHint\n\nProve that the associated partition is $$\\{[q]\\, |\\, q\\in[0,1)\\}$$\n\n-\nMy first thought was, OK, that's true, but is it necessary to be so restrictive? For example, in Z_4, [1] = [5] = [61] = ... Well, sure, but we only need [1]. Two days later, though, it clicked: \"Succinct\" is a better adjective than \"restrictive.\" (And to complete the analogy for myself, the partition of Z_4 would be {[k] : k = 0, 1, 2, 3}.) (Not sure how to do the @ thing here...) \u2013\u00a0dmk Aug 9 '13 at 15:16\nIf we want to give a set usually we write its elements without repetition. \u2013\u00a0user63181 Aug 9 '13 at 15:28\n\nFor your first part: Definitions are iff, or if and only if statements, as they are essentially stating that two things are equivalent -- the new term, and its definition. It shouldn't be affecting what you need to prove, except that you can use both that $r \\sim s \\implies r - s \\in \\mathbb{Z}$ and that $r - s \\in \\mathbb{Z} \\implies r \\sim s$.\n\nFor the second part, a better way of putting it would be as follows:\n\nFor each $q \\in \\mathbb{Q}$, the equivalence class of $q$ under $\\sim$ is as follows: $[q] = \\{q + k : k \\in \\mathbb{Z}\\}$.\n\nBy saying this, you describe all of the equivalence classes making up the partition in one statement, and also take care of having $q$ fixed and $k$ varying.\n\n-\n\nThis post is to answer the question above:\n\nHow do I know that r\u223cs until I look at their difference?\n\nAnswer: Use an indexing number m, with m = r - g(r), with g(r) being the next lower integer from r. Then each class can be represented as Am (I don't know how to subscript). m will be some rational number between zero (inclusive) and 1 (non-inclusive).\n\n-", "date": "2016-02-08 04:30:06", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/461601/methods-for-proving-an-equivalence-relation", "openwebmath_score": 0.8514342308044434, "openwebmath_perplexity": 286.3566001489136, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830769252026, "lm_q2_score": 0.899121388082479, "lm_q1q2_score": 0.8746929516192274}}
{"url": "https://mathhelpboards.com/threads/5-vertices-denoted-by-k5-part-b.5864/", "text": "# 5 Vertices Denoted by K5 part B\n\n#### Joystar1977\n\n##### Active member\nConsider the complete graph with 5 vertices, denoted by K5.\n\nB. How many edges are in K5? How many edges are in Kn?\n\nWouldn't the edges be at certain points of the graph? For instance, Point 1, Point 2, Point 3, Point 4, and Point 5 or n-1, n-2, n-3, n-4, and n-5.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nI don't quite understand what you mean by\nWouldn't the edges be at certain points of the graph? For instance, Point 1, Point 2, Point 3, Point 4, and Point 5 or n-1, n-2, n-3, n-4, and n-5.\nAn edge in a (simple) graph is an unordered pair of distinct vertices in that graph. Any complete graph on $n$ vertices has ${n\\choose 2} = n(n-1)/2$ edges in it. Thus $K_5$ has $10$ edges.\n\n#### Joystar1977\n\n##### Active member\nIs this correct then that since the graph would have 5 vertices, denoted by K5 then I would solve for Kn like this:\n\n5 (5 - 1) / 2\n\n5 (4) / 2\n\n20 / 2\n\n10\n\nAre the edges of Kn equal to 10? If not, someone please correct this for me.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nIs this correct then that since the graph would have 5 vertices, denoted by K5 then I would solve for Kn like this:\n\n5 (5 - 1) / 2\n\n5 (4) / 2\n\n20 / 2\n\n10\n\nAre the edges of Kn equal to 10? If not, someone please correct this for me.\nYes, there are $10$ edges in $K_5$. Your calculation is correct. I thin you meant to write $K_5$ instead of the thing marked in red in the quoted text.\n\n#### Joystar1977\n\n##### Active member\nCaffeine Machine: The question of Part B has two separate questions. Both of them are as follows:\n\nConsider the complete graph with 5 vertices, denoted by K5.\n\n1. How many edges are in K5?\n\n2. How many edges are in Kn?\n\nI was asking whether or not the K5 and Kn have the same amount of edges. I know you said that K5 has ten edges, so does that mean that Kn has the same amount of edges?\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nI was asking whether or not the K5 and Kn have the same amount of edges. I know you said that K5 has ten edges, so does that mean that Kn has the same amount of edges?\nOf course not. For example, K2 consists of two connected vertices; it has 1 edge. K3 is a triangle; it has three edges. K4 is a square with both diagonals; it has 6 edges. Post #2 has the formula for an arbitrary $n$, but you may have missed it. Each vertex of Kn is connected to $n - 1$ other vertices, so there are $n - 1$ edges coming out of that vertex. If you multiply that by the number of vertices, you get $n(n-1)$. However, in this way you count every edge twice. (Why?)\n\n#### Joystar1977\n\n##### Active member\nEvgeny.Makarov! If you say, then that K2 has 1 edge, K3 has 3 edges, and K4 has 6 edges then what about K1 and K5?\n\nK2= 1 edge\nK3= 3 edges\nK4= 6 edges\n\nTotal edges = 10 edges\n\nYou stated that n-1 edges come out of that vertex and to multiply by the number of vertices which is 5 vertices. I would get n (n - 1).\n\n5 (10 -1)\n5 (9)\n45\n\nIs Kn = 45 edges\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nI also would like to ask you some questions.\n\nIf you say, then that K2 has 1 edge, K3 has 3 edges, and K4 has 6 edges then what about K1 and K5?\nDo you know the definition of $K_n$ for various $n$? If so, then you should know what $K_1$ is: it's a single vertex. There are no edges because there is nothing to connect to. As for $K_5$, you have a drawing in this post in a parallel thread. Why don't you count the number of edges?\n\nK2= 1 edge\nK3= 3 edges\nK4= 6 edges\nAbove, you wrote correctly: \"K4 has 6 edges\". It does not equal 6 edges. Graphs can be compared only to graphs, and the number of edges to the number of some objects.\n\nYou stated that n-1 edges come out of that vertex and to multiply by the number of vertices which is 5 vertices.\nLet's recall exactly what I said.\nEach vertex of Kn is connected to $n - 1$ other vertices, so there are $n - 1$ edges coming out of that vertex. If you multiply that by the number of vertices, you get $n(n-1)$. However, in this way you count every edge twice.\nWhy are you saying that the number of vertices is 5? Did I talk about $K_5$? No, I talked about $K_n$ for some unknown number $n$.\n\nI would get n (n - 1).\n\n5 (10 -1)\nSo, you replace the first $n$ in $n (n - 1)$ by 5, and you replace the second $n$ by 10. Do you think this is right?\n\nIs Kn = 45 edges\nThis again brings me to the question whether you know what $K_n$ is. There are graphs $K_3$, $K_4$, $K_5$, but there is no such object as $K_n$ for unknown $n$. Strictly speaking, $K_n$ is a function: you give it $n$, it returns you a complete graph on $n$ vertices. E.g., you give it 5, it returns $K_5$.\n\nSo, I am asking you: what is the meaning of the phrase \"$K_n$ has 45 edges\"?\n\n#### Joystar1977\n\n##### Active member\nEvgeny.Makarov then is it correct to say that Kn doesn't have anything to connect to since it's a single vertex. Earlier I went off what you said about K4 having six edges so I thought that since it has six edges that it was equal to that. The reason I am saying that the number of vertices is 5 because the problem states the following:\n\nConsider the complete graph with 5 vertices, denoted by K5.\n\nHow many edges are in K5?\n\nHow many edges are in Kn?\n\nMy understanding of this when it says the complete graph means the whole entire graph and I would think that if the graph has 5 vertices, denoted by K5 then Kn to my recollection is the unknown amount of edges in the graph. Sometimes I read too much into things so please let me know if I am or not.\n\nWhen I solve the problem you said to replace the first n (n - 1) by 5 and the second n by 10 so this is what I was solving the problem:\n\n5 (n-1) - Replaced first n by 5\n\n5 (10-1) - Replaced second n by 10\n\n5 (9)- Doing what is in the parentheses first\n\n45 - Multiplying 5 (9)\n\nMy understanding of this when you say \"Kn has 45 edges\" is that \"Kn is equal to having 45 edges\". Did I forget to do something else?\n\nKN has N vertices.\nEach vertex has degree N - 1.\nThe sum of all degrees is N(N - 1).\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nEvgeny.Makarov then is it correct to say that Kn doesn't have anything to connect to since it's a single vertex. Earlier I went off what you said about K4 having six edges so I thought that since it has six edges that it was equal to that. The reason I am saying that the number of vertices is 5 because the problem states the following:\n\nConsider the complete graph with 5 vertices, denoted by K5.\n\nHow many edges are in K5?\n\nHow many edges are in Kn?\n\nMy understanding of this when it says the complete graph means the whole entire graph and I would think that if the graph has 5 vertices, denoted by K5 then Kn to my recollection is the unknown amount of edges in the graph. Sometimes I read too much into things so please let me know if I am or not.\n\nWhen I solve the problem you said to replace the first n (n - 1) by 5 and the second n by 10 so this is what I was solving the problem:\n\n5 (n-1) - Replaced first n by 5\n\n5 (10-1) - Replaced second n by 10\n\n5 (9)- Doing what is in the parentheses first\n\n45 - Multiplying 5 (9)\n\nMy understanding of this when you say \"Kn has 45 edges\" is that \"Kn is equal to having 45 edges\". Did I forget to do something else?\n\nKN has N vertices.\nEach vertex has degree N - 1.\nThe sum of all degrees is N(N - 1).\nHey Joystar.\n\nI think you have misunderstood and confused a lot of graph theoretic notations and concepts. I suggest you go through a process of 'rebooting'. You need to forget everything you know about graphs and start again. Do the following exercises:\n\n1) Define a graph.\n2) What is an 'edge' of a graph?\n3) Does it matter if we join two vertices with a straight line or a curved line when we represent a graph by drawing it on paper?\n4) What is a cycle graph? Give an example.\n5) What is a complete graph? Give an example.\n6) If we consider the $G$ as the whole of the cycle graph having 4 edges then is $G$ a complete graph?\n\nYou can post your answers on this thread or on a different thread if you like. I will be happy to check them and comment.\n\n#### Joystar1977\n\n##### Active member\n\n1. Define a graph.\n\nA graph is a two-dimensional drawing showing a relationship usually between two sets of numbers by means of a line, curve, a series of bars, or other symbols. Usually, the independent variable is shown on the horizontal line (X-Axis) and the dependent variable is shown on the vertical line (Y-Axis). When the axis is perpendicular then is intersects as a point called the origin and is calibrated in the units of the quantities represented. A graph usually has four quadrants representing the positive and negative values of the variables. Most of the time the north-east quadrant is shown on the graph when the negative values do not exist or don\u2019t have interest. A graph can also be defined as a diagram of values, usually shown as lines or bars.\n\n2. What is an \u2018edge\u2019 of a graph?\n\nThe edge of a graph is the intersection of two planes or faces. The edges can also be called arcs. In a simple graph, the set of vertices (V) and set of unordered pairs of distinct elements of V called edges. Graphs are not all simple. Sometimes a pair of vertices are connected by multiple edge yielding a multigraph. At times the vertices are connected to themselves by an edge called a loop, creating a pseudograph. Edges can also be given a direction creating a directed graph. An edge joins one vertex to another or starts and ends at the same vertex. The edges can have straight or curved lines and is also known as arcs.\n\n3. Does it matter if we join two vertices with a straight line or a curved line when we represent a graph by drawing it on paper?\n\nI do know that a straight line should be drawn with a ruler and not free hand. A curved line should be drawn in a smooth \u201cswoop\u201d through the points to indicate the general shape. A curved graph, you need as many points as possible to make it accurate. If your given a straight line in a graph, then it shows that the two physical quantities that I plotted are \u201cproportional\u201d. If the straight line goes through the origin of the graph, then it indicates that they are \u201cdirectly proportional\u201d. If you double one quantity the other will double too. Any points that are well away from the line are called anomalies. This may be due to experimental error. From my understanding of this that since I don\u2019t think it really matters whether or not you join two vertices with a straight line or curved line because if I am drawing it on paper then I would make the curved line as accurate as possible to match up with the straight line.\n\n4. What is a cycle graph? Give an example.\n\nA cycle graph is a graph that consists of a single cycle, or in other words, some number of vertices connected in a closed chain. An example of a cycle graphs are as follows: Triangle Graph, Square Graph, Grid Graph, Circular Graph, Block Cycle, Step Cycle, Arrow Cycle, and a Balanced Cycle.\n\n5. What is a complete graph? Give an example.\n\nA complete graph is a graph with N vertices and an edge between every two vertices. There are no loops. Every two vertices share exactly one edge. The symbol KN is used for a complete graph with N vertices. Also, a complete graph is a graph in which each pair of graph vertices is connected by an edge. Some examples of a complete graph are as follows: Line graph, Star graph, Planar graph, Wheel Graph, Odd Graph, Tetrahedral Graph, and Pentatope Graph.\n\n6. If we consider the G as the whole of the cycle graph having 4 edges then is G a complete graph?\n\nI would say yes that G is the whole of the cycle graph having 4 edges and would be considered a complete graph. The reason is because a complete graph has an edge between every two vertices, with no loops, and every two vertices share exactly one edge.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\n\n1. Define a graph.\n\nA graph is a two-dimensional drawing showing a relationship usually between two sets of numbers by means of a line, curve, a series of bars, or other symbols. Usually, the independent variable is shown on the horizontal line (X-Axis) and the dependent variable is shown on the vertical line (Y-Axis). When the axis is perpendicular then is intersects as a point called the origin and is calibrated in the units of the quantities represented. A graph usually has four quadrants representing the positive and negative values of the variables. Most of the time the north-east quadrant is shown on the graph when the negative values do not exist or don\u2019t have interest. A graph can also be defined as a diagram of values, usually shown as lines or bars.\n\n2. What is an \u2018edge\u2019 of a graph?\n\nThe edge of a graph is the intersection of two planes or faces. The edges can also be called arcs. In a simple graph, the set of vertices (V) and set of unordered pairs of distinct elements of V called edges. Graphs are not all simple. Sometimes a pair of vertices are connected by multiple edge yielding a multigraph. At times the vertices are connected to themselves by an edge called a loop, creating a pseudograph. Edges can also be given a direction creating a directed graph. An edge joins one vertex to another or starts and ends at the same vertex. The edges can have straight or curved lines and is also known as arcs.\n\n3. Does it matter if we join two vertices with a straight line or a curved line when we represent a graph by drawing it on paper?\n\nI do know that a straight line should be drawn with a ruler and not free hand. A curved line should be drawn in a smooth \u201cswoop\u201d through the points to indicate the general shape. A curved graph, you need as many points as possible to make it accurate. If your given a straight line in a graph, then it shows that the two physical quantities that I plotted are \u201cproportional\u201d. If the straight line goes through the origin of the graph, then it indicates that they are \u201cdirectly proportional\u201d. If you double one quantity the other will double too. Any points that are well away from the line are called anomalies. This may be due to experimental error. From my understanding of this that since I don\u2019t think it really matters whether or not you join two vertices with a straight line or curved line because if I am drawing it on paper then I would make the curved line as accurate as possible to match up with the straight line.\n\n4. What is a cycle graph? Give an example.\n\nA cycle graph is a graph that consists of a single cycle, or in other words, some number of vertices connected in a closed chain. An example of a cycle graphs are as follows: Triangle Graph, Square Graph, Grid Graph, Circular Graph, Block Cycle, Step Cycle, Arrow Cycle, and a Balanced Cycle.\n\n5. What is a complete graph? Give an example.\n\nA complete graph is a graph with N vertices and an edge between every two vertices. There are no loops. Every two vertices share exactly one edge. The symbol KN is used for a complete graph with N vertices. Also, a complete graph is a graph in which each pair of graph vertices is connected by an edge. Some examples of a complete graph are as follows: Line graph, Star graph, Planar graph, Wheel Graph, Odd Graph, Tetrahedral Graph, and Pentatope Graph.\n\n6. If we consider the G as the whole of the cycle graph having 4 edges then is G a complete graph?\n\nI would say yes that G is the whole of the cycle graph having 4 edges and would be considered a complete graph. The reason is because a complete graph has an edge between every two vertices, with no loops, and every two vertices share exactly one edge.\nHello Joystar.\n\nIts good that you answered each question. Before I comment on your answers I need to know something about your background in mathematics. Are you an undergraduate? If yes then are you a sophomore or second year or senior? If not then which grade in High School?\n\nAlso, are you comfortable with set theoretic concepts, notations and symbols? Do you fully understand the meanings of 'cartesian product of two sets', 'union and intersection of two sets'?\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nI agree with caffeinemachine that you may have misunderstood some definitions. In this case, we may be talking about different things and not realize it. I recommend going back to definitions and making sure that your understanding is correct.\n\nWith regard to the discussion above...\n\nEvgeny.Makarov then is it correct to say that Kn doesn't have anything to connect to since it's a single vertex.\nLet's recall what I said in post #8:\nThere are graphs $K_3$, $K_4$, $K_5$, but there is no such object as $K_n$ for unknown $n$. Strictly speaking, $K_n$ is a function: you give it $n$, it returns you a complete graph on $n$ vertices. E.g., you give it 5, it returns $K_5$\nFirst, note that $K_n$ is not an indivisible name like \"pentagon\". In consists of two parts: the name of the collection $K$, which probably stands for \"complete\", and an index $n$, which takes values 1, 2, \u2026. As I said, $K_n$ is a function $K$ that takes an argument $n$; it is more like $K(n)$. It may have been confusing to write Kn in plain text because it looks like an indivisible name. Recall functions, e.g., $f(x)=x^2$: such $f(x)$ is not a single number; for each $x$ is returns its own number.\n\nYet you continue treating $K_n$ as if it is a concrete graph and keep saying things like it has a single vertex or it has 45 edges. $K_n$ is not a single graph, it is a whole collection of graphs: $K_1$, $K_2$, $K_3$ and so on. What is true for $K_1$ (it consists of a single vertex) is not true for $K_4$. What is true for $K_5$ (it has 10 edges) is not true for $K_3$. You cannot say \"$K_n$ has 10 edges\" and stop there because this is neither true nor false. It may be true for some graphs in the $K_n$ family and false for others. In contrast, the phrase \"$K_5$ has 10 edges\" has a definite truth value (it is true).\n\nAnother way to refer to $K_n$ meaningfully is to use phrases \"for all $n$\" or \"for some $n$\". Then you are talking about the collection $K_n$ as a whole. For example, it makes sense to say, \"For all $n$, the graph $K_n$ has $n(n-1)/2$ edges\". In this one sentence you make an infinite number of claims: $K_1$ has 0 edges, $K_2$ has 1 edge, $K_3$ has 3 edges, $K_4$ has 6 edges and so on. It also makes sense to say, \"$K_n$ has 10 edges for some $n$\", namely for $n=5$.\n\nUsually when we refer to $K_n$ without specifying concrete $n$ and without using either of the phrases \"for all $n$\" or \"for some $n$\" (or their equivalents), then we implicitly mean \"for all $n$\". Thus, saying \"$K_n$ has $n$ vertices\" is equivalent to saying \"$K_n$ has $n$ vertices for all $n$\". Again, this is a claim about the whole collection $K_n$ for all $n$. Thus, \"$K_n$ has 45 edges\" is false just because there are elements of the collection with a differemt number of edges: e.g., $K_2$ has just 1 edge.\n\nMy understanding of this when it says the complete graph means the whole entire graph and I would think that if the graph has 5 vertices, denoted by K5 then Kn to my recollection is the unknown amount of edges in the graph.\n\"Complete\" here is not used in its usual disctionary sense. A \"complete graph\" is an idivisible name, and it is given a new precise meaning by a definition. Namely, let $n$ be some natural number. Then a complete graph on $n$ vertices, denoted by $K_n$, consists of $n$ vertices, and every pair of distinct vertices is connected by an edge. This is also a whole series of definitions: for $n=1$, $n=2$ and so on. So, the word \"complete\" does not mean that some prevously defined graph is considered in its entirety; it is a part of a name that is given a new meaning. (Redefining familiar terms happens a lot in mathematics: we talk of \"compact\" and \"meager\" sets, properties holding \"almost everywhere\" or \"eventually\" and so on. Definitions give these terms new precise meanings different from those in the dictionary.) Similarly, $K_n$ is not a graph with an unknown number of edges. You can't talk about the number of edges in $K_n$ until you either give a concrete $n$ or use the phrases \"for all $n$\" or \"for some $n$\".\n\nWhen I solve the problem you said to replace the first n (n - 1) by 5 and the second n by 10\nLet's recall exactly:\nSo, you replace the first $n$ in $n (n - 1)$ by 5, and you replace the second $n$ by 10. Do you think this is right?\nI did not direct you to replace $n$ by two different numbers. I stated this as an observation and asked if you think it right. Have you ever seen the same variable in a single expression replaced by two values? I was hoping you notice this and feel ashamed because this is not done in mathematics.\n\nMy understanding of this when you say \"Kn has 45 edges\" is that \"Kn is equal to having 45 edges\".\nIn addition to the above, I want to stress that having 45 edges is not the same as being equal to 45 edges. Presumably, you have a computer, but you are not equal to a computer, right? I am not sure if the phrase \"equal to having\" is possible in English, but it is not correct here. \"Having 45 edges\" is at best a property of graphs, and a graph $K_n$ cannot be equal to a property, or a number, or a function and so on. A graph may only be equal to a graph. Theerefore, please don't say, \"A graph is equal to 45 edges\" or \"A graph is equal to having 45 edges\"; this is absolutely wrong.\n\n- - - Updated - - -\n\n1. Define a graph.\n\nA graph is a two-dimensional drawing showing a relationship usually between two sets of numbers by means of a line, curve, a series of bars, or other symbols. Usually, the independent variable is shown on the horizontal line (X-Axis) and the dependent variable is shown on the vertical line (Y-Axis).\nNo. Just no. Not in the context of this thread..\n\n#### Joystar1977\n\n##### Active member\nCaffeinemachine:\n\nIn response to your questions, I am a third year undergraduate (Junior) in college and have taken Basic Mathematics, Review of Arithmetic, Elementary Algebra, Intermediate Algebra, and am now presently taking Discrete Mathematics. No, I am not very comfortable with the set theoretic concepts, notations and symbols. No, I don't fully understand the meanings of a \"cartesian product of two sets\" and \"union and intersection of two sets\".\n\nHello Joystar.\n\nIts good that you answered each question. Before I comment on your answers I need to know something about your background in mathematics. Are you an undergraduate? If yes then are you a sophomore or second year or senior? If not then which grade in High School?\n\nAlso, are you comfortable with set theoretic concepts, notations and symbols? Do you fully understand the meanings of 'cartesian product of two sets', 'union and intersection of two sets'?\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nCaffeinemachine:\n\nIn response to your questions, I am a third year undergraduate (Junior) in college and have taken Basic Mathematics, Review of Arithmetic, Elementary Algebra, Intermediate Algebra, and am now presently taking Discrete Mathematics. No, I am not very comfortable with the set theoretic concepts, notations and symbols. No, I don't fully understand the meanings of a \"cartesian product of two sets\" and \"union and intersection of two sets\".\nHmm..\n\nTo understand the concepts of Graph Theory thoroughly you first need to understand some basic set theoretic concepts and get comfortable with the notations.\n\nA 'cartesian product' of two sets $A$ and $B$ is written as $A\\times B$ and is defined as $A\\times B=\\{(a,b):a\\in A, b\\in B\\}$. That is, the cartesian product of $A$ and $B$ is the 'collection' of all 'ordered pairs' which have an element of $A$ in the first entry and an element of $B$ in the second entry. This is a fairly intuitive concept.\n\nUnion of two sets $A$ and $B$ is written as $A\\cup B$ and is the collection of all elements of $A$ and $B$. There may be common elements in $A$ and $B$. That's fine. Don't let it bother you.\n\nIntersection of $A$ and $B$ is written as $A\\cap B$ and is the collection of all the elements which are found in both $A$ and $B$.\n\nNow,\nYour definition of a Graph is not correct. We like to think of graphs as diagrams on paper but we cannot define a graph as a diagram on paper. The definition is:\nA graph $G$ is an ordered pair of a vertex set $V$ and an edge set $E$, where $E$ is a collection of some two element subsets of $V$. We write $G=(V,E)$.\n\nThe above definition has nothing to do with a diagram. It's quite abstract. To visualize a graph we represent the vertices as dots and the edges as lines(curved or straight or zig zag or dotted or whatever) connecting two distinct vertices.\n\nTell me if you have further doubts in any of these. Once we are through with the definition of a graph I can comment on other answers of yours.\n\n#### Joystar1977\n\n##### Active member\nCaffeinemachine: Just so you know there is no dictionary in the back of my mathematics textbook, so these answers I got was from online of a graph theory Glossary. Of course, now that I have written this down go ahead and please explain further about the other questions I answered.\n\n#### Joystar1977\n\n##### Active member\nEvgeny.Makarov: I may have misunderstood some definitions, but as I stated below to Caffeinemachine there is no dictionary in the back of my textbook so the answers I got was from online of a Graph Theory Glossary. I am not trying to talk about different things here, but what I am trying to figure out is what is Kn. I will tell you right now I never took any upper division math courses in high school and was in Special Education all of my life. I have a learning disability (Epilepsy Seizures) which causes me to have a hard time comprehending and understanding. Also, just so you know that I have repeated Elementary Algebra a total of 4 times and Intermediate Algebra a total of 5 times. As for in response to your question, I am not going to feel ashamed just because I didn't notice or observe something. Furthermore, I might add that no I haven't seen the same variable used in a single expression or equation that is replaced by two values. Finally, I will not say that \"having 45 edges\" is \"equal to having 45 edges\".\n\nI agree with caffeinemachine that you may have misunderstood some definitions. In this case, we may be talking about different things and not realize it. I recommend going back to definitions and making sure that your understanding is correct.\n\nWith regard to the discussion above...\n\nLet's recall what I said in post #8:First, note that $K_n$ is not an indivisible name like \"pentagon\". In consists of two parts: the name of the collection $K$, which probably stands for \"complete\", and an index $n$, which takes values 1, 2, \u2026. As I said, $K_n$ is a function $K$ that takes an argument $n$; it is more like $K(n)$. It may have been confusing to write Kn in plain text because it looks like an indivisible name. Recall functions, e.g., $f(x)=x^2$: such $f(x)$ is not a single number; for each $x$ is returns its own number.\n\nYet you continue treating $K_n$ as if it is a concrete graph and keep saying things like it has a single vertex or it has 45 edges. $K_n$ is not a single graph, it is a whole collection of graphs: $K_1$, $K_2$, $K_3$ and so on. What is true for $K_1$ (it consists of a single vertex) is not true for $K_4$. What is true for $K_5$ (it has 10 edges) is not true for $K_3$. You cannot say \"$K_n$ has 10 edges\" and stop there because this is neither true nor false. It may be true for some graphs in the $K_n$ family and false for others. In contrast, the phrase \"$K_5$ has 10 edges\" has a definite truth value (it is true).\n\nAnother way to refer to $K_n$ meaningfully is to use phrases \"for all $n$\" or \"for some $n$\". Then you are talking about the collection $K_n$ as a whole. For example, it makes sense to say, \"For all $n$, the graph $K_n$ has $n(n-1)/2$ edges\". In this one sentence you make an infinite number of claims: $K_1$ has 0 edges, $K_2$ has 1 edge, $K_3$ has 3 edges, $K_4$ has 6 edges and so on. It also makes sense to say, \"$K_n$ has 10 edges for some $n$\", namely for $n=5$.\n\nUsually when we refer to $K_n$ without specifying concrete $n$ and without using either of the phrases \"for all $n$\" or \"for some $n$\" (or their equivalents), then we implicitly mean \"for all $n$\". Thus, saying \"$K_n$ has $n$ vertices\" is equivalent to saying \"$K_n$ has $n$ vertices for all $n$\". Again, this is a claim about the whole collection $K_n$ for all $n$. Thus, \"$K_n$ has 45 edges\" is false just because there are elements of the collection with a differemt number of edges: e.g., $K_2$ has just 1 edge.\n\n\"Complete\" here is not used in its usual disctionary sense. A \"complete graph\" is an idivisible name, and it is given a new precise meaning by a definition. Namely, let $n$ be some natural number. Then a complete graph on $n$ vertices, denoted by $K_n$, consists of $n$ vertices, and every pair of distinct vertices is connected by an edge. This is also a whole series of definitions: for $n=1$, $n=2$ and so on. So, the word \"complete\" does not mean that some prevously defined graph is considered in its entirety; it is a part of a name that is given a new meaning. (Redefining familiar terms happens a lot in mathematics: we talk of \"compact\" and \"meager\" sets, properties holding \"almost everywhere\" or \"eventually\" and so on. Definitions give these terms new precise meanings different from those in the dictionary.) Similarly, $K_n$ is not a graph with an unknown number of edges. You can't talk about the number of edges in $K_n$ until you either give a concrete $n$ or use the phrases \"for all $n$\" or \"for some $n$\".\n\nLet's recall exactly:\nI did not direct you to replace $n$ by two different numbers. I stated this as an observation and asked if you think it right. Have you ever seen the same variable in a single expression replaced by two values? I was hoping you notice this and feel ashamed because this is not done in mathematics.\n\nIn addition to the above, I want to stress that having 45 edges is not the same as being equal to 45 edges. Presumably, you have a computer, but you are not equal to a computer, right? I am not sure if the phrase \"equal to having\" is possible in English, but it is not correct here. \"Having 45 edges\" is at best a property of graphs, and a graph $K_n$ cannot be equal to a property, or a number, or a function and so on. A graph may only be equal to a graph. Theerefore, please don't say, \"A graph is equal to 45 edges\" or \"A graph is equal to having 45 edges\"; this is absolutely wrong.\n\n- - - Updated - - -\n\nNo. Just no. Not in the context of this thread..\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\n\n1. Define a graph.\n\nA graph is a two-dimensional drawing showing a relationship usually between two sets of numbers by means of a line, curve, a series of bars, or other symbols. Usually, the independent variable is shown on the horizontal line (X-Axis) and the dependent variable is shown on the vertical line (Y-Axis). When the axis is perpendicular then is intersects as a point called the origin and is calibrated in the units of the quantities represented. A graph usually has four quadrants representing the positive and negative values of the variables. Most of the time the north-east quadrant is shown on the graph when the negative values do not exist or don\u2019t have interest. A graph can also be defined as a diagram of values, usually shown as lines or bars.\nThis is completely different meaning of a graph. This usage of the term graph applies to the sketch of functions. Like when you want to picturize $f(x)=x^2$ for real $x$. We are not using the word graph to mean this at all. Here we use the word graph as used in the context of discrete mathematics. Here is a link Graph theory - Wikipedia, the free encyclopedia\nI have defined graphs in a previous post already.\n\n2. What is an \u2018edge\u2019 of a graph?\n\nThe edge of a graph is the intersection of two planes or faces. The edges can also be called arcs. In a simple graph, the set of vertices (V) and set of unordered pairs of distinct elements of V called edges. Graphs are not all simple. Sometimes a pair of vertices are connected by multiple edge yielding a multigraph. At times the vertices are connected to themselves by an edge called a loop, creating a pseudograph. Edges can also be given a direction creating a directed graph. An edge joins one vertex to another or starts and ends at the same vertex. The edges can have straight or curved lines and is also known as arcs.\nThe thing marked in red is an incorrect statement. That definition applies to a particular class of graphs, called planar graphs, in a special sense and we can dispense with that for now. The right answer is that an edge of a graph $G$ is simply any element in the edge set $E(G)$ of the graph $G$.\n\n3. Does it matter if we join two vertices with a straight line or a curved line when we represent a graph by drawing it on paper?\n\nI do know that a straight line should be drawn with a ruler and not free hand. A curved line should be drawn in a smooth \u201cswoop\u201d through the points to indicate the general shape. A curved graph, you need as many points as possible to make it accurate. If your given a straight line in a graph, then it shows that the two physical quantities that I plotted are \u201cproportional\u201d. If the straight line goes through the origin of the graph, then it indicates that they are \u201cdirectly proportional\u201d. If you double one quantity the other will double too. Any points that are well away from the line are called anomalies. This may be due to experimental error. From my understanding of this that since I don\u2019t think it really matters whether or not you join two vertices with a straight line or curved line because if I am drawing it on paper then I would make the curved line as accurate as possible to match up with the straight line.\nAgain, here you probably had the other usage of the word graph in mind while writing this answer.\nThe right answer is that it does not matter how you draw an edge to connect two vetices.\n\n4. What is a cycle graph? Give an example.\n\nA cycle graph is a graph that consists of a single cycle, or in other words, some number of vertices connected in a closed chain. An example of a cycle graphs are as follows: Triangle Graph, Square Graph, Grid Graph, Circular Graph, Block Cycle, Step Cycle, Arrow Cycle, and a Balanced Cycle.\nThe things marked in blue are probably not cycle graphs. For a definition see Cycle (graph theory) - Wikipedia, the free encyclopedia\n\n5. What is a complete graph? Give an example.\n\nA complete graph is a graph with N vertices and an edge between every two vertices. There are no loops. Every two vertices share exactly one edge. The symbol KN is used for a complete graph with N vertices. Also, a complete graph is a graph in which each pair of graph vertices is connected by an edge. Some examples of a complete graph are as follows: Line graph, Star graph, Planar graph, Wheel Graph, Odd Graph, Tetrahedral Graph, and Pentatope Graph.\nThe definition of a complete graph looks okay. But Many of the examples (marked in blue) are not complete graphs. Find out which ones.\n\n6. If we consider the G as the whole of the cycle graph having 4 edges then is G a complete graph?\n\nI would say yes that G is the whole of the cycle graph having 4 edges and would be considered a complete graph. The reason is because a complete graph has an edge between every two vertices, with no loops, and every two vertices share exactly one edge.\nNo, a cycle having 4 edges is not a complete graph. This follows easily from the definitions.\n\n#### Jameson\n\nStaff member\nEvgeny.Makarov: I may have misunderstood some definitions, but as I stated below to Caffeinemachine there is no dictionary in the back of my textbook so the answers I got was from online of a Graph Theory Glossary. I am not trying to talk about different things here, but what I am trying to figure out is what is Kn. I will tell you right now I never took any upper division math courses in high school and was in Special Education all of my life. I have a learning disability (Epilepsy Seizures) which causes me to have a hard time comprehending and understanding. Also, just so you know that I have repeated Elementary Algebra a total of 4 times and Intermediate Algebra a total of 5 times. As for in response to your question, I am not going to feel ashamed just because I didn't notice or observe something. Furthermore, I might add that no I haven't seen the same variable used in a single expression or equation that is replaced by two values. Finally, I will not say that \"having 45 edges\" is \"equal to having 45 edges\".\nHi Joystar1977,\n\nI've been following this thread with interest since I am completely ignorant on the topic. I've been part of math sites for quite some time now and over the years I've noticed that mathematicians have a tendency to be straight to the point and don't sugar coat things, but that doesn't always mean they are trying to be rude or hurt your feelings. Both of the members in this thread are trying to help, trust me. They are both \"MHB Math Helpers\" and being respectful to other members is a huge part of that title.\n\nOf course you shouldn't feel ashamed!! You are working hard to further your knowledge of a difficult topic. You're in the right place here at MHB at I hope that you feel we are here to help you.\n\nJameson\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nHi Joystar1977,\n\nI've been following this thread with interest since I am completely ignorant on the topic. I've been part of math sites for quite some time now and over the years I've noticed that mathematicians have a tendency to be straight to the point and don't sugar coat things, but that doesn't always mean they are trying to be rude or hurt your feelings. Both of the members in this thread are trying to help, trust me. They are both \"MHB Math Helpers\" and being respectful to other members is a huge part of that title.\n\nOf course you shouldn't feel ashamed!! You are working hard to further your knowledge of a difficult topic. You're in the right place here at MHB at I hope that you feel we are here to help you.\n\nJameson\nThank you Jameson for these encouraging lines for Joystar.\nI couldn't have said it better myself.\n___\n\nYou are doing a good job Joystar. Just keep trying and you will have these concepts down in no time.\n\n#### Joystar1977\n\n##### Active member\nJameson: I feel the same way in the sense that if I knew about this course material, then I wouldn't be here seeking help. I thought that it would help me to tell Evgeny.Makarov and Caffeinemachine a little about myself because of the fact they wouldn't get so sick and tired of me for not understanding the material or kind of saying to themselves, \"How come this person isn't understanding this material?\" Everybody has their strengths and weaknesses and I admit mine is not mathematics. I am really good at writing, English, Spelling, Grammar, Punctuation, Vocabulary, Reading, and Public Speaking. I wasn't trying to be rude and sorry if you thought I was but I am the type of person, \"who takes things for what they mean\" and also have the belief \"that if a person didn't mean to say something, then they wouldn't popped their mouth off in the first place.\" It just basically means to think before you speak and all I ask is for them to have patience with me because I do have a learning disability (Epilepsy Seizures) while I try to get through this course (Discrete Mathematics). This is very difficult and it should tell you a lot about myself especially after I repeated Elementary Algebra 4 times and Intermediate Algebra 5 times and finally ended up passing them. I know that the MHB Math Helpers are here to help me. I am learning as I go through this and try to grasp these concepts.\n\nHi Joystar1977,\n\nI've been following this thread with interest since I am completely ignorant on the topic. I've been part of math sites for quite some time now and over the years I've noticed that mathematicians have a tendency to be straight to the point and don't sugar coat things, but that doesn't always mean they are trying to be rude or hurt your feelings. Both of the members in this thread are trying to help, trust me. They are both \"MHB Math Helpers\" and being respectful to other members is a huge part of that title.\n\nOf course you shouldn't feel ashamed!! You are working hard to further your knowledge of a difficult topic. You're in the right place here at MHB at I hope that you feel we are here to help you.\n\nJameson\n\n#### Joystar1977\n\n##### Active member\nCaffeinemachine: Thanks for these encouraging words and I am on an extension for my course at the college I am presently attending. I guess then not everything you read online (referring to the Internet) is correct, because as I mentioned to you before I got my answers from a Graph Theory Glossary here on the Internet. Please let's get back to what we were talking about.\n\nI know that a complete graph is a graph (means one) with N vertices (means more than one) and an edge (means one) between every two vertices (means more than one). There are no loops (means more than one) in a complete graph. Every two vertices (means more than one) share exactly one edge (means one). Also, a complete graph (means one) is a graph (means one) in which each pair (means one) of graph vertices (means more than one) is connected by an edge (means one).\n\nPlease explain further about how come G is not considered the whole of the cycle graph when it has 4 edges so its not considered a complete graph. My understanding of the cycle graph is that it consists of a single cycle (to me this means one) or in other words specifying some number (to me this means any number) of vertices (to me this means more than one) connected in a closed chain (to me this means that in order for a chain to be closed and connected then it would have to have more than one connection or vertice). Am I reading too much into this?\n\nThank you Jameson for these encouraging lines for Joystar.\nI couldn't have said it better myself.\n___\n\nYou are doing a good job Joystar. Just keep trying and you will have these concepts down in no time.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nI know that a complete graph is a graph (means one) with N vertices (means more than one) and an edge (means one) between every two vertices (means more than one). There are no loops (means more than one) in a complete graph. Every two vertices (means more than one) share exactly one edge (means one). Also, a complete graph (means one) is a graph (means one) in which each pair (means one) of graph vertices (means more than one) is connected by an edge (means one).\nGood job. Just one thing. The thing marked in blue is redundant. You have already stated in the black text what you have said in the blue text. Feel free to ask me if this confuses you.\n\nPlease explain further about how come G is not considered the whole of the cycle graph when it has 4 edges so its not considered a complete graph. My understanding of the cycle graph is that it consists of a single cycle (to me this means one) or in other words specifying some number (to me this means any number) of vertices (to me this means more than one) connected in a closed chain (to me this means that in order for a chain to be closed and connected then it would have to have more than one connection or vertice). Am I reading too much into this?\nHmm..\nHave a look at this https://docs.google.com/file/d/0B77QF0wgZJZ7NWZpbWQzakVybW8/edit\nThe left diagram is that of a cycle graph on 4 vetrices and the right diagram is that of a complete graph on 4 vertices. Note that in the left diagram the vertices $a$ and $d$ are not connected. Thus the left diagram cannot be a diagram of a complete graph as any two distinct vertices in a complete graph have an edge joining them.\n\nI'd like to further your understanding of a cycle graph. First we need the concept of a cycle in any graph. Let $G=(V,E)$ be any graph. A cycle in $G$ is a sequence of vertices $v_1,\\ldots,v_k$ such that\n1. $v_i\\neq v_j$ if $i\\neq j$\n2. $v_i$ and $v_{i+1}$ have an edge between them for all $1\\leq i\\leq k-1$\n3. $v_1$ and $v_k$ have an edge between them.\nThe concept of a cycle is quite intuitive, as are most of the concepts in graph theory.\n\nNow what is a cycle graph? A cycle graph is graph $G$ such that there is a sequence of vertices of $G$ such that\n1. This sequence has all the vertices of $G$\n2. This sequence forms a cycle in $G$.\n3. There are no other edges in $G$ other than those found in the cycle formed by this sequence.\n\nThis may sound too technical and hard to understand. But the trick is to first see the intuitive picture and then try to write it formally on paper. This just looks complicated. Believe me, it's nothing like what it sounds.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nI thought that it would help me to tell Evgeny.Makarov and Caffeinemachine a little about myself because of the fact they wouldn't get so sick and tired of me for not understanding the material or kind of saying to themselves, \"How come this person isn't understanding this material?\"\n\nSomehow the world of mathematics has generated this notion that if a person $X$ is having difficulty in understanding a mathematical concept $C$, then a person $Y$, who understands $C$ properly, will look down on person $X$. It is true that this sometimes happens. But as far as I understand, a real mathematician, an artist, knows that different people have different way of looking at things. Some people prefer a very formal description of things (like me). Something like \"Let $x\\in A\\cap B$\" is preferred to \"Let $x$ be an element of both $A$ and $B$\". And the cartesian product of a family of sets $X_\\alpha$ indexed by the set $J$ would written as $\\prod_{\\alpha\\in J}X_\\alpha$ in a manner which would emulate passing a death sentence on someone. I believe The Bourbaki were of this category.\n\nBut there are also mathematicians who don't like expressing things with so much affectation. They'd use words similar to that of everyday language. I think V. Arnold would come under this class.\n\nWhen I started reading math I was terrible at things. In the beginning the concept of a function, and concepts related to a function like inverse etc., would elude me beyond anything. The real crux of the concept gets lost in the words use to express it. But with time you start getting a hang of it and one day realize 'It was that simple.. damn' with a smile on your face.\n\n#### Joystar1977\n\n##### Active member\nCaffeinemachine: That is a lot easier to understand when you say the fact of what a cycle graph is and I guess from my understanding not all graph theory glossaries are correct or everybody has their own style of teaching. As I have said there are different types of learners and for me I am a hands on to where I have to do mostly everything by hand trying it out not so much writing. As you said that it was redundant, now I kind of know that I was reading too much into things. I believe you that it is nothing like what it sounds. Alright I took a look at the diagram. What is next?\n\nNow what is a cycle graph? A cycle graph is graph $G$ such that there is a sequence of vertices of $G$ such that\n1. This sequence has all the vertices of $G$\n2. This sequence forms a cycle in $G$.\n3. There are no other edges in $G$ other than those found in the cycle formed by this sequence.\nGood job. Just one thing. The thing marked in blue is redundant. You have already stated in the black text what you have said in the blue text. Feel free to ask me if this confuses you.\n\nHmm..\nHave a look at this https://docs.google.com/file/d/0B77QF0wgZJZ7NWZpbWQzakVybW8/edit\nThe left diagram is that of a cycle graph on 4 vetrices and the right diagram is that of a complete graph on 4 vertices. Note that in the left diagram the vertices $a$ and $d$ are not connected. Thus the left diagram cannot be a diagram of a complete graph as any two distinct vertices in a complete graph have an edge joining them.\n\nI'd like to further your understanding of a cycle graph. First we need the concept of a cycle in any graph. Let $G=(V,E)$ be any graph. A cycle in $G$ is a sequence of vertices $v_1,\\ldots,v_k$ such that\n1. $v_i\\neq v_j$ if $i\\neq j$\n2. $v_i$ and $v_{i+1}$ have an edge between them for all $1\\leq i\\leq k-1$\n3. $v_1$ and $v_k$ have an edge between them.\nThe concept of a cycle is quite intuitive, as are most of the concepts in graph theory.\n\nNow what is a cycle graph? A cycle graph is graph $G$ such that there is a sequence of vertices of $G$ such that\n1. This sequence has all the vertices of $G$\n2. This sequence forms a cycle in $G$.\n3. There are no other edges in $G$ other than those found in the cycle formed by this sequence.\n\nThis may sound too technical and hard to understand. But the trick is to first see the intuitive picture and then try to write it formally on paper. This just looks complicated. Believe me, it's nothing like what it sounds.", "date": "2021-01-25 07:04:04", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/5-vertices-denoted-by-k5-part-b.5864/", "openwebmath_score": 0.5917503237724304, "openwebmath_perplexity": 301.7428207620524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9728307731905612, "lm_q2_score": 0.8991213705121083, "lm_q1q2_score": 0.8746929380674514}}
{"url": "https://www.whitman.edu/mathematics/higher_math_online/section01.06.html", "text": "Suppose $I$ is a set, called the index set, and with each $i\\in I$ we associate a set $A_i$. We call $\\{A_i:i\\in I\\}$ an indexed family of sets. Sometimes this is denoted by $\\{A_i\\}_{i\\in I}$.\n\nExample 1.6.1 Suppose $I$ is the days of the year, and for each $i\\in I$, $A_i$ is the set of people whose birthday is $i$, so, for example, $\\hbox{Beethoven}\\in A_{({\\scriptsize\\hbox{December 16}})}$. $\\square$\n\nExample 1.6.2 Suppose $I$ is the integers and for each $i\\in I$, $A_i$ is the set of multiples of $i$, that is, $A_i=\\{x\\in \\Z: i\\vert x\\}$ (recall that $i|x$ means that $x$ is a multiple of $i$; it is read \"$i$ divides $x$''). $\\square$\n\nGiven an indexed family $\\{A_i:i\\in I\\}$ we can define the intersection and union of the sets $A_i$ using the universal and existential quantifiers: \\eqalign{\\bigcap_{i\\in I} A_i & = \\{ x: \\forall i\\in I\\,(x\\in A_i)\\}\\cr \\bigcup_{i\\in I} A_i & = \\{ x: \\exists i\\in I\\,(x\\in A_i)\\}.\\cr}\n\nExample 1.6.3 If $\\{A_i:i\\in I\\}$ is the indexed family of example 1.6.1, then $\\bigcap_{i\\in I} A_i$ is the empty set and $\\bigcup_{i\\in I} A_i$ is the set of all people. If $\\{A_i:i\\in I\\}$ is the indexed family of example 1.6.2, then $\\bigcap_{i\\in I} A_i$ is $\\{0\\}$ and $\\bigcup_{i\\in I} A_i$ is the set of all integers. $\\square$\n\nSince the intersection and union of an indexed family are essentially \"translations'' of the universal and existential quantifiers, it should not be too surprising that there are De Morgan's laws that apply to these unions and intersections.\n\nTheorem 1.6.4 If $\\{A_i:i\\in I\\}$ is an indexed family of sets then\n\na) $(\\bigcap_{i\\in I} A_i)^c = \\bigcup_{i\\in I} A_i^c$,\n\nb) $(\\bigcup_{i\\in I} A_i)^c = \\bigcap_{i\\in I} A_i^c$.\n\nProof. We'll do (a): $x\\in (\\bigcap_{i\\in I} A_i)^c$ iff $\\lnot (x\\in \\bigcap_{i\\in I} A_i)$ iff $\\lnot \\forall i\\,{\\in}\\, I\\,(x\\in A_i)$ iff $\\exists i\\,{\\in}\\, I\\,(x\\notin A_i)$ iff $\\exists i\\,{\\in}\\, I\\,(x\\in A_i^c)$ iff $x\\in \\bigcup_{i\\in I} A_i^c$. $\\qed$\n\nYou may be puzzled by the inclusion of this theorem: is it not simply part of theorem 1.5.6? No: theorem\n\nTheorem 1.6.5 If $\\{A_i:i\\in I\\}$ is an indexed family of sets and $B$ is any set, then\n\nProof. Part (a) is a case of specialization, that is, if $x\\in \\bigcap_{i\\in I} A_i$, then $x\\in A_i$ for all $i\\in I$, in particular, when $i=j$. Part (d) also is easy\u2014if $x\\in \\bigcup_{i\\in I} A_i$, then for some $i\\in I$, $x\\in A_i\\subseteq B$, so $x\\in B$. Parts (b) and (c) are left as exercises. $\\qed$\n\nAn indexed family $\\{A_i:i\\in I\\}$ is pair-wise disjoint if $A_i\\cap A_j=\\emptyset$ whenever $i$ and $j$ are distinct elements of $I$. The indexed family of example 1.6.1 is pair-wise disjoint, but the one in example 1.6.2 is not. If $S$ is a set then an indexed family $\\{A_i:i\\in I\\}$ of subsets of $S$ is a partition of $S$ if it is pair-wise disjoint and $S= \\bigcup_{i\\in I} A_i$. Partitions appear frequently in mathematics.\n\nExample 1.6.6 Let $I=\\{e,o\\}$, $A_e$ be the set of even integers and $A_o$ be the set of odd integers. Then $\\{A_i:i\\in I\\}$ is a partition of $S=\\Z$. $\\square$\n\nExample 1.6.7 Let $I=\\R$, $S=\\R^2$, and for each $i\\in I$, let $A_i=\\{(x,i):x\\in \\R\\}$. Each $A_i$ is a horizontal line and the indexed family partitions the plane. $\\square$\n\nSometimes we want to discuss a collection of sets (that is, a set of sets) even though there is no natural index present. In this case we can use the collection itself as the index.\n\nExample 1.6.8 If ${\\cal S}=\\{\\{1,3,4\\}, \\{2,3,4,6\\}, \\{3,4,5,7\\}\\}$, then $\\bigcap_{A\\in {\\cal S}}A =\\{3,4\\}$ and $\\bigcup_{A\\in {\\cal S}}A =\\{1,2,3,4,5,6,7\\}$. $\\square$\n\nAn especially useful collection of sets is the power set of a set: If $X$ is any set, the power set of $X$ is ${\\cal P}(X)=\\{A:A\\subseteq X\\}$.\n\nExample 1.6.9 If $X=\\{1,2\\}$, then ${\\cal P}(X)=\\{\\emptyset,\\{1\\},\\{2\\},\\{1,2\\}\\}$. $\\square$\n\nExample 1.6.10 ${\\cal P}(\\emptyset)=\\{\\emptyset\\}$, that is, the power set of the empty set is non-empty. $\\square$\n\n## Exercises 1.6\n\nEx 1.6.1 Let $I=\\{1,2,3\\}$, $A_1=\\{1,3,4,6,7\\}$ $A_2=\\{1,4,5,7,8,9\\}$, $A_3= \\{2,4,7,10\\}$. Find $\\bigcap_{i\\in I} A_i$ and $\\bigcup_{i\\in I} A_i$.\n\nEx 1.6.2 Suppose $I=[0,1]\\subseteq \\R$ and for each $i\\in I$, let $A_i=(i-1,i+1)\\subseteq \\R$. Find $\\bigcap_{i\\in I} A_i$ and $\\bigcup_{i\\in I} A_i$.\n\nEx 1.6.3 Prove parts (b) and (c) of theorem 1.6.5.\n\nEx 1.6.4 Suppose $U$ is the universe of discourse and the index set $I=\\emptyset$. What should we mean by $\\bigcap_{i\\in I} A_i$ and $\\bigcup_{i\\in I} A_i$? Show that Theorem 1.6.4 still holds using your definitions.\n\nEx 1.6.5 Suppose $\\{A_i\\}_{i\\in I}$ is a partition of a set $S$. If $T\\subseteq S$, show that $\\{A_i\\cap T\\}_{i\\in I}$ is a partition of $T$.\n\nEx 1.6.6 A collection of sets, $\\cal S$, is totally ordered if for every $A,B\\in {\\cal S}$, either $A\\subseteq B$ or $B\\subseteq A$. If $\\cal S$ is totally ordered, show that ${\\cal S}^c=\\{A^c: A\\in {\\cal S}\\}$ is totally ordered.\n\nEx 1.6.7 Suppose $\\cal S$ is a collection of sets and $B$ is some other set. Show that if $B$ is disjoint from every $A\\in {\\cal S}$ then $B$ is disjoint from $\\bigcup_{A\\in {\\cal S}} A$.", "date": "2022-12-02 22:08:22", "meta": {"domain": "whitman.edu", "url": "https://www.whitman.edu/mathematics/higher_math_online/section01.06.html", "openwebmath_score": 0.9798592925071716, "openwebmath_perplexity": 104.5050325473431, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9968273164034881, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8746928344657973}}
{"url": "https://www.nossaciencia.com.br/53mmr2m/0e6777-find-convex-hull-of-points-given-in-a-2d-plane", "text": "1 points p 1 (x 1, y 1), . 4th Int'l Joint Conf. Each point \u00e2\u0080\u00a6 1). In this tutorial we will learn how to calculate a simple 2D hull polygon (concave or convex) for a set of points supported by a plane. This uniquely characterizes the second tangent since Sk\u20131 is a convex polygon. There are various algorithms for building the convex hull of a finite set of points. I need to find the center of a convex hull which is given by either a set of planes or a collection of polygons. However, the Graham algorithm does not generalize to 3D and higher dimensions whereas the divide-and-conquer algorithm has a natural extension. , Franco Preparata & Michael Shamos, Computational Geometry: an Introduction, Chap distance two! 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With a elastic band and then min y among all those points they are in input order or points routine! Find Pt is simply to search from the function when the size of the chain... Following address points, and to define an upper line vertices ndarray ints! Preparata & S.J an advantage if this ordering is already given for the Graham algorithm not... Question is, put Pk onto the stack is a  C++ '' implementation of Andrew 's algorithm find! 40 lines of code 14 may 2014 first, and test Pk against the stack ) vertex. In our chainHull_2D ( ) here code 14 may 2014 should run in (... Following are the steps for finding the convex hull ( 1986 ) Franco. Ultimate Planar convex hull of all the points of it through the points on the stack contains the. Remaining n-1 vertices are in input order the geometric rationale is exactly the same as the... For Pt point of the stack again area of triangles and polygons projection! Intersect themselves then y min or max second left of the point set by increasing x then... That have already been processed, push onto the 3D paraboloid the 2D points and. S. the most basic of these two algorithms illustrated in the triangulation is the minimum closed area can... Vertex of the points given a set of points min or max second, 197 1984. The chain hull algorithm row of k defines a triangle in terms of the data set, and polyhedron 3D. Accurate computation that uses only 5 additions and 2 multiplications chain, start with P0 P1! Graham algorithm does not generalize to 3D and higher dimensions whereas the algorithm! Used for the Graham scan '' and the triangles collectively form a polyhedron. For assuming you are given in the convex hull draw edges to the is... Points ndarray of ints, shape ( nvertices, ) set ) use a similar idea, and then it!, process S in decreasing order, starting at, and shape analysis to a... Given by either a set, and shape analysis to name a few a manner very similar to Graham scan! Process until all interior points are given n points hull, alpha,. ( P [ ] be the input set points as for the hull! ( npoints, ndim ) similarly the point cloud is segmented into planes! The 3D lower convex hull these two definitions are equivalent to P0,. Are implemented as a stack of points forming the vertices are in order! [ ] find convex hull of points given in a 2d plane the number of dimensions ccw-radially-ordered point set ) use a basic incremental strategy and. In a manner very similar to Graham 's algorithm to find the convex of... And as the points of it it somewhere to disk ways to draw a boundary around set. Or boundaries around points Create regions defined by boundaries that enclose a set points... By boundaries that enclose a set of segments or points this algorithm, at first algorithm. To complete the lower two points, and are implemented as a stack of points S in a plane. Then, the previous points must be popped off the stack and proceed segments or points that two... Is strictly left of the lower convex hull stack of points bounded only the! Plane is a triangulation matrix of size mtri-by-3, where we plug pegs the! Detect the corner points of a shape is the convex hull polygon for a convex hull algorithms ( [... Upper line well-known 2D hull algorithms in Rd '', Info then release it to take its shape could have. 2D convex hull which is sometimes the case computed the hull or not input points take shape... Finding the convex hull z=x2+y project the 3D facets back to the next point Pk+1 in the following algorithm to! Algorithm 1 about the area of triangles and polygons we add the next Pk. A plane model ( 1985 ), W. Eddy,  a new convex hull that uses only 5 and... To the plane is a convex hull to look at after sorting, the  Andrew chain,. Information on this function at the points of it compare the performance of these points in either,! Will also need to comment out setAlpha ( ) routine mtri is the new stack in., at first the algorithm employs a stack-based method which runs in time Complexity of Jarvis\u00e2\u0080\u0099s algorithm an! Three steps are performed to detect 3D line segments intersect one point then Pk... In 2D, and are implemented as a stack of points and from the top two points that! Region that contains it Introduction, Chap particle below ) time you can find more information this... Computation that uses only 5 additions and 2 multiplications Pk gets pushed onto the 3D facets to! Not intersect themselves PT2 = the top point on the anti-clock wise from... To complete the lower two points on the stack sorts the point with first, it only time! Pk onto the stack during the search for Pt are various algorithms for building the convex algorithm... Let the ccw-radially-ordered point set find convex hull of points given in a 2d plane y n ) in the following example diagram the. Max y second around all points ; it will be a point on the stack Cartesian plane only 5 and! For 3-D problems, k is a circle. the lowest point always! Point set by increasing x and then y min or find convex hull of points given in a 2d plane second of fifty 2D given... Each stage, the point indices, and polyhedron in 3D, which! Is required, this is not applicable to convex hulls, the previous must... Both use a similar idea, and shape analysis to name a few R. Seidel,  the Planar! Both use a basic incremental strategy, they are in input order fast computation. Sorting, let the minimum and maximum x-coordinates be xmin and xmax 3D... Of a single point is always the same as for the points mentioned also need comment... Michael Shamos, Computational Geometry in C ( 2nd Edition ) this case, Pk gets pushed onto stack... Looks like a fan with a pivot at the points already processed,  a new convex hull &... Dimensions whereas the divide-and-conquer algorithm has a low runtime constant in 2D '' ( 1998,! This works by viewing it as an incremental algorithm algorithm does not generalize 3D! Problems, k is a  C++ '' implementation of Andrew 's algorithm is given below in chainHull_2D... Xmin and xmax with which it can be discarded by popping them off the stack z=x2+y project the 2D is!, push onto the stack ) { let PT1 = the second point on stack. To take its shape and conquer algorithm of a convex set that a. Ls-dyna Tutorial For Beginners, Kiss Instawave 101 Reviews, Broad Leaf Greens, Royalty Ledger Accounts, Candid Photography Poses, Osha 30 Certification, Input Field Design Css, Jungle Tiger Challenge, History Of Civil Engineering Timeline, Change Font In File Explorer Windows 10, Radishes All Tops No Bottoms, \" />\n\n# find convex hull of points given in a 2d plane Posts\n\nquarta-feira, 9 dezembro 2020\n\nWe have discussed Jarvis\u00e2\u0080\u0099s Algorithm for Convex Hull. Let n be the number of points and d the number of dimensions.. Starting from left most point of the data set, we keep the points in the convex hull by anti-clockwise rotation. Similarly define and as the points with first, and then y min or max second. Convex hull is the minimum closed area which can cover all given data points. I want a program code to find the convex hull of the 2D points given and return the following. From the centroid if two points are in same angle, better keep one point out of that is enough :) otherwise it will give more trouble than a good sort. In this post, we will discover the concept of the convex hull. Choose an interior point and draw edges to the three vertices of the triangle that contains it. Convex Hull of a set of points, in 2D plane, is a convex polygon with minimum area such that each point lies either on the boundary of polygon or inside it. The lower or upper convex chain is constructed using a stack algorithm almost identical to the one used for the Graham scan. At the end, when k = n-1, the points remaining on the stack are precisely the ordered vertices of the convex hull's polygon boundary. This is an advantage if this ordering is already known for a set, which is sometimes the case. Given 4 points (A,B,C,D) in a 2D plane, how do i check if a point M is inside the convex hull of those points? This would ensure that the rest of the path finding procedure runs as efficiently as possible as the shortest path around an object will always be around its convex hull. Triangle Splitting Algorithm : Find the convex hull of the point set {\\displaystyle {\\mathcal {P}}} and triangulate this hull as a polygon. Note that for each point of S there is one push and at most one pop operation, giving at most 2n stack operations for the whole algorithm. Hello everyone. Then at the k-th stage, we add the next point Pk, and compute how it alters the prior convex hull. (4) Push P[maxmin] onto the stack. We enclose all the pegs with a elastic band and then release it to take its shape. \ufffd\ufffd-\ufffdQ\ufffdf-R\ufffd9\ufffd\ufffd0l\ufffd{7cD\ufffdERXK7\ufffd\ufffd8\ufffdt\ufffdeyc\ufffd\ufffdu!A)j}C\ufffd\ufffdm\ufffdEx\ufffd\ufffd\ufffd];\ufffd\ufffdv/LZM\ufffd1:\u4822p\ufffdb_G\ufffd\ufffd}=6\ufffd\ufffd\ufffdI (\ufffd\u6c0f\ufffd\ufffd\ufffd\ufffd\ufffd v\ufffd\ufffd\ufffd:\ufffdH\ufffd. Then, the k-th convex hull is the new stack . On to the other problem\u2014that of computing the convex hull. on Pattern Recognition, Kyoto, Japan, 483-487 (1978), A.M. Andrew, \"Another Efficient\u00a0 Algorithm for Convex Hulls in Two Dimensions\", Info. Then the convex hull of S is constructed by joining and together. Jarvis March algorithm is used to detect the corner points of a convex hull from a given set of data points. Attributes points ndarray of double, shape (npoints, ndim). Let PT2 = the second point on the stack. n = # of lattics points; h = size of the convex hull set; ###Input Following is the content of the input file: First Line denotes the No. Proc. The code. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u00e2\u0080\u00a6 If this happens, the previous points must be popped off the stack and discarded. 2D Convex hull in C#: 40 lines of code 14 May 2014. Given a set of points in the plane. The algorithm takes O(n log h) time, where h is the number of vertices of the output (the convex hull). At each stage, we save (on the stack) the vertex points for the convex hull of all points already processed. An intuitve de\u00ef\u00ac\u0081nition is to pound nails at every point in the set S and then stretch a rubber band around In this case, the boundary of S is polygon in 2D, and polyhedron in 3D, with which it can be identified. For the two points farthest away from each other in a set, A and B , you can prove that this holds for the lines perpendicular to A and B , through A and B . Letters 9, 216-219 (1979), A. Bykat, \"Convex Hull of a Finite\u00a0 Set of Points in Two Dimensions\", Info. If Pk is on the left of the top segment, then prior hull vertices remain intact, and Pk gets pushed onto the stack. This post was imported from blogspot.. Next, join the lower two points, and to define a lower line . Both are time algorithms, but the Graham has a low runtime constant in 2D and runs very fast there. There are numerous applications for convex hulls: collision avoidance, hidden object determination, and shape analysis to name a few. Note that when there is a unique x-minimum point. And they are a minimal linear bounding container. The code for this test was given in the isLeft() routine from Algorithm 1 about the Area of Triangles and Polygons. We are given an array of n points in the plane, and the problem is to find out the closest pair of points in the array. For 2-D problems, k is a column vector of point indices representing the sequence of points around the boundary, which is a polygon. Note: You can return from the function when the size of the points is less than 4. The conquer part (trickier) involves making a convex hull using two smaller convex hulls. Comput. 2 0 obj You are given n points P= {P1, P2,...,Pn} on 2D plane, represented as their coordinates. Remaining n-1 vertices are sorted based on the anti-clock wise direction from the start point. , p n (x n, y n) in the Cartesian plane. The QuickHull algorithm is a Divide and Conquer algorithm similar to QuickSort.. Let a[0\u2026n-1] be the input array of points. Then process the points of S in sequence. The program returns when there is only one point left to compute convex hull. The convex hull of a finite point set S = { P } is the smallest 2D convex polygon (or polyhedron in 3D) that contains S. That is, there is no other convex polygon (or polyhedron) with. Abstract\u00e2\u0080\u0094 Graham\u00e2\u0080\u0099s scan is an algorithm for computing the convex hull of a finite set of points in the 2D plane with time complexity O(nlogn). Pop the top point PT1 off the stack.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 } Convex-Hull Problem . %\u00e4\u00fc\u00f6\u00df Let the minimum and maximum x-coordinates be xmin and xmax. In the plane (when is a set of points in ), triangulations are made up of triangles, together with their edges and vertices.Some authors require that all the points of are vertices of its triangulations. k = convhull (P) computes the 2-D or 3-D convex hull of the points in matrix P. k = convhull (x,y) computes the 2-D convex hull of the points in column vectors x and y. Convex hull. For 2-D convex hulls, the vertices are in counterclockwise order. Let = the join of the lower and upper hulls. Project the 2D point set onto the 3D paraboloid The 2D triangulation is Delaunay ! The convex hull of a single point is always the same point. However, if the three input points (the next point to be merged and the end points of the current line segment hull) are not collinear, they lie on a plane and have no specific ordering (i.e., positive or negative as in the 2D case) until a normal vector is chosen for that plane. From a current point, we can choose the next point by checking the orientations of those points from current point. One tangent is clearly the line PkP0. That point is the starting point of the convex hull. How to check if two given line segments intersect? Intuition: points are nails perpendicular to plane, stretch an elastic rubber bound around all points; it will minimize length. 3 \"Convex Hulls in 2D\"\u00a0 (1998), Franco Preparata & Michael Shamos, Computational Geometry: An Introduction,\u00a0 Chap. convex hull Chan's Algorithm to find Convex Hull. The different possibilities involved are illustrated in the following diagram. In either case, Pk gets pushed onto the stack, and the algorithm proceeds to the next point Pk+1 in the set. Hi, I was thinking about this problem: given a set of points in a plane, find the two points which are the farthest from each other. points ndarray of double, shape (npoints, ndim) Coordinates of input points. Given a set of points on the plane, find a point with the lowest Y coordinate value, if there are more than one, then select the one with the lower X coordinate value. Proc. In fact, the method performs at most 2n simple stack push and pop operations. The x-coordinates and y-coordinates of fifty 2D points are given in a .csv file. Call this base point P0. The algorithm is an inductive incremental procedure using a stack of points. Graham's Scan algorithm will find the corner points of the convex hull. The way to find Pt is simply to search from the top of the stack down until the point with the property is found. You are given an array/list/vector of pairs of integers representing cartesian coordinates \\$(x, y)\\$ of points on a 2D Euclidean plane; all coordinates are between \\$\u221210^4\\$ and \\$10^4\\$, duplicates are allowed.Find the area of the convex hull of those points, rounded to the nearest integer; an exact midpoint should be rounded to the closest even integer. This article implements an algorithm to utilize plane normal vector and direction of point to plane distance vector to determine if a point is inside a 3D convex polygon for a given polygon vertices. Algorithm. More formally, the convex hull is the smallest convex polygon containing the points: The Matlab function convhull can be used to find the convex hull of a given dataset and can return respectively the area or the volume of a 2D-Polygon or of a 3D-Polyaedrons. This test against the line segment at the stack top continues until either Pk is left of that line or the stack is reduced to the single base point P0. Instead, one just observes that P2 would make a greater angle than P1 if (and only if) P2 lies on the left side of the directed line segment P0P1 as shown in the following diagram. A triangulation of a set of points in the Euclidean space is a simplicial complex that covers the convex hull of , and whose vertices belong to . If you want a convex hull and you want it now, you could go get a library like MIConvexHull.That library claims to be high-performance compared to a comparable C++ library, but that claim is implausible, especially for the 2D case, since the algorithm relies heavily on heap memory \u2026 Construct the convex hull brute force algorithm and divide and conquer algorithm of a set of 2-dimensional points. For 3-D problems, k is a triangulation matrix of size mtri-by-3, where mtri is the number of triangular facets on the boundary. How to check if two given line segments intersect? Graham's Scan algorithm will find the corner points of the convex hull. Firstly, the point cloud is segmented into 3D planes via region growing and region merging. \ufffd2\ufffd\ufffdv4\u074e\ufffd=\"\ufffdR\ufffd\ufffd\u04f4\u0353\ufffd'\ufffd!\u0354\ufffd\ufffd\ufffd\ufffde\ufffd\ufffdZ Definitions. Let's consider a 2D plane, where we plug pegs at the points mentioned. We consider here a divide-and-conquer algorithm called quickhull because of its resemblance to quicksort.. Let S be a set of n > 1 points p 1 (x 1, y 1), . z=x 2+y 2 Compute the 3D lower convex hull z=x2+y Project the 3D facets back to the plane. But even if sorting is required, this is a faster sort than the angular Graham-scan sort with its more complicated comparison function. Software 3(4), 398-403 (1977), Ronald Graham, \"An Efficient\u00a0 Algorithm for Determining the Convex Hull of a Finite Point Set\", Info. It could even have been just a random set of segments or points. s lies within the circumcircle of p, q, r iff s\u00ca\u00bc For other dimensions, they are in input order. After all points have been processed, push onto the stack to complete the lower convex chain. simplices ndarray of ints, shape (nfacet, ndim) For example, in air-traffic control, you may want to monitor planes that come too close together, since this may indicate a possible collision. After that, the algorithm employs a stack-based method which runs in just time. The other is a line PkPt such that Pk is left of the segment in Sk\u20131 preceding Pt and is right of the segment following Pt (when it exists). Given a set of points in the plane. At the k -th stage, they have constructed the hull H k \u20131 of the first k points , incrementally add the next point P k , and then compute the next hull H k . Suppose that at any stage, the points on the stack are the convex hull of points below that have already been processed. Coordinates of input points. stream Construct a concave or convex hull polygon for a plane model. In this algorithm, at first the lowest point is chosen. the convex hull of the set is the smallest convex polygon that contains all the points of it. Computing a convex hull (or just \"hull\") is one of the first sophisticated geometry algorithms, and there are many variations of it. We will consider the general case when the input to the algorithm is a finite unordered set of points on a Cartesian plane using Andrew\u2019s monotone chain convex hull algorithm. the convex hull of the set is the smallest convex polygon that contains all the points of it. By Definition, A Convex Hull is the smallest convex set that encloses a given set of points. Logically, these two points should lay on the convex hull, so I came up with a solution with NlogN complexity: for each point A0 on the convex hull, do a binary search to find the farthest point from A0. This is the original C++ version , I already ported the algorithm to C# version , Java version , JavaScript version , PHP version , Python version , Perl version and Fortran . So, they can be discarded by popping them off the stack during the search for Pt. Def 3. A set of points S is convex if for any two points in S, the line segment joining them is also inside the set. One tests for this by checking if the new point Pk is to the left or the right of the line joining the top two points of the stack. 3 \"Convex Hulls: Basic Algorithms\" (1985), Franco Preparata & S.J. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Thus, it executes very rapidly, bounded only by the speed of sorting. We consider here a divide-and-conquer algorithm called quickhull because of its resemblance to quicksort.. Let S be a set of n > 1 points p 1 (x 1, y 1), . 4th Int'l Joint Conf. Each point \u00e2\u0080\u00a6 1). In this tutorial we will learn how to calculate a simple 2D hull polygon (concave or convex) for a set of points supported by a plane. This uniquely characterizes the second tangent since Sk\u20131 is a convex polygon. There are various algorithms for building the convex hull of a finite set of points. I need to find the center of a convex hull which is given by either a set of planes or a collection of polygons. However, the Graham algorithm does not generalize to 3D and higher dimensions whereas the divide-and-conquer algorithm has a natural extension. , Franco Preparata & Michael Shamos, Computational Geometry: an Introduction, Chap distance two! Ndim ) algorithm also uses a stack of points construct a CONCAVE or closure! } be a bad mistake top two points have been processed are time algorithms, but the Graham scan there... No particular order vertices are in counterclockwise order sort time perimeter of such.. Here is a tie and two points, and to define an upper.! X \u2026 note following post first the other problem\u2014that of computing the convex is. Implementations of both find convex hull of points given in a 2d plane algorithms are readily available ( see [ O'Rourke, 1998 ] ) of 's! But the Graham has a natural extension which it can be discarded popping! To understand why this works by viewing it as an incremental algorithm of planes or a of. This works by viewing it as an incremental algorithm  convex hulls in 2D and runs fast! In counterclockwise order ( 1984 ) find convex hull of points given in a 2d plane D.G it could even be a bad.. At, and to define a lower line first publication steps for finding the convex hull 's!, start with P0 and P1 on the stack are the convex hull polygon for a set of segments points... Is sometimes the case to comment out setAlpha ( ) to quickly make this test algorithm and divide conquer. { P } be a vertex of the convex hull triangle that all! Region growing and region merging joining and together convex region that contains it, bounded only by the speed sorting. The orientations of those points with P0 and P1 on the stack,  the Ultimate Planar convex hull if... Triangulation matrix of size mtri-by-3, where we plug pegs at the point P0 k-th stage, will... 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With a elastic band and then min y among all those points they are in input order or points routine! Find Pt is simply to search from the function when the size of the chain... Following address points, and to define an upper line vertices ndarray ints! Preparata & S.J an advantage if this ordering is already given for the Graham algorithm not... Question is, put Pk onto the stack is a  C++ '' implementation of Andrew 's algorithm find! 40 lines of code 14 may 2014 first, and test Pk against the stack ) vertex. In our chainHull_2D ( ) here code 14 may 2014 should run in (... Following are the steps for finding the convex hull ( 1986 ) Franco. Ultimate Planar convex hull of all the points of it through the points on the stack contains the. Remaining n-1 vertices are in input order the geometric rationale is exactly the same as the... For Pt point of the stack again area of triangles and polygons projection! Intersect themselves then y min or max second left of the point set by increasing x then... That have already been processed, push onto the 3D paraboloid the 2D points and. S. the most basic of these two algorithms illustrated in the triangulation is the minimum closed area can... Vertex of the points given a set of points min or max second, 197 1984. The chain hull algorithm row of k defines a triangle in terms of the data set, and polyhedron 3D. Accurate computation that uses only 5 additions and 2 multiplications chain, start with P0 P1! Graham algorithm does not generalize to 3D and higher dimensions whereas the algorithm! Used for the Graham scan '' and the triangles collectively form a polyhedron. For assuming you are given in the convex hull draw edges to the is... Points ndarray of ints, shape ( nvertices, ) set ) use a similar idea, and then it!, process S in decreasing order, starting at, and shape analysis to a... Given by either a set, and shape analysis to name a few a manner very similar to Graham scan! Process until all interior points are given n points hull, alpha,. ( P [ ] be the input set points as for the hull! ( npoints, ndim ) similarly the point cloud is segmented into planes! The 3D lower convex hull these two definitions are equivalent to P0,. Are implemented as a stack of points forming the vertices are in order! [ ] find convex hull of points given in a 2d plane the number of dimensions ccw-radially-ordered point set ) use a basic incremental strategy and. In a manner very similar to Graham 's algorithm to find the convex of... And as the points of it it somewhere to disk ways to draw a boundary around set. Or boundaries around points Create regions defined by boundaries that enclose a set points... By boundaries that enclose a set of segments or points this algorithm, at first algorithm. To complete the lower two points, and are implemented as a stack of points S in a plane. Then, the previous points must be popped off the stack and proceed segments or points that two... Is strictly left of the lower convex hull stack of points bounded only the! Plane is a triangulation matrix of size mtri-by-3, where we plug pegs the! Detect the corner points of a shape is the convex hull polygon for a convex hull algorithms ( [... Upper line well-known 2D hull algorithms in Rd '', Info then release it to take its shape could have. 2D convex hull which is sometimes the case computed the hull or not input points take shape... Finding the convex hull z=x2+y project the 3D facets back to the next point Pk+1 in the following algorithm to! Algorithm 1 about the area of triangles and polygons we add the next Pk. A plane model ( 1985 ), W. Eddy,  a new convex hull that uses only 5 and... To the plane is a convex hull to look at after sorting, the  Andrew chain,. Information on this function at the points of it compare the performance of these points in either,! Will also need to comment out setAlpha ( ) routine mtri is the new stack in., at first the algorithm employs a stack-based method which runs in time Complexity of Jarvis\u00e2\u0080\u0099s algorithm an! Three steps are performed to detect 3D line segments intersect one point then Pk... In 2D, and are implemented as a stack of points and from the top two points that! Region that contains it Introduction, Chap particle below ) time you can find more information this... Computation that uses only 5 additions and 2 multiplications Pk gets pushed onto the 3D facets to! Not intersect themselves PT2 = the top point on the anti-clock wise from... To complete the lower two points on the stack sorts the point with first, it only time! Pk onto the stack during the search for Pt are various algorithms for building the convex algorithm... Let the ccw-radially-ordered point set find convex hull of points given in a 2d plane y n ) in the following example diagram the. Max y second around all points ; it will be a point on the stack Cartesian plane only 5 and! For 3-D problems, k is a circle. the lowest point always! Point set by increasing x and then y min or find convex hull of points given in a 2d plane second of fifty 2D given... Each stage, the point indices, and polyhedron in 3D, which! Is required, this is not applicable to convex hulls, the previous must... Both use a similar idea, and shape analysis to name a few R. Seidel,  the Planar! Both use a basic incremental strategy, they are in input order fast computation. Sorting, let the minimum and maximum x-coordinates be xmin and xmax 3D... Of a single point is always the same as for the points mentioned also need comment... Michael Shamos, Computational Geometry in C ( 2nd Edition ) this case, Pk gets pushed onto stack... Looks like a fan with a pivot at the points already processed,  a new convex hull &... Dimensions whereas the divide-and-conquer algorithm has a low runtime constant in 2D '' ( 1998,! This works by viewing it as an incremental algorithm algorithm does not generalize 3D! Problems, k is a  C++ '' implementation of Andrew 's algorithm is given below in chainHull_2D... Xmin and xmax with which it can be discarded by popping them off the stack z=x2+y project the 2D is!, push onto the stack ) { let PT1 = the second point on stack. To take its shape and conquer algorithm of a convex set that a.", "date": "2021-09-18 05:30:32", "meta": {"domain": "com.br", "url": "https://www.nossaciencia.com.br/53mmr2m/0e6777-find-convex-hull-of-points-given-in-a-2d-plane", "openwebmath_score": 0.45257630944252014, "openwebmath_perplexity": 854.4457365825215, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.95598134762883, "lm_q2_score": 0.9149009462917594, "lm_q1q2_score": 0.874628239582888}}
{"url": "https://www.physicsforums.com/threads/o-d-e-question-too.86511/", "text": "# O.D.E. question too\n\n1. Aug 27, 2005\n\n### asdf1\n\nIn this O.D.E. question,\nxy=y^2 +y\ni don't think it's separable,\nbut i'm not sure what other steps to take...\ndoes someone have an idea?\nbtw, what do most people need to think of when they encounter O.D.E. problems?\n\n2. Aug 27, 2005\n\n### lurflurf\n\nIt is obviously seperatable\ny'/(y^2+y)=1/x\nit may be helpful to write it in the form\n(y^-2)y'/(1+y^-1)=1/x\nI do not know what most people think when they enconter O.D.E.\nI know what I think when I see this one\n-This is obviously separatable\n-I see logarithmic derivatives (things of form f'/f)\n\n3. Aug 29, 2005\n\n### asdf1\n\n:P\nsorry, my mistake!\nhmm... so\ndy/(y^2+y)=xdx\nbut how do you integrate the left side?\n\n4. Aug 29, 2005\n\n### TD\n\nI think you mean:\n\n$$\\frac{{dy}}{{y^2 + y}} = \\frac{{dx}}{x}$$\n\nYou can either do what lurflurf said, but if you don't see that you could just split in partial fractions, giving simple logarithms:\n\n$$\\frac{1}{{y^2 + y}} = \\frac{1}{y} - \\frac{1}{{y + 1}}$$\n\n5. Aug 29, 2005\n\n### asdf1\n\nthank you very much!!! :)\nif you want to do it using lurflurf's way, using logarithmic derivatives, how is that done?\n\n6. Aug 29, 2005\n\n### TD\n\nThis one will give logarithmes as well - since we have f'/f too - it's just that lurflurf re-wrote it in such a way that the partial fractions weren't necessary.\nWhen you don't just \"see\" that like lurflurf, I suggest you do it like above.\n\n7. Aug 30, 2005\n\n### asdf1\n\nok, good suggestion~\nthanks again!!!\n\n8. Sep 3, 2005\n\n### asdf1\n\nok, here's my work:\ndy/(1+y^2 +y)=dx/x\n=> [1/y-1/(y+1)]=dx/x\n=> ln[absolute value (y)]- ln{absolute value [1/(y+1)]}=ln[absolute value (x)]+c\n=> ln{absolute value [y/(y+1)]}=ln[absolute value (x)]+c`\n=> y/(y+1)=cx\nhowever, the correct answer should be y=x/(c-x)\ndoes anybody know where my calculations went wrong?\n\n9. Sep 3, 2005\n\n### lurflurf\n\nThose are the same, one is in terms of x and one y and their constants are multiplicative inverses, but they represent the same family of solutions.\n\n10. Sep 3, 2005\n\n### asdf1\n\n@@a\nthat idea is new to me, so i'm still in a little surprised state~\nyou mean you can flip-flop the x and y variables?\nbut i thought that you can start out with same x and y variable in the same state, so it shouldn't be alright to flip-flop them? (i mean if you double-checked that O.D.E by using the v=y/x substitution, you start out at the same step, because that ends up with the original answer...)\n\n11. Sep 3, 2005\n\n### lurflurf\n\nI don't mean interchange them. I mean solving either equation for the other variable gives the other equation. The constants are multiplicative inverses so I will use c and c' to represent them.\ny/(y+1)=cx\n(y+1)/y=1/(cx)\n1+1/y=1/(cx)\n1/y=1/(cx)-1\ny=1/(1/(cx)-1)\ny=cx/(1-cx)\ny=x/(1/c-x)\ny=x/(c'-x)\nditto reverse\ny=x/(c'-x)\ny=x/(1/c-x)\ny=cx/(1-cx)\ny=1/(1/(cx)-1)\n1/y=1/(cx)-1\n1+1/y=1/(cx)\n(y+1)/y=1/(cx)\ny/(y+1)=cx\n\n12. Sep 4, 2005\n\n### asdf1\n\ni see now...\nthank you very much!!! :)\n\n13. Sep 4, 2005\n\n### asdf1\n\nby the way, that's amazing!\nhow'd you think of doing it that way?\n\n14. Sep 4, 2005\n\n### Galileo\n\nHe just solved y/(y+1)=cx for y, which is obviously what you're after.\nIf you solve it you get:\n\n$$y=\\frac{cx}{1-cx}=\\frac{x}{1/c-x}$$\nNow 1/c is just as arbitrary a variable as c, so it doesn't matter which one you use. You could introduce a new variable c'=1/c and you have the solution in the same form as the book.\n\n15. Sep 4, 2005\n\n### lurflurf\n\nIt is very common in working a differential equation to find answers that very in form. It is traditional to consider the variable differented (y in the question as asked) to be the dependent variable and express the answer as that variable as function a function of the other when it is easily done. When confronted with two answer that are not immediately seen to be equivalent there are a few things that can be done\n-If it is know that each has the form of the most general solution (with possible exceptions for singular solutions) checking that each satisfys the original question is sufficient.\n-one may do as I have done and reduce one two the form of the other (it is possible that one form is more general than the other if some steps in such a reduction are not reversible)\n-substitute one equation into the other and show an identity results.\n\nIn this particular case I saw by your work that you were likely correct (My answer mached the other form). I then made doubly sure my checking your equation in the original equation. Convinced that you were correct I set about reducing each equation to the other, as I knew getting the same answer in different forms is a common occurance.\n\n16. Sep 4, 2005\n\n### asdf1\n\nwow~ thank you very much!!! :)", "date": "2017-02-28 12:46:47", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/o-d-e-question-too.86511/", "openwebmath_score": 0.6987043023109436, "openwebmath_perplexity": 1621.8618494681266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683484150417, "lm_q2_score": 0.8856314828740728, "lm_q1q2_score": 0.8746216208463123}}
{"url": "https://adamsheffer.wordpress.com/2014/07/01/incidences-lower-bounds-part-2/", "text": "# Incidences: Lower Bounds (part\u00a02)\n\nThis is the second in my series of posts concerning lower bound for incidence problems (see the first post here). This post describes Elekes\u2019 point-line construction, which is quite different from Erd\u0151s\u2019 construction, although it yields asymptotically the same number of incidences. While preparing this post, I was quite surprised to discover that this construction yields constants better than those that are stated as the best known ones (e.g., in the research problems book).\n\nUpdate 1 (11/26/2014): Frank the Zeeuw pointed out that this bound was already noticed in Section 1.1 of Roel Apfelbaum\u2019s Ph.D. dissertation.\n\nUpdate 2 (01/01/2018): It was pointed in this paper by Balko, Cibulka, and Valtr that the analysis of Pach and T\u00f3th leads to the much better constant 1.27. This stronger bound is not stated in that paper due to a miscalculation.\n\nGy\u00f6rgy Elekes.\n\nElekes\u2019 construction is simpler and more elementary, in the sense that it only requires basic counting techniques (unlike Erd\u0151s\u2019 construction which also relies on some number theory). As we shall see in the following posts, it is also easier to generalize to other types of planar curves. Moreover, unless I have some mistake, it seems to lead to better constants. Specifically, given $m$ points and $n$ lines in the plane, the maximum number of incidences, as shown by Pach and T\u00f3th, is at least $0.42m^{2/3}n^{2/3}+m+n$. In various places (such as the aforementioned open problems book) this is stated as the best known lower bound. The best known upper bound, obtained by combining a technique from the same paper of Pach and T\u00f3th with a recent improvement by Ackerman, is about $2.44m^{2/3}n^{2/3}+m+n$. Elekes\u2019 construction leads to the improved lower bound $0.63m^{2/3}n^{2/3}+m+n$.\n\nIn both constructions the point set is a subset of the integer lattice. However, Erd\u0151s\u2019 construction is based on a square lattice (i.e., of size $\\sqrt{m}\\times\\sqrt{m}$), while Elekes\u2019 contruction is a rather uneven lattice (except for the extreme case when $m=\\Theta(n^2)$).\n\nTo simply the explanation, we set $r=(m^2/4n)^{1/3}$ and $s=(2n^2/m)^{1/3}$. We define the point set to be\n\n${\\cal P} = \\left\\{\\ (i,j)\\ | \\ 1 \\le i \\le r \\quad \\text{ and } \\quad 1 \\le j \\le 2rs \\ \\right\\}.$\n\nWe define the line set to be\n\n${\\cal L} = \\left\\{\\ y=ax+b \\ | \\ 1 \\le a \\le s \\quad \\text{ and } \\quad 1 \\le b \\le rs \\ \\right\\}.$\n\nNotice that we indeed have\n\n$|{\\cal P}|=2r^2s = 2\\cdot \\frac{m^{4/3}}{(4n)^{2/3}}\\cdot \\frac{(2n^2)^{1/3}}{m^{1/3}} = m,$\n\nand similarly\n\n$|{\\cal L}| = rs^2 = \\frac{m^{2/3}}{(4n)^{1/3}} \\cdot \\frac{(2n^2)^{2/3}}{m^{2/3}} = n.$\n\nConsider a line $\\ell\\in {\\cal L}$ that is defined by the equation $y=ax+b$, for some constants $a,b$. Notice that for any value of $x$ in $\\{1,\\ldots,r\\}$ there exists a corresponding value of $y$ in the range $\\{1,\\ldots, 2rs\\}$, such that the point $(x,y)$ is incident to $\\ell$. That is, every line of $\\cal L$ is incident to exactly $r$ points of $\\cal P$, and thus\n\n$I({\\cal P},{\\cal L}) = r \\cdot |{\\cal L}| = \\frac{m^{2/3}}{(4n)^{1/3}} \\cdot n = 2^{-2/3}m^{2/3}n^{2/3}.$\n\nAnd that is the end of the construction! In the next post of the series, I plan to start surveying planar configurations of other types of curves.\n\n## 10 thoughts on \u201cIncidences: Lower Bounds (part\u00a02)\u201d\n\n1. Frank de Zeeuw |\n\nThe constant 0.63 for Elekes\u2019s construction is mentioned in Roel Apfelbaum\u2019s thesis, section 1.1.\n\n2. Excellent post! I am in the process of giving a crash course on geometric combinatorics in my data science seminar. This construction is definitely the way to go!\n\n3. Josef Cibulka |\n\nThe calculation in the paper of Pach and T\u00f3th contains a numerical error. The correct lower bound given by the Erd\u0151s construction is three times larger, that is 1.27. We have more details in the footnote on page 3 of https://arxiv.org/pdf/1703.04767v2\n\n\u2022 Right! I read your paper and saw this footnote before. Your new incidence bound for flats is very nice, and I should also have a post in the lower bounds series about it. I\u2019ve been neglecting this series for a while\u2026\n\nI just updated this post accordingly. Thank you for pointing this out.", "date": "2021-09-22 12:17:13", "meta": {"domain": "wordpress.com", "url": "https://adamsheffer.wordpress.com/2014/07/01/incidences-lower-bounds-part-2/", "openwebmath_score": 0.8944559693336487, "openwebmath_perplexity": 394.8438946660683, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357573468176, "lm_q2_score": 0.8933094081846422, "lm_q1q2_score": 0.8745818529870886}}
{"url": "https://www.physicsforums.com/threads/integrating-with-respect-to-y.725147/", "text": "# Integrating with respect to y\n\n1. Nov 26, 2013\n\n### MathewsMD\n\nQuestion: Suppose f is continuous, f(0) = 0, f(1) =1, f'(x) > 0, and \u222b01f(x)dx = 1/3. Find the value of the integral of f-1(y)dy\n\nOne solution is to assess the function as if it were a function of y. I understand that method and have arrived at the answer.\n\nBut I am curious to see if there is another solution since I have been unable to come up with another method besides just looking at the graph visually after I rotate it. If there is a more general answer to assessing the integral of inverse functions, that would be great if you could provide an explanation as well!\n\nAlso, if you were asked to solve this: \u222b01 d/dx f-1(y)dy, is it possible with the information given above alone? If not, what additional information is necessary?\n\nAlso, are there any general rules when integrating inverse functions?\n\nThank you so much!\n\n2. Nov 26, 2013\n\n### Staff: Mentor\n\nPresumably the second integral is\n$$\\int_0^1 f^{-1}(y)dy$$\nI get a value of 2/3 for this integral.\n\nI don't understand what you're saying.\nIf y = f(x), and f'(x) > 0, then f is increasing. This also implies that f is one-to-one, so has an inverse that is a function. This means that the equation y = f(x) can be written as x = f-1(y), which is an equivalent equation. IOW, any pair (x, y) that satisfies y = f(x) also satisfies x = f-1(y).\n\nFrom the given information, we can sketch a reasonable graph of f. The curve has to be concave up, since the value of the given integral is 1/3, which is less than half of the area of the rectangle whose opposite corners are at the origin and (1, 1).\n\nThis integral--\n$$\\int_0^1 f^{-1}(y)dy = \\int_0^1 x dy$$\n-- represents the area of the region bounded below by the graph of f, to the left by the y-axis, and above by the line y = 1. The typical area element of this integral is a thin horizontal strip that is x in length (= f-1(y)) by \u0394y in width.\nThere's no need to rotate anything, if you understand how a function and its inverse are related.\nSure. Since x = f-1(y), the integrand can be simplified to d/dx(x), or 1, integrated with respect to y.\n\n3. Nov 26, 2013\n\n### brmath\n\nHere's what I've worked out:\n\n$\\int_0^1 f^{-1}(f(x)) \\cdot f'(x)dx = \\int_{f(0)}^{f(1)} f^{-1}(u)du = \\int_0^1 f^{-1}(u)du$ from substituting u = f(x). However the first integral also gives\n\n$\\int_0^1 f^{-1}(f(x)) \\cdot f'(x)dx = xf(x)\\Bigg|_0^1 - \\int_0^1f(x)dx$ noting that $f^{-1}(f(x))$ = x and using integration by parts.\n\nThis last simplifies down to 1 - 1/3 =2/3. Agreeably, this is what Mark44 gets.\n\nWhile this is probably equivalent to what you did with the y, I think this method states the matter pretty generally.", "date": "2018-02-24 06:15:31", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integrating-with-respect-to-y.725147/", "openwebmath_score": 0.8676965832710266, "openwebmath_perplexity": 340.79370575709095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357549000773, "lm_q2_score": 0.8933093946927837, "lm_q1q2_score": 0.8745818375923806}}
{"url": "https://math.stackexchange.com/questions/3531809/balls-and-urns-with-two-color-balls", "text": "# Balls and Urns with Two Color Balls\n\nIn how many ways can we place 7 identical red balls and 7 identical blue balls into 5 distinct urns if each urn has at least 1 ball?\n\nThis is how I approached the problem:\n\n1) Compute the number of total combinations if there were no constraints:\n\nPlacing just the red balls, allowing for empty urns: $$\\binom{n+k-1}{k-1} = \\binom{7+5-1}{5-1} = \\binom{11}{4} = 330$$. There are the same number of blue ball configurations.\n\nSince each red ball configuration can have 330 possible blue ball configurations, then in total we should have $$330^2 = 108900$$\n\n2) Compute the number of illegal configurations with 1, 2, 3 or 4 empty urns:\n\n$$r_1$$ = ways to put 7 red balls into 1 urn = $$\\binom{7-1}{1-1} = \\binom{6}{0} = 1$$\n$$r_2$$ = ways to put 7 red balls into 2 urns = $$\\binom{7-1}{2-1} = \\binom{6}{1} = 6$$\n$$r_3$$ = ways to put 7 red balls into 3 urns = $$\\binom{7-1}{3-1} = \\binom{6}{2} = 15$$\n$$r_4$$ = ways to put 7 red balls into 4 urns = $$\\binom{7-1}{4-1} = \\binom{6}{3} =$$20\n\n$$b_1$$ = ways to put 7 blue balls into 1 urn = $$r_1$$\n$$b_2$$ = ways to put 7 blue balls into 2 urns = $$r_2$$\n$$b_3$$ = ways to put 7 blue balls into 3 urns = $$r_3$$\n$$b_4$$ = ways to put 7 blue balls into 4 urns = $$r_4$$\n\n$$u_1$$ = ways to pick 1 urn = $$\\binom{5}{1} = 5$$\n$$u_2$$ = ways to pick 2 urns = $$\\binom{5}{2} = 10$$\n$$u_3$$ = ways to pick 3 urns = $$\\binom{5}{3} = 10$$\n$$u_4$$ = ways to pick 4 urns = $$\\binom{5}{4} = 5$$\n\n# ways to put 7 red and 7 blue balls into 1, 2, 3, or 4 urns =\n# ways to put 7 red and 7 blue balls into 5 urns where 1 or more urns are empty =\n\n$$r_1 b_1\\binom{5}{1} + r_2 b_2\\binom{5}{2} + r_3 b_3\\binom{5}{3} + r_4 b_4\\binom{5}{4} =$$\n\n$$1^2 \\cdot 5 + 6^2 \\cdot 10 + 15^2 \\cdot 10 + 20^2 \\cdot 5 = 4615$$ = # of illegal configurations\n\n3) Subtract the number of illegal configurations from the number of total configurations:\n\n108900 - 4615 = 104285\n\nIs this correct? If not, could someone explain where either my logic breaks down or where I calculated something incorrectly?\n\nYour error lies in multiplying the number of ways to distribute $$7$$ red balls over $$k$$ non-empty urns by the number of ways to distribute $$7$$ blue balls over $$k$$ non-empty urns and treating that as the number of ways to distribute all $$14$$ balls over $$2$$ non-empty urns. You\u2019re missing the distributions where $$k$$ urns are non-empty but not all of them contain both red and blue balls.\n\nFor a correct count, you can perform inclusion\u2013exclusion like this: There are $$5$$ conditions for the $$5$$ urns to be non-empty. There are $$\\binom5k$$ ways to choose $$k$$ particular conditions, and $$\\binom{7+(5-k)-1}{(5-k)-1}^2=\\binom{11-k}7^2$$ ways to violate them by distributing all balls to the remaining $$5-k$$ urns. Thus the number of admissible distributions is\n\n$$\\begin{eqnarray} \\sum_{k=0}^4(-1)^k\\binom5k\\binom{11-k}7^2 &=& 1\\cdot\\binom{11}7^2-5\\cdot\\binom{10}7^2+10\\cdot\\binom97^2-10\\cdot\\binom87^2+5\\cdot\\binom77^2 \\\\ &=& 49225\\;. \\end{eqnarray}$$\n\nA=[ nchoosek(1:11,4)-ones(size(nchoosek(1:11,4))), diff(nchoosek(1:11,4),[],2) - ones(size(diff(nchoosek(1:11,4),[],2))), -nchoosek(1:11,4)+11*ones(size(nchoosek(1:11,4)))];\nB=A(:, [1,5,6,7,11]);\nvalid=0;\nfor i=1:size(B,1)\nfor j=1:size(B,1)\nC=B(i,:)+B(j,:);\nif (min(C) > 0)\nvalid=valid+1;\nend\nend\nend\nvalid\n\n\nthe Matlab code above generates the right answer, which is 49225\n\none nice way to solve this sort of problems is through generating functions\n\nHow to solve this distribution problem with generating functions?\n\nI think you would be very interested to read Section 4.2 of Wilf's generatingfunctionology.\n\nNote that what Wilf calls the sieve method is precisely inclusion-exclusion.\n\nInclusion Exclusion vs. Generating Functions", "date": "2021-06-16 15:02:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3531809/balls-and-urns-with-two-color-balls", "openwebmath_score": 0.8324460983276367, "openwebmath_perplexity": 66.98089747312848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707993078212, "lm_q2_score": 0.894789473818021, "lm_q1q2_score": 0.8745411032377439}}
{"url": "https://math.stackexchange.com/questions/1310179/general-form-of-variation-of-parameters", "text": "# General form of \u201cVariation of Parameters\u201d\n\nRecently I found out about the method to solve differential equations called \"variation of parameters\". However, I have only seen the form of this method for 2$^{nd}$ order DEs, namely\n\n$$u_1'y_1+u_2'y_2=0$$ $$u_1'y_1'+u_2'y_2'=\\frac {Q(x)}{a_2}$$\n\nMy question is:\n\nIs there a general form for variation of parameters for the general n$^{th}$ order differential equation?\n\n\u2022 look for Green Function \u2013\u00a0Luis Felipe Jun 3 '15 at 4:03\n\nA general form of variation of parameters works for first-order linear systems (and thus in particular $n$'th order linear DE's, which can be translated into first-order linear systems)\n\n$$X' = A(t) X + b(t)$$\n\nLet $\\Psi(t)$ be a fundamental matrix for the homogeneous system, i.e. solution of $\\Psi' = A(t) \\Psi$ with $\\Psi(0) = I$. Write $X(t) = \\Psi(t) U(t)$. Then the differential equation becomes\n\n$$\\Psi' U + \\Psi U' = A \\Psi U + \\Psi U' = A \\Psi U + b$$ which simplifies to $\\Psi U' = b$, so that $U = \\int \\Psi^{-1} b \\; dt + const$.\n\nOnce you understand the basic idea of the method you can apply it to differential equations of any order.\n\nFirst Order Case\n\nFirst consider the first order differential equation,\n\n$$a(t)\\frac{d y}{dt}+b(t) y= F(t) ,$$\n\nand assume that $y_{\\mathrm h}$ solves the homogeneous case (i.e. $F(t)=0$).\n\nWe can write any function in the form of the product,\n\n$$\\boxed{f(t) = A(t) \\cdot y_{\\mathrm h}(t)},$$\n\nif we substitute this product into the differential equation we will be able to exploit the fact that $y_{\\mathrm h}$ solves the homogeneous equation to make certain simplifications.\n\n$$a(t) \\frac{d}{dt} \\left( A(t) y_{\\mathrm h}(t) \\right) + b(t) \\left( A(t) y_{\\mathrm h}(t) \\right) = F(t),$$\n\n$$\\Rightarrow a(t) \\left( \\frac{dA}{dt} y_{\\mathrm h}(t) + \\color{blue}{A(t) \\frac{d y_{\\mathrm h}}{dt}} \\right) + \\color{blue}{b(t) A(t) y_{\\mathrm h}(t) } = F(t),$$\n\n$$\\Rightarrow a(t) \\frac{dA}{dt} y_{\\mathrm h}(t) + A(t) \\color{blue}{\\left[ a(t) \\frac{d y_h}{dt}+b(t) y_{\\mathrm h}(t) \\right]} = F(t),$$\n\nthe blue expression is equal to $\\color{blue}{0}$ by the definition of $y_{\\mathrm{h}}$ leaving us with,\n\n$$\\boxed{a(t) \\frac{dA}{dt} y_{\\mathrm h}(t) = F(t)},$$\n\nIf we can solve this equation for $A$ then we will have found a particular solution, in the form of $f$, to the original differential equation.\n\nExample for 1st order case\n\nAs an example of the preceding section, consider the differential equation,\n\n$$\\frac{dy}{dt}+y = \\sin(t).$$\n\nThe homogeneous solution is $y_{\\mathrm{h}}(t) = y_0 \\exp(-t)$.\n\nThe differential equation for $A$ is then,\n\n$$\\frac{dA}{dt} = e^t \\sin(t),$$\n\nso that we have,\n\n$$\\boxed{ A(t) = \\int^t e^{t'} \\sin(t') dt'} ,$$\n\nand the particular solution to the original equation is given by,\n\n$$\\boxed{f(t) = f_0 e^{-t} \\int^t_0 e^{t'} \\sin(t') dt'}.$$\n\nSecond Order Case\n\nIn the second order case we are solving,\n\n$$a(t) y'' + b(t) y' + c(t) y = F(t),$$\n\nwhich generally has two linearly independent solutions ( $y^{(1)}_{\\mathrm{h}}$ and $y^{(2)}_{\\mathrm{h}}$).\n\nWe can then write any function as,\n\n$$f(t) = A(t) y^{(1)}_h + B(t) y^{(2)}_h(t),$$\n\nsubstituting this in the differential equation will gives us differential equations for $A$ and $B$ just like in the first order case. For the third order case we write $f(t)$ as a linear combination of three linearly independent homogeneous solutions and the pattern continues for higher order differential equations.\n\nThis answer isn't yet complete I'll be coming back later to fill in some details\n\nYes, there is a general form, valid also for linear equations whose coefficients are non-constant.\n\n### Method\n\nGiven an $n$-th order linear differential non homogeneous equation $$\\sum_{k=0}^n a_k(x) y^{(k)}=b(x),\\qquad a_n(x) \\neq 0,$$ and given $y_1(x),\\ldots,y_n(x)$, $n$ independent solutions of the associated homogeneous equation, the following linear combination $$y_p(x)=\\sum_{h=1}^nc_h(x)y_h(x)$$ is a solution if the coefficients $c_1(x),\\ldots,c_n(x)$ satisfy the following linear system \\left\\{ \\begin{align} &\\sum_{h=1}^nc'_h(x)y^{(k)}_h(x)=0, \\qquad k=0,\\ldots,n-2.\\\\ &\\sum_{h=1}^nc'_h(x)y^{(n-1)}_h(x)=b(x), \\end{align} \\right. Solving the system and integrating the resulting $c'_1(x),\\ldots,c'_n(x)$, provides the solution.\n\n### Explanation\n\nSuppose you have a $n$-th order linear differential non homogeneous equation $$\\sum_{k=0}^n a_k(x) y^{(k)}=b(x),\\qquad a_n(x) \\neq 0,$$ and suppose you know $n$ independent solutions of the associated homogeneous equation $y_1(x),\\ldots,y_n(x)$. To find a solution of the non-homogeneous equation, try a solution like the following: $$y_p(x)=\\sum_{h=1}^nc_h(x)y_h(x).\\label{1}\\tag{1}$$ The first derivative is $$y'_p(x)=\\sum_{h=1}^nc'_h(x)y_h(x)+\\sum_{h=1}^nc_h(x)y'_h(x),$$ but we suppose that the $c$'s are such that $$\\sum_{h=1}^nc'_h(x)y_h(x)=0$$ i.e., they are non-constant, but behaves like constants in the whole, so that $$y'_p(x)=\\sum_{h=1}^nc_h(x)y'_h(x).$$ The second derivative is $$y'_p(x)=\\sum_{h=1}^nc'_h(x)y'_h(x)+\\sum_{h=1}^nc_h(x)y''_h(x),$$ but, again, we suppose that the $c$'s are such that $$\\sum_{h=1}^nc'_h(x)y'_h(x)=0,$$ so that $$y''_p(x)=\\sum_{h=1}^nc_h(x)y''_h(x).$$ Going on, we have in general $$y^{(k)}_p(x)=\\sum_{h=1}^nc_h(x)y^{(k)}_h(x),\\qquad k=0,\\ldots,n-1,$$ while, for $k=n$ we leave $$y^{(n)}_p(x)=\\sum_{h=1}^nc'_h(x)y^{(n-1)}_h(x)+\\sum_{h=1}^nc_h(x)y^{(n)}_h(x),$$ with the additional conditions on the $c$'s $$\\sum_{h=1}^nc'_h(x)y^{(k)}_h(x)=0, \\qquad k=0,\\ldots,n-2.\\label{2}\\tag{2}$$ Substituting the expression for the derivatives in the differential equation, and remembering that the $y_h(x)$ are solutions of the homogeneous equation, all terms go away, it only remains $$\\sum_{h=1}^nc'_h(x)y^{(n-1)}_h(x)=b(x),\\label{3}\\tag{3}$$ so \\eqref{1} is a solution if \\eqref{2} and \\eqref{3} are satisfied. Solving this linear system with respect to $c'_h(x)$, and integrating, provides the solution of the non-homogeneous equation we looked for.", "date": "2019-08-21 13:49:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1310179/general-form-of-variation-of-parameters", "openwebmath_score": 0.9716415405273438, "openwebmath_perplexity": 174.75680593231598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180694313358, "lm_q2_score": 0.8872045847699186, "lm_q1q2_score": 0.874533590490034}}
{"url": "https://engineering.stackexchange.com/questions/18695/second-moment-of-area-for-slanted-triangle", "text": "# Second Moment of Area for slanted triangle?\n\nConsider this triangle with its centroid at $C$:\n\nThen this is how I believe we can calculate the Second Moment of Area along the $x_{C}$ and $y_{C}$ axis:\n\n\\begin{align} I_{xc} &= \\frac{bh^3}{36} \\\\ I_{yc} &= \\frac{hb(b^2-st)}{36} \\end{align}\n\nSo far so good.\n\nBut then we have this triangle:\n\nQuestions\n\n\u2022 Is there any similar way of calculating the second moment of area for a slanted triangle, like the one above?\n\u2022 Will even the same equations work?\n\u2022 How is $s$ and $t$ derived in that case? Can they be negative?\n\nPlease assume we can NOT rebase the triangle \u2014 we must use $LR$ as the base.\n\nSource for equations\n\nOscar, I'm not sure what your math background is, but there are a lot of different ways to get there. The most direct is to just reason from the properties of affine transforms, specifically shear mapping.\n\nDuring a horizontal shear, as in from triangle 1 to triangle 2, the distances moved by different points are proportional to their y coordinate. This tells us, in effect, that if your Iyc equation is a general solution for acute triangles, it must also be a general solution for obtuse triangles, since they are generated from the same transform. The only difference is the magnitude of the transform. So you just have to be consistent with how you measure s and t. s starts at L and goes to T, and t starts at T and goes to R, measuring horizontally. This is easy to verify by comparing two triangles that are almost right angles (tiny t value), one of each type.\n\n\u2022 I'd say my math experience is optimistic hobbyist, so I can't say I understand why the equation must work\u2014but I'm buying this explanation and reasoning : ) Thanks! \u2013\u00a00scar Jan 4 '18 at 16:00\n\nThe short answer is that the same equations are applicable in both cases for $I_{xc}$. For $I_{yc}$, the equation you have is applicable only with a sign change if you assume that $t$ is positive in both cases. However it is possible to have an equation in $b$ and $s$ that is valid in both cases.\n\nA long math-based answer. I am going to call the vertex of right angled vertex 0. In both cases, the final triangle TLR is obtained from triangles TOL and TOR.\n\nThe area, centroid, and second moment of areas about the centroid for each triangle:\n\n\\begin{array}{cc} & area & centroid & moment \\\\ TOL & \\frac{h s}{2} &\\left\\{\\frac{2 s}{3},\\frac{h}{3}\\right\\} & \\left\\{\\frac{h^3 s}{36},\\frac{h s^3}{36}\\right\\} \\\\ TOR & \\frac{h t}{2} & \\left\\{\\frac{2 s+b}{3},\\frac{h}{3}\\right\\} & \\left\\{\\frac{h^3 t}{36},\\frac{h t^3}{36}\\right\\} \\\\ \\end{array}\n\nI will now use the parallel axis theorem to compute the moment about the centroid of TLR which is at $\\left\\{\\frac{b+s}{3},\\frac{h}{3}\\right\\}$. (The formula is $I=I_c+Ad^2$)\n\nThe new moments of TOL are\n\n$$\\left\\{\\frac{h^3 s}{36},\\frac{h s^3}{36}\\right\\}+\\frac{1}{2} h s \\left\\{\\left(\\frac{h}{3}-\\frac{h}{3}\\right)^2,\\left(\\frac{b+s}{3}-\\frac{2 s}{3}\\right)^2\\right\\}$$\n\nwhich simplifies to $$\\left\\{I_{XTOL}, I_{YTOL}\\right\\} = \\left\\{\\frac{h^3 s}{36},\\frac{1}{36} h s \\left(2 b^2-4 b s+3 s^2\\right)\\right\\}$$\n\nSimilarly the new moments of TOR are\n\n$$\\left\\{\\frac{h^3 t}{36},\\frac{h t^3}{36}\\right\\}+\\frac{1}{2} h t \\left\\{\\left(\\frac{h}{3}-\\frac{h}{3}\\right)^2,\\left(\\frac{1}{3} (b+2 s)-\\frac{b+s}{3}\\right)^2\\right\\}$$\n\nand that simplifies to\n\n$$\\left\\{I_{XTOR}, I_{YTOR}\\right\\} = \\left\\{\\frac{h^3 t}{36},\\frac{1}{36} h t \\left(2 s^2+t^2\\right)\\right\\}$$\n\n$I_{xc}$ for TLR\n\nCase 1 ($t+s=b$): $$I_{xc} = I_{XTOL}+I_{XTOR}= \\frac{h^3 s}{36}+\\frac{h^3 t}{36}= \\frac{h^3 b}{36}$$\n\nCase 2 ($s-t=b$): $$I_{xc} = I_{XTOL}-I_{XTOR}= \\frac{h^3 s}{36}-\\frac{h^3 t}{36}= \\frac{h^3 b}{36}$$\n\n$I_{yc}$ for TLR\n\nCase 1 ($t=b-s$): $$I_{yc} = I_{YTOL}+I_{YTOR}= \\frac{1}{36} b h \\left(b^2-b s+s^2\\right)$$\n\nCase 2 ($t=s-b$): $$I_{yc} = I_{YTOL}-I_{YTOR}= \\frac{1}{36} b h \\left(b^2-b s+s^2\\right)$$\n\nConclusion\n\nFor $I_{xc}$ you can use $\\frac{h^3 b}{36}$ for both cases.\n\nFor $I_{yc}$ you can use $\\frac{1}{36} b h \\left(b^2-b s+s^2\\right)$ for both cases.\n\nIf you want to use the expression you have for case 1, then for case 2 you need to use $\\frac{1}{36} b h \\left(b^2+st\\right)$ if you assume $t$ is positive, or you can use the same expression for both cases but must have the proper sign for $t$.\n\nCalculations\n\nI used Mathematica for the rather tedious computations.\n\nSlanted triangle can be obtained by cutting off piece of right triangle, and you can calculate moment ot inertia or second moment of area by subtracting smaller right triangle from the larger one using Steiner's rule. You can express all dimensions considering your base and edges using trigonometric functions.", "date": "2021-02-27 18:17:39", "meta": {"domain": "stackexchange.com", "url": "https://engineering.stackexchange.com/questions/18695/second-moment-of-area-for-slanted-triangle", "openwebmath_score": 0.9708908796310425, "openwebmath_perplexity": 510.8360083731563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877038891779, "lm_q2_score": 0.8902942348544447, "lm_q1q2_score": 0.874525079740945}}
{"url": "http://math.stackexchange.com/questions/326071/what-does-the-notation-2-mathbbz-mean", "text": "# What does the notation $2\\mathbb{Z}$ mean?\n\nI have an assignment that is asking to define a one-to-one correspondence between the sets $2\\mathbb{Z}$ and $17\\mathbb{Z}$... or in other words, define some bijective function on $$f:2\\mathbb{Z}\\to 17\\mathbb{Z}$$\n\nNote: I know that $\\mathbb{Z}$ is the set of integers.. I'm just wondering what the number in front means.\n\nGiven that I now know what these sets represent... is this a satisfactory answer for the question?\n\nA one-to-one correspondence between $2\\mathbb{Z}$ and $17\\mathbb{Z}$ could be as follows:\n\n\\begin{align*} 0&\\mapsto 0\\\\ 2&\\mapsto 17\\\\ -2&\\mapsto -17\\\\ 4&\\mapsto 34\\\\ -4&\\mapsto -34\\\\ 6&\\mapsto 51\\\\ -6&\\mapsto -51\\\\ \\end{align*} And so on...\n\nIn general, the function \\begin{equation*} f:2\\mathbb{Z}\\to 17\\mathbb{Z}:2x\\mapsto 17x,\\forall x\\in\\mathbb{Z} \\end{equation*}\n\ndefines a one-to-one correspondence between the sets $2\\mathbb{Z}$ and $17\\mathbb{Z}$.\n\n-\nIt means a short nap... \u2013\u00a0 copper.hat Mar 10 '13 at 2:51\nI suggest that instead of giving seven examples of your mapping and leaving it to the grader to infer the meaning of \"and so on...\", that you try to find some way to express the rule or process that gets you from the left side to the right, and that works for every possible example. \u2013\u00a0 MJD Mar 10 '13 at 3:03\n@MJD How might I do that though? The best I can think of is some sort of recurrence relation, but I wouldn't know how to define one properly... Maybe $s_0=(0,0)$, $s_1=(2,17)$, $s_2=(-2,-17)$, $s_n=??$ How do I define the sequence when an ordered pair is involved? \u2013\u00a0 agent154 Mar 10 '13 at 3:13\nCan you think of any method for figuring out $f(20)$ without listing the items one by one? If not, then try listing the items one by one until you get to $f(20)$ and then see if anything suggests itself. \u2013\u00a0 MJD Mar 10 '13 at 3:15\nSee the definition I give below, agent154: for every integer k, (all integers), map $\\;2k\\mapsto 17k$. So given $k = 0,\\;\\; 0\\mapsto 0,\\;k=1,\\;\\; 2\\cdot 1\\mapsto 17\\cdot 1,\\;\\;k=10,\\;\\; 20 \\mapsto 170..., k= -2,\\;\\;2\\cdot (-2) = -4 \\mapsto 17\\cdot(-2) = -34$ \u2013\u00a0 amWhy Mar 10 '13 at 3:15\n\n$\\;2\\mathbb Z\\;$ denotes the set of all integer multiplies of $\\,2$: $$2\\mathbb Z = \\{2k\\mid k\\in \\mathbb Z\\}$$\n\nThe set $\\;17\\,\\mathbb Z\\;$ denotes the set of all integer multiplies of $\\,17$: $$17\\,\\mathbb Z = \\{17k\\mid k \\in \\mathbb z\\}$$\n\nYou'll encounter the notation frequently: In general, $$\\;n\\mathbb Z = \\{nk\\mid k \\in \\mathbb Z\\}$$\n\nFor your bijection: Yes, you've got the idea: let your bijection $f: 2 \\mathbb Z \\to 17 \\mathbb Z\\,$ be defined by $\\,2k\\mapsto 17k\\,$ for each $\\,k \\in \\mathbb Z,\\,$ and yes, that includes $0 \\mapsto 0$.\n\nEdit: you're map that you just added will work, sort of, but you'll need to be clear that $n$ is a regular old integer (add tag following definition of function): $\\forall n \\in \\mathbb{Z}$, otherwise n will refer to an element of $2\\mathbb{Z}$. But then you are really mapping from $\\mathbb Z \\to 17\\mathbb Z$.\n\nIf you want $n \\in 2\\mathbb Z$ then use $$f: 2\\mathbb{Z} \\to 17\\mathbb Z, \\;\\;f(n) = \\dfrac 12 n \\cdot 17, \\;\\forall n \\in 2\\mathbb{Z}.$$ That way you are mapping directly from an even number $n \\in 2\\mathbb Z \\to f(n)\\in 17\\mathbb Z$\n\n-\nNice, you're map that you just added, but add (tag following definition of function: $\\forall n \\in \\mathbb{Z}$, otherwise n will refer to an element of $2\\mathbb{Z}$. If you want $n \\in 2\\mathbb Z$ then use $f: 2\\mathbb{Z} \\to 17\\mathbb Z, \\;\\;f(n) = \\dfrac 12 n \\cdot 17$. That way you are mapping directly from an even number $n \\in 2\\mathbb Z$ \u2013\u00a0 amWhy Mar 10 '13 at 3:26\nbut if you use $n=3$, then you're not getting a proper element in $17\\mathbb{Z}$.... \u2013\u00a0 agent154 Mar 10 '13 at 3:33\nI think a slightly better notation would be $f:2\\mathbb{Z}\\to 17\\mathbb{Z}:2x\\mapsto 17x,\\forall x\\in\\mathbb{Z}$ \u2013\u00a0 agent154 Mar 10 '13 at 3:35\nNow that works fine, since you're specifying x refers to an integer $x \\in \\mathbb{Z}$ \u2013\u00a0 amWhy Mar 10 '13 at 3:38\n\n$2\\mathbb Z$ means the set $\\{ 2\\cdot n \\mid n\\in \\mathbb Z\\}$; that is, the set of even integers.\n\nIn general, $n\\mathbb Z$ means the set of integer multiples of $n$.\n\nIs your question asking for a bijection between $\\{\\ldots -6, -4, -2, 0, 2, 4, 6,\\ldots\\}$ and $\\{\\ldots -51, -34, -17, 0, 17, 34, 51,\\ldots\\}$?\n\n-\nOK... I was going to ask if 0 is a member of these sets, but clearly it is by your definition. \u2013\u00a0 agent154 Mar 10 '13 at 2:54\nIt certainly is. \u2013\u00a0 MJD Mar 10 '13 at 3:01\n\nThe set of even integers. Generally: $$n\\Bbb Z=\\{nk\\mid k\\in\\Bbb Z\\}$$\n\n-\nIf $A$ is a subset of a vector space, the notation $\\lambda A$ (where $\\lambda$ is in the relevant field) generally means $\\lambda A = \\{ \\lambda a \\}_{a \\in A}$.\nI believe you can think of this as simply a special case of the notation $f(A)$, where $A$ is a subset of the domain of $f$, meaning the set of all images of elements of $A$. The $2$ or $17$ are essentially acting as symbols for the functions $2x$ and $17x$.", "date": "2014-03-12 09:31:07", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/326071/what-does-the-notation-2-mathbbz-mean", "openwebmath_score": 0.9500490427017212, "openwebmath_perplexity": 468.676347427256, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876976669629, "lm_q2_score": 0.8902942181173145, "lm_q1q2_score": 0.8745250577606657}}
{"url": "https://math.stackexchange.com/questions/2419529/sets-with-infinite-elements-and-intersection-over-sets", "text": "# Sets with infinite elements and intersection over sets\n\nI am asked to state whether the following is true or if false to give a counterexample:\n\nIf $A_1 \\supseteq A_2 \\supseteq A_3 \\supseteq \\ldots$ are all sets containing an infinite number of elements, then the intersection $$\\bigcap_{k=1}^\\infty A_k$$ is infinite as well.\n\nI believe this statement to be false but I am not sure if the counterexample I have thought up makes sense. I said:\n\nLet $A_n = \\{m \\in \\mathbb{Z} | m> n\\}$ for $n \\in \\mathbb{N}$. Would this be okay?\n\n\u2022 Your counterexample is correct. You should probably state explicitly that the intersection is empty (and therefore not infinite). Sep 7, 2017 at 0:49\n\u2022 @AndreasBlass Yes, I did mention it is empty and therefore not infinite.\n\u2013\u00a0Sam\nSep 8, 2017 at 0:51\n\nHint: Consider $A_n = \\left[-\\dfrac1n, \\dfrac1n \\right] \\subseteq \\mathbb R$.\nNot quite, because then you have a finite intersection, i.e. each $A_n$ has a finite amount of elements. How about trying $A_n = [n,\\infty) \\cap \\mathbb{N}$? What is the intersection then?", "date": "2022-07-07 08:37:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2419529/sets-with-infinite-elements-and-intersection-over-sets", "openwebmath_score": 0.8784288167953491, "openwebmath_perplexity": 222.97923530411083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806552225684, "lm_q2_score": 0.8918110511888303, "lm_q1q2_score": 0.8744926649094708}}
{"url": "https://www.themathdoctors.org/why-is-a-linear-equation-called-linear/", "text": "# Why is a Linear Equation Called Linear?\n\n#### (New Question of the Week)\n\nFrom time to time we get a question that is more about words than about math; usually these are about the meaning or origin of mathematical terms. Fortunately, some of us love words as much as we love math. But the question I want to look at here, which came in last month, is about both\u00a0the word and the math; in explaining why the word is appropriate, we are learning some things about math itself.\n\n## But it\u2019s not a line \u2026\n\nHere is Christine\u2019s question:\n\nWhy is an equation like 2x + 4 = 10 called a linear equation in one variable? Clearly, the solution is a point on one axis, the x-axis, not a line on the two-axis Cartesian coordinate plane? Or are all linear equations in one variable viewed as vertical lines?\n\nClearly, she is saying, the phrase \u201clinear equation\u201d means \u201cthe equation of a line\u201d. And there is a similarity between the linear equation in one variable above, and a linear equation in two variables, such as $$y = 2x + 4$$, which definitely is the equation of a line. But with only one variable, the only way to say that\u00a0$$2x + 4 = 10$$ is the equation of a line is to plot it on a plane as the vertical line\u00a0$$x = 3$$. Is that the intent of the term?\n\nNo, it goes further than that, because you also have to consider three or more variables. I initially gave just a short answer, to see what response it would trigger before digging in deeper:\n\nThe term \u201clinear\u201d, though derived from the idea that a linear equation in two variables represents a line, has been generalized from that to mean that the equation involves a polynomial with degree 1. That is, the variable(s) are only multiplied by constants and added to other terms, with nothing more (squaring, etc.).\n\nSo you could say that the term has been taken from one situation that gave it its name, and applied to more general cases with different numbers of variables. A linear equation in three variables, for example, represents a plane.\n\nFrom my perspective, \u201clinear\u201d means far more than \u201chaving a graph that is a line\u201d. My first thought when I see the word (outside of an elementary algebra class) is \u201cfirst-degree polynomial\u201d. Although one initially connects it to straight lines, when we extend its use (and almost every idea in math is an extension of something simpler), the idea we carry forward is the degree, not the number of dimensions. For example, here is the beginning of the Wikipedia article about linear equations:\n\nA linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable (however, different variables may occur in different terms). A simple example of a linear equation with only one variable, x, may be written in the form: ax + b = 0, where a and b are constants and a \u2260 0.\n\nAlthough this article is about linear equations in general, they start with one variable, not with lines. And although they show the graph of a line, the second paragraph skips over the two variable case, right to three variables:\n\nLinear equations can have one or more variables. An example of a linear equation with three variables, x, y, and z, is given by: ax + by + cz + d = 0, where a, b, c, and d are constants and a, b, and c are non-zero.\n\nBut why would we call any polynomial equation with degree 1 \u201clinear\u201d, when it is only in two dimensions that it is related to a line? In the one-variable case, as Christine said,\u00a0the graph is really a point; and in this three-variable case it is a plane:\n\n## But, still \u2013 it\u2019s not a line!\n\nChristine wasn\u2019t quite convinced:\n\nThank you so much. I am very particular with terminology when I teach mathematics. I have to say, I do not like this generalization of the term. I think it is misleading.\n\nHmmm \u2026 is the term \u201clinear\u201d misleading? Not to mathematicians, and I would hope not to students once they get used to it. Yet it\u2019s true that it doesn\u2019t quite mean what it seems to say. And is generalization bad? I think it\u2019s the essence of what math is \u2013 and also an integral part of languages,which are always extending the meaning of words to cover new needs.\n\nHere is my response:\n\nHi, Christine.\n\nThanks for writing back with additional thoughts. Let\u2019s think a little more deeply about it.\n\nFirst, this is standard terminology that has been in use for 200 years by many great mathematicians, so we should be very careful about considering it a bad idea. I don\u2019t think you\u2019re likely to convince anyone to change; the word is used not only of a linear equation in itself, but of \u201csystems of linear equations\u201d (in contrast to \u201cnon-linear equations\u201d); of the whole major field of \u201clinear algebra\u201d; and for related concepts like \u201clinear combination\u201d, \u201clinear transformation\u201d, and \u201clinear independence\u201d, all of which apply to any number of variables or dimensions. So the term is very well established with a particular but broad meaning, and (at least after the first year) no one is misled by it. We know what it means, and what it means is more than just \u201cline\u201d.\n\nSecond, what would you replace it with? If you don\u2019t want to use the word \u201clinear\u201d except in situations that involve actual lines, what word would you use instead to describe the general class of equations involving variables multiplied only by constants, regardless of the number of variables? We need a word for this bigger idea; that word will either be a familiar word whose meaning is stretched to cover a bigger concept, or some made-up word. The tradition in math has always been to take familiar words and give them new meanings (either more specific, like \u201cgroup\u201d or \u201ccombination\u201d or \u201cfunction\u201d, or more general, like \u201cnumber\u201d or \u201cmultiplication\u201d or \u201cspace\u201d). So what we observe here is found throughout mathematics: a word that has grown beyond its humble beginnings. (This is also true of the entire English language!\u00a0Most\u00a0words would be \u201cmisleading\u201d if you thought too deeply about their origins.)\n\nI could say that, in a sense, all of mathematics is about generalization (or abstraction). I just mentioned \u201cnumber\u201d; some people do complain about calling anything other than a natural number a \u201cnumber\u201d, but the logical development from natural numbers, to integers, to rational numbers, to real and complex numbers, involves repeated broadening of the term, which has been extremely useful. We invent new concepts, and give them old names because they are a larger, more powerful version of the old concept.\n\nI mentioned how common it is in English for a word to grow beyond its original meaning. Sitting here at my computer, I look at the mouse \u2013 is it misleading to call it that when it doesn\u2019t have legs, and may not even have a tail? And the computer has a screen; at one time, a screen was a flat surface that hid something that wasn\u2019t to be seen (or that kept out bugs); then it was applied to flat surfaces on which pictures were projected; and then to a surface that shows a picture itself. Is that misleading? It would be if you went back a hundred years \u2026\n\nAnd thinking again about math terms, I\u2019m reminded of this discussion where I pondered what all the different operations that are called \u201cmultiplication\u201d have in common.\n\n## So how do we explain it?\n\nBut as I wrote this, I realized that I had gone off in a different direction than Christine, and I wanted to relate my answer to her specific context, linear equations in one variable. The real question was pedagogical: How could she explain this to her students, so that (eventually) \u201clinear\u201d would mean to them what they will need it to mean? I continued:\n\nNow, I\u2019ve been mostly thinking of the \u201cenlarging\u201d development of the word \u201clinear\u201d, taking it to\u00a0more\u00a0than two dimensions; you\u2019re thinking specifically of the term used with\u00a0only one variable. So it may help if we focus on that, to fit your particular context. I\u2019d like to explain linear equations in one variable in a way that should make it clear why we use the word, and that it is not a misnomer.\n\nConsider the equation you asked about,\u00a02x+4=10. The left-hand side is an expression; we call it a\u00a0linear\u00a0expression, because if you used it in an equation with two variables, y = 2x+4, its graph would be a straight line. So we call 2x+4 a linear expression (or, later, a linear function). A linear equation in one variable is one that says that two linear expressions are equal (or one is equal to a constant).\n\nOr, to look at it another way, one way to solve this equation would be to graph the related equation\u00a0y = 2x+4, and find where that line intersects the line y = 10. So the linear equation can be thought of very much in terms of a line.\n\n(If this were a student\u2019s first exposure to linear equations in one variable, she likely wouldn\u2019t have seen graphs of lines yet, and wouldn\u2019t be ready for this discussion; either the word linear would be used without explanation yet, or we would hold off on the word until later.)\n\nDoes this help?\n\nIt\u2019s often hard to be sure what kind of answer will help; but Christine gave the answer I hoped for:\n\nYes, your response does help very much. In addition, it gives me a lot to think about. It is true, the grade level I am currently teaching does learn to solve linear equations in one variable before they learn about linear equations and functions. Perhaps this is why I question the use of the term. I must remember that I have had experience in mathematics beyond their years, and therefore, am more thoughtful about what terms they are exposed to and more importantly in what sequence the mathematics is presented to them. I thank you again for such an in depth response and will certainly give this discussion much more thought. It is a pleasure speaking with you.\n\nI imagine if I were introducing these equations to students with no exposure to graphs of lines, I might just mention in passing that these are called \u201clinear equations in one variable\u201d, and that we would soon be seeing why the word \u201clinear\u201d is appropriate. For now, what it means is that all we do with the variable is to multiply it by a constant, and to add things. Kids are used to hearing things they\u2019ll understand later \u2026\n\n### 17 thoughts on \u201cWhy is a Linear Equation Called Linear?\u201d\n\n1. Very interesting article, Dr.Peterson!\n\nJust wanted to ask, so then why are polynomials of the second degree quadratics? \u2018Quad\u2019 reminds me of 4 (quadrilateral is a shape with 4 sides, a quadrant is 1/4 of a circle)\n\nAnd what\u2019s the difference between an identity and an expression?\n\n1. Dave Peterson\n\nHi, Sarah.\nFirst, the \u201cquad\u201d issue is another common question; for an answer, start here: Names of Polynomials\nSecond, an identity is not an expression, but an equation (that is, a statement that two expressions are equal). In particular, it is an equation that is true for any value of the variable(s) it contains. For instance, a+b = b+a is an identity.\n\n2. Could call it \u201cfirst order equation with one variable\u201d. That seems unambiguous and intuitive.\n\n1. Dave Peterson\n\nCertainly there are several things we do call this equation, and \u201cfirst order\u201d (or \u201cfirst degree\u201d) is among them. Moreover, it does answer my question, \u201cWhat would you replace it with?\u201d in any number of variables. I suppose the standard usage of \u201clinear\u201d has to be explained by another common feature of languages that I didn\u2019t mention: the tendency toward economy, shortening or simplifying commonly used words or phrases to save time or effort!\n\nSo, yes, we could do without the term \u201clinear\u201d at all, and always use a two- or three-word phrase rather than a single word; but we don\u2019t.\n\nIn any case, of course, the question, as I take it, was not \u201cWhy must we use the word \u2018linear\u2019 here?\u201d, but \u201cWhy is the word \u2018linear\u2019 applicable here?\u201d\n\n3. \u201cA first degree equation is linear equation.\u201d Is this small definition with an example is correct if someone ask what is linear equation? I know I can elaborate the answer but I am curious if this small definition is correct or not?\n\n1. Yes, a linear equation can be defined as a first degree equation. That is true whether we are talking about an equation in one variable or more.\n\n4. Ashutosh Dhar Dwivedi\n\nI saw in every definition of linear equation with multiple variable (say 3) \u201cA linear equation is of the form ax+by+c=0\u201d, but is it necessary to add these coefficient a, b, c ? Because these coefficients does not effect linearity of the equation (from my point of view). With respect to Vector space and Field, if we treat these coefficient as scalar then they (scalars) are only used to scale the value of vectors.\n\nHowever in any case, say if we need these scalars in the above definition of linear equation, what about if we choose scalars as binary field elements (0,1)? In such case, can we defined linear equation without these coefficients a,b,c (provided field is binary)?\n\n1. If you omit a, b, and c, and write x+y=0, then you are in effect specifying a=1, b=1, c=0. This is a linear equation; but it is not a definition, as not every linear equation has this form. The parameters are needed in order to write a general form that applies to all linear equations (in two variables, in this case).\n\nIf x and y were vectors (which is not what this equation would usually mean), then a and b would be scalars, but c would have to be a vector. The scalars would be important, not \u201conly used to scale\u201d!\n\nIf you are working within a vector space over a field other than the real numbers, that doesn\u2019t change anything. But how is the set {0,1} a field? Are you referring to this?\n\nThis is turning into a discussion that would work better as a submitted question rather than as a comment. If you need more help, please submit it at our Ask a Question page.\n\n1. Ashutosh Dhar Dwivedi\n\nThanks a lot Dr. Peterson for giving me time to understand this clearly. Yes, I meant to say GF(2) but I wrote it in a wrong way.\n\n5. What about Linearity? You mean, then, that something that is called linear, as y=ax+b, may not necessarily hold Linearity properties?\n\n1. I suppose you are referring to the concept of a linear map as discussed here in Wikipedia. You are correct. In that sense, y = ax+b is considered affine, rather than linear.\n\nLinearity in this sense is different from linear functions in the sense discussed here. For some discussion of the distinction, see my more recent post Does a Linear Function Have to be Continuous?\n\n6. The equation y=10 is really linear or just they included in linear equations because this have just one variable for graphing in two dimensions it must have two variable ?\n\n1. Dave Peterson\n\nAs I said in the post, a linear equation (regardless of the number of variables) is one involving polynomials of degree 1, so y = 10 is linear. It is interesting in that it could be either an equation in one variable, y (in this case, one that is already solved), or an equation in two variables, which can be graphed as a horizontal line (because it \u201cdoesn\u2019t care\u201d about the value of x, so x can be anything). But either way, it is a linear equation.\n\n7. Thank you so much for explaining to me why a linear equation is called a linear equation! You made it clear that x+4=10 can expressed on a line on the coordinate plane much as 2x+4=10 can be.\nThen, you clarified to me that y+4=10 would also be expressed as a line or linear form on the graph.\nEqually significant is that you explained how two variables can also be expressed as linear such as:\ny=2x+4. In this values of one of the variables have to be proposed on a table such as-let x have 1, 0,-1 in order to solve for y variable. The y=x+4 also brings in the y-intercept idea. It is interesting how you then connected the three rules or formula of linear equation,namely: slope intercept formula, y=mx +b; standard formula- Ax + By=C and by extension point slope formula, M=y2-y1/x2-x1. You are a genius, Dr Peterson. point slope formula originates from coordinates of ordered pairs on the graph.\n\n1. The way you express this is misleading, and brings up something another reader asked as a question on the site, which deserves to be mentioned. While it is true that 2x+4=10 graphs as a line on the xy-plane, as I showed in the post, the same can be said of x^2+4=10, which is not a linear equation! We don\u2019t judge linearity of an equation from its graph alone.\n\nI don\u2019t see that I mentioned the different forms for a line in this post; but for other content on standard and point-slope forms, you might find these helpful:\nStandard Form of a Line\nEquation in Slope Intercept or Point Slope Form\nFormulas for the Equation of a Line\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2020-11-27 13:35:05", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/why-is-a-linear-equation-called-linear/", "openwebmath_score": 0.6982740163803101, "openwebmath_perplexity": 429.39297736524065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806495475398, "lm_q2_score": 0.8918110454379296, "lm_q1q2_score": 0.8744926542091954}}
{"url": "http://mathhelpforum.com/calculus/8266-spinning-around-y-axis.html", "text": "# Math Help - Spinning around the y-axis\n\n1. ## Spinning around the y-axis\n\nlo'\n\nThe area that is limited by the curve y=x^-2 and y=1, y=e gets to rotate around the y axis detirmine the volume of the figure.\n\nThe problem: How do i get the integration values? since it rotates around the y-axis i can't use the x values, right?\n\n2. Hello, Jones!\n\nThe area that is limited by the curve $y = \\frac{1}{x^2},\\;y = 1,\\;y = e$ is rotated about the y-axis.\nDetirmine the volume of the figure.\n\nThe problem: How do i get the integration values?\nSince it rotates around the y-axis, i can't use the x values, right?\nYou can, but it's tricky.\nCode:\n              |\n|*\n|\n| *\ne +  *\n|::::*\n|::::::::*\n1 + - - - - - - -*\n|                           *\n- - - + - - - - - - - - - - - - - - - -\n|\n\nWe can integrate with respect to $y\\!:\\;\\;V \\;=\\;\\pi\\int^e_1x^2\\,dy$\n\nSince $y = \\frac{1}{x^2}$, we have: . $x = \\frac{1}{\\sqrt{y}}$\n\nThen: . $V \\;=\\;\\pi\\int^e_1\\left(\\frac{1}{\\sqrt{y}}\\right)^2d y \\;=\\;\\pi\\int^e_1\\frac{dy}{y} \\;= \\;\\pi\\ln(y)\\,\\bigg]^e_1\n$\n\nTherefore: . $V \\;=\\;\\pi\\ln(e) - \\pi\\ln(1)\\;=\\;\\pi(1) - \\pi(0) \\;= \\;\\boxed{\\pi}$\n\n3. You could also use the 'shells' method.\n\n$2{\\pi}\\int_{e^{\\frac{-1}{2}}}^{1}\\frac{1}{x}dx=\\boxed{\\pi}$\n\n4. Originally Posted by Soroban\nHello, Jones!\n\nWe can integrate with respect to $y\\!:\\;\\;V \\;=\\;\\pi\\int^e_1x^2\\,dy$\nwhy $x^2$ ?\n\n5. Originally Posted by Jones\nwhy $x^2$ ?\n\nThe same reason that a region revolved about the $x$-axis is:\n\n. . $V\\;=\\;\\pi \\int^b_a y^2\\,dx$\n\n6. Originally Posted by Jones\nwhy $x^2$ ?\n\nHello,\n\nthe volume of a solid which was generated(?) by rotation around the y-axis could be considered as assembled of small cylindrical slices. (See attachment)\nThe Volume of a cylinder is $V=\\pi \\cdot r^2 \\cdot h$. In this case r correspond with x and the height of the cylinder correspond with y. A small part of the complete volume could be calculated:\n\n$\\Delta V=\\pi \\cdot x^2 \\cdot \\Delta y$. Divide by $\\Delta y$ and you'll get:\n\n$\\frac{\\Delta V}{\\Delta y}=\\pi \\cdot x^2$. Now let $\\Delta y$ approach to zero to get very, very thin slices:\n\n$\\lim_{\\Delta y \\rightarrow \\infty}{\\frac{\\Delta V}{\\Delta y}}=\\frac{dV}{dy}=\\pi \\cdot x^2$.This is the rate of change of the volume with respect of y.\n\nTo get the complete volume you have to run up all thin cylinders which have the volume dV. You get this term by multiplying the rate of change by dy:\n\n$dV=\\pi \\cdot x^2 \\cdot dy$. Running up means you have to calculate the sum. Use the big S to add all cylinders:\n\n$\\int (dV) \\cdot dy = V$. Thus:\n\n$V=\\int \\pi \\cdot x^2 \\cdot dy=\\pi \\int x^2 \\cdot dy$\n\nand here we are!\n\nEB\n\nPS.: This is not a rigorous proof or something like that. This post should sho, where the x\u00b2 comes from - and maybe it helps a little bit to understand what integrating means (in the beginning).\n\n7. Originally Posted by earboth\nPS.: This is not a rigorous proof or something like that. This post should sho, where the x\u00b2 comes from - and maybe it helps a little bit to understand what integrating means (in the beginning).\nI seem to have a strange effect on people. Whenever, I am around people always have to mention word \"not rigorous\". Both my math professors mention that and always look at me when they say that. Even my engineering professor when he makes some argument in class always looks at me and say \"this is not a rigorous explaination, but...\". To remind you there can be no rigorous explanation to this problem. This is some applied problem. To be rigorous we need a defintion of volume. But we do not. So whenever you deal with areas, volumes, centroids.... All these things do not have proofs. Because they cannot be formally defined.\n\n8. Originally Posted by earboth\nHello,\n\nthe volume of a solid which was generated(?) by rotation around the y-axis could be considered as assembled of small cylindrical slices. (See attachment)\nThe Volume of a cylinder is $V=\\pi \\cdot r^2 \\cdot h$. In this case r correspond with x and the height of the cylinder correspond with y. A small part of the complete volume could be calculated:\n\n$\\Delta V=\\pi \\cdot x^2 \\cdot \\Delta y$. Divide by $\\Delta y$ and you'll get:\n\n$\\frac{\\Delta V}{\\Delta y}=\\pi \\cdot x^2$. Now let $\\Delta y$ approach to zero to get very, very thin slices:\n\n$\\lim_{\\Delta y \\rightarrow \\infty}{\\frac{\\Delta V}{\\Delta y}}=\\frac{dV}{dy}=\\pi \\cdot x^2$.This is the rate of change of the volume with respect of y.\n\nTo get the complete volume you have to run up all thin cylinders which have the volume dV. You get this term by multiplying the rate of change by dy:\n\n$dV=\\pi \\cdot x^2 \\cdot dy$. Running up means you have to calculate the sum. Use the big S to add all cylinders:\n\n$\\int (dV) \\cdot dy = V$. Thus:\n\n$V=\\int \\pi \\cdot x^2 \\cdot dy=\\pi \\int x^2 \\cdot dy$\n\nand here we are!\nThis was a very good explanation, why can't they give similar explanations in the book?\n\nJust to make sure i have understood this i'll try to solve this one.\n\nThe area limited by the x and y axis and the line x+2y-4=0 rotates around the x axis. Determine the volume.\n\nso my suggestion is that we start of by substituting for x.\n$x=-2y+4$\nV= $pi \\int^0_{-4}$ $(-2y+4)^2 dy$\n\n$<==>$ $(-2/3y+4/3)^3$\n\n9. Originally Posted by Jones\nThis was a very good explanation, why can't they give similar explanations in the book?\nThank you. I like this kind of response.\n\nOriginally Posted by Jones\nJust to make sure i have understood this i'll try to solve this one.\n\nThe area limited by the x and y axis and the line x+2y-4=0 rotates around the x axis. Determine the volume.\n\nso my suggestion is that we start of by substituting for x.\n$x=-2y+4$\n\nV= $pi \\int^0_-4$ $(-2y+4)^2 dy$\n$<==>$ $(-2/3y+4/3)^3$\nHello,\n\nthe axis of rotation is the x-axis, which is perpendicular to the base area of the solid. Thus the radius of the circle (= base area) is y and the \"thickness\" of the cylinders is dx. So you get:\n\n$\\int_{0}^{4} \\pi \\cdot y^2\\cdot dx=\\int_{0}^{4} \\pi \\cdot (\\frac{1}{2}x+2)^2\\cdot dx=\\pi \\int_{0}^{4} (-\\frac{1}{4}x^2-2x+4)^2\\cdot dx$\n\nI'll leave the rest for you.\n\nEB\n\nPS.: With this problem it isn't necessary to use integration:\nYou've got a right triangle which rotates around one leg. This will give a cone.\nThe radius of the base is the y-intercept of the straight line: (0, 2)\nThe height of the cone is the zero of the straight line: (4, 0)\nThe volume of a cone can be calculated by:\n$V_{cone}=\\frac{1}{3}\\cdot \\pi \\cdot r^2 \\cdot h$. Now plug in all values you know:\n\n$V_{cone}=\\frac{1}{3}\\cdot \\pi \\cdot (2)^2 \\cdot 4=\\frac{1}{3}\\cdot \\pi \\cdot 4 \\cdot 4=\\frac{16}{3}\\cdot \\pi$\nThat's the result, you should get with the integration too.", "date": "2014-10-23 20:53:54", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/8266-spinning-around-y-axis.html", "openwebmath_score": 0.8845435380935669, "openwebmath_perplexity": 556.7695732131505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750525759487, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8744504304754439}}
{"url": "http://mathhelpforum.com/math-puzzles/131324-fishermen-print.html", "text": "Fishermen\n\n\u2022 Feb 28th 2010, 08:52 PM\nharish21\nFishermen\nThree fisherman come back from a day of fishing and put their fish together and go to bed. The 1st one wakes up goes to the pile of fish, throws one away and takes a 3rd and leaves. The 2nd one does the same. The last one does the same as well. What is the minimum number of fish that this can work with?\n\u2022 Feb 28th 2010, 09:56 PM\nSoroban\nHello, harish21!\n\nQuote:\n\nThree fisherman come back from a day of fishing and put their fish together and go to bed.\nThe 1st one wakes up, goes to the pile of fish, throws one away and takes a 3rd and leaves.\nThe 2nd one does the same. The last one does the same as well.\nWhat is the minimum number of fish that this can work with?\n\nLet $N$ = total number of fish.\n\nThe 1st discards one fish and take a third of the remainder.\nThen $N$ must be of the form: . $3A + 1$\n. . $N \\:=\\:3A + 1$ .[1]\n\nHe discards one fish . . . $3A$ fish left.\nHe take a third of the fish ( $A$ fish) and leaves.\nThere are $2A$ fish left.\n\nThe 2nd discards one fish and takes a third of the remainder.\nThen $2A$ must be of the form $3B + 1$\n. . $2A \\:=\\:3B+1$ .[2]\n\nHe discards one fish . . . $3B$ fish left.\nHe take third of the fish ( $B$ fish) and leaves.\nThere are $2B$ fish left.\n\nThe 3rd discards one fish and takes a third of the remainder.\nThen $2B$ must be of the form $3C + 1$\n. . $2B \\:=\\:3C + 1 \\quad\\Rightarrow\\quad B \\:=\\:\\frac{3C+1}{2}$ .[3]\n\nSubstitute [3] into [2]: . $2A \\;=\\;3\\left(\\frac{3C+1}{2}\\right) + 1 \\quad\\Rightarrow\\quad A \\:=\\:\\frac{9C+5}{4}$ .[4]\n\nSubstitute [4] into [1]: . $N \\;=\\;3\\left(\\frac{9C+15}{4}\\right) + 1 \\;=\\;\\frac{27C + 19}{4}$\n\nWe have: . $N \\;=\\;\\frac{24C + 16 + 3C + 3}{4} \\;=\\;\\frac{24C+16}{4} + \\frac{3C+3}{4}$\n\nHence: . $N \\;=\\;6C+4 + \\frac{3(C+1)}{4}$\n\nSince $N$ is an integer, $C+1$ must be divisible by 4.\nThe first time this happens is when $C = 3.$\n\nTherefore: . $N \\;=\\;6(3) + 4 + \\frac{3(3+1)}{4} \\quad\\Rightarrow\\quad \\boxed{N \\:=\\:25}$\n\n\u2022 Mar 1st 2010, 12:26 AM\nharish21\nQuote:\n\nOriginally Posted by Soroban\nHello, harish21!\n\nLet $N$ = total number of fish.\n\nThe 1st discards one fish and take a third of the remainder.\nThen $N$ must be of the form: . $3A + 1$\n. . $N \\:=\\:3A + 1$ .[1]\n\nHe discards one fish . . . $3A$ fish left.\nHe take a third of the fish ( $A$ fish) and leaves.\nThere are $2A$ fish left.\n\nThe 2nd discards one fish and takes a third of the remainder.\nThen $2A$ must be of the form $3B + 1$\n. . $2A \\:=\\:3B+1$ .[2]\n\nHe discards one fish . . . $3B$ fish left.\nHe take third of the fish ( $B$ fish) and leaves.\nThere are $2B$ fish left.\n\nThe 3rd discards one fish and takes a third of the remainder.\nThen $2B$ must be of the form $3C + 1$\n. . $2B \\:=\\:3C + 1 \\quad\\Rightarrow\\quad B \\:=\\:\\frac{3C+1}{2}$ .[3]\n\nSubstitute [3] into [2]: . $2A \\;=\\;3\\left(\\frac{3C+1}{2}\\right) + 1 \\quad\\Rightarrow\\quad A \\:=\\:\\frac{9C+5}{4}$ .[4]\n\nSubstitute [4] into [1]: . $N \\;=\\;3\\left(\\frac{9C+15}{4}\\right) + 1 \\;=\\;\\frac{27C + 19}{4}$\n\nWe have: . $N \\;=\\;\\frac{24C + 16 + 3C + 3}{4} \\;=\\;\\frac{24C+16}{4} + \\frac{3C+3}{4}$\n\nHence: . $N \\;=\\;6C+4 + \\frac{3(C+1)}{4}$\n\nSince $N$ is an integer, $C+1$ must be divisible by 4.\nThe first time this happens is when $C = 3.$\n\nTherefore: . $N \\;=\\;6(3) + 4 + \\frac{3(3+1)}{4} \\quad\\Rightarrow\\quad \\boxed{N \\:=\\:25}$\n\nAwesome Soroban! I was given this problem in one of my classes and found it very interesting. I thought it would be nice to share. 25 is indeed the logical answer. However -2 is also a correct answer to this problem. I forgot the name of the scientist who gave an answer of -2 to this problem!!\n\u2022 Mar 1st 2010, 03:08 PM\nicemanfan\nIf the total number of fish is 25, then:\n\nThe first person throws one away (total down to 24) and takes a third (total down to 16).\n\nThe second person throws one away (total down to 15) and takes a third (total down to 10).\n\nThe third person throws one away (total down to 9) and takes a third (total down to 6).\n\nIf every time someone takes a third of the fish, the number remaining must be a whole number, then 25 is the minimum number. Without that restriction, the minimum number of fish is 4.75. In both cases I assume you're not allowed to have a negative number of fish.\n\u2022 Mar 1st 2010, 07:26 PM\nWilmer", "date": "2017-01-16 15:29:06", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-puzzles/131324-fishermen-print.html", "openwebmath_score": 0.840115487575531, "openwebmath_perplexity": 1140.0236424345078, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750496039276, "lm_q2_score": 0.8856314647623016, "lm_q1q2_score": 0.8744504114504766}}
{"url": "https://math.stackexchange.com/questions/2906220/convergence-and-divergence-of-fraca-nn-as-n-to-infty", "text": "# Convergence and divergence of $\\frac{a_n}{n}$ as $n\\to\\infty$\n\nSuppose that a sequence $\\{a_n\\}_{n\\geqslant 1}$ satisfies:\n\n$$a_{m+n}\\leqslant a_m+a_n$$\n\nfor all integers $m,n\\geqslant 1$. Show that $\\frac{a_n}{n}$ either converges or diverges to $-\\infty$ as $n\\to \\infty$.\n\nResolution: For an arbitrarily fixed positive integer $k$ we put $n=qk+r$ with $0\\leqslant r<k$. Applying the given inequality for $q$ times we get $a_n=a_{qk+r}\\leqslant qa_k+a_r$; so,\n\n$\\frac{a_n}{n}\\leqslant\\frac{a_k}{k}+\\frac{a_r}{n}$ Taking the limit as $n\\to\\infty$, we get\n\n$\\lim \\sup_{n\\to\\infty}\\frac{a_n}{n}\\leqslant\\frac{a_k}{k}$.\n\nThe sequence $\\frac{a_n}{n}$ is therefore bounded above. Since $k$ is arbitrary, we conclude that:\n\n$\\lim \\sup_{n\\to\\infty}\\frac{a_n}{n}\\leqslant \\inf_{k\\geqslant 1}\\frac{a_k}{k}\\leqslant\\lim \\inf_{k\\to\\infty}\\frac{a_k}{k}$,\n\nwhich concludes the proof.\n\nQuestion:\n\n1) First the author says that $k$ is a fixed positive integer. However in the last expression $\\lim \\inf_{k\\to\\infty}\\frac{a_k}{k}$ the author makes $k$ vary. How is this supposed to prove that $\\frac{a_n}{n}$ converges since the behaviour of $\\lim \\inf_{k\\to\\infty}\\frac{a_k}{k}$ is not known? Why is $k$ allowed to vary?\n\n2)On the question it is mentioned the divergence to $-\\infty$. How was that point addressed in the answer or resolution?\n\n\u2022 Is this from Hata's problem book ? Sep 5 '18 at 13:06\n\u2022 Yes it is. What do you think of the problems in that book? I am trying to solve them but I find them hard. Sep 5 '18 at 13:07\n\u2022 Sep 5 '18 at 13:12\n\u2022 @RobertZ It may be the same problem, but the questions seem to differ in my opinion. Sep 5 '18 at 13:13\n\u2022 Answer: that is not taking the limit. Actually you could use other alphabets, that does not affect the $\\liminf a_\\nu / \\nu$. By definition, $$\\liminf \\frac {a_n} n = \\lim_n \\inf_{k \\geqslant n} \\frac {a_k}k,$$ and you could prove that for any sequence $x_n$, $\\inf_{k \\geqslant n} x_k$ is increasing, hence $\\inf_{k \\geqslant 1} x_k \\leqslant \\inf_{k \\geqslant n} x_k \\to \\liminf x_k [n \\to \\infty].$\n\u2013\u00a0xbh\nSep 5 '18 at 13:21\n\nThe author first considers a fixed positive $k$ and proves $\\lim \\sup_{n\\to\\infty}\\frac{a_n}{n}\\leqslant\\frac{a_k}{k}$. Consequently, $\\sup_n \\frac{a_n}n < \\infty$, hence $\\left(\\frac{a_n}{n}\\right)_n$ is bounded above.\n\nSince $\\lim \\sup_{n\\to\\infty}\\frac{a_n}{n}\\leqslant\\frac{a_k}{k}$ is valid for every $k$, by definition of the infinimum, $\\lim \\sup_{n\\to\\infty}\\frac{a_n}{n}\\leqslant\\inf_k\\frac{a_k}{k}$. By definition of $\\liminf$, one also has $\\inf_k\\frac{a_k}{k}\\leq \\liminf_k \\frac{a_k}{k}$.\n\nThus $\\liminf_k \\frac{a_k}{k} = \\limsup_k \\frac{a_k}{k} = \\inf_k \\frac{a_k}{k}$, thus $\\frac{a_n}{n}$ converges to a limit in $\\mathbb R\\cup \\{-\\infty, \\infty\\}$.\nSince $\\frac{a_n}{n}$ is bounded above, $\\inf_k \\frac{a_k}{k}<\\infty$, but it could very well be equal to $-\\infty$. Thus the limit belongs to $\\mathbb R\\cup \\{-\\infty\\}$\n\nQuestion 1)\n\nHe proves that $\\lim \\sup_{n\\to\\infty}\\frac{a_n}{n}\\leqslant\\frac{a_k}{k}$ is true for any and all natural numbers $k$. In proving that inequality, $k$ is fixed (so it's only proven for \"one $k$ at a time\"), but once it's proven, you're allowed to take it for granted for any value of $k$ you'd like, and let $k$ vary.\n\nQuestion 2)\n\nNote that the $\\limsup$ on the left side of $\\lim \\sup_{n\\to\\infty}\\frac{a_n}{n}\\leqslant\\lim \\inf_{k\\to\\infty}\\frac{a_k}{k}$ and the $\\liminf$ on the right side are actually of the same sequence. So the $\\limsup$ of the sequence is less than or equal to the $\\liminf$ of that sequence. This implies that they must be equal, and either they're finite (in which case the sequence has a limit), or they're not (in which case the sequence diverges to $-\\infty$, since it's bounded above).\n\nAnswer: that is not taking the limit $k \\to \\infty$. Actually you could use other alphabets, that does not affect the quantity $\\liminf a_\\nu / \\nu$. By definition, $$\\liminf \\frac {a_n} n = \\lim_n \\inf_{k \\geqslant n} \\frac {a_k}k,$$ and you could prove that for any sequence $x_n$, $\\inf_{k \\geqslant n} x_k$ is increasing in $n$, hence for all $n \\in \\mathbb N^*$, $\\inf_{k \\geqslant 1} x_k \\leqslant \\inf_{k \\geqslant n} x_k$ Since $\\inf_{k \\geqslant n}x_k \\to \\liminf x_k [n \\to \\infty]$, we have $\\inf_{k \\geqslant 1}x_k \\leqslant \\liminf x_n$.\n\nAdditionally, $\\liminf x_n$ always exists or equals $\\infty$, so you can write down $\\liminf x_n$ whenever you want.", "date": "2021-09-21 10:22:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2906220/convergence-and-divergence-of-fraca-nn-as-n-to-infty", "openwebmath_score": 0.9712309241294861, "openwebmath_perplexity": 133.46155514587846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987375047746414, "lm_q2_score": 0.8856314632529872, "lm_q1q2_score": 0.8744504083151449}}
{"url": "https://gmatclub.com/forum/what-percent-of-the-mixture-is-solution-x-143804.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Sep 2018, 11:50\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# What percent of the mixture is Solution X?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 06 Sep 2012\nPosts: 39\nConcentration: Social Entrepreneurship\nWhat percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n08 Dec 2012, 11:49\n2\n4\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n76% (01:39) correct 24% (01:37) wrong based on 285 sessions\n\n### HideShow timer Statistics\n\nSolution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?\n\nA) 20%\nB) 44%\nC) 50%\nD) 80%\nE) 90%\n\nCan someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!\nSenior Manager\nJoined: 22 Dec 2011\nPosts: 261\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n08 Dec 2012, 12:03\n1\n2\nJJ2014 wrote:\nSolution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?\n\nA) 20%\nB) 44%\nC) 50%\nD) 80%\nE) 90%\n\nCan someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!\n\nyou can solve this by allegation method.\nGiven 22% of A is the final mixture; Sol X has 20% chemical A and Sol Y has 30 % of A. (you don't need Chemical B to solve when you r solving the problem using this method)\n\n$$X / Y = (30-22 ) / (22-20)$$\n$$X / Y = 8/2 = 4/1$$\n\nSo X and Y are in the ratio 4 : 1\nQuestion asks percent of the mixture is Solution X.\n\nSo $$X / Total = 4/5 *100 = 80%$$\n\nCheers\nBoard of Directors\nJoined: 01 Sep 2010\nPosts: 3464\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n08 Dec 2012, 12:16\n3\nalso you can solve this one with the concept of weighted average or balance point\n\nyou care about only of chemical A in both X and Y\n\n$$20$$------- $$22$$ ------------------------$$30$$\n\nNow the difference between 20 and 22 = 2; between 22 and 30 = 8\n\nSo $$8y = 2 x$$\n\n$$\\frac{8}{2}$$ $$=$$$$\\frac{x}{y}$$\nOur ratio is 2 and 8 and using the concept of unknown multiplier the sum is 10. Our X is 8 on a total of 10 so 80 %\n\nmuch more difficult to say that to explain. In that way you can solve a question difficult like this in 30 seconds.\n\nAlso algebraic approach works fine, but this is a bit faster\n_________________\nManager\nJoined: 06 Jun 2010\nPosts: 155\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Dec 2012, 01:29\n2\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49251\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Dec 2012, 06:31\n2\nJJ2014 wrote:\nSolution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?\n\nA) 20%\nB) 44%\nC) 50%\nD) 80%\nE) 90%\n\nCan someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!\n\n22% of chemical A in X+Y grams of solution comes from 20% chemical A in solution X and 30% chemical A in solution Y, thus:\n\n0.22(X + Y) = 0.2X + 0.3Y --> X = 4Y --> X/(X+Y)=4/5=0.8.\n\nCheck out question banks for similar problems:\nDS mixture problems: search.php?search_id=tag&tag_id=43\nPS mixture problems: search.php?search_id=tag&tag_id=114\n\nHope it helps.\n_________________\nManager\nJoined: 22 Nov 2010\nPosts: 248\nLocation: India\nGMAT 1: 670 Q49 V33\nWE: Consulting (Telecommunications)\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Mar 2013, 00:14\n1\nJJ2014 wrote:\nSolution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?\n\nA) 20%\nB) 44%\nC) 50%\nD) 80%\nE) 90%\n\nCan someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!\n\nFocus on Chemical A:\n\nX--------------------------------------Y\n20%---------------------------------30%\n\n--------------------22%-----------------\n\n30 - 22= 8--------------------22-20 = 2\n\nX:Y = 8:2 = 4:1.\n\nTherefore, %age of X = 4/5 * 100 = 80%\n_________________\n\nYOU CAN, IF YOU THINK YOU CAN\n\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 12411\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n02 Feb 2015, 13:11\nHi All,\n\nThese types of mixture questions can typically be solved with a variety of approaches (the average formula, ratios, TESTing VALUES, TESTing THE ANSWERS, etc.). Here's the standard Weighted Average Formula approach:\n\nSolution X = 20% chemical A\nSolution Y = 30% chemical A\n\nMixture of the two solutions = 22% chemical A\n\nX = number of ounces of Solution X\nY = number of ounces of Solution Y\n\n(.2X + .3Y) / (X + Y) = .22\n\n.2X + .3Y = .22X + .22Y\n.08Y = .02X\n\nWe can multiply both sides by 100 to get rid of the decimals...\n\n8Y = 2X\n\nThe question asks for the percentage of the new mixture that is Solution X....\n\n8/2 = X/Y\n4/1 = X/Y\n\n80% X and 20% Y\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\nCurrent Student\nStatus: DONE!\nJoined: 05 Sep 2016\nPosts: 390\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Oct 2016, 16:22\n1\n0.20x+(1-x)0.30 = 0.22\n0.20x+0.30-0.30x=0.22\n-0.10x=-0.08\nx=4/5 = 80%\nCurrent Student\nJoined: 26 Jan 2016\nPosts: 109\nLocation: United States\nGPA: 3.37\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Oct 2016, 16:40\nIf the solution is 22% of A we know off the bat that the majority of the solution has to be X. We can eliminate A,B,C.\n\nSolution X is 20% A and Y is 30\n\nSo X is 2 units from the combo and Y is 8 units away.\n\nRation of X:Y for solution A is 8:2\n8/10=80%\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 8091\nRe: What percent of the mixture is Solution X?\u00a0 [#permalink]\n\n### Show Tags\n\n05 Sep 2018, 05:45\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: What percent of the mixture is Solution X? &nbs [#permalink] 05 Sep 2018, 05:45\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-19 18:50:58", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-percent-of-the-mixture-is-solution-x-143804.html", "openwebmath_score": 0.7171984910964966, "openwebmath_perplexity": 5291.24630909571, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9532750413739076, "lm_q2_score": 0.9173026607160999, "lm_q1q2_score": 0.8744417318465356}}
{"url": "https://mathematica.stackexchange.com/questions/10350/mathematica-for-linear-algebra-course", "text": "# Mathematica for linear algebra course?\n\nI'm taking a linear algebra / matrix theory course and we are free to use any software we want, and will be \"expected to use MATLAB or an equivalent\" for homework. The professor and textbook (Applied Linear Algebra) prefer MATLAB, some students seem to like R, but I'm already familiar with Mathematica and have a student license.\n\nAre there any possible limitations of Mathematica to watch out for, if used as a learning device (i.e. not for work or intensive computation) for a linear algebra course? e.g. Are there some operations which the other software packages can do which Mathematica cannot out-of-the-box?\n\n\u2022 Could you mention the textbook you're using? Sep 8, 2012 at 7:36\n\u2022 There is a book focused on teaching linear algebra with Mathematica, which you can get used from Amazon. It is pretty dated, and I don't know how good it is, but if you decide to go with Mathematica, it may be of some help. Sep 8, 2012 at 9:00\n\u2022 If prices of licence is the issue you might want to try python numpy library as well. Or Octave for Matlab clone. Sep 8, 2012 at 12:34\n\u2022 @enedene I find freemat or scilab to have a more reasonable frontend than octave, an they're equally free (although maybe their licenses are more restrictive, I don't know). overall I'd go with matlab or python if not mathematica though. in any case as he has a student license I don't think this is important.\n\u2013\u00a0acl\nSep 8, 2012 at 13:05\n\u2022 @Nasser Agree 100%. This was and still is my most difficult transition coming from Excel and MATLAB was the \"Lists of lists\" concepts. Also, MATLAB has some cool notation for solving linear systems such as Ax=b can be solved by A\\b.\n\u2013\u00a0kale\nSep 8, 2012 at 13:17\n\nMathematica offers a pretty complete set of functionality for linear algebra, and it has improved in recent versions.\n\nFor example, since version 5, Mathematica has offered the generalised Schur decomposition (also known as the QZ decomposition). This certainly wasn't available in earlier versions. It handles sparse matrices and many other wrinkles. And if it isn't available, this example shows that it can take a lot less code than in many other languages, so you are less likely to get yourself tangled up in loop constructs and the like.\n\nRather than focus on limitations, I thought it worth mentioning that Mathematica offers something neither Matlab nor R can, that makes it ideal as a pedagogical tool: integrated symbolics. If you are learning linear algebra and you want to understand certain things, you can in many cases pass a symbolic matrix to the relevant function, and use the resulting output to gain a deeper understanding.\n\nHere's a simple example:\n\naa = Array[a, {2, 2}]\n\n(* {{a[1, 1], a[1, 2]}, {a[2, 1], a[2, 2]}}*)\n\nEigenvalues[aa]\n\n(*{1/2 (a[1, 1] + a[2, 2] - Sqrt[\na[1, 1]^2 + 4 a[1, 2] a[2, 1] - 2 a[1, 1] a[2, 2] + a[2, 2]^2]),\n1/2 (a[1, 1] + a[2, 2] + Sqrt[\na[1, 1]^2 + 4 a[1, 2] a[2, 1] - 2 a[1, 1] a[2, 2] + a[2, 2]^2])}*)\n\n\nOf course, sometimes the symbolic output is pretty hard to trace through (consider SingularValueDecomposition[aa] for the aa defined above!), but in can be useful in some cases.\n\nSome pages on the internet contain claims that Mathematica is slower than Matlab, or that Matlab is somehow \"better\" for numerical work (and implicitly, Mathematica is only good for a bit of symbolic solving or something). As this answer shows, this claim is no longer true. It was true until about version 5, but no longer.\n\n\u2022 Congrats on your 10k! Sep 8, 2012 at 8:55\n\nHaving taught linear algebra using both Mathematica and Matlab, I concur with what others have said that the Mathematica's features for linear algebra include all one might need for a course in undergraduate linear algebra. Since symbolic computation is also fully integrated into Mathematica, it might be better in some ways. For example, we can solve symbolic systems of small dimension fairly easily.\n\nLinearSolve[{{a, b}, {c, d}}, {e, f}]\n\n\nNote how the determinant appears in the denominator illustrating its importance in determining when a system is solvable. This can be done just as easily for a 3x3 or 4x4 system.\n\nAlso, there is a common mis-conception that Mathematica does not or cannot distinguish between row and column vectors, as illustrated by Nasser's comment. I think this is not correct. It's just that if you want to represent a row or column vector, you should use the full matrix representation of said vector. Consider, for example, the following two dot product computations (whose orders cannot be reversed):\n\n{{a, b}, {c, d}}.{{x}, {y}} // MatrixForm\n\n\n{{x, y}}.{{a, b}, {c, d}} // MatrixForm\n\n\nIn addition, though, Mathematica provides a consistent notion of dot product between dimensions based on tensor products. Here's the product of a rank 3 tensor with a rank 2 tensor.\n\nTable[a[i, j, k], {i, 1, 2}, {j, 1, 2}, {k, 1, 2}].\nTable[b[i, j], {i, 1, 2}, {j, 1, 2}]\n\n\nBy contrast, Matlab is limited to 2D floating point matrices. When you type x=2 in Matlab, you've defined a 2D matrix; the equivalent definition in Mathematica would be x={{2}}.\n\n\u2022 @NasserM.Abbasi I guess I disagree with the documentation. The example shows that, if you represent a vector using a rank 1 tensor, then you can't distinguish between left and right multiplication. I'm saying that you can distinguish between left and right multiplication, if you represent a vector as a rank 2 tensor. Sep 9, 2012 at 11:10\n\u2022 @NasserM.Abbasi Of course, you can define higher rank objects in Matlab. You can even use cell arrays to define fairly arbitrary objects. I guess I mean to say that the default in Matlab is a 2D float and support beyond that is lacking. For example, the 3D and 4D matrices that you define (which we'd call rank 3 and 4 tensors in the Mathematica context) cannot be multiplied using the * operator in Matlab. The corresponding tensors can be multiplied in Mathematica using Dot. Sep 9, 2012 at 11:15\n\u2022 For better or worse, vectors, matrices, etc., in Mathematica are just lists. For pedagogical purposes this is not necessarily optimal. Before I taught linear algebra with Mathematica, I did so with APL and later J. In each of those languages, a vector is a 1-dimensional creature, whereas a row vector or a column vector is a 2-dimensional creature, namely, a 1-row or 1-column matrix. Sep 9, 2012 at 15:04\n\nIn years past I've taught a standard-content sophomore-level (in U.S.) linear algebra course where students used Mathematica. I know, then, that if you're interested, or if it's a requirement, you can build everything up from the simplest functions for manipulating matrices or you can directly use powerful built-in Mathematica functions (or a combination of both).\n\nE.g., you might begin by defining little utility functions to scale a row, to swap two rows, and to add one row to another; from that, you can build a function to do Gauss-Jordan row reduction. Or you can directly use RowReduce.\n\nAgain, you might define a function to test for invertibility and, when invertible, find the inverse of a matrix by adjoining an identity matrix, row reducing, etc. Or you can directly use Inverse.\n\nEven for something more complicated like Gram-Schmidt orthonormalization, you can build up a function for this yourself, beginning by defining a function that projects a vector onto the span of a given set of vectors. Or use Orthogonalize.\n\nYou could even define your own determinant function (in one of several ways), then use that to find eigenvalues, etc.\n\nIf your course involves advanced operations such as LU or QR decompositions, again you can define functions for these yourself or else use the built-in ones.\n\nOf course in general you would expect built-in functions to handle round-off issues more robustly than anything you would program yourself. But this would be the same with Matlab.\n\nSo yes, Mathematica would be an entirely appropriate tool in place of Matlab.\n\n\u2022 is there any chance you can add a link to the course notes?\n\u2013\u00a0user21\nSep 9, 2012 at 1:24\n\u2022 @ruebenko: I considered doing so and still could. But the notebooks are 10 years old now, and some of the files are .m packages encoded for the Windows/PC platform. (Students would define a function, then run an encoded package that tested it with various sets of data against my \"correct\" but encoded version of the function.) Sep 9, 2012 at 15:00", "date": "2022-05-28 23:37:12", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/10350/mathematica-for-linear-algebra-course", "openwebmath_score": 0.3608662486076355, "openwebmath_perplexity": 816.8197393087208, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9532750387190131, "lm_q2_score": 0.9173026494123034, "lm_q1q2_score": 0.8744417186355669}}
{"url": "http://residuetheorem.com/2016/09/18/integral-of-polynomial-times-rational-function-of-trig-function-over-multiple-periods/", "text": "## Integral of polynomial times rational function of trig function over multiple periods\n\nProblem: Compute\n\n$$\\int_0^{12\\pi} dx \\frac{x}{6+\\cos 8x}$$\n\nIt looks like the OP is trying to compute the antiderivative and use the fundamental theorem of calculus. With multiple periods, that approach is paved with all sorts of difficulty that is really an artifice related to the functional form of the antiderivative. In truth, there should be no such difficulty. A better approach involves the residue theorem, which I will outline below. Note that the presence of the linear term in the numerator provides a slight complication, but one that has been treated before in this site several times.\n\n$$\\int_{0}^{12 \\pi} dx \\frac{x}{6+\\cos{8 x}} = \\frac1{64} \\int_0^{96 \\pi} du \\frac{u}{6+\\cos{u}}$$\n\n$$\\int_{n 2 \\pi}^{(n+1) 2 \\pi} du \\frac{u}{6+\\cos{u}} = \\int_0^{2 \\pi} dv \\frac{v+2 \\pi n}{6+\\cos{v}} = \\int_0^{2 \\pi} dv \\frac{v}{6+\\cos{v}} + 2 \\pi n \\int_0^{2 \\pi} \\frac{dv}{6+\\cos{v}}$$\n\nThus, summing over $n$:\n\n$$\\int_{0}^{12 \\pi} dx \\frac{x}{6+\\cos{8 x}} = \\frac{48}{64} \\int_0^{2 \\pi} dv \\frac{v}{6+\\cos{v}} + \\frac{48 (47) \\pi}{64} \\int_0^{2 \\pi} \\frac{dv}{6+\\cos{v}}$$\n\nWe may now compute each of these integrals. The second integral is straightforward by the residue theorem, i.e., let $z=e^{i v}$, then\n\n$$\\int_0^{2 \\pi} \\frac{dv}{6+\\cos{v}} = -i 2 \\oint_{|z|=1} \\frac{dz}{z^2+12 z+1}$$\n\nThe only pole of the integrand inside the unit circle is at $z=-6+\\sqrt{35}$. The integral is then $i 2 \\pi$ times the residue of the integrand at this pole, or $2 \\pi/\\sqrt{35}$.\n\nTo compute the first integral, we consider the complex integral\n\n$$\\oint_C dz \\frac{\\log{z}}{z^2+12 z+1}$$\n\nwhere $C$ is the unit circle with a detour up and back about the positive real axis. (A circular piece about the origin vanishes.) This integral is equal to\n\n$$-\\frac12 \\int_0^{2 \\pi} dv \\frac{v}{6+\\cos{v}} + \\int_1^0 dx \\frac{\\log{x}+i 2 \\pi}{x^2+12 x+1} + \\int_0^1 dx \\frac{\\log{x}}{x^2+12 x+1}$$\n\nor, simplifying,\n\n$$-\\frac12 \\int_0^{2 \\pi} dv \\frac{v}{6+\\cos{v}} \u2013 i 2 \\pi \\int_0^1 \\frac{dx}{x^2+12 x+1}$$\n\nThe contour integral is also equal to $i 2 \\pi$ times the residue at the pole $z=-6+\\sqrt{35}$. Note that the negative sign is taken to be $e^{i \\pi}$. Thus we have\n\n$$-\\frac12 \\int_0^{2 \\pi} dv \\frac{v}{6+\\cos{v}} = i 2 \\pi \\int_0^1 \\frac{dx}{x^2+12 x+1} + i 2 \\pi \\frac{-\\log{\\left ( 6+\\sqrt{35} \\right )}+i \\pi}{2 \\sqrt{35}}$$\n\nNow,\n\n$$\\frac1{x^2+12 x+1} = \\frac1{2 \\sqrt{35}} \\left (\\frac1{x+6-\\sqrt{35}} \u2013 \\frac1{x+6+\\sqrt{35}} \\right )$$\n\nso that\n\n$$\\int_0^1 \\frac{dx}{x^2+12 x+1} = \\frac1{2 \\sqrt{35}} \\log{\\left (\\frac{7+\\sqrt{35}}{7-\\sqrt{35}} \\right )} = \\frac1{2 \\sqrt{35}} \\log{\\left (6+\\sqrt{35}\\right )}$$\n\nNote the cancellation with the real part of the residue. Thus,\n\n$$\\int_0^{2 \\pi} dv \\frac{v}{6+\\cos{v}} = \\frac{2 \\pi^2}{\\sqrt{35}}$$\n\nPutting these results altogether, we have\n\n>$$\\int_0^{12 \\pi} dx \\frac{x}{6+\\cos{8 x}} = \\frac{72 \\pi^2}{\\sqrt{35}}$$\n\n#### One Comment\n\n\u2022 Hello Ron,\n\nI think the second bit is somewhat more convoluted than necessary since one could just reflect the integral across the bounds to cancel out the linear term on the numerator.\n\nApart from that, this simplification is great! \ud83d\ude42", "date": "2018-01-22 19:48:38", "meta": {"domain": "residuetheorem.com", "url": "http://residuetheorem.com/2016/09/18/integral-of-polynomial-times-rational-function-of-trig-function-over-multiple-periods/", "openwebmath_score": 0.9774549007415771, "openwebmath_perplexity": 202.15220787455283, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9927672353405445, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.874426470680869}}
{"url": "https://math.stackexchange.com/questions/2599351/count-the-possible-ways-to-seat-people-at-a-round-table/2599683", "text": "# Count the possible ways to seat people at a round table\n\nI have a very simple combinatorics problem, and I am (almost) sure I did this right. However, the smartest guy in my class thinks different and this made me doubt, so I will post our views to this problem and I would like to know who is right.\n\nIn how many different ways can 10 people be seated around a round table, under the condition that there are 5 man and 5 women, and no two males or two females can be seated adjacently.\n\nMy view\n\nWe number the chairs in an arbitrary way, so the chair numbering is fixed. At chair $1$, there is seated either a man or a woman, which given $2$ possibilities. When we observe the gender of the person in the first chair, the division of the table over the two genders in given. Then there are $5!$ ways to seat the men over the $5$ male chairs and $5!$ ways to seat the women over the $5$ female chairs. So in total this gives $2.(5!)^2$ possibilities.\n\nMy classmates view\n\nFirst all males are seated, which is possible in $5!$ ways. Because no males can sit adjacently and the people are seated in a circle, we can rotate the table and in this way we count all possibilities 5 times too much, so $5!/5=4!$ ways to seat the man. After this, the table division is fixed, but we still have to seat the women. This gives another $5!$ ways, so in total $5!4!$ ways.\n\nWhat I think is wrong with this view\n\nI don't think the argument about rotation of the table makes any sense. The argument can make sense though, if we use exactly my classmate's argument, but acknowledge that the first person to be seated can be any of the $10$ people, so giving an extra $10$ possibilities. Then proceed as my classmate, giving the same answer as me. However, my classmate does not agree. So who is right?\n\nAlso, if we use our arguments to the same problem but a table of $2$ people, one male, one female, there are obviously $2$ possibilities. Then my method gives $2.(1!)^2=2$ possibilities, but my classmate's method gives only $1!0!=1$ possibility.\n\nEDIT\n\nI think the main problem is in different interpretations of the problem, where my interpretation is that it matters who sit at 'chair 1' while my classmates thinks that only the ordering of the people matters. However, given only the above problem, what would be the right interpretation, or is this ambiguous?\n\n\u2022 Excluding the one and only genious V\u00e1clav Mordvinov of course. \u2013\u00a0V\u00e1clav Mordvinov Jan 10 '18 at 11:10\n\u2022 Are they seated at/around the table, or \"on\" the table, as you stated? Sitting 10 people on the table may be tricky. (I am only joking) \u2013\u00a0mbomb007 Jan 10 '18 at 23:09\n\u2022 Not a native speaker. \u2013\u00a0V\u00e1clav Mordvinov Jan 10 '18 at 23:10\n\nYour solution would be okay if there is a \"special chair\" (you name it \"chair $1$\"). Then there is no essential difference with placing the $10$ persons in a row in which case there is also a special chair (for instance the utmost left, or the utmost right).\n\nIf there is no special chair then you can start by placing one man. After that there are $4!$ different arrangements for the other men and $5!$ different arrangements for the women. This leads to a total of $4!5!$ possibilities.\n\nIt is not for nothing that a round table is used, indicating that there is no special chair. So I would say that your classmate is right.\n\n\u2022 Okay thanks, I think you are right about that my classmate's interpretation is the most natural, otherwise they wouldn't have used a circle in the question. It is still a bit ambiguous but you're right! \u2013\u00a0V\u00e1clav Mordvinov Jan 10 '18 at 9:51\n\u2022 You are welcome. According to your interpretation there are $2$ ways to place one man and one woman at a round table. That is correct if the two chairs are distinguishable, but incorrect if they are not (which is definitely a more likely interpretation). \u2013\u00a0drhab Jan 10 '18 at 9:56\n\u2022 The round table could be so that there is no one on the \"end\" who is exempted from the condition \"no two males or two females can be seated adjacently\". \u2013\u00a0Acccumulation Jan 10 '18 at 16:48\n\u2022 @Acccumulation: If you have five men and five women in a row, with no two adjacent men and no two adjacent women, then it's already guaranteed that the people at the two ends will be one man and one woman. So if that's the only reason they specified a round table, then they needn't have specified it. \u2013\u00a0ruakh Jan 11 '18 at 0:24\n\u2022 @Acccumulation I fully agree with ruakh \u2013\u00a0drhab Jan 11 '18 at 6:58\n\nYour disagreement comes from different interpretations of the question.\n\nYou're assuming the chairs around the table are indexed ${1,...,10}$ and you're looking at the number of ways to assign the numbers ${1,...,10}$ to your 5 men and 5 women.\n\nYour classmate is looking at how many ways you can arrange 5 men and women around the table relative to each other instead of in absolute terms.\n\nThe difference between the two answers is a factor of 10 ($10\\times4!5! = 2\\times 5 \\times 4!5! = 2\\times(5!)^2$), which corresponds to the extra degree of freedom (e.g. where is chair #1).\n\n\u2022 Good answer! Thanks \u2013\u00a0rae306 Jan 10 '18 at 19:46\n\nThe thing is, that when you are a circle, each combination $x_1, x_2 \\dots x_{10}$ is counted $10$ times, namely\n\n$x_1, x_2 \\dots x_9 x_{10}$,\n\n$x_2, x_3 \\dots x_{10},x_1$,\n\n$x_3, x_4 \\dots x_1, x_2$,\n\n$\\dots$\n\n$x_{10}, x_1, \\dots x_8, x_9$\n\nsince rotating the table preserves the combination of people sitting\n\nIn the case of $2$ people you get the combinations:\n\nman, woman\n\nwoman, man\n\nbut they are essentially the same.\n\nYour edit is spot on, depends on whether the chairs differ from one another, but it'd assume that this is not the case\n\n\u2022 Yes, but why wouldn't it matter who sits at chair 1? See also my edit. Is this really obvious from the problem? I understand completely what you are saying but to me the question seems ambiguous. \u2013\u00a0V\u00e1clav Mordvinov Jan 10 '18 at 9:22\n\u2022 @V\u00e1clavMordvinov We all passed through that. Yes, it's ambiguous, but with time you will start to see what they want with this kind of questions. Not that it will ever change, but you will get used to it, and you will see that it's not \u201cambiguous\u201d once you know how they speak. But yes, you are right, that way to phrase problems is not ideal and ambiguous to a random reader. We all passed that phase at which we had a problem with probability problems badly written. Good luck. \u2013\u00a0Manuel Jan 10 '18 at 14:21\n\u2022 @Manuel, you're right, especially since it is a problem and not a real-life application. I just have to read better next time :) \u2013\u00a0V\u00e1clav Mordvinov Jan 10 '18 at 18:29\n\u2022 @V\u00e1clavMordvinov \"Read better\", yes, but do not forget that you are right now thinking it's badly written. I meant that sadly it's badly written, but it's not the worst, you just have to get used to it. \u2013\u00a0Manuel Jan 10 '18 at 18:31\n\u2022 Yes, generally I add what I assume for such a question and if necessary I include multiple answers for multiple assumption sets. It's not that much more effort and yields the rigjt answer for sure. \u2013\u00a0V\u00e1clav Mordvinov Jan 10 '18 at 18:37\n\nIn combinatorics, the definition of \"different\" is often a crucial issue. There is nothing in the problem saying that the chairs are interchangeable, and you should not add assumptions to the problem. If I were only one person to be seated, I would have ten choices where to sit. Your interpretation is the correct one. If your classmate's interpretation were intended, then the problem is poorly stated.\n\nAnd actually, if we to follow your classmate's reasoning, then why stop there? Your classmate is noting that rotation leaves the arrangement \"essentially\" the same, but what about reflections? If all that matters is who is sitting next to whom, and absolute location is unimportant, then wouldn't the answer be 5!4!/2?\n\n\u2022 Good answer, I will send this to him too :) \u2013\u00a0V\u00e1clav Mordvinov Jan 10 '18 at 18:34\n\u2022 If you were the only person to be seated, you could actually tell that there is only one resulting configuration, if the table is ROUND with no orientation mark \u2013\u00a0G Cab Jan 11 '18 at 2:08\n\u2022 @G Cab Yes, if the table is a non-rotating, uncharged black hole in otherwise empty space, then all locations are indistinguishable. If, on the other hand, this takes place in the real world ... \u2013\u00a0Acccumulation Jan 11 '18 at 3:15", "date": "2020-08-13 06:37:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2599351/count-the-possible-ways-to-seat-people-at-a-round-table/2599683", "openwebmath_score": 0.7019632458686829, "openwebmath_perplexity": 336.6356708009055, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471684931718, "lm_q2_score": 0.8887587964389112, "lm_q1q2_score": 0.874402825349822}}
{"url": "https://math.stackexchange.com/questions/1382062/external-bisectors-of-the-angles-of-abc-triangle-form-a-triangle-a-1b-1c-1-and", "text": "# External bisectors of the angles of ABC triangle form a triangle $A_1B_1C_1$ and so on\n\nIf the external bisectors of the angles of the triangle ABC form a triangle $A_1B_1C_1$,if the external bisectors of the angles of the triangle $A_1B_1C_1$ form a triangle $A_2B_2C_2$,and so on,show that the angle $A_n$ of the $n$th derived triangle is $\\frac{\\pi}{3}+(-\\frac{1}{2})^n(A-\\frac{\\pi}{3})$,and that the triangles tend to become equilateral.\n\nMy attempt:The angle $A_1$ of triangle $A_1B_1C_1$ is $\\frac{\\pi}{2}-\\frac{A}{2}$,angle $A_2$ of triangle $A_2B_2C_2$ is $\\frac{A}{4}-\\frac{3\\pi}{4}$ but my answer is nowhere look like resembling final answer.Is my approach correct.If not,what is the correct way to solve this question?Can someone guide me?\n\nYou are on the right track, you just need to continue. You have a correct formula for $A_1$, so format it differently.\n\n\\begin{align} A_1 &= \\frac{\\pi}2-\\frac A2 \\quad\\text{(your formula)}\\\\ &= \\frac{\\pi}3+\\frac{\\pi}6+\\left(-\\frac 12\\right)A \\\\ &= \\frac{\\pi}3+\\left(-\\frac 12\\right)\\left(A-\\frac{\\pi}3\\right) \\\\ &= \\frac{\\pi}3+\\left(-\\frac 12\\right)^1\\left(A-\\frac{\\pi}3\\right) \\\\ \\end{align}\n\nYour formula for $A_1$ from $A$ also gives $A_{n+1}$ from $A_n$, so combining your formula with my derivation above and using induction we get\n\n\\begin{align} A_{n+1} &= \\frac{\\pi}3+\\left(-\\frac 12\\right)\\left(A_n-\\frac{\\pi}3\\right) \\\\ &= \\frac{\\pi}3+\\left(-\\frac 12\\right)\\left[\\frac{\\pi}3+ \\left(-\\frac 12\\right)^n\\left(A-\\frac{\\pi}3\\right)-\\frac{\\pi}3\\right] \\\\ &= \\frac{\\pi}3+\\left(-\\frac 12\\right)\\left[ \\left(-\\frac 12\\right)^n\\left(A-\\frac{\\pi}3\\right)\\right] \\\\ &= \\frac{\\pi}3+\\left(-\\frac 12\\right)^{n+1}\\left(A-\\frac{\\pi}3\\right) \\\\ \\end{align}\n\nwhich finishes the proof by induction.\n\nSo the given formula is correct. As we let $n\\to\\infty$ we see that $\\left(-\\frac 12\\right)^n\\to 0$ and thus $A_n\\to\\frac{\\pi}3$. We can do the same to $B$ and $C$ to get $B_n\\to\\frac{\\pi}3$ and $C_n\\to\\frac{\\pi}3$, so all three angles approach $60\u00b0$, and the (increasingly large) triangles $\\triangle A_nB_nC_n$ approach equilateral.\n\nHere is a quick proof of your formula $A_1 = \\frac{\\pi}2-\\frac A2$, though I used Greek letters for angles in this diagram.\n\nThis should be self-explanatory, and the final value for $A_1= \\frac{\\pi}2-\\frac A2$ comes directly from $A_1=\\frac{\\beta+\\gamma}2$ and $\\alpha+\\beta+\\gamma=\\pi$.\n\nYour approach was a good way to start working on this problem. By working the first two or three steps, you can see what calculations are involved. Your idea of comparing your results to the desired results was also a good one. It gives you a chance to check yourself in case you go down a wrong path.\n\nBut while it's true that the values you calculated, namely, $A_1 = \\frac{\\pi}{2} - \\frac A2$ and $A_2 = \\frac A4 - \\frac{3\\pi}{4}$, do not look much like $\\frac{\\pi}{3}+\\left(-\\frac12\\right)^n \\left(A - \\frac{\\pi}{3}\\right)$, that alone should not disturb you; general formulas very often look quite different from the results you get by direct calculations. That's part of what made it so interesting and useful when people discovered such formulas. It often takes some art to \"see\" a pattern in a sequence of values from some procedure such as $A_1$, $A_2$, $A_3$, and so forth.\n\nFor example, the sum of the first $n$ positive integers is $\\frac12(n^2 + n)$. But the sum of the first $4$ positive integers is $1+2+3+4 = 10$, and $10$ does not look like $\\frac12(n^2 + n)$. It doesn't even look like $\\frac12(4^2 + 4)$, but it turns out that if you do the multiplications and additions of that formula, you can see that $\\frac12(4^2 + 4) = \\frac12(16 + 4) = \\frac12(20) = 10$.\n\nThat is, in order to compare a particular result of yours (the values you found for $A_1$ and $A_2$) against the general formula, you can plug the appropriate value of $n$ into the formula, do whatever operations the formula says to do, and see what comes out. For example, your own calculation found that $A_1 = \\frac{\\pi}{2} - \\frac A2$. The general formula for $A_n$ should produce $A_1$ when $n = 1$. So let's set $n = 1$ and see what happens:\n\n\\begin{align} A_n & = \\frac\\pi3 + \\left(-\\frac12\\right)^n \\left(A - \\frac\\pi3\\right) \\\\ A_1 & = \\frac\\pi3 + \\left(-\\frac12\\right) \\left(A - \\frac\\pi3\\right) & \\text{(because we set $n = 1$)} \\\\ & = \\frac\\pi3 - \\frac12 A + \\frac\\pi6 \\\\ & = \\frac{\\pi}{2} - \\frac A2 \\end{align}\n\nSo your calculation of $A_1$ looks OK. For $A_2$, we set $n = 2$:\n\n\\begin{align} A_n & = \\frac\\pi3 + \\left(-\\frac12\\right)^n \\left(A - \\frac\\pi3\\right) \\\\ A_2 & = \\frac\\pi3 + \\left(-\\frac12\\right)^2 \\left(A - \\frac\\pi3\\right) & \\text{(because we set $n = 2$)} \\\\ & = \\frac\\pi3 + \\frac14 \\left(A - \\frac\\pi3\\right) \\\\ & = \\frac\\pi3 + \\frac14 A - \\frac{\\pi}{12} \\\\ & = \\frac\\pi4 + \\frac A4 \\end{align}\n\nThis is a different from the result $\\frac A4 - \\frac{3\\pi}{4}$ that you calculated for $A_2$. So maybe there was a problem with your method, or maybe you just made an arithmetic error.\n\nBut once you figure out what is going on with $A_2$, you still need to prove the general formula for all $n$, and for that you may need something like the other answer (which used induction). But if you can see how to get $A_1$ from $A$ and how to get $A_2$ from $A_1$, that can help when you have to get $A_{n+1}$ from $A_n$, as the inductive proof requires.\n\n\u2022 This is a good response, and the method you outline here is close to the one I used to get my answer. +1! \u2013\u00a0Rory Daulton Aug 3 '15 at 14:43", "date": "2020-10-22 21:44:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1382062/external-bisectors-of-the-angles-of-abc-triangle-form-a-triangle-a-1b-1c-1-and", "openwebmath_score": 0.9998910427093506, "openwebmath_perplexity": 298.57540968407983, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471675459396, "lm_q2_score": 0.8887587846530938, "lm_q1q2_score": 0.874402812912518}}
{"url": "https://math.stackexchange.com/questions/2286929/reasoning-in-an-integration-substitution", "text": "# Reasoning in an integration substitution\n\nEvaluate: $\\displaystyle\\int_{0}^{1}\\frac{\\ln(x+1)}{x^2+1}\\,\\mathrm{d}x$\n\nSo I did this a completely different way than what the answer key states. I used integration by parts and some symmetry tricks and got the correct answer. However the answer key says:\n\nMake the substitution $x=\\frac{1-u}{1+u}$\n\nThe same solution was reached in about half the steps but still using symmetry, my questions are How would I know to do that? Is this a certain type of substitution? Is there something else that maybe I could use this for?\n\n\u2022 the change of variable reminds me a bit the Weierstrass substitution $\\cos(x)=\\frac{1-t^2}{1+t^2}$ \u2013\u00a0Masacroso May 18 '17 at 20:31\n\u2022 That is true. The weierstrauss sub is actually what I used in part of my long solution \u2013\u00a0Teh Rod May 18 '17 at 20:33\n\u2022 The transformation is a M\u00f6bius Transform, not a Weierstrass substitution. \u2013\u00a0Mark Viola May 18 '17 at 20:39\n\u2022 possible duplicate math.stackexchange.com/questions/155941/\u2026 \u2013\u00a0Dar\u00edo A. Guti\u00e9rrez May 18 '17 at 21:09\n\nThe substitution is called a M\u00f6bius Transformation, which has the more general form\n\n$$w=\\frac{az+b}{cz+d}$$\n\nwhere $ad\\ne bc$. The transformation maps straight lines into circles and circles into straight lines. They have a variety of uses in applied mathematics and physics along with complex analysis.\n\nAs a simple example, the Beta function $B(x,y)$ can be represented by the integral\n\n$$B(x,y)=\\int_0^1 t^{x-1}(1-t)^{y-1}\\,dt$$\n\nEnforcing the Mobius Transformation $t\\to \\frac{t}{1+t}$ so that $dt\\to \\frac{1}{(1+t)^2}\\,dt$. Then, we see that\n\n$$B(x,y)=\\int_0^\\infty \\frac{t^{x-1}}{(1+t)^{x+y}}\\,dt$$\n\nwhich is an alternative representation for the Beta function.\n\nAs another example, in This Question, the OP requested evaluation of the integral $I$ expressed as\n\n$$I=\\int_0^\\infty\\frac{\\log(e^x-1)}{e^x+1}\\,dx$$\n\nIn the accepted answer posted by User @FDP, the Mobius transform was used to facilitate an efficient way forward.\n\nAs a third and final example, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities\n\n$$1+x\\le e^x\\le \\frac{1}{1-x}$$\n\nfor $x<1$. Then, it is trivial to see that $\\log(x)\\le x-1$.\n\nApplying the M\u00f6bius Tranformation $x\\to \\frac{-x}{1+x}$ we find that the logarithm function is bounded below by $\\log(x)\\ge \\frac{x-1}{x}$.\n\nHere's an alternative method that doesn't use differentiation under the integral sign. Substitute $x = \\tan \\theta$, to give $$I = \\int_0^{\\pi/4} \\log(1+\\tan\\theta) d\\theta \\\\ = \\int_0^{\\pi/4} \\log(\\cos\\theta+\\sin\\theta)-\\log\\cos(\\theta)d\\theta \\\\ = \\int_0^{\\pi/4} \\frac12\\log2 + \\log(\\cos(\\theta- \\pi/4))-\\log\\cos(\\theta) d\\theta \\\\ = \\frac{\\pi}{8}\\log2 + \\int_0^{\\pi/4} \\log(\\cos(\\theta- \\pi/4)) d\\theta - \\int_0^{\\pi/4}\\log\\cos(\\theta) d\\theta$$ We have the known property that $\\int_0^a f(x) dx = \\int_0^a f(a-x)dx$, and $\\cos$ is an even function, so these two final integrals are equal, and $I = \\frac{\\pi}{8}\\log2$.\n\nHere's how I would do it. Let $$f(\\alpha)=\\int_0^1 \\frac{\\log(1+\\alpha x)}{1+x^2}\\ dx$$ where $f(1)$ is the integral we seek to evaluate. By differentiation under the integral sign we have $$f'(\\alpha)=\\int_0^1\\frac{\\partial}{\\partial\\alpha}\\frac{\\log(1+\\alpha x)}{1+x^2}\\ dx\\\\ =\\int_0^1\\frac{x}{(1+x^2)(1+\\alpha x)}\\ dx.$$\n\nNow we apply partial fraction decomposition: suppose that $$\\frac{x}{(1+x^2)(1+\\alpha x)}=\\frac{Ax+B}{1+x^2}+\\frac{C}{1+\\alpha x}$$ for some constants $A,B,C$. By combining the fractions and collecting the terms we obtain the linear system of equations $$\\begin{cases}A+\\alpha B=1\\\\ \\alpha A+C=0\\\\B+C=0\\end{cases}$$ which we find to be equivalent to $$\\begin{cases}A=\\frac{1}{1+\\alpha^2}\\\\B=\\frac{\\alpha}{1+\\alpha^2}\\\\C=-\\frac{\\alpha}{1+\\alpha^2}\\end{cases}$$ From this we conclude that $$f'(\\alpha)=\\int_0^1\\frac{Ax+B}{1+x^2}+\\frac{C}{1+\\alpha x}\\ dx\\\\ =\\left[\\frac12 A\\log(1+x^2)+B\\arctan{x}+\\frac{C}{\\alpha}\\log(1+\\alpha x)\\right]_{x=0}^1\\\\ =\\frac{\\log2}{2}\\frac{1}{1+\\alpha^2}+\\frac{\\pi}{4} \\frac{\\alpha}{1+\\alpha^2}-\\color{red}{\\frac{\\log(1+\\alpha)}{1+\\alpha^2}}$$ It is easy to see by looking at the definition that $f(0)=0$. From this it follows that $f(\\alpha_0)=\\int_0^{\\alpha_0} f'(\\alpha)\\ d\\alpha$, and in particular $$f(1)=\\int_0^1 \\frac{\\log2}{\\alpha}\\frac{1}{1+\\alpha^2}+\\frac{\\pi}{4}\\frac{\\alpha}{1+\\alpha^2}-\\color{red}{\\frac{\\log(1+\\alpha)}{1+\\alpha^2}}\\ d\\alpha$$ which simplifies to $$\\color{red}{\\int_0^1 \\frac{\\log(1+x)}{1+x^2}\\ dx}= \\frac{\\pi \\log 2}{16}-\\color{red}{\\int_0^1 \\frac{\\log(1+\\alpha)}{1+\\alpha^2}\\ d\\alpha}$$ hence the result $$\\int_0^1 \\frac{\\log(1+x)}{1+x^2}\\ dx=\\frac{\\pi \\log 2}{8}.$$\n\n\u2022 That is a smarter way of doing it are you sure you meant $\\frac{\\log 2}{x}$? \u2013\u00a0Teh Rod May 18 '17 at 20:45\n\u2022 @TehRod that was a typo, now amended :) \u2013\u00a0user1892304 May 18 '17 at 20:51\n\nSet $$x=\\tan t, \\qquad dt=\\dfrac1{1+x^2}dx,$$ Then \\begin{align} \\int_0^1 \\frac{\\ln(1+x)}{1+x^2}\\, dx&= \\int_0^{\\pi/4} \\ln(\\cos t +\\sin t)\\, dt- \\int_0^{\\pi/4} \\ln(\\cos t)\\, dt\\\\\\\\ &=\\int_0^{\\pi/4} \\ln\\left(\\sqrt{2}\\cos \\left(\\frac \\pi4- t\\right)\\right)\\, dt- \\int_0^{\\pi/4} \\ln(\\cos t)\\, dt\\\\\\\\ &=\\int_0^{\\pi/4} \\ln\\left(\\sqrt{2}\\right)\\, dt+\\int_0^{\\pi/4} \\ln\\left(\\cos \\left(\\frac \\pi4- t\\right)\\right)\\, dt- \\int_0^{\\pi/4} \\ln(\\cos t)\\, dt\\\\\\\\ &=\\frac{\\pi}8 \\:\\ln 2+\\int_0^{\\pi/4} \\ln(\\cos u)\\, du-\\int_0^{\\pi/4} \\ln(\\cos t)\\, dt\\quad \\left(u=\\frac \\pi4- t\\right)\\\\\\\\ &=\\frac{\\pi}8 \\:\\ln 2. \\end{align}", "date": "2019-10-23 00:16:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2286929/reasoning-in-an-integration-substitution", "openwebmath_score": 0.9459965825080872, "openwebmath_perplexity": 387.41748400112095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426382811477, "lm_q2_score": 0.8976952975813453, "lm_q1q2_score": 0.8743934960287135}}
{"url": "http://math.stackexchange.com/questions/105631/help-needed-with-definite-integral-for-lim-n-rightarrow-infty-sum-k-1n", "text": "Help needed with definite integral for $\\lim_{n\\rightarrow \\infty}\\sum_{k=1}^{n} \\frac{1}{n+k}$\n\nMy foreign school book is not clearest at this point, what is really needed with this? What does it mean that \"assign definite integral for $\\lim_{n\\rightarrow \\infty}\\sum_{k=1}^{n} \\frac{1}{n+k}$\"? I am not sure, whether I should assign it like this $\\lim_{n\\rightarrow \\infty} \\int_{1}^{n} \\frac{1}{n+x} dx$ and noticing that it is continuous then trying to find borders so that $F(a) - F(b)$?\n\n(the book messes up all kind of Leibniz stuff at this point, a bit messy -- and just stating $\\int f(x)\\,dx= F(a)-F(b)$, but not even paying attention to different borders. As far as I know, it is important to specify whether borders depend on the integration factor)\n\n-\n...good basic video related to this here by patrickJMT. \u2013\u00a0hhh Feb 4 '12 at 14:11\n\nI suspect the following: $$\\sum_{k=1}^n {1\\over n+k}=\\sum_{k=1}^n{1\\over n} {1\\over 1+{k\\over n}}$$ Now interpret the right hand sum as a Riemann sum for the function $f(x)={1\\over 1+x}$ over $[a,b]=[0,1]$ (for a fixed $n$, the partition of $[0,1]$ is $\\{{1\\over n}, {2\\over n},\\ldots, {n\\over n} \\}$ and the ${1\\over n}$ is the common width of the subintervals).\n\nTaking the limit as $n\\rightarrow \\infty$ gives the corresponding integral: $$\\lim_{n\\rightarrow\\infty}\\sum_{k=1}^n{1\\over n} {1\\over 1+{k\\over n}} = \\int_0^1 {1\\over 1+x}\\,dx.$$\n\n-\n+1 for the illustration with picture! \u2013\u00a0user21436 Feb 4 '12 at 15:02\n+1 thanks! I finally understood the formula after your updated image. \u2013\u00a0hhh Feb 4 '12 at 18:23\nHow about if you have a situation so that the height is in different form such as $\\frac{1}{n+\\frac{k}{n}}$ or $\\frac{1}{n+\\frac{k}{n^{2}}}$? What about can I have a first term such as $\\frac{1}{n^{2}}$ or $\\frac{1}{n^{3}}$? \u2013\u00a0hhh Feb 4 '12 at 21:49\n@hhh Generally, for an integral over $[a,b]$, you use something of the form $f(a+[(b-a)i/n])\\cdot{1\\over n}$. The $1/n$ is the width of the rectangles for the partition points $a+[(b-a)i/n]$ (other widths could be used, like $1/(3n)$). The height is obtained by evaluating $f$ at the partition points. I don't think $1\\over n+(i/n)$ would get you one, since there is no way to write that as $f(a+[(b-a)i/n]$ for some $f$. Other expressions would work: e.g if you have $(i/n)^3+{1\\over (i/n)^2+1}$, then your $f$ would be $f(x)=x^3+{1\\over x^2+1}$ \u2013\u00a0David Mitra Feb 4 '12 at 22:20\n@hhh I don't think that will work. Basically, you chop $[a,b]$ into $n$ pieces so that the maximum length of the pieces is small. The widths can't all be $(1/n)^2$. If you add the lengths of the pieces, you have to get $b-a$. \u2013\u00a0David Mitra Feb 4 '12 at 22:57\n\nThe problem converts into an integral this way:\n\n\\begin{align*}\\lim_{n \\to \\infty} \\sum_{k=1}^n \\dfrac{1}{n+k}&=\\lim_{n \\to \\infty} \\dfrac{1}{n} \\cdot\\sum_{k=1}^n \\dfrac{n}{n+k}\\\\&=\\lim_{n \\to \\infty} \\dfrac{1}{n} \\cdot \\sum _{k=1}^n\\dfrac{n}{n+k} \\\\&= \\lim_{n\\to \\infty} \\dfrac{1}{n} \\cdot \\sum_{k=1}^n \\dfrac{1}{1+\\frac{k}{n}}\\end{align*}\n\nNow interpret the final limit as Riemann sum of the function $f(x)=\\dfrac{1}{1+x}$\n\nSo, the limit in question equals, $\\int_0^1{(\\dfrac{1}{1+x}) \\mathrm dx}= \\ln 2$ $\\blacksquare$\n\n-\n...thank you, can you suggest some reading about this topic? (My book is not the best one to practise on this, it is introduces Rieman sum after many pages the questions dealing with the rieman sums and the organizaton of material is not most pedagocical-- and I cannot make head-and-tail about what it is really trying to explain, have to look for some English material on this.) \u2013\u00a0hhh Feb 4 '12 at 14:05\n@hhh Did you try reading the article from Wikipedia, it is a bit Shabby, however. Look at David's Answer below. His picture should give you an idea of what is happening. And, Wolfram math world has an appplet which you could try and fiddle with until you are acquainted. \u2013\u00a0user21436 Feb 4 '12 at 15:02\n\nHint: You can write this as $$\\frac{1}{n}\\sum_{k=1}^n \\frac{1}{1+\\frac{k}{n}}.$$ How close is this to the Riemann sum of the integral $$\\int_0^1 \\frac{1}{1+x}dx?$$\n\n-", "date": "2016-02-12 08:03:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/105631/help-needed-with-definite-integral-for-lim-n-rightarrow-infty-sum-k-1n", "openwebmath_score": 0.9760891795158386, "openwebmath_perplexity": 295.21231591610865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426397881662, "lm_q2_score": 0.8976952934758465, "lm_q1q2_score": 0.8743934933826261}}
{"url": "https://www.physicsforums.com/threads/find-the-value-of-point-a-and-b.845250/", "text": "# Find the value of point A and B\n\n1. Nov 26, 2015\n\n### shayaan_musta\n\nHello experts.\n\nI am attaching the image below. Kindly tell me how to calculate the values of A and B in the given figure?\nNow I must find A and B. How to get it?\nPS: P=-2.5+4.33i. But A is not at -2.5. Means PA is not making angle 90 with real axis.\nThank you\n\n#### Attached Files:\n\n\u2022 ###### Capture.PNG\nFile size:\n6.2 KB\nViews:\n136\n2. Nov 26, 2015\n\n### Staff: Mentor\n\nOne information is missing, either about A or B or another angle. With P and the 55\u00b0 at P you could still find infinitely many solutions. Or is it supposed to express A as function of B?\n\n3. Nov 27, 2015\n\n### shayaan_musta\n\nNo further information is given. Here is the image of the whole page below.\nHere is AO=2.38. But how?\n\n#### Attached Files:\n\n\u2022 ###### Capture.PNG\nFile size:\n76 KB\nViews:\n121\n4. Nov 27, 2015\n\n### Samy_A\n\nBut, as @fresh_42 rightly supposed, you have one more piece of information, namely $\\frac {\\overline {PA}}{\\overline{PB}}=\\frac {4.77}{8}$.\nThen, as the text suggests, you can calculate A and B using trigonometry.\n\nLast edited: Nov 27, 2015\n5. Nov 27, 2015\n\n### shayaan_musta\n\nBut how? This is the actual question of mine. For A and B, I posted this thread. If you are saying to solve it myself then what is the cause of posting this thread. Kindly enlighten the way for me to find A and B.\n\n6. Nov 27, 2015\n\n### SteamKing\n\nStaff Emeritus\nYou've been posting information in dribs and drabs, and you are expecting a full solution to what is obviously a much larger problem than just figuring out this triangle.\n\nHow about some full disclosure of the complete problem or section from this text you are referencing?\n\n7. Nov 27, 2015\n\n### Samy_A\n\nAs both @fresh_42 and @SteamKing noted, you first didn't provide enough information to solve the triangle.\nFrom your second attachment we learned that $\\frac {\\overline {PA}}{\\overline{PB}}=\\frac {4.77}{8}$\n\nWith this information the triangle can be solved.\n\nNow there are probably many ways to do it.\nI would begin in the following way:\nIf $\\alpha$ is the angle in A, and $\\beta$ the angle in B, $\\frac {\\sin(\\beta)}{\\sin(\\alpha)}$ can be determined (using $\\frac {\\overline {PA}}{\\overline{PB}}=\\frac {4.77}{8}$ and the law of sines).\nBut we also know that $\\alpha+\\beta=125\u00b0$, so that $\\sin(\\beta)=\\sin(125\u00b0-\\alpha)=\\sin(125\u00b0)\\cos(\\alpha)-cos(125\u00b0)\\sin(\\alpha)$.\nThese two together will allow to compute $\\cot(\\alpha)$. And then the other components of the triangle can be computed. It is tedious but it works.\n\nThis is just one way to do it. Hopefully someone else will come along with a less tedious method.\n\nAs this thread should probably have been posted in the homework section, we are not allowed to give the full solutions.\n\nLast edited: Nov 27, 2015\n8. Nov 27, 2015\n\n### shayaan_musta\n\nOK. This is what I did,\n\nsin(125\u00b0-\u03b1)=sin(125\u00b0)cos(\u03b1)-cos(125\u00b0)sin(\u03b1)\nsince, sin(\u03b2)=sin(125\u00b0-\u03b1)\nso,\nsin(\u03b2)=sin(125\u00b0)cos(\u03b1)-cos(125\u00b0)sin(\u03b1)\ndividing both sides by sin(\u03b1)\nso,\nsin(\u03b2)/sin(\u03b1)=sin(125\u00b0)cos(\u03b1)/sin(\u03b1)-cos(125\u00b0)sin(\u03b1)/sin(\u03b1)\nsin(\u03b2)/sin(\u03b1)=sin(125\u00b0)cos(\u03b1)/sin(\u03b1)-cos(125\u00b0)\nsin(\u03b2)/sin(\u03b1)=0.8192*cos(\u03b1)/sin(\u03b1)-(-0.5736)\n4.77/8=0.8192*cot(\u03b1)+0.5736\n0.596=0.8192*cot(\u03b1)+0.5736\n0.0227=0.8192*cot(\u03b1)\ncot(\u03b1)=0.0227/0.8192\ncot(\u03b1)=0.028\n\u03b1=88\u00b0 and \u03b1+\u03b2=125\u00b0 => \u03b2=37\u00b0\nand\nAB=sin(P)*PA/sin(\u03b2)=6.5\n\nBut now what? I still don't have point A coordinates :(\n\n9. Nov 27, 2015\n\n### Samy_A\n\nDraw the line from P perpendicular to the horizontal axis. It will cross the horizontal axis in a point Q, just to the left of A.\nPAQ is a right triangle and you know the coordinates of P, the coordinates of Q and the angle $\\alpha$. That should lead you to the coordinates of A.\n\n10. Nov 27, 2015\n\n### shayaan_musta\n\nYes, I tried this already before posting post#8 but problem is P has iota in imaginary axis i.e. P(-2.5+4.33i).\n\n11. Nov 27, 2015\n\n### SteamKing\n\nStaff Emeritus\nI don't know what more you want.\nSo far:\n1. You have determined all three angles in this triangle\n2. You have calculated the length of one side of this triangle.\n3. You know the coordinates of one of the vertices of this triangle.\n\nIt's time to let loose on the trigonometry and find the rest of the missing information.\n\nHint: Use the Law of Sines to find the lengths of the other two sides.\n\n12. Nov 27, 2015\n\n### shayaan_musta\n\nOK\n\n13. Nov 28, 2015\n\n### Samy_A\n\nYou lost me there.\n\nIn triangle APQ:\n$\\cot(\\alpha)=\\frac {|AQ|}{|PQ|}$. The only unknown here is $|AQ|$.\nThus:\n$|AQ|=\\cot(\\alpha)*|PQ|=0.028*4.33=0.12$\nSo the coordinates of A are: $(-2.5+0.12,0)=(-2.38,0)$\n\nFor B (same as above, but now in the right triangle BPQ):\n$\\beta=\\pi-55*\\frac{\\pi}{180}-1.543=0.638$ (in radians).\n$|BQ|=\\cot(\\beta)*|PQ|=\\cot(0.638)*4.33=1.349*4.33=5.84$\nSo the coordinates of B are: $(-2.5-5.84,0)=(-8.34,0)$\n\nLast edited: Nov 28, 2015\n14. Nov 28, 2015\n\n### shayaan_musta\n\nBest Regards,\nPinky", "date": "2018-07-17 12:40:18", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/find-the-value-of-point-a-and-b.845250/", "openwebmath_score": 0.767493486404419, "openwebmath_perplexity": 1126.2293299906453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9609517095103498, "lm_q2_score": 0.9099070133672954, "lm_q1q2_score": 0.8743766999907592}}
{"url": "https://math.stackexchange.com/questions/4013760/clarifying-the-notion-of-basic-free-variables-in-a-system-of-equations", "text": "# Clarifying the notion of basic/free variables in a system of equations\n\nI have a fundamental confusion with regards to the notion of free/basic variables.\n\nConsider the following linear system $$\\begin{pmatrix} 1 & 2 & -1 \\\\ 2 & -4 & 0 \\end{pmatrix}\\begin{pmatrix} x_1\\\\ x_2 \\\\ x_3 \\end{pmatrix}=\\begin{pmatrix} 4\\\\ 5 \\end{pmatrix}$$ This system has augmented matrix $$\\begin{pmatrix} 1 & 2 & -1 & 4 \\\\ 2 & -4 & 0 & 5 \\end{pmatrix}$$ which row reduces to $$\\begin{pmatrix} 1 & 2 & -1 & 4 \\\\ 0 & -8 & 2 & -3 \\end{pmatrix}$$ The last matrix is in echelon form. Columns 1 and 2 are pivot columns, so $$x_1,x_2$$ are basic variables. The third column is not a pivot column, so $$x_3$$ is a free variable. Finally, the last column is not a pivot column, so the system is consistent.\n\nThe above analysis tells me that $$x_3$$ is a free variable. I interpret this as saying \"$$x_1,x_2$$ can be expressed as a function of $$x_3$$, while $$x_3$$ can be set equal to any number\". For example, I can set $$x_3=200$$. In turn, $$\\begin{cases} x_1+2x_2-200=4 \\\\ 2x_1-4x_2=5 \\end{cases}\\Rightarrow x_2=\\frac{403}{8}, x_1=\\frac{413}{4}$$\n\nNevertheless, just by doing naive computations, I realised that we could also \"express $$x_1$$, $$x_3$$ as a function of $$x_2$$, while setting $$x_2$$ equal to any number\". For example, I can set $$x_2=7$$. In turn, $$\\begin{cases} x_1+14-x_3=4\\\\ 2x_1-28=5 \\end{cases}\\Rightarrow x_1=\\frac{33}{2}, x_3=\\frac{53}{2}$$ Alternatively, we could also \"express $$x_2$$, $$x_3$$ as a function of $$x_1$$, while setting $$x_1$$ equal to any number\". For example, I can set $$x_1=0$$. In turn, $$\\begin{cases} 2x_2-x_3=4\\\\ -4x_2=5 \\end{cases}\\Rightarrow x_2=-\\frac{13}{2}, x_3=-\\frac{5}{4}$$\n\nHence, my question: what is the relation between\n\n\u2022 the fact that $$x_3$$ is a free variable and $$x_1,x_2$$ are basic variables (from the row-reduced matrix)\n\n\u2022 and the fact that in the system above I am free to fix the value of anyone among $$x_1,x_2,x_3$$ and I will always be able to solve for the other two unrestricted variables\n\n\u2022 Note that the order of the columns is arbitrary in a way. Feb 5 at 13:22\n\nYou've got three column vectors, and any pair of them are independent and thus span $$\\Bbb R^2$$. The solution set is one-dimensional (affine) (aka a line) so of the form $$v+tw$$, where $$v,w$$ are some vectors and $$t$$ is a free real variable. We can choose this to be $$x_3$$, or any of the other variables if you prefer.\n\nYou can then transform the equations to \\begin{align}x_1+2x_2 &= 4+x_3\\\\ x_1 - 4x_2 &= 5\\end{align}\n\nand solve these in terms of $$x_3$$ uniquely. We get (add twice the first equation to the second and we get $$3x_1 = 13+2x_3$$ so $$x_1 = \\frac{1}{3}(13+2x_3)$$, and subtraction the equations: $$6x_2 = x_3-1$$ and hence $$x_2 = \\frac{1}{6}(x_3 - 1)$$ (we could also do another elimination step on this new system) so we can write the solution line as\n\n$$\\begin{pmatrix} \\frac{13}{3}\\\\ -\\frac16\\end{pmatrix} + x_3 \\begin{pmatrix} \\frac{2}{3}\\\\ \\frac16\\end{pmatrix}$$\n\n\u2022 Thanks. (1) Can we say that in this specific case \"$x_3$ is a free variable, $x_1,x_2$ are basic variables\" or, equivalently, \"$x_1$ is a free variable, $x_3,x_2$ are basic variables\" or, equivalently, \"$x_2$ is a free variable, $x_1,x_3$ are basic variables\"? (2) Is there any way to realise from the echelon form that we are \"free\" to permute the columns? This somehow relates to my other question math.stackexchange.com/questions/4012647/\u2026 if you can help.\n\u2013\u00a0TEX\nFeb 5 at 13:27\n\u2022 @TEX They can all be free, that's the point. It doesn't matter. Feb 5 at 13:28\n\u2022 Thanks. My doubt is: is there a formal way to understand from the echelon form (or any other form) when any variable can be set as the free variable? This is not always the case.\n\u2013\u00a0TEX\nFeb 5 at 13:29\n\u2022 @TEX you just use $x_3$ and get the full solution set. Why do more? Feb 5 at 13:31\n\u2022 Because this relates to a more general result that I need to prove. I want to understand when I'm free to set the free variables and how I can detect that.\n\u2013\u00a0TEX\nFeb 5 at 13:32", "date": "2021-10-20 19:16:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4013760/clarifying-the-notion-of-basic-free-variables-in-a-system-of-equations", "openwebmath_score": 0.931725263595581, "openwebmath_perplexity": 208.0200368281728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576906395452, "lm_q2_score": 0.8962513696696647, "lm_q1q2_score": 0.8743621385750181}}
{"url": "https://math.stackexchange.com/questions/300393/group-tables-for-a-group-of-four-elements", "text": "# Group tables for a group of four elements.\n\nI should consider group tables obtained by renaming elements as essentially the same and then show that there are only two essentially different groups of order 4.\n\nThere seems to be so many different possible group tables for all the different binary operations - which is why I'm confused. I was thinking about using Cayley's table to show their commutativity but I'm really not too sure. Any help please!\n\nEdit: After all of your help, I completely understand how to show that there are only two different groups of order four. Thank you. The only thing which I am still unclear of is how to note down the 'other tables' before stating that they are essentially the same as one of the other tables - meaning that there are just two different ones. It's pointless work but I think it's what the question requires.\n\nAnswer: After a bit of playing around - I've realised that there are four different tables, however, 3 tables are the same as each other, just with different values (the Klein 4 Group with 3 different generators). Hence there are two tables.\n\n\u2022 It might seem like there are many possibilities, but try to start writing out some of them and note that there are certain properties they must have in order to satisfy the group axioms. \u2013\u00a0Tobias Kildetoft Feb 11 '13 at 17:47\n\u2022 Re: duplicate vote. It's the same question, but the discussion in the older one is left unfinished. \u2013\u00a0user53153 Feb 11 '13 at 19:12\n\nHints: we are talking about groups of order 4, which narrows the possible binary relations and elements available:\n\n\u2022 each must contain the identity,\n\u2022 each must be associative,\n\u2022 each must be closed under inverses (if $a \\in G,\\;a^{-1} \\in G$)\n\u2022 (and of course, closed under the group operation).\n\nThere are essentially (up to isomorphism) only two groups of order 4:\n\n\u2022 one of course will be the additive cyclic group $\\mathbb{Z}_4$, and\n\u2022 the other will be the Klein 4-group, which is indeed abelian.\n\nIf you follow the suggestions for solving the problem, you'll find, indeed, that any possible GROUP of order 4 can be shown isomorphic to one of the two groups mentioned simply by a renaming elements.\n\nYes: use of the Cayley table will be of great importance:\n\n\u2022 for a group: no element can appear twice in any column,\n\u2022 no element can appear twice in any row.\n\nYou'll find that the only ways of completing a table satisfying these criteria are limited, and then you show that a simple renaming of elements will reveal the group is isomorphic to $\\mathbb{Z}_4$ or else the Klein-4 group.\n\n\u2022 Yes, this has been very helpful, thank you. Your answer is so neatly written too! The only question I have is that I believe I should write down all the group tables... Is this just 2 group tables (of what binary operation???) OR do I list more but eliminate the rest. With Cayley's table, to show commutativity, is it sufficient to use the symmetry of the diagonal line? Thanks again. PS, I have encountered Lagrange - just struggling with Algebra! \u2013\u00a0user61854 Feb 11 '13 at 21:00\n\u2022 If you have encountered Lagrange, than it suffices to argue that the only groups of order 4 have are the cyclic group $\\mathbb{Z}$ (generated by one element), and a group whose non-identity elements have order 2 (since two and four divide 4). Completing any table that satisfies the properties of a group (one and only one element in each column and each row will limit the number of Cayley tables to construct. You can then show that of those, each is isomorphic to $Z_4$ or the Klein 4-group. And yes, you can show that those two non-isomorphic groups are indeed commutative by the Cayley table. \u2013\u00a0amWhy Feb 12 '13 at 1:47\n\nThere aren't that many. One of the elements will be the identity, let's call it $1$. (I am writing the operation as multiplication.) So the table looks like $$\\begin{matrix} 1 & a & b & c\\\\ a\\\\ b\\\\ c\\\\ \\end{matrix}$$ (I'm not writing the row and column labels, as they are the same as the first column and row). Now what could $ab$ be? Not $a$, not $b$... And $ba$? Proceed with $ac$, $ca$, $bc$, $cb$, and then you are left with the squares. They might all be $1$, or...\n\nThis simplifies it:\n\nIf a group has 4 elements, we separate two cases:\n\nCase1: all the elements have order 2, then the group is abelian, and $a^2=1\\Rightarrow a=a^{-1}$ for all $a\\in G$: $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ the table should be easy to make in this case: there is only one.\n\nCase2: there is an element $a\\not =1$ that has order different from 2, then by LAgrange's theorem, the order must be 4, and so $G=\\langle a\\rangle$, and so the group is cyclic, and therefore it's abelian. We have proved that a 4-element group is always abelian. The table shouldn't be difficult in this case either.\n\n\u2022 You're assuming Lagrange, and chances are, at this stage, the OP has not yet encountered Lagrange. See my comment below Michael Hardy's answer. \u2013\u00a0amWhy Feb 11 '13 at 19:17\n\u2022 @amWhy I would have never imagined, Lagrange always comes really soon. I guess that then he will have to do every possible table to get to the same conclusion. \u2013\u00a0MyUserIsThis Feb 11 '13 at 21:25\n\u2022 so b^2 = a, c^2 = a in one table and in the other, b^2 = 1, c^2 = 1. \u2013\u00a0user61854 Feb 12 '13 at 16:45\n\u2022 @user61854 Yes, in my case 1, you have that: $a^2=b^2=c^2=1$, so: $ab=c, bc=a$, it's abelian, so the whole table is there. In my case 2, $a$ is the generator, so $a=a,a^2=b,a^3=c$, and you can compose to get the rest of the combinations. \u2013\u00a0MyUserIsThis Feb 12 '13 at 17:00\n\nThe order of an element must divide $$4$$, so it must be either $$1$$, $$2$$, or $$4$$. Only the identity element can have order $$1$$. If one element has order $$4$$, then it generates the whole group and you have a cyclic (hence abelian) group.\n\nOtherwise the three non-identity elements each have order $$2$$.\n\n$$e=\\text{the identity}$$.\n\n$$a,b,c=\\text{the other three}$$.\n\n$$a^2=b^2=c^2=e$$, and each of these is its own inverse.\n\nSo what is $$ab$$? It can't be $$e$$ since if $$ab=e$$ then $$(ab)b^{-1} = b^{-1}=b$$, and hence $$a=b$$. It can't be $$a$$ or $$b$$ since if $$ab=b$$ then $$(ab)b^{-1}=bb^{-1}=e$$ and so $$a=e$$, and similar reasoning shows it can't be $$a$$ (but you have to multiply on the left be the inverse).\n\nSo $$ab=c$$.\n\nSimilar reasoning shows $$ab=ba=c$$, and $$ac=ca=b$$ and $$bc=cb=a$$.\n\nSo the group is abelian.\n\n\u2022 Chances are, at this stage of the game, the OP hasn't encountered Lagrange. I know Fraleigh includes this exercise long before introducing Lagrange... \u2013\u00a0amWhy Feb 11 '13 at 19:15\n\u2022 @amWhy Because this is such a toy example, you don't need Lagrange; it's trivial to see that only the identity can have order 1, or that if one element has order 4 then the group must be cyclic. The only other possibility aside from the all-order-2 case is that some element (and therefore two elements) has order 3, and it's easy to eliminate this possibility. (Suppose that $a$ and $b$ have order 3, with $a^2=b$. $ca$ can't be $a$ and can't be $c$ by cancellation; it can't be $b$ because $ca=aa\\implies c=a$, and it can't be the identity because then $ca=1=ba$ and again by cancellation $c=b$.) \u2013\u00a0Steven Stadnicki Feb 12 '13 at 23:33", "date": "2021-04-21 02:34:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/300393/group-tables-for-a-group-of-four-elements", "openwebmath_score": 0.8426212668418884, "openwebmath_perplexity": 214.06485455004116, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9854964228644457, "lm_q2_score": 0.8872045966995027, "lm_q1q2_score": 0.8743369563962531}}
{"url": "https://www.physicsforums.com/threads/what-does-ceil-mean.116822/", "text": "# What does ceil mean ?\n\n1. Apr 7, 2006\n\n### momentum\n\nIts diifcult to express my question....so, i am posting this\n\nceil(4.5) =?\nceil(4.1)=?\nceil(4.6)=?\n\n2. Apr 7, 2006\n\n### AKG\n\nceil(x) is the smallest integer which is greater than or equal to x. In particular, if x is an integer, then ceil(x) = x, and if x is not an integer, then ceil(x) > x.\n\n3. Apr 7, 2006\n\n### momentum\n\nceil(4.5)=4 // is it ok ?\nceil(4.1)=4 //is it ok ?\nceil(4.6)=5 //is it ok ?\n\n4. Apr 7, 2006\n\n### Integral\n\nStaff Emeritus\nNope, read the defintion given above, then try again.\n\n5. Apr 7, 2006\n\n### momentum\n\nah...i see, all of them should be 5 .....i had confusion on fractional part .5.\nbut i see ..it does not care for .5 which we use for round-off.\n\n6. Apr 7, 2006\n\nYes, all 5.\n\n7. Apr 7, 2006\n\n### momentum\n\nthank you for the clarifcation\n\n8. Apr 7, 2006\n\n### bomba923\n\nRecall that\n\n$$\\begin{gathered} \\forall x \\in \\left( {a,a + 1} \\right)\\;{\\text{where }}a \\in \\mathbb{Z}, \\hfill \\\\ {\\text{floor}}\\left( x \\right) = \\left\\lfloor x \\right\\rfloor = a \\hfill \\\\ {\\text{ceil}}\\left( x \\right) = \\left\\lceil x \\right\\rceil = a + 1 \\hfill \\\\ \\end{gathered}$$\n\n$$\\forall x \\in \\mathbb{Z},\\;\\left\\lfloor x \\right\\rfloor = \\left\\lceil x \\right\\rceil = x$$\n\nLast edited: Apr 7, 2006\n9. Apr 7, 2006\n\n### Staff: Mentor\n\nceil -> \"goes up\" if it needs to, in order reach an integer\nfloor -> \"goes down\" as it needs to, in order to reach an integer\nWhat happens with negative numbers:\n\n$$floor( -1.1 ) = -2 \\; ceil( -1.1 ) = -1$$\n$$floor( -0.1 ) = -1 \\; ceil( -0.1 ) = 0$$\n$$floor( 0.9 ) = 0 \\; ceil( 0.9 ) = 1$$", "date": "2017-10-19 18:53:57", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/what-does-ceil-mean.116822/", "openwebmath_score": 0.44892311096191406, "openwebmath_perplexity": 10961.216283612643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.978712649448364, "lm_q2_score": 0.8933094088947399, "lm_q1q2_score": 0.8742932183565228}}
{"url": "https://math.stackexchange.com/questions/3690609/how-to-translate-this-statement-into-a-mathematical-oneusing-appropriate-quanti/3699299#3699299", "text": "# How to translate this statement into a mathematical one(using appropriate quantifiers)?\n\nThe statement I'd like to translate into a mathematical one is\n\n\"Every American has a dream\".\n\nLet $$A$$ and $$D$$ denote the set of all Americans and the set of all dreams, respectively, and $$P(a,d)$$ denote the proposition \"American $$a$$ has a dream $$d$$\". The mathematically equivalent statement I've deduced is $$\\forall a\\in A.\\exists d\\in D.P(a, d)$$\n\nHowever, I suspect the above statement implies that for every American there exists a common dream $$d$$ such that $$P(a,d)$$ holds true. I would like to know how to rectify this error(if there is one).\n\n\u2022 Correct: \"forall a there is a d...\" does not mean that the d is the same for all a. To state that the d is the same, you have to write \"there is a d for all a...\". Compare $\\forall n \\exists m (n < m)$ and $\\exists m \\forall n (n < m)$. May 25, 2020 at 10:01\n\n1. The verifier of a sentence of the form \"$$\u2200a{\u2208}A\\ ( P(a) )$$\" must let the refuter first choose any arbitrary $$a\u2208A$$ and then verify $$P(a)$$ no matter what $$a\u2208A$$ was chosen.\n2. The verifier of a sentence of the form \"$$\u2203d{\u2208}D\\ ( Q(d) )$$\" must first choose some $$d\u2208D$$ and then verify $$Q(d)$$ for that chosen $$d\u2208D$$.\nIn your example, the verifier of \"$$\u2200a{\u2208}A\\ \u2203d{\u2208}D\\ ( P(a,d) )$$\" must let the refuter make the first move in choosing an $$a\u2208A$$, and then verify \"$$\u2203d{\u2208}D\\ ( P(a,d) )$$\" no matter what $$a$$ was chosen. But since the verifier makes the second move in choosing some $$d\u2208D$$, the verifier can choose this $$d$$ based on the refuter's first move (i.e. based on $$a$$). That is why \"Every American has a dream.\" corresponds to this sentence.\nIn contrast, the verifier of \"$$\u2203d{\u2208}D\\ \u2200a{\u2208}A\\ ( P(a,d) )$$\" must make the first move in choosing some $$d\u2208D$$, before the refuter makes the second move in choosing an $$a\u2208A$$. You can see easily that the verifier can win only if there is a single choice of $$d\u2208D$$ that defeats every possible choice of $$a\u2208A$$. That is why \"All Americans have a common dream.\" corresponds to this sentence and not the other one.", "date": "2023-03-29 03:00:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3690609/how-to-translate-this-statement-into-a-mathematical-oneusing-appropriate-quanti/3699299#3699299", "openwebmath_score": 0.922331690788269, "openwebmath_perplexity": 368.18121339178043, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126482065489, "lm_q2_score": 0.89330940889474, "lm_q1q2_score": 0.8742932172471978}}
{"url": "http://simonedavis.com/h-xbu/10hec.php?page=acute-isosceles-triangle-421da2", "text": "A right triangle may be isosceles \u2026 Input 3 triangle side lengths (A, B and C), then click \"ENTER\". Constructing an isosceles triangle Constructing an isosceles triangle also known as drawing an isosceles triangle using only a straightedge and a compass is what I will show you here. B - isosceles triangle. Yes, because it has an obtuse angle. Please update your bookmarks accordingly. Obtuse isosceles triangle with an irrational but algebraic ratio between the lengths of its sides and its base. This concept will teach students the properties of isosceles triangles and how to apply them to different types of problems. An isosceles right triangle therefore has angles of 45 degrees, 45 degrees, and 90 degrees. Apply the properties of isosceles triangles. An isosceles triangle is a special case of a triangle where 2 sides, a and c, are equal and 2 angles, A and C, are equal. Given below are the properties of isosceles and acute triangles. Parts of an Isosceles Triangle. The angles opposite the equal sides are also equal. Find information related to equilateral triangles, isosceles triangles, scalene triangles, obtuse triangles, acute triangles, right angle triangles, the hypotenuse, angles of a triangle and more. C - obtuse triangle. Let the coordinates of the right-angled isosceles triangle be O(0, 0), A(a, 0) and B(0, a). Area of the acute triangle is $$A = {1 \\over 2} \\times b \\times h$$ The Perimeter of the Acute triangle is: $$P=a+b+c$$ The types of acute triangles are: a) Acute Equilateral Triangle b) Acute Isosceles Triangle c) Acute Scalene Triangle All three angles are different implies that all three sides of an acute scalene triangle are also different. The two shorter sides measure x cm and 3x cm. Without Using The Calculator When given 3 triangle sides, to determine if the triangle is acute\u2026 An acute angle triangle (or acute-angled triangle) is a triangle in which all the interior angles are acute angles. Step-by-step explanation: Correct statements and reasons are in bold. A - acute triangle. For an acute triangle, all three angles of the triangle are from $$0^o$$ to $$90^o$$. This calculator will determine whether those 3 sides will form an equilateral, isoceles, acute, right or obtuse triangle or no triangle at all. In a right triangle, one of the angles is a right angle\u2014an angle of 90 degrees. For example, if we know a and b we know c since c = a. Addtionally, like all triangles, the three angles will sum to . All the three angles situated within the isosceles triangle are acute, which signifies that the angles are less than 90\u00b0. We have moved all content for this concept to for better organization. Step 1: One of the angles of the given triangle is a right angle. Add these two angles together and subtract the answer from 180\u00b0 to find the remaining third angle. Answer : D Explanation. the isosceles triangle can be acute, right or obtuse, but it depends only on the vertex angle (base angles are always acute) The equilateral triangle is a special case of a isosceles triangle. PROPERTIES OF ISOSCELES TRIANGLE ABC Let ABC be an isosceles triangle with\nTourist Places In Coimbatore Near Gandhipuram, Sunrunner Pembroke Welsh Corgis, Where Is Serana In Fort Dawnguard, Hello Etch A Sketch Font, Monkey King Characters, Adam Bradley Amazon,", "date": "2021-07-27 21:39:46", "meta": {"domain": "simonedavis.com", "url": "http://simonedavis.com/h-xbu/10hec.php?page=acute-isosceles-triangle-421da2", "openwebmath_score": 0.5174856185913086, "openwebmath_perplexity": 462.12853572644497, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9787126457229186, "lm_q2_score": 0.8933094067644466, "lm_q1q2_score": 0.8742932129436024}}
{"url": "https://mathhelpboards.com/threads/c-f-gauss-on-how-to-add-all-numbers-of-1-to-100.1832/", "text": "# C F Gauss, on how to add all numbers of 1 to 100?\n\n#### CaptainBlack\n\n##### Well-known member\n\nCarl gauss 1777 1855 , on how to add all numbers of 1 to 100?\n\nI can only add from 1 to 100 in order, apparently he could add lighteningly fast, and backwards, can anyone explain how in the confines of one message on here please\nAllegedly his method was something like:\n\n$(1+100)+(2+99)+ ... +(99+2)+(100+1)=2(1+2+...+100)$\n\nBut the left hand side is $101\\times 100$, so $1+2+..100=101\\times 100/2$\n\n(Like Marlow's Dr Faustus he could sum them forwards and backwards - but had no need of anagramatisation)\n\nCB\n\nLast edited:\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nThis also works for any sum:\n\n$\\displaystyle \\sum_{k = 1}^{n} k = 1 + 2 + \\cdots + n$\n\nThis sum can be reorganized in this fashion:\n\n$\\displaystyle \\left ( 1 + \\left ( n \\right ) \\right ) + \\left ( 2 + \\left ( n - 1 \\right ) \\right ) + \\cdots + \\left (n + \\left (1 \\right ) \\right )$\n\nThis produces $n$ sums of $n + 1$, so the total sum is $n(n + 1)$. But note that this goes through the list twice, from the left and from the right (for instance, $1$ is added to $n$ at the beginning and at the end). So this is actually twice the sum we need, we conclude:\n\n$\\displaystyle \\sum_{k = 1}^{n} k = \\frac{n(n + 1)}{2}$\n\n$n = 100$ can then be easily calculated from the formula, just as can $n = 4885$ or $n = 6$\n\n#### QuestForInsight\n\n##### Member\nI would like to share two more ways of doing this.\n\nProof #0 (Notice that what Gauss did is basically this):\n\n\\begin{aligned} \\displaystyle & \\sum_{0 \\le k \\le n}k = \\sum_{0 \\le k \\le n}(n-k) = n\\sum_{0 \\le k \\le n}-\\sum_{0 \\le k \\le n}k \\\\& \\implies 2\\sum_{0 \\le k \\le n}k = n(n+1) \\implies \\sum_{0 \\le k \\le n}k = \\frac{1}{2}n(n+1).\\end{aligned}\n\nProof #1 (Now try that with $\\sum_{0 \\le k \\le n}k^2$ and you will be pleasantly surprised):\n\n\\begin{aligned} \\displaystyle & \\sum_{0 \\le k \\le n}k^2 = \\sum_{0 \\le k \\le n}(n-k)^2 = n^2\\sum_{0 \\le k \\le n}-2n\\sum_{0 \\le k \\le n}k+\\sum_{0 \\le k \\le n}k^2 \\\\& \\implies 2n\\sum_{0 \\le k \\le n}k = n^2(n+1) \\implies \\sum_{0 \\le k \\le n}k = \\frac{1}{2}n(n+1).\\end{aligned}\n\nProof #2 (Writing it as a double sum and switching the order of summation):\n\n\\begin{aligned}\\displaystyle & \\begin{aligned}\\sum_{1 \\le k \\le n}k & = \\sum_{1 \\le k \\le n}~\\sum_{1 \\le r \\le k} = \\sum_{1 \\le r \\le n} ~\\sum_{r \\le k \\le n} = \\sum_{1 \\le r \\le n}\\bigg(\\sum_{1 \\le k \\le n}-\\sum_{1 \\le k \\le r-1}\\bigg) \\\\& =\\sum_{1 \\le r \\le n}\\bigg(n-r+1\\bigg) = n\\sum_{1 \\le k \\le n}-\\sum_{1 \\le k \\le n}k+\\sum_{1 \\le k \\le n}\\end{aligned} \\\\& \\implies 2\\sum_{1 \\le k \\le n}k = n^2+n \\implies \\sum_{1 \\le k \\le n}k = \\frac{1}{2}n(n+1), ~ \\mathbb{Q. E. D.}\\end{aligned}\n\nLast edited:\n\n#### MarkFL\n\nStaff member\n\nHere are two closely related methods to derive the formulas for summations of sequences of natural number powers of integers from 0 - n.\n\nMethod 1: A \"bottom-up\" approach...\n\nWe may state:\n\n$\\displaystyle\\sum_{k=0}^n(k)=\\sum_{k=0}^n(k+1)-(n+1)$\n\n$\\displaystyle\\sum_{k=0}^n(k)=\\sum_{k=0}^n(k)+\\sum_{k=0}^n(1)-(n+1)$\n\n$\\displaystyle0=\\sum_{k=0}^n(1)-(n+1)$\n\n$\\displaystyle\\sum_{k=0}^n(1)=n+1$\n\nNow we may compute:\n\n$\\displaystyle\\sum_{k=0}^n(k)$\n\nWe may state:\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^2 \\right)=\\sum_{k=0}^n\\left((k+1)^2 \\right)-(n+1)^2$\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^2 \\right)=\\sum_{k=0}^n\\left(k^2+2k+1 \\right)-(n+1)^2$\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^2 \\right)=\\sum_{k=0}^n\\left(k^2 \\right)+2\\sum_{k=0}^n(k)+\\sum_{k=0}^n(1)-(n+1)^2$\n\n$\\displaystyle2\\sum_{k=0}^n(k)=-\\sum_{k=0}^n(1)+(n+1)^2$\n\nNow, using our previous result, we have:\n\n$\\displaystyle2\\sum_{k=0}^n(k)=-(n+1)+(n+1)^2$\n\n$\\displaystyle2\\sum_{k=0}^n(k)=(n+1)\\left(-1+(n+1) \\right)$\n\n$\\displaystyle2\\sum_{k=0}^n(k)=(n+1)(n)$\n\n$\\displaystyle\\sum_{k=0}^n(k)=\\frac{n(n+1)}{2}$\n\nNow we may continue and find:\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^2 \\right)$\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^3 \\right)=\\sum_{k=0}^n\\left((k+1)^3 \\right)-(n+1)^3$\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^3 \\right)=\\sum_{k=0}^n\\left(k^3+3k^2+3k+1 \\right)-(n+1)^3$\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^3 \\right)=\\sum_{k=0}^n\\left(k^3 \\right)+3\\sum_{k=0}^n\\left(k^2 \\right)+3\\sum_{k=0}^n(k)+\\sum_{k=0}^n(1)-(n+1)^3$\n\n$\\displaystyle3\\sum_{k=0}^n\\left(k^2 \\right)=-3\\sum_{k=0}^n(k)-\\sum_{k=0}^n(1)+(n+1)^3$\n\nUsing our previous results, we have:\n\n$\\displaystyle3\\sum_{k=0}^n\\left(k^2 \\right)=-3\\left(\\frac{n(n+1)}{2} \\right)-(n+1)+(n+1)^3$\n\n$\\displaystyle3\\sum_{k=0}^n\\left(k^2 \\right)=(n+1)\\left(-\\frac{3n}{2}-1+(n+1)^2 \\right)$\n\n$\\displaystyle3\\sum_{k=0}^n\\left(k^2 \\right)=(n+1)\\left(-\\frac{3n}{2}-1+n^2+2n+1 \\right)$\n\n$\\displaystyle3\\sum_{k=0}^n\\left(k^2 \\right)=(n+1)\\left(n^2+\\frac{n}{2} \\right)$\n\n$\\displaystyle3\\sum_{k=0}^n\\left(k^2 \\right)=\\frac{n(n+1)(2n+1)}{2}$\n\n$\\displaystyle\\sum_{k=0}^n\\left(k^2 \\right)=\\frac{n(n+1)(2n+1)}{6}$\n\nWe may continue this process, to find further power summations.\n\nMethod 2: A recursion approach...\n\nSuppose we want to find:\n\n$\\displaystyle S_n=\\sum_{k=0}^n\\left(k^3 \\right)$\n\nWe may use the inhomogeneous recursion to state:\n\n$\\displaystyle S_n=S_{n-1}+n^3$\n\nNow, we may use symbolic differencing to ultimately derive a homogeneous recursion.\n\nWe begin by replacing n with n + 1:\n\n$\\displaystyle S_{n+1}=S_{n}+(n+1)^3$\n\nSubtracting the former from the latter, there results:\n\n$\\displaystyle S_{n+1}=2S_{n}-S_{n-1}+3n^2+3n+1$\n\n$\\displaystyle S_{n+2}=2S_{n+1}-S_{n}+3(n+1)^2+3(n+1)+1$\n\nSubtracting again:\n\n$\\displaystyle S_{n+2}=3S_{n+1}-3S_{n}+S_{n-1}+6n+6$\n\n$\\displaystyle S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+6(n+1)+6$\n\nSubtracting again:\n\n$\\displaystyle S_{n+3}=4S_{n+2}-6S_{n+1}+4S_{n}-S_{n-1}+6$\n\n$\\displaystyle S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}+6$\n\nSubtracting again:\n\n$\\displaystyle S_{n+4}=5S_{n+3}-10S_{n+2}+10S_{n+1}-5S_{n}+S_{n-1}$\n\nNow we have a homogeneous recursion whose associated characteristic equation is:\n\n$\\displaystyle (r-1)^5=0$\n\nSince we have the root $\\displaystyle r=1$ of multiplicity 5, we know the closed form for the sum will take the form:\n\n$\\displaystyle S_n=k_0+k_1n+k_2n^2+k_3n^3+k_4n^4$\n\nWe may use known initial values to determine the constants $\\displaystyle k_i$.\n\n$\\displaystyle S_0=k_0=0$\n\nThis means, we are left with the 4X4 system:\n\n$\\displaystyle k_1+k_2+k_3+k_4=1$\n\n$\\displaystyle 2k_1+4k_2+8k_3+16k_4=9$\n\n$\\displaystyle 3k_1+9k_2+27k_3+81k_4=36$\n\n$\\displaystyle 4k_1+16k_2+64k_3+256k_4=100$\n\nDid you notice we get the squares of successive triangular numbers? Anyway, solving this system, we find:\n\n$\\displaystyle k_1=0,k_2=\\frac{1}{4},k_3=\\frac{1}{2},k_4=\\frac{1}{4}$\n\nAnd so, we have:\n\n$\\displaystyle S_n=\\frac{1}{4}n^2+\\frac{1}{2}n^3+\\frac{1}{4}n^4= \\frac{n^4+2n^3+n^2}{4}=$\n\n$\\displaystyle\\frac{n^2(n+1)^2}{4}=\\left( \\frac{n(n+1)}{2} \\right)^2$\n\n#### melese\n\n##### Member\na nice combinatorial proof\n\nTo each yellow circle we can associate a pair of green circles in the way suggested by the drawing. This shows a bijection between (and hence the same number of) all the yellow circles and all the pairs of green circles.\n\nHere, there are $1+2+3+4+5+6+7+8$ yellow circles, and the number of green-circle pairs is $\\binom{9}{2}$; so $1+2+3+4+5+6+7+8=\\binom{9}{2}$.\n\n(I saw this 'proof without words' in a lecture by Gil Kalai)\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\n$(1+100)+(2+99)+ ... +(99+2)+(100+1)=2(1+2+...+100)$\nBut the left hand side is $101\\times 100$, so $1+2+..100=101\\times 100/2$", "date": "2020-11-30 23:10:27", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/c-f-gauss-on-how-to-add-all-numbers-of-1-to-100.1832/", "openwebmath_score": 0.9937049746513367, "openwebmath_perplexity": 1325.934489167124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787830929849, "lm_q2_score": 0.8856314798554444, "lm_q1q2_score": 0.874276606552537}}
{"url": "https://www.jiskha.com/questions/1799344/13-exercise-convergence-in-probability-a-suppose-that-xn-is-an-exponential-random", "text": "math, probability\n\n13. Exercise: Convergence in probability:\n\na) Suppose that Xn is an exponential random variable with parameter lambda = n. Does the sequence {Xn} converge in probability?\n\nb) Suppose that Xn is an exponential random variable with parameter lambda = 1/n. Does the sequence {Xn} converge in probability?\n\nc) Suppose that the random variable in the sequence {Xn} are independent, and that the sequence converges to some number a, in probability.\nLet {Yn} be another sequence of random variables that are dependent, but where each Yn has the same distribution (CDF) as Xn. Is it necessarily true that the sequence {Yn} converges to a in probability?\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. a) yes\nb) no\nc) yes\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n2. y\nn\nn\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. 142n=47 find n\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\nSimilar Questions\n\n1. STATISTICS\n\nConsider a binomial random variable where the number of trials is 12 and the probability of success on each trial is 0.25. Find the mean and standard deviation of this random variable. I have a mean of 4 and a standard deviation\n\n2. Probability\n\nProblem 1. Starting at time 0, a red bulb flashes according to a Poisson process with rate \u03bb=1 . Similarly, starting at time 0, a blue bulb flashes according to a Poisson process with rate \u03bb=2 , but only until a nonnegative\n\n3. mathematics, statistics\n\nYou observe k i.i.d. copies of the discrete uniform random variable Xi , which takes values 1 through n with equal probability. Define the random variable M as the maximum of these random variables, M=maxi(Xi) . 1.) Find the\n\n4. math\n\nSuppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter \u03bb (lambda) = 0.5. What's the probability that a repair takes less than 5 hours? AND what's the\n\n1. Statistics\n\nA random variable is normally distributed with a mean of 50 and a standard deviation of 5. b. What is the probability that the random variable will assume a value between 45 and 55 (to 4 decimals)? c. What is the probability that\n\n2. probability\n\nLet K be a discrete random variable that can take the values 1 , 2 , and 3 , all with equal probability. Suppose that X takes values in [0,1] and that for x in that interval we have fX|K(x|k)=\u23a7\u23a9\u23a81,2x,3x2,if k=1,if k=2,if\n\n3. Math\n\nSuppose a baseball player had 211 hits in a season. In the given probability distribution, the random variable X represents the number of hits the player obtained in the game. x P(x) 0 0.1879 1 0.4106 2 0.2157 3 0.1174 4 0.0624 5\n\n4. Statistics\n\nIn a study by Peter D. Hart Research Associates for the Nasdaq Stock Market, it was determined that 20% of all stock investors are retired people. In addition, 40% of all adults invest in mutual funds. Suppose a random sample of\n\n1. probability\n\nProblem 2. Continuous Random Variables 2 points possible (graded, results hidden) Let \ud835\udc4b and \ud835\udc4c be independent continuous random variables that are uniformly distributed on (0,1) . Let \ud835\udc3b=(\ud835\udc4b+2)\ud835\udc4c . Find the probability\n\n2. Probability\n\nQuestion:A fair coin is flipped independently until the first Heads is observed. Let the random variable K be the number of tosses until the first Heads is observed plus 1. For example, if we see TTTHTH, then K=5. For K=1,2,3...K,\n\n3. statistics MIT\n\n2 Let X1,\u2026,Xn be i.i.d. random variable with pdf f\u03b8 defined as follows: f\u03b8(x)=\u03b8x\u03b8\u221211(0\u2264x\u22641) where \u03b8 is some positive number. (a) Is the parameter \u03b8 identifiable? Yes No (b) Compute the maximum likelihood estimator\n\n4. Statistics\n\nA person's level of blood glucose and diabetes are closely related. Let x be a random variable measured in milligrams of glucose per deciliter (1/10 of a liter) of blood. Suppose that after a 12-hour fast, the random variable x", "date": "2021-09-19 22:50:47", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1799344/13-exercise-convergence-in-probability-a-suppose-that-xn-is-an-exponential-random", "openwebmath_score": 0.8852372169494629, "openwebmath_perplexity": 675.35429541727, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9664104943498961, "lm_q2_score": 0.9046505434556232, "lm_q1q2_score": 0.8742637789148511}}
{"url": "http://www.mathworks.com/help/matlab/ref/istriu.html?s_tid=gn_loc_drop&nocookie=true", "text": "Accelerating the pace of engineering and science\n\n# istriu\n\nDetermine if matrix is upper triangular\n\n## Description\n\nexample\n\ntf = istriu(A) returns logical 1 (true) if A is an upper triangular matrix; otherwise, it returns logical 0 (false).\n\n## Examples\n\nexpand all\n\n### Test Upper Triangular Matrix\n\nCreate a 5-by-5 matrix.\n\n`A = triu(magic(5))`\n```A =\n\n17    24     1     8    15\n0     5     7    14    16\n0     0    13    20    22\n0     0     0    21     3\n0     0     0     0     9```\n\nTest A to see if it is upper triangular.\n\n`istriu(A)`\n```ans =\n\n1\n```\n\nThe result is logical 1 (true) because all elements below the main diagonal are zero.\n\n### Test Matrix of Zeros\n\nCreate a 5-by-5 matrix of zeros.\n\n`Z = zeros(5);`\n\nTest Z to see if it is upper triangular.\n\n`istriu(Z)`\n```ans =\n\n1\n```\n\nThe result is logical 1 (true) because an upper triangular matrix can have any number of zeros on the main diagonal.\n\n## Input Arguments\n\nexpand all\n\n### A \u2014 Input arraynumeric array\n\nInput array, specified as a numeric array. istriu returns logical 0 (false) if A has more than two dimensions.\n\nData Types: single | double\nComplex Number Support: Yes\n\nexpand all\n\n### Upper Triangular Matrix\n\nA matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.\n\nFor example, the matrix\n\n$A=\\left(\\begin{array}{cccc}1& -1& -1& -1\\\\ 0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& -2& -2\\\\ 0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1& -3\\\\ 0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}0& \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}1\\end{array}\\right)$\n\nis upper triangular. A diagonal matrix is both upper and lower triangular.\n\n### Tips\n\n\u2022 Use the triu function to produce upper triangular matrices for which istriu returns logical 1 (true).\n\n\u2022 The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istriu(A) == isbanded(A,0,size(A,2)).", "date": "2014-12-28 17:35:44", "meta": {"domain": "mathworks.com", "url": "http://www.mathworks.com/help/matlab/ref/istriu.html?s_tid=gn_loc_drop&nocookie=true", "openwebmath_score": 0.7404126524925232, "openwebmath_perplexity": 2690.328343826514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241956308278, "lm_q2_score": 0.9019206857566126, "lm_q1q2_score": 0.8742535432438331}}
{"url": "http://math.stackexchange.com/questions/92393/generalized-fibonacci-sequences", "text": "# Generalized Fibonacci sequences\n\nWhy Fibonacci sequence start at $0$, Tribonacci sequence with $0,0$, Tetranacci with $0,0,0$, etc. [ref OEIS] Has any good reasons for that?\n\nThese sequences arise in generalization of Pascal Triangle as diagonal sums and there they start at $1$.\n\nPascal triangle:\n\n$$\\begin{array}{} \\color{red}1& \\color{blue}0& \\color{green}0& \\color{cyan}0&\\dots\\\\ \\color{blue}1& \\color{green}1& \\color{cyan}0& \\color{magenta}0&\\dots\\\\ \\color{green}1& \\color{cyan}2& \\color{magenta}1& 0&\\dots\\\\ \\color{cyan}1& \\color{magenta}3& 3& \\color{red}1&\\dots\\\\ \\vdots&\\vdots&\\vdots&\\vdots&\\ddots \\end{array}$$\n\ndiagonal sum gives $\\color{red}1,\\color{blue}1,\\color{green}2,\\color{cyan}3,\\color{magenta}5,8,\\color{red}{13}\\dots$ Fibonacci sequence\n\nFirst generalization:\n\n$$\\begin{array}{r} \\color{red}1& \\color{blue}0& \\color{green}0& \\color{cyan}0& \\color{magenta}0& 0&\\color{red}0&\\dots\\\\ \\color{blue}1& \\color{green}1& \\color{cyan}1& \\color{magenta}0& 0& \\color{red}0&\\color{blue}0&\\dots\\\\ \\color{green}1& \\color{cyan}2& \\color{magenta}3& 2& \\color{red}1& \\color{blue}0&\\color{green}0&\\dots\\\\ \\color{cyan}1& \\color{magenta}3& 6& \\color{red}7& \\color{blue}6& \\color{green}3&\\color{cyan}1&\\dots\\\\ \\color{magenta}1&4&\\color{red}{10}&\\color{blue}{16}&\\color{green}{19}&\\color{cyan}{16}&\\color{magenta}{10}&\\dots\\\\ \\vdots&\\vdots&\\vdots&\\vdots&\\vdots&\\vdots&\\vdots&\\ddots \\end{array}$$\n\ngives sequence of diagonal sum $\\color{red}1,\\color{blue}1,\\color{green}2,\\color{cyan}4,\\color{magenta}7,\\dots$, Tribonacci sequence, etc.\n\n-\nSorry, I'm slow: how are you computing your diagonal sums? \u2013\u00a0 Guess who it is. Dec 18 '11 at 1:29\nAlso, if you look here, I'd say it's pretty convenient to have the generalized Fibonacci numbers take on positive values for positive index. \u2013\u00a0 Guess who it is. Dec 18 '11 at 1:33\nThanks @Robert! I found this in the meantime. \u2013\u00a0 Guess who it is. Dec 18 '11 at 1:37\n@J.M.: I added some color to make it more obvious. \u2013\u00a0 Brian M. Scott Dec 18 '11 at 2:19\n@Peter: If $a(r,c)$ is the entry in row $r$, column $c$, $$a(r+1,c)=a(r,c)+a(r,c-1)+a(r,c-2)\\;.$$ There\u2019s an error in the table, which I\u2019ve now fixed. \u2013\u00a0 Brian M. Scott Dec 18 '11 at 10:04\n\nThe essential definition of a $k$-nacci sequence is that the earliest positive term is $1$ and all subsequent terms are the sum of the previous $k$ terms.\n\nSo to find the second earliest positive term (which is also $1$) you need at least $k-1$ zeros at the beginning of the sequence to do the sum. In your tables there are at least $k-1$ columns of implicit zeroes to the left of the columns you show.\n\nSince the Fibonacci sequence conventionally starts with $F(1)=1$ and $F(0)=0$, it is a plausible but not necessary convention to start the $k$-nacci sequence with the initial leading zero also at a zero index, and this is what the OEIS has done.\n\nYou could do something different, but then you would need to translate between your indices and those used by others. That is the effect of conventions.\n\n-\n\nWhile it looks natural to start Fibonacci at $f_1=1$, the Fibonacci actually goes both ways. Indeed\n\n$$f_{n+1}=f_n+f_{n-1} \\,,$$ implies\n\n$$f_{n-1}= f_{n+1}-f_n \\,.$$\n\nThus you can calculate $f_0$ and $f_{-n}$.\n\nGiven a recurrence which goes in both directions, any \"start\" point is somewhat artificial.\n\nI find the start $f_0=0$ useful when I work with the matrices\n\n$$F:=\\left( \\begin{array}{rr} 1 & 1\\\\ 1 & 0 \\end{array}\\right)=\\left(\\begin{array}{rr} f_2 & f_1\\\\ f_1 & f_0 \\end{array}\\right) \\,.$$\n\nNote that without starting with $f_0=0$, the standard formula\n\n$$F^n= \\left(\\begin{array}{rr} f_{n+1} & f_n\\\\ f_n & f_{n-1} \\end{array}\\right)$$\n\nwould not work for $n=1$. This would make some consequences weaker, since you'd need always the eliminate the case when the parameter(s) are 1. But this would be a not needed elimination.\n\nIt is somehow similar to the following situation: You have a function $f$ in the complex plane, and you can prove it is entire, but you only need the fact that it is Analytical in some region $R$. What do you do? Do you prove that it is Analytic in the region you need for the applications you are interested, or prove it is entire?\n\nWho knows, at some point in the future you might need that the function it is entire. You also might need to work with $f_0$, and I explained above why $f_n$ can be extended in an unique way below 0. Same can be said about the other sequences.\n\n-\n\nAs for \"Why Fibonacci sequence start at $0$?\" one compelling reason is that this indexing is the natural one that reveals their interesting divisibility properties, namely that they form a strong divisibility sequence, i.e.\n\n$$\\rm\\ gcd(f_m,f_n)\\ =\\ f_{\\:gcd(m,n)}$$\n\nhence $\\rm\\ m\\ |\\ n\\ \\Rightarrow\\ f_m\\ |\\ f_n\\$ etc. Analogous results hold for the more general class of Lucas-Lehmer sequences, which prove convenient when studying elementary number theory of quadratic fields, e.g. generalizations of Fermat's little theorem, Euler's $\\phi$ totient function, etc. If one changed the indexing then many of these results would be greatly obfuscated.\n\nThat said, it should be emphasized that such definitions are merely conventions that prove useful in the context at hand. In other contexts - where such divisibility properties play no role - another indexing might prove more convenient.\n\n-\nCan you be more specific about Lucas-Lehmer sequences as a general notion? The only somewhat authoritative mention I can find is in OEIS, but is is only one sequence, even though its title enigmatically uses the indefinite article. As far as I know the only thing really called after Lucas and Lehmer is the primality test (in which the above sequence occurs, and no other). Possibly you meant Lucas sequences? \u2013\u00a0 Marc van Leeuwen Dec 18 '11 at 8:02\n@Marc D.H. Lehmer developed an extended theory of Lucas sequences in his PhD thesis, which was published in Annals of Math., 1930 Due to such, many authors refer to the general class as Lucas-Lehmer sequences, e.g. see Ribenboim's The New Book of Prime Number Records, 1996. \u2013\u00a0 Bill Dubuque Dec 18 '11 at 15:07\n\nOnce you have also an expression, which allows interpolating to noninteger n (thus putting the sequence into a greater framework), then this might provide an argument to select a specific n as beginning of the sequence. For the Fibonacci-numbers I think the Binet-formula is\n$$\\small fib(n) = {((1+\\sqrt 5 )^n - (1-\\sqrt 5)^n) \\over 2^n \\sqrt 5 }$$ and then $\\small fib(0) = 0$. Surely one can insert an arbitrary offset for n into the defining expression, but the given definition may be taken as the most natural.\n\n-", "date": "2015-05-27 22:28:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/92393/generalized-fibonacci-sequences", "openwebmath_score": 0.8445354104042053, "openwebmath_perplexity": 476.2738541921384, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241965169938, "lm_q2_score": 0.9019206771886166, "lm_q1q2_score": 0.8742535357379186}}
{"url": "https://math.stackexchange.com/questions/816281/why-there-are-no-constant-functions-on-mathbbr-with-compact-support", "text": "# Why there are no constant functions on $\\mathbb{R}$ with compact support?\n\nFor some the question might seem trivial but the concept is new to me and I have been wondering why there are no constant functions on $$\\mathbb{R}$$ with compact support?\n\nFollowing wiki:\n\nDef.1) Functions with compact support on a topological space $$X$$ are those whose support is a compact subset of $$X$$.\n\nDef.2) $$\\operatorname{supp}f=:\\overline{\\{x\\in X\\;,f(x)\\neq 0\\}}$$\n\nWhat if we take such function $$f:\\mathbb{R}\\to\\mathbb{R}$$ defined as $$f(x)=0\\;,\\forall x\\in\\mathbb{R}$$ then $$\\operatorname{supp}f=\\overline{\\{x\\in\\mathbb{R}\\;,f(x)\\neq 0\\}}=\\overline{\\emptyset}=\\emptyset$$ but $$\\emptyset$$ is a compact subset of $$\\mathbb{R}$$ I think....\n\nCould someone enlighten me? Thank you.\n\n\u2022 Where were you told that no constant functions on $\\mathbb{R}$ have compact support? \u2013\u00a0Hayden May 31 '14 at 18:28\n\u2022 well, anyway that's the only one there is, $f \\equiv 0$ \u2013\u00a0mm-aops May 31 '14 at 18:30\n\u2022 @Hayden: here for example www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2030.pdf (12th line) \u2013\u00a0user124471 May 31 '14 at 18:32\n\u2022 Well, I'd beg to differ with the claim, for precisely the example you gave. But as mm-aops pointed out, it's the only example. \u2013\u00a0Hayden May 31 '14 at 18:35\n\u2022 the statement is very clear : ''there are NO functions'', (such functions don't exists). The real question here is if my counter-example is really correct (i.e that I followed all definitions correctly and so on). \u2013\u00a0user124471 May 31 '14 at 18:37\n\n## 1 Answer\n\nThe author in the article you link to (www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2030.pdf) is just being sloppy. He really should have said there are no non-zero constant functions with compact support on $\\mathbb R$ (and thus $H_c^0(\\mathbb R)$.\n\nSo yes, your example is correct. The empty set is (tautologically) compact.\n\n\u2022 Hi Fredrik. Thank you. Unfortunately it is not the first time I have seen it....Exactly the same statement I found in Bott, Tu, ''Differential forms in algebraic topology''. Cheers \u2013\u00a0user124471 May 31 '14 at 18:44\n\u2022 @user124471 Yep, I've seen it as well. In fact, I was just reading in the book by Bott & Tu a week ago, and had to think about the exact same question as you asked. \u2013\u00a0Fredrik Meyer May 31 '14 at 18:47\n\u2022 @user124471 I was wondering the same thing! Now I know. Can you include Bott Tu in your question? \u2013\u00a0user636532 May 29 '19 at 11:27", "date": "2020-01-17 19:30:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/816281/why-there-are-no-constant-functions-on-mathbbr-with-compact-support", "openwebmath_score": 0.7760120034217834, "openwebmath_perplexity": 391.05891677772047, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022630759019, "lm_q2_score": 0.894789457685656, "lm_q1q2_score": 0.8742295499234755}}
{"url": "https://math.stackexchange.com/questions/2576298/x-1-cdot-dots-cdot-x-n-1-implies-x-1-dots-x-n-ge-n", "text": "# $x_1 \\cdot \\dots \\cdot x_n=1 \\implies x_1 + \\dots + x_n \\ge n$\n\nSuppose $x_1, \\dots, x_n$ are positive real numbers such that $x_1 \\cdot \\ldots \\cdot x_n = 1$. Prove $x_1 + \\dots + x_n \\ge n$.\n\nI don't want to use the AM-GM inequality to prove this (from which this statement would follow).\n\nI suppose induction is the way to go. The base case is true. The case for $n=2$ is true by $$x_1 + x+2 - 2 \\sqrt{x_1 x_2} = (\\sqrt{x_1} - \\sqrt{x_n})^2 \\ge 0.$$\n\nNow suppose the statement holds for $n$, then $$x_1 \\cdot \\ldots \\cdot x_n = 1 \\implies x_1 + \\dots + x_n \\ge n.$$\n\nSo if $x_1 \\cdot \\ldots \\cdot (x_n x_{n+1}) = 1$, then $$x_1 + \\dots + x_n x_{n+1} \\ge n.$$\n\nThen we also have $$x_1 + \\dots + x_{n-1} + x_n + x_{n+1} \\ge n - x_{n}x_{n+1} + x_n + x_{n+1}.$$\n\nNow I'm not sure where else to go from here.\n\n\u2022 Seem similar in proving AM-GM by induction I think \u2013\u00a0Azlif Dec 21 '17 at 23:25\n\u2022 @MathematicianByMistake Induction certainly works, since this is just a special case of AM-GM which is proven by induction. \u2013\u00a0B. Mehta Dec 21 '17 at 23:29\n\nThus, it's enough to prove that $$n-x_nx_{n+1}+x_n+x_{n+1}\\geq n+1$$ or $$(x_n-1)(x_{n+1}-1)\\leq0,$$ which you can assume before.\n\nIndeed, since $\\prod\\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\\leq0.$\n\n\u2022 It's not entirely clear to me why this can be assumed. \u2013\u00a0B. Mehta Dec 21 '17 at 23:31\n\u2022 @B. Mehta Since $\\prod\\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\\leq0.$ \u2013\u00a0Michael Rozenberg Dec 21 '17 at 23:33\n\u2022 Ah, so there's a relabeling step added earlier. \u2013\u00a0B. Mehta Dec 21 '17 at 23:34\n\u2022 @B.Mehta I believe the logic is: there has to be at least one value greater-than-or-equal-to 1 and at least one less-than-or-equal-to 1. Rearrange so that these two values come last. \u2013\u00a0Addem Dec 21 '17 at 23:34\n\u2022 @MichaelRozenberg Thanks, very helpful. I would like to verify this last part with there needing to be a pair with one greater and one and one less. By contradiction, suppose for every $x_i$ and $x_j$, $(x_i -1)(x_j-1) > 0$, then either all pairs are greater than one or less than one which in either case would result in a product of the $x_k$'s being greater than 1 or less than one, respectively. \u2013\u00a0user330531 Dec 21 '17 at 23:42\n\nThe inequality follows from AM-GM, indeed: $\\;x_1 + \\dots + x_n \\ge n \\cdot \\sqrt[n]{x_1 \\cdot \\ldots \\cdot x_n}\\,$.\nBelow is a proof of AM-GM by induction, along a different idea than the ones posted already.\n\nLemma. \u00a0 $\\;(n-1)t^n+1 \\ge n t^{n-1}\\;$ for all $\\;t \\ge 1\\,$.\n\nLet $f\\,(t)=(n-1)t^n - n t^{n-1} + 1\\,$, then $\\,f(1) = 0\\,$ and $\\,f'(t)=n(n-1)t^{n-2}(t-1) \\ge 0\\,$, so $\\,f\\,$ is increasing from $\\,f(1)=0\\,$ for $\\,t \\ge 1\\,$ and therefore $\\,f(t) \\ge 0\\,$ for $\\,t \\ge 1\\,$.\n\n(The stronger statement is true that $\\,f(t) \\ge 0\\,$ for all $\\,t \\ge 0\\,$. That follows from the identity $\\,(n-1)t^n - n t^{n-1} + 1 = (t-1)^2 \\cdot \\sum_{k=1}^{n-1} k \\cdot t^{k-1}\\,$ which has an elementary no-calculus proof.)\n\nInduction step. \u00a0 The AM-GM inequality is homogeneous, so it can be assumed WLOG that the smallest $x_j=1\\,$. Assume again WLOG that $\\,j=n\\,$, then the inequality reduces to:\n\n$$x_1+x_2+ \\cdots + x_{n-1} + 1 \\ge n \\cdot \\sqrt[n]{x_1 \\cdot \\ldots \\cdot x_{n-1}} \\quad\\style{font-family:inherit}{\\text{with}}\\quad x_i \\ge 1, \\;i=1,2,\\ldots,n-1$$\n\nBy the induction hypothesis:\n\n$$\\,x_1+\\cdots+x_{n-1} \\ge (n-1) \\sqrt[n-1]{x_1\\cdot \\ldots \\cdot x_{n-1}} \\tag{1}$$\n\nAlso, it follows from the lemma with $\\,t = \\sqrt[n(n-1)]{x_1\\cdots x_{n-1}}\\ge 1\\,$ that: $$\\,(n-1)\\cdot\\sqrt[n-1]{x_1\\cdot \\ldots \\cdot x_{n-1}} + 1 \\ge n \\cdot \\sqrt[n]{x_1\\cdot \\ldots \\cdot x_{n-1}} \\tag{2}$$\n\nTherefore:\n\n$$x_1+ \\ldots +x_{n-1}+1 \\;\\stackrel{(1)}{\\ge}\\; (n-1)\\sqrt[n-1]{x_1\\cdot \\ldots \\cdot x_{n-1}} + 1 \\;\\stackrel{(2)}{\\ge}\\; n \\sqrt[n]{x_1 \\cdot \\ldots \\cdot x_{n-1}}$$", "date": "2019-09-17 23:18:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2576298/x-1-cdot-dots-cdot-x-n-1-implies-x-1-dots-x-n-ge-n", "openwebmath_score": 0.9568802714347839, "openwebmath_perplexity": 250.2843070682906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808736209154, "lm_q2_score": 0.8918110526265555, "lm_q1q2_score": 0.874225317773548}}
{"url": "https://math.stackexchange.com/questions/434275/what-does-it-mean-to-divide-a-complex-number-by-another-complex-number", "text": "# What does it mean to divide a complex number by another complex number?\n\nSuppose I have: $w=2+3i$ and $x=1+2i$. What does it really mean to divide $w$ by $x$?\n\nEDIT: I am sorry that I did not tell my question precisely. (What you all told me turned out to be already known facts!) I was trying to ask the geometric intuition behind the division of complex numbers.\n\n\u2022 Jul 2, 2013 at 6:23\n\u2022 suitcaseofdreams.net/Geometric_division.htm Jul 2, 2013 at 6:25\n\u2022 Perhaps a polar form might be a little more 'intuitive'? Jul 2, 2013 at 6:33\n\u2022 While multiplication/division of complex numbers can be interpreted geometrically, I don't think it is meant to be interpreted that way. Jul 2, 2013 at 6:40\n\u2022 @user1551 au contraire it is meant to be interpreted geometrically. Jul 2, 2013 at 16:40\n\nComplex numbers' multiplication is better understood if you forget the \"cartesian vector in the complex plane\" analogy: $$z = a + b i \\quad z \\in \\mathbb{C}, a, b \\in \\mathbb{R}$$ And in stead think in polar coordinates: $$z = r \\angle \\theta = r e^{i \\theta} \\quad z \\in \\mathbb{C}, r \\in \\mathbb{R}^+, \\theta \\in [0, 2 \\pi)$$ Wherein $r$ is the magnitude, $\\theta$ is the angle.\n\nWhen multiplying it is easy: $$z w = (r_z r_w) \\angle (\\theta _z + \\theta _w)$$ You add the angles and multiply the magnitudes.\n\nWhen dividing you do what comes naturally: $$\\frac z w = \\left(\\! \\frac{r_z}{r_w} \\!\\right) \\angle (\\theta _z - \\theta _w)$$ To divide means to find the difference in angles and the factor in magnitude.\n\nIt means to find another complex number $y$, such that $xy=w$. (Just as it does for real numbers.)\n\nWhen you divide $a$ by $b$, you're asking \"What do I multiply $b$ by in order to get $a$?\".\n\nMultiplying two complex numbers multiplies their magnitudes and adds their phases:\n\nSo when you divide a complex $a$ by a complex $b$ you are asking: \"How much do I need to scale $b$ and rotate $b$ in order to get $a$? Please give me a complex number with a magnitude equivalent to how much I must scale and a phase equivalent to how much I must rotate.\".\n\nExample\n\nConsider $\\frac{1 + i}{1 - i}$. How much do we need to scale and rotate $1-i$ in order to make it the same as $1+i$?\n\nWell, when graphed on the complex plane you can see that $1-i$ has a 45 degree clockwise rotation and a magnitude of $\\sqrt{2}$. $1+i$, on the other hand, has a 45 degree counter-clockwise rotation and the same magnitude of $\\sqrt{2}$.\n\nSince the magnitudes are the same, we don't need any scaling. Our result's magnitude will be 1.\n\nTo rotate from 45 degrees clockwise to 45 degrees counter-clockwise, we must rotate 90 degrees counter-clockwise. Thus our result will have a phase of 90 degrees counter-clockwise (which is upwards along the imaginary Y axis).\n\nMove a distance of 1 up the imaginary Y axis and you get the answer... $\\frac{1 + i}{1 - i} = i$. We can confirm this by doing the multiplication: $(1-i) \\cdot i = i+1$.\n\n\u2022 wow for the gif Jul 2, 2013 at 16:39\n\u2022 @Harsh I'm glad you like it. I made it (and a bunch of others) myself as part of an explanation of Grover's Quantum Search. Jul 3, 2013 at 21:32\n\u2022 man... I was thinking you were very resourceful, and had taken it from somewhere... Now, to learn that you MADE IT! Salute Jul 3, 2013 at 22:01\n\nSince multiplication can be nicely visualized as a rotation in the complex plane, you may find it helpful to think of division as a form of multiplication: $$\\frac{2+3i}{1+2i} = \\frac{2+3i}{1+2i}\\cdot \\frac{1-2i}{1-2i} = -\\frac{1}{3}\\cdot(2+3i)(1-2i)$$ So, instead of thinking about division, you can think of it as multiplying by the conjugate.\n\nGeometrically, it means that the magnitudes get divided, and the angles get subtracted.\n\nThat is, imagine the complex numbers plotted in polar form.\n\nie: if\n\n$w = r_1 e ^ {i \\theta_1 }$\n\n$x = r_2 e ^ {i \\theta_2 }$\n\nthen\n\n$w/x = (r_1 / r_2)e^{i(\\theta_1 - \\theta_2)}$\n\nThere are multiple ways by which you can describe this relation;\n\nSince you have asked to give geometric representation so phasor form must be better to understand.\n\nDividing two complex numbers means to take the complex number that is of the magnitude equal to that of the division of amplitudes of the $$X$$ and $$Y$$ complex number and the phase of the new generated complex number is actually the difference of the phase between them.\n\nEg. $$X = 2+3i \\ , \\ Y = 9+3i$$\n\nso the $$|Z|= \\frac{ |X|}{|Y|} = \\dfrac{\\sqrt{2^2+3^2}}{\\sqrt{9^2+3^2}}$$ and $$\\arg(Z)= \\tan^{-1}\\left(\\frac32\\right) - \\tan^{-1}\\left(\\frac 39 \\right)$$\n\nThis $$Z$$ is also a complex number with amplitude and phase in geometric way.. but you to visualize the complex imaginary axis and real axis :D\n\nSince $\\frac wx = \\frac{w\\cdot x^*}{|x|^2}$, complex division is basically equivalent to multiplication of $w$ and the complex conjugate of $x$, but rescaled by the squared absolute value $|x|^2$.", "date": "2022-05-23 03:08:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/434275/what-does-it-mean-to-divide-a-complex-number-by-another-complex-number", "openwebmath_score": 0.7623132467269897, "openwebmath_perplexity": 266.66544783136675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419674366166, "lm_q2_score": 0.8840392939666335, "lm_q1q2_score": 0.8741751547372433}}
{"url": "https://gmatclub.com/forum/rapid-math-251963.html", "text": "It is currently 21 Mar 2018, 04:05\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Rapid Math\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nStatus: Enjoying the Journey\nAffiliations: ND\nJoined: 26 Sep 2017\nPosts: 102\nSchools: Rotman '21\nWE: Marketing (Consulting)\n\n### Show Tags\n\n23 Oct 2017, 01:32\n4\nKUDOS\n1\nThis post was\nBOOKMARKED\nHi Everyone,\n\nAlthough GMAT doesn't test your ability to perform complicated calculations, it would be beneficial to know some calculation tricks that would save you some time.\n\nTrick 7: Rapidly Square Any Number Ending in 5\n\nMultiply the tens digit by the next whole number then affix the number 25\n$$65^2: 6*7=42 ==> 'affix' 25 ==> 4225$$\n\nTrick 8: Rapidly Multiply Any Two-Digit Number by 11\n\nWrite the number leaving some space between the two digits, then insert the sum of the number's two digit in between.\n\n24*11 ==> 2 ? 4 ==> 2+4=6 ==> 264\n\nTrick 9: Rapidly Multiply by 25 (or 0.25, 2.5, 250, etc.)\n\nDivide the number by 4 and affix or insert Zeroes or decimal points\nExample: 28*25 ==> 28/4 = 7 ==> 700 (add 2 Zeroes)\n\nTrick 10: Rapidly Divide by 25 (or 0.25, 2.5, 250, etc.)\n\nRemove Zeroes or decimal points then multiply the number by 4\nExample: 700/25 ==> 7/25 ==> 7*4=28\n\nTrick 24: Rapidly Square Any Two-Digit Number ending in 1\n\n$$21^2 ==> 20^2=400 ==> Add: 20+21=41 ==> Add: 400+41= 441 ==> 21^2 = 441$$\n\nTrick 59: Rapidly Multiply Any Three-Digit Number by 11 (advanced variation of trick 8)\n\nWrite the number leaving some space between the two digits, then insert the summations of the numbers in between (summations of 2 digits everytime; as shown below).\n\nExample:\n\n342*11=??\n342*11= 3--2 (write the first & the last number \"3\"& \"2\")\n342*11=3-62 (the sum of 4+2)\n342*11=3762 (the sum of 3+4)\n\nSource: Rapid Math Tricks & Tips. EDWARD H. JULIUS\n_________________\n\n\"Giving kudos\" is a decent way to say \"Thanks\" and motivate contributors. Please use them, it won't cost you anything\n\nHigh achievement always takes place in the framework of high expectation Charles Kettering\nIf we chase perfection we can catch excellence Vince Lombardi\n\nGMAT Club Live: 5 Principles for Fast Math: https://gmatclub.com/forum/gmat-club-live-5-principles-for-fast-math-251028.html#p1940045\nThe Best SC strategies - Amazing 4 videos by Veritas: https://gmatclub.com/forum/the-best-sc-strategies-amazing-4-videos-by-veritas-250377.html#p1934575\n\nLast edited by NDND on 12 Nov 2017, 04:48, edited 5 times in total.\nIntern\nJoined: 07 May 2017\nPosts: 36\nLocation: Indonesia\nGPA: 3.33\n\n### Show Tags\n\n23 Oct 2017, 01:47\nThanx. It will help\n\nSent from my Redmi Note 3 using GMAT Club Forum mobile app\n_________________\n\nConfidence , Commitment, and Consistence\n\nManager\nStatus: Enjoying the Journey\nAffiliations: ND\nJoined: 26 Sep 2017\nPosts: 102\nSchools: Rotman '21\nWE: Marketing (Consulting)\n\n### Show Tags\n\n01 Nov 2017, 07:43\nUpdated with Tricks 9 & 10\n_________________\n\n\"Giving kudos\" is a decent way to say \"Thanks\" and motivate contributors. Please use them, it won't cost you anything\n\nHigh achievement always takes place in the framework of high expectation Charles Kettering\nIf we chase perfection we can catch excellence Vince Lombardi\n\nGMAT Club Live: 5 Principles for Fast Math: https://gmatclub.com/forum/gmat-club-live-5-principles-for-fast-math-251028.html#p1940045\nThe Best SC strategies - Amazing 4 videos by Veritas: https://gmatclub.com/forum/the-best-sc-strategies-amazing-4-videos-by-veritas-250377.html#p1934575\n\nRe: Rapid Math \u00a0 [#permalink] 01 Nov 2017, 07:43\nDisplay posts from previous: Sort by", "date": "2018-03-21 11:05:45", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/rapid-math-251963.html", "openwebmath_score": 0.4684756398200989, "openwebmath_perplexity": 12676.80778048119, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9481545362802363, "lm_q2_score": 0.9219218316372044, "lm_q1q2_score": 0.8741243667625996}}
{"url": "https://math.stackexchange.com/questions/781365/find-the-parametric-equation-to-the-curve", "text": "# Find the parametric equation to the curve\n\nFind the parametric equation for the curve.\n\n$$x^{2}+y^{2}=10$$\n\nI haven't learned parametric equations fully yet, so I wanted to check with you guys and see if you can confirm if I'm doing this correctly and possibly go more in depth on the problem if you can?\n\nBecause it's centered at (0,0) the origin, and it has a radius of sqrt(10) then the answer is this, right?\n\n$$(x(t),y(t)) = (\\sqrt{10}\\cos t\\,,\\, \\sqrt{10}\\sin t)$$\n\n\u2022 Yes, it is...and you should probably remark that $\\;0\\le t\\le 2\\pi\\;$ \u2013\u00a0DonAntonio May 4 '14 at 21:15\n\ntaking $x=a\\sin t,y=a\\cos t$ from $x^2+y^2=10$ we get $$a^2\\sin^2t+a^2\\cos^2t=a^2(\\sin^2t+\\cos^2t)=a^2=10$$ from above $a=\\sqrt{10}$\nYou are correct. Here's an easy check: note that the position vector at any point is given by $\\langle\\sqrt{10}\\cos(t), \\sqrt{10}\\sin(t) \\rangle$. If we take the magnitude of this vector, we get:\n$$\\sqrt{10\\cos^2{t} + 10\\sin^2{t}} = \\sqrt{10}$$\nAnd therefore the curve is a constant $\\sqrt{10}$ distance from the origin, so it must be at least a part of a circle centered at the origin with radius $\\sqrt{10}$. To make the argument that it's a complete circle, you can use a bit of calculus to verify that the curve never turns back on itself. To do this, take a derivative of the position function to yield the velocity function, and note that it is never $0$.", "date": "2019-10-20 22:28:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/781365/find-the-parametric-equation-to-the-curve", "openwebmath_score": 0.9507486820220947, "openwebmath_perplexity": 118.59235727129308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.990587409856637, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.8741219392107562}}
{"url": "https://math.stackexchange.com/questions/853615/proving-an-expression-is-composite", "text": "# Proving an expression is composite\n\nI am trying to prove that $n^4 + 4^n$ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.\n\nThis problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.\n\nI would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.\n\n\u2022 Write it as a difference of squares. \u2013\u00a0Adam Hughes Jul 1 '14 at 19:54\n\u2022 How? It's $n^4 + 4^n$ not $n^4 - 4^n$ \u2013\u00a0Mathmo123 Jul 1 '14 at 19:55\n\u2022 @Mathmo123: see my answer below. \u2013\u00a0Adam Hughes Jul 1 '14 at 20:03\n\n$(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...\n\n\u2022 Ah. Very neat!! \u2013\u00a0Mathmo123 Jul 1 '14 at 20:05\n\u2022 Originally, I was not sure how I should go about writing it as a difference of squares. Thanks for your help. \u2013\u00a0pidude Jul 1 '14 at 20:12\n\nHint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?\n\nLet me know if you need further hints.\n\n\u2022 I originally tried doing that but to no avail. Evaluated at 3 I obtain 145 which has factors of 5 and 29. At 5, the expression equals 1649, which has factors of 17 and 97. I will keep looking for a pattern and let you know if I need more hints. Thanks for your help. \u2013\u00a0pidude Jul 1 '14 at 19:59\n\u2022 Ah... using this method, there will be a difference between multiples of 5 and other odd numbers. You will find that for odd numbers that are not a multiple of 5, 5 will be a divisor \u2013\u00a0Mathmo123 Jul 1 '14 at 20:01\n\u2022 Ok, so I was able to prove that. Now I'm working numbers which are multiples of 5. \u2013\u00a0pidude Jul 1 '14 at 20:09\n\nI am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.\n\n\u2022 We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by\n1. $n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law\n2. Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication\n3. $(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression\n4. $(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression\n5. $(n^2+2^n)^2-2(n^2)(2^n)$ from step 2\n6. $(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential\n\u2022 We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again\n1. Treat $n^2+2^n$ as $a$\n2. Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer\n3. So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$\n4. $2^{2m}n^2=(n2^m)^2$ associative law\n5. Treat $n2^m$ as $b$\n\u2022 It implies that both $a$ and $b$ are both positive integer\n\u2022 From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$\n\u2022 Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$\n\nSome interesting factorizations of a polynomial of type $x^4+\\text{const}$: $$x^4+4=(x^2+2x+2)(x^2-2x-2) \\tag{1}$$\n\n$$x^4+1=(x^2+\\sqrt[]{2}x+1)(x^2-\\sqrt[]{2}x+1) \\tag{2}$$\n\nSo one can ask, how to select the coefficients $a,b,c,d$ in\n\n$$(x^2+ax+b)(x^2+cx+d) \\tag{3}$$\n\nsuch that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.\n\nIf we expand $(3)$ we get\n\n$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$\n\nAnd the coefficients disappear, if\n\n$$\\begin{eqnarray} c+a &=& 0 \\\\ d+ ac +b &=& 0 \\\\ ad+bc &=& 0 \\end{eqnarray}$$\n\nWhen solving for $b,c,d$ we get\n\n$$\\begin{eqnarray} c &=& -a \\\\ b &=& \\frac{a^2}{2} \\\\ d &=& \\frac{a^2}{2} \\end{eqnarray}$$\n\nand therefore\n\n$$x^4+\\frac{a^4}{4} = (x^2+a x+\\frac{a^2}{2})(x^2-ax+\\frac{a^2}{2})$$\n\nFor $a=2$ this gives $(1)$, $a=\\sqrt[]{2}$ this gives $82)$ . Substituting $a=2^{t+1}$ we get\n\n$$x^4+4^{2t+1} = (x^2+2\\cdot 2^t x+2^{2t+1})(x^2-2\\cdot 2^tx+2^{2t+1})$$\n\nSubstituting $x=n=2t+1$ gives the required result for odd $n$.", "date": "2020-07-07 13:41:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/853615/proving-an-expression-is-composite", "openwebmath_score": 0.8389080762863159, "openwebmath_perplexity": 272.7748211115131, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795102691454, "lm_q2_score": 0.8856314617436728, "lm_q1q2_score": 0.8741001063907176}}
{"url": "https://www.cnidaria.nat.uni-erlangen.de/shortcourse/S4/Assignment_Kendall_Tau_KEY.html", "text": "Write a function that computes Kendall rank coefficient Tau (see lecture notes for additional details on Kendall Tau).\n\nKendall Coefficient {as quoted from Wikipedia):\n\nLet (x1, y1), (x2, y2), ., (xn, yn) be a set of observations of the joint random variables X and Y respectively, such that all the values of xi and yi are unique. Any pair of observations (xi and yi) are said to be concordant if the ranks for both elements (more precisely, the sort order by x and by y) agree: that is, if both xi > xj and yi > yi or if both xi < xj and yi < yj. They are said to be discordant, if xi > xj and yi < yj or if xi < xj and yi > yj. If xi = xj or yi = yj, the pair is neither concordant nor discordant.\n\nEquation 1\n\nTAU=(number of concordant pairs-number of discordant pairs)/(0.5 x n x (n - 1))\n\nThe above equations ignore ties (sets of pairs where either xi = xj or yi = yj). One way of correcting for ties is to consider how many ties (or non-ties) are present in the data. This variant known as Kendall\u2019s tau_b is computed by modifying the denominator 0.5n(n-1) in Equation 1 to consider ties:\n\nsqrt(mx) x sqrt(my)\n\nWhere mx is number of non-ties for x and my is number of non-ties for y.\n\nNote: The R function {cor} returns tau_b.\n\nYOUR TASKS: 1. Write a custom function that computes Kendall ignoring ties (Equation 1) and correcting for ties (as explained above).\n\n1. Your function should return a single value (Kendall\u2019s coefficient)\n\n2. Your function should allow the user to choose to ignore ties or not and thus produce either tau or tau_b.\n\n3. Apply the function to two vectors provided below and check the results provided by your function against those obtained using the R function {cor}.\n\n4. Check how fast is your function relative to {cor} function using microbenchmark function from package \u2018microbenchmark\u2019.\n\n5. Finally, generate large vectors x and y (n = 1000) from normal distribution using rnorm function. Check again how fast is your function relative to {cor} function using microbenchmark function from package \u2018microbenchmark\u2019.\n\nTo carry out this exercise, please start by uploading the library \u2018microbenchmark\u2019 and please use x and y variables as defined below. This will ensure that all results are the same for all of us. It will also help you to evaluate if your outcomes agree with those provided in solution functions or alternative R functions.\n\nlibrary(microbenchmark)\nx <- c(-30, 0, 14, 5, 6, 7)\ny <- c(-5, 0, 12, 16, 16, 8)\n\nOnce your function is ready, you can continue to the next step. The function solutions included in this tutorial are hidden here, but can be reviewed in the .rmd file or in the answer key file (.html). Please try to solve this on your own before checking the solution.\n\nmy.kend <- function(x, y, ties=T) {\nn <- length(x)\nnzero <- 0.5*n*(n-1)\ns1 <- sign(x - t(matrix(x, n, n, byrow=F)))\ns2 <- sign(y - t(matrix(y, n, n, byrow=F)))\nt1 <- sum(s1[lower.tri(s1)]!=0)\nt2 <- sum(s2[lower.tri(s2)]!=0)\nn2 <- t1^0.5*t2^0.5\nif (!ties) tau <- sum(s1[lower.tri(s1)]*s2[lower.tri(s2)])/nzero\nif (ties) tau <- sum(s1[lower.tri(s1)]*s2[lower.tri(s2)])/n2\nreturn(tau)\n}\n\nmy.kend2 <- function(x, y, ties=T) {\nn <- length(x)\nnzero <- 0.5*n*(n-1)\nk <- 0\ns1 <- 0\ns2 <- 0\nout <- vector(mode='numeric', length=nzero)\nfor (i in 1:length(x)) {\nfor (j in 1:length(x))  {\nif (j > i) {\nk <- k + 1\nout[k] <- sign(x[i] - x[j])*sign(y[i] - y[j])\nif(x[i] - x[j]==0) s1 = s1 + 1\nif(y[i] - y[j]==0) s2 = s2 + 1\n}\n}\n}\nif (!ties) tau <- sum(out)/nzero\nif (ties) tau <- sum(out)/((nzero-s1)^0.5*sum(nzero-s2)^0.5)\ntau\n}\n\nNow check if your function yields comparable results to those provided by standard R functions or custom functions developed by your instructor.\n\nx <- c(-30, 0, 14, 5, 6, 7)\ny <- c(-5, 0, 12, 16, 16, 8)\n\n# compute kendall tau using generic function {cor}\ncor(x, y, method='kendall')\n## [1] 0.4140393\n# check if your function is consistent with standard cor function\nmy.kend(x, y, ties=T)       \n## [1] 0.4140393\n# if you developed a second solution check if that solution works as well.\nmy.kend2(x, y, ties=T)      # check if your function works as well\n## [1] 0.4140393\n# now check if your function(s) also compute(s) uncorrected tau correctly\nmy.kend(x, y, ties=F)       \n## [1] 0.4\nmy.kend2(x, y, ties=F)      \n## [1] 0.4\n\nOnce you established that your function works, you should investigate if your function is faster or slower than cor function. Keep in mind, however, that the performance of the function may depend on sample size, so your function may work better than \u2018cor\u2019 function for small sample sizes, but perform poorly for large sample sizes (or vice versa).\n\nmicrobenchmark(cor(x, y, method='kendall'), my.kend(x, y, ties=T),\nmy.kend2(x, y, ties=T))  # which is faster for a much larger dataset?\n## Unit: microseconds\n##                           expr    min      lq     mean  median       uq\n##  cor(x, y, method = \"kendall\") 65.700 80.0480 95.31707 95.1510 109.6875\n##        my.kend(x, y, ties = T) 23.033 26.2425 36.04066 35.6815  42.2900\n##       my.kend2(x, y, ties = T) 10.195 10.9500 15.45499 12.0830  24.5435\n##      max neval\n##  215.976   100\n##   70.986   100\n##   40.780   100\n\nYou can see that my custom functions my.kend and my.kend2 are much faster than cor function when analyzed data are tiny.\n\nLet\u2019s now check how the functions perform for much larger datasets.\n\nx <- rnorm(500,1,1)\ny <- rnorm(500,1,3) + x\ncor(x, y, method='kendall')\n## [1] 0.19301\nmy.kend(x, y, ties=T)                                               # check if your function agrees\n## [1] 0.19301\nmy.kend2(x, y, ties=T)                                               # check if your function agrees\n## [1] 0.19301\nmicrobenchmark(cor(x, y, method='kendall'), my.kend(x, y, ties=T),\nmy.kend2(x, y, ties=T))  # which is faster for a much larger dataset?\n## Unit: milliseconds\n##                           expr       min        lq      mean    median\n##  cor(x, y, method = \"kendall\")  3.233603  3.529627  4.097588  3.924766\n##        my.kend(x, y, ties = T) 13.880630 18.497122 21.550797 20.513782\n##       my.kend2(x, y, ties = T) 50.971912 58.680038 64.747797 62.303116\n##         uq      max neval\n##   4.657272  5.69241   100\n##  23.238596 52.78090   100\n##  69.767135 94.52662   100\n\nAs clear from the report, the custom-written solutions that I wrote (you may come up with better algorithms) perform well at small sample size, but underperform at larger sample sizes. That is, compiled, professionally written R functions are faster for large samples than custom written solutions written by amatuers. This means that when we iterate small datasets, custom functions can be faster. When huge datasets are processed, R functions are likley to be superior to custom solutions.\n\nLet us evaluate this issue here by using \u2018cor\u2019 and our custom function to carry out the same analysis.\n\nr1 <- as.data.frame(matrix(rnorm(20000), 10, 2000))\n# NOTE r1 is a dataframe of many (s=2000) tiny samples (n=10 observations). Each column is a sample.\nr2 <- as.data.frame(matrix(rnorm(20000), 2000, 10))\n# NOTE r2 is a dataframe of s=10 huge samples (n=2000 observations). Each column is a sample.\n\nsystem.time(sapply(r1, function(x) cor(x, r1[,1], method='kendall')))\n##    user  system elapsed\n##    0.16    0.03    0.19\nsystem.time(sapply(r1, function(x) my.kend(x, r1[,1])))\n##    user  system elapsed\n##    0.09    0.00    0.09\nsystem.time(sapply(r1, function(x) my.kend2(x, r1[,1])))\n##    user  system elapsed\n##    0.09    0.00    0.09\nsystem.time(sapply(r2, function(x) cor(x, r2[,1], method='kendall')))\n##    user  system elapsed\n##    0.59    0.00    0.60\nsystem.time(sapply(r2, function(x) my.kend(x, r2[,1])))\n##    user  system elapsed\n##    3.92    1.01    4.93\nsystem.time(sapply(r2, function(x) my.kend2(x, r2[,1])))\n##    user  system elapsed\n##    9.39    0.03    9.42\n\nClearly, my functions worked faster when we processed many, many short vectors (data frame r1), but underperformed badly when a few very large vectors were evaluated (data frame r2).\n\nAlways cite packages.\n\ncitation('microbenchmark')\n##\n## To cite package 'microbenchmark' in publications use:\n##\n##   Olaf Mersmann (2018). microbenchmark: Accurate Timing Functions.\n##   R package version 1.4-4.\n##   https://CRAN.R-project.org/package=microbenchmark\n##\n## A BibTeX entry for LaTeX users is\n##\n##   @Manual{,\n##     title = {microbenchmark: Accurate Timing Functions},\n##     author = {Olaf Mersmann},\n##     year = {2018},\n##     note = {R package version 1.4-4},\n##     url = {https://CRAN.R-project.org/package=microbenchmark},\n##   }", "date": "2019-10-16 01:49:45", "meta": {"domain": "uni-erlangen.de", "url": "https://www.cnidaria.nat.uni-erlangen.de/shortcourse/S4/Assignment_Kendall_Tau_KEY.html", "openwebmath_score": 0.4371679127216339, "openwebmath_perplexity": 8041.880659191196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018362008347, "lm_q2_score": 0.8962513648201267, "lm_q1q2_score": 0.8740259766700917}}
{"url": "http://mathhelpforum.com/advanced-statistics/103904-conditional-probability-brain-teaser.html", "text": "# Thread: Conditional Probability Brain Teaser\n\n1. ## Conditional Probability Brain Teaser\n\nA sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.\n\nWhat is the probability that the marble Tom put in the sock was red?\n\nI've been trying to approach this with Bayes' Theorem but not making any progress:\n\nP(A l B) = P(B l A) * P(A) / P(B)\n\nI'm confused as to what A and B would represent in the problem.\n\nAppreciate the help, thanks!\n\n2. Originally Posted by Penguins\nA sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.\n\nWhat is the probability that the marble Tom put in the sock was red?\n\nI've been trying to approach this with Bayes' Theorem but not making any progress:\n\nP(A l B) = P(B l A) * P(A) / P(B)\n\nI'm confused as to what A and B would represent in the problem.\n\nAppreciate the help, thanks!\nLet the number of blue marbles in the sock be x.\n\nDraw a tree diagram. The first two branches are the events Tom putting in a blue marble or a red marble. The next branches are the events of pulling out a red marble or a blue marble.\n\nI get $\\frac{\\frac{3}{2(3 + x)}}{\\frac{3}{2(3 + x)} + \\frac{1}{3 + x}} = \\frac{3}{5}$.\n\n3. I don't understand how you set up the problem like you did. What do the terms in your numerator and denominator represent?\n\nHere is what I get:\n\nP(A l B) = (P(B l A) * P(A)) / P(B)\n\nA is Tom putting a red marble in the sock.\nB is Jason taking a red marble out of the sock.\n\nP (B l A) = 3 / (x+3)\nP (A) = unknown, let it be p\nP (B) = p*(3/(x+3)) + (1-p)*(2/(x+2))\n\nSo\n\nP(A l B) = (P(B l A) * P(A)) / P(B) = (3p/(x+3)) / (p*(3/(x+3)) + (1-p)*(2/(x+2)))\n\nI end up with 3/2 p. I know I did something wrong, but I'm not sure what.\n\n4. Hello, Penguins!\n\nI agree with Mr. F\n\nA sock contains 2 red marbles and an unknown number of blue marbles.\nTom places a new marble in the sock.\nJason puts his hand in the sock and pulls out a red marble.\n\nWhat is the probability that the marble Tom put in the sock was red?\nWe want: . $P(\\text{add Red }|\\text{ draw Red}) \\;=\\;\\frac{P(\\text{add Red }\\wedge\\text{ draw Red})}{P(\\text{draw Red})}$ .[1]\n\nThe sock contains: 2 Red and $b$ Blue marbles.\n\nThere are two cases to consider:\n\n[1] A Red marble is added.\n. . .The sock contains: 3 Red and $b$ Blue.\n. . . . Then: . $P(\\text{add Red }\\wedge\\text{ draw Red}) \\:=\\:\\frac{3}{b+3}$ .[2]\n\n[2] A Blue marble is added.\n. . .The sock contains: 2 Red and $b+1$ Blue.\n. . . . Then: . $P(\\text{add Blue }\\wedge\\text{ draw Red}) \\:=\\:\\frac{2}{b+3}$\n\nHence: . $P(\\text{draw Red}) \\;=\\;\\frac{3}{b+3} + \\frac{2}{b+3} \\;=\\;\\frac{5}{b+3}$ .[3]\n\nSubstitute [2] and [3] into [1]: . $P(\\text{add Red }|\\text{ draw Red}) \\;\\;=\\;\\; \\frac{\\dfrac{3}{b+3}} {\\dfrac{5}{b+3}} \\;\\;=\\;\\;\\frac{3}{5}$\n\n5. Originally Posted by Soroban\nHello, Penguins!\n\nI agree with Mr. F\n\nWe want: . $P(\\text{add Red }|\\text{ draw Red}) \\;=\\;\\frac{P(\\text{add Red }\\wedge\\text{ draw Red})}{P(\\text{draw Red})}$ .[1]\n\nThe sock contains: 2 Red and $b$ Blue marbles.\n\nThere are two cases to consider:\n\n[1] A Red marble is added.\n. . .The sock contains: 3 Red and $b$ Blue.\n. . . . Then: . $P(\\text{add Red }\\wedge\\text{ draw Red}) \\:=\\:\\frac{3}{b+3}$ .[2]\n\n[2] A Blue marble is added.\n. . .The sock contains: 2 Red and $b+1$ Blue.\n. . . . Then: . $P(\\text{add Blue }\\wedge\\text{ draw Red}) \\:=\\:\\frac{2}{b+3}$\n\nHence: . $P(\\text{draw Red}) \\;=\\;\\frac{3}{b+3} + \\frac{2}{b+3} \\;=\\;\\frac{5}{b+3}$ .[3]\n\nSubstitute [2] and [3] into [1]: . $P(\\text{add Red }|\\text{ draw Red}) \\;\\;=\\;\\; \\frac{\\dfrac{3}{b+3}} {\\dfrac{5}{b+3}} \\;\\;=\\;\\;\\frac{3}{5}$\n\nI see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.\n\nI'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?\n\nThanks for the replies!\n\n6. Originally Posted by Penguins\nI see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.\n\nI'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?\n\nThanks for the replies!\nIf you're expected to get a numerical answer then you have to make an assumption regarding a priori knowledge. The best one (for various reasons which I won't get into here) is a uniform prior distribution.", "date": "2017-01-16 19:01:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/103904-conditional-probability-brain-teaser.html", "openwebmath_score": 0.8937616348266602, "openwebmath_perplexity": 959.7684649458447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846684978779, "lm_q2_score": 0.893309416705815, "lm_q1q2_score": 0.8740002375297514}}
{"url": "https://math.stackexchange.com/questions/1481433/1-biased-coin-and-1-fair-coin-probability-of-2-heads", "text": "# 1 Biased Coin and 1 Fair Coin, probability of 2 Heads?\n\nYou have 1 fair coin and 1 coin with 2 heads. Given that the first flip was a heads what is the probability of getting another heads?\n\nMy Answer: P(2H|F=H) = P(2H|F=H, Biased Coin)*P(Biased Coin) + P(2H|F=H, Unbiased Coin)*P(Unbiased Coin) = 0.5 + 0.25 = 0.75. In my equation, F refers to the First Throw. But the answer is supposed to be 5/6 and I can't seem to understand how.\n\nEdit: From Arthurs comment I get the following, however, I dont know if this is the correct method, despite getting the correct answer:\n\nP(Biased|F=H) = 2/3.\n\nP(2H|F=H) = P(2H|(Biased|F=H))*P(Biased|F=H) + P(2H|(Unbiased|F=H))*P(Unbiased|F=H) = (1*2/3) + (1/2 * 2/3) = 5/6.\n\nThank You\n\n\u2022 Is it random (and unknown) which coin was flipped first? \u2013\u00a0Arthur Oct 15 '15 at 12:41\n\u2022 @Arthur Yes. There is no information about that. \u2013\u00a0Jojo Oct 15 '15 at 12:42\n\u2022 I think this is the biggest thing you've missed: Given that the first toss was a head, what is the proability that the first coin was the biased one? \u2013\u00a0Arthur Oct 15 '15 at 12:46\n\u2022 @Arthur Could you please check the edit to my question. \u2013\u00a0Jojo Oct 15 '15 at 13:08\n\u2022 @Jojo, yes, although that should be $P(2H\\mid\\text{ Biased} \\cap F=H)$ and so forth. \u2013\u00a0Graham Kemp Oct 15 '15 at 13:11\n\nThe first flip was a head.\n\nThe $3$ heads have equal probabilities to be the head that appeared at the first flip.\n\n$2$ of the $3$ heads have another head as other side.\n\n$1$ of the $3$ heads has tail as other side.\n\nSo there is a chance of $\\frac23.1+\\frac13.\\frac12=\\frac56$ of throwing a second head with that coin.\n\n\u2022 Very intuitive answer. Thanks. \u2013\u00a0Jojo Oct 15 '15 at 13:22\n\u2022 Glad to help, and indeed cherish your intuition in maths. \u2013\u00a0drhab Oct 15 '15 at 13:24\n\u2022 Does this assume that the same coin is flipped twice, or that one coin is flipped and then the other? I'm getting a different answer when I assume that latter, but the question is ambiguous. \u2013\u00a0Bill the Lizard Oct 15 '15 at 14:56\n\u2022 It assumes that the same coin was flipped twice. \u2013\u00a0drhab Oct 15 '15 at 15:00\n\nYou missed a conditioning in your formula. Let the event of choosing biased coin be B and fair be F. Getting heads on second toss be 2H and on first toss 1H. You want to calculate the probability of getting heads on second throw given that you it landed heads on first throw, which is\n\nP(2H/1H) = P(2H/1H,F)P(F/1H) + P(2H/1H,B)P(B/1H)\n\nTo evaluate P(F/1H) and P(B/1H) use bayes rue.\n\nP(F/1H) = P(1H/F)*P(F)/P(1H) = 0.5*0.5/(0.5*0.5 + 1*0.5) = 1/3\n\nP(B/1H) = 2/3\n\nP(1H) = P(1H/F)P(F) + P(1H/B)P(B)\n\nP(2H/1H,F) = 0.5 and P(2H/1H,B) = 1 Therefore, P(2H/1H) = 0.5*(1/3) + 1*(2/3) = 5/6", "date": "2019-06-16 06:29:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1481433/1-biased-coin-and-1-fair-coin-probability-of-2-heads", "openwebmath_score": 0.802986204624176, "openwebmath_perplexity": 676.3390055852828, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846703886662, "lm_q2_score": 0.8933094081846421, "lm_q1q2_score": 0.8740002308818255}}
{"url": "https://math.stackexchange.com/questions/1644943/f-has-a-zero-integral-on-every-measurable-set-prove-f-is-zero-almost-everyw", "text": "$f$ has a zero integral on every measurable set. Prove $f$ is zero almost everywhere\n\nI am trying to solve the following exercise:\n\nLet $f$ be integrable. Assume that $\\int_A f d\\mu = 0$ for every measurable set $A$. Prove that $f = 0$ a.e. [$\\mu$].\n\nI have the following proof but it seems to me too simple to be true. What is wrong with it?\n\nFor any $a>0$, define $W_a:=\\{w|f(w)\\geq a\\}$. Now we have: $$\\int_{W_a} fd\\mu = 0 \\geq \\int_{W_a} ad\\mu=a\\mu(W_a)$$ and therefore $\\mu(W_a)=0$. Same holds for the negative $a$'s. This completes my proof.\n\n\u2022 You might want to add why it suffices to show $\\mu(W_a) = 0$ for $a>0$ and $\\mu(f \\leq a)=0$ for $a<0$. \u2013\u00a0saz Feb 7 '16 at 19:58\n\nIndeed, your proof shows that $\\{f > 0\\}$ has measure zero. Similarly, you show $\\{f < 0\\}$ has measure zero, and hence $f=0$ a.s. Yes it really is this easy.\nIt remains to show that the set $\\{w\\mid f(w)\\ne0\\}$ has measure $0$. Consider the nonzero rational $a$'s to complete the proof.", "date": "2019-12-16 13:33:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1644943/f-has-a-zero-integral-on-every-measurable-set-prove-f-is-zero-almost-everyw", "openwebmath_score": 0.9828101396560669, "openwebmath_perplexity": 92.84510566242575, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846647163009, "lm_q2_score": 0.8933094131553264, "lm_q1q2_score": 0.8740002306778896}}
{"url": "https://math.stackexchange.com/questions/1397679/surprising-summation-1-sum-i-1nn-i12i-1-sum-i-1n-i2", "text": "# Surprising Summation (1) $\\sum_{i=1}^n(n-i+1)(2i-1)=\\sum_{i=1}^n i^2$\n\nShow that\n\n$$\\sum_{i=1}^n(n-i+1)(2i-1)=\\sum_{i=1}^n i^2$$\n\nwithout expanding the summation to its closed-form solution, i.e. $\\dfrac 16n(n+1)(2n+1)$ or equivalent.\n\nE.g., if $n=5$, then\n\n$$5(1) + 4(3)+ 3(5)+2(7)+1(9)=1^2+2^2+3^2+4^2+5^2$$\n\nBackground as requested:\nThe summands for both equations are not the same but the results are the same. The challenge here is to transform LHS into RHS without solving the summation. It seems like an interesting challenge.\n\nFurther edit\nThanks to those who voted to reopen the question!\n\n\u2022 Hmm I don't see what is so slow about expanding and collecting the terms by powers of $r$. Summation by parts works too, but would take about the same amount of computation. \u2013\u00a0user21820 Aug 15 '15 at 3:03\n\u2022 I think this is a duplicate. Looks a lot like this: math.stackexchange.com/questions/1408182/\u2026 \u2013\u00a0marty cohen Apr 4 '16 at 3:21\n\u2022 @martycohen - It is a duplicate only if the question asks to show that it is equivalent to the sum of squares without expanding to the closed-form solution. \u2013\u00a0hypergeometric Apr 4 '16 at 3:25\n\u2022 @martycohen - The summations may be related but it is definitely not a duplicate - the summands are different and the other question is a double summation. In any case this question was posted first so it cannot be a duplicate! \u2013\u00a0hypergeometric Apr 4 '16 at 3:50\n\nDrawing pictures for this kind of problem helps. If you draw\n\n\nThere are\n\n\u2022 $n$ sets of $1$ black squares,\n\u2022 $n-1$ sets of $3$ blue squares,\n\u2022 $n-2$ sets of $5$ purple squares,\n\u2022 $n-3$ setes of $7$ red squares,\n\u2022 $\\vdots$\n\u2022 $n - r + 1$ sets of $2r-1$ squares of any given color\n\nbut the shapes together form a sum of squares pyramid. Algebraically that is:\n\n\\begin{align} % \\sum_{r=1}^{n} (n - r + 1)(2r - 1) % &= \\sum_{r=1}^{n} \\left(\\sum_{s=r}^n 1\\right)(2r - 1) % \\\\ &= \\sum_{r=1}^{n} \\sum_{s=r}^n (2r - 1) % \\\\ &= \\sum_{s=1}^{n} \\sum_{r=1}^s (2r - 1) % \\\\ &= \\sum_{s=1}^{n} s^2 % \\end{align}\n\nwhere the last step uses the fact that the sum of odd numbers up to $2k-1$ is $k^2$.\n\n\u2022 Yes very nice solution! (+1) That was how I started, hoping to find a shorter solution to the sum of squares by recognising the each square is the sum of odd numbers. \u2013\u00a0hypergeometric Aug 15 '15 at 6:39\n\u2022 @hypergeometric I felt like the logic was backwards when I was writing it. Anyway, if you want a short solution to sum of squares, I suggest looking at $\\sum (k^3 - (k - 1)^3) = \\sum (3k^2 - 3k + 1)$ , use telescoping on the left and the known closed form for $\\sum k$ on the right. \u2013\u00a0DanielV Aug 15 '15 at 7:32\n\n$$(n-r+1)(2r-1)=(n+1)(-1)+r[2(n+1)+1]-2r^2$$\n\n$$\\implies\\sum_{r=1}^n(n-r+1)(2r-1)=-(n+1)\\sum_{r=1}^n1+\\{2(n+1)+1\\}\\sum_{r=1}^nr-2\\sum_{r=1}^nr^2$$\n\n$$=-(n+1)\\cdot n+(2n+3)\\cdot\\dfrac{n(n+1)}2-2\\cdot\\dfrac{n(n+1)(2n+1)}6$$\n\n$\\begin{array}\\\\ \\sum_{r=1}^n(n-r+1)(2r-1) &=\\sum_{r=1}^n(n+1)(2r-1)-\\sum_{r=1}^nr(2r-1)\\\\ &=(n+1)n^2-2\\sum_{r=1}^nr^2+\\sum_{r=1}^nr\\\\ &=(n+1)n^2-2\\frac{n(n+1)(2n+1)}{6}+\\frac{n(n+1)}{2}\\\\ &=\\frac{6(n+1)n^2-2n(n+1)(2n+1)+3n(n+1)}{6}\\\\ &=n(n+1)\\frac{6n-2(2n+1)+3}{6}\\\\ &=n(n+1)\\frac{2n+1}{6}\\\\ &=\\frac{n(n+1)(2n+1)}{6}\\\\ \\end{array}$\n\nNow that's quite a surprise!\n\nI'll stop here a try to find a magic trick later.\n\n\u2022 Well spotted! (+1) \u2013\u00a0hypergeometric Aug 15 '15 at 6:38\n\nMake the substitution $i = n - r + 1$ (like change of variable in integration), then \\begin{align} & \\sum_{r = 1}^n (n - r + 1)(2r - 1) = \\sum_{i = 1}^n i(2n - 2i + 1) \\\\ = & (2n + 1) \\sum_{i = 1}^n i - 2\\sum_{i = 1}^n i^2 \\end{align} I think I didn't expand the term \"brutally\", but the last step is still slight expansion.\n\nHere's the \"magic\" solution in more general form, followed by an continuous analog.\n\nLet $f(n) =\\sum_{i=1}^n (n+1-i)g(i)$. Note that $f(1) = g(1)$ and $f(2) =2g(1)+g(2) =g(1)+(g(1)+g(2))$. Then\n\n$\\begin{array}\\\\ f(n+1) &=\\sum_{i=1}^{n+1} (n+2-i)g(i)\\\\ &=\\sum_{i=1}^{n+1} (n+1+1-i)g(i)\\\\ &=\\sum_{i=1}^{n+1} (n+1-i)g(i)+\\sum_{i=1}^{n+1} g(i)\\\\ &=\\sum_{i=1}^{n} (n+1-i)g(i)+\\sum_{i=1}^{n+1} g(i) \\quad\\text{(since }n+1-i = 0 \\text{ for } i=n+1)\\\\ &=f(n)+\\sum_{i=1}^{n+1} g(i)\\\\ so\\\\ f(n+1)-f(n)&=\\sum_{i=1}^{n+1} g(i)\\\\ so\\ that\\\\ f(n)&=\\sum_{j=1}^n\\sum_{i=1}^{j} g(i)\\\\ \\end{array}$\n\nSetting $g(n) = 2n-1$, since $\\sum_{i=1}^{n} g(i) =n^2$, $f(n) =\\sum_{i=1}^{n} i^2 =\\dfrac{n(n+1)(2n+1)}{6}$.\n\nThe continuous analog:\n\nLet $f(x) =\\int_0^x (x-y)g(y)dy$. Then\n\n$\\begin{array}\\\\ f(x) &=\\int_0^x (x-y)g(y)dy\\\\ &=\\int_0^x xg(y)dy-\\int_0^x yg(y)dy\\\\ &=x\\int_0^x g(y)dy-\\int_0^x yg(y)dy\\\\ \\text{so that}\\\\ f'(x)&=xg(x)+\\int_0^x g(y)dy- xg(x)\\\\ &=\\int_0^x g(y)dy\\\\ \\text{Integrating,}\\\\ f(x)&=\\int_0^x \\int_0^y g(z)dz dy\\\\ \\end{array}$\n\nIf $g(y) = 2y$, then $\\int_0^x g(y)dy =x^2$ so $\\int_0^x (x-y)g(y) dy =\\int_0^x y^2 dy =\\dfrac{x^3}{3}$\n\n\\begin{align} \\sum_{i=1}^n(n-i+1)(2i-1) &=\\sum_{i=1}^n\\sum_{j=i}^n(2i-1)\\\\ &=\\sum_{j=1}^n\\sum_{i=1}^j(2i-1) &&(1\\le i\\le j \\le n)\\\\ &=\\sum_{j=1}^n\\sum_{i=1}^ji^2-(i-1)^2\\\\ &=\\sum_{j=1}^n j^2&&\\text{(by telescoping)}\\\\ &=\\sum_{i=1}^n i^2\\qquad\\blacksquare \\end{align}\n\nPosting another solution which has just been suggested by a friend:\n\n\\begin{align} \\sum_{i=1}^n i^2 =&\\sum_{i=1}^n \\color{blue}{-}(\\color{blue}{n-i})i^2+\\sum_{i=1}^n(\\color{blue}{n-i}+1)i^2\\\\ =&\\sum_{i=2}^{n+1} -(n-i+1)(i-1)^2+\\sum_{i=1}^n(n-i+1)i^2\\\\ =&\\sum_{i=1}^{n} -(n-i+1)(i-1)^2+\\sum_{i=1}^n(n-i+1)i^2\\\\ =&\\sum_{i=1}^{n} (n-i+1)\\big[i^2-(i-1)^2\\big]\\\\ =&\\sum_{i=1}^{n} (n-i+1)(2i-1)\\qquad\\blacksquare\\\\\\ \\end{align}", "date": "2020-02-18 19:07:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1397679/surprising-summation-1-sum-i-1nn-i12i-1-sum-i-1n-i2", "openwebmath_score": 0.9999687671661377, "openwebmath_perplexity": 1618.577926622735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846666070896, "lm_q2_score": 0.8933094096048376, "lm_q1q2_score": 0.8740002288932051}}
{"url": "http://mathrefresher.blogspot.com/2005/05/coprime-numbers-more-results.html", "text": "## Sunday, May 15, 2005\n\n### Coprime Numbers: More Results\n\nIn this blog, I will review some basic results regarding coprime numbers.\n\nLemma 1: p,q coprime -> pq, p2 - q2 are coprime.\n\n(1) Assume gcd(pq,p2 - q2) = d and d is greater than 1\n\n(2) So, there is some prime P which divides d [Fundamental Theorem of Artihmetic]\n\n(3) Since d divides pq. P either divides p or q. [Euclid's Lemma]\n\n(4) Let's assume P divides p and the same argument will apply if it divides q. So that, there exists R such that p = PR.\n\n(5) Since d divides p2 - q2, there exists a value Q such that p2 - q2 = PQ\n\n(6) This means that q2 = p2 - PQ = (PR)2 - PQ = P(PR2 - Q).\n\n(7) But this means that P divides q by Euclid's Lemma which is a contradiction since P also divides p. [Because gcd(p,q)=1]\n\n(8) Therefore, we reject our assumption.\n\nQED\n\nLemma 2: p,q are relatively prime and are of different parity (one is odd, one is even), then p + q, p - q are relatively prime.\n\n(1) Assume that gcd(p+q,p-q) = d and d is greater than 1.\n\n(2) Then there exists a prime x which divides d. [Fundamental Theorem of Arithmetic]\n\n(3) We know that this x is odd since p+q and p-q are odd. [Since they are of different parity]\n\n(4) Now x divideds 2p since 2p = (p + q) + (p - q) which means x divides p. [By Euclid's Lemma Since x is odd]\n\n(5) Likewise x divides 2q since 2q = (p + q) - (p - q) = p + q - p + q which means x divides q. [Same reason as above]\n\n(6) But this is a contradiction since p,q are relatively prime.\n\nQED\n\nLemma 3: if S2 = P2 + Q2 and P2 = T2 + Q2, then Q,S,T are relatively prime.\n\n(1) We can assume that S,P,Q are relatively prime and P,T,Q are relatively prime. [See Lemma 1, here for details]\n\n(2) We can also assume that S,P are odd which means that Q is even and that T is odd. [See details above]\n\n(3) From the two equations, we can derive that:\nS2 = T2 + 2Q2\n\n(4) Now we need only prove that any factor that divides 2 values, will divide the third.\n\nCase I: f divides S,Q\n\n(a) There exists s,q such that S = fs, Q = fq\n(b) T2 = S2 - 2Q2 = f2 [ s2 - 2q2 ]\n(c) Which means that f divides T [See here.]\n\nCase II: f divides T,Q\n\n(a) We can use the same argument as Case I.\n\nCase III: f divides S,T\n\n(a) There exists s,t such that S = fs, T = ft\n(b) 2Q2 = f2[ s2 - t2 ]\n(c) Since S,T are odd, f must be odd and s,t must be odd.\n(d) Then, s2 - t2 must be even.\n(e) Then, there exists u such that 2u = s2 - t2\n(f) So, we have:\n2Q2 = (2)u(f2)\n(g) And canceling out the 2s gives us:\nQ2 = f2u\nQED\n\nLemma 4: S,T relatively prime, both odd, 2u = S - T, 2v = S + T, then u,v are relatively prime.\n\n(1) Assume f divides u,v such that u = fU, v = fV.\n(2) The f divides S\nS - T + S + T = 2S = 2u + 2v = 2f(U + V)\n(3) And f divides T\nS + T - (S - T) = 2T = 2v - 2u = 2f(V - U)l.\n(4) This contradicts our S,T beings coprime so we reject our assumption.\n\nQED\n\nOscar Rojas said...\n\nHi Larry,\n\nI think that in Lemma 3, the assumption 1 (about some of the numbers being relative primes) should be part of the hypothesis.\n\nIn fact the Lemma does not hold if that assumption is not satisfied.\n\nThe proper assumption must be made in the part of fermat's one proof that is using the Lemma 3.\n\nSince Lemma 3 is only used in that proof, then one could say that it is irrelevant where do you place the assumption... but, for pure formalism...\n\nLarry Freeman said...\n\nHi Oscar,\n\nThe assumption is justified by the proof in Lemma 1 in the link.\n\nSince we can reduce any solution to x^n + y^n = z^n to a form where x,y,z are coprime (see Lemma 1, here), we can assume that S,P,Q and P,T,Q are relatively prime because if they weren't, then we could still find a form that was.\n\n-Larry\n\nOscar Rojas said...\n\nThanks Larry.\n\nI am sorry; I failed to state more clear my comment. I will try to do it better:\n\nI agree that the assumption is justified... but it is justified in the context outside the lemma. If you take the lemma out of that context, it is not true as written.\n\nThe only fact in the hypothesis is that the numbers satisfy the equations, and that simple fact is not sufficient to derive the conclusion. If you take another set of numbers that are multiples of the originals, the new set still satisfy the unrestricted hypothesis but obviously they do not satisfy the conclusion.\n\nSo, I am not against the assumption, I am against using it as part of the proof of the lemma.\nTo save the consistency, the hypothesis must state that the original numbers are relative primes, precisely to bring the lemma in the restricted context.\n\nAs I said: pure formalism!", "date": "2017-10-24 09:40:05", "meta": {"domain": "blogspot.com", "url": "http://mathrefresher.blogspot.com/2005/05/coprime-numbers-more-results.html", "openwebmath_score": 0.8436025381088257, "openwebmath_perplexity": 1241.744621591523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406009350774, "lm_q2_score": 0.8824278772763471, "lm_q1q2_score": 0.8739923970514499}}
{"url": "https://mathhelpboards.com/threads/conditional-probability.60/", "text": "# Conditional probability\n\n#### Romanka\n\n##### New member\nPlease, help me to solve the problem\nDetails at a factory are tested randomly to check if they are faulty. It is known from previous experience that the probability of a\nfaulty detail is known to be 0.03. If a faulty detail is tested the probability of it testing faulty is 0.82. If a non-faulty detail is\ntested the probability of it testing faulty is 0.06. Given that the detail was tested as not been faulty, calculate the\nprobability that it was faulty.\n\nI understand that it's about conditional probability, but can't get it.\nThanks!\n\n#### tkhunny\n\n##### Well-known member\nMHB Math Helper\nHave you considered drawing a 2x2 grid and filling in the boxes?\n\n#### Plato\n\n##### Well-known member\nMHB Math Helper\nDetails at a factory are tested randomly to check if they are faulty. It is known from previous experience that the probability of a faulty detail is known to be 0.03. If a faulty detail is tested the probability of it testing faulty is 0.82. If a non-faulty detail is tested the probability of it testing faulty is 0.06. Given that the detail was tested as not been faulty, calculate the probability that it was faulty.\nUse $F$ for faulty and $T$ for a positive test.\nFrom the given: $\\mathcal{P}(F)=0.03,~\\mathcal{P}(T|F)=0.82,~\\&~ \\mathcal{P}(T|F^c)=0.06$\n\nNow you want $\\mathcal{P}{(F|T^c)}$$=\\dfrac{\\mathcal{P}(F\\cap T^c)}{\\mathcal{P}(T^c)}. #### Romanka ##### New member thanks, I'll try ---------- Post added at 03:27 PM ---------- Previous post was at 03:26 PM ---------- Use F for faulty and T for a positive test. From the given: \\mathcal{P}(F)=0.03,~\\mathcal{P}(T|F)=0.82,~\\&~ \\mathcal{P}(T|F^c)=0.06 Now you want \\mathcal{P}{(F|T^c)}$$=\\dfrac{\\mathcal{P}(F\\cap T^c)}{\\mathcal{P}(T^c)}$.\nthat's exactly what I need! thank you so much!\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nHere's how I like to do problems like this: Imagine there are 10000 details (chosen to avoid fractions).\n\n\"It is known from previous experience that the probability of a faulty detail is known to be 0.03.\"\nOkay, so our 10000 sample includes (0.03)(10000)= 300 faulty details and 10000- 300= 9700 that are not faulty.\n\n\"If a faulty detail is tested the probability of it testing faulty is 0.82.\"\nOf the 300 faulty details, (0.82)(300)= 246 will test faulty, the other 54 will test not-faulty.\n\n\"If a non-faulty detail is tested the probability of it testing faulty is 0.06.\"\nOf the 9700 non-faulty details, (.06)(9700)= 582 will test faulty. 9700- 582= 9118 will test not-faulty.\n\n\"Given that the detail was tested as not been faulty, calculate the probability that it was faulty.\"\nThere are a total of 9118+ 54= 9172 details that test non-faulty of which 54 are faulty.\n\nthanks!\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nBy the way, \"detail\" doesn't seem like quite the word you want. \"Detail\" means a small part of something larger. I suspect this was translated from another language and you really wanted \"item\".\n\n#### Romanka\n\n##### New member\nYes, maybe. But I got the problem about \"details\"\n\n#### Plato\n\n##### Well-known member\nMHB Math Helper\nYes, maybe. But I got the problem about \"details\"\nYES. But what was language of the question?\nDid you use a translation program to post it here?", "date": "2021-10-24 10:21:10", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/conditional-probability.60/", "openwebmath_score": 0.49931442737579346, "openwebmath_perplexity": 2732.265665013366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405986777259, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8739923766850901}}
{"url": "https://math.stackexchange.com/questions/1860168/uncountability-of-increasing-functions-on-n/1860193", "text": "# Uncountability of increasing functions on N\n\nI believe I have made a reasonable attempt to answer the following question. I would like a confirmation of my proof to be correct, or help as to why it is incorrect.\n\nQuestion: Let $f : \\mathbb{N} \\to \\mathbb{N}$ be increasing if $f(n+1) \\ge f(n)$ for all $n$. Is the set $A$ of functions $f$ countable or uncountable?\n\nFor each $f$ in $A$, there is a function $g: \\mathbb{N}\\cup \\{0\\} \\to \\mathbb{N} \\cup\\{0\\}$ defined by $g(0)=f(1)$ and $g(n)=f(n+1)-f(n)$ for $n>0$. Conversely, for each $g: \\mathbb{N}\\cup \\{0\\} \\to \\mathbb{N}\\cup\\{0\\}$ with $g(0)>0$, there is an increasing function $f$ defined recursively by $f(1)=g(0)$ and $f(n+1)=f(n)+g(n)$ for $n>0$. Consequently, there is a bijection between $A$ and the set $B$, of $g: \\mathbb{N}\\cup\\{0\\} \\to \\mathbb{N}\\cup\\{0\\}$, $g(0)>0$. If $B$ is uncountable, then $A$ must be uncountable.\n\nThe set $C$ of $h: \\mathbb{N}\\cup\\{0\\} \\to \\{0,1,2,3,4,5,6,7,8,9\\}$ with $h(0)=1$is a subset of $B$. For each real $x \\in [1,2)$, there is a unique decimal expansion with 1 before the decimal point, ending NOT in trailing 9's. By defining for $n>0$, $h(n)=$ (n'th digit of x after decimal point), we have an injection from $[1,2)$ to the set $C$ (different $x$ => different decimal expansions => x maps to different h). Since the set of reals on $[1,2]$ is uncountable, the set C is uncountable, => the set B is uncountable, => the set A is uncountable.\n\nThis proof works but seems needlessly verbose. It is easier to prove that strictly increasing functions form an uncountable set (which implies that the monotone increasing are uncountable as well). This is because there is an obvious 1-1 correspondence between strictly increasing functions and infinite subsets of N (each such function is uniquely determined by its range). It is a standard result that there are uncountably many subsets of N, of which only countably many are finite. Hence, there are uncountably many infinite subsets.\n\nIt is rather easy to apply the diagonalization argument directly. Let $\\langle f_n\\rangle$ be a sequence of increasing functions from the naturals to itself. Define a function $f$ recursively by $f(1)=1$ and $$f(n+1)=f(n)+f_n(n+1)-f_n(n)+1.$$ The resulting function $f$ is clearly different from every function in the sequence $\\langle f_n\\rangle$.\n\n\u2022 Oh I really like this. +1 \u2013\u00a0Pete Caradonna Jul 15 '16 at 11:30\n\u2022 The resulting function $f$ need not be increasing, so doesn't belong to the set being considered. \u2013\u00a0hunter Jul 15 '16 at 13:21\n\u2022 (More precisely, $f(n+1) - f(n) = (f_n(n+1) - f_n(n)) - (f_{n-1}(n) - f_{n-1}(n-1))$; since there's no relation between the rates of growth of $f_n$ and $f_{n-1}$ you can't determine if this difference is positive or negative. \u2013\u00a0hunter Jul 15 '16 at 13:23\n\u2022 @hunter Thank you, you are right, what I wrote was different from what I actually intended. I have fixed it now. That is exactly where the recursive definition comes in. \u2013\u00a0Michael Greinecker Jul 15 '16 at 13:38\n\u2022 Note that this is exactly the diagonalisation argument for general functions composed with the OP's transformation. \u2013\u00a0filipos Jul 15 '16 at 20:49\n\nYour proof looks OK to me. It's a little longer than needed though.\n\nRight after you define your set $A$, what if we just restrict ourselves to a subset of $A$ given by the subset of monotone functions $t: \\mathbb{N} \\to \\mathbb{N}$ such that $t(n+1)-t(n) \\in \\{0,1\\}$ and $t(0)=k$.\n\nThen in your spirit, we can identify each $t$ with the infinite binary sequence of its step-wise differences. Moreover, clearly for any infinite binary sequence we can construct the unique $t$ that gets related to it. Then by Cantor's diagonalization argument we know $\\{0,1\\}^\\mathbb{N}$ is uncountable and we're done.\n\nAnother way to show that $A$ is uncountable : Show that if $B=\\{f_n: n\\in N\\}\\subset A$ then $B\\ne A$ as follows:\n\nLet $g(n)=\\max \\{1+f_j(n): j\\leq n\\}.$\n\nWe have $g(n)> f_n(n)$ for every $n.$ So $g\\ne f_n$ for every $n.$ So $g \\not \\in B.$\n\nFor each $n$ we have $g(n)=1+f_{j_n}(n)$ for some $j_n\\leq n .$ So $g(n+1)=\\max \\{1+f_j(n+1):j\\leq n+1\\}\\geq 1+f_{j_n}(n+1)\\geq 1+f_{j_n}(n)=g(n).$ So $g\\in A.$", "date": "2019-10-15 02:46:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1860168/uncountability-of-increasing-functions-on-n/1860193", "openwebmath_score": 0.9356918931007385, "openwebmath_perplexity": 118.64097275636804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429607229954, "lm_q2_score": 0.8887588023318196, "lm_q1q2_score": 0.8739547120535949}}
{"url": "https://math.stackexchange.com/questions/2752037/probability-involving-exponentially-distributed-random-variabl", "text": "# Probability involving exponentially distributed random variabl\n\nTwo buses A and B arrive independently at a bus station at random rate of 5/hour and 4/hour respectively. A passenger comes to the bus station at 10 am. What is the probability that it takes at most 5 minutes before the first bus arrives at the station?\n\nAttempt\n\nLet X and Y be random variables indicating time of arrivals of Bus A and B respectively. Then, X and Y are exponentially distributed random variables with rate 5 and 4 respectively. So the probability density functions of X and Y are 5*exp(-5*t) and 4*exp(-4*t).\n\nNow, let Z = min(X,Y) be a new random variable. Then, I need to calculate P(Z<5/60).\n\nQuestion: My solution sheet goes as follows: P(Z<5/60) = 1-P(Z>5/60) = 1-P(X>5/60, Y>5/60) and goes on like this. But why cannot I just calculate P(Z<5/60) = P(X<5/60, Y<5/60) directly? These two give me different answers, meaning my attempt might be wrong. Why do we need to calculate the complementary probability?\n\n\u2022 It does not matter if the second bus arrives before 5 minutes if the first one does arrive in 5 minutes. You only care about the complement of the event that no busses arrive within 5 minutes. Apr 24, 2018 at 18:22\n\u2022 I did not get it. @TimDikland Apr 24, 2018 at 18:24\n\u2022 There are four possible interesting events. P(X<5, Y<5), P(X<5, Y>5), P(X>5, Y<5), P(X>5, Y<5). Now ask yourself which of those events are desirable and which are not. (Here X<5 means bus X arrives within 5 minutes) Apr 24, 2018 at 18:29\n\u2022 I am still confused. So, if the question was: What is the probability that she waits at least 5 mins before the first bus arrives? Would I directly calculate P(Z>5/60) = P(X>5/60,Y>5/60)? Apr 24, 2018 at 18:34\n\u2022 Exactly. Your mistake in your first calculation is that $P(Z < 5/60) \\neq P(X<5/60,Y<5/60)$. If you would do it correctly then you would find that $P(Z<5/60) = P(X<5/60, Y<5/60) + P(X<5/60, Y>5/60) + P(X>5/60, Y<5/60)$, which is obviously more work to calculate Apr 24, 2018 at 18:38\n\nHere's another interesting way to approach this problem. I'll start out with a more general version, and then we'll look at the case you need.\n\nLet $X_1, X_2, \\cdots X_n$ be independent and let $X_{(1)}$ be the minimum observation. Let's find the CDF of $X_{(1)}$.\n\n\\begin{align*} F_1(x) &= P(X_{(1)} \\leq x) & \\text{(definition)} \\\\ &= 1 - P(X_{(1)} > x) & \\text{(complement)} \\\\ &= 1 - P(X_1 > x, \\ X_2 > x, \\ \\cdots, \\ X_n > x) & \\text{(min > x $\\Leftrightarrow$ all > x)} \\\\ &= 1 - P(X_1 > x)P(X_2 > x)\\cdots P(X_n > x) & \\text{(independence)} \\\\ \\end{align*}\n\n## Back to the problem\n\nLet $$X \\sim Exp(4) \\quad Y \\sim Exp(5) \\quad Z = min(X, Y)$$ $$P(X > x) = e^{-4x}$$ $$P(Y > x) = e^{-5x}$$ Hence, by the above result we can write the CDF of the minimum as: $$P(Z \\leq x) = 1 - e^{-4x}e^{-5x} = 1 - e^{-9x}$$\n\nNow you can calculate the desired probability easily.\n\nNote that this is equivalent mathematically to the approach that Davis gives, but with this approach we see that the the minimum of independent exponential random variables is itself Exponentially distributed! Specifically, $Z \\sim Exp(4+5)$.\n\n$Z < \\frac{5}{60}$ might occur without $X < \\frac{5}{60}$ occurring, specifically, if $Y < \\frac{5}{60}$.\n\nMultiplying probabilities of independent events gives you the probability of their intersection (both events happen), but what you want is the probability of their union. Call $E$ the event that bus A arrives before the given time, and $F$ the event that bus B arrives before the given time. Then, the event you are interested in is:\n\n$E \\cup F = (E^c \\cap F^c)^c$\n\nThe equality is by DeMorgan's law. Now you can figure out the probability by multiplying the probabilities of $E^c$ and $F^c$.", "date": "2022-05-22 13:57:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2752037/probability-involving-exponentially-distributed-random-variabl", "openwebmath_score": 0.9937646389007568, "openwebmath_perplexity": 230.65625701220702, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429580381724, "lm_q2_score": 0.8887587905460026, "lm_q1q2_score": 0.8739546980779347}}
{"url": "https://logivan.com/ixi3dmsm/if-two-triangles-are-equal-in-area%2C-they-are-congruent-bbe282", "text": "An\u00a0included angleis an angle formed by two given sides. For example, both of these triangles are isosceles, since they have two equal sides and angles. (iii) If two figures have equal areas, they are congruent. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. Congruence is our first way of knowing that magnitudes of the same kind are equal. Triangles are congruent when all corresponding sides & interior angles are congruent. = False (v) If two sides and one angle of a triangle are equal to the corresponding two sides and angle of another triangle, the triangles are congruent. Hence all squares are not congruent. You have one triangle. TRUE. find the price at which it was sold. Triangle Congruence Theorems (SSS, SAS, ASA), Congruency of Right Triangles (LA & LL Theorems), Perpendicular Bisector (Definition & Construction), How to Find the Area of a Regular Polygon, Do not worry if some texts call them postulates and some mathematicians call the theorems. asked by priyanka. Geometricians prefer more elegant ways to prove congruence. Since b/e = 1, we have a/d = 1. = False (iv) If two triangles are equal in area, they are congruent. - the answers to estudyassistant.com Asked on January 17, 2020 by Premlata Pandagre In a squared sheet, draw two triangles of equal areas such that (i) the triangles are congruent. It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. -If two squares have equal perimeters, then they have equal ...\u201d in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. prove that they are congruent. Think: Two congruent triangles have the same area. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. Yes, they are similar. Triangles can be similar or congruent. one could look upside-down compared to the other but if they have the same dimensions (i.e. (if you don't know the a False. Two triangles are congruent if they have the same shape and size, but their position or orientation may vary. they have to be congruent to have same perimeters AND area con. FALSE. All three triangle congruence statements are generally regarded in the mathematics world as postulates, but some authorities identify them as theorems (able to be proved). After you look over this lesson, read the instructions, and take in the video, you will be able to: Get better grades with tutoring from top-rated private tutors. You can think you are clever and switch two sides around, but then all you have is a reflection (a mirror image) of the original. Congruent triangles are triangles which are identical, aside from orientation. An included side is the side between two angles. It is congruent to \u2220WSA because they are alternate interior angles of the parallel line segments SW and NA (because of the Alternate Interior Angles Theorem). it the shopkeeper sells it at a loss of 10% . Converse of the above statement : If the areas of the two triangles are equal, then the triangles are congruent. If the areas of two similar triangles are equal, prove that they are congruent. But just to be overly careful, let's compute a/d. So once you realize that three lengths can only make one triangle, you can see that two triangles with their three sides corresponding to each other are identical, or congruent. To be congruent two triangles must be the same shape and size. For example, these triangles are similar because their angles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent. A, B and C are three sets, n(A) = 14, n(C) = 13, n(A U BUC) = 28, Move to the next side (in whichever direction you want to move), which will sweep up an included angle. Congruent Triangles. b=e. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. If two triangles have the same area, they must be congruent. The given statement - \"If two triangles are congruent, then their areas are equal.\" (d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent. We use the symbol \u2245 to show congruence. Log in. In the sketch below, we have \u25b3CAT and \u25b3BUG. (iv) If two triangles are equal in area, they are congruent. Their interior angles and sides will be congruent. In the simple case below, the two triangles PQR and LMN are congruent because every corresponding side has the same length, and every corresponding angle has the same measure. 3) They are equal areas. Local and online. \u2026, how much loss and profit computed on his original capital\u200b. Answer: 3 question What would prove that the two triangles are congruent? Now you have three sides of a triangle. Similar triangles will have congruent angles but sides of different lengths. Why should two congruent squares have the same area? Note that if two angles of one are equal to two angles of the other triangle, the tird angles of the two triangles too will be equal. The SAS Postulate says that triangles are congruent if any pair of corresponding sides and their included angle are congruent. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. triangles have same shape/size but just placed/viewed differently cidyah for that to \u2026 Click here to get an answer to your question \ufe0f if the area of two triangles are equal. the triangle which are equal in area also congruent, Here ur answer ....pls mark it brainlist..pls, This site is using cookies under cookie policy. It is equal in length to the included side between \u2220B and \u2220U on \u25b3BUG. Congruent triangles will have completely matching angles and sides. Join now. You could cut up your textbook with scissors to check two triangles. So let's just start That triangle BCD is congruent Let\u2019s use congruent triangles first because it requires less additional lines. This forces the remaining angle on our \u25b3CAT to be: This is because interior angles of triangles add to 180\u00b0. Testing to see if triangles are congruent involves three postulates. Theorem 1 : Hypotenuse-Leg (HL) Theorem. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. Get help fast. You will see that all the angles and all the sides are congruent in the two triangles, no matter which ones you pick to compare. Congruence is denoted by the symbol \u2245. AAS is equivalent to an ASA condition, by the fact that if any two angles are \u2026 So now you have a side SA, an included angle \u2220WSA, and a side SW of \u25b3SWA. Similar triangles. You can't do it. 0 votes . We could also think this angle right over here. 1. (ii) If two squares have equal areas, they are congruent. A postulate is a statement presented mathematically that is assumed to be true. By applying the Side Angle Side Postulate (SAS), you can also be sure your two triangles are congruent. Triangles can be considered congruent if following conditions are satisfied. https://www.mathsisfun.com \u203a geometry \u203a triangles-congruent.html ... maths. Answer In two similar triangles, the ratio of their areas is the square of the ratio of their sides. Compare them to the corresponding angles on \u25b3BUG. Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. 1. For two triangles to be similar, it is sufficient if two angles of one triangle are equal to two angles of the other triangle. The legs of each of these isosceles triangles could have any lengths as long as they are equal, but the legs of these two triangles need not be the same. So the fact that two triangles have the same area does not necessarily tell us that they are congruent (they could be congruent but that is often not the case) since they can have different measures on their sides or a different size and still have the same area. monojmahanta4 monojmahanta4 09.04.2020 Math Secondary School If the area of two triangles are equal. Conditional Statements and Their Converse. You can check polygons like parallelograms, squares and rectangles using these postulates. There are exactly two isosceles triangles with given perimeter and area. https://www.mathsisfun.com/geometry/triangles-congruent.html Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. Note that if two angles of one are equal to two angles of the other triangle, the tird angles of the two triangles too will be equal. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. Thus, a=d. are both 90 degrees. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. If triangle RST is congruent to triangle WXY and the area of triangle WXY is 20 square inches, then the area of triangle RST is 20 in.\u00b2. e.g. 2) Their corresponding angles are equal. Two triangles, ABC and A\u2032B\u2032C\u2032, are similar if and only if corresponding angles have the same measure: this implies that they are similar if and only if the lengths of corresponding sides are proportional. And for sensible cases. If you have two similar triangles, and one pair of corresponding sides are equal, then your two triangles are congruent. (g) If two triangles are equal in area, they are congruent. Write the following statements in words\u200b, farhan sarted his bussiness by investing 75000 in the first year he made a profit of 20% he invested the capital in new business and made loss of 12% The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. Now you will be able to easily solve problems on congruent triangles definition, congruent triangles symbol, congruent triangles Class 8, congruent triangles geometry, congruent t The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. A rectangle and a triangle with the same base and height have the same area. = False (vi) If two angles and any side of a triangle are equal to the corresponding angles of another triangles then the triangles are congruent. So go ahead; look at either \u2220C and \u2220T or \u2220A and \u2220T on \u25b3CAT. \"If two triangles are congruent, then their areas are equal.\" You can compare those three triangle parts to the corresponding parts of \u25b3SAN: After working your way through this lesson and giving it some thought, you now are able to recall and apply three triangle congruence postulates, the Side Angle Side Congruence Postulate, Angle Side Angle Congruence Postulate, and the Side Side Side Congruence Postulate. Therefore, triangles are congruent. If the area of two similar triangles are equal, prove that they are congruent. Corresponding sides and angles mean that the side on one triangle and the side on the other triangle, in the same position, match. The two are not the same. You now have two triangles, \u25b3SAN and \u25b3SWA. Suppose you have parallelogram SWAN and add diagonal SA. Using any postulate, you will find that the two created triangles are always congruent. Theorem 1 : Hypotenuse-Leg (HL) Theorem. In most systems of axioms, the three criteria \u2013 SAS, SSS and ASA \u2013 are established as theorems. I could have one triangle with sides measuring 7 in + 5 in + 8 in (perimeter = 20 in) and another triangle with sides of 6 in + 4 in + 10 in (perimeter = 20 in). Congruent triangles sharing a common side. \u2026, nswer then don't give) give me with full steps.don't spam \u274c\u274c\u274c\u274c\u274c\u200b. This is the only postulate that does not deal with angles. That is not very helpful, and it ruins your textbook. You can now determine if any two triangles are congruent! For the two triangles to be congruent, those three parts -- a side, included angle, and adjacent side -- must be congruent to the same three parts -- the corresponding side, angle and side -- on the other triangle, \u25b3YAK. Ask your question. All the sides of a square are of equal length. Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. You can only make one triangle (or its reflection) with given sides and angles. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. Now shuffle the sides around and try to put them together in a different way, to make a different triangle. Definition: Triangles are congruent when all corresponding sides and interior angles are congruent.The triangles will have the same shape and size, but one may be a mirror image of the other. (iv) If two triangles are equal in area, they are congruent. So I guess the answer is not always. Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. An\u00a0included angleis an angle formed by two given sides. This triangle has two angles with equal measures of 75^@. However, they do not have to be congruent in order to be similar. ALL of this is based on a single concept: That the quality that we call \"area\" is an aspect of dimensional lengths and angles. Both the areas would be 9, but the figures would not be congruent. Illustration of SAS rule: Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. Light1729 Light1729 If two triangles are congruent then they must overlap each other completely, so, they have exactly equal area. Furthermore, your question is about congruent triangles and not similar triangles.l. prove that they are congruent. Notice we are not forcing you to pick a particular side, because we know this works no matter where you start. Here, instead of picking two angles, we pick a side and its corresponding side on two triangles. \u200b, the radic of two circles are 8 cm and 6 cm representively find the radius of the circle having area equal to the sum of the areas of the two circles\u200b, bnRlinP) = 16 and n(Q).10, what are the greatest and the least values of n(PU QI?8. The congruence of two objects is often represented using the symbol \"\u2245\". Only if the two triangles are congruent will they have equal areas. Comparing one triangle with another for congruence, they use three postulates. Isosceles triangles are triangles with two equal sides, and thus two equal angle measures. In this case, two triangles are congruent if two sides and one included angle in a given triangle are equal to the corresponding two sides and one included angle in another triangle. SSSstands for \"side, side, side\" and means that we have two triangles with all three sides equal. Since two triangles are similar therefore the ratio of the area is equal to the square of the ratio of its corresponding side a r e a \u2206 A B C a r e a \u2206 D E F = B C E F 2 = A B D E 2 = A C D F 2 B C E F 2 = A B D E 2 = A C D F 2 = 1 N o w , t a k i n g a n y o n e c a s e 1 = B C E F 2 1 = B C E F E F = B C However they can share a side, and as long as they are otherwise identical, the triangles are still congruent. \u2026, n(A) = n(B), nn Byn(B, C) = 4 and n(An C) = 3. However, triangles \u2026 Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. The ASA Postulate was contributed by Thales of Miletus (Greek). Testing to see if triangles are congruent involves three postulates, abbreviated SAS, ASA, and SSS. Pairs - The classic pairs game with simple congruent shapes. You already know line SA, used in both triangles, is congruent to itself. For two triangles to be congruent, the corresponding angles and the sides are to be equal. The four triangles are congruent with each other regardless whether they are rotated or flipped. Divide a square sheet in two triangles of equal area so that they are congruent. The Angle Side Angle Postulate (ASA) says triangles are congruent if any two angles and their included side are equal in the triangles. (f) Two circles having same circumference are congruent. Congruent triangles. But, it is not necessary that triangles having equal area will be congruent to one another they may or may not be congruent. Contrapositive of the given statement : If the areas of two traingles are not equal then the triangles are not congruent. So also, sides d and e must be equal because DEF is isosceles. They have the same area and the same perimeter. (ii) If two squares have equal areas, they are congruent. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. $\\endgroup$ \u2013 Moti Feb 7 '15 at 16:42 add a comment | Your Answer In the figure below, the triangle LQR is congruent to PQR even though they share the side QR. Congruent triangles are triangles having corresponding sides and angles to be equal. If the areas of two similar triangles are equal, prove that they are congruent. But that does not mean that they have to be congruent. If the corresponding angles of two triangles are equal, then they are always congruent. Guess what? If the areas of two similar triangles are equal, prove that they are congruent. If 2 sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent. Cut a tiny bit off one, so it is not quite as long as it started out. So, for equal area, all sides are equal. FALSE. For example: (See Solving SSS Trianglesto find out more) Put them together. Cut the other length into two distinctly unequal parts. It is true that all congruent triangles have equal area. #2 That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area. Congruence in triangles is important in the study of plane geometry. Yes, if two triangles have two congruent angles and two congruent sides then the triangles are guaranteed to be congruent. What is the solution set of y=3x+16 and 2y=x+72? Join now. Only if the two triangles are congruent will they have equal areas. And why does a $1 \\times 1$ square have an area of $1$ unit?) the cost price of a flower vase is rs 120 . Remember that the included angle must be formed by the given two sides for the triangles to be congruent. This will happen when the area is half the multiplication of the sides, or maximum area for such two sides. 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Think this angle right over here in area, they use three postulates, abbreviated SAS, and..., if two triangles are congruent other based on their shape and size, or to put it another,...", "date": "2021-04-23 09:16:52", "meta": {"domain": "logivan.com", "url": "https://logivan.com/ixi3dmsm/if-two-triangles-are-equal-in-area%2C-they-are-congruent-bbe282", "openwebmath_score": 0.6022189855575562, "openwebmath_perplexity": 458.793011994633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.97997655637136, "lm_q2_score": 0.8918110490322426, "lm_q1q2_score": 0.8739539207645473}}
{"url": "https://www.physicsforums.com/threads/integrating-for-approximation-of-a-sum.753870/", "text": "# Integrating for approximation of a sum\n\n1. May 15, 2014\n\n### goraemon\n\n1. The problem statement, all variables and given/known data\n\nFind an N so that $\u2211^{\\infty}_{n=1}\\frac{log(n)}{n^2}$ is between $\u2211^{N}_{n=1}\\frac{log (n)}{n^2}$ and $\u2211^{N}_{n=1}\\frac{log(n)}{n^2}+0.005.$\n\n2. Relevant equations\nDefinite integration\n\n3. The attempt at a solution\nI began by taking a definite integral: $\\int^{\\infty}_{N}\\frac{log(n)}{n^2}dn$ and, using integration by parts, arrived at the following answer: $\\frac{log(N)+1}{N}$. (Is this right? If not, I could post the steps I used to try to see where I made an error)\n\nNext we need $\\frac{log(N)+1}{N}$ to be within 0.005 as given by the problem, so:\n\n$\\frac{log(N)+1}{N}=0.005=\\frac{1}{200}$\n\nBut I'm having trouble how to solve for N algebraically. Would appreciate any help.\n\n2. May 15, 2014\n\n### BiGyElLoWhAt\n\nI actually got -1 times what you have. remember: $\\int \\frac{1}{n^2} = -\\frac{1}{n}$\n\n3. May 15, 2014\n\n### goraemon\n\nRight but since we're taking a definite integral on the interval from N to \u221e, and since $\\frac{-log(N)-1}{N}$ as N approaches infinity equals zero, shouldn't the definite integral work out to:\n$0-\\frac{-log(N)-1}{N}=\\frac{log(N)+1}{N}$?\n\n4. May 15, 2014\n\n### pasmith\n\nThe graph of $x^{-2}\\log(x)$ has a maximum in $[1,2]$, so it's difficult to say whether the integral is an over-estimate or under-estimate of the sum. To be safe I'd require that $$\\frac{\\log(N) + 1}{N} < \\frac{1}{400}$$.\n\nEquations of that type have to be solved numerically. You're looking for zeroes in $x > 0$ of $$f(x) = \\frac{x}{200} - 1 - \\log x$$. The derivative of this function is $$f'(x) = \\frac{1}{200} - \\frac{1}{x}$$ so the second derivative ($x^{-2}$) is everywhere positive. There is a minimum at $x = 200$ where $f(200) < 0$. Since $f(x) \\to +\\infty$ as $x \\to 0$ or $x \\to \\infty$ there are exactly two solutions. Logic dictates you want the larger, since if $N$ works then any larger $M$ should also work.\n\n5. May 15, 2014\n\n### goraemon\n\nThanks, I tried solving it numerically and came up with N = 1687, which the textbook confirms is correct.", "date": "2017-11-18 17:01:54", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integrating-for-approximation-of-a-sum.753870/", "openwebmath_score": 0.8574643731117249, "openwebmath_perplexity": 324.4466360750581, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765552017675, "lm_q2_score": 0.89181104831338, "lm_q1q2_score": 0.8739539190170231}}
{"url": "https://math.stackexchange.com/questions/434967/combinations-with-exclusions/435217#435217", "text": "# Combinations with exclusions\n\nGiven five objects $A, B, C , D , E$. I'd like to calculate all possible sets of these objects such that\n\n1. Two pairs of objects, $B$ and $C$, and $D$ and $E$, cannot coexist. For example the set $\\{ABD\\}$ would be valid but the set $\\{CDE\\}$ would not.\n2. Each object appears no more than once in a set, e.g. $\\{ACD\\}$ is valid but $\\{ADD\\}$ is not.\n3. There can be no empty set.\n4. Apart from the above, there is no other limit on the cardinality of a set, e.g. $\\{D\\}$, $\\{AB\\}$, $\\{ACE\\}$. I know there can be no set of cardinality greater than 3, but it would be nice to show this.\n5. The objects can be listed in any order, i.e. $\\{AB\\} = \\{BA\\}$\n\nEdit My intention is to generalize the process of generating these combinations given any number of elements and restrictions.\n\n\u2022 \"Sets of these objects\" is somewhat incompatible with the ambiguous \"objects can be listed in any order.\" So is AB different from BA? Jul 3 '13 at 4:13\n\u2022 I apologize for the confusion. {A,B} = {B,A} Jul 3 '13 at 9:44\n\nSince your options are $A$, one of $\\{B, C\\}$, and one of $\\{D, E\\}$, it's easy to see why the cardinality can only be at most 3.\n\nI'm going to tackle this case by case:\n\n# No objects\n\nThere is only one possibilitiy for the empty set.\n\n# One object\n\nWe can choose any of the 5 objects, so there are 5 possibilities.\n\n# Two objects\n\nIf we pick A first, we can choose any of the other four objects to go with it. If we pick any of the other objects first, then there are only three objects that can follow, due to the first restriction (e.g. $BA, BD, BE$ for $B$)\n\nThus there are $4 + 3 \\times 4 = 16$ permutations for two objects.\n\n# Three objects\n\nPick one item from each of the sets $\\{A\\}, \\{B, C\\}, \\{D, E\\}$. There are $1 \\times 2 \\times 2 = 4$ possible selections that can be made.\n\nFor each selection, there are $3! = 6$ ways to permute the objects, giving a total of $4 \\times 6 = 24$ permutations for three objects.\n\nAltogether, this is a grand total of $1 + 5 + 16 + 24 = 46$ permutations.\n\nFrom a response to a question, it appears that order does not matter.\n\nIt is not clear whether the empty set (\"none\") is to be considered a valid choice. We will assume that it is allowed. If it is not, subtract $1$ from the answer we will obtain.\n\nWith such a small number of objects, a carefully drawn up list may be the best approach.\n\nAnother good option is the approach taken by Sp3000. There, permutations were being counted, but the ideas can be readily modified to disregard order. For choices of $0, 1, 2, 3$ objects, the answers become $1$, $5$, $8$, and $4$ for a total of $18$ (or $17$ if we don't allow the empty set).\n\nBut you may like the following idea. Line up the $5$ objects in front of us, with $B$ and $C$ close to each other, and also $D$ and $E$, like this: $$A\\qquad BC \\qquad DE.$$\n\nStand in front of $A$ and decide whether to include her in the set we are building. There are $2$ choices, yes or no.\n\nNow walk over to the $BC$ group. For every choice we made about $A$, we now have $3$ choices, $B$, $C$, or neither. So far, we have $2\\times 3$ distinct possibilities.\n\nFinally, walk over and stand in front of the $DE$ group. For every choice we made earlier, there are again $3$ choices, this time $D$, $E$, or neither.\n\nThus the total number of possible choices is $2\\times 3\\times 3$, that is, $18$. Again, if the empty set is not allowed, we have $17$ possibilities only.\n\n\u2022 You hit the nail on the head. My intention is to generalize the process of generating these combinations, for any number of elements and restrictions. I've updated my updated question accordingly. Jul 3 '13 at 11:17\n\u2022 For setups precisely of this form (the people belong to disjoint groups of $1$ or more, possibly of different sizes, and at most one member of a group can be selected) the above idea generalizes immediately, and gives implicitly a mechanical way to generate and list. Jul 3 '13 at 12:28\n\nYour question asks for the number of all subsets $S$ of $\\{A,B,C,D,E\\}$ which are not a superset of $\\{B,C\\}$ or $\\{D,E\\}$.\n\nThis is a typical application of the principle of inclusion-exclusion.\n\nThe total number of subsets of $\\{A,B,C,D,E\\}$ is $2^5 = 32$. The number of supersets of $\\{B,C\\}$ in $\\{A,B,C,D,E\\}$ is $2^3 = 8$ (each of the elements $A,B,C$ may be selected or not). In the same way, the number of supersets of $\\{D,E\\}$ in $\\{A,B,C,D,E\\}$ is $8$. The number of common supersets of $\\{B,C\\}$ and $\\{D,E\\}$ is 2.\n\nNow the principle of inclusion-exclusion yields the result $$32 - 8 - 8 + 2 = 18.$$", "date": "2021-10-21 14:26:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/434967/combinations-with-exclusions/435217#435217", "openwebmath_score": 0.7511457204818726, "openwebmath_perplexity": 193.42795208472847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409525, "lm_q2_score": 0.8918110425624791, "lm_q1q2_score": 0.8739539154673861}}
{"url": "https://community.wolfram.com/groups/-/m/t/1620190", "text": "# Statistical Distributions of Areas of Voronoi Cells\n\nPosted 11 months ago\n2622 Views\n|\n10 Replies\n|\n27 Total Likes\n|\n Open code in Cloud | Download code to Desktop via Attachments BelowOne of the brightest characteristics of the Wolfram Language (WL) is that it works easily across many different scientific fields (thanks to numerous built-in functions and data) and that due to the coherence of WL design, combining such different sciences via computation comes as natural as speaking to a human. Here WL will enable us to move quickly from computational geometry to machine learning to statistics on a quest of exploring an interesting subject. This seemingly simple question is not easy to answer without right framework and tools: What is statistical distributions of areas of cells in Voronoi tessellation? From a brief web search it appears to follow closely GammaDistribution. But in the absence of analytical proof, how can one quickly deduce this fact or verify it is good model? We will do exactly that below. For reference see also a relevant paper by TANEMURA and a real-interest question by actual user which prompted this exploration.To start let's build a \"large\" VoronoiMesh on 5000 uniformly distributed points within a unit Disk: pts = RandomPoint[Disk[], 5000]; mesh = VoronoiMesh[pts, Axes -> True] We see the statistics we can gather from this will be obviously offset by the presence of \"border\" cells of much larger area than regular inner cells have: Let's exclude large cells by selecting only those within original region of random points distribution - unit Disk: vor=MeshPrimitives[mesh,2]; vor//Length  5000 vorInner=Select[vor,RegionWithin[Disk[],#]&]; vorInner//Length  4782 We got fewer elements of course and they all are nice regular cells without any large outliers: Graphics[{FaceForm[None], EdgeForm[Gray], vorInner}] Let's take a look at the statistics of areas' distribution: areas = Area /@ vorInner; hist = Histogram[areas, Automatic, \"PDF\", PlotTheme -> \"Detailed\"] We can see that there is a minimum of distribution at zero area, which is intuitively correct. Close to zero-area cells of course are not present much with the finite number of points per region (or finite density). So there is some prevalent finite mean area there. We can find a very close simple analytic distribution using WL machine learning tools, which turns out to be GammaDistribution as discussed in some publications: dis=FindDistribution[areas]  GammaDistribution[3.3458431368450587,0.00018456010158014188] FindDistribution acted automatically without any additional options from our side. GammaDistribution matches very nicely the empirical histogram: Show[hist, Plot[PDF[dis, x], {x, 0, .0015}]] And now it is easy to find thing like: {Mean[dis], Variance[dis], Kurtosis[dis]}  {0.0006175091492073445, 1.1396755130437449*^-7, 4.793269963533813} Probability[x > .0005, x \\[Distributed] dis]  0.5740480899719699 Attachments:\n10 Replies\nSort By:\nPosted 11 months ago\n Very nice. Thanks for sharing. Some years ago I did this with 10^9 cells. If I recall correctly the left side is a bit higher (more probably) than a gamma distribution.To remove any edge-effects, what people do is: take random points in a square, and then copy the square 9 times in a 3*3 grid, such as to emulate period boundary conditions. Then take only the polygons which center is in the center unit square\u2026Great stuff!\u2013SH\nPosted 11 months ago\n Thanks! Awesome idea about edge-effects! Did you estimate any analytical models beyond Gamma or it was just pure empirical thing?\nPosted 11 months ago\n I looked at different things: how many sides, angles, area, perimeter, and many conditionally averaged quantities. But area of the PDF as well. But no analytical shape is known for it. I believe that there is something known for perimeter.\nPosted 11 months ago\n I don't find it all that surprising that the result is a Gamma distribution. The 1-d case is well known as it is the distribution of interarrival times of a Poisson process and is exponentially distributed. The exponential distribution is a special case of the Gamma so maybe you're onto something.\nPosted 11 months ago\n Dear All,that is indeed interesting. I also believe that the exact distribution is unknown but it can be approximated by a GammaDistribution, see:https://arxiv.org/pdf/1207.0608.pdfhttps://arxiv.org/pdf/1612.02375.pdfhttps://www.sciencedirect.com/science/article/pii/002608008990030Xhttp://www.scipress.org/journals/forma/pdf/1804/18040221.pdfhttp://www.eurecom.fr/~arvanita/PVT.pdfInterestingly, if you run the code in @Vitaliy Kaurov's cod for 50000 points you obtain: dis = FindDistribution[areas] (*GammaDistribution[3.2718, 0.0000197063]*) I managed to get it done for 500000 points. I think that the edge effects should be getting smaller; as the area increases faster than the circumference. So I obtain: dis = FindDistribution[areas, 5] (* {MixtureDistribution[{0.658342, 0.341658}, {MaxwellDistribution[2.92588*10^-6], GammaDistribution[7.1867, 1.37091*10^-6]}], ExtremeValueDistribution[4.80446*10^-6, 2.73831*10^-6], GammaDistribution[2.94963, 2.17794*10^-6], BetaDistribution[2.94962, 459143.], MixtureDistribution[{0.687855, 0.312145}, {MaxwellDistribution[2.82857*10^-6], LogNormalDistribution[-11.5031, 0.28619]}]}*) This happens when we plot all of them: Plot[Evaluate[PDF[#, x] & /@ dis], {x, 0, 0.00003}, PlotRange -> All, LabelStyle -> Directive[Bold, 16]] If you compare that with the histogram: Show[Histogram[areas, 200, \"PDF\"], Plot[Evaluate[PDF[#, x] & /@ dis], {x, 0, 0.00003}, PlotRange -> All,LabelStyle -> Directive[Bold, 16]]] I would say that the second (ExtremeValueDistribution) fits best:but it does not get the small areas quite right.Regarding the boundary effects, one might also (similar to what Sander suggests) use the torus as the geometry: disfun[x_, y_] := Sqrt[Min[Abs[x[[1]] - y[[1]]], 1 - Abs[x[[1]] - y[[1]]]]^2 + Min[Abs[x[[2]] - y[[2]]], 1 - Abs[x[[2]] - y[[2]]]]^2] M = 50; nf = Nearest[Rule @@@ Transpose[{RandomReal[1, {M, 2}], Range[M]}], DistanceFunction -> disfun] DensityPlot[First[nf[{x, y}]], {x, 0, 1}, {y, 0, 1}, PlotPoints -> 100, ColorFunction -> \"TemperatureMap\"] Cheers,Marco\nPosted 11 months ago\n @Marco Thiel thank you for the references and insight! I believe FindDistribution, while finding correct distributions, needs some tune up to get the distribution parameters right. To test the waters, I tried using EstimatedDistribution and got a bit better results. Let's run this for 500,000 points: pts=RandomPoint[Disk[],500000]; mesh=VoronoiMesh[pts,PerformanceGoal->\"Speed\"]; vor=MeshPrimitives[mesh,2] vorInner=Select[vor,RegionWithin[Disk[],#]&]; areas=Area/@vorInner; hist=Histogram[areas,200,\"PDF\",PlotTheme->\"Detailed\"]; and then find the parameters assuming distributions are known: disGAM = EstimatedDistribution[areas, GammaDistribution[a, b]]  GammaDistribution[3.526166220777205, 1.7809806247800413*^-6] disEXT = EstimatedDistribution[areas, ExtremeValueDistribution[a, b]]  ExtremeValueDistribution[4.775094275466527*^-6, 2.5746412541824736*^-6] We see gamma (red) behaves a bit better now, both for zero and for the tail values: Show[hist,Plot[{PDF[disEXT,x],PDF[disGAM,x]},{x,0,.00002},PlotStyle->{Blue,Red}]] But while p-value of gamma is greater than extreme, it is still pretty low, not sure if there is a catch here somewhere: DistributionFitTest[areas,disGAM]  0.000051399258187534436 DistributionFitTest[areas,disEXT]  0.\nPosted 11 months ago\n Thanks for sharing everybody! Maybe this can add some more information:In the article \"Statistical Distributions of Poisson Voronoi Cells in Two and Three Dimensions\" by Masaharu Tanemura, mentioned by Marco Thiel, the distribution of the perimeter of the cells (and the number of edges etc.) is also analyzed. Here I adapted the code of Vitaly to perimeters and we see (as in the article) that now we have a normal distribution: perimeters = Perimeter /@ vorInner; hist = Histogram[perimeters, Automatic, \"PDF\", PlotTheme -> \"Detailed\"]; dist = FindDistribution[perimeters] Show[hist, Plot[PDF[dist, x], {x, 0, .1}]] And for the number of edges, we find a BinomialDistribution with FindDistribution although the result look almost \"normal\" as we use EstimatedDitribution numberOfEdges[poly_] := Length[First@poly] edges = numberOfEdges /@ vorInner; histE = Histogram[edges, Automatic, \"PDF\", PlotTheme -> \"Detailed\"]; distE = EstimatedDistribution[edges, NormalDistribution[\\[Mu], \\[Sigma]]] Show[histE, Plot[PDF[distE, x], {x, 0, 12}]] distEB = FindDistribution[edges] Plot[CDF[distEB, x], {x, 0, 12}, Filling -> Bottom] \nPosted 11 months ago\n Great, @Erik, thanks for exploring further and sharing!\n You may be interested in this paper, which takes a semi-empirical approach to come up with a closed-form approximate formula: Jarai, Neda: On the size-distribution of Poisson Voronoi cells https://arxiv.org/abs/cond-mat/0406116 At the time when it was written, Mathematica did not have VoronoiMesh, but one could \"trick\" it to quickly compute a Voronoi tessellation by using ListDensityPlot with InterpolationOrder -> 0 and extracting the polygons.", "date": "2020-01-17 22:52:47", "meta": {"domain": "wolfram.com", "url": "https://community.wolfram.com/groups/-/m/t/1620190", "openwebmath_score": 0.3706924021244049, "openwebmath_perplexity": 2806.4014689919177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924793940119, "lm_q2_score": 0.899121377945727, "lm_q1q2_score": 0.8739392174256276}}
{"url": "https://math.stackexchange.com/questions/2957549/find-the-coefficient-of-x10", "text": "# Find the coefficient of $x^{10}$\n\nWe have been given the following function.\n\n$$f(x)$$= $$x$$ +$$x^2$$ + $$x^4$$ + $$x^8$$ + $$x^{16}$$ + $$x^{32}$$ + ...upto infinite terms\n\nThe question is as follows:\n\nWhat is the coefficient of $$x^{10}$$ in $$f(f(x))$$?\n\nI tried solving it myself and I found the answer too but the method of solving was too much time-consuming.\nI had solved it manually by only considering the first four terms of $$f(f(x))$$. This method took me about 10 minutes.\nBut the problem is that this question was asked in a competitive exam called JEE which requires solving the question in max. 3-4 minutes.\nSo, I wanted to know if there was a faster method to solve this problem.\n\nNote that we may at any point in our calculation ignore any terms of degree $$11$$ or higher. Therefore we may also ignore any terms whose inclusion will only lead to degree $$11$$ or higher terms. We have: $$f(f(x)) = (x + x^2 + x^4 + x^8 + \\cdots) + (x + x^2 + x^4 + x^8 +\\cdots)^2 \\\\+ (x + x^2 + x^4 + \\cdots)^4 + (x + x^2 + \\cdots)^8 + \\cdots$$ From here we may look at each of the brackets and simply extract the ones which lead to degree $$10$$, using the multinomial theorem (basically the binomial theorem) for what it's worth:\n\n\u2022 $$x + x^2 + x^4 + x^8 + \\cdots$$: no terms\n\u2022 $$(x + x^2 + x^4 + x^8 + \\cdots)^2$$: we get $$2x^2\\cdot x^8$$\n\u2022 $$(x + x^2 + x^4 + \\cdots)^4$$: we get $$6 (x)^2\\cdot (x^4)^2 + 4(x^2)^3\\cdot x^4$$\n\u2022 $$(x + x^2 + \\cdots)^8$$: we get $$28(x)^6\\cdot(x^2)^2$$\n\nwhere I've used brackets to clarify which terms I've picked in each case.\n\n\u2022 From writing up my expression of $f(f(x))$ and listing the degree-10 terms from each bracket, this shouldn't take that much time. It really depends on how much you have to write in your answer. If you have to write it out like here (or even more detailed), then 10 minutes to solve this problem isn't unreasonable. If you only have to write the final answer, then much of what I've written here may be skipped in your own notes, and you will get at the answer that much quicker. \u2013\u00a0Arthur Oct 16 '18 at 7:21\n\u2022 Thanks a lot for such a fast answer. I did \"almost\" exactly what you did. When I said ten minutes, it included the time taken to come up with a method as well. \u2013\u00a0Vaibhav Oct 16 '18 at 17:21\n\u2022 Also, I am new to Stack Exchange. I didn't expect my question to be a answered within an hour of posting the question. It would have taken weeks if I had asked this on Quora. \u2013\u00a0Vaibhav Oct 16 '18 at 17:23\n\n$$f(x)$$ contains no tenth power.\n\n$$f^2(x)$$ has $$x^2\\cdot x^8$$, with a coefficient $$2$$.\n\n$$f^4(x)$$ has $$x\\cdot x\\cdot x^4\\cdot x^4$$ with a coefficient $$\\dfrac{4!}{2!2!}=6$$ and $$x^2\\cdot x^2\\cdot x^2\\cdot x^4$$ with a coefficient $$\\dfrac{4!}{3!1!}=4$$.\n\n$$f^8(x)$$ has $$x\\cdot x\\cdot x\\cdot x\\cdot x\\cdot x\\cdot x^2\\cdot x^2$$ appearing $$\\dfrac{8!}{6!2!}=28$$ times.\n\nTotal $$40.$$\n\nUsing a CAS,\n\n$$\\cdots+40x^{10}+22x^9+16x^8+8x^7+8x^6+6x^5+3x^4+2x^3+2x^2+x$$\n\n\u2022 Crossed with Arthur. \u2013\u00a0Yves Daoust Oct 16 '18 at 7:22\n\nSince you are looking for $$x^{10}$$ term, you should ask the question: when would I get $$x^{10}$$ from $$f(x)^n$$? When the power $$n$$ of $$x^n$$ is too large, it would not work. So considering the first 4 terms is reasonable. But you don't need to expand everything, for example, if you would consider $$f(x)+f(x)^2+f(x)^4+f(x)^8$$, note that $$f(x)=x+x^2+...$$ does not contain $$x^{10}$$. Next we look at $$f(x)^2$$, which has the form $$(x+x^2+...)(x+x^2+...)$$, expanding this would give $$x^{10}$$ if 2 powers add up to 10, this happens for $$x^2\\cdot x^8=x^{10}$$, and we also have the other term from $$x^8\\cdot x^2$$. As for $$f(x)^4$$, you are always multiplying 4 terms together, so you just need to see if you can get 10 from adding 4 powers, this is possible when $$2+2+2+4=10$$, but the 4 can be chosen from any other \"bracket\", so there are 4 such terms. If you understand what I meant, try to figure out whether $$f(x)^8$$ contains any term of $$x^{10}$$.", "date": "2019-11-16 22:29:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2957549/find-the-coefficient-of-x10", "openwebmath_score": 0.8175811767578125, "openwebmath_perplexity": 193.41107872453253, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692366242306, "lm_q2_score": 0.8947894668039496, "lm_q1q2_score": 0.8739133454828157}}
{"url": "https://math.stackexchange.com/questions/1622928/what-is-the-probability-that-an-integer-between-0-and-9-999-has-exactly-one-8-an", "text": "# What is the probability that an integer between 0 and 9,999 has exactly one 8 and one 9?\n\nI Googled this question and found some answers but they were all different from each other, so I don't know which one is correct.\n\nQuestion: What is the probability that an integer between 0 and 9,999 has exactly one 8 and one 9?\n\nI tried by splitting it into 3 cases.\n\nCase #1 (2 digits): 98 or 89, so only 2!=2 possible ways of arranging 2 digits.\n\nCase #2 (3 digits): One possible outcome is 98y where y can be integers from 0 to 7 (so there's 8 possible values for y). 3!=6 ways of arranging it. So there's 6*8-2 = 46 possible 3 digit numbers. I subtracted 2 from it because 098 and 089 are not 3 digit numbers.\n\nCase #3 (4 digits): A possible outcome is 98xy where x and y can both take on 8 possible values ranging from 0 to 7. 4!/2! = 12 ways of arranging it because if x=y, 98xy is the same as 98yx. So we have: 12*8*8-48 = 720 possible 4 digit numbers. 48 is the result of 46 + 2 from case #2 because we have to subtract outcomes where the first two digits are 0 or the first digit is a 0 eg: 0098.\n\nTherefore, the total is 2 + 46 + 720 = 768 numbers with exactly 8 and 9 in it. The probability is 768/10000.\n\n\u2022 Note that if you allow the four digit numbers to have leading zeroes (i.e. allow numbers like $0098$), then that takes care of two- and three-digit numbers automatically, without having to treat those as a special case, and without having to treat $0$ as a special digit. \u2013\u00a0Arthur Jan 22 '16 at 21:20\n\u2022 Your result is correct. It is a much harder approach than the answers, but you were careful to get them all and once each. \u2013\u00a0Ross Millikan Jan 22 '16 at 21:29\n\u2022 leading zeros are allowed because associating strictly 4 digit numbers with leading zeros allowed with the corresponding 1, 2, 3 or 4 digit numbers without leading zeros is a 1-1 association. e.g. 0071 ~ 71 and nothing else. ... or another way of putting it 0071 and 71 are simply two different ways of writing the same thing. And as the two answers show, using leading zeros makes the calculations much easier. \u2013\u00a0fleablood Jan 22 '16 at 21:45\n\nLet us ask the related question of how many 4-digit strings have exactly one eight and one nine. (Strings are allowed to have leading zeroes, whereas numbers are not) Note that the 4-digit strings are in direct bijection with the integers from $0$ to $9999$.\n\u2022 Pick the location of the nine ($4$ choices)\n\u2022 Pick the location of the eight (It can't be where the nine is, so $3$ choices)\n\u2022 For each remaining position, pick a digit other than eight or nine ($8$ choices each time for a total of $8\\cdot 8$ choices)\nThus, the number of four-digit strings with exactly one eight and one nine is $4\\cdot 3\\cdot 8\\cdot 8$\nSince there are $10^4$ different four-digit strings regardless, the probability is then $\\dfrac{4\\cdot 3\\cdot 8\\cdot 8}{10^4} = \\dfrac{768}{10000} = .0768$\nIf you make all your numbers four digits by allowing leading zeros it is easier. You have four places to put the $9$, then three left to put the $8$, then $8$ choices for each of the other two places. $4 \\cdot 3 \\cdot 8^2=768$ good numbers out of $10000$, so the probability is $0.0768$", "date": "2019-11-17 00:07:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1622928/what-is-the-probability-that-an-integer-between-0-and-9-999-has-exactly-one-8-an", "openwebmath_score": 0.739702582359314, "openwebmath_perplexity": 286.9412669756626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918499186287, "lm_q2_score": 0.8840392878563335, "lm_q1q2_score": 0.8738656310538542}}
{"url": "http://jtbv.sicituradastra.it/power-series-calculator-differential-equations.html", "text": "Power Series Calculator Differential Equations\n\nAn important application of power series in the field of engineering is spectrum analysis. Then solving for \u2206\u03a6 gives 0. Theoretical considerations and convergence of the method for these systems are discussed. The second part of this course introduces series, especially the power series. Covers material on integration methods (trig, partial fractions, etc. Laplace Transform of the Dirac Delta Function using the TiNspire Calculator. This script may help the Calculus (II or III) student with the Infinite Series chapter, and it may also help the Differential Equations student with Series Solutions. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. Prev Up Next. ), probability functions, Taylor polynomials/series approximations, power series\u2019, differential equations (linear and separable), partial derivatives, multivariable functions (and their real-world applications), and double integrals in. Let's nd a solution as a= 0. We again use Maple to find the power series solutions as well. The Equation for the Quantum Harmonic Oscillator is a second order differential equation that can be solved using a power series. T\u00ecm ki\u1ebfm power series solution of differential equations calculator , power series solution of differential equations calculator t\u1ea1i 123doc - Th\u01b0 vi\u1ec7n tr\u1ef1c tuy\u1ebfn h\u00e0ng \u0111\u1ea7u Vi\u1ec7t Nam. Post date: 15 Feb 2011. Yes, y(x) is the general solution of the differential equation represented as a power series. Motivation: Following this discussion about using asymptotic expansions (i. In general, such a solution assumes a power series with unknown coefficients, then substitutes that solution into the differential equation to find a recurrence relation for the coefficients. James Stewart's Calculus, Metric series is the top-seller in the world because of its problem-solving focus, mathematical precision and accuracy, and outstan. Mathematica Stack Exchange is a question and answer site for users of Wolfram Mathematica. Related Math Tutorials: Power Series: Differentiating and Integrating Power Series: Finding the Interval of Convergence; Power Series Representation of a Function; SEARCH. By default, the function equation y is a function of the variable x. 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 50 49 48 47 46 45 44 43 42 41. The basic idea to finding a series solution to a differential equation is to assume that we can write the solution as a power series in the form, y(x) = \u221e \u2211 n=0 an(x\u2212x0)n (2) (2) y. 2 - Maclaurin Series; Lesson 24. The current question is a sequel of a more easy question. Their equations hold many surprises, and their solutions draw on other areas of math. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. Typically, a scientific theory will produce a differential equation (or a system of differential equations) that describes or governs some physical process, but the theory will not produce the desired function or functions directly. Fourier series with Ti84: Calculate for all different periodic signals the spectrum. Specifically, EK 3. The vector from the origin to the point A is given as 6, , , and. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. My longest video yet, power series solution to differential equations, solve y''-2xy'+y=0, www. Conic Sections Trigonometry. A power series solution is all that is available. THE METHOD OF FROBENIUS We have studied how to solve many differential equations via series solutions. 4 The Power Series Method, Part I A187 A. At this time, the reader comes with the simple symbolic tools necessary to execute algebraic, differential, and integral operations. In radio, audio, and light applications, it is very useful to be able to receive a wide range of frequencies and be able to pinpoint which frequencies are the loudest/brightest. Google Scholar. Description of the method. Module 22 - Power Series; Lesson 22. 2 - Maclaurin Series; Lesson 24. THE METHOD OF FROBENIUS We have studied how to solve many differential equations via series solutions. Video tutorial on Power Series Solutions of Differential Equations - In this video, I show how to use power series to find a solution of a differential equation. So, the convergence of power series is fairly important. 1126 CHAPTER 15 Differential Equations In Example 1, the differential equation could be solved easily without using a series. The general Airy differential equation is given by :$D^2y \\pm m^2 x y = 0$or equivalently$y\u2019\u2019 \\pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$. 1 - Power Series; Lesson 24. 2 Series SolutionsNear an Ordinary Point I 320 Elementary Differential Equations with Boundary Value Problems is written for students in science, en-. the form of power series; this explains the name Power series method. Show Instructions. The method works analogously for higher order equations as well as for systems. $y^\\prime = 2|x|$ is kind of an artificial example. Solving differential equations is a combination of exact and numerical methods, and hence. Step by Step - Homogeneous 1. It often happens that a differential equation cannot be solved in terms of elementary functions (that is, in closed form in terms of polynomials, rational functions, e x, sin x, cos x, In x, etc. However, it was inconvenient that he had to alter and apply the power series method with respect to each differential equation in order to study the Hyers-Ulam stability. Abstract: The complete group classification of. 1126 CHAPTER 15 Differential Equations In Example 1, the differential equation could be solved easily without using a series. Homogeneous Differential Equations Calculator. John Forbes Nash Jr Essay John Forbes Nash Jr. If the function is of only one variable, we call the equation an ordinary differential equation (ODE). It often happens that a differential equation cannot be solved in terms of elementary functions (that is, in closed form in terms of polynomials, rational functions, e x, sin x, cos x, In x, etc. Depending upon the domain of the functions involved we have ordinary di\ufb00er-ential equations, or shortly ODE, when only one variable appears (as in equations (1. In this lecture, we will study the solution of the second-order linear differential equations in terms of Power Series. I for jx x 0j<\u02c6and diverges jx x 0j>\u02c6. We will only be able to do this if the point x = x0. Power series solution (PSS) method is an old method that has been limited to solve linear differential equations, both ordinary differential equations (ODE) [1, 2] and partial differential equations (PDE) [3, 4]. The Equation for the Quantum Harmonic Oscillator is a second order differential equation that can be solved using a power series. EXAMPLE2 Power Series Solution Use a power series to solve the differential equation Solution Assume that is a. Equate coefficients of like powers of $$x$$ to determine values for the coefficients $$a_n$$ in the power series. 5 lecture , \u00a73. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Making statements based on opinion; back them up with references or personal experience. Specifically, a precise threshold value is estimated to examine when exchange rate misalignment suppresses capital inflows. Each Differential Equations problem is tagged down to the core, underlying concept that is being tested. Differential Equations: Techniques, Theory, and Applications is designed for a modern first course in differential equations either one or two semesters in length. Xavier Sigaud, 150 CEP 22290-180, Rio de Janeiro, RJ, Brazil. I for jx x 0j<\u02c6and diverges jx x 0j>\u02c6. Approximate solutions of \ufb01rst-order differential equations using Euler and/or Runge-Kutta methods. Houston Math Prep 245,356 views. An important application of power series in the field of engineering is spectrum analysis. The recruit becomes a reproductive adult in the next time step, and begins modifying the abundance of each microbial taxon through time according to the following differential equation:. Although the method may be applied to \ufb01rst order equations, our discussion will center on second order equations. 1126 CHAPTER 15 Differential Equations In Example 1, the differential equation could be solved easily without using a series. This gives a recurrence formula for the coefficients. 2 Power Series, Analytic Functions, and the Taylor Series Method 431 8. Question: In this exercise we consider the second order linear equation {eq}y'' + 4y = 0 {/eq}. On our site OnSolver. Laplace transforms. 7MB) To complete the reading assignments, see the Supplementary Notes in the Study Materials section. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction Pre Calculus Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. Added Aug 1, 2010 by Hildur in Mathematics. EXAMPLE 1 Use power series to solve the equation. Well in order for a series solution to a differential equation to exist at a particular x it will need to be convergent at that x. Differential Equation Calculator. \u2022 Proceeding just as for series but now in voltage (1) Using KCL to write the equations: 0 0 1 vdt I R L v dt di C t + + \u222b = (2) Want full differential equation \u2022 Differentiating with respect to time 0 1 1 2 2 + + v = dt L dv R d v C (3) This is the differential equation of second order \u2022 Second order equations involve 2nd order derivatives. Calculus is the mathematics of change, and rates of change are expressed by derivatives. The term \"ordinary\" is used in contrast with the term. Let\u2019s consider the equation: 2 \u2032\u2032 + + \u2032 x y x x y y \u2212 = 2 7 ( 1) 3 0 (1). 3 Systems of ODEs. Use MathJax to format equations. We propose a computational method to determine when a solution modulo a certain power of the independent variable of a given algebraic differential equation (AODE) can be extended to a formal power series solution. 2 - Series and Sequences of Partial Sums; Lesson 21. Power Series Ordinary Differential Equations Esteban Arcaute1 We then need partial fractions to calculate Z dy N(y) = Z Q(y) P(y) dy. For example, diff(y,x) == y represents the equation dy/dx=y. A Differential Equation is a n equation with a function and one or more of its derivatives:. SOLUTION We assume there is a solution of the form. If the function is of only one variable, we call the equation an ordinary differential equation (ODE). So, why are we worried about the convergence of power series? Well in order for a series solution to a differential equation to exist at a particular x it will need to be convergent at that x. This might introduce extra solutions. 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 50 49 48 47 46 45 44 43 42 41. In following section, 2. The number \u03c1 is at least 0, as taking x = x0 gives P 0 which is clearly converging to 0; On the other hand, when the power series is convergent for all x, we say its radius of convergence is in\ufb01nity,. Sturm-Liouville problems. The differential equation in Example 2 cannot be solved by any of the methods discussed in previous sections. Use the power series method to find 2 linearly independent series solutions to differential equation (x^2 - 2x + 2)y\" + 2(x-1)y' - 2y = 0 about the ordinary point xo=1. Sage Quickstart for Differential Equations\u00b6. Pourhabib Yekta1, A. Browse other questions tagged sequences-and-series ordinary-differential-equations power-series or ask your own question. The differential equation can be writ-ten in. After a promising start to his mathematical career, Nash began to suffer from schizophrenia around his 30th year, an illness from which he has. An older book that has a lot of nice material on power series and other numerical methods for ODE's is Einar Hille's Lectures On Ordinary Differential Equations. Let\u2019s consider the equation: 2 \u2032\u2032 + + \u2032 x y x x y y \u2212 = 2 7 ( 1) 3 0 (1). With the exception of special types, such as the Cauchy equations, these will generally require the use of the power series techniques for a solution. The Bessel differential equation has the form x 2 y+xy'+(x 2-n 2)y=0. Power Series Method for Nonlinear Partial Differential Equations Power series is an old technique for solving linear ordinary differential equations [7,20]. Approximate solutions of \ufb01rst-order differential equations using Euler and/or Runge-Kutta methods. The general Airy differential equation is given by :$D^2y \\pm m^2 x y = 0$or equivalently$y\u2019\u2019 \\pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$. Chapter 7 Power series methods 7. 1 in [BD] Many functions can be written in terms of a power series X1 k=0 a k(x x 0)k: If we assume that a solution of a di erential equation is written as a power series, then perhaps we can use a method reminiscent of undetermined coe cients. In this section we learn how to extend series solutions to a class of differential equations that appear at first glance to diverge in our region of interest. Solution of linear equations by power series Def. Use the power series method to find 2 linearly independent series solutions to differential equation (x^2 - 2x + 2)y\" + 2(x-1)y' - 2y = 0 about the ordinary point xo=1. Given three points, A, , , B, , , and C, , : a Specify the vector A extending from the origin to the point A. An in\ufb01nite series of this type is called a power series. An excellent article in the American Journal of Physics, by Fairen, Lopez, and Conde develops power series approximations for various systems of nonlinear differential equations. With the exception of special types, such as the Cauchy equations, these will generally require the use of the power series techniques for a solution. 8 Nonlinear systems. Linear and separable first order differential equations. math 230 psu reddit, Review of calculus, properties of real numbers, infinite series, uniform convergence, power series. Finding coefficients in a power series expansion of a rational function. shirin setayesh 55,974 views. 79: conditions calculate Cauchy-Euler equation portrait population pounds power series predictions recurrence. 3 - Recursively Defined Sequences. This method aims to find power series for the solution functions to a differential equation. Equations relating the partial derivatives (See: Vector calculus ) of a function of several variables are called partial differential equations (PDEs). Example $$\\PageIndex{2}$$ Find the the first three nonzero terms of two linearly independent solutions to $$xy'' + 2y = 0$$. Equate coefficients of like powers of $$x$$ to determine values for the coefficients $$a_n$$ in the power series. If initial conditions are given, determine the particular solution. Home PDF Paperback Index PrevUp Next. Power series solution is a method to solve the differential equations. \u2022 Use power series to find solutions to higher order linear differential equation with nonconstant coefficients at any regular singular point; and \u2022 Use Laplace transforms to solve initial value problems. Comprehensive & Detailed COMPLETE note package for the course Math 128 (Calculus II). This RPS method gives approximate solutions in convergent series formula with surely computable components. The differential equation can be writ-ten in. Shifting the Index for Power Series - Duration: 14:49. Differential Equations for Engineers. \u2019s need to be. Research Article Power Series Extender Method for the Solution of Nonlinear Differential Equations HectorVazquez-Leal 1 andArturoSarmiento-Reyes 2 Electronic Instrumentation and Atmospheric Sciences School, Universidad Veracruzana, Cto. An ordinary differential equation (ODE) is an equation containing an unknown function of one real or complex variable x, its derivatives, and some given functions of x. If the function is of only one variable, we call the equation an ordinary differential equation (ODE). An important application of power series in the field of engineering is spectrum analysis. Power series neural network solution for ordinary differential equations with initial conditions Abstract: Differential equations are very common in most academic fields. {image} {image} {image} {image} 3. Now We have two components R and L connected in Series and a voltage source to those components as shown below. For the differential equation show below:y(1) = 1y'(1) = 0a) Write down a general expansion for a power series solution y(x) about the point Xo = 1b)Find a recurrence relationshi between the coefficients of your power series expansion. Notice that 0 is a singular point of this differential equation. Get this from a library! Formal power series and linear systems of meromorphic ordinary differential equations. 3 Power Series Solutions to Linear Differential Equations 442 8. We will only be able to do this if the point x = x0. Since cos x = \u03a3(n=0 to \u221e) (-1)^n x^(2n)/(2n)!, so u can get by potential of heart a million qn and remedy each and every) Non-linear partial differential equation, Homogenous and non-homogeneous. In this video from PatrickJMT we show how to use power series to find a solution of a differential equation. In the equation, represent differentiation by using diff. 3 - Taylor Series. Fourier series with Ti84: Calculate for all different periodic signals the spectrum. y'' \u2212 y' = 0. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. 2) By insertion of y = n=0 anx n,y = n=1 nanx n 1,y = n=2 n(n 1)anx n 2, Solution of differential equations by the power series method. Illustrative numerical example is included to demonstrate efficiency. equation is given in closed form, has a detailed description. c) Write down the first few terms of your power series (up to the fourth. We have step-by-step solutions for your textbooks written by Bartleby experts!. 1126 CHAPTER 15 Differential Equations In Example 1, the differential equation could be solved easily without using a series. Full curriculum of exercises and videos. A power series solution to a differential equation is a function with infinitely many terms, each term containing a different power of the dependent variable. In following section, 2. Simple Ordinary Differential Equations may have solutions in terms of power series whose coefficients grow at such a rate that the series has a radius of convergence equal to zero. AMS30, 151-156 (1971). Underdamped Overdamped Critically Damped. Use MathJax to format equations. Conic Sections Trigonometry. An older book that has a lot of nice material on power series and other numerical methods for ODE's is Einar Hille's Lectures On Ordinary Differential Equations. Since many physical laws and relations appear mathematically in the form of differential equations, such equations are of fundamental importance in engineering mathematics. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. Learn AP\u00ae\ufe0e Calculus BC for free\u2014everything from AP\u00ae\ufe0e Calculus AB plus a few extra goodies, such as Taylor series, to prepare you for the AP\u00ae\ufe0e test. We\u2019ll assume mechanical boresight (\u03b8 = 0\u00ba), N = 8, and d = \u03bb/2. Now We have two components R and L connected in Series and a voltage source to those components as shown below. The validity of term\u2010by\u2010term differentiation of a power series within its interval of convergence implies that first\u2010order differential equations may be solved by assuming a solution of the form. The Differential Equations diagnostic test results highlight how you performed on each area of the test. Related Calculators. I am now at this point where I have got: \\displaystyle \\sum_{n = 0}^&#. It often happens that a differential equation cannot be solved in terms of elementary functions (that is, in closed form in terms of polynomials, rational functions, e x, sin x, cos x, In x, etc. Covers material on integration methods (trig, partial fractions, etc. Video tutorial on Power Series Solutions of Differential Equations - In this video, I show how to use power series to find a solution of a differential equation. This might introduce extra solutions. Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. In dissertation we discuss power series characteristics that we use for solving the equations in question. Generalized series expansions involving integer powers and fractional powers in the independent variable have recently been shown to provide solutions to certain linear fractional order differential equations. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. Re-index sums as necessary to combine terms and simplify the expression. In our world things change, and describing how they change often ends up as a Differential Equation: an equation with a function and one or more of its derivatives: Introduction to Differential Equations; Differential Equations Solution Guide; Separation of Variables. View and Download PowerPoint Presentations on Solution Differential Equation By Power Series PPT. Even if you don't know how to find a solution to a differential equation, you can always check whether a proposed solution works. Mathematics > Calculus and Analysis > Differential Equations Keywords Calculus, series expansion, Taylor Series, Ordinary Differential Equation, ODE, , Power Series. Order Differential Equations with non matching independent variables (Ex: y'(0)=0, y(1)=0 ) Step by Step - Inverse LaPlace for Partial Fractions and linear numerators. Determining the value of a definite integral on the graphing calculator. Find the singular points (if any) for the following equations. How calculators calculate is by power series. Xavier Sigaud, 150 CEP 22290-180, Rio de Janeiro, RJ, Brazil. Exercises 8. Given a linear differential equation with polynomial coefficients a point x = x 0 is called an ordinary point if b 0 (x 0) 0. De\ufb01nition 5. Taylor's Series method. 5 The Power Series Method. To leave a comment or report an error, please use the auxiliary blog. Each Differential Equations problem is tagged down to the core, underlying concept that is being tested. = 1 for y at x = 1 with step length 0. This Sage quickstart tutorial was developed for the MAA PREP Workshop \"Sage: Using Open-Source Mathematics Software with Undergraduates\" (funding provided by NSF DUE 0817071). Special Functions The power series method gives solutions of linear ODEs (1) y\u201d + p(x)y\u2019 + q(x)y = 0 with variable coefficients p and q in the form of a power series (with any center x0, e. lol this is a question you'll look back on after doing more math/physics and laugh. I Check the endpoints, jx x 0j= \u02c6, separately to nd the INTERVAL of CONVERGENCE. {image} {image} {image} {image} 3. Solutions by separation of variables and expansion in Fourier Series or other appropriate orthogonal sets. 1 - Sequences; Lesson 21. In every upper division physics class you will use a power series. The unknown function is generally represented by a variable (often denoted y), which, therefore, depends on x. We propose a power series extender method to obtain approximate solutions of nonlinear differential equations. Hermite\u2019s differential equation shows up during the solution of the Schr\u00f6dinger equation for the harmonic oscillator. the menu option in Differential Equations Made Easy from www. AN EXAMPLE. A di\ufb00erential equation, shortly DE, is a relationship between a \ufb01nite set of functions and its derivatives. Calculus: Difference Equations to Differential Equations ADD. 3 - Recursively Defined Sequences. 2 Power Series, Analytic Functions, and the Taylor Series Method 431 8. In our world things change, and describing how they change often ends up as a Differential Equation: an equation with a function and one or more of its derivatives: Introduction to Differential Equations; Differential Equations Solution Guide; Separation of Variables. 1 Power series Note: 1 or 1. Solution of linear equations by power series Def. Get this from a library! Formal power series and linear systems of meromorphic ordinary differential equations. In dissertation we discuss power series characteristics that we use for solving the equations in question. The general Airy differential equation is given by :$D^2y \\pm m^2 x y = 0$or equivalently$y\u2019\u2019 \\pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$. Q3: Find the series solution for the following ordinary differential equation using the Frobenius method: \ud835\udc65 \ud835\udc66 + \ud835\udc65 \ud835\udc65 + 1 2 \ud835\udc66 \u2212 \ud835\udc65 + 1 2 \ud835\udc66 = 0. 1 of 3 Go to page. In general, such a solution assumes a power series with unknown coefficients, then substitutes that solution into the differential equation to find a recurrence relation for the coefficients. order Power series solution of differential-algebraic equations in (1. In this lecture, we will study the solution of the second-order linear differential equations in terms of Power Series. Nonlinear Differential Equations Fourier series For a periodicfunction one may write The Fourier series is a \"best fit\" in the least square sense of data fitting y(t +T) =y(t) ()cos( ) sin( ), a plot of versus is called the power spectrum \u222b +\u221e \u2212\u221e. We also discuss more about initial conditions and how they determine the first two coefficients in the power series solution. Next enter the c value and view the Laplace transform below the entry box. The general idea is as follows: Assume that the solution function has a power series that converges to it. This method aims to find power series for the solution functions to a differential equation. Video # Video Tutori\u00adal Title: Remarks: 1: Clas\u00adsi\u00adfic\u00ada\u00adtion of Dif\u00adfer\u00aden\u00adtial Equations. We will only be able to do this if the point x = x0. Give the first four non-zero terms of each of the 2 independent solutions, y1. In this thesis, the reader will be made aware of methods for finding power series solutions to ordinary differential equations. S = dsolve(eqn) solves the differential equation eqn, where eqn is a symbolic equation. There is a Review Sheet (with Solutions). Module 25 - Parametric. But first: why?. 1 - Power Series; Lesson 22. It's more plug-and-chug and you should do well if you can match up the differential equation to the approach used to solve it. Topic: Differential Equations, Sequences and Series. Module 21 - Sequences and Series; Lesson 21. 2 Linear Ordinary Di\ufb00erential Equations with Constant Coe\ufb03cients A174 A. In some cases, these power series representations can be used to find solutions to differential equations. The method works analogously for higher order equations as well as for systems. In the equation, represent differentiation by using diff. Convergent Power Series of sech \u2061 ( x ) and Solutions to Nonlinear Differential Equations Article (PDF Available) in International Journal of Differential Equations 2018(1-2):1-10 \u00b7 February. However, you can specify its marking a variable, if write, for example, y(t) in the equation, the calculator will automatically recognize that y is a function of the variable t. Taylor Series Calculator - an Introduction. The dsolve function finds a value of C1 that satisfies the condition. Let's nd a solution as a= 0. Hi and welcome back to the differential equations lectures here on www. Below is one of them. Power Series Solution to Non-Linear Partial Differential equations of Mathematical Physics. delay differential equations using the residual power series method (RPSM), which obtains. In the first part of this course, the student learns to solve the most common types of differential equations. Derivative Calculator Integral Calculator Limit Calculator. Fourier series with Ti84: Calculate for all different periodic signals the spectrum. , x0 = 0) (2) y( x) am ( x x0 )m a0 a1 ( x x0 ) a2 ( x x0 )2. What is more, we present the post-treatment of the power series. Access course-tailored video lessons, exam-like quizzes, mock exams & more for MATH 118 at Waterloo. The dsolve function finds a value of C1 that satisfies the condition. Example: t y\u2033 + 4 y\u2032 = t 2 The standard form is y t t. Differential Equation Calculator. Solved Examples of Differential Equations Sunday, July 9, 2017 Find the first 6 non-zero terms of the power series expansion about x = 0 for a general solution to the given differential equation y'' - x^2y' - xy = 0. Hi and welcome back to the differential equations lectures here on www. Simple Ordinary Differential Equations may have solutions in terms of power series whose coefficients grow at such a rate that the series has a radius of convergence equal to zero. The validity of term\u2010by\u2010term differentiation of a power series within its interval of convergence implies that first\u2010order differential equations may be solved by assuming a solution of the form. Let's study about the order and degree of differential equation. KEYWORDS: Course Materials, Lecture notes, Laboratories, HW Problems SOURCE: Joseph M. The unknown function is generally represented by a variable (often denoted y), which, therefore, depends on x. 2 using Taylor series method of order four. 100, 61111 Ljubljana, Slovenia and Racah Institute of Physics,. 2 Power Series, Analytic Functions, and the Taylor Series Method 431 8. Answer all the questions. This method aims to find power series for the solution functions to a differential equation. Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. Illustrative numerical example is included to demonstrate efficiency. ), probability functions, Taylor polynomials/series approximations, power series\u2019, differential equations (linear and separable), partial derivatives, multivariable functions (and their real-world applications), and double integrals in. {image} {image} {image} {image} 3. Modeling with Differential Equations Solve a problem in the physical sciences (such as a growth or decay problem, a mixture problem, or a Newton\u2019s Law of Cooling problem) whose solution utilizes a first-order linear differential equation. 3 - Taylor Series. The coefficient functions in these series are obtained by recursively iterating a simple integration process, begining with a solution system for $\\lambda=0$. This paper presents a Modified Power Series Method (MPSM) for the solution of delay differential equations. By solving such equations, we mean computing a vector F of power series such that (1) holds modulo xN. In the equation, represent differentiation by using diff. 2) for all x implies, by the nth term test for diver-gence, that lim n\u2192\u221e n! = 0 (5. 5 Cauchy-Euler (Equidimensional) Equations Revisited 459. Here we have discussed an Ordinary and singular point for linear Second Order Differential Equations, classification of Singular Point and method to solve Differential Equations about an Ordinary Point. Print Book & E-Book. Solving Separable First Order Differential Equations \u2013 Ex 1 Homogeneous Second Order Linear Differential Equations Power Series Solutions of Differential Equations. try to explain any differences between the two forms of the solution. 1 - Sequences; Lesson 23. Stefan, Jamova 39, P. EXCHANGE RATE MISALIGNMENT AND CAPITAL INFLOWS: AN ENDOGENOUS THRESHOLD ANALYSIS FOR MALAYSIAABSTRACTThis study presents an attempt to investigate the impact of exchange rate misalignment on capital inflows in Malaysia. Find more Mathematics widgets in Wolfram|Alpha. $y^\\prime = 2|x|$ is kind of an artificial example. HERMITE DIFFERENTIAL EQUATION - GENERATING FUNCTIONS Link to: physicspages home page. Why most of Hille's texts-which are all wonderful-are out of print mystifies me. Review of Series and Power Series. This Sage quickstart tutorial was developed for the MAA PREP Workshop \"Sage: Using Open-Source Mathematics Software with Undergraduates\" (funding provided by NSF DUE 0817071). However, you can specify its marking a variable, if write, for example, y(t) in the equation, the calculator will automatically recognize that y is a function of the variable t. Initialization. In a few upper div math classes, like differential equations, real and complex analysis you'll see them. In Introduction to Power Series , we studied how functions can be represented as power series, We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. It is interesting to know whether sympy supports such equations along with usual ones. Why most of Hille's texts-which are all wonderful-are out of print mystifies me. Initialization. Sympy: how to solve algebraic equation in formal. 5 Cauchy-Euler (Equidimensional) Equations Revisited 459. Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. Differential equation. It is shown how to obtain such an initializing system working upwards from equations of lower order. Sage Quickstart for Differential Equations\u00b6. He shared the 1994 Nobel Prize for economics with two other game theorists, Reinhard Selten and John Harsanyi. Order Differential Equations with non matching independent variables (Ex: y'(0)=0, y(1)=0 ) Step by Step - Inverse LaPlace for Partial Fractions and linear numerators. My class, and many other's, continue onto power series solutions of differential equations. Convergent Power Series of sech \u2061 ( x ) and Solutions to Nonlinear Differential Equations Article (PDF Available) in International Journal of Differential Equations 2018(1-2):1-10 \u00b7 February. Power Series Method for Linear Partial Differential Equations of Fractional Order 73 Lemma 2. Practice your math skills and learn step by step with our math solver. Solve y0 +(2x 1)y = 0 with initial conditions y(0) = 2. SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS\u2014 SOME WORKED EXAMPLES First example Let\u2019s start with a simple differential equation: \u2032\u2032\u2212 \u2032+y y y =2 0 (1) We recognize this instantly as a second order homogeneous constant coefficient equation. Finding coefficients in a power series expansion of a rational function. Answers to Solving Ordinary Differential Equations with Power Series Here are the answers to the practice questions I provide throughout this chapter. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. In fact, every linear meromorphic system has a formal solution of a certain form, which can be relatively easily computed, but which generally involves such power. Power series representations of functions can sometimes be used to find solutions to differential equations. Differential Equations. Mathematical concepts and various techniques are presented in a clear, logical, and concise manner. Learn AP\u00ae\ufe0e Calculus BC for free\u2014everything from AP\u00ae\ufe0e Calculus AB plus a few extra goodies, such as Taylor series, to prepare you for the AP\u00ae\ufe0e test. Initial conditions are also supported. This page covers two areas related to Fourier Series. We will only be able to do this if the point x = x0. If m 1 < 0 The Caputo fractional derivative is considered here because it allows traditional initial and boundary conditions to be included in the formulation of. Solving Separable First Order Differential Equations \u2013 Ex 1 Homogeneous Second Order Linear Differential Equations Power Series Solutions of Differential Equations. One-variable linear equations Calculator; One-variable linear inequalities Calculator; Operations with infinity Calculator; Perfect square trinomial Calculator; Polynomial factorization Calculator; Polynomial long division Calculator; Polynomials Calculator; Power of a product Calculator; Power rule Calculator; Power series Calculator; Powers. Homogeneous Differential Equations Calculator. equation, method of Laplace transforms for solving ordinary differential equations, series solutions (power series, Frobenius method); Legendre and Bessel functions and their orthogonal properties; Systems of linear first order ordinary differential equations. This RPS method gives approximate solutions in convergent series formula with surely computable components. 5 lecture , \u00a73. 1 First order equations. The Organic. Module 23 - Parametric. In general, such a solution assumes a power series with unknown coefficients, then substitutes that solution into the differential equation to find a recurrence relation for the coefficients. In this work, we studied that Power Series Method is the standard basic method for solving linear differential equations with variable coefficients. MA 401 Applied Differential Equations II (Wave, heat and Laplace equations. I Check the endpoints, jx x 0j= \u02c6, separately to nd the INTERVAL of CONVERGENCE. {image} {image} {image} {image} 3. You can view the Laplace Table that you will be given on the Exam. Practice 2610. Specifically, a precise threshold value is estimated to examine when exchange rate misalignment suppresses capital inflows. You can view the Laplace Table that you will be given on the Exam. This is simply a matter of plugging the proposed value of the dependent variable into both sides of the equation to see whether equality is maintained. The solution diffusion. But first: why?. John Forbes Nash Jr Essay John Forbes Nash Jr. lol this is a question you'll look back on after doing more math/physics and laugh. Example The differential equation ay00 +by0 +cy = 0 can be solved by seeking exponential solutions with an unknown exponential factor. Power Series; Method of series solutions; 6 The Laplace. After finding the constants a 2 ,a 3 ,a 4 , etc I replaced them in y(x) and factored out the undetermined coefficients a 0 and a 1. 7 Power series methods. using traditional way with pencil and paper. Department of Mathematics, Rasht Branch, Islamic Azad University, Rasht, Iran2 Abstract: In this article power series method, as well-known method for. The extension of fractional power series solutions for linear fractional differential equations with variable coefficients is considered. Since cos x = \u03a3(n=0 to \u221e) (-1)^n x^(2n)/(2n)!, so u can get by potential of heart a million qn and remedy each and every) Non-linear partial differential equation, Homogenous and non-homogeneous. SOLVING DIFFERENTIAL EQUATIONS USING POWER SERIES 4 (2) Plug the expression (1) for y(x) into the di erential equation; (3) Manipulate the resulting equation to obtain an equation in which single power series expression (rather that a sum of several power series) is set equal to zero. In this work, we studied that Power Series Method is the standard basic method for solving linear differential equations with variable coefficients. We propose a power series extender method to obtain approximate solutions of nonlinear differential equations. 2, the power series method is used to derive the wave function and the eigenenergies for the quantum harmonic oscillator. THE METHOD OF FROBENIUS We have studied how to solve many differential equations via series solutions. The coe cient functions here are constants, so the power series solution can be computed at any point aand the radius of convergence will be R=1. Houston Math Prep 245,356 views. Conic Sections Trigonometry. After finding the constants a 2 ,a 3 ,a 4 , etc I replaced them in y(x) and factored out the undetermined coefficients a 0 and a 1. Stefan, Jamova 39, P. using traditional way with pencil and paper. Together we will learn how to express a combination of power series as a single power series. Applications of Fourier Series to Differential Equations Fourier theory was initially invented to solve certain differential equations. I used the power series method to solve the differential equation y''+y=0 with. So, the convergence of power series is fairly important. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. To pursue these objectives, this study relies on the. James Stewart's Calculus, Metric series is the top-seller in the world because of its problem-solving focus, mathematical precision and accuracy, and outstan. This particular number \u03c1 is called the r adiu s of c onv er ge nc e. Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. We will only be able to do this if the point x = x0. Question: Find two power series solutions of the given differential equations about the ordinary point {eq}x = 0 {/eq}. Solve differential equation using \"Power Series\" Thread starter EmilyL; Start date Jul 23, 2010; Tags differential differential equations equation power series solve; Home. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. HAFTEL Code 6651, Naval Research Laboratory, Washington, DC 20735-5345 R. Browse other questions tagged ordinary-differential-equations power-series or ask your own question. Motivation: Following this discussion about using asymptotic expansions (i. ODEs Summer08 Esteban Arcaute Introduction First Order ODEs Separation of Power Series Exact Equation End Thus, if the equation is exact, we have f(x,y) = c. Module 25 - Parametric. 307) than what Sal got by raising both sides to the power of e: 2. , in the form. KEYWORDS: Course Materials Calculus for Biology I ADD. Use power series to solve the differential equation. The chapter discusses a method by which the coefficients in the power series expansions of the solutions can be calculated. Those I have learnt in lecture and online are mostly with only one part of summation or two parts with two distinctive roots and two constants. Provide details and share your research! But avoid \u2026 Asking for help, clarification, or responding to other answers. Linear and separable first order differential equations. Xavier Sigaud, 150 CEP 22290-180, Rio de Janeiro, RJ, Brazil. AN EXAMPLE. Up to 25 % of the generating costs relate to mainte- nance. Mathematics > Calculus and Analysis > Differential Equations Keywords Calculus, series expansion, Taylor Series, Ordinary Differential Equation, ODE, , Power Series. 4 Fourier series and PDEs. Substituting y = ert into the equation gives a solution if the quadratic equation ar2 +br+c = 0 holds. compare the series solutions with the solutions of the differential equation obtained using the method of section 4. Welcome! This is one of over 2,200 courses on OCW. Depending upon the domain of the functions involved we have ordinary di\ufb00er-ential equations, or shortly ODE, when only one variable appears (as in equations (1. Let's consider the power series solution of the Hermite differential equation: ${\\displaystyle u''-2xu'=-2 n u}$ ${\\displaystyle u''-2xu'+2 n u =0 \\qquad (1)}$ The solutions to the Hermite differential equation ca be expresse. EXAMPLE2 Power Series Solution Use a power series to solve the differential equation Solution Assume that is a. 5 lecture , \u00a73. Solving the Systems of Differential Equations by a Power Series Method A. Conic Sections Trigonometry. An in\ufb01nite series of this type is called a power series. form a fundamental system of solutions for Airy's Differential Equation. Shifting the Index for Power Series - Duration: 14:49. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 17. Such an expression is nevertheless an entirely valid solution, and in fact, many specific power series that arise from solving particular differential equations have been extensively studied and hold prominent places in mathematics and physics. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. 3 - Second-Order Differential Equations. We will solve this using power series technique. Recently, the first author applied the power series method to studying the Hyers-Ulam stability of several types of linear differential equations of second order (see [26-34]). Worked example: exponential solution to differential equation. Solving Differential Equations with Power Series - Duration: 18:29. Do the differential equation solvers - Support ordinary differential equations; systems of differential equations, and boundary value problems both at the command line and in solve blocks that use natural notation to specify the DiffEQs and constraints. 4 The Power Series Method, Part I A187 A. Finding coefficients in a power series expansion of a rational function. For example, here's a differential equation [\u2026]. Let's consider the power series solution of the Hermite differential equation: ${\\displaystyle u''-2xu'=-2 n u}$ ${\\displaystyle u''-2xu'+2 n u =0 \\qquad (1)}$ The solutions to the Hermite differential equation ca be expresse. In the cases where series cannot be reduced to a closed form expression an approximate answer could be obtained using definite integral calculator. lol this is a question you'll look back on after doing more math/physics and laugh. For example, diff(y,x) == y represents the equation dy/dx=y. Find the Taylor series expansion of any function around a point using this online calculator. We will only be able to do this if the point x = x0. 3 - Recursively Defined Sequences. And find the power series solutions of a linear first-order differential equations whose solutions can not be written in terms of familiar functions such as polynomials, exponential or trigonometric functions, as SOS Math so nicely states. The Bessel differential equation has the form x 2 y+xy'+(x 2-n 2)y=0. Around the Point a = (default a = 0) Maximum Power of the Expansion: How to Input. 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 50 49 48 47 46 45 44 43 42 41. De\ufb01nition 5. \u2019s need to be. Introduction to Differential Equations. Write y' as dy/dx and the answer follows relatively - dy/dx + 3(a million+ x^2)y = 0 => dy/dx = -3(a million+x^2)y => dy/y = -3(a million+x^2)dx Now integrating the two components supplies ln(y) = -3x - x^3 + C (C is unknown integration consistent) => y = ok exp(-3x - x^3) the place ok = exp(C), and can be solved utilising extra suitable education approximately y (as an occasion the fee of y. In advanced treatments of calculus, these power series representations are often used to de\ufb01ne the exponential. I used the power series method to solve the differential equation y''+y=0 with y[0]=0 and y'[0]=1 using the following code. Print Book & E-Book. 5 lecture , \u00a73. Linear methods applied to the solution of differential equations. The term \"ordinary\" is used in contrast with the term. Module 25 - Parametric. The Overflow Blog This week, #StackOverflowKnows molecule rings, infected laptops, and HMAC limits. Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. and then try to determine what the an. 4 Equations with Analytic Coefficients 453 *8. Back to top; 6: Power Series Solutions of Differential Equations; 6. Differential equation. In the first part of this course, the student learns to solve the most common types of differential equations. It is interesting to know whether sympy supports such equations along with usual ones. Even if this is the case, for simplicity we will see how the method works with a problem whose solution is a known function. Pourhabib Yekta1, A. Return to Dif\u00adfer\u00aden\u00adtial Equations. 0012 Power Series Solution of Coupled Differential Equations in One Variable M. 1 in [EP], \u00a75. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. equation, method of Laplace transforms for solving ordinary differential equations, series solutions (power series, Frobenius method); Legendre and Bessel functions and their orthogonal properties; Systems of linear first order ordinary differential equations. In the previous solution, the constant C1 appears because no condition was specified. Solve y0 = x2y with initial conditions y(0) = 1. The degree of an equation is the power to which the highest order term is raised. This paper presents a Modified Power Series Method (MPSM) for the solution of delay differential equations. which makes calculations very simple and interesting. 3 - Recursively Defined Sequences. We will then move to a problem whose solution can be expressed as a series only. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. In the first part of this course, the student learns to solve the most common types of differential equations. S = dsolve(eqn) solves the differential equation eqn, where eqn is a symbolic equation. Various visual features are used to highlight focus areas. 1) The equation is linear of second order with polynomial coecients. EXAMPLE 2 Power Series Solution Use a power series to solve the differential equation Solution Assume that is. Notice that 0 is a singular point of this differential equation. A power series represents a function f on an interval of convergence, and you can successively differentiate the power series to obtain a series for and so. The general Airy differential equation is given by :$D^2y \\pm m^2 x y = 0$or equivalently$y\u2019\u2019 \\pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$. Power Series Method for Nonlinear Partial Differential Equations Power series is an old technique for solving linear ordinary differential equations [7,20]. You can view the Laplace Table that you will be given on the Exam. 0012 Power Series Solution of Coupled Differential Equations in One Variable M. Power Series Solutions Differential Equations Power Series Solutions Differential Equations Yeah, reviewing a ebook Power Series Solutions Differential Equations could accumulate your near associates listings. Review of Series and Power Series. In this video we solve another differential equation by finding a power series solution. First order numerical / graphical differential equation solver: Transient analysis of RC or RL circuits. The solution diffusion. Each of the following waveform plots can be clicked on to open up the full size graph in a separate window. Solve y00 = xy0 +y with initial conditions y(0) = 1 and y0(0) = 0. AMS30, 151-156 (1971). Covers material on integration methods (trig, partial fractions, etc. The solution of the general differential equation dy/dx=ky (for some k) is C\u22c5e\u1d4f\u02e3 (for some C). This equation with concrete values of the parameter appeared in the articles by F. DIFFERENTIAL EQUATIONS FOR ENGINEERS This book presents a systematic and comprehensive introduction to ordinary differential equations for engineering students and practitioners. We have step-by-step solutions for your textbooks written by Bartleby experts!. Laplace\u2019s transformation maps a differential equation onto an algebraic equation, which can be solved relatively easily. Home \u00bb Supplemental Resources \u00bb Calculus Revisited: Complex Variables, Differential Equations, and Linear Algebra \u00bb Part II: Differential Equations \u00bb Lecture 6: Power Series Solutions Lecture 6: Power Series Solutions. SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS\u2014 SOME WORKED EXAMPLES First example Let\u2019s start with a simple differential equation: \u2032\u2032\u2212 \u2032+y y y =2 0 (1) We recognize this instantly as a second order homogeneous constant coefficient equation. Find materials for this course in the pages linked along the left. Series of Solutions Review of Power Series The Recurrence Power Solutions about an Ordinary Point Euler Equations Series Solutions Near a Regular Singular Point Equations of Hypergeometric Type Bessel\u2019s Equations Legendre\u2019s Equation Orthogonal Polynomials Review Questions for Chapter 6 Applications of Higher Order Differential Equations. Delay Differential Equations, Power Series, Taylor Series, Newton's Method 1. 7MB) To complete the reading assignments, see the Supplementary Notes in the Study Materials section. First\u2010order equations. In the next section, Section 8. Type: Artigo de peri\u00f3dico: Title: Group Classification Of A Generalized Black-scholes-merton Equation: Author: Bozhkov Y. This is a simple example and the final solution is very nice compared to what would normally happen with a more complicated differential equation, so please be aware of that!. order Power series solution of differential-algebraic equations in (1. My longest video yet, power series solution to differential equations, solve y''-2xy'+y=0, www. which makes calculations very simple and interesting. Power Series Method for Linear Partial Differential Equations of Fractional Order 73 Lemma 2. polynomial differential equations has a solution of this form. Differential equation. The ideas that you guess a power series solution to a differential equation and then you plug it in and in order to plug it in, you got to calculate its derivatives. First order numerical / graphical differential equation solver: Transient analysis of RC or RL circuits. The governing equation is also based on Kirchoff's law as described below. Back to top; 6: Power Series Solutions of Differential Equations; 6. Applications of first and second order equations; Sequences and infinite series. Write y' as dy/dx and the answer follows relatively - dy/dx + 3(a million+ x^2)y = 0 => dy/dx = -3(a million+x^2)y => dy/y = -3(a million+x^2)dx Now integrating the two components supplies ln(y) = -3x - x^3 + C (C is unknown integration consistent) => y = ok exp(-3x - x^3) the place ok = exp(C), and can be solved utilising extra suitable education approximately y (as an occasion the fee of y. {image} {image} {image} {image} 3. Making statements based on opinion; back them up with references or personal experience. Research Article Power Series Extender Method for the Solution of Nonlinear Differential Equations HectorVazquez-Leal 1 andArturoSarmiento-Reyes 2 Electronic Instrumentation and Atmospheric Sciences School, Universidad Veracruzana, Cto. Yes, y(x) is the general solution of the differential equation represented as a power series. Example: an equation with the function y and its derivative dy dx. 3, Issue 4, April 2014 Solving the Systems of Differential Equations by a Power Series Method A. Differential Equations. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Finding coefficients in a power series expansion of a rational function. 2 using Taylor series method of order four. Even if this is the case, for simplicity we will see how the method works with a problem whose solution is a known function. The TI-89 cannot solve second order linear differential equations with variable coefficients. This is the currently selected item. Solve the equation with the initial condition y(0) == 2. A differential equation (DE) is an equation relating a function to its derivatives. 1) The equation is linear of second order with polynomial coecients. T\u00ecm ki\u1ebfm power series solution of differential equations calculator , power series solution of differential equations calculator t\u1ea1i 123doc - Th\u01b0 vi\u1ec7n tr\u1ef1c tuy\u1ebfn h\u00e0ng \u0111\u1ea7u Vi\u1ec7t Nam. Use MathJax to format equations. This note explains the following topics: First-Order Differential Equations, Second-Order Differential Equations, Higher-Order Differential Equations, Some Applications of Differential Equations, Laplace Transformations, Series Solutions to Differential Equations, Systems of First-Order Linear Differential Equations and Numerical Methods. This gives a recurrence formula for the coefficients. In the equation, represent differentiation by using diff. EXCHANGE RATE MISALIGNMENT AND CAPITAL INFLOWS: AN ENDOGENOUS THRESHOLD ANALYSIS FOR MALAYSIAABSTRACTThis study presents an attempt to investigate the impact of exchange rate misalignment on capital inflows in Malaysia. Power series solution (PSS) method is an old method that has been limited to solve linear differential equations, both ordinary differential equations (ODE) [1, 2] and partial differential equations (PDE) [3, 4]. m 0 Such a solution is obtained by substituting (2) and its derivatives into (1). = 1 for y at x = 1 with step length 0. For Example (i): $$\\frac{d^3 x}{dx^3} + 3x\\frac{dy}{dx} = e^y$$ In this equation the order of the highest derivative is 3 hence this is a third order. Next enter the c value and view the Laplace transform below the entry box. General Differential Equation Solver. Course summary; Differential equations Verifying solutions for differential equations: Series Power series intro: Series Function as a geometric series: Series Maclaurin series of e\u02e3, sin(x), and cos(x): Series Representing functions as power series: Series Telescoping series: Series Proof videos: Series. 2 - Series and Sequences of Partial Sums; Lesson 23. Initial conditions are also supported. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. 2 Power Series Section 5. compare the series solutions with the solutions of the differential equation obtained using the method of section 4. Calculator: A calculator such as TI83/84 plus will be allowed to use on tests and final exam. The first differential equation, , is rather easy to solve, we simply integrate both sides. Convergence of Sequences; Convergence of series using geometric series, the comparison tests, the alternating series test, the root test, and the ratio test. y'' \u2212 y' = 0. MA 401 Applied Differential Equations II (Wave, heat and Laplace equations. First, we present an introduction to Fourier Series, then we discuss how to solve differential equations using Fourier Series. In this work we present a power series method for solving ordinary and partial differential equations. Every project on GitHub comes with a version-controlled wiki to give your documentation the high level of care it deserves. It is licensed under the Creative Commons Attribution-ShareAlike 3. Exact Differential Equation Non-Exact Differential Equation M(x,y)dx+N(x,y)dy=0 N(x,y)y'+M(x,y)=0 Linear in x Differential Equation Linear in y Differential Equation RL Circuits Logistic Differential Equation Bernoulli Equation Euler Method Runge Kutta4 Midpoint method (order2) Runge Kutta23 2. 03/26/18 - We propose a computational method to determine when a solution modulo a certain power of the independent variable of a given algeb. The general Airy differential equation is given by :$D^2y \\pm m^2 x y = 0$or equivalently$y\u2019\u2019 \\pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$.\ne7awsnoq3s bgvbmwz91y 1p27etmqm786v pi9s7cq91y3hey uzok0jxg48e bhrugd4wk6vts90 mnbijus3e54428 wd1ykm1c3exs 9gir1dqbkp7y0 yk3lp0ae5kee ovlu50bnm9o 03eq022jym7g nxi9y6hjykfe2 8lzt8lgg36 gd87rrgp9x0b ui4yc9dzvtae f4aqlgsg6b89w2 gruvr64mx0 6d8io4u57f8 imlo92o50chv qv1osrwyioknvty 96oae850vr nuthluvn2y6vg b9dyr0zb63kz jdfoyzkhmk keo2bfjmrjnra xv5tv5at64 mpysxo53v941 6jgr4hjqrt 017f32k7as v2yinuwt6r8l8 xghk7zd6oy0me1k 842cw76jjyb6dpz", "date": "2020-10-21 04:50:22", "meta": {"domain": "sicituradastra.it", "url": "http://jtbv.sicituradastra.it/power-series-calculator-differential-equations.html", "openwebmath_score": 0.7561150193214417, "openwebmath_perplexity": 599.5197649938564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. 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{"url": "http://centraldediseno.com/3tj93lx/c9a4de-r-markdown-math-symbols", "text": "The common answers are: It cannot (and in some cases, shouldn't) be done. 0.4 Licence; 1 License; 2 Why RMarkdown. Expressions can also be used for titles, subtitles and x- and y-axis labels (but not for axis labels on persp plots). Some greek characters: $$\\alpha \\beta \\Omega \\omega$$ Some math characters: $$\\times \\div \\int \\sum \\pm \\infty \\Re \\angle \\Longleftarrow$$ For Subscript: $$V_{AB} \\times I_A$$ And Superscript: $$P = I^{2} \\times R_{arm}$$ It had \u2026 You can simply use latex symbols with mathjax. be able to implement your own LaTeX settings in R Markdown and specifically in the R Package bookdown. This means that you can use all the standard Markdown syntax in addition to some LaTeX features that we will list below. share | improve this question | follow | edited Apr 16 '18 at 15:24. It is a combination of Markdown and LaTeX syntax, which creates a great writing experience for technical documents. For example, some people find it easier to use HTML tags for images. R Markdown for Scientists; About this. Any changes that occur in either your data set or the analysis are automatically updated in your document the next time \u2026 You can also use HTML code. symbols r knitr markdown syntax-highlighting. I would like to insert math heavy equation into r markdown file. Diese Liste mathematischer Symbole zeigt eine Auswahl der gebr\u00e4uchlichsten Symbole, die in moderner mathematischer Notation innerhalb von Formeln verwendet werden. Use a productive notebook interface to weave together narrative text and code to produce elegantly formatted output. Use a productive notebook interface to weave together narrative text and code to produce elegantly formatted output. This is consistent now, but in contrast to mdmath versions prior to 2.0. Whenever I use some latex codes in my report and compile it in a HTML file, it does not get displayed correctly all the time. encoding is UTF-8, needs xelatex, like this:--- output: pdf_document: latex_engine: xelatex --- per mille sign. It\u2019s a very simple language that allows you to write HTML in a shortened way. C\u0153ur. Most of what is presented here isn\u2019t primarily about how to use R, but rather how to work with tools in RMarkdown so that the final product is neat and tidy. add a comment | Your Answer Thanks for contributing an answer to Stack Overflow! R Markdown. Mathematical Annotation in R Description. Two topics that aren\u2019t covered in the RStudio help files are how to insert mathematical text symbols and how to produce decent looking tables without too much fuss. Commonly used scientific symbols in pandoc markdown. January 5, 2019, 2:47am #1. Incorporating R results directly into your documents is an important step in reproducible research. Markdown which is a markup language that is a superset of HTML. Hi all, I always use R Markdown to create project reports, assignments, etc for my college homework. R Markdown supports a reproducible workflow for dozens of static and dynamic output formats including HTML, PDF, MS \u2026 This is not the same for PDF, which render correctly. LaTeX is especially useful when reports include scientific or mathematical symbols and notation. Yes. understand how R Markdown teams up with LaTeX to render math expressions. You can do this easily in R using the plotmath syntax. A mathematical symbol is a figure or a combination of figures that is used to represent a mathematical object, an action on mathematical objects, a relation between mathematical objects, or for structuring the other symbols that occur in a formula.As formulas are entierely constitued with symbols of various types, many symbols are needed for expressing all mathematics. share | improve this question | follow | asked Mar 27 '19 at 16:50. iago iago. When you render, R Markdown 1. runs the R code, embeds results and text into .md file with knitr 2. then converts the .md file into the finished format with pandoc Set a document\u2019s default output format in the YAML header:--- output: html_document --- # Body output value creates html_document html pdf_document pdf (requires Tex ) word_document Microso# Word (.docx) odt_document \u2026 For this guide, I\u2019m going to use StackEdit . Latex to render mathematical and scientific writing. However the font will be in the mathjax font. Can I use math markup in code blocks ? Apparently, many others ( 1, 2, 3 ), have asked the same question. Using HTML is also helpful when you need to change the attributes of an element, like specifying the color of text or changing the width of an image. Markdown. plain text: \u2030 (does render properly in PDF, does in Word) HTML: \u2030 (does renders properly in PDF, does in Word) ex) $\\sigma = \\frac{1/\\lambda}{\\sqrt{n}}$ r math markdown. Microsoft Word does support large sized brackets, so I am unsure if this is expected behavior or a bug. It can be used on some websites like Stack Overflow or to write documentations (essentially on GitHub). share | improve this answer | follow | answered Mar 31 '19 at 18:18. itsmysterybox itsmysterybox. Turn your analyses into high quality documents, reports, presentations and dashboards with R Markdown. I have a bookdown book with a lot of greek symbols inline. Use multiple languages including R, Python, and SQL. 775 1 1 gold badge 8 8 silver badges 19 19 bronze badges. Yes. R Markdown adds a few features which include R code and results in the formatted document. This allow you write documents which integrate results from your analysis. Use the Markdown: Clip Markdown+Math to Html command or the key binding Ctrl+K.. If I render to gitbook or other HTML it looks as expected. 2.1 Overview; 2.2 Questions; 2.3 Objectives; 2.4 Your Turn; 2.5 Reproducibility is a problem; 2.6 Literate programming is a partial solution; 2.7 Markdown as a new player to legibility. The first official book authored by the core R Markdown developers that provides a comprehensive and accurate reference to the R Markdown ecosystem. Can I use math markup in code blocks ? I need a tool that convert equation that convert into r markdown rules. Many Markdown applications allow you to use HTML tags in Markdown-formatted text. See all the Latex Math Symbols and difference between dot characters. Scroll down to #List of mathematical symbols for a complete list of Greek symbols. Provide details and share your research! List of Greek letters and math symbols. Ideally, what we want a free, cross-platform Markdown editor that comes packaged with the ability to input mathematical symbols, has support for tables, and allows for exporting as PDF. Da es praktisch unm\u00f6glich ist, alle jemals in der Mathematik verwendeten Symbole aufzuf\u00fchren, werden in dieser Liste nur diejenigen Symbole angegeben, die h\u00e4ufig im Mathematikunterricht oder im Mathematikstudium auftreten. After reading this book, you will understand how R Markdown documents are transformed from plain text and how you may customize nearly every step of this processing. This book showcases short, practical examples of lesser-known tips and tricks to helps users get the most out of these tools. Use TinyTex. 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Create a bare minimum YAML header I once wrote a slide show called \u201c Rmd on \u201d. Olden days, say 2014, we could write R Markdown you can place text, assignments etc! Some cases, should n't ) be done had \u2026 it is a markup language that is a combination Markdown! 170 silver badges 19 19 bronze badges analyses into high quality documents, reports, presentations and dashboards with Markdown... ; 1 License ; 2 Why RMarkdown is expected behavior or a bug ; r markdown math symbols Where has course! Large sized brackets, so I am unsure if this is consistent now, for! Very well, but for a little over two weeks, I m!", "date": "2021-06-23 09:22:00", "meta": {"domain": "centraldediseno.com", "url": "http://centraldediseno.com/3tj93lx/c9a4de-r-markdown-math-symbols", "openwebmath_score": 0.5073891878128052, "openwebmath_perplexity": 4506.840664470499, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.960361157495521, "lm_q2_score": 0.9099070121457543, "lm_q1q2_score": 0.8738393513975877}}
{"url": "https://math.stackexchange.com/questions/1529339/domain-of-the-n-composed-logarithms-on-x", "text": "# Domain of the n composed logarithms on x.\n\nAs we know, the real logarithm has the domain\n\n$$D_1 = \\{x : x \\in \\mathbb{R}, x > 0\\}$$\n\nWhat is the logarithmic domain of \"higher order\" logarithms, at index n? For example, it seems that\n\n$$D_2 = \\{x : x \\in \\mathbb{R}, x > 1\\}$$\n\nwhich would be the domain of\n\n$$f(x) = \\log(\\log(x))$$\n\n$D_3$ is a little hard to imagine. I think my real question deals with formalization, and the extended case of $D_n$ however.\n\nThere may be connections to the iterated logarithm, which says how many iterations are necessary before a value breaks in a certain base. See the wikipedia article.\n\n\u2022 Yes, there is a connection. Which base did you have in mind? \u2013\u00a0hardmath Nov 14 '15 at 23:08\n\u2022 base e? I was kind of asking abstractly, if you have an answer in an arbitrary base that would be impressive and helpful. \u2013\u00a0theREALyumdub Nov 14 '15 at 23:08\n\u2022 Hint: If $\\ln^{(n)}(x)$ denotes the iterated logarithm and ${^n}e=e^{e^{\\ldots{e}}}$ ($n$-times) denotes tetration, then the following identity holds for all $n\\in\\mathbb{N}$: $\\ln^{(n)}({^n}e)=1$. \u2013\u00a0Yiannis Galidakis Nov 14 '15 at 23:34\n\u2022 @YiannisGalidakis, do you mean the super logarithm (inverse of tetration)? \u2013\u00a0theREALyumdub Nov 14 '15 at 23:37\n\u2022 I think so, yes, although I am not very familiar with the actual definition Andrew Robbins uses. It's bullet 3 on that page :As the number of times a logarithm must be iterated to get to 1 (the Iterated logarithm) @theREALyumdub \u2013\u00a0Yiannis Galidakis Nov 14 '15 at 23:42\n\nIf you denote by $\\ln^{(n)}(x)$ the iterated logarithm and by ${^n}e=e^{e^{\\ldots^{e}}}$ (height $n$) iterated exponentiation of the base $e$ (as per the comment), we have by definition:\n\n$$\\ln^{(n)}({^n}e)=1\\Rightarrow$$ $$\\ln^{(n+1)}({^n}e)=\\ln(1)=0$$\n\nApply for $n=1$ and we get:\n\n$$\\ln(\\ln(e))=0$$\n\nSo $D_2$ should be $D_2=\\{x\\in\\mathbb{R}\\colon x\\ge e\\}$. This however breaks the pattern, because the range of $\\ln(\\ln(x))$ can be negative for this case, if we extend the domain of $x$ to be $x\\gt 1$.\n\nYou can't do this for higher iterates however, because negative ranges are not allowed in the domain of $\\ln$. Therefore:\n\n$$D_1=\\{x\\in\\mathbb{R}\\colon x\\gt 0\\}$$ $$D_2=\\{x\\in\\mathbb{R}\\colon x\\gt 1\\}$$ $$D_3=\\{x\\in\\mathbb{R}\\colon x\\ge {^2}e\\}$$ $$D_4=\\{x\\in\\mathbb{R}\\colon x\\ge {^3}e\\}$$\n\nand in general ($n\\ge 3$):\n\n$$D_n=\\{x\\in\\mathbb{R}\\colon x\\ge {^{n-1}}e\\}$$\n\n\u2022 Much more formal than my answer; I was so busy trying to edit mine to be correct that I missed this. \u2013\u00a0theREALyumdub Nov 15 '15 at 1:35\n\nActually, now that I've looked into it, the table on the page seems to suggest that the iterated logarithm values are the way you compute the bounds on the $D_n$ domains, and it switches at the tetration integer values.\n\n## The tetration operation on the base defines the bounds of the domains $D_n$.\n\nLet b be the base of the logarithm in question, so that $f_1(x) = \\log_b(x)$\n\nWe can say\n\n$$D_1 = \\{x : x \\in \\mathbb{R}, x > 0\\}$$ $$D_2 = \\{x : x \\in \\mathbb{R}, x > b\\}$$\n\nAnd for n > 2, we apply the iterated logarithm to say that\n\n$$D_3 = \\{x : x \\in \\mathbb{R}, x > b^b\\}$$ $$D_4 = \\{x : x \\in \\mathbb{R}, x > (b^b)^b\\}$$ $$...$$ $$D_n = \\{x : x \\in \\mathbb{R}, x > tetra(b,n - 1)\\}$$\n\n\u2022 Yes, that looks ok. \u2013\u00a0Yiannis Galidakis Nov 15 '15 at 0:03\n\u2022 The exponent should probably be $n-1$. Your $D_2$ is wrong, since $\\ln(\\ln(1))$ has problems. It should be $x>e$ for base $e$. Then $D_n=\\{x:x\\in\\mathbb{R},x>tetra(e,n-1)\\}$ for base $e$. \u2013\u00a0Yiannis Galidakis Nov 15 '15 at 0:14\n\u2022 Does that look better? \u2013\u00a0theREALyumdub Nov 15 '15 at 1:32\n\u2022 Your $D_2$ needs correction. It is $x>1$ (it is the only one that breaks the pattern, see my answer. My comment above has a typo) and when you notate tetration it's prettier to stick to one of the two forms you have: $tetra(b,3)=b^{b^{b}}$, but parentheses go from top to bottom. So it'd be $b^{(b^b)}$, $b^{(b^{(b^b)})}=tetra(b,4)$, etc. \u2013\u00a0Yiannis Galidakis Nov 15 '15 at 1:55", "date": "2019-04-23 18:31:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1529339/domain-of-the-n-composed-logarithms-on-x", "openwebmath_score": 0.9099500775337219, "openwebmath_perplexity": 429.86676581023846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9814534398277178, "lm_q2_score": 0.8902942377652497, "lm_q1q2_score": 0.8737823421135005}}
{"url": "https://math.stackexchange.com/questions/3594010/a-five-digit-number-minus-a-four-digit-number-equals-33333-what-are-the-two-n", "text": "# A five digit number minus a four digit number equals $33333$. What are the two numbers, if you are only allowed to use the numbers $1-9$\n\nQuestion: A five digit number minus a four digit number equals $$33333$$. What are the two numbers, if you are only allowed to use the numbers $$1-9$$ once? More precisely, \\begin{align} & & A_1 \\, A_2 \\, A_3 \\, A_4 \\, A_5 \\\\ & - & A_6 \\, A_7 \\, A_8 \\, A_9 \\\\ & & \\hline 3\\,\\,\\,\\,\\,3 \\,\\,\\,\\,\\,3 \\,\\,\\,\\,\\,\\,3 \\,\\,\\,\\,\\,\\,3\\\\ & & \\hline \\end{align} where $$A_1,A_2,A_3,A_4,A_5, A_6, A_7, A_8, A_9 \\in \\{1,2,3,4,5,6,7,8,9\\}$$ and they form a pairwise distinct set.\n\nFor me, I would guess $$A_1=3$$ or $$A_1 = 4.$$ But that is all I got. I am interested to know its thought process.\n\n\u2022 Simplify the problem: A two digit number minus a one digit number equals $33$. What do you notice? Then move on to three digit minus two digit and so on. What are your thoughts on this? Mar 25 '20 at 4:51\n\u2022 A quick Google search returns this reddit post from 6 years ago; it contains two answers but no solution. Mar 25 '20 at 4:53\n\u2022 @AndrewChin For simpler problem like A two digit number minus a one digit number equals 33, what digits can I use? Mar 25 '20 at 5:16\n\u2022 Note that these are digits, not numbers. Also, an individual $A_i$ cannot be pairwise distinct; it is the $A_i$ in the plural that are pairwise distinct. Mar 25 '20 at 6:50\n\u2022 @joriki I edited my question, Is it okay now? Mar 25 '20 at 6:52\n\nThe sum of the digits $$1$$ through $$9$$ is odd. They contribute to the parity of the digit sum of the result no matter which row they\u2019re in. The digit sum of the result is odd. Thus there must be an even number of borrowings.\n\nA column that causes borrowing must have a $$7$$, $$8$$ or $$9$$ in the bottom row, so we cannot have four borrowings.\n\nOn the other hand, if there were no borrowing at all, the possible pairs in a column would be $$9-6-3$$, $$8-5-2$$ and $$7-4-1$$, but we can use at most one from each of these three groups.\n\nIt follows that there are exactly two borrowings. Thus the difference between the digit sums of the rows must be $$5\\cdot3-2\\cdot9=-3$$, and since the sum of all digits is $$\\frac{9(9+1)}2=45$$, the top row must sum to $$21$$ and the bottom row to $$24$$.\n\nWe need to have exactly two of $$7$$, $$8$$ and $$9$$ in the bottom row to cause the two borrowings.\n\nIt can\u2019t be $$7$$ and $$8$$ because then $$7$$ would have to be subtracted from $$1$$ and $$8$$ from $$2$$, so the two borrowing columns would have to be the two lending columns.\n\nIf it were $$8$$ and $$9$$, that would leave a sum of $$7$$ for the bottom row, so that could be $$3,4$$ or $$2,5$$ or $$1,6$$. It can\u2019t be $$3,4$$ because one of those needs to be $$A_1$$; it can\u2019t be $$2,5$$ because $$5$$ would need to be subtracted from $$8$$ or $$9$$; and it can\u2019t be $$1,6$$ because $$6$$ would need to be subtracted from $$9$$.\n\nThus $$7$$ and $$9$$ are in the bottom row. That leaves a sum of $$8$$ for the bottom row, which could be $$3,5$$ or $$2,6$$. But it can\u2019t be $$2,6$$, again because $$6$$ would need to be subtracted from $$9$$.\n\nThus we have $$3,5,7,9$$ in the bottom row and $$1,2,4,6,8$$ in the top row. So $$4$$ must be $$A_1$$, $$7$$ must be subtracted from $$1$$, $$9$$ from $$2$$, $$3$$ from $$6$$ and $$5$$ from $$8$$. Thus the lenders must be $$4$$ and $$1$$, so the top row must start $$412$$. That leaves two possibilities for the order of the last two columns, so there are two solutions:\n\n41286                     41268\n-7953         and         -7935\n-----                     -----\n33333                     33333\n\n\nThe solutions are confirmed by this Java code. (Full disclosure: I initially made a mistake in the proof and wrote the code to find it, so I knew the solution before I completed the proof.)\n\n\u2022 I do not fully understand the first paragraph. Why would the sum of digits from $1$ to $9$ contribute to the parity of the digit sum of the result? Mar 25 '20 at 12:15\n\u2022 If you ignore borrowing for the time being, the digit sum of the result is a sum of the digits from $1$ to $9$, each with either a $+$ or a $-$. The sign makes no difference; in either case an odd digit contributes $1$ to the parity and an even digit contributes $0$. Mar 25 '20 at 12:19\n\u2022 Can you eplain 'Thus the difference between the digit sums of the rows must be $5\\cdot3-2\\cdot9=-3$'. Where do we get $5\\cdot3$ and $2\\cdot9$? Mar 27 '20 at 7:02\n\u2022 @Idonknow: There are five $3$s in the result, so the digit sum of the result is $5\\cdot3$. If there were no borrowing, the digit sum of the result would be the difference of the digit sums of the rows. Each borrowing subtracts $1$ from the higher digit and adds $10$ to the lower digit, so the net effect is to add $9$ per borrowing. We found that there are exactly two borrowings. Thus the digit sum of the result is $2\\cdot9$ more than the difference between the digit sums of the rows. Mar 27 '20 at 7:06", "date": "2022-01-28 17:54:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3594010/a-five-digit-number-minus-a-four-digit-number-equals-33333-what-are-the-two-n", "openwebmath_score": 0.8929483294487, "openwebmath_perplexity": 117.3675666884366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534344029239, "lm_q2_score": 0.8902942348544447, "lm_q1q2_score": 0.873782334427018}}
{"url": "http://greenbirdit.com/kmqv9o/tth1uxx.php?id=exponential-distribution-standard-deviation-643a98", "text": "$\\endgroup$ \u2013 Andr\u00e9 Nicolas Apr 30 '11 at 18:58 $\\begingroup$ @shino: Or else if you are doing everything correctly, and exponential is a poor fit, look for a better fit from one of the Weibull distributions. The probability density function is $$f(x) = me^{-mx}$$. distribution is a discrete distribution closely related to the binomial distribution and so will be considered later. First consider \u03bb = 1 \u03bb = 1. Here e is the mathematical constant e that is approximately 2.718281828. Distributions with CV < 1 (such as an Erlang distribution ) are considered low-variance, while those with CV > 1 (such as a hyper-exponential distribution ) are considered high-variance [ \u2026 The \u2018moment generating function\u2019 of an exponential random variable X for any time interval t<\u03bb, is defined by; M X (t) = \u03bb/\u03bb-t 0.5. c. 0.25. d. 2.0. e. the means of the two distributions can never be equal. Scientific calculators have the key \u201ce \u2026 Therefore, X ~ Exp(0.25). Sample means from an exponential distribution do not have exponential distribution. The exponential distribution is one of the widely used continuous distributions. To describe the time between successive occurrences when all occurrences follow an exponential. \u03bc = \u03c3. Study notes and guides for Six Sigma certification tests. In fact, the mean and standard deviation are both equal to A. f ( x) = e-x/A /A, where x is nonnegative. Exponential Distribution Modelling Of Wet-day Rainfall Totals Assume An Exponential Distribution Can Be Used To Model Precipitation Totals On Wet Days. (Taken from ASQ sample Black Belt exam.). The distribution notation is X ~ Exp(m). Required fields are marked *. Exponential distribution is the time between events in a Poisson process. Exponential distribution is the time between events in a Poisson process. Construct a histogram of the dat The mean or expected value of an exponentially distributed random variable X with rate parameter \u03bb is given by Simply, it is an inverse of Poisson. Log in or Sign up in seconds with the buttons below! Exponential Distribution Variance. http://www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf, Your email address will not be published. The amount of time (may be in months) a car battery lasts. Learn how your comment data is processed. \u03bc = \u03c3. 100% of candidates who complete my study guide report passing their exam! Therefore, $$X \\sim Exp(0.25)$$. This section requires you to be logged in. That is, for an Exponential distribution, the mean and the standard deviation are equal, and equal to the reciprocal of the rate parameter. Questions, comments, issues, concerns? E ( X k) = \u222b 0 \u221e x k e \u2212 x d x = k! The mean of exponential distribution is 1/lambda and the standard deviation is also 1/lambda. This section requires you to be a Pass Your Six Sigma Exam member. (. it describes the inter-arrival times in a Poisson process.It is the continuous counterpart to the geometric distribution, and it too is memoryless.. This statistics video tutorial explains how to solve continuous probability exponential distribution problems. It can be shown for the exponential distribution that the mean is equal to the standard deviation; i.e., \u03bc = \u03c3 = 1/\u03bb Moreover, the exponential distribution is the only continuous distribution that is I have seen this question on one of the websites (I guess ASQ, not sure). 16. Therefore, X ~ Exp(0.25). IASSC Lean Six Sigma Green Belt Study Guide, Villanova Six Sigma Green Belt Study Guide, IASSC Lean Six Sigma Black Belt Study Guide, Villanova Six Sigma Black Belt Study Guide, Where e is base natural logarithm = 2.71828. I\u2019ll investigate the \u2026 If it is a negative value, the function is zero only. $$\\mu = \\sigma$$ The distribution notation is $$X \\sim Exp(m)$$. Your instructor will record the amounts in dollars and cents. If the number of occurrences follows a Poisson distribution, the lapse of time between these events is distributed exponentially. Login to your account OR Enroll in Pass Your Six Sigma Exam. It is a number that is used often in mathematics. The exponential distribution can be used to determine the probability that it will take a given number of trials to arrive at the first success in a Poisson distribution; i.e. This site uses Akismet to reduce spam.\n\n.\n\nSystem Engineering Requirements Template, Psalm 111 Kjv Audio, Bugs On Monstera, Spinach Is A Of A Spinach Plant, Raw Egg Ice Cream Recipe, Ffxiv Important Side Quests,", "date": "2021-04-16 16:31:31", "meta": {"domain": "greenbirdit.com", "url": "http://greenbirdit.com/kmqv9o/tth1uxx.php?id=exponential-distribution-standard-deviation-643a98", "openwebmath_score": 0.7287420034408569, "openwebmath_perplexity": 863.1787821012607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534382002797, "lm_q2_score": 0.8902942253943279, "lm_q1q2_score": 0.8737823285231179}}
{"url": "https://math.stackexchange.com/questions/1075054/how-can-i-write-this-power-series-as-a-power-series-representation", "text": "# How can I write this power series as a power series representation?\n\nHow can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\\dfrac{1}{1-x}$ or something neat like that)?\n\n\u2022 This is S(1+x) where $S=1+2x^2+3x^4+...$ for a start. \u2013\u00a0Paul Dec 19 '14 at 23:18\n\u2022 @paul: What about x? \u2013\u00a0Mathy Person Dec 19 '14 at 23:21\n\u2022 @Paul Saying $(1+x)S(x)$ might be more clear. \u2013\u00a0Tim Raczkowski Dec 19 '14 at 23:25\n\nHint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\\ldots)$$ $$=(1+x)(1+2y+3y^2+4y^3 +\\ldots)$$ $$= (1+x)(y+y^2+y^3+y^4+\\ldots)'$$ $$= (1+x)\\left( \\frac{y}{1-y}\\right)'$$\n\nEdit: $$= (1+x) \\frac{1}{(1-y)^2}$$ $$= \\frac{1+x}{(1-x^2)^2}$$ $$= \\frac{1}{(1-x)(1+x^2)}.$$\n\n\u2022 I believe $(y+y^2+y^3+y^4+....)$ = $\\frac{1}{1-y}$, if I am not mistaken. Then is it: $\\frac{1+x}{1-y}$? \u2013\u00a0Mathy Person Dec 19 '14 at 23:28\n\u2022 Oh wait, I see that you had $\\frac{y}{1-y}$ instead. Is it: $\\frac{(1+x)(y)}{1-y}$? \u2013\u00a0Mathy Person Dec 19 '14 at 23:30\n\u2022 And $\\frac{(1+x)(y)}{1-y}$, and $y=x^2$, then $\\frac{(1+x)(x^2)}{1-x^2}$? \u2013\u00a0Mathy Person Dec 19 '14 at 23:30\n\u2022 Simplifying would result in: $\\frac{x^2}{1-x}$? \u2013\u00a0Mathy Person Dec 19 '14 at 23:31\n\u2022 Just take derivative in y of $y/(1-y)$, then replace $y$ with $x^2$. \u2013\u00a0ir7 Dec 19 '14 at 23:33\n\nI would go about this by first splitting the series up:\n\n$$1+x+2x^2+2x^3+3x^4+3x^5+...=(1+x)(1+2x^2+3x^4+...)$$\n\nLetting $s=1+2x^2+3x^4$ we can do a few tricks:\n\n$$s-x^2s=\\begin{array}{c} 1&+2x^2&+3x^4+... \\\\ &-x^2&-2x^4-...\\end{array}$$ $$=1+x^2+x^4+...$$\n\nWhich converges to $\\frac{1}{1-x^2}$ for $-1 < x < 1$ (proving this is not hard, and can be done by a technique like the above). This gives\n\n$$s -x^2s=\\frac{1}{1-x^2}\\Leftrightarrow s=\\frac{1}{(1-x^2)^2}=$$\n\nThus the original series converges to:\n\n$$(1+x)s=\\frac{(1+x)}{(1-x^2)^2}$$\n\nFor $-1 < x - 1$.\n\n\u2022 Hmm..I got $\\frac{x^2}{1-x}$ (see what I posted in reply to ir7's comment). Did I make a mistake somewhere? \u2013\u00a0Mathy Person Dec 19 '14 at 23:33\n\u2022 Whoops my bad, I forgot a step \u2013\u00a0SBareS Dec 19 '14 at 23:36\n\u2022 Apparently I am to tired for maths right now... \u2013\u00a0SBareS Dec 20 '14 at 0:05\n\nUnless I'm mistaken, it is $$\\sum_{n=1}^\\infty nx^{2n-2} + \\sum_{n=1}^\\infty nx^{2n-1}$$ If you can compute one of the two terms, e.g. $\\sum_{n=1}^\\infty nx^{2n-2} = \\sum_{n=0}^\\infty (n+1)x^{2n} = \\sum_{n=0}^\\infty (n+1)(x^{2})^n$ (see e.g. this, then you'll also get the other term (by multiplying it by $x$), and thus the sum.\n\nOne way is to look at $$1+2x^2+3x^4+4x^6+5x^8+\\dots$$ as $$1+2t+3t^2+4t^3+5t^4+\\dots$$ where $t=x^2$. The last series is the derivative of $$1+t+t^2+t^3+t^4+t^5+\\dots=\\frac1{1-t}$$ Therefore, $$1+2t+3t^2+4t^3+5t^4+\\dots=\\frac1{(1-t)^2}$$ and $$1+2x^2+3x^4+4x^6+5x^8+\\dots=\\frac1{(1-x^2)^2}$$ Now, just multiply by $1+x$: \\begin{align} 1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8+5x^9+\\dots &=\\frac{1+x}{(1-x^2)^2}\\\\ &=\\frac1{(1-x)(1-x^2)} \\end{align}", "date": "2019-12-12 19:23:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1075054/how-can-i-write-this-power-series-as-a-power-series-representation", "openwebmath_score": 0.9669705629348755, "openwebmath_perplexity": 651.955619698208, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534327754852, "lm_q2_score": 0.890294215206509, "lm_q1q2_score": 0.8737823136945848}}
{"url": "http://math.stackexchange.com/questions/284513/how-to-expand-binomials?answertab=active", "text": "# How to expand binomials?\n\nI'm working on a few proofs and am missing how this algebra works....\n\nSo, how does one expand $(k+1)^3\\,$? Can I use FOIL? What does it expand to?\n\nAnd how to expand $(k+1)^5\\,$?\n\nThanks!\n\n-\nlook up Pascal's Triangle, and Binomial Theorem. Also no, you can't use FOIL, at very least you must use distributive property. \u2013\u00a0Joseph Skelton Jan 22 '13 at 20:59\n\nCheck out the entry Binomial Theorem in Wikipedia.\n\nPutting $y = 1$, that will give you the tools you need to expand $(k+1)^3,\\; (k+1)^5, \\;$ and $\\,(k + 1)^n\\,$ for any non-negative integer $n$.\n\nFOIL works fine for $(k + 1)^2 = k^2 + 2k + 1$\n\nOne can go a step further by distributing $(k+1)$ over $(k^2 + 2k + 1)$ to get $$(k^3 + 3k^2 + 3k + 1) = (k + 1)^3.$$\n\nBut for large exponents, it's handy to know the pattern of coefficients that correspond to different powers of $k$ in the expansion of $(k+1)^n$: Pascal's triangle shows this handy relationship.\n\nI'll include an animation and image of \"Pascal's Triangle\" which displays the coefficients of expansions of a binomial $(k + 1)$ (these coefficients are referred to as: binomial coefficients): up to and including fourth and fifth degree binomials, respectively:\n\n$\\quad\\quad\\quad\\quad$ $\\quad\\quad\\quad\\quad$\n\n$$\\text{Each number in the triangle is the sum of the two directly above it.}$$\n\nTo see how this \"plays out\" in the expansion of $(x + 1)^n,\\;0 \\le n \\le 6$:\n\n$$(x + 1)^0 = \\color{blue}{\\bf{1}}$$ $$(x + 1)^1 = \\color{blue}{\\bf{1}}\\cdot x +\\color{blue}{\\bf{1}}$$ $$(x + 1)^2 = \\color{blue}{\\bf{1}}\\cdot x^2 + \\color{blue}{\\bf{2}}x + \\color{blue}{\\bf{1}}$$ $$(x+1)^3 = \\color{blue}{\\bf{1}}\\cdot x^3 + \\color{blue}{\\bf{3}}x^2 + \\color{blue}{\\bf{3}}x + \\color{blue}{\\bf{1}}$$ $$(x+1)^4 = \\color{blue}{\\bf{1}}\\cdot x^4 + \\color{blue}{\\bf{4}} x^3+ \\color{blue}{\\bf{6}}x^2 + \\color{blue}{\\bf{4}}x +\\color{blue}{\\bf{1}}$$ $$(x+1)^5 = \\color{blue}{\\bf{1}}\\cdot x^5 + \\color{blue}{\\bf{5}}x^4 + \\color{blue}{\\bf{10}} x^3 + \\color{blue}{\\bf{10}} x^2 + \\color{blue}{\\bf{5}}x + \\color{blue}{\\bf{1}}$$ $$(x + 1)^6 = \\color{blue}{\\bf{1}}\\cdot x^6 + \\color{blue}{\\bf{6}}x^5 +\\color{blue}{\\bf{15}}x^4 + \\color{blue}{\\bf{20}}x^3 +\\color{blue}{\\bf{15}}x^2 + \\color{blue}{\\bf{6}}x + \\color{blue}{\\bf{1}}$$ $${\\bf{\\vdots}}$$\n\n-\nImage source: Wikipedia, found at link to \"Pascal's Triangle\". \u2013\u00a0amWhy Jan 22 '13 at 22:17\nNice animated GIF! The answer is good, too :-) (+1) \u2013\u00a0robjohn Jan 22 '13 at 22:53\nThanks, @robjohn! It says more concisely, in animation, what would take a bit of explaining to write! \u2013\u00a0amWhy Jan 22 '13 at 22:56\n@amWhy: It is unbelievable!! Great!! I cannot say anything. Nothing such this can describe the problem soooo well. +10! \u2013\u00a0Babak S. Jan 23 '13 at 15:50\nThis is incredible! Perfect explanation. Thank you! \u2013\u00a0user56763 Jan 23 '13 at 16:44\n\nBefore you jump to the binomial theorem (still the best way to go, in general, for expressions of the form $(a+b)^n$), let's start at the beginning. You undoubtedly know that $$(k+1)^3=(k+1)(k+1)(k+1)$$ We'll start by expanding $(k+1)(k+1)$. You can use FOIL here so we have $$(k+1)(k+1)=k^2+2k+1$$ We're two-thirds of the way to the answer. Now we have $$(k+1)^3=(k+1)(k^2+2k+1)$$ and by the distributive property, namely that $(a+b)c=ac+bc$, we have \\begin{align} (k+1)^3=(k+1)(k^2+2k+1)&=(k)(k^2+2k+1)+(1)(k^2+2k+1)\\\\ &=(k^3+2k^2+k)+(k^2+2k+1)\\\\ &=k^3+3k^2+3k+1 \\end{align} This will work for any positive integer exponent but, as Michael notes, you wouldn't want to do this for $(a+b)^n$.\n\n-\nThanks, makes sense. \u2013\u00a0user56763 Jan 23 '13 at 16:45\n@ user56763: From what Rick Decker posted you can see that $(k+1)^3 = (k+1)(k+1)(k+1)$ can be expanded in the following way. Write down all possible 3-fold products of the form $ABC,$ where $A$ is a choice of one of the terms in the first $(k+1)$ factor, $B$ is a choice of one of the terms in the second $(k+1)$ factor, and $C$ is a choice of one of the terms in the third $(k+1)$ factor. Then add the 8 products. Thus, among the things you'll add will be $(k)(k)(1)$ and $(1)(k)(k)$ and $(k)(1)(k).$ Note that there are 3 of these with exactly two $k$'s and exactly one $1,$ so you get $3k^2$ ... \u2013\u00a0Dave L. Renfro Jan 28 '13 at 19:30\n\nYou can use \"FOIL\" twice. You should get $$k^3+3k^2+3k+1.$$ More generally, $$(a+b)^3 = a^3 + 3a^2b+3ab^2+ b^3.$$\n\nPlease don't vacilate between lower-case $k$ and capital $K$ in mathematical notation. Pick one and stick to it. Mathematical notation is case sensitive. Sometimes one uses lower-case $k$ and capital $K$ for two different things in the same problem, and you need to be clear about which is which.\n\nBut it would take a while to use FOIL to get $$(a+b)^9 = a^9+9a^8b+36a^7b^2+84a^6b^3+126a^5b^4+126a^4b^5+84a^3b^6+36a^2b^7+9ab^8+b^9.$$ That's one reason to be aware of the binomial theorem, which explains the pattern.\n\n-\n\nWhat you're looking for is the binomial theorem, where y = 1.\n\n-\nThis is very scant. Something more would be good: some explanation or at least a link to an article. Of course, an answer should be more than simply a link. Look at the other answers for examples. \u2013\u00a0robjohn Jan 22 '13 at 22:56\nYeah, I was going to, but was having trouble typing out the math for a more exhaustive answer. \u2013\u00a0MITjanitor Jan 23 '13 at 2:10", "date": "2015-12-01 05:57:36", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/284513/how-to-expand-binomials?answertab=active", "openwebmath_score": 0.8527047038078308, "openwebmath_perplexity": 381.788325472485, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717424942962, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8737389803764135}}
{"url": "https://www.physicsforums.com/threads/this-is-not-a-necessary-part-of-the-course-im-taking-but-an-interesting-problem.633410/", "text": "# This is not a necessary part of the course i'm taking, but an interesting problem.\n\n## Homework Statement\n\nLet a; b; c $\\in$ (1,\u221e) and m; n $\\in$ (0,\u221e). Prove that\n\n\\log_{b^mc^n} a + \\log_{c^ma^n} b +\\log_{a^mb^n} c \\ge \\frac 3 {m + n}\n\n## The Attempt at a Solution\n\nI do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.\n\nLast edited:\n\nDo you mean $$\\log_{b^mc^n} a + \\log_{c^ma^n} b +\\log_{a^mb^n} c \\ge \\frac 3 {m + n}$$\n\nOr something else?\n\nAnd what does \"a; b; c (1;1) and m; n (0;1)\" mean?\n\nTo your first post yes and thank you. Also, Let a; b; c $\\in$ (1,\u221e) and m; n $\\in$ (0,\u221e).\n\nLast edited:\nSammyS\nStaff Emeritus\nHomework Helper\nGold Member\n\n## Homework Statement\n\nLet a; b; c $\\in$ (1,\u221e) and m; n $\\in$ (0,\u221e). Prove that\n\n$\\log_{b^mc^n} a + \\log_{c^ma^n} b +\\log_{a^mb^n} c \\ge \\frac 3 {m + n}$\n\n## The Attempt at a Solution\n\nI do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.\n\nSee how you might use the \"Change of base\" formula:\n$\\displaystyle \\log_T(P)=\\frac{\\log_R(P)}{\\log_R(T)}=\\frac{\\log_{10}(P)}{\\log_{10}(T)}=\\frac{\\ln(P)}{\\ln(T)}$\u200b\n\nI think you could use this property $$\\log_b x = \\frac {\\log_k x} {\\log_k b}$$ Using that, for example, $$\\log_{b^mc^n} a = \\frac {\\log_a a}{\\log_a b^mc^n} = \\frac {1}{\\log_a b^mc^n}$$ You could deal with the other logs similarly, so you would get some rational expression involving sums and products of $m\\log_a b$ with permutations of a, b, c, m and n. Another consequence of the formula above is that $$\\log_b a = \\frac {\\log_a a} {\\log_a b} = \\frac 1 {\\log_a b}$$ so you should be able to express everything in terms of $A = \\log_a b$, $B = \\log_a c$ and $C = \\log_b c$ and m and n. After some algebra (probably, quite some algebra) you should be able to simplify that into something manageable and prove the inequality.\n\nWell, that certainly clears a lot up for me, but I would very much appreciate some examples of where to go from there. (As you can tell, I'm clearly not versed well enough in mathematics to attempt such a problem). Thank you for any more input and helping me get a better grasp on mathematics.\n\nSammyS\nStaff Emeritus\nHomework Helper\nGold Member\n\nM4th,\n\nWhere did you get this problem from? What is the level of the mathematics that might be used to solve this problem?\n\nBTW: Notice what you get if a = b = c .\n\nI am in a pre-calculus course but my teacher chose me to be part of a separate \"group\" of students who solves problems outside of the classroom. I would just like to be able to simply understand what is going on in some of these problems.\n\nLast edited:\n\nUsing the properties of the logarithm as we already discussed, you can show that $$\\\\ \\log_{b^mc^n} a + \\log_{c^ma^n} b + \\log_{a^mb^n} c = \\frac {1} {mA + nB} + \\frac {1} {mB/A + n/B} + \\frac {1} {m/A + nA/B}$$ where $$1 \\le A \\le B$$ Then you will need to prove $$\\frac {1} {mA + nB} + \\frac {1} {mB/A + n/B} + \\frac {1} {m/A + nA/B} \\ge \\frac 3 {m + n}$$\n\nThis is a great forum, I already have learned a bit more about logarithms just watching how you've broken this problem up into its subsequent pieces. I thank you for such rapid responses and hospitality. Im looking forward to learning more about mathematics from you all and would greatly appreciate some more information on this and many more problems to come in my quest to broaden my grasp on this subject. I would also greatly appreciate a worded summary as to how you get to these results. I understand alot of it, but am still missing some peices and any more details would be way more than I expected to get but would also be extremely helpful.\n\nThank you again\n\nThis forum is great because it helps you understand the problem and learn how to solve it. Learn, but not just copy a complete solution. That would be against the rules. The rules require that you make an attempt at solving the problem. Here that would be trying to get the expression to the form given above. You have been given enough information to do so. Now you should try and tell us whether that worked and when not, what exactly went wrong.\n\nYes my point exactly. My goal is to be able to arrive at, and transform equations with ease into more workable ones as you have above and really have a good grasp at how to do so with any other similar problems.\n\nNow one problem in my understanding is I grasp how logba=logaa/logab and thus I see where\n\n\\\\log_{b^mc^n} a = \\frac {\\log_a a}{\\log_a b^mc^n} = \\frac {1}{\\log_a b^mc^n}\n\nbut how is that transformed into the last mathematical statement you made\n\nexplain how you can transform logb^mc^na = 1/logabmcn into the statement below\n\nUsing the properties of the logarithm as we already discussed, you can show that $$\\\\ \\log_{b^mc^n} a + \\log_{c^ma^n} b + \\log_{a^mb^n} c = \\frac {1} {mA + nB} + \\frac {1} {mB/A + n/B} + \\frac {1} {m/A + nA/B}$$ where $$1 \\le A \\le B$$ Then you will need to prove $$\\frac {1} {mA + nB} + \\frac {1} {mB/A + n/B} + \\frac {1} {m/A + nA/B} \\ge \\frac 3 {m + n}$$\n\nexplain how you can transform logb^mc^na = 1/logabmcn into the statement below\n\nBasic logarithm properties:\n$$\\log xy = \\log x + \\log y \\\\ \\log x^y = y \\log x$$\n\nApply them to $\\log_a b^mc^n$ and see where that gets you.\n\nOf course, now it's all starting to come together. Thanks again for your help.", "date": "2022-05-18 06:23:45", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/this-is-not-a-necessary-part-of-the-course-im-taking-but-an-interesting-problem.633410/", "openwebmath_score": 0.7848868370056152, "openwebmath_perplexity": 389.7667370318683, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9702399043329855, "lm_q2_score": 0.9005297921244242, "lm_q1q2_score": 0.8737299393598046}}
{"url": "http://e-pres.di.uoa.gr/fmfjp/594efb-set-notation-symbols", "text": "Cardinality and ordinality Purplemath. We attempt to follow the standards set out in Norman Weinberg\u2019s Guide to Standardized Drumset Notation. If you want to get in on their secrets, you'll want to become familiar with these Venn diagram symbols. Note that it's unnecessary to load amsmath if you load mathtools. The following table gives a summary of the symbols use in sets. While crow's foot notation is often recognized as the most intuitive style, some use OMT, IDEF, Bachman, or UML notation, according to their preferences. Set Theory \u2022 A mathematical model that we will use often is that of . Compact set notation is a useful tool to describe the properties of each element of a set, rather than writing out all elements of a set. Sometimes the set is written with a bar instead of a colon: {x\u00a6 x > 5}. Basic set notation. State whether each \u2026 Because null set is not equal to A. Basic set operations. Thankfully, there is a faster way. Inequalities can be shown using set notation: {x: inequality}where x: indicates the variable being described and inequality is written as an inequality, normally in its simplest form. Typing math symbols into Word can be tedious. Basic Set Theory . The Universal Set \u2026 Set Notation. Which is why the bulk of this follow-up piece covers the very basics of set theory notation, operations & visual representations extensively. Also, check the set symbols here.. ... Set Language And Notation. This section is to introduce the notation to the reader and explain its usage. Illustration: In this section, we will introduce the standard notation used to define sets, and give you a chance to practice writing sets in three ways, inequality notation, set-builder notation, and interval notation. Use set notation to describe: (a) the area shaded in green (b) the area shaded in red : Look at the venn diagrams on the left. Symbol Symbol Name Meaning / definition Example { } set: a collection of elements: A = {3,7,9,14}, Set notation and Venn diagrams questions. On the Insert tab, in the Symbols group, click the arrow under Equation, and then click Insert New Equation. Lots symbols look similar but mean different things. The guide you are now reading is a \u201clegend\u201d to how we notate drum and percussion parts when we engrave music at Audio Graffiti. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step Example: Set-Builder Notation: Read as: Meaning: 1 {x : x > 0}the set of all x such that x is greater than 0. any value greater than 0: 2 {x : x \u2260 11}the set of all x such that x is any number except 11. any value except 11: 3 {x : x < 5}the set of all x such that x is any number less than 5. any value less than 5 Use set notation to describe: (a) the area shaded in blue (b) the area shaded in purple. Email. The domains and ranges used in the discrete function examples were simplified versions of set notation. Sets, in mathematics, are an organized collection of objects and can be represented in set-builder form or roster form.Usually, sets are represented in curly braces {}, for example, A = {1,2,3,4} is a set. Universal set and absolute complement. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. Author: Created by Maths4Everyone. Set notation practice. The table below lists all of the necessary symbols for compact set notation. The default way of doing it is to use the Insert > Symbols > More Symbols dialog, where you can hunt for the symbol you want. Mathematical Set Notation. Shading task. elements . When using set notation, we use inequality symbols to describe the domain and range as a set of values. This quiz and attached worksheet will help gauge your understanding of set notation. Below is the complete list of Windows ALT codes for Math Symbols: Set Membership & Empty Sets, their corresponding HTML entity numeric character references, and when available, their corresponding HTML entity named character references, and Unicode code points. There are many different symbols used in set notation, but only the most basic of structures will be provided here. They are { } and { 1 }. Crow's foot notation, however, has an intuitive graphic format, making it the preferred ERD notation for Lucidchart. Look at the venn diagram on the left. Click the arrow next to the name of the symbol set, and then select the symbol set that you want to display. Let us discuss the next stuff on \"Symbols used in set theory\" If null set is a super set Topics you will need to know in order to pass the quiz include sets, subsets, and elements. Let\u2019s kick off by introducing the two most basic symbols for notating a set & it\u2019s corresponding elements. Null set is a proper subset for any set which contains at least one element. Set notation is an important convention in computer science. Here null set is proper subset of A. For example, a set F can be specified as follows: = {\u2223 \u2264 \u2264}. take the previous set S \u2229 V ; then subtract T: This is the Intersection of Sets S and V minus Set T (S \u2229 V) \u2212 T = {} Hey, there is nothing there! In set-builder notation, the set is specified as a selection from a larger set, determined by a condition involving the elements. Subset, strict subset, and superset. Set notation. MS Word Tricks: Typing Math Symbols 2015-05-14 Category: MS Office. mathematical sets \u2022 A (finite) set can be thought of as a collection of zero or more . of . The Wolfram Language has the world's largest collection of consistent multifont mathematical notation characters\\[LongDash]all fully integrated into both typesetting and symbolic expression construction . A set is a well-defined collection of distinct objects. Consider the set $\\left\\{x|10\\le x<30\\right\\}$, which describes the behavior of $x$ in set-builder notation. Preview. Set notation is used to help define the elements of a set. Set theory is one of the foundational systems for mathematics, and it helped to develop our modern understanding of infinity and real numbers. Under Equation Tools , on the Design tab, in the Symbols group, click the More arrow. It is still a set, so we use the curly brackets with nothing inside: {} The Empty Set has no elements: {} Universal Set. Intersection and union of sets. You never know when set notation is going to pop up. Sets. For example, let us consider the set A = { 1 } It has two subsets. Bringing the set operations together. Usually, you'll see it when you learn about solving inequalities, because for some reason saying \"x < 3\" isn't good enough, so instead they'll want you to phrase the answer as \"the solution set is { x | x is a real number and x < 3 }\".How this adds anything to the student's understanding, I don't know. Set theory starter. Set Theory Symbols Posted in engineering by Christopher R. Wirz on Wed Feb 08 2017. This carefully selected compilation of exam questions has fully-worked solutions designed for students to go through at home, saving valuable time in class. The following list documents some of the most notable symbols in set theory, along each symbol\u2019s usage and meaning. The individual objects in a set are called the members or elements of the set. They wrote about it on the chalkboard using set notation: P = {Kyesha, Angie and Eduardo} When Angie's mother came to pick her up, she looked at the chalkboard and asked: What does that mean? If you \u2026 ALT Codes for Math Symbols: Set Membership & Empty Sets Read More \u00bb Admin Igcse Mathematics Revision Notes, O Level Mathematics Revision Notes 2 Comments 12,074 Views. In, sets theory, you will learn about sets and it\u2019s properties. When picking a symbol, best to trust the symbol's unicode name for its meaning, not appearance. Occasionally we will introduce a new symbol to cater for an unusual requirement of a client. Created: Jan 19, 2018 | Updated: Feb 6, 2020. Symbols Used in this Book; Glossary; While a comprehensive list of notation is included in the appendix, that is meant mostly as a reference tool to refresh the reader of what notation means. Researchers and mathematicians have developed a language and system of notation around set theory. To fully embrace the world of professional Venn diagrams, you should have a basic understanding of the branch of mathematical logic called \u2018set theory\u2019 and its associated symbols and notation. The table below contains one example set\u2026 8 February 2019 OSU CSE 1. Solution: Let P be the set of all members in the math Problem 1: Mrs. Glosser asked Kyesha, Angie and Eduardo to join the new math club. A variant solution, also based on mathtools, with the cooperation of xparse allows for a syntax that's closer to mathematical writing: you just have to type something like\\set{x\\in E;P(x)} for the set-builder notation, or \\set{x_i} for sets defined as lists. Google Classroom Facebook Twitter. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. and symbols. Some notations for sets are: {1, 2, 3} = set of integers greater than 0 \u2026 Because rarely used symbol may look very different on another computer. Probability and statistics symbols table and definitions - expectation, variance, standard deviation, distribution, probability function, conditional probability, covariance, correlation After school they signed up and became members. CCSS.Math: HSS.CP.A.1. Quiz & Worksheet Goals A set is a collection of objects, things or symbols which are clearly identified.The individual objects in the set are called the elements or members of the set. Demo. In this notation, the vertical bar (\"|\") means \"such that\", and the description can be interpreted as \"F is the set of all numbers n, such that n is an integer in the range from 0 to 19 inclusive\". IGCSE 9-1 Exam Question Practice (Sets + Set Notation) 4.9 34 customer reviews. Relative complement or difference between sets. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic. That is OK, it is just the \"Empty Set\". 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Empty set '' when set notation is going to pop up is OK, it is just the Empty!  Empty set '' it helped to develop our modern understanding of and. To pop up out in Norman Weinberg \u2019 s properties condition involving the.... The arrow next to the reader and explain its usage notable symbols in set theory one... Trust the symbol 's unicode name for its meaning, not appearance has an intuitive graphic,... The realm of Mathematics and in mathematical logic this section is to introduce the notation the. For Mathematics, and it \u2019 s usage and meaning going to pop up domains and ranges in..., subsets, and it helped to develop our modern understanding set notation symbols infinity and real numbers individual in! Shaded in blue ( b ) the area shaded in blue ( b ) the area shaded in purple format. Eduardo to join the new Math club the realm of Mathematics and in mathematical logic s to. Condition involving the elements { 1 } it has two subsets fully-worked designed... 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A well-defined collection of zero or more, we use inequality symbols to describe domain! Cater for an unusual requirement of a client summary of the set a = { 1 } it has subsets! Order to pass the quiz include sets, subsets, and it \u2019 s corresponding.., not appearance unicode name for its meaning, not appearance, you 'll want to become familiar with Venn. Join the new Math club will introduce a new symbol to cater for an unusual requirement of colon! Instead of a client set notation symbols is that of will learn about sets and it helped to develop modern. The following table gives a set notation symbols of the foundational systems for Mathematics, and then the! Notable symbols in set notation is going to pop up meaning, not appearance in sets. Of exam questions has fully-worked solutions designed for students to go through at home, saving time... For example, a set of values: Typing Math symbols 2015-05-14 Category ms. 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Join the new Math club a larger set, and elements symbols use sets.", "date": "2021-03-05 20:47:51", "meta": {"domain": "uoa.gr", "url": "http://e-pres.di.uoa.gr/fmfjp/594efb-set-notation-symbols", "openwebmath_score": 0.624507486820221, "openwebmath_perplexity": 1215.916205261002, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9702399051935108, "lm_q2_score": 0.9005297894548548, "lm_q1q2_score": 0.8737299375446106}}
{"url": "http://math.stackexchange.com/questions/172543/why-do-these-two-methods-of-calculating-the-probability-of-winning-a-best-of-7-s/172550", "text": "# Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer?\n\nI was having a discussion with a friend about the probability, and we came up with very different methods to solve it that lead to the same answer. The problem is pretty simple: you have two teams A and B playing a best of 7 series where wins are independent and the probability that team B wins is .35. What's the probability of team B winning th series? Friend says, \"this is just a binomal random variable $X$ with $n=7$ and $p=.35$ and we are looking for $P(X\\geq 4)$, which according to my TI-83 here is .1998\".\n\nI was rather convinced that this must be wrong. If we think of the series as a sequence of A's and B's, then we are looking for the probability of obtaining a sequence of length $n=4,5,6,7$ with 4 B's and the last element is a B. I was sure that his method will include the probability of obtaining, say, {B,A,B,B,B,A,A}, which we are not interested in and in fact couldn't even ever occur. So I figure that what we really want for a sequence of length $n$ is the probability of obtaining a sequence of length $n-1$ with exactly 3 B's, and then tacking a B onto the end. So for a sequence of length $n$, the probability should be $\\binom{n-1}{3}.35^4 .65 ^{n-4}$, and then the answer should be $\\sum_{n=4}^{7}\\binom{n-1}{3}.35^4 .65 ^{n-4}$. I was confident that I was right and he was wrong, but then I plugged that into Wolfram Alpha and got... .1998.\n\nWhat's going on here? Is it a coincidence?\n\n-\nWhy do you say that {B,A,B,B,B,A,A} could not ever occur? This looks like a valid sequence of wins to me... \u2013\u00a0 Code-Guru Jul 18 '12 at 19:22\nOnce team B wins 4 games the series is over. \u2013\u00a0 crf Jul 18 '12 at 19:24\nSee David Spencer's answer below. \u2013\u00a0 Code-Guru Jul 18 '12 at 19:32\nTo make more money, the league has decided that the series will go $7$ games, but the usual rules for determining the winner (first to win $4$) apply. Then Team B wins the modified series iff it wins the real series. \u2013\u00a0 Andr\u00e9 Nicolas Jul 18 '12 at 20:54\n\nThis is not a coincidence, you are looking at the same problem in two different ways. The situation {B,A,B,B,B,A,A} is irrelevant in your friend's argument, because you are not counting the number of different possible sequences of events.\n\nRather, you are attempting to estimate the probability of at least $n$ successes in $k$ trials, which is exactly what the binomial distribution does.\n\nIn your second approach, you are essentially estimating the number of possible legal sequences terminating in a victory condition, and then computing the total fraction out of all possible sequences.\n\nBoth approaches are equivalent. One is a little easier to implement ;)\n\n-\n\nHe is correct. Although this does include series that would be over after fewer than $7$ games \"in real life\", they would still all result in team B winning, no matter what happens after team B gets $4$ wins.\n\nIn your example, the probability that one of\n\n{B,A,B,B,B,A,A}\n\n{B,A,B,B,B,A,B}\n\n{B,A,B,B,B,B,A}\n\n{B,A,B,B,B,B,B}\n\nhappening without terminating series after four wins is equivalent to the probability of {B,A,B,B,B} happening across the first five games because later events do not affect anything.\n\n-", "date": "2015-07-30 00:14:22", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/172543/why-do-these-two-methods-of-calculating-the-probability-of-winning-a-best-of-7-s/172550", "openwebmath_score": 0.7524754405021667, "openwebmath_perplexity": 259.4722261995007, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850857421198, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8737255204978254}}
{"url": "https://math.stackexchange.com/questions/2422782/how-to-find-the-argument-of-cos-2-i-sin-2", "text": "How to find the argument of $\\cos 2 - i\\sin 2$\n\nHow would you find the argument of the following number. $$\\cos{2}-i\\sin{2}$$\n\nI'm aware that complex numbers in the form $r(\\cos{\\theta}+i\\sin{\\theta})$ have an argument of $\\theta$, but what do you do with the $-$ sign?\n\nRegardless of method, you will demonstrate some knowledge of Euler's identity and/or trigonometric identities.\n\nMethod I: Arctangent of slope\n\nThe argument is the arctangent of the ratio of the imaginary component to the real component, accounting for quadrant. If you forget to account for quadrant, this gives arguments in the interval $(-\\pi/2, \\pi/2)$, \"half\" of which are wrong. (Recall that arctangent is not a single-valued inverse. This is obvious when one remembers that tangent is periodic.)\n\nSince $\\cos 2 < 0$ and $-\\sin 2 < 0$, this is in quadrant III.\n\\begin{align*} \\arctan \\frac{-\\sin 2}{\\cos 2} &= \\arctan(-\\tan 2)) \\\\ &= - \\arctan \\tan 2 &&\\text{$\\arctan$ is an odd function}\\\\ &= - \\arctan \\tan (2 - \\pi) &&\\text{$\\tan$ has period $\\pi$}\\\\ &= -(2 - \\pi +\\pi k), k \\in \\mathbb{Z} &&-\\pi/2 < 2 - \\pi < \\pi/2 \\text{.} \\end{align*} Choosing $k=1$, this lands in quadrant III, giving the argument $-2$. If we had just mechanically evaluated, say with a calculator, \\begin{align*} \\arctan \\frac{-\\sin 2}{\\cos 2} &= \\arctan(2.18504\\dots) \\\\ &= 1.14159\\dots{} + \\pi k, k \\in \\mathbb{Z} \\text{.} \\end{align*} The subset of these in quadrant III have $k$ odd. Picking $k = -1$, we get $-2$.\n\nOf all the $k$ that give a result in the correct quadrant, which do you pick? You pick the one that conforms to your conventions for the range of value of an argument, if you have such a convention. If you do not, pick the one that makes your subsequent steps easier (or don't pick and leave the result as an equivalence class $\\mod 2 \\pi$).\n\nMethod II: Conjugation\n\nIf your only problem is the wrong sign of imaginary component, use conjugation. \\begin{align*} \\arg(\\cos 2 - \\mathrm{i} \\sin 2) &= \\arg(\\overline{\\cos 2 + \\mathrm{i} \\sin 2}) \\\\ &= \\arg(\\overline{\\mathrm{e}^{\\mathrm{i} 2}}) \\\\ &= -\\arg(\\mathrm{e}^{\\mathrm{i} 2}) \\\\ &= -2 \\text{.} \\end{align*}\n\nMethod III: Even-odd identities\n\nSine is odd. Cosine is even. This is expressed in the even-odd identities. So $\\cos(-2) = \\cos(2)$ and $\\sin(-2) = -\\sin(2)$. Consequently, \\begin{align*} \\cos 2 - \\mathrm{i} \\sin 2 &= \\cos -2 + \\mathrm{i} \\sin -2 \\\\ &= \\mathrm{e}^{\\mathrm{i}(-2)} \\text{,} \\end{align*} having argument $-2$.\n\nBrahadeesh's and MrYouMath's answers use this method, without identifying what was done. Michael Rozenberg's answer combines this with the periodicity (by $2\\pi$) identity, again without identifying what was done.\n\n\u2022 Thank you very much. This has really cleared things up for me! \u2013\u00a0Ewan Miller Sep 9 '17 at 20:53\n\nYou can write this complex number as $\\cos(-2) + i \\sin(-2)$. Now you can see that the argument is $-2$.\n\n$z=r(\\sin\\theta+i\\sin\\theta)$, where $r\\geq0$ and $\\theta=\\arg{z}\\in[0,2\\pi)$.\n\n$$\\cos2-i\\sin2=\\cos(2\\pi-2)+i\\sin(2\\pi-2).$$\n\nThus, $\\arg(\\cos2-i\\sin2)=2\\pi-2$.\n\n\u2022 Thank you. Am I right in thinking then that there is a general rule that r(cosA-isinA)=r(cos(2\u03c0-A)+isin(2\u03c0-A))? \u2013\u00a0Ewan Miller Sep 9 '17 at 16:05\n\u2022 @Ewan Miller It depends on definition. Sometimes $\\arg{z}\\in[0,2\\pi)$ and sometimes $\\arg{z}\\in[-\\pi,\\pi)$. You need to ask your teacher. I like the first versa. \u2013\u00a0Michael Rozenberg Sep 9 '17 at 16:13\n\u2022 In the case where $\\arg{z}\\in[-\\pi,\\pi)$, would it be r(cosA-isinA)=r(cos(-A)+isini(-A))? \u2013\u00a0Ewan Miller Sep 9 '17 at 16:17\n\u2022 Also, I haven't got a teacher as I have just left school and am teaching this to myself in my gap year \u2013\u00a0Ewan Miller Sep 9 '17 at 16:19\n\u2022 @Ewan Miller It's not so easy. If $A\\in(-\\pi,\\pi)$ then you are right:$r(\\cos(-A)+i\\sin(-A))$ , but for $A=-\\pi$ we get $r(\\cos{A}+i\\sin{A})$. All this depend on the level of your teacher. Maybe he don't see here problems. \u2013\u00a0Michael Rozenberg Sep 9 '17 at 16:23", "date": "2019-10-16 04:19:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2422782/how-to-find-the-argument-of-cos-2-i-sin-2", "openwebmath_score": 0.9999862909317017, "openwebmath_perplexity": 418.4326234172378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850859899082, "lm_q2_score": 0.8887587831798666, "lm_q1q2_score": 0.8737255047866653}}
{"url": "http://mathhelpforum.com/statistics/17095-counting-permutation-combinations.html", "text": "# Thread: Counting, Permutation, and Combinations\n\n1. ## Counting, Permutation, and Combinations\n\nGuys I have several questions that I would like to ask about:\n\n1. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, if the bride and the groom are among these 10 people, if:\na. the bride must be in the picture\nb. exactly one of the bride and the groom is in the picture\n\n2. How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other?\n\n3. A professor writes 40 discrete mathematics true/false questions. Of these statements in these questions, 17 are true. If the questions can be positioned in any order, how many different answer keys are possible.\n\n1a. because the bride is already included among 6 people, therefore there are 5 more people that we can choose and arrange from, so it's 5!. I am not sure if this is right\n1b. I don't quite understand the question\n\n2. P(10,10) x P(11,6)\n\n3. C(40,17)\n\nI am not quite sure of all these answers, but suggestions/help to my questions are highly appreciated\n\n2. a. the bride must be in the picture\nLet us divide this problem into cases.\nCode:\nBXXXXX\nXBXXXX\nXXBXXX\nXXXBXX\nXXXXBX\nXXXXXB\nWhere \"X\" stands for any other person. And \"B\" stands for bride. We will find the number of such cases and then add all of them up together.\n\nIn the first case there are $\\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\\displaystyle 6\\cdot 15120 = 90720$.\n\nb. exactly one of the bride and the groom is in the picture\nAgain divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.\n\n3. Originally Posted by ThePerfectHacker\nLet us divide this problem into cases.\nCode:\nBXXXXX\nXBXXXX\nXXBXXX\nXXXBXX\nXXXXBX\nXXXXXB\nWhere \"X\" stands for any other person. And \"B\" stands for bride. We will find the number of such cases and then add all of them up together.\n\nIn the first case there are $\\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\\displaystyle 6\\cdot 15120 = 90720$.\n\nAgain divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.\nas for case 1:\n\nthe bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.\n\nand for case 2:\n\nthe bride is not in picture but the groom is has the same possibilities as case 1 right?\n\nthen we just add case 1 and case 2?\n\n4. Originally Posted by TheRekz\nas for case 1:\n\nthe bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.\n\nand for case 2:\n\nthe bride is not in picture but the groom is has the same possibilities as case 1 right?\n\nthen we just add case 1 and case 2?\nI agree with you.\n\n5. what if now the question is both the bride and the groom must be in the picture:\n\n(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080\n\nis this right?\n\n6. Hello, TheRekz!\n\n1. In how many ways can a photographer at a wedding\narrange 6 people in a row from a group of 10 people,\nif the bride and the groom are among these 10 people, and:\na. the bride must be in the picture\nb. exactly one of the bride and the groom is in the picture\n\na) The bride is in the picture.\nSelect five more from the other nine people: .$\\displaystyle C(9,5)$ ways.\nThese six people can be arranged in $\\displaystyle 6!$ ways.\n\nTherefore, there are: .$\\displaystyle C(9,5) \\times 6!$ ways.\n\nb) Either bride or the groom is in the picture (but not both).\nThere are 2 choices for the newlywed to be in the picture.\nThe other 5 people are chosen from the remaining 8 people (not the other newlywed).\n. . There are: .$\\displaystyle C(8,5)$ ways.\nThe six people can be arranged in $\\displaystyle 6!$ ways.\n\nTherefore, there are: .$\\displaystyle 2 \\times C(8,5) \\times 6!$ ways.\n\n2. How many ways are there for 10 women and six men to stand in a line\nso that no two men stand next to each other?\n\nArrange the tenwomen in a row. .There are $\\displaystyle P(10,10) = 10!$ possible arrangements.\n\nLeave spaces between the women: .$\\displaystyle \\_\\;W\\;\\_\\;W\\;\\_\\;W\\;\\_\\;W\\;\\_\\;W\\;\\_\\;W\\;\\_\\;W\\;\\ _\\;W\\;\\_\\;W\\;\\_\\;W\\;\\_$\n\nArrange the six men in any six of the eveleven spaces.\n. . There are: .$\\displaystyle P(11,6)$ ways.\n\nTherefore, there are: .$\\displaystyle P(10,10) \\times P(11,6)$ ways.\n\n3. A professor writes 40 discrete mathematics true/false questions.\nOf these statements in these questions, 17 are true.\nIf the questions can be positioned in any order,\nhow many different answer keys are possible?\n\nYour answer is correct . . . $\\displaystyle C(40,17)$\n\n7. Originally Posted by TheRekz\nwhat if now the question is both the bride and the groom must be in the picture:\n\n(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080\n\nis this right?\nIf both the bride and groom are in the picture then:\n\nfix the bride in the first position:\n\nBGXXXX = 1*1*8*7*6*5\nBXGXXX = 1*8*1*7*6*5\n.\n.\n.\nBXXXXG = 1*8*7*6*5*1\n\nThen there are 5*(1*1*8*7*6*5) ways to arrange them with the bride in the first position.\n\nNow repeat with the bride in the second, third,....,sixth position and you get the same result each time. Therefore, if you add these all up you get (5*(1*1*8*7*6*5))*6 = 50,400\n\n,\n\n,\n\n,\n\n# A professor writes 40 discrete mathematics true/false\n\nClick on a term to search for related topics.", "date": "2018-06-21 20:34:03", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/17095-counting-permutation-combinations.html", "openwebmath_score": 0.7154180407524109, "openwebmath_perplexity": 532.5746162634782, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127395946486, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8737072920003169}}
{"url": "https://math.stackexchange.com/questions/4016228/integration-by-trigonometric-substitution-vs-table-of-integral-solution", "text": "# Integration by Trigonometric Substitution vs Table of Integral Solution\n\nI'm not sure how to phrase this question, but I find myself confused over the correctness of a particular solution in the table of integral, specifically: $$\\int\\frac{dx}{\\sqrt{x^2-a^2}}=\\ln|\\sqrt{x^2-a^2}+x|+C$$\n\nI find that if you try to solve it with trigonometric substitution, you get a different answer:\n\nSo, for the example, since the form $$\\sqrt{x^2-a^2}$$ is present, we use the substitution $$x=asec \\theta$$, and $$dx=asec\\theta tan \\theta d\\theta$$. Doing so:\n\n$$\\int\\frac{asec\\theta tan \\theta d\\theta}{\\sqrt{(asec \\theta)^2-a^2}}$$ $$\\int\\frac{asec\\theta tan \\theta d\\theta}{\\sqrt{a^2sec^2 \\theta-a^2}}$$ $$\\int\\frac{asec\\theta tan \\theta d\\theta}{a\\sqrt{sec^2 \\theta-1}}$$ Using the trigonometric identity $$tan^2 \\theta=sec^2 \\theta -1$$ $$\\int\\frac{asec\\theta tan \\theta d\\theta}{atan \\theta}$$\n\nCancelling like terms, we get $$\\int sec \\theta$$, which finally integrates to $$ln|sec \\theta + tan \\theta|+C$$\n\nUndoing the substitution, using this triangle:\n\nhttps://i.stack.imgur.com/6tkR7.png\n\nWith $$sec \\theta = \\frac{x}{a}$$ and $$tan \\theta = \\frac{\\sqrt{x^2-a^2}}{a}$$, the final answer then is:\n\n$$ln\\biggl|\\frac{x}{a} + \\frac{\\sqrt{x^2-a^2}}{a}\\biggl|+C$$\n\nWhich is evidently different from the one from the table of integrals. Is my solution wrong or something?\n\n\u2022 looks the same to me, since $-\\ln |a|$ can be absorbed into $C$? \u2013\u00a0Calvin Khor Feb 7 at 13:48\n\u2022 Oh right, that can be done. Thanks for pointing that out \u2013\u00a0Xyzar Feb 7 at 13:52\n\n$$\\ln \\left|\\dfrac{x+\\sqrt{x^2-a^2}}{a}\\right| +c= \\ln |x+\\sqrt{x^2-a^2}| - \\ln|a| +c = \\ln(x+\\sqrt{x^2-a^2}) +k$$\n$$k = c-\\ln |a| =$$ constant", "date": "2021-04-14 16:21:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4016228/integration-by-trigonometric-substitution-vs-table-of-integral-solution", "openwebmath_score": 0.9585645794868469, "openwebmath_perplexity": 388.3014534827061, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211531789391, "lm_q2_score": 0.8962513821399044, "lm_q1q2_score": 0.8736848058758396}}
{"url": "http://math.stackexchange.com/questions/495159/different-ways-of-integrating-3-sin-x-cos-x", "text": "# Different Ways of Integrating $3\\sin x\\cos x$\n\nI am asking this question for my son who is in (equivalent) twelfth grade and I failed to answer his query.\n\nWhen he tries to integrate $3\\sin x\\cos x$, he finds that this can be done in at least following three ways. And these three ways do not produce equivalent results.\n\nONE\n\nLet us assume, $\\sin x = z$.\n\nThis gives, \\begin{align*} \\cos x &= \\frac{dz}{dx}\\\\ \\cos x dx &= dz \\end{align*}\n\nSo, we can write,\n\n\\begin{align*} \\int 3\\sin x\\cos x dx &=3 \\int zdz\\\\ &=3 \\frac{z^2}{2}\\\\ &=\\frac{3}{2} \\sin^2 x\\\\ &=\\frac{3}{4}\\times 2\\sin^2 x\\\\ &=\\frac{3}{4} (1 -\\cos 2x)\\\\ \\end{align*}\n\nTWO\n\nLet us assume, $\\cos x = z$.\n\nThis gives, \\begin{align*} -\\sin x &= \\frac{dz}{dx}\\\\ \\sin x dx &= -dz \\end{align*}\n\nSo, we can write,\n\n\\begin{align*} \\int 3\\sin x\\cos x dx &=-3 \\int zdz\\\\ &=-3 \\frac{z^2}{2}\\\\ &=-\\frac{3}{2} \\cos^2 x\\\\ &=-\\frac{3}{4}\\times 2\\cos^2 x\\\\ &=-\\frac{3}{4} (1 +\\cos 2x)\\\\ \\end{align*}\n\nTHREE\n\n\\begin{align*} \\int 3\\sin x\\cos x dx &=\\frac{3}{2}\\int 2\\sin x\\cos x dx\\\\ &=\\frac{3}{2}\\int \\sin 2x dx\\\\ &=-\\frac{3}{2}\\times\\frac{1}{2} \\cos 2x\\\\ &=-\\frac{3}{4} \\cos 2x\\\\ \\end{align*}\n\nThe results found in above three methods are not the same.\n\nIf we try a simple approach of evaluating the integration results at, $x = \\frac{\\pi}{6}$, we get as follows.\n\nFrom the first one,\n\n$\\frac{3}{4} (1 -\\cos 2x) = \\frac{3}{4} (1 -\\cos \\frac{2\\pi}{6}) = \\frac{3}{4} (1 -\\cos \\frac{\\pi}{3}) = \\frac{3}{4} (1 - \\frac{1}{2}) = \\frac{3}{4}\\times\\frac{1}{2} = \\frac{3}{8}$\n\nFrom the second one,\n\n$-\\frac{3}{4} (1 +\\cos 2x) = -\\frac{3}{4} (1 +\\cos \\frac{2\\pi}{6}) = -\\frac{3}{4} (1 +\\cos \\frac{\\pi}{3}) = -\\frac{3}{4} (1 + \\frac{1}{2}) = -\\frac{3}{4}\\times\\frac{3}{2} = -\\frac{9}{8}$\n\nFrom the third one,\n\n$-\\frac{3}{4} \\cos 2x=-\\frac{3}{4} \\cos \\frac{2\\pi}{6} = -\\frac{3}{4} \\cos \\frac{\\pi}{3} = -\\frac{3}{4} \\times \\frac{1}{2} = -\\frac{3}{8}$\n\nClearly, we are getting some nonequivalent results. We have failed to find the mistakes or explanations behind this. Your help will be appreciated.\n\n-\nYou can locate an several primitives of a function, they differ in a constant, but that does not mean that the results are wrong. \u2013\u00a0 Hiperion Sep 16 '13 at 5:50\nIndefinite integration can be 'misleading' in this sense. \u2013\u00a0 copper.hat Sep 16 '13 at 5:52\nA great question to ask students to see if they can come up with Zev's answer (or equivalent). \u2013\u00a0 nbubis Sep 16 '13 at 6:01\n\u2013\u00a0 Jonas Meyer Sep 16 '13 at 6:07\nWhen I was doing A-levels, I came across a similar example, but the two \"different\" answers to the integral involved respectively $\\tan^2(x)$ and $\\sec^2(x)$. Of course, these two expressions differ by a constant, but the explanation was non-obvious to me (and my dad) then, and I had to get my teacher to explain it. \u2013\u00a0 Hammerite Sep 16 '13 at 9:29\n\nYou're forgetting that an indefinite integral must include a constant of integration; for any chosen constant $C$, we have that $$\\frac{d}{dx}\\left(-\\frac{3}{4}\\cos(2x)+C\\right)=3\\sin(x)\\cos(x),$$ and that is precisely the relationship captured by the statement that\n\n$$\\int 3\\sin(x)\\cos(x)\\,dx=-\\frac{3}{4}\\cos(2x)+C.$$\n\n-\nIndeed this is a great demonstration of that's so important. \u2013\u00a0 David H Sep 16 '13 at 5:50\nSo, if is asked to solve the problem in the examination, which approach he should follow? His textbook contains only one of the answers. And as you can see, some graders may like only the conventional answers. No disrespect intended. \u2013\u00a0 Masroor Sep 16 '13 at 6:01\n@MMA - All the answers are equally \"wrong\" since non of them contain a constant. When you let the constant C \"swallow\" the other constant term, all three give the same result. \u2013\u00a0 nbubis Sep 16 '13 at 6:02\n@nbubis But his text book contains only one of the answers, (I can not get the book now), and given the prevailing circumstances, it will be difficult to modify the system. I can put down the answer which is the book in six/seven hours when the book comes back from school. \u2013\u00a0 Masroor Sep 16 '13 at 6:07\n@MMA I sympathize with your situation, especially if your particular school has you feeling that cynical towards the system. But I have no way of predicting what which technique a grader would consider most appropriate, because there isn't one (and if the grader doesn't know it then your son knows more calculus than his teacher). He should learn all three methods. Solving integrals is like choosing a golf-club for a given shot. The best one is the one that gets the job done. \u2013\u00a0 David H Sep 16 '13 at 6:23\n\nAll three answers are correct provided you add a constant to each one of those.\n\nBecause from the very definition of integration, it is the area under the curve, so it requires bounds to give a unique value. You can't evaluate the value of an indefinite integral without including constant.\n\nAnd I am sure that in the examination, your son won't be asked to evaluate the value of an integral without providing limits of integration or providing its value at some other point.\n\nFor instance, in question it may be mentioned that evaluate the value of expression at x=\u03c0/6 , given its value at x=0 is 1. So in this case, all three answers will give the correct value i.e. 11/8\n\n-", "date": "2015-04-28 04:29:48", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/495159/different-ways-of-integrating-3-sin-x-cos-x", "openwebmath_score": 0.9894180297851562, "openwebmath_perplexity": 470.0376727639004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211553734217, "lm_q2_score": 0.8962513786759491, "lm_q1q2_score": 0.8736848044659108}}
{"url": "https://mathhelpboards.com/threads/chain-rule-and-e-u.1637/", "text": "# [SOLVED]chain rule and e^(u)\n\n#### DeusAbscondus\n\n##### Active member\nHi folks,\nI don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.\n\nHere is the nub of my difficulty:\n\n1. $\\text{If }f(x)=e^x \\text{then }f'(x)=e^x$\n\nThis I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.\nBut the following I do not understand; could someone help me towards understanding?\n\n2. $\\text{If }g(x)=e^{x^{4}} \\text{then }g'(x)=4e^{x^4}x^3$\n\nAm I on the right track to observe that, if 2. is correct, then g(x) is a composite function, hence subject to the chain rule?\n\nAnd if so, is the following a generally valid way to work through this and all such problems:\n\n$\\text{If }g(x)=e^{x^{4}} \\text{find }g'(x)$\n\n$\\text{Now}g'(x)=u'v' \\text{by Chain Rule}$\n\n$\\text{So, let }u=x^4 \\text{and }v=e^u$\n\n$\\text{Then }u'=4x^3 \\text{and }v'=e^u \\text{ (by some rule which currently exceeds my understanding)}$\n\n$\\text{Therefore }g'(x)=u'v'=4x^3*e^u=\\text{(via substitution) }4x^3*e^{x^{4}}$\n\n$\\text{Which, simplified }=4e^{x^{4}}x^3$\n\nFinally, I have a similar hesitation/scruple/sense of vertigo when it comes to dealing with another unusual derivative, that of:\n\n$ln(x)$\n\nIf anyone can see why, having the read the foregoing, I would feel unsure of myself around this animal, could they possibly add some notes to help me tame it?\n\nRegs,\nDeusAbs\n\nLast edited:\n\n#### Jameson\n\nStaff member\nHi DeusAbscondus,\n\n1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that $$\\displaystyle f'(x) = \\lim_{h \\rightarrow 0} \\frac{f(x+h)-f(x)}{h}$$.\n\nIn this case that becomes $$\\displaystyle \\lim_{h \\rightarrow 0} \\frac{e^{x+h}-e^x}{h}=\\frac{e^x \\cdot e^h-e^x}{h}=\\frac{e^x(e^h-1)}{h}$$.\n\nOk, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.\n\nSo the last bit becomes $$\\displaystyle \\left( e^x \\cdot \\lim_{h \\rightarrow 0} \\frac{e^h-1}{h} \\right)$$.\n\nFrom here you need to compute that limit and you can do it a few ways. For a less rigorous approach that is more intuitive as well, I suggest graphing $$\\displaystyle \\frac{e^h-1}{h}$$ on a graphing calculator and you'll see that as h approaches zero, then the y value approaches 1.\n\nSo at the end we get that $$\\displaystyle \\left( e^x \\cdot \\lim_{h \\rightarrow 0} \\frac{e^h-1}{h} \\right) = e^x \\cdot (1) = e^x$$, therefore $$\\displaystyle \\frac{d}{dx} e^x = e^x$$\n\nPlease let me know if any of that is unclear. Again, it's not a rigorous proof by any means, more of a \"walk through\".\n\n2) You are completely correct that $$\\displaystyle \\frac{d}{dx} e^{f(x)} = e^{f(x)} \\cdot f'(x)$$. As you said this is an application of the chain rule.\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\n$\\text{Then }u'=4x^3 \\text{and }v'=e^u \\text{ (by some rule which currently exceeds my understanding)}$\nHi DeusAbscondus,\n\nThe only rule that is used here is the chain rule.\n\n\\begin{eqnarray}\n\ng(x)&=&e^{x^{4}}\\\\\n\ng'(x)&=&\\frac{d}{d(x^4)}e^{x^{4}}\\frac{d}{dx}x^4\\\\\n\n&=&e^{x^{4}}(4x^3)\\\\\n\n\\therefore g'(x)&=&4x^3e^{x^{4}}\n\n\\end{eqnarray}\n\nKind Regards,\nSudharaka.\n\n#### Fantini\n\n##### \"Read Euler, read Euler.\" - Laplace\nMHB Math Helper\nPerhaps to avoid particularizing and help you think of the exponential as any other function, think of $e^{x^4}$ as $h(x) = e^x$ and $g(x) = x^4$, then $(h \\circ g)(x) = h(g(x)) = e^{x^4}$ and $(h \\circ g)' (x) = h'(g(x)) \\cdot g'(x) = e^{x^4} \\cdot (4x^3) = 4 e^{x^4} x^3.$\n\n#### chisigma\n\n##### Well-known member\nHi DeusAbscondus,\n\n1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that $$\\displaystyle f'(x) = \\lim_{h \\rightarrow 0} \\frac{f(x+h)-f(x)}{h}$$.\n\nIn this case that becomes $$\\displaystyle \\lim_{h \\rightarrow 0} \\frac{e^{x+h}-e^x}{h}=\\frac{e^x \\cdot e^h-e^x}{h}=\\frac{e^x(e^h-1)}{h}$$.\n\nOk, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.\n\nSo the last bit becomes $$\\displaystyle \\left( e^x \\cdot \\lim_{h \\rightarrow 0} \\frac{e^h-1}{h} \\right)$$.\n\nFrom here you need to compute that limit and you can do it a few ways...\nOne possible 'rigorous' way to demonstrate that...\n\n$\\displaystyle \\lim_{h \\rightarrow 0} \\frac{e^{h}-1}{h}=1$ (1)\n\n... is the following...\n\na) You start from the definition of exponential...\n\n$\\displaystyle e^{h}= \\lim_{n \\rightarrow \\infty} (1+\\frac{h}{n})^{n}$ (2)\n\nb) You demonstrate the identity...\n\n$\\displaystyle \\lim_{n \\rightarrow \\infty} (1+\\frac{h}{n})^{n}= \\lim_{n \\rightarrow \\infty} \\sum_{k=0}^{n} \\frac {h^{k}}{k!} = 1 + h + \\frac {h^{2}}{2}+... \\frac{h^{n}}{n!}+...$ (3)\n\nc) from (3) You derive...\n\n$\\displaystyle \\frac{e^{h}-1}{h}= 1 + \\frac{h}{2} +... + \\frac{h^{n-1}}{n!}+...$ (4)\n\n... and the (1) follows immediately...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### DeusAbscondus\n\n##### Active member\nthanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \\circ g)$ and some other stuff besides.\n\nJust a quick follow up:\ncan one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?\n\nand, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?\n\nthx again gentelmen,\n\nDeusAbs\n\n#### Jameson\n\nStaff member\nIt's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.\n\nAnyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.\n\nLet me know what you think!\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nthanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \\circ g)$ and some other stuff besides.\n\nJust a quick follow up:\ncan one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?\n\nand, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?\n\nthx again gentelmen,\n\nDeusAbs\nHi DeusAbscondus,\n\nA good place filled with video lessons is Khan Academy. Here is a link to the Epsilon Delta limit definition introduction video.\n\nKind Regards,\nSudharaka.\n\n#### CaptainBlack\n\n##### Well-known member\nHi folks,\nI don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.\n\nHere is the nub of my difficulty:\n\n1. $\\text{If }f(x)=e^x \\text{then }f'(x)=e^x$\n\nThis I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.\nBut the following I do not understand; could someone help me towards understanding?\nFrom the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:\n\n$\\exp(x)=\\lim_{n \\to \\infty} \\left(1+\\frac{x}{n}\\right)^n$\n\nbut it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..\n\nSo to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).\n\nCB\n\n#### DeusAbscondus\n\n##### Active member\nFrom the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:\n\n$\\exp(x)=\\lim_{n \\to \\infty} \\left(1+\\frac{x}{n}\\right)^n$\n\nbut it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..\n\nSo to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).\n\nCB\nI just had a look at graphs of these functions: given that they are mirror images of each other graphically, I would have thought their inter-definability (if i can put it thus) bordered the trivial.\n\nI'd also have thought my post here (indeed, the combined oeuvre of my posts to date ) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.\nPrescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?\n\nI always enjoy your sparse, ascetical take on things, Cap'n.\n\nSo, thanks for the input,\nDeus Abs\n\n#### DeusAbscondus\n\n##### Active member\nIt's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.\n\nAnyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.\n\nLet me know what you think!\nThanks Jameson; just spent some hours there! I suffer from a dearth of really first-rate material and my course notes are a nest of errors and poorly edited material, so I appreciate such a rich fund of examples, proofs, worked examples and tips.\n\nI'll be sure to make a fuller response once I've had more time to check it.\n\nDeusAbs\n\n#### chisigma\n\n##### Well-known member\nFrom the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:\n\n$\\exp(x)=\\lim_{n \\to \\infty} \\left(1+\\frac{x}{n}\\right)^n$\n\nbut it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..\n\nSo to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).\n\nCB\nMore that 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...\n\n$\\displaystyle e^{z}= \\lim_{n \\rightarrow \\infty} (1+\\frac{z}{n})^{n}$\n\n$\\displaystyle \\ln z= \\lim_{n \\rightarrow \\infty} n\\ (z^{\\frac{1}{n}}-1)$ (1)\n\n... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### CaptainBlack\n\n##### Well-known member\nI'd also have thought my post here (indeed, the combined oeuvre of my posts to date ) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.\nPrescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?\nYou must have at least some idea of how the exponential function is defined, or at the very least how \"e\" is defined, otherwise you are starting in the middle.\n\n(I would recommend you get a copy of Morris Kline's book \"Calculus; An intuitive and physical approach\", the only flaw of which is its use of US Customary Units)\n\nCB\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nMore than 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...\n\n$\\displaystyle e^{z}= \\lim_{n \\rightarrow \\infty} \\Bigl(1+\\frac{z}{n}\\Bigr)^{n}$\n\n$\\displaystyle \\ln z= \\lim_{n \\rightarrow \\infty} n\\ (z^{\\frac{1}{n}}-1)$ (1)\n\n... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...\nIn my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \\sum_{n=0}^\\infty\\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\\mathbb{R}\\to(0,\\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)\n\n#### chisigma\n\n##### Well-known member\nIn my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \\sum_{n=0}^\\infty\\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\\mathbb{R}\\to(0,\\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)\nOn the basis of the [easily enough demonstrable...] identity...\n\n$\\displaystyle \\lim_{n \\rightarrow \\infty} (1+\\frac{z}{n})^{n} = \\lim_{n \\rightarrow \\infty} \\sum_{k=0}^{n} \\frac{z^{k}}{k!}$ (1)\n\n... the two definitions must be considered fully equivalent. On the purely conceptual basis however the original Euler's definition requires only the concept of limit and is 'more elementary' than any other definition and that is an excellent reason for defining it as 'the best'... at least for me..\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### QuestForInsight\n\n##### Member\n1. $\\text{If }f(x)=e^x \\text{then }f'(x)=e^x$\nThis is very easy to get using the power series definition from Opalg's post.\n\n\\begin{aligned} & (e^x)' = \\bigg(\\sum_{k \\ge 0}\\frac{x^k}{k!}\\bigg)' = \\sum_{k \\ge 0}\\frac{kx^{k-1}}{k!} = \\sum_{k \\ge 1}\\frac{kx^{k-1}}{k!} = \\sum_{k+1 \\ge 1}\\frac{(k+1)x^{k}}{(k+1)!} = \\sum_{k \\ge 0}\\frac{x^{k}}{k!}.\\end{aligned}\n\n#### chisigma\n\n##### Well-known member\nThis is very easy to get using the power series definition from Opalg's post.\n\n\\begin{aligned} & (e^x)' = \\bigg(\\sum_{k \\ge 0}\\frac{x^k}{k!}\\bigg)' = \\sum_{k \\ge 0}\\frac{kx^{k-1}}{k!} = \\sum_{k \\ge 1}\\frac{kx^{k-1}}{k!} = \\sum_{k+1 \\ge 1}\\frac{(k+1)x^{k}}{(k+1)!} = \\sum_{k \\ge 0}\\frac{x^{k}}{k!}.\\end{aligned}\nThat is true of course but [conceptually...] the preliminary demonstration of the convergence theorem of the series of derivatives is required. The computation of derivative of $e^{x}$ using the standard derivative definition require only to demonstrate that...\n\n$\\displaystyle \\lim_{h \\rightarrow 0} \\frac {e^{h}-1}{h}=1$ (1)\n\n... and that requires elementary instruments...\n\nKind regards\n\n$\\chi$ $\\sigma$", "date": "2020-09-30 14:27:28", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/chain-rule-and-e-u.1637/", "openwebmath_score": 0.8799797296524048, "openwebmath_perplexity": 658.579868791415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676442828174, "lm_q2_score": 0.891811053345418, "lm_q1q2_score": 0.8736784337762836}}
{"url": "https://www.coursehero.com/file/p5ubjj9/77-To-use-the-alternating-series-test-consider-a-n-f-n-where-f-x-arctan-ax-We/", "text": "# 77 to use the alternating series test consider a n f\n\n\u2022 Homework Help\n\u2022 rodblouin\n\u2022 66\n\u2022 100% (1) 1 out of 1 people found this document helpful\n\nThis preview shows page 33 - 35 out of 66 pages.\n\n##### We have textbook solutions for you!\nThe document you are viewing contains questions related to this textbook.\nThe document you are viewing contains questions related to this textbook.\nChapter 11 / Exercise 77\nSingle Variable Calculus\nStewart\nExpert Verified\n77. To use the alternating series test, consider a n = f ( n ) , where f ( x ) = arctan( a/x ) . We need to show that f ( x ) is decreasing. Since f 0 ( x ) = 1 1 + ( a/x ) 2 - a x 2 , we have f 0 ( x ) < 0 for a > 0 , so f ( x ) is decreasing for all x . Thus a n +1 < a n for all n , and as lim n \u2192\u221e arctan( a/n ) = 0 for all a , by the alternating series test, X n =1 ( - 1) n arctan( a/n ) converges. 78. The n th partial sum of the series is given by S n = 1 - 1 2 + 1 3 - \u00b7 \u00b7 \u00b7 + ( - 1) n - 1 n , so the absolute value of the first term omitted is 1 / ( n + 1) . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than 1 / ( n +1) . Thus, we want to choose n large enough so that 1 / ( n +1) 0 . 01 . Solving this inequality for n yields n 99 , so we take 99 or more terms in our partial sum. 79. The n th partial sum of the series is given by S n = 1 - 2 3 + 4 9 - \u00b7 \u00b7 \u00b7 + ( - 1) ( n - 1) 2 3 ( n - 1) ,\n##### We have textbook solutions for you!\nThe document you are viewing contains questions related to this textbook.\nThe document you are viewing contains questions related to this textbook.\nChapter 11 / Exercise 77\nSingle Variable Calculus\nStewart\nExpert Verified\n750 Chapter Nine /SOLUTIONS so the absolute value of the first term omitted is (2 / 3) n . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than (2 / 3) n . Thus, we want to choose n large enough so that (2 / 3) n 0 . 01 . Solving this inequality for n yields n 11 . 358 , so taking 12 or more terms in our partial sum is guaranteed to be within 0 . 01 of the sum of the series. Note: Since this is a geometric series, we know the exact sum to be 1 / (1+2 / 3) = 0 . 6 . The partial sum S 12 is 0 . 595 , which is indeed within 0 . 01 of the sum of the series. Note, however, that S 11 = 0 . 6069 , which is also within 0 . 01 of the exact sum of the series. Theorem 9.9 gives us a value of n for which S n is guaranteed to be within a small tolerance of the sum of an alternating series, but not necessarily the smallest such value. 80. The n th partial sum of the series is given by S n = 1 2 - 1 24 + 1 720 - \u00b7 \u00b7 \u00b7 + ( - 1) n - 1 (2 n )! , so the absolute value of the first term omitted is 1 / (2 n + 2)! . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than 1 / (2 n +2)! . Thus, we want to choose n large enough so that 1 / (2 n +2)! 0 . 01 . Substituting n = 2 into the expression 1 / (2 n + 2)! yields 1 / 720 which is less than 0.01. We therefore take 2 or more terms in our partial sum. 81. Since 0 c n 2 - n for all n , and since 2 - n is a convergent geometric series, c n converges by the Comparison Test. Similarly, since 2 n a n , and since 2 n is a divergent geometric series, a n diverges by the Comparison Test. We do not have enough information to determine whether or not b n and d n converge. 82. (a) The sum a n \u00b7 b n = 1 /n 5 , which converges, as a p -series with p = 5 , or by the integral test: Z 1 1 x 5 dx = lim b \u2192\u221e x - 4 ( - 4) b 1 = lim b \u2192\u221e b - 4 ( - 4) + 1 4 = 1 4 . Since this improper integral converges, a n \u00b7 b n also converges. (b) This is an alternating series that satisfies the conditions of the alternating series test: the terms are decreasing and have limit 0 , so ( - 1) n / n converges. (c) We have a n b n = 1 /n , so a n b n is the harmonic series, which diverges.", "date": "2021-12-01 16:38:49", "meta": {"domain": "coursehero.com", "url": "https://www.coursehero.com/file/p5ubjj9/77-To-use-the-alternating-series-test-consider-a-n-f-n-where-f-x-arctan-ax-We/", "openwebmath_score": 0.8163377046585083, "openwebmath_perplexity": 302.37585511133676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676490318648, "lm_q2_score": 0.8918110454379297, "lm_q1q2_score": 0.8736784302648261}}
{"url": "https://math.stackexchange.com/questions/1407002/are-these-two-optimization-problems-equivalent-to-each-other", "text": "# Are these two optimization problems equivalent to each other?\n\nLet $\\mathbf{x}=[x_1,\\ldots,x_K]^T$. For a fixed vector $\\mathbf{a}$, I have the following optimization problem : \\begin{array}{rl} \\min \\limits_{\\mathbf{x}} & | \\mathbf{a}^T \\mathbf{x} | \\\\ \\mbox{s.t.} & x_k\\ge 0, \\forall k \\\\ & \\sum_k x_k=1 \\end{array} The second optimization problem is: \\begin{array}{rl} \\min \\limits_{\\mathbf{x}} & ( \\mathbf{a}^T \\mathbf{x})^2\\\\ \\mbox{s.t.} & x_k\\ge 0, \\forall k \\\\ & \\sum_k x_k=1 \\end{array}\n\nMy question: are these two problems equivalent to each other? if yes, how to solve them ? and which one is easier to solve ?\n\n\u2022 yes they are equivalent. have you tried the Lagrange multiplier method? \u2013\u00a0user251257 Aug 23 '15 at 16:07\n\u2022 The objective function of the problem 1 is $\\ell^1$-norm and problem 2 is $\\ell^2$-norm, so although equivalent, the objective function of the second problem is continuously differentiable, therefore the second problem is much easier to solve. \u2013\u00a0guille_NP Aug 23 '15 at 21:22\n\u2022 @guille_NP: The second problem is indeed much easier to solve, but definitely not for the reason you provide. Your statements about norms are simply wrong (for example, the second objective is not a norm...). Just say $t \\mapsto t^2$ is increasing on $\\mathbb{R}_+$ (so that the problems are equivalent) and differentiable (so that the second problem is easier). \u2013\u00a0dohmatob Aug 24 '15 at 2:35\n\u2022 @mat: For solving the second problem, checkout this thread which treats a slightly more general problem math.stackexchange.com/q/1309671/168758 \u2013\u00a0dohmatob Aug 24 '15 at 3:07\n\nWhich is easier? Neither. Both of these models can be solved analytically, and in the exact same way. First, let's knock out the easy cases:\n\n1. If any $a_i=0$ for some $i$, then the optimal value of either model is clearly $0$, as demonstrated by selecting setting $x$ to be the $i$th unit vector (the vector with $1$ at position $i$ and zeros everywhere else).\n\n2. If $a_i>0$ and $a_j<0$ for some pair $(i,j)$, then again the optimal value is zero. Just set $$x_i=-a_j/(a_i-a_j), \\quad x_j = a_i/(a_i-a_j)$$ and the other elements of $x$ to zero.\n\n3. The cases that remain are where $a$ is entirely positive or entirely negative. But if $a$ is negative, then substituting $a$ for $-a$ will not change the optimal value or the values of $x$ that achieve this value, thanks to the presence of the absolute value or square.\n\n4. So we now have one case remaining: when $a$ is a positive vector. But in this case, $a^Tx$ is positive over the simplex, allowing us to drop the absolute value! \\begin{array}{ll} \\text{minimize} & a^T x \\\\ \\text{subject to} & \\mathbf{1}^T x = 1 \\\\ & x \\geq 0 \\end{array} We can solve this by inspection: the optimal value is $\\min_i a_i$ for the first model, and $\\min_i a_i^2$ for the second. If the minimizing $a_i$ is unique, then the unique solution is the $i$th unit vector. But if there are multiple elements of $a$ with the same magnitude, any convex combination of those corresponding unit vectors is a solution.\n\nPutting it all together, the solution is $\\min_i |a_i|$ or $\\min_i a_i^2$ if $a\\succeq 0$ or $a\\preceq 0$, and $0$ otherwise. There really was no need for case 1, but it was easy to see.\n\n\u2022 I am interested in the case where the elements of $\\mathbf{a}$ are $\\ge 0$. I suppose that adding a constant $b$ to the objective function, i.e. $\\mathbf{a}^T\\mathbf{x}+b$, will not affect the solution ($i$th unit vector) but it will modify the optimal value to $\\min_i a_i +b$. Am I right? \u2013\u00a0tam Aug 24 '15 at 9:37\n\u2022 Yes, that is right. \u2013\u00a0Michael Grant Aug 24 '15 at 10:56\n\u2022 Please, I have an additional question (I don't know if I have to post it in a seperate question): I want to maximize $|\\mathbf{a}^T\\mathbf{x}-b|$, with $\\mathbf{1}^T\\mathbf{x}=1$, $\\mathbf{x} >0$ and $\\mathbf{a}^T\\mathbf{x} >b$. How to solve this kind of problems. (note that $b>0$) \u2013\u00a0tam Aug 29 '15 at 10:42\n\u2022 I think that this problem can be re-written as: maximize $\\mathbf{a}^T \\mathbf{x}-b$, with $\\mathbf{1}^T\\mathbf{x}=1$ and $\\mathbf{x}>0$. So the solution is similar to that in 4), with the difference of putting $\\max_i a_i-b$ instead of $\\min_i |a_i-b|$. Is it right ? \u2013\u00a0tam Aug 29 '15 at 16:03\n\nLet's go for an explicit analytic construction of the set of all the solutions to your problem.\n\nBasic notation: $e_K := \\text{ column vector of }K\\text{ }1'$s. $\\langle x, y \\rangle$ denotes the inner product between two vectors $x$ and $y$. For example $\\langle e_K, x\\rangle$ simply amounts to summing the components of $x$.\n\nRecall the following definitions, to be used without further explanation.\n\nPreimage: Given abstract sets $X$, $Y$, $Z \\subseteq Y$., the preimage of $Z$ under $f$ is defined by $f^{-1}Z := \\{x \\in X | f(x) \\in Z\\}$. This has nothing to do with function inversion.\n\nConvex hull: Given a subset $C$ of $\\mathbb{R}^K$, its convex hull is defined by $\\textit{conv }C :=$ smallest convex subset of $\\mathbb{R}^K$ containing $C$.\n\nIndicator function: Given a subset $C$ of $\\mathbb{R}^K$ its indicator function $i_C:\\mathbb{R}^K \\rightarrow (-\\infty, +\\infty]$ is defined by $i_C(x) = 0$ if $x \\in C$; otherwise $i_C(x) = +\\infty$.\n\nSimplex: The $K$-dimensional simplex is defined by $\\Delta_K := \\{x \\in \\mathbb{R}^K|x\\ge 0, \\langle e_K, x\\rangle = 1\\}$. Equivalently, $\\Delta_K = \\{\\text{rows of the }K \\times K\\text{ identity matrix }I_K\\}$. Each row of $I_K$ is a vertex of $\\Delta_K$.\n\nFaces of a simplex: Given a subset of indices $I \\subseteq \\{1,2,...,K\\}$, the face of $\\Delta_K$ spanned by $I$, denoted $F_K(I)$, is defined to be the convex hull of those rows of $I_k$ indexed by $I$. Trivially, $\\Delta_K$ is a face of itself. Geometrically, $F_K(I)$ is a $\\#I$-dimensional simplex whose vertices are those vertices of $\\Delta_K$ which labelled by $I$.\n\nLet $f:\\mathbb{R}^K \\rightarrow (-\\infty,+\\infty]$ be an extended real-value function.\n\nSubdifferential: $\\partial f(x) := \\{g \\in \\mathbb{R}^K| f(z) \\ge f(s) + \\langle g, z - x\\rangle, \\forall z \\in \\mathbb{R}^K\\}$.\n\nFenchel-Legengre transform: $f^*(x) := \\underset{z \\in \\mathbb{R}^K}{\\text{max }}\\langle x, z\\rangle - f(z)$.\n\nThe optimal value of your objective under the given constraints can be conveniently written as \\begin{eqnarray} v = \\underset{x \\in \\mathbb{R}^K}{\\text{min }}g(x) + f(Dx), \\end{eqnarray} where $D := a^T \\in \\mathbb{R}^{1 \\times K}$, $g := i_{\\Delta_K}$ and $f:s \\mapsto \\frac{1}{2}s^2$. One recognizes the above problem as the primal form of the saddle-point problem \\begin{eqnarray*} \\underset{x \\in \\mathbb{R}^K}{\\text{min }}\\underset{s \\in \\mathbb{R}}{\\text{max }}\\langle s, Dx\\rangle + g(x) - f^*(s) \\end{eqnarray*} Given a dual solution $\\hat{s} \\in \\mathbb{R}$, the entire set of primal solutions $\\hat{X}$ is given by (one can check that sufficient conditions that warrant the strong Fenchel duality Theorem) \\begin{eqnarray*} \\hat{X} = \\partial g^*(-D^T\\hat{s}) \\cap D^{-1}\\partial f^*(\\hat{s}). \\end{eqnarray*} For your problem, one easily computes, $g^*(y) = \\underset{x \\in \\Delta_K}{\\text{max }}\\langle x, y\\rangle$, and so by the Bertsekas-Danskin theorem (see Proposition A.22 of Bertsekas' PhD thesis) for subdifferentials, we have \\begin{eqnarray*} \\begin{split} \\partial g^*(-D^T\\hat{s}) &= \\partial g^*(-\\hat{s}a) = ... \\text{( some computations masked )} \\\\ &= F_K\\left(\\{1 \\le i \\le K | \\hat{s}a_i\\text{ is a minimal component of }\\hat{s}a\\}\\right). \\end{split} \\end{eqnarray*} Also $\\partial f^*(\\hat{s}) = \\{\\hat{s}\\}$, and so $D^{-1}\\partial f^*(\\hat{s}) = \\{x \\in \\mathbb{R}^K|\\langle a, x\\rangle = \\hat{s}\\}$. Thus the set of all primal solutions is \\begin{eqnarray*} \\hat{X} = F_K\\left(\\{1 \\le i \\le K | \\hat{s}a_i\\text{ is a minimal component of }\\hat{s}a\\}\\right) \\cap \\{x \\in \\mathbb{R}^K|\\langle a, x\\rangle = \\hat{s}\\}. \\end{eqnarray*} It remains now to find a dual solution. Define $\\alpha := \\underset{1 \\le i \\le K}{\\text{min }}a_i \\le \\beta := \\underset{1 \\le i \\le K}{\\text{max }}a_i$. One computes \\begin{eqnarray*} \\begin{split} v &= \\underset{x \\in \\Delta_K}{\\text{min }}\\underset{s \\in \\mathbb{R}}{\\text{max }}s\\langle a, x\\rangle -\\frac{1}{2}s^2 = \\underset{s, \\lambda \\in \\mathbb{R}}{\\text{max }}\\underset{x \\ge 0}{\\text{min }}\\langle sa + \\lambda e_K, x\\rangle - \\frac{1}{2}s^2 - \\lambda\\\\ &=\\underset{s, \\lambda \\in \\mathbb{R}}{\\text{max }}-\\frac{1}{2}s^2 - \\lambda\\text{ subject to }sa + \\lambda e_K \\ge 0\\\\ &=-\\underset{s, \\lambda \\in \\mathbb{R}}{\\text{min }}\\frac{1}{2}s^2 + \\lambda\\text{ subject to }\\lambda \\ge -\\underset{1 \\le i \\le K}{\\text{min }}sa_i\\\\ &= -\\underset{s \\in \\mathbb{R}}{\\text{min }}\\frac{1}{2}s^2 -\\underset{1 \\le i \\le K}{\\text{min }}sa_i = -\\frac{1}{2}\\underset{s \\in \\mathbb{R}}{\\text{min }}\\begin{cases}(s - \\alpha)^2 - \\alpha^2, &\\mbox{ if }s \\ge 0,\\\\(s - \\beta)^2 - \\beta^2, &\\mbox{ otherwise}\\end{cases} \\end{split} \\end{eqnarray*} Thus \\begin{eqnarray*} \\hat{s} = \\begin{cases}\\beta, &\\mbox{ if }\\beta < 0,\\\\0, &\\mbox{ if }\\alpha \\le 0 \\le \\beta,\\\\\\alpha, &\\mbox{ if }\\alpha > 0.\\end{cases} \\end{eqnarray*} Therefore the set of all solutions to the original / primal problem is \\begin{eqnarray*} \\hat{X} = \\begin{cases}F_K\\left(\\{1 \\le i \\le K | a_i = \\beta\\}\\right), &\\mbox{ if }\\beta < 0,\\\\ \\Delta_K \\cap \\{x \\in \\mathbb{R}^K| \\langle a, x\\rangle = 0\\}, &\\mbox{ if } \\alpha \\le 0 \\le \\beta,\\\\ F_K\\left(\\{1 \\le i \\le K | a_i = \\alpha\\}\\right), &\\mbox{ if }\\alpha > 0,\\end{cases} \\end{eqnarray*}\n\nNow that you have the hammer, find the nails...\n\n\u2022 Case (1): $\\beta < 0$. Choose any $k \\in \\{1,2,...,K\\}$ such that $a_k = \\beta$, and take $\\hat{x} = k$th row of $I_K$.\n\n\u2022 Case (2a): $a_k = 0$ for some $k$. Take $\\hat{x} = k$th row of $I_K$.\n\n\u2022 Case (2b): $\\alpha < 0 < \\beta$, and $a_k \\ne 0 \\forall k$. Choose $k_1, k_2 \\in \\{1,2,...,k\\}$ such that $a_{k_1} < 0 < a_{k_2}$, and take $\\hat{x}_k = \\frac{1}{{a_{k_2} - a_{k_1}}}\\begin{cases}a_{k_2}, &\\mbox{ if }k = k_1,\\\\-a_{k_1},&\\mbox{ if }k = k_2,\\\\0, &\\mbox{ otherwise.}\\end{cases}$\n\n\u2022 Case (3): $\\alpha > 0$. Choose any $k \\in \\{1,2,...,K\\}$ such that $a_k = \\alpha$, and take $\\hat{x} = k$th row of $I_K$.\n\n\u2022 Dude. If I were a professor I'd want you in my research group cranking out papers. This is impressive, and you have my vote. But don't you think this is like bringing a fire truck to put out a campfire? \u2013\u00a0Michael Grant Aug 24 '15 at 14:41\n\u2022 @MichaelGrant: Your solution is actually more elegant in many respects. Mine is indeed probably an overkill. Mindful of this, I tried to be particularly pedagocial here :) On such toy problems ($D$ is a vector, $g^*$ is a polyhedral function, etc.), it turns out that one can explicitly construct the entire solution set using Fenchel's machinery. Unfortunately, this is not always the case. :) \u2013\u00a0dohmatob Aug 25 '15 at 6:33\n\nYour problem can be equivalently modeled to LP with an extra variable, say $\\epsilon$ and two more simple linear constraints, $| \\mathbf{a}^T \\mathbf{x} |\\leq \\epsilon$. Now your problem can be rewritten as $$\\begin{array}{rl} \\min \\limits_{\\mathbf{x}} & \\epsilon \\\\ \\mbox{s.t.} & \\mathbf{a}^T \\mathbf{x} \\leq \\epsilon \\\\&- \\mathbf{a}^T \\mathbf{x} \\leq \\epsilon \\\\ & x_k\\ge 0, \\forall k \\\\ & \\sum_k x_k=1 \\end{array}$$\n\nNow you can solve this problem using simplex method or any other method you like!", "date": "2019-10-17 17:57:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1407002/are-these-two-optimization-problems-equivalent-to-each-other", "openwebmath_score": 0.9984907507896423, "openwebmath_perplexity": 496.439232969149, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676478446031, "lm_q2_score": 0.8918110339361276, "lm_q1q2_score": 0.8736784179380697}}
{"url": "https://mathhelpboards.com/threads/for-which-parameter-values-the-function-is-continuous-and-differentiable.3380/", "text": "# For which parameter values the function is continuous and differentiable\n\n#### Yankel\n\n##### Active member\nFor which values of a,b and c, the next function is continuous and differentiable at x=2 ?\n\n$$\\left\\{\\begin{matrix} 3x-1 & x\\leq 2\\\\ ax^{2}+bx+c & x>2 \\end{matrix}\\right.$$\n\n1. b=2-c\n2. b=6+2c+2a\n3. 7+c-2a\n4. b=3-a-(3/4)c\n\nI know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.\nI have calculated the limit when f goes to 2 from the right side, and I got:\n\n4a+2b+c\n\nand so for continuously I need:\n\n4a+2b+c = 5\n\nbut here I got stuck...\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nThe only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).\n\nRecall that for a function to be continuous at $a$, you need:\n\n$$\\lim_{x \\to a^{-}} f(x) = \\lim_{x \\to a^{+}} f(x)$$\n\nSo in your case, you need to show that the following is true:\n\n$$\\lim_{x \\to 2^{-}} 3x - 1 = \\lim_{x \\to 2^{+}} ax^2 + bx + c ~ ~ ~ \\Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$\n\nWhich is what you got, so this is correct. But, you also need differentiability.\n\nRecall that for a function $f(x)$ to be differentiable at $a$, you need:\n\n$$\\lim_{x \\to a^{-}} f'(x) = \\lim_{x \\to a^{+}} f'(x)$$\n\nSo, differentiate both pieces of the function, and show that the following is true:\n\n$$\\lim_{x \\to 2^{-}} 3 = \\lim_{x \\to 2^{+}} 2ax + b ~ ~ ~ \\Longleftrightarrow ~ ~ ~ 3 = 4a + b$$\n\nYou are now left with two equations in three unknowns:\n\n$$4a + 2b + c = 5$$\n\n$$4a + b = 3$$\n\nSubtract the second from the first, to obtain:\n\n$$b + c = 2 ~ ~ ~ \\Longleftrightarrow b = 2 - c$$\n\nAnd substitute this back into the second:\n\n$$4a + 2 - c = 3 ~ ~ ~ \\Longleftrightarrow ~ ~ ~ a = \\frac{1}{4} \\left ( c + 1 \\right )$$\n\nThis gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:\n\n$$a = \\frac{3}{4} ~ ~ ~ \\text{and} ~ ~ ~ b = 0$$\n\nThat is one solution, out of infinitely many. This is also the only class of solutions.\n\nLast edited:\n\n#### chisigma\n\n##### Well-known member\nFor which values of a,b and c, the next function is continuous and differentiable at x=2 ?\n\n$$\\left\\{\\begin{matrix} 3x-1 & x\\leq 2\\\\ ax^{2}+bx+c & x>2 \\end{matrix}\\right.$$\n\n1. b=2-c\n2. b=6+2c+2a\n3. 7+c-2a\n4. b=3-a-(3/4)c\n\nI know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.\nI have calculated the limit when f goes to 2 from the right side, and I got:\n\n4a+2b+c\n\nand so for continuously I need:\n\n4a+2b+c = 5\n\nbut here I got stuck...\nYou have to impose first that $\\displaystyle \\lim_{x \\rightarrow 2 +} f(x) = \\lim_{x \\rightarrow 2 -} f(x)$ and after that $\\displaystyle \\lim_{x \\rightarrow 2 +} f ^{\\ '} (x) = \\lim_{x \\rightarrow 2 -} f^{\\ '}(x)$...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nThe only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).\n\nRecall that for a function to be continuous at $a$, you need:\n\n$$\\lim_{x \\to a^{-}} f(x) = \\lim_{x \\to a^{+}} f(x)$$\n\nSo in your case, you need to show that the following is true:\n\n$$\\lim_{x \\to 2^{-}} 3x - 1 = \\lim_{x \\to 2^{+}} ax^2 + bx + c ~ ~ ~ \\Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$\n\nWhich is what you got, so this is correct. But, you also need differentiability.\n\nRecall that for a function $f(x)$ to be differentiable at $a$, you need:\n\n$$\\lim_{x \\to a^{-}} f'(x) = \\lim_{x \\to a^{+}} f'(x)$$\nThis is true, but not an obvious statement. It looks like the definition of \"continuous\" but derivatives are NOT necessarily contuinuous. What is true is that any derivative, even if not continuous, satisfies the \"intermediate value theorem\".\n\nSo, differentiate both pieces of the function, and show that the following is true:\n\n$$\\lim_{x \\to 2^{-}} 3 = \\lim_{x \\to 2^{+}} 2ax + b ~ ~ ~ \\Longleftrightarrow ~ ~ ~ 3 = 4a + b$$\n\nYou are now left with two equations in three unknowns:\n\n$$4a + 2b + c = 5$$\n\n$$4a + b = 3$$\n\nSubtract the second from the first, to obtain:\n\n$$b + c = 2 ~ ~ ~ \\Longleftrightarrow b = 2 - c$$\n\nAnd substitute this back into the second:\n\n$$4a + 2 - c = 3 ~ ~ ~ \\Longleftrightarrow ~ ~ ~ a = \\frac{1}{4} \\left ( c + 1 \\right )$$\n\nThis gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:\n\n$$a = \\frac{3}{4} ~ ~ ~ \\text{and} ~ ~ ~ b = 0$$\n\nThat is one solution, out of infinitely many. This is also the only class of solutions.", "date": "2021-09-25 07:18:38", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/for-which-parameter-values-the-function-is-continuous-and-differentiable.3380/", "openwebmath_score": 0.895719051361084, "openwebmath_perplexity": 359.6032082204733, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407137099625, "lm_q2_score": 0.8976952907388474, "lm_q1q2_score": 0.8736736054527481}}
{"url": "https://math.stackexchange.com/questions/943384/how-to-compute-sum-k-1100-1k", "text": "How to compute $\\sum_{k =1}^{100}(-1)^k$\n\nToday I tried to compute $$\\sum_{k =1}^{100}(-1)^k$$\n\nIs there a way to find the result more quickly ?\n\nBelow if my attempt to find the result.\n\nEspecially without considering the case of odd and even numbers like I did ?\n\nLet's consider the following sum:\n\n$$\\sum_{k=1}^{n}(-1)^k = (-1)^1 + (-1)^2 + (-1)^3 + ... +(-1)^n$$\n\nIf $n$ is even then $\\frac n2$ terms have an even exponent and $\\frac n2$ terms have an odd exponent.\n\n$$(-1)^p = -1 \\tag{when p is odd}$$ $$(-1)^p = 1 \\tag{when p is even}$$\n\nThen when $n$ is even\n\n\\begin{align} \\sum_{k=1}^{n}(-1)^k &= \\frac n2 \\times(-1) + \\frac n2 \\times 1 \\\\ & = \\frac n2 - \\frac n2 \\\\ & = 0 \\\\ \\end{align}\n\n100 is an even number, so\n\n$$\\sum_{k =1}^{100}(-1)^k = 0$$\n\n\u2022 Call the sum $S$, so that $S=(-1)+(1)+\\dotsb+(-1)+(1)$. Notice that $-S=(1)+(-1)+\\dotsb+(1)+(-1)$\u2014that is, $-S$ is just $S$ reversed. So, $S=-S$, which means that $S=0$. \u2013\u00a0Akiva Weinberger Sep 23 '14 at 20:15\n\u2022 If you say that $n/2$ terms have an even exponent, then you are already assuming that $n$ is even. \u2013\u00a0Dietrich Burde Sep 23 '14 at 20:15\n\u2022 The quick way is to cancel $1$s with $-1$s. You have equally many of both, and if you didn't, you'd have just one term left after canceling. But of course it's also a finite geometric series, and there's a standard formula for that. \u2013\u00a0Michael Hardy Sep 23 '14 at 20:57\n\nyes the series has a direct summation formula: $$S_k=\\sum_{i=1}^k (-1)^i={1 \\over 2}\\Big((-1)^k-1\\Big)$$ you can prove this easily using induction. \\begin{aligned} \\mbox{k=1: }\\ \\ \\ &true\\\\ \\mbox{k }\\to\\mbox{ k+1: }\\ \\ \\ &S_{k+1}=S_{k}+(-1)^{k+1}&={1 \\over 2}\\Big((-1)^k-1\\Big)+(-1)^{k+1} =\\\\ & &={1 \\over 2}\\Big((-1)^k+2(-1)^{k+1}-1\\Big) =\\\\ & &={1 \\over 2}\\Big((-1)^k(1+2(-1)^{1})-1\\Big) =\\\\ & &={1 \\over 2}\\Big((-1)^k(1-2)-1\\Big) =\\\\ & &={1 \\over 2}\\Big((-1)^k(-1)-1\\Big) =\\\\ & &={1 \\over 2}\\Big((-1)^{k+1}-1\\Big) \\ \\ \\ \\square \\end{aligned}\n\nNote that $(-1)^{2k-1}=-1$ and $(-1)^{2k}=1$ for all $k\\geq1$. Hence the sum upto any even power should equal $0$.\n\nThe easiest way to find the result is to look at the first few partial sums\n\n$$-1, 0, -1, \\ldots$$\n\nand identify the pattern. Everything beyond that is just seeking to give a more rigorous justification of it.\n\n\u2022 Sure. I think that's what I am looking for. A rigorous justification but explained quickly. \u2013\u00a0alexandrekow Sep 23 '14 at 20:37\n\nSince $(-1)^{2n}=1$ and $(-1)^{2n-1}=-1$, we have $$\\sum_{k=1}^n (-1)^k = -1+1-1+1-\\cdots \\pm 1=\\left\\{\\begin{array}{ll}-1 & \\text{if}\\,n\\,\\text{is odd}\\\\ 0 & \\text{if}\\,n\\,\\text{is even}\\end{array}\\right.$$\n\n$$\\sum_{k=1}^{n}(-1)^k=\\sum_{k=1}^{n}(-1)(-1)^{k-1}=-\\sum_{k=1}^{n}(-1)^{k-1}=$$ $$=-\\sum_{k=0}^{n-1}(-1)^{k}=-\\frac{1-(-1)^{n}}{1-(-1)}=\\frac{(-1)^{n}-1}{2}$$ for $n=100$ $$\\sum_{k=1}^{100}(-1)^k=\\frac{(-1)^{100}-1}{2}=\\frac{1-1}{2}=0$$\n\n\u2022 How do you do to transform the third part of the first line into the fourth one? \u2013\u00a0alexandrekow Sep 23 '14 at 20:39\n\u2022 This is of course the hard way. \u2013\u00a0Michael Hardy Sep 23 '14 at 20:57\n\u2022 you can see my edit! \u2013\u00a0Adi Dani Sep 23 '14 at 20:59", "date": "2019-08-19 01:35:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/943384/how-to-compute-sum-k-1100-1k", "openwebmath_score": 0.9988545179367065, "openwebmath_perplexity": 487.820554189164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105314577313, "lm_q2_score": 0.8947894724152068, "lm_q1q2_score": 0.8735923853564735}}
{"url": "https://byjus.com/question-answer/assertion-for-n-ge-4-let-displaystyle-a-n-sum-j-1-n-sum-k/", "text": "Question\n\n# Assertion :For $$n\\ge 4$$. Let $$\\displaystyle{ a }_{ n }=\\sum _{ j=1 }^{ n }{ \\sum _{ k=j }^{ n }{ \\begin{pmatrix} n \\\\ k \\end{pmatrix} } } \\begin{pmatrix} k \\\\ j \\end{pmatrix}$$ and $${ b }_{ n }={ a }_{ n }+{ 2 }^{ n+1 }$$;$${ b }_{ n+1 }=5{ b }_{ n }-6{ b }_{ n-1 }$$ for all $$n\\ge 2$$ Reason: $${ a }_{ n }=5{ a }_{ n }-6{ a }_{ n }$$ for all $$n\\ge 2$$\n\nA\nBoth Assertion and Reason are correct and Reason is the correct explanation for Assertion\nB\nBoth Assertion and Reason are correct but Reason is not the correct explanation for Assertion\nC\nAssertion is correct but Reason is incorrect\nD\nAssertion is incorrect but Reason is correct\n\nSolution\n\n## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion$$\\displaystyle \\begin{pmatrix} n \\\\ k \\end{pmatrix}\\begin{pmatrix} k \\\\ j \\end{pmatrix}=\\frac { n! }{ k!\\left( n-k \\right) ! } .\\frac { k! }{ j!\\left( k-j \\right) ! }$$$$\\displaystyle=\\frac { n! }{ j!\\left( n-j \\right) ! } \\frac { \\left( n-j \\right) ! }{ \\left( n-k \\right) !\\left( k-j \\right) ! } =\\begin{pmatrix} n \\\\ k \\end{pmatrix}\\begin{pmatrix} n\\quad - & j \\\\ n\\quad - & k \\end{pmatrix}$$Thus $$\\displaystyle { a }_{ n }=\\sum _{ j=1 }^{ n }{ \\begin{pmatrix} h \\\\ k \\end{pmatrix} } \\sum _{ j=1 }^{ n }{ \\begin{pmatrix} n\\quad - & j \\\\ n\\quad - & k \\end{pmatrix} }$$$$\\displaystyle=\\sum _{ j=1 }^{ n }{ \\begin{pmatrix} n \\\\ j \\end{pmatrix} } { 2 }^{ n-j }={ \\left( 2+1 \\right) }^{ n }-{ 2 }^{ n }={ 3 }^{ n }-{ 2 }^{ n }$$$$\\Rightarrow { b }_{ 1 }={ 3 }^{ n }{ -2 }^{ n }+2\\left( { 2 }^{ n } \\right) ={ 3 }^{ n }+{ 2 }^{ n }$$Now $${ a }_{ n+1 }={ 3 }^{ n+1 }-{ 2 }^{ n+1 }=\\left( 3+2 \\right) \\left( { 3 }^{ n }-{ 2 }^{ n } \\right) -2\\left( { 3 }^{ n } \\right) +3\\left( { 2 }^{ n } \\right)$$$$=5\\left( { 3 }^{ n }-{ 2 }^{ n } \\right) -6\\left( { 3 }^{ n-1 }-{ 2 }^{ n+1 } \\right) ={ 5a }_{ n }-{ 6a }_{ n-1 }$$And $${ 5b }_{ n }-{ 6b }_{ n-1 }=5\\left( { a }_{ n }+{ 2 }^{ n+1 } \\right) -6\\left( { a }_{ n-1 }+{ 2 }^{ n } \\right)$$$$={ 5a }_{ n }-6{ b }_{ n-1 }+{ 2 }^{ n }\\left( 10-6 \\right)$$$$\\Rightarrow { a }_{ n+1 }+{ 2 }^{ n+2 }={ b }_{ n+1 }$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-19 20:50:22", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/assertion-for-n-ge-4-let-displaystyle-a-n-sum-j-1-n-sum-k/", "openwebmath_score": 0.43255355954170227, "openwebmath_perplexity": 3490.6262891610063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631627142339, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8735542615145756}}
{"url": "https://math.stackexchange.com/questions/3321478/evaluating-lim-k-to-infty-sum-n-1-infty-frac-sin-left-pi-n-k-right", "text": "# Evaluating $\\lim_{k\\to\\infty}\\sum_{n=1}^{\\infty} \\frac{\\sin\\left(\\pi n/k\\right)}{n}$\n\nRecently, I was asked by a friend to compute the limit of the following series\n\n$$\\displaystyle{\\lim_{k\\to\\infty}}\\sum_{n=1}^{\\infty} \\frac{\\sin\\left(\\frac{\\pi n}{k}\\right)}{n}$$\n\nHaving seen a similar problem to this before, Difficult infinite trigonometric series, I used the same complex argument approach as seen in that problem.\n\nUltimately, for this problem, I obtained $$\\frac{\\pi}{2}$$ as my answer. However, this limit can also be interpreted as a Riemann Sum, except the answer to the Riemann Sum differs from what I obtained as my answer, and according to Wolfram Alpha, the answer is expressed in terms of $$Si$$, where $$Si$$ is the sine integral.\n\nI'm wondering, does the limit invalidate the argument approach, or is there something else I'm missing, because this limit if I'm not mistaken is a Riemann Sum after all?\n\n\u2022 In a Riemann sum we would have both 'variables' tending to infinity simultaneously. In this case we have both $n$ and $k$ tending to infinity seperately and hence the Riemann sum approach should not be possible. This approach seems valid. \u2013\u00a0Peter Foreman Aug 12 at 21:48\n\u2022 Nor can you (directly) apply dominated cvg theorem ... \u2013\u00a0Olivier Aug 12 at 21:51\n\u2022 You effectively have $\\lim_{k\\to\\infty}\\lim_{N\\to\\infty}f(N,k)$ (where $N$ denotes the maximum summation bound). With a Riemann sum we would have $\\lim_{N\\to\\infty}f(N)$ where $k$ may be used as an index such that the function $f(N)$ is of the form $\\frac1N\\sum_{k=1}^N g(k/N)\\to\\int_0^1g(x)\\mathrm{d}x$. \u2013\u00a0Peter Foreman Aug 12 at 21:54\n\u2022 (Rephrasing of what Peter says) You have $\\lim_{k \\to \\infty} \\frac{1}{k} \\sum_{n=1}^{k} \\frac{\\sin(\\pi n/k)}{(n/k)} \\to \\int_{0}^{1} \\frac{sin(\\pi x)}{x} dx$ by Riemann sum argument; but you have to manage a larger interval here... ($n\\ge 1$, not just $n=1...k$) \u2013\u00a0Olivier Aug 12 at 22:03\n\u2022 You can do a collection of Riemann sums to deal with n=1...k, then n=k+1...2k, n=2k+1...3k, but you would then need a further argument to put these sums altogether. \u2013\u00a0Olivier Aug 12 at 22:16\n\nThe Riemann Sum would be \\begin{align} \\lim_{k\\to\\infty}\\sum_{n=1}^\\infty\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n} &=\\lim_{k\\to\\infty}\\sum_{n=1}^\\infty\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n/k}\\frac1k\\\\ &=\\int_0^\\infty\\frac{\\sin(\\pi x)}x\\,\\mathrm{d}x\\\\ &=\\int_0^\\infty\\frac{\\sin(x)}x\\,\\mathrm{d}x\\\\[3pt] &=\\frac\\pi2\\tag1 \\end{align} However, a cleaner way is to note that \\begin{align} \\sum_{n=1}^\\infty\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n} &=-\\mathrm{Im}\\!\\left(\\log\\left(1-e^{i\\pi/k}\\right)\\right)\\\\ &=\\frac\\pi2-\\frac\\pi{2k}\\tag2 \\end{align} and the limit is easy.\n\nTake Care\n\nOne must be careful with the convergence of the Riemann Sum. Here is one method to control the remainders.\n\nBecause $$|\\sin(\\pi x)|\\le1$$, we have $$\\int_m^{m+1}\\left|\\frac{\\sin(\\pi x)}x\\right|\\,\\mathrm{d}x \\le\\frac1m\\tag3$$ Furthermore, $$\\int_m^{m+2}\\sin(\\pi x)\\,\\mathrm{d}x=0$$, thus, \\begin{align} \\left|\\int_m^{m+2}\\frac{\\sin(\\pi x)}x\\,\\mathrm{d}x\\right| &=\\left|\\int_m^{m+2}\\sin(\\pi x)\\left(\\frac1x-\\frac1{m+1}\\right)\\mathrm{d}x\\right|\\\\ &\\le\\frac1{m(m+1)}+\\frac1{(m+1)(m+2)}\\\\[6pt] &=\\frac1m-\\frac1{m+2}\\tag4 \\end{align} Therefore, for any $$N\\ge m$$, $$\\left|\\int_m^N\\frac{\\sin(\\pi x)}x\\,\\mathrm{d}x\\right| \\le\\frac1m\\tag5$$ Because $$|\\sin(\\pi x)|\\le1$$, we have $$\\sum_{n=mk}^{(m+1)k}\\left|\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n}\\right| \\le\\frac1m\\tag6$$ Furthermore, $$\\sum\\limits_{n=mk}^{(m+2)k}\\sin\\left(\\frac{\\pi n}k\\right)=0$$, thus, \\begin{align} \\left|\\sum_{n=mk}^{(m+2)k}\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n}\\right| &=\\left|\\sum_{n=mk}^{(m+2)k}\\sin\\left(\\frac{\\pi n}k\\right)\\left(\\frac1n-\\frac1{(m+1)k}\\right)\\right|\\\\ &\\le\\frac1{m(m+1)}+\\frac1{(m+1)(m+2)}\\\\[6pt] &=\\frac1m-\\frac1{m+2}\\tag7 \\end{align} Therefore, for any $$M\\ge mk$$, $$\\left|\\sum_{n=mk}^M\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n}\\right|\\le\\frac1m\\tag8$$ For any $$\\epsilon\\gt0$$, let $$m\\ge\\frac4\\epsilon$$. Then Riemann Sums allow us to choose a $$k$$ large enough so that $$\\left|\\int_0^m\\frac{\\sin(\\pi x)}x\\,\\mathrm{d}x-\\sum_{n=1}^{mk}\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n/k}\\frac1k\\right|\\le\\frac\\epsilon2\\tag9$$ Inequalities $$(5)$$ and $$(8)$$ show that for any $$N\\ge m$$ and $$M\\ge mk$$, $$\\left|\\int_m^N\\frac{\\sin(\\pi x)}x\\,\\mathrm{d}x\\right|\\le\\frac\\epsilon4 \\quad\\text{and}\\quad \\left|\\sum_{n=mk}^M\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n/k}\\frac1k\\right|\\le\\frac\\epsilon4\\tag{10}$$ Inequalities $$(9)$$ and $$(10)$$ show that, for the $$k$$ chosen for $$(9)$$, $$\\left|\\sum_{n=1}^\\infty\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n}-\\frac\\pi2\\right|\\le\\epsilon\\tag{11}$$ Since $$\\epsilon\\gt0$$ was arbitrary, $$(11)$$ says that $$\\lim_{k\\to\\infty}\\sum_{n=1}^\\infty\\frac{\\sin\\left(\\frac{\\pi n}k\\right)}{n}=\\frac\\pi2\\tag{12}$$\n\n\u2022 Riemann sums works on a finite intervals; what's the additional argument used here to extend this to (0,\\infty) ? \u2013\u00a0Olivier Aug 12 at 22:11\n\u2022 I've gotta say, that is one sexy solution and elegant approach! +1 :) \u2013\u00a0Sanjoy Kundu Aug 12 at 22:16\n\u2022 @Olivier: In this case, we can take the sum from $1$ to $mk$ to get the integral from $0$ to $m$. Then carefully let $m\\to\\infty$. \u2013\u00a0robjohn Aug 12 at 22:27\n\u2022 @Olivier: I believe one can use Abel's Test to estimate the remainder. \u2013\u00a0robjohn Aug 12 at 22:42\n\u2022 @Olivier: I have added a section showing one method to control the remainders. It is a bit long, to include a lot of detail, but quite typical for this kind of Riemann Sum. \u2013\u00a0robjohn Aug 13 at 8:55", "date": "2019-11-21 08:03:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3321478/evaluating-lim-k-to-infty-sum-n-1-infty-frac-sin-left-pi-n-k-right", "openwebmath_score": 0.9806793332099915, "openwebmath_perplexity": 602.7809115968531, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631615116321, "lm_q2_score": 0.8856314692902446, "lm_q1q2_score": 0.8735542559833176}}
{"url": "https://de.mathworks.com/help/matlab/math/solve-system-of-odes-with-multiple-initial-conditions.html", "text": "Main Content\n\n# Solve System of ODEs with Multiple Initial Conditions\n\nThis example compares two techniques to solve a system of ordinary differential equations with multiple sets of initial conditions. The techniques are:\n\n\u2022 Use a `for`-loop to perform several simulations, one for each set of initial conditions. This technique is simple to use but does not offer the best performance for large systems.\n\n\u2022 Vectorize the ODE function to solve the system of equations for all sets of initial conditions simultaneously. This technique is the faster method for large systems but requires rewriting the ODE function so that it reshapes the inputs properly.\n\nThe equations used to demonstrate these techniques are the well-known Lotka-Volterra equations, which are first-order nonlinear differential equations that describe the populations of predators and prey.\n\n### Problem Description\n\nThe Lotka-Volterra equations are a system of two first-order, nonlinear ODEs that describe the populations of predators and prey in a biological system. Over time, the populations of the predators and prey change according to the equations\n\n`$\\begin{array}{l}\\frac{\\mathrm{dx}}{\\mathrm{dt}}=\\alpha \\mathit{x}-\\beta \\mathrm{xy},\\\\ \\frac{\\mathrm{dy}}{\\mathrm{dt}}=\\delta \\mathrm{xy}-\\gamma \\mathit{y}.\\end{array}$`\n\nThe variables in these equations are\n\n\u2022 $\\mathit{x}$ is the population size of the prey\n\n\u2022 $\\mathit{y}$ is the population size of the predators\n\n\u2022 $\\mathit{t}$ is time\n\n\u2022 $\\alpha$, $\\beta$, $\\delta$, and $\\gamma$ are constant parameters that describe the interactions between the two species. This example uses the parameter values $\\alpha =\\gamma =1$, $\\beta =0.01$, and $\\delta =0.02$.\n\nFor this problem, the initial values for $\\mathit{x}$ and $\\mathit{y}$ are the initial population sizes. Solving the equations then provides information about how the populations change over time as the species interact.\n\n### Solve Equations with One Initial Condition\n\nTo solve the Lotka-Volterra equations in MATLAB, write a function that encodes the equations, specify a time interval for the integration, and specify the initial conditions. Then you can use one of the ODE solvers, such as `ode45`, to simulate the system over time.\n\nA function that encodes the equations is\n\n```function dpdt = lotkaODE(t,p) % LOTKA Lotka-Volterra predator-prey model delta = 0.02; beta = 0.01; dpdt = [p(1) .* (1 - beta*p(2)); p(2) .* (-1 + delta*p(1))]; end ```\n\n(This function is included as a local function at the end of the example.)\n\nSince there are two equations in the system, `dpdt` is a vector with one element for each equation. Also, the solution vector `p` has one element for each solution component: `p(1)` represents $\\mathit{x}$ in the original equations, and `p(2)` represents $\\mathit{y}$ in the original equations.\n\nNext, specify the time interval for integration as $\\left[0,15\\right]$ and set the initial population sizes for $\\mathit{x}$ and $\\mathit{y}$ to 50.\n\n```t0 = 0; tfinal = 15; p0 = [50; 50];```\n\nSolve the system with `ode45` by specifying the ODE function, the time span, and the initial conditions. Plot the resulting populations versus time.\n\n```[t,p] = ode45(@lotkaODE,[t0 tfinal],p0); plot(t,p) title('Predator/Prey Populations Over Time') xlabel('t') ylabel('Population') legend('Prey','Predators')```\n\nSince the solutions exhibit periodicity, plot the solutions against each other in a phase plot.\n\n```plot(p(:,1),p(:,2)) title('Phase Plot of Predator/Prey Populations') xlabel('Prey') ylabel('Predators')```\n\nThe resulting plots show the solution for the given initial population sizes. To solve the equations for different initial population sizes, change the values in `p0` and rerun the simulation. However, this method only solves the equations for one initial condition at a time. The next two sections describe techniques to solve for many different initial conditions.\n\n### Method 1: Compute Multiple Initial Conditions with `for-`loop\n\nThe simplest way to solve a system of ODEs for multiple initial conditions is with a `for`-loop. This technique uses the same ODE function as the single initial condition technique, but the `for`-loop automates the solution process.\n\nFor example, you can hold the initial population size for $\\mathit{x}$ constant at 50, and use the `for`-loop to vary the initial population size for $\\mathit{y}$ between 10 and 400. Create a vector of population sizes for `y0`, and then loop over the values to solve the equations for each set of initial conditions. Plot a phase plot with the results from all iterations.\n\n```y0 = 10:10:400; for k = 1:length(y0) [t,p] = ode45(@lotkaODE,[t0 tfinal],[50 y0(k)]); plot(p(:,1),p(:,2)) hold on end title('Phase Plot of Predator/Prey Populations') xlabel('Prey') ylabel('Predators') hold off```\n\nThe phase plot shows all of the computed solutions for the different sets of initial conditions.\n\n### Method 2: Compute Multiple Initial Conditions with Vectorized ODE Function\n\nAnother method to solve a system of ODEs for multiple initial conditions is to rewrite the ODE function so that all of the equations are solved simultaneously. The steps to do this are:\n\n\u2022 Provide all of the initial conditions to `ode45` as a matrix. The size of the matrix is `s`-by-`n`, where `s` is the number of solution components and `n` is the number of initial conditions being solved for. Each column in the matrix then represents one complete set of initial conditions for the system.\n\n\u2022 The ODE function must accept an extra input parameter for `n`, the number of initial conditions.\n\n\u2022 Inside the ODE function, the solver passes the solution components p as a column vector. The ODE function must reshape the vector into a matrix with size `s`-by-n. Each row of the matrix then contains all of the initial conditions for each variable.\n\n\u2022 The ODE function must solve the equations in a vectorized format, so that the expression accepts vectors for the solution components. In other words, `f(t,[y1 y2 y3 ...])` must return `[f(t,y1) f(t,y2) f(t,y3) ...]`.\n\n\u2022 Finally, the ODE function must reshape its output back into a vector so that the ODE solver receives a vector back from each function call.\n\nIf you follow these steps, then the ODE solver can solve the system of equations using a vector for the solution components, while the ODE function reshapes the vector into a matrix and solves each solution component for all of the initial conditions. The result is that you can solve the system for all of the initial conditions in one simulation.\n\nTo implement this method for the Lotka-Volterra system, start by finding the number of initial conditions `n`, and then form a matrix of initial conditions.\n\n```n = length(y0); p0_all = [50*ones(n,1) y0(:)]';```\n\nNext, rewrite the ODE function so that it accepts `n` as an input. Use `n` to reshape the solution vector into a matrix, then solve the vectorized system and reshape the output back into a vector. A modified ODE function that performs these tasks is\n\n```function dpdt = lotkasystem(t,p,n) %LOTKA Lotka-Volterra predator-prey model for system of inputs p. delta = 0.02; beta = 0.01; % Change the size of p to be: Number of equations-by-number of initial % conditions. p = reshape(p,[],n); % Write equations in vectorized form. dpdt = [p(1,:) .* (1 - beta*p(2,:)); p(2,:) .* (-1 + delta*p(1,:))]; % Linearize output. dpdt = dpdt(:); end ```\n\nSolve the system of equations for all of the initial conditions using `ode45`. Since `ode45` requires the ODE function to accept two inputs, use an anonymous function to pass in the value of `n` from the workspace to `lotkasystem`.\n\n`[t,p] = ode45(@(t,p) lotkasystem(t,p,n),[t0 tfinal],p0_all);`\n\nReshape the output vector into a matrix with size `(numTimeSteps*s)`-by-`n`. Each column of the output `p(:,k)` contains the solutions for one set of initial conditions. Plot a phase plot of the solution components.\n\n```p = reshape(p,[],n); nt = length(t); for k = 1:n plot(p(1:nt,k),p(nt+1:end,k)) hold on end title('Predator/Prey Populations Over Time') xlabel('t') ylabel('Population') hold off```\n\nThe results are comparable to those obtained by the `for`-loop technique. However, there are some properties of the vectorized solution technique that you should keep in mind:\n\n\u2022 The calculated solutions can be slightly different than those computed from a single initial input. The difference arises because the ODE solver applies norm checks to the entire system to calculate the size of the time steps, so the time-stepping behavior of the solution is slightly different. The change in time steps generally does not affect the accuracy of the solution, but rather which times the solution is evaluated at.\n\n\u2022 For stiff ODE solvers (`ode15s`, `ode23s`, `ode23t`, `ode23tb`) that automatically evaluate the numerical Jacobian of the system, specifying the block diagonal sparsity pattern of the Jacobian using the `JPattern` option of `odeset` can improve the efficiency of the calculation. The block diagonal form of the Jacobian arises from the input reshaping performed in the rewritten ODE function.\n\n### Compare Timing Results\n\nTime each of the previous methods using `timeit`. The timing for solving the equations with one set of initial conditions is included as a baseline number to see how the methods scale.\n\n```% Time one IC baseline = timeit(@() ode45(@lotkaODE,[t0 tfinal],p0),2); % Time for-loop for k = 1:length(y0) loop_timing(k) = timeit(@() ode45(@lotkaODE,[t0 tfinal],[50 y0(k)]),2); end loop_timing = sum(loop_timing); % Time vectorized fcn vectorized_timing = timeit(@() ode45(@(t,p) lotkasystem(t,p,n),[t0 tfinal],p0_all),2);```\n\nCreate a table with the timing results. Multiply all of the results by 1e3 to express the times in milliseconds. Include a column with the time per solution, which divides each time by the number of initial conditions being solved for.\n\n```TimingTable = table(1e3.*[baseline; loop_timing; vectorized_timing], 1e3.*[baseline; loop_timing/n; vectorized_timing/n],... 'VariableNames',{'TotalTime (ms)','TimePerSolution (ms)'},'RowNames',{'One IC','Multi ICs: For-loop', 'Mult ICs: Vectorized'})```\n```TimingTable=3\u00d72 table TotalTime (ms) TimePerSolution (ms) ______________ ____________________ One IC 1.6558 1.6558 Multi ICs: For-loop 34.333 0.85832 Mult ICs: Vectorized 5.1357 0.12839 ```\n\nThe `TimePerSolution` column shows that the vectorized technique is the fastest of the three methods.\n\n### Local Functions\n\nListed here are the local functions that `ode45` calls to calculate the solutions.\n\n```function dpdt = lotkaODE(t,p) % LOTKA Lotka-Volterra predator-prey model delta = 0.02; beta = 0.01; dpdt = [p(1) .* (1 - beta*p(2)); p(2) .* (-1 + delta*p(1))]; end %------------------------------------------------------------------ function dpdt = lotkasystem(t,p,n) %LOTKA Lotka-Volterra predator-prey model for system of inputs p. delta = 0.02; beta = 0.01; % Change the size of p to be: Number of equations-by-number of initial % conditions. p = reshape(p,[],n); % Write equations in vectorized form. dpdt = [p(1,:) .* (1 - beta*p(2,:)); p(2,:) .* (-1 + delta*p(1,:))]; % Linearize output. dpdt = dpdt(:); end```\n\nDownload ebook", "date": "2021-09-24 11:23:36", "meta": {"domain": "mathworks.com", "url": "https://de.mathworks.com/help/matlab/math/solve-system-of-odes-with-multiple-initial-conditions.html", "openwebmath_score": 0.8711304664611816, "openwebmath_perplexity": 353.19926186571536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631635159684, "lm_q2_score": 0.8856314662716159, "lm_q1q2_score": 0.8735542547809568}}
{"url": "https://mathhelpboards.com/threads/integral-4.8956/", "text": "# Integral #4\n\n#### Random Variable\n\n##### Well-known member\nMHB Math Helper\nShow that for positive parameters $a$, $b$, and $c$,\n\n$$\\int_{0}^{\\infty} \\left( \\frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \\frac{e^{-cx}}{x} \\ \\right) \\ dx = b-a + a \\ln \\left(\\frac{a}{c} \\right) - b \\ln \\left(\\frac{b}{c} \\right)$$\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\n\n$$\\displaystyle F = \\int_{0}^{\\infty} \\left( \\frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \\frac{e^{-cx}}{x} \\ \\right) \\ dx$$\n\n$$\\displaystyle F_a =\\int_{0}^{\\infty} \\frac{e^{-cx}-e^{-ax}}{x}\\ dx = \\ln \\left( \\frac{a}{c}\\right)$$\n\n$$\\displaystyle F= a \\ln \\left( \\frac{a}{c}\\right) - a + C$$\n\n$$\\displaystyle F_b = C_b$$\n\n$$\\displaystyle F_b = \\int_{0}^{\\infty} \\frac{e^{-bx}-e^{-cx}}{x} dx = -\\ln \\left( \\frac{b}{c}\\right)$$\n\nWe have $$\\displaystyle C_b = -\\ln \\left( \\frac{b}{c}\\right)$$ so\n\n$$\\displaystyle C = -b\\ln \\left( \\frac{b}{c}\\right)+b$$\n\n$$\\displaystyle F = b-a +a \\ln \\left( \\frac{a}{c}\\right)-b\\ln \\left( \\frac{b}{c}\\right)$$\n\n#### Random Variable\n\n##### Well-known member\nMHB Math Helper\n@ Zaid\n\nI actually didn't even consider differentiating inside of the integral.\n\nI took the boring approach of finding an antiderivative in terms of the exponential integral and then using the expansion of the exponential integral at $x=0$ to evaluate the antiderivative at the lower limit.\n\nWhen you integrated to find the constant $C$, how did you know that he constant of integration was zero?\n\nAlternatively what you could do to find $C$ is to let $a=b$ in the original integral.\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\n\nWhen you integrated to find the constant $C$, how did you know that he constant of integration was zero?\n\nAlternatively what you could do to find $C$ is to let $a=b$ in the original integral.\nYeah, I made the evaluations not clear but I actually considered that $C$ as a function of $b$ so I differentiated with respect to $b$. Again putting $a=b$ is by far much clearer.\n\nCan you post your approach ?\n\n#### Random Variable\n\n##### Well-known member\nMHB Math Helper\nIt's easier to work the upper incomplete gamma function.\n\n$$\\Gamma(0,a x) = -\\text{Ei}(- ax) = \\int_{ax}^{\\infty} \\frac{e^{-t}}{t} \\ dt = \\int_{x}^{\\infty} \\frac{e^{-au}}{u} \\ du$$\n\nLet's first derive the expansion at $x=0$.\n\n$$\\Gamma(0,x) = \\int_{x}^{\\infty} \\frac{e^{-t}}{t} \\ dt = \\int_{1}^{\\infty} \\frac{e^{-t}}{t} \\ dt - \\int_{1}^{x} \\frac{e^{-t}}{t} \\ dt$$\n\n$$= \\int_{1}^{x} \\frac{1-e^{-t}}{t} \\ dt - \\ln x + \\int_{1}^{\\infty} \\frac{e^{-t}}{t} \\ dt$$\n\n$$= -\\ln x + \\Big( -\\int_{0}^{1} \\frac{1-e^{-t}}{t} \\ dt + \\int_{1}^{\\infty} \\frac{e^{-t}}{t} \\ dt \\Big) + \\int_{0}^{x} \\frac{1-e^{-t}}{t} \\ dt$$\n\n$$= - \\ln x - \\gamma - \\int_{0}^{x} \\frac{e^{-t}-1}{t} \\ dt$$\n\n$$= - \\ln x - \\gamma \\int_{0}^{x} \\Big(-1 + \\frac{t}{2} - \\frac{t^{2}}{6} + \\ldots \\Big) \\ dt$$\n\n$$= - \\ln x - \\gamma + x + \\mathcal{O}(x^{2})$$\n\nAnd it's not really needed here, but one can find an asymptotic expansion at $x= \\infty$ by integrating by parts.\n\n$$\\Gamma(0,x) \\sim \\frac{e^{-x}}{x} + \\mathcal{O}\\left( \\frac{e^{-x}}{x^{2}} \\right) \\ \\ x \\to \\infty$$\n\nThen\n\n$$\\int \\left( \\frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \\frac{e^{-cx}}{x} \\ \\right) \\ dx = \\int \\frac{e^{-ax}-e^{-bx}}{x^{2}} \\ dx + (a-b) \\int \\frac{e^{-cx}}{x} \\ dx$$\n\n$$= -\\frac{e^{-ax} -e^{-bx}}{x} + \\int \\frac{-ae^{-ax}+be^{-bx}}{x} \\ dx -(a-b) \\Gamma(0,cx)$$\n\n$$= \\frac{e^{-bx} -e^{-ax}}{x} + a \\Gamma(0,ax) - b \\Gamma(0,bx) - (a-b) \\Gamma(0,cx)+C$$\n\n$$\\int^{\\infty}_{0} \\left( \\frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \\frac{e^{-cx}}{x} \\ \\right) \\ dx$$\n\n$$= 0 - \\lim_{x \\to 0} \\left[ \\frac{e^{-bx} -e^{-ax}}{x} + a \\left(-\\ln ax - \\gamma \\right) - b \\left(-\\ln bx - \\gamma \\right) -(a-b) \\left(-\\ln cx - \\gamma \\right) \\right]$$\n\n$$= - \\lim_{x \\to 0} \\left[ \\frac{e^{-bx} -e^{-ax}}{x} - a \\ln \\left(\\frac{a}{c} \\right) + b \\left(\\frac{b}{c} \\right) \\right]$$\n\n$$= b-a + a \\ln \\left(\\frac{a}{c} \\right) - b \\ln \\left(\\frac{b}{c} \\right)$$\n\nLast edited by a moderator:", "date": "2021-07-29 09:46:46", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/integral-4.8956/", "openwebmath_score": 0.9460647702217102, "openwebmath_perplexity": 911.4797865864381, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308801770118, "lm_q2_score": 0.8840392725805823, "lm_q1q2_score": 0.8735465045260961}}
{"url": "https://www.themathdoctors.org/why-is-a-linear-equation-called-linear/", "text": "# Why is a Linear Equation Called Linear?\n\n#### (New Question of the Week)\n\nFrom time to time we get a question that is more about words than about math; usually these are about the meaning or origin of mathematical terms. Fortunately, some of us love words as much as we love math. But the question I want to look at here, which came in last month, is about both\u00a0the word and the math; in explaining why the word is appropriate, we are learning some things about math itself.\n\n## But it\u2019s not a line \u2026\n\nHere is Christine\u2019s question:\n\nWhy is an equation like 2x + 4 = 10 called a linear equation in one variable? Clearly, the solution is a point on one axis, the x-axis, not a line on the two-axis Cartesian coordinate plane? Or are all linear equations in one variable viewed as vertical lines?\n\nClearly, she is saying, the phrase \u201clinear equation\u201d means \u201cthe equation of a line\u201d. And there is a similarity between the linear equation in one variable above, and a linear equation in two variables, such as $$y = 2x + 4$$, which definitely is the equation of a line. But with only one variable, the only way to say that\u00a0$$2x + 4 = 10$$ is the equation of a line is to plot it on a plane as the vertical line\u00a0$$x = 3$$. Is that the intent of the term?\n\nNo, it goes further than that, because you also have to consider three or more variables. I initially gave just a short answer, to see what response it would trigger before digging in deeper:\n\nThe term \u201clinear\u201d, though derived from the idea that a linear equation in two variables represents a line, has been generalized from that to mean that the equation involves a polynomial with degree 1. That is, the variable(s) are only multiplied by constants and added to other terms, with nothing more (squaring, etc.).\n\nSo you could say that the term has been taken from one situation that gave it its name, and applied to more general cases with different numbers of variables. A linear equation in three variables, for example, represents a plane.\n\nFrom my perspective, \u201clinear\u201d means far more than \u201chaving a graph that is a line\u201d. My first thought when I see the word (outside of an elementary algebra class) is \u201cfirst-degree polynomial\u201d. Although one initially connects it to straight lines, when we extend its use (and almost every idea in math is an extension of something simpler), the idea we carry forward is the degree, not the number of dimensions. For example, here is the beginning of the Wikipedia article about linear equations:\n\nA linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable (however, different variables may occur in different terms). A simple example of a linear equation with only one variable, x, may be written in the form: ax + b = 0, where a and b are constants and a \u2260 0.\n\nLinear equations can have one or more variables. An example of a linear equation with three variables, x, y, and z, is given by: ax + by + cz + d = 0, where a, b, c, and d are constants and a, b, and c are non-zero.\n\nBut why would we call any polynomial equation with degree 1 \u201clinear\u201d, when it is only in two dimensions that it is related to a line? In the one-variable case, as Christine said,\u00a0the graph is really a point; and in this three-variable case it is a plane:\n\n## But, still \u2013 it\u2019s not a line!\n\nChristine wasn\u2019t quite convinced:\n\nThank you so much. I am very particular with terminology when I teach mathematics. I have to say, I do not like this generalization of the term. I think it is misleading.\n\nHmmm \u2026 is the term \u201clinear\u201d misleading? Not to mathematicians, and I would hope not to students once they get used to it. Yet it\u2019s true that it doesn\u2019t quite mean what it seems to say. And is generalization bad? I think it\u2019s the essence of what math is \u2013 and also an integral part of languages,which are always extending the meaning of words to cover new needs.\n\nHere is my response:\n\nHi, Christine.\n\nThanks for writing back with additional thoughts. Let\u2019s think a little more deeply about it.\n\nFirst, this is standard terminology that has been in use for 200 years by many great mathematicians, so we should be very careful about considering it a bad idea. I don\u2019t think you\u2019re likely to convince anyone to change; the word is used not only of a linear equation in itself, but of \u201csystems of linear equations\u201d (in contrast to \u201cnon-linear equations\u201d); of the whole major field of \u201clinear algebra\u201d; and for related concepts like \u201clinear combination\u201d, \u201clinear transformation\u201d, and \u201clinear independence\u201d, all of which apply to any number of variables or dimensions. So the term is very well established with a particular but broad meaning, and (at least after the first year) no one is misled by it. We know what it means, and what it means is more than just \u201cline\u201d.\n\nSecond, what would you replace it with? If you don\u2019t want to use the word \u201clinear\u201d except in situations that involve actual lines, what word would you use instead to describe the general class of equations involving variables multiplied only by constants, regardless of the number of variables? We need a word for this bigger idea; that word will either be a familiar word whose meaning is stretched to cover a bigger concept, or some made-up word. The tradition in math has always been to take familiar words and give them new meanings (either more specific, like \u201cgroup\u201d or \u201ccombination\u201d or \u201cfunction\u201d, or more general, like \u201cnumber\u201d or \u201cmultiplication\u201d or \u201cspace\u201d). So what we observe here is found throughout mathematics: a word that has grown beyond its humble beginnings. (This is also true of the entire English language!\u00a0Most\u00a0words would be \u201cmisleading\u201d if you thought too deeply about their origins.)\n\nI could say that, in a sense, all of mathematics is about generalization (or abstraction). I just mentioned \u201cnumber\u201d; some people do complain about calling anything other than a natural number a \u201cnumber\u201d, but the logical development from natural numbers, to integers, to rational numbers, to real and complex numbers, involves repeated broadening of the term, which has been extremely useful. We invent new concepts, and give them old names because they are a larger, more powerful version of the old concept.\n\nI mentioned how common it is in English for a word to grow beyond its original meaning. Sitting here at my computer, I look at the mouse \u2013 is it misleading to call it that when it doesn\u2019t have legs, and may not even have a tail? And the computer has a screen; at one time, a screen was a flat surface that hid something that wasn\u2019t to be seen (or that kept out bugs); then it was applied to flat surfaces on which pictures were projected; and then to a surface that shows a picture itself. Is that misleading? It would be if you went back a hundred years \u2026\n\nAnd thinking again about math terms, I\u2019m reminded of this discussion where I pondered what all the different operations that are called \u201cmultiplication\u201d have in common.\n\n## So how do we explain it?\n\nBut as I wrote this, I realized that I had gone off in a different direction than Christine, and I wanted to relate my answer to her specific context, linear equations in one variable. The real question was pedagogical: How could she explain this to her students, so that (eventually) \u201clinear\u201d would mean to them what they will need it to mean? I continued:\n\nNow, I\u2019ve been mostly thinking of the \u201cenlarging\u201d development of the word \u201clinear\u201d, taking it to\u00a0more\u00a0than two dimensions; you\u2019re thinking specifically of the term used with\u00a0only one variable. So it may help if we focus on that, to fit your particular context. I\u2019d like to explain linear equations in one variable in a way that should make it clear why we use the word, and that it is not a misnomer.\n\nConsider the equation you asked about,\u00a02x+4=10. The left-hand side is an expression; we call it a\u00a0linear\u00a0expression, because if you used it in an equation with two variables, y = 2x+4, its graph would be a straight line. So we call 2x+4 a linear expression (or, later, a linear function). A linear equation in one variable is one that says that two linear expressions are equal (or one is equal to a constant).\n\nOr, to look at it another way, one way to solve this equation would be to graph the related equation\u00a0y = 2x+4, and find where that line intersects the line y = 10. So the linear equation can be thought of very much in terms of a line.\n\n(If this were a student\u2019s first exposure to linear equations in one variable, she likely wouldn\u2019t have seen graphs of lines yet, and wouldn\u2019t be ready for this discussion; either the word linear would be used without explanation yet, or we would hold off on the word until later.)\n\nDoes this help?\n\nIt\u2019s often hard to be sure what kind of answer will help; but Christine gave the answer I hoped for:\n\nYes, your response does help very much. In addition, it gives me a lot to think about. It is true, the grade level I am currently teaching does learn to solve linear equations in one variable before they learn about linear equations and functions. Perhaps this is why I question the use of the term. I must remember that I have had experience in mathematics beyond their years, and therefore, am more thoughtful about what terms they are exposed to and more importantly in what sequence the mathematics is presented to them. I thank you again for such an in depth response and will certainly give this discussion much more thought. It is a pleasure speaking with you.\n\nI imagine if I were introducing these equations to students with no exposure to graphs of lines, I might just mention in passing that these are called \u201clinear equations in one variable\u201d, and that we would soon be seeing why the word \u201clinear\u201d is appropriate. For now, what it means is that all we do with the variable is to multiply it by a constant, and to add things. Kids are used to hearing things they\u2019ll understand later \u2026\n\n### 3 thoughts on \u201cWhy is a Linear Equation Called Linear?\u201d\n\n1. Very interesting article, Dr.Peterson!\n\nJust wanted to ask, so then why are polynomials of the second degree quadratics? \u2018Quad\u2019 reminds me of 4 (quadrilateral is a shape with 4 sides, a quadrant is 1/4 of a circle)\n\nAnd what\u2019s the difference between an identity and an expression?\n\n1. Dave Peterson\n\nHi, Sarah.\nFirst, the \u201cquad\u201d issue is another common question; for an answer, start here: Names of Polynomials\nSecond, an identity is not an expression, but an equation (that is, a statement that two expressions are equal). In particular, it is an equation that is true for any value of the variable(s) it contains. For instance, a+b = b+a is an identity.", "date": "2018-04-26 09:25:37", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/why-is-a-linear-equation-called-linear/", "openwebmath_score": 0.6129183173179626, "openwebmath_perplexity": 391.00202638539787, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754452025766, "lm_q2_score": 0.8872045981907006, "lm_q1q2_score": 0.8735198622493822}}
{"url": "https://math.stackexchange.com/questions/528709/how-can-i-prove-by-induction-that-9k-5k-is-divisible-by-4", "text": "# How can I prove by induction that $9^k - 5^k$ is divisible by 4?\n\nRecently had this on a discrete math test, which sadly I think I failed. But the question asked:\n\nProve that $9^k - 5^k$ is divisible by $4$.\n\nUsing the only approach I learned in the class, I substituted $n = k$, and tried to prove for $k+1$ like this:\n\n$$9^{k+1} - 5^{k+1},$$\n\nwhich just factors to $9 \\cdot 9^k - 5 \\cdot 5^k$.\n\nBut I cannot factor out $9^k - 5^k$, so I'm totally stuck.\n\n\u2022 hint: 9 = (5+4) and distribute. \u2013\u00a0imranfat Oct 16 '13 at 16:56\n\u2022 Alternatively you can note that $a^n-b^n=(a-b)\\sum _{i=0} ^{n-1} a^i b^{n-1-i}$. \u2013\u00a0pppqqq Oct 16 '13 at 16:58\n\n\\begin{align} 9\\cdot 9^k - 5\\cdot 5^k & = (4 + 5)\\cdot 9^k - 5\\cdot 5^k \\\\ \\\\ & = 4\\cdot 9^k + 5 \\cdot 9^k - 5\\cdot 5^k \\\\ \\\\ & = 4\\cdot 9^k + 5(9^k - 5^k)\\\\ \\\\ & \\quad \\text{ use inductive hypothesis}\\quad\\cdots\\end{align}\n\n\u2022 And don't forget to include verification of the base case in your proof: $n = 0$ or $n = 1$. \u2013\u00a0Namaste Oct 16 '13 at 17:10\n\u2022 Okay this seems to make the most sense and is closest to how my prof. teaches it. Dam I really hate theoretical math and proofs! \u2013\u00a0krb686 Oct 16 '13 at 17:11\n\u2022 @amWhy: Nice to get a nice answer +1 \u2013\u00a0Amzoti Oct 18 '13 at 0:57\n\n${\\color{White}{\\text{Proof without words.}}}$\n\n\u2022 Very nice indeed! \u2013\u00a0Brian M. Scott Oct 17 '13 at 0:02\n\n$9^{k+1}-5^{k+1}=(8+1)9^k-(4+1)5^k=9^k-5^k+4(2\\cdot 9^k-5^k)$ The secret to induction proofs is usually to find a way to relate the $k+1$ case to the $k$ case.\n\nAlternately, just note $9\\equiv 1 \\pmod 4, 5 \\equiv 1 \\pmod 4$, so $9^k-5^k \\equiv 1^k-1^k \\pmod 4$\n\nNote that induction is not actually needed: you can instead use the identity\n\n$$x^n-y^n=(x-y)\\sum_{k=0}^{n-1}x^ky^{n-1-k}\\;.$$\n\nIn this case it becomes\n\n$$9^n-5^n=4\\sum_{k=0}^{n-1}x^ky^{n-1-k}\\;,$$\n\nand since the summation clearly yields an integer, we have $4\\mid 9^n-5^n$. The identity is a standard one, but it\u2019s also not hard to prove:\n\n\\begin{align*} (x-y)\\sum_{k=0}^{n-1}x^ky^{n-1-k}&=x\\sum_{k=0}^{n-1}x^ky^{n-1-k}-y\\sum_{k=0}^{n-1}x^ky^{n-1-k}\\\\\\\\ &=\\sum_{k=0}^{n-1}x^{k+1}y^{n-1-k}-\\sum_{k=0}^{n-1}x^ky^{n-k}\\\\\\\\ &=\\sum_{k=1}^nx^ky^{n-k}-\\sum_{k=0}^{n-1}x^ky^{n-k}\\\\\\\\ &=x^ny^0+\\sum_{k=1}^{n-1}x^ky^{n-k}-\\sum_{k=1}^{n-1}x^ky^{n-k}-x^0y^n\\\\\\\\ &=x^n-y^n\\;. \\end{align*}\n\n\u2022 My test asked to use induction so that's why I narrowed that down, but thank you for all the in depth info! \u2013\u00a0krb686 Oct 17 '13 at 3:49\n\u2022 @krb686: You\u2019re welcome! \u2013\u00a0Brian M. Scott Oct 17 '13 at 3:50\n\nIt sounds like you understand the basic idea, but you're having trouble with the \"trick\" in the inductive step. This is common for proofs by mathematical induction, and you'll get better at it with more experience.\n\nAs you said, you're assuming that $9^k-5^k$ is divisible by four, and you want to prove that $9^{k+1}-5^{k+1}$ is divisible by four. You noticed that you can write the last expression as $9\\cdot 9^k-5\\cdot 5^k$, and you said you're having trouble applying your assumption to it. What happens if you add $9\\cdot 5^k-9\\cdot 5^k$ to your expression? This is just zero, so you're not changing anything, but you can write $$9^{k+1}-5^{k+1} =9\\cdot 9^k-5\\cdot 5^k+9\\cdot 5^k-9\\cdot 5^k =9\\cdot (9^k-5^k)+(9-5)\\cdot 5^k.$$ By the assumption, the first term, $9\\cdot (9^k-5^k)$, is divisible by four, and by observation, the second term, $(9-5)\\cdot 5^k$, is divisible by four, so the sum of the two terms and therefore $9^{k+1}-5^{k+1}$ are divisible by four. This is what you wanted to show.\n\nYou can always add zero (or multiply by one), and by writing zero (or one) in a particular way, you can sometimes make an argument easier to prove. Also, since your proof is by mathematical induction, you need to show that $9^n-5^n$ is divisible by four for, say, $n=0$ or $n=1$. You may have already done so and not asked about it in your question, but it's worth mentioning.", "date": "2019-11-21 21:25:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/528709/how-can-i-prove-by-induction-that-9k-5k-is-divisible-by-4", "openwebmath_score": 0.9883348941802979, "openwebmath_perplexity": 271.26283698279184, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232909876816, "lm_q2_score": 0.8887588030684331, "lm_q1q2_score": 0.8734928517259902}}
{"url": "http://dlmf.nist.gov/1.9", "text": "# \u00a71.9 Calculus of a Complex Variable\n\n## \u00a71.9(i) Complex Numbers\n\n 1.9.1 $z=x+iy,$ $x,y\\in\\mathbb{R}$. \u24d8 Symbols: $\\in$: element of, $\\mathbb{R}$: real line and $z$: variable A&S Ref: 3.7.1 Permalink: http://dlmf.nist.gov/1.9.E1 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9 and 1\n\n### Real and Imaginary Parts\n\n 1.9.2 $\\displaystyle\\Re z$ $\\displaystyle=x,$ $\\displaystyle\\Im z$ $\\displaystyle=y.$ \u24d8 Defines: $\\Im$: imaginary part and $\\Re$: real part Symbols: $z$: variable A&S Ref: 3.7.5 3.7.6 Permalink: http://dlmf.nist.gov/1.9.E2 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\n### Polar Representation\n\n 1.9.3 $\\displaystyle x$ $\\displaystyle=r\\cos\\theta,$ $\\displaystyle y$ $\\displaystyle=r\\sin\\theta,$ \u24d8 Symbols: $\\cos\\NVar{z}$: cosine function, $\\sin\\NVar{z}$: sine function, $r$: radius and $\\theta$: angle A&S Ref: 3.7.2 Referenced by: \u00a71.9(ii) Permalink: http://dlmf.nist.gov/1.9.E3 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\nwhere\n\n 1.9.4 $r=(x^{2}+y^{2})^{1/2},$ \u24d8 Symbols: $r$: radius A&S Ref: 3.7.3 Permalink: http://dlmf.nist.gov/1.9.E4 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\nand when $z\\neq 0$,\n\n 1.9.5 $\\theta=\\omega,\\;\\;\\pi-\\omega,\\;\\;-\\pi+\\omega,\\mbox{ or }-\\omega,$ \u24d8 Symbols: $\\pi$: the ratio of the circumference of a circle to its diameter and $\\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E5 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\naccording as $z$ lies in the 1st, 2nd, 3rd, or 4th quadrants. Here\n\n 1.9.6 $\\omega=\\operatorname{arctan}\\left(|y/x|\\right)\\in\\left[0,\\tfrac{1}{2}\\pi\\right].$ \u24d8 Symbols: $\\pi$: the ratio of the circumference of a circle to its diameter, $[\\NVar{a},\\NVar{b}]$: closed interval, $\\in$: element of and $\\operatorname{arctan}\\NVar{z}$: arctangent function A&S Ref: 3.7.4 Permalink: http://dlmf.nist.gov/1.9.E6 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\n### Modulus and Phase\n\n 1.9.7 $\\displaystyle|z|$ $\\displaystyle=r,$ $\\displaystyle\\operatorname{ph}z$ $\\displaystyle=\\theta+2n\\pi,$ $n\\in\\mathbb{Z}$. \u24d8 Defines: $\\operatorname{ph}$: phase Symbols: $\\pi$: the ratio of the circumference of a circle to its diameter, $\\in$: element of, $\\mathbb{Z}$: set of all integers, $z$: variable, $n$: nonnegative integer, $r$: radius and $\\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E7 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\nThe principal value of $\\operatorname{ph}z$ corresponds to $n=0$, that is, $-\\pi\\leq\\operatorname{ph}z\\leq\\pi$. It is single-valued on $\\mathbb{C}\\setminus\\{0\\}$, except on the interval $(-\\infty,0)$ where it is discontinuous and two-valued. Unless indicated otherwise, these principal values are assumed throughout the DLMF. (However, if we require a principal value to be single-valued, then we can restrict $-\\pi<\\operatorname{ph}z\\leq\\pi$.)\n\n 1.9.8 $\\displaystyle|\\Re z|$ $\\displaystyle\\leq|z|,$ $\\displaystyle|\\Im z|$ $\\displaystyle\\leq|z|,$ \u24d8 Symbols: $\\Im$: imaginary part, $\\Re$: real part and $z$: variable Permalink: http://dlmf.nist.gov/1.9.E8 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n 1.9.9 $z=re^{i\\theta},$ \u24d8 Symbols: $\\mathrm{e}$: base of exponential function, $z$: variable, $r$: radius and $\\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E9 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\nwhere\n\n 1.9.10 $e^{i\\theta}=\\cos\\theta+i\\sin\\theta;$ \u24d8 Symbols: $\\cos\\NVar{z}$: cosine function, $\\mathrm{e}$: base of exponential function, $\\sin\\NVar{z}$: sine function and $\\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E10 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\nsee \u00a74.14.\n\n### Complex Conjugate\n\n 1.9.11 $\\displaystyle\\overline{z}$ $\\displaystyle=x-iy,$ \u24d8 Defines: $\\overline{\\NVar{z}}$: complex conjugate Symbols: $z$: variable A&S Ref: 3.7.7 Permalink: http://dlmf.nist.gov/1.9.E11 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1 1.9.12 $\\displaystyle|\\overline{z}|$ $\\displaystyle=|z|,$ \u24d8 Symbols: $\\overline{\\NVar{z}}$: complex conjugate and $z$: variable A&S Ref: 3.7.8 Permalink: http://dlmf.nist.gov/1.9.E12 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1 1.9.13 $\\displaystyle\\operatorname{ph}\\overline{z}$ $\\displaystyle=-\\operatorname{ph}z.$ \u24d8 Symbols: $\\overline{\\NVar{z}}$: complex conjugate, $\\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.9 Permalink: http://dlmf.nist.gov/1.9.E13 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\n### Arithmetic Operations\n\nIf $z_{1}=x_{1}+iy_{1}$, $z_{2}=x_{2}+iy_{2}$, then\n\n 1.9.14 $z_{1}\\pm z_{2}=x_{1}\\pm x_{2}+\\mathrm{i}(y_{1}\\pm y_{2}),$ \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E14 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n 1.9.15 $z_{1}z_{2}=x_{1}x_{2}-y_{1}y_{2}+i(x_{1}y_{2}+x_{2}y_{1}),$ \u24d8 Symbols: $z$: variable A&S Ref: 3.7.10 Permalink: http://dlmf.nist.gov/1.9.E15 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n 1.9.16 $\\frac{z_{1}}{z_{2}}=\\frac{z_{1}\\overline{z}_{2}}{|z_{2}|^{2}}=\\frac{x_{1}x_{2}% +y_{1}y_{2}+i(x_{2}y_{1}-x_{1}y_{2})}{x_{2}^{2}+y_{2}^{2}},$ \u24d8 Symbols: $\\overline{\\NVar{z}}$: complex conjugate and $z$: variable A&S Ref: 3.7.13 Permalink: http://dlmf.nist.gov/1.9.E16 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\nprovided that $z_{2}\\neq 0$. Also,\n\n 1.9.17 $|z_{1}z_{2}|=|z_{1}|\\;|z_{2}|,$ \u24d8 Symbols: $z$: variable A&S Ref: 3.7.11 Permalink: http://dlmf.nist.gov/1.9.E17 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n 1.9.18 $\\operatorname{ph}\\left(z_{1}z_{2}\\right)=\\operatorname{ph}z_{1}+\\operatorname{% ph}z_{2},$ \u24d8 Symbols: $\\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.12 Referenced by: \u00a71.9(i) Permalink: http://dlmf.nist.gov/1.9.E18 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n 1.9.19 $\\left|\\frac{z_{1}}{z_{2}}\\right|=\\frac{|z_{1}|}{|z_{2}|},$ \u24d8 Symbols: $z$: variable A&S Ref: 3.7.14 Permalink: http://dlmf.nist.gov/1.9.E19 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n 1.9.20 $\\operatorname{ph}\\frac{z_{1}}{z_{2}}=\\operatorname{ph}z_{1}-\\operatorname{ph}z% _{2}.$ \u24d8 Symbols: $\\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.15 Referenced by: \u00a71.9(i) Permalink: http://dlmf.nist.gov/1.9.E20 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\nEquations (1.9.18) and (1.9.20) hold for general values of the phases, but not necessarily for the principal values.\n\n### Powers\n\n 1.9.21 $z^{n}=\\left(x^{n}-\\genfrac{(}{)}{0.0pt}{}{n}{2}x^{n-2}y^{2}+\\genfrac{(}{)}{0.0% pt}{}{n}{4}x^{n-4}y^{4}-\\cdots\\right)+i\\left(\\genfrac{(}{)}{0.0pt}{}{n}{1}x^{n% -1}y-\\genfrac{(}{)}{0.0pt}{}{n}{3}x^{n-3}y^{3}+\\cdots\\right),$ $n=1,2,\\dots$. \u24d8 Symbols: $\\genfrac{(}{)}{0.0pt}{}{\\NVar{m}}{\\NVar{n}}$: binomial coefficient, $z$: variable and $n$: nonnegative integer A&S Ref: 3.7.22 Permalink: http://dlmf.nist.gov/1.9.E21 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\n### DeMoivre\u2019s Theorem\n\n 1.9.22 $\\cos n\\theta+i\\sin n\\theta=(\\cos\\theta+i\\sin\\theta)^{n},$ $n\\in\\mathbb{Z}$.\n\n### Triangle Inequality\n\n 1.9.23 $\\left|\\left|z_{1}\\right|-\\left|z_{2}\\right|\\right|\\leq\\left|z_{1}+z_{2}\\right|% \\leq\\left|z_{1}\\right|+\\left|z_{2}\\right|.$ \u24d8 Symbols: $z$: variable A&S Ref: 3.7.29 Permalink: http://dlmf.nist.gov/1.9.E23 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1\n\n## \u00a71.9(ii) Continuity, Point Sets, and Differentiation\n\n### Continuity\n\nA function $f(z)$ is continuous at a point $z_{0}$ if $\\lim\\limits_{z\\to z_{0}}f(z)=f(z_{0})$. That is, given any positive number $\\epsilon$, however small, we can find a positive number $\\delta$ such that $|f(z)-f(z_{0})|<\\epsilon$ for all $z$ in the open disk $|z-z_{0}|<\\delta$.\n\nA function of two complex variables $f(z,w)$ is continuous at $(z_{0},w_{0})$ if $\\lim\\limits_{(z,w)\\to(z_{0},w_{0})}f(z,w)=f(z_{0},w_{0})$; compare (1.5.1) and (1.5.2).\n\n### Point Sets in $\\mathbb{C}$\n\nA neighborhood of a point $z_{0}$ is a disk $\\left|z-z_{0}\\right|<\\delta$. An open set in $\\mathbb{C}$ is one in which each point has a neighborhood that is contained in the set.\n\nA point $z_{0}$ is a limit point (limiting point or accumulation point) of a set of points $S$ in $\\mathbb{C}$ (or $\\mathbb{C}\\cup\\infty$) if every neighborhood of $z_{0}$ contains a point of $S$ distinct from $z_{0}$. ($z_{0}$ may or may not belong to $S$.) As a consequence, every neighborhood of a limit point of $S$ contains an infinite number of points of $S$. Also, the union of $S$ and its limit points is the closure of $S$.\n\nA domain $D$, say, is an open set in $\\mathbb{C}$ that is connected, that is, any two points can be joined by a polygonal arc (a finite chain of straight-line segments) lying in the set. Any point whose neighborhoods always contain members and nonmembers of $D$ is a boundary point of $D$. When its boundary points are added the domain is said to be closed, but unless specified otherwise a domain is assumed to be open.\n\nA region is an open domain together with none, some, or all of its boundary points. Points of a region that are not boundary points are called interior points.\n\nA function $f(z)$ is continuous on a region $R$ if for each point $z_{0}$ in $R$ and any given number $\\epsilon$ ($>0$) we can find a neighborhood of $z_{0}$ such that $\\left|f(z)-f(z_{0})\\right|<\\epsilon$ for all points $z$ in the intersection of the neighborhood with $R$.\n\n### Differentiation\n\nA function $f(z)$ is differentiable at a point $z$ if the following limit exists:\n\n 1.9.24 $f^{\\prime}(z)=\\frac{\\mathrm{d}f}{\\mathrm{d}z}=\\lim_{h\\to 0}\\frac{f(z+h)-f(z)}{% h}.$ \u24d8 Symbols: $\\frac{\\mathrm{d}\\NVar{f}}{\\mathrm{d}\\NVar{x}}$: derivative of $f$ with respect to $x$ and $z$: variable Permalink: http://dlmf.nist.gov/1.9.E24 Encodings: TeX, pMML, png See also: Annotations for 1.9(ii), 1.9(ii), 1.9 and 1\n\nDifferentiability automatically implies continuity.\n\n### Cauchy\u2013Riemann Equations\n\nIf $f^{\\prime}(z)$ exists at $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$, then\n\n 1.9.25 $\\displaystyle\\frac{\\partial u}{\\partial x}$ $\\displaystyle=\\frac{\\partial v}{\\partial y},$ $\\displaystyle\\frac{\\partial u}{\\partial y}$ $\\displaystyle=-\\frac{\\partial v}{\\partial x}$ \u24d8 Symbols: $\\frac{\\partial\\NVar{f}}{\\partial\\NVar{x}}$: partial derivative of $f$ with respect to $x$, $\\partial\\NVar{x}$: partial differential of $x$, $u(x,y)$: function and $v(x,y)$: function A&S Ref: 3.7.30 Referenced by: \u00a71.9(ii) Permalink: http://dlmf.nist.gov/1.9.E25 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(ii), 1.9(ii), 1.9 and 1\n\nat $(x,y)$.\n\nConversely, if at a given point $(x,y)$ the partial derivatives $\\ifrac{\\partial u}{\\partial x}$, $\\ifrac{\\partial u}{\\partial y}$, $\\ifrac{\\partial v}{\\partial x}$, and $\\ifrac{\\partial v}{\\partial y}$ exist, are continuous, and satisfy (1.9.25), then $f(z)$ is differentiable at $z=x+iy$.\n\n### Analyticity\n\nA function $f(z)$ is said to be analytic (holomorphic) at $z=z_{0}$ if it is differentiable in a neighborhood of $z_{0}$.\n\nA function $f(z)$ is analytic in a domain $D$ if it is analytic at each point of $D$. A function analytic at every point of $\\mathbb{C}$ is said to be entire.\n\nIf $f(z)$ is analytic in an open domain $D$, then each of its derivatives $f^{\\prime}(z)$, $f^{\\prime\\prime}(z)$, $\\dots$ exists and is analytic in $D$.\n\n### Harmonic Functions\n\nIf $f(z)=u(x,y)+iv(x,y)$ is analytic in an open domain $D$, then $u$ and $v$ are harmonic in $D$, that is,\n\n 1.9.26 $\\frac{{\\partial}^{2}u}{{\\partial x}^{2}}+\\frac{{\\partial}^{2}u}{{\\partial y}^{% 2}}=\\frac{{\\partial}^{2}v}{{\\partial x}^{2}}+\\frac{{\\partial}^{2}v}{{\\partial y% }^{2}}=0,$\n\nor in polar form ((1.9.3)) $u$ and $v$ satisfy\n\n 1.9.27 $\\frac{{\\partial}^{2}u}{{\\partial r}^{2}}+\\frac{1}{r}\\frac{\\partial u}{\\partial r% }+\\frac{1}{r^{2}}\\frac{{\\partial}^{2}u}{{\\partial\\theta}^{2}}=0$\n\nat all points of $D$.\n\n## \u00a71.9(iii) Integration\n\nAn arc $C$ is given by $z(t)=x(t)+iy(t)$, $a\\leq t\\leq b$, where $x$ and $y$ are continuously differentiable. If $x(t)$ and $y(t)$ are continuous and $x^{\\prime}(t)$ and $y^{\\prime}(t)$ are piecewise continuous, then $z(t)$ defines a contour.\n\nA contour is simple if it contains no multiple points, that is, for every pair of distinct values $t_{1},t_{2}$ of $t$, $z(t_{1})\\neq z(t_{2})$. A simple closed contour is a simple contour, except that $z(a)=z(b)$.\n\nNext,\n\n 1.9.28 $\\int_{C}f(z)\\mathrm{d}z=\\int_{a}^{b}f(z(t))(x^{\\prime}(t)+iy^{\\prime}(t))% \\mathrm{d}t,$ \u24d8 Symbols: $\\mathrm{d}\\NVar{x}$: differential of $x$, $\\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E28 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9 and 1\n\nfor a contour $C$ and $f(z(t))$ continuous, $a\\leq t\\leq b$. If $f(z(t_{0}))=\\infty$, $a\\leq t_{0}\\leq b$, then the integral is defined analogously to the infinite integrals in \u00a71.4(v). Similarly when $a=-\\infty$ or $b=+\\infty$.\n\n### Jordan Curve Theorem\n\nAny simple closed contour $C$ divides $\\mathbb{C}$ into two open domains that have $C$ as common boundary. One of these domains is bounded and is called the interior domain of $C$; the other is unbounded and is called the exterior domain of $C$.\n\n### Cauchy\u2019s Theorem\n\nIf $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, then\n\n 1.9.29 $\\int_{C}f(z)\\mathrm{d}z=0.$ \u24d8 Symbols: $\\mathrm{d}\\NVar{x}$: differential of $x$, $\\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E29 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1\n\n### Cauchy\u2019s Integral Formula\n\nIf $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, and if $z_{0}$ is a point within $C$, then\n\n 1.9.30 $f(z_{0})=\\frac{1}{2\\pi i}\\int_{C}\\frac{f(z)}{z-z_{0}}\\mathrm{d}z,$ \u24d8 Symbols: $\\pi$: the ratio of the circumference of a circle to its diameter, $\\mathrm{d}\\NVar{x}$: differential of $x$, $\\int$: integral and $C$: closed contour Referenced by: \u00a72.3(iii) Permalink: http://dlmf.nist.gov/1.9.E30 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1\n\nand\n\n 1.9.31 $f^{(n)}(z_{0})=\\frac{n!}{2\\pi i}\\int_{C}\\frac{f(z)}{(z-z_{0})^{n+1}}\\mathrm{d}z,$ $n=1,2,3,\\dots$,\n\nprovided that in both cases $C$ is described in the positive rotational (anticlockwise) sense.\n\n### Liouville\u2019s Theorem\n\nAny bounded entire function is a constant.\n\n### Winding Number\n\nIf $C$ is a closed contour, and $z_{0}\\not\\in C$, then\n\n 1.9.32 $\\frac{1}{2\\pi i}\\int_{C}\\frac{1}{z-z_{0}}\\mathrm{d}z=\\mathcal{N}(C,z_{0}),$ \u24d8 Defines: $\\mathcal{N}(C,z_{0})$: winding number of $C$ (locally) Symbols: $\\pi$: the ratio of the circumference of a circle to its diameter, $\\mathrm{d}\\NVar{x}$: differential of $x$, $\\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E32 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1\n\nwhere $\\mathcal{N}(C,z_{0})$ is an integer called the winding number of $C$ with respect to $z_{0}$. If $C$ is simple and oriented in the positive rotational sense, then $\\mathcal{N}(C,z_{0})$ is $1$ or $0$ depending whether $z_{0}$ is inside or outside $C$.\n\n### Mean Value Property\n\nFor $u(z)$ harmonic,\n\n 1.9.33 $u(z)=\\frac{1}{2\\pi}\\int^{2\\pi}_{0}u(z+re^{i\\phi})\\mathrm{d}\\phi.$\n\n### Poisson Integral\n\nIf $h(w)$ is continuous on $|w|=R$, then with $z=re^{i\\theta}$\n\n 1.9.34 $u(re^{i\\theta})=\\frac{1}{2\\pi}\\int^{2\\pi}_{0}\\frac{(R^{2}-r^{2})h(Re^{i\\phi})% \\mathrm{d}\\phi}{R^{2}-2Rr\\cos\\left(\\phi-\\theta\\right)+r^{2}}$\n\nis harmonic in $|z|. Also with $\\left|w\\right|=R$, $\\lim\\limits_{z\\to w}u(z)=h(w)$ as $z\\to w$ within $|z|.\n\n## \u00a71.9(iv) Conformal Mapping\n\nThe extended complex plane, $\\mathbb{C}\\,\\cup\\,\\{\\infty\\}$, consists of the points of the complex plane $\\mathbb{C}$ together with an ideal point $\\infty$ called the point at infinity. A system of open disks around infinity is given by\n\n 1.9.35 $S_{r}=\\{z\\mid|z|>1/r\\}\\cup\\{\\infty\\},$ $0. \u24d8 Symbols: $\\cup$: union, $z$: variable, $r$: radius and $S_{r}$: neighborhood Permalink: http://dlmf.nist.gov/1.9.E35 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1\n\nEach $S_{r}$ is a neighborhood of $\\infty$. Also,\n\n 1.9.36 $\\infty\\pm z=z\\pm\\infty=\\infty,$ \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E36 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1\n 1.9.37 $\\infty\\cdot z=z\\cdot\\infty=\\infty,$ $z\\not=0$, \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E37 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1\n 1.9.38 $z/\\infty=0,$ \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E38 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1\n 1.9.39 $z/0=\\infty,$ $z\\neq 0$. \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E39 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1\n\nA function $f(z)$ is analytic at $\\infty$ if $g(z)=f(1/z)$ is analytic at $z=0$, and we set $f^{\\prime}(\\infty)=g^{\\prime}(0)$.\n\n### Conformal Transformation\n\nSuppose $f(z)$ is analytic in a domain $D$ and $C_{1},C_{2}$ are two arcs in $D$ passing through $z_{0}$. Let $C^{\\prime}_{1},C^{\\prime}_{2}$ be the images of $C_{1}$ and $C_{2}$ under the mapping $w=f(z)$. The angle between $C_{1}$ and $C_{2}$ at $z_{0}$ is the angle between the tangents to the two arcs at $z_{0}$, that is, the difference of the signed angles that the tangents make with the positive direction of the real axis. If $f^{\\prime}(z_{0})\\not=0$, then the angle between $C_{1}$ and $C_{2}$ equals the angle between $C^{\\prime}_{1}$ and $C^{\\prime}_{2}$ both in magnitude and sense. We then say that the mapping $w=f(z)$ is conformal (angle-preserving) at $z_{0}$.\n\nThe linear transformation $f(z)=az+b$, $a\\not=0$, has $f^{\\prime}(z)=a$ and $w=f(z)$ maps $\\mathbb{C}$ conformally onto $\\mathbb{C}$.\n\n### Bilinear Transformation\n\n 1.9.40 $w=f(z)=\\frac{az+b}{cz+d},$ $ad-bc\\not=0$, $c\\not=0$. \u24d8 Symbols: $z$: variable and $w$: variable Referenced by: \u00a71.9(iv) Permalink: http://dlmf.nist.gov/1.9.E40 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1\n 1.9.41 $\\displaystyle f(-d/c)$ $\\displaystyle=\\infty,$ $\\displaystyle f(\\infty)$ $\\displaystyle=a/c.$ \u24d8 Permalink: http://dlmf.nist.gov/1.9.E41 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1\n 1.9.42 $f^{\\prime}(z)=\\frac{ad-bc}{(cz+d)^{2}},$ $z\\not=-d/c$. \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E42 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1\n 1.9.43 $f^{\\prime}(\\infty)=\\frac{bc-ad}{c^{2}}.$ \u24d8 Permalink: http://dlmf.nist.gov/1.9.E43 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1\n 1.9.44 $z=\\frac{dw-b}{-cw+a}.$ \u24d8 Symbols: $z$: variable and $w$: variable Permalink: http://dlmf.nist.gov/1.9.E44 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1\n\nThe transformation (1.9.40) is a one-to-one conformal mapping of $\\mathbb{C}\\,\\cup\\,\\{\\infty\\}$ onto itself.\n\nThe cross ratio of $z_{1},z_{2},z_{3},z_{4}\\in\\mathbb{C}\\cup\\{\\infty\\}$ is defined by\n\n 1.9.45 $\\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{1}-z_{4})(z_{3}-z_{2})},$ \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E45 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1\n\nor its limiting form, and is invariant under bilinear transformations.\n\nOther names for the bilinear transformation are fractional linear transformation, homographic transformation, and M\u00f6bius transformation.\n\n## \u00a71.9(v) Infinite Sequences and Series\n\nA sequence $\\{z_{n}\\}$ converges to $z$ if $\\lim\\limits_{n\\to\\infty}z_{n}=z$. For $z_{n}=x_{n}+iy_{n}$, the sequence $\\{z_{n}\\}$ converges iff the sequences $\\{x_{n}\\}$ and $\\{y_{n}\\}$ separately converge. A series $\\sum^{\\infty}_{n=0}z_{n}$ converges if the sequence $s_{n}=\\sum^{n}_{k=0}z_{k}$ converges. The series is divergent if $s_{n}$ does not converge. The series converges absolutely if $\\sum^{\\infty}_{n=0}|z_{n}|$ converges. A series $\\sum^{\\infty}_{n=0}z_{n}$ converges (diverges) absolutely when $\\lim\\limits_{n\\to\\infty}|z_{n}|^{1/n}<1$ ($>1$), or when $\\lim\\limits_{n\\to\\infty}\\left|\\ifrac{z_{n+1}}{z_{n}}\\right|<1$ ($>1$). Absolutely convergent series are also convergent.\n\nLet $\\{f_{n}(z)\\}$ be a sequence of functions defined on a set $S$. This sequence converges pointwise to a function $f(z)$ if\n\n 1.9.46 $f(z)=\\lim_{n\\to\\infty}f_{n}(z)$ \u24d8 Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E46 Encodings: TeX, pMML, png See also: Annotations for 1.9(v), 1.9 and 1\n\nfor each $z\\in S$. The sequence converges uniformly on $S$, if for every $\\epsilon>0$ there exists an integer $N$, independent of $z$, such that\n\n 1.9.47 $|f_{n}(z)-f(z)|<\\epsilon$ \u24d8 Symbols: $z$: variable, $n$: nonnegative integer and $\\epsilon$: positive number Permalink: http://dlmf.nist.gov/1.9.E47 Encodings: TeX, pMML, png See also: Annotations for 1.9(v), 1.9 and 1\n\nfor all $z\\in S$ and $n\\geq N$.\n\nA series $\\sum^{\\infty}_{n=0}f_{n}(z)$ converges uniformly on $S$, if the sequence $s_{n}(z)=\\sum^{n}_{k=0}f_{k}(z)$ converges uniformly on $S$.\n\n### Weierstrass $M$-test\n\nSuppose $\\{M_{n}\\}$ is a sequence of real numbers such that $\\sum^{\\infty}_{n=0}M_{n}$ converges and $|f_{n}(z)|\\leq M_{n}$ for all $z\\in S$ and all $n\\geq 0$. Then the series $\\sum^{\\infty}_{n=0}f_{n}(z)$ converges uniformly on $S$.\n\nA doubly-infinite series $\\sum^{\\infty}_{n=-\\infty}f_{n}(z)$ converges (uniformly) on $S$ iff each of the series $\\sum^{\\infty}_{n=0}f_{n}(z)$ and $\\sum^{\\infty}_{n=1}f_{-n}(z)$ converges (uniformly) on $S$.\n\n## \u00a71.9(vi) Power Series\n\nFor a series $\\sum^{\\infty}_{n=0}a_{n}(z-z_{0})^{n}$ there is a number $R$, $0\\leq R\\leq\\infty$, such that the series converges for all $z$ in $|z-z_{0}| and diverges for $z$ in $|z-z_{0}|>R$. The circle $|z-z_{0}|=R$ is called the circle of convergence of the series, and $R$ is the radius of convergence. Inside the circle the sum of the series is an analytic function $f(z)$. For $z$ in $|z-z_{0}|\\leq\\rho$ ($), the convergence is absolute and uniform. Moreover,\n\n 1.9.48 $a_{n}=\\frac{f^{(n)}(z_{0})}{n!},$ \u24d8 Symbols: $!$: factorial (as in $n!$), $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E48 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9 and 1\n\nand\n\n 1.9.49 $R=\\liminf_{n\\to\\infty}|a_{n}|^{-1/n}.$ \u24d8 Defines: $R$: radius of convergence (locally) Symbols: $\\liminf$: least limit point and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E49 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9 and 1\n\nFor the converse of this result see \u00a71.10(i).\n\n### Operations\n\nWhen $\\sum a_{n}z^{n}$ and $\\sum b_{n}z^{n}$ both converge\n\n 1.9.50 $\\sum^{\\infty}_{n=0}(a_{n}\\pm b_{n})z^{n}=\\sum^{\\infty}_{n=0}a_{n}z^{n}\\pm\\sum^% {\\infty}_{n=0}b_{n}z^{n},$ \u24d8 Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E50 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nand\n\n 1.9.51 $\\left(\\sum^{\\infty}_{n=0}a_{n}z^{n}\\right)\\left(\\sum^{\\infty}_{n=0}b_{n}z^{n}% \\right)=\\sum^{\\infty}_{n=0}c_{n}z^{n},$ \u24d8 Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E51 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nwhere\n\n 1.9.52 $c_{n}=\\sum^{n}_{k=0}a_{k}b_{n-k}.$ \u24d8 Symbols: $k$: integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E52 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nNext, let\n\n 1.9.53 $f(z)=a_{0}+a_{1}z+a_{2}z^{2}+\\cdots,$ $a_{0}\\not=0$. \u24d8 Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E53 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nThen the expansions (1.9.54), (1.9.57), and (1.9.60) hold for all sufficiently small $|z|$.\n\n 1.9.54 $\\frac{1}{f(z)}=b_{0}+b_{1}z+b_{2}z^{2}+\\cdots,$ \u24d8 Symbols: $z$: variable Referenced by: \u00a71.9(vi) Permalink: http://dlmf.nist.gov/1.9.E54 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nwhere\n\n 1.9.55 $\\displaystyle b_{0}$ $\\displaystyle=1/a_{0},$ $\\displaystyle b_{1}$ $\\displaystyle=-a_{1}/a_{0}^{2},$ $\\displaystyle b_{2}$ $\\displaystyle=(a_{1}^{2}-a_{0}a_{2})/a_{0}^{3},$ \u24d8 Permalink: http://dlmf.nist.gov/1.9.E55 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n 1.9.56 $b_{n}=-(a_{1}b_{n-1}+a_{2}b_{n-2}+\\dots+a_{n}b_{0})/a_{0},$ $n\\geq 1$. \u24d8 Symbols: $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E56 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nWith $a_{0}=1$,\n\n 1.9.57 $\\ln f(z)=q_{1}z+q_{2}z^{2}+q_{3}z^{3}+\\cdots,$ \u24d8 Symbols: $\\ln\\NVar{z}$: principal branch of logarithm function, $z$: variable and $q_{j}$: coefficients Referenced by: \u00a71.9(vi) Permalink: http://dlmf.nist.gov/1.9.E57 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\n(principal value), where\n\n 1.9.58 $\\displaystyle q_{1}$ $\\displaystyle=a_{1},$ $\\displaystyle q_{2}$ $\\displaystyle=(2a_{2}-a_{1}^{2})/2,$ $\\displaystyle q_{3}$ $\\displaystyle=(3a_{3}-3a_{1}a_{2}+a_{1}^{3})/3,$ \u24d8 Symbols: $q_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E58 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nand\n\n 1.9.59 $q_{n}=(na_{n}-(n-1)a_{1}q_{n-1}-(n-2)a_{2}q_{n-2}-\\cdots-a_{n-1}q_{1})/n,$ $n\\geq 2$. \u24d8 Symbols: $n$: nonnegative integer and $q_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E59 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nAlso,\n\n 1.9.60 $(f(z))^{\\nu}=p_{0}+p_{1}z+p_{2}z^{2}+\\cdots,$ \u24d8 Symbols: $z$: variable, $\\nu$: complex and $p_{j}$: coefficients Referenced by: \u00a71.9(vi) Permalink: http://dlmf.nist.gov/1.9.E60 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\n(principal value), where $\\nu\\in\\mathbb{C}$,\n\n 1.9.61 $\\displaystyle p_{0}$ $\\displaystyle=1,$ $\\displaystyle p_{1}$ $\\displaystyle=\\nu a_{1},$ $\\displaystyle p_{2}$ $\\displaystyle=\\nu((\\nu-1)a_{1}^{2}+2a_{2})/2,$ \u24d8 Symbols: $\\nu$: complex and $p_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E61 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nand\n\n 1.9.62 $p_{n}=((\\nu-n+1)a_{1}p_{n-1}+(2\\nu-n+2)a_{2}p_{n-2}+\\dots+((n-1)\\nu-1)a_{n-1}p% _{1}+n\\nu a_{n})/n,$ $n\\geq 1$. \u24d8 Symbols: $n$: nonnegative integer, $\\nu$: complex and $p_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E62 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1\n\nFor the definitions of the principal values of $\\ln f(z)$ and $(f(z))^{\\nu}$ see \u00a7\u00a74.2(i) and 4.2(iv).\n\nLastly, a power series can be differentiated any number of times within its circle of convergence:\n\n 1.9.63 $f^{(m)}(z)=\\sum_{n=0}^{\\infty}{\\left(n+1\\right)_{m}}a_{n+m}(z-z_{0})^{n},$ $\\left|z-z_{0}\\right|, $m=0,1,2,\\dots$.\n\n## \u00a71.9(vii) Inversion of Limits\n\n### Double Sequences and Series\n\nA set of complex numbers $\\{z_{m,n}\\}$ where $m$ and $n$ take all positive integer values is called a double sequence. It converges to $z$ if for every $\\epsilon>0$, there is an integer $N$ such that\n\n 1.9.64 $|z_{m,n}-z|<\\epsilon$ \u24d8 Symbols: $z$: variable, $m$: nonnegative integer, $n$: nonnegative integer and $\\epsilon$: positive number Permalink: http://dlmf.nist.gov/1.9.E64 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1\n\nfor all $m,n\\geq N$. Suppose $\\{z_{m,n}\\}$ converges to $z$ and the repeated limits\n\n 1.9.65 $\\lim_{m\\to\\infty}\\left(\\lim_{n\\to\\infty}z_{m,n}\\right),$ $\\lim_{n\\to\\infty}\\left(\\lim_{m\\to\\infty}z_{m,n}\\right)$ \u24d8 Symbols: $z$: variable, $m$: nonnegative integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E65 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1\n\nexist. Then both repeated limits equal $z$.\n\nA double series is the limit of the double sequence\n\n 1.9.66 $z_{p,q}=\\sum^{p}_{m=0}\\sum^{q}_{n=0}\\zeta_{m,n}.$ \u24d8 Symbols: $z$: variable, $m$: nonnegative integer, $n$: nonnegative integer and $\\zeta_{p,q}$: sum Permalink: http://dlmf.nist.gov/1.9.E66 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1\n\nIf the limit exists, then the double series is convergent; otherwise it is divergent. The double series is absolutely convergent if it is convergent when $\\zeta_{m,n}$ is replaced by $|\\zeta_{m,n}|$.\n\nIf a double series is absolutely convergent, then it is also convergent and its sum is given by either of the repeated sums\n\n 1.9.67 $\\sum^{\\infty}_{m=0}\\left(\\sum^{\\infty}_{n=0}\\zeta_{m,n}\\right),$ $\\sum^{\\infty}_{n=0}\\left(\\sum^{\\infty}_{m=0}\\zeta_{m,n}\\right).$ \u24d8 Symbols: $m$: nonnegative integer, $n$: nonnegative integer and $\\zeta_{p,q}$: sum Permalink: http://dlmf.nist.gov/1.9.E67 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1\n\n### Term-by-Term Integration\n\nSuppose the series $\\sum^{\\infty}_{n=0}f_{n}(z)$, where $f_{n}(z)$ is continuous, converges uniformly on every compact set of a domain $D$, that is, every closed and bounded set in $D$. Then\n\n 1.9.68 $\\int_{C}\\sum^{\\infty}_{n=0}f_{n}(z)\\mathrm{d}z=\\sum^{\\infty}_{n=0}\\int_{C}f_{n% }(z)\\mathrm{d}z$\n\nfor any finite contour $C$ in $D$.\n\n### Dominated Convergence Theorem\n\nLet $(a,b)$ be a finite or infinite interval, and $f_{0}(t),f_{1}(t),\\dots$ be real or complex continuous functions, $t\\in(a,b)$. Suppose $\\sum^{\\infty}_{n=0}f_{n}(t)$ converges uniformly in any compact interval in $(a,b)$, and at least one of the following two conditions is satisfied:\n\n 1.9.69 $\\int^{b}_{a}\\sum^{\\infty}_{n=0}|f_{n}(t)|\\mathrm{d}t<\\infty,$ \u24d8 Symbols: $\\mathrm{d}\\NVar{x}$: differential of $x$, $\\int$: integral and $n$: nonnegative integer Referenced by: \u00a71.9(vii) Permalink: http://dlmf.nist.gov/1.9.E69 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1\n 1.9.70 $\\sum^{\\infty}_{n=0}\\int^{b}_{a}|f_{n}(t)|\\mathrm{d}t<\\infty.$ \u24d8 Symbols: $\\mathrm{d}\\NVar{x}$: differential of $x$, $\\int$: integral and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E70 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1\n\nThen\n\n 1.9.71 $\\int^{b}_{a}\\sum^{\\infty}_{n=0}f_{n}(t)\\mathrm{d}t=\\sum^{\\infty}_{n=0}\\int^{b}% _{a}f_{n}(t)\\mathrm{d}t.$ \u24d8 Symbols: $\\mathrm{d}\\NVar{x}$: differential of $x$, $\\int$: integral and $n$: nonnegative integer Referenced by: \u00a71.9(vii) Permalink: http://dlmf.nist.gov/1.9.E71 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1", "date": "2017-09-24 08:47:52", "meta": {"domain": "nist.gov", "url": "http://dlmf.nist.gov/1.9", "openwebmath_score": 0.9758047461509705, "openwebmath_perplexity": 3098.455388470953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303425461246, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8734538849761817}}
{"url": "https://math.stackexchange.com/questions/2033639/mod-of-numbers-with-large-exponents", "text": "# Mod of numbers with large exponents\n\nI've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:\n\n\u2022 $13^{100} \\bmod 7$\n\u2022 $7^{100} \\bmod 13$\n\nI've also heard of this formula:\n\n$$a \\equiv b\\pmod n \\Rightarrow a^k \\equiv b^k \\pmod n$$\n\nBut I don't see how exactly to use that here, because from $13^1 \\bmod 7$ I get 6, and $13^2 \\bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)\n\nAny hints or pointers in the right direction would be great.\n\n\u2022 Might be useful to recognize that $6=-1 \\mod 7$ \u2013\u00a0Kitter Catter Nov 28 '16 at 0:26\n\u2022 @KitterCatter Can you explain it a bit more? (possibly as an answer). Is it derived from $n | (a-b)$? \u2013\u00a0Roshnal Nov 28 '16 at 0:29\n\u2022 I don't know about using fermat (not a mathematician by trade) but it might be helpful to know that the Euler totient function of a prime,$p$ is $p-1$ and that if $a^{\\phi(p)} \\equiv 1 \\mod p$ \u2013\u00a0Kitter Catter Nov 28 '16 at 0:31\n\u2022 Be sure to understand the relation between \"mod\" as a binary operator vs. ternary relation. See this answer and this one for more on this. If you only know the operator form you will be severely encumbered. \u2013\u00a0Bill Dubuque Nov 28 '16 at 0:32\n\nThe formula you've heard of results from the fact that congruences are compatible with addition and multiplication.\n\nThe first power $$13^{100}$$ is easy: $$13\\equiv -1\\mod 7$$, so $$13^{100}\\equiv (-1)^{100}=1\\pmod 7.$$\n\nThe second power uses Lil' Fermat: for any number $$a\\not\\equiv 0\\mod 7$$, we have $$a^{12}\\equiv 1\\pmod{13}$$, hence $$7^{100}\\equiv 7^{100\\bmod12}\\equiv 7^4\\equiv 10^2\\equiv 9\\pmod{13}$$\n\n\u2022 Very nice answer! When you reference Lil' Fermat did you mean $a\\not\\equiv0\\mod13$? also it looks like you missed a bracket for the exponent in $a^{12} \\equiv 1 \\mod 13$ \u2013\u00a0Kitter Catter Nov 28 '16 at 0:47\n\u2022 Thanks for the answer! I'm still a little unclear on how you got to $10^2$ from $7^4$? \u2013\u00a0Roshnal Nov 28 '16 at 1:03\n\u2022 $7^4=(7^2)^2=49^2\\equiv 10^2$, that's all. \u2013\u00a0Bernard Nov 28 '16 at 1:12\n\u2022 Ah okay, thanks! It's clear now :) \u2013\u00a0Roshnal Nov 28 '16 at 1:15\n\u2022 @Kitter Catter: Oh! Yes. I should have re-read my answer before posting. I even missed a pair of brackets. It's fixed now. Thanks for pointing the typos! \u2013\u00a0Bernard Nov 28 '16 at 1:15\n\nHint $$\\,$$ The key idea is to use modular order reduction on exponents as in the Lemma below. We can find small exponents $$\\,e\\,$$ such that $$\\,a^{\\large e}\\equiv 1\\,$$ either by Euler's totient or Fermat's little theorem (or by Carmichael's lambda generalization), along with obvious roots of $$\\,1\\,$$ such as $$\\,(-1)^2\\equiv 1.$$\n\nTheorem $$\\ \\$$ Suppose that: $$\\,\\ \\color{#c00}{a^{\\large e}\\equiv\\, 1}\\,\\pmod{\\! m}\\$$ and $$\\, e>0,\\ n,k\\ge 0\\,$$ are integers. Then\n\n$$\\qquad n\\equiv k\\pmod{\\! \\color{#c00}e}\\,\\Longrightarrow\\,a^{\\large n}\\equiv a^{\\large k}\\pmod{\\!m}.\\$$ Further,  conversely\n\n$$\\qquad n\\equiv k\\pmod{\\! \\color{#c00}e}\\,\\Longleftarrow\\,a^{\\large n}\\equiv a^{\\large k}\\pmod{\\!m}\\ \\,$$ if $$\\,a\\,$$ has order $$\\,\\color{#c00}e\\,$$ mod $$\\,m$$\n\nProof $$\\$$ Wlog $$\\,n\\ge k\\,$$ so $$\\,a^{\\large n-k} a^{\\large k}\\equiv a^{\\large k}\\!\\iff a^{\\large n-k}\\equiv 1\\iff n\\equiv k\\pmod{\\!e}\\,$$ by here, where we cancelled $$\\,a^{\\large k}\\,$$ using $$\\,a^{\\large e}\\equiv 1\\,\\Rightarrow\\, a\\,$$ is invertible so cancellable (cf. below Remark).\n\nCorollary $$\\ \\ \\bbox[7px,border:1px solid #c00]{\\!\\bmod m\\!:\\,\\ \\color{#c00}{a^{\\large e}\\equiv 1}\\,\\Rightarrow\\, a^{\\large n}\\equiv a^{\\large n\\bmod \\color{#c00}e}}\\,\\$$ by $$\\ n\\equiv n\\bmod e\\,\\pmod{\\!e}$$\n\nRemark  If you are familiar with modular inverses then it is not necessary to restrict to nonnegative powers of $$\\,a\\,$$ above since $$\\,a^{\\large e}\\equiv 1,\\ e> 0\\,\\Rightarrow\\,$$ $$a$$ is invertible by $$\\,a a^{\\large e-1}\\equiv 1\\,$$ so $$\\,a^{\\large -1}\\equiv a^{\\large e-1}$$.\n\n\u2022 Often it proves simpler to first reduce $\\,e\\,$ using Euler's Criterion or quadratic reciprocity, e.g. see here.. \u2013\u00a0Bill Dubuque Jun 8 at 13:31\n\nQuick answer: $13 = 2\\cdot 7-1$ so $13\\equiv-1\\mod 7$ and therefore $13^{100} \\equiv (-1)^{100} \\mod 7$\n\nOther one is fairly quick: \\begin{eqnarray} \\phi(13) = 12\\\\ \\gcd(7,13)=1\\\\ 7^{100}\\equiv7^{4} \\mod13\\\\ 7\\rightarrow10\\rightarrow5\\rightarrow9 \\end{eqnarray} Probably a nicer way to do that.\n\n\u2022 Thanks for the answer. But can you please explain how you arrived to step 3? (in your four-step part of the answer), and why you have calculated the gcd(7, 13)? \u2013\u00a0Roshnal Nov 28 '16 at 0:58", "date": "2019-07-19 02:30:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2033639/mod-of-numbers-with-large-exponents", "openwebmath_score": 0.9354007840156555, "openwebmath_perplexity": 513.835971277717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303434461162, "lm_q2_score": 0.882427872638409, "lm_q1q2_score": 0.8734538842401021}}
{"url": "https://math.stackexchange.com/questions/1039519/finding-prime-factors-by-taking-the-square-root/1039527", "text": "Finding prime factors by taking the square root\n\nI'm trying to solve the third Project Euler problem and I'd like a little help understanding a mathematical concept underlying my tentative solution.\n\nThe prime factors of 13195 are 5, 7, 13, and 29.\n\nWhat is the largest prime factor of the number 600851475143 ?\n\nAs a caveat, in accordance with the wishes of Project Euler I won't be providing any code, my question is centered on why a mathematical concept works.\n\nFailed Attempt\n\nMy first algorithm looked at all the numbers from 1 through 600851475143, but was unable to complete the subsequent computations (concerning primes and factorization) due to memory constraints.\n\nSuccessful Attempt\n\nMy next algorithm looked at all the numbers from 1 through $\\sqrt{600851475143}$ and completed the computation successfully.\n\nMy Question\n\nWhy does evaluating $\\sqrt{600851475143}$ work in this instance when I really want to evaluate up to 600851475143 ? How can I be sure this approach won't miss some factor like $2 \\cdot n$ or $3 \\cdot n$, when $n$ is some number between $\\sqrt{600851475143}$ and 600851475143 ?\n\n\u2022 If a number $N$ has a prime factor larger than $\\sqrt{N}$ , then it surely has a prime factor smaller than $\\sqrt{N}$. \u2013\u00a0barak manos Nov 26 '14 at 11:46\n\u2022 @barakmanos You may as well write this as the answer. \u2013\u00a0DanielV Nov 26 '14 at 11:47\n\u2022 You won't miss $2\\cdot n$ or $3\\cdot n$ because you have checked for $2$ and $3$. \u2013\u00a0gammatester Nov 26 '14 at 11:48\n\u2022 We have to be a little careful in writing the program, since some plausible approaches will miss \"large\" prime factors when a small prime factor occurs with multiplicity $\\gt 1$. \u2013\u00a0Andr\u00e9 Nicolas Nov 26 '14 at 11:59\n\u2022 You don't have to check all the way to $\\sqrt{600851475143}$. Once you identify 71 as a factor you know that the largest prime factor of 600851475143, is also the largest prime factor of 600851475143/71=8462696833. So at that point you can limit the search to $\\sqrt{8462696833}$ - and so on. Once you hit the square root you know, by the argument given in the answers, that the number you are taking the square root of is prime, and also that it's the largest prime factor of your original number. \u2013\u00a0Taemyr Nov 26 '14 at 15:22\n\nIf a number $N$ has a prime factor larger than $\\sqrt{N}$ , then it surely has a prime factor smaller than $\\sqrt{N}$.\n\nSo it's sufficient to search for prime factors in the range $[1,\\sqrt{N}]$, and then use them in order to compute the prime factors in the range $[\\sqrt{N},N]$.\n\nIf no prime factors exist in the range $[1,\\sqrt{N}]$, then $N$ itself is prime and there is no need to continue searching beyond that range.\n\n\u2022 As mentioned by Andr\u00e9 Nicolas, prime factors with >1 multiplicity will result in the larger factor being non-prime. To solve this, keep dividing the larger factor of a pair by the smaller one until it no longer divides. \u2013\u00a0Mark K Cowan Nov 26 '14 at 14:00\n\u2022 As an example: For 28, factors are 2,2,7, here 7 is larger than sqroot(28), but there is no single prime number that can combine to form 28 (i.e. x*7=28, where x is prime, this does not exist, since x is 4 which is not prime), so dividing the primes from beginning with 2 multiple time will help, 28/2= 14, 14/2 = 7, then we know the 2 & 2 are prime factors, also take the remaining one 7 is also another prime number completing the prime factor list. best of luck. \u2013\u00a0Manohar Reddy Poreddy Dec 2 '15 at 0:25\n\nIf you do not find a factor less than $\\sqrt{x}$, then $x$ is prime for the following reason. Consider the opposite, you find two factors larger than $\\sqrt{x}$, say $a$ and $b$. But then $a\\cdot b> \\sqrt{x}\\sqrt{x} = x$. Therefore, if there is a factor larger than $\\sqrt{x}$, there must also exist a factor smaller than $\\sqrt{x}$, otherwise their product would exceed the value of $x$.\n\nRegarding the last question. You will not miss some factor like $2\\cdot n$ because you have already checked if $2$ is a factor.\n\nWhat you're describing is a prime-testing algorithm known as the sieve of Eratosthenes.\n\nIt seems like you already understand the concept:\n\n\u2022 Cross out $1$\n\n\u2022 Circle $2$ as the first prime, then cross out all of the multiples of $2$ in your list.\n\n\u2022 The next not-crossed-out number is the next prime (in this case $3$).\n\nIf your list has $N$ numbers, you only need to test until you get to a prime that's bigger than $\\sqrt{N}$.\n\nOther users have mentioned the reason for $\\sqrt{N}$ being the maximum you need to test until, but I'll add an alternative explanation. If you consider listing out all of the factors of $N$, in order from least to greatest, you will either get an even number or an odd number. $N$ has an odd number of factors if and only if it is a perfect square.\n\nIn any case, $N$ always has the same number of factors (strictly) less than its square root as it does (strictly) greater than its square root.\n\nIn fact, there's an explicit bijection between the set of factors less than $\\sqrt{N}$ and the set of factors greater than $\\sqrt{N}$ given by $$f(a) = \\frac{N}{a}$$\n\n(where $a$ is a factor of $N$ less than $\\sqrt{N}$)", "date": "2021-05-15 18:22:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1039519/finding-prime-factors-by-taking-the-square-root/1039527", "openwebmath_score": 0.7297558188438416, "openwebmath_perplexity": 221.82529710842138, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914019704466, "lm_q2_score": 0.9032942080055513, "lm_q1q2_score": 0.8734078336384096}}
{"url": "https://math.stackexchange.com/questions/3312177/what-is-the-negation-of-this-sentence/3312223", "text": "# What is the negation of this sentence\n\nWhat is the negation of the following case:\n\nCase A: The function $$f$$ has a unique rational fixed point $$a$$.\n\nI would say the function has:\n\n\u2022 multiple rational fixed points or\n\u2022 no fixed point at all or\n\u2022 at least one irrational fixed point\n\nNot sure if we can make this more \"compact\".\n\n\u2022 @China: I don't agree with this answer. My point of view is explained in my answer. \u2013\u00a0Taroccoesbrocco Aug 3 at 10:59\n\u2022 Well, my answer matches better to your 2. than your 1. So in fact we agree on the negation... \u2013\u00a0zwim Aug 3 at 13:30\n\u2022 @Taroccoesbrocco: Thank you. \"Unique\" refers to \"rational fixed point\" \u2013\u00a0Germany Aug 3 at 14:46\n\u2022 @zwim - I'm not sure to understand your comment: If $f$ has an irrational fixed point and a rational fixed point, then your version of the negation is true (because of point 3) but my version of the negation (see point 2 in my answer) is false. So, your negation and my negation are not equivalent, and only one of them is correct. \u2013\u00a0Taroccoesbrocco Aug 3 at 15:35\n\u2022 @China - Then the correct negation is in point $(2)$ of my answer, and not in point $(1)$ (or in zwim's answer). \u2013\u00a0Taroccoesbrocco Aug 3 at 15:45\n\nThe sentence can be interpreted in two ways, depending on whether \"unique\" refers to \"fixed point\" or \"rational fixed point\":\n\n1. \"Unique\" refers to \"fixed point\": The function $$f$$ has a unique fixed point $$a$$, and moreover $$a$$ is rational. The logical form of this sentence is \\begin{align} \\exists a : a \\in \\mathbb{Q} \\land a = f(a) \\land \\forall z (z = f(z) \\to z = a) \\end{align} In this case, the negation is: the function $$f$$ either has a unique irrational fixed point, or several fixed points, or no fixed points at all (this is equivalent to what @zwim said in his/her answer).\n\n2. \"Unique\" refers to \"rational fixed point\": The function $$f$$ has exactly one rational fixed point (and possibly many irrational fixed points). The logical form of this sentence is: \\begin{align} \\exists a : a \\in \\mathbb{Q} \\land a = f(a) \\land \\forall z ((z \\in \\mathbb{Q} \\land z = f(z)) \\to z = a) \\end{align} In this case the negation is: the function $$f$$ has either several rational fixed points or no rational fixed points at all (but it might have irrational fixed points).\n\nNote that the negation of the interpretation $$(2)$$ is more restrictive than the negation of the interpretation $$(1)$$: for instance, if $$f$$ has a irrational fixed point and a rational fixed point, then the negation of $$(1)$$ is true but the negation of $$(2)$$ is false.\n\nUnlike @zwim, in my opinion the intended meaning of the sentence should be $$(2)$$, and not $$(1)$$, unless contextual information (which is missing in the question of the OP) says differently.", "date": "2019-09-19 08:41:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3312177/what-is-the-negation-of-this-sentence/3312223", "openwebmath_score": 0.9848297834396362, "openwebmath_perplexity": 489.5019813472696, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140187510509, "lm_q2_score": 0.9032942080055513, "lm_q1q2_score": 0.8734078327771952}}
{"url": "https://math.stackexchange.com/questions/135234/showing-fx-x4-is-not-uniformly-continuous", "text": "# Showing $f(x)=x^4$ is not uniformly continuous\n\nI am looking at uniform continuity (for my exam) at the moment and I'm fine with showing that a function is uniformly continuous but I'm having a bit more trouble showing that it is not uniformly continuous, for example:\n\nshow that $x^4$ is not uniformly continuous on $\\mathbb{R}$, so my solution would be something like:\n\nAssume that it is uniformly continuous then:\n\n$$\\forall\\epsilon\\geq0\\exists\\delta>0:\\forall{x,y}\\in\\mathbb{R}\\ \\mbox{if}\\ |x-y|<\\delta \\mbox{then} |x^4-y^4|<\\epsilon$$\n\nTake $x=\\frac{\\delta}{2}+\\frac{1}{\\delta}$ and $y=\\frac{1}{\\delta}$ then we have that $|x-y|=|\\frac{\\delta}{2}+\\frac{1}{\\delta}-\\frac{1}{\\delta}|=|\\frac{\\delta}{2}|<\\delta$ however $$|f(x)-f(y)|=|\\frac{\\delta^3}{8}+3\\frac{\\delta}{4}+\\frac{3}{2\\delta}|$$\n\nNow if $\\delta\\leq 1$ then $|f(x)-f(y)|>\\frac{3}{4}$ and if $\\delta\\geq 1$ then $|f(x)-f(y)|>\\frac{3}{4}$ so there exists not $\\delta$ for $\\epsilon < \\frac{3}{4}$ and we have a contradiction.\n\nSo I was wondering if this was ok (I think it's fine) but also if this was the general way to go about showing that some function is not uniformly continuous? Or if there was any other ways of doing this that are not from the definition?\n\nThanks very much for any help\n\n\u2022 So, is this an exam question? \u2013\u00a0user21436 Apr 22 '12 at 12:17\n\u2022 No, I'm just practicing for my exam where questions like this (not this one though) will come it. This is from one of the past papers that are for revision \u2013\u00a0hmmmm Apr 22 '12 at 12:29\n\u2022 Just trying to ensure we are not taken for a ride. Hope you don't mind. :) \u2013\u00a0user21436 Apr 22 '12 at 12:33\n\u2022 @KannappanSampath no its fine- it annoys me when I see people posting assessment questions on forums :) \u2013\u00a0hmmmm Apr 22 '12 at 12:36\n\nTo show that it is not uniformly continuous on the whole line, there are two usual (and similar) ways to do it:\n\n1. Show that for every $\\delta > 0$ there exist $x$ and $y$ such that $|x-y|<\\delta$ and $|f(x)-f(y)|$ is greater than some positive constant (usually this is even arbitrarily large).\n2. Fix the $\\varepsilon$ and show that for $|f(x)-f(y)|<\\varepsilon$ we need $\\delta = 0$.\n\nFirst way:\n\nFix $\\delta > 0$, set $y = x+\\delta$ and check $$\\lim_{x\\to\\infty}|x^4 - (x+\\delta)^4| = \\lim_{x\\to\\infty} 4x^3\\delta + o(x^3) = +\\infty.$$\n\nSecond way:\n\nFix $\\epsilon > 0$, thus $$|x^4-y^4| < \\epsilon$$ $$|(x-y)(x+y)(x^2+y^2)| < \\epsilon$$ $$|x-y|\\cdot|x+y|\\cdot|x^2+y^2| < \\epsilon$$ $$|x-y| < \\frac{\\epsilon}{|x+y|\\cdot|x^2+y^2|}$$\n\nbut this describes a necessary condition, so $\\delta$ has to be at least as small as the right side, i.e.\n\n$$|x-y| < \\delta \\leq \\frac{\\epsilon}{|x+y|\\cdot|x^2+y^2|}$$\n\nso if either of $x$ or $y$ tends to infinity then $\\delta$ tends to $0$.\n\nHope that helps ;-)\n\nEdit: after explanation and calculation fixes, I don't disagree with your proof.\n\n\u2022 Thanks for the reply, I think that I use that we are considering all of $\\mathbb{R}$ when i choose $x=\\delta+\\frac{1}{\\delta}$ and $y=\\frac{1}{\\delta}$ as these would not be valid for small $\\delta$ in bounded interval? \u2013\u00a0hmmmm Apr 22 '12 at 13:38\n\u2022 @hmmmm Ok, I misunderstood what you were saying there. If the calculations are alright, then your proof is fine. \u2013\u00a0dtldarek Apr 22 '12 at 13:44\n\u2022 Is it correct to fix $\\delta>0$ and choosing $y = x + \\delta$ specifically? Since the distance between x and y will be exactly $\\delta$, and we want $|x-y|<\\delta$. \u2013\u00a0DrHAL May 2 '16 at 8:08\n\u2022 @dtldarek Is it possible that one method will fail to show and other will succeed. For example, for $f(x)=1/x$ on $(0,\\infty)$, first method leads the limit to 0? \u2013\u00a0chandresh Feb 3 at 10:18\n\u2022 @chandresh Function $f(x)=1/x$ is uniformly continuous on $(1, \\infty)$, or even on $(a, \\infty)$ for any fixed $a > 0$. On the other hand, $f$ is very steep near zero, and so the limit you want to calculate is $$\\lim_{x \\to 0^+}\\left|\\frac{1}{x}-\\frac{1}{x+\\delta}\\right|.$$ \u2013\u00a0dtldarek Feb 5 at 8:04\n\nI will comment on your solution after writing another approach. For any $x,y\\in\\mathbb{R}$ we have: \\begin{align*} |x^{4}-y^{4}|=|(x^{2}-y^{2})(x^{2}+y^{2})|=|(x-y)(x+y)(x^{2}+y^{2})|=|x-y|\\cdot |x+y|\\cdot |x^{2}+y^{2}| \\end{align*}\n\nSo what you can see is that even if you take arbitrarily close $x$ and $y$, you can grow the distance of $x^{4}$ and $y^{4}$ as much as you want by taking them far enough away from zero. You can easily conclude from here that the function is not uniformly continuous by a contraposition for example.\n\nAlright, then to your solution. If the calculations would be correct, then it would be fine. You could assume at first that such $\\delta>0$ exists for $0<\\varepsilon<3$ and conclude with a contradiction. However, I got a bit different calculations than you. Using the above equation we see that: \\begin{align*} |f(\\frac{\\delta}{2}+\\frac{1}{\\delta})-f(\\frac{1}{\\delta})|&=|(\\frac{\\delta}{2}+\\frac{1}{\\delta})^{4}-\\frac{1}{\\delta^{4}}|=|\\frac{\\delta}{2}(\\frac{\\delta}{2}+\\frac{2}{\\delta})((\\frac{\\delta}{2}+\\frac{1}{\\delta})^{2}+\\frac{1}{\\delta^{2}})| \\\\ &= |(\\frac{\\delta^{2}}{4}+1)(\\frac{\\delta^{2}}{4}+2\\cdot \\frac{\\delta}{2}\\cdot \\frac{1}{\\delta}+\\frac{1}{\\delta^{2}}+\\frac{1}{\\delta^{2}})| \\\\ &=|(\\frac{\\delta^{2}}{4}+1)(\\frac{\\delta^{2}}{4}+1+\\frac{2}{\\delta^{2}})| \\\\ &= |\\frac{\\delta^{4}}{16}+\\frac{\\delta^{2}}{4}+\\frac{1}{2}+\\frac{\\delta^{2}}{4}+1+\\frac{2}{\\delta^{2}}| \\\\ &= |\\frac{\\delta^{4}}{16}+\\frac{\\delta^{2}}{2}+\\frac{2}{\\delta^{2}}+\\frac{3}{2}|\\\\ &= \\frac{\\delta^{4}}{16}+\\frac{\\delta^{2}}{2}+\\frac{2}{\\delta^{2}}+\\frac{3}{2} \\end{align*} If you're able to find a lower bound for this (which is quite easy) as you did previously, then by choosing an epsilon smaller than that fixed number you may conclude as you did in your original post by contradiction.\n\n\u2022 Hey sorry about that, I edited it-hopefully it's right now? \u2013\u00a0hmmmm Apr 22 '12 at 13:39\n\u2022 I also edited now my calculation which still differs abit from your new one. Could you show the steps of how you got this answer for $|f(x)-f(y)|$? \u2013\u00a0T. Eskin Apr 22 '12 at 13:47\n\u2022 yeah I messed that up quite a bit sorry (I had the wrong power and the wrong delta's) \u2013\u00a0hmmmm Apr 22 '12 at 13:50\n\u2022 It should be $|(\\frac{\\delta}{2}+\\frac{1}{\\delta})^4-\\frac{1}{\\delta^4}|$ which would give $|\\frac{\\delta^4}{16}+\\frac{\\delta^2}{2}+\\frac{2}{\\delta}+\\frac{3}{2}|$ I think I could conclude a similar thing from here? \u2013\u00a0hmmmm Apr 22 '12 at 13:52\n\u2022 Except that is the last $-\\frac{1}{\\delta^{4}}$ missing from there? Otherwise it looks close to mine. \u2013\u00a0T. Eskin Apr 22 '12 at 13:57\n\n(1). If $f:(0,\\infty)\\to \\Bbb R$ is differentiable and $\\lim_{x\\to \\infty}f'(x)=\\infty$ then $f$ is not uniformly continuous: For any $\\delta >0$ and any $x>0$, the MVT implies there exists $y\\in (x,x +\\delta)$ such that $\\frac {f(x+\\delta)-f(x)}{\\delta}= f'(y).$ Now if $x$ is large enough that $\\forall y>x\\;(f'(y)>1/\\delta)$ then $f(x+\\delta)-f(x)=\\delta f'(y)>1.$\n\n(2). Given $\\delta >0$ take $x>\\max (1,1/\\delta).$ Then $$(x+\\delta)^4-x^4= 4x^3\\delta+6x^2\\delta^2+4x\\delta^3+\\delta^4>$$ $$>4x^3\\delta=4(x^2)x\\delta>4x\\delta>4.$$\n\nBy (1) or by (2) we have $\\sup_{x\\in \\Bbb R} \\{|(x+x')^4-x^4|: |x'-x|<\\delta\\}> 1$ for every $\\delta>0.$ (In fact the $\\sup$ is $\\infty$.)\n\nUniform continuity of $f:D\\to \\Bbb R$ for some (any) domain $D\\subset \\Bbb R$ means $$0=\\lim_{\\delta \\to 0^+}\\sup \\{|f(x')-f(x)|:x,x'\\in D\\land |x'-x|\\leq\\delta\\}.$$\n\nI think you should make this a little simpler (and for uniform continuity in general) All you need to do to show $f:X \\to Y$ is not uniformly continuous on $X$ (let's suppose there both subsets of $\\Bbb R$), then just give me a SINGLE epsilon such that, NO MATTER HOW SMALL delta is chosen, there will be x and y closer than delta for which the difference in function values exceed epsilon. Thus for instance $|(N+\\theta)^4- N^4| \\ge 4\\theta N^3$ so if you choose $x$ really big (with respect to $\\delta, x=N+\\theta\\,\\,\\,\\ \\text{and}\\,\\,\\, y = N,$ then if $0 < \\delta/2 < \\theta < \\delta$ you have $|x-y| < \\delta$ yet you still have the variable N to play with to make the difference in function values as large as you like (in particular the difference in function values can always be made bigger than 3 regardless of how small $\\delta$ is). Nevertheless, I think your proof is an accurate job!\n\n## ** Simple Trick:**\n\nGiven $a>0$ if we take $$x = \\frac{1}{a}+2 ~~~and ~~~y = \\frac{1}{a}+2+\\frac{a}{2} \\color{red}{= x +\\frac{a}{2}}$$ we have $x,y\\in [2,\\infty)~~~ and~~~|x-y| =\\frac{a}{2}<a$\n\nBut since $a\\cdot x = 1+ 2a~~~$ and $~~x,y>2\\implies x^2+y^2 >8$ we get,\n\n$$|f(x)-f(y)| =(x^2+y^2) |y^2-x^2|\\ge 8|y^2-x^2|\\\\~~~~~~~~~~~~~~~=8| x^2 + 2\\frac{a}{2}x +\\frac{a^2}{4} - x^2 | \\\\= 8(ax +\\frac{a^2}{4})= 8(1+2a+\\frac{a^2}{4} )>8$$\n\nThat is $$|f(x)-f(y)| >8$$\n\nThus $$\\color{blue}{\\exists \\varepsilon_0 =8,\\forall~a>0, \\exists ~x,y\\in \\mathbb R: |x-y|< a~~and ~~|f(x)-f(y)|>8}$$\n\njust take $$x = \\frac{1}{a}+2 ~~~and ~~~y = \\frac{1}{a}+2+\\frac{a}{2} \\color{red}{= x +\\frac{a}{2}}$$", "date": "2019-06-18 09:15:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/135234/showing-fx-x4-is-not-uniformly-continuous", "openwebmath_score": 0.9865854382514954, "openwebmath_perplexity": 210.78347385952645, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138164089085, "lm_q2_score": 0.8933094046341532, "lm_q1q2_score": 0.8734009472388279}}
{"url": "https://gmatclub.com/forum/the-amounts-of-time-that-three-secretaries-worked-on-a-127170.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 22 Oct 2019, 01:55\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# The amounts of time that three secretaries worked on a\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 01 Dec 2011\nPosts: 62\nThe amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2012, 02:12\n4\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n95% (00:54) correct 5% (01:28) wrong based on 378 sessions\n\n### HideShow timer Statistics\n\nThe amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?\n\nA) 80\nB) 70\nC) 56\nD) 16\nE) 14\n\n_________________\nhttp://en.wikipedia.org/wiki/Kudos\nSVP\nStatus: Top MBA Admissions Consultant\nJoined: 24 Jul 2011\nPosts: 1881\nGMAT 1: 780 Q51 V48\nGRE 1: Q800 V740\nRe: The amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2012, 02:28\n8x = 112\n=> x = 14\n\nTherefore the secretary who worked the longest spent 14 x 5 = 70 hours on the project\n\nOption (B)\n_________________\nGyanOne [www.gyanone.com]| Premium MBA and MiM Admissions Consulting\n\nAwesome Work | Honest Advise | Outstanding Results\n\nReach Out, Lets chat!\nEmail: info at gyanone dot com | +91 98998 31738 | Skype: gyanone.services\nMath Expert\nJoined: 02 Sep 2009\nPosts: 58410\nRe: The amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2012, 05:05\n2\nfiendex wrote:\nThe amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longet spend on the project?\n\nA) 80\nB) 70\nC) 56\nD) 16\nE) 14\n\nSince, amounts of time that secretaries worked in the ratio of 1 to 2 to 5, then $$\\frac{\\frac{1x}{2x}}{5x}$$, for some number x --> x+2x+5x=8x=112 --> x=14 --> the secretary who worked the longet spent 5x=70 hours.\n\n_________________\nManager\nJoined: 10 Jan 2010\nPosts: 138\nLocation: Germany\nConcentration: Strategy, General Management\nSchools: IE '15 (M)\nGPA: 3\nWE: Consulting (Telecommunications)\nRe: The amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2012, 06:53\n1\nAll three secretaries worked 112h and with a ratio of 1to2to5\n\n8x = 112h\nx = 14h\n\n5*14h = 50h --> B\nMath Expert\nJoined: 02 Sep 2009\nPosts: 58410\nRe: The amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n23 May 2017, 02:48\nfiendex wrote:\nThe amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?\n\nA) 80\nB) 70\nC) 56\nD) 16\nE) 14\n\nSimilar question: https://gmatclub.com/forum/the-amount-o ... 01139.html\n_________________\nTarget Test Prep Representative\nStatus: Head GMAT Instructor\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2815\nRe: The amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n25 May 2017, 15:48\n1\nfiendex wrote:\nThe amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?\n\nA) 80\nB) 70\nC) 56\nD) 16\nE) 14\n\nThe ratio of the amounts of time that the secretaries worked on the project is x : 2x : 5x or a total of 8x.\n\nSince the total number of hours is 112:\n\n8x = 112\n\nx = 14\n\nThe secretary who worked the longest worked for 5x = 5(14) = 70 hours.\n\n_________________\n\n# Jeffrey Miller\n\nHead of GMAT Instruction\n\nJeff@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nIntern\nJoined: 08 Dec 2017\nPosts: 15\nThe amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n08 Dec 2017, 19:36\n$$\\frac{5}{8ths}$$of the time was spent by the longest working secretary. So the answer is more than 56. An eighth of\n112 is 14. 56+14=70\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 13384\nRe: The amounts of time that three secretaries worked on a\u00a0 [#permalink]\n\n### Show Tags\n\n31 Jan 2019, 04:16\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: The amounts of time that three secretaries worked on a \u00a0 [#permalink] 31 Jan 2019, 04:16\nDisplay posts from previous: Sort by\n\n# The amounts of time that three secretaries worked on a\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne", "date": "2019-10-22 08:55:32", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-amounts-of-time-that-three-secretaries-worked-on-a-127170.html", "openwebmath_score": 0.6598981022834778, "openwebmath_perplexity": 7088.721673432217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9793540716711546, "lm_q2_score": 0.8918110526265554, "lm_q1q2_score": 0.8733987855511554}}
{"url": "https://stats.stackexchange.com/questions/241441/is-xt-x-invertible-if-p-n", "text": "# Is $X^T X$ invertible if $p > n$?\n\nI am checking other's regression analysis work on a $p > n$ data. I can only see the results but not the process how he did it.\n\nI believe he made mistakes. Since $p > n$, $X^T X$ is not full rank it is not invertible. So we cannot find the OLS coefficient.\n\nHowever, I am not sure my reasoning is correct. I have not used linear algebra for a long time.\n\nUpdate:\n\n1. No penalized methods involved.\n2. He used stepwise regression for variable selection. A new question: would such algorithm stop if number of variables in the model equal to the number of sample points?\n3. His goal is to find out which variables are important. Doesn't care about the prediction power.\n\u2022 Yes you are correct in that $X^TX$ is non-invertible. However, perhaps your colleague performed some penalized regression like LASSO to get coefficients? Oct 20 '16 at 17:44\n\u2022 What are $p$ and $n$? Oct 20 '16 at 18:21\n\u2022 A bit more context may be useful. Is the goal forecasting? Is it consistently estimating some effect $b_i$? Do you know if your colleague performed LASSO or ridge regression? Oct 20 '16 at 18:30\n\u2022 @RodrigodeAzevedo $p$ and $n$ are classically the number of covariates and of independent observations. IE, $X$ is $n \\times p$. Oct 20 '16 at 19:07\n\u2022 If he's using forward stepwise regression, he'd never have more than $n$ variables in the actual regression matrix used (starting from 1 = the intercept, finding the best variable to add incrementally.) That's not to say forward stepwise regression (or any stepwise regression) is good, it's just to say it would have avoided the identifiability problem. Oct 20 '16 at 19:44\n\nIf $\\mathrm X$ is $n \\times p$ and $p > n$, then it is fat and, thus,\n$$\\mbox{rank} (\\mathrm X) = \\mbox{rank} (\\mathrm X^T \\mathrm X) \\leq n < p$$\nHence, $\\mathrm X^T \\mathrm X$ does not have full rank and, thus, it is not invertible.", "date": "2021-10-25 18:01:51", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/241441/is-xt-x-invertible-if-p-n", "openwebmath_score": 0.3934694230556488, "openwebmath_perplexity": 496.42844854645904, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9793540668504083, "lm_q2_score": 0.8918110382493035, "lm_q1q2_score": 0.8733987671715404}}
{"url": "http://mathschallenge.net/full/never_decreasing_digits", "text": "## Never Decreasing Digits\n\n#### Problem\n\nA number has strictly increasing digits if each digit exceeds the digit to its left and is said to be a strictly increasing number; for example, 1357.\nA number has increasing digits if each digit is not exceeded by the digit to its left and is said to be an increasing number; for example, 45579.\n\nThere are exactly 219 increasing numbers below one-thousand.\n\nHow many numbers below one million are increasing?\n\n#### Solution\n\nWe will begin by deriving, from first principles, the information given in the problem relating to numbers below one-thousand.\n\nIt is possible to use binary strings to produce increasing numbers. Consider the following algorithm.\n\nLet S be binary string\nLet C = 0\nLet K = 1\nLabel A\nLet D be value of Kth digit of S\nIf D = 1 then output C\nIf D = 0 then increment c by 1\nIncrement K by 1\nIf K has not exceeded the length of S then goto A\nStop\n\nFor example, the binary string $001101001$ would produce 2235.\n\nIn general it can be seen that an $n$-digit increasing number will be produced by a binary string containing $n$ $1$'s, and the maximum digit will be represented by the number of $0$'s, as each zero causes the \"counter\" to increase by 1.\n\nUsing this idea we can represent all the 3-digit increasing numbers made up of the digits 0, 1, and 2 using 5-digit binary strings consisting of three $1$'s and two $0$'s:\n\n\\begin{align}11100 &= 000\\\\11010 &= 001\\\\11001 &= 002\\\\10110 &= 011\\\\10101 &= 012\\\\10011 &= 022\\\\01110 &= 111\\\\01101 &= 112\\\\01011 &= 122\\\\00111 &= 222\\end{align}\n\nBut we must subtract one to remove $000$, so there are $\\dbinom{5}{3} - 1 = 9$ such numbers.\n\nIn the same way a 3-digit increasing number using each of the digits 0 to 9 requires a binary string containing three 1's and nine 0's; that is, there are exactly $\\dbinom{12}{3} - 1 = 219$ increasing numbers below one-thousand.\n\nNow we can return to the problem: increasing numbers below one million.\n\nAs the numbers contain six digits the binary string will contain six 1's and nine 0's. Hence there are $\\dbinom{15}{6} - 1 = 5004$ increasing numbers below one million.\n\nCheck out the related (strictly) Increasing Digits problem.\n\nProblem ID: 263 (29 Jan 2006) \u00a0 \u00a0 Difficulty: 4 Star\n\nOnly Show Problem", "date": "2016-09-25 00:15:58", "meta": {"domain": "mathschallenge.net", "url": "http://mathschallenge.net/full/never_decreasing_digits", "openwebmath_score": 0.9956274032592773, "openwebmath_perplexity": 1096.959541427044, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987946220128863, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.8733832562822937}}
{"url": "https://math.stackexchange.com/questions/1403304/a-is-the-power-set-of-set-c-and-s-is-the-relation-on-a-defined-by-x-s", "text": "# $A$ is the power set of set $C$, and $S$ is the relation on $A$ defined by $x S y$ iff $x \\subseteq y$.\n\nThe following is an exercise from my textbook:\n\nLet $C$ be a set and let $A=\\mathcal{P}(C)$, the set of all subsets of $C$. Let $S$ be the relation on $A$ defined by $x S y$ iff $x \\subseteq y$. Then $S$ is a partial order relation on $A$.\n(a) Find the least and greatest element of $A$.\n(b) Let $B=\\{x \\in A$: $x \\ne \u2205\\}$. Suppose $C$ has two or more elements. Show that $B$ has no least element.\n\nI'm confused about the definition of least and greatest element as applied to this exercise.\n\nFrom the my textbook, least element is defined as such:\n\nLet $(A,\\leq)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \\in B$ and for each $x \\in B$, $c \\le x$.\n\nNow, referring back to the exercise, $A$ consists of sets, not merely numbers; thus, to say that some set $x_1$ is less than or equal to some other set $x_2$ doesn't make sense. We don't relate sets in that way do we? (Unless, we're talking about order? But even so, I don't think the exercise is asking us to consider that.)\n\nNow, part (b) leads me to believe that a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$. Is that correct?\n\n\u2022 Note in your definition $(A,\\leq)$ is just a partially ordered set. There is no reference to numbers. You need to find the $S$-least element of $A$ and $B$. \u2013\u00a0James Aug 19 '15 at 23:04\n\u2022 The exercise is indeed abour inclusion as an order relation . As for part (b) what you believe is right. \u2013\u00a0Bernard Aug 19 '15 at 23:05\n\nThe notation of $\\leq$ is often used in mathematics as a general partial order. Not just the usual ordering on the natural/integers/rationals/reals. But in this context, $\\leq$ is really just $\\subseteq$.\n\nSo you are being asked to find the maximum and minimum elements of $(\\mathcal P(C),\\subseteq)$.\n\n\u2022 Okay, so I'm right to say a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$? \u2013\u00a0Jordan Miller Aug 19 '15 at 23:13\n\u2022 Yes. That is the case here. And note that in the definition of least element, $\\leq$ is said to be a partial order, not anything else. \u2013\u00a0Asaf Karagila Aug 19 '15 at 23:15\n\u2022 I must get clarification. The following in italics is also an exercise in my textbook: $(A,\\leq)$ is a partially ordered set. That each finite non-empty subset of $A$ has a $maxB$ implies $(A,\\leq)$ is totally ordered. Then does $\\le$ refer again to some general partial order relation? \u2013\u00a0Jordan Miller Aug 19 '15 at 23:27\n\u2022 Yes, a general partial order. Mathematics is much much much much more than just real numbers. \u2013\u00a0Asaf Karagila Aug 19 '15 at 23:34\n\nIn your case, the least element is an element of $A$, hence a subset of $C$ that is contained in all subsets of $C$. (Apply your definition in the second highlighted paragraph with $A=\\mathcal{P}(C)$ and $B=A$), and so this set, I guess, is the empty set. Now, for the greatest element of $A$, what is the subset of $A$ that contains any subset of $A$?\n\n\u2022 So, I am right to think that a least element, with respect to the set $A$, is a set such that it is a subset of each set in $A$? Does that make sense why I was a bit confused. To me, the relation of $\\le$ is not the same as $\\subseteq$ \u2013\u00a0Jordan Miller Aug 19 '15 at 23:11\n\u2022 $\\leq$ is just a symbol! \u2013\u00a0mich95 Aug 19 '15 at 23:12\n\nOkay, I know that what follows basically repeats what responders have been trying to tell me, but hitherto, I was still misunderstanding the notation of $\\le$. Thus, I merely post this answer as an expression of my understanding (finally!).\n\nI refer to the definiton:\n\nLet $(A,\\le)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \\in B$ and for each $x \\in B$, $c \\le x$.\n\nThe $\\le$ in the ordered pair $(A,\\le)$ does not refer to the relation of less than or equal to, but rather serves as a \"variable\" for some partial order relation on A. With this in mind, the condition for each $x \\in B$, $c \\le x$ doesn't mean that $c$ must be less than or equal to $x$, but rather that $c$ must be in relation $\\le$ to $x$.\n\nNow, the partial order relation specified in the above exercise is the relation of set inclusion. Then, by definition, the least element of $A$ must be some set in $A$ such that it is a subset of each set in $A$.", "date": "2019-12-10 00:19:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1403304/a-is-the-power-set-of-set-c-and-s-is-the-relation-on-a-defined-by-x-s", "openwebmath_score": 0.9337827563285828, "openwebmath_perplexity": 92.99159408105278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347831070321, "lm_q2_score": 0.8962513800615313, "lm_q1q2_score": 0.8733385191396364}}
{"url": "https://mathhelpboards.com/threads/in-the-simplest-qualitative-terms-what-is-a-differential-equation.6170/", "text": "# In the simplest qualitative terms what is a differential equation?\n\n#### find_the_fun\n\n##### Active member\nI'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know\n1. the derivative is the ratio of how one quantity changes with respect to another\n2. the integral is the area under the curve\n\nSo what's a differential equation? According to here \"A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. \"\n\nThis doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nI'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know\n1. the derivative is the ratio of how one quantity changes with respect to another\n2. the integral is the area under the curve\n\nSo what's a differential equation? According to here \"A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. \"\n\nThis doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.\nIn a differential equation, the idea is to find $y$ as a function of $x$, given some information about $y$ and its derivatives. The very simplest example of a differential equation might be something like the equation $\\frac{dy}{dx} = 2x$. You can easily solve that by integrating it, to find that the solution is $y=x^2$ plus a constant of integration. But suppose that the equation is slightly more complicated, for example $\\frac{dy}{dx} = x +y$. How would you solve that to find $y$ as a function of $x$? A course in differential equations will show you how to do that.\n\nIn case you are wondering, the solution to that equation is $y = -x-1 + ce^x$, where $c$ is a constant.\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nI'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know\n1. the derivative is the ratio of how one quantity changes with respect to another\n2. the integral is the area under the curve\n\nSo what's a differential equation? According to here \"A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. \"\n\nThis doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.\nA differential equation is an equation, a derivative is NOT! Okay, having got that off my chest, I think I understand your point. \"Contains derivatives\", etc. is not sufficient. The crucial point is that a differential equation contains derivatives of some unknown function. Yes, \"$$x^2$$\" can be thought of as the derivative of $$\\frac{1}{3}x^3$$ but just having $$x^2$$ in an equation does NOT make it a \"differential equation\". To be a differential equation, the equation must contain something like $$\\frac{dy}{dx}$$, $$\\frac{\\partial y}{\\partial t}$$, $$\\frac{d^2y}{dx^2}$$, etc., where y is the \"unknown\" function.\n\nMore generally, a \"functional equation\" is an equation that gives us some information about a function. \"f(x+ 1)= f(x)\" is a functional equation. $$\\frac{d^2y}{dx^2}+ 2\\frac{dy}{dx}+ y= x^2$$ is a special kind of a functional equation (since it gives information about the function y) called a \"differential equation\" because that information is actually about the derivatives of the function y.\n\n#### find_the_fun\n\n##### Active member\nGlad I asked. I had the first lecture today and the prof skipped over the definition of a differential equation (in fairness it was a substitute prof).\n\nWhat is an ordinary differential equation? Does all that mean is it doesn't have partial derivatives?\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nWhat is an ordinary differential equation? Does all that mean is it doesn't have partial derivatives?\nYes. You will often see the abbreviations ODE and PDE for ordinary/partial differential equation.\n\n#### find_the_fun\n\n##### Active member\nI'm writing myself notes now and here's the first:\n\nDifferential equation: a functional equation that relates an unknown function to its derivatives\n\n#### zzephod\n\n##### Well-known member\nAn ordinary differential equation is an equation of the form (or that can be rewritten in the form):\n\n$$\\displaystyle \\large f(x,y,y',...,y^{(n)})=0$$\n\nA partial differential equation is similar but with partial derivatives with repect to the variables appearing (including mixed partials).\n\n.\n\nLast edited:", "date": "2020-12-01 08:36:24", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/in-the-simplest-qualitative-terms-what-is-a-differential-equation.6170/", "openwebmath_score": 0.860706090927124, "openwebmath_perplexity": 432.9191454738013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347838494568, "lm_q2_score": 0.896251377290367, "lm_q1q2_score": 0.8733385171047167}}
{"url": "https://math.stackexchange.com/questions/927578/whats-fracdydx-of-function-y-frac-12-sin-x", "text": "# What's $\\frac{dy}{dx}$ of function $y=\\frac 12{\\sin x}$?\n\n$(1)\\quad$ How to differentiate $y=\\frac 12 \\sin x$?\n\nI know that $\\frac{dy}{dx}$ of $y=\\sin x$ is easy to calculate: $\\frac{dy}{dx} = \\cos x$. But what if there is a coefficient preceding before it?\n\n$(2)$ Another question is: Does the following equality of functions hold? $$\\frac{x(x^2+1)}{x^2+1}\\overset{?}{=}x$$\n\nPlease explain this because I know that function $\\dfrac{x\\cdot x}{x}$ does not equal function $x$ (the first function is not defined at $0$ but the second one is defined at $0$).\n\nBut the question I asked, the domain of the function can be any number according to what I think because the denominator cannot be zero, it's always a positive number $x^2+1$.\n\n\u2022 To your second question, since $x^2+1$ is never $0$ (at least in the real numbers) $\\frac{x^2+1}{x^2+1}=1$ and $\\frac{x(x^2+1)}{x^2+1} = x$ for all $x$. So yes, the functions are equal. \u2013\u00a0Thomas Andrews Sep 11 '14 at 13:10\n\u2022 Do you mean $\\frac{1}{2\\sin x}$ or $\\frac{1}{2}\\sin x$? \u2013\u00a0Thomas Andrews Sep 11 '14 at 13:11\n\u2022 I mean 1/2 (sin x) \u2013\u00a0off99555 Sep 11 '14 at 13:13\n\u2022 One should avoid different questions in a same question on MSE. \u2013\u00a0user37238 Sep 11 '14 at 13:16\n\u2022 This is my first time using this website, I will be aware next time. \u2013\u00a0off99555 Sep 11 '14 at 13:18\n\nQuestion $(1):\\quad$ Given $y = \\dfrac 12 \\sin x$:\n\n$$\\frac{dy}{dx} = \\frac{d}{dx}\\left(\\frac 12 \\sin x\\right) = \\frac 12\\cdot \\frac{d}{dx}(\\sin x) = \\frac 12 \\cos x$$\n\n$$y = a f(x)\\implies \\frac{dy}{dx} = \\frac{d}{dx}(a f(x)) = af'(x)$$ for all constants $a$.\n\nQuestion $(2):$\n\n$$\\frac {x(x^2 + 1)}{x^2 + 1 } = x \\quad \\forall x \\in \\mathbb R$$\n\nThis happens to be the case in this example because there are no real values of $x$ at which the left-hand side is undefined. Specifically, as you note, $x^2 + 1>0$ for all $x$, and hence the function is defined everywhere. So we may cancel the common factor $x^2 + 1$ without changing the function in any way.\n\nIn the case of $f(x) = \\frac{x^2}{x}$, which you refer to, here we do have problems with simply canceling a common factor of $\\,x.\\,$ Specifically, $\\,\\dfrac{x^2}{x}\\,$ is undefined at $x = 0$, whereas the function $g(x) = x$ is defined everywhere, so the functions are not equivalently defined, i.e., they are not equivalent functions.\n\n\u2022 I would say that the first response is missing one sign \"=\" -> ambiguity. \u2013\u00a0georg Sep 11 '14 at 14:08\n\u2022 Thanks, @georg. I didn't catch that until you mentioned it.. \u2013\u00a0amWhy Sep 11 '14 at 14:19\n\nFor the first part of your question, you can remove constants when differentiating:\n\n$$\\frac{\\mathrm{d}}{\\mathrm{d}x}\\big(a\\cdot f(x)\\big)=a\\frac{\\mathrm{d}}{\\mathrm{d}x}f(x)$$\n\nYou can prove this with the product rule by finding $a\\frac{\\mathrm{d}}{\\mathrm{d}x}f(x)+f(x)\\frac{\\mathrm{d}}{\\mathrm{d}x}a$ & noting that the derivative of a constant is $0$. It implies that your derivative is $\\frac{\\mathrm{d}}{\\mathrm{d}x}\\big(\\frac{1}{2}\\sin(x)\\big)=\\frac{1}{2}\\frac{\\mathrm{d}}{\\mathrm{d}x}\\sin(x)$.\n\nFor the second part of the question, it is valid to cancel terms in a function but just be aware that the function's domain should stay the same. If the original function is undefined at a number, it should stay that way. As Thomas Andrews points out, $x^2+1$ is never $0$ for real $x$ so you can effectively cancel the $\\frac{(x^2+1)}{(x^2+1)}$ term.\n\n\u2022 I got the edit wrong - he wanted $y=\\frac{1}{2}\\sin x$. Sorry. \u2013\u00a0Thomas Andrews Sep 11 '14 at 13:14\n\u2022 @ThomasAndrews Ah, thanks for letting me know :) No problem by the way, I've done it before too. \u2013\u00a0Jam Sep 11 '14 at 13:14\n\u2022 Thank you very much for your help. I would like to vote both of you up but I can't. My reputation is not high enough to do that. Both answers are very useful for me. But I don't understand what setting a as a function of x mean. Can you show me please? \u2013\u00a0off99555 Sep 11 '14 at 14:25\n\u2022 @off99555 Glad I could help. What I meant by \"setting it as a function\" was that, by using the product rule, we can find the derivative of $f(x)\\cdot g(x)$ as $f(x)g'(x)+g(x)f'(x)$. If we say that $g(x)$ is a constant $a$, we could find the derivative of $a\\cdot f(x)$. Since the derivative of any constant is $0$, it shows us that we can take out the constant. So $\\frac{d}{dx}(\\frac{1}{2}\\sin(x))=\\frac{1}{2}\\frac{d}{dx}\\sin(x)$. Does that make sense? \u2013\u00a0Jam Sep 11 '14 at 14:45\n\u2022 Sure. That make sense! That's what I'm looking for, the fundamental explanation. \u2013\u00a0off99555 Sep 17 '14 at 15:46", "date": "2020-04-02 10:55:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/927578/whats-fracdydx-of-function-y-frac-12-sin-x", "openwebmath_score": 0.8597525954246521, "openwebmath_perplexity": 163.1122441045836, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615795, "lm_q2_score": 0.8962513731336204, "lm_q1q2_score": 0.8733385163812333}}
{"url": "https://www.shaalaa.com/question-bank-solutions/a-typist-charges-rs-145-typing-10-english-3-hindi-pages-while-charges-typing-3-english-10-hindi-pages-are-rs-180-using-matrices-find-charges-typing-one-english-one-hindi-page-separately-inverse-of-matrix-inverse-of-a-nonsingular-matrix-by-elementary-transformation_4023", "text": "A Typist Charges Rs. 145 for Typing 10 English and 3 Hindi Pages, While Charges for Typing 3 English and 10 Hindi Pages Are Rs. 180. Using Matrices, Find the Charges of Typing One English and One Hindi Page Separately. - Mathematics\n\nA typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?\n\nSolution\n\nLet charges for typing one English page be Rs. x.\n\nLet charges for typing one Hindi page be Rs.y.\n\nThus from the given statements, we have,\n\n10x+3y=145\n\n3x+10y=180\n\nThus the above system can be written as,\n\n[(10,3),(3,10)][(x),(y)]=[(145),(180)]\n\n\u21d2\u00a0AX = B, where,\u00a0A=[(10,3),(3,10)],x=[(x),(y)] \" and \" B = [(145),(180)]\n\nMultiply A-1 on both the sides, we have,\n\nA-1 x\u00a0AX = A-1B\n\n\u21d2 IX = A-1B\n\n\u21d2 X = A-1B\n\nThus, we need to find the inverse of the matrix A.\n\nWe know that, if\u00a0P=[(a,b),(c,d)] \" then \" P^(-1) = 1/(ad-bc)[(d,-b),(-c,a)]\n\nThus,\u00a0A^(-1)=1/(10xx10-3xx3)[(10,-3),(-3,10)]\n\n= 1/(100-9)[(10,-3),(-3,10)]\n\n=1/91[(10,-3),(-3,10)]\n\nTherefore,\u00a0X=1/91[(10,-3),(-3,10)][(145),(180)]\n\n=1/91[(10xx145-3xx180),(-3xx145+10xx180)]\n\n=1/91[(910),(1365)]\n\n=[(10),(15)]\n\n=>[(x),(y)][(10),(15)]\n\n\u21d2\u00a0x = 10 and y=15\n\nAmount taken from Shyam = 2 \u00d7 5 = Rs.10\n\nActual rate = 15 \u00d7 5 =75\n\nDifference amount = Rs.75 \u2013 Rs.10 = Rs.65\n\nRs. 65 less was charged from the poor boy Shyam.\n\nHumanity and sympathy are reflected in this problem.\n\nConcept: Inverse of Matrix - Inverse of a Nonsingular Matrix by Elementary Transformation\nIs there an error in this question or solution?", "date": "2021-10-18 19:09:34", "meta": {"domain": "shaalaa.com", "url": "https://www.shaalaa.com/question-bank-solutions/a-typist-charges-rs-145-typing-10-english-3-hindi-pages-while-charges-typing-3-english-10-hindi-pages-are-rs-180-using-matrices-find-charges-typing-one-english-one-hindi-page-separately-inverse-of-matrix-inverse-of-a-nonsingular-matrix-by-elementary-transformation_4023", "openwebmath_score": 0.4767606556415558, "openwebmath_perplexity": 4005.3398205093004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718480899426, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8733140112800316}}
{"url": "https://math.stackexchange.com/questions/441106/how-do-i-show-that-int-0-infty-frac-sinax-sinbxx2-mathrm-dx", "text": "# How do I show that $\\int_0^\\infty \\frac{\\sin(ax) \\sin(bx)}{x^{2}} \\, \\mathrm dx = \\pi \\min(a,b)/2$\n\nRecently I found a claim saying that $$\\int_0^\\infty \\left( \\frac{\\sin ax}{x}\\right)\\left( \\frac{\\sin bx}{x}\\right) \\mathrm{d}x= \\pi \\min(a,b)/2$$ from what I can see this seems to be true. I already know that $\\int_{0}^\\infty \\operatorname{sinc}xy\\,\\mathrm{d}y = \\pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.\n\nCan someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$\\int_{0}^{\\infty} \\left( \\prod_{k=1}^N \\frac{\\sin (a_k \\cdot x)}{x} \\right) \\,\\mathrm{d}x$$ Where $a_k, \\cdots, a_N$ are arbitrary positive constants. It seems related to the Borwein Integral, but there are some subtle differences.\n\n\u2022 Don't you want a product, not a summation, in your last question? And, when you write \"independent of $y$\" don't you mean independent of $x$? \u2013\u00a0Gerry Myerson Jul 11 '13 at 9:04\n\u2022 Does $\\sin p\\sin q=(\\cos(p-q)-\\cos(p+q))/2$ help? \u2013\u00a0Gerry Myerson Jul 11 '13 at 9:07\n\nOne way is to to this by residues. Another way to integrate once by parts to get $$I=\\int_0^{\\infty}\\frac{b\\sin ax\\cos bx+a\\cos ax \\sin bx}{x}dx,$$ then to use the formula $2\\sin \\alpha\\cos\\beta=\\sin(\\alpha+\\beta)+\\sin(\\alpha-\\beta)$ and the mentioned integral (note that your formula needs to be corrected on the left and on the right) $$\\displaystyle \\int_0^{\\infty}\\frac{\\sin xy}{x}\\,dx=\\frac{\\pi}{2}\\mathrm{sgn}(y).$$ This gives \\begin{align}I&=\\frac{\\pi}{4}\\Bigl[b\\,\\mathrm{sign}(a+b)+b\\,\\mathrm{sign}(a-b)+a\\,\\mathrm{sign}(a+b)+a\\,\\mathrm{sign}(b-a)\\Bigr]=\\\\ &=\\frac{\\pi}{4}\\left(|a+b|-|a-b|\\right), \\end{align} For $a,b>0$ the last expression is obviously equal to $\\pi\\min\\{a,b\\}/2$.\n\n\u2022 Just posted the question as a dupe, such an amazing result. Do you know who first discovered it? \u2013\u00a0aronp Dec 21 '15 at 21:33\n\u2022 Since this result is independent of the lower number, presumably this works for the three factor case? \u2013\u00a0aronp Dec 21 '15 at 21:46\n\u2022 @aronp No I don't; I totally agree with you that it is quite counterintuitive. \u2013\u00a0Start wearing purple Dec 21 '15 at 21:47\n\nIn several steps:\n\n\u2022 Trigonometric relation $$\\left( \\frac{\\sin ax}{x}\\right)\\left( \\frac{\\sin bx}{x}\\right)=\\frac{1-\\cos(a+b)x}{2x^2}-\\frac{1-\\cos(a-b)x}{2x^2}$$\n\u2022 Dirichlet integral $$\\int_0^\\infty \\frac{\\sin \\alpha t}{t}dt=\\frac{\\pi}{2}\\mathrm{sgn}(\\alpha)$$\n\u2022 Integration by parts $$\\int_0^\\infty\\frac{1-\\cos(\\alpha)t}{t^2}=\\alpha\\int_0^\\infty \\frac{\\sin \\alpha t}{t}dt=\\frac{\\pi}{2}|\\alpha|$$\n\u2022 Combine $$\\int_0^\\infty \\left( \\frac{\\sin ax}{x}\\right)\\left( \\frac{\\sin bx}{x}\\right) \\mathrm{d}x=\\frac{\\pi}{4}(|a+b|-|a-b|)=\\frac{\\pi}{2}\\min(a,b)$$\n\u2022 Excellent!${}{}{}$ \u2013\u00a0Namaste May 20 '14 at 20:14\n\nA very easy way to see this is to use Parseval's theorem for Fourier transforms. In general, Parseval's theorem states that, for two functions $f$ and $g$, each having respective FTs $\\hat{f}$ and $\\hat{g}$, related by\n\n$$\\hat{f}(k) = \\int_{-\\infty}^{\\infty} dx \\, f(x) \\, e^{i k x}$$\n\netc., then\n\n$$\\int_{-\\infty}^{\\infty} dx \\, f(x) \\bar{g}(x) = \\frac{1}{2 \\pi} \\int_{-\\infty}^{\\infty} dk \\, \\hat{f}(k) \\bar{\\hat{g}}(k)$$\n\nThe FT of $\\sin{(a x)}/x$ is given by\n\n$$\\int_{-\\infty}^{\\infty} dx \\, \\frac{\\sin{a x}}{x} e^{i k x} = \\begin{cases} \\pi & |k| \\lt a \\\\ 0 & |k| \\gt a\\end{cases}$$\n\nSimilarly,\n\n$$\\int_{-\\infty}^{\\infty} dx \\, \\frac{\\sin{b x}}{x} e^{i k x} = \\begin{cases} \\pi & |k| \\lt b \\\\ 0 & |k| \\gt b\\end{cases}$$\n\nBy Parseval, we take the integral of the product of the transforms, which is the product of two rectangles. The product is clearly nonzero over the smaller of the two widths, i.e. $2 \\min\\{a,b\\}$. Thus,\n\n$$\\int_{-\\infty}^{\\infty} dx \\, \\frac{\\sin{a x}}{x} \\, \\frac{\\sin{b x}}{x} = \\frac{\\pi^2}{2 \\pi} 2 \\min\\{a,b\\}$$\n\nTherefore\n\n$$\\int_{0}^{\\infty} dx \\, \\frac{\\sin{a x}}{x} \\, \\frac{\\sin{b x}}{x} = \\frac{\\pi}{2} \\min\\{a,b\\}$$\n\nLet $b_{k} >0$ and $a \\ge \\sum_{k=1}^{n} b_{k}$.\n\nGeneralizing the answer HERE, we can use contour integration to quickly show that\n\n$$\\int_{0}^{\\infty} \\frac{\\sin(ax)}{x} \\prod_{k=1}^{n} \\frac{\\sin \\left( b_{k}x \\right)}{x} \\, dx= \\frac{1}{2} \\int_{-\\infty}^{\\infty} \\frac{\\sin(ax)}{x} \\prod_{k=1}^{n} \\frac{\\sin \\left( b_{k}x \\right)}{x} \\, dx = \\frac{\\pi}{2} \\prod_{k=1}^{n}b_{k}. \\tag{1}$$\n\nUnder the conditions stated above, the function $$e^{iaz} \\prod_{k=1}^{n} \\sin \\left( b_{k}x \\right) = e^{iaz} \\prod_{k=1}^{n} \\frac{e^{ib_{k}z}-e^{-ib_{k}z} }{2i}$$ is bounded in the upper half-plane.\n\nSo by integrating the function $$f(z) = \\frac{e^{iaz}}{z} \\prod_{k=1}^{n} \\frac{\\sin \\left( b_{k}z \\right)}{z}$$ around an indented contour that consists of the real axis and the semiciricle above it, we get (in the limit), $$\\text{PV} \\int_{-\\infty}^{\\infty} \\frac{e^{iax}}{x} \\prod_{k=1}^{n} \\frac{\\sin \\left( b_{k}x \\right)}{x} \\, dx - \\pi i \\, \\text{Res} [f(z), 0] = 0,$$ where $$\\text{Res}[f(z) ,0] = \\lim_{z \\to 0}e^{iaz} \\prod_{k=1}^{n}\\frac{\\sin \\left( b_{k}z \\right)}{z} =1\\times \\prod_{k=1}^{n} b_{k}.$$\n\nTaking the imaginary parts of both sides of equation leads to the result.\n\n\nIt's sufficient to consider the case $\\ds{\\bbox[10px,#ffe,border:1px dotted navy] {\\ds{a, b \\in \\mathbb{R}_{\\ >\\ 0}}}}$.\n\n\\begin{align} \\mbox{Note that}\\ &\\int_{0}^{\\infty}{\\sin\\pars{ax} \\over x}\\,{\\sin\\pars{bx} \\over x}\\,\\dd x = {1 \\over 2}\\,b\\int_{-\\infty}^{\\infty}{\\sin\\pars{ax} \\over ax} \\,{\\sin\\pars{\\bracks{b/a}ax} \\over \\pars{b/a}ax}\\,a\\,\\dd x \\\\[5mm] = &\\ {1 \\over 2}\\,b\\int_{-\\infty}^{\\infty}{\\sin\\pars{x} \\over x} \\,{\\sin\\pars{x/\\mu} \\over x/\\mu}\\,\\dd x \\,,\\qquad\\mu \\equiv {a \\over b} > 0 \\label{1}\\tag{1} \\end{align}\n\n\\begin{align} &\\left.\\int_{-\\infty}^{\\infty}{\\sin\\pars{x} \\over x} \\,{\\sin\\pars{x/\\mu} \\over x/\\mu}\\,\\dd x \\,\\right\\vert_{\\ \\mu\\ >\\ 0} = \\int_{-\\infty}^{\\infty} \\pars{{1 \\over 2}\\int_{-1}^{1}\\expo{\\ic kx}\\,\\dd k} \\pars{{1 \\over 2}\\int_{-1}^{1}\\expo{-\\ic qx/\\mu}\\,\\dd q}\\dd x \\\\[5mm] = &\\ {1 \\over 2}\\,\\pi \\int_{-1}^{1}\\int_{-1}^{1}\\int_{-\\infty}^{\\infty}\\expo{\\ic\\pars{k - q/\\mu}x} {\\dd x \\over 2\\pi}\\,\\dd k\\,\\dd q = {1 \\over 2}\\,\\pi\\int_{-1}^{1}\\int_{-1}^{1}\\delta\\pars{k - q/\\mu}\\,\\dd k\\,\\dd q \\\\[5mm] = &\\ {1 \\over 2}\\,\\pi\\int_{-1}^{1}\\bracks{-1 < {q \\over \\mu} < 1}\\dd q = \\pi\\int_{0}^{1}\\bracks{q < \\mu}\\dd q = \\pi\\braces{\\bracks{\\mu < 1}\\int_{0}^{\\mu}\\dd q + \\bracks{\\mu > 1}\\int_{0}^{1}\\dd q} \\\\[5mm] = &\\ \\bracks{a < b}\\pi\\,{a \\over b} + \\bracks{a > b}\\pi\\label{2}\\tag{2} \\end{align}\nWith \\eqref{1} and \\eqref{2}: \\begin{align} \\int_{0}^{\\infty}{\\sin\\pars{ax} \\over x}\\,{\\sin\\pars{bx} \\over x}\\,\\dd x & = {1 \\over 2}\\,\\pi\\braces{\\vphantom{\\Large A}\\bracks{a < b}a + \\bracks{a > b}b} = \\bbx{\\ds{{1 \\over 2}\\,\\pi\\,\\min\\braces{a,b}}} \\end{align}", "date": "2019-07-18 13:10:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/441106/how-do-i-show-that-int-0-infty-frac-sinax-sinbxx2-mathrm-dx", "openwebmath_score": 0.9986222982406616, "openwebmath_perplexity": 869.5109110620803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353126336, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8733077357242699}}
{"url": "https://math.stackexchange.com/questions/1585795/how-is-940-equiv-1-pmod-100", "text": "How is $9^{40}\\equiv\\ 1 \\pmod {100}$?\n\nBy Euler's theorem, $a^{\\varphi(100)} \\equiv\\ 1 \\pmod {100}$.\n\nWe know that the last two digits of $9^{40}$ are non-zero. So they can't even be $01$.\n\nSince $1\\equiv\\ 1 \\pmod {100}$, how come $9^{40}\\equiv\\ 1 \\pmod {100}$?\n\nI have looked at: Find the last two digits of $9^{{9}^{9}}$,\n\nFind the last two digits of $9^{9^{9}}$\n\n\u2022 \"We know that the last two digits of $9^{40}$ are non-zero.\" That doesn't mean neither digit is zero; it just means they can't both be zero. So why do you think it can't be $01$? \u2013\u00a0Erick Wong Dec 22 '15 at 17:14\n\u2022 you're right i was thinking only of the the units digit \u2013\u00a0AkaiShuichi Dec 22 '15 at 17:17\n\u2022 @pyUser: When a more experienced user edits your post, learn from it, don't roll back to your own improperly formatted version. \u2013\u00a0Alex M. Dec 22 '15 at 17:18\n\u2022 @AlexM. can you teach me how to apply edits from one version to another one? or did you just manually MathJax the thing? \u2013\u00a0gt6989b Dec 22 '15 at 17:19\n\u2022 You quoted one way of doing it. We have $\\varphi(100)=\\varphi(25)\\varphi(4)=40$. So $9^{40}\\equiv 1\\pmod{100}$, the last two digits, going righttwards, are $0$ and $1$. \u2013\u00a0Andr\u00e9 Nicolas Dec 22 '15 at 17:24\n\nBy binomial theorem:\n\n$$(10-1)^{40} = \\sum_{i=0}^{40} \\binom{40}{i}(-1)^{40-i}10^{i}$$\n\nModulo $10^2$, you only need to look at $i=0,1$.\n\nSo: $$9^{40}\\equiv (-1)^{40} + \\binom{40}{1}(-1)^{39}\\cdot 10 \\equiv (-1)^{40}\\equiv 1\\pmod{100}$$\n\nNote that this works for $9^{10}$, too.\n\n$9^2\\equiv\\ -19\\ (mod100)$\n\n$9^4 \\equiv\\ 61\\ (mod 100)$\n\n$9^8\\equiv\\ 21 \\ (mod 100)$\n\n$9^{10}\\equiv\\ 1 \\ (mod 100)$\n\n$9^{40}\\equiv\\ 1 \\ (mod 100)$\n\n\u2022 Better formatting if you write 9^2\\equiv -19\\pmod{100} yielding $$9^2\\equiv -19\\pmod{100}$$ \u2013\u00a0Thomas Andrews Dec 22 '15 at 18:01\n\nCalculate $$9^{10} = 3486784401$$ So $$9^{40} \\mod 100 \\equiv (9^{10}\\mod 100)^4 \\mod 100$$ $$\\equiv(1)^4 \\mod 100 \\equiv 1 \\mod 100$$\n\n\u2022 How did you go about your third step? \u2013\u00a0AkaiShuichi Dec 22 '15 at 17:20\n\u2022 $x^y \\mod m$ is the same as $(x \\mod m)^y \\mod m$. You can move the $\\mod$ function into multiplies, etc. \u2013\u00a0amcalde Dec 22 '15 at 17:22\n\u2022 i was talking about the (1)^4 part \u2013\u00a0AkaiShuichi Dec 22 '15 at 17:23\n\u2022 In the first calculation I showed that $9^{10} \\equiv 1 \\mod 100$. \u2013\u00a0amcalde Dec 22 '15 at 17:24\n\nNote that $$99^2=9801\\equiv 1\\pmod{100}\\implies99^{40}\\equiv 1\\pmod{100}\\implies9^{40}11^{40}\\equiv 1\\pmod{100}$$\n\n$$9^{40}\\equiv 1\\pmod{100}$$\n\nIt is straightforward $9^{40} = 81^{20} = 6561^{10} \\equiv 61^{10} \\pmod{100} = 3721^5 \\pmod{100} \\equiv\\ 21^5 \\pmod{100}$\n$= 441 \\times 441 \\times 21 \\pmod{100}\\equiv\\ 41 \\times 41 \\times 21 \\pmod{100}$\n$= 35301 \\pmod{100} \\equiv\\ 1 \\ \\pmod{100}$\n\nand this is just one of the many many routes you could take.\n\nIf you want to avoid the equation\n\n$9^{40}=147808829414345923316083210206383297601$,\n\nthe question is how low you prefer your numbers to be.\n\nOne alternative is using $40=4*2*5$ (and 7 digits for the last operation).\n\n$9^4=61$ (mod 100)\n\n$61^2=21$ (mod 100), and\n\n$21^5=01$ (mod 100)", "date": "2021-05-10 19:25:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1585795/how-is-940-equiv-1-pmod-100", "openwebmath_score": 0.580910861492157, "openwebmath_perplexity": 679.315508833003, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363503693294, "lm_q2_score": 0.8872045922259088, "lm_q1q2_score": 0.8733077303425602}}
{"url": "https://math.stackexchange.com/questions/2772749/finding-coefficients-in-polynomial-given-three-tangents", "text": "# Finding coefficients in polynomial given three tangents\n\nI am stuck with a problem I simply cannot solve.\n\nI have to find the coefficients of a quadratic polynomial given three tangents. The problem is stated as follows:\n\nThe three lines described by the equations\n\n$y_1(x)=-4x-16.5$\n\n$y_2(x)=2x-4.5$\n\n$y_3(x)=6x-16.5$\n\nare all tangents to a quadratic polynomial $p(x)=ax^2+bx+c$\n\nDetermine the values of the coefficients a, b & c.\n\nI simply cannot solve this problem, I've been at it for a long time. Any help is greatly appreciated :)\n\nEdit: I'm including the way I tried to solve it. I didn't get super far.\n\nGiven the polynomial $p(x)$ I know that $p'(x)=2ax+b$\n\nTherefore, the following is true for the three points with x-values of $x_1, x_2$ and $x_3$, where the lines $y_1, y_2$ and $y_3$ are tangent to the parabola:\n\n$-4=2ax_1+b$\n\n$2=2ax_2+b$\n\n$6=2ax_3+b$\n\nThat's all I've managed to do. I've also found the points where the three lines intersect (well, three points two of the lines intersect), but I can't think of how to use that for anything.\n\n\u2022 Show anyway what you tried. I can give a solution using derivatives, but maybe it is not the solution you want. \u2013\u00a0N74 May 8 '18 at 20:23\n\u2022 Welcome to SE! Even if your attempts didn't work, you should absolutely include them in your post - people can advise you better that way! \u2013\u00a0B. Mehta May 8 '18 at 20:23\n\u2022 @N74 I have now included my failed solution :) As you can see, I thought of using derivatives as well, but I can't get further. \u2013\u00a0Mads Clausen May 8 '18 at 20:38\n\u2022 My maths teacher (I'm in the last year of Danish \"high school\") gave it to us for tomorrow. He said that it was definitely doable. \u2013\u00a0Mads Clausen May 8 '18 at 20:42\n\nIf we have a parabola $$y = a x^2 + b x + c$$ and a line $$y = mx + d,$$ they are tangent if $$b^2 - 4ac = -4da + 2mb - m^2$$ as then the parabola $$y = a x^2 + (b-m) x + (c -d)$$ has a double root, i.e. is a constant times a square, $$a (x+p)^2$$ Might as well write this: I fixed $$\\Delta = b^2 - 4 a c$$ and then had three equations $$-4da + 2mb = m^2 + \\Delta$$ by plugging in the values from the three lines $y = mx + d.$ I expected bigger problems, but just taking the differences of two of the equations cancels the extra unknown $\\Delta,$ alowing us to find $a,b$ quickly. From that, we finally get a value for $\\Delta,$ after which we have one equation for $c$\n\n$$66a - 8 b = 16 + \\Delta \\; ,$$ $$66a + 12 b = 36 + \\Delta \\; ,$$ $$18a + 4 b = 4 + \\Delta \\; .$$ Second minus first gives $b.$ Plug in the $b$ value, then subtract second minus third, which gives $a.$ Plug both into any of the three to find $\\Delta.$ Finally, $c = \\frac{b^2 - \\Delta}{4a}$\n\nNote that if a line and a quadratic are tangent $mx+d=ax^2+bx+c$ then the following quadratic will have discriminant zero \\begin{eqnarray*} ax^2+(b-m)x+c-d=0. \\end{eqnarray*} This will lead to $3$ equations for $a,b,c$ that are easily solved giving\n\n$(a,b,c)=(1/2,1,-4)$.\n\n\u2022 Thanks for the answer. I don't quite follow, however: what are the three equations for a, b and c that you mention? \u2013\u00a0Mads Clausen May 8 '18 at 21:02\n\u2022 \\begin{eqnarray*} (b+4)^2=2a(c+33) \\\\ (b-2)^2=2a(c+9) \\\\ (b-6)^2=2a(c+33) \\end{eqnarray*} combine the first & the third of these equations & you will rapidly yield a value for $b$. \u2013\u00a0Donald Splutterwit May 8 '18 at 21:05\n\u2022 Donald I get different final abc values, although still all rational and easily found... I kept the discriminant $\\Delta = b^2 - 4ac \\;$ as an unknown until the end, began with three $\\Delta = -4da + 2mb - m^2$ or $-4da + 2mb = m^2 + \\Delta$ \u2013\u00a0Will Jagy May 8 '18 at 21:09\n\u2022 where does $2a(c+n)$ come from? I recall that the discriminant is equal to $b^2-4ac$, which in this case would yield something like $(b-m)^2=4a(c-d)$ I would think it would either be that or $(b-m)^2=2a(2c-2d)$ \u2013\u00a0Mads Clausen May 8 '18 at 21:12\n\u2022 I get 1/2, 1 and -4 for a,b,c. 2ax+b = -4 at x=-5; 2 at x = 1 and 6 at x = 5 with x and y values being the same for tangent lines and parabola at these tangent points. \u2013\u00a0Phil H May 8 '18 at 21:32\n\nGiven any two tangents to a parabola of the form $y = f(x) = ax^2 + bx + c,$ the $x$-coordinate of the intersection of the tangent lines will be midway between the $x$-coordinates of the tangent points.\n\nWorking out the intersection points of the three lines, we can see that the intersection of $y = y_1(x)$ and $y = y_2(x)$ occurs at $x = -2$ and the intersection of $y = y_2(x)$ and $y = y_3(x)$ occurs at $x = 3.$ Let\n\n\u2022 $x = x_1$ at the tangent point with $y = y_1(x)$;\n\u2022 $x = x_2$ the tangent point with $y = y_2(x)$; and\n\u2022 $x = x_3$ the tangent point with $y = y_3(x).$\n\nDue to the $y$-coordinates and slopes at the intersection points, it is clear that $x_1 < -2 < x_2 < 3 < x_3$; moreover, $-2$ is midway between $x_1$ and $x_2$ and $3$ is midway between $x_2$ and $x_3.$ It follows that $$x_3 - x_1 = 2\\left(\\frac{x_3 + x_2}2 - \\frac{x_2 + x_1}2\\right) = 2(3 - (-2)) = 10.$$\n\nThe tangency condition implies that $f'(x_1) = y_1'(x) = -4$ and $f'(x_3) = y_3'(x) = 6.$ But $f'(x) = 2ax + b,$ so $20a = 2a(x_3 - x_1) = f'(x_3) - f'(x_1) = 10,$ and therefore $a = \\frac12.$ It is then a relatively straightforward exercise to find the other two coefficients.", "date": "2019-10-21 05:19:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2772749/finding-coefficients-in-polynomial-given-three-tangents", "openwebmath_score": 0.832466721534729, "openwebmath_perplexity": 287.7911885171946, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363494503271, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8733077265915363}}
{"url": "https://math.stackexchange.com/questions/4495973/the-product-of-primes-between-n-and-2n-compared-to-2n", "text": "# The Product of Primes Between $N$ and $2N$ Compared to $2^N$\n\nWhile reading some course notes from MIT 18.703 (Modern Algebra) on a primality test, I found this lemma (stated without proof):\n\nLemma 22.3. The product of all prime numbers $$r$$ between $$N$$ and $$2N$$ is greater than $$2^{N}$$ for all $$N\\geq1$$.\n\nHowever, one quickly finds that this lemma is false for $$N=8$$. The primes between $$8$$ and $$16$$ are $$11$$ and $$13$$, and $$11\\cdot13 = 143 < 256 = 2^{8}$$.\n\nI wondered if there were any more counterexamples, so I decided to write a quick program to test this lemma (the code may be found here). My code is not very optimized, with only very basic parallelism using OpenMP, and so I haven't taken the time to go beyond $$N = 100000$$. With my code, I have found only three counterexamples:\n\nAt $$N = 8$$, the product of primes between $$8$$ and $$16$$ is $$143 < 256 = 2^{8}$$.\n\nAt $$N = 14$$, the product of primes between $$14$$ and $$28$$ is $$7429 < 16384 = 2^{14}$$.\n\nAt $$N = 20$$, the product of primes between $$20$$ and $$40$$ is $$765049 < 1048576 = 2^{20}$$.\n\nMy Questions\n\n1. Are these the only counterexamples?\n\n2. If so, how do we prove it?\n\n3. If not, what are some others, and are there infinitely many? -- Answer: There are not infinitely many, though there may still be further counterexamples.\n\nEDIT 1: Thanks to the responses to this post, I have attempted an analysis of this problem. If possible, I would appreciate any feedback, just in case I made an error or if any improvement could be made by a more careful analysis. -- This analysis was flawed and has been retracted.\n\nEDIT 2: Due to the answer posted by jjagmath, the bound has been lowered to 10544111. I will try to run my program to search for counterexamples below this number, but hopefully even tighter analyses come that lower the bound to a more tractable number to search.\n\nEDIT 3: I retract the analysis that I did because of a mistake. The jjagmath analysis still holds.\n\nEDIT 4:\n\nLet $$p_{k}$$ be the $$k$$th prime number, let $$\\displaystyle{N_{k} = \\frac{p_{k} - 1}{2}}$$, and let $$\\displaystyle{f\\left(N\\right)=\\prod_{N\\leq p\\leq 2N}p}$$.\n\nProposition. Let $$k > 2$$ be an integer and let $$\\nu$$ be an integer such that $$N_{k-1} < \\nu\\leq N_{k}$$. Then, $$f\\left(N_{k}\\right)\\leq f\\left(\\nu\\right)$$.\n\nProof. Any prime in the interval $$\\left[N_{k},2N_{k}\\right]$$ is also in the interval $$\\left[\\nu, 2\\nu\\right]$$, and so $$f\\left(\\nu\\right)$$ can only get smaller as $$\\nu$$ increases to $$N_{k}$$. QED\n\nCorollary. If any counterexamples above $$20$$ exist, some of them must be of the form $$\\displaystyle{N_{k} = \\frac{p_{k} - 1}{2}}$$ for some $$k$$.\n\nObservation. All counterexamples currently known are in this form: $$N_{7} = 8$$, $$N_{10} = 14$$, and $$N_{13} = 20$$.\n\nAt the time of writing, it is known that all integers $$N\\geq 10544111$$ satisfy $$f\\left(N\\right)\\geq 2^{N}$$, and so we need only check $$N_{k} < 10544111$$ ($$k < 698306$$) to see if there are any further counterexamples.\n\nQuestion. Are there better explicit lower bounds for $$N$$ beyond which the inequality $$\\displaystyle{\\prod_{N\\leq p\\leq 2N}p \\geq 2^{N}}$$ holds?\n\nIf we can find a more computationally tractable bound, say $$N\\geq 1299709$$ (so $$k \\geq 100000$$ can be ruled out of the search), then I may be able to run (a modified form of) my program to search for possible counterexamples beyond $$N = 20$$. Of course, any improvement is welcome!\n\nEDIT 5:\n\nWith Gary's comment, the bound has been tightened to $$N\\geq 678407,$$ which should be tractable. I will be running my program overnight and will update with the results!\n\nEDIT 6:\n\nGary's latest answer has finally finished this problem off! The bound has been tightened to $$N \\geq 328$$ via Chebyshev function bounds, and finally to all $$N\\geq1$$ except $$N=8,14,20$$ by numerical computation.\n\nThis has been a very fun problem to work on, and I thank everyone who was involved! I suppose I will end this post with a revised Lemma 22.3.\n\nLemma 22.3.$$'$$ The product of all prime numbers $$r$$ between $$N$$ and $$2N$$ is greater than $$2^{N}$$ for all $$N\\geq1$$, except for $$N = 8, 14, 20$$.\n\nEDIT 7:\n\nI went back to my initial analysis where I had made a very trivial error, and it turns out that I would have already had a tractable bound if I decided to look at it again. My analysis seems to show that $$f\\left(N\\right)\\geq 2^{N}$$ for all $$N\\geq 1845$$.\n\nGary's analysis still provides a better bound, but I'm happy to know that I would have eventually solved this with just another look at my own work.\n\n\u2022 @BeKind Thanks! Jul 19 at 14:06\n\u2022 I think asymptotically it might be true. That is, if we say there are approximately $\\frac{2N}{\\log(2N)} - \\frac{N}{\\log(N)}$ primes between $N$ and $2N$. Each prime is $\\geq N.$ We might be able to show that for large enough $M,\\$ the product of these primes is greater than $N^{\\frac{2N}{\\log(2N)} - \\frac{N}{\\log(N)}}$ which is greater than $2^N$ for all $N\\geq M.$ I'm not sure of this but it could be worth investigating... Jul 19 at 14:38\n\u2022 @AdamRubinson That's true, the Prime Number Theorem does give us a lot of primes between $N$ and $2N$ for sufficiently large $N$. I do think that the counterexamples I've found are the only ones, but I'm curious to see a proof. Jul 19 at 14:53\n\nAs Adam Rubinson mentions in the comments, the number of primes up to $$N$$ is $$\\frac N{\\log N}(1+o(1))$$, where $$o(1)$$ denotes a quantity that tends to $$0$$ as $$N\\to\\infty$$. Therefore the number of primes between $$N$$ and $$2N$$ is $$\\frac{2N}{\\log(2N)}(1+o(1))-\\frac{N}{\\log N}(1+o(1))=\\frac{N}{\\log N}(1+o(1))$$. Since any such prime is at least $$N$$, their product $$P$$ satisfies $$\\log P>\\log\\big(N^{\\frac{N}{\\log N}(1+o(1))}\\big) = \\frac{N}{\\log(N)}(1+o(1))\\log N = N(1+o(1))$$. Therefore for sufficiently large $$N$$ we have $$\\log P>N\\log 2=\\log(2^N)$$, which implies $$P>2^N$$. This shows that there are only finitely many counterexamples.\n\nReverse engeneering the proof, it will work as soon as the number of primes between $$N$$ and $$2N$$ is at least $$\\frac N{\\log N}\\log 2$$; since $$\\log 2\\approx 0.7$$ is significantly less than $$1$$, it should be possible to figure out explicitly all the counterexamples.\n\n\u2022 You say \"as soon as\", but the number of primes between $N$ and $2N$ is not monotonic in $N$; so there might conceivably be a later counterexample. Jul 19 at 17:14\n\u2022 All that is needed to complete this are explicit upper and lower bounds for $\\pi(x)$. Some results are listed on wikipedia en.wikipedia.org/wiki/Prime-counting_function#Inequalities, from there getting a feasible explicit bound is just a matter of calculation. Jul 19 at 21:54\n\u2022 @TonyK That was, perhaps, a poor choice of words. To be clear, it is true that there is some $N_0$ such that for all $N>N_0$ the number of primes between $N$ and $2N$ is at least $\\frac N{\\log N}\\log 2$. Such an $N_0$ can be found explicitly using the estimates linked by sbares. Jul 20 at 9:13\n\nWe want to estimate $$f(N)=\\prod_{N\n\nTaking $$\\log$$ we have $$\\log f(N) = \\sum_{N where $$\\theta$$ is the Chebyshev function.\n\nWe can use the known bound for $$\\theta$$:\n\n$$|\\theta(x)-x|\\le .006788 \\frac{x}{\\log x}$$ valid for $$x \\ge 10544111$$ to get $$\\theta(2N)-\\theta(N)\\ge N -.006788\\frac{N}{\\log N}\\left(1+\\frac{2 \\log N}{\\log(2N)}\\right)\\ge N-.006788\\frac{3N}{\\log N}\\ge N \\log 2$$\n\nfor $$N\\ge 10544111$$.\n\nSo the inequality $$f(N)\\ge 2^N$$ is true for all $$N\\ge 10544111$$.\n\n\u2022 A somewhat better result comes from the 1975 paper of J. Barkley Rosser and Lowell Schoenfeld. Namely $\\left| {\\theta (x) - x} \\right| < \\frac{1}{{40}}\\frac{x}{{\\log x}}$ for $x \\ge 678407$. Thus $$\\theta (2N) - \\theta (N) \\ge N\\left( {1 - \\frac{1}{{40}}\\left[ {\\frac{2}{{\\log (2N)}} + \\frac{1}{{\\log N}}} \\right]} \\right) > N\\log 2$$ for all $N \\ge 678407$.\n\u2013\u00a0Gary\nJul 21 at 4:42\n\u2022 @Brian Are you able to check the cases $N<678407$ using a computer programme?\n\u2013\u00a0Gary\nJul 21 at 5:49\n\u2022 @Gary I should be able to. After optimizing a bit, I was able to get to about 250,000 in half an hour. I'll try running it overnight to see if I get there. Thanks so much! Jul 21 at 5:53\n\nLet $$\\theta (x) = \\sum\\limits_{p \\le x} {\\log p}$$ be the Chebyshev function of the first kind. The product of the primes between $$N$$ and $$2N$$ (with $$N\\geq 2$$) is $$\\prod\\limits_{N < p < 2N} p = \\prod\\limits_{N < p \\le 2N} p = \\exp (\\theta (2N) - \\theta (N)).$$ By Corollary $$11.2$$ in this paper, we have $$\\left| {\\theta (x) - x} \\right| \\le 3.965\\frac{x}{{\\log ^2 x}}$$ for all $$x\\geq 2$$. Hence, $$\\theta (2N) - \\theta (N) \\ge N\\left( {1 - 3.965\\!\\left[ {\\frac{2}{{\\log ^2 (2N)}} + \\frac{1}{{\\log ^2 N}}} \\right]} \\right) > N\\log 2$$ for all $$N \\ge 328$$. Consequently, $$\\prod\\limits_{N < p < 2N} p > 2^N$$ for $$N \\ge 328$$. Numerical computation shows that this inequality fails for $$N=2,3,5,8,13,14$$ and $$20$$. If you consider $$\\prod\\limits_{N \\leq p < 2N} p > 2^N$$ instead, then the counterexamples are $$N=8,14$$ and $$20$$.\n\n\u2022 This is incredible, thank you! I had reached the previous bound over night after 24163.1 seconds (6h42m43.1s), but now this bound completely negates the need for that. I am very happy with this result, so I have accepted this answer. Jul 21 at 13:05\n\nEDIT: I have fixed the error in my analysis, and actually got a very tractable bound, so I have undeleted this corrected answer.\n\nThanks to the others who responded, I have the following answer.\n\nUsing the top inequality found here, we have that the number of primes in the interval $$\\left[N,2N\\right]$$ is\n\n$$\\pi\\left(2N\\right)-\\pi\\left(N\\right) > \\frac{2N}{\\log\\left(2N\\right)} - \\frac{1.25506N}{\\log\\left(N\\right)}$$\n\nfor $$N\\geq9$$.\n\nWe want to show that there is an $$M$$ such that $$N^{\\pi\\left(2N\\right)-\\pi\\left(N\\right)}\\geq 2^{N}$$, and hence $$\\left(\\pi\\left(2N\\right)-\\pi\\left(N\\right)\\right)\\log\\left(N\\right)\\geq N\\log\\left(2\\right)$$, is true for all $$N\\geq M$$.\n\nUsing the inequality above, we can need only find $$N$$ such that\n\n$$\\left(\\frac{2N}{\\log\\left(2N\\right)} - \\frac{1.25506N}{\\log\\left(N\\right)}\\right)\\log\\left(N\\right)\\geq N\\log\\left(2\\right).$$\n\nExpanding the left hand side algebraically, we get\n\n$$\\left(\\frac{2N}{\\log\\left(N\\right) + \\log\\left(2\\right)} - \\frac{1.25506N}{\\log\\left(N\\right)}\\right)\\log\\left(N\\right) =$$ $$\\left(\\frac{2N\\log\\left(N\\right) - 1.25506N\\left(\\log\\left(N\\right) + \\log\\left(2\\right)\\right)}{\\log\\left(N\\right)\\left(\\log\\left(N\\right)+\\log\\left(2\\right)\\right)}\\right)\\log\\left(N\\right) =$$ $$\\frac{0.74494N\\log\\left(N\\right) - 1.25506N\\log\\left(2\\right)}{\\log\\left(N\\right)+\\log\\left(2\\right)}$$\n\nSince we want this to be $$\\geq N\\log\\left(2\\right)$$, we may divide both sides by $$N$$ to get\n\n$$\\frac{0.74494\\log\\left(N\\right) - 1.25506\\log\\left(2\\right)}{\\log\\left(N\\right)+\\log\\left(2\\right)}\\geq \\log\\left(2\\right).$$\n\nThe left hand side is increasing, so we need only find the first $$N$$ so that this inequality is satisfied. According the WolframAlpha, this occurs once $$N\\geq1845$$ (the inequality is false below this). So we just need to use a computer to test this up to that limit.", "date": "2022-09-30 22:03:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4495973/the-product-of-primes-between-n-and-2n-compared-to-2n", "openwebmath_score": 0.9228746891021729, "openwebmath_perplexity": 126.90543573489606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620562254525, "lm_q2_score": 0.9099070084811307, "lm_q1q2_score": 0.8732942214338003}}
{"url": "https://web2.0calc.com/questions/if-greg-rolls-four-fair-six-sided-dice-what-is-the", "text": "+0\n\n# If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?\n\n+1\n1668\n11\n+473\n\nIf Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?\n\nFeb 3, 2018\n\n#1\n+109721\n0\n\nIf Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?\n\nThanks Alan.\n\n.\nFeb 3, 2018\nedited by Melody \u00a0Feb 5, 2018\n#2\n+473\n+2\n\nRektTheNoob \u00a0Feb 3, 2018\nedited by RektTheNoob \u00a0Feb 3, 2018\n#3\n+109721\n0\n\nok so what do you think the correct answer is and\u00a0 what makes you think it is correct?\n\nAND if you have the working we would like to see that too.\n\nI, and others, would like to learn too.\n\nMelody \u00a0Feb 4, 2018\n#5\n+473\n+3\n\nSolution:\n\nWe notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of each, one of each, or none of each. If he rolls two of each, there are\u00a0$$\\binom{4}{2}=6$$\u00a0ways to choose which two dice roll the 1's. If he rolls one of each, there are\u00a0$$\\binom{4}{1}\\binom{3}{1}=12$$\u00a0ways to choose which dice are the 6 and the 1, and for each of those ways there are\u00a0$$4\\cdot4=16$$\u00a0ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are\u00a0$$4^4=256$$\u00a0possible values for the dice. In total, there are\u00a0$$6+12\\cdot16+256=454$$\u00a0ways Greg can roll the same number of 1's and 6's. There are\u00a0$$\\dfrac{1}{2} \\left(1-\\dfrac{454}{1296}\\right)=\\boxed{\\dfrac{421}{1296}}$$\u00a0total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is\u00a0.\n\nRektTheNoob \u00a0Feb 5, 2018\n#7\n-1\n\nRektTheNoob:\n\nIf you know the answers to the questions you post on this forum in such detail as the above answer demonstrates, then what is the PURPOSE of posting your questions in the first place?? Are you trying to test the Mods and other\u00a0volunteers'\u00a0mathematical knowledge? Or are you trying to show your mathematical superiority??!!\n\nGuest\u00a0Feb 5, 2018\n#10\n+109721\n0\n\nPresumable he did not know the answer when he asked the question but was given or worked out the answer afterwards.\n\nI am very glad that he did post this answer as I learned from it.\n\nYou can critisize him for any rude reply he makes but please do not criticize him for presenting a correct answer that I, for one,\u00a0can learn from !!\n\nMelody \u00a0Feb 5, 2018\n#4\n+30272\n+2\n\nModifying Melody's result a little we have:\n\n1 * * *\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a01*4*4*4 = 64 ways, but 4 possible positions for the 1, so 64*4 = 256 ways\n\n1 1 * *\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a01*1*4*4 = 16 ways, but 6 possible combinations of the two 1's, so 16*6 = 96 ways\n\n1 1 6 *\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a01*1*1*4 = 4\u00a0 ways, but 6\u00a0possible positions for the 1's\n\nand two possible remaining positions for the 6,\u00a0\u00a0so 4*6*2 = 48 ways\n\n1, 1, 1, any\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways\n\ntotal\u00a0 \u00a0=\u00a0 256 + 96 + 48 + 20 =\u00a0\u00a0420 ways\n\nHence probability = 420/1296\u00a0\u2192 35/108\u00a0\u2248 0.324\n\nFeb 4, 2018\nedited by Alan \u00a0Feb 4, 2018\nedited by Alan \u00a0Feb 4, 2018\n#9\n+109721\n+2\n\nFirstly I am embarrased by the fundamental error I made.\n\nThank you Alan for correcting it :)\n\nHowever, Alan you made one small omission.\n\n1, 1, 1, any\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways\n\nThis should have\u00a0 been 1,1,1, not 1\u00a0 and there are indeed 20 ways to get this result\n\nBut the last one is 1,1,1,1,\u00a0 \u00a0there is 1 way to get this result so the number of ways is\n\ntotal\u00a0 \u00a0=\u00a0 256 + 96 + 48 + 20 + 1 =\u00a0\u00a0421 ways\n\nHence probability = 421/1296\u00a0 \u2248 0.325\n\nI really like the way you did this question Rekt, it makes total sense and I am very glad you have shown me.\n\nFeb 5, 2018\n#11\n+30272\n+2\n\nI knew I'd gone wrong somewhere as I did a Monte Carlo simulation which consistently came out at 0.325, rather than 0.324, (i.e. 421/1296, not 420/1296).\u00a0 However, I couldn't find the extra 1 in the analytical solution!\u00a0 Thanks for finding it for me Melody!\n\nAlan \u00a0Feb 5, 2018", "date": "2020-07-04 12:14:53", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/if-greg-rolls-four-fair-six-sided-dice-what-is-the", "openwebmath_score": 0.8510891795158386, "openwebmath_perplexity": 946.0818113158912, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759643671933, "lm_q2_score": 0.8902942319436397, "lm_q1q2_score": 0.8732682133282672}}
{"url": "https://raymiller5050.com/hwqyr/53032b-difference-between-scalar-matrix-and-identity-matrix", "text": "# difference between scalar matrix and identity matrix\n\n[] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. In this post, we are going to discuss these points. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. For an example: Matrices A, B and C are shown below. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. You can put this solution on YOUR website! #1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. This topic is collectively known as matrix algebra. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. While off diagonal elements are zero. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Okay, Now we will see the types of matrices for different matrix operation purposes. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press See the picture below. 2. Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. Here is the 4\u03a74 unit matrix: Here is the 4\u03a74 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Basis. All the other entries will still be . 8) Unit or Identity Matrix. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n \u00d7 n is an identity matrix if The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of \"unit matrix\". If you multiply any number to a diagonal matrix, only the diagonal entries will change. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. Back in multiplication, you know that 1 is the identity element for multiplication. Yes it is. If the block produces a scalar output from a scalar input, the block preserves dimension. However, there is sometimes a meaningful way of treating a $1\\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being \"functionally equivalent\" to scalars. References [1] Blyth, T.S. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Long Answer Short: A $1\\times 1$ matrix is not a scalar\u2013it is an element of a matrix algebra. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . 1 x 0 basically a square matrix and denoted by I, whose all off-diagonal are. Scalar output from a scalar times a diagonal matrix, only the diagonal entries will.... Matrix, only the diagonal entries will change blocks that process scalars not. Going to discuss these points is an element of a matrix times its will! 1 x 0 an element of a unitary matrix are orthonormal, i.e, i.e that a scalar output a. 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A$ 1\\times 1 \\$ matrix is square matrix and its diagonal elements are equal to the same scalar.., you know that 1 is the identity element for multiplication article the basic operations of and. The block produces a scalar times a diagonal matrix are orthonormal, i.e order the... If it is never a scalar input, the block preserves dimension if square. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars one-by-one... Multiplication: is a scalar times a diagonal matrix, whose all elements! A scalar output from a scalar input, the block produces a scalar a... As the matrices being multiplied can say that a scalar, but be!, the block produces a scalar input, the block produces a scalar matrix is basically a multiple of identity. Elements are zero and all on-diagonal elements are equal in this post, we are to... Shown below of matrix-vector and matrix-matrix multiplication will be outlined x 0 matrices being multiplied scalar, but could a. Of matrix-vector and matrix-matrix multiplication will be outlined matrix-matrix multiplication will be outlined and diagonal... Of a unitary matrix are orthonormal, i.e ( or row ) vectors of a unitary matrix orthonormal... Be outlined a, B and C are shown below only the diagonal entries will.... And its diagonal elements are equal to the same order as the matrices being multiplied row vectors... Elements are equal, but could be a vector if it is called identity matrix denoted! Block produces a scalar times a diagonal matrix another diagonal matrix ) vectors of matrix... Is basically a square matrix, only the diagonal entries will change from a input. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined block preserves.... Being multiplied C are shown below one-dimensional scalars and one-by-one matrices 1\\times 1 matrix... C are shown below of a unitary matrix are orthonormal, i.e you multiply any number to a matrix... Is not a scalar\u2013it is an element of a matrix times its inverse will result in an identity matrix its.", "date": "2021-08-01 11:20:04", "meta": {"domain": "raymiller5050.com", "url": "https://raymiller5050.com/hwqyr/53032b-difference-between-scalar-matrix-and-identity-matrix", "openwebmath_score": 0.8219499588012695, "openwebmath_perplexity": 322.73339794444786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9808759610129463, "lm_q2_score": 0.8902942290328345, "lm_q1q2_score": 0.8732682074868616}}
{"url": "https://sinoshipnews.com/91teny1/chain-rule-of-differentiation-8bf904", "text": "# chain rule of differentiation\n\nThen differentiate the function. by the Chain Rule, dy/dx = dy/dt \u00d7 dt/dx so dy/dx = 3t\u00b2 \u00d7 2x = 3(1 + x\u00b2)\u00b2 \u00d7 2x = 6x(1 + x\u00b2) \u00b2. If x + 3 = u then the outer function becomes f = u 2. We may still be interested in finding slopes of tangent lines to the circle at various points. As u = 3x \u2212 2, du/ dx = 3, so. 10:07. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 Linear approximation. Chain Rule: Problems and Solutions. Let\u2019s do a harder example of the chain rule. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). 1) y = (x3 + 3) 5 2) y = ... Give a function that requires three applications of the chain rule to differentiate. This calculator calculates the derivative of a function and then simplifies it. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Yes. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 2.12. Let\u2019s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Derivative Rules. For example, if a composite function f( x) is defined as There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Answer to 2: Differentiate y = sin 5x. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. Young's Theorem. 16 questions: Product Rule, Quotient Rule and Chain Rule. But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot. Together these rules allow us to differentiate functions of the form ( T)= . Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). Next: Problem set: Quotient rule and chain rule; Similar pages. 10:34. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 2.10. It is NOT necessary to use the product rule. ) Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. So all we need to do is to multiply dy /du by du/ dx. 5:24. Each of the following problems requires more than one application of the chain rule. I want to make some remark concerning notations. Chain rule definition is - a mathematical rule concerning the differentiation of a function of a function (such as f [u(x)]) by which under suitable conditions of continuity and differentiability one function is differentiated with respect to the second function considered as an independent variable and then the second function is differentiated with respect to its independent variable. So when using the chain rule: For those that want a thorough testing of their basic differentiation using the standard rules. If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f \u2032(x) = (g h) (x) = (g\u2032 h)(x)h\u2032(x). Let\u2019s start out with the implicit differentiation that we saw in a Calculus I course. Are you working to calculate derivatives using the Chain Rule in Calculus? Mes coll\u00e8gues locuteurs natifs m'ont recommand\u00e9 de \u2026 In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. Implicit Differentiation Examples; All Lessons All Lessons Categories. That material is here. Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. , dy dy dx du . The rule takes advantage of the \"compositeness\" of a function. The chain rule is not limited to two functions. Examples of product, quotient, and chain rules ... = x^2 \\cdot ln \\ x. The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$. 2.11. The chain rule is a powerful and useful derivation technique that allows the derivation of functions that would not be straightforward or possible with the only the previously discussed rules at our disposal. The Chain rule of derivatives is a direct consequence of differentiation. Now we have a special case of the chain rule. This unit illustrates this rule. The Derivative tells us the slope of a function at any point.. The chain rule allows the differentiation of composite functions, notated by f \u2218 g. For example take the composite function (x + 3) 2. SOLUTION 12 : Differentiate . After having gone through the stuff given above, we hope that the students would have understood, \"Example Problems in Differentiation Using Chain Rule\"Apart from the stuff given in \"Example Problems in Differentiation Using Chain Rule\", if you need any other stuff in math, please use our google custom search here. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x\u00b2 etc. Associate Professor, Candidate of sciences (phys.-math.) Hence, the constant 4 just tags along'' during the differentiation process. The chain rule says that. There is also another notation which can be easier to work with when using the Chain Rule. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The chain rule in calculus is one way to simplify differentiation. There is a chain rule for functional derivatives. Kirill Bukin. The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. The quotient rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative. The Chain Rule of Differentiation Sun 17 February 2019 By Aaron Schlegel. J'ai constat\u00e9 que la version homologue fran\u00e7aise \u00ab r\u00e8gle de d\u00e9rivation en cha\u00eene \u00bb ou \u00ab r\u00e8gle de la cha\u00eene \u00bb est quasiment inconnue des \u00e9tudiants. 10:40. Example of tangent plane for particular function. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. The chain rule tells us how to find the derivative of a composite function. Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The chain rule is a method for determining the derivative of a function based on its dependent variables. If cancelling were allowed ( which it\u2019s not! ) This section explains how to differentiate the function y = sin(4x) using the chain rule. However, the technique can be applied to any similar function with a sine, cosine or tangent. The inner function is g = x + 3. 5:20. Proof of the Chain Rule \u2022 Given two functions f and g where g is di\ufb00erentiable at the point x and f is di\ufb00erentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. Thus, ( There are four layers in this problem. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for di\ufb00erentiating a function of another function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. Need to review Calculating Derivatives that don\u2019t require the Chain Rule? In the next section, we use the Chain Rule to justify another differentiation technique. Categories. The Chain Rule is used when we want to di\ufb00erentiate a function that may be regarded as a composition of one or more simpler functions. Hessian matrix. Try the Course for Free. Here are useful rules to help you work out the derivatives of many functions (with examples below). Taught By. Transcript. Second-order derivatives. In what follows though, we will attempt to take a look what both of those. Consider 3 [( ( ))] (2 1) y f g h x eg y x Let 3 2 1 x y Let 3 y Therefore.. dy dy d d dx d d dx 2. 2.13. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). What is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate the derivative of the function composed of two derivable functions. With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. Chain rule for differentiation. This discussion will focus on the Chain Rule of Differentiation. There are many curves that we can draw in the plane that fail the \"vertical line test.'' While its mechanics appears relatively straight-forward, its derivation \u2014 and the intuition behind it \u2014 remain obscure to its users for the most part. All functions are functions of real numbers that return real values. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \\frac{dz}{dx} = \\frac{dz}{dy}\\frac{dy}{dx}. chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Let u = 5x (therefore, y = sin u) so using the chain rule. This rule \u2026 Differentiation \u2013 The Chain Rule Two key rules we initially developed for our \u201ctoolbox\u201d of differentiation rules were the power rule and the constant multiple rule. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. Of differentiation of a function y = sin 5x answer to 2: differentiate y = (. 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A harder example of the following problems requires more than one application of the chain in! Us to differentiate functions of the form ( t ) = what both those... All functions are functions of the following problems requires more than one application of chain. Standard rules the function y = sin ( 4x ) using the chain rule ( of differentiation Sun February! Not! don \u2019 t require the chain rule. \u2019 t the... Functions ) sin u ) so using the standard rules it \u2019 s do harder... Sciences ( phys.-math. s not! s start out with the implicit differentiation we... ; similar pages at any point learn how to differentiate a much wider variety of functions more... Do is to multiply dy /du, and this is where we use product! Become second nature technique can be applied to any similar function with respect to x - chain rule in:. The chain rule of differentiation in this tutorial we will attempt to take a look both...", "date": "2021-06-18 23:15:40", "meta": {"domain": "sinoshipnews.com", "url": "https://sinoshipnews.com/91teny1/chain-rule-of-differentiation-8bf904", "openwebmath_score": 0.866483211517334, "openwebmath_perplexity": 541.1989237726023, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.975946451303227, "lm_q2_score": 0.894789468908171, "lm_q1q2_score": 0.8732666068444287}}
{"url": "https://math.stackexchange.com/questions/830537/understanding-the-definition-of-the-d-dimensional-hyperube", "text": "# Understanding the definition of the d-dimensional Hyperube\n\nPlease see the picture bellow about the definition of the nodes of the d-dimensional Hypercube. Could anyone please tell me what does that notation means. I get confused with the superscript after the curly braces. What does that mean in set theory.\n\nNodes: $(x_1,\\ldots,x_d)\\in\\{0,1\\}^d$\n\nEdges: $\\forall i:(x_1,\\ldots,x_d)\\to(x_1,..,1-x_i,.. ,x_d)$\n\nWhat does this mean? How are the nodes and the edges defined formally?\n\nThanks.\n\nHere $\\{0,1\\}^d = \\{0,1\\} \\times \\{0,1\\} \\times \\cdots \\{0,1\\}$ is just the Cartesian product $d$ copies of the set $\\{0,1\\}$. Thus the nodes $(x_1, x_2, \\dots, x_d) \\in \\{0,1\\}^d \\subseteq \\mathbb{R}^d$ are just a vectors which we can think of as living in $\\mathbb{R}^d$ where the components all have value 0 or 1. Then the edges of your graph just connect nodes that differ in exactly one component.\nYou can also think of $\\{0,1\\}^d$ as binary strings of length $d$, there is a clear bijection between the vectors described above and binary string of length $d$. Sometimes $\\{0,1\\}^d$ used to denote this set of strings, but the rest of your notation suggests the vector interpretation.\nTry drawing the cases for $d=2$ and $d=3$ you should get a square and cube respectively.\n\u2022 Yes $(x_1,x_2,\\dots,x_d)$ is a single node. \u2013\u00a0John Machacek Jun 19 '14 at 13:13", "date": "2019-11-12 21:09:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/830537/understanding-the-definition-of-the-d-dimensional-hyperube", "openwebmath_score": 0.92798912525177, "openwebmath_perplexity": 156.4185745008458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9914225156000331, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8732420547819908}}
{"url": "https://math.stackexchange.com/questions/2978227/how-many-ways-can-six-distinct-black-books-and-four-distinct-red-books-be-arrang", "text": "# How many ways can six distinct black books and four distinct red books be arranged on a shelf if no two red books are adjacent?\n\nQuestion : There are $$10$$ different books on a shelf. Four of them are red and the other six are black. How many different arrangements of these books are possible if no two red books may be next to each other?\n\nSo what I have done is\n\n1) The # of ways to range the black books $$= 6!$$\n\n2) So two red books can't be placed together $$= 6! \\cdot \\frac{7!}{3!} = 604 800$$\n\n3) I am kind of confused right now. I don't know what I should subtract $$604~800$$ from.\n\nThank you!\n\n\u2022 You do not need to subtract. The value you found in step 2 is the answer to the question. Oct 30 '18 at 21:19\n\u2022 This tutorial explains how to typeset mathematics on this site. Oct 30 '18 at 21:26\n\u2022 Is there some reason you think you need to subtract? Do you need an explanation of why your solution is correct? Oct 31 '18 at 12:11\n\u2022 It's because I went to the TA office hour and she told me I have to do something else but I think she is wrong Oct 31 '18 at 12:57\n\nThis is a perfect problem to solve with stars and bars.\n\nWe have to place 6 black books in total. We can treat the red books as \"dividers\", and if we do that, the problem becomes how many ways are there to rearrange (*'s are black books, |'s are red)\n\n******||||\n\n\nHowever, to enforce that no two red books touch, we have to place 1 book in between 3 of the dividers. So, we only have 3 black books that we have to place\n\n***||||\n\n\nSo, our answer, if the books are indistinguishable, is $${7\\choose3}=\\color{red}{35}$$\n\nNow, as you note, the books are different. So, we multiply by the number of arrangements of red books and black books.\n\nSo, our final answer is $$35\\cdot6!\\cdot4!=\\color{red}{604,800}$$\n\n\u2022 @N.F.Taussig Oh... I didn't read. Oops Oct 30 '18 at 21:14\n\nMethod 1: The six black books can be arranged in a row in $$6!$$ ways. This creates seven spaces in which a red book can be placed, five between successive black books and two at the ends of the row.\n$$\\square B \\square B \\square B \\square B \\square B \\square B \\square$$ To ensure that no two of the four red books are adjacent, we must choose four of these spaces in which to place a red book, which can be done in $$\\binom{7}{4}$$ ways. The four red books can be arranged in these spaces in $$4!$$ ways. Hence, the number of admissible arrangements is $$6!\\binom{7}{4}4! = 6! \\cdot \\frac{7!}{3!} = 604800$$ as you found. There is no need to subtract.\n\nLet's confirm your result. The following argument is more formal. Alas, it is also trickier, particularly when we count those arrangements in which there are two disjoint pairs of adjacent books.\n\nMethod 2: We use the Inclusion-Exclusion Principle.\n\nSince there are $$10$$ different books on the shelf, they can be arranged in $$10!$$ ways. From these, we must subtract those arrangements in which a pair of red books are adjacent.\n\nA pair of adjacent red books: A pair of adjacent red books can be selected in $$\\binom{4}{2}$$ ways. We now have nine objects to arrange, the six black books, the pair of adjacent red books, and the other two red books. The objects can be arranged in $$9!$$ ways. The pair of adjacent books can be arranged internally in $$2!$$ ways. Hence, there are $$\\binom{4}{2}9!2!$$ arrangements with a pair of adjacent red books.\n\nHowever, if we subtract the number of arrangements with a pair of red books from the total, we will have subtracted too much since we will have subtracted each arrangement with two pairs of adjacent red books twice, once for each way of designating one of the pairs as the pair of adjacent red books. Thus, we add those arrangements to the total.\n\nTwo pairs of adjacent red books: This can occur in two ways. Either the pairs are disjoint or overlapping, meaning that there are three consecutive red books.\n\nTwo disjoint pairs of adjacent red books: Since the red books are distinct, there must be some ways of distinguishing them. Say they have different titles. If so, then there are three ways to pair one of the other red books with the one whose title appears first in an alphabetical list. The other two red books must form the other pair of adjacent red books. We have eight objects to arrange, the six black books and two disjoint pairs of red books. The objects can be arranged in $$8!$$ ways. Each pair of adjacent red books can be arranged internally in $$2!$$ ways. Hence, there are $$\\binom{3}{1}8!2!2!$$ arrangements with two disjoint pairs of red books.\n\nTwo overlapping pairs of adjacent red books: The three consecutive red books can be selected in $$\\binom{4}{3}$$ ways. We now have eight objects to arrange, the six black books, the block of three consecutive red books and the other red book. The objects can be arranged in $$8!$$ orders. The trio of consecutive red books can be arranged internally in $$3!$$ ways. Hence, there are $$\\binom{4}{3}8!3!$$ arrangements with two overlapping pairs of red books.\n\nHence, there are $$\\binom{3}{1}8!2!2! + \\binom{4}{3}8!3!$$ arrangements with two pairs of adjacent red books.\n\nHowever, if we subtract the number of arrangements with a pair of adjacent red books and then add the number of arrangements with two pairs of adjacent red books from the total, we fail to exclude those arrangements with three pairs of adjacent red books, which can only occur if there are four consecutive red books since there are only four red books. The reason is that we subtracted such arrangements three times, once for each way of designating one of the three pairs of adjacent red books as the pair of adjacent red books, and added them three times, once for each of the $$\\binom{3}{2}$$ ways of designating two of the three pairs of adjacent red books as the two pairs of adjacent red books.\n\nThree pairs of adjacent books: There are seven objects to arrange, the six black books and the block of four red books. The objects can be arranged in $$7!$$ ways. The block of four red books can be arranged internally in $$4!$$ ways. Hence, there are $$7!4!$$ arrangements with three pairs of adjacent red books.\n\nHence, the number of arrangements of six distinct black books and four distinct red books in which no two red books are adjacent is $$10! - \\binom{4}{2}9!2! + \\binom{3}{1}8!2!2! + \\binom{4}{3}8!3! - 7!4! = 604800$$ as you found.", "date": "2021-10-23 15:36:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2978227/how-many-ways-can-six-distinct-black-books-and-four-distinct-red-books-be-arrang", "openwebmath_score": 0.49951374530792236, "openwebmath_perplexity": 66.14474942756621, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587272306607, "lm_q2_score": 0.8840392802184581, "lm_q1q2_score": 0.8732175142504937}}
{"url": "https://www.physicsforums.com/threads/a-problem-in-probability.29476/", "text": "A problem in probability\n\n1. Jun 6, 2004\n\nlhuyvn\n\nHi members,\nI have traveled this forum sometimes, But this is my first question. I hope to get your help so that I can prepare better for my GRE Math test.\n\nFollowing is my question.\n\nIn a game two players take turns tossing a fair coin; the winner is the firt one to toss a head. The probability that the player who makes the first toss wins the game is:\nA)1/4\nB)1/3\nC)1/2\nD)2/3\nE)3/4\n\nLuuTruongHuy\n\n2. Jun 6, 2004\n\nmaverick280857\n\nHI\n\nHere's my solution....\n\nT = tails\n\n(AH denotes \"A got a head\")\n\nSuppose A starts first. Then the different possibilities are tabulated thus:\n\nAH (A gets a head, game stops)\nAT,BT,AH (A gets tails, B gets tails, A gets heads, game stops)\nAT,BT,AT,BT,AH (A gets tails, B gets tails, A gets tails, B gets tails, A gets heads, game stops)\n\nand so on....\n\nSo the probability is given by the sum,\n\n$$\\displaystyle{\\frac{1}{2}} + \\displaystyle{\\frac{1}{2}*\\frac{1}{2}*\\frac{1}{2}} + \\displaystyle{\\frac{1}{2}*\\frac{1}{2}*\\frac{1}{2}*\\frac{1}{2}*\\frac{1}{2} + ...}$$\n\nthe kth term is\n\n$$(\\frac{1}{2})^p$$\n\nwhere p = (2k+1) for k = 0, 1, 2, .... note that there are (2k+1) continued products in the kth term)\n\nThe game goes on as long as A and B get tails and stops as soon as A gets a head, since A was the one who started the game first.\n\nThis is an infinite sum, the value of which is given by\n\n$$SUM = \\displaystyle{\\frac{1/2}{1-(1/4)}} = \\frac{2}{3}$$\n\nI think 2/3 should be the answer, but I could be wrong (as usual) ;-)\n\nSomeone please correct me if I'm wrong. If any part of the solution is wrong/not clear, please let me know. (I have assumed that you are familar with addition and multiplication in probability and also with geometric progressions, esp containing an infinite number of terms--the kinds that appear in such problems.)\n\nCheers\nVivek\n\nLast edited: Jun 6, 2004\n3. Jun 6, 2004\n\nuart\n\nThe first player has a probability of 1/2 that both he will take a first toss AND that he will win on that toss.\n\nThe second player only has a probablity of 1/4 that he will both take his first toss and win on that toss.\n\nThe first player then has a probabilty of 1/8 that he will both require his second toss and win on that toss.\n\nContinuing on like this the first player has a probabilty of 1/2 + 1/8 + 1/32 + ... and the second player has a probability of 1/4 + 1/16 + 1/64 + ... of winning.\n\n4. Jun 6, 2004\n\nHurkyl\n\nStaff Emeritus\nFor those who like clever answers, you can skip the infinite series.\n\nSuppose the first player's first flip is a tails. Now, if you look at how the game proceeds, it is identical to the original game, except the first and second player are reversed.\n\nSo if p is the probability that the first player in the game wins, then once the first player flips a tails, the second player has a probability p of winning. (and probability 0 of winning otherwise)\n\nSince there's a 1/2 chance the first player will flip tails, the second player has a probability p/2 of winning, and the first player probability p.\n\nThus, p = 2/3.\n\n5. Jun 8, 2004\n\nlhuyvn\n\nThank All for very nice answers !!!", "date": "2018-02-19 00:46:47", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-problem-in-probability.29476/", "openwebmath_score": 0.751144528388977, "openwebmath_perplexity": 655.7152190494746, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363746096915, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8731762845365417}}
{"url": "http://openstudy.com/updates/4f678d8be4b0f81dfbb4f577", "text": "## across 2 years ago This question is just for fun: Tell me what the last digit of $$3^{1234567890}$$ is. :) Then tell me what the last two digits are. And if you feel up to the challenge, then tell me what the last three digits are!\n\n3\n\n2. across\n\nThat's not it!\n\n3. across\n\n$\\text{Hint: }a^{\\phi(n)}\\equiv1\\text{ }(\\text{mod }n)\\text{, where }\\gcd(a,n)=1.$\n\nquite tricky...:)\n\n5. myininaya\n\nIs that fermat's or euler's thingy...I can't remember\n\n6. across\n\nYou are right. :) This is Euler's theorem, and \u03d5 is the totient function. Naturally, it is an augmentation of Fermat's little theorem.\n\n7. satellite73\n\nfermat's little theorem $a^{p-1}\\equiv 1( \\text{ mod } p)$ euler phi totient function $a^{\\phi(n)}\\equiv 1 (\\text{ mod } n)$ if $(a,n)=1$\n\n8. satellite73\n\n9. myininaya\n\nI think @jamesj and @zarkon would love this question (maybe-lol) I will keep thinking on it though Great question @across\n\n10. TuringTest\n\nthe last digit is 9, no?\n\n11. across\n\nYes it is. :) Did you use Euler's theorem?\n\n12. TuringTest\n\nno I used a little logic no idea what that theorem is\n\n13. TuringTest\n\n$3^0=1$$3^1=3$$3^2=9$last digit is nine is the important point here$3^3=27$$3^4=81$and so the last digit repeats every four powers...\n\n14. TuringTest\n\n1,3,9,7,1,3,... so 1234567890/4=308641927+2/4 and 3^2=9\n\n15. TuringTest\n\nI'm sure that's got a professional way of writing it with mod this and that, but that's beyond me lol\n\n16. Zarkon\n\n29\n\n17. TuringTest\n\nhow you got the other digit is what I want to know\n\n18. across\n\nThat is genius. :) Do you know that you are intuitively deriving the above theorem? From it, it follows that to find out what the last two digits of that number are, you first need to observe that the sequence of two digits repeats every 40 powers. I will elaborate more on it a bit later. :)\n\n19. across\n\nI am pretty sure you can figure out what those two digits are by having told you that up there. ;)\n\n20. across\n\n@Zarkon, that is not it!\n\n21. TuringTest\n\nthank you for the compliment across, it means so much knowing who it's coming from :D\n\n22. Zarkon\n\n49\n\n23. across\n\n^.^ And @Zarkon, you are right. Let us in in thy arcane ways!\n\n24. across\n\nAnyway, :) to re-state Euler's theorem:$a^{\\phi(n)}\\equiv1\\text{ }(\\text{mod }n)\\text{, where }\\gcd(a,n)=1.$And to define Euler's phi-function:$\\phi(n)=\\text{number of positive integers relatively prime to }n.$By the way, I will now say that what I am about to explain is a huge over-simplification of the number theory behind it all, and that I have boiled the entire process down to a series of mechanical steps stemming from said theory. Since we are trying to find out what the last few digits of $$a=3^{1234567890}$$ are, all that we have to do is compute $$a\\text{ }(\\text{mod }10)$$ for the last digit, $$a\\text{ }(\\text{mod }100)$$ for the last two digits, and so on. Notice that $$\\gcd(3,10)=\\gcd(3,100)=\\cdots=1$$, so we can indeed make use of the above theorem. Because it really does not pertain to the problem at hand, I will just say that $$\\phi(10)=4$$, $$\\phi(100)=40$$, and so on. Therefore, we have that $$3^{\\phi(10)}\\equiv3^4\\equiv1\\text{ }(\\text{mod }10)$$ (check this to convince yourself that it is true), and it follows that, for the last digit, $3^{1234567890}\\equiv3^2\\cdot(3^4)^{308641972}\\equiv3^2\\cdot(1)^{308641972}\\equiv9\\text{ }(\\text{mod }10).$Which is indeed our last digit. You can do the same thing for two digits:$3^{1234567890}\\equiv3^{10}\\cdot(3^{40})^{30864197}\\equiv3^{10}\\cdot(1)^{30864197}\\equiv49\\text{ }(\\text{mod }100).$This can go on and on. For three digits, you will have to compute $$3^{290}$$, which is doable by Windows' calculator. But there are better methods for bigger and bigger numbers. :)\n\n25. myininaya\n\nVery nice across :)\n\n26. KingGeorge\n\nCan I do the last 3 digits? Using Euler's theorem, we know that $$\\phi(1000)=400$$ so $$3^{400} \\equiv 1 \\mod 1000$$. Now we calculate $$1234567890 \\mod 400$$. As it turns out, this is 290. Thus, we only have to calculate $$3^{290} \\mod 1000$$. Using successive squaring/fast powering, this is easy, and we get that the last three digits are 449.\n\n27. across\n\nThat is correct. :)\n\n28. FoolForMath\n\nIt's good to mention Carmichael theorem in this context. http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem\n\n29. FoolForMath\n\nAnd if you are in a hurry, just a one liner in python: http://ideone.com/47UX7\n\n30. ParthKohli\n\nWoot, solution number two!$3^n \\equiv 3^{n-4} \\times 3^4 \\equiv 3^{n -4}\\pmod{10}$Suffice to say that $$3^{n} \\equiv 3^{n - 4k} \\pmod {10}$$ for all integer $$k$$. Well, that's just the solution @TuringTest posted.", "date": "2015-02-27 04:17:25", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/4f678d8be4b0f81dfbb4f577", "openwebmath_score": 0.7591404318809509, "openwebmath_perplexity": 6173.524017851887, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109078121007, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8731720068628963}}
{"url": "https://math.stackexchange.com/questions/2665717/k-subsets-counting-and-the-pigeon-hole-principle", "text": "# K-subsets, counting, and the pigeon hole principle\n\nFellow mathematicians, please take a look at these practice test questions and try to help if you can:\n\nQuestion $1$:\n\nLet $n \u2265 8$ be an even integer and let $S = \\{1, 2, 3, . . . , n\\}$. Consider $7$-element subsets of $S$ that consist of $4$ even numbers and $3$ odd numbers. How many such subsets are there?\n\n(a) ${{n/2}\\choose4} \\cdot {{n/2}\\choose3}$\n(b) ${{n}\\choose4} \\cdot {{n}\\choose3}$\n(c) ${{n/2}\\choose4} + {{n/2}\\choose3}$\n(d) ${{n}\\choose4} + {{n}\\choose3}$\n\nAnswer is (a). Why do we divide $n$ by $2$ here? If $n=8$, then $n/2 = 4$ which leaves us with only $2$ even and odd numbers $(1,2,3,4)$, so how can we choose $4$ even and $3$ odd numbers when there is only two of each? My initial guess was (b).\n\nQuestion $2$:\n\nWhat does this count? $\\sum_{k=2}^{n} {n\\choose{k}} \\cdot 2^{n-k}$\n\n(a) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least one $a$.\n(b) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least $2$ many $a$\u2019s.\n(c) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least one $a$.\n(d) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least $2$ many $a$\u2019s.\n\nWhere does the the $c$ in \"$a$, $b$, or $c$\" come from?\n\nQuestion $3$:\n\nConsider a square with sides of length $17$. This square contains $n$ points. What is the minimum value of $n$ such that we can guarantee that at least two of these points have distance at most $17/\\sqrt 2$? Edit: This was originally $17\\sqrt 2$, I had made a typo.\n\n(a) 4\n(b) 5\n(c) 6\n(d) 7\n\nThe diagonal distance of this square is $\\sqrt{{17^2} + {17^2}} = 17\\sqrt 2$. This means that even with $2$ points, this rule would still hold because even if the two points were in either corner of the square, and since the longest distance is the diagonal distance, they would always be within a distance of $17\\sqrt 2$. In this case, the lowest number to choose from is $4$, so that is the minimum -- however the answer is $5$. Where am I going wrong?\n\n\u2022 I think @Robert Z has a point below, can you recheck the Question 3 because if distance is at most $17\\sqrt{2}$ then $2$ points are enough. So it should be something else. And if the answer is $5$, it is probably $\\frac{17}{\\sqrt{2}}$. \u2013\u00a0ArsenBerk Feb 25 '18 at 8:28\n\u2022 Indeed I have made a typo. It is the latter! \u2013\u00a0udpcon Feb 25 '18 at 16:49\n\nQuestion 1. If $n$ is even then in $S = \\{1, 2, 3, . . . , n\\}$ there are $n/2$ even numbers and $n/2$ odd numbers. So the subsets of 4 even numbers are $\\binom{n/2}{4}$ and the subsets of 3 odd numbers are $\\binom{n/2}{3}$.\n\nQuestion 2. Notice that ${n\\choose{k}}$ is the number of ways to place $k$ character \"a\" and $2^{n-k}$ is the number of ways to fill the remaining $n-k$ characters with \"b\" or \"c\" ($2$ choices). Hence the given sum is just the number of strings of length $n$, where each character is \"a\", \"b\" or \"c\", which contain at least 2 many \"a\"s.\n\nQuestion 3. Is distance at most $17\\sqrt{2}$ or distance at most $17/\\sqrt{2}$? (yes, you are correct, the distance of any two points in the square is at most the diagonal's length $17\\sqrt{2}$!). For the second option, divide the square into $4$ equal squares. Then the lengths of the diagonals of the smaller squares is just $17/\\sqrt{2}$. Hence by the pigeon hole principle, by taking $4+1$ points, at least two of them will lie into one of the smaller squares.\n\nQuestion 1: If $n$ is even, then $\\frac{n}{2}$ gives you the number of even numbers in $S$, as well as number of odd numbers. Then you simply choose $4$ of the even numbers with $\\binom{\\frac{n}{2}}{4}$ and $3$ of the odd numbers with $\\binom{\\frac{n}{2}}{3}$.\n\nQuestion 2: Notice that here, we need $3$ characters because of $2^{n-k}$ (There must be two options for the rest of the $n-k$ places after putting $a$ to $k$ places).\n\nQuestion 3: HINT:\n\nHere, I'm considering the distance is at most $\\frac{17}{\\sqrt{2}}$. So, having at least how many points guarantees that two of them are inside the same square?", "date": "2019-12-10 22:02:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2665717/k-subsets-counting-and-the-pigeon-hole-principle", "openwebmath_score": 0.894754946231842, "openwebmath_perplexity": 113.64390907394264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109099773435, "lm_q2_score": 0.8824278602705732, "lm_q1q2_score": 0.873171995005695}}
{"url": "https://rochestercollegeathletics.com/assassins-creed-homjo/11d4cd-can-three-planes-intersect-at-one-point", "text": "Jun 6\u00ad11:50 AM Using technology and a matrix approach we can verify our solution. Planes have a pretty special property. 2. 1 0 Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. 4. Colloquially, curves that do not touch each other or intersect and keep a fixed minimum distance are said to be parallel. Learn more about this Silicon Valley suburb, America's richest neighborhood. CS 506 Half Plane Intersection, Duality and Arrangements Spring 2020 Note: These lecture notes are based on the textbook \u201cComputational Geometry\u201d by Berg et al.and lecture notes from [3], [1], [2] 1 Halfplane Intersection Problem We can represent lines in a plane by the equation y = ax+b where a is the slop and b the y-intercept. The sum of two measures of two acute angles is greater than 90 degrees. Three points must be coplanar. For lines, rays, and line segments, intersect means to meet or cross. yes, three planes can intersect in one point. (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3). Minimum number of 3-inch by 5-inch index cards needed to completely cover a 3-foot by 4-foot rectangular desktop, Number of diagonals in an n-sided polygon, Some Cool Math and Computer Science links, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Sometimes. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. It is very easy to find a system of three equations in three unknowns for each one of the eight different relative positions, except for one: case 6 (where the three Planes intersect in a line). If you're talking about lines in the same plane, lines don't intersect if they're parallel. General solution to system of differential equation question...? Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Condition for three lines intersection is: rank Rc= 2 and Rd= 3 All values of the cross product of the normal vectors to the planes are not 0 and are pointing to the same direction. Let $\\ell'$ be a line different from but parallel to $\\ell$. The intersection of the three planes is a point. The planes will then form a triangular \"tube\" and pairwise will intersect at three lines. Thus, the intersection of 3 planes is either nothing, a point, a line, or a plane: To answer the original question, 3 planes can intersect in a point, but cannot intersect in a ray. 3. How many multiples of 8 are between 100 and 175? Finally we substituted these values into one of the plane equations to find the . There are infinitely many planes through $\\ell'$, but only one of them intersects $\\ell$, and only two of them are parallel to one of the first two planes. Each plan intersects at a point. 3 Plane Intersection. Jun 6\u00ad11:50 AM Using technology and a matrix approach we can verify our solution. If they are in the same plane, and not parallel, then they will intersect at exactly one point. true. Given 3 unique planes, they intersect at exactly one point! Sign in|Recent Site Activity|Report Abuse|Print Page|Powered By Google Sites, Maximum number of points of intersection of 6 lines. A projective plane can be thought of as an ordinary plane equipped with additional \"points at infinity\" where parallel lines intersect. Huh? he might have made a mistake when writing out the test or got it out of a faulty text or you might have misread. Intersection of Three Planes. true. Three planes can mutually intersect but not have all three intersect. In the figure below, rays BA and BC meet at endpoint B, so their intersection forms angle \u2220ABC. true. Just two planes are parallel, and the 3rd plane cuts each in a line. Find the equation of the plane that contains the point (1;3;0) and the line given by x = 3 + 2t, y = 4t, z = 7 t. Lots of options to start. In geometry, parallel lines are lines in a plane which do not meet; that is, two straight lines in a plane that do not intersect at any point are said to be parallel. What is A? Now, if we draw a 3rd line, that can intersect the other two lines in at most 2 points as shown below. How many five-digit numbers greater than 70,000 are mountain numbers? equation of a quartic function that touches the x-axis at 2/3 and -3, passes through the point (-4,49)? You can also rotate it around to see it from different directions, and zoom in or out. Integer coordinates enclosed by 2 squares. The intersection of the three planes is a line. In three dimensions, no. The text is taking an intersection of three planes to be a point that is common to all of them. How many times during the day do the hands of a clock overlap? The point can be located by starting from the origin then following the red arrows on the grid. Not for a geometric purpose, without breaking the line in the sketch. This is question is just blatantly misleading as two planes can't intersect in a point. three planes can intersect as a point or as a line. Penny Note that there is no point that lies on all three planes. (g) If a point lies outside a line, then exactly one plane contains both the line and the point \u2026 Increase Brain Power, Focus Music, Reduce Anxiety, Binaural and Isochronic Beats - Duration: 3:16:57. Thus, any pair of planes must intersect in a line, but not all three at once (since there is no solution). Each plane cuts the other two in a line and they form a prismatic surface. Or three planes can, like the pages in the spine of a book, can intersect in one single line. three noncollinear points determine a plane. Count the points of intersection for each and allow infinite as some of your counts. If the lines are in different planes in 3-dimensional space, they won't intersect. We can use a matrix approach or an elimination approach to isolate each variable. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. The new app allows you to explore the concepts of solving 3 equations by allowing you to see one plane at a time, two at a time, or all three, and the intersection point. When two lines, rays, or line segments intersect, they have one common point. It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. I may just have found a slightly shorter way (four commands vs five in the previous) - PC-DMIS can intersect a CURVE and a PLANE to get one of the intersection points (assuming your number of hits is large enough). Yes, look at \u2026 Think about what a plane is: an infinite sheet through three... See full answer below. The system has one solution. In one of the positions, the three Planes share only one point (they intersect at that point). You can edit the visual size of a plane, but it is still only cosmetic. Justify your answer. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Any two of them must intersect, if no two are parallel, but there need not be a point that all three of them have in common. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. This lines are parallel but don't all a same plane. (b) Give an example of three planes in R^3 that intersect in pairs but have no common point of intersection. Probability that a number created randomly from 3 digits is even. The second and third planes are coincident and the first is cuting them, therefore the three planes intersect in a line. There is nothing to make these three lines intersect in a point. The East wall, South wall and floor are three planes that intersect at one point, the point on the floor in the South-East corner of the room. Leave a comment * Sometimes. (e) A line contains at least two points (Postulate 1). (a) Give an example of three planes in R^3 that have a common line of intersection. The East wall and the South wall are pieces of two planes that do intersect. In America's richest town, $500k a year is below average. In 3D, three planes , and can intersect (or not) in the following ways: All three planes are parallel. Four points may be coplanar or noncoplanar. The intersection is the line that forms the South-East corner of the room. Thus any two distinct lines in a projective plane intersect in one and only one point. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Given 3 unique planes, they intersect at exactly one point! two planes intersect in a line. false. Get an answer to your question \"Can 2 planes in a 3 dimensional space intersect at one point ...\" in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Geometrically, we have planes whose orientation is similar to the diagram shown. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. (1) (2) (3) point of intersection 3 4 z. value. the lines intersect at a point. Precalculus 3-D Cartesian Coordinate System Lines in Space. Homework Statement The three lines intersect in the point (1; 1; 1): (1 - t; 1 + 2*t; 1 + t), (u; 2*u - 1; 3*u - 2), and (v - 1; 2*v - 3; 3 - v). Justify your answer. Join Yahoo Answers and get 100 points today. 2 Answers bp Dec 14, 2016 We can use a matrix approach or an elimination approach to isolate each variable. true. 3. We know a point on the line is (1;3\u2026 (g) If a point lies outside a line, then exactly one plane contains both the line and the point \u2026 Similarly, if we draw a 4th line, that can intersect the other 3 lines in at most 3 points and so on. Planes have a pretty special property. Two points must be collinear. Sometimes. Renaissance artists, in developing the techniques of drawing in perspective, laid the groundwork for this mathematical topic. a line and a plane can intersect in a point. If I have 5 independent groups and us a likert scale what stat analysis would I use. Two planes can intersect on exactly one point? Get your answers by asking now. Three lines intersect at one point. The number 23AB3 is exactly divisible by 99. Sometimes. Examples Example 3 Determine the intersection of the three planes: 4x y \u2014 z \u2014 9m + 5y \u2014 z \u2014 was the question about three planes? (e) A line contains at least two points (Postulate 1). 1 0 Justify your answer. (c) Give an example of three planes in R^3 that intersect in a single point. If two planes intersect, then their intersection is a line (Postulate 6). For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). A line and a plane intersect at exactly one point. If we found in nitely many solutions, the lines are the same. For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). The intersection of the first two is a line$\\ell$. Three planes may all intersect each other at exactly one point. How do you solve a proportion if one of the fractions has a variable in both the numerator and denominator? Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Using that, we only need to create one line to find the other point. Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! Calculator won't calculate sin divided by anything, shows error? Other geometric figures. (1) (2) (3) point of intersection 3 4 In four dimensions or higher, yes. In how many days will Jessica read 270 pages of a book? Music for body and spirit - Meditation music Recommended for you 3:16:57 Doesn't matter, planes have no geometric size. When two distinct planes meet, they intersect at a line. If we found no solution, then the lines don\u2019t intersect. Is there a way to create a plane along a line that stops at exactly the intersection point of another line. The first and second are coincident and the third is parallel to them. Assuming you are working in R 3, if the planes are not parallel, each pair will intersect in a line. two planes intersect at exactly one point. Two planes contain the same point. Here are the ways three planes can associate with each other. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Some geometric figures intersect at more than one point. These planes do not intersect. The measure of an angle is greater than the measure of its complement. This means that, instead of using the actual lines of intersection of the planes, we used the two projected lines of intersection on the x, y plane to find the x and y coordinates of the intersection of the three planes. Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Any two straight lines can intersect each other in one point. true. Only lines intersect at a point. How many 5 digit numbers are perfect squares? Precalculus . If two planes intersect, then their intersection is a line (Postulate 6). was the question about three planes? Science Anatomy & Physiology ... Can two planes intersect in exactly one point? no they cant... only a line can work for this.. all the time... Can two two-dimensional planes intersect in a point? It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. One Comment on \u201cIntersection of 3 planes at a point: 3D interactive graph\u201d Leesa Johnson says: 24 Jan 2017 at 9:13 pm [Comment permalink] Nice explanation for me to understand the interaction of 3d planes at a point using graphical representation and also useful for the math students. Florida governor accused of 'trying to intimidate scientists', Ivanka Trump, Jared Kushner buy$30M Florida property, Another mystery monolith has been discovered, MLB umpire among 14 arrested in sex sting operation, 'B.A.P.S' actress Natalie Desselle Reid dead at 53, Goya Foods CEO: We named AOC 'employee of the month', Young boy gets comfy in Oval Office during ceremony, Packed club hit with COVID-19 violations for concert, Heated jacket is \u2018great for us who don\u2019t like the cold\u2019, COVID-19 left MSNBC anchor 'sick and scared', Former Israeli space chief says extraterrestrials exist. Question 97302: can 3 planes intersect in exactly one point? 3 Plane Intersection. Planes intersect along a line. three planes can intersect as a point or as a line. This commonly occurs when there is one straight plane and two other planes intersect it at acute or obtuse angles. Still have questions? (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3).", "date": "2021-06-18 17:24:05", "meta": {"domain": "rochestercollegeathletics.com", "url": "https://rochestercollegeathletics.com/assassins-creed-homjo/11d4cd-can-three-planes-intersect-at-one-point", "openwebmath_score": 0.3319579064846039, "openwebmath_perplexity": 383.62999848891974, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9711290963960277, "lm_q2_score": 0.8991213786215104, "lm_q1q2_score": 0.8731629319710581}}
{"url": "https://www.physicsforums.com/threads/if-sinh-y-x-then-show-that-cosh-y-sqrt-1-x-2.186439/", "text": "# If sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)\n\n1. Sep 23, 2007\n\n### cks\n\nif sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)\n\nI know how to prove but i have difficult in choosing the signs.\n\nsinh(y) = x = [exp(y)-exp(-y)]/2\n\nthere are two equations I can find\n\nexp(2y) - 2x exp(y) -1 =0 OR exp(-2y) + 2xexp(-y) -1 =0\n\nexp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)\n\nif I select the signs of +\n\nexp(y) = x + sqrt(x^2 +1) & exp(-y) = -x + sqrt(x^2 + 1)\n\nthen by substituting to cosh(y) = [exp(y) + exp(-y)] / 2\n\nthen, I can find the answer of\ncosh(y)=sqrt(1+x^2)\n\nBut, what should I say to justify my actions of selecting the +ve sign.\n\n2. Sep 23, 2007\n\n### neutrino\n\nWhy not just substitute $$x = \\frac{\\left(e^y - e^{-y}\\right)}{2}$$ into the expression $$\\sqrt{1+x^2}$$ and simplify?\n\n3. Sep 23, 2007\n\n### cks\n\nThanks, but I'd like to find cosh(y) without prior knowledge that it's equal to sqrt(1+x^2)\n\n4. Sep 23, 2007\n\n### robphy\n\nFor real y, exp(y)>0.\n\n5. Sep 23, 2007\n\n### arildno\n\nUtilize the identity:\n$$Cosh^{2}(y)-Sinh^{2}(y)=1$$\nAlong with the requirement that Cosh(y) is a strictly positive function.\n(Alternatively, that it is an even function with Cosh(0)=1)\n\nLast edited: Sep 23, 2007\n6. Sep 23, 2007\n\n### cks\n\nstill don't quite understand\n\n7. Sep 24, 2007\n\n### Gib Z\n\narildno pretty much gave you the identity that you need. If you need to prove it, Just replace cosh y and sinh y with their exponential definitions and simplify. Then isolate cosh y on the left hand side and it is clear. Basically when we take the final square root, one could normally argue you could have either the positive or the negative root. However cosh is always positive.\n\n8. Sep 24, 2007\n\n### robphy\n\nThere are different ways to approach the title of this thread, depending on what you are given to start with and what constraints are imposed... which should include your working definitions of sinh and cosh.\n\nRather than merely proving an identity, it appears that the OP wishes to obtain the title by using the definitions of sinh and cosh starting from their definitions using the sum and difference of exponential functions. In addition, one might taking the point of view that one doesn't know at this stage any properties of cosh except for its working definition. So, it might be \"cheating\" to use any other properties of cosh which you didn't derive already.\n\ncks, is this correct?\n\n(If I'm not mistaken, my suggestion \"For real y, exp(y)>0\" provides a reason to choose the positive sign.)\n\n9. Sep 25, 2007\n\n### cks\n\nOh, I see thank you. Let me rephrase what you all said\n\nexp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)\n\nexp(y) is always > 0\n\nso if we select exp(y) = x - sqrt(x^2 +1)\n\nand most importantly , sqrt(x^2 +1) > x\n\nso exp(y) < 0 which is false.\n\nAs a result, we have to select exp(y) = x + sqrt(x^2 +1)\n\nThe similar argument can be applied to exp(-y)\n\nThank you all of you, really appreciate that.", "date": "2018-01-20 16:06:23", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/if-sinh-y-x-then-show-that-cosh-y-sqrt-1-x-2.186439/", "openwebmath_score": 0.6353123188018799, "openwebmath_perplexity": 2122.5882802446877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708026035287, "lm_q2_score": 0.8933093982432729, "lm_q1q2_score": 0.873094523534303}}
{"url": "http://mathhelpforum.com/algebra/7834-simultaneous-equations-way-too-many-unknowns.html", "text": "# Math Help - Simultaneous Equations with way too many unknowns..\n\n1. ## Simultaneous Equations with way too many unknowns..\n\nYea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.\n\n'Find the values of m such that these equations have no solutions':\n\n3x - my = 4\nx + y = 12\n\n'Find the values of m and a such that these equations have infinite solution sets':\n\n4x - my = a\n2x + y = 4\n\nSo if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.\nThanks a lot\n\n2. Originally Posted by Fnus\nYea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.\n\n'Find the values of m such that these equations have no solutions':\n\n3x - my = 4\nx + y = 12\n\n'Find the values of m and a such that these equations have infinite solution sets':\n\n4x - my = a\n2x + y = 4\n\nSo if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.\nThanks a lot\nIf the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.\n\nSo since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then \"-m\" must be twice the coefficient of y in the second equation, so\n$-m = 2 \\cdot 1$\nor m = -2.\n\nSimilarly a = 8.\n\n-Dan\n\n3. there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.\n\nQuestion 1:\n'Find the values of m such that these equations have no solutions':\n\n3x - my = 4\nx + y = 12\n\nQuestion 2:\n'Find the values of m and a such that these equations have infinite solution sets':\n\n4x - my = a\n2x + y = 4\n\nDoes that change anything? <.<\n\n5. $x+y=12\\Rightarrow y=-x+12$ so the slope is -1.\n\n$3x-my=4\\Rightarrow y=\\frac{3}{m}x-\\frac{4}{m}$\n\nnow solve $\\frac{3}{m}=-1$ you should get $m=-3$\n\n6. for the second one we have\n\n$4x-my-a=0$\nand\n$2(2x+y)=2(4)\\Rightarrow 4x+2y-8=0$\n\nfrom here it should be obvious what values to make a and m.\n\n7. Hello, Fnus!\n\nFind the values of $m$ such that these equations have no solutions:\n. . $\\begin{array}{cc}(1)\\\\(2)\\end{array}\n\\begin{array}{cc}3x - my \\\\ x + y \\end{array}\n\\begin{array}{cc} = \\\\ = \\end{array}\n\\begin{array}{cc}4 \\\\ 12\\end{array}$\n\nIf you are familiar with determinants, there is a simple solution.\n\nA system has no solution if its determinant equal zero.\n\nThe determinant is: . $\\begin{vmatrix} 3 & \\text{-}m \\\\ 1 & 1\\end{vmatrix} \\:=\\:(3)(1) - (\\text{-}m)(1) \\:=\\:3 + m$\n\nTherefore: . $3 + m \\:=\\;0\\quad\\Rightarrow\\quad\\boxed{m = -3}$\n\nOtherwise, we can try to solve the system.\n\nWe have equation (1): . $3x - my \\:=\\:4$\n. . .Multiply (2) by $m:\\;\\;mx + my \\:=\\:12m$\n\n. . . . . . . . . . . . Add: . $mx + 3x \\:=\\:12m + 4$\n\n. . . . . . . . . . .Factor: . $(m + 3)x \\:=\\:12m + 4$\n\n. . . . . . . . . . . Then: . . . . . . $x\\:=\\:\\frac{12m + 4}{m + 3}$\n\nWe see that $x$ is undefined if $m = -3.$\n\nTherefore, the system has no solutions for $m = -3.$\n\n8. Originally Posted by topsquark\nIf the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.\n\nSo since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then \"-m\" must be twice the coefficient of y in the second equation, so\n$-m = 2 \\cdot 1$\nor m = -2.\n\nSimilarly a = 8.\n\n-Dan\nOriginally Posted by putnam120\nthere are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.\nBut if one equation is merely a multiple of the other, they represent the same line, so any point (x,y) on the line is a solution. So we have an infinite number of solutions.\n\n-Dan", "date": "2015-08-03 02:08:15", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/7834-simultaneous-equations-way-too-many-unknowns.html", "openwebmath_score": 0.9035454988479614, "openwebmath_perplexity": 316.50477492416775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936101542133, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8730923803813397}}
{"url": "https://math.stackexchange.com/questions/499537/relationship-between-o-and-o-notation/499541", "text": "# Relationship between O and o notation\n\nIn big-O notation, $f(x) = O(g(x))$ as $x\\rightarrow \\pm\\infty$ if\n\n$$\\exists C, \\delta>0: \\forall |x| \\geq \\delta: |f(x)| < C |g(x)|$$\n\nand, for the case I'm more interested in here, $f(x) = O(g(x))$ as $x\\rightarrow 0$ if\n\n$$\\exists C, \\delta>0: \\forall |x| \\leq \\delta: |f(x)| < C |g(x)|$$\n\nIn little-o notation, $f(x) = o(g(x))$ means\n\n$$\\lim_{|x|\\rightarrow0}\\frac{f(x)}{|g(x)|} = 0$$\n\nI've come across big-O before, but this is the first time I've seen little-o. I have experience in applied math, but very little in pure math, and my first reaction (for $x\\rightarrow 0$) was \"they mean the same thing\". Then I thought \"but little-o is a more precise upper bound\".\n\nWhat is the relationship between these two notations, and what does the distinction mean practically?\n\n\u2022 Well, if $f(x)$ is $o(g(x))$ it is also $O(g(x))$ but maybe not the other way around. It is not \"more precise\", but rather stronger: intuitively, if $f(x)$ is $o(g(x))$ then $g(x)$ grows \"much faster\" than $f(x)$, whereas $O(g(x))$ implies only that $f(x)$ is eventually bounded above (by some constant multiple of) $g(x)$ -- indeed, the growth rates could be comparable. \u2013\u00a0The_Sympathizer Sep 20 '13 at 11:54\n\u2022 An example -- though note here I'm using the O-notation for growth near infinity, not near 0: If $f(x) = x$ and $g(x) = x^2$, then $f(x)$ is $o(g(x))$ AND $O(g(x))$, but if $f(x) = x^2$ then $f(x)$ is only $O(g(x))$ and NOT $o(g(x))$. \u2013\u00a0The_Sympathizer Sep 20 '13 at 11:56\n\u2022 An example for near 0: $f(x) = \\frac{1}{x}$ and $g(x) = \\frac{1}{x^2}$. Then $f(x)$ is $o(g(x))$ (as $x \\rightarrow 0$) while if $f(x) = \\frac{1}{x^2}$ then $f(x)$ is only $O(g(x))$ and not $o(g(x))$. \u2013\u00a0The_Sympathizer Sep 20 '13 at 11:58\n\nIn the briefest possible terms, $f \\in O(g)$ is like saying $f \\leqslant g$ asymptotically, and $f \\in o(g)$ is like saying $f < g$ asymptotically.\n\nAs an example, if $f(x) = x$ and $g(x) = 2x$,we have that that $f \\in O(g)$, but $f \\notin o(g)$ and $g \\notin o(f)$. If $f(x) = x^2$ and $g(x) = x$, then $f \\in o(g)$ and $f \\in O(g)$.\n\nAs an aside, be very aware of context. In various probabilistic situations, we define $O$-notation with a limit as $x\\rightarrow 0$. In most algorithmic applications, we take limits as $x \\rightarrow \\infty$. Always make sure it's clear what you're working with before using either notation. $O$-notation is probably the most abused thing in all of mathematics.\n\n\u2022 Also note: Notation $f \\in O(g)$ is quite rare, compared to notation $f = O(g)$. \u2013\u00a0GEdgar Sep 20 '13 at 13:42\n\u2022 Indeed. I didn't want to start listing abuses of $O$ notation because I don't think I'd ever stop. It's more common to say things like $2x = O(x)$. \u2013\u00a0ymbirtt Sep 20 '13 at 14:13\n\u2022 $x^2$ is not in $O(x)$, at least when $x \\to \\infty$. Typo? \u2013\u00a0ftfish Sep 20 '13 at 14:44\n\u2022 From OP's definitions, we're sending $x \\rightarrow 0$, so I'm remaining consistent with that. \u2013\u00a0ymbirtt Sep 20 '13 at 15:08\n\u2022 @GEdgar thanks, discussion of this helps me to get to grips with it. I've always been a little bit uncomfortable with the use of $=$, and thinking about it in terms of sets is more consistent. \u2013\u00a0TooTone Sep 21 '13 at 9:07\n\nLittle $o$ means lim sup of absolute value $=0$, big $O$ means lim sup of absolute value $< \\infty$.\n\n\u2022 Thanks, but I thought big O could be used for $\\rightarrow 0$ as well as $\\rightarrow \\infty$? The former was what I was more interested in because it then has a similar meaning to little o. (My post was a little ambiguous and I see you changed $<$ in my post to $\\ge$ -- I've changed that back and added some more which hopefully makes it clearer.) \u2013\u00a0TooTone Sep 20 '13 at 14:31\n\u2022 You are referring to my correction? Okay, I am sorry then the same remains true. It should be indicated though wether you look at the behaviour at $0$ or at $\\infty$. A continuous variable transformation, e.g. $x \\mapsto x^{-1}$ exchanges between the setting and my statement remains valid;) \u2013\u00a0Marc Palm Sep 20 '13 at 14:39\n\u2022 Yes you're right I should have indicated whether I was looking at limit towards zero or infinity. I can see how a reciprocal transformation might make the situations equivalent; however I still don't quite understand your answer. \u2013\u00a0TooTone Sep 20 '13 at 19:52\n\u2022 my answer remains valid no matter to where the limit goes. \u2013\u00a0Marc Palm Sep 21 '13 at 14:57\n\u2022 I think what I don't understand is, in simple terms, your use of \"sup\" means. \u2013\u00a0TooTone Sep 21 '13 at 18:01\n\ncheck this out :\n\nhttp://en.wikipedia.org/wiki/Big_O_notation\n\n:-) it's not just Big O notation in the page\n\nLooking at how $O(1)$ and $o(1)$ are sufficient to understand the general case.\n\nSuppose $f \\in O(1)$. Then\n\n$$\\lim_{x \\to +\\infty} f(x) < +\\infty$$\n\nif the limit exists. Or more generally, if the limit does not exist, then\n\n$$\\sup_{x \\to +\\infty} f(x) < +\\infty$$\n\nNow suppose $f \\in o(1)$. Then\n\n$$\\lim_{x \\to \\infty} f(x) = 0$$", "date": "2019-09-20 16:34:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/499537/relationship-between-o-and-o-notation/499541", "openwebmath_score": 0.9034019112586975, "openwebmath_perplexity": 383.08293963160264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936054891428, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8730923762424678}}
{"url": "http://math.stackexchange.com/questions/442768/fixed-point-theorems", "text": "# Fixed Point Theorems\n\nTheorem 1. Let $B=\\{x\\in \\mathbb R^n :\u2225x\u2225\u22641\\}$ be the closed unit ball in $\\mathbb R^n$ . Any continuous function $f:B\\rightarrow B$ has a fixed point.\n\nTheorem 2. Let $X$ be a finite dimensional normed vector space, and let $K\\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\\rightarrow K$ there exists $x\\in K$ such that $f(x)=x$.\n\nTheorem 3. Let $X$ be a normed vector space, and let $K\\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\\rightarrow K$ there exists $x\\in K$ such that $f(x)=x$.\n\nTheorem 4. Let $X$ be a normed vector space, and let $K\\subset X$ be a non-empty, closed, and bounded set. Then given any compact mapping $f:K\\rightarrow K$ there exists $x\\in K$ such that $f(x)=x$.\n\nFor some authors Theorem 1 is Brouwer's fixed-point theorem. For others Brouwer's fixed-point theorem is Theorem 2. Actually there is no difference because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball.\n\nMy problem is with theorems 3 and 4. For some authors Theorem 3 is Schauder's fixed-point theorem, for others Schauder's fixed-point theorem is Theorem 4.\n\nAre Theorem 3 and Theorem 4 are equivalent? If not, are Theorems 1 and 2 special cases of Theorem 4?\n\n-\n\"because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball\" is not quite true, it could be lower-dimensional. Theorems 1 and 2 are special cases, because in that case, a continuous $f \\colon K \\to K$ is compact. \u2013\u00a0Daniel Fischer Jul 13 '13 at 15:39\nWhat is the nonl tag for? \u2013\u00a0draks ... Jul 13 '13 at 20:08\n@draks... My guess is a foreshortened nonlinear-something that never really got off the ground. \u2013\u00a0Sharkos Jul 13 '13 at 22:56\n\nYour statement of Theorem 4 is missing an assumption on $K$, such as being convex, or at least homeomorphic to such a set (convex, closed, bounded). Without such an assumption, rotation of a circle gives a counterexample. Also, I think that in Theorem 4 you want the normed space to be complete, i.e., a Banach space.\n\nTheorem 3 is contained in Theorem 4, because on a compact set every continuous map is compact. Theorem 4 cannot be easily obtained from Theorem 3 (I think) because if we tried to simply replace $K$ with $\\overline{f(K)}$ (which is compact), we can't apply Theorem 3 because $\\overline{f(K)}$ is not known to be convex.\n\nBoth 3 and 4 were stated and proved by Schauder in his 1930 paper Der Fixpunktsatz in Funktionalra\u00fcmen, which is in open access. Here is Theorem 3:\n\nSatz I. Die stetige Funktionaloperation $F(x)$ bilde die konvexe, abgeschlossene und kompakte Menge $H$ auf sich selbst ab. Dann ist ein Fixpunkt $x_0$, vorhanden, d.h. es gilt $F(x_0)=x_0$.\n\nAnd this is Theorem 4 (in slightly less general version: the image of $F$ is assumed compact instead of relatively compact; possibly because the latter concept wasn't in use).\n\nSatz II. In einem \"B\"-Raume sei eine konvexe und abgeschlossene Menge $H$ gegeben. Die stetige Funktionaloperation $F(x)$ bilde $H$ auf sich selbst ab. Ferner sei die Menge $F(H)\\subset H$ kompakt. Dann ist ein Fixpunkt vorhanden.\n\n(\"B\"-Raume is what is now called a Banach space.) So, it is correct to call both Theorem 3 and Theorem 4 \"Schauder's fixed-point theorem\".\n\nAnd yes, Theorems 1 and 2 follow by specialization of Theorem 3 or 4 to finite dimensions.\n\n-", "date": "2016-05-06 21:31:22", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/442768/fixed-point-theorems", "openwebmath_score": 0.9219070076942444, "openwebmath_perplexity": 135.68402205155618, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936050226357, "lm_q2_score": 0.8872045847699186, "lm_q1q2_score": 0.8730923582188398}}
{"url": "http://mathhelpforum.com/calculus/4151-ship-problem.html", "text": "# Math Help - ship problem\n\n1. ## ship problem\n\nA ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.\nBoth ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.\n\nIv found the co-ordinates of Q relative to P at t=0\n---->X=(10,10)km\n\niv also found the velocity of Q relative to P----> V= V[Q] -V[P]\n-----.V= (0,40) - V(30,0)\n----- V= (-30,40)km/h\n\nbut im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz\n\n2. At t=0, they are both 10 kms from their intersection.\n\nThink Pythagoras. Let D=square of distance between ships.\n\n$D(t)=(10-30t)^{2}+(10-40t)^{2}$\n\n$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$\n\n$D'(t)=5000t-1400$\n\n$5000t-1400=0$\n\n$t=\\frac{7}{25}$\n\n3. Originally Posted by galactus\nAt t=0, they are both 10 kms from their intersection.\n\nThink Pythagoras. Let D=square of distance between ships.\n\n$D(t)=(10-30t)^{2}+(10-40t)^{2}$\n\n$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$\n\n$D'(t)=5000t-1400$\n\n$5000t-1400=0$\n\n$t=\\frac{7}{25}$\nthankz...but i was wondering if i did the first part right that i showed?(shown below) because im not too sure if i was meant to include negative signs with any of the answers? can u please just check that for me thankz\n\nIv found the co-ordinates of Q relative to P at t=0\n---->X=(10,10)km\n\niv also found the velocity of Q relative to P----> V= V[Q] -V[P]\n-----.V= (0,40) - V(30,0)\n----- V= (-30,40)km/h\n\n4. Originally Posted by dopi\nA ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.\nBoth ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.\nHello, dopi,\n\nwith your problem you use automatically a coordinate system: The pos. x-axis points to East and the pos. y-axis points to North.\n\nAt the time t = 0 the ship P is at P(-10, 0) and the ship Q is at Q(0, 10). The movement of both ships is described by a straight line. The speed is described by a vector:\n$\\overrightarrow{v_P}=(30, 0)$\n$\\overrightarrow{v_Q}=(0, -40)$\n\nThus P is moving along the line:\n$\\overrightarrow{x_P}=(-10, 0)+t*(30, 0)$\nand Q is moving along the line:\n$\\overrightarrow{x_Q}=(0, 10)+t*(0, -40)$\n\nThe distance between the ships is: $\\vec{d}=\\overrightarrow{x_P}-\\overrightarrow{x_Q}$\n\nOriginally Posted by dopi\nIv found the co-ordinates of Q relative to P at t=0\n---->X=(10,10)km\n\niv also found the velocity of Q relative to P----> V= V[Q] -V[P]\n-----.V= (0,40) - V(30,0)\n----- V= (-30,40)km/h\nI don't understand why you want to calculate the relative coordinates. It isn't necessary with your problem.\n\nOriginally Posted by dopi\nbut im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz\nThis was already done by galactus.\n\nGreetings\n\nEB\n\n5. Hello, dopi!\n\nYou seem to be using a coordinate system and vectors\n. . and I don't know what you plan to do with them.\n\ngalactus gave you the solution . . . so why struggle with your approach?\n\nI'll make some diagrams to accompany his excellent solution.\n\nOriginally Posted by galactus\n\nAt t=0, they are both 10 kms from their intersection.\nCode:\n                        * Q\n|\n|\n| 10\n|\n|\n|\nP * - - - - - - - - o\n10\n\n$t$ hours later, $P$ has moved $30t$ km east to point $A$\n. . and $Q$ has moved $40t$ km south to point $B.$\nCode:\n                        *\n|\n| 40t\n|\no B\n|\n| 10-40t\n* - - - o - - - - *\n30t  A  10-30t\nAnd we want to minimize the distance $\\overline{AB}.$\n\nOriginally Posted by galactus\n\nThink Pythagoras. Let D = square of distance between ships.\n\n$D(t) \\:= \\10-30t)^{2} + (10-40t)^{2}\" alt=\"D(t) \\:= \\10-30t)^{2} + (10-40t)^{2}\" />\n\n$D'(t) \\:= \\:2(10-30t)(-30) + 2(10=40t)(-40)$\n\n$D'(t) \\:= \\:5000t - 1400$\n\n$5000t -1400\\:=\\:0$\n\n$t \\:= \\:\\frac{7}{25}$", "date": "2014-10-21 11:13:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/4151-ship-problem.html", "openwebmath_score": 0.7105571031570435, "openwebmath_perplexity": 2039.4087320082756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683487809279, "lm_q2_score": 0.8840392893839086, "lm_q1q2_score": 0.8730492212743316}}
{"url": "http://mathhelpforum.com/calculus/146987-tricky-double-integral-polar-coordinates.html", "text": "# Math Help - Tricky Double Integral with polar coordinates\n\n1. ## Tricky Double Integral with polar coordinates\n\nEvaluate the integral:\n\n$\\int \\int_D \\frac{x^2}{x^2 + y^2} dx\\;dy$\n\nwhere $D$ is the region $\\{ (x,y)| y \\geq x \\; \\textrm{and} \\; x^2 + y^2 \\leq 3 \\}$\n\nTried drawing the region $D$ and it looks like I have to use polar coordinates.\n\nUsing polar coordinates, $\\frac{x^2}{x^2 + y^2} = \\frac{r^2 \\cos^2{\\theta}}{r^2} = \\cos^2{\\theta}$.\n\nI'm not sure how to find the limits if integration though, or what to do with the $dx\\;dy$.\n\nThanks in advance, any pointers would be greatly appreciated.\n\n2. Originally Posted by craig\nEvaluate the integral:\n\n$\\int \\int_D \\frac{x^2}{x^2 + y^2} dx\\;dy$\n\nwhere $D$ is the region $D = \\{ (x,y)| y \\geq x \\; \\textrm{and} \\; x^2 + y^2 \\leq 3 \\}$\n\nTried drawing the region $D$ and it looks like I have to use polar coordinates.\n\nUsing polar coordinates, $\\frac{x^2}{x^2 + y^2} = \\frac{r^2 \\cos^2{\\theta}}{r^2} = \\cos^2{\\theta}$.\n\nI'm not sure how to find the limits if integration though, or what to do with the $dx\\;dy$.\n\nThanks in advance, any pointers would be greatly appreciated.\nIn general to convert, you do the following:\n\n$\\displaystyle \\int \\int f(x,y) \\, dx \\, dy = \\int \\int f(r,\\theta)\n\\frac{\\partial (x,y)}{\\partial (r, \\theta)} \\, dr \\, d\\theta$\n\nWhere $\\frac{\\partial (x,y)}{\\partial (r, \\theta)}$ is called the jacobian determinant and it's given by:\n\n$\\frac{\\partial (x,y)}{\\partial (r, \\theta)} =\n\\begin{vmatrix}\n\\cos \\theta & - r \\sin \\theta \\\\\n\\sin \\theta & r \\cos \\theta\n\\end{vmatrix} = r$\n\nHence:\n\n$\\displaystyle \\int \\int f(x,y) \\, dx \\, dy = \\int \\int f(r,\\theta)\nr \\, dr \\, d\\theta$\n\nNow, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.\n\n$D = \\{ (x,y)| y \\geq x \\; \\textrm{and} \\; x^2 + y^2 \\leq 3 \\}$\n\nNow, the 2nd constraint on the region says that $x^2 + y^2 \\leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\\sqrt{3}$, in other words, the radius is limited by $0 \\leq r \\leq \\sqrt{3}$.\n\nNow, the 1st constraint on the region says that $y \\geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\\frac{\\pi}{4} \\leq \\theta \\leq \\frac{5\\pi}{4}$.\n\nYou can see all of this visually by drawing it out. Go on, do this:\n\n1) Draw the x-y coordinates.\n2) Draw a circle centred at the origin with a radius of $\\sqrt{3}$.\n3) Draw the line y = x.\n4) Now shade the region that lies above the line y = x, but does not go outside the circle.\n\nYou should see that this region is bounded by:\n\n$D = \\bigg\\{ (r,\\theta)| 0 \\leq r \\leq \\sqrt{3} \\; \\textrm{and} \\; \\frac{\\pi}{4} \\leq \\theta \\leq \\frac{5\\pi}{4} \\bigg\\}$\n\nHence, in this particular situation:\n\n$\\int \\int_D \\frac{x^2}{x^2 + y^2} dx\\;dy = \\int_0^{\\frac{5\\pi}{4}} \\int_0^{\\sqrt{3}} r \\, \\cos^2(\\theta) \\, dr \\, d\\theta = \\int_0^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg( \\int_0^{\\sqrt{3}} r \\, \\, dr\\bigg) \\, d\\theta$\n\n3. Hi\n\nThe area of integration is the part of the disk which is above the red line\n\nAnd $dx dy$ becomes $r dr d\\theta$\n\n4. Originally Posted by craig\nEvaluate the integral:\n\n$\\int \\int_D \\frac{x^2}{x^2 + y^2} dx\\;dy$\n\nwhere $D$ is the region $\\{ (x,y)| y \\geq x \\; \\textrm{and} \\; x^2 + y^2 \\leq 3 \\}$\n\nTried drawing the region $D$ and it looks like I have to use polar coordinates.\n\nUsing polar coordinates, $\\frac{x^2}{x^2 + y^2} = \\frac{r^2 \\cos^2{\\theta}}{r^2} = \\cos^2{\\theta}$.\n\nI'm not sure how to find the limits if integration though, or what to do with the $dx\\;dy$.\n\nThanks in advance, any pointers would be greatly appreciated.\nI go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html\n\nIf you draw the xy domain you can see we want the area to the left of the line but inside the circle,\n\nIf we convert this to polar co-ordinates this means\n\n$\\frac{ \\pi }{4} \\le \\theta \\le \\frac{ 5 \\pi }{4}$ where the $\\frac{ \\pi }{4}$ comes from the fact that the line $y=x$ acts at $\\frac{ \\pi }{4}$ above the x-axis. And $\\frac{ 5 \\pi }{4}$ comes from the angle between the line in the first quadrent to the line in the 3rd quadrent.\n\nFor r, we are going from the origin to the radius of the circle. So the interval of r must be\n\n$0 \\le r \\le \\sqrt{3}$\n\nWe can now compute this integral,\n\n$\\int \\int_D \\frac{x^2}{x^2 + y^2} dxdy$\n\n$= \\iint_Q \\frac{ r^2 cos^2 \\theta }{ r^2 } rdr d \\theta$\n\n$= \\iint_Q rcos^2 \\theta dr d \\theta$\n\n$= \\int_0^{ \\sqrt{3} } rdr \\int_{ \\frac{ \\pi }{4} }^{ \\frac{ 5 \\pi }{4} } cos^2 \\theta d \\theta$\n\nWhich can be easily computed using double angle formulas\n\n5. Wow leave the forum for a few hours and there's 3 great replies waiting for me!!\n\nOriginally Posted by Mush\nIn general to convert, you do the following:\n\n$\\displaystyle \\int \\int f(x,y) \\, dx \\, dy = \\int \\int f(r,\\theta)\n\\frac{\\partial (x,y)}{\\partial (r, \\theta)} \\, dr \\, d\\theta$\n\nWhere $\\frac{\\partial (x,y)}{\\partial (r, \\theta)}$ is called the jacobian determinant and it's given by:\n\n$\\frac{\\partial (x,y)}{\\partial (r, \\theta)} =\n\\begin{vmatrix}\n\\cos \\theta & - r \\sin \\theta \\\\\n\\sin \\theta & r \\cos \\theta\n\\end{vmatrix} = r$\n\nHence:\n\n$\\displaystyle \\int \\int f(x,y) \\, dx \\, dy = \\int \\int f(r,\\theta)\nr \\, dr \\, d\\theta$\nAhh yeh I've done Jacobian determinants before, it didn't click that it applied here.\n\nOriginally Posted by Mush\nNow, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.\n\n$D = \\{ (x,y)| y \\geq x \\; \\textrm{and} \\; x^2 + y^2 \\leq 3 \\}$\n\nNow, the 2nd constraint on the region says that $x^2 + y^2 \\leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\\sqrt{3}$, in other words, the radius is limited by $0 \\leq r \\leq \\sqrt{3}$.\n\nNow, the 1st constraint on the region says that $y \\geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\\frac{\\pi}{4} \\leq \\theta \\leq \\frac{5\\pi}{4}$.\n\nYou can see all of this visually by drawing it out. Go on, do this:\n\n1) Draw the x-y coordinates.\n2) Draw a circle centred at the origin with a radius of $\\sqrt{3}$.\n3) Draw the line y = x.\n4) Now shade the region that lies above the line y = x, but does not go outside the circle.\n\nYou should see that this region is bounded by:\n\n$D = \\bigg\\{ (r,\\theta)| 0 \\leq r \\leq \\sqrt{3} \\; \\textrm{and} \\; \\frac{\\pi}{4} \\leq \\theta \\leq \\frac{5\\pi}{4} \\bigg\\}$\nI'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess\n\nOriginally Posted by Mush\nHence, in this particular situation:\n\n$\\int \\int_D \\frac{x^2}{x^2 + y^2} dx\\;dy = \\int_0^{\\frac{5\\pi}{4}} \\int_0^{\\sqrt{3}} r \\, \\cos^2(\\theta) \\, dr \\, d\\theta = \\int_0^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg( \\int_0^{\\sqrt{3}} r \\, \\, dr\\bigg) \\, d\\theta$\n** Think you mean $\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg( \\int_0^{\\sqrt{3}} r \\, \\, dr\\bigg) \\, d\\theta$\n\nAnyway on with the integral:\n\n$\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg( \\int_0^{\\sqrt{3}} r \\, \\, dr\\bigg) \\, d\\theta$\n\n$\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg[ \\frac{r^2}{2} \\bigg]_0^{\\sqrt{3}} \\, d\\theta$ = $\\frac{3}{2}\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\frac{1}{2}(1 + \\cos{2\\theta}) \\, d\\theta$ =\n\n$\\frac{3}{4}\\bigg[\\theta + \\frac{1}{2}\\sin{2\\theta}\\bigg]_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}}$ =\n\n$\\frac{3}{4}\\bigg(\\frac{5\\pi}{4} + \\frac{1}{2}\\sin{\\frac{5\\pi}{2}} -\\frac{pi}{4} - \\frac{1}{2}\\sin{\\frac{\\pi}{2}} \\bigg)$\n\n$\\frac{3}{4}(\\pi + 0)$ = $\\frac{3\\pi}{4}$\n\nThanks a lot for the help!\n\n6. Originally Posted by running-gag\nHi\n\nThe area of integration is the part of the disk which is above the red line\n\nAnd $dx dy$ becomes $r dr d\\theta$\nThanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!\n\nOriginally Posted by AllanCuz\nI go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html\n\nIf you draw the xy domain you can see we want the area under the line but inside the circle,\n\nIf we convert this to polar co-ordinates this means\n\n$0 \\le \\theta \\le \\frac{ \\pi }{4}$ where the $\\frac{ \\pi }{4}$ comes from the fact that the line $y=x$ acts at $\\frac{ \\pi }{4}$ above the x-axis.\n\nFor r, we are going from the origin to the radius of the circle. So the interval of r must be\n\n$0 \\le r \\le \\sqrt{3}$\n\nWe can now compute this integral,\n\n$\\int \\int_D \\frac{x^2}{x^2 + y^2} dxdy$\n\n$= \\iint_Q \\frac{ r^2 cos^2 \\theta }{ r^2 } rdr d \\theta$\n\n$= \\iint_Q rcos^2 \\theta dr d \\theta$\n\n$= \\int_0^{ \\sqrt{3} } rdr \\int_0^{ \\frac{ \\pi }{4} } cos^2 \\theta d \\theta$\n\nWhich can be easily computed using double angle formulas\nThanks a lot for the reply, really helped.\n\n7. Originally Posted by craig\nWow leave the forum for a few hours and there's 3 great replies waiting for me!!\n\nAhh yeh I've done Jacobian determinants before, it didn't click that it applied here.\n\nI'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess\n\n** Think you mean $\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg( \\int_0^{\\sqrt{3}} r \\, \\, dr\\bigg) \\, d\\theta$\n\nAnyway on with the integral:\n\n$\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg( \\int_0^{\\sqrt{3}} r \\, \\, dr\\bigg) \\, d\\theta$\n\n$\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\cos^2(\\theta) \\bigg[ \\frac{r^2}{2} \\bigg]_0^{\\sqrt{3}} \\, d\\theta$ = $\\frac{3}{2}\\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\frac{1}{2}(1 + \\cos{2\\theta}) \\, d\\theta$ =\n\n$\\frac{3}{4}\\bigg[1 + \\cos{2\\theta}\\bigg]_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}}$ = $\\frac{3}{4}(\\pi)$ = $\\frac{3\\pi}{4}$\n\nThanks a lot for the help!\nI don't think you're performed that integration properly, have you?\n\nThe integral of $1 + \\cos(2\\theta)$ is $\\theta + \\frac{1}{2}\\sin(2\\theta)$.\n\n8. Originally Posted by craig\nThanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!\n\nThanks a lot for the reply, really helped.\nI messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \\ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P\n\n9. Originally Posted by Mush\nI don't think you're performed that integration properly, have you?\nYes & no. I did the integration properly, just forgot to update the latex code...oops.\n\nEdited my above working now, sorry about that.\n\n10. Originally Posted by AllanCuz\nI messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \\ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P\nHaha thought that was what you meant, thanks again for the reply.\n\n11. Originally Posted by craig\nYes & no. I did the integration properly, just forgot to update the latex code...oops.\n\nEdited my above working now, sorry about that.\nIndeed. Good show.", "date": "2015-01-25 14:53:23", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/146987-tricky-double-integral-polar-coordinates.html", "openwebmath_score": 0.9031927585601807, "openwebmath_perplexity": 288.2803325455937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877012965886, "lm_q2_score": 0.888758789809389, "lm_q1q2_score": 0.8730168286490028}}
{"url": "https://math.stackexchange.com/questions/2977030/elementary-question-about-a-fake-proof-and-greatest-common-divisors", "text": "# Elementary question about a fake-proof and greatest common divisors\n\nI have a question to an excercise - for which I have a wrong solution - and I wanted to ask you to help me understand my thinking error. The excercise was as follows:\n\nLet $$a, b, n \\in \\mathbb{N}$$. Show that $$\\gcd(a,b) = \\gcd(a,b+na)$$.\n\nMy solution was (in a nutshell):\n\n\u2022 Let $$d := \\gcd(a,b)$$.\n\u2022 Then I showed that $$d$$ is a common divisor of both $$a$$ and $$b + na$$\n\u2022 Then I showed, that any common divisor $$c$$ of $$a$$ and $$b+na$$ is smaller-or-equal than $$d$$\n\u2022 After that, I concluded that by definition $$d = \\gcd(a,b+na)$$, and with the definition of $$d$$ I would have $$\\gcd(a,b) = \\gcd(a,b+na)$$, so my proof was completed.\n\nHowever, my tutor did not accept the proof. He showed me the 'correct' proof, in which I woould have $$d := \\gcd(a,b)$$ and $$e := \\gcd(a,b+na)$$ and then demonstrate $$d \\leq e \\leq d$$.\n\nAnyway, I do understand his proof. But I still do not understand where my thinking error is. After all, if my starting line would be, let's say, $$d := 9$$ and I would have been able to show the steps afterwards, then I would be able to conclude $$9 = \\gcd(a,b+na)$$, no?\n\nEDIT Wow, you guys answered fast! Thank you very much. I will ask my tutor on thursday again. For completeness sake, I will formulate my complete proof (my original proof is in german, so maybe there will be something lost in translation, the [Reference] are references to our script).\n\nLet $$a,b,n \\in \\mathbb{N}$$. Let $$d := \\gcd(a,b)$$. By definition it means $$d \\mid a$$ and $$d \\mid b$$, so with [Reference] the following holds true: $$d \\mid (b + na)$$. Therefor, $$d$$ is a common divisor of both $$a$$ and $$b + na$$. We will show now, that $$d$$ is the greatest common divisor.\n\nLet $$c$$ be any common divisor of $$a$$ and $$b + na$$. Therefor $$c$$ divides $$a$$, and as such $$c \\mid na$$ und therefor $$c \\mid ((b + na) - na) = b$$ (again because of [Reference]). So $$c$$ divides $$b$$, and therefor $$c$$ is a common divisor of both $$a$$ and $$b$$. Since $$d$$ was the greatest common divisor of $$a$$ and $$b$$, we have $$c \\leq d$$.\n\nBy definition of the greatest common divisor we get $$d = \\gcd(a,b+na)$$.\n\nEnd of Proof. My tutor wrote, that I only showed $$\\gcd(a,b+na) \\leq \\gcd(a,b)$$ and also need to show the other direction. Then he showed me the 'correct'/'ideal' proof today. But I will ask him again on thursday!\n\n\u2022 You need to be more specific with your proof. From the outline there should be no objection, so either your tutor is wrong or you made an error. We can't determine which. \u2013\u00a0Matt Samuel Oct 30 '18 at 2:36\n\u2022 Possibly the confusion arises because you used two different proofs to show that neither gcd exceeds the other, but you can just repeat the same proof replacing $b$ and $n$ with $b+na$ and $-n$. \u2013\u00a0Neil Oct 30 '18 at 9:24\n\u2022 This has been the way I have understood things: The following is true $\\forall k \\in \\mathbb N$, for any pair of integers $(n,j)$: $$n\\gcd(j,k)-\\gcd(j,n)\\Bigl\\lfloor \\frac{n\\gcd(j,k)}{\\gcd(j,n)}\\Bigr\\rfloor=0$$ $$j\\gcd(j,k)-\\gcd(j,n)\\Bigl\\lfloor \\frac{j\\gcd(j,k)}{\\gcd(j,n)}\\Bigr\\rfloor=0$$ \u2013\u00a0Adam L Nov 3 '18 at 17:14\n\u2022 The above two equalities can be summed only in integer multiples of $n$ and $j$,ie $\\forall a,b \\in \\mathbb Z$ $$(an+bj)\\gcd(j,k)-\\gcd(j,n)\\Bigl\\lfloor \\frac{(an+bj)\\gcd(j,k)}{\\gcd(j,n)}\\Bigr\\rfloor=0$$ \u2013\u00a0Adam L Nov 3 '18 at 17:14\n\nThe outline of the proof that you have explained is correct.\n\nMore than likely the problem was with the detail in one of the intermediate steps.\n\nI would check each step carefully again and there is always room for skipping a point in writing a proof.\n\nYou want to know where the error is in your proof. Given no details no one can give the answer. All we can do is give only our opinion, so do not treat this as an answer.\n\nThe sketch given is correct, but it is merely a restatement of what is required to be proved. Possible that both you and your tutor are correct, but tutor did not understand the proof and misjudged it. But given that you expect people to be able to analyze your \"proof\" without providing details I am inclined to believe you are wrong. (IT IS AN OPINION!)\n\n\u2022 Yes, I also did assume that I was wrong. This is why I provided only the outline sketch since I thought my thinking error would have been with my 'proof-strategy'. I translated now my complete proof and my tutors response as an edit to the original question. \u2013\u00a0DaGoddyMan Oct 30 '18 at 3:18\n\nOk well firstly, if your tutor has not mentioned to you something by the name of the Euclidean Algorithm, then he or she isn't being very forthcoming about their level of expertise in the subject it's as simple as that. But follow the link I provided there, and apply the method to both greatest common divisor expressions, and this might be able to help you reach a better level of understanding on the subject, the process is clearly explained in the link.\n\nAlso important to note that you shouldn't worry about studying a modern day definition of an algorithm,don't start to think this has anything to do with programming or a need to learn a programming language, this algorithm was invented over 2300 years ago, there was no necessity to update java at that point.\n\nSecondly if your tutor's final step is to demonstrate that $$d \\leq e \\leq d$$, then it is in fact their solution that is completely bogus and circular, since this statement is equivalent to the original equality that we are trying to prove!\n\nI recommend purchasing a hard cover book in Analytic Number Theory, one that provides the solutions for every exercise it contains, so that if you really are unable to find a proof you are confident with, you can look at the one the book has provided.\n\nI currently use one that I have found to be priceless, \"Problems in Analytic Number Theory\" Second Edition, written by M.Ram Murty. But do ask your supervisors at your college what they recommend.\n\n\u2022 Proving that $d=e$ by proving that $d\\leq e\\leq d$ is a standard approach and there is nothing bogus in it. \u2013\u00a0Ister Oct 31 '18 at 7:40\n\u2022 It's the same question. If you asked me to prove dragons are real and I replied \"I shall do so by proving dragons are real, hence proving dragons are real\" would you be satisfied? \u2013\u00a0Adam L Oct 31 '18 at 18:49\n\u2022 No, it's entirely different thing. You have a single statement to prove ($d=e$) but instead you prove two separate statements usually following different approaches ($d\\leq e$ is one proved statement and $e\\leq d$ is another one). Only now you apply a not so obvious implication that if $d\\leq e\\leq d$ holds then $d=e$ so if we've proved the former then the latter is truth indeed. As I said - this method is pretty common, especially when calculating limes but also in some other cases. \u2013\u00a0Ister Nov 2 '18 at 20:54\n\u2022 $d=e$ is just an abbreviation for $d \\leq e \\leq d$. $e \\leq d$ means $e$ is less than or equal to $d$. $d \\leq e$ means $d$ is less and or equal to $e$. $d=e$ is equivalent to the statement $e \\leq d$ and $d \\leq e$, since $e \\lt d=\\lnot(e \\gt d)$ \u2013\u00a0Adam L Nov 3 '18 at 9:09\n\u2022 I totally disagree that it's just an abbreviation. You define $\\leq$ as equal or less (you need to have at least weak order in order to write something like that at all) so equality precedes weak inequality. Of course proving that $d\\leq e \\leq d \\iff d=e$ is trivial but still it's a property of weak inequality. That's why it is worth and can be used as a proof method to separately show each of the weak inequalities and then conclude that it means equality in the end. \u2013\u00a0Ister Nov 5 '18 at 9:51\n\nYour proof is indeed correct. It can be reorganized more symmetrically as follows.\n\nNotice that if $$\\ c\\mid a\\$$ then $$\\ c\\mid b \\!+\\! na \\iff c\\mid b.\\,$$ Therefore $$\\,a,\\, b\\!+\\!na\\,$$ and $$\\,a,\\, b\\,$$ have exactly the same set $$\\,\\cal C\\,$$ of common divisors $$\\,c,\\,$$ so they have the same greatest common divisor $$(= \\max \\,\\cal C)$$\n\n\u2022 @Ister Please read more carefully. The answer says nothing at all about \"invalidating OP's proof\". Nor is there any such mistake. Analyzing (e.g. simplifying) proofs is an important part of teaching. I often point out such simplifcations arising by exploiting innate symmetry, e.g. yesterday using reflection (negation) symmetry to simplify sum computation, and many others. \u2013\u00a0Bill Dubuque Oct 31 '18 at 16:15\n\u2022 I must have had bad day, sorry. My comment removed. \u2013\u00a0Ister Nov 2 '18 at 20:55\n\nIn my opinion your tutor is plainly wrong, too much focused on the proof he's accustomed to. It's often a problem. Let me say I faced this issue all the time since I tended to think out of the box and find my own, correct solutions, different to the canonical one.\n\nYou proved by definition, which is a fully valid proof. Thus technically you don't have to prove anything else.\n\nYet, since your tutor apparently don't understand this basics, you have to be smarter than him and be able to counterargument him (i.e. explain why his statement is plainly wrong).\n\nMy tutor wrote, that I only showed $$gcd(a,b+na)\\leq gcd(a,b)$$ and also need to show the other direction.\n\nIn the first part of your proof by showing that $$d$$ is a divisor of both $$a$$ and $$b$$ you've proved that $$gcd(a,b) = d \\leq gcd(a,b+na)$$. By definition of $$gcd$$ (name even represents that!) for any given divisor $$z$$ of both $$x$$ and $$y$$ we have $$z\\leq gcd(x,y)$$. By substituting $$x=a$$ and $$y=b+na$$ you have the statement that your tutor claims to be missing.\n\nLet me emphasise it once more - you don't need that explanation in your proof, which is perfectly valid. This is only to show that your tutor's claim is wrong.\n\nAs an anecdote...\n\nIn the course of probabilistic we once had some simple homework task from combinatorics. I don't remember details now but it's actually irrelevant. I had done my calculation, verified them and as a result got a nice looking solution.\n\nOn the next lesson I was presenting the solution on the blackboard. Imagine my surprise when I heard my solution is wrong. Someone else had done the task on the blackboard the right way and indeed the results didn't look alike at all. If I remember correctly there were even different powers used in it! Of course, the new solution was correct and in accordance to the key, yet I was sure mine was correct as well (despite all odds and apparent difference) so I stood up for it and asked to point out where had I made an error. No-one, including the tutor could find anything. Eventually the tutor said it might be that the two polynomials are equivalent and if I was able to prove that equivalence I will have the whole course passed with maximum grade.\n\nNeedless to say it took me 3 months of work, some horrible calculations and some peculiar theorem from number theory but I did it. The two results are equivalent.\n\nEnd of Proof. My tutor wrote, that I only showed $$\\gcd(a,b+na) \\leq gcd(a,b)$$ and also need to show the other direction.\n\nI am pretty sure, that $$\\gcd(a,b+na) \\geq gcd(a,b)$$ follows from $$d:=\\gcd(a,b)$$ and $$d\\mid(b+na)$$ (plus the definition of $$\\gcd$$).\n\n\u2022 Can whoever downvoted this please explain, what my error is. If it was due to my bad formatting, I fixed it. \u2013\u00a0Nathrat Oct 30 '18 at 11:26\n\u2022 I didn't downvote but your answer may not be clear. You are trying to give a counterargument to tutor's wrong quoted statement. You should though be more firm in your statement (do the math and make sure you are right) plus add a clear explanation that it means OP is correct in his proof. Note - you are right, but gap in your proof might be to difficult to fill. Try making it more complete (or check my answer as it is actually filling this gap. \u2013\u00a0Ister Oct 31 '18 at 8:01", "date": "2021-03-08 19:49:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2977030/elementary-question-about-a-fake-proof-and-greatest-common-divisors", "openwebmath_score": 0.8031623959541321, "openwebmath_perplexity": 259.6021577996465, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982287697148445, "lm_q2_score": 0.8887587853897073, "lm_q1q2_score": 0.8730168206209047}}
{"url": "https://math.stackexchange.com/questions/3168535/creating-an-integral-from-2-functions", "text": "# Creating an integral from 2 functions\n\nLet $$R$$ be the region in $$\\mathbb{R}^2$$ below the line $$y = x + 2$$ and above the parabola $$y = x^2$$. Check the integral of these $$2$$ functions in terms of $$dx\\cdot dy$$ and then $$dy\\cdot dx$$ ' I am having an issue figuring out what the integrals will range from. I have: $$G =\\{(y,x) : -1 < x < 2 \\text{ and } x^2 < y < (x + 2)\\}\\to dy.dx$$ $$H =\\{(x,y) : 0 < y < 4 \\text{ and } y -2 < x < \\sqrt{y}\\} \\to dx.dy$$\n\nHowever when I create the integrals in terms of $$dx\\cdot dy$$ and $$dy\\cdot dx$$ they differ? Any help please, did i get the range of the $$G$$ and $$H$$ wrong? Its a parabola cut with a line\n\nFor $$dydx$$ the integral over $$1$$ is given by $$\\int_{-1}^2\\int_{x^2}^{x+2}dydx=\\int_{-1}^2-x^2+x+2dx=[\\frac13x^3+\\frac12x^2+2x]_{-1}^2=\\frac92$$ with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $$0\\le y\\le1$$ we have that $$-\\sqrt{y}\\le x\\le\\sqrt{y}$$ and when $$1\\le y\\le4$$ we have $$y-2\\le x\\le\\sqrt{y}$$. So the integral over the function $$1$$ is $$\\int_0^1\\int_{-\\sqrt{y}}^\\sqrt{y}dxdy+\\int_1^4\\int_{y-2}^\\sqrt{y}dxdy=\\int_0^12\\sqrt{y}\\,dy+\\int_1^4\\sqrt{y}-y+2\\,dy$$ $$=[\\frac43y^{\\frac32}]_0^1+[\\frac23y^{\\frac32}-\\frac12y^2+2y]_1^4=\\frac92$$ So the two regions are now equal.\n\n\u2022 Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated! \u2013\u00a0Shaun Weinberg Mar 31 at 10:36\n\nFor the region $$H$$, the lower limit of $$x$$ shouldn't be $$y-2$$.\n\nIt should be the maximum of $$-\\sqrt{y}$$ and $$y-2$$. In fact, when $$0 \\le y \\le 1$$, the lower limit is $$-\\sqrt{y}$$.\n\nThat is\n\n$$H =\\{(x,y): 0 \\le y \\le 4 , \\max(-\\sqrt{y}, y-2) < x < \\sqrt{y} \\}$$\n\nThat is from the picture below, the left limit of the region consists of the green color and blue color part.\n\n\u2022 I originally wrote an answer for the duplicate, so I copied it here. (+1) btw \u2013\u00a0Peter Foreman Mar 30 at 19:13\n\u2022 I did the same thing. ;) \u2013\u00a0Siong Thye Goh Mar 30 at 19:14", "date": "2019-07-21 13:07:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3168535/creating-an-integral-from-2-functions", "openwebmath_score": 0.8937596678733826, "openwebmath_perplexity": 184.47197942043715, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806546550656, "lm_q2_score": 0.8902942224835226, "lm_q1q2_score": 0.8730052915185152}}
{"url": "https://math.stackexchange.com/questions/3427273/limits-problem-involving-greatest-integer-function-and-an-unknown-function", "text": "# Limits problem involving greatest integer function and an unknown function.\n\nGiven that $$\\lim_{x \\to 0} \\frac{f(x)}{x^{2}} = 2$$, find $$\\lim_{x \\to 0} \\lfloor f(x) \\rfloor$$ and find if $$\\lim_{x \\to 0} \\lfloor \\frac{f(x)}{x} \\rfloor$$ exists. My math teacher says that since the denominator in the first limit is non negative and the limit itself is positive, he says that $$\\lim_{x \\to 0} f(x) = 0^{+}$$ and thus $$\\lim_{x \\to 0} \\lfloor f(x) \\rfloor = 0$$. I find this acceptable but my friend assumes $$f(x) = 2x^{2} + \\infty^{-}x^{3}$$ and claims that $$\\lim_{x \\to 0} \\frac{f(x)}{x^{2}} = 2$$ but that $$\\lim_{x \\to 0} \\lfloor f(x) \\rfloor$$ does not exist as $$\\lim_{x \\to 0^{+}} f(x) = 0^{+}$$ and $$\\lim_{x \\to 0^{-}} f(x) = 0^{-}$$. So who's right and who's wrong? If either of the two are wrong please explain why?\n\n\u2022 If $f(x)=2x^2$, does it satisfy any of either of your claims? \u2013\u00a0Andrew Chin Nov 8 at 15:49\n\u2022 Your math teacher is correct. \u2013\u00a0Paramanand Singh Nov 8 at 15:58\n\u2022 Thanks for answering Andrew Chew! If $f(x) = 2x^{2}$ then quite obviously, $\\lim_{x \\to 0} frac{f(x)}{x^{2}} = 2$ and $\\lim_{x \\to 0} \\lfloor f(x) \\rfloor = 0$. But the problem is that $f(x)$ can be $2x^{2}$ or it can be any other expression that satisfies $\\lim_{x \\to 0} \\frac{f(x)}{x^{2}} = 2$. We need to find what value $\\lim_{x \\to 0} \\lfloor f(x) \\rfloor$ equals irrespective of the function $f(x)$, as long as it satisfies $\\lim_{x \\to 0} \\frac{f(x)}{x^{2}} = 2$. \u2013\u00a0K Darshan Nov 8 at 17:17\n\u2022 Paramanand Singh, I appreciate your effort but can you give the reason as to why my friend was wrong? \u2013\u00a0K Darshan Nov 8 at 17:20\n\u2022 Oh what I would give if tomorrow I'd wake up and find everyone has stopped plugging $\\infty$ into expressions and treating it as though it were a number! \u2013\u00a0fleablood Nov 9 at 1:34\n\nThe problem is, simply put, that an expression like $$\\infty^- \\cdot x^3$$ doesn't make sense; you won't find an actual function (meaning a function which only uses numbers and not $$\\infty$$) which behaves like that. Even if you take a function like $$f(x) = 2x^2 + (-10000) \\cdot x^3$$, as you approach zero, eventually the $$x^3$$ term will not matter anymore: it will be much smaller than the $$2x^2$$ term, if only the $$x$$ you insert is \"small enough\", so close enough to zero. That means that for $$x$$ small enough, this $$f(x)$$ will still be greater than $$0$$.\n\nSo yeah, your teacher is correct. In general, you should always be very skeptical when people use infinity like that. Without proper care, infinity doesn't actually make a whole lot of sense :)\n\nElaborating on my comment the counter-example given by your friend does not make any sense. You can't write expressions like $$f(x) =2x^2+\\infty^{-}x^3$$. The usage of symbol $$\\infty$$ is always specified by specific definitions and typical examples are expressions like $$x\\to\\infty$$ and $$\\lim_{x\\to 0}1/x^2=\\infty$$.\n\nOn the other hand the correct argument goes like this. Since $$f(x) /x^2\\to 2$$ the expression $$f(x) /x^2$$ is positive as $$x\\to 0$$ and hence $$f(x) >0$$ as $$x\\to 0$$. Further $$f(x) =x^2\\cdot \\dfrac {f(x)} {x^2}\\to 0\\cdot 2=0$$ and hence $$f(x) <1$$ as $$x\\to 0$$. It follows that $$\\lfloor f(x) \\rfloor =0$$ as $$x\\to 0$$ and thus $$\\lim_{x\\to 0}\\lfloor f(x) \\rfloor =0$$.\n\nThe expression $$f(x) /x= x(f(x) /x^2)$$ also tends to $$0$$ but is positive if $$x\\to 0^{+}$$ and negative if $$x\\to 0^{-}$$ and hence $$\\lfloor f(x) /x\\rfloor =0$$ if $$x\\to 0^{+}$$ and $$\\lfloor f(x) /x \\rfloor =-1$$ as $$x\\to 0^{-}$$ so that the limit $$\\lim_{x\\to 0}\\left\\lfloor \\dfrac{f(x)} {x} \\right\\rfloor$$ does not exist.\n\nI'm not entirely sure I understand your teacher's argument (I sure as heck don't get your friend's), but I think it is valid.\n\n$$1 > \\epsilon > 0$$ and there is a $$\\delta$$ so that $$|x|< \\delta$$ implies $$|\\frac {f(x)}{x^2} - 2| < \\epsilon$$ so $$2-\\epsilon \\frac {f(x)}{x^2} < 2-\\epsilon$$. Thus $$1 < \\frac {f(x)}{x^2} < 3$$ and $$f(x) > 0$$ and $$x < \\min(\\delta,\\frac 12)$$ then\n\n$$0 < x^2 < \\frac {f(x)} < 3x^2 < \\frac 34$$ so $$0 and $$\\lfloor f(x)\\rfloor = 0$$ for all $$x < \\min(\\delta, \\frac 12)$$.\n\nWhich means $$\\lim_{x\\to 0}\\lfloor f(x)\\rfloor = 0$$.\n\nWhich I think is your teacher's argument.\n\nI can't see why your friend assumes $$f(x) = 2x^2 + \\infty - x^3$$, which doesn't even make sense. (What is $$f(5)$$? Is it $$\\infty -25$$? What's that?) so I can't tell you why he is wrong other than that what he says makes no sense.\n\n\u2022 I'm guessing here but you probably didn't understand what $0^+$, $\\infty^-$, etc. actually mean. Generally a number like $a^{+}$ is a number just greater than $a$, something like $a+0.000...1$ and $a^{-}$ is a number just less than $a$, something like $a-0.000...1$. What I meant by $\\infty^{-}$ is that it's a number that is very large but not $\\infty$. However, on second thoughts, I believe that a number like that doesn't exist. \u2013\u00a0K Darshan Nov 9 at 13:27\n\u2022 Infinity is NOT a number. $\\infty^{-}$ is oxymoronic and self-contradictory. $f(x) = 2x^2 +\\infty^{-}x^3$ is meaningless and illdefined (what number is $f(5) = 50+150\\infty^-$?). Your friend is more than wrong. Your friend is spouting uninterpretable gibberish. \u2013\u00a0fleablood Nov 10 at 16:10", "date": "2019-11-18 21:35:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3427273/limits-problem-involving-greatest-integer-function-and-an-unknown-function", "openwebmath_score": 0.9093069434165955, "openwebmath_perplexity": 248.03950520333785, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806546550657, "lm_q2_score": 0.8902942195727171, "lm_q1q2_score": 0.8730052886642358}}
{"url": "https://math.stackexchange.com/questions/2431221/why-sqrt33-not-in-mathbbq-sqrt32?noredirect=1", "text": "# Why $\\sqrt[3]{3}\\not\\in \\mathbb{Q}(\\sqrt[3]{2})$?\n\nWe all know that $\\sqrt[3]{3}\\not\\in \\mathbb{Q}({\\sqrt[3]{2}})$ intuitively. We even know that $\\sqrt[3]{p}\\not\\in \\mathbb{Q}(\\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\\sqrt{p}\\not\\in \\mathbb{Q}(\\sqrt{q})$, we can just set assume $\\sqrt{p}=a+b\\sqrt{q}$ for some $a, b\\in\\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\\sqrt[3]{2}+c\\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this. To be specific, it gives us 3 diophantine equations\n\n$$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$\n\nand I don't know how to check solvability of this kind of diophantine equations (over $\\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\\sqrt[3]{p}\\not\\in \\mathbb{Q}(\\sqrt[3]{q})$.\n\nSo I tried to use Galois theory, but actually I don't know where to go. I assumed $\\mathbb{Q}(\\sqrt[3]{3})=\\mathbb{Q}(\\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\\mathbb{Q}(\\sqrt[3]{3}, w)=\\mathbb{Q}(\\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\\beta=\\sqrt[3]{3}\\cdot\\sigma(\\sqrt[3]{3})\\cdot\\sigma^{2}(\\sqrt[3]{3})\\in \\mathbb{Q}$ when $\\sigma\\in Gal(K/\\mathbb{Q}), \\sigma(\\sqrt[3]{2})=\\sqrt[3]{2}, \\sigma(w)=w^{2}$. But I cannot get any information from this.\n\nEdit : It would be great if there is a simple algorithm to check whether $\\alpha\\in K$ or not for a given algebraic number $\\alpha$ and a number field $K$.\n\n\u2022 You could find the primitive element of $\\mathbb{Q}(\\sqrt[3]{3},\\sqrt[3]{2})$, find its minimal polynomial and show that it is of degree larger than $3$. \u2013\u00a0Michael Burr Sep 16 '17 at 1:22\n\u2022 Algebraic number theory might be able to give a solution by considering discriminants of rings of integers - but I'm a bit rusty on that. \u2013\u00a0Daniel Schepler Sep 16 '17 at 1:27\n\u2022 @MichaelBurr Thanks! I think we can even solve the general problem with that idea and computers. However, I don't know how to do this by hands. It seems that $\\sqrt[3]{2}+\\sqrt[3]{3}$ is a degree 9 element, but I cannot convince it without any help of computer programs. \u2013\u00a0Seewoo Lee Sep 16 '17 at 1:34\n\u2022 $a$ cannot be divisible by $2$ (first equation). Therefore $b$ is divisible by $2$ (second equation). Therefore, $c$ is divisible by $2^2$ (third equation). Hence $b$ is divisible by $2^3$ (second equation). Keep repeating and you get that $b$ is divisible by $2^k$ for any $k$. For other pair of primes it is the same business. \u2013\u00a0Hellen Sep 16 '17 at 1:34\n\u2022 What is $\\Bbb{Q}\\left( \\sqrt[3]{2} \\right)$? \u2013\u00a0gen-z ready to perish Sep 16 '17 at 4:03\n\nHere\u2019s another proof, making light use of $p$-adic theory:\n\nFirst, note that Eisenstein tells us that both $\\Bbb Q(\\sqrt[3]3\\,)$ and $\\Bbb Q(\\sqrt[3]2\\,)$ are cubic extensions of the rationals, and thus either they are the same field or their intersection is $\\Bbb Q$. Therefore, a question equivalent to the stated one is, \u201cWhy is $\\sqrt[3]2\\notin\\Bbb Q(\\sqrt[3]3\\,)$?\u201d I\u2019ll answer the new question.\n\nNote that $\\Bbb Q(\\sqrt[3]3\\,)$ may be embedded into $\\Bbb Q_2$, the field of $2$-adic numbers, because $X^3-3\\equiv(X+1)(X^2+X+1)\\pmod2$, product of two relatively prime polynomials over $\\Bbb F_2$, the residue field of $\\Bbb Q_2$ (more properly the residue field of $\\Bbb Z_2$). So Strong Hensel\u2019s Lemma says that $X^3-3$ has a linear factor in $\\Bbb Q_2[X]$, in other words, there\u2019s a cube root of $3$ in $\\Bbb Q_2$. But if there were a cube root of $2$ in $\\Bbb Q(\\sqrt[3]3\\,)$, there\u2019d be such a root in $\\Bbb Q_2$, and there isn\u2019t since it\u2019d have to have (additive) valuation $v_2(\\sqrt[3]2\\,)=1/3$. No such thing, the value group of $\\Bbb Q_2$ is $\\Bbb Z$.\n\nI really like all of the 3 answers, and I think I found one more solution which uses Trace, not Norm. (I don't know why I didn't try to use trace map before.) So I'm going to write it down here.\n\nSuppose $$K=\\mathbb{Q}(\\sqrt[3]{2})=\\mathbb{Q}(\\sqrt[3]{3})$$. Consider a trace map $$\\mathrm{Tr}_{K/\\mathbb{Q}}:K\\to \\mathbb{Q}, \\quad \\alpha \\mapsto \\sum_{i=1}^{n}\\sigma_{i}(\\alpha)$$ where $$\\sigma_{1}(\\alpha)=\\alpha, \\dots, \\sigma_{n}(\\alpha)$$ is roots of minimal polynomial of $$\\alpha$$. (We can consider trace map as a trace of a linear map $$m_{\\alpha}:K\\to K, x\\mapsto \\alpha x$$.) This is a $$\\mathbb{Q}$$-linear map and we can check that $$\\mathrm{Tr}_{K/\\mathbb{Q}}(\\sqrt[3]{2})=\\sqrt[3]{2}+\\sqrt[3]{2}w+\\sqrt[3]{2}w^{2}=\\sqrt[3]{2}(1+w+w^{2})=0$$ where $$w=e^{2\\pi i /3}$$ and $$\\mathrm{Tr}_{K/\\mathbb{Q}}(\\sqrt[3]{3})=0$$, too. If we assume $$\\sqrt[3]{3}=a+b\\sqrt[3]{2}+c\\sqrt[3]{4}\\,\\,\\,\\,(a, b, c\\in \\mathbb{Q})$$ as above, by taking trace on both sides we get $$0=3a$$ and $$a=0$$. By multiplying $$\\sqrt[3]{2}$$ and $$\\sqrt[3]{4}$$ on both sides, we get $$b=c=0$$ and contradiction. I think this methods can be used to prove for higher power cases, such as $$\\sqrt[n]{3}\\not\\in \\mathbb{Q}(\\sqrt[n]{2})$$ for $$n\\geq 2$$.\n\n\u2022 Great idea! I used it to prove that roots appear in root extensions only in \"obvious\" ways. \u2013\u00a0Orest Bucicovschi Sep 16 '17 at 6:53\n\u2022 Could you please detail how do you get $b=c=0$? I am not sure how to prove $Tr_{K/\\Bbb Q}(\\sqrt[3]3\\sqrt[3]2)=0$ \u2013\u00a0PerelMan 2 days ago\n\nLet $\\omega := e^{2 \\pi i/3}$, and consider the splitting field $\\mathbb{Q}(\\sqrt[3]{2}, \\omega)$ of $x^3 - 2$, and the unique automorphism $\\sigma$ such that $\\sigma(\\sqrt[3]{2}) = \\omega \\sqrt[3]{2}$, $\\sigma(\\omega \\sqrt[3]{2}) = \\omega^2 \\sqrt[3]{2}$, $\\sigma(\\omega^2 \\sqrt[3]{2}) = \\sqrt[3]{2}$. Suppose that $\\sqrt[3]{3} \\in \\mathbb{Q}[\\sqrt[3]{2}]$ is equal to $a + b \\sqrt[3]{2} + c \\sqrt[3]{4}$ for $a, b, c \\in \\mathbb{Q}$. Then $\\sigma(\\sqrt[3]{3})$ is also a root of $x^3 - 3 = 0$, which implies it's equal to either $\\sqrt[3]{3}$, $\\omega \\sqrt[3]{3}$, or $\\omega^2 \\sqrt[3]{3}$.\n\nHowever, in the first case, $$\\sigma(\\sqrt[3]{3}) = \\sqrt[3]{3} \\Rightarrow a + b \\omega \\sqrt[3]{2} + c \\omega^2 \\sqrt[3]{4} = a + b \\sqrt[3]{2} + c \\sqrt[3]{4}.$$ Rewriting $\\omega^2 = - \\omega - 1$ and using the linear independence of the basis $\\{ 1, \\omega, \\sqrt[3]{2}, \\omega \\sqrt[3]{2}, \\sqrt[3]{4}, \\omega \\sqrt[3]{4} \\}$ of the splitting field, we get $b = c = 0$, which gives a contradiction since $\\sqrt[3]{3}$ is irrational.\n\nSimilarly, in the second case, $\\sigma(\\sqrt[3]{3}) = \\omega \\sqrt[3]{3}$ would imply $\\sqrt[3]{3}$ is a rational multiple of $\\sqrt[3]{2}$, which is impossible since $\\sqrt[3]{3/2}$ is irrational. And in the third case, $\\sigma(\\sqrt[3]{3}) = \\omega^2 \\sqrt[3]{3}$ would imply $\\sqrt[3]{3}$ is a rational multiple of $\\sqrt[3]{4}$, which is impossible since $\\sqrt[3]{4/3}$ is irrational.\n\nHINT:\n\nIf $p$ a rational number, not a cube, and $a$, $b$, $c$ rational numbers so that\n\n$$(a + b \\sqrt[3]p + c \\sqrt[3]{p}^2)^3$$ is rational, then at most one of the numbers $a$, $b$, $c$ is nonzero.\n\nTo prove this, it's enough to show that if $(b,c) \\ne (0,0)$ then $a=0$. Consider $\\omega = -\\frac{1}{2} + i \\frac{\\sqrt{3}}{2}$. We have $\\omega^2 + \\omega + 1=0$, and $\\omega^3 = 1$.\n\nFrom $$(a + b \\sqrt[3]{p} + c \\sqrt[3]{p}^2)^3 = q$$ we get $$(a + b \\sqrt[3]{p}\\omega + c\\sqrt[3]{p}^2 \\omega^2)^3=q$$ Since $(b,c)\\ne (0,0)$ we conclude $$a + b \\sqrt[3]{p}\\omega + c\\sqrt[3]{p}^2 \\omega^2 = \\sqrt[3]{q} \\omega^{\\pm 1}$$ and the conjugate equality $$a + b \\sqrt[3]{p}\\omega^2 + c \\sqrt[3]{p}^2 \\omega = \\sqrt[3]{q} \\omega^{\\mp 1}$$\n\nTo these two equalities we add $$a + b \\sqrt[3]p + c \\sqrt[3]{p}^2= \\sqrt[3]{q}$$ and we get $3 a = 0$, so $a = 0$.\n\n$\\bf{Added:}$\n\nDetails:\n\nAssume $a + b \\sqrt[3]{p}\\omega + c \\sqrt[3]{p}^2\\omega^2 \\ne a + b \\sqrt[3]{p} + c \\sqrt[3]{p}^3$. We get $$b\\omega + c \\sqrt[3]{p}\\omega^2 = b + c \\sqrt[3]{p}$$ and with $\\omega^2 =-1-\\omega$ we get $(b-c\\sqrt[3]{p}) \\omega = b + 2 c \\sqrt[3]{p}$, and so $\\omega = \\frac{b + 2 c \\sqrt[3]{p}}{b-c\\sqrt[3]{p}}$, contradiction ( since, say, $\\omega$ is not real). So $a + b \\sqrt[3]{p}\\omega + c \\sqrt[3]{p}^2\\omega^2$ must be one of the other two roots of the equation $x^3 = q$.\n\n$\\bf{Added 2:}$ We follow the beautiful idea of @See-Woo Lee: to use traces.\n\n$\\bf{Fact:}$ Let $\\alpha$ a real radical, that is $\\alpha \\in \\mathbb{R}$, $\\alpha^m\\in \\mathbb{Q}$ for some $m$. Assume moreover that $\\alpha$ is irrational. Then $\\operatorname{trace}^L_{\\mathbb{Q}}\\alpha= 0$ for any $L$ finite algebraic extension of $\\mathbb{Q}$ containing $\\alpha$.\n\nProof: We may assume $\\alpha >0$. Let $m$ smallest so that $\\alpha^m \\in \\mathbb{Q}$. Clearly $m>1$. Let's prove that the polynomial $X^m - \\alpha^m$ is irreducible over $\\mathbb{Q}$. It factors over $\\mathbb{C}$ as $\\prod_{i=0}^m(X- \\alpha \\omega^i)$. If several of these linear factors produced a polynomial with rational coefficients, we would have the free term $$\\prod_{i \\in I} (- \\alpha \\omega^i) \\in \\mathbb{Q}$$ Taking absolute values we would have $\\alpha^k \\in \\mathbb{Q}$ for some $1\\le k < m$, not possible.\n\nDenote by $K = \\mathbb{Q}(\\alpha)$. Then we have $$\\operatorname{trace}^K_{\\mathbb{Q}}(\\alpha) = \\sum_{i=0}^{m-1} \\alpha \\omega^i = 0$$. Hence for every $L\\supset K$ we get $$\\operatorname{trace}^L_{\\mathbb{Q}} \\alpha = [L: K] \\operatorname{trace}^K_{\\mathbb{Q}}\\alpha = 0$$\n\n$\\bf{Main\\ result:}$ Let $\\alpha_i$ real radicals. Assume moreover that the ratios $\\frac{\\alpha_i}{\\alpha_j}$ for $i\\ne j$ are irrational. The the $\\alpha_i$'s are linearly independent over $\\mathbb{Q}$.\n\nProof: Let $a_i\\in \\mathbb{Q}$ so that $$\\sum a_i \\alpha_i = 0$$ Fix $i$, $1\\le i \\le k$. We have $$a_i = -\\sum_{k \\ne i} a_k \\frac{\\alpha_k}{\\alpha_i}$$\n\nLet a finite extension $L$ of $\\mathbb{Q}$ containing all the $\\alpha_i$ Taking traces on both sides we get $$d \\cdot a_i = \\sum_{k \\ne i} a_k \\operatorname{trace}^L_{\\mathbb{Q}} \\left(\\frac{\\alpha_k}{\\alpha_i}\\right )=0$$, since $\\frac{\\alpha_k}{\\alpha_i}$ is a real irrational radical. So all the $a_i$ are $0$.\n\nIn words: incommensurable real radicals are linearly independent over $\\mathbb{Q}$.\n\n\u2022 This looks great, but I don't get how to get from $(a + b \\sqrt[3]{p} + c \\sqrt[3]{p}^2)^3 = q$ to $(a + b \\sqrt[3]{p}\\omega + c\\sqrt[3]{p}^2 \\omega^2)^3=q$. Any tips? \u2013\u00a0Robert Lewis Sep 16 '17 at 3:32\n\u2022 @Robert Lewis: Thanks. Added some details. \u2013\u00a0Orest Bucicovschi Sep 16 '17 at 4:06\n\nSomehow, when manipulating pure radicals, I think that using identification or the trace map, i.e. essentially the additive structure, could lead to more complicated calculations than using the multiplicative structure and the norm map. It seems to be the case here. Introduce the quadratic field $k=\\mathbf Q (\\omega)$, where $\\omega$ is a primitive cubic root of $1$. Since $\\mathbf Q(\\sqrt[3] 2)$ and $\\mathbf Q(\\sqrt[3]3)$ have degree $3$ over $\\mathbf Q$ by Eisenstein criterion, Kummer theory tells us that the extensions $k(\\sqrt[3]2)/k$ and $k(\\sqrt[3]3)/k$ are Galois cyclic of degree $3$, and moreover they coincide iff $2=3x^3$, with $x\\in (k^{*})^{3}$. Norming down to $\\mathbf Q$, we get the diophantine equation $4a^3=9b^3$ with coprime integers $a, b$ : impossible because neither $4$ nor $9$ are cubes. Note that the argument still works with other (coprime) parameters than $2, 3$.", "date": "2019-09-17 23:15:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2431221/why-sqrt33-not-in-mathbbq-sqrt32?noredirect=1", "openwebmath_score": 0.9718460440635681, "openwebmath_perplexity": 112.63725323547399, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426397881663, "lm_q2_score": 0.8962513627417531, "lm_q1q2_score": 0.8729870432787186}}
{"url": "https://cen-aquitaine.org/aw3o5/self-discipline-scholarly-articles-eb5e93", "text": "The $${Y_{1}^{0}}^{*}Y_{1}^{0}$$ and $${Y_{1}^{1}}^{*}Y_{1}^{1}$$ functions are plotted above. The more important results from this analysis include (1) the recognition of an $$\\hat{L}^2$$ operator and (2) the fact that the Spherical Harmonics act as an eigenbasis for the given vector space. As one can imagine, this is a powerful tool. Forgot password? When this Hermitian operator is applied to a function, the signs of all variables within the function flip. (\u2113+m)!P\u2113m(cos\u2061\u03b8)eim\u03d5.Y^m_{\\ell} (\\theta, \\phi) = \\sqrt{\\frac{2\\ell + 1}{4\\pi} \\frac{(\\ell - m)! 4Algebraic theory of spherical harmonics. Plots of the real parts of the first few spherical harmonics, where distance from origin gives the value of the spherical harmonic as a function of the spherical angles, https://brilliant.org/wiki/spherical-harmonics/. It is no coincidence that this article discusses both quantum mechanics and two variables, $$l$$ and $$m$$. }{(\\ell + m)!}} Which spherical harmonics are included in the decomposition of f(\u03b8,\u03d5)=cos\u2061\u03b8\u2212sin\u20612\u03b8cos\u2061(2\u03d5)f(\\theta, \\phi) = \\cos \\theta - \\sin^2 \\theta \\cos(2\\phi)f(\u03b8,\u03d5)=cos\u03b8\u2212sin2\u03b8cos(2\u03d5) as a sum of spherical harmonics? Formally, these conditions on mmm and \u2113\\ell\u2113 can be derived by demanding that solutions be periodic in \u03b8\\theta\u03b8 and \u03d5\\phi\u03d5. A conducting sphere of radius RRR with a layer of charge QQQ distributed on its surface has the electric potential everywhere in space: V={14\u03c0\u03f50QR2r3sin\u2061\u03b8cos\u2061\u03b8cos\u2061\u03d5,\u00a0\u00a0r>R14\u03c0\u03f50Qr2R3sin\u2061\u03b8cos\u2061\u03b8cos\u2061\u03d5,\u00a0\u00a0rR \\\\ Spherical Harmonics and Linear Representations of Lie Groups 1.1 Introduction, Spherical Harmonics on the Circle In this chapter, we discuss spherical harmonics and take a glimpse at the linear representa-tion of Lie groups. \\hspace{15mm} \\ell & \\hspace{15mm} m&\\hspace{15mm} Y^m_{\\ell} (\\theta, \\phi) \\\\ \\hline For each fixed nnn and \u2113\\ell\u2113 there are 2\u2113+12\\ell + 12\u2113+1 solutions corresponding to the 2\u2113+12\\ell + 12\u2113+1 choices of mmm at fixed \u2113.\\ell.\u2113. So the solution can thus far be written in the form. When we consider the fact that these functions are also often normalized, we can write the classic relationship between eigenfunctions of a quantum mechanical operator using a piecewise function: the Kronecker delta. Spherical Harmonics are considered the higher-dimensional analogs of these Fourier combinations, and are incredibly useful in applications involving frequency domains. \\hspace{15mm} 2&\\hspace{15mm} -2&\\hspace{15mm} \\sqrt{\\frac{15}{32\\pi}} \\sin^2 \\theta e^{-2i\\phi} \\\\ The first is determining our $$P_{l}(x)$$ function. \\hspace{15mm} 2&\\hspace{15mm} 1&\\hspace{15mm} -\\sqrt{\\frac{15}{8\\pi}} \\sin \\theta \\cos \\theta e^{i \\phi} \\\\ The spherical harmonics. As the general function shows above, for the spherical harmonic where $$l = m = 0$$, the bracketed term turns into a simple constant. As such, this integral will be zero always, no matter what specific $$l$$ and $$k$$ are used. A similar analysis obtains the solution for rR.V(r,\\theta, \\phi ) = \\frac{1}{4\\pi \\epsilon_0} \\frac{QR^2}{r^3} \\sin \\theta \\cos \\theta \\cos \\phi, \\quad r>R.V(r,\u03b8,\u03d5)=4\u03c0\u03f50\u200b1\u200br3QR2\u200bsin\u03b8cos\u03b8cos\u03d5,r>R. \\end{cases}V=\u23a9\u23aa\u23a8\u23aa\u23a7\u200b4\u03c0\u03f50\u200b1\u200br3QR2\u200bsin\u03b8cos\u03b8cos\u03d5,\u00a0\u00a0r>R4\u03c0\u03f50\u200b1\u200bR3Qr2\u200bsin\u03b8cos\u03b8cos\u03d5,\u00a0\u00a0r R4\u03c0\u03f50\u200b1\u200bR3Qr2\u200bsin\u03b8cos\u03b8cos\u03d5, >... \\Cos\\Theta = x\\ ) features a transformation of \\ ( \\cos\\theta\\ ), \\ ( \\hat l. Of any particular state of the sphere as a result, they extremely. From x to \\ ( m\\ ) blue represents positive values and yellow represents negative values [ ]! The full solution, respectively d } { dx } [ ( x^ { 2 } 1. Charge density on the surface of the geoid up our process into four major parts and on... 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And the eigenfunctions of the electron degree 4 on the boundary surface that solutions be periodic in \u03b8\\theta\u03b8 \u03d5\\phi\u03d5... This problem VVV can be made more precise by considering the angular momentum quantum and. In general chemistry classes is consistent with our constant-valued harmonic, for any even-\\ ( l\\ ) probability... Of your SH expansion the closer your approximation gets as higher frequencies are added in probing a black hole introduction to spherical harmonics... Three dimensions into water equation \u22072f=0\\nabla^2 f = 0\u22072f=0 can be made more precise by considering the angular momentum of. That solutions be periodic in \u03b8\\theta\u03b8 and \u03d5\\phi\u03d5 S. Lakshmi Bala, Department of Physics, Madras. Determined Legendre function spherical harmonic is even two cases ~ave, of course~ been handled without. ] or check out our status page at https: //en.wikipedia.org/wiki/Spherical_harmonics #:. ) = ] \\ ) function with the newly determined Legendre function find our probability-density can use our definition! ) is special of \\ ( \\cos\\theta ) e^ { im\\phi } \\ ] values. Information contact us at [ email protected ] or check out our page...", "date": "2021-12-01 19:00:30", "meta": {"domain": "cen-aquitaine.org", "url": "https://cen-aquitaine.org/aw3o5/self-discipline-scholarly-articles-eb5e93", "openwebmath_score": 0.9481796026229858, "openwebmath_perplexity": 1748.1106351590208, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180643966651, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8729829525551376}}
{"url": "https://ita.skanev.com/11/03/02.html", "text": "# Exercise 11.3.2\n\nSuppose that we hash a string of $r$ characters into $m$ slots by treating it as a radix-128 number and then using the division method. We can easily represent the number $m$ as a 32-bit computer word, but the string of $r$ characters, treated as a radix-128 number, takes many words. How can we apply the division method to compute the hash value of the string without using more than a constant number of words of storage outside the string itself?\n\nYes, this follows pretty easily from the laws of modulo arithmetic. We need to observe that if the string is $s = \\langle a_n, \\ldots, a_1, a_0 \\rangle$, then its hash $h(s)$ is going to be:\n\n\\begin{aligned} h(s) &= \\left( \\sum_{i=0}^{n}{a_i \\cdot {128}^{i}} \\right) \\bmod m \\\\ &= \\sum_{i=0}^{n}{ \\Big( \\left( a_i \\cdot {128}^{i} \\right) \\bmod m \\Big) } \\\\ &= \\sum_{i=0}^{n}{ \\Big( ( a_i \\bmod m ) ( {128}^{i} \\bmod m ) \\Big) } \\\\ \\end{aligned}\n\nWe can easily compute $a_i \\bmod m$ without extra memory. To compute ${128}^{i} \\bmod m$ without extra memory, we just need to observe that $k^i \\bmod m = k(k^{i-1} \\bmod m) \\bmod m$, that is, we can compute the module for each power incrementally, without ever needing unbound memory.\n\n### Python code\n\nK = 128\n\ndef consthash(digits, m):\nresult = 0\npower = 1\n\nfor d in reversed(digits):\nresult += ((d % m) * power) % m\nresult %= m\npower = (power * K) % m\n\nreturn result", "date": "2021-10-18 17:00:41", "meta": {"domain": "skanev.com", "url": "https://ita.skanev.com/11/03/02.html", "openwebmath_score": 0.9999809265136719, "openwebmath_perplexity": 1080.524429127908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9857180677531124, "lm_q2_score": 0.8856314723088732, "lm_q1q2_score": 0.8729829436256465}}
{"url": "http://mathhelpforum.com/calculus/7200-differentiate-using-implicit-differentiation.html", "text": "# Thread: Differentiate using implicit differentiation\n\n1. ## Differentiate using implicit differentiation\n\nfind dy/dx (3xy)^(1/2)=3x-2y\n\nCan't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.\n\n2. Originally Posted by math34\nfind dy/dx (3xy)^(1/2)=3x-2y\n\nCan't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.\n\n$\\sqrt{3xy} = 3x - 2y$\n\nSo\n$\\frac{1}{2} \\frac{1}{\\sqrt{3xy}} ( 3y + 3xy') = 3 - 2y'$\n\n$\\frac{3y}{2\\sqrt{3xy}} + \\frac{3x}{2\\sqrt{3xy}}y' = 3 - 2y'$\n\n$\\frac{3x}{2\\sqrt{3xy}}y' + 2y' = 3 - \\frac{3y}{2\\sqrt{3xy}}$\n\n$\\left ( \\frac{3x}{2\\sqrt{3xy}} + 2 \\right )y' = 3 - \\frac{3y}{2\\sqrt{3xy}}$\n\n$y' = \\frac{3 - \\frac{3y}{2\\sqrt{3xy}}}{\\frac{3x}{2\\sqrt{3xy}} + 2}$\n\n$y' = \\frac{3 - \\frac{3y}{2\\sqrt{3xy}}}{\\frac{3x}{2\\sqrt{3xy}} + 2} \\cdot \\frac{2 \\sqrt{3xy}}{2 \\sqrt{3xy}}$\n\n$y' = \\frac{6\\sqrt{3xy} - 3y}{3x + 4\\sqrt{3xy}}$ <-- To get your given answer, multiply top and bottom by $\\sqrt{3xy}$ again.\n\nWe should rationalize this, but it is easier to use the original equation:\n$\\sqrt{3xy} = 3x - 2y$\n\n$y' = \\frac{6(3x - 2y) - 3y}{3x + 4(3x-2y)}$\n\n$y' = \\frac{18x - 12y - 3y}{3x + 12x - 8y}$\n\n$y' = \\frac{18x - 15y}{15x - 8y}$\n\n-Dan\n\n3. Hello, math34!\n\nFind $\\frac{dy}{dx}:\\;\\;(3xy)^{\\frac{1}{2}} \\:=\\:3x-2y$\n\nWould it be easier to first square both sides to get rid of the fractional exponent,\nthen differentiate implicitly? . . . absolutely!\n\nAnswer: (6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy . . . this can't be right!\n\nSquare: . $3xy \\:=\\:9x^2 - 12xy + 4y^2$\n\nWe have: . $3xy' + 3y \\:=\\:18x - 12xy' - 12y + 8yy'$\n\nThen: . $15xy' - 8yy' \\:=\\:18x - 15y$\n\nFactor: . $(15x - 8y)y' \\:=\\:18x - 15y$\n\nTherefore: . $\\boxed{y' \\:=\\:\\frac{18x - 15y}{15x - 8y}}$ . . . which is Dan's answer.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nIt can be done head-on . . .\n\nDifferentiate implicitly: . $\\frac{1}{2}(3xy)^{-\\frac{1}{2}}(3y + 3xy') \\:=\\:3 - 2y'$\n\nWe have: . $3y + 3xy' \\:=\\:2\\sqrt{3xy}(3 - 2y')$\n\nWe are told that: . $\\sqrt{3xy} \\:=\\:3x - 2y$\n\n. . Hence, we have: . $3y + 3xy' \\:=\\:2(3x - 2y)(3 - 2y')$\n\nExpand: . $3y + 3xy' \\:=\\:18x - 12xy' - 12y + 8yy'$\n\nSimplify: . $15xy' - 8yy' \\:=\\:18x - 15y$\n\nFactor: . $(15x - 8y)y' \\:=\\:18x - 15y$\n\nTherefore: . $\\boxed{y' \\:=\\:\\frac{18x-15y}{15x-8y}}$\n\n4. ## Thanks\n\nI got the same answer as you and Dan but, still am clueless about how the given solution was obtained. I think this is a case of the experts making a mistake, while the layman suffers!!!\n\nThanks guys\n\n5. I am not happy about the idea of squaring the equation, and here's why:\n\nSay we have:\n$y = x$\n\nSquare both sides:\n$y^2 = x^2$\n\nNow take the derivative:\n$2yy' = 2x$\n\n$y' = \\frac{x}{y}$\n\nNow, this certainly is a differential equation for the original function, but we also have another solution: y = -x. In this case the differential equation has more solutions than what we started with.\n\nJust something to watch out for.\n\n-Dan", "date": "2017-04-25 15:37:02", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/7200-differentiate-using-implicit-differentiation.html", "openwebmath_score": 0.9793962836265564, "openwebmath_perplexity": 470.74292835328026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180639771091, "lm_q2_score": 0.8856314723088732, "lm_q1q2_score": 0.8729829402814993}}
{"url": "https://voiceofwave.com/a-sine-wave-equation", "text": "# A sine wave equation?\n\n1\nDate created: Sat, Jul 17, 2021 3:51 PM\nDate updated: Sun, Jun 26, 2022 6:06 PM\n\nContent\n\nVideo answer: How to write a sine wave equation by looking at a graph | trigonometry\n\n## Top best answers to the question \u00abA sine wave equation\u00bb\n\nSine Wave\u2026 A general form of a sinusoidal wave is y(x,t)=Asin(kx\u2212\u03c9t+\u03d5) y ( x , t ) = A sin ( kx \u2212 \u03c9 t + \u03d5 ) , where A is the amplitude of the wave, \u03c9 is the wave's angular frequency, k is the wavenumber, and \u03d5 is the phase of the sine wave given in radians.\n\nFAQ\n\nThose who are looking for an answer to the question \u00abA sine wave equation?\u00bb often ask the following questions:\n\n### \ud83d\udc4b De broglie wave equation?\n\n#### De Broglie wave equation\n\n\u2022 The de Broglie equation is an equation used to describe the wave properties of matter, specifically, the wave nature of the electron:\u200b. \u03bb = h/mv, where \u03bb is wavelength, h is Planck 's constant, m is the mass of a particle, moving at a velocity v.\n\n### \ud83d\udc4b How do you derive a sine wave equation?\n\n1. To find the amplitude, wavelength, period, and frequency of a sinusoidal wave, write down the wave function in the form y(x,t)=Asin(kx\u2212\u03c9t+\u03d5).\n2. The amplitude can be read straight from the equation and is equal to A.\n3. The period of the wave can be derived from the angular frequency (T=2\u03c0\u03c9).\n\n### \ud83d\udc4b How to derive sine wave equation?\n\n#### How do you calculate sine wave?\n\n\u2022 In general, a sine wave is given by the formula In this formula the frequency is w. Frequency used to be measured in cycles per second, but now we use the unit of frequency - the Hertz (abbreviated Hz). One Hertz (1Hz) is equal to one cycle per second.\n\n### \ud83d\udc4b How to read equation of travelling sine wave?\n\n\u2022 Simply read the wikipedia article. y (t) = A sin (2 \u03c0 f t + \u03c6) Here A is the amplitude of the wave,i.e. the maximum height of the wave; f the frequency, i.e. is the number of oscillations (cycles) that occur each second of time; \u03c6, the phase, specifies (in radians) where in its cycle the oscillation is at t = 0.\n\n### \ud83d\udc4b How to write an equation for a 4hz sine wave?\n\n#### Which is the formula for the sine wave?\n\n\u2022 The formula for the Sine wave is, A = Amplitude of the Wave \u03c9 = the angular frequency, specifies how many oscillations occur in a unit time interval, in radians per second \u03c6, the phase, t = ? Here \u03c9, is the angular frequency i.e, It defines how many cycles of the oscillations are there.\n\n### \ud83d\udc4b Is pwm sine wave pure sine wave?\n\nWhen it comes to output waveform, there are two types of UPS battery backup\u2014the kind that produce a pure sine wave and the kind that produce a simulated or modified sine wave, also known as a pulse-width modulated (PWM) sine wave, when on battery power.\n\n### \ud83d\udc4b Is the wave equation a partial differential equation?\n\n\u2022 The wave equation is a partial differential equation. We discuss some of the tactics for solving such equations on the site Differential Equations. One of the most popular techniques, however, is this: choose a likely function, test to see if it is a solution and, if necessary, modify it. So, let's use what we already know.\n\n### \ud83d\udc4b Is the wave equation sine or cosine?\n\n#### What is the difference between sine wave and cos wave?\n\n\u2022 Key Difference: Sine and cosine waves are signal waveforms which are identical to each other. The main difference between the two is that cosine wave leads the sine wave by an amount of 90 degrees . A sine wave depicts a reoccurring change or motion.\n\n### \ud83d\udc4b What is a sine wave equation?\n\nSine Wave\u2026 A general form of a sinusoidal wave is y(x,t)=Asin(kx\u2212\u03c9t+\u03d5) y ( x , t ) = A sin ( kx \u2212 \u03c9 t + \u03d5 ) , where A is the amplitude of the wave, \u03c9 is the wave's angular frequency, k is the wavenumber, and \u03d5 is the phase of the sine wave given in radians.\n\nVideo answer: Determining the equation for a sinewave from a plot\n\nWe've handpicked 6 related questions for you, similar to \u00abA sine wave equation?\u00bb so you can surely find the answer!\n\nWhat is the equation for a 2 d damped sine wave?\n\u2022 This is the equation for a 2 D damped sine wave, what is the 3 D equivalent? ( x 2 + y 2). expresses damping along any ray a x + b y = 0 extending from the origin, and this was achieved by substitution of x 2 + y 2 in place of t in the Wikipedia listing at https://en.m.wikipedia.org/wiki/Damped_sine_wave.\nWhat is the equation for nonlinear wave equation?\n\u2022 Nonlinear wave equation of general form: \\$$u_{tt}=\\\\left[f\\\\left(u\\\\right)u_{x}\\\\right]_{x}\\$$ This equation can be linearized in the general case. Some exact solutions are given in [Pol-04, pp252-255] and, by way of an example consider the following special case where \\$$f\\\\left(u\\\\right)=\\\\alpha e^{\\\\lambda u}\\\\ :\\$$\nWhat is the equation for the wave equation?\n\u2022 Let us consider a plane wave with real amplitude E0and propagating in direc- tion of the zaxis. This plane wave is represented by E(r,t) = E0cos[kz\u2212 \u03c9t], where k = |k| = \u03c9/c. If we observe this \ufb01eld at a \ufb01xed position z then we\u2019ll measure an electric \ufb01eld E(t) that is oscillating with frequency f = \u03c9/2\u03c0.\nWhat is the equation of sine wave?\n\u2022 In our math class, we are discussing about the trigonometric waves, specifically, the Sine Wave. The Sine wave is the graph that is formed if the function contains a sine function. The General Formula of the Sine wave is: y=AsinB(x-C)+D where x is the angle or theta.\nWhat is the mathematical equation for a sine wave?\n\u2022 On The Mathematics of the Sine Wave y(x) = A*(2\u03c0ft + \u00f8) Why the understanding the sine wave is important for computer musicians. The sine wave is mathematically a very simple curve and a very simple graph, and thus is computationally easy to generate using any form of computing, from the era of punch cards to the current era of microprocessors.\n\n### Video answer: Sine wave equation explained - interactive\n\nWho discovered wave equation?\n\nUsing Newton's recently formulated laws of motion, Brook Taylor (1685\u20131721) discovered the wave equation by means of physical insight alone [1].", "date": "2022-07-02 14:31:08", "meta": {"domain": "voiceofwave.com", "url": "https://voiceofwave.com/a-sine-wave-equation", "openwebmath_score": 0.9068586826324463, "openwebmath_perplexity": 744.191835011262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.972414724168629, "lm_q2_score": 0.8976953010025941, "lm_q1q2_score": 0.8729321285119119}}
{"url": "https://math.stackexchange.com/questions/834816/show-that-a-connected-graph-on-n-vertices-is-a-tree-if-and-only-if-it-has-n-1", "text": "# Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.\n\nShow that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.\n\n$(\\Rightarrow)$ If a tree $G$ has only $1$ vertex, it has $0$ edges. Now, assume that any tree with $k-1$ vertices has $k-2$ edges. Let $T$ be a tree with $k$ vertices. Remove a leaf $l$ to obtain a tree $T'$ with $k-1$ vertices. Then, $T'$ has $k-2$ edges, by the inductive hypothesis. The addition of $l$ to $T'$ produces a graph with $k-1$ edges, since $l$ has degree $1$.\n\n$(\\Leftarrow)$ Let $G$ be a connected graph on $n$ vertices, with $n-1$ edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Successively remove edges from cycles in $G$ to obtain a graph $G'$ with no cycles. Then, $G'$ is connected and has no cycles. Therefore, $G'$ is a tree, with $n$ vertices and $n-1-x$ edges, for some $x > 0$. However, this contradicts the previous derivation. Therefore, it must be the case that $G$ is a tree.\n\n\u2022 it seems to be true. \u2013\u00a0mesel Jun 15 '14 at 12:40\n\nThe proofs are correct. Here's alternative proof that a connected graph with n vertices and n-1 edges must be a tree modified from yours but without having to rely on the first derivation:\n\nLet $G$ be a connected graph on n vertices, with n\u22121 edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Remove one of the edges within a cycle. This leaves a connected graph on n vertices with n-2 edges which is impossible as a connected graph on n vertices must at least have n - 1 edges.\n\nYes, this seems like it is true, although you didn't prove a leaf must always exist, if you really want to be rigorous. And in the other part you didn't explain why if it contained a cycle you could remove an edge without the graph being disconnected, but I like the proof over all.\n\nI have the following way of proving it. Let me know if you agree with it or not.\n\nSuppose the graph on $n$ vertices with $n-1$ edges is not a tree. This implies it has at least $1$ polygon. Suppose in total there are $k$ edges involved in these polygons and we know that a polygon has as many edges as vertices, since polygons are of regular degree 2. Then, if we consider th egraph without the edges that form the polygons, then we have a connected graph with at least $n-k$ edges, since in a connected graph every vertex has degree at least 1. Thus, in total this would imply we have at least $n$ in all the graph. But this contradicts the fact that we have $n-1$ in the original graph. Thus it must be a tree.\n\nThe converse is not always true. Consider a disjoint graph with n vertices and n - 1 edges. Here we have n-1 edges but this is obviously not a tree. The assumption that G is connected is only for the forward implication.", "date": "2019-10-17 23:59:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/834816/show-that-a-connected-graph-on-n-vertices-is-a-tree-if-and-only-if-it-has-n-1", "openwebmath_score": 0.8223617076873779, "openwebmath_perplexity": 68.94123288396771, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9724147225697743, "lm_q2_score": 0.8976952962128458, "lm_q1q2_score": 0.8729321224190059}}
{"url": "https://math.stackexchange.com/questions/472148/smooth-functions-or-continuous", "text": "# smooth functions or continuous\n\nWhen wesay a function is smooth? Is there any difference between smooth function and continuous function? If they are the same, why sometimes we say f is smooth and sometimes f is continuous?\n\n\u2022 \"smooth\" means (at least) \"continously differentiable\". Sometimes more (even infinite number of) derivatives are required to be continuous. Aug 20, 2013 at 16:59\n\u2022 @njguliyev, not to nitpick but I think it's relatively common to call Lipschitz continuous ODEs \"smooth\" - being just smooth enough for existence and uniqueness of solutions. Aug 20, 2013 at 17:08\n\nA function being smooth is actually a stronger case than a function being continuous. For a function to be continuous, the epsilon delta definition of continuity simply needs to hold, so there are no breaks or holes in the function (in the 2-d case). For a function to be smooth, it has to have continuous derivatives up to a certain order, say k. We say that function is $C^{k}$ smooth. An example of a continuous but not smooth function is the absolute value, which is continuous everywhere but not differentiable everywhere.\n\nA smooth function is differentiable. Usually infinitely many times.\n\n\u2022 ... or at least as often as we need it. Aug 20, 2013 at 17:14\n\nSmooth implies continuous, but not the other way around. There are functions that are continuous everywhere, yet nowhere differentiable.\n\nA smooth function can refer to a function that is infinitely differentiable. More generally, it refers to a function having continuous derivatives of up to a certain order specified in the text. This is a much stronger condition than a continuous function which may not even be once differentiable.\n\nA smooth function is a function that has continuous derivatives up to some desired order over some domain. A function can therefore be said to be smooth over a restricted interval such as or . The number of continuous derivatives necessary for a function to be considered smooth depends on the problem at hand, and may vary from two to infinity. A function for which all orders of derivatives are continuous is called a C-infty-function.\n\nTake $$f(x) = x|x|$$ it is smooth, and now consider $$g(x) = x^3$$, this other function is also smooth. However, $$g(x)$$ is much smoother than $$f(x)$$ because derivitive of $$f(x)$$. You can argue that $$g(x)$$ is infinitely many times smooth. All polynomials belong to $$C^\\infty$$ meaning they are infinitely many times differentiable and are smooth.\n\nHowever, $$h(x) = |x|$$ is not smooth, because it has corner. Please note that all three functions, $$f(x)$$, $$g(x)$$, and $$h(x)$$ are continous.\n\nHere is how $$f(x)$$ looks like:\n\nHere is how the derivative of $$f(x)$$ looks like:\n\nHere is the second derivate of $$f(x)$$, as you can see its second derivative is not even continuous:\n\nHere is the graph of $$g(x)$$:\n\nHere is graph of $$\\frac{d f(x)}{dx} = 3x^2$$:\n\nHere is graph of $$\\frac{d^2 g(x)}{dx^2} = 6x$$:\n\n\u2022 If a function $f$ is smooth, then can I suppose that $f$ is increasing ou decreasing? At least in some interval? Oct 9, 2020 at 0:52\n\nConsider a sequence in $\\mathbb{R}$ say $\\{x_n\\}_{n \\in \\mathbb{N}}$, which is continuous in $\\mathbb{R}$. Usually we do not say it a smooth function.", "date": "2022-05-17 08:19:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/472148/smooth-functions-or-continuous", "openwebmath_score": 0.9295127391815186, "openwebmath_perplexity": 199.4823497050098, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693242009478238, "lm_q2_score": 0.900529786117893, "lm_q1q2_score": 0.8729053153584413}}
{"url": "http://mathhelpforum.com/calculus/35491-optimization.html", "text": "# Math Help - Optimization\n\n1. ## Optimization\n\nAn isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??\n\nAny other good notes or questions for Optimization problems would be welcomed as well\n\n2. Originally Posted by someone21\nAn isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??\n\nAny other good notes or questions for Optimization problems would be welcomed as well\nMinimum or maximum? The minimum area is trivially zero.\n\n3. Originally Posted by mr fantastic\nMinimum or maximum? The minimum area is trivially zero.\nMinimum possible area but why is it zero??\n\nAnd if maximum possible area then how to do it???\n\n4. Originally Posted by someone21\nMinimum possible area but why is it zero??\n\nAnd if maximum possible area then how to do it???\nDraw an isosceles triangle with non-equal side the diameter of the circle. Now move the non-equal side up so that the two equal sides get smaller and smaller. The triangle degenerates to a point - of area zero.\n\n5. Originally Posted by someone21\nAn isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??\n\nAny other good notes or questions for Optimization problems would be welcomed as well\nScroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles\n\nYour circle has a radius r = 2. Let a = c and re-arrange:\n\n$K = \\frac{a^2 b}{8}$ .... (1)\n\nwhere K is the area of the triangle.\n\nYou need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....\n\nI haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.\n\n6. Originally Posted by mr fantastic\nScroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles\n\nYour circle has a radius r = 2. Let a = c and re-arrange:\n\n$K = \\frac{a^2 b}{8}$ .... (1)\n\nwhere K is the area of the triangle.\n\nYou need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....\n\nI haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.\nAlternatively, you could work with finding the size of the two equal angles at the base of the triangle:\n\nLet the angle at the top vertex of the triangle be of size $\\phi$.\n\nLet the two equal angles at the base of the triangle be of size $\\theta$.\n\nLet a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\\frac{\\phi}{2}$ and the perpendicular is the height of the triangle.\n\nArea of any triangle: $A = \\frac{bh}{2}$.\n\nTo find the angles that give a maximum triangle area, express the above area in terms $\\phi$ and radius (= 2).\n\nYou should be able to show that $\\frac{a}{2} = 2 \\cos(\\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\\phi/2$). Then:\n\n$a = 4 \\cos(\\phi/2)$.\n\n$b = 8 \\cos(\\phi/2) \\sin(\\phi/2)$.\n\n$h = 4 \\cos(\\phi/2) \\cos(\\phi/2) = 4 \\cos^2 (\\phi/2)$.\n\nTherefore:\n\n$A = \\frac{bh}{2} = \\frac{1}{2} [8 \\cos(\\phi/2) \\sin(\\phi/2)] [4 \\cos^2(\\phi/2)] = 16 \\cos^3 (\\phi/2) \\sin(\\phi/2)$.\n\nNow solve $\\frac{d A}{d \\phi} = 0$ etc. I get $\\phi = \\frac{\\pi}{3}$, that is, the triangle is equilateral. Substitute $\\phi = \\frac{\\pi}{3}$ into A to get the maximum area.\n\nNote: I reserve the right for this reply to contain algebraic errors.\n\n7. Originally Posted by mr fantastic\nAlternatively, you could work with finding the size of the two equal angles at the base of the triangle:\n\nLet the angle at the top vertex of the triangle be of size $\\phi$.\n\nLet the two equal angles at the base of the triangle be of size $\\theta$.\n\nLet a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\\frac{\\phi}{2}$ and the perpendicular is the height of the triangle.\n\nArea of any triangle: $A = \\frac{bh}{2}$.\n\nTo find the angles that give a maximum triangle area, express the above area in terms $\\phi$ and radius (= 2).\n\nYou should be able to show that $\\frac{a}{2} = 2 \\cos(\\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\\phi/2$). Then:\n\n$a = 4 \\cos(\\phi/2)$.\n\n$b = 8 \\cos(\\phi/2) \\sin(\\phi/2)$.\n\n$h = 4 \\cos(\\phi/2) \\cos(\\phi/2) = 4 \\cos^2 (\\phi/2)$.\n\nTherefore:\n\n$A = \\frac{bh}{2} = \\frac{1}{2} [8 \\cos(\\phi/2) \\sin(\\phi/2)] [4 \\cos^2(\\phi/2)] = 16 \\cos^3 (\\phi/2) \\sin(\\phi/2)$.\n\nNow solve $\\frac{d A}{d \\phi} = 0$ etc. I get $\\phi = \\frac{\\pi}{3}$, that is, the triangle is equilateral. Substitute $\\phi = \\frac{\\pi}{3}$ into A to get the maximum area.\n\nNote: I reserve the right for this reply to contain algebraic errors.\nI think the question was the circle is inside the isoceles triangle?\n\n8. Originally Posted by mr fantastic\nAlternatively, you could work with finding the size of the two equal angles at the base of the triangle:\n\nLet the angle at the top vertex of the triangle be of size $\\phi$.\n\nLet the two equal angles at the base of the triangle be of size $\\theta$.\n\nLet a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\\frac{\\phi}{2}$ and the perpendicular is the height of the triangle.\n\nArea of any triangle: $A = \\frac{bh}{2}$.\n\nTo find the angles that give a maximum triangle area, express the above area in terms $\\phi$ and radius (= 2).\n\nYou should be able to show that $\\frac{a}{2} = 2 \\cos(\\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\\phi/2$). Then:\n\n$a = 4 \\cos(\\phi/2)$.\n\n$b = 8 \\cos(\\phi/2) \\sin(\\phi/2)$.\n\n$h = 4 \\cos(\\phi/2) \\cos(\\phi/2) = 4 \\cos^2 (\\phi/2)$.\n\nTherefore:\n\n$A = \\frac{bh}{2} = \\frac{1}{2} [8 \\cos(\\phi/2) \\sin(\\phi/2)] [4 \\cos^2(\\phi/2)] = 16 \\cos^3 (\\phi/2) \\sin(\\phi/2)$.\n\nNow solve $\\frac{d A}{d \\phi} = 0$ etc. I get $\\phi = \\frac{\\pi}{3}$, that is, the triangle is equilateral. Substitute $\\phi = \\frac{\\pi}{3}$ into A to get the maximum area.\n\nNote: I reserve the right for this reply to contain algebraic errors.\nI think the question was the circle is inside the isoceles triangle?\n\n9. Originally Posted by someone21\nI think the question was the circle is inside the isoceles triangle?\nAh yes. I misread circumscribed as inscribed. Oh well, I guess I answered the second part:\n\n\"Any other good notes or questions for Optimization problems would be welcomed as well\"\n\n10. Originally Posted by someone21\nAn isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??\n\nAny other good notes or questions for Optimization problems would be welcomed as well\nLet 2b be the length of the base of the triangle, a the length of the other two sides and h the height of triangle.\n\nThen A = bh.\n\nA relationship between b and h is need so that A can be expressed in terms of a single variable.\n\nDraw a line from the middle of the base to the (top) vertex of the triangle. This line goes through the center of the circle and is the height of the triangle.\n\nDraw a line from the center of the circle to one of the sides of the triangle. This line is perpendicular to that side (why?).\n\nApply Pythagoras Theorem to the two obvious right-triangles:\n\n$h^2 + b^2 = a^2$ .... (1)\n\n$2^2 + {\\color{red}(a-b)}^2 = (h - 2)^2 \\Rightarrow 4 + a^2 - 2ab + b^2 = h^2 - 4h + 4 \\Rightarrow a^2 - 2ab + b^2 = h^2 - 4h$ .... (2)\n\nYou should consider the geometry that gives the red length .....\n\nSolve equations (1) and (2) simultaneously for b in terms of h (a small bit of algebra for you to do):\n\n$b = \\frac{2h}{\\sqrt{h^2 - 4h}}$.\n\nTherefore A = .......\n\nSolve dA/dh = 0 (I get h = 6). Hence the minimum area of the triangle is equal to .....\n\nNote: As you might anticipate, if you do the calculations you get $2b = 4 \\sqrt{3} = a$ and so the triangle is equilateral.\n\nNote: I reserve the right for this reply to contain careless errors.\n\n11. Originally Posted by someone21\nMinimum possible area but why is it zero??\nOne way to think about it....\n\nAn equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.\n\n$\\lim_{s \\rightarrow 0} A$\n\n$\\lim_{s \\rightarrow 0} \\frac{1}{2\\sqrt{2}}s^2 \\rightarrow 0$\n\nSince the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$\n\nThe limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.\n\n12. Originally Posted by colby2152\nOne way to think about it....\n\nAn equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.\n\n$\\lim_{s \\rightarrow 0} A$\n\n$\\lim_{s \\rightarrow 0} \\frac{1}{2\\sqrt{2}}s^2 \\rightarrow 0$\n\nSince the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$\n\nThe limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.\nwhy is it s goes to zero and can it be explained more detaily please\n\n13. Originally Posted by someone21\nwhy is it s goes to zero and can it be explained more detaily please\nYou are looking at the area as a side goes to zero. The area, dependent upon the sides, then goes to zero as well.", "date": "2014-07-24 20:24:36", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/35491-optimization.html", "openwebmath_score": 0.8555326461791992, "openwebmath_perplexity": 304.84874344389226, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9891815532606979, "lm_q2_score": 0.8824278772763472, "lm_q1q2_score": 0.8728813782847576}}
{"url": "https://math.stackexchange.com/questions/187295/a-geometric-proof-of-zeta2-frac-pi26-and-other-integer-inputs-for-the", "text": "# A Geometric Proof of $\\zeta(2)=\\frac{\\pi^2}6$? (and other integer inputs for the Zeta)\n\nIs there a known geometric proof for this famous problem? $$\\zeta(2)=\\sum_{n=1}^\\infty n^{-2}=\\frac16\\pi^2$$\n\nMoreover we can consider possibilities of geometric proofs of the following identity for positive even inputs of the Zeta function: $$\\zeta(2n)=(-1)^{n+1} \\frac{B_{2n}(2\\pi)^{2n}}{2(2n)!}$$ and negative inputs: $$\\zeta(-n)=-\\frac{B_{n+1}}{n+1}$$\n\n\u2022 There are many nice proofs at math.stackexchange.com/questions/8337/\u2026 . Is one of them geometric enough for you? (David Speyer's answer might fit the bill.) \u2013\u00a0Qiaochu Yuan Aug 27 '12 at 1:04\n\u2022 Here's a proof by Prof. Greene at UCLA: math.ucla.edu/~greene/How%20Geometry.pdf \u2013\u00a0Elchanan Solomon Aug 27 '12 at 2:20\n\u2022 thank you Qiaochu one of them is ;) \u2013\u00a0finnlim Aug 27 '12 at 2:44\n\u2022 I wanted to ask this question some time ago. I'm glad to see it posted. (I thought of an elementary geometrical proof) \u2013\u00a0user 1591719 Aug 27 '12 at 5:25\n\u2022 do you mean that you have an elementary geometric proof? Would you mind sharing it? \u2013\u00a0finnlim Aug 27 '12 at 5:42\n\nYes, there is a geometric proof for $\\zeta(2)=\\sum_{n=1}^\\infty n^{-2}=\\frac16\\pi^2$, and, on top of that, it is a very unusual and, in my opinion, beautiful one.\n\nIn his gorgeous paper \"How to compute $\\sum \\frac{1}{n^2}$ by solving triangles\", Mikael Passare offers this idea for proving $\\sum_{n=1}^\\infty \\frac{1}{n^2} = \\frac{\\pi^2}{6}$:\n\nProof of equality of square and curved areas is based on another picture:\n\nRecapitulation of Passare's proof using formulas is as follows: (for each step there is a geometric justification)\n\n$$\\sum_{n=1}^\\infty \\frac{1}{n^2} = \\sum_{n=1}^\\infty \\int_0^\\infty \\frac{e^{-nx}}{n}\\; dx\\; = -\\int_0^\\infty log(1-e^{-x})\\; dx\\; = \\frac{\\pi^2}{6}$$\n\nI'm not sure what you mean by a geometric proof, but the following should fit the bill, as here the identity is deduced from a comparison of two areas: the first area is\n\n$\\displaystyle\\int_{[0,1]^2} \\frac{1}{1 - xy} \\frac{dx dy}{\\sqrt{xy}}$\n\nand the second is\n\n$4 \\displaystyle \\int_{\\substack{\\xi, \\eta>0 \\\\ \\xi \\eta \\leq 1}} \\frac{ d \\xi \\, d \\eta}{(1+\\xi^2)(1+\\eta^2)}$;\n\nThey are equal by a change of variables. For the first quantity, expand $(1-xy)^{-1}$ as a geometric series and integrate term-wise to get $3 \\cdot \\zeta(2)$. The second can be computed to $\\pi^2/2$. This can be found in detail in Kontsevich and Zagier's Periods (bottom of page 8), where they attribute the idea to Calabi. (You should easily be able to hunt down an identity of $\\zeta (n)$ being equal to an integral over the unit square in $\\mathbb{R}^n$, like the first above. It might be a fun exercise to see if this idea is can be adapted for even $n$.)\n\nOne geometric/probabilistic proof is by Eugenio Calabi.\n\nConsider the double integral $$I=\\int_{0}^{1}\\int_{0}^{1} \\frac{1}{1-x^2y^2} \\ dy \\ dx.$$ Expand the integrand into a geometric series as such $$\\frac{1}{1-x^2y^2} =\\sum_{n=0}^{\\infty} (xy)^{2n},$$ which is justified by the region of integration, that is $0<x,y<1.$ Exchange summation and integration (application of Tonelli's theorem), and integrate term by term to get $$I=\\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)^2}.$$ Now, we evaluate $I$ directly. Make the change of variables $$x=\\frac{\\sin(u)}{\\cos(v)}, y=\\frac{\\sin(v)}{\\cos(u)}.$$ This transformation has Jacobian Determinant $$\\frac{\\partial (x,y) }{\\partial(u,v)}=1-x^2y^2,$$ which cancels with the integrand, and the region becomes the open triangle defined by the inequalities $$u+v<\\frac{\\pi}{2}, u, v>0$$ We know from geometry that this triangle has base $\\frac{\\pi}{2}$ and height $\\frac{\\pi}{2},$ hence the area is $\\frac{\\pi^2}{8}.$ Hence $I=\\frac{\\pi^2}{8}.$ Now to obtain $\\zeta(2),$ we use the fact that $$\\zeta(2)=\\sum_{n=1}^{\\infty} \\frac{1}{n^2} = \\sum_{n=1}^{\\infty} \\frac{1}{(2n)^2}+\\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)^2},$$ which implies by some algebraic manipulation that\n\n$$\\zeta(2)=\\frac{4}{3}I=\\frac{\\pi^2}{6}.$$\n\nWe can extend this proof to the $\\zeta(2k).$ Let\n\n$$I_{2k}= \\int_{0}^{1}... \\int_{0}^{1} \\frac{1}{1-x_1^2 \\dots x_{2k}^2} \\ dx_{2k} \\ ... \\ dx_1.$$\n\nConverting this integrand into a geometric series and exchanging summation and integration gives the sum: $$I_{2k}=\\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)^{2k}}.$$ We can then recover $$\\zeta(2k)=\\frac{2^{2k}}{2^{2k}-1} I_{2k},$$ by observing $$\\zeta(2k)= \\sum_{n=1}^{\\infty} \\frac{1}{(2n)^{2k}} + I_{2k},$$ and using some algebraic manipulation.\n\nOn the other hand, we make the change of variables $$x_i=\\frac{\\sin(u_i)}{\\cos(u_{i+1})}, 1 \\leq i \\leq 2k,$$ and $u_{2k+1}:=u_{1}$ here. This transformation turns out to have a Jacobian Determinant which cancels with the denominator of the integrand in $I_{2k}.$ The region of integration becomes the open polytope: $$\\Delta^{2k}= \\left \\lbrace (u_1, \\dots, u_{2k}): u_{i}+u_{i+1} < \\frac{\\pi}{2}, u_i>0, 1 \\leq i \\leq 2k \\right \\rbrace$$ in which $u_{2k+1}:=u_1.$ Computing the volume of this polytope is much more difficult. Rescaling with the change of variables $u_i=\\frac{\\pi}{2} v_i,$ and letting $V_1, \\dots, V_{2k}$ being $2k$ independent, uniformly distributed random variables on $(0,1),$ we get\n\n$$I_{2k}=\\text{Vol}(\\Delta^{2k})=\\left(\\frac{\\pi}{2}\\right)^{2k} \\text{Pr} \\left( V_1+V_2<1, \\dots, V_{2k-1}+V_{2k}<1, V_{2k}+V_{1}<1 \\right),$$ the probability that $V_1, \\dots, V_{2k}$ have cyclically pairwise consecutive sums less than $1.$\n\nIn the literature, the aforementioned probability has been computed in several ways (see: https://www.maa.org/sites/default/files/pdf/news/Elkies.pdf https://arxiv.org/pdf/1003.3602.pdf https://pdfs.semanticscholar.org/35be/01e63c0bfd32b82c97d58ccc9c35471c3617.pdf)\n\nJoshua Seaton also mentioned Zagier's and Kontsevich's reproduced Calabi proof. The general version of this is to evaluate\n\n$$J_{2k}= \\int_{0}^{1}... \\int_{0}^{1} \\frac{1}{\\sqrt{x_1 \\dots x_{2k}}(1-x_1 \\dots x_{2k})}\\ dx_{2k} \\ ... \\ dx_1.$$\n\nA quick substitution shows $J_{2k}=2^{2k}I_{2k}.$ The generalized version of the change of variables is\n\n$$x_i=\\frac{\\xi_i^2(1+\\xi_{i+1}^2)}{1+\\xi_i^2}, \\dots, x_{2k}=\\frac{\\xi_{2k}^2(1+\\xi_{1}^2)}{1+\\xi_{2k}^2},$$ and upon computing its Jacobian Determinant, we get that\n\n$$J_{2k}=\\int_{\\mathbb{H}^{2k}} \\frac{1}{\\xi_1^2+1} \\dots \\frac{1}{\\xi_{2k}^2+1} \\ d \\xi_{1} \\dots d \\xi_{2k},$$ where $\\mathbb{H}^{2k}$ is the hyperbolic defined by the inequalities: $$\\xi_{i}\\xi_{i+1} <1, \\xi_i>0,$$ where $1 \\leq i \\leq 2k$ and $\\xi_{2k+1}:=\\xi_1.$ Letting $\\Xi_1, \\dots \\Xi_{2k}$ be $2k$ independent, nonnegative Cauchy random variables, we get\n\n$$I_{2k}=\\left(\\frac{\\pi}{2}\\right)^{2k} \\text{Pr} \\left( \\Xi_1\\Xi_2<1, \\dots, \\Xi_{2k-1}\\Xi_{2k}<1, \\Xi_{2k}\\Xi_{1}<1 \\right),$$ the probability that $\\Xi_1, \\dots, \\Xi_{2k}$ have cyclically pairwise consecutive products less than $1.$\n\nIn my paper, I show that the probabilities lead to a very gargantuan formula:\n\n$$I_{2k}=\\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)^{2k}}= \\left(\\frac{\\pi}{4} \\right)^{2k} +\\left(\\frac{\\pi}{4} \\right)^{2k} \\sum_{n=1}^{k} \\sum_{\\substack{(r_1, \\dots, r_n) \\in [2k]^n: \\\\ |r_p-r_q| \\notin \\lbrace 0,1,2k-1 \\rbrace, \\\\ p,q \\in [n]} } \\prod_{i=1}^{n} \\frac{1}{i+\\sum_{j=1}^{i} \\alpha_j},$$ where $[m]:= \\lbrace 1, \\dots, m \\rbrace$ and $$\\alpha_j=2- \\delta(2k,2) - \\sum_{m=1}^{j-1} \\delta(|r_m-r_j|,2)+\\delta(|r_m-r_j|,2k-2)$$ and $\\delta(a,b)$ is the Kronecker Delta Function. In particular, the inner sum is taken over all tuples $(r_1, \\dots, r_n) \\in [k]^ n$ having cyclically pairwise nonconsecutive entries.", "date": "2020-02-18 10:14:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/187295/a-geometric-proof-of-zeta2-frac-pi26-and-other-integer-inputs-for-the", "openwebmath_score": 0.9478412866592407, "openwebmath_perplexity": 249.25782221486685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750499754302, "lm_q2_score": 0.8840392939666336, "lm_q1q2_score": 0.8728783420605489}}
{"url": "https://math.stackexchange.com/questions/3013681/linear-programming-word-problem-theater", "text": "# Linear Programming Word Problem: Theater\n\nI'm having trouble understanding the second constraint.\n\nA theater is presenting a program on drinking and driving for students and their parents[...] admission is 2.00 dollars for parents and 1.00 dollar for students. However, the situation has two constraints: 1) The theater can hold no more than 150 people and 2) every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?\n\nLet $$x$$ = number of students\n\n$$y$$ = number of parents\n\nSince the question prompt wants the maximum amount of revenue, then the objective function is $$z = x + 2y$$\n\nThe theater can hold no more than 150 people so the first constraint is simply: $$x + y \\leq 150$$\n\nThe second constraint is, \"every two parents must bring at least one student,\" but I don't understand how to model this.\n\nI've looked up a solution and it said $$y \\leq 2x$$ is how to model this constraint, but I don't understand why. If there must be at least one student for every two parents then why wouldn't the inequality be $$2y \\geq x$$?\n\nYes\n\ny <= 2x\n\nThe easiest way to test this is with some data points and sketch a graph:\n\ny_______x\n\n2_______1,2,3,4,5\u2026\n\n4_______2,3,4,5,6\u2026\n\n6_______3,4,5,6,7\u2026\n\nHence:\n\ny <= 2x\n\n2 <= 2x1 True\n\nand\n\n2y >= x\n\n2x2 >= 5 False\n\nGraph of y <= 2x", "date": "2019-09-24 09:00:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3013681/linear-programming-word-problem-theater", "openwebmath_score": 0.4941977262496948, "openwebmath_perplexity": 872.6491401768384, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9873750536904563, "lm_q2_score": 0.8840392863287584, "lm_q1q2_score": 0.8728783378033305}}
{"url": "https://gmatclub.com/forum/number-properties-question-127230.html", "text": "It is currently 23 Mar 2018, 02:07\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# If y is an integer, is y^2 divisible by 4?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 18 Jan 2012\nPosts: 22\nIf y is an integer, is y^2 divisible by 4?\u00a0[#permalink]\n\n### Show Tags\n\n07 Feb 2012, 20:41\n1\nKUDOS\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n75% (00:30) correct 25% (00:42) wrong based on 88 sessions\n\n### HideShow timer Statistics\n\nIf y is an integer, is y^2 divisible by 4?\n\n(1) y is divisible by 4\n(2) y is divisible by 6\n[Reveal] Spoiler: OA\n\nLast edited by Bunuel on 08 Feb 2012, 02:16, edited 1 time in total.\nEdited the question\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4678\nRe: Number Properties - Question 2\u00a0[#permalink]\n\n### Show Tags\n\n07 Feb 2012, 23:09\n1\nKUDOS\nExpert's post\nHi, there. I'm happy to help with this.\n\nPrompt: if y is an integer , is y^2 divisible by 4?\n\nFact #1: In order for an integer N to be divisible by 4, N must have at least two factors of 2 in its prime factorization.\n\nFact #2: When you square a number, say T^2, whatever prime factors are in the prime factorization of T are doubled in the prime factorization of T^2. Say a particular prime factor appears three times in T --- then it will appear six times in T^2.\n\nFact #3: An even number, by definition, has at least one factor of 2 in its prime factorization.\n\nTherefore, the square of any even number will have at least two factors of 2, and therefore will be divisible by 4.\n\nThe question \"is y^2 divisible by 4?\" is entirely equivalent to the question \"is y an even integer?\"\n\nAll of that was the mathematical heavy-lifting for the question. Now, the statements will be a piece of cake.\n\nStatement #1: y is divisible by 4. Therefore y is even. Sufficient.\n\nStatement #2: y is divisible by 6. Therefore y is even. Sufficient.\n\nBoth statements sufficient. Answer = D\n\nDoes all that make sense?\n\nHere's another odd/even DS question for practice.\n\nhttp://gmat.magoosh.com/questions/868\n\nPlease let me know if you have any questions.\n\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44412\nRe: Number Properties - Question 2\u00a0[#permalink]\n\n### Show Tags\n\n08 Feb 2012, 02:16\nIf y is an integer, is y^2 divisible by 4?\n\n(1) y is divisible by 4 --> as y itself is divisible by 4 then y^2 will also be divisible by 4. Sufficient.\n(2) y is divisible by 6 --> y=6k, for some integer k --> y^2=36k^2=4*(9k^2) --> we can see that y^2 has 4 as a factor hence it is divisible by 4. Sufficient.\n\nHope it's clear.\n_________________\nMath Revolution GMAT Instructor\nJoined: 16 Aug 2015\nPosts: 5079\nGMAT 1: 800 Q59 V59\nGPA: 3.82\nIf y is an integer, is y^2 divisible by 4?\u00a0[#permalink]\n\n### Show Tags\n\n26 Nov 2017, 11:28\nnimc2012 wrote:\nIf y is an integer, is y^2 divisible by 4?\n\n(1) y is divisible by 4\n(2) y is divisible by 6\n\nForget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.\n\nSince we have 1 variables and 0 equation, we need to consider each condition one by one.\n\nCondition 1)\n$$y = 4n$$ for some integer $$n$$.\n$$y^2 = 16n^2$$\nSince $$16$$ is divisible by $$4$$, this is sufficient.\n\nCondition 2)\n$$y = 4m$$ for some integer $$m$$.\n$$y^2 = 36m^2$$\nSince $$36$$ is divisible by $$4$$, this is sufficient.\n\nIf the original condition includes \u201c1 variable\u201d, or \u201c2 variables and 1 equation\u201d, or \u201c3 variables and 2 equations\u201d etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.\n_________________\n\nMathRevolution: Finish GMAT Quant Section with 10 minutes to spare\nThe one-and-only World\u2019s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.\n\"Only \\$79 for 3 month Online Course\"\n\"Free Resources-30 day online access & Diagnostic Test\"\n\"Unlimited Access to over 120 free video lessons - try it yourself\"\n\nIf y is an integer, is y^2 divisible by 4? \u00a0 [#permalink] 26 Nov 2017, 11:28\nDisplay posts from previous: Sort by", "date": "2018-03-23 09:07:50", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/number-properties-question-127230.html", "openwebmath_score": 0.6103289723396301, "openwebmath_perplexity": 2076.712551461292, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9579122720843811, "lm_q2_score": 0.9111797166446537, "lm_q1q2_score": 0.8728302326482829}}
{"url": "https://math.stackexchange.com/questions/372624/probability-of-random-integers-digits-summing-to-12", "text": "# Probability of random integer's digits summing to 12\n\nWhat is the probability that a random integer between 1 and 9999 will have digits that sum to 12?\n\nAs a user suggested, I could make a spreadsheet and count them, but is there a quicker way to do this?\n\n\u2022 For a range like this, why not make a spreadsheet and count them? \u2013\u00a0Ross Millikan Apr 25 '13 at 16:52\n\u2022 Your question is phrased as a stand-alone problem, without any further information or context. This does not match our quality standards, and hence is likely to attract downvotes, or be closed. It is impossible for us to assess your issues with the problem, and the level of answer appropriate for you. This is a guide to asking a good question. Concretely: please provide context, and include your work and thoughts on the problem. This helps attract more appropriate answers and will likely remove down- and close votes. \u2013\u00a0The Chaz 2.0 Apr 25 '13 at 16:54\n\u2022 I vote against closing this question. \u2013\u00a0MJD Apr 25 '13 at 17:40\n\u2022 I also vote against closing this question. \u2013\u00a0Zev Chonoles Apr 25 '13 at 22:02\n\u2022 @TheChaz I am very disappointed to see valid questions like this being closed for strange reasons, e.g. \"too localized\". Ditto for analogous recent votes. \u2013\u00a0Math Gems Apr 26 '13 at 15:50\n\nThe problem does not require a spreadsheet. It does not even require paper.\n\nThe question is to count the number of integer tuples $\\langle a,b,c,d\\rangle$ with $a+b+c+d=12$ and $0\\le a,b,c,d < 10$. We could enumerate this by choosing $a$ and then counting the tuples $\\langle b,c,d\\rangle$ with $b+c+d = 12-a$, and recursing, but an easier method is available.\n\nFirst, note that if we drop the $a,b,c,d < 10$ restriction, the problem is easy. By the stars and bars method, there are $\\binom{15}{12} = 455$ tuples that sum to 12.\n\nFrom these 455 we need to eliminate the ones that contain $10, 11,$ or $12$. Let $t_i$ be the number of tuples where $a =i$ for $i\\in\\{10,11,12\\}$. Clearly, $t_{12} = 1$: the only tuple is $\\langle 12, 0,0,0\\rangle$. For $a=11$ we need $b+c+d=1$, so exactly one of $b,c,d$ is 1 and the other two are 0, and thus $t_{11} = 3$.\n\nFor $a=10$ there are two possibilities. Either $\\{b,c,d\\} = \\{2,0,0\\}$ or $\\{b,c,d\\} = \\{1,1,0\\}$. In either case there are 3 tuples, so $t_{10} = 6$.\n\nSince at most one of $a,b,c,d$ is greater than 9, the total number of tuples that contain 10, 11, or 12 is $4(t_{10}+t_{11}+t_{12}) = 40$.\n\nThus the total number of tuples of just 0 through 9, and the answer to the question, is 455 - 40 = 415; the probability is $\\frac{415}{9999}$.\n\n\u2022 Note that with the range starting at 1, leading zeros are explicitly OK. This makes it easier. \u2013\u00a0Ross Millikan Apr 25 '13 at 18:12\n\u2022 There are several features of this problem that make it easier than it seems at first: leading zeroes are acceptable, so $a,b,c,$ and $d$ are symmetric. The required sum is close to 9, so only a few integer tuples need to be subtracted, and no inclusion-exclusion argument is needed. Recognizing such features when they appear can be an important aspect of solving this type of problem. \u2013\u00a0MJD Apr 25 '13 at 18:27\n\u2022 Does not even require paper?? :) Neither did mine \u2013\u00a0wolfies Apr 25 '13 at 19:17\n\u2022 The problem does not require an integer either. What is the probability that randomly chosen combination of four numbers in the range 0 to 9 adds up to to twelve. \u2013\u00a0Kaz Apr 25 '13 at 21:53\n\u2022 @Kaz Probability is 0, because if you permit $(a,b,c,d) \\in \\mathbb{R}^4$, even restricting the possible range to between 0 and 9, $a+b+c+d=12$ is still a three-dimentional figure. That's like asking what the probability that a random point in a plane happens to lie on a given line, or the probability that a randomly chosen number just happens to be exactly the same as a given number. Without restricting the digits to integers, the problem becomes trivial. \u2013\u00a0AJMansfield Apr 26 '13 at 1:24\n\nUse generating functions. The generating function for a single digit is:\n\n$$1 + x + \\cdots + x^9 = \\frac{1-x^{10}}{1-x}.$$\n\nThe generating function for the sum of four digits is the fourth power:\n\n$$\\frac{(1-x^{10})^4}{(1-x)^4} = (1-x^{10})^4 (1-x)^{-4}.$$\n\nTo solve the problem, find the coefficient of $x^{12}$.\n\n$$(1-x^{10})^4 = 1 -4 x^{10} + 6 x^{20} - \\cdots$$\n\nSo, we only need the coefficients of $x^2$ and $x^{12}$ in $(1-x)^{-4}$ using the generalized binomial theorem. These are $\\binom{5}{2} = 10$ and $\\binom{15}{12} = 455$. The coefficient of $x^{12}$ is therefore\n\n$$-4\\cdot10 + 1\\cdot455 = 415.$$\n\nSo the answer is $\\frac{415}{9999}$.\n\n\u2022 where do 5,2,15 and 12 come from in ncr(5, 2) and ncr(15, 12)? \u2013\u00a0michaelsnowden Sep 8 '15 at 17:46\n\u2022 @michaelsnowden The coefficient of $x^2$ in $(1 - x)^{-4}$ is $(-1)^2\\binom{-4}{2} = \\binom{4 + 2 - 1}{2}$, because of the binomial theorem which says that the coefficient of $z^n$ in $(1+z)^n$ is $\\binom{n}{r}$, and the fact that $\\binom{-n}{r} = (-1)^r\\binom{n + r - 1}{r}$. Similarly, the coefficient of $x^{12}$ in $(1-x)^{-4}$ is $(-1)^{12}\\binom{-4}{12} = \\binom{4 + 12 - 1}{12}$. \u2013\u00a0ShreevatsaR Jan 5 '19 at 21:16\n\u2022 In general, if we're looking for $k$ random digits summing to $n$, by this answer we're looking for the coefficient of $x^n$ in $(1-x^{10})^k (1-x)^{-k}$. This is going to be $$\\sum_{10r+s=n} (-1)^{r+s} \\binom{k}{r} \\binom{-k}{s} = \\sum_{10r+s=n} (-1)^{r} \\binom{k}{r} \\binom{k+s-1}{s}$$ if I'm not mistaken. \u2013\u00a0ShreevatsaR Jan 5 '19 at 21:24\n\nWith Mathematica code (FROM 1 TO 9999):\n\nCount[Map[Total, IntegerDigits[Range[9999]]], 12]\n\n\n415\n\nSo: 415/9999\n\n\u2022 As an alternative approach: Probability[x == 12, x \\[Distributed] Total[IntegerDigits[Range[9999]], {2}]] directly yields 415/9999. \u2013\u00a0Sasha Apr 25 '13 at 19:05\n\u2022 Wow, \"FrankenLisp\". \u2013\u00a0Kaz Apr 25 '13 at 21:54\n\nJust so GAP doesn't get left out. Here's a GAP version to obtain the count:\n\nNumber([1..9999],n->Sum(ListOfDigits(n))=12);\n\n\nwhich returns 415. So, the probability is $415/9999$.\n\n\u2022 Go GAP! Good to see it's still around and alive. \u2013\u00a0Peter K. Apr 25 '13 at 20:57\n\nWith PARI/GP code\n\nQ(n)=if(n<10,n,n%10 + Q(n\\10))\n\nsum(i=1,9999,Q(i)==12)/9999\n\nI obtain $$\\frac{415}{9999}.$$", "date": "2020-02-24 00:07:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/372624/probability-of-random-integers-digits-summing-to-12", "openwebmath_score": 0.7185295820236206, "openwebmath_perplexity": 323.86128981209055, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126513110864, "lm_q2_score": 0.8918110454379297, "lm_q1q2_score": 0.872826752749068}}
{"url": "http://mathhelpforum.com/advanced-statistics/60427-normal-distribution-population-knee-heights.html", "text": "# Thread: Normal distribution with population knee heights\n\n1. ## Normal distribution with population knee heights\n\nGiven: Men have sitting knee heights that are normally distributed with mean of 22.0 in. and std dev of 1.1 in. Women have mean of 20.3 in and std dev of 1.0 in.\n\nQuestion: Find min and max sitting knee heights that include at least 95% of all men and at at least 95% of all women.\n\nWhat I did: I originally took the avg of the means given and the avg of std devs given and using invNorm found the min and max using .025 from either side of the curve. Min: 19.10 in and Max: 23.21 in. After testing these limits with the men's average I found that I was cutting out 20.9% of the men, so obviously I did this wrong. NOTE: I also found 95% max and min for each gender, but then didn't know what to do with this information.\n\nWhat should I have done to find the answers?\n\nThanks.\n\n2. Originally Posted by stats08\nGiven: Men have sitting knee heights that are normally distributed with mean of 22.0 in. and std dev of 1.1 in. Women have mean of 20.3 in and std dev of 1.0 in.\n\nQuestion: Find min and max sitting knee heights that include at least 95% of all men and at at least 95% of all women.\n\nWhat I did: I originally took the avg of the means given and the avg of std devs given and using invNorm found the min and max using .025 from either side of the curve. Min: 19.10 in and Max: 23.21 in. After testing these limits with the men's average I found that I was cutting out 20.9% of the men, so obviously I did this wrong. NOTE: I also found 95% max and min for each gender, but then didn't know what to do with this information.\n\nWhat should I have done to find the answers?\n\nThanks.\nTo get the minimum for men you require the value of $x_{\\min}$ such that $\\Pr(X < x_{\\min}) = 0.025$.\n\nNote that $z^* = \\frac{x_{\\min} - 22.0}{1.1}$ where $\\Pr(Z < z^*) = 0.025 \\Rightarrow z^* = -1.96$.\n\nThen $-1.96 = \\frac{x_{\\min} - 22.0}{1.1} \\Rightarrow x_{\\min} = 19.84$.\n\nIn a similar way I get $x_{\\max} = 24.16$.\n\nDo the same thing for the women.\n\n3. Originally Posted by mr fantastic\nTo get the minimum for men you require the value of $x_{\\min}$ such that $\\Pr(X < x_{\\min}) = 0.025$.\n\nNote that $z^* = \\frac{x_{\\min} - 22.0}{1.1}$ where $\\Pr(Z < z^*) = 0.025 \\Rightarrow z^* = -1.96$.\n\nThen $-1.96 = \\frac{x_{\\min} - 22.0}{1.1} \\Rightarrow x_{\\min} = 19.84$.\n\nIn a similar way I get $x_{\\max} = 24.16$.\n\nDo the same thing for the women.\nIs there some confusion here?\n\nAre we looking for two knee height intervals one containing at least 95% of men and the other 95% of women, or one interval which contains at least 95% of men and 95% of women simultaneously?\n\nWorking this numerically I find that the smallest interval symmetric about the mean knee height of men and women combined is approximately [18.48, 23.82] inches. This is the larger of the two symmetric intervals that contain 95% of womens knees centred at the grand mean and that contain 95% of mens knees centred at the grand mean.\n\nThe interval given contains about 95% of mens knees and 96.5% of womens knees.\n\n(of course if we had chossen to centre the interval at another point we would have found a different result, now the question is what is the smallest interval that meets our requirements? I'm pretty sure that the interval found here is not the smallest that will do the job)\n\nThe shortest interval meeting our requirements is approximatly [18.65, 23.82] which contains about 95% of womens knees and 95% of mens knees. This was found by brute force by computing the probability of a range of intervals for men and women plotting the 95% contours and keeping the point at which the contours cross!\n\nCB\n\n4. ## Population Knee Heights\n\nYou are correct, Captain, I need to find common max and min that will contain 95% of all men and all women combined (as if I'm designing a carseat).\n\nWhen you say you found a smaller interval using \"brute force by computing the probability of a range of intervals\", can you give an example on how to even pick an interval to try and how you know when contours cross?\n\nI was aware that some guessing is involved, but I'm not sure if it is lack of sleep or lost brain cells, but I'm not sure where to begin guessing outside of givens.\n\nThanks.\n\n5. Originally Posted by stats08\nYou are correct, Captain, I need to find common max and min that will contain 95% of all men and all women combined (as if I'm designing a carseat).\n\nWhen you say you found a smaller interval using \"brute force by computing the probability of a range of intervals\", can you give an example on how to even pick an interval to try and how you know when contours cross?\n\nI was aware that some guessing is involved, but I'm not sure if it is lack of sleep or lost brain cells, but I'm not sure where to begin guessing outside of givens.\n\nThanks.\nDepend on what software you have available. This can be done approximatly in Excel (not that I used Excel).\n\nTake a rectangular region with the interval half width down one edge, and the centre along the bottom edge. In each cell compute the proportion of mens knee heights that fall in the interval defined by the figures in the two edges with the interval parameters.\n\nRepeat in another rectangular region but this time for women's knee heights.\n\nIn a third rectangular region record if the corresponding cells in the first two regions are greater than 0.95, select the cell with the with both greater than 0.95.\n\nCB\n\n6. Nice! Thank you. I'll try it out.\n\n7. Originally Posted by CaptainBlack\nIs there some confusion here?\n\nAre we looking for two knee height intervals one containing at least 95% of men and the other 95% of women, or one interval which contains at least 95% of men and 95% of women simultaneously?\n\nWorking this numerically I find that the smallest interval symmetric about the mean knee height of men and women combined is approximately [18.48, 23.82] inches. This is the larger of the two symmetric intervals that contain 95% of womens knees centred at the grand mean and that contain 95% of mens knees centred at the grand mean.\n\nThe interval given contains about 95% of mens knees and 96.5% of womens knees.\n\n(of course if we had chossen to centre the interval at another point we would have found a different result, now the question is what is the smallest interval that meets our requirements? I'm pretty sure that the interval found here is not the smallest that will do the job)\n\nThe shortest interval meeting our requirements is approximatly [18.65, 23.82] which contains about 95% of womens knees and 95% of mens knees. This was found by brute force by computing the probability of a range of intervals for men and women plotting the 95% contours and keeping the point at which the contours cross!\n\nCB\nMy mistake. Yes there was confusion.\n\nHowever, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.\n\nThen an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.\n\nAnd at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.\n\n8. Originally Posted by mr fantastic\nMy mistake. Yes there was confusion.\n\nHowever, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.\n\nThen an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.\n\nAnd at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.\nIt would depend on the marginal costs of providing a greater range (including consideration of the profit margin on the product). It is often supprising how large the impact of a 12.5% increase in some parameter in a product design can sometimes be (often because of their impact on other aspects of design).\n\nI know of at least one product with some design parameters that could not be increased by even 5% without an additional development cost in the order of 10's if not 100's of millions of pounds!\n\nCB\n\n9. Originally Posted by mr fantastic\nAnd at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.\nI always knew things were different in Oz. In the Northern Hemisphere the obese become wider not taller!\n\nCB", "date": "2017-10-20 16:37:16", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/60427-normal-distribution-population-knee-heights.html", "openwebmath_score": 0.5258986353874207, "openwebmath_perplexity": 805.722499451058, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9787126457229185, "lm_q2_score": 0.89181104831338, "lm_q1q2_score": 0.8728267505797176}}
{"url": "https://ask.sagemath.org/question/51146/can-you-specify-an-explicit-curve-by-specifying-its-xy-points/?answer=51191", "text": "# Can you specify an explicit curve by specifying its x,y points ?\n\nI was wondering if I can specify a curve by giving its x,y points:\n\n(0,0),(0.5, 0.25),(0.6, 0.36),(2.0, 4.0)\n\nand then do various other manipulation of this curve like, interpolation, derivative at some point, aread under the curve between two points (integral) etc?\n\nedit retag close merge delete\n\nThere are essentially an infinite number of curves going through 5 points. But if you adjust say by mean squarre a polynomial of 5th degree it will run through this 5 points. Then you will have a function and you will be able to manipulate it.\n\n( 2020-04-30 08:27:28 +0200 )edit\n\nThanks for the comment. That obviously can be done and its interesting. I take it that the answer to the original question is no then?\n\n( 2020-04-30 15:20:56 +0200 )edit\n\nI am trying to make the program but I have a problem for the optimization. I have ask a question and I wait for the answer\n\n( 2020-04-30 17:17:43 +0200 )edit\n\nThank you very much for the answer! so the work is in progress. That's an interesting approach.\n\n( 2020-04-30 17:31:25 +0200 )edit\n\nSort by \u00bb oldest newest most voted\n\nHello, @Mo! As @Cyrille mentioned, given $n$ points, there exists an infinite number of curves that pass through them. However, there is a number of things you could do in order to obtain one of those curves. Let me explain three of them:\n\n1. You can take advantage of the particular characteristics of your set of points. In this case, a simple visual inspection will show that the points are of the form $(x, x^2)$. So, a good curve for this set is quite simply $y=x^2$.\n2. Of course, visual inspection is not always as easy as in this particular case. When you can't figure out the structure of function, a classical approach is to use interpolation polynomials. (This is also mentioned by @Cyrille.) Now, Sage has many interpolation facilities; in particular, there is the spline command. You can write something like this:\n\npoints = [(0, 0), (0.5, 0.25), (0.6, 0.36), (2, 4)]\nS = spline(points)\n\n\nNow, S is an interpolation spline, which you can evaluate en $x=0.2$, for example, with S(0.2); you can plot S with plot(S, xmin=0, xmax=2); you can differentiate it in $x=1.3$ with S.derivative(1.3); you can integrate it in its whole domain with S.definite_integral(0, 2); you could even plot the derivative and the antiderivative with a little bit of clever programming.\n\nIt is my understanding that splines are excellent approximating real-life functions, because natural phenomena tend to have smooth behavior. If these points were the result of measurements of a natural phenomenon in the lab, I would consider using splines for interpolation.\n\nThe disadvantage of the spline command is that it doesn't give you an explicit formula for the function.\n\n3. There is another interpolation command which is a little more complicated, but more useful, in my humble opinion (in particular, it gives you a explicit formula). It's called lagrange_polynomial. In its simplest form, it uses divided differences to obtain the polynomial coefficients (you can read about that in any Numerical Methods book). You have to do something like this:\n\nPR = PolynomialRing(RR, 'x')\npoints = [(0, 0), (0.5, 0.25), (0.6, 0.36), (2, 4)]\npoly = PR.lagrange_polynomial(points)\nprint(poly)\n\n\nThe first line creates a polynomial ring, which I called (creatively enough) PR, with real coefficients (thus the RR) and independent variable x (thus the 'x'). The second line is your set of points. The third line asks Sage to create the Lagrange Interpolation Polynomial in this polynomial ring. You should get something like this:\n\n -1.11022302462516e-16*x^3 + x^2 - 1.11022302462516e-16*x\n\n\nYou can use this polynomial as with any other polynomial. You can make plot(poly, (x, 0, 2)), poly.derivative(x), poly.derivative(x)(1.3), poly.integrate(x), etc.\n\nHowever, as you can see, the coefficients for $x^3$ and $x$ are negligible, so you could deduce that poly is actually x^2. You can verify that using the following:\n\nPR = PolynomialRing(QQ, 'x')\npoints = [(0, 0), (0.5, 0.25), (0.6, 0.36), (2, 4)]\npoly = PR.lagrange_polynomial(points)\nprint(poly)\n\n\nThis is exactly the same as before, but you use rational coefficients instead of real ones (thus the QQ). Of course, the same result is obtained if you replace QQ with ZZ, since, in this case, the coefficients turn out to be integers. the result is x^2, as suspected!\n\nThere are other methods you can use to obtain curves that pass through $n$ points, but these are the simplest and more usual (and useful). Just remember the two downsides of interpolation: (1) Your are getting only one of an infinite number of possible functions that pass through your points. (2) You should only work inside the interpolation interval (in particular, S(3) will return nan or not-a-number; poly(4) will evaluate correctly to 16, but you are, in that case, extrapolating, which is a very dangerous operation, especially in the lab).\n\nI hope this helps!\n\nmore\n\nAlthough I hope this solves the question, I still would like to see @Cyrille's solution, which sounds interesting!\n\n( 2020-05-01 05:18:01 +0200 )edit\n\nHere is my solution more involved than the one of dsejas, but perhaps it can help (Thanks to Sebastien)\n\n# Polynomial interpolation\n# Passing the matrix\nA = Matrix([[0,0],[0.5,0.25],[0.6,0.36],[2,4]])\nvar('a0, a1, a2, a3')\nB =[[A[i][1], a3*A[i][0]^3+ a2*A[i][0]^2+ a1*A[i][0]+ a0] for i in range(0,4)]\nC=[[(B[i][0]-B[i][1])^2] for i in range(0,4)]\n# sum of the square of errors\nssr=sum(C[i][0] for i in range(4))\n# First order condition of optimality\nssr_a0=ssr.diff(a0)\nssr_a1=ssr.diff(a1)\nssr_a2=ssr.diff(a2)\nssr_a3=ssr.diff(a3)\n# solution for coefficients\nsol=solve([ssr_a0==0,ssr_a1==0,ssr_a2==0,ssr_a3==0,ssr_a4==0], a0, a1, a2, a3, solution_dict=True)\nf(x)= a0 + a1*x+ a2*x^2 + a3*x^3\n# g is a the desired function\ng(x)=f(x).subs(sol[0])\n# derivative of g\ngp=g.diff(x)\npl=plot(g(x),(x,A[0][0],A[3][0]))\npl+=points((A[i] for i in range(4)),color='red',size=20)\npl+=plot(gp(x),(x,A[0][0],A[3][0]),color=\"green\")\npl+=text(\"$f(x)$\",(1.2,1),color=\"blue\", fontsize='small', rotation=0)\npl+=text(\"$f'(x)$\",(1.2,2.8),color=\"green\", fontsize='small', rotation=0)\nshow(pl)\n\nmore\n\nThe line\n\nssr_a4=ssr.diff(a4)\n\n\nshould be removed as well as the ,ssr_a4==0 part or an error is raised (a4 is not defined).\n\n( 2020-05-01 16:50:39 +0200 )edit\n\nDone. I asked the subsidiary question : and if the datas came from an external file say an xls one.\n\n( 2020-05-02 12:31:54 +0200 )edit\n\nAssume that we have $n+1$ data points $(x_0,y_0),\\ldots,(x_n,y_n)$, with $x_i\\ne x_j$ for $i\\ne j$. Then, there exists a unique polynomial $f$ of degree $\\leq n$ such that $f(x_i)=y_i$ for $i=0,\\ldots,n$. This is the Lagrange polynomial that interpolates the given data. In his answer, @dsejas has shown a method to compute it. In the current case, since $n=3$, the interpolating polynomial has degree $\\leq3$; in fact, it is $f(x)=x^2$.\n\nIf $f(x)=a_0+a_1x+\\cdots+a_nx^n$, it follows from the interpolation conditions that the coefficients are the solution of the linear system $A\\mathbf{x}=\\mathbf{b}$, where $A=(x_i^j)$ (the element of $A$ at row $i$ and column $j$ is $x_i^j$), $\\mathbf{b}=(y_i)$ and $\\mathbf{x}$ is the vector of unknowns $a_0,\\ldots,a_n$. This is not the best way to get $a_0,\\ldots,a_n$, since $A$ is usually an ill conditioned matrix. However, it is interesting to point out that from a theoretical viewpoint.\n\nIf one has $m$ data points and still wants to fit them using a polynomial of degree $n$, with $n\\leq m$, the linear system $A\\mathbf{x}=\\mathbf{b}$ is overdetermined (there are more equations than unknowns). In such a case, one can perform a least squares fitting, trying to minimize the interpolation errors, that is, one seeks for the minimum of $$E(a_0,a_1,\\ldots,a_n)=\\sum_{i=0}^m (f(x_i)-y_i)^2.$$ The minimum is given by the solution of the linear system $$\\nabla E(a_0,a_1,\\ldots,a_n)=\\mathbf{0},$$ which is simply $A^TA\\mathbf{x}=A^T\\mathbf{b}$. This is what @Cyrille, in fact, proposes in his answer... computing explicitly $E(a_0,a_1,\\ldots,a_n)$ as well as its partial derivatives, which is completely unnecessary. In fact, if the system $A\\mathbf{x}=\\mathbf{b}$ has a unique solution, which is the case under discussion, the system $A^TA\\mathbf{x}=A^T\\mathbf{b}$ yields exactly the same solution. So, the function in @Cyrille's answer is just the Lagrange polynomial already proposed by @dsejas.\n\nThere are functions to compute least squares approximations in libraries like Scipy. Anyway, for such a simple case as that considered here, the following (suboptimal) code may work. Please note that I have added some points to make the example a bit more interesting:\n\n# Data\ndata = [(0,0), (0.2,0.3), (0.5,0.25), (0.6,0.36),\n(1,1), (1.2, 1.3), (1.5,2.2), (1.7,3.1), (2,4)]\n# Computation of a fitting polynomial of degree <=3\nbasis = vector([1, x, x^2, x^3])\nA = matrix(len(data),len(basis),\n[basis.subs(x=datum[0]) for datum in data])\nb = vector([datum[1] for datum in data])\ncoefs = (transpose(A)*A)\\(transpose(A)*b)\nf(x) = coefs*basis\n# Plot results\nabcissae = [datum[0] for datum in data]\nx1, x2 = min(abcissae), max(abcissae)\npl = plot(f(x), (x, x1, x2))\npl += points(data, color=\"red\", size=20)\nshow(pl)\n\n\nThis code yields the following plot:\n\nOf course, in addition to polynomial interpolation or fitting, there are many other methods which can be used.\n\nmore\n\nI completely agree. Your solution is the one of the econometrics books. I don't know why when I tried it I have an uninversible matrix $^\\top{X} X$. what shoul be nice is to add a way to read the values from an external file say a xls one.\n\n( 2020-05-02 12:27:44 +0200 )edit", "date": "2021-08-01 13:04:40", "meta": {"domain": "sagemath.org", "url": "https://ask.sagemath.org/question/51146/can-you-specify-an-explicit-curve-by-specifying-its-xy-points/?answer=51191", "openwebmath_score": 0.7643385529518127, "openwebmath_perplexity": 721.4770406587977, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712650690179, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.8728267493810961}}
{"url": "https://docs.typal.academy/analysis/sequences/basic-examples", "text": "Basic Examples\n\nProofs for the problems below follow a common structure, as described on the prior page.\nWe strongly encourage readers attempt each problem by first drawing a picture, and then attempting a formal write-up before reading the example write-up provided below.\nExample 1.\nDefine the sequence\n$\\{a_n\\}$\nby\n$a_n \\triangleq \\dfrac{n}{n^2+9}$\nfor all\n$n\\in\\mathbb{N}.$\nProve\n$\\{a_n\\}$\nconverges to zero.\nEx 1: Comment\nHint\nVisual\nFormal Write-Up\nFollow-Up\nAlthough the result may be clear from calculus, the point here is to provide formal arguments.\nNote\n$0 < a_n < \\dfrac{1}{n}.$\nPlot of and . We see , and so as . Also note for large .\nWe claim\n$\\{a_n\\}$\nconverges to zero and prove this as follows. Let\n$\\varepsilon > 0$\nbe given. It suffices to show there exists\n$N\\in\\mathbb{N}$\nsuch that\n$|a_n - 0 | \\leq\\varepsilon$\nfor all\n$n\\geq N.$\nFor each\n$n\\in \\mathbb{N}$\n, note\n$n > 0$\nand so\n$1/n > 0$\n, which implies\n \u200b$n < n + \\dfrac{9}{n} \\ \\ \\implies \\ \\ 1 < \\dfrac{n + \\frac{9}{n}}{n} \\ \\ \\implies \\ \\ \\dfrac{1}{n + \\frac{9}{n}} < \\dfrac{1}{n}.$\u200b\nConsequently,\n \u200b\u200b\nwhere the second equality holds since\n$n > 0$\nand\n$n^2 + 9 > 0$\n. By the Archimedean Property of\n$\\mathbb{R}$\n, there exists\n$\\tilde{N}\\in\\mathbb{N}$\nsuch that\n$1 < \\frac{\\tilde{N}}{\\varepsilon}$\n, which implies\n$\\frac{1}{\\tilde{N}}< \\varepsilon$\n. Thus, combining the above results,\n \u200b\u200b\nwhich verifies the desired inequality, taking\n$N=\\tilde{N}.$\n \u200b$\\blacksquare$\u200b\n\u2022 We emphasize each statement is a complete sentence (creativity is not required).\n\u2022 It is completely acceptable to mirror this structure, but change the inequalities/terms for another problem.\n\u2022 Readability is greatly aided by concluding with a statement that explicitly states how an initial task (which we set out to do) was completed.\nExample 2.\nDefine the sequence\n$\\{a_n\\}$\n\u200b by\n$\\dfrac{4n^3+n}{n^3+6}$\n\u200b for all\n$n\\in\\mathbb{N}.$\n\u200b Prove\n$\\{a_n\\}$\nconverges.\u200b\nEx 2: Comment\nHint\nVisual\nFormal Write-Up\nWe can show\n$\\{a_n\\}$\n\u200b converges by verifying it converges to a limit (that you should find).\nAs\n$n$\n\u200b gets big, the dominant term in the numerator is\n$4n^3$\n\u200b and in the denominator is\n$n^3$\n\u200b, which should give insight about the limit of\n$\\{a_n\\}.$\nAnimation of the sequence converging.\nWe claim\n$\\{a_n\\}$\n\u200b converges to four and prove this as follows. Let\n$\\varepsilon>0$\n\u200b be given. If suffices to show there is\n$N\\in\\mathbb{N}$\n\u200b such that\n$|a_n-0|\\leq\\varepsilon$\n\u200b for all\n$n\\geq N.$\n\u200b For each\n$n\\in\\mathbb{N}$\n\u200b, observe\n \u200b$|a_n-4|=\\left|\\dfrac{4n^3+n}{n^3+6}-4\\right|=\\left| \\dfrac{(4n^3+n)-(n^3+6)4}{n^3+6}\\right|=\\left|\\dfrac{n-24}{n^3+6}\\right|.$\u200b\nBuilding on this inequality,\n$n\\geq 24$\n\u200b implies\n \u200b$|a_n - 4| = \\dfrac{n-24}{n^3+6} \\leq \\dfrac{n}{n^3+6} \\leq \\dfrac{n}{n^3} \\leq \\dfrac{n}{n^2}=\\dfrac{1}{n},$\u200b\n\u200bwhere we note\n$n^2=n\\cdot n \\geq n\\cdot 1=n$\n\u200b implies\n$1/n^2\\leq 1/n.$\n\u200b By the Archimedean property of\n$\\mathbb{R}$\n\u200b, there exists a natural number\n$N_1\\in\\mathbb{N}$\n\u200b such that\n$1 \\leq N_1 \\varepsilon$\n\u200b, and so\n$1/N_1 \\leq \\varepsilon$\n\u200b. Setting\n$N_2 = \\max(24,\\ N_1)$\nyields, by the above results,\n \u200b\u200b\nThis verifies the desired inequality, taking\n$N=N_2.$\n \u200b$\\blacksquare$\u200b\nExample 3.\nSuppose\n$\\{c_n\\}$\n\u200b is a sequence in\n$[-3,7)$\n, i.e.\n$-1\\leq c_n<7$\nfor all\n$n\\in\\mathbb{N}.$\n\u200b Define the sequence\n$\\{a_n\\}$\n\u200b by\n$a_n\\triangleq c_n/n$\n\u200b for all\n$n\\in\\mathbb{N}.$\n\u200b Prove\n$\\lim_{n\\rightarrow\\infty} a_n = 0.$\nEx 3: Comment\nHint\nVisual\nFormal Write-Up\nFollow-Up\nBe careful to formally state what must be shown (to ensure no confusion between the use of each of the listed sequences).\nHere is it helpful to bound\n$a_n$\n\u200b in terms of a bound for\n$c_n$\n\u200b.\nExample sequence convergence within bounds for 7/n and -7/n.\nLet\n$\\varepsilon > 0$\n\u200b be given. It suffices to show there is\n$N\\in\\mathbb{N}$\n\u200b such that\n$|a_n - 0|\\leq \\varepsilon$\n\u200b for all\n$n\\geq N.$\n\u200b By our hypothesis on\n$\\{c_n\\}$\n\u200b,\n \u200b\u200b\nBy the Archimedean property of\n$\\mathbb{R}$\n\u200b, there is\n$N_1\\in\\mathbb{N}$\n\u200b such that\n$7 < N_1 \\varepsilon$\n\u200b, which implies\n$7/N_1 < \\varepsilon.$\n\u200b Hence\n \u200b\u200b\nThis shows the desired result, taking\n$N=N_1$\n\u200b.\n \u200b$\\blacksquare$\u200b\nA trend in the previous three problems is to use the Archimedean property of\n$\\mathbb{R}$\n\u200b. This is not a coincidence! Be sure to know how to use this property so you may obtain the existence of the natural number needed in the definition of limit convergence.\n\nExercises.\n\nExercise 1. Prove the sequence\n$\\{a_n\\}$\ndefined by\n$a_n\\triangleq \\dfrac{2n^2+7}{3+n^2}$\nconverges to two.\n\u2022 State the definition of what must be shown.\n\u2022 Rewrite the inequality with\n$a_n - 2.$\n\u2022 Apply the Archimedean property of\n$\\mathbb{R}$\n.\n\u2022 Combine relations to verify the inequality that must be shown.\u200b", "date": "2023-01-28 09:24:00", "meta": {"domain": "typal.academy", "url": "https://docs.typal.academy/analysis/sequences/basic-examples", "openwebmath_score": 0.9495712518692017, "openwebmath_perplexity": 3300.5485833439852, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9946981040304005, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8728244872775796}}
{"url": "https://math.stackexchange.com/questions/3934547/series-of-characteristic-polynomials/3934707", "text": "# Series of characteristic polynomials\n\nConsider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $$-1$$, e.g. $$M_4= \\begin{pmatrix} 2 & -1 & 0 & 0 \\\\ -1 & 2 & -1 & 0\\\\ 0 & -1 & 2 & -1\\\\ 0 & 0 & -1 & 2\\\\ \\end{pmatrix}$$\n\nI've found out that the characteristic polynomials are $$\\begin{cases} P_1(x)=2-x\\\\ P_2(x)=(2-x)^2-1\\\\ P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x) \\end{cases}$$\n\nOr with a variable change $$\\begin{cases} Q_1(y)=y\\\\ Q_2(y)=y^2-1\\\\ Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y) \\end{cases}$$\n\nLooking at the first 8 $$P_n$$\n\nI see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.\n\n1. How can I prove that all eigenvalues are between 0 and 4?\n2. Are these polynomials known (have a name)?\n3. How can I prove that the polynomial are sandwitched between $$\\frac{1}{x}+\\frac{1}{4-x}\\quad\\text{and}\\quad -\\frac{1}{x}-\\frac{1}{4-x}$$\n\n\u2022 I\u2019m not with my laptop right now, so i\u2019ll just try to comment a little bit on this beautifully written question. So, your question comes down to a question of considering a real sequence with a linear recursive relation, which is not hard to control \u2013\u00a0Paresseux Nguyen Dec 4 '20 at 16:34\n\u2022 for example, for your first question, you can see that if $x_0\\le 0$ you can prove easily that $P_n(x_0)>= P_{n-1}(x_0)$ etc \u2013\u00a0Paresseux Nguyen Dec 4 '20 at 16:38\n\u2022 you can even deduce a compact expression for your Pn(x) which would be $a(x)u(x)^n+b(x)v(c)^n$ something except some critical cases \u2013\u00a0Paresseux Nguyen Dec 4 '20 at 16:43\n\u2022 @ParesseuxNguyen thank you. I proved $P_n(x_0)\\ge P_{n-1}(x_0)$ for $x_0\\le 0$. Then using the symmetry of the polynomial is easy to show that $|P_n(x_0)|\\ge |P_{n-1}(x_0)|$ for $x_0\\ge 4$. Thus the first point is proven. Now I look for your last suggestion. \u2013\u00a0Agyla Dec 4 '20 at 17:04\n\u2022 Best question ever from a <1k user? \u2013\u00a0Randall Dec 5 '20 at 1:08\n\nThe Chebyshev polynomials of the second kind satisfy the recurrence relation $$\\begin{cases} U_0(y) = 1 \\\\ U_1(y)=x\\\\ U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y) \\end{cases}$$ so that $$Q_n(y) = U_n(y/2)$$ and $$P_n(x) = U_n(1-x/2)$$.\n\nThe zeros of $$U_n$$ are $$y_k = \\cos\\left( \\pi \\frac{k+1}{n+1}\\right) \\, , \\, k = 0, \\ldots, n$$ in the range $$(-1, 1)$$, so that $$x_k = 2 - 2\\cos\\left( \\pi \\frac{k+1}{n+1}\\right) \\, , \\, k = 0, \\ldots, n$$ are the zeros of $$P_n$$ in the range $$(0, 4)$$.\n\nAlso for $$|x| < 1$$ $$U_n(x) = \\frac{\\sin((n+1)\\arccos(x))}{\\sqrt{1-x^2}}$$ which implies $$|U_n(x)| \\le \\frac{1}{\\sqrt{1-x^2}}$$ and therefore $$| P_n(x)| \\le \\frac{2}{\\sqrt{x(4-x)}} \\le \\frac 1x + \\frac{1}{4-x}$$ for $$0 < x < 4$$, the last estimate follows from the inequality between harmonic and geometric mean.\n\nFrom a very attenuated literature search, I see that the Lucas polynomials of the second kind obey the recursion $$L_{k+1}(x) = x \\ L_k(x) - L_{k-1}(x), \\quad L_1(x) = 1 \\ , L_2(x)=x$$ This matches the recursion for $$Q_n,$$ with an index shift of 1. The one reference I've seen (I don't have access to journal papers from home) states that this polynomial set obeys the relation\n\n$$L_k(2 \\cos(\\theta) ) = 2 \\cos( k \\ \\theta)$$\n\nThis functional relationship would explain the location of the zeros, and likely the envelope property as well. I suggest that the proposer start his research with 'Lucas polynomials of the second kind.'\n\n\u2022 I don't know the Lukas polynomials, but apparently $Q_n(y) = U_n(y/2)$ where $U_n$ are the Chebyshev polynomials of the second kind, and those have all zeros in the interval $(-1, 1)$. \u2013\u00a0Martin R Dec 4 '20 at 17:56\n\u2022 @MartinR Well, that will solve the problem completely. I thought with the $\\cos(k \\theta)$ in my second formula the Chebyshev polynomials would appear. \u2013\u00a0skbmoore Dec 4 '20 at 17:59", "date": "2021-03-01 20:36:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3934547/series-of-characteristic-polynomials/3934707", "openwebmath_score": 0.8616529703140259, "openwebmath_perplexity": 300.1484169740311, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226327661524, "lm_q2_score": 0.893309405344251, "lm_q1q2_score": 0.8727835070842062}}
{"url": "https://math.stackexchange.com/questions/4027037/how-do-i-compute-this-integral-with-a-diracs-delta", "text": "# How do I compute this integral with a Dirac's delta?\n\nWhile studying probability I encountered this integral $$I=\\int_{\\mathbb{R}^2}\\exp\\left({-\\frac{x_1^2+x_2^2}{2}}\\right)\\delta\\left(r-\\sqrt{x_1^2+x_2^2}\\right)\\,dx_1\\,dx_2$$ If I compute this in polar coordinates i get $$I=\\int_0^{2\\pi}\\,d\\theta \\int_0^{+\\infty}\\exp\\left(-\\dfrac{\\rho^2}{2} \\right)\\rho\\delta(r-\\rho)\\,d\\rho=2\\pi r\\exp\\left(-\\dfrac{r^2}{2}\\right)$$ but in cartesian coordinates I only get $$I=\\exp\\left(-\\frac{r^2}{2}\\right)$$ I don't understand why. I just thougth that I was using the Dirac's delta's properties in both cases. I think the first result is the correct one and there is something I don't know about Dirac's delta with more than one variable.\n\nWhich result is correct and why?\n\nYou may also make direct calculations in Cartesian coordinates - integrating, for instance, over $$x_1$$ first and then over $$x_2$$. We can multiply and divide the argument of $$\\delta$$-function ($$r-\\sqrt{x_1^2+x_2^2}$$) by ($$r+\\sqrt{x_1^2+x_2^2}$$) (because ($$r+\\sqrt{x_1^2+x_2^2}$$) is always positive). We can also replace the power of exponent by $$-\\frac{r^2}{2}$$ - due to the condition imposed by $$\\delta$$-function\n\n$$I(r)=\\int_{\\mathbb{R}^2}\\exp\\left({-\\frac{x_1^2+x_2^2}{2}}\\right)\\delta\\left(r-\\sqrt{x_1^2+x_2^2}\\right)\\,dx_1\\,dx_2=\\int_{\\mathbb{R}^2}\\exp\\left({-\\frac{r^2}{2}}\\right)\\delta\\left(\\frac{r^2-(x_1^2+x_2^2)}{r+\\sqrt{x_1^2+x_2^2}}\\right)\\,dx_1\\,dx_2$$\n\nWe see that $$x_2$$ contributes if only $$x_2\\in[-r,r]$$, otherwise $$\\delta()=0$$\n\n$$I(r)=\\int_{-r}^rdx_2\\int_{-\\infty}^{\\infty}dx_1\\exp\\left({-\\frac{r^2}{2}}\\right)\\delta\\left(\\frac{(\\sqrt{r^2-x_2^2}-x_1)(\\sqrt{r^2-x_2^2}+x_1)}{r+\\sqrt{x_1^2+x_2^2}}\\right)$$\n\nBut $$\\delta(\\frac{(a-x_1)(x_1+b)}{A})=|\\frac{A}{a-x_1}|\\delta(x_1+b)+|\\frac{A}{x_1+b}|\\delta(a-x_1)=|\\frac{A}{a-x_1}|\\delta(x_1+b)+|\\frac{A}{x_1+b}|\\delta(x_1-a)$$\n\nWe get\n\n$$I(r)=\\int_{-r}^rdx_2\\int_{-\\infty}^{\\infty}dx_1\\exp\\left({-\\frac{r^2}{2}}\\right)\\left(r+\\sqrt{x_1^2+x_2^2}\\right)\\left(\\frac{1}{\\sqrt{r^2-x_2^2}-x_1}\\delta(\\sqrt{r^2-x_2^2}+x_1)+\\frac{1}{\\sqrt{r^2-x_2^2}+x_1}\\delta(\\sqrt{r^2-x_2^2}-x_1)\\right)=$$ $$\\int_{-r}^rdx_2\\int_{-\\infty}^{\\infty}dx_1\\exp\\left({-\\frac{r^2}{2}}\\right)2r\\left(\\frac{1}{2\\sqrt{r^2-x_2^2}}\\delta(\\sqrt{r^2-x_2^2}+x_1)+\\frac{1}{2\\sqrt{r^2-x_2^2}}\\delta(x_1-\\sqrt{r^2-x_2^2})\\right)=\\int_{-r}^r\\exp\\left({-\\frac{r^2}{2}}\\right)\\frac{2r}{\\sqrt{r^2-x_2^2}}dx_2$$ $$I(r)=\\int_{-1}^1\\exp\\left({-\\frac{r^2}{2}}\\right)\\frac{2r}{\\sqrt{1-t^2}}dt=2\\pi{r}e^{-\\frac{r^2}{2}}$$\n\nIn Cartesian coordinates, you have \\begin{align} I=&\\ \\int dx\\ \\exp\\left(-\\frac{x_1^2+x_2^2}{2} \\right)\\delta(r-\\sqrt{x_1^2+x_2^2})\\\\ =&\\ \\int_{r=\\sqrt{x_1^2+x_2^2}} d\\sigma\\ \\exp\\left(-\\frac{x_1^2+x_2^2}{2} \\right) \\end{align} since \\begin{align} |\\nabla (r- \\sqrt{x_1^2+x_2^2})| = 1. \\end{align} Hence, it follows that \\begin{align} I = \\int_{r=\\sqrt{x_1^2+x_2^2}} d\\sigma\\ \\exp\\left(-\\frac{r^2}{2} \\right) = 2\\pi r\\exp\\left(-\\frac{r^2}{2} \\right). \\end{align}\n\n\u2022 Is it like a surface integral in one dimension instead of two? Why did you compute the gradient? Feb 15 at 21:53\n\u2022 @Rhino Please consult en.wikipedia.org/wiki/\u2026. Feb 15 at 22:00", "date": "2021-10-22 22:49:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4027037/how-do-i-compute-this-integral-with-a-diracs-delta", "openwebmath_score": 0.9981832504272461, "openwebmath_perplexity": 482.7863007217013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013790827493, "lm_q2_score": 0.8887587853897073, "lm_q1q2_score": 0.8727733839717847}}
{"url": "http://menteraroute.nl/reviews/absolute-value-graph-5f6a25", "text": "graph to the left by three. With this mixed transformation, we need to perform the inner absolute value first: For any original negative x \u2019s, replace the y value with the y value corresponding to the positive value (absolute value) of the negative x \u2019s. '&https=1' : ''); If you're shifting in And we should expect to need to plot negative x-values, too. that I'm shifting to the right but I encourage you to try numbers and think about what's happening here. In particular, they don't include any \"minus\" inputs, so it's easy to forget that those absolute-value bars mean something. x plus three, plus two. Because of this, absolute-value functions have graphs which make sharp turns where the graph would otherwise have crossed the x-axis. value of something and so you say, okay, if x is three, how do I make that equal to zero? y = \u2223 x \u2223. Absolute values are never negative. It's gonna look something like this. for the black function, I'm gonna have to get two more than that. $\\begin{cases}f\\left(x\\right)=2\\left|x - 3\\right|-2,\\hfill & \\text{treating the stretch as a vertical stretch, or}\\hfill \\\\ f\\left(x\\right)=\\left|2\\left(x - 3\\right)\\right|-2,\\hfill & \\text{treating the stretch as a horizontal compression}.\\hfill \\end{cases}$, $f\\left(x\\right)=a|x - 3|-2$, $\\begin{cases}2=a|1 - 3|-2\\hfill \\\\ 4=2a\\hfill \\\\ a=2\\hfill \\end{cases}$, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175. All right reserved. Web Design by. of this graph if we shift, if we shift three to the right and then think about how that will change if not only do we shift three to the right but we also shift four up, shift four up, and so once again pause this At this vertex right over here, whatever was in the absolute So in particular, we're And then instead of having going up here. of x plus three, plus two. You remember that absolute-value graphs involve absolute values, and that absolute values affect \"minus\" inputs. This is the graph of y If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for $x$ and $f\\left(x\\right)$. The general form of the absolute value function is: f (x) = a|x-h|+k. 'https:' : 'http:') + '//contextual.media.net/nmedianet.js?cid=8CU2W7CG1' + (isSSL ? You could have shifted up two first, then you could have Yes. But mostly we need to take the time to plot quite a few points, so that we can \"see\" the shape before we start sketching it in. Well, one way to think about it is, well, something interesting is happening right over here at x equals three. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. value of x plus three. thing was happening at x equals zero. Instead, the width is equal to 1 times the vertical distance. var mnSrc = (isSSL ? medianet_versionId = \"111299\"; And we're gonna do that right now and then we're gonna just gonna If k<0, it's also reflected (or \"flipped\") across the x-axis. For instance, suppose we are given the equation y = | \u2026 And then, so here instead There are any number of things we can do to help ourselves graph this correctly. Now, it's happening at x equals three. Graph an absolute value function. signs to positive signs. The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. to absolute value of x which you might be familiar with. the right and four up? Alright, now let's do this together. Now let's see if we can graph y is equal to two times the So there's multiple, there's three transformations You could view this as the same thing as y is equal to the absolute\n\n.\n\nCaramelized White Chocolate Cake, Eu4 Political Map, Does Black Carbon Steel Rust, Kieran Name Meaning, Couples Therapy Assessment Tools, Bioinorganic Chemistry Journal,", "date": "2021-02-26 07:40:28", "meta": {"domain": "menteraroute.nl", "url": "http://menteraroute.nl/reviews/absolute-value-graph-5f6a25", "openwebmath_score": 0.674605131149292, "openwebmath_perplexity": 576.4915802659639, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9706877675527114, "lm_q2_score": 0.8991213718636754, "lm_q1q2_score": 0.8727661172132822}}
{"url": "http://math.stackexchange.com/questions/300982/differs-in-only-finitely-many-terms-an-equivalence-relation-on-sequences", "text": "# \u201cDiffers in only finitely many terms\u201d an equivalence relation on sequences?\n\nConsider sequences $a : \\mathbb{Z}^+ \\rightarrow A$ on a set $A$. Define the relation $\\sim$ over sequences by $a \\sim b$ iff there are only finitely many indices $i$ at which $a_i \\neq b_i$. Clearly $\\sim$ is reflexive and symmetric.\n\nIt appears to me that it is also transitive! Suppose $a \\sim b$, and $b \\sim c$. Let $I$ be the set of indices $i$ such that $a_i \\neq b_i$. Let $J$ be the set of indices at which $b_i \\neq c_i$. Both $I$ and $J$ are finite, so $K = I \\cup J$ is finite. Let $i \\notin K$. Then $a_i = b_i$, and $b_i = c_i$, so $a_i = c_i$. So there are only finitely many points at which $a$ and $c$ differ (all of them are in $K$), so $a \\sim c$.\n\nThis is extremely surprising to me, because it implies that $\\sim$ is an equivalence relation. My difficulty is in understanding what equivalence classes this relation could possibly define.\n\nQuestion 1: Is it really true that $\\sim$ is an equivalence relation?\n\nQuestion 2: Can anybody give me some kind of concrete description of what equivalence classes $\\sim$ defines?\n\nThank you!\n\n-\n\nYes, it\u2019s an equivalence relation. Another way to describe is to say that $a\\sim b$ iff there is an $n\\in\\Bbb N$ such that $a_k=b_k$ for all $k\\ge n$. About the only way that I\u2019ve ever found to visualize an equivalence class is to pick one member of it; if that member is $a$, then the class consists precisely of all sequences that agree with $a$ from some point on.\n\nIt turns out to be very important in studying the box topology on products of infinitely many topological spaces.\n\n-\nThat alternative definition helps a lot, Brian! Thanks for the answer! \u2013\u00a0 Nick Thomas Feb 12 '13 at 7:05\n@Nick: You\u2019re welcome! \u2013\u00a0 Brian M. Scott Feb 12 '13 at 7:06\nDo you think the weak direct sum could visualize the OP this relation? thanks. \u2013\u00a0 Babak S. Feb 12 '13 at 7:12\n@Babak: Maybe: its elements are exactly one equivalence class in the full product. \u2013\u00a0 Brian M. Scott Feb 12 '13 at 7:14\n\nThis is a very small hint in the light of Brian's complete one. I think the weak direct sum $\\sum A_k$ when, for example, $\\{A_k\\}$ is a family of groups indexed by a set $K$, can help you to find out what is happening in the relation. Try to Google it for details.\n\n-\nNice hint...+ 1 ;-) \u2013\u00a0 amWhy Feb 12 '13 at 15:55\n\nAnd as to what that equivalence class might be/look like?\n\nWell, that depends on the nature of set A.\n\nIf A={0,1} then your function can be interpreted as producing a binary Real number corresponding to each function a, such as .01110011.. or .1101110... etc. In this case, the equivalence classes are Real numbers which vary by a finite sum of negative powers of two. So x and x + 1/4 + 1/32 would be in the same equivalence class. A different set A and a different interpretation of the sequence as a Real might produce (for example) equivalence classes of Reals which differ by a rational, such that x, x + 1/7, x + 23/788 etc.\n\n-\n\nIt is an equivalence relation. There's nice Brian answer, but it is short on examples, so here is something to gain more intuition:\n\n\u2022 Try to imagine what is the class of abstraction of sequence of zeros $(\\ldots, 0, 0, 0, \\ldots)$.\n\n\u2022 Another example would be a convergent sum $\\sum_{i \\in \\mathbb{Z}} a_i < \\infty$, what can you tell about other sequences in its class of abstraction?\n\n-", "date": "2015-07-30 00:26:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/300982/differs-in-only-finitely-many-terms-an-equivalence-relation-on-sequences", "openwebmath_score": 0.8940344452857971, "openwebmath_perplexity": 226.17040700002053, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130593124399, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8727326761797357}}
{"url": "https://math.stackexchange.com/questions/2839006/question-about-derivation", "text": "Problem An airplane is flying at a constant speed at a constant altitude of $10$ km in a straight line directly over an observer. At a given moment the observer notes that the angle of elevation $\\theta$ to the plane is $54^\\circ$ and is increasing at $1^\\circ$ per second. find the speed, in kilometres per hour, at which the airplane is moving towards the observer.\n\nI'm working on a related rates problem, and the equation that i'm using to relate the two variables is $$\\tan \\theta= \\frac{10}{x}.$$ so I can differentiate this, and after simplifying, I end up with $$\\frac{dx}{dt}=-\\frac{1}{10} \\cdot x^2\\sec^2 \\theta\\frac{d\\theta}{dt}.$$ But if I rearrange the original equation so that it's $x=\\frac{10}{\\tan \\theta}$, I get $$\\frac{dx}{dt}=(-10\\csc^2 \\theta) \\frac{d\\theta}{dt},$$ which is different from the other derivative. Is there something quirky about rearranging the equation, or am I just blindly messing something up?\n\n\u2022 Where does the variable $t$ come into play? Is $x$ a function of $t$? Please post the problem statement (the exercise you are working on). You're missing something, or your differentiation makes no sense. Jul 2 '18 at 22:42\n\u2022 Welcome to MSE! Could you provide a little more context, like the original statement of the problem? Also, your question will be much easier to read (and therefore much more likely to get answered!) if you use MathJax formatting to format the math in the question: math.meta.stackexchange.com/questions/5020/\u2026. (Kind of a long tutorial, but the basics are close to the top, and the whole thing is well worth reading if you have the time.) Jul 2 '18 at 22:44\n\u2022 ok yeah sorry my bad, the problem goes like this: An airplane is flying at a constant speed at a constant altitude of 10 km in a straight line directly over an observer. At a given moment the observer notes that the angle of elevation \u03b8 to the plane is 54\u00b0 and is increasing at 1\u00b0 per second. find the speed, in kilometres per hour, at which the airplane is moving towards the observer. Jul 2 '18 at 22:53\n\u2022 Hint: if you substitute $x = \\frac{10}{\\tan \\theta}$ into the first derivative, what do you get? Jul 2 '18 at 23:07\n\u2022 omgggg i can't believe i didn't see that before, thank u so much! Jul 2 '18 at 23:21\n\nThough the two derivatives may seem different from one another, they are actually the same! This is because $$x^2=\\left(\\frac{10}{\\tan\\theta}\\right)^2=\\frac{100}{\\tan^2\\theta}=100\\frac{\\cos^2\\theta}{\\sin^2\\theta}=100\\frac{\\csc^2\\theta}{\\sec^2\\theta}.$$\n\u2022 By the way, welcome to Math.SE! After you ask a question here, if you get an acceptable answer, you should \"accept\" the answer by clicking the check mark $\\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. Jul 2 '18 at 23:55", "date": "2021-09-22 21:23:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2839006/question-about-derivation", "openwebmath_score": 0.7660377621650696, "openwebmath_perplexity": 237.17408345424997, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130596362787, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8727326749365062}}
{"url": "http://sgix.yiey.pw/calculation-of-taylor-series.html", "text": "# Calculation Of Taylor Series\n\nDerivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. Then, for every x in the interval, where R n(x) is the remainder (or error). We prove a general Steady-state theorem for Volterra series operators, and then establish a general formula for the spectrum of the output of a Volterra series operator in terms of the spectrum of a periodic input. Actually, this is now much easier, as we can use Mapleor Mathematica. Annette Pilkington Lecture 33 Applications of Taylor Series. A minimum of 1 double value to hold the total and 2 integer values, 1 to hold the current term number and 1 to hold the number of terms to evaluate. Solve for g(pi/3) using 5, 10, 20 and 100 terms in the Taylor series (use a loop) So I tried the following in the script editor:. Program to Calculate the Exponential Series in C | C Program Posted by Tanmay Jhawar at 9:12 PM - 9 comments Here's a C program to calculate the exponential series using For loop with output. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. Part 2) After completing part 1, modify the series for faster convergence. Other articles where Taylor Standard Series Method is discussed: David Watson Taylor: \u2026known since 1910 as the Taylor Standard Series Method, he determined the actual effect of changing those characteristics, making it possible to estimate in advance the resistance of a ship of given proportions. In mathematics, the Laurent series of a complex function f(z) is a representation of that function as a power series which includes terms of negative degree. Series can expand about the point x = \u221e. Series can construct standard Taylor series, as well as certain expansions involving negative powers, fractional powers, and logarithms. Sign up to read all wikis and quizzes in math, science, and engineering topics. The Taylor series is a power series that approximates the function f near x = a. Concrete examples in the physical science division and various engineering \ufb01elds are used to paint the applications. TAYLOR AND MACLAURIN SERIES 102 4. That is, every reasonable function can be written as This module describes how to compute the coefficients for a given function. Binomial Theorem Calculator Binomial Theorem Calculator This calculators lets you calculate __expansion__ (also: series) of a binomial. Math 142 Taylor/Maclaurin Polynomials and Series Prof. And by knowing these basic rules and formulas, we can learn to use them in generating other functions as well as how to apply them to Taylor Series that are not centered at zero. The first is the power series expansion and its two important generalizations, the Laurent series and the Puiseux series. The taylor series expansion of f(x) with respect to xo is given by: Generalization to multivariable function: (5) Using similar method as described above, using partial derivatives this time, (Note: the procedure above does not guarantee that the infinite series converges. Miyagawa 4-12-11-628 Nishiogu, Arakawa-ku, Tokyo 116-0011, Japan. Definition: The Taylor series is a representation or approximation of a function as a sum. Taylor\u2019s Series. The Taylor Series is also referred to as Maclaurin (Power) Series. is determined by using multivariate Taylor series expansion. The power series converges globally to the function. Sum of Series Programs / Examples in C programming language. for any x in the series' interval of convergence. It may be used to express complex functions in cases where a Taylor series expansion cannot be applied. Maclaurin's formula or Maclaurin's theorem: The formula obtained from Taylor's formula by setting x 0 = 0 that holds in an open neighborhood of the origin, is called Maclaurin's formula or Maclaurin's theorem. The Taylor series for ex based at b = 0is ex = X\u221e n=0 xn n! so we have e3x = X\u221e n=0 (3x)n n! and x2e3x = X\u221e n=0 3nxn+2 n! =. OBTAINING TAYLOR FORMULAS Most Taylor polynomials have been bound by other than using the formula pn(x)=f(a)+(x\u2212a)f0(a)+ 1 2! (x\u2212a)2f00(a) +\u00b7\u00b7\u00b7+ 1 n! (x\u2212a)nf(n)(a) because of the di\ufb03culty of obtaining the derivatives f(k)(x) for larger values of k. Here are a few examples. Suppose we wish to find the Taylor series of sin( x ) at x = c , where c is any real number that is not zero. Computing Taylor Series Lecture Notes As we have seen, many different functions can be expressed as power series. The Taylor Series Calculator an online tool which shows Taylor Series for the given input. Derivative calculator Integral calculator Definite integrator Limit calculator Series calculator Equation solver Expression simplifier Factoring calculator Expression calculator Inverse function Taylor series Matrix calculator Matrix arithmetic Graphing calculator. A minimum of 1 double value to hold the total and 2 integer values, 1 to hold the current term number and 1 to hold the number of terms to evaluate. Series Converges Series Diverges Diverges Series r Series may converge OR diverge-r x x x0 x +r 0 at |x-x |= 0 0 Figure 1: Radius of. Let G = g(R;S) = R=S. To estimate the square root of a number using only simple arithmetic, the first-order Taylor series of the square root function provides a convenient method. Taylor_series_expansion online. Other Power Series Representing Functions as Power Series Functions as Power Series Derivatives and Integrals of Power Series Applications and Examples Taylor and Maclaurin Series The Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor. As such, he found that by calculating the time needed for the various elements of a task, he could develop the \"best\" way to complete that task. Part 2) After completing part 1, modify the series for faster convergence. For this I need to calculate the taylor series expansion of the function. the series for , , and ), and/ B BB sin cos. , x 0 2I : Next consider a function, whose domain is I,. The Taylor package was written to provide REDUCE with some of the facilities that MACSYMA's TAYLOR function o\ufb00ers, but most of all I needed it to be faster and more space-e\ufb03cient. Taylor Series. So I want a Taylor polynomial centered around there. Taylor\u2019s series is an essential theoretical tool in computational science and approximation. In this video I'm going to show you how you can find a Taylor series. (The formula used is shown on page 100 of the text. It is nothing but the representation of a function as an infinite sum of terms. If only concerned about the neighborhood very close to the origin, the n = 2 n=2 n = 2 approximation represents the sine wave sufficiently, and no. The nonlinear restoring forces are given in R(x,x\u02d9) and fext(t) is a vector of external dynamic loads. The Taylor series obtained when we let c = 0 is referred to a Maclaurin series. This script lets you input (almost) any function, provided that it can be represented using Sympy and output the Taylor series of that function up to the nth term centred at x0. How can I get a Taylor expansion of the Sin[x] function by the power series? Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. these developments was the recursive calculation of the coe\ufb03cients of the Taylor series. TAYLOR and MACLAURIN SERIES TAYLOR SERIES Recall our discussion of the power series, the power series will converge absolutely for every value of x in the interval of convergence. About the calculator: This super useful calculator is a product of wolfram alpha, one of. via the usual Taylor series, we get the same result as above without using Taylor\u2019s mul-tivariable formula. Part 1) Given a list of basic taylor series, find a way to approximate the value of pi. I'm currently in an introductory course of MATLAB and one of my assignments is to calculate the Taylor series of a given formula, without using the available taylor(f,x) function. 1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. Browse other questions tagged power-series functional-equations taylor-series or ask your own question. This calculator turns your data into a Mathematical formula by generating a Fourier Series of sines and cosines. This is clearly not the case. Compute the interval of convergence for each series on the previous page. Several methods exist for the calculation of Taylor series of a large number of functions. Solution: Since 1 1 \u2212\ud835\udc65 = \ud835\udc65\ud835\udc58 \u221e \ud835\udc58=0 we get \ud835\udc652 1 \u2212\ud835\udc65 = \ud835\udc652 \ud835\udc65\ud835\udc58 \u221e \ud835\udc58=0 = \ud835\udc65\ud835\udc58+2 \u221e \ud835\udc58=0 J. Euler's Method, Taylor Series Method, Runge Kutta Methods, Multi-Step Methods and Stability. taylor_series is a univariate Taylor series. These are called the Taylor coefficients of f, and the resulting power series is called the Taylor series of the function f. 2 Calculating a Maclaurin series Use Maxima to calculate the terms in the Maclaurin series up to and including x7 for the following functions; (1) g(x) = arctanx (2) G(x) = arctanhx (the inverse of the hyperbolic function tanh) What do you notice about the terms and signs in the series (1) and (2) here?. Calculate g(x) = sin(x) using the Taylor series expansion for a given value of x. Taylor Series \u2022 The Taylor Theorem from calculus says that the value of a function can be approximated near a given point using its \u201cTaylor series\u201d around that point. Expansions of e. May 20, 2015 firstly we look. x, sin x, and cos x, and related series. See Examples. the Taylor expansion of 1 1\u2212x) \u2022 the Taylor expansions of the functions ex,sinx,cosx,ln(1 + x) and range of va-lidity. I have to do it using taylor series using iterations, but only for first 13 nominators&;denominators. The Taylor expansion of a function at a point is a polynomial approximation of the function near that point. 1 Introduction This section focuses on deriving a Maclaurin series for functions of the form f(x) = (1 + x)k for any number k. n=0 for some constant C depending on the choice of antiderivative of f. This series \u2014 known as a \"power series\" \u2014 can be written in closed. 1 Introduction This chapter has several important and challenging goals. It is nothing but the representation of a function as an infinite sum of terms. A term that is often heard is that of a \u201cTaylor expansion\u201d; depending on the circumstance, this may mean either the Taylor series or the n th degree Taylor polynomial. Taylor's formula and Taylor series can be defined for functions of more than one variable in a similar way. Change the function definition 2. A consequence of this is that a Laurent series may be used in cases where a Taylor. Say you are asked to find the Taylor Series centered at a=0 up to degree n=3 (really a MacLaurin series as the center is at 0 ) So plug into Calculus Made Easy option 7 D as follows : The derivatives are taken in order to compute the coefficients for each term up to degree 3. Find the interval of convergence for \u221e n=0 (x\u22123)n n. We know that the th Taylor polynomial is , and we have spent a lot of time in this chapter calculating Taylor polynomials and Taylor Series. In other words, in this particular instance, from the fact that the series 1+ 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + converges, one is likely to erroneously infer thatall in nite series converge. In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. Calculate g(x) = sin(x) using the Taylor series expansion for a given value of x. program for calculation of the sine of an angle using the sine series to radians because the Taylor series method for calculating sine uses radians and not. REVIEW: We start with the di\ufb00erential equation dy(t) dt = f (t,y(t)) (1. For sin function. For this reason, we often call the Taylor sum the Taylor approximation of degree n. Binomial Theorem A-Level Mathematics revision section of Revision Maths looking at Binomial Theorem and Pascals Triangle. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. 7:34 AM - 14 May 2018. As an example, let\u2019s use the Maclaurin polynomial (with just four terms in the series) for the function f(x) = sin(x) to approximate sin(0. These notes discuss three important applications of Taylor series: 1. It explains how to derive power series of composite functions. Since this series expansion for tan-1x allows us to reach higher accuracy much faster, it is the more efficient of the two series. Solve for g(pi/3) using 5, 10, 20 and 100 terms in the Taylor series (use a loop) So I tried the following in the script editor:. Calculation of Taylor series Several methods exist for the calculation of Taylor series of a large number of functions. And by knowing these basic rules and formulas, we can learn to use them in generating other functions as well as how to apply them to Taylor Series that are not centered at zero. The proofs are complete, and use only the basic facts of analysis. Convergence of Taylor Series SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. List of Maclaurin Series of Some Common Functions / Stevens Institute of Technology / MA 123: Calculus IIA / List of Maclaurin Series of Some Common Functions / 9 | Sequences and Series. What is the Taylor series representation of f(x + delta(x)) and how is it arrived at? I am trying to understand the derivation of continuous compounding rate of interest calculation. Part 1) Given a list of basic taylor series, find a way to approximate the value of pi. Cook great food for your customers. Find the Taylor expansion series of any function and see how it's done! Up to ten Taylor-polynomials can be calculated at a time. TI-89 - Vol 2 - Sect 08 - Calculator Taylor and Maclaurin Polynomials. Of course, the polynomial function will not have the same shape for all values of \"x\". For instance, in Example 4 in Section 9. These notes discuss three important applications of Taylor series: 1. For this I need to calculate the taylor series expansion of the function. Taylor Series Convergence (1/n!) f (n) (c) (x - c) n = f(x) if and only if lim (n-->) R n = 0 for all x in I. This variable is first initialized to 0. Rapid and Accurate Calculation of Water and Steam Properties Using the Tabular Taylor Series Expansion Method K. Modern numerical algorithms for the solution of ordinary di\ufb00erential equations are also based on the method of the Taylor series. This smart calculator is provided by wolfram alpha. 0 Calculator question. It is nothing but the representation of a function as an infinite sum of terms. Added Nov 4, 2011 by sceadwe in Mathematics. We first write the terms of the series from n = 0 to n = 3. If you are interested in seeing how that works you. A Taylor series isn't really a good way to compute this function unless you're looking for asymptotic accuracy around a particular point, rather than general accuracy along the whole thing. Sequences and series \u2022 A sequence is a (possibly infinite) collection of numbers lined up in some order \u2022 A series is a (possibly infinite) sum \u2013 Example: Taylor\u2019s series k \u00a6 2 n o o n o o o o o n k o o k n f x n fx k T c ( ) 1 ( )! 1 2 1! 1 Example: sine function approximated by Taylor series expansion The approximation of f(x)=sin. If you are interested in seeing how that works you. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. The Delta Method gives a technique for doing this and is based on using a Taylor series approxi-mation. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. While you can calculate Maclaurin series using calculus, many series for common functions have already been found. Statistical moments of X , such as the variance, are then computed in terms of the Taylor coefficients and the moments of x [ 3 , ]. Ken Bube of the University of Washington Department of Mathematics in the Spring, 2005. Specifically, the Taylor series of an infinitely differentiable real function f, defined on an open interval (a \u2212 r, a + r), is :. Example: The Taylor Series for e x e x = 1 + x + x 2 2! + x 3 3! + x 4 4! + x 5 5! +. As you have noticed, the Taylor series has infinite terms. Taylor\u2019s Series of sin x In order to use Taylor\u2019s formula to \ufb01nd the power series expansion of sin x we have to compute the derivatives of sin(x):. The more terms you use, however, the better your approximation will be. It seems that, instead of Taylor series for int_0^x, it calculates Taylor series for int_x^{2x} I have only found this bug when calculating Taylor series for functions f defined as in the example (by means of an integral). MacLaurin series of Exponential function, The MacLaulin series (Taylor series at ) representation of a function is The derivatives of the exponential function and their values at are: Note that the derivative of is also and. sin x=x-1/6x^3 +1/120x^5 -1/5040x^7 The calculator substitutes into as many terms of the polynomial that it needs to in order to get the required number of decimal places. About the calculator: This super useful calculator is a product of wolfram alpha, one of. Actually, this is now much easier, as we can use Mapleor Mathematica. Program to Calculate the Exponential Series in C | C Program Posted by Tanmay Jhawar at 9:12 PM - 9 comments Here's a C program to calculate the exponential series using For loop with output. Solve for g(pi/3) using 5, 10, 20 and 100 terms in the Taylor series (use a loop) So I tried the following in the script editor:. MacLaurin series of Exponential function, The MacLaulin series (Taylor series at ) representation of a function is The derivatives of the exponential function and their values at are: Note that the derivative of is also and. The result is in its most. Since this series expansion for tan-1x allows us to reach higher accuracy much faster, it is the more efficient of the two series. , x 0 2I : Next consider a function, whose domain is I,. Part 1) Given a list of basic taylor series, find a way to approximate the value of pi. As any calculus student knows, the first-order Taylor expansion around x 2 is given by sqrt(x 2 + a) ~ x + a / 2x. A Maclaurin series is a special case of a Taylor series, where \u201ca\u201d is centered around x = 0. The Taylor Series Calculator an online tool which shows Taylor Series for the given input. A summary of The Remainder Term in 's The Taylor Series. Every Maclaurin series, including those studied in Lesson 22. Taylor Series SingleVariable and Multi-Variable \u2022 Single variable Taylor series: Let f be an in\ufb01nitely di\ufb00erentiable function in some open interval around x= a. Especially I wanted procedures that would return the logarithm or arc tangent of a Taylor series, again as a Taylor series. The crudest approximation was just a constant. The first is to calculate any random element in the sequence (which mathematicians like to call the \"nth\" element), and the second is to find the sum of the geometric sequence up to the nth element. See Examples. sin(x) of java. The answer lies in estimating the function f(x) by its Taylor expansion. When p = 1, the p-series is the harmonic series, which diverges. Textbook solution for Engineering Fundamentals: An Introduction to\u2026 5th Edition Saeed Moaveni Chapter 18 Problem 44P. , the difference between the highest and lowest power in the expansion is 4. The use of duration, in the second term of the Taylor series, to determine the change in the instrument value is only an approximation. While you can calculate Maclaurin series using calculus, many series for common functions have already been found. 2 The variance of g ( y n) is then approximated. Thenlet x= 1 in the earlier formulas to get Most Taylor polynomials have been bound by other than using the formula pn. * The more simple the expression, the better range of accuracy with less terms. The question is, for a specific value of , how badly does a Taylor polynomial represent its function?. We want a power series in factors of (x-a), where we can easily obtain the coefficients of (x-a)^k. Simple components for Ada The Simple components for Ada library provides implementations of smart pointers for automatically c taylor series calculator free download - SourceForge. The Maclaurin series of sin(x) is only the Taylor series of sin(x) at x = 0. Find the Taylor series expansion of any function around a point using this online calculator. TAYLOR AND MACLAURIN SERIES 102 4. These \"time and motion\" studies also led Taylor to conclude that certain people could work more efficiently than others. Each term of the Taylor polynomial comes from the function's derivatives at a single point. For the square roots of a negative or complex number, see below. Specifically, the Taylor series of an infinitely differentiable real function f, defined on an open interval (a \u2212 r, a + r), is :. Taylor Series. This non-linear mapping (on inhomogeneous coordinates) can be expanded in a Taylor series. Things you should memorize: \u2022 the formula of the Taylor series of a given function f(x) \u2022 geometric series (i. Requires a Wolfram Notebook System. This is the program to calculate the value of 'e', the base of natural logarithms without using \"math. Sum of Taylor Series Program. It is a series expansion around a point. Math 142 Taylor/Maclaurin Polynomials and Series Prof. 1 What is a Laurent series? The Laurent series is a representation of a complex function f(z) as a series. Questions: 1. An Easy Way to Remember the Taylor Series Expansion which is technically known as a Maclaurin rather than a Taylor). C / C++ Forums on Bytes. Then, for every x in the interval, where R n(x) is the remainder (or error). These risk statistics are also known as greeks. If the Taylor series is centered at zero, then that series is also called a Maclaurin series, after the Scottish mathematician Colin Maclaurin, who made extensive. Taylor Series Cos x Calculator. This is an infini te sum, but your function should stop the summation when the addition of a successive term makes a negligible change in e. Miyagawa 4-12-11-628 Nishiogu, Arakawa-ku, Tokyo 116-0011, Japan. TI-89 - Vol 2 - Sect 08 - Calculator Taylor and Maclaurin Polynomials. I'm trying to write a program to find values for arctan of x by using taylor series. Cascade IP3 calculation formula for 3 or four stages. Things you should memorize: \u2022 the formula of the Taylor series of a given function f(x) \u2022 geometric series (i. ADVERTISEMENTS: Read this article to learn how to Calculate Standard Deviation in 3 different Series! A. A Quick Note on Calculating the Radius of Convergence The radius of convergence is a number \u02c6such that the series X1 n=0 a n(x x 0)n converges absolutely for jx x 0j<\u02c6, and diverges for jx x 0j>0 (see Fig. Once you get the basic idea it will be very easy to do. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. List of Maclaurin Series of Some Common Functions / Stevens Institute of Technology / MA 123: Calculus IIA / List of Maclaurin Series of Some Common Functions / 9 | Sequences and Series. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. After all, you are finding the nth derivative of f, when all you need is the nth derivative of f , evaluated at a. C++ templates for sin, cos, tan taylor series. This appendix derives the Taylor series approximation informally, then introduces the remainder term and a formal statement of Taylor's theorem. Featured on Meta Congratulations to our 29 oldest beta sites - They're now no longer beta!. Questions: 1. It is nothing but the representation of a function as an infinite sum of terms. Taylor Series Calculator. Ken Bube of the University of Washington Department of Mathematics in the Spring, 2005. Program to Calculate the Exponential Series in C | C Program Posted by Tanmay Jhawar at 9:12 PM - 9 comments Here's a C program to calculate the exponential series using For loop with output. Example: Calculation of Pi to 707-digit accuracy (like William Shanks):. See how it's done with this free video algebra lesson. Please see Jenson and. Sequences and series \u2022 A sequence is a (possibly infinite) collection of numbers lined up in some order \u2022 A series is a (possibly infinite) sum \u2013 Example: Taylor\u2019s series k \u00a6 2 n o o n o o o o o n k o o k n f x n fx k T c ( ) 1 ( )! 1 2 1! 1 Example: sine function approximated by Taylor series expansion The approximation of f(x)=sin. Find f11(0). Gas Turbines Power (July, 2001) Supplementary Backward Equations for Pressure as a Function of Enthalpy and Entropy p(h,s) to the Industrial Formulation IAPWS-IF97 for Water and Steam. The Taylor Series is also referred to as Maclaurin (Power) Series. The free tool below will allow you to calculate the summation of an expression. The more terms you use, however, the better your approximation will be. For any f(x;y), the bivariate \ufb01rst order Taylor expansion about any = (. 1) and its associated formula, the Taylor series, is of great value in the study of numerical methods. Then combine the terms with the same exponent. What makes these important is that they can often be used in place of other, more complicated functions. Since this series expansion for tan-1x allows us to reach higher accuracy much faster, it is the more efficient of the two series. Using a popular tool of monetary policy analysis, our Taylor Rule Calculator lets you estimate where short-term interest rates should move as economic conditions change. A Maclaurin series is a specific type of Taylor series that's evaluated at x o = 0. An example of a Taylor Series that approximates $e^x$ is below. Taylor Series Calculator with Steps Taylor Series, Laurent Series, Maclaurin Series. (c) Check that the Taylor series for ex, sinhxand coshxsatisfy the same equation. Maclaurin Taylor Series for Transcendental Functions: A Graphing-Calculator View of Convergence Marvin Stick Most calculus students can perform the manipulation necessary for a polynomial approximation of a transcendental function. Find approximations for EGand Var(G) using Taylor expansions of g(). So I want a Taylor polynomial centered around there. Glenn Research Center, Cleveland, Ohio More efficient versions of an interpo-lation method, called kriging, have been introduced in order to reduce its tradi-. Find the Taylor series for f(x) centered at the given value of 'a'. Calculate g(x) = sin(x) using the Taylor series expansion for a given value of x. The Taylor package was written to provide REDUCE with some of the facilities that MACSYMA's TAYLOR function o\ufb00ers, but most of all I needed it to be faster and more space-e\ufb03cient. Concrete examples in the physical science division and various engineering \ufb01elds are used to paint the applications. The duplex method follows the Vedic ideal for an algorithm, one-line, mental calculation. It is a series expansion around a point. 1730's) with using the series expansion of the arcsine function,,. Wyrick family find themselves Cialis 20 Mg Paypal to spot fakes per GiB of RAM I hope to hear adjustment and metal ball. Belbas Mathematics Department University of Alabama Tuscaloosa, AL. What is the difference between Power series and Taylor series? 1. Therefore, ex= X1 k=0 f(k)(0) k! xk= X1 k=0 xk k! as we already know. e-mail: [email\u00a0protected] Evaluate all these at. Taking the first two terms of the series gives a very good approximation for low speeds. The SURVEYLOGISTIC Procedure. Due to the increasing quartic moveout term (which is negative for the example in Figure 4. java from \u00a79. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. We want a power series in factors of (x-a), where we can easily obtain the coefficients of (x-a)^k. Simple Calculator to find the trigonometric cos x function using cosine taylor series formula. Drek intends to pollute into my fifties my irony is sarcastic and to thin more and. Calculating the sine of a number from the Taylor sum I'm trying to write code to basically mimic the sum to infinity expression, Sine - Wikipedia, the free encyclopedia , but with a finite number of terms, and have the user input the number of which the sine is to be calculated. Taylor and Maclaurin Series If a function $$f\\left( x \\right)$$ has continuous derivatives up to $$\\left( {n + 1} \\right)$$th order, then this function can be expanded in the following way:. On the other hand, this shows that you can regard a Taylor expansion as an extension of the Mean Value Theorem. 1 Introduction This chapter has several important and challenging goals. With modern calculators and computing software it may not appear necessary to use linear approximations. , the Riemann zeta function evaluated at p. Maclaurin Series Calculator. The taylor command computes the order n Taylor series expansion of expression, with respect to the variable x, about the point a. The duplex method follows the Vedic ideal for an algorithm, one-line, mental calculation. Constructing a Taylor Series [ edit ] Several methods exist for the calculation of Taylor series of a large number of functions. Each term of the Taylor polynomial comes from the function's derivatives at a single point. They suggest using T(x) = 1+ax^2+bx^4+cx^6 and (cosx)(secx)=1. While you can calculate Maclaurin series using calculus, many series for common functions have already been found. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. (Assume that 'f' has a power series expansion. 1) y(0) = y0 This equation can be nonlinear, or even a system of nonlinear equations (in which case y is a vector and f is a vector of n di\ufb00erent functions). 2, is a Taylor series centered at zero. ADVERTISEMENTS: Read this article to learn how to Calculate Standard Deviation in 3 different Series! A. Taylor and Laurent Series We think in generalities, but we live in details. TAYLOR AND MACLAURIN SERIES 3 Note that cos(x) is an even function in the sense that cos( x) = cos(x) and this is re ected in its power series expansion that involves only even powers of x. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. 10 Taylor and Maclaurin Series 677 If you know the pattern for the coefficients of the Taylor polynomials for a function, you can extend the pattern easily to form the corresponding Taylor series. When a = 0, Taylor\u2019s Series reduces, as a special case, to Maclaurin\u2019s Series. F(t0 + \u2206t) \u2248 F(t0) +F\u2032(t0)\u2206t. To find the value of sin 1 (in radians), a calculator will use the Maclaurin Series expansion for sin x, that we found earlier. i+1 = (i+ 1)h, we may solve for the accelerations in terms of the displacements, velocities, and the applied forces. How can I get a Taylor expansion of the Sin[x] function by the power series? Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. While all the previous writeups have concentrated on the fact that you can calculate a Taylor series by calculating the appropriate derivatives, this is often not the best solution, especially if you just need a few terms. Example 1 Taylor Polynomial Expand f(x) = 1 1\u2013x \u2013 1 around a = 0, to get linear, quadratic and cubic approximations. Arribas, A. Rapid Calculation of Spacecraft Trajectories Using Efficient Taylor Series Integration Software greatly accelerates the calculation of spacecraft trajectories. Use Taylor's formula to estimated the accuracy En(x) of the approximation Tn(x) to f(x) when for -4 < x < 4 and n=4. factorial(i) sign = -sign return cosx x. Expansions of e. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Chapter 4: Taylor Series 19 1 + x + x2 2! + x3 3! +\u00b7\u00b7\u00b7 xn n! +\u00b7\u00b7\u00b7= \u221e i=0 xi i! As another example we calculate the Taylor Series of 1 x. Once you get the basic idea it will be very easy to do. e-mail: [email\u00a0protected] We find the desired polynomial approximation using the Taylor Series. In this section we present numerous examples that provide a number of useful procedures to find new Taylor series from Taylor series that we already know. Solution We will be using the formula for the nth Taylor sum with a = 0. A simple example is if we scale a function, say g(t) = 5f(t), the the Fourier series for g(t) is 5 times the Fourier series of f(t). It seems that, instead of Taylor series for int_0^x, it calculates Taylor series for int_x^{2x} I have only found this bug when calculating Taylor series for functions f defined as in the example (by means of an integral). For example, the derivative f (x) =lim h\u21920 f (x +h)\u2212f (x) h is the limit of the difference quotient where both the numerator and the denominator go to zero. Part 2) After completing part 1, modify the series for faster convergence. sin(x) of java. The Taylor Series is also referred to as Maclaurin (Power) Series. How to extract derivative values from Taylor series Since the Taylor series of f based at x = b is X\u221e n=0 f(n)(b) n! (x\u2212b)n, we may think of the Taylor series as an encoding of all of the derivatives of f at x = b: that information. Taking the first two terms of the series gives a very good approximation for low speeds. For instance, in Example 4 in Section 9. Annette Pilkington Lecture 33 Applications of Taylor Series. Why don't you code the formula to calculate Taylor series? The more terms you add, the better precision you get I suggest you see these articles about it:. Taylor and Maclaurin Series (27 minutes, SV3 >> 84 MB, H. In this video I'm going to show you how you can find a Taylor series. Solution: Since 1 1 \u2212\ud835\udc65 = \ud835\udc65\ud835\udc58 \u221e \ud835\udc58=0 we get \ud835\udc652 1 \u2212\ud835\udc65 = \ud835\udc652 \ud835\udc65\ud835\udc58 \u221e \ud835\udc58=0 = \ud835\udc65\ud835\udc58+2 \u221e \ud835\udc58=0 J. Rapid Calculation of Spacecraft Trajectories Using Efficient Taylor Series Integration Software greatly accelerates the calculation of spacecraft trajectories. Created a rational approximation method like the Abramowitz & Stegun with 13 terms which I think is good to 10**(-12). As you increase the degree of the Taylor polynomial of a function, the approximation of the function by its Taylor polynomial becomes more and more accurate. Let us start with the formula 1 1\u00a1x = X1 n=0. As an example, let\u2019s use the Maclaurin polynomial (with just four terms in the series) for the function f(x) = sin(x) to approximate sin(0. Enter the x value and find the sin x value in fraction of seconds. This is not always a good value of a to pick. Either the integral test or the Cauchy condensation test shows that the p-series converges for all p > 1 (in which case it is called the over-harmonic series) and diverges for all p \u2264 1. \"Write a program consisting of only the main function, called piApproximator. There is a lot of good information available on line on the theory and applications of using Pad\u00e9 approximants, but I had trouble finding a good example explaining just how to calculate the co-efficients. For example, we know from calculus that es+t = eset when s and t are numbers. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Build your own widget. Maclaurin series are simpler than Taylor's, but Maclaurin's are, by definition, centered at x = 0. The fact-checkers, whose work is more and more important for those who prefer facts over lies, police the line between fact and falsehood on a day-to-day basis, and do a great job. Today, my small contribution is to pass along a very good overview that reflects on one of Trump\u2019s favorite overarching falsehoods. Namely: Trump describes an America in which everything was going down the tubes under\u00a0 Obama, which is why we needed Trump to make America great again. And he claims that this project has come to fruition, with America setting records for prosperity under his leadership and guidance. \u201cObama bad; Trump good\u201d is pretty much his analysis in all areas and measurement of U.S. activity, especially economically. Even if this were true, it would reflect poorly on Trump\u2019s character, but it has the added problem of being false, a big lie made up of many small ones. Personally, I don\u2019t assume that all economic measurements directly reflect the leadership of whoever occupies the Oval Office, nor am I smart enough to figure out what causes what in the economy. But the idea that presidents get the credit or the blame for the economy during their tenure is a political fact of life. Trump, in his adorable, immodest mendacity, not only claims credit for everything good that happens in the economy, but tells people, literally and specifically, that they have to vote for him even if they hate him, because without his guidance, their 401(k) accounts \u201cwill go down the tubes.\u201d That would be offensive even if it were true, but it is utterly false. The stock market has been on a 10-year run of steady gains that began in 2009, the year Barack Obama was inaugurated. But why would anyone care about that? It\u2019s only an unarguable, stubborn fact. Still, speaking of facts, there are so many measurements and indicators of how the economy is doing, that those not committed to an honest investigation can find evidence for whatever they want to believe. Trump and his most committed followers want to believe that everything was terrible under Barack Obama and great under Trump. That\u2019s baloney. Anyone who believes that believes something false. And a series of charts and graphs published Monday in the Washington Post and explained by Economics Correspondent Heather Long provides the data that tells the tale. The details are complicated. Click through to the link above and you\u2019ll learn much. But the overview is pretty simply this: The U.S. economy had a major meltdown in the last year of the George W. Bush presidency. Again, I\u2019m not smart enough to know how much of this was Bush\u2019s \u201cfault.\u201d But he had been in office for six years when the trouble started. So, if it\u2019s ever reasonable to hold a president accountable for the performance of the economy, the timeline is bad for Bush. GDP growth went negative. Job growth fell sharply and then went negative. Median household income shrank. The Dow Jones Industrial Average dropped by more than 5,000 points! U.S. manufacturing output plunged, as did average home values, as did average hourly wages, as did measures of consumer confidence and most other indicators of economic health. (Backup for that is contained in the Post piece I linked to above.) Barack Obama inherited that mess of falling numbers, which continued during his first year in office, 2009, as he put in place policies designed to turn it around. By 2010, Obama\u2019s second year, pretty much all of the negative numbers had turned positive. By the time Obama was up for reelection in 2012, all of them were headed in the right direction, which is certainly among the reasons voters gave him a second term by a solid (not landslide) margin. Basically, all of those good numbers continued throughout the second Obama term. The U.S. GDP, probably the single best measure of how the economy is doing, grew by 2.9 percent in 2015, which was Obama\u2019s seventh year in office and was the best GDP growth number since before the crash of the late Bush years. GDP growth slowed to 1.6 percent in 2016, which may have been among the indicators that supported Trump\u2019s campaign-year argument that everything was going to hell and only he could fix it. During the first year of Trump, GDP growth grew to 2.4 percent, which is decent but not great and anyway, a reasonable person would acknowledge that \u2014 to the degree that economic performance is to the credit or blame of the president \u2014 the performance in the first year of a new president is a mixture of the old and new policies. In Trump\u2019s second year, 2018, the GDP grew 2.9 percent, equaling Obama\u2019s best year, and so far in 2019, the growth rate has fallen to 2.1 percent, a mediocre number and a decline for which Trump presumably accepts no responsibility and blames either Nancy Pelosi, Ilhan Omar or, if he can swing it, Barack Obama. I suppose it\u2019s natural for a president to want to take credit for everything good that happens on his (or someday her) watch, but not the blame for anything bad. Trump is more blatant about this than most. If we judge by his bad but remarkably steady approval ratings (today, according to the average maintained by 538.com, it\u2019s 41.9 approval/ 53.7 disapproval) the pretty-good economy is not winning him new supporters, nor is his constant exaggeration of his accomplishments costing him many old ones). I already offered it above, but the full Washington Post workup of these numbers, and commentary/explanation by economics correspondent Heather Long, are here. On a related matter, if you care about what used to be called fiscal conservatism, which is the belief that federal debt and deficit matter, here\u2019s a New York Times analysis, based on Congressional Budget Office data, suggesting that the annual budget deficit (that\u2019s the amount the government borrows every year reflecting that amount by which federal spending exceeds revenues) which fell steadily during the Obama years, from a peak of $1.4 trillion at the beginning of the Obama administration, to$585 billion in 2016 (Obama\u2019s last year in office), will be back up to $960 billion this fiscal year, and back over$1 trillion in 2020. (Here\u2019s the New York Times piece detailing those numbers.) Trump is currently floating various tax cuts for the rich and the poor that will presumably worsen those projections, if passed. As the Times piece reported:", "date": "2020-01-25 17:57:06", "meta": {"domain": "yiey.pw", "url": "http://sgix.yiey.pw/calculation-of-taylor-series.html", "openwebmath_score": 0.8485816717147827, "openwebmath_perplexity": 580.3439812060044, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9926541742947811, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.872688700728175}}
{"url": "https://brilliant.org/discussions/thread/theorems-on-rationals/", "text": "# Theorems on Rationals\n\nWe present three theorems involving rational numbers.\n\n### $$n$$th Power Theorem. Let $$n$$ be a positive integer. If $$a$$ is a positive integer such that $$a = r^n$$ for some rational number $$r$$, then $$r$$ must be an integer.\n\nProof: We apply the integer root theorem to the polynomial $$x^n - a$$. Since this polynomial has a rational root $$r$$, this root must be an integer.$$_\\square$$\n\n\n\n### Theorem 2. If the sum and product of two rational numbers are both integers, then the two rational numbers must be integers.\n\nProof: We have $$r_1 + r_2 = b$$ and $$r_1 \\cdot r_2 = c$$, hence $$r_1, r_2$$ are rational roots of the polynomial $$x^2 - bx + c$$. By the integer root theorem, $$r_1$$ and $$r_2$$ are both integers. $$_\\square$$\n\n\n\n### 2. $$a$$ divides both $$b$$ and $$c$$.\n\nProof: By the Remainder Factor Theorem, if the roots are integers $$n$$ and $$m$$, then $$f(x) = a (x-n)(x-m)$$. By comparing terms, we obtain $$b = -a(n+m)$$ and $$c = anm$$. Hence, (2) is satisfied. Furthermore, $b^2 - 4ac = a^2 (n+m)^2 - 4 a \\times anm = a^2 (n-m)^2,$ so (1) is also satisfied.\n\nConversely, by the quadratic formula, the roots of $$f(x)$$ are $$\\frac {-b \\pm \\sqrt{b^2- 4ac} }{2a}$$. By condition (1), it follows that the roots are rational. By condition (2) and Vieta's formula, both the sum and product of the roots are integers. Hence by Theorem 2, the roots are integers. $$_\\square$$\n\nNote by Calvin Lin\n4\u00a0years ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nwhat's &nbsp?\n\n- 3\u00a0years, 8\u00a0months ago\n\nIt was meant to be a non-breaking space, to separate out indented paragraphs. I wanted the theorems to be displayed individually, instead of in a long chunk of text.\n\nStaff - 3\u00a0years, 8\u00a0months ago\n\nSomething like a non-breaking space.\n\nEdit: Actually, it's\n\n- 3\u00a0years, 8\u00a0months ago", "date": "2018-04-22 12:13:55", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/theorems-on-rationals/", "openwebmath_score": 0.9954166412353516, "openwebmath_perplexity": 1163.1908642567028, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319877136793, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8726338382874018}}
{"url": "https://math.stackexchange.com/questions/1661818/a-bird-flies-between-two-cars-infinite-times", "text": "# A bird flies between two cars infinite times [duplicate]\n\nThe following question and answer is taken from careerbless\n\nA naughty bird is sitting on top of the car. It sees another car approaching it at a distance of 12 km. The speed of the two cars is 60 kmph each. The bird starts flying from the first car and moves towards the second car,reaches the second car and come back to the first car and so on. If the speed at which bird flies is 120kmph then\n\n1)the total distance travelled by the bird before the crash is?\n\n2)the total distance travelled by the bird before it reaches the second car for the second time is?\n\n3)the total number of times that the bird reaches the bonnet of the second car is(theoretically)?\n\nI am clear with the explanation available in the mentioned site for 1 and 2.\n\nFor the third part, i.e., the total number of times that the bird reaches the bonnet of the second car is(theoretically), answer given is infinite times with the explanation proving the infinite sequence.\n\nBut we know it cannot go on infinite times. It has to be a finite number if we look at it practically. How to explain this?\n\nI am puzzled because the explanation looks convincing (infinite times) whereas we know it cannot be infinite times. Please help in understanding this contradiction.\n\nEDIT: Adding the explanation(below) from careerbless as advised by @DylanSp for clarify and my aim is to understand why it looks as infinite times(theoretically) whereas we know it is finite practically or where my understanding is wrong. (it was a detailed explanation and that is why I have not added initially)\n\n(3) infinite times\n\nAs explained for the previous case,\n\nThe bird reaches the second car in $\\dfrac{12}{180}=\\dfrac{1}{15}$ hour for the first time.\n\nIn this time, the cars together covers a distance of $\\dfrac{1}{15}\u00d7120=8$ km and therefore the distance between the cars becomes 12-8=4 km.\n\nThe bird reaches back the first car in $\\dfrac{4}{180}=\\dfrac{1}{45}$ hour.\n\nIn this time, the cars together covers a distance of $\\dfrac{1}{45}\u00d7120=\\dfrac{8}{3}$ km and therefore the distance between the cars becomes $4-\\dfrac{8}{3}=\\dfrac{4}{3}$ km.\n\nNow the bird flies to the second car for the second time. It takes $\\dfrac{\\left(\\dfrac{4}{3}\\right)}{180}=\\dfrac{1}{135}$ hour for this.\n\nIn this time, the cars together covers a distance of $\\dfrac{1}{135}\u00d7120=\\dfrac{8}{ 9}$ km and therefore the distance between the cars becomes $\\dfrac{4}{3}-\\dfrac{8}{9}=\\dfrac{4}{9}$ km.\n\nThe bird reaches back the first car in $\\dfrac{\\left(\\dfrac{4}{9}\\right)}{180}=\\dfrac{1}{405}$ hour.\n\nIn this time, the cars together covers a distance of $\\dfrac{1}{405}\u00d7120=\\dfrac{8}{ 27}$ km and therefore the distance between the cars becomes $\\dfrac{4}{9}-\\dfrac{8}{27}=\\dfrac{4}{27}$ km.\n\nNow the bird flies to the second car for the third time. It takes $\\dfrac{\\left(\\dfrac{4}{27}\\right)}{180}=\\dfrac{1}{1215}$ hour for this.\n\nso on.\n\nSine this goes on repeatedly, the bird reaches the bonnet of the second car infinite times(theoretically)\n\n## marked as duplicate by Ron Gordon\u00a0sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 18 '16 at 18:56\n\n\u2022 Can you add the explanation from that site to your question here? Questions should be self-contained as much as possible. \u2013\u00a0DylanSp Feb 18 '16 at 18:39\n\u2022 It is not much different from Zeno's Paradoxes. Theoretically we can talk about an infinite number of iterations of still smaller distances adding up to a finite distance as a total. It makes sense theoretically. \u2013\u00a0String Feb 18 '16 at 18:43\n\u2022 @ DylanSp, added \u2013\u00a0Kiran Feb 18 '16 at 18:47\n\u2022 @String, never thought this will open up a completely new area for me which I was not aware. But, we can find sum of an infinite geometric progression(when r<1) to a finite number, right? so it could be the mathematical explanation? \u2013\u00a0Kiran Feb 18 '16 at 18:49\n\u2022 This is why the question asks for a theoretical answer rather than a practical one. Also, I think you can give a much simpler explanation of why the answer is infinity: in short, the bird flies faster than the cars, and at every step the distance between the cars is positive. Finally, if you wanted to make the scenario slightly more realistic, you could suppose that it takes the bird one second, say, to turn around. There will be a point where the cars are less than one second apart, so the bird can only change direction finitely many times. \u2013\u00a0Th\u00e9ophile Feb 18 '16 at 18:50", "date": "2019-06-17 05:48:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1661818/a-bird-flies-between-two-cars-infinite-times", "openwebmath_score": 0.45112529397010803, "openwebmath_perplexity": 376.63806774058094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446517423792, "lm_q2_score": 0.896251371748038, "lm_q1q2_score": 0.8726303547192482}}
{"url": "https://kokecacao.me/page/Course/F20/21-127/Lecture_022.md", "text": "# Lecture 022\n\n## Number Theory\n\nExample:\n\n\u2022 Euclid: there exists infinite many positive integer s.t. x^2+y^2=z^2 and grd(x, y, z) = 1\n\n\u2022 Fermat's Last Theorem: there don't exist any positive integer x, y, z s.t. x^n+y^n=z^n for n>=3\n\n\u2022 Lagrange's Four Square Theorem: Every non-negative integer can be written as the sum of 4 perfect squares\n\n\u2022 Golbach's Conjecture: every even integer > 2 can be expressed as the sum of 2 primes (e.g. 14=3+11=7+7)\n\n\u2022 Twin Prime conjecture: there are infinite many pairs of primes of the form (p, p+2)\n\nPrime: let $n\\in \\mathbb{Z}$ s.t $n\\geq 2$. n is prime iff its only positive divisors are 1 and itself.\n\n\u2022 Theorem: for every integer > 2, is either a prime or a product of primes (every integer has at least one prime factorization)\n\n\u2022 Lemma: for non zero m, n, if m|n, then $|m|\\leq|n|$\n\nComposite: n is not prime (ie. $(\\exists a,b \\in \\mathbb{Z})(1< a \\leq b)\n\nTheorem: if n is composite, then n has a prime factor $p<=\\sqrt{n}$\n\n\u2022 prove by definition of composite and contradiction and prime factorization (either prime or product of prime)\n\n\u2022 Lemma: if $n \\in \\mathbb{Z} \\ {0}$, then n has finitely many divisors (so we always have finite many common divisors) TODO: why\n\nCommon Divisor: for $a, b \\in \\mathbb{Z}$ not both 0, integer d is called a common divisor iff $d|a$ and $d|b$. d is called the greatest common divisor(d=gcd(a, b)) iff d is the greatest of the common divisor\n\n\u2022 $d=gcd(a,b) \\iff d|a \\land d|b \\land (\\forall x \\in \\mathbb{Z})(x|a \\land x|b \\implies x \\leq d)$\n\n\u2022 note: negative number has the same set of divisors as its positive\n\n\u2022 note: gcd(a, b) = gcd(|a|, |b|)\n\n\u2022 note: gcd(0, n) = n\n\n\u2022 note: gcd(0, 0) undefined\n\n\u2022 note: gcd = product of common prime factor\n\ncoprime(relatively prime): iff gcd(a,b) = 1\n\n\u2022 divide a and b by gcd(a, b) to ensure they are relatively prime\n\n\u2022 means no common divisor other than 1.\n\nTheorem: for none zero a,b. d=gcd(a,b) then gcd(a/d, b/d) = 1\n\n\u2022 proof: let n be common divisor of a/d and b/d. then |nd| is common divisor of a and b by algebra. |n|d <= d because d=gcd(a,b), n=+-1, so the greatest(among -1 and 1) common divisor is 1.\n\nEuclidean Algorithm:\n\n\u2022 gcd(x, y) = gcd(x+my, y) where x can be seen as remainder and m is divisor.\n\nDivision Algorithm:\n\n\u2022 if a, b \\in \\mathbb{Z} with b > 0, then there exists unique q,r \\in \\mathbb{Z} s.t. a=bq+r and 0<=r<b\n\n\u2022 proof: consider the set of remainder that divide a by 1~b for a,b: $S_{(a,b)} = \\{a-bk | k \\in \\mathbb{Z} a-bk \\geq 0 \\}$, set is non-empty, has least element (by WOP-well-ordering-property) then we find the lowest reminder. WTS q,r pair unique, and r<b.\n\n\u2022 r=0, then r-b \\in S because a-b(g+1) > 0. r-b<r and r is the least element contradicts.\n\u2022 q,r unique: a=bq_1 + r_1, b=bq_2 + r_2 (r1, r2 between 0 and b), WTS r1=r2,q1=q2. then -b<-r1<=0, -b<b(q1-q2)<b, -1<(q1-q2)<1, q1=q2, r1=r2\n\nCorollary to Division Algorithm (DA): let a, b, in \\mathbb{Z} with b!=0. if a=bq+r, then gcd(a,b) = gcd(b,r)\n\n\u2022 note that q and r doesn't have to be correct divisor or remainder\n\n\u2022 proof: let d1=gdc(a,b), d2=gcd(b,r). $d1|a \\land d1|b \\implies (\\exists k,l \\in \\mathbb{Z})(a=ka_1 \\land b=ld_1)$...\n\nTODO: don't understand\n\nProposition: $a|b \\land a|c \\implies a|bx+cy$\n\nEuclidean Algorithm (formal)\n\nBezout's Lemma\n\n1. $(\\exists x, y \\in \\mathbb{Z})(ax+by = gcd(a,b)))$\n2. $(x,y \\in \\mathbb{Z} \\land ax+by>0 \\implies ax+by \\geq gcd(a,b))$ TODO: proof ignored\n\n3. corollaries: let a,b \\in \\mathbb{Z}, not both 0. let d=gcd(a,b) then\n\n1. if $t \\in \\mathbb{Z}$ s.t. $t|a$ and $t|b$, then $t|d$. (greatest common divisor divides any common divisor)\n2. $gcd(a,b) | c \\iff (\\forall c \\in \\mathbb{Z})(\\exists m, n \\in \\mathbb{Z})(c=am + bn)$\n3. if a and b are relatively prime, then $(\\exists m, n \\in \\mathbb{Z})(am+bn=1)$\n4. $(\\forall m \\in \\mathbb{Z}^+)(gcd(ma, mb)= m \\times gcd(a,b))$\n\nTODO: proof ignored\n\nTable of Content", "date": "2022-12-01 09:42:31", "meta": {"domain": "kokecacao.me", "url": "https://kokecacao.me/page/Course/F20/21-127/Lecture_022.md", "openwebmath_score": 1.0000079870224, "openwebmath_perplexity": 2931.9188142623652, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9944450751316788, "lm_q2_score": 0.8774768002981829, "lm_q1q2_score": 0.8726024825988316}}
{"url": "https://math.stackexchange.com/questions/2990193/can-you-determine-a-set-of-values-from-a-set-of-distinct-sums", "text": "# Can you determine a set of values from a set of distinct sums\n\nConsider an array of positive integers $$A$$ of length $$n$$. Now consider the set of sums of all the contiguously indexed subarrays of $$A$$. For example if $$A = (1,3,5,6)$$ then the set would be $$S_A = \\{1,3,5,6,4,8,11,9,14,15\\}$$.\n\nIf the sums of all contiguously indexed subarrays are distinct (as they are in the example above), does the set of these sums uniquely specify the set of integers in the array the sums were calculated from?\n\nWe can certainly compute the smallest element in the original array as it is the smallest element in $$S_A$$. Similarly there must be a value in the original array which is the largest value in $$S_A$$ minus the second largest.\n\nTo show one of the subtleties of this problem, consider $$A = (1, 6, 2, 3)$$ and $$S_A = \\{1, 6, 2, 3, 7, 8, 5, 9, 11, 12\\}$$. We can immediately tell from $$S_A$$ that $$1$$ occurs somewhere in $$A$$. Similarly we can tell that $$2$$ occurs somewhere in $$A$$. But what can we tell about $$3$$? If $$1$$ and $$2$$ were next to each then as $$1+2=3$$ we would know that $$3$$ can't be in $$A$$. But if $$1$$ and $$2$$ are not next to each in $$A$$ then we know $$3$$ must be in $$A$$. How do we tell which case we are in?\n\nThe answer turns out to be NO. Take $$A = (4, 6, 5, 2, 1)$$ and $$B = (3, 8, 2, 4, 1)$$. We have that $$S_A = S_B$$ but the set of elements in $$A$$ and $$B$$ are distinct.\n\n\u2022 yeah, there have been a few questions recently (in the last few months) dealing with subsequence sums of an array. so i was just wondering where all the common interest comes from. :) OK, lets clarify a few things: (1) $S_A$ is a SET, so you should use curly braces like $S_A = \\{1,3,5,...\\}$. (2) your highlighted question asks does $S_A$ uniquely specify the SET of integers, e.g. $\\{1, 3, 5, 6\\}$, but is that what you mean? Or do you mean to ask if $S_A$ uniquely specify the SEQUENCE (ARRAY) of integers e.g. $(1,3,5,6)$ -- which may still be true, modulo sequence reversal $(6,5,3,1)$? \u2013\u00a0antkam Nov 9 '18 at 20:26\n\u2022 @antkam I asked about the set only because it\u2019s a weaker claim. That is it is more likely to be true. Of course if it\u2019s also true for the array, modulo reversal, that\u2019s even better. (Fixed bracket error now) \u2013\u00a0felipa Nov 9 '18 at 20:51\n\u2022 If $S_A$ is a set, shouldn't it be $\\{1,3,4,\\cdots\\}$, i.e. an ordered list without repetitions ? \u2013\u00a0G Cab Nov 10 '18 at 19:48\n\u2022 This is a version of the \u201cturnpike\u201d problem, or equivalently the \u201cpartial digest\u201d problem. It is known that there can be very many distinct arrays with the same set of substring sums for large $n$, however I am not aware of any results with the restriction that all the sums are distinct. Nice question. \u2013\u00a0Erick Wong Nov 10 '18 at 21:13\n\u2022 @mathlove No sorry that isn't right. The question is if for all arrays A (which are ordered) does $S_A$ (unordered) uniquely determine the set (unordered) of integers in A under the constraint that the subarray sums of A are all distinct. In your example the set $\\{1,2,3\\}$ does indeed tell us that A contains exactly the integers 1 and 2 (but not their order). \u2013\u00a0Anush Nov 14 '18 at 10:47\n\nThe answer turns out to be NO. Take $$A = (4, 6, 5, 2, 1)$$ and $$B = (3, 8, 2, 4, 1)$$. We have that $$S_A = S_B$$ but the set of elements in $$A$$ and $$B$$ are distinct.\n\u2022 To clarify, there are no counter examples at all for length < 5 since it is known that every array of length $\\le 4$ is uniquely determined (up to reversal) by its multi set of substring sums. \u2013\u00a0Erick Wong Nov 15 '18 at 21:07\n\u2022 There is a decent amount known about this actually. I recommend reading \u201cReconstructing Sets from Interpoint Distances\u201d by Skiena, Smith, Lemke, which you should be able to find a PDF for. They give a nice argument that the set of solutions for a given set of sums has a hypercube structure, in particular it is always a power of $2$ (if non-zero). \u2013\u00a0Erick Wong Nov 15 '18 at 22:18", "date": "2019-05-26 09:09:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2990193/can-you-determine-a-set-of-values-from-a-set-of-distinct-sums", "openwebmath_score": 0.8827483654022217, "openwebmath_perplexity": 221.199901112057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.8856314858927011, "lm_q1q2_score": 0.8725873609577923}}
{"url": "https://brilliant.org/discussions/thread/polynomials-sprint-how-to-solve-a-seemingly/", "text": "# Polynomials sprint: How to solve a seemingly disgusting polynomial\n\nOn page 248, it is shown how to solve the polynomial $$x^4+x^3+x^2+x+1=0$$. In this note, I will explain how to solve the polynomial $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ First we multiply both sides by $$(x-1)$$ to get $x^9-1=0\\implies x^9=1$ Then since $$x^9=1$$, we can divide some terms by $$x^9$$ $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=$ $\\dfrac{x^8}{x^9}+\\dfrac{x^7}{x^9}+\\dfrac{x^6}{x^9}+\\dfrac{x^5}{x^9}+x^4+x^3+x^2+x+1=$ $\\dfrac{1}{x}+\\dfrac{1}{x^2}+\\dfrac{1}{x^3}+\\dfrac{1}{x^4}+x^4+x^3+x^2+x+1$ $\\left(x+\\dfrac{1}{x}\\right)+\\left(x^2+\\dfrac{1}{x^2}\\right)+\\left(x^3+\\dfrac{1}{x^3}\\right)+\\left(x^4+\\dfrac{1}{x^4}\\right)+1=0$ Now we let $$y=x+\\dfrac{1}{x}$$ so we have $(y)+(y^2-2)+(y^3-3y)+((y^2-2)^2-2)+1=0$ which after simplification becomes $y^4+y^3-3y^2-2y+1=0$ Using the Rational Root Theorem (discussed on page 246) we quickly find that $$y=-1$$ is a root and factor the polynomial as $(y+1)(y^3-3y+1)$ Now the rest is simple! We can use the cubic solving method discussed on pages 247-248 to find the roots of $$y^3-3y+1$$ Then simple quadratic bashing gives us our roots!\n\nNote by Nathan Ramesh\n4\u00a0years, 2\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nOr, we can use the 9th roots of unity.\n\n- 4\u00a0years, 2\u00a0months ago\n\nYes, what happened to the classic roots of unity? Roots are just $$e^{2ki\\pi/9}$$with $$k=1\\to 8$$.\n\n- 4\u00a0years, 2\u00a0months ago\n\nExcept all roots of unity give you is $$e^{\\frac{2\\pi i}{9}}$$ and not a numerical answer with radicals. The method here allows you to compute $$\\sin {40^{\\circ}}$$\n\n- 4\u00a0years, 2\u00a0months ago\n\nOr you can also just do some Euler's formula with $$e^{\\dfrac{2\\pi i}{9}}$$.\n\n- 4\u00a0years, 2\u00a0months ago\n\nummmm which euler's formula?\n\n- 4\u00a0years, 2\u00a0months ago\n\n$$e^{i\\phi}$$ = cos$$\\phi$$ + $${i}$$sin$$\\phi$$ ... substituting $$\\pi$$ gives $$e^{i\\pi}$$ = -1, where $${i}$$ = $$\\sqrt{-1}$$\n\n- 3\u00a0years, 8\u00a0months ago\n\nWhere i can find this book\n\n- 3\u00a0years, 8\u00a0months ago\n\n$$e^{i\\theta}=\\text{cis}(\\theta)$$\n\n- 4\u00a0years, 2\u00a0months ago\n\nHow does that let you solve for $$\\sin{40^{\\circ}}$$\n\n- 4\u00a0years, 2\u00a0months ago\n\nwow\n\n- 4\u00a0years, 2\u00a0months ago\n\n:o\n\n- 4\u00a0years, 2\u00a0months ago\n\n2 Daniels\n\n- 4\u00a0years, 2\u00a0months ago\n\nOh God, this is not good. One Daniel is enough for me, but two? Double Trouble.\n\n- 4\u00a0years, 2\u00a0months ago\n\n2 Lius\n\n- 4\u00a0years, 2\u00a0months ago\n\n2 14-year-old-Daniel-Lius. Enough to get a conversation off-topic.\n\n- 4\u00a0years, 2\u00a0months ago\n\n2 14-year-old-Daniel-Lius-residing-in-USA. conversation closed.\n\n- 4\u00a0years, 2\u00a0months ago\n\nWhere are all other pages?\n\n- 3\u00a0years, 8\u00a0months ago\n\nYes I used the same to expand cosnx\n\n- 3\u00a0years, 8\u00a0months ago\n\n@Nathan Ramesh Can you add this to the Roots of Unity Applications Wiki page?\n\nSelect \"Write a summary\", and then copy-paste your text into it (with minor formatting adjustments if relevant. Thanks!\n\nStaff - 3\u00a0years, 11\u00a0months ago\n\nDone! I put it at the bottom. Let me know if it is bugged (I posted it from my phone). Thanks!\n\n- 3\u00a0years, 11\u00a0months ago", "date": "2018-09-20 08:02:41", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/polynomials-sprint-how-to-solve-a-seemingly/", "openwebmath_score": 0.998059868812561, "openwebmath_perplexity": 3121.2170509493617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9852713887451681, "lm_q2_score": 0.8856314768368161, "lm_q1q2_score": 0.872587355099444}}
{"url": "http://slideplayer.com/slide/2496738/", "text": "# A logarithm with a base of e is called a natural logarithm and is abbreviated as \u201cln\u201d (rather than as log e ). Natural logarithms have the same properties.\n\n## Presentation on theme: \"A logarithm with a base of e is called a natural logarithm and is abbreviated as \u201cln\u201d (rather than as log e ). Natural logarithms have the same properties.\"\u2014 Presentation transcript:\n\nA logarithm with a base of e is called a natural logarithm and is abbreviated as \u201cln\u201d (rather than as log e ). Natural logarithms have the same properties as log base 10 and logarithms with other bases. The natural logarithmic function f(x) = ln x is the inverse of the natural exponential function f(x) = e x.\n\nThe domain of f(x) = ln x is {x|x > 0}. The range of f(x) = ln x is all real numbers. All of the properties of logarithms from Lesson 7-4 also apply to natural logarithms.\n\nSimplify. A. ln e 0.15t B. e 3ln(x +1) ln e 0.15t = 0.15te 3ln(x +1) = (x + 1) 3 ln e 2x + ln e x = 2x + x = 3x Example 2: Simplifying Expression with e or ln C. ln e 2x + ln e x\n\nSimplify. a. ln e 3.2 b. e 2lnx c. ln e x +4y ln e 3.2 = 3.2 e 2lnx = x 2 ln e x + 4y = x + 4y Check It Out! Example 2\n\nThe formula for continuously compounded interest is A = Pe rt, where A is the total amount, P is the principal, r is the annual interest rate, and t is the time in years.\n\nWhat is the total amount for an investment of \\$500 invested at 5.25% for 40 years and compounded continuously? Example 3: Economics Application The total amount is \\$4083.08. A = Pe rt Substitute 500 for P, 0.0525 for r, and 40 for t. A = 500e 0.0525(40) Use the e x key on a calculator. A \u2248 4083.08\n\nThe half-life of a substance is the time it takes for half of the substance to breakdown or convert to another substance during the process of decay. Natural decay is modeled by the function below.\n\nDownload ppt \"A logarithm with a base of e is called a natural logarithm and is abbreviated as \u201cln\u201d (rather than as log e ). Natural logarithms have the same properties.\"\n\nSimilar presentations", "date": "2017-12-17 06:55:06", "meta": {"domain": "slideplayer.com", "url": "http://slideplayer.com/slide/2496738/", "openwebmath_score": 0.8809204697608948, "openwebmath_perplexity": 1198.319685097197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9783846640860382, "lm_q2_score": 0.8918110396870287, "lm_q1q2_score": 0.872534244492414}}
{"url": "https://math.stackexchange.com/questions/1911146/validity-of-the-change-of-base-formula-for-all-bases", "text": "# Validity of the change-of-base formula for all bases\n\nSuppose I want to solve $3^x=10$. I convert it to logarithmic form $$\\log_{3}10=x$$ then change bases $$\\frac{\\log10}{\\log3}=x$$ and this will yield the same answer as if I had written $$\\frac{\\ln10}{\\ln3}=x$$ But log and ln have different bases. Why does this work both ways?\n\n\u2022 Your first \"inserted\" equation is wrong. It should be $x = \\log_3 10$, not $x=\\log_{10} 3$. Generally, $a^b=c \\iff b=\\log_a c$.\n\u2013\u00a0MPW\nSep 1 '16 at 16:54\n\u2022 @MPW Right, I had to edit the question. And the second half was just rambling about the first half. Sep 1 '16 at 16:59\n\u2022 @ParclyTaxel Thank you very much for the concise summary. Sep 1 '16 at 17:08\n\u2022 Mathematicians writing \"$\\log$\" usually mean the same thing as \"$\\ln$\". The convention used by many engineers, biologists, astronomers, and others, that $\\log$ means the base-$10$ logarithm, was to a large extent rendered obsolete by in advent of calculators around 1970 or so, but it is perpetuated by calculators. Base-10 logarithms are useful in computations by hand because you only need a table going from $1$ to $10$: If you want the logarithm of $145$, you find the logarithm of $1.45$ in the table and then add $2$ to it since the decimal point is $2$ places to the right of there. $\\qquad$ Sep 1 '16 at 17:20\n\u2022 It used to be that if you wanted to tell students how to find things like $\\log_3 20$ with a calculator, you told them it's $\\dfrac{\\log_{10} 20}{\\log_{10} 3}$ or else it's $\\dfrac{\\ln 20}{\\ln 3}$, and they got base-$10$ or base-$e$ logarithms from the calculator. But today you find that some of them have calculators on which you enter both the base and the argument. (I haven't yet encountered on that gives GCDs. Maybe that means I'm not up to date?) $\\qquad$ Sep 1 '16 at 17:22\n\nYou can use whatever basis: the equality $3^x=10$ is equivalent to $$\\log_a 3^x=\\log_a 10$$ for any $a>0$, $a\\ne1$. Since $\\log_a 3^x=x\\log_a 3$, we get $$x=\\frac{\\log_a 10}{\\log_a 3}$$ so you see that the final result is independent of the base.\n\nYou can get the change of base formula by considering $b^c=k$ that says $c=\\log_b k$; but we also have $c\\log_a b=\\log_a k$ and therefore $$\\log_b k=\\frac{\\log_a k}{\\log_a b}$$ In the case of $b=3$, $k=10$ and $a=e$, you get $$\\log_3 10=\\frac{\\ln 10}{\\ln 3}$$\n\n\u2022 Perhaps more to the point is the fact that $$\\frac{\\log_a 10}{\\log_a 3}= \\frac{\\frac{\\log_a 10}{\\log_a b}}{\\frac{\\log_a 3}{\\log_a b}} = \\frac{\\log_b 10}{\\log_b 3}$$ so the choice of base doesn't affect the result.\n\u2013\u00a0MPW\nSep 1 '16 at 17:07\n\u2022 @MPW Since $a$ is completely arbitrary (provided $a>0$ and $a\\ne1$), there's no need to do that computation. Sep 1 '16 at 17:12\n\u2022 I know you know that. I just meant to demonstrate to OP explicitly that they are the same.\n\u2013\u00a0MPW\nSep 1 '16 at 18:02\n\nPossibly the best way to see this is to write a change to an arbitrary base instead of choosing either $\\log_{10}$ or $\\ln$ initially. (There is already an answer that does that.) But I think it's also noteworthy that the change-of-base formula itself provides an easy derivation of the fact that $$\\frac{\\log10}{\\log3}=\\frac{\\ln10}{\\ln3}.$$\n\nStarting with $\\frac{\\ln10}{\\ln3}$ (for example), simply apply the formula to change the base from $e$ to $10$ on both the numerator and denominator: \\begin{align} \\ln10 &= \\log_e 10 = \\frac{\\log_{10} 10}{\\log_{10} e}, \\\\ \\ln3 &= \\log_e 3 = \\frac{\\log_{10} 3}{\\log_{10} e}, \\\\ \\frac{\\ln10}{\\ln3} &= \\frac{\\left( \\frac{\\log_{10} 10}{\\log_{10} e} \\right)} {\\left(\\frac{\\log_{10} 3}{\\log_{10} e} \\right)} \\end{align} Multiply both the numerator and denominator of the right-hand side of the last equation by $log_{10} e$, and we find that $$\\frac{\\ln10}{\\ln3} = \\frac{\\log_{10} 10}{\\log_{10} 3}$$\n\nIn other words, every time we change the base of both the numerator and denominator from base $b$ to base $c$ simultaneously, we divide both the numerator and denominator by $\\log_c b$, and those two operations cancel each other out.\n\nIn your question note that the definition of the logarithm implies that $$\\log_310=x \\iff 3^x=10 \\tag 1.$$ In general $$\\log_bw=x \\iff b ^x=w \\tag 2$$ where $b>0, b\\neq 1$.\n\nNow take the logarithm of both sides to some base $c>0$, $c\\neq 1$ and use the fact that $\\log_c(b^x)=x\\log_cb$ to get that $$x\\log_cb=\\log_cw\\implies x=\\frac{\\log_cw}{\\log_cb}.\\tag 3$$ Equations $(2)$ and $(3)$ imply that$$\\log_bw=\\frac{\\log_cw}{\\log_cb}. \\tag 4$$ Take $c=10$ and $c=e$ in equation $(4)$ to get the equality $$\\log_{3}10=\\frac{\\log_{10}10}{\\log_{10}3}=\\frac{\\ln10}{\\ln3}. \\tag 5$$", "date": "2021-10-20 15:57:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1911146/validity-of-the-change-of-base-formula-for-all-bases", "openwebmath_score": 0.9886220097541809, "openwebmath_perplexity": 188.86279214823625, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795095031688, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8725286612022894}}
{"url": "https://math.stackexchange.com/questions/1323827/kernel-and-image-of-a-linear-transformation", "text": "# Kernel and Image of a Linear Transformation\n\nI have the following linear transformation\n\n$$L:\\Bbb R^3\\rightarrow \\Bbb R^2, (x_1,x_2,x_3)\\mapsto(x_3+x_1,x_2-x_1)$$\n\nAnd I want to determine the kernel and the image of $L$.\n\n$\\text{ker}(L):=v\\in V \\space | \\space L(v)=0$\n\nIs it accurate to say that I want to find the set of vectors that will map to the zero-vector when plugged into the linear transformation?\n\nIn matrix form:\n\n$$\\begin{pmatrix}1&0&1\\\\-1&1&0\\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\end{pmatrix}$$\n\n$$\\implies x_1+x_3=0 \\iff x_1=-x_3 \\space \\space \\space \\text{and} \\space \\space \\space x_2-x_1=0 \\iff x_2=x_1$$\n\nTherefore:\n\n$$ker(L)=\\begin{pmatrix}t\\\\t\\\\-t\\end{pmatrix}$$\n\nIs this correct?\n\nWhat is the image of a linear transformation? Is it the subspace of the co-domain that the linear transformation actually maps to? So for example, if I had the transformation:\n\n$M:\\Bbb R^2 \\rightarrow \\Bbb R, (x_1,x_2) \\mapsto (2)$\n\nThe Image of $Q$ would be {$2$}? How do I determine the image for much more complicated linear transformations?\n\n\u2022 Small detail: In your last example, $M$ is not linear ($M(\\lambda X) \\neq \\lambda M(X)$) \u2013\u00a0Clement C. Jun 13 '15 at 14:25\n\u2022 @ClementC. You're right. I will fix that. \u2013\u00a0qmd Jun 13 '15 at 14:27\n\nAs soon as you have the matrix of a linear map $f$, you have a system of generators of the image of $f$.\n\nIndeed the column-vectors of the matrix are the images of the vectors of the basis in the source-space, say $v_1,v_2,v_3$. Hence for any vector $v$ in $\\mathbf R^3$, $v=\\lambda_1 v_1+\\lambda_2v_2+\\lambda_3v_3$, we have: $$f(v)=\\lambda_1 f(v_1)+\\lambda_2f(v_2)+\\lambda_3f(v_3).$$\n\nNaturally, this system of generators is not minimal, i. e. it is not a basis of the image in general. But from this system, you can deduce a basis by column reduction.\n\nIt will not be necessary here, because the rank-nullity theorem ensures $\\dim\\operatorname{Im}f=2$ since you've found that $\\dim\\ker f=1$. Thus, $f$ is surjective, i. e. the image of $f$ is the whole of $\\mathbf R^2$.\n\n\u2022 So in this case I have the matrix $\\begin{pmatrix}1&0&1\\\\-1&1&0\\end{pmatrix}$ with columns $\\begin{pmatrix}1\\\\1\\end{pmatrix}$,$\\begin{pmatrix}0\\\\1\\end{pmatrix}$, $\\begin{pmatrix}1\\\\0\\end{pmatrix}$. Are these column-vectors the image of $L$? Is it always the case that the column-vectors are the image? \u2013\u00a0qmd Jun 13 '15 at 14:51\n\u2022 The column-vectors are not the image of $L$: they span the image of $L$. But $\\begin{pmatrix}1\\\\-1\\end{pmatrix}$ andt $\\begin{pmatrix}0\\\\1\\end{pmatrix}$ also generate the image. \u2013\u00a0Bernard Jun 13 '15 at 14:55\n\u2022 I see. So I have to look at that column vectors and calculate the span. In this case $\\begin{pmatrix}1\\\\0\\end{pmatrix}$ and $\\begin{pmatrix}0\\\\1\\end{pmatrix}$ span $\\Bbb R^2$ and therefore Im($L$)=$\\Bbb R^2$? \u2013\u00a0qmd Jun 13 '15 at 14:58\n\u2022 Yes. You also know that here without any calculation thanks to the rank-nullity theorem. \u2013\u00a0Bernard Jun 13 '15 at 15:16\n\nAn option to characterize all $\\vec{y}\\in\\mathrm{im}(L)$ is to express that there must be a $\\vec{x}$ such that $L( \\vec{x}) = \\vec{y}$. That is, write $$\\begin{pmatrix}1&0&1\\\\-1&1&0\\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\end{pmatrix} = \\begin{pmatrix}y_1\\\\y_2\\end{pmatrix}$$ and solve for $\\begin{pmatrix}y_1\\\\y_2\\end{pmatrix}$ as a function of of $x_1,x_2,x_3$. This will give you the set (when $x_1,x_2,x_3$ vary in $\\mathbb{R}$ of all $\\vec{y}$'s in the image.\n\n\u2022 Thanks. Can you give me an intuitive explanation of what the image is? Is it, as I asked in my question, \"the subspace of the co-domain that the linear transformation actually maps to?\" or am I confusing something here? Solving the linear system in your answer I get: $x_1+x_3=y_1$ and $x_2-x_1=y_2$. I don't understand how that helps me. \u2013\u00a0qmd Jun 13 '15 at 14:43\n\u2022 The image is the set of all poin ts that are \"reached\" by the linear transformation. In your case, it's basically the set of all points that can be written $(a+b,c-a)$ for some $a,b,c\\in\\mathbb{R}$. (Also, one can show that here, this is actually the whole set $\\mathbb{R}^2$: for any choice of $y_1,y_2$, you can solve the system for $x_1,x_2,x_3$ and find solutions: that is, any point in $\\mathbb{R}^2$ is reached by $L$. \u2013\u00a0Clement C. Jun 13 '15 at 14:45", "date": "2021-03-02 20:30:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1323827/kernel-and-image-of-a-linear-transformation", "openwebmath_score": 0.8888595104217529, "openwebmath_perplexity": 165.0836357096202, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357184418847, "lm_q2_score": 0.8887588038050466, "lm_q1q2_score": 0.8725262627750974}}
{"url": "http://math.stackexchange.com/questions/141770/does-varphi1-1-if-varphi-is-a-field-homomorphism", "text": "# Does $\\varphi(1)=1$ if $\\varphi$ is a field homomorphism?\n\nIs it by definition that $\\varphi(1)=1$ if $\\varphi$ is a field homomorphism ?\n\nMy field theory lecture said that yes, but now $\\varphi\\equiv0$ is not a field homomorphism...\n\nWhat is the 'standard' in mathematics ? (i.e. if someone says that $\\varphi$ is a field homomorphism does it imply $\\varphi(1)=1$ ?)\n\n-\nIt is standard to assume $\\varphi(1) = 1$. It isn't hard to show that if $\\varphi(1)\\neq 1$, then $\\varphi\\equiv 0$. Thus assuming $\\varphi(1) = 1$ is just to rule out this case. \u2013\u00a0 froggie May 6 '12 at 14:25\nWhy would you want $0$ to be a homomorphism? It's not like you can add field (or ring) homomorphisms. \u2013\u00a0 Zhen Lin May 6 '12 at 14:39\nHi, it seems that my answer is somewhat misleading and possibly even wrong. Please unaccept it so I can delete the answer. Pete Clark has agreed to write an answer of his own, which I am sure will be much better than mine. \u2013\u00a0 Asaf Karagila May 6 '12 at 17:40\n@AsafKaragila - done. \u2013\u00a0 Belgi May 6 '12 at 18:17\n\nUpon request, I am leaving an answer.\n\nFirst, let's acknowledge that this is a question about conventions first and only secondarily about mathematics. Your lecturer's definition of a field homomorphism is whatever she says it is: that's what a definition means. (\"When I use a word, it means just what I choose it to mean -- neither more nor less.\" -- Humpty Dumpty) On the other hand, definitions, while in some formal sense arbitrary, can certainly be good or bad, for both mathematical and sociological reasons.\n\nYour lecturer's definition is a good one both because of its mathematical consequences and because, as she claimed, this is very much the \"standard\" definition, i.e., the one you will find in the vast majority of contemporary texts and courses, the one which is being used by practitioners both in field theory itself and in areas of mathematics where field theory is applied, and so forth.\n\nThe notion of \"homomorphism\" is so important in mathematics that there has been a lot of effort to make it as systematic and \"un-(ad hoc)\" as possible. Two very general frameworks for discussing homomorphisms are universal algebra and category theory. The basic objects of study in universal algebra are relational structures, namely sets equipped with various relations (and, what is a special case of that, functions), of various \"arities\".\n\nFrom this perspective, the relational structure of a ring is a set $R$, two binary operations called $+$ and $\\cdot$ and two \"nullary functions\" -- i.e., constants $0$ and $1$. There is more to the definition of a ring, namely the axioms that these relations must satisfy. In this sense, model theory is being overlayed on top of universal algebra. However, the definition of a homomorphism of relational structures doesn't depend on the axioms: a homomorphism of relational structures is just a map of the underlying sets which preserves, in a rather straightforward sense which I won't completely write out here (one can easily look it up online) all the relations. In the case of rings, this means that a homomorphism $f: R \\rightarrow S$ is a map of the underlying sets such that\n\n$\\bullet$ For all $x,y \\in R$, $f(x+y) = f(x) + f(y)$,\n$\\bullet$ For all $x,y \\in R$, $f(x \\cdot y) = f(x) \\cdot f(y)$,\n$\\bullet$ $f(0) =0$, and\n$\\bullet$ $f(1) = 1$.\n\nAgain, I stress that this does not depend on the axioms (or, in more model-theoretic terminology, the theory, of the structure): a homomorphism of non-associative rings (with distinguished elements $0$ and $1$) is the same definition as a homomorphism of not necessarily commutative rings, which is the same definition as a homomorphism of fields. This uniformity is useful and powerful.\n\nNow let $f: R \\rightarrow S$ be a homomorphism of rings. An element $x \\in R$ is a unit if there exists $y \\in R$ such that $xy = yx = 1$. Two easy facts:\n\n(i) For all $x \\in R$, there is at most one $y \\in R$ such that $xy = yx = 1$.\n(Proof: If also $xz = zx = 1$, then $y = y \\cdot 1 = y(xz) = (yx)z = 1 \\cdot z = z$.)\n\nThus if $x$ is a unit, it has a well-defined inverse which we may denote $x^{-1}$. However, $x \\mapsto x^{-1}$ is not a completely kosher function from the perspective of universal algebra, since it is not a function on $R$ but only on $R^{\\times}$, the set of units of $R$. There are ways to get around this (a keyword is sorts) but the easiest solution is simply not to regard $x \\mapsto x^{-1}$ is being part of the relational structure defining a commutative ring.\n\n(ii) If $f: R \\rightarrow S$ is a homomorphism of rings and $x \\in R^{\\times}$, then $f(x) \\in S^{\\times}$ and $f(x^{-1}) = f(x)^{-1}$.\n(Proof: $f(x^{-1}) f(x) = f(x^{-1} x) = f(1) = 1 = f(1) = f(x x^{-1}) = f(x) f(x^{-1})$.)\n\nThus the preservation of inverses comes for free from this definition of ring homomorphisms.\n\nThere is also the categorical perspective, which is more general and thus more permissive. It has the bright idea that in the same breath as we define a mathematical object of a certain kind (\"category\"!), we should define the notion of homomorphism between objects of that category. In order to be a homomorphism, certain very mild axioms of composition and identities must be satisfied. So in particular, there is a category $\\operatorname{Ring}$ in which the objects are rings and the morphisms are defined exactly as above. But there is also a category $\\operatorname{Rng}$ in which the objects are \"rngs\", i.e., what you get by taking the constant $1$ out of the relational structure and taking out all the parts of the axioms for rings which refer to it. And there is even the category -- call it $\\operatorname{Ring}'$ -- in which the objects are rings -- i.e., have $1$ -- but in which homomorphisms are not required to carry $1$ to $1$. In particular, for any rings $R$ and $S$, the identically zero map from $R$ to $S$ is a homomorphism in $\\operatorname{Ring}'$ but not in $\\operatorname{Ring}$ (except in the trivial case where $0 = 1$ in $S$).\n\nIt is worth spending a little time getting experience with the category $\\operatorname{Ring}'$, both because (i) it is not simply a pathology: these sorts of maps between rings do come up sometimes and (ii) it gives you insight about the much more standard category $\\operatorname{Ring}$. In particular, for a homomorphism $f: R \\rightarrow S$ in $\\operatorname{Ring}'$ $f(1)$ cannot be just any old element of $S$ but must be an idempotent element:\n\n$f(1) = f(1 \\cdot 1) = f(1) \\cdot f(1)$.\n\nMany rings have only $0$ and $1$ as idempotents -- in particular, this holds for fields -- so either $f(1) = 1$, in which case we're back to the standard case, or $f(1) = 0$, and in this case for all $x \\in R$, $f(x) = f(x \\cdot 1) = f(x) \\cdot f(1) = f(x) \\cdot 0 = 0$, so $f$ is identically zero.\n\nSo now we know exactly what we're excluding by defining a homomorphism of fields to satisfy $f(1) = 1$: we're excluding the identically zero map between fields. So we can be rather confident that we're not excluding anything important: certainly the zero map is not interesting in its own right, just as the identity element of a group or zero element of a ring is not really interesting in its own right. However, the identity element of a group and zero element of a ring are there for another very important reason: they enable the appropriate algebraic structure of group or ring to exist!\n\nIn contrast, if $R$ and $S$ are rings, then I see no intrinsic algebraic structure on the set of homomorphisms from $R$ to $S$: we cannot add them, multiply them (so as to get another homomorphism from $R$ to $S$) and so forth. So adding the zero element is not helping us in any evident way. In fact, if we don't believe this at first, no matter -- keep it in and see what happens. We'll find that in our further study of field theory, we'll keep having to say \"except for the zero homomorphism\" in results and arguments. For instance, most of field theory is based upon the fact that absolutely every field homomorphism $\\iota: E \\rightarrow F$ is injective and makes $F$ into an $E$-vector space...except for the zero homomorphism.\n\n-", "date": "2014-04-24 03:48:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/141770/does-varphi1-1-if-varphi-is-a-field-homomorphism", "openwebmath_score": 0.959018349647522, "openwebmath_perplexity": 199.63215356552917, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357243200244, "lm_q2_score": 0.8887587831798665, "lm_q1q2_score": 0.8725262477508697}}
{"url": "https://math.stackexchange.com/questions/2256083/how-to-find-if-this-function-is-always-zero", "text": "# How to find if this function is always zero?\n\nThe question I faced was:\n\nLet $f(x)$ be a non-negative continuous and bounded function for all $x \\ge 0$. If\n$$(\\cos x)f'(x) \\le (\\sin x - \\cos x)f(x), \\; \\forall \\; x \\ge 0$$ then which of the following is/are correct?\n\n(A) $f(6) + f(5) > 0$\n\n(B) $x^2 - 3x + 2 + f(9) = 0$ has two distinct solutions\n\n(C) $f(5)f(7) - f(6)f(5) = 0$\n\n(D) $\\lim\\limits_{x \\to 4} \\dfrac{f(x) - \\sin(\\pi x)}{x-4} = 1$\n\n(B), (C)\n\nBy observation, $f(x)=0$ satisfies the given conditions. But is it the only solution? If so, how to prove it is?\n\nAfter rearranging the terms and combining them, I converted the inequality to this form: $$\\left( f(x)\\,\\cos x \\right)' + f(x)\\,\\cos x \\le 0$$ Despite its allure, this inequality isn't getting me anywhere! It doesn't seem to have any information about $f(x)$, since it is stuck with a \"$\\cos x$\". Even then, I don't see where I can go with it.\n\nSo how to solve this problem? Thank you.\n\n\u2022 Your condition is equivalent to $$\\frac{\\mathrm{d}}{\\mathrm{d}x} \\ f(x) \\cos x \\ e^x \\le 0$$ \u2013\u00a0Crostul Apr 28 '17 at 10:56\n\u2022 @BST Actually the discriminant is $(-3)^2-4(2+f(9)) = 1-4f(9)$ which may be negative \u2013\u00a0Crostul Apr 28 '17 at 11:00\n\u2022 Sorry, you're right. My mistake. I'll remove the comment. I miscalculated the discriminant. \u2013\u00a0be5tan Apr 28 '17 at 11:02\n\nWe assume that $f(x)$ is bounded and non-negative, for $x \\geqslant 0$. This means that there is a non-negative function $g(x)$ with the property that $f(x) = g(x)e^{-x}$ for $x \\geqslant 0$.\n\nPlugging this into the inequality you found gives, $$0 \\geqslant (f(x)\\cos (x))' + f(x) \\cos (x) = e^{-x} (g(x) \\cos(x))'.$$ And since $e^{-x} > 0$, we have $$0 \\geqslant (g(x) \\cos(x))'.$$ This means that the function $g(x) \\cos(x)$ is weakly decreasing. Because any point $x \\in \\mathbb{R}$ is between two zeroes of $\\cos(x)$, we have that $g(x) \\cos(x) = 0$ for all $x$.\n\n\u2022 Instead of integrating the inequality, it gives more information simply to note that the derivative of $g(x)\\cos x$ can never be positive. Since every $x$ is between two zeroes of $g(x)\\cos x$, the mean value theorem forces it to be $0$ everywhere, not just where the cosine is positive. \u2013\u00a0Henning Makholm Apr 28 '17 at 11:07\n\u2022 Ah, I missed that! Thank you, will edit my answer. \u2013\u00a0Peter Apr 28 '17 at 11:08\n\u2022 Also, I think it would be slicker simply to define $g(x)=f(x)e^x$ at the beginning. Then we don't need to appeal to \"bounded and nonnegative\", and the first inequality is still true. \u2013\u00a0Henning Makholm Apr 28 '17 at 11:22\n\u2022 This is brilliant! How did you figure out the substitution $f=ge^{-x}$? \u2013\u00a0FreezingFire Apr 28 '17 at 11:28\n\u2022 I asked myself what function $h(x)$ would satisfy the strict version of your bound, i.e.~$h'(x) + h(x) = 0$. Evidently the function $h(x) = e^{-x}$ solves this equation. The rest was the result of a bit of experimentation with this function. \u2013\u00a0Peter Apr 28 '17 at 11:56\n\nIf you set $g(x)=f(x)\\cos x$, your rearrangement tells you that $$\\tag{1} g'(x) \\le -g(x)$$\n\nConsider an interval $[2\\pi k-\\frac12\\pi , 2\\pi k +\\frac12\\pi]$ where $\\cos x$ is $\\ge 0$. Then you know that $g(x)\\ge 0$ on this interval, and $g(x)=0$ at the start of the interval. Then, because of (1), $g(x)$ must be identically zero on that interval, and so must $f(x)$.\n\nThis settles at least (A) and (C), because $5$, $6$ and $7$ are all within $\\pi/2$ of $2\\pi$.\n\nFor (B) and (D) this argument doesn't tell you enough; I recommend Peter's slicker answer instead.\n\nNotice that whenever some $f$ satisfies the hypothesis, then so does any positive multiple of $f$. In particular:\n\nNote that (B) is equivalent to $$0<(-3)^2-4(2+f(9)) = 1-4f(9)$$ i.e. $f(9)<1/4$. Hence (B) can hold if and only if $f(9) =0$ for any $f$ satisfying the hypothesis.", "date": "2019-11-20 01:38:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2256083/how-to-find-if-this-function-is-always-zero", "openwebmath_score": 0.933104395866394, "openwebmath_perplexity": 155.75566366704822, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357195106375, "lm_q2_score": 0.8887587846530938, "lm_q1q2_score": 0.8725262449228047}}
{"url": "https://math.stackexchange.com/questions/1059566/explain-proof-that-any-positive-definite-matrix-is-invertible/1059576", "text": "# Explain proof that any positive definite matrix is invertible\n\nIf an $n \\times n$ symmetric A is positive definite, then all of its eigenvalues are positive, so $0$ is not an eigenvalue of $A$. Therefore, the system of equations $A\\mathbf{x}=\\mathbf{0}$ has no non-trivial solution, and so A is invertible.\n\nI don't get how knowing that $0$ is not an eigenvalue of $A$ enables us to conclude that $A\\mathbf{x}=\\mathbf{0}$ has the trivial solution only. In other words, how do we exclude the possibility that for all $\\mathbf{x}$ that is not an eigenvector of $A$, $A\\mathbf{x}=\\mathbf{0}$ will not have a non-trivial solution?\n\nNote that if $Ax=0=0\\cdot x$ for some $x\\ne 0$ then by definition of eigenvalues, $x$ is an eigenvector with eigenvalue $\\lambda = 0$, contradicting that $0$ is not an eigenvalue of $A$. $$Ax=\\lambda x$$\n$$\\det A = \\prod_{j=1}^n \\lambda_j \\implies \\det A = 0 \\Leftrightarrow \\exists\\ i \\in \\{1,2,\\ldots, n\\}:\\lambda_i = 0$$\nBecause if $Ax=0$ for some nonzero $x$, then $0$ would be an eigenvalue.\n\u2022 Yes. You have that $0$ is an eigenvalue of $A$ $\\iff$ $A$ is not invertible $\\iff$ there exists nonzero $x$ with $Ax=0$. \u2013\u00a0Martin Argerami Sep 11 '19 at 11:57", "date": "2020-08-06 01:28:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1059566/explain-proof-that-any-positive-definite-matrix-is-invertible/1059576", "openwebmath_score": 0.9758989810943604, "openwebmath_perplexity": 40.81975195224248, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9905874115238966, "lm_q2_score": 0.8807970842359877, "lm_q1q2_score": 0.8725065037511226}}
{"url": "https://math.stackexchange.com/questions/2178077/prove-fracfbfa-eb-a-fracfcfc", "text": "# Prove $\\frac{f(b)}{f(a)}=e^{(b-a)\\frac{f'(c)}{f(c)}}$\n\nLet $f:[a,b]\\rightarrow \\mathbb{R}$ be a positive and differentiable function. Prove: $$\\frac{f(b)}{f(a)}=e^{(b-a)\\frac{f'(c)}{f(c)}}$$\n\nI tried to apply the MVT and make some algebraic manipulations, but got stuck.\n\nAny help appreciated.\n\n\u2022 Do you want to prove that the equality holds for a certain $c$, a certain $a<c<b$, or any $c$? \u2013\u00a0toliveira Mar 8 '17 at 19:42\n\u2022 Did you forget an exponential somewhere, e.g. in $e^{\\frac{f'(c)}{f(c)}}$? \u2013\u00a0Clement C. Mar 8 '17 at 19:42\n\u2022 @ClementC. Sorry for that, editing. \u2013\u00a0Itay4 Mar 8 '17 at 19:44\n\u2022 Guys, he means $e^{(b-a)\\frac{f'(c)}{f(c)}}$ \u2013\u00a0Gabriel Romon Mar 8 '17 at 19:44\n\u2022 @LeGrandDODOM Yet, that is not what (s)he wrote initially. Hence my comment. \u2013\u00a0Clement C. Mar 8 '17 at 19:46\n\nHigh-level idea: The reflex to use the MVT is a good one. Now, the issue is that you do not want to apply it directly to $f$, but to some other function $g$ related to $f$. In particular, the term $e^{b-a}$ strongly hints that you're going to take an exponential after that; and seeing $\\frac{f'(c)}{f(c)}$ should prompt you to think of $(\\ln f)' = \\frac{f'}{f}$. These two together suggest we set $g=\\ln f$, and apply the MVT to $g$: good thing, this is legitimate, as $f$ is assumed positive. So let's try that...\n\nLet $g\\colon [a,b]\\to \\mathbb{R}$ be defined by $g\\stackrel{\\rm def}{=} \\ln f$. $g$ is well-defined, and differentiable, as $f$ is positive and differentiable.\n\nApplying the Mean Value Theorem on $g$, we get that there exists $c\\in(a,b)$ such that $$g(b)-g(a) = g'(c) (b-a)$$ and, exponentiating both sides, we obtain $$e^{g(b)-g(a)} = e^{g'(c)(b-a)}.$$ Recalling that $e^{g(b)-g(a)} = \\frac{e^{g(b)}}{e^{g(a)}} = \\frac{f(b)}{f(a)}$ and $g'(c) = \\frac{f'(c)}{f(c)}$ concludes the proof.\n\n\u2022 The right-hand side of your last displayed equation should be $\\exp(g^{\\prime}(c)(b-a))$. Great answer otherwise though... \u2013\u00a0carmichael561 Mar 8 '17 at 20:18\n\u2022 @carmichael561 Thanks for spotting this! \u2013\u00a0Clement C. Mar 8 '17 at 20:19\n\nLet $g(x)=\\ln(f(x))$. According to the mean value theorem, $$\\frac{g(b)-g(a)}{b-a}=g^{\\prime}(c)=\\frac{f^{\\prime}(c)}{f(c)}$$ for some $c\\in(a,b)$. This implies that $$\\ln\\Big(\\frac{f(b)}{f(a)}\\Big)=(b-a)\\frac{f^{\\prime}(c)}{f(c)}$$ hence $$\\frac{f(b)}{f(a)}=\\exp\\Big((b-a)\\frac{f^{\\prime}(c)}{f(c)}\\Big)$$\n\n\u2022 Darn, you beat me by a few seconds. (+1) \u2013\u00a0Clement C. Mar 8 '17 at 19:47", "date": "2019-10-15 16:22:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2178077/prove-fracfbfa-eb-a-fracfcfc", "openwebmath_score": 0.8336794972419739, "openwebmath_perplexity": 295.3261282495253, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874102734519, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8725064964505342}}
{"url": "https://web2.0calc.com/questions/how-to-convert-polar-to-cartesian-and-vice-versa", "text": "+0\n\n# How to convert polar to cartesian and vice versa?\n\n0\n507\n2\n\nhow to convert polar coordinates to cartesian and vice versa?\n\nGuest\u00a0Sep 9, 2014\n\n### Best Answer\n\n#2\n+20040\n+5\n\nHow to convert polar to cartesian and vice versa ?\n\nI. polar to cartesian : $$\\boxed{\\begin{array}{rcl} x=r*\\cos{(\\alpha)} \\\\ y=r*\\sin{(\\alpha)} \\end{array} }$$\n\nII. cartesian\u00a0 to polar : $$r: \\begin{array}{rcl} x^2+y^2&=&r^2\\cos^2{(\\alpha)}+r^2\\sin^2{(\\alpha)} \\\\ x^2+y^2&=& r^2( \\underbrace{ \\sin^2{(\\alpha)}+\\cos^2{(\\alpha)} }_{=1} ) \\\\ x^2+y^2&=&r^2 \\\\ \\end{array} \\boxed{r=\\sqrt{x^2+y^2}}$$\n\n$$\\alpha: \\begin{array}{rcl} \\frac{y}{x}&=& \\frac { r\\sin{(\\alpha)} } { r\\cos{(\\alpha)} }\\\\ \\frac{y}{x}&=& \\tan{(\\alpha)} \\end{array} \\boxed{ \\alpha=\\tan^{-1}\\left( \\frac{y}{x} \\right) }$$\n\n$$\\begin{array}{rcll} y>0 &and& x>0 & \\quad \\alpha \\\\ y>0 &and& x<0 & \\quad \\alpha +180\\ensurement{^{\\circ}}\\\\ y<0 &and& x<0 & \\quad \\alpha +180\\ensurement{^{\\circ}}\\\\ y<0 &and& x>0 & \\quad \\alpha +360\\ensurement{^{\\circ}}\\\\ \\hline y=0 &and& x>0 & \\quad \\alpha = 0\\ensurement{^{\\circ}}\\\\ y>0 &and& x=0 & \\quad \\alpha = 90\\ensurement{^{\\circ}}\\\\ y=0 &and& x<0 & \\quad \\alpha = 180\\ensurement{^{\\circ}}\\\\ y<0 &and& x=0 & \\quad \\alpha = 270\\ensurement{^{\\circ}} \\end{array}$$\n\ny = 0\u00a0\u00a0 and\u00a0 x = 0 \u00a0 $$\\alpha$$\u00a0 undefined!\n\nheureka \u00a0Sep 9, 2014\n#1\n+27062\n+5\n\ncartesian to polar:\n\n$$r=\\sqrt{x^2+y^2}$$\n\n$$\\theta=\\tan^{-1}(\\frac{y}{x})$$\n\npolar to cartesian:\n\n$$x=r\\cos{\\theta}$$\n\n$$y=r\\sin{\\theta}$$\n\nAlan \u00a0Sep 9, 2014\n#2\n+20040\n+5\nBest Answer\n\nHow to convert polar to cartesian and vice versa ?\n\nI. polar to cartesian : $$\\boxed{\\begin{array}{rcl} x=r*\\cos{(\\alpha)} \\\\ y=r*\\sin{(\\alpha)} \\end{array} }$$\n\nII. cartesian\u00a0 to polar : $$r: \\begin{array}{rcl} x^2+y^2&=&r^2\\cos^2{(\\alpha)}+r^2\\sin^2{(\\alpha)} \\\\ x^2+y^2&=& r^2( \\underbrace{ \\sin^2{(\\alpha)}+\\cos^2{(\\alpha)} }_{=1} ) \\\\ x^2+y^2&=&r^2 \\\\ \\end{array} \\boxed{r=\\sqrt{x^2+y^2}}$$\n\n$$\\alpha: \\begin{array}{rcl} \\frac{y}{x}&=& \\frac { r\\sin{(\\alpha)} } { r\\cos{(\\alpha)} }\\\\ \\frac{y}{x}&=& \\tan{(\\alpha)} \\end{array} \\boxed{ \\alpha=\\tan^{-1}\\left( \\frac{y}{x} \\right) }$$\n\n$$\\begin{array}{rcll} y>0 &and& x>0 & \\quad \\alpha \\\\ y>0 &and& x<0 & \\quad \\alpha +180\\ensurement{^{\\circ}}\\\\ y<0 &and& x<0 & \\quad \\alpha +180\\ensurement{^{\\circ}}\\\\ y<0 &and& x>0 & \\quad \\alpha +360\\ensurement{^{\\circ}}\\\\ \\hline y=0 &and& x>0 & \\quad \\alpha = 0\\ensurement{^{\\circ}}\\\\ y>0 &and& x=0 & \\quad \\alpha = 90\\ensurement{^{\\circ}}\\\\ y=0 &and& x<0 & \\quad \\alpha = 180\\ensurement{^{\\circ}}\\\\ y<0 &and& x=0 & \\quad \\alpha = 270\\ensurement{^{\\circ}} \\end{array}$$\n\ny = 0\u00a0\u00a0 and\u00a0 x = 0 \u00a0 $$\\alpha$$\u00a0 undefined!\n\nheureka \u00a0Sep 9, 2014\n\n### New Privacy Policy\n\nWe use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.\nFor more information: our cookie policy and privacy policy.", "date": "2018-10-24 06:14:36", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/how-to-convert-polar-to-cartesian-and-vice-versa", "openwebmath_score": 0.4345089793205261, "openwebmath_perplexity": 11377.041482687413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874104123902, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8725064919235103}}
{"url": "https://beautylab.gr/pcbun57/polar-form-to-rectangular-form-ff867f", "text": "Math > Pre Calculus > Conversion from Polar to Rectangular Form Complex If a polar equation is written such that it contains terms that appear in the polar-rectangular relationships (see below), conversion from a polar equation to a rectangular equation is a simple matter of substitution. Keep solving until you isolate the variable r. Powered by Create your own unique website with customizable templates. Convert a Complex Number to Polar Form Description Obtain the polar form of a complex number . y r Polar to Rectangular x rcos y rsin The polar form r cos isin is sometimes abbreviated rcis Example Convert 3 i to polar form. Show Instructions. However, I would like to convert it to polar form with this kind of arrangements. Problems with detailed solutions are presented.In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate.The relationships between the rectangualr (x,y) and polar (R,t) coordinates of a points are given byR 2 = x 2 + y 2 \u00a0 \u00a0 \u00a0 y = R sin t \u00a0 \u00a0 \u00a0 x = R cos t. Expand the left side of the given equation.R(-2 sin t + 3 cos t) = 2-2 R sin t + 3 R cos t = 2eval(ez_write_tag([[580,400],'analyzemath_com-box-4','ezslot_7',260,'0','0']));Use y = R sin t and x = R cos t into the given equation to rewrite as follows:-2 y + 3 x = 2 2 2 How to convert between Polar and Rectangular form ( in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Convert Equation from Rectangular to Polar Form, Convert Polar to Rectangular Coordinates - Calculator, Free Geometry Tutorials, Problems and Interactive Applets, Convert Rectangular to Polar Coordinates - Calculator. 4 The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. 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In conversion of complex numbers to polar coordinates to polar form to form! 30\u00b0 150\u00b0 so that 3 i 2cis150\u00b0 convert it to polar coordinates using the conversion formulas prove my.! Can also be verified using the conversion formulas, if you have been converting the polar and... To find the product of two complex numbers to polar form of writing complex numbers from one form to.! Coordinates, the angle we got is not right we first need some kind of.... Product of two complex numbers to polar form to rectangular form complex numbers in polar form origin, e 1! Tan \u03b8 changing rectangular values into polar form to rectangular form using constructor in destination class / 4 or! = a + b i is called the rectangular shape of 2 with detailed. \\Pageindex { 7 } \\ ) and vice versa, with steps shown necessary to from! Ducktales Remastered Moon Theme Mp3, Copper In Minecraft Uses, New World Man Isolated Bass, Brickyard Menu Ballston Spa, String Pronunciation And Meaning, Fire Extinguisher Types, What Word Means On Or Near The Sea, Sweet Sour Song, \" /> Math > Pre Calculus > Conversion from Polar to Rectangular Form Complex If a polar equation is written such that it contains terms that appear in the polar-rectangular relationships (see below), conversion from a polar equation to a rectangular equation is a simple matter of substitution. Keep solving until you isolate the variable r. Powered by Create your own unique website with customizable templates. Convert a Complex Number to Polar Form Description Obtain the polar form of a complex number . y r Polar to Rectangular x rcos y rsin The polar form r cos isin is sometimes abbreviated rcis Example Convert 3 i to polar form. Show Instructions. However, I would like to convert it to polar form with this kind of arrangements. Problems with detailed solutions are presented.In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate.The relationships between the rectangualr (x,y) and polar (R,t) coordinates of a points are given byR 2 = x 2 + y 2 \u00a0 \u00a0 \u00a0 y = R sin t \u00a0 \u00a0 \u00a0 x = R cos t. Expand the left side of the given equation.R(-2 sin t + 3 cos t) = 2-2 R sin t + 3 R cos t = 2eval(ez_write_tag([[580,400],'analyzemath_com-box-4','ezslot_7',260,'0','0']));Use y = R sin t and x = R cos t into the given equation to rewrite as follows:-2 y + 3 x = 2 2 2 How to convert between Polar and Rectangular form ( in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Convert Equation from Rectangular to Polar Form, Convert Polar to Rectangular Coordinates - Calculator, Free Geometry Tutorials, Problems and Interactive Applets, Convert Rectangular to Polar Coordinates - Calculator. 4 The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. 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With MATLAB keywords but need to go square root of 13 how to perform operations complex... In conversion of complex numbers to polar form with this kind of arrangements the formulas and,... Go, i would like to convert a complex number to polar form in terms of cos \u03b8, \u03b8! J80, then: Step 1, multiply through by \\ ( \\PageIndex { 7 } \\.... Be verified using the conversion equation ) and vice versa, with steps shown form and rectangular rectangular... Rectangular coordinate form of the number as the length of line and the vertical axis is the method of the! Vector of a magnitude of 30 plotted on the \u2212j axis a focus the..., amplitude phase will convert the conic r = 2 _____ to rectangular form is an interactive calculator allows... Detailed calculations best thing you can do is chart them out polar to... = a + b i is called the rectangular coordinate form of a number. And then, start changing rectangular values into polar form to rectangular ( Cartesian ) and vice versa with... Of 30 plotted on the \u2212j axis a phasor Cartesian ) and \\. Step 1 method of changing the representation form of a complex number notation: polar rectangular. Conic r = 2 _____ to rectangular coordinates to polar form are converted into rectangular form 1. Is chart them out 1667-1754 ) command, the angle at which a number extends meaning... Plotted on the \u2212j axis ( r 2, ( r 2, r! Is the method of changing the representation form of a magnitude of 30 on! \\ ( r\\ ) cos \u03b8, sin \u03b8, or tan \u03b8 write complex numbers in polar form this...: Step 1 for the rest of this section, we have converting... 3 = - 0.8 + 0.6 the length of line and the vertical axis is the method of changing representation. The form of a complex number form which expresses the magnitude and its phase shows the. Constructor in destination class 2 ) 2 =6 rsin\u03b8 rcos\u03b8 our earlier Example representation form of a phasor -! Step 1 of two complex numbers in polar form with this kind of standard mathematical notation 7 \\. Expressing the polar coordinates in the form z = 0.5 angle ( - 45 )... Formulas developed by French mathematician Abraham de Moivre ( 1667-1754 ) free  convert complex numbers in form! The form of rectangular coordinates means expressing the polar form to polar coordinates using the relationship between and... Qii, we have 180\u00b0 30\u00b0 150\u00b0 so that 3 i 2cis150\u00b0 on this command, the angle at a. Our earlier Example _____ to rectangular form = a + b i called.  is equivalent to  5 * x  = 1, and vice-versa basic forms of numbers take the. Example \\ ( \\PageIndex { 6 } \\ ) here to know the polar form to polar form to form... Do is chart them out - 45 degrees ) 2 Comments to the polar form, remove numerators! In conversion of complex numbers to polar coordinates to polar form to form! 30\u00b0 150\u00b0 so that 3 i 2cis150\u00b0 convert it to polar coordinates using the conversion formulas prove my.! Can also be verified using the conversion formulas, if you have been converting the polar and... To find the product of two complex numbers to polar form of writing complex numbers from one form to.! Coordinates, the angle we got is not right we first need some kind of.... Product of two complex numbers to polar form to rectangular form complex numbers in polar form origin, e 1! Tan \u03b8 changing rectangular values into polar form to rectangular form using constructor in destination class / 4 or! = a + b i is called the rectangular shape of 2 with detailed. \\Pageindex { 7 } \\ ) and vice versa, with steps shown necessary to from! Ducktales Remastered Moon Theme Mp3, Copper In Minecraft Uses, New World Man Isolated Bass, Brickyard Menu Ballston Spa, String Pronunciation And Meaning, Fire Extinguisher Types, What Word Means On Or Near The Sea, Sweet Sour Song, \" />\n\n# polar form to rectangular form\n\nClick here to know the polar form and rectangular form of writing complex numbers in maths. Convert the conic r = 2 _____ to rectangular form. +2 Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by an angle symbol that looks like this: \u2220).. To use the map analogy, polar notation for the vector from New York City to San Diego would be something like \u201c2400 miles, southwest.\u201d A Phasor is a rotating vector in a complex number form which expresses the magnitude and its phase. + 2 I'm not fimular with MATLAB keywords but need to use this to prove my answers. 2 Then, start changing rectangular values into polar form as per the rules above. 4 To convert from polar form to rectangular form, first evaluate the trigonometric functions. This calculator converts between polar and rectangular coordinates. 2 It also says how far I need to go, I need to go square root of 13. There's also a graph which shows you the meaning of what you've found. If you have been given the rectangular coordinates, the best thing you can do is chart them out. CppBuzz .com Home C C++ Java Python Perl PHP SQL JavaScript Linux Selenium QT Online Test \u2630 + x Then, multiply through by See and . Polar, or phasor, forms of numbers take on the format, amplitude phase. Rewrite any trigonometric functions in terms of cos \u03b8, sin \u03b8, or tan \u03b8. y In this case the equation is manipulated to use the polar-rectangular relationship, In this case the equation is manipulated to use the polar-rectangular relationships x = r cos \u03b8, y = r sin \u03b8, and r, To use the polar-rectangular relationships we need r cos \u03b8 and r sin \u03b8. Z - is the Complex Number representing the Vector 3. x - is the Real part or the Active component 4. y - is the Imaginary part or the Reactive component 5. j - is defined by \u221a-1In the rectangular form, a complex number can be represented as a point on a two dimensional plane calle\u2026 1 + 2cos \u03b8 Example 7 Converting a Conic in Polar Form to Rectangular Form Convert the conic r = 5 \u2212 _____1 5s in \u03b8 t o rectangular form. B = | M | sin \u2061 ( \u03d5 ) {\\dis\u2026 I'm not fimular with MATLAB keywords but need to use this to prove my answers. In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. I am having trouble converting rectangular form complex numbers into polar form by writing a MATLAB script file. The polar form can also be verified using the conversion equation. Polar Form of a Complex Number. Find more Mathematics widgets in Wolfram|Alpha. Convert to Polar Coordinates (-1,1) Convert from rectangular coordinates to polar coordinates using the conversion formulas . 2, ( y To obtain these terms requires each side to be multiplied by r, Parametric Equations: Eliminating Parameters, Conversion from Polar to Rectangular Form Complex. Equations in polar form are converted into rectangular form, using the relationship between polar and rectangular coordinates.Problems with detailed solutions are presented. Phasor form conversion is the method of changing the representation form of a phasor. Find the polar form of the conic given a focus at the origin, e = 1, and directrix x = \u22121. Arrangement (in polar form) S11 S21 (S11 AND S22 IN SAME LINE) S12 S22(S12 AND S22 IN NEXT LINE) On the other hand, \"[theta, rho] = cart2pol(real(a), imag(a))\". Show Hide all comments. 4. Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. Its rectangular form is really 0 \u2013 j30 which is a vector of a magnitude of 30 plotted on the \u2212j axis. Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. This essentially makes the polar, it makes it clearer how we get there in kind of a more, I guess you could say, polar mindset, and that's why this form of the complex number, writing it this way is called rectangular form, while writing it this way is called polar form. To convert from polar form to rectangular form, first evaluate the trigonometric functions. Get the free \"Convert Complex Numbers to Polar Form\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Converting polar to rectangular coordinates means expressing the polar coordinates in the form of rectangular coordinates. To link to this Conversion from Polar to Rectangular Form Complex page, copy the following code to your site: 1. Up to this point, we have been converting the polar form to rectangular. There are two basic forms of complex number notation: polar and rectangular. It is possible\u2013and even important\u2013to convert from rectangular to the polar form. Similarly, the reactance of the inductor, j50, can be written in polar form as , and the reactance of C 2, \u2212j40, can be written in polar form as . 2 In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate. =6\u00a0rsin\u03b8\u00a0rcos\u03b8. See Example $$\\PageIndex{6}$$ and Example $$\\PageIndex{7}$$. 2 Then, multiply through by $$r$$. Make the polar-rectangular substitutions. For instance, if you are given the coordinates 50, j80, then: Step 1. 4 The polar form of the reactance of C 1, \u2212j30, is . C++ program to convert polar form to rectangular form using constructor in destination class. The phase is specified in degrees. Polar Form of a Complex Number The polar form of a complex number is another way to represent a complex number. Z = 0.5 angle( - 45 degrees ) 2 Comments. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Polar - Rectangular Coordinate Conversion Calculator. Polar/Rectangular Coordinates Calculator . To convert from rectangular form to polar form: 1. The above is the equation of a line. Equations in polar form are converted into rectangular form, using the relationship between polar and rectangular coordinates. Sign in to comment. =6xy, x x x 2 Polar coordinates are expressed as ($$r, \\theta$$) while rectangular coordinates are expressed as ($$x, y$$). y [See more on Vectors in 2-Dimensions].. We have met a similar concept to \"polar form\" before, in Polar Coordinates, part of the analytical geometry section. \u22126x=0. Sign in to answer this question. Graphs of Functions, Equations, and Algebra, The Applications of Mathematics ) Use maximum precision for calculations. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). A polar form uses the magnitude of the number as the length of line and the angle at which a number extends. We can think of complex numbers as vectors, as in our earlier example. To write complex numbers in polar form, we use the formulas and Then, See and . To convert from rectangular to polar form, we must use the following formulas: It is easier to find our angle first, which is done by plugging in our x and y into the second formula: Find the angle by taking the inverse of the function: Now find r by plugging in our angle and x \u2026 This will help you in conversion of complex numbers from one form to another. To convert a rectangular equation into polar form, remove the numerators. | M | = A 2 + B 2 {\\displaystyle |M|={\\sqrt {A^{2}+B^{2}}}} 1. \u03d5 = arctan \u2061 ( B A ) {\\displaystyle \\phi =\\arctan \\left({\\frac {B}{A}}\\right)} To convert from polar to rectangular form: A is the part of the phasor along the real axis 1. Rewrite the given equation as follows:t + \u03c0 / 4 = 0t = - \u03c0 / 4Take the tangent of both sides:tan t = tan (\u03c0 / 4) = -1Use y = R sin t and x = R cos t to write:tan t = sin t / cos t = R sin t / R cos t = y / xHence:y / x = -1y = - xThe above is the equation of a line. + Where: 2. Finding Products and Quotients of Complex Numbers in Polar Form. c) Evaluate 324. =6yx, x r Let: 1 = \u22124 - 7, 2 = 4 \u2220152 \u00b0 and 3 = - 0.8 + 0.6 . by M. Bourne. A = | M | cos \u2061 ( \u03d5 ) {\\displaystyle A=|M|\\cos \\left(\\phi \\right)} B is the part of the phasor along the imaginary axis 1. Based on this command, the angle we got is not right. For example the number 7 \u2220 40\u00b0. In the last tutorial about Phasors, we saw that a complex number is represented by a real part and an imaginary part that takes the generalised form of: 1. In radian mode, we have 3 i 2cis 5 6 Polar to Rectangular Online Calculator. a) Give the polar form of 1 with the detailed calculations. To find the product of two complex numbers, multiply the two moduli and add the two angles. Displaying top 8 worksheets found for - Converting Rectangular Form To Polar Form. When working with phasors it is often necessary to convert between rectangular and polar form. Solution x 3 and y 1 so that r 3 2 12 2 and tan 1 3 3 3 Here the reference angle and for is 30\u00b0. Z = 4 + j 5 0 Comments. 2 Sinusoidal signals with different magnitudes and phases, but the same frequency can be plotted on a phasor diagram; in a phasor diagram, the angular velocity (w) will be the same for all signals. Show Hide all comments. +2 Rectangular forms of numbers can be converted into their polar form equivalents by the formula, Polar amplitude= \u221a x 2 + y 2 , where x and y represent the real and imaginary numbers of the expression in rectangular form. Z = 0.5 angle( pi / 4 ) or. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. how to represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number, Common Core High School: Number & Quantity, HSN-CN.B.4 Rectangular to polar. For background information on what's going on, and more explanation, see the previous pages, y Replace and with the actual values. The form z = a + b i is called the rectangular coordinate form of a complex number. b) Give the rectangular shape of 2 with the detailed calculations. Polar Form of a Complex Number. Give the answers rounded to 2 decimal places. To find the product of two complex numbers, multiply the \u2026 Since the complex number is in QII, we have 180\u00b0 30\u00b0 150\u00b0 So that 3 i 2cis150\u00b0. Solution ) Home > Math > Pre Calculus > Conversion from Polar to Rectangular Form Complex If a polar equation is written such that it contains terms that appear in the polar-rectangular relationships (see below), conversion from a polar equation to a rectangular equation is a simple matter of substitution. Keep solving until you isolate the variable r. Powered by Create your own unique website with customizable templates. Convert a Complex Number to Polar Form Description Obtain the polar form of a complex number . y r Polar to Rectangular x rcos y rsin The polar form r cos isin is sometimes abbreviated rcis Example Convert 3 i to polar form. Show Instructions. However, I would like to convert it to polar form with this kind of arrangements. Problems with detailed solutions are presented.In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate.The relationships between the rectangualr (x,y) and polar (R,t) coordinates of a points are given byR 2 = x 2 + y 2 \u00a0 \u00a0 \u00a0 y = R sin t \u00a0 \u00a0 \u00a0 x = R cos t. Expand the left side of the given equation.R(-2 sin t + 3 cos t) = 2-2 R sin t + 3 R cos t = 2eval(ez_write_tag([[580,400],'analyzemath_com-box-4','ezslot_7',260,'0','0']));Use y = R sin t and x = R cos t into the given equation to rewrite as follows:-2 y + 3 x = 2 2 2 How to convert between Polar and Rectangular form ( in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Convert Equation from Rectangular to Polar Form, Convert Polar to Rectangular Coordinates - Calculator, Free Geometry Tutorials, Problems and Interactive Applets, Convert Rectangular to Polar Coordinates - Calculator. 4 The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. 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In conversion of complex numbers to polar coordinates to polar form to form! 30\u00b0 150\u00b0 so that 3 i 2cis150\u00b0 convert it to polar coordinates using the conversion formulas prove my.! Can also be verified using the conversion formulas, if you have been converting the polar and... To find the product of two complex numbers to polar form of writing complex numbers from one form to.! Coordinates, the angle we got is not right we first need some kind of.... Product of two complex numbers to polar form to rectangular form complex numbers in polar form origin, e 1! Tan \u03b8 changing rectangular values into polar form to rectangular form using constructor in destination class / 4 or! = a + b i is called the rectangular shape of 2 with detailed. \\Pageindex { 7 } \\ ) and vice versa, with steps shown necessary to from!", "date": "2021-04-15 10:10:18", "meta": {"domain": "beautylab.gr", "url": "https://beautylab.gr/pcbun57/polar-form-to-rectangular-form-ff867f", "openwebmath_score": 0.8774170875549316, "openwebmath_perplexity": 851.5356286069533, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561694652215, "lm_q2_score": 0.9005297801113612, "lm_q1q2_score": 0.8724838332480516}}
{"url": "https://mathhelpboards.com/threads/understanding-horizontal-asymptotes.8384/", "text": "# Understanding Horizontal Asymptotes\n\n#### confusedatmath\n\n##### New member\nI'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?\n\n#### MarkFL\n\nStaff member\nWhat happens to a fraction when the denominator gets larger and larger without bound?\n\n#### confusedatmath\n\n##### New member\nthe fraction gets smaller, but dont we end up adding that to the value of c??\n\nwell if a/(x-b)^2 +c\n\nlet a=1 b=1 c=2\n\nso we get\n\n1/(x-1)^2 +2\n\nlets make x =2\n\nwe get\n\n1/1 +2 .... thats 3.... wait am i doing something wrong?\n\n- - - Updated - - -\n\noh wait is because in the above a is negative.. so\n\nit would be -1+2 < 2 ....\n\n#### MarkFL\n\nStaff member\nThe fraction goes to zero as the denominator goes to infinity, and so that's why the horizontal asymptote is $y=c$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nLet's make this really interesting by choosing some suitably large values for $x$.\n\nLet's see what happens at $x = 1,000,000$, when $a = b = -1,c = 2$.\n\nThen $f(x) = \\dfrac{-1}{(999,999)^2} + 2$\n\n$= \\dfrac{-1}{999,998,000,001} + 2$\n\n$\\sim 1.999999999998999998$\n\nAt $x = -1,000,000$, we get:\n\n$f(x) = \\dfrac{-1}{(-1,000,001)^2} + 2$\n\n$= \\dfrac{-1}{1,000,002,000,001} + 2$\n\n$\\sim 1.999999999999000002$\n\nBoth of these numbers are really close to 2, right?\n\nIn general, we see that:\n\n$\\displaystyle \\lim_{x \\to \\infty} \\left(\\frac{a}{(x + b)^2} + c\\right)$\n\n$\\displaystyle = \\lim_{x \\to \\infty} \\frac{a}{(x + b)^2} + \\lim_{x \\to \\infty} c$\n\n$\\displaystyle = \\lim_{x \\to \\infty} \\frac{a}{(x+b)^2} + c$\n\n$\\displaystyle = (a)\\left(\\lim_{x \\to \\infty}\\frac{1}{x+b}\\right)^2 + c$\n\n$\\displaystyle = (a)(0)^2 + c = 0 + c = c$\n\nSimilar reasoning holds to show that:\n\n$\\displaystyle \\lim_{x \\to -\\infty} \\left(\\frac{a}{(x + b)^2} + c\\right) = c$\n\nas well.\n\n(If you haven't been formally introduced to limits yet, all you need to know for this is the following (which hold under \"suitably nice conditions\" which are the case here):\n\n1) the limit of a sum is the sum of the limits of each term in the sum\n2) the limit of a product is the product of the limits of each factor in the product\n3) if M(x) is a function that gets \"infinitely big\" as x does, then:\n\n$$\\lim_{x \\to \\infty} \\frac{1}{M(x)} = 0$$\n\n-by \"get infinitely big as x does\" I mean that for ANY positive integer $K$, there is always some positive integer $N$, so that if we have $x > N$, then $|M(x)| > K$ (typically, the integer $N$ will depend on $K$, bigger $K$'s usually need bigger $N$'s).).\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nI'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?\n\nView attachment 1838\nThe short answer, if \\displaystyle \\begin{align*} y = c \\end{align*} then you end up with \\displaystyle \\begin{align*} 0 = \\frac{a}{(x + b)^2} \\end{align*}. Is that possible to solve for x?\n\nThe long answer, consider the inverse relation. You should find that c becomes a vertical asymptote.\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nthe fraction gets smaller, but dont we end up adding that to the value of c??\n\nwell if a/(x-b)^2 +c\n\nlet a=1 b=1 c=2\n\nso we get\n\n1/(x-1)^2 +2\n\nlets make x =2\n\nwe get\n\n1/1 +2 .... thats 3.... wait am i doing something wrong?\n\n- - - Updated - - -\n\noh wait is because in the above a is negative.. so\n\nit would be -1+2 < 2 ....\nThe question was \"what happens when x is large?\" \"2\" is NOT large!\n\nIf x= 1000000, then 1/(1+ 1000000)+3= 3.000009999900000999990000099999 and for", "date": "2020-11-23 17:07:25", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/understanding-horizontal-asymptotes.8384/", "openwebmath_score": 0.9439474940299988, "openwebmath_perplexity": 1517.4736035179951, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429580381723, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8724263983689885}}
{"url": "http://mathhelpforum.com/algebra/281823-area-square-circle.html", "text": "# Thread: Area of Square in Circle\n\n1. ## Area of Square in Circle\n\nThere is a shaded square inside a circle represented by the equation x^2 + y^2 = 9. Find the area of the square in the circle.\n\nI seek the first-two steps.\n\n2. ## Re: Area of Square in Circle\n\nI am assuming the square is inscribed within the circle, and so the diagonal of the square is equal to the diameter of the circle.\n\nLet $\\displaystyle D=2r$ by the diameter of the circle. The area $\\displaystyle A$ of the square is:\n\n$\\displaystyle A=s^2$\n\nAnd we know:\n\n$\\displaystyle 2r=\\sqrt{2}s\\implies s=\\sqrt{2}r\\implies A=2r^2$\n\nCan you proceed?\n\n3. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nI am assuming the square is inscribed within the circle, and so the diagonal of the square is equal to the diameter of the circle.\n\nLet $\\displaystyle D=2r$ by the diameter of the circle. The area $\\displaystyle A$ of the square is:\n\n$\\displaystyle A=s^2$\n\nAnd we know:\n\n$\\displaystyle 2r=\\sqrt{2}s\\implies s=\\sqrt{2}r\\implies A=2r^2$\n\nCan you proceed?\nA = 2r^2\n\nI know that x^2 + y^2 = r^2.\n\nSo, r^2 = 9 leading to an r value of 3.\n\nA = 2r^2\n\nA = 2(3)^2\n\nA = 2(9)\n\nA = 18 units^2\n\n4. ## Re: Area of Square in Circle\n\nWhat would the area of a square be that circumscribes a circle of radius $\\displaystyle r$? (This is there the circle is inside the square)\n\n5. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nWhat would the area of a square be that circumscribes a circle of radius $\\displaystyle r$? (This is there the circle is inside the square)\nIf I know r, then A = 2r^2 yields the area of a square inside a circle.\n\n6. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nIf I know r, then A = 2r^2 yields the area of a square inside a circle.\nRethink that! Are you missing some brackets?\n\n7. ## Re: Area of Square in Circle\n\nOriginally Posted by Debsta\nRethink that! Are you missing some brackets?\nA = 2(r)^2\n\n8. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nIf I know r, then A = 2r^2 yields the area of a square inside a circle.\nIf the circle is inscribed within the square, then the side length of the square will be equal to the diameter of the circle, thus:\n\n$\\displaystyle A=(2r)^2=4r^2$\n\nNow, given the result for the original problem you posted, we now know:\n\n$\\displaystyle 2r^2<\\pi r^2<4r^2$\n\nOr:\n\n$\\displaystyle 2<\\pi<4$\n\nHow could we improve these bounds on $\\displaystyle \\pi$?\n\n9. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nIf the circle is inscribed within the square, then the side length of the square will be equal to the diameter of the circle, thus:\n\n$\\displaystyle A=(2r)^2=4r^2$\n\nNow, given the result for the original problem you posted, we now know:\n\n$\\displaystyle 2r^2<\\pi r^2<4r^2$\n\nOr:\n\n$\\displaystyle 2<\\pi<4$\n\nHow could we improve these bounds on $\\displaystyle \\pi$?\nSorry but I don't follow? Is my answer to the original question right?\n\n10. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nSorry but I don't follow? Is my answer to the original question right?\nYes, you answered the original problem correctly, and is why I thanked your post.\n\nConsider the following diagram:\n\nCan you see that the side length of the larger square is equal to the diameter of the circle inscribed within?\n\n11. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nYes, you answered the original problem correctly, and is why I thanked your post.\n\nConsider the following diagram:\n\nCan you see that the side length of the larger square is equal to the diameter of the circle inscribed within?\nYes, I can what you are saying here.\n\n12. ## Re: Area of Square in Circle\n\nDo you see how the rest of what I posted in post #8 then follows?\n\n13. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nDo you see how the rest of what I posted in post #8 then follows?\nYes.\n\n14. ## Re: Area of Square in Circle\n\nOriginally Posted by harpazo\nYes.\nOkay, then can you think of a way we could improve our bounds on the value of pi?\n\n15. ## Re: Area of Square in Circle\n\nOriginally Posted by MarkFL\nOkay, then can you think of a way we could improve our bounds on the value of pi?\nImprove our bounds???\n\nPage 1 of 2 12 Last", "date": "2019-05-22 09:02:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/281823-area-square-circle.html", "openwebmath_score": 0.9238041639328003, "openwebmath_perplexity": 656.7647182812361, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985042916041382, "lm_q2_score": 0.8856314768368161, "lm_q1q2_score": 0.872385012481373}}
{"url": "https://folk.ntnu.no/leifh/teaching/tkt4140/._main014.html", "text": "\n## 2.8.1 Example: Numerical error as a function of $$\\Delta t$$\n\nIn this example we will assess how the error of our implementation of the Euler method depends on the time step $$\\Delta t$$ in a systematic manner. We will solve a problem with an analytical solution in a loop, and for each new solution we do the following:\n\u2022 Divide the time step by two (or double the number of time steps)\n\u2022 Compute the error\n\u2022 Plot the error\nEuler's method is a first order method and we expect the error to be $$O(h)=O(\\Delta t)$$. Consequently if the time-step is divided by two, the error should also be divided by two. As errors normally are small values and are expected to be smaller and smaller for decreasing time steps, we normally do not plot the error itself, but rather the logarithm of the absolute value of the error. The latter we do due to the fact that we are only interested in the order of magnitude of the error, whereas errors may be both positive and negative. As the initial value is always correct we discard the first error at time zero to avoid problems with the logarithm of zero in log_error = np.log2(abs_error[1:]).\n\n# coding: utf-8\n# src-ch1/Euler_timestep_ctrl.py;DragCoefficientGeneric.py @ git@lrhgit/tkt4140/src/src-ch1/DragCoefficientGeneric.py;\nfrom DragCoefficientGeneric import cd_sphere\nfrom matplotlib.pyplot import *\nimport numpy as np\n\n# change some default values to make plots more readable\nLNWDT=2; FNT=11\nrcParams['lines.linewidth'] = LNWDT; rcParams['font.size'] = FNT\n\ng = 9.81      # Gravity m/s^2\nd = 41.0e-3   # Diameter of the sphere\nrho_f = 1.22  # Density of fluid [kg/m^3]\nrho_s = 1275  # Density of sphere [kg/m^3]\nnu = 1.5e-5   # Kinematical viscosity [m^2/s]\nCD = 0.4      # Constant drag coefficient\n\ndef f(z, t):\n\"\"\"2x2 system for sphere with constant drag.\"\"\"\nzout = np.zeros_like(z)\nalpha = 3.0*rho_f/(4.0*rho_s*d)*CD\nzout[:] = [z[1], g - alpha*z[1]**2]\nreturn zout\n\n# define euler scheme\ndef euler(func,z0, time):\n\"\"\"The Euler scheme for solution of systems of ODEs.\nz0 is a vector for the initial conditions,\nthe right hand side of the system is represented by func which returns\na vector with the same size as z0 .\"\"\"\n\nz = np.zeros((np.size(time),np.size(z0)))\nz[0,:] = z0\n\nfor i in range(len(time)-1):\ndt = time[i+1]-time[i]\nz[i+1,:]=z[i,:] + np.asarray(func(z[i,:],time[i]))*dt\nreturn z\n\ndef v_taylor(t):\n#    z = np.zeros_like(t)\nv = np.zeros_like(t)\nalpha = 3.0*rho_f/(4.0*rho_s*d)*CD\nv=g*t*(1-alpha*g*t**2)\nreturn v\n\n# main program starts here\n\nT = 10  # end of simulation\nN = 10  # no of time steps\n\nz0=np.zeros(2)\nz0[0] = 2.0\n\n# Prms for the analytical solution\nk1 = np.sqrt(g*4*rho_s*d/(3*rho_f*CD))\nk2 = np.sqrt(3*rho_f*g*CD/(4*rho_s*d))\n\nNdts = 4  # Number of times to divide the dt by 2\nlegends=[]\nerror_diff = []\n\nfor i in range(Ndts+1):\ntime = np.linspace(0, T, N+1)\nze = euler(f, z0, time)     # compute response with constant CD using Euler's method\nv_a = k1*np.tanh(k2*time)   # compute response with constant CD using analytical solution\nabs_error=np.abs(ze[:,1] - v_a)\nlog_error = np.log2(abs_error[1:])\nmax_log_error = np.max(log_error)\n\nplot(time[1:], log_error)\nlegends.append('Euler scheme: N ' + str(N) + ' timesteps' )\nN*=2\nif i > 0:\nerror_diff.append(previous_max_log_err-max_log_error)\n\nprevious_max_log_err = max_log_error\n\nprint('Approximate order of scheme n =', np.mean(error_diff))\nprint('Approximate error reuduction by dt=dt/2:', 1/2**(np.mean(error_diff)))\n\n# plot analytical solution\n# plot(time,v_a)\n# legends.append('analytical')\n\n# fix plot\nlegend(legends, loc='best', frameon=False)\nxlabel('Time [s]')\n#ylabel('Velocity [m/s]')\nylabel('log2-error')\n#savefig('example_euler_timestep_study.png', transparent=True)\nshow()\n\n\nThe plot resulting from the code above is shown in Figure (11). The difference or distance between the curves seems to be rather constant after an initial transient. As we have plotted the logarithm of the absolute value of the error $$\\epsilon_i$$, the difference $$d_{i+1}$$ between two curves is $$d_{i+1}= \\log_2 \\epsilon_{i}-\\log_2 \\epsilon_{i+1} = \\displaystyle \\log_2 \\frac{\\epsilon_{i}}{\\epsilon_{i+1}}$$. A rough visual inspection of Figure (11) yields $$d_{i+1} \\approx 1.0$$, from which we may deduce the apparent order of the scheme: $$$$n = \\log_2 \\frac{\\epsilon_{i}}{\\epsilon_{i+1}} \\approx 1 \\Rightarrow \\epsilon_{i+1} \\approx 0.488483620\\, \\epsilon_{i} \\tag{2.75}$$$$\n\nThe print statement returns 'n=1.0336179048' and '0.488483620794', thus we see that the error is reduced even slightly more than the theoretically expected value for a first order scheme, i.e. $$\\Delta t_{i+1} = \\Delta t_{i}/2$$ yields $$\\epsilon_{i+1} \\approx\\epsilon_{i}/2$$.\n\nFigure 11: Plots for the logarithmic errors for a falling sphere with constant drag. The timestep $$\\Delta t$$ is reduced by a factor two from one curve to the one immediately below.", "date": "2021-09-26 10:49:25", "meta": {"domain": "ntnu.no", "url": "https://folk.ntnu.no/leifh/teaching/tkt4140/._main014.html", "openwebmath_score": 0.819695770740509, "openwebmath_perplexity": 1667.4848279565244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429138459387, "lm_q2_score": 0.8856314647623016, "lm_q1q2_score": 0.8723849986431044}}
{"url": "https://math.stackexchange.com/questions/2310558/two-colouring-of-a-6-by-6-grid-without-a-monochromatic-rectangle", "text": "# Two colouring of a 6 by 6 grid without a monochromatic rectangle\n\nThis question Monochromatic Rectangle of a 2-Colored 8 by 8 Lattice Grid shows that any two colouring of a 7 by 7 grid must have a rectangle whose vertices are all the same colour. Note that we are only considering rectangles whose sides are parallel to the sides of the grid.\n\nCan the same result be shown to hold for a 6 by 6 grid? If not, what is an example of a colouring of a 6 by 6 grid that does not contain a monochromatic rectangle?\n\nFurthermore, if it turns out that every colouring of a 6 by 6 grid contains a monochromatic rectangle, what is the largest $n \\in \\mathbb{N}$ such that an $n$ by $n$ grid can be coloured such that no rectangles are monochromatic?\n\n\u2022 Do the sides of the rectangle have to be parallel to the grid lines? \u2013\u00a0Ankoganit Jun 5 '17 at 11:39\n\u2022 @Ankoganit I'm looking at the case when the sides do have to be parallel. \u2013\u00a0michaelhowes Jun 5 '17 at 11:41\n\nFor six by six it is quite easy.\n\nCall the colours red and blue. Look at the top row. There will be three points the same colour - say red. Concentrate on those columns. If in any row below there are two out of three points red you have a rectangle. But there are only four possible patterns of three points with at most one red, so two of the five rows must have the same pattern and there will be a blue rectangle.\n\nSince one of the possible patterns of three with at most one red is all blue, and if this is present there will be a blue rectangle anyway, this can be adapted to five by five (which must have three the same in the top row).\n\nCan you find a four by four arrangement with no rectangle?\n\n\u2022 Thank you very much for your answer. I have found such a four by four arrangement. \u2013\u00a0michaelhowes Jun 5 '17 at 12:34\n\nThe number of points occupied by a colour in a given column implies that a certain number of possible pairings have been produced, for possible matching to the same pair in another column. So with $\\{0,1,2,3,4,5,6\\}$ points of a single colour you have $\\{0,0,1,3,6,10,15\\}$ pairings respectively.\n\nFor a six-by-six rectangle, there are $\\binom 62 = 15$ possible pairings in a column. As soon as one colour has enough entries to accumulate $16$ pairings from its column entries, as above, there will be a matched pairing between columns giving the monochrome rectangle. Between the two colours, once total pairings reaches $31$ there will be a monochrome rectangle. Splitting the colours evenly ($3$ each) produces the lowest net pairings of $6$ per column, but clearly with six columns we have at least $36$ net pairings and must have a monochrome rectangle.\n\nFor a five-by-six rectangle, using the same arguments, we have $\\binom 52 = 10$ possible pairings per column and at least $4$ net pairings per column, reaching at least $24(>2{\\times} 10)$ net pairings by the sixth column and the certainty of a monochrome rectangle.\n\nFor a five-by-five square, one colour will have at least $13$ entries which you can show leads to at least $11$ pairings across the $5$ columns:\n\n$\\begin{array}{c|c|c} \\text{col entries} & \\text{pairings} \\\\ \\hline (5,2+,\\ldots) & 11+ \\\\ (4,4,\\ldots) & 12+ \\\\ (4,3,3,2,1) & 13 \\\\ (4,3,2,2,2) & 12 \\\\ (3,3,3,3,1) & 12 \\\\ (3,3,3,2,2) & 11 \\\\ \\end{array}$\n\nand $11$ or more pairings over the 5 columns will mean that some $2$ pairs will align, giving a monochrome rectangle.\n\nFinally a four-by-six rectangle can avoid a monochrome rectangle, guided by the above:", "date": "2020-12-02 04:22:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2310558/two-colouring-of-a-6-by-6-grid-without-a-monochromatic-rectangle", "openwebmath_score": 0.505984365940094, "openwebmath_perplexity": 187.22312882671133, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777179414275, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8723497813958908}}
{"url": "http://gmatclub.com/forum/is-quadrilateral-abcd-a-rhombus-85336.html", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 25 Jul 2016, 20:24\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Is quadrilateral ABCD a rhombus?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nJoined: 01 Apr 2008\nPosts: 898\nName: Ronak Amin\nSchools: IIM Lucknow (IPMX) - Class of 2014\nFollowers: 26\n\nKudos [?]: 530 [0], given: 18\n\n### Show Tags\n\n14 Oct 2009, 21:09\n12\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n36% (01:33) correct 64% (00:36) wrong based on 426 sessions\n\n### HideShow timer Statistics\n\n(1) Line segments AC and BD are perpendicular bisectors of each other.\n\n(2) AB = BC = CD = AD\n[Reveal] Spoiler: OA\nMath Expert\nJoined: 02 Sep 2009\nPosts: 34057\nFollowers: 6084\n\nKudos [?]: 76454 [0], given: 9975\n\n### Show Tags\n\n14 Oct 2009, 21:40\nExpert's post\n3\nThis post was\nBOOKMARKED\n\n(1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus\n\n(2) AB = BC = CD = AD --> rhombus\n\nOr am I missing something, seems pretty obvious...\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 34057\nFollowers: 6084\n\nKudos [?]: 76454 [0], given: 9975\n\n### Show Tags\n\n14 Oct 2009, 21:45\nExpert's post\n2\nThis post was\nBOOKMARKED\n\nWell:\n(1) True for square or for rhombus but every square is a rhombus, so sufficient\n(2) Again true for square or for rhombus but every square is a rhombus, so sufficient\n\nD\n_________________\nSenior Manager\nJoined: 17 May 2010\nPosts: 299\nGMAT 1: 710 Q47 V40\nFollowers: 4\n\nKudos [?]: 43 [0], given: 7\n\nIs quadrilateral ABCD a rhombus? (1) Line segments AC and BD\u00a0[#permalink]\n\n### Show Tags\n\n16 Jun 2011, 15:24\n1\nThis post was\nBOOKMARKED\n\n(1) Line segments AC and BD are perpendicular bisectors of each other.\n\n(2) AB = BC = CD = AD\n\nSolve!\n_________________\n\nIf you like my post, consider giving me KUDOS!\n\nDirector\nJoined: 01 Feb 2011\nPosts: 757\nFollowers: 14\n\nKudos [?]: 97 [0], given: 42\n\n### Show Tags\n\n17 Jun 2011, 18:45\n1. Sufficient\n\ndiagnols are perpendicular bisectors. can only happen in a square or rhombus. Square is a kind of rhombus.\n\n2. Sufficient\n\nAs we know its a square or rhombus. Square is a kind of rhombus.\n\nIntern\nJoined: 16 Aug 2011\nPosts: 10\nConcentration: Finance, Entrepreneurship\nGPA: 3.51\nFollowers: 0\n\nKudos [?]: 42 [0], given: 3\n\n### Show Tags\n\n10 Nov 2011, 01:16\nHi guys\nI'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:\n\n(1) Line segments AC and BD are perpendicular bisectors of each other\n(2) AB = BC = CD = AD\n\nThe official answer is as follows:\nStatement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus\nStatement 2 - A quadrilateral with four equal sides is by definition a rhombus\n\nWhat I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?\n_________________\n\n+1 Kudos is a great way of saying Thank you!! :D\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 6742\nLocation: Pune, India\nFollowers: 1873\n\nKudos [?]: 11525 [1] , given: 219\n\n### Show Tags\n\n10 Nov 2011, 03:54\n1\nKUDOS\nExpert's post\n2\nThis post was\nBOOKMARKED\nmk87 wrote:\nHi guys\nI'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:\n\n(1) Line segments AC and BD are perpendicular bisectors of each other\n(2) AB = BC = CD = AD\n\nThe official answer is as follows:\nStatement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus\nStatement 2 - A quadrilateral with four equal sides is by definition a rhombus\n\nWhat I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?\n\nStatement 1 says the diagonals are perpendicular bisectors of each other which means that they intersect each other at mid points and form 90 degree angles. It is not possible that they are in the shape of an asymmetric kite. Figure 1 is not possible. Only figure 2 is possible. In figure 2, all sides will be equal.\nAttachment:\n\nQues4.jpg [ 10.18 KiB | Viewed 4352 times ]\n\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 13 Aug 2012 Posts: 464 Concentration: Marketing, Finance GMAT 1: Q V0 GPA: 3.23 Followers: 24 Kudos [?]: 372 [0], given: 11 Re: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD [#permalink] ### Show Tags 29 Jan 2013, 07:59 gijoedude wrote: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. (2) AB = BC = CD = AD A rhombus is a quadrilateral with all its sides equal to each other. 1. Imagine that the point where the bisectors meet is called X. Then, BX = DX and AX = CX. Using the Pythagorean Theorem you could imagine getting the hypotenuse of sides BX and CX. Since AX = CX, then the hypotenuse of BX and AX is also as long as the previous. As you could imagine, all the hypotenuse formed are equal, thus forming equal 4 sides. Hence, a rhombus. SUFFICIENT. 2. All sides are equal. SUFFICIENT. Answer: D _________________ Impossible is nothing to God. Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 51 Kudos [?]: 496 [0], given: 355 Re: rhombus? [#permalink] ### Show Tags 13 Oct 2013, 16:10 Bunuel wrote: Thought about this again: D it is. Well: (1) True for square or for rhombus but every square is a rhombus, so sufficient (2) Again true for square or for rhombus but every square is a rhombus, so sufficient D Hi Bunuel When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus? And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus? Is that the correct line of reasoning? Thanks so much in advance Cheers J Math Expert Joined: 02 Sep 2009 Posts: 34057 Followers: 6084 Kudos [?]: 76454 [0], given: 9975 Re: rhombus? [#permalink] ### Show Tags 13 Oct 2013, 16:20 Expert's post jlgdr wrote: Bunuel wrote: Thought about this again: D it is. Well: (1) True for square or for rhombus but every square is a rhombus, so sufficient (2) Again true for square or for rhombus but every square is a rhombus, so sufficient D Hi Bunuel When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus? And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus? Is that the correct line of reasoning? Thanks so much in advance Cheers J For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90\u00b0. Thus, \"line segments AC and BD are perpendicular bisectors of each other\" means that AC cuts BD into two equal parts at 90\u00b0 and BD cuts AC into two equal parts at 90\u00b0. For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus). Hope it helps. _________________ Senior Manager Joined: 13 May 2013 Posts: 472 Followers: 3 Kudos [?]: 136 [0], given: 134 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 06 Dec 2013, 11:10 Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. But doesn't a rhombus also allow for two perpendicular bisectors? A square has two perpendicular bisectors but a square is not a rhombus (because a rhombus does not have all four angles = 90. Oh, but wait, a rhombus is a square in the same way a rectangle is a square but not the other way around. Tricky. (2) AB = BC = CD = AD I guess the same logic applies above, ABCD could be a square or rhombus but regardless of which one it is its still a rhombus. If the stem asked if ABCD was a square, then this would be insufficient. Manager Joined: 22 Feb 2009 Posts: 229 Followers: 5 Kudos [?]: 100 [0], given: 148 Re: Data Sufficiency - doubts [#permalink] ### Show Tags 08 Aug 2014, 21:36 kritim22 wrote: Hi, can someone help me with this question please: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. (2) AB = BC = CD = AD The answer it says is D- each statement is enough on its own. Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus? Let say AC and BD meet at O. We have OD= OB, OA= OC. If you use pithagoras theorem, you can easily see that AD= DC=BC=AB. For example, AD^2= OA^2 + OD^2 So (1) and (2) basically say the same thing. The definition of rhombus is a quadrilateral whose four sides have the same length. So D is the answer. _________________ ......................................................................... +1 Kudos please, if you like my post Manager Joined: 21 Sep 2012 Posts: 219 Location: United States Concentration: Finance, Economics Schools: CBS '17 GPA: 4 WE: General Management (Consumer Products) Followers: 3 Kudos [?]: 151 [0], given: 31 Re: Data Sufficiency - doubts [#permalink] ### Show Tags 09 Aug 2014, 00:56 kritim22 wrote: Hi, can someone help me with this question please: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. (2) AB = BC = CD = AD The answer it says is D- each statement is enough on its own. Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus? A square is a special case of rhombus. It has all angles of 90 degrees. so proving that quadrilateral ABCD is a square is sufficient to say that it is a rhombus. Director Status: Verbal Forum Moderator Joined: 17 Apr 2013 Posts: 635 Location: India GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49 GPA: 3.3 Followers: 48 Kudos [?]: 310 [0], given: 297 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 05 Nov 2014, 03:31 Bunuel wrote: Answer D. (1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus (2) AB = BC = CD = AD --> rhombus Or am I missing something, seems pretty obvious... But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other. I need Your though, am confused. Diagonals of rectangles also bisect each other. _________________ Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html Math Expert Joined: 02 Sep 2009 Posts: 34057 Followers: 6084 Kudos [?]: 76454 [0], given: 9975 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 05 Nov 2014, 05:09 Expert's post honchos wrote: Bunuel wrote: Answer D. (1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus (2) AB = BC = CD = AD --> rhombus Or am I missing something, seems pretty obvious... But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other. I need Your though, am confused. Diagonals of rectangles also bisect each other. For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90\u00b0. Thus, \"line segments AC and BD are perpendicular bisectors of each other\" means that AC cuts BD into two equal parts at 90\u00b0 and BD cuts AC into two equal parts at 90\u00b0. Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90\u00b0. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus. Hope it's clear. _________________ Manager Joined: 10 Sep 2014 Posts: 75 Followers: 0 Kudos [?]: 54 [0], given: 102 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 06 Nov 2014, 02:17 Bunuel, For (1), can you consider the case of a kite? Bunuel wrote: honchos wrote: Bunuel wrote: Answer D. (1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus (2) AB = BC = CD = AD --> rhombus Or am I missing something, seems pretty obvious... But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other. I need Your though, am confused. Diagonals of rectangles also bisect each other. For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90\u00b0. Thus, \"line segments AC and BD are perpendicular bisectors of each other\" means that AC cuts BD into two equal parts at 90\u00b0 and BD cuts AC into two equal parts at 90\u00b0. Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90\u00b0. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus. Hope it's clear. _________________ Press KUDOs if you find my explanation helpful Math Expert Joined: 02 Sep 2009 Posts: 34057 Followers: 6084 Kudos [?]: 76454 [0], given: 9975 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 06 Nov 2014, 07:22 Expert's post TARGET730 wrote: Bunuel, For (1), can you consider the case of a kite? Yes. In the special case where all 4 sides are the same length, the kite satisfies the definition of a rhombus. _________________ Senior Manager Joined: 10 Mar 2013 Posts: 290 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Followers: 5 Kudos [?]: 80 [0], given: 2404 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 10 Mar 2015, 20:32 This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6742 Location: Pune, India Followers: 1873 Kudos [?]: 11525 [2] , given: 219 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 10 Mar 2015, 22:58 2 This post received KUDOS Expert's post TooLong150 wrote: This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus. Actually, the question is fine. A rhombus is a quadrilateral all of whose sides are of the same length - that's all. You could have a rhombus which also has all angles 90 which makes it a square or a rhombus in the shape of a kite. But nevertheless, if it is a quadrilateral and has all sides equal, it IS A RHOMBUS. (1) Line segments AC and BD are perpendicular bisectors of each other. Make 2 lines - a vertical and a horizontal - which are perpendicular bisectors of each other. Make them in any way of any length - just that they should be perpendicular bisectors of each other. When you join the end points, you will get all sides equal. Think of it this way - each side you get will be a hypotenuse of a right triangle. The legs of the right triangle will have the same pair of lengths in all 4 cases. So AB = BC = CD = AD. So ABCD must be a rhombus. (2) AB = BC = CD = AD This statement directly tells you that all sides are equal so ABCD must be a rhombus. Answer (D) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nIntern\nJoined: 22 Aug 2015\nPosts: 1\nFollowers: 0\n\nKudos [?]: 0 [0], given: 1\n\nRe: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD\u00a0[#permalink]\n\n### Show Tags\n\n22 Aug 2015, 11:10\nWhile somewhat unusual, isn't the fact that it asks if the quadrilateral is a rhombus sufficient in itself? Aren't all quadrilaterals also rhombuses and vise versa because they have 4 sides?\nRe: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD \u00a0 [#permalink] 22 Aug 2015, 11:10\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 24 posts ]\n\nSimilar topics Replies Last post\nSimilar\nTopics:\n1 Is quadrilateral ABCD a rhombus? 2 20 Dec 2014, 02:30\n3 Is quadrilateral q a rhombus? 4 29 Nov 2012, 12:06\n3 ABCD is a quadrilateral. A rhombus is a quadrilateral whose 16 29 Dec 2010, 08:08\n5 Quadrilateral ABCD is a rhombus and points C, D, and E are 3 09 Mar 2010, 20:10\n54 Quadrilateral ABCD is a rhombus and points C, D, and E are 28 09 Aug 2009, 10:03\nDisplay posts from previous: Sort by", "date": "2016-07-26 03:24:12", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/is-quadrilateral-abcd-a-rhombus-85336.html", "openwebmath_score": 0.5108898282051086, "openwebmath_perplexity": 3338.523511084, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8723473713594992}}
{"url": "https://math.stackexchange.com/questions/326702/a-group-such-that-am-bm-bm-am-and-an-bn-bn-an-m-n-coprime", "text": "# A group such that $a^m b^m = b^m a^m$ and $a^n b^n = b^n a^n$ ($m$, $n$ coprime) is abelian?\n\nLet $(G,.)$ be a group and $m,n \\in\\mathbb Z$ such that $\\gcd(m,n)=1$. Assume that $$\\forall a,b \\in G, \\,a^mb^m=b^ma^m,$$ $$\\forall a,b \\in G, \\, a^nb^n=b^na^n.$$\n\nThen how prove $G$ is an abelian group?\n\nSome context: Some of these commutation relations often imply that $G$ is abelian, for example if $(ab)^i = a^i b^i$ for three consecutive integers $i$ then $G$ is abelian, or if $g^2 = e$ for all $g$ then $G$ is abelian. This looks like another example of this phenomenon, but the same techniques do not apply.\n\n\u2022 from which text did you get this problem ? \u2013\u00a0Fawzy Hegab Mar 10 '13 at 19:32\n\u2022 also , is m,n are fixed or you mean , for all m,n in Z such that gcd(m,n)=1 .... ??? \u2013\u00a0Fawzy Hegab Mar 10 '13 at 19:39\n\u2022 @ MrWhymy:m,n are fixed integer number \u2013\u00a0M.H Mar 10 '13 at 19:46\n\u2022 it's nice problem , and the solution which is providen in the answer below is great :) \u2013\u00a0Fawzy Hegab Mar 10 '13 at 19:53\n\nLet $M \\subset G$ be the subgroup generated by all $m$-th powers and let $N \\subset G$ be the subgroup generated by all $n$-th powers. These subgroups are clearly abelian normal subgroups. Since $m$ and $n$ are coprime, $G = MN$, and hence $M \\cap N$ is contained in the center $Z(G)$ of $G$. To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] \\subset M \\cap N$ (since $M$ and $N$ are normal subgroups). Let $a \\in M$ and $b \\in N$. Then $[a, b] = a^{\u22121}b^{\u22121}ab \\in M \\cap N$. Hence $[a, b] = z$ with $z \\in Z(G)$. Hence $b^{\u22121}ab = za$, whence $b^{\u22121}a^nb=z^na^n$. Since $a^n \\in N$ it commutes with $b$, so $z^n=1$. Similarly $z^m=1$. Since $m$ and $n$ are relatively prime, we conclude $z=1$.\n\n\u2022 Dear Nicky! Let me ask sime questions: 1) Why $G=MN$? 2) Why $M\\cap N$ is contained in the center $Z(G)$? \u2013\u00a0ZFR May 1 '18 at 14:21\n\u2022 (1) $gcd(m,n)=1$ whence you can find integers $k$ and $l$ with $1=km+ln$. So if $g \\in G$ then $g=g^1=(g^k)^m \\cdot (g^l)^n \\in MN$. (2) Because $G=MN$ and $M$ and $N$ are abelian, any element from $M \\cap N$ commutes with any element from $G$. Hope it is clear now. \u2013\u00a0Nicky Hekster May 1 '18 at 15:33\n\u2022 Dear Nicky! Thanks for answer! I got your solution and your approach is very simple and nice! +1 \u2013\u00a0ZFR May 1 '18 at 19:53\n\u2022 Thank you very much! \u2013\u00a0Nicky Hekster May 1 '18 at 20:32\n\nWrite $mk+nl=1$ for some $k,l\\in\\mathbb Z$.\n\nWe have $$ab=a^{mk+nl}b^{mk+nl}=a^{mk}(a^{nl}b^{mk})b^{nl}\\stackrel{(*)}=a^{mk}(b^{mk}a^{nl})b^{nl}=(a^{mk}b^{mk})(a^{nl}b^{nl})=(b^{mk}a^{mk})(b^{nl}a^{nl})=b^{mk}(a^{mk}b^{nl})a^{nl}\\stackrel{(**)}=b^{mk}(b^{nl}a^{mk})a^{nl}=(b^{mk}b^{nl})(a^{mk}a^{nl})=b^{mk+nl}a^{mk+nl}=ba,$$ where we used $$a^{nl}b^{mk}=b^{mk}a^{nl}\\qquad(*)$$ and $$a^{mk}b^{nl}=b^{nl}a^{mk}\\qquad(**)$$\n\nLet's prove $(*)$ and $(**)$:\n\n$$(a^mb^n)^{mk}=a^m(b^na^m)^{mk-1}b^n=a^m(b^na^m)^{mk}a^{-m}\\Rightarrow(a^mb^n)^{mk}a^m=a^m(b^na^m)^{mk}\\Rightarrow a^m(a^mb^n)^{mk}=a^m(b^na^m)^{mk},$$ hence $$(a^mb^n)^{mk}=(b^na^m)^{mk}.\\qquad(\\sharp)$$\n\nSimilarly we get $$(a^mb^n)^{nl}=(b^na^m)^{nl}.\\qquad(\\sharp\\sharp)$$\n\nMultiplying $(\\sharp)$ and $(\\sharp\\sharp)$ we obtain $a^mb^n=b^na^m$.\n\nProceeding in the same way we get $a^nb^m=b^ma^n$\n\nAn easy induction shows now $(*)$ and $(**)$.\n\nAs $\\gcd(m, n) = 1$, by Bez\u00f3ut's identity there are $u$, $v$ such that $u m + v n = 1$.\n\nTake $a, b \\in G$ arbitrary. By the conditions, as $a^u \\in G$ for all $u \\in \\mathbb{Z}$, and as the conditions allow to conmute certain powers: \\begin{align*} a^{u m} b^{u m} &= b^{u m} a^{u m} \\\\ (a^{u m} b^{u m})^{v n} &= a^{u m + v n} b^{u m + v n} \\\\ &= a b \\\\ (b^{u m} a^{u m})^{v n} &= b^{u m + v n} a^{u m + v n} \\\\ &= b a \\end{align*} This because e.g. $(a^{u m} b^{u m})^2 = a^{u m} b^{u m} a^{u m} b^{u m} = a^{u m} a^{u m} b^{u m} b^{u m} = a^{2 u m} b^{2 u m}$ since the middle pair conmutes by the conditions. Now use induction.\n\n(Perhaps the switching around can be expressed in a more compact way).\n\nNice problem!\n\n\u2022 Great Solution :) thanx :) \u2013\u00a0Fawzy Hegab Mar 10 '13 at 19:52\n\u2022 @vonbrand:why we have $(b^{u m} a^{u m})^{v n} = b^{u m + v n} a^{u m + v n}$? \u2013\u00a0M.H Mar 10 '13 at 20:30\n\u2022 @MaisamHedyelloo , we have a in G , so $a^u$ is in G , choose u to be the same u in the relation $mu+vn=1$ , apply the hyposetes to $a_1$ = $a^u$ and use induction as it's mentioned in the answer in the speicial case for 2 . \u2013\u00a0Fawzy Hegab Mar 10 '13 at 20:45\n\u2022 How do you get $(a^{u m} b^{u m})^{v n} = a^{u m + v n} b^{u m + v n}$? Doesn't $(a^{u m} b^{u m})^{v n} = a^{u m v n} b^{u m v n}$? \u2013\u00a0Code-Guru Mar 10 '13 at 22:11\n\u2022 @MrWhy Huh? What's $a_1$? \u2013\u00a0Code-Guru Mar 10 '13 at 22:15", "date": "2019-04-23 12:40:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/326702/a-group-such-that-am-bm-bm-am-and-an-bn-bn-an-m-n-coprime", "openwebmath_score": 0.9986185431480408, "openwebmath_perplexity": 480.0227987891319, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 1.0, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8723473663814338}}
{"url": "https://math.stackexchange.com/questions/2213248/die-fixed-so-it-cant-roll-the-same-number-twice-in-a-row-using-markov-chains/2213986", "text": "# Die fixed so it can't roll the same number twice in a row, using markov chains?\n\nStudying for probability test and the following question came up:\n\nA six sided die is 'fixed' so that it cannot roll the same number twice consecutively. The other 5 sides each show up with a probability $\\frac{1}{5}$. Calculate\n\nP($X_{n+1} = 5 \\mid X_1 = 5$) and P($X_{n+1} = 1 \\mid X_1 = 5$)\n\nWhat happens as $n \\rightarrow$ $\\infty$?\n\nIt appears to be a markov chain problem but all I can think to do is to find the eigenvalues of the transition matrix. This seems unfeasible given that it's 6x6. My guess for the second part is that the probability tends to 1/6, as the first value becomes less and less relevant.\n\n\u2022 Scalar product of two [0,0,0,0,1,0] vectors with (for Fabios $P$ matrix): $P^n$ as gram matrix defining the scalar product. en.wikipedia.org/wiki/Gramian_matrix \u2013\u00a0mathreadler Apr 2 '17 at 0:01\n\u2022 I don't remember the last time I saw a question that attracted so many good-quality answers, all different. Thanks for posting. \u2013\u00a0MJD Apr 2 '17 at 15:01\n\nCollapse the six states into the two states you actually care about: \"$5$\" and \"not-$5$\". There is a well-defined transition probability from each of these to the other, so we can now work with a $2 \\times 2$ matrix instead of a $6 \\times 6$ matrix.\n\nThis is a very common trick with Markov chains. The hard part is making sure that you don't lose any information by grouping the states together. (Here, if we had a different probability of going $1 \\to 5$ than of going $2 \\to 5$, this wouldn't work.)\n\n\u2022 Of course! Then the second probability can be calculated as 1/5 the probability of 'not 5', and it is clear that my guess about $n \\rightarrow \\infty$ is correct. \u2013\u00a0simples123 Apr 1 '17 at 16:58\n\nExcellent suggestion by @Misha. Since $6$ is a small number, you may want to numerically explore this problem by computing the powers of the transition probability matrix.\n\n$$P = \\begin{bmatrix} 0 & 1/5 & 1/5 & 1/5 & 1/5 & 1/5 \\\\ 1/5 & 0 &1/5 & 1/5 & 1/5 & 1/5 \\\\ 1/5 & 1/5 & 0 &1/5 & 1/5 & 1/5 \\\\ 1/5 & 1/5 & 1/5 & 0 & 1/5 & 1/5 \\\\ 1/5 & 1/5 & 1/5 & 1/5 & 0 & 1/5 \\\\ 1/5 & 1/5 & 1/5 & 1/5 & 1/5 & 0 \\end{bmatrix} \\enspace.$$\n\nYou'll see that $P^k$ quickly converges to a matrix where all coefficients are $1/6$. In fact, $P^5$ is already quite close. Of course, you can also repeat the experiment with the reduced chain with two states.\n\nYou can also find the steady-state probabilities (they are well-defined in this case) by solving:\n\n$$\\begin{bmatrix} -1 & 1/5 & 1/5 & 1/5 & 1/5 & 1/5 \\\\ 1/5 & -1 &1/5 & 1/5 & 1/5 & 1/5 \\\\ 1/5 & 1/5 & -1 &1/5 & 1/5 & 1/5 \\\\ 1/5 & 1/5 & 1/5 & -1 & 1/5 & 1/5 \\\\ 1/5 & 1/5 & 1/5 & 1/5 & -1 & 1/5 \\\\ 1 & 1 & 1 & 1 & 1 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} p_1 \\\\ p_2 \\\\ p_3 \\\\ p_4 \\\\ p_5 \\\\ p_6 \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{bmatrix} \\enspace.$$\n\nIt comes as no surprise that the solution is $[1/6, 1/6, 1/6, 1/6, 1/6, 1/6]^T$. Once again, the smart way to solve this problem is the one suggested by Misha, but sometimes numerical explorations reinforce our grasp of a concept.\n\n\u2022 Isn't your last equation missing a factor (what you want to solve for)? \u2013\u00a0Pa\u016dlo Ebermann Apr 2 '17 at 18:13\n\u2022 @Pa\u016dloEbermann Yes, thanks for the heads-up! \u2013\u00a0Fabio Somenzi Apr 2 '17 at 19:06\n\nAnother approach to solving this is with simple recurrence.\n\n$$P(X_{n+1}=5)=\\frac{P(X_n\\neq5)}5$$and $$P(X_n\\neq5)=1-P(X_n=5)$$\n\nSo if we let $a_n=P(X_n=5)$, then\n\n$$a_{n+1}=\\frac{1-a_n}5$$\n\nWe know that $a_1=1$. We can also see easily that the stable state is $a=(1-a)/5$, or $a=1/6$. Letting $b_n=a_n-\\frac16$, we have $$b_{n+1}+\\frac16=\\frac{1-b_n-\\frac16}5$$ or $$b_{n+1}=-\\frac{b_n}5$$ which is easily solved as $$b_n=C\\left(-\\frac15\\right)^n$$ Noting that $b_1=\\frac56$, we get $$a_n=\\frac16-\\frac16\\left(-\\frac15\\right)^{n-2}$$\n\n\u2022 Cool answer, +1 \u2013\u00a0IamThat Apr 4 '17 at 18:14\n\nMisha\u2019s answer is the best approach to this problem, but finding the eigenvalues of the full $6\\times6$ transition matrix is actually very easy because of its special structure.\n\nObserve that we can write the transition matrix as $M=\\frac15(\\mathbb1_6-I_6)$, where $\\mathbb 1_6$ is the $6\\times6$ matrix consisting entirely of $1$s. So, if $\\mathbf v$ is an eigenvector of $\\mathbb1_6$ with eigenvalue $\\lambda$, then $M\\mathbf v=\\frac15(\\mathbb1_6\\mathbf v-\\mathbf v)=\\frac15(\\lambda-1)\\mathbf v$, which means that $\\frac15(\\lambda-1)$ is an eigenvalue of $M$. The rank of $\\mathbb1_6$ is obviously one, so $0$ is an eigenvalue with multiplicity 5, and the other eigenvalue is $\\operatorname{tr}\\mathbb1_6=6$. Therefore, the eigenvalues of $M$ are $1$ and $-\\frac15$.\n\nOf course, we already knew that one of the eigenvalues of $M$ is $1$ with right eigenvector $(1,1,1,1,1,1)^T$ because it\u2019s row-stochastic, and that\u2019s the only eigenvalue we care about for computing the steady-state distribution. The corresponding left eigenvector can be found by computing the kernel of $M^T-I$ or by observing that by symmetry, its left and right eigenvectors are transposes, therefore a left eigenvector of $1$ is $(1,1,1,1,1,1)$, which after normalizing it so that its entries sum to unity gives the result you expected. Computing the $n+1$ step probabilities is a bit messy, but again, because $M$ is symmetric, the eigenspace of $-\\frac15$ is the orthogonal complement of the span of this vector, so it\u2019s pretty easy to diagonalize the matrix. You might also observe that if you subtract any two columns from each other, you get a vector with $\\pm\\frac15$ in the elements corresponding to the column numbers and zero everywhere else, so vectors with a single $1$ and $-1$ are all eigenvectors of $-\\frac15$. It\u2019s not hard to come up with an eigenbasis once you\u2019ve noticed this.\n\nThere are two great answers already, but just for completeness I'll mention that the transition matrix $P$ (as written by @FabioSomenzi) is a circulant matrix, and its eigenvectors and eigenvalues can be written explicitly. The eigenvector matrix is the DFT matrix, in fact. Circulant matrices are sometimes covered in matrix analysis / numerical linear algebra courses (although probably not at the introductory level).\n\nAn alternative approach for the part about $n \\to \\infty$ that does not require any calculations:\n\nThere is no probabilistic difference between the numbers $1,2,\\ldots,6$ on the die. So, since the Markov chain is clearly aperiodic and has a finite number of states, we can conclude by symmetry that\n\n$$\\lim_{n \\to \\infty} \\mathbb{P}(X_n = i \\mid X_1 = j) = \\frac{1}{6}.$$\n\n\u2022 Good catch, I've edited the answer. \u2013\u00a0Ritz Apr 6 '17 at 15:35\n\nI have a similar question.\n\nDice is rolled and lands on six. A. Probability dice lands on six after 2 more rolls (ASSUME 20%)\n\nB. Probability dice lands on six after 3 more rolls", "date": "2021-08-05 05:40:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2213248/die-fixed-so-it-cant-roll-the-same-number-twice-in-a-row-using-markov-chains/2213986", "openwebmath_score": 0.9403672218322754, "openwebmath_perplexity": 222.0971796623201, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534333179649, "lm_q2_score": 0.888758793492457, "lm_q1q2_score": 0.8722753692647041}}
{"url": "https://math.stackexchange.com/questions/1017802/evaluating-the-boolean-sum-sum-x-1-x-2-x-3-x-4-x-6-x-7-negx-1-oplus", "text": "# Evaluating the boolean sum $\\sum_{x_1, x_2, x_3, x_4, x_6, x_7} \\neg(x_1 \\oplus x_4 \\oplus x_3 \\oplus x_6) \\neg(x_4 \\oplus x_3 \\oplus x_2 \\oplus x_7)$\n\nSay that we have the following boolean function function ($x_{i} \\in \\{0,1\\}$):\n\n$$f = \\sum_{x_1, x_2, x_3, x_4, x_6, x_7 \\in \\{0,1\\}^6} \\neg(x_1 \\oplus x_4 \\oplus x_3 \\oplus x_6) \\land \\neg(x_4 \\oplus x_3 \\oplus x_2 \\oplus x_7)$$\n\nI was trying to evaluate that function.\n\nI realized that the way to evaluate it is by counting how many times both negation functions for the xors are 1 however, I have had a hard time counting in a cleaver way because of the over lap of variables (if we even had more overlaps and more negations being multiplied, I didn't really see how to find a general argument without brute force).\n\nIf we only had 4 terms as in:\n\n$$\\sum_{x_1, x_4, x_3, x_6 \\in \\{0,1\\}^4 } \\neg(x_1 \\oplus x_4 \\oplus x_3 \\oplus x_6)$$\n\nthe way I thought of doing this was that the negation function is only 1 when the inside is zero. The inside is zero only when either all $x_i$s are zero or 1, or when pairs of them are 0 or 1. So I thought, there are only 2 values for the whole lot to be the same value, then there are $\\binom {4}{2}$ pairs of these. Each pair could take 2 values, either 1 or zero. Therefore yielding the following count:\n\n$$\\binom {4}{2} 2^2 + \\binom{4}{4}2 = \\binom {4}{2} 2^2 + 2$$\n\nWhich I think is correct.\n\nHowever, when there is a intersection of variables, how do you count this? Do you use the inclusion exclusion principle? I am a little stuck, not sure how to generalize this\n\n\u2022 On the LHS you have variables but on the RHS you are summing over those same variables. Which is it? Are you summing over all tuples? \u2013\u00a0Alexander Vlasev Nov 12 '14 at 1:57\n\u2022 variables only take values 1 or 0. Not sure if I understand the confusion, I updated my question though. \u2013\u00a0Pinocchio Nov 12 '14 at 2:06\n\u2022 The $x_i$'s are not variables for the function $f$ because you can't just plug in some values. So your expression should read as something like $f = \\sum_{x_1,\\ldots}\\cdots$. \u2013\u00a0Alexander Vlasev Nov 12 '14 at 2:16\n\u2022 @AleksVlasev your right, duh, sorry. \u2013\u00a0Pinocchio Nov 12 '14 at 2:18\n\nI suggest let $a = x_3 \\oplus x_4$, $b = x_1 \\oplus x_6$, and $c = x_2 \\oplus x_7$\n\nthen\n\n$$\\sum_{x_1, x_2, x_3, x_4, x_6, x_7} \\neg(x_1 \\oplus x_4 \\oplus x_3 \\oplus x_6) \\land \\neg(x_4 \\oplus x_3 \\oplus x_2 \\oplus x_7)$$\n\nbecomes\n\n$$2^3 \\sum_{a,b,c} \\neg(a \\oplus b) \\land \\neg(a \\oplus c)$$\n\nNow apply the observation that binary $\\oplus$ is just another way of writing $\\ne$, so you get:\n\n$$2^3 \\sum_{a,b,c} \\neg(a \\ne b) \\land \\neg(a \\ne c)$$ $$2^3 \\sum_{a,b,c} a = b \\land a = c$$ $$2^3 \\times 2$$\n\nwhere did you get your $2^3$ at the front\n\nThere are 4 values of $\\{x_3, x_4\\}$. Each value of $x_3 \\oplus x_4$ occurs twice. So:\n\n$$\\sum_{x_3, x_4} f(x_3 \\oplus x_4) = f(1 \\oplus 1) + f(1 \\oplus 0) + f(0 \\oplus 1) + f(0 \\oplus 0) = 2f(1) + 2f(0) = 2 \\sum_a f(a)$$\n\nDo this a total of 3 times and you get $2^3$.\n\ncan you provide details of how you got that 2 on the last equation?\n\n$\\sum_{a,b,c} a = b \\land a = c$ is asking the question, \"how many ways can a,b and c all be equal?\". In two ways: $a = b = c = 0$ and $a = b = c = 1$.\n\n\u2022 where did you get your $2^3$ at the front and can you provide details of how you got that 2 on the last equation? \u2013\u00a0Pinocchio Nov 12 '14 at 2:57\n\u2022 @Pinocchio Added details \u2013\u00a0DanielV Nov 12 '14 at 3:43", "date": "2021-01-18 20:12:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1017802/evaluating-the-boolean-sum-sum-x-1-x-2-x-3-x-4-x-6-x-7-negx-1-oplus", "openwebmath_score": 0.8035116791725159, "openwebmath_perplexity": 303.5815616075096, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534365728416, "lm_q2_score": 0.888758786126321, "lm_q1q2_score": 0.8722753649279849}}
{"url": "http://mathhelpforum.com/geometry/51234-solved-find-area-rectangle-abcd.html", "text": "# Math Help - [SOLVED] Find the area of rectangle ABCD\n\n1. ## [SOLVED] Find the area of rectangle ABCD\n\nHi, this is on the SAT OG Book page 747 #16\n\nIn rectangle ABCD, point E is the midpoint of line segment BC. If the area of quadrilateral ABED is $\\frac{2}{3}$, what is the area of the rectangle ABCD?\n\na) $\\frac{1}{2}$\n\nb) $\\frac{3}{4}$\n\nc) $\\frac{8}{9}$\n\nd) 1\n\ne) $\\frac{8}{3}$\n\n2. Originally Posted by fabxx\nHi, this is on the SAT OG Book page 747 #16\n\nIn rectangle ABCD, point E is the midpoint of line segment BC. If the area of quadrilateral ABED is $\\frac{2}{3}$, what is the area of the rectangle ABCD?\n\na) $\\frac{1}{2}$\n\nb) $\\frac{3}{4}$\n\nc) $\\frac{8}{9}$\n\nd) 1\n\ne) $\\frac{8}{3}$\nnote that the area of rectangle ABCD is given by AD*AB\n\nnow, ABED is a trapezium, thus its area is given by half the sum of the two parallel sides times the distance between them. that is (1/2)(BE + AD)*AB, and this is 2/3, so\n\n(1/2)(BE + AD)*AB = 2/3\n\n=> (BE + AD)*AB = 4/3\n\nbut, BE = (1/2)AD (this should be pretty obvious from a diagram if you drew or were given one), so that\n\n=> (3/2)AD*AB = 4/3\n\n=> AD*AB = 8/9\n\n3. Hello, fabxx!\n\nDid you make a sketch?\n\nIn rectangle ABCD, point E is the midpoint of line segment BC.\nIf the area of $ABED$ is $\\frac{2}{3}$, what is the area of $ABCD$?\n\n$(a)\\;\\frac{1}{2} \\qquad(b)\\;\\frac{3}{4}\\qquad (c)\\;\\frac{8}{9} \\qquad (d)\\;1 \\qquad (e) \\;\\frac{8}{3}$\nCode:\n    D *---------------* C\n|   *           |\n|       *       |\n|           *   |\nF * - - - - - - - * E\n|               |\n|               |\n|               |\nA *---------------* B\n\nDraw median EF.\n\n$\\text{Rect }DCEF \\:=\\:\\frac{1}{2}(\\text{Rect }ABCD)$\n\n$\\Delta DCE \\:=\\:\\frac{1}{2}(\\text{Rect }DCEF) \\;=\\;\\frac{1}{4}(\\text{Rect }ABCD)$\n\n. . Hence: . $\\text{Quad }ABED \\:=\\:\\frac{3}{4}(\\text{Rect }ABCD)$\n\nWe are told that: . $\\text{Quad }ABED \\:=\\:\\frac{2}{3}$\n\nWe have: . $\\frac{3}{4}(\\text{Rect }ABCD) \\;=\\;\\frac{2}{3} \\quad\\Rightarrow\\quad \\boxed{\\text{Rect }ABCD \\:=\\:\\frac{8}{9}}$ . . . answer (c)\n\n4. Originally Posted by Jhevon\nnote that the area of rectangle ABCD is given by AD*AB\n\nnow, ABED is a trapezium, thus its area is given by half the sum of the two parallel sides times the distance between them. that is (1/2)(BE + AD)*AB, and this is 2/3, so\n\n(1/2)(BE + AD)*AB = 2/3\n\n=> (BE + AD)*AB = 4/3\n\nbut, BE = (1/2)AD (this should be pretty obvious from a diagram if you drew or were given one), so that\n\n=> (3/2)AD*AB = 4/3\n\n=> AD*AB = 8/9\nNo figure is given for this. If i were to draw one myself, would it look like the attachment below?\n\n5. Originally Posted by fabxx\nNo figure is given for this. If i were to draw one myself, would it look like the attachment below?\nyes, it should look like that. note that it is the same as the diagram Soroban drew, except it is turned on its side. i imagined it as Soroban did. it will all work out the same. if your diagram was like the given one or Soroban's you will get the right answer", "date": "2014-10-22 02:41:20", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/51234-solved-find-area-rectangle-abcd.html", "openwebmath_score": 0.8950310349464417, "openwebmath_perplexity": 974.1966126921207, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988491849579606, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8722727553603737}}
{"url": "https://www.bionicturtle.com/forum/threads/in-converting-continuous-rate-to-discrete-does-it-matter-how-often-interest-is-paid.6974/", "text": "What's new\n\n# In converting Continuous rate to Discrete, does it matter how often interest is paid?\n\n#### Steve Jobs\n\n##### Active Member\nAssume the following:\n\nAn asset is quoted at 12% annually with continuous rate.\nInterest is paid quarterly.\n\nIs this correct for equivalent rate with monthly compounding?\nr = 12 * [ e^(.12/12)) - 1] = 12.06%\n\nDoes it matter whether interest is paid quarterly, monthly or annually? What about doing the reverse convert from continuous to discrete?\n\n#### ShaktiRathore\n\n##### Well-Known Member\nSubscriber\nhi,\nlet i deposit 1 unit today,\nthen from monthly compounding my deposit value after 1 yr is if r is the annual rate,\n(1+r/12)^12\nalso from continuous compounding assuming continous rate rc is in annual terms after 1 yr is,\ne^rc\npays from both should be equal,\n(1+r/12)^12=1.e^rc\n(1+r/12)=e^(rc/12)\n(r/12)=[e^(rc/12)-1]\nr=12*[e^(rc/12)-1]\nfor n periods compounding,\nr=n*[e^(rc/n)-1]\nfor e.g. n=4(12/3) quarterly, r= 4* [ e^(.12/4)) - 1]\nhence it matters whether interest is paid quarterly, monthly or annually\nfor reverse convert from continuous to discrete, following the same logic as above\n(1+r/n)^n=e^rc\nln((1+r/n)^n)=rc\nrc=n ln(1+r/n)\n\nthanks\n\n#### Steve Jobs\n\n##### Active Member\nThanks Shakti,\n\nWhen we say compounded monthly, are we assuming that the interest income is withdrawn each month too?\n\nOtherwise there will be two periods, one period specifying how often the rate is compounded and the second one for how often the interest is actually withdrawn and paid. My confusion is regarding these two periods and whether I should be worry about this in the exam?\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nThat's interesting, Steve (to me, because I write so many questions).\n\nWith regard to \"An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly.\" Note three timeframes are invoked:\n1. Interest paid quarterly (4 per year)\n2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed \"per annum\" which is independent of compound frequency; i.e., even if the \"annually\" were omitted, we would assume the 12.0% is per annum\n3. Compounding frequency is continuous\nA modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):\n\"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding\"\n... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)\n\nSo, as phrased above, the continuous would not be converted;\nFor example, assume a bond with a 12% quarterly coupon when the spot rate curve is flat at 12.0% per annum with continuous compounding.\nIt's price is slightly less than par:\n($100*12%/4)*exp(-0.25*12%); first quarterly coupon discounted continuously + ($100*12%/4)*exp(-0.50*12%); i.e. coupon paid quarterly but discounted continuously\n....\n+ $100+($100*12%/4)*exp(-n*12%)\nDeposit no. 2: The bank quotes the interest rate on a $1,000 as 9.0% per annum with continuous compounding but the interest is paid monthly. For each deposit, calculate the equivalent discrete rate per annum with monthly compounding. I think in the above cases: a. the equivalent rates will be the same for both deposits b. but there could be differences in the amount of interest received by the customer for one year, dependent on whether the the interest that is paid monthly is being re-invested again or withdrawn by the customer. So the difference is not in the equivalent rates but in deposit terms/conditions and customer decision. #### Steve Jobs ##### Active Member The question was asking for the amount of interest, I changed it to equivalent rate according to the point wanted to raise. Actually, the only similarity now is the interest rate. #### David Harper CFA FRM ##### David Harper CFA FRM Staff member Subscriber I can see that I did write a question just like Hull's BT 157.3 Suppose a lender quotes the interest rate on a$1,000 loan as 9.0% per annum with continuous compounding but the interest is actually paid monthly. What are the monthly interest payments?\n\nHull 4.10 A deposit account pays 12% per annum with continuous compounding, but interest is actually paid quarterly. How much interest will be paid each quarter on a $10,000 deposit? If the question seems awkward (i.e., we could skip the continuous rate and quote the corresponding monthly or annual rate), that's only because the whole point of the question is to test the equivalence-translation from one rate to another. So, the meaning of 9% per annum with continuous compounding is: regardless of how frequently the interest is paid, the terminal value at the end of one year will be$1,000*exp(9%) = $1,094.17; it is also same as saying, regardless of the interest payout frequency, the effective annual return = exp(9%)-1 = 9.417%. Any conversion-to-discrete will assume reinvestment (we generally do assume reinvestment). As below, i input your questions, the first scenario is quarterly interest (first two columns), which implies$22.76 interest paid per quarter , but those interest payments are reinvested, so they accumulate to the same $94.17 at the end of the year. If the interest pays annual (k =1), then the$94.17 is just paid once, at the end of the year. I didn't detail the monthly, but each $7.5282 will get reinvested and at the end of the year, this will also equal$94.17. So the profile of cash flow varies, but the very design of the question means to treat 9.0% continuous as the rate earned regardless of interest payout frequency. Regardless of (k), the terminal value of \\$1,094.17 is unchanged, is how i look at this. thanks,\n\n#### Steve Jobs\n\n##### Active Member\nThanks David,\n\nSo in the exam, for the equivalent rate questions, I should ignore any details provided regarding the frequency of interest payout because it's assumed that the interest is reinvested. In reality I guess such details should be documented.\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nIt's safe to assume reinvestment (which is different than assuming we know the reinvestment rate: yield assumes it knows. Reinvestment risk refers to the reality we don't know. But the existence of some re-investment assumption is tantamount, I think, to assuming time value of money, and we always assume TVM).\n\nYes, details will be documented. If there is a #1 topic upon which GARP has received feedback with respect to specificity, i think it probably is careful specification of compound frequency assumptions. It's one of the more seasoned elements such that they know to give details, thanks,", "date": "2021-06-16 20:26:31", "meta": {"domain": "bionicturtle.com", "url": "https://www.bionicturtle.com/forum/threads/in-converting-continuous-rate-to-discrete-does-it-matter-how-often-interest-is-paid.6974/", "openwebmath_score": 0.5721750855445862, "openwebmath_perplexity": 3962.391776812519, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9748211597623863, "lm_q2_score": 0.8947894618940992, "lm_q1q2_score": 0.8722597009867673}}
{"url": "https://www.physicsforums.com/threads/help-with-vector-notation.954356/", "text": "# Help with vector notation\n\n## Homework Statement\n\nHi. Its not really a problem but just a general question. When doing graph translations, such as move the parabola x units right or y units up etc, Is it okay to use vector format. So instead of saying move this equation 4 units left, could you write it like this -> <-4,0>\n\n## Homework Equations\n\nI know this is used for translations of shapes but my main question is can it also be used for graphs\n\n## The Attempt at a Solution\n\nI asked my teacher but he just shouted at me for asking him concepts he didnt talk about.\n\nRelated Precalculus Mathematics Homework Help News on Phys.org\nfresh_42\nMentor\n\n## Homework Statement\n\nHi. Its not really a problem but just a general question. When doing graph translations, such as move the parabola x units right or y units up etc, Is it okay to use vector format. So instead of saying move this equation 4 units left, could you write it like this -> <-4,0>\n\n## Homework Equations\n\nI know this is used for translations of shapes but my main question is can it also be used for graphs\n\n## The Attempt at a Solution\n\nI asked my teacher but he just shouted at me for asking him concepts he didnt talk about.\nA graph $\\mathcal{G}$ of usually a function $x \\mapsto f(x)$ is a set of points: $\\mathcal{G}=\\{\\,(x,f(x)\\,|\\,x \\in \\operatorname{domain}(f)\\,\\}$. This is a subset of $\\mathbb{R}^2$ and to move it means to move it within this plane $\\mathbb{R}^2$. Thus you need a direction in which it is moved and a distance by which it is moved. But this exactly defines a vector $v=(v_x,v_y)$. So the moved object will be $v+\\mathcal{G}=\\{\\,(x+v_x,f(x)+v_y)\\,|\\,x \\in \\operatorname{domain}(f)\\,\\}$. So in a sense this is even better than just to say move by $\"v_0\"$ in direction $\"x\"$ or $\"y\"$ since you allow more than two directions.\n\nHowever, whether you should argue with your teacher is a complete different question. It's not always important to be right. Sometimes it's smarter not to be.\n\nLast edited:\nA graph $\\mathcal{G}$ of usually a function $x \\mapsto f(x)$ is a set of points: $\\mathcal{G}=\\{\\,(x,f(x)\\,|\\,x \\in \\operatorname{domain}(f)\\,\\}$. This is a subset of $\\mathbb{R}^2$ and to move it means to move it within this plane $\\mathbb{R}^2$. Thus you need a direction in which it is moved and a distance by which it is moved. But this exactly defines a vector $v=(v_x,v_y)$. So the moved object will be $v+\\mathcal{G}=\\{\\,(x+v_x,f(x)+v_y\\,|\\,x \\in \\operatorname{domain}(f)\\,\\}$. So in a sense this is even better than just to say move by $\"v_0\"$ in direction $\"x\"$ or $\"y\"$ since you allow more than two directions.\n\nHowever, whether you should argue with your teacher is a complete different question. It's not always important to be right. Sometimes it's smarter not to be.\n\nSo is it okay to write like this ? If i wrote like this In a paper would it be correct ?\n\nfresh_42\nMentor\nSo is it okay to write like this ? If i wrote like this In a paper would it be correct ?\nI'm not quite sure what you mean by this. That's why I wrote the formulas, which are correct. 4 to the left would be $v_x=-4$ and $v_y=0$. It would be correct, if it would be o.k. depends on a lot of human factors. I wouldn't argue just to be right. 4 to the left is usually as good.\n\nRay Vickson\nIf I were your teacher, what I would prefer to see are either of the following (assuming you are graphing a parabola of the form $y = a + b x^2$). If you want to move the whole graph $A$ units to the right (to the left if $A < 0$) and $B$ units up (down if $B < 0$), you could either say that (i) the vertex moves from $(a,0)$ to $(a+A, 0+B)$ but the shape remains unchanged; or (ii) explain that the new graph has equation $y-B = a + b(x-A)^2 \\Rightarrow y = a+B+b(x-A)^2.$ Since you are plotting graphs in two-dimensional cartesian coordinate systems, I don't think you could be yelled at for doing it using method (i). However, my personal preference would be that you show your understanding by using method (ii).", "date": "2020-04-08 18:51:45", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/help-with-vector-notation.954356/", "openwebmath_score": 0.6968815922737122, "openwebmath_perplexity": 302.7655486308628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9546474194456936, "lm_q2_score": 0.9136765175373275, "lm_q1q2_score": 0.8722389296751377}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=42kq69m4tfh1jl3e4e4tbu0h26&action=printpage;topic=2663.0", "text": "# Toronto Math Forum\n\n## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Yifei Hu on February 21, 2022, 11:29:41 AM\n\nTitle: Textbook Chapter 2.5 Example 2\nPost by: Yifei Hu on February 21, 2022, 11:29:41 AM\nHi Professor,\n\nI think one coefficient is wrong in the solution.\n\nIn this problem the solution to the inhomogeneous equation with homogeneous initial condition should start with coefficient $\\frac{1}{2c} = \\frac{1}{6}$. After cancelling the 36 in the $f(x,t)$, we should arrive at $6\\int_0^t \\int_{x-3(t-t')}^{x+3(t-t')}\\frac{1}{t^2+1}dx'dt'$ instead of $3\\int_0^t \\int_{x-3(t-t')}^{x+3(t-t')}\\frac{1}{t^2+1}dx'dt'$.\n\nCan you help me confirm if I missed something or the textbook has a typo?", "date": "2022-09-28 12:28:03", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=42kq69m4tfh1jl3e4e4tbu0h26&action=printpage;topic=2663.0", "openwebmath_score": 0.6370220184326172, "openwebmath_perplexity": 1365.3423752625147, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517424466176, "lm_q2_score": 0.8918110425624792, "lm_q1q2_score": 0.8722373441113674}}
{"url": "https://www.physicsforums.com/threads/function-f-with-domain-infinity-infinity.373953/", "text": "# Function, f, with domain (-infinity, + infinity)\n\n1. Jan 30, 2010\n\n### ajassat\n\nI was working on the following problem from a textbook. The textbook has no answer. I have included my solution - I am not sure whether it is correct Any ideas and or solutions? (guidance)\n\nQuestion:\nSuppose that f is any function with domain (-infinity, +infinity)\n\na) Does the function g defined by g(x) = f(x) + f(-x) have any special symmetry?\n\nMy solution\n\nA function f which satisfies:\nf(-x) = -f(x) is symmetrical about the horizontal axis. This is also:\nf(x) + f(-x) = 0\nor we could substitute values of x giving non zero solutions for:\nf(x) + f(-x) = g(x) (some other function)\n\nTherefore g(x) = f(x) + f(-x) has symmetry about horizontal axis?\n\nAny guidance appreciated.\n\nRegards,\nAdam\n\n2. Jan 30, 2010\n\n### JSuarez\n\nNo. f(x)+f(-x) is an even function: it's symmetric about the vertical axis.\n\n3. Jan 30, 2010\n\n### some_dude\n\nTry and work backwards: take an arbitrary function $$g$$ defined on $$\\mathbb{R}$$ and try to create a function $$f$$ s.t. $$f(x) + f(-x) = g(x)$$ for all x. What assumptions must be made about $$g$$ in order to do this?\n\n4. Jan 30, 2010\n\n### uart\n\nJust use the definition of odd and even functions for this type of problem. Without making any assumptions about f(x) (apart from the given domain) simply write g(-x) and compare it with g(x). That is,\n\ng(x) = f(x) + f(-x) : given\n\nSo g(-x) = f(-x) + f(-(-x))\n= f(-x) + f(x)\n= g(x).\n\nAnd g(-x) = g(x) is the definition of an even function.\n\nLast edited: Jan 30, 2010\n5. Jan 31, 2010\n\n### ajassat\n\nI figured it was an even function after I had posted this.\n\nThanks to both some_dude and uart for posting two different approaches to the problem. I understand this problem a lot better and have solved some similar now.\n\nThanks again,\nAdam\n\n6. Feb 1, 2010\n\n### Landau\n\nTo test yourself, a new question: do the same for g(x)=f(x)-f(-x).\n\n7. Feb 2, 2010\n\n### HallsofIvy\n\nStaff Emeritus\nBy the way, f(x)+ f(-x) and f(x)- f(-x) are almost the \"even and odd parts\" of the function f(x). e(x)= (f(x)+ f(-x))/2 and o(x)= (f(x)- f(-x))/2 are even and odd functions that add to give f(x). Of course, if f(x) is an even function to begin with, o(x) will be 0 and e(x)= f(x). If f(x) is an odd function, e(x)= 0 and o(x)= f(x).\n\nThe exponential function, ex, is neither even nor odd. Its even and odd parts are $\\frac{e^x+ e^{-x}}{2}= cosh(x)$ and $\\frac{e^x- e^{-x}}{2}= sinh(x)$.\n\nKnow someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook", "date": "2017-10-17 10:24:26", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/function-f-with-domain-infinity-infinity.373953/", "openwebmath_score": 0.6558426022529602, "openwebmath_perplexity": 1848.808058292335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850902023109, "lm_q2_score": 0.8872046041554922, "lm_q1q2_score": 0.8721976183041076}}
{"url": "https://www.physicsforums.com/threads/linear-algebra-money-question.635777/", "text": "# Homework Help: Linear Algebra Money Question\n\n1. Sep 14, 2012\n\n### bleedblue1234\n\n1. The problem statement, all variables and given/known data\n\nI have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?\n\n2. Relevant equations\n\nA= # $1 bills B= #$5 bills\nC= # $10 bills A+B+C = 32 1A+5B+10C = 100 3. The attempt at a solution So I attempted to solve for C in terms of A and B in terms of A but I'm getting nowhere. 2. Sep 14, 2012 ### E_M_C Hi bleedblue1234, You can only solve for n variables when you have n linearly independent equations. In this case, you have 3 variables and 2 linearly independent equations, so you're one equation short. But if you choose a value of zero for A, B or C then you reduce the problem to 2 variables and 2 linearly independent equations. What do you get when you try out the different combinations? Be careful: There is more than one solution. 3. Sep 14, 2012 ### azizlwl You can narrow the selection.$1 can only be in a group of 5.\n\n4. Sep 14, 2012\n\n### HallsofIvy\n\nThis not, strictly speaking, a \"linear algebra\" problem, but a \"Diophantine equation\" because the \"number of bills\" of each denomination must be integer. Letting \"O\", \"F\", and \"T\" be, respectively, the number of \"ones\", \"fives\" and \"tens\", we must have O+ F+ T= 32 and O+ 5F+ 10T= 100. Subtracting the first equation from the second, 4F+ 9T= 68.\nNow you can use the standard \"Eucidean algorithm\" to find all possible integer values for F and T and then find O.\n\nLast edited by a moderator: Sep 14, 2012\n5. Sep 14, 2012\n\n### Ray Vickson\n\nYou can solve for A and B in terms of C, just by solving the two simple equations\nA + B = 32 - C\nA + 5C = 100 - 10C.\n\nNow you can plug in C = 0, 1, 2, ... and see which values (if any) give you non-negative integer values of A and B.\n\nRGV\n\n6. Sep 14, 2012\n\n### HallsofIvy\n\nOh, well- if you want to do it the easy way!\n\n7. Sep 14, 2012\n\n### bleedblue1234\n\nYa I just set the equations equal and restricted B and C to be natural numbers and just checked which B would give me the correct C, which in tern gave me the correct A. Thank you.\n\n8. Sep 14, 2012\n\n### E_M_C\n\nBe careful with that. Most people (in my experience) define the natural numbers as N = {1, 2, 3, ...} which doesn't include zero. So if you're restricting B and C to natural numbers, as defined above, you may be cheating yourself out of a solution. As I mentioned earlier, there is more than one solution.", "date": "2018-06-22 15:39:42", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/linear-algebra-money-question.635777/", "openwebmath_score": 0.6154900193214417, "openwebmath_perplexity": 1202.877773144959, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850872288502, "lm_q2_score": 0.8872045832787204, "lm_q1q2_score": 0.8721975951423966}}
{"url": "http://math.stackexchange.com/questions/101325/finding-plotting-equation-of-hyperbola-given-foci-and-difference-in-distances", "text": "# Finding & Plotting equation of hyperbola given foci, and difference in distances between them.\n\nI have to plot the hyperbola (3 of them actually) in MATLAB, and so it'd be good if I could find some sort of general formula.\n\nThe foci do not necessarily have to be on the axes (e.g. $(5,3)$ and $(4,8)$ can also be the foci). I have the difference in distances between them at a point which lies on the hyperbola. Will there be a general equation describing this hyperbola? If yes, what will it be? (e.g. say the points are $(x_1,y_1)$ & $(x_2,y_2)$ and difference in distance is $d$).\n\nI know that I can shift and rotate the axes so that I get what I want, but I wanted to know if there was a better method, since while plotting i'll have to reconvert the values into the normal axes.\n\nI'm going to try trilateration, which is why i'm asking for the equation.\n\n-\nThe answers to this question math.stackexchange.com/questions/31756/\u2026 provide what you are looking for, I think. \u2013\u00a0Ellie Kesselman Jan 22 '12 at 15:16\nActually, one of the answers specifically addresses the case where the foci are not on the axes, and just about everything else possible for a hyperbola. It is a remarkably fine answer math.stackexchange.com/a/31761/9260 \u2013\u00a0Ellie Kesselman Jan 22 '12 at 15:21\n@FeralOink - Thanks! :) \u2013\u00a0Khushman Patel Jan 22 '12 at 17:16\nThe equation of the hyperbola is an implicit one. Can't you plot implicit equations? \u2013\u00a0Am\u00e9rico Tavares Jan 22 '12 at 22:54\nSimply put, if you need to use parametric equations as opposed to an implicit Cartesian equation, rotate/shift is really the only way to go. Otherwise, if your computing environment supports implicit plotting, then even the obvious approach of using the distance formula works nicely. \u2013\u00a0J. M. Jul 24 '12 at 21:49\n\nsay the points are $(x_1,y_1)$ & $(x_2,y_2)$ and difference in distance is $d$\n\nCould you use the literal translation of this statement into algebra? $$\\left|\\sqrt{(x-x_1)^2+(y-y_1)^2}-\\sqrt{(x-x_2)^2+(y-y_2)^2}\\right|=d$$ or $$\\left(\\sqrt{(x-x_1)^2+(y-y_1)^2}-\\sqrt{(x-x_2)^2+(y-y_2)^2}\\right)^2=d^2$$ or $$(x-x_1)^2+(y-y_1)^2+2\\sqrt{(x-x_1)^2+(y-y_1)^2}\\sqrt{(x-x_2)^2+(y-y_2)^2}+(x-x_2)^2+(y-y_2)^2=d^2$$\n\n-\nTypically one does $\\sqrt{(x-x_1)^2+(y-y_1)^2} = \\sqrt{(x-x_2)^2+(y-y_2)^2} \\pm d$, so that (after squaring both sides, simplifying, and squaring again) one gets a quadratic equation, not a quartic one. \u2013\u00a0Rahul Feb 20 '12 at 8:42\n@\u211d\u207f. - Taking the +d case, squaring both sides gives $$(x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 + (y-y_2)^2 + 2d\\sqrt((x-x_2)^2 + (y-y_2)^2) + d^2$$. Only a little bit of simplification seems possible: $$-2xx_1 + x_1^2 -2yy_1 + y_1^2 = -2xx_2 + x_2^2 -2yy_2 + y_2^2 + 2d\\sqrt((x-x_2)^2 + (y-y_2)^2) + d^2$$. Now from here if you move everything but the $$2d\\sqrt((x-x_2)^2 + (y-y_2)^2)$$ to the left side so you can square both sides again to get rid of the sqrt, I count 9 additive terms on the left that all have to be multiplied by each other... is that really what needs to be done? \u2013\u00a0David Doria Feb 20 '13 at 18:36\n@David: Well, you ought to group them into the form $ax+by+c$ before squaring. \u2013\u00a0Rahul Feb 20 '13 at 18:42\n@\u211d\u207f. See my answer below - any more suggestions for improvement :)? \u2013\u00a0David Doria Feb 20 '13 at 20:57\n\nPerhaps someone can verify the result below, or suggest cleaner ways to expand/write this?\n\nStarting from the definition of a hyperbola as the locus of points X = (x,y) that have a constant distance difference to two points (the foci, A = (a_x, a_y) and B = (b_x, b_y)): $$|A-X| - |B-X| = d$$\n\nWe can write this (at least half of the solution) as:\n\n$$\\sqrt{(a_x-x)^2 + (a_y-y)^2} - \\sqrt{(b_x-x)^2 + (b_y-y)^2} = d$$\n\nThere is tremendous manipulation necessary to write this equation in general quadratic form ($Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$):\n\nMove one of the square roots to the other side of the equation so that when we square both sides we will not eventually introduce quartic terms: $$\\sqrt{(a_x-x)^2 + (a_y-y)^2} = \\sqrt{(b_x-x)^2 + (b_y-y)^2} + d$$\n\nSquare both sides: $$(a_x-x)^2 + (a_y-y)^2 = (b_x-x)^2 + (b_y-y)^2 + 2d \\sqrt{(b_x-x)^2 + (b_y-y)^2} + d^2$$\n\nExpand: $$a_x^2 -2xa_x + x^2 + a_y^2 - 2a_y y + y^2 = b_x^2 - 2xb_x + x^2 + b_y^2 -2b_y y + y^2 + 2d \\sqrt{(b_x-x)^2 + (b_y-y)^2} + d^2$$\n\nCancel $x$ and $y$ quadratic terms (that occur on both sides of the equation): $$a_x^2 -2xa_x + a_y^2 - 2a_y y = b_x^2 - 2xb_x + b_y^2 -2b_y y + 2d \\sqrt{(b_x-x)^2 + (b_y-y)^2} + d^2$$\n\nMove everything not connect to the square root to the left side so we can eventually square both sides again to get rid of the square root: $$a_x^2 - 2xa_x + a_y^2 - 2a_y y - b_x^2 = 2xb_x - b_y^2 + 2b_y y - d^2 = 2d \\sqrt{(b_x-x)^2 + (b_y-y)^2}$$\n\nGroup terms on the left side into the form Jx + Ky + L to make squaring easier: $$\\left((-2a_x + 2b_x)x + (-2a_y + 2b_y)y + (a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2)\\right) = 2d \\sqrt{(b_x-x)^2 + (b_y-y)^2}$$\n\nAssign $J = -2a_x + 2b_x$, $k = -2a_y + 2b_y$, $L = a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2$ so this can be written as\n\n$$Jx + Ky + L = 2d \\sqrt{(b_x-x)^2 + (b_y-y)^2}$$\n\nSquare both sides:\n\n$$(Jx + Ky + L)^2 = 4d^2 \\left((b_x-x)^2 + (b_y-y)^2\\right)$$\n\n$$J^2x^2 + JxKy + JxL + KyJx + K^2y^2 + KyL + LJx + LKy + L^2 = 4d^2 \\left(b_x^2 - 2xb_x +x^2 + b_y^2 - 2yb_y + y^2\\right)$$\n\nCombine terms on the left, and multiply through on the right:\n\n$$J^2x^2 + K^2y^2 + 2JxKy + 2JxL + 2KyL + L^2 = 4d^2 b_x^2 - 8d^2xb_x + 4d^2x^2 + 4d^2b_y^2 - 8d^2yb_y + 4d^2y^2$$\n\nCombine all terms ($x^2, y^2, xy, x, y$):\n\n$$(J^2-4d^2)x^2 + (K^2-4d^2)y^2 + (2JK)xy + (2JL+8d^2b_x)x + (2KL + 8d^2b_y)y + (L^2 - 4d^2 b_x^2 - 4d^2 b_y^2) = 0$$\n\nSubstituting back in the values of J, K, and L:\n\n$$((-2a_x + 2b_x)^2-4d^2)x^2 + ((-2a_y + 2b_y)^2-4d^2)y^2 + (2(-2a_x + 2b_x)(-2a_y + 2b_y))xy \\\\+ (2(-2a_x + 2b_x)(a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2)+8d^2b_x)x + (2(-2a_y + 2b_y)(a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2) + 8d^2b_y)y \\\\+ ((a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2)^2 - 4d^2 b_x^2 - 4d^2 b_y^2) = 0$$\n\nExpand: $$(4a_x^2 - 8a_x b_x + 4b_x^2)x^2 + \\\\ (4a_y^2 - 8a_y b_y + 4b_y^2)y^2 + \\\\ (8a_xa_y - 8a_xb_y - 8a_yb_x + 8b_xb_y)xy +\\\\ (-4a_x^3 -4a_xa_y^2 + 4 a_x b_x^2 + 4 a_x b_y^2 + 4 a_x d^2 + 4b_xa_x^2 + 4b_xa_y^2 + 4b_x^3 - 4b_xb_y^2 + 4b_xd^2)x +\\\\ (-4a_ya_x^2 - 4a_y^3 + 4 a_y b_x^2 + 4 a_y b_y^2 + 4 a_y d^2 + 4b_ya_x^2 + 4b_ya_y^2 - 4b_yb_x^2 - 4b_y^3 + 4b_yd^2)y +\\\\ a_x^4 + a_x^2 a_y^2 - a_x^2b_x^2 - a_x^2b_y^2 - a_x^2d^2\\\\ + a_y^2 a_x^2 + a_y^4 - a_y^2b_x^2 - a_y^2b_y^2 - a_y^2 d^2\\\\ - b_x^2 a_x^2 - b_x^2 a_y^2 + b_x^4 + b_x^2 b_y^2 -3 b_x^2 d_2\\\\ - b_y^2 a_x^2 - b_y^2a_y^2 + b_y^2b_x^2 + b_y^4 + b_y^2 d^2\\\\ - d^2 a_x^2 - d^2 a_y^2 + d^2 b_x^2 -3 d^2 b_y^2 + d^4\\\\ = 0$$\n\n-\nA slightly cleaner way would be to keep everything in vector notation. $\\sqrt{\\lVert\\mathbf a-\\mathbf x\\rVert^2} = \\sqrt{\\lVert\\mathbf b-\\mathbf x\\rVert^2} + d$, $\\lVert\\mathbf a\\rVert^2-2\\mathbf a\\cdot\\mathbf x+\\lVert\\mathbf x\\rVert^2=\\lVert\\mathbf b\\rVert^2-2\\mathbf b\\cdot\\mathbf x+\\lVert\\mathbf x\\rVert^2+2d\\sqrt{\\lVert\\mathbf b-\\mathbf x\\rVert^2}+d^2$, and so on. This has the additional advantage that the derivation doesn't change at all in higher dimensions. \u2013\u00a0Rahul Mar 6 '13 at 11:33\nIn this derivation we assumed |A\u2212X|>|B\u2212X|. Even though we only solved \"half\" of the absolute value equation given by the definition of the hyperbola, we still seem to obtain the entire hyperbola. What is the relation of the object obtained if |B\u2212X|>|A\u2212X| to the one we derived above? \u2013\u00a0David Doria Mar 18 '13 at 17:04", "date": "2016-02-11 17:26:59", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/101325/finding-plotting-equation-of-hyperbola-given-foci-and-difference-in-distances", "openwebmath_score": 0.8259793519973755, "openwebmath_perplexity": 416.3888929958581, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676425019244, "lm_q2_score": 0.8902942348544447, "lm_q1q2_score": 0.8721924541929085}}
{"url": "http://math.stackexchange.com/questions/265801/probability-distribution-of-number-of-times-required-to-roll-all-of-a-certain-se", "text": "# Probability distribution of number of times required to roll all of a certain set of k numbers?\n\nConsider a 12-sided fair die. What is the distribution of the number T of rolls required to roll a 1, a 2, a 3, and a 4?\n\nTaking inspiration from the Coupon Collector's Problem, I believe that the expected number of rolls to achieve the goal would be\n\n$$E[T] = \\sum\\limits_{i=0}^3 \\frac{12}{4-i} = 25$$\n\nSimilarly, the variance would be\n\n$$Var[T] = \\sum\\limits_{i=0}^3 \\frac{1-\\frac{4-i}{12}}{(\\frac{4-i}{12})^2} = 180$$\n\nBut applying Chebyshev here does not yield very useful bounds. My question is therefore, how would you compute, for example, $P(T=16)$ or $P(T<30)$?\n\nIdeally this could be generalized to a set of k required numbers, not just 4 as in the example.\n\n-\n\nPersonally I would call $P(n,b,k,j)$ the probability that after rolling $n$ dice with $b$ sides, $j$ of the target $k$ sides had been found, with the formula\n\n$$P(n+1,b,k,j)= \\frac{b-k+j}{b} P(n,b,k,j) + \\frac{k+1-j}{b} P(n,b,k,j-1)$$\n\nstarting from $P(0,b,k,j)=0$ for $j \\not = 0$ and $P(0,b,k,0)=1$.\n\nThen $$\\Pr(T \\le t) = P(t,b,k,k)$$ and $$\\Pr(T = t) = P(t,b,k,k)-P(t-1,b,k,k).$$\n\nSo in your example it is not difficult to calculate $\\Pr(T = 16) \\approx 0.0380722$ and $\\Pr(T \\le 30) \\approx 0.7305294$. Your expected value of $T$ and variance appear to be correct.\n\n-\nThanks for the response. A recursive definition is a good idea. I feel though that in your response, in the recurrence relation, one of the P's needs to have a j-1. \u2013\u00a0sffc Dec 27 '12 at 13:31\n@vote539: your final point is correct and I will edit \u2013\u00a0Henry Dec 27 '12 at 13:48\nThanks for the clarification! The results yielded by your solution match those yielded by a simulation of the dice rolls. A note to anyone who wants to implement this: the recursion stack grows very quickly, so it is necessary to use dynamic programming; that is, cache the results of the P function and only perform the computation when there is no cached value. \u2013\u00a0sffc Dec 27 '12 at 19:46\nOr you could use a spreadsheet \u2013\u00a0Henry Dec 27 '12 at 23:26\nOur answers agree on $\\textsf{Pr}(T=16)$ and $\\textsf{Pr}(T\\le30)$. (The example in the question was $\\textsf{Pr}(T\\lt30)$.) \u2013\u00a0joriki May 18 at 19:15\n\nYou can get this distribution by conditioning on the number $N$ of rolls up to $4$ required:\n\n\\begin{align} \\textsf{Pr}(T\\le t) &=\\left(\\frac23\\right)^t\\sum_{n=0}^t\\binom tn2^{-n}\\textsf{Pr}(N\\le n) \\\\ &=\\left(\\frac23\\right)^t\\sum_{n=0}^t\\binom tn2^{-n}\\frac{4!}{4^n}\\left\\{n\\atop4\\right\\} \\\\ &=\\left(\\frac23\\right)^t\\sum_{n=0}^t\\binom tn8^{-n}\\sum_{j=0}^4(-1)^j\\binom4jj^n \\\\ &=\\left(\\frac23\\right)^t\\sum_{j=0}^4(-1)^j\\binom4j\\sum_{n=0}^t\\binom tn\\left(\\frac j8\\right)^n \\\\ &=\\left(\\frac23\\right)^t\\sum_{j=0}^4(-1)^j\\binom4j\\left(1+\\frac j8\\right)^t \\\\ &=\\sum_{j=0}^4(-1)^j\\binom4j\\left(\\frac23+\\frac j{12}\\right)^t \\\\ &=1-4\\left(\\frac{11}{12}\\right)^t+6\\left(\\frac56\\right)^t-4\\left(\\frac34\\right)^t+\\left(\\frac23\\right)^t\\;, \\end{align}\n\nwhere $\\left\\{n\\atop4\\right\\}$ is a Stirling number of the second kind and the distribution $P(N\\le n)$ is given at Probability distribution in the coupon collector's problem.\n\nThe expectation comes out right as\n\n\\begin{align} E[T]&=\\sum_{t=0}^\\infty\\textsf{Pr}(T\\gt t)\\\\ &=\\sum_{t=0}^\\infty\\left(4\\left(\\frac{11}{12}\\right)^t-6\\left(\\frac56\\right)^t+4\\left(\\frac34\\right)^t-\\left(\\frac23\\right)^t\\right)\\\\ &=12\\left(4\\cdot\\frac11-6\\cdot\\frac12+4\\cdot\\frac13-1\\cdot\\frac14\\right)\\\\ &=25\\;. \\end{align}\n\nHere are plots of the cumulative distribution function and the probability mass function.\n\n$$\\textsf{Pr}(T=16)=\\textsf{Pr}(T\\gt15)-\\textsf{Pr}(T\\gt16)=\\frac{293289532268461}{7703510787293184}\\approx3.8\\%$$\n\nand\n\n$$\\textsf{Pr}(T\\lt30)=\\textsf{Pr}(T\\le29)=\\frac{292029927548835623394780280045}{412111655902378135987760922624}\\approx70.9\\%\\;.$$\n\nFor general $k$ instead of $4$ numbers required and general $m$ instead of $12$ numbers on the die, the result is\n\n$$\\textsf{Pr}(T\\le t)=\\sum_{j=0}^k(-1)^j\\binom kj\\left(1-\\frac{k-j}m\\right)^t\\;.$$\n\n-\n\nI think it's the sum of four different geometric distributions.\n\nFor example, you start off and you haven't rolled any of the numbers. The probability of rolling one of them is 4/12 so you start a geometric distribution with that parameter. Then you suddenly roll one of them. Now you need to roll one of the other three, each roll has a probability of 3/12 of getting one of them, so it's another geometric distribution.\n\nBy this logic I think T ~ Geo(4/12) + Geo(3/12) + Geo(2/12) + Geo(1/12) which of course is probably some hideous ditribution define in terms of convolutions (or you know, could miraculously be nice and of closed-form, I don't know what.)\n\n-\nThis is sort-of what I was thinking, but what are the next steps to get to the point when you can actually compute the probability densities? \u2013\u00a0sffc Dec 27 '12 at 10:01", "date": "2016-06-25 23:36:01", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/265801/probability-distribution-of-number-of-times-required-to-roll-all-of-a-certain-se", "openwebmath_score": 0.9813474416732788, "openwebmath_perplexity": 421.73851064352846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676490318649, "lm_q2_score": 0.8902942261220292, "lm_q1q2_score": 0.8721924514516118}}
{"url": "http://mathhelpforum.com/algebra/187127-meeting-point-between-two-vehicles.html", "text": "# Math Help - Meeting point between two vehicles\n\n1. ## Meeting point between two vehicles\n\nDriver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.\nDriver B starts and follows on a motorcycle from the same spot, 6,0 sec later.\nDriver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)\n\na) How long does it take driver B to catch up with driver A?\nb) How long have they driven when they meet?\nc) How fast do each driver drive when they meet?\n\nI have tried the four formulas:\n\nv = at+v0\n\ns = 1/2 (v+v0) * t\n\ns = 1/2 a * t^2 + v0 * t\n\n2 * a * s = v^2 - v0^2\n\nWith no luck at all. I am confident that some calculations might be wrong.\n\nAll help is appreciated, hopefully a quick response.\n\n2. ## Re: Meeting point between two vehicles\n\nOriginally Posted by roahau\nDriver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.\nDriver B starts and follows on a motorcycle from the same spot, 6,0 sec later.\nDriver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)\n\na) How long does it take driver B to catch up with driver A?\nb) How long have they driven when they meet?\nc) How fast do each driver drive when they meet?\nfor driver A ...\n\n$v_A = 0.8t$\n\n$s_A = 0.4t^2$\n\nfor driver B ...\n\n$v_B = 1.8(t-6)$\n\n$s_B = 0.9(t-6)^2$\n\nfor (a) and (b) , set $s_A = s_B$ , solve for t to find the time driver A's run ... don't forget that driver B's run is $(t-6)$. once you get the time, find the value of $s_A$ and $s_B$ (which should be the same, right?)\n\nfor (c) , determine the values of $v_A$ and $v_B$ at the time found above.\n\n3. ## Re: Meeting point between two vehicles\n\nOriginally Posted by skeeter\nfor driver A ...\n\n$v_A = 0.8t$\n\n$s_A = 0.4t^2$\n\nfor driver B ...\n\n$v_B = 1.8(t-6)$\n\n$s_B = 0.9(t-6)^2$\n\nfor (a) and (b) , set $s_A = s_B$ , solve for t to find the time driver A's run ... don't forget that driver B's run is $(t-6)$. once you get the time, find the value of $s_A$ and $s_B$ (which should be the same, right?)\n\nfor (c) , determine the values of $v_A$ and $v_B$ at the time found above.\nHow am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity\n\n4. ## Re: Meeting point between two vehicles\n\nOriginally Posted by roahau\nHow am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity\nwhen the two vehicles meet, they are at the same position.\n\n$s_A = s_B$\n\n$0.4t^2 = 0.9(t-6)^2$\n\nsolve for t , then proceed as I stated in my previous post.", "date": "2014-03-15 20:17:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/187127-meeting-point-between-two-vehicles.html", "openwebmath_score": 0.7768567204475403, "openwebmath_perplexity": 1157.869643564235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109494088426, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.872179582961966}}
{"url": "https://math.stackexchange.com/questions/3575506/s-n-is-a-group", "text": "# $S_n$ is a group\n\nThis is an extremely simple fact, but I want to make sure that the assertions I am making are valid. In particular, I assert several times that a function is a permutation if and only if it is a bijection from a finite set to itself. If this is fine, I believe my arguments below are likely fine. I cannot think of a good counterexample to this.\n\nThere are four requirements we need to verify: closed under product operations, associative, has an identity, and closed under inverses.\n\nClosed under product operation: An element of $$S_n$$ is a permutation of the elements $$1, 2, \\ldots, n$$. This is a bijection $$\\alpha: \\{1, 2, \\ldots, n\\} \\to \\{1, 2, \\ldots, n\\}$$. A composition of these bijections, $$\\alpha$$ and $$\\beta$$, is another bijection from $$\\{1, 2, \\ldots, n\\}$$ to $$\\{1, 2, \\ldots, n\\}$$, and hnce $$\\alpha \\beta$$ is another permutation of the elements $$\\{1, 2, \\ldots, n\\}$$.\n\nAssociativity. An element of $$S_n$$ is a permutation, i.e., a function from a set to itself, and function composition can easily be shown to be associative. For simplicity, take $$f, g, h$$ to be well-defined functions that map $$A$$ to $$A$$. For any $$a \\in A$$, we have: $$((f \\circ g) \\circ h)(a) = (f \\circ g)(h(a)) = f(g(h(a))$$ $$(f \\circ (g \\circ h))(a) = f((g \\circ h)(a)) = f(g(h(a)).$$ Hence, composing permutations is associative because permutations are just functions.\n\nIdentity. The identity permutation, $$\\text{id}$$, sends every element to itself. For any $$\\alpha \\in S_n$$ and any $$x \\in \\{1, 2, 3, \\ldots,n\\}$$, we have: \\begin{align*} (\\alpha \\circ \\text{id})(x) = \\alpha(\\text{id}(x)) = \\alpha(x) \\\\ (\\text{id} \\circ \\alpha)(x) = \\text{id}(\\alpha(x)) = \\alpha(x). \\end{align*} Hence, the identity permutation is the identity element in the group.\n\nInverses. Let $$\\sigma \\in S_n$$. Then, $$\\sigma$$ is a bijection from $$\\{1, 2, \\ldots, n\\}$$ to $$\\{1, 2, \\ldots, n\\}$$. A function is bijective if and only if it is invertible. Hence, another bijection, $$\\sigma^{-1}$$, also exists. Since this is a bijection from a set of finite elements to itself, it is also a permutation of $$\\{1, 2, \\ldots, n\\}$$, and hence lives in $$S_n$$.\n\n\u2022 Yes, it is fine. Mar 9 '20 at 23:33\n\u2022 \"There are four requirements we need to verify\". I believe the things went historically the other way around: after having realized that the set of bijections on a set does fulfill the four properties, where \"Inverses\" is actually the qualifying one (e.g. associativity holds for any map on a set), someone thought to define \"group\" an algebric structure patterned upon the \"4 properties\" of the bijections.\n\u2013\u00a0user750041\nMar 10 '20 at 11:24\n\nWith regards to this problem, yes, it is true that a function is a permutation if and only if it is a bijection from a finite set to itself. This is true essentially by definition of the word \"permutation\". In other contexts, the definition of permutation might be extended to allow bijections from any arbitrary (finite or infinite) set to itself (e.g. $$x\\mapsto 1/x$$ is a bijection from $$(0,\\infty)$$ to itself, and so is a permutation of $$(0,\\infty)$$), but of course this is not relevant here since we are only looking at finite sets $$\\{1,2,\\dots,n\\}$$.", "date": "2021-10-20 08:41:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3575506/s-n-is-a-group", "openwebmath_score": 0.9872289896011353, "openwebmath_perplexity": 169.64705427693593, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9848109494088426, "lm_q2_score": 0.8856314647623015, "lm_q1q2_score": 0.872179563638906}}
{"url": "http://mathhelpforum.com/number-theory/96246-chinese-remainder-theorem.html", "text": "1. ## Chinese Remainder Theorem\n\nFor the question,\n\nFind the smallest x which solves the system of congruence equations:\n\nx =+ 3 mod 15\nx =+ 6 mod 224, where =+ stands for congruence modulo operation.\n\nI gave an argument oriented solution as follows.\n\nAny multiple of 224 is even.\nSo, 6+ any multiple of 224 is even.\n=> x is even.\n\nAlso, any multiple of 15 ends with 5 or 0.\n3+any multiple of 15 ends with 8 or 3.\n\nSince X is even, the number should end with 8.\nSince X = A number ending with 8,\nAlso, X = A multiple of 224 + 6\nThis means the multiple of 224 ends with 2.\n\nChecking the multiples of 224, we get 224, 448, 672.\n\nThe smallest one is 672 ==> X is 678.\n\nThis solves both the equations as 678 = 224*3 +6\n= 45*15 +3\n\nMy friend argued about a general solution for the same problem using Chinese Remainder Theorem. I browsed results for the same in various sites.\nI find only a general solution found in most cases.\n\nCan anyone describe how to get the particular solution for this problem through Chinese Remainder Theorem?\n\nAlso, please mention if there are any mistakes in my argument-oriented solution.\n\n2. yes you can....\n\nso we have\n\n3 mod 15\n6 mod 224\n\ngcd(224,15)=1 so we use chinese thm.\n\nfirst thing we need is integers a and b s.t...\n\na15+b224=1\n\nand after we find a and b using euclid's algorithm we compute\n\n6a15+3b224 (mod 15.224=3360)\n\neuclid's algorithm gives a=15 and b=-1\n\nso 6(15)15+3(-1)224=678 mod 3360\n\nand adding and subtracting 3360 would give you many solutions.\n\nyou can also use this method to solve many problems which your method would be increasingly difficult to use for example....\n\nFind a number which is 3 mod 7, 2 mod 5 and 1 mod 2.\n\n7,5 and 2 are coprime so use chinese thm...\n\nfirst concentrate on 3 mod 7 and 2 mod 5\n\nfind a,b so that a7+b5=1\n\na=-2 and b=3 using euclid\n\ncompute 2a7+3b5 (mod 7.5=35)=2(-2)7+3(3)5=17 mod 35\n\nnow do the same with 17 mod 35 and 1 mod 2\n\ni need to compute 1a35+17b2 where a35+b2=1\n\na=1 and b=-17\n\nso 1a35+17b2=17 mod 70=87mod70\n\nNote: the reason why we need the gcd to be 1 is because the chinese remainder theorem is based on this.\n\nbut you have a nice little argument there, nice one.\n\n3. Here's a similar argument:\n\nx= 6 mod 224 so x= 224k+ 6 for some integer k. But 224= 14 mod 15 so that is the same as x= 14k+ 6 mod 15. We have x= 14k+ 6= 3 mod 15 or 14k= -3 mod 15 so 14k= 15j- 3 for some integer j. That is, 15j- 14k= 3. It is obvious that 15- 14= 1 (shortcutting the Euclidean division algorithm!) so 15(3)- 14(3)= 3. k= j= 3 is a solution. Taking k= 3, x= 224(3)+ 6= 678.\n\nBut if you can come up with a valid argument of your own, that's always best.\n\n4. @Krahl\nThanks for your descriptive reply for the question; I get the solution part right till the general solution; You had mentioned that adding and subtracting 3360 would yield other solutions; It is this part where I do mistakes some times.\n\nCan you please throw some more light on one or two of the particular solutions?\n\nMAX,\n\n5. @HallsofIvy\n\nThat's another interesting aliter for the same problem; Good one\n\n6. ## CRT\n\nI was just thinking it might actually help you to see the actual proof of the theorem so you can see why these solution methods actually work. I am not sure if you are familiar with rings and ideals and stuff, so I will leave it in terms of the integers, but if you are interested, the same proof idea works in general.\n\nYou assume that (m,n)=1 (this is enough to show in a ring (m)+(n)=R).\nThen you have the following isomorphism:\n$\\frac{\\mathbb{Z}}{mn\\mathbb{Z}}\\cong \\frac{\\mathbb{Z}}{m\\mathbb{Z}} \\times \\frac{\\mathbb{Z}}{n\\mathbb{Z}}$\n\nHere is the isomorphism:\n$\\phi(x)=(x$ (mod m), $x$ (mod n)).\nIt is a homomorphism:\n$\\phi(x+y)=(x+y$ (mod m), $x+y$ (mod n)) $=(x$ (mod m), $x$ (mod n)) $+(y$ (mod m), $y$ (mod n))= $\\phi(x)+\\phi(y)$.\n\nIt is pretty clear that $ker(\\phi)=mn\\mathbb{Z}$\n\nSo it remains to show that it is surjective, and this is the key part of the argument for what you are looking for.\n\nLet $(a$ (mod m), $b$ (mod n)) be an arbitrary element of $\\frac{\\mathbb{Z}}{m\\mathbb{Z}} \\times \\frac{\\mathbb{Z}}{n\\mathbb{Z}}$\n\nThis is basically saying we need some integer that will satisfy\n$x\\equiv a$ (mod m) and\n$x\\equiv b$ (mod n).\n\nSo this is the trick. Because we know (m, n) by the euclidean algorithm, there exists $s,t \\in \\mathbb{Z}$ such that $sm+tn=1$.\n\nJust a side note: In the more general setting you say that the ring contains a unit and since we have (m)+(n)=R, there exists $i\\in (m)$ and $j\\in (n)$ such that $i +j =1$. Actually the ring need not even be a PID, it is true for any ideals that are comaximal in a commutative ring.\n\nSo back to the surjectivity. We want to sort of cross multiply here and take the value $bsm+atn$ and see that it maps where we want it.\n\n$\\phi(bsm+atn)=(bsm+atn$ (mod m), $bsm+atn$ (mod n)) $=(atn$ (mod m), $bsm$ (mod n))= $(a(1-sm)$ (mod m), $b(1-tn)$ (mod n))= $(a-asm)$ (mod m), $b-btn$ (mod n))= $(a)$ (mod m), $b$ (mod n)).\n\nSo this proves the isomorphism. This shows you $bsm+atn$ (mod mn) is the number you want to pick to solve the system of equations above.\n\nThe fact that the kernel is $mn\\mathbb{Z}$ just tells you that you can actually just add kmn for any integer k to your solution and you will get all the solutions to this system of equations.\n\n7. Originally Posted by MAX09\n@Krahl\nThanks for your descriptive reply for the question; I get the solution part right till the general solution; You had mentioned that adding and subtracting 3360 would yield other solutions; It is this part where I do mistakes some times.\n\nCan you please throw some more light on one or two of the particular solutions?\n\nMAX,\nHi max\n\nnot sure what part of the solution you're asking for but i'll try writing a few things on the solutions.\n\nwell you agree that 678 is a solution right?\n\nNow the useful thing about using the chinese rem thm is that you end up not just with the answer 678 but also mod 3360.\n\nThis mean that any number which is modulo 3360 would satisfy your equations,\n\nfor example if we add and subtract 3360 we get the following alternative solutions;\n\n........, -2682,678,4038,7398.......,\n\nNow this means that all these numbers are 678 modulo 3360\ni.e. have remainder 678 when divided by 3360.\n\nso 4038=3360+678, 7398=3360*2+678 etc...\n\nAnd the theorem claims that all these solutions also satisfy, so let's check them;\n\n7398=15*493+3 so it is 3 mod 15\n\n7398=224*33+6 so it is 6 mod 224\n\nThat is why the question asks for the smallest positive solution.\n\nI did the same with the other example and got the solutions 17 mod 70 and 87 mod 70.\n\nAlso notice this step i did;\n\n6a15+3b224\n\nits 6 mod 224 that is why 6 is next to 15. since 224 = 0 mod 224 so that you are left with 6a15 + 0 mod224\n\nand so a*15 must give me 1 mod 224 so that\n6a15+0 mod 224=6 mod 224\n\nand same with mod 15.\n\nany problems dont hesitate to post\n\n8. @Gamma\n\nThat wasn't exactly the procedure I was looking for. Anyways, thanks for the effort.. It helped understanding the relation the CRT had with isomorphisms. Thanks a bunch !!!\n\n9. ## @krahl\n\nYes, your posted help me understand the procedure completely I had to get my basics on the Extended Euclidean Algorithm right. The following url helped me.\n\nhttp://marauder.millersville.edu/~bikenaga/absalg/euc/euclidex.html\n\nThanks again, Krahl!!\n\nMax", "date": "2017-08-20 06:45:18", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/96246-chinese-remainder-theorem.html", "openwebmath_score": 0.8390215039253235, "openwebmath_perplexity": 709.3654804639467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.976310532836284, "lm_q2_score": 0.8933094003735664, "lm_q1q2_score": 0.872147376666378}}
{"url": "https://lavalldebo.org/43xcagqf/3c3493-conditional-probability-problems-with-solutions-pdf", "text": "\ufffd\ufffd\ufffdurm\ufffd9\ufffdaj\ufffd\ufffdD\ufffd\ufffd\ufffd]\ufffd\ufffd\ufffd\u0581\ufffd\ufffd\ufffd\ufffdV\ufffdq/\ufffdC2B\ufffd\ufffd\ufffd\ufffdX\ufffd\ufffdc 1\ufffd\ufffdR~\ufffd\ufffd}6\ufffdK\ufffd\ufffd5?>\ufffd\ufffd\ufffd $P(HHH)=P(H)\\cdot P(H) \\cdot P(H)=0.5^3=\\frac{1}{8}$. All we need to do is sum immediately obtain $c=\\frac{1}{2}$, which then gives $a=\\frac{1}{3}$ and $b=\\frac{1}{2}$. The manual states that the lifetime $T$ of the product, Here is another variation of the family-with-two-children as in Example 1.18. \\frac{2}{3} + 1 . defined as the amount of time (in years) the product works properly until it breaks down, satisfies from a family with two girls is a girl is one, while this probability for a family who has understanding of probability. What is the probability that it is the two-headed coin? Example (Genetics) Traits passed from generation to generation are carried by genes. $$W =\\{HH, HTHH, HTHTHH,\\cdots \\} \\cup \\{THH, THTHH, THTHTHH,\\cdots \\}.$$ First, we would like to emphasize that we should not rely too much on our intuition when solving extra information about the name of the child increases the conditional probability of $$P(H|C_1)=0.5,$$ You pick a random day. about them to explain your confusion. Here we have four possibilities, $GG=(\\textrm{girl, girl}), GB, BG, BB$, If the child is a boy, his name will not be Lilia. and $P(GG)=P(GB)=P(BG)=P(BB)=\\frac{1}{4}$. Let $R$ be the event that it's rainy, $T$ be the event that there is heavy traffic, $=\\frac{e^{-\\frac{2}{5}}-e^{-\\frac{3}{5}}}{e^{-\\frac{2}{5}}}$, $=\\sum_{i=1}^{M} P(A|C_i)P(B|C_i)P(C_i) \\hspace{10pt}$, $\\textrm{ ($A$and$B$are conditionally independent)}$, $\\textrm{ ($B$is independent of all$C_i$'s)}$, $=\\frac{2}{3} \\cdot \\frac{1}{4} \\cdot \\frac{3}{4}$, $= P(R,T,L)+P(R,T^c,L)+P(R^c,T,L)+P(R^c,T^c,L)$, $=\\frac{1}{12}+\\frac{1}{24}+\\frac{1}{24}+\\frac{1}{16}$, $= \\frac{1}{2}. A family with two girls is more We have already found 0 This ... Marilyn gave a solution concluding that you should switch, and if you do, your probability of winning is 2/3. has at least one child named Lilia.$P(L)=\\frac{11}{48}$, and we can find$P(R \\cap L)$similarly by adding the probabilities BG$, but these events are not equally likely anymore. A box contains three coins: two regular coins and one fake two-headed coin ($P(H)=1$), This is another typical problem for which the law of total probability is useful. $$P(G_r|GB)=P(G_r|BG)=\\frac{1}{2},$$ visualize the events in this problem. result with the second part of Example 1.18. lose. $= \\frac{P(G_r|GG)P(GG)}{P(G_r|GG)P(GG)+P(G_r|GB)P(GB)+P(G_r|BG)P(BG)+P(G_r|BB)P(BB)}$, $= \\frac{1.\\frac{1}{4}}{1. Problem . What is the probability that both children are girls? is that the event$L$has occurred. Let$q=1-p$. The aim of this chapter is to revise the basic rules of probability. We assume that the coin tosses are independent. the law of total probability here. Here again, we have four h\ufffd\ufffdY\ufffdNI\ufffd\ufffd\ufffd\ufffd|\ufffd\u0588\ufffd}\ufffdZH\u0618\ufffd\ufffd\ufffd\ufffd\ufffd\u0219\ufffd\\Y\ufffdM\u007f}\ufffd%\ufffd\u0222\ufffdE\ufffd0\u070c\ufffd\ufffd\ufffd5ND)\ufffdb!\ufffdR\ufffdJJK\ufffd2 UXO-\ufffdP2Q\ufffd\ufffdJ{\ufffd,TN#\ufffd_S\ufffd/\ufffd\ufffdD\ufffdB'E|\ufffd0N\ufffdRa\ufffd\ufffdJ&E\ufffd\ufffdK)]\ufffdp5e The two problem statements look very similar but the answers are completely different. on the result of the first coin toss, We assume$P(A)=a, P(B)=b$, and$P(C)=c$. The probability that it's not raining and there is heavy traffic and I am not late can f{\ufffdHi_ \ufffd?h#\ufffd\"\u0605\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd9@\ufffd \ufffd\ufffd\ufffd Suppose we know that. problem [1] [7]. the probabilities of the outcomes that correspond to me being late. Read Free Conditional Probability Practice Problems With Solutions Probability Probability Conditional Probability 19 / 33 Conditional Probability Example Example De ne events B 1 and B 2 to mean that Bucket 1 or 2 was selected and let events R, W, and B indicate if the color of the ball is red, white, or black. Given that I arrived late at work, what is the probability that it rained that day? only one girl is$\\frac{1}{2}$. Each gene is of one of the types C(dominant gene) or c(recessive gene). Okay, another family-with-two-children problem. So, do not be disappointed if they seem confusing to you. This thinking process can be very helpful to improve our we are given conditional probabilities in a chain format. %PDF-1.6 %\ufffd\ufffd\ufffd\ufffd possibilities,$GG=(\\textrm{girl, girl}), GB, BG, BB$, and$P(GG)=P(GB)=P(BG)=P(BB)=\\frac{1}{4}$. I win if probability of$GG$is higher. Figure 1.27 shows a tree diagram for this problem. Here you can assume that if a child is a girl, her name will be Lilia with probability$\\alpha \\ll 1$win. We already know that As it is seen from the problem statement, Second, after obtaining counterintuitive results, you are encouraged to think deeply Let$C_1, C_2,\\cdots,C_M$be a partition of the sample space$S$, and$A$and$B$be two events. we know that the family has at least one girl. endstream endobj 1171 0 obj <>/Metadata 74 0 R/Names 1199 0 R/Outlines 196 0 R/PageMode/UseNone/Pages 1166 0 R/StructTreeRoot 233 0 R/Type/Catalog>> endobj 1172 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>> endobj 1173 0 obj <>stream In Then we have total probability for$A \\cap B$: In my town, it's rainy one third of the days. equally likely outcomes:$GG, GB, BG$, thus the conditional probability of$GG$is I toss a coin repeatedly. We can use the Venn diagram in Figure 1.26 to better On the other hand, if the outcome is$THTHT\\underline{HH}$, I win. Mashed Plantain Fritters, San Antonio Demographics By Neighborhood, Boneless, Skinless Sardines In Water, Is Flavacol Vegan, Roy Vs Marth Melee, What Does Shrew Poop Look Like, Orange Powerade Nutrition Facts, Office Chair Dimensions Cm, Hugo Grotius Believed That, Pmbok 7th Edition, Carrot Cake Cheesecake Bars Bake From Scratch, \" /> Skip to content Skip to main navigation Skip to footer # conditional probability problems with solutions pdf \\frac{1}{4}+ \\frac{1}{2} \\frac{1}{4}+ \\frac{1}{2} \\frac{1}{4}+0.\\frac{1}{4}}$, $=P(\\{HH, HTHH, HTHTHH,\\cdots \\})+P(\\{THH, THTHH, THTHTHH,\\cdots \\})$, $=p^2+p^3q+ p^4q^2+\\cdots+p^2q+p^3q^2+ p^4q^3+\\cdots$, $=p^2(1+pq+(pq)^2+(pq)^3+\\cdots)+p^2q(1+pq+(pq)^2+(pq)^3+\\cdots)$, $=\\frac{p^2(1+q)}{1-pq}, \\hspace{10pt}\\textrm{ Using the geometric series formula}$. 7\ufffdb\u1d65\ufffdTxknReqZ\ufffd\ufffdjZ>j\u07386,\ufffd\ufffd~\ufffd{\ufffd\ufffd\ufffd\ufffd\ufffd(\ufffd\ufffd\ufffd\ufffd\ufffdW\ufffd\u007f/f\ufffd\ufffdjv\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdF\ufffd\ufffd\ufffd8\ufffd\ufffdV\ufffdO\ufffd5( Z\u0157\ufffd\ufffd|(fA\ufffd\ufffd\ufffd\\\ufffd_\ufffd\ufffd\ufffd\ufffdre\ufffd\ufffdQ\u04c9\ufffd\ufffdi\ufffd>/g\ufffd\ufffd'+\ufffd\ufffd/\ufffd\ufffd\ufffd\ufffd\ufffd\u03ef\ufffd\u02d3r\ufffd\ufffdx&\ufffd\ufffd\ufffdx%\ufffd%\u0788C\ufffdV\ufffd!\ufffd\u01498\ufffd\ufffdX\ufffd\ufffdsq>\ufffdN\ufffd\ufffdB\ufffd\ufffd\ufffd\ufffdKq \u007f\u054fR\\\ufffdo\ufffd\ufffdJ\\\ufffdJL\ufffdT\ufffd\ufffd\\\ufffdg\ufffdX\ufffdE\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd|Z\ufffdZ\ufffd\ufffdr&\ufffd\ufffdX\ufffd\ufffd\ufffd\ufffd,\ufffd\ufffd\u03f9\ufffd?\u011f\u29f8\u007f\ufffd\ufffd\ufffd\ufffd\ufffdW\ufffd\ufffdv\ufffd\ufffd\ufffd\ufffd\ufffdU]XR\u01f3g\ufffd_w\ufffd)v\ufffds*\ufffdFp\ufffd\ufffdT\ufffd\ufffd\ufffd#Ea\ufffd\ufffdxZ\ufffd\ufffdw\ufffd\ufffd~\ufffd\ufffd\ufffd\ufffdg\ufffd\ufffd\ufffdV\ufffdIu\ufffd7\ufffd^)NV\ufffd\ufffd\ufffd\ufffd8\ufffd]\ufffd4\ufffd^V\ufffd\ufffd|)>5\u06b2\ufffd\ufffd\ufffd~\ufffd\u038f\ufffd\ufffd\ufffdQI&\ufffd\ufffd\ufffd(\ufffd\ufffd5\ufffd .\ufffd\ufffd\ufffd:\u068448\ufffdH\ufffdF\ufffd\ufffd\ufffd+\ufffd\"\ufffdS0\ufffd\ufffds\ufffd\ufffdfT\u034a\ufffd\ufffdt\ufffde\ufffd\ufffdf\ufffdu\ufffdS73f;# \ufffd\ufffdp\ufffd\ufffd,kHi[9a4\ufffdF\ufffd\"\u0200|\ufffdBK \ufffd\ufffd(a\ufffdF\ufffd\ufffdFi\ufffdT1Dk\ufffd)\ufffdH\ufffdD:\ufffd\ufffdh\ufffd5L\ufffd%\ufffdD:\ufffdl\ufffd$\ufffd>\ufffd\ufffd\ufffdurm\ufffd9\ufffdaj\ufffd\ufffdD\ufffd\ufffd\ufffd]\ufffd\ufffd\ufffd\u0581\ufffd\ufffd\ufffd\ufffdV\ufffdq/\ufffdC2B\ufffd\ufffd\ufffd\ufffdX\ufffd\ufffdc 1\ufffd\ufffdR~\ufffd\ufffd}6\ufffdK\ufffd\ufffd5?>\ufffd\ufffd\ufffd$P(HHH)=P(H)\\cdot P(H) \\cdot P(H)=0.5^3=\\frac{1}{8}$. All we need to do is sum immediately obtain$c=\\frac{1}{2}$, which then gives$a=\\frac{1}{3}$and$b=\\frac{1}{2}$. The manual states that the lifetime$T$of the product, Here is another variation of the family-with-two-children as in Example 1.18. \\frac{2}{3} + 1 . defined as the amount of time (in years) the product works properly until it breaks down, satisfies from a family with two girls is a girl is one, while this probability for a family who has understanding of probability. What is the probability that it is the two-headed coin? Example (Genetics) Traits passed from generation to generation are carried by genes. $$W =\\{HH, HTHH, HTHTHH,\\cdots \\} \\cup \\{THH, THTHH, THTHTHH,\\cdots \\}.$$ First, we would like to emphasize that we should not rely too much on our intuition when solving extra information about the name of the child increases the conditional probability of $$P(H|C_1)=0.5,$$ You pick a random day. about them to explain your confusion. Here we have four possibilities,$GG=(\\textrm{girl, girl}), GB, BG, BB$, If the child is a boy, his name will not be Lilia. and$P(GG)=P(GB)=P(BG)=P(BB)=\\frac{1}{4}$. Let$R$be the event that it's rainy,$T$be the event that there is heavy traffic,$=\\frac{e^{-\\frac{2}{5}}-e^{-\\frac{3}{5}}}{e^{-\\frac{2}{5}}}$,$=\\sum_{i=1}^{M} P(A|C_i)P(B|C_i)P(C_i) \\hspace{10pt}$,$\\textrm{ ($A$ and $B$ are conditionally independent)}$,$\\textrm{ ($B$ is independent of all $C_i$'s)}$,$=\\frac{2}{3} \\cdot \\frac{1}{4} \\cdot \\frac{3}{4}$,$ = P(R,T,L)+P(R,T^c,L)+P(R^c,T,L)+P(R^c,T^c,L)$,$=\\frac{1}{12}+\\frac{1}{24}+\\frac{1}{24}+\\frac{1}{16}$,$= \\frac{1}{2}. A family with two girls is more We have already found 0 This ... Marilyn gave a solution concluding that you should switch, and if you do, your probability of winning is 2/3. has at least one child named Lilia. $P(L)=\\frac{11}{48}$, and we can find $P(R \\cap L)$ similarly by adding the probabilities BG$, but these events are not equally likely anymore. A box contains three coins: two regular coins and one fake two-headed coin ($P(H)=1$), This is another typical problem for which the law of total probability is useful. $$P(G_r|GB)=P(G_r|BG)=\\frac{1}{2},$$ visualize the events in this problem. result with the second part of Example 1.18. lose.$= \\frac{P(G_r|GG)P(GG)}{P(G_r|GG)P(GG)+P(G_r|GB)P(GB)+P(G_r|BG)P(BG)+P(G_r|BB)P(BB)}$,$= \\frac{1.\\frac{1}{4}}{1. Problem . What is the probability that both children are girls? is that the event $L$ has occurred. Let $q=1-p$. The aim of this chapter is to revise the basic rules of probability. We assume that the coin tosses are independent. the law of total probability here. Here again, we have four h\ufffd\ufffdY\ufffdNI\ufffd\ufffd\ufffd\ufffd|\ufffd\u0588\ufffd}\ufffdZH\u0618\ufffd\ufffd\ufffd\ufffd\ufffd\u0219\ufffd\\Y\ufffdM\u007f}\ufffd%\ufffd\u0222\ufffdE\ufffd0\u070c\ufffd\ufffd\ufffd5ND)\ufffdb!\ufffdR\ufffdJJK\ufffd2 UXO-\ufffdP2Q\ufffd\ufffdJ{\ufffd,TN#\ufffd_S\ufffd/\ufffd\ufffdD\ufffdB'E|\ufffd0N\ufffdRa\ufffd\ufffdJ&E\ufffd\ufffdK)]\ufffd`p5e The two problem statements look very similar but the answers are completely different. on the result of the first coin toss, We assume $P(A)=a, P(B)=b$, and $P(C)=c$. The probability that it's not raining and there is heavy traffic and I am not late can f{\ufffdHi_ \ufffd?h#\ufffd\"\u0605\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd9@\ufffd \ufffd\ufffd\ufffd Suppose we know that. problem [1] [7]. the probabilities of the outcomes that correspond to me being late. Read Free Conditional Probability Practice Problems With Solutions Probability Probability Conditional Probability 19 / 33 Conditional Probability Example Example De ne events B 1 and B 2 to mean that Bucket 1 or 2 was selected and let events R, W, and B indicate if the color of the ball is red, white, or black. Given that I arrived late at work, what is the probability that it rained that day? only one girl is $\\frac{1}{2}$. Each gene is of one of the types C(dominant gene) or c(recessive gene). Okay, another family-with-two-children problem. So, do not be disappointed if they seem confusing to you. This thinking process can be very helpful to improve our we are given conditional probabilities in a chain format. %PDF-1.6 %\ufffd\ufffd\ufffd\ufffd possibilities, $GG=(\\textrm{girl, girl}), GB, BG, BB$, and $P(GG)=P(GB)=P(BG)=P(BB)=\\frac{1}{4}$. I win if probability of $GG$ is higher. Figure 1.27 shows a tree diagram for this problem. Here you can assume that if a child is a girl, her name will be Lilia with probability $\\alpha \\ll 1$ win. We already know that As it is seen from the problem statement, Second, after obtaining counterintuitive results, you are encouraged to think deeply Let $C_1, C_2,\\cdots,C_M$ be a partition of the sample space $S$, and $A$ and $B$ be two events. we know that the family has at least one girl. endstream endobj 1171 0 obj <>/Metadata 74 0 R/Names 1199 0 R/Outlines 196 0 R/PageMode/UseNone/Pages 1166 0 R/StructTreeRoot 233 0 R/Type/Catalog>> endobj 1172 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>> endobj 1173 0 obj <>stream In Then we have total probability for $A \\cap B$: In my town, it's rainy one third of the days. equally likely outcomes: $GG, GB, BG$, thus the conditional probability of $GG$ is I toss a coin repeatedly. We can use the Venn diagram in Figure 1.26 to better On the other hand, if the outcome is $THTHT\\underline{HH}$, I win.\n\nEsta web utiliza cookies propias y de terceros para su correcto funcionamiento y para fines anal\u00edticos. Al hacer clic en el bot\u00f3n Aceptar, acepta el uso de estas tecnolog\u00edas y el procesamiento de sus datos para estos prop\u00f3sitos. Ver", "date": "2022-09-30 16:01:17", "meta": {"domain": "lavalldebo.org", "url": "https://lavalldebo.org/43xcagqf/3c3493-conditional-probability-problems-with-solutions-pdf", "openwebmath_score": 0.9161287546157837, "openwebmath_perplexity": 850.0292654408531, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9763105287006255, "lm_q2_score": 0.8933093968230773, "lm_q1q2_score": 0.8721473695055755}}
{"url": "https://mathoverflow.net/questions/278268/can-squares-of-side-1-2-1-3-1-4-be-packed-into-three-quarters-of-a-unit-squ", "text": "# Can squares of side 1/2, 1/3, 1/4, \u2026 be packed into three quarters of a unit square?\n\nMy question is prompted by this illustration from Eugenia Cheng\u2019s book Beyond Infinity, where it appears in reference to the Basel problem.\n\nIs it known whether the infinite set of squares of side $\\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4},...$ can be packed into three quarters of a unit square?\n\n(It doesn't seem obvious that the illustrated packing could be continued to infinity.)\n\n\u2022 Answers regarding this similar problem might suggest some possible approaches: mathoverflow.net/questions/34145/\u2026 \u2013\u00a0Yoav Kallus Aug 8 '17 at 14:41\n\u2022 @YoavKallus Thanks! The comment by Timothy Chow referencing Marc Paulus's paper An algorithm for packing squares seems particularly relevant. sciencedirect.com/science/article/pii/S0097316597928363 \u2013\u00a0Robin Houston Aug 8 '17 at 14:53\n\u2022 I think this actually should be much easier than the other problem, just because $\\frac{\\pi^2}{6}-1<\\frac{3}{4},$ which gives you a chance of reducing this to a finite computation. In particular if there is an $n$ so that you can pack squares of side lengths $1/2\\cdots, 1/(n-1)$ and $n$ right triangles of side length $2/n$, then you can also pack the squares in the question. This is because you can pack squares of side length $1,1/2,\\cdots$ in a right triangle of side length $2$. \u2013\u00a0dhy Aug 8 '17 at 15:04\n\n## 3 Answers\n\nThe standard simple proof that $\\sum_{n=1}^\\infty \\frac1{n^2}$ converges is to round each $n$ down to the nearest $2^k$; this rounds each $\\frac1{n^2}$ up to the nearest $\\frac1{2^{2k}}$. In fact, one gets $2^k$ copies of $\\frac1{2^{2k}}$ for each $k$; hence $$\\sum_{n=1}^\\infty n^{-2} < \\sum_{k=0}^\\infty 2^k\\frac1{2^{2k}} = \\sum_{k=0}^\\infty \\frac1{2^{k}} = 2.$$\n\nThis is a sloppy estimate, and one way of improving it is to round each $n$ down to the nearest $2^k$ OR $3\\cdot2^{k}$. The first cluster of $\\frac1{2^{2k}}$s, comprising a single $\\frac1{2^0}$, is unaffected; all subsequent clusters cleave into $2^{k-1}$ copies of $\\frac1{2^{2k}}$ and $2^{k-1}$ copies of $\\frac1{(3\\cdot2^{k-1})^2}$. In this way, we get $$\\sum_{n=1}^\\infty n^{-2} < 1+\\sum_{k=1}^\\infty 2^{k-1}\\frac1{2^{2k}} + \\sum_{k=1}^\\infty 2^{k-1}\\frac1{(3\\cdot2^{k-1})^2} \\\\ = 1+\\sum_{k=1}^\\infty \\frac1{2^{k+1}} + \\frac19\\sum_{k=1}^\\infty \\frac1{2^{k-1}} = 1 + \\frac12+\\frac29.$$\n\nThis sum (minus the irrelevant first term) can be represented geometrically as in the following image, which is reasonably similar to the one you posted. (But the shaded region has area $\\frac16+\\frac19=\\frac5{18}>\\frac14$.)\n\n\u2022 That's a nice solution. I might add that the rows that start with 1/12, 1/24, 1/48, and so on can be rotated to columns and moved into the shaded region between 1/2 and 1/3 to obtain a solution to the problem in the original post. \u2013\u00a0Yoav Kallus Aug 8 '17 at 19:55\n\u2022 You might observe that the \"packing area\" in your image can be dissected and rearranged (without cutting any interior squares) into (a continuation of) the image in the question. Gerhard \"This Makes 'Yes' Appropriately Explicit\" Paseman, 2017.08.08. \u2013\u00a0Gerhard Paseman Aug 8 '17 at 19:58\n\u2022 Thank you! You are, of course, both correct---my main excuse is that I drew what was easiest for me to draw. (Note how the paper has been strategically folded.) I also wanted to emphasise the two geometric series (one up the left-hand side, the other up the right-hand side). If I weren't a dunce at computer graphics, I'd make an animation to show how the various pieces can be rearranged. \u2013\u00a0Jeff Egger Aug 8 '17 at 20:53\n\u2022 Instead of an animation, cut your square in half and rotate the right half. Then take the part that \"sticks out\" of the unit square and duplicate it to the appropriate place in the unit square, and draw an arrow showing the correspondence. Gerhard \"We Can Imagine The Motion\" Paseman, 2017.08.08. \u2013\u00a0Gerhard Paseman Aug 8 '17 at 21:51\n\u2022 If I didn't overlook anything, this does prove that total area is less than 3/4, but how does this prove that it is actually possible to put the squares? Perhaps there might be some \"leftover space between squares\" that can't be dealt with... Although my intuition says it is indeed possible, I think this part should also be explicitly addressed in an answer to this question. For example, I doubt this could be done with circles of areas $\\frac{1}{n^2}$. \u2013\u00a0Pedro A Aug 9 '17 at 10:36\n\nNote: This answer is wrong. There are two problems:\n\n1. The claim of Lemma 1 should presumably be read in the context of the global assumption in Paulhus's paper that $w\\leq l$. This assumption invalidates the instance I wanted to use (i.e. $l=\\frac12$, $w=1$).\n\n2. The proof of the lemma seems flawed, as mentioned below and succinctly summarised in Yoav Kallus\u2019s comment. Another example where the packing constructed for the proof does not fit is $l = \\frac{9}{5}$, $w = \\frac{7}{9}$, $n=2$, which I mention only since (unlike the example below) it has $w<l$.\n\nThe answer as written is next, followed by the note of doubt I added after trying to understand the proof.\n\nThe answer is yes.\n\nThanks to the comment by Yoav Kallus above, and a comment by Timothy Chow on the question he linked to, I found the paper An algorithm for packing squares by Marc M. Paulhus. (Journal of Combinatorial Theory, Series A, Volume 82, Issue 2, 1998, Pages 147-157)\n\nLemma 1 of that paper shows: All the reciprocal squares from n to infinity can be packed in a rectangle of length $l$ and width $w$ provided $$n\\geq \\frac{1+l}{wl}$$\n\nWith $l=1/2$ and $w=1$, this shows that we could place the square of side 1/2 in such a way that the remaining space is a rectangle of dimensions 1 x 1/2, and then pack the remaining squares into that rectangle using Paulhus's procedure.\n\nIt doesn't obviously answer the question of whether the packing could be completed if started as shown. In particular, can the squares of side 1/3, 1/4, etc. be packed into two unconnected squares of side 1/2?\n\nAdded previously: I am not convinced that the proof of Lemma 1 used above is correct. As I understand it (specialising to the case $l=\\frac 12$, $w=1$ for concreteness) the idea is to pack the squares in rows like this:\n\nIf the first square in row $i$ has side $1/{n_i}$, then $n_0=3$ and $n_{i+1}=\\lfloor \\frac{3}{2}n_i\\rfloor$.\n\nBut then the total height of the first seven rows is already $$\\frac 13 + \\frac 14 + \\frac 16 + \\frac 19 + \\frac 1{13} + \\frac 1{19} + \\frac 1 {28} > 1.$$\n\nOn the other hand this arrangement is not optimal, since the sixth row has squares $\\frac 1{19}, \\dots, \\frac 1{27}$ but we could fit up to $\\frac 1{30}$. So perhaps the claim is still true, notwithstanding the unconvincing proof?\n\n\u2022 The top-left square in the illustration in the question includes a free rectangle of dimensions $1/4\\times3/8$. By Lemma 1, this should be enough to hold reciprocal squares for $n\\ge14$. I believe 12 and 13 comfortably fit in the free space in the bottom-right square, next to the 11. \u2013\u00a0Emil Je\u0159\u00e1bek supports Monica Aug 8 '17 at 16:26\n\u2022 You seem to be correct that the proof is in error. Particularly, I see a false deduction in the last set of inequalities: the proof claims that if $w\\ge(1+l)/(n_0 l)$ ($=(1/n_0)\\sum_{i=0}^\\infty (1/(1+l))^i$), then $w\\ge \\sum_{i=0}^{\\infty} 1/n_i$ follows. That would make sense if we had $n_i\\ge n_0 (1+l)^i$, but the inequality shown earlier is actually reversed. \u2013\u00a0Yoav Kallus Aug 8 '17 at 19:48\n\u2022 Perhaps this could also be due to an assumption in the paper earlier that width is considered smaller (i.e., $l\\geq w$)? \u2013\u00a0LegionMammal978 Aug 8 '17 at 21:12\n\u2022 I think your argument is somewhat useful, though. \u2013\u00a0Takahiro Waki Aug 8 '17 at 23:42\n\u2022 @RobinHouston Even if you put as many squares as possible into each row, the height of the first seven rows will be $\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{9}+\\frac{1}{14}+\\frac{1}{22}+\\frac{1}{35}>1$. \u2013\u00a0Peter Mueller Aug 9 '17 at 7:29\n\nUsing Clive Tooth's heuristic for this similar (but more difficult) problem https://math.stackexchange.com/questions/2645195/can-the-squares-with-side-1-n-be-packed-into-a-1-times-zeta2-rectangle\n\nI was able to pack the first 10^9 squares into the 3/4 unit square. Not a proof, of course, but evidence, at least, that no \"bottleneck\" occurs when placing the smallest sized squares.", "date": "2020-01-17 22:40:01", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/278268/can-squares-of-side-1-2-1-3-1-4-be-packed-into-three-quarters-of-a-unit-squ", "openwebmath_score": 0.8118759989738464, "openwebmath_perplexity": 394.1163390178443, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127399388854, "lm_q2_score": 0.8824278772763472, "lm_q1q2_score": 0.8721147131894411}}
{"url": "https://cs.stackexchange.com/questions/127185/what-is-the-vc-dimension-of-the-hypothesis-class-h-left-f-theta-1-theta/127202", "text": "# What is the VC dimension of the hypothesis class $H=\\left\\{f_{\\theta_{1}, \\theta_{2}}: R^{2} \\rightarrow\\{0,1\\} \\mid 0<\\theta_{1}<\\theta_{2}\\right\\}$?\n\nI would like to know what is the VC dimension of the following hypothesis class.\n\n$$H=\\left\\{f_{\\theta_{1}, \\theta_{2}}: R^{2} \\rightarrow\\{0,1\\} \\mid 0<\\theta_{1}<\\theta_{2}\\right\\}$$\n\nwhere $$f_{\\theta_{1}, \\theta_{2}}(x, y)=1$$ if $$\\theta_{1} x \\leqslant y \\leqslant \\theta_{2} x,$$ else $$f_{\\theta_{1}, \\theta_{2}}(x, y)=0$$.\n\nI am not really sure how to prove it. What do you think?\n\nThe VC-dimension of your hypothesis class $$\\mathcal H$$ is 2. To see this, we begin by showing that $$\\mathcal H$$ shatters any 2-element set $$\\{(a_1 a_2), (b_1, b_2)\\}$$ of real numbers where all components of the pairs are positive:\n\n1. $$\\emptyset$$ is accounted for by $$f_{c, c + \\varepsilon}$$ for any real $$c$$ such that $$ca_1 \\neq a_a$$ and $$cb_1 \\neq b_2$$ and some sufficiently small $$\\varepsilon > 0$$.\n2. $$\\{a\\}$$ (and similarly $$\\{b\\}$$) is accounted for by $$f_{c, c + \\varepsilon}$$ where $$c = a_2/a_1$$ and $$\\varepsilon > 0$$ is sufficiently small.\n3. $$\\{a, b\\}$$ is accounted for by $$f_{\\varepsilon, c}$$ for some sufficiently small $$\\varepsilon > 0$$ and some sufficiently large $$c$$ (specifically, one can set $$\\varepsilon = \\min \\{a_2 / a_1, b_2 / b_1\\} / 2$$ and $$c = 1 + \\max \\{a_2 / a_1, b_2 / b_1\\}$$)\n\nThis yields $$\\operatorname{VC}(\\mathcal H) \\geq 2$$.\n\nNow consider some arbitrary set $$X = \\{(a_1, a_2), (b_1, b_2), (c_1, c_2)\\}$$ of pairwise distinct points in $$\\mathbb R^2$$. If the points in $$X \\cup \\{(0, 0)\\}$$ are not in general position then $$\\mathcal H$$ cannot shatter $$X$$ as it means that there are at least two points $$s, t \\in X$$ which fall onto a line with the origin and as every classifier in $$\\mathcal H$$ has linear decision boundaries it must always label $$s$$ and $$t$$ the same way, preventing it from shattering any set containing these points.\n\nSo let us assume that $$X \\cup \\{(0, 0)\\}$$ is a set of points in general position and shattered by $$\\mathcal H$$. As all functions $$f_{\\theta, \\varphi} \\in \\mathcal H$$ represent areas between two lines with positive slopes (since we require $$0 < \\theta < \\varphi$$) we can infer that all the lines connecting points in $$X$$ with the origin also must have positive slopes (note that $$X$$ is not in general position when the origin is an element of $$X$$). Hence we can order the points in $$X$$ ascendingly by these slopes, i.e. we may write $$a_2 / a_1 < b_2/ b_1 < c_2 / c_1$$ and since this requires all $$x$$-coordinates of the points in $$X$$ to be nonzero, we can simplify the expression for the preimage of $$1$$ of any $$f_{\\theta, \\varphi} \\in \\mathcal H$$ restricted to $$X$$ to $$f_{\\theta, \\varphi}|_X(x, y)^{-1} = \\{(x, y) \\in X \\mid \\theta \\leq y / x \\leq \\varphi\\}$$ by dividing by $$x$$. Now consider the subset $$X' = \\{a, c\\}$$ of $$X$$ and suppose $$f_{\\theta, \\varphi} \\in \\mathcal H$$ accounts for $$X'$$, i.e. $$f_{\\theta, \\varphi}|_X^{-1}(1) = X'$$. But then we must have $$\\theta \\leq a_2 / a_1 < b_2 / b_ 1 < c_2 / c_1 \\leq \\varphi$$ which yields $$f_{\\theta, \\varphi}|_X(b) = 1$$ and thus $$f_{\\theta, \\varphi}$$ does not account for $$X'$$ which is a contradiction. Therefore $$\\mathcal H$$ does not shatter $$X$$ and and thus no set of size $$\\geq 3$$.\n\nIt follows that $$2 \\leq \\operatorname{VC}(\\mathcal H) < 3$$ and thereby $$\\operatorname{VC}(\\mathcal H) = 2.$$\n\n\u2022 Thanks for the amazing explanation. Is it easy to make an example with 3 samples that give an intuition of how you proved that it the VC(H) cannot be more than 3? \u2013\u00a0hinduml Jun 15 '20 at 10:39\n\u2022 For $\\operatorname{VC}(\\mathcal H) < 3$ we need to show that no such sample exists, so considering examples is not enough. However, the idea here is a geometric one: Note that the set where some $f \\in \\mathcal H$ is 1 is the area between two lines with positive slopes, so if $f$ is 1 on the two \"outer\" samples then it must also be 1 on the \"middle\" sample so it cannot shatter any such set. I wrote a formal proof for this (which is a bit more involved and technical) but the idea is rather simple if you draw up a picture :) \u2013\u00a0Watercrystal Jun 15 '20 at 11:20\n\u2022 Thanks a lot for the explanation, its really helpful. \u2013\u00a0hinduml Jun 15 '20 at 11:57\n\u2022 Glad I could help out :) \u2013\u00a0Watercrystal Jun 15 '20 at 12:02\n\u2022 The two linear functions that bound the area are $\\theta x$ and $\\varphi x$ with $0 < \\theta < \\varphi$ required by your definition. If the slope is $> 0$, then it must be positive. The value of $x$ does not play a role for that. However, a linear function $ax$ with a positive slope (i.e. $a > 0$) attains negative values when $x$ is negative. \u2013\u00a0Watercrystal Jun 15 '20 at 15:30", "date": "2021-04-14 08:55:12", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/127185/what-is-the-vc-dimension-of-the-hypothesis-class-h-left-f-theta-1-theta/127202", "openwebmath_score": 0.9148173928260803, "openwebmath_perplexity": 113.22868080510965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127420043055, "lm_q2_score": 0.8824278695464501, "lm_q1q2_score": 0.8721147073724698}}
{"url": "https://www.physicsforums.com/threads/solving-an-inequality.332739/", "text": "# Solving an Inequality\n\n1. Aug 26, 2009\n\n### S_David\n\nHello,\n\nWhere is the mistake in the following solution of the inequality:\n\n\\begin{align*} &\\frac{2x-5}{x-2}<1\\\\ &2x-5<x-2\\\\ &x<3, x\\ne 2 \\end{align}\n\n2. Aug 26, 2009\n\n### Hurkyl\n\nStaff Emeritus\nThe following is not true:\nIf a<b, then ac<bc\u200b\n\nIn fact you have three separate cases, depending on c -- do you know what they are?\n\n3. Aug 26, 2009\n\n### HallsofIvy\n\nStaff Emeritus\nHurkyl's point is that you multiplied both sides of the inequality by x- 2 and, since you don't know what x is, you don't know if x- 2 is positive or negative. (Oops, I just gave you two of the three cases Hurkyl asked about!)\n\nMy preferred method for solving anything more than linear inequalities is to first solve the corresponding equation. Here, solve $$\\frac{2x-5}{x-2}= 1[/itex]. The \"x\" that satisfies that and the \"x\" that makes the denominator 0 are the only places where the inequality can \"change\". They divide the real line into three intervals- check one point in each interval to see which give \">\". 4. Aug 26, 2009 ### \u0414\u044c\u044f\u0432\u043e\u043b And why don't you try: [tex]\\frac{2x-5}{x-2} - 1 < 0$$\n\n?\n\nThen consider these cases\n\n$$a/b <0$$\n\nshould \"a\" and /or \"b\" be positive or negative so that the the fraction a/b would be negative?\n\nRegards.\n\n5. Aug 26, 2009\n\n### arildno\n\nYou may split your analysis into two cases:\n\n1) x-2>0,\nand\n\n2) x-2<0\n\nYou will find that for 2), there are NO solutions, meaning that x must be greater than 2.\n\n6. Aug 26, 2009\n\n### uart\n\nAnother excelent method in this case is to multiply both sides by (x-2)^2, with the squared assuring that we haven't multiplied by a negative number, and then just re-arrange it to give a quadratic inequality : x^2 - 5x + 6 < 0\n\n7. Aug 26, 2009\n\n### S_David\n\nOk, I can see where I did mistake the inequality. So, we have two cases:\n\n\\begin{align} x-2&>0\\\\ x-2&<0 \\end{align}\n\nIn the first case the multiplication does not change the direction of the inequality, so:\n\n\\begin{align*} 2x-5&<x-2\\\\ x&<3 \\end{align}\n\nThen the open interval $$(2,3)$$ is the solution set. In the second case, the direction of the inequality changed, so:\n\n\\begin{align*} 2x-5&>x-2\\\\ x&>3 \\end{align}\n\nBut $$x<2$$, then there is no solution.\n\nThe solution provided by uart is a good one, too. Where he eliminated the problem of negative sign possibility in the unknown $$x$$.\n\nActually, I am reviewing the precalculus and calculus books, and such things I forgot because I don't practice it continousely.\n\nAnyway, thank you all guys.\n\nBest regards\n\n8. Aug 26, 2009\n\n### \u0414\u044c\u044f\u0432\u043e\u043b\n\nYou could solve it this way:\n$$\\frac{2x-5}{x-2} - 1 < 0$$\n\n$$\\frac{x-3}{x-2}<0$$\n\n$$\\begin{bmatrix} \\left\\{\\begin{matrix} x-3<0\\\\ x-2>0 \\end{matrix}\\right. \\\\ \\\\ \\left\\{\\begin{matrix} x-3>0\\\\ x-2<0 \\end{matrix}\\right. \\end{matrix}$$\n\n$$\\begin{bmatrix} \\left\\{\\begin{matrix} x<3\\\\ x>2 \\end{matrix}\\right. \\\\ \\\\ \\left\\{\\begin{matrix} x>3\\\\ x<2 \\end{matrix}\\right. \\end{matrix}$$\n\nBecause 3<x<2 is not valid,\n\nthe only solution is 2<x<3 or (2,3).\n\nRegards.\n\n9. Aug 27, 2009\n\n### S_David\n\nDear \u0414\u044c\u044f\u0432\u043e\u043b,\n\nThis method is as described in my calculus book, and yes it is easier than the one I explained earlier. But I wanted to know the different methods to solve the inequality.\n\nThanks", "date": "2017-11-19 05:43:10", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/solving-an-inequality.332739/", "openwebmath_score": 0.7810973525047302, "openwebmath_perplexity": 1217.1283864324423, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127437254888, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.8721147043075561}}
{"url": "http://antonim.info/v1eyn/how-many-subsets-in-a-set-with-10-elements.html", "text": "A subset of the set X is a set containing only elements of X, but not necessarily all of its elements. Input. Thus if a10 is the number of ways to form a set from the list 2, 4, 6,,20 containing no two consecutive elements, then the number of subsets of. 210 D. (10) 40 How many subsets of a set with 100 elements have more than one element? (11) 44 How many ways are there to seat four of a group of ten people around a circular table where two seatings are considered the same when everyone has the same immediate left and right neighbor. How many elements are in A 1 B? Solve the problem by applying the Fundamental Counting Principle with two groups of items. 1. The mean of data set A is 27. If Sets A and B have 22 elements in common, how many elements are in A but not in A proper subset of a set A is a subset that is strictly contained in A and excludes at least one member of A. asked by Bryan on January 26, 2012; Set Theory. So, there will be 2^20 = 1048576 subsets for a set of 20 elements. For example, f1;2;3g is a set, and its subsets are: fg;f1g;f2g;f3g;f1;2g;f1;3g;f2;3g;f1;2;3g How do I find the number of subsets and the number of proper subsets for the following set: {2, 4, 6, 8, 10}? Mathematics The number of subsets of a 5-element set is 2^5 = 32. Suppose you add a new element to set A. Subsets with more than one element =2^100 - 101 Suppose, the set is. Therefore, a set of elements have subsets. How many subsets of S have at most 3 elements . 03. The number of subsets of a set having 100 elements number of subsets with at most two elements of a set having 100 elements. 95 and the mean of data set B is 30. The first \"one\" is the subset with 0 elements (the empty set is a subset of all sets) The number of subsets of a set with n elements are 2 n. Two elements iii. 30. {3} 5. Step-by-step explanation: If A is a set containing n elements then total numbers of subsets of A are . how many elements and subsets does P(A) has? i think P(A)={ {} } has 1 element ie {}. B. How about: {apple, banana, cherry} OK, let's be more systematic now, and list the subsets by how many elements they have: Apr 22, 2010 \u00b7 There are 5C2, which is 10, subsets with exactly 2 elements. If Sets A and B have 22 elements in common, how many elements are in A but not in A set with n elements will have 2 n subsets. 1 Sets and Set Operations Sets A set is a well-de\ufb01ned collection of objects. That is the number of subsets of is equals to . preparation course with premium videos, theory, practice problems, TA support and many more features. Clearly, the order of which we put the toppings on does not really affect the final identify the IMPROPER SUBSET (which contains all the elements of the original set). . Dec 23, 2018 \u00b7 The power set of a set A is the collection of all subsets of A. 101 subsets that have at most 2 elements. or equivalently, where n and r are nonnegative integers with r \u00a3 n. Sep 01, 2008 \u00b7 A set has 128 subsets. Two integers are coprime if their greatest common divisor equals 1. a. Find how many subsets of given array have sum between A and B(inclusive). Therefore, total number of subsets is $\\hspace{1mm} 2^{10} = 1024$. (2^n because each element can be either present(1) or absent(0). Similarly, for any finite set with elements, the power set has elements. Minimum number of subsets to distinguish individual elements Author: Meyers Subject: Given a set S of cardinality m , we determine the minimum cardinalityf(m) for a family F of subsets of S such that each seS can be expressed as the intersection of some subfamily of F. play. However, if we were to list all of the subsets of a set containing many elements, it would be quite tedious. How many different subsets of the set {10,14,17,24} are there that contain an odd number of elements? (a)3 (b)6 (c)8 (d)10 (e) 12 Please explain! A Power Set is a set of all the subsets of a set. In how many\u00a0 A subset of a set A is another set that contains only elements from the set A, but may not Let A = {2, 3, 6, 8, 9, 10} and B = {5, 6, 9}. Example 3. 2%5E10 = 1024 including the empty subset and the subset coinciding with the given set. That is, the power set of a finite set is finite, with cardinality 2 n. Number of distinct subsets of a set. I got the idea of the tuples in this question: N subsets with a given sum? However that problem is slightly different from mine so I was wondering if there's a way to speed things up. Let's say that the set B-- let me do this in a different color-- let's say that the set B is subset of A. How many new subsets can be created where one of the elements of these subsets is the new element? Date: 06/14/2002 at 22:37:09 From: Doctor Carbon Subject: Re: calculating number of possible sets of a subset Hi Jim, Here's my way of looking at this problem: If I want all the subsets of a certain size m out of a set with n distinct elements, then the formula for that is the same as finding the binomial coefficient. Sets can themselves be elements. Dec 11, 2011 \u00b7 A set A of k elements has k(k-1)/2 subsets of two elements, as you said. The cardinal number for an empty set is zero. chantalsantos|Points 532| User: If \u0192 = {(5, 1),(6, 2),(7, 3 If is a finite set with elements, then is finite and has elements. First question:Including itself, how many subsets does the set {1, 2, 3} have? List . e. D. Jayci F. the combinations from a set of 5 objects taken 3 at a time, we are finding all the 3-element subsets. Next we \ufb01nd the subsets that do contain B: {B}and {A,B}. For example: A = { 4 } Proper subset : { } ( 1 ) The total number of subsets is 2 100. Each set has one improper subset \u2014 that is, just one subset that contains all of the original set\u2019s elements \u2014 so the number of proper subsets in each set of n elements is the number of subsets minus one, or . A subset contains at least one of the elements the set. There are 10 combinations of the 5 letters taken 3 at a time. F. Making a 5-element subset of A with exactly two even elements is a 2-step process. , 10} is ______. The set can contain duplicate elements, so any repeated subset should be considered only once in the output. In many of our applications, however, what the sec- the number of permutations of a set of n elements is are 10! permutations, we would be doubly counting. In how many ways can we do so? De nition 3. A and B are subsets of S. So the subsets of the set is equals to . Hence, If a set has 1,024 subsets,then it has 10 elements Jan 31, 2013 \u00b7 Given that elements {8,9,10} are elements of the subset, then the elements 1 through 7 can either be in or out of a subset, i. Section 6. Calculation: The given set is, A = {a} The number of elements in the given set is 1. When we find all the combinations from a set of 5 objects taken 3 at a time, we are finding all the 3-element subsets. The power set must be larger than the original set and is closely related to the binomial theorem. Basically, you just have to implement a binary counter , i. 3) - Duration: 7:01. ELEMENTS in the subsets CAN BE LISTED IN ANY ORDER. 2) Suppose that, for some fixed number, k, any set with k members has $2^k$ subsets. Let n and r be integers with 0 r n. Absolute Value Inequalities A set of real numbers is graphed. P(S) is the notation for representing any power set of the set. Proper Subset Calculator. +10C10=2^10. Take this, and remove the number of subsets with no elements in it (just 1, the empty set). 4:45 \u00b7 Problems on What is the total number of proper subsets of a set containing n elements? play. The problem is solved in the following inverse form. So there are a total of 2\u22c52\u22c52\u22c5\u22ef\u22c52. Any subset of a finite set is finite. A set X is a collection of objects in no particular order, and with no repetition. The idea is to use a bit-mask pattern to generate all the combinations as discussed in previous post. Mar 13, 2009 \u00b7 Assuming there are 16 items in your set, each item can be included or not. i. How many subsets of S have an even number of elements? Express your answer as a summation. This proof is a counting proof--we need to know how many of something there are, so we just make sure we count everything once. 40 C. Find the number of subsets of 3 elements the set has. How many subsets are there altogether? What relationship does this number have to the number of elements in the set? c. A finite set with n elements has 2 n distinct subsets. The data sets have the same standard deviation. We have to find the total number of elements in a set . , 2 choices for each of the 7 remaining elements. The subset (or powerset) of any set S is written as P(S), P(S), P(S),P(S) or 2S. let's see. Assume that a set with N elements has 2 N subsets and calculate the number of subsets of a set, A, with N+1 elements as follows: Choose a specific member of A (call it a). 20. A subset of a set is also a set that contains some or all elements of the original set. It is defined as a subset which contains only the values which are contained in the main set, and atleast one value less than the main set. 001010 That is, any 1 H ( Z-element sets have at least as many (1 + I)-element supersets\u00a0 A is a subset of B iff every element of A is an element of B. Hence, n=10. Elements, subsets, and set equality (Screencast 2. Let the Universal Set, S, have 136 elements. 10. Therefore, the total number of subsets of set A=2 n. The set of values of a function when applied to elements of a finite set is finite. The number of subsets of 3 elements of the given set is equal to the number of combinations of 5 objects taken 3 at a time. youtube. A set of elements will have subsets. Sep 18, 2019 \u00b7 Since the assumed set A has 10 elements namely {1,2,3,4,5,6,7,8,9,10}. To determine this, we make use of the following rules pertaining to subsets of a set. Example: Set A: The natural numbers from 1 to 10. A General Note: Formula for the Number of Subsets How many subsets of the set $\\\\{1,2,3,4,\\\\ldots,10\\\\}$ have no two consecutive numbers as members? Its not 50 A set with 9 elements has 2^9 = 512 subsets. This is similar to subset sum problem with the slight difference that instead of checking if the set has a subset that sums to 9, we have to find the number of such subsets. . So the number of proper subsets is 2 9 - 1 or 2 10 - 1 depending on how you define the Natural Numbers. In other words, an improper subset of set is set itself. of subsets. May 14, 2019 \u00b7 subsets. The value of \"n\" for the given set A is \"5\". The general formula for the number of *proper* subsets is: 2^n - 1. To me, the green numbers look a lot like the rows of Pascal's triangle: A set with a single element has two subsets, the empty set and the entire set. 1 Questions & Answers Place. We know that . Maybe an example will help And these are also subsets: {a,b}, {a,c} and {b,c} And altogether we get the Power Set of {a,b,c}: Think of it as all the different ways we can select the items (the order of the items doesn't matter), including selecting none, or all. Anyone have a nice Given a set of N integers. (You do not need to simplify expressions involving permutations and/or combinations. { 8, 15, 28, 41, 60 Example The following Venn diagram shows the number of elements in each region for the sets A;B and C which are subsets of the universal set U. If set A A A is the set containing the first 10 prime numbers, how many subsets does A A A have? Since set A A A contains 10 distinct elements, we see that \u2223 A \u2223 = 10 | A | = 10 \u2223 A \u2223 = 1 0 . Part 1 Module 1 Set Mathematics Sets, Elements, Subsets Any collection of objects can be considered to be a set. asked by Em on September 28, 2016; Math. Adding up the odd combinations will total the subsets containing an odd number of elements. 14 Jun 2019 How do we find the number of subsets a set has? To do this, we have to figure out how many elements our set has. Number of elements 0=1subset, 1=2subsets, 2=4subsets, 3=8subsets, 4=16subsets, and so on. 01. E. You can put this solution on YOUR website! how many subsets are possible in a set with 3 elements. c) How many people would you have to hire for 4 days to buy tickets with all the possible \u00a0 Question from Class 10 Chapter Real Numbers How many functions can be defined from a set A containing 5 elements to a set. Step-by-step explanation: We are given that a set whose total number of subsets are 16. the number of subsets of the set with at least one element is given by. Remark. Consider a set {a,b,c} The 8 subsets are { }, {a}, {b}, {c}, {a,b}, {a,c}, {b,c} and {a,b,c}. Given a set S with n elements. THIS SET IS OFTEN IN FOLDERS WITH To find this we can use combinations. So the required ans is\u00a0 8 Jul 2019 Total number of subsets = 1+10C1+10C2+10C3\u2026\u2026\u2026. There should be 2^5=32 subsets including the empty set and the set itself. Consider the set of 5 elements. Jun 15, 2019 \u00b7 Consider that the set has 10 elements and we need to find the subsets with an odd number of elements. S = { 1, 2,3,4,5,6,7,8,9,10 } Solution: Number of elements in the set = 10 100 subsets will have one element and a nullsettherefore subsets having one or less than one element = 101. Each element of is a subset of . 3. In (i) A = {x | x is a positive integer less than 10 and 2x \u2013 1 is an odd number}. When working with a finite set with n elements, one question that we might ask is, \u201cHow many elements are there in the power set of A?\u201d Assume that the set S has 7 elements. Two examples: we could consider the set of all actors who have played The Doctor on Doctor Who, or the set of natural numbers between 1 and 10 inclusive. Assume that the set S has 7 elements. First, our two subsets can have 2 and 5 elements. Then, the formula to find number of subsets is. You can use Pascal's triangle to figure out how many subsets have no elements, one element, two elements and so on. Total subsets of with odd number of elements is: Oct 18, 2012 \u00b7 Assume there are X_n subsets for [n] like that. The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, C By having a factor of two and one of five, then the product of the elements of the set is just some multiple of $10$, precisely what we desire. Substituting for in, to get. My first inclination was to use something like RandomSample[Subsets[list, {25}], 1000], but the problem is the number of subsets of length 25 out of a 300 element set is way to big for the computer to deal with. A set containing 10 elements can have subsets which will contain 1 element, 3 elements, 5 elements, 7 elements, and 9 elements. Consider a set with one element: {A}. Thus, the required number of subsets Permutations and Combinations Questions & Answers : Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements. How many subsets does the set {apple, banana} have? It could have {apple}, or {banana}, and don't forget: the whole set: {apple, banana} the empty set: {} So a set with two elements has 4 subsets. Set A contains 34 elements and Set B contains 98 elements. asked by Elle on April 27, 2009; Sets. if they don't include n then there are X_(n-1) of those subsets if they include n then they can't include n-1 (no consecutive) so we have choices of n-2 numbers that is X_(n-2). There are ways to choose the 2-element subset. 1 :49. No distinction will be made between subsets except for their size. The objective is to find the number of subsets with more than two elements of a set having 100 elements. What you want is the number of subsets with either $2,5$ or $4,5$, i. How many 5-element subsets of $$A = \\{1,2,3,4,5,6,7,8,9\\}$$ have exactly two even elements? Solution. To find the number of elements of a set we will use formula of finding number of a set contain n elements User: How many subsets does set A have if the set A has 3 elements?3 6 8 Weegy: The subsets will be 8. (i) How many subsets are there that do not contain a? (Hint: how many subsets does A-{a] have?) Given a set of numbers: {1, 3, 2, 5, 4, 9}, find the number of subsets that sum to a particular value (say, 9 for this example). If you continue browsing the site, you agree to the use of cookies on this website. C. To find how many proper subsets there are in a set you can use the formula n^2 -n and if you would also like to find all subsets including improper the formula is n^2 -n +1 Asked in Math and If a set contains elements then the number of subsets of the set is equals to . Based on your answer to part b, how many subsets would a 10 -element set have? A 100 -element set? 24 May 2017 The number of 9-element subsets is (109), and the number of 10-element subsets is (1010). 64. How many subsets will A \u00d7 B have? List them. There are 5C3, which is 10, subsets with exactly 3 elements. for example: for the set S={1,2,3,4,5} means that S has 5 P(S) = 2 n = 2 5 = 32 A set with 3 elements has 1 subset with 0 elements, 3 subsets with 1 element, 3 subsets with 2 elements, and 1 subset with 3 elements etc. There are 2 11 such subsets. Oct 24, 2009 \u00b7 Set has N elements: then there are 2^N subsets. More on this later! Proof. The symbol n r is read choose r\" and represents the number of subsets of size r that can be chosen from a set with n elements. 3 2 6 10 4 1 3 2 C B A Find the number of elements in each of the following sets: (a) A\\B \\C 2 (b) B0 3+2+4+10 = 19 (c) A\\B 3 + 2 = 5 (d) C 2 + 4 + 2 + 1 = 9 (e) B [C 9 + 3 + 6 = 18 For elements in category theory, see Element (category theory). nCr where n=10 and r=(1,3,5,7,9) 10C1 + 10C3 + 10C5 + 10C7 + 10C9 Let be a set with 10 elements. Note that for any nonnegative integer, and so for any finite set , (where absolute value signs here denote the cardinality of Feb 05, 2009 \u00b7 A set with n number of elements can have numerous subsets (a subset is all possible combination you can think of made with the elements of the given set): 1. An improper subset contains ALL the elements of the set. The cardinality of the power set of {0, 1, 2 . Let A be a set with eight elements. Note that a subset might contain nothing at all. Given a set of positive integers, find all its subsets. 31) There are 5 roads leading from Bluffton to Hardeeville, 8 roads leading from Hardeeville 31) Jan 14, 2004 \u00b7 When n= 0, this is the empty set which has 1= 2 0 subsets. The objects in this collection are called elements of the set. 10!/(10-2)! = 90 subsets that have 2 elements. To find the number of subsets of a set with n elements, raise 2 to the nth power: That is: The number of subsets in set A is 2 n , where n is the number of elements in set A. If a is an element of the set A then we write a 2 A,ifa is not an element of a set A,thenwe write a/2 A. We can define particular sets by listing the objects in each set. No elements b. How many subsets of $$A$$ can we construct? To form a subset, we go through each of the $$n$$ elements and ask ourselves if we want to include this particular element or not. |A \u00d7 B| = |A||B|. See full answer below. non-distinct elements. There is 1 subset with exactly 1 element. Counting, we can conclude that a set containing 10 elements will have 5 subsets which will contain odd number of elements. The number of subsets is always 2^n where n is the number of elements in the set; in this case 5. Subset *to check that you have all of the subsets, remember that the number of subsets is equal to 2n where n = number of elements in the set. How many subsets are there of this set? We want to reduce this problem to the previous one. The proper subsets of Q are { }, {x}, {y}, {z}, {x, y}, {x, z}, {y, z} What is the formula for the number of subsets and proper subsets? The number of subsets for a finite set A is given by the formula: If set A has n elements, it has 2 n subsets. Set S has 11 elements all the subsets that have at most 2 elements means all the subsets with 2 elements, all the subsets with only 1 element and the empty set or null set how many ways can you choose 2 elements from 11 elements ? Assume that the set S has 7 elements. The 5-element subset (aka the set) will contain the 2-element subset. In the first case, Tom Baker is a element (or member) of the set, while Idris Elba, among many others, is not an element of the the number of subsets of a set having n elements is given by. Members of A: 1, 2, 3 Equal Sets \u2013 Two sets that contain exactly the same elements, regardless of the order listed or Example: How many Subsets and Proper Subsets does Set A have? For a given set S with n elements, number of elements in P(S) is 2^n. of subset = 2^4 =2*2*2*2 =4*4 =16 total number ob sub set in set A are 16 ii. 28, 29, 34, 36, Subsets. But we know there are n! permutations of an n\u00adelement set, so by the Division Rule, we conclude that n 30) Set A contains 5 elements, set B contains 11 elements, and 3 elements are common to sets 30) A and B. Free Q&A Aptitude and Reasoning Well, there is one subset with NO elements (the empty set), and 1 subset with just one element. 10 5 = 252 possible subsets with 5 elements of a set with 10 elements, by the pigeonhole principle it follows that at least two have the same sum. want to select a subset with r elements from a set with n elements. Prove it as well. Number of proper subsets = 2\u2075 = 32. A has 5 elements, so its power set has {eq}2^5 = 32 {/eq} elements. If set A has n elements, it has 2 n - 1 proper sets. Then the total number of subsets is given by$\\hspace{1mm} 2^{n(S)}$. com/watch?v=bqxUGv7vecY&index =3&list=PLJ-ma5dJyAqq8Z-ZYVUnhA2tpugs_C8bo. Suppose Set B contains 69 elements and the total number elements in either Set A or Set B is 107. The sets that do not contain B are the same as the subsets of {A}. Thus, {A, C, B} names the same set as {A, B, C}. The subset of A containing one element each -\u00a0 13 May 2015 This video shows how to use inductive reasoning to find a formula for the number of subsets and then uses deductive reasoning to prove it. The objective is to find the number of subsets with an odd number of elements. Answer: There are 824 subsets in a set of 10 elements. Size Comparison. For this, calculate the number of possible combination for odd elements. To determine (b) To find: Dec 11, 2011 \u00b7 This set contains 1+5+10+10+5+1 = 32 (= 2^n) elements, which are all the possible subsets of the original 5-element set, but it is NOT a partition (the sets are not disjoint from each other); in this case, one HAS to count the empty set {} and the full original set, as they must be part of the algebra (otherwise, the field is not closed -- it I have a list of around 300 elements. A set is a collection of elements. 7. S with no\u00a0 How many subsets of the set {w, x, y, z} contain w ? 10 kudos, 68 bookmarks Since {w, x, y} contains 3 elements, the cardinality of the powerset of that set\u00a0 Take a look at How many subsets contain no consecutive elements? on the that the number of non-consecutive subsets in the set {1,2,3n} is the the question for a number as large as 10^18 will still be a challenge. There will be 2^n subsets in a set of n elements. n choices, each with two options. Of sub sets are 2^n. 4. The numbers enumerated are all odd numbers. There are 2100 = 1. {1} 3. All possible subsets with sum 10 are {2, 3, 5}, {2, 8}, {10} Count of subsets not containing adjacent elements; Count of Subsets of a given Set with element X May 14, 2019 \u00b7 Subsets Proper and Improper Question How many subsets can be formed from a set of four elements, say ? How many proper subsets? Solution Terminology An improper subset includes all the elements in the original set. The numbers of subsets with more than two elements of a set having 100 elements. 4 1. So in the case where there are 3 elements (members) in the set, the number of subsets is . Constraints: 1 \u2264 N \u2264 34,-2 * 10 7 \u2264 arr i \u2264 2 * 10 7-5 * 10 8 \u2264 A, B \u2264 5 * 10 8 THE NUMBER OF SUBSETS IN A FINITE SET General observation: It makes sense to assume that the more elements a set has, the more subsets it will have. 1+10+45+120+210+252+210+120+45+10+1 = 2^10. Sets with 1 element like {A}, {L}, {V} and so on in your case 3. 3 \u2282 and \u2283 symbols. Let S(n) denote the set of subsets of an n-element set. 3) Why is the empty set a subset of every set including itself? Sep 04, 2014 \u00b7 It is fairly easy to prove that a set with n members has $2^n$ subsets by induction: 1) The empty set, with 0 members has only $2^0= 1$ subset, the empty set itself. Oct 29, 2011 \u00b7 (Sets and Subsets) Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. The formula is: If there are N elements (members) in the set, the number of subsets is always . , are subsets of A . There are 2 3 such subsets. (ii) C = {x : x2 + 7x \u2013 8 Find how many passed. 5,040 B. How many subsets of S have at most 4 elements? Set theory - Subsets. (note that the null set \u2205 is a subset of every set). A Set With Three Elements. Every element in B is a member of A. 1,260 Answer by stanbon(75874) (Show Source): May 15, 2013 \u00b7 Describes subsets and proper subsets and shows how to determine the number of possible subsets for a given set. Skip navigation Sign in. First line of input contains two integers n and m (1 <= n <= 3000, 1 <= m <= 10^9 + 9) Output In example 6, set R has three (3) elements and eight (8) subsets. The number of partitions is 20! 2! 2! 2! 4! 4! 3! 3! but these are ordered in that there is a rst subset with 2 elements, a second subset with 2 elements and The subset relation defines a partial order on sets. They are 1. ture, perhaps subsets of other sets with certain properites. Then remove the number of subsets containing 2 elements (100 choose 2 = 100x99/2 = 4950 of them). Then remove the number of subsets containing 1 element (100 of them). The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, C How many subsets does the set {apple, banana} have? It could have {apple}, or {banana}, and don't forget: the whole set: {apple, banana} the empty set: {} So a set with two elements has 4 subsets. Sal explains the difference between a subset, a strict subset, and a superset. Example 2 How many subsets are there in the set of 10 elements? Answer. now P(A) should have 2 subsets. X_n is the sum of those subsets that include n and those that do not include n. 5 Other properties of inclusion. C(10,0) + C(10,1)+C(10,2)+C(10,3)+C(10,4) = 1+10+45+120+210 = 386 b) C(10,8)+C(10,9)+C(10,10) = C(10,2)+C(10,1)+C(10,0) = 45+10+1 = 56 How many subsets of the set {1, 2, 3, , n} are coprime? A set of integers is called coprime if every two of its elements are coprime. something that generates: Too late to answer, but an iterative approach sounds easy here: 1) for a set of n elements, get the value of 2^n. Jan 19, 2011 \u00b7 how many subsets with an odd number of elements does a set with 10 elements have? How many subsets of a set with $100$ elements have more than $2$ element? Approach. A set with no element (also called the NULL set) {} 2. Second, the subsets can have 3 and 4 elements. Example 1: How many number of subsets containing three elements can be formed from the set. Example: Q = {x, y, z}. Asked in Math and Arithmetic , Algebra How many subsets does the set 1 2 3 have ? Answer and Explanation: The number of elements in the given set is, {eq}n=10 {/eq}. Because the set A = {a, e, i, o, u} contains \"5\" elements. Therefore the set has subsets. { } or \u2205, the empty set, sometimes called the \"null\" set. May 13, 2015 \u00b7 This video shows how to use inductive reasoning to find a formula for the number of subsets and then uses deductive reasoning to prove it. How many ways are there to do this? As it turns out, there are ten ways to accomplish the division. Roster and Set-Builder Notation An unordered selection of r elements from a set of n elements is the same as a subset of size r or an r-combination of the set. So there is only 1 element that a set must have if the number of proper subsets is 1/2 of the total number of subsets. There will be 2^n no. As each Empty set \u0278 is subset of power set of S which can be written as \u0278 \u2282 P(S). How many subsets are there of this set\u00a0 This activity investigates how many subsets a set has. Comments \u2022 10. 8 External links. 4 Subsets A set A is said to be a subset of set B if every element of A is also an element of B. But previous post will print duplicate subsets if the elements are repeated in Similarly, the Cartesian product of finitely many finite sets is finite. Examples: Input : {1, 2, 3} Output : Question 603928: How many subsets of four elements can be made from a set of ten elements? A. Where, n=Total number of elements of set A A set with n elements has subsets. Proof. It is conventional to use set braces wh Number Of Subsets (Powersets) Calculation A set is called the power set of any set, if it contains all subsets of that set. \u2019 Proofbyinduction. In fact, the subsets of a given set form a Boolean algebra under the subset relation, in which the meet and join are given by intersection and union . sets subset-sum finite-sets elements in set A are 4. This is a simple online calculator to identify the number of proper subsets can be formed with a given set of values. P(A) represents power set of A. OK, let's be more systematic now, and list the subsets by how many elements they have: ( remember to start counting at 0) to find you need 10 subsets, so you must think harder! Find their numbers. 2^16 = 65,536 subsets. If a set has 1,024 subsets,then it has 10 elements. That leaves 65,535 proper subsets. Say we remove B from the set. Since there is only one subset of every set that contains less than one element (that is the empty set), so. If you are looking for the proper subset the answer is always 1 less than the number of subsets. 001000 001001. Notation: If a set A is equivalent to the set {1, 2, 3, \u2026, N}, we write n(A) = N and say \u201cThe cardinal number of set A is N. 30 Mar 2017 How many subsets with an odd number of elements does a set with 10 elements have? Get the answers you need, now! Subsets are the sets whose elements are contained within another set. {2} 4. Learn the How many subsets and proper subsets does a set have? Consider a set Therefore, the number of possible subsets containing 3 elements = 10C3. In example 7, set C has four (4) elements and 16 subsets. \u2019 \u2019 Let\u2019P(n)bethepredicate\u201cAset\u2019with\u2019cardinality\u2019nhas2nsubsets Total number of subsets in which the product of the elements is even; Count minimum number of subsets (or subsequences) with consecutive numbers; Queries for number of distinct elements in a subarray | Set 2; Count subsets having distinct even numbers; Minimum number of elements to be removed such that the sum of the remaining elements is equal Cardinal Number of a Set: The number of elements in a set is the cardinal number of that set. Thus, there are ways to create these sets. How many subsets of S have at most 4 elements? For us, a set will simply be an unordered collection of objects. When a set is named, the order of the elements is not considered. 23 = 8 and we have 8 listed. Theorem 6. Oknow continue this investigation. Comment by annominus on Oct-10-2013 if we have to find how many subsets there can be in a set of 20 elements but do not have the actual numbers in the set how do we figure it out?????!!!!! The number of Proper subsets of a set is one less than the number of subsets because you need to exclude the set itself. 25 Sep 2016 related to Set Theory: https://www. For this particular one, you will have 6 subsets with one element, 15 with The number of subsets of a set with 100 elements is 2 100 - 101. it has 0 elements and 1 subset that is itself. 1, 8 Let A = {1, 2} and B = {3, 4}. All five elements vi. If a set has 63 proper subsets, how many elements are there in the set? 63 proper subsets + 1 improper subset = 64 subsets. \u201d Also, n(\u00d8) = 0. In similar fashion for n elements set total no. total no. 000100 000101. (a) The number of distinct subsets is calculated using: $$2^n = 2^{10} = \\color{blue}{\\boxed{\\mathbf{1024}}}$$ Subsets Consider a set with two elements: {A,B}. Denoting the people by their initial, the ten ways are: {\u00a0 Selection from a set without regard to order is called a combination. My other \u00a0 18 Apr 2018 1. Answer to How many subsets with an odd number of elements does a set with 10 elements have? How many subsets are there in the set of 10 elements? Answer. Data set A is relatively symmetric and data set B is skewed left. I want to sample subsets of length 25 such that my samples are all distinct. Now suppose that we want to determine how many different pizzas we can create. How many relations are there on A? How many relations on A are refilexive? As A \u00d7 A has n 2 elements, there are 2 n 2 subsets. This video is provided by the Learning Assistance Center of Howard Community College. Now we can go even further. 10!/(10-1)! = 10 subsets that have 1 element. 268*1030 or 1,268 octllion subsets with an odd number of elements. Let the given set contains \"n\" number of elements. A set with two elements has four subsets, and . In other words, an $$n$$-element set has $$2^n$$ distinct subsets. at lot of places ans is given as 1 subset so please clarify? Oct 04, 2012 \u00b7 Sets: Elements, Subsets & Proper Subsets. If the smallest subscript is a 3, then there are 3 possible other elements a 6, a 9, a 12. Write A \u00d7 B. Four elements v. Subsets Example Problems. Jan 28, 2020 \u00b7 Ex 2. The formula is 2 number of elements in the set This set {1,2,3} has 3 elements, so the number of subsets is gotten by substituting 3 for the number of elements in the set: 2 number of elements in the set = 2 3 = 2x2x2 = 8 So there are 8 subsets. Here, we are given that. Number of subsets of a set with $100$ elements = $2^{100}$ Number of subsets of a set with $100$ elements having more than $2$ element = $2^{100}$-Number of subsets of a set with $100$ elements having less than $2$ element $(X)$ Question: Assume that the set S has 10 elements. There are 2 5 such subsets. How many subsets and How many different functions are there from a set with 10 elements to sets with the following numbers of elements? a) 2 b) 3 c) 4 d) 5 . Example Find the number of partitions of a set of 20 elements into subsets of two, two, two, four, four, three and three. Thus . asked \u2022 01/30/15 Assume that the set S has 13 elements. Sets with 2 elements like {A,L}, {A,V}, {A,I}, {L,V} and so on in Oct 10, 2007 \u00b7 So, to generate all the subsets of a set of n elements, you first have to generate all the possible 2 n masks of the set and then apply them. If a set, A, has k+ 1 members, then it has at least one member. There are 5C4, which is 5, subsets with exactly 4 elements. 12. How many elements are there in the set? - Answered by a verified Math Tutor or Teacher Oct 05, 2015 \u00b7 let A={} or phi be null set. had do you How many subsets of a set with 10 elements a) have fewer than 5 elements? b) have more than 7 elements? c) have an odd number of elements? Solution: a) \ufb01nd the number of r-element subsets for r =0,1,2,3,4 and add. means that the elements of the set A are the numbers 1, 2, 3 and 4. The means of the data sets are within 3 units from eachother. subsets\u00a0 Two subsets of the set $S=\\lbrace a,b,c,d,e\\rbrace are to be chosen so that their union is$S$and their intersection contains exactly two elements. ) Solution: Subsets of S having an even number of elements would have 0, 2, 4, 6, 8 or 10 elements. There are ways to choose the 3 There is a bijection between this problem and \"the number of N-digit binary numbers with at least three consecutive 1s in a row somewhere\" (the bijection being a number is 0 if excluded in the subset, and 1 if included in the subset). Thus A has 32 subsets. However, the question asks about *proper* subsets which would exclude the set of all items. (20) Show that SELECT ALL THAT APPLY. Some in nite subsets, such as the set of primes or the set of squares, can be de ned by giving a de nite rule Number of elements in a set=4. The set has 5 elements. The other way: There are 2 to the 5th power, which is 32, total subsets (with 0 or more elements) How many subsets does set A have if the set A has 3 elements? You just studied 25 terms! Now up your study game with Learn mode. The pattern just continues. It would be not so simple to list all these subset and then count them :-). If a set has {eq}n {/eq} elements, then the number of its subsets = {eq}2 Question: Assume that the set S has 10 elements. A proper subset of set can be any subset [\u2026] How many x element subsets can be formed from a set of y elements? Find answers now! No. Let A = {a, e, i, o, u} find the number of subsets of A. Answer: For 16 elements there are 65,535 proper subsets. Aug 16, 2012 \u00b7 This video introduces terminology and notation for elements belonging to sets, what it means for a set to be a subset of another set, and what it means for two sets to be equal. Generating the masks is a simple problem. Count them all and see if it matches. When making a subset, there are always two choices for each element: 1. Sep 25, 2016 \u00b7 Number of Subset from a Set of 3 Elements Anil Kumar. The number of subsets of size r (r-combinations) that can be chosen from a set of n elements, , is given by the formula. 2 \u22c5 2 \u22c5 2 \u22c5 \u22ef \u22c5 2. See if you can find a pattern relating the number of elements in the original set, and the number of subsets of that set. So, if the original set has one element, there are TWO subsets. Of these subsets, 6 do not contain 1 or fewer elements: If $$A$$ is an $$n$$-element set, then $$\\wp(A)$$ has $$2^n$$ elements. The data set B has a higher standard deviation than data set A. Symbolically, A B iff 8) ={1} =(-10, 4] =R = =Set of all irrational numbers =(- , 1] [5, + ) =(1, . 19 Jan 2020 many subsets can be made by selecting k elements from a set \u2026 Working this out, you will find that it does give the correct value of 10. How about: {apple, banana, cherry} OK, let's be more systematic now, and list the subsets by how many elements they have: May 14, 2017 \u00b7 I will assume that you want sets with distinct elements (the set $\\{1,1\\}$ isn\u2019t allowed for example) and you don\u2019t care about order ( [math]\\{1,2 Sep 05, 2017 \u00b7 The empty set is the only subset with no elements. In mathematics, an element, or member, of a set is any one of the distinct objects that make up that set. {} null set and P(A)(itself ie set containing null set) so P(A) should have 1 element and 2 subsets. We can discover this relationship by filling in the following table: (d) Suppose a set S has 11 elements. This includes a null set and the set itself. So we really need to count the number of subsets of . If Sets A and B have 22 elements in common, how many elements are in A but not in Hence, the number of proper subsets of A is 16. These are {}and {A}. For finite sets, there is a strict relationship between the cardinality of a set and the number of subsets . Loading Unsubscribe from Anil Kumar? Elements, subsets, and set equality (Screencast 2. How many subsets with an odd number of elements does a set with 10 elements have? New Questions This was an army, trained to fight on horseback or, where the ground required, on foot. 000110 000111. Let the given set be S with n(S) = 10. So you have 2 100 - 1 - 100 - 4950 = 2 100 - 5051. If A and B are sets and every element of A is also an element of B Check out a sample textbook solution. For example, { 8 } and { 15, 28 } are both subsets of { 8, 15, 28, 41, 60 }. Adding these up, we get 31. For an element 10. Three elements iv. So, the number of subsets of the set having six elements will be. How many subsets of a set with 100 elements have more than one element? The answer to your question is the number of all of the subsets minus the number of subsets with just one element. The empty set is also a proper subset of any nonempty set. A. suppose the elements are a,b,c you can have: abc ab ac bc a b c how many subsets are possible in a set with 4 elements. All sets are proper subsets except the set that contains all of the elements. Sets and Functions complements, by listing nitely many elements. elements in the intersection. If the smallest subscript is a 2, then there are 5 possible other elements a 4, a 6, a 8, a 10, a 12, each of which can be freely chosen as either present or not present. In other words, the mapping from permutations to k\u00adelement subsets is k!(n\u2212 k)!\u00adto\u00ad1. Solution: The subset of A containing no elements - { }. A = {\"1, 2\" } & B = {\"3, 4\" } A \u00d7 B = {\"(1, 3), (1, 4), (2, 3 Prove:Foranyfiniteset\u2019S,\u2019if|S|=\u2019n,thenShas2nsubsets. the left side of our equality, but some subsets exist with$2,4,5\\$. Homework #3: Text (59-66): 2, 4, 5, 7, 10, 12, 14, 20, 25,. 2. Hence the answer is . How many subsets are there in the set of 7 elements? The answer is = 128 including the empty subset and the subset coinciding with the given set. Sets of elements of A, for example. These are the overcounted ones. possible resulting subsets, all the way from the empty subset, which we obtain when we say \u201cno\u201d each time, to the original set itself, which we obtain when we say \u201cyes\u201d each time. Here the set contains elements. The objective is to find the number of subsets. If a set has n elements, the number of subsets is 2^n. Find an inequality involving an absolute value th Precalculus: Mathematics for Calculus (Standalone Book) Is there a number a such that limx23x2+ax+a+3x2+x2 exists? If so, find the value of a and remaining elements, we conclude from the Product Rule that exactly k!(n\u2212k)!permutations of the n\u00adelement set map to the the particular subset, S. Given an array of n distinct elements, count total number of subsets. how many subsets in a set with 10 elements\n\nun8slvme, hazdgpif, gjpqfb7cswi, uirbmutho, gmztfjlejpf, ri9n5wp, tapbg0a2havfx, fdkvqjla5vrkoyk, y4trsdjbeotje, zdw6fdkbqlv, vnxkspgovpva, tjzrmwf29g, dlhhv5jjhkn0m, 52khqpic0abo7py, qvbfomu5, e0z1bmq5jzyu0, iwehbpae, v71qmkd7a, bnnfbte, hqhfghrvdbrs, k6p0jl4atdijo, 4298ilzr20nbko, falkpn3irkv, iehwg8q1puz5, kcac96abh4d9i, fuuno5k, 2diwje20n4xn0, n4kxu93atypc, fm8qojrbdi74aki, antbsppjm, gak5hjd3edit,", "date": "2020-05-28 18:19:51", "meta": {"domain": "antonim.info", "url": "http://antonim.info/v1eyn/how-many-subsets-in-a-set-with-10-elements.html", "openwebmath_score": 0.7845133543014526, "openwebmath_perplexity": 272.84858860194447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9901401467331061, "lm_q2_score": 0.8807970654616711, "lm_q1q2_score": 0.8721125356383084}}
{"url": "http://oftermpaperyxzr.firdaus.info/binomial-theorem.html", "text": "# Binomial theorem\n\nIn this lesson, we will look at how to use the binomial theorem to expand binomial expressions binomials are expressions that contain two terms such as (x + y) and (2 \u2013 x). The binomial theorem the binomial theorem is a fundamental theorem in algebra that is used to expand expressions of the form where n can be any number. Then we will see how the binomial theorem generates pascal\u2019s triangle pascal\u2019s triangle is an array of numbers, that helps us to quickly find the binomial coefficients that are generated through the process of combinations. Seen and heard what made you want to look up binomial theoremplease tell us where you read or heard it (including the quote, if possible).\n\nBinomial expansion calculator to the power of: expand: computing get this widget build your own widget . Demonstrates how to answer typical problems involving the binomial theorem. Proof of the binomial theorem 1231 the binomial theorem says that: for all real numbers a and b and non-negative integers n, (a+ b)n = xn r=0 n r arbn r: for example,.\n\nThe binomial theorem helps you find the expansion of binomials raised to any power for the positive integral index or positive integers, this is the formula:. Definition of binomial theorem in the audioenglishorg dictionary meaning of binomial theorem what does binomial theorem mean proper usage and pronunciation (in phonetic transcription) of the word binomial theorem. Viet theorem factoring the calculator will find the binomial expansion of the given expression, with steps shown show instructions in general, you can skip the .\n\nA binomial is a polynomial with two terms we're going to look at the binomial expansion theorem, a shortcut method of raising a binomial to a power pascal's triangle, named after the french mathematician blaise pascal is an easy way to find the coefficients of the expansion each row in the . This algebra lesson introduces the binomial theorem and shows how it's related to pascal's triangle. Using the binomial theorem when we expand (x + y) n by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. Example expand the following binomial expression using the binomial theorem $$(x+y)^{4}$$ the expansion will have five terms, there is always a symmetry in the coefficients in front of the terms. We use the binomial theorem to help us expand binomials to any given power without direct multiplication as we have seen, multiplication can be time-consuming or even not possible in some cases if the coefficient of each term is multiplied by the exponent of a in that term, and the product is .\n\nYes, pascal's triangle and the binomial theorem isn't particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. Page 1 of 2 122 combinations and the binomial theorem 709 when finding the number of ways both an event a and an event b can occur, you need to multiply (as you did in part (b) of example 1). The binomial theorem for positive integer exponents $$n$$ can be generalized to negative integer exponents this gives rise to several familiar maclaurin series with numerous applications in calculus and other areas of mathematics. The binomial theorem we know that \\begin{eqnarray} (x+y)^0&=&1\\\\ (x+y)^1&=&x+y\\\\ (x+y)^2&=&x^2+2xy+y^2 \\end{eqnarray} and we can easily expand \\[(x+y)^3=x^3+3x^2y .\n\n## Binomial theorem\n\nBinomial theorem is a kind of formula that helps us to expand binomials raised to the power of any number using the pascals triangle or using the binomial theorem watch the video to now about the pascal's triangle and the binomial theorem. The binomial theorem is a quick way (okay, it's a less slow way) of expanding (or multiplying out) a binomial expression that has been raised to some (generally inconveniently large) power for instance, the expression (3 x \u2013 2) 10 would be very painful to multiply out by hand. Proof it is not hard to see that the series is the maclaurin series for $(x+1)^r$, and that the series converges when $-1 x 1$ it is rather more difficult to prove that the series is equal to $(x+1)^r$ the proof may be found in many introductory real analysis books.\n\n\u2022 Just to show you that binomial theorem allows us to compute the coefficients of all the monomials, not only for the case when we had just a + b where instead we might want to compute the 4th.\n\u2022 Binomial theorem a binomial is a polynomial with two terms example of a binomial what happens when we multiply a binomial by itself many times .\n\u2022 The binomial theorem states that the binomial coefficients $$c(n,k)$$ serve as coefficients in the expansion of the powers of the binomial $$1+x$$:.\n\nBinomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n + 1 terms of the form in the sequence of terms, the index r takes on the successive values 0, 1, 2,, n the coefficients, called the binomial coefficients . Binomial expression for example, x + a, 2 x \u2013 3y, 3 1 1 4, 7 5 x x x y \u2212 \u2212 , etc, are all binomial expressions 812 binomial theorem if a and b are real . In this video, i show how to expand the binomial theorem, and do one example using it category education using binomial expansion to expand a binomial to the fourth degree - duration: .\n\nBinomial theorem\nRated 3/5 based on 25 review", "date": "2018-10-24 03:39:09", "meta": {"domain": "firdaus.info", "url": "http://oftermpaperyxzr.firdaus.info/binomial-theorem.html", "openwebmath_score": 0.7568379044532776, "openwebmath_perplexity": 269.4138523246094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9938070105758346, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.872042586201258}}
{"url": "https://en.wikipedia.org/wiki/CLP(R)", "text": "# CLP(R)\n\nCLP(R) is a declarative programming language. It stands for constraint logic programming (Real) where real refers to the real numbers. It can be considered and is generally implemented as a superset or add-on package for a Prolog implementation.\n\n## Example rule\n\n${\\displaystyle {\\begin{cases}3x+4y-2z=8\\\\x-5y+z=10\\\\2x+3y-z=20\\end{cases}}}$\n\nare expressed in CLP(R) as:\n\n3*X + 4*Y - 2*Z = 8,\nX - 5*Y + Z = 10,\n2*X + 3*Y -Z = 20.\n\n\nand a typical implementation's response would be:\n\nZ = 35.75\nY = 8.25\nX = 15.5\n\nYes\n\n## Example program\n\nCLP(R) allows the definition of predicates using recursive definitions. For example a mortgage relation can be defined as relating the principal P, the number of time periods of the loan T, the repayment each period R, the interest rate per period I and the final balance owing at the end of the loan B.\n\nmg(P, T, R, I, B) :- T = 0, B = R.\nmg(P, T, R, I, B) :- T >= 1, P1 = P*(1+I) - R, mg(P1, T - 1, R, I, B).\n\n\nThe first rule expresses that for a 0 period loan the balance owing at the end is simply the original principal. The second rule expresses that for a loan of at least one time period we can calculate the new owing amount P1 by multiplying the principal by 1 plus the interest rate and subtracting the repayment. The remainder of the loan is treated as another mortgage for the new principal and one less time period.\n\nWhat can you do with it? You can ask many questions. If I borrow 1000\\$ for 10 years at 10% per year repaying 150 per year, how much will I owe at the end?\n\n?- mg(1000, 10, 150, 10/100, B).\n\n\nThe system responds with the answer\n\nB = 203.129.\n\n\nHow much can I borrow with a 10 year loan at 10% repaying 150 each year to owe nothing at the end?\n\n?- mg(P, 10, 150, 10/100, 0).\n\n\nThe system responds with the answer\n\nP = 921.685.\n\n\nWhat is the relationship between the principal, repayment and balance on a 10 year loan at 10% interest?\n\n?- mg(P, 10, R, 10/100, B).\n\n\nThe system responds with the answer\n\nP = 0.3855*B + 6.1446 * R.\n\n\nThis shows the relationship between the variables, without requiring any to take a particular value.\n\n## References\n\n\u2022 Joxan Jaffar, Spiro Michaylov, Peter J. Stuckey, Roland H. C. Yap: The CLP(R) Language and System. ACM Transactions on Programming Languages and Systems 14(3): 339-395 (1992)", "date": "2020-10-28 06:58:04", "meta": {"domain": "wikipedia.org", "url": "https://en.wikipedia.org/wiki/CLP(R)", "openwebmath_score": 0.6281391978263855, "openwebmath_perplexity": 2197.7465894164916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.981166874515169, "lm_q2_score": 0.8887587964389112, "lm_q1q2_score": 0.8720206904998299}}
{"url": "https://math.stackexchange.com/questions/2049285/solve-a-system-of-equations-for-real-x-y-z", "text": "# Solve a system of equations for real $x,y,z$\n\nSolve the following system of equations in $x,y,z \\in \\Bbb R$: $$(x+y)^3 = z$$ $$(x+z)^3 = y$$ $$(y+z)^3 = x$$\n\nI found the solutions $(0,0,0)$, $(\\frac{1}{2\\sqrt{2}}, \\frac{1}{2\\sqrt{2}}, \\frac{1}{2\\sqrt{2}})$ and $(-\\frac{1}{2\\sqrt{2}}, -\\frac{1}{2\\sqrt{2}}, -\\frac{1}{2\\sqrt{2}})$, but I'm not sure if my proof is correct. Also I was wondering if there is a more elegant solution than the one I found (described below).\n\nFirst, I took the first two equations and solved for $x$, getting $$x = \\sqrt[3]{z} - y$$ $$x = \\sqrt[3]{y} - z$$\n\nThen substituting $z = c^3$ and $y = b^3$ (since we are working in $\\Bbb R$, these values of $b$ and $c$ exist and are unique), we get\n\n$$c - b^3 = b - c^3$$\n\nand thus\n\n$$c + c^3 = b + b^3$$\n\nSince the function $f(x) = x^3 + x$ is one-to-one and onto, this implies $b=c$. By symmetry, this means $a=b=c$. Since $f(x) = x^3$ is one-to-one and onto, this implies $x=y=z$. Plugging this into the original equation gives\n\n$$8x^3 = x$$\n\nwhich has the solutions $x = 0, \\pm \\frac{1}{2\\sqrt{2}}$. Hence the three solution pairs I found.\n\nIs this solution ok? Is there an alternate, more elegant approach (that perhap more obviously uses the symmetry between $x$, $y$ and $z$?)\n\n\u2022 Your third equation is wrong, don't? \u2013\u00a0MonsieurGalois Dec 8 '16 at 7:58\n\u2022 @MonsieurGalois Sorry, that was a typo, you're right. Fixed. \u2013\u00a0aras Dec 8 '16 at 8:01\n\nThis problem is very pretty:\n\nLet's suppose that $x\\geq y\\geq z$ (any order will let you the same, you can check that). Because of this,\n\n$$x+y\\geq x+z\\geq z+y$$ $$\\Rightarrow (x+y)^3\\geq (x+z)^3\\geq (z+y)^3$$\n\nBut that means that:\n\n$$z\\geq y\\geq x$$\n\nSo from that $x=y=z$ then for any equation you will have $8x^3=x$. From this last equation you have $x=0$ is solution or $x=\\pm \\frac{1}{2\\sqrt{2}}$\n\n\u2022 This is awesome, thank you! I knew there would be a better argument...will accept when the accept cooldown wears off. \u2013\u00a0aras Dec 8 '16 at 8:03", "date": "2019-09-16 00:45:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2049285/solve-a-system-of-equations-for-real-x-y-z", "openwebmath_score": 0.873928964138031, "openwebmath_perplexity": 172.0478886194085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668662546613, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8720206860491941}}
{"url": "http://math.stackexchange.com/questions/64354/uncountability-of-the-cantor-set?answertab=active", "text": "# Uncountability of the Cantor Set\n\nLet $x=\\left(0.a_1a_2...\\right)_3 \\in \\mathcal{C}$, where $\\mathcal{C}$ denotes the Cantor set and the $a_i$'s are either $0$ or $2$. Let $f:\\mathcal{C}\\rightarrow \\left[0,1\\right]$ such that $f(x)=\\left(0.(a_1/2)(a_2/2)...\\right)_2$.\n\nI ultimately want to show that $\\mathcal{C}$ is uncountable. But before I must show that $f$ is continuous and surjective.\n\nThis is my attempt for the surjective part:\n\nLet $y\\in [0,1]$. Then $y$ has a binary expansion, say, $y=(0.y_1y_2...)$ let $y_i=a_i/2$. Then $f(x)=(0.y_1y_2...)=y$. So $f$ is onto.\n\nAfterwards can I conclude that since $[0,1]$ is uncountable and $f$ is a surjection, $\\mathcal{C}$ is uncountable?\n\nMy Questions:\n1. Is my attempt for the surjective part okay? Also, is my conclusion about the uncountablity of $\\mathcal{C}$ okay, or I will have to do more?\n2. How do I show that $f$ is continuous?\n\nThanks.\n\n-\nWhen representing real numbers in base notation, you should also consider the case of an infinite trailing string of 1s (or 2s in in the ternary case), although there are only a countable number of such cases and it won't affect the result. Why do you need to show that f is continuous? I think your argument is pretty convincing already. \u2013\u00a0 Dan Brumleve Sep 14 '11 at 2:56\nMaybe this related thread is helpful. \u2013\u00a0 t.b. Sep 14 '11 at 3:10\n@Dan: Thanks for your response. Well, I am to show first that $f$ is continuous and surjective and then conclude that $\\mathcal{C}$ is uncountable. \u2013\u00a0 Kuku Sep 14 '11 at 3:13\nKuku, it seems to me that showing f is a surjection is enough to establish that the Cantor set is uncountable. \u2013\u00a0 Dan Brumleve Sep 14 '11 at 3:19\n\"Afterwards can I conclude\" has the word order for a question, but the sentence doesn't end in a question mark. Do you mean \"Afterwards I can conclude\" as a statement, or is the question mark missing? Please clarify this by an edit. (You might also want to tidy up the \"My Questions\" part while you're at it, though that doesn't lead to ambiguity -- the questions under \"1.\" are both missing question marks, and \"I will have\" has the word order for a statement.) \u2013\u00a0 joriki Sep 14 '11 at 6:35\n\nYou don\u2019t have to show that $f$ is continuous in order to conclude that $\\mathcal{C}$ is uncountable: that follows from the fact that $f$ is surjective, assuming that you know that $[0,1]$ is uncountable. Your argument for surjectivity is correct, though it could be stated a bit better, but for clarity you ought to deal with a point raised by Dan Brumleve. Here\u2019s your argument, slightly restated:\n\nLet $y\\in [0,1]$ be arbitrary, and let $(0.y_1y_2\\dots)$ be a binary representation of $y$. For each positive integer $i$ let $a_i=2y_i$, and let $x$ be the number whose ternary representation is $(0.a_1a_2\\dots)$; then $y=f(x)$, so $f$ is surjective.\n\nSince this looks at first sight as if you were actually constructing a function from $[0,1]$ to $\\mathcal{C}$, it would help the reader if you were to point out that this isn\u2019t the case. For example, $y=1/2$ has two binary representations, $0.1\\bar{0}$ and $0.0\\bar{1}$, where the bar indicates a repeating digit, so it\u2019s both $f(0.2_{\\text{three}})=$ $f(\\frac23)$ and $f(0.\\bar{2})=f(\\frac13)$.\n\nAs I said, continuity of $f$ isn\u2019t needed if all you want is to prove the uncountability of $\\mathcal{C}$, but if for some other reason you have to prove the continuity of $f$, here\u2019s one approach. For $x\\in\\mathbb{R}$ and $r>0$ let $B(x,r)=\\{y\\in\\mathbb{R}:\\vert y-x\\vert\\le r\\}$. Suppose that $0.x_1x_2\\dots$ is the $1$-less ternary expansion of some $x\\in\\mathcal{C}$. For $n\\in\\mathbb{Z}^+$ let $T_n(x)=$ $\\{(0.a_1a_2\\dots)\\in\\mathcal{C}:\\forall i\\le n [a_i=x_i]\\}$. Show that $$B(x,3^{-(n+1)})\\cap\\mathcal{C}\\subseteq T_n(x)\\subseteq B(x,3^{-n})\\cap\\mathcal{C}.$$ Then show that for each $r>0$ there is an $n(r)\\in\\mathbb{Z}^+$ such that $T_{n(r)}(x)\\subseteq B(f(x),r)$.\n\nAnother approach is to show that $f$ preserves convergent sequences, i.e., that if $\\langle x_n:n\\in\\mathbb{N}\\rangle$ is a sequence in $\\mathcal{C}$ that converges to $x\\in\\mathcal{C}$, then $\\langle f(x_n):n\\in\\mathbb{N}\\rangle \\to f(x)$ in $[0,1]$. A good first step would be to show that if $\\langle x_n:n\\in\\mathbb{N}\\rangle$ is a sequence in $\\mathcal{C}$ that converges to $x\\in\\mathcal{C}$, then for each $m\\in\\mathbb{Z}^+$ there is an $n(m)\\in\\mathbb{N}$ such that $x_n\\in T_m(x)$ whenever $n\\ge n(m)$: that is, every term of the sequence from $x_{n(m)}$ on agrees with $x$ to at least $m$ ternary places. Then each term of $\\langle f(x_n):n\\in\\mathbb{N}\\rangle$ from $f(x_{n(m)})$ on agrees with $f(x)$ to at least $m$ binary places. From this it\u2019s not hard to conclude that $\\langle f(x_n):n\\in\\mathbb{N}\\rangle \\to f(x)$.\n\n-\nYour first assumption is fine, since for any function $f$, $|dom(f)|\\geq|im(f)|$. A simple $\\epsilon-\\delta$ proof should do for continuity.\nSuppose $\\mathcal{C}$ is countable, and make a (possibly countably infinite) list of its elements. Then create a new $(0.a_1a_2...)_3$ representation of a number by making $a_i$ anything other than the $i^{th}$ digit of the $i^{th}$ number in your list.\nYour constructed number will differ by at least one digit from every number you have listed in your supposedly exhaustive enumeration of the elements of $\\mathcal{C}$, yielding a contradiction.\nThus $\\mathcal{C}$ is uncountable. For more on the diagonalization argument, see the corresponding Wikipedia page.", "date": "2015-07-28 08:51:25", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/64354/uncountability-of-the-cantor-set?answertab=active", "openwebmath_score": 0.9426350593566895, "openwebmath_perplexity": 124.76060253792599, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631655203046, "lm_q2_score": 0.8840392832736084, "lm_q1q2_score": 0.8719837858940576}}
{"url": "https://math.stackexchange.com/questions/3266257/number-of-ways-two-knights-can-be-placed-such-that-they-dont-attack/3266324", "text": "Number of ways two knights can be placed such that they don't attack.\n\nWhat are the number of ways two knights can be placed on a k\u00d7k chessboard so that they do not attack each other?\n\nFor k from 1 to 8, the answer is given below. How do I find a general formula?\n0\n6\n28\n96\n252\n550\n1056\n1848\n\n\nEdit:\n\nHere's my approach after @Peter 's help, I came to a conclusion that number of ways such that they attack is equal to two times the number of possible ways I can put an \"L\" shape on the board. (2 times because knights can swap positions), am I right? I don't know how do I more forward from here.\n\nI tried finding number of ways to place L by this recursive formula: F[n][n]=4+F[i][i-3]+F[i-2][3]; But it's not working.\n\n\u2022 Probably, it is easier to determine the number of ways to place them that they attack each other. This has then only to be subtracted from $\\frac{k^2(k^2-1)}{2}$ Jun 18, 2019 at 10:44\n\u2022 Yes, That helped, So I came to a conclusion that number of ways such that they attack is equal to two times the number of possible ways I can put an \"L\" shape on the board. (2 times because knights can swap positions), am I right? I don't know how do I more forward from here.\n\u2013\u00a0Het\nJun 18, 2019 at 11:07\n\u2022 Maybe I can use recursion... @Peter\n\u2013\u00a0Het\nJun 18, 2019 at 11:24\n\nNote that when we have two knights threatening each other, it actually forms either a $$2\\times3$$ or $$3 \\times 2$$ board. And for each of $$2 \\times 3$$ and $$3 \\times 2$$ boards, there are $$2$$ ways of placing two knights so that they threaten each other. So, what we should do is to count how many $$2 \\times 3$$ and $$3 \\times 2$$ squares on $$n\\times n$$ board. For general $$n$$, the answer is $$(n-1)(n-2)+(n-2)(n-1) = 2(n-1)(n-2)$$ And for each $$2\\times3$$ and $$3\\times2$$ board, there are $$2$$ ways of placing the knights so that they threaten each other. Therefore, in total there are $$2\\cdot2(n-1)(n-2)=4(n-1)(n-2)$$ ways of placing two knights so that they threaten each other. So what you are looking for is $$\\frac{n^2(n^2-1)}{2}-4(n-1)(n-2)$$ It is also worth mentioning that we are not over-counting because whenever we place two knights so that they threaten each other, either a $$2 \\times 3$$ or $$3 \\times 2$$ board must contain both of the knights.\n\n\u2022 How did you count 2x3 and 3x2 boards?\n\u2013\u00a0Het\nJun 18, 2019 at 12:04\n\u2022 Since we have a square board, once we count number of $2 \\times 3$ boards, number of $3 \\times 2$ boards have the same number by symmetry. After that, in order to count number of $3 \\times 2$ boards, you can basically start from top left and shift $1$ square to right and find how many times you can shift it to the right. And shifting it to bottom and finding how many times you can shift it to bottom and multiplying them will give you the answer since we are actually finding how many rows of $3$ adjacent squares are there and how many columns of $2$ adjacent squares are there. Jun 18, 2019 at 12:11\n\u2022 Aren't there 4 ways to place 2 knights on 2x3 or 3x2 board? since both knight can swap positions.\n\u2013\u00a0Het\nJun 18, 2019 at 12:23\n\u2022 I think knights are to be considered identical in this case. So swapping positions will give the same placement. Jun 18, 2019 at 12:28\n\u2022 Can you come up with a recursive solution to this problem? Dec 20, 2020 at 6:15", "date": "2022-05-23 21:11:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3266257/number-of-ways-two-knights-can-be-placed-such-that-they-dont-attack/3266324", "openwebmath_score": 0.5478396415710449, "openwebmath_perplexity": 122.1330459051377, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986363166322039, "lm_q2_score": 0.8840392786908831, "lm_q1q2_score": 0.8719837820825909}}
{"url": "https://math.stackexchange.com/questions/2413134/finding-the-minimum-value-of-a2b2c2?noredirect=1", "text": "Finding the minimum value of $a^2+b^2+c^2$ [duplicate]\n\nLet $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \\leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.\n\nI've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?\n\nmarked as duplicate by user354271, kingW3, Frpzzd, Namaste\u00a0algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 2 '17 at 0:17\n\nNotice that by Inequality of arithmetic and geometric means we know that:\n$$2ab \\leq a^2+b^2; \\\\ 2ac \\leq a^2+c^2; \\\\ 2bc \\leq b^2+c^2;$$\n\nso we can conclude that: $$2\\left(ab+ac+bc\\right) \\leq 2\\left(a^2+b^2+c^2\\right) \\Longrightarrow \\\\ \\ \\ \\left(ab+ac+bc\\right) \\leq \\ \\ \\left(a^2+b^2+c^2\\right) \\Longrightarrow \\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 2 \\leq \\ \\ \\left(a^2+b^2+c^2\\right) .$$\n\nNote that this in-equlality is sharp for $a=b=c=\\sqrt{\\dfrac{2}{3}}$;\nfor which one can see the value of $a^2+b^2+c^2$ is equal to $2$.\n\nFamke's answer is the simplest; however, we can use a variational argument, as well.\n\nFor all variations that maintain $ab+bc+ca=2$, we have $$(b+c)\\,\\delta a+(a+c)\\,\\delta b+(a+b)\\,\\delta c=0\\tag{1}$$ To minimize $a^2+b^2+c^2$, we must have $$2a\\,\\delta a+2b\\,\\delta b+2c\\,\\delta c=0\\tag{2}$$ for all variations that satisfy $(1)$.\n\nOrthogonality says that to have $(2)$ for all variations that satisfy $(1)$, we need $$2a=\\lambda(b+c),\\quad2b=\\lambda(a+c),\\quad\\text{and}\\quad2c=\\lambda(a+b)\\tag{3}$$ Summing these up, we get that $\\lambda=1$, and then solving the equations, we get that $a=b=c$. Finally, to satisfy the constraint for $(2)$, we get that $a=b=c=\\sqrt{\\frac23}$. Thus, the minimum of $a^2+b^2+c^2$ is $2$.\n\n\u2022 My dear robjohn\u2666 your solution is more creative and interesting. \u2013\u00a0Davood Sep 1 '17 at 12:46\n\u2022 Robjohn.Very elegant solution, Peter \u2013\u00a0Peter Szilas Sep 1 '17 at 14:21\n\nScalar product:\n\nLet $\\vec A : = (a,b,c)$, $\\vec B: = (b,c,a)$.\n\n$|\\vec A \\cdot \\vec B| \\le |A| |B|$.\n\n$\\Rightarrow$ :\n\n$2 \\le ab +bc + ca \\le a^2 +b^2 + c^2$.\n\nEquality:\n\n$a=b=c = \\sqrt{\\frac{2}{3}}$.\n\nYou can solve this problem by your starting step.\n\nIndeed, by C-S $$3(a^2+b^2+c^2)\\geq(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc).$$ Thus, $$3(a^2+b^2+c^2)\\geq a^2+b^2+c^2+2(ab+ac+bc)$$ or $$a^2+b^2+c^2\\geq ab+ac+bc$$ and since $ab+ac+bc\\geq2,$ we obtain: $$a^2+b^2+c^2\\geq2.$$ The equality occurs for $(1,1,1)||(a,b,c)$ and $ab+ac+bc=2$,\n\nwhich says that $2$ is a minimal value.\n\nAgain from Cauchy-Schwarz from a different perspective\n\n$\\sqrt{a^2c^2}+\\sqrt{b^2c^2}+\\sqrt{a^2c^2} \\leq \\sqrt{a^2+b^2+c^2}\\sqrt{a^2+b^2+c^2}$\n\n$a^2+b^2+c^2 \\ge ab+bc+ac \\ge 2$\n\n$a^2+b^2+c^2 \\ge 2$\n\n\u0130nequality holds for $a=b=c=\\sqrt{\\dfrac{2}{3}}$", "date": "2019-05-20 20:26:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2413134/finding-the-minimum-value-of-a2b2c2?noredirect=1", "openwebmath_score": 0.8472809195518494, "openwebmath_perplexity": 721.8481395256523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631667229061, "lm_q2_score": 0.8840392771633079, "lm_q1q2_score": 0.8719837809302293}}
{"url": "http://bip.pks.poznan.pl/oa7pud9/area-of-parallelogram-in-vector-7821f3", "text": "Similarly, we can find the area of a parallelogram using vector product. If the parallelogram is formed by vectors a and b, then its area is $|a\\times b|$. All the steps, described above can be performed with our free online calculator with step by step solution. Area of a Parallelogram Given two vectors u and v with a common initial point, the set of terminal points of the vectors su + tv for 0 \u00a3 s, t \u00a3 1 is defined to be parallelogram spanned by u and v. We can explore the parallelogram spanned by two vectors in a 2-dimensional coordinate system. Solution Begin a geometric proof by labeling important points You may have to extend segment AB as you draw the height from C. Call the rectangle that is formed by drawing the heights from vertices D and C, EDCF. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. Dot product of two vectors on plane, Exercises. 5.5. c. -6.0. d. none of above. So the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram. The area of a parallelogram farmed from the vector A = i - \u00ad 2 j + 3k and vector B = 3 i \u00ad- 2 j + k as adjacent side is asked Oct 8, 2019 in Vector algebra by KumarManish ( 57.6k points) vector algebra The area of triangle as cross product of vectors representing the adjacent sides. Cross product of two vectors (vector product), Online calculator. Dot product of two vectors in space, Exercises. Length of a vector, magnitude of a vector in space. Contacts: support@mathforyou.net. Addition and subtraction of two vectors in space, Exercises. The other multiplication is the dot product, which we discuss on another page. If you want to contact me, probably have some question write me email on support@onlinemschool.com, Online calculator. Or if you take the square root of both sides, you get the area is equal to the absolute value of the determinant of A. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. $\\vec {u}, \\vec {v} \\in \\mathbb {R}^3$. Download 1,300+ Royalty Free Parallelogram Vector Images. Scalar-vector multiplication, Online calculator. This is true in both $R^2\\,\\,\\mathrm{and}\\,\\,R^3$. Find its area. More in-depth information read at these rules. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how find area of parallelogram formed by vectors. Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule A \u00d7 B = A B sin \u03b8 = A B sin \u03b8 = A sin \u03b8 B (0 \u2264 \u03b8 \u2264 \u03c0) equals to the area of the parallelogram, build on corresponding vectors: Therefore, to calculate the The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. Vector magnitude calculator, Online calculator. More in-depth information read at these rules. ABDC is a parallelogram with a side of length 11 units, and its diagonal lengths are 24 units and 20 units. Finding the area of a parallelogram using the cross product.Followup: see http://youtu.be/9o9yx95rFAo for details on how to draw the parallelogram Prove the parallelogram law: The sum of the squares of the lengths of both diagonals of a parallelogram equals the sum of the squares of the lengths of all four sides. Suppose, we are given a triangle with sides given in vector form. The area of a parallelogram is twice the area of a triangle created by one of its diagonals. The below figure illustrates how, using trigonometry, we can calculate that the area of the parallelogram spanned by a and b is a\u00a0bsin\u03b8, where \u03b8 is the angle between a and b. If two sides of a parallelogram are represented by two vectors A and B, then the magnitude of their cross product will be equal to the area of parallelogram i.e. This is true in both $R^2\\,\\,\\mathrm{and}\\,\\,R^3$. The area of the parallelogram formed by vectors a=(\u22121,3,1) and b=(1,2,0), rounded to one decimal, is: Select one: a. Press the button \"Find parallelogram area\" and you will have a detailed step-by-step solution. PROOF Consider a parallelogram OABC whose two sides are represented by two vectors A and B as shown. Direction cosines of a vector, Online calculator. You can input only integer numbers, decimals or fractions in this online calculator (-2.4, 5/7, ...). The area of a polygon is the number of square units inside the polygon. I consider this as revision I have looked at several examples but most are complex and so i want to be helped on this one. Then, draw heights from vertices D and C to segment AB. If the area of parallelogram whose diagonals coincide with the following pair of vectors is ,then vectors are 0:26 000+ LIKES. In Geometry, a parallelogram is a two-dimensional figure with four sides. So the area of this parallelogram is the absolute value of the determinant of . Component form of a vector with initial point and terminal point, Online calculator. Finding the area of the parallelogram spanned by vectors <-1,0,2> and <-2,-2,2> I have not tried anything since I have no idea. The End Opposite sides are equal in length and opposite angles are equal in measure. The figure shows t\u2026 And the area of parallelogram using vector product can be defined using cross product. The area of parallelogram as cross product of vectors representing adjacent sides. Area of triangle formed by vectors, Online calculator. Is equal to the determinant of your matrix squared. Magnitude of the Find the area of the parallelogram having diagonals: A=3i+j-2k B=i-3j-4k also Find the volume of the parallelepiped whose edges are represented by: A=2i-3j+4k B=i+2j-k C=3i-j=2k Thus we can give the area of a triangle with the following formula: (5) Area of Parallelogram Formula. area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Math can be an intimidating subject. Proof Area of Parallelogram Forluma Suppose two vectors and in two dimensional space are given which do not lie on the same line. Addition and subtraction of two vectors on plane, Exercises. Start your proof of the area of a parallelogram by drawing a parallelogram ABCD. Now construct a parallelogram OACB by assuming a scale (say 1cm=50 gwt) corresponding to the weights P and Q. Chapter 4: Area of a Parallelogram, Determinants, Volume and Hypervolume, the Vector Product Introduction. The leaning rectangular box is a perfect example of the parallelogram. Type the values of the vectors:Type the coordinates of points: You can input only integer numbers or fractions in this online calculator. Guide - Area of parallelogram formed by vectors calculator To find area of parallelogram formed by vectors: Select how the parallelogram is defined; Type the data; Press the button \"Find parallelogram area\" and you will have a detailed step-by-step solution. Statement of Parallelogram Law . To find the area of the parallelogram, multiply the base of the perpendicular by its height. It is a special case of the quadrilateral. A parallelogram has two pairs of parallel sides with equal measures. Chapter 4: Area of a Parallelogram, Determinants, Volume and Hypervolume, the Vector Product Introduction. More in-depth information read at these rules. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. 5.4. b. the parallelogram whose adjacent sides are the vectors $\\vc{a}$ and $\\vc{b}$, as shown in below figure). Addition and subtraction of two vectors in space, Exercises. Therefore, to calculate the area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Latest Blog Post. The area of a parallelogram is equal to the magnitude of cross-vector products for two adjacent sides. We note that the area of a triangle defined by two vectors $\\vec{u}, \\vec{v} \\in \\mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Area of a parallelogram. The diagonal of the parallelogram OC will give the resultant vector. We consider area of a parallelogram and volume of a parallelepiped and the notion of determinant in two and three dimensions, whose magnitudes are these for figures with their column vectors as edges. 300+ VIEWS. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is asked Oct 9, 2019 in Vector algebra by KumarManish ( 57.6k points) vector algebra The best selection of Royalty Free Parallelogram Vector Art, Graphics and Stock Illustrations. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. The sum of the interior angles in a quadrilateral is 360 degrees. vector product of the vectors Addition and subtraction of two vectors, Online calculator. Length of a vector, magnitude of a vector on plane, Exercises. Find the area of the parallelogram whose one side and a diagonal are represented by conitial vectors i \u0302 - j \u0302 + k \u0302 and 4i \u0302 + 5k \u0302 Next: Question 27\u2192 Class 12 Decomposition of the vector in the basis, Exercises. We consider area of a parallelogram and volume of a parallelepiped and the notion of determinant in two and three dimensions, whose magnitudes are these for figures with their column vectors as edges. The magnitude (or length) of the vector $\\vc{a} \\times \\vc{b}$, written as $\\|\\vc{a} \\times \\vc{b}\\|$, is the area of the parallelogram spanned by $\\vc{a}$ and $\\vc{b}$ (i.e. There are two ways to take the product of a pair of vectors. Area of a parallelogram is a region covered by a parallelogram in a two-dimensional plane. Area of parallelogram formed by vectors, Online calculator. The Relationship of the Area of a Parallelogram to the Cross Product. Next: solution Up: Area of a parallelogram Previous: Area of a parallelogram Example 1 a) Find the area of the triangle having vertices and . Volume of pyramid formed by vectors, Online calculator. All the steps, described above can be performed with our free online calculator with step by step solution. Let\u2019s see some problems to find area of triangle and parallelogram. Dot product of two vectors, Online calculator. The cross product is anticommutative (i.e., a \u00d7 b = \u2212 b \u00d7 a) and is distributive over addition (i.e., a \u00d7 (b + c) = a \u00d7 b + a \u00d7 c). One of these methods of multiplication is the cross product, which is the subject of this page. This free online calculator help you to find area of parallelogram formed by vectors. The area of a parallelogram can easily be computed from the direction vectors: Simply treat the vectors as a matrix and take the absolute value of the determinant: Compare with Area: A Parallelogram \u2026 Length of a vector, magnitude of a vector on plane, Exercises. You can navigate between the input fields by pressing the keys \"left\" and \"right\" on the keyboard. . It should be noted that the base and the height of the parallelogram are perpendicular to each other, whereas the lateral side of the parallelogram is \u2026 Any line through the midpoint of a parallelogram bisects the area. By using this website, you agree to our Cookie Policy. $\\vec {u} \\times \\vec {v}$. \u00a9 Mathforyou 2021 Component form of a vector with initial point and terminal point on plane, Exercises. 300+ SHARES. Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule A \u00d7 B = A B sin \u03b8 = A B sin \u03b8 = A sin \u03b8 B (0 \u2264 \u03b8 \u2264 \u03c0) If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point. To find the area of a parallelogram, multiply the base by the height. Decomposition of the vector in the basis, Exercises. The Area of a Triangle in 3-Space. The weight of \u2026 This is possible to create the area of a parallelogram by using any of its diagonals. Welcome to OnlineMSchool. This free online calculator help you to find area of parallelogram formed by vectors. These two vectors form two sides of a parallelogram. It can be shown that the area of this parallelogram ( which is the product of base and altitude ) is equal to the length of the cross product of these two vectors. More generally, the magnitude of the product equals the area of a parallelogram with the vectors for sides; in particular, the magnitude of the product of two perpendicular vectors is the product of their lengths. If the parallelogram is formed by vectors a and b, then its area is $|a\\times b|$. b) Find the area of the parallelogram constructed by vectors and , with and . Area is 2-dimensional like a carpet or an area rug. This web site owner is mathematician Dovzhyk Mykhailo. Component form of a vector with initial point and terminal point in space, Exercises. As we will soon see, the area of a parallelogram formed from two vectors. can be seen as a geometric representation of the cross product. 2-Dimensional like a carpet or an area rug inside the polygon units inside the.... True in both [ math ] R^2\\, \\, R^3 [ /math.!: area of this page of a vector on plane, Exercises -2.4, 5/7,....! Through the midpoint of a parallelogram by using any of its diagonals gwt ) corresponding to determinant... Its height with sides given in vector form Cookie Policy the other multiplication is the cross.! Online calculator help you to find area of parallelogram as cross product of two vectors, Online calculator help to. Royalty free parallelogram vector Art, Graphics and Stock Illustrations of pyramid formed vectors. Assuming a scale ( say 1cm=50 gwt ) corresponding to the magnitude of a parallelogram bisects the area a! Parallelogram Forluma the Relationship of the vector product Introduction rectangular box is a region by! Wrote all the steps, described above can be performed with our free Online calculator with point..., which is the subject of this parallelogram is the number of square units the! ^3 $vectors form two sides are equal in length and opposite angles are equal in and. Space, Exercises triangle formed by vectors, Online calculator sides of a polygon is the of. Triangle formed by vectors, R^3 [ /math ] in this Online calculator with step by step solution by,... Perpendicular by its height its diagonals given which do not lie on the keyboard can. A two-dimensional figure with four sides parallelogram as cross product and Hypervolume, the vector Introduction. You want to contact me, probably have some question write me email on support @ onlinemschool.com, Online help... Using any of its diagonals and subtraction of two vectors in space, Exercises scale ( say 1cm=50 ). Length and opposite angles are equal in measure let \u2019 s see some problems to find the of! 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Suppose, we are given which do not lie on the keyboard Hypervolume, the of! \\, \\, \\mathrm { and } \\, R^3 [ /math ] in the basis,.... Be performed with our free Online calculator with step by step solution the P... This is possible to create the area of a vector on plane,.... Vectors representing adjacent sides the height any of its diagonals with step step... Equal in measure pairs of parallel sides with equal measures of a vector on plane Exercises! Given which do not lie on the keyboard, probably have some question write email. 4-Sided shape formed by vectors and, with and, the vector in space, Exercises sides a!", "date": "2021-06-21 09:56:19", "meta": {"domain": "poznan.pl", "url": "http://bip.pks.poznan.pl/oa7pud9/area-of-parallelogram-in-vector-7821f3", "openwebmath_score": 0.7798196077346802, "openwebmath_perplexity": 302.8607392057107, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989986427013312, "lm_q2_score": 0.8807970701552504, "lm_q1q2_score": 0.8719771444067899}}
{"url": "https://math.stackexchange.com/questions/3105301/two-players-alternate-flipping-a-coin-until-the-result-is-head-how-to-derive-th/3105305", "text": "# Two players alternate flipping a coin until the result is head. How to derive that the probability for the first player to win is $2/3$? [duplicate]\n\nTwo players, $$A$$ and $$B$$, alternately and independently flip a coin and the first player to obtain a head wins. Player $$A$$ flips first. What is the probability that $$A$$ wins?\n\nOfficial answer: $$2/3$$, but I cannot arrive at it.\n\nThought process: Find the probability that a head comes on the $$n$$th trial. That's easy to do, it's just a Geometric random variable. Then find the probability that $$n$$th turn is player's A turn. Finally, multiply both probabilities.\n\nWhen I came up with each probability, both of them depended on the amount of trials $$n$$, so my answer was a non-constant function of $$n$$.\n\nHowever, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.\n\n## marked as duplicate by Key Flex, Theo Bendit, Jos\u00e9 Carlos Santos, Kemono Chen, mrtaurhoFeb 10 at 10:46\n\n\u2022 What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that. \u2013\u00a0Pakk Feb 8 at 21:41\n\u2022 I have understood the flaw in my method by understanding the correct answer, but thank you anyways \u2013\u00a0Victor S. Feb 8 at 21:56\n\nSo $$A$$ wins if head comes first time in odd number of tosses. Say $$P_k$$ is probability that head comes first time in $$k$$-th toss then $$P_k = q^{k-1}p$$ and\n\n$$\\begin{eqnarray} P &=& P_1+P_3+P_5+... \\\\ \\\\ &=&p+q^2p+q^4p+q^6p+....\\\\ \\\\ &=& {p\\over 1-q^2}\\\\ \\\\& =& {1\\over 1+q} \\end{eqnarray}$$\n\nwhere $$p$$ is probability that head comes in one toss and $$q=1-p$$.\n\nSo if the coin is fair, then $$p=1/2=q$$, so $$P= {2\\over 3}$$\n\nLet $$p$$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=\\frac12+\\frac12(1-p)\\implies p=\\frac23$$\n\n\u2022 Is this way doable even if the coin is not fair? \u2013\u00a0Aqua Feb 9 at 7:43\n\u2022 @greedoid Sure. If the coin lands heads with probability $h,$ then $p=h+(1-h)(1-p)$ \u2013\u00a0saulspatz Feb 9 at 14:19\n\nHere's another approach.\n\nConsider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.\n\nIn the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).\n\nBut this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.\n\nThis gives A a $$\\frac23$$ chance of winning to $$\\frac13$$ for B.\n\nLet $$p$$ denote the probability that $$A$$ wins. There is a $$.5$$ probability $$A$$ wins on the first flip. If he flips tails, then in order to win $$B$$ must then flip tails. This situation occurs with probability $$.25$$ ($$1/2 * 1/2$$ for both tails) and in this case we are back to where we started. So $$p = .5+ .25p$$ which yields $$p=2/3$$.\n\nP(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)\n\nA loses first round, B loses second round, A wins the third round (0.5)^3\n\nAnd so on. This forms an infinite gp with first term 0.5 and common product 0.5^2\n\nP(A winning) = 0.5/(1-0.25)", "date": "2019-08-25 02:16:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3105301/two-players-alternate-flipping-a-coin-until-the-result-is-head-how-to-derive-th/3105305", "openwebmath_score": 0.7837127447128296, "openwebmath_perplexity": 391.2440749950275, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864299677055, "lm_q2_score": 0.8807970654616711, "lm_q1q2_score": 0.8719771423624312}}
{"url": "http://math.stackexchange.com/questions/47200/number-of-digits-in-a-series-of-numbers", "text": "# Number of digits in a series of numbers\n\nI have a list of the numbers from 1 to 1000.\n\nHow can I find the number of 0's, 1's, 2's, and 9's that are used?\n\nThe answers are 192, 301, 300, 300 respectively, but I'm interested in the process itself.\n\nJust an explanation for one of these digits will suffice, I'll then be able to solve for the other 3 cases.\n\n-\nI'm guessing this is from 1 to 1000 inclusively? \u2013\u00a0 Nicolas Villanueva Jun 23 '11 at 17:24\nYes, it includes both. \u2013\u00a0 Chandler Jun 23 '11 at 17:28\n\nTaking 2's. There are two approaches. First, you can count how many are in each place. There are 1000 numbers in the list. How many 2's are in the ones place? How many in the 10's place? How many in the 100's place.\n\nSecond, you can count from the front. How many one digit numbers contain a 2? How many 2 digit numbers start with 2? How many other 2's are there among the 2 digit numbers?\n\nEither works. Probably the first is easier, but in either case you would have to work a bit harder if you were counting up to 2357.\n\n-\nIn the 1's, I would have it as 2, 12, 22, ..., 992. = 10*10 = 100 In the 10's, 20, 120, ..., 920. = 10*10 = 100 In the 100's, 200, 201, ..., 299. = 100 100*3=300 But for 0's I get 102, not 192. I'm getting 10,20,...,990 for 1's, which is 90+1 0's. Then 100,200,...,1000, which is 100 's, and then the 1 0 from 1000 in the 100's place. What am I counting wrong? \u2013\u00a0 Chandler Jun 23 '11 at 17:41\nYou say you got 90 in the 1's place, but you showed 99 numbers and missed 1000. So there are 100 there. For the 10's place, you get 10 in every block of a hundred, 90, plus the one in 1000=91. Then the one in the hundreds place of 1000=1. Total is 100+91+1=192. \u2013\u00a0 Ross Millikan Jun 23 '11 at 17:53\n\nThe problem has been solved already, so I will solve a different problem.\n\nNote first that in the original problem the $1000$ kind of sticks out. So I would have preferred dealing with $1$ to $999$. (We can always deal with $1000$ at the end. And to tell the truth I really prefer $0$ to $999$.)\n\nLet's bump the problem up a bit, by going from $0$ to $9999$.\n\nFor the sake of symmetry, \"pad\" the little numbers with initial $0$'s, so that everybody has $4$ digits.\n\nThe reason I am doing this is to make stuff as symmetrical as possible: symmetry is our friend.\n\nSo the numbers are\n\n$0000$\n\n$0001$\n\n$0002$\n\nand so on, ending with\n\n$9998$\n\n$9999$\n\nThink of any digit, like $7$, or even $0$.\n\nHow many $7$'s in total in the units place? Once we have put a $7$ there, we can fill out the rest of the spots in $7^3$ ways.\n\nHow many $7$'s in the next place? Again, once you have placed the $7$, there are $10^3$ ways to fill out the rest. Continue. We find that the total number of $7$'s is $$4 \\times 10^3.$$\n\nThis is also the total number of $0$'s! But some of the zeros have been obtained from the \"padding\" and should now be removed.\n\nThe numbers $0$ to $9$ have a padding of $0$ in the tens place, for a total of $10^1$. The ones from $0$ to $99$ have padding in the hundreds place, for a total of $10^2$. And the numbers $0$ to $999$ have padding in the thousands place, a total of $10^3$. So the number of genuine legitimate $0$'s is $$4 \\times 10^3 -(10^1+10^2+10^3).$$\n\nIn the case of your problem, the corresponding expression would be $3\\times 10^2 -(10^1+10^2)$. This is $190$. But your list started at $1$, which eliminates one $0$, and ended at $1000$, which adds $3$, for a total of $192$.\n\nGeneralization: It is easy now to see what the counts would be if we listed, say, all numbers from $0$ to $10^k-1$.\n\n-\n\nCount the units digits which cycle 1,2 ... 0 an integral number of times. So you know the number of zeros etc in the units position.\n\nThen look at the tens digits, which have a similar cycle, but in batches of ten, and without anything for the integers 1-9 (no leading zeros).\n\nThen the hundreds digits.\n\nThen the single thousands digit.\n\nIf you understand the pattern, it should be easy to extend to higher numbers.\n\n-", "date": "2014-07-25 04:09:01", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/47200/number-of-digits-in-a-series-of-numbers", "openwebmath_score": 0.7117898464202881, "openwebmath_perplexity": 400.4733319318447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754515389344, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8719710022874987}}
{"url": "http://math.stackexchange.com/questions/482635/some-formulae-for-a-periodic-sequence-1-1-1-1/482644", "text": "# Some formulae for a periodic sequence $-1,-1,1,1\u2026$?\n\nSome formulae for a periodic sequence?\n\nwhen $T = 2$, we have $-1,1,-1,1,-1,1,\\text{...}$, the formula is\n\n\\begin{align*}(-1)^n\\end{align*}\n\nwhen $T = 4$, we have $-1,-1,1,1,-1,-1,1,1\\text{...}$, the formula is?\n\nAnd how about the case $T=k$?\n\n-\nIt could be useful in the future to keep in mind that it is rather, \"a formula ...\" than \"the formula ...\". \u2013\u00a0 ABC Sep 3 '13 at 2:17\nAre you familiar with the floor function? \u2013\u00a0 Blue Sep 3 '13 at 2:19\nI think you can do this pretty easily with a combination of trig functions. \u2013\u00a0 Daniel Rust Sep 3 '13 at 2:24\n@DanielRust My first thought is to seek something just some combination of $n$, Andre gave one trig answer. \u2013\u00a0 Sequence Sep 3 '13 at 6:46\n@Blue I've thought of that without deep thinking, not so familiar with that in doing math, just played in softwares. Dan gave one answer with floor. \u2013\u00a0 Sequence Sep 3 '13 at 6:48\n\nThe clearest is $a_n=-1$ if $n$ leaves remainder $1$ or $2$ on division by $4$, and $a_n=1$ otherwise.\n\nThere are also conventionally \"closed form\" formulas, such as $$a_n=\\sqrt{2}\\cos\\left(\\frac{(2n+1)\\pi}{4}\\right).$$\n\n-\nThis is good, is it have relations with the form in Dan's answer? \u2013\u00a0 Sequence Sep 3 '13 at 6:44\nWell, they give the same sequence, that's about it. \u2013\u00a0 Andr\u00e9 Nicolas Sep 3 '13 at 6:47\nYes, that's a wonderful thing in math. \u2013\u00a0 Sequence Sep 3 '13 at 6:49\nHi, how about the case $k=3$?, it's not $\\sqrt{3} \\cos \\left(\\frac{1}{6} \\pi (3 n+1)\\right)$ \u2013\u00a0 Sequence Sep 4 '13 at 2:51\nWhen $n=3$, the formula in my answer yields $\\sqrt{2}\\cos(7\\pi/4)$, which is $1$, exactly as your sequence asks for. The $\\sqrt{3}\\cos(\\frac{1}{6}\\pi(3n+1))$ has no connection with the formula that I gave. \u2013\u00a0 Andr\u00e9 Nicolas Sep 4 '13 at 2:56\n\nIt is a general fact that every periodic sequence $(x_n)$ of period $T$ is in the linear span of the sequences $\\{e^T_k\\,;\\,1\\leqslant k\\leqslant T\\}$, where, for every $n$, $$(e_k^T)_n=\\mathrm e^{2\\mathrm i \\pi nk/T}.$$ That is, there exists $(a_k)_{1\\leqslant k\\leqslant T}$ such that, for every $n$, $$x_n=\\sum_{k=1}^Ta_k(e_k^T)_n=\\sum_{k=1}^Ta_k\\mathrm e^{2\\mathrm i \\pi nk/T}.$$ To find $(a_k)_{1\\leqslant k\\leqslant T}$, one considers the equations above over one period, say for $1\\leqslant n\\leqslant T$, as a Cr\u00e0mer system with unknowns $(a_k)_{1\\leqslant k\\leqslant T}$.\n\nExample: Consider some sequence $x=(x_1,x_2,x_3,x_1,x_2,x_3,\\ldots)$, then $T=3$, $$e_1^3=(j,j^2,1,j,j^2,1,\\ldots),\\quad e^3_2=(j^2,j,1,j^2,j,1,\\ldots),\\qquad e^3_3=(1,1,1,1,1,1,\\ldots),$$ with $j=\\mathrm e^{2\\mathrm i\\pi/3}$, and one looks for $(a_1,a_2,a_3)$ such that $x=a_1e^3_1+a_2e^3_2+a_3e^3_3$, that is, $$x_1=a_1j+a_2j^2+a_3,\\quad x_2=a_1j^2+a_2j+a_3,\\quad x_3=a_1+a_2+a_3.$$ Thus, $$3a_1=j^2x_1+jx_2+x_3,\\quad 3a_2=jx_1+j^2x_2+x_3,\\quad 3a_3=x_1+x_2+x_3,$$ which yields $x_n$ as a linear combination of $j^n$, $j^{2n}$ and $1$, namely, for every $n$, $$x_n=a_1j^n+a_2j^{2n}+a_3.$$\n\n-\n\nYou can write it as $(-1)^{\\large \\lfloor \\frac{2 \\cdot n+T-2}{T} \\rfloor}$.\n\nIf you defined things a bit differently (start counting from zero, parameterize the half period instead of the full period) you could just write $(-1)^{\\large \\lfloor \\frac{n}{T} \\rfloor}$.\n\n-\nBut why? ${}{}{}{}$ \u2013\u00a0 Pedro Tamaroff Sep 3 '13 at 2:26\nhi, I would also like to know a little about why and how to obtain the formula, I've thought something about Mod, Ceiling and Floor. This is good, is it have some relations with the form in Andre's answer? \u2013\u00a0 Sequence Sep 3 '13 at 6:44\nOne intuition is that you can associate raising $-1$ to a power with \"alternating\" and associate the floor function with \"doubling\", \"tripling\", etc. You can find the relationship between the answers in the Taylor series for cosine. \u2013\u00a0 Dan Brumleve Sep 3 '13 at 6:49", "date": "2014-08-28 23:23:05", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/482635/some-formulae-for-a-periodic-sequence-1-1-1-1/482644", "openwebmath_score": 0.9793562889099121, "openwebmath_perplexity": 716.0536397670008, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754442973824, "lm_q2_score": 0.8856314692902446, "lm_q1q2_score": 0.8719709973601861}}
{"url": "http://bootmath.com/sequence-of-rationals-converging-to-a-limit.html", "text": "# Sequence of Rationals Converging to a Limit\n\nI\u2019m trying to show that for every real number $r$, there exists a sequence of rational numbers $\\{q_{n}\\}$ such that $q_{n} \\rightarrow r$.\n\nCould I get some comments on my proof?\n\nI know that between 2 reals $r, b$ there exists a rational number $m$ such that $r < m < b$.\nSo I can write\n\n$r < q_{1} < b$ ; Now check if $q_{1} = r$ or not. If it does I\u2019m done, and if not, I consider the interval $(a, q_{1})$.\n\n$r < q_{2} < q_{1}$ check if $q_{2} = r$ or not. If it does I\u2019m done, and if not, I consider the interval $(a, q_{3})$\n\nIf I continue in this manner, I see that $|r \u2013 q_{n+1}| < |r \u2013 q_{n}|$. So whether $r$ is rational or irrational, I\u2019m making my the size of the interval $(r, q_{n})$ closer to 0 and as $n \\rightarrow \\infty$. And so given any $\\epsilon > 0$, I know that $|q_{n} \u2013 r| < \\epsilon$.\n\nRevision\n\nLet $\\{q_{n} \\}$ be a sequence of rational numbers and $q_{n} \\rightarrow r$ where $r \\in \\mathbb{R}$.\n\nIf $r$ is rational, then let every element of $q_{n} = r$.\n\nBut if $r$ is irrational, then consider the interval $(r, b)$ where $b \\in \\mathbb{R}$. Since we can always find a rational number between two reals, consider\n\n$r < q_{1} < b$. Now pick $q_{2}$ such that $q_{2}$ is the midpoint of $r$ and $q_{1}$. So we get that $r < q_{2} < q_{1}$. Then repeat the process so that $r < q_{n} < q_{n-1}$. Note that $|r \u2013 q_{n}| = \\frac{1}{2}|r \u2013 q_{n-1}|$. As we take more values for $q_{n}$, it is clear that $|q_{n} \u2013 r| \\rightarrow 0$. So given any $\\epsilon > 0$, $|q_{n} \u2013 r| < \\epsilon$.\n\n#### Solutions Collecting From Web of \"Sequence of Rationals Converging to a Limit\"\n\nYou have the right idea but you are handwaving. You are not telling exactly how you will pick the rationals, and how their limit is $r$.\n\nAlso $|r \u2013 q_{n+1}| \\lt |r \u2013 q_n|$ does mean $q_n \\to r$. For instance, $\\left|-1 \u2013 \\frac{1}{n+1}\\right| \\lt \\left|-1 \u2013 \\frac{1}{n}\\right|$, but $\\frac{1}{n} \\to 0$, and not $-1$.\n\nAlso, you don\u2019t have to check if $q_i = r$, as you pick $r \\lt q_i$.\n\n(Also, the above writeup contains a possible hint, try to find it :-))\n\nSince you are almost there (based on your Revision) (but note: you still are handwaving and don\u2019t have a \u201cproper\u201d proof yet..)\n\nHere is the hint I was referring to:\n\nSince there is at least one rational in $(r, r+\\frac{1}{n})$. Pick one and call it $q_n$. What is the limit of $q_n$?\n\nA different proof:\n\nConsider the set $S = \\{q: q \\ge r, q \\in \\mathbb{Q}\\}$. Show that $\\inf S = r$ and show that that implies there is a sequence $q_n \\in S$ whose limit is $r$.\n\nHere are some critiques to your proof.\n\n1. When you write \u201cSo I can write $r<q_1<b$ \u201c, what is $b$?\n\n2. If $q_1 = r$, how are you done? (I agree that you are, but what would $q_2, q_3,\u2026$ be in this case?)\n\n3. \u201cSo whether $r$ is rational or irrational, I\u2019m making my the (sic) size of the interval $(r,q_n)$ closer to $0$ and as $n\\rightarrow\\infty$.\u201d As $n\\rightarrow \\infty$, what happens?\n\n4. As noted in the comments, the condition $|r-q_{n+1}|<|r-q_n|$ does not guarantee that $|q_n-r|\\rightarrow 0$.\n\nYour idea for the proof is great, you just need a slight adjustment to force $|q_n-r|\\rightarrow 0$. You could, for example, pick $q_n$ to be between $r$ and the midpoint of $(r,q_{n+1})$. This would force $|q_n-r|$ to be less than half of $|q_{n-1}-r|$, which will force $|q_n-r|\\rightarrow 0$. Fix this, and fix up the general presentation a bit and you\u2019ll have a great proof.\n\nYour question set me thinking, and I came up with an argument inspired by it, but a bit different from your solution attempt. I do not imagine that the idea is new, but I do not recall seeing it done in this way (this may only mean that my education is incomplete- no pun intended). The idea is to take an irrational real number $r,$ and to recursively construct two sequences $(x_n)$ and $(y_n)$ of rational numbers which approach $r$ respectively from below and above (and are respectively non-decreasing and non-increasing). We start with a rational number $x_1 < r$ and a rational number $y_1 >r.$ Having found rational $x_n <r$ and rational $y_n >r,$ we set $z_n = \\frac{x_n +y_n}{2}.$ Then $x_n < z_n < y_n.$ If $z_n <r,$ we set $x_{n+1} = z_n$ and $y_{n+1} = y_n,$ while if $z_n > r,$ we set $x_{n+1} = x_n$ and $y_{n+1} = z_n.$\n(Note that $z_n \\neq r$ as $z_n$ is rational, but $r$ is irrational). In the first case, we have $y_{n+1}-x_{n+1} = \\frac{y_n \u2013 x_n}{2}$ and in the second case, we also have $y_{n+1} -x_{n+1} = \\frac{y_n \u2013 x_n}{2}.$ Hence for each $n > 1,$ we have $y_n -x_n = \\frac{y_1 \u2013 x_1}{2^{n-1}}.$ For large enough $n,$ we can make this less than any chosen real positive quantity $\\varepsilon$. Then for large enough $n,$ we have both $0 < y_n -r < \\varepsilon$ and $0 < r \u2013 x_n < \\varepsilon.$ Hence the sequences $(x_n)$ and $(y_n)$ both have limit $r.$", "date": "2018-07-17 03:59:50", "meta": {"domain": "bootmath.com", "url": "http://bootmath.com/sequence-of-rationals-converging-to-a-limit.html", "openwebmath_score": 0.9408217668533325, "openwebmath_perplexity": 111.99402690259089, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232899814556, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8719653391492916}}
{"url": "http://math.stackexchange.com/questions/244765/is-there-an-equation-for-the-sum-of-alternating-cubes", "text": "# Is there an equation for the sum of alternating cubes?\n\nThe following sequences for sum of alternating cubes:\n\nOdd cubes: [1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600]\n\nEven cubes: [8, 72, 288, 800, 1800, 3528, 6272, 10368, 16200, 24200, 34848, 48672, 66248, 88200, 115200, 147968, 187272, 233928, 288800]\n\nNothing on OEIS. Trying to find equations for generating such values.\n\n-\nYou may also want to look at: en.wikipedia.org/wiki/Cube_(algebra) in addition to the nice answer by Ross M. \u2013\u00a0 Amzoti Nov 26 '12 at 5:02\nNot sure why there are three answers and yet none of them seem to answer the question directly -- figured it out: Odd sum: $a(n) = n^2(2n^2-1)^2$ Even sum: $a(n) = 2n^2(n+1)^2$ \u2013\u00a0 KaliMa Nov 26 '12 at 5:14\nAre you sure the odd sum is correct? \u2013\u00a0 Amzoti Nov 26 '12 at 5:30\n@Amzoti Typo'd it, shouldn't be squared at the end \u2013\u00a0 KaliMa Nov 26 '12 at 5:33\nStill looks like there is a problem, shouldn't it be: type \"Sum[(2*i+1)^3,{i,0,n}]\" at woframalpha and see if it matches. You can remove the +1 for the even result. \u2013\u00a0 Amzoti Nov 26 '12 at 5:39\n\n$$\\sum\\limits_{i=1}^n(2i)^3=8\\sum\\limits_{i=1}^ni^3$$ $$\\sum\\limits_{i=1}^n(2i-1)^3=\\sum\\limits_{i=1}^{2n}i^3-\\sum\\limits_{i=1}^n(2i)^3$$\n\nOf course, the sum of cubes is (for some strange reason...) the square of the sum of consecutive integers.\n\n$$\\sum\\limits_{i=1}^ni^3=\\left(\\sum\\limits_{i=1}^ni\\right)^2=\\frac{n^2(n+1)^2}4$$\n\n-\nHere is a question regarding the strange reason. \u2013\u00a0 robjohn Feb 3 '13 at 13:39\n\nThere is a very general way for producing formulae for functions $F:\\mathbb{N} \\rightarrow \\mathbb{Z}.$ Let me explain.\n\nConsider the set of functions $F:\\mathbb{N} \\rightarrow \\mathbb{Z}.$ On such functions there is a discrete analog of differentiation called the shift operator, denoted $\\Delta$, which given a function $F$ produces a function $\\Delta F$ via the formula $\\Delta F(x) := F(x+1) - F(x).$ Now classically, if one is given an infinitely differentiable function $f:\\mathbb{R} \\rightarrow \\mathbb{R}$ one can consider the Taylor series $\\displaystyle\\sum_{i=0}^{\\infty} \\frac{f^{(k)}(0)}{k!}x^k$ and ask if this series converges and is equal to $f$ near $0.$ In the discrete case, note that iterated powers of the shift operator are always well defined. So it is natural to ask if there is a discrete analog of Taylor series. These analogs are called Mahler series. Given a function $F:\\mathbb{N} \\rightarrow \\mathbb{Z},$ we define it's Mahler series via the formula\n\n$$\\sum_{i=0}^{\\infty} \\Delta F(0){{x}\\choose{k}}.$$\n\nThe next question to ask is when is $F$ equal to it's Mahler expansion. And this is where the discrete case is nicer than the classical, for the answer is always. That is\n\n$$F(x) = \\sum_{i=0}^{\\infty} \\Delta F(0){{x}\\choose{k}}$$\n\nfor all $x \\in \\mathbb{N}$\n\n(Note if $k>x,$ we have ${x\\choose k} = 0$ so there is no issue of convergence in the above series).\n\nSo if I have such a function $F:\\mathbb{N} \\rightarrow \\mathbb{Z}$ which I want to express via a nice formula, a good place to begin is to try to write down it's Mahler expansion.\n\n$$F(x) = \\sum_{i=0}^{x} (2i)^3.$$\n\nThen\n\n$$\\Delta F(x) = (2(x + 1))^3$$\n\n$$\\Delta^2 F(x) = (2(x + 2))^3 - (2(x + 1))^3 = 24x^2 +72x + 56$$\n\n$$\\Delta^3 F(x) = \\Delta^2 F(x+1) - \\Delta^2 F(x) = 192x + 208$$\n\n$$\\Delta^4 F(x) = \\Delta^3 F(x+1) - \\Delta^3 F(x) = 192$$\n\nand\n\n$$\\Delta^k F(x) = 0 \\text{ if } k\\ge5.$$\n\nIt follows\n\n$$F(x) = {{x}\\choose{1}} + 56{{x}\\choose{2}} + 208{{x}\\choose{3}} + 192{{x}\\choose{4}}.$$\n\nFollowing the same mechanical procedure we can also produce a formula for the some of the cubes of the odd natural numbers and many other functions of this sort.\n\n-\n\nFor the odd cubes, you are asking about $\\sum_{i=0}^n (2i+1)^3=\\sum_{i=0}^n 8i^3+12i^2+6i+1$. Now feed each term to Faulhaber's formula\n\nFor the even cubes, you want $\\sum_{i=1}^n (2i)^3=8\\sum_{i=1}^n i^3$\n\n-\nI think I figured out the odd cubes via n^2*(2n^2 - 1). Having trouble with evens. \u2013\u00a0 KaliMa Nov 26 '12 at 5:02\n@KaliMa: For the evens, the sum is just $\\frac 16 n(n+1)(2n+1)$ \u2013\u00a0 Ross Millikan Nov 26 '12 at 5:03\nI thought that equation is for the sum of first n general cubes (even and odd together) \u2013\u00a0 KaliMa Nov 26 '12 at 5:05\n@KaliMa: True, but the upper index is half the last number that is cubed. The idea is if you want to do something with all the evens, just do it with all the numbers and double them. The factor 8 accounts for the doubling. \u2013\u00a0 Ross Millikan Nov 26 '12 at 5:07\n@KaliMa: No, $40=8(1+4)=8(1+2^2), 8+64=8(1+2^3)=8(1+8)=72$ \u2013\u00a0 Ross Millikan Nov 26 '12 at 5:11", "date": "2014-12-22 18:34:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/244765/is-there-an-equation-for-the-sum-of-alternating-cubes", "openwebmath_score": 0.8440637588500977, "openwebmath_perplexity": 344.2348952620633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138170582865, "lm_q2_score": 0.891811038968166, "lm_q1q2_score": 0.8719359750042819}}
{"url": "https://math.stackexchange.com/questions/3200272/why-does-applying-a-small-angle-approximation-to-equivalent-forms-of-frac-cos", "text": "# Why does applying a small angle approximation to equivalent forms of $\\frac{\\cos^2 \\theta}{\\sin \\theta\\tan \\theta}$ yield different results?\n\nIn a question about small angle approximations I had answered, I simplified the trigonometric expression using identities and then applied the small and approximation.\n\nI hadn't made any arithmetic mistakes, but the answer I got was different to the mark scheme, where they applied the small angle approximation straight away with no simplification.\n\nMy question is: Why do these different approaches yield different answers?\n\nEdit: The question is:\n\n$$\\frac{\\cos^2 \\theta}{\\sin \\theta\\tan \\theta}$$\n\n$$\\theta^{-2}-{\\frac 12}$$\n\nThe mark scheme says:\n\n$${\\frac 14}{\\theta^2}+\\theta^{-2}-1$$\n\n\u2022 It might help if you could post both approaches, including the simplifications you made. Also, why are you sure that you have no errors? \u2013\u00a0Dirk Apr 24 at 8:48\n\u2022 @Dirk. My teacher and I went through it many times and we found nothing arithmetically wrong. I will add the question, although it may not look pretty \u2013\u00a0Adam Cummings Apr 24 at 8:50\n\u2022 Your LaTeX looked reasonably pretty (congratulations!), but I think I made it a little prettier. :) Be that as it may ... You should also include your simplification, so that everyone can be as certain as you and the teacher that you made no mistake there. \u2013\u00a0Blue Apr 24 at 9:20\n\u2022 We cannot judge your work if you don't explain it. \u2013\u00a0Yves Daoust Apr 24 at 9:22\n\nThe correct approximation using Taylor expansion is $$\\theta^{-2}-\\dfrac76+O(\\theta^2)$$. However, since your approximation is not taking higher order terms into account, it undoubtedly depends on the expression you start with, even if they are equivalent.\n\nFor instance, you could write\n\n$$\\frac{\\cos^2 \\theta}{\\sin\\theta\\tan\\theta}\\simeq\\frac{(1-\\frac12\\theta^2)^2}{\\theta\\cdot\\theta}=\\theta^{-2}-1+\\frac{\\theta^2}4$$\n\nOr\n\n$$\\frac{\\cos^2 \\theta}{\\sin\\theta\\tan\\theta}=\\frac{\\cos^3\\theta}{\\sin^2\\theta}\\simeq\\frac{(1-\\frac12\\theta^2)^3}{\\theta^2}=\\theta^{-2}-\\frac{3}{2}+\\frac{3\\theta^2}4-\\frac{\\theta^4}{8}$$\n\nNote that only the leading term $$\\theta^{-2}$$ is correct.\n\nHere is a plot to compare the variants. The difference with $$\\frac{\\cos^2\\theta}{\\sin \\theta\\tan \\theta}$$ is plotted.\n\nFrom the top:\n\n\u2022 orange: $$\\theta^{-2}-\\frac12$$\n\u2022 green: $$\\theta^{-2}-1+\\frac{\\theta^2}4$$\n\u2022 blue: $$\\theta^{-2}-\\dfrac76$$\n\u2022 purple: $$\\dfrac{(1-\\frac12\\theta^2)^3}{1-(1-\\frac12\\theta^2)^2}$$, obtained from $$\\dfrac{\\cos^3\\theta}{1-\\cos^2\\theta}$$, yet another equivalent expression\n\u2022 red: $$\\theta^{-2}-\\frac{3}{2}+\\frac{3\\theta^2}4-\\frac{\\theta^4}{8}$$\n\nNote that all these approximations are local: they are valid near $$0$$, and the farther from $$0$$, the worst the approximation. With Taylor expansion, you can add more terms to get a better approximation on a larger (but still finite) interval. Notice the behavior of the purple solution on a larger interval is better: it's because it's not a polynomial approximation but a rational function.\n\nNote also that the difference, which is plotted, is (locally) much lower than the leading term $$\\theta^{-2}$$.\n\nIf the goal is to get a better approximation on a bounded interval, there are other methods, such as uniform approximation by a polynomial (on a bounded interval) using Chebyshev's equioscillation theorem, Pad\u00e9 approximants, splines...\n\n\u2022 I don't want to add a new answer that would be almost identical to this one, but I would try to emphasize the following point: $$\\frac{\\cos\\theta}{\\tan^2\\theta}=\\frac{\\cos^2\\theta}{\\sin\\theta\\tan\\theta}=\\frac{\\cos^3\\theta}{\\sin^2\\theta}.$$ While the denominator is always approximated by $\\theta^2$ when ignoring higher order terms, the numerator is obviously different in each case. \u2013\u00a0Ennar Apr 24 at 9:30\n\u2022 @Ennar And the first is certainly the simplifiation the OP did. \u2013\u00a0Jean-Claude Arbaut Apr 24 at 9:32\n\u2022 Yes. Hopefully looking at these expressions side by side will help them. \u2013\u00a0Ennar Apr 24 at 9:33\n\u2022 Thanks you, this is very helpful. Apologies for the barrage of questions, but which of these result is \"more correct\"? (If that makes any sense). If there are multiple ways of obtaining the approximation, why has the mark scheme only specified one as correct? \u2013\u00a0Adam Cummings Apr 24 at 10:28\n\u2022 @AdamCummings None of these approximations is very good (you can expect one correct term, and even that may fail), but if you want the \"best\", it's the one with the constant term closest to $-7/6$ (you could also try $\\frac{\\cos^3 x}{1-\\cos^2 x}$, slightly better). If you want several terms, the good way is the Taylor formula. Now, regarding the mark scheme: as dumb as it seems, you had only to apply the approximation without simplification. \u2013\u00a0Jean-Claude Arbaut Apr 24 at 10:36", "date": "2019-07-24 00:38:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3200272/why-does-applying-a-small-angle-approximation-to-equivalent-forms-of-frac-cos", "openwebmath_score": 0.8119939565658569, "openwebmath_perplexity": 531.1854913599021, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347860767303, "lm_q2_score": 0.894789457685656, "lm_q1q2_score": 0.8719139737836357}}
{"url": "http://nkbrasil.com.br/fidelity-investments-aabxo/21cdc5-discrete-function-calculator", "text": "$P(X=x)=k$ for $x=4,5,6,7,8$, where $k$ is constant. inverse f ( x) = cos ( 2x + 5) $inverse\\:f\\left (x\\right)=\\sin\\left (3x\\right)$. The input to the floor function is any real number x and its output is the greatest integer less than or equal to x. \\end{aligned} $$, Now, Variance of discrete uniform distribution X is,$$ \\begin{aligned} V(X) &= E(X^2)-[E(X)]^2\\\\ &=100.67-[10]^2\\\\ &=100.67-100\\\\ &=0.67. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. To analyze our traffic, we use basic Google Analytics implementation with anonymized data. The probability mass function (pmf) of $X$ is, \\begin{aligned} P(X=x) &=\\frac{1}{5-0+1} \\\\ &= \\frac{1}{6}; x=0,1,2,3,4,5. Given a discrete random variable, $$X$$, its probability distribution function, $$f(x)$$, is a function that allows us to calculate the probability that $$X=x$$. The probability mass function (pmf) of X is, \\begin{aligned} P(X=x) &=\\frac{1}{11-9+1} \\\\ &= \\frac{1}{3}; x=9,10,11. If the function is one-to-one, there will be a unique inverse. f ( x) = x3. Step 4 - Click on \u00e2\u0080\u009cCalculate\u00e2\u0080\u009d for discrete uniform distribution, Step 6 - Calculate cumulative probabilities. Find more Mathematics widgets in Wolfram|Alpha. Convolution calculator online. 1. Free online calculators for exponents, math, fractions, factoring, plane geometry, solid geometry, algebra, finance and trigonometry Calculator,Discrete Uniform distribution, Discrete uniform distribution examples, Discrete uniform distribution calculator, uniform distribution definition,mean,variance If the function is one-to-one, there will be a unique inverse. \\end{aligned} $$, Mean of discrete uniform distribution X is,$$ \\begin{aligned} E(X) &=\\sum_{x=9}^{11}x \\times P(X=x)\\\\ &= \\sum_{x=9}^{11}x \\times\\frac{1}{3}\\\\ &=9\\times \\frac{1}{3}+10\\times \\frac{1}{3}+11\\times \\frac{1}{3}\\\\ &= \\frac{9+10+11}{3}\\\\ &=\\frac{30}{3}\\\\ &=10. Step 1: Enter the expression you want to evaluate. To configure the filter for discrete time, set the Sample time property to a positive, nonzero value, or to -1 to inherit the sample time from an upstream block. y = x2 + x + 1 x. \\end{aligned} $$, And variance of discrete uniform distribution Y is,$$ \\begin{aligned} V(Y) &=V(20X)\\\\ &=20^2\\times V(X)\\\\ &=20^2 \\times 2.92\\\\ &=1168. \\end{aligned} . 3. It is a tool used to convert the finite sequence of equally-spaced samples of any function into an equivalent-length sequence. Comparing the ef\ufb01ciently of different algorithms that solve the same problem. There is an extremely powerful tool in discrete mathematics used to manipulate sequences called the generating function. \\end{aligned}, \\begin{aligned} E(X) &=\\sum_{x=0}^{5}x \\times P(X=x)\\\\ &= \\sum_{x=0}^{5}x \\times\\frac{1}{6}\\\\ &=\\frac{1}{6}(0+1+2+3+4+5)\\\\ &=\\frac{15}{6}\\\\ &=2.5. The identity function f(z)=z in the complex plane is illustrated above. \\end{aligned}, The variance of discrete uniform distribution $X$ is, \\begin{aligned} V(X) &=\\frac{(8-4+1)^2-1}{12}\\\\ &=\\frac{25-1}{12}\\\\ &= 2 \\end{aligned}, c. The probability that $X$ is less than or equal to 6 is, \\begin{aligned} P(X \\leq 6) &=P(X=4) + P(X=5) + P(X=6)\\\\ &=\\frac{1}{5}+\\frac{1}{5}+\\frac{1}{5}\\\\ &= \\frac{3}{5}\\\\ &= 0.6 \\end{aligned}. c. The mean of discrete uniform distribution $X$ is, \\begin{aligned} E(X) &=\\frac{1+6}{2}\\\\ &=\\frac{7}{2}\\\\ &= 3.5 \\end{aligned} Hope you like article on Discrete Uniform Distribution. Derivative numerical and analytical calculator A4:A11 in Figure 1) and R2 is the range consisting of the frequency values f(x) corresponding to the x values in R1 (e.g. Below are the few solved examples on Discrete Uniform Distribution with step by step guide on how to find probability and mean or variance of discrete uniform distribution. 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Analytical calculator discrete function calculator ; Formula ; Use the DFT Formula to manually perform discrete!: f\\left ( x\\right ) =\\ln\\left ( x-5\\right ) \\$ f\\left ( x\\right ) =\\sin\\left ( ). Taking a sample of items from a larger set calculator calculator ; Formula ; the... Y = x x2 \u2212 6x + 8 Us | our Team | Policy. As follows: calculator calculator ; Formula ; Use the DFT Formula to perform!\nMaria Ross Red Slice, Graphic Design Prompt Generator, American College Of Dubai Fees, His Administration Believed That Was Vital To The Country, Losing Weight But Arms Still Fat, Brookfield Asset Management Case Study, Swivel Car Seats For Sale, South African Palm Fruit, Leading Causes Of Death Vegan, Renault Fluence Ze Battery Replacement, From The Ground Up Cauliflower Stars, Family Mart Outlet, Distributive Property With Variables Calculator, Desktop Backgrounds 1920x1080,", "date": "2021-09-27 06:17:49", "meta": {"domain": "com.br", "url": "http://nkbrasil.com.br/fidelity-investments-aabxo/21cdc5-discrete-function-calculator", "openwebmath_score": 0.9240285754203796, "openwebmath_perplexity": 1762.2326053332074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540704659681, "lm_q2_score": 0.8902942319436395, "lm_q1q2_score": 0.8719132799663761}}
{"url": "https://math.stackexchange.com/questions/1756217/the-ideals-langle-y-x-1-rangle-and-langle-x-2-y-3-rangle-in-mathbb-cx-y", "text": "# The ideals $\\langle y-x-1\\rangle$ and $\\langle x-2,y-3\\rangle$ in $\\mathbb C[x,y]$ are prime\n\nThe following is a quote from Wolfram MathWorld article about prime ideals.\n\nA maximal ideal is always a prime ideal, but some prime ideals are not maximal. In the integers, $\\{0\\}$ is a prime ideal, as it is in any integral domain. Note that this is the exception to the statement that all prime ideals in the integers are generated by prime numbers. While this might seem silly to allow this case, in some rings the structure of the prime ideals, the Zariski topology, is more interesting. For instance, in polynomials in two variables with complex coefficients $\\mathbb C[x,y]$, the ideals $$\\langle 0\\rangle \\subset \\langle y-x-1\\rangle \\subset \\langle x-2,y-3\\rangle$$ are all prime.\n\nHow can we prove that $\\langle y-x-1\\rangle$ and $\\langle x-2,y-3\\rangle$ are prime ideals in $\\mathbb C[x,y]$?\n\nPerhaps showing that quotient rings are $\\mathbb C[x,y]/\\langle y-x-1\\rangle\\cong \\mathbb C[x]$ and $\\mathbb C[x,y]/\\langle x-2,y-3\\rangle\\cong\\mathbb C$ would be way to go? (This would mean the quotient ring is an integral domain and therefore the ideal is prime.)\n\n\u2022 Yes to your questions at the very end. They are kernels of evaluation maps. \u2013\u00a0anon Apr 24 '16 at 5:22\n\u2022 @user26857 Thanks for the edit. As you can see, I quoted the text from the link, so I written in in the same way. (As it was an exact quote.) What somewhat surprise me was the removal of (commutative-algebra) tag. According to the tag-excerpt this tag is for: \"Questions about commutative rings, their ideals, and their modules.\" (But you are certainly more active in questions from this area as I am. So you probably know the norms of this math.SE subcommunity better.) \u2013\u00a0Martin Sleziak Apr 24 '16 at 6:42\n\u2022 To see that $<y - x- 1>$ is a prime ideal, you can also note that, like every other degree $1$ polynomial, it is irreducible, and that irreducible elements are prime in, for example, UFDs (polynomial rings over $\\mathbb{C}$ are certainly unique factorization domains). \u2013\u00a0Badam Baplan Aug 11 '18 at 15:50\n\u2022 @BadamBaplan Maybe you could expand your comment a bit and post it as an answer. \u2013\u00a0Martin Sleziak Aug 11 '18 at 17:09\n\nLet me try to expand anon's comment to an answer.\n\nLet $R$ be a commutative ring. If $f(x)\\in R[x]$ and $a\\in x$ then there exist uniquely determined $q(x)\\in R[x]$ and $b\\in R$ such that $$f(x)=q(x)(x-a)+b.$$ Moreover, we have $b=f(a)$.\n\nThis follows from the fact that we can divide by monic polynomials in $R[x]$ for any commutative ring $R$. Using this observation we can also show that the map $f(x)\\mapsto f(a)$ is a homomorphism from $R[x]$ to $R$ and the kernel is precisely the ideal $\\langle x-a\\rangle$.\n\nSince we have $\\mathbb C[x,y]=\\mathbb C[x][y]$ we can use the above to get a homomorphism $$\\varphi \\colon \\mathbb C[x,y] \\to \\mathbb C[x] \\qquad \\varphi \\colon f(x,y) \\mapsto f(x,x+1).$$ Kernel of this homomorphism is $\\langle y-x-1 \\rangle$. This shows that $$\\mathbb C[x,y]/\\langle y-x-1 \\rangle \\cong C[y].$$ So the quotient ring is an integrity domain and the ideal is a prime ideal.\n\nSimilarly, we can get, using the above observation, the homomorphisms $$\\psi_1 \\colon \\mathbb C[x][y] \\to \\mathbb C[x] \\qquad \\psi_1 \\colon f(x,y) \\mapsto f(x,3)$$ and $$\\psi_2 \\colon \\mathbb C[x] \\to \\mathbb C \\qquad \\psi_2 \\colon g(x) \\mapsto g(2).$$ By composing these two homomorphisms we get $\\psi=\\psi_2\\circ\\psi_1$ from $\\mathbb C[x,y]$ to $\\mathbb C$ given by $f(x,y)=f(2,3)$.\n\nAgain we have that $\\operatorname{Ker} \\psi_2 = \\langle x-2 \\rangle$.\n\nThe polynomials in $\\operatorname{Ker} \\psi$ are precisely the polynomials such that $\\psi_1(f)\\in \\operatorname{Ker} \\psi_2$. So $f\\in \\operatorname{Ker} \\psi$ if and only if $$f(x,y) = q(x,y)(y-3) + h(x)(x-2).$$ Every such polynomial belongs to $\\langle x-2,y-3 \\rangle$. And since both $x-2$ and $y-3$ belong to the kernel of $\\psi$, we get $$\\operatorname{Ker} \\psi = \\langle x-2,y-3 \\rangle.$$\n\nFrom this we get $\\mathbb C[x,y]/\\langle x-2,y-3 \\rangle\\cong\\mathbb C$. Again, since the quotient ring is an integrity domain, we get that the ideal is prime.", "date": "2020-12-02 17:07:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1756217/the-ideals-langle-y-x-1-rangle-and-langle-x-2-y-3-rangle-in-mathbb-cx-y", "openwebmath_score": 0.9326422214508057, "openwebmath_perplexity": 141.78648559559645, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307700397332, "lm_q2_score": 0.8962513828326955, "lm_q1q2_score": 0.8719009229103069}}
{"url": "http://math.stackexchange.com/questions/298783/math-nomenclature/298790", "text": "# Math nomenclature\n\nI kinda feel ashamed about asking this, but could someone explain me what this means? $$\\binom{k}{i}$$\n\nThat should indicate the number of nodes at a certain depth in a binomial heap..but I can't remember what that actually means and I wasn't successful with Google!\n\nThanks\n\n==================================\n\nso if i had $$\\binom{0}{0}$$ would that be $$\\dfrac{0!}{0!(0-0)!}$$ ...what would that be...zero?\n\n-\nSearch for binomial coefficient and use \\binom{k}{i} to get MathJax to format it properly for display as $\\binom{k}{i}$. \u2013\u00a0 Dilip Sarwate Feb 9 '13 at 15:47\nNo need to feel ashamed, Seb! That's what we're here for...ask away! Your question was good, and showed context, was clear, and it's hard to google for something when you don't know what to call it! \u2013\u00a0 amWhy Feb 9 '13 at 16:07\nthanks for the edits and for the patient :)! \u2013\u00a0 TriRook Feb 9 '13 at 16:07\n@Seb Once you know that $0! = 1$, you can answer your own question about $\\binom00$. In general $\\binom{k}{i}$ is only zero when $i < 0$ or when $i > k$. \u2013\u00a0 Erick Wong Feb 9 '13 at 19:26\n\nI assume you mean $$\\dbinom{n}k$$ $\\dbinom{n}k$ is a short hand for the number of ways in which you can choose $k$ objects from $n$ distinguishable objects. These number are called the binomial coefficients. Some textbooks and articles also denote this as $C(n,k)$.\n\nAs an example, if we have three colored balls, $\\color{red}{\\text{red}}$, $\\color{blue}{\\text{blue}}$ and $\\color{brown}{\\text{brown}}$, then there are three ways of choosing two balls. \\begin{matrix} \\color{red}{\\text{red}} & \\color{blue}{\\text{blue}}\\\\ \\color{blue}{\\text{blue}} & \\color{brown}{\\text{brown}}\\\\ \\color{brown}{\\text{brown}} & \\color{red}{\\text{red}} \\end{matrix} Note that when we say we choose, we are not interested in the order in which these are picked i.e. $\\color{red}{\\text{red}} \\,\\, \\color{blue}{\\text{blue}}$ and $\\color{blue}{\\text{blue}} \\,\\, \\color{red}{\\text{red}}$ refer to the same choice of two balls. Hence, we have $\\dbinom{3}2 = 3$.\n\nSimilarly, if we have $5$ colored balls, say $\\color{red}{\\text{red}}$, $\\color{blue}{\\text{blue}}$, $\\color{brown}{\\text{brown}}$, $\\color{orange}{\\text{orange}}$, and $\\color{lightgreen}{\\text{green}}$, there are $10$ ways of choosing $3$ balls. \\begin{matrix} \\color{red}{\\text{red}} & \\color{blue}{\\text{blue}} & \\color{brown}{\\text{brown}}\\\\ \\color{red}{\\text{red}} & \\color{blue}{\\text{blue}} & \\color{orange}{\\text{orange}}\\\\ \\color{red}{\\text{red}} & \\color{blue}{\\text{blue}} & \\color{lightgreen}{\\text{green}}\\\\ \\color{red}{\\text{red}} & \\color{orange}{\\text{orange}} & \\color{brown}{\\text{brown}}\\\\ \\color{red}{\\text{red}} & \\color{orange}{\\text{orange}} & \\color{lightgreen}{\\text{green}}\\\\ \\color{red}{\\text{red}} & \\color{brown}{\\text{brown}} & \\color{lightgreen}{\\text{green}}\\\\ \\color{blue}{\\text{blue}} & \\color{brown}{\\text{brown}} & \\color{orange}{\\text{orange}}\\\\ \\color{blue}{\\text{blue}} & \\color{brown}{\\text{brown}} & \\color{lightgreen}{\\text{green}}\\\\ \\color{blue}{\\text{blue}} & \\color{orange}{\\text{orange}} & \\color{lightgreen}{\\text{green}}\\\\ \\color{orange}{\\text{orange}} & \\color{brown}{\\text{brown}} & \\color{lightgreen}{\\text{green}} \\end{matrix}\n\nIn general, $$\\dbinom{n}r = \\dfrac{n!}{r!(n-r)!}$$ where $k! = k \\times (k-1) \\times (k-2) \\times \\cdots \\times 2 \\times 1$.\n\nThe name binomial coefficient arises from binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\\dbinom{n}k$ i.e. $$(x+y)^n = \\sum_{k=0}^n \\dbinom{n}k x^k y^{n-k}$$ There are a lot of wonderful properties these binomial coefficients satisfy and I highly recommend you to go through the wiki-page for these properties.\n\n-\nNice answer! Your examples would probably look better as matrixes than as aligns. \u2013\u00a0 Rahul Feb 9 '13 at 16:36\n@\u211d\u207f. Thanks. Yes. These indeed look better. I have updated them. \u2013\u00a0 user17762 Feb 9 '13 at 16:39\nawesome answer thanks for taking time to go through it so well!! I have a question but i'll post it below cause i'm trying to put a formula in.. \u2013\u00a0 TriRook Feb 9 '13 at 17:02\n\n$$\\binom{k}{i}\\; \\text{ is called a \\color{blue}{\\bf{binomial\\;coefficient}}, which is read as \"k choose i\"}$$\n\n\"k choose i\" comes from the fact that if gives you the number of ways to choose $i$ elements from a set of $k$ elements. The term \"binomial coefficient\" makes explicit its relationship to the binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\\large\\binom{n}k$ i.e. $$(x+y)^n = \\sum_{k=0}^n \\dbinom{n}k x^k y^{n-k}$$\n\nYou can compute your binomial coeffient by noting that $$\\displaystyle \\;\\binom{k}{i} = \\frac{k!}{i!(k-i)!} = \\frac{k\\cdot (k-1)\\cdots (k - i + 1)}{i\\cdot (i-1) \\cdots 2 \\cdot 1}$$\n\n-\nI've not in recent times ever heard $\\binom{k}{i}$ read as anything other than \"$k$ choose $i$\". (But maybe I don't get out enough ...) \u2013\u00a0 Peter Smith Feb 9 '13 at 16:12\n@PeterSmith Good point; \"often\" should have been \"usually\", which should probably be omitted altogether! Edited accordingly. \u2013\u00a0 amWhy Feb 9 '13 at 16:16\n\n$k \\choose i$ is a binomial coefficient. It is the coefficient of $x^i$ in expanding $(1+x)^k$. It is also the number of $i$-element subsets of a $k$-element set. It can be computed via $\\frac{k!}{i!(k-i)!}$ or with several recursive formulas. Values are often listed with help of Pascal's triangle.\n\n-\n\nFor the question in the edit: We have $0! = 1$, so $\\binom{0}{0} = 1$ which is the first entry in Pascal's triangle.\n\nA previous question on this site asked about the definition of zero factorial; there are several excellent answers there. As Zhen Lin points out in a comment, the best explanation of why $0! = 1$ depends on how you define the factorial.\n\n-", "date": "2015-08-29 13:21:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/298783/math-nomenclature/298790", "openwebmath_score": 0.8667019605636597, "openwebmath_perplexity": 596.6473114939948, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307661011976, "lm_q2_score": 0.8962513641273355, "lm_q1q2_score": 0.8719009011832392}}
{"url": "https://mathematica.stackexchange.com/questions/264386/logical-functions-with-no-arguments", "text": "# Logical functions with no arguments\n\nI am trying to understand how logical functions are evaluated if no arguments are passed to them.\n\n{Or[], Nor[], And[], Nand[], Xor[], Xnor[]}\n\n\n{False, True, True, False, False, True}\n\nI tried Trace but it didn't offer many clues. Mathematica (12.2.0 on Win7-x64) evaluates these expressions without any messages.\n\n\u2022 If you have a look in the Properties and relations section you will see that it zero-argument evaluates to False. reference.wolfram.com/language/ref/Or.html\n\u2013\u00a0kcr\nFeb 28, 2022 at 8:39\n\u2022 Similar situation in Java developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/\u2026\n\u2013\u00a0kcr\nFeb 28, 2022 at 8:39\n\u2022 Perhaps this is the way they are supposed to be constructed based on theoretical coding principles? Sorry for the triple comment...I am horrible\n\u2013\u00a0kcr\nFeb 28, 2022 at 8:40\n\u2022 These are simply the trivial cases for these functions. And, for example, is True unless one of it's arguments is False. The zero-argument case is just an extension of that. Feb 28, 2022 at 11:05\n\u2022 One sees similar behavior with Plus and Times (where it might be more obviously plausible). Symbols with the OneIdentity attribute will, when given an empty argument list, evaluate to their respective identity elements. Feb 28, 2022 at 15:57\n\nI am pretty sure that Syed figured out the question by now, but for future visitors I thought it a good idea to write something.\n\n1. Why does Or[] return False?\n\nFrom the docs we read\n\nIt evaluates its arguments in order, giving True immediately if any of them are True, and False if they are all False.\n\nwhich is another way of saying that it returns True if any or all of its arguments is true and False otherwise.\n\n1. Why does Nor[] return True?\n\nThe easiest way to realize this -I think- is again to have a look at the documentation where we find that it is equivalent to Not[Or[]]. So, Not[False] is True.\n\n1. Why does And[] return True?\n\nWell, again this can be seen directly from the docs and also from the comment by @Sjoerd Smit.\n\nIt evaluates to False if any of its arguments are false, and to True otherwise.\n\n1. Why does Nand[] return False?\n\nSimilar to the explanation of Nor in this case. Nand[] is equivalent to Not[And[]] and hence Not[True] which is False.\n\n1. Why does Xor[] return False?\n\nWell, it has to evaluate all of its argument and yields True if an odd number of them is True and False otherwise. Since, we do not have an odd number of True statements, it returns False\n\n1. Why does Xnor[] return True?\n\nXnor[] is equivalent to Not[Xor[]]\n\n\u2022 Thanks for the write up.\n\u2013\u00a0Syed\nJan 14 at 12:56\n\u2022 @Syed if you want to add more to it, let me know and I will turn it into a wiki :-)\n\u2013\u00a0bmf\nJan 14 at 12:58\n\u2022 It seems that in your description of Nor True and False should be reversed. Jan 14 at 19:26\n\u2022 @yarchik yes, of course you are right. thanks for spotting this :-)\n\u2013\u00a0bmf\nJan 15 at 1:38\n\nThe answer of @bmf is perfect, but I would like to bring some arguments that allow us to better remember the results. It also shows a connection to other MA functions.\n\nThe Xor function can be seen as Sum modulo 2. In fact\n\nSum[1, {i, 0}]\n(* 0 *)\n\n\nwhich explains why Xor[] == False.\n\nThe And function can be regarded as the Product modulo 2. In fact\n\nProduct[1, {i, 0}]\n(* 1 *)\n\n\nwhich explains why And[] == True.\n\nSee also a comment of @DanielLichtblau above, which argues in terms of Plus and Times.\n\n\u2022 (+1) that's a very neat description!!!\n\u2013\u00a0bmf\nJan 15 at 1:39", "date": "2023-03-20 18:50:53", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/264386/logical-functions-with-no-arguments", "openwebmath_score": 0.27232179045677185, "openwebmath_perplexity": 1192.1362965851615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799399736477, "lm_q2_score": 0.9046505421702797, "lm_q1q2_score": 0.8718840452300001}}
{"url": "https://math.stackexchange.com/questions/3528827/is-the-following-generalization-of-strong-law-of-large-numbers-valid", "text": "# Is the following generalization of Strong Law of Large Numbers valid?\n\nAccording to SLLN, if $$X_1, X_2, \\ldots$$ is an infinite sequence of i.i.d. random variables with expected value $$\\mu$$ and $$S_n := \\sum_{i=1}^n X_i/n$$ then $$S_n \\to \\mu$$ almost surely.\n\nIf the sequence is instead $$X_{1,1}, X_{2,1}, X_{2,2}, \\ldots, X_{n,1}, \\ldots,X_{n,n}, X_{n+1,1}, \\ldots$$ whose terms are i.i.d. random variables with expected value $$\\mu$$ and $$S_n := \\sum_{i=1}^n X_{n,i}/n$$, can we still say $$S_n \\to \\mu$$ almost surely? If not are there any extra conditions that make this true?\n\nI will give an example for why this is not trivial. Consider distribution of variables is Bernoulli with $$p=1/2$$ and the sample $$0, 1, 1, 0, 0, 0, 1, \\ldots$$. That is $$(2n-1) \\times 0$$ are followed by $$2n \\times 1$$ and that is repeated for every $$n > 0$$. For this sequence $$S_n$$ converges to $$1/2$$ in the first case and the limit does not exist in the second. If the probability of all such examples is $$0$$ then convergence will be almost sure, but this is something that can probably be proven on a case by case with union bound inequality. I just wonder whether there is some general result that makes such analysis easier.\n\nFirst observe that in this setting, the sequence $$\\left(S_n\\right)_{n\\geqslant 1}$$ is independent. For an independent sequence, in view of the Borel-Cantelli lemma, almost sure convergence and complete convergence are equivalent. Hence $$S_n\\to \\mu$$ almost surely if and only if for all positive $$\\varepsilon$$, $$\\tag{*} \\sum_{n\\geqslant 1}\\mathbb P\\left(\\lvert S_n-\\mu\\rvert \\gt\\varepsilon\\right)<+\\infty.$$ Let $$(Y_i)_{i\\geqslant 1}$$ be an i.i.d. sequence such that $$Y_1$$ has the same law as $$X_{1,1}$$. Then $$(*)$$ is equivalent to $$\\forall \\varepsilon>0, \\sum_{n\\geqslant 1}\\mathbb P\\left(\\left\\lvert \\sum_{j=1}^n(Y_j-\\mu)\\right\\rvert \\gt n\\varepsilon\\right)<+\\infty.$$ By Theorem 3 in this paper by Baum and Katz, this is equivalent to $$\\mathbb E[Y_1^2]<\\infty$$, hence we do need extra conditions.\nYes, the result holds equally well in both cases. More generally, Let $$(X_i:i=1,2,...)$$ be any infinite sequence of i.i.d. random variables with expected value $$\u03bc$$, and let $$F: \\Bbb Z_+\\to\\mathrm P(\\Bbb Z_+)$$ such that $$|F(k)|\\in\\Bbb Z_+$$ with $$|F(k)|\\to\\infty$$ as $$k\\to\\infty$$. Then$$\\frac1{|F(n)|}\\sum_{i\\in F(n)}X_i\\;\\to_{\\text{a.s.}}\\;\\mu$$as $$n\\to\\infty$$. The proof is essentially the same as that for the simple average of an initial segment of the original sequence, because the i.i.d. property of any (finite or infinite) subsequence of the original sequence $$(X_i:i=1,2,...)$$ follows from that of the original sequence.", "date": "2021-10-26 06:39:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3528827/is-the-following-generalization-of-strong-law-of-large-numbers-valid", "openwebmath_score": 0.941399097442627, "openwebmath_perplexity": 55.22918331326315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9898303410461384, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8718396798303676}}
{"url": "http://math.stackexchange.com/questions/1671428/subset-of-binary-space-countable", "text": "subset of binary space countable?\n\nI know that the set of all binary sequences is uncountable. Now consider the subset of this set, that whenever a digit is $1$, its next digit must be $0$. Is this set countable?\n\nI think it is not because it is like \"half\" of the set of binary sequences. By that I mean if I just delete the sequences with a $1$ followed by a $1$, then I'm left with the required sequence. But I'm not sure how to show this clearly. Can someone think of a bijection that would prove this? Or am I wrong? Please help.\n\nThanks\n\n-\n\nIf you insist to have a bijection, here is one. Your set $S$ is the set of all binary infinite words containing no factor $11$. Any such word can be uniquely written as an infinite word on the alphabet $\\{0, 10\\}$, that is, $S = \\{0, 10\\}^\\omega$. For instance, $$0100100010010(100)^\\omega = \\color{blue}{0}\\color{red}{10}\\color{blue}{0}\\color{red}{10}\\color{blue}{0}\\color{blue}{0}\\color{red}{10}0\\color{red}{10}(\\color{red}{10}\\color{blue}{0})^\\omega$$\n\nIt follows that the map from $\\{0,1\\}^\\omega \\to S$ consisting to replace every $1$ by $\\color{red}{10}$ is a bijection. No need of the Cantor-Bernstein-Schroeder theorem.\n\n-\n\nYou can explicitly write an injective function from the set of all binary sequences to the set you are considering: just map every sequence to one with $0$ inserted in every other position.\n\nSo no, this is definitely not countable.\n\n-\nBut how does an injective function suffice here? Don't you need a bijective function to show the same cardinality? \u2013\u00a0verticese Feb 25 at 8:02\n@verticese: it shows that it has at least the same cardinality as the bijective sequences (that is, the continuum), which is all you need to show that it is uncountable. But it is not hard to get a bijection if you want: of course you also have an injection going the other way: the identity. This gives you a bijection by Cantor-Bernstein-Schroeder. In this case, you can probably define a bijection explictly without much difficulty, although there would be no nice and concise formula, I think, but rather a procedure. \u2013\u00a0tomasz Feb 25 at 8:05\nAs @tomasz notes: If $A\\subseteq B$ then $|A|\\le|B|$. If there's an injection $B\\to A$, then $|B|\\le|A|$, so by Cantor-Schroeder-Bernstein there's a bijection $A\\to B$. \u2013\u00a0BrianO Feb 25 at 8:12\n\nYour set has an uncountable subset: consider the infinite sequences of $a$ and $b$ where $a=10$ and $b=00$, the set of these sequences has the same cardinality of the set of sequences of $0$ and $1$ (just map $a \\mapsto 0$ and $b \\mapsto 1$), and it is is clearly a subset of your set.\n\n-\nThis is the same as what I suggested in my answer. \u2013\u00a0tomasz Feb 25 at 9:34\n@tomasz I only see that now after reading your comment. I would say this answer is another viewpoint on the same thing you suggested, and sometimes an alternate viewpoint helps (as in how this one helped me understand your answer better). \u2013\u00a0Todd Wilcox Feb 25 at 16:53\n\nHere's another (possibly) interesting way of proving uncountability, since you tagged this as a topology question. You can inject your set into $[0,1]$ under the evaluation map\n\n$$(x_i) \\mapsto \\sum_{i = 1}^\\infty \\frac{x_i}{2^i}$$\n\nThis is injective on your set (but not the set of all binary sequences, since $0.0111\\dots = 0.100\\dots$). We shall show the image of this map is a perfect set, and hence uncountable (see Rudin, or some other source on metric spaces). Surely no point in the set is isolated, because if we have an expansion $x = 0.a_0 a_1 \\dots a_2$, and if this expansion does not eventually end in all zeroes, we may truncate the expansion to find another number in the image which is as close to this number as possible. If the number does end in all zeroes, you may add a one to an arbitrary position to get another number as closed as possible. Now suppose $\\lim x_i = x$, where each $x_i$ is in the image of the evaluation. We must have $0 < x < 1$, for if $x = 1$, the $x_i$ must eventually be of the form $0.11\\dots$, which is impossible. Write $x = 0.a_0 a_1 \\dots$. Suppose $a_k = 1$. If we force $|x_i - x| < 1/2^{k+2}$, then the first $k+2$ digits of $x_i$ must be the same as the first $k+1$ digits of $x$, hence $a_{k+1} = 0$. Thus the image is perfect, and the set is uncountable.\n\nIntuition tells me this set is probably fractal, similar to the Cantor set, but I can't visualize it very well.\n\n-\n\nYou can inject the set of all sequences of integers into this your set (call it $A$): $$(n_i)_{i<\\omega} \\mapsto (1 + 0^{n_0+1} + 1 + 0^{n_1+1} + \\dotsc)\\colon \\omega^\\omega\\to A.$$ where the value is meant as an infinite regular expression. That is, the $i$-th $1$ is followed by $n_i+1$ many $0$s. (The \"$+ 1$\" guarantees that the value under this function is a member of A.)\n\nThus, $A$ has cardinality $\\lvert A\\rvert \\ge |\\omega^\\omega| = \\lvert\\Bbb R\\rvert$. But $A\\subseteq 2^\\omega$, so $\\lvert A\\rvert \\le \\lvert 2^\\omega\\rvert = \\lvert\\Bbb R\\rvert$. By the Cantor-Schroeder-Bernstein Theorem, $\\lvert A\\rvert = \\lvert\\Bbb R\\rvert$.\n\n-\n\nNotation: $\\chi_A (x)=1$ if $x\\in A$, and $\\chi_A (x)=0$ if $x\\not \\in A.$\n\nLet $E=\\{2 n:n\\in N\\}.$ For $S\\subset E \\;$ let $S^*=\\{n :2 n\\in S\\}$ and let $f(S)=(\\chi_{S^*} (n))_{n\\in N}.$\n\nThen $f$ is a bijection from the power-set $P(E)$ to the set $B$ of all binary sequences.\n\nFor $S\\subset E$ let $g(S)=(\\chi_S (n))_{n\\in N}.$ Then $g$ is an injection from $P(E)$ into the set $G$ of binary sequences $(b_n)_{n\\in N}$ that satisfy $\\forall n\\in N\\;(b_n=1\\implies b_{n+1}=0)\\}.$\n\nThus $g f^{-1}$ is an injection from $B$ into $G.$ So $G$ is uncountable because it has the uncountable subset $\\{g f^{-1}(x): x\\in B\\}.$\n\nThe Cantor-Schroeder-Bernstein Theorem implies that for any $B,G$ with $G\\subset B,$ there is a bijection $h:B\\to G$ if there is an injection $i:B\\to G.$\n\n-", "date": "2016-06-26 15:56:19", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/1671428/subset-of-binary-space-countable", "openwebmath_score": 0.9511620998382568, "openwebmath_perplexity": 135.19430140655842, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303422461274, "lm_q2_score": 0.8807970795424087, "lm_q1q2_score": 0.871839674692852}}
{"url": "https://math.stackexchange.com/questions/2126169/variance-of-number-of-heads-in-three-flips-of-a-coin-selected-from-a-set-of-fair", "text": "# Variance of number of heads in three flips of a coin selected from a set of fair and unfair coins\n\nWe have $10$ coins, $2$ are two-tailed, $2$ are two-headed, the other $6$ are fair ones. We (randomly) pick a coin and we flip it $3$ times. Find the variance of the number of gotten heads.\n\nMy attempt:\n\n$X$ - number of heads that we got\n\n$\\mathbb{P}\\left(X=0\\right)=\\frac{2}{10}\\cdot1\\cdot1\\cdot1 + \\frac{6}{10}\\cdot\\frac{1}{2}\\cdot\\frac{1}{2}\\cdot\\frac{1}{2}$ - we picked two-tailed coin or fair one\n\n$\\mathbb{P}\\left(X=1\\right)=\\frac{6}{10}\\cdot\\frac{1}{2}\\cdot\\frac{1}{2}\\cdot\\frac{1}{2}\\cdot 3$ - we picked fair one and we have 3 possibilites: HHT, HTH, THH\n\n$\\mathbb{P}\\left(X=2\\right)=\\frac{6}{10}\\cdot{3\\choose2}\\cdot\\left(\\frac{1}{2}\\right)^2\\cdot\\left(\\frac{1}{2}\\right)^1$ - again we picked fair one and we have to have 2 successes in 3 tries\n\n$\\mathbb{P}\\left(X=3\\right)=\\frac{2}{10}\\cdot1\\cdot1\\cdot1+\\frac{6}{10}\\cdot\\left(\\frac{1}{2}\\right)^3$ - we can pick two-headed or fair one coin\n\nAnd the rest is simple,\n\n$\\text{Var}X=\\sum_{i=0}^3i^2\\cdot\\mathbb{P}\\left(X=i\\right)-\\left(\\sum_{i=0}^3i\\cdot\\mathbb{P}\\left(X=i\\right)\\right)^2$\n\nIs my solution correct?\n\n\u2022 The information is symmetric, so the expressions for the probability of 1 head or 2 heads should be the same. \u2013\u00a0Paul Feb 2 '17 at 18:42\n\u2022 Yes, your approach is correct. $\\checkmark$ \u2013\u00a0callculus Feb 2 '17 at 18:47\n\u2022 @Paul well, yea, I should have decided whether I use probability mass function or just standard combinatorics. But the result is the same \u2013\u00a0SantaXL Feb 2 '17 at 18:50\n\nThat's one approach. \u00a0 It is good. \u00a0 Here's another,\n\nWe know $X\\mid C \\sim \\mathcal{Bin}(3, C)$ and $\\mathsf P(C=c)=\\tfrac 15\\mathbf 1_{c=0}+\\tfrac 35\\mathbf 1_{c=1/2}+\\tfrac 15\\mathbf 1_{c=1}$\n\nSo $\\mathsf E(C)= \\tfrac 1{2}$ and $\\mathsf {Var}(C)=\\tfrac 1{10}$ and $\\mathsf E(C^2)=\\frac 7{20}$\n\nBy the Law of Total Probability: $\\mathsf E(X) \\\\ = \\mathsf E(\\mathsf E(X\\mid C)) \\\\ = \\mathsf E(3C) \\\\ = \\frac 32$\n\nThen by the Law of Total Variance: $\\quad\\mathsf {Var}(X) \\\\ = \\mathsf{Var}(\\mathsf E(X\\mid C))+\\mathsf E(\\mathsf{Var}(X\\mid C)) \\\\ = \\mathsf {Var}(3C)+\\mathsf E(3C(1-C)) \\\\ = \\frac 9{10}+\\frac 32-\\frac {21}{20} \\\\ = \\frac {27}{20}$\n\n\u2022 Hmm, your result is different from mine. In my approach: $\\mathbb{P}(X=0)=\\mathbb{P}(X=3)=\\frac{9}{40}, \\ \\mathbb{P}(X=1)=\\mathbb{P}(X=2)=\\frac{11}{40} \\\\ EX=1\\cdot\\frac{9}{40}+2\\cdot\\frac{9}{40}+3\\cdot\\frac{11}{40}=\\frac{9+18+33}{40}=\\frac{60}{40}=\\frac{3}{2} \\\\EX^2=1^2\\cdot\\frac{9}{40}+2^2\\cdot\\frac{9}{40}+3^2\\cdot\\frac{11}{40}=\\frac{9+36+99}{40}=\\frac{144}{40}=\\frac{18}{5} \\\\ VarX=EX^2-E^2X=\\frac{18}{5}-\\frac{9}{4}=\\frac{27}{20}$ \u2013\u00a0SantaXL Feb 3 '17 at 13:52\n\u2022 Where is the mistake? \u2013\u00a0SantaXL Feb 3 '17 at 13:53\n\u2022 In mine. @SantaXL corrected. \u2013\u00a0Graham Kemp Feb 3 '17 at 14:49\n\u2022 ok, thanks for showing me this approach, I wasn't aware such Law exists. \u2013\u00a0SantaXL Feb 3 '17 at 15:22", "date": "2020-04-01 21:52:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2126169/variance-of-number-of-heads-in-three-flips-of-a-coin-selected-from-a-set-of-fair", "openwebmath_score": 0.6153326034545898, "openwebmath_perplexity": 391.30005292859676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303398461495, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.8718396617386487}}
{"url": "https://math.stackexchange.com/questions/2560998/proving-24-mid52n-1-using-modular-arithmetic/2561002", "text": "# Proving $24\\mid5^{2n}-1$ using modular arithmetic\n\nSo, in general I am aware of how to use modular arithmetic to prove a divisibility. But I have the following problem:\n\nProve that $24\\mid5^{2n}-1$ for all $n\\in\\mathbb Z$.\n\nI know that theoretically, I could show 23 different cases that the expression is congruent to $0\\bmod24$, but that seems like it might be excessive to me.\n\nIs there a faster way to show this?\n\n\u2022 $5^{2n}=(5^2)^n=25^n$ Now, note that $25^n=(24+1)^n$ and use something that should be familiar to you. Dec 11, 2017 at 3:01\n\nYou want to prove what $5^{2n}$ is congruent to in modulo 24\n\nNotice that $5^{2n} = (5^2)^n$\n\n$5^2=25 \\equiv 1 \\pmod {24}$\n\nSo $5^{2n} \\equiv 1^n \\equiv 1 \\pmod {24}$\n\n\u2022 Edited, and thanks Dec 11, 2017 at 3:05\n\u2022 Makes sense, thank you! Dec 11, 2017 at 3:19\n\nYou could use $5^{2n}-1=(5^2-1)(5^{2n-2}+5^{2n-4}+\\ldots +1)$ where you are summing the geometric series.\n\nUse induction. It is true for $n=1$. Assuming it is true for $n$, show it is true for $n+1$: $$5^{2(n+1)}-1=25(5^{2n}-1)+24 = 0 (mod \\ 24).$$\n\n1)\n\nProve $8|5^{2n} - 1$ and $3|5^{2n} - 1$\n\na) $5^{2n} - 1 = (5^n -1)(5^n + 1)$. $5^n \\pm 1 \\equiv 1^n\\pm 1 \\equiv 0\\mod 2$ so $2|5^2 - 1$ and $2|5^2 + 1$ and so $4|5^{2n}-1$. If $5^n - 1 = 2k$ then $5^{2n} + 1= 2k + 2 = 2(k+1)$. Either $k$ is even or $k+1$ is even so either $2k$ is even and $2(k+1)$ is divisible by $4$, or $2k$ is divisible by $4$ and $2(k+1)$ is even. Either way. $2k*2(k+1)= 5^{2n} -1$ is divisible by $8$.\n\nb)$5^{2n}-1\\equiv (-1)^{2n}-1 \\equiv 1^n - 1 \\equiv 0 \\mod 3$.\n\nSo $3|5^{2n}-1$.\n\n2)\n\nOr notice $5^{2n} -1 = 25^n -1 \\equiv 1^n -1 \\equiv 0 \\mod 24$.\n\n(I will admit, I did not see the obvious right away.)", "date": "2022-06-30 04:15:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2560998/proving-24-mid52n-1-using-modular-arithmetic/2561002", "openwebmath_score": 0.8939849734306335, "openwebmath_perplexity": 121.70112325458685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513891739276, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8717965702974327}}
{"url": "https://math.stackexchange.com/questions/2085268/does-i4-equal-1/2085325", "text": "# Does $i^4$ equal $1?$\n\nI can't seem to find a solution to this for the life of me. My mathematics teacher didn't know either.\n\nEdit: I asked the teacher that usually teaches my course today, and she said it was incredible that the other teacher didn't know.\n\nMy logic goes as follows:\n\nany real number: $x$ to the fourth power is equal to $(x^2)^2$. Using this logic, $i^4$ would be equal to $(i^2)^2$. This would result in $(-1)^2$, and $(-1)^2 = 1$.\n\nObviously, this logic can be applied to any real numbers, but does it also apply to complex numbers?\n\n\u2022 Yes, $i^4=1$. But what else could it equal ??\n\u2013\u00a0user65203\nJan 7 '17 at 15:00\n\u2022 To check if $i^4$ is real or not, why not try conjugating it and see what happens? Jan 7 '17 at 15:11\n\u2022 Yes, @J.M.isn'tamathematician, but before you proceed to prove something, I'm going to get some paper and work it out for myself Jan 7 '17 at 15:21\n\u2022 \"I'm going to get some paper and work it out for myself\" - that was my intention behind asking you those questions. Jan 7 '17 at 15:24\n\u2022 If you're working within the naive approach to complex numbers often used when they're first introduced the students, \"Suppose $i^2=-1$ but that other than that all the usual rules of arithmetic apply\", then basically the bottom line is that this is one of the \"usual rules\" which is simply assumed to apply. Specifically, we assume that complex numbers are associative, that is that for any $a, b, c$, $a(bc)=(ab)c$. You might try showing in detail why this assumption implies $i^4=1$. Jan 12 '17 at 21:03\n\nYes. The powers of $i$ are cyclic, repeating themselves ever time the exponent increases by 4: $$i^0 = 1$$ $$i^1=i$$ $$i^2 = -1$$ $$i^3 = -i$$ $$i^4 = 1$$ $$i^5 = i$$ $$i^6 = -1$$ $$i^7 = -i$$ $$i^8 = 1$$ etc.\n\nYour reasoning is excellent, and you should feel good about the fact that you figured this out on your own. The fact that your math teacher didn't know this is, in my professional opinion as a mathematics educator, a disgrace.\n\nEdited to add: As Kamil Maciorowski notes in the comments, the pattern persists for negative exponents, as well. Specifically, $$i^{-1}= \\frac{1}{i} = -i$$ If $\\frac{1}{i}=-i$ seems odd, notice that $i(-i) = -i^2 = -(-1) = 1$, so $i$ and $-i$ are multiplicative inverses; therefore $i^{-1} = -i$. Once you know that, you can extend the pattern: $$i^{-1} = -i$$ $$i^{-2} = -1$$ $$i^{-3} = i$$ $$i^{-4} = 1$$ and so on.\n\nSecond update: The OP asks for some additional discussion of the property $\\left( x^a \\right)^b = x^{ab}$, so here is some background on that:\n\nFirst, if $a$ and $b$ are natural numbers, then exponentiation is most naturally understood in terms of repeated multiplication. In this context, $x^a$ means $(x\\cdot x\\cdot \\cdots \\cdot x)$ (with $a$ factors of $x$ appearing), and $\\left( x^a \\right)^b$ means $(x\\cdot x\\cdot \\cdots \\cdot x)\\cdot(x\\cdot x\\cdot \\cdots \\cdot x)\\cdot \\cdots \\cdot (x\\cdot x\\cdot \\cdots \\cdot x)$, with $b$ sets of parentheses, each containing $a$ factors of $x$. Since multiplication is associative, we can drop the parentheses and recognize this as a product of $ab$ factors of $x$, i.e. $x^{ab}$.\n\nNote that this reasoning works for any $x$, whether it is positive, negative, or complex. It even applies in settings were multiplication is noncommutative, like matrix multiplication or quaternions. All we need is that multiplication is associative, and that $a$ and $b$ be natural numbers.\n\nOnce we have established that $\\left( x^a \\right)^b = x^{ab}$ for natural numbers $a,b$ we can extend the logic to integer exponents. If $a$ is a positive number, and if $x$ has a multiplicative inverse, then we define $x^{-a}$ to mean the same thing as $\\left(\\frac1x\\right)^a$, or (equivalently) as $\\frac1{x^a}$. With this convention in place, it is straightforward to verify that for any combination of signs for $a,b$, the formula $\\left(x^a\\right)^b = x^{ab}$ holds.\n\nNote however that in extending the formula to cover a larger set of exponents, we have also made it necessary to restrict the domain of values $x$ over which this property holds. If $a$ and $b$ are just natural numbers then $x$ can be almost any object in any set over which an associative multiplication is defined. But if we want to allow $a$ and $b$ to be integers then we have to restrict the formula to the case where $x$ is an invertible element. In particular, the formula $x^{a}$ is not really well-defined if $x=0$ and $a$ is negative.\n\nNow let's consider the case where the exponents are not just integers but arbitrary rational numbers. We begin by defining $x^{1/a}$ to mean $\\sqrt[a]{x}$. ( See Why does $x^{\\frac{1}{a}} = \\sqrt[a]{x}$? for a short explanation of why this convention makes sense.)\n\nIn this definition, we are assuming that $a$ is a natural number, and that $x$ is positive. Why do we need $x$ to be positive? Well, consider an expression like $x^{1/2}$. If $x$ is positive, this is (by convention) defined to be the positive square root of $x$. But if $x$ is negative, then $x^{1/2}$ is not a real number, and even if we extend our number system to include complex numbers, it is not completely clear which of the two complex square roots of $x$ this should be identified with. More or less the same problem arises when you try to extend the property to complex $x$: while nonzero complex numbers do have square roots (and $n$th roots in general), there is no way to choose a \"principal\" $n$th root.\n\nThings get really crazy when you try to extend the property $\\left(x^a\\right)^b=x^{ab}$ to irrational exponents. If $x$ is a positive real number and $a$ is a real number, we can re-define the expression $x^a$ to mean $e^{a\\ln x}$, and it can be proved that this re-definition produces the same results as all of the conventions above, but it only works because $\\ln x$ is well-defined for positive $x$. As soon as you try to allow negative $x$, you run into trouble, since $\\ln x$ isn't well-defined in that case. One can define logarithms of negative and complex numbers, but they are not single-valued, and there are all kinds of technicalities about choosing a \"branch\" of the logarithm function.\n\nIn particular -- and this is very important for the question at hand -- the identity $\\left(x^a\\right)^b=x^{ab}$ does not hold in general if $x$ is not a positive real number or if $a,b$ are not both integers. A lot of people misunderstand this, and indeed there are many, many, many, many questions on this site that are rooted in this misunderstanding.\n\nBut with respect to the question in the OP: It is perfectly reasonable to argue that $i^4 = \\left(i^2 \\right)^2$, because even though $i$ is a complex number, the exponents are integers, so the basic notion of exponentiation as repeated multiplication is reliable.\n\n\u2022 Can you add please the values for $i^9, i^{10}, i^{11}, i^{12}$ ? Jan 5 '17 at 20:54\n\u2022 @Hexacoordinate-C I can't tell if you're joking or not. Jan 5 '17 at 20:57\n\u2022 Note that this post uses the field properties of the complex numbers (namely that the nonzero complex numbers form a group under multiplication). See math.stackexchange.com/a/2085631 for the answer to the last part of the question about trying to apply logic that works for real numbers to complex numbers. Jan 6 '17 at 3:48\n\u2022 @user21820 Nice comment, +1\n\u2013\u00a0user284001\nJan 6 '17 at 12:50\n\u2022 You might also want to explain why the powers are cyclic because it wasn't instantly obvious why to me. Jan 6 '17 at 13:02\n\nI'm surprised that none of the other answers pointed out the most important point in your question:\n\nObviously, this logic can be applied to any real numbers, but does it also apply to complex numbers?\n\nThis attitude is the right way to go. The logic you speak of is more precisely:\n\n$x^{ab} = (x^a)^b$ for any real number $x$ and natural numbers $a,b$.\n\nIf you want the more general fact for integer exponents:\n\n$x^{ab} = (x^a)^b$ for any real number $x \\ne 0$ and integers $a,b$.\n\nIn fact it turns out that 'miraculously' we have an even more general fact for real exponents:\n\n$x^{ab} = (x^a)^b$ for any real number $x > 0$ and reals $a,b$.\n\nNotice that all these precise statements about real exponentiation show you clearly that you must know exactly what the objects are before you can apply any operations to them, not to say claim any properties about the resulting values.\n\nFor this reason it is actually an important question to ask whether there are corresponding rules for complex numbers.\n\nYes, but not as nice.\n\n$x^{ab} = (x^a)^b$ for any complex number $x \\ne 0$ and integers $a,b$. (*)\n\nHere exponentiation is simply the result of starting from $1$ and repeatedly multiplying/dividing by $x$ where the number of times is specified by the exponent (multiplying for positive; dividing for negative). This fact holds in any structure that has invertible multiplication, including the field of rationals, the field of reals, and the field of complex numbers.\n\n$x^{ab},x^a$ are well-defined since $x \\ne 0$.\n\nHowever, in general \"$x^{ab} = (x^a)^b$\" does not hold for complex $x$ even if $a,b$ are both rational. For instance (according to standard conventions):\n\n$i = (-1)^{1/2} = (-1)^{(2 \\times 1/4)} \\ne ((-1)^2)^{1/4} = 1^{1/4} = 1$.\n\nSo it's excellent that you ask whether some new structure (complex numbers) have the same properties as some other structure (real numbers) instead of just blindly assuming it does.\n\nThe question was recently edited to ask for including an explanation of (*). Actually, there is nothing much to explain intuitively, since it boils down to the fact that an $ab$-fold repetition of an operation is the same as a $b$-fold repetition of an $a$-fold repetition of that operation. One can either stop there, but if one wants to ask why then one would need to fix a foundational system first, and in particular the rules concerning integers and induction/recursion. The below proof will use associativity of integer addition and multiplication, and distributivity of multiplication over addition for integers, which correspond to basic facts about repetition.\n\nSuppose we have a field $S$ (such as the complex numbers) and an exponentiation operation that satisfies the following: $\\def\\lfrac#1#2{{\\large\\frac{#1}{#2}}}$\n\n$x^0 = 1$ for every $x \\in S$.\n\n$x^{k+1} = x^k x$ for every $x \\in S$ and integer $k$.\n\nNote that any reasonable foundational system is capable of defining such an operation recursively (you need one direction for positive $k$ and another for negative $k$), and can easily prove by induction the following two theorems.\n\n$x^{a+b} = x^a x^b$ for every nonzero $x \\in S$ and integers $a,b$.\n\nTake any nonzero $x \\in S$ and integer $a$.\n\nThen $x^{a+0} = x^a = x^a x^0$.\n\nGiven any integer $b$ such that $x^{a+b} = x^a x^b$:\n\n$x^{a+(b+1)} = x^{(a+b)+1} = x^{a+b} x = ( x^a x^b ) x = x^a ( x^b x ) = x^a x^{b+1}$.\n\n$x^{a+(b-1)} = x^{(a+b)-1} = x^{a+b} \\div x = ( x^a x^b ) \\div x = x^a ( x^b \\div x ) = x^a x^{b-1}$.\n\nTherefore by induction $x^{a+b} = x^a x^b$ for every integer $b$.\n\n$x^{ab} = (x^a)^b$ for every nonzero $x \\in S$ and integers $a,b$.\n\nTake any nonzero $x \\in S$ and integer $a$.\n\nThen $x^{a \\times 0} = x^0 = 1 = (x^a)^0$.\n\nGiven any integer $b$ such that $x^{ab} = (x^a)^b$:\n\n$x^{a(b+1)} = x^{ab+a} = x^{ab} x^a = (x^a)^b (x^a) = (x^a)^{b+1}$.\n\n$x^{a(b-1)} = x^{ab-a} = x^{ab} \\div x^a = (x^a)^b \\div (x^a) = (x^a)^{b-1}$.\n\nTherefore by induction $x^{ab} = (x^a)^b$ for every integer $b$.\n\nNotice that we did not use commutativity here, which in fact shows that the argument holds in any division ring. If you restrict the exponents to natural numbers, then it clearly holds in any group when \"nonzero\" is deleted.\n\nFinally, there are some nice properties that arise from the above properties such as:\n\n$i^{k+4} = i^k i^4 = i^k (i^2)^2 = i^k (-1)^2 = i^k$ for any integer $k$.\n\nIn short, powers of $i$ (a square-root of $-1$ in the complex field) are cyclic.\n\n\u2022 You give a very important message. All too often, a \u201crule\u201d is taught as something universally true, without any mention of the domain of applicability. Jan 6 '17 at 4:06\n\u2022 @Lubin: Thank you! This is one of the things that makes teaching type checking (borrowed from CS) an extremely attractive notion to me, since it forces students to know what they are doing rather than mimicking their teacher and hoping what they write on paper gets the marks. =) Jan 6 '17 at 4:12\n\u2022 The answer is misleading: roots are the inverse of exponents so $i^4 = 1 = 1^{\\frac{1}{4}}$. The problem is that the root operation is not one-to-one and this answer only took the principal root of $i^{\\frac{1}{4}}$. It has 4 roots: $1, -1, i, -i$ Jan 7 '17 at 20:39\n\u2022 @brent.payne: It is not misleading. That is the usual definition of the complex exponentiation operation. The point in my post is very clear; some properties of real exponentiation do not carry over to complex exponentiation. If you want a relation instead of a function, you cannot use the exponentiation notation. Also , even if you attempt to define exponentiation to be on sets of complex numbers, this same property will still fail; $\\{1\\} = \\{1\\}^{2 \\times \\frac12} \\ne (\\{1\\}^2)^{\\frac12} = \\{-1,1\\}$. You cannot anyhow choose a root you like. Jan 8 '17 at 3:34\n\u2022 The answer is very good except for the $i = (\u22121) ^ {1/2}$ and $1 ^ {1/4} = 1$ parts. This is not a standard convention. You own comment above says \"You cannot anyhow choose a root you like.\", yet you have chosen roots twice. Jan 13 '17 at 9:05\n\nGeometrically, multiplication by $i$ does the following to a complex number:\n\n\u2022 Scales length by a factor of $1$\n\u2022 Rotates 90 degrees\n\nIf you rotate by 90 degrees four times in the same direction, where do you end up?\n\n\u2022 You should clarify this with a complex number plotted on the complex plane. Jan 5 '17 at 23:47\n\u2022 Obviously depending on where you start the rotation, so no one knows what $i^4$ is actually Jan 13 '17 at 4:56\n\u2022 @BradPitt Neal is thinking of multiplication by a number as a function. Multiplication by $i^4$ is the identity function- scale by 1 and rotate by 0. Sep 12 '17 at 7:15\n\u2022 @BradPitt The answer is, \"The same way you were pointing when you started.\" In other words, nothing changed, so you multiplied by $1$. :)\n\u2013\u00a0Neal\nSep 12 '17 at 10:26\n\nSeeing as the current best answer that I understand doesn't contain proof at my level (and likely the level of the people this question would help), I'm going to answer.\n\nThe solutions of $i^n$ are repeating in a simple pattern. The pattern goes as follows\n\n$$i^0 = 1$$ $$i^1 = i$$ $$i^2 = i\\cdot i = -1$$ $$i^3 = (i\\cdot i)\\cdot i = -i$$ $$i^4 = i\\cdot i\\cdot i\\cdot i = (i\\cdot i)(i\\cdot i) = (-1)^2 = 1$$\n\nThis pattern is repeated infinitely.\n\nedit: As tomazs pointed out, this only works because multiplication by pure and simplified complex numbers is associative.\n\n$$i\\cdot i\\cdot i\\cdot i = (i\\cdot i)(i\\cdot i) = (i\\cdot i\\cdot i)\\cdot i = ((i\\cdot i)\\cdot i)\\cdot i$$ etc.\n\n\u2022 Good answer. Just be exact with parenthesis. In the last step it is $(-1)^2=1$. Since otherwise one reads it as $-(1)^2=-1$ and this can lead to confusion. Jan 5 '17 at 21:39\n\u2022 @MarvinF, thanks for pointing that out, didn't even think about it. Jan 5 '17 at 21:42\n\u2022 I think you were setting up a nice pattern with $i^2$ and $i^3$: To get to the next power of $i$, multiply the previous one by $i$. But you didn't continue that pattern for $i^4$, and compute it as $i^3 \\cdot i$. There's nothing wrong with what you have, but \"thematically,\" you might consider the \"multiply by another copy of $i$\" approach. Jan 6 '17 at 0:41\n\u2022 @pjs36 I had that originally, but it looked even messier than my current proof of $i^4$ Jan 6 '17 at 0:56\n\u2022 A fair point, and a rather sophisticated one :) I just wanted to mention it Jan 6 '17 at 0:58\n\nAlternatively:\n\n$$i^4=i^{2+2}=i^2i^2=(-1)(-1)=1$$\n\nRaising to positive integer power is the same as repeated multiplication, thus you don't even have to think whether $i^4=(i^2)^2$ is true, just expand the power:\n\n$$i^4=i\\cdot i\\cdot i\\cdot i=(i\\cdot i)(i\\cdot i)=(-1)(-1)=1.$$\n\nThe second equality works due to associativity of multiplication for complex numbers.\n\n\u2022 I prefer @Tomasz answer, that explicitly invokes associativity. Your parenthesing isn't justified.\n\u2013\u00a0user65203\nJan 7 '17 at 15:02\n\u2022 @YvesDaoust you're right. I was thinking of mentioning it, but for some reason decided to not do when I was writing the answer. Anyway, tomasz's answer is way more complete than mine, and it was written later, so I couldn't have had it as an example :). Jan 7 '17 at 15:18\n1. $$i^4=(i^2)^2=(-1)^2=1$$\n2. $$i^4=\\left(|i|e^{\\arg(i)i}\\right)^4=\\left(e^{\\frac{\\pi i}{2}}\\right)^4=e^{\\frac{4\\pi i}{2}}=e^{2\\pi i}=1$$\n\u2022 Pretty sure that if OP is at the level of wondering whether $i^4 = 1$, they won't be familiar with the complex exponential. I'm quite often wrong though, so who knows... Jan 5 '17 at 20:54\n\u2022 Proofing this with exponentiation looks like a cyclic proof to me Jan 5 '17 at 21:43\n\nThe property of multiplication of real numbers which you used and referred to, but did not precisely quote, is associativity (i.e. $z_1\\cdot (z_2\\cdot z_3)=(z_1\\cdot z_2)\\cdot z_3)$). Multiplication of complex numbers is associative, so you have: $$i^4=i\\cdot(i\\cdot(i\\cdot i))=(i\\cdot i)\\cdot(i\\cdot i)=(-1)\\cdot (-1)=1.$$ (Note that if multiplication was not associative, an expression like $i^4$ would not immediately make sense: it is not at all obvious if it means $i\\cdot(i\\cdot(i\\cdot i))$ or $((i\\cdot i)\\cdot i)\\cdot i$. There are expansions of complex numbers with non-associative multiplication for which the rule you are using here would not apply.)\n\nYou can also verify it directly using the definition of multiplication of complex numbers: \\begin{align} i\\cdot(i\\cdot(i\\cdot i))&=(0+1i)\\cdot ((0+1i)\\cdot(-1+0i))\\\\ &=(0+1i)\\cdot(0+(-1)i)\\\\ &=-(-1)+0i\\\\ &=1 \\end{align}\n\nChecking associativity in general is a bit more troublesome (in fact, it is almost always tiresome to check directly), but still workable. Recall that the definition of multiplication in complex numbers is $$(a_1+b_1i)\\cdot(a_2+b_2i)=(a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2).$$ Now you can just compute directly \\begin{multline*} (a_1+b_1i)\\cdot \\left( (a_2+b_2i)\\cdot (a_3+b_3i)\\right)=(a_1+b_1i)\\cdot((a_2a_3-b_2b_3)+i(a_2b_3+b_2a_3))=\\\\ =(a_1a_2a_3-a_1b_2b_3-(b_1a_2b_3+b_1b_2a_3))+i(a_1a_2b_3+a_1b_2a_3+b_1a_2a_3-b_1b_2b_3). \\end{multline*} On the other hand, \\begin{multline*} ((a_1+b_1i)\\cdot (a_2+b_2i))\\cdot (a_3+b_3i)=((a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2))\\cdot(a_3+b_3i)=\\\\ =(a_1a_2a_3-b_1b_2a_3-(a_1b_2b_3+b_1a_2b_3))+i(a_1a_2b_3-b_1b_2b_3+a_1b_2a_3+b_1a_2b_3).\\end{multline*} Finally, $$a_1a_2a_3-a_1b_2b_3-(b_1a_2b_3+b_1b_2a_3)=a_1a_2a_3-b_1b_2a_3-(a_1b_2b_3+b_1a_2a_3),$$ and $$a_1a_2b_3+a_1b_2a_3+b_1a_2a_3-b_1b_2b_3=a_1a_2b_3-b_1b_2b_3+a_1b_2a_3+b_1a_2a_3,$$ so, since two complex numbers are equal when their real and complex parts are equal, the multiplication is associative. (This argument can be made to be a little bit shorter if you apply commutativity of multiplication of complex numbers.)\n\nYes you are right. Rotation by 90 degree in complex plane and multiplication by $i$ are very much same. If you raise $i$ to 4,8,12th ... powers you get 1. Raising to 1, 5, 9 .. you get back $i$; raising to 2,6,10 powers gives you $-1$ and so on.\n\nThe property $$(x^{m})^n=x^{mn}\\tag{1}$$ for $m$ and $n$ nonnegative integers holds whenever we are dealing with an associative operation with a neutral element $1$, like for complex multiplication. Indeed it holds for $n=1$, because $(x^{m})^0=1=x^{m0}$ by definition.\n\nRecall that $x^n$ is defined recursively: $x^0=1$, $x^{k+1}=x^k\\cdot x$.\n\nSuppose property $(1)$ holds for $n$; then \\begin{align} (x^{m})^{n+1}&= (x^m)^n\\cdot x^m &&\\text{definition of powers}\\\\ &=x^{mn}\\cdot x^m &&\\text{induction hypothesis}\\\\ &=x^{mn+m} && \\text{rule of powers $(2)$}\\\\ &=x^{m(n+1)} && \\text{property of integers} \\end{align} The mentioned rule of powers is where associativity is used: $$x^{h+k}=x^h\\cdot x^k\\tag{2}$$\n\nAgain, this is true by definition when $k=1$. Suppose it holds for $k$; then \\begin{align} x^{h+(k+1)}&=x^{(h+k)+1} && \\text{property of integers}\\\\ &=x^{h+k}\\cdot x &&\\text{definition of powers}\\\\ &=\\bigl(x^h\\cdot x^k\\bigr)\\cdot x &&\\text{induction hypothesis}\\\\ &=x^h\\cdot\\bigl(x^k\\cdot x\\bigr) &&\\text{associativity}\\\\ &=x^h\\cdot x^{k+1} &&\\text{definition of powers} \\end{align}\n\nTherefore you are certainly allowed to say that $$i^4=(i^2)^2=(-1)^2=1$$\n\nLight comes from the definition of the complex multiplication, which only involves real arithmetic:\n\n$$(a+ib)(c+id):=ac-bd+(ad+bc)i.$$\n\nThen\n\n$$(0+1i)^2=(0+1i)(0+1i)=(\\bar1+0i),\\\\ (0+1i)^3=(\\bar1+0i)(0+1i)=(0+\\bar1i),\\\\ (0+1i)^4=(0+\\bar1i)(0+1i)=(1+0i).$$\n\n\u2022 I'm not the downvoter, so I've no idea about the reason for the downvote, but if you ask me, you definitely should not use \"$\\overline{1}$\" for \"$-1$\" since in complex analysis \"$\\overline{z}$\" denotes the conjugate of $z$. Also, I don't think your answer quite addresses the underlying question of what complex exponentiation satisfies. You can make it relevant by stating that in your answer the integer exponent is defined to mean repeated multiplication, and so you are simply evaluating the powers one by one. Makes sense? Jan 13 '17 at 15:56\n\u2022 $\\bar 1$ makes a much nicer alignment than $-1$. Purely cosmetic. I didn't think of the confusion with conjugation, which I use to denote $^*$. The accepted answer just says that the powers are cyclically $i,-1,-i,1$ without any justification. IMO, it is the one that deserves a downvote.\n\u2022 I also use \"$z^*$\" to denote the conjugate of $z$, but the overline notation is actually very common. As I said, I didn't downvote your answer, but I do think it doesn't address the underlying question. And I don't know who downvoted my answer either..... Not you right? Jan 14 '17 at 8:52", "date": "2021-10-24 21:50:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2085268/does-i4-equal-1/2085325", "openwebmath_score": 0.9037195444107056, "openwebmath_perplexity": 278.31217136572945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513889704252, "lm_q2_score": 0.8840392848011833, "lm_q1q2_score": 0.8717965686111082}}
{"url": "https://math.stackexchange.com/questions/2241622/eigenvectors-of-harmonic-series-matrix?noredirect=1", "text": "# Eigenvectors of harmonic series matrix\n\nI have a finite upper triangular $n$ by $n$ matrix $A$ which has the following form:\n\n$$A = \\begin{pmatrix} 1 & 1\\over 2 & 1\\over 3 & \\cdots & 1\\over n \\\\ 0 & 1\\over 2 & 1\\over 3 & \\cdots & 1\\over n \\\\ 0 & 0 & 1\\over 3 & \\cdots & 1\\over n \\\\ \\vdots &\\vdots &\\vdots &\\ddots &\\vdots \\\\ 0 & 0 & 0 & \\cdots& 1\\over n \\end{pmatrix}$$\n\nOr alternatively $A =TH$, where $T$ is the all-ones upper $n$ by $n$ triangular matrix and $H$ is the first $n$ terms of the harmonic series on the diagonal.\n\nI'm interested in diagonalizing this matrix, as I'm interested in understanding and computing $A^k$ (specifically the final column). Since all elements on the diagonal are distinct the eigenvalues are simply the first $n$ terms of the harmonic series.\n\nDoing some numerical analysis using numpy gave fairly opaque numbers, however Mathematica produced very interesting eigenvectors. E.g. for $n = 5$ the matrix with the eigenvectors as columns is:\n\n$$P = \\begin{pmatrix} 1 & -1 & 1 & -1 & 1 \\\\ 0 & 1 & -2 & 3 & -4 \\\\ 0 & 0 & 1 & -3 & 6 \\\\ 0 & 0 & 0 & 1 & -4 \\\\ 0 & 0 & 0 & 0 & 1 \\\\ \\end{pmatrix}, P^{-1} = \\begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\\\ 0 & 1 & 2 & 3 & 4 \\\\ 0 & 0 & 1 & 3 & 6 \\\\ 0 & 0 & 0 & 1 & 4 \\\\ 0 & 0 & 0 & 0 & 1 \\\\ \\end{pmatrix}$$\n\nThis seems to be some alternating sign Pascal triangle. In fact, $P^{-1}$ seems the be the actual Pascal triangle! And indeed $PHP^{-1} = A$ (no PHP pun intended), for the cases I've tried.\n\n### I have two questions:\n\n1. Is there a good explanation for this seeming connection between the eigenvectors of the truncated harmonic series matrix and Pascal's triangle, assuming my numerical exploration is not a coincidence?\n\n2. Can we use this to say anything interesting about (the final column of) $A^k$? A closed form that does not require constructing the Pascal triangles perhaps?\n\n\u2022 For curiosity, it is the probability matrix of a markov chain for standing on a stairs $n-1$ steps left to the top and with equal $1/n$ probability to stay on the same stage or go to any other step (but only upwards). Apr 19 '17 at 10:51\n\u2022 Maybe there is something with the equidistribution of uniform variables and pascals triangle making successive better approximations to normal (law of large numbers). Apr 19 '17 at 10:54\n\u2022 @mathreadler Yes! The origin is a study of the probability distribution of a $k$-throws dice game where the size of the dice you throw is determined by your previous throw (and initially $n$).\n\u2013\u00a0orlp\nApr 19 '17 at 10:55\n\u2022 This earlier question seems relevant: math.stackexchange.com/q/69925/137524. Apr 19 '17 at 16:00\n\nLet's indicate your matrix as $${\\bf A}_{\\,h} = \\left\\| {\\;a_{\\,n,\\,m} \\quad \\left| {\\;0 \\le n \\le m \\le h} \\right.\\;} \\right\\| = \\left\\| {\\;{{\\left[ {n \\le m} \\right]} \\over {m + 1}}\\;} \\right\\|_{\\,h}$$ so that it is a $(h+1) \\times (h+1)$ matrix, indexed from $0$, and where the square brackets indicate the Iverson's bracket.\n\nThen let's define \\eqalign{ & {\\bf E}_{\\,h} = \\left\\| {\\;\\left[ {n + 1 = m} \\right]\\;} \\right\\|_{\\,h} \\cr & {\\bf S}_{\\,h} = \\left\\| {\\;\\left[ {n \\le m} \\right]\\;} \\right\\|_{\\,h} = {{{\\bf I}_{\\,h} } \\over {{\\bf I}_{\\,h} - {\\bf E}_{\\,h} }}\\quad \\cr & \\left( {f(n) \\circ {\\bf I}_{\\,h} } \\right) = \\left\\| {\\;f(n)\\left[ {n = m} \\right]\\;} \\right\\|_{\\,h} \\cr} so that\n${\\bf E}_{\\,h}$ is the Shifting matrix, with $1$ on the first upper sub-diagonal and null otherwise;\n${\\bf S}_{\\,h}$ is the Summing matrix, upper triangular with all $1$ 's;\n$\\left( {f(n) \\circ {\\bf I}_{\\,h} } \\right)$ is the diagonal matrix, with diagonal elements $f(0),f(1), \\cdots, f(h)$.\n\nThen, as you correctly pointed out, we can write: $${\\bf A}_{\\,h} = {\\bf S}_{\\,h} \\left( {{1 \\over {n + 1}} \\circ {\\bf I}_{\\,h} } \\right)$$ which means $${\\bf A}_{\\,h} ^{\\; - \\,{\\bf 1}} = \\left( {\\left( {n + 1} \\right) \\circ {\\bf I}_{\\,h} } \\right)\\left( {{\\bf I}_{\\,h} - {\\bf E}_{\\,h} } \\right)$$\n\nIt is not difficult to demonstrate that \\eqalign{ & \\left( {f(n) \\circ {\\bf I}_{\\,h} } \\right)\\;{\\bf E}_{\\,h} = {\\bf E}_{\\,h} \\;\\left( {f(n - 1) \\circ {\\bf I}_{\\,h} } \\right) = \\cr & = \\left( {\\prod\\limits_{0\\, \\le \\,k\\, \\le \\,n - 1} {f(k)} \\circ {\\bf I}_{\\,h} } \\right)^{\\; - \\,{\\bf 1}} \\;{\\bf E}_{\\,h} \\left( {\\prod\\limits_{0\\, \\le \\,k\\, \\le \\,n - 1} {f(k)} \\circ {\\bf I}_{\\,h} } \\right) \\cr} and that, given in general $${\\bf A}_{\\,h} (f) = {\\bf S}_{\\,h} \\left( {f(n) \\circ {\\bf I}_{\\,h} } \\right)\\quad \\left| {\\;f(m) \\ne f(n)} \\right.$$ the relevant matrix of eigenvectors $\\mathbf W$ is : $${\\bf W}_{\\,h} (f) = \\left\\| {\\;\\left[ {n \\le m} \\right]{{f(m)^{\\,m - n} } \\over {\\prod\\limits_{n\\, \\le \\,k\\, \\le \\,m - 1} {\\left( {f(m) - f(k)} \\right)} }}\\;} \\right\\|_{\\,h} = \\left\\| {\\;\\left[ {n \\le m} \\right]\\prod\\limits_{n\\, \\le \\,k\\, \\le \\,m - 1} {\\left( {{{f(m)} \\over {f(m) - f(k)}}} \\right)} \\;} \\right\\|_{\\,h}$$ while the inverse is: \\eqalign{ & {\\bf W}_{\\,h} ^{\\; - \\,{\\bf 1}} (f)\\quad \\left| {\\;0 \\ne f(0)} \\right.\\quad = \\cr & = \\left\\| {\\;\\left[ {n \\le m} \\right]{{f(m)f(n)^{\\,m - n - 1} } \\over {\\prod\\limits_{n + 1\\, \\le \\,k\\, \\le \\,m} {\\left( {f(n) - f(k)} \\right)} }}\\;} \\right\\|_{\\,h} = \\left\\| {\\;\\left[ {n \\le m} \\right]{{f(m)} \\over {f(n)}}\\prod\\limits_{n + 1\\, \\le \\,k\\, \\le \\,m} {{{f(n)} \\over {\\left( {f(n) - f(k)} \\right)}}} \\;} \\right\\|_{\\,h} \\cr}\n\nNow, in the present case, we obtain \\eqalign{ & {\\bf W}_{\\,h} = \\left\\| {\\;\\left[ {n \\le m} \\right]\\prod\\limits_{n\\, \\le \\,k\\, \\le \\,m - 1} {\\left( {{{k + 1} \\over {k - m}}} \\right)} \\;} \\right\\|_{\\,h} = \\cr & = \\left\\| {\\;\\left[ {n \\le m} \\right]{{m!} \\over {n!}}\\prod\\limits_{1\\, \\le \\,j\\, \\le \\,m - n} {\\left( { - {1 \\over j}} \\right)} \\;} \\right\\|_{\\,h} = \\left\\| {\\;\\left( { - 1} \\right)^{\\,m - n} \\left( \\matrix{ m \\cr n \\cr} \\right)\\;} \\right\\|_{\\,h} = \\overline {\\bf B} _{\\,h} ^{\\; - \\,{\\bf 1}} \\cr} where $\\overline {\\bf B} _{\\,h}$ is the transpose of the Lower Triangular Pascal array.\n\nFinally $${\\bf A}_{\\,h} = {\\bf S}_{\\,h} \\left( {{1 \\over {n + 1}} \\circ {\\bf I}_{\\,h} } \\right) = \\overline {\\bf B} _{\\,h} ^{\\; - \\,{\\bf 1}} \\;\\left( {{1 \\over {n + 1}} \\circ {\\bf I}_{\\,h} } \\right)\\;\\overline {\\bf B} _{\\,h}$$ which demonstrates your assumption.\n\nConcerning your last question instead, we have of course that $${\\bf A}_{\\,h} ^{\\;{\\bf q}} = \\overline {\\bf B} _{\\,h} ^{\\; - \\,{\\bf 1}} \\;\\left( {{1 \\over {\\left( {n + 1} \\right)^{\\;q} }} \\circ {\\bf I}_{\\,h} } \\right)\\;\\overline {\\bf B} _{\\,h}$$ whose single components are: \\eqalign{ & a^{\\left( q \\right)} _{\\,n,\\,m} = \\sum\\limits_{0\\, \\le \\,\\left( {n\\, \\le } \\right)\\,k\\,\\left( { \\le \\,m} \\right)\\, \\le \\,h} {\\left( { - 1} \\right)^{\\,k - n} \\left( \\matrix{ k \\cr n \\cr} \\right){1 \\over {\\left( {k + 1} \\right)^{\\;q} }}\\left( \\matrix{ m \\cr k \\cr} \\right)} = \\cr & = \\sum\\limits_{0\\, \\le \\,\\left( {n\\, \\le } \\right)\\,k\\,\\left( { \\le \\,m} \\right)\\, \\le \\,h} {\\left( { - 1} \\right)^{\\,k - n} \\left( \\matrix{ m \\cr n \\cr} \\right)\\left( \\matrix{ m - n \\cr k - n \\cr} \\right)\\left( {k + 1} \\right)^{\\; - \\,q} } = \\cr & = \\left( \\matrix{ m \\cr n \\cr} \\right)\\sum\\limits_{0\\, \\le \\,\\left( {n\\, \\le } \\right)\\,k\\,\\left( { \\le \\,m} \\right)\\, \\le \\,h} {\\left( { - 1} \\right)^{\\,k - n} \\left( \\matrix{ m - n \\cr k - n \\cr} \\right)\\left( {k - n + n + 1} \\right)^{\\; - \\,q} } = \\cr & = {{m!} \\over {n!}}\\sum\\limits_{\\,0\\, \\le \\,j\\, \\le \\,m - n} {{{\\left( { - 1} \\right)^{\\,j} } \\over {j!\\left( {m - n - j} \\right)!}}\\left( {j + n + 1} \\right)^{\\; - \\,q} } \\cr} Cannot find at the moment a \"more compact\" expression.\n\nSince the given matrix is upper triangular, the eigenvalues are trivially $\\frac{1}{1},\\frac{1}{2},\\frac{1}{3},\\ldots$ and the shown decomposition is just an instance of the following identity: $$\\frac{1}{x(x+1)(x+2)\\cdots(x+n)} = \\frac{1}{n!}\\sum_{k=0}^{n}\\frac{(-1)^k}{x+k}\\binom{n}{k}$$ that follows from the residue theorem.\n\n\u2022 Could you elaborate a bit on how the decomposition is an instance of the identity? I did manage to find a closed form for the last column of $A^k$ though, if $x$ goes from $1$ to $n$ denoting the row number, then $f_{k, n}(x) = \\frac{x}{n} \\left | \\sum_{i=1}^n \\frac{(-1)^i}{i^k}\\binom{n}{i}\\binom{i}{x} \\right |$.\n\u2013\u00a0orlp\nApr 19 '17 at 12:47", "date": "2021-12-05 21:09:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2241622/eigenvectors-of-harmonic-series-matrix?noredirect=1", "openwebmath_score": 0.9990427494049072, "openwebmath_perplexity": 1681.5298036583429, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513865283938, "lm_q2_score": 0.8840392848011834, "lm_q1q2_score": 0.8717965664522567}}
{"url": "http://riosite.com.br/v4hyp/zero-polynomials.html", "text": "## Zero polynomials\n\nIn general, two polynomials are only considered equal if they have the same coe cients. are all constants. 1. What is the degree of zero polynomial? 0 is considered as constant polynomial. Given that 2 is a zero of the cubic polynomial 6x3 + 2 x2 \u2013 10x \u2013 4 2, find its other two zeroes. Example. You can put this solution on YOUR website! If c is a zero of the polynomial P, which of the following statements must be true? True/false (a) P(c) = 0 (b) P(0) = c (c) c is the y-intercept of the graph of P Polynomials of odd degrees have at least one real zero. So, before we get into that we need to get some ideas out of the way regarding zeroes of polynomials that will help us in that process. Zero Polynomials The constant polynomial whose coefficients are all equal to 0. 1 De\ufb01nitions A complex polynomial is a function of the form P (z) = n k =0 a k z k, (1. 2, 4 Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 Putting p(x) = 0 x + 5 = 0 x = \u2212 5 So, x = \u22125 is a zero of the given polynomial\u00a0 Step2. Polynomial Exponents Lessons. +. Repeat the process using Q (x) this time instead of P (x). Constant & Linear Polynomials Constant polynomials A constant polynomial is the same thing as a constant function. Driver, P. If an equation has a zero then you will have factors with no remainders when you divide them. The roots function calculates the roots of a single-variable polynomial represented by a vector of coefficients. Zero Product Rule. Symbolic Computation (1992) 13, 117-131 Solving Zero-dimensional Algebraic Systems D. Please see explanation below. The z i terms are the zeros of the transfer function; as s\u2192z i the numerator polynomial goes to zero, so the transfer function From the section on polynomials, we know that the root of a polynomial P(x) is defined as the value of x for which the polynomial is equal to zero. A unique zero of a system of polynomials is a zero of a finite system. As we\u2019ve seen, long division of polynomials can involve many steps and be quite cumbersome. Find all zeros. This is because the function value never changes from a , or is constant. If the leading coefficient is not a The algebra of polynomials 1. Because the extended Euclidean algorithm works for polynomials, the same proof used to prove unique factorization for integers also Negative Exponents and Zero Exponents. Zeros of the derivative of a p-adic meromorphic function and applications 0]) = (-1, 0, 3) (we use the convention that the zero polynomial has degree -1). You were taught long division of polynomials in Intermediate Algebra. If P(x) is a polynomial with integer coefficients and if is a zero of P(x) (P() = 0), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x). The previous lesson explained how to simplify exponents of a single term inside parentheses, like the problem below. The graph of a quadratic polynomial is a parabola which opens up if a > 0, down if a < 0. A polynomial can have any number of terms. These results follow from a general theorem which models such polynomials by Hermite polynomials. The form of a monomial is an expression is where n is a non-negative integer. Find f(x) and if you get zero --- then its a factor of f(x)=x^3 - 5x^2 + 2x + 8. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial. Y. Factoring a polynomial is the opposite process of multiplying polynomials. We prove that a given subset of the vector space of all polynomials of degree three of less is a subspace and we find a basis for the subspace. Determining a This is just a constant term (b 0 /a 0) multiplied by a ratio of polynomials which can be factored. If you choose, you could then multiply these factors together, and you should get the original polynomial (this is a great way to check yourself on your factoring skills). We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial. Polynomials are a type of function that you will see regularly as you study mathematics. Hey, our polynomial buddies have caught up to us, and they seem to have calmed down a bit. For example, the polynomial x^3 - 4x^2 + 5x - 2 has zeros x = 1 and x = 2. Thus, there can be only one positive zero. What is the Root of this constant polynomial? The answer is a Non-zero constant polynomial has no zero. Because of our restriction above, that a transfer function must not have more zeros than poles, we can state that the polynomial order of D(s) must be greater than or equal to the polynomial order of N(s). Ascending order is basically when the power of a term increases for each succeeding term. Factoring polynomials free answersheets, quadratic word problems, 2 variable polynomials addition solver, cheats for my geometry home work, elementary operation algebraic fractions, how to do seventh grade algebra online, algebra solving equations, checking solutions, literal equations worksheets. In our study of mathematics, we\u2019ve found that some functions are easier to work with than others. Lojasiewicz, Triangulation of semi-analytic sets, Ann. If you're seeing this message, it means we're having trouble loading external resources on our website. Factor the following polynomial functions completely. The graph of a linear polynomial is a straight line. Ask Question 2. When x = 1 or 2, the polynomial equals zero. Boyer and William M. Such a repre-sentation is frequently called a canonical form. Viewed 438 times 22. Pritsker Abstract We study global distribution of zeros for a wide range of ensembles of random poly-nomials. 1 and N. This was the key idea in Euler\u2019s method. 6 Zeros of a Polynomial Fundamental Theorem of Algebra Every complex polynomial function fx( ) of degree n \u22651 has at least one complex zero. Polynomials\u00b6 Polynomials in NumPy can be created, manipulated, and even fitted using the Using the Convenience Classes of the numpy. is a polynomial with integer coefficients, the polynomial. It can only be a monomial which is equal to a constant . For example, you can use synthetic division to divide by x + 3 or x \u2013 6, but you cannot use synthetic division to divide by x 2 + 2 or 3x 2 \u2013 x + 7. PRELIMINARIES . Binomial theorem. 12. Now we are going to study two more aspects of monomials: those that have negative exponents and those that have zero as an exponent. Find the Roots of a Polynomial Equation. Spend time developing the purpose of the zero product property so that young mathematicians understand why the equations should be set equal to zero and how that The degree of the zero polynomial is either left undefined, or is defined to be negative (usually \u22121 or\u00a0 whose coefficients are all equal to 0. is a polynomial with integer coefficients \u00a0 There is only one zero of this polynomial, and it is easy to find out that zero. \u2022 a standard form: 0. Sometimes you will not know it is prime until you start looking for factors of it. Example \u2013 3: Divide the polynomial 4x 2-3x +x 3 +10 by x+4 and verify the remainder with zero of the divisor. Free printable worksheets with answer keys on Polynomials (adding, subtracting, multiplying etc. This is due to the fact that imaginary roots come in pairs (Conjugates). [Fed69] Herbert Federer, Geometric measure theory,\u00a0 31 Jul 2013 Zero Polynomials The constant polynomial whose coefficients are all equal to 0. Well, if you a polynomial is factorable then its roots/zeroes can be easily found by setting it to zero and using the zero factor property. Geometry of poles. It can also be said as the roots of the polynomial equation. Now we've gotta find factors and roots of polynomials. =. Polynomials have much in common with integers. Section 2 deals with almost sure convergence of the zero counting measures for polynomials with random coe cients that satisfy only weak log-integrability assumptions. B. This section\u00a0 Descartes' rule of sign is used to determine the number of real zeros of a polynomial function. Able to display the work process and the detailed explanation. does not have only integer coefficients! You will learn how to find all those roots of such polynomials, which are rational numbers, such as. LAZARD ' LIT_P, Institut Blaise Pascal, Boite 168, 4, place Jussieu, F-75252 Paris Cedex 05, France (Received 27 June 1989) It is shown that a good output for a solver of algebraic systems of dimension zero consists of a family of \"triangular sets of polynomials\". You can add them, subtract them and multiply them together and the result is another polynomial. This section is where we look at some concepts in graph and group theories that aid in computing the independence and clique polynomials of zero-divisor graphs of the ring of integers modulo n It is also possible for the same zero to occur more than once. Dividing Polynomials by Binomials. Free trial available at KutaSoftware. Find the zeros of an equation using this calculator. A product is zero if and only if one of the factors is zero, so to find the roots of we need to look only at the roots of the individual factors. When we are given a list of the zeros of a polynomial, we can conclude the polynomial must have certain factors, which gives us information about the equation of the polynomial. are all real and simple, andthat a zero of Ek-1 lies strictly between every two con-secutive zeros ofEk, 2 <_ k <m. Secondly, and probably more importantly, in order to use the zero factor property we MUST have a zero on one side of the equation. The numbers a0, a1\u00a0 POLYNOMIALS AND THEIR ZEROS. 15. To find the zeros of a polynomial by grouping, we first equate the polynomial to 0 and then use our Process for Finding Rational Zeroes Use the rational root theorem to list all possible rational zeroes of the polynomial P (x). POLYNOMIALS 15 3. A zero with an even multiplicity, like (x + 3) 2, doesn't go through the x-axis. That is, a constant polynomial is a function of the form p(x)=c for some number c. Roots in a Specific Interval. x^2 - 3x + x + 2x - x^2 = zero Zero of a Linear Polynomial in Polynomials with Definition, Examples and Solutions. A core concept in algebra, polynomials are used in calculus and throughout all areas of mathematics. Any polynomial can be converted to a monic polynomial by dividing all the terms by the coefficient of the highest order term. Problems related to polynomials with real coefficients and complex solutions are also in These results follow from a general theorem which models such polynomials by Hermite polynomials. PDF | We provide two examples of complex homogeneous quadratic polynomials P on Banach spaces of the form l_1(I). There are two very important things you need to know when working with Zero Power or Negative Exponents. Determine if the set of polynomials is closed under division. In the examples, C is set equal to zero. 2 Multiply Polynomials My Powerpoint Connecting the 3 different ways of finding roots of a quadratic equation. Real Zeros 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. ZERO DISTRIBUTION OF COMPOSITE POLYNOMIALS AND POLYNOMIALS BIORTHOGONAL TO EXPONENTIALS D. Why do we care about zero in particular? Well, we don't, but we do care about equality: it seems worthwhile to be able to figure out when f(x) = g(x). Determining the equation of a polynomial function. Inequalities for the derivative and for the maximum modulus on a larger circle of a polynomial with a given zero on the unit circle are obtained in terms of its degree and maximum modulus on the unit circle; examples are given Factoring. Find k so that x 2 + 2x + k is a factor of 2x 4 + x 3 \u2013 14 x 2 + 5x + 6. A value of x that makes the equation equal to 0 is termed as zeros. Determining a subspace of polynomials with degree 3 or whether it means that U contains a polynomial equal to zero specifically for x=0. where it is proven that real algebraic (actually, even\u00a0 7 Aug 2002 A method to determine the distribution of the zeros of a polynomial with respect to the unit-circle, proposed by this author in the past, is revisited\u00a0 11 Jan 2013 Zero sets throughout mathematics. Zeros of Polynomials Number of Zeros Theorem. This holds for k 1 since E1 2Alx and 0 _< x0. If is a rational number written in lowest terms, and if is a zero of , a polynomial function with integer coefficients, then p is a factor of the. Weproceedto prove byinduction on k that if 1 <k <m, thenthe largest zero of Ek is _< x0. In the next couple of sections we will need to find all the zeroes for a given polynomial. N\u00f8rsett2 AbstractThe authors have presented in [6] a technique to generate transfor-mations T of the set Pn of nth degree polynomials to itself such that if p\u2208 Pn has all its zeros in (c,d) then T {p} has all its zeros in (a,b), where (a,b) Using Synthetic Division to Divide Polynomials. Degree 3, 4, and 5 polynomials also have special names: cubic, quartic, and quintic functions. If ever you have to have assistance on percents or perhaps exponents, Factoring-polynomials. Zero Degree Polynomials. The integral of any polynomial is the sum of the integrals of its terms. A polynomial with one term is called a monomial. Doing this serves two purposes. Zero degree polynomial functions are also known as constant functions. For all n \u2265 2 and 1 \u2264 k < d, there exists a constant \u03b4n,d,k > 0 The limits of the numerator and denominator follow from Theorems 1, 2, and 4. 4, numpy. Let\u2019s show that this is irreducible over Q. They have a polynomial for us. e. In the case of the Riemann zeta function, this proves the Gaussian unitary ensemble random matrix model prediction in derivative aspect. The list must contain an element for each nonzero monomial of the polynomial. So, why find the root? Well, let's say that we know we have the problem, $x^2+2x+100=99$ So, what can x be? The power of the constant polynomial is Zero. 11. The Zero Product Principle says that if there is a product of two number that is equal to zero, than or the first, or the second (or both) has to be zero. The degree of a polynomial is the maximum of the degrees of each\u00a0 See. Explain why or why not. Theorem 1: Let f, g [member of] A(K) be such that W(f,g) is a non-identically zero polynomial. Long Division of Polynomials. In fact, the process of factoring is so important that very little of algebra beyond this point can be accomplished without understanding it. If degree of remainder is equal to or\u00a0 Rational Zeros of Polynomials. In other words, if the product of two things is zero then one of those two things must be zero, because the only way to multiply something and get zero is to multiply it by zero. di Pisa, 18 (1964), 449-474. Algebra 1 Notes Review Real Numbers and Closure Real Numbers and Closurerev Notes Page 2 of 20 9/4/2014 Sample Questions* (A. We also use the terms analytic polynomial (re\ufb02ecting the fact that Introduction. Introduction to Algebraic Expressions and Polynomials An algebraic expression is an expression formed from any combination of numbers and variables by using the operations of addition, subtraction, multiplication, division, exponentiation (raising to powers), or extraction of roots. Factoring Cubic Polynomials March 3, 2016 A cubic polynomial is of the form p(x) = a 3x3 + a 2x2 + a 1x+ a 0: The Fundamental Theorem of Algebra guarantees that if a 0;a 1;a 2;a 3 are all real numbers, then we An important problems associated with polynomials is factoring. If the remainder is 0 or degree of remainder is less than divisor, then we cannot continue the division any further. Pascal's triangle. Goh Abstract. If the constant term = 4, then the polynomial form is given by f(x)= 4x 0. A polynomial having value zero (0) is known as zero polynomial. If a polynomial has no variable, it is called polynomial of zero variable. 4. Real Zeros. 5) becomes Hermite\u2019s ODE and H n(x) are the Hermite polynomials. If f(x) = P n i=0 c ix i with c n 6= 0, then c n is the leading coe\ufb03cient and c 0 is the constant term. Now we can plot the equation y=p(x) on the Cartesian plane by taking various values of x and y obtained by putting the values. This section deals with polynomials which have integer coefficients only. If a polynomial is not factorable we say that it is a prime polynomial. First, any number to the Zero Power always equals one. The output of a constant polynomial does not depend on the input (notice Classification of polynomials vocabulary defined. 3x to the 4th power minus 75x to the second power = 0. The zeros of a polynomial function of x are the values of x that make the function zero. S. Find a polynomial given its graph. A-APR. \u2022 Polynomials of degree 3: Cubic polynomials P(x) = ax3 +bx2 + cx+d. The Degree of a Polynomial with one variable is POLYNOMIALS 9 Sample Question 2: Given that two of the zeroes of the cubic polynomial ax3 + bx2 + cx + d are 0, the third zero is (A) \u2013b a (B) b a (C) c a (D) \u2013 d a Solution : Answer (A). For example \u2013 5. The zero-power property can be used to solve an equation when one side of the equation is a product of polynomial factors and the other side is zero. How to Divide Polynomials. Exact answers only!!! No Decimal approximations allowed! 8) FACTOR: f(x)=x5+2x4\u2212184x32\u2212+\u2212xx4930 Polynomials usually are arranged in one of two ways. 1 Roots of\u00a0 polynomial,. Multiple Zeros: If a zero is repeated an even number of times it is called a double root/zero; If a zero is repeated an odd number of times, it is called a triple root/zero; If a zero is not repeated it is called a simple root/zero Let p be the set of polynomials let a,b,c be elements ofn0 such that b and d are non zero elements of p which of the following is true regarding the sum - 11944\u2026 The same thing can occur with polynomials. All of these are polynomials of \u00a0 [Voiceover] So, we have a fifth-degree polynomial here, p of x, and we're asked to do several things. 2 Addition, subtract are not polynomials, because the powers of x are not positive integers or zero. Calculator returns the roots (zeroes) of any polynomial. Using the Remainder Theorem -- we can try and guess for one of the zeros and then divide. Let me show you two examples: f(x)= 2(x+3) and x 1(x+10). 3 - Real Zeros of Polynomial Functions. A polynomial is a series of terms, each of which is the product of a constant coefficient and an integer power of the independent variable. Basic Shapes - Even Degree (Intro to Zeros) Basic Shapes - Odd Degree (Intro to Zeros) Important Rules About Zeros. Polynomials of even degrees may have no real zeros. Odd-degree Polynomials Another type of application of the Intermediate Zero Theorem is not to find a root but to simply show that a root exists. -(existence theorem; tells us that a zero exists, but not where; it is between the two points a and b where sign changes, has to cross the x axis somewhere to change signs) This zero point energy is an aspect of the uncertainty principle, a genuine quantum phenomenon. But 0 is the only term here. Click here for K-12 lesson plans, family activities, virtual labs and more! Home. RAHMAN ABSTRACT. Example 4. It happened In other words, the usual arrangement of a monic polynomial is in descending powers. Roots of a Polynomial. Polynomials are stored in Maxima either in General Form or as Canonical Rational Expressions (CRE) form. Algebra -> Polynomials-and-rational-expressions-> SOLUTION: Solve the polynomial equation by factoring and then using the zero product principle. Here is the twist. They include many other families of multivariable orthogonal polynomials as special cases, including the Jack polynomials, the Hall\u2013Littlewood polynomials, the Heckman\u2013Opdam polynomials, and the Koornwinder polynomials. Isolate the variable term. Prior to NumPy 1. Constant functions have these be a polynomial function with real coefficients: The number of positive real zeros is either equal to the number of sign changes of f (x) or is less than the number of sign changes by an even integer. . A polynomial of degree n has at most n distinct zeros. One reason why factoring is important is that we know that a product is zero precisely when one or more of its factors are zero. where a and C are constants. We start with our new Polynomials Class 9 Maths Notes with Formulas Download in pdf. A polynomial has coefficients: The terms are in order from highest to lowest exponent (Technically the 7 is a constant, but here it is easier to think of them all as coefficients. If x+a is a zero of a polynomial f, then the following three statements are true: A. The number of negative real zeros is either equal to the number of sign changes of f ( \u2212 x) or is One method is to use synthetic division, with which we can test possible polynomial function zeros found with the rational roots theorem. Enter the polynomial expression: FACTOR: Computing Get this widget. Purplemath. The results after simpli\ufb01cation are shown in Table 4. 1 )( axa xa xaxf n n n n. Factoring-polynomials. \u2022 Polynomials of degree 1: Linear polynomials P(x) = ax+b. Every linear polynomial has one and only one zero. SIDI2 Abstract. Section 3 is devoted to the discrepancy results, and establishes ex-pected rates of convergence of the zero counting measures to the equilibrium mea-sures. It is clear from that the roots of our particular polynomial are , , and . New videos every week. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1. You can even di vide a polynomial by another polynomial, as you will see in Section 1. polynomial functions zeros factored form expanded form the factor theorem the rational roots theorem potential zeros synthetic division the reduced polynomial the quadratic formula. An expression that is a real number, a variable, or a product of a real number and a variable with whole- A multiple zero has Section 2-1: Polynomials Polynomials are expressions whose exponents are non-negative integers and the coefficients are real numbers. A polynomial in a single variable x can always be written in the form. There is a very nice relationship between the zeros of a polynomial and the factors of a polynomial. Appell Polynomials and Their Zero Attractors Robert P. 8. The polynomial doesn't change signs at a zero of even multiplicity. Let p (x) be a polynomial function with real coefficients. If g(t) is an entire function, g(0) 6= 0, with at least one zero, the asymptotics of linearly scale d polynomials Classification of Polynomials Polynomial equations are the equation that contains monomial, binomial, trinomial and also the higher order polynomial. Biorthogonal polynomials and zero-mapping transformations Arieh Iserles1 and Syvert P. For instance, the quadratic ( x + 3)( x \u2013 2) has the zeroes x = \u20133 and x = 2 , each occuring once. A monic polynomial is one with leading coe\ufb03cient 1. displaymath151. Therefore the zero Factoring Polynomials \u2013 Finding Zeros of Polynomials - 1 7) Find a polynomial function (factored form) of degree 3 which has the corresponding table of values to the right. But when As it turns out, the zero knowledge (ie. 7. Deriving zero bounds for real and complex zeros of polynomials is a classical problem that has been proven essential in various disciplines such as engineering, mathematics, and mathematical chemistry \u2013. This example shows several different methods to calculate the roots of a polynomial. For example, p(x)=5 3 or q(x)=7. In theory, root finding for multi-variate polynomials can be transformed into that for single-variate polynomials. This program runs exceptionally well for lacunary polynomials with not too many non-zero terms. (13. The limit of the fraction follows from Theorem 3. 2) in Quarteroni, Sacco, and Saleri, but their presentation focusses on orthogonal polynomials. In other words it divides through the Polynomials The polynomial 2x 4 + 3x 3 \u2212 10x 2 \u2212 11x + 22 is represented in Matlab by the array [2, 3, -10, -11, 22] (the coefficients of the polynomial are starting with the highest power and ending with the constant term, which means power zero). 2 Addition, subtract the work, section 3 is where the zero-divisor graphs are constructed, the independence and clique polynomials are computed in section 4. We will add, subtract, multiply, and even start factoring polynomials. It is a constant polynomial whose all the\u00a0 1 Oct 2018 As the zero polynomial does not have a variable and equating it with zero will give no result. privacy) guarantee is (relatively!) easy to provide; there are a bunch of ways to convert any computation into an instance of something like the three color graph problem, where a three-coloring of the graph corresponds to a solution of the original problem, and then use a traditional zero knowledge Unit 6: Polynomials. Imaginary zeros of polynomials. We analyze polynomials Pn that are biorthogonal to exponentials INEQUALITIES FOR POLYNOMIALS WITH A PRESCRIBED ZERO BY A. Note that (x-3) is a factor of . The latter is a standard form, and is used internally by operations such as factor, ratsimp, and so on. 13. Basically, the procedure is carried out like long division of real numbers. journal of the american mathematical society volume 00, number 0, pages 000\u2013000 s 0894-0347(xx)0000-0 on the number of zero-patterns of a sequence of polynomials Complex Zeros of Polynomials \u2014 5. Geometric Meaning of the Zero's of the polynomials Lets us assume y= p(x) where p(x) is the polynomial of any form. Well, you can write any constant with a variable having an exponential power of zero. To know more about Polynomials, please visit https://DontMemorise. The next section explains the topic remainder theorem. Any time you get a zero remainder, the divisor is a factor of the dividend. Also, every real number is a zero of the Zero Polynomial. Numeric Roots. Solve for the variable. real and complex numbers. d. Question 7. Although we would almost always like to find a basis in which the matrix representation of an operator is The graph of a polynomial with roots meets the axis at those roots At a simple root the curve crosses the axis at an angle At a multiple root the axis is tangent to The analytic theory of orthogonal polynomials is well documented in a number of treatises; for classical orthogonal polynomials on the real line as well as on the circle, see [25], for those on the real line also [24]. It tells us that the number of positive real zeroes in a polynomial\u00a0 where n is a nonnegative integer and a0, a1,,an\u22121, an are real numbers with an = 0. Improve your math knowledge with free questions in \"Solve a quadratic equation using the zero product property\" and thousands of other math skills. The solutions \u03c8 n (Fig. 14. Polynomial Long Division Calculator - apply polynomial long division step-by-step. Jensen polynomials of each degree. The terms are in order from highest to lowest exponent (Technically the 7 is a constant, but here it is easier to think of them all as coefficients. 1 Complex polynomials 1. \u2022 Polynomials of degree 2: Quadratic polynomials P(x) = ax2 +bx+c. In this post, we Roots and zeros When we solve polynomial equations with degrees greater than zero, it may have one or more real roots or one or more imaginary roots. Zeros of polynomials and their importance in combinatorics and\u00a0 It is called the zero polynomial (or the zero function. In particular, an empty list results in the zero polynomial. Follow Byju's and understand things easily. For example, the polynomial f(x) = x7 x over Z 7 has the property that f(a) = 0 for all a 2Z 7, but this does not mean that f is equal to zero polynomial. and the indefinite integral of that term is. Zeros at x = ~+mn~ \u221a5, correspond to the factors. Polynomials. The only possible choices are 1 and 2. Solution: Say f(x) = 4x 2-3x +x 3 +10 and g(x)= x+4. . (a,0) is an ----- of the graph f A summary of Complex Zeros and the Fundamental Theorem of Algebra in 's Algebra II: Polynomials. 4. We're finding the zeros of polynomial functions. $$\\PageIndex{5}$$ Find a third degree polynomial with real coefficients that has zeros of $$5$$ and $$\u22122i$$ such that $$f (1)=10$$. 1) of Eq. Click on the lesson below that interests you, or follow the lessons in order for a complete study of the unit. 1. Ken Bube of the University of Washington Department of Mathematics in the Spring, 2005. A. (Since the interval of integration is symmetric about the origin, the integral of an odd monomial is zero. Zeroes/Roots of Polynomials. The names of different polynomial functions are The zero is very useful because it helps find the root. What must be added to f(x) = 4x 4 + 2x 3 \u2013 2x 2 + x \u2013 1 so that the resulting polynomial is divisible by g(x) = x 2 + 2x -3? Question 6. I. The Macdonald polynomials are orthogonal polynomials in several variables, depending on the choice of an affine root system. The zeros of a polynomial are the values of x for which the value of the polynomial is zero. For this division, we have using the steps as follows Polynomial definition is - a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a nonnegative integral power (such as a + bx + cx2). The expression applies for both positive and negative values of n except for the special case of n= -1. Using Remainder Theorem, we can redefine a root as a value a for which the factor (x - a) divides through the polynomial P(x) to get a remainder of zero. Rational zero theorem. com offers great facts on zero product property calculator, trigonometric and two variables and other algebra topics. f (x) (x ) Create your own worksheets like this one with Infinite Precalculus. It has the determinant and the trace of the matrix as coefficients. A solution of that equation is called a root or zero of . 1) where the a k are complex numbers not all zero and where z is a complex variable. Once we find a zero we can partially factor the polynomial and then find the polynomial function zeros of a reduced polynomial. ) Alternatively, you can say that the degree of the zero polynomial is undefined; in that case, you will need to make minor changes to some of the results below. Polynomial zeros or polynomial roots, factoring polynomials, properties, definition , examples and solved problems. This page will show you how to raise a polynomial to some power, or exponent. Factor the quadratic expression x 2 + 7 x + 12. To find the degree of a polynomial we need the highest degree of individual terms with non-zero coefficient. Rational Zeros Theorem. Polynomials can be divided the same as numeric constants, either by factoring or by long division. A \"root\" (or \" zero\") is where the polynomial is equal to zero: Graph of Inequality. Two polynomials are said to be equal if reduction of all similar terms makes them identical (except for order, and terms with zero coefficients). addition and subtraction: Adding and subtracting polynomials is the same as the procedure used in combining like terms. Graphing Polynomials. We'll also work through some sample problems using 3. Cuemath material for JEE & CBSE, ICSE board to understand Zero of a Linear Polynomial better. q f zM ba Kdje o RwJiAtNhG eIBn4fbi hn DiFt 4eh zA El9g BeIb jr TaH U1h. ) A polynomial also has roots: A \"root\" (or \"zero\") is where the polynomial is equal to zero. Also, with 2\u03bb n \u22121 = 2n, Eq. polynomial package, introduced in NumPy 1. p q If the answer is no, it is probably because zero is that of a one-term monomial (not polynomial), not a sum. Back We have already seen degree 0, 1, and 2 polynomials which were the constant, linear, and quadratic functions, respectively. \u00a9n p2C031 B2f tK au GtDaF bS Ao5f ptlw Gaur meI 4LbLSCt. 12 and 4. Multiplicities of polynomials . Descartes' rule of signs. When adding polynomials, simply drop the parenthesis and combine like terms. Given that x \u2013 5 is a factor of the cubic polynomial x3 \u2013 3 5x2 + 13x \u2013 3 5, find all the NCERT Solutions for Class 10 Mathematics CBSE, 2 Polynomials. a. It is useful if an equation has to be solved. i. If we know the function value at some point (say f (a)) and the value of the derivative at the same NCERT Exemplar Problems Class 9 Maths \u2013 Polynomials August 19, 2019 by Mereena 5 Comments CBSETuts. Functions are provided to evaluate the polynomials, determine their zeros, produce their polynomial coefficients, produce related quadrature rules, project other functions onto these polynomial bases, and integrate double and triple products of the polynomials. The corresponding polynomial function is the constant function with value 0, also called the zero map. 13. First, it puts the quadratics into a form that can be factored. Learn exactly what happened in this chapter, scene, or section of Algebra II: Polynomials and what it means. Tutorial and problems with detailed solutions on finding polynomial functions given their zeros and/or graphs and other information. You don't have to worry about the\u00a0 zeros, of polynomials in one variable. APR. The degree of a term pk, lxkyl is defined as k + l if pk, l is non-zero. Explore the Science of Everyday Life . 2, we found that we can use synthetic division to determine if a given real number is a zero of a polynomial function. The degree of the polynomial is the largest exponent of x which appears in the polynomial -- it is also the subscript on the leading term. \u2022 recognize the typical shapes of the graphs of polynomials, of degree up to 4, The function f(x)=0 is also a polynomial, but we say that its degree. 5. In the latter case the polynomial is called identically zero and is denoted by the symbol 0. It is easy to check that none of these are zeroes of x2 2. One way to find the zeros of a polynomial is to write in its factored form. Here are some examples you could try: (x+1)^2 (5x^2+10x-3)^2 (x^2-7)^5 Poly1 - Division Algorithm: If polynomial f(x) divided by polynomial D(x) results in quotient Q(x) with with remainder R(x), then we may write. Lets see if 0, 1, or 2 is a zero. Build your own widget In the last section, we saw how to determine if a real number was a zero of a polynomial. Did you know that polynomials are used in every walk of life from shopping to engineering? Learn this vital topic with ease using these polynomials worksheets, featuring key skills like recognizing polynomials, identifying the degree and type of polynomials, performing arithmetic operations on polynomials and more. The nonnegative integer n is called the degree of P. And also we can say that the reminder is not zero, 3x is not a factor of 3x 3 + 9x 2 + 5. A general term of a polynomial can be written. For example, x + x 2 + x 3 or 5 x + 2 x 2 \u2013 3 x 3 + x 5 are arranged in ascending order. One of the nice things about polynomials is that you can add or subtract two polynomials to get a third one. For typical polynomials, one can expect that f(x) and its non-reciprocal part are the same (this, however, is not really the case when one considers 0-1 polynomials with a limited number of non-zero terms). Thus x2 2 is irreducible over Q. Thus, having a factorization allows us to find the zeros of a polynomial. And let's sort of remind ourselves\u00a0 Zeros of Polynomial | Relation Between Zeros & Coefficients of polynomial equations. An important problems associated with polynomials is factoring. (2) Polynomials : The expression which contains one or more terms with non-zero coefficient is called a polynomial. A Question for You Multiply Polynomials - powered by WebMath. A root or a zero of a polynomial in one variable, say $p(x)$, is a number $a$ such that substituting $x=a$ in the polynomial gives zero, i. Roots: The roots of a polynomial f(x) are the values of x where f is equal to zero. Duren / Zero distribution of hypergeometric polynomials as n!1, without giving a detailed proof. Engaging math & science practice! Improve your skills with free problems in 'Using the Zero-Product Property Given the Product of Two Binomials' and thousands of other practice lessons. Factoring Polynomials. Note that a polynomial is de ned to be a formal sum, not a function. Continuing our discussion of analytic geometry and trigonometry, in this segment, I want to talk about polynomials and conics. The bisection method is an application of the Intermediate Zero Theorem. The calculator will show you the work and detailed explanation. Plus examples of polynomials. Answered by Yasmeen Khan | 1st Oct, 2018,\u00a0 Zeroes of Polynomial are the real values of the variable for which the value of the polynomial becomes zero. 2 is based on the same concept. 16) Write a polynomial function of degree ten that has two imaginary roots. We are commited to providing you with factoring help in areas such as quadratic expressions complete squares 2 - lots of lessons in factoring polynomials can be found on our free site. Set the equation to equal zero. Before going to start other sections of Polynomials, try to solve the below-given question. Article Summary: Polynomial is also a vital concept in algebra and all through the science and Math. Consequently x=3 is a root of the polynomial . Also find the other zero of the polynomial Question 5. Let k > 1. solution. The student might think that to evaluate a limit as x approaches a value, all we do is evaluate the function at that value. In other words, is a root or zero of a polynomial if it is a solution to the equation . It must go from to so it must cross the x-axis. Conjugate Zeros Theorem. So f(0) = 8 --- thus 0 is not a factor and not For polynomials, the role of primes in integer factorization is taken by irreducible polynomials, where a polynomial p is irreducible if p(x) = a(x)b(x) holds only if at lest one of a(x) or b(x) has degree zero. Still, degree of zero polynomial is not 0. 1 Linear Approximations We have already seen how to approximate a function using its tangent line. For Polynomials of degree less than or equal to 4, the exact value of any roots (zeros) of the polynomial are returned. CHEBYSHEV_POLYNOMIAL is a MATLAB library which considers the Chebyshev polynomials T(i,x), U(i,x), V(i,x) and W(i,x). This polynomial has only one term, which is constant. So, real numbers, 'm' and 'n' are zeroes of\u00a0 Polynomials: Sums and Products of Roots. A polynomial family {pn(x)} is Appell if it is given by e xt g(t) = P\u221e n=0 pn(x)tn or, equivalently, p\u2032n (x) = pn\u22121(x). following property of zero sets of harmonic polynomials: Hn,k points can be detected in zero sets of harmonic polynomials of degree d (1 \u2264 k \u2264 d) by \ufb01nding a single, su\ufb03ciently good approximation at a coarse scale. Moreover, we establish hyperbolicity for all d 8. In particular, the zero measures of such random polynomials converge almost surely to normalized Lebesgue measure on the unit circle if and only if the underlying coefficient distribution satisfies a particular moment condition. Courses The zero of a polynomial in the variable x, is a value of x for which the polynomial is zero. Math scholars and students use polynomials to create polynomial equations, which inscribe a huge range of Math difficulties ranging from basic to intricate problems in both science and Math field. In other words, p 2 is irrational. Polynomials can approximate some functions. -If P(x) is a polynomial function and P(a) and P(b) have opposite signs, then there is a real number c between a and b such that P(c)=0. Factoring Polynomials Calculator. (This is necessary in order to make the degree formulas work out. NCERT Exemplar Problems Class 10 Maths \u2013 Polynomials. 14. Also find all the zeroes of the two polynomials. If we don\u2019t have a zero on one side of the equation we won\u2019t be able to use the zero factor property. An intimately related concept is that of a root, also called a zero, of a polynomial. Polynomial of one variable. In this case, there is a possibility that the coefficients of the lower order terms will appear as fractions. The procedure is explained in the textbook if you're not familiar with it. If ab = 0 then either a = 0 or b = 0 (or both). To stay The zeros of a polynomial equation are the solutions of the function f(x) = 0. ) Each sheet includes visual aides, model problems and many practice problems Polynomial Worksheets- Free pdf's with answer keys on adding,subtracting, dividing polynomials Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. If we have a polynomial, and you know, for example, that $2$ is a zero, then $(x - 2)$ is a factor of that polynomial! That means if you were to factor the polynomial completely, you would see $(x-2)$ somewhere in there. Concept Chapter 2 Class 10 Maths POLYNOMIALS Ncert Solutions Degree of the polynomial :-If p(x) is a polynomial in terms of x, the highest power of x in p(x) is called the degree of the polynomial p(x). Page 1 of 23 1. RN. cos(\u03c0/8), the coef\ufb01cient polynomials are tedious to \ufb01nd (but this can be done by a computer). Zero-free regions. Example : 7, 3, -2, 3/7, etc. The set of polynomials is closed under division. Don\u2019t Memorise brings learning to life through its captivating FREE educational videos. A number x=a is called a root of the polynomial f(x), if Once again consider the polynomial Let's plug in x=3 into the polynomial. Symbolic Roots. So, first of all, a polynomial equation is an equation that contains variables which we sometimes call indeterminates and coefficients. f(x) = D(x) Q(x) + R(x). Find k so that x2 + 2x + k is a factor of 2x4 + x3 \u2013 14 x2 + 5x + 6. The solutions to a polynomial equation are called roots. Two main directions are related to almost sure limits of the zero counting measures, and to quantitative results on the expected number of zeros in various sets. The degree of the zero polynomial is undefined, but many authors conventionally set it equal to -1 or \u221e. $p(a)=0$. x=a is a ----- of the polynomial equation f(x)=0 B. It just \"taps\" it, and then goes back the way it came. A polynomial has coefficients:. s O LARljl g DrPi zg 5hvt Ss1 mrNeusfe mrEvDexdt. In mathematics, the fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. Write 3. coefficients tend to concentrate near the unit circle. In the last post, we talked about how to multiply polynomials. Basically, the\u00a0 5 Jun 2019 In Section 3. For convenience, we\u2019ll usually that we write our polynomials so that c n 6= 0. ----- is a factor of the polynomial f(x) C. The Factor Theorem. Fundamental theorem of algebra. Now, let\u2019s look at a constant polynomial \u20185\u2019. In the days before graphing technology was commonplace, mathematicians discovered a lot of clever tricks for determining the likely locations of zeros. Therefore, you must use sparse input involving only nonzero terms. In this equation the constant k=b 0 /a 0. Degree. Put simply: a \u00a0 Ex 2. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. poly1d was the class of choice and it is still available in order to maintain backward compatibility. a) If x = 1 is a zero of multiplicity 2, then (x - 1) 2 is a factor of p (x) Rational Zeros Theorem: If p Polynomials: Bounds on Zeros. All the solutions of Polynomials - Mathematics explained in detail by experts to help students prepare for their CBSE exams. How to Find Zeros of Polynomials Questions with Detailed Solutions solution. A polynomial all of whose coefficients are zero is called an identical zero polynomial and is denoted by 0. 0 may be zero of a polynomial. com is the right site to pay a visit to! In the case of the above polynomial division, the zero remainder tells us that x + 1 is a factor of x 2 \u2013 9x \u2013 10, which you can confirm by factoring the original quadratic dividend, x 2 \u2013 9x \u2013 10. Zero Distribution of Random Polynomials Igor E. 13, respectively. Evaluate the polynomial at the numbers from the first step until we find a zero. The first step in solving a polynomial is to find its degree. 148 K. 8. There are two cases for dividing polynomials: either the \"division\" is really just a simplification and you're just reducing a fraction (albeit a fraction containing polynomials), or else you need to do long polynomial division (which is explained on the next page). Zeros with an odd multiplicity, like x and (x \u2013 4) 3, pass right through the x-axis and change signs. In this section, we will learn how to find good candidates to test using synthetic division. This call is the fastest method to create polynomials of the type DOM_POLY because the input already has the form that MuPAD uses internally. This is not merely an esoteric exercise. Roots Using Substitution. factor the polynomial. com . Polynomial with only one variable is called Polynomial of one variable. So far in this unit, you've learned how to simplify monomial expressions with positive exponents. You can write this as 5x 0. 9. Introduction. It is well known that the roots of a random polynomial with i. The zeros of a polynomial equation are the solutions of the function f(x) = 0. Simplifying Expressions with Zero Exponents A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such Roots of polynomials. The corresponding polynomial function is the constant function\u00a0 Polynomials with quaternionic coefficients located on only one side of the powers (we call them simple polynomials) may have two different types of zeros:\u00a0 Common Core Standard. You will agree that degree of any constant polynomial is zero. The leading term is the term with the highest power. ) Equation is related to Equations (10. The general theorem also allows us to prove a conjecture of Chen, Jia, and Wang on the partition function. Scu. GIROUX AND Q. ) Its degree is undefined, $-1$ , or $-\\infty$ , depending on the author. These always graph as horizontal lines, so their slopes are zero, meaning that there is no vertical change throughout the function. com A zero of a polynomial need not to be 0. Limits of polynomials. Polynomials with degree n > 5 are just called n th degree polynomials. 5 $\\begingroup$ There is an extensive theory of the real and complex Linear Transformations and Polynomials We now turn our attention to the problem of finding the basis in which a given linear transformation has the simplest possible representation. 3) 1. For example (-50) 0 = 1 There is one number that CANNOT be raised to the Zero Power, 0 0 does not exist! When dealing with Negative Exponents there is a simple trick. 2 $\\begingroup$ The zero polynomial is defined by convention to have degree . ) Two polynomials are called equal if, after reduction, all terms with non-zero coefficients are pairwise identical (but, possibly, written in a different order), and also if all the coefficients of both of these polynomials turn out to be zero. First, find the real roots. A polynomial of degree n may always be written in. If f (c) = 0, then x - c is a factor of f (x). 1 Taylor Polynomials The tangent line to the graph of y = f(x) at the point x = a is the line going through the point ()a, f (a) that has slope f '(a). 10. 6 = 2 \u00d7 3 , or 12 = 2 \u00d7 2 \u00d7 3. In mathematics, factorization or factoring is the breaking apart of a polynomial into a product of other smaller polynomials. 3 - Real Zeros of Polynomial Functions Long Division of Polynomials. The method you use depends upon how complex the polynomial dividend and divisor are. How to Solve Polynomials - Solving a Linear Polynomial Determine whether you have a linear polynomial. In this lesson, we'll tackle polynomial equations and learn how to solve for their zero values. Welcome to the Algebra 1 Polynomials Unit! This unit is a brief introduction to the world of Polynomials. 27 Aug 2015 A polynomial function on Rn to R, is either identically 0, or non-zero almost everywhere. polynomials of degree 0, together with the zero polynomial, are called constant polynomials. Norm. of all polynomials over F. The Fundamental Theorem of Algebra tells us that every polynomial can be written as a product of complex linear factors. Polynomial Roots Calculator. Zero Factor Theorem Remember the Fundamental Theorem of Algebra which states that whatever the degree of the polynomial, that is exactly the number of zeros (roots or x-intercepts) we will get, as Paul\u2019s Online Notes so accurately states. 3 Identify zeros of polynomials when suita- ble factorizations are available, and use the zeros to construct a rough graph of the\u00a0. For instance, if you are doing calculus, typically polynomials are \u201ceasy\u201d to work with because they are easy to differentiate and integrate. Exercise 2. Once you have exhausted all possibilities, then you can call it prime. Rational Zeros Theorem: If the polynomial ( ) 1 11 nn Px ax a x ax ann \u2212 = +++ \u2212 +0 has integer coefficients, then every rational zero of P is of the form . Consider f(x) = x4 + 3x2 7x+ 1 2Q[x]. LUBINSKY1 AND A. Recall that when we factor a number, we are looking for prime factors that multiply together to give the number; for example . The process of factoring is essential to the simplification of many algebraic expressions and is a useful tool in solving higher degree equations. In linear algebra, the characteristic polynomial of a square matrix is a polynomial which is invariant under matrix similarity and has the eigenvalues as roots. Rational Zeros of Polynomials: The next theorem gives a method to determine all possible candidates for rational zeros of a polynomial function with integer coefficients. 1) and (10. The precise statement is as follows. Video notes on connecting the 3 ways of how to find roots of quadratic equations: Quadratic Formula, Square Root Method, Factoring The term a n is assumed to be non-zero and is called the leading term. Actually, the term 0 is itself zero polynomial. Find the degree and classify them by degree and number of terms. one rational zero. It is a value where the graph of the polynomial intersects the x-axis. The Euclidean algorithm for polynomials is similar to the Euclidean algorithm for finding the greatest common divisor of nonzero integers. P(x) = a 0 x n + a 1 x n-1 + \u2026 OR 1 is the Zero of the Polynomial equation p(x) = x \u2013 1 = 0. where pk, l are constants. A Brief Review. y = x^8 + 8 has no roots. Constants : A symbol having a fixed numerical value is called a constant. Use synthetic division to evaluate a given possible zero by synthetically dividing Repeat step two using the quotient found with A zero has a \"multiplicity\", which refers to the number of times that its associated factor appears in the polynomial. Example: 5x+2, 2x^2+x+3, etc. The first polynomial P has both separable and nonseparable maximal zero subspaces. Descartes' rule of signs for positive zeros may be stated as follows: When a polynomial function is written in standard form, the number of changes in sign of the coefficients is the maximum number of positive zeros of the function. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Zero curves of Tutte Polynomials? Ask Question Asked 3 years, 10 months ago. How do we solve polynomials? That depends on the Degree!. RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. The zero polynomial \u00a0 5 Aug 2016 A polynomial with degree of zero must have no unknowns, only constants. 4) are proportional to the Hermite polynomials1H n(x). We simply equate this polynomial to 0 and find out the corresponding value of x:. Applications of polynomial functions. \\$x=1 Zero in on your pupils' understanding of solving quadratic equations. In order to divide polynomials using synthetic division, you must be dividing by a linear expression and the leading coefficient (first number) must be a 1. Polynomial of zero variable. Compare the Lagrange polynomials of degree N =3forf (x) =ex that are obtained by using the coef\ufb01cient polynomials in Tables 4. By is a quadratic polynomial then it would have a zero in Z and this zero would divide 2. STRUCTURE OF SETS WHICH ARE WELL APPROXIMATED BY ZERO SETS OF HARMONIC POLYNOMIALS MATTHEW BADGER, MAX ENGELSTEIN, AND TATIANA TORO Abstract. D Worksheet by Kuta Software LLC Adding and Subtracting Polynomials Activity For example, I would probably never ask students to find the sum of two polynomials that summed to zero. Use the Rational Zero Theorem to list all possible rational zeros of the function. TAYLOR POLYNOMIALS AND TAYLOR SERIES The following notes are based in part on material developed by Dr. com provides you Free PDF download of NCERT Exemplar of Class 9 Maths chapter 2 Polynomials solved by expert teachers as per NCERT (CBSE) book guidelines. In this section, the solved example is given in which each step of the solution is Ch 9. Roots of Polynomials. A degree 0 polynomial is a constant. Simple examples: y = x^7 + 8 has one root. A polynomial can have more than one zero. If you would like to see the details, take this side trail. Poles are the roots of D(s) (the denominator of the transfer function), obtained by setting D(s) = 0 and solving for s. Since there are several ways to sum zero in a polynomial even without 0 as a coefficient (when terms of common power have opposite signs), you can always simplify the polynomial and make it say 0. Note that this is the same result that applies to zero degree polynomials, i. Auxiliary results exploit the connection be-tween hypergeometric polynomials and associated functions. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. By induction hypothesis, the largest zero of Ek-1 is <_ x0, so byseparation of zeros are not polynomials, because the powers of x are not positive integers or zero. General orthogonal polynomials are dealt with in [5] and more recently in [22], especially with regard to nth-root asymptotics. Rational Zeros of Polynomials. Characteristics of polynomial graphs. 1 Multiple Choice Questions (MCQs) Question 1: Solution: (a) Given that, one of the zeroes of the quadratic polynomial say p(x) = (k- 1)x 2 + kx + 1 All right, we've trekked a little further up Polynomial Mountain and have come to another impasse. Polynomials are mathematical expressions that contain a sum of powers of indeterminate variables multiplied by coefficients. 1 Introduction to Polynomials . Theorem 1. The zero sets of harmonic polynomials play a crucial role in the study of the Polynomials. How To: Given a polynomial function f, use synthetic division to find its zeros. Example 17. This step reduces the degrees of the polynomials involved, and so repeating the procedure leads to the greatest common divisor of the two polynomials in a finite number of steps. zero polynomials\n\n0tw, yxqfss5, 5t5k, kjwo6xeqew, xqnb, 0po, kb3fx, bhwrxgo, i6xhbw, c19, qy6jro,", "date": "2020-06-03 19:25:58", "meta": {"domain": "com.br", "url": "http://riosite.com.br/v4hyp/zero-polynomials.html", "openwebmath_score": 0.7238003611564636, "openwebmath_perplexity": 364.1644106487573, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9861513873424044, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8717965566269317}}
{"url": "http://crawfordcountyhealthdepartment.com/r7b53/archive.php?page=d5dd1c-application-of-differential-equation-in-pharmacy", "text": "Vedantu Simple harmonic motion: Simple pendulum: Azimuthal equation, hydrogen atom: Velocity profile in fluid flow. Applications of Laplace Transforms Circuit Equations. This review focuses on the basics and principle of centrifugation, classes of centrifuges, \u00e2\u0080\u00a6 Introduction to differential equations View this lecture on YouTube A differential equation is an equation for a function containing derivatives of that function. l n m m 0 = k t. when t = 1 , m = 1 2 m 0 gives k = \u00e2\u0080\u0093 ln 2. l n m m 0 = \u00e2\u0088\u0092 2 ( l n 2) t. Now when the sheet loses 99% of the moisture, the moisture present is 1%. YES! Biologists use differential calculus to determine the exact rate of growth in a bacterial culture when different variables such as temperature and food source are changed. The forum of differential calculus also enables us to introduce, at this point, the contraction mapping principle, the inverse and implicit function theorems, a discussion of when they apply to Sobolev spaces, and an application to the prescribed mean curvature equation. In this paper we discussed about first order linear homogeneous equations, first order linear non homogeneous equations and the application of first order differential equation in electrical circuits. , and allowing the well-stirred solution to flow out at the rate of 2 gal/min. The Langmuir adsorption model explains adsorption by assuming an adsorbate behaves as an ideal gas at isothermal conditions. There are two (related) approaches: Derive the circuit (differential) equations in the time domain, then transform these ODEs to the s-domain;; Transform the circuit to the s-domain, then derive the circuit equations in the s-domain (using the concept of \"impedance\"). Differential Equation Applications. 1 INTRODUCTION . Studies of various types of differential equations are determined by engineering applications. OF PHARMACEUTICAL CHEMISTRY ISF COLLEGE OF PHARMACY WEBSITE: - WWW.ISFCP.ORG EMAIL: RUPINDER.PHARMACY@GMAIL.COM ISF College of Pharmacy, Moga Ghal Kalan, GT Road, Moga- 142001, Punjab, INDIA Internal Quality Assurance Cell - (IQAC) One of the common applications of differential equations is growth and decay. How to Solve Linear Differential Equation? Considering, the number of height derivatives in a differential equation, the order of differential equation we have will be \u20133\u200b. This is an introductory course in mathematics. And the amazing thing is that differential equations are applied in most disciplines ranging from medical, chemical engineering to economics. Polarography DR. RUPINDER KAUR ASSOCIATE PROFESSOR DEPT. In applications of differential equations, the functions represent physical quantities, and the derivatives, as we know, represent the rates of change of these qualities. Local minima and maxima. Differential equations have a remarkable ability to predict the world around us. Calculating stationary points also lends itself to the solving of problems that require some variable to be maximised or minimised. 4 B. Applications of Differential Equations Anytime that we a relationship between how something changes, when it is changes, and how much there is of it, a differential equations will arise. Application of Differential Equation: mixture problem Submitted by Abrielle Marcelo on September 17, 2017 - 12:19pm A 600 gallon brine tank is to be cleared by piping in pure water at 1 gal/min. APPLICATIONS TO ORDINARY DIFFERENTIAL EQUATIONS Our aim is to find the solution of the ordinary differential equation Lt = m=O 1 u,(x) m dmt/dxm= 7, (4) where z is an arbitrary known distribution. This book may also be consulted for In order to solve this we need to solve for the roots of the equation. Newton\u00e2\u0080\u0099s and Hooke\u00e2\u0080\u0099s law. 6.7 Applications of differential calculus (EMCHH) Optimisation problems (EMCHJ) We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. The degree of a differentiated equation is the power of the derivative of its height. How Differential equations come into existence? The degree of the differential equation is: A. with an initial condition of h(0) = h o The solution of Equation (3.13) can be done by separating the function h(t) and the Differential equations are of basic importance in molecular biology mathematics because many biological laws and relations appear mathematically in the form of a differential equation. So this is a homogenous, first order differential equation. In applications, the functions usually denote the physical quantities whereas the derivatives denote their rates of alteration, and the differential equation represents a relationship between the two. The Laplace transform and eigenvalue methods were used to obtain the solution of the ordinary differential equations concerning the rate of change of concentration in different compartments viz. ; We will use the first approach. Here, we have stated 3 different situations i.e. The constant r will alter based on the species. formula. NCERT Exemplar Class 12 Maths Chapter 9 Differential Equations Solutions is given below. blood and tissue medium. Objectives: Upon completion of the course the student shall be able to: Know the theory and their application in Pharmacy 2 SOLUTION OF WAVE EQUATION. endstream endobj 72 0 obj <> endobj 73 0 obj <> endobj 74 0 obj <>stream Course: B Pharmacy Semester: 1st / 1st Year Name of the Subject REMEDIAL MATHEMATICS THEORY Subject Code: BP106 RMT Units Topics (Experiments) Domain Hours 1 1.1 1.2 1.3 Partial fraction Introduction, Polynomial, Rational fractions, Proper and Improper fractions, Partial [\u00e2\u0080\u00a6] The rate constants governing the law of mass action were used on the basis of the drug efficacy at different interfaces. 1. ln m = kt + ln m 0. ln m \u00e2\u0080\u0093 ln m 0 = kt. There are basically 2 types of order:-. Electrical and Mechanical) Sound waves in air; linearized supersonic airflow Aug 29, 2020 differential equations with applications and historical notes third edition textbooks in mathematics Posted By Ann M. Aug 29, 2020 differential equation analysis in biomedical science and engineering partial differential equation applications Posted By Yasuo UchidaPublishing TEXT ID c111c3f6b Online PDF Ebook Epub Library. is a function of x alone, the differential equation has . This might introduce extra solutions. Oscillations naturally occur in virtually every area of applied science including, e.g., mechanics, electrical, radio engineering, and vibrotechnics. Since . Abstract Mathematical models in pharmacodynamics often describe the evolution of phar- macological processes in terms of systems of linear or nonlinear ordinary dierential equations. \u00e2\u0080\u009cPharmaceutical Mathematics with Application to Pharmacy\u00e2\u0080\u009d authored by Mr. Panchaksharappa Gowda D.H. Application in Physics. The mathematical description of various processes in chemistry and physics is possible by describing them with the help of differential equations which are based on simple model assumptions and defining the boundary conditions [2, 3].In many cases, first-order differential equations are completely describing the variation dy of a function y(x) and other quantities. 3/4 C. not defined D. 2 example is the equation used by Nash to prove isometric embedding results); however many of the applications involve only elliptic or parabolic equations. By means of DSC, the melting range can be determined for a substance, and based on the equation of Van\u00e2\u0080\u0099t Hoff (Ca-notilho et al., 1992, Bezjak et al., 1992) (Equation 1) it is So, since the differential equations have an exceptional capability of foreseeing the world around us, they are applied to describe an array of disciplines compiled below;-, explaining the exponential growth and decomposition, growth of population across different species over time, modification in return on investment over time, find money flow/circulation or optimum investment strategies, modeling the cancer growth or the spread of a disease, demonstrating the motion of electricity, motion of waves, motion of a spring or pendulums systems, modeling chemical reactions and to process radioactive half life. If we can get a short list which contains all solutions, we can then test out each one and throw out the invalid ones. Generally, $\\frac{dQ}{dt} = \\text{rate in} \u00e2\u0080\u0093 \\text{rate out}$ Typically, the resulting differential equations are either separable or first-order linear DEs. This book describes the fundamental aspects of Pharmaceutical Mathematics a core subject, Industrial Pharmacy and Pharmacokinetics application in a very easy to read and understandable language with number of pharmaceutical examples. As defined in Section 2.6, the fundamental solution is the solution for T = 6(x). Actuarial Experts also name it as the differential coefficient that exists in the equation. For many nonlinear systems in our life, the chaos phenomenon generated under certain conditions in special cases will split the system and result in a crash-down of the system. d m d t = k m. \u00e2\u0087\u0092 ln m = kt + c. initially when t = 0, m = m 0 thus substituting we get. They can describe exponential growth and decay, the population growth of \u00e2\u0080\u00a6 We begin by multiplying through by P max P max dP dt = kP(P max P): We can now separate to get Z P max P(P max P) dP = Z kdt: The integral on the left is di cult to evaluate. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Within mathematics, a differential equation refers to an equation that brings in association one or more functions and their derivatives. APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS . Therefore, this equation is normally taught to second- or third-year students in the schools of medicine and pharmacy. A differential equationis an equation which contains one or more terms which involve the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable) dy/dx = f(x) Here \u00e2\u0080\u009cx\u00e2\u0080\u009d is an independent variable and \u00e2\u0080\u009cy\u00e2\u0080\u009d is a dependent variable For example, dy/dx = 5x A differential equation that contains derivatives which are either partial derivatives or ordinary derivatives. Order of a differential equation represents the order of the highest derivative which subsists in the equation. Index References Kreyzig Ch 2 during infusion t = T so, \u00ef\u0080\u00a8 kt\u00ef\u0080\u00a9 e t Background of Study. the solution of the differential equation is The presence of oxygen in the atmosphere has a profound effect on the redox properties of the aquatic environment\u00e2\u0080\u0094 that is, on natural waters exposed directly or indirectly to the atmosphere, and by extension, on organisms that live in an aerobic environment.This is due, of course, to its being an exceptionally strong oxidizing agent and thus a low \u00e2\u0080\u00a6 An alternative approach is presented that uses a computer algebra system to calculate a limit and allows one to bypass the use of differential equations. The differential equation for the mixing problem is generally centered on the change in the amount in solute per unit time. Modeling is an appropriate procedure of writing a differential equation in order to explain a physical process. In this chapter we will cover many of the major applications of derivatives. - Could you please point me out to some successful Medical sciences applications using partial differential equations? First order differential equations have an applications in Electrical circuits, growth and decay problems, temperature and falling body problems and in many other fields\u00e2\u0080\u00a6 This subject deals with the introduction to Partial fraction, Logarithm, matrices and Determinant, Analytical geometry, Calculus, differential equation and Laplace transform. This model even explains the effect of pressure i.e at these conditions the adsorbate's partial pressure, , is related to the volume of it, V, adsorbed onto a solid adsorbent. \u00f0\u009d\u0091\u0091 2 \u00f0\u009d\u0091\u00a6 \u00f0\u009d\u0091\u0091\u00f0\u009d\u0091\u00a5 2 + \u00f0\u009d\u0091\u009d(\u00f0\u009d\u0091\u00a5) \u00f0\u009d\u0091\u0091\u00f0\u009d\u0091\u00a6 \u00f0\u009d\u0091\u0091\u00f0\u009d\u0091\u00a5 + \u00f0\u009d\u0091\u009e(\u00f0\u009d\u0091\u00a5)\u00f0\u009d\u0091\u00a6= \u00f0\u009d\u0091\u0094(\u00f0\u009d\u0091\u00a5) APPLICATION OF DIFFERENTIAL EQUATION IN PHYSICS . Now let\u2019s know about the problems that can be solved using the process of modeling. \"Functional differential equation\" is the general name for a number of more specific types of differential equations that are used in numerous applications. e.g. differential scanning calorimetry (DSC) method has been satisfactorily used as a method of evaluating the degree of purity of a compound (Widmann, Scherrer, 1991). This equation of motion may be integrated to find $$\\mathbf{r}(t)$$ and $$\\mathbf{v}(t)$$ if the initial conditions and the force field $$\\mathbf{F}(t)$$ are known. This paper discusses the stable control of one class of chaotic systems and a control method based on the accurate exponential solution of a differential equation is used. Logistic Differential Equation Let\u00e2\u0080\u0099s recall that for some phenomenon, the rate of change is directly proportional to its quantity. Recall the equation dC dt = \u00e2\u0088\u0092k Rearranging dC = - kdt We now need to integrate (to remove the differential and obtain an equation for C). Applications in Pharmacy Functions of several variables: graphical methods, partial derivatives and their geometrical meaning. Differential equations in Pharmacy: basic properties, vector fields, initial value problems, equilibria. \u00e2\u0080\u009csolve the differential equation\u00e2\u0080\u009d). Let P (t) be a quantity that increases with time t and the rate of increase is proportional to the same quantity P as follows. Oxygen and the Aquatic Environment. dp/dt = rp represents the way the population (p) changes with respect to time. Models such as these are executed to estimate other more complex situations. Applications include population dynamics, business growth, physical motion of objects, spreading of rumors, carbon dating, and the spreading of a pollutant into an environment to name a few. 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{"url": "https://math.stackexchange.com/questions/1719162/does-the-limit-exist", "text": "# Does the limit exist?\n\nI had a question about the definition of a limit. I know for a limit to exist the right hand limit must equal the left hand limit but what if the graph of a specific function has the domain from [0,5] and if you wanted to find the limit at x=5 you know the left hand limit exists but the right hand doesn't.\n\nSo in this situation does the limit exist or not since only the right-hand limit exists?\n\n\u2022 The left hand limit exist only \u2013\u00a0imranfat Mar 29 '16 at 18:42\n\u2022 No, the limit does exist. \u2013\u00a0D_S Mar 29 '16 at 18:43\n\u2022 If the right hand limit is not defined, then THE limit does not exist, unless the domain can be extended beyond 5 and both right hand and left hand limit exist and are equal \u2013\u00a0imranfat Mar 29 '16 at 18:44\n\u2022 Well, only left limit exists (since to the right the function is not defined), but that's ok: the limit coincides with the left limit. \u2013\u00a0Crostul Mar 29 '16 at 18:44\n\u2022 \"The limit coincides with the left limit, so the limit exists? Ehhh...?? \u2013\u00a0imranfat Mar 29 '16 at 18:47\n\nYou can only discuss limits in neighborhoods where the function is defined.\n\nIf the function is only defined in $[0, 5]$, then it makes no sense to talk about a right-hand limit at $5$. Therefore, when you talk about a limit at $5$, you can only mean left-hand limit.\n\n\u2022 That a better way of wording it +1 \u2013\u00a0imranfat Mar 29 '16 at 18:47\n\u2022 In this case, the left-hand limit and the limit are the same thing though. \u2013\u00a0D_S Mar 29 '16 at 18:58\n\u2022 So does the Limit exist overall? \u2013\u00a0Asker123 Mar 29 '16 at 19:07\n\u2022 @Asker123 I think it is better to come up with particular examples... \u2013\u00a0imranfat Mar 29 '16 at 19:13\n\u2022 For my question above, after reading all of the comments and the answers I think the limit does exist assuming that the limit can exist only on a given domain with boundaries. \u2013\u00a0Asker123 Mar 29 '16 at 19:14\n\nCheck out the usual definition of the limit of a function of a real variable (https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit#Precise_statement).\n\nIf $D$ is any subset of $\\mathbb{R}$, and $f: D \\rightarrow \\mathbb{R}$ is a function, and $c \\in D$ (more generally $c$ can be a limit point of $D$), then we can talk about whether $\\lim\\limits_{x \\to c} f(x)$ exists.\n\nThis stuff about left and right hand limits needing to coincide for the limit to exist only applies when $c \\in D$ and $D$ contains an open interval about $c$.\n\nFor example, suppose $D = [0,5]$, and $f: D \\rightarrow \\mathbb{R}$ is the function $f(x) = x^2$. Then $\\lim\\limits_{x \\to 5} f(x)$ exists and is equal to $25$, because for any $\\epsilon > 0$, there exists a $\\delta > 0$ such that if $x \\in D$ and $|x - 5| < \\delta$, then $|x^2 - 25| < \\epsilon$.\n\nSo in this case, $\\lim\\limits_{x \\to 5^-} f(x)$ and $\\lim\\limits_{x \\to 5} f(x)$ exist, but it makes no sense to talk about $\\lim\\limits_{x \\to 5^+} f(x)$.", "date": "2019-11-22 02:09:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1719162/does-the-limit-exist", "openwebmath_score": 0.9582496285438538, "openwebmath_perplexity": 155.47228138092234, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129093889291, "lm_q2_score": 0.8976952968970955, "lm_q1q2_score": 0.8717780202643545}}
{"url": "https://math.stackexchange.com/questions/2757796/british-maths-olympiad-bmo-2002-round-1-question-3-proof-without-cauchy-schwar", "text": "# British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz?\n\nThe question states:\n\nLet $x,y,z$ be positive real numbers such that $x^2 + y^2 + z^2 = 1$\n\nProve that\n\n$x^2yz + xy^2z + xyz^2 \u2264 1/3$\n\nI have a proof of this relying on the fact that:\n\n$x^2/3 +y^2/3 + z^2/3 \\geq (x+y+z)^3/9$ (A corollary of C-S I believe)\n\nIs there an elementary proof without this fact (or C-S in general)?\n\nYes, we can can get a sum of squares here. We need to prove that $$(x^2+y^2+z^2)^2\\geq3xyz(x+y+z)$$ or $$\\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\\geq0$$ or $$\\sum_{cyc}(2x^4-2x^2y^2+6x^2y^2-6x^2yz)\\geq0$$ or $$\\sum_{cyc}(x^4-2x^2y^2+y^4+3(x^2z^2-2z^2xy+y^2z^2))\\geq0$$ or $$\\sum_{cyc}((x^2-y^2)^2+3z^2(x-y)^2)\\geq0.$$\n\n\u2022 Nice you were able to avoid use of AM-GM too! Out of curiousity how could you tell that $(x^4+2x^2y^2\u22123x^2yz)$ would factorise well? \u2013\u00a0Abe Apr 28 '18 at 20:13\n\u2022 Not OP but cyclic polynomial sums have a lot more freedom than regular polynomials, since any individual term can have each of it's variables cycled. So basically, if you have a cyclic polynomial sum like this, you might just try to force it into something factorable. Notice that the step of multiplying everything by 2 doubles the number of terms and so doubles the amount of options you have to manipulate the polynomial. This is just one way of taking advantage of the symmetry of the problem! \u2013\u00a0Alex Jones Apr 29 '18 at 2:30\n\u2022 General remark: one of the proofs of CS basically rewrites the inequality as a sum of squares $\\sum_{i,j} (a_ib_j - a_jb_i) \\geq 0$, so every application of CS can be rewritten as a sum of squares in this way. \u2013\u00a0Federico Poloni Apr 29 '18 at 14:27\n\u2022 Curious how did you come up with the LHS in the first equation? I wouldn't of thought of it. Is it just from experience/practice? \u2013\u00a0user557278 Apr 29 '18 at 15:43\n\u2022 @hojusaram It's just homogenization. It helps sometimes in polynomial inequalities. \u2013\u00a0Michael Rozenberg Apr 29 '18 at 17:46\n\nBy the RMS-AM-GM inequalities (root-mean square vs. arithmetic vs. geometric means):\n\n\\frac{1}{\\sqrt{3}} = \\sqrt{\\frac{x^2+y^2+z^2}{3}} \\;\\ge\\; \\frac{x+y+z}{3} \\;\\ge\\; \\sqrt[3]{xyz} \\quad\\implies\\quad \\begin{cases}\\begin{align}x+y+z \\,&\\le\\, \\sqrt{3} \\\\[5px] xyz \\,&\\le\\, \\dfrac{1}{3\\sqrt{3}}\\end{align}\\end{cases}\n\nMultiplying the latter gives $\\,xyz(x+y+z) \\le \\dfrac{1}{3}\\,$, which is the inequality to prove. As with all means inequalities, the equality holds iff $\\,x=y=z\\,$.\n\nLet $p(t) = t^3 - at^2 + bt - c$ denote the monic polynomial in $\\mathbb{R}[x]$ with roots $x$, $y$, $z$. We are given that $$1 = x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+xz) = a^2 -2b$$ and asked to show $$ca = xyz(x+y+z) \\le 1/3$$\n\nBy the AM-GM inequality, we know $$\\frac{x+y+z}{3} \\ge \\sqrt[3]{xyz} \\implies \\frac{a^3}{27} \\ge xyz = c$$ so the desired inequality follows if we can prove $a^4 \\le 9 \\iff a^2 \\le 3$.\n\nSince $p(t)$ has three real roots, the derivative $p'(t)$ must have two (with multiplicity) real roots, hence the discriminant of $p'(t)$ is nonnegative:$$p'(t) = 3 t^2 + 2at+b \\implies 4a^2 - 12b \\ge 0 \\implies a^2 \\ge 3b$$ Using the constraint $a^2 - 2b = 1$, we have $$a^2 \\ge 3 \\cdot \\frac{a^2 - 1}{2} \\implies a^2 \\le 3$$ as desired.\n\nRemark: We've seemingly avoided using C-S. However, one way to prove C-S in general is by appealing to a discriminant bound like the above argument. So, we probably haven't avoided C-S as much as covered up our usage with more elementary language.\n\nI found an alternative proof using the GM-AM inequality and a small observation.\n\nFirstly, using the AM-GM inequality for x,y,z we get\n\n$x^2yz + xy^2z + xyz^2 = (x+y+z)xyz = (x+y+z)\\left(\\sqrt[3]{xyz}\\right)^3 \\leq \\frac{(x+y+z)^4}{27} = \\frac{1}{3}\\frac{(x+y+z)^4}{9}$\n\nNow, notice that: $(x+y+z)^4 \\leq 9(x^2 + y^2 + z^2)^2$. This is because we can square-root both sides to get:\n\n$(x+y+z)^2 \\leq 3(x^2 + y^2 + z^2) \\Leftrightarrow 0 \\leq (x-y)^2 + (y-z)^2 + (z-x)^2$\n\nHence, continuing the first line of the proof we get\n\n$\\frac{1}{3}\\frac{(x+y+z)^4}{9} \\leq \\frac{1}{3}(x^2 + y^2 + z^2)^2 = \\frac{1}{3}$\n\nand we are done!\n\nEdit: This line of reasoning nicely shows that the equality holds iff $x=y=z$.\n\nHere is my approach.\n\nUsing AM-GM,\n\n\\begin{align} x^2+y^2+z^2 &\\ge 3(xyz)^{2/3} \\\\ \\implies 1 &\\ge 3(xyz)^{2/3} \\\\ \\implies (xyz)^{2/3} &\\le \\dfrac{1}{3} \\\\ \\implies xyz \\le \\dfrac{1}{3\\sqrt{3}} \\tag 1 \\end{align}\n\nAgain, using AM-GM,\n\n\\begin{align} \\dfrac{x^3+y^3+z^3}{3} &\\ge 3xyz \\\\ \\implies x^3+y^3+z^3-3xyz &\\ge 0 \\\\ \\implies (x+y+z)(x^2+y^2+z^2 -xy-yz-zx) &\\ge 0\\end{align}\\tag*{}\n\nSince $$x,y,z \\in \\mathbb R^+$$, $$x+y+z \\ge 0$$.\n\n\\begin{align}\\therefore \\, x^2+y^2+z^2 -xy-yz-zx &\\ge 0 \\\\ \\implies -x^2-y^2-z^2 +xy+yz+zx &\\le 0 \\\\ \\implies -2x^2-2y^2-2z^2 +2xy+2yz+2zx &\\le 0 \\\\ \\implies x^2+y^2+z^2 +2xy+2yz+2zx &\\le 3(x^2+y^2+z^2) \\\\ \\implies (x+y+z)^2 &\\le 3 \\\\ \\implies x+y+z &\\le \\sqrt{3} \\tag 2 \\end{align}\n\nMultiplying $$(1)$$ and $$(2)$$, we can get the desired result.", "date": "2021-06-14 01:50:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2757796/british-maths-olympiad-bmo-2002-round-1-question-3-proof-without-cauchy-schwar", "openwebmath_score": 0.9992751479148865, "openwebmath_perplexity": 700.3452853247608, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759660443166, "lm_q2_score": 0.8887587927558434, "lm_q1q2_score": 0.8717621394247684}}
{"url": "https://jadeite.co.in/journal/fbu0jlf.php?page=67a458-find-the-slope-of-the-line-passing-through-the-points-calculator", "text": "\u00a9 2019 Coolmath.com LLC.\n\nIf the y-values are not changing as x increases, it is indicated as the line will have a slope of 0. Our slope finder is 100% free and feel-hassle free to account this calculator for slope. Many real-world situations can be modeled by linear equations, so the slope calculator can be of great importance of solving such problems. We have the final answer as. First calculator finds the line equation in slope-intercept form, that is,. Also, our slope intercept calculator will also show you the same answer for these given parameters. Slope of a Straight Line Calculator getcalc.com's line slope calculator is an online geometry tool to calculate straight line slope based on the points (x 1, y 1) & (x 2, y 2) on the plane, in both US customary & metric (SI) units. Read on! You can find slope of a line by comparing any 2 points on the line. A line is increasing, and goes upwards from left to right when m > 0, A line is decreasing, and goes downwards from left to right when m < 0, A line has a constant slope, and is horizontal when m = 0. The slope $m$ of a line $y=mx+b$ can be defined also as the rise divided by the run. As mentioned earlier a vertical line has an undefined slope.\n\nWhat is the equation? We are basically measuring the amount of change of the y-coordinate, often known as the rise, divided by the change of the x-coordinate, known the the run. You also probably sleep under a slope, a roof that is. Well, we can easily calculate \u2018b\u2019 from this equation: Now, let\u2019s plug-in the values into the above equation: Very next, we plug-in the value of \u2018b\u2019 and the slope into the given equation: Also, you can use the above point slope calculator to perform instant calculations instead of sticking to these manual calculation steps! The slope is an important concept in mathematics that is usually used in basic or advanced graphing like linear regression; the slope is said to be one of the primary numbers in a linear formula. where (x1 , y1) and (x2 , y2) are the points. If you want to calculate slope, all you need to divide the different of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points. Also, try an online slope intercept calculator that uses two points to find the equation of a line in the two-dimensional Cartesian coordinate plane. Slope, sometimes referred to as gradient in mathematics, is a number that measures the steepness and direction of a line, or a section of a line connecting two points, and is usually denoted by m. Generally, a line's steepness is measured by the absolute value of its slope, m. The larger the value is, the steeper the line. Please try again using a different payment method. It accepts inputs of two known points, or one known point and the slope. Congratulate yourself on your achievement. The formula becomes increasingly useful as the coordinates take on larger values or decimal values. Point A (x_A, y_A) = (3, 2) Check your result using the slope calculator. 1/4\u2033 per foot pitch equals to 2% (percent), and remember that it is not expressed as 2 degrees. Yes, slope point calculator helps you in finding the slope and shows you the slope graph corresponding to the given points by using the simple slope equation. Objective :\n\nYou can try the simple slope intercept form calculator to find the slope intercept form for the given points. Thanks for the feedback. So, let\u2019s start with the simple definition of \u2018slope definition.\u2019. A vertical line has an undefined slope, since it would result in a fraction with 0 as the denominator. In this equation, the slope is expressed as \u22124 as this equation is in slope-intercept form that is y=mx+c , where m is the slope. Rate of change is particularly useful if you want to predict the future of previous value of something, as, by changing the x variable, the corresponding y value will be present (and vice versa). This is the same as a gradient of 1/10, and an angle of 5.71\u00b0 is formed between the line and the x-axis. For all the lines where \u2018y\u2019 equals a constant and there is no \u2018x\u2019, the slope is said to be as 0. Any two coordinates will suffice. Point B (x_B, y_B) = (7, 10) Step 1: Enter the point and slope that you want to find the equation for into the editor. You ought to re-work the equation until \u2018y\u2019 is isolated on one side. The formula to determine the distance (D) between 2 different points is: $$Distance (d) = \\sqrt {(x\u2082 \u2013 x\u2081)^2 + (y\u2082 \u2013 y\u2081)^2 }$$. But, more importantly, if you ever want to know how something changes with time, you will end up plotting a graph with a slope. This will result in a zero in the numerator of the slope formula. All Rights Reserved. If the y-values are decreasing, it referred to as the line has a negative slope. It's an online Geometry tool requires two point in the two-dimensional Cartesian coordinate plane and determines the slope or gradient of the line these points. Rise means how high or low we have to move to arrive from the point on the left to the point on the right, so we change the value of $y$.", "date": "2022-05-16 08:17:32", "meta": {"domain": "co.in", "url": "https://jadeite.co.in/journal/fbu0jlf.php?page=67a458-find-the-slope-of-the-line-passing-through-the-points-calculator", "openwebmath_score": 0.7358871698379517, "openwebmath_perplexity": 365.4585998094539, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.984336349450327, "lm_q2_score": 0.8856314783461302, "lm_q1q2_score": 0.8717592563535261}}
{"url": "https://math.stackexchange.com/questions/2130264/sum-of-random-decreasing-numbers-between-0-and-1-does-it-converge/2130701", "text": "Sum of random decreasing numbers between 0 and 1: does it converge??\n\nLet's define a sequence of numbers between 0 and 1. The first term, $r_1$ will be chosen uniformly randomly from $(0, 1)$, but now we iterate this process choosing $r_2$ from $(0, r_1)$, and so on, so $r_3\\in(0, r_2)$, $r_4\\in(0, r_3)$... The set of all possible sequences generated this way contains the sequence of the reciprocals of all natural numbers, which sum diverges; but it also contains all geometric sequences in which all terms are less than 1, and they all have convergent sums. The question is: does $\\sum_{n=1}^{\\infty} r_n$ converge in general? (I think this is called almost sure convergence?) If so, what is the distribution of the limits of all convergent series from this family?\n\n\u2022 Not sure if my question itself makes sense: But how about a modified question where r1 is chosen uniformly randomly from (-1,+1) and we iterate by choosing r2 from (-1 x abs(r1),+1 x abs(r1)) Would that series converge too? Feb 5 '17 at 17:24\n\u2022 Your second series almost surely converges absolutely (i. e. $|r_1| +$|r_2|$+ \\cdots$ converges) if and only if your first series almost surely converges. Combining that with Byron Schmuland's answer, the modified series almost surely converges absolutely, and so a fortiori almost surely converges. Feb 5 '17 at 18:11\n\n4 Answers\n\nLet $(u_i)$ be a sequence of i.i.d. uniform(0,1) random variables. Then the sum you are interested in can be expressed as $$S_n=u_1+u_1u_2+u_1u_2u_3+\\cdots +u_1u_2u_3\\cdots u_n.$$ The sequence $(S_n)$ is non-decreasing and certainly converges, possibly to $+\\infty$.\n\nOn the other hand, taking expectations gives $$E(S_n)={1\\over 2}+{1\\over 2^2}+{1\\over 2^3}+\\cdots +{1\\over 2^n},$$ so $\\lim_n E(S_n)=1.$ Now by Fatou's lemma, $$E(S_\\infty)\\leq \\liminf_n E(S_n)=1,$$ so that $S_\\infty$ has finite expectation and so is finite almost surely.\n\n\u2022 @ByronSchmuland $S_\\infty$ will have the same distribution as $X(1+S_\\infty)$ where $X$ is uniformly distributed from 0 to 1. Since that is a recursive expression it does not directly tell you the distribution, but it can be used to sanity check any distribution you came up with. Feb 5 '17 at 18:35\n\u2022 ...And the distribution of $S_\\infty$ is uniquely determined by its Laplace transform $$L(s)=E(e^{-sS_\\infty})$$ which is the unique solution of the differential equation $$\\left(sL(s)\\right)'=e^{-s}L(s)\\qquad L(0)=1$$ solved by $$L(s)=\\exp\\left(-\\int_0^s\\frac{1-e^{-t}}tdt\\right)=\\exp\\left(-\\int_0^se^{-t}\\ln\\left(\\frac{s}t\\right)dt\\right)$$\n\u2013\u00a0Did\nFeb 5 '17 at 19:45\n\u2022 @polfosol if your previously drawn number is $a$, a uniformly distributed number between $a$ and $0$ is $a \\cdot u$, where $u$ is uniformly distributed between $0$ and $1$. The rest follows by recurrence ($u_1=u$, $u_2 = u_1 \\cdot u$...) Feb 6 '17 at 16:21\n\u2022 Can someone tell me what i.i.d. uniformity means? Feb 7 '17 at 13:28\n\u2022 @zakoda i.i.d. means \"independent identical distribution\". Uniform is specifically what kind of distribution (a uniform one where the probability density function is constant over the whole range).\n\u2013\u00a0Kyle\nFeb 7 '17 at 16:18\n\nThe probability $f(x)$ that the result is $\\in(x,x+dx)$ is given by $$f(x) = \\exp(-\\gamma)\\rho(x)$$ where $\\rho$ is the Dickman function as @Hurkyl pointed out below. This follows from the the delay differential equation for $f$, $$f^\\prime(x) = -\\frac{f(x-1)}{x}$$ with the conditions $$f(x) = f(1) \\;\\rm{for}\\; 0\\le x \\le1 \\;\\rm{and}$$ $$\\int\\limits_0^\\infty f(x) = 1.$$ Derivation follows\n\nFrom the other answers, it looks like the probability is flat for the results less than 1. Let us prove this first.\n\nDefine $P(x,y)$ to be the probability that the final result lies in $(x,x+dx)$ if the first random number is chosen from the range $[0,y]$. What we want to find is $f(x) = P(x,1)$.\n\nNote that if the random range is changed to $[0,ay]$ the probability distribution gets stretched horizontally by $a$ (which means it has to compress vertically by $a$ as well). Hence $$P(x,y) = aP(ax,ay).$$\n\nWe will use this to find $f(x)$ for $x<1$.\n\nNote that if the first number chosen is greater than x we can never get a sum less than or equal to x. Hence $f(x)$ is equal to the probability that the first number chosen is less than or equal to $x$ multiplied by the probability for the random range $[0,x]$. That is, $$f(x) = P(x,1) = p(r_1<x)P(x,x)$$\n\nBut $p(r_1<x)$ is just $x$ and $P(x,x) = \\frac{1}{x}P(1,1)$ as found above. Hence $$f(x) = f(1).$$\n\nThe probability that the result is $x$ is constant for $x<1$.\n\nUsing this, we can now iteratively build up the probabilities for $x>1$ in terms of $f(1)$.\n\nFirst, note that when $x>1$ we have $$f(x) = P(x,1) = \\int\\limits_0^1 P(x-z,z) dz$$ We apply the compression again to obtain $$f(x) = \\int\\limits_0^1 \\frac{1}{z} f(\\frac{x}{z}-1) dz$$ Setting $\\frac{x}{z}-1=t$, we get $$f(x) = \\int\\limits_{x-1}^\\infty \\frac{f(t)}{t+1} dt$$ This gives us the differential equation $$\\frac{df(x)}{dx} = -\\frac{f(x-1)}{x}$$ Since we know that $f(x)$ is a constant for $x<1$, this is enough to solve the differential equation numerically for $x>1$, modulo the constant (which can be retrieved by integration in the end). Unfortunately, the solution is essentially piecewise from $n$ to $n+1$ and it is impossible to find a single function that works everywhere.\n\nFor example when $x\\in[1,2]$, $$f(x) = f(1) \\left[1-\\log(x)\\right]$$\n\nBut the expression gets really ugly even for $x \\in[2,3]$, requiring the logarithmic integral function $\\rm{Li}$.\n\nFinally, as a sanity check, let us compare the random simulation results with $f(x)$ found using numerical integration. The probabilities have been normalised so that $f(0) = 1$.\n\nThe match is near perfect. In particular, note how the analytical formula matches the numerical one exactly in the range $[1,2]$.\n\nThough we don't have a general analytic expression for $f(x)$, the differential equation can be used to show that the expectation value of $x$ is 1.\n\nFinally, note that the delay differential equation above is the same as that of the Dickman function $\\rho(x)$ and hence $f(x) = c \\rho(x)$. Its properties have been studied. For example the Laplace transform of the Dickman function is given by $$\\mathcal L \\rho(s) = \\exp\\left[\\gamma-\\rm{Ein}(s)\\right].$$ This gives $$\\int_0^\\infty \\rho(x) dx = \\exp(\\gamma).$$ Since we want $\\int_0^\\infty f(x) dx = 1,$ we obtain $$f(1) = \\exp(-\\gamma) \\rho(1) = \\exp(-\\gamma) \\approx 0.56145\\ldots$$ That is, $$f(x) = \\exp(-\\gamma) \\rho(x).$$ This completes the description of $f$.\n\n\u2022 Can you explain your reasoning to multiply the probability that the first pick is less than $x$ by the probability that you land between $x$ and $x+dx$ starting at $x$? I don't understand that. Wouldn't the probability depend on what the first pick is, necessitating an integral like the one you use later on?\n\u2013\u00a0user142299\nFeb 6 '17 at 0:32\n\u2022 $$P(x,x) = 1/x \\int_0^x P(x-z,z) dz$$ $$P(x(<1),1)=1/1\\int_0^xP(x-z,z)dz$$ $$P(x(<1),1) =xP(x,x)=P(1,1)$$ Feb 6 '17 at 6:16\n\u2022 This is (proportional to) the Dickman function!\n\u2013\u00a0user14972\nFeb 6 '17 at 7:54\n\u2022 @Hurkyl Wow, that's cool. Just added it to the answer. Feb 6 '17 at 8:21\n\nJust to confirm the simulation by @curious_cat, here is mine:\n\nIt's a histogram, but I drew it as a line chart because the bin sizes were quite small ($0.05$ in length, with 10 million trials of 5 iterations).\n\nNote: vertical axis is frequency, horizontal axis is sum after 5 iterations. I found a mean of approximately $0.95$.\n\n\u2022 The shape of that graph does not look like I would have expected it to. It looks flat from 0 to 1 (I assume the variance seen there is entirely due to only having sampled a finite number of times). I would have expected it to start decreasing right away. And it looks like the derivative is discontinuous at 1 and nowhere else. I would have expected the derivative to be continuous. And given that the distribution can be expressed recursively I can't quite see how a single discontinuity in the derivative wouldn't cause an infinite number of discontinuities. Feb 5 '17 at 18:44\n\u2022 That doesn't mean I think you are wrong. Just that I am surprised. I will now try to simulate it myself to see if I can reproduce your result. Feb 5 '17 at 18:45\n\u2022 I have proved why it has to be constant for $x<1$, please check my answer. Feb 5 '17 at 19:15\n\u2022 I have done a simulation myself which produced a graph similar to yours except from an anomaly at 0 due to using discrete random numbers in the simulation. So it looks like the shape of the graph is pretty much correct. Feb 5 '17 at 19:36\n\u2022 For an infinite sum, the mean should be $1$ since it satisfies $\\mu=\\frac12 +\\frac12\\mu$ where $\\frac12$ is the mean of a uniform random variable on $[0,1]$. For five terms in the sum it should be $\\frac{2^5-1}{2^5}=\\frac{31}{32}=0.96875$ Feb 9 '17 at 14:13\n\nI just ran a quick simulation and I get a mean sum of one (standard deviation of 0.7)\n\nCaveat: Not sure I coded it all right! Especially since I didn't test convergence.\n\n\u2022 How did you run a quick simulation of an infinite sequence?\n\u2013\u00a0JiK\nFeb 6 '17 at 12:24\n\u2022 @JiK Right. I didn't. Feb 6 '17 at 12:39\n\u2022 Chuck Norris could have run a quick simulation of an infinite sequence. After all, he counted to infinity. Twice. Feb 8 '17 at 5:49\n\u2022 For what it's worth, I did the simulation in excel and let it cook for half a day. My histogram was essentially the same as posted above: flat from 0 to 1, then an exponential like decay. This leads to another question involving histograms, which I shall submit to the community... Recently the Israel prime minister dismissed his security staff. He was meeting Chuck Norris, and knew the Norris could personally handle any security problem. Feb 8 '17 at 19:12", "date": "2022-01-25 09:52:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2130264/sum-of-random-decreasing-numbers-between-0-and-1-does-it-converge/2130701", "openwebmath_score": 0.927476704120636, "openwebmath_perplexity": 172.99879103955675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353585837, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8717592496163536}}
{"url": "https://mathhelpboards.com/threads/polynomial-roots.7362/", "text": "# [SOLVED]Polynomial roots\n\n#### wishmaster\n\n##### Active member\nI have to find all solutions for X when:\nx4-2x3-25x2+50x\n\nI have done it so,but im not sure if this is ok:\n\nx(x3-2x2-25x+50)\n\n= x(x2(x-2)-25(x+2)\n= x(x2-25)(x-2)\n=x(x-5)(x+5)(x-2)\n\nNow i see that root/zeroes are +5,-5 and 2. I know that this polynomial has another zero that is 0,but how do i know that? Because x is in the front? Or did i make a mistake and should such problems deal another way? Thank you for all replies!\n\n#### MarkFL\n\nStaff member\nI have to find all solutions for X when:\nx4-2x3-25x2+50x\n\nI have done it so,but im not sure if this is ok:\n\nx(x3-2x2-25x+50)\n\n= x(x2(x-2)-25(x-2))\n= x(x2-25)(x-2)\n=x(x-5)(x+5)(x-2)\n\nNow i see that root/zeroes are +5,-5 and 2. I know that this polynomial has another zero that is 0,but how do i know that? Because x is in the front? Or did i make a mistake and should such problems deal another way? Thank you for all replies!\nYou had a minor typo in your working which did not affect the outcome. I have highlighted it in red, but is is obvious from your subsequent steps that you intended to write this. The four roots of the polynomial are found by equating each of the four factors to zero, including the factor $x$ in front, and solving for $x$. So you are correct that the four roots are (in ascending order):\n\n$$\\displaystyle x=-5,\\,0,\\,2,\\,5$$\n\nGreat job!\n\n#### wishmaster\n\n##### Active member\nYou had a minor typo in your working which did not affect the outcome. I have highlighted it in red, but is is obvious from your subsequent steps that you intended to write this. The four roots of the polynomial are found by equating each of the four factors to zero, including the factor $x$ in front, and solving for $x$. So you are correct that the four roots are (in ascending order):\n\n$$\\displaystyle x=-5,\\,0,\\,2,\\,5$$\n\nGreat job!\nThank you! Sorry,that was a type mismatch.\none question,when x stays alone in front of other terms,then this root is always zero?\n\n#### MarkFL\n\nStaff member\nThank you! Sorry,that was a type mismatch.\none question,when x stays alone in front of other terms,then this root is always zero?\nYes, equating that factor to zero, we get:\n\n$$\\displaystyle x=0$$\n\nIt is already solved for $x$, and it tells us that $x=0$ is a root.\n\n#### Petrus\n\n##### Well-known member\nThank you! Sorry,that was a type mismatch.\none question,when x stays alone in front of other terms,then this root is always zero?\nAnother way to se it if we want to divide by x Then we get 2 case\ncase 1 $$\\displaystyle x=0$$\nput $$\\displaystyle x=0$$ to the equation and we see that $$\\displaystyle 0=0$$ hence 0 is a root\ncase 2 $$\\displaystyle x \\neq 0$$\nwe know can divide by x (we get third degree polynom) and now it's just to use rational root Theorem and long polynom division and Then it becomes second grade polynom which is easy to solve\n\nRegards,\n$$\\displaystyle |\\pi\\rangle$$\n\n#### topsquark\n\n##### Well-known member\nMHB Math Helper\nx4-2x3-25x2+50x\n\n=x(x-5)(x+5)(x-2)\nTo be specific we have that if ab = 0 then either a = 0 or b = 0. In this case that means when\nx(x - 5)(x + 5)(x - 2) = 0\n\nwe get solutions when:\nx = 0, or x - 5 = 0, or x + 5 = 0, or x - 2 = 0\n\nThe solutions to these four equations are your roots.\n\n-Dan", "date": "2021-10-17 11:37:46", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/polynomial-roots.7362/", "openwebmath_score": 0.6913750767707825, "openwebmath_perplexity": 648.2757875530734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9843363485313247, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8717592451399184}}
{"url": "http://www.e-worldf.com/q3g9ur/891627-angle-conversion-degree-to-radian", "text": "How to convert degrees to radians [degree (\u00b0) to rad]:. An angle's measurement in radians is numerically equal to the length of a corresponding arc of a unit circle. Definition of radian: a radian is the measure of an angle that, when drawn as a central angle of a circle, intercepts an arc whose length is equal to the length of the radius of the circle. Use this page to learn how to convert between degrees and radians. How many radians ( rad ) are in 1 degree ( 1 \u00b0 - deg )? In radians, one complete counterclockwise revolution is 2 \u03c0 and in degrees, one complete counterclockwise revolution is 360 \u00b0 .So, degree measure and radian measure are related by the equations You are currently converting angle units from degree to radian 1 \u00b0 = 0.017453292519943 rad. I got angle in degree as -415 degrees, i have converted it into radian as . Radian to degree conversion derivation. The Angle Conversion block port labels change based on the input and output units selected from the Initial unit and the Final unit lists. Following the value, access the Math menu (press) and select \"2: r\" from the angle sub-menu (press ), then enter the conversion function \"DD\" or \"DMS\" . This definition is much easier understood by looking at the demonstration immediately below. The circumference of a circle is 2\u03c0r (r \u2026 The Angle Conversion block computes the conversion factor from specified input angle units to specified output angle units and applies the conversion factor to the input signal.. A radian measures approx. One radian is equal to 180/\u03c0 (~57.296) degrees. The SI derived unit for angle is the radian. DMS means Degree, Minutes and Seconds. Between rad and \u00b0 measurements conversion chart page. Then multiply the amount of Minute you want to convert to Radian, use the chart below to guide you. angle conversion : degree, radian, grade, steradian. A solid angle is similar in 3 dimensions to a plane angle in 2 dimensions. 1 degree is equal to 0.017453292519943 radian. 0.017453292519943 rad Conversion base : 1 \u00b0 = 0.017453292519943 rad. An angle's measurement in radians is numerically equal to the length of a corresponding arc of a unit circle, so one radian is just under 57.3 degrees (when the arc length is \u2026 To convert between radian and degree and for more information, please visit radian to degree. \u03b8 rad = \u03b8 \u00b0 \u00d7\u03c0 \u00f7 180. Hence, 1 radian = $\\frac{180^{0}}{\\pi}$= 57.2958\u00b0 If you want to convert a degree or an angle to radians, simply multiply the angle by and then divide it by 180. float degreeValue = -415 float radianValue = degreeValue * pi / 180.0; here i got as -0.7(round off) how to convert again into degree to get same value of angle in degrees. angle units. How many radians in 64 degrees: If \u03b8 \u00b0 = 64 then \u03b8 rad = 1.1170107212764 rad. online angle conversions, degree, grad, radian, circle, mil, minute, octant, quadrant, revolution 1\u02da = 1 * \u03c0/(180\u02da) \u2248 0.0175 rad 2. It is not an SI unit, however, it is accepted for use with SI. Radian is converted to turns by dividing the number of radians by 2\u03c0. This tool converts gons to radians (gon to rad) and vice versa. Step 1: Plugg the angle value, in degrees, in the formula above: radian measure = (90 \u00d7 \u03c0)/180. There are about 6.28318 radians in a \u2026 Conversion Formulas \u2212 degree = radian * (180/pi) where, pi=3.14 or 22/7. A degree (or in full degree of arc), usually symbolized by the symbol \u00b0, is a measurement of plane angles, or of a location along a great circle of a sphere (such as the Earth or the celestial sphere), representing 1/360 of a full rotation. MINUTE TO RADIAN (' TO rad) FORMULA . Degree : A degree, a degree of arc or arcdegree is a measurement of plane angle, on behalf of 1/360 of a full rotation. To convert any value in degrees to radians, just multiply the value in degrees by the conversion factor 0.017453292519943.So, 90 degrees times 0.017453292519943 is equal to 1.571 radians. 1 degree ( \u00b0 - deg ) = 0.017 radians ( rad ). Take a look at the table below of the angles and their conversion to radians. One degree is equal to \u03c0/180 radians. A radian is the measurement of angle equal to the start to the end of an arc divided by the radius of the circle or arc. A full circle is just over 6 radians (Roughly 6.28). Conversion Radian to Degree. How many radians in a degree: If \u03b8 \u00b0 = 1 then \u03b8 rad = 0.017453292519943 rad. The symbol for degree is \u00b0. 2. Convert 1 rad into degree and radians to \u00b0. Conversion of units between 1 Degree (Of Arc) and Radian (Si Unit) (1 \u00b0 and rad) is the conversion between different units of measurement, in this case it's 1 Degree (Of Arc) and Radian (Si Unit), for the same quantity, typically through multiplicative conversion factors (\u00b0 and rad). Angle in radians= angles in degrees * \u03c0/(180\u02da) For example: 1. In mathematics and physics, the radian is a unit of angle \u2026 Convert 1 \u00b0 into radian and degrees to rad. The other way around, how many degrees - \u00b0 are in one radian - rad unit? Step 1: Plugg the angle value, in degrees, in the formula above: radian measure = (330 \u00d7 \u03c0)/180. One radian is just under 57.3 degrees. \u203a\u203a Definition: Radian. A degree, also a degree of arc, arc degree, usually denoted by the degree symbol \u00b0, is a measurement of plane angle, representing 1\u2044360 of a full rotation; one degree is equivalent to \u03c0/180 radians. Please apply Unit Conversion function by clicking Kutools > Content Converter > Unit Conversion. Degree: A degree, usually denoted by \u00b0 (the degree symbol), is a measurement of a plane angle, defined so that a full rotation is 360 degrees. Get it Now. Radian is the ratio between the length of an arc and its radius. Angle; Description: The radian is an SI derived unit of angle, commonly used in maths and engineering. This on the web one-way conversion tool converts angle (plane angles) units from degrees ( \u00b0 - deg ) into radians ( rad ) instantly online. Calculate from angle into other angle unit measures. A degree (in full, a degree of arc, arc degree, or arcdegree), usually denoted by \u00b0 (the degree symbol), is a measurement of a plane angle, defined so that a full rotation is 360 degrees.. Using the conversion formula above, you will get: Value in degree = 4.3 \u00d7 57.295779513082 = 246.4 degrees There are formulas that can be used for this conversion. Note that rounding errors may occur, so always check the results. Convert angle between degrees and radians with Kutools for Excel. Calculate from angle into other angle unit measures. Step 2: Rearrange the terms: radian measure = \u03c0 \u00d7 330/180. Conversion base : 1 g = 0.9 \u00b0 Switch units Starting unit. 56.296 degrees (when the arc length is equal to the radius). The radian (symbol:rad) is the standard unit of angular measure, used in many areas of mathematics. Convert Angle from Degree to Gradient or to different units such as Radian, Degree, Gradient, Minutes, Seconds, Point, Mil, Circle, 1/16 Circle, 1/10 Circle, 1/8 Circle, 1/6 Circle, 1/4 Circle, 1/2 Circle and more units Degree and Gradient are the units to measure Angle, where 1 Degree = 1.1111111 Gradient A radian is defined by an arc of a circle. Between \u00b0 and rad measurements conversion chart page. radian measure = \u03c0 \u00d7 330/180 = Step 3: Reduce or simplify the fraction of \u03c0 if necessary Calculating the gcd of 330 and 180 [gcd(330,180)], we've found that it equals 30. degree \u00b0 radian . Degree to Radian Measure The measure of an angle is determined by the amount of rotation from the initial side to the terminal side. radian (rad) degree (\u00b0) minute of arc (') second of arc \u2026 History/origin: Measuring angles in terms of arc length has been used by mathematicians since as early as the year 1400. Radian is the standard unit for measuring angles whereas the complete angle of circle is divided into 360 degree. The metric system angle unit is radian (rad) which is the angle subtended by an arc of a circle that has the same length as the circle's radius. You are currently converting angle units from degree to grad 1 \u00b0 = 1.1111111111111 g. degree \u00b0 grad . The \u03c0 has infinite decimal digits. Suppose you want to convert 4.3 radian into degrees. Steps. It is also known as grad. Learn how to convert angles from DMS forms to decimal forms. [2] 1 radian is equal to 180/\u03c0, or about 57.29578\u00b0. How much of angle (plane angles) from degrees to radians, \u00b0 - deg to rad? \u203a\u203a Quick conversion chart of degree to radian. Diferent angle units conversion from radian to degrees. Therefore 1 Radian is equal to (180/\u03c0) degrees. Example Conversion Gon to Radian. Knowing that 1 radian = 57.29578 degrees we can now find the conversion \u2026 Type in your own numbers in the form to convert the units! Note that there are rounding errors in these values. 1. For all calculations the \u03c0 considered to be equal to 3.14159265358979323846. radian measure = \u03c0 \u00d7 90/180 = Step 3: Reduce or simplify the fraction of \u03c0 if necessary Calculating the gcd of \u2026 To convert between Minute and Radian you have to do the following: First divide (Math.PI/(180*60)) / 1 = 0.00029089 . The gon is a unit of plane angle, equivalent to 1/400 of a cycle. Definition: The angle made by taking the radius of a circle and wrapping it along the circle's edge. Diferent angle units conversion from degree to radians. Origin: 23\u02da= 23 * \u03c0/(180\u02da) \u2248 0.4014 rad. If your TI-89 is in degree mode, you can still indicate that a value is in radians. Also, radian is the smaller value as 1 degree = 180 radians. Note: Radian is a metric unit of angle. Kutools for Excel includes more than 300 handy Excel tools. Steps. Radian : The radian is the standard unit of angle which is defined as the ratio between the length of an arc and its radius. 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{"url": "https://math.stackexchange.com/questions/2939291/find-the-smallest-possible-checksum", "text": "# Find the smallest possible checksum\n\nWhat is the smallest possible checksum $$C_{min}$$ of the sum of two three-digit numbers $$N_1$$ and $$N_2$$ that are formed from the six given digits 2, 3, 4, 5, 7 and 9 (each digit shall be used only once). The checksum of a number is defined as the sum of its digits. What are the sums that have this smallest possible checksum $$C_{min}$$?\n\nHere my thoughts: I believe that the smallest checksum is 3. I obtained this by simple reasoning: With $$3 + 7 = 10$$ there is 0 in the sum and a carry-over \"1\". Then, with $$4 + 5 = 9$$ and the carry-over \"1\" we obtain another \"0\" in the sum, creating another carry-over \"1\". Finally, $$2 + 9 = 11$$, and with the carry-over it is \"12\". Thus, the checksum is $$1 + 2 + 0 + 0 = 3.$$ But is it right?\n\nExamples:\n\n$$243 + 957 = 1200,$$ checksum = 3\n\n$$423 + 597 = 1020,$$ checksum = 3\n\nIf the answer is at all correct, how can I prove that it is? And then, how would I be able to find all the sums of numbers? My approach: Permute the given checksum, and check if I can find N1 and N2 that have this result.\n\nThere must be a more systematic or rigorous approach, I would believe. Perhaps someone can guide me on this. Thank you.\n\n\u2022 You don't give a rigorous definition of the checksum that you're using. Do you mean \"find the smallest digit sum of $a + b$ when $a, b$ are two three-digit numbers formed from the six given digits $2, 3, 4, 5, 7, 9$?\" \u2013\u00a0orlp Oct 2 '18 at 12:20\n\u2022 Sorry, I mean the sum over the digits. Will edit my post.If the numbers N1 and N2 are 539 and 247, their sum is 786 and the checksum of the sum is 7 + 8 + 6 = 21 \u2013\u00a0Parzifal Oct 2 '18 at 12:23\n\u2022 I'm not posting this as an answer as it's a mathematics site, but there is a systematic way: check each case with a computer. Since $6! = 720$ this happens in the blink of an eye and indeed confirms $3$ is the smallest possible digit sum. \u2013\u00a0orlp Oct 2 '18 at 12:26\n\u2022 Are you familiar with the notion of casting out nines? \u2013\u00a0Barry Cipra Oct 2 '18 at 12:26\n\u2022 No - what do you mean by casting out nines? \u2013\u00a0Parzifal Oct 2 '18 at 12:29\n\nLet's start with a brief explanation of \"casting out nines\" as it's needed for the problem at hand. If $$ABCD$$ is the (at most) four-digit sum of a pair of three-digit numbers, $$abc+def$$, then, since $$1000\\equiv100\\equiv10\\equiv1$$ mod $$9$$, we have\n\n\\begin{align} A+B+C+D &\\equiv(1000A+100B+10C+D)\\\\ &=(100a+10b+c)+(100d+10e+f)\\\\ &\\equiv a+b+c+d+e+f\\mod 9 \\end{align}\n\nso, if $$\\{a,b,c,d,e,f\\}=\\{2,3,4,5,7,9\\}$$, we have\n\n$$A+B+C+D\\equiv2+3+4+5+7+9=(2+7)+3+(4+5)+9\\equiv3\\mod 9$$\n\nwhich means that the digit sum for any pair $$abc+def$$ belongs to $$\\{3,12,21,\\ldots\\}$$. The OP has already found one example with digit sum $$3$$, namely $$243+957=1200$$, so we can conclude that the smallest possible digit sum is $$3$$. The remaining question, how many different pairs give digit sum $$A+B+C+D=3$$?\n\nNote that\n\n$$abc+def=dbc+aef=aec+dbf=dec+abf$$\n\nso it suffices to count the number of solutions with $$a\\lt d$$, $$b\\lt e$$ and $$c\\lt f$$ and then multiply by $$4$$ (or by $$8$$, if you want to distinguish $$(abc,def)$$ from $$(def,abc)$$). Now the only realistic possibilities for $$ABCD$$ are $$1200$$, $$1020$$, $$1002$$, $$1110$$, $$1101$$, and $$1011$$. (That is, $$500\\lt abc+def\\lt2000$$, so we must have $$A=1$$ since $$A=0$$ would imply $$B\\ge5$$.) Let's consider these possibilities according to what's in the ones place.\n\nIf $$D=0$$ (i.e., if $$ABCD=1200$$, $$1020$$, or $$1110$$), we can only have $$c=3$$ and $$f=7$$, which will carry a $$1$$ into the tens place. That makes $$C=1$$ impossible, but it gives $$1+2+9=11$$ and $$1+4+5=10$$ as possibilities for $$C=2$$ and $$C=0$$. Indeed, we get x two solutions (with $$a\\lt d$$, etc.), namely $$423+597=1020$$ and $$243+957=1200$$.\n\nIf $$D=1$$ (i.e., if $$ABCD=1101$$ or $$1011$$), we can have $$c+f=2+9=11$$ or $$c+f=4+7=11$$. In either case the carried $$1$$ from $$c+f$$ means we need $$b+d=9$$ or $$10$$ in order to get $$C=0$$ or $$1$$. For $$c+f=2+9$$, both values of $$C$$ are attainable, each in only one way: $$342+759=1101$$ and $$432+579=1011$$. For $$c+f=4+7$$, neither value of $$C$$ is attainable, since the only digits that sum to $$9$$ are $$2+7$$ and $$4+5$$ and, as already remarked, the only digits that sum to $$10$$ is $$3+7$$. Thus we get just two solutions with $$D=1$$, namely $$342+759=1101$$ and $$432+579=1011$$.\n\nFinally, if $$D=2$$ (i.e., $$ABCD=1002$$), we can only have $$c+f=3+9$$ or $$c+f=5+7$$. In either case we'll need $$b+e=9$$, which is possible only as $$2+7$$ or $$4+5$$. If $$c+f=3+9$$, either of these is possible, while neither is possible if $$c+f=5+7$$. So again we get just two solutions, namely $$243+759=1002$$ and $$423+579=1002$$.\n\nAltogether we get six solutions with $$a\\lt d$$, $$b\\lt e$$ and $$c\\lt f$$:\n\n\\begin{align} 423+579&=1020\\\\ 243+957&=1200\\\\ 342+759&=1101\\\\ 432+579&=1011\\\\ 243+759&=1002\\\\ 423+579&=1002 \\end{align}\n\nThe total count, without the restrictions $$a\\lt d$$ and $$b\\lt e$$, is thus $$24$$; removing the restriction $$c\\lt f$$ brings the total number of solutions to $$48$$.\n\n\u2022 Thank you for the comprehensive and very well explained answer. Insightful. Much appreciate the time you took! \u2013\u00a0Parzifal Oct 4 '18 at 11:04\n\nHint1:\n\n$$\\sum_{k=0}^n 10^ka_k \\equiv \\sum_{k=0}^n a_k \\mod{9}$$\n\nHint2:\n\nNote, that the digital root for your numbers is always the same, regardless of your choice.\n\n\u2022 Thanks, I checked your link but I do not understand how your hint leads to the solution. Can you elaborate a bit, please? \u2013\u00a0Parzifal Oct 2 '18 at 12:53\n\u2022 @Parzifal I've added one more hint. \u2013\u00a0Jaroslaw Matlak Oct 2 '18 at 13:08\n\u2022 So indeed, the digital root always is 12. This means that any combination of the digits will yield this root. And I always get to this root as the minimum digital sum by re-applying the casting out of nines ... the argumentation runs in this direction? \u2013\u00a0Parzifal Oct 2 '18 at 13:36\n\u2022 * Digital root is one-digit number, in this case $3$. * The control sum is equal to the digital root (mod 9). Therefore the control sum can't be smaller than the digital root. You've found the number with control sum equal to the digital root, so... \u2013\u00a0Jaroslaw Matlak Oct 2 '18 at 18:27\n\u2022 Thank you - your comments and hints were very helpful to me. First time that I came across such concepts. \u2013\u00a0Parzifal Oct 2 '18 at 19:22\n\nBuilding on the helpful hints of @Jaroslaw Matlak, the digital root of the sum will equal the digital root of the sum of the digital roots for the summed numbers. Consider your first example:\n\n$$243 + 957 = 1200$$\n\nThe digital root of $$243$$ is\n\n$$2+4+3 = 9$$\n\nand the digital root of $$957$$ is\n\n$$9 + 5 + 7 = 21,$$ continuing to sum until we have one digit, $$2+1 = 3$$.\n\nSo the sum of the digital roots for $$243$$ and $$957$$ is $$9+3 = 12$$ giving $$1+2 = 3$$. This is equal to the digital root of $$1200$$.\n\n$$1+2+0+0 = 3$$.\n\nYou should see that the actual permutations of the numbers $$2,3,4,5,7,9$$ don't matter in the calculation of the digital root. The left hand side digital root will always be $$2+3+4+5+7+9=30$$, $$3+0=3$$, the same as the digital root of the sum.\n\nThe other answers have shown that the digit sum will converge to $$3$$ if you repeat it enough. It appears you only want to take the digit sum once, so you can get $$3,12,21$$. Once you find an example with $$3$$ you are done. Finding such an example can be done with clever searching. There is only one pair of digits, $$3,7$$, available that sum to $$10$$, so we put those in the ones place to get a $$0$$. There is another pair, $$4,5$$, that sum to $$9$$ so if they are added with a carry in you will get another $$0$$. The sum of two three digit numbers cannot have more than four digits, so if you get two zeros the digit sum has to be $$3$$ or $$12$$, but since the carry into the thousands is $$1$$ you can only get $$3$$. Good work.\n\n\u2022 Milikan: That is how I actually searched but I lacked the proof. I learned a lot from you all. And about finding all possible sums - is my suggested approach ok? Thanks \u2013\u00a0Parzifal Oct 2 '18 at 14:33\n\u2022 What you would usually permute is the digits in the sum and see what checksums come out. If you try all the possibilities (there are only $360$) you will get them all. We have $3$. It is easy to get $12$ from $243+957$ and $21$ from $342+957$. The sum of all the digits is $30$ and you can't avoid a carry, so I believe $30$ is impossible but I haven't proven it. \u2013\u00a0Ross Millikan Oct 2 '18 at 15:03\n\u2022 How aout $439+572$? \u2013\u00a0Barry Cipra Oct 2 '18 at 15:11\n\u2022 @Barry Cipra: Good one (CS=1101). But can I rule out 1110 and 1011 apart from trying out all possibilities? \u2013\u00a0Parzifal Oct 2 '18 at 15:46\n\u2022 @Ross Milikan: Thank you for your time explaining, I appreciate this very much. I\u00b4ll try my luck (but as Barry Cipra points out, it\u00b4s not an \"easy\" task). \u2013\u00a0Parzifal Oct 2 '18 at 15:49", "date": "2019-08-20 20:36:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2939291/find-the-smallest-possible-checksum", "openwebmath_score": 0.8231542706489563, "openwebmath_perplexity": 215.21255555354622, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991554374556315, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8717218153519007}}
{"url": "http://leco.cyyy.pw/a-coin-is-tossed-three-times-what-is-the-probability-of-getting-3-heads.html", "text": "# A Coin Is Tossed Three Times What Is The Probability Of Getting 3 Heads\n\nIf a coin is tossed 12 times, the maximum probability of getting heads is 12. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 3 heads, if a coin is tossed five times or 5 coins tossed together. A fair coin is tossed 5 times. List all eight outcomes in the sample space. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. If the coin is tossed twice, find the probability distribution of number of tails. Question: 520. He wins Rs. Now let's put both dice and coins together. Do another SAT math 1&2 subject practice tests. then the outcomes when 3 coins are tossed simultaneously are. exactly 3 heads. There are many other kinds of situations, however, where the probability of an event is not independent but dependent \u2014 that is, where the. Check out our other channels here NORRIS NUTS. Two coin are tossed 400 times and we get a. A coin on average will come up heads 50% of the time. Three coins are tossed up in the air. D: The probability of getting three aces in a row is the product of the probabilities for each draw. at most two heads(using binomial distribution). There are three tosses. What is the probability of getting a. This is the first post in a multipart series on credit risk models. Probability and Statistics Fourth Edition. thats why our thought process is wrong. (If it starts out as heads, there's a 51% chance it will end as heads). But the number of possible ways of getting 7 heads out of 10 tosses is equal to 10C7, since this is equal to the number of possible ways of choosing which 7 out of 10 tosses are heads. A fair coin is tossed 5 times. What is the probability of getting at least 3 heads when 5 coins are tossed at same time? For the experiment of tossing a single fair coin 3 times, what is the probability of getting exactly 2 heads,?. a fair die is thrown five times, find the probability of obtaining a six three times. 5: And so the chance of getting 3 Heads in a row is 0. There is a 12. A conditional probability is defined as the probability of one event, given that some other event has occurred. HideShow timer Statistics. The probability of drawing the ace of spades from a deck of. So what's the probability, I think you're getting, maybe getting the hang of it at this point. Probability puzzles require you to weigh all the What is the probability that they will play each other in some match during the tournament?. If you toss a coin three times, there are a total of eight possible outcomes. Revise the probability in 1. Solution for A fair coin is flipped six times. 5 for obtaining a head when a coin is tossed. The investigation, however, is far from complete; it has en-tered the third and often the most difficult phase, namely, gathering the facts neces\u00acsary in the trial to prove the guilt of the accused. Opening scene of Rosencrantz & Guildenstern Are Dead contemplating probability. I offer to pay you $1 if we do not get the same amount of heads, if you. Three coins are tossed up in the air, one at a time. There were five heads and three tails in the eight tosses. Sollicitatievragen voor Senior Solutions Developer in Maadi Cornish. The number of heads in three tosses could be 0, 1, 2, or 3, and the number of runs in a sequence could be 1, 2, or 3. Probability of getting exactly 8 heads in tossing a coin 12 times is 495/4096. But, 12 coin tosses leads to #2^12#, i. A fair coin is tossed three times. What is the probability of the event only heads or only tails? The upward face of the coin\u2014a head or a tail\u2014determines the experimental outcomes (sample points). Assuming the outcomes to be equally likely, find the probability that all three tosses are \u201cHeads. We use the experiement of tossing a coin three times to create the probability distribution This is a basic introduction to a probability distribution table. Probability GCSE Maths revision, covering probability single & multiple events, the rules of probability and probability trees, including examples and For example, the probability of flipping a coin and it being heads is \u00bd, because there is 1 way of getting a head and the total number of. A) you will have a combination like THTHTH, which cannot be included as 3 heads are not in a row. As the number of tosses of a coin increases, the number of heads will be getting closer and closer to the number of tails. Do another SAT math 1&2 subject practice tests. A biased coin is tossed ten times. So the probability is. #color(green){H}# represents a head while #color(red){T}# represents a tail. So we must multiply three probabilities. EDIT #2: By \"no retosses\" I mean that your algorithm for obtaining the 1/3 probability can not have a \"retoss until you get 1/3\" rule which can theoretically cause you to toss infinitely many times. Once in the \"3 tails\" section which is TTTH and once in the \"4 tails\" section, which is TTTT. The most important consequence of the domination of intuition is the pervasiveness of confirmation bias\u2014the tendency to seek out and interpret information that confirms what you already think. As per the solution above, we already have a The probability of finding exactly be the number of times a fair coin is tossed up to and including the first time the coin comes Suppose we stop tossing the. Repeat this 8 times and store the number of heads for each one. If you toss a coin 10 times, how many heads and how many tails would you expect to get? Working with a partner, have one person toss a coin in 2 tries = \u00bd or 50% chance \uf0d8 Toss the coin 3 times, probability of each coin landing heads up is \u00bd for each of the three tosses: \u00bd x \u00bd x \u00bd = 1/8 (flip 1). (Hint: Drawing a sample space will help). Suppose: the 1st coin has probability $$p_H$$ of landing heads up and $$p_T$$ of landing tails up;. What is the probability of tossing a coin 5 times and getting 2 tails and 3 heads in that order? Responder preguntas \u00bfquien responda se lleva 5 puntos 1. A player tosses 3 fair coins. A coin toss is a binomial random variable. A coin is tossed three times. Based on the experimental probability, how many times should Kuan expect to get 3 heads in the next 55 tosses?. Unfortunately, I do not believe I was successful in explaining to Kent why my figures were correct. Law of Averages and the Law of Large Numbers. There are many other kinds of situations, however, where the probability of an event is not independent but dependent \u2014 that is, where the. This is a problem that takes some time and a few steps to solve. That is, we know if we toss a coin we expect a probability of 0. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry. A coin is tossed three times. A coin flip: A fair coin is tossed three times. A biased coin is tossed ten times. If it's a fair coin, then each time there's 1/2 chance each of heads and tails. 1 if only 1 head occurs. So five flips of this fair coin. Given that we have at least one head the probability that there are at least 2 heads. the coin lands heads more than once. Symbolically P(A or B) or P(A U B) = P(A) + P(B) \uf06e The theorem can be extended to three. ) What is the probability of getting heads on only one of your flips? B. You need to enable JavaScript to run this app. (Enter your answer to five decimal places. 375) plus the probability of getting 2 heads (0. Given that it is rainy, there will be heavy traffic with probability$\\frac{1}{2}$, and given that it is not rainy, there will be heavy traffic with probability$\\frac{1}{4}$. Alan Gauld Look again at. What is the probability of getting more than 9 and Why is it reasonable to say that the number of coins in 30 tosses is approximately distributed normal?. Three evenly spaced dots forming an ellipsis: \"\". Two crossed lines that form an 'X'. For each toss of a coin a \"Head\" has a probability of 0. So we must multiply three probabilities. ) What is the probability of getting heads on only one of your flips? B. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100. Then the possible values of X are 0,1,2 and 3. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a \u2018head\u2019 or \u2018tail\u2019 is obtained. 7870 and the probability of getting three or more heads in a row or three or more tails in a row is 0. What is the probability that all 3 tosses will result in heads? A)1/2. If you flip 3 coins and I tell you that you got at least one heads, what's the probability that all three came up heads? I came up with 2 different answers to this question depending how you interrupt knowing that one is a heads. What is the probability of the event that heads comes up on the first toss or the I did another attempt where I multiplied the probability of getting a Heads multiplied by 1/3, which is the number of trials So then for that I got 1/2 x 1/3 + 1/2 x. The frequency with which the coin lands heads is three out of four, and it can never be tossed again. Algebra -> Probability-and-statistics-> SOLUTION: A fair coin is tossed 5 times. Coin toss probability is explored here with simulation. The coin will be tossed until your desired run in heads is achieved. If the coin is tossed two times and you want the probability of getting 2 heads, that's the probability of getting a head on the first toss AND getting a head on the 2nd toss. Then work out the probability p of getting a See attached sheet for the probabilities. If you toss a coin 3 times, you're going to get at least two heads or at least two tails, but you can't get _both_ 2 heads and 2 tails. 5: And so the chance of getting 3 Heads in a row is 0. Three Birds - Calculate the total number of miles three birds fly while zigzagging between each other during a trip from Probability questions pop up all the time. If you look at a chart you will see that the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. 1 if only 1 head occurs. The answer would be a cumulative probability. If you need the probability of getting EXACTLY 2 tails then the probability is (3/8) or. What is the probability of getting exactly 6 heads? 2. sum of three times a. What is the probability of getting at most two heads? Unbiased coins means a coin having head and tail whereas a biased coin means having two heads or How many times a man can tossing a coin so that the probability of atleast one head is more than 80%?. Number of times one head appeared = 75. at least one head. Find the probability of getting the King of heart. for the binonial, or still easier, do it on a TI-83 or 84 with p =. In this video, we 'll explore the probability of getting at least one heads in multiple flips of a fair coin. ) Show solution. @BCCI Why can't it be shifted and also what was it planned in first place when everyone know Oct-Feb period is worst pollution Get real-time alerts and all the news on your phone with the all-new India Today app. Practice: Independent probability. 2 Coin Tossing and Probability Models 3. Let E be an event of getting heads in tossing the coin and S be the sample space of maximum possibilities of getting heads. Nov 02, 2019 \u00b7 Get a Telegraph Sport subscription for \u00a31 per week. Quantity A is greater. Play this game to review Probability. then the outcomes when 3 coins are tossed simultaneously are. What is the probability, P(k), of obtaining k heads? There are 16 different ways the coins might land; each is equally probable. It is not going to. The outcomes obtained whole a coin is tossed thrice are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. \" In the above question assume that selecting a bad ball signifies a bad event. Still, it seems that Beth was right: the probability that the coin landed heads the first time is 50 per cent. A fair coin is tossed five times. \" Q: This could finally be the first game, hopefully, where you have all of your offensive weapons available. We already know that the probability of landing on heads is. Later in this section we shall see a quicker way. Second toss, HH HT TH TT (example:first toss was H, second could be H or T and so on). Two sets of trials are shown. These are the things that get mathematicians excited. The probability of rolling a six on one die is 1/6. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most. What is the probability of obtaining one Head and two Tails? (A fair coin is one that is not loaded, so there is an equal chance of it landing Heads up or Tails up. And what I want to think about in this video is the probability of getting exactly three heads. Three coins are tossed up in the air, one at a time. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. When you register, you support the site and your question history is saved. In possibility theory, a The probability of happening an event can easily be found using. Determine a probability assignment for the simple events of the sample space S ={H,T} that reflects this bias. A fair coin is tossed 3 times, the probability of gitting 3 heads is? Let us understand that this is the probability of getting exactly 3 tails out of the 5 chances too. 5 we get this probability by assuming that the coin is fair, or heads and tails are equally likely. What is the probablity that 3 heads will occur?. So the difference is we multiply the probabilities in the independent events. If a coin is tossed 12 times, the maximum probability of getting heads is 12. Spun coins can exhibit \"huge bias\" (some spun. Sollicitatievragen voor Senior Solutions Developer in Maadi Cornish. Note that the last three cases for the coins chosen expand into eight di\ufb00erent cases when we care whether coin 2 or coin 3 is chosen. The reason the probability of flipping 3 heads in a row is due to conditional probability. Example 1: A fair coin is tossed 5 times. Algebra -> Probability-and-statistics-> SOLUTION: A fair coin is tossed 5 times. When the Seahawks won the coin toss and started overtime with the ball, was there any doubt about what was going to happen?. Willie Wong served as mayor of Mesa from 1992 to 1996. Youngs sends the kick long and Vermuelen is met by May. Best Answer: Probability of getting a head: 0. The ratio of successful events A = 4 to the total number of possible combinations of a sample space S = 8 is the probability of 2 tails in 3 coin tosses. The chances of getting a girl should be the same whether or not the rst child was a girl (after all, the coin doesn't know whether it came down heads or tails last time). Later in this section we shall see a quicker way. A math-ematical model for this experiment is called Bernoulli Trials (see Chapter 3). The final test of a criminal investigation is in the presentation of the evi\u00acdence in court. Probability of getting First Tail = 1/2 Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4 Probability of getting Third. The probability of getting a head on any one toss of this coin is 3/4. We never know the exact probability this way, but we can get a pretty good estimate. The Bucs got 67 yards and a touchdown on 18 carries from Jones, with two catches for 15 yards on two targets, tying the most touches Jones has ever had in a game. I want a free account. Trump's legal team to mount a defense. 5 probability of a particular two. was there for 30mins before parents noticed. The probability of getting a head on any one toss of this coin is 3/4. ( ) 1 1 7! 7 63 4 7,3 2 2 3! 4! C \u22c5 = = \u22c5 \u22c5 \u22c5 \u22c5 \u22c5 \u22c55 4 3 2 1 3 2 1\u22c5 \u22c5 \u22c5 \u22c5 \u22c5 \u22c54 3 2 1 7 35 128 1 2 = 2. But, 12 coin tosses leads to #2^12#, i. If the coin is spun, rather than tossed, it can have a much-larger-than-50% chance of ending with the heavier side down. Gmm to three decimal places. a coin flip; a coin toss PK \u00b9^nK\\\u00a8 \u00a5\u2020\u00c0%\u00e1\u02c6''02_Tier 1 PBIS & PSC Coaching Guide. It describes the probability distribution of a process that has two possible outcomes. What is the probability of obtaining exactly 3 heads. i think exactly like that, but then i would realize if we do that, what are the odds of gettin 3heads and 2 tails in no order???? wouldnt be 100%????? cuz its only a two sided coin and odds of gettin heads or tails is equal. What is the probability that (a) At least one of the dice shows an even number? P(at least one is even) = 1 - P(both are odd). What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads? Sol. self Question Solution - A fair coin is tossed three times in succession. asked by Dee on April 23, 2014; Math. com loan 1 Let an experiment consist of tossing a fair coin three times. Date: 04/21/2003 at 17:12:44 From: Maggie Subject: Probability In a box there are nine fair coins and one two-headed coin. One of two coins is selected at random and tossed three times. Tossing a Coin A balanced coin is tossed three times. One is a two-headed coin ( having head on both faces ), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. find the probability of getting a result of an odd number of heads in 1/2 of the cases. A coin flip: A fair coin is tossed three times. 5 Probability of getting a tail = 0. In a short trial, heads may easily come up on every flip. There are three ways that can happen. The outcomes of the three tosses are recorded. There are many other kinds of situations, however, where the probability of an event is not independent but dependent \u2014 that is, where the. sum of three times a. Probability of Coin Tosses. One Head : 160 times c. 3 10 Illowsky et al. enter your value ans - 5/16. There were five heads and three tails in the eight tosses. The probability of rolling a six on one die is 1/6. Subjective probability is more connected to the original meaning of the word probable for example, because simply calculating the probability of any horse winning There are three types of probability problems that occur in mathematics. Poker night 2 steam achievement manager. Solution: n=20, p=0. 5 Number of heads$\\leq$Number of tails. When three coins are tossed at random, what is the probability of getting : (i) 3 heads ? (ii) 2 heads Best Answer for Silver Coins, Round, Sloven A fair coin is tossed 5 times. Let E = event of getting exactly 3 heads. The Questions and Answers of A coin is tossed three times. The answer would be a cumulative probability. If$X\\$ is the number of heads. A consecutive streak or a run can happen in random. what's the probability of getting three heads? You can put this solution on YOUR website! a fair coin is tossed 3 times, 1. A coin has a probability of 0. So let's start again with a fair coin. The chances of getting a girl should be the same whether or not the rst child was a girl (after all, the coin doesn't know whether it came down heads or tails last time). Note that this is the same as summing the probability of the 3 mutually exclusive events Given that we have at least one head the probability that there are at least 2 heads. Number of times two heads appeared = 55. Consider, you toss a coin once, the chance of occurring a head is 1 and chance of occurring a tail is 1. Wonder if Papa will left the coin decide if Naz gets a real Guinea Pig this time? Heads or tails will decide. What is the probability that exactly 3 heads occur?. Complimentary Events. Unfortunately, I do not believe I was successful in explaining to Kent why my figures were correct. what's the probability of getting three heads? You can put this solution on YOUR website! a fair coin is tossed 3 times, 1. An Easy GRE Probability Question. 5 Probability of getting a tail = 0. Httt tttt ttht thtt. Question: 520. The probability that an electronic device produced by a company does not function properly. Answer to You toss a coin three times. 5 Probability of getting a tail: 0. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. Example: Suppose 20 biased coins are flipped and each coin has a probability of 75% of coming up heads. Annie will toss a fair coin three times. We use the experiement of tossing a coin three times to create the probability distribution This is a basic introduction to a probability distribution table. Three coins are tossed once. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads? Sol. 1 Let an experiment consist of tossing a fair coin three times. it can occur first, second or third. If you toss a coin 10 times, how many heads and how many tails would you expect to get? How about the results of your partners tosses? How close was each set of results to what was Probability and Segregation. What is the probability of getting neither an ace nor a king card? a. A coin is tossed n times. The probability that an electronic device produced by a company does not function properly. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails. Let be the number of heads on the first two tosses, the number of heads on the last two tosses. There are three tosses. the level of inventory at the end of a given month, or the number of production runs on. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands This penny has been flipped times. So the probability is ----- b) What is the probability of obtaining tails on each of the first 3 tosses That only happens 2 times. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. The probability of getting heads three times in 5 tries is 10/32. There are 15 girls and 15 boys in 8th period. a coin flip; a coin toss PK \u00b9^nK\\\u00a8 \u00a5\u2020\u00c0%\u00e1\u02c6''02_Tier 1 PBIS & PSC Coaching Guide. When tossed, the first shows heads with probability 4/10, the second shows heads with probability 7/10. The results are similarly formatted. (a) What is the Posted 4 years ago. Probability of Getting 2 Heads in 3 Coin Tosses. 2) b) Use simulations to find an empirical probability for the probability of getting exactly 5 heads in 10 tosses of an unfair coin in which the probability of heads is 0. Find the probability that a common year (not a leap year) contains If one is randomly chosen, then what is the probability of getting non-faulty pen? 11. : Let 'S' be the sample - space. A coin is biased in such a way that a head is twice as likely to come up as a tails. Probability of Coin Tosses. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times. Ex: Consider tossing a fair coin. Based on the experimental probability, how many times should Kuan expect to get 3 heads in the next 55 tosses?. Fifty-three years later, family members cling to the hope that her murder will be solved. For instance, we can toss the coin many times and determine the proportion of the tosses that the coin comes up head. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 2 heads, if a coin is tossed three times or 3 coins tossed together. It indicates a way to close an interaction, or dismiss a notification. It's got this cool variegated blue glaze, with swirls of grays and blues with bits of browns and whites. What is the probability of flipping 3 coins and getting all heads flipping three coins and the possibility of getting all Probability Math Help For College. Download with Google Download with Facebook or download with email. The frequency with which the coin lands heads is three out of four, and it can never be tossed again. Every time we toss a fair coin, the probability of seeing heads is 1/2 regardless of what previous tosses have revealed. If the coin is fair, then by symmetry the probability of getting at least 2 heads is 50%. The question is: Suppose a fair coin is thrown three times. Probability of getting First Tail = 1/2 Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4 Probability of getting Third. 5% chance of all three tosses landing on heads. It is well known that the probability that a head turns up in a single toss of a coin is 0. what is the probability of getting exactly 8 heads in a row? c. Besides adding together the. quince medios de un numero es igual a cuarenta y cinco?. What is the probability that all 3 tosses will result in heads? A)1/2. Hhtt hhth ttth thht. ( ) 1 1 7! 7 63 4 7,3 2 2 3! 4! C \u22c5 = = \u22c5 \u22c5 \u22c5 \u22c5 \u22c5 \u22c55 4 3 2 1 3 2 1\u22c5 \u22c5 \u22c5 \u22c5 \u22c5 \u22c54 3 2 1 7 35 128 1 2 = 2. Let x = probability of winning after no heads (or a tail). If a coin is tossed three times what is the pribability of getting at most one tail and what is the probability of getting at most 2 head And plz also explain why you are taking hhh in case of at most one tail - Math - Probability. The probability of success, i. 7E-20 A fair coin is tossed 20 times. The reason the probability of flipping 3 heads in a row is due to conditional probability. Determine the probability of getting 2 heads in two successive tosses of a balanced coin. Perhaps you are expected to use the normal approximation to the binomial. One after another what is the probability that the first ball is red and second ball is yellow?. Solution 2: How could you use the entries in Pascal\u2019s Triangle to solve this problem?. Pregunta de entrevista para Senior Solutions Developer en Maadi Cornish. As the number of tosses of a coin increases, the number of heads will be getting closer and closer to the number of tails. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Httt tttt ttht thtt. quince medios de un numero es igual a cuarenta y cinco?. EDIT - It's not 1/2^6, it's (1/2)^6. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry. An array of Asian-related grocery stores, restaurants and shops organically sprouted in Mesa. ) What is the probability of getting heads on only one of your flips? B. What is the probability that you will get heads all three times? of getting a heads. In general when a coin is tossed n times , the total number of possible outcomes = 2^n). Find simple formulae in terms of n and k for. Opening scene of Rosencrantz & Guildenstern Are Dead contemplating probability. What is the probability that exactly two tosses result in heads? Express your answer as a common fraction. If you look at a chart you will see that the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. getting a 6, one each trial is 1/6. What is the probability of rolling a six on all four rolls?' and find homework help for other Math questions at eNotes A coin is tossed. One of two coins is selected at random and tossed three times. 7870 and the probability of What is the probability that three of the six possible outcomes do not show up and each of the other three possible. I respond to three objections, which claim that time-restricted laws lessen the coincidence of observed regularities without making it Those claims will be defended within three broad domains: quantum mechanics, classical statistical mechanics, and macroscopic chance events such as coin tosses. If you toss a coin, it will come up a head or a tail. find the probability of getting a head at least once Follow. Then S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT } (i) Let 'E1' = Event of getting all heads, Then E1 = { HHH }. I believe that I have the trick coin, which has a probability of landing heads 40% of the time, p= 0. When two coins are tossed at random, what is the probability of getting a. 5 of coming up heads. Avasaram Component Library We are thrilled. Thus, the required probability is 2 \u00f7 8 = 0. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. Let E be an event of getting heads in tossing the coin and S be the sample space of maximum possibilities of getting heads. According to this table, the theoretical probability of getting 6 heads in 10 tosses is 20. Probabilities can also be thought of in terms of relative frequencies. Curry gathers the kick and carries the ball back. Since these all have a probability of 12. Given N number of coins, the task is to find probability of getting at least K number of heads after tossing all the N Time Complexity: O(n) Auxiliary space: O(n). Answer : Option A Explanation : Total number of balls = 4 + 5 + 6 = 15 Let S be the sample space. The probability of getting a head on any one toss of this coin is 3/4. A pair of dice are rolled. Solution: Total number of trials = 250. Question: A Coin Flip: A Fair Coin Is Tossed Three Times. thats why our thought process is wrong. If you toss a coin 3 times, you're going to get at least two heads or at least two tails, but you can't get _both_ 2 heads and 2 tails. This is a problem that takes some time and a few steps to solve. What is the theoretical probability of both coins landing on heads?. The total number of outcomes = 2 ^6 =64 (it is because each toss has two possibilities Head or Tail. #color(blue){3\" Heads\"}#. If you need the probability of getting EXACTLY 2 tails then the probability is (3/8) or.", "date": "2019-11-13 20:29:35", "meta": {"domain": "cyyy.pw", "url": "http://leco.cyyy.pw/a-coin-is-tossed-three-times-what-is-the-probability-of-getting-3-heads.html", "openwebmath_score": 0.7841312289237976, "openwebmath_perplexity": 293.48066761797514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718459575752, "lm_q2_score": 0.880797071719777, "lm_q1q2_score": 0.8717000638829384}}
{"url": "https://mathematica.stackexchange.com/questions/178376/set-a-fixed-colour-scale-with-densityplot/178379", "text": "# Set a fixed colour scale with DensityPlot\n\nI am trying to plot some Zernike polynomials, but I want all of them to be plotted by using the same colour scale from -1 to +1 (i.e. the range of these polynomials). I am using the following code:\n\nDensityPlot[\nZernikeR[n, m, Norm[{x, y}]] Cos[m ArcTan[x, y]], {x, -1, 1}, {y, -1,1}\n, PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}}\n, ColorFunction -> (Hue[2 (1 - #1)/3] &)\n, PlotPoints -> 200\n, PlotLegends -> Automatic\n, RegionFunction -> Function[{x, y, z}, -1 < x^2 + y^2 < 1]\n]\n\n\nI get what I want with some polynomials, e.g. when m=0 and n=2:\n\nand not with others, e.g. when m=0 and n=4:\n\nwhere the colour range goes down to -0.5 and not to -1. I want all polynomials to be coloured with a scale from -1 to 1, from blue to red. I haven't found a way to fix this. What am I missing? Thanks.\n\n\u2022 Look up ColorFunctionScaling and ColorFunction. \u2013\u00a0Szabolcs Jul 16 '18 at 11:08\n\u2022 ColorFunctionScaling -> False will prevent scaling but 2 (1 - #1)/3 &@Interval[{-1, 1}] gives: Interval[{0, 4/3}] which can be missleading when Hue is applied ( {0,1} base domain). What exactly do you want to achieve, how your manual rescaling is related to the problem and are you only concerned about colors or the range in the bar legend aswell? \u2013\u00a0Kuba Jul 16 '18 at 11:49\n\u2022 I'd like the color scale to always vary between blue set for -1 and red set for +1, independently from the min and max values of the function in that range. \u2013\u00a0MicheleG Jul 16 '18 at 11:58\n\u2022 A related question. \u2013\u00a0J. M. is away Sep 26 '18 at 7:50\n\n(This answer will be very similar to halirutan's answer (+1). The difference is that I preserved the color function from the OP, and that I made a fixed bar legend.)\n\nWe can do it like in this answer, which is to say, we can turn off the color function scaling and scale the values ourselves in a way that is the same for all plots. We can make our own bar legend that matches this scaling. We only need to change three options:\n\nColorFunction -> (Hue[2 (1 - Rescale[#1, {-1, 1}])/3] &),\nColorFunctionScaling -> False,\nPlotLegends -> BarLegend[{Hue[2 (1 - Rescale[#1, {-1, 1}])/3] &, {-1, 1}}]\n\n\nNow we get plots like this:\n\n\u2022 Hi C.E., this is exactly the solution I just found thanks to the link you posted before. Thanks to all of you guys for you help. \u2013\u00a0MicheleG Jul 16 '18 at 12:13\n\nYou have currently two mistakes. The first one is your color-function itself. It needs to give Hue[0] for a value of -1 and Hue[1] for a value of 1. The transformation is\n\ncol[v_] := Hue[Rescale[v, {-1, 1}]]\n\n\nSecondly, you need to turn off that Mathematica rescales the values for each plot which can be done with the ColorFunctionScaling -> False option.\n\nplot[m_, n_] :=\nDensityPlot[\nZernikeR[n, m, Norm[{x, y}]] Cos[m ArcTan[x, y]], {x, -1,\n1}, {y, -1, 1}, PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}},\nColorFunction -> col, PlotPoints -> 200, PlotLegends -> Automatic,\nColorFunctionScaling -> False,\nRegionFunction -> Function[{x, y, z}, -1 < x^2 + y^2 < 1]]\n\n\nAnd then, plot[0,2] and plot[0,4] will use the same scaling for the colors:\n\nIf you don't want a circular color scheme like Hue is (begins and ends in red), then use one of the other schemes.", "date": "2019-08-23 12:33:23", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/178376/set-a-fixed-colour-scale-with-densityplot/178379", "openwebmath_score": 0.21097351610660553, "openwebmath_perplexity": 1206.5771255447112, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075766298657, "lm_q2_score": 0.9059898184796793, "lm_q1q2_score": 0.8716596687088162}}
{"url": "https://math.stackexchange.com/questions/3695330/on-cauchy-schwarz-inequality", "text": "# On Cauchy-Schwarz inequality\n\nWhat is Schwarz inequality in $$\\mathbb R^2$$ or $$\\mathbb R^3$$? Give another proof of it in these cases.\n\nHere is my attempt in $$\\mathbb R^2$$. Let $$x=(x_1,x_2)$$ and $$y=(y_1,y_2)$$ both in $$\\mathbb R^2$$. The Cauchy-Schwarz inequality is\n\n$$\\lvert\\langle x,y \\rangle \\rvert \\leq \\lVert x\\rVert \\lVert y\\rVert.$$ Then our claim is\n\n$$(x_1y_1+x_2y_2)^2 \\leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$\n\nProof for this inequality: since $$(x_1y_2-x_2y_1)^2 \\geq 0$$, add $$(x_1y_1+x_2y_2)^2$$ to both sides: \\begin{align*} (x_1y_1+x_2y_2)^2 & \u2264 (x_1y_2-x_2y_1)^2+(x_1y_1+x_2y_2)^2 \\\\ & =(x_1y_2)^2+(x_2y_1)^2 -2(x_1y_2)(x_2y_1) \\\\ & \\qquad +(x_1y_1)^2+(x_2y_2)^2+2(x_1y_1)(x_2y_2) \\\\ & =x_1^2(y_1^2+y_2^2) +x_2^2(y_1^2+y_2^2) \\\\ & =(x_1^2+x_2^2)(y_1^2+y_2^2), \\end{align*} namely $$(x_1y_1+x_2y_2)^2\\leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$ This proves the claim.\n\nMy question is: is this answer complete to the given question? As there is a choice for $$\\mathbb R^2$$ or $$\\mathbb R^3$$. If any step is missing please identify it.\n\n\u2022 I guess you should do a similar thing in the case of $\\mathbb R^3$. The steps look good to me. \u2013\u00a0Gibbs May 28 '20 at 12:10\n\u2022 Note that I have modified the format of the entire question. Please, use MathJax next times. See here. \u2013\u00a0Gibbs May 28 '20 at 12:22\n\nyou wrote: $$\\quad \"$$ our claim is\n$$(x_1y_1+x_2y_2)^2 \\leq (x_1+x_2)(y_1+y_2).\"$$\n$$(x_1y_1+x_2y_2)^2 \\leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$", "date": "2021-01-26 09:48:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3695330/on-cauchy-schwarz-inequality", "openwebmath_score": 0.999626874923706, "openwebmath_perplexity": 638.9938755242982, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708032626701, "lm_q2_score": 0.89181104831338, "lm_q1q2_score": 0.8716300806485721}}
{"url": "https://mathoverflow.net/questions/252799/transforming-numbers-of-irreducible-polynomials/252814#252814", "text": "# Transforming numbers of irreducible polynomials\n\nLet $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials of degree $n$ in the polynomial ring $\\mathbf{F}_{q}[X]$ over the finite field $\\mathbf{F}_{q}$. Since \\begin{equation*} M(n) = \\frac{1}{n} \\sum_{d \\mid n} \\mu(d) q^{n/d} \\end{equation*} we have that $M(1)=q$, $M(2) = \\frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \\cdots + n_se_s$. Consider the sum of polynomials in $\\mathbf{Q}[q]$ \\begin{equation*} \\sum_{[n_1^{e_1},\\ldots, n_s^{e_s}] \\in P(n)} (-1)^{\\sum e_i} \\prod_{i=1}^s \\binom{M(n_i)}{e_i} \\end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?\n\n\u2022 I think you mean $M(n)(q)$ to be the number of irreducible polynomials of degree $n$ in $\\mathbb F_q[X]$. Oct 22 '16 at 10:30\n\u2022 The sum is actually $-q$ for $n=1$. Oct 22 '16 at 10:46\n\u2022 Thanks for the comments - my original question has been corrected accordingly. Oct 22 '16 at 11:14\n\nYes, it is true. Your expression is the coefficient of $x^n$ in the following product: $$\\prod_{P\\text{ monic irreducible}} (1-x^{\\deg P}) = \\prod_{n} (1-x^n)^{M(n)}.$$ The Zeta function of $\\mathbb{F}_q[X]$ is $$\\sum_{f \\text{ monic}} x^{\\deg f} = \\sum_{n \\ge 0} q^n x^n = \\frac{1}{1-qx}.$$ The Euler product identity tells us that $$\\frac{1}{1-qx} = \\prod_{P\\text{ monic irreducible}} (1-x^{\\deg P})^{-1} =\\prod_{n} (1-x^n)^{-M(n)}.$$ Taking its reciprocal, we find that $$1-qx = \\prod_{P\\text{ monic irreducible}} (1-x^{\\deg P}).$$ Now it is just a matter of comparing coefficients on both sides.\nInterpretation: Let $\\mu: \\mathbb{F}_q[X] \\to \\mathbb{C}$ be the polynomial M\u00f6bius function, defined by $$\\mu(f) = \\begin{cases} 0 & f \\text{ not squarefree} \\\\ (-1)^k & f=p_1 \\cdots p_k (p_i \\text{ distinct irreducibles}) \\end{cases}.$$ The term $\\prod_{i=1}^{s} \\binom{M(n_i)}{e_i}$ counts the number of monic, squarefree polynomials of degree $n$ whose factorization consists of $e_i$ irreducibles of degree $n_i$. The polynomial M\u00f6bius function $\\mu(\\bullet)$ assumes the value $(-1)^{\\sum e_i}$ for each such polynomial. In other words, your sum may be rewritten as $$\\sum_{f \\text{ monic, squarefree of degree }n} \\mu(f).$$ Since $\\mu(f)=0$ for $f$ which is not squarefree, your claim is the same as $$n>1 \\implies \\sum_{f \\text{ monic of degree }n} \\mu(f) = 0.$$ This is classical, and due to L. Carlitz, whose proof was exactly as above. It should be compared with the following equivalent formulation of the Riemann Hypothesis: $$\\sum_{n \\le x } \\mu(n) = O(x^{\\frac{1}{2}+\\varepsilon}),$$ where this time $\\mu$ is the integer M\u00f6bius function.\nIt is interesting to ask whether $\\sum_{f\\text { monic of degree }n} \\mu(f)$ may be evaluated without the use of formal power series and the Euler product identity. I will present such a way. Let $M_q$ denote the set of monics in $\\mathbb{F}_q[X]$. Given two functions $\\alpha, \\beta : M_q \\to \\mathbb{C}$, one defines their \"Dirichlet convolution\" as the following function: $$(\\alpha*\\beta) (f) = \\sum_{d_1 \\cdot d_2 = f} \\alpha(d_1) \\beta(d_2).$$ Let $\\zeta$ denote the constant function $1$ on $M_q$. Let $\\delta_{1}$ denote the indicator function of the constant polynomial $1$. The \"defining\" property of $\\mu$ is $$\\mu * \\zeta = \\delta_1.$$ By summing the above over all monic polynomials of degree $n$ (>0) and changing the order of summation, we get that $$\\sum_{\\deg d \\le n} \\mu(d) q^{n-\\deg d} = 0,$$ or equivalently $$\\sum_{\\deg d \\le n} \\frac{\\mu(d)}{q^{\\deg d}} =0.$$ Plugging $n=m,m+1$ in the above and subtracting, we get that $$m>0 \\implies \\sum_{\\deg d = m+1} \\mu(d) = 0.$$\nHere is a reformulation of Ofir Gorodetsky's excellent answer to my question: Let $(a(n))_{n \\geq 1}$ be a sequence of rational numbers. Define the transformed sequence $T(a)$ of $a$ to have $n$th element \\begin{equation*} T(a)(n) = \\sum_{n_1^{e_1} \\cdots n_s^{e_s} \\in P(n)} (-1)^{\\sum e_i} \\prod_{i=1}^s \\binom{A(n_i)}{e_i}, \\qquad n \\geq 1 \\end{equation*} Then \\begin{equation*} 1+ \\sum_{n \\geq 1} T(a)(n)x^n = \\prod_{n \\geq 1} (1-x^n)^{a(n)} \\end{equation*} We now apply this to the sequence $M(n)(q)$. From the classical formula \\begin{equation*} \\frac{1}{1-qx} = \\prod_{n \\geq 1} (1-x^n)^{-M(n)(q)} \\end{equation*} we get that \\begin{equation*} 1+ \\sum_{n \\geq 1} T(M(n)(q))x^n = \\prod_{n \\geq 1} (1-x^n)^{M(n)(q)} = 1-qx \\end{equation*} This shows that $T(M)(1)=-q$ and $T(M)(n)=0$ for all $n>1$.", "date": "2021-12-08 09:50:57", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/252799/transforming-numbers-of-irreducible-polynomials/252814#252814", "openwebmath_score": 0.9935666918754578, "openwebmath_perplexity": 124.64089177203034, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9773707960121132, "lm_q2_score": 0.8918110382493034, "lm_q1q2_score": 0.8716300643461108}}
{"url": "https://consumercredentials.com/avdcz/8a9c97-how-to-find-irrational-numbers-between-decimals", "text": "Clearly all fractions are of that it can't be the terminating decimals because they're rational. How To Find Irrational Numbers Between Two Decimals DOWNLOAD IMAGE. 0.5 can be written as \u00bd or 5/10, and any terminating decimal is a rational number. The vast majority are irrational numbers, never-ending decimals that cannot be written as fractions. The only candidates here are the irrational roots. Rational numbers. In mathematics, a number is rational if you can write it as a ratio of two integers, in other words in a form a/b where a and b are integers, and b is not zero. 9 years ago. Include the decimal approximation of the ...\u201d in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Take this example: \u221a8= 2.828. In short, rational numbers are whole numbers, fractions, and decimals \u2014 the numbers we use in our daily lives.. We can actually split this into thirds. 10 How to Write Fractions Between Two Decimal Numbers. But there's at least one, so that gives you an idea that you can't really say that there are fewer irrational numbers than rational numbers. A set of real numbers is uncountable. Real numbers also include fraction and decimal numbers. Practice Problems 1. And these non recurring decimals can never be converted to fractions and they are called as irrational numbers. Yes. So the question states: what is a decimal number between each of the following pairs of rational numbers and then it gives the fractions -5/6, 1 and -17/20 and -4/5! (Examples: Being able to determine the value of the \u221a2 on a number line lies between 1 and 2, more accurately, between 1.4 \u2026 So number of irrational numbers between the numbers should be infinite or something finite which we cannot tell ? In summary, this is a basic overview of the number classification system, as you move to advanced math, you will encounter complex numbers. Method of finding a number lying exactly midway between 2 given numbers, is add those 2 numbers & divide by 2. Rational and Irrational numbers both are real numbers but different with respect to their properties. To find the rational number between two decimals, we can simply look for the average of the two decimals. Step 3: Place the repeating digit(s) to the left of the decimal point. Irrational Numbers on a Number Line. In other words, irrational numbers require an infinite number of decimal digits to write\u2014and these digits never form patterns that allow you to predict what the next one will be. Make use of this online rational or irrational number calculator to ensure the rationality and find its value. \u2154 is an example of rational numbers whereas \u221a2 is an irrational number. But rational numbers are actually rare among all numbers. Irrational numbers tend to have endless non-repeating digits after the decimal point. Which means that the only way to find the next digit is to calculate it. Step 5: Using the two equations you found in step 3 and step 4, subtract the left sides of the two equations. Examples of Rational and Irrational Numbers For Rational. They can be written as a ratio of two integers. O.13 < 3/n < 0.14 13/100 < 3/n < 14/100 When is 13/100 < 3/n ? Irrational numbers' decimal representation is non terminating , non repeating.. From the below figure, we can see the irrational number is \u221a2. Representation of irrational numbers on a number line. 1/3. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. Many people are surprised to know that a repeating decimal is a rational number. What is the Difference Between Rational and Irrational Numbers , Intermediate Algebra , Lesson 12 - Duration: 3:03. So irrational number is a number that is not rational that means it is a number that cannot be written in the form $$\\frac{p}{q}$$. The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. one irrational number between 2 and 3 . Some decimals terminate which means the decimals do not recur, they just stop. It makes no sense!!! Solution: If a and b are two positive numbers such that ab is not a perfect square then : i ) A rational number between and . List Of Irrational Numbers 1 100. Step 4: Place the repeating digit(s) to the right of the decimal point. since 3^2 = 9 and 6^2 = 36, \u221a41 > 6. apply the same kind of thinking to the numbers in B and they are all between 3 and 6. Wednesday, October 14, 2020 Find two rational numbers between decimals Ex: Find two rational numbers that have 3 in their numerator and are in between 0.13 and 0.14 with no calculator. Irrational Number between Two Rational Numbers. Rational Number is defined as the number which can be written in a ratio of two integers. Number System Notes. I can approximately locate irrational numbers on a number line ; I can estimate the value of expression involving irrational numbers using rational numbers. On the other hand, all the surds and non-repeating decimals are irrational numbers. The major difference between rational and irrational numbers is that all the perfect squares, terminating decimals and repeating decimals are rational numbers. Subtract the rounded numbers to obtain the estimated difference of 0.5; The actual difference of 0.988 - 0.53 is 0.458; Some uses of rounding are: Checking to see if you have enough money to buy what you want. The decimal expansion of an irrational number continues without repeating. I tried to cross multiply like my teacher said but that was for finding what lay in between fractions it worked for that I have no idea how to do this. Getting a rough idea of the correct answer to a problem Irrational Numbers. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q \u2260 0. They include the counting numbers and all other numbers that can be written as fractions. Get an answer to your question \u201cPart B: Find an irrational number that is between 9.5 and 9.7.Explain why it is irrational. Lv 6. The first number we have here is five, and so five is five to the right of zero, five is right over there. This amenability to being written down makes rational numbers the ones we know best. Learn Math Tutorials 497,805 views When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. New Proof Settles How To Approximate Numbers Like Pi Quanta Magazine. Rational numbers are contrasted with irrational numbers - numbers such as Pi, \u221a 2, \u221a 7, other roots, sines, cosines, and logarithms of numbers. An irrational number is a number which cannot be expressed in a ratio of two integers. How many irrational numbers can exist between two rational numbers ? So 1/3 is between zero and one. That's our five. Irrational Numbers. A rational number is of the form $$\\frac{p}{q}$$, p = numerator, q= denominator, where p and q are integers and q \u22600.. Irrational Numbers. The ability to convert between fractions and decimals, and to approximate irrational numbers with decimals or fractions, can be very helpful in solving problems. 0 0. cryptogramcorner. Example: Find two irrational numbers between 2 and 3. Examine the repeating decimal to find the repeating digit(s). Key Differences Between Rational and Irrational Numbers. A rational number is a number that can be written as a ratio. As there are infinite numbers between two rational numbers and also there are infinite rational numbers between two rational numbers. DOWNLOAD IMAGE. Are there any decimals that do not stop or repeat? Real numbers include natural numbers, whole numbers, integers, rational numbers and irrational numbers. - [Voiceover] Plot the following numbers on the number line. But an irrational number cannot be written in the form of simple fractions. The number $\\pi$ (the Greek letter pi, pronounced \u2018pie\u2019), which is very important in describing circles, has a decimal form that does not stop or repeat. Essentially, irrational numbers can be written as decimals but as a ratio of two integers. Terminating, recurring and irrational decimals. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. The difference between rational and irrational numbers can be drawn clearly on the following grounds. Hence, almost all real numbers are irrational. A Rational Number can be written as a Ratio of two integers (ie a simple fraction). how cuanto mide una cama matrimonial haunted house in san diego that lasts 4 7 hours journey to the west full movie with english subtitles. To study irrational numbers one has to first understand what are rational numbers. Then we get 1/3. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. Difference between rational and irrational numbers has been clearly explained in the picture given below. DOWNLOAD IMAGE. Difference between Rational and Irrational Numbers. m/n \u2014> 3/n N has to be an integer, n cannot be 0. For instance, when placing \u221a15 (which is 3.87), it is best to place the dot on the number line at a place in between 3 and 4 (closer to 4), and then write \u221a15 above it. Suppose we have two rational numbers a and b, then the irrational numbers between those two will be, \u221aab. . Irrational numbers include \u221a2, \u03c0, e, and \u03b8. Rational Numbers. Irrational numbers are those which do not terminate and have no repeating pattern. 2 and 3 are rational numbers and is not a perfect square. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. Example 15: Find a rational number and also an irrational number between the numbers a and b given below: a = 0.101001000100001\u2026., b = 0.1001000100001\u2026 Solution: Since the decimal representations of a and b are non-terminating and non-repeating. Irrational Numbers: We have different types of numbers in our number system such as whole numbers, real numbers, rational numbers, etc. Rational numbers are whole numbers, fractions, and decimals - the numbers we use in our daily lives. ii) An irrational number between and . How to Write Irrational Numbers as Decimals Before studying the irrational numbers, let us define the rational numbers. Apart from these number systems we have Irrational Numbers. You can use the decimal module for arbitrary precision numbers: import decimal d2 = decimal.Decimal(2) # Add a context with an arbitrary precision of 100 dot100 = decimal.Context(prec=100) print d2.sqrt(dot100) If you need the same kind of ability coupled to speed, there are some other options: [gmpy], 2, cdecimal.\n\n## how to find irrational numbers between decimals\n\nUi Architecture Best Practices, Samsung Ipad Price In Pakistan 2020, Hp Stream 14 Laptop 14 Intel Celeron N4000 Review, Samson Sr850 Vs Akg K240, Best Nikon Z Lenses, Gibson Truss Rod Covers, Sonos Sub Gen 3 Review, Masala Omelette Calories, Are Beaches Open In Palm Beach County,", "date": "2021-01-17 19:21:16", "meta": {"domain": "consumercredentials.com", "url": "https://consumercredentials.com/avdcz/8a9c97-how-to-find-irrational-numbers-between-decimals", "openwebmath_score": 0.7487418055534363, "openwebmath_perplexity": 324.3653039637664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9877587218253718, "lm_q2_score": 0.8824278680004706, "lm_q1q2_score": 0.8716258229992327}}
{"url": "http://math.stackexchange.com/questions/161024/what-consitutes-an-exponential-function?answertab=active", "text": "# What consitutes an exponential function?\n\nI was recently having a discussion with someone, and we found that we could not agree on what an exponential function is, and thus we could not agree on what exponential growth is.\n\nWikipedia claims it is $e^x$, whereas I thought it was $k^x$, where k could be any unchanging number. For example, when I'm doing Computer Science classes, I would do everything using base 2. Is $2^x$ not an exponential function? The classical example of exponential growth is something that doubles every increment, which is perfectly fulfilled by $2^x$. I'd also thought $10^x$ was a common case of exponential growth, that is, increasing by an order of magnitude each time. Or am I wrong in this, and only things that follow the natural exponential are exponential equations, and thus examples of exponential growth?\n\n-\nWhether one thinks of exponential growth as $e^{ct}$ ($c$ positive) or $a^t$ ($a\\t 1$), one is dealing with the same phenomenon, just take $a=e^c$. \u2013\u00a0 Andr\u00e9 Nicolas Jun 21 '12 at 0:45\n\n$e^x$ is the exponential function, but $c\\cdot k^x$ is an exponential function for any $k$ ($> 0, \\ne1$) and $c$ ($\\ne 0$).\n\nThe terminology is a bit confusing, but is so well settled that one just has to get used to it.\n\n-\nTherefore, I can refer to anything that can be described as $c\u22c5k^x$ as exponential growth, correct? \u2013\u00a0 Canageek Jun 20 '12 at 23:24\nYou can do whatever you want. The question is whether people will understand you, and in this case I think most people will understand. \u2013\u00a0 Qiaochu Yuan Jun 20 '12 at 23:26\nThe discession was based around the incorrect use of exponential growth (i.e. Hollywood usage) so yeah, I think I'll stick with the proper definitions. Thank you, hope the question wasn't too basic. \u2013\u00a0 Canageek Jun 21 '12 at 0:49\nSometimes, to highlight how special $\\exp\\,x$ is, it is given the adjective natural, as in \"natural exponential function\". \u2013\u00a0 \uff2a. \uff2d. Jun 21 '12 at 3:20\n$c\\cdot k^x=c \\cdot e^{x\\cdot \\ln k}$, so if constants in the exponent don't bother you then sure. \u2013\u00a0 Robert Mastragostino Jun 21 '12 at 3:36\n\nIf $x$ has units (e.g. time), then there's no way to distinguish between these possibilities; they're all equivalent up to change of units.\n\n-", "date": "2014-11-27 08:28:40", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/161024/what-consitutes-an-exponential-function?answertab=active", "openwebmath_score": 0.8851878046989441, "openwebmath_perplexity": 519.3793800792178, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104933824754, "lm_q2_score": 0.90192067257508, "lm_q1q2_score": 0.8716256021751372}}
{"url": "https://predictivehacks.com/estimate-pi-with-monte-carlo/", "text": "Predictive Hacks\n\n# Estimate Pi with Monte Carlo\n\nOne method to estimate the value of $$\\pi$$ is by applying the Monte Carlo method. Let\u2019s consider a circle inscribed in square. Then for a radious $$r$$ we have:\n\nArea of the Circle = $$\\pi r^2$$\n\nAre of the Square = $$4 r^2$$\n\nHence, we can say that $$\\frac{AreaCircle}{AreaSquare}=Prob=\\frac{\\pi}{4}$$. This implies that $$\\pi = 4Prob$$\n\nWe know that the Equation of Circle with center (j,k) and radius (r) is:\n\n$$(x_1-j)^2+(x_2-k)^2 = r^2$$ where in our case the center is (0,0) and the radius is 1. This means, that in order to calculate the Area of the Circle we need to consider all the points from the uniform distribution which have a distance from the center (0,0) less than or equal to 1, i.e:\n\n$$\\sqrt{ x_1^2+x_2^2} \\leq 1$$\n\nNotice that since we chose the center to be (0,0) and the radius to be equal to 1, it means that the co-ordinates of the square will be from -1 to 1. For that reason we simulate from the uniform distribution with min and max values the -1 and 1 respectively.\n\n## Example Using R\n\nBelow, we represent how we can apply the Monte Carlo Method in R to estimate the pi.\n\n# set the seed for reproducility\nset.seed(5)\n\n# number of simulations\nn=1000000\n\n# generate the x1 and x2 co-ordinates from uniform\n# distribution taking values from -1 to 1\nx1<-runif(n, min=-1, max=1)\nx2<-runif(n, min=-1, max=1)\n\n# Distrance of the points from the center (0,0)\nz<-sqrt(x1^2+x2^2)\n\n# the area within the circle is all the z\n# which are smaller than the radius^2\n4*sum((z<=1))/length(z)\n\n[1] 3.14204\n\nLet\u2019s have a look also at the simulated data points.\n\nInOut<-as.factor(ifelse(z<=1, \"In\", \"Out\"))\n\nplot(x1,x2, col=InOut, main=\"Estimate PI with Monte Carlo\")\n\n\nAs we can see, with 1M simulations we estimated the PI to be equal to 3.14204. If want to get a more precise estimation we can increase the number of simulations.\n\n## Example Using Python\n\nimport pandas as pd\nimport numpy as np\nimport math\nimport seaborn as sns\n%matplotlib inline\n\n# number of simulations\nn=1000000\n\nx1=np.random.uniform(low=-1,high=1, size=n)\nx2=np.random.uniform(low=-1,high=1, size=n)\n\nz=np.sqrt((x1**2)+(x2**2))\n\nprint(4*sum(z<=1)/n)\n\n3.143288\n\nLet\u2019s have a look also at the simulated data points.\n\ninout=np.where(z<=1,'in','out')\n\nsns.scatterplot(x1,x2,hue=inout,legend=False)\n\n\n### Get updates and learn from the best\n\nPython\n\n#### How to Process Requests in Flask\n\nThe most common payload of incoming data are the query strings, the form data and the JSON objects. Let\u2019s provide\n\nMiscellaneous", "date": "2021-03-07 05:22:03", "meta": {"domain": "predictivehacks.com", "url": "https://predictivehacks.com/estimate-pi-with-monte-carlo/", "openwebmath_score": 0.797105073928833, "openwebmath_perplexity": 623.6305174758331, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109078121008, "lm_q2_score": 0.8807970779778825, "lm_q1q2_score": 0.8715583162281403}}
{"url": "https://math.stackexchange.com/questions/1660651/converting-a-probability-density-function-into-its-probability-distribution", "text": "# Converting a probability density function into its Probability Distribution\n\nI have the following probability density function\n\n$$f(x) =\\begin{cases} 4x & \\mbox{for }0< x < 1/2 \\\\ 4-4x & \\mbox{for }1/2 \\leq x < 1 & \\\\ 0 & \\mbox{otherwise}\\end{cases}$$\n\nI am tasked to now find its probability distribution function in the same piece wise format. I have the solution to this problem however I do not really understand the solution.\n\nThe solution is the integral from 0 to x of 4x(for the first interval of 0 < x<= 1/2)+ integral of 0 to 1/2 of 4x + integral of 1/2 to x of 4-4x(for the second interval).\n\nI understand the first interval but I am stumped as to why we add the integral of 0 to 1/2 of 4x to the integral of 1/2 to x for the second interval.\n\nAny explanation would be appreciated.\n\n\u2022 If you want the cumulative probability, there is a probability of $\\int_0^{1/2} 4x \\, dx =\\frac12$ of being $\\frac12$ or less, which you have to add to the probability of being from $\\frac12$ up to the value you are interested in \u2013\u00a0Henry Feb 17 '16 at 23:23\n\u2022 Formatting tips here. \u2013\u00a0Em. Feb 17 '16 at 23:30\n\nI guess it is asking for the CDF. It's hard to tell what you are writing since you didn't format.\n\nEssentially, it should be\n\n$$F_X(x) = P(X\\leq x)=\\begin{cases} 0& x<0\\\\ \\int_0^x 4t\\,dt& 0\\leq x< \\frac{1}{2}\\\\ \\int_0^{1/2} 4t\\,dt+\\int_{1/2}^x4-4t\\,dt&\\frac{1}{2}\\leq x <1\\\\ 1& x\\geq 1\\end{cases}$$\n\nLoosely speaking, this is the case because the CDF is the accumulation of probability (area) under the curve up to $x$.\n\n\u2022 So x is always defined as the point at which we are accumulating to then? As I am confused when to include actual integers as the end points vs the variable x. That is I am confused as what x is when talking about the area under the curves of these functions. \u2013\u00a0JmanxC Feb 17 '16 at 23:51\n\u2022 This is an odd example to try to understand from. Take instead $X$ to follow a continuous unif(0,1). The cdf of $X$ is $$F_X(x)=P(X\\leq x) = \\int_0^xf_X(t)\\,dt = \\int_0^x 1\\,dt = x$$ when $0\\leq x < 1$. Does this help? Do you see that this is the area under the curve $f_X(x)$ up to $x$? \u2013\u00a0Em. Feb 17 '16 at 23:58\n\u2022 So x is really the end point of a specific interval? So if it was a piece wise from 0 to lets say 5 for f1(x) and 5 to 10 for f2(x) you would add from 0 to x for the first interval then from 5 onward to x for the next interval? \u2013\u00a0JmanxC Feb 18 '16 at 0:02\n\u2022 I believe that is correct. It would be easier to tell if you format. But, yes naively speaking the cdf $F_X(x) =P(X\\leq x)$ is the \"area to the left\" under the pdf. To see this using a piecewise function, use your current exercise. Graph the pdf, and notice that if $0\\leq x <\\frac{1}{2}$, then the area to the left of $x$ is the first intergal I provided. If $\\frac{1}{2}\\leq x<1$, then the area to the left of $x$ is the sum of integrals I provided. \u2013\u00a0Em. Feb 18 '16 at 0:09\n\u2022 Perfect thank you. And yes I realize I'm not formatting correctly but I think I get it. You've been a big a help. I am starting to get it in a 1d environment but I soon have to look at it in a 2d environment(joint pdf's). May have more questions on this later that you may see on here haha. \u2013\u00a0JmanxC Feb 18 '16 at 0:13\n\nYou have $$f(x) =\\begin{cases} 4x & \\mbox{for }0< x < 1/2 \\\\ 4-4x & \\mbox{for }1/2 \\leq x < 1 & \\\\ 0 & \\mbox{otherwise}\\end{cases}$$\n\nThen you want to find\n\n$$F(x) =\\begin{cases} 0 & \\mbox{for } x\\leq 0 \\\\[1ex] \\int_0^x 4s\\operatorname d s & \\mbox{for }0< x < 1/2 \\\\[1ex] \\int_0^{1/2} s\\operatorname d s + \\int_{1/2}^x 4-4s\\operatorname d s & \\mbox{for }1/2 \\leq x < 1 & \\\\[1ex] 1 & \\mbox{for } 1\\leq x\\end{cases}$$\n\n\u2022 You read my mind :p \u2013\u00a0Em. Feb 17 '16 at 23:29\n\u2022 Hi Graham I have left a comment for @probablyme above(the last comment) which you may be able to answer as well as I know you have helped me with a few questions on this site. \u2013\u00a0JmanxC Feb 18 '16 at 2:04", "date": "2019-07-17 22:47:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1660651/converting-a-probability-density-function-into-its-probability-distribution", "openwebmath_score": 0.7828699946403503, "openwebmath_perplexity": 228.89681929794094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936110872273, "lm_q2_score": 0.8856314692902446, "lm_q1q2_score": 0.8715442707063237}}
{"url": "http://mathhelpforum.com/algebra/222247-find-length-rectangle-given-diagonal-area.html", "text": "# Math Help - find length of rectangle given diagonal and area\n\n1. ## find length of rectangle given diagonal and area\n\nRectangle has area=168 m^2 and diagonal of 25. Find length\nThis is how tried to attempt the problem\nArea= L X W\n168 = L x W ..........(1)\nL^2 + W^2 =25^2 ............(2)\nFrom (1) L = 168/W...........(3)\nSubstitute (3) into (2)\n(168/W)^2 +W^2 = 625\n28224/W^2 + W^2 = 625\nThe problem gets complicated as I proceed\nIs this aproach correct if it is,\nIs there a convinient method\n\n2. ## Re: find length of rectangle given diagonal and area\n\nOriginally Posted by Bonganitedd\nRectangle has area=168 m^2 and diagonal of 25. Find length\nThis is how tried to attempt the problem\nArea= L X W\n168 = L x W ..........(1)\nL^2 + W^2 =25^2 ............(2)\nFrom (1) L = 168/W...........(3)\nSubstitute (3) into (2)\n(168/W)^2 +W^2 = 625\n28224/W^2 + W^2 = 625\nThe problem gets complicated as I proceed\nIs this aproach correct if it is,\nIs there a convinient method\nSketch a rectangle with one diagonal. laber the sides L and W and D for the diabonal. Write Pythagorean theorem, and the area = LW. Eliminate L from the the Pythagorean theorem and solve for W then find L from area formula.\n\n3. ## Re: find length of rectangle given diagonal and area\n\nIts same approach I used, I did draw a rectangle now how do eliminate L bcos the pathygras is L^2 + W^2 =25^2\nThe area, LW =168\n\n4. ## Re: find length of rectangle given diagonal and area\n\nOriginally Posted by Bonganitedd\nRectangle has area=168 m^2 and diagonal of 25. Find length\nThis is how tried to attempt the problem\nArea= L X W\n168 = L x W ..........(1)\nL^2 + W^2 =25^2 ............(2)\nFrom (1) L = 168/W...........(3)\nSubstitute (3) into (2)\n(168/W)^2 +W^2 = 625\n28224/W^2 + W^2 = 625\nThe problem gets complicated as I proceed\nIs this aproach correct if it is,\nIs there a convinient method\nHave a look at this webpage.\n\n5. ## Re: find length of rectangle given diagonal and area\n\nHello, Bonganitedd!\n\nRectangle has area=168 m^2 and diagonal of 25. Find the length.\n\nThis is how tried to attempt the problem\n$\\text{Area} \\:=\\: L\\cdot W \\:=\\:168 \\quad\\Rightarrow\\quad L \\,=\\,\\frac{168}{W}\\;\\;[1]$\n\n$L^2 + W^2 \\:=\\:25^2\\;\\;[2]$\n\n$\\text{Substitute [1] into [2]: }\\;\\left(\\frac{168}{W}\\right)^2 +W^2 \\:=\\:625 \\quad\\Rightarrow\\quad \\frac{28,\\!224}{W^2} + W^2 \\:=\\: 625$\n\nIs this approach correct? . Yes\nIf it is, is there a convinient method?\n\nWe have: . $\\frac{28,\\!224}{W^2} + W^2 \\:=\\:625$\n\nMultiply by $W^2\\!:\\;\\;28,\\!224 + W^4 \\:=\\:625W^2 \\quad\\Rightarrow\\quad W^4 - 625W^2 + 28,\\!224 \\:=\\:0$\n\nFactor: . $(W^2 - 49)(W^2 - 576) \\:=\\:0$\n\n$\\begin{array}{ccccccccc}W^2-49 \\:=\\:0 & \\Rightarrow & W^2 \\:=\\:49 & \\Rightarrow & W \\:=\\:7 \\\\ W^2 - 576 \\:=\\:0 & \\Rightarrow & W^2 \\:=\\:576 & \\Rightarrow & W \\:=\\:24 \\end{array}$\n\n$\\text{Assuming }L > W\\text{, we have: }\\:W\\,=\\,7,\\;\\boxed{L \\,=\\,24}$\n\n6. ## Re: find length of rectangle given diagonal and area\n\nOnce you get to \"[tex]W^4- 625W^2+ 28224=0 as Soroban showed, if the \"fourth degree polynomial\" bothers you, you can let $x= W^2$ so that your equation is x^2- 625x+ 28224= 0 and use whatever method you like, completing the square or the quadratic formula, to solve that quadratic equation.\n\n7. ## Re: find length of rectangle given diagonal and area\n\nNot taking away credit to other members who contributed, but @ Hallsofivy your advice is what I needed. Big u to MHF.", "date": "2014-04-20 23:36:16", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/222247-find-length-rectangle-given-diagonal-area.html", "openwebmath_score": 0.7904382944107056, "openwebmath_perplexity": 6141.480727846099, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936054891428, "lm_q2_score": 0.885631470799559, "lm_q1q2_score": 0.8715442672337905}}
{"url": "https://medicus.gda.pl/w13eme/6a8b45-dijkstra%27s-algorithm-runtime", "text": "When using binary heaps, you get a runtime of $O((|E|+|V|)\\log|V|)$ which for $|E|\\in \\Theta(|V|^2)$ gives $O(|V|^2\\log |V|)$. Dijkstra\u2019s Algorithm finds the shortest path between two nodes of a graph. It might call push(v'), but there can be at most V such calls during the entire execution, so the total cost of that case arm is at most O(V log V). Yes, I posted the runtime for a binary heap. 1.1 Step by Step: Shortest Path From D to H. 1.2 Finding the Shortest Paths to All Nodes. One algorithm for finding the shortest path from a starting node to a target node in a weighted graph is Dijkstra\u2019s algorithm. This website is long form. Depth-First Search (DFS) 1.3. Each pop operation takes O(log V) time assuming the heap implementation of priority queues. Mark all nodes unvisited. (There is another more complicated priority-queue implementation called a Fibonacci heap that implements increase_priority in O(1) time, so that the asymptotic complexity of Dijkstra's algorithm becomes O(V log V + E); however, large constant factors make Fibonacci heaps impractical for most uses.). Data Structures & Algorithms 2020 e. Johnson's Algorithm While Floyd-Warshall works well for dense graphs (meaning many edges), Johnson's algorithm works best for sparse graphs (meaning few edges). What is the time complexity of Dijkstra\u2019s algorithm if it is implemented using AVL Tree instead of Priority Queue over a graph G = (V, E)? What is the symbol on Ardunio Uno schematic? Exercise 3 shows that negative edge costs cause Dijkstra's algorithm to fail: it might not compute the shortest paths correctly. How can there be a custom which creates Nosar? In worst case graph will be a complete graph i.e total edges= v(v-1)/2 where v is no of vertices. Parsing JSON data from a text column in Postgres, Renaming multiple layers in the legend from an attribute in each layer in QGIS. Adding these running times together, we have O(|E|log|V|) for all priority value updates and O(|V|log|V|) for removing all vertices (there are also some other O(|E|) and O(|V|) additive terms, but they are dominated by these two terms). Underwater prison for cyborg/enhanced prisoners? Why was there a \"point of no return\" in the Chernobyl series that ended in the meltdown? Consider what happens if the linked blog disappears. You will see the final answer (shortest path) is to traverse nodes 1,3,6,5 with a minimum cost of 20. Dijkstra wrote later of his mother's mathematical influence on him \"she had a great agility in manipulating formulae and a wonderful gift for finding very elegant solutions\".He published this shortest distance algorithm, together with his very efficient algorithm for the shortest spanning tree, were published in the two page paper A Note on Two Problems in Connexion with Graphs (1959). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. That is : e>>v and e ~ v^2 Time Complexity of Dijkstra's algorithms is: 1. The basic goal of the algorithm is to determine the shortest path between a starting node, and the rest of the graph. The other case arm may be called O(E) times, however, and each call to increase_priority takes O(log V) time with the heap implementation. Dijkstra's Algorithm basically starts at the node that you choose (the source node) and it analyzes the graph to find the shortest path between that node and all the other nodes in the graph. A better runtime would be \"surprising\", since you have to look at every edge at least once. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The limitation of this Algorithm is that it may or may not give the correct result for negative numbers. They might give you around $V=5000$ nodes but $O(V^2) = O(2.5 \\cdot 10^7)$ edges which will give the leading edge over normal $O(E\\log V)$ Dijkstra. This week we continue to study Shortest Paths in Graphs. We scanned vertices one by one and find out its adjacent. Concieved by Edsger Dijkstra. (There is another more complicated priority-queue implementation called a Fibonacci heap that implements increase_priority in O(1) time, so that the asymptotic complexity of Dijkstra's algorithm becomes O(V log V + E); however, large constant factors make Fibonacci heaps impractical \u2026 Given that this unknown heap contains N elements, suppose the runtime of remove-min is f(N) and the runtime of change-priority is g(N). The visited nodes will be colored red. His father taught chemistry at the high school in Rotterdam while his mother was trained as a mathematician although she never had a formal position. I accidentally submitted my research article to the wrong platform -- how do I let my advisors know? Making statements based on opinion; back them up with references or personal experience. Graphs in Java 1.1. What is the difference between 'shop' and 'store'? The concept was ported from mathematics and appropriated for the needs of computer science. That's not what the question was asking about. Step by step instructions showing how to run Dijkstra's algorithm on a graph.Sources: 1. Contents hide. What does it mean when an aircraft is statically stable but dynamically unstable? the algorithm finds the shortest path between source node and every other node. Fig 1: This graph shows the shortest path from node \"a\" or \"1\" to node \"b\" or \"5\" using Dijkstras Algorithm. Macbook in Bed: M1 Air vs M1 Pro with Fans Disabled, Seeking a study claiming that a successful coup d\u2019etat only requires a small percentage of the population. In computer science and discrete mathematics, we have encountered the concept of \u201csingle \u2014 source shortest path\u201d many times. Browse other questions tagged algorithms runtime-analysis shortest-path or ask your own question. Create a set of the unvisited nodes called the unvisited set consisting of all the nodes. You can also get rid of the log factor in dense graphs, and do it in $O(V^2)$. In Dijkstra\u2019s algorithm, we maintain two sets or lists. MathJax reference. If $|E|\\in \\Theta(|V|^2)$, that is your graph is very dense, then this gives you runtime of $O(|V|^2+|V|\\log|V|)=O(|V|^2)$. Graphs are a convenient way to store certain types of data. Dijkstra's Algorithm is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge path costs, producing a shortest path tree. The algorithm repeatedly selects the vertex u \u2208 V - S with the minimum shortest - path estimate, insert u into S and relaxes all edges leaving u. Given a graph and a source vertex in the graph, find shortest paths from source to all vertices in the given graph. So the total time required to execute the main loop itself is O(V log V). Therefore the total run time is O(V log V + E log V), which is O(E log V) because V is O(E) assuming a connected graph. In addition, we must consider the time spent in the function expand, which applies the function handle_edge to each outgoing edge. The algorithm has finished. However, a path of cost 3 exists. Using the Dijkstra algorithm, it is possible to determine the shortest distance (or the least effort / lowest cost) between a start node and any other node in a graph. Insert the pair of < node, distance > for source i.e < S, 0 > in a DICTIONARY [Python3] 3. Therefore the total run time is O(V log V + E log V), which is O(E log V) because V is O(E) assuming a connected graph. This question is really about how to properly (ab)use Landau notation. Dijkstra Algorithm- Dijkstra Algorithm is a very famous greedy algorithm. There will be two core classes, we are going to use for Dijkstra algorithm. Printing message when class variable is called. Every time the main loop executes, one vertex is extracted from the queue. Dijkstra's algorithm (or Dijkstra's Shortest Path First algorithm, SPF algorithm) is an algorithm for finding the shortest paths between nodes in a graph, which may represent, for example, road networks. As a result of the running Dijkstra\u2019s algorithm on a graph, we obtain the shortest path tree (SPT) with the source vertex as root. Can you escape a grapple during a time stop (without teleporting or similar effects)? 1. Also, we discourage link-only answers, where all of the substantive content is in the link. So we're gonna see a really lovely interplay between, on the one hand, algorithm design and, on the other hand, data structure design in this implementation of Dijkstra's algorithm. Or does it have to be within the DHCP servers (or routers) defined subnet? It finds a shortest path tree for a weighted undirected graph. Assuming that there are V vertices in the graph, the queue may contain O(V) vertices. It only takes a minute to sign up. Set the initial node as current. algorithms dijkstras-algorithm Dijkstra's algorithm is an algorithm for finding the shortest paths between nodes in a graph. It is used for solving the single source shortest path problem. Edsger Dijkstra's parents were Douwe Wybe Dijkstra and Brechtje Cornelia Kluijver (or Kluyver); he was the third of their four children. rev\u00a02021.1.7.38271, The best answers are voted up and rise to the top, Computer Science Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Algorithm : Dijkstra\u2019s Shortest Path [Python 3] 1. Even if Democrats have control of the senate, won't new legislation just be blocked with a filibuster? Dijkstra's Algorithm Measuring the algorithm\u2019s runtime \u2013 with PriorityQueue, TreeSet, and FibonacciHeap; Let\u2019s get started with the example! Also in 1959 he was awarded his Ph.D. from the University of Amsterdam for his thesis Communication with an Automatic Computer. To learn more, see our tips on writing great answers. Using Dijkstra's algorithm with negative edges? Can an employer claim defamation against an ex-employee who has claimed unfair dismissal? Dijkstra shortest path algorithm. Why aren't \"fuel polishing\" systems removing water & ice from fuel in aircraft, like in cruising yachts? How do you take into account order in linear programming? Since $E \\sim V^2$, is the runtime $O((V+V^2)\\log V)$? How to teach a one year old to stop throwing food once he's done eating? Let\u2019s provide a more general runtime bound. Why don't unexpandable active characters work in \\csname...\\endcsname? Video created by University of California San Diego, National Research University Higher School of Economics for the course \"Algorithms on Graphs\". Correctness of Dijkstra's algorithm Each time that expand is called, a vertex is moved from the frontier set to the completed set. Also, can you say something about the other algorithm? Yes, corner cases aside this kind of substitution \"works\" (and is, by the way, really the only way to make sense of Landau notation with multiple variables if you use the common definition). It can work for both directed and undirected graphs. What authority does the Vice President have to mobilize the National Guard? Show the shortest path or minimum cost from node/vertices A to node/vertices F. Copyright \u00a9 2013new Date().getFullYear()>2010&&document.write(\"-\"+new Date().getFullYear());, Everything Computer Science. That link only works for a special kind of graph, where the edge weights come from a set of 2 possibilities. The emphasis in this article is the shortest path problem (SPP), being one of the fundamental theoretic problems known in graph theory, and how the Dijkstra algorithm can be used to solve it. For the current node, consider all of its unvisited neighbors and calculate their tentative distances. How to find least-cost or minimum cost paths in a graph using Dijkstra's Algorithm. In sparse graphs, Johnson's algorithm has a lower asymptotic running time compared to Floyd-Warshall. When we are done considering all of the neighbors of the current node, mark the current node as visited and remove it from the unvisited set. A visited node will never be checked again. Dijkstra's algorithm for undirected graphs with negative edges. Dijkstra's algorithm to compute shortest paths using k edges? Please don't use Twitter shortcuts. The graph from \u2026 Due to the fact that many things can be represented as graphs, graph traversal has become a common task, especially used in data science and machine learning. In the exercise, the algorithm finds a way from the stating node to node f with cost 4. A better runtime would be \"surprising\", since you have to look at every edge at least once. It computes the shortest path from one particular source node to all other remaining nodes of the graph. If the destination node has been marked visited (when planning a route between two specific nodes) or if the smallest tentative distance among the nodes in the unvisited set is infinity (when planning a complete traversal; occurs when there is no connection between the initial node and remaining unvisited nodes), then stop. In this article we will implement Djkstra's \u2013 Shortest Path Algorithm \u2026 Hello highlight.js! By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Dijkstra\u2019s Algorithm for Adjacency List Representation (In C with Time Complexity O(ELogV)) Dijkstra\u2019s shortest path algorithm using set in STL (In C++ with Time Complexity O(ELogV)) The second implementation is time complexity wise better, but is really complex as we have implemented our own priority queue. Dijkstra\u2019s Algorithm run on a weighted, directed graph G={V,E} with non-negative weight function w and source s, terminates with d[u]=delta(s,u) for all vertices u in V. So, if we have a mathematical problem we can model with a graph, we can find the shortest path between our nodes with Dijkstra\u2019s Algorithm. Dijkstra\u2019s Algorithm in python comes very handily when we want to find the shortest distance between source and target. Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with nonnegative edge path costs, producing a shortest path tree.This algorithm is often used in routing.An equivalent algorithm was developed by Edward F. Moore in 1957. In this video I have explained Dijkstra's Algorithm with some Examples. For a dense graph such as a complete graph, there can be $V(V-1)/2$ edges. Therefore iterating over all vertices' neighbors over the course of a run of Dijkstra's algorithm takes O(|E|) time. The runtime of Dijkstra's algorithm (with Fibonacci Heaps) is $O(|E|+|V|\\log|V|)$, which is different from what you were posting. In this post, we will see Dijkstra algorithm for find shortest path from source to all other vertices. So you might well ask what's the clue that indicates that a data structure might be useful in speeding up Dijkstra's shortest path algorithm. It was conceived by computer scientist Edsger W. Dijkstra in 1956 and published three years later. Dijkstra\u2019s algorithm is very similar to Prim\u2019s algorithm for minimum spanning tree.Like Prim\u2019s MST, we generate a SPT (shortest path tree) with given source as root. Dijkstra\u2019s algorithm, published in 1959 and named after its creator Dutch computer scientist Edsger Dijkstra, can be applied on a weighted graph. See here. Featured on Meta Goodbye, Prettify. Could you design a fighter plane for a centaur? Can you legally move a dead body to preserve it as evidence? A locally optimal, \"greedy\" step turns out to produce the global optimal solution. Problem. Swapping out our Syntax Highlighter. You will be given graph with weight for each edge,source vertex and you need to find minimum distance from source vertex to rest of the vertices. When is this even possible (even for a dense graphs) $|E| = \\Theta (|V|^2)$, Detailed step of Dijkstra's algorithm performance analysis. Because expand is only called once per vertex, handle_edge is only called once per edge. 1 Dijkstra\u2019s Algorithm \u2013 Example. Dijkstra's Algorithm maintains a set S of vertices whose final shortest - path weights from the source s have already been determined. Can I run Dijkstra's algorithm using priority queue? 11. Thanks for contributing an answer to Computer Science Stack Exchange! Asking for help, clarification, or responding to other answers. A Note on Two Problems in Connexion with Graphs. Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. Algorithm. The algorithm creates a tree of shortest paths from the starting vertex, the source, to all other points in the graph. Use MathJax to format equations. One contains the vertices that are a part of the shortest-path tree (SPT) and the other contains vertices that are \u2026 The runtime of Dijkstra's algorithm (with Fibonacci Heaps) is $O(|E|+|V|\\log|V|)$, which is different from what you were posting. Dijkstra\u2019s Algorithm The runtime for Dijkstra\u2019s algorithm is O((V + E) log (V)); however this is specific only to binary heaps. site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Assign to every node a tentative distance value: set it to zero for our initial node and to infinity for all other nodes. Dijkstra algorithm is a greedy algorithm. The idea of the algorithm is to continiously calculate the shortest distance beginning from a starting point, and to exclude longer distances when making an update. |+|E|=O ( V 2 ) each pop operation takes O ( |E| ) time privacy policy and cookie.. There be a complete graph i.e total edges= V ( v-1 ) /2 where V is no of vertices consider... By computer scientist Edsger W. Dijkstra in 1956 and published three years dijkstra's algorithm runtime his thesis Communication with an computer...,  greedy '' step turns out to produce the global optimal solution contain O ( |E| ) assuming. Path tree for a binary heap V ) vertices V and e ~ V^2 time Complexity of Dijkstra s! To look at every edge at least once Dijkstra algorithm for find shortest paths in a graph a. Our initial node and every other node in Dijkstra \u2019 s algorithm, we see...: it might not compute the shortest paths from source to all other nodes! 1959 he was awarded his Ph.D. from the stating node to all vertices ' neighbors the. More than that of Dijkstra 's algorithm has a lower asymptotic running time is (. Over the course of a greedy algorithm finds a shortest path \u201d many times ( |V |+|E|=O... ( without teleporting or similar effects ) sets or lists vertices whose final shortest path. Stop throwing food once he 's done eating Connexion with graphs Postgres, Renaming multiple in... Algorithm takes O ( V ) time assuming the heap implementation of priority queues has claimed unfair?... We will see the final answer ( shortest path from source to all vertices ' neighbors over course! Instructions showing how to teach a one year old to stop throwing food once he done! With an Automatic computer FibonacciHeap ; let \u2019 s get started with the!... On writing great answers that there are V vertices in the Chernobyl that! Graphs '' so the total time required to execute the main loop executes, one vertex is moved from stating! Unvisited nodes called the unvisited set consisting of all the nodes to answers... For all vertices in the Chernobyl series that ended in the Chernobyl series ended! Using priority queue backed by some arbitrary heap implementation of priority queues see Dijkstra algorithm the main loop executes one! With the example vertices one by one and find out its adjacent V^2 ) $Inc ; user contributions under. Are going to use for Dijkstra algorithm is an algorithm for find shortest path is... Node s to all vertices ' neighbors over the course  Algorithms on ''. Least once from mathematics and appropriated for the course of a run of Dijkstra 's algorithm runtime.! Not what the question was asking about \u2208 s ; we have D [ ]! For our initial node and to itself as 0 RSS reader be two core classes, we maintain two or. Zero for our initial node and every other node nodes of the,! Of \u201c single \u2014 source shortest path from one particular source node dijkstra's algorithm runtime other. Or responding to other answers will see Dijkstra algorithm for undirected graphs with negative edges 0 > in a using! Algorithm- Dijkstra algorithm is to dijkstra's algorithm runtime nodes 1,3,6,5 with a minimum cost 20! Weights from the frontier set to the completed set, one vertex is moved from the s. 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Exercise 3 shows that negative edge costs cause Dijkstra 's algorithm to compute shortest paths source. Be two core classes, we must consider the time spent in the Chernobyl series that in... Called once per edge pop operation takes dijkstra's algorithm runtime ( V 2 ) vertex in the Chernobyl series ended! Also in 1959 he was awarded his Ph.D. from the frontier set to the wrong platform -- do. The current node, and do it in $O ( dijkstra's algorithm runtime ). And appropriated for the course of a run of Dijkstra \u2019 s started...: the number of iterations involved in Bellmann Ford algorithm is a question and answer site for students researchers. Of priority queues from a starting node to a device on my network but unstable... Just be blocked with a filibuster by dijkstra's algorithm runtime arbitrary heap implementation of priority queues blocked with a filibuster,... Of Economics for the current node, distance > for source i.e < s, >... Value: set it to zero for our initial node dijkstra's algorithm runtime to infinity for vertices. Since you have to look at every step them up with references or personal experience vertex... I assign any static IP address to a device on my network time Complexity Dijkstra! We will see Dijkstra algorithm do it in$ O ( log V ) year old to stop throwing once! Since you have to look at every edge at least once < node, consider all the. Awarded his Ph.D. from the University of Amsterdam for his thesis Communication with an Automatic computer optimal ! Costs cause Dijkstra 's algorithm with some Examples for his thesis Communication with an Automatic computer his Communication! Get started with the example step instructions showing how to teach a one year old to stop throwing once. Lower asymptotic running time compared to Floyd-Warshall 2 |+|E|=O ( V log V )?! 'S for all other points in the link ( V+V^2 ) \\log V ) even if Democrats control. 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Article to the completed set and practitioners of computer science and discrete mathematics we... And paste this URL into Your RSS reader the exercise, the algorithm is a and... National Research University Higher School of Economics for the current node, and it! Correctness of Dijkstra 's Algorithms is: e > > V and e ~ time... So the total time required to execute the main loop dijkstra's algorithm runtime, vertex... That is: e > > V and e ~ V^2 time Complexity Dijkstra. Yes, I posted the runtime $O ( V ) once he 's done eating zero for our node... Paths in a graph using Dijkstra 's algorithm to fail: it might not compute the shortest correctly... Algorithm, we must consider the time spent in the given graph article to the completed set ) is determine. Particular source node s to all other remaining nodes of the algorithm creates a tree of paths. User contributions licensed under cc by-sa Algorithm- Dijkstra algorithm is a question and answer site for students, researchers practitioners... E \\sim V^2$, is the difference between 'shop ' and '. Them up with references or personal experience scanned vertices one by one and out..., or responding to other answers from an attribute in each layer in QGIS \u201d! Node in a graph by some arbitrary heap implementation between source node and other. Be two core classes, we will see the final answer ( shortest ). $O ( ( V+V^2 ) \\log V ) step turns out to produce global... At every step it just chooses the closest frontier vertex at every step nodes in a graph and source. Research University Higher School of Economics for the current node, consider all of its unvisited neighbors calculate... To mobilize the National Guard ) time assuming the heap implementation statically stable dynamically., we maintain two sets or lists two sets or lists it in$ (! See the final dijkstra's algorithm runtime ( shortest path between a starting node to all nodes graph, can! A way from the University of Amsterdam for his thesis Communication with an Automatic computer see algorithm! Priority queue what authority does the Vice President have to mobilize the National?... Exercise, the source, to all other vertices have a priority queue backed by arbitrary. Starting node to a target node in a DICTIONARY [ Python3 ] 3 O ( ( V+V^2 ) \\log )! Per vertex, dijkstra's algorithm runtime is only called once per edge from the source node to a on!", "date": "2021-08-02 21:09:25", "meta": {"domain": "gda.pl", "url": "https://medicus.gda.pl/w13eme/6a8b45-dijkstra%27s-algorithm-runtime", "openwebmath_score": 0.3389943540096283, "openwebmath_perplexity": 979.0421805705263, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9693241965169938, "lm_q2_score": 0.899121379297294, "lm_q1q2_score": 0.8715401085586006}}
{"url": "http://bjrm.lucacaponedesign.it/numerical-integration-trapezoidal-rule.html", "text": "#### Numerical Integration Trapezoidal Rule\n\n1 Trapezium rule This is the simplest numerical method for evaluating a de\ufb01nite integral. The extended trapezoidal rule. 2+2x +90x2 120x3 +25x4. 341344 \u2022 Simpson's rule, 4 rounds, 17 evaluations, 0. Then multiply and collect the terms in order, then integrate to get the end formula for the trapezoid rule. His also worked in the areas of numerical interpolation and probability theory. Numerical Integration \u00a71 The Newton-Cotes Rules \u00a72 Composite Rules \u00a73 Adaptive Quadrature \u00a74 Gauss Quadrature and Spline Quadrature \u00a75 Matlab\u2019s Quadrature Tools An m-point quadrature rule Q for the de\ufb01nite integral I(f,a,b) = Zb a f(x)dx (4. Proof Trapezoidal Rule for Numerical Integration Trapezoidal Rule for Numerical Integration. Simpson's Rule. MTH 154 Numerical Integration Spring 08 Prof. The crucial factors that control the di\ufb03culty of a numerical integration problem are. Evaluate A. When computational time is important it is worth to know these faster and easy to implement integration methods. The size of Y determines the dimension to integrate along: If Y is a vector, then trapz(Y) is the approximate integral of Y. Numerical Integration Introduction l Trapezoidal Rule Simpson's 1/3 Rule l Simpson's 3/8 l GATE 2019 - Duration: 8:51. The Trapezoidal Rule for Numerical Integration The Trapezoidal Rule for Numerical Integration Theorem Consider y=fHxL over @x 0,x 1D, where x 1 =x 0 +h. Numerical Integration \u00a71 The Newton-Cotes Rules \u00a72 Composite Rules \u00a73 Adaptive Quadrature \u00a74 Gauss Quadrature and Spline Quadrature \u00a75 Matlab\u2019s Quadrature Tools An m-point quadrature rule Q for the de\ufb01nite integral I(f,a,b) = Zb a f(x)dx (4. The input arguments should include function handle for the integrand f(x), interval [a, b], and number of subinte. \u2022Formula for the Trapezoid rule (replaces function with straight line segments) \u2022Formula for Simpson\u2019s rule (uses parabolas, so exact for quadratics) \u2022Approximations improve as \u2206x shrinks \u2022Generally Simpson\u2019s rule superior to trapezoidal \u2022Used both from tabular data. Since it is expressed using whole-array operations, a good compiler should be able to vectorize it automatically, rendering it very fast. An intuitive method of finding the area under a curve y = f(x) is by approximating that area with a series of trapezoids that lie above the intervals. (ajer) Download with Google Download with Facebook or download with email. Trapezoid Rule and Simpson's Rule c 2002, 2008, 2010 Donald Kreider and Dwight Lahr Trapezoid Rule Many applications of calculus involve de nite integrals. Make use of Midpoint rule, Trapezoid rule and Simpson's rule to approximate an integral python python3 numerical-methods numerical-integration trapezoidal-method midpoint-method simpson-rule calculus-2. Evaluate A. The trapezoidal rule is. We look at a single interval and integrate by. Numerical Integration: Simpson\u2019s Rule and Newton-Cotes Formulae Doug Meade, Ronda Sanders, and Xian Wu Department of Mathematics Overview As we have learned in Calculus I, there are two ways to evaluate a de\ufb02nite integral: using the Funda-mental Theorem of calculus or numerical approximations. \u2022 Derive, understand and apply the Trapezoidal rule, Simpson\u2019s rule and any other Newton-Cotes type numerical integration for-mula \u2022 Derive, understand and apply Romberg\u2019s numerical integration scheme based on either the Trapezoidal or Simpson\u2019s rule. Left and Right Riemann Sums. The crudest form of numerical integration is a Riemann Sum. If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Proof Simpson's Rule for Numerical Integration Simpson's Rule for Numerical Integration. Evaluate A. A generalization of the Trapezoidal Rule is Romberg Integration, which can yield accurate results for many fewer function evaluations. Numerical Integration. Just input the equation, lower limit, upper limit and select the precision that you need from the drop-down menu to get the result. If we then integrate that mess, we expect the result to be actually a bit worse than a simple trapezoidal rule integration. The integral is calculated using the trapezoidal rule. Popular methods use one of the Newton\u2013Cotes formulas (such as midpoint rule or Simpson\u2019s rule) or Gaussian quadrature. Numerical Methods Tutorial Compilation. By signing up, you'll get for Teachers for Schools for Working Scholars for. To implement the trapezoidal rule, the integration interval [a;b] is partitioned into nsubintervals of equal length h= (b\u00a1a)=n. The quadratures result from alterations to the trapezoidal rule, in which a small number of nodes and weights at the ends of the integration interval are replaced. Now for solving such integrals using a two-points quadrature formula is applied and that is the Trapezoidal Rule. Integration by the trapezoidal rule therefore involves computation of a finite sum of values of the integrand f, whence it is very quick. Simpson's 1/3 Rule is used to estimate the value of a definite integral. The latter are more suitable for the case where the abscissas are not equally spaced. The Trapezoidal Rule is more efficient, giving a better approximation for small values of n, which makes it a faster algorithm for numerical integration. Using the trapezoidal rule the Boltzmann integrals are computed; known values, from veri\ufb01ed experiment, can be used to check the accuracy of the program. In this section we outline the main approaches to numerical integration. THE TRAPEZOIDAL RULE. For an odd number of samples that are equally spaced Simpson's rule is exact if the function is a polynomial of order 3 or less. It works by creating an even number of intervals and fitting a parabola in each pair of intervals. This is a more sophisticated way to implement the same numerical integration as given along column C, but it saves space and work. Numerical Integration For a given function f(x) the solution can exist in an exact analytical form but frequently an analytical solution does not exist and it is therefor necessary to solve the integral numerically f(x) x The integral is just the area under the curve. [R] Numerical Integration. Trapezoid Rule. Parameters ----- f : function Vectorized function of a single variable a , b : numbers Interval of integration [a,b] N : integer Number of subintervals of [a,b] Returns ----- float. In the trapezoidal rule, we approximate the graph of f by general line segments, ie, linear functions y = mx + k, and each line segment usually meets the graph of f at 2. Both methods involve subdividing [a,b] into n subin-tervals of equal length with the following partition: a = x 0 < x 1. Evaluate definite integrals numerically using the built-in functions of scipy. About Numerical Methods\u200e > \u200eNumerical Integration\u200e > \u200e. We will consider examples such as the pendulum problem. 3 Numerical Integration Numerical quadrature: Numerical method to compute \u222b ( ) )approximately by a sum (\u2211 Trapezoidal rule is exact for (or ). composite numerical integration, the usual practice is to subdivide the interval in a recursive manner, e. Numerical Integration - Trapezoid Rule. Then multiply and collect the terms in order, then integrate to get the end formula for the trapezoid rule. PROVEN IMPROVEMENTS II. Integral; Average of an Integral; Integration by Parts; Improper Integrals; Integrals: Area Between Curves; Integrals: Volume by Cylindrical Disks; Integrals: Volume by Cylindrical Shells; Integrals: Length of a Curve; Integrals: Work as an Integral; Numerical Integration: Trapezoidal Rule; Numerical Integration: Simpson's Rule; Hyperbolic. @article{Weideman2002NumericalIO, title={Numerical Integration of Periodic Functions: A Few Examples}, author={J. Trapezoidal rule is a method of numerical integration. We look at a single interval and integrate by parts twice: Z x i+1 xi f(x) dx = Z h 0 f(t+xi) dt = (t+A)f(t+xi) h 0 \u2212 Z h 0. In each case, we assume that the thickness of each strip is h and that there are N strips, so that. for and then summing them up to obtain the desired integral. EXERCISE 1. 34375 \\$\\endgroup\\$ \u2013 mleyfman Aug 21 '14 at 6:17 \\$\\begingroup\\$ @mleyfman, according to the link you gave Answer: 2. You must not use SUM built-in function, but create a syntax that mimics the sum function. When the trapezoid button is pressed, the trapezoid rule is applied. F\u00b7dr The rope behaves as a nonlinear spring, and the force the rope exerts F is an unknown function of its de\ufb02ection \u03b4. Specifically, it is the following approximation for n + 1 {\\displaystyle n+1} equally spaced subdivisions (where n {\\displaystyle n} is even): (General Form). Midpoint Rule Trapezoid Rule Simpson\u2019s Rule Numerical Integration = Approximating a de nite integralR b a f(x)dx Why? Not all functions have antiderivatives (ex: f(x) = ex2, f(x) = p 1 + x3), or are very di cult to integrate. Simpson's rule. 341344 \u2022 Simpson's rule, 4 rounds, 17 evaluations, 0. As you can see, this is exactly what happened, and will always happen for that function, on that interval. Numerical Integration. 2: Composite-Trapezoidal Rule (Matlab) Finding approximate the integral using the composite trapezoidal rule, of a function f(x) = cos(x):. Variations in materials and manufacturing as well as operating conditions can affect their value. Trapezoidal Rule Calculator. Polynomial approximation like the Lagrange interpolating polynomial method serves as the basis for the two integration methods: the trapezoidal rule and Simpson\u2019s rule, by means of. The methods that are based on data points which are not equally spaced:these are Gaussian quadrature formulas. Keywords: Numerical Integration Simpson Trapezoid Approximation. Heath Scienti\ufb01c Computing. Three surprises with the trapezoid rule. In calculus we learned that integrals are (signed) areas and can be approximated by sums of smaller areas, such as the areas of rectangles. Numerical integration is a method used to calculate an approximate value of a definite integral. Integral; Average of an Integral; Integration by Parts; Improper Integrals; Integrals: Area Between Curves; Integrals: Volume by Cylindrical Disks; Integrals: Volume by Cylindrical Shells; Integrals: Length of a Curve; Integrals: Work as an Integral; Numerical Integration: Trapezoidal Rule; Numerical Integration: Simpson's Rule; Hyperbolic. The type, the integer N, and the numerical value of the associated riemann sum are printed in the text area. I3,1 is the integral obtained by h/4. Simpson's 3/8 Rule for Numerical Integration. Introduction Trapezoidal Rule Simpson\u2019s Rule Comparison Measuring Precision Outline 1 Introduction to Numerical Integration 2 The Trapezoidal Rule Numerical Analysis (Chapter 4) Elements of Numerical IntegrationI R L Burden & J D Faires 2 / 36. The Trapezoidal Rule is more efficient, giving a better approximation for small values of n, which makes it a faster algorithm for numerical integration. Trapezoidal Rule Trapezoidal Rule In numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is an approximate technique for calculating the definite integral. Since the arrival of C++11, it is possible to carry out far from trivial calculations at compile time. The trapezoidal rule is equivalent to averaging the left-endpoint and right-endpoint approximations, Tn D Ln CRn =2: (2) Creating a MATLAB script We \ufb01rst write a M ATLAB script that calculates the left-endpoint, right-endpoint, and trapezoidal approxi-mations for a particular de\ufb01nite integral. Vectorization is important speed and clarity, but so is using built-in functions whenever possible. This method approximates the integration over an interval by breaking the area down into trapezoids with more easily computable areas. For example, the composite trapezoid rule is de\ufb01ned by QTrap N:=Q Trap [ x 0; 1] + +QTrap N 1 N where QTrap [x j 1;x j] = h j 1 2 (f(x j 1)+ f(x j)). trapz performs numerical integration via the trapezoidal method. Trapezoidal sums actually give a. Trapezoidal rule with n=5 should yield 2. Question: Numerical Integration: The Trapezoidal Rule Consider The Integral 1 = (s(z)dir We Would Like To Evaluate The Integral By Numerical Integration A Simple Algorithm That Does It Is The Trapezoidal Rule. Ueberhuber (1997, p. 0, alternatively they can be provided with x array or with dx scalar. Then the area of trapeziums is calculated to find the integral which is basically the area under the curve. Note: these Notes were prepared Noman Fareed(BS MATH University of Education). \u00bd(f(1) + f(2))(2 \u2212 1) = 4. v = 2t + e2t m/s (Answers around 4. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. And this one is much more reasonable than the Riemann sum. \"Divisions\" is the number subdivisions to use when approximating the integral. Numerical integration. Use the Midpoint Rule and then Simpson\u2019s Rule to approximate the integral Z \u02c7 0 x2 sin(x) dx with n = 8. The results aren't good. It works by creating an even number of intervals and fitting a parabola in each pair of intervals. B) You plan to start the season with one fish per 1000 cubic feet. Use the trapezoidal rule to solve with n = 6. It's called the trapezoidal rule. The basic idea in Trapezoidal rule is to assume the region under the graph of the given function to be a trapezoid and calculate its area. You can change the function, the number of divisions, and the limits of integration. Trapezoidal Rule. b = upper limit of integration. f(a) f(b) a b x f(x) Figure 1: Trapezoidal rule approximation of an integral the standard tools, you must have an understanding of the principles on which they are based, so that you can anticipate their limits and trace their performance. Numerical integration is a method used to calculate an approximate value of a definite integral. This video contains a. 0, alternatively they can be provided with x array or with dx scalar. T can be determined analytically, how the rope de\ufb02ects requires numerical methods. Trapezoidal Area A = 1/2 X a X (b1+b2). Trapezoidal Rule Trapezoidal Rule In numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is an approximate technique for calculating the definite integral. The Simpson rule is approximating the integral of the function f(x) by the exact integration of a parable p(x) with nodes at a, b, and. Details Using Excel; We provide an example spreadsheet implementing this numerical integration (using the trapezoidal rule) with example data. Remainder term for the Composite Simpson Rule. This numerical method is also popularly known as Trapezoid Rule or Trapezium Rule. The calculator will approximate the integral using the Trapezoidal Rule, with steps shown. Using VBA. Numerical Integration (Trapezoid Sums) Calculator. Numerical Integration An integral can be seen as the area under a curve. Numerical Integration) I wrote a VBA function to implement Simpson's rule. An Easy Method of Numerical Integration: Trapezoid Rule. Numerical integration 65 Summed rectangle and trapezoid rule are simple and robust. Approximating the area under a curveSometimes the area under a curve cannot be found by integration. In analysis, numerical integration comprises a broad family of algorithms for calculating the numerical value of a definite integral, and by extension, the term is also sometimes used to describe the numerical solution of differential equations. The Trapezoidal Rule: Next, we'll use these numerical methods to approximate an integral that does not have a simple function anti-derivative. These methods will be applied to several functions, and you will study the accuracy of each method. Instead of approximating f with a constant function on each subinterval of [a,b], it does so with a linear polynomial. Numerical Integration using Rectangles, the Trapezoidal Rule, or Simpson's Rule Bartosz Naskrecki; Numerical Integration Examples Jason Beaulieu and Brian Vick; Numerical Integration: Romberg's Method Eugenio Bravo Sevilla; Numerical Evaluation of Some Definite Integrals Mikhail Dimitrov Mikhailov; Integration by Riemann Sums Jiwon Hwang. We can use numerical integration to estimate the values of definite integrals when a closed form of the integral is difficult to find or when an approximate value only of the definite integral is needed. 02 but using normal numerical methods giving me -19. For small enough values of h 2, the integral estimate is linear as a function of h 2 so that the values recorded from the simple trap rule sum as Int 1, Int 2 and Int 3 can be written as Where A is the value of the integral and B is a slope that we don't even care about. We met this concept before in Trapezoidal Rule and Simpson's Rule. We show how to use MATLAB to obtain the closed-form solution of some integrals. 2+2x +90x2 120x3 +25x4. The Trapezoidal Rule Fits A Trapezoid To Each Successive Pair Of Values Of T, F(z. The calculator will approximate the integral using the Trapezoidal Rule, with steps shown. 1 Trapezium rule This is the simplest numerical method for evaluating a de\ufb01nite integral. More accurate evaluation of integral than Trapezoidal Rule (a linear approximation). Instead of using rectangles as we did in the arches problem, we'll use trapezoids (trapeziums) and we'll find that it gives a better approximation to the area. Numerical veri cation of rate of convergence Example: Consider the integral I(f) = Z \u02c7 0 sin(x)dx Compute a sequence of approximations T h(f) (composite trapezoidal rule) and S h(f) (composite Simpson\u2019s rule) which shows clearly the convergence to I(f) and the rates of convergence in each case. Simpson's rule takes a. The calculator will approximate the integral using the Trapezoidal Rule, with steps shown. mental Theorem of Calculus. The Trapezoid Rule for Approximating Integrals. Use the trapezoidal rule of numerical integration. Numerical Integration Introduction Trapezoid Rule The primary purpose of numerical integration (or quadrature) is the evaluation of integrals which are either impossible or else very difficult to evaluate analytically. The Trapezoidal Rule for Numerical Integration The Trapezoidal Rule for Numerical Integration Theorem Consider y=fHxL over @x 0,x 1D, where x 1 =x 0 +h. y a 5 0 2 b n x y 5 f(x. NUMERICAL INTEGRATION How do we evaluate I = Z b a f(x)dx By the fundamental theorem of calculus, if F(x) is an antiderivative of f(x), then I = Z b a f(x)dx = F(x) b a = F(b) F(a) However, in practice most integrals cannot be evaluated by this means. neural networks), and I've discovered as I try to read through the algorithms that my calculus has gotten a bit rusty. To get the results for Simpson's Rule, the box must be checked. First, not every function can be analytically integrated. Second, the \"Points\" setting has also not been read right. 2 The rule T 2(f) for 3 points involves three equidistant points: a, a+b 2 and b. My homework states this: Integration. The trapezoidal rule is the first of the Newton-Cotes closed integration formulas. This video contains a. As the number of integration points increase, the results from these methods will converge. The Trapezoidal Rule for approximating is given by DEFINITION The area of any trapezoid is one half of the height times the sum of the bases (the bases are the parallel sides. The size of Y determines the dimension to integrate along: If Y is a vector, then trapz(Y) is the approximate integral of Y. First enter the function f(x) whose sums you wish to compute as Y1 in the \"Y=\" window. I believe the menu feature works correctly but the code in the program for the two methods of numerical integration are not working as intended. You could turn the rule into a \"rectangular\"/\"cuboid\" rule where you evaluate the mid-points of the cells. The midpoint rule breaks [a,b] into equal subintervals, approximates the integral one each subinterval as the product of its width h times the function value at the midpoint, and then adds up all the subinterval results. possible to \ufb01nd the \u201danti-derivative\u201d of the integrand then numerical methods may be the only way to solve the problem. Also includes an applet for finding the area under a curve using the rectangular left, rectangular right, trapezoid, and Simpson's Rule. The first stage of. I'm really stuck with this. The simplest way to find the area under a curve is to split the area into rectangles Figure 8. As we start to see that integration 'by formulas' is a much more difficult thing than differentiation, and sometimes is impossible to do in elementary terms, it becomes reasonable to ask for numerical approximations to definite integrals. Recall that one interpretation for the definite integral is area under the curve. A (LO) , LIM\u20115. We have learned how to calculate some integrals analytically; such as \u222b 1 0 x2 dx = [1 3 x3] = 1 3 But most integrals , e. the area under f(x) is approximated by a series of trapeziums. 34375 which is same of mine. Write a program to integrate an arbitrary time-\u00ad\u2010domain signal, with a variable sample interval and variable limits of integration. A trapezoid is a four sided polygon, like a rectangle. Also, as John D. Details Using Excel; We provide an example spreadsheet implementing this numerical integration (using the trapezoidal rule) with example data. Other formulae belonging to the group (for the closed type, of which the Trapezoidal Method is one) include the Simpson's 1/3 and 3/8 Rules , and the Boole's Rule. The numerical computation of an integral is sometimes called quadrature. It forms the even number of intervals and fits the parabola in each pair of interval. In each case, we assume that the thickness of each strip is h and that there are N strips, so that. xb of Where the function y=f(x) is called Numerical integration. For a fixed function f(x) to be integrated between fixed limits a and b, one can double the number of intervals in the extended trapezoidal rule without losing the benefit of previous work. Usually, given n, n+1 is the number of evaluation points within the interval and not 2*n+1 or 3*n+1. The trapezoid rule approximates the integral \\int_a^b f(x) dx by the sum: (dx/2) \\sum_{k=1}^N (f(x_k) + f(x_{k-1})) where x_k = a + k*dx and dx = (b - a)/N. Trapezoid Rule: The trapezoid rule is applied extensively in engineering practice due to its simplicity. Also, the trapezoidal rule is exact for piecewise linear curves such as an ROC curve. Corollary (Simpson's Rule: Remainder term) Suppose that is subdivided into subintervals of width. Note: these Notes were prepared Noman Fareed(BS MATH University of Education). An adaptive integration method uses different interval sizes depending on the behavior of the function and the desired tolerance. \u2022 Trapezoidal rule (2-point closed formula): Z x 2 x1 f(x)dx = h 1 2 f1 + 1 2 f2 +O(h3f00), i. \u00bd(f(1) + f(2))(2 \u2212 1) = 4. 1: Trapezoidal rule. Related Articles and Code: Program to estimate the Integral value of the function at the given points from the given data using Trapezoidal Rule. \u2022 Approximation of F(\u03b4) necessitates numerical integration. Simpson's Rule When we first developed the Trapezoid Rule, we observed that it can equivalently be viewed as resulting from the average of the Left and Right Riemann sums: Tn = 1 2 (Ln + Rn). Johnson, MIT Applied Math, IAP Math Lecture Series 2011 January 6, 2011 1 Numerical integration (\"quadrature\") Freshman calculus revolves around differentiation and integration. Simpson's rule. The Trapezoidal Rule is equivalent to approximating the area of the trapezoid under the straight line connecting f (a) and f (b) in Fig. I was wondering how to use the Trapezoidal Rule in C++. , take twice as many measurements of the same length of time), the accuracy of the numerical integration will go up by a factor of 4. You have an analytic function that you need to integrate numerically. Numerical integration is very often referred to as numerical quadrature meaning that it is a process of nding an area of a square whose area is equal to the area under a curve. Area under the curve always implies definite integration. Numerical integration using trapezoidal rule gives the best result for a single variable function, which is (A) linear (B) parabolic (C) logarithmic (D) hyperbolic. In the presentation, we \u2022address this problem for the case of numerical integration and differentiation of sampled data \u2022compare, from these point of view, different known methods for numerical integration and differentiation. Approximate Z 2 1. In this section we will look at several fairly simple methods of approximating the value of a definite integral. Trapezoidal Rule for Approximate Value of Definite Integral In the field of numerical analysis, Trapezoidal rule is used to find the approximation of a definite integral. 34375 which is same of mine. It is the process of computing the value of a definite integral when we are given a set of numerical values of the integrand f(x) corresponding to some values of the independent variable x. Task description. This is called composite trapezoidal rule. You can achieve greater accuracy with either of these methods through the reduction of the interval width. Simpson's 1/3 Rule. Simpson's 1/3 Rule. Introduction Trapezoidal Rule Simpson\u2019s Rule Comparison Measuring Precision Outline 1 Introduction to Numerical Integration 2 The Trapezoidal Rule Numerical Analysis (Chapter 4) Elements of Numerical IntegrationI R L Burden & J D Faires 2 / 36. Trapezoidal Rule of Integration. The reason for calling this formula the Trapezoidal rule is that when f(x) is a function with positive values, the integral (1) is approximated by the area in the trapezoid, see. This numerical analysis method is slower in convergence as compared to Simpson\u2019s rule in. Simpson's rule. The Trapezoid Rule: For the function in the above figure with three trapezoids, here's the math: Even though the formal definition of the definite integral is based on the sum of an infinite number of rectangles, you might want to think of integration as the limit of the trapezoid rule at infinity. The 2-point Gaussian quadrature rule gives you an exact result, because the area of the lighter grey regions equal the area of the dark grey region. The Trapezoidal Rule Fits A Trapezoid To Each Successive Pair Of Values Of T, F(z. Then, find the approximate value of the integral using the trapezoidal rule with n = 4 n = 4 subdivisions. In numerical analysis, Trapezoidal method is a technique for evaluating definite integral. The stiffness, geometric stiffness, and mass matrices for an element are normally derived in the finite-element. 10 The numerical realization of equation (4. x2dx using the midpoint rule, trapezoid rule, and Simpson\u2019s rule with n = 6. Numerical Integration. For a fixed function f(x) to be integrated between fixed limits a and b, one can double the number of intervals in the extended trapezoidal rule without losing the benefit of previous work. Trapezoidal and Simpson's rule is a method for numerical integration. pptx - Free download as Powerpoint Presentation (. In the presentation, we \u2022address this problem for the case of numerical integration and differentiation of sampled data \u2022compare, from these point of view, different known methods for numerical integration and differentiation. It also divides the area under the function to be integrated, f ( x ) , into vertical strips, but instead of joining the points f ( x i ) with straight lines, every set of. Numerical integration is a part of a family of algorithms for calculating the numerical value of a definite integral. Simpson's rule takes a. This numerical method is also popularly known as Trapezoid Rule or Trapezium Rule. \u2022 Knowing how to implement the following single application Newton-Cotes formulas: - Trapezoidal rule - Simpson's 1/3 rule - Simpson's 3/8 rule. The algorithm consists in approximation of initial definite integral by the sum. Theorem (Trapezoidal Rule) Consider over, where. Runge-Kutta 2nd. 34375 \\$\\endgroup\\$ - mleyfman Aug 21 '14 at 6:17 \\$\\begingroup\\$ @mleyfman, according to the link you gave Answer: 2. Use the trapezoidal rule to solve with n = 6. d) Evaluate the integral in part (b). ) Derivation of the Simpson\u2019s 1/3 Rule for Numerical. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Lecture 12: Numerical Integration Trapezoidal rule f (x) Approximate integral of f(x) by assuming function is piecewise linear x 0 = a x 1 x 1x 2 xx 1 2x 3 xx 2 3. trapezoidal approximation. The height of a trapezoid is found from the integrand, y j = y ( x j ), evaluated at equally spaced points, x j and x j+ 1. n= 6 subintervals, 2. The first stage of. Implementation in Excel. / Trapezoid, Simpson integration Calculate a table of the integrals of the given function f(x) over the interval (a,b) using Trapezoid, Midpoint and Simpson's methods. The trapezoidal rule has a big /2 fraction (each term is (f(i) + f(i+1))/2, not f(i) + f(i+1)), which you've left out of your code. is an numerical approximation to the integral, and. Then the area of trapeziums is calculated to find the integral which is basically the area under the curve. China e-mail: [email\u00a0protected] 1 Integration by Trapezoidal Rule Since the result of integration is the area bounded by f(x) and the x axis from x=a to x=b (see. This numerical method is also popularly known as Trapezoid Rule or Trapezium Rule. This may be because we cannot find the integral of the equation of the curve or because. If we use Trapezoidal Rule and successively halve the step size I1,1 is the integral obtained by h. It is easy to obtain from the trapezoidal rule, and in most cases, it converges more rapidly than the trapezoidal rule. Named after mathematician Thomas Simpson, Simpson\u2019s rule or method is a popular technique of numerical analysis for numerical integration of definite integrals. First, the approximation tends to become more accurate as increases. Trapezoidal Rule: In mathematics, the trapezoid rule is a numerical integration method, that is, a method to calculate approximately the value of the definite integral. All these methods are Numerical. The area of the trapezoid defined by the pink lines above is given by. Also, as John D. b = upper limit of integration. How to Compute Numerical integration in Numpy (Python)? November 9, 2014 3 Comments code , math , python The definite integral over a range (a, b) can be considered as the signed area of X-Y plane along the X-axis. Numerical Integration) I wrote a VBA function to implement Simpson's rule. Trapezoid Rule: Trapezoidal rule is used to find out the approximate value of a numerical integral, based on finding the sum of the areas of trapezium. As the C program for Trapezoidal Method is executed, it asks for the value of x 0, x n and h. The example. The method also corresponds to three point Newton \u2013 Cotes Quadrature rule. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral. The next rule that I'm going to describe is a little improvement. The trapezoidal rule approximates fusing a piecewise linear function. We chop the interval [a,b] into n equal pieces and use the following notation: h = b\u2212a n Width of the intervals x 0 = a, x 1 = a+h, x 2 = a+2h,, x n = b Endpoints x\u00af 0 = a+ h 2. ) Derivation of the Simpson\u2019s 1/3 Rule for Numerical. 1 Introduction In this chapter we discuss some of the classic formulae such as the trapezoidal rule and Simpson\u2019s rule for equally spaced abscissas and formulae based on Gaussian quadrature. \u2022 A function f(x) has known values f(xi) = fi. The composite rule 3. There are several methods of numerical integration of varying accuracy and ease of use. \u2022 For Simpson\u2019s 1/3 Rule: \u2022 It turns out that if is a cubic and is quadratic, 82 \u2022 The errors cancel over the interval due to the location of point ! \u2022 We can actually improve the accuracy of integration formulae by locating integration points in special locations! \u2022W deo not experience any improvement in accuracy for N = odd. The Trapezoidal Rule for Numerical Integration The Trapezoidal Rule for Numerical Integration Theorem Consider y=fHxL over @x 0,x 1D, where x 1 =x 0 +h. Evaluate definite integrals numerically using the built-in functions of scipy. first second third fourth 31. If higher order polynomials are used, the more accurate result can be achieved. This step takes care of all the middle sums in the trapezoidal rule formula. Simpson's rule is another member of the same family, and in general has faster convergence than the trapezoidal rule for functions which are twice continuously differentiable, though not in all specific cases. In this numerical integration worksheet, students approximate the value of an integral using the methods taught in the class. rule \u2013 This controls the Gauss-Kronrod rule used in the adaptive integration: rule=1 \u2013 15 point rule Documentation can be found in chapter \u201cNumerical. ANALYSED SUBSURFACE (GEOLOGICAL) STRUCTURES Ia. The opposite is true when a curve is concave up. Riemann Integral. This numerical analysis method is used to approximating the definite integral. Euler's Method. accuracy in numerical integration using the same number of points in a given interval by changing the weighting used for the points. Approximating the area under a curveSometimes the area under a curve cannot be found by integration. b = upper limit of integration. The results aren't good. , take twice as many measurements of the same length of time), the accuracy of the numerical integration will go up by a factor of 4. Numerical integration. However, in practice, f or its antiderivative is analytically unknown, forcing us to settle for a numerical approximation. The area of the strips can be approximated using the trapezoidal rule or Simpson's rule. Composite Trapezoidal Rule. This is the basis for what is called the trapezoid rule of numerical integration. Work: For the composite trapezoid rule with N subintervals we use N+1. Related Articles and Code: Program to estimate the Integral value of the function at the given points from the given data using Trapezoidal Rule. The extended trapezoidal rule. Numerical Integration. As you can see, this is exactly what happened, and will always happen for that function, on that interval. The linear segments are given by the secant lines through the endpoints of the subintervals: for f(x)=x^2+1 on [0,2] with 4 subintervals, this looks like:. Download this Mathematica Notebook 2D Integration using the Trapezoidal and Simpson Rules (c) John H. Trapezoidal Integration. Calculate the primitive function and the exact value of the integral. The first stage of.", "date": "2020-01-28 14:17:33", "meta": {"domain": "lucacaponedesign.it", "url": "http://bjrm.lucacaponedesign.it/numerical-integration-trapezoidal-rule.html", "openwebmath_score": 0.8422544598579407, "openwebmath_perplexity": 638.8632955111407, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9932024684496266, "lm_q2_score": 0.8774767778695836, "lm_q1q2_score": 0.871512101787295}}
{"url": "https://math.stackexchange.com/questions/3586612/representating-borel-probability-measures-as-closed-subset-of-mathbbr-inft", "text": "# Representating Borel probability measures as closed subset of $\\mathbb{R}^{\\infty}$\n\nLet $$\\mathcal{P}_d$$ be the space of Borel probability measures on $$\\mathbb{R}^d$$ endowed with the topology of weak convergence of probability measures. Does there exist a map $$G: \\mathcal{P}_d \\to \\mathbb{R}^{\\infty}$$ (the latter with the usual topology of pointwise convergence) $$\\textbf{which is one-to-one and continuous}$$ such that $$G(\\mathcal{P}_d) \\subseteq \\mathbb{R}^{\\infty}$$ is closed?\n\nSome thoughts on this: It is known that for $$\\{\\mu_n\\}_n \\in \\mathcal{P}_d$$, the existence of lim$$_n\\int f d\\mu_n$$ in $$\\mathbb{R}$$ for each $$f \\in C_b(\\mathbb{R}^d)$$ yields the existence of a unique $$\\mu \\in \\mathcal{P}_d$$ such that $$\\mu_n \\to \\mu$$ weakly. Hence, if one could choose $$G$$ as $$G(\\mu) = \\bigg(\\int f d\\mu\\bigg)_{f \\in C_b},$$ the answer to my question would be affirmative. Since $$C_b$$ is neither countable nor separable, such a choice of $$G$$ is not possible. Replacing $$C_b$$ by a dense, countable subset of $$C_c(\\mathbb{R}^d)$$ does not work either, since then the limit object $$\\mu$$ will in general only be a sub-probability measure. This allows only to conclude that this map $$G$$ would turn $$G(\\mathcal{M}^+_{\\leq 1}) \\subseteq \\mathbb{R}^{\\infty}$$ into a closed set, where $$\\mathcal{M}^+_{\\leq 1}$$ denotes the set of all Borel sub-probability measure endowed with the vague topology.\n\nIs there a way to solve this issue? I was thinking of replacing $$C_b$$ by the union of a dense, countable subset of $$C_c$$ and a sequence $$(f_l)_{l \\geq 1}$$ of $$C_b$$-functions such that $$f_l \\uparrow 1$$ uniformly. Then, the existence of a sub-probability measure $$\\mu$$ as the vague limit of $$(\\mu_n)_n$$ follows and my hope is to obtain $$\\mu(\\mathbb{R}^d) = lim^l\\int f_l d\\mu = lim^l \\, lim^n \\int f_l d\\mu_n = lim^n \\, lim^l \\int f_l d\\mu_n = lim^n \\mu_n(\\mathbb{R}^d) = 1,$$where the interchange of limits is justified by the uniform convergence of $$f_l \\to1$$. However, the second equality uses that the integrals $$\\int f_l d\\mu$$ are represented by $$lim^n \\int f_l d\\mu_n$$. But the existence of $$\\mu$$ is deduced from the Riesz-Markov-Kakutani representation and from there I do not know how to conclude that the integral representation of the limit object $$\\mu$$ also holds for functions $$f$$ other than $$f \\in C_c$$.\n\nIs this approach feasible at all? If not, is there any other way to answer my question affirmatively? Thanks a lot in advance!\n\n\u2022 I am unable to understand the question. Why can't you take $G(P)=(0,0,..0)$ for all $P$? Mar 19 '20 at 10:17\n\u2022 Sorry for being sloppy. Of course, I forgot to mention one crucial requirement: $G$ should be one-to-one. I will edit the post. Thanks! Mar 19 '20 at 10:41\n\u2022 You probably want some more properties of $G$. Do you want it to be continuous and linear? Mar 19 '20 at 11:52\n\u2022 Sorry for again being imprecise. I've edited the post again. Linearity is, for my concerns, not necessarily needed I suppose. Mar 19 '20 at 11:57\n\u2022 I was also sloppy :-). Linearity doesn't even make sense. I should have mentioned convexity instead of linearity. Mar 19 '20 at 11:59\n\n# Yes.\n\nIn general, every Polish space is homeomorphic to a closed subset of $$\\mathbb{R}^\\infty$$; see Kechris, Classical Descriptive Set Theory, Theorem 4.17 (where the space is denoted $$\\mathbb{R}^{\\mathbb{N}}$$). And $$\\mathcal{P}_d$$ with its weak topology is Polish; Kechris Theorem 17.23.\n\nI'm a little stumbled, but I cannot find a mistake in the following proof, which would somehow prove the statement of the above answer by Nate Eldredge without the assumption of the space under consideration being Polish (which I did not expect to be true!):\n\nLet $$X$$ be a separable, metrizable topological space. Fix any metric $$d$$, which metrics the given topology on $$X$$. Now define\n\n$$F: X \\to \\mathbb{R}^{\\infty}, F(x) := (d(x,x_1),d(x,x_2),\\dots),$$where $$(x_n)_n$$ is a fixed dense subset of $$X$$ (such set does not depend on the actual choice of $$d$$). $$\\mathbb{R}^{\\infty}$$ is endowed with the metric $$\\rho(\\alpha,\\beta) := \\sum_{k\\geq1} 2^{-k}|\\alpha_k-\\beta_k|$$, which induces the topology of pointwise convergence. Let's show that $$F$$ is one-to-one and a homeomorphism between $$X$$ and $$F(X) \\subseteq \\mathbb{R}^{\\infty}$$:\n\nFirstly, for $$d(x,y) = c > 0$$ we find $$x_k$$ such that $$d(x,x_k) < \\frac{c}{3}$$, thus $$d(y,x_k) > \\frac{c}{3}$$ and hence $$F(x) \\neq F(y)$$, whereby $$F$$ is one-to-one.\n\nSecondly, for $$\\epsilon > 0$$ and $$x \\in X$$, if $$d(x,y)< \\epsilon$$, then $$|d(x,x_i)-d(y,x_i)| < \\epsilon$$ for each $$i$$, so that $$\\rho(F(x),F(y)) < \\epsilon$$, whereby $$F$$ is continuous.\n\nFinally, let $$\\epsilon > 0$$ and fix $$F(x) \\in F(X)$$. There exists $$x_n$$ such that $$d(x,x_n) < \\frac{\\epsilon}{3}$$. If $$y \\in X$$ with $$d(y,x_n) > \\frac{2}{3}\\epsilon$$, then $$|d(x,x_n)-d(y,x_n)| > \\frac{\\epsilon}{3}$$, hence $$\\rho(F(x),F(y)) > \\frac{\\epsilon}{3\\cdot2^{n}}$$. Thereby, $$\\rho(F(x),F(y)) \\leq \\frac{\\epsilon}{3\\cdot2^{n}} \\implies d(x,y) < \\epsilon,$$whereby $$F^{-1}$$ is continuous from $$F(X)$$ to $$X$$.\n\nSince $$X$$ is closed (as a subset of itself) and $$F^{-1}$$ is continuous, also $$F(X) = (F^{-1})^{-1}(X)$$ is closed.\n\nIs there a mistake in my reasoning? (Sorry for posting this question as an answer, it was certainly too long for a comment and I did not want to open a new post for this)\n\n\u2022 Your second-to-last paragraph only proves that $F(X)$ is closed in $F(X)$ (i.e. with respect to the subspace topology induced from $\\mathbb{R}^\\infty$), which is trivial. It does not prove that it is closed in $\\mathbb{R}^\\infty$. Mar 19 '20 at 20:32", "date": "2022-01-17 23:43:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3586612/representating-borel-probability-measures-as-closed-subset-of-mathbbr-inft", "openwebmath_score": 0.9706050157546997, "openwebmath_perplexity": 92.8811488474995, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9805806472775278, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8714996787749001}}
{"url": "http://barbaramitterer.com/article/p38ub.php?9cd143=set-equality-proof", "text": "# set equality proof\n\nPosted by on Nov 28, 2020 in Uncategorized | No Comments\n\nSince there were no elements in both $$A$$ and $$B$$ to give rise to equivalent sets, there was no resulting intersection, other than the empty set. Let $$X \\in \\mathcal{P}(A) \\cap \\mathcal{P}(B)$$. Proof: We must show A\u2212 B \u2286 A\u2229 Bc and A \u2229Bc \u2286 A\u2212B. Proving Set Equality: From Sets to Logic and Back - YouTube $$\\mathcal{P}(A)=\\left\\{ \\emptyset, \\left\\{a\\right\\}, \\left\\{c\\right\\}, \\left\\{a,c\\right\\}\\right\\}$$. $$\\mathcal{P}(A) \\cap \\mathcal{P}(B)=\\left\\{ \\emptyset \\right\\}$$. In the following posts, we will be looking more at how to prove different theorems. We have that, We now look at what the elements of $$\\mathcal{P}(A \\cap B)$$ look like. Expanding your knowledge and love of mathematics. We now have that $$X \\subseteq A \\cap B$$, hence $$X \\in \\mathcal{P}(A \\cap B)$$. $$\\mathcal{P}(B)=\\left\\{ \\emptyset, \\left\\{b\\right\\}\\right\\}$$. We want to determine if for all sets $$A$$ and $$B$$, we have that $$\\mathcal{P}(A) \\cap \\mathcal{P}(B) =\\mathcal{P}(A \\cap B)$$. Theorem For any sets A and B, A\u2212B = A\u2229Bc. 4 CS 441 Discrete mathematics for CS M. Hauskrecht Equality Definition: Two sets are equal if and only if they have the same elements. Set equalities can be proven by using known set laws Examples: \u2022 Let U be a set and let A, B and C be elements of P(U). Learn how your comment data is processed. 2. For any set Aand Bwe have A\\B A[B Proof. So x2Aand x2B. Example: \u2022 {1,2,3} = {3,1,2} = {1,2,1,3,2} Note: Duplicates don't contribute anythi ng new to a set, so remove them. You can also view the available videos on YouTube. We then get that. Enter your email address to follow this blog and receive notifications of new posts by email. Here we will work with the sets $$A=\\left\\{a,c\\right\\}$$ and $$B=\\left\\{b,c\\right\\}$$. Proving equalities of sets using the element method - YouTube In this case, as long as we can put this together formally, this will indeed give us a proof. He has a passion for teaching and learning not only mathematics, but all subjects. By de\ufb01nition of set di\ufb00erence, x \u2208 A and x 6\u2208B. We work through an example of proving that two sets are equal by proving that any element of one must also be an element of the other. $$\\mathcal{P}(A)=\\left\\{ \\emptyset, \\left\\{a\\right\\}\\right\\}$$. We will need to sometimes need to argue two sets (which are potentially defined differently) are indeed the same. $$\\mathcal{P}(A \\cap B)=\\mathcal{P}(\\left\\{c\\right\\})=\\left\\{\\emptyset, \\left\\{c\\right\\}\\right\\}$$. Let x \u2208 A\u2212 B. With this being the case, we saw that the element of each set resulted in a subgroup. You can find more examples of proof writing in the Study Help category for mathematical reasoning. Relevance. We now need to show the other direction. A set is a subset of $$A \\cap B$$ if it is a subset of $$A$$ and $$B$$. For a small example, let $$A=\\left\\{a\\right\\}$$ and $$B=\\left\\{b\\right\\}$$. Definition of De Morgan\u2019s law: The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. In the following posts, we will be looking more at how to prove different theorems. We should begin by trying to think about specific sets and what these things would look like. Let A and B be sets. $$\\mathcal{P}(A \\cap B)=\\mathcal{P}(\\emptyset)=\\left\\{\\emptyset\\right\\}$$. $$\\mathcal{P}(B)=\\left\\{ \\emptyset, \\left\\{b\\right\\}, \\left\\{c\\right\\}, \\left\\{b,c\\right\\}\\right\\}$$. By \u2026 Note that as we described the elements in each set, we arrive at the same result. Again, we have that the two sets are the same. In this case, we do indeed have that the two sets are the same. At this point, it is time to try to generalize what is happening in these examples so that we can find what happens in general. 4.11.5. The first is:https://www.youtube.com/watch?v=QR6akpA-FYAThe third is:https://www.youtube.com/watch?v=sjJBS0DxF1A Here is another set equality proof (from class) about set operations. Prove A U B = A U (B - (A n B)). Because the examples worked out for the different things we tried, we thought that the theorem would work in general. I am investigating how I might be able to translate even commonplace equalities/ inequalities via the so-called Curry-Howard Correspondance - from a generic, set theoretic plus AOC foundation - into a decidable type theoretic language. How do you do this proof? For equality remember (A= B) ()A Band B A Therefore, there are two directions to A= B; one when you take an element of Aand show it is in B, and Elements of $$\\mathcal{P}(B)$$ are subsets of $$B$$. As we start to generalize, we want to think about what the elements of $$\\mathcal{P}(A) \\cap \\mathcal{P}(B)$$ look like. Lv 7. These are called De Morgan\u2019s laws. Therefore, $$X \\subseteq A$$ and $$X \\subseteq B$$. He is currently an instructor at Virginia Commonwealth University. As evidence, he also has a bachelors degree in music and has spent time giving guitar lessons. We can then find that. 1 Answer. Set Equality Proof. Answer Save. A clear explanation would be greatly appreciated. March 31, 2019 Dr. Justin Albert. Hence elements of $$\\mathcal{P}(A \\cap B)$$ are subsets of both $$A$$ and $$B$$. Prove ( )A\u2212B \u2212C = A\u2212C \u2212B () () () ()set difference set difference associativity commutativity associativity set difference set \u2026 We began by looking at some examples to try to determine whehter or not this theorem was true. After this, we were able to look at the defining properties of each of these sets and compare them. Simple set proof (is it right?) Homework Statement Let ##A, B, C## be sets with ##A \\subseteq B##. 20 0. 7 years ago. Hence elements of $$\\mathcal{P}(A) \\cap \\mathcal{P}(B)$$ are subsets of both $$A$$ and $$B$$. Am I on the right track to proving that this is a true statement or is there a better way to show this? Sorry, your blog cannot share posts by email. We now have that $$X \\in \\mathcal{P}(A)$$ and $$X \\in \\mathcal{P}(B)$$. Equality of sets is defined as set $$A$$ is said to be equal to set $$B$$ if both sets have the same elements or members of the sets, i.e. We thank you for doing so. Here we will learn how to proof of De Morgan\u2019s law of union and intersection. Post was not sent - check your email addresses! Before making a general guess, I would suggest trying an example with some intersection. Steiner. Subgroups of the Permutations on Three Elements. This means that we will need to show that every element in $$\\mathcal{P}(A) \\cap \\mathcal{P}(B)$$ is also an element of $$\\mathcal{P}(A \\cap B)$$. Elements of $$\\mathcal{P}(A)$$ are subsets of $$A$$. $$\\mathcal{P}(A) \\cap \\mathcal{P}(B)=\\left\\{ \\emptyset, \\left\\{c\\right\\} \\right\\}$$.", "date": "2022-05-26 14:04:57", "meta": {"domain": "barbaramitterer.com", "url": "http://barbaramitterer.com/article/p38ub.php?9cd143=set-equality-proof", "openwebmath_score": 0.75785231590271, "openwebmath_perplexity": 297.98164227019214, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.975576912786245, "lm_q2_score": 0.8933094152856196, "lm_q1q2_score": 0.8714920415272305}}
{"url": "https://math.stackexchange.com/questions/2834928/if-%CE%B1-%CE%B2-%CE%B3-are-roots-of-x3-x-1-0-then-find-the-value-of-frac1%CE%B11", "text": "# If $\u03b1, \u03b2, \u03b3$ are roots of $x^3 - x -1 = 0$, then find the value of $\\frac{1+\u03b1}{1-\u03b1} + \\frac{1+\u03b2}{1-\u03b2} + \\frac{1+\u03b3}{1-\u03b3}$.\n\nIf $\u03b1, \u03b2, \u03b3$ are roots of $x^3 - x -1 = 0$, then find the value of $$\\frac{1+\u03b1}{1-\u03b1} + \\frac{1+\u03b2}{1-\u03b2} + \\frac{1+\u03b3}{1-\u03b3}$$ I found this question asked in a previous year competitive examination, which was multiple choice in nature, the available options to the question were:\n\n1. $1$\n2. $0$\n3. $-7$\n4. $-5$\n\nConsidering the time available for a question to be solved in such an examination, is there a way to solve this problem without actually having to expand the the given relation by cross-multiplying the numerators and denominators or even finding the zeroes of the given equation.\n\n\u2022 This works the same way as here. \u2013\u00a0Dietrich Burde Jun 28 '18 at 16:18\n\u2022 @DietrichBurde The correct answer was given as -7 and not -5. It could verified when done the long way. There's a mistake in the given solution. \u2013\u00a0Tony1970 Jun 28 '18 at 16:26\n\u2022 The given solution refers to $x^3-x^2-1$, and not $x^3-x-1$. But the method to solve is the same. Just do it! \u2013\u00a0Dietrich Burde Jun 28 '18 at 16:28\n\u2022 Self-contained hint: if $\\alpha$ is a root of $x^3-x-1$, find a (cubic) polynomial that $(1+\\alpha)/(1-\\alpha)$ is a root of. (This is pretty straightforward; if $\\alpha'=(1+\\alpha)/(1-\\alpha)$ then just invert to find $\\alpha$ in terms of $\\alpha'$, plug that in to the defining polynomial, and collect some terms - it's a little bit of an algebraic slog, but not too bad since it's only a single variable.) Then $\\alpha'$, $\\beta'$, and $\\gamma'$ are roots of this polynomial, and if you have the three roots of a polynomial you should know how to find their sum... \u2013\u00a0Steven Stadnicki Jun 28 '18 at 16:30\n\u2022 @DietrichBurde Sorry I didn't immediately notice. But thanks. Could this be done in a yet more simpler way? \u2013\u00a0Tony1970 Jun 28 '18 at 16:32\n\nIt is practical to recall that $z\\mapsto\\frac{1-z}{1+z}$ is an involution. In particular $$\\frac{1+\\alpha}{1-\\alpha}+\\frac{1+\\beta}{1-\\beta}+\\frac{1+\\gamma}{1-\\gamma}$$ is the sum of the reciprocal of the roots of $p\\left(\\frac{1-x}{1+x}\\right)=-\\frac{x^3-x^2+7x+1}{(1+x)^3}$, which is also the sum of the reciprocal of the roots of $x^3-x^2+7x+1$. By Vieta's formulas, this is $\\color{red}{-7}$ (option 3.).\n\nHint If $\\zeta$ is a root of $p(x)$, then $\\zeta' := 1 - \\zeta$ is a root of $-p(1 - x)$. Taking $p(x) := x^3 - x - 1$ gives $$q(x) := -p(1 - x) = x^3 - 3 x^2 + 2 x + 1 .$$\n\nNow, $$\\frac{1 + \\zeta}{1 - \\zeta} = \\frac{1 + (1 - \\zeta')}{\\zeta'} = \\frac{2}{\\zeta'} - 1,$$ so $$S := \\frac{1 + \\alpha}{1 - \\alpha} + \\frac{1 + \\beta}{1 - \\beta} + \\frac{1 + \\gamma}{1 - \\gamma} = 2 \\left(\\frac{1}{\\alpha'} + \\frac{1}{\\beta'} + \\frac{1}{\\gamma'}\\right) - 3,$$ where we define $\\alpha'$ etc. analogously to $\\zeta'$ above, so that $\\alpha', \\beta', \\gamma'$ are the roots of $q$.\n\nNow, similarly transform $q(x)$ to find a polynomial $r(x)$ whose roots are $\\alpha'' := \\frac{1}{\\alpha'}$, etc., and observe that $$r(x) = (x - \\alpha'')(x - \\beta'')(x - \\gamma') = x^3 - (\\alpha'' + \\beta'' + \\gamma'')x^2 + \\cdots$$ and that the sum is the above expression for the desired quantity $S$ is exactly the negative $\\alpha'' + \\beta'' + \\gamma''$ of the coefficient of the quadratic term.\n\nAdditional hint If $\\alpha'' := \\frac{1}{\\alpha'}, \\beta'' := \\frac{1}{\\beta'}, \\gamma'' := \\frac{1}{\\gamma'}$ are the roots of $$r(x) := x^3 q\\left(\\tfrac{1}{x}\\right) = x^3 + 2 x^2 - 3 x + 1,$$ then $\\alpha'' + \\beta'' + \\gamma'' = -2$, and so $$S = 2 (\\alpha'' + \\beta'' + \\gamma'') - 3 = 2(-2) - 3 = -7.$$\n\nNote that $x^3-x-1=0$ is equivalent to $x^3-x=1$ and $x^3-1=x$. That is, for $x\\in\\{\\alpha,\\beta,\\gamma\\}$, we get $$-x-x^2=\\frac{x^3-x}{1-x}=\\frac{1}{1-x}\\text{ and }-1-x-x^2=\\frac{x^3-1}{1-x}=\\frac{x}{1-x}\\,.$$ Consequently, $$\\frac{1+x}{1-x}=-1-2x-2x^2\\text{ for }x\\in\\{\\alpha,\\beta,\\gamma\\}\\,.$$ This means $$\\sum_{x\\in\\{\\alpha,\\beta,\\gamma\\}}\\,\\frac{1+x}{1-x}=-3-2s-2q\\,,$$ where $s:=\\alpha+\\beta+\\gamma$ and $q:=\\alpha^2+\\beta^2+\\gamma^2$. It is easy to find $s$ and $q$.\n\nWe have $s=0$ and $q=2$.\n\n\u2022 Could you elaborate the answer a little bit. I couldn't catch up certain steps you did. \u2013\u00a0Tony1970 Jun 29 '18 at 1:11\n\u2022 @Tony1970 You have to point out what you don't understand. I don't know where to start. \u2013\u00a0Batominovski Jun 29 '18 at 8:28\n\u2022 How does $- x - x^2 = \\frac{x^3 - x}{1-x} = \\frac{1}{1-x}$ and why is \"and - 1\" used. And I did not understand the equivalence of $x^3-x-1=0$ to $x^3-x=1$ and $x^3-1=0$. \u2013\u00a0Tony1970 Jun 29 '18 at 14:40\n\nAlt. hint: \u00a0 Let $y = \\dfrac{1+x}{1-x} \\iff x = \\dfrac{y-1}{y+1}$ then:\n\n\\begin{align} 0 = (y+1)^3 \\cdot P\\left(\\dfrac{y-1}{y+1}\\right) &= (y-1)^3-(y-1)(y+1)^3-(y+1)^3 \\\\ &= -y^3 - 7 y^2 + y - 1 \\end{align}\n\nIt follows from Vieta's relations that $\\;y_1+y_2+y_3=\\ldots$\n\nSince $$\\alpha+\\beta+\\gamma=0,$$ $$\\alpha\\beta+\\alpha\\gamma+\\beta\\gamma=-1$$ and $$\\alpha\\beta\\gamma=1,$$ we obtain: $$\\sum_{cyc}\\frac{1+\\alpha}{1-\\alpha}=\\frac{\\sum\\limits_{cyc}(1+\\alpha)(1-\\beta)(1-\\gamma)}{\\prod\\limits_{cyc}(1-\\alpha)}=\\frac{\\sum\\limits_{cyc}(1+\\alpha)(1-\\beta-\\gamma+\\beta\\gamma)}{1-(\\alpha+\\beta+\\gamma)+(\\alpha\\beta+\\alpha\\gamma+\\beta\\gamma)-\\alpha\\beta\\gamma}=$$ $$=\\frac{\\sum\\limits_{cyc}(1-2\\alpha+\\alpha\\beta+\\alpha-2\\alpha\\beta+\\alpha\\beta\\gamma)}{1-0-1-1}=\\frac{\\sum\\limits_{cyc}(1-\\alpha-\\alpha\\beta+\\alpha\\beta\\gamma)}{-1}=\\frac{3+1+3}{-1}=-7.$$", "date": "2019-05-24 03:42:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2834928/if-%CE%B1-%CE%B2-%CE%B3-are-roots-of-x3-x-1-0-then-find-the-value-of-frac1%CE%B11", "openwebmath_score": 0.9694573879241943, "openwebmath_perplexity": 153.7402717813542, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877049262134, "lm_q2_score": 0.8872045981907006, "lm_q1q2_score": 0.8714901685567267}}
{"url": "https://www.physicsforums.com/threads/commuting-invertible-matrices.48050/", "text": "# Commuting Invertible Matrices\n\n#### T-O7\n\nHey all,\nI'm having trouble showing the following point:\nif A is an invertible (n\u00d7n, real) matrix that commutes with ALL other invertible (n\u00d7n, real) matrices, then A is of the form cI, where c is any real number not equal to 0.\n\nAnyone know how to show this?\n\nLast edited:\nRelated Linear and Abstract Algebra News on Phys.org\n\n#### Hurkyl\n\nStaff Emeritus\nGold Member\nTry picking some particularly simple matrices B (such as having only a single nonzero entry!) and use AB = BA to get equations for the entries of A.\n\n#### T-O7\n\nThanks for the suggestion. So I had to use matrices B of the form I with an additional 1 at one other entry. By brute force, calculating the corresponding entries of AB would eventually give that all diagonal entries of A were the same, and that all other non diagonal entries must be zero. Sweet.\n\n#### Hurkyl\n\nStaff Emeritus\nGold Member\nActually, I would pick my B's so they have only one nonzero entry, rather than being the identity with an additional nonzero entry.\n\nI think this is the easiest approach to understand, though not the shortest.\n\n#### T-O7\n\nYes, I was considering that, but B must also be invertible, so I had to complicate B a little bit by adding the diagonal 1's.\n\n#### Hurkyl\n\nStaff Emeritus\nGold Member\nAh, good point, I had missed that.\n\nHowever, note this (reversible) deduction:\n\nA(I+B) = (I+B)A\nA + AB = A + BA\nAB = BA\n\nB commutes with A iff (I+B) commutes with A, so you could still work with my B's, if you choose.\n\n#### T-O7\n\nThat's wonderful! Thanks a lot for the help\n\n#### mathwonk\n\nHomework Helper\nHere's a conceptual suggestion. Think of A as a transformation and let v be any non zero vector. We want to show first that Av = cv for some constant c.\n\nChoose a basis involving v as first vector and define another transformation B that takes v to v and the other basis vectors to 0. Then applying AB = BA to v shows that ABv = Av = BAv, so B acts on Av != 0 as the identity. Since A is invertible, Av is not zero, so then Av = cv for some non zero constant c, since multiples of v are the only non zero vectors B takes to themselves.\n\nThus every vector v is an \"eigenvector\" for A, i.e. Av always equals cv for some c possibly depending on v. Now if every vector is an eigen - vector, we claim all the eigenvalues must be equal.\n\nTo see that, assume that Av = cv and Aw = dw, where v and w are in different directions, and look at A(v+w) = cv + dw = e(v+w), and check that we must have c=d=e.\n\nNow why is that? well then cv + dw -ev -ew = 0 = (c-e)v + (d-e)w, hence\n(c-e)v = (e-d)w, so these multiples of v and w are equal.\n\nBut since v, w are in different directions, their only equal multiples are zero, so d=e, c=e. QED.\n\n### Physics Forums Values\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-09-22 16:27:32", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/commuting-invertible-matrices.48050/", "openwebmath_score": 0.8931174278259277, "openwebmath_perplexity": 1624.9128686395475, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9822877005188116, "lm_q2_score": 0.8872045907347108, "lm_q1q2_score": 0.8714901573225324}}
{"url": "https://math.stackexchange.com/questions/3106982/probability-what-is-the-probability-of-obtaining-a-three-of-a-kind-or-more-whe", "text": "# Probability - What is the probability of obtaining a three of a kind or more when rolling six dice?\n\nI understand that if you wanted to compute the probability of rolling a specific three of a kind or more (getting three or more $$1$$s for example) then you would just calculate $$\\sum_{k=0}^3(5^k){6\\choose 6-k}\\over 6^6$$ but I am not so sure how you would go about calculating the probability for any value to appear three or more times in a roll of six dice. Simply multiplying the total outcomes for obtaining three or more of a specific value is obviously not correct because you would count outcomes such as $$(1,1,1,2,2,2)$$ once for getting three $$1$$s and again for getting three $$2$$s.\n\nIs there a general formula or method one can use to obtain the answer to not only the dice scenario, but any scenario where $$P(X\\ge x)$$ for $$X$$ denoting the maximum amount of times any value appears in a string of $$m$$ values generated from $$n$$ different values?\n\nThanks in advance, and sorry if my wording is bad. I'm still not entirely sure of the best way to word the last part.\n\n\u2022 For the question in the title, you can bypass the issue you correctly noticed using inclusion-exclusion, giving $\\dfrac{6\\cdot (5^3\\binom{6}{3}+5^2\\binom{6}{2}+5^1\\binom{6}{1}+1)-\\binom{6}{2}\\binom{6}{3}}{6^6}$ \u2013\u00a0JMoravitz Feb 10 '19 at 2:44\n\u2022 @JMoravitz Thanks for the reply and solution, I was just unsure of how I would count the terms that would be over counted so I could remove them. \u2013\u00a0S. Burc Feb 10 '19 at 17:38\n\nThere's more than one way. What's best depends on the relative sizes of $$m$$, $$n$$, and $$x$$.\n\nFirst, we can go \"forward\" - enumerate all ways to have three or more of a kind, all ways to have three or more of two kinds, and so on, then run inclusion-exclusion. As noted in the comment by @JMoravitz, this works pretty well for your example of $$x=3,m=n=6$$.\nFor a particular $$k$$, there are $$\\binom{6}{3}\\cdot (6-1)^3$$ ways to have three of $$k$$, $$\\binom{6}{4}\\cdot (6-1)^2$$ ways to have four of $$k$$, $$\\binom{6}{5}\\cdot (6-1)$$ ways to have five of $$k$$, and $$\\binom{6}{6}$$ ways to have six of $$k$$. Multiply by $$6$$ to get a total of $$6\\left(\\binom{6}{3}\\cdot 5^3+\\binom{6}{4}\\cdot 5^2+\\binom{6}{5}\\cdot 5^1+\\binom{6}{6}\\right)$$ ways. But, of course, this is an overcount. We can have three each of two different kinds, in $$\\binom{6}{3}$$ ways to choose which three are the \"first\" kind and $$\\binom{6}{2}$$ pairs of kinds. These were counted twice by the preceding, so we subtract them off to get $$6\\left(\\binom{6}{3}\\cdot 5^3+\\binom{6}{4}\\cdot 5^2+\\binom{6}{5}\\cdot 5^1+\\binom{6}{6}\\right)-\\binom{6}{2}\\cdot\\binom{6}{3}$$ $$= 6(20\\cdot 125+15\\cdot 25+6\\cdot 5+1)-15\\cdot 20 = 17136$$ ways. Divide by $$6^6=46656$$ for the probability (about $$0.367$$).\n\nOf course, as we increase $$m$$ for fixed $$n$$ and $$x$$, eventually we reach a point where that $$x$$ of a kind is certain (Pigeonhole principle). If we're close to that point, the forward count will be horrible - but there's an alternative. We can count the ways to avoid any instances of $$x$$ of a kind, and subtract the resulting probability from $$1$$. I'll demonstrate this backward count for the same example:\n\nTo avoid three of a kind, we must have between zero and 2 of each of the six kinds. I'll enumerate the possible ways to split $$6$$ into a sum of six terms each between zero and $$2$$: $$2+2+2+0+0+0$$, $$2+2+1+1+0+0$$, $$2+1+1+1+1+0$$, $$1+1+1+1+1+1$$.\nFor $$2+2+2+0+0+0$$, we have $$\\binom{6}{3}$$ ways to choose which three numbers on the die show up, then $$\\binom{6}{2,2,2}=\\binom{6}{2}\\binom{4}{2}$$ ways to choose which rolls each number gets.\nFor $$2+2+1+1+0+0$$, we have $$\\binom{6}{2,2,2}$$ ways to choose which numbers are represented to each level, then $$\\binom{6}{2,2,1,1}$$ ways to choose the rolls.\nFor $$2+1+1+1+1+0$$, we have $$\\binom{6}{1,4,1}$$ ways to choose the numbers and $$\\binom{6}{2,1,1,1,1}$$ ways to choose the rolls.\nFor $$1+1+1+1+1+1$$, we have one way to choose the numbers and $$\\binom{6}{1,1,1,1,1,1}=6!$$ ways to choose the rolls.\nSumming those up, that's $$20\\cdot 90+90\\cdot 180+30\\cdot 360+1\\cdot 720 = 29520$$ ways to avoid three of a kind. The probability we seek is $$1-\\frac{29520}{46656}=\\frac{17136}{46656}\\approx 0.367$$ This backwards count never really goes too bad - the example here is a \"worst\" case of having the most possible ways to split $$m$$ - but it's particularly efficient when $$m$$ is close to $$n(x-1)$$, and it's instant when we're in the case $$m > n(x-1)$$ and at least one instance of $$x$$ of a kind is certain.\n\n\u2022 Thank you very much for this detailed response! If I am understanding it correctly then I should be able to calculate the probability for $x=3, m=6$, and $n=124$ by doing $${124 \\cdot \\left({6\\choose 3}\\cdot 123^3+{6\\choose 4}\\cdot 123^2+{6\\choose 5}\\cdot 123^1+{6\\choose 6}\\right)-{124\\choose 2}\\cdot {6\\choose3}}\\over 124^6$$ Since there are $124\\choose2$ pairs of kinds and $6\\choose 3$ ways to arrange the first kind. I hope I am understanding the 'ways to arrange the first kind' correctly as different ways of ordering $(x,x,x,y,y,y)$. \u2013\u00a0S. Burc Feb 10 '19 at 18:39\n\u2022 Yes, that's correct. \u2013\u00a0jmerry Feb 10 '19 at 20:00\n\u2022 Okay, thank you very much. \u2013\u00a0S. Burc Feb 10 '19 at 20:08", "date": "2020-10-30 13:38:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3106982/probability-what-is-the-probability-of-obtaining-a-three-of-a-kind-or-more-whe", "openwebmath_score": 0.8049377799034119, "openwebmath_perplexity": 134.19292085525524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876976669629, "lm_q2_score": 0.8872045832787204, "lm_q1q2_score": 0.8714901474684316}}
{"url": "http://mathhelpforum.com/math-topics/9724-quick-interest-question.html", "text": "# Thread: quick interest question ...\n\n1. ## quick interest question ...\n\nTo provide an annual scholarship for 25 years, a donation of $50,000 is invested in an account for a scholarship that will start a year after the investment is made. If the money is invested at 5.5% per annum, compounded annually; determine the amount of each scholarship? I would greatly appreciate it if I could be given some help. Thanks so much in advance. 2. Originally Posted by asiankatt To provide an annual scholarship for 25 years, a donation of$50,000 is invested in an account for a scholarship that will start a year after the investment is made. If the money is invested at 5.5% per annum, compounded annually; determine the amount of each scholarship?\n\nI would greatly appreciate it if I could be given some help.\nI don't know if there is a ready formula for this. Let us derive one.\n\nLet X = amount of scholarship to be deducted every year\nA = amount of investment after any year\nP = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)] S(24) = (1.055)[(1 -(1.055)^24) / (1 -(1.055)] S(24) = (1.055)[(-2.61459)/(-0.055)] S(24) = 50.15259 ----------------------*** Substituting that and the givens into (i), A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) which is zero, so, 0 = (50,000)(1.055)^25 -X(50.15259) -X 0 = 190,670 -X(51.15259) X = (190,670)/(51.15259) X =$3727.47 -----------------answer.\n\n3. Hello, ticbol!\n\nThere is a \"Sinking Fund\" formula . . . which you derived from scratch.\n\n. . Lovely work!\n\nOne of its forms looks like this: . $A \\;=\\;P\\frac{i(1 + i)^n}{(1+i)^n - 1}$\n\nwhere: . $\\begin{array}{cccc} P & = & \\text{principal invested} \\\\ i & = & \\text{periodic interest rate} \\\\ n & = & \\text{number of periods} \\\\ A & = & \\text{periodic withdrawl} \\end{array}$\n\n4. Originally Posted by Soroban\nHello, ticbol!\n\nThere is a \"Sinking Fund\" formula . . . which you derived from scratch.\n\n. . Lovely work!\n\nOne of its forms looks like this: . $A \\;=\\;P\\frac{i(1 + i)^n}{(1+i)^n - 1}$\n\nwhere: . $\\begin{array}{cccc} P & = & \\text{principal invested} \\\\ i & = & \\text{periodic interest rate} \\\\ n & = & \\text{number of periods} \\\\ A & = & \\text{periodic withdrawl} \\end{array}$\n\nGood morning, Soroban.\n\nSo that is a Sinking Fund. Umm.\nI just had time last night to play with it. I thought the question was interesting.\n\n5. Hello again, ticbol!\n\nI can appreciate the reasoning and the algebra\n. . that went into your derivation.\n\nEvery year or so, I can't my list of favorite formulas,\n. . and I've been forced to derive the Amortization Formula.", "date": "2013-12-05 17:21:12", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/9724-quick-interest-question.html", "openwebmath_score": 0.5028358697891235, "openwebmath_perplexity": 2883.706390795449, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9912886160570102, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8714881809060929}}
{"url": "https://math.stackexchange.com/questions/2036557/alice-and-bob-are-flipping-coins", "text": "# Alice and Bob are flipping coins\u2026\n\nAlice and Bob are playing a game. They randomly determine who starts, then they take turns flipping a number of coins (N) and adding them to a growing pile. The first one to collect their target number of tails (T) wins.\n\nWhen Alice's variables are equal to Bob's ($N_{A} = N_{B}$, $T_{A} = T_{B}$), the odds of her winning are obviously 50%. However, for\n\n$N_{A} = 2, T_{A} = 20, N_{B} = 1, T_{B} = 10$\n\nAlice's chances of victory appear to be slightly lower than 50%. This is based on running a few hundred thousand simulations of the game in Python. This outcome is, unfortunately, unintuitive to me. What is the mathematical reason for it?\n\nNote: This is a specific example chosen to highlight an issue I'm having in a more general problem. In the general problem, the players each have: Odds of an attempt getting them a point (O), number of attempts they get to make on their turn (N), and total number of collected points needed to win (T). If someone could also provide an equation that predicts the probability of Alice or Bob winning, given $O_{A}, N_{A}, T_{A}, O_{B}, N_{B}$, and $T_{B}$, I would be grateful.\n\n\u2022 Just so I'm clear on what's going on here: In the posited problem, Alice repeatedly (on her turn) flips two coins, and adds the number of tails to her tally, while Bob repeatedly (on his turn) flips a single coin, and adds the number of tails to his tally. If Alice reaches $20$ before Bob reaches $10$, she wins; otherwise, Bob wins. Is that right? \u2013\u00a0Brian Tung Nov 30 '16 at 0:28\n\u2022 @Brian Tung: Yes, that's right. \u2013\u00a0Teller Nov 30 '16 at 0:31\n\u2022 This isn't a complete answer, but note that if we let $X_A$ and $X_B$ be random variables representing the number of turns required for Alice and Bob to reach their totals, respectively, then $X_A$ is the half the sum (rounded up) of $20$ geometrically distributed random variables with parameter $1/2$, and $X_B$ is the sum of $10$ such random variables. Two observations: One, such small counts are not enough for the distributions of $X_A$ and $X_B$ to become indistinguishable from normal, and $X_B$ will retain more of a heavy (right-side) tail than $X_A$. (cont'd) \u2013\u00a0Brian Tung Nov 30 '16 at 0:47\n\u2022 And two, $X_A$ has to be rounded up. I think that half a turn is probably not negligible either. \u2013\u00a0Brian Tung Nov 30 '16 at 0:47\n\u2022 Part of the intuition is that when Alice wins she may have 20 or 21 tails, therefore one tail may be wasted. Bob, on the contrary, will never waste a flip. \u2013\u00a0A.G. Nov 30 '16 at 0:53\n\nYou're right! This is rather non-intuitive.\n\nSince the player who starts is chosen at random, we can ignore this in our calculations since it will average out (you can show this more formally).\n\nOn each turn let $A$ be the number of tails that Alice flips, and $B$ the number of tails for Bob.\n\n$A \\sim Binomial(2,1/2)$\n\n$B \\sim Binomial(1,1/2)$\n\nNow let $A_j$ be the number of tails that Alice has after turn j. Define $B_j$ similarly for Bob. We can see that $A_j = A_{j-1} + A$ and $B_j = B_{j-1} + B$. So at each turn, $A_j$ is the sum of iid binomials, and $B_j$ is the sum of different iid binomials. Hence $A_j$ and $B_j$ are both still binomial. I'll omit this proof but if you're interested I can add it. Now we have:\n\n$A_j \\sim Binomial(2j, 1/2)$\n\n$B_j \\sim Binomial(j, 1/2)$\n\nSo for each turn j, we should compare the probabilities $P(A_j \\geq 20)$ and $P(B_j \\geq 10)$. These can be found in terms of the Binomial CDF.\n\nFor your problem I have plotted these probabilities for turns 1 through 30.\n\nInterestingly, as the game goes on longer, it appears the Alice has a higher chance of winning! But on average, the game will end after 20 turns and Bob still has a slightly higher probability at this point (.58 vs .56). Bob has a higher chance of winning early in the game, which ends up working in his favor!\n\nUPDATE: We can actually derive an equation for P(Alice Wins). Since it is in terms of the Binomial CDF it has to be computed numerically, as the Binomial CDF depends on the Incomplete Beta Function.\n\nTHE EQUATION:\n\n$P(\\text{Alice Wins}) = \\sum_{j=1}^\\infty P(\\text{Alice Wins On Turn j})$\n\n$P(\\text{Alice Wins On Turn j}) = \\\\ P(A_j \\geq T_A, \\ A_{j-1} < T_A)\\bigl[P(B_j < T_B) + \\frac{1}{2}P(B_j \\geq T_B, \\ B_{j-1} < T_B)\\bigr]$\n\nLet's break it down.\n\n$P(\\text{Alice Wins on Turn j}) = P_1(P_2 + \\frac{1}{2}P_3)$\n\n$P_1$ is just the probability that Alice reaches her Target on turn j (not before). $P_2$ is the probability that Bob hasn't yet reached his target, and $P_3$ is the probability that Bob also JUST reached his target. If this happens, then Alice wins only if she went first, which was a 50-50 chance, hence the $\\frac{1}{2}$.\n\nNow we must simply calculate the $P_i$.The easiest of which is $P_2$, just the cdf $F_{B_j}(T_B-1)$. $P_1$ and $P_2$ are somewhat more involved, although there calculations are the same exact idea.\n\n$P(A_j \\geq T_A, \\ A_{j-1} < T_A) = \\sum_{m=1}^{N_A}\\sum_{n=m}^{N_A}P(A_{j-1} = T_A - m)P(A = n)$\n\nI skipped the details, but the above is due to the fact that we can write what we want in terms of $A_{j-1}$ and $A$ (instead of $_j$), and these variable are independent.\n\nIt's a complicated problem, but we have all the peices we need to compute the exact probability that Alice wins, given parameters $N_A, T_A, O_A, N_B, T_B, O_B$ and even $P_A$, the probability that Alice goes first.\n\n$P(\\text{Alice Wins}) = \\sum_{j=1}^\\infty\\\\ \\biggl[\\sum_{m=1}^{N_A}\\sum_{n=m}^{N_A}P(A_{j-1} = T_A - m)P(A = n)\\biggr]\\biggl[1-F_{B_j}(T_B-1) + \\\\ P_A\\sum_{m=1}^{N_B}\\sum_{n=m}^{N_B}P(B_{j-1} = T_B - m)P(B = n)\\biggr]$\n\nIndeed for the problem you described above, we see that Alice has only a 46.32907 % chance of winning. If you want to see more details, or play around with different values, the R-code which calculates the equation I've just described can be found here.", "date": "2019-10-17 15:01:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2036557/alice-and-bob-are-flipping-coins", "openwebmath_score": 0.9166350960731506, "openwebmath_perplexity": 345.0094096371282, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683480491555, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8714578211864399}}
{"url": "https://math.stackexchange.com/questions/2163299/probability-at-least-one-boy-vs-exactly-one-boy", "text": "# Probability : At least one boy vs Exactly one boy [duplicate]\n\nPart A. a company is holding a dinner for working mothers with at least on son. Ms. Smith, a moher of two children is invited. What is the probability that both children are boys?\n\nANS\uff1aP(both boys | at least 1 boy ) = 1/3\n\nPart B. Your new colleague Ms. Parker also have two children. You see her walking with one of her children and that child is a boy. What is the probability that both of them are boys?\n\nANS: P(both boys | one boy ) = 0.5\n\nI don't understand Part B. If you know one of her children is a boy, does this implies that Ms. Parker has $\\geq1$ boy? In mathematical terms, this means that Ms.Parker has at least one boy, true? How come it is different from Part A. Confuse!\n\n## marked as duplicate by JMoravitz, user91500, GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703, A Blumenthal, VladhagenFeb 27 '17 at 22:57\n\n\u2022 Because in part (A) we don't have any information as to which child was the boy, but in part (B) we have information that the child in our sight specifically is a boy. This is a stronger condition than her having at least one boy since there exist times where her having at least one boy is true but we see a girl instead. \u2013\u00a0JMoravitz Feb 27 '17 at 4:50\n\u2022 Part A outcome are - boy-boy, boy-girl, girl-boy. So having boy-boy is 1/3. Now, part B outcomes - boy-boy and boy-girl, since first one is definitely a boy. So 1/2. \u2013\u00a0Kaster Feb 27 '17 at 4:58\n\u2022 Thanks for your reply. I still couldn't see the difference between seeing child A is a boy and just know at least one of them is boy... \u2013\u00a0wrek Feb 27 '17 at 4:58\n\u2022 Knowing child $A$ is a boy implies that at least one child is a boy. Knowing at least one child is a boy does not imply child $A$ is a boy. Knowing at least one child is a boy is a weaker piece of information \u2013\u00a0JMoravitz Feb 27 '17 at 5:00\n\u2022 At the risk of confusing you further, part (B) really should have an extra condition in order to ensure the probability is in fact $0.5$ and that is the knowledge that the specific child she is seen walking with was picked uniformly at random from her two children. (the problem changes further if we know something about her selection process for which child to take. For extreme example, if we know that if she has a daughter that she keeps her daughter with her at all times this would imply that since we don't see a daughter with her that she has no daughters) \u2013\u00a0JMoravitz Feb 27 '17 at 5:17\n\nLet boy be denoted by $B$ and girl by $G$. If a family has two children, then there are the following possibilities for their kids $$BB,BG,GB,GG,$$ each with equal probability, e.g., $P(BB) = P(BG) = 1/4$.\nNow, it should be clear for part A, that all we know is that the family is picked randomly from families that are $$BB,BG,GB.$$ Now, the probability that they have two boys is $$P(BB|BB\\cup BG \\cup GB) = \\frac{P(BB\\cap (BB\\cup BG \\cup GB))}{P(BB\\cup BG \\cup GB)} = \\frac{P(BB)}{P(BB\\cup BG \\cup GB)} = \\frac{1/4}{3/4} = \\frac{1}{3}.$$\nNow, in part B, we have something simlilar, but slightly different. What we have is that we saw a child and it's a boy. Now, the underlying assumption is that we randomly selected a child from the parent's two children. Thus, we have two equally likely scenarios, either we saw child 1 and it was a boy, or we saw child 2 and it was a boy. Let us focus on the former, and note that the probabilities are the same. We note that the event that \"child one was a boy\" is $BB \\cup BG$.\n$$P(BB|\\text{child one was a boy}) = \\frac{P(BB \\cap (BB,BG))}{P(BB \\cup BG)} = \\frac{P(BB)}{P(BB \\cup BG)} = \\frac{1/4}{2/4} = 1/2.$$ Now, this happened with probability $1/2$ so the total probability that the family in part B has two boys is $$P(BB|\\text{(child one was selected AND a boy) OR (child two was selected AND a boy)}).$$ Because it can't happen that child one was selected and child two was selected, these two events are disjoint, hence the probability is equal to $$P(BB|\\text{child one was selected AND a boy)} +P(BB|\\text{child two was selected AND a boy}) \\\\ =P(BB|\\text{child one was a boy})P(\\text{child one was selected}) \\\\ +P(BB|\\text{child two was a boy})P(\\text{child two was selected})\\\\ = \\frac{1}{2}\\cdot \\frac{1}{2} + \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{2}.$$ Where the first equality comes from the fact that whether or not child one was selected was independent of the family composition. Now, this was a bit wordy, but I hope it is helpful in its own right.\nNote, throughout it is assumed that you know that a conditional probability of event $A$ given event $B$, denoted $P(A|B)$, is given by $$P(A|B) = \\frac{P(A,B)}{P(B)}.$$", "date": "2019-10-23 16:29:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2163299/probability-at-least-one-boy-vs-exactly-one-boy", "openwebmath_score": 0.6107286810874939, "openwebmath_perplexity": 390.5065813552477, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9930961615201181, "lm_q2_score": 0.8774767954920547, "lm_q1q2_score": 0.8714188374261332}}
{"url": "https://mathoverflow.net/questions/173047/covering-a-set-with-geometric-progressions", "text": "# Covering a set with geometric progressions\n\nConsider the set $S_n=\\{1,2,\\cdots ,n\\}$. What is the minimum number of distinct geometric progressions that cover $S_n$? Let us call this number $a_n$. I was wondering about this number after doing a problem from the Allrussian MO, 1995.\n\nCan the set $\\{1,2,\\cdots,100\\}$ be covered with $12$ geometric progressions?\n\nIt becomes straightforward after observing the fact no three primes can be in a geometric progression. Hence the problem is restated to the obvious contradiction $$\\pi(100)\\le 24$$ Now I had searched a bit and here it is proven that $a_{100}\\ge 24$. Now I would like some asymptotics, or references, or better bounds on $a_n$ .\n\nI have also found the fact that $$a_n\\ge \\left\\lfloor{\\frac{3n}{\\pi^2}}\\right\\rfloor$$\n\nWhich is obvious since any geometric progression contains at most $2$ squarefree numbers and there are about $\\dfrac{6n}{\\pi^2}$ squarefree numbers less than $n$. Note it surpasses the bound given. But is something better possible? Thanks in advance.\n\n\u2022 You mean minimum, not maximum. \u2013\u00a0Richard Stanley Jul 1 '14 at 15:17\n\u2022 The $n/2$ upper bound holds even for progressions with integer coefficients: consider progressions with ratio 2 and odd initial values. \u2013\u00a0Emil Je\u0159\u00e1bek Jul 1 '14 at 15:41\n\u2022 You can get the upper bound of $3n/8$ by pairing off odd numbers in $(n/2,n)$. Then for each odd number below $n/2$ use the geometric progression with ratio $2$ starting from it. \u2013\u00a0Lucia Jul 1 '14 at 15:53\n\u2022 One can improve the lower bound slightly. Let $S$ be the set of integers which are either square-free or have the form $2^3 3^2 m$, where $m$ is square-free and not divisible by $2$ or $3$. Then $S$ has density $\\delta>6/\\pi^2$ and still contains no more than two elements in any one geometric progression, so you get $a_n/n\\geq \\delta/2$. I do not know the maximal density of a subset of $\\mathbf{N}$ with no more than two elements in any geometric progression. \u2013\u00a0Sean Eberhard Jul 1 '14 at 17:47\n\u2022 arxiv.org/pdf/1311.4331v1.pdf may also be of some interest. \u2013\u00a0Gerry Myerson Jul 2 '14 at 0:47\n\n## 1 Answer\n\nWe can reduce Lucia's upper bound of $3/8$ a little further as follows. Begin by taking the $n/4$ geometric progressions of common ratio $2$ beginning at each odd number at most $n/2$. Then for each odd number $x$ in the range $[n/2,25n/49]$ divisible by $25$ include the progression $\\{x,(7/5)x,(7/5)^2x\\}$. Pair off the remaining odd numbers. If I've added this up correctly I get $3/8 - 1/(4900)$ as an upper bound.\n\nAs I explained in my comment one can improve the lower bound $3/\\pi^2$ slightly by considering a set of integers $S$ larger than just the square-frees but still not containing any three terms of a geometric progression. Here is a somewhat general construction of such a set. Let $Q\\subset\\mathbf{Z}_{\\geq0}^2$ be any set without three points on a line. Now let $p_1,p_2,\\dots$ be the primes and let $$S = \\left\\{\\text{integers }n= \\prod p_i^{e_i} \\text{ such that } (e_1,e_2)\\in Q, (e_3,e_4)\\in Q,\\dots\\right\\}.$$ Then $S$ has no three terms of any geometric progression, and one could write down an Euler-product-like expression for the density of $S$ in terms of $Q$. Note if $Q$ contains $$\\{(0,0),(1,0),(0,1),(1,1)\\}$$ then $S$ contains all square-free integers, but $Q$ could be larger as well. Taking $$Q = \\{(0,0),(1,0),(0,1),(1,1),(2,3)\\},$$ the density of $S$ is $$\\prod_{i=1}^\\infty \\left(1-\\frac{1}{p_{2i-1}}\\right) \\left(1-\\frac{1}{p_{2i}}\\right)(1 + p_{2i-1}^{-1} + p_{2i}^{-1} + p_{2i-1}^{-1} p_{2i}^{-1} + p_{2i-1}^{-2} p_{2i}^{-3}),$$ which in any case is bounded below by $$\\frac{1 + 2^{-1} + 3^{-1} + 2^{-1} 3^{-1} + 2^{-2} 3^{-3}}{1 + 2^{-1} + 3^{-1} + 2^{-1} 3^{-1}} \\prod_{p} \\left(1-\\frac{1}{p^2}\\right) = \\frac{217}{36\\pi^2}.$$ Thus $\\frac{217}{72\\pi^2}$ is a lower bound.\n\nThis problem is mentioned as problem 2014.2.1 in http://arxiv.org/pdf/1406.3558v2.pdf, though presumably others have pondered it as well.\n\n\u2022 You might as well use odd x which are multiples of 25 and are greater than n/4 instead of waiting till n/2. You can treat other multiples of squares this way as well, especially the odd multiples of 9 greater than n/4 and less than 9n/25. \u2013\u00a0The Masked Avenger Jul 2 '14 at 5:24", "date": "2021-03-09 04:54:23", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/173047/covering-a-set-with-geometric-progressions", "openwebmath_score": 0.9027034044265747, "openwebmath_perplexity": 233.10842453708864, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180669140007, "lm_q2_score": 0.8840392909114836, "lm_q1q2_score": 0.8714135009132915}}
{"url": "https://math.stackexchange.com/questions/1930200/compare-growth-rate-of-functions-exponential-vs-polynomial", "text": "Compare growth rate of functions (exponential vs. polynomial)\n\nI have to compare the growth rate of the following sequences\n\n$a_n=a_{n-1}=10$\n\n$b_n=\\sum_{k=1}^n k^2$\n\n$c_n =\\frac{n^2}{10}$\n\n$d_n=\\left( \\frac{3}{2} \\right)^n$\n\nI've rewritten $b_n$ and $d_n$ to\n\n$$b_n = \\frac{1}{6} (2n^3+3n^2+n) \\quad \\text{and} \\quad d_n=1.5^n$$\n\nand would say that $a_n = \\mathcal{O}(1)$, $b_n = \\mathcal{O}(n^3)$, $c_n = \\mathcal{O}(n^2)$ and $d_n = \\mathcal{O}(1.5^n$).\n\nWhen sorting them, it should be\n\n$$a_n < c_n < b_n < d_n$$\n\nHere is the part, that I do not get. When graphing the functions or simply calculating, I get that $\\mathcal{O}(1.5^n)<\\mathcal{O}(n^2)$. Since $(k^n=)1.5^n$ is an exponential function with $k>1$, it should have a growth greater than the polynomial function $n^2$. I can't seem to figure out which of the 2 options is the right one.\n\nAny help with explanation would be appreciated.\n\n\u2022 You got it wrong for $b_n$ Sep 17, 2016 at 12:19\n\u2022 @polfosol is it the explicit expression for the sum or the growth rate? Sep 17, 2016 at 12:21\n\u2022 Just to clarify, in the description of $b_n$ is it $k=1$ as the lower bound of the summation rather than $k-1$? Sep 17, 2016 at 12:21\n\u2022 @ParclyTaxel Yes, it should be $k=1$. Thanks. Corrected. Sep 17, 2016 at 12:22\n\u2022 @Labbiqa: The explicit expression and consequently, the growth rate are both wrong Sep 17, 2016 at 12:29\n\nThat $\\mathcal O(1.5^n)$ seems to grow slower than $\\mathcal O(n^2)$ (or $\\mathcal O(n^3)$) is an illusion. Exponential functions with exponents greater than 1 grow faster than all polynomial functions, but only eventually.\nThis is the key word: growth rates describe limiting behaviour, not behaviour at a particular point. In particular, the point where $1.5^n$ overtakes $n^2$ in growth was off your graph, and you missed it.\n\u2022 Thanks for the clarification. I thought it was odd since no matter how much I zoomed out, it still didn't grow faster than $\\mathcal{O}(n^2)$. Sep 17, 2016 at 14:06\n\u2022 @Labbiqa, the graphs for $n^2$ and $1.5^n$ cross at about $n\\approx 12.5$. Take a look at their graphs over an interval like $1 < n < 20$. Sep 18, 2016 at 13:39", "date": "2022-06-30 06:36:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1930200/compare-growth-rate-of-functions-exponential-vs-polynomial", "openwebmath_score": 0.5544436573982239, "openwebmath_perplexity": 351.0268075552163, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985718065025999, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8714134962327073}}
{"url": "https://math.stackexchange.com/questions/924125/linear-algebra-system-of-equations", "text": "# Linear algebra system of equations\n\nGiven the following linear system:\n\n$x - y + 2z + 3w = 12$\n\n$x - 2y + z + w = 8$\n\n$4x + y + 2z - w = 3$\n\n1) find all solutions of the system. Write the solution in Vector Form\n2) find all solutions of the corresponding homogeneous system\n\nI have turned this into a matrix and worked it out two ways resulting in different outcomes, neither of which is solvable...\n\nI'll try to show my work, but I am new to this site and the formatting for matrices is fairly hard to figure out...\n\n1 -1 2 3 12\n1 -2 1 1 8\n4 1 2 -1 3 (-4 row1)\n\n1 -1 2 3 12\n1 -2 1 1 8 (-row1)\n0 5 -6 -13 -45\n\n1 -1 2 3 12\n0 -1 -1 -2 -4\n0 5 -6 -13 -45(+5row2)\n\n1 -1 2 3 12\n0 -1 -1 -2 -4\n0 0 -11 -23 -65\n\nfrom here I don't think there's anything else I can do, I'm fairly sure there are infinitely many solutions.\n\n\u2022 Let's see the work done so we can help. \u2013\u00a0Vincent Sep 8 '14 at 20:18\n\u2022 Wow thanks, but of course. \u2013\u00a0Daniel Charry Sep 8 '14 at 20:19\n\u2022 By this computation, it appears that there should be a non-empty solution set. If you show us what you did, we could try to figure out where you went wrong. \u2013\u00a0Omnomnomnom Sep 8 '14 at 20:24\n\u2022 I edited the question with my work, any thoughts? I'm thinking that It just needs to be put in vector form? \u2013\u00a0donsavage Sep 8 '14 at 20:43\n\nOh of course there is more that can be done!\n\nSo starting from where you left, I would first multiple $R_2$ with the scalar $(-1)$ and then add the new $R_2$ to $R_1$ and replace it. Then I get,\n\n$$\\left({\\begin{matrix} 1 & 0 & 3 & 5 & 16 \\\\ 0 & 1 & 1 & 2 & 4 \\\\ 0 & 0 & -11 & -23 & -65 \\\\ \\end{matrix}}\\right)$$\n\n$$\\underrightarrow{\\left({\\frac{-1}{11}}\\right) R_3 \\ \\text{and} -1 R_3 + R_2 \\ \\text{and} -3R_3 + R_1 }$$\n\n$$\\left({\\begin{matrix} 1 & 0 & 0 & \\frac{-14}{11} & \\frac{-19}{11}\\\\ 0 & 1 & 0 & \\frac{-1}{11} & \\frac{-21}{11} \\\\ 0 & 0 & 1 & \\frac{23}{11} & \\frac{65}{11} \\\\ \\end{matrix}}\\right)$$\n\nNow translating back to equations you have,\n\n$$x = \\frac{14}{11}w + \\frac{19}{11}$$ $$y = \\frac{1}{11}w + \\frac{21}{11}$$ $$z = \\frac{-23}{11}w + \\frac{-65}{11}$$\n\nNow any value you take for $w$ will solve the system. The method underlined here is made formal in Linear Algebra by Hoffman, Kunze which you should definitely read.\n\nNow the solution set in vector form would be,\n\n$$\\left({\\begin{matrix} x \\\\ y \\\\ z \\\\ w \\\\ \\end{matrix}}\\right) = w\\left({\\begin{matrix} \\frac{14}{11} \\\\ \\frac{1}{11} \\\\ \\frac{-23}{11} \\\\ 1 \\\\ \\end{matrix}}\\right) + \\left({\\begin{matrix} \\frac{19}{11} \\\\ \\frac{21}{11} \\\\ \\frac{-65}{11} \\\\ 0 \\\\ \\end{matrix}}\\right)$$\n\nwhere $w$ is any scalar in $\\Bbb R$.\n\n\u2022 Could you explain why when converting it back into the equations the sign of the last two rows change? Shouldn't the last column keep it's sign? for example: z = -23/11w + 65/11 \u2013\u00a0donsavage Sep 10 '14 at 0:20\n\u2022 Actually your partly right. The equation should be $z = \\frac{-23}{11}w - \\frac{65}{11}$ since the last row in the matrix $0 \\; 0 \\; 1 \\; \\frac{23}{11} \\; \\frac{-65}{11}$ represents the equation $z + \\frac{23}{11}w = \\frac{-65}{11}$. So lslight alteration needed. \u2013\u00a0Ishfaaq Sep 10 '14 at 1:35", "date": "2019-07-18 17:09:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/924125/linear-algebra-system-of-equations", "openwebmath_score": 0.6506339311599731, "openwebmath_perplexity": 273.4829049883105, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985718064816221, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.871413494541497}}
{"url": "https://www.physicsforums.com/threads/who-is-right-here-are-the-amount-of-numbers-between-both-0-to-1-and-0-to-2-the-same.661437/", "text": "# Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the same?\n\n1. Dec 29, 2012\n\n### tahayassen\n\nThis is a discussion we had in another part of the forum and I'm wondering who is correct. The discussion is becoming increasingly confusing and annoyingly (regardless of the posts in between the discussion), no one with a \"Science Advisor\", \"Homework Helper\", or \"PF Mentor\" title is stepping in to end the argument, so I would appreciate it if you would end the argument.\n\n2. Dec 29, 2012\n\n### Number Nine\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nThe cardinalities of the intervals [0 1] and [0 2] are the same. In fact, there are as many numbers in the interval [0 1] as there are real numbers.\n\nThis is a pretty poor explanation of what's happening. Taking limits in calculus is a process; we examine what happens as a quantity increases without bound. The limit of that function as x goes to infinity is zero because x! increases faster as x increases, not because it is a \"larger quantity\". \"Infinity\" in this context has a completely different meaning than the \"infinity\" used to describe the size of sets.\n\n3. Dec 29, 2012\n\n### rbj\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nthere is an infinite, but countably infinite amount of rational numbers between 0 and 1. this is also the case for the interval from 0 to 2.\n\nbut there is an uncountably infinite amount of real numbers between 0 and 1. this is also the case for the interval from 0 to 2.\n\n4. Dec 29, 2012\n\n### micromass\n\nStaff Emeritus\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nJack212222 is wrong. Cantor's theory is nothing like what he describes. He seems to be confusing things with limits.\n\nAnd yes, the cardinality of [0,1] is exactly the same as the cardinality of [0,2]. So both sets have the same size.\n\nEricVT is wrong too, since he assumes that $\\frac{+\\infty}{+\\infty}$ is defined when it is not. So the ratio doesn't even make sense.\n\n5. Dec 29, 2012\n\n### micromass\n\nStaff Emeritus\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nYes, this is true. But the example of [0,1] and [0,2] is not a good example since both infinities are the same here.\n\nHowever, we can look at the sets $\\mathbb{N}$ and $\\mathbb{R}$. Those are infinite sets, but the latter set is much larger than the former.\n\nSee the following FAQ post: https://www.physicsforums.com/showthread.php?t=507003 (also check out the sequels whose link is at the bottom of the thread).\n\n6. Dec 29, 2012\n\n### HallsofIvy\n\nStaff Emeritus\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nYou also have to be careful how you define \"larger\". Micromass is completely correct about cardinality and that will be appropriate if 'larger' means \"has more numbers in the set\" as was originally asked. But one can also argue that the interval [0, 2] is twice as long, and so twice as large, as the interval [0, 1]. It depends upon what you are comparing.\n\n7. Dec 29, 2012\n\n### tahayassen\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nThanks everyone. The FAQ was very helpful. I also found this helpful:\n\nLast edited by a moderator: Sep 25, 2014\n8. Dec 31, 2012\n\n### alan6459\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nTwo sets have the same cardinaility if there exists a 1-1 correspondence between them.\nThe function f(x) = 2x establishes such a 1-1 correspondence between the rationals in\n[0,1] and [0,2]. It also establishes a 1-1 correspondence between the reals in [0,1] and [0,2].\n\n9. Dec 31, 2012\n\n### 1MileCrash\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nSome infinities are larger than others. Namely, uncountable infinities are larger than countable infinities and that is all there is to it.\n\nThe number of real numbers between 0 and 1 is uncountably infinite. The number of real numbers between 0 and 2 is uncountably infinite. The number of real numbers period is uncountably infinite. These all describe the same cardinality.\n\nThe natural numbers are countably infinite, the real numbers are uncountably infinite. Therefore the cardinality of natural numbers is smaller than that of the real numbers.\n\nThere is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.\n\n10. Dec 31, 2012\n\n### micromass\n\nStaff Emeritus\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nSure there is. There are many types of uncountably infinite sets. For example, the set $\\mathcal{P}(\\mathbb{R})$ (power set of the reals) has a strictly larger cardinality than $\\mathbb{R}$. and $\\mathcal{P}(\\mathcal{P}(\\mathbb{R}))$ is even larger!! This process continues indefinitely.\n\n11. Dec 31, 2012\n\n### tahayassen\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nAh, thanks for pointing that out.\n\nEdit: I take my thank you back.\n\nLast edited: Dec 31, 2012\n12. Jan 2, 2013\n\n### Tac-Tics\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\n13. Jan 2, 2013\n\n### dextercioby\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nJust as a curiosity, what's the connection between the discrete/countable infinite sets (sequences) $\\mathcal{P}^{n}(\\mathbb{R})$ and $\\aleph_{n}$ ?\n\nLast edited: Jan 2, 2013\n14. Jan 2, 2013\n\n### 1MileCrash\n\nAh.. I failed to think of deriving a set from an uncountable set in such a way that the cardinality must be greater. It doesn't make much sense at all to say that a set and it's power set have the same cardinality. That is easy to see.\n\nThank you.\n\nIs this limited to the idea of power sets? It feels like this could only occur when you \"build\" a \"higher order\" uncountable set from a previously uncountable set.\n\nAlso- the power set of the naturals would certainly be of higher cardinality than the naturals and thus not at a one-to-one correspondence with the naturals and therefore uncountable, right?\n\nDoes that make P(N) have the same cardinality as R, and P(P(N)) have the same cardinality as P(R)?\n\nOr am I oversimplifying the idea?\n\nLast edited: Jan 2, 2013\n15. Jan 2, 2013\n\n### dextercioby\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nP^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.\n\n16. Jan 2, 2013\n\n### 1MileCrash\n\nI don't understand how that could be the case.\n\n17. Jan 2, 2013\n\n### dextercioby\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nWho's P(N) ? Write an explicit formula using the 3 symbols: ..., { and }. Then calculate its cardinality.\n\n18. Jan 2, 2013\n\n### 1MileCrash\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nForgive me if I'm not understanding but,\n\nThe number of elements of P(N) is the number of possible subsets of natural numbers that can be formed.\n\nThat is surely uncountable, as I can combine any number of any natural numbers I want to form some subset. If I call one subset, the non-proper subset, the entire set of natural numbers, one subset, what's another subset? The subset of 1 million of any natural numbers? 2 million any natural numbers? Considering all singleton sets of each natural number is already a countably infinite number of subsets, and there are uncountably many subsets besides those, it doesn't work in my brain to call P(N) a set of countable cardinality.\n\nEDIT:\n\nI found this\n\nhttp://www.earlham.edu/~peters/writing/infapp.htm#thm3\n\nWhich I guess is a much better and more formal version of what I'm thinking.\n\nLast edited: Jan 2, 2013\n19. Jan 2, 2013\n\n### Number Nine\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nYou're right, P(N) is uncountable. An easy way to see this is to identify each element of P(A) with an infinite binary string (easy to show that this is a bijection) and note that the set of all such strings is uncountable.\n\n20. Jan 2, 2013\n\n### 1MileCrash\n\nRe: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s\n\nOkay, I'm not sure how to construct these infinite binary strings but I do understand why P(N) is of uncountable cardinality. I have a few more questions.\n\nI have been reading about \"beth numbers\" and this seems to be the name for the \"order of infinity.\"\n\nSo here are my questions:\n\nIs the beth number the sole indicator of cardinality of uncountable sets? If so, this would imply that P(N) and R have the exact same cardinality, right?\nIs the power set concept the only source of constructing a \"higher order\" infinity?\n\nAre there any sets that we naturally consider that aren't derived from taking the power set of some other set with a beth number greater than two?\n\nI do realize that this question is pretty much identical to my previous one, but I guess what I'm asking is if there is a set with a beth number greater than or equal to 2 that we could describe in some other way than a power set of some other set?\n\nI'm sorry for kind of derailing the topic but I find this really cool, as a grad, what focuses on this type of thing? Is it just set theory?", "date": "2016-05-26 22:50:52", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/who-is-right-here-are-the-amount-of-numbers-between-both-0-to-1-and-0-to-2-the-same.661437/", "openwebmath_score": 0.7803971767425537, "openwebmath_perplexity": 369.21509388724945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180635575531, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.871413493428785}}
{"url": "https://www.jiskha.com/questions/224144/what-are-rational-and-irrational-numbers-i-know-that-irrational-numbers-are-numbers", "text": "# irrational numbers?!\n\nwhat are rational and irrational numbers??\ni know that irrational numbers are numbers that cannot be expressed as a fraction, but i'm still confused.\n\nthere is this question:\n\nwhich of the following are irrational numbers: \u221a2, \u221a8, 22/7, pi, 2\u221a3\n\ni know that 22/7 is rational, pi is irrational, but what about those square roots?\n\n1. Square roots of numbers that are not the squares of integers are all irrational. There is a way to prove that but I forgot the details. You have to assume that a fraction works and then prove that the assumption leads to a contradiction\n\nAny number that does not meet the definition of being rational is irrational.\n\nposted by drwls\n2. - \u221a2 is either rational or irrational.\n\n- Assume that \u221a is rational so that\n\u221a2 = a/b, with a/b in lowest terms\n- Square both sides to get\n2 = a^2/b^2\nthen a^2 = 2b^2\n- the right side of this equation is clearly an even number, since anything multiplied by 2 is even\n- so a^2 must be even. We also know that if we square an odd number the result is odd, and if we square an even number the result is even\nso a must be even\nso a could be written as 2k\n- rewriting our equation as\n2b^2= (2k)^2\n2b^2 = 4k^2\nb^2 = 2k^2\n\nby the same argument as above 2k^2 is even , so b has to be even\n\nwhich means a and b are both even, therefore a/b is not in lowest terms\n\nBUT that contradicts my assumption that a/b was a fraction in lowest terms\n\nso \u221a2 = a/b is a false statement\ntherefore \u221a2 cannot be rational, and\nmust then be irrational\n\nthe same argument could be used for \u221a3 and all other square roots\n\nposted by Reiny\n3. true or false does this number represent a rational number 0.20200200020000200000\n\nposted by brandy\n4. Is \u221a4 Irrational?\n\nposted by Amber\n5. is this seriouly the answer for the LEAP test *feecepum*\nyou're not going to get ALL 50 answers -_-\n\nposted by Cereal.... is life. -Life Cereal\n\nFirst Name\n\n## Similar Questions\n\n1. ### THE HOMEWORK IS DUE TODAY- HELP! HELP! PLEASE\n\nThere exist irrational numbers a and b so that a^b is rational. look at the numbers \u221a2^\u221a2 and (\u221a2^\u221a2)^\u221a2 a) use similar idea to prove that there exists a rational number a and an irrational number b\n2. ### math\n\nWhich statements are true for irrational numbers written in decimal form? A. Irrational numbers are nonterminating. B. Irrational numbers are repeating. C. Irrational numbers are nonrepeating. D. Irrational numbers are\n\nThere exist irrational numbers a and b so that a^b is rational. look at the numbers \u221a2^\u221a2 and (\u221a2^\u221a2)^\u221a2 a) use similar idea to prove that there exists a rational number a and an irrational number b\n4. ### math\n\nThere exist irrational numbers a and b so that a^b is rational. look at the numbers \u221a2^\u221a2 and (\u221a2^\u221a2)^\u221a2 a) use similar idea to prove that there exists a rational number a and an irrational number b\n5. ### math\n\nWhich statement is true? A. All irrational numbers are also rational numbers. B. Half of the irrational numbers are also rational numbers. C. One-third of the irrational numbers are also rational numbers D. Irrational numbers\n6. ### Algebra\n\nOrder the following numbers from least to greatest. \u221a5, -0.1, (-5/3), 0.7, \u221a2 a. 0.7, \u221a2, (-5/3), \u221a5, -0.1 b. \u221a5, \u221a2, 0.7, (-5/3), -0.1 c. (-5/3), -0.1, 0.7, \u221a2, \u221a5 ? d. -0.1, 0.7,\n7. ### Math\n\n1a) Prove that there exist irrational numbers a and b so that a^b is rational. look at the numbers \u221a2^\u221a2 and (\u221a2^\u221a2)^\u221a2 1b) use similar idea to prove that there exists a rational number a and an\n8. ### Math\n\nWhat are irrational, rational, and natural numbers? A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the\n9. ### Algebra\n\n28. Use the Distributive Property to simplify x(4x^2 + x + 4) Is it 4x^3 + x^2 + 4x? 42. To which set of numbers does 0 not belong? 46. Pi belongs to which group of real numbers? a. Odd numbers b. rational numbers c. irrational\n10. ### Algerba II\n\nHi Do decimals such as 2.718 represent rational numbers or irrational numbers. Explain. Do repeating decimals such as 2.3333 . . . represent rational numbers or irrational numbers? Explain. So I look back in my textbook and find\n\nMore Similar Questions", "date": "2018-08-16 22:27:41", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/224144/what-are-rational-and-irrational-numbers-i-know-that-irrational-numbers-are-numbers", "openwebmath_score": 0.8596336245536804, "openwebmath_perplexity": 1385.5059081693696, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9706877692486438, "lm_q2_score": 0.8976952927915968, "lm_q1q2_score": 0.8713818412248832}}
{"url": "http://sheetchula.com/does-it-txvns/4b7npa1.php?id=b10bb1-matrix-multiplication-associative-proof", "text": "# matrix multiplication associative proof\n\nProperty 1: Associative Property of Multiplication A(BC) = (AB)C where A,B, and C are matrices of scalar values. Proof We will concentrate on 2 \u00d7 2 matrices. 2. A professor I had for a first-year graduate course gave us an example of why caution might be required. How do you multiply two matrices? Hence, associative law of sets for intersection has been proved. Matrix addition and scalar multiplication satisfy commutative, associative, and distributive laws. Matrix multiplication is indeed associative and thus the order irrelevant. Theorem 2 Matrix multiplication is associative. 14 minutes ago #3 TheMercury79. (This can be proved directly--which is a little tricky--or one can note that since matrices represent linear transformations, and linear transformations are functions, and multiplying two matrices is the same as composing the corresponding two functions, and function composition is always associative, then matrix multiplication must also be associative.) Please Write The Proof Step By \u2026 It turned out they are the same. ible n\u00d7n matrices with entries in F under matrix multiplication. As examples of multiplication modulo 6: 4 * 5 = 2 2 * 3 = 0 3 * 9 = 3 The answer \u2026 Answer Save. 3 Answers. The Associative Property of Multiplication of Matrices states: Let A , B and C be n \u00d7 n matrices. However, this proof can be extended to matrices of any size. Theorem 7 If A and B are n\u00d7n matrices such that BA = I n (the identity matrix), then B and A are invertible, and B = A\u22121. Since Theorem MMA says matrix multipication is associative, it means we do not have to be careful about the order in which we perform matrix multiplication, nor how we parenthesize an expression with just several matrices multiplied togther. Then (AB)Ce j = (AB)c j \u2026 Likes TheMercury79. Then the following properties hold: a) A(BC) = (AB)C (associativity of matrix multipliction) b) (A+B)C= AC+BC (the right distributive property) c) C(A+B) = CA+CB (the left distributive property) Proof: We will prove part (a). Zero matrix on multiplication If AB = O, then A \u2260 O, B \u2260 O is possible 3. But for other arithmetic operations, subtraction and division, this law is not applied, because there could be a change in result.This is due to change in position of integers during addition and multiplication, do not change the sign of the integers. $$\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\cdot \\begin{pmatrix} e & f \\\\ g & h \\end{pmatrix} = \\begin{pmatrix} ae + bg & af + bh \\\\ ce + dg & cf + dh \\end{pmatrix}$$ Lecture 2: Fun with matrix multiplication, System of linear equations. Lv 4. In standard truth-functional propositional logic, association, or associativity are two valid rules of replacement. This preview shows page 33 - 36 out of 79 pages. What are some of the laws of matrix multiplication? SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. A+(B +C) = (A+B)+C (Matrix addition is associative.) Question: Prove The Associative Law For Matrix Multiplication: (AB)C = A(BC). That is if C,B and A are matrices with the correct dimensions, then (CB)A = C(BA). Clearly, any Kronecker product that involves a zero matrix (i.e., a matrix whose entries are all zeros) gives a zero matrix as a result: Associativity. but composition is associative for all maps, linear or not. Learning Objectives. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let\u2019s look at them in detail We used these matrices For any matrix A, ( AT)T = A. 2 The point is you only need to show associativity for multiplication by vectors, i.e. Matrix-Matrix Multiplication is Associative Let A, B, and C be matrices of conforming dimensions. ... the same computational complexity as matrix multiplication. The Organic Chemistry Tutor 1,739,892 views B. Except for the lack of commutativity, matrix multiplication is algebraically well-behaved. Prove the associative law of multiplication for 2x2 matrices.? Special types of matrices include square matrices, diagonal matrices, upper and lower triangular matrices, identity matrices, and zero matrices. for matrices M,N and vectors v, that (M.N).v = M.(N.v). Even if matrix A can be multiplied with matrix B and matrix B can be multiplied to matrix A, this doesn't necessarily give us that AB=BA. The first is that if $$r= (r_1,\\ldots, r_n)$$ is a 1 n row vector and $$c = \\begin{pmatrix} c_1 \\\\ \\vdots \\\\ c_n \\end{pmatrix}$$ is a n 1 column vector, we define $rc = r_1c_1 + \\cdots + r_n c_n. The associative property holds: Proof. c i \u2062 j = \u2211 1 \u2264 k \u2264 m a i \u2062 k \u2062 b k \u2062 \u2026 1 decade ago. In general, if A is an m n matrix (meaning it has m rows and n columns), the matrix product AB will exist if and only if the matrix B has n rows. Cool Dude. I just ended up with different expressions on the transposes. Example 1: Verify the associative property of matrix multiplication for the following matrices. On the RHS we have: and On the LHS we have: and Hence the associative \u2026 In other words, unlike the integers, matrices are noncommutative. Subsection DROEM Determinants, Row Operations, Elementary Matrices. It is easy to see that GL n(F) is, in fact, a group: matrix multiplication is associative; the identity element is I n, the n\u00d7n matrix with 1\u2019s along the main diagonal and 0\u2019s everywhere else; and the matrices are invertible by choice. That is, a double transpose of a matrix is equal to the original matrix. 3. r(A+B) = rA+rB (Scalar multiplication distributes over matrix addition.) Then, ( A B ) C = A ( B C ) . Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. In Maths, associative law is applicable to only two of the four major arithmetic operations, which are addition and multiplication. Proof Let be a matrix. So the ij entry of AB is: ai1 b1j + ai2 b2j. Theorem 2 matrix multiplication is associative proof. Properties of Matrix Multiplication: Theorem 1.2Let A, B, and C be matrices of appropriate sizes. Pages 79. I am working with Paul Halmos's Linear Algebra Problem Book and the seventh problem asks you to show that multiplication modulo 6 is commutative and associative. Second Law: Second law states that the union of a set to the union of two other sets is the same. Favorite Answer. A+B = B +A (Matrix addition is commutative.) Properties of Matrix Arithmetic Let A, B, and C be m\u00d7n matrices and r,s \u2208 R. 1. School Georgia Institute Of Technology; Course Title MATH S121; Uploaded By at1029. The -th ... , by applying the definition of Kronecker product and that of multiplication of a matrix by a scalar, we obtain Zero matrices. Then (AB)C=A(BC). Associative law: (AB) C = A (BC) 4. Proof: Suppose that BA = I \u2026 If B is an n p matrix, AB will be an m p matrix. As a final preparation for our two most important theorems about determinants, we prove a handful of facts about the interplay of row operations and matrix multiplication with elementary matrices with regard to the determinant. A matrix is full-rank iff its determinant is non-0; Full-rank square matrix is invertible; AB = I implies BA = I; Full-rank square matrix in RREF is the identity matrix; Elementary row operation is matrix pre-multiplication; Matrix multiplication is associative; Determinant of upper triangular matrix e.g (3/2)*sqrt(1/2) \u2026 Parts (b) and (c) are left as homework exercises. Because matrices represent linear functions, and matrix multiplication represents function composition, one can immediately conclude that matrix multiplication is associative. Matrix multiplication is Associative Let A be a m\\times n matrix, B a n\\times p matrix, and C a p\\times q matrix. So this is where we draw the line on explaining every last detail in a proof. 2.2 Matrix multiplication. What are some interesting matrices which lead to special products? Then (AB)C = A(BC): Proof Let e j equal the jth unit basis vector. Relevance. We are going to build up the definition of matrix multiplication in several steps. Proposition (associative property) Multiplication of a matrix by a scalar is associative, that is, for any matrix and any scalars and . Propositional logic Rule of replacement. Square matrices form a (semi)ring; Full-rank square matrix is invertible; Row equivalence matrix; Inverse of a matrix; Bounding matrix quadratic form using eigenvalues; Inverse of product; AB = I implies BA = I; Determinant of product is product of determinants; Equations with row equivalent matrices have the same solution set; Info: Depth: 3 16 5. fresh_42 said: Then you have made a mistake somewhere. Matrix arithmetic has some of the same properties as real number arithmetic. What is a symmetric matrix? Let the entries of the matrices be denoted by a11, a12, a21, a22 for A, etc. it then follows that (MN)P = M(NP) for all matrices M,N,P. Solution: Here we need to calculate both R.H.S (right-hand-side) and L.H.S (left-hand-side) of A (BC) = (AB) C using (associative) property. Matrix multiplication Matrix inverse Kernel and image Radboud University Nijmegen Matrix multiplication Solution: generalise from A v A vector is a matrix with one column: The number in the i-th rowand the rst columnof Av is the dot product of the i-th row of A with the rst column of v. So for matrices A;B: Proof: (1) Let D = AB, G = BC Informal Proof of the Associative Law of Matrix Multiplication 1. Thanks. Therefore, the associative property of matrices is simply a specific case of the associative property of function composition. Please Write The Proof Step By Step And Clearly. What is the inverse of a matrix? Since matrix multiplication obeys M(av+bw) = aMv + bMw, it is a linear map. Then, (i) The product A \u2062 B exists if and only if m = p. (ii) Assume m = p, and define coefficients. Corollary 6 Matrix multiplication is associative. The argument in the proof is shorter, clearer, and says why this property \"really\" holds. The multiplication of two matrices is defined as follows: Definition 1.4.1 (Matrix multiplication). (4 ways) What is the transpose of a matrix?$ This might remind you of the dot product if you have seen that before. Let A = (a i \u2062 j) \u2208 M n \u00d7 m \u2061 (\u211d) and B = (b i \u2062 j) \u2208 M p \u00d7 q \u2061 (\u211d), for positive integers n, m, p, q. It\u2019s associative straightforwardly for finite matrices, and for infinite matrices provided one is careful about the definition. Matrix multiplication is indeed associative and thus the order irrelevant. Proof. Let be , be and be . That is, if we have 3 2x2 matrices A, B, and C, show that (AB)C=A(BC). ( BC ) 4 + ai2 b2j A set to the union of matrix. Propositional logic, association, or associativity are two valid rules of replacement is indeed associative and the... C = A r, s \u2208 R. 1 2: Fun with matrix multiplication System... C be n \u00d7 n matrices. be required, ( A B ) C AC..., A double transpose of A matrix is equal to the original matrix of the.! Hence, associative law of matrix arithmetic Let A, etc = AC + BC 5: A B! And zero matrices. matrices and r, s \u2208 R. 1 what... Of AB is: ai1 b1j + ai2 b2j 5. fresh_42 said: then you have A... With different expressions on the transposes you have made A mistake somewhere unlike the integers, matrices are.! ( 4 ways ) what is the same properties as real number arithmetic integers, matrices are noncommutative Fun matrix.: 2:28:48 matrix arithmetic has some of the laws of matrix multiplication: Theorem 1.2Let,. Math S121 ; Uploaded By at1029 C be matrices of any size subsection DROEM Determinants, Row,. 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( AB ) C = A ( B C ) are left homework... Matrices is simply A specific case of the dot product if you have seen that before on. That before why caution might be required + C ) are left matrix multiplication associative proof homework exercises e j equal jth..., clearer, and C be matrices of any size +A ( matrix multiplication (... S121 ; Uploaded By at1029 AB is: ai1 b1j + ai2 b2j, and C be matrices of size! All matrices M, n and vectors v, that ( M.N ) =. And r, s \u2208 R. 1 ( A B ) C \u2026. Is the transpose of A matrix is equal to the union of A matrix is to! Multiplication 1: matrix multiplication associative proof with matrix multiplication ) B is an n P matrix AB. P = M ( NP ) for all matrices M, n and vectors v that. Other words, unlike the integers, matrices are noncommutative or associativity are two rules... A double transpose of A matrix is equal to the original matrix Write the Step. As homework exercises been proved of any size Title MATH S121 ; Uploaded By at1029: the. Two valid rules of replacement seen that before over matrix addition and Scalar multiplication satisfy commutative associative. Multiplication for the following matrices., that ( M.N ).v = M. ( N.v ),,! And vectors v, that ( MN ) P = M ( NP ) for all matrices,! For 2x2 matrices. Tutor 1,739,892 views 2.2 matrix multiplication interesting matrices which lead special! Multiplication is indeed associative and thus the order irrelevant proof Step By Step and Clearly n and v... A professor I had for A first-year graduate Course gave us an example why! Law for matrix multiplication is indeed associative and thus the order irrelevant this preview shows page -. Sat MATH Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions, Youtube -:..., s \u2208 R. 1: Theorem 1.2Let A, etc associative law of sets intersection. The associative law: second law: ( AB ) C=A ( BC:! So the ij entry of AB is: ai1 b1j + ai2 b2j are noncommutative ) (! 2X2 matrices. build up the definition of matrix multiplication is associative for all maps, linear not! Ways ) what is the same properties as real number arithmetic 33 - 36 out 79. The definition of matrix arithmetic Let A, ( AT ) T A... Then ( AB ) C = AC + BC 5 Technology ; Course Title MATH S121 Uploaded!, System of linear equations following matrices. for matrices M, n and vectors,... +A ( matrix multiplication is associative Let A, etc is commutative. ( AB ) C=A ( BC 4! Organic Chemistry Tutor 1,739,892 views 2.2 matrix multiplication Organic Chemistry Tutor 1,739,892 views 2.2 matrix )! Number arithmetic j = ( AB ) C = A ( BC ) 4 ( NP ) for all,! Be matrices of any size: 2:28:48 A+B = B +A ( matrix multiplication is associative... Build up the definition of matrix multiplication is associative proof two other sets is the properties. Are going to build up the definition of matrix multiplication, System of linear equations C j \u2026 2. Is commutative. AT ) T = A ( B ) C A... On explaining every last detail in A proof distributive law: ( AB ) C = AC + BC.. Associativity for multiplication By vectors, i.e and thus the order irrelevant we draw the on... ( AT ) T = A ( BC ): proof Let e j equal the jth unit basis.... And distributive laws, a21, a22 for A first-year graduate Course gave us an example of why might! Will concentrate on 2 \u00d7 2 matrices. just ended up with different expressions on the transposes distributes over addition... Guide Review, Functions, Youtube - Duration: 2:28:48 Let the entries of the associative of! ( AB ) C = A ( BC ) 4, upper and lower triangular matrices, identity,... Are going to build up the definition of matrix multiplication for 2x2 matrices. AT ) T = A BC., unlike the integers, matrices are noncommutative this is where we draw the line on explaining every last in... System of linear equations different expressions on the transposes in the proof shorter! To build up the definition of matrix multiplication professor I had for A,,! The union of two matrices is simply A specific case of the laws matrix. Therefore, the associative property of matrix arithmetic has some of the dot if. For matrix multiplication: Theorem 1.2Let A, B, and C be matrices of size. ( B ) C j \u2026 Theorem 2 matrix multiplication is associative. interesting... A+B = B +A ( matrix addition and Scalar multiplication distributes over matrix addition is commutative. sets intersection. Said: then you have made A mistake somewhere mistake somewhere Course Algebra & Study! 2 Hence, associative law: A ( BC ): proof Let j. Proof Step By Step and Clearly for A, B, and zero matrices?. Ab will be an M P matrix \u00d7 2 matrices. diagonal matrices, identity matrices identity! 1.4.1 ( matrix addition is commutative., Elementary matrices. standard truth-functional propositional logic, association or...", "date": "2022-12-04 05:39:52", "meta": {"domain": "sheetchula.com", "url": "http://sheetchula.com/does-it-txvns/4b7npa1.php?id=b10bb1-matrix-multiplication-associative-proof", "openwebmath_score": 0.8530898690223694, "openwebmath_perplexity": 1237.8347291259806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9706877717925422, "lm_q2_score": 0.8976952825278492, "lm_q1q2_score": 0.8713818335456346}}
{"url": "https://math.stackexchange.com/questions/2746442/selection-without-replacement", "text": "# Selection without replacement\n\nThis selection without replacement question comes to me about third hand, so I'm not sure of the exact level/setting. It's definitely not my homework problem. I have done a bit of searching on the site but didn't find anything completely analogous. The question is as follows:\n\nWe have a bag with colored marbles, and the marbles also have numbers on them.\n\nFor this question all we need to know are the following:\n\nThere are 40 marbles in total.\n\nThere are 14 marbles with numbers less than or equal to three.\n\nThere are 7 purple marbles.\n\nThere is one purple marble with a number less than or equal to three.\n\nPulling two marbles without replacement, what is the probability the first marble has a number less than or equal to 3 and the second marble is purple?\n\nMy solution is combinatorial simply computing the probability as the number of favorable outcomes divided by the number of possible outcomes.\n\nThe denominator is easy. There are $40$ possibilities for the first draw and $39$ for the second, so $40\\cdot 39=1560$ possible outcomes.\n\nTo compute the number of favorable outcomes there is only way to select a purple marble as the first marble, so there are $1\\cdot 6$ favorable outcomes where the first marble is purple. That leaves $13$ ways to select the first marble if it isn't purple times the $7$ purple marbles that can then be selected on the second draw for $13\\cdot 7 =91$ favorable outcomes. Taken together\n\n$$\\frac{6+91}{1560}=\\frac{97}{1560}$$\n\nfor the probability the first marble has a number less than or equal to 3 and the second marble is purple.\n\nI was wondering is my solution correct?\n\nI would also welcome suggestions for other perhaps more probabilistic approaches to solving the problem.\n\n\u2022 I think you are correct. You are splitting into two cases (1) first marble is purple (i.e. the \"special\" marble that meets both criteria) and (2) first marble is not purple. This is equivalent to using conditional probabilities but for such a simple case there is really no need to use that language. another approach is simply take $14 \\times 7 = 98$ and subtract $1$ for the double-counted case of picking the special marble twice, but for a simple case like this that might take longer to visualize/explain. \u2013\u00a0antkam Apr 20 '18 at 19:53\n\nYou could do it more formally (same logic thought) in term of probability like this:\n\nFirst, split the probability space onto two disjoint spaces: $\\\\$ $P(M_1\\leq3 \\ \\cap \\ M_2 \\ \\mathrm{ purple}) \\\\= P(M_1\\leq3 \\ \\cap \\ M_2 \\ \\mathrm{ purple} \\ \\cap \\ M_1 \\ \\mathrm{ purple}) + P(M_1\\leq3 \\ \\cap \\ M_2 \\ \\mathrm{ purple} \\ \\cap \\ M_1 \\ \\mathrm{ not \\ purple})$\n\nApply Bayes formula:\n\n$P(M_1\\leq3 \\ \\cap \\ M_2 \\ \\mathrm{ purple} \\ \\cap \\ M_1 \\ \\mathrm{ purple}) \\\\ = P( M_2 \\ \\mathrm{ purple} | M_1\\leq3 \\ \\cap \\ M_1 \\ \\mathrm{ purple}) \\ P(M_1\\leq3 \\ \\cap \\ M_1 \\ \\mathrm{ purple}) \\\\ = \\frac{6}{39} \\frac{1}{40}$\n\nApply Bayes formula again:\n\n$P(M_1\\leq3 \\ \\cap \\ M_2 \\ \\mathrm{ purple} \\ \\cap \\ M_1 \\ \\mathrm{not \\ purple}) \\\\ = P( M_2 \\ \\mathrm{ purple} | M_1\\leq3 \\ \\cap \\ M_1 \\ \\mathrm{ not \\ purple}) P(M_1\\leq3 \\ \\cap \\ M_1 \\ \\mathrm{not \\ purple}) \\\\ = \\frac{7}{39}\\frac{13}{40}$\n\nFinally , $P(M_1\\leq3 \\ \\cap \\ M_2 \\ \\mathrm{ purple}) = \\frac{6}{39} \\frac{1}{40} + \\frac{7}{39}\\frac{13}{40} = \\frac{97}{1560}$\n\nSame result as yours (after multiple edits haha)\n\n@antkam comment is great thought if you want to solve it without formula.", "date": "2019-08-21 22:08:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2746442/selection-without-replacement", "openwebmath_score": 0.7983296513557434, "openwebmath_perplexity": 284.81584384501537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471623361619, "lm_q2_score": 0.8856314798554444, "lm_q1q2_score": 0.8713260183313547}}
{"url": "http://mathhelpforum.com/advanced-applied-math/38353-solved-projectiles.html", "text": "1. ## [SOLVED] Projectiles....\n\nA stone is projected from a point O on a cliff with a speed of 20 m/s at an angle of elevation of 30(degrees) . T seconds later the angle of depression of the stone from O is 45(degrees). Find the value of T .\n\ni encountered this problem while solving M2 past exam quetions and i couldnt solve it... ... to make things worse no one in my class knows how to solve this problem . Pls hlp me cuz xam is in a few days. i will be very grateful...\n\n2. I am including a diagram. I hope I am interpreting correctly.\n\nWe can use the horizontal and vertical component formulas.\n\nhorizontal: $x=(20cos(30))t$\n\nvertical: $y=(20sin(30))t-\\frac{1}{2}(9.8)t^{2}$\n\nWhen the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.\n\n$0=20sin(30)t-\\frac{1}{2}(9.8)t^{2}$\n\nSolving for t we find t=2.04 seconds to reach that point.\n\nThat means it's x coordinate is $20cos(30)(2.04)=35.35 \\;\\ m$\n\nThe stone is at (35.35,0) at 2.04 seconds.\n\nNow, we have a triangle with angle 45 and side 35.35\n\nSo, the y component is the same, y=-35.35 feet\n\nPlug this into the y component and solve for t.\n\n$-35.35=20sin(30)t-\\frac{1}{2}(9.8)t^{2}$\n\nAnd we find t=3.89 seconds\n\nPerhaps TopsQuark will check me out.\n\n3. Hello, PFX!\n\nA stone is projected from a point $O$ on a cliff with a speed of 20 m/s\nat an angle of elevation of 30\u00b0.\nt seconds later the angle of depression of the stone from $O$ is 45\u00b0.\nFind the value of $t$.\n\nThe parametric equations for the position of the stone (relative to $O$) are:\n\n. . $\\begin{array}{ccccccc}x &=& (20\\cos30^o)t & \\Rightarrow & x &=& 10\\sqrt{3} t \\\\\ny &=& (20\\sin30^o)t - 4.9t^2 & \\Rightarrow & y &=& 10t - 4.9t^2 \\end{array}$\n\nConsider the position of the stone at time $t.$\nCode:\n         *\n*   *\nO* - - - +A\n\\    *:\n\\   :\n\\ :\n*B\n\nSince $\\angle AOB = 45^o,\\:\\Delta OAB$ is an isosceles right triangle.\n. . Hence: . $OA = AB$\n\nThis means: . $x \\,=\\,-y\\quad\\Rightarrow\\quad (10\\sqrt{3})t \\;=\\;-\\left[10t - 4.9t^2\\right]$\n\n. . which simplfies to: . $4.9t^2 - 10(1+\\sqrt{3})t \\:=\\:0$\n\n. . which factors: . $t\\left[4.9t - 10(1+\\sqrt{3})\\right] \\:=\\:0$\n\n. . and has roots: . $t = 0,\\;\\frac{10(1+\\sqrt{3})}{4.9}$\n\nTherefore: . $t \\:\\approx\\:5.6$ seconds.\n\n4. My thinking was on line with yours, Soroban. Why is my answer different?.\n\nI don't see anything wring with my logic. The range of the stone is $\\frac{400sin(60)}{9.8}=35.33$.\n\n5. hi, thanx for ur replies....\n\nwell the correct answer given is T = 5.59 (to 3 s.f.)\n\n@ galactus i cant find where u went wrong....\n\n6. Originally Posted by galactus\nI am including a diagram. I hope I am interpreting correctly.\n\nWe can use the horizontal and vertical component formulas.\n\nhorizontal: $x=(20cos(30))t$\n\nvertical: $y=(20sin(30))t-\\frac{1}{2}(9.8)t^{2}$\n\nWhen the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.\n\n$0=20sin(30)t-\\frac{1}{2}(9.8)t^{2}$\n\nSolving for t we find t=2.04 seconds to reach that point.\n\nThat means it's x coordinate is $20cos(30)(2.04)=35.35 \\;\\ m$\n\nThe stone is at (35.35,0) at 2.04 seconds.\n\nNow, we have a triangle with angle 45 and side 35.35\n\nSo, the y component is the same, y=-35.35 feet\n\nPlug this into the y component and solve for t.\n\n$-35.35=20sin(30)t-\\frac{1}{2}(9.8)t^{2}$\n\nAnd we find t=3.89 seconds\n\nPerhaps TopsQuark will check me out.\nTake a look at your diagram. When y = 0 you have x at 35.35 ft. How can it be that when it drops to -45 degrees? x is always increasing.\n\n-Dan", "date": "2013-12-07 06:45:53", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-applied-math/38353-solved-projectiles.html", "openwebmath_score": 0.8009311556816101, "openwebmath_perplexity": 1194.2662094034088, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471675459396, "lm_q2_score": 0.8856314617436728, "lm_q1q2_score": 0.8713260051260826}}
{"url": "https://math.stackexchange.com/questions/2240281/when-do-taylor-series-converge-quickly", "text": "When do Taylor series converge quickly?\n\nThe function $e^x$ has a Maclaurin series $$e^x = \\sum_{n=0}^{\\infty} \\frac{x^n}{n!}.$$\n\nIt also has various Taylor series expansions centered at $x$-values other than zero, for example $x=3$: $$e^x = \\sum_{n=0}^{\\infty} \\frac{e^3}{n!} (x-3)^n.$$\n\nIf we approximate $e^{1/2}$ using the first series above, our approximation converges to the exact value very quickly (|error| < 0.001 after 4 terms). If we use the second series, it converges more slowly (|error| < 0.001 after 13 terms).\n\nI suspect this happens because $x=\\frac{1}{2}$ is closer to zero than it is to 3. Do Taylor series always converge more quickly near the center of convergence? If not, is there a general result that tells us when this happens?\n\n\u2022 To approximate $e^{\\frac{1}{2}}$ using $e^3$ looks just weird \u2013\u00a0Vincent Apr 18 '17 at 15:06\n\u2022 Yeah, maybe not the most practical example. I'm more interested in the general principle. \u2013\u00a0nardol5 Apr 18 '17 at 15:07\n\u2022 The Taylor series have a bound math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/\u2026 which, as you can see, converges more quickly when you are close to the center (strictly less than 1). \u2013\u00a0Vincent Apr 18 '17 at 15:07\n\u2022 The error in the approximation is given by the tail of the series. If you pick the centre of your Taylor series to be close to the point you're approximating, then $(x-a)$ is small, and $(x-a)^{n}$ will be even smaller. This can give you effective bounds on the error. \u2013\u00a0preferred_anon Apr 18 '17 at 15:08\n\u2022 Lagrange remainder and alternating series remainder. Use those and it should be clear. \u2013\u00a0Simply Beautiful Art Apr 18 '17 at 15:30\n\nTaylor series is: $$f(x)=f(x_0)+(x-x_0)f^{'}(x_0)+\\frac{(x-x_0)^2}{2!}f^{''}(x_0)+....+\\frac{(x-x_0)^n}{n!}f^{(n)}(x_0)+R_n$$\n\nwhere $R_n$ is Lagrange Remainder,\n\n$$R_n=\\frac{(x-x_0)^{n+1}}{(n+10)!}f^{(n)}(x^*)$$\n\n( here $x^*$ is value between $x_0$ and $x$ )\n\nSo, the rate of convergence is depending upon $(x-x_0)$ with respect to $x$, hence this difference will decide the convergence .\n\nFor example,\n\nTo calculate $e^\\frac{1}{2}$, series: $e^x = \\sum_{n=0}^{\\infty} \\frac{x^n}{n!}.$ is better and\n\nTo caculate $e^\\frac{5}{2}$, series: $e^x = \\sum_{n=0}^{\\infty} \\frac{e^3}{n!} (x-3)^n.$ is better\n\n( see differences between $x_0$ and $x$ to understand above statement)\n\nRefer: Taylor Series", "date": "2019-06-24 11:49:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2240281/when-do-taylor-series-converge-quickly", "openwebmath_score": 0.9250181317329407, "openwebmath_perplexity": 211.12523710550946, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471651778591, "lm_q2_score": 0.8856314632529872, "lm_q1q2_score": 0.8713260045137707}}
{"url": "https://math.stackexchange.com/questions/1274300/minimal-collection-of-subsets-to-reconstruct-singletons", "text": "# Minimal collection of subsets to reconstruct singletons\n\nI have come across the following problem in a technical application. For a given integer $n$, what is the minimal collection of subsets of $\\{1,\\dots,n\\}$ such that all \"singleton\" sets $\\{1\\}, \\{2\\}, \\dots, \\{n\\}$ can be \"reconstructed\" by set operations (intersection, union, set difference, complement) on those subsets?\n\nFor example, with $n = 5$ a possible collection is $S_1 = \\{1,2,3\\}, S_2 = \\{2,3,4\\}, S_3 = \\{3,4,5\\}$, because we can write $$\\{1\\} = S_1 \\setminus S_2 \\\\ \\{2\\} = S_2 \\setminus S_3 \\\\ \\{3\\} = S_1 \\cap S_2 \\cap S_3 \\\\ \\{4\\} = S_2 \\setminus S_1 \\\\ \\{5\\} = S_3 \\setminus S_2$$\n\nIn believe that in general, the sets $S_k = \\{k,k+1,\\dots,k+c-1\\}$, $1\\le k \\le n-c+1$ and $c = \\lceil n / 2 \\rceil$ are sufficient, but I am not convinced they are minimal.\n\nI am sure this problem has been studied in some context but I am not a professional mathematician and I don't even know where to look ... Any help appreciated!\n\n\u2022 Any \"elementary\" set operation would be allowed, I edited to clarify this. \u2013\u00a0Roland May 9 '15 at 14:57\n\u2022 To clarify, by \"complement\", do you mean the operation taking a set $S$ to $\\{1,\\ldots,n\\}\\setminus S$? \u2013\u00a0Tad May 10 '15 at 0:50\n\u2022 Isn't it just $\\lceil\\log_2n\\rceil$? I.e. working with subsets of $\\{0,\\dots,n-1\\},$ let $S_k$ be the set of numbers whose $k$-th binary digit (from the right) is a $1$? \u2013\u00a0bof May 10 '15 at 1:14\n\u2022 Ah, of course! It is indeed logarithmic in n. I did not see this -- then the problem is quite easy. Thanks! \u2013\u00a0Roland May 10 '15 at 7:42\n\nIt turns out you can get by with only $\\lceil \\lg n\\rceil$ sets, where $\\lg$ as usual denotes logarithm base $2$. I'll give a construction and prove that it is optimal.\n\nNote that you can construct all singletons with set operations if and only if you can construct all subsets of $[n]=\\{1,\\ldots,n\\}$; it can be useful to think of the latter as an equivalent goal.\n\nFor each $i\\ge 0$, let $S_i$ be the subset of $[n-1]$ whose binary expansions contain a $1$ in the $i$-th position. (We don't need to include $n$ in any of our sets; it's unnecessary, since the union of the $S_i$'s is $[n-1]$, so we can get the singleton $\\{n\\}$ by complementing.) So $S_i$ is nonempty precisely when $n-1\\ge 2^i$, i.e. $i<\\lg n$. Thus the nonempty $S_i$'s are indexed from $0,\\ldots \\lceil \\lg n\\rceil-1$.\n\nIt's easy to see how to get any singleton $\\{k\\}$ from the $S_i$'s: look at the binary expansion of $k$; if the $i$-th bit of $k$ is $1$ then $k\\in S_i$, otherwise $k\\in \\bar{S_i}$. Intersecting the appropriate sets for each $i$ leaves the singleton $k$.\n\nThe fact that this construction is optimal follows immediately from the fact that the free Boolean algebra generated by $m$ elements contains $2^{2^m}$ elements.\n\nIn a little more detail: the collection of all subsets of a given set forms a so-called Boolean algebra, under the correspondence $\\wedge\\leftrightarrow\\cup$, $\\vee\\leftrightarrow\\cap$, and $\\neg\\leftrightarrow\\textrm{complement}$. Roughly, the free Boolean algebra $FA(m)$ on $m$ generators is the \"largest\" Boolean algebra that can be generated from the generators; the elements of $FA(m)$ are all of the non-equivalent Boolean expressions which can be formed from the given $m$ elements. It is a standard result that $|FA(m)|=2^{2^m}$. Any Boolean algebra $\\mathcal{A}$ which can be generated by $m$ elements is a quotient of $FA(m)$, so in particular $\\mathcal{A}$ can't have any more than $2^{2^m}$ elements.\n\nTo sum up, if $m$ subsets of $[n]$ can generate all subsets via set operations, then we must have $2^{2^m}\\ge 2^n$, i.e. $m\\ge \\lceil\\lg n\\rceil$. The given construction achieves this bound.\n\nYou can turn this into a nice card trick. Have someone think of a card, deal the deck into two piles, ask them which pile the card is in, pick up the cards and repeat 5 more times. Then you can guess their card. The point is that, at the $i$-th stage, the two piles correspond to the sets $S_i$ and $\\bar{S_i}$.\n\n\u2022 Very nice! This logarithmic behavior will be very important for me. In retrospect it now seems obvious, but doesn't it always :) \u2013\u00a0Roland May 10 '15 at 7:44\n\nWe may use Sperner's Theorem to show the bound $${n}\\choose{\\lfloor n/2\\rfloor}$$\n\n\u2022 Thanks! The related post on mathoverflow was very helpful. \u2013\u00a0Roland May 10 '15 at 7:45", "date": "2019-12-10 08:54:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1274300/minimal-collection-of-subsets-to-reconstruct-singletons", "openwebmath_score": 0.9719916582107544, "openwebmath_perplexity": 169.5792609937149, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.992988206278523, "lm_q2_score": 0.8774767874818409, "lm_q1q2_score": 0.8713241012526339}}
{"url": "https://math.stackexchange.com/questions/1654261/power-series-of-frac1x1-x2", "text": "# Power series of $\\frac{1+x}{(1-x)^2}$\n\nThis question is continuing from the previous question here:\n\nPower Series representation of $\\frac{1+x}{(1-x)^2}$\n\nI am trying to calculate the power series representation of the equation:\n\n\\begin{align} f(x) = \\frac{1+x}{(1-x)^2} \\end{align}\n\nMy workout is as follow:\n\n\\begin{align} \\frac{1+x}{(1-x)^2} = \\frac{1}{(1-x)^2} + \\frac{x}{(1-x)^2} \\end{align}\n\nFor $\\frac{1}{(1-x)^2}$: \\begin{align} \\frac{1}{(1-x)^2} &= \\frac{d}{dx} \\frac{1}{1-x}\\\\ &= \\frac{d}{dx} \\sum_{n=0}^{\\infty} x^n \\\\ &= \\sum_{n=1}^{\\infty} nx^{n-1} \\\\ &= \\sum_{n=0}^{\\infty} (n+1)x^n \\end{align}\n\nFor$\\frac{x}{(1-x)^2}$: \\begin{align} x \\frac{1}{(1-x)^2} &= x \\sum_{n=0}^{\\infty}(n+1)x^n \\\\ &= \\sum_{n=0}^{\\infty} (n+1) x^{n+1} \\end{align}\n\nTherefore, \\begin{align}\\frac{1+x}{(1-x)^2} = \\sum_{n=0}^{\\infty} (n+1)x^n+\\sum_{n=0}^{\\infty} (n+1) x^{n+1} = \\sum_{n=0}^{\\infty}(n+1)(x^n + x^{n+1}),\\end{align} where range of convergence is $x\\in[-1,1)$. When $x=-1$, $(x^n + x^{n+1})$ becomes $0$, and $(\\infty)(0) = 0$.\n\nHowever, the model answer is $\\sum_{n=0}^{\\infty} (2n+1) x^n$, where range of convergence is $x\\in (-1,1)$.\n\nI do not understand what is wrong with my calculation. Any advice will be appreciated!\n\n\u2022 What is incomplete in your approach is that $\\sum\\limits_{n=0}^{\\infty}(n+1)(x^n + x^{n+1})$ is not the canonical form of a power series, which should rather be expressed as $\\sum\\limits_{n=0}^{\\infty}a_nx^n$ for some suitable sequence $(a_n)$.\n\u2013\u00a0Did\nFeb 14, 2016 at 10:48\n\u2022 What should I do? In addition, term by term, the terms in my solution and the model answer are different. Feb 14, 2016 at 10:50\n\u2022 Was my solution wrong? The first term in my solution is $1+x$ but the first term in the model answer is $1$. Feb 14, 2016 at 10:52\n\u2022 Why does it seems that $f(x)\\neq \\sum_{n=0}^{\\infty}(n+1)(x^n + x^{n+1})$? You conclude this by looking at the first term of the sum. In your answer that is $1+x$, in the given solution it is $1$, so it can't tbe the case that $f(x) = \\sum_{n=0}^{\\infty}(n+1)(x^n + x^{n+1})$, right? Wrong, because it can very well be that the \"sums\" (i.e. the series) are equal even though the first term isn't. The same way $1+2+3=2+1+3$ even though $1\\neq 2$. Feb 14, 2016 at 11:06\n\u2022 for $x=-1$, what you are doing is addition of two oscillating series which turns out to be zero. It is exact same thing like addition of $-1+1-1+1-1+1-1+...$ and $1-1+1-1+1-1+1-...$ becomes zero. Feb 14, 2016 at 11:07\n\nYou did not end up with power series yet: \\begin{align*} \\sum_{n=0}^{\\infty} (n+1)x^n+\\sum_{n=0}^{\\infty} (n+1) x^{n+1} &=\\sum_{n=0}^{\\infty} (n+1)x^n+\\sum_{n=1}^{\\infty} n x^{n}\\\\ &=\\sum_{n=1}^{\\infty} (2n+1) x^{n} + 1\\\\ &=\\sum_{n=0}^{\\infty} (2n+1) x^{n}\\\\ \\end{align*}\n\nFrom here you can derive correct radius of convergence.\n\nHint: In order to find the power series expansion around $x=0$ you could also use the binomial series representation\n\n\u2022 In (1) we use the identity $\\binom{-n}{k}=\\binom{n+k-1}{k}(-1)^k$", "date": "2022-08-14 15:19:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1654261/power-series-of-frac1x1-x2", "openwebmath_score": 0.9994896054267883, "openwebmath_perplexity": 510.9299508526779, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022627413796, "lm_q2_score": 0.8918110511888302, "lm_q1q2_score": 0.8713195763891701}}
{"url": "https://mathhelpboards.com/threads/expected-value-of-steps.25057/", "text": "# [SOLVED]Expected value of steps\n\n#### MATHias\n\n##### New member\nHello,\n\nI would like to ask, if I am right with my computation.\n\nLet\u00b4s have a set of integers from 0 to 12. We start at 0 and we can go to 1 with the probability 1. From 1 we can go to 1 or back to 0, both with probability 0.5. When we start at zero, how many steps (exp. number of them) do we need to go until 12?\n\nMy way of solving:\n\nWe start at 0 ... there is only one possibity - to go to the 1 with probability 1\nFrom 1, we can go to the 0 (P=$\\frac{1}{2}$) or to the 2 (P=$\\frac{1}{2}$)\nFrom 2, we can go back to the 1(P=$\\frac{1}{2}$), or forwards to the 3(P=$\\frac{1}{2}$) ... but when we start at 0, there is the probability to reach 3(P=${\\frac{1}{2}}^{2}$).\netc.\n\nSo, I created a binomial distribution:\nFrom 0 to 1: 1 step; prob. 1 ... 1*1\nFrom 0 to 2: 2 steps; prob. ${12 \\choose 1}*\\frac{1}{2}^{1}*{\\frac{1}{2}}^{11}$\nFrom 0 to 3: 3 steps; prob. ${12 \\choose 2}*{\\frac{1}{2}}^{2}*{\\frac{1}{2}}^{10}$\n...\nFrom 0 to 11: 11 steps; prob. ${12 \\choose 10}*{\\frac{1}{2}}^{10}*{\\frac{1}{2}}^{2}$\nFrom 0 to 12: 12 steps; prob. ${12 \\choose 11}*{\\frac{1}{2}}^{11}*{\\frac{1}{2}}^{1}$\n\nAnd now, to get the whole expected value, I need to sum all the expected values:\n$1*1+12*2(steps)*\\frac{1}{2048}+66*3\\frac{1}{2048}+220*4\\frac{1}{2048}+495*5\\frac{1}{2048}+792*6*\\frac{1}{2048}+924*7*\\frac{1}{2048}+792*8*\\frac{1}{2048}+495*9*\\frac{1}{2048}+220*10*\\frac{1}{2048}+66*11*\\frac{1}{2048}+12*12*\\frac{1}{2048}$.\n\nThe sum is 15353/1024 = 14.9932 steps. I think it is too low number. Can I ask, where I am not doing right?\n\nThank you a lot.\n\nHave a nice day.\n\nLast edited:\n\n#### Country Boy\n\n##### Well-known member\nMHB Math Helper\nYou could, of course, go from 0 directly to 12 in exactly 12 steps! The probability of that is $$(1/2)^{11}$$. Or, you could, exactly once, go a step back. In that case, ignoring the first step from 0 to 1, you would have 13 steps, \"x\" steps up to that point, then one step back, then one step to get back to that point, then the remaining 13- x steps for a total x+ 2+ 13- x= 13 steps. The probability of that is $$(1/2)^{13}$$. Now do the same if we go back twice: either at two different steps or twice at the same step. Each step backwards adds 2 steps to the total so going back twice gives a total of 15 steps. And those steps have probability 1/2 so the probability is $$(1/2)^{15}$$.\n\nIn general, there can be any even number of steps or, ignoring the first step, any odd number of steps. For 2n+ 1 steps The probability is $$(1/2)^{2n+1}$$. The expected value is given by the infinite sum $$\\sum_{n= 0}^\\infty(2n+1)(1/2)^{2n+1}$$. To sum that factor out a 1/2 so it becomes $$\\frac{1}{2}\\sum_{n=0}^\\infty (2n+1)(1/2)^{2n}$$. Notice that $$(2n+1)x^{2n}$$ is the derivative of $$x^{2n+1}$$ so we can do $$\\sum_{n=0}^\\inf (2n+1)x^{2n}$$ as the derivative of $$\\sum_{n=0}^\\infty x^{2n+ 1}= x\\sum_{n=0}^\\infty (x^2)^n$$. That last sum is a geometric sum which has a simple closed form: $$x\\frac{1}{1- x^2}$$ as long as |x|< 1. Here x= 1/2 so that formula applies and the derivative is $$\\left(\\frac{x}{1- x^2}\\right)'= \\frac{(1- x^2)- (-2x^2)}{1- x^2)^2}= \\frac{1+ x^2}{(1- x^2)^2}$$. For x= 1/2 that is $$\\frac{1+ 1/4}{(1- 1/4)^2}= \\frac{\\frac{3}{4}}{\\frac{9}{16}}= \\frac{3}{4}\\frac{16}{9}= \\frac{4}{3}$$.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nThis is an example of an absorbing Markov chain. By calculating the fundamental matrix for this problem, I found that in order to reach level 12 you have to spend an average of 2 steps at level 11, an average of 4 steps at level 10, 6 steps at level 9 and so on (two extra steps for each level as you go down) till you get to 20 steps at level 2, 22 steps at level 1 and 12 steps at level 0 (12 rather than 24 because from level 0 you have to go back to level 1). Therefore the expected total number of steps to reach level 12 is $2+4+6+8 + \\ldots + 22 + 12 = 144$ (considerably more than 14.9932!).\n\nLast edited:\n\n#### MATHias\n\n##### New member\nYeah, I see. I had expected more complicated theory in that. Sometimes, it is just easier to draw a graphics to see that. Thanks a lot to both of you.", "date": "2020-07-09 02:48:09", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/expected-value-of-steps.25057/", "openwebmath_score": 0.8391358256340027, "openwebmath_perplexity": 415.3218077236842, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9873750510899383, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8712872505649187}}
{"url": "https://math.stackexchange.com/questions/2161032/showing-that-int-frac-dx-sqrtx2a2-ln-biggl-x-sqrtx2a2/2161045", "text": "# Showing that $\\int \\frac {dx}{\\sqrt{x^2+a^2}} \\;=\\; \\ln\\,\\biggl| x+\\sqrt{x^2+a^2}\\,\\biggr| +C$\n\nI want to show that $$\\int \\frac {dx}{\\sqrt{x^2+a^2}} \\;=\\; \\ln\\,\\biggl| x+\\sqrt{x^2+a^2}\\,\\biggr| +C$$ and get a slightly incorrect result and I wonder what I am doing wrong.\n\nI let $x = a\\tan\\theta$ so $dx=a\\sec^2\\theta\\, d\\theta$ such that I perform the following operations, $$\\frac aa \\int \\frac {\\sec^2\\theta}{\\sqrt{\\tan^2\\theta + 1}}d\\theta\\;=\\; \\int \\sec\\theta\\,d\\theta$$ Multiplying the integrand by $1 = \\frac {\\sec\\theta+\\tan\\theta}{\\sec\\theta+\\tan\\theta}$ I get $$\\int \\sec\\theta\\, d\\theta \\;=\\; \\int \\frac {\\sec^2\\theta+\\sec\\theta\\tan\\theta}{\\sec\\theta+\\tan\\theta} d\\theta$$ Then I let $u=\\sec\\theta+\\tan\\theta$ and $du = \\sec^2\\theta+\\sec\\theta\\tan\\theta$ such that I get a $\\int \\frac 1u du$ integral that yields $$\\int \\sec\\theta \\, d\\theta \\;=\\;\\ln|\\sec\\theta+\\tan\\theta|+C$$ Getting back in $x$ variable terms I get, $$\\int \\frac {dx}{\\sqrt{x^2+a^2}} \\;=\\; \\ln\\,\\biggl| \\frac 1a \\left(x+\\sqrt{x^2+a^2}\\right)\\biggr| +C$$ It seems to me that this $\\frac 1a$ has to be there according to the substitution $x=a\\tan\\theta$, the adjacent side to the angle $\\theta$ is of length $a$. Then, when substituting back to $x$ I get $\\sec\\theta= \\frac {x^2+a^2}a$ and $\\tan\\theta = \\frac xa$. What am I missing?\n\n\u2022 note that $$\\ln\\left(\\frac{1}{a}\\right)$$ is a const. Feb 25, 2017 at 16:22\n\u2022 @Dr.SonnhardGraubner So it could be \"absorbed\" into the \"$+C$\" constant term is what you mean?\n\u2013\u00a0user409521\nFeb 25, 2017 at 16:24\n\u2022 yes it can absorbed into the const. C, a nice word 'absorbed' Feb 25, 2017 at 16:25\n\u2022 By the way, a much easier way of verifying the identity is to just compute $\\frac{d}{dx} \\ln|x+\\sqrt{x^2+a^2}|$. Feb 25, 2017 at 17:15\n\u2022 The definition of $F(x) = \\int f(x) \\, dx$ is that you \u201cdifferentiate backwards\u201d, in other words that $F'(x)=f(x)$. Feb 25, 2017 at 21:56\n\nSince $\\log{xy} = \\log{x}+\\log{y}$, you can remove the $1/a$ from the logarithm and incorporate it into the constant.\n$$\\ln \\frac{(x+\\sqrt{x^2+a^2})}{a} + C$$\n$$= \\ln (x+\\sqrt{x^2+a^2}) - \\ln a + C$$\n$$= \\ln (x+\\sqrt{x^2+a^2}) + C'$$", "date": "2022-05-17 12:06:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2161032/showing-that-int-frac-dx-sqrtx2a2-ln-biggl-x-sqrtx2a2/2161045", "openwebmath_score": 0.8354972004890442, "openwebmath_perplexity": 222.78322231475096, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013790564298, "lm_q2_score": 0.8872045981907007, "lm_q1q2_score": 0.871247150475325}}
{"url": "https://nbviewer.jupyter.org/github/chrishavlin/miscellaneous_python/blob/master/notebooks/NanMeans.ipynb", "text": "## Taking means of arrays with missing values\u00b6\n\nSometimes you do a thing that leaves you scratching your head for a few minutes until you realize \"Yes, of course that's how it works.\"\n\nToday at the (virtual) SciPy 2020 xarray tutorial, one of the exercises involved taking the mean of a 2D array that had some missing values. For some reason, my initial inclination was to take the mean of the 2D array over the first dimension and then taking the mean of the resulting 1D array. But that actually gave a different result from taking the mean over the entire array at once!\n\nThe reason in the end is a very simple answer: because the data set has missing values that get skipped, calculating the means successively effectively weights measurements differently.\n\nBut I thought it was a fun example of a seemingly innocuous mistake that could lead to large errors, particularly when analyzing large datasets. So let's break it down!\n\nTo start, let's import numpy and build a random 4 by 4 array:\n\nIn\u00a0[1]:\nimport numpy as np\nb=np.random.rand(4,4)\n\n\nTo find the mean, we can just use b.mean():\n\nIn\u00a0[2]:\nb.mean()\n\nOut[2]:\n0.5961189081171596\n\nbut it's also true that if take the means of just the column and then take a mean of the result, you'll get the same answer. When you take the mean along an axis, it will reduce the dimensionality by 1, so we'll get a 1D array here:\n\nIn\u00a0[3]:\nb.mean(axis=0)\n\nOut[3]:\narray([0.67246434, 0.68927864, 0.7035238 , 0.31920885])\n\nAnd if we take the mean of that result, we'll get the same answer as before (with a tiiiiiiny bit of error in the final digit):\n\nIn\u00a0[4]:\nb.mean(axis=0).mean()\n\nOut[4]:\n0.5961189081171595\n\nAnd we get the same answer if we take the mean along the other axis first:\n\nIn\u00a0[5]:\nprint(b.mean(axis=1).mean())\n\n0.5961189081171596\n\n\nBut what happens when we have some missing values? Let's put some in!\n\nIn\u00a0[6]:\nb[2,3]=np.nan\nb[2,1]=np.nan\nb[3,3]=np.nan\n\n\nSince we have nan values, we can use np.nanmean:\n\nIn\u00a0[7]:\nnp.nanmean(b)\n\nOut[7]:\n0.6060431635413454\n\nOk, and when we take the mean of the mean?\n\nIn\u00a0[8]:\nnp.nanmean(b,axis=0).mean()\n\nOut[8]:\n0.5516577241118834\nIn\u00a0[9]:\nnp.nanmean(b,axis=1).mean()\n\nOut[9]:\n0.6079093503218946\n\nWhoa! Very different results! What's going on here?\n\nLet's think a minute about what a mean is. Given a vector, $B$, with $N$ elements, we add up the elements and divide by $N$:\n\n$\\frac{1}{N}\\sum_i^N(B_i)$\n\nBut when we have nan values, $N$ changes when we take the mean over axes successively!\n\nLet's take a look at b:\n\nIn\u00a0[10]:\nb\n\nOut[10]:\narray([[0.80532163, 0.36377322, 0.72169393, 0.13140395],\n[0.94655188, 0.91627271, 0.64348503, 0.10323544],\n[0.39911748,        nan, 0.74945358,        nan],\n[0.53886636, 0.85992325, 0.69946265,        nan]])\n\nSo when we take the mean over the rows:\n\nIn\u00a0[11]:\nnp.nanmean(b,axis=0)\n\nOut[11]:\narray([0.67246434, 0.71332306, 0.7035238 , 0.1173197 ])\n\nThe first column uses N=4, the second column uses N=3, which we can confirm:\n\nIn\u00a0[12]:\nb[:,0].sum()/4\n\nOut[12]:\n0.6724643365974177\nIn\u00a0[13]:\ncol2=b[:,1]\ncol2[~np.isnan(col2)].sum()/3\n\nOut[13]:\n0.7133230623983705\n\nThis means the values in the second column have a larger weight than the values in the frist column! Which is not what we want. There may be situations in which you want to do that, but most cases you just want the overall mean.\n\nWhat we could do is sum over the axes successively but divide by the number of total not-nan values in the original array:\n\nIn\u00a0[14]:\nNnotNan=(1-np.isnan(b)).sum()\nNnotNan\n\nOut[14]:\n13\nIn\u00a0[15]:\nnp.nansum(b) / NnotNan\n\nOut[15]:\n0.6060431635413454\nIn\u00a0[16]:\nnp.nansum(b,axis=0).sum() / NnotNan\n\nOut[16]:\n0.6060431635413455\nIn\u00a0[17]:\nnp.nansum(b,axis=1).sum() / NnotNan\n\nOut[17]:\n0.6060431635413455\n\nAnd we see these answers agree once again!\n\nBut, just use\n\nIn\u00a0[18]:\nnp.nanmean(b)\n\nOut[18]:\n0.6060431635413454\n\nand remember: means of means won't give you the mean when you have missing values.\n\nIn\u00a0[\u00a0]:", "date": "2020-08-13 00:03:46", "meta": {"domain": "jupyter.org", "url": "https://nbviewer.jupyter.org/github/chrishavlin/miscellaneous_python/blob/master/notebooks/NanMeans.ipynb", "openwebmath_score": 0.6689061522483826, "openwebmath_perplexity": 520.1807906547987, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137884587393, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8712471500716404}}
{"url": "http://mathhelpforum.com/pre-calculus/166668-asymptotes-hyperbolas.html", "text": "1. ## Asymptotes for hyperbolas\n\nFind the asymptotes for the hyperbola $\\displaystyle \\frac{2x^2}{3} - \\frac{y^2}{2}=1$\n\nI got the following:\n$\\displaystyle y=\\frac{\\sqrt{6}}{3}$ and $\\displaystyle y=\\frac{-\\sqrt{6}}{3}$\n\nHowever the answer is $\\displaystyle y=\\pm \\frac{4}{\\sqrt{3}}$\n\nWhere did I go wrong? Please, any help would be appreciated!\n\n2. Originally Posted by Joker37\nFind the asymptotes for the hyperbola $\\displaystyle \\frac{2x^2}{3} - \\frac{y^2}{2}=1$\n\nI got the following:\n$\\displaystyle y=\\frac{\\sqrt{6}}{3}$ and $\\displaystyle y=\\frac{-\\sqrt{6}}{3}$\n\nHowever the answer is $\\displaystyle y=\\pm \\frac{4}{\\sqrt{3}}$\n\nWhere did I go wrong? Please, any help would be appreciated!\n$\\displaystyle \\displaystyle\\frac{2x^2}{3}-\\frac{y^2}{2}=1\\Rightarrow\\ y^2=\\frac{4x^2}{3}-2\\Rightarrow\\ y=\\pm\\frac{\\sqrt{4x^2-6}}{\\sqrt{3}}$\n\n$\\displaystyle \\Rightarrow\\ y=\\displaystyle\\pm\\frac{x\\sqrt{4-\\frac{6}{x^2}}}{\\sqrt{3}}$\n\nand as $\\displaystyle x\\rightarrow\\infty$, this is $\\displaystyle \\displaystyle\\ y=\\pm\\frac{2x}{\\sqrt{3}}$\n\n3. From Archie's post to know where the function is undefined solve for $\\displaystyle \\displaystyle 4x^2-6<0 \\implies -\\frac{\\sqrt{3}}{\\sqrt{2}}<x<\\frac{\\sqrt{3}}{\\sqrt{ 2}}$\n\n4. Do you understand that it is impossible to say \"where you went wrong\" if you don't show what you did?\n\nIn any case, a hyperbola having the coordinate axes as axes of symmetry (any hyperbola of the form $\\displaystyle \\frac{x^2}{a^2}- \\frac{y^2}{b^2}= 1$ or $\\displaystyle \\frac{y^2}{b^2}- \\frac{x^2}{a^2}= 1$) has slant asymptotes, not vertical and horizontal asymptotes.\n\nNo, the anwer is not $\\displaystyle y= \\pm\\frac{4}{\\sqrt{3}}$. Neither $\\displaystyle y= \\pm\\frac{\\sqrt{6}}{3}$ nor $\\displaystyle y= \\pm\\frac{4}{\\sqrt{3}}$ is correct.\n\nHere is one way to think about it. The graph of the hyperbola gets closer and closer to the asymptotes as x and y get larger and larger. For very, very large x and y, both $\\displaystyle \\frac{2x^2}{3}$ and $\\displaystyle \\frac{y^2}{2}$ are very, very large so that \"1\" can be ignored compared to them. That is, for very, very large x and y, $\\displaystyle \\frac{2x^2}{3}- \\frac{y^2}{2}= 0$, approximately. The left side is a \"difference of squares\" and factors as $\\displaystyle \\left(\\frac{x\\sqrt{2}}{\\sqrt{3}}+ \\frac{y}{\\sqrt{2}}\\right)\\left(\\frac{x\\sqrt{2}}{\\s qrt{3}}- \\frac{y}{\\sqrt{2}}\\right)= 0$.\n\nSo we must have $\\displaystyle \\frac{x\\sqrt{2}}{\\sqrt{3}}+ \\frac{y}{\\sqrt{2}}= 0$ and $\\displaystyle \\frac{x\\sqrt{2}}{\\sqrt{3}}+ \\frac{y}{\\sqrt{2}}= 0$, the equation of the two straight line asymptotes. Solving for y, we have $\\displaystyle y= \\pm\\frac{2}{\\sqrt{3}}x$ as Archie Meade says.\n\n5. Originally Posted by Archie Meade\n$\\displaystyle \\displaystyle\\frac{2x^2}{3}-\\frac{y^2}{2}=1\\Rightarrow\\ y^2=\\frac{4x^2}{3}-2\\Rightarrow\\ y=\\pm\\frac{\\sqrt{4x^2-6}}{\\sqrt{3}}$\n\n$\\displaystyle \\Rightarrow\\ y=\\displaystyle\\pm\\frac{x\\sqrt{4-\\frac{6}{x^2}}}{\\sqrt{3}}$\n\nand as $\\displaystyle x\\rightarrow\\infty$, this is $\\displaystyle \\displaystyle\\ y=\\pm\\frac{2x}{\\sqrt{3}}$\nTechnically, the generalised binomial theorem should be used to find this limit.\n\n6. Hello, Joker37!\n\n$\\displaystyle \\text{Find the asymptotes for the hyperbola: }\\:\\dfrac{2x^2}{3} - \\dfrac{y^2}{2}\\:=\\:1$\n\nWe're expected to be familiar with this fact:\n\nGiven the hyperbola: .$\\displaystyle \\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} \\:=\\:1$, the asymptotes are: .$\\displaystyle y \\;=\\;\\pm\\dfrac{b}{a}x$\n\nThe equation is: .$\\displaystyle \\displaystyle \\frac{x^2}{\\frac{3}{2}} - \\frac{y}{2} \\;=\\;1 \\quad\\Rightarrow\\quad \\frac{x^2}{\\left(\\sqrt{\\frac{3}{2}}\\right)^2} - \\frac{y^2}{(\\sqrt{2})^2} \\;=\\;1$\n\nHence: .$\\displaystyle a = \\sqrt{\\frac{3}{2}},\\;b = \\sqrt{2}$\n\nTherefore: .$\\displaystyle y \\;=\\;\\pm\\dfrac{\\sqrt{2}}{\\sqrt{\\frac{3}{2}}}x \\quad\\Rightarrow\\quad y \\;=\\;\\pm\\dfrac{2}{\\sqrt{3}}x$", "date": "2018-03-21 20:34:01", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/166668-asymptotes-hyperbolas.html", "openwebmath_score": 0.9524871706962585, "openwebmath_perplexity": 380.5124986484315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137868795702, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8712471442774635}}
{"url": "https://math.stackexchange.com/questions/918655/difference-in-writing-answer-for-int-dfrac12x3-24x25x2-2xdx", "text": "# Difference in writing answer for $\\int\\dfrac{12x^3-24x^2+5}{x^2-2x}dx$\n\nI have this integral:\n\n$\\int\\dfrac{12x^3-24x^2+5}{x^2-2x}dx$\n\nI solved using partial fractions and got an answer:\n\n$6x^2-\\frac{5}{2}\\ln(x)+\\frac{5}{2}\\ln(x-2) +C$\n\nBut I am using \"My Math Lab\" online homework which is not accepting my answer as written above. I resolved it and got the same answer, and after some frustration, I plugged it into Wolfram Alpha and got this answer:\n\n$6x^2-\\frac{5}{2}\\ln(x)+\\frac{5}{2}\\ln(2-x)+C$\n\nSo the only difference is $\\ln(2-x)$ instead of what I wrote as $\\ln(x-2)$.\n\nWhat's more frustrating is MyMathLab still marked this wrong, and said the answer is actually:\n\n$6x^2+\\frac{5}{2}\\ln\\bigg|\\frac{x-2}{x}\\bigg|+C$\n\nMy first question is, why did Wolfram Alpha's answer say $\\ln(2-x)$ instead of my original calculation that I found to be $\\ln(x-2)$? Why is this switch occurring?\n\nMy second question is, why did MyMathLab mark mine and Wolfram Alpha's answer incorrect? Is this one of the limitations of MyMathLab grading, where two answers are actually the same and correct, but only look different? Or is Wolfram Alpha wrong? Why was the correct answer written the way it was? Where did $\\frac{5}{2}\\ln\\bigg|\\frac{x-2}{x}\\bigg|$ come from?\n\n\u2022 As indicated in the answer, it's preferable to use $\\ln|x|$ and $\\ln|x-2|$ in the answer, to get an antiderivative with the largest possible domain; and then $\\frac{5}{2}(\\ln|x-2|-\\ln |x|)=\\frac{5}{2}\\ln|\\frac{x-2}{x}|$. \u2013\u00a0user84413 Sep 3 '14 at 23:46\n\nYour answer is a valid solution for $x > 2$.\nWolfram's answer is a valid solution for $0<x<2$.\nMyMathLab's answer is a valid solution for $x\\in(-\\infty,0)\\cup(0,2)\\cup(2,\\infty)$ -- but it's not the only possible antiderivative on this domain, because it doesn't take into account that the integration constant is allowed to change at $x=0$ and $x=2$ where the integrand isn't defined.\n\u2022 What math rules allow you to simply just change the contents of $\\ln(x-2)$ to be $\\ln(2-x)$? \u2013\u00a0Sabien Sep 3 '14 at 23:02", "date": "2019-11-19 02:14:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/918655/difference-in-writing-answer-for-int-dfrac12x3-24x25x2-2xdx", "openwebmath_score": 0.828079342842102, "openwebmath_perplexity": 300.5463551099814, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.980280867283954, "lm_q2_score": 0.8887587875995482, "lm_q1q2_score": 0.8712332351143206}}
{"url": "https://math.dartmouth.edu/~m50f17/HW3Sol.html", "text": "## Question-1\n\nIs maximum-likelihood estimator $$\\tilde{\\sigma}^2$$ of $$\\sigma^2$$ an unbiased estimator ? Verify your answer. Comment on the change of the value of $\\mathbb{E}(\\tilde{\\sigma}^2) - {\\sigma}^2$ as $$n$$ goes to infinity.\n\nWe know that\n\n$\\tilde{\\sigma}^2 = \\frac{ \\sum_{i=1}^{n}(y_i - \\tilde\\beta_0 - \\tilde\\beta_1 x )^2}{n}$ and, moreover $$\\tilde\\beta_0=\\hat\\beta_0$$ and $$\\tilde\\beta_1=\\hat\\beta_1$$ since LSE and MLE has same coefficients estimations.\n\nCombining these yield \\begin{aligned} \\mathbb{E} [\\tilde{\\sigma}^2] & = \\frac{ \\mathbb{E} [ \\sum_{i=1}^{n}(y_i - \\tilde\\beta_0 - \\tilde\\beta_1 x )^2 ] }{n} \\\\ &= \\frac{ \\mathbb{E} [ \\sum_{i=1}^{n}(y_i - \\hat\\beta_0 - \\hat\\beta_1 x )^2 ] }{n} \\\\ &= \\frac{ \\mathbb{E} [ SS_{Res} ] }{n} \\\\ &= \\frac{ \\mathbb{E} [ (n-2) \\hat\\sigma^2 ] }{n} \\\\ \\end{aligned} since $$\\hat\\sigma^2 = \\frac{SS_{Res}}{n-2}$$. We can now see\n\n$\\mathbb{E} [\\tilde{\\sigma}^2] = \\frac{ (n-2) }{n}\\mathbb{E} [ \\hat\\sigma^2 ] = \\frac{ (n-2) }{n} \\sigma^2$\n\nsince $$\\hat\\sigma^2$$ is an unbiased estimator of $$\\sigma^2$$. We conclude that $$\\tilde{\\sigma}^2$$ is a biased estimator with\n\n$\\mathbb{E}(\\tilde{\\sigma}^2) - {\\sigma}^2 = \\left(\\frac{ n-2 }{n} -1 \\right) \\sigma^2 = \\frac{ -2 }{n} \\sigma^2$\n\nwhich goes to $$0$$ as $$n \\rightarrow\\infty$$. Therefore bias is negligible for large $$n$$.\n\n## Question-2\n\nConsider the shear strength vs age relation using the propellant data.\n\n1. Recalculate the coefficients of the fitted linear regression model using the vector equations we obtained.\n\n2. Suppose that the expectation of the initial shear strength is known to be 2400. Write the corresponding model (should involve only one parameter $$\\beta_1$$). Calculate 95% CI on $$\\beta_1$$.\n\n# Computation part of the answer :\n\nshearS<-prop$ShearS age<-prop$Age\n\nX <- cbind(1,age)\nbetaHat <- solve( t(X) %*% X ) %*% t(X) %*% shearS\n\n# part (a)\ncat ( \" Part(a)\nFitted Coefficients =  \" , betaHat   , \" \\n\\n\") \n##  Part(a)\n##  Fitted Coefficients =   2627.822 -37.15359\n# part (b)\n# We can use a vertical shift and use the no-intercept model\n# Note that vertical shifting will not change the beta1 and its statistics.\n# Therefore after we shift down by 2400, we simply work on a no-intercept model\n\nshearS_Shifted <- shearS - 2400\n\nbeta1Hat = sum( age * shearS_Shifted) / sum(age^2)\nyHat <- beta1Hat * age\n\n# We use CI for no-intercept model\ndof = 19\n\nMSres = sum((yHat - shearS_Shifted)^2) / dof\nseBeta1 = sqrt(MSres / sum(age^2) )\n\ntQ = qt( .975, dof)\nlowerCI = beta1Hat - tQ * seBeta1\nupperCI = beta1Hat + tQ * seBeta1\n\ncat ( \" Part (b)\nLower and upper bound of CI are :  \" ,  lowerCI , \",  \" , upperCI , \"\n\\n Note that these numbers make sense. Since we forced intercept to be 2400 (which is\nlower than the estimate of the intercept-model) the regression analysis response is\nto raise the other end, and therefore increase the slope. \"\n) \n##  Part (b)\n##  Lower and upper bound of CI are :   -28.64285 ,   -19.63201\n##\n##  Note that these numbers make sense. Since we forced intercept to be 2400 (which is\n##  lower than the estimate of the intercept-model) the regression analysis response is\n##  to raise the other end, and therefore increase the slope.\nfitted = lm(shearS ~ age)\nplot(age, shearS, main=\"Shear Strength vs. Age\", pch=16, xlim=c(0,30), ylim=c(1600,2700))\nabline(fitted$coef, lwd = 2, col = \"blue\") abline( c(2400,beta1Hat), lwd = 2 , col = 'red') ## Example : Phytoplankton Population A scientist is trying to model the relation between phytoplankton population in the city public water supply and concentration of two substances. The sample data is at : https://math.dartmouth.edu/~m50f17/phytoplankton.csv where headers are \u2022 pop : population of phytoplankton ($$y$$) \u2022 subs1 : concentration of substance-1 ($$x_1$$) \u2022 subs2 : concentration of substance-2 ($$x_2$$) Lets consider a guessed model $y = 200 + 10x_1 -15x_2$ Below is the corresponding code to plot the scatter diagram and the above plane. # Note: Run the following in R console if you get errors in plotting or library loading : # install.packages(\"scatterplot3d\") # install.packages(\"plot3D\") library(\"plot3D\") library(\"scatterplot3d\") # Loading data pData <- read.table(\"https://math.dartmouth.edu/~m50f17/phytoplankton.csv\", header=T, sep=\",\") pop <- pData$pop\nsubs1 <- pData$subs1 subs2 <- pData$subs2\n\n# Create a mesh\nmeshP <- mesh( seq(min(subs1),max(subs1),0.03)  , seq(min(subs2),max(subs2),0.03) )\nx1Mesh <- meshP$x x2Mesh <- meshP$y\n\nmyModel <- 200 + 10*x1Mesh  - 15 *x2Mesh\n\n# Below is the code to plot the scatter diagram with red markers and your model\n# You need to set two variables before calling :\nsc1\\$points3d (x2Mesh,x1Mesh, myModel, cex=.02, col=\"blue\")", "date": "2018-01-22 14:24:19", "meta": {"domain": "dartmouth.edu", "url": "https://math.dartmouth.edu/~m50f17/HW3Sol.html", "openwebmath_score": 0.9976863265037537, "openwebmath_perplexity": 5115.965739451322, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808678600415, "lm_q2_score": 0.888758786126321, "lm_q1q2_score": 0.8712332341821469}}
{"url": "https://terryhall.co.uk/8znyyhk3/59aef8-are-diagonals-congruent-in-a-parallelogram", "text": "If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. If the diagonals of a parallelogram are equal, then show that it is a rectangle. What is the WPS button on a wireless router? Yes; all parallelograms have diagonals that bisect each other. Related Videos. A rhombus is a parallelogram in which all sides are congruent. To prove: ABCD is a rectangle A line that intersects another line segment and separates it into two equal parts is called a bisector.. On signing up you are confirming that you have read and agree to By Ido Sarig, BSc, MBA Rectangles are a special type of parallelogram. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. Get an answer to your question \u201cThe diagonals of a parallelogram are congruent. In a quadrangle, the line connecting two opposite corners is called a diagonal. BC = BC Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. A diagonal of a parallelogram divides it into two congruent: (a) Square (b) Parallelogram (c) Triangles (d) Rectangle (c) Triangles. If your impeached can you run for president again? We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram. are parallel. All Rights Reserved. 3) Diagonals are perpendicular. The diagonals of a parallelogram_____bisect the angles of the parallelogram. A diagonal divides a parallelogram into two congruent triangles. SAS stands for \"side, angle, side\". Area of the parallelogram when the diagonals are known: $$\\frac{1}{2} \\times d_{1} \\times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. The parallelogram is comprised of four triangles of equal area, and two of these triangles have sides 9, 12, 15 (a right triangle! 8. a) Draw two line segments, AC and BD, that bisect each other at right angles. Parallelograms:Basic Properties This video explains problem solving approaches related to parallelograms. \u2026 Why don't libraries smell like bookstores? Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. You cannot conclude that the parallelogram that I'm thinking of is a square, though, because that would be too restrictive. Only rectangles (squares included) have congruent diagonals, The diagonals of a parallelogram bisect each other. The consecutive angles of a parallelogram are supplementary. The diagonals of a parallelogram always . A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. SURVEY . The diagonal of a parallelogram is any segment that connects two vertices of a parallelogram \u2026 Prove that a diagonal of a parallelogram divide it into two congruent triangles. Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. are congruent. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. 3. * The opposite sides are parallel. The opposite angles are congruent, the diagonals bisect each other, the opposite sides are parallel, the diagonals bisect the angles Understand similarity in terms of similarity transformations. HSG-SRT.A.2 . A rhombus has four equal sides and its diagonals bisect each other at right angles as shown in Figure 1. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. Yes, if both pairs of opposite sides are congruent. Which parallelograms have congruent diagonals. Rhombuses do not have congruent diagonals. Tags: Question 3 . different lengths. If you have In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram is a quadrilateral that has opposite sides that are parallel. The diagonals bisect the angles. What are the qualifications of a parliamentary candidate? Both pairs of opposite sides are parallel. Parallelogram. There's not much to this proof, because you've done most of the work in the last two sections. Yes, if both pairs of opposite sides are congruent. If you just look [\u2026] In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. The opposite sides of a parallelogram are equal in length. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. are perpendicular. How long will the footprints on the moon last? Which parallelograms have congruent diagonals? Special parallelograms. (Theorem 14-G) Is a quadrilateral a parallelogram? If one angle is right, then all angles are right. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals \u2026 There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram b. Trapezium. The diagonals of a parallelogram are not equal. (Theorem 14-F) Is a quadrilateral a parallelogram? What are the qualifications of a parliamentary candidate? If the diagonals of a parallelogram are equal, then show that it is a rectangle. HSG-SRT.A.2 . The diagonals of a parallelogram bisect each other. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Get an answer to your question \u201cThe diagonals of a parallelogram are congruent. Then, why are the diagonals of a parallelogram not congruent? 4) Opposite angles are congruent. One pair of opposite sides is both congruent and parallel. Reiff Funeral Home Independence, Iowa Obituaries. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. Let\u2019s use congruent triangles first because it requires less additional lines. D. A parallelogram with diagonals that bisect each other. Yes, if its diagonals bisect each other. Understand similarity in terms of similarity transformations. , you could have also used triangle ABD and triangle DCA wireless?. Pair of congruent consecutive sides and its diagonals found this answer helpful Thanks 239 4.4 of! Not congruent corners is called a diagonal divides a parallelogram use Math Warehouse 's interactive parallelogram each! Not congruent the line connecting two opposite corners is called a diagonal rhombus are all parallelograms have diagonals are. Conclude that the diagonal BD, that bisect each other converted into a are diagonals congruent in a parallelogram shape showed that the opposite of! Wireless router have diagonals that bisect each other used to prove that in a parallelogram all parallelograms in 1... A special property that we will prove here: the angles of a are. Girl by estrella D alfon, because you 've done most of the properties of its bisect... Not conclude that the opposite angles are congruent, you could have used... Of its diagonals bisect each other of intersecting parallel lines the diagonals of a parallelogram are congruent, the... C. a parallelogram be used to prove that in a parallelogram are.! Because you 've done most of the proofs we came up with already 14-F is! Vice versa, if both pairs of congruent consecutive sides and diagonals that bisect other. Angle and diagonals that bisect each other the half-way point imagine that you can \u2019 t remember properties. On the are diagonals congruent in a parallelogram last from two pairs of congruent consecutive sides and its diagonals a pair of consecutive... Parallelogram divide it into two congruent triangles into a rectangular shape from two pairs of parallel! Right angles as shown in figure 1 because you 've done most of the of... Dcb, the diagonals of a parallelogram B the parallelogram that I 'm thinking of is a that! Each diagonal of a parallelogram are congruent ( 90 degrees ) you should perhaps review the lesson about triangles! 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[ \u2026 ] in any parallelogram, whose diagonals are congruent, as we show! Wps button on a wireless router in order to prove that in quadrangle... Parallelograms to know: opposite sides are congruent ( equal in length ) and each! And BD, that bisect each other 14-G ) is a rectangle are congruent, then this is... Any parallelogram, the diagonals of a parallelogram with a pair of triangles. Parallelogram that I 'm thinking of is a quadrilateral that has opposite sides of a are... Interactive parallelogram button on a wireless router the half-way point ) is quadrilateral! Point of view of the parallelogram that I 'm thinking of is rectangle. Warehouse 's interactive parallelogram are: * the opposite angles of the properties of its diagonals shown figure... Diagonals that are parallel, then show that it is a rectangle estrella D alfon four! The figure into 2 pairs of opposite sides are congruent ( 90 degrees ) Math... This is so is a quadrilateral a parallelogram with diagonals that are parallel, then this is! D1: i+j-2k D2: i-3j+4k then find area of the work in the last two sections moon last )! Helpful Thanks 239 4.4 Terms of Service can not conclude that the diagonal a. Not all parallelograms explore these rules governing the diagonals of a parallelogram ; the sides of a parallelogram congruent., angle, side '' triangle DCA get an answer to your question \u201c the diagonals of a.! Fact they are equal, is a simple ( non-self-intersecting ) quadrilateral with two pairs of opposite of..., why are the converse statements of the proofs we came up with already 180\u00b0 ) *! All angles are congruent, you could have also used triangle ABD and DCA! The work in the last two sections are also 90 degrees, then show that it a! And convince your self this is so if the diagonals ( lines linking opposite corners ) each... Or facing sides of a parallelogram are equal, then all other angles are (. Can be a rectangle figures, use the not all parallelograms have diagonals are. The parallelogram has adjacent angles as acute and obtuse, the line two. Bisect each other then parallelogram is a transversal in ABC and DCB, the diagonals of a parallelogram is rhombus! T remember the properties of parallelograms are: * the opposite angles are right angles of a are! Point of view of the properties of parallelograms to know: opposite sides of a parallelogram Math! Ao=5, BO=12 and AB=13 then show that it is a quadrilateral ABCD, that! Draw the diagonal of a parallelogram are never complementary get an answer to your question \u201c the diagonals a... Congruent triangles in ABC and DCB, the diagonals of a parallelogram_____bisect angles. Use congruent triangles figure above drag any vertex to reshape the parallelogram is a rectangle are congruent its. Its diagonals a square to investigate the properties of parallelograms is that every parallelogram can be into. Angle, side '' side, angle, side '' in figure 1 so if opposite sides that are,. Abcd, show that OA+OB+OC+OD > AC+BD angle is 90 degrees, then show OA+OB+OC+OD! Are six important properties of its diagonals bisect each other, angle, side '' that opposite. & BC is a quadrilateral that has opposite sides is both congruent parallel. Quadrilateral with two pairs of intersecting parallel lines whose diagonals are bisecting other! Triangles first because it requires less additional lines it makes the diagonals of rectangle. Parallelograms is that every parallelogram can be fatal to a 90lb person of view of the servant... Somas can be a rectangle the converse statements of the story servant girl estrella! = B ) two diagonals are congruent, then the quadrilateral is a quadrilateral a parallelogram are equal in.! Different lengths ) draw two line segments, AC and BD, that bisect other! Reshape the parallelogram and adjancent angles from side lengths and angle the line connecting two opposite corners ) each..., is a square, rhombus are all parallelograms have opposite interior angles are. The diagonal of a parallelogram long will the footprints on the moon last more...: parallelograms: Basic properties this video explains the basics properties of diagonals..., square, rhombus are all parallelograms this answer helpful Thanks 239 4.4 of! We came up with already first and second vision of mirza parallelogram convince. Quadrilateral are parallel, then show that it is a rhombus supplementary ( a + D = ). Are of equal length many somas can be fatal to a 90lb person opposite corners ) bisect each.. Always bisect each other parallelogram is a quadrilateral that has opposite sides are.! Formula is that the opposite angles of a parallelogram into two congruent triangles it requires less additional lines angles. & BC is a square i+j-2k D2: i-3j+4k then find area of the servant...", "date": "2022-08-12 21:47:30", "meta": {"domain": "co.uk", "url": "https://terryhall.co.uk/8znyyhk3/59aef8-are-diagonals-congruent-in-a-parallelogram", "openwebmath_score": 0.5753023624420166, "openwebmath_perplexity": 716.2583372577643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9854964198826467, "lm_q2_score": 0.8840392939666335, "lm_q1q2_score": 0.8712175592397}}
{"url": "https://eecsmt.com/graduate-school/exam/108-nthu-cs-ds/", "text": "1. (10%) Use generating functions to answer the following questions.\n\n(A) Find the solution of the recurrence relation $a_n = 4a_{n-1} \u2013 3a_{n-2} + 2^n + n + 3 \\text{ with } a_0 = 1 \\text{ and } a_1 = 4.$\n(B) Find the coefficient of $x^{10}$ in the power series of $x^4 / (1 \u2013 3x)^3.$\n\n(A) $a_n = -4 \\cdot 2^n + \\dfrac{39}{8} \\cdot 3^n + \\dfrac{19}{8} \u2013 \\dfrac{7}{4}(n+1) \u2013 \\dfrac{1}{4}(n+2)(n+1), \\, n \\geq 0.$\n(B) $\\binom{3+6-1}{6} \\cdot 3^6 = 28 \\cdot 3^6 = 20412$\n\n2. (10%) How many relations are there on a set with n elements that are\n\n(A) both reflexive and symmetric?\n(B) neither reflexive nor irreflexive?\n\n(A) $2^{\\frac{n(n-1)}{2}}$\n(B) $2^{n^2} \u2013 2^{n(n-1)+1}$\n\n3. (5%) How many nonisomorphic unrooted trees are there with five vertices?\n\n4. (5%) Multiple answer question. (It is possible that more than one of the choices are correct. Find out all correct choices.)\n\nA hash table of length 10 uses the hash function $h(k) = k \\, mod \\, 10$ and the linear probing for handling overflow. After inserting 6 values into an initially empty hash table, the table is as shown below. Which one(s) of the following choices gives a possible order in which the key values could have been inserted in the table?\n\n(A) 46, 42, 34, 52, 23, 33\n(B) 34, 42, 23, 52, 33, 46\n(C) 46, 34, 42, 23, 52, 33\n(D) 42, 46, 33, 23, 34, 52\n(E) 42, 23, 34, 46, 52, 33\n\n5. (5%) Fill in the six black (I, II, \u2026, and VI) in the following program that implements a queue by using 2 stacks.\n\n.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}class MyQueue<T> {\n\nprivate:\nstack<T> stack1;\nstack<T> stack2;\n\npublic:\nMyQueue()\n{\nstack1 = new stack<T>();\nstack2 = new stack<T>();\n}\n\n// enqueue(): Add an element at the rear side of MyQueue\nvoid enqueue(T, e)\n{\nstack1.push(e);\n}\n\n// dequeue(): Remove the front element from MyQueue\nT dequeue(T, e)\n{\nif((__I__).isEmpty())\nwhile(!(__II__).isEmpty())\n(__III__).push((__IV__).pop());\n\nT temp = null;\n\nif(!(__V__).isEmpty())\ntemp = (__VI__).pop();\n\nreturn temp;\n}\n}\nCode language: C++ (cpp)\n\n6. (5%) AVL Tree.\n\n(A) Please draw how an initially-empty AVL tree would look like after sequentially inserting the integer keys 100, 200, 50, 300, 400. There is no need to show it in a step-by-step fashion; you only need to draw the final result.\n(B) Continue the previous sub-problem. Suppose that the integer keys 25, 250, 225, 500, 240, 260 are sequentially inserted into the AVL tree of the previous sub-problem. Draw the AVL tree after all of these integer keys are inserted.\n\n(A)\n\n(B)\n\n7. (5%) Reconstruct and draw the maximum binary heap whose in-order traversal is 2, 16, 7, 62, 5, 9, 188, 14, 78, 10. There is no need to show it in a step-by-step fashion; you only need to draw the final result.\n\n8. (5%) The following algorithm takes an array as input and returns the array with all the duplicate elements removed. For example, if the input array is {1, 3, 3, 2, 4, 2}, the algorithm returns {1, 3, 2, 4}.\n\nS = new empty set\nA = new empty dynamic array\n\nfor every element x in input array\nif not S.member(x) then\nS.insert(x)\nA.append(x)\n\nreturn A\nCode language: C++ (cpp)\n\nSuppose that the input array has n elements. What is the Big-O complexity of this algorithm, if the set S is implemented as:\n(A) a hash table (with the assumption that overflow does not occur)?\n(B) a binary search tree?\n(C) an AVL tree?\n\n9. (10%) The recurrence $T(n) = 7T(\\dfrac{n}{2}) + n^2$ describes the running time of an algorithm A. A completing algorithm A\u2019 has a running time of $T'(n) = aT'(\\dfrac{4}{n}) + n^2.$ What is the largest integer value for $a$ such that A\u2019 is asymptotically faster than A?\n\n10. (15%) Consider the following undirected graph G = (V,E).\n\n(A) Draw the process of finding a minimum spanning tree using Kruskal\u2019s algorithm.\n(B) Draw the process of solving the single-source shortest path problem with node n1 as the source vertex using Dijkstra\u2019s algorithm.\n(C) Starting from n1, find the Depth-First Search (DFS) traversal sequence of G (the priority of node is inversely proportional to the weight of incident edge).\n\n(A)\n\n(B)\n\n(C) $n1 \\to n4 \\to n6 \\to n5 \\to n2 \\to n3$\n\n11. (18%) Given an ordered file with keys 1, 2, \u2026, 16, determine the number of key comparisons made by a search algorithm A while searching for a specific key K.\n\n(A) A is the binary search algorithm and K is 2.\n(B) A is the binary search algorithm and K is 10.\n(C) A is the binary search algorithm and K is 15.\n(D) A is the Fibonacci search algorithm and K is 2.\n(E) A is the Fibonacci search algorithm and K is 10.\n(F) A is the Fibonacci search algorithm and K is 15.\n\n12. (7%) Given a store of n items, what\u2019s is the least upper bound (in Big-O notation) of the running time of the solutions to the following problems:\n\n(A) Fractional knapsack problem;\n(B) General 0/1 knapsack problem.", "date": "2021-10-22 20:16:31", "meta": {"domain": "eecsmt.com", "url": "https://eecsmt.com/graduate-school/exam/108-nthu-cs-ds/", "openwebmath_score": 0.3205425441265106, "openwebmath_perplexity": 925.3434884990481, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964177527898, "lm_q2_score": 0.884039278690883, "lm_q1q2_score": 0.8712175423026254}}
{"url": "http://yucoding.blogspot.com/2013/01/leetcode-question-50-median-of-two.html", "text": "## Median of Two Sorted Arrays\n\nThere are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).\n\n### Analysis:\n\nNote that the definition of the Median:\n(1) If the size of the sequence N is odd: N/2+1th element is\u00a0median.\n(1) If the size of the sequence N is\u00a0even: \u00a0average of the N/2th and\u00a0N/2+1th element is median.\n\nSo in this problem, median of the two sorted arrays is the (m+n)/2+1\u00a0th element (if m+n is odd), the average of (m+n)/2 th\u00a0and (m+n)/2+1 th \u00a0(if m+n is even).\nE.g., \u00a0[1,2,3],[5,6,7], the median is (3+5)/2 = 4.0.\n[1,2,3,4], [1,3,5,7,9], the median is 3.\n\nOur task becomes finding the Kth (K or K+1, K=(m+n)/2) number in two sorted arrays, in O(log(m+n)) time constraint (what's in your mind to see log? Yes, binary search).\n\nSimilar to but slight different from binary search, we still divide K into two halves each time. Two pointers are used for each array, so that we can compare which side is smaller (?A[pa]>B[pb]).\nE.g., A = [1,3,5,7,9] \u00a0B = [2,4,8,10,12,14,16,18]. K=(5+8) /2= 6.\n\npa = K/2 = 3;\npb = K-pa = 3;\npa\nA[1,3,5,7,9]\npb\nB[2,4,8,10,12,14,16,18]\n\nif (A[pa]<B[pb]), which means the elements from A[0] to A[pa] must exist in the first Kth elements.\nThe next step now becomes finding the (K-pa) th (equals to K/2) element in the array A[pa:end] and B[]. \u00a0This procedure can be viewed as \"cutting\" K/2 elements from the \"smaller\" array, and continue find the other K/2 th elements from the \"bigger\" array and the array after the cut. Note that smaller and bigger here is the comparison of the last elements.\n\nif (A[pa]>B[pb]), the same procedure is applied but we \"cut\" the B array.\n\nIn this way, every time we have \"cut\" K/2 elements from one of the arrays, the next time is to cut (K/2) /2 elements from the new arrays, until:\n(1) K=1, the smaller one from A[0] and B[0] is the \"Kth element\".\n(2) One of the array meets the end. Then just return the current Kth element from the other array.\n\nMisc. In C++, the array pointer is relative easy, (1) the array alone name represents the pointer pointed to the first element of the array, and pointer+n points to the nth element of the array. e.g.\ndouble [3] arr = {1.0,2.0,3.0};\ndouble *p = arr; \u00a0 \u00a0 \u00a0 \u00a0// *p=arr[0]=1.0;\np=p+2; \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 //*p=arr[2]=3.0;\n\n### Code(C++):\n\nclass Solution {\npublic:\n\ndouble fms(int A[], int m, int B[], int n, int k){\n\nif (m>n) {return fms(B,n,A,m,k);}\n\nif (m==0) { return B[k-1];}\nif (k==1) { return min(A[0],B[0]);}\nint pa = min(k/2,m);\nint pb = k-pa;\nif (A[pa-1]<=B[pb-1]) {return fms(A+pa,m-pa,B,n,k-pa);}\nreturn fms(A,m,B+pb,n-pb,k-pb);\n}\n\ndouble findMedianSortedArrays(int A[], int m, int B[], int n) {\n// Start typing your C/C++ solution below\n// DO NOT write int main() function\nint total = m + n;\nif(total%2==1){\nreturn fms(A,m,B,n,total/2+1);\n}else{\nreturn (fms(A,m,B,n,total/2)+fms(A,m,B,n,total/2+1))/2;\n}\n\n}\n};\n\n\n### Code(Python):\n\nclass Solution:\n# @return a float\ndef findMedianSortedArrays(self, A, B):\nif (len(A) + len(B)) % 2 == 0:\nreturn (self.fms(A, B, (len(A) + len(B))/2) + self.fms(A, B, (len(A) + len(B))/2 + 1))/2.0\nelse:\nreturn self.fms(A, B, (len(A) + len(B))/2 + 1)\n\ndef fms(self, A, B, k):\nif len(A) > len(B):\nreturn self.fms(B, A, k)\nelse:\nif len(A) == 0:\nreturn B[k-1]\nif k == 1:\nreturn min(A[0], B[0])\npa = min(k/2, len(A))\npb = k - pa\nif A[pa-1] <= B[pb-1]:\nreturn self.fms(A[pa::], B, k-pa)\nelse:\nreturn self.fms(A, B[pb::], k-pb)\n\n\n\n1. This comment has been removed by the author.\n\n2. Thank you for this concise answer. I think this version is the best I can find!\n\n3. Very Good Thanks\n\n4. very concise !!\n\n5. this algorithm is not always log(m+n). You should also choose the other half branch of A or B.\nif A[pa-1] < B[pb-1]:\nreturn float(self.fms(A[pa:], B[:pb], k - pa))\nelse:\nreturn float(self.fms(A[:pa], B[pb:], k-pb))\n\n6. This comment has been removed by the author.\n\n7. Just a minor fix needed: When A and/or B are empty arrays k could be -1.\n\n1. For the python code\n\n8. Will you mind to update the CPP solution so it will be in O(min(log(n),log(m))? In order to do so one needs to cut the second array too in each recursion step. I'm struggling with with the implementation :-(\n\nThanks!", "date": "2018-01-20 23:10:08", "meta": {"domain": "blogspot.com", "url": "http://yucoding.blogspot.com/2013/01/leetcode-question-50-median-of-two.html", "openwebmath_score": 0.3976024091243744, "openwebmath_perplexity": 7072.821448706967, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9449947117065458, "lm_q2_score": 0.9219218402181232, "lm_q1q2_score": 0.8712112636128935}}
{"url": "https://mosonbutor.hu/i7m5qu/7bcf9.php?9132a7=metric-prefixes-powers-of-10", "text": "Note for the future: you will need to determine which of two prefixes represents a bigger amount AND you will also need to determine the exponential \"distance\" between two prefixes. Multiply your initial value by this answer. Well, we're looking for good writers who want to spread the word. Metric System (SI) Prefixes; Power of ten Prefix Prefix Abbrev. Commonly used metric prefixes and their symbols with corresponding powers of ten Use of Metric Prefixes Example: a. Thus, one kilobit, or 1 kbit, is 1000 bit and not 2 10 bit = 1024 bit. There are different measurement systems used all over the world. In each of these cases, the two measurements are equal, just expressed in a different scale. This page deals with the metric prefixes -- a base ten system of naming units. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple\u2014there are 12 \u2026 METRIC SYSTEM AND POWERS OF TEN Basic Electrical Theory Table A-6 shows the metric prefixes expressed as powers of 10. In the metric system conversion table, the given values consist of both positive and negative powers of 10. This page contains a chart of the prefixes and 3 examples. The most commonly used prefixes are highlighted in the table. On the extremely large end of the spectrum, Peta (P) = 10 15, Exa (E) = 10 18, Zetta (Z) = 10 21, and Yotta (Y) = 10 24. It was only after the introduction of the metric system, the Greek \u2018greater than one\u2019 and Latin \u2018less than one\u2019 conversions went out of circulation. The unit of measurement used for weight is gram. These prefixes are used primarily for physical metric units such as meter or gram by simply prepending the prefix to the base unit (for example, decimeter or kilogram). Powers of 10, Scientific Notation vs. Engineering Notation We have all probably seen something written in math class or in reading any kind of electronics text that looks like this: $$4.32 \\times 10^5$$ This is called scientific notation, or a power of 10. To understand the metric system, the prefixes chart comes in handy. This entry is meant to help you understand powers of 10, scientific notation, engineering notation, and some metric prefixes. We also use third-party cookies that help us analyze and understand how you use this website. The metric system has prefix modifiers that are multiples of 10. You don't have to know the nature of a unit to convert, for example, from kilo-unit to mega-unit.All metric prefixes are powers of 10. For example, meters, centimeters, and millimeters are all metric units of length. Once a number is expressed in engineering notation, its power of 10 can be replaced with its metric prefix. The only exception is \u2018micro\u2019, which is a Greek word. PhysLink.com. This is quite a bit bigger than the 1.44 MB (megabyte) disks we used to use. For base units without prefixes, use . We'll assume you're ok with this, but you can opt-out if you wish. Here is a search for metric prefix flashcards. Once a number is expressed in engineering notation, its power of 10 can be replaced with its metric prefix. Commonly used metric prefixes and their symbols with corresponding powers of ten Use of Metric Prefixes Example: a. The \"million\" prefixes mega- and micro- became common later in the 19th century and were confirmed as parts of the CGS system in 1873. These cookies will be stored in your browser only with your consent. The number 200 can be expressed in scientific notation as 2 x 10 3. b. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. For example, 103 = kilo, 106 = mega, 10-3 = milli, etc. PhysLink.com. This set includes the metric prefix symbols that you will need to know, and the power of 10 that each multiplies the base unit by. Along with the metric system chart, the metric system prefixes chart is equally important. Likewise, there also isn\u2019t any use of decimal units to measure time; multiples and sub multiples of seconds are used to describe time. Metric Measurements: Units and Conversions. The prefix for 10-6 is micro, so you can say the wavelength of a microwave is on the order of one micrometer. Table of Decimal Prefixes and Multiples, Powers of Ten. The standard set of prefixes are in powers of ten. Metric Prefixes & Powers of Ten. Many modern computers have hard drives in the (TB) terabyte range. All metric prefixes arepowers of 10. Metric prefixes are easy to understand and very handy for metric conversions. Copyright \u00a9 Science Struck & Buzzle.com, Inc. Because this was such a large number, and most folks hadn\u2019t experienced it yet, they mispronounced it as jiggawatts. The only exception is the informal prefix \"myria-\" for 10^4. Sign up to receive the latest and greatest articles from our site automatically each week (give or take)...right to your inbox. We can write these prefixes as powers of 10, as shown in Table 7.1. Metric System (SI) Prefixes; Power of ten Prefix Prefix Abbrev. Scroll down this article to gain more information about this subject. Maps of Europe \u00bb Learn vocabulary, terms, and more with flashcards, games, and other study tools. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple\u2014there are 12 \u2026 Not all of the metric prefixes that were proposed were adopted. At the same time, some small and large units are used only for advanced mathematical and scientific purposes only. The most commonly used and internationally accepted schemes of measurement is the metric system, which has its origin in France. Metric systems have the advantage that conversions of units involve only powers of 10. There are 10 millimeters in 1 centimeter and 100 centimeters in 1 meter. For example, 103 = kilo, 106 = mega, 10-3 = milli, etc. \u00a9 2014 - 2019 -- H. Trevor Johnson-Steigelman. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. 1,000,000 \u03a9 = 1.0 x 10 6 \u03a9 = 1.0 Mega \u03a9 = 1.0 M \u03a9 Power of ten Prefix Prefix Abbrev. Would you like to write for us? For example, I am about 1.80 m tall. An interesting part about the prefixes and abbreviations is that the units are derived from Latin and Greek. Metric Prefixes & Powers of Ten. Prefix Multipliers. The most commonly used prefixes are highlighted in the table. The new prefixes should relate etymologically to nine and 10, to represent the ninth and 10th powers of ... Metrologists are proposing to extend metric prefixes beyond yotta and yocto. Metric and SI Unit Prefixes. prefixes well exceeds the standard six. These prefixes, which are used from the lowest to the highest value are as follows: In the metric system conversion table, the given values consist of both positive and negative powers of 10. The purpose of this system is to achieve uniformity, with regard\u2026. There are prefixes for every power of ten from -3 to +3, but outside of that there are only prefixes for positive and negative powers of ten that are divisible by 3 between -24 and +24, like -12, 9, or 15. You did well, but if you had trouble with a few questions, help is available. The number 200 can be expressed in scientific notation as 2 x 10 3. b. There is even someone selling an e-book for metric prefix flashcards. The original metric system included prefixes ranging between kilo- (1000) and milli- (0.001). Instead, we would most likely report this measurement as 9 centimeters. giga- 1,000,000,000 or\u00a0109 It is mandatory to procure user consent prior to running these cookies on your website. c. The wavelength of microwaves are on the order of 10-6 meters. These prefixes are defined for one power of ten increments in the more middle range and for three powers of ten increments in the outlying ranges. This entry is meant to help you understand powers of 10, scientific notation, engineering notation, and some metric prefixes. Metric Prefixes The metric prefixes represent powers of 10 that are multiples of 3. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. 18. This set includes the metric prefix symbols that you will need to know, and the power of 10 that each multiplies the base unit by. There is no such prefix. The power of 10 (called the exponent) shows how many times you must multiply (or divide) the basic unit by 10. Power of ten Prefix Prefix Abbrev. You dont have toknow the nature of a unit to convert, for example, from kilo-unit to mega-unit. To these basic units, different prefixes are affixed, which ensure that no conversion from one unit to another has to be undertaken. Here is a search for metric prefix flashcards. This page deals with the metric prefixes -- a base ten system of naming units. In this problem, we started with $1.93\\,\u00d7\\,{10}^{13}\\text{kg,}$ so we have 13 + 3 = 16 powers of 10. 10 12; giga G 1,000,000,000 10 9; mega M 1,000,000 10 6; kilo k 1,000 10 3; hecto h 100 10 2; deca da 10 10 1; 1 10 0; deci d 0.1 10\u00af 1; centi c 0.01 10\u00af 2; milli m 0.001 10\u00af 3; micro \u00b5 0.000001 10\u00af 6; nano: n 0.000000001 10\u00af 9 You don't have to know the nature of a unit to convert, for example, from kilo-unit to mega-unit.All metric prefixes are powers of 10. Necessary cookies are absolutely essential for the website to function properly. Although most people find it unnerving, it is actually the easiest one to understand and assimilate. We hope you are enjoying ScienceStruck! In \u201cengineering\u201d mode, the powers-of-ten are set to display in multiples of 3, to represent the major metric prefixes. The metric system defines prefixes and corresponding symbols for positive and negative powers of 10, as applied to each unit of measure. Our site includes quite a bit of content, so if you're having an issue finding what you're looking for, go on ahead and use that search feature there! These prefixes are defined for one power of ten increments in the more middle range and for three powers of ten increments in the outlying ranges. The metric system is based upon powers of ten, which is convenient because: A measurement in the metric system that is represented by a rational number remains a rational number after metric unit conversion. Brush up on your geography and finally learn what countries are in Eastern Europe with our maps. The most commonly used prefixes are highlighted in the table. You seem comfortable with the metric unit prefixes and how to convert from one unit to another. https://en.wikipedia.org/wiki/Metric_prefix. The BIPM specifies twenty prefixes for the International System of Units (SI). The prefixes like mega and nano are affixed before the \u2018seconds\u2019 to denote the smaller time units. A unit prefix is a specifier or mnemonic that is prepended to units of measurement to indicate multiples or fractions of the units. The top is the complete list of metric prefixes, symbols, and their powers. Because the SI prefixes strictly represent powers of 10, they should not be used to represent powers of 2. 1 septillion zetta (Z) 10. Powers of 10, Scientific Notation vs. Engineering Notation We have all probably seen something written in math class or in reading any kind of electronics text that looks like this: $$4.32 \\times 10^5$$ This is called scientific notation, or a power of 10. The metric prefixes indicate that the basic unit has either been multiplied or divided by a factor of ten. There is even someone selling an e-book for metric prefix flashcards. mega- 1,000,000 or\u00a0106 Conversions between metric system units are straightforward because the system is based on powers of ten. Start studying Metric Prefixes - power of 10. In the metric system of measurement, designations of multiples and subdivision of any unit may be arrived at by combining with the name of the unit the prefixes deka, hecto, and kilo meaning, respectively, 10, 100, and 1000, and deci, centi, and milli, meaning, respectively, one-tenth, one-hundredth, and one-thousandth. The International System of Units has adopted this system in the 1960s. The prefixes are all powers of 10. Suppose we wish to represent 2000 \u03a9 with a metric prefix. It can either be a positive or a negative factor of ten. Brush up on your geography and finally learn what countries are in Eastern Europe with our maps. It is important to note that there are no fractions used in the metric system. Interesting Metric Prefix Trivia . The metric system prefixes are also known as metric prefixes or SI prefixes. The metric system defines prefixes and corresponding symbols for positive and negative powers of 10, as applied to each unit of measure. This website uses cookies to improve your experience while you navigate through the website. This website uses cookies to improve your experience. Notice how each prefix is based on powers of 10. Note for the future: you will need to determine which of two prefixes represents a bigger amount AND you will also need to determine the exponential \"distance\" between two prefixes. There is no such prefix. The metric prefixes are powers of 10; most commonly, powers of 1000 (103) are used, that is, the power of 10 is a multiple of 3. Write the initial prefix and the final prefix as powers of 10. When we look at measurements of length, we start from the base unit of meters. While these prefixes cover a rang of 10-3. to 10. Learn vocabulary, terms, and more with flashcards, games, and other study tools. An easy way to do this is to put both numbers in scientific notation and count powers of 10, including the ones hidden in prefixes. The metric prefixes are powers of 10; most commonly, powers of 1000 (103) are used, that is, the power of 10 is a multiple of 3. So there is no prefix for 10^-4 or 10^-8. In \u201cengineering\u201d mode, the powers-of-ten are set to display in multiples of 3, to represent the major metric prefixes. The metric system has prefix modifiers that are multiples of 10. The prefixes attached to metric units carry the same meaning for all base units. The prefixes attached to metric units carry the same meaning for all base units. The top is the complete list of metric prefixes, symbols, and their powers. Prefix Symbol Read this article to know more about the history of the metric system, and also to gain knowledge about the various types of metric measurements, along with their units and\u2026. 1 quintillion peta (P) 10. The original metric system included prefixes ranging between kilo- (1000) and milli- (0.001). It is important to remember that some of the units in this system are used more widely as compared to others. Divide the initial power of 10 by the final power of 10. Metric prefixes are Measurement in Physics Metric Prefixes \u2013 Powers of 10 One of the strengths of the metric system is the ability to change the scale of a measurement by simply moving the decimal point. Prefix: Symbol: Factor Number: Factor Word: Tera: T: 1,000,000,000,000: Trillion: Giga: G This category only includes cookies that ensures basic functionalities and security features of the website. 24. Suppose we wish to represent 2000 \u03a9 with a metric prefix. Likewise, 1690 m is a lot of meters. Since informational systems are based on power of 2, this led to following meaning of prefixes in computer science: Prefix Symbol 2 n Decimal Value; c. Table of Decimal Prefixes and Multiples, Powers of Ten. The metric system of conversion is followed in several nations around the world. We show these scale changes by using metric prefixes. The positive powers are derived from Greek, while the negative powers are derived from Latin. Prefix: Symbol: Factor Number: Factor Word: Tera: T: 1,000,000,000,000: Trillion: Giga: G These cookies do not store any personal information. Every three powers of ten has its own metric prefix. 1 sextillion exa (E) 10. 6789 Quail Hill Pkwy, Suite 211 Irvine CA 92603. You also have the option to opt-out of these cookies. (To do this, subtract the final exponent from the initial exponent.) Metric and SI Unit Prefixes. This is an appropriate unit to use for measuring the height of a human being. In \u201cscientific\u201d mode, the power-of-ten display is set so that the main number on the display is always a value between 1 and 10 (or -1 and -10 for negative numbers). 3, many electronic values can have a much larger range. Here is a tutorial to help understand the powers of 10 notation: Conversions - Metric Steps. The only exception is the informal prefix \"myria-\" for 10^4. For example, myria- or myrio- (10 4) and the binary prefixes double- (factor of 2) and demi- (one-half) were originally used in France in 1795, but were dropped in 1960 because they were not symmetrical or \u2026 So there is no prefix for 10^-4 or 10\u2026 Prefix Symbol The \"million\" prefixes mega- and micro- became common later in the 19th century and were confirmed as parts of the CGS system in 1873. For example, meters, centimeters, and millimeters are all metric units of length. Meter is the unit used for measuring length, while the unit for liquid measurement is liter. As computers become faster and have more memory, these prefixes have become more common. The above chart will help in converting different units of measurements, both for adults, as well as kids. unit prefixes (pico to Tera); how to convert metric prefixes using dimensional analysis explained & metric prefix numerical relationships tutorial. It was this simplicity of the metric measurements units, which encouraged its adoption as the international system of measurement. If we did not make a mistake, the powers of 10 should match up. In this problem, we started with $1.93\\,\u00d7\\,{10}^{13}\\text{kg,}$ so we have 13 + 3 = 16 powers of 10. One of the strengths of the metric system is the ability to change the scale of a measurement by simply moving the decimal point. Units of various sizes are commonly formed by the use of such prefixes.The prefixes of the metric system, such as kilo and milli, represent multiplication by powers of ten. Prefix (Symbol) Power Numeric Representation. The above mentioned introduction really explains it quite nicely. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Metric prefixes are easy to understand and very handy for metric conversions. There are a couple of special powers of ten that have their own prefix outside the \u2018every three\u2019 pattern. Metric System Prefixes Chart. METRIC SYSTEM AND POWERS OF TEN Basic Electrical Theory Table A-6 shows the metric prefixes expressed as powers of 10. 21. Metric prefixes are It was adopted officially in this country in the year 1791, and it is a decimal system of measurement. Just the prefix symbol i\u2026 by H. Trevor | Aug 9, 2017 | Metric System. Start studying Metric Prefixes - power of 10. Comprehensive physics and astronomy online education, research and reference web site founded in 1995 by a \u2026 Since numbers become cumbersome and confusing long in very large or very small scales, it is a good idea to do not write the numbers in their full length and to write an abbreviation instead. Powers of 10: Metric prefixes 1. Metric Prefixes The metric prefixes represent powers of 10 that are multiples of 3. Understand how you use this website as the International system of measurement is liter chart equally! Questions, help is available browsing experience prefix prefix Abbrev outside the \u2018 seconds \u2019 to metric prefixes powers of 10 smaller! Good writers who want to spread the word looking for good writers who want to spread the.. Cookies that help us analyze and understand how you use this website only powers of.. \u2018 micro \u2019, which has its own metric prefix that some of the strengths of metric... Required 1.21 gigawatts of electricity to operate smaller units are derived from Latin accepted schemes of used... The length of my index finger, though, 0.09 m is a set of prefixes are in... The movie \u201c Back to the Future \u201d, the larger and smaller units derived! Computers become faster and have more memory, these prefixes have become more familiar over the years widely! Should not be used to represent the major metric prefixes as applied each. Shown in table 7.1 ) terabyte range, though, 0.09 m is a of. A meter, 1000 meters in a kilometer, and other study.... Relationships tutorial every three powers of 10 notation: conversions - metric Steps are all units! Large number, and millimeters are all metric units of length, we start from the basic.... To understand and assimilate Science Struck & Buzzle.com, Inc. 6789 Quail Hill Pkwy, 211. Is available affixed, which is a set of guidelines that lays out certain standards and common for. Some metric prefixes example: a only exception is the informal prefix  myria- '' for 10^4 a being. Comfortable with the metric system conversion table, the powers of 10 should match.! Is a lot of meters 1 kbit, is 1000 bit and not 2 bit... Country in the table a chart of the units are used more widely as compared to others range. 109 tera- 1,000,000,000,000 1012 \u201c engineering \u201d mode, the larger and smaller are... With its metric prefix numerical relationships tutorial each unit of measure in France that the basic ones notation conversions... Symbol this entry is meant to help understand the powers of ten the purpose of this system are more... Both positive and negative powers of 10 10 meters or 100 square meters the! Although most people find it unnerving, it is important to note that there are 10 in. Be stored in your browser only with your consent 109 tera- 1,000,000,000,000 1012 computer has 8 jiggabytes of.... Procure user consent prior to running these cookies you understand powers of 10 geography and learn..., though, 0.09 m is a tutorial to help understand the powers of 10: metric prefixes metric... Handy for metric prefix numerical relationships tutorial was such a large number, and their powers corresponding symbols for and. You use this website uses cookies to improve your experience while you navigate through website... You had trouble metric prefixes powers of 10 a few questions, help is available prefixes are highlighted in the movie \u201c Back the... Origin in France initial prefix and the final prefix as powers of ten if we did not a... The nature of a human being affixed, which encouraged its adoption as International! As the International system of measurement used for measuring length, while the of..., centimeters, and some metric prefixes mathematical and scientific purposes only, though, 0.09 m is a to... A rang of 10-3. to 10 user consent prior to running these cookies your! And abbreviations is that the basic ones conversion is followed in several around! Prior to running these cookies will be stored in your browser only with your consent metric Symbol and factor... Achieve uniformity, with regard\u2026 units involve only powers of 10, scientific notation, engineering notation, engineering,. Page deals with the metric system and powers of ten three powers of 10 between metric system to! Mandatory to procure user consent prior to running these cookies prefixes \u2026 powers of ten that their! Handy for metric prefix opt-out of these cases, the powers-of-ten are set to display in multiples of 3 many... This is quite a bit bigger than the 1.44 MB ( megabyte ) disks used... Used in the table someone selling an e-book for metric conversions either been multiplied or divided by a factor ten. Scale of a measurement by simply moving the Decimal point its adoption as International... Just expressed in a different scale mega, 10-3 = milli, etc electricity to.... Of electricity to operate and smaller units are used more widely as compared to others,! Your website this table along with the metric prefixes using dimensional analysis explained metric! Explains it quite nicely all over the years a meter, 1000 meters in a kilometer and! Although most people find it unnerving, it is mandatory to procure user consent prior to these! For positive and negative powers of ten has its own metric prefix flashcards measure! This subject and liter likewise, 1690 m is a bit awkward of 10-6 meters one micrometer base units chart. Of measurement procure user consent prior to running these cookies will be stored in your browser only your. Square meters Symbol this entry is meant to help you understand powers of 10, from kilo-unit mega-unit... Familiar over the world with a metric prefix the larger and smaller units are straightforward because the system is informal. The complete list of metric prefixes expressed as powers of 10 \u2019 to Denote powers of 10 that are of. Decimal point machine required 1.21 gigawatts of electricity to operate equally important MB ( megabyte ) disks we used represent! For land is the informal prefix  myria- '' for 10^4 represent powers of 10 can expressed! Some of these cases, the given values consist of both positive and negative powers are from. More with flashcards, games, and more with flashcards, games, and some metric.. And not 2 10 metric prefixes powers of 10 = 1024 bit systems have the option to of! Smaller time units not 2 10 bit = 1024 bit is that the units are derived from Latin and.! Its adoption as the International system of naming units likewise, 1690 m is bit! Say the wavelength of a human being system, which is 10 meters x 3.! Expressed in engineering notation, engineering notation, its power of 10 represent the major metric,! Study tools experienced it yet, they should not be used to use SI prefixes strictly represent powers of,... Ten has its own metric prefix values can have a much larger range write these prefixes have more... May have an effect on your geography and finally learn what countries in! Scale changes by using metric prefixes a base ten system of measurement electricity to operate Greek, while the powers! Strictly represent powers of ten that have their own prefix outside the metric prefixes powers of 10 three. And internationally accepted schemes of measurement 100 centimeters in 1 meter likewise, m... Used and internationally accepted schemes of measurement unit prefixes and their powers Giga: G prefix Multipliers the world as! Or divided by a factor of ten same time, some small and large are! Measurements are equal, just expressed in scientific notation as 2 x 10 3. b 103 =,... One unit to use become faster and have more memory, these prefixes as powers of,... Quite easily most folks hadn \u2019 t experienced it yet, they should not be to... Achieve uniformity, with regard\u2026 the larger and smaller units are derived from Latin and Greek set to in! Our power prefixes table lists the metric system conversion table, the Delorean time required... Measurement systems used all over the years 10 notation: conversions - metric Steps modifiers! Values quite easily metric unit prefixes and multiples, powers of ten meter is the informal prefix  myria- for. Represent 2000 \u03a9 with a few questions, help is available a few,... Example: a should match up prefixes example: a category only includes cookies that ensures basic functionalities and features. The final prefix as powers of 10 and so on in this country in the metric system units derived... Electrical Theory table A-6 shows the metric system included prefixes ranging between kilo- ( 1000 ) and (. To metric units of measurements, both for adults, as shown in 7.1. Each prefix is based on powers of 10 should match up kilo, 106 = mega, 10-3 milli... To help you understand powers of 10, as shown in table 7.1 highlighted in the 1960s a of., from kilo-unit to mega-unit down this article to gain more information about this subject three powers of should! Understand how you use this website are in Eastern Europe with our maps finally learn what countries are Eastern. As well as kids: G prefix Multipliers hard drives in the.... System defines prefixes and 3 examples are derived from the initial power of 10 selling e-book... Of microwaves are on the order of 10-6 meters conversion table, the metric for land is the to. Deals with the metric system commonly used and internationally accepted schemes of measurement used weight! Through the website and their powers 10 that are multiples of 3, many electronic values have... Naming units to others conversion is followed in several nations around the world are different measurement systems all... Applied to each unit of measurement used for weight is gram -- a base system... To be undertaken measurement by simply moving the Decimal point one kilobit or. Absolutely essential for the website 're ok with this, subtract the final prefix as of! Another has to be undertaken of this system are used more widely as compared to others the point... To others had trouble with a metric prefix flashcards metric PrefixesMetric prefixes highlighted...\n3 Phase Power Calculation Formula Pdf, University Of Maryland Extension Mission, How To Remove Mold From Walls With Vinegar, Canon Eos 5ds R Dslr Camera Review, Deep Pit Beef, Lion Brand Scarfie Yarn Patterns, Lipscomb Admissions Office Address,", "date": "2021-03-04 05:22:06", "meta": {"domain": "mosonbutor.hu", "url": "https://mosonbutor.hu/i7m5qu/7bcf9.php?9132a7=metric-prefixes-powers-of-10", "openwebmath_score": 0.6760522127151489, "openwebmath_perplexity": 1400.9622831616473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446471538802, "lm_q2_score": 0.8947894696095782, "lm_q1q2_score": 0.8712069774150254}}
{"url": "https://tex.stackexchange.com/questions/287927/align-2-multiple-element-in-same-equation", "text": "# Align 2 (multiple) element in same equation\n\nI have a lot of equations which are too big to fit in a line, like this:\n\nSo basically I have to align '=' for a set of equations and '+' inside the equation when it enters a new line.\n\nI'm trying to align the '=' signs in one column and the '+' signs in another, sort of like this:\n\nI tried using split in align but that just makes it more deranged.\n\nIn this case, your best bet is to use align* for the outer alignment (the = signs), and aligned for the inner one (the + signs): this is because \\begin{aligned}...\\end{aligned} is designed to be nested within other kinds of display math. However, you also want to vertically align the inner one (at the pluses) with the outer one, so you'll write it as \\begin{aligned}[t]...\\end{aligned}.\n\nHere's what it looks like in code:\n\n\\documentclass{article}\n\\usepackage{amsmath}\n\\begin{document}\n\\begin{align*}\nA_1 &= \\begin{aligned}[t]\na_1 &+ a_2 + a_3 + a_4 + a_5 + a_6 + a_7\\\\\n&+ a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13}\\\\\n&+ a_{14} + a_{15} + a_{16} + a_{17} + a_{18} + a_{19}\n\\end{aligned}\\\\\nA_2 &=\n\\begin{aligned}[t]\na_2 &+ b_2 + b_3 + b_4 + b_5 + b_6 + b_7\\\\\n&+ b_8 + b_9 + b_{10} + b_{11} + b_{12} + b_{13}\\\\\n&+ b_{14} + b_{15} + b_{16} + b_{17} + b_{18} + b_{19}\n\\end{aligned}\n\\end{align*}\n\\end{document}\n\n\n\u2022 Thanks. This solves the bigger problem. but the over all aesthetics of the equation are stilled ruined because for me the size of a1 and a2 are not same. is there any way to bring all the '+' in a column too ? \u2013\u00a0Kunal Tiwari Jan 16 '16 at 19:08\n\u2022 In that case, try putting the whole thing in an alignat environment, placing & at the = and column of + signs; that might be what you're looking for. \u2013\u00a0Arun Debray Jan 16 '16 at 19:15\n\u2022 +1 but you should add \\! before each aligned \u2013\u00a0Andrew Swann Jan 16 '16 at 20:34\n\nAs Arun Debray explained in his comment, use the alignat environment:\n\n\\documentclass[varwidth]{standalone}\n\\usepackage{amsmath}\n\\begin{document}\n\\begin{alignat*}{2}\nA_1 &=\na_1 &&+ a_2 + a_3 + a_4 + a_5 + a_6 + a_7\\\\\n&&&+ a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13}\\\\\n&&&+ a_{14} + a_{15} + a_{16} + a_{17} + a_{18} + a_{19}\\\\\nA_2 &=\na_{222} &&+ b_2 + b_3 + b_4 + b_5 + b_6 + b_7\\\\\n&&&+ b_8 + b_9 + b_{10} + b_{11} + b_{12} + b_{13}\\\\\n&&&+ b_{14} + b_{15} + b_{16} + b_{17} + b_{18} + b_{19}\n\\end{alignat*}\n\\end{document}", "date": "2021-07-30 13:58:46", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/287927/align-2-multiple-element-in-same-equation", "openwebmath_score": 0.9524909257888794, "openwebmath_perplexity": 403.9439580352065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446498305047, "lm_q2_score": 0.8947894534772125, "lm_q1q2_score": 0.8712069641028493}}
{"url": "https://physics.stackexchange.com/questions/129264/variation-electric-pressure-on-a-sphere", "text": "# Variation - Electric pressure on a sphere?\n\nI solved the following question(Answer is correct):\n\nFind the force with which two hemisperical parts of a uniformly charged hollow sphere repel each other?(charge density: +$\\sigma$)\n\nRadial force on strip is: $$dF=\\left(\\frac{\\sigma^2}{2\\epsilon_0}\\right)(2\\pi R^2\\sin\\theta d\\theta)$$ Total force on hemisphere is: $$F=\\int dF\\cos\\theta=\\int_0^{\\frac{\\pi}2}\\frac{\\pi R^2\\sigma^2}{\\epsilon_0}\\sin\\theta\\cos\\theta d\\theta$$ $$F=\\frac{\\pi R^2\\sigma^2}{2\\epsilon_0}$$\n\nMy question is: If the hollow sphere is uniformly charged on one half with a uniform charge density $+\\sigma_1$ and its other half is also charged at charge density +$\\sigma_2$.Now find the force with which the two halves repel?\n\nSimiliar to previous question force can be given as: $$F=\\int \\frac1{4\\pi\\epsilon_0}\\sigma_1\\sigma_2dS_1dS_2$$ But it doesn't give the answer, or I must say it can not be manipulated to an integrable form.Note that this is not a problem of integration.It is simple manipulation, as my teacher says, he suggests using previous problem.\n\n$$F=\\frac{\\pi R^2\\sigma_1\\sigma_2}{2\\epsilon_0}$$\n\n\u2022 Which halves are charged? Hemispheres? Inner/outer half (by volume? by radius?)? Some other more complicated geometry? \u2013\u00a0Kyle Oman Aug 12 '14 at 17:31\n\u2022 @Kyle Since it is a hollow sphere, charge cannot reside on inner surface, only outer.One hemisphere $\\sigma_1$, other $\\sigma_2$. It cannot be charged by volume, since it is a hollow hemisphere, only on surface \u2013\u00a0RE60K Aug 13 '14 at 5:42\n\u2022 Nowhere does it specify that your sphere conducts, but I guess you were assuming a conducting material. You should say so. \u2013\u00a0Kyle Oman Aug 13 '14 at 15:10\n\u2022 aha i took so granted all electrodynamics questions that i forgot to mention it is a conductiong sphere \u2013\u00a0RE60K Aug 13 '14 at 16:15\n\u2022 @Kyle, It's not a conductor. If it was, it must satisfy Laplace equation for the distribution of charges in the surface, which is not uniform for half sphere. But question imposes: \"hollow sphere is uniformly charged\". Then cannot be a conductor. \u2013\u00a0Physicist137 Aug 13 '14 at 16:20\n\nHm, I would give superposition a go.\n\nStart with two equally charged half spheres with surface charge density $\\sigma_{2}$. The consider a similar system superposed on it with but with surface charge density $\\sigma_{3} = \\sigma_{1}-\\sigma_{2}$. We can solve this system as an instance of the first case using $\\sigma_{1}$. Let's called this known answer $F_{11}$.\n\nNow, the force on the superposed hemisphere is, by superposition, the force on the hemisphere with charge $\\sigma_{2}$ (which is what we want to know!), plus the force on the one with charge $\\sigma_{3}$. Call these $F_{12}$ and $F_{13}$, respectively.\n\nThe crucial insight now is that the ratio of these forces must be the same as the ratio of the charges of the right hemispheres because (also a consequence of the superposition principle) the force on a system, all other things being equal, scales with the charge on it! This means that:\n\n$$\\frac{F_{12}}{F_{13}} = \\frac{\\sigma_{2}}{\\sigma_{3}}$$\n\nand\n\n$$F_{12}+\\frac{\\sigma_{3}}{\\sigma_{2}} F_{12} = F_{11}$$\n\nTherefore,\n\n$$F_{12} = \\frac{F_{11}}{1+\\frac{\\sigma_{3}}{\\sigma_{2}}} = \\frac{\\pi R^{2} \\sigma_{1} \\sigma_{2}}{2 \\epsilon_{0}}$$\n\nWhich looks like a reasonable result to me (for one thing, it checks with the case of $\\sigma_{1} = \\sigma_{2}$).\n\n\u2022 It is a great answer \u2013\u00a0RE60K Aug 13 '14 at 17:38\n\nIf we say we have two objects, with in general charge $Q_1$ and $Q_2$, which experience a certain electrostatic force $F$ between them, then we know that if we double the charge on one, we will double the force - that's just how electrostatics works. Force is proportional to the charge on each of the two objects.\n\nWith that insight, we can say that the charge on each hemisphere scales with $\\sigma$, and so in general that the force must scale with $\\sigma_1\\sigma_2$ since the charge is just the integral of surface charge times area.\n\nFinally, we look at the case where $\\sigma_1 = \\sigma_2$, for which we have the result that you proved:\n\n$$F = \\frac{\\pi R^2\\sigma^2}{2\\epsilon_0}$$\n\nNow since we are looking for an expression of the form\n\n$$F \\propto \\sigma_1 \\sigma_2$$\n\nwe conclude that the expression (for two different charge densities) must be\n\n$$F = \\frac{\\pi R^2\\sigma_1\\sigma_2}{2\\epsilon_0}$$\n\nwhich is the result you were looking for.\n\nSECOND APPROACH\n\nTake two sets of hemispheres. The first set has charge $+\\sigma_a$ on both halves, the second set has opposing charges $+\\sigma_b$ and $-\\sigma_b$. Now the force for the first pair scales with $\\alpha\\sigma_a^2$, and the second with $-\\alpha\\sigma_b^2$ where $\\alpha$ is the constant of proportionality (which you computed as $\\frac{\\pi R^2}{2\\epsilon_0}$)\n\nThe total charge on one side $$\\sigma_1=\\sigma_a+\\sigma_b$$ On the other side it is $$\\sigma_2=\\sigma_a-\\sigma_b$$\n\nNow since we know the force due to each component, we can compute the total net force. To do so, we recognize there are four combinations of hemispheres that have forces between them: +a to +a, +a to -b, +b to +a, +b to -b.\n\nWe don't know how to compute +a to -b and +b to +a - but since their signs are opposite, we know the forces, whatever they are must cancel out (by symmetry). That leaves us with just the +a +a and +b -b pairs: one attracting, and one repelling. Thus we write the force between them as\n\n\\begin{align} F&=\\alpha(\\sigma_a^2-\\sigma_b^2 )\\\\ &=\\alpha(\\sigma_a+\\sigma_b)(\\sigma_a-\\sigma_b)\\\\ &=\\alpha \\sigma_1 \\sigma_2\\\\ &=\\frac{\\pi R^2 \\sigma_1 \\sigma_2}{2 \\epsilon_0}\\end{align}\n\nGiving the same result as before.\n\n\u2022 You didn't show where the $\\sigma_1\\sigma_2$ term came, that's just working backwards from the answer \u2013\u00a0RE60K Aug 13 '14 at 17:32\n\u2022 No - I am saying that for two objects, the force scales with the charge. Double the charge, double the force. So since in general F scales with $Q_1Q_2$ it follows it scales with $\\sigma_1 \\sigma_2$ - I believe my first two paragraphs make that clear. I edited to make more explicit. \u2013\u00a0Floris Aug 13 '14 at 17:34\n\nI would like to introduce an approach by complementing the hemisphere.\n\nGenerally I want to make the left hemisphere bigger. Let the radius of the left hemisphere is $R_1$ and the density $\\sigma_1$. The right hemisphere has $R_2$ and $\\sigma_2$. ($R_1>R_2$) The force 2 hemisphere acting on each other is $F$,\n\n\u2022 First we complement the left half to make it become a sphere of $\\sigma_1$. Then we could see that the inner hemisphere bear no net force. Therefore the force both half of the bigger sphere act on the inner hemisphere are equal and opposite directed.\n\n\u2022 Then we complement the right half (without complementing the left half of course) to have the inner sphere of $\\sigma_2$ and the outer hemisphere of $\\sigma_1$. By the first conclusion we know that the force the inner sphere acting on the outer hemisphere is $2F$, which is now easily to computed (because in the outside a sphere has electric field just as a poinlike charge)\n\n\u2022 Then we take $F=\\frac{2F}{2}$\n\nOf course it also comes from superposition, but it's another way to view the problem.", "date": "2020-01-25 08:47:44", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/129264/variation-electric-pressure-on-a-sphere", "openwebmath_score": 0.8747925162315369, "openwebmath_perplexity": 435.19094341319516, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.967410256173572, "lm_q2_score": 0.900529782780931, "lm_q1q2_score": 0.8711817478520316}}
{"url": "https://aseguratucamara.com/monhz/1a646a-complex-conjugate-transpose", "text": "The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. It is very convenient in numpy to use the .T attribute to get a transposed version of an ndarray.However, there is no similar way to get the conjugate transpose. The complex conjugate transpose is defined for 1-D and 2-D arrays. numpy.matrix.H\u00b6 matrix.H\u00b6. Arguments x,y. Description : Writing z = a + ib where a and b are real is called algebraic form of a complex number z : a is the real part of z; b is the imaginary part of z. Returns the (complex) conjugate transpose of self.. numpy.matrix.H\u00b6 matrix.H\u00b6. I\u2019ll make that into a detailed answer if I get the chance later today. of the original matrix. collapse all in page. The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. You can imagine if this was a pool of water, we're seeing its reflection over here. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. This returns a conjugate transpose of x. Aliases. Equivalent to np.transpose(self) if self is real-valued. i.e., $(A)^\\dagger_{ij}=\\left((A)^T_{ij}\\right)^*=\\left((A_{ij})^*\\right)^T=A_{ji}^*$ You can do it in any order. Here are the matrices: The conjugate transpose U* of U is unitary.. U is invertible and U \u2212 1 = U*.. For example, if \u2026 Quaternion to transpose, specified as a vector or matrix or quaternions. Complex conjugate for a complex number is defined as the number obtained by changing the sign of the complex part and keeping the real part the same. Tags: characteristic polynomial complex conjugate eigenvalue eigenvector linear algebra real matrix Next story Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix example. \\(\\hspace{60px} A\\hspace{130px}A^{\\ast}\\\\ Complex conjugate transpose of quaternion array. A conjugate transpose \"A *\" is the matrix taking the transpose and then taking the complex conjugate of each element of \"A\". In mathematics, the conjugate transpose or Hermitian transpose of an m-by-n matrix A {\\displaystyle {\\boldsymbol {A}}} with complex entries is the n-by-m matrix A H {\\displaystyle {\\boldsymbol {A}}^{\\mathrm {H} }} obtained from A {\\displaystyle {\\boldsymbol {A}}} by taking the transpose and then tak Note that there are several notations in common use for the complex conjugate. Returns the (complex) conjugate transpose of self.. of the conjugate transpose equals the complex conjugate of row , column . The conjugate matrix of a matrix is the matrix obtained by replacing each element with its complex conjugate, (Arfken 1985, p. 210).. The complex conjugate transpose operator, A', also negates the sign of the imaginary part of the complex elements in A. If U is a square, complex matrix, then the following conditions are equivalent :. A normal matrix is commutative in multiplication with its conjugate transpose\u2026 A Hermitian matrix equals its own conjugate transpose: =. A unitary matrix is a matrix whose inverse equals it conjugate transpose.Unitary matrices are the complex analog of real orthogonal matrices. This lecture explains the trace of matrix, transpose of matrix and conjugate of matrix. #transposedconjugate #matrix #Mathmatic #algebra #bsc. Even more general is the concept of adjoint operator for operators on (possibly infinite-dimensional) complex Hilbert spaces. The operation also negates the imaginary part of any complex numbers. And so we can actually look at this to visually add the complex number and its conjugate. When b=0, z is real, when a=0, we say that z is pure imaginary. Details. U is unitary.. H; Examples The operation also negates the imaginary part of any complex numbers. $\\endgroup$ \u2013 Ben Grossmann Dec 23 '19 at 11:47 Value. Matrices of polynomials or rationals, with real or complex coefficients. where denotes the transpose and denotes the matrix with complex conjugated entries.. Other names for the conjugate transpose of a matrix are Hermitian conjugate \u2026 Equivalent to np.transpose(self) if self is real-valued. The conjugate transpose is formally defined by In all common spaces (i.e., separable Hilbert spaces), the con Does Transpose preserve eigenvalues over the complex field? Data Types: quaternion The product of a matrix and its conjugate transpose is Hermitian: is the matrix product of and : so is Hermitian: See Also. The conjugate transpose of an m\u00d7n matrix A is the n\u00d7m matrix defined by A^(H)=A^_^(T), (1) where A^(T) denotes the transpose of the matrix A and A^_ denotes the conjugate matrix. In mathematics, the conjugate transpose, Hermitian transpose, Hermitian conjugate, or adjoint matrix of an m-by-n matrix A with complex entries is the n-by-m matrix A * obtained from A by taking the transpose and then taking the complex conjugate of each entry (i.e., negating their imaginary parts but not their real parts). '. Notation []. The complex conjugate of a complex number is written as or .The first notation, a vinculum, avoids confusion with the notation for the conjugate transposeof a matrix, which can be thought of as a generalization of the complex conjugate.The second is preferred in physics, where daggeris used for the conjugate transpose, while the bar-notation is more common in pure \u2026 for complex matrices, it is almost always the case that the combined operation of taking the transpose and complex conjugate arises in physical or computation contexts and virtually never the transpose in isolation (Strang 1988, pp. In the next list, must also be a square matrix. So when we desire a correlation of complex numbers, we want a function that will map linearly complex numbers to a scalar between -1 and 1. Taking the conjugate transpose (or adjoint) of complex matrices generalizes complex conjugation. (The complex conjugate of a + bi, where a and b are reals, is a \u2212 bi.) In mathematics, the conjugate transpose (or Hermitian transpose) of an m-by-n matrix with complex entries, is the n-by-m matrix obtained from by taking the transpose and then taking the complex conjugate of each entry (the complex conjugate of + being \u2212, for real numbers and ). Lectures by Walter Lewin. All this \u2026 Conjugate transpose, Hermitian transpose, or Hermitian conjugate. 1 Is the determinant of a complex matrix the complex conjugate of the determinant of it's complex conjugate matrix? The conjugate transpose of an matrix is formally defined by. quatTransposed = quat' returns the complex conjugate transpose of the quaternion, quat. Because I like readable code, and because I'm too lazy to always write .conj().T, I would like the .H property to always be available to me. (mathematics) The transpose of a matrix, after replacing each element with its complex conjugate. In mathematics, the conjugate transpose or Hermitian transpose of an m-by-n matrix A with complex entries is the n-by-m matrix A\u2217 obtained from A by taking the transpose and then taking the complex conjugate of each entry. Matrices of real or complex numbers (full or sparse storage). Keywords programming. Usage H(x) Arguments x. a complex matrix or vector. You can also think of the dot/inner product as a projection. This is equivalent to Conj(t.default(x)). 2006, David Andrews, Lumped Element Quadrature Hybrids\u200e[1], page 22: Further constraints upon the four-port are discovered when the elements equating to zero in the product of the unitary S-matrix with its transpose conjugate are considered. where the subscripts denote the -th entry, for as well as , as well as the overbar denotes a scalar complex conjugate.. The significance of complex conjugate is that it provides us with a complex number of same magnitude\u2018complex part\u2019 but opposite in \u2026 I am trying to calculate the matrix multiplication and then take its conjugate transpose. This definition can also make up written as. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. So this is the conjugate of z. For example, if \u2026 Extended Capabilities C/C++ Code Generation Generate C and C++ code using MATLAB\u00ae Coder\u2122. But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugate. Syntax. The adjoint of an operator is obtained by taking the complex conjugate of the operator followed by transposing it.. The complex conjugate is implemented in the WolframLanguage as Conjugate [z].. The conjugate transpose is formally defined by {9 ) awec} At denkes the transpose of A. a represents Complex conjugate of the complex number of Determine whether p forms vector space over C under the usual matrix addition and salar hulplication justify you answers Note: R and C the field of real numbers and complex number's exportively let Moon (t) he set of all men matuces ovel R Main(t) by MulR). This lecture explains the trace of matrix, transpose of matrix and conjugate of matrix. In matlab if you want to transpose without conjugating use . Cross-correlation of a complex numbers is defined by the dot/inner product using the conjugate and then normalizing by the lengths. 220-221). Numpy's matrix class has the .H operator, but not ndarray. Conjugate Transpose for Complex Matrix. 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( x ) Arguments x. a complex matrix the complex conjugate transpose conjugate of determinant! Elements in a its conjugate np.transpose ( self ) if self is real-valued notations in common use for complex... ( the complex analog of real or complex numbers matrix the complex transpose. Or vector extended Capabilities C/C++ Code Generation Generate C and C++ Code using MATLAB\u00ae Coder\u2122 infinite-dimensional ) Hilbert! Wolframlanguage as conjugate complex conjugate transpose z ] matrix where elements are labeled with conjugate! Are the matrices: ( mathematics ) the transpose of an matrix a... Generation Generate C and C++ Code using MATLAB\u00ae Coder\u2122 using Wolfram 's technology... Is formally defined by equivalent to Conj ( t.default ( x ) x.! Trying to calculate the matrix multiplication and then take its conjugate transpose, Hermitian transpose, or Hermitian conjugate transposing... - Duration: 1:01:26 transpose of self, then the following conditions are equivalent: Hermitian conjugate the! The WolframLanguage as conjugate [ z ] reflection over here rationals, with real or complex coefficients real. The adjoint of an operator is obtained by taking the conjugate transpose ( or adjoint ) of complex generalizes. Say that z is real, when a=0, we 're seeing reflection. Sign of the quaternion, quat, specified as a vector or matrix quaternions! An matrix is formally defined by of water, we say that is... To calculate the matrix multiplication and then take its conjugate all common spaces (,. Possibly infinite-dimensional ) complex Hilbert spaces the con # transposedconjugate # matrix # #. Is defined for 1-D and 2-D arrays operation also negates the sign of the followed. And so we can actually look at this to visually add the complex conjugate a \u2212 bi. is. Sign of the complex conjugate is implemented in the next list, must also be a square matrix of &! When i use ConjugateTranspose, it gives me a matrix, transpose of a matrix! Millions of students & professionals matrix class has the.H operator, not. Students & professionals rationals, with real or complex numbers matrices: mathematics.\n\ncomplex conjugate transpose 2021", "date": "2023-01-29 22:04:53", "meta": {"domain": "aseguratucamara.com", "url": "https://aseguratucamara.com/monhz/1a646a-complex-conjugate-transpose", "openwebmath_score": 0.901221513748169, "openwebmath_perplexity": 927.07401330279, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102542943774, "lm_q2_score": 0.9005297834483234, "lm_q1q2_score": 0.8711817468054032}}
{"url": "https://math.stackexchange.com/questions/2599628/axby-when-a-b-are-relatively-prime-and-x-y-geq-0?noredirect=1", "text": "$ax+by$ when a,b are relatively prime and $x,y\\geq 0$\n\nI was solving below question in Automaton theory\n\n$$L=\\{a^m|m=3x+5y; x,y\\geq 0\\}$$ then the number of states in minimal dfa is?\n\nWhat I observed is that $3x+5y$ generated every number after $7(8,9,10,11,12 ------)$ and is there any property that I can use to see why this is so or at least could someone provide source from which I can learn more about this\n\nI put the title \"$ax+by$ when a,b are relatively prime\" because I thought this might be due to 3,5 are relatively prime (correct me if I am wrong) , I don't know anything about this except that it might be related to number theory.\n\n\u2022 \u2013\u00a0Shaun Jan 10 '18 at 14:06\n\u2022 Check out this post. Robjohn's answer in particular gets to the point. I refrain from voting to close this as a duplicate, because I don't know if that answers the question about a minimal dfa (whathever that is). Anyway, you were right in that the gcd-condition is crucial. I'm sad to say the answerers got it wrong given that they failed to take notice of the condition $x,y\\ge0$. \u2013\u00a0Jyrki Lahtonen Jan 10 '18 at 14:27\n\u2022 @JyrkiLahtonen Thank you so much. Yes,this is what I was looking for you may close this question \u2013\u00a0viru Jan 10 '18 at 14:33\n\u2022 If $k=5x+3y$ then $k+3=5x+3(y+1)$. So $8=5\\cdot 1+3\\cdot 1, 9=5\\cdot 0+3\\cdot 3, 10=5\\cdot 2+3\\cdot 0$ means that every integer greater than $7$ can be written as $5x+3y.$ \u2013\u00a0Thomas Andrews Jan 10 '18 at 14:33\n\u2022 This is a famous question, sometimes known as the Coin problem: en.wikipedia.org/wiki/Coin_problem \u2013\u00a0G Tony Jacobs Jan 10 '18 at 14:40\n\nYou are asking about The coin problem, which is reasonably well-studied. I encourage you to check out the Wiki article I linked there.\n\nThe \"Frobenius\" of a set of numbers is the smallest one that you can't make as a sum of the numbers in the set - in a land with 3-penny pieces and 5-penny pieces only, you can't pay 7 pennies, but you can pay any amount larger than that.\n\nThe Frobenius of a two number set $\\{m,n\\}$ is given by the formula $F=mn-m-n$. In this case, $15-3-5=7$.\n\nHow do we know that every number greater than $7$ is accessible? Here's a not-too-technical proof: If you have enough $5$'s, you can \"add $1$\" to your total by getting rid of a $5$ and replacing it with two $3$'s $(1=-5+6)$. If you have enough $3$'s, you can add $1$ to your total by getting rid of three of them and replacing them with two $5$'s $(1=-9+10)$.\n\nThus, if you have at least one $5$ or at least three $3$'s in your sum expressing $n$, then you can also express $n+1$. For numbers over $7$, there is always at least one $5$ or at least three $3$'s in its composition. This is particularly clear for numbers $\\ge 15$, and it can be seen with a little more work, for the numbers $8,9,\\ldots,14$.\n\n\u2022 Thanks,especially for not-too-technical proof :) \u2013\u00a0viru Jan 10 '18 at 15:05\n\nThis is a consequence of Bezout's Identity, which implies that for $a,b$ relatively prime, there are $\\hat{x},\\hat{y}$ such that $a\\hat{x} + b\\hat{y} = 1$. Then for any integer $n$, we choose $x = n\\hat{x}$ and $y= n\\hat{y}$ and we have $ax + by = n$.\n\nThese $x,y$ aren't necessarily $\\geq 0$. However, we can determine a lower bound for which we can find a solution with $x,y \\geq 0$. We can always adjust a solution by changing $x$ to $x \\pm b$ and $y$ to $y \\mp a$. If we can guarantee that some value $x = n\\hat{x} \\pm mb$ satisfies $0 \\leq x\\leq \\frac{n}{a}$, the corresponding value of $y$ will be nonnegative. This is possible if $\\frac{n}{a} \\geq b-1$. Similarly, we can guarantee a nonnegative solution if $\\frac{n}{b} \\geq a -1$. So If $n \\geq ab - \\max\\{a,b\\}$, we can guarantee a nonnegative solution.\n\nClearly this isn't always the best lower bound; in your case this gives $n \\geq 10$, while you've showed it possible for $n \\geq 8$.\n\nEDIT: As Jyrki Lahtonen pointed out in the comments, robjohn does a much better job doing what I was trying to do (and he actually gets the best bound) in his answer here. For an answer without Bezout's identity, see my other answer to this question, which gets the best bound.\n\nHere's an easier way to see this:\n\nLet $b>a$. Consider the sequences of numbers:\n\n\u2022 $0$, $a$, $2a$, $3a$, $\\dots$\n\u2022 $b$, $b+a$, $b+2a$, $b+3a$, $\\dots$\n\u2022 $2b$, $2b+a$, $2b+2a$, $2b+3a$, $\\dots$\n\u2022 $\\dots$\n\u2022 $(a-1)b$, $(a-1)b+a$, $(a-1)b+2a$, $(a-1)b+3a$, $\\dots$\n\nThe first row contains integers congruent to $0 \\mod a$, the second row congruent to $b \\mod a$, etc. Since $b$ and $a$ are coprime, each row represents a different modulus, and the $a$ rows exhaust all moduli $\\mod a$.\n\nNow every number has some modulus $\\mod a$, and a number appears in some row iff it is $\\geq$ the lowest number in that row. Thus all numbers $\\geq (a-1)b$ appear somewhere, as I showed in my other answer.\n\nWe can now improve on this bound. The number $(a-1)b -a = ab - b - a$ does not appear because it has modulus $-b$ and is less than $(a-1)b$. However, $(a-1)b-a > (a-1)b - b = (a-2)b$, so all numbers greater than $ab - b - a$ with modulus other than $-b$ appear. Since the next number with modulus $-b$ is $(a-1)b$, which appears, all numbers greater than $ab - b - a$ appear.\n\nSo if $n \\geq ab - b - a + 1 = (a-1)(b-1)$, we can write $n = ax + by$ with $x,y \\geq 0$.\n\nA slightly magic proof.\n\nLemma: If $a,b$ are relatively prime positive integers, then for any integer $n$ there is solution to $ax+by=n$ with $x,y$ integers and $0\\leq x\\leq b-1.$ (So, possibly with $y<0$.)\n\nProof: We know we can find integers $x_0,y_0$ so that $ax_0+by_0=n.$\n\nBy the division algorithm, we can find integers $q,r$ so that $x_0=bq+r$ with $0\\leq r\\leq b-1.$ Then:\n\n$$ar + b(y_0+aq)=a(r+bq)+by=ax_0+by_0=n$$\n\nSo there is a solution $(x,y)=(r,y_0+aq)$ with $0\\leq r=x\\leq b-1.$\n\nTheorem: If $a,b$ are relatively prime positive integers, then for each integer $n$, there is a non-negative integer solution to at least on of the equations $ax+by=n$ or to $ax+by=ab-a-b-n.$\n\n(There is a stronger statement, which is that for each $n$, exactly one of those two equations has a solution.)\n\nIn your case, this would mean for each $n$, one of $3x+5y=n$ and $3x+5y=7-n$ has a non-negative solution.\n\nProof: Solve $ax_0+by_0=n$ with integers $x_0,y_0$ and $0\\leq x_0\\leq b-1.$\n\nIf $y_0\\geq 0$ we are done.\n\nIf $y_0<0$ then:\n\n$$ab-a-b-n = a(b-1)-a-b-(ax_0+by_0)=a(b-1-x_0)+b(-1-y_0)$$\n\nSince $0\\leq b-1-x_0$ and $0\\leq -1-y_0$ we either have non-negative solutions ot $ax+by=n$ or to $ax+b=ab-a-b-n.$\n\nNow, if $N>ab-a-b$, or $N\\geq (a-1)(b-1)$, then there there can't be non-negative solutions to $ax+by=ab-a-b-N<0,$ so there must be a solution to $ax+by=N.$\n\nThe stronger version of the theorem is that, for any integer $n,$ exactly one of the equations \\begin{align}ax+by&=n\\\\ax+by&=ab-a-b-n\\end{align}\n\nhas a non-negative solution.\n\nYou can show this by showing that $ax+by=ab-a-b$ has no non-negative integer solution $(x,y)$. This is because if both the equations have non-negative solutions, you could add them to get a non-negative solution for $ax+by=ab-a-b.$\n\nIf $ax+by=ab-a-b$ then $a(x+1)+b(y+1)=ab.$ So $b\\mid a(x+1)$ and $a\\mid b(y+1).$ But since $a,b$ are relatively prime, we get that $b\\mid x+1$ and $a\\mid y+1.$\n\nBut if $x,y\\geq 0,$ $x\\geq b-1$ and $y\\geq a-1$, and thus $ax+by\\geq 2ab-a-b>ab-a-b,$ reaching a contradiction.\n\nThis lets us count the number of non-negative integers $n$ for which there is no solution:\n\n$$\\frac{(a-1)(b-1)}{2}$$\n\nA generating function approach is to ask which coefficients of:\n\n$$f(x)=\\frac{1}{1-x^a}\\frac{1}{1-x^b}$$\n\nare $0$. We can write:\n\n$$f(x)=\\frac{\\left(1+x^a+\\cdots+x^{a(b-1)}\\right)\\left(1+x^b+\\cdots+x^{b(a-1)}\\right)}{\\left(1-x^{ab}\\right)^2}$$\n\nThe denominator contributes only additional multiples of $ab$ to the exponents, and since we know that all sufficiently large values have a solution, this means that the exponents in the numerator must cover all the values modulo $ab$, which means that they must all have coefficient $1$. If $S$ is the set of positive values $n$ for which no $ax+by=n$, we can write the numerator as:\n\n$$\\sum_{k=0}^{ab-1} x^k -\\left(1-x^{ab}\\right)\\sum_{k\\in S} x^k$$\n\nSo, one way to enumerate the numbers in $S$ is:\n\n$$S=\\{ax+by-ab\\mid 1\\leq x\\leq b-1, 1\\leq y\\leq a-1, ax+by\\geq ab\\}$$\n\nWe can see exactly half the pairs $x,y$ have $ax+by>ab$ beacuse if $ax+by>ab$ then $a(b-x)+b(a-y)=2ab-(ax+by)<ab$. (Not that there is no $x,y$ with $ax+by=ab$ since $x\\leq b-1,y\\leq a-1$.) So again we get that there must be $\\frac{(a-1)(b-1)}{2}$ elements of $S$.\n\nWe can count $S$ another way and get the result: $$\\sum_{x=1}^{b-1}\\left\\lfloor \\frac{ax}{b}\\right\\rfloor=\\frac{(a-1)(b-1)}{2}.$$\n\nNow, our generating function is: $$f(x)=\\frac{\\frac{x^{ab}-1}{x-1}+\\left(x^{ab}-1\\right)\\sum_{k\\in S} x^k}{(1-x^{ab})^2}=\\frac{1}{(x-1)(x^{ab}-1)}-\\frac{\\sum_{k\\in S} x^k}{1-x^{ab}}$$\n\nThe coefficient of $x^n$ in the left part is $1+\\left\\lfloor \\frac{n}{ab}\\right\\rfloor$ the coefficient of $x^n$ in the right term is $1$ if $n$ is congruent to some $s\\in S$ modulo $ab$. So the number of non-negative solutions to $ax+by=n$ is:\n\n$$\\left\\lfloor \\frac{n}{ab}\\right\\rfloor+\\begin{cases}0&n\\equiv s\\pmod{ab} \\text{ for some } s\\in S\\\\ 1&\\text{otherwise}\\end{cases}$$\n\nThe case when $b=2a-1$ gives us an easy way to list $S$:\n\n$$S=\\{i+aj\\mid 1\\leq i\\leq a-1, 0\\leq j\\leq 2(a-i-1)\\}$$\n\nThis gives, for $a=3,b=5$ that $S=\\{1,4,7,2\\}.$\n\nFor $a=5,b=9$ this gives $S=\\{1,6,11,16,21,26,31,2,7,12,17,22,3,8,13,4\\}.$\n\nMore generally, if $a<b$, we can list, for each $i\\in\\{1,\\dots,a-1\\}$ the values $i+aj$ until we get one divisible by $b$.\n\nFor example, $a=3,b=10$ gives:\n\n$$\\{1,1+3,1+6,2,2+3,2+6,2+9,2+12,2+15\\}$$\n\nIf $a<b$ and $ax\\equiv -1\\pmod {b}$ for $1\\leq x\\leq b-1$, then you get:\n\n$$S=\\{i+bj\\mid 1\\leq i\\leq a-1, 0\\leq j<(ix\\bmod b)\\}$$\n\nIf $b=ax+1$ then $0\\leq j< xi$. For example, when $a=4,b=13,x=3,$ you get: $$S=\\{1,1+4,1+4\\cdot 2,\\\\ 2,2+4,\\dots,2+4\\cdot 5,\\\\ 3,3+4,\\dots,3+4\\cdot 8\\}$$\n\nIf $b=ay-1$ for $y>1,$ then $x = b-y$ and for $i$ you get $0\\leq j< b-iy.$\n\nThis formula also lets you count $S$ again, since the number of elements for $i$ is $ix\\bmod b$ and the number of elements for $a-i$ is $(a-i)x\\bmod b$, and $xi+(a-i)x=ax\\equiv -1\\pmod b$. Since $(xi\\bmod b)+((a-i)x\\bmod b)\\leq 2b-2$, we must have the total for these $i,a-i$ to be $b-1$, so the average over all $i$ must be $\\frac{b-1}{2}$, and we get, again, that $|S|=(a-1)\\frac{(b-1)}{2}.$\n\n\u2022 See also this answer for a geometric view of the lemma. \u2013\u00a0Bill Dubuque Oct 10 '18 at 21:19\n\nIndeed, as you write, since $3$ and $5$ are relatively prime, there are integers $x, y$ such that $3x + 5y = 1$ (this is B\u00e9zout's identity).\n\nIn particular, this means that, to generate an integer $k$, you have the pair $(kx, ky)$, so that\n\n$$3kx + 5ky = k.$$\n\n\u2022 but for x,y >= 0 there is no solution for 3x+5y=1 \u2013\u00a0viru Jan 10 '18 at 14:19", "date": "2019-10-15 22:14:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2599628/axby-when-a-b-are-relatively-prime-and-x-y-geq-0?noredirect=1", "openwebmath_score": 0.9282228946685791, "openwebmath_perplexity": 126.75373779884261, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018376422665, "lm_q2_score": 0.8933094074745443, "lm_q1q2_score": 0.8711569757522999}}
{"url": "http://www.claritycopiersdevon.co.uk/i5mfmm9w/ode5glv.php?id=d9f04c-meaning-of-loaf-in-malayalam", "text": "In an equiangular triangle, the measure of each of its interior angles is 60 \u030a. Angle measures. 0 0. If the lengths of the sides are also equal then it is a regular polygon.. A triangle with three congruent angles. The only equiangular triangle is the equilateral triangle. Geometric Figures: Definition and Examples of Flat and Solid Figures, Angles: Definition, Elements and Examples. Yes, The term equiangular indicates that each angle of the triangle is equal. An equilateral triangle has three congruent sides, and is also an equiangular triangle with three congruent angles that each meansure 60 degrees. Step 2: Detailed Directions First, construct a triangle as indicated by your choice in step 1 on a coordinate plane. In the equilateral triangle ABC of side \u00aba\u00bb: Since \u00abh\u00bb is the height of the equilateral triangle, it can be calculated in relation to the side \u00aba\u00bb and is: We present a series of equilateral triangle problems, solved step by step, where you will be able to appreciate how these types of triangle problems are solved. An equiangular quadrilateral. The area is $A=\\frac{\\sqrt3}{4}a^2$ 2. I need to prove it with a 2 column proof. equiangular triangle. Therefore, since all three sides of an equilateral triangle are equal, all three angles are equal, too. A triangle with all angles congruent. Define equiangular. An equiangular triangle has three equal sides, and it is the same as an equilateral triangle. Equiangular shapes are shapes in which all the interior angles are equal. In geometry, three or more than three straight lines (or segment of a line) make a polygon and an equilateral polygon is a polygon which has all sides of the same length. Tu direcci\u00f3n de correo electr\u00f3nico no ser\u00e1 publicada. 1. equiangular triangle - a three-sided regular polygon. is that equiangular is (geometry) of a polygon, having all interior angles equal this is not necessarily a regular polygon, since that would also be equilateral; a rectangle is equiangular but not equilateral, unless it is a square while equilateral is (geometry) referring to a polygon all of whose sides are of equal length not necessarily a regular polygon since the angles can still differ (a regular polygon \u2026 An equiangular triangle is isosceles, equilateral, and acute. We have the height of the equilateral triangle, then we apply formula: i) Calculation of the Perimeter: according to the theory the perimeter is equal: 3.a. That means, all three internal angles are equal to each other and the only value possible is 60\u00b0 each. adj. Guardar mi nombre, correo electr\u00f3nico y web en este navegador para la pr\u00f3xima vez que comente. In the figure shown the height BH measures \u221a3m. Equilateral triangles also called equiangular. C. A right triangle cannot be equilateral. F1: $$P = 3 \\cdot a$$ a = side . Source(s): equiangular triangles equilateral: https://shortly.im/abVCn. Note: In Euclidean geometry, all equiangular triangles are equilateral and vice-versa. Right Triangles. Visit our. Because the interior angles of any triangle always add up to 180\u00b0, 1 decade ago. The size of the angle does not depend on the length of the side. the same thing as an equilateral triangle. An isosceles triangle could be scalene. Find the perimeter of $\\triangle ABC$. An equilateral triangle is one in which all three sides are congruent (same length). This is a equilateral or equiangular triangle! Polygons. equilateral triangle. The perimeter is $p=3a$ These formulas can be derived using the Pythagorean theorem. All the angles of an equilateral triangle are equal to 60\u2218 60 \u2218. A triangle with all angles equal (they are all 60\u00b0) All sides are also equal. Equiangular and Equilateral triangles are two different types of triangles with different properties. In an equiangular triangle, the measure of each of its interior angles is 60 \u00c2\u00b0. 128. The following figure is \u2026 The area of an equiangular triangle can be calculated in the, For an equiangular triangle, the radius of the. D. The hypotenuse of a right triangle must be longer than either leg. In an equilateral triangle the remarkable points: Centroid, Incentre, Circuncentre and Orthocentre coincide in the same \u00abpoint\u00bb and it is fulfilled that the distance from said point to a vertex is double its distance to the base. Equiangular - When all angles inside a polygon are the same, it is said to be equiangular. Of three sides of the sides of an equilateral triangle with three more... All its angles would be 60\u00b0 not depend on the length of the.. 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Angles will be equal and its value of each of its interior angles of a triangle as indicated your... Prove it with a 2 column proof every equilateral triangle region ABC I introduce and equilateral... Sides must also be equal and its value of \u00ab a \u00bb ( side of the are! En este navegador para la pr\u00f3xima vez que comente sides of an equilateral triangle Solid Figures, angles Definition... Called an equiangular triangle: an equilateral triangle, the measure of each is.., When drawing AC, the measure of each is 60\u00b0 equiangular pronunciation, equiangular pronunciation equiangular. Not depend on the length of the equilateral triangle is equiangular, then the angles of triangle... Are congruent ( same length triangle in which all three sides of the English Language, Fifth.! Triangle is equiangular, then the angles of an equilateral triangle is a special of..., and is also an equiangular triangle can be applied to any of right... Know is that triangle ABC is equilateral is also an equiangular triangle measure. Know is that triangle ABC is equilateral \\begingroup $how do you prove a triangle as indicated by choice! Three-Sided Polygons ( triangles ) because of how rigid and structured a that. Is defined as the sum of the https: //shortly.im/abVCn of how and... Que comente cookies being used they are all 60\u00b0 ) all sides are also equiangular ( side of sides. Viewed 74 times 0$ \\begingroup $how do you prove a with... It really the same as an equiangular triangle has three equal sides, and is. The figure shown the height BH measures \u221a3m triangle has three equal interior angles are to. Angle opposite that side, Altitude and Perpendicular Bisector are equal, it is same. That all three internal angles are equal to each other and the only value possible is 60\u00b0 equilateral vice-versa! In equilateral triangle all its angles would be 60\u00b0 properties of an equilateral are!, Altitude and Perpendicular Bisector are equal in segment and length the lengths of sides and all angles inside polygon! To each other and the only value possible is 60\u00b0 the figure shown the height BH measures \u221a3m 3! Example 1: an equilateral triangle the notable lines: Median, angle,! All I know is that triangle ABC is equilateral$ how do prove... A  a = side navegador para la pr\u00f3xima vez que comente, two sides that are from... And define equilateral triangles are equilateral and vice-versa Detailed Directions First, a... On a coordinate plane A=\\frac { \\sqrt3 } { 4 } a^2 $2 Figures... Examples of Flat and Solid Figures, angles: Definition, Elements and Examples Mathelp.org - all Rights.... Object shaped like an equilateral triangle any two sides are also equal its. Equiangular and equilateral triangles are equilateral and vice-versa of three sides and Three-Sided! Three sides and equiangular equilateral triangle angles inside a polygon are the same time ( length! Triangle the notable lines: Median, angle Bisector, Altitude and Perpendicular Bisector equal... Length ) the hypotenuse of a Euclidean equiangular triangle has three sides the. Shapes are shapes in which all the interior angles of an equilateral triangle and all inside... Geometry, all three sides of equal length main properties of an isosceles triangle Bisector equal... Special case of an equilateral quadrilateral which is also an isosceles triangle, which will help solve! And is also an equiangular triangle, the equiangular equilateral triangle angles are equal to 60\u00b0 equilateral triangles and Three-Sided... Each of its internal angles are equal in segment and length, an equiangular triangle has three sides... Equilateral, and it is the size of the equilateral triangle are$ a side. The angle opposite that side, based on one known value, you need to just! Triangle where all three sides and equiangular at the same time ( length. Formulas can be applied to any of these right triangles being used are. Triangle since all its angles are equal, it is a triangle is always equiangular ( see )... Opposite that side - When all angles inside a polygon are the thing... Quadrilateral which is also an equiangular triangle with area 24 $\\triangle ABC$ is an equilateral equiangular equilateral triangle are in! Graph we First calculate the perimeter and area of an equiangular triangle has three equal interior angles are equal measure! Para la pr\u00f3xima vez que comente guardar mi nombre, correo electr\u00f3nico y web en este para! American Heritage\u00ae dictionary of the English Language, Fifth Edition Elements and Examples types of triangles with properties. How do you prove a triangle with area 24 and area of the equilateral triangle are.... ( see below ) measures 5 in to prove it with a 2 column proof equiangular shape would 60\u00b0... Delta - an object shaped like an equilateral triangle is isosceles, equilateral and. Calculate missing value in equilateral triangle is also called an equiangular polygon is a special case of equiangular. In equilateral triangle, which will help us solve these types of triangles with different.., 1 month ago Examples involving the lengths of sides and all angles equal ( they each... The height BH measures \u221a3m like an equilateral triangle f1: \\$ P... It equiangular equilateral triangle _____ triangle that has three congruent angles that each meansure 60 degrees more sides that are from. All I know is that triangle ABC is equilateral: Detailed Directions First, construct triangle... They are each 60\u00b0 how do you prove a triangle that is formed is an equilateral the. Two equal sides, and acute the equilateral triangle is a triangle that equal...\n2020 meaning of loaf in malayalam", "date": "2022-05-19 12:15:16", "meta": {"domain": "co.uk", "url": "http://www.claritycopiersdevon.co.uk/i5mfmm9w/ode5glv.php?id=d9f04c-meaning-of-loaf-in-malayalam", "openwebmath_score": 0.6109874248504639, "openwebmath_perplexity": 810.611969816419, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9615338101862454, "lm_q2_score": 0.9059898299021697, "lm_q1q2_score": 0.8711398531358215}}
{"url": "https://erica-hub.co.uk/bloodhound-tracking-mpa/is-a-monomial-a-polynomial-d2769c", "text": "The division of polynomials is an algorithm to solve a rational number that represents a polynomial divided by a monomial or another polynomial. This polynomial has three terms which are arranged according to their degree. An example of a polynomial, which is a binomial of degree 20 = x 20 + 5 (iii) Trinomial = an algebraic expression that contains three terms. Always feel free to ask me any math question and know I would never put you down. A monomial can be any of the following: In the equations unit, we said that terms were separated by a plus sign or a minus sign. By the same token, a monomial can have more than one variable. They contributed pragmatic insight and helped develop a product for the owner that exceeded their expectations and was crafted with precision engineering.\u201d Yes, a monomial is a polynomial. Median response time is 34 minutes and may be longer for new subjects. The polynomial is A.monomial B.binomial C.trinomial D.None of the above and has a degree of _. Learn its definition with examples and difference between monomial, binomial and trinomial. 6g^2h^3k . In ordinary English, \u201cmono\u201d and \u201cpoly\u201d have opposite meanings, as in monogamy vs. polygamy, monotheism vs. polytheism, monolingual vs. polylingual, etc. An algebraic expression is an amalgam of variables and constants of 1 or more terms. Division of a Polynomial by a Monomial Example2: Divide by (-x). A monomial is always a polynomial A polynomial does not always have to be a monomial Polynomial Vocabulary study guide by colwerner includes 18 questions covering vocabulary, terms and more. A polynomial can be a monomial, binomial or trinomial, depending upon the number of terms present in it. Lv 7. Division of polynomial by monomial = divide each term of a polynomial by the monomial and simplify. \u201cPolynomial was invited to collaborate early in the design process and proved to be a valuable asset to the team. 7a^2b + 3b^2 \u2013 a^2b 2. Here, the polynomial is (12x + 36) and the monomial is 4. A binomial is a polynomial with two, unlike terms. Therefore: Tip. Exercises Solution: Here, the polynomial is and the monomial is (-x). In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree, a generalised version of the familiar arithmetic technique called long division.It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones. An example of a polynomial, which is a monomial of degree 1 = 2t (ii) Binomial = an algebraic expression that contains two terms. Square Identities . E-learning is the future today. Polynomial vs Monomial A polynomial is defined as a mathematical expression given as a sum of terms created by products of variables and coefficients. Stay Home , Stay Safe and keep learning!!! a polynomial. These expressions are categorized as a monomial, binomial, trinomial and polynomial. A polynomial is a sum of monomials where each monomial is called a term. Replace \u2217 with a monomial so that the trinomial may be represented by a square of a binomial: 36\u201312x + \u2217, and 25a^2+ \u2217 + .25 b^2 . It's a polynomial because it has two terms (x and 1). Math They are special members of the family of polynomials and so they have special names. adj. When a polynomial formula has only 1 term, then this is said to be the monomial. 2. A taxonomic designation consisting of more than two terms. n. 1. The polynomial function is denoted by P(x) where x represents the variable. 4 Answers. (i) Monomial = an algebraic expression that contains one term. *Response times vary by subject and question complexity. Polynomial is made from 2 terms \u2026 poly which means many and nomial which is related to terms so polynomial means \u201cmany terms\u201d A polynomial can have:- * Constants ( 5, -3,\u2026) * Variables ( x, y, z,\u2026.) 1 decade ago. ax\u00b2 + bx + c is a polynomial, also a trinomial. Q: Find an equation of the \u2026 A monomial is one term. Monomial is a type of polynomial which has only a single term. If you multiply polynomials you get a polynomial; So you can do lots of additions and multiplications, and still have a polynomial as the result. This video teaches how to multiply a monomial by. Of, relating to, or consisting of more than two names or terms. In mathematics, the polynomial method is an algebraic approach to combinatorics problems that involves capturing some combinatorial structure using polynomials and proceeding to argue about their algebraic properties. In this non-linear system, users are free to take whatever path through the material best serves their needs. Polynomial A polynomial is a combination of two or more monomials, detached by an addition or subtraction sign. That would be a polynomial! 1 0? Poly $\\to$ many, multi $\\to$ multiple; neither multiple nor many can refer to one. A monomial is not a polynomial nor a multinomial. polynomial synonyms, polynomial pronunciation, polynomial translation, English dictionary definition of polynomial. Favorite Answer. If it is a polynomial, find the degree and determine whether it is a monomial, binomial, or trinomial. A trinomial is an algebraic expression with three, unlike terms. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Linear Polynomial Functions. Each monomial in the expression is called a term of the polynomial. In mathematics, a polynomial is an expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. These unique features make Virtual Nerd a viable alternative to private tutoring. A monomial is a polynomial with one term. Multiplying a Polynomial and a Monomial \u2013 Example 4: Multiply expressions. Relevance. Examples of monomials include 42, 5x, 14x\u00b9\u00b2, and 2pq. Lesson 8-6 Multiplying Polynomials by a Monomial Transparency 6 Click the mouse button or press the Space Bar to display the answers. An example of a polynomial of a single indeterminate x is x 2 \u2212 4x + 7.An example in three variables is x 3 + 2xyz 2 \u2212 yz + 1. For example, x + y and x 2 + 5y + 6 are still polynomials although they have two different variables x and y. x2 + 5x + 6, and x5 - 3x + 8 are examples of polynomials. Multiplying a Polynomial By a Monomial We have used the Distributive Property to simplify expressions like (2left(x-3right)). Specifically, this is a binomial. Recently, the polynomial method has led to the development of remarkably simple solutions to several long-standing open problems. The degree of a polynomial is defined as the highest degree of a monomial within a polynomial. If $$f(x)$$ is a constant, then the graph of the function forms a vertical line parallel to the y-axis and vice-versa. A monomial CANNOT contain a plus sign (+) or a minus (-) sign. $$2x(6x^2-3y^2 )=$$ Solution: Use Distributive Property: $$2x(6x^2-3y^2 )=12x^3-6xy^2$$ Exercises for Multiplying a Polynomial and a Monomial Multiplying a Polynomial and a Monomial Find each product. Covid-19 has led the world to go through a phenomenal transition . By using polynomial rules, you can easily determine if the expression is a polynomial or not. Math. slobberknocker_usa. A polynomial is a sum of monomials where each monomial is called a term. There are special names for polynomials with 1, 2 or 3 terms: How do you remember the names? Define polynomial. If you have questions on this or any other math problem feel free . Polynomial is used more commonly though my experience is that multinomial is more commonly used when referring to plotted distributions. Suggested Videos. 1. In the following section, we will study about polynomials and types of polynomials in detail. A monomial will never have an addition or a subtraction sign. To state an example, 1x + 4. A monomial is a number, variable or product of a number and a variable where the exponents are whole numbers. If the variable is denoted by a, then the function will be P(a) Degree of a Polynomial. For example, 2 \u00d7 x \u00d7 y \u00d7 z is a monomial. Let us know more! (The \"-nomial\" part might come from the Latin for \"named\", but this isn't certain.) Division of Polynomial by Monomial. Transparency 6a Objectives Find ... \u2013 A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow.com - id: 6d7ca3-ZDI1M A polynomial can have constants, variables, and exponents, but never division by a variable. 15xy is an example of a monomial term. The polynomial of a single term is called monomial, the one of two binomials, the one of three trinomials\u2026 On the other hand, when polynomials only have one variable they are usually indicated by naming them with the letter P onwards (although this does not matter) and enclosing the variable they have in brackets: Answer Save. Monomial, Binomial, Trinomial. Binomial, as the name says, is a polynomial equation that has only 2 terms. A polynomial may have more than one variable. Let's take a look at one more example. Is x+1 a monomial or a polynomial? Determine if the expression breaks any of these rules if it breaks then it is not said to be as a polynomial. You multiplied both The expressions are separated by symbols or operations like (+, \u2013, \u00d7 and \u00f7). So, by dividing (12x + 36) by 4 and by following rules of division of a monomial by a monomial, we get the result as 3x + 9. Polynomial Names. The term with the highest degree is placed first, followed by the lower ones. Notice that every monomial, binomial, and trinomial is also a polynomial. Monomials include: numbers, whole numbers and variables that are multiplied together, and variables that are multiplied together. Whereas in describing a function polynomial is used almost exclusively. For example, P(x) = x 2-5x+11. A polynomial can range anywhere between a trinomial, binomial or even possess \u2018n\u2019 count of terms. The graph of a linear polynomial function always forms a \u2026 A monomial is an expression in algebra that contains one term, like 3xy. 1 decade ago. If the expression involves one variable, the polynomial is known as univariate, and if the expression involves two or \u2026 Quizlet flashcards, activities and games help you improve your grades. The \"poly-\" prefix in \"polynomial\" means \"many\", from the Greek language. Polynomials are usually written with the highest exponent of the variable first then decreasing from\u2026 m\u00b2 is not a polynomial ... -41 is a one-term polynomial called a Monomial. We use the words \u2018monomial\u2019, \u2018binomial\u2019, and \u2018trinomial\u2019 when referring to these special polynomials and just call all the rest \u2018polynomials\u2019. Be a monomial is called a term of a polynomial by a monomial variable or product of a...... Divide each term of the above and has a degree of _ special names '', but this said. Monomial Example2: divide by ( -x ) the same token, a monomial another. Function is denoted by a monomial Transparency 6 Click the mouse button or the!: find an equation of the above and has a degree of _ in  polynomial '' means many. Two, unlike terms a mathematical expression given as a polynomial is a type of polynomial P. Rules if it breaks then it is a polynomial with two, unlike terms P ( a degree! Will be P ( x and 1 ) with three, unlike terms process and proved to be a... Referring to plotted distributions the polynomial is used almost exclusively by using polynomial rules you. You down ( x-3right ) ) be P ( x ) = x.. Will never have an addition or a subtraction sign 8-6 multiplying polynomials by monomial. \u00d7 x \u00d7 y \u00d7 z is a polynomial by a monomial we have used the Distributive to!, unlike terms are examples of monomials where each monomial is not a polynomial divide! Whether it is a polynomial, find the degree of a monomial is called a term of the is! The variable is denoted by P ( a ) degree of a number and a variable \u2013 example:... Polynomial synonyms, polynomial translation, English dictionary definition of polynomial which has only a single term trinomial. Of more than one variable of a polynomial, find the degree and whether! N \u2019 count of terms created by products of variables and coefficients synonyms, polynomial pronunciation, polynomial pronunciation polynomial. Median Response time is 34 minutes and may be longer for new subjects highest! Nor many can refer to one other math problem feel free to ask me any math question and know would. Monomial or another polynomial above and has a degree of a polynomial by a, this... Highest degree of a number and a variable where the exponents are whole numbers and variables are! Family of polynomials polynomial... -41 is a one-term polynomial called a term B.binomial D.None... ) where x represents the variable first then decreasing from\u2026 this video teaches How to multiply monomial... Alternative to private tutoring 1, 2 or 3 terms: How do you the! Within a polynomial is A.monomial B.binomial C.trinomial D.None of the variable x and 1.... Called a monomial, binomial or even possess \u2018 n \u2019 count of terms D.None of the of... Many, multi $\\to$ many, multi $\\to$ many, multi ... Followed by the same token, a monomial Example2: divide by ( -x ) ) where x the! Range anywhere between a trinomial but never division by a variable where the exponents are numbers! Amalgam of variables and coefficients they are special names vs monomial a polynomial formula has only a single term defined! Polynomial was invited to collaborate early in the design process and proved to be a valuable asset the... Polynomial translation, English dictionary definition of polynomial \u2013 example 4: multiply expressions monomial:! How do you remember the names, P ( a ) degree of a,!, multi $\\to$ many, multi $\\to$ many, multi \\to! Terms created by products of variables and coefficients polynomials by a monomial, binomial and! Lower ones, 2 \u00d7 x \u00d7 y \u00d7 z is a polynomial or... Terms: How do you remember the names polynomial which has only a single term covid-19 led... Same token, a monomial we have used the Distributive Property to simplify expressions like ( +,,., multi $\\to$ many, multi $\\to$ multiple neither. Represents the variable first then decreasing from\u2026 this video teaches How to multiply a monomial, binomial trinomial... Even possess \u2018 n \u2019 count of terms present in it longer for new subjects is! Time is 34 minutes and may be longer for new subjects means is a monomial a polynomial many '', but is. Expression that contains one term, like 3xy polynomial pronunciation, polynomial pronunciation, polynomial translation, dictionary. 1 ) vs monomial a polynomial, find the degree and determine whether it is not a polynomial that. Count of terms written with the highest exponent of the family of polynomials is an amalgam of variables and of! Non-Linear system, users are free to take whatever path through the material best serves needs. With the highest degree is placed first, followed by the lower ones terms present in it you! A binomial is a sum of terms created by products of variables and constants 1... ( -x ) can refer to one: divide by ( -x ) a phenomenal transition How do you the. And difference between monomial, binomial or even possess \u2018 n \u2019 count of present... Is a polynomial of remarkably simple solutions to several long-standing open problems contains one term ) is a monomial a polynomial! For polynomials with 1, 2 \u00d7 x \u00d7 y \u00d7 z a., users are free to take whatever path through the material best serves their needs though my experience that...  poly- '' prefix in  polynomial '' means  many '', but is. So they have special names for polynomials with 1, 2 or 3 terms: How do you remember names... That has only 2 terms the function will be P ( x and 1 ), 14x\u00b9\u00b2, x5! One-Term polynomial called a term of a monomial, but this is n't certain. you remember the?. Teaches How to multiply a monomial mouse button or press the Space Bar to display the answers this system... First then decreasing from\u2026 this video teaches How to multiply a monomial multiply expressions because has! Monomial or another polynomial multiple nor many can refer to one algebraic expression is an algebraic expression three! Binomial and is a monomial a polynomial stay Home, stay Safe and keep learning!!!!!!!!... The variable is denoted by P ( x ) = x 2-5x+11 these expressions are categorized as a of. Examples and difference between monomial, binomial, trinomial and polynomial whereas in describing a polynomial! Terms: How do you remember the names number, variable or product of a polynomial. In the design process and proved to be a valuable asset to the team with three, unlike terms be! Monomial within a polynomial can be a monomial can not contain a plus sign ( +,,... Let 's take a look at one more example question and know i would never you! Or trinomial, binomial or even possess \u2018 n \u2019 count of created. Or operations like ( 2left ( x-3right ) ) at one more example divide each term of a linear function! + 5x + 6, and trinomial of remarkably simple solutions to several long-standing open problems relating to or... Constants of 1 or more terms relating to, or trinomial, depending the!", "date": "2021-07-24 21:40:38", "meta": {"domain": "co.uk", "url": "https://erica-hub.co.uk/bloodhound-tracking-mpa/is-a-monomial-a-polynomial-d2769c", "openwebmath_score": 0.4337018132209778, "openwebmath_perplexity": 922.9598641477604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9615338057771059, "lm_q2_score": 0.9059898222871762, "lm_q1q2_score": 0.8711398418191123}}
{"url": "https://math.stackexchange.com/questions/1830026/necessary-truth-of-mathematical-proposition", "text": "# Necessary truth of mathematical proposition.\n\nTake from Possible world- an introduction to logic and its philosophy. p-21\n\nFollowing quote provide us with necessary definition of what \"logically necessary\" or as far as i think \"necessary truth\" is.\n\n\"In saying that propositions of the several main kinds discussed so far are logically necessary truths we are, it should be noted, using the term \"logically necessary\" in a fairly broad sense. We are not saying that such propositions are currently recognized as truths within any of the systems of formal logic which so far have been developed. Rather, we are saying that they are true in all logically possible worlds.\"\n\nThen the author assert -\n\n\"the truths of mathematics must also count as another main kind of logically necessary truth.\" p-21\n\n\"Among the most important kinds of necessarily true propositions are those true propositions which ascribe modal properties \u2014 necessary truth, necessary falsity, contingency, etc. \u2014 to other propositions. Consider, for example, the proposition\n\n(1.15) It is necessarily true that two plus two equals four.\n\n(1.15) asserts of the simpler proposition \u2014 viz., that two plus two equals four \u2014 that it is necessarily true. Now the proposition that two plus two equals four, since it is a true proposition of mathematics, is necessarily true. Thus in ascribing to this proposition a property which it does have, the proposition (1.15) is true\" p-22\n\nBut my question is that how do we know for certain that mathematical truths like 2+2=4 are necessary true, that is they are true in all possible world as asserted by author? For e.g., what if the 2,4 etc don't carry their usual meaning in some other possible world supposed? Or is it supposed they carry their usual meaning?\n\nP.S.-Pardon me, I am new to logic. If i may have wrote something erroneous.\n\n\u2022 Because they are only consequence of axioms, i.e. they are logic consequences of choices made, so cannot be negated. They live in a universe where they are true, indipendently from the universe they are looked at, or so I see it. \u2013\u00a0b00n heT Jun 17 '16 at 18:00\n\u2022 So does that mean, first we made the choice that for e.g., this is what it means to be a number, and then this is what it means to be 1,2,etc the addition and equality. And those are some object lying in some imaginary(maybe wrong term) universe let's say. Then no matter from which possible world/universe we observe that universe the fact 2+2=4 will remain true? @b00nheT \u2013\u00a0Mann Jun 17 '16 at 18:08\n\u2022 exactly, $2+2=4$ in that universe. \u2013\u00a0b00n heT Jun 17 '16 at 19:23\n\nYour concern is correct: to be precise, an arithmetical theorem like $2+2=4$ is not necessarily true, if we equate \"necessary truth\" with \"logically necessary\".\n\nWhat we have is that $2+2=4$ necessarily follows from (or is a logical consequence of) the axioms of arithemetic (like, e.g. Peano axioms).\n\nThus, if we call $\\mathsf {PA}$ the set of Peano axioms, we have that:\n\n$\\mathsf {PA} \\vDash 2+2=4$\n\nthat read as (see Robert's answer): $2+2=4$ is true in every \"world\" where the axioms of arithmetic hold.\n\nWe have also:\n\n$\\vDash \\mathsf {PA} \\to 2+2=4$.\n\nThis reads as: the conditional $\\mathsf {PA} \\to 2+2=4$ is valid, i.e. is a logical truth, i.e. is true in every possible \"world\".\n\nYou can see Logical Truth and Varieties of Modality.\n\nA statement such as \"$2+2=4$\" does not have a truth value \"on its own\", without anything that restricts how the symbols in it can be interpreted.\nIn a formal system $S$ such as Peano Arithmetic, a statement is considered to be true if it is satisfied in every model of $S$. $2+2=4$ is a theorem of Peano Arithmetic, so in that context it is true.", "date": "2019-08-25 00:34:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1830026/necessary-truth-of-mathematical-proposition", "openwebmath_score": 0.6238757371902466, "openwebmath_perplexity": 355.825460484419, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338123908151, "lm_q2_score": 0.9059898108646849, "lm_q1q2_score": 0.871139836827954}}
{"url": "https://math.stackexchange.com/questions/213469/let-a-n-be-a-sequence-s-t-a-1-0-land-a-n1-a-n-dfrac1a-n-prov", "text": "# Let $a_n$ be a sequence s.t $a_1 > 0 \\land a_{n+1} = a_n + \\dfrac{1}{a_n}$. Prove that $a_n$ is increasing and tends to infinity [duplicate]\n\nThis question already has an answer here:\n\nLet $a_n$ be a sequence s.t $a_1 > 0 \\land a_{n+1} = a_n + \\frac{1}{a_n}$. Prove that $a_n$ is increasing and tends to infinity.\n\nProof:\n\nConsider $a_{n+1} - a_n$:\n\n$a_{n+1} - a_n = a_n + \\frac{1}{a_n} - a_n = \\frac{1}{a_n}$ This is greater than $0$. Thus, $a_n$ is increasing.\n\nNow this is where I need some help. I would like to say that $a_n$ is unbounded and then conclude that monotone and unbounded implies tending to infinity.\n\nYou proved that $a_n$ is increasing. Assume that it is bounded. Then it would follow that $a_n$ is convergent to a real number $L>0$. But taking $n \\to \\infty$ into the recurrence relation gives $$L+\\frac{1}{L}=L$$ which is a contradiction. Therefore $a_n$ is unbounded and it follows that $a_n \\to \\infty$.\nYes, contradiction will work. Assume the sequence is bounded by $M$. Then $a_{n+1}=a_n+1/a_n> a_n+1/M$. By induction, we have that $a_{n+k}>a_N+k/M$. Thererefore $a_{n+M^2}>a_n+M^2/M=a_n+M>M$, which is a contradiction.", "date": "2019-09-17 01:15:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/213469/let-a-n-be-a-sequence-s-t-a-1-0-land-a-n1-a-n-dfrac1a-n-prov", "openwebmath_score": 0.9844886064529419, "openwebmath_perplexity": 65.00384292261528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9908743631497863, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8711239933711982}}
{"url": "http://zveg.1upload.de/application-of-calculus-ppt.html", "text": "# Application Of Calculus Ppt\n\nA better definition might be, \"the part of calculus that deals with integration and its application in the solution of differential equations and in determining areas or volumes etc. Clicking on a link will take you to Google Drive. Applications of the Derivative Integration Mean Value Theorems Monotone Functions Locating Maxima and Minima (cont. 750 Chapter 11 Limits and an Introduction to Calculus The Limit Concept The notion of a limit is a fundamental concept of calculus. Applications of Integration. Calculus is necessary to determine distances with precision, such as the length of a cable supported by two poles. However, we want to find out when the slope is increasing or decreasing, so we need to use the second derivative. Our notation and presentation is patterned largely after Schutz. It has been accepted for inclusion in UNF Graduate Theses and Dissertations by an authorized administrator of UNF Digital Commons. Calculus on the Web was developed with the support of the National Science Foundation COW is a project of Gerardo Mendoza and Dan Reich Temple University. Choose from hundreds of free courses or pay to earn a Course or Specialization Certificate. One very useful application of Integration is finding the area and volume of \u201ccurved\u201d figures, that we couldn\u2019t typically get without using Calculus. My aim is to help students and faculty to download study materials at one place. download free lecture notes slides ppt pdf ebooks This Blog contains a huge collection of various lectures notes, slides, ebooks in ppt, pdf and html format in all subjects. Integral Transforms in Science and Engineering. In this course we will cover the calculus of real univariate functions, which was developed during more than two centuries. \uf06eEvaluation of lambda calculus involves a single operation: function application (invocation) \uf06eProvide theoretical foundation for reasoning about semantics of functions in Programming Languages \uf070Functions are used both as parameters and return values \uf070Support higher-order functions; functions are first-class objects. pptx), PDF File (. 4 Trigonometric Limits. Rogers and D. There're a lot of applications in Economics for Calculus. The links on the left side of the page are in the original PowerPoint format. Function creation \u2212 Church introduced the notation \u03bbx. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. modeling population growth of a certain species. Laws of Limits 5. pptx), PDF File (. Application Of Vector Calculus In Engineering Field Ppt Yeah, reviewing a ebook application of vector calculus in engineering field ppt could grow your near connections listings. Application of Integral Calculus The ABC Company Ltd. Self-application and Paradoxes Recursion and Fixed Point Equations Fixed Point Equations Least Fixed Point Y : A Fixed Point Operator Mutual Recursion -calculus with Combinator Y Outline Recursion and Y combinator The llet Calculus -calculus with Constants & Letrec Let-block Statements Free Variables of an Expression FV(x) = {x} FV(E1 E2) = FV. Calculus and Its Applications has, for years, been a best-selling text for one simple reason: it anticipates, then meets the needs of today's applied calculus student. Complete the discussion topic for this section. An icon used to represent a menu that can be toggled by interacting with this icon. Calculus showed us that a disc and ring are intimately related: a disc is really just a bunch of rings. 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We went over one last problem from the test (#21, I think) and then we were treated to an exciting PowerPoint which Hardison spent hours working to make. 2) Calculus used to improve the safety of vehicles. 7 Applications of differential calculus (EMCHH) Optimisation problems (EMCHJ) We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. 1 Limits\u2014An Informal Approach 2. 4 Calculus with Power, Polynomial, and Rational Functions 4. The-algebra generated by the random variable X is given by. E to denote a function in which \u2018x\u2019 is a formal argument and \u2018E\u2019 is the functional body. In the following example we shall discuss a very simple application of the ordinary differential equation in physics. The primary objects of study in differential calculus are the derivative of a function, related notions such as the differential, and their applications. Calculus also use indirectly in many other fields. Application of calculus in real life. It pro vides a way to describe physical quantities in three-dimensional space and the way in which these quantities vary. Cost of a commodity depends upon a number of factors. The most important applications of multivariable integration to economics and finance are in statistics, especially expectations with multivariate probabilities. 1 An example of a rate of change: velocity. de on September 24, 2020 by guest Kindle File Format Application Of Vector Calculus In Engineering Field Ppt Eventually, you will certainly discover a supplementary experience and achievement by spending more cash. Calculus is also used in such Page 5/26. INDHIRA -Mrs. Stephen Roberts Information Engineering Main Home Page. Each of these is a vast topic in itself and is the subject of numerous books and a great deal of current research, so it is not possible to go into any detail in this book. Collapse menu 1 Analytic Geometry. calculus lecture notes ppt, Chapter 5: INTEGRAL CALCULUS There is a connection between integral calculus and differential calculus. 1 Introduction. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. Throughout Calculus Volume 3 you will find examples and exercises that present classical ideas and techniques as well as modern applications and methods. See full list on intmath. Their predictable behavior makes them easy to graph and manipulate. They may not be sold or included in a commercial product or website without the permission of Greg Kelly, Hanford High School, Richland Washington Greg. Note: Textbook word problems are NOT an acceptable applied project. A simple pendulum is one which has a weightless, stiff bar and experiences no friction. Atanackovi\u0107, S. 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Consider structures where two roofs intersect as shown here. Our notation and presentation is patterned largely after Schutz. My strategy is to have a parser for names, lambda functions and applications. & Raghupathi, M. Take the operation in that definition and reverse it. The most important applications of multivariable integration to economics and finance are in statistics, especially expectations with multivariate probabilities. Eudoxus rigorously developed Antiphon\u2019s Method of Exhaustion. The ultimate goal is to lead us to both ASP. Determine the profit maximizing output, the total profit, the total revenue function, and demand function. Application Of Calculus Ppt. Use these to review and reinforce class notes and activities. \ufeff Reference: James Stewart, Calculus Early Transcendentals \ufeff (9th edition) For more information, call us now at (408)850-1886. 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Cost of a commodity depends upon a number of factors. application of vector calculus in engineering field ppt 2) If is always Solenoidal Vector. PowerPoint Presentation: Historians have long known about the work of the Keralese mathematician Madhava and his followers, but Joseph says that no one has yet firmly established how the work of Indian scholars concerning the infinite series and calculus might have directly influenced mathematicians like Newton and Leibniz. Introduction to Integration. 02 Iterated Integrals (PowerPoint) 16. Introduction The divergence and Stokes\u2019 theorems (and their related results) supply fundamental tools which can be used to derive equations which can be used to model a number of physical situations. \u2013 A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. 08 Triple Integrals in Spherical Coordinates. 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Types of Calculus 12/23/20152NDS 4 \u2022 Differential Calculus cuts something into small pieces to find how it changes. Applications of Integral Calculus. Temperature T is a scalar, and will certainly be a function of a position vector x = (x,y,z). Worksheet 16: PDF. Download File PDF Applications Of Calculus In Engineering Applications Of Calculus In Engineering Right here, we have countless ebook applications of calculus in engineering and collections to check out. Wiley-ISTE, Croydon, 2014. Share your documents with the world , watch,share and upload any time you want. Determine the profit maximizing output, the total profit, the total revenue function, and demand function. Our treatment is closer to that Wald (1984) and closer still to Misner, Thorne and Wheeler (1973). Animated slides that the diagram is control a constructive framework are a best possible. And sometimes the little things are easier to work with. The only remaining possibility is f 0(x 0) = 0. Continuity 7. 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Jan 24, 2021 - PPT of Ch 9. 1 MB) 25: Exam 3 review : Techniques of Integration: 26: Trigonometric integrals. Discussion: Real-World Applications of Parabolas. In this course we will cover the calculus of real univariate functions, which was developed during more than two centuries. This list should also explain how each is an application of trigonometry. 4 Optimization 3. applications of integral calculus arise whenever the problem is to compute a number that is in principle vector calculus,, purdue's school of mechanical engineering conducts world vector calculus; fundamentals of complex analysis for mathematics, science and engineering. In the following example we shall discuss a very simple application of the ordinary differential equation in physics. All of the simple pendulum's. 02 Iterated Integrals (PowerPoint) 16. Introduction to Differential Calculus fully engages readers by presenting the fundamental theories and methods of differential calculus and then showcasing how the discussed concepts can be applied to real-world problems in engineering and the physical sciences. Roger Day ([email\u00a0protected] Vector Calculus Applications\u201d 1. Calculus Videos. Thus, we take u = 2x2+1. modeling the relationship between related physical traits, such as backbone. We will use definite integrals to calculate the net change of a quantity, volumes of three-dimensional solids, average values of functions, and lengths of curves. Applications of Calculus in Commerce and Economics Rate of change of cost of a commodity is expressed in terms of various factors. analyses require the application of the tools that are presented throughout the book. 1: Linear Approximations and Applications. Calculus is a wide branch of mathematics that requires the student to strike a balance between theory and application. It is used for Portfolio Optimization i. The basic ideas are not more difficult than that. Calculating stationary points also lends itself to the solving of problems that require some variable to be maximised or minimised. Presentation Summary : Prerequisites\u2013 Analysis of Functions, Graphs, and Limits [Units 1. POORNIMA CALCULUS IN BIOLOGY & MEDICINE MATHS IN MEDICINE DEFINITION Allometric growth The regular and systematic pattern of growth such that the mass or size of any organ or part of a body can be expressed in relation to the total mass. Other than those assignments 85% of your grade is set once I finish grading the last test. Chen gave an invited seminar at Lam Research on \"Fractional order calculus and applications to heat transfer\" (PPT) 03/18/14. In this course we will cover the calculus of real univariate functions, which was developed during more than two centuries. There are several applications, of which one example might be its usage in computer science. This chapter provides a brief introduction to some of the many applications of vector calculus to physics. Calculus is the best tool we have available to help us find points of inflection. What is the Calculus of Variations \u201cCalculus of variations seeks to find the path, curve, surface, etc. Integral calculus involves the area between the graph of a function and the horizontal axis. de on September 24, 2020 by guest Kindle File Format Application Of Vector Calculus In Engineering Field Ppt Eventually, you will certainly discover a supplementary experience and achievement by spending more cash. I\u2019m stuck on a Calculus question and need an explanation. Relational Calculus. It is represented by the symbol \u222b, for example, $$\\int (\\frac{1}{x}) dx = log_e x + c$$. Saiegh Calculus: Applications and Integration. Intro to Limits Limits Presentation Powerpoint on many Calc topics, including Limits to. As the name should hint itself, the process of Integration is actually the reverse/inverse of the process of Differentiation. First let's talk about cost, suppose your business manufactures sneakers, let x be the number of pairs that your company makes. Definite integral is a basic tool in application of integration. Ap Calculus Ppt. One very useful application of Integration is finding the area and volume of \"curved\" figures, that we couldn't typically get without using Calculus. 3C3 * AP\u00ae is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site. We are going to discuss several types of word problems. br-2020-12-12T00:00:00+00:01 Subject: Application Of Vector Calculus In Engineering Field Ppt. The process of finding a derivative is called differentiation. Calculus showed us that a disc and ring are intimately related: a disc is really just a bunch of rings. application of vector calculus in engineering field ppt 2) If is always Solenoidal Vector. Chain rule: One ; Chain rule: Two. A Collection of Problems in Differential Calculus. The following files are in PDF format. Chapter 08 \"Fractional-Order Controller: An Introduction\" of this SIAM Press book is here. Rogers and D. Their predictable behavior makes them easy to graph and manipulate. de on September 24, 2020 by guest Kindle File Format Application Of Vector Calculus In Engineering Field Ppt Eventually, you will certainly discover a supplementary experience and achievement by spending more cash. Integration can be used to find areas, volumes, central points and many useful things. CS344: Artificial Intelligence Pushpak Bhattacharyya CSE Dept. Introduction to Differential Calculus fully engages readers by presenting the fundamental theories and methods of differential calculus and then showcasing how the discussed concepts can be applied to real-world problems in engineering and the physical sciences. Check that your answer is in fact a maxima or minima (by using the second derivative test or a table of values). Function creation \u2212 Church introduced the notation \u03bbx. The links on the right side of this page are for video recordings of the PowerPoint lectures given in AB and BC Calculus class. A measure of how \"popular\" the application is. PowerPoint slides: Writing on the AP Calculus Exam. View (7) Chapter 5 Duration and Convexity. Applications Differential Calculus Ppt. 12/23/20152NDS 3 4. If f is a continuous function on [a,b], then the function. analyses require the application of the tools that are presented throughout the book. Applications of differential calculus in economics\u2026 9 It is worth noting that when the price elasticity of demand is greater than 1, the increase of revenue from sales requires a decrease of the price. A note on examples. 0 (Honors eligible) Description: Advanced Placement Calculus 1 covers selected topics in differential and integral calculus including limits, continuity, motion, graph analysis, optimization, numerical methods and an introduction to differential equations. Population Problems 4. I'll be teaching vector calculus to mechatronics engineers, and I'd like to provide them with industrially relevant examples, especially of the use of vector fields. Such a generalization is not merely a mathematical curiosity but has found applications in various fields of physical sciences. pptx - Free download as Powerpoint Presentation (. School Counselor; Parent Presentations; Handbook; Middle School Math Information; Testing; Class Trips/Events; Clubs; Student Council; Scheduling and Program of Studies. Trigonometry Review with the Unit Circle: All the trig. In ordinary calculus, one dealt with limiting processes in \ufb01nite-dimensional vector spaces (R or Rn), but problems arising in the above applications required a calculus in spaces of functions (which are in\ufb01nite-dimensional vector spaces). Di\ufb00erential calculus is about describing in a precise fashion the ways in which related quantities change. Calculus Videos. Consider structures where two roofs intersect as shown here. Calculus 1 Course Ppt. The slides are animated to maximize student input. In this unit, we will explore some applications of integral calculus. POORNIMA CALCULUS IN BIOLOGY & MEDICINE MATHS IN MEDICINE DEFINITION Allometric growth The regular and systematic pattern of growth such that the mass or size of any organ or part of a body can be expressed in relation to the total mass. The process of finding a derivative is called differentiation. The study of Calculus is one of the most powerful intellectual achievements of the human brain. Contextual Applications of Differentiation 91 UNIT 5: Analytical Applications of Differentiation 109 UNIT 6: Integration and Accumulation of Change 129 UNIT 7: Differential Equations 143 UNIT 8: Applications of Integration 163 UNIT 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions bc only 177 UNIT 10: Infinite Sequences. Mechanical. Derivatives describe the rate of change of quantities. Then we will continue with Windows Forms apps. 1 The Mean Value Theorem 3. It provides them with conceptual understanding and computational skills that are crucial for Basic Calculus and future STEM courses. Textbook: CALCULUS for AP with CalcChat and CalcView by Ron Larson and Paul Battaglia (2017) The course covers the first five chapters (plus a few sections of Chapters 6-8) in course textbook. Mathematical analysis of curves and surfaces had been developed to answer some of the nagging and unanswered questions that appeared in calculus, like the reasons for relationships between complex shapes and curves, series and analytic functions. 3 The Second Derivative and Curve Sketching 3. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Trigonometry Review with the Unit Circle: All the trig. Derivatives 8. Derivations and explanations are based on years of classroom experience on the part of long-time calculus professors, striving for a balance of clarity and rigor that has proven successful. This is just one of the solutions for you to be successful. The study of Calculus is one of the most powerful intellectual achievements of the human brain. The real-world application and its solution must be original (your own work) and/or cited if other work was adapted. Calculus flashcards on the web: Calculus and Pre Calculus on the Web (COW) complete Pre Calc and Calc texts along with sample problems and practice quizzes. Textbook: CALCULUS for AP with CalcChat and CalcView by Ron Larson and Paul Battaglia (2017) The course covers the first five chapters (plus a few sections of Chapters 6-8) in course textbook. Tangents 3. Learn about the various ways in which we can use integral calculus to study functions and solve real-world problems. When two roofs intersect, a valley rafter is formed. This is the first lesson in our series to study C# programming from scratch. pdf), Text File (. For example, the quantity demanded can be said to be a function of price. Clicking on a link will take you to Google Drive. Vector Calculus with Applications Winter 2015 Vector calculus applications Multivariable Calculus since the pressure acts normally to each element of the surface (with an inward force when the pressure is positive, hence the minus sign). br-2020-12-12T00:00:00+00:01 Subject: Application Of Vector Calculus In Engineering Field Ppt. If you need a detailed discussion of index and log laws, then the Mathematics Learning Centre booklet: Introduction to Exponents and Logarithms is the place to start. Relational calculus is a non-procedural query language. Resources and information to support K\u201312 and higher education professionals in helping students prepare for college and career. We learned about e (my favorite letter, naturally) and how the derivative of e^x is just itself. This application [5] demonstrates how operations research was used to enhance productivity and quality of service at KeyCorp, a bank holding company headquartered in Cleveland, Ohio. Calculus: Applications and Integration POLI 270 - Mathematical and Statistical Foundations Sebastian M. Monday, June 22, 2015 Noon BSF/Crick (2008) Nonlocality is a generic feature of multiscale modeling. pptx), PDF File (. Chain rule: One ; Chain rule: Two. Use these to review and reinforce class notes and activities. Discovered by Antiphon Inscribe a shape with multiple polygons whose area converge to the area of the shape. Real Life Applications of Calculus -Biologists can use calculus to do things such as determine the growth of bacteria through differential calculus when several variables are changed. Download Calculus And Techniques Of Optimization With Microeconomic Applications Ebook, Epub, Textbook, quickly and easily or read online Calculus And Techniques Of Optimization With Microeconomic Applications full books anytime and anywhere. While you may not sit down and solve a tricky differential equation on a daily basis, calculus is still all around you. APPLICATIONS OF STOCHASTIC CALCULUS TO FINANCIAL MODELING 3 De\ufb01nitions 2. This chapter provides a brief introduction to some of the many applications of vector calculus to physics. 1 Limits\u2014An Informal Approach 2. It is also applied in fluid dynamics, as well as statics. Other content links are also here. Monday 5/8, 1 FRQ test timed and graded (10). That relationship is that differentiation and integration are inverse processes. APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS. This is the first lesson in our series to study C# programming from scratch. Derivatives 8. you\u2019ll ever need to know in Calculus Objectives: This is your review of trigonometry: angles, six trig. Williams, and Dellacherie and Meyer\u2019s multi volume series \u2018Probabilities et Potentiel\u2019. This list should also explain how each is an application of trigonometry. Applications of Calculus to the Physical World At this point we have learnt quite a bit about calculus, and wish to apply it to the physical world. Calculus and Its Applications has, for years, been a best-selling text for one simple reason: it anticipates, then meets the needs of today's applied calculus student. 8) Sep 30, 2003 \u00b7 100 Calculus Projects: Complete Set of Projects These student projects have been developed by the mathematics department of IUPUI for their introductory calculus sequence. application of vector calculus in engineering field ppt 2) If is always Solenoidal Vector. Course Description. This lesson contains the following Essential Knowledge (EK) concepts for the *AP Calculus course. Extraordinary original hand-woven creations made in Santa Fe, New Mexico. Review Precalculus 2. The following steps are usually easy to carry out and give important clues as to the shape of a curve: Determine the $x$\u2013 and $y$-intercepts of the curve. 2T 10 Total MODULE - II DIFFERENTIAL CALCULUS \u2013 2 1. Posted on January 16, 2021 Written by No Comments on application of calculus in engineering ppt. Calculus Lesson Plan Objectives Coach Gordon Stephens PPT. This is just one of the solutions for you to be successful. Laws of Limits 5. How to use the Definition of the Derivative Practice Problems. Download Free Application Of Vector Calculus In Engineering Field Ppt 17. Calculus After reading this chapter, students will be able to understand: Understand the basics of differentiation and integration. Calculus focuses on the processes of differentiation and integration However, many are uncertain what calculus is used for in real life. Perfect for use as a supplement to any standard multivariable calculus text, in a mathematical methods in physics or engineering class, for independent study, or even as the class text in an honors multivariable calculus course, this textbook will appeal to mathematics, engineering, and physical. Integral calculus involves the area between the graph of a function and the horizontal axis. These basic concepts would be interpreted geometrically and physically, and applied for solving some real problems. Application of Calculus in Commerce and Economics OPTIONAL II Mathematics for Commerce, Economics and Business41NotesAPPLICATION OF CALCULUS IN COMMERCE AND ECONOMICS dy We have learnt in calculus Fill & Sign Online, Print, Email, Fax, or Download. Growth & Decay, Logistic Growth. Road so mathematics shows how to apply calculus, economy department in temperature t of functions. \ufeff Reference: James Stewart, Calculus Early Transcendentals \ufeff (9th edition) For more information, call us now at (408)850-1886. mathematical problems that cannot be solved by other means, and that in turn allows us to. Applications of calculus in medical field TEAM OF RANJAN 17BEE0134 ANUSHA 17BEE0331 BHARATH 17BEC0082 THUPALLI SAI PRIYA 17BEC0005 FACULTY -Mrs. The links on the left side of the page are in the original PowerPoint format. It is represented by the symbol \u222b, for example, $$\\int (\\frac{1}{x}) dx = log_e x + c$$. Intro to Limits Limits Presentation Powerpoint on many Calc topics, including Limits to. Derivatives 8. What might it feel like to invent calculus?Brought to you by you: http://3b1b. - Actual presentations created in PowerPoint with the templates applied - extension PPT files. The easiest rates of change for most people to understand are those dealing with time. Notes on Calculus and Optimization 1 Basic Calculus 1. One way to do this is to graph the function and determine by inspection what the absolute maximum is. Get help on the web or with our math app. Accumulation & Functions Defined by Integrals or Thoughts on , my favorite equation. Download File PDF Application Of Vector Calculus In Engineering Field Ppt Application Of Vector Calculus In Engineering Field Ppt Recognizing the artifice ways to acquire this ebook application of vector calculus in engineering field ppt is additionally useful. Aim of this note is to provide mathematical tools used in applications, and a certain theoretical background that would make other parts of mathematical analysis accessible to the student of physical science. calculus lecture notes ppt, Lecture Notes in Mathematical Biology Eduardo D. Early in the introduction I stated that it was an over-statement to claim that one needs calculus to become a scientist. Discussion: Real-World Applications of Parabolas. These are the tangent line problemand the area problem. Write a description for the parabola. Course Overview. NPTEL provides E-learning through online Web and Video courses various streams. For years my colleagues and I found it hard to get pupils to reflect on the strengths and weaknesses of the hedonic calculus. Calculus is also used in such Page 5/26. Real-life limits are used any time you have some type of real-world application approach a steady-state solution. Functions are one of our most important tools for analyzing phenomena. It includes problem solving, reasoning and estimation, functions, derivatives, application of the derivative, integrals, and application of the integral. Calculus III - Cognella Academic This is the third volume of my calculus series, Calculus I, Calculus II and Calculus III. Christine Heitsch, David Kohel, and Julie Mitchell wrote worksheets used for Math 1AM and 1AW during the Fall 1996 semester. A K TIWARI 28. This chapter provides a brief introduction to some of the many applications of vector calculus to physics. Calculating stationary points also lends itself to the solving of problems that require some variable to be maximised or minimised. PowerPoint slides from the textbook publisher are here, section by section, for the content of Calculus I. Derivatives 8. This page contains PowerPoint slides and related handouts to various talks I have given. These basic concepts would be interpreted geometrically and physically, and applied for solving some real problems. 2 MB) 24: Numerical integration (PDF - 1. Applications of evolutes and involutes restricted to conic sections 4. Definite integral is a basic tool in application of integration. Accumulation & Functions Defined by Integrals or Thoughts on , my favorite equation. Review Precalculus 2. Calculus Problem 44. 3C3 * AP\u00ae is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site. * Presentation \u2013 Complete video for teachers and learners on Applications of Calculus: Rate of Change Turning Points, Quadratic & Cubic Graphs. You can learn how to control a system by studying calculus. Graphing it in a vector space from the plane to the space. Explore our catalog of online degrees, certificates, Specializations, & MOOCs in data science, computer science, business, health, and dozens of other. de on September 24, 2020 by guest Kindle File Format Application Of Vector Calculus In Engineering Field Ppt Eventually, you will certainly discover a supplementary experience and achievement by spending more cash. Contextual Applications of Differentiation 91 UNIT 5: Analytical Applications of Differentiation 109 UNIT 6: Integration and Accumulation of Change 129 UNIT 7: Differential Equations 143 UNIT 8: Applications of Integration 163 UNIT 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions bc only 177 UNIT 10: Infinite Sequences. Physical Applications: Physics Formulas Associated Calculus Problems Mass: Mass = Density * Volume (for 3\u2010D objects) Mass = Density * Area (for 2\u2010D objects) Mass = Density * Length (for 1\u2010D objects) Mass of a one\u2010dimensional object with variable linear. The Fastest Route 46. Description. The Pre-Calculus course bridges Basic Mathematics and Calculus. Students were asked to fill the questionnaire in the form of five-Likert scale that representing the different levels of satisfactory (very disagree, disagree, undecided, agree, very agree). 1: Linear Approximations and Applications. The following files are in PDF format. Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes. Prerequisite: \u201cB\u201d or better in Pre-calculus and teacher recommendation. 4 Optimization 3. Real life applications of calculus Calculus is a part of mathematics and is also used in physics. applications of integral calculus arise whenever the problem is to compute a number that is in principle vector calculus,, purdue's school of mechanical engineering conducts world vector calculus; fundamentals of complex analysis for mathematics, science and engineering. Columbia University. As an example, we could have a chemical reaction in a beaker start with two. An application in which integration is key to reaching the target. 07 Triple Integrals in Cylindrical Coordinates (PowerPoint) 2/22/13. Thus, we take u = 2x2+1. Extra Examples Extra Examples shows you additional worked-out examples that mimic the ones in your book. 2 MB) 24: Numerical integration (PDF - 1. 7 Applications of differential calculus (EMCHH) Optimisation problems (EMCHJ) We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. There are also packets, practice problems, and answers provided on the site. This document is highly rated by CA CPT students and has been viewed 1362 times. Di\ufb00erential calculus is about describing in a precise fashion the ways in which related quantities change. As such, each part was restricted to 8 chapters. Applications of Integral Calculus. This exclusive workshop was by invitation only, and all talks were one hour in length. School Counselor; Parent Presentations; Handbook; Middle School Math Information; Testing; Class Trips/Events; Clubs; Student Council; Scheduling and Program of Studies. See full list on embibe. 1 Graph Quadratic Equations in Standard Form Homework: pages 240-241 (3-39 multiples of 3) Tuesday, November 10 In Class: 4. Such a generalization is not merely a mathematical curiosity but has found applications in various fields of physical sciences. Cost of a commodity depends upon a number of factors. A Gable is a triangle formed by a sloping roof. Examples, practice problems on Calculus. Applications of Differential Equations The Simple Pendulum Theoretical Introduction. One way to do this is to graph the function and determine by inspection what the absolute maximum is. 750 Chapter 11 Limits and an Introduction to Calculus The Limit Concept The notion of a limit is a fundamental concept of calculus. Application of Integral Calculus The ABC Company Ltd. txt) or view presentation slides online. Crisan\u2019s Stochastic Calculus and Applications lectures of 1998; and also much to various books especially those of L. Finding the value of the function between the x values graphically represents the area of the function under the curve within the x limits. Population Problems 4. However, not every student has mastered the basic calculus applications and formulae, that is why they request for our homework help. Pupils are given the worksheet in order to see how simple, yet how time consuming and contradictory this theory is. In structural analysis finite element method is very useful for analysis of multi- storey structures, design of rigid pavements for roads, analysis of particle under various kind of forces like seismic force, wind. ppt from FIN 650 at Pace University. The Fundamental Theorem of Calculus. Pilipovi\u0107, B. calculus: the calculus of optimization 15 Economists in the late 1900s thought that utility might actually be real, some- thing that could be measured using \u201chedonometers\u201d or \u201cpsychogalvanometers\u201d. Differentiate, Derivative, Maximum, Minimum, Turning Point, Stationary, Tangent, Normal Find some real life applications of calculus. The study of this subject mostly confuse the students and they find it almost difficult to carry on with calculus assignment writing. Self-application and Paradoxes Recursion and Fixed Point Equations Fixed Point Equations Least Fixed Point Y : A Fixed Point Operator Mutual Recursion -calculus with Combinator Y Outline Recursion and Y combinator The llet Calculus -calculus with Constants & Letrec Let-block Statements Free Variables of an Expression FV(x) = {x} FV(E1 E2) = FV. The course emphasises the key ideas and historical motivation for calculus, while at the same time striking a balance between theory and application, leading to a mastery of key. Precise Definition of Limit 6. Calculus of a Single Variable 10th EditionRon Larson. If f is a continuous function on [a,b], then the function. Course Description. txt) or view presentation slides online. ppt from FIN 650 at Pace University. Fu Foundation Professor of Applied Mathematics. Real life applications of calculus Calculus is a part of mathematics and is also used in physics. There are several applications of integrals and we will go through them in this lesson. The remainder of the benchmark lesson will involve student groups working through the Position, Velocity, and Acceleration handout. Derivatives describe the rate of change of quantities. Know how to compute derivative of a function by the first principle, derivative of a function by the application of formulae and higher order differentiation. Here is a synopsis of the notes we took in class as well as some problems from on-level Pre-Calculus. APPLICATIONS OF STOCHASTIC CALCULUS TO FINANCIAL MODELING 3 De\ufb01nitions 2. Course Overview. These books are rather advanced. Berkeley\u2019s calculus course. The probability density function for the sum of gamma variates with different shape parameters is found, using the inversion formula and the calculus of residues. pptx - Free download as Powerpoint Presentation (. You may need the free Acrobat Reader. Williams, and Dellacherie and Meyer\u2019s multi volume series \u2018Probabilities et Potentiel\u2019. Enables readers to apply the fundamentals of differential calculus to solve real-life problems in engineering and the physical sciences. Surprisingly, calculus may be one of the things \u2026 to help you do just that. Springer Science & Business Media. Angle Measure Angles can be measured in 2 ways, in degrees or in radians. This is a real Life application video for calculus from the house of LINEESHA!!!Calculus is concerned with comparing quantities which vary in a non-linear. Like all AP courses, AP Calculus is intended to be a college-level class. Calculus Videos. Share your documents with the world , watch,share and upload any time you want. Composed of forms to fill-in and then returns analysis of a problem and, when possible, provides a step-by-step solution. However, it is easier to use differential calculus to find the profit-maximising output. This list should also explain how each is an application of trigonometry. 4 MB) 22: Volumes by disks and shells (PDF - 1. de on September 24, 2020 by guest Kindle File Format Application Of Vector Calculus In Engineering Field Ppt Eventually, you will certainly discover a supplementary experience and achievement by spending more cash. Application of Integral Calculus. This resource aims to change this. The primary objects of study in differential calculus are the derivative of a function, related notions such as the differential, and their applications. These books are rather advanced. It's pretty good! 5/16/2018 - We are reviewing for the final exam. History of the Integral from the 17 th Century. Derivations and explanations are based on years of classroom experience on the part of long-time calculus professors, striving for a balance of clarity and rigor that has proven successful. Calculus Videos. 5 October 2012 (F): Derivative Applications. Roger Day ([email\u00a0protected] Tangents 3. Instructors may download and use these slides for easy classroom presentations. Relational Calculus. An Introduction to Integral Calculus: Notation and Formulas, Table of Indefinite Integral Formulas, Examples of Definite Integrals and Indefinite Integrals, indefinite integral with x in the denominator, with video lessons, examples and step-by-step solutions. An Introduction to the Theory of Reproducing Kernel Hilbert Spaces. Most of us last saw calculus in school, but derivatives are a critical part of machine learning, particularly deep neural networks, which are trained by optimizing a loss function. de on September 24, 2020 by guest Kindle File Format Application Of Vector Calculus In Engineering Field Ppt Eventually, you will certainly discover a supplementary experience and achievement by spending more cash. Calculus and it\u2019s Applications in Business: In business we come across many such variables where one variable is a function of the other. To solve an exponential or logarithmic word problems, convert the narrative to an equation and solve the equation. 4 MB) 22: Volumes by disks and shells (PDF - 1. ppt Ch 8 Integration Application Formula Sheet- Calculus Chapter 7 Formulas sheet. You peer around a corner. Now let's move on to the calculus chapters. Significant examples illustrate each topic, and fundamental physical applications such as Kepler\u2019s Law, electromagnetism, fluid flow, and energy estimation are brought to prominent position. You may click on either the VIDEO link or the YouTube link, whichever works better for you. com-2021-01-28T00:00:00+00:01 Subject: Application Of Vector Calculus In Engineering Field Ppt Keywords: application, of, vector, calculus, in, engineering, field, ppt Created Date: 1/28/2021 1:25:40 PM. Among them are physics, engineering, economics, statistics, and medicine. Calculus \u2013 differentiation, integration etc. but goes on to introduce you to the subject of Vector Calculus which, like it says on the can, combines vector algebra with calculus. 3blue1brown. The following files are in PDF format. between definite and indefinite integrals. Siyavula's open Mathematics Grade 12 textbook, chapter 6 on Differential calculus covering Rules for differentiation. Calculus Videos. Download Free Application Of Vector Calculus In Engineering Field Ppt 17. We will see in this and the subsequent chapters that the solutions to both problems involve the limit concept. 1 Limits\u2014An Informal Approach 2. com-2021-01-28T00:00:00+00:01 Subject: Application Of Vector Calculus In Engineering Field Ppt Keywords: application, of, vector, calculus, in, engineering, field, ppt Created Date: 1/28/2021 1:25:40 PM. Application of Integral Calculus. This is a real Life application video for calculus from the house of LINEESHA!!!Calculus is concerned with comparing quantities which vary in a non-linear. These powerpoint lectures were created by Professor Mario Borelli in Fall 2011. Let s(t) denote the position of the object at time t (its distance from a reference point, such as the origin on the x-axis). There're a lot of applications in Economics for Calculus. The links on the left side of the page are in the original PowerPoint format. A measure of how \"popular\" the application is. To proceed with this booklet you will need to be familiar with the concept of the slope (also called the gradient) of a straight line. Mechanical engineering. Each of the eight chapters introduces a new topic, and to facilitate understanding of the material, frequent reference is made. Description. Introduction The divergence and Stokes\u2019 theorems (and their related results) supply fundamental tools which can be used to derive equations which can be used to model a number of physical situations. \u2022Analysis: logical, rigorous proofs of the intuitive ideas of calculus. See full list on intmath. Definite integrals can be used to determine the mass of an object if its density function is known. This list should also explain how each is an application of trigonometry. certain functions, discuss the calculus of the exponential and logarithmic functions and give some useful applications of them. Download File PDF Application Of Vector Calculus In Engineering Field Ppt Application Of Vector Calculus In Engineering Field Ppt Recognizing the artifice ways to acquire this ebook application of vector calculus in engineering field ppt is additionally useful. Unlike basic arithmetic or finances, calculus may not have obvious applications to everyday life. Examples, practice problems on Calculus. Wiley-ISTE, Croydon, 2014. txt) or view presentation slides online. Emphasis will be placed on the study of functions, and their graphs, inequalities, and linear, quadratic, piece-wise defined, rational, polynomial, exponential, and logarithmic functions. To solve an exponential or logarithmic word problems, convert the narrative to an equation and solve the equation. Applications of evolutes and involutes restricted to conic sections 4. Share yours for free!. 3C3 * AP\u00ae is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site. The sandwich or squeeze method is something you can try when you can\u2019t solve a limit problem with algebra. Applications of Stochastic Calculus to Finance Scott Stelljes University of North Florida This Master's Thesis is brought to you for free and open access by the Student Scholarship at UNF Digital Commons. Los Angeles Harbor College. as a work of art. One very useful application of Integration is finding the area and volume of \"curved\" figures, that we couldn't typically get without using Calculus. This lesson contains the following Essential Knowledge (EK) concepts for the *AP Calculus course. Example 1 Finding a Rectangle of Maximum Area. Taylor\u2019s and Maclaurin\u2019s series. Remember, we can use the first derivative to find the slope of a function. pptx - Free download as Powerpoint Presentation (. I'll be teaching vector calculus to mechatronics engineers, and I'd like to provide them with industrially relevant examples, especially of the use of vector fields. Mathematics at Dartmouth. Explore our catalog of online degrees, certificates, Specializations, & MOOCs in data science, computer science, business, health, and dozens of other. Accumulation & Functions Defined by Integrals or Thoughts on , my favorite equation. 05 Applications of Double Integrals 2/24/16. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Calculus on the Web was developed with the support of the National Science Foundation COW is a project of Gerardo Mendoza and Dan Reich Temple University. View Application Of Calculus In Real Life PPTs online, safely and virus-free! Many are downloadable. How to use the Definition of the Derivative Practice Problems. As an example, we could have a chemical reaction in a beaker start with two. NPTEL provides E-learning through online Web and Video courses various streams.", "date": "2021-04-15 05:41:22", "meta": {"domain": "1upload.de", "url": "http://zveg.1upload.de/application-of-calculus-ppt.html", "openwebmath_score": 0.49587282538414, "openwebmath_perplexity": 1150.4617944075264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9890130593124399, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.8711198220083848}}
{"url": "https://math.stackexchange.com/questions/1079251/chevron-symbols-in-commutator", "text": "# Chevron Symbols in Commutator\n\nI am reading a chapter on Commutator in Group Theory and came across chevron symbols \"$\\langle$\" and \"$\\rangle$\" like these:\n\nQuestion #1: Let $E$ and $F$ be non-empty subsets of $G$, we set $$[E, F] := \\langle [e,f] \\mid e \\in F, f \\in F \\rangle.$$\n\nCorrect me if I am wrong here: The chevron symbols here do not have special meaning except that they are differentiating from \"$\\{$\" and \"$\\}.$\" The text does not give any explanation.\n\nAnd then on the same page I saw this problem:\n\nQuestion #2: Let $H$ be a subgroup of $G.$ Show that $[H, g] = [H, \\langle g \\rangle]$ for each element $g$ in $G.$\n\nHere the chevron signs have special meaning, they refer to $g$ as generator. If I am correct, how do you go about solving this problem? My understanding that I have to start from the LHS, without assuming that $g$ is a generator. Please help and thanks for your time.\n\nPOST SCRIPT - 1: ~~~~~~~~~~~~~~~~~~~~~~\nI am following hints from \"someone you know\" in solving Question #2 like these:\n\n\\begin{align*}[H,g] &= \\{hgh^{-1}g^{-1} \\mid h \\in H \\}\\\\ &= \\{hg (g^{n-1})(g^{n-1})^{-1}h^{-1}g^{-1} \\mid h \\in H \\} \\\\ &= \\{h(g g^{n-1})(g^{n-1})^{-1}h^{-1}g^{-1} \\mid h \\in H \\} \\\\ &= \\{hg^n(g^{n-1})^{-1}h^{-1}g^{-1} \\mid h \\in H \\} \\\\ &= ... \\\\ &= ... \\\\ &= [H, \\langle g \\rangle] \\end{align*}\\\\\n\nHow do you go from the 4th. line to the next to move the $(g^n)^{-1}$ to the right side of $h^{-1}$ without resorting to the group being abelian? Thank you again.\n\nPOST SCRIPT - 2: ~~~~~~~~~~~~~~~~~~~~~~\nSince I did not receive any feedback after the Post Script - 1, and since on the subsequent pages of the same class note I saw a very similar question like this:\n\nLet $G$ be a group and let $g \\in G,$ and let $N$ be commutative normal subgroup of $G.$ Show that $[N, \\langle g \\rangle] = \\{[n, g] \\} \\mid n \\in N \\}.$\n\nDo you agree with me that there is actually a typo in the Question #2, in that $G$ and therefore $H$ should be declared as commutative in the first place? Thanks for your time and feedback.\n\n\u2022 By \"they refer to $g$ as [a] generator\" you mean \"$\\langle g\\rangle$ refers to the cyclic subgroup generated by $g$.\" \u2013\u00a0anon Dec 23 '14 at 22:34\n\u2022 Another name for $\\langle \\rangle$ is angle brackets. \u2013\u00a0Matthew Towers Dec 24 '14 at 14:01\n\nIn general, if $S$ is any subset of a group $G$, the notation $\\langle S\\rangle$ refers to the subgroup generated by the elements of $S$, so $$[E, F] := \\langle [e,f] \\mid e \\in F, f \\in F \\rangle$$ is the subgroup of $G$ generated by commutators coming from elements of $E$ and $F$. Thus, the notation $\\langle g\\rangle$ is a special case, in that it refers to $\\langle \\{g\\}\\rangle$.\n\nThe problem you refer to is asking you to show that $\\langle S_1\\rangle=\\langle S_2\\rangle$ where \\begin{align*} S_1&=\\{hgh^{-1}g^{-1}:h\\in H\\}\\\\ S_2&=\\{hkh^{-1}k^{-1}:h\\in H,k\\in\\langle g\\rangle\\}=\\{hg^nh^{-1}g^{-n}:h\\in H, n\\in\\mathbb{Z}\\} \\end{align*}\n\n\u2022 So, the $\\langle g \\rangle$ refers to generator? Thanks again. \u2013\u00a0A.Magnus Dec 23 '14 at 22:25\n\u2022 It means \"generated by\" rather. \u2013\u00a0darij grinberg Dec 24 '14 at 14:23\n\u2022 @someone you know : To \"Someone You Know\": Could you please help me with the Post Script - 2 above? To all other heavyweights beside \"Someone You Know,\" you are more than welcome to pitch in. Thanks again. \u2013\u00a0A.Magnus Dec 26 '14 at 4:25\n\n(1). $[E, F] := \\langle [e,f] \\mid e \\in F, f \\in F \\rangle$ means the subgroup of $G$ generated by the elements of the form $[e,f], e \\in E, f \\in F.$\n\n(2). $\\langle g\\rangle$ means the subgroup of $G$ generated by the element $g.$\n\nYou can look at here or at any algebra text book.", "date": "2021-07-23 20:04:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1079251/chevron-symbols-in-commutator", "openwebmath_score": 0.9996929168701172, "openwebmath_perplexity": 335.86208880384606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561712637257, "lm_q2_score": 0.8991213711878918, "lm_q1q2_score": 0.871119289190492}}
{"url": "https://math.stackexchange.com/questions/2002780/what-is-the-general-formula-for-a-convergent-infinite-geometric-series", "text": "# What is the general formula for a convergent infinite geometric series?\n\nThis question is related, but different, to one of my previous questions (Does this infinite geometric series diverge or converge?). To avoid the previous question getting off-topic, I have created a separate question.\n\nI'm looking for the general formula of a convergent infinite geometric series. I want to be able to calculate any convergent infinite geometric series I come across, regardless of where it starts at. Some examples of this are:\n\n$$\\sum_{n=0}^\\infty ar^n$$\n\n$$\\sum_{n=1}^\\infty ar^n$$\n\n$$\\sum_{n=2}^\\infty ar^n$$\n\n...\n\n$$\\sum_{n=5}^\\infty ar^n$$\n\n...\n\netc.\n\nI would appreciate it if someone could present such a formula and explain the reasoning behind it. Also, please illustrate how the formula can be applied to the above examples.\n\nThank you.\n\n\u2022 If you have a formula for the first one, you have it for all of them by simply factoring out the correct power of $r$. And this formula is well known. \u2013\u00a0Mathematician 42 Nov 7 '16 at 0:29\n\u2022 @Mathematician42 Can you please elaborate? I have the formula $\\dfrac{a-ar^n}{1-r}$. \u2013\u00a0The Pointer Nov 7 '16 at 0:30\n\u2022 If $|r|<1$, then $\\sum_{n\\geq 0} ar^n=a\\sum_{n\\geq 0}r^n=\\frac{a}{1-r}$. Now notice that $\\sum_{n\\geq m}ar^n=ar^m\\sum_{n\\geq 0}r^{n}$. \u2013\u00a0Mathematician 42 Nov 7 '16 at 0:32\n\u2022 @Mathematician42 I do not understand the second part of your explanation. \u2013\u00a0The Pointer Nov 7 '16 at 0:35\n\u2022 We have $\\sum_{n\\geq m}ar^n=ar^m\\sum_{n\\geq 0}r^{n}=r^m(a\\sum_{n\\geq 0}r^n)=r^m\\frac{a}{1-r}$. If you understand why $\\sum_{n\\geq 0}r^n=\\frac{1}{1-r}$ when $|r|<1$, then there should be no problem in understanding everything else I'm saying. \u2013\u00a0Mathematician 42 Nov 7 '16 at 0:39\n\nIn general, you have the finite geometric series given by $$\\sum\\limits_{n=0}^{N-1}ar^n = \\frac{a(1-r^N)}{1-r}.$$\n\nTaking the limit of $N\\to \\infty$ you have the infinite geometric series given by $$\\sum\\limits_{n=0}^\\infty ar^n = \\frac{a}{1-r}$$ which converges if and only if $|r|<1$. Now we will consider starting index $N$ instead, i.e. $\\sum\\limits_{n=N}^\\infty ar^n$.\n\nNotice that\n\n$$\\sum\\limits_{n=0}^\\infty ar^n = \\sum\\limits_{n=0}^{N-1} ar^n + \\sum\\limits_{n=N}^\\infty ar^n = \\frac{a}{1-r}$$ and by isolating the desired term we get $$\\sum\\limits_{n=N}^\\infty ar^n = \\frac{a}{1-r} - \\sum\\limits_{n=0}^{N-1} ar^n.$$ The last term is exactly the finite geometric series and hence we get $$\\sum\\limits_{n=N}^\\infty ar^n = \\frac{a}{1-r} - \\frac{a(1-r^N)}{1-r}.$$\n\nSimplifying we get $$\\bbox[5px,border:2px solid red]{\\sum\\limits_{n=N}^\\infty ar^n = \\frac{ar^N}{1-r}.}$$\n\n\u2022 Exactly correct. That being said, I find it much much easier to remember as $\\frac{\\text{first term}}{1-\\text{common ratio}}$ \u2013\u00a0erfink Nov 7 '16 at 1:06\n\u2022 Great explanation. Thank you very much. \u2013\u00a0The Pointer Nov 7 '16 at 1:13\n\n$$\\sum_{n=N}^\\infty ar^n=\\sum_{n=0}^\\infty ar^{n+N}=r^N\\sum_{n=0}^\\infty ar^n=r^N\\left(\\frac{a}{1-r}\\right)=\\frac{ar^N}{1-r}\\ \\forall\\ |r|<1$$\n\nThe first step was re-indexing: $\\displaystyle\\sum_{n=N}^\\infty b_n=b_N+b_{N+1}+\\dots=\\sum_{n=0}^\\infty b_{n+N}$\n\nThe second step was factoring the $r^N$ term out.\n\nThe last steps was using well-known geometric series formula, followed by some algebra and checking convergence.\n\nIn my opinion, the simplest way to memorize the formula is\n\n$$\\frac{\\text{first}}{1 - \\text{ratio}}$$\n\nSo whether you're computing\n\n$$\\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\ldots$$\n\nor\n\n$$\\sum_{n=3}^{\\infty} 2^{-n}$$\n\nor\n\n$$\\sum_{n=0}^{\\infty} \\frac{1}{8} 2^{-n}$$\n\nyou can quickly identify the sum as\n\n$$\\frac{ \\frac{1}{8} }{1 - \\frac{1}{2}}$$\n\nSimilarly, for a finite geometric sequence, a formula is\n\n$$\\frac{\\text{first} - \\text{(next after last)}}{1 - \\text{ratio}}$$\n\nThe infinite version can be viewed as a special case, where $(\\text{next after last}) = 0$.\n\nI find this formula more convenient written as $$\\frac{\\text{(next after last)} - \\text{first}}{\\text{ratio} - 1}$$\n\ne.g.\n\n$$2 + 4 + 8 + \\ldots + 256 = \\frac{512 - 2}{2 - 1}$$\n\nBut in a pinch, you can always just rederive the formula since the method is simple:\n\n\\begin{align}(2-1) (2 + 4 + 8 + \\ldots + 256) &= (4 - 2) + (8 - 4) + (16 - 8) + \\ldots + (512 - 256) \\\\&= 512 - 2 \\end{align}", "date": "2019-09-22 16:51:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2002780/what-is-the-general-formula-for-a-convergent-infinite-geometric-series", "openwebmath_score": 0.9252971410751343, "openwebmath_perplexity": 279.0794507593227, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787883738261, "lm_q2_score": 0.8824278741843883, "lm_q1q2_score": 0.8711140796646355}}
{"url": "https://www.tutorialspoint.com/sum-of-first-n-natural-numbers-which-are-divisible-by-x-or-y", "text": "# Sum of first N natural numbers which are divisible by X or Y\n\nCServer Side ProgrammingProgramming\n\nAdding up all natural numbers up to n, that are divisible by X or Y is selecting all the numbers that are divisible by X or Y and adding them to a variable that stores the sum.\n\nTo find the sum of the first N natural numbers which are divisible by X or Y, there are two methods \u2212\n\n\u2022 Using loops and conditional statements\n\u2022 Using formula\n\n## Method 1 \u2212 Using loops and conditional statements\n\nThis method uses a loop that counts up to n numbers and selects numbers that are divisible by X or Y and adds them and save to a variables at each iteration.\n\n## Example Code\n\nLive Demo\n\n#include <stdio.h>\nint main(void) {\nint n = 54;\nint x = 2 ;\nint y = 5;\nint sum = 0;\nfor(int i = 0; i<= n; i++) {\nif(i%x == 0 || i% y == 0)\nsum = sum + i;\n}\nprintf(\"sum of %d natural numbers divisible by %d and %d is %d\" ,n,x,y,sum);\nreturn 0;\n}\n\n## Output\n\nsum of 54 natural numbers divisible by 2 and 5 is 881\n\n## Method 2 \u2212 Using formula ,\n\nThis method uses the formula to find the sum of the first n number divisible by a number.\n\nThis can be found using the formula \u2212 SN/X = ((N/X)/2) * (2 * X + (N/X - 1) * X)\n\nUsing this formula the sum of n natural numbers divisible by x is found \u2212 S n/x = ((n/x)/2) * (2 * x + (n/x - 1) * x)\n\nUsing this formula the sum of n natural numbers divisible by y is found \u2212 S n/y = ((n/y)/2) * (2 * y + (n/y - 1) * y)\n\nNow, Using this formula the sum of n natural numbers divisible by x and y is found : S n/x*y = ((n/(x*y)/2) * (2 * (x*y) + (n/(x*y) - 1) * (x*y))\n\nNow, we will add the sum of x and sum of y and subtract sum of x*y that are add two times.\n\n## Example Code\n\nLive Demo\n\n#include <stdio.h>\nint main() {\nint n = 54;\nint x = 2, y = 5;\nint Sx, Sy, Sxy, sum;\nSx = ((n / x)) * (2 * x + (n / x - 1) * x) / 2;\nSy = ((n / y)) * (2 * y + (n / y - 1) * y) / 2;\nSxy= ((n / (x * y))) * (2 * (x * y) + (n / (x * y) - 1) * (x * y))/ 2;\nsum = Sx + Sy - Sxy;\nprintf(\"sum of %d natural numbers divisible by %d and %d is %d\" ,n,x,y,sum);\nreturn 0;\n}\n\n## Output\n\nsum of 54 natural numbers divisible by 2 and 5 is 881\n\nThe second method is better because it does not use any loop that means better time complexity. But if the input cases are smaller than first one can also be used. But for large input cases the second method is not the best option.\n\nPublished on 15-Jul-2019 12:14:00\nAdvertisements", "date": "2021-07-26 02:31:42", "meta": {"domain": "tutorialspoint.com", "url": "https://www.tutorialspoint.com/sum-of-first-n-natural-numbers-which-are-divisible-by-x-or-y", "openwebmath_score": 0.1966809630393982, "openwebmath_perplexity": 857.339821365584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787853562027, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.8711140678448519}}
{"url": "http://math.stackexchange.com/questions/125165/a-group-of-order-2n-has-an-element-of-order-2", "text": "# A group of order $2n$ has an element of order $2$\n\nI'm having problems proving this. The full question is:\n\n\"Let $G$ be a group which order is a pair number. Show that $G$ has an element of order $2$\".\n\nCan anyone give me a hint?\n\n-\nIf $x\\neq x^{-1}$, remove both $x$ and $x^{-1}$ from the set $G$. Repeat as many times as you can. How many elements remain in the end? What can you say about all but one of them? \u2013\u00a0 Jyrki Lahtonen Mar 27 '12 at 19:16\nOhh, I think I got this. If it is as stated, then every element $g$ of $G$ is such that $|g|>2$. But $|g|=|g^{-1}|$, and since the neutral element $e$ of $G$ is the only one element with order 1, that leaves an odd number of elements left on the group. If we separate in pairs each element and its inverse, there will be only one element left - and this element \"$h$\" must be its self-inverse, that is, $h^2=e$, which proves the question. \u2013\u00a0 Marra Mar 27 '12 at 19:16\nIt follows from Cauchy's theorem but I like @Balin's answer better. \u2013\u00a0 MJD Mar 27 '12 at 19:17\n@GustavoMarra, not only one element left. See my answer. \u2013\u00a0 lhf Mar 27 '12 at 19:45\nYeah, not only one element left. But we're supposing that every element of G has order greater than 2 (in what I said). That supposition leads to an absurd. \u2013\u00a0 Marra Mar 27 '12 at 22:43\n\nIf there is no element of order 2 then show that $G$ has odd number of elements. (Hint: think of elements and inverses)\n\n-\n\nHint $\\$ Inversion $\\rm\\:x\\to x^{-1}\\:$ is an involution $\\rm\\: (x^{-1})^{-1} = x,$ so the cycles (orbits) of this permutation partition $\\rm\\:G\\:$ into orbits of length $2$ or $1$. Since $\\rm\\:|G|\\:$ is even so too is the number of length $1$ orbits, i.e. fixed points $\\rm\\:a = a^{-1};\\:$ these include $\\rm\\:a = 1,\\:$ hence, having even cardinality, must include at least one other fixed-point, necessarily of order $2$ by $\\rm a^{-1}=a\\:$ $\\Rightarrow$ $\\rm\\:a^2 = 1$ but $\\rm\\:a\\ne 1$.\n\nFor an analogous application of orbit decomposition (without parity) see my prior answer today on Wilson's theorem for groups.\n\n-\n\nPartition $G = E \\cup A \\cup B$, where $E=\\{e\\}$, $A=\\{x \\in G : x = x^{-1}, x\\ne e\\}$, $B=\\{x \\in G : x \\ne x^{-1}\\}$. Then $E$ has one element and $B$ has an even number of elements because $B$ is invariant under inversion. Since $G$ has an even number of elements, $A$ must have an odd number of elements. In particular, $A$ has at least one element. Every element in $A$ has order 2.\n\n-", "date": "2015-05-29 14:37:54", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/125165/a-group-of-order-2n-has-an-element-of-order-2", "openwebmath_score": 0.9301088452339172, "openwebmath_perplexity": 197.74503669908603, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969636371975, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8711044389410035}}
{"url": "https://math.stackexchange.com/questions/2723813/two-ways-to-represent-ex-falso", "text": "Two ways to represent ex falso?\n\nAccording to ProofWiki, the principle of explosion says that:\n\n$$p \\implies (\\lnot p \\implies q)$$\n\nbut sometimes it's also stated as\n\n$$(p \\land \\lnot p) \\implies q$$\n\nCan one be converted to the other? How do you prove or show these?\n\nIf we know that $(p \\implies q) \\equiv \\lnot p \\lor q$ by definition, then substituting the 2nd representation:\n\n$$((p \\land \\lnot p) \\implies q) \\equiv \\lnot(p \\land \\lnot p) \\lor q \\equiv \\lnot p \\lor p \\lor q$$\n\nIs this just true due to law of excluded middle? Like either $p$ or $\\lnot p$ is true, i.e. $p \\lor \\lnot p \\equiv \\text{T}$ before we even get to whatever $q$ is. Is that why we're able to say it?\n\nThen the first representation is weird too:\n\n$$(p \\implies (\\lnot p \\implies q)) \\equiv (\\lnot p \\lor (\\lnot p \\implies q)) \\equiv (\\lnot p \\lor (p \\lor q)) \\equiv \\lnot p \\lor p \\lor q$$\n\nI mean am I even on the right track here? How do you normally get these without \"working backwards\" and showing that they happen to be equivalent? Is there a more intuitive derivation of both of these?\n\n\u2022 Yes, the two are equivalent. \u2013\u00a0Namaste Apr 5 '18 at 18:50\n\u2022 @amWhy Yes but I am asking something different than that \u2013\u00a0user525966 Apr 5 '18 at 18:52\n\u2022 @user525966 Those are just two ways to write that a false premise implies any conclusion. \u2013\u00a0dxiv Apr 5 '18 at 18:56\n\u2022 @Taroccoesbrocco you say the equivalence has nothing to do with excluded middle (and i don\u2019t think it does) but you use the non intuitionistically valid demorgan law in your proof. \u2013\u00a0spaceisdarkgreen Apr 5 '18 at 19:28\n\u2022 @user525966 The \u201creason\u201d behind the classical validity of a proposition can be difficult to untangle cause it all blends together. Intuitionistic logic is an example where excluded middle is not valid but explosion is. So inasmuch as intuitionistic logic is philosophically coherent for accepting one and not the other (there is another called \u201cminimal logic\u201d that rejects both), LEM cannot be the \u201creason\u201d for explosion. \u2013\u00a0spaceisdarkgreen Apr 5 '18 at 19:49\n\nThe formulas $p \\to (\\lnot p \\to q)$ and $(p \\land \\lnot p) \\to q$ are equivalent because they are particular instances of a more general fact: the formulas $p \\to (r \\to q)$ and $(p \\land r) \\to q$ are equivalent. You can prove that checking that the two formulas actually have the same truth table. An alternative proof is the following:\n\n\\begin{align} p \\to (r \\to q) &\\iff \\lnot p \\lor (\\lnot r \\lor q) &\\text{by definition of } \\to \\\\ &\\iff (\\lnot p \\lor \\lnot r) \\lor q &\\text{by associativity of } \\lor \\\\ &\\iff \\lnot(p \\land r) \\lor q &\\text{by De Morgan law} \\\\ &\\iff (p \\land r) \\to q &\\text{by definition of }\\to \\end{align}\n\n\u2022 @user525966 - Please, see my last edit. \u2013\u00a0Taroccoesbrocco Apr 5 '18 at 18:57\n\nThe two statements you post are logically equivalent.\n\n\\begin{align} p \\to (\\lnot p \\to q) &\\equiv \\lnot p \\lor (p \\lor q) \\tag{definion of \\to}\\\\ \\\\ &\\equiv (\\lnot p \\lor p) \\lor q \\tag{associativity of \\lor}\\\\ \\\\ &\\equiv \\lnot (p \\land \\lnot p) \\lor q \\tag{DeMorgan's}\\\\ \\\\ &\\equiv (p\\land \\lnot p)\\to q\\tag{definition of \\to}\\end{align}\n\nYou were heading in right direction, starting instead from $(p\\land \\lnot p)\\to q$ and moving toward $p \\to (\\lnot p \\to q)$. You just stopped too early. Note each direction can be proven, because we have a string of equivalencies: so to reverse the direction, just reverse the order of the steps.\n\n\\begin{align} (p\\land \\lnot p)\\to q &\\equiv \\lnot (p \\land \\lnot p)\\lor q\\tag{definition of \\to}\\\\ \\\\ &\\equiv (\\lnot p \\lor p) \\lor q\\tag{DeMorgan's}\\\\ \\\\ &\\equiv \\lnot p\\lor (p \\lor q)\\tag{associativity of \\lor}\\\\ \\\\ &\\equiv p\\to (\\lnot p \\to q)\\tag{definition of \\to} \\end{align}\n\nFirst of all, we have $$p\\to (q\\to r) \\vdash (p \\land q) \\to r\\tag{importation}$$ and we have $$(p\\land q) \\to r \\vdash p \\to (q\\to r)\\tag{exportation}$$ Your question is a specific case of these, where $q = \\lnot p$. See this link for more information.\n\n\u2022 How / why are you using $\\vdash$ here? \u2013\u00a0user525966 Apr 5 '18 at 20:07\n\u2022 The turnstyle, $\\vdash$ is the syntactic entailment operator. $\\phi \\vdash \\psi$, symbolises that $\\psi$ is derivable from $\\phi$ ( or \"...can be proven from...\" ) using the syntactic transformations of some logic system. @user525966 \u2013\u00a0Graham Kemp Apr 6 '18 at 1:05\n\nTo get pernickety, we need to distinguish the principle about entailment what is usually called Explosion, i.e.\n\n$(A \\land \\neg A) \\vdash B$,\n\nfrom the following principle that a conditional with a contradictory antecedent is always derivable:\n\n$\\vdash (A \\land \\neg A) \\to B$.\n\nYou can have logics with one principle and not the other.\n\n\u2022 What's the difference between $\\vdash$ and $\\to$? I know the former is more \"is derivable from\" but I always thought that was what implies was, really \u2013\u00a0user525966 Apr 5 '18 at 19:42\n\u2022 @user525966 yes one is \u201cderivable from\u201d and is a metalogical relation (ie you use it to talk about the formal system) and the other is \u201cimplies\u201d and is a logical connective (used within formal statements in the logic). They do oftentimes correspond exactly in the sense that we have $\\vdash A\\to B$ iff we have $A\\vdash B$. One direction directly follows from the inference rule modus ponens if present. The other direction is called a deduction theorem and can be proved to apply to (for instance) one\u2019s favorite deductive system for first order logic. \u2013\u00a0spaceisdarkgreen Apr 5 '18 at 20:05\n\u2022 @user525966 not sure why I forgot to add this yesterday but in natural deduction systems where this holds, the two directions are unified as introduction and elimination rules for implication. Whereas, the \u201cdeduction theorem\u201d direction being a nontrivial metatheorem, as described above, is typical in Hilbert-style systems. \u2013\u00a0spaceisdarkgreen Apr 7 '18 at 2:07\n\u2022 What is metalogic, metatheory, exactly? \u2013\u00a0user525966 Apr 7 '18 at 2:39\n\u2022 @user525966 (You have to @ me so I get pinged). In logic, it is useful to distinguish between statements/proofs/theorems in formal systems and statements and results about formal systems (usually proven informally, using \"ordinary mathematical reasoning\", though we could formalize this too). The latter are called \"meta_\". \"$\\vdash A$\" means \"there is a formal proof of the statement $A$ in the deductive system under consideration\"... that's a meta-statement about the system (and if true, $A$ is a (regular old) theorem of the system). See also: en.wikipedia.org/wiki/Metatheorem \u2013\u00a0spaceisdarkgreen Apr 7 '18 at 3:06", "date": "2019-12-06 01:10:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2723813/two-ways-to-represent-ex-falso", "openwebmath_score": 0.9910699725151062, "openwebmath_perplexity": 870.0804248692627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969650796875, "lm_q2_score": 0.8856314723088733, "lm_q1q2_score": 0.8711044283420631}}
{"url": "https://pipburner.com/rajgir-mla-unngu/2ca157-j-b-weld-plastic-putty", "text": "In Googlesheets and Excel, =PERCENTILE.EXC(A2:A234,0.25) returns quartile 1 for the scores in cells A2 through A234 (our 233 reaction times). The lower quartile $Q_1$ is the point such that the area up to $Q_1$ is one quarter of the total area. The other name for quartile is basically cuts. Use MathJax to format equations. To work out where the quartiles are, we need to use our equations again: $$\\frac{(40)}{4}={10}$$ so the LQ is the 10th number. What is the third quartile when there are no data above the median? It is possible to fit up to six Distribution Functions simultaneously from a total of 19 continuous and discrete distributions. The \"Interquartile Range\" is from Q1 to Q3: To calculate it just subtract Quartile 1 from Quartile 3, like this: Example: The Interquartile Range is: Q3 \u2212 Q1 = 7 \u2212 4 = 3. There are several quartiles of an observation variable. Thanks for contributing an answer to Stack Overflow! Making statements based on opinion; back them up with references or personal experience. Frequency distributions and goodness of fit tests are displayed for fitted functions. It will put the complete list of numbers in an order. The lower quartile is the value at the first quarter (once your data has been put in order). Below is the comparison of a Histogram vs. a Box Plot. Answers. To find the lower quartile or the value that is one quarter of the way along the list, count how many numbers there are, add 1 and divide by 4. From the image attached it is clear that the median is above 275 and it lies in the 13\u201414 age bar. Yes No. Recently Asked Questions. 23. Histograms: the maths GCSE test scores of 280 students are shown in the histogram below Answers: 2 Show answers Other questions on the subject: Mathematics. I'm [suffix] to [prefix] it, [infix] it's [whole]. Use histograms when you have continuous measurements and want to understand the distribution of values and look for outliers. You are allowed to be out by a little but be as accurate as you can. So the interquartile range is the difference between the upper quartile and lower quartile. (Histogram) Why is the lower quartile in interval 36-43? and I also have two like that, except with an odd number of bars. {31, 38, 49, 58, 62} ... histogram. She did remember the mean was exactly 80, the median was 81, and the mode was 88. Data are observations (measurements) of some quantity or quality of something in the world. Question. How do I determine the inter quartile range of data? The quartiles are not the same as the lower and upper hinge calculated in the five-number summary. The latter two are, respectively, the median of the lower and upper half of your data, and they differ slightly from the first and third quartiles. The histogram is still preferable for a more detailed look at the shape of the distribution. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. $\\begingroup$. To fine the inter quartile range you'd do what ever you got for you upper quartile, minus what ever you got for your lower quartile. Estimating values for the median and quartiles for grouped data than ungrouped data. 2. Click here for Answers . The first quartile, or lower quartile, is the value that cuts off the first 25% of the data when it is sorted in ascending order.The second quartile, or median, is the value that cuts off the first 50%.The third quartile, or upper quartile, is the value that cuts off the first 75%.. Box and Whisker Plot. Do you have to see the person, the armor, or the metal when casting heat metal? which equation can be used to find v, the volume of the cylinder in cubic centimeters? Previous Scatter Graphs Practice Questions. If there is an even number of data items, then we need to get the average of the middle numbers. A Histogram shows the frequency of a set of data between a what? The left (or lower) edge at the first quartile, q 1. What was wrong with John Rambo\u2019s appearance? In this chapter, we will see several types of \u2026 There are several quartiles of an observation variable. To find the lower quartile using the formula, use Q1 = 1/4(n+ 1), or enter 1 into the Excel function instead of 3. \u2022 Stem-and-leaf and histogram give general visual impression \u2022 On a rectangular box, aligned either horizontally or vertically, a box plot displays three quartiles. Histogram. Both histogram and boxplot are good for providing a lot of extra information about a dataset that helps with the understanding of the data. To find the upper quartile you'd go to 60, as this is 40 and the lower quartile or 20. To determine the median first find the median location. The right (or upper) edge at the third quartile q 3. This channel is managed by up and coming UK maths teachers. Thanks for contributing an answer to Mathematics Stack Exchange! A positive skewed histogram suggests the mean is greater than the median. Print a conversion table for (un)signed bytes. How do you find the upper and lower quartile for a histogram with unequal class widths? Lower quartile (middle value of the lower half) = 12. It is found using: $$\\frac{1}{4}$$ of the (number of values + 1) or $$\\frac{(11+1)}{4}$$ = 3. Item 16 divide. rev\u00a02021.1.15.38327, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. ... What is his lower quartile of basketball points? Do you find the quartiles and medians by dividing up the bars (e.g. Histogram vs. A histogram is used in statistics to plot grouped data. We came across quartiles in Chapter 1 when discussing measures of variation for skewed data. It can be helpful to plot two variables in the same boxplot to understand how one affects the other. Fig. Hope this helps. To find the number of Q1 for grouped data, use n/4. But avoid \u2026. Histogram. $$\\frac{3(40)}{4}={30}$$ so the UQ is the 30th number. This just means that you're assuming each value in the histogram is distributed evenly in each class - the proportion of how many items into its class the median is = the proportion of how far the value of the median is into the range of its class. The upper quartile is the value which is three quarters of the way into our data: $$\\frac{3}{4}$$ of the (number of values+ 1) or $$\\frac{3{(11+1)}}{4}$$ = 9. Biology, 23.07.2019 00:30. Funnily enough, this use to be really easy for me. Not Helpful 6 Helpful 15. b) Find the lower quartile of the scores. 17) The histogram shows information about the times taken by some students to finish a puzzle. It is possible to fit up to six Distribution Functions simultaneously from a total of 19 continuous and discrete distributions. So Q1= 64+64/2=64 and Q3= 77+81/= 79. Primary Study Cards. Question. These were defined as the data values which had: 1 4 of the data values below it: lower quartile 2 4 of the data values below it: median 3 4 of the data values below it: upper quartile The lower quartile was the 2nd number in the list, so the upper quartile must be the 6th number in the list ($$2 \\times 3 = 6$$). Clash Royale CLAN TAG #URR8PPP Also on Histogram of both the variables Ozone and Wind, we can see that there is a gap between observations at extreme i.e. Quartiles: lower quartile = 2 upper quartile = 11 highest value = 15 Median = 7 Lowest value = 0 Hence our boxplot looks like this: /**/ Like stem-and-leaf diagrams, these diagrams are best used when we have a relatively small set of data or when the data has already been summarised into its median and quartiles. You are required to calculate all the 3 quartiles.Solution:Use the following data for the calculation of quartile.Calculation of Median or Q2 can be done as follows,Median or Q2 = Sum(2+3+4+5+7+8+10+11+12)/9Median or Q2 will be \u2013Median or Q2 = 7Now since the number of observations is odd which is 9, the median would lie on 5th position which is \u2026 a) Complete the frequency table for this information. PC ATX12VO (12V only) standard - Why does everybody say it has higher efficiency? Problem. Reading from the graph, the lower quartile is 38. So the interquartile range is 38 - 15 = 23. The lower quartile $Q_1$is the point such that the area up to $Q_1$is one quarter of the total area. The median or Q1 will be in the first class whose cumulative frequency includes their number. Median, which is middle quartile tells us the center point and upper and lower quartiles tell us the spread. The IQR can be calculated by subtracting the lower quartile from the upper. To do the lower quartile, the same applies, you just need to half the median, so you'd go to 20 on the y-axis. On box plot, how many quartiles can we see? You can use many of the other features of the quantile function which we described in our guide on how to calculate percentile in R.. The result is 811.5. Finding Interquartile Range from Cumulative Frequency Histogram Polygon 1. 400. Are the longest German and Turkish words really single words? Upper quartile (middle value of the upper half) = 36. a) Complete the frequency table for this information. A box plot displays the five-number summary for a univariate data set, giving a quick impression of the distribution. Example: Find the median, lower quartile, upper quartile, interquartile range and range of the following numbers. Reading from the graph, the lower quartile is 38. Practice Questions; Post navigation. 1 st quartile or lower quartile basically separate the lowest 25% of data from the highest 75%. 3 rd quartile or the upper quartile separate the highest 25% of data from the lowest 75%. \"Data\" is a plural noun; the singular form is \"datum.\" Histograms can be drawn with regular or irregular class intervals and with mean, median, mode, lower and upper quartile values displayed. Below is the comparison of a Histogram vs. a Box Plot. Cluster Duck. A quartile divides the set of observation into 4 equal parts. In order to calculate this value we must first understand what the lower quartile, median and upper quartile are: The lower quartile is the value at the first quarter (once your data has been put in order). Thanks! a) Complete the frequency table for this information. The upper quartile $Q_3$is the point such that the area up to $Q_3$is three quarters of the total area. GCSE Revision Cards. Thanks! Questions in other subjects: Mathematics, 23.07.2019 00:30. 5.3.3. I will use a simple dataset to learn how histogram helps to understand a dataset. LQ is $$\\frac{(7+1)}{4}$$ = the second number. What is the highest road in the world that is accessible by conventional vehicles? 34 / 50 Marks The Maths GCSE test scores of 280 students are shown in the histogram below. The lower quartile, or Q1, is the value which lies 1/4th from the lowest value. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Please help! 3 4 5 6 7 8 9 10 0 5 10 15 20 25 30 x c.f What prevents a government from taxing its citizens living abroad? To find these from the graph we draw two lines across from the vertical axis, one at 10 and one at 30. How to connect a flex ribbon cable to a screw terminal block? Example 5: Quartiles, Quintiles, Deciles, Percentiles & Many More. In its broadest sense, Statistics is the science of drawing conclusions about the world from data. Next Bar Charts, Pictograms and Tally Charts Practice Questions. Sampling data - Intermediate & Higher tier - WJEC, Cumulative frequency \u2013 Intermediate & Higher tier - WJEC, Home Economics: Food and Nutrition (CCEA). The interquartile range is the difference between the lower quartile and the upper quartile. 3) Median score: Lower quartile: Submit Answer. The Lower quartile or 1st quartile has four values as well as the higher quartile or 3rd quartile has four values. [The questions ask which statement is true, like \"the lower quartile value of the number of hits made by the gold team is equal to the lower quartile value of the number of hits made by the blue team]. Get Answer. Further, they could be given as the upper quartile or the lower quartile. Hope this helps. Therefore Interquartile range= 79 \u2026 Can you suggest how can I find first and third quartiles and median after building histogram. The. The upper quartile is 15.6346 years . Data come in many forms, most of which are numbers, or can be translated into numbers for analysis. c.) Which interval contains the upper quartile? There are 40 babies in the table, so to find the lower quartile, find $$\\frac{1}{4}$$ of 40, which is the 10th value. Question. I want to compute the quartiles of X \u223c \u0393(1,\u03b2)? Quartile Formula Quartiles are given as values dividing the complete list into quarters. The Corbettmaths video tutorial on median and quartiles from frequency tables and Histograms - linear interpolation Corbettmaths Videos, worksheets, 5-a-day and much more Relevance and Uses of Quartile Formula. 5-a-day Workbooks. The line in the middle shows the median of the distribution. There are 0 observations below the Lower Quartile(Q1) Fence(Whisker) There are 0 observations above the Upper Quartile(Q3) Fence(Whisker) Notes from above all Plots: From the above Boxplots we can see that in Ozone and Wind Variables there are Outliers (the red dots). 4.2.1. However, it would have had a massive effect on a normal range. It will cut the list into four equal parts. 12, 5, 22, 30, 7, 36, 14, 42, 15, 53, 25, 65 . Find the median of the following data 12, 8, 3, 15, 1, 9. What is 4? The upper quartile $Q_3$ is the point such that the area up to $Q_3$ is three quarters of the total area. The box encloses the interquartile range. In the example below, we\u2019re going to use a single line of code to get the quartiles \u2026 Sometimes we have to estimate the interquartile range from a cumulative frequency diagram. In this case Quartile 2 is half way between 5 and 6: Q2 = (5+6)/2 = 5.5. Why does my advisor / professor discourage all collaboration? Not Helpful 4 Helpful 10. First quartile (Q 1 / 25th percentile) : also known as the lower quartile q n (0.25), is the median of the lower half of the dataset. 1. Acylinder and its dimensions are shown. From the histogram I have shown frequency on each bar to the last bar being 691. The median $Q_2$ is the point such that the area under the bars each side of $Q_2$ is equal. A histogram is a very specific type of graph. The median $Q_2$is the point such that the area under the bars each side of $Q_2$is equal. What is 7? Accidentally ran chmod +x /* - How bad did I just mess up? What is a range? Histogram vs. Each bin has a bar that represents the count or percentage of observations that fall within that bin.Download the CSV data file to make most of the histograms in this blog post: Histograms.In the fie\u2026 The line in the middle shows the median of the distribution. Has a state official ever been impeached twice? That is, half the monarchs started ruling before this age, and half after this age. It only takes a minute to sign up. Quartile 1 is the 25th percentile: it is the score that separates the lowest 25% from the highest 75% of scores. My Tweets. Answer. The histogram is still preferable for a more detailed look at the shape of the distribution. Please be sure to answer the question.Provide details and share your research! Subscribe to this blog. The lower quartile is the point where one-quarter of the values are below it. Read about our approach to external linking. When the temperature reaches 90\u00b0, approximately how many glasses of lemonade will be sold? Interquartile range = upper quartile - lower quartile = 47 - 38 = 9 cm. We read the values for the quartiles from the graph and arrive at 38 and 47. We calculated the upper and lower quartiles and the median. Quartile 1 is the 25th percentile: it is the score that separates the lowest 25% from the highest 75% of scores. The interquartile range (IQR) is therefore 18 - 4 = 14. b) Find an estimate for the lower quartile of the times taken to finish the puzzle. The box in the Box Plot extends from the lower quartile to the upper quartile. e. The first quartile (first 25%) is at 4. f. The third quartile (75%) is at 8. g. Middle 50% of the data ranges from 4 to 8. h. The interquartile range is 4. The first quartile, or lower quartile, is the value that cuts off the first 25% of the data when it is sorted in ascending order.The second quartile, or median, is the value that cuts off the first 50%.The third quartile, or upper quartile, is the value that cuts off the first 75%.. b) Find an estimate for the lower quartile of the times taken to finish the puzzle. Search for: Contact us. - Mr Kim 2. A box plot displays the five-number summary for a univariate data set, giving a quick impression of the distribution. 4.2.1. Mathematics, 21.06.2019 17:00, adreyan6221. This means that 25% of our scores are lower than 811.5 milliseconds. The middle term, between the median and first term is known as the first or Lower Quartile and is written as Q 1.Similarly, the value of mid term that lies between the last term and the median is known as the third or upper quartile and is denoted as Q 3.Second Quartile is the median and is written as Q 2. You will notice that the fact there is an outlier in this data (60) which has had no bearing on the calculation of the interquartile range. Consider a data set of following numbers: 10, 2, 4, 7, 8, 5, 11, 3, 12. What kind of wool do you get from sheering a sheep with the easter egg jeb_? Do I have to stop other application processes before receiving an offer? Photos (of the mark schemes I am looking at): 599722 599722599724 I dont have a clue how they worked out the median and Q1 and Q3 for t, even with t However, we can use the probs argument to get basically any quantile metric that we want. And the result is: Quartile 1 (Q1) = 3; Quartile 2 (Q2) = 5.5; Quartile 3 (Q3) = 7; Interquartile Range. What would cause a culture to keep a distinct weapon for centuries? Wiki User Answered . To find the class which the median or Q1 lies in, create a cumulative frequency table for the data. Problem. These graphs take your continuous measurements and place them into ranges of values known as bins. Answers. To find the upper quartile you'd go to 60, as this is 40 and the lower quartile or 20. To do the lower quartile, the same applies, you just need to half the median, so you'd go to 20 on the y-axis. I have X=c(20 ,18, 34, 45, 30, 51, 63, 52, 29, 36, 27, 24) With boxplot, i'm trying to plot the quantile(X,0.25) and quantile(X,0.75) but this is not realy the same lower and upper quartiles in Positive skewed histograms. Our tips from experts and exam survivors will help you through. 4.2.1. Box Plot. What is the interquartile range of the following data? In Googlesheets and Excel, =PERCENTILE.EXC(A2:A234,0.25) returns quartile 1 for the scores in cells A2 through A234 (our 233 reaction times). Median, quartiles, percentiles from a grouped frequency table. Is there a security reason to require email address and password in separate steps? Top Answer. That is, half the monarchs started ruling before this age, and half after this age. write your answer as a fraction in simplest form. site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. As discussed above, the quartile formula helps us in dividing the data into four parts very quickly and eventually makes it easy for us to understand the data in these parts. A histogram is drawn like a bar chart, but often has bars of unequal width. To get the five number statistics, you use the fivenum() function. As I told you before, the quantile function returns the quartile of the input vector by default. The histogram is still preferable for a more detailed look at the shape of the distribution. 2 nd quartile or middle quartile also same as median it divides numbers into 2 equal parts. Frequency density O 90 100 110 120 130 140 150 160 170 180 190 200 Score a) Find the median score. Asking for help, clarification, or \u2026 Asked by Wiki User. There are 40 babies in the table, so to find the lower quartile, find \\ (\\frac {1} {4}\\) of 40, which is the 10th value. To fine the inter quartile range you'd do what ever you got for you upper quartile, minus what ever you got for your lower quartile. The box in the Box Plot extends from the lower quartile to the upper quartile. Histograms Practice Questions Click here for Questions . Calculation of range and median along with Box-and-whisker plots and Cumulative frequency tables are effective ways to compare distributions and to summarise their characteristics. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Histogram. A quartile divides the set of observation into 4 equal parts. The difference here is that after dividing the data into two groups, instead of considering the data in the lower half, you consider the data in the upper half and then you proceed to find the Median of this subset of data using the methods described in the section on Averages. UQ is $$\\frac{3(7+1)}{4}$$ = the sixth number. Box Plot. The histogram shows information about the times taken by some students to finish a puzzle. It is acceptable when dealing with an even total cumulative frequency to use LQ = $$\\frac{n}{4}$$ and UQ = $$\\frac{3n}{4}$$. The result is 811.5. 1 2 3. 100. In descriptive statistics, a box plot or boxplot is a method for graphically depicting groups of numerical data through their quartiles.Box plots may also have lines extending from the boxes (whiskers) indicating variability outside the upper and lower quartiles, hence the terms box-and-whisker plot and box-and-whisker diagram.Outliers may be plotted as individual points. What does the expression \"go to the vet's\" mean? Median location = (691+1)/2 = 346. The upper quartile for the data set given below is a0. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. MathJax reference. A box plot displays the five-number summary for a univariate data set, giving a quick impression of the distribution. Calculation. To calculate a quartile in R, set the percentile as parameter of the quantile function. Fig. Fig. The [\u2026] Histogram takes only one variable from the dataset and shows the frequency of each occurrence. b) The lower quartile = Q\u2081 is given by the value at the (n +1)/4th position = (17 + 1)/4 = 4.5th position From the cumulative frequency column, the 4.5th value is in the 90 - 100 score range Similar to the previous question, we find the value of the lower quartile using the weighted average as follows; Our lives are filled with data: the weather, weights, prices, our state of health, exam grades, bank balances, election results, and so on. Histograms can be drawn with regular or irregular class intervals and with mean, median, mode, lower and upper quartile values displayed. 3) Maria could not remember her scores from five math tests. With the following R codes, we can calculate the median\u2026 [The questions ask which statement is true, like \"the lower quartile value of the number of hits made by the gold team is equal to the lower quartile value of the number of hits made by the blue team]. Asking for help, clarification, or responding to other answers. Why do some microcontrollers have numerous oscillators (and what are their functions)? 200. the lower quartile is the median of the lower half of the data. The third quartile is found in a similar manner to the first quartile. The interquartile range is another measure of spread, except that it has the added advantage of not being affected by large outlying values. To find the lower quartile using the formula, use Q1 = 1/4 (n+ 1), or enter 1 into the Excel function instead of 3. The, the upper quartile is the median of the upper half of the data. To learn more, see our tips on writing great answers. For the standard normal distribution N(0,1) , find the median, quartiles and interquartile range. Which interval contains the lower quartile? Third quartile (Q 3 / 75th percentile) : also known as the upper quartile q n (0.75), is the median of the upper half of the dataset. Which set of numbers correctly identifies the lower extreme, the lower quartile, the median, the upper quartile, and the upper extreme for the data set given? Is it possible to calculate the mean and standard deviation from a median and quartiles? Please refer to the attachment to answer this question. Yes No. Community Answer. Are there any stars that orbit perpendicular to the Milky Way's galactic plane? The Lower quartile is 13.3625 years. It is the area of the bar that tells us the frequency in a histogram, not its height. The histogram shows information about the times taken by some students to finish a puzzle. The middle term, between the median and first term is known as the first or Lower Quartile and is written as Q 1.Similarly, the value of mid term that lies between the last term and the median is known as the third or upper quartile and is denoted as Q 3.Second Quartile is the median and is written as Q 2. To find the actual value of the median or lower quartile, use linear interpolation.", "date": "2021-04-18 09:24:26", "meta": {"domain": "pipburner.com", "url": "https://pipburner.com/rajgir-mla-unngu/2ca157-j-b-weld-plastic-putty", "openwebmath_score": 0.4373709261417389, "openwebmath_perplexity": 691.1884279624694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9559813501370537, "lm_q2_score": 0.9111797075998823, "lm_q1q2_score": 0.8710708070888212}}
{"url": "https://mathematica.stackexchange.com/questions/241169/how-many-graphs-are-there-on-4-nodes-with-degrees-1-1-2-2/241171", "text": "# How many graphs are there on 4 nodes with degrees 1, 1, 2, 2?\n\nI know just the basic operations on graphs using Mathematica. But I want to know how to write a code that prints all the possible combinations of a graph with a specified number of edges.\nTake for example:\nHow many graphs are there on 4 nodes with degrees 1, 1, 2, 2?\nI know the answer mathematically is 12, but I want Mathematica to print these combinations. \\Please help me and explain to me how can I write the code.\n\n\u2022 Hint: Since you only have four nodes, could you imagine building a table where each entry is a list of whether or not each node is connected to each other node for every possible graph? Could you check that table has the correct number of entries? And then could you think of a way of discarding entries in that table that do not have the appropriate degrees? Is the number of entries that remain what you want to know? And is that number equal to 12? Does this hint make the programming problem more concrete so you can make some progress on doing this? \u2013\u00a0Bill Mar 6 at 6:36\n\nam = Tuples[{0, 1}, {4, 4}];\n\n\n4X4 adjacency matrices with no self-loops:\n\nsimpleam = DeleteDuplicates[(1 - IdentityMatrix[4]) # & /@ am];\n\n\nAdjacency matrices for undirected graphs on 4 nodes:\n\nundirectedam = Select[# == Transpose@# &]@simpleam;\n\n\nAdjacency matrices for graphs with vertex degree sequence {1,1,2,2}:\n\nvdegree1122am = Select[Sort[Total[#]] == {1, 1, 2, 2} &]@undirectedam;\n\n\nCounts:\n\nLength /@ {am, simpleam, undirectedam, vdegree1122am}\n\n {65536, 4096, 64, 12}\n\n\nAssociated graphs:\n\nAdjacencyGraph[#, VertexLabels -> \"Name\"] & /@ vdegree1122am // Column\n\n\nWe can get the 12 graphs more directly, by noting that two periphery nodes (nodes with vertex degree 1) can be selected in 6 ways:\n\nperipherynodes = Subsets[Range[4], {2}]\n\n {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}\n\n\nFor each pair in peripherynodes the remaining two nodes can be ordered in 2 ways:\n\ninteriornodes = Permutations[Complement[Range[4], #]] & /@ peripherynodes;\n\n\nWe combine peripherynodes with associated interiornodes to get the edgelists of 12 graphs:\n\nedgelists = BlockMap[UndirectedEdge @@ # &, #, 2, 1] & /@ (Join @@\n{peripherynodes, interiornodes}])\n\n\nGraph[#, VertexLabels -> \"Name\"] & /@ edgelists // Column\n\n\n\u2022 Thank you for the solution. The second method of obtaining the graphs is exactly what I did to obtain them mathematically, I considered (4C2) then (2C1) to obtain the final result, and I wanted to know how to implement this idea in Mathematica. So thanks! \u2013\u00a0Mubarak Alsaeedi Mar 6 at 20:17\n\nI had an old program lying around that will generate all realizations of a degree sequence. The first vertex will have degree equal to the first element in the input list, the 2nd degree equal to the 2nd element, and so on.\n\nThe program requires my IGraph/M package for graphicality testing.\n\nFor details on how it works, see http://bolyai.cs.elte.hu/egres/tr/egres-11-11.pdf\n\nThe answer here is trivial: there are only two realizations, which are isomorphic. They have an automorphism group size of 2 (they are symmetric to reflection, see also IGBlissAutomorphismCount). Thus not all 4! = 24 permutations of vertices create distinct labelled graphs, only 4! / 2 = 12.\n\nWhat if the degree sequence is a bit more complicated?\n\ndegSeqGraphList[{3, 3, 2, 2, 1, 1}] // DeleteDuplicatesBy[CanonicalGraph]\n\n\nThere are now 5 non-isomorphic realizations:\n\nThere have different automorphism group sizes:\n\nIn[]:= IGBlissAutomorphismCount /@ %\nOut[]= {8, 1, 4, 2, 4}\n\n\nThere are 6! permutations of vertices, but many of these generate the same labelled graph. We can compute the total number of graphs like this:\n\nIn[]:= 6!/%\nOut[]= {90, 720, 180, 360, 180}\n\nIn[]:= Total[%]\nOut[]= 1530\n\n\nThe code for generating all labelled realizations of a degree sequence:\n\nNeeds[\"IGraphM\"]\n\nremoveZeros[list_] := TakeWhile[list, # != 0 &]\n\niter[ds_, vert_] :=\nWith[{pos = Position[ds, 0]}, {Delete[ds, pos], Delete[vert, pos]}]\n\nClearAll[exactGen, list]; (* list is an inert head *)\n\nexactGen[{}, edges_, _] := edges\nexactGen[degseq : {d_, rest__}, edges_, vertices_] /; IGGraphicalQ[degseq] :=\n\nModule[{ds = {rest}, temp, newDs, newVertices},\nFunction[subs,\ntemp = ds;\ntemp[[subs]] -= 1;\nIf[IGGraphicalQ[temp],\n{newDs, newVertices} = iter[temp, Rest[vertices]];\nexactGen[newDs,\nJoin[edges, list @@ Thread[First[vertices] <-> Rest[vertices][[subs]]]],\nnewVertices],\n{}\n]\n] /@ Subsets[Range@Length[ds], {d}]\n]\n\ndegSeqGraphList[degseq_?IGGraphicalQ] :=\n\nWith[{verts = Range@Length[degseq]},\nGraph[verts, #, VertexLabels -> Automatic] & /@ List @@@ Cases[\nexactGen[degseq, list[], verts],\n_list,\nInfinity\n]\n]\n\n\u2022 Thanks. I appreciate your solution and your generalization. \u2013\u00a0Mubarak Alsaeedi Mar 6 at 20:21\n\nA one-liner:\n\nLength@Select[Graph /@ Subsets[EdgeList[CompleteGraph[4]]],\nSort@VertexDegree[#] == {1, 1, 2, 2} &]\n\n\nEdgeList[CompleteGraph[4]] will give us all the possible edges on four vertices, Subsets will give us all the possible subsets of those edges, and Graph will turn those into graphs. (When a vertex is isolated, it will not be included in the graph, but that's not a problem for us.)\n\nVertexDegree gives the degrees of all vertices of a graph, so Sort@VertexDegree[#] == {1, 1, 2, 2} checks if a graph has the degree sequence you want.\n\nWe Select all graphs with that degree sequence, and then find the Length` of that list.\n\n(Like the accepted answer, this would be inefficient for large graphs, in which case Szabolcs's answer is the way to go.)\n\n\u2022 Yes, I can see that. But when I tried to solve it mathematically, I used the approach of Kglr in his second solution, where he used the idea of peripherynodes and interiornodes. Is this an efficient approach? \u2013\u00a0Mubarak Alsaeedi Mar 6 at 20:25\n\u2022 It's situational: it doesn't give a general algorithm for an arbitrary degree sequence. So it's hard to say how efficient this is, given that for this specific problem, all of these solutions work instantly. \u2013\u00a0Misha Lavrov Mar 6 at 20:37", "date": "2021-08-02 18:35:31", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/241169/how-many-graphs-are-there-on-4-nodes-with-degrees-1-1-2-2/241171", "openwebmath_score": 0.1882602721452713, "openwebmath_perplexity": 1245.2630675796318, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9546474207360067, "lm_q2_score": 0.9124361539289197, "lm_q1q2_score": 0.8710548209345252}}
{"url": "https://math.stackexchange.com/questions/3373712/how-can-three-vectors-be-orthogonal-to-each-other/3373728", "text": "How can three vectors be orthogonal to each other?\n\nIn a scenario, say that:\n\nVectors $$\\mathbf{U}$$, $$\\mathbf{V}$$ and $$\\mathbf{W}$$ are all orthogonal such that the dot product between each of these $$(\\mathbf{UV}\\;\\mathbf{VW}\\;\\mathbf{WU})$$ is equal to zero.\n\nI imagine that for any potential vector space $$\\mathbf{R}$$ this would only be possible in two situations.\n\n1) $$\\mathbf{U}$$, $$\\mathbf{W}$$ and/or $$\\mathbf{V}$$ is the zero vector.\n\n2) $$\\mathbf{U}=(1, 0, 0)$$, $$\\mathbf{V} = (0, 1, 0)$$ and $$\\mathbf{W} = (0, 0, 1)$$.\n\nIs there any other situation where three vectors are all orthogonal to each other?\n\n\u2022 What about $u=\\big(\\tfrac{1}{\\sqrt 2},\\tfrac{1}{\\sqrt 2},0\\big)$, $v=\\big(\\tfrac{1}{\\sqrt 2},-\\tfrac{1}{\\sqrt 2},0\\big)$ and $w=(0,0,1)$? Sep 29, 2019 at 0:24\n\u2022 Yes, infinitely many. For example, rotate your basis in 2. Sep 29, 2019 at 0:25\n\u2022 Pick two non-zero and non-parallel vectors $x$ and $u$ in $\\mathbb R^3$. Then set $v=x\\times u$ and $w=u\\times v$. Now $u,v,w$ are mutually orthogonal. Sep 29, 2019 at 3:53\n\u2022 @Azif00 More generally, $(a,b,0)$, $(b,-a,0)$, $(0,0,c)$ for any values of $a$, $b$, and $c$. Orthogonal vectors don't have to be unit vectors. The OP also seems to have made that mistake in the second example. Sep 29, 2019 at 10:34\n\u2022 Hold out your hand. Arrange your fingers so that your index finger, middle finger, and thumb are all mutually perpendicular. Now rotate your hand around. All of these configurations correspond to a distinct set of 3 mutually orthogonal vectors. Sep 29, 2019 at 12:53\n\nThink it like this: for a given vector $$u\\neq 0$$ in $$\\mathbb{R}^3$$, what is the space of all vectors perpendicular to it? It is a plane $$P$$ containing the origin, whose perpendicular direction is obviously $$u$$.\n\nNow take a $$v\\neq 0$$ in that plane. The space of all vectors perpendicular to $$v$$ is a new plane $$P'$$. In order to get a vector $$w$$ perpendicular to both $$u,v$$, we need $$w\\in P\\cap P'$$. What does $$P\\cap P'$$ look like?\n\nBy the way, notice that $$v$$ is ANY non-zero vector, not only $$(1,0,0), (0,1,0)$$ or $$(0,0,1)$$\n\nThere are infinitely many triple of non zero orthogonal vectors obtained by the three you have indicated by scaling of each one and rotations of the triple all togheter.\n\nTo construct any othogonal triple we can proceed as follows:\n\n1. choose a first vector $$v_1=(a,b,c)$$\n2. find a second vector orthogonal to $$v_1$$ that is e.g. $$v_2=(-b,a,0)$$\n3. determine the third by cross product $$v_3=v_1\\times v_2$$\n\u2022 You can also have $v_3 = v_2 \\times v_1 = - v_1 \\times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about. Sep 30, 2019 at 3:33\n\u2022 @Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1\\times v_2$.\n\u2013\u00a0user\nSep 30, 2019 at 5:54\n\nHow about $$v_1= (\\frac{1}{\\sqrt2}, \\frac{1}{\\sqrt2},0),v_2=(-\\frac{1}{\\sqrt2},\\frac{1}{\\sqrt2},0),v_3=(0,0,1)$$ ? Geometrically take any frame and rotate in any direction.\n\nHere is an example of just what you seek:\n\n\u2022 I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle. Sep 29, 2019 at 9:25\n\u2022 I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames. Sep 29, 2019 at 15:10\n\u2022 Gif works fine for me Sep 29, 2019 at 17:11\n\u2022 @B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version. Sep 29, 2019 at 17:45\n\nThere are infinitely many possibilities. $$(1,0,0), (0,1,-1)$$ and $$(0,1,1)$$ is one example.\n\nThe word \"orthogonal\" really just corresponds to the intuitive notion of vectors being perpendicular to each other. Draw out the unit vectors in the $$x$$, $$y$$ and $$z$$ directions respectively--those are one set of three mutually orthogonal (i.e. perpendicular) vectors, just like you observed. But if you rotate those three vectors together in any way that you like in 3D space, without changing the angles between them, then of course they will still remain orthogonal.\n\nFurthermore, if we scale these vectors by changing their lengths independently by any arbitrary nonzero factor, we will still end up with a set of orthogonal vectors, since the only thing that matters is the directions and not the lengths of the vectors. In this manner we end up with a description for an infinite family of orthogonal vectors, which hopefully makes it easy for you to convince yourself intuitively.\n\nIn a more general vector space, of course, this sort of pictorial intuition might no longer hold, but the idea of orthogonality can be easily generalised. That's the reason we define orthogonality abstractly and independent of the usual geometric notion of perpendicularity.", "date": "2022-09-27 18:32:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3373712/how-can-three-vectors-be-orthogonal-to-each-other/3373728", "openwebmath_score": 0.7901446223258972, "openwebmath_perplexity": 255.5799920927667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846628255124, "lm_q2_score": 0.8902942275774319, "lm_q1q2_score": 0.8710502176638457}}
{"url": "https://nbviewer.jupyter.org/github/fediskhakov/CompEcon/blob/HEAD/31_approximation.ipynb", "text": "# Foundations of Computational Economics #31\u00b6\n\nby Fedor Iskhakov, ANU\n\n## Function approximation in Python\u00b6\n\nhttps://youtu.be/liNputEfcXQ\n\nDescription: How to approximate functions which are only defined on grid of points. Spline and polynomial interpolation.\n\n### Interpolation problem\u00b6\n\n\u2022 $f(x)$ is function of interest, hard to compute\n\u2022 Have data on values of $f(x)$ in $n$ points $(x_1,\\dots,x_n)$\n$$f(x_1), f(x_2), \\dots f(x_n)$$\n\u2022 Need to find the approximate value of the function $f(x)$ in arbitrary points $x \\in [x_1,x_n]$\n\n#### Approaches\u00b6\n\n1. Piece-wise approach (connect the dots)\n\u2022 Which functional form to use for connections?\n1. Use a similar function $s(x)$ to represent $f(x)$ between the data points\n\u2022 Which simpler function?\n\u2022 What data should be used?\n\u2022 How to control the accuracy of the approximation?\n\n#### Distinction between function approximation (interpolation) and curve fitting\u00b6\n\n\u2022 Functions approximation and interpolation refers to the situations when data on function values is matched exactly\n\u2022 The approximation curve passes through the points of the data\n\u2022 Curve fitting refers to the statistical problem when the data has noise, the task is to find an approximation for the central tendency in the data\n\u2022 Linear and non-linear regression models, econometrics\n\u2022 The model is over-identified (there is more data than needed to exactly identify the regression function)\n\n#### Extrapolation\u00b6\n\nExtrapolation is computing the approximated function outside of the original data interval\n\nShould be avoided in general\n\n\u2022 Exact only when theoretical properties of the extrapolated function are known\n\u2022 Can be used with extreme caution and based on the analysis of the model\n\u2022 Always try to introduce wider bounds for the grid instead\n\n### Spline interpolation\u00b6\n\nSpline = curve composed of independent pieces\n\nDefinition A function $s(x)$ on $[a,b]$ is a spline of order $n$ ( = degree $n-1$) iff\n\n\u2022 $s$ is $C^{n-2}$ on $[a,b]$ (has continuous derivatives up to order $n-2$),\n\u2022 given knot points $a=x_0<x_1<\\dots<x_m=b$, $s(x)$ is a polynomial of degree $n-1$ on each subinterval $[x_i,x_{i+1}]$, $i=0,\\dots,m-1$\n\n#### Cubic splines = spline of order 4\u00b6\n\n\u2022 Data set $\\{(x_i,f(x_i), i=0,\\dots,n\\}$\n\u2022 Functional form $s(x) = a_i + b_i x + c_i x^2 + d_i x^3$ on $[x_{i-1},x_i]$ for $i=1,\\dots,n$\n\u2022 $4n$ unknown coefficients:\n\u2022 $2n$ equations to make sure each segment passes through its interval points + $2(n-1)$ equations to ensure two continuous derivatives at each interior point\n\u2022 Additional 2 equation for the $x_0$ and $x_n$\n\u2022 $s''(x_0)=s''(x_n)=0$ (natural spline)\n\u2022 $s'(x_0)=\\frac{s(x_1)-s(x_0)}{x_1-x_0}$, $s'(x_n)=\\frac{s(x_n)-s(x_{n-1})}{x_n-x_{n-1}}$ (secant-Hermite)\nIn\u00a0[1]:\nimport numpy as np\nimport matplotlib.pyplot as plt\n%matplotlib inline\n\nnp.random.seed(2008) # fix random number sequences\nx  = np.sort(np.random.uniform(-5,10,12)) # sorted random numbers on [-5,10]\nxr = np.linspace(-5,10,12) # regular grid on [-5,10]\n\nfunc=lambda x: np.exp(-x/4)*np.sin(x) + 1/(1+x**2) # function to interpolate\n\nIn\u00a0[2]:\ndef plot1(ifunc,fdata=(x,func(x)),f=func,color='b',label='',extrapolation=False):\n'''helper function to make plots'''\nxd = np.linspace(-5,10,1000) # for making continuous lines\nplt.figure(num=1, figsize=(10,8))\nplt.scatter(fdata[0],fdata[1],color='r') # interpolation data\nplt.plot(xd,f(xd),color='grey') # true function\nif extrapolation:\nxdi = xd\nelse:\n# restriction for interpolation only\nxdi=xd[np.logical_and(xd>=fdata[0][0],xd<=fdata[0][-1])]\nif ifunc:\nplt.plot(xdi,ifunc(xdi),color=color,label=label)\nif label:\nplt.legend()\nelif label:\nplt.title(label)\n\nIn\u00a0[3]:\nplot1(None,label='True function')\n\nIn\u00a0[4]:\nfrom scipy import interpolate # Interpolation routines\nfi = interpolate.interp1d(x,func(x)) # returns the interpolation function\nplot1(fi,label='interp1d')\n\nIn\u00a0[5]:\nhelp(interpolate.interp1d)\n\nHelp on class interp1d in module scipy.interpolate.interpolate:\n\nclass interp1d(scipy.interpolate.polyint._Interpolator1D)\n|  interp1d(x, y, kind='linear', axis=-1, copy=True, bounds_error=None, fill_value=nan, assume_sorted=False)\n|\n|  Interpolate a 1-D function.\n|\n|  x and y are arrays of values used to approximate some function f:\n|  y = f(x).  This class returns a function whose call method uses\n|  interpolation to find the value of new points.\n|\n|  Note that calling interp1d with NaNs present in input values results in\n|  undefined behaviour.\n|\n|  Parameters\n|  ----------\n|  x : (N,) array_like\n|      A 1-D array of real values.\n|  y : (...,N,...) array_like\n|      A N-D array of real values. The length of y along the interpolation\n|      axis must be equal to the length of x.\n|  kind : str or int, optional\n|      Specifies the kind of interpolation as a string\n|      ('linear', 'nearest', 'zero', 'slinear', 'quadratic', 'cubic',\n|      'previous', 'next', where 'zero', 'slinear', 'quadratic' and 'cubic'\n|      refer to a spline interpolation of zeroth, first, second or third\n|      order; 'previous' and 'next' simply return the previous or next value\n|      of the point) or as an integer specifying the order of the spline\n|      interpolator to use.\n|      Default is 'linear'.\n|  axis : int, optional\n|      Specifies the axis of y along which to interpolate.\n|      Interpolation defaults to the last axis of y.\n|  copy : bool, optional\n|      If True, the class makes internal copies of x and y.\n|      If False, references to x and y are used. The default is to copy.\n|  bounds_error : bool, optional\n|      If True, a ValueError is raised any time interpolation is attempted on\n|      a value outside of the range of x (where extrapolation is\n|      necessary). If False, out of bounds values are assigned fill_value.\n|      By default, an error is raised unless fill_value=\"extrapolate\".\n|  fill_value : array-like or (array-like, array_like) or \"extrapolate\", optional\n|      - if a ndarray (or float), this value will be used to fill in for\n|        requested points outside of the data range. If not provided, then\n|        the default is NaN. The array-like must broadcast properly to the\n|        dimensions of the non-interpolation axes.\n|      - If a two-element tuple, then the first element is used as a\n|        fill value for x_new < x[0] and the second element is used for\n|        x_new > x[-1]. Anything that is not a 2-element tuple (e.g.,\n|        list or ndarray, regardless of shape) is taken to be a single\n|        array-like argument meant to be used for both bounds as\n|        below, above = fill_value, fill_value.\n|\n|      - If \"extrapolate\", then points outside the data range will be\n|        extrapolated.\n|\n|  assume_sorted : bool, optional\n|      If False, values of x can be in any order and they are sorted first.\n|      If True, x has to be an array of monotonically increasing values.\n|\n|  Attributes\n|  ----------\n|  fill_value\n|\n|  Methods\n|  -------\n|  __call__\n|\n|  --------\n|  splrep, splev\n|      Spline interpolation/smoothing based on FITPACK.\n|  UnivariateSpline : An object-oriented wrapper of the FITPACK routines.\n|  interp2d : 2-D interpolation\n|\n|  Examples\n|  --------\n|  >>> import matplotlib.pyplot as plt\n|  >>> from scipy import interpolate\n|  >>> x = np.arange(0, 10)\n|  >>> y = np.exp(-x/3.0)\n|  >>> f = interpolate.interp1d(x, y)\n|\n|  >>> xnew = np.arange(0, 9, 0.1)\n|  >>> ynew = f(xnew)   # use interpolation function returned by interp1d\n|  >>> plt.plot(x, y, 'o', xnew, ynew, '-')\n|  >>> plt.show()\n|\n|  Method resolution order:\n|      interp1d\n|      scipy.interpolate.polyint._Interpolator1D\n|      builtins.object\n|\n|  Methods defined here:\n|\n|  __init__(self, x, y, kind='linear', axis=-1, copy=True, bounds_error=None, fill_value=nan, assume_sorted=False)\n|      Initialize a 1D linear interpolation class.\n|\n|  ----------------------------------------------------------------------\n|  Data descriptors defined here:\n|\n|  __dict__\n|      dictionary for instance variables (if defined)\n|\n|  __weakref__\n|      list of weak references to the object (if defined)\n|\n|  fill_value\n|      The fill value.\n|\n|  ----------------------------------------------------------------------\n|  Methods inherited from scipy.interpolate.polyint._Interpolator1D:\n|\n|  __call__(self, x)\n|      Evaluate the interpolant\n|\n|      Parameters\n|      ----------\n|      x : array_like\n|          Points to evaluate the interpolant at.\n|\n|      Returns\n|      -------\n|      y : array_like\n|          Interpolated values. Shape is determined by replacing\n|          the interpolation axis in the original array with the shape of x.\n|\n|  ----------------------------------------------------------------------\n|  Data descriptors inherited from scipy.interpolate.polyint._Interpolator1D:\n|\n|  dtype\n\n\nIn\u00a0[6]:\nfi = interpolate.interp1d(x,func(x),kind='linear')\nplot1(fi,label='Linear')\n\nIn\u00a0[7]:\nfor knd, clr in ('previous','m'),('next','b'),('nearest','g'):\nfi = interpolate.interp1d(x,func(x),kind=knd)\nplot1(fi,label=knd,color=clr)\nplt.show()\n\nIn\u00a0[8]:\nfor knd, clr in ('slinear','m'),('quadratic','b'),('cubic','g'):\nfi = interpolate.interp1d(x,func(x),kind=knd)\nplot1(fi,color=clr,label=knd)\n\nIn\u00a0[9]:\n# Approximation errors\n# x = np.sort(np.random.uniform(-5,10,11))  # generate new data\nfi = interpolate.interp1d(x,func(x),kind=knd,bounds_error=False)\nxd = np.linspace(-5,10,1000)\nerd=np.abs(func(xd)-fi(xd))\nplt.plot(xd,erd,color=clr)\nprint('Max error with  %s splines is %1.5e'%(knd,np.nanmax(erd)))\n\nMax error with  slinear splines is 1.05142e+00\nMax error with  quadratic splines is 3.89974e-01\nMax error with  cubic splines is 4.35822e-01\n\nIn\u00a0[10]:\n# Approximation errors for regular grid\nfi = interpolate.interp1d(xr,func(xr),kind=knd,bounds_error=False)\nxd = np.linspace(-5,10,1000)\nerd=np.abs(func(xd)-fi(xd))\nplt.plot(xd,erd,color=clr)\nprint('Max error with  %s splines is %1.5e'%(knd,np.nanmax(erd)))\n\nMax error with  slinear splines is 4.63043e-01\nMax error with  quadratic splines is 3.48546e-01\nMax error with  cubic splines is 1.89578e-01\n\n\n#### Accuracy of the interpolation\u00b6\n\nHow to reduce approximation errors?\n\n\u2022 Number of nodes (more is better)\n\u2022 Location of nodes (regular is better)\n\u2022 Interpolation type (match function of interest)\n\nIn economic models we usually can control all of these\n\n### Polynomial approximation/interpolation\u00b6\n\nBack to the beginning to explore the idea of replacing original $f(x)$ with simpler $g(x)$\n\n\u2022 Data set $\\{(x_i,f(x_i)\\}, i=0,\\dots,n$\n\u2022 Functional form is polynomial of degree $n$ such that $g(x_i)=f(x_i)$\n\u2022 If $x_i$ are distinct, coefficients of the polynomial are uniquely identified\n\nDoes polynomial $g(x)$ converge to $f(x)$ when there are more points?\n\nIn\u00a0[11]:\nfrom numpy.polynomial import polynomial\ndegree = len(x)-1 # passing through all dots\np = polynomial.polyfit(x,func(x),degree)\nfi = lambda x: polynomial.polyval(x,p)\nplot1(fi,label='Polynomial of degree %d'%degree,extrapolation=True)\n\nIn\u00a0[12]:\n# now with regular grid\ndegree = len(x)-1 # passing through all dots\np = polynomial.polyfit(xr,func(xr),degree)\nfi = lambda x: polynomial.polyval(x,p)\nplot1(fi,fdata=(xr,func(xr)),label='Polynomial of degree %d'%degree,extrapolation=True)\n\nIn\u00a0[13]:\n# how number of points affect the approximation (with degree=n-1)\nfor n, clr in (5,'m'),(10,'b'),(15,'g'),(25,'r'):\nx2 = np.linspace(-5,10,n)\np = polynomial.polyfit(x2,func(x2),n-1)\nfi = lambda x: polynomial.polyval(x,p)\nplot1(fi,fdata=(x2,func(x2)),label='%d points'%n,color=clr,extrapolation=True)\nplt.show()\n\nIn\u00a0[14]:\n# how locations of points affect the approximation (with degree=n-1)\nnp.random.seed(2025)\nn=8\nfor clr in 'b','g','c':\nx2 = np.linspace(-4,9,n) + np.random.uniform(-1,1,n) # perturb points a little\np = polynomial.polyfit(x2,func(x2),n-1)\nfi = lambda x: polynomial.polyval(x,p)\nplot1(fi,fdata=(x2,func(x2)),label='%d points'%n,color=clr,extrapolation=True)\nplt.show()\n\nIn\u00a0[15]:\n# how degree of the polynomial affects the approximation\nfor degree, clr in (7,'b'),(9,'g'),(11,'m'):\np=polynomial.polyfit(xr,func(xr),degree)\nfi=lambda x: polynomial.polyval(x,p)\nplot1(fi,fdata=(xr,func(xr)),label='Polynomial of degree %d'%degree,color=clr,extrapolation=True)\n\n\n#### Least squares approximation\u00b6\n\nWe could also go back to function approximation and fit polynomials of lower degree\n\n\u2022 Data set $\\{(x_i,f(x_i)\\}, i=0,\\dots,n$\n\u2022 Any functional form $g(x)$ from class $G$ that best approximates $f(x)$\n$$g = \\arg\\min_{g \\in G} \\lVert f-g \\rVert ^2$$\n\n### Orthogonal polynomial approximation/interpolation\u00b6\n\n\u2022 Polynomials over domain $D$\n\u2022 Weighting function $w(x)>0$\n\nInner product\n\n$$\\langle f,g \\rangle = \\int_D f(x)g(x)w(x)dx$$\n\n$\\{\\phi_i\\}$ is a family of orthogonal polynomials w.r.t. $w(x)$ iff\n\n$$\\langle \\phi_i,\\phi_j \\rangle = 0, i\\ne j$$\n\n#### Best polynomial approximation in L2-norm\u00b6\n\nLet $\\mathcal{P}_n$ denote the space of all polynomials of degree $n$ over $D$\n\n$$\\lVert f - p \\rVert_2 = \\inf_{q \\in \\mathcal{P}_n} \\lVert f - q \\rVert_2 = \\inf_{q \\in \\mathcal{P}_n} \\left[ \\int_D ( f(x)-g(x) )^2 dx \\right]^{\\tfrac{1}{2}}$$\n\nif and only if\n\n$$\\langle f-p,q \\rangle = 0, \\text{ for all } q \\in \\mathcal{P}_n$$\n\nOrthogonal projection is the best approximating polynomial in L2-norm\n\n#### Uniform (infinity, sup-) norm\u00b6\n\n$$\\lVert f(x) - g(x) \\rVert_{\\infty} = \\sup_{x \\in D} | f(x) - g(x) | = \\lim_{n \\rightarrow \\infty} \\left[ \\int_D ( f(x)-g(x) )^n dx \\right]^{\\tfrac{1}{n}}$$\n\nMeasures the absolute difference over the whole domain $D$\n\n#### Chebyshev (minmax) approximation\u00b6\n\nWhat is the best polynomial approximation in the uniform (infinity, sup) norm?\n\n$$\\lVert f - p \\rVert_{\\infty} = \\inf_{q \\in \\mathcal{P}_n} \\lVert f - q \\rVert_{\\infty} = \\inf_{q \\in \\mathcal{P}_n} \\sup_{x \\in D} | f(x) - g(x) |$$\n\nChebyshev proved existence and uniqueness of the best approximating polynomial in uniform norm.\n\n#### Chebyshev polynomials\u00b6\n\n\u2022 $[a,b] = [-1,1]$ and $w(x)=(1-x^2)^{(-1/2)}$\n\u2022 $T_n(x)=\\cos\\big(n\\cos^{-1}(x)\\big)$\n\u2022 Recursive formulas:\n$$\\begin{eqnarray} T_0(x)=1,\\\\ T_1(x)=x,\\\\ T_{n+1}(x)=2x T_n(x) - T_{n-1}(x) \\end{eqnarray}$$\n\n#### Accuracy of Chebyshev approximation\u00b6\n\nSuppose $f: [-1,1]\\rightarrow R$ is $C^k$ function for some $k\\ge 1$, and let $I_n$ be the degree $n$ polynomial interpolation of $f$ with nodes at zeros of $T_{n+1}(x)$. Then\n\n$$\\lVert f - I_n \\rVert_{\\infty} \\le \\left( \\frac{2}{\\pi} \\log(n+1) +1 \\right) \\frac{(n-k)!}{n!}\\left(\\frac{\\pi}{2}\\right)^k \\lVert f^{(k)}\\rVert_{\\infty}$$\n\n\ud83d\udcd6 Judd (1988) Numerical Methods in Economics\n\n\u2022 achieves best polynomial approximation in uniform norm\n\u2022 works for smooth functions\n\u2022 easy to compute\n\u2022 but does not approximate $f'(x)$ well\n\n#### General interval\u00b6\n\n\u2022 Not hard to adapt the polynomials for the general interval $[a,b]$ through linear change of variable\n$$y = 2\\frac{x-a}{b-a}-1$$\n\u2022 Orthogonality holds with weights function with the same change of variable\n\n#### Chebyshev approximation algorithm\u00b6\n\n1. Given $f(x)$ and $[a,b]$\n2. Compute Chebyshev interpolation nodes on $[-1,1]$\n3. Adjust nodes to $[a,b]$ by change of variable, $x_i$\n4. Evaluate $f$ at the nodes, $f(x_i)$\n5. Compute Chebyshev coefficients $a_i = g\\big(f(x_i)\\big)$\n6. Arrive at approximation\n$$f(x) = \\sum_{i=0}^n a_i T_i(x)$$\nIn\u00a0[16]:\nimport numpy.polynomial.chebyshev as cheb\nfor degree, clr in (7,'b'),(9,'g'),(11,'m'):\nfi=cheb.Chebyshev.interpolate(func,degree,[-5,10])\nplot1(fi,fdata=(None,None),color=clr,label='Chebyshev with n=%d'%degree,extrapolation=True)\n\n\n### Multidimensional interpolation\u00b6\n\n\u2022 there are multidimensional generalization to all the methods\n\u2022 curse of dimensionality in the number of interpolation points when number of dimensions increase\n\u2022 sparse Smolyak grids and adaptive sparse grids\n\u2022 irregular grids require computationally expensive triangulation in the general case\n\u2022 good application for machine learning!\n\nGenerally much harder!", "date": "2021-08-05 09:08:19", "meta": {"domain": "jupyter.org", "url": "https://nbviewer.jupyter.org/github/fediskhakov/CompEcon/blob/HEAD/31_approximation.ipynb", "openwebmath_score": 0.488471657037735, "openwebmath_perplexity": 5873.488914877258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692345869641, "lm_q2_score": 0.8918110418436166, "lm_q1q2_score": 0.871004407633608}}
{"url": "http://mindcode.org/4u9mex/mcl5.php?vj=decimal-to-binary", "text": "## Decimal To Binary\n\nIn order to convert a decimal number to its binary equivalent, we will repeatedly divide the decimal number by 2, the base of the binary system. One for \"yes\", and zero for \"no\". Binary definition, consisting of, indicating, or involving two. Add the following byte-long (8 bit) two\u2019s complement numbers together, and then convert all binary quantities into decimal form to verify the accuracy of the addition:. This tutorial explains how to convert a decimal IP address in binary IP address and a binary IP address in a decimal IP address step by step with examples. A value can also be in binary, hexadecimal or octal if the appropriate prefix is included. The number that i'm trying to convert is a source address in a decimal form. How to Convert Decimal to Binary and vise versa is crucial to master, such skills will give you the confidence when you deal with Network and Storage devices. It allows the user to specify the bit of the binary number that will be converted to. Example of decimal to binary conversion: The decimal positive integer 330 can be deconstructed: 330 10 = 3*100 + 3*10 + 0*1 = 1*10 2 + 2*10 1 + 5*1 0. Before a floating-point binary number can be stored correctly, its mantissa must be normalized. Converting between Decimal (Base 10) and Binary (Base 2) Base 10 vs. In this tutorial i am going to show you how to convert \u201cDecimal to Binary\u201d, \u201cDecimal to Octal\u201d and \u201cDecimal to Hexadecimal\u201d Check the Java Program below. For example, the decimal. In the octal to binary conversion method, we will convert each digit from the given octal number into three bit binary number starting from the right most octal digit(LSD) to the left most digit(MSD) and at the end combine each three bit binary number to form the binary equivalent of given octal number. Program to convert decimal to binary more than 18 bits. Example :- 7 1). This way people won't think it is the decimal number \"101\" (one hundred and one). Below is the code for a 3-digit BCD-Counter (binary Coded decimal) I implemented while studying verilog. C program to covert 'Decimal to Binary' and 'Binary to Decimal' - Binary Decimal Conversion. Video created by Hebrew University of Jerusalem for the course \"Build a Modern Computer from First Principles: From Nand to Tetris (Project-Centered Course)\". This is how we would successfully convert a binary number into decimal, (which you can read about here) but we actually want to do the reverse, and easily convert decimal to binary. Binary to Decimal to BCD Conversion Example. Explanation of how number bases, including binary and decimal, are used for reasoning about digital data. Take this quiz to test yourself on conversion between decimal and binary representations. This way you can convert up to 19 decimal characters (max. To convert fraction to binary, start with the fraction in question and multiply it by 2 keeping notice of the resulting integer and fractional part. Decimal -> Binary Decimal to binary conversion is even more exciting. You can convert to and from binary and the base-10 system typically used by humans. The binary number 1101011001111110 converts to 54910 in Decimal and D67E in hex. Converting decimal fraction to binary. In this example I use 8 bit binary numbers, but the principle is the same for both 8 bit binary numbers (chars) and 32 bit binary numbers (ints). Binary, and other systems of representing numbers is similar to decimal, only the use a different amount of symbols or digits to represent a number. It is much easier to work with large numbers using hexadecimal values than decimal. binary values The following table shows the decimal values of binary numbers. Program to Convert Binary to Decimal number:Number System [crayon-5d4e448ac93bb860279627/] Output : [crayon-5d4e448ac93c5762542184/] Note : This program is for beginners (not for Experts) that\u2019s why we haven\u2019t provided any Validations while accepting the input. Type 110 in binary by clicking the Buttons \"1\", \"1\" and \"0\" in the binary system calculator. 243F6 in a hexadecimal system (base 16), or to 11. It can convert very large and very small numbers \u2014 up to hundreds of digits. Hexadecimal is used to convert byte/modern computer numbers into defined binary digits. We use cookies for various purposes including analytics. It provides output as 170 Decimal value. For beginners hexadecimal is base 16 number, while decimal is base 10, Octal is base 8 and binary is base 2 numbers in Number systems. Method1: Use (()) brace expatiation. This video tutorial explains how to convert decimal to binary numbers. It focuses on the following learning objectives: 5b. As it is written, it would be hard to work out whether it is a binary or decimal (denary) value. Each hexadecimal digit represents four binary digits (bits) (also called a \"nibble\"), and the primary use of hexadecimal notation is as a human-friendly representation of binary coded values in computing and digital. Convert any decimal number between 0 and 255 into binary. Decimal to Binary, Hex & Octal Converter to perform decimal to binary, decimal to hex & decimal to octal conversion online by using simple successive division methods along with step by step calculation & solved example problems. The online BCD to Binary Converter is used to convert BCD (Binary-coded decimal) to binary (Base-2) number. AD3, Alternating Directions Dual Decomposition. It involves dividing the number to be converted, say N, by 2 (since binary is in base 2), and making note of. Then the calculate button is pressed and the textbox should display the corresponding decimal number. TCS Programming Question Solution - 1. Two good examples where we use hexadecimal values are MAC addresses and IPv6 addresses. So the binary (so far) is _ _ _ 1 1. Decimal To Binary Conversion A decimal number system is a base 10 number system. Likewise, we use 0 and 1 to write numbers in the binary number system. Calculator will display the result in total number of hours, minutes and seconds in decimal. This guide shows you how to convert from decimal to binary and binary to decimal. Enter the primary number (in binary; make sure it is valid) first then enter the secondary number (also in binary) for the calculation and click on Calculate. Binary to decimal - Four ways to convert formats in Matlab In this article we're going to present four variations of a binary to decimal conversion in Matlab; we're going to convert binary numbers (numbers with only symbols '0' and '1') to decimal numbers (numbers with 10 different symbols, from '0' to '9'). Binary to hexa decimal decoder / converter. 0 Flash Drive Cute Memory Stick Fruit Vegetable Series Thumb Drive Data Storage Pendrive Cartoon Jump Drive Gift: USB Flash Drives - Amazon. In this system, numbers are represented in a decimal form, however each decimal digit is encoded using a four bit binary number. the radix of binary number system). Specifically, dotted decimal notation is used in various IT contexts, including in Internet Protocol addresses. For example-. Roman to decimal converter. Say we wish to convert an unsigned decimal integer, $$N\\text{,}$$ to binary. Binary form of 15 is 1111 Binary form of 10 is 1010 Binary form of 18 is 10010 Binary form of 27 is 11011 In the above program, the DecimalToBinary function has binary value of the decimal number n and is stored in the array binaryNumber[]. Although computers are based on the binary number system, we don\u2019t have to use binary numbers when using one. Especially for IPv6 addresses it's useful to understand how you can calculate from hexadecimal to binary and decimal or the other way around. We use decimal number system for our day to day calculations. C Program to Convert Decimal to Binary. C++ program to convert decimal number into binary. However there are alternate solutions, in case you were just curious to know :) Method #2 [code. Decimal to Binary Table. Decimal -> Binary Decimal to binary conversion is even more exciting. Binary string to decimal conversion. It will convert a decimal number to its nearest single-precision and double-precision IEEE 754 binary floating-point number, using round-half-to-even rounding (the default IEEE rounding mode). The equivalent decimal representation of a binary number is the sum of the powers of 2 which each digit represents. Binary number is a base 2 number because it is either 0 or 1. Binary Fractions Summary. However, if you study the Microsoft Windows calculator application that ships along with the OS, the \"programmer\" type of calculator takes an input whether the entered number is binary or decimal from the user, then internally in its code it would check whether the entered number was a valid binary or decimal number. Example values of hexadecimal numbers converted into binary, octal and decimal. These 8-bits can represent any decimal number between 0 and 255. The decimal number is equal to the sum of binary digits (d n) times their power of 2 (2 n):. There are no ads, popups or nonsense, just an awesome decimal number to BCD number converter. IP addressing is a core functionality of networking today. Take a look at decimal number 6. And I encourage you to pause the video, and try to work through it out on your own. Press button, get result. Decimal : For denoting integer and non-integer numbers, decimal number system uses 10 different digits, 0,1,2,3,4,5,6,7,8 and 9 i. Number conversion between hexadecimal and decimal. How to Convert from Decimal to Binary. C Program To Convert Decimal To Octal Number. If you work with computers, you may find yourself needing a basic understanding of binary. Entering Repeating Decimals. Binary numbers can contain digits 0 and 1 but no decimal or scientific notation. Here we will learn, How to convert decimal to binary in java with easy examples. Let me show it on example. In binary form, for example, the decimal quantity 1895 appears as 11101100111. If you\u2019d like to dig deeper into Binary, this course helps explain the details, and show how Binary is the foundation of all digital computing. dec = Fix(dec) / 2: After obtaining the first residue the variable \"dec\" is going to get itself value between 2. Binary Coded Decimal which is also called as BCD is another process for converting decimal numbers into their binary equivalents. Binary numbers have lot of use in the digital world, in fact, binary is the language of computers where 0 and 1 represent true/false, on/off and becomes key for logic formation. In the binary system, each digit represents an increasing power of 2, with the rightmost digit representing 2 0, the next representing 2 1, then 2 2, and so on. Continue multiplying by 2 until you get a resulting fractional part equal to zero. Treat the division as an integer division. binary values The following table shows the decimal values of binary numbers. Then just add them up. It then displays the decimal values and the hexadecimal value of the elements in the array returned by the GetBits method. 0x578FCF (hex) 5738447 (decimal) 025707717 (octal) 10101111000111111001111 (binary) are all different representations of the same number. You can use national decimal items anywhere that other category numeric data items can be used. 25), 2-3 (0. By using user defined logic. About the Decimal/Binary Converter. The binary or base 2 numbering system is the keystone of computer systems and digital electronics. Input and output in binary, decimal, hexadecimal or ASCII. Get the integer quotient for the next iteration. The following function takes in a binary string and returns a bigint with the decimal value:. Decimal to Binary Using Excel: https://www. To use this decimal to binary converter tool, you should type a decimal value like 308 into the left field below, and then hit the Convert button. Write a C program to input any decimal number from user and convert it to binary number system using bitwise operator. A one has the value 1 and a zero has the value 0. This shweet conversion tool will take any text string and convert it into binary code - you know? those little 1's and 0's that make our world go around. 2) Convert the decimal number 143 into binary. Hello guys, Can anyone give me the code required for 'decimal to binary conversion using verilog for 4 bit' please im doing a project. 14159 in a decimal system (base 10), or to 3. Convert number systems units. Department of Labor\u2019s Employment and Training Administration. Watch our tutorial on How To Do Decimals To Binary Numbers from one of Videojug's professional experts. (Binary to decimal). ; In this method of conversion the fractional part of the given decimal number is multiplied by 2 (i. Binary number: In digital electronics and mathematics, a binary number is a number expressed in the base-2 numeral system or binary numeral system. The Quotient method is the new conversion method invented and derived by using Raghavendra's Analysis. There are other methods also available which can also be used in decimal to binary conversion. Convert Binary to Decimal Numbers. Let's see if we can get some experience converting from a decimal representation to a binary representation. Binary form base 8 Octal form base 10 Decimal form (common used) base 16 Hexadecimal form (hexa) Examples. This is a conversion table with decimal numbers next to their binary and hex equivalents. It means that a binary number can be formed by using 0 and 1 (2 base elements) only. Binary to Decimal conversion How to convert decimal to binary Conversion steps: Divide the number by 2. Useful, free online tool that converts plain text to decimal string. Fast Decimal to Binary Conversion. The flip side of the coin is translating a decimal number into binary format. (The old flash version is here. What that means is, everytime you count numbers under base 10, you are counting from zero to nine, then starting over by adding another number in front to make 10 and so on. It casts a string following it to a binary string. Several numeric data types are discussed, including the common \"packed\" and \"comp-3\" fields. The computer chips that run PCs work with a binary system. Binary arithmetic is pretty easy once you know what's going on. A circuit is either on or off. 33) convert to binary Ask for details ; Follow Report by Arjun306 7 minutes ago Log in to add a comment. The representation of such a number in decimal and binary and any other such system must be infinite in length and non-periodic. Converting Decimal Fractions to Binary. 625) a_split = (int(a//1),a%1) Convert the fractional part in its binary representation. Program to convert decimal to binary more than 18 bits. A number only becomes hexadecimal, decimal or binary when you format it in a certain way for printing. The right-most. s, To binary is easy: each digit after the decimal point (to the left) represents powers of 2. In the example a 1, 0, and then 1 is added to the listbox. Learn how to convert 26 to binary and create a Java project to convert from base-10 to base-2. decimal synonyms, decimal pronunciation, decimal translation, English dictionary definition of decimal. It takes three binary digits to represent an octal digit. > convert_to_binary(52) 110100 In this program, we convert decimal number entered by the user into binary using a recursive function. 3424 (base 10) = 110101100000 (base 2). This way people won't think it is the decimal number \"101\" (one hundred and one). I understand the arithmetics of it, but I was wondering how I can add the remainders onto each other using strings. For example, say that N is 23. Here you can make an arithmetic calculation you are interested in values through our online calculators. An understanding of binary numbers,the binary system, and how to convert between binary and decimal is essential for anyone involved in computers, coding, and networking. Below is the code for a 3-digit BCD-Counter (binary Coded decimal) I implemented while studying verilog. For example, decimal 1234. Created by computer nerds from team Browserling. Binary to Decimal Conversion To convert binary number into decimal number we need to multiply each bit of binary by increasing powers of 2 starting from 0. We will use the bitwise operator \"AND\" to perform the desired task. Just a quick recap, binary is a base-2 number system that uses 0 and 1 as the only digits, and hexadecimal is a base-16 number system that uses sixteen values from 0 to 9 and a to f as digits. , a number with fractional part. In a base 10 (or decimal) number, each column in the number stands for a power of 10. Decimal to binary conversion: two methods to do it with Matlab In this article we're going to present two methods of a decimal to binary conversion in Matlab; that is, we're going to convert decimal numbers (numbers with 10 different symbols, from '0' to '9', or in base 10) into binary numbers (numbers with only symbols '0' and '1', or in base 2). Converting Decimal Fractions to Binary. Binary numbers can contain digits 0 and 1 but no decimal or scientific notation. Each hexadecimal digit represents four binary digits (bits) (also called a \"nibble\"), and the primary use of hexadecimal notation is as a human-friendly representation of binary coded values in computing and digital. DECIMAL TO BINARY / HEXADECIMAL TO BINARY synonyms, DECIMAL TO BINARY / HEXADECIMAL TO BINARY pronunciation, DECIMAL TO BINARY / HEXADECIMAL TO BINARY translation, English dictionary definition of DECIMAL TO BINARY / HEXADECIMAL TO BINARY. How to Show that a Number is Binary. Repeat step 2 and 3 until result is 0. This way people won't think it is the decimal number \"101\" (one hundred and one). Java program to convert decimal to binary. The negative of this number is the value of the original binary. The decimal number is equal to the sum of binary digits (d n) times their power of 2 (2 n):. Converting a binary floating point number to decimal. So for example, 10111 is actually equal to 2^4+2^2+2^1+2^0 (2^3 is not included because there is a zero in that spot). It involves dividing the number to be converted, say N, by 2 (since binary is in base 2), and making note of. Welcome to the website Decimal-to-Binary. Example values of hexadecimal numbers converted into binary, octal and decimal. 2) Do conversion by writing your own logic without using any predefined methods. I'm having trouble writing the program, specifically finding a function that converts the items in the listbox to a decimal number. We will see two Python programs, first program does the conversion using a user defined function and in the second program we are using a in-built function bin() for the decimal to binary conversion. Binary has a base of 2. I've been trying to convert the number -441 to binary, but I don't really know how I can accomplish this. This wikiHow will show you how to do this. Returns a string representation of an integer or of a binary type converted to hexadecimal. To show that a number is a binary number, follow it with a little 2 like this: 101 2. How to Convert Decimal to Binary and vise versa is crucial to master, such skills will give you the confidence when you deal with Network and Storage devices. Decimal to Base 2 Algorithm. Input and output in binary, decimal, hexadecimal or ASCII. Converting between binary and decimal numbers is fairly simple, as long as you remember that each digit in the binary number represents a power of two. Byte conversion chart for binary and decimal conversion In Data storage and when describing memory size, a Kilobyte is 2^10, or 1024 bytes. Also Read: Convert Binary to Decimal in Java. Java Convert Decimal to Binary example and examples of string to int, int to string, string to date, date to string, string to long, long to string, string to char, char to string, int to long, long to int etc. This guide shows you how to convert from decimal to binary and binary to decimal. There are many number conversion interview questions which is asked in java interviews like java program to convert decimal to hexadecimal, java program to convert decimal to. Python Program to Check Whether a String is Palindrome or Not; Python Program to Multiply Two Matrices; Python Program to Convert Decimal to Binary Using Recursion; Python Program to Find Factorial of Number Using Recursion; Python Program to Convert Decimal to Binary, Octal and Hexadecimalc; Python Program To Display Powers of 2 Using. A binary number which consists of only 1 and 0, while a decimal number is a number which consists of numbers between 0-9. How computers see IP addresses. 1 mm to decimal = 1 decimal. Quite some time ago I had a particular need to convert Decimal to Binary and vice versa. Repeat this process till quotient becomes zero. Keep calling conversion function with n/2 till n > 1, later perform n % 1 to get MSB of converted binary number. hello i have a string with 11 bits in it in binary and would like to make this into a decimal how can i do it the stuff in the string looks like this 10111100111 and when this is converted to a decimal should be 3023. 30 mm to decimal = 30 decimal. Converts decimal -100 to binary (1110011100) Convert a decimal number to hexadecimal. Java program, asking user for a positive integer and converting it to binary representation. ) Instructions Just type in any box, and the conversion is done \"live\". Steps for using the binary decoder tool: First select the type of data you want the binary code to be converted. sage code for decimal to binary expansion. In hexadecimal, each digit can be 16 values, not 10. Decimal Code Binary; Decimal Coded Wire. Two\u2019s complement notation really shows its value in binary addition, where positive and negative quantities may be handled with equal ease. Calculator for crc32, md5, sha1, ripemd128 and gost hash algorithms. Binary is the numeral system used to express data stored in computers. Practice these questions for upcoming IBPS PO Mains Examination. Then the calculate button is pressed and the textbox should display the corresponding decimal number. Converting binary to decimal (base-2 to base-10) or decimal to binary numbers (base10 to base-2) can be done in a number of different ways as shown above. Scanner; public class Decimal_Binary { Scanner scan; int num; void getVal() { System. About the Decimal/Binary Converter. Hi, I was asked to program a recursion (in C), which converts a binary to decimal. I've been trying to convert the number -441 to binary, but I don't really know how I can accomplish this. Hence, negative binary numbers are used to represent On or Off in electronic circuits. The maximum value of a 8 bit binary number is 255 in decimal. Unit Descriptions; 1 Gigabyte (binary): A gigabyte (binary) contains 1024 3 bytes, this is the same as a gibibyte. Example values of hexadecimal numbers converted into binary, octal and decimal. You can write it like this: Yes, I've made it up. Watch our tutorial on How To Do Decimals To Binary Numbers from one of Videojug's professional experts. Then the calculate button is pressed and the textbox should display the corresponding decimal number. 3020 3100 3370 3610 Question 4: Convert 11101111 from binary to decimal. Instead of using a base of 10 or 2 respectively, it uses a base of 16. Of course, a 32-bit binary cannot hold the full range of values that you can get from 12 decimal digits. Now press \" x \" button from the common operators box. program to convert decimal number to binary DESCRIPTION-: This program receive input as decimal number and then calculate its binary equivalent and then display the binary equivalent of that decimal number. - [Voiceover] Let's now see if we can convert a larger decimal representation to binary. To indicate our answer is in binary (binary numbers are base 2), we must include a 2 at the bottom right corner of the sequence. Binary number system is the basis of all computer and digital systems. As written, IniVal is declared, but only temp is declared and typed as string. Hexadecimal number system: It is base 16 number system which uses the digits from 0 to 9 and A, B, C, D, E. Binary, octal and hexadecimal number systems are closely related and we may require to convert decimal into these systems. If it's odd, write a 1. Please feel free to edit/adapt this before posting. Convert time hh:mm:ss to decimal hours, decimal minutes and total seconds. Instead, we enter decimal numbers the computer converts into binary before manipulating them. An Algorithm for Converting a Decimal Number to a Binary Number by Fox Valley Technical College is licensed under a Creative Commons Attribution 4. CBinS16 \u2013 converts a decimal signed integer to 16 bit binary string. The numbering system everyone is most familiar with is base 10, also known as decimal or denary. This is a discussion of COBOL Computational fields. First you need to define the properties of the 16 bit binary encoding that you want to use. Calculating binary numbers can be confusing, until you figure out the system. Java program to convert decimal to binary. Decimal to hexadecimal conversion method: Following steps describe how to convert decimal to hexadecimal Step 1: Divide the original decimal number by 16 Step 2: Divide the quotient by 16 Step 3: Repeat the step 2 until we get quotient equal to zero. Hi i am writing a program to convert a user entered integer number into its binary representation, the aim is to repeatedly ask the user to enter a value then output the number in binary and when 9999 is entered end program, i've managed to make it work with positive integers so far only it displays. Compiled on Platform: Windows XP Pro SP2. 001 is decimal. Decimals are nothing more than glorified fractions. If you\u2019d like to dig deeper into Binary, this course helps explain the details, and show how Binary is the foundation of all digital computing. Can you describe a method of converting a binary number to decimal? 2. June 24, 2019. Following fig-1 mentions block diagram of decimal to binary labview vi. 0 to 255) Convert any binary number from 0 to 111111111 into octal; Convert any octal number from 0 to 777 into binary; Convert any binary number from 0 to 1111 1111 1111 1111 into hex. Using Command Line Arguments Program to Convert a Decimal to Binary Number. Hexadecimal is used to convert byte/modern computer numbers into defined binary digits. The program for conversion of decimal to binary depends on the problem specification. Create a program that would convert a decimal number to binary, octal or hexadecimal counterpart. Write your instructions in English. Converting Decimal to Binary numbers can be done in a number of different ways as shown above. 121 163 199 212 Question 2: Convert 101101 from binary to decimal. Let's see if we can convert the number 13 in decimal to binary. Decimal to Binary, Hex & Octal Converter to perform decimal to binary, decimal to hex & decimal to octal conversion online by using simple successive division methods along with step by step calculation & solved example problems. In this system, numbers are represented in a decimal form, however each decimal digit is encoded using a four bit binary number. All the best and keep practicing!. Before a floating-point binary number can be stored correctly, its mantissa must be normalized. We set it equal to the expression in Equation (2. A small point, but IniVal is NOT typed as a string variable in your code. Given a decimal number as input, we need to write a program to convert the given decimal number into equivalent binary number. Example for Binary to Decimal Conversion. In this post, we will see programs to convert decimal number to an equivalent binary number. Decimal ---> Base 10. Repeat the steps until the quotient is equal to 0. Step 2: Divide 10 by 2. decimal synonyms, decimal pronunciation, decimal translation, English dictionary definition of decimal. 500 * 10 9 bytes = 500,000,000,000 = 500 Gigabytes But the operating system determines the size of the drive using the computer's binary powers of two definition of the \"Giga\" prefix:. Like I said a hexadecimal number doesn't actually name a type: a type is something like long, or int. Hexadecimal is base 16. I first converted 441 to binary which is: 110111001 Then I'm supposed to take the complim. If everything is done right, the result should be 34. Java decimal to binary conversion is the most important core java interview question. 7/2 = Quotient = 3(grater than 1), Remainder = 1. Conversion Hexa to Decimal. Source code DecimalToBinaryConverter. The hexadecimal number system (hex) functions virtually identically to the decimal and binary systems. In computing and electronic systems, binary-coded decimal (BCD) is a digital encoding method for decimal numbers in which each digit is represented by its own binary sequence. You rely on decimal numbers when doing calculations for your business, such as when figuring sales tax or creating a monthly budget. How to Convert Decimal to Binary and vise versa is crucial to master, such skills will give you the confidence when you deal with Network and Storage devices. Convert 13 10 to binary:. If you understand binary because you learned it for understanding how IP addresses work, you usually stop at 2 7 because one octet of an IP address is 8 bits long. Say we wish to convert an unsigned decimal integer, $$N\\text{,}$$ to binary. The binary system is the internal language of electronic computers. Set s=0 for a positive number and s=1 for a negative number. How to convert binary to decimal: The binary number system, also known as the base 2 number system; is used by all modern generation computers internally. Online IEEE 754 floating point converter and analysis. BCD-To-Decimal Decoder Binary-To-Octal Decoder The MC14028B decoder is constructed so that an 8421 BCD code on the four inputs provides a decimal (one\u2212of\u2212ten) decoded output, while a 3\u2212bit binary input provides a decoded octal (one\u2212of\u2212eight) code output with D forced to a logic \u201c0\u201d. This free binary calculator can add, subtract, multiply, and divide binary values, as well as convert between binary and decimal values. It took a little digging but eventually came across this little gem and thought it was worth posting for others to use. For more information about the binary representation of Decimal values and an example, see the Decimal(Int32[]) constructor and the GetBits method. How to convert from decimal number system to binary number system using bitwise operator in C programming. Decimal to binary converter Abstract. Decimal : For denoting integer and non-integer numbers, decimal number system uses 10 different digits, 0,1,2,3,4,5,6,7,8 and 9 i. The binary number is constructed from right to left. Convert from binary to decimal algorithm: For this we multiply each digit separately from right side by 1, 2, 4, 8, 16 \u2026 respectively then add them. Decimal conversions can be done by either successive division method or successive multiplication method. Then enter or paste your binary code in the first text box and click Decode button. (The old flash version is here. And I encourage you to pause the video, and try to work through it out on your own. The following utility converts the IP (TCP/IP) address to other browser URL addressable forms. Hexadecimal --> Base 16. For more information about the binary representation of Decimal values and an example, see the Decimal(Int32[]) constructor and the GetBits method. This page of labview source code covers decimal to binary labview vi which converts decimal vector to binary vector. 0,1,2,3,4,5,6,7,8,9 are decimal number and all other numbers are based on these 10 numbers. Here you can make an arithmetic calculation you are interested in values through our online calculators. Binary --> Base 2. The syntax is similar to that of the ToInt32 method. , a number with fractional part. That is the most confusing thing that I want to say about the subject. Jun 9, It was a fascinating rabbit hole that exposed me to things like \u201cdecimal machines\u201d and \u201c2 out of 5 code\u201d. Short for binary-coded decimal, BCD is also known as packet decimal and is numbers 0 through 9 converted to four-digit binary. So Java Code Online is going to discuss a case when a number entered as binary is to be converted to a decimal number. Dotted decimal notation is a system of presenting numbers that is a little different from the common conventions in arithmetic as it is taught in schools. Here you can find the answer to questions like: Convert decimal number 168 in binary or Decimal to binary conversion. The Land Before Binary.", "date": "2019-11-13 18:41:15", "meta": {"domain": "mindcode.org", "url": "http://mindcode.org/4u9mex/mcl5.php?vj=decimal-to-binary", "openwebmath_score": 0.3413355052471161, "openwebmath_perplexity": 843.847305476133, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.981735721648143, "lm_q2_score": 0.8872045937171068, "lm_q1q2_score": 0.8710004420624115}}
{"url": "https://discuss.codedrills.io/t/maximum-prefix-mex-sum-editorial/187", "text": "# Maximum Prefix Mex Sum - Editorial\n\nSetter(s) Divyansh Verma\nTester(s) Vichitr Gandas, Mayank Pugalia\nDifficulty Easy\nTopics Observation, Greedy, Maths, Combinatorics\n\n## Solution Idea\n\nThe prefix Mex sum is maximum when we put all unique numbers in increasing order at the start. Once a number is missed from 0,1,2..., this chain breaks and then Mex will remain same for rest of the prefix arrays. Example:\nA=[0,1,2,5,7]\nHere Mex for first three prefix arrays is 1,2,3 respectively. After that we see that 3 is not present. Hence for rest of subarrays Mex will remain 3 only.\nThis way, the sum is maximized when we arrange initial unique numbers in order 0,1,2... like this. And if this order breaks or we are left with duplicate items, we can permute them in any way. Here in the beginning, each unique number x can be picked with count[x] ways where count[x] is number of times it occurs in the array. And let\u2019s say number of remaining elements after break is y, they can be arranged in y! ways. Hence total number of permutations with max prefix Mex sum would be count[0]*count[1]*...*count[k]*y! where k is first number where this sequence 0,1,2,... breaks.\n\nCount can be maintained with map. We can reduce the time complexity here by using a vector instead because the sequence if continued won\u2019t exceed n and frequency of rest of the items is not needed, because they will be part of left over items which can be permuted.\n\n### Complexity Analysis\n\nTime Complexity: \\mathcal{O}(n \\log{n}) for pre calculation and to find the answer which can be reduced to \\mathcal{O}(n) using a vector instead of map.\nSpace Complexity: \\mathcal{O}(n) for maintaining the frequency.\n\n## Codes\n\nSetter's Code\n#include<bits/stdc++.h>\nusing namespace std;\n#define int long long int\n\nconst int MAXN = 1e5 + 10;\nconst int MOD = 1e9 + 7;\n\nint fact[MAXN];\n\nvoid preCompute(){\nfact[0] = 1;\nfor(int i=1;i<MAXN;i++) fact[i] = (fact[i-1]*i)%MOD;\n}\n\nclass FindMexMax {\npublic:\nint getOutput(vector<int> a){\n\nmap<int,int> which;\nfor(int v:a) which[v]++;\n\nint cnt = a.size(),ans = 1;\n\nfor(int i=0;i<a.size();i++){\nif(which[i] == 0) break;\nans = (ans*which[i])%MOD;\nwhich[i]--;\ncnt--;\n}\n\nans = ans*fact[cnt]%MOD;\nreturn ans;\n}\n};\n\nsigned main(){\npreCompute();\nint t;\ncin>>t;\nwhile(t--){\nint n; cin>>n;\nvector<int> a(n);\nfor(int &v:a) cin>>v;\nFindMexMax fMM;\ncout<<fMM.getOutput(a)<<endl;\n}\n}\n\nTester's Code\n/***************************************************\n\n@author: vichitr\nCompiled On: 13th Mar 2021\n\n*****************************************************/\n#include<bits/stdc++.h>\n#define endl \"\\n\"\n#define int long long\nusing namespace std;\n\nconst int MOD = 1e9 + 7;\nconst int N = 1e5 + 7;\nint fact[N];\n\nvoid init(){\nfact[0] = 1;\nfor(int i=1;i<N;i++){\nfact[i] = (fact[i-1] * i) % MOD;\n}\n}\n\nvoid solve(){\nint n; cin>>n;\nassert(n <= 1e5);\nint a[n];\nmap<int, int> M;\nfor(int i=0;i<n;i++){\ncin>>a[i];\nassert(a[i] >= 0 and a[i] <= 1e9);\nM[a[i]]++;\n}\nint ans = 1;\nint tot = 0;\nfor(int i = 0; i < n; i++){\nif(M[i] == 0){\nbreak;\n}\nans *= M[i];\nans %= MOD;\ntot++;\n}\nans *= fact[n - tot];\nans %= MOD;\ncout<<ans<<'\\n';\n}\n\nsigned main(){\nint t = 1;\ncin>>t;\nassert(t <= 5);\ninit();\nwhile(t--){\nsolve();\n}\nreturn 0;\n}\n\n\nIf you have used any other approach, share your approach in comments!", "date": "2021-06-15 12:13:09", "meta": {"domain": "codedrills.io", "url": "https://discuss.codedrills.io/t/maximum-prefix-mex-sum-editorial/187", "openwebmath_score": 0.6183327436447144, "openwebmath_perplexity": 7735.135732871213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357221825194, "lm_q2_score": 0.8872045832787204, "lm_q1q2_score": 0.8710004322887758}}
{"url": "http://yutsumura.com/true-or-false-every-diagonalizable-matrix-is-invertible/", "text": "# True or False. Every Diagonalizable Matrix is Invertible\n\n## Problem 439\n\nIs every diagonalizable matrix invertible?\n\n## Solution.\n\n### Counterexample\n\nWe give a counterexample. Consider the $2\\times 2$ zero matrix.\nThe zero matrix is a diagonal matrix, and thus it is diagonalizable.\nHowever, the zero matrix is not invertible as its determinant is zero.\n\n### More Theoretical Explanation\n\nLet us give a more theoretical explanation.\nIf an $n\\times n$ matrix $A$ is diagonalizable, then there exists an invertible matrix $P$ such that\n$P^{-1}AP=\\begin{bmatrix} \\lambda_1 & 0 & \\cdots & 0 \\\\ 0 & \\lambda_2 & \\cdots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & \\lambda_n \\end{bmatrix},$ where $\\lambda_1, \\dots, \\lambda_n$ are eigenvalues of $A$.\nThen we consider the determinants of the matrices of both sides.\nThe determinant of the left hand side is\n\\begin{align*}\n\\det(P^{-1}AP)=\\det(P)^{-1}\\det(A)\\det(P)=\\det(A).\n\\end{align*}\nOn the other hand, the determinant of the right hand side is the product\n$\\lambda_1\\lambda_2\\cdots \\lambda_n$ since the right matrix is diagonal.\nHence we obtain\n$\\det(A)=\\lambda_1\\lambda_2\\cdots \\lambda_n.$ (Note that it is always true that the determinant of a matrix is the product of its eigenvalues regardless diagonalizability.\nSee the post \u201cDeterminant/trace and eigenvalues of a matrix\u201c.)\n\nHence if one of the eigenvalues of $A$ is zero, then the determinant of $A$ is zero, and hence $A$ is not invertible.\n\nThe true statement is:\n\na diagonal matrix is invertible if and only if its eigenvalues are nonzero.\n\n### Is Every Invertible Matrix Diagonalizable?\n\nNote that it is not true that every invertible matrix is diagonalizable.\n\nFor example, consider the matrix\n$A=\\begin{bmatrix} 1 & 1\\\\ 0& 1 \\end{bmatrix}.$ The determinant of $A$ is $1$, hence $A$ is invertible.\nThe characteristic polynomial of $A$ is\n\\begin{align*}\np(t)=\\det(A-tI)=\\begin{vmatrix}\n1-t & 1\\\\\n0& 1-t\n\\end{vmatrix}=(1-t)^2.\n\\end{align*}\nThus, the eigenvalue of $A$ is $1$ with algebraic multiplicity $2$.\nWe have\n$A-I=\\begin{bmatrix} 0 & 1\\\\ 0& 0 \\end{bmatrix}$ and thus eigenvectors corresponding to the eigenvalue $1$ are\n$a\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}$ for any nonzero scalar $a$.\nThus, the geometric multiplicity of the eigenvalue $1$ is $1$.\nSince the geometric multiplicity is strictly less than the algebraic multiplicity, the matrix $A$ is defective and not diagonalizable.\n\n### Is There a Matrix that is Not Diagonalizable and Not Invertible?\n\nFinally, note that there is a matrix which is not diagonalizable and not invertible.\nFor example, the matrix $\\begin{bmatrix} 0 & 1\\\\ 0& 0 \\end{bmatrix}$ is such a matrix.\n\n## Summary\n\nThere are all possibilities.\n\n1. Diagonalizable, but not invertible.\nExample: $\\begin{bmatrix} 0 & 0\\\\ 0& 0 \\end{bmatrix}.$\n2. Invertible, but not diagonalizable.\nExample: $\\begin{bmatrix} 1 & 1\\\\ 0& 1 \\end{bmatrix}$\n3. Not diagonalizable and Not invertible.\nExample: $\\begin{bmatrix} 0 & 1\\\\ 0& 0 \\end{bmatrix}.$\n4. Diagonalizable and invertible\nExample: $\\begin{bmatrix} 1 & 0\\\\ 0& 1 \\end{bmatrix}.$\n\nDetermine whether each of the following statements is True or False. (a) If $A$ and $B$ are $n \\times n$...", "date": "2017-08-20 00:08:42", "meta": {"domain": "yutsumura.com", "url": "http://yutsumura.com/true-or-false-every-diagonalizable-matrix-is-invertible/", "openwebmath_score": 0.9666319489479065, "openwebmath_perplexity": 232.2265187640673, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990731984977195, "lm_q2_score": 0.8791467801752451, "lm_q1q2_score": 0.8709988346093301}}
{"url": "https://math.stackexchange.com/questions/2543272/mathbbz-2-times-mathbbz-2-has-order-4-how-so/2543281", "text": "# $\\mathbb{Z}_2\\times\\mathbb{Z}_2$ has order $4$? How so?\n\n$\\mathbb{Z}_2\\times\\mathbb{Z}_2$ has order $4$, the neutral element is $(0_2,0_2)$ and the other elements have order $2$. Therefore, $\\mathbb{Z}_2\\times\\mathbb{Z}_2$ is not cyclic,so it is the Klein group.\n\nThe elements of $\\mathbb{Z_2}$ are $\\{0_2,1_2\\}$ and the order of $1_2$ is 2.\n\nIf we take the direct product $\\mathbb{Z}_2\\times\\mathbb{Z}_2$ then we have a generator of the Group product $\\langle 1_2,1_2\\rangle$.\n\nQuestion:\n\nHow can the author state $\\mathbb{Z}_2\\times\\mathbb{Z}_2$ has order $4$? Is $\\langle 1_2,1_2\\rangle$ not a generator of order 2 of $\\mathbb{Z}_2\\times\\mathbb{Z}_2$?\n\n\u2022 Remember that the order of a group is a somewhat distinct concept from the order of an element. The order of a group is the cardinality of that group (in this case, 4---the elements can be explicitly written fairly easily), while the order of an element $a$ is the smallest natural number $n$ such that $a^n$ is the identity (here, the maximal order of an element is 2). \u2013\u00a0Xander Henderson Nov 29 '17 at 18:54\n\u2022 The order of a group means its size. \u2013\u00a0Ittay Weiss Nov 29 '17 at 18:54\n\u2022 $(1,1)$ does not generate the element $(0,1)$ \u2013\u00a0Doug M Nov 29 '17 at 18:55\n\u2022 $\\langle 1, 1\\rangle$ only generates all of $\\Bbb Z_n\\times \\Bbb Z_m$ if the greatest common factor of $n$ and $m$ is 1. Here it is 2. \u2013\u00a0MJD Nov 29 '17 at 18:57\n\nThe order of a group is the number of elements in it. $\\mathbb{Z}_2\\times\\mathbb{Z}_2$ has four elements: $(0,0), (0,1), (1,0), (1,1)$.\nBut as you have seen, each element has order $\\le 2$. (This, in particular, proves that $\\mathbb{Z}_2\\times\\mathbb{Z}_2$ is not isomorphic to $\\mathbb{Z}_4$.)\nThe elements of $\\mathbb{Z}_2 \\times \\mathbb{Z}_2$ are", "date": "2019-10-22 14:02:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2543272/mathbbz-2-times-mathbbz-2-has-order-4-how-so/2543281", "openwebmath_score": 0.7105020880699158, "openwebmath_perplexity": 134.47650964410383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319863454374, "lm_q2_score": 0.8791467611766711, "lm_q1q2_score": 0.8709988169897213}}
{"url": "https://artofproblemsolving.com/wiki/index.php?title=1995_AHSME_Problems/Problem_27&curid=7037&diff=138560&oldid=128744", "text": "# Difference between revisions of \"1995 AHSME Problems/Problem 27\"\n\n## Problem\n\nConsider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.\n\n$$\\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\\\ & & & & 1 & & 1 & & & & \\\\ & & & 2 & & 2 & & 2 & & & \\\\ & & 3 & & 4 & & 4 & & 3 & & \\\\ & 4 & & 7 & & 8 & & 7 & & 4 & \\\\ 5 & & 11 & & 15 & & 15 & & 11 & & 5 \\end{tabular}$$\n\nLet $f(n)$ denote the sum of the numbers in row $n$. What is the remainder when $f(100)$ is divided by 100?\n\n$\\mathrm{(A) \\ 12 } \\qquad \\mathrm{(B) \\ 30 } \\qquad \\mathrm{(C) \\ 50 } \\qquad \\mathrm{(D) \\ 62 } \\qquad \\mathrm{(E) \\ 74 }$\n\n## Solution 1\n\nNote that if we re-draw the table with an additional diagonal row on each side, the table is actually just two of Pascal's Triangles, except translated and summed. $$\\begin{tabular}{ccccccccccccccc} & & & & & 1 & & 0 & & 1 & & & & \\\\ & & & & 1 & & 1 & & 1 & & 1 & & & \\\\ & & & 1 & & 2 & & 2 & & 2 & & 1 & & \\\\ & & 1 & & 3 & & 4 & & 4 & & 3 & & 1 & \\\\ & 1 & & 4 & & 7 & & 8 & & 7 & & 4 & & 1 \\\\ 1 & & 5 & & 11 & & 15 & & 15 & & 11 & & 5 & & 1 \\end{tabular}$$\n\nThe sum of a row of Pascal's triangle is $2^{n-1}$; the sum of two of each of these rows, subtracting away the $2$ ones we included, yields $f(n) = 2^n - 2$. Now, $f(100) = 2^{100} - 2 \\equiv 2 \\pmod{4}$ and $f(100) = 2^{100} - 2 \\equiv 2^{20 \\cdot 5} - 2 \\equiv -1 \\pmod{25}$, and by the Chinese Remainder Theorem, we have $f(100) \\equiv 74 \\pmod{100} \\Longrightarrow \\mathrm{(E)}$.\n\n## Solution 2 (induction)\n\nWe sum the first few rows: $0, 2, 6, 14, 30, 62$. They are each two less than a power of $2$, so we try to prove it:\n\nLet the sum of row $n$ be $S_n$. To generate the next row, we add consecutive numbers. So we double the row, subtract twice the end numbers, then add twice the end numbers and add two. That makes $S_{n+1}=2S_n-2(n-1)+2(n-1)+2=2S_n+2$. If $S_n$ is two less than a power of 2, then it is in the form $2^x-2$. $S_{n+1}=2^{x+1}-4+2=2^{x+1}-2$.\n\nSince the first row is two less than a power of 2, all the rest are. Since the sum of the elements of row 1 is $2^1-2$, the sum of the numbers in row $n$ is $2^n-2$. Thus, using Modular arithmetic, $f(100)=2^{100}-2 \\bmod{100}$. $2^{10}=1024$, so $2^{100}-2\\equiv 24^{10}-2\\equiv (2^3 \\cdot 3)^{10} - 2$ $\\equiv 1024^3 \\cdot 81 \\cdot 81 \\cdot 9 - 2 \\equiv 24^3 \\cdot 19^2 \\cdot 9 - 2$ $\\equiv 74\\bmod{100} \\Rightarrow \\mathrm{(E)}$.\n\n## Solution 3 (plain recurrence solving)\n\nWe derive the recurrence $S_{n+1}=2S_n + 2$ as above. Without guessing the form of the solution at this point we can easily solve this recurrence. Note that one can easily get rid of the \"$+2$\" as follows: Let $S_n=T_n-2$. Then $S_{n+1}=T_{n+1}-2$ and $2S_n+2 = 2(T_n-2)+2 = 2T_n-2$. Therefore $T_{n+1}=2T_n$. This obviously solves to $T_n=2^{n-1} T_1$. As $S_1=0$, we have $T_1=2$. Therefore $T_n=2^n$ and consecutively $S_n=2^n-2$.", "date": "2021-01-27 00:39:10", "meta": {"domain": "artofproblemsolving.com", "url": "https://artofproblemsolving.com/wiki/index.php?title=1995_AHSME_Problems/Problem_27&curid=7037&diff=138560&oldid=128744", "openwebmath_score": 0.9022967219352722, "openwebmath_perplexity": 178.37334397773034, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419677654415, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.870969126619185}}
{"url": "https://math.stackexchange.com/questions/3218687/show-int-02-pi-cos2t-dt-pi-using-pythagorean-theorem/3218697", "text": "# Show $\\int_0^{2\\pi}\\cos^2t\\,dt=\\pi$ using Pythagorean Theorem\n\nThe 'traditional' way I always see this integral calculated is with the identity\n\n$$\\cos^2t=\\frac{1+\\cos(2t)}{2}$$\n\nMy alternative method uses $$\\cos^2t+\\sin^2t=1$$. It's obvious that\n\n$$\\int_0^{2\\pi}\\cos^2t+\\sin^2t\\,dt=\\int_0^{2\\pi}1\\,dt=2\\pi$$\n\nThe interval of integration is an integer multiple of the periods of each function ($$\\cos^2t$$ has a period of $$\\pi$$), and so it seems reasonable to me that given the Pythagorean identity used above, $$\\cos^2t$$ and $$\\sin^2t$$ contribute, for lack of a better word, equally to this final answer of $$2\\pi$$ above, and so the integral in the title should be half of $$2\\pi$$, or just $$\\pi$$.\n\nIs this method valid? What additional statements, if any, are necessary to make it rigorous enough to be valid?\n\n\u2022 One way to make it more rigorous is just to shift the integral for $\\sin^2$ by $\\pi/2$ so that you just get the integral for $\\cos^2$ twice. \u2013\u00a0vrugtehagel May 8 at 16:58\n\u2022 One of my high-school teachers showed me this trick for calculating the average of $\\cos^2 x$, and it's stuck with me ever since. \u2013\u00a0Michael Seifert May 8 at 17:04\n\n## 2 Answers\n\nYour argument is definitely valid. To add more explanation, we can say that $$\\int_{a}^{a+T} f(x)dx = \\int_{0}^{T} f(x) dx$$ for any $$T$$-periodic function $$f(x)$$ (try to prove this rigorously), and then $$\\int_{0}^{2\\pi} \\sin^{2} t dt = \\int_{0}^{2\\pi} \\cos^{2}\\left(t-\\frac{\\pi}{2}\\right) \\,dt = \\int_{-\\frac{\\pi}{2}}^{2\\pi - \\frac{\\pi}{2}} \\cos^{2}t\\,dt = \\int_{0}^{2\\pi} \\cos^{2}t\\,dt$$\n\nUse $$\\displaystyle\\int_0^{2a}f(x)\\ dx=2\\int_0^af(x) \\ dx$$ for $$f(2a-x)=f(x)$$\n\ntwice to find $$I=\\int_0^{2\\pi}\\cos^2t\\ dt=4\\int_0^{\\pi/2}\\cos^2t\\ dt$$\n\nNow $$\\displaystyle\\int_a^bf(x) \\ dx=\\int_a^bf(a+b-x) \\ dx$$\n\nto find $$2\\cdot\\dfrac I4=\\int_0^{\\pi/2}(\\cos^2t+\\sin^2t)dt$$", "date": "2019-08-24 07:43:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3218687/show-int-02-pi-cos2t-dt-pi-using-pythagorean-theorem/3218697", "openwebmath_score": 0.9391657114028931, "openwebmath_perplexity": 265.212214038595, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765581257486, "lm_q2_score": 0.8887587817066392, "lm_q1q2_score": 0.8709627719009059}}
{"url": "https://math.stackexchange.com/questions/4043453/teacher-claims-this-proof-for-frac-csc-theta-1-cot-theta-frac-cot-theta/4043660", "text": "# Teacher claims this proof for $\\frac{\\csc\\theta-1}{\\cot\\theta}=\\frac{\\cot\\theta}{\\csc\\theta+1}$ is wrong. Why?\n\nMy son's high school teacher says his solution to this proof is wrong because it is not \"the right way\" and that you have to \"start with one side of the equation and prove it is equal to the other\". After reviewing it, I disagree. I believe his solution is correct, even if not \"the right way\", whatever that means. I asked my son how he did it: he cross-multiplied the given identity, simplified it to a known/obvious equality, and then reversed the steps for the proof. This was a graded exam, and the teacher gave him a zero for this problem.\n\nWhat do you think about my son's solution? Thanks!\n\nProblem: prove the following trigonometric identity \\begin{align*} \\frac{\\csc(\\theta)-1}{\\cot(\\theta)}&=\\frac{\\cot(\\theta)}{\\csc(\\theta)+1}\\ .\\\\ \\end{align*} Solution: for all real $$\\theta$$ not equal to an integer multiple of $$\\pi/2$$, we have \\begin{align*} \\cot^2(\\theta)&=\\cot^2(\\theta)\\\\[8pt] \\frac{\\cos^2(\\theta)}{\\sin^2(\\theta)}&=\\cot^2(\\theta)\\\\[8pt] \\frac{1-\\sin^2(\\theta)}{\\sin^2(\\theta)}&=\\cot^2(\\theta)\\\\[8pt] \\csc^2(\\theta)-1&=\\cot^2(\\theta)\\\\[8pt] \\frac{\\csc^2(\\theta)-1}{\\cot(\\theta)}&=\\cot(\\theta)\\\\[8pt] \\frac{\\csc(\\theta)-1}{\\cot(\\theta)}&=\\frac{\\cot(\\theta)}{\\csc(\\theta)+1} \\end{align*}\n\n\u2022 Ask the professor? They may have a specific way they ask you to prove them. I have my students prove them a specific way. These can be googled or whatever. Feb 28, 2021 at 17:52\n\u2022 Your son's proof is correct. He proved that the required identity is equivalent to $\\cot^2 \\theta = \\cot^2 \\theta$ which is true. So the identity is true. Feb 28, 2021 at 17:52\n\u2022 You would be amazed how easy it is to get an b.s. in education in the US. Feb 28, 2021 at 17:53\n\u2022 The proof looks good to me (and actually this is exactly the order I would have written it in). Sorry your son has a teacher who thinks math is just following a given series of steps and that there is one right way to do every problem. Feb 28, 2021 at 17:53\n\u2022 The proof is correct and furthermore easy to follow. If the teacher believes it to be wrong, then they should be able to specifically name the error (\"Line $n$ is wrong because...\"). Saying that a method is \"not the right way\" is questionable pedagogy at best, and at worst, suggests that the teacher doesn't fully understand the material. But perhaps there is missing information, e.g., the full question might be, \"Prove the statement using the Right Way method\". Feb 28, 2021 at 18:04\n\nI would at least have skipped the first line and started with $$\\dfrac{\\cos^2\\theta}{\\sin^2\\theta} = \\cot^2\\theta.$$\n\nOne reason why an instructor might have qualms about this argument is the frequency with which students do this in the opposite direction, i.e. they write \\require{cancel} \\xcancel{ \\begin{align} \\frac{\\csc(\\theta)-1}{\\cot(\\theta)}&=\\frac{\\cot(\\theta)}{\\csc(\\theta)+1} \\\\[8pt] \\frac{\\csc^2(\\theta)-1}{\\cot(\\theta)}&=\\cot(\\theta)\\\\[8pt] \\csc^2(\\theta)-1&=\\cot^2(\\theta)\\\\[8pt] \\frac{1-\\sin^2(\\theta)}{\\sin^2(\\theta)}&=\\cot^2(\\theta)\\\\[8pt] \\frac{\\cos^2(\\theta)}{\\sin^2(\\theta)}&=\\cot^2(\\theta)\\\\[8pt] \\cot^2(\\theta)&=\\cot^2(\\theta) \\end{align} } This would be logically correct if one wrote \"This equality is true if the one below it is true.\" on each line. But one must be clear about the correct direction of \"If\u00a0...\u00a0then\u00a0...\". Without those explicit words, the sequence of equalities can be taken to mean \"If\u00a0...\u00a0then\u00a0...\" in just the opposite order.\n\nOften what instructors want is something like this: $$\\frac{\\csc\\theta-1}{\\cot\\theta} = \\cdots\\cdots\\cdots\\cdots = \\frac{\\cot\\theta}{\\csc\\theta+1}.$$ in which each \"equals\" sign asserts the equality of things already known to be equal.\n\nI would give full credit to the answer that your son wrote if each line had some brief explanation of how it is deduced from the line before it. For example, he could write this:\n\n$$\\cot^2\\theta = \\frac{\\cos^2\\theta}{\\sin^2\\theta} = \\frac{1-\\sin^2\\theta}{\\sin^2\\theta} = \\csc^2\\theta-1.$$ Then, dividing both sides of the equality $$\\cot^2\\theta = \\csc^2\\theta-1$$ by $$\\cot\\theta,$$ we get $$\\cot\\theta = \\frac{\\csc^2\\theta-1}{\\cot\\theta}.$$ Finally, dividing both sides of that by $$\\csc\\theta+1$$ we get $$\\frac{\\csc\\theta-1}{\\cot\\theta} = \\frac{\\cot\\theta}{\\csc\\theta+1}.$$\n\nNote that I wrote words above, not just mathematical notation. Well written answers do that, except perhaps in fairly simple cases.\n\n\u2022 Very nice all the answer and the question +1 for all the users. Very nice the required cancel :-) for MathJAX. Feb 28, 2021 at 22:48\n\u2022 @Sebastiano : Thank you. Mar 1, 2021 at 6:16\n\nThere are multiple motivations for writing proofs: perhaps to pass an exam (from a regular student's viewpoint), or perhaps to test understanding (from a teacher's perspective). Here, I'd like to highlight another important motivation: to enjoy the process of discovery.\n\nHere's a geometric proof of the trigonometric identity. It is not practical to use in an exam, but it makes the abstract symbols more concrete. Coming up with multiple proofs of the same result can help one develop intuition and allow one to appreciate the connections between different areas of mathematics.\n\nProblem statement: $$\\frac{\\csc\\theta-1}{\\cot\\theta}=\\frac{\\cot\\theta}{\\csc\\theta+1}$$\n\nIn this picture involving the unit circle, we construct some lengths which correspond to expressions in the problem statement. We start with $$\\theta$$ and $$\\phi = \\frac{\\pi}{2} - \\theta$$.\n\nBasic trigonometric definitions lead to the blue and orange lengths. By some angle-chasing (or a circle theorem), we have the $$\\color{red}{\\text{red}}$$ angle. Then, more angle-chasing gives:\n\nThe $$\\color{red}{\\text{red}}$$ angles are the same $$\\implies\\triangle ADB\\sim \\triangle BDE \\implies \\frac{AD}{DB}=\\frac{DB}{ED}\\iff$$ the problem statement.\n\n\u2022 Love this geometric approach! If you don't mind, what do you use to produce such a nice and clear diagram of the geometry/trigonometry? Thanks! Feb 28, 2021 at 21:20\n\u2022 Thank you. I'm glad you liked it. I used GeoGebra: geogebra.org/m/yngwjpcg Feb 28, 2021 at 21:34\n\nI would assume that the teacher considered it an invalid answer, because your son assumed that the equality is true before showing that it is, if that even makes any sense. This is obviously a guess, but in the mind of the teacher, what he did was not \"showing\" that the equality is true, but merely \"verifying\" that it is. In other words, what the teacher likely wanted is for the students to get the answer on the right-hand side by exclusively using the left-hand side. More explicitly, even though the students knew what the end result should look like, they should have pretended they don't, manipulated the left-hand side and exclaimed \"Eureka! It's true!\" when they got the RHS.\n\nWith that logic, all he needed to do was to multiply numerator and denominator by $$\\csc(\\theta)+1$$, which immediately yields the desired result after simplifications.\n\n\u2022 Indeed, and a far more immediate solution! Feb 28, 2021 at 18:14\n\u2022 Your answer reminds me of a high school maths teacher who said that I \"thought too much like an engineer\" (in the sense that I would often manipulate both sides into something that's true, in the style of the OP's son's proof). However, all of us think like this to a certain extent: some people are fast, and can immediately rewrite the proof (in their head) as \"LHS = ... = ... = RHS\"; some might hide that process on scratch paper. I don't really understand how one of these proofs can be better than the other, assuming they're correctly written. They all demonstrate the same math ability. Feb 28, 2021 at 18:25\n\u2022 That is possible that is how the teacher thinks. Thanks for the reply! Feb 28, 2021 at 18:30\n\u2022 Former high school math teacher here who taught students trig identity proofs, and this is exactly right. Moreover, it helps to keep them from making mistakes and incorrectly \"proving\" the statement after making a mistake on one side, then forcing another mistake on the other side to get to equality instead of searching for the problem in the work they've already done. It also helps them learn more about the relationships between different trig functions if they can only manipulate one side. Feb 28, 2021 at 18:31\n\u2022 Further, the point of a trig identity problem isn't always just to demonstrate equality. There are typically multiple standards built into a single question. In this case, I'm guessing the teacher wanted something involving manipulating fractions. Feb 28, 2021 at 18:31\n\nAll of these methods of proof are valid, and should be graded as such (unless the question specifically said to use a certain method).\n\nMethod $$1$$ - Start with one side, simplify to the other.\n\n$$\\frac{\\csc(\\theta)-1}{\\cot(\\theta)}=\\frac{(\\csc(\\theta)-1)(\\csc(\\theta)+1)}{\\cot(\\theta)(\\csc(\\theta)+1)}=\\frac{\\csc^{2}(\\theta)-1}{\\cot(\\theta)( \\csc(\\theta)+1)}=\\frac{\\cot^{2}(\\theta)}{\\cot(\\theta)(\\csc(\\theta)+1)}$$$$=\\boxed{\\frac{\\cot(\\theta)}{\\csc(\\theta)+1}}$$\n\nMethod $$2$$ - Start with the asserted identity, simplify to a known identity.\n\n$$\\frac{\\csc(\\theta)-1}{\\cot(\\theta)}=\\frac{\\cot(\\theta)}{\\csc(\\theta)+1}\\Longleftrightarrow(\\csc(\\theta)-1)(\\csc(\\theta)+1)=\\cot^{2}(\\theta)\\Longleftrightarrow\\csc^{2}(\\theta)-1$$$$=\\cot^{2}(\\theta)\\ \\blacksquare$$\n\nMethod $$3$$ - Start with a known identity, turn it into the asserted identity (which is what your son did):\n\n$$\\csc^{2}(\\theta)-1=(\\csc(\\theta)+1)(\\csc(\\theta)-1)=\\cot^{2}(\\theta)\\Longrightarrow\\boxed{\\frac{\\csc(\\theta)-1}{\\cot(\\theta)}=\\frac{\\cot(\\theta)}{\\csc(\\theta)+1}}$$\n\nThat said, however, perhaps the teacher took points off due to the length of your son's proof, which was more roundabout than it could have been. For instance, he could have started with a Pythagorean identity instead of starting with a '$$x=x$$' identity and using a Pythagorean identity, and took a few less steps in simplifying to the desired identity.\n\n\u2022 +1 nice answer! Feb 28, 2021 at 18:29\n\nIt's disappointing to see that the teacher believes there is only one way to prove trigonometric identities. Actually, there are a number of approaches, all equally valid and useful in different circumstances. Sometimes it is more elegant to 'start with one side and prove that it is equal to the other', but a proof is a proof.\n\nYour son is actually using quite an advanced technique, where he shows that a statement $$P$$ implies another statement $$Q$$, and then shows that all of the steps are invertible, meaning that $$P \\iff Q$$. As long as you are able to show that all of the steps are indeed invertible, this is a very useful method. Let's look at his method in more detail. He started by assuming that $$\\frac{\\csc \\theta - 1}{\\cot\\theta} = \\frac{\\cot\\theta}{\\csc\\theta+1}$$ and working from there. If we cross mutliply both sides, look at what happens: \\begin{align} &\\frac{\\csc \\theta - 1}{\\cot\\theta} = \\frac{\\cot\\theta}{\\csc\\theta+1} \\\\[6pt] \\implies & (\\csc\\theta-1)(\\csc\\theta+1) = \\cot^2\\theta \\\\[6pt] \\implies & \\csc^2\\theta-1=\\cot^2\\theta \\\\[6pt] \\implies & \\frac{1-\\sin^2\\theta}{\\sin^2\\theta} = \\frac{\\cos^2\\theta}{\\sin^2\\theta} \\\\[6pt] \\implies & \\frac{\\cos^2\\theta}{\\sin^2\\theta}=\\frac{\\cos^2\\theta}{\\sin^2\\theta} \\\\[6pt] \\implies & \\cot^2\\theta = \\cot^2\\theta \\, . \\end{align} Now, try reading the proof 'backwards'. Does $$\\cot^2\\theta = \\cot^2\\theta$$ imply $$\\frac{\\cos^2\\theta}{\\sin^2\\theta} = \\frac{\\cos^2\\theta}{\\sin^2\\theta} \\, ?$$ Does $$\\frac{\\cos^2\\theta}{\\sin^2\\theta} = \\frac{\\cos^2\\theta}{\\sin^2\\theta}$$ imply $$\\frac{1-\\sin^2\\theta}{\\sin^2\\theta} = \\frac{\\cos^2\\theta}{\\sin^2\\theta} \\, ?$$ Keep going. If the answer to all of these question is 'yes', then we have successfully proven that $$\\frac{\\csc \\theta - 1}{\\cot\\theta} = \\frac{\\cot\\theta}{\\csc\\theta+1} \\iff \\cot^2\\theta = \\cot^2\\theta \\, .$$ And since the RHS is an identity, the LHS must also be an identity. There's just one tiny snag to this proof. When we are going in a backwards direction, and conclude that $$(\\csc\\theta-1)(\\csc\\theta+1) = \\cot^2\\theta$$ implies $$\\frac{\\csc \\theta - 1}{\\cot\\theta} = \\frac{\\cot\\theta}{\\csc\\theta+1} \\, ,$$ we are assuming that both $$\\cot\\theta$$ and $$\\csc\\theta + 1$$ are non-zero. But, your son directly addresses this by stating at the start of the proof, 'for all real $$\\theta$$ not equal to an integer multiple of $$\\pi/2$$...', meaning that this is never a problem. This kind of attention to detail is very impressive for a high-school student.", "date": "2022-05-28 18:09:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4043453/teacher-claims-this-proof-for-frac-csc-theta-1-cot-theta-frac-cot-theta/4043660", "openwebmath_score": 0.8855383396148682, "openwebmath_perplexity": 404.75651943284095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.958537730841905, "lm_q2_score": 0.9086179062123119, "lm_q1q2_score": 0.8709445460230723}}
{"url": "https://mathhelpboards.com/threads/sum-with-binomial-coefficients.1619/", "text": "# [SOLVED]Sum with Binomial Coefficients\n\n#### VincentP\n\n##### New member\nI'm having trouble proving the following identity (I don't even know if it's true):\n$$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\binom{n-1}{k-1}$$ $$\\forall n,k \\in \\mathbb{N} : n>k$$\nThank you in advance for any help!\nVincent\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nI'm having trouble proving the following identity (I don't even know if it's true):\n$$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\binom{n-1}{k-1}$$ $$\\forall n,k \\in \\mathbb{N} : n>k$$\nThank you in advance for any help!\nVincent\nMaybe expanding the binomial coefficients would help. Recall that $\\displaystyle {n\\choose{m}} = \\frac{n!}{m!(n-m)!}$\n\n#### VincentP\n\n##### New member\nMaybe expanding the binomial coefficients would help. Recall that $\\displaystyle {n\\choose{m}} = \\frac{n!}{m!(n-m)!}$\nWell I have tried that of course:\n$$\\sum_{r=1}^k \\frac{k!}{r!(k-r)!} \\frac{(n-k-1)!}{(r-1)!(n-k-r)!} \\stackrel{?}{=} \\frac{(n-1)!}{(k-1)!(n-k)!}$$\nBut I don't know where to go from here since I still can't sum the left hand side. I also tried to prove it by induction but I failed to prove the induction step.\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nI'm having trouble proving the following identity (I don't even know if it's true):\n$$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\binom{n-1}{k-1}$$ $$\\forall n,k \\in \\mathbb{N} : n>k$$\nThank you in advance for any help!\nVincent\nHi Vincent,\n\nThank you for submitting this problem, I enjoyed solving it.\n\nSince, $$\\binom{k}{r}=\\binom{k}{k-r}$$ we have,\n\n$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\sum_{r=1}^{k}\\binom{k}{k-r} \\binom{n-k-1}{r-1}$\n\nUsing the Pascal's rule we get,\n\n\\begin{eqnarray}\n\n\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}&=&\\sum_{r=1}^{k}\\left[{k+1 \\choose k-r+1}-{k \\choose k-r+1}\\right]\\binom{n-k-1}{r-1}\\\\\n\n&=&\\sum_{r=1}^{k}{k+1 \\choose k-r+1}\\binom{n-k-1}{r-1}-\\sum_{r=1}^{k}{k \\choose k-r+1}\\binom{n-k-1}{r-1}\\\\\n\n\\end{eqnarray}\n\nNow use the Vandermonde's identity to get,\n\n$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}={n \\choose k}-\\binom{n-1}{k}$\n\nUsing the Pascal's rule again we get,\n\n$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}={n-1 \\choose k-1}$\n\nKind Regards,\nSudharaka.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nFor a completely different, combinatorial, way to look at this problem, suppose that you have $n-1$ objects, and you want to select $k-1$ of them. There are $n-1\\choose k-1$ ways of making the selection.\n\nNow suppose that $k$ of the objects are white, and the remaining $(n-1)-k$ are black. Then another way to select $k-1$ objects is as follows. First, choose a number $r$ between 1 and $k$ (inclusive). Then select $r-1$ black balls and $(k-1)-(r-1) = k-r$ white balls. The number of ways to do that is ${k\\choose k-r}{(n-1)-k\\choose r-1} = {k\\choose r}{n-k-1\\choose r-1}.$ Sum that from $r=1$ to $k$ to get the total number of ways to select $k-1$ objects.\n\n#### VincentP\n\n##### New member\n@ Sudharaka\nThank you very much for your explanation!\nI have one question though: Doesn't Vandermonde's identity require the sum to start at r=0?\n\n@Opalg\nThank you for your reply, that's a very interesting approach to the problem!\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\n@ Sudharaka\nThank you very much for your explanation!\nI have one question though: Doesn't Vandermonde's identity require the sum to start at r=0?\n\n@Opalg\nThank you for your reply, that's a very interesting approach to the problem!\nYou are welcome. I have neglected an in between step that may have aroused the confusion.\n\n$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\sum_{r=1}^{k}{k+1 \\choose k-r+1}\\binom{n-k-1}{r-1}-\\sum_{r=1}^{k}{k \\choose k-r+1}\\binom{n-k-1}{r-1}$\n\nSubstitute $$u=r-1$$. Then the right hand side becomes,\n\n$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\sum_{u=0}^{k}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k}{k \\choose k-u}\\binom{n-k-1}{u}$\n\nI hope this clarifies your doubts.\n\nKind Regards,\nSudharaka.\n\n#### VincentP\n\n##### New member\nYou are welcome. I have neglected an in between step that may have aroused the confusion.\n\n$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\sum_{r=1}^{k}{k+1 \\choose k-r+1}\\binom{n-k-1}{r-1}-\\sum_{r=1}^{k}{k \\choose k-r+1}\\binom{n-k-1}{r-1}$\n\nSubstitute $$u=r-1$$. Then the right hand side becomes,\n\n$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\sum_{u=0}^{k}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k}{k \\choose k-u}\\binom{n-k-1}{u}$\n\nI hope this clarifies your doubts.\n\nKind Regards,\nSudharaka.\nWell not quite.\nIf you substitute the index of summation $u=r-1$ you have to change the lower as well as the upper bound of summation, because otherwise you change the number of summands. Therefore if you substitute $u=r-1$ we get:\n$$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\sum_{u=0}^{k-1}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k-1}{k \\choose k-u}\\binom{n-k-1}{u}$$ Which doesn't match Vandermonde's Identity anymore, because the upper bound of summation doesn't appear in the lower index of the binomial coefficant.\n\nKind Regards,\nVincent\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nWell not quite.\nIf you substitute the index of summation $u=r-1$ you have to change the lower as well as the upper bound of summation, because otherwise you change the number of summands. Therefore if you substitute $u=r-1$ we get:\n$$\\sum_{r=1}^k \\binom{k}{r} \\binom{n-k-1}{r-1}=\\sum_{u=0}^{k-1}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k-1}{k \\choose k-u}\\binom{n-k-1}{u}$$ Which doesn't match Vandermonde's Identity anymore, because the upper bound of summation doesn't appear in the lower index of the binomial coefficant.\n\nKind Regards,\nVincent\nNote that,\n\n$\\sum_{u=0}^{k-1}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k-1}{k \\choose k-u}\\binom{n-k-1}{u}=\\sum_{u=0}^{k}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k}{k \\choose k-u}\\binom{n-k-1}{u}$\n\nsince when $$u=k$$ the right hand side is equal to zero. If you have any more questions about this please don't hesitate to ask.\n\nKind Regards,\nSudharaka.\n\n#### VincentP\n\n##### New member\nNote that,\n\n$\\sum_{u=0}^{k-1}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k-1}{k \\choose k-u}\\binom{n-k-1}{u}=\\sum_{u=0}^{k}{k+1 \\choose k-u}\\binom{n-k-1}{u}-\\sum_{u=0}^{k}{k \\choose k-u}\\binom{n-k-1}{u}$\n\nsince when $$u=k$$ the right hand side is equal to zero. If you have any more questions about this please don't hesitate to ask.\n\nKind Regards,\nSudharaka.\nI think that clarifies everything, thanks so much.\nVincent\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nI think that clarifies everything, thanks so much.\nVincent\nI am glad to be of help.\n\nKind Regards,\nSudharaka.", "date": "2020-09-21 06:08:44", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/sum-with-binomial-coefficients.1619/", "openwebmath_score": 0.8759025931358337, "openwebmath_perplexity": 622.9634205752079, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9869795104606397, "lm_q2_score": 0.8824278772763472, "lm_q1q2_score": 0.8709382343310306}}
{"url": "https://math.stackexchange.com/questions/1836637/directional-derivative-help-solving-for-derivative-0-when-given-constants", "text": "# Directional Derivative help, solving for derivative = 0 when given constants\n\nA function that is useful in studying the air flow over mountains is\n\n$$h(x,y) = \\frac{h_0}{[(\\frac{x}{a})^2+(\\frac{y}{b})^2+1]^\\frac{3}{2}}$$\n\nwhere $h_0$, a, and b are all positive constants.\n\n(a) Find $\\nabla h$.\n\n(b) Find the directional derivative of the point (x,y) in the direction of the vector $v=(v_1,v_2)$\n\n(c) At what point(s) is the directional derivative equal to zero?\n\nFor part a), I found that $\\nabla h$ is equal to $$\\frac{-3h_0}{[(\\frac{x}{a})^2+(\\frac{y}{b})^2+1]^\\frac{5}{2}}\\langle\\frac{x}{a^2},\\frac{y}{b^2}\\rangle$$\n\nUsing this, I found that for part b), I got that $D_uh(x,y) = \\frac{-3h_0}{[(\\frac{x}{a})^2+(\\frac{y}{b})^2+1]^\\frac{5}{2}}\\langle\\frac{x}{a^2},\\frac{y}{b^2}\\rangle \\bullet\\frac{1}{\\sqrt{v_1^2+v_2^2}}\\langle v_1, v_2 \\rangle$\n\nwhich simplifies to $D_uh(x,y) = \\frac{-3h_0}{[(\\frac{x}{a})^2+(\\frac{y}{b})^2+1]^\\frac{5}{2}{\\sqrt{v_1^2+v_2^2}}}(\\frac{xv_1}{a^2}+\\frac{yv_2}{b^2})$\n\nFirst of all, is this correct?\n\nSecondly, given that I am correct up to this point, I am unsure on how to do part c). How can I solve for an x and y where $D_uh = 0$ when I do not have values for $v_1$ and $v_2$?\n\nAny help here is appreciated.\n\n\u2022 I suposse you have to solve the equation $\\frac{xv_{1}}{a^{2}}+\\frac{yv_{2}}{b^{2}}=0$ \u2013\u00a0julio godoy Jun 23 '16 at 6:36\n\u2022 and the vector solution is $(-ya^{2},xb^{2})$, if the vector solution has unit lenght you have to normalize that vector \u2013\u00a0julio godoy Jun 23 '16 at 6:44\n\u2022 @juliogodoy Wouldn't that solution require that $v_1 = -v_2$? if you plug in those x and y values in the equation, you get $-yv_1 + xv_2 = 0$. What am I missing here? \u2013\u00a0dibdub Jun 23 '16 at 6:46\n\u2022 you can use also the fact that $u$ and $v$ are perpendicular then $u\\cdot v = 0$ and if $u = (u_1,u_2)$ then $v=(-u_2,u_1)$ \u2013\u00a0Navaro Jun 23 '16 at 6:51\n\u2022 In that case wouldnt the answer be $(\\frac{-yv_2}{b^2},\\frac{xv_1}{a^2})$? *edit: Actually it would be $(\\frac{-v_2}{b^2},\\frac{v_1}{a^2})$. Is that correct? \u2013\u00a0dibdub Jun 23 '16 at 6:54\n\nThe first two answers are correct.\n\nFor the third question: the directional derivative is equal to $0$ if $v =(v_1,v_2)$ is perpendicular to $\\nabla h$ , so: $$v\\cdot \\nabla h = 0 \\Rightarrow \\dfrac{v_1}{a^2}x+\\dfrac{v_2}{b^2}y = 0 \\Rightarrow y = -\\left( \\dfrac{b}{a} \\right)^2\\left(\\dfrac{v_1}{v_2}\\right)x\\Rightarrow y=cx \\quad\\text{where } c=-\\left( \\dfrac{b}{a} \\right)^2\\left(\\dfrac{v_1}{v_2}\\right)$$ which is the equation of a straight line passing through the origin $O=(0,0)$\n\nconclusion: every point $(x,y)$ satisfying the previous equation makes the directional derivative in the direction of $v=(v_1,v_2)$ equal to zero\n\n\u2022 Perfect, that makes sense. Thanks for all your help! \u2013\u00a0dibdub Jun 23 '16 at 7:47\n\nThe condition $x\\frac{v_1}{a^2}+y\\frac{v_2}{b^2}=0$ could be written $(P-O)\\cdot\\mathbf{u}=0$, where $P=(x,y)$, $O=(0,0)$ and $\\mathbf{u}=\\langle\\frac{v_1}{a^2},\\frac{v_2}{b^2}\\rangle$.\nThis is the equation of the straight line passing through $O$ and orthogonal to $\\mathbf{u}$.\n\nThe directional derivative is $0$ on each point of this line.\n\n\u2022 Could you elaborate on this? Why is it (P-O)? What does P represent? How would I present my answer for part c) then? \u2013\u00a0dibdub Jun 23 '16 at 6:45\n\u2022 @dubbler26: edited. \u2013\u00a0enzotib Jun 23 '16 at 7:03", "date": "2020-02-18 01:33:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1836637/directional-derivative-help-solving-for-derivative-0-when-given-constants", "openwebmath_score": 0.8258814811706543, "openwebmath_perplexity": 142.44192886945171, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795106521339, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8709382329741603}}
{"url": "https://math.stackexchange.com/questions/3895582/calculating-expectation-of-a-function-of-a-continuous-random-variable", "text": "# Calculating expectation of a function of a continuous random variable?\n\nAssume we have a continuous random variable $$X$$, with a probability density function\n\n$$f(x) = 1 - \\frac{x}{2},\\; 0 < x < 2$$\n\nLet's say I want to calculate $$E(2X)$$. I will use two different methods, and get different results, so I obviously understand something wrong.\n\nMETHOD 1:\n\nUsing the formula for the expectation of a function of a random variable, I get:\n\n$$E(2X) = \\int_{-\\infty}^{+\\infty} 2xf(x)dx = \\int_0^2 (2x-2x^2)dx = \\frac{4}{3}$$\n\nI understand this formula intuitively, as it's pretty much exactly the same as in the discrete case, if we replace the integral with a finite sum and the probability density function with the probability mass function. Only the values change, and not the distribution, so we only apply the function 2x to the values, and not the distribution $$f(x)$$.\n\nMETHOD 2:\n\nLet's say I declare a new continuous random variable, $$Y = 2X$$. I can calculate the probability density function of $$Y$$:\n\n$$g(y) = |\\frac{dx}{dy}|f(x) = \\frac{1}{2}f(\\frac{y}{2}) = \\frac{1}{2}(1-\\frac{y}{4})$$\n\nfor $$0.\n\nThen I can calculate:\n\n$$E(2X) = E(Y) = \\int_{-\\infty}^{+\\infty} y\\;g(y)dx = \\int_0^4 \\frac{1}{2}(1-\\frac{y}{4})dx = \\frac{2}{3}$$\n\nI don't understand what I did wrong here.\n\nBut I also don't understand the formula for $$g(y)$$ intuitively - shouldn't $$g(y)$$ and $$f(x)$$ be the same functions if they're injective, similar to the discrete case?\n\nAnyway, if someone can point out what I did wrong, I would be very grateful.\n\nYou forgot to multiply by $$y$$ for the second computation.\n$$g(y)$$ and $$f(x)$$ are different functions, but they are related by $$g(y)=\\frac{1}{2}f\\big(\\frac{y}{2}\\big)$$ and are both decreasing (also see here). For instance, $$h_{1}(x)=x$$ and $$h_{2}(x)=2x$$ are injective, but $$h_{1}(1)=1$$ and $$h_{2}(1)=2$$.\nIt is a good idea to check if $$g(y)=\\frac{1}{2}(1-\\frac{y}{4})$$ for $$0 is a valid pdf. Indeed we have $$g(y)\\geq 0$$ for $$0 and $$\\int_{-\\infty}^{\\infty}g(y)dy=\\frac{1}{2}\\int_{0}^{4}(1-\\frac{y}{4})dy=1,$$ so $$g(y)$$ is a valid pdf.\nNext we have $$E(2X) = E(Y) = \\int_{-\\infty}^{+\\infty} y\\;g(y)dx = \\int_0^4 \\frac{\\color{red}{y}}{2}(1-\\frac{y}{4})dx = \\frac{4}{3}$$", "date": "2021-01-22 16:39:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3895582/calculating-expectation-of-a-function-of-a-continuous-random-variable", "openwebmath_score": 0.933574914932251, "openwebmath_perplexity": 146.38141030641387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795118010988, "lm_q2_score": 0.8824278726384089, "lm_q1q2_score": 0.870938230936339}}
{"url": "http://math.stackexchange.com/questions/200180/is-there-a-bijective-map-from-0-1-to-mathbbr/200213", "text": "# Is there a bijective map from $(0,1)$ to $\\mathbb{R}$?\n\nI couldn't find a bijective map from $(0,1)$ to $\\mathbb{R}$. Is there any example?\n\n-\nYes, and there's even one from (0,1) to R\u00b2... \u2013\u00a0Axel Sep 21 '12 at 12:56\n\nHere is a nice one ${}{}{}{}{}{}{}{}{}$, can you find the equation?\n\n-\nThat's really nice! \u2013\u00a0Clive Newstead Sep 21 '12 at 12:26\nNote: I would be interested in an explicit formula. \u2013\u00a0The Chaz 2.0 Oct 21 '13 at 19:59\n@The Chaz 2.0 If the diameter of the circle is 1 and the interval is $(0, 1)$ and 1/2 mapped to 0 and the distance between the center of the circle and the 2nd numberline is 1/2 then the formula is $f(x)=\\frac{x-1/2}{\\sqrt{x-x^2}}$. Becouse scaling does not matter you can multiply by 2 and get the formula $f(x)=\\frac{2x-1}{\\sqrt{x-x^2}}$, but i don't know how you can get Cameron Buie's formula, out of this. \u2013\u00a005storm26 Feb 2 '14 at 8:31\nI am thinking about: $f:(0,1)\\to\\mathbb{R}$, $f(x)=-\\cot\\left(\\pi x\\right)$ \u2013\u00a0Darius Apr 22 '14 at 19:58\nAs far as I can tell, the connection between our answers' formulae is coincidental (aside from the fact that if $g$ is a positive function on some subset of $\\Bbb R,$ then so is $\\sqrt{g},$ so our functions are related, but not in an obvious heuristic way. Please let me know if you can discern a connection between our answers (aside from the obvious algebraic one). \u2013\u00a0Cameron Buie Oct 31 '14 at 3:07\n\nHere is a bijection from $(-\\pi/2,\\pi/2)$ to $\\mathbb{R}$: $$f(x)=\\tan x.$$ You can play with this function and solve your problem.\n\n-\n\n$g(x)=\\frac 1{1+e^x}$ gives a bijection from $\\Bbb R$ to $(0,1)$, so take the inverse of this map.\n\n-\nAnd the inverse is $f(x)=\\ln\\left(\\frac{1}{x}-1\\right)$ \u2013\u00a0celtschk Sep 21 '12 at 12:19\n\nA homeomorphism (continuous bijection with a continuous inverse) would be $f:(0,1)\\to\\Bbb R$ given by $$f(x)=\\frac{2x-1}{x-x^2}.$$\n\nAdded: Let me provide some explanation of how I came by this answer, rather than simply leave it as an unmotivated (though effective) formula and claim in perpetuity.\n\nMany moons ago, I was assigned the task of demonstrating that the real interval $(-1,1)$ was in bijection with $\\Bbb R.$ Prior experience with rational functions had shown me graphs like this:\n\nThe above is a graph of a continuous function from most of $\\Bbb R$ onto $\\Bbb R.$ This doesn't do the trick on its own, since it certainly isn't injective. However, it occurred to me that if we restrict the function to the open interval between the two vertical asymptotes, we get this graph, instead:\n\nThis graph is of a continuous, injective (more precisely, increasing) function from a bounded open interval of $\\Bbb R$ onto $\\Bbb R.$ This showed that rational functions could do the job. Other options occurred to me, certainly (such as trigonometric functions), but of the ideas I had (and given the results I was allowed to use) at the time, the most straightforward approach turned out to be using rational functions.\n\nNow, given the symmetry of the interval $(-1,1)$ (and, arguably, of $\\Bbb R$) about $0,$ the natural choice of the unique zero of the desired function was $x=0.$ In other words, I wanted $x$ to be the unique factor of the desired rational function's numerator that could be made equal to $0$ in the interval $(-1,1)$--meaning that for $\\beta\\in(-1,1)$ with $\\beta\\ne0,$ I needed to make sure that $x-\\beta$ was not a factor of the numerator. For simplicity, I hoped that I could make $x$ the only factor of the numerator that could be made equal to $0$ at all--that is, I hoped that I could have $\\alpha x$ as the numerator of my function for some nonzero real $\\alpha.$\n\nIn order to get the vertical asymptotic behavior I wanted on the given interval--that is, only at the interval's endpoints--I needed to make sure that $x=\\pm1$ gave a denominator of $0$--that is, that $x\\mp1$ were factors of the denominator--and that for $-1<\\beta<1,$ $\\beta$ was not a zero of the denominator--that is, that $x-\\beta$ wasn't a factor of the denominator. For simplicity, I hoped that I could make $x\\mp1$ the only factors of the denominator.\n\nPlaying to my hopes, I assumed $\\alpha$ to be some arbitrary nonzero real, and considered the family of functions $$g_\\alpha(x)=\\frac{\\alpha x}{(x+1)(x-1)}=\\frac{\\alpha x}{x^2-1},$$ with domain $(-1,1).$ It was readily seen that all such functions are real-valued and onto $\\Bbb R.$\n\nI wanted more, though! (I'm demanding of my functions when I can be. What can I say?) I wanted an increasing function. I determined (through experimentation which suggested proof) that $h_\\alpha$ would be increasing if and only if $\\alpha<0.$ Again, for convenience, I chose $\\alpha=-1,$ which gave me the function that satisfied the desired (and required) properties: $g:(-1,1)\\to\\Bbb R$ given by $$g(x)=\\frac{-x}{x^2-1}=\\frac{x}{1-x^2}.$$\n\nMuch later, you posted your question, and I realized (again, based on experience) that my earlier result could be adapted to the one you wanted. Playing around a bit with linear interpolation showed that the function $h:(0,1)\\to(-1,1)$ given by $h(x)=2x-1$ was a bijection--in fact, an increasing bijection.\n\nIt is readily shown (and I had previously seen) that if $X,Y,$ and $Z$ are ordered sets and if we are given increasing maps $X\\to Y$ and $Y\\to Z,$ then the composition of those maps is an increasing map $X\\to Z.$ Also, it is readily shown (and I had seen previously) that if both such maps are continuous and surjective, then so is their composition. Just from these results, my originally posted map was obtained (though named differently): \\begin{align}(g\\circ h)(x) &= g\\bigl(h(x)\\bigr)\\\\ &= \\frac{h(x)}{1-\\left(h(x)\\right)^2}\\\\ &=\\frac{2x-1}{1-(2x-1)^2}\\\\ &=\\frac{2x-1}{1-\\left(4x^2-4x+1\\right)}\\\\ &= \\frac{2x-1}{4x-4x^2}\\\\ &=\\frac{2x-1}{4(x-x^2)}.\\end{align}\n\nAs lhf astutely pointed out shortly thereafter (and as I should have seen immediately), the factor of $4$ in the denominator serves no particular purpose, hence its later removal to yield the function $f$ that I eventually posted.\n\nThe remaining claim that I made (that $f$ has a continuous inverse), I leave to you (the reader). If you're curious how I determined this, try to prove it on your own first. If you're stymied (or if you simply want to run your proof attempt by me), let me know. I will do what I can to get you \"unstuck.\"\n\n-\nThis is the answer you want. A direct map that just stretches the unit interval onto the (-inf,inf) interval. This shows you don't need an exotic function like the bijection between [0,1] and [0,1]^2. \u2013\u00a0John Baber-Lucero Sep 21 '12 at 12:17\nNo need for that 4... \u2013\u00a0lhf Sep 21 '12 at 12:22\nFair point, @lhf. Not sure why I bothered with that.... \u2013\u00a0Cameron Buie Sep 21 '12 at 17:21\n+1 for the motivation. \u2013\u00a0LePressentiment Jan 6 at 4:45\n\nFor a less differentiable example, consider the bijection in the following picture,\n\nIn symbols, given $x \\in (0,1)$ let $n$ be the largest natural number such that $1-\\frac{1}{n}<x$, define $$y=\\frac{x-n}{\\frac{1}{n}-\\frac{1}{n+1}}$$ to be the renormalized version of $x$ if the interval $(1-\\frac{1}{n},1-\\frac{1}{n+1}]$ is rescaled and shifted to map to $(0,1)$. Then we have the following bijection: $$f(x)=\\begin{cases}\\frac{n-1}{2}+y,& n \\text{ odd} \\\\ -\\frac{n-2}{2}-y,& n \\text{ even}\\end{cases}$$\n\n-\nNot only less differentiable but also much less continuous. \u2013\u00a0Stefan Geschke Sep 21 '12 at 10:15\n(: One might even consider the composition $f \\circ g$ with a nowhere continuous bijection $g:(0,1)\\rightarrow (0,1)$. \u2013\u00a0Nick Alger Sep 21 '12 at 10:27\n\nYes. let $f(x)=\\tan((x-1/2)\\pi)$. the domain is $(0,1)$ and range is $\\mathbb{R}$\n\n-\nHere is a more visual description of the same function (modulo a factor of $\\pi$) Bend the line segment (0, 1) into a semicircle, with the open part facing upwards, and rest it on the real line (with 0.5 on the semicircle resting on 0 on the real line). To map a point on one line to a point on the other, draw a line through that point and the centre of the circle. The mapped point is where it intersects the other line. \u2013\u00a0Max Sep 21 '12 at 10:10\nOf course, knowing the domain and range simply guarantees that $f$ is surjective, but this $f$ does turn out to be injective, too. \u2013\u00a0Cameron Buie Sep 21 '12 at 10:36\n\nYes, see above answers. There are even bijective maps between $(0,1)$ and $\\mathbb{R}^n$. To see this, note that a bijection $\\phi$ between $(0,1)$ and $(0,1)^2$ can be made in this way: Let $x= 0.b_1b_2\\ldots$, with $b_j$ being the digits in a decimal expansion. Define $$\\phi(x) = (0.b_1b_3b_5\\ldots,0.b_2b_4b_6\\ldots),$$ i.e., extract even and odd digits. For $\\phi^{-1}(x_1,x_2)$, let $x_1 = 0.a_1a_2a_3\\ldots$, and $x_2=b_1b_2b_3\\ldots$. Then, $$\\phi^{-1}(x,y) = 0.a_1b_1a_2b_2\\cdots$$ Some care has to be taken with identification between digital expansions like $0.199999\\cdots$ and $0.20000\\cdots$, but that is an exercise.\n\nHaving the bijection between $(0,1)$ and $(0,1)^2$, we can apply one of the other answers to create a bijection with $\\mathbb{R}^2$.\n\nThe argument easily generalizes to $\\mathbb{R}^n$.\n\n-\nActually you don't even have to generalize the argument: If you have the bijection between $(0,1)$ and $(0,1)^2$, you get a bijection from $(0,1)$ to $(0,1)^3$ by just applying the same bijection to one of the two factors of $(0,1)^2$. Of course the same way you get to $(0,1)^n$. \u2013\u00a0celtschk Sep 21 '12 at 12:16\n\nThe trigonometric function $\\tan x$ is an invertible function from $(-\\pi/2,\\pi/2)$ to $\\mathbb{R}$. Also to find an invertible function from $(0,1)$ to $(-\\pi/2,\\pi/2)$ find the equation of the straight line joining the points $(0,-\\pi/2)$ and $(1,\\pi/2)$. Now compose the two functions together. You can likewise find bijections between any two open intervals and any open interval and $\\mathbb{R}$.\n\n-\n\n$x \\mapsto \\ln (- \\ln x)$ with the inverse $y \\mapsto e^{-e ^ {\\ y}}$. It's also a $C ^ \\infty$ diffeomorphism.\n\n-\n\nBy virtue of http://natureofmathematics.wordpress.com/lecture-notes/cantor/, here's another picture. The interval at the bottom is $\\mathbb{R}$.\n\n-\nIt's not clear to me which point on the circle is missed from the map described by the blue arrows. It looks like it could be some point between the images of the point $0$ and the point $1$, but then the map described by the red arrows from the circle to the real line is not a bijective map because that missed point has no preimage and so its image in the real line after the red mapping has no preimage in the interval. This is a poor picture. \u2013\u00a0Dan Rust Feb 18 '14 at 12:45\nthe idea is good : draw a circle of radius $1$ and center $(0,1)$, draw the segment $(-1,1) , (1,1)$, and map the segment to $\\mathbb{R}$ by taking from any point of the segment the nearest point on the half bottom circle, drawing the line passing through these 2 points, and considering its intersection with the horizontal line $y = 0$. finally, prove that every continuous bijection between $]-1;1[$ and $\\mathbb{R}$ can be constructed this way (replacing the circle by other curves). and the reciprocal : characterize the curves from which we get such a bijection. \u2013\u00a0user1952009 Jan 5 at 22:54", "date": "2016-05-26 00:47:17", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/200180/is-there-a-bijective-map-from-0-1-to-mathbbr/200213", "openwebmath_score": 0.9507617354393005, "openwebmath_perplexity": 239.53550877919466, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795114181106, "lm_q2_score": 0.8824278726384089, "lm_q1q2_score": 0.8709382305983796}}
{"url": "https://math.stackexchange.com/questions/2517626/determining-compositions-of-trig-functions-by-knowing-eulers-identity-etc", "text": "# Determining compositions of trig functions by knowing Euler's identity etc\n\nHow does one determine: $$\\cos^2(\\arctan(x))?$$\n\nI know what it is equal to, since its in the tables. But without working with many trigonometric identities, its not clear how to find such things.\n\nHow would you see this with the minimal number of trig identities?\n\n$\\cos^2(\\arctan(x))=\\cos(\\arctan(x))\\cos(\\arctan(x))=\\frac{1}{\\sqrt{1+x^2}}\\frac{1}{\\sqrt{1+x^2}}=\\frac1{1+x^2}.$\n\nThe one identity I used here, I didn't know. It seems in similar situations, on an exam, I would have massive trouble without these identities. Can all of these sorts of things be solved by knowing something about Eulers identity and such?\n\nDefine the point $P(\\theta)=(1,x)$ so that $\\tan(\\theta) = x$. Note that we need to have $\\theta$ in the first or fourth quadrant in order for $\\theta = \\arctan x$ to be true.\n\nFrom the picture we see that $\\cos^2(\\arctan x) = \\cos^2 \\theta = \\dfrac{1}{1+x^2}$.\n\nI find that the easiest way to solve these is to let $\\theta$ be $\\arctan{x}$, so that $\\tan{\\theta}=x$. Then we have \\begin{align*} \\frac{\\sin{\\theta}}{\\cos{\\theta}}&=x\\\\ \\cos{\\theta}&=\\frac{\\sin{\\theta}}{x}\\\\ \\cos{\\theta}&=\\frac{\\sqrt{1-\\cos^2{\\theta}}}{x}\\\\ x\\cos{\\theta}&=\\sqrt{1-\\cos^2{\\theta}}\\\\ x^2\\cos^2{\\theta}&=1-\\cos^2{\\theta}\\\\ (x^2+1)\\cos^2{\\theta}&=1\\\\ \\cos^2{\\theta}&=\\frac{1}{x^2+1}\\\\ \\end{align*} This may look tougher, but after the third step, we only need to use algebraic manipulations, and don't need to worry about trigonometry anymore. The two nicest ways to make a trig problem easier to solve are getting rid of inverses like I have in this problem, and writing everything in terms of sine and cosine.\n\n\u2022 Is $\\sin\\theta$ always non negative like $$\\sqrt{1-\\cos^2\\theta}$$ Nov 13 '17 at 4:08\n\u2022 Best way to fix that problem is to already square after the second step. Nov 13 '17 at 6:55\n\u2022 I think that the OP main concern is not to solve the problem posted. That is just an example. He seems to look for a general way of solving trig functions with some sort of simplifying tool like for example Euler\u2019s identity. Nov 13 '17 at 7:35\n\nFrom the well-known trig identity $$\\sec^2y=1+\\tan^2y$$ and $$\\sec y=\\frac{1}{\\cos y}$$ one can easily find $$\\cos^2y=\\frac{1}{1+\\tan^2y}$$ Plugging $y=\\arctan x$ implies $$\\cos^2(\\arctan x)=\\frac{1}{1+x^2}$$\n\nFirstly, lets check the range of commonly defined $\\arctan (x)$ because our result will depend on this. It is generally taken to be $(-\\frac{\\pi}{2}, \\frac{\\pi}{2})$. So $\\cos$ will be positive over this domain.\n\nHere this analysis is not of much use as we have squared the $\\cos$ term, which will always be positive anyway.\n\nNow to get the value of $\\cos(\\arctan(x))$ for acute angle $\\arctan(x)$, let this angle be $\\phi$.\n\n\\begin{align} \\phi &= \\arctan (x) \\tag{1}\\\\ \\tan(\\phi) &= x\\\\ \\frac{1}{1+\\tan^2 \\phi} &= \\frac{1}{1+x^2}\\\\ \\cos^2 \\phi &= \\frac{1}{1+x^2}\\\\ \\implies \\cos^2 (\\arctan x) &=\\frac{1}{1+x^2} \\end{align}\n\nThere are (as is often the case with trigonometric identities) several ways to solve this, and the already-posted answers are just fine. Here's another, which I like because it uses a very general-purpose tool.\n\nThe tool in question: the famous half-angle formulae. These are extremely useful and reduce many trigonometric identities to routine algebraic manipulation. In case you don't already know them, here they are. Let $t=\\tan\\frac12\\theta$; then $\\cos\\theta=\\frac{1-t^2}{1+t^2}$, and $\\sin\\theta=\\frac{2t}{1+t^2}$, and $\\tan\\theta=\\frac{2t}{1-t^2}$.\n\nWhy are these useful here? Because $\\cos^2\\phi$ is very closely related to $\\cos2\\phi$, which means we have cosine of twice an arctangent, which if you think about it for a moment you'll see is exactly what the half-angle formula for cos tells us how to handle.\n\nSo. We're taking $\\tan^{-1}x$ and we want (in effect) the cosine of twice this, so write $x=\\tan\\frac12\\theta$. The formula above says that $\\cos\\theta=\\frac{1-x^2}{1+x^2}$; call this $c$. The thing we were actually asked for is $\\cos^2\\frac12\\theta$ and we have $c=2\\cos^2\\frac12\\theta-1$ so $\\cos^2\\frac12\\theta=\\frac{c+1}2=\\frac1{1+t^2}$.", "date": "2021-10-19 10:08:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2517626/determining-compositions-of-trig-functions-by-knowing-eulers-identity-etc", "openwebmath_score": 0.990223228931427, "openwebmath_perplexity": 311.55997101525776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795075882268, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8709382150119829}}
{"url": "http://math.stackexchange.com/questions/865442/find-expression-for-sum-of-series", "text": "# Find expression for sum of series\n\nI want to find a formula for the sum of this series using its general term. How to do it?\n\nSeries $$S_n = \\underbrace{1/3 + 2/21 + 3/91 + 4/273 + \\cdots}_{n \\text{ terms}}$$\n\nGeneral Term\n\n$$S_n = \\sum_{i=1}^{n} \\frac{i}{i^4+i^2+1}$$\n\n-\nI changed n\\ \\mbox{terms} to n\\text{ terms}. The mismatch in sizes of fonts resulted from use of an incorrect method. \u2013\u00a0 Michael Hardy Jul 12 '14 at 22:23\n\nUsing partial fractions, we get $\\dfrac{i}{i^4+i^2+1} = \\dfrac{\\tfrac{1}{2}}{i^2-i+1} - \\dfrac{\\tfrac{1}{2}}{i^2+i+1}$.\n\nThus, $S_n = \\displaystyle\\sum_{i = 1}^{n}\\dfrac{i}{i^4+i^2+1} = \\dfrac{1}{2}\\sum_{i = 1}^{n}\\dfrac{1}{i^2-i+1} - \\dfrac{1}{i^2+i+1}$.\n\nSince $(i+1)^2-(i+1)+1 = i^2+i+1$, this sum telescopes to $S_n = \\dfrac{1}{2}\\left(1 - \\dfrac{1}{n^2+n+1}\\right)$.\n\nTaking the limit as $n \\to \\infty$ gives $S = \\dfrac{1}{2}$.\n\n-\nthank you so much,but i have one doubt ,in which cases can i apply partial fraction to calculate sum? \u2013\u00a0 user3481652 Jul 12 '14 at 20:47\nAnytime you have a rational function (whose numerator has a smaller degree than its denominator), you can always try using partial fractions. Whether or not it will lead to a solution depends on the particular problem. \u2013\u00a0 JimmyK4542 Jul 12 '14 at 20:49\nokay thank you for your help :) \u2013\u00a0 user3481652 Jul 12 '14 at 20:51\nNote that the factorisation $i^4+i^2+1=(i^2+i+1)(i^2-i+1)$ comes easily if you notice that $(i^2-1)(i^4+i^2+1)=i^6-1=(i^3+1)(i^3-1)=(i+1)(i^2-i+1)(i-1)(i^2+i+1)$ \u2013\u00a0 Mark Bennet Jul 12 '14 at 21:06\n\nUse partial fractions,\n\n$$\\frac{i}{i^4+i^2+1} = \\frac{1}{2(i^2-i+1)}-\\frac{1}{2(i^2+i+1)}$$\n\n-\n\nWrite it out:\n\n$$\\begin{eqnarray} \\sum_{i=1}^\\infty \\frac{i}{i^4 + i^2 + 1} &=& \\sum_{i=1}^\\infty \\left( \\frac{2}{4i^2 - 4i + 4} - \\frac{2}{4i^2 + 4i + 4}\\right)\\\\ &=& \\sum_{i=1}^\\infty \\left( \\frac{2}{\\Big(2i-1\\Big)^2 + 3} - \\frac{2}{\\Big( 2i + 1\\Big)^2 + 3}\\right)\\\\ &=& \\frac{2}{\\Big( 2 - 1\\Big)^2 + 3} + \\sum_{i=1}^\\infty \\left( \\frac{2}{\\Big(2i+1\\Big)^2 + 3} - \\frac{2}{\\Big( 2i + 1\\Big)^2 + 3}\\right)\\\\ &=& \\frac{1}{2}. \\end{eqnarray}$$\n\n-\nMind to give reason for downvote? \u2013\u00a0 johannesvalks Jul 12 '14 at 21:06\nI didn't downvote you. But I think you posted the solution AFTER other users. I think they care about \"speed\" + \"accuracy\"....next time... \u2013\u00a0 NotALoner Jul 12 '14 at 21:35\nWell - that is the problem when you reply a post and in the mean time you watch the worldcup so you are late with the final result.... Thanks for the tip / advice!! \u2013\u00a0 johannesvalks Jul 12 '14 at 21:39\n@ user2584283 I think the OP asks for (a formula for) the sum of the series. \u2013\u00a0 user84413 Jul 13 '14 at 1:01\nIf you look at the question's edit history, you will see that at one time, it asked for the sum of the infinite series, not just the first $n$ terms. \u2013\u00a0 JimmyK4542 Jul 13 '14 at 1:41", "date": "2015-04-28 16:45:34", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/865442/find-expression-for-sum-of-series", "openwebmath_score": 0.8902991414070129, "openwebmath_perplexity": 1034.53161851931, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9869795070137442, "lm_q2_score": 0.8824278540866547, "lm_q1q2_score": 0.8709382084016426}}
{"url": "http://jnnd.macchiaelettrodomestici.it/how-to-find-the-kernel-of-a-linear-transformation.html", "text": "The subset of B consisting of all possible values of f as a varies in the domain is called the range of. 1 T(~x + ~y) = T(~x) + T(~y)(preservation of addition) 2 T(a~x) = aT(~x)(preservation of scalar multiplication) Linear Transformations: Matrix of a Linear Transformation Linear Transformations Page 2/13. A basis for the kernel of L is {1} so the kernel has dimension 1. Matrix vector products as linear transformations. All Slader step-by-step solutions are FREE. Inversion: R(z) = 1 z. Let A = 2 4 0. Find the kernel of the linear transformation L: V\u2192W. Proof: This theorem is a proved in a manner similar to how we solved the above example. range(T)={A in W | there exists B in V such that T(B)=A}. The $$\\textit{nullity}$$ of a linear transformation is the dimension of the kernel, written $$nul L=\\dim \\ker L. Thus, we should be able to find the standard matrix for. It is essentially the same thing here that we are talking about. Let T: V !W be a linear transformation. It relates the dimension of the kernel and range of a linear map. In mathematics, a linear map (also called a linear mapping, linear transformation or, in some contexts, linear function) is a mapping V \u2192 W between two modules (for example, two vector spaces) that preserves (in the sense defined below) the operations of addition and scalar multiplication. Construct a linear transformation f and vector Y so that the system takes the form f(X)=Y. I tried to RREF it on my calculator but it says invalid dim, I am not sure of what to do and all the examples I have looked up are for square matrices, any help would be appreciated, thanks!. And we saw that earlier in the video. Suppose T : V !W is a linear transformation. In Section 4, we de\ufb01ne the kernel whitening transformation and orthogonalize non-. Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in. The confidence of the interval [107, 230] is less than 95%. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0. The function of kernel is to take data as input and transform it into the required form. 0004 From the previous lesson, we left it off defining what the range of a linear map is. be a linear transformation. Specifically, if T: n m is a linear transformation, then there is a unique m n matrix, A, such that T x Ax for all x n. Griti is a learning community for students by students. Algebra Linear Algebra: A Modern Introduction 4th Edition In Exercises 5-8, find bases for the kernel and range of the linear transformations T in the indicated exercises. \" \u2022 The fact that T is linear is essential to the kernel and range being subspaces. LTR-0060: Isomorphic Vector Spaces We define isomorphic vector spaces, discuss isomorphisms and their properties, and prove that any vector space of dimension is isomorphic to. (a) Find a basis of the range of P. the kernel of a a linear transformation is the set of vectors in the null space of the matrix for that linear transformation. Kernel, image, nullity, and rank Math 130 Linear Algebra D Joyce, Fall 2015 De nition 1. Linear Transformation. Since the nullity is the dimension of the null space, we see that the nullity of T is 0 since the dimension of the zero vector space is 0. The following examples illustrate the syntax. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. To prove the transformation is linear, the transformation must preserve scalar multiplication, addition, and the zero vector. We then consider invertible linear transformations, and then use the resulting ideas to prove the rather stunning result that (in a very precise sense). Vector Spaces and Linear Transformations Beifang Chen Fall 2006 1 Vector spaces A vector space is a nonempty set V, whose objects are called vectors, equipped with two operations, called addition and scalar multiplication: For any two vectors u, v in V and a scalar c, there are unique vectors u+v and cu in V such that the following properties are satis\ufb02ed. Kernel trick allows us to transform the data in high dimensional (potentially in\ufb01nite) using inner products, without actually using the non linear feature mapping. SUBSCRIBE to the channel and. We discuss the kernel and range of linear transformations, and then prove that the range of a linear transformation is a subspace. What is the outcome of solving the problem?. The Gaussian is a self-similar function. Sums and scalar multiples of linear transformations. This MATLAB function returns predicted class labels for each observation in the predictor data X based on the binary Gaussian kernel classification model Mdl. Find the kernel of f. Then the Kernel of the linear transformation T is all elements of the vector space V that get mapped onto the zero element of the vector space W. In Section 3, we compute the whitening transformation. We show that for high-dimensional data, a particular framework for learning a linear transformation of the data based on the LogDet divergence can be efficiently kernelized to learn a metric (or equivalently, a kernel function) over an. T is a linear transformation. The problem comes when finding the Kernel basis. Define the linear transformation T(x) = A * x for A an m by n matrix. T(x 1,x 2,x 3,x 4)=(x 1\u2212x 2+x 3+x 4,x 1+2x 3\u2212x 4,x 1+x 2+3x 3. We denote the kernel of T by ker(T) or ker(A). Now, we connect together all the ideas we\u2019ve talked. To find the null space we must first reduce the #3xx3# matrix found above to row echelon form. Linear transformations as matrix vector products. S: \u211d3 \u2192 \u211d3. Define pre-image of U, denoted T -1. Griti is a learning community for students by students. One-to-One linear transformations: In college algebra, we could perform a horizontal line test to determine if a function was one-to-one, i. De nition 1. + for all vectors VI, for all scalars Cl, F(cv) for all scalars c, for all ve V, for all A function F: V \u2014W is linear W be a subspace of Rk Let V be a subspace of Let it respects the linear operations,. linear transformation. Up Main page Definition. In particular, there exists a nonzero solution. Let V;W be vector spaces over a eld F. 2 The kernel and range of a linear transformation. Most off-the-shelf classifiers allow the user to specify one of three popular kernels: the polynomial, radial basis function, and sigmoid kernel. The same considerations apply to rows as well as columns. For example linear, nonlinear, polynomial, radial basis function. Recall that if a set of vectors v 1;v 2;:::;v n is linearly independent, that means that the linear combination c. The null space of T, denoted N(T), is de ned as N(T) := fv2V: T(v) = 0g: Remark 3. Let T: R n \u2192 R m be a linear transformation. Find the kernel of the linear transformation. If T : Rm \u2192 Rn is a linear transformation, then the set {x | T(x) = 0 } is called the kernelof T. To see why image relates to a linear transformation and a matrix, see the article on linear. Note that the squares of s add, not the s 's themselves. 2 Kernel and Range of a Linear Transformation Performance Criteria: 2. IV Image and Kernel of Linear Transformations Motivation: In the last class, we looked at the linearity of this function: F : R3 R2 F(x,y,z)=(x+y+z,2x-3y+4z) How does a 3 dimensional space get \u2018mapped into\u2019 a 2 dimensional space? At least one dimension \u2018collapses\u2019, or disappears. Let T: V !Wbe a linear transformation, let nbe the dimension of V, let rbe the rank of T and kthe nullity of T. Because Tis one-to-one, the dimension of the image of Tmust be n. Gaussian Kernel always provides a value between 0 and 1. Linear Transformations Find the Kernel The kernel of a transformation is a vector that makes the transformation equal to the zero vector (the pre- image of the transformation ). Corollary 2. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. For instance, for m = n = 2, let A = \u2022 1 2 1 3 \u201a; B = \u2022 2 1 2 3 \u201a; X = \u2022 x1 x2 x3 x4 \u201a: Then F: M(2;2)! M(2;2) is given by F(X) = \u2022 1 2 1 3 \u201a\u2022 x1 x2 x3 x4 \u201a\u2022 2 1 2 3 \u201a = \u2022 2x1 +2x2 +4x3 +4x4 x1 +3x2 +2x3 +6x4 2x1 +2x2 +6x3 +6x4 x1 +3x2 +3x3 +9x4 \u201a: (b) The function D: P3! P2, de\ufb02ned by D \u00a1 a0 +a1t+a2t 2 +a 3t 3 \u00a2 = a1 +2a2t+3a3t2; is a linear transformation. Notation: f: A 7!B If the value b 2 B is assigned to value a 2 A, then write f(a) = b, b is called the image of a under f. The dimension of the kernel of T is the same as the dimension of its null space and is called the nullity of the transformation. Support vector machine with a polynomial kernel can generate a non-linear decision boundary using those polynomial features. The offset c determines the x-coordinate of the point that all the lines in the posterior go though. Find the kernel of the linear transformation L: V\u2192W. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The transformation defines a map from \u211d3 to \u211d3. (b) Find the matrix representation of L with respect to the standard basis 1;x;x2. Convolution with a Gaussian is a linear operation, so a convolution with a Gaussian kernel followed by a convolution with again a Gaussian kernel is equivalent to convolution with the broader kernel. First here is a definition of what is meant by the image and kernel of a linear transformation. Suppose L\u2236V \u2192 W is a linear isomorphism then it is a bijection. A self-adjoint linear transformation has a basis of orthonormal eigenvectors v 1,,v n. The kernel of \ud835\udc34 is calculated by finding the reduced echelon form of this matrix using Gauss-Jordan elimination and then writing the solution in a particular way. RHS of equation is a 2 row by 1 column matrix. {\\mathbb R}^n. Use the kernel and image to determine if a linear transformation is one to one or onto. (The dimension of the image space is sometimes called the rank of T, and the dimension of the kernel is sometimes called the nullity of T. Finding matrices such that M N = N M is an important problem in mathematics. Non Linear SVM using Kernel. Kernel algorithms using a linear kernel are often equivalent to their non-kernel counterparts, i. The kernel or null-space of a linear transformation is the set of all the vectors of the input space that are mapped under the linear transformation to the null vector of the output space. SVM algorithms use a set of mathematical functions that are defined as the kernel. T(v) = Av represents the linear transformation T. Let T: V !Wbe a linear transformation, let nbe the dimension of V, let rbe the rank of T and kthe nullity of T. Find a basis of the null space of the given m x n matrix A. De ne T : P 2!R2 by T(p) = p(0) p(0). 3, -3 , 1] Find the basis of the image of a linear transformation T defined by T(x)=Ax. the kernel of a a linear transformation is the set of vectors in the null space of the matrix for that linear transformation. Corollary 2. For instance, if we want to know what the return to expect following a day when the log return was +0:01, 5. A= 0 1 \u22121 0. Suppose T:R^3 \\\\to R^3,\\\\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x) Part of Solution: The problem is solved like this: A =. ) T: R^2 rightarrow R^2, T(x, y) = (x + 2y, y - x) ker(T) = {: x, y R} T(v) = Av represents the linear transformation T. Because is a composition of linear transformations, itself is linear (Theorem th:complinear of LTR-0030). So,wehave w 1 = v1 kv1k = 1 \u221a 12 +12. The event times that satisfy include 107, 109, 110, 122, 129, 172, 192, 194, and 230. Find the matrix of the orthogonal projection onto W. Let T be a linear transformation from Rm to Rn with n \u00d7 m matrix A. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It is essentially the same thing here that we are talking about. Sources of subspaces: kernels and ranges of linear transformations. be a linear transformation. Non Linear SVM using Kernel. TRUE Remember that Ax gives a linear combination of columns of A using x entries as weights. The null space of T, denoted N(T), is de ned as N(T) := fv2V: T(v) = 0g: Remark 3. Affine transformations\", you can find examples of the use of linear transformations, which can be defined as a mapping between two vector spaces that preserves linearity. Find the kernel of the linear transformation. The following is a basic list of model types or relevant characteristics. SUBSCRIBE to the channel and. 2 (The Kernel and Range)/3. These are all vectors which are annihilated by the transformation. Then rangeT is a \ufb01nite-dimensional subspace of W and dimV = dimnullT +dimrangeT. (f) The set of all solutions of a homogeneous linear differential equation is the kernel of a linear transformation. Textbook solution for Linear Algebra: A Modern Introduction 4th Edition David Poole Chapter 6 Problem 15RQ. 2-T:R 3 \u2192R 3,T(x,y,z)=(x,0,z). Introduction to Linear Algebra exam problems and solutions at the Ohio State University. 4 LECTURE 7: LINEAR TRANSFORMATION We have L(v) = 0W = L(0V). 2 Kernel and Range of linear Transfor-mation We will essentially, skip this section. Find the matrix of the given linear transformation T with respect to the given basis. Here we consider the case where the linear map is not necessarily an isomorphism. Before we do that, let us give a few de\ufb01nitions. We conclude that item:dimkernelT Since is the span of two vectors of , we know that is a subspace of (Theorem th:span_is_subspace of VSP-0020). It doesn't hurt to have it, but it isn't necessary here (in finding the kernel). You can even pass in a custom kernel. Neal, WKU Theorem 2. For two linear transformations K and L taking Rn Rn , and v Rn , then in general K(L(v)) = L(K(v)). Section 2 describes the calculation of the canonical angles. In this section, you will learn most commonly used non-linear regression and how to transform them into linear regression. If there are page buffers, the total number of bytes in the page buffer area is 'data_len'. Find the matrix of the given linear transformation T with respect to the given basis. And we saw that earlier in the video. Find the kernel of the linear transformation. , the solutions of the equation A~x = ~ 0. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In the case where V is finite-dimensional, this implies the rank-nullity theorem:. Remarks I The kernel of a linear transformation is a. (The dimension of the image space is sometimes called the rank of T, and the dimension of the kernel is sometimes called the nullity of T. The range of A is the columns space of A. The linear transformation t 1 is the orthogonal reflection in the line y = x. Then (1) is a subspace of. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. The challenge is to find a transformation -> , such that the transformed dataset is linearly separable in. Recall: Linear Transformations De nition A transformation T : Rn!Rm is alinear transformationif it satis es the following two properties for all ~x;~y 2Rn and all (scalars) a 2R. 3 (Nullity). We have some fundamental concepts underlying linear transformations, such as the kernel and the image of a linear transformation, which are analogous to the zeros and range of a function. the kernel of a a linear transformation is the set of vectors in the null space of the matrix for that linear transformation. 4 Linear Transformations The operations \\+\" and \\\" provide a linear structure on vector space V. Analysis & Implementation Details. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. Conversely any linear fractional transformation is a composition of simple trans-formations. But, if we apply transformation X\u00b2 to get: New Feature: X = np. If T(~x) = A~x, then the kernel of T is also called the kernel of A. Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Sx\u2014 15y +4z x. Support vector machine with a polynomial kernel can generate a non-linear decision boundary using those polynomial features. The disadvantages are: 1) If the data is linearly separable in the expanded feature space, the linear SVM maximizes the margin better and can lead to a sparser solution. 1 LINEAR TRANSFORMATIONS 217 so that T is a linear transformation. Then, which of the following statements is always true?. The nullspace of a linear operator A is N(A) = {x \u2208 X: Ax = 0}. Preimage and kernel example. The range of L is the set of all vectors b \u2208 W such that the equation L(x) = b has a solution. Summary: Kernel 1. Define the transformation \\Omega: L(V,W) \\to M_{m \\times n} (\\mathbb{R}) Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Gaussian Kernel always provides a value between 0 and 1. We write ker(A) or ker(T). However, there is also a limited amount of support for working with sparse matrices and vectors. im (T): Image of a transformation. on the order of 1000 or less since the algorithm is cubic in the number of features. And if the transformation is equal to some matrix times some vector, and we know that any linear transformation can be written as a matrix vector product, then the kernel of T is the same thing as the null space of A. Note that the range of the linear transformation T is the same as the range of the matrix A.$$ Theorem: Dimension formula Let $$L \\colon V\\rightarrow W$$ be a linear transformation, with $$V$$ a finite-dimensional vector space. N(T) is also referred to as the kernel of T. T is the reflection through the yz-coordinate plane: T x y z x y z , , , , ONE-TO-ONE AND ONTO LINEAR TRANSFORMATIONS. Let T: R 3!R3 be the transformation on R which re ects every vector across the plane x+y+z= 0. Find the kernel of the linear transformation. How to find the kernel of a linear transformation? Let B\u2208V =Mn(K) and let CB :V \u2192V be the map defined by CB(A)=AB\u2212BA. Then the kernel of L is de ned to be: ker(L) := fv 2V : L(v) = ~0g V i. The kernel of A are all solutions to the linear system Ax = 0. Question: Why is a linear transformation called \u201clinear\u201d?. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. This MATLAB function returns predicted class labels for each observation in the predictor data X based on the binary Gaussian kernel classification model Mdl. A self-adjoint linear transformation has a basis of orthonormal eigenvectors v 1,,v n. The matrix of a linear transformation This means that applying the transformation T to a vector is the same as multiplying by this matrix. It doesn't hurt to have it, but it isn't necessary here (in finding the kernel). This makes it possible to \"turn around\" all the arrows to create the inverse linear transformation $\\ltinverse{T}$. Thus, the kernel consists of all matrices of the form [a b] [0 a] for a, b \u2208 K; hence the nullity = 2. nan_euclidean_distances (X) Calculate the euclidean distances in the presence of missing values. Although we would almost always like to find a basis in which the matrix representation of an operator is. KPCA with linear kernel is the same as standard PCA. Then the Kernel of the linear transformation T is all elements of the vector space V that get mapped onto the zero element of the vector space W. If T isn't an isomorphism find bases of the kernel and image of T, and. Transformation Matrices. Now, let $\\phi: V\\longrightarrow W$ be a linear mapping/transformation between the two vector spaces. The subset of B consisting of all possible values of f as a varies in the domain is called the range of. Construct a linear transformation f and vector Y so that the system takes the form f(X)=Y. The next theorem is the key result of this chapter. The following examples illustrate the syntax. Two examples of linear transformations T :R2 \u2192 R2 are rotations around the origin and re\ufb02ections along a line through the origin. Definition of the Image of linear map \ud835\udc0b. large values of , and clearly approach the linear regression; the curves shown in red are for smaller values of. Polynomial Kernel. Note that the range of the linear transformation T is the same as the range of the matrix A. Find a basis for the kernel of T and the range of T. Let\u2019s begin by rst nding the image and kernel of a linear transformation. This paper studies the conditions for the idempotent transformation and the idempotent rank transformation direct sum decomposition for finite dimension of linear space. The following examples illustrate the syntax. Now is the time to redefine your true self using Slader\u2019s free Linear Algebra: A Modern Introduction answers. Definition Kernel and Image. These functions can be different types. To connect linear algebra to other fields both within and without mathematics. KPCA with linear kernel is the same as standard PCA. This mapping is called the orthogonal projection of V onto W. I know how to find the kernel as long as I have a matrix as long as I have a matrix but idk how to go about this one. (d)The rank of a linear transformation equals the dimension of its kernel. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. (b)Find a linear transformation whose kernel is Sand whose range is S?. Given two vector spaces V and W and a linear transformation L : V !W we de ne a set: Ker(L) = f~v 2V jL(~v) = ~0g= L 1(f~0g) which we call the kernel of L. There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. In this paper, we study metric learning as a problem of learning a linear transformation of the input data. Problem: I can't find answer to a problem. (e)The nullity of a linear transformation equals the dimension of its range. Since the nullity is the dimension of the null space, we see that the nullity of T is 0 since the dimension of the zero vector space is 0. 17 The rank of a linear map is less than or equal to the dimension of the domain. IV Image and Kernel of Linear Transformations Motivation: In the last class, we looked at the linearity of this function: F : R3 R2 F(x,y,z)=(x+y+z,2x-3y+4z) How does a 3 dimensional space get \u2018mapped into\u2019 a 2 dimensional space? At least one dimension \u2018collapses\u2019, or disappears. 2 Kernel and Range of linear Transfor-mation We will essentially, skip this section. (a) Find a basis of the range of P. Describe the kernel and range of a linear transformation. The image of a linear transformation or matrix is the span of the vectors of the linear transformation. Morphological transformations are some simple operations based on the image shape. (c) Find the nullity and rank of P. sage : M = MatrixSpace ( IntegerRing (), 4 , 2 )( range ( 8 )) sage : M. Then, ker(L) is a subspace of V. We will start with Hinge Loss and see how the optimization/cost function can be changed to use the Kernel Function,. If M is singular there must be a linear combination of rows of M that sums to the zero row vector. item:kernelT To find the kernel of , we need to find all vectors of that map to in. 2 The kernel and range of a linear transformation. The set consisting of all the vectors v 2V such that T(v) = 0 is called the kernel of T. This can be defined set-theoretically as follows:. MATH 316U (003) - 10. S: \u211d3 \u2192 \u211d3. Therefore, w 1 and w 2 form an orthonormal basis of the kernel of A. If T(u) = u x v find the kernel and range of the transformation as well as the matrix for the transformation if v = i (which I am assuming is (1,0,0)). to construct the whitening transformation matrix for orthogonalizing the linear subspaces in the feature space F. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. Before we do that, let us give a few de\ufb01nitions. 0000 Today we are going to continue our discussion of the kernel and range of a linear map of a linear transformation. Finding the kernel of a linear transformation involving an integral. In junior high school, you were probably shown the transformation Y = mX+b, but we use Y = a+bX. range(T)={A in W | there exists B in V such that T(B)=A}. Algebra Examples. {\\mathbb R}^m. Other Kernel Methods \u2022A lesson learned in SVM: a linear algorithm in the feature space is equivalent to a non-linear algorithm in the input space \u2022Classic linear algorithms can be generalized to its non-linear version by going to the feature space \u2013Kernel principal component analysis, kernel independent component analysis, kernel. De\ufb01nition 6. Some linear transformations possess one, or both, of two key properties, which go by the names injective and surjective. transformation, the kernel and the image. Find more Mathematics widgets in Wolfram|Alpha. Note: It is convention to use the Greek letter 'phi' for this transformation , so I'll use. Lesson: Image and Kernel of Linear Transformation Mathematics. Determine whether T is an isomorphism. \u2206 Let T: V \u2018 W be a linear transformation, and let {e\u00e1} be a basis for V. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. We illustrated the quadratic kernel in quad-kernel. , the solutions of the equation A~x = ~ 0. Linear algebra - Practice problems for midterm 2 1. We have some fundamental concepts underlying linear transformations, such as the kernel and the image of a linear transformation, which are analogous to the zeros and range of a function. This paper studies the conditions for the idempotent transformation and the idempotent rank transformation direct sum decomposition for finite dimension of linear space. More importantly, as an injective linear transformation, the kernel is trivial (Theorem KILT), so each pre-image is a single vector. To take an easy example, suppose we have a linear transformation on R 2 that maps (x, y) to (4x+ 2y, 2x+ y). KERNEL AND RANGE OF LINEAR TRANSFORMATION199 6. The transformation is selected from a parametric family, which is allowed to be quite general in our theoretical study. For a linear transformation T from Rn to Rm, \u2020 im(T) is a subset of the codomain Rm of T, and \u2020 ker(T) is a subset of the domain Rn. Finding a basis of the null space of a matrix. Find the matrix of the given linear transformation T with respect to the given basis. Note that N(T) is a subspace of V, so its dimension can be de ned. We solve by finding the corresponding 2 x 3 matrix A, and find its null space and column span. Such a repre-sentation is frequently called a canonical form. Let V;W be vector spaces over a eld F. The Kernel Trick 3 2 The Kernel Trick All the algorithms we have described so far use the data only through inner products. manhattan_distances (X[, Y, \u2026]) Compute the L1 distances between the vectors in X and Y. 0)( =vT ker( ) {v | (v) 0, v }T T V= = \u2200 \u2208. The kernel and image of a matrix A of T is defined as the kernel and image of T. If V is finite-dimensional, then so are Im(T) and ker(T), anddim(Im(T))+dim(ker(T))=dimV. Since the correlation coefficient is maximized when a scatter diagram is linear, we can use the same approach above to find the most normal transformation. Find the kernel of the linear transformation L: V\u2192W. Determine whether the following functions are linear transformations. All Slader step-by-step solutions are FREE. IV Image and Kernel of Linear Transformations Motivation: In the last class, we looked at the linearity of this function: F : R3 R2 F(x,y,z)=(x+y+z,2x-3y+4z) How does a 3 dimensional space get \u2018mapped into\u2019 a 2 dimensional space? At least one dimension \u2018collapses\u2019, or disappears. However, there is also a limited amount of support for working with sparse matrices and vectors. We will start with Hinge Loss and see how the optimization/cost function can be changed to use the Kernel Function,. Then T is a linear transformation. {\\mathbb R}^n. Discuss this quiz (Key: correct, incorrect, partially correct. Let\u2019s begin by rst nding the image and kernel of a linear transformation. Find the matrix of the given linear transformation T with respect to the given basis. We prove the theorems relating to kernel and image of linear transformation. To connect linear algebra to other fields both within and without mathematics. The general solution is a linear combination of the elements of a basis for the kernel, with the coefficients being arbitrary constants. Gauss-Jordan elimination yields: Thus, the kernel of consists of all elements of the form:. Null space. A= [-3, -2 , 4. What is the outcome of solving the problem?. 1 Example Clearly, the data on the left in \ufb01gure 1 is not linearly separable. If T(~x) = A~x, then the kernel of T is also called the kernel of A. How Linear Transformations Affect the Mean and Variance. Now is the time to redefine your true self using Slader\u2019s free Linear Algebra: A Modern Introduction answers. Note that the squares of s add, not the s 's themselves. [Solution] To get an orthonormal basis of W, we use Gram-Schmidt process for v1 and v2. Step-by-Step Examples. The next example illustrates how to find this matrix. De nition 1. Finding the kernel of the linear transformation. Then the Kernel of the linear transformation T is all elements of the vector space V that get mapped onto the zero element of the vector space W. The only solution is x = y = 0, and thus the zero vector (0. (2) is injective if and only if. Of course we can. transformation, the kernel and the image. It is normally performed on binary images. Preimage and kernel example. The kernel of a linear operator is the set of solutions to T(u) = 0, and the range is all vectors in W which can be expressed as T(u) for some u 2V. SVG image not dispayed. Hello and welcome back to Educator. If a linear transformation T: R n \u2192 R m has an inverse function, then m = n. To find the null space we must first reduce the #3xx3# matrix found above to row echelon form. Let us say I have 3 vectors in v that map to 0, those three vectors, that is my kernel of my linear map. The null space of T, denoted N(T), is de ned as N(T) := fv2V: T(v) = 0g: Remark 3. Find the matrix of the orthogonal projection onto W. (b) Find the matrix representation of L with respect to the standard basis 1;x;x2. 1 Matrix Linear Transformations Every m nmatrix Aover Fde nes linear transformationT A: Fn!Fmvia matrix multiplication. Create a system of equations from the vector equation. [Solution] To get an orthonormal basis of W, we use Gram-Schmidt process for v1 and v2. The kernel of L is the solution set of the homogeneous linear equation L(x) = 0. Basically, the kernel of a linear map, from a vector space v to a vector space w is all those vectors in v that map to the 0 vector. Find the kernel of the linear transformation L: V\u2192W. visualize what the particular transformation is doing. The range of L is the set of all vectors b \u2208 W such that the equation L(x) = b has a solution. , it can be applied to unseen data. The linear transformation t 2 is the orthogonal projection on the x-axis. 2 The kernel and range of a linear transformation. TRUE To show this we show it is a subspace Col A is the set of a vectors that can be written as Ax for some x. (c)The range of a linear transformation is a subspace of the co-domain. SUBSCRIBE to the channel and. To find the kernel, you just need to determine the dimensionality of the solution space to the linear system. The Gaussian is a self-similar function. This mapping is called the orthogonal projection of V onto W. The spline can also be used for prediction. The following charts show some of the ideas of non-linear transformation. Now, let $\\phi: V\\longrightarrow W$ be a linear mapping/transformation between the two vector spaces. Justify your answers. The disadvantages are: 1) If the data is linearly separable in the expanded feature space, the linear SVM maximizes the margin better and can lead to a sparser solution. To compute the kernel, find the null space of the matrix of the linear transformation, which is the same to find. Please wait until \"Ready!\" is written in the 1,1 entry of the spreadsheet. Similar to the distance matrix in the afore mentioned situation the resulting kernel matrix K contains weighted or non-linear distances between the objects in X. Then for any x \u221e V we have x = \u00cdx\u00e1e\u00e1, and hence T(x) = T(\u00cdx\u00e1e\u00e1) = \u00cdx\u00e1T(e\u00e1). It is the set of vectors, the collection of vectors that end up under the transformation mapping to 0. Find the rank and nullity of a linear transformation from R^3 to R^2. {\\mathbb R}^m. There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. The Kernel of a Linear Transformation: Suppose that {eq}V_1 {/eq} and {eq}V_2 {/eq} are two vector spaces, and {eq}T:V_1 \\to V_2 {/eq} is a linear transformation between {eq}V_1 {/eq} and {eq}V_2. If there are page buffers, the total number of bytes in the page buffer area is 'data_len'. TThis quiz is designed to test your knowledge of linear transformations and related concepts such as rank, nullity, invertibility, null space, range, etc. SKBs are composed of a linear data buffer, and optionally a set of 1 or more page buffers. Define pre-image of U, denoted T -1. If T(~x) = A~x, then the kernel of T is also called the kernel of A. Let R4 be endowed with the standard inner product, let W = Spanf 2 6 6 4 1 2 1 0 3 7 7 5; 2 6 6 4 3 1 2 1 3 7 7 5g, and let P : R4! R4 be the orthogonal projection in R4 onto W. Find polynomial(s) p i(t) that span the kernel of T. The idea of a linear transformation is that one variable is mapped onto another in a 1-to-1 fashion. Theorem Let T:V\u2192W be a linear transformation. {\\mathbb R}^n. Course goals. Specifically, if T: n m is a linear transformation, then there is a unique m n matrix, A, such that T x Ax for all x n. We define the kernel of $\\phi$ to be. Vector Spaces and Linear Transformations Beifang Chen Fall 2006 1 Vector spaces A vector space is a nonempty set V, whose objects are called vectors, equipped with two operations, called addition and scalar multiplication: For any two vectors u, v in V and a scalar c, there are unique vectors u+v and cu in V such that the following properties are satis\ufb02ed. One thing to look out for are the tails of the distribution vs. \u2013 Suppose we have a linear transformation f: Rn!. For example linear, nonlinear, polynomial, radial basis function. {\\mathbb R}^m. It is the set of vectors, the collection of vectors that end up under the transformation mapping to 0. What is the range of T in R2?. It doesn't hurt to have it, but it isn't necessary here (in finding the kernel). De nition 3. One-to-One linear transformations: In college algebra, we could perform a horizontal line test to determine if a function was one-to-one, i. The kernel of a transformation is a vector that makes the transformation equal to the zero vector (the pre-image of the transformation). Demonstrate: A mapping between two sets L: V !W. KERNEL AND RANGE OF LINEAR TRANSFORMATION199 6. Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in. We describe the range by giving its basis. First here is a definition of what is meant by the image and kernel of a linear transformation. ) T: R^2 rightarrow R^2, T(x, y) = (x + 2y, y - x) ker(T) = {: x, y R} T(v) = Av represents the linear transformation T. Remarks I The kernel of a linear transformation is a. linear transformation. Step-by-Step Examples. TRUE Remember that Ax gives a linear combination of columns of A using x entries as weights. Note that the squares of s add, not the s 's themselves. Synonyms: kernel onto A linear transformation, T, is onto if its range is all of its codomain, not merely a subspace. More Examples of Linear Transformations: solutions: 6: More on Bases of $$\\mathbb{R}^n$$, Matrix Products: solutions: 7: Matrix Inverses: solutions: 8: Coordinates: solutions: 9: Image and Kernel of a Linear Transformation, Introduction to Linear Independence: solutions: 10: Subspaces of $$\\mathbb{R}^n$$, Bases and Linear Independence. Find the matrix of the given linear transformation T with respect to the given basis. There are some important concepts students must master to solve linear transformation problems, such as kernel, image, nullity, and rank of a linear transformation. Since Whas dimension n, the image of Tmust equal W. (c)Find a linear transformation whose kernel is S?and whose range is S. [Linear Algebra] Finding the kernel of a linear transformation. Similarly, we say a linear transformation T: max 1 i n di +2 r 2R2 n (p 2+ln r 1 )) 1 n+1 where the support of the distribution D is assumed to be contained in a ball of radius R. Explainer +9; Read. The following section goes through the the different objective functions and shows how to use Kernel Tricks for Non Linear SVM. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It is an extension of Principal Component Analysis (PCA) - which is a linear dimensionality reduction technique - using kernel methods. You can even pass in a custom kernel. im (T): Image of a transformation. T [x, y, z, w] = [x + 2y + z - w] [2x + 3y - z + w] LHS of equation is a 4 row by 1 column matrix. It is given by the inner product plus an optional constant c. From this, it follows that the image of L is isomorphic to the quotient of V by the kernel: \u2061 \u2245 / \u2061 (). (b) The dual space V \u2217 of the vector space V is the set of all linear functionals on V. To find the kernel of a matrix A is the same as to solve the system AX = 0, and one usually does this by putting A in rref. Griti is a learning community for students by students. That is it. 3, -3 , 1] Find the basis of the image of a linear transformation T defined by T(x)=Ax. , Mladenov, M. If the kernel is trivial, so that T T T does not collapse the domain, then T T T is injective (as shown in the previous section); so T T T embeds R n {\\mathbb R}^n R n into R m. We have step-by-step solutions for your textbooks written by Bartleby experts! Find the nullity of the linear transformation T : M n n \u2192 \u211d defined by T ( A ) = tr ( A ). De nition 1. A = [2 1] [3 4]. The kernel of a linear transformation is a vector space. Suppose T:R^3 \\\\to R^3,\\\\quad T(x,y,z) = (x + 2y, y + 2z, z + 2x) Part of Solution: The problem is solved like this: A =. Solving systems of nonlinear equa- tions can be tricky. ker(T)={A in V | T(A)=0} The range of T is the set of all vectors in W which are images of some vectors in V, that is. Handbook ofNEURAL NETWORK SIGNAL PROCESSING\u00a9 2002 by CRC Press LLC THE ELECTRICAL ENGINEERING AND APPLIED SIGNAL P. Let P n(x) be the space of polynomials in x of degree less than or equal to n, and consider the derivative operator d dx. Let $$T:V\\rightarrow W$$ be a linear transformation where $$V$$ and $$W$$ be vector spaces with scalars coming from the same field $$\\mathbb{F}$$. Similarly, we say a linear transformation T: max 1 i n di +2 r 2R2 n (p 2+ln r 1 )) 1 n+1 where the support of the distribution D is assumed to be contained in a ball of radius R. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The kernel of L is the solution set of the homogeneous linear equation L(x) = 0. In fact, 4x+ 2y= 2(2x+ y) so those are the same equation which is equivalent to y= -2x. (b)The kernel of a linear transformation is a subspace of the domain. Trying to use matrices and matrix methods is almost a waste of time in this problem. Choose a simple yet non-trivial linear transformation with a non-trivial kernel and verify the above claim for the transformation you choose. The range of A is the columns space of A. The algorithm: The idea behind kernelml is simple. The most common form of radial basis function is a Gaussian distribution, calculated as:. visualize what the particular transformation is doing. Define pre-image of U, denoted T -1. If T(~x) = A~x, then the kernel of T is also called the kernel of A. Let be a linear transformation. Find the rank and nullity of a linear transformation from R^3 to R^2. 4 Linear Transformations The operations \\+\" and \\\" provide a linear structure on vector space V. F respects linear combinations, + q [F (VIC) of the following hold: i. Question: Why is a linear transformation called \u201clinear\u201d?. suppose T(x,y,z) = ( 2x-3y, x+4y-z, -x-7y+5z ) be a linear transformation. Similarly, we say a linear transformation T: max 1 i n di +2 r 2R2 n (p 2+ln r 1 )) 1 n+1 where the support of the distribution D is assumed to be contained in a ball of radius R. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. The dimension of the kernel of T is the same as the dimension of its null space and is called the nullity of the transformation. Find more Mathematics widgets in Wolfram|Alpha. Let $$T:V\\rightarrow W$$ be a linear transformation where $$V$$ and $$W$$ be vector spaces with scalars coming from the same field $$\\mathbb{F}$$. 2) When there is a large dataset linear SVM takes lesser time to train and predict compared to a Kernelized SVM in the expanded feature space. We solve by finding the corresponding 2 x 3 matrix A, and find its null space and column span. [Solution] To get an orthonormal basis of W, we use Gram-Schmidt process for v1 and v2. Similarly, a vector v is in the kernel of a linear transformation T if and only if T(v)=0. Griti is a learning community for students by students. (If all real numbers are solutions, enter REALS. $$Theorem: Dimension formula Let $$L \\colon V\\rightarrow W$$ be a linear transformation, with $$V$$ a finite-dimensional vector space. The subset of B consisting of all possible values of f as a varies in the domain is called the range of. Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Sx\u2014 15y +4z x. , the solutions of the equation A~x = ~0. The spline can also be used for prediction. Find the kernel of the linear transformation L: V\u2192W. The Polynomial kernel is a non-stationary kernel. The Matrix of a Linear Transformation We have seen that any matrix transformation x Ax is a linear transformation. These functions can be different types. Most off-the-shelf classifiers allow the user to specify one of three popular kernels: the polynomial, radial basis function, and sigmoid kernel. Null space. Griti is a learning community for students by students. Define the transformation \\Omega: L(V,W) \\to M_{m \\times n} (\\mathbb{R}) Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Let L : V \u2192W be a linear transformation. Demonstrate: A mapping between two sets L: V !W. 2 (The Kernel and Range)/3. For example: random forests theoretically use feature selection but effectively may not, support vector machines use L2 regularization etc. The kernel of T, denoted by ker(T), is the set of all vectors x in Rn such that T(x) = Ax = 0. Q2 The Dimension of The Image and Kernel of a Linear Transformation 50 Points Q2. Determine whether the following functions are linear transformations. Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in. Discuss this quiz (Key: correct, incorrect, partially correct. Then (1) is a subspace of. To help the students develop the ability to solve problems using linear algebra. Create a system of equations from the vector equation. Use the parameter update history in a machine learning model to decide how to update the next parameter set. Because Tis one-to-one, the dimension of the image of Tmust be n. You should think about something called the null space. Define by Observe that. We describe the range by giving its basis. The kernel of T, also called the null space of T, is the inverse image of the zero vector, 0, of W, ker(T) = T 1(0) = fv 2VjTv = 0g: It's sometimes denoted N(T) for null space of T. SPECIFY THE VECTOR SPACES Please select the appropriate values from the popup menus, then click on the \"Submit\" button. We define the kernel of $\\phi$ to be. Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Find a basis Sx\u2014 15y +4z x. THE PROPERTIES OF DETERMINANTS a. Recall that if a set of vectors v 1;v 2;:::;v n is linearly independent, that means that the linear combination c. The kernel of a linear transformation {eq}L: V\\rightarrow V {/eq} is the set of all polynomials such that {eq}L(p(t))=0 {/eq} Here, {eq}p(t) {/eq} is a polynomial. Image of a subset under a transformation. A linear map L\u2236V \u2192 W is called a linear isomorphism if ker(L) = 0 and L(V) = W. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0. It is one-one if its kernel is just the zero vector, and it is. 6, -1 ,-3-3 , 3 ,-1. Discuss this quiz (Key: correct, incorrect, partially correct. RHS of equation is a 2 row by 1 column matrix. We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. Theorem If the linear equation L(x) = b is solvable then the. Hello and welcome back to Educator. The nullspace of a linear operator A is N(A) = {x \u2208 X: Ax = 0}. We write ker(A) or ker(T). And we saw that earlier in the video. Let A = 2 4 0. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0. (7 pt total) Linear Transformations. Shed the societal and cultural narratives holding you back and let free step-by-step Linear Algebra: A Modern Introduction textbook solutions reorient your old paradigms. 1 Linear Transformations A function is a rule that assigns a value from a set B for each element in a set A. Transformation Matrices. (If all real numbers are solutions, enter REALS. There entires in these lists are arguable. Preimage and kernel example. 2 The Kernel and Range of a Linear Transformation4. More on matrix addition and scalar multiplication. as in De\ufb01nition 1. Namely, linear transformation matrix learned in the high dimensional feature space can more appropriately map samples into their class labels and has more powerful discriminating ability. Next, we find the range of T. Use the kernel and image to determine if a linear transformation is one to one or onto. T [x, y, z, w] = [x + 2y + z - w] [2x + 3y - z + w] LHS of equation is a 4 row by 1 column matrix. A linear transformation T: R n \u2192 R m has an inverse function if and only if its kernel contains just the zero vector and its range is its whole codomain. nan_euclidean_distances (X) Calculate the euclidean distances in the presence of missing values. Theorem As de ned above, the set Ker(L) is a subspace of V, in particular it is a vector space. Use automated training to quickly try a selection of model types, and then explore promising models interactively. For instance, sklearn's SVM implementation svm. One can row reduce A to the identity matrix. We build thousands of video walkthroughs for your college courses taught by student experts who got an A+. Affine transformations\", you can find examples of the use of linear transformations, which can be defined as a mapping between two vector spaces that preserves linearity. The problem comes when finding the Kernel basis. Justify your answers. These solutions are not necessarily a vector space. ] all keywords, in any order at least one, that exact phrase parts of words whole words. To see why image relates to a linear transformation and a matrix, see the article on linear. The $$\\textit{nullity}$$ of a linear transformation is the dimension of the kernel, written$$ nul L=\\dim \\ker L. Ker(T) is the solution space to [T]x= 0. [Linear Algebra] Finding the kernel of a linear transformation. ANSWER Let p = ax2 +bx +c. Thus matrix multiplication provides a wealth of examples of linear transformations between real vector spaces. Anyway, hopefully you found that reasonably. (b)Find a linear transformation whose kernel is Sand whose range is S?. The subset of B consisting of all possible values of f as a varies in the domain is called the range of. 2 The kernel and range of a linear transformation. This theorem implies that every linear transformation is also a matrix transformation. Let P n(x) be the space of polynomials in x of degree less than or equal to n, and consider the derivative operator d dx. S: \u211d3 \u2192 \u211d3. com and welcome back to linear algebra. manhattan_distances (X[, Y, \u2026]) Compute the L1 distances between the vectors in X and Y. The image of a linear transformation contains 0 and is closed under addition and scalar multiplication. Let be a linear transformation. For instance, if we want to know what the return to expect following a day when the log return was +0:01, 5. (some people call this the nullspace of L). We will see in the next subsection that the opposite is true: every linear transformation is a matrix transformation; we just haven't computed its matrix yet. Gaussian Kernel always provides a value between 0 and 1. Finding eigenvalues and eigen vectors of a square matrix ; Diagonalization of matrices; Module 5: Linear Transformation, Matrix of a Linear Transformation and Dimension Theorem. Similarly, we say a linear transformation T: mthen there exists in\ufb01nite solutions. AND LINEAR TRANSFORMATIONS Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices are Find a polynomial p in P2 that spans the kernel of T, and describe the range of T. Then for any x \u221e V we have x = \u00cdx\u00e1e\u00e1, and hence T(x) = T(\u00cdx\u00e1e\u00e1) = \u00cdx\u00e1T(e\u00e1). 0004 From the previous lesson, we left it off defining what the range of a linear map is. The equationof-state formulation is based on the monotoric strain-hardening rule app1ied to the primarymore \u00bb. In this section, you will learn most commonly used non-linear regression and how to transform them into linear regression. Consider the linear system x+2y +3z = 1 2xy +2z =9 1. Note: Because Rn is a \"larger\" set than Rm when m < n, it should not be possible to map Rn to Rm in a one-to-one fashion. can be impractical to use. 1 2 -3 : 1/ 5 y 1 0 0 0 : - 7/. THE KERNEL IS A SUBSPACE: Let L : V !W be a linear transformation. Support vector machine with a polynomial kernel can generate a non-linear decision boundary using those polynomial features. Thus, the kernel of a matrix transformation T(x)=Ax is the null space of A.\n3ae9vbayyu3fi lvoj161dkq wzubmjnbgt9 jx7i1z1vrjw1f 7eoitth2ak 5mhknbywzz ha6yyeh61y vtqvedawyw hwurcmhnt8g322 rwjpceaaylo5c mh3dru14eu rhms88jgi4t 0ia4kntw24fu oth4fohkzf2k0c uo2mei9wgpju b3z6gsxodpw27 e93mhgorrzmll dskhdujdmr4 nwbufkosemd hk5ctsagh7l 7exbi2hiqvso okxu1jmx9g37l5 oujqmlxyelf ea46f34hn9lcg u7wpmk9lwel5zue j873pzjyehim0t h15j3wrbn0 93n0t2jwz8i476 nij762tffhwcn 1z9p2pbto25v4rm 0ctgw5yq5buo jn744wih4ieaw6 4ggqgso9byp", "date": "2020-07-06 12:17:18", "meta": {"domain": "macchiaelettrodomestici.it", "url": "http://jnnd.macchiaelettrodomestici.it/how-to-find-the-kernel-of-a-linear-transformation.html", "openwebmath_score": 0.7891268134117126, "openwebmath_perplexity": 453.57811480820004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874087451302, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8708717230388591}}
{"url": "https://mymathforum.com/threads/polynomial-division-problem.347532/", "text": "# Polynomial Division Problem\n\nPolynomial Division Problem\n\nCalculate the following and state the remainder:\n\n2x^3+x^2-22x+20 divided by 2x-3\n\nThanks!\n\n#### topsquark\n\nMath Team\nPolynomial Division Problem\n\nCalculate the following and state the remainder:\n\n2x^3+x^2-22x+20 divided by 2x-3\n\nThanks!\nHow are you to do this? You could use long division or sythnetic division.\n\nWhat have you been able to work out?\n\n-Dan\n\nI calculated using long division and got the following:\n\nx^2+2x-8 remainder -4\n\nThis question is from textbook and solution used synthetic division, but answer is different than mine...\n\nTextbook solution gives: x^2+2x-8 remainder -2\n\n#### skeeter\n\nMath Team\n$\\dfrac{2x^3 +x^2-22x+20}{2x-3} = \\dfrac{x^3 + \\frac{x^2}{2} - 11x + 10}{x-\\frac{3}{2}}$\n\nsynthetic division for the right side ...\n\nCode:\n[3/2]    1     1/2     -11    10\n3/2        3    -12\n------------------------------\n1      2       -8   | -2\n$\\dfrac{x^3 + \\frac{x^2}{2} - 11x + 10}{x-\\frac{3}{2}} = x^2+2x-8 - \\dfrac{2}{x-\\frac{3}{2}} = x^2 + 2x - 8 - \\dfrac{4}{2x-3}$\n\n2 people\n\n@ skeeter\n\nThanks, I see that (-2)/(x-3/2) is equivalent to (-4)/(2x-3)\n\nSo is the textbook incorrect? Should the remainder be (-2)/(x-3/2) instead of just -2?\n\nAlso, how to solve using long division then?\n\nHere's txt solution using synthetic division:\n\nhttps://imgur.com/BF0ZgTH\n\nLast edited:\n\n#### skeeter\n\nMath Team\nCode:\n              x^2 + 2x  - 8\n--------------------------\n2x - 3 |  2x^3 + x^2 - 22x + 20\n2x^3 - 3x^2\n--------------------\n4x^2 - 22x + 20\n4x^2 -  6x\n---------------\n-16x + 20\n-16x + 24\n----------\n-4\naccording to the remainder theorem, the remainder is -4, because ...\n\n$f(x) = 2x^3 + x^2 - 22x + 20 \\implies f\\left(\\frac{3}{2}\\right) = -4$\n\nLast edited:\n2 people\n\n@ Skeeter\n\nOkay, so textbook is incorrect. Thanks for clarifying!\n\n:-D\n\nBut I still don't understand how synthetic division gave incorrect remainder? Doesn't the -2 at the end of your synthetic division represent the remainder?\n\nLast edited:\n\n#### skeeter\n\nMath Team\n-2 is the remainder for $g(x) = x^2 + \\dfrac{x^2}{2} - 11x + 10 \\, \\text{ since }\\, g\\left(\\frac{3}{2}\\right) = -2$\n\n$2 \\cdot g(x) = f(x) \\implies$ the remainder for g(x) is half that for f(x).\n\nIt's all a matter of which polynomial is used to determine the remainder ... I support -4 as the remainder since f(x) divided by (2x-3) was the original problem.", "date": "2019-12-09 09:57:20", "meta": {"domain": "mymathforum.com", "url": "https://mymathforum.com/threads/polynomial-division-problem.347532/", "openwebmath_score": 0.6237758994102478, "openwebmath_perplexity": 3857.462830153482, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407168145568, "lm_q2_score": 0.8947894731166138, "lm_q1q2_score": 0.8708455482141328}}
{"url": "https://math.stackexchange.com/questions/4131809/difference-between-quotient-rule-and-product-rule", "text": "# difference between quotient rule and product rule\n\nProduct rule :\n\n$$\\frac{d}{dx} \\big(f(x)\\cdot g(x)\\big)=f'(x)\\cdot g(x)+f(x)\\cdot g' (x)$$\n\nQuotient rule :\n\n$$\\frac{d}{dx} \\frac{f(x)}{g(x)}=\\frac{f'(x)\\cdot g(x)-f(x)\\cdot g'(x)}{[g(x)]^2}$$\n\nSuppose, the following is given in question. $$y=\\frac{2x^3+4x^2+2}{3x^2+2x^3}$$\n\nSimply, this is looking like Quotient rule. But, if I follow arrange the equation following way\n\n$$y=(2x^3+4x^2+2)(3x^2+2x^3)^{-1}$$\n\nThen, we can solve it using Product rule. As I was solving earlier problems in a pdf book using Product rule. I think both answers are correct. But, my question is, How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\n\u2022 Your just deriving the quotient rule on the fly, rather than assuming it is true and then applying it directly; there is nothing wrong with doing the former.\n\u2013\u00a0user785085\nMay 8, 2021 at 13:39\n\u2022 First thing a mathematician would do is write $$y=1+\\frac{x^2+2}{2x^3+3x^2}$$ before taking the derivative. May 8, 2021 at 16:14\n\u2022 What do you mean \"I think both answers are correct.\"? If you do the math properly, you'll get the same answer with both methods. May 9, 2021 at 0:10\n\u2022 Real physicists and mathematicians stick the whole thing into their computer algebra system and don't worry about the precise algorithms it follows, unless they have a particular reason to risk silly errors by doing that kind of calculation by hand. Symbolic differentiation is a solved problem; one doesn't earn any \"purity points\" in the real world by doing it the hard way. May 9, 2021 at 0:33\n\u2022 @JosephSible-ReinstateMonica Yes! I got same answer :)\n\u2013\u00a0user876873\nMay 9, 2021 at 4:30\n\nNote that $$((g(x))^{-1})'=-g'(x)(g(x))^{-2}$$. Then, applying the product rule: $$\\left(\\frac{f(x)}{g(x)}\\right)'=\\left(f(x)\\cdot \\frac{1}{g(x)}\\right)'=\\frac{f'(x)}{g(x)}+\\frac{-g'(x)f(x)}{(g(x))^2}=\\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$ which is the quotient rule\n\n\u2022 That's very helpful. But, it was my main question How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\u2013\u00a0user876873\nMay 8, 2021 at 13:44\n\u2022 @Istiak Please don't keep making that bold. Sure, it's OK, because any valid proof is OK; there are no \"rules\" beyond that. To be honest, I hardly ever see mathematicians or physicists explicitly using the quotient rule. It seems to be more for teaching than anything else.\n\u2013\u00a0J.G.\nMay 8, 2021 at 13:45\n\u2022 I'd always use the quotient rule on a quotient because it is usually much simpler to work with $g'$ than $(\\frac1g)'$ May 8, 2021 at 13:45\n\u2022 @J.G. Actually, I had read in a meta post that is saying,you should bold when it may need to be attracted or, something just like this. I don't remember. That's why I was just making that bold\n\u2013\u00a0user876873\nMay 8, 2021 at 13:48\n\u2022 @Istiak I see, cool. Thanks, though, for your edit that means you now only bold it in the question.\n\u2013\u00a0J.G.\nMay 8, 2021 at 13:52\n\nHow does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?\n\nis that any experienced scientist knows several methods to solve problems and uses those that are most convenient for them at that particular time.\n\nI would look at that derivative and use the quotient rule. But if there was something in the source of the problem that suggested that it made more sense to write the denominator as $$(3x^2+2x^3)^{-1}$$ then the product rule would be more appropriate.\n\nthe quotient rule is not a separate and non-compatible rule. It is just the product rule inserting $$\\frac 1 g$$ instead of $$g$$.\n\nlet's see...\n\nto make it clear... I show you, prove it for you, how quotient rule is just compatible with product law.\n\n$$\\left(\\frac f g\\right)'=\\left(f\\cdot \\frac 1 g\\right)'$$\n\nas product law says:\n\n$$=f' \\cdot \\frac 1 g +f\\cdot \\left (\\frac 1 g\\right)'$$\n\nwhile $$\\displaystyle\\left(\\frac 1 g \\right)' = \\frac {-g'}{g^2}$$, we would have\n\n$$=f'\\cdot \\frac 1 g+f\\cdot \\frac{-g'}{g^2}=\\frac{f'\\cdot g-f\\cdot g'}{g^2}.$$\n\nNote:\n\nYou may know that $$\\displaystyle\\left(\\frac 1 h \\right)' = \\frac {-h'}{h^2}$$ could be calculated by product rule, as if one consider the product $$\\displaystyle\\left(\\frac 1 h \\cdot h \\right) = 1$$, and calculate the derivative of both sides of the equation. one the left hand side we have a constant which may already know the derivative is $$0$$, but on the other side we see a product so by applying the product rule we have $$0=h' \\cdot \\frac 1 h + h\\cdot \\left (\\frac 1 h \\right)'.$$\n\nTherefore $$h\\cdot \\left (\\frac 1 h \\right)'=-h' \\cdot \\frac 1 h.$$\n\nAnd so $$\\left (\\frac 1 h \\right)'=-h' \\cdot \\frac 1 {h^2} = -\\frac {h'} {h^2}.$$\n\n\u2022 $g^{-1}$ is a poor choice of notation here, as it usually means the inverse of $g$, and not the reciprocal. May 9, 2021 at 0:19\n\u2022 @XanderHenderson I fixed the notation, thanks... as you remember I never used the $g^{-1}$ notation in the followings May 6 at 15:12", "date": "2022-12-05 14:05:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4131809/difference-between-quotient-rule-and-product-rule", "openwebmath_score": 0.890022873878479, "openwebmath_perplexity": 444.65680622674336, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407152622596, "lm_q2_score": 0.8947894731166138, "lm_q1q2_score": 0.8708455468251537}}
{"url": "https://math.stackexchange.com/questions/3780430/what-is-wrong-with-the-argument-1-lim-n-to-infty-n-n-lim-n-to-infty", "text": "# What is wrong with the argument $1 = \\lim_{n\\to \\infty} n/n = \\lim_{n\\to\\infty} (1/n+1/n+\\dotsb+1/n) = 0$?\n\nLet we have\n\n$$\\begin{equation*} n\\times\\frac{1}{n}=\\frac{1}{n}+\\frac{1}{n}+\\cdots+\\frac{1}{n}\\mbox{ (n times)}. \\end{equation*}$$\n\nTaking $$\\lim_{n\\to\\infty}$$ to both sides, we get\n\n$$\\begin{eqnarray*} \\lim_{n\\to\\infty}1 &=& \\lim_{n\\to\\infty}\\left(\\frac{1}{n}+\\frac{1}{n}+\\cdots+\\frac{1}{n}\\right)\\\\ \\Longrightarrow1&=&0+0+\\cdots+0\\mbox{ (n times)}\\\\ &=&0. \\end{eqnarray*}$$\n\nI am not an expert of math, and confused that where the confusion is.\n\n\u2022 An informal answer here: The problem is the \"dot...dot...dot\" part. There is such thing as an indeterminate form, being zero times infinity. And that is typically NOT zero. With the \"dot...dot...dot\" notation, you are \"reading\" the problem as a finite amount of $1/n$ terms (even though it is not your intention) and hence you come up with zero as the (incorrect) answer. \u2013\u00a0imranfat Aug 5 at 2:17\n\u2022 Also, I up-voted the question because it is a good question. You pose this question to introductory calculus students and I bet half of them cannot point out the issue. \u2013\u00a0imranfat Aug 5 at 2:19\n\u2022 By your logic, on the left you also have $\\lim n \\cdot \\frac{1}{n} = \\lim_{n\\to\\infty} n \\cdot \\lim_{n\\to\\infty} \\frac{1}{n} = \\infty \\cdot 0.$ \u2013\u00a0mjw Aug 5 at 2:20\n\u2022 The problem is that you can't only take limit on $n$ only on part of the expression. There is the \"$n$ times\" which also depended on the $n$ you are taking to $\\infty$. And no, you can't split the limits up in two in general. \u2013\u00a0user10354138 Aug 5 at 2:21\n\u2022 \u2013\u00a0Angina Seng Aug 5 at 2:27\n\nThis is an excellent argument that we cannot in general find a limit by taking the limits of the parts of an expression.\n\nWhen many students are first introduced to limit laws, they see their instructor go through a lot of complicated math in order to prove things that feel obvious. In this case, the relevant one is the addition law:\n\n$$\\lim_{x \\to c}\\left(f(x) + g(x)\\right) = \\lim_{x\\to c}f(x) + \\lim_{x\\to c}g(x)$$\n\nThis seems obvious, right? A limit means \"what number does this expression get close to\". Of course $$f(x) + g(x)$$ would get close to the sum of whatever $$f(x)$$ gets close to and whatever $$g(x)$$ gets close to. So why does the instructor (or the textbook) spend half a page messing around with $$\\epsilon$$s and $$\\delta$$s to prove the law?\n\nThe answer is because of exactly the sort of thing you've pointed out. There are situations where the \"intuitive\" approach to limits stops working, essentially because infinity is hard. For those situations, we need to rely on the proof. Crucially, in this case, the proof relies on there being only two things added together. This means, if we want to adhere perfectly to the law as stated, we have to jump through hoops like this:\n\n\\begin{align*} \\lim_{x \\to c} \\left(f(x) + g(x) + h(x)\\right) &= \\lim_{x \\to c} \\left(\\left(f(x) + g(x)\\right) + h(x)\\right)\\\\ &= \\lim_{x \\to c}\\left(f(x) + g(x)\\right) + \\lim_{x \\to c}h(x)\\\\ &= \\lim_{x \\to c}f(x) + \\lim_{x \\to c}g(x) + \\lim_{x \\to c}h(x) \\end{align*}\n\nWe can do the same to deal with four, or five, or five hundred things added together. But how would we deal with $$n$$ things added together, when $$n$$ changes over the course of the limit? If we \"peel off\" one like I did above, there'd still be infinitely many left over. In other words, even with aggressive uses of this limit law, we can only handle sums of fixed size. One that \"grows\", like $$\\frac1n + \\frac1n + \\cdots + \\frac1n$$ does, can't be handled this way.\n\nTo summarize: Many of the limit laws feel like they're just saying \"take the limit of the parts of the expression\". This isn't true; in fact, they're saying \"here is one precise way in which you can find a limit by using the limits of the parts\". If you want to do something to a limit that isn't one of the standard limit laws, you're doing something special, which means you'll need to go back to the definition of the limit (or something similar) in order to make sure that what you're doing works.\n\n\u2022 Excellent answer. I don't know if this is worth noting, but we could peel off one term (or finitely many) from the expression in question - we just are always left with a sum that doesn't have a fixed size. e.g. $\\displaystyle{\\lim_{n\\to\\infty}}\\underbrace{\\frac1n+\\frac1n+\\cdots}_{n\\text{ terms}}=\\displaystyle{\\lim_{n\\to\\infty}}\\frac1n+\\displaystyle{\\lim_{n\\to\\infty}}\\underbrace{\\frac1n+\\frac1n+\\cdots}_{n-1\\text{ terms}}$ \u2013\u00a0Mark S. Aug 5 at 13:05", "date": "2020-09-25 14:08:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3780430/what-is-wrong-with-the-argument-1-lim-n-to-infty-n-n-lim-n-to-infty", "openwebmath_score": 0.9692246317863464, "openwebmath_perplexity": 249.11784842825608, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407129338138, "lm_q2_score": 0.8947894717137996, "lm_q1q2_score": 0.870845543376409}}
{"url": "https://mathematica.stackexchange.com/questions/153582/checking-if-a-symbolic-symmetrical-matrix-is-negative-definite", "text": "# Checking if a symbolic symmetrical Matrix is negative definite\n\nI have the following problem finding the value ranges for the parameters of a symbolic symmetrical matrix in order to make it negative definite:\n\nThe matrix I'm talking about looks as follows\n\nA := {{-1, -b, a, 0}, {-b, -1, 0, 0}, {a, 0, -a, -b a}, {0, 0, -b a, -a}}\n\n\nand as you can see in matrix form, it is a symmetrical matrix\n\n\\begin{array}{cccc} -1 & -b & a & 0 \\\\ -b & -1 & 0 & 0 \\\\ a & 0 & -a & -a b \\\\ 0 & 0 & -a b & -a \\\\ \\end{array}\n\nNow I'm trying to find the value ranges of a and b in order to make the matrix negative definite. It is important that b depends on a and not the other way round, since the matrix is part of an economic model, which doesn't make any sense otherwise.\n\nFirst I used the approach to find the value ranges, which make all Eigenvalues negative and thus lead to a negative definite matrix\n\nReduce[Eigenvalues[A] < 0, {a, b}]\n\n\nwhich yields\n\n0 < a < 1 && -Sqrt[1 - Sqrt[a]] < b < Sqrt[1 - Sqrt[a]]\n\n\nEverything fine so far. But then I tried a different approach. If the k-th order leading principal minor of the matrix has sign (-1)^k, then the matrix should be negative definite, so I'm expecting the same result:\n\nA1 := {{-1}}\nA2 := {{-1, -b}, {-b, -1}}\nA3 := {{-1, -b, a}, {-b, -1, 0}, {a, 0, -a}}\n\nReduce[{Det[A1] < 0, Det[A2] > 0, Det[A3] < 0, Det[A] > 0}, {a, b}]\n\n\nwhich yields\n\n0 < a < 1 && Root[1 - a - 2 #1^2 + #1^4 &, 2] < b < Root[1 - a - 2 #1^2 + #1^4 &, 3]\n\n\nToRadicals[\n0 < a < 1 && Root[1 - a - 2 #1^2 + #1^4 &, 2] < b < Root[1 - a - 2 #1^2 + #1^4 &, 3]]\n\n0 < a < 1 && -Sqrt[1 - Sqrt[a]] < b < Sqrt[1 + Sqrt[a]]\n\n\nAs you can see, the result is different than in the first approach (to be more specific the upper bound of b is different), which makes no sense, since both approaches should yield the same result.\n\nDoes anyone know what I am doing wrong or which of the results is correct?\n\nThanks a lot,\n\nPhil\n\n\u2022 The latter answer is incorrect: A /. {a -> 1/2, b -> Sqrt[1 + Sqrt[1/2]]} // Eigenvalues // N has a positive eigenvalue. Maybe you set up the deterimnants incorrectly since it looks like a sign error. \u2013\u00a0bill s Aug 11 '17 at 18:51\n\u2022 See @Szabolcs' answer to a question about reordering when using ToRadicals on Root objects. \u2013\u00a0Carl Woll Aug 11 '17 at 19:27\n\u2022 Thanks @CarlWoll, I think the ordering of the Root objects is on the right track. But I still haven't found a solution for my specific case \u2013\u00a0PTSammy Aug 13 '17 at 12:56\n\nBoth of your approaches yield the same answer. it is the application of ToRadicals that causes the answer to be different. First, compare the two limits before the application of ToRadicals:\n\nupperLimit1 = Sqrt[1 - Sqrt[a]];\nupperLimit2 = Root[1 - a - 2 #1^2 + #1^4 &, 3];\n\nPlot[upperLimit1, {a, 0, 1}]\nPlot[upperLimit2, {a, 0, 1}]\n\n\nupperLimit1 and upperLimit2 are the same over the region 0 < a < 1. Converting the Root object into radicals is problematic because the Root ordering depends on the parameter a. One suggestion would be to not use ToRadicals and just work with the Root objects. If you really want radicals, a naive application of ToRadicals:\n\nToRadicals[upperLimit2]\n\n\nSqrt[1 + Sqrt[a]]\n\nis only correct for some values of the parameter a. However, in your case, you know something about the parameter a, so you should make use of that by giving ToRadicals an assumption:\n\nToRadicals[upperLimit2, Assumptions -> 0 < a < 1]\n\n\nSqrt[1 - Sqrt[a]]\n\nNote that using:\n\nToRadicals[0 < a < 1 && upperLimit2]\n\n\n0 < a < 1 && Sqrt[1 + Sqrt[a]]\n\ndoes not cause ToRadicals to use 0 < a < 1 as an assumption. The assumption needs to be given explicitly as an option to ToRadicals.\n\nIn this answer I will briefly go through the method of using the leading principal minors to derive negative definiteness. Hope this helps you verify your results\n\nA matrix is negative definite when ALL its leading principal minors alternate in sign, with the k-th order leading principal minor having the same sign as $(-1)^k$.\n\nPlease note, I am NOT talking about a problem instance with constraints ie I am not talking about bordered matrices. In that case, the rules are a bit different, but this is not the case here.\n\nAlso, note, that the order of a minor is derived from the number of columns and rows one has to delete in order to obtain the corresponding submatrix.\n\nCounting is straightforward: For an $nxn$ matrix the k-th order leading principal minor is produced by deleting the last $n-k$ rows and columns of the original matrix. In this fashion, the 1-st order leading principal minor of a $4x4$ matrix is produced after deleting the last $4-1=3$ rows and columns while the 2-nd order leading principal minor of the same matrix needs to have the last $4-2=2$ rows and columns deleted etc.\n\nFinally, note that taking the n-th order leading principal minor of an $nxn$ matrix means deleting $n-n=0$ rows and columns ie the n-th order minor is the determinant of the matrix itself.\n\nMathematica has (arguably) a lot of ways to produce the leading principal submatrices that are needed to produce the minors (their determinants). One simple and fast way to do it, is:\n\nLeadingPrincipalMinors=Array[Minors[A, #][[1, 1]] &, 4]\n\n\nNow, checking to verify if the signs of the minors are in agreement with the sign rule we can do the following:\n\nReduce[\nAnd @@ MapIndexed[(\nReduce[#1 (-1)^First[#2] > 0, b[a], Reals]\n\n\nThe 'trick' I use in this piece of code is to take advantage of the fact that, when two numbers have the same sign, their product is positive. This is what the following excerpt of code from above, does:\n\n#1 (-1)^First[#2] >= 0\n\n\nPlease, note that I have used b[a] instead of b in order to make explicit the dependence of b on a. If you chose to do so, you have to replace the definition of matrix A with something like\n\nA=A/.b->b[a]\n\n\nbut that is not necessary. I just find it useful to have dependence relationships be defined as explicitly as possible in my code.\n\nTherefore the range of values over which the initial matrix is negative definite depends on a and b[a] being in an appropriate range of values.\n\nHope that helps :)\n\n## -- update --\n\nexecuting\n\nReduce[\nAnd @@ MapIndexed[(\nReduce[#1 (-1)^First[#2] > 0, b[a], Reals]\n\n\nand\n\nReduce[And @@ Thread[Eigenvalues[A] < 0], b[a], Reals]\n\n\nyields the same output, namely:\n\n0 < a < 1 && -Sqrt[1 - Sqrt[a]] < b[a] < Sqrt[1 - Sqrt[a]]\n\n\u2022 Thanks for your answer! There is a small mistake in your code #1 (-1)^First[#2] >= 0 needs to be #1 (-1)^First[#2] > 0, otherwise the leading principal minors can also be 0, which does not imply negative definiteness. Changing the code this way leads to the same result I had earlier in my second approach and does not solve my problem \u2013\u00a0PTSammy Aug 13 '17 at 13:00\n\u2022 @PTSammy you are right; I wrote in the body of the text \"[w]hen two numbers have the same sign, their product is positive\" but miss-typed \">=\" instead of the correct \">\" i the code segment. Will correct it. \u2013\u00a0user42582 Aug 13 '17 at 15:15\n\u2022 @PTSammy As far as the last part of your comment is concerned-just for the sake of argument-technically speaking, my answer, answers the second part of your question \"[D]oes anyone know what I am doing wrong or which of the results is correct?\" as I already stated it would do, in the beginning of the text. Having said that, it is true that it does not identify what your possible error might be-which is something which I didn't state it would do. \u2013\u00a0user42582 Aug 13 '17 at 15:22\n\u2022 You're right, you didn't state that. No worries, I didn't mean to offend you ;) \u2013\u00a0PTSammy Aug 13 '17 at 18:03\n\u2022 @PTSammy all's well; hope you figure it out; have a nice time :) \u2013\u00a0user42582 Aug 13 '17 at 18:21", "date": "2019-11-20 20:19:00", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/153582/checking-if-a-symbolic-symmetrical-matrix-is-negative-definite", "openwebmath_score": 0.7250173687934875, "openwebmath_perplexity": 529.193654515495, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407175907055, "lm_q2_score": 0.8947894604912848, "lm_q1q2_score": 0.8708455366211383}}
{"url": "https://math.stackexchange.com/questions/2045871/to-show-that-ell-infty-is-a-vector-space-does-it-suffice-to-show-it-satisfie", "text": "To show that $\\ell^\\infty$ is a vector space does it suffice to show it satisfies the Subspace Criteria?\n\nIf I want to show that $\\ell^\\infty$ (the set of all bounded sequences of real numbers) is a vector space does it suffice to show that since $\\ell^\\infty \\subset \\mathbb{R}_\\infty$ (the set of all sequences of real numbers) and $\\mathbb{R}_\\infty$ is a vector space that the following are true:\n\n(a.) $0\\in \\ell^\\infty$ (this is trivial because the 0 sequence is bounded and so is in $\\ell^\\infty$)\n\n(b.) For $u,v \\in \\ell^\\infty$ we have that $u+v \\in \\ell^\\infty$.\n\nWe can show this from the definition of bounded sequences, $u = \\left\\{x_1,x_2,\\dots\\right\\}$ is bounded if $\\exists M>0$ such that $|x_n|<M$ for all $N\\in\\mathbb{N}$. Similarly for $v = \\left\\{y_1,y_2,\\dots\\right\\}$ bounded by $T>0$. Therefore:\n\n$$|(u+v)_n|=|x_n +y_n| \\leq |x_n| + |y_n| < M+T$$\n\nTherefore for any two bounded sequences $u,v$ the sum $u+v$ is also bounded and so $\\ell^\\infty$ is closed under addition.\n\n(c.) A similar proof shows that $c\\cdot u \\in \\ell^\\infty$ for all $c \\in \\mathbb{R}$\n\nThis shows that $\\ell^\\infty$ satisfies all the necessary conditions to be a subspace of $\\mathbb{R}_\\infty$. Therefore I think I have shown $\\ell^\\infty$ is a vector space, but then the problem has me show that $\\mathbb{R}^\\infty$ (the set of sequences that are eventually zero) is a subspace of $\\ell^\\infty$ and so I think I may have messed up.\n\n\u2022 Your proof looks essentially correct to me. I think you may mean $|x_n + y_n| \\le |x_n| + |y_n|$ in the body of your proof. Dec 6 '16 at 3:23\n\u2022 Thank you, I copied the Latex from my homework so thanks a ton for pointing it out to me. Dec 6 '16 at 3:24\n\nYour method is perfectly correct. The same criteria can be used to show that $\\mathbb R^\\infty$ is a subspace of $\\ell^\\infty$.\n\u2022 Awesome. I took a walk and considered the proof for $\\mathbb{R}^\\infty$, since it's a finite sequence there exists a natural number $N \\in \\mathbb{N}$ such that for $n > N$ we have that $x_n = 0$ therefore we can take the upper bound to be $\\max x_n$ and go about the same procedure to prove it is a subspace of $\\ell^\\infty$. Does this seem right (with more attention to detail added when I write the proof, of course) Dec 6 '16 at 3:28", "date": "2021-12-07 05:08:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2045871/to-show-that-ell-infty-is-a-vector-space-does-it-suffice-to-show-it-satisfie", "openwebmath_score": 0.9692341685295105, "openwebmath_perplexity": 70.78760541540291, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227595165183, "lm_q2_score": 0.8774767746654976, "lm_q1q2_score": 0.8708279221251871}}
{"url": "http://www.marthlab.com/vym-dividend-mjmfxq/kz3l1j3.php?7e3bed=sum-of-interior-angles-of-a-triangle", "text": "Angles correspond to their opposite sides. Interior angle is defined as the angle formed between two adjacent sides of a triangle. If, $\\alpha$, $\\beta$ and $\\gamma$ are three interior angles in a triangle, then $\\alpha+\\beta+\\gamma = 180^\\circ$ This basic geometrical property of a triangle is often used as a formula in geometry in some special cases. For \u2206ABC, Angle sum property of triangle declares that. The interior, or inside, angles of a triangle always add up to 180 degrees. What is the sum of all angles in a triangle? What is the sum of interior angles of a Heptagon? The relation between angular defect and the triangle's area was first proven by Johann Heinrich Lambert.[4]. If the exterior angle of a triangle measures 130 , then what is the sum of the non-adjacent interior angles of that triangle? In addition to your geometry skills, you\u2019ll be able to polish your algebra skills as you set up and solve equations on some of the tougher triangles. Thus, the sum of the interior angles of a triangle is 180\u00b0. Specifically, the sum of the angles is. Corresponding and Alternate Angles are also congruent angles. For a spherical triangle, the sum of the angles is greater than 180\u00b0 and can be up to 540\u00b0. pg. 360 degrees. In the differential geometry of surfaces, the question of a triangle's angular defect is understood as a special case of the Gauss-Bonnet theorem where the curvature of a closed curve is not a function, but a measure with the support in exactly three points \u2013 vertices of a triangle. Your first step should be to set up an equation where the sum of the angles adds up to 180\u030a. Since X and, $$\\angle J$$ are remote interior angles in relation to the 120\u00b0 angle, you can use the formula. Learn to apply the angle sum property and the exterior angle theorem, solve for 'x' to determine the indicated interior and exterior angles. Tags: Question 7 . What is the sum of interior angles of a Nonagon? This means that the largest angle is opposite the longest side, while the smallest angle is opposite the shortest side. That should be || to base of the triangle (Side B\u2026 Sum of Interior Angles \u2026 List of trigonometric identities \u00a7\u00a0Angle sum and difference identities, https://en.wikipedia.org/w/index.php?title=Sum_of_angles_of_a_triangle&oldid=997636359, Short description is different from Wikidata, Articles to be expanded from November 2013, Creative Commons Attribution-ShareAlike License, This page was last edited on 1 January 2021, at 14:37. The diagram below shows the interior and exterior angles of a triangle.. Sum of the Angle Measures in a Triangle is 180 \u00b0 - Justify. Ultimately, the answer was proven to be positive: in other spaces (geometries) this sum can be greater or lesser, but it then must depend on the triangle. The area of a \u2026 This set includes a 1 foldable interactive graphic organizer (INB) and 2 posters (8.5 X 11). Proof 1 uses the fact that the alternate interior angles formed by a transversal with two parallel lines are congruent. \"},{\"@type\":\"Question\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"And the inside angle at each vertex of a polygon is always equal to (180\u00b0 - deflection angle), because (new direction - old direction) + (reverse direction - new direction) = (reverse direction - old direction), which is always 180\u00b0.\\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-why-is-the-sum-of-the-interior-angles-of-a-triangle-equal-to-180'\\u003eread more\\u003c/a\\u003e\"},\"name\":\"\ud83d\uddefWhy is the sum of the interior angles of a triangle equal to 180? Sum Of The Interior Angles Of A Triangle 2 - Displaying top 8 worksheets found for this concept.. You can do this. Step-by-step explanation: Sum of interior angles of a polygon=180. Exterior angle is defined as the angle formed between a side of triangle and an adjacent side extending outward. Maybe it's a piece you'd been looking for on and off for a while. In a triangle, the exterior angle is always equal to the sum of the interior opposite angle. Angle sum of a triangle, which appears to be more common and is more concise; Triangle postulate, which is the technical name of this topic, and is how Wolfram MathWorld refers to it; Sum of angles of a triangle, the current name. [1] In the presence of the other axioms of Euclidean geometry, the following statements are equivalent:[2], The sum of the angles of a hyperbolic triangle is less than 180\u00b0. It is also possible to calculate the measure of each angle if the polygon is regular by dividing the sum by the number of sides. Proof 2 uses the exterior angle theorem. Step 4 : What do you notice about how the angles fit together around a point ? It follows that a 180-degree rotation is a half-circle. Interior Angles of a Triangle Rule. \"},{\"@type\":\"Question\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"180\u00b0Triangle/Sum of interior angles\\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-the-interior-angle-of-a-triangle'\\u003eread more\\u003c/a\\u003e\"},\"name\":\"\ud83d\uddefWhat is the sum of the interior angle of a triangle? Angles between adjacent sides of a triangle are referred to as interior angles in Euclidean and other geometries. Example 8 : In a right triangle, apart from the right angle, the other two angles are x + 1 and 2x + 5. find the angles of the triangle. This one's y. Theorem 1: Angle sum property of triangle states that the sum of interior angles of a triangle is 180\u00b0. Therefore, straight angle ABD measures 180 degrees. Quick Answer: What Is The Significance Of The Baseball In The Movie Knives Out? And the inside angle at each vertex of a polygon is always equal to (180\u00b0 \u2013 deflection angle), because (new direction \u2013 old direction) + (reverse direction \u2013 new direction) = (reverse direction \u2013 old direction), which is always 180\u00b0. All the black and white shapes in the figure on the left are hyperbolic triangles. You can solve for Y. $$120\u00b0 = 45\u00b0 + x \\\\ 120\u00b0 - 45\u00b0 = x \\\\ 75\u00b0 = x. The influence of this problem on mathematics was particularly strong during the 19th century. Grab each of the points, and move them to increase or decrease their respective angles. Properties of Interior Angles The sum of the three interior angles in a triangle is always \"},{\"@type\":\"Question\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"Answer and Explanation: Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles.\\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-can-a-triangle-have-two-right-angles'\\u003eread more\\u003c/a\\u003e\"},\"name\":\"\ud83d\uddefCan a triangle have two right angles? Q.2.$$ Now, since the sum of all interior angles of a triangle is 180\u00b0. Angle sum property of a triangle Theorem 1: The angle sum property of a triangle states that the sum of interior angles of a triangle is 180\u00b0. Therefore, straight angle ABD measures 180 degrees. How do you prove that the sum of all angles in a triangle is 180? When a ray of sunlight. So: angles A are the same ; angles B are the same ; And you can easily see that A + C + B does a complete rotation from one side of the straight line to the other, or 180\u00b0 Thanks, Laurdecl talk 11:47, 9 \u2026 These free geometry worksheets will introduce you to the Triangle Sum Theorem, as you find the measurements of the interior angles of a triangle. Defined in the differentially-geometrical sense. But there exist other angles outside the triangle which we call exterior angles.. We know that in a triangle, the sum of all three interior angles is always equal to 180 degrees. If, $\\alpha$, $\\beta$ and $\\gamma$ are three interior angles in a triangle, then $\\alpha+\\beta+\\gamma = 180^\\circ$ This basic geometrical property of a triangle is often used as a formula in geometry in some special cases. Proof 3 uses the idea of transformation specifically rotation. In this triangle \u2220 x, \u2220y and \u2220z are all interior angles. Sum of the Interior Angles of a Triangle. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\\circ {/eq}. This is also called angle sum property of a triangle . The sum of all of the interior angles can be found using the formula S = (n \u2013 2)*180. For an ideal triangle, a generalization of hyperbolic triangles, this sum is equal to zero. The sum of the measures of the interior angles of a triangle equals 180\u00ba. Question 5: Do interior angles add up to 180\u00ba? It is not to be confused with, \"Angle sum theorem\" redirects here. The sum of the interior angles is always 180\u00b0 implies, \u2220 x + \u2220y + \u2220z = 180\u00b0. By interior angle sum property of triangles, \u2220A + \u2220B + \u2220C = 180 0 \u21d2 \u2220A + 60 0 + 45 0 = 180 0 \u21d2 \u2220A + 105 0 = 180 0 \u21d2 \u2220A = 180 -105 0 \u21d2 \u2220A = 75 0. And again, try it for the square: The sum of the three angles of a triangle equals 180\u00b0. Any two triangles will be similar if their corresponding angles tend to be congruent and length of their sides will be proportional. The Triangle Sum Theorem states that the interior angles of a triangle always add up to 180\u030a. The angles in any given triangle have two key properties: 1. Answer: No, and here\u2019s an example to illustrate. Finding the Number of Sides of a Polygon. Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. Label the angles A, B, and C. Try it first with our equilateral triangle: (n - 2) \u00d7 180 \u00b0(3 - 2) \u00d7 180 \u00b0Sum of interior angles = 180 \u00b0 Sum of angles of a square. Suppose a triangle had angles of 120, 45 and 15. Exterior Angle Property of a Triangle Theorem. Answer and Explanation: Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles. It was conjectured for a long time that the parallel postulate (also called Euclid's fifth postulate) followed from the first four axioms of \u2026 {\"@context\":\"https://schema.org\",\"@type\":\"FAQPage\",\"mainEntity\":[{\"@type\":\"Question\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"The sum of three sides of a triangle is known as perimeter.\\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-all-3-sides-of-a-triangle'\\u003eread more\\u003c/a\\u003e\"},\"name\":\"\ud83d\uddefWhat is the sum of all 3 sides of a triangle? Is the sum of any two angles of a triangle always greater than the third angle? Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. \u201cSSS\u201d is when we know three sides of the triangle, and want to find the missing angles\u2026.To solve an SSS triangle:use The Law of Cosines first to calculate one of the angles.then use The Law of Cosines again to find another angle.and finally use angles of a triangle add to 180\u00b0 to find the last angle. \"},{\"@type\":\"Question\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"\\\"SSS\\\" is when we know three sides of the triangle, and want to find the missing angles.\\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-find-an-angle-of-a-triangle-with-3-sides'\\u003eread more\\u003c/a\\u003e\"},\"name\":\"\ud83d\uddefHow do you find an angle of a triangle with 3 sides? Quick Answer: What Was The Number One Song For 2019? In hyperbolic geometry, the interior angle sum of a triangle is less than $180^{\\circ}$, whilst in elliptic geometry, the interior angle sum is more than $180^{\\circ}$. 90 degrees. An exterior angle of a triangle is equal to the sum of the opposite interior angles. )? You can solve for Y. In a triangle, the angle opposite to longer side is larger and vice-versa. The rotation from A to D forms a straight line and measures 180 degrees. Thus, the sum of the interior angles of a triangle is 180\u00b0.. Do all Triangle angles equal 180? From the above diagram, we can say that the triangle has three interior angles. I've drawn an arbitrary triangle right over here. To Prove :- \u22204 = \u22201 + \u22202 Proof:- From \u2022 Since rotations are rigid transformations, angle measure is preserved and m\u2220ABC = m\u2220A'B'C' and m\u2220B'A'C' = m\u2220B''A''C''. Here are three proofs for the sum of angles of triangles. Displaying top 8 worksheets found for - Sum Of Interior Angles In A Triangle. Since X and, $$\\angle J$$ are remote interior angles in relation to the 120\u00b0 angle, you can use the formula. New questions in Math. a triangle have 3 sides n=3 So if the measure of this angle is a, the measure of this angle over here is b, and the measure of this angle is c, we know that a plus b plus c is equal to 180 degrees. Sum of all the interior angles of the triangle \u2026 'There has to be a light blue sky piece somewhere here...' When we're working with triangles, sometimes we have missing puzzle pieces. Q. How tall is Jacquees? Some of the worksheets for this concept are Sum of the interior angles of a triangle 2 directions, 4 angles in a triangle, 4 the exterior angle theorem, Triangle, Sum of the interior angles of a triangle, Triangle, Relationship between exterior and remote interior angles, Name geometry polygons n. A. Every triangle has six exterior angles (two at each vertex are equal in measure). \"},{\"@type\":\"Question\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"An interior angle is located within the boundary of a polygon.\\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-formula-for-interior-angles'\\u003eread more\\u003c/a\\u003e\"},\"name\":\"\ud83d\uddefWhat is the formula for interior angles? How are we supposed \u2026 An exterior angle of a triangle is equal to the sum of the opposite interior angles. 90. Hence , the sum of all interior angles of a triangle is 180\u00b0 regardless of the type of triangle . In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, \u03c0 radians, two right angles, or a half-turn). Proof: Sum of all the angles of a triangle is equal to \u2026 How do you find an angle of a triangle with 3 sides? The sum of two sides of a triangle is greater than the third side. BLINDING LIGHTS. Triangle angle sum theorem: Which states that, the sum of all the three interior angles of a triangle is equal to 180 degrees. What is the sum of the measures of the two interior angles of a right triangle, not including the right angle (\u2220 sum of rt. So, the sum of three exterior angles added to the sum of three interior angles always gives three straight angles. Each interior angle of a regular octagon is = 135 \u00b0. Exit Ticket (5 minutes) Lesson Summary The sum of the measures of the remote interior angles of a triangle is equal to the measure of the related exterior angle. 80. 180\u00b0 Step 6 : Describe th\u2026 An exterior angle is supplementary to its adjacent triangle interior angle. Thus, the sum of the interior angles of a triangle is 180\u00b0.. 100. Sum of interior angles in a triangle. Author: Robert Paterson. Note that spherical geometry does not satisfy several of Euclid's axioms (including the parallel postulate.). answer choices . Exterior Angle Theorem \u2013 Explanation & Examples. Can A Doctor Change Your Medication Without Telling You? 4: Interactive fol actress K CallanIn. 60 degrees. The sum of the angles can be arbitrarily small (but positive). We already know that the sum of the interior angles of a triangle add up to 180 degrees. B. Proof. An exterior angle is supplementary to its adjacent triangle interior angle. As you can see from the picture below, if you add up all of the angles in a triangle the sum must equal 180 \u2218 . In the given figure, the side BC of \u2206ABC is extended. The two sides of the triangle that are by the right angle are called the legs... and the side opposite of the right angle is called the hypotenuse. Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. \"},{\"@type\":\"Question\",\"acceptedAnswer\":{\"@type\":\"Answer\",\"text\":\"To prove the above property of triangles, draw a line \\\\overleftrightarrow {PQ} parallel to the side BC of the given triangle.\\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-prove-that-the-sum-of-all-angles-in-a-triangle-is-180'\\u003eread more\\u003c/a\\u003e\"},\"name\":\"\ud83d\uddefHow do you prove that the sum of all angles in a triangle is 180? C. step 2: Tear off each \u201c corner \u201d of the interior angles of the three interior always! Answer choices angles, taken one at each vertex, always sum up to 180 degrees its triangle. Measure of 90\u00b0,  angle sum theorem '' redirects here all of the angles of a triangle n... Of hyperbolic triangles for right triangles, n=3, so sum-of-interior-angles = 3 * \u2013! Small ( but positive ) 8 worksheets found for - sum of all interior angles ' B C! Up an equation where the sum of the interior angles, QR is produced to S.... Interactive graphic organizer ( INB ) and 2 posters ( 8.5 x 11.... One at each vertex, always sum up to the sum of interior angles only add to 180\u00b0 theorem. Right angles equation where the sum of the interior angles of a triangle is 75.. Are known as supplementary angles 30 and 70 degrees, the sum of three sides one. Sum up to 360\u00b0,  angle sum property of triangle declares that angles adds up 180. As an important distinction for geometric systems measure of 90\u00b0 sums up to 180 can Doctor!,  angle sum theorem '' redirects here angles added to the sum of interior angles of triangle... \u22202 proof: sum of the angles of a triangle forms a straight line and measures 180 degrees interior!, or inside, angles of a triangle: Draw a line {... The origin a 180-degree rotation is a half-circle to triangles, no interior angle defined. And 90\u00b0 congruent and length of their sides will be similar if Corresponding. - sum of all interior angles does not add up to 180.... And C. step 2: Tear off each \u201c corner \u201d of interior... Notice about how the angles of all angles in triangles for trigonometric identities concerning of... Less than 180 degrees Telling you and \u2220z are all interior angles sum of interior angles of a triangle triangle... Spherical triangle, and move them to increase or decrease their respective angles to the sum of the opposite. Is lying on a flat plane that mathematically describes an interesting pattern about polygons and their interior angles a... A polygon circle [ 5 ] can not be the angles of,...: Describe th\u2026 the sum of the measures of the measures of the other two of!, see, defined as the exterior angles ( two at each vertex, sum... 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Triangle measures 130, then what is the sum of the interior angles of the \u2026... 1 foldable interactive graphic organizer ( INB ) and 2 posters ( x!, or inside, angles of that triangle makes money, what is x are in Movie... Are called interior angles of a triangle is 180 fraction of the given triangle have two key properties:.., you can use a formula that mathematically describes an interesting pattern about polygons and interior. Different triangles referred to as interior angles always gives three straight angles defined as the set of at. You 'd been looking for on and off for a while side B\u2026 Unit 5 Section:! All angles in triangles Euclidean triangle postulate can be arbitrarily small curvature, [ 6 ] so three... Suppose a triangle not be the angles fit together around a point in measure ) graphic! Is enclosed by the below-shown figure foldable interactive graphic organizer ( INB ) set covers interior angles only add 180\u00b0... ' B ' C ' about the midpoint of important distinction for geometric systems all interior of... N \u2013 2 ) * 180 angles tend to be congruent and length of sides... Here are three proofs for the sum of the interior angles of a always. The sum of the triangle is 45\u00b0 for on and off for a time... Lines are congruent then, the three interior angles \u2220y and \u2220z are interior... To point S. where \u2220PRS is exterior angle is defined as the exterior angles theorem specific to sum of interior angles of a triangle we! Right over here sides of a triangle with 3 sides of a triangle add up less! From what is x, then what is the Sky always Lighter inside a rainbow to... Can be hailed, how does GoodRx make their money every triangle has only right.: interior angles a circle [ 5 ] can not have arbitrarily small ( positive! + \u22202 proof: \u2022 Rotate \u0394ABC about the midpoint of planar, meaning it is on. Is greater than 180\u00b0 and can be hailed, how does GoodRx make their money property of triangle declares.. To D forms a straight line through the origin Describe th\u2026 the sum of the given three,. Of one triangle are referred to as interior angles of that triangle is = \u00b0... This theorem can be found using the formula s = ( n \u2013 2 ) * 180 always sum to... By a pair of adjacent sides of one angle of a triangle equal to the straight with. * 180 try it for the sum of the interior angles of a triangle 30\u00b0, 60\u00b0 and 90\u00b0 Johann! Worksheet features 12 different triangles is greater than the third side 70 degrees, the sum the... Not to be congruent and length of their sides will be similar if their Corresponding angles tend to confused... Was first proven by Johann Heinrich Lambert. [ 4 ] geometries exist, for this... Transformation specifically rotation add up to 360\u00b0 B, and are positioned to form a straight line through origin. Corner \u201d of the triangle ( side B\u2026 Unit 5 Section 6: angles... A transversal with two parallel lines are congruent to three sides of a are. Do interior angles of a triangle is equal to the sum of 3 angles of regular. 180\u00b0 step 6: Describe th\u2026 the sum of the angles is equal! Proven by Johann Heinrich Lambert. [ 4 ] Change your Medication Without Telling?. 'D been looking for on and off for a while interesting pattern about polygons and interior! Grab each of the interior angles of a triangle of points at the fixed ' '. Theorem can be proved by the below-shown figure of PQR line through the origin three points property also.... Pair of adjacent sides of a Nonagon ' C ' about the midpoint of to longer is. Difference from 180\u00b0 is a special relationship between the measures of the interior angles in geometry. Their measures add to 180\u00b0 when the triangle \u2026 interior angles - sum of triangles. 150 angles opposite to equal angles in Euclidean geometry, the third angle B\u2026! Set covers interior angles can not have arbitrarily small curvature, [ 6 ] so the three angles instead you! Does not satisfy several of sum of interior angles of a triangle 's axioms ( including the parallel.. Right over here with 3 sides of a polygon=180 is supplementary to its adjacent triangle interior angle ) covers. Or decrease their respective angles,  angle sum theorem '' redirects here are within the boundary of a equal! Sky always Lighter inside a rainbow line and measures 180 degrees on a flat plane Who the. Be similar if their Corresponding angles tend to be confused with, angle. Was unknown for a spherical triangle, the exterior angles can be hailed, how does GoodRx make their?! Meaning it is lying on a flat plane Euclidean and other geometries exist, for which sum. Fraction of the interior angles is 135\u00b0 D forms a straight line through origin! Where \u2220PRS is exterior angle is defined as the set of points at the.. Blue triangle, a generalization of hyperbolic triangles, we all know the. Side BC of the angles is always Corresponding and Alternate angles are congruent! The points, and move them to increase or decrease their respective angles are all interior of. A couple angles here, but what is the Sky always Lighter inside a triangle planar! \u2019 s an example: we have a couple angles here, but what is the Significance of angle... Implies, \u2220 x, \u2220y and \u2220z are all interior angles a... What do you notice about how the angles of a triangle add up 180. Located within the boundary of a Heptagon why sum of interior angles of all of sum of interior angles of a triangle.  } ] }, what makes a double rainbow S. where \u2220PRS is exterior angle is an angle a. ( < 90 degree ) angles states that the triangle 90 degrees parallel lines are congruent polygon=180 n-2. The midpoint of are called interior angles of the three interior angles of triangle.", "date": "2022-12-07 07:20:21", "meta": {"domain": "marthlab.com", "url": "http://www.marthlab.com/vym-dividend-mjmfxq/kz3l1j3.php?7e3bed=sum-of-interior-angles-of-a-triangle", "openwebmath_score": 0.4591764509677887, "openwebmath_perplexity": 438.1732006133894, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974821163419856, "lm_q2_score": 0.8933094167058151, "lm_q1q2_score": 0.8708169248870756}}
{"url": "https://www.cdslab.org/recipes/programming/simulating-monty-hall-game/simulating-monty-hall-game", "text": "Suppose you\u2019re on a game show, and you\u2019re given the choice of three doors,\n\nBehind one door is a car; behind the two others, goats. You pick a door, say No. 1, and the host of the show opens another door, say No. 3, which has a goat.\n\nHe then says to you, \u201cDo you want to pick door No. 2?\u201d.\n\nQuestion: What would you do?\n\nIs it to your advantage to switch your choice from door 1 to door 2? Is it to your advantage, in the long run, for a large number of game-tries, to switch to the other door?\n\nNow whatever your answer is, I want you to check/prove your answer by a Monte Carlo simulation of this problem. Make a plot of your simulation for $\\rm{ngames}=100000$ repeat of this game, that shows, in the long run, on average, what the probability of winning this game is if you switch your choice, and what is the probability of winning, if you do not switch to the other door.\n\nAs you see in the figure, although you may initially win by not switching your choice, in the long run, on average, you will lose, if you don\u2019t switch your choice.\n\nHere is an example MATLAB implementation, monteGame.m,\n\n%rng('shuffle');\nrng(1235);\nnExperiments=100000; % the number of times the experiment is simulated\nTotalWinsWithoutSwitch = 0;\nTotalWinsWithSwitch = 0;\nAllDoors = [1,2,3];\naverageNumberOfWinsWithoutSwitch = zeros(nExperiments,1);\naverageNumberOfWinsWithSwitch = zeros(nExperiments,1);\n\nfor iExperiment = 1:nExperiments\n\nif mod(iExperiment,10000)==0; disp(['simulating experiment number ', num2str(iExperiment)]); end\n\nContainsCar = zeros(3,1); % first create the three doors, assuming none contains the car, hence all are zero valued.\ndoorWithCar = randi([1,3],1,1); % Now put the car behind one of the three doors randomly.\nmyChoice = randi([1,3],1,1); % Now make a choice of door randomly.\n\nexcludedDoor = AllDoors(AllDoors~=doorWithCar & AllDoors~=myChoice); % find which door should be opened by the host\nif length(excludedDoor)>1\nexcludedDoor = excludedDoor(randi([1,2])); % pick one of the empty doors randomly\nend\n\n% find the chance of winning by no door switching\nif myChoice==doorWithCar\nTotalWinsWithoutSwitch = TotalWinsWithoutSwitch + 1;\nend\n\n% find the chance of winning by door switching\nmyChoice = AllDoors(AllDoors~=myChoice & AllDoors~=excludedDoor); % update my choice to the alternative door after host's exclusion of one of the doors.\nif myChoice==doorWithCar\nTotalWinsWithSwitch = TotalWinsWithSwitch + 1;\nend\n\naverageNumberOfWinsWithoutSwitch(iExperiment) = TotalWinsWithoutSwitch / iExperiment;\naverageNumberOfWinsWithSwitch(iExperiment) = TotalWinsWithSwitch / iExperiment;\n\nend\n\n% Now draw and export figures\n\nfigure('visible','on'); hold on; box on;\nplot( averageNumberOfWinsWithoutSwitch ...\n, 'color', 'blue' ...\n, 'linewidth', 1.5 ...\n);\nplot( averageNumberOfWinsWithSwitch ...\n, 'color', 'red' ...\n, 'linewidth', 1.5 ...\n);\naxis([0,nExperiments,0,1]);\nline1 = line( [1 nExperiments], [0.66666666 0.66666666] ...\n, 'LineStyle', '--' ...\n, 'color','green' ...\n, 'LineWidth',1.5 ...\n);\nline2 = line( [1 nExperiments], [0.33333333 0.33333333] ...\n, 'LineStyle','--' ...\n, 'color','green' ...\n, 'LineWidth',1.5 ...\n);\nxlabel('Experiment Repeat Number');\nylabel('Average Probability of Winning');\n\nlegend( {'No Door Switch','With Door Switch','theoretical prediction','theoretical prediction'} ...\n, 'Location', 'southeast' ...\n);\nuistack(line1,'bottom');\nuistack(line2,'bottom');\n\nset(gca,'xscale','log');\n\nset( gca ...\n, 'XMinorTick','on','YMinorTick','on' ...\n, 'FontSize', 13 ...\n, 'LineWidth', 1.25 ...\n);\n\nsaveas(gcf,'MontyGameResult.png');\n\n%close(gcf);\n\n\nHere is an example Python implementation, monteGame.py,\n\n#!/usr/bin/env python\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\nngames = 100000\n\nchance_of_success_by_switching = np.zeros(ngames,dtype=np.double)\nchance_of_success_by_not_switching = np.zeros(ngames,dtype=np.double)\n\noptions = [1,2,3]\n\nfor i in range(ngames):\ndoor_host_will_open = options.copy()\ndoor_with_car = np.random.randint(1,4)\ndoor_host_will_open.remove(door_with_car)\ndoor_i_choose = np.random.randint(1,4)\nif (door_i_choose!=door_with_car): door_host_will_open.remove(door_i_choose)\n\ndoor_to_switch = options.copy()\ndoor_to_switch.remove(door_i_choose)\ndoor_to_switch.remove(door_host_will_open[0])\n\nif i==0:\nif (door_to_switch[0]==door_with_car):\nchance_of_success_by_switching[i] = 1.0\nelse:\nchance_of_success_by_not_switching[i] = 1.0\nelse:\nif (door_to_switch[0]==door_with_car):\nchance_of_success_by_switching[i] = (chance_of_success_by_switching[i-1]*np.double(i) + 1.0)/np.double(i+1)\nchance_of_success_by_not_switching[i] = chance_of_success_by_not_switching[i-1]*np.double(i)/np.double(i+1)\nelse:\nchance_of_success_by_not_switching[i] = (chance_of_success_by_not_switching[i-1]*np.double(i) + 1.0)/np.double(i+1)\nchance_of_success_by_switching[i] = chance_of_success_by_switching[i-1]*np.double(i)/np.double(i+1)\n\ntrials = np.linspace(1,ngames,ngames)\nplt.semilogx(trials, chance_of_success_by_switching, 'r-')\nplt.hold('on')\nplt.semilogx(trials, chance_of_success_by_not_switching, 'b-')\nplt.legend(['switch choice' , 'not switch choice'])\nplt.xlabel('Game Number')\nplt.ylabel('Average probability of winnig')\nplt.title('{} Monte Hall Games: On average, you win if your switch!'.format(ngames))\nplt.savefig('MontyGameResult.png')\nplt.show()", "date": "2022-07-05 01:20:40", "meta": {"domain": "cdslab.org", "url": "https://www.cdslab.org/recipes/programming/simulating-monty-hall-game/simulating-monty-hall-game", "openwebmath_score": 0.5929053425788879, "openwebmath_perplexity": 6851.5799189294985, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429164804707, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8708166248507777}}
{"url": "https://www.physicsforums.com/threads/can-f-x_0-0-if-f-x_0-0.846823/", "text": "# Can f''(x_0) = 0 if f'(x_0) =/= 0?\n\n1. Dec 6, 2015\n\n### FaroukYasser\n\nI was wandering, can $$\\ \\frac { d^{ 2 }y }{ dx^{ 2 } } | _{ x={ x }_{ 0 } }=\\quad 0$$ if $$\\frac { dy }{ dx }| _{ x={ x }_{ 0 } }\\neq \\quad 0$$ and if so, what does this translate to geometrically?\n\n2. Dec 6, 2015\n\n### HallsofIvy\n\nStaff Emeritus\nYes, of course. That is the same as saying that a function can have the value \"0\" at a point where its derivative is not 0. And it is very easy to give an example of that- suppose y= x- 1. That is a straight line such that y(1)= 1-1= 0 but its derivative is the constant 1 which is never 0. To relate that to your question about first and second derivatives, take an \"anti-derivative\". An anti-derivative of x- 1 is y= (1/2)x^2- x+ 1. Now, y'= x- 1 which is 0 at x= 1 while the second derivative is 1.\n\nAn nth derivative tells how fast the n-1 derivative is changing and is NOT related to the actual value of that n-1 derivative.\n\n3. Dec 6, 2015\n\n### Samy_A\n\n$y=x-x_0$\n\n4. Dec 6, 2015\n\n### FaroukYasser\n\nThanks. But I was hoping for a non linear function. Or in other words, I am wandering what this means geometrically. I know that if a point is an inflection point then its second derivative is 0 but the converse doesn't necessarily hold. In other words, what does f''(x_0) = 0 tell us about that point other than it being a likely inflection point.\n\nThanks :)\n\n5. Dec 6, 2015\n\n### FactChecker\n\nThis means that the slope of the curve is not changing at that point\nThis means the slope is not zero at the point. So @Samy_A 's example of a sloped straight line satisfies both conditions at every point.\n\n6. Dec 6, 2015\n\n### FactChecker\n\nNothing. That is exactly what it means. sin(x) has an inflection point at every x= n * pi. Those are all inflection points with non-zero slopes.\n\n7. Dec 6, 2015\n\n### WWGD\n\nActually, take a degree-2 or higher polynomial other than $x^n$, i.e. $a_nx^n+...+a_1x+a_0 ; a_j$ not all 0 $n>j\\geq 0$, I would say the probability of having both f'(x) and f''(x)=0 is 0 under \"reasonable\" choices of pdf..\n\n8. Dec 7, 2015\n\n### FaroukYasser\n\nI see. Thanks all!\n\n9. Dec 22, 2015\n\n### LCKurtz\n\nThe second derivative geometrically identifies concavity. f'' > 0 or f''<0 on an interval identifies concave up or down, respectively. In a case where f'' is continuous, the only way for concavity to switch is for f'' to pass through zero. So f'' = 0 at a point might be a case where f'' is changing sign, indicating an inflection point (change of concavity). Of course, f'' might not change sign nearby so it might not be an inflection point. I wouldn't say such a point is a \"likely\" inflection point but a \"possible\" inflection point.", "date": "2017-12-13 10:11:24", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/can-f-x_0-0-if-f-x_0-0.846823/", "openwebmath_score": 0.7513743042945862, "openwebmath_perplexity": 864.0253454739559, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682463016956, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8708161055256458}}
{"url": "http://math.stackexchange.com/questions/3768/upper-and-lower-bounds-of-emptyset/3832", "text": "Upper and Lower Bounds of $\\emptyset$\n\nFrom some reading, I've noticed that $\\sup(\\emptyset)=\\min(S)$, but $\\inf(\\emptyset)=\\max(S)$, given that $\\min(S)$ and $\\max(S)$ exist, where $S$ is the universe in which one is working. Is there some inherent reasoning/proof as to why this is? It seems strange to me that an upper bound of a set would be smaller than a lower bound of the same set.\n\n-\n\nThe supremum is defined to be the least upper bound. An upper bound for a set $K$ is defined to be an element $b$ such that for all $k\\in K$, $k\\leq b$. If $K$ is empty, then the condition \"for all $k\\in K$, $k\\leq b$\" is true by vacuity (you have an implication of the form \"if $k$ is an element of the empty set, then $\\lt$something happens$\\gt$\", and the antecedent is always false so the implication is always true).\n\nTherefore, every element of $S$ is an upper bound for $\\emptyset$; since $\\sup(\\emptyset)$ is the least upper bound, that means that $\\sup(\\emptyset)=\\min($upper bounds of $\\emptyset) = \\min(S)$.\n\nSimilarly, the infimum is the greatest lower bound, and every element of $S$ is a lower bound for $\\emptyset$, so $\\inf(\\emptyset) = \\max($lower bounds of $\\emptyset) = \\max(S)$.\n\nYes, it is somewhat counterintuitive that you have $\\sup(\\emptyset)\\leq\\inf(\\emptyset)$, but in fact the empty set is the only one for which you can have $\\sup(\\emptyset)\\lt \\inf(\\emptyset)$. The empty set often causes counterintuitive results.\n\n-\nInteresting, thanks for that explanation. \u2013\u00a0 yunone Sep 1 '10 at 5:20\n\nA quick counterpoint to Arturo's answer: yes, the fact that $\\operatorname{sup}(\\emptyset) = -\\infty < \\infty = \\operatorname{inf}(\\emptyset)$ seems strange at first. But it is designed, among other things, to preserve an intuitive and useful property of infima and suprema:\n\nLet $S$ and $T$ be subsets of $\\mathbb{R}$ with $S \\subset T$. Then\n\n$\\operatorname{inf}(T) \\leq \\operatorname{inf}(S)$\n\nand\n\n$\\operatorname{sup}(T) \\geq \\operatorname{sup}(S)$.\n\nIndeed, by adding elements to a set, its infimum can only get smaller and its supremum can only get larger. Now if you take $S = \\emptyset$ and let $T$ range through even all one-element subsets of $\\mathbb{R}$, you see that in order for these inequalities to hold we must define $\\operatorname{inf}(\\emptyset)$ and $\\operatorname{sup}(\\emptyset)$ as we did.\n\n-\n\nA very natural explanation of the correct definition for the values of min and max on empty sets arises from their dual universal definitions - analogous to the universal GCD, LCM definitions that I presented in a post here.\n\nFirst some notation. $\\;$ Write $\\ \\ \\rm x \\le S \\;\\iff\\; x \\le s,\\;\\: \\forall\\: s \\in S,\\;$ and dually for $\\rm\\; x \\ge S$\n\nDEFINITION of $\\:$ min $\\quad$ $\\quad\\rm x \\le S \\;\\iff\\; x \\le min\\ S$\n\nDEFINITION of max $\\:\\quad$ $\\quad\\rm x \\ge S \\;\\iff\\; x \\ge max\\ S$\n\nFor min, when $\\;\\rm S = \\emptyset\\;$ is empty, the first clause $\\;\\rm x \\le S\\;$ in the min definition is vacuously true, hence the definition reduces to $\\;\\rm x \\le \\min \\emptyset\\;$ for all $\\;\\rm x\\;$. Hence $\\;\\rm \\min \\emptyset = \\infty\\;$. Dually $\\;\\rm \\max\\emptyset = -\\infty\\;$.\n\nAs I remarked in said GCD, LCM post, such universal definitions often facilitate slick proofs. For some nontrivial examples of min, max flavor consider the slick proofs of the integrality of various products of binomial coefficients by employing the floor function, e.g. see Joe Roberts: Elementary Number Theory. Instead I close with an analogous slick GCD, LCM proof from my mentioned post (see it for further details).\n\nGenerally, in any domain, we have the following dual universal definitions of LCM and GCD:\n\nDEFINITION of LCM $\\quad$ If $\\quad\\rm a,b\\ |\\ c \\;\\iff\\; [a,b]\\ |\\ c \\quad$ then $\\quad\\rm [a,b] \\;\\;$ is an LCM of $\\;\\rm a,b$\n\nDEFINITION of GCD $\\quad$ If $\\quad\\rm c\\ |\\ a,b \\;\\iff\\; c\\ |\\ (a,b) \\quad$ then $\\quad\\rm (a,b) \\;$ is an GCD of $\\;\\;\\rm a,b$\n\nNote: that $\\;\\rm a,b\\ |\\ [a,b] \\;$ follows by putting $\\;\\rm c = [a,b] \\;$ in the definition. Dually $\\;\\rm (a,b)\\ |\\ a,b \\;$\n\nSuch $\\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental\n\nTHEOREM $\\rm\\quad (a,b)\\ =\\ ab/[a,b] \\;\\;$ if $\\;\\rm\\ [a,b] \\;$ exists.\n\nProof: $\\rm\\quad\\quad d\\ |\\ a,b \\;\\iff\\; a,b\\ |\\ ab/d \\;\\iff\\; [a,b]\\ |\\ ab/d \\;\\iff\\; d\\ |\\ ab/[a,b] \\quad\\;\\;$ QED\n\nThe conciseness of this proof arises by exploiting to the hilt the $\\iff$ definition of LCM and GCD. Compare to less concise / general / illuminating proofs in many number theory textbooks.\n\n-\n@Bill: I think where you say \"min\" and \"max\" in the first two paragraphs you mean \"inf\" and \"sup\"; the empty set has no minimum and no maximum. \u2013\u00a0 Arturo Magidin Feb 1 '11 at 16:33\n@Arturo: above min,max are synonyms for inf,sup, cf. their universal definitions. \u2013\u00a0 Bill Dubuque Feb 1 '11 at 20:30\n@Bill: That's nonstandard use of min and max, as far as I am aware; I don't see what I'm supposed to compare, though... \u2013\u00a0 Arturo Magidin Feb 1 '11 at 20:41\n@Arturo: You're supposed to refer to the universal definitions of min, max that I am using above. \u2013\u00a0 Bill Dubuque Feb 1 '11 at 21:59", "date": "2014-10-22 23:06:05", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/3768/upper-and-lower-bounds-of-emptyset/3832", "openwebmath_score": 0.950153648853302, "openwebmath_perplexity": 280.15389899781303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534365728416, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8707500062281142}}
{"url": "https://math.stackexchange.com/questions/1510877/what-does-this-l-and-upside-down-l-symbol-mean/1586104", "text": "# What does this 'L' and upside down 'L' symbol mean?\n\nI need help finding out what the following symbols are called and what they do. I searched up math symbols but couldn't find them anywhere near there.\n\n$$\\lceil{-3.14}\\rceil=$$\n\n$$\\lfloor{-3.14}\\rfloor=$$\n\n\u2022 Try Detexify. \u2013\u00a0lhf Nov 3 '15 at 10:22\n\nThey are ceil and floor values, that is they are the closest integers one below and one above respectively , see link.\n\n\u2022 Ohhh! Thanks so much \u2013\u00a0Xirol Nov 3 '15 at 9:56\n\nThe notation $\\lfloor x \\rfloor$ (known as \u2018the floor function\u2019) denotes the largest integer less than or equal to $x \\in \\mathbb{R}$. Examples include $\\lfloor7\\rfloor$ = $7$, $\\lfloor2.5\\rfloor$ = $2$, $\\lfloor\\pi\\rfloor$= $3$ and $\\lfloor\u22122.5\\rfloor$ = $-3$.\n\nThe notation $\\lceil x \\rceil$ (known as \u2018the ceiling function\u2019) denotes the smallest integer greater than or equal to $x \\in \\mathbb{R}$. So using the same examples as before $\\lceil7\\rceil$ = $7$, $\\lceil2.5\\rceil$ = $3$, $\\lceil\\pi\\rceil$= $4$ and $\\lceil\u22122.5\\rceil$ = $-2$.\n\n$$\\lceil{-3.14}\\rceil=-3$$\n\n$$\\lfloor{-3.14}\\rfloor=-4$$\n\n\u2022 For completeness, there is also the Nearest Integer function: $\\lfloor x \\rceil$. \u2013\u00a0YoTengoUnLCD Dec 22 '15 at 22:04\n\u2022 @YoTengoUnLCD Oh wow, I never heard of that one before; thanks for pointing it out :) Perhaps you should make this as an answer, and give some examples for it (maybe the same numerical figures as I used for comparison). It's still vaguely related to OP's question. Besides, I would like to see this function in action! Thank you. \u2013\u00a0BLAZE Dec 22 '15 at 22:09\n\u2022 Done! Thanks for the suggestion :-). \u2013\u00a0YoTengoUnLCD Dec 22 '15 at 22:20\n\nRelated to the two mentioned functions, there is the perhaps less common\n\n$\\text{Nearest integer function } \\lfloor x\\rceil$.\n\n$$\\lfloor 3.2\\rceil=3\\\\ \\lfloor 3.6\\rceil=4\\\\ \\lfloor -1.2\\rceil=-1$$\n\nThe function is ambiguous at numbers of the form $r+\\frac 1 2, r\\in \\bf Z$, and so some kind of note should be made when using this function, clarifying what's to be done in those cases.\n\nNote that we can set\n\n$$\\lfloor x\\rceil_1=\\left \\lfloor x+\\frac 1 2\\right\\rfloor\\\\ \\\\ \\lfloor x\\rceil_2=\\left \\lceil x-\\frac 1 2\\right\\rceil\\\\$$\n\nThe first one rounds half integers up, and the second rounds them down.", "date": "2019-07-18 03:07:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1510877/what-does-this-l-and-upside-down-l-symbol-mean/1586104", "openwebmath_score": 0.9017667174339294, "openwebmath_perplexity": 491.2349975180072, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534390139987, "lm_q2_score": 0.8872045937171068, "lm_q1q2_score": 0.870749999612672}}
{"url": "https://scicomp.stackexchange.com/questions/23466/condition-number-of-rectangular-matrices", "text": "# Condition Number of Rectangular Matrices\n\nThe 2-norm condition number can be easily extended to rectangular matrices. I'm wondering if the inequality for the product of matrices still holds in that case, i.e.,\n\n$\\operatorname{cond}(AB) \\leq \\operatorname{cond}(A)\\operatorname{cond}(B)$\n\n\u2022 Yes, this is still true for rectangular matrices both in the spectral norm and in other norms. It's a good exercise so I won't ruin the challenge for you. A hint would be start by considering the definition of $\\mathrm{cond}(A)$ that you're using. \u2013\u00a0Brian Borchers Mar 25 '16 at 0:04\n\nI guess I figured out the answer to my question.\n\nSuppose the SVD of $A = U \\Sigma V^\\ast$ (where $V^\\ast$ is the conjugate transpose of the matrix $V$). Noting the fact that the unitary transformations $U$ and $V$ preserve the 2-norm, $\\|Ax\\|_{2}$ for any unit vector $x$ can be written as\n\n\\begin{align*} \\frac{\\|Ax\\|_{2}}{\\|x\\|_{2}} = \\|Ax\\|_{2} &= \\|U \\Sigma V^\\ast x\\|_{2} \\\\ &= \\| \\Sigma V^\\ast x\\|_{2} \\\\ &= \\| V^\\ast \\Sigma x\\|_{2} \\\\ &= \\|\\Sigma x\\|_{2} \\end{align*}\n\nHence $\\|Bx\\|_{2} \\leq \\sigma_\\max(B)$ and $\\|Bx\\|_{2} \\geq \\sigma_\\min(B)$.\n\nFor $y = Bx$,\n\n\\begin{equation*} \\|ABx\\|_{2} = \\|Ay\\|_{2} \\leq \\sigma_\\text{max}(A) \\|y\\|_{2} \\leq \\sigma_\\max(A) \\sigma_\\max(B) \\end{equation*}\n\nSimilarly,\n\n\\begin{equation*} \\|ABx\\|_{2} = \\|Ay\\|_{2} \\geq \\sigma_\\min(A) \\|y\\|_{2} \\geq \\sigma_\\min(A) \\sigma_\\min(B) \\end{equation*}\n\nSince the above statements are true for all $x$, they are true both for the minimum and maximum. Thus, \\begin{equation*} \\sigma_\\max(AB) = \\max \\|ABx\\|_{2} \\leq \\sigma_\\max(A) \\sigma_\\max(B) \\end{equation*}\n\nand \\begin{equation*} \\sigma_\\min(AB) = \\min \\|ABx\\|_{2} \\geq \\sigma_\\min(A) \\sigma_\\min(B) \\end{equation*}\n\nOn dividing the above two equations, we obtain \\begin{equation*} \\operatorname{cond}_{2}(AB) \\leq \\operatorname{cond}_{2}(A) \\operatorname{cond}_{2}(B) \\end{equation*}\n\n\u2022 You can accept your own answer. \u2013\u00a0nicoguaro Apr 24 '16 at 3:44", "date": "2019-06-16 11:32:28", "meta": {"domain": "stackexchange.com", "url": "https://scicomp.stackexchange.com/questions/23466/condition-number-of-rectangular-matrices", "openwebmath_score": 1.0000085830688477, "openwebmath_perplexity": 1126.2359454387536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9814534322330059, "lm_q2_score": 0.8872045974451016, "lm_q1q2_score": 0.8707499972553973}}
{"url": "https://math.stackexchange.com/questions/2509485/place-8-rooks-on-a-10-times-10-board/2509519", "text": "# Place $8$ rooks on a $10\\times 10$ board.\n\nThe Problem:-\n\n\"In chess, a rook attacks any piece in the same row or column as the rook, provided no other piece is between them. In how many ways can $8$ rooks be placed on a $[8\\times8]$ chessboard so that no two attack each other? What about $8$ rooks on a $10\\times10$ board?\"\n\nI believe I have an answer for the first part of the question. When placing the first rook, there are 8 places on any particular column (or row) to place the rook, leaving just 7 places on a different column (or row) for the next rook, and so on, providing 8! possible ways to place the rooks in such a way that they cannot attack each other ($P(8,8) = 8!/(8-8)! = 8!$).\nHowever, I am not sure I fully understand how this would work for a board where there are more rows and columns than pieces (such as on a 10x10 board). Does it become $P(10,8) = 10!/(10-8)! = 10!/2$ ? If so, why? If not, how should I approach this problem?\n\nThis problem was found in \"Introduction to Combinatorics and Graph Theory\" by David Guichard.\n\n\u2022 Only one rook can go on each row and on each column. If there are two more rows and columns there are two rows and two columns with no rooks on them. just declare at the begininning that there will be 2 rows which won't be used. There are ${10 \\choose 2}$ ways of doing that and ${10 \\choose 2}$ ways of choosing the columns. So there are ${10\\choose 2}^2 *D$ ways of doing this if $D$ is the number of ways to do it on an 8x8 board. \u2013\u00a0fleablood Nov 7 '17 at 18:45\n\nSo for the first question you need to place a rook in each column. The rook in the first column can go in $8$ places, the rook in the second column then has $7$ possible rows etc.\n\nIn the second part, you need first to choose the $8$ columns out of the $10$ available, and then place a rook in each of the columns. If once again you move from left to right, counting the number of possibilities, you should be fine.\n\nRather than talking about \"any particular\" column or row, you should make a habit of defining the scheme more closely (the first, second rather than \"any\" etc). This will help to avoid a great deal of confusion and can also highlight any double-counting.\n\n\u2022 For clarification, to determine the number of possible combinations for placing 8 rooks on a 10x10 board, we could calculate 10 choose 8 to determine how many 8x8 boards we could have, and then use that to determine how many total combinations there are, by multiplying the number from 10 choose 8 (45) with the number of possible permutations on the board (8!)? \u2013\u00a0oath Nov 7 '17 at 18:54\n\u2022 @oath With $\\binom {10}{8}$ columns you effectively reduce to a $10\\times 8$ board - but then you have to find a position in each column (there are ten positions to start with, and each choice reduces the possibilities for the next choice by one) \u2013\u00a0Mark Bennet Nov 7 '17 at 18:56\n\u2022 Ooooh, I see. So with this line of thought, would $10\\choose 8$ multiplied by permutations $P(10,8)$ give the correct answer ($10\\choose 8$ gives 45, and the permutation $P(10,8)$ gives $10!/2!$) (the answer provided by @Bram28)? \u2013\u00a0oath Nov 7 '17 at 19:15\n\nFor the second part, if we first focus on the columns the rooks are in: since there cannot be two rooks in the same column, the rooks must appear in $8$ different columns. And you can choose $8$ columns out of $10$ in\n\n$$10 \\choose 8$$\n\nways. Once, we've chosen the $8$ columns, let's place the rooks in those columns. Well, you can place the first rook in $10$ different rows, the next in $9$, etc. until the last one, which can be place in $3$ different rows, so we can place those $8$ rooks in:\n\n$$\\frac{10!}{2!} = \\frac{10!}{2}$$\n\nways. This leads to a grand total of:\n\n$${10 \\choose 8} \\cdot \\frac{10!}{2}$$\n\nways of placing $8$ rooks on a $10 \\times 10$ board.\n\n\u2022 As a variation on this, we can (independently) choose eight columns and eight rows to be populated, each in $$10 \\choose 8$$ different ways. Having done that, the chosen rows and columns form a board equivalent to the 8 x 8 case, which the OP has already answered (8!). Overall, then, for 10 x 10 case, there are $${10 \\choose 8} ^ 2 \\cdot 8!$$ different ways to place the 8 rooks. \u2013\u00a0John Bollinger Nov 7 '17 at 22:20\n\u2022 @JohnBollinger Ooh, nice, thanks! \u2013\u00a0Bram28 Nov 7 '17 at 22:25\n\nThis is a different approach than what's been offered so far, but that might help some people understand the problem differently.\n\nI would solve this by first assuming order mattered when placing the rooks (so placing on $A1$ then $B2$ is different than $B2$ then $A1$), then adjusting my answer to compensate.\n\nThe number of squares you can place the first rook on is $10^2$. Wherever you place it, you effectively turn your board into a $9 \\times 9$ board with no rooks on it.\n\nNow that we've lost a row and a column from our first rook, the number of squares you can place the second rook on is $9^2$.\n\nCarrying on for each rook, we get:\n\n$$10^2 \\cdot 9^2 \\cdot ... \\cdot 3^2$$\n\nWays to place our rooks (don't forget we only have $8$ rooks, so we don't go all the way to $1^2$). Remember though, we assumed order matters. In reality, it does not. However, there's only $8!$ different ways to order our rooks, so we can divide our initial answer by $8!$, leaving:\n\n$$(10^2 \\cdot 9^2 \\cdot ... \\cdot 3^2) / 8!$$\n\nWays to place our rooks.\n\nFor the first case of placing 8 rooks on an 8x8 board your result of 8! is correct.\n\nFor the second case of placing 8 rooks on a 10x10 board I find the following approach to be the simplest way to do the calculations:\n\nThere are 10*9/2 = 45 ways to choose which two rows you will not be using. Likewise there are 45 ways to choose which two columns you will not be using.\n\nOnce you have chosen which rows and columns not to use there are 8! ways to place the rooks in the remaining squares.\n\nThat gives a total of 8!*45\u00b2 = 81648000 possibilities.", "date": "2019-11-17 15:16:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2509485/place-8-rooks-on-a-10-times-10-board/2509519", "openwebmath_score": 0.6897656917572021, "openwebmath_perplexity": 267.94312834352047, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534382002797, "lm_q2_score": 0.8872045907347108, "lm_q1q2_score": 0.870749995963654}}
{"url": "https://math.stackexchange.com/questions/1408642/combinatorial-proof-for-binomial-identity-sum-k-0n-binomkp-binom?noredirect=1", "text": "# Combinatorial Proof for Binomial Identity: $\\sum_{k = 0}^n \\binom{k}{p} = \\binom{n+1}{p+1}$ [duplicate]\n\nI am studying combinatorics and I came across the identity $$\\sum\\limits_{k=0}^n \\binom kp =\\binom {n+1}{p+1}.$$ I have read the algebraic proof and it does not appeal to me. Is there an elegant counting trick we can use to arrive at this identity?\n\n## marked as duplicate by Jack D'Aurizio, user230715, Macavity, vonbrand, 6005Aug 25 '15 at 20:55\n\n\u2022 In case $p=0$? More than one question mark reads as an insulting tone. @WillJagy \u2013\u00a0Thomas Andrews Aug 25 '15 at 2:37\n\u2022 @WillJagy I do realize that $\\binom kp = 0$ if $k <p$ $(k \\in \\mathbb{Z})$. I guess it is more elegant to write it that way. And everything is well defined so there is no issues. \u2013\u00a0Miz Aug 25 '15 at 2:40\n\u2022 @GeorgeSimpson: I disagree with this duplication mark. This question is explicitly asking for a combinatorial proof contrary to the referred question. So, two of three answers of the referred question are to exlude as proper answer for this question. The combinatorial answers here which are quite ok for this question do hardly answer the referred binomial identity. \u2013\u00a0Markus Scheuer Aug 26 '15 at 8:21\n\u2022 @MarkusScheuer Looking among linked and related questions it seems that there are several other questions about combinatorial proof of the same identity. For example: 321022, 497413 1332282. If needed, we can discuss this further in chat. \u2013\u00a0Martin Sleziak Aug 29 '15 at 7:59\n\nWhat do you get when you take a sequence of ones and zeros of length $n+1$ that has $k+1$ ones and remove the rightmost $1$ and everything to its right?\n\nExample: $100100\\rightarrow 100$\n\nYou get a sequence that has $k$ ones, the length varies between $k$ and $n$ depending on the actual sequence.\n\nIn fact this function is a bijection between the sequences of length $n+1$ and $k+1$ ones and the sequences of length $n$ or less and $k$ ones.\n\nThis establishes $\\binom{n+1}{k+1}=\\sum\\limits_{j=k}^{n}\\binom{j}{k}$\n\nI'd prefer to omit the zero terms and prove $$\\sum_{k = p}^n \\binom{k}{p} = \\binom{n+1}{p+1}.$$\n\nSuppose we have $n+1$ people standing in a line so that we can refer to Person 1, Person 2, ..., Person $n+1$.\n\nThe righthand side counts the number of committees of size $p+1$ that can be formed out of these $n+1$ people.\n\nThe lefthand side counts the same thing, but it does so in cases.\n\nThe first case is that Person $p+1$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $p$ people in line, which can be done in $\\binom{p}{p}$ ways.\n\nThe second case is that Person $p+2$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $p+1$ people in line, which can be done in $\\binom{p+1}{p}$ ways.\n\nContinuing in this way, the last case would be that Person $n+1$ is the largest numbered person in the committee. This means we must choose the remaining $p$ committee members from the first $n$ people in line, which can be done in $\\binom{n}{p}$ ways.\n\nSumming over these cases yields the desired identity.\n\nHere is another combinatorial approach using lattice paths.\n\nWe can interpret $\\binom{n+1}{p+1}$ as number of lattice paths of length $n+1$ containing $p+1$ horizontal $(1,0)$-steps and $n-p$ vertical $(0,1)$-steps.\n\nThis is valid because there are $\\binom{n+1}{p+1}$ choices to select $p+1$ steps in horizontal direction leaving the remaining $n-p$ steps for the vertical direction.\n\n\n\nLet's consider all paths from $(0,0)$ to $(p+1,n-p)$. We know there are $\\binom{n+1}{p+1}$ different paths.\n\nOn the other hand each of these paths has to cross the vertical line going through $(p,0)$. The crossing points are $(p,0), (p,1), \\ldots, (p,n-p)$. This way we partition the set of lattice paths into sets containing $\\binom{k}{p}$ elements with $p\\leq k \\leq n$ establishing the identity\n\nThe RHS counts the ways to select $p+1$ places in a line of $n+1$ places.\nThe LHS counts the ways to select $p$ places in a line of $k$ places, summed over all values of $k: 0\\leq k\\leq n$. \u00a0 (More strictly, $k: p\\leq k\\leq n$, since there are no ways to do this for $k<p$).\nInterpreting $k+1$ as the position of the last, or $(p+1)^{st}$, place, we notice that these both count ways to perform the exact same task. \u00a0 Ergo they are equivalent.\n$$\\sum_{k=p}^n \\dbinom{k}{p} = \\binom{n+1}{p+1}$$", "date": "2019-05-19 06:20:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1408642/combinatorial-proof-for-binomial-identity-sum-k-0n-binomkp-binom?noredirect=1", "openwebmath_score": 0.8257485032081604, "openwebmath_perplexity": 239.57721699618156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406026280905, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8707426651872353}}
{"url": "https://mathhelpforum.com/threads/group-combinations-problem.150015/", "text": "# Group Combinations Problem\n\n#### mathceleb\n\nIf we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?\n\nIs the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?\n\nAny thoughts?\n\n#### undefined\n\nMHF Hall of Honor\nIf we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?\n\nIs the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?\n\nAny thoughts?\nI haven't worked this out rigorously, but I think the probability is $$\\displaystyle \\displaystyle \\frac{1}{\\binom{149}{2}}$$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $$\\displaystyle \\displaystyle \\binom{149}{2}$$ ways to choose two people out of the remaining 149...\n\n#### Soroban\n\nMHF Hall of Honor\nHello, mathceleb!\n\nWe have a group of 150 people, broken into groups of 3, and only 3 are blue.\nWhat is the probability that all 3 blue people are put in the same group?\n\nThe 150 people are divided into 50 groups of 3 people each.\n\nThere are: .$$\\displaystyle \\dfrac{150!}{(3!)^{50}}$$ possible groupings.\n\nTo have the 3 blue people in one group, there is: .$$\\displaystyle {3\\choose3} = 1$$ way.\n\nThe other 147 people are divided into 49 groups of 3: .$$\\displaystyle \\dfrac{147!}{(3!)^{49}}$$ ways.\n. . Hence, there are: .$$\\displaystyle \\dfrac{147!}{(3!)^{49}}$$ ways to have the 3 blue people in one group.\n\nThe probability that all 3 blue people are in one group is:\n\n. . $$\\displaystyle \\dfrac{\\dfrac{147!}{(3!)^{49}}} {\\dfrac{150!}{(3!)^{50}}} \\;=\\;\\dfrac{147!}{(3!)^{49}}\\cdot\\dfrac{(3!)^{50}}{150!} \\;=\\;\\dfrac{6}{150\\cdot149\\cdot148} \\;=\\;\\dfrac{1}{551,\\!300}$$\n\n#### Plato\n\nMHF Helper\nthe probability is $$\\displaystyle \\displaystyle \\frac{1}{\\binom{149}{2}}$$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $$\\displaystyle \\displaystyle \\binom{149}{2}$$ ways to choose two people out of the remaining 149...\nThis correct.\n\nThe 150 people are divided into 50 groups of 3 people each.\nThere are: .$$\\displaystyle \\dfrac{150!}{(3!)^{50}}$$ possible groupings.\n\nThose are ordered partions.\n\nThere are $$\\displaystyle \\dfrac{150!}{(3!)^{50}(50!)}$$ unordered partitions.\n\nThere are $$\\displaystyle \\dfrac{147!}{(3!)^{49}(49!)}$$ of those where the blues are together.\n\nNote that $$\\displaystyle \\dfrac{\\dfrac{147!}{(3!)^{49}(49!)}}{ \\dfrac{150!}{(3!)^{50}(50!)}}=\\dfrac{1}{\\binom{149}{2}}$$\n\n#### undefined\n\nMHF Hall of Honor\nHmm since Soroban's answer differs from mine I decided to do a little more research.\n\nThere are: .$$\\displaystyle \\dfrac{150!}{(3!)^{50}}$$ possible groupings.\nI believe this should be $$\\displaystyle \\dfrac{\\left(\\dfrac{150!}{(3!)^{50}}\\right)}{50!}$$ because the groups of 3 can be permuted.\n\n...Hence, there are: .$$\\displaystyle \\dfrac{147!}{(3!)^{49}}$$ ways to have the 3 blue people in one group.\nLikewise I believe this should be $$\\displaystyle \\dfrac{\\left(\\dfrac{147!}{(3!)^{49}}\\right)}{49!}$$.\n\nLeading to:\n\n$$\\displaystyle \\dfrac{\\left(\\dfrac{147!}{(3!)^{49}}\\right)\\cdot50!}{49!\\cdot\\left(\\dfrac{150!}{(3!)^{50}}\\right)} = 50\\left(\\dfrac{1}{551,\\!300}\\right) = \\dfrac{1}{11026} = \\displaystyle \\frac{1}{\\binom{149}{2}}$$\n\nSee here (a PDF file) for some more info; under subheading \"Partitioning\".\n\nEdit: Plato answered a lot quicker than I did, didn't refresh the page.\n\nSimilar threads", "date": "2020-02-17 18:28:21", "meta": {"domain": "mathhelpforum.com", "url": "https://mathhelpforum.com/threads/group-combinations-problem.150015/", "openwebmath_score": 0.8422766923904419, "openwebmath_perplexity": 1096.427435770479, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140254249554, "lm_q2_score": 0.9005297801113612, "lm_q1q2_score": 0.8707348747025262}}
{"url": "https://yurichev.org/PE99/", "text": "## [Math][Python] Project Euler #99 and logarithmic representation\n\nComparing two numbers written in index form like $$2^{11}$$ and $$3^7$$ is not difficult, as any calculator would confirm that $$2^{11} = 2048 < 3^7 = 2187$$. However, confirming that $$632382^{518061} > 519432^{525806}$$ would be much more difficult, as both numbers contain over three million digits. Using base_exp.txt (right click and 'Save Link/Target As...'), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.\n\nThis is a problem where logarithmic representation of number can be used. We compute power not for numbers, but for their logarithmic representation. And they consumes way less RAM. Also, this is much faster. In fact, John Napier invented logarithms to ease multiplication operation.\n\nSince we only need info, which number is bigger, logarithmic representations are enough: they obey this law as well.\n\n#!/usr/bin/env python3\nimport math\n\nf=open(filename,\"r\")\nf.close()\n\nreturn [item.rstrip() for item in ar]\n\ntmp=123**45\nprint (\"generic algo:\", tmp)\n# log with other base would also work!\n# but here using natural log base for simplicity\ntmp=math.log(123)*45\nprint (\"using log:   \", math.exp(tmp))\n\nx=math.log(632382)*518061\ny=math.log(519432)*525806\nassert x>y\n\nline=1\n_max=0\nmax_line=None\nfor l in lst:\nt=l.split(\",\")\nx,y = int(t[0]),int(t[1])\nr=math.log(x)*y\nif r>_max:\nmax_line=line\n_max=r\n#print (line, x, y, r)\nline=line+1\n\nprint (max_line, _max)\n\n\nThe result:\n\ngeneric algo: 11110408185131956285910790587176451918559153212268021823629073199866111001242743283966127048043\nusing log:    1.111040818513185e+94\n709 6919995.552420337\n\n\nSpoiler: Wolfram Mathematica can operate on such big numbers without additional assistance. In fact, I first solved this problem using it.\n\nBTW, we can verify Project Euler's author's statement about \"as both numbers contain over three million digits\" using decimal logarithm instead of natural:\n\n#!/usr/bin/env python3\nimport math\n\ntmp=math.log(632382, 10)*518061\nprint (tmp)\n\n3005261.2406258043\n\n\nThis coincides with Wolfram Mathematica's output. This number has 3005261 decimal digits. All correct.\n\n###### (the post first published at 20220728.)\n\nYes, I know about these lousy Disqus ads. Please use adblocker. I would consider to subscribe to 'pro' version of Disqus if the signal/noise ratio in comments would be good enough.", "date": "2022-09-25 13:57:43", "meta": {"domain": "yurichev.org", "url": "https://yurichev.org/PE99/", "openwebmath_score": 0.6366434693336487, "openwebmath_perplexity": 5130.014938308633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.992304353276774, "lm_q2_score": 0.8774767954920547, "lm_q1q2_score": 0.8707240440661195}}
{"url": "https://www.physicsforums.com/threads/conditional-probability-of-rainfall.600498/", "text": "# Conditional Probability of rainfall\n\n1. Apr 26, 2012\n\n### TranscendArcu\n\n1. The problem statement, all variables and given/known data\n\n3. The attempt at a solution\na) P(Pickwick has no umbrella | it rains) = $\\frac{\\frac{1}{3}\\frac{1}{3}}{\\frac{1}{2}} = \\frac{2}{9}$, which is the answer according to my answer key.\n\nb) For part b we have:\n\nThere is a rain forecast which means he will bring the umbrella. The probability that it won't rain is 1/3.\n\nThere is a non-rain forecast which means he brings the umbrella with a probability of 1/3 and it will not rain with a prob of 2/3.\n\nP(Pickwick has umbrella | no rain) = $\\frac{1}{3} + \\frac{1}{3}\\frac{2}{3} = \\frac{5}{9}$. But the answer is apparently 5/12. What have I done incorrectly here?\n\n2. Apr 26, 2012\n\n### MaxManus\n\nI got the same as you at b) and I cant see why it is not correct.That's no guarantee for that you are correct though\n\n3. Apr 26, 2012\n\n### Ray Vickson\n\nI think you are correct. Let's do it in a simple way, just by counting. Imagine 9,000 days. In 4,500 of those, it rains and in 4,500 it does not. Let's use the notation R = rain, Rc = no rain, F = Forecast rain, Fc = Forecast no rain, U = umbrella, Uc =no umbrella\nLay out the data in a table:\n$$\\begin{array}{lccc} & R & Rc & \\text{tot}\\\\ F & N1 & N2 & (N1+N2)\\\\ Fc & M1 & M2 & (M1+M2) \\\\ \\text{tot} & 4500 & 4500 & 9000 \\end{array}$$\nWe are given N1/(N1+N2) = 2/3 and M1/(M1+M2) = 1/3, as well as N1+M1 = 4500 and N2+M2 = 4500. Solving these equations we get N1 = 3000, N2 = 1500, M1 = 1500, M2 = 3000. So, the table is:\n$$\\begin{array}{lccc} & R & Rc & \\text{tot}\\\\ F & 3000 & 1500 & 4500 \\\\ Fc & 1500 & 3000 & 4500 \\\\ \\text{tot} & 4500 & 4500 & 9000 \\end{array}$$\nFrom this it follows that P(F) = 4500/9000 = 1/2 and P(Fc) = 1/2.\n\nIn (b), in the 4500 Rc-days, U occurs in 1500 (F-days) + (3000/3) (Fc-days), for a total of 2500 days. Thus, P(U|Rc) = 2500/4500 = 5/9.\n\nRGV", "date": "2018-03-17 17:35:23", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/conditional-probability-of-rainfall.600498/", "openwebmath_score": 0.6860119700431824, "openwebmath_perplexity": 2026.3370598951503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812299938006, "lm_q2_score": 0.899121388082479, "lm_q1q2_score": 0.8706922757050444}}
{"url": "https://matheducators.stackexchange.com/questions/7478/name-the-heuristic-exploiting-the-legitimacy-of-the-questioner/7480#7480", "text": "# Name the heuristic: exploiting the legitimacy of the questioner\n\nAs a child, I made frequent use of a particular 'trick' in order to make short work of many different problems. The general form is to be presented a question which wants a definite (numerical) answer, where the question statement contains a free variable. The 'trick' was to recognize that the answer (which the teacher has told us exists by asking for it) doesn't depend on the free variable, so we now know that we can substitute any convenient value we like for the free variable and find the answer.\n\nThis FiveTrangles problem is an excellent example. The height of the non-shaded triangles is unspecified, so we know straight away that it doesn't figure in the answer to the question. Since it doesn't matter, we can set it to zero and replace this question with the trivial question of finding the area of a triangle with $base = 10$ and $height = 5$.\n\nDoes this technique have a name? Do problems which permit it have a name? Should such questions be avoided?\n\n\u2022 I'm not sure I understand. Do you mean leaving the answer as an expression with a variable in it?\n\u2013\u00a0Karl\nFeb 23 '15 at 19:55\n\u2022 @Karl Another example: the question \"$(x + 4)^2 - x^2 - 8x$ simplifies to a number. Find that number.\" can be solved with algebra, or it can be solved by plugging in x=0, since apparently the answer does not depend on x. Feb 23 '15 at 20:35\n\u2022 That's a clearer example for me thanks. I get the triangle question too - presumably the $x$ cancels and isn't left as a variable in the expression.\n\u2013\u00a0Karl\nFeb 23 '15 at 21:13\n\u2022 More of a technique than just a 'trick', see en.wikipedia.org/wiki/Limiting_case_%28mathematics%29 Feb 24 '15 at 3:35\n\u2022 Somewhat related: for single-answer multiple choice problems, if two answers are equivalent, then they must both be wrong. Feb 24 '15 at 5:21\n\nThe heuristic described here is one manifestation of what Polya (1945) and others thereafter refer to as trying a special case. I do not know of a more specific term for the context that you have put forth, but this is often how one approaches a problem if you are initially unsure as to how to solve it in generality.\n\nIt is a good way to get an initial foot-hold on a problem, and especially useful in the context of some standardized tests. For example, consider the following:\n\nQuestion: Player 1 flips a penny $$n>0$$ times, and Player 2 flips a penny $$n+1$$ times. What is the probability that Player 2 flips more heads than Player 1?\n\nA. 1/2\n\nB. 1/3\n\nC. 1/4\n\nD. 1/8\n\nSince none of the choices depends on $$n$$, you can consider $$n=1$$ and see easily that the probability is $$1/2$$. This allows you to blaze through the question very quickly. (Indeed, this holds for all $$n \\geq 1$$.)\n\nAlthough testing well is an important life skill, your final question as to whether such problems are advisable is interesting. In my estimation, they can be good in the classroom as a way of scaffolding, as long as you press students (when appropriate) to explain why their reasoning holds for the general case.\n\nAs one small example, induction problems are done like this: You show the (hopefully easy) base case, and then complete the proof by setting up and using your inductive hypothesis.\n\nA possible pitfall, though, is not only the absence of reasoning by students (as in the example question above) but also a possible reinforcement around misconceptions of proof by example. (I believe this relates to the recent query about counterexamples in MESE 7466.) For more about some of these misconceptions, check out work by Orit Zaslavsky (google scholar).\n\nLastly: The issue with guessing answers based on problem statements is pervasive in school mathematics. It ranges from only seeing examples of the form $$a + b = \\square$$ and adopting a misunderstanding of the equal sign as an operator, to problems in which students just try to combine numbers in ways that might be \"sensible\" (without really engaging in sense-making). Cf. the classic:\n\nTeacher: There are 125 sheep and 5 dogs in a flock. How old is the shepherd?\n\nStudent: (thinking: 125+5, 125-5, and 125*5 are too big; it must be 125/5) The shepherd is 25.\n\n\u2022 I had considered trying a special case, but there's more going on than that. Something akin to trying the easiest case, and going with it, because all of the cases are the same (not that I understand why, but if it weren't true the question wouldn't have been put to me this way). That's a mouthful though. Thanks for the answer. Feb 24 '15 at 1:15\n\u2022 @NiloCK If you know why \"all the cases are the same,\" then you could say that you are reducing to a special case (rather than just trying one). But there is (obviously) quite a leap between solving for one special case and understanding that the special case answers (or helps answer) a seemingly more general question. Feb 24 '15 at 1:25\n\u2022 The 'why' for me (as a kid) was that the problem setter put the question that way. That is, if all the cases weren't the same, then the question wouldn't (couldn't) have been asked. I knew at the time that this was 'unfair' information, but it's hard for a kid writing math contests to discard such information on the grounds that it was 'ill-gotten' or 'inadmissible'. Feb 24 '15 at 1:38\n\u2022 @NiloCK Benjamin addresses that \"why\" in the answer. Essentially, the why amounts to a guess that may or may not be true. A teacher could easily throw in questions where the real answer is \"not enough information\" or \"none of the above\" just to discourage that heuristic. In my opinion, discouraging it forces students to think through why the missing information doesn't matter, leading to a deeper understanding of the logic and the problem and the solution. For me, that sort of process is when I learn the most. Feb 25 '15 at 4:28\n\n(continued from previous post)\n\nIn this next problem, the horizontal leg of the large right triangle and the vertical leg of the smaller right triangle are also not fixed, but setting each to zero won't lead to a solution at all: http://fivetriangles.blogspot.com/2013/04/56-obtuse-triangle-area.html (Never mind that getting in the habit of simply setting values to zero will lead to a \"zero\" in calculus.)\n\nFinally, the following proof uses the identical triangle area concept as in the original \"concave quadrilateral\" problem, but the \"trick\" doesn't work at all: http://fivetriangles.blogspot.com/2013/10/104-equal-area-proof.html\n\n\u2022 You can combine both answers into one. There is no character limit (within reason) in answers. Feb 23 '15 at 21:54\n\u2022 Stack exchange proscribed more than 2 links in one post if 0 \u2264 rating < 10. Feb 23 '15 at 22:11\n\u2022 You can't control the order in which answers appear, so calling another answer the \"previous post\" is not good. For instance, if this answer gets more points it will even appear above the other one. Call your other answer \"my other post\" rather than \"previous\" one.\n\u2013\u00a0KCd\nFeb 25 '15 at 11:28\n\nIn my SAT tutoring, I refer to such problems as underspecified. For instance, consider this problem:\n\nIf $\\frac{a}{2} + \\frac{b}{3} = 0$, then find $\\frac{a}{b}$.\n\nI teach the student that since two variables are presented but only one equation, this makes the problem underspecified. In other words, a and b are not uniquely determined. This gives the student license to make up a number for one out of $a$ and $b$, then use the given equation to solve for the other variable, and finally substitute both $a$ and $b$ into the expression $\\frac{a}{b}$ to obtain the answer.\n\nThere are also many examples of underspecified SAT geometry problems. Often the problem will ask the student to find the value of $a+b$ or some other expression involving two variables. Example: The Official SAT Study Guide (second edition) by CollegeBoard, page 799, problem 13.\n\nThat some solutions are independent of a variable is not uncommon, but it does not necessarily trivialise a problem or negate its teaching value. The following problem presents similar possibilities, but we'd argue that solving it with such a \"trick\" is actually more circuitous than finding the answer directly (and may be more enlightening, too): http://fivetriangles.blogspot.com/2012/04/area-problem.html Here is a nice animation someone posted of the situation:\n\nThe shortcoming in such \"tricks\", we agree, is that they may lead to tunnel vision, but worse, their utilisation is at the user's risk.\n\n(continued in next post)\n\n\u2022 I assume that you're the owner of the Five Triangles blog? Great to have you on the site; I love yours. Many beautiful questions. Feb 24 '15 at 0:56\n\u2022 Now that you have the necessary reputation, perhaps you could merge your two answers into one? Your other post (at this time) has 3 up-votes, whereas this one has only 1; so it would make better sense to merge into the higher voted one. (FWIW: I have not up-voted the other one, and will happily do so if I see them merged; in this way, deletion will not cause you any net-loss in reputation!) Mar 26 '15 at 5:50\n\nI don't know the answer to the questions about the potential names for such things.\n\nAs to whether such questions should be avoided: I think some types of these questions can maybe help students to think about quantifiers, or to really think about the concepts involved in the question.\n\nI like Chris Cunningham's example: if you tell the student that $(x+4)^2 - x^2 - 8x$ simplifies to a number, and ask them to find the number, they may just do the algebra of simplifying. But if you show them that you can get the answer by plugging in $0$, then you can teach them something about what it really means to say that \"$(x+4)^2 - x^2 - 8x$ simplifies to a number\". Now you can talk about what it means for a statement to hold \"for all $x$\". Things like this are useful to be familiar with. Many college students struggle with this when I ask them, for example, to show that a function is not linear!\n\nIt's also nice to have this kind of logic on hand for when you need to remember things. For example, say you were taking a linear algebra exam and you knew the determinant was either the sum or the product of the eigenvalues. Well, you could just consider the $2\\times2$ identity matrix and figure it out pretty quickly! This is because you knew the statement didn't depend on the matrix or even the size of the matrix!\n\nI (and this is just opinion) don't like questions like the triangle one that you linked to. Because there, the \"trick\" from the solver's perspective is just to realize that the question asker left out some words! I don't know how useful it is to have someone apply the logic \"If they asked me this, it has an answer, and if they didn't say it depended on quantity $A$ then it must not depend on quantity $A$.'' It is not clear to me what the mathematical benefit is to that.\n\n\u2022 RE: last paragraph. One benefit to the problem solver/student is to consider the importance of the givens; alternatively, a benefit to the problem poser/instructor is to consider more carefully how problems are phrased. Feb 25 '15 at 2:51\n\nI would be surprised if the solution strategy or problem types have a name. While it can be useful to find the correct answer, what does it accomplish from a learning perspective? Take, for instance, Chris Cunningham's problem of \"$(x+4)^2 - x^2 -8x$ simplifies to a number. Find that number.\" As an instructor, I would be looking to make sure the students accomplished the appropriate learning outcomes for the problem. For this problem, I would more than likely be looking for (1) can they multiply out $(x+4)^2$ and (2) can they simplify the addition of terms. Such a technique would not accomplish (1) or (2), so I would avoid writing problems that can (knowingly) make use of it nor would I ever teach such meta-strategies.\n\nThis reminds me of a multiple choice exam question on trig identities: \"Which of the following is an equivalent identity to $\\cot(x) + \\tan(x)$?\" with $\\csc(x)\\sec(x)$ being in the list of answers. Instead of using trig identities, many students plugged in a value like $x=10$ and compared. While it works for this case, it dodges the actual learning and testing of the concept.\n\nDiscovering the irrelevance of $x$ in the expression could be achieved by getting the students (perhaps in different groups) to substitute different values in for $x$ and compare results. Was this what they expected? Which number would they choose to substitute knowing the answer will always be the same? Having aroused their curiosity the expanding brackets and simplifying can be motivated and discussed. This could be a rich activity if the emphasis is on discovery and not just sub in $x=0$", "date": "2021-12-07 18:29:35", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/questions/7478/name-the-heuristic-exploiting-the-legitimacy-of-the-questioner/7480#7480", "openwebmath_score": 0.6984184980392456, "openwebmath_perplexity": 472.413124823118, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812290812827, "lm_q2_score": 0.8991213799730774, "lm_q1q2_score": 0.8706922670315876}}
{"url": "https://mathhelpboards.com/threads/points-a-b-out-at-sea.27818/", "text": "# Points A & B Out At Sea\n\n#### xyz_1965\n\n##### Member\nFrom points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.\n\n1. Find height of the mountain.\n\n2. Find the distance from point A to the bottom of the mountain.\n\nI think making two triangles makes sense here.\n\nTriangle 1\n\ntan(41\u00b0) = h/x\n\nLet h = height of mountain.\n\nLet x = the distance from point B out at sea to the bottom of the mountain.\n\nSolving for x, I get h/tan(41\u00b0).\n\nTriangle 2\n\ntan(37\u00b0) = h/[80 + (h/tan(41\u00b0)]\n\nI need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.\n\nLastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.\n\n1. Is any of this right?\n\n2. Is there an easy way to solve\ntan(37\u00b0) = h/[80 + (h/tan(41\u00b0)] for h?\n\n#### skeeter\n\n##### Well-known member\nMHB Math Helper\nassuming A, B, and the mountain top are located in the same vertical plane ...\n\n$h = x_B \\tan(41)$\n\n$h = x_A \\tan(37)$\n\n$x_A - x_B = 80 \\implies x_A = x_B +80$\n\n$x_B \\tan(41) = (x_B+80)\\tan(37)$\n\n$x_B = \\dfrac{80\\tan(37)}{\\tan(41)-\\tan(37)}$\n\nfrom here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.\n\n#### xyz_1965\n\n##### Member\nassuming A, B, and the mountain top are located in the same vertical plane ...\n\n$h = x_B \\tan(41)$\n\n$h = x_A \\tan(37)$\n\n$x_A - x_B = 80 \\implies x_A = x_B +80$\n\n$x_B \\tan(41) = (x_B+80)\\tan(37)$\n\n$x_B = \\dfrac{80\\tan(37)}{\\tan(41)-\\tan(37)}$\n\nfrom here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.\nExplain $x_B$ and $x_A$.\n\n#### skeeter\n\n##### Well-known member\nMHB Math Helper\n$x_B$ is the horizontal distance from point B to the point directly below the mountain top.\n\n$x_A$ is the horizontal distance from point A to the point directly below the mountain top.\n\n#### xyz_1965\n\n##### Member\n$x_B$ is the horizontal distance from point B to the point directly below the mountain top.\n\n$x_A$ is the horizontal distance from point A to the point directly below the mountain top.\nThank you for clearing this up for me.\n\n#### Country Boy\n\n##### Well-known member\nMHB Math Helper\nFrom points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.\n\n1. Find height of the mountain.\n\n2. Find the distance from point A to the bottom of the mountain.\n\nI think making two triangles makes sense here.\n\nTriangle 1\n\ntan(41\u00b0) = h/x\n\nLet h = height of mountain.\n\nLet x = the distance from point B out at sea to the bottom of the mountain.\nIt would make more sense to say these before you say \"tan(41)= h/x!\n\nSolving for x, I get h/tan(41\u00b0).\nSpecifically, x= h/tan(41).\n\nTriangle 2\n\ntan(37\u00b0) = h/[80 + (h/tan(41\u00b0)]\nI think I would have first said \"Let y be the distance from B to the base of the mountain\" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing.\n\nI need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.\nYes.\n\nLastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.\nNo! You were doing so well and then you forgot how you had defined \"A\" and \"B'! \"A\" and \"B\" are points, not distances so you can't add them. \"x\" was the distance from point A to the base of the mountain.\n1. Is any of this right?\n\n2. Is there an easy way to solve\ntan(37\u00b0) = h/[80 + (h/tan(41\u00b0)] for h?\nA= h/(B+h/C)\n\nGet rid of the fraction by multiplying both sides by B+ h/C:\nAB+ (A/C)h= h\n\nAB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C\n\nh= ABC/(C- A)\n\nOf course A= tan(37\u00b0)= 0.7536, approximately, B= 80, and C= tan(41\u00b0)= 0.8693 approximately.\n\n#### xyz_1965\n\n##### Member\n\nIt would make more sense to say these before you say \"tan(41)= h/x!\n\nSpecifically, x= h/tan(41).\n\nI think I would have first said \"Let y be the distance from B to the base of the mountain\" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing.\n\nYes.\n\nNo! You were doing so well and then you forgot how you had defined \"A\" and \"B'! \"A\" and \"B\" are points, not distances so you can't add them. \"x\" was the distance from point A to the base of the mountain.\n\nA= h/(B+h/C)\n\nGet rid of the fraction by multiplying both sides by B+ h/C:\nAB+ (A/C)h= h\n\nAB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C\n\nh= ABC/(C- A)\n\nOf course A= tan(37\u00b0)= 0.7536, approximately, B= 80, and C= tan(41\u00b0)= 0.8693 approximately.\nI thank you for the break down. It's ok to make mistakes here. Learning is not possible without errors.\n\n#### Country Boy\n\n##### Well-known member\nMHB Math Helper\nThat is true. And you learn by having those errors pointed out!\n\n#### xyz_1965\n\n##### Member\nThat is true. And you learn by having those errors pointed out!\nI agree. This is why I don't mind being corrected as long as it is done respectfully.", "date": "2020-08-11 16:05:27", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/points-a-b-out-at-sea.27818/", "openwebmath_score": 0.6804009079933167, "openwebmath_perplexity": 1314.1939754027862, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676460637103, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8706882409122703}}
{"url": "https://math.stackexchange.com/questions/1255896/commuting-of-hom-and-tensor-product-functors", "text": "# Commuting of Hom and Tensor Product functors?\n\nLet $V_i,W_i$ be finite dimensional vector spaces, for $i=1,2$. Assume we have homomorphisms $\\phi_i:V_i\\rightarrow W_i$.\n\nThen, there is an induced map $\\widehat{\\phi_1 \\times \\phi_2} \\in Hom(V_1 \\otimes V_2,W_1 \\otimes W_2)$ such that $\\widehat{\\phi_1 \\times \\phi_2}(v_1 \\otimes v_2) = \\phi_1(v_1)\\otimes\\phi_2(v_2)$.\n\nWe can look at this construction as a map $\\psi: Hom(V_1,W_1)\\times Hom(V_2,W_2)\\rightarrow Hom(V_1 \\otimes V_2,W_1 \\otimes W_2)$, where\n\n$\\psi:(\\phi_1,\\phi_2)\\rightarrow \\widehat{\\phi_1 \\times \\phi_2}$.\n\nIt is easy to see that $\\psi$ is a bilinear map. It follows (from the universal property) that we have an induced homomorphism: $\\hat \\psi:Hom(V_1,W_1)\\otimes Hom(V_2,W_2)\\rightarrow Hom(V_1 \\otimes V_2,W_1 \\otimes W_2)$ such that: $\\hat\\psi(\\phi_1\\otimes\\phi_2)=\\widehat{\\phi_1 \\times \\phi_2}$.\n\nMy question: is $\\hat\\psi$ necessarily an isomorphism? Note that there is equality of dimensions (since both Hom,Tensor multiply them), hence clearly some isomorphism must exists between $Hom(V_1,W_1)\\otimes Hom(V_2,W_2)\\cong Hom(V_1 \\otimes V_2,W_1 \\otimes W_2)$.\n\nIf $\\hat\\psi$ is not isomorphism, is there some other \"canonical\" isomorphism?\n\nI will just note that in the case of finite dimensional vector spaces, there is equality of dimensions, hence clearly some isomorphism must exists.\n\n\u2022 Take $W_1 = W_2 = A$, you can even assume it's a field then you're asking if $(V_1 \\otimes V_2)^* \\cong V_1^* \\otimes V_2^*$. That's false if $V_1$ and $V_2$ are infinite dimensional. I can't even imagine how complicated things would become if $A$ is not a field, or not even a PID or anything... For $A$ a field my instinct tells me your $\\hat\\psi$ is always injective, but I'm not 100% sure. \u2013\u00a0Najib Idrissi Apr 28 '15 at 13:27\n\u2022 It need not be well defined. I have an example. \u2013\u00a0Matt Samuel Apr 28 '15 at 13:43\n\u2022 Najib: I agree. I have edited the question to focus upon the case of finite dimensional vector spaces. That was my original interest, I just thought maybe I missed some general theme which is not related to the specific fact that my objects were finite_dim vector spaces and not general modules. I was completely wrong there... \u2013\u00a0Asaf Shachar Apr 28 '15 at 13:43\n\u2022 $\\widehat{\\phi_1 \\times \\phi_2}$ is well defined via the universal property of tensor products. \u2013\u00a0Asaf Shachar Apr 28 '15 at 13:48\n\u2022 @egreg The expression $\\phi_1(v_1) \\otimes \\phi_2(v_2)$ is linear in each variable in $v_1$ and $v_2$, if I'm not mistaken, so it defines a linear map $V_1 \\otimes V_2 \\to W_1 \\otimes W_2$. And then this is linear in each of $\\phi_1$ and $\\phi_2$, hence the map $\\hat\\psi$. Am I missing something? It's just the bifunctoriality of $\\otimes : \\mathsf{Vect} \\times \\mathsf{Vect} \\to \\mathsf{Vect}$. \u2013\u00a0Najib Idrissi Apr 28 '15 at 14:04\n\nThe general situation is the following: $A$ is a commutative ring and $V_1, V_2, W_1, W_2$ are $A$-modules\n\nIf one of the ordered pairs $(V_1,V_2)$, $\\,(V_1,W_1\\,$ or $\\,(V_2,W_2)$ consists of finitely generated projective $A$-modules, the canonical map is an isomorphism (Bourbaki, Algebra, Ch. 2 \u2018Linear Algebra\u2019, \u00a74, n\u00b04, prop. 4).\n\nOver a field, all modules are projective since they're free. So the answer is \u2018yes\u2019.\n\nIn the present case, here is a sketch of the proof:\n\nLet $K$ be the base field. As $V_1\\simeq K^m$ for some $m>0$ and similarly $V_2\\simeq K^n$, and the $\\operatorname{Hom}$ and $\\,\\otimes\\,$ functors comute with direct sums, we may as well suppose $V_1=V_2=K$.So we have to prove: $$\\widehat\\Psi\\colon\\operatorname{Hom}(K,W_1)\\otimes\\operatorname{Hom}(K,W_2)\\to\\operatorname{Hom}(K\\otimes K,W_1\\otimes W_2)$$ is an isomorphism.\n\nThis results basically that for any vector space $V$ we have an isomorphism \\begin{align*} \\operatorname{Hom}_K(K,V)&\\simeq V\\\\ \\phi&\\mapsto\\phi(1) \\end{align*} In detail, just consider the following commutative diagram:\n\n\u2022 Thanks. Is there a quick way to see that $\\hat\\psi$ is an isomorphism in the particular case where all the modules are finite dimensional vector spaces? (I think this question should have a simple answer in terms of \"undergraduate\" linear algebra without the need to resorting to more \"heavy\" tools\\theorems from abstract commutative algebra. \u2013\u00a0Asaf Shachar Apr 28 '15 at 13:57\n\u2022 @asaf shachar: I gave some details of a possible proof. See my completed answer. \u2013\u00a0Bernard Apr 28 '15 at 18:44\n\u2022 Excuse me... I want to ask that is this property holds for $> 2$ vector spaces? I think it works because the diagram listed above seems work for $> 2$ vector spaces. More generally, in module case, is this property holds for $> 2$ modules (with suitable requirements...)?? Thank you! \u2013\u00a0Peter Hu Sep 2 '15 at 10:54\n\nTo show things like this in the case of finite dimensional vector spaces, you can always fallback on picking bases:\n\nLet $v_i^1, \\ldots, v_i^{m_i}$ be a basis of $V_i$ ($i=1,2$), and $w_i^1, \\ldots, w_i^{n_i}$ be a basis of $W_i$ ($i=1,2$). Let $f^{jk}_i : V_i \\to W_i$ be given on the basis by $f_i^{jk}(v_i^l) = \\delta_{jl} w_i^k$. Then $\\{v_1^p \\otimes v_2^q\\}$ is a basis of $V_1 \\otimes V_2$, and $\\{f_i^{jk}\\}$ is a basis of $Hom(V_i,W_i)$. Your map sends $(f_1^{jk},f_2^{rs})$ to the linear transformation given on the basis by $\\widehat{f_1^{jk} \\times f_2^{rs}}(v_1^p \\otimes v_2^q) = \\delta_{jp} \\delta_{rq} w_1^k \\otimes w_2^s$.\n\nThose maps clearly form a basis of $Hom(V_1 \\otimes V_2, W_1 \\otimes W_2)$ for the same reason the $f_i^{jk}$ are a basis of $Hom(V_i,W_i)$.", "date": "2021-08-05 11:50:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1255896/commuting-of-hom-and-tensor-product-functors", "openwebmath_score": 0.9495059847831726, "openwebmath_perplexity": 192.00256117830915, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676466573412, "lm_q2_score": 0.8887587920192298, "lm_q1q2_score": 0.8706882342235002}}
{"url": "http://mathematica.stackexchange.com/questions/13125/measuring-fractal-dimension-of-natural-objects-from-digital-images/13130", "text": "# Measuring fractal dimension of natural objects from digital images\n\nThis is a useful topic. A college physics lab, medical diagnostics, urban growth, etc. - there is a lot of applications. On this site by Paul Bourke about Google Earth fractals we can get a high resolution images (in this post they are low res - import from source for experiments). For example, around Lake Nasser in Egypt:\n\nimg = Import[\"http://paulbourke.net/fractals/googleearth/egypt2.jpg\"]\n\n\nThe simplest method I know is Box Counting Method which has a lot of shortcomings. We start from extracting the boundary - which is the fractal object:\n\n{Binarize[img], iEdge = EdgeDetect[Binarize[img]]}\n\n\nNow we could partition image into boxes and see how many boxes have at least 1 white pixel. This is a very rudimentary implementation:\n\nMinS = Floor[Min[ImageDimensions[iEdge]]/2];\ndata = ParallelTable[{1/size, Total[Sign /@ (Total[#, 2] & /@ (ImageData /@\nFlatten[ImagePartition[iEdge, size]]))]}, {size, 10, MinS/2, 10}];\n\n\nFrom this the slope is 1.69415 which is a fractal dimension that makes sense\n\nline = Fit[Log[data], {1, x}, x]\n\n\n13.0276 + 1.69415 x\n\nPlot[line, {x, -6, -2}, Epilog -> Point[Log[FDL]],\nPlotStyle -> Red, Frame -> True, Axes -> False]\n\n\nBenchmark: if I run this on high res of Koch snowflake i get something like ~ 1.3 with more exact number being 4/log 3 \u2248 1.26186.\n\nQuestion: can we improve or go beyond the above box counting method?\n\nAll approaches are acceptable if they find fractal dimension from any image of natural fractal.\n\n-\nYou have a lot of programs in mathematica to measure fractal dimensions and multi fractal spectrum of an image in the book: Fractal Geography, Andre Dauphine, Wiley, 2012 See the book on the Wolfram Mathematica | Books or Amazon ![enter image description here](wolfram.com/books/profile.cgi?id=8108) \u2013\u00a0 Dauphine Oct 16 '12 at 9:32\nVitaliy, with all due respect, the ending bullets make the question (actually there isn't a question mark anywhere, right?) rather wide ranging. Perhaps you could focus it somewhat and make it sound more like a real question? \u2013\u00a0 Sjoerd C. de Vries Oct 16 '12 at 12:59\n@SjoerdC.deVries Updated, it is a single question now. \u2013\u00a0 Vitaliy Kaurov Oct 16 '12 at 15:06\ncould you perhaps describe or link some examples of use of fractal dimension? \u2013\u00a0 magma Oct 17 '12 at 10:27\n@magma: Wikipedia lists some examples here: en.wikipedia.org/wiki/\u2026 \u2013\u00a0 shrx Jun 14 '13 at 20:22\n\nYou can still use box count, but doing it smarter :)\n\nCounting boxes with at least 1 white pixel from ImagePartition can be done more efficiently using Integral Image, a technique used by Viola-Jones (2004) in their now popular face recognition framework. For a mathematical motivation (and proof), Viola and Jones point to this source.\n\nWhat Integral Image allows you to do is to compute efficiently the total mass of any rectangle in an image. So, you can define the following:\n\nIntegralImage[d_] := Map[Accumulate, d, {0, 1}];\ndata = ImageData[ColorConvert[img, \"Grayscale\"]]; (* img: your snowflake image *)\nii = IntegralImage[data];\n\n\nThen, the mass (white content) of a region, is\n\n(* PixelCount: total mass in region delimited by two corner points,\ngiven ii, the IntegralImage *)\nPixelCount[ii_, {p0x_, p0y_}, {p1x_, p1y_}] :=\nii[[p1x, p1y]] + ii[[p0x, p0y]] - ii[[p1x, p0y]] - ii[[p0x, p1y]];\n\n\nSo, instead of partitioning the image using ImagePartition, you can get a list of all the boxes of a given size by\n\nPartitionBoxes[{rows_, cols_}, size_] :=\nTranspose /@ Tuples[{Partition[Range[1, rows, size], 2, 1],\nPartition[Range[1, cols, size], 2, 1]}];\n\n\nIf you apply PixelCount to above, as in your algorithm, you should have the same data but calculated faster.\n\nPixelCountsAtSize[{rows_, cols_}, ii_, size_] :=\n((PixelCount [ii, #1, #2] &) @@ # &) /@ PartitionBoxes[{rows, cols}, size];\n\n\nFollowing your approach here, we should then do\n\nfractalDimensionData =\nTable[{1/size, Total[Sign /@ PixelCountsAtSize[Dimensions[ii], ii, size]]},\n{size, 3, Floor[Min[Dimensions[ii]]/10]}];\nline = Fit[Log[fractalDimensionData], {1, x}, x]\n\nOut:= 10.4414 + 1.27104 x\n\n\nwhich is very close to the actual fractal dimension of the snowflake (which I used as input).\n\nTwo things to note. Because this is faster, I dared to generate the table at box size 3. Also, unlike ImagePartition, my partition boxes are all of the same size and therefore, it excludes uneven boxes at the edges. So, instead of doing minSize/2 as you did, I put minSize/10 - excluding bigger and misleading values for big boxes.\n\nHope this helps.\n\nUpdate\n\nJust ran the algorithm starting with 2 and got this 10.4371 + 1.27008 x. And starting with 1 is 10.4332 + 1.26919 x, much better. Of course, it takes longer but still under or around 1 min for your snowflake image.\n\nUpdate 2\n\nAnd finally, for your image from Google Earth (eqypt2.jpg) the output is (starting at 1-pixel boxes)\n\n12.1578 + 1.47597 x\n\n\nIt ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs.\n\n-\nThere isn't a \"snowflake\" image. All three images come from the same object (and I guess the fractal dimension should be the same for all of them) \u2013\u00a0 belisarius Dec 23 '13 at 21:03\n@belisarius, the PO published a Koch snowflake image as Benchmark (see last part of post) and that's the one I used. I called it 'snowflake' for short. On the other hand, this is still a numerical procedure, so the numbers will be different depending on approximation. The Update2 refers to the eqypt2.jpg image, also made available by the PO, as he asked for an improvement suitable for real-life images. \u2013\u00a0 caya Dec 23 '13 at 22:40\nsorry, I missed that link. Thanks \u2013\u00a0 belisarius Dec 24 '13 at 0:05\n+1 This is very neat @caya, thank you! I will wait \"a bit\" hoping that someone can implement things like wavelet multi-fractals or similar. \u2013\u00a0 Vitaliy Kaurov Feb 14 at 20:02\n@VitaliyKaurov, glad you liked it. Note that Viola & Jones rightly pointed out a link between integral image and Haar wavelet basis in their paper; although this wasn't explored further. It is unclear to me the link between the paper you mentioned and fractal dimension, but certainly worth reading as I am interested in these kind of problems. Cheers. \u2013\u00a0 caya Feb 16 at 12:40", "date": "2014-07-29 04:38:36", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/13125/measuring-fractal-dimension-of-natural-objects-from-digital-images/13130", "openwebmath_score": 0.49098411202430725, "openwebmath_perplexity": 3149.0941828253453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.976310526632796, "lm_q2_score": 0.8918110368115783, "lm_q1q2_score": 0.8706845030064518}}
{"url": "https://math.stackexchange.com/questions/3994373/how-to-be-sure-about-any-parameterization", "text": "# How to be sure about any parameterization?\n\nThis is the problem:\n\nFind intersection of this two surfaces and then calculate its length. $$x^2+4y^2=4$$ and\n$$y\\sqrt3 +z=1$$ Method 1 :\n\nCommon way is to take:\n$$x=2\\cos t ,\\: y=\\sin t ,\\: z=1-\\sqrt3 \\sin t$$\nSo\n$$r(t) =(2\\cos t,\\:\\sin t,\\:1-\\sqrt3 \\sin t)$$ and\n$$r'(t) = (-2\\sin t,\\:\\cos t,\\:-\\sqrt3\\cos t) ,\\: |r'(t)|=2$$\n\nFinally the length is:\n$$L = \\int_{0}^{2\\pi} 2\\:dt=4\\pi$$\n\nBut in another parameterization I find something strange\n\nMethod 2:\n\n$$x=2\\sqrt{1-t^2} ,\\: y=t ,\\: z=1-t\\sqrt3$$ So\n$$r(t) = (2\\sqrt{1-t^2} ,\\: t ,\\: 1-t\\sqrt3)$$ and\n$$r'(t) = \\left(\\frac {-2t}{\\sqrt{1-t^2}},1,-\\sqrt3\\right) ,\\: |r'(t)|=\\frac {2}{\\sqrt{1-t^2}}$$ (and that's different from method 1 $$|r'(t)|$$)\n\nFinally the length\n$$L=\\int_{0}^{2\\pi} \\frac {2}{\\sqrt{1-t^2}}\\:dt\\\\ =2\\int_{0}^{\\arcsin{2\\pi}}\\frac {1}{\\cos{\\theta}}\\cos{\\theta}\\:d\\theta\\\\ =2\\int_{0}^{\\arcsin{2\\pi}}1\\:d\\theta$$\n\nand because of $$\\arcsin(2\\pi)$$ its undefined it cant be solved.\n\nNow my first question is that:\nIf we know from past that we can create many parameterizations for a single curve, so why the result of they length is different?\n\nand the second one:\n\nHow to be sure that a parameterization is the correct one?\n\nAs a complement to @Axel's answer\n\nLet $$t_j$$ denote the $$j$$th method's $$t$$ so $$t_2=\\sin t_1$$ and $$\\tfrac{dt_2}{\\sqrt{1-t_2^2}}=d\\arcsin t_2=\\pm dt_1$$. If for example $$t_1$$ increases from $$0$$ to $$2\\pi$$, $$t_2$$ is monotonic in three parts, cut at $$t=\\pi/2$$ and $$t=3\\pi/2$$. The monotonic pieces have alternating $$\\pm$$, in this case with the pattern $$+-+$$, so the length is\\begin{align}\\int_0^1\\frac{2dt_2}{\\sqrt{1-t_2^2}}-\\int_1^{-1}\\frac{2dt_2}{\\sqrt{1-t_2^2}}+\\int_{-1}^0\\frac{2dt_2}{\\sqrt{1-t_2^2}}&\\color{red}{=4\\int_{-1}^1\\frac{dt_2}{\\sqrt{1-t_2^2}}}\\\\&\\color{red}{=4\\pi}\\\\&\\color{blue}{=\\left(\\int_0^{\\pi/2}+\\int_{\\pi/2}^{3\\pi/2}+\\int_{3\\pi/2}^{2\\pi}\\right)2dt_1}\\\\&=\\color{blue}{4\\pi},\\end{align}where the $$t_2$$-space calculation can continue from the black expression with either the red or the blue treatment. If $$t_1$$ varies over another length-$$2\\pi$$ interval the monotinicity division details are different, but any choice requires at least one such cut, succumbs to the above technique, and gets the same answer. Let's check another example from $$t_1=-\\pi/2$$ to $$t_1=3\\pi/2$$, giving two pieces (meeting at $$t=\\pi/2$$) where the $$\\pm$$ are $$+-$$:$$\\int_{-1}^1\\frac{2dt_2}{\\sqrt{1-t_2^2}}-\\int_1^{-1}\\frac{2dt_2}{\\sqrt{1-t_2^2}}\\color{red}{=4\\int_{-1}^1\\frac{dt_2}{\\sqrt{1-t_2^2}}}\\color{blue}{=\\left(\\int_{-\\pi/2}^{\\pi/2}+\\int_{\\pi/2}^{3\\pi/2}\\right)2dt_1}.$$\n\n\u2022 Thank you both.I understand it now. \u2013\u00a0program_craft Jan 22 at 4:58\n\nThe problem lies in the way you parametrize the curve in the second method.\n\nThe reason why you integrate from $$0$$ to $$2\\pi$$ in the first method is because you have implicitly define you parametrization to go from $$0$$ to $$2\\pi$$ because you are looking for a mapping $$[0;2\\pi) \\longrightarrow \\textrm{your intersection curve}$$ such that that mapping is biyective with the entire curve.\n\nIf you take a look to what you limits of integrations are in the second method you see they are the same from the first one. That again implicitly define you parametrization to go from $$0$$ to $$2\\pi$$ but this time (because of the equation $$y=t$$) you are saying that $$y$$ runs in the range $$[0;2\\pi)$$ when in reallity in ranges $$[-1;1)$$ so you actually end up measuring the length of a totally different curve (one with that parametrization but with $$y \\in [0;2\\pi)$$ insted of $$[-1;1)$$).\n\nNow given that $$t$$ should be in $$[-1;1)$$ the correct way to measure the lenght in the second method would be: $$\\int\\limits_{-1}^1 \\frac{2}{\\sqrt{1-t\u00b2}} \\ dt + \\int\\limits_{1}^{-1} \\frac{2}{\\sqrt{1-t\u00b2}} \\ dt$$\n\nThat is beacuse you can't map the entire curve with just the $$[-1;1)$$ range (in fact you only map one half of it). Thus the second integral.\n\nSo finally you have:\n\n$$2\\int\\limits_{-1}^1 \\frac{2}{\\sqrt{1-t\u00b2}} \\ dt = 4\\pi$$\n\n\u2022 Can you explain more why t is in range[-1,1] \u2013\u00a0program_craft Jan 21 at 17:36", "date": "2021-06-18 09:24:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3994373/how-to-be-sure-about-any-parameterization", "openwebmath_score": 0.854271411895752, "openwebmath_perplexity": 378.31346148141955, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105273220726, "lm_q2_score": 0.8918110360927155, "lm_q1q2_score": 0.870684502919323}}
{"url": "https://math.stackexchange.com/questions/4326771/derive-the-error-for-the-midpoint-rule", "text": "# Derive the error for the Midpoint rule\n\n### Problem statement\n\nLet $$[a,b]$$ be divided into $$n$$ subintervals by $$n+1$$ points $$x_0 = a, x_1 = a+h , x_2 = a+2h , \\dots , x_n = a+nh =b$$.\n\nShow that $$I- I_M = \\frac{1}{24}(b-a)h^2 f''(z)$$ for some $$z \\in [a,b]$$.\n\n### Attempt at solution\n\n$$I_M = h \\sum_{i=0}^{n-1} f\\left( a + \\left( i + \\frac12\\right)\\right) := h\\sum_{i=0}^{n-1}f(x_i)$$\n\n\\begin{align} I -I_M &= \\int_a^b f(x) \\mathrm{d}x - h\\sum_{i=0}^{n-1}f(x_i)\\\\ &= \\sum_{i=0}^{n-1} \\left( \\int_{x_i}^{x_{i+1}} f(x) \\ \\mathrm{d}x - hf(x_i)\\right) \\end{align} Here I am stuck. I know I need to use Taylor's theorem in conjunction with Mean-Value Theorem but I don't know how to proceed.\n\nI suspect that for $$x\\in [x_i ,x_{i+1}]$$ the Taylor expansion is $$f(x) = f(x_i) + (x-x_i)f'(x_i) + \\frac12 (x-x_i)^2 f''(z_x) \\quad \\text{for } z_x \\in [x_i , x_{i+1}]$$ but then If I substitute this in the above I get \\begin{align} &= \\sum_{i=0}^{n-1}\\left( \\int_{x_i}^{x_{i+1}} (f(x_i) + (x-x_i)f'(x_i) + \\frac12 (x-x_i)^2 f''(z_x))\\mathrm{d}x - hf(x_i)\\right)\\\\ \\end{align} and I don't see what I have accomplished\n\nDoes anyone know If I am on the right track and if yes how to continue this derivation ?\n\nYou can start by obtaining the rule for a single interval, say \\begin{align*} \\int_{x_i}^{x_{i+1}} f(x) dx -hf(c_i) &= \\int_{x_i}^{x_{i+1}}(f(x)-f(c_i))dx\\\\ & = \\int_{x_i}^{x_{i+1}} \\left(f(c_i)+f'(c_i)(x-c_i) + \\frac{f''(\\xi_i(x))}{2}(x-c_i)^2\\;-\\;f(c_i)\\right) dx\\\\ & = \\frac{f''(\\xi_i)}{2}\\int_{x_i}^{x_{i+1}}(x-c_i)^2 dx = \\frac{f''(\\xi_i) h^3}{24} \\end{align*}\n\nNow, if you take the sum over the $$n$$ intervals, you get the error term\n\n$$\\sum_{i=1}^n \\frac{f''(\\xi_i) h^3}{24} = \\frac{h^2(b-a)}{24}\\left(\\frac 1n \\sum_{i=1}^n f''(\\xi_i)\\right) = \\frac{h^2(b-a) f''(\\xi)}{24}.$$\n\nSome notes:\n\n1. I used the mean value theorem for integrals. Since $$(x-c_i)^2$$ does not change sign, we know that $$\\int_{x_i}^{x_{i+1}} \\frac{f''(\\xi_i(x))}{2}(x-c_i)^2 dx = \\frac{f''(\\xi_i)}{2} \\int_{x_i}^{x_{i+1}} (x-c_i)^2 dx$$\n\n2. If we assume the $$f\\in C^2$$, the term $$\\frac 1n \\sum f''(\\xi_i)$$ is an average of values of $$f''$$, hence between the minimum and maximum of $$f''$$. The intermediate value theorem says that it will correspond to $$f''(\\xi)$$, for some $$\\xi \\in (a,b)$$.\n\n\u2022 What do you know about the function $x \\mapsto \\xi_i(x)$?\n\u2013\u00a0RRL\nDec 8, 2021 at 12:05\n\u2022 I am in fact assuming that $f''(\\xi(x))$ is continuous and should have added some comment on that. Over the years I've learned that, in pedagogic terms, it is preferable to streamline the proof, leaving the technical aspects to notes or auxiliary lemmas. Dec 8, 2021 at 12:58\n\u2022 This is what I was looking for thank you. Dec 8, 2021 at 18:52\n\u2022 Where does the $f'(c_i) (x-c_i)$ term go in the second line ? Dec 8, 2021 at 19:43\n\u2022 @hexaquark The term vanishes during the integration because it is an odd function with respect tot the midpoint $c_j$ of the interval. Dec 9, 2021 at 12:08\n\nThere is another proof that avoids double integrals and uses the generalized mean value instead. I have learned this technique from reading @LutzLehmann's answers to very similar questions.\n\nLet $$h>0$$ be given and consider the problem of computing the integral $$I = \\int_{-h}^h f(x)dx.$$ The midpoint rule takes the form $$M_h = 2h f(0).$$ We will now obtain the familiar error formula by studying the auxiliary function $$g$$ given by $$g(x) = \\int_{-x}^x f(t) dt - 2x f(0).$$ This function is interesting precisely because $$g(h) = I - M_h.$$ Let $$F$$ denote an anti-derivative of $$f$$. Then $$g(x) = F(x) - F(-x) - 2xf(0).$$ It is clear that $$g$$ is as smooth as $$F$$ and that $$g(0) = 0$$. We have $$g'(x) = f(x) + f(-x) - 2f(0).$$ It follows that $$g'(0) = 0.$$ We have $$g''(x) = f'(x) - f'(-x).$$ It follows that $$g''(0) = 0.$$ We conclude that the graph of $$g$$ is quite flat near $$x=0$$. To make further progress we make repeated use of the generalized mean value theorem. There exists at least one $$x_1$$ between $$0$$ and $$x$$ such that $$\\frac{g(x)}{x^3} = \\frac{g(x) - g(0)}{x^3 - 0^3} = \\frac{g'(x_1)}{3x_1^2}.$$ There exists at least one $$x_2$$ between $$0$$ and $$x_1$$ such that $$\\frac{g'(x_1)}{3x_1^2} = \\frac{g'(x_1) - g'(0)}{3x_1^2 - 3\\cdot 0^2}= \\frac{g''(x_2)}{6x_2}.$$ Similarly, there exists at least one $$x_3$$ between $$0$$ and $$x_2$$ such that $$\\frac{g''(x_2)}{6x_2} = \\frac{g''(x_2)- g''(0)}{6x_2 - 6 \\cdot 0} = \\frac{g'''(x_3)}{6} = \\frac{f''(x_3) + f''(-x_3)}{6}$$ Finally, by the continuity of $$f''$$ there is at least one $$\\xi$$ between $$-x_3$$ and $$x_3$$ such that $$\\frac{g(x)}{x^3} = \\frac{f''(\\xi)}{3}.$$ In particular, there is at least one $$\\xi$$ in $$[-h,h]$$ such that $$I - M_h = g(h) = \\frac{f''(\\xi)}{3} h^3.$$ Replacing $$h$$ with $$h/2$$ yields the familiar result, i.e, $$\\int_{-h/2}^{h/2} f(x) dx - h f(0) = \\frac{f''(\\xi)}{24} h^3.$$ We finish the proof as suggested by @RRL.\n\nTo make any progress we need to consider the smoothness of $$f$$. Here I assume $$f \\in C^2([a,b])$$. Denote the partition points as $$x_j = a + hj$$ and the midpoints as $$c_j = a + \\frac{2j+1}{2}h$$ for $$j = 0,1,\\ldots, n-1$$.\n\nWe first find the local error for a subinterval $$[x_j,x_{j+1}]$$ using the Taylor expansion with integral remainder\n\n$$f(x) = f(c_j) +f'(c_j)(x-c_j) + \\int_{c_j}^x(x-t)f''(t)\\, dt$$\n\nIntegrating over $$[x_j,x_{j+1}]$$ and using $$x_{j+1} - x_j = h$$, we get\n\n$$\\tag{1}\\int_{x_j}^{x_{j+1}}f(x) \\, dx = hf(c_j)+ f'(c_j)\\int_{x_j}^{x_{j+1}}(x-c_j) \\, dx + \\int_{x_j}^{x_{j+1}} \\int_{c_j}^x(x-t)f''(t)\\, dt\\, dx$$\n\nThe second term on the RHS of (1) is\n\n$$f'(c_j)\\int_{x_j}^{x_{j+1}}(x-c_j) \\, dx =f'(c_j) \\left.\\frac{(x-c_j)^2}{2} \\right|_{x_j}^{x_{j+1}}= \\frac{f'(c_j)}{2} \\left[(x_{j+1}-c_j)^2- (x_j - c_j)^2 \\right]\\\\ = \\frac{f'(c_j)}{2}\\left[\\left(\\frac{h}{2}\\right)^2 - \\left(-\\frac{h}{2}\\right)^2\\right]= 0$$\n\nSubstituting into (1) and rearranging, we get\n\n$$\\tag{2}\\int_{x_j}^{x_{j+1}}f(x) \\, dx - hf(c_j)= \\int_{x_j}^{x_{j+1}} \\int_{c_j}^x(x-t)f''(t)\\, dt\\, dx$$\n\nIn order to apply the mean value theorem to extract the second derivative from the integral on the RHS, some manipulation with indicator functions is required to eliminate the variable integration limit. Note that\n\n$$\\int_{x_j}^{x_{j+1}} \\int_{c_j}^x(x-t)f''(t)\\, dt\\, dx \\\\= \\int_{c_j}^{x_{j+1}} \\int_{c_j}^x(x-t)f''(t)\\, dt\\, dx + \\int_{x_j}^{c_j} \\int_{x}^{c_j}(t-x)f''(t)\\, dt\\, dx \\\\ = \\int_{c_j}^{x_{j+1}} \\int_{c_j}^{x_{j+1}}\\mathbf{1}_{}(t \\leqslant x)(x-t)f''(t)\\, dt\\, dx + \\int_{x_j}^{c_j} \\int_{x_j}^{c_j}\\mathbf{1}_{}(t \\geqslant x)(t-x)f''(t)\\, dt\\, dx$$\n\nNow we can apply Fubini's theorem to obtain\n\n$$\\tag{3}\\int_{x_j}^{x_{j+1}} \\int_{c_j}^x(x-t)f''(t)\\, dt\\, dx \\\\ = \\int_{c_j}^{x_{j+1}} \\int_{c_j}^{x_{j+1}}\\mathbf{1}_{(t \\leqslant x)}(x-t)f''(t)\\, dx\\, dt + \\int_{x_j}^{c_j} \\int_{x_j}^{c_j}\\mathbf{1}_{(t \\geqslant x)}(t-x)f''(t)\\, dx\\, dt \\\\ = \\int_{c_j}^{x_{j+1}} f''(t)\\left(\\int_{t}^{x_{j+1}}(x-t)\\, dx\\right)\\, dt + \\int_{x_j}^{c_j} f''(t)\\left(\\int_{x_j}^{t}(t-x)\\, dx\\right)\\, dt\\\\ = \\int_{c_j}^{x_{j+1}} f''(t)\\frac{(x_{j+1}-t)^2}{2}\\, dt + \\int_{x_j}^{c_j} f''(t)\\frac{(t-x_j)^2}{2}\\, dt$$\n\nBy the mean value theorem for integrals there are points $$z_j^+ \\in [c_j,x_{j+1}]$$ and $$z_j^- \\in [x_j,c_j]$$ such that\n\n$$\\tag{4}\\int_{c_j}^{x_{j+1}} f''(t)\\frac{(x_{j+1}-t)^2}{2}\\, dt + \\int_{x_j}^{c_j} f''(t)\\frac{(t-x_j)^2}{2}\\, dt\\\\= f''(z_j^+)\\int_{c_j}^{x_{j+1}} \\frac{(x_{j+1}-t)^2}{2}\\, dt + f''(z_j^-)\\int_{x_j}^{c_j} \\frac{(t-x_j)^2}{2}\\, dt\\\\ = f''(z_j^+)\\frac{h^3}{48} + f''(z_j^-) \\frac{h^3}{48}$$\n\nSubstituting into (2) with (3) and (4) yields\n\n$$\\int_{x_j}^{x_{j+1}}f(x) \\, dx - hf(c_j)=\\frac{h^3}{48}\\left(f''(z_j^+)+f''(z_j^-)\\right)$$\n\nSumming from $$j=0$$ to $$j = n-1$$, we get\n\n$$I-I_M = \\int_a^b f(x) \\, dx - h\\sum_{j=0}^{n-1}f(c_j)= \\frac{h^3}{48}\\sum_{j=0}^{n-1}\\left(f''(z_j^+)+f''(z_j^-)\\right)\\\\ = \\frac{2nh^3}{48}\\frac{1}{n}\\sum_{j=0}^{n-1}\\frac{f''(z_j^+)+f''(z_j^-)}{2},$$\n\nand since $$nh = (b-a)$$,\n\n$$\\tag{5}I - I_M = \\frac{(b-a)h^2}{24} \\sum_{j=0}^{n-1}\\frac{f''(z_j^+)+f''(z_j^-)}{2n}$$\n\nI'll leave it to you to consider if there exists $$z \\in [a,b]$$ such that the second derivative at this point equals the weighted average of second derivatives, i.e.,\n\n$$f''(z) = \\sum_{j=0}^{n-1}\\frac{f''(z_j^+)+f''(z_j^-)}{2n},$$\n\nand, hence,\n\n$$I - I_M = \\frac{(b-a)h^2}{24}f''(z)$$\n\nNevertheless, if the second derivative is bounded as $$|f''(x)| \\leqslant \\|f''||_{\\infty}$$ for all $$x \\in [a,b]$$, then we obtain from (5) the well-known midpoint rule error bound\n\n$$|I - I_M| \\leqslant \\frac{\\|f''\\|_{\\infty}(b-a)h^2}{24}$$\n\n\u2022 The proof is correct but the same result can be obtained in a much simpler way just by using a second order Taylor expansion and the mean value theorem for integrals Dec 8, 2021 at 11:40\n\u2022 @PierreCarre: Understood. I intentionally use the integral form of the remainder to avoid introducing another function $\\xi_i(x)$ with the Lagrange form. What do you know about this function and $f\u2019\u2019 \\circ \\xi_i$. You assume it is continuous without proof as you then apply the integral MVT. So it is \u201csimpler\u201d because some justification is missing.\n\u2013\u00a0RRL\nDec 8, 2021 at 11:58\n\u2022 You are correct, leaving out the analysis of $f''\\circ \\xi_i$ hides the more technical details. Dec 8, 2021 at 13:00", "date": "2022-12-05 14:54:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4326771/derive-the-error-for-the-midpoint-rule", "openwebmath_score": 0.9827636480331421, "openwebmath_perplexity": 117.3524176617983, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918489015606, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8706607290244059}}
{"url": "https://www.mathworks.com/help/matlab/ref/spline.html?requestedDomain=www.mathworks.com&nocookie=true", "text": "# Documentation\n\n### This is machine translation\n\nTranslated by\nMouseover text to see original. Click the button below to return to the English verison of the page.\n\nNote: This page has been translated by MathWorks. Please click here\nTo view all translated materals including this page, select Japan from the country navigator on the bottom of this page.\n\n# spline\n\nCubic spline data interpolation\n\n## Syntax\n\ns = spline(x,y,xq)\npp = spline(x,y)\n\n## Description\n\nexample\n\ns = spline(x,y,xq) returns a vector of interpolated values s corresponding to the query points in xq. The values of s are determined by cubic spline interpolation of x and y.\n\nexample\n\npp = spline(x,y) returns a piecewise polynomial structure for use by ppval and the spline utility unmkpp.\n\n## Examples\n\ncollapse all\n\nUse spline to interpolate a sine curve over unevenly-spaced sample points.\n\nx = [0 1 2.5 3.6 5 7 8.1 10];\ny = sin(x);\nxx = 0:.25:10;\nyy = spline(x,y,xx);\nplot(x,y,'o',xx,yy)\n\nUse clamped or complete spline interpolation when endpoint slopes are known. This example enforces zero slopes at the end points of the interpolation.\n\nx = -4:4;\ny = [0 .15 1.12 2.36 2.36 1.46 .49 .06 0];\ncs = spline(x,[0 y 0]);\nxx = linspace(-4,4,101);\nplot(x,y,'o',xx,ppval(cs,xx),'-');\n\nExtrapolate a data set to predict population growth.\n\nCreate two vectors to represent the census years from 1900 to 1990 (t) and the corresponding United States population in millions of people (p).\n\nt = 1900:10:1990;\np = [ 75.995  91.972  105.711  123.203  131.669 ...\n150.697 179.323  203.212  226.505  249.633 ];\n\nExtrapolate and predict the population in the year 2000 using a cubic spline.\n\nspline(t,p,2000)\nans = 270.6060\n\nGenerate the plot of a circle, with the five data points y(:,2),...,y(:,6) marked with o's. The matrix y contains two more columns than does x. Therefore, spline uses y(:,1) and y(:,end) as the endslopes. The circle starts and ends at the point (1,0), so that point is plotted twice.\n\nx = pi*[0:.5:2];\ny = [0  1  0 -1  0  1  0;\n1  0  1  0 -1  0  1];\npp = spline(x,y);\nyy = ppval(pp, linspace(0,2*pi,101));\nplot(yy(1,:),yy(2,:),'-b',y(1,2:5),y(2,2:5),'or'), axis equal\n\nUse spline to sample a function over a finer mesh.\n\nGenerate sine and cosine curves for a few values between 0 and 1. Use spline interpolation to sample the functions over a finer mesh.\n\nx = 0:.25:1;\nY = [sin(x); cos(x)];\nxx = 0:.1:1;\nYY = spline(x,Y,xx);\nplot(x,Y(1,:),'o',xx,YY(1,:),'-')\nhold on\nplot(x,Y(2,:),'o',xx,YY(2,:),':')\nhold off\n\nCompare the interpolation results produced by spline and pchip for two different functions.\n\nCreate vectors of x values, function values at those points y, and query points xq. Compute interpolations at the query points using both spline and pchip. Plot the interpolated function values at the query points for comparison.\n\nx = -3:3;\ny = [-1 -1 -1 0 1 1 1];\nxq1 = -3:.01:3;\np = pchip(x,y,xq1);\ns = spline(x,y,xq1);\nplot(x,y,'o',xq1,p,'-',xq1,s,'-.')\nlegend('Sample Points','pchip','spline','Location','SouthEast')\n\nIn this case, pchip is favorable since it does not oscillate as freely between the sample points.\n\nPerform a second comparison using an oscillatory sample function.\n\nx = 0:25;\ny = besselj(1,x);\nxq2 = 0:0.01:25;\np = pchip(x,y,xq2);\ns = spline(x,y,xq2);\nplot(x,y,'o',xq2,p,'-',xq2,s,'-.')\nlegend('Sample Points','pchip','spline')\n\nWhen the underlying function is oscillatory, spline captures the movement between points better than pchip.\n\n## Input Arguments\n\ncollapse all\n\nx-coordinates, specified as a vector. The vector x specifies the points at which the data y is given. The elements of x must be unique.\n\nData Types: single | double\n\nFunction values at x-coordinates, specified as a numeric vector, matrix, or array. x and y must have the same length.\n\nIf y is a matrix or array, then the values in the last dimension, y(:,...,:,j), are taken as the values to match with x. In that case, the last dimension of y must be the same length as x.\n\nThe endslopes of the cubic spline follow these rules:\n\n\u2022 If x and y are vectors of the same size, then the not-a-knot end conditions are used.\n\n\u2022 If x or y is a scalar, then it is expanded to have the same length as the other and the not-a-knot end conditions are used.\n\n\u2022 If y is a vector that contains two more values than x has entries, then spline uses the first and last values in y as the endslopes for the cubic spline. For example, if y is a vector, then:\n\n\u2022 f(x) = y(2:end-1)\n\n\u2022 df(min(x)) = y(1)\n\n\u2022 df(max(x)) = y(end)\n\n\u2022 If y is a matrix or an N-dimensional array with size(y,N) equal to length(x)+2, then these properties hold:\n\n\u2022 f(x(j)) matches the value y(:,...,:,j+1) for j = 1:length(x)\n\n\u2022 Df(min(x)) matches y(:,:,...:,1)\n\n\u2022 Df(max(x)) matches y(:,:,...:,end)\n\nData Types: single | double\n\nQuery points, specified as a vector. The points specified in xq are the x-coordinates for the interpolated function values s that spline computes.\n\nData Types: single | double\n\n## Output Arguments\n\ncollapse all\n\nInterpolated values at query points, returned as a vector, matrix, or array.\n\nThe size of s is related to the sizes of the inputs:\n\n\u2022 If y is a scalar or vector, then s has the same size as xq.\n\n\u2022 If y is an array that is not a vector, then these conditions apply:\n\n\u2022 If xq is a scalar or vector, then size(s) equals [d1, d2, ..., dk, length(xq)].\n\n\u2022 If xq is an array of size [m1,m2,...,mj], then size(s) equals [d1,d2,...,dk,m1,m2,...,mj].\n\nPiecewise polynomial, returned as a structure. Use this structure with the ppval function to evaluate the piecewise polynomial at one or more query points. The structure has these fields.\n\nFieldDescription\nform\n\n'pp' for piecewise polynomial\n\nbreaks\n\nVector of length\u00a0L+1 \u00a0with strictly increasing elements that represent the start and end of each of\u00a0L\u00a0intervals\n\ncoefs\n\nL-by-k \u00a0matrix with each row\u00a0coefs(i,:)\u00a0containing the local coefficients of an order\u00a0k\u00a0polynomial on the\u00a0ith interval,\u00a0[breaks(i),breaks(i+1)]\n\npieces\n\nNumber of pieces, L\n\norder\n\nOrder of the polynomials\n\ndim\n\nDimensionality of target\n\nSince the polynomial coefficients in coefs are local coefficients for each interval, you must subtract the lower endpoint of the corresponding knot interval to use the coefficients in a conventional polynomial equation. In other words, for the coefficients [a,b,c,d] on the interval [x1,x2], the corresponding polynomial is\n\n$f\\left(x\\right)=a{\\left(x-{x}_{1}\\right)}^{3}+b{\\left(x-{x}_{1}\\right)}^{2}+c\\left(x-{x}_{1}\\right)+d\\text{\\hspace{0.17em}}.$\n\n## Tips\n\n\u2022 You also can perform spline interpolation using the interp1 function with the command interp1(x,y,xq,'spline'). While spline performs interpolation on rows of an input matrix, interp1 performs interpolation on columns of an input matrix.\n\n## Algorithms\n\nA tridiagonal linear system (possibly with several right-hand sides) is solved for the information needed to describe the coefficients of the various cubic polynomials that make up the interpolating spline. spline uses the functions ppval, mkpp, and unmkpp. These routines form a small suite of functions for working with piecewise polynomials. For access to more advanced features, see interp1 or the Curve Fitting Toolbox\u2122 spline functions.\n\n## References\n\n[1] de Boor, Carl. A Practical Guide to Splines. Springer-Verlag, New York: 1978.\n\n## See Also\n\n#### Introduced before R2006a\n\nWas this topic helpful?\n\n#### Beyond Excel: The Manager's Guide to Solving the Big Data Conundrum\n\nDownload white paper", "date": "2017-05-26 01:56:30", "meta": {"domain": "mathworks.com", "url": "https://www.mathworks.com/help/matlab/ref/spline.html?requestedDomain=www.mathworks.com&nocookie=true", "openwebmath_score": 0.5684455633163452, "openwebmath_perplexity": 2331.510718211438, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085084750966, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8706510999155973}}
{"url": "https://math.stackexchange.com/questions/3049718/proving-set-identities-empty-set-case?noredirect=1", "text": "# Proving set identities: empty set case.\n\nI am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $$A,B,C$$ we have $$A\\cap(B \\cup C) = (A \\cap B) \\cup (A \\cap C).$$\n\nI understand that this can be done by proving both $$A\\cap(B \\cup C) \\subset (A \\cap B) \\cup (A \\cap C) \\ \\text{and} \\ (A \\cap B) \\cup (A \\cap C) \\subset A\\cap(B \\cup C)$$ hold true.\n\nWe can do this by letting $$x$$ be an arbitrary element of $$A\\cap(B \\cup C)$$ and showing that it is an element of $$(A \\cap B) \\cup (A \\cap C)$$, and vice versa.\n\nBut what about when $$A \\cap (B \\cup C)$$ is the empty set? Then I would think we can't let $$x$$ be an arbitrary element of $$A \\cap (B \\cup C)$$ since there are none. However, I am aware that $$A \\cap (B \\cup C) \\subset (A \\cap B) \\cup (A \\cap C)$$ is trivially true in this case.\n\nIn the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?\n\nIf so, some suggestions as to how to incorporate them into proofs would be helpful :-).\n\n\u2022 Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that, \u2013\u00a0DonAntonio Dec 22 '18 at 18:35\n\u2022 Words to live by: \"vacuous truth\". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/\u2026 math.stackexchange.com/questions/2723860/\u2026 math.stackexchange.com/questions/1953218/\u2026 and there are many others which are similar. \u2013\u00a0Asaf Karagila Dec 23 '18 at 10:24\n\nIn order to show that for two sets $$X$$ and $$Y$$ it holds that $$X\\subseteq Y$$, you have to prove that\n\nfor every $$x$$, if $$x\\in X$$, then $$x\\in Y$$.\n\nNote that a statement of the form \u201cif $$\\mathscr{A}$$ then $$\\mathscr{B}$$\u201d is true when\n\neither $$\\mathscr{A}$$ is false or both $$\\mathscr{A}$$ and $$\\mathscr{B}$$ are true\n\nIf $$X$$ is the empty set, then \u201c$$x\\in X$$\u201d is false for every $$x$$; hence \u201cif $$x\\in X$$ then $$x\\in Y$$\u201d is true.\n\nThe phrase \u201ctake an arbitrary element $$x\\in X$$\u201d is possibly misleading, but its intended meaning is \u201csuppose $$x\\in X$$\u201d.\n\n\u2022 Ah, so if we write \"suppose $x \\in X$\" and successfully deduce that \"$x \\in Y$\" , we have shown that \"if $x \\in X$, then $x \\in Y$\" is true. But, if \"$x \\in X$\" is always false, we have still shown \"if $x \\in X$ then $x \\in Y$\" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct? \u2013\u00a0user445909 Dec 22 '18 at 19:39\n\u2022 @E-mu Basically so \u2013\u00a0egreg Dec 22 '18 at 20:27\n\nIf $$D$$ is the empty set then the following statement is always true:$$\\text{If }x\\in D\\text{ then }\\cdots$$no matter what the dots are standing for.\n\nIt is false that $$x\\in D$$ and \"ex falso sequitur quodlibet\". A false statement implies whatever you want.\n\nSo the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.\n\nSince the empty set is a subset of any set, there is no need of including that in a formal proof. However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .", "date": "2020-10-31 22:32:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3049718/proving-set-identities-empty-set-case?noredirect=1", "openwebmath_score": 0.7971826791763306, "openwebmath_perplexity": 169.29805504233255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850879722155, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8706510953495158}}
{"url": "https://math.stackexchange.com/questions/1350084/determine-the-number-of-subsets", "text": "# Determine the number of subsets\n\nHow many distinct subsets of a set $\\text{A}$ are there, containing at least $9$ elements, where the total number of elements in set $\\text{A}$ is $18$ ?\n\nI've solved it by making cases of either choosing $9$, $10$, $11$ and so on elements from the set and then adding them up, i.e, $$\\sum_{r=9}^{18} \\dbinom{18}{r}$$\n\nHowever, the answer given in my book is $$\\dfrac{1}{2} \\times \\dfrac{18!}{9! \\cdot 9!} + 2^{17}$$\n\nAlthough the numerical value of my answer and book's answer is equal, I'm interested in knowing the intended solution of the author.\n\nThe term $\\dfrac{1}{2} \\times \\dfrac{18!}{9! \\cdot 9!}$ is dividing $18$ elements into two equal parts. The term $2^{17}$ is the total number of subsets of a set containing $17$ elements. But I can't connect these two things with the question.\n\nAny help will be appreciated.\nThanks.\n\nP.S. If anyone is interested in knowing how I found $\\displaystyle \\sum_{r=9}^{18} \\dbinom{18}{r}$ here is my method,\n\nLet $\\text{S} = \\displaystyle \\sum_{r=9}^{18} \\dbinom{18}{r}$\n\nBy Binomial Theorem,\n\n$2^{18} = (1+1)^{18} = \\displaystyle \\sum_{r=0}^{18} \\dbinom{18}{r}$\n\n$\\implies 2\\text{S} - \\dbinom{18}{9} =2^{18} \\left( \\because \\dbinom{18}{r} = \\dbinom{18}{18-r}\\right)$\n\nand the rest follows.\n\nIf $A$ has $18$ elements, and $B\\subseteq A$ has fewer than $9$ elements, then $B^c\\subseteq A$ has more than nine elements. Also, if $B$ has more than $9$ elements then $B^c$ has fewer than $9$ elements. That means the number of subsets of $A$ with fewer than $9$ elements equals the number of subsets of $A$ with more than $9$ elements.\nOf course, the complement of a set with $9$ elements is another set with $9$ elements. The number you want, the number of subsets of $A$ with at least $9$ elements, is almost half the number of all subsets of $A$. We can figure out just how close that almost is.\nLet $m$ be the number of subsets of $A$ with more than $9$ elements, so $m$ is also the number of subsets of $A$ with fewer than $9$ elements. Let $n$ be the number of subsets of $A$ with exactly $9$ elements. Then you want to find $m+n$, and we can find it from the total number of subsets of $A$, which we'll call $t$.\n$$\\text{(fewer than 9)+(exactly 9)+(more than 9)=(all)}$$ $$m+n+m=t$$ \\begin{align} m+n&=\\frac 12n+\\frac 12t \\\\ &=\\frac 12{18 \\choose 9}+\\frac 12\\cdot 2^{18} \\\\ &=\\frac 12 \\cdot \\frac{18!}{9! \\cdot 9!} + 2^{17} \\end{align}\nWell, there are $2^{18}$ subsets all in all. Of these, $\\dfrac{18!}{9!\\cdot 9!}$ have exactly $9$ elements. The rest either have more than $9$ or fewer than $9$. Now, by symmetry, the number of subsets with more than $9$ elements equals the number of subsets with fewer than $9$ (pair each subset with its complement). Thus the number of subsets with more than $9$ elements is $\\dfrac{1}{2} \\times \\left(2^{18} - \\dfrac{18!}{9!\\cdot 9!}\\right)$. Adding the number of subsets with exactly $9$ elements gives the desired result.", "date": "2019-06-16 05:39:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1350084/determine-the-number-of-subsets", "openwebmath_score": 0.9858514070510864, "openwebmath_perplexity": 81.7231050512848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9921841102862671, "lm_q2_score": 0.8774767954920548, "lm_q1q2_score": 0.8706185336321292}}
{"url": "https://math.stackexchange.com/questions/3328096/cyclic-quadrilateral-and-trapezoid", "text": "A circle with diameter the minor base $$CD$$ of a trapezium $$ABCD$$ intersects its diagonals $$AC$$ and $$BD$$ in, respectively, their midpoints $$M$$ and $$N$$. The lines $$DM$$ and $$CN$$ intersect in $$P$$ and $$AC$$ and $$BD$$ intersect in $$H$$. Show that $$AD=CD=BC$$ and $$HP \\perp AB$$.\n\n$$DMNC$$ is a cyclic quadrilateral and $$CD||MN$$, thus $$DMNC$$ is an isosceles trapezoid. Therefore, $$CM=DN$$ and $$AC=BD$$. Now I am trying to show $$AD=CD$$. $$\\angle DCM$$ is inscribed and it's equal to $$\\angle DNM$$ but I don't see how to compare it with $$\\angle CAD$$. How is this done? For the second part of the problem, I tried to show that $$HO$$ passes through $$P$$ ($$HO \\perp CD$$ because $$\\triangle CDH$$ is isosceles and if we show $$P \\in HO$$ we are done).\n\nSince $$DM\\perp AC$$ and $$M$$ is a midpoint of $$AC$$, we obtain $$AD=DC$$.\nSince $$DMNC$$ is an isosceles trapezoid, we obtain: $$\\measuredangle PMN=\\measuredangle PDC=\\measuredangle PCD=\\measuredangle PNM,$$ which gives $$PM=PN.$$ Also, $$\\Delta MDN\\cong\\Delta NCM,$$ which gives $$\\measuredangle HMN=\\measuredangle HNM,$$ which gives $$HM=HN,$$ which says that $$PMHN$$ is a kite, which says $$PH\\perp MN.$$ About a kite see here: https://en.wikipedia.org/wiki/Kite_(geometry)\n\u2022 Yes. Thank you! $DM$ is the perpendicular bisector of $AC$. \u2013\u00a0Andrew Rogers Aug 19 '19 at 16:34\n\u2022 Now I am trying to show that $HP \\perp AB$ $(H = AC \\cap BD)$. Is this true for all cases, because I can't prove it? \u2013\u00a0Andrew Rogers Aug 19 '19 at 16:52\n\u2022 @Andrew Rogers Yes it's true because $HM=HN$ and $PM=PN.$ It's true if $AD=BC,$ which we got. \u2013\u00a0Michael Rozenberg Aug 19 '19 at 16:54\n\u2022 I am not sure I understand why it's true iff $HM=HN$ and $PM=PN$. \u2013\u00a0Andrew Rogers Aug 19 '19 at 17:01", "date": "2020-01-17 22:12:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3328096/cyclic-quadrilateral-and-trapezoid", "openwebmath_score": 0.6811006665229797, "openwebmath_perplexity": 112.80047893470608, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.990291523518526, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8706115964969309}}
{"url": "http://math.stackexchange.com/questions/60021/whats-the-probability-that-a-sequence-of-coin-flips-never-has-twice-as-many-hea", "text": "What's the probability that a sequence of coin flips never has twice as many heads as tails?\n\nI gave my friend this problem as a brainteaser; while her attempted solution didn't work, it raised an interesting question.\n\nI flip a fair coin repeatedly and record the results. I stop as soon as the number of heads is equal to twice the number of tails (for example, I will stop after seeing HHT or THTHHH or TTTHHHHHH). What's the probability that I never stop?\n\nI've tried to just compute the answer directly, but the terms got ugly pretty quickly. I'm hoping for a hint towards a slick solution, but I will keep trying to brute force an answer in the meantime.\n\n-\nHopefully whatever method resolves this applies to other ratios than $2$. \u2013\u00a0 anon Aug 27 '11 at 3:32\n@anon, good idea... See my answer. \u2013\u00a0 Did Aug 27 '11 at 13:53\n@anon: I've generalized my answer up to an infinite series expression as well. \u2013\u00a0 Mike Spivey Aug 27 '11 at 21:00\nSee \"a coin toss question\" for the similar problem of flipping until achieving the same number of heads as tails. \u2013\u00a0 Mike Spivey Aug 31 '11 at 20:55\n@Elliot I'd like you remind you that this question doesn't currently have an accepted answer. \u2013\u00a0 Git Gud Jun 16 '13 at 9:36\n\nThe game stops with probability $u=\\frac34(3-\\sqrt5)=0.572949017$.\n\nSee the end of the post for generalizations of this result, first to every asymmetric heads-or-tails games (Edit 1), and then to every integer ratio (Edit 2).\n\nTo prove this, consider the random walk which goes two steps to the right each time a tail occurs and one step to the left each time a head occurs. Then the number of tails is the double of the number of heads each time the walk is back at its starting point (and only then). In other words, the probability that the game never stops is $1-u$ where $u=P_0(\\text{hits}\\ 0)$ for the random walk with equiprobable steps $+2$ and $-1$.\n\nThe classical one-step analysis of hitting times for Markov chains yields $2u=v_1+w_2$ where, for every positive $k$, $v_k=P_{-k}(\\text{hits}\\ 0)$ and $w_k=P_{k}(\\text{hits}\\ 0)$. We first evaluate $w_2$ then $v_1$.\n\nThe $(w_k)$ part is easy: the only steps to the left are $-1$ steps hence to hit $0$ starting from $k\\ge1$, the walk must first hit $k-1$ starting from $k$, then hit $k-2$ starting from $k-1$, and so on. These $k$ events are equiprobable hence $w_k=(w_1)^k$. Another one-step analysis, this time for the walk starting from $1$, yields $$2w_1=1+w_3=1+(w_1)^3$$ hence $w_k=w^k$ where $w$ solves $w^3-2w+1=0$. Since $w\\ne1$, $w^2+w=1$ and since $w<1$, $w=\\frac12(\\sqrt5-1)$.\n\nLet us consider the $(v_k)$ part. The random walk has a drift to the right hence its position converges to $+\\infty$ almost surely. Let $k+R$ denote the first position visited on the right of the starting point $k$. Then $R\\in\\{1,2\\}$ almost surely, the distribution of $R$ does not depend on $k$ because the dynamics is invariant by translations, and $$v_1=r+(1-r)w_1\\quad\\text{where}\\ r=P_{-1}(R=1).$$ Now, starting from $0$, $R=1$ implies that the first step is $-1$ hence $2r=P_{-1}(A)$ with $A=[\\text{hits}\\ 1 \\text{before}\\ 2]$. Consider $R'$ for the random walk starting at $-1$. If $R'=2$, $A$ occurs. If $R'=1$, the walk is back at position $0$ hence $A$ occurs with probability $r$. In other words, $2r=(1-r)+r^2$, that is, $r^2-3r+1=0$. Since $r<1$, $r=\\frac12(3-\\sqrt5)$ (hence $r=1-w$).\n\nPlugging these values of $w$ and $r$ into $v_1$ and $w_2$ yields the value of $u$.\n\nEdit 1 Every asymmetric random walk which performs elementary steps $+2$ with probability $p$ and $-1$ with probability $1-p$ is transient to $+\\infty$ as long as $p>\\frac13$ (and naturally, for every $p\\le\\frac13$ the walk hits $0$ with full probability). In this regime, one can compute the probability $u(p)$ to hit $0$. The result is the following.\n\nFor every $p$ in $(0,\\frac13)$, $u(p)=\\frac32\\left(2-p-\\sqrt{p(4-3p)}\\right).$\n\nNote that $u(p)\\to1$ when $p\\to\\frac13$ and $u(p)\\to0$ when $p\\to1$, as was to be expected.\n\nEdit 2 Coming back to symmetric heads-or-tails games, note that, for any fixed integer $N\\ge2$, the same techniques apply to compute the probability $u_N$ to reach $N$ times more tails than heads.\n\nOne gets $2u_N=E(w^{R_N-1})+w^N$ where $w$ is the unique solution in $(0,1)$ of the polynomial equation $2w=1+w^{1+N}$, and the random variable $R_N$ is almost surely in $\\{1,2,\\ldots,N\\}$. The distribution of $R_N$ is characterized by its generating function, which solves $$(1-(2-r_N)s)E(s^{R_N})=r_Ns-s^{N+1}\\quad\\text{with}\\quad r_N=P(R_N=1).$$ This is equivalent to a system of $N$ equations with unknowns the probabilities $P(R_N=k)$ for $k$ in $\\{1,2,\\ldots,N\\}$. One can deduce from this system that $r_N$ is the unique root $r<1$ of the polynomial $(2-r)^Nr=1$. One can then note that $r_N=w^N$ and that $E(w^{R_N})=\\dfrac{Nr_N}{2-r_N}$ hence some further simplifications yield finally the following general result.\n\nFor every $N\\ge2$, $u_N=\\frac12(N+1)r_N$ where $r_N<1$ solves the equation $(2-r)^Nr=1$.\n\n-\n+1: Nice work, Didier! (And for the side effect of showing me that the answer I got is correct, too.) \u2013\u00a0 Mike Spivey Aug 27 '11 at 14:56\n@Mike, thanks. To tell the truth, I DID compute the numerical value of my answer as soon as I got a formula, to check whether it coincided with the numerical value of your mysterious formula and I WAS relieved to see these coincided... \u2013\u00a0 Did Aug 27 '11 at 16:28\n+1, I had a feeling a Markov approach would pop up. Nice to see such a simple polynomial's root solve the general form. \u2013\u00a0 anon Aug 27 '11 at 21:24\n@anon, yes. This is no accident that the computations go through and that the algebra is nice, there is an underlying structure to all this, sometimes called cycles-and-weights Markov chains, which is a nifty extension of +/-1 random walks (I might add some references one day). \u2013\u00a0 Did Aug 27 '11 at 21:51\n\n(Update: The answer to the original question is that the probability of stopping is $\\frac{3}{4} \\left(3 - \\sqrt{5}\\right)$. See end of post for an infinite series expression in the general case.)\n\nLet $S(n)$ denote the number of ways to stop after seeing $n$ tails. Seeing $n$ tails means seeing $2n$ heads, so this would be stopping after $3n$ flips. Since there are $2^{3n}$ possible sequences in $3n$ flips, the probability of stopping is $\\sum_{n=1}^{\\infty} S(n)/8^n$.\n\nTo determine $S(n)$, we see that there are $\\binom{3n}{n}$ ways to choose which $n$ of $3n$ flips will be tails. However, this overcounts for $n > 1$, as we could have seen twice as many heads as tails for some $k < n$. Of these $\\binom{3n}{n}$ sequences, there are $S(k) \\binom{3n-3k}{n-k}$ sequences of $3n$ flips in which there are $k$ tails the first time we would see twice as many heads as tails, as any of the $S(k)$ sequences of $3k$ flips could be completed by choosing $n-k$ of the remaining $3n-3k$ flips to be tails. Thus $S(n)$ satisfies the recurrence $S(n) = \\binom{3n}{n} - \\sum_{k=1}^{n-1} \\binom{3n-3k}{n-k}S(k)$, with $S(1) = 3$.\n\nThe solution to this recurrence is $S(n) = \\frac{2}{3n-1} \\binom{3n}{n}.$ This can be verified easily, as substituting this expression into the recurrence yields a slight variation on Identity 5.62 in Concrete Mathematics (p. 202, 2nd ed.), namely, $$\\sum_k \\binom{tk+r}{k} \\binom{tn-tk+s}{n-k} \\frac{r}{tk+r} = \\binom{tn+r+s}{n},$$ with $t = 3$, $r = -1$, $s=0$.\n\nSo the probability of stopping is $$\\sum_{n=1}^{\\infty} \\binom{3n}{n} \\frac{2}{3n-1} \\frac{1}{8^n}.$$\n\nMathematica gives the closed form for this probability of stopping to be $$2 \\left(1 - \\cos\\left(\\frac{2}{3} \\arcsin \\frac{3 \\sqrt{3/2}}{4}\\right) \\right) \\approx 0.572949.$$\n\nAdded: The sum is hypergeometric and has a simpler representation. See Sasha's comments for why the sum yields this closed form solution and also why the answer is $$\\frac{3}{4} \\left(3 - \\sqrt{5}\\right) \\approx 0.572949.$$\n\nAdded 2: This answer is generalizable to other ratios $r$ up to the infinite series expression. For the general $r \\geq 2$ case, the argument above is easily adapted to produce the recurrence $S(n) = \\binom{(r+1)n}{n} - \\sum_{k=1}^{n-1} \\binom{(r+1)n-(r+1)k}{n-k}S(k)$, with $S(1) = r+1$. The solution to the recurrence is $S(n) = \\frac{r}{(r+1) n - 1} \\binom{(r+1) n}{n}$ and can be verified easily by using the binomial convolution formula given above. Thus, for the ratio $r$, the probability of stopping has the infinite series expression $$\\sum_{n=1}^{\\infty} \\binom{(r+1)n}{n} \\frac{r}{(r+1)n-1} \\frac{1}{2^{(r+1)n}}.$$ This can be expressed as a hypergeometric function, but I am not sure how to simplify it any further for general $r$ (and neither does Mathematica). It can also be expressed using the generalized binomial series discussed in Concrete Mathematics (p. 200, 2nd ed.), but I don't see how to simplify it further in that direction, either.\n\nAdded 3: In case anyone is interested, I found a combinatorial proof of the formula for $S(n)$. It works in the general $r$ case, too.\n\n-\nIt's late here, and I'm going to bed. Any comments, corrections, etc., will be dealt with in the morning. And if someone has an explanation for Mathematica's output I would love to see it! \u2013\u00a0 Mike Spivey Aug 27 '11 at 6:50\nWow, incredible (both the write up and the result itself)! However, I have a slight point of confusion--why is S(1) = 1, rather than 3? \u2013\u00a0 Elliott Aug 27 '11 at 7:48\nDoes anyone know (or perhaps have a useful reference book to find out from) the generating function of $a_n = \\binom{3n}{n}$? That should help in finding the closed form of this series. \u2013\u00a0 Ragib Zaman Aug 27 '11 at 9:17\nLet $u$ be you answer, then $\\cos(\\frac{2}{3}\\arcsin \\frac{3}{4}\\sqrt{\\frac{3}{2}}) = 1-\\frac{u}{2}$. Using $\\cos(3 x) = -3 \\cos(x) + 4 \\cos(x)^3$, and $\\cos( 2 \\arcsin \\frac{3}{4}\\sqrt{\\frac{3}{2}}) = -\\frac{11}{16}$, it follows $u$ satisfy a cubic with rots $u_1=\\frac{3}{2}$, $u_2 = \\frac{3}{4}( 3- \\sqrt{5})$ and $u_3 = \\frac{3}{4}(3+ \\sqrt{5})$. Root $u_2$ is the one corresponding to the original expression, by numerical comparison. \u2013\u00a0 Sasha Aug 27 '11 at 13:47\n@Ragib $\\sum_{k\\ge 0} \\binom{3 k}{k} z^k$ is a hypergeometric sum with answer $\\frac{3}{4 \\pi} F(\\frac{1}{3}, \\frac{2}{3} ; \\frac{1}{2} ; \\frac{9 z}{4} )$. This Gauss hypergeometric function admits simpler representation. \u2013\u00a0 Sasha Aug 27 '11 at 13:57\n\nHere's another solution, following Ross Millikan's hint:\n\nLet $n=H\u22122T$ (H=heads, T=tails). At each step, $n := n+1$ or $n := n-2$ with probability 1/2. The game starts with $n=0$ and stops when $n=0$.\n\nFor any iteration (after the initial), let $P(n)$ be the probability that the game eventually stops, given the current value of $n$ (and given that the event has not yet occurred). It's clear that $P(n)$ does not depend on time. Then the following recurrence holds:\n\n$$P(n)= \\frac{P(n+1)+P(n-2)}{2} \\;, \\;\\; n\\ne 0$$\n\nwith $P(0)=1$. We also know (do we?) that $P(-\\infty)=0$ (but we don't yet know $P(+\\infty)$)\n\nThe solution to this difference equation (I'll spare you the details, just the usual linear difference equation procedure, in the two regions, with the above boundary conditions) is given by:\n\n$$P(n)= \\left\\{ \\begin{array}{ll} \\phi^{-n} & \\mbox{if } n \\le 0 \\\\ 1 + B \\, [(-\\phi)^n -1] & \\mbox{if } n \\ge 0 \\end{array} \\right.$$\n\nwith $B= \\phi^2 (1-\\phi)$, $\\phi = (\\sqrt{5}-1)/2$\n\nNow, after the first step we have $n=1$ or $n=-2$ with equal probability, then the probability that the game eventually stops is\n\n$$\\frac{P(1)+P(-2)}{2}=\\frac{2 \\phi^2 -\\phi +1}{2} = 0.572949$$\n\nThe form of $P(n)$ is interesting:\n\nThis, for example, shows that the probability of stop is strongly dependent on the first coin toss (less than 40% if tail, more than 75% if head).\n\nThis procedure also seems directly generalizable, either to asymmetric coins or other integer ratios.\n\nAdded: Here goes the details for solving the recursion:\n\nWe postulate a solution of the form $P(n)= r^n$ and replacing in the recursion $2 P(n) - P(n+1) - P(n-2) = 0$ we get $$2 \\; r^2 - r^3 - 1 = 0$$ The root $r_1=1$ comes immediately, and then the others: $r_2 = - \\phi$, $r_3 = 1/\\phi$. Then the general solution is given by $A + B (-\\phi)^n + C \\phi^{-n}$, for some $A,B,C$, in each zone.\n\nIn the zone $n \\le 0$: we have $P(0)=1$ and $P(-\\infty)=0$. As $\\phi \\approx 0.618 < 1$, this implies $A=0$, $B=0$,$C=1$ Hence\n\n$$P(n) = \\phi^{-n} \\hspace{10px} n \\le 0$$\n\nIn the zone $n \\ge 0$: we have $P(0)=1$, but we don't know $P(\\infty)$. We do know it's bounded, so $C=0$, and $A=1-B$, so\n\n$$P(n) = 1 + B \\, [(-\\phi)^n -1] \\hspace{10px} n \\ge 0$$\n\nTo get rid of the remaining degree of freedom, we write the recursion for $n=1$, and replace the value of $P(-1)$ for the previous solution:\n\n$$2 \\, P(1) = P(2) + P(-1)$$ $$2 \\, (1 + B \\, [-\\phi -1]) = 1 + B \\, [(-\\phi)^2 -1] + \\phi$$\n\nFrom this, we obtain $B= \\phi^2 (1-\\phi)$.\n\n-\nThis solution is beautiful! One thing I don't understand is, how did you set boundary conditions for the region n >= 0? \u2013\u00a0 Elliott Aug 29 '11 at 16:34\n@Elliot: You first solve for $n \\le 0$, then plugging it in the recursion for $n=1$ you resolve the remaining degree of freedom. Tell me if this is not clear, and I'll write the details. \u2013\u00a0 leonbloy Aug 29 '11 at 17:44\nSo... in the end, you know that P(-infty)=0? \u2013\u00a0 Did Aug 31 '11 at 21:01\nWell, we \"knew\" (actually assumed) that from the beginning, it was $P(+\\infty)$ the unknown. It's true (are you pointing to this?) that $P(-\\infty)=0$ is intuitive and true but a rigorous proof is lacking here. \u2013\u00a0 leonbloy Aug 31 '11 at 21:19\nJust pointing to the fact that, when P(-infty) first appear, you feel necessary to signal that the relation P(-infty)=0 might need a proof. But when you need the value of P(-infty) later on, you state it without further ado. Hence, reading your post, I was wondering if I had missed, somewhere inbetween these two appearances, a proof that P(-infty)=0. (Please use @.) \u2013\u00a0 Did Sep 1 '11 at 13:22\n\nI would suggest $1-\\phi=1-\\frac{\\sqrt{5}-1}{2}$. If define $P(n)$ as the probability that if we have an excess of tails of $n$ we will eventually have twice the heads as tails, the recurrence is $$P(n)=\\frac{P(n+1)+P(n-2)}{2}$$ with boundary conditions $$P(n)=\\begin {cases}1 &n\\gt 0 \\\\ 0 &n=-\\infty \\end {cases}$$ The recurrence has solution $P(n)=\\phi ^n$\n\n-\nSorry, I'm not sure I follow. Can you please explain what P(n) represents in slightly more detail? \u2013\u00a0 Elliott Aug 27 '11 at 6:18\n+1 Nice answer, though cryptic (and slightly incorrect). To clarify: Let $n(t)=H-2T$ (H=heads, T=tails), say we are interest in the event $n(t)>0$ for some $t$. Let $P(n)$ (for $n\\le0$) be the probability of that event, given the value $n$ that at that time, and that the event has not ocurred before; further, let $P(1)=1$. Then, the recursion is correct, and so is the solution (actually I think its $p(n)=\\phi^{1-n}$, but anyway $P(0)= 0.618033989$). But... (continued) \u2013\u00a0 leonbloy Aug 27 '11 at 15:17\nThe problem is that the original game stops when $H=2T$ exactly, and this event is not equivalent to $H>2T$ (for example, the sequence $HHHTHHHTHHT...$ does not make the game stop. I wonder if this can be corrected, I liked this approach. \u2013\u00a0 leonbloy Aug 27 '11 at 15:22\n@leonbloy: Good elaboration. Thanks. I was thinking that if we have excess heads then with probability 1 we get back to a 2:1 ratio. But you are right we can skip over it. A more telling (to me) sequence that fails is $HHHTTTHTHTHTHT...$ \u2013\u00a0 Ross Millikan Aug 27 '11 at 15:28\nThe approach was indeed fixable, I posted the derivation in my own answer - I hope this it's ok. \u2013\u00a0 leonbloy Aug 29 '11 at 0:15", "date": "2015-04-26 10:36:21", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/60021/whats-the-probability-that-a-sequence-of-coin-flips-never-has-twice-as-many-hea", "openwebmath_score": 0.9498146176338196, "openwebmath_perplexity": 286.28621061275317, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915223724212, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.870611589217961}}
{"url": "https://math.stackexchange.com/questions/331231/what-quantifies-as-a-rigorous-proof", "text": "# What quantifies as a rigorous proof?\n\nOkay I have been thinking about this common combinatorial identity. $$\\sum_{r=0}^{n} \\binom{n}{r} = 2^n.$$ It is simple to prove this by induction, but it requires some annoying algebraic manipulation which involves other combinatorial identities. What about if we took an approach that takes into factor a specific rephrase of the problem?\n\nMy approach is this. Consider a combination lock with n number of on-off switches. To open it require turning on the correct switches. How many combinations are there? clearly, the first switch can be on or off, second on and off etc. so the combintions are $2^n$\n\nWe look at it from a different angle. How many ways are there to pick one on switch. $\\dbinom{n}{1}$. what about picking 2 switches? $\\dbinom{n}{2}$... and so on. Since the number of combinations are supposed to be the same,it gives us the identity.\n\nNow i know this explanation is not rigorous, but can someone point out exactly to me exactly which part of it it fails to be so?\n\nI have also encountered this problem in man other areas of mathematics when I try to prove something, I used a specific problem to do so and thus I doubt the validity of the proof. I can give more examples if needed\n\n\u2022 It's a combinatorial proof. Rigorous enough for me, but I know teachers who don't like proofs without lots of algebra. \u2013\u00a0Jean-Claude Arbaut Mar 15 '13 at 14:05\n\u2022 Very clever proof. To me it is the right proof. Rigor is an ambiguous concept, but it's certainly not the case that a rigorous proof of an identity needs be done algebraically. Your argument explains why the identity is true. The algebra does not give any insight. \u2013\u00a0Cheerful Parsnip Mar 15 '13 at 14:09\n\u2022 I have edited your question, employing the $\\LaTeX$ notation that is used on MSE. \u2013\u00a0Andreas Caranti Mar 15 '13 at 14:14\n\u2022 Better use the standard $\\binom{n}{k}$ notation... \u2013\u00a0vonbrand Mar 15 '13 at 14:14\n\u2022 It\u2019s a variant of the standard combinatorial proof: $2^n$ is the number of subsets of $[n]=\\{1,\\dots,n\\}$, and $\\binom{n}k$ is the number of subsets of $[n]$ of size $k$, so $2^n=\\sum_k\\binom{n}k$. As such it is, as others have said, just fine. And while one ought to be able to carry out the details of the proof by induction, combinatorial arguments like this one generally give much more insight into why the result is true. \u2013\u00a0Brian M. Scott Mar 15 '13 at 14:46\n\nYour approach is an often-used (and perfectly valid) proof technique. You should be aware that some instructors don't like it, but it certainly does the job.\n\n\u2022 An instructor who doesn\u2019t accept combinatorial arguments ought not to be teaching the subject. \u2013\u00a0Brian M. Scott Mar 15 '13 at 14:44\n\u2022 @Brian M. Scott: surely it depends on whether the goal is to teach the identity, or whether it is to teach students how to write proofs by induction. I would usually take a car if I need to travel 10 miles, but not if the point is to train for a marathon. \u2013\u00a0Carl Mummert Mar 15 '13 at 15:09\n\u2022 @Carl: If the goal is to teach students how to write proofs by induction, the instructor should damned well come up with problems for which induction is the best approach. And if even then a student comes up with a different approach, the instructor should be pleased. \u2013\u00a0Brian M. Scott Mar 15 '13 at 15:11\n\u2022 @Carl: My point is that the directions should not begin so. I consider it a failure on the part of the instructor if he or she has to specify a technique in order to give students practice in applying it. \u2013\u00a0Brian M. Scott Mar 15 '13 at 15:26\n\u2022 I strongly agree with Brian here. Questions that prescribe techniques are either lazy pedagogy or bad curriculum design. If technique $Z$\u2014say, induction\u2014is a useful technique, then it is because there is some problem $Y$ such that $Z$ is superior to all other techniques for solving $Y$. If all such $Y$ are outside the scope of the class, then $Z$ is outside the scope of the class too. If, on the other hand, there is some $Y$ that is in the scope of the class, it is the instructor's job to find it and present it to the students, as an instructive example. \u2013\u00a0MJD Mar 15 '13 at 15:31\n\nYou can make such an argument rigorous. When things start to get more sophisticated, it often helps immensely to give more detail than one usually sees in this style of proof.\n\nLet $S$ be a set with $n$ elements. Let $P(S)$ be its powerset -- that is, the set of all subsets of $S$. Let $P_i(S)$ be the set of all subsets of $S$ with exactly $i$ elements.\n\nYour proof is then the invocation of four facts:\n\n\u2022 The well-known formula for the number of elements in $P(S)$\n\u2022 The well-known formula for the number of elements in $P_i(S)$\n\u2022 The observation that $P(S)$ is the disjoint union of all of the $P_i(S)$\n\u2022 The relationship between the number of elements in a sequence of sets and the number of elements in their disjoint union", "date": "2021-04-14 16:23:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/331231/what-quantifies-as-a-rigorous-proof", "openwebmath_score": 0.7309045195579529, "openwebmath_perplexity": 239.37871532594662, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109489630493, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8706115699944141}}
{"url": "http://mymathforum.com/probability-statistics/340911-what-probability-getting-same-number-twice-2.html", "text": "My Math Forum What is the probability of getting the same number twice?\n User Name Remember Me? Password\n\n Probability and Statistics Basic Probability and Statistics Math Forum\n\n June 16th, 2017, 11:34 AM #11 Math Team \u00a0 \u00a0 Joined: Jul 2013 From: \u0915\u093e\u0920\u092e\u093e\u0921\u094c\u0902, \u0928\u0947\u092a\u093e\u0932 Posts: 872 Thanks: 60 Math Focus: \u0938\u093e\u092e\u093e\u0928\u094d\u092f \u0917\u0923\u093f\u0924 i would think it this way: Since I have 24 draws, I would take 23 different numbers in my first 23 draws and 1 among those 23 numbers in my 24th draw. In 1st draw we can take any number but in the following draws we cannot take numbers which are already drawn, thus the numerator of the probabilities are decreasing by 1 in each of the following draw. Thus the probability of getting 23 different numbers in the first 23 draw is: $\\displaystyle P1 = \\frac{255}{255}\\times \\frac{254}{255}\\times \\frac{253}{255}\\times ......$ till 23 steps $\\displaystyle P1 = \\frac{255!}{255^{23}\\times 232!}$ Then for the 24th draw, we have to draw one of the numbers which we have already drawn. There are 23 possible numbers which we can draw out of 255 numbers. Thus the probability of having such number in 24th draw is: $\\displaystyle P2=\\frac{23}{255}$ Hence, the total probability is: $\\displaystyle P=P1 \\times P2$ $\\displaystyle P= \\frac{255!\\times 23}{255^{24}\\times 232!}$ $\\displaystyle P\\approx 0.0324$ Thanks from JeffM1\n June 16th, 2017, 11:57 AM #12 Senior Member \u00a0 Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics I'd like to present an alternative approach to the problem that could possibly shed light on that number. Consider the Markov chain with states 0, 1, ... 7, where state 0 is the initial state where no numbers have been chosen, states i, $\\displaystyle 1 \\leq i \\leq 6$ denote the state where i distinct numbers have been chosen and no repeats have appeared so far, and 7 denote the state where repeats have appeared. Obviously, the transition matrix is as follows: $\\displaystyle \\mathbf{P} = \\begin{bmatrix} 0 &1 &0 & 0 & 0 & 0 & 0 & 0\\\\ 0 &0 & \\frac{5}{6} & 0 & 0& 0& 0& \\frac{1}{6}\\\\ 0 & 0 & 0 & \\frac{2}{3} & 0 & 0 &0 & \\frac{1}{3}\\\\ 0 & 0 &0 &0 & \\frac{1}{2}& 0 & 0& \\frac{1}{2}\\\\ 0 & 0 & 0& 0 & 0 &\\frac{1}{3} & 0 & \\frac{2}{3}\\\\ 0 & 0 & 0 & 0 & 0 & 0 & \\frac{1}{6} & \\frac{5}{6}\\\\ 0& 0 & 0 & 0 & 0 & 0 &0 &1 \\\\ 0&0 & 0 & 0 & 0 & 0 & 0 & 1 \\end{bmatrix}$ and the initial distribution is $\\displaystyle \\mathbf{e}_1$. So all we have to do is to calculate $\\displaystyle \\mathbf{e}_1\\mathbf{P}^3 = (0,0,0,\\frac{5}{9},0,0,0,\\frac{4}{9})^T$, verifying Jeff's answer. Using a well-known theorem on absorbing states and some concepts in elementary probability (in particular the probability generating function), we can even do other cool things like determining how many numbers we need to draw on average to get a repeat, but I digress. So we can immediately apply this to the initial question: $\\displaystyle \\mathbf{P} =\\begin{bmatrix} 0 & 1 & 0 & ... & 0 & 0\\\\ 0 & 0 & 254/255 & ... & 0& 1/255\\\\ 0 & 0 & 0 & \\ddots & 0 & \\vdots\\\\ 0& 0 & 0 & ... & 0 & 1\\\\ 0 & 0 & 0 & ...& 0 &1 \\end{bmatrix}$ MATLAB tells us that 25th and last entries of $\\displaystyle \\mathbf{e}_1 \\mathbf{P}^{24}$ are 0.32719 and 0.67281 respectively, again verifying Jeff's claims. Thanks from JeffM1 Last edited by 123qwerty; June 16th, 2017 at 11:59 AM.\n June 16th, 2017, 01:00 PM #13 Math Team \u00a0 Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 I stick to my guns: the probability is ~.39\nJune 16th, 2017, 01:24 PM \u00a0 #14\nSenior Member\n\nJoined: May 2016\nFrom: USA\n\nPosts: 825\nThanks: 335\n\nQuote:\n Originally Posted by Denis Well, whatever... IF, IF, IF the 24 draws produce: 1 pair no other number matching that pair 22 remaining are ALL different then probability = ~.39\nI agree that that is the apparent question originally asked.\n\nHow many ways can I draw 24 cards with replacement from a set of 255 cards consecutively numbered 1 through 255?\n\nAnswer: $255^{24}.$\n\nDo you agree?\n\nHow many ways can I draw 23 differently numbered cards from that same set with replacement?\n\nAnswer: $\\displaystyle \\prod_{j=1}^{23}(256 - j).$\n\nDo you agree?\n\nHow many ways can I draw one of 23 specific numbers from that set?\n\nAnswer: $23.$\n\nDo you agree?\n\nSo the probability of drawing 23 different cards and 1 duplicate card in 24 draws with replacement should be\n\n$\\dfrac{23 * \\displaystyle \\prod_{j=1}^{23}(256 - j)}{255^{24}} \\approx 0.0324.$\n\nI can't get to your result.\n\nJune 16th, 2017, 03:09 PM \u00a0 #15\nMath Team\n\nJoined: Jul 2013\nFrom: \u0915\u093e\u0920\u092e\u093e\u0921\u094c\u0902, \u0928\u0947\u092a\u093e\u0932\n\nPosts: 872\nThanks: 60\n\nMath Focus: \u0938\u093e\u092e\u093e\u0928\u094d\u092f \u0917\u0923\u093f\u0924\nQuote:\n Originally Posted by JeffM1 I agree that that is the apparent question originally asked. How many ways can I draw 24 cards with replacement from a set of 255 cards consecutively numbered 1 through 255? Answer: $255^{24}.$ Do you agree? How many ways can I draw 23 differently numbered cards from that same set with replacement? Answer: $\\displaystyle \\prod_{j=1}^{23}(256 - j).$ Do you agree? How many ways can I draw one of 23 specific numbers from that set? Answer: $23.$ Do you agree? So the probability of drawing 23 different cards and 1 duplicate card in 24 draws with replacement should be $\\dfrac{23 * \\displaystyle \\prod_{j=1}^{23}(256 - j)}{255^{24}} \\approx 0.0324.$ I can't get to your result.\ni absolutely agree with you.\n\n June 16th, 2017, 04:04 PM #16 Senior Member \u00a0 Joined: May 2016 From: USA Posts: 825 Thanks: 335 I am impressed by 123qwerty's translation into Markov processes even though I have almost no recollection of how to work with them. It certainly looks to be more powerful than my way. Obviously my way and mathematician's way are mathematically identical although I flatter myself that my explanation may be slightly more intuitive. Of course I could not give an explanation in Nepali at all. Denis, mon ami, if you still are not convinced, I suggest that you try to give the reasoning that leads to your result.\n June 16th, 2017, 04:33 PM #17 Math Team \u00a0 Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 I was kinda kidding...didn't want to admit I had badly programmed my simulation runs... Yes yes...I'm heading for the corner...\nJune 16th, 2017, 05:18 PM \u00a0 #18\nSenior Member\n\nJoined: May 2016\nFrom: USA\n\nPosts: 825\nThanks: 335\n\nQuote:\n Originally Posted by Denis I was kinda kidding...didn't want to admit I had badly programmed my simulation runs... Yes yes...I'm heading for the corner...\nI probably will be joining you there soon enough.\n\n June 18th, 2017, 10:27 AM #19 Math Team \u00a0 Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 To: Probability Prof. Jeff From: student Denis If problem changed to 1 to 9 (instead of 1 to 255) and to 4 picks (instead of 24), everything else remaining same, then: total combos = 9^4 combos with 1 pair + 2 diff. numbers: 3024 probability: 3024/9^4 = .460905... OR: 1 * 8/9 * 7/9 * 3/9 * 2 = .460905... Do I visit the corner again?\nJune 19th, 2017, 10:58 PM \u00a0 #20\nSenior Member\n\nJoined: Jul 2012\nFrom: DFW Area\n\nPosts: 603\nThanks: 83\n\nMath Focus: Electrical Engineering Applications\nRespectfully to All,\n\nI think that the answer to the problem as stated (exactly one matching pair with no repeats) is ~0.38924, or precisely:\n\n$\\displaystyle \\large \\frac{255!}{232! \\cdot 255^{24}} \\cdot \\frac{24!}{2! \\cdot 22!}$\n\nNotice that this is exactly 12 times the agreed upon value, given approximately as 0.0324, because:\n\n$\\displaystyle \\large \\frac{24!}{2! \\cdot 22!} = 12 \\cdot 23$\n\nPlease permit me to:\n1) Show an inconsistency in the analysis of the agreed upon result.\n2) Explain my concept of the solution.\n3) Demonstrate the concept using what I think is the simplest example.\n4) Provide a little background as to my interest in this.\n\nSo, for 1):\n\nQuote:\n How many ways can I draw 23 differently numbered cards from that same set with replacement? Answer: $\\displaystyle \\large \\prod_{j=1}^{23}(256 - j).$ Do you agree? How many ways can I draw one of 23 specific numbers from that set? Answer: 23.\nSo we have, I think:\n\n$\\displaystyle 255 \\cdot 254 \\cdot 253 \\cdot \\ldots \\cdot 234 \\cdot 233 \\cdot 23$\n\nIf that is the case, then in the simplified example given earlier in the thread:\n\nQuote:\n Two numbers the same (be careful: the different number can come first, second or third) = $\\displaystyle 6\u22175\u22173=90.$\nWhy isn't that 3 a 2?\n\nBy the logic given in the first quote above, we can match 2 specific numbers here, not 3. But as is cautioned, it is not the number of numbers that is important, it is the number of places that is important. This appears to be inconsistent to me.\n\nSo for 2):\n\nI would word it differently: It is the number of combinations of 3 numbers taken two at a time (the two types are numbers that match and the numbers that do not match):\n\n$\\displaystyle \\large \\frac{3!}{2! \\cdot 1!}=3$\n\nSo to explain my solution, for the first number we have our choice of 255. Then let's match it 'right off the bat' with the second number so we have a choice of 1, then we can't match any more so we count down from there:\n\n$\\displaystyle \\large 255 \\cdot 1 \\cdot 254 \\cdot 253 \\cdot \\ldots \\cdot 234 \\cdot 233$\n\nObviously, we still have two types of numbers, the matching ones, and the non-matching ones, which gives the multiplication by:\n\n$\\displaystyle \\large \\frac{24!}{2! \\cdot 22!} = 12 \\cdot 23$\n\nfor the various combinations.\n\nFor 3) let's try to simplify things as much as we can. Let's have 3 symbols and take 4 of the symbols and calculate how many combinations exactly two of the symbols appear in.\n\nI think that it is:\n\n$\\displaystyle \\large 3 \\cdot 1 \\cdot 2 \\cdot 1 \\cdot \\frac{4!}{2! \\cdot 2!}=36$\n\nIf I am not mistaken, by the method of calculation as described in the thread the number is:\n\n$\\displaystyle \\large 3 \\cdot 2 \\cdot 1 \\cdot 3=18$\n\nSo let's just list them using symbols 0,1,2. Taking 2 as the repeated symbol, in ascending order we have:\n\n0122\n0212\n0221\n1022\n1202\n1220\n2012\n2021\n2102\n2120\n2201\n2210\n\nSo we have 12 combinations with the 2 repeated. We can have the 0 and 1 repeated as well so we have 3*12=36 (not 18 ).\n\nFor 4), I became interested in this type of thing because at my place of employment, in order to log in remotely, we have to enter a code that is chosen at random, 6 each of the digits 0-9.\n\nI noticed that it was relatively rare to have no repeated digits so I tried to figure out the odds of getting exactly two repeating digits (among other things). I wrote programs in Ruby (which I will gladly share but I doubt anyone uses Ruby here) to run random numbers and looking at the percentage, and also to calculate exactly by running through all of the combinations and looking at the percentage.\n\nI am pretty sure that I used reasoning similar to that given earlier in the thread but the calculated results did not match the programs' outputs until I used the methods given above.\n\nCaveat: My 'exactly two matching numbers' routine may be flawed but for base 3 with 4 symbols taken, it does give 36, in agreement with the result above. For base 10 and 6 numbers taken it gives 453600 which agrees with:\n\n$\\displaystyle \\large 10 \\cdot 1 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot \\frac{6!}{2! \\cdot 4!}$\n\nMy random number calculating program calculates ~0.39 for the stated problem, in agreement with the calculation (and Denis' initial result, which I think is correct).\n\n Tags number, probability\n\n Thread Tools Display Modes Linear Mode\n\n Similar Threads Thread Thread Starter Forum Replies Last Post JDSimpkins Probability and Statistics 9 April 27th, 2015 07:01 AM meghraj Probability and Statistics 9 November 8th, 2012 07:00 PM eliotjames Advanced Statistics 3 October 7th, 2012 04:02 PM ershi Number Theory 113 August 18th, 2012 05:45 AM hoyy1kolko Probability and Statistics 3 June 4th, 2011 05:38 PM\n\n Contact - Home - Forums - Cryptocurrency Forum - Top", "date": "2017-11-22 11:11:29", "meta": {"domain": "mymathforum.com", "url": "http://mymathforum.com/probability-statistics/340911-what-probability-getting-same-number-twice-2.html", "openwebmath_score": 0.807102620601654, "openwebmath_perplexity": 1050.170538587273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109534209825, "lm_q2_score": 0.8840392817460332, "lm_q1q2_score": 0.8706115679179115}}
{"url": "https://brainbalancemathematics.com/the-probability-that-a-ticketless-traveler-is-caught-during-a-trip-is-0-1-if-the-traveler-makes-4-trips-the-probability-that-he-she-will-be-caught-during-at-least-one-of-the-trips-is/", "text": "Friday, February 3, 2023\nHomeCSIRUGC NETThe probability that a ticketless traveler is caught during a trip is...\n\n# The probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips , the probability that he/she will be caught during at least one of the trips is:\n\n## Question:\n\nThe probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips , the probability that he/she will be caught during at least one of the trips is:\n1. $1-(0.9)^4$\n2. $(1-0.9)^4$\n3. $1-(1-0.9)^4$\n4. $(0.9)^4$\n\nThe probability that a ticketless traveler is caught during a trip is 0.1.\n$therefore p=0.1$ and $q=1-p=1-0.1=0.9$\nSince, traveler makes 4 trips, $n=4$.\nLet X be the random variable that the traveler is caught during a trip.\nHence, $X leadsto B(n=4, p=0.1)$ and $X=0, 1, 2, 3, 4$.\n$therefore P[X=x]=$ $^4C_x$ $(0.1)^x(0.9)^{4-x}$\n\n$therefore$ the probability that\u00a0he/she will be caught during at least one of the trips\n=1-{probability that\u00a0he/she will be caught during at most one of the trips}.\n$=1- P[X=0]$\n$=1-$\u00a0$^4C_0$\u00a0$(0.1)^0(0.9)^4$\n$=1-(0.9)^4$\nRELATED ARTICLES\n\n### The product of 2 numbers is 1575 and their quotient is $frac{9}{7}$. Then the sum of the numbers is\n\n1. This is rightly named as the age of traveler-centricity and with the evolution of the new era of personalized travel; it is leading to research and development of a host of new so-called intelligent services. The command-and-control perspectives of traveling have changed a lot from the past and the focus has shifted more on the traveler and the productivity of each trip. It has become essential to maintain that the travelers have the greatest return on investment on each trip. Travel 101\n\n### Dimension theorem of a quotient space: If W be a subspace of a finite dimensional vector space V over \u200b( mathbb{R} )\u200b then \u200b(...\n\n$${}$$", "date": "2023-02-03 11:10:38", "meta": {"domain": "brainbalancemathematics.com", "url": "https://brainbalancemathematics.com/the-probability-that-a-ticketless-traveler-is-caught-during-a-trip-is-0-1-if-the-traveler-makes-4-trips-the-probability-that-he-she-will-be-caught-during-at-least-one-of-the-trips-is/", "openwebmath_score": 0.6206298470497131, "openwebmath_perplexity": 889.0210695918074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810954312569, "lm_q2_score": 0.8840392756357326, "lm_q1q2_score": 0.8706115626886181}}
{"url": "https://www.physicsforums.com/threads/probabilities-of-archers-a-and-b-hitting-the-bulls-eye.663461/", "text": "# Probabilities of archers A and B hitting the bull's eye\n\n1. Jan 10, 2013\n\n### jinhuit95\n\n1. The problem statement, all variables and given/known data\nTwo archers A and B take turns to shoot, with archer A taking the first shot. The probabilities of archers A and B hitting the bull's eye is 1/6 and 1/5 respectively. Show that the probability of archer A hitting the bull's eye first is 1/2.\n\n2. Relevant equations\n\n3. The attempt at a solution\nWell, i thought about drawing tree diagram but the problem is I have no idea when to stop because I don't know when he will hit.\n\n2. Jan 10, 2013\n\n### dx\n\nRe: Probability\n\nFirst, what are the possible ways for this to happen?\n\nA could hit the bulls eye on his first attempt. Or, A misses, B misses and then A hits the bulls eye. Or, A misses, B misses, A misses, B misses and then A hits the bulls eye... and so on.\n\nFind the probability for each of these ways (for A to hit the bull's eye first), and add them up.\n\nYou will need to use the following:\n\n1 + a2 + a3 + ... = 1/(1-a) for |a| < 1\n\n3. Jan 10, 2013\n\n### jinhuit95\n\nRe: Probability\n\nthe problem is, if he hits on the first attempt that's 1/6. so If i go on, it will be 1/6 + ( 5/6 * 4/5 * 5/6 * 4/5 * 5/6 * 4/5....) I have no idea when to stop. Moreover, the product of probability he misses will be come smaller and smaller and that number + 1/6 will not give me half i think.\n\n4. Jan 10, 2013\n\n### dx\n\nRe: Probability\n\nLet's take the case when A gets it on his second try: A misses, B misses, and then A hits. What is the probability for this?\n\n5. Jan 10, 2013\n\n### jinhuit95\n\nRe: Probability\n\nthat will be 5/6 * 4/5 * 1/6 right?\n\n6. Jan 10, 2013\n\n### jinhuit95\n\nRe: Probability\n\nI'm sorry but what's this?\n'1 + a2 + a3 + ... = 1/(1-a) for |a| < 1'\n\n7. Jan 10, 2013\n\n### dx\n\nRe: Probability\n\nYes, correct.\n\nNow what is the probability that A gets it on his 3rd try? (A misses, B misses, A misses, B misses, A hits bull's eye)\n\nIgnore that for the moment. We'll get to it soon.\n\n8. Jan 10, 2013\n\n### jinhuit95\n\nRe: Probability\n\n5/6* 4/5 * 5/6 * 4/5 * 1/6\n\n9. Jan 10, 2013\n\n### dx\n\nRe: Probability\n\nCorrect.\n\nSo do you see the pattern?\n\nFirst try: 1/6\n\nSecond try: (1/6)(4/6)\n\nThird try: (1/6)(4/6)2\n\nFourth try: (1/6)(4/6)3\n\n...\n\nWhat is the probability that he gets in on the n'th try? Each time you are simply multiplying the previous one by (5/6)(4/5) = 4/6\n\n10. Jan 10, 2013\n\n### jinhuit95\n\nRe: Probability\n\n1/6 * 4/6 ^ n-1!\n\n11. Jan 10, 2013\n\n### dx\n\nRe: Probability\n\nExactly.\n\nNow you simply add them up. Let a = 4/6\n\n(1/6) + (1/6)a + (1/6)a2 + (1/6)a3...\n\nwhich is (1/6)(1 + a + a2 + a3 + a4...)\n\nThis is where you use the formula I mentioned before.\n\n12. Jan 10, 2013\n\n### jinhuit95\n\nRe: Probability\n\nGot it! Thanks alot!!\n\n13. Jan 10, 2013\n\n### dx\n\nRe: Probability\n\nNo problem.\n\n14. Jan 10, 2013\n\n### Ray Vickson\n\nRe: Probability\n\nHere is another approach: let x = the probability that A will win, given that A is just about to shoot.\n\nIf A hits the target on his first shot, he wins; the probability of this is 1/6. If A misses his first shot (prob. = 5/6), then for A to win it must be the case that B misses his first shot (prob = 4/5); at that point, A is just about to shoot and we are back to the starting point (win prob. = x at that point). So, altogether we have\n$$x = \\frac{1}{6} + \\frac{5}{6} \\frac{4}{5}\\, x = \\frac{1}{6} + \\frac{2}{3} x.$$\nSolving, we get x = 1/2.\n\n15. Jan 10, 2013\n\n### CAF123\n\nRe: Probability\n\n@Ray\nIs there a reason why doing: $$x = \\frac{1}{6} + \\frac{2}{3}x + \\frac{5}{6} \\frac{4}{5} \\frac{5}{6} \\frac{4}{5}x,$$ and solving is incorrect? (I am just considering the possibility of both A and B losing until A hits on his third go).\n\n16. Jan 10, 2013\n\n### Joffan\n\nRe: Probability\n\nMy friend Kim and I had a similar competition, but we're much worse. I am slightly worse than Kim, so I went first. On any one shot, my chance of hitting the bull is 1/39, and Kim's chance is 1/38.\n\nShow that, once again, this is an even competition - my chance of getting the bull's eye first is 1/2.\n\n17. Jan 10, 2013\n\n### CAF123\n\nRe: Probability\n\nAre you stuck? If so, just use exactly the same method as you used to solve the first one. Alternatively, say you win on your $i$th shot. This means you (or Kim) cannot have hit the bull on the previous $i-1$ attempts.\nWhat is the prob of you not hitting the bull on the $i-1$ attempts?\nWhat is the prob of Kim not hitting the bull on her $i-1$ attempts?\nWhat is the prob of you hitting on the $i$th trial given that you and Kim failed on the $i-1$ trials?\nBy the independence of trials you can then multiply these together and sum.\n(This is really the same method by dx just reworded a bit differently)\n\n18. Jan 10, 2013\n\n### haruspex\n\nRe: Probability\n\nJust use Ray's method again. It's much the simplest way to solve these problems. Did you understand Ray's method?\n\n19. Jan 10, 2013\n\n### Joffan\n\nRe: Probability\n\nSorry for the misunderstanding guys. I was giving an extension question to illustrate the reasoning; I'm not stuck.\n\n20. Jan 10, 2013\n\n### Ray Vickson\n\nRe: Probability\n\nWe can see it is incorrect by solving it and getting the wrong answer (x = -3/2 instead of x = 1/2). To understand WHY it is wrong, go back and look at the definition of x and the argument I gave for the equation satisfied by x. The point is that x already includes all the possibilities of winning on the first, third, fifth, .... go.\n\nHowever, it would be correct to write\n$$x = \\frac{1}{6} + \\frac{2}{3}\\frac{1}{6} + \\frac{2}{3}\\frac{2}{3}\\,x$$\nfor similar reasons. This equation does have the correct solution x = 1/2.\n\nLast edited: Jan 10, 2013", "date": "2017-12-13 06:18:26", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/probabilities-of-archers-a-and-b-hitting-the-bulls-eye.663461/", "openwebmath_score": 0.7214272618293762, "openwebmath_perplexity": 1557.7460466744242, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109503004294, "lm_q2_score": 0.884039278690883, "lm_q1q2_score": 0.8706115621504746}}
{"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Oct 2018, 20:07\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# A farmer has an apple orchard consisting of Fuji and Gala\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 23 Apr 2010\nPosts: 7\nA farmer has an apple orchard consisting of Fuji and Gala\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 12 Aug 2012, 07:01\n3\n12\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n70% (02:56) correct 30% (03:10) wrong based on 696 sessions\n\n### HideShow timer Statistics\n\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\nA. 22\nB. 33\nC. 55\nD. 77\nE. 88\n\n_________________\n\n\"Choose to chance the rapids and dance the tides\"\n\nOriginally posted by iamseer on 26 Apr 2010, 15:13.\nLast edited by Bunuel on 12 Aug 2012, 07:01, edited 1 time in total.\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8386\nLocation: Pune, India\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n02 Apr 2011, 07:04\n3\n3\njay121 wrote:\nPlease tell me how should I calculate 187*0.85=220 within 30 seconds?\n\nIs there any specific trick for that?\n\nEven if i split it up and say: 0.1x +0.75x = 187 I have difficulties to solve it fast? Any tipps?\n\nThanks\n\n10% are cross pollinated and 75% are pure Fuji so 85% are Fuji which gives you\n\n$$(\\frac{85}{100})*x = 187$$\n\nNow it is obvious that number of trees has to be an integer so 85 and 187 need to have come common factor.\n187 isn't divisible by 2 (not even), by 3(1+8+7 = 16 so not divisible by 3), by 7 (since 7*2 = 14, you have 47 left which will not go by 7) but it is divisible by 11 (11*1 = 11 and 11*7 = 77). So you split 187 into 11*17.\nNow 85 is 17*5.\n\nNow the equation becomes:\n$$(\\frac{17*5}{100})*x = 11*17$$\n17 gets canceled and 5 gets canceled with 100 leaving a 20. So you get x = 220\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\n##### General Discussion\nManager\nJoined: 12 Jan 2010\nPosts: 246\nSchools: DukeTuck,Kelogg,Darden\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n26 Apr 2010, 15:17\n3\n1\niamseer wrote:\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\nLet the total trees be x\n\n3/4 are pure Fuji = 3x/4\n\n10% cross pollinated = x/10\n\nnow The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187\n\n3x/4 + x/10 = 187\nsolve this x = 220\n\n220-187 = 33 are the pure Gala trees.\n_________________\n\nRun towards the things that make you uncomfortable daily. The greatest risk is not taking risks\nhttp://gmatclub.com/forum/from-690-to-730-q50-v38-97356.html\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2606\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 26 Apr 2010, 15:21\n1\n2\niamseer wrote:\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\nLet f = pure fuji , g = pure gala and c - cross pollinated.\n\nc = 10% of x where x is total trees.\nc = .1x\n\nalso 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220\n\n220 - 187 = pure gala = 33.\n\nPS: While posting the questions please post options and source.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\nhttp://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nOriginally posted by gurpreetsingh on 26 Apr 2010, 15:19.\nLast edited by gurpreetsingh on 26 Apr 2010, 15:21, edited 1 time in total.\nManager\nJoined: 12 Jan 2010\nPosts: 246\nSchools: DukeTuck,Kelogg,Darden\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n26 Apr 2010, 15:21\ngurpreetsingh wrote:\niamseer wrote:\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\nLet f = pure fuji , g = pure gala and c - cross pollinated.\n\nc = 10% of x where x is total trees.\nc = .1x\n\nalso 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220\n\n220 - 187 = pure gala = 33.\n\nHi shouldnt it be 33, since 187 includes the fuji and the ones that cross pollinated???\n_________________\n\nRun towards the things that make you uncomfortable daily. The greatest risk is not taking risks\nhttp://gmatclub.com/forum/from-690-to-730-q50-v38-97356.html\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2606\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n26 Apr 2010, 15:23\nSilvers wrote:\ngurpreetsingh wrote:\niamseer wrote:\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\nLet f = pure fuji , g = pure gala and c - cross pollinated.\n\nc = 10% of x where x is total trees.\nc = .1x\n\nalso 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220\n\n220 - 187 = pure gala = 33.\n\nHi shouldnt it be 33, since 187 includes the fuji and the ones that cross pollinated???\n\nIts 33 only check again what you have quoted.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\nhttp://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nManager\nJoined: 22 Oct 2009\nPosts: 235\nGMAT 1: 760 Q49 V44\nGPA: 3.88\n\n### Show Tags\n\n04 Jul 2010, 11:16\n1\nFirst - I want to point out the fact that the Google ad on this particular thread is an advertisement that says, \"Apple trees are on sale now!\"\nClassic.\n\nI'm not sure how to do this without simply setting up the formulas and hacking your way through the algebra. I'm guessing from your post (\"need to understand the easiest way to solve this one\") that you don't want this method.\n\nAnyone else?\nManager\nJoined: 16 Feb 2010\nPosts: 181\n\n### Show Tags\n\n04 Jul 2010, 11:52\nyup....no need for the 4 equations method\nf+g= T\n0.1T = c\n0.75T = F\nf+c= 187\nyou can figure out the rest.....i hope there is a simpler way of doing it....\nIntern\nJoined: 03 Jun 2010\nPosts: 21\n\n### Show Tags\n\n05 Jul 2010, 13:10\n2\nGala apples - G\nFuji apples - F\nC - Cross Pollinated apples\nTotal apples - X\nGiven:\nF = 3/4X\nX = G + F + cross pollinated\nFuji + Cross pollinated (10 % of all apples) = 187\n\nSolution:\n3/4X + 1/10X = 187\nHence X = 220\nX = G + F + cross pollinated\n220 = G + 187\nHence G = 33.\nDirector\nJoined: 21 Dec 2009\nPosts: 534\nConcentration: Entrepreneurship, Finance\n\n### Show Tags\n\n05 Jul 2010, 14:05\nzisis wrote:\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\n(A) 22\n(B) 33\n(C) 55\n(D) 77\n(E) 88\n\nOA:\n\nplease explain method.....need to understand the easiest way to solve this one\n\nI got confused with the highlighted part; how would the question be framed if the part in bold where to mean\npure fuji plus cross-pollinated fuji(not including cross-pollinated Gala)? Am I reading too much meaning to the question?\n_________________\n\nKUDOS me if you feel my contribution has helped you.\n\nIntern\nJoined: 18 Dec 2010\nPosts: 4\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n01 Apr 2011, 21:04\nPlease tell me how should I calculate 187*0.85=220 within 30 seconds?\n\nIs there any specific trick for that?\n\nEven if i split it up and say: 0.1x +0.75x = 187 I have difficulties to solve it fast? Any tipps?\n\nThanks\nRetired Moderator\nJoined: 16 Nov 2010\nPosts: 1437\nLocation: United States (IN)\nConcentration: Strategy, Technology\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n01 Apr 2011, 21:35\n1\nLet x be the number of apple trees\n\n0.1x of apple trees are cross-pollinated\n\n3x/4 = Fuji\n\n3x/4 + 0.1x = 187\n\n=> 0.85x = 187\n\n=> 0.05x = 11\n\n=> x = 1100/5 = 220\n\nNow x/4 - 0.1x = Pure Gala\n\n= 55 - 22 = 33\n_________________\n\nFormula of Life -> Achievement/Potential = k * Happiness (where k is a constant)\n\nGMAT Club Premium Membership - big benefits and savings\n\nDirector\nJoined: 01 Feb 2011\nPosts: 668\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n02 Apr 2011, 13:24\nLets assume F stands for pure Fuji, G for pure Gala , CP for cross pollinated and T for total trees.\n\nF + CP = 187\n\n=> 3T/4 + 10T/100 = 187 = > T = 220\n\nT = F+G+CP = > G = 33\nRetired Moderator\nJoined: 20 Dec 2010\nPosts: 1835\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n02 Apr 2011, 13:48\niamseer wrote:\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\n10% of his trees cross pollinated i.e. Cross Pollinated = 0.1*Total\nThe number of his trees that are pure Fuji plus the cross-pollinated ones totals 187 i.e. Pure Fuji+ Cross Pollinated=187\n3/4 of all his trees are pure Fuji i.e. Pure Fuji=0.75*Total\n\nPure Fuji+ Cross Pollinated=0.75*Total+0.1*Total=0.85*Total\n\n0.85*Total=187\nTotal=187/0.85\n\nPure Gala = Total - (Pure Fuji+ Cross Pollinated) = 187/0.85-187=187((1/0.85)-1)=187(0.15/0.85)=187(3/17)=11*3=33\n\nAns: 33 Pure Gala Apples\n_________________\nVP\nStatus: There is always something new !!\nAffiliations: PMI,QAI Global,eXampleCG\nJoined: 08 May 2009\nPosts: 1055\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n01 May 2011, 23:35\n0.1x + 0.75x = 187\nx = 220\n\nnumber of G trees = 220-187 = 33\nHence B\nIntern\nJoined: 15 Nov 2011\nPosts: 20\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n12 Aug 2012, 05:53\nHello,\n\nI got this question today on my MGMAT CAT and tried first to solve it through a chart for overlapping sets, but then realized that this is not possible. I solved the question afterwards as in the posts here equation system). How can I identify that the chart won't work here?\n\nThanks!\nIntern\nJoined: 31 May 2012\nPosts: 10\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n22 Sep 2012, 14:17\nrocketscience wrote:\nHello,\n\nI got this question today on my MGMAT CAT and tried first to solve it through a chart for overlapping sets, but then realized that this is not possible. I solved the question afterwards as in the posts here equation system). How can I identify that the chart won't work here?\n\nThanks!\n\n2X2 works as well. You just need to set \"neither\" as 0 because apples have to be either G or F.\n\n-------F -----nF\nG----.1x-----y\nnG---.75x---0\n------187----?---x\n\nFrom here solve for x and y will be the difference between x and 187.\nManager\nJoined: 02 Nov 2009\nPosts: 112\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n04 Oct 2012, 15:48\nthis is the explanatio i understood\n\ncan anyone explain y it should be 55-22\n\nThank\n\nsubhashghosh wrote:\nLet x be the number of apple trees\n\n0.1x of apple trees are cross-pollinated\n\n3x/4 = Fuji\n\n3x/4 + 0.1x = 187\n\n=> 0.85x = 187\n\n=> 0.05x = 11\n\n=> x = 1100/5 = 220\n\nNow x/4 - 0.1x = Pure Gala\n\n= 55 - 22 = 33\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8386\nLocation: Pune, India\nRe: A farmer has an apple orchard: MGMAT, PS\u00a0 [#permalink]\n\n### Show Tags\n\n04 Oct 2012, 21:16\nvenmic wrote:\nthis is the explanatio i understood\n\ncan anyone explain y it should be 55-22\n\nThank\n\nsubhashghosh wrote:\nLet x be the number of apple trees\n\n0.1x of apple trees are cross-pollinated\n\n3x/4 = Fuji\n\n3x/4 + 0.1x = 187\n\n=> 0.85x = 187\n\n=> 0.05x = 11\n\n=> x = 1100/5 = 220\n\nNow x/4 - 0.1x = Pure Gala\n\n= 55 - 22 = 33\n\n3/4 (i.e. 75%) are pure Fuji and 10% are cross so 15% are pure Gala.\n\nSo once you get x, you can calculate pure Gala as 15% of x = 15/100 * 220 = 33\n\nor you can say Pure Gala = 25% of x (Gala apples) - 10% of x (cross) = 55 - 22 = 33 (as done above)\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nIntern\nJoined: 12 Jun 2012\nPosts: 36\nRe: A farmer has an apple orchard consisting of Fuji and Gala\u00a0 [#permalink]\n\n### Show Tags\n\n05 Oct 2012, 00:45\n2\n1\nA farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?\n\nA. 22\nB. 33\nC. 55\nD. 77\nE. 88\n\nTHE QUICK METHOD...\n\nFuji + Cross = 187\n10% are cross\n75% are Fuji\n\nso 85% = 187\n\nWe want to know what the 15% is\n\nDivide our percent by 10, 8.5% = 18.7\nDouble it, 17% = 38\n\nWe need 15 percent and it is pretty obvious 33 fits the bill\n_________________\n\nRe: A farmer has an apple orchard consisting of Fuji and Gala &nbs [#permalink] 05 Oct 2012, 00:45\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 28 posts ]\n\nDisplay posts from previous: Sort by", "date": "2018-10-18 03:07:17", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html", "openwebmath_score": 0.5062430500984192, "openwebmath_perplexity": 9131.965340582265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8705972768020107, "lm_q1q2_score": 0.8705972768020107}}
{"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 31 May 2016, 06:03\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# The probability that a visitor at the mall buys a pack of\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSenior Manager\nJoined: 05 Oct 2008\nPosts: 272\nFollowers: 3\n\nKudos [?]: 278 [4] , given: 22\n\nThe probability that a visitor at the mall buys a pack of\u00a0[#permalink]\n\n### Show Tags\n\n20 Oct 2009, 06:44\n4\nKUDOS\n16\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n(N/A)\n\nQuestion Stats:\n\n68% (02:04) correct 32% (01:01) wrong based on 137 sessions\n\n### HideShow timer Statistics\n\nThree for comparison: How do we decide what formula to use? Can someone explain the logic please..\n\nThe probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?\n\n* 0.343\n* 0.147\n* 0.189\n* 0.063\n* 0.027\n\nHow do we decide whether the solution is\n\n3/10* 3/10 * 7/10 = .063\nOR\n3/10* 3/10 * 7/10 *3 = 0.189\n\nLikewise the following question:\n\nA certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior.If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?\n\nA: 3/40000\nB: 1/3600\nC: 9/2000\nD: 3/20000\nE: 1/15\n\nHow do we decide whether the solution is\n\n60/1000 * 1/800 = 3/40,000\nOR\n60/1000 * 1/800 * 2 = 3/20,000\n\nThere are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?\n\n* 1\n\n* $$\\frac{2}{3}$$\n\n* $$\\frac{1}{2}$$\n\n* $$\\frac{2}{5}$$\n\n* $$\\frac{1}{3}$$\n\nThis has yet another method to calculate!!\nManager\nJoined: 25 Aug 2009\nPosts: 175\nLocation: Streamwood IL\nSchools: Kellogg(Evening),Booth (Evening)\nWE 1: 5 Years\nFollowers: 11\n\nKudos [?]: 161 [8] , given: 3\n\n### Show Tags\n\n20 Oct 2009, 13:46\n8\nKUDOS\nAnswer to the first question -\n\n0.3*0.3*0.7 is equivalent to saying the the first person picks candy and second person picks candy and third person doesn't pick candy.\n\nhowever this is a different case compared to\n0.3*0.7*0.3 is equivalent to saying the the first person picks candy and second person doesn't pick candy and third person picks candy.\n\nwhich in turn is a fifferent case compared to\n0.7*0.3*0.3 (I'll skip the verbose comments)\n\nAnswer to the second question -\n\nProbability of picking a sibling from the junior class is 60/1000\nand Probability of picking the corresponding sibling from the senior class is 1/800.\nHowever in this case if we flip the case we still end up with the same sibling pair since the order is not important. Hence we don't multiply by 2\n\nAnswer to the third question -\nWe just list the various cases where the sum will be 8\n\n6 2\n2 6\n5 3\n3 5\n4 4\n\nTotal = 5\n5 occurs in 2 cases.\nhence prob is 2/5.\n_________________\n\nRock On\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 33103\nFollowers: 5782\n\nKudos [?]: 70969 [19] , given: 9857\n\n### Show Tags\n\n20 Oct 2009, 21:06\n19\nKUDOS\nExpert's post\n8\nThis post was\nBOOKMARKED\nThe probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?\n* 0.343\n* 0.147\n* 0.189\n* 0.063\n* 0.027\n\nSolution: P(B=2)=3!/2!*0.3^2*0.7=0.189\n\nThe point here is not about to multiply or not by 3, the point is to find # of ways favorable scenario to occur: in our case we are asked to find the probability of 2 out of 3 visitors to buy the candy.\n\n3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren\u2019t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn\u2019t.\n\nNOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn\u2019t.\n\nLet\u2019s consider some similar examples:\n1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?\n\nThe same here favorable scenarios are: NNB, NBN, BNN \u2013 total of three. 3!/2! because again two visitors who didn\u2019t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B \u2013 first two visitors didn\u2019t buy the candy and the third one did.\n\nSo, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441\n\nNOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn\u2019t buy, means that exactly one did.\n\n2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?\n\nAt least ONE buys, means that buys exactly one OR exactly two OR exactly three:\n\nP(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657\n\nP(B=1) --> 0.3*0.7^2 (one bought, two didn\u2019t) multiplied by combinations of BNN=3!/2!=3 (Two identical N\u2019s)\n\nP(B=2) --> 0.3^2*0.7 (two bought, one didn\u2019t) multiplied by combinations of BBN=3!/2!=3 (Two identical B\u2019s)\n\nP(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B\u2019s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.\n\nBUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it\u2019s better to solve it as below:\n\nP(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.\n\n3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?\n\nP(B=2)=5!/2!3!*0.3^2*0.7^3\n\nWe want to count favorable scenarios possible for BBNNN (two bought the candy and three didn\u2019t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.\n\nHope now it's clear.\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 33103\nFollowers: 5782\n\nKudos [?]: 70969 [12] , given: 9857\n\n### Show Tags\n\n20 Oct 2009, 21:09\n12\nKUDOS\nExpert's post\n3\nThis post was\nBOOKMARKED\nLet\u2019s move to the second problem:\n\nA certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?\nA: 3/40000\nB: 1/3600\nC: 9/2000\nD: 3/20000\nE: 1/15\n\nThis one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.\n\nWhat is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).\n\nWhat is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).\n\nSo the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000\n\nI see no argument of multiplying this by two.\n\nThis problem can be solved in another way and maybe this way shows that no need of multiplication:\n\nIn how many ways we can choose 1 person from 1000=1C1000=1000\nIn how many ways we can choose 1 person from 800=1C800=800\nSo total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.\n\nLet\u2019s count favorable outcomes: 1 from 60=60C1=60\nThe pair of the one chosen=1C1=1\nSo total favorable outcomes=60C1*1C1=60\n\nProbability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000\n\nLet\u2019s consider another example:\nA certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I\u2019m not sure whether it\u2019s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?\n\nThe same way here:\n\nWhat is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).\nWhat is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).\n\nSo the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000\n\nAnother way:\nIn how many ways we can choose 1 person from 1000=1C1000=1000\nIn how many ways we can choose 1 person from 800=1C800=800\nSo total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.\n\nFavorable outcomes:\n1 from 120=120C1=120\nThe pair of the one chosen=1C2=2\nSo total favorable outcomes=120C1*1C2=240\n\nProbability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 33103\nFollowers: 5782\n\nKudos [?]: 70969 [8] , given: 9857\n\n### Show Tags\n\n20 Oct 2009, 21:10\n8\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nThird problem:\n\nThere are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?\n* 1\n* 2/3\n* 1/2\n* 2/5\n* 1/3\n\nIn this case we know that drawing has already happened and the sum of two cards was 8.\n\nWhat combinations of two cards are possible to total 8?\n(first card, second card)\n(6,2) (2,6) (5,3) (3,5) (44) \u2013 only 5 possible ways sum to be 8. One from this 5 has already happened.\n\nFrom this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.\n\nIf the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?\n\nWhat combinations of two cards are possible to total 8 AND one of them to be 5?\n(5,3) (3,5) \u2013 two favorable outcomes.\n\nTotal number of outcomes=6*6=36.\nP(one card 5, sum 8)=2/36=1/18\n_________________\nSenior Manager\nJoined: 05 Oct 2008\nPosts: 272\nFollowers: 3\n\nKudos [?]: 278 [0], given: 22\n\n### Show Tags\n\n21 Oct 2009, 06:44\nGreat Explanations and examples - couldn't have been more clear!!\n\nKudos given for all of them..\n\nThanks!!\nManager\nJoined: 06 Oct 2009\nPosts: 69\nFollowers: 2\n\nKudos [?]: 33 [1] , given: 5\n\n### Show Tags\n\n21 Oct 2009, 23:51\n1\nKUDOS\nProblem 1:\nAs there are three visitors, scenarios can be as below\nso Probability= (3/10*3/10*7/10)+(3/10*7/10*3/10)+(7/10*3/10*3/10) = 0.189\n\nProblem 2:\nProbability of selecting 1 sibling pair= (60/1000*60/800*1/60*1/60)= 1/800000\nThis can happen 60 times so probability= (1/800000)*60= 3/40000\n\nProblem 3:\npossible pairs are (2,6),(3,5),(4,4),(5,3),(6,2) so total 5 possibilities\ntwo pairs are having 5\nso probability = 2/5\nIntern\nJoined: 01 Dec 2008\nPosts: 5\nFollowers: 0\n\nKudos [?]: 0 [0], given: 4\n\n### Show Tags\n\n08 May 2010, 03:32\nBunuel,\nExcellent explanations. I really loved the way you explain with examples.\nI am really glad that i started using gmatclub.. Thanks!\n\n-Vinod\nManager\nJoined: 05 Mar 2010\nPosts: 221\nFollowers: 1\n\nKudos [?]: 28 [0], given: 8\n\n### Show Tags\n\n09 May 2010, 02:02\ngood questions and excellent explanations\n_________________\n\nSuccess is my Destiny\n\nCurrent Student\nJoined: 15 Jul 2010\nPosts: 256\nGMAT 1: 750 Q49 V42\nFollowers: 8\n\nKudos [?]: 143 [0], given: 65\n\n### Show Tags\n\n12 Nov 2010, 20:21\nThanks for the question and explanations.\n\nKudos for both!\n_________________\n\nConsider KUDOS if my post was helpful.\n\nMy Debrief: 750-q49v42-105591.html#p825487\n\nManager\nJoined: 23 Jan 2011\nPosts: 127\nFollowers: 1\n\nKudos [?]: 49 [0], given: 13\n\n### Show Tags\n\n24 May 2011, 10:30\nThere are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?\n\nWhat combinations of two cards are possible to total 8 AND one of them to be 5?\n(5,3) (3,5) \u2013 two favorable outcomes.\n\nTotal number of outcomes=6*6=36.\nP(one card 5, sum 8)=2/36=1/18\n\nCan you please explain why number of outcomes will be 36?\nDirector\nStatus: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.\nAffiliations: University of Chicago Booth School of Business\nJoined: 03 Feb 2011\nPosts: 920\nFollowers: 13\n\nKudos [?]: 294 [1] , given: 123\n\n### Show Tags\n\n24 May 2011, 22:12\n1\nKUDOS\nWelcome to the club !\n\nThis interpretation is wrong. Probability = Favourables / Total. But this question total is constrained if you read it carefully - \"the sum of two cards would be 8\". Total is not the absolute ways in which you will pick the cards i.e. 6 * 6 = 36 absolute ways. Total ways to get a sum of 8 using two cards is 5. And that's why the ultimate probability is 2/5\n\nChetangupta wrote:\nThere are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?\n\nWhat combinations of two cards are possible to total 8 AND one of them to be 5?\n(5,3) (3,5) \u2013 two favorable outcomes.\n\nTotal number of outcomes=6*6=36.\nP(one card 5, sum 8)=2/36=1/18\n\nCan you please explain why number of outcomes will be 36?\nManager\nJoined: 23 Jan 2011\nPosts: 127\nFollowers: 1\n\nKudos [?]: 49 [0], given: 13\n\n### Show Tags\n\n24 May 2011, 22:55\nthank you. makes sense.\nManager\nJoined: 13 May 2010\nPosts: 124\nFollowers: 0\n\nKudos [?]: 10 [0], given: 4\n\n### Show Tags\n\n08 Jan 2012, 10:44\nRegarding the below listed explanation from funnel -\n\nA certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?\nA: 3/40000\nB: 1/3600\nC: 9/2000\nD: 3/20000\nE: 1/15\n\nThis one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.\n\nWhat is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).\n\nWhat is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).\n\nSo the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000\n\nI see no argument of multiplying this by two.\n\n-------------------------------------------------------------------------------------------------------------\n\nI have a question regarding this one -\n\nSo basically you are considering the case when the first sibling selected is from the junior class and the second sibling is from the senior class.....but why does it have to be only in that order.....why would you not consider the case when first sibling is from senior class and the second is from the junior class?\n\nManager\nJoined: 29 Jul 2011\nPosts: 107\nLocation: United States\nFollowers: 5\n\nKudos [?]: 48 [0], given: 6\n\n### Show Tags\n\n08 Jan 2012, 20:47\n1. (0.3)(0.3)(0.7) x 4C3 = 0.189\n2. 60/1000 x 1/800 = 3/20000\n_________________\n\nI am the master of my fate. I am the captain of my soul.\nPlease consider giving +1 Kudos if deserved!\n\nDS - If negative answer only, still sufficient. No need to find exact solution.\nPS - Always look at the answers first\nCR - Read the question stem first, hunt for conclusion\nSC - Meaning first, Grammar second\nRC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min\n\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 9719\nFollowers: 466\n\nKudos [?]: 120 [0], given: 0\n\nRe: The probability that a visitor at the mall buys a pack of\u00a0[#permalink]\n\n### Show Tags\n\n21 Sep 2013, 12:02\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nSenior Manager\nJoined: 15 Aug 2013\nPosts: 328\nFollowers: 0\n\nKudos [?]: 38 [0], given: 23\n\n### Show Tags\n\n20 Apr 2014, 17:28\nBunuel wrote:\nThird problem:\n\nThere are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?\n* 1\n* 2/3\n* 1/2\n* 2/5\n* 1/3\n\nIn this case we know that drawing has already happened and the sum of two cards was 8.\n\nWhat combinations of two cards are possible to total 8?\n(first card, second card)\n(6,2) (2,6) (5,3) (3,5) (44) \u2013 only 5 possible ways sum to be 8. One from this 5 has already happened.\n\nFrom this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.\n\nIf the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?\n\nWhat combinations of two cards are possible to total 8 AND one of them to be 5?\n(5,3) (3,5) \u2013 two favorable outcomes.\n\nTotal number of outcomes=6*6=36.\nP(one card 5, sum 8)=2/36=1/18\n\nHi Bunuel,\n\nThanks for the great explanations. I'm stuck on #3 -- first two worked out great.\n\n-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?\n\n-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 33103\nFollowers: 5782\n\nKudos [?]: 70969 [1] , given: 9857\n\n### Show Tags\n\n21 Apr 2014, 00:34\n1\nKUDOS\nExpert's post\nruss9 wrote:\nBunuel wrote:\nThird problem:\n\nThere are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?\n* 1\n* 2/3\n* 1/2\n* 2/5\n* 1/3\n\nIn this case we know that drawing has already happened and the sum of two cards was 8.\n\nWhat combinations of two cards are possible to total 8?\n(first card, second card)\n(6,2) (2,6) (5,3) (3,5) (44) \u2013 only 5 possible ways sum to be 8. One from this 5 has already happened.\n\nFrom this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.\n\nIf the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?\n\nWhat combinations of two cards are possible to total 8 AND one of them to be 5?\n(5,3) (3,5) \u2013 two favorable outcomes.\n\nTotal number of outcomes=6*6=36.\nP(one card 5, sum 8)=2/36=1/18\n\nHi Bunuel,\n\nThanks for the great explanations. I'm stuck on #3 -- first two worked out great.\n\n-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?\n\n-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?\n\nThe number of total outcomes is 5, not 10: (6,2) (2,6) (5,3) (3,5) (4, 4). Only 5 possible ways sum to be 8.\n\nSo, the probability that one of the cards drawn was a 5 is 2/5.\n\nAs for your second question: 6 options for the first card and 6 options for the second card, thus total 6*6=35 options.\n_________________\nIntern\nJoined: 01 Aug 2010\nPosts: 14\nFollowers: 0\n\nKudos [?]: 0 [0], given: 41\n\nRe: The probability that a visitor at the mall buys a pack of\u00a0[#permalink]\n\n### Show Tags\n\n14 Apr 2015, 22:08\nBunuel wrote:\nLet\u2019s move to the second problem:\n\nA certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?\nA: 3/40000\nB: 1/3600\nC: 9/2000\nD: 3/20000\nE: 1/15\n\nThis one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.\n\nWhat is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).\n\nWhat is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).\n\nSo the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000\n\nI see no argument of multiplying this by two.\n\nThis problem can be solved in another way and maybe this way shows that no need of multiplication:\n\nIn how many ways we can choose 1 person from 1000=1C1000=1000\nIn how many ways we can choose 1 person from 800=1C800=800\nSo total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.\n\nLet\u2019s count favorable outcomes: 1 from 60=60C1=60\nThe pair of the one chosen=1C1=1\nSo total favorable outcomes=60C1*1C1=60\n\nProbability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000\n\nLet\u2019s consider another example:\nA certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I\u2019m not sure whether it\u2019s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?\n\nThe same way here:\n\nWhat is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).\nWhat is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).\n\nSo the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000\n\nAnother way:\nIn how many ways we can choose 1 person from 1000=1C1000=1000\nIn how many ways we can choose 1 person from 800=1C800=800\nSo total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.\n\nFavorable outcomes:\n1 from 120=120C1=120\nThe pair of the one chosen=1C2=2\nSo total favorable outcomes=120C1*1C2=240\n\nProbability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000\n\nI understand your explanation but have a doubt. This is for the second question.\nAfter 60/1000 * 1/800 = 3/40000. But shouldn't we add 59 to it, to include the different ways of choosing the first sibling pair? I don't think I am clear on when to count these different ways and when not to. For e.g., when we count the number of shoppers in the first question, we count the different arrangements of shoppers, then why shouldn't the different sibling pairs be counted here?\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 9719\nFollowers: 466\n\nKudos [?]: 120 [0], given: 0\n\nRe: The probability that a visitor at the mall buys a pack of\u00a0[#permalink]\n\n### Show Tags\n\n17 Apr 2016, 01:21\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: The probability that a visitor at the mall buys a pack of \u00a0 [#permalink] 17 Apr 2016, 01:21\nSimilar topics Replies Last post\nSimilar\nTopics:\n1 At the wholesale store you can buy an 8-pack of hot dogs for \\$1.55, a 7 22 Dec 2015, 21:36\n1 The probability that a visitor at the mall buys a pack of candy is 30% 4 03 Apr 2011, 16:14\nProbability 9 23 Jan 2010, 15:49\n24 The probability that a visitor at the mall buys a pack of 19 16 Nov 2007, 09:27\n19 The probability that a visitor at the mall buys a pack of 23 13 Nov 2007, 09:20\nDisplay posts from previous: Sort by", "date": "2016-05-31 13:03:38", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html", "openwebmath_score": 0.6559093594551086, "openwebmath_perplexity": 1810.2808087801413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8705972751232809, "lm_q1q2_score": 0.8705972751232809}}
{"url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html", "text": "It is currently 23 Mar 2018, 17:45\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n### Show Tags\n\n12 Dec 2012, 05:29\n30\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n66% (01:53) correct 34% (01:59) wrong based on 1656 sessions\n\n### HideShow timer Statistics\n\nA pharmaceutical company received $3 million in royalties on the first$20 million in sales of the generic equivalent of one of its products and then $9 million in royalties on the next$108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first $20 million in sales to the next$108 million in sales?\n\n(A) 8%\n(B) 15%\n(C) 45%\n(D) 52%\n(E) 56%\n[Reveal] Spoiler: OA\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44419\n\n### Show Tags\n\n14 Dec 2012, 03:18\n3\nKUDOS\nAns:\n\nfor this kind of percentage change questions we apply the formula (change/original)x100, so here we have initial ratio=3/20 final ratio=1/12 . now change = 3/20-1/12=1/15 , putting these values in the formula we get the answer as (C).\n_________________\n\nwww.mnemoniceducation.com\n\nIntern\nJoined: 04 Dec 2012\nPosts: 2\n\n### Show Tags\n\n26 Feb 2013, 11:34\n1\nKUDOS\ncarloswn wrote:\nTo Brunuel:\n\nThanks for the sharing, I'm wondering whether your formula shouldn't be (9/108-3/20) / (3/20) ? This doesn't change the final answer anyway.\n\nI don't think it'll make a difference because in the numerator we are just looking for the difference between the two amounts. Just make sure that in the denominator you have the original value.\n\nFor better understanding:\n\n$$Percent Change = \\frac{Change in Value}{Original Value}$$\n\nHence, the \"change in value\" is just the difference. I wouldn't worry about a negative here.\n_________________\n\nIf my post has contributed to your learning or teaching in any way, feel free to hit the kudos button ^_^\n\nManager\nJoined: 07 Apr 2014\nPosts: 130\n\n### Show Tags\n\n29 May 2015, 02:57\nFor me, percentage questions seem time consuming ..not sure if I am the only one feel this way..\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 11313\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: 340 Q170 V170\nRe: A pharmaceutical company received $3 million in royalties [#permalink] ### Show Tags 29 May 2015, 15:53 3 This post received KUDOS Expert's post Hi katzzzz, Percent questions come from the broader family of 'ratio-based' questions and you're going to see a bunch of those on Test Day, so you have to make sure that you're ready for them. While some of these questions can be wordier/longer than average, the 'key' to answering these types of questions quicker is to organize information in the most effective way possible (for the question that is asked and for the answer choices that are given). For example, ALL of the following examples mean the same thing, so you have to decide which would be easiest to work with... Men/Women = 1 to 10 = 1:10 = 1/10 = 0.1 = 10% 10M = 1W GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 882\n\n### Show Tags\n\n23 Nov 2015, 15:55\n1\nKUDOS\n1\nThis post was\nBOOKMARKED\n% change also equals :\n$$\\frac{new}{old}$$$$- 1$$\n\n$$\\frac{9}{108}$$*$$\\frac{20}{3}$$$$- 1$$\n\n= $$\\frac{-4}{9}$$ ==> since 1/9 = 0,11 ==> -44%\n_________________\n\nNew Application Tracker : update your school profiles instantly!\n\nManager\nJoined: 30 Dec 2015\nPosts: 90\nGPA: 3.92\nWE: Engineering (Aerospace and Defense)\n\n### Show Tags\n\n08 Mar 2016, 00:30\nTo Bunuel:\n\nI tend to get confused when estimating in certain scenarios and select the wrong answer. In this case I took denominator 20 as 18 so that 3/20 becomes 3/18 = 1/6 and 9/108 = 1/12 so 1/6-1/12 =1/12 and now (1/12)/(1/6)*100 = 50% so I would have selected (D) - 52% which is the closest. In another case had I taken 20 as 21(more convenient) the fraction would have been 3/21 = 1/7 and to balance the round off I would have taken 12 as 10 (one increase other decrease) so (1/7-1/10)/(1/7) = (3/70)*7 = 30% its quite far apart from the options so either 15 or 45 although we are actually reducing the 108 mn to 90 mn which is a huge difference so the overall result should be higher so 45 should be the case. In the next scenario I took 20 mn as 21 mn and then ratio became (1/7) so ({1/7-1/12}/(1/7))*100 = 5/12*100 = 42% approx so again option C (45%). So in these cases the answers are different so then how to reach the correct answer in quick time if choices are quite close, there is a high chance of error. Kindly advise the extent to which we can approximate values of fractions as answers can be a huge difference in the answers arrived by taking multiple scenarios as if one scenario leads us to one choice we will quickly select it and move on as there is the time factor.\n_________________\n\nGive kudos and appreciate if you think its worthwhile\n\nSVP\nJoined: 12 Sep 2015\nPosts: 2159\nRe: A pharmaceutical company received $3 million in royalties [#permalink] ### Show Tags 10 Apr 2016, 10:43 2 This post received KUDOS Expert's post Walkabout wrote: A pharmaceutical company received$3 million in royalties on the first $20 million in sales of the generic equivalent of one of its products and then$9 million in royalties on the next $108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first$20 million in sales to the next $108 million in sales? (A) 8% (B) 15% (C) 45% (D) 52% (E) 56% First$20 million: royalties/sales ratio = 3/20 = 36/240\nNext $108 million: royalties/sales ratio = 9/108 = 1/12 = 20/240 Noticed that I rewrote both with the SAME DENOMINATOR. So, now all we need to is determine the percent change from 36 to 20. To do so, we could use some more lengthy calculations [e.g., 100(36-20)/36] HOWEVER, notice that, if we start at 36, a 50% decrease would give us 18. So going from 36 to 20, must be a decrease that's LESS THAN 50% (but also pretty close to 50%) Only one answer choice works. Answer: C Related Resource The following free video covers the concepts/strategies that are useful for answering this question: Cheers, Brent _________________ Brent Hanneson \u2013 Founder of gmatprepnow.com Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2166 Re: A pharmaceutical company received$3 million in royalties\u00a0[#permalink]\n\n### Show Tags\n\n03 May 2016, 09:27\n2\nKUDOS\nExpert's post\nA pharmaceutical company received $3 million in royalties on the first$20 million in sales of the generic equivalent of one of its products and then $9 million in royalties on the next$108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first $20 million in sales to the next$108 million in sales?\n\n(A) 8%\n(B) 15%\n(C) 45%\n(D) 52%\n(E) 56%\n\nSolution:\n\nThis is a percent decrease problem. We will use the formula: percent change = (new \u2013 old)/old x 100 to calculate the final answer.\n\nWe first set up the ratios of royalties to sales. The first ratio will be for the first 20 million in sales, and the second ratio will be for the next 108 million in sales. Because all of the sales are in millions, we do not have to express all the trailing zeros in our ratios.\n\nFirst 20 Million\n\nroyalties/sales = 3/20\n\nNext 108 Million\n\nroyalties/sales = 9/108 = 1/12\n\nBecause each ratio is not an easy number to use, we can simplify each one by multiplying each by the LCM of the two denominators, which is 60. Keep in mind that we are able to do this only because our answer choices are expressed in percents.\n\nFirst 20 Million\n\nroyalties/sales = (3/20) x 60 = 9\n\nNext 108 Million\n\nroyalties/sales = 9/108 = (1/12) x 60 = 5\n\nWe can plug 9 and 5 into our percent change formula:\n\n(new \u2013 old)/old x 100\n\n[(5 \u2013 9)/9] x 100\n\n-4/9 x 100\n\nAt this point we can stop and consider the answer choices. Since we know that 4/9 is just a bit less than \u00bd, we know that -4/9 x 100 is about a 45% decrease.\n\n_________________\n\nJeffery Miller\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nManager\nJoined: 03 Jan 2017\nPosts: 191\n\n### Show Tags\n\n11 Jun 2017, 04:15\nI did this by \"Long hand\", Of course it eat up more time!\n\nJust sharing in case someone interested.\n\nFor the first 20 Million, $$\\frac{Royalty}{sales}= \\frac{3}{20}$$\n\nFor the next 108 Million, $$\\frac{Royalty}{sales} = \\frac{9}{108}$$\n\nRequired percentage\n\n= $$(\\frac{3}{20}- \\frac{9}{108})/\\frac{3}{20} * 100$$\n\n= $$(\\frac{3}{20} - \\frac{9}{108})* \\frac{20}{3} * 100$$\n\n= $$3* (\\frac{1}{20}-\\frac{3}{108}) * \\frac{20}{3} * 100$$\n\n= $$(1 - \\frac{5}{9}) * 100$$\n\n= $$\\frac{4}{9} * 100$$\n\n= 44.44444\n\n_________________\n\nMy Best is yet to come!\n\nIntern\nJoined: 03 Jun 2017\nPosts: 15\n\n### Show Tags\n\n24 Sep 2017, 22:33\nPercentage change = (3/20*100 -9/108*100) / (3/20*100)\n= only percentage change to near to 50 and less that 50 is 45% hence answer must be C\n_________________\n\nRegards,\nNaveen\nemail: nkmungila@gmail.com\nPlease press kudos if you like this post\n\nStudy Buddy Forum Moderator\nJoined: 04 Sep 2016\nPosts: 780\nLocation: India\nWE: Engineering (Other)\n\n### Show Tags\n\n14 Jan 2018, 00:22\n1\nKUDOS\nExpert's post\nBunuel chetan2u amanvermagmat niks18\n\nQuote:\n\n$$=\\frac{\\frac{3}{20}-\\frac{9}{108}}{\\frac{3}{20}}*100\\approx{44%}$$.\n\nIs below approach the most efficient for simplification:\nTaking 1/4 common after simplifying(3/20 - 1/12)\nin numerator which finally simplifies to 2/3 and then multiplying by 20/3\nwhich approx to 40/9. Now since denominator is slightly less than 10\nand 40/10 is 4 so we shall get fraction as slightly more than 4.xx as a value.\n\nI'd suggest another way:\n\n$$\\frac{\\frac{3}{20}-\\frac{9}{108}}{\\frac{3}{20}}=(\\frac{3}{20}-\\frac{1}{12})*\\frac{20}{3}=1 -\\frac{1}{12}*\\frac{20}{3}=1-\\frac{5}{9}=\\frac{4}{9}=0.444....$$\n_________________", "date": "2018-03-24 00:45:39", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html", "openwebmath_score": 0.6309756636619568, "openwebmath_perplexity": 3873.8104959385996, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.870597270087091}}
{"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 22 Feb 2019, 06:25\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. 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Saturday, February 23rd at 7 AM PT\n\u2022 ### FREE Quant Workshop by e-GMAT!\n\nFebruary 24, 2019\n\nFebruary 24, 2019\n\n07:00 AM PST\n\n09:00 AM PST\n\nGet personalized insights on how to achieve your Target Quant Score.\n\n# If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2597\nIf x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 07 Aug 2018, 05:17\n5\n15\n00:00\n\nDifficulty:\n\n75% (hard)\n\nQuestion Stats:\n\n59% (02:21) correct 41% (02:32) wrong based on 901 sessions\n\n### HideShow timer Statistics\n\nIf x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0, what is the least possible value of the product xy?\n\n(A) -12\n(B) -3\n(C) 0\n(D) 2\n(E) None of the above\n\nThis is Question 5 of\n\nProvide your solution below. Kudos for participation. Happy Solving!\n\nBest Regards\nThe e-GMAT Team\n\n_________________\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nOriginally posted by EgmatQuantExpert on 20 May 2015, 09:28.\nLast edited by EgmatQuantExpert on 07 Aug 2018, 05:17, edited 3 times in total.\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2597\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n22 May 2015, 06:55\n2\n4\nOfficial Explanation\n\n|y + 3| \u2264 3\n\nmeans that the distance of y from the point -3 on the number line is less than, or equal to, 3\n\nThis gives us, -6 \u2264 y \u2264 0\n\nSo, possible values of y (keeping in mind the constraint that y is an integer): -6, -5, -4, -3, -2, -1, 0\n\nWe\u2019re given 2y \u2013 3x = -6\n\nSo, $$x= \\frac{(2y+6)}{3}$$\n\nFor x to be an integer, y must be a multiple of 3.\n\nThis means, possible values of y: -6, -3, 0\nCorresponding values of x: -2, 0, 2\nCorresponding values of xy: 12, 0, 0\n\nThus, minimum possible value of xy: 0\n_________________\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\n##### General Discussion\nRetired Moderator\nJoined: 29 Apr 2015\nPosts: 839\nLocation: Switzerland\nConcentration: Economics, Finance\nSchools: LBS MIF '19\nWE: Asset Management (Investment Banking)\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 21 May 2015, 13:20\n2\n2\nEgmatQuantExpert wrote:\nIf x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0, what is the least possible value of the product xy?\n\n(A) -12\n(B) -3\n(C) 0\n(D) 2\n(E) None of the above\n\nThis is Question 5 of\n\nProvide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 22nd May.\n\nTill then, Happy Solving!\n\nBest Regards\nThe e-GMAT Team\n\nHow to deal with inequalities involving absolute values? First example shows us the so called \"number case\"\nIn this case we have |y + 3| \u2264 3 which is generalized |something| \u2264 some number. First we solve as if there were no absolute value brackets:\n\ny + 3 \u2264 3\ny \u2264 0\nSo y is 0 or negative\n\nSecond scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:\n\ny + 3 >= -3\ny >= -6\nTherefore we have a possible range for y: -6=<y<=0\n\nOk, so far so good, we're half way through. What about x?\n\nHere's the formula: 2y \u2013 3x + 6 = 0, rewrite it as 2y + 6 = 3x.\n\nYou can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers (-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value.\n\nHence, Answer Choice C is the one to go.\n_________________\n\nSaving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!\n\nPS Please send me PM if I do not respond to your question within 24 hours.\n\nOriginally posted by reto on 20 May 2015, 13:19.\nLast edited by reto on 21 May 2015, 13:20, edited 2 times in total.\nManager\nJoined: 18 Nov 2013\nPosts: 78\nConcentration: General Management, Technology\nGMAT 1: 690 Q49 V34\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n20 May 2015, 21:43\n3\nEq: 1 |y + 3| \u2264 3 and Eq: 2 2y \u2013 3x + 6 = 0, x & y are integers\nSolve Eq: 1 |y + 3| \u2264 3\n\n$$+(y + 3) \u2264 3$$\n$$y\u2264 0$$\n\n$$-(y + 3) \u2264 3$$\n$$y\\geq{-6}$$\n\nSolve Eq: 2 2y \u2013 3x + 6 = 0\n\n$$x = \\frac{(2y + 6)}{3}$$\n\nas range of y: $$-6=<y<=0$$\n\nwe get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}\n\nlowest xy = 0\n\nAns : C\n_________________\n\n_______\n- Cheers\n\n+1 kudos if you like\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2597\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n21 May 2015, 03:39\n2\nreto wrote:\n\nHow to deal with inequalities involving absolute values? First example shows us the so called \"number case\"\nIn this case we have |y + 3| \u2264 3 which is generalized |something| \u2264 some number. First we solve as if there were no absolute value brackets:\n\ny + 3 \u2264 3\ny \u2264 0\nSo y is 0 or negative\n\nSecond scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:\n\ny + 3 >= -3\ny >= -6\n\nTherefore we have a possible range for y: -6=<y<=0\n\nOk, so far so good, we're half way through. What about x?\n\nHere's the formula: 2y \u2013 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.\n\nDear reto\n\nVery well explained! Kudos for that.\n\nJust want to ask 2 questions:\n\n1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not?\n2. Is the question asking you to find the value of the product xy?\n\nBased on the answers to these 2 questions, you may feel a need to edit the last part of your solution\n\nBest Regards\n\nJapinder\n_________________\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nManager\nJoined: 09 Nov 2013\nPosts: 70\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n21 May 2015, 05:23\n\nAs per the equation |y+3| <=3, y can have value from -6 to 0.\nNow as per the equation x=(2y+6)/3; x will have negative values for all y in {-6,-5,-4,-3} hence the xy will be (-x)*(-y) a positive value. for y in { -2,-1,0}, x will have { 2/3,4/3,2}. Hence x*y will have minimum at x=2/3,y=-2 or x=4/3,y=-1\nIntern\nJoined: 23 Nov 2014\nPosts: 3\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n22 May 2015, 03:28\n1\n-6<=y<=0\nTo get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.\n\nNote : It is clearly mentioned that both x and y are integers.\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 2597\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n22 May 2015, 07:03\ncsirishac wrote:\n-6<=y<=0\nTo get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.\n\nNote : It is clearly mentioned that both x and y are integers.\n\nDear csirishac\n\nYou bring up an interesting point by saying the part in red.\n\nThis part is not true. As you yourself saw, when you minimized the values of both x and y (both negative), you got a positive product.\n\nHowever, if x = +2 and y = 0 (both are actually the maximum values of x and y in this question), the product = 0, far lesser than the product 12 that you got above.\n\nSo, be careful about what you need to minimize.\n\nBest Regards\nJapinder\n_________________\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nSenior Manager\nJoined: 27 Jul 2014\nPosts: 251\nSchools: ISB '15\nGMAT 1: 660 Q49 V30\nGPA: 3.76\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n10 Nov 2015, 01:50\nIf x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0, what is the least possible value of the product xy?\n\n(A) -12\n(B) -3\n(C) 0\n(D) 2\n(E) None of the above\n\nsolving |y+3|<=3\ngives\n-6<=Y<=0\n\nfurther equation 2y-3x+6=0\nIn order to fiind least value of x we nned to maximize y i,e 0\nthis gives x =2\nAnd least value of y=-6\n\nSo product will be -12\nBunuel @VeritasPrepKarsihma Engr2012\nExperts please let me know where I am missing something?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 53069\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n10 Nov 2015, 02:07\n3\nkanigmat011 wrote:\nIf x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0, what is the least possible value of the product xy?\n\n(A) -12\n(B) -3\n(C) 0\n(D) 2\n(E) None of the above\n\nsolving |y+3|<=3\ngives\n-6<=Y<=0\n\nfurther equation 2y-3x+6=0\nIn order to fiind least value of x we nned to maximize y i,e 0\nthis gives x =2\nAnd least value of y=-6\n\nSo product will be -12\nBunuel @VeritasPrepKarsihma Engr2012\nExperts please let me know where I am missing something?\n\nThe relationship between the values of x and y are given by the formula 2y \u2013 3x + 6 = 0. You cannot take the values of x and y separately, meaning that you cannot take x= 2 and y = -6, because if x = 2, then y = 0, and xy = 0.\n_________________\nIntern\nJoined: 22 Nov 2012\nPosts: 21\nLocation: United States\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n03 Dec 2015, 09:38\nEgmatQuantExpert wrote:\nreto wrote:\n\nHow to deal with inequalities involving absolute values? First example shows us the so called \"number case\"\nIn this case we have |y + 3| \u2264 3 which is generalized |something| \u2264 some number. First we solve as if there were no absolute value brackets:\n\ny + 3 \u2264 3\ny \u2264 0\nSo y is 0 or negative\n\nSecond scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:\n\ny + 3 >= -3\ny >= -6\n\nTherefore we have a possible range for y: -6=<y<=0\n\nOk, so far so good, we're half way through. What about x?\n\nHere's the formula: 2y \u2013 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.\n\nDear reto\n\nVery well explained! Kudos for that.\n\nJust want to ask 2 questions:\n\n1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not?\n2. Is the question asking you to find the value of the product xy?\n\nBased on the answers to these 2 questions, you may feel a need to edit the last part of your solution\n\nBest Regards\n\nJapinder\n\nJapinder,\nI'm getting back to yours qs:\n1) x=2/3 must be rejected as per costraints in the question stem.\n2) we are asked to find the least product of xy. This happens when y=0 x=2 or y=-3 x=0.\n\nThanks.\n_________________\n\nGMAT,\nIt is not finished untill I win!!!\n\nManager\nJoined: 03 Dec 2014\nPosts: 100\nLocation: India\nGMAT 1: 620 Q48 V27\nGPA: 1.9\nWE: Engineering (Energy and Utilities)\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n07 Dec 2015, 19:38\nUJs wrote:\nEq: 1 |y + 3| \u2264 3 and Eq: 2 2y \u2013 3x + 6 = 0, x & y are integers\nSolve Eq: 1 |y + 3| \u2264 3\n\n$$+(y + 3) \u2264 3$$\n$$y\u2264 0$$\n\n$$-(y + 3) \u2264 3$$\n$$y\\geq{-6}$$\n\nSolve Eq: 2 2y \u2013 3x + 6 = 0\n\n$$x = \\frac{(2y + 6)}{3}$$\n\nas range of y: $$-6=<y<=0$$\n\nwe get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}\n\nlowest xy = 0\n\nAns : C\n\nis it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12\n\npls. clear the doubt. thank in advance.\nMath Expert\nJoined: 02 Aug 2009\nPosts: 7334\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n07 Dec 2015, 20:17\nrobu wrote:\nUJs wrote:\nEq: 1 |y + 3| \u2264 3 and Eq: 2 2y \u2013 3x + 6 = 0, x & y are integers\nSolve Eq: 1 |y + 3| \u2264 3\n\n$$+(y + 3) \u2264 3$$\n$$y\u2264 0$$\n\n$$-(y + 3) \u2264 3$$\n$$y\\geq{-6}$$\n\nSolve Eq: 2 2y \u2013 3x + 6 = 0\n\n$$x = \\frac{(2y + 6)}{3}$$\n\nas range of y: $$-6=<y<=0$$\n\nwe get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}\n\nlowest xy = 0\n\nAns : C\n\nis it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12\n\npls. clear the doubt. thank in advance.\n\nHi,\nthe eq ly+3l<=3 gives following values of y..\ny=0 ,-1,-2,-3,-4,-5,-6..\nfor these values x=2,-,-,0,-,-,-2...as x has to be an integer..\nyou have to take the corresponding values only..\ni)when y=-6, then x =-2\nii)and when y=0, x=2..\niii) when y=-3, x=0..\n\nofcourse x as 0 and y as 0 is wrong inputs\nso you get xy as 12 in (i) case and 0 in (ii) and (iii) case..\nhope it helps\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html\n\nGMAT Expert\n\nBoard of Directors\nJoined: 17 Jul 2014\nPosts: 2587\nLocation: United States (IL)\nConcentration: Finance, Economics\nGMAT 1: 650 Q49 V30\nGPA: 3.92\nWE: General Management (Transportation)\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n28 Oct 2016, 06:11\nEgmatQuantExpert wrote:\nIf x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0, what is the least possible value of the product xy?\n\n(A) -12\n(B) -3\n(C) 0\n(D) 2\n(E) None of the above\n\nwe are given that\n-3<= y <= -1.\ny can take 3 values:\n-3, -2, -1, and 0.\n\n2y-3x +6 = 0\n2y+6 = 3x\nwe can test values, but we can clearly see that if y is not -3/0, then x can't be an integer, but we need it to be an integer.\ntherefore, the only cases that actually work are y=-3, x=0 or y=0, and x=2\nxy = 0 in any case\n\nSenior Manager\nJoined: 15 Jan 2017\nPosts: 353\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n26 Jul 2017, 04:33\nA simple algebraic approach.\nWe have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.\n2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2\n\nIf we take y = -6; then we have :\n- 6 = 3x - 6/ 2\n-12 = 3x - 6\n-6 = 3x\nThere fore x = -2.\nX * Y = -2*-6 = positive 12 --> not the least value\n\nNow lets take y = 0\n0 = 3x - 6/2\n3x = -6 => 6 = 3x so x =2\nX*Y = 2*0 = 0\n\nWould this approach be right? Expert comment would be welcome!\nManager\nJoined: 11 Feb 2017\nPosts: 188\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n26 Jul 2017, 04:46\nA simple algebraic approach.\nWe have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.\n2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2\n\nIf we take y = -6; then we have :\n- 6 = 3x - 6/ 2\n-12 = 3x - 6\n-6 = 3x\nThere fore x = -2.\nX * Y = -2*-6 = positive 12 --> not the least value\n\nNow lets take y = 0\n0 = 3x - 6/2\n3x = -6 => 6 = 3x so x =2\nX*Y = 2*0 = 0\n\nWould this approach be right? Expert comment would be welcome!\n\nYes Indeed....This approach is correct\nManager\nJoined: 23 Oct 2017\nPosts: 64\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n09 Dec 2017, 08:33\nIf you visualize |y+3|<=3 on the number line you will get to know that y can take values in the range [-6,0]\n\nNext express the give relation in terms of x i.e. y= (2/3)*(y-3).\nand since y is in the range of [-6,0] and not in (0,3) .. x will follow y's sign.\nThus, the product xy has to be >=0.\nMinumum being 0 .. check whether one can be 0 .. y=0 falls in the range. min. value of the product =0\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 9889\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0,\u00a0 [#permalink]\n\n### Show Tags\n\n20 Jan 2019, 04:37\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: If x and y are integers such that |y + 3| \u2264 3 and 2y \u2013 3x + 6 = 0, \u00a0 [#permalink] 20 Jan 2019, 04:37\nDisplay posts from previous: Sort by", "date": "2019-02-22 14:25:56", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html", "openwebmath_score": 0.7057603597640991, "openwebmath_perplexity": 1847.3903594093272, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8705972684083609}}
{"url": "https://gmatclub.com/forum/root-80-root-139868.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 16 Jul 2018, 00:01\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Root(80)+root(125)\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47006\n\n### Show Tags\n\n01 Oct 2012, 05:02\n2\n5\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n92% (00:38) correct 8% (01:14) wrong based on 1872 sessions\n\n### HideShow timer Statistics\n\n$$\\sqrt{80}+\\sqrt{125} =$$\n\n(A) $$9\\sqrt{5}$$\n(B) $$20\\sqrt{5}$$\n(C) $$41\\sqrt{5}$$\n(D) $$\\sqrt{205}$$\n(E) 100\n\nPractice Questions\nQuestion: 52\nPage: 159\nDifficulty: 600\n\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47006\n\n### Show Tags\n\n01 Oct 2012, 05:03\n1\n3\nSOLUTION\n\n$$\\sqrt{80}+\\sqrt{125} =$$\n\n(A) $$9\\sqrt{5}$$\n(B) $$20\\sqrt{5}$$\n(C) $$41\\sqrt{5}$$\n(D) $$\\sqrt{205}$$\n(E) 100\n\n$$\\sqrt{80}+\\sqrt{125} =\\sqrt{16*5}+\\sqrt{25*5}=4\\sqrt{5}+5\\sqrt{5}=9\\sqrt{5}$$.\n\n_________________\nBoard of Directors\nJoined: 01 Sep 2010\nPosts: 3428\n\n### Show Tags\n\n01 Oct 2012, 07:43\n2\nWhe we deal with this problems may be the best thing to do is to break the number into factors\n\n$$\\sqrt{80}$$$$=$$$$\\sqrt{2*5*2*2*2}$$ ---> $$\\sqrt{4*4*5}$$ ---> $$4$$$$\\sqrt{5}$$\n\n$$\\sqrt{125}$$ $$=$$ $$\\sqrt{25 *5}$$ -----> $$5$$ $$\\sqrt{5}$$\n\n$$4$$$$\\sqrt{5}$$ $$+ 5$$ $$\\sqrt{5}$$ $$=$$ $$9$$$$\\sqrt{5}$$\n\n_________________\nIntern\nJoined: 30 Aug 2011\nPosts: 17\nLocation: United States\nSchools: ISB '15\nGMAT 1: 680 Q46 V37\nWE: Project Management (Computer Software)\n\n### Show Tags\n\n04 Oct 2012, 00:50\n2\n$$\\sqrt{80} = \\sqrt{16*5} = 4\\sqrt{5}$$ ........(1)\n\n$$\\sqrt{125} = \\sqrt{25*5} = 5\\sqrt{5}$$ .......(2)\n\nThus $$4\\sqrt{5} + 5\\sqrt{5} = 9\\sqrt{5}$$\n\nManager\nJoined: 12 Jan 2013\nPosts: 188\n\n### Show Tags\n\n18 Dec 2013, 14:04\nBunuel wrote:\n$$\\sqrt{80}+\\sqrt{125} =$$\n\n(A) $$9\\sqrt{5}$$\n(B) $$20\\sqrt{5}$$\n(C) $$41\\sqrt{5}$$\n(D) $$\\sqrt{205}$$\n(E) 100\n\nPractice Questions\nQuestion: 52\nPage: 159\nDifficulty: 600\n\nThe only thing I know is that the answer is just under 24.. And it is definitely not D since D falls between 14 and 15.. And it's clearly not E. And out of the 3 left, B and C are waay too high above 24, and I figured sqrrt of 5 is around \"2.something\" so A made most sense.\n\nClearly this approach is very shaky but this time it worked.\nManager\nJoined: 13 Feb 2011\nPosts: 94\n\n### Show Tags\n\n20 Aug 2014, 23:32\n$$\\sqrt{80}$$ is approx. 9 and $$\\sqrt{125}$$ is approx. 11 so the answer should be close to 20. Choice A is the closest.\nManager\nJoined: 26 Feb 2015\nPosts: 119\n\n### Show Tags\n\n05 Mar 2015, 07:51\nI see this question is from a subcategory called \"Practice Questions\", where can I find that sub?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47006\n\n### Show Tags\n\n05 Mar 2015, 07:58\nerikvm wrote:\nI see this question is from a subcategory called \"Practice Questions\", where can I find that sub?\n\nThose are practice questions from OG13. You can find all questions HERE.\n_________________\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 2916\nLocation: United States (CA)\n\n### Show Tags\n\n04 May 2016, 10:03\n1\nBunuel wrote:\n$$\\sqrt{80}+\\sqrt{125} =$$\n\n(A) $$9\\sqrt{5}$$\n(B) $$20\\sqrt{5}$$\n(C) $$41\\sqrt{5}$$\n(D) $$\\sqrt{205}$$\n(E) 100\n\nSolution:\n\nTo solve the problem we must simplify the radicals. Radicals should be simplified whenever possible. Since the square root of a perfect square produces integers, it will often be helpful to locate and simplify perfect squares within a radical expression. Thus, we first locate the perfect squares that divide evenly into 80 and 125, making the simplification of each radical straightforward.\n\n\u221a80 = \u221a16 x \u221a5 = 4\u221a5\n\n\u221a125 = \u221a25 x \u221a5 = 5\u221a5\n\nNow we can add these two results together. Remember to keep the value inside the radical constant and add together the values on the outside.\n\n4\u221a5 + 5\u221a5 = 9\u221a5\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nIntern\nJoined: 01 Jan 2015\nPosts: 31\nLocation: India\nGPA: 3.71\nWE: Consulting (Retail Banking)\n\n### Show Tags\n\n04 May 2016, 10:06\nA\n80=4*4*5\n125=5*5*5\nRoot(80)=4Root(5)\nSimilarly 125\nhence 9Root(5)\n_________________\n\n_________________\n\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 3622\nLocation: India\nGPA: 3.5\n\n### Show Tags\n\n04 May 2016, 10:19\nBunuel wrote:\n$$\\sqrt{80}+\\sqrt{125} =$$\n\n(A) $$9\\sqrt{5}$$\n(B) $$20\\sqrt{5}$$\n(C) $$41\\sqrt{5}$$\n(D) $$\\sqrt{205}$$\n(E) 100\n\nPractice Questions\nQuestion: 52\nPage: 159\nDifficulty: 600\n\n$$\\sqrt{80}+\\sqrt{125} =$$ $$4\\sqrt{5}+5\\sqrt{5}$$ = $$9\\sqrt{5}$$\n\n_________________\n\nThanks and Regards\n\nAbhishek....\n\nPLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS\n\nHow to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )\n\nManager\nStatus: On a 600-long battle\nJoined: 22 Apr 2016\nPosts: 138\nLocation: Hungary\nSchools: Erasmus '19\nGMAT 1: 410 Q18 V27\nGMAT 2: 490 Q35 V23\n\n### Show Tags\n\n22 Mar 2017, 23:29\nPrime factorization can really help you and so can converting the roots to fraction exponents.\n\n$$\\sqrt { 80 } +\\sqrt { 125 } =\\\\ { ({ 2 }^{ 4 } }\\times { 5 }^{ 1 })^{ \\frac { 1 }{ 2 } }+{ ({ 5 }^{ 2 }\\times { 5 }^{ 1 }) }^{ \\frac { 1 }{ 2 } }=\\\\ ({ 2 }^{ \\frac { 4 }{ 2 } }\\times { 5 }^{ \\frac { 1 }{ 2 } })+({ 5 }^{ \\frac { 2 }{ 2 } }\\times { 5 }^{ \\frac { 1 }{ 2 } })\\quad =\\\\ { 2 }^{ 2 }\\sqrt { 5 } +5\\sqrt { 5 } =\\\\ 4\\sqrt { 5 } +5\\sqrt { 5 } =\\\\ \\Rightarrow 9\\sqrt { 5 } \\\\$$\n_________________\n\n\"When the going gets tough, the tough gets going!\"\n\n|Welcoming tips/suggestions/advices (you name it) to help me achieve a 600|\n\nManager\nJoined: 07 Jun 2017\nPosts: 177\nLocation: India\nConcentration: Technology, General Management\nGMAT 1: 660 Q46 V38\nGPA: 3.6\nWE: Information Technology (Computer Software)\n\n### Show Tags\n\n14 Sep 2017, 22:49\n\u221a80 ~~ 9\n\u221a 125 ~~ 11\n9+11 = 20\n\u221a5 ~~2.1\n9*2.1~~20\nall other answer not near to this\n_________________\n\nRegards,\nNaveen\nemail: nkmungila@gmail.com\nPlease press kudos if you like this post\n\nRe: Root(80)+root(125) \u00a0 [#permalink] 14 Sep 2017, 22:49\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-07-16 07:01:11", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/root-80-root-139868.html", "openwebmath_score": 0.7343714833259583, "openwebmath_perplexity": 13453.318723910088, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8705972684083609}}
{"url": "http://math.stackexchange.com/questions/353028/is-there-an-idempotent-element-in-a-finite-semigroup", "text": "# Is there an idempotent element in a finite semigroup?\n\nLet $(G,.)$ be a finite semigroup. Is there any $a\\in G$ such that: $$a^2=a$$\n\nIt seems to be true in view of theorem 2.2.1 page 97 of this book (I'm not sure). But is there an elementary proof?\n\nTheorem 2.2.1. [R. Ellis] Let $S$ be a compact right topological semigroup. Then there exists an idempotent in it.\n\n-\n\"Pages 77 to 99 are not shown in this preview\". Do you have another link? \u2013\u00a0 1015 Apr 6 '13 at 15:42\n\nNote first that it suffices to prove that $a^k = a$ for some $k \\geq 2$. If $k = 2$ we are done. Otherwise $k > 2$ and multiplying both sides by $a^{k-2}$ gives $(a^{k-1})^2 = a^{k-1}$.\n\nFix $x \\in G$ and consider the sequence\n\n$$x, x^2, x^4, x^8, x^{16}, \\ldots$$\n\nSince $G$ is finite, there is repetition in this sequence. That is, $x^{2^t} = x^{2^s}$ for some integers $t > s \\geq 1$. Thus $x^{2^t} = (x^{2^s})^{2^{t-s}} = x^{2^s}$, so choosing $a = x^{2^{s}}$ and $k = 2^{t-s}$ gives $a^k = a$. Note that $k \\geq 2$ since $t > s$.\n\n-\n\nPick an arbitrary element and start iterating $x\\mapsto x^2$. Since the semigroup is finite you will eventually hit a cycle. This gives you a $b$ such that $b^k=b$ for some $k\\ge 2$. Now set $a=b^{k-1}$.\n\n-\n\nWhen I first saw the question, I remembered there was a proof on MO using Ramsey theory, but couldn't remember how the argument went, so I came up with the following, that I first posted as a comment:\n\nA cute proof using Schur's theorem: Fix $a$ in your semigroup $S$, and color $n$ and $m$ with the same color whenever $a^n=a^m$. By Schur's theorem (and the fact that the semigroup is finite) there are $n\\le m$ such that $n$, $m$, and $n+m$ have the same color. That is, $a^n=a^m=a^{n+m}=(a^n)^2$.\n\n(Today I finally found the thread on MO with the Ramsey theory proof, using Ramsey's theorem directly rather than Schur's theorem.)\n\n-\n\nThis is a very good point proved by E.H.Moore that says:\n\nSome power of every element of a finite semigroup is idempotent.\n\nTrans. Amer. Math. Soc. 3 1902\n\n-\nGood day, @Babak! How is your's going? ;-) \u2013\u00a0 amWhy Apr 7 '13 at 12:13\nAhhh, maybe you need a nice, sweet nap? ;-) \u2013\u00a0 amWhy Apr 7 '13 at 12:19\nOh yes Amy. Teaching Maths makes everyone tired. I am so glad I could see you at this time. ;-) \u2013\u00a0 B. S. Apr 7 '13 at 12:21\n8^) ${}{}{}{}{}{}{}{}$ \u2013\u00a0 amWhy Apr 7 '13 at 12:23\n\nYou can find a proof $^1$ of the following theorem, from which your assertion follows, at Proof Wiki\n\nTheorem\n\nLet $(S,\\circ)$ be a finite semigroup.\n\nFor every element in $(S,\\circ),$ there is a power of that element which is idempotent. That is: $$\\forall x \\in S:\\exists i \\in \\mathbb N:x^i=x^i\\circ x^i$$\n\nEssentially, then, for your purposes: you can simply set $a = x^i$,\nand you have the existence of an idempotent element $a \\in S$ such that $\\;a^2 = a$.\n\n$1.$ Source of proof: Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978).\n\n-\nN.B. Proof Wiki is a great resource when looking for a proof, or an alternative proof, to mathematical proofs. (Link is to \"main page\"). \u2013\u00a0 amWhy Apr 6 '13 at 16:20\nI do not agree with \"stronger, and more general\". Because it's a simple consequence of the question. \u2013\u00a0 user59671 Apr 7 '13 at 17:53\n@CutieKrait: Actually, no, this theorem talks about every element of a finite semigroup, not just the existence of some element such that... Your question and its answer follow from the theorem above. Besides, the point of my post was not to argue the merits of the theorem or your question, the point of my post was to simply offer help. \u2013\u00a0 amWhy Apr 7 '13 at 18:06\nassume we know every finite element in a semigroup $S$ has an idempotent. Then for each $a$ in $S$ , $\\{a^n \\mid n\\in \\Bbb N \\}$ is a semigroup. So it has an idempotent, say $a^k$. $a^ka^k=a^k$ for some $k$. \u2013\u00a0 user59671 Apr 7 '13 at 18:19\nYou asked about whether there exists an element in each semigroup...such that it is idempotent. We don't know that for each x $x \\circ x = x$, we only know (from the theorem) that for each $x$, exists a power $i$ of $x$ such that $x^i\\circ x^i = x^i$...that's what the theorem above states. From that we know that exists $y = x^i$ such that $y\\circ y = y$, we don't know that for all $a$ in the semigroup, $a\\circ a = a$. The assumption that \"every finite element in a semigroup has [a power that is] an idempotent\" is NOT to say ever finite element in a semigroup is idempotent. \u2013\u00a0 amWhy Apr 7 '13 at 18:24", "date": "2014-07-23 06:14:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/353028/is-there-an-idempotent-element-in-a-finite-semigroup", "openwebmath_score": 0.9014335870742798, "openwebmath_perplexity": 389.4375717154165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717472321277, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.870578406589985}}
{"url": "http://bstermpaperwcen.allstarorchestra.info/the-natural-logarithmic-function-essay.html", "text": "# The natural logarithmic function essay\n\nThe natural logarithmic function essay, The history of logarithms is the story of a correspondence around 1730, leonhard euler defined the exponential function and the natural logarithm by = \u2192.\n\nHome math, popular an intuitive guide to exponential functions & e the mathematical constant e is the base of the natural logarithm. 32 logarithmic functions and their graphs the definition above implies that the natural logarithmic function and the natural exponential function are inverse. Introduction to exponents and logarithms introduction to exponential and logarithmic functions the natural logarithm [latex]ln(x. 32 the natural logarithm function brian e veitch this means the value of lnx, x the the negative of the area shown above this should make sense from one of. Introduction to logarithms mathematicians use log (instead of ln) to mean the natural logarithm this can lead to confusion: example engineer thinks.\n\nThe common and natural logarithms (page since the log function is the inverse of the exponential function, the graph of the log is the flip of the graph of the. A summary of the number e and the natural log in 's inverse, exponential, and logarithmic functions learn exactly what happened in this chapter, scene, or section of. In fact, another approach to introduce the exponential and logarithmic functions in calculus is to present the natural logarithm first, defined exactly as we did l (x. Yes, exponential and logarithmic functions isn\u00e2\u20ac\u2122t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong.\n\nNatural logarithms used with base e logarithmic function: definition & examples related study materials essay prompts. College essay financial the number e and the natural logarithm function of an exponential function natural logarithms are special types of. A-1 natural exponential function in lesson 21, we explored the world of logarithms in base 10 the natural logarithm has a base of \u201ce\u201d \u201ce\u201d is approximately. 51 the natural logarithmic function and differentiation and the definition of the natural logarithmic function: to this point later in this essay.\n\nThe natural logarithmic function-integration name _____ (no trigonometry) homework find the indefinite integral of the following functions. Open document below is an essay on ma1310: week 1 exponential and logarithmic functions from anti essays, your source for research papers, essays, and term paper. The graph of an exponential function logarithmic and exponential logarithmic and exponential functions here is the graph of the natural logarithm. The integration of rational functions, resulting in logarithmic or arctangent functions where represents the natural (base e) logarithm of x and.\n\nWe have all seen the natural logarithm appear in why does the natural logarithm appear so much in its inverse function, the natural logarithm will crop up a. Read this essay on ma1310: week 1 exponential and logarithmic functions come browse our large digital warehouse of free sample essays get the knowledge you need in.\n\nFrom the early 1600's there have been two kinds of logarithms\u2014natural logarithms and common logarithms common logarithms are base 10, denoted [math]\\log_{10. Natural logarithm functiongraph of natural logarithmalgebraic properties of ln(x) i applying the natural logarithm function to both sides of the equation ex 4. Evaluating natural logarithm with and so the reason why you wouldn't see log base e written this way is log base e is referred to as the natural logarithm. The natural logarithm of a number is its logarithm to the base of the plots of the natural logarithm function on the complex plane (principal branch.\n\nThe natural logarithmic function essay\nRated 4/5 based on 39 review\n\n2017. bstermpaperwcen.allstarorchestra.info", "date": "2018-08-20 18:05:17", "meta": {"domain": "allstarorchestra.info", "url": "http://bstermpaperwcen.allstarorchestra.info/the-natural-logarithmic-function-essay.html", "openwebmath_score": 0.8453141450881958, "openwebmath_perplexity": 709.9886028109532, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.986571742099477, "lm_q2_score": 0.882427872638409, "lm_q1q2_score": 0.8705784035860106}}
{"url": "https://math.stackexchange.com/questions/541978/can-the-inverse-of-a-function-be-the-same-as-the-original-function?noredirect=1", "text": "# Can the inverse of a function be the same as the original function?\n\nI was wondering if the inverse of a function can be the same function.\n\nFor example when I try to invert\n\n$g(x) = 2 - x$\n\nThe inverse seems to be the same function. Am I doing something wrong here?\n\n\u2022 No, you're not doing anything wrong. You can certainly have functions such that $f(f(x))=x$. Such functions are called involutions. \u2013\u00a0mjqxxxx Oct 27 '13 at 20:54\n\u2022 Consider $f(x) = x$. \u2013\u00a0Thomas Oct 28 '13 at 1:35\n\u2022 There is also a simple graphical interpretation: any function that when graphed is symmetrical over $y=x$ (that is, a diagonal line at 45\u00b0 angle splits the graph in two mirror images of each other) will have this property. It's easy to see that you can draw boatloads of functions like that. \u2013\u00a0Euro Micelli Oct 28 '13 at 1:40\n\u2022 In general, given an invertible function $g$, $f(x)=g^{-1}(-g(x))$ is an involution. \u2013\u00a0Workaholic Sep 23 '16 at 17:23\n\nYou're correct. A function that's its own inverse is called an involution.\n\nEdit: Oh let's have some fun. :) What are some other functions that are easy to check are involutions? I've cherry picked some of my favorites in what follows, both from memory and also the references I provide below.\n\n\u2022 First, note that there's an easy test to determine whether or not $f$ is an involution. Namely, since $f^{-1} = f$, you just need to double check that $f(f(x)) = x$ for all $x$ in the domain of $f$. This can be used to verify all three of the following examples are actually involutions.\n\n\u2022 Your function $g(x)$ generalizes to a whole class of involutions! Namely, $$f(x) = a-x$$ is an involution for any real number $a$. In particular, $f(x) = 0 - x = -x$ is an involution (as is $f(x) = x$, of course).\n\n\u2022 As someone already pointed out, $f(x) = 1/x$ (defined for all real $x \\neq 0$) is also an involution. More generally, for any real $a$ and $b$ the function $$f(x) = a + \\frac{b}{x-a} = \\frac{ax + (b-a^2)}{x-a}$$ satisfies $$f(f(x)) = a + \\frac{b}{a + \\frac{b}{x-a} - a} = a + (x-a) = x$$ for all real $x \\neq a$, and as such is also an involution on this domain.\n\n\u2022 Here's a less obvious (but cool) example. Consider the function $f(x) = (a - x^3)^{1/3}$. You can check this directly that $f(f(x)) = (a - ((a-x^3)^{1/3})^3)^{1/3} = x$. This is an example of a large class of involutions generated by a special type of symmetric function $F(x,y)$ (as explained here).\n\n\u2022 Fun Fact: The only continuous, odd ($f(-x) = -f(x)$ for all $x$) involutions with domain $(-\\infty,\\infty)$ are $f(x) = \\pm x$. (A short proof of this fact is given here.)\n\n\u2022 There are many, many, more of these functions, and they occur naturally/are useful tools in many branches of mathematics.\n\n\u2022 For \"fun fact\", that is the only non-trivial involution, but the trivial involution is also odd and continuous. \u2013\u00a0Dietrich Epp Oct 28 '13 at 3:25\n\u2022 @DietrichEpp: Good catch! I've edited the post. Thanks! \u2013\u00a0Dan Oct 28 '13 at 5:26\n\nThat's perfectly fine, and your answer is correct. For another function that is its own inverse, see: $$f(x) = \\frac{1}{x} = f^{-1}(x)$$\n\n$g(x): y=2-x$\n\n$g^{-1}(x): x=2-y\\implies x-2=-y\\implies y=2-x$\n\nSo you're correct. It is possible that a function can be an inverse of itself.\n\nComing from theory of coding (LDPC codes decoding), you have another involution: $x\\mapsto-\\log(\\tanh(x/2))$.\n\nYes, you are correct, a function can be it's own inverse. However, I noticed no one gave a graphical explanation for this.\n\nThe inverse for a function of $x$ is just the same function flipped over the diagonal line $x=y$ (where $y=f(x)$). So, if you graph a function, and it looks like it mirrors itself across the $x=y$ line, that function is an inverse of itself.\n\nThis graphical representation can be used to understand a lot of really cool ideas about inverse functions.\n\nFor example, any straight line that's perpendicular to the line $x=y$ will of course be the exact same on both sides of the line $x=y$. Any function that is a straight line perpendicular to $x=y$ can be written in the form $f(x)=a\u2212x$, which as Dan mentioned is an entire class of functions that your function just happens to be one of. Note that the $a$ just determines how high up on the $y$ axis the function starts, but it will always mirror itself across $x=y$.\n\nYou can also see some interesting facts about functions that are their own inverse that might be fun to take some time to prove. For example, since it needs to mirror itself over the line $x=y$, any function that is its own inverse that has any points where $f(x) = x$ must have a slope of either 1 or -1 at that point (or have an undefined slope).\n\nEdit: I must have skipped over Euro Micelli's comment, which explained the graphical aspect of this literally years before I did.\n\nThere are many examples for such types of function's Y=1/x X^2+Y^2=1,2,3,4,5,6,7.....(any other positive number) Simply the fact behind this is that the graph of the function should be symmetric about line Y=X While calculating inverse what we actually calculate is image of that function with respect to line Y=X\n\n\u2022 $x^2 + y^2 = 1$ is not a function. \u2013\u00a0mlc Aug 2 '17 at 6:01\n\u2022 Welcome to the math.stachexchange. Try using mathjax for formulas to make your answers/questions more readable. Extensive how-to found on site. \u2013\u00a0Thanassis Aug 2 '17 at 6:04", "date": "2019-10-18 11:20:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/541978/can-the-inverse-of-a-function-be-the-same-as-the-original-function?noredirect=1", "openwebmath_score": 0.8139545321464539, "openwebmath_perplexity": 252.32056974090685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717484165853, "lm_q2_score": 0.8824278633625322, "lm_q1q2_score": 0.870578400009085}}
{"url": "https://mathematica.stackexchange.com/questions/86126/dft-why-are-frequencies-mirrored-around-the-frequency-axis", "text": "# DFT: why are frequencies mirrored around the frequency axis?\n\nI want to apply Mathematica's DFT procedure, Fourier[], to a Gaussian. Why a Gaussian? Because I know the analytical result.\n\nSo I define\n\na = 0.5;\nh[t_] := Exp[-(t^2/a^2)];\n\n\nThe analytic FourierTransform of h using FourierParameters -> {0, -2 Pi} is:\n\nHAnalytic[v] := Sqrt[Pi]*a*Exp[-a^2*Pi^2*v^2];\n\n\nThen I create time values:\n\ndt = 1/100;\ntList = Table[t, {t, 0, 20000 dt, dt}];\n\n\nand I create sample data:\n\ninputData = h[tList]\n\n\nI then create the frequency values:\n\ndv = 1/((Length[tList] - 1)*dt);\nvList = dv*Table[i, {i, 0, Length[tList] - 1}];\n\n\nand carry out the DFT calculation:\n\nHDFT = Re[Fourier[inputData]];\n\n\nNow I plot it:\n\nShow[\nPlot[HAnalytic[v], {v, 0, 1000*dv}, PlotRange -> {{0, 1.5}, {-1, 1}},\nPlotStyle -> Green],\n]\n\n\nSo far no problem. Only the scaling is different.\n\nBUT NOW MY QUESTION!\n\nWhy do I get\n\ni.e. the Fourier-transformed values are mirrored around the frequency axis, when using\n\ntListSym = Table[t, {t, -10000 dt, 10000 dt, dt}]\n\n\nI mean h[tListSym] gives really kind of the whole function to the DFT, as h is symmetric around $t = 0$, whereas h[tList] reproduces only the right hand side of h.\n\nIf I look at the values of HDFT, I can see that the values are alternatively positive / negative.\n\nCan someone explain to me what is going on?\n\n\u2022 Does this help ? dsp.stackexchange.com/questions/431/\u2026 \u2013\u00a0image_doctor Jun 17 '15 at 11:07\n\u2022 Not really... as I am not wondering about negative frequenies. I am wondering why the amplitude of the frequencies is alternating for t =ListSym = Table[t, {t, -10000 dt, 10000 dt, dt}]. \u2013\u00a0newandlost Jun 17 '15 at 12:24\n\u2022 @Hugh has recently written up an excellent primer on DFT in Mathematica. Maybe something in there might be of help to you. In particular, I think you might want to plot the Abs[] of your DFT values, rather than the Re[]. \u2013\u00a0MarcoB Jun 17 '15 at 14:45\n\u2022 Does not matter, as the fouriertransform of a real even function is a real even function. \u2013\u00a0newandlost Jun 17 '15 at 15:15\n\nThe answer is that Fourier does not know your time values and does not see an even function. Thus if we generate your new input data and plot we get\n\ninputData2 = h[tListSym];\nListLinePlot[inputData2, PlotRange -> All]\n\n\nThe problem is thus how to tell Fourier that the first half of your input is for negative x values. We could change the phase of the values created by Fourier or we can put your input into the usual Fourier form of positive values then negative values - a shifted version. Thus\n\nnn = Length[tListSym]\ninputData3 =\nJoin[inputData2[[(nn - 1)/2 + 1 ;; -1]],\ninputData2[[1 ;; (nn - 1)/2]]];\nListLinePlot[inputData3, PlotRange -> All]\n\n\nThus we are starting with the positive values and then putting the negative values after. Now when we do the Fourier transform we get what you want.\n\nHDFT2 = Fourier[inputData2];\nHDFT3 = Fourier[inputData3];\n\n\nHere I have not taken the real parts as you did. We can compare the two versions\n\nListPlot[{Re[HDFT2][[1 ;; 500]], Re[HDFT3][[1 ;; 500]]},\nPlotRange -> All]\n\n\nListPlot[{Im[HDFT2][[1 ;; 500]], Im[HDFT3][[1 ;; 500]]},\nPlotRange -> All]\n\n\nThe imaginary part of the shifted version is zero because we have constructed an even function. We now get what you want.\n\nShow[\nPlot[HAnalytic[v], {v, 0, 1000*dv},\nPlotRange -> {{0, 1.5}, {-1, 1}}, PlotStyle -> Green],\nListPlot[{Thread[{vList, Re[HDFT2]}][[1 ;; 500]]}, PlotStyle -> Blue],\nListPlot[{Thread[{vList, HDFT3}][[1 ;; 500]]}, PlotStyle -> Brown]\n]\n\n\nOn a secondary point I think you are off by one point in defining your frequency values. The increment should be 1/(Length[] dt).\n\nI suppose a deeper question is why does Fourier do positive values then negative values? There is an assumption of a periodic function so it does not matter. However, it is often a nuisance as you have found.\n\nWith regard to getting the scaling correct you need to work out your version of Parseval's Theorem using FourierParameters.\n\n\u2022 I was hoping you might be induced to weigh in on this one :-) Let me also add a link to a post in the DSP stackexchange site dealing explicitly with the math behind the even/odd function problem, just for reference. (+1) \u2013\u00a0MarcoB Jun 17 '15 at 18:09\n\u2022 @Hugh Thanks you :) \"I suppose a deeper question is why does Fourier do positive values then negative values?\" Yes ... But I think no I understand it. You have to think about it like this: It is a N periodic function, hence 0 = N, AND the first value given to the DFT defines 0. This is the same as sying the function lives on a ring. Therefore for getting a even function we have fist hand it the positive and then the negative values. I will make a small drawing: drive.google.com/file/d/0B1ecQW5sX1LJLWhfUUt6ckktY1U/edit \u2013\u00a0newandlost Jun 18 '15 at 11:15\n\u2022 @newandlost Unfortunately I can't download or see your video -I will try when I get to another computer. However, I think I can see what is happening and I agree with your comment. There is also as solution that would involve adjusting the phase of the Fourier transform after the calculation i.e. making a transformation in time. Thus by writing Exp[- I 2 Pi f (t-t0)] in the Fourier integral choosing a t0 and then taking this phase outside the integral it would be possible to correct the spectrum. Glad I could help. \u2013\u00a0Hugh Jun 18 '15 at 14:07", "date": "2020-10-28 23:52:31", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/86126/dft-why-are-frequencies-mirrored-around-the-frequency-axis", "openwebmath_score": 0.638904869556427, "openwebmath_perplexity": 1364.7799658351746, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.963779943094681, "lm_q2_score": 0.9032942138630786, "lm_q1q2_score": 0.8705768460347125}}
{"url": "https://gmatclub.com/forum/paul-sells-encyclopedias-door-to-door-he-earns-160-on-ever-166375.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Nov 2018, 19:59\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. 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Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)\n\u2022 ### Free lesson on number properties\n\nNovember 23, 2018\n\nNovember 23, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nPractice the one most important Quant section - Integer properties, and rapidly improve your skills.\n\n### Show Tags\n\nUpdated on: 22 Jan 2014, 06:15\n3\n00:00\n\nDifficulty:\n\n45% (medium)\n\nQuestion Stats:\n\n72% (01:55) correct 28% (01:51) wrong based on 128 sessions\n\n### HideShow timer Statistics\n\nPaul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->\n\nHe does not earn \"double commission.\" That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?\n\nA)13,200\nB)14,800\nC)16,400\nD)15,800\nE)19,600\n\nOriginally posted by guerrero25 on 20 Jan 2014, 21:16.\nLast edited by guerrero25 on 22 Jan 2014, 06:15, edited 1 time in total.\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8574\nLocation: Pune, India\nRe: Paul sells encyclopedias door-to-door.\u00a0 [#permalink]\n\n### Show Tags\n\n20 Jan 2014, 22:00\n1\n1\nguerrero25 wrote:\nPaul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->\n\nHe does not earn \"double commission.\" That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?\n\nA)13,200\n\nB)14,800\n\nC)16,400\n\nD)15,800\n\nE)19,600\n\nI will post the OA soon\n\nHis pay check was $1320. Out of this,$160 was his fixed salary so the total commission he earned was $1320 -$160 = $1160 He earns 10% on the sales of first$10,000 which gives a commission of $1000. He earns 5% on every subsequent dollar. Since he earns another$160, he must have had sales of another 160*(100/5) = 3200\nSo his total sales must have been $10,000 +$3200 = $13,200 _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 04 Oct 2013 Posts: 152 Location: India GMAT Date: 05-23-2015 GPA: 3.45 Re: Paul sells encyclopedias door-to-door. He earns$160 on ever\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jan 2014, 00:11\nPaul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->\nHe does not earn \"double commission.\" That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?\n\nUsing the equation y = mx + C, where, y = Paycheck amount; x= amount of sales exceeding $10000; m = 5% or 1/20; C = 160 + 10000(1/10) = 1160; $$1320 = \\frac{x}{20} + 1160$$ Or,$$x = (1320-1160)(20) = 3200$$ So, Total sales $$= 10000+3200=13200$$ Answer: (A) Senior SC Moderator Joined: 22 May 2016 Posts: 2117 Paul sells encyclopedias door-to-door. He earns$160 on ever\u00a0 [#permalink]\n\n### Show Tags\n\n26 Aug 2018, 18:31\n1\nguerrero25 wrote:\nPaul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->\n\nHe does not earn \"double commission.\" That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?\n\nA)13,200\nB)14,800\nC)16,400\nD)15,800\nE)19,600\n\nPaycheck = fixed pay + commissions\n$$1,320 = 160$$ + commissions\n$$(1,320 -160)=1,160$$ in commissions\n\nAll of the answer choices are greater than $10,000 $$x=$$ first$10,000 in sales\n$$y$$ = sales amount after the first $10,000 $$x+y=$$ total sales amount $$.10x + .05y=$$ total commissions amount $$.10x+.05y=1,160$$ $$.10(10,000)+.05(y)=1,160$$ $$1,000+0.05y=1,160$$ $$.05y=160$$ $$y=\\frac{160}{.05}=\\frac{16,000}{5}=3,200$$ Total sales = $$(x+y)=(10,000+3,200)=13,200$$ Answer A Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4170 Location: United States (CA) Re: Paul sells encyclopedias door-to-door. He earns$160 on ever\u00a0 [#permalink]\n\n### Show Tags\n\n12 Nov 2018, 07:22\nguerrero25 wrote:\nPaul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->\n\nHe does not earn \"double commission.\" That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?\n\nA)13,200\nB)14,800\nC)16,400\nD)15,800\nE)19,600\n\nSince he earns $160 as base salary and 10% of$10,000 is $1,000, if he sells no more than$10,000 worth of encyclopedias, the maximum he earns is $1,160. However, since he earns$1,320, he must have sold more than $10,000 worth of encyclopedias. Therefore, if we let x = his sales for the paycheck of$1,320, we can create the equation:\n\n160 + 0.1(10,000) + 0.05(x - 10,000) = 1,320\n\n160 + 1,000 + 0.05x - 500 = 1,320\n\n660 + 0.05x = 1,320\n\n0.05x = 660\n\nx = 13,200\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 12891\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: Paul sells encyclopedias door-to-door. He earns $160 on ever [#permalink] ### Show Tags 13 Nov 2018, 11:15 Hi All, We're told that Paul sells encyclopedias door-to-door. He earns$160 on every paycheck, regardless of how many sets he sells and a commission as follows:\n\n10% on the first $10,000.00 and 5% for any additional sales beyond that. His largest paycheck of the year was$1,320. We're asked for his sales for that pay period. This question can be approached in a couple of different ways. Based on the given answer choices, we can take advantage of a couple of built-in patterns and avoid some of the extra 'math steps.'\n\nTo start, we can see that all of the answers are ABOVE $10,000, so we know that part of his commissions was 10% of$10,000 = $1,000. Paul received$160 on top of that piece of his commission, so we have accounted for $1,160. Now, we just have to account for the extra$1320 - $1160 =$160 he earned...\n\nThat extra commission was based on 5% of the sales ABOVE $10,000. 5% of$1,000 = $50$160 is a little more than (3)($50), so the extra sales had to be a little more than (3)($1000). Thus, his total sales for that period were a LITTLE MORE than $10,000 +$3000 = $13,000... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****", "date": "2018-11-21 03:59:25", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/paul-sells-encyclopedias-door-to-door-he-earns-160-on-ever-166375.html", "openwebmath_score": 0.45608147978782654, "openwebmath_perplexity": 13958.469966398803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9637799378929587, "lm_q2_score": 0.9032942106088968, "lm_q1q2_score": 0.8705768381997117}}
{"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal", "text": "prove that opposite angles of a parallelogram are equal\n\nStrategy: how to prove that opposite sides of a parallelogram are equal. Always [\u2026] if angle AOB=118 degree, find i) angle ABO, ii) angle ADO, iii) angle \u2026 The reflexive property refers to a number that is always equal to itself. Join now. Opposite Angles of a Parallelogram. Most questions answered within 4 hours. Prove that. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean parallel postulate and neither condition can be proven without appealing to the Euclidean parallel postulate or one of its equivalent formulations. (iv) In quadrilateral ACFD, AD || CF and AD = CF | From (iii) \u2234 quadrilateral ACFD is a parallelogram. A good way to begin a proof is to think through a game plan that summarizes your basic argument or chain of logic. Answer Save. Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. 12 Answers. Relevance? There is a theorem in a parallelogram where opposite angles are equal. ... angle ABD\u2245angle BDC --> Alternate Interior Angles Theorem. One of the properties of parallelograms is that the opposite angles are congruent, as we will now show. To Prove that the Opposite Angles of a Parallelogram are equal in measure 00:06:23 undefined Related concepts Properties of a Parallelogram - Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram. Property: The Opposite Angles of a Parallelogram Are of Equal Measure. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram. In a parallelogram , 1. prove that in a parallelogram opposite angles are equal. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. how_to_reg Follow . For example, z = z or 1000 = 1000 are examples of the reflexive property. Class-IX . Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! Answer: 3 question Given: SV || TU and SVX = UTX Prove: VUTS is a parallelogram. 1. Log in. In the adjoining figure abcd is a rectangle and diagonals intersect at O point. | \u2235 A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length (v) \u2235 ACFD is a parallelogram | \u2026 State the given then use CPCTC to say all their parts match and say what specifically makes it a parallelogram. 1 decade ago. Favourite answer. Prove that opposite sides of a parallelogram are congruent.? if one pair of opposite sides are parallel and one pair of opposite angles are congruent? This video is highly rated by Class 8 students and has been viewed 322 times. Since each pair of alternate interior angles are congruent, the sum of them must be as well. Why are the opposite angles of a parallelogram equal? Its diagonals bisect each other.. or i need both pairs for it to be a parallelogram? For the other opposite angles, we can prove that the angles are equal by drawing another diagonal line and proving that the triangles are congruent. Parallel Lines Transversals Angle. - Get the answer to this question and access a vast question bank that is tailored for students. Prove that A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel. In any case, in a parallelogram, the opposite angles are always equal. Yes if diagonals of a parallelogram are equal then it is a rectangle. Given, opposite angles of a quadrilateral are equal. Opposite angles of a parallelogram are equal (or congruent) Consecutive angles are supplementary angles to each other (that means they add up to 180 degrees) Read more: Area of Parallelogram. There can be more than one community in a society. Click hereto get an answer to your question \ufe0f Prove that in a parallelogram the opposite angles are equal . Area = base $$\\times$$ height = b $$\\times$$ h. It includes every relationship which established among the people. A parallelogram has two parallel pairs of opposite sides. The diagonals of a parallelogram are not of equal length. 2. asked Jan 19, 2019 in Mathematics by Bhavyak ( 67.3k points) quadrilaterals So as you can see parallelogram has opposite sides parallel. - the answers to estudyassistant.com So adjacent angles must sum up to 180\u00b0(because if the angle is more or less the lines would converge on either side) Is it possible to have a regular polygon with measure of each interior angle of 45 degree 3. Angles \"a\" and \"b\" add up to 180\u00b0, so they are supplementary angles. Update: simple geometry proof required. Lv 6. Construction: Draw a parallelogram ABCD. Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram. don't use any co-ordinate fundamentals. Prove theorem: If a quadrilateral is a parallelogram, then opposite angles are congruent. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. 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'' and  b '' add up to 180\u00b0, so: its opposite sides parallel each other O... Has opposite sides are equal and parallel opposite angles are equal and parallel think a... We will now show community in a parallelogram the opposite angles are equal opposite angles are equal property refers a. Z or 1000 = 1000 are examples of the following reasons would complete the in. Why are the opposite angles are equal quadrilateral are equal answers you need keep... In line 6 equal and its diagonal bisects the parallelogram 2 its diagonals bisect each other the properties of is. 1000 are examples of the properties of parallelogram proofs show game plans followed by the resulting proofs! Click hereto Get an answer to a number that is always equal for example, z z... Ad = CF need to keep in mind when you prove that opposite sides of a parallelogram are,. From ( 1 ) and ( 2 ), we obtain AD || CF AD... Why are the opposite angles are congruent. is always equal to.... One community in a quadrilateral is a parallelogram where opposite angles of a quadrilateral each! Parallelogram are of equal measure the interior angle of 45 degree 3 we... Side, in addition to the angles, to show congruency interior angle of parallelogram... As well your basic argument or chain of logic should be aware of about the proof above proves. So they are supplementary angles is the number of sides of a parallelogram not... A question for free Get a free answer to a quick problem ( 2,! To prove that opposite sides of a quadrilateral is equal, then it is a parallelogram equal... This question and access a vast question bank that is always equal to itself not of equal.... \u2026 prove that in a quadrilateral is equal, then opposite angles in a parallelogram If one pair opposite!, as we will now show of 45 degree 3 ( \\times\\ ) h. it includes every which. And say what specifically makes it a parallelogram opposite sides are equal and.! All their parts match and say what specifically makes it a parallelogram If one of! Pairs for it to be a parallelogram are equal is it possible to have regular... Get the answer to your question \ufe0f prove that in a quadrilateral is parallelogram... Think through a game plan that summarizes your basic argument or chain of logic, in a parallelogram show... Get the answer to your question \ufe0f prove that in a parallelogram and we know,! Intersect at O point to have a regular polygon exit exterior angle by 100.! Pairs for it to be a parallelogram and we know that, a. Parallelogram has opposite sides parallel and one pair of opposite sides are also equal diagonal bisects the 2! ( 105k points ) in a society sides are equal parallel and equal, then is... Diagonals of a parallelogram need to keep in mind when you prove that in a parallelogram equal! Use CPCTC to say all their parts match and say what specifically makes it a parallelogram, the opposite of! Get a free answer to a quick problem or 1000 = 1000 examples... Has two parallel pairs of opposite angles of a parallelogram, so they supplementary... Angles is equal, then it is a flat shape with opposite sides a. Good way to begin a proof is to think through a game plan how!", "date": "2021-09-23 11:50:33", "meta": {"domain": "eventfullevents.com", "url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal", "openwebmath_score": 0.6664319038391113, "openwebmath_perplexity": 507.28567752899374, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9697854146791214, "lm_q2_score": 0.8976952996340946, "lm_q1q2_score": 0.8705718084111486}}
{"url": "https://math.stackexchange.com/questions/1461604/direct-proof-instead-of-an-indirect-proof-if-b-is-the-set-of-lower-bounds-of", "text": "# Direct Proof instead of an Indirect Proof: If $B$ is the set of lower bounds of $A$, then $\\sup B \\in B$\n\nThe author proved the following exercise using a proof by contradiction, but I think it can be accomplished using a direct proof, so I was wondering if someone would verify that my argument is sound.\n\nExercise\n\nSuppose $A$ is a nonempty set of reals that is bounded below. Let $B$ be the set of lower bounds for $A$, and assume further that $B$ is not empty and bounded above.\n\nProve that $\\sup B \\in B$.\n\nBefore proving the statement, I present the following lemma.\n\nLemma Let $x,y \\in \\mathbb{R}$ such that $x \\leq y + \\epsilon$ for every $\\epsilon > 0$. Then $x \\leq y$.\n\nProof\n\nSince $A$ is not empty and bounded below, the completeness axiom implies that there exists a real number $\\psi$ such that $\\psi = \\inf A$. Further, because $B$ is not empty and bounded above, the completeness axiom entails the existence of a real number $\\phi$ such that $\\phi = \\sup B$.\n\nNow choose any $\\epsilon > 0$. Because $\\phi$ is the least upper bound of B, it follows that $\\phi - \\epsilon$ is not an upper bound of B. Thus there exists an $x \\in B$ such that\n\n$$\\phi - \\epsilon < x$$\n\nSince $x \\in B$, it follows by definition of $B$ that $x$ is a lower bound of A. Due to the fact that $\\psi$ is the greatest lower bound of A, it follows that\n\n$$x \\leq \\psi$$\n\nBy transitivity, we deduce that $\\phi - \\epsilon < \\psi$, which implies that\n\n$$\\phi < \\psi + \\epsilon$$\n\nFrom the lemma we conclude that\n\n$$\\phi \\leq \\psi$$ From the last inequality, we conclude that $\\phi$ is a lower bound of A, i.e. $\\phi = \\sup B \\in B$, as desired.\n\n\u2022 Note: I understand that $\\phi$ and $\\psi$ are typically used to denote either angle degrees or as functions due to convention, but I like using them because of how they look haha. Oct 2 '15 at 21:18\n\u2022 \"Let B be the set of lower bounds for A, and assume further that B is not empty and bounded above.\" You can not make that assumption you must demonstrate that, which is relatively easy to do. (Anything less than inf A is a lower bound and anything in A is greater than all lower bounds). Your lemma is a direct consequence of axioms of inequalities and doesn't need stating. Otherwise your proof seems sound and valid. (Not sure why your professor choice a proof by contradiction as your proof is the standard and quite direct.) Oct 2 '15 at 21:41\n\u2022 @fleablood The exercise that is in my post is the second part of a three part exercise, where part one is to prove that $B \\neq \\emptyset$ and $B$ is bounded above. So, I included the assumption without proof for convenience. Oct 2 '15 at 21:56\n\u2022 Votaire. Never mind then :) Oct 2 '15 at 22:58\n\u2022 Instead of saying $\\phi - \\epsilon$ just say \"for any y < $\\phi$\". We know R isn't bounded below so we know y exists. This is simpler, more direct, easier to read and means you don't need to present the lemma. Oct 2 '15 at 23:05\n\nThe proof is correct except a possible minor modification. So, $x\\in B$ which happens to be the set of all lower bounds of $A$ and $\\psi$ is the greatest lower bound of $A$. This implies $x\\leq \\psi$ (contrary to strict inequality in your proof.) However, this won't change your proof because we have $\\phi -\\epsilon <x\\leq \\psi \\implies \\phi-\\epsilon<\\psi$.\n\n\u2022 ah, good point. Thank you! Oct 2 '15 at 21:34\n\u2022 Also, you can use a stronger lemma which states that x=y with the same hypothesis. This will also prove that sup (B) =inf (A)! Oct 2 '15 at 21:40", "date": "2021-12-08 08:02:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1461604/direct-proof-instead-of-an-indirect-proof-if-b-is-the-set-of-lower-bounds-of", "openwebmath_score": 0.9573087692260742, "openwebmath_perplexity": 112.7183002591206, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370156, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8705029715565851}}
{"url": "http://mathhelpforum.com/statistics/274308-discs-bag.html", "text": "# Thread: discs in a bag\n\n1. ## discs in a bag\n\na bag contains 20 discs marked with numbers from 1 to 20 inclusive. if two discs are picked at random from the bag whats the probability of getting one even numbered disc and one odd numbered disc.\n\nthere are 10 even and odd numbers so i was thinking the answer would be 10/20 *10/20 =1/4 the answer at the back of the book is 10/19\n\n2. ## Re: discs in a bag\n\nHey markosheehan.\n\nYou should think about the order of picking things.\n\nIf you pick one even one then the next even one has a lower chance.\n\nTry listing the different combinations of frequencies for picking even and odd in different orders and add them up.\n\n3. ## Re: discs in a bag\n\nsorry yes so the answer would be 10/20*10/19=5/19 this is not the right answer either\n\n4. ## Re: discs in a bag\n\nThe first disc chosen is even or odd ... probability of choosing one or the other is 1. Choosing the first leaves 19 in the bag.\n\nYou're looking for the second disc to be different than the first ... odd if the first was even or even if the first was odd. In either case, there are 10 discs of the remaining 19 in the bag that are different than the first choice. The probability of pulling out odd/even disc different than the first is 10/19.\n\nthe probability that the fisrt two discs drawn are odd/even or even/odd is 1 times 10/19\n\n5. ## Re: discs in a bag\n\nThere are initially 20 disks in the bag, 10 of them even. The probability the first disk drawn is even is 10/20= 1/2. There are then 19 disks in the bag, 10 of them odd. The probability the second disk drawn is odd is 10/19. The probability the first disk drawn is even and the second odd, in that order, is (1/2)(10/19)= 5/19.\n\nSimilarly, since there were also 10 odd disks in the bag initially, the probability the first disk drawn is odd is 1/2. Then there are 19 disks left in the bag, 10 of them even. The probability the second disk drawn is even is 10/19. The probability the first disk drawn is odd and the second even, in that order, is (1/2)(10/19)= 5/19.\n\nThe probability one disk is even and the other odd, in either order, is 5/19+ 5/19= 10/19.\n\n(It is always the case, in problems like this, that the probabilities of the two orders are the same. We could have just calculated the first probability, 5/19, then multiplied by 2 for the two orders.)\n\n6. ## Re: discs in a bag\n\nanother way of doing it is using combinations.\nchoose 2 from 20 = 20ncr2=190 total number\n\n(choose 1 from 20 even)and(chose 1 from 10 odd)\n10ncr1*10ncr1=100\n\n100/190=10/19\n\ni do not like this method as i am finding it hard to see why it works even though it gives the right answer", "date": "2018-01-19 16:21:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/274308-discs-bag.html", "openwebmath_score": 0.8258749842643738, "openwebmath_perplexity": 593.9153418589925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127406273587, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8705029709804077}}
{"url": "https://math.stackexchange.com/questions/2089393/how-to-prove-that-gcdm1-n1-divides-mn-1", "text": "# How to prove that gcd(m+1, n+1) divides (mn-1)\n\nI'm learning about divisors and the gcd, but now I'm stuck at proving:\n\ngcd(m+1, n+1) divides (mn-1) for all m,n in the set of Integers\n\nHelp is appreciated on how to prove this! Thanks\n\n\u2022 $mn-1=(m+1)(n+1)-(m+1)-(n+1)$ \u2013\u00a0Wojowu Jan 8 '17 at 20:29\n\nA common strategy to solve this type of problem is using the following simple fact:\n\nIf $d$ divides $a$ and $b$, then d divides $$a\\cdot r+b\\cdot s$$ for all $r,s \\in \\mathbb{Z}$.\n\nThus if $d=\\gcd(m+1,n+1)$, then obviously $d$ divides $m+1$ and $n+1$, and so d divides $$(m+1)\\cdot r+(n+1)\\cdot s$$ for all $r,s \\in \\mathbb{Z}$. In particular, for $r=n$ and $s=-1$, we have that $d$ divides $$(m+1)\\cdot n+(n+1)\\cdot (-1)=mn-1.$$\n\n\u2022 Thanks for this clarification! One of my attempts stranded at (m+1)\u22c5r + (n+1)\u22c5s. Could you maybe explain why in particular r = n and s = -1? Or do you simply take these to get to mn-1 as we can state that (m+1)\u22c5r+(n+1)\u22c5s for all r and s in Z? \u2013\u00a0Superkuuk Jan 8 '17 at 21:10\n\u2022 Exactly! The suitable choices you make for $r$ and $s$ is completely based on your target \"$mn-1$\". \u2013\u00a0MathChat Jan 8 '17 at 21:13\n\u2022 @Superkuuk You hit the nail on the head.The problem with this approach is that it pulls the answer out of a hat, i.e. there is no motivation for the choice of $\\,r,s.\\,$ But if you use the remainder-based approach as in my answer it is much more more natural, i.e. it is just a special case of the Polynomial Remainder Theorem, i.e that $\\,P(x) \\equiv P(c)\\pmod{x-c}.\\$ Good luck trying as above to find those coefficients for more complex polynomials $\\,P(m,n).\\,$ OTOH, evaluating $\\,P(m,n)\\,$ at $\\,n,m = -1\\,$ is utterly trivial. \u2013\u00a0Bill Dubuque Jan 8 '17 at 21:36\n\u2022 @BillDubuque - I disagree that it's \"pulling the answer out of a hat\". Mathematics is, first and foremost, a pursuit of pattern-matching. As neither $m^2$ nor $m^2$ appear in the target term, we can see that, for the simplest solution, $r$ does not contain $m$ and $s$ does not contain $n$. So we have $(m+1)(an+b)+(n+1)(cm+d)=mn-1$. Expanding the left and equating, we get $a+c=1$, $b+c=0$, $a+d=0$, and $b+d=-1$. It can easily be seen that one equation is redundant, and so we can set, for example, $b=0$, which gives $a=1$, $c=0$, $d=-1$, which is MathChat's solution. \u2013\u00a0Glen O Jan 9 '17 at 8:45\n\u2022 @GlenO But no such derivation is given in the answer. Further, as I emphasized above, ad hoc methods like that do not generalize, but the remainder based methods do. In fact they lead to very very powerful algorithms such as the Grobner basis algorithm. Ad-hoc attempts at \"pattern matching\" cannot compete with algorithms. \u2013\u00a0Bill Dubuque Jan 9 '17 at 14:57\n\nHint $\\$ mod \\,\\gcd(m\\!+\\!1,n\\!+\\!1)\\!:\\ \\ \\begin{align} &m\\!+\\!1\\equiv0\\equiv n\\!+\\!1\\\\ \\Rightarrow\\ &\\ \\ \\,m\\equiv -1\\equiv n\\\\ \\Rightarrow\\ &mn\\equiv (-1)(-1)\\equiv 1\\end{align}\n\nRemark $\\$ More generally, as above, using the Polynomial Congruence Rule we deduce\n\n$$P(m,n)\\equiv P(a,b)\\,\\ \\pmod{\\gcd(m\\!-\\!a,m\\!-\\!b)}$$\n\nfor any polynomial $\\,P(x,y)\\,$ with integer coefficients, and for any integers $\\,m,n,a,b.$\n\nOP is special case $\\, a,b = -1\\$ and $\\ P(m,n) = mn\\$ (or $\\ mn-1)$\n\nAlthough it's less elegant than the other approaches, you can also prove this through direct substitution. First observe that \\begin{align} \\gcd(m+1, n+1)=d & \\implies m+1=pd,n+1=qd \\\\ & \\implies m=pd-1,n=qd-1 \\end{align} for some $p,q\\in\\mathbb{Z}$. Then \\begin{align} mn-1 & = (pd-1)(qd-1)-1 \\\\ & = (pqd^2-(p+q)d+1)-1 \\\\ & = (pqd-(p+q))d, \\end{align} which is an integer multiple of $d$.\n\nUse the fact that\n\n$$(m+1)(n+1) = mn + m + n + 1 = (mn - 1) + (m+1) + (n+1)$$\n\nTHen it is straightforward", "date": "2019-09-16 00:00:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2089393/how-to-prove-that-gcdm1-n1-divides-mn-1", "openwebmath_score": 0.9780576825141907, "openwebmath_perplexity": 504.38805445444433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668701095653, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8704957558273562}}
{"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Nov 2018, 20:01\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\nYour Progress\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in November\nPrevNext\nSuMoTuWeThFrSa\n28293031123\n45678910\n11121314151617\n18192021222324\n2526272829301\nOpen Detailed Calendar\n\u2022 ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!\n\nNovember 22, 2018\n\nNovember 22, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nMark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)\n\u2022 ### Free lesson on number properties\n\nNovember 23, 2018\n\nNovember 23, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nPractice the one most important Quant section - Integer properties, and rapidly improve your skills.\n\n# A lottery is played by selecting X distinct single digit numbers from\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 11 Aug 2012\nPosts: 116\nSchools: HBS '16, Stanford '16\nA lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n05 Feb 2013, 07:40\n2\n1\n23\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n69% (01:00) correct 31% (01:05) wrong based on 923 sessions\n\n### HideShow timer Statistics\n\nA lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?\n\n(1) Players must match at least two numbers with machine to win.\n\n(2) X = 4\n##### Most Helpful Expert Reply\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4488\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2017, 12:07\n6\n2\nTheMastermind wrote:\nExcellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.\n\nThank you.\n\nDear TheMastermind,\n\nI'm happy to respond.\n\nJust so we are clear, here would be the PS version of the problem:\nA lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery?\n\nThink about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use \"at least\" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution.\n\nCase I: the machine's four choices don't overlap at all with mine.\nThere are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits:\nfirst choice P = 6/10 = 3/5\nsecond choice P = 5/9\nthird choice P = 4/8 = 1/2\nfourth choice = 3/7\n\n$$P_1 = \\dfrac{3}{5} \\times \\dfrac{5}{9} \\times \\dfrac{1}{2} \\times \\dfrac{3}{7}$$\n\n$$P_1 = \\dfrac{1}{1} \\times \\dfrac{1}{1} \\times \\dfrac{1}{2} \\times \\dfrac{1}{7} = \\dfrac{1}{14}$$\n\nThat's the Case I probability.\n\nCase II: the machine's four choices overlap with exactly one digit of mine.\n\nFirst, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc.\n\n$$P_2 = \\dfrac{4}{10} \\times \\dfrac{6}{9} \\times \\dfrac{5}{8} \\times \\dfrac{4}{7}$$\n$$+ \\dfrac{6}{10} \\times \\dfrac{4}{9} \\times \\dfrac{5}{8} \\times \\dfrac{4}{7}$$\n$$+ \\dfrac{6}{10} \\times \\dfrac{5}{9} \\times \\dfrac{4}{8} \\times \\dfrac{4}{7}$$\n$$+ \\dfrac{6}{10} \\times \\dfrac{5}{9} \\times \\dfrac{4}{8} \\times \\dfrac{4}{7}$$\n\n$$P_2 = \\dfrac{1}{1} \\times \\dfrac{2}{3} \\times \\dfrac{1}{1} \\times \\dfrac{1}{7}$$\n$$+ \\dfrac{2}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{1}{7}$$\n$$+ \\dfrac{1}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{2}{7}$$\n$$+ \\dfrac{1}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{2}{7}$$\n\n$$P_2 = \\dfrac{4*2}{3*7} = \\dfrac{8}{21}$$\n\nThat's the Case II probability.\n\nAdd those two:\n\n$$P_{1,2} = \\dfrac{1}{14} + \\dfrac{8}{21} = \\dfrac{3}{42} + \\dfrac{16}{42} = \\dfrac{19}{42}$$\n\nThat's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question.\n\n$$P = 1 - \\dfrac{19}{42} = \\dfrac{23}{42}$$\n\nThat's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown.\n\nDoes all this make sense?\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\n##### General Discussion\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4488\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n05 Feb 2013, 10:12\n5\ndanzig wrote:\nA lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?\n(1) Players must match at least two numbers to win.\n(2) X = 4\n\nI'm happy to help.\n\nThis is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you.\n\nSo, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a \"single digit number\" ----- let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning.\n\nStatement #1: Players must match at least two numbers to win.\nNow, at least we know what constitutes winning. The trouble is --- we don't know the how many digits are picked. If X = 9 ---- the lottery picks all the digits from 1-9, then I also pick all the digits from 1-9 --- then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3 --- the lottery picks three, and then I pick three --- well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient.\n\nStatement #2: X = 4\nNow, we know how many digits are picked ---- lottery picks 4, then I pick 4 --- but now I have no idea what constitutes \"winning\". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient.\n\nCombined statements:\nNow, we know --- the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient.\n\nAnswer = C\n\nDoes all this make sense?\n\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\nManager\nJoined: 27 Jan 2013\nPosts: 228\nGMAT 1: 780 Q49 V51\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n05 Feb 2013, 11:49\n1\nHi,\n\nI think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9\n\nWhat does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just \"the digits\" but could include the single digit decimals .1,.2,.3.... Anyone?\n\nI guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question.\n\nHG.\n_________________\n\n\"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years.\" -Dr. Edwin Land\n\nGMAT vs GRE Comparison\n\nIf you found my post useful KUDOS are much appreciated.\n\nIMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:\n\nHere is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html\n\nIntern\nJoined: 09 Oct 2012\nPosts: 7\nLocation: United States\nConcentration: Strategy, Finance\nGMAT 1: 590 Q34 V38\nGPA: 3.56\nWE: Analyst (Consumer Products)\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2013, 08:22\nHerrGrau wrote:\nHi,\n\nI think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9\n\nWhat does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just \"the digits\" but could include the single digit decimals .1,.2,.3.... Anyone?\n\nI guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question.\n\nHG.\n\nIt's tagged as a grockit problem, so I would guess that this is not an official question. It is also tagged as a 600-700 lvl question, which seems high to me for this question...I don't have any experience with Grockit, but MGMAT questions that are in the 600-700 range tend to be more difficult for me to solve.\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4488\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2013, 10:09\nHerrGrau wrote:\nHi,\nThe wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9 What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just \"the digits\" but could include the single digit decimals .1,.2,.3.... Anyone?\n\nDear HerrGrau\nThank you for your kind words.\n\nIn my experience, I have never heard \"single digit numbers\" apply to anything other than non-negative integers. I have never seen any book or hear anyone ever refer to, say, -5 or 0.008 or 3 x 10^8 as a \"single digit number. Yes, technically, each of these is written with a single non-zero digit, but I have never heard the term \"single digit number\" used for them. (We could say that each of these has \"one significant figure\", but that carries us far afield into measurement theory, well beyond GMAT territory.)\n\nIn my mind, the only ambiguity in this question is whether zero is included --- I guess I was assuming, whatever their understanding of the term, this understanding would be fixed and wouldn't change as we moved through the statements of the DS ---- therefore, with both statements, and with whatever convention they are following, the question can be definitely answered.\n\nI agree with you ---- this question is not written with the tight precision so characteristic of official GMAT questions. The very fact that there's any ambiguity at all makes this a woefully substandard question.\n\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\nManager\nJoined: 27 Jan 2013\nPosts: 228\nGMAT 1: 780 Q49 V51\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n07 Feb 2013, 14:47\nHi Mike,\n\nIn general I agree with everything that you're saying. Just to clarify, I think that .008 is a three digit number with 2 unique digits and so would be excluded. The numbers in question are .1, .2, .3, .4, .5, .6, .7, .8, and .9. They are numbers that have only one digit and hence are \"single digit numbers\". I have never seen this distinction be an actual issue on a GMAT question but am curious as to what the general consensus is.\n\nHG.\n_________________\n\n\"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years.\" -Dr. Edwin Land\n\nGMAT vs GRE Comparison\n\nIf you found my post useful KUDOS are much appreciated.\n\nIMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:\n\nHere is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html\n\nManager\nJoined: 30 Mar 2013\nPosts: 109\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n16 Oct 2014, 11:04\njust for the sake of understanding, how would we solve this question?\n1- probability of matching none - prob of matching 1?\nnone= (5/9)*4/9*3/9*2/9\n\nprobability of matching one: 1*5/9*4/9*3/9\n\nis this the way?\n\nThank you!\nMath Revolution GMAT Instructor\nJoined: 16 Aug 2015\nPosts: 6531\nGMAT 1: 760 Q51 V42\nGPA: 3.82\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n05 Dec 2015, 11:47\n2\nForget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.\n\nA lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?\n\n1) Players must match at least two numbers to win.\n\n(2) X = 4\n\nThere are 2 variables (x and how many to be win) and 2 equations are given, so there is high chance (C) will be the answer.\nLooking at the conditions together,\nthe probability to win is 1-(10C4/10^4), which is unique and makes the condition sufficient,\nand the answer becomes (C).\n\nFor cases where we need 2 more equations, such as original conditions with \u201c2 variables\u201d, or \u201c3 variables and 1 equation\u201d, or \u201c4 variables and 2 equations\u201d, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.\n_________________\n\nMathRevolution: Finish GMAT Quant Section with 10 minutes to spare\nThe one-and-only World\u2019s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.\n\"Only \\$99 for 3 month Online Course\"\n\"Free Resources-30 day online access & Diagnostic Test\"\n\"Unlimited Access to over 120 free video lessons - try it yourself\"\n\nManager\nJoined: 02 Feb 2016\nPosts: 89\nGMAT 1: 690 Q43 V41\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2017, 04:06\nmikemcgarry wrote:\ndanzig wrote:\nA lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?\n(1) Players must match at least two numbers to win.\n(2) X = 4\n\nI'm happy to help.\n\nThis is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you.\n\nSo, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a \"single digit number\" ----- let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning.\n\nStatement #1: Players must match at least two numbers to win.\nNow, at least we know what constitutes winning. The trouble is --- we don't know the how many digits are picked. If X = 9 ---- the lottery picks all the digits from 1-9, then I also pick all the digits from 1-9 --- then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3 --- the lottery picks three, and then I pick three --- well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient.\n\nStatement #2: X = 4\nNow, we know how many digits are picked ---- lottery picks 4, then I pick 4 --- but now I have no idea what constitutes \"winning\". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient.\n\nCombined statements:\nNow, we know --- the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient.\n\nAnswer = C\n\nDoes all this make sense?\n\nMike\n\nExcellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.\n\nThank you.\nIntern\nJoined: 12 Sep 2018\nPosts: 1\nRe: A lottery is played by selecting X distinct single digit numbers from\u00a0 [#permalink]\n\n### Show Tags\n\n14 Nov 2018, 17:47\nmikemcgarry wrote:\nTheMastermind wrote:\nExcellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.\n\nThank you.\n\nDear TheMastermind,\n\nI'm happy to respond.\n\nJust so we are clear, here would be the PS version of the problem:\nA lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery?\n\nThink about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use \"at least\" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution.\n\nCase I: the machine's four choices don't overlap at all with mine.\nThere are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits:\nfirst choice P = 6/10 = 3/5\nsecond choice P = 5/9\nthird choice P = 4/8 = 1/2\nfourth choice = 3/7\n\n$$P_1 = \\dfrac{3}{5} \\times \\dfrac{5}{9} \\times \\dfrac{1}{2} \\times \\dfrac{3}{7}$$\n\n$$P_1 = \\dfrac{1}{1} \\times \\dfrac{1}{1} \\times \\dfrac{1}{2} \\times \\dfrac{1}{7} = \\dfrac{1}{14}$$\n\nThat's the Case I probability.\n\nCase II: the machine's four choices overlap with exactly one digit of mine.\n\nFirst, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc.\n\n$$P_2 = \\dfrac{4}{10} \\times \\dfrac{6}{9} \\times \\dfrac{5}{8} \\times \\dfrac{4}{7}$$\n$$+ \\dfrac{6}{10} \\times \\dfrac{4}{9} \\times \\dfrac{5}{8} \\times \\dfrac{4}{7}$$\n$$+ \\dfrac{6}{10} \\times \\dfrac{5}{9} \\times \\dfrac{4}{8} \\times \\dfrac{4}{7}$$\n$$+ \\dfrac{6}{10} \\times \\dfrac{5}{9} \\times \\dfrac{4}{8} \\times \\dfrac{4}{7}$$\n\n$$P_2 = \\dfrac{1}{1} \\times \\dfrac{2}{3} \\times \\dfrac{1}{1} \\times \\dfrac{1}{7}$$\n$$+ \\dfrac{2}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{1}{7}$$\n$$+ \\dfrac{1}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{2}{7}$$\n$$+ \\dfrac{1}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{2}{7}$$\n\n$$P_2 = \\dfrac{4*2}{3*7} = \\dfrac{8}{21}$$\n\nThat's the Case II probability.\n\nAdd those two:\n\n$$P_{1,2} = \\dfrac{1}{14} + \\dfrac{8}{21} = \\dfrac{3}{42} + \\dfrac{16}{42} = \\dfrac{19}{42}$$\n\nThat's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question.\n\n$$P = 1 - \\dfrac{19}{42} = \\dfrac{23}{42}$$\n\nThat's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown.\n\nDoes all this make sense?\nMike\n\nHi Mike,\n\nI know this is DS, so we do not need to solve, but I was able to come up with the same solution, using the following method - is there anything done incorrectly?\n\n1. Total possible machine selections: 10C4 = 210\n\n2. Total possible human selections where no numbers match (choose any 4 from the remaining 6 numbers the machine did not select): 6C4 = 15\n\n3. Total possible human selections where 1 number matches (choose one of the 4 numbers selected by machine x 3 of the unselected numbers): 4 x 6C3 = 4 x 20 = 80\n\nProbability that human does not win: 1- (combined losing selections/total possible machine selections) = 1 - ((15+80)/210) = 1 - 19/42 = 23/42\nRe: A lottery is played by selecting X distinct single digit numbers from &nbs [#permalink] 14 Nov 2018, 17:47\nDisplay posts from previous: Sort by\n\n# A lottery is played by selecting X distinct single digit numbers from\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-11-21 04:01:08", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html", "openwebmath_score": 0.585705041885376, "openwebmath_perplexity": 1063.8596555375786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9811668679067631, "lm_q2_score": 0.8872045907347108, "lm_q1q2_score": 0.8704957494836778}}
{"url": "http://mathhelpforum.com/discrete-math/35974-question-cambridge-step-paper.html", "text": "# Math Help - Question from Cambridge STEP Paper.\n\n1. ## Question from Cambridge STEP Paper.\n\nHi everyone, I was looking through STEP Papers because I'm considering doing Mathematics at University. On the University of Cambridge website, they have past STEP papers, I tried a question and I was wondering if anyone could confirm the method for doing this. The question was:\n\nHow many integers between 10 000 and 100 000 (inclusive) contain exactly 2 different digits? (23 332 contains exactly 2 different digits but neither of 33 333 or 12 331 does.)\n\nMy calculation gave me an answer of 1 217.\n\nCan someone try this question and let me know what they got and how they got it?\n\nI'll explain my method too after there has been confirmation of whether my answer is right or wrong.\n\n2. Originally Posted by AAKhan07\nHi everyone, I was looking through STEP Papers because I'm considering doing Mathematics at University. On the University of Cambridge website, they have past STEP papers, I tried a question and I was wondering if anyone could confirm the method for doing this. The question was:\n\nHow many integers between 10 000 and 100 000 (inclusive) contain exactly 2 different digits? (23 332 contains exactly 2 different digits but neither of 33 333 or 12 331 does.)\n\nMy calculation gave me an answer of 1 217.\n\nCan someone try this question and let me know what they got and how they got it?\n\nI'll explain my method too after there has been confirmation of whether my answer is right or wrong.\nHere's my naive counting method for doing this. Start by counting only the five-digit numbers, but then we must remember to add 1 at the end to allow for the number 100000.\n\nIf both digits are nonzero then one of them, say x, occurs once or twice, and the other one, y, occurs 3 or 4 times. If x occurs once, there are 5 positions where it can occur. If it occurs twice, there are $\\textstyle{5\\choose2}=10$ possible positions. In each case, there are 9 ways to choose x and 8 ways to choose y. This gives a total of (5+10)\u00d79\u00d78 = 1080 possible numbers.\n\nTo those we must add the numbers containing one or more zeros. But a zero cannot occur as the first digit. The zeros can occur in any of the four other positions, and there can be one, two, three or four of them. There are 4\u00d79 numbers with one zero, 6\u00d79 with two zeros, 4\u00d79 with three zeros, and 9 with four zeros. Total: 15\u00d79 = 135.\n\nMy final answer is thus 1080 + 135 + 1 = 1216 (the \"1\" is for the number 100000 that we had to remember to account for).\n\nSo, where did number 1217 come from?\n\nThe papers are brilliant they cover come really good topics, Cambrdige do set high offers based on the papers though, I have my test in two months\n\nLet me know if you want more resources.\n\nBobak\n\n4. I agree with the 1216. Here is my approach.\n$\\binom{9}{2}\\sum\\limits_{k = 1}^4 \\binom{5}{k}+ 9\\sum\\limits_{k = 1}^4 \\binom{4}{k} +1$\n\n5. Originally Posted by bobak\nThe papers are brilliant they cover come really good topics, Cambrdige do set high offers based on the papers though, I have my test in two months\nYou are probably going to do well.\n\n6. ## My mistake, it was 1,216\n\nYes, sorry, you were all right, I meant to say 1,216. I remember my initial calculation gave me 1,215 so I added 1 for 100,000 but I think I added one again because I forgot I had already done it.\n\nMy method was 9(4C1+4C2+4C3+4C4)+36(5C1+5C2+5C3+5C4), same as Plato's I think although I could not express it as elegantly as Plato.\n\nThanks to bobak for the link, I'll be happy to hear about any more Cambridge STEP resources you have, and good luck for your test. Are you hoping to get into Cambridge? Which college? Are you doing AS Mathematics now?\n\n7. ## my crap approach :)\n\nThis is my approach to the problem:\n\n(I found the total possible values of \"x\" and \"y\", excluding 0 for now, by finding the sum of one to 8 which is 36)\n\nThen calculating the number of posible possitions of \"x\" and \"y\" by doing 2^5 -2 = 30 (the -2 is because the digits can not all be the same)\n\nSo the number of combinations of the digits excluding 0 is therefore 36 * 30 = 1080\n\nThe total posible values of \"x\" and \"y\", when \"x\" = 0 is 9, and as the first digit can't be 0, you half the number of posible possitions making it 15, 15 * 9 = 135\n\nso 1080 + 135 + 1 = 1216 ....", "date": "2015-04-25 22:27:53", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/35974-question-cambridge-step-paper.html", "openwebmath_score": 0.6888198852539062, "openwebmath_perplexity": 556.3250619187047, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9728307700397331, "lm_q2_score": 0.8947894654011352, "lm_q1q2_score": 0.8704787246496275}}
{"url": "http://mathhelpforum.com/trigonometry/89376-general-solution-trig-problem.html", "text": "# Math Help - general solution to trig problem\n\n1. ## general solution to trig problem\n\nHi,\n\nI'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!\n\nFind the general solution to\n\n$\\sin 2x + \\sin 4x = \\cos 2x + \\cos 4x$\n\nRegards,\n\nStonehambey\n\n2. Originally Posted by Stonehambey\nHi,\n\nI'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!\n\nFind the general solution to\n\n$\\sin 2x + \\sin 4x = \\cos 2x + \\cos 4x$\n\nRegards,\n\nStonehambey\n\nUse the sum to product identities\n\nList of trigonometric identities - Wikipedia, the free encyclopedia\n\n$2\\sin\\left(\\frac{2x+4x}{2} \\right)\\cos\\left(\\frac{2x-4x}{2} \\right)=2\\cos\\left(\\frac{2x+4x}{2} \\right)\\cos\\left(\\frac{2x-4x}{2} \\right)$\n\n$2\\sin(3x)\\cos(-2x)=2\\cos(3x)\\cos(-2x)$\n\n$2\\sin(3x)\\cos(2x)=2\\cos(3x)\\cos(2x)$\n\n$2\\sin(3x)\\cos(2x)-2\\cos(3x)\\cos(2x)=0$\n\n$2\\sin(3x)\\cos(2x)-2\\cos(3x)\\cos(2x)=0$\n\n$2\\cos(2x)[\\sin(3x)-\\cos(3x)]=0$\n\nYou should be able to finish from here\n\n3. sum to product! Now it's obvious :P\n\nThanks!\n\n4. I'm assuming you are looking for all solutions on interval [0,2pi)\n\nI include thr graph just to show we can expect 8 solutions\n\nI've included the solutions and method in the attachment .\n\nBut the basic method i used was to rearrange as\n\nsin(2x)-cos(2x) = cos(4x)-sin(4x)\n\nThen I squared both sides and got 2 fairly simple equations\n\nhowever there are 16 possibilities of which 8 are solutions--this is the tedious part.\n\nIf someone has an easier way I'd like to know it.\n\nshould have waited\n\n5. hmm, I tried the way theEmptySet suggested and got a little stuck again\n\nI applied the formulas and got to the same bit that you did. When I solved for\n\n$2\\cos(2x)=0$\n\nI get $x=\\frac{\\pi}{4} + n\\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong?\n\n6. Originally Posted by Stonehambey\nhmm, I tried the way theEmptySet suggested and got a little stuck again\n\nI applied the formulas and got to the same bit that you did. When I solved for\n\n$2\\cos(2x)=0$\n\nI get $x=\\frac{\\pi}{2} + n\\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong?\nYou are really close you should have gotten\n\n$2x = \\frac{\\pi}{2}+n\\pi \\iff x =\\frac{\\pi}{4}+n\\frac{\\pi}{2}$\n\n7. Originally Posted by TheEmptySet\nYou are really close you should have gotten\n\n$2x = \\frac{\\pi}{2}+n\\pi \\iff x =\\frac{\\pi}{4}+n\\frac{\\pi}{2}$\neek, sorry that was a typo on my part What I meant to type was\n\n$\\cos(2x) = 0$\n\n$2x = \\pm \\frac{\\pi}{2} + 2\\pi n$\n\n$\\implies x = \\pm \\frac{\\pi}{4} + n \\pi$\n\nIf we let n = 0 and sub back into the original equation we have\n\n$\\sin\\left(\\frac{\\pi}{2}\\right) + \\sin(\\pi) = \\cos\\left(\\frac{\\pi}{2}\\right) + \\cos(\\pi)$\n\n$1 = -1$\n\nwhich is obviously wrong but I can't see where the error is\n\nEDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P\n\n8. Originally Posted by Stonehambey\neek, sorry that was a typo on my part What I meant to type was\n\n$\\cos(2x) = 0$\n\n$2x = \\pm \\frac{\\pi}{2} + 2\\pi n$\n\n$\\implies x = \\pm \\frac{\\pi}{4} + n \\pi$\n\nIf we let n = 0 and sub back into the original equation we have\n\n$\\sin\\left(\\frac{\\pi}{2}\\right) + \\sin(\\pi) = \\cos\\left(\\frac{\\pi}{2}\\right) + \\cos(\\pi)$\n\n$1 = -1$\n\nwhich is obviously wrong but I can't see where the error is\n\nEDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P\nCosine has a zero every Pi units 2pi is too much see my above post.\n\n9. Originally Posted by TheEmptySet\nCosine has a zero every Pi units 2pi is too much see my above post.\nYou are right it only works for odd n.\n\n10. Originally Posted by TheEmptySet\nCosine has a zero every Pi units 2pi is too much see my above post.\nyeah, I accounted for that when I gave the general solution as\n\n$2x = \\pm\\frac{\\pi}{2} + 2n\\pi$\n\nstarting at $\\frac{\\pi}{2}$ and going up by $2\\pi$ each time gives $\\frac{\\pi}{2}, \\frac{5\\pi}{2}, \\frac{9\\pi}{2}, ...$\n\nStarting at $-\\frac{\\pi}{2}$ and going up by $2\\pi$ each time gives us $\\frac{3\\pi}{2}, \\frac{7\\pi}{2}, \\frac{11\\pi}{2}, ...$\n\nSo all the zeros are covered by these two sets of solutions\n\n(obviously n could be negative and we get zeros going the other way as well)\n\nBut how do I know which ones will hold true for the original equation (when I divide them by 2 to get x on its own)?\n\n11. Originally Posted by TheEmptySet\nYou are really close you should have gotten\n\n$2x = \\frac{\\pi}{2}+n\\pi \\iff x =\\frac{\\pi}{4}+n\\frac{\\pi}{2}$\nSo can any one explain why this only works for odd n? This problems is driving me up the wall!!\n\nRegards,\n\nStonehambey", "date": "2015-04-26 07:15:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/89376-general-solution-trig-problem.html", "openwebmath_score": 0.9131166338920593, "openwebmath_perplexity": 380.09546312716003, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307668889048, "lm_q2_score": 0.8947894611926921, "lm_q1q2_score": 0.8704787177361966}}
{"url": "https://www.physicsforums.com/threads/need-to-determine-function-based-on-graph.776914/", "text": "# Need to determine function based on graph\n\n1. Oct 18, 2014\n\n### TheExibo\n\nThere is this graph: http://i.imgur.com/BdwhN3b.jpg where the equation of the function must be determined. Down below is some thinking I've done, where I've found that the oblique asymptote is -0.5x+2, and the v-asymptote=6. I know how to solve something similar with 2 x-int's, but I'm having difficulty understanding what to do with 3 x-int's. Can anyone help me please? Thanks!\n\n2. Oct 19, 2014\n\n### zoki85\n\nplugin 3 values of (x,f(x)) from the graph of the eq. in order to get system of equations and solve it to find a,b,c.\n\nLast edited: Oct 19, 2014\n3. Oct 19, 2014\n\n### NTW\n\nI believe I'm saying the same as zoki85, but I'll try to elaborate a bit more...\n\nIt seems that you have come to the conclusion that your function has the form\n\ny = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)\n\nIn my opinion, and to find a, b, c you should take three values of x, y from the drawing, reasonably chosen among the left and right branches. For the left branch, one point. For the right branch, and always IMHO, one for the maximum, and another to the right, away from the maximum.\n\nThus, you'll have a 3 x 3 system and solve it for the coefficients a, b, c...\n\n4. Oct 19, 2014\n\n### TheExibo\n\nI got it using only two of the intercepts: (-40,0) and (0.5,0) and putting them into -0.2x+50+(ax+b)/-(x+6)2\n\nGraphing it in desmos.com, it shows the exact same graph. I must explain how I got this function, so how can I explain the third x-intercept? Is it true that I only need two points, one on each side of the asymptote, to find the exact function of the graph that is shown?\n\n5. Oct 19, 2014\n\n### NTW\n\nI now see that you needed only to find two coefficients, a and b. I saw the numerator in x2, but it's true that I misquoted the denominator in your notes, writing (y = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)\n\n6. Oct 19, 2014\n\n### TheExibo\n\nBut by doing that, I've come across the problem where I can't explain how the graph ended up with all 3 correct x-intercepts, when I only used the two. Anyone have any ideas?\n\n7. Oct 20, 2014\n\n### ehild\n\nYou also used the asymptote at infinity where the function behaves as 50-0.2x.\n\nehild\n\n8. Oct 20, 2014\n\n### tycoon515\n\nBeing that your graph has an oblique asymptote, an improper rational function should be the kind of function that comes to mind. By the looks of the graph, as we let $x$ tend toward either $\\infty$ or $-\\infty$, the function gets closer and closer in value to a linear function, so we should consider an improper rational function whose numerator is $1$ greater in degree than its denominator. Also, a rational function is equal to $0$ wherever its numerator, but not its denominator, is equal to $0$. Since you have three $x$-intercepts, the numerator should be a polynomial that is the product of three factors, each one corresponding to a different $x$-intercept. A preliminary construction might look like the following:\n\n$f(x)=k\\frac{(x+40)(x-.5)(x-277.4)}{?}$\n\nThe numerator, when expanded, appears to be a polynomial whose dominant term is $x^3$. Therefore, we should consider a denominator that is a polynomial of degree $2$. Also, the graph is undefined and has a vertical asymptote at only one value of $x$: $-6$. Notice also that the graph of $f(x)$ tends toward $-\\infty$ as $x$ approaches $-6$ from either side. Therefore, the denominator should be a quadratic with at least one $(x+6)$ factor. To find the other factor, observe either that there is no sign change of the function across the vertical asymptote at $x=-6$, or that for any factor other than $(x+6)$ to be present in the denominator, the graph would have to have another vertical asymptote that isn't at $x=-6$, so the denominator clearly has repeated $(x+6)$ factors. Our construction now takes the following form:\n\n$f(x)=k\\frac{(x+40)(x-.5)(x-277.4)}{(x+6)^2}=k\\frac{(x+40)(\\frac{2x-1}{2})(\\frac{5x-1387}{5})}{(x+6)^2}=k\\frac{(x+40)(2x-1)(5x-1387)}{10(x+6)^2}$\n\n$=k\\frac{(2x^2+79x-40)(5x-1387)}{10(x^2+12x+36)}=k\\frac{10x^3-2379x^2-109773x+55480}{10x^2+600x+1800}$\n\nA function of this form is EXACTLY equal to $0$ at all three of the intercepts shown on the graph provided. And, if we let $k=-\\frac{1}{5}$, it has an oblique asymptote of $-\\frac{1}{5}x+\\frac{2379}{50}$, which has the same slope as the asymptote on the graph but NOT the same $y$-intercept. Also, the equation you came up with has the correct slope and $y$-intercept of the oblique asymptote, but it is NOT EXACTLY equal to $0$ when $x=277.4$, so one of two things must be going on here: either the $x$-intercepts displayed on the graph are slightly off (rounded), or the actual equation of the oblique asymptote is slightly different than $-\\frac{1}{5}x+50$.\n\n9. Oct 20, 2014\n\n### TheExibo\n\nAwesome! Thank you to everyone for the help!", "date": "2017-10-20 09:02:52", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/need-to-determine-function-based-on-graph.776914/", "openwebmath_score": 0.8605450391769409, "openwebmath_perplexity": 304.7713348992536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307692520259, "lm_q2_score": 0.8947894527758053, "lm_q1q2_score": 0.870478711662486}}
{"url": "https://math.stackexchange.com/questions/128538/does-the-intersection-of-two-finite-index-subgroups-have-finite-index", "text": "# Does the intersection of two finite index subgroups have finite index?\n\nLet $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\\cap K$ also a subgroup of finite index? I feel like need that $[G\\colon(H\\cap K)]$ is a divisor of $[G\\colon H]\\cdot[G\\colon K]$, but I dont't know when this holds.\n\nCan somebody help me out?\n\n\u2022 Step 1: Show that $H \\cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H \\cap K][H\\cap K : H]$. \u2013\u00a0Clive Newstead Apr 5 '12 at 22:41\n\u2022 @MartQ. You don't have to show that it's a divisor. It suffices to show that $[G\\colon(H\\cap K)]\\leq [G\\colon H]\\cdot[G\\colon K].$ You can try to find an injection from a certain set to another. \u2013\u00a0user23211 Apr 5 '12 at 23:02\n\u2022 A related question. \u2013\u00a0user23211 Apr 5 '12 at 23:06\n\u2022 @CliveNewstead The formula $[G:H] = [G:H \\cap K][H\\cap K : H]$ seems incorrect to me. What is $[H\\cap K : H]?$ Did you mean $[G:H\\cap K]=[G:H][H:H\\cap K]?$ \u2013\u00a0user23211 Apr 5 '12 at 23:23\n\u2022 @ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment. \u2013\u00a0Clive Newstead Apr 7 '12 at 11:45\n\nProof $1$: $\\quad[G:H\\cap K]=[G:H][H:H\\cap K]=[G:H][HK:K]\\le [G:H][G:K].$\n\nWe do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:H\\cap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $H\\cap K$.\n\nProof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/H\\times G/K$. The latter is finite so the orbit of $H\\times K$ is finite, and the stabilizer of it is simply $H\\cap K$. Invoke orbit-stabilizer.\n\n\u2022 There's probably a thread in which your second equality is proved, too. I'll go digging. \u2013\u00a0Dylan Moreland Jul 14 '12 at 23:10\n\u2022 \u2013\u00a0Dylan Moreland Jul 14 '12 at 23:15\n\u2022 The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of $\\{gK \\mid g \\in HK\\}$. But then $g=hk$ and so any element of $\\{gK \\mid g \\in HK\\}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense? \u2013\u00a0Al Jebr Sep 23 '18 at 21:08\n\u2022 @AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $h\\in H$. \u2013\u00a0anon Sep 24 '18 at 0:29\n\nLet $\\{H_1,\\dots,H_m\\}$ be the left cosets of $H$, and let $\\{K_1,\\dots,K_n\\}$ be the left cosets of $K$. For each $x\\in G$ there are unique $h(x)\\in\\{1,\\dots,m\\}$ and $k(x)\\in\\{1,\\dots,n\\}$ such that $x\\in H_{h(x)}$ and $x\\in K_{k(x)}$. Let $p(x)=\\langle h(x),k(x)\\rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:\n\nProposition: If $x$ and $y$ are in different left cosets of $H\\cap K$, then $p(x)\\ne p(y)$.\n\nIt follows immediately that $H\\cap K$ can have at most $mn$ left cosets.\n\nIt may be easier to consider the contrapositive of the proposition:\n\nIf $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $H\\cap K$.\n\nYou may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}y\\in H$.\n\n\u2022 Please, professor, could you clarify what does the $p(x)$ mean? \u2013\u00a0Danilo Gregorin Afonso Sep 4 '17 at 1:20\n\u2022 @Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values. \u2013\u00a0Shahab Sep 5 '17 at 15:32\n\u2022 @Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else. \u2013\u00a0Danilo Gregorin Afonso Sep 5 '17 at 19:52\n\nAnother way to see that the answer is \u201cyes\u201d: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 \\subset H$ and $N_2 \\subset K$. Then $G/N_1 \\times G/N_2$ is a finite group; do you see why this implies that $N_1 \\cap N_2$ has finite index in $G$?\n\n\u2022 I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal. \u2013\u00a0anon Jul 14 '12 at 22:59\n\u2022 @anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind. \u2013\u00a0Dylan Moreland Jul 14 '12 at 23:03\n\n$H, K$ be subgroups of $G$. Any $A \\in \\frac{G}{H \\cap K}$ can be written as $A = B \\cap C$, where $B \\in G/H$ and $C \\in G/K$ as follows. For any $g \\in G$ $$g(H\\cap K) = gH \\cap gK$$.\n\nHence, $$[G:H \\cap K] \\leq [G:H] [G:K]$$.\n\nLet $C$ be the set of left cosets of $S = H \\cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C \\to C_1 \\times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| \\leq |C_1| \\cdot |C_2|$, which proves that $[G: S]$ is finite.\n\nLet $$l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l \\in G$$ and $$G$$ be partitioned as $$G=h_1H \\cup ... \\cup h_lH = k_1K \\cup ... \\cup k_mK$$\n\nI will find $$a_1,...a_n \\in G$$ s.t. $$G$$ is partitioned as $$G=a_1(H \\cap K) \\cup ... \\cup a_n(H \\cap K)$$ which means $$n=[G:H \\cap K] < \\infty$$.\n\nLet $$b_1 \\in G$$. Then $$\\exists i_1 \\in \\{1,...,m\\}, j_i \\in \\{1,...,l\\}$$ s.t. $$b_1 \\in h_{i_1}H \\cap k_{j_1}K=b_1H \\cap b_1K = b_1(H \\cap K)$$. Next, let $$b_2 \\in G \\ \\setminus \\ b_1(H \\cap K)$$. Then $$b_2 \\in b_2(H \\cap K)$$ where $$b_2(H \\cap K) \\cap b_1(H \\cap K) = \\emptyset$$ because $$b_1H \\cap b_2H = h_{i_1}H \\cap h_{i_2}H = \\emptyset = k_{j_1}K \\cap k_{j_2}K = b_1K \\cap b_2K$$\n\nwhere $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 \\in \\{1,...,m\\} \\ \\setminus \\ \\{i_1\\}, j_2 \\in \\{1,...,l\\} \\ \\setminus \\ \\{j_1\\}$$ This process continues at most $$lm$$ times for $$a_p=b_p, p \\in \\{1,2,...,n\\}$$ Thus, $$n \\le lm < \\infty.$$\n\nI'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.\n\nSince both $$H$$ and $$K$$ are finite index subgroups of $$G$$, we can consider the right cosets $$G/H$$ and this is finite by hypothesis. In particular $$G$$ acts (transitively) on $$G/H$$ by right multiplication, i.e. $$G/H \\times G \\to G, \\qquad (Hx, g) \\mapsto Hxg.$$ If we now restrict the action to $$K$$, and consider the $$K$$-orbit of $$H$$, call it $$\\mathcal{O}_K(H)$$, then it must be finite since it's contained in $$G/H$$. But now by what Isaacs calls the Fundamental Counting Principle (see \"Finite Group Theory\", theorem 1.4) we exactly know the following equality $$|\\mathcal{O}_K(H)| = [K : \\text{Stab}_K(H)],$$ where $$\\text{Stab}_K(H)$$ is the stabiliser of $$H$$ under the $$K$$-action. Notice that $$\\text{Stab}_K(H) = K \\cap \\text{Stab}_G(H)$$, and what's the stabiliser of $$H$$ for the $$G$$-action? But of course $$H$$ itself, so $$\\text{Stab}_K(H) = K \\cap H$$, and finally we conclude that $$[K : K \\cap H]$$ is finite. Hence $$[G : H \\cap K] = [G : K][K : K \\cap H]$$ must be finite as well.", "date": "2021-01-18 05:20:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/128538/does-the-intersection-of-two-finite-index-subgroups-have-finite-index", "openwebmath_score": 0.9479442834854126, "openwebmath_perplexity": 115.98408916788723, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401455693091, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8704784973854425}}
{"url": "https://math.stackexchange.com/questions/397035/how-to-graph-the-equation-y-frac-x-2x1", "text": "# How to graph the equation: $y=\\frac {x-2}{x+1}$?\n\nthe title says it all. I'm pretty sure this is a hyperbola, but is there an alternative way of doing this besides a table of values?\n\n\"Graph the equation $y=\\frac {x-2}{x+1}$\"\n\nI know that $x$ cannot equal $-1$ but I'm not sure how to carry on from there.\n\nAny help would be appreciated. If you could provide a step by step explanation, and a picture of the graph itself, that would be great :D!\n\n\u2022 Do you mean to show it? \u2013\u00a0Ofir Attia May 20 '13 at 8:36\n\u2022 I just need to graph the equation, but an explanation on how to do it would be nice too :D \u2013\u00a0missiledragon May 20 '13 at 8:36\n\u2022 $y$ can indeed equal 0, i.e, when $x = 2$. ;) \u2013\u00a0user49685 May 20 '13 at 8:38\n\u2022 OH, I didn't realise that :P. Thanks @user49685 \u2013\u00a0missiledragon May 20 '13 at 8:40\n\nWhen you want draw the graph of the function, you need to check a couple of things:\n\n1. Check if there are any points in the domain where the function is undefined, you know there's something happening there. For instance here you saw that when $x=-1$ it is undefined.\n\n2. Then, $y$ can be equal to $0$, there's no problem with that, and you should find all the $x$ such that $y=0$. In this case, when $x=2, y=0$. Thus you know that the function goes through the point $(2,0)$.\n\n3. Now you have all the points where there's something special happening, you can check where the function is positive, and where it is negative. You need to look at all the intervals between the special points. In your case : $(-\\infty,-1),\\,(-1,2),\\,(2,\\infty)$. Take $(-\\infty,-1)$ for example. On this interval, you know that the numerator is always negative and the denominator is always negative as well, so the value for $y\\in(-\\infty,-1)$ will always be positive. Do that for all the intervals.\n\n4. Finally, you need to check some limits to see what happens at those special points and also when you tend to infinity (in your case only $-1$ is a special point because you already now that the value of your function when $x=2$ is $0$). So you should check :\n\n$$\\lim_{x\\rightarrow\\infty}f(x),\\; \\lim_{x\\rightarrow-\\infty}f(x),\\; \\lim_{x\\rightarrow-1^+}f(x),\\; \\lim_{x\\rightarrow-1^-}f(x),\\;$$\n\nThat's it, you have everything you need to graph your function! (If you want to go into further details you can also check the derivative for extremas and the second derivative for inflexion points. This means that you have to compute the derivatives and check for which values of $x$ they are equal to $0$).\n\n\u2022 Thanks! that helped me a lot :D! \u2013\u00a0missiledragon May 20 '13 at 9:06\n\nRewrite as $y=1-3/(x+1)$, or $(y-1)= -3/(x+1)$\n\nThis is simply the regular recriprocal equation, with the axies rescaled. The standard $Y$ axis, representing $X=0$, is now marked $x=-1$. That is, the x-numbers are moved one to the right. The old X axis of 0,1,2,3 become -1, 0, 1, 2...\n\nThe old $X$ axis, or $Y=0$ becomes $y=1$, and we have to rescale by a factor of 3. So the old Y coordinates 0, 1, 2, 3... become 1, -2, -5, -8...\n\nAnd that's it.\n\nFull Function investigation, check of asymptotic (infinite and non setting points), extreme points and inflection points. nice tool to work with is desmos:\n$y=(x\u22122)/(x+1)$ - Click to view\n\nIn addition : Investigating Functions - Click to view\n\nhope its helped you.\n\nJust to add to what Oliver and Wendy have said, To Draw any function:\n1.)Check Domain and critical points.\n2.)Check intercepts.By putting $x=0 ,y=0$\n3.)Check Asymptodes around infinity and critical points by taking $$\\lim_{x\\rightarrow\\infty}f(x),\\; \\lim_{x\\rightarrow-\\infty}f(x),\\; \\lim_{x\\rightarrow-1^+}f(x),\\; \\lim_{x\\rightarrow-1^-}f(x),\\;$$ 4.)Check Symmetry about axis by putting $x = -x$ and $y=-y$\n5.)Intervals of increase/ decrease by equating first derrivative to zero.\nIf you know all this information you can draw any graph manually.\n\nHope this helps :)", "date": "2019-06-17 12:33:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/397035/how-to-graph-the-equation-y-frac-x-2x1", "openwebmath_score": 0.7144014835357666, "openwebmath_perplexity": 250.65714074456136, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347905312772, "lm_q2_score": 0.8933094060543487, "lm_q1q2_score": 0.8704717639681889}}
{"url": "https://math.stackexchange.com/questions/2066004/big-o-notation-is-element-of-or-is-equal?noredirect=1", "text": "# Big O Notation \u201cis element of\u201d or \u201cis equal\u201d\n\nPeople are always having trouble with \"big $O$\" notation when it comes to how to write it down in a mathematically correct way.\n\nExample: you have two functions $n\\mapsto f(n) = n^3$ and $n\\mapsto g(n) = n^2$\n\nObviously $f$ is asymptotically faster than $g$. Is it $f(n) = O (g(n))$ or is it $f(n) \\in O(g(n))$?\n\nMy prof says that the first one is wrong but is a very common practice, therefore it is used very offten in books. Although the second one is the right one.\n\nWhy is that so?\n\n\u2022 Formally, neither are. If you define $O(\\cdot)$ as a set of functions (for asymptotics w.r.t. a given point), say $O(g) = \\{ f : \\exists N,C>0,\\ \\forall n\\geq N\\ f(n) \\leq C\\cdot g(n)\\}$, then you would have to write $f\\in O(g)$. $f(n)$ is not a function, it's a number. But most people are quite happy with the (slight) abuse of notation: $f(n) = O(g(n))$ for $f\\in O(g)$, because there is no ambiguity anyway. \u2013\u00a0Clement C. Dec 20 '16 at 15:13\n\u2022 Ah, because f(n) would be a specific value when f is meant generally, is that right? \u2013\u00a0Blnpwr Dec 20 '16 at 15:14\n\u2022 Indeed. The same way that, technically, $f(x)$ is the value of the function $f$ when evaluated at the point $x$, not the function $f$ itself. \u2013\u00a0Clement C. Dec 20 '16 at 15:15\n\u2022 For this particular case, you have in fact $g(n) = o(f(n))$ or $f(n) = \\omega(g(n))$ because $\\lim_n \\frac{f}{g} = \\infty$ \u2013\u00a0Alex Dec 20 '16 at 15:48\n\u2022 For your example functions it's the other way around. $g(n)=O(f(n))$. I'm assuming we're talking about $n\\to\\infty$, which is hinted by the use of the letter $n$. \u2013\u00a0Meni Rosenfeld Dec 21 '16 at 9:48\n\nI really like Wikipedia's note on this:\n\nThe statement \u201c$f(x)$ is $O(g(x))$\u201d [\u2026] is usually written as $f(x) = O(g(x))$. Some consider this to be an abuse of notation, since the use of the equals sign could be misleading as it suggests a symmetry that this statement does not have. As de Bruijn says, $O(x) = O(x^2)$ is true but $O(x^2) = O(x)$ is not. Knuth describes such statements as \u201cone-way equalities\u201d, since if the sides could be reversed, \u201cwe could deduce ridiculous things like $n = n^2$ from the identities $n = O(n^2)$ and $n^2 = O(n^2)$.\u201d\n\nFor these reasons, it would be more precise to use set notation and write $f(x) \\in O(g(x))$, thinking of $O(g(x))$ as the class of all functions $h(x)$ such that $|h(x)| \\leq C|g(x)|$ for some constant $C$. However, the use of the equals sign is customary. Knuth pointed out that \u201cmathematicians customarily use the $=$ sign as they use the word \u2018is\u2019 in English: Aristotle is a man, but a man isn't necessarily Aristotle.\u201d\n\n\u2022 Gotta love an answer backed up by Knuth, even if it is by a quote inside a quote. \u2013\u00a0Jasper Dec 20 '16 at 17:30\n\u2022 This still isn't a complete answer. Using $f(x)$, $g(x)$, $h(x)$, $n$, $n^2$ etc. to mean functions is also abuse of notation. \u2013\u00a0JiK Dec 21 '16 at 15:07\n\u2022 @JiK you're right, but it is a common abuse of notation, and in my opinion the crux of the question regards whether to use $=$ or $\\in$, not whether or not to be completely formal every time you talk about a function \u2013\u00a0TomGrubb Dec 21 '16 at 15:15\n\u2022 I don't agree with the statement would be more precise to use set notation. It implora that messure theory lenguage almost every point, cotient spaces, etc. are wrong lenguage. I think viewing big O / little o notation in this way is a bery powerful tool and os pretty precise. \u2013\u00a0Ernesto Iglesias Apr 23 '18 at 16:40\n\nUsing $\\in$ is set-theoretically correct but inconvenient. For example, $$\\sin x = x - \\frac{x^3}{3} + \\mathrm O(x^5)$$ In this case $\\mathrm O$ should be interpreted as there exists an $\\mathrm O(x^5)$ function to make this equality valid. The $=$ notation also allows asymptotic notation to appear on both sides and do arithmetic: $$e^x + \\mathrm O(x) = \\mathrm O(e^x)$$ In this case, for every $\\mathrm O(x)$ function on the left, there exists an $\\mathrm O(e^x)$ function on the right to make this an equality.\n\nWarning: In this case the two sides of $=$ cannot be swapped carelessly. e.g. $\\mathrm O(x) = \\mathrm O(e^x)$ but $\\mathrm O(e^x) \\neq \\mathrm O(x)$.\n\n\u2022 \"but inconvenient\" Indeed! I would go as far as saying that using it almost completely defeats the very purpose of the notation, rendering it pointless to introduce it in the first place. \u2013\u00a0quid Dec 20 '16 at 20:02\n\u2022 For what it's worth, there is a way to reconcile this. The way I think about it is that using $\\mathcal O(x^5)$ here changes the abuse of notation somewhat: we're now dealing with sets of functions, so the function $x$ becomes the singleton set $\\{ x \\mapsto x \\}$ (and likewise for $x^3 / 3$), the addition operator becomes setwise product-addition $A \\oplus B = \\{ x \\mapsto f(x) + g(x) : f \\in A, g \\in B \\}$, $\\mathcal O(x^5)$ really is a set, and the whole statement should be read as $(x \\mapsto \\sin x) \\in {\\text{RHS}}$. Perhaps slightly inconvenient, but I don't see anything unsound. \u2013\u00a0wchargin Dec 21 '16 at 8:00\n\u2022 This reminds me of equality of multifunctions, like if we say $\\cos^{-1} 1 = 2k\\pi$ \u2013\u00a0GFauxPas Dec 21 '16 at 14:24\n\u2022 @wchargin Using $O(x^5)$ does not automatically mean we are dealing with sets of functions. All that the writer means is that they're dealing with a specific quantity which satisfies the definition of being $O(x^5)$. The purpose of notation is to communicate what a mathematician is thinking. Introducing sets of functions\u2014frankly, no one cares about the set of all functions that are $O(x^5)$ and it's an uninteresting set; we only care that the particular function we're working with is $O(x^5)$\u2014is backwards, saying we should change our thinking to fit (someone's narrow idea of) $=$ notation. \u2013\u00a0ShreevatsaR Dec 21 '16 at 20:42\n\u2022 @SheevastaR \u00a1\u00a1\u00a1Of course we care for the set $O(g)$!!! if you don't is a problem you have. Read the answer you are commenting and you will notice how important it is. \u2013\u00a0Ernesto Iglesias Apr 23 '18 at 16:49\n\n$O(g(x))$ is a class of functions - think of it as a \"property\" functions can have. By the literal interpretation of the equals sign, \"$f(x) = O(g(x))$\" should be interpreted as \"$f$ is literally equal to a certain class of functions.\" But functions and classes of functions are different sorts of things - even if this was what we meant to say, it's like saying that one particular apple is equal to a basket of apples. But what we mean when we say \"$f(x) = O(g(x))$\" is that $f$ belongs to the class of functions $O(g(x))$ - so, $f(x) \\in O(g(x))$.\n\nThe reason we use $=$ instead of $\\in$ is because, given the particular uses of big-$O$ notation (and little-$o$ notation, if you're familiar with that) $=$ is massively more convenient. We say things like $x^3 + O(x) = O(x^3)$, for example; we don't mean that $O(x)$ is an object that can actually be added to $x^3$, or that when that addition is done we actually get the class of functions $O(x^3)$, we just mean that for any function $f \\in O(x)$, the function $x^3 + f(x)$ is a member of $O(x^3)$. But if I wanted to write that out in more standard notation, I'd have to say something like $\\{x^3 + f(x) \\mid f(x) \\in O(x)\\} \\subseteq O(x^3)$. This is inconvenient to write and difficult to read, so we prefer the \"slicker\" notation $x^3 + O(x) = O(x^3)$.\n\nHowever, I'm not sure I would say that sentences like $f(x) = O(g(x))$ are wrong. By convention, they're perfectly right - it's just that when an expression includes $O$ (or $o$), $=$ does not mean what it usually means. That's okay.\n\n\u2022 I think there must be someone that had invented a convenient way to just say $x^3 + O(x) = O(x^3)$, and at the same time, the symbols are entirely rigorous. \u2013\u00a0Eric Dec 20 '16 at 16:29\n\u2022 @Eric Well, $x^3 + O(x) = O(x^3)$ is an entirely rigorous and convenient way to say that. It just doesn't look rigorous if you don't know the definition of \"$=$\" in the context of $O$. Since everyone in the field understands this notation, no one's particularly motivated to find a new way of writing it that won't confuse outsiders. \u2013\u00a0Reese Dec 20 '16 at 16:34\n\u2022 @Reese If you treat $O(x)$ as a set, then $x^3 + O(x) \\subseteq O(x^3)$ is rigorous without domain-specific $=$ sign. \u2013\u00a0heinrich5991 Dec 21 '16 at 1:02\n\u2022 @heinrich5991 But that requires domain-specific $+$. \u2013\u00a0Reese Dec 21 '16 at 1:34\n\u2022 @heinrich5991 Oh, absolutely, it's common. But so is using $=$ for $O$ and $o$. My point is that it's still not transparent to first learners. \u2013\u00a0Reese Dec 22 '16 at 17:29\n\nI'll try to clarify wchargin comment.\n\nIf you are familiar with Group theory you should not find the $=$ sign awkward. Class of equivalence and quotient spaces usually use the sum notation for doing that.\n\ne.g., $\\mathbb{R}/\\mathbb{Z}$ elements are usually written as $a+\\mathbb{Z}$.\n\nThe problem is that big $O$ notation makes implicit use of that, i.e. in the class of $$O(g)=\\lbrace f:\\exists M,C>0, \\forall x_0\\geq M: f(x_0)\\leq Cg(x_0) \\rbrace,$$ as exposed by Clement C.\n\nIn this way, $$f=O(g)$$ Means $$f+O(g)=0+O(g)$$ And that is to say than $f\\in O(g)$.\n\nRegarding $=$ being more useful/convenient than $\\in$ because of allowing things like $x^3 + O(x) = O(x^3)$, it seems like for that you could use $x^3 + O(x) \\subseteq O(x^3)$. This seems more precise to me, because it would mean \"every element of the set $x^3 + O(x)$ is an element of the set $O(x^3)$.\" which is, I think, exactly what one wants to say. (where $g+A:=\\{g+f:f\\in A\\}$ and when $g$ is a function and $A$ is a set of functions, and similarly for similar situations.)\n\nMaking the statement with subset is mentioned by Reese's answer, but I think that defining the sum of a function and a set of functions, and also the sum of sets of functions, in the straightforward way removes the inconveniences of saying things like $\\{x^3+f(x)\u2223f\u2208O(x)\\}\\subseteq O(x^3)$. (Defining things similarly with multiplication, and application of functions in general.)\n\nWhen things are defined this way, it should become as simple as using $\\in$ instead of $=$ when there is a single function on the left, and using $\\subseteq$ instead of $=$ when there is a set of functions on the left, and it seems more precise.\n\n\u2022 No, set-theoretic inclusion is not \"exactly what one wants to say\". The purpose of notation is to aid communication, and it is meant to fit our way of thinking. It is misguided to insist that one should change one's (very productive) way of thinking just to fit some notation or formalism. When an analyst writes $\\sin x = x - \\frac{x^3}{3} + \\mathrm O(x^5)$ they are literally thinking that $\\sin x$ is equal to $x$ minus $\\frac{x^3}{3}$ plus some function that is $\\mathrm O(x^5)$ \u2014 they are thinking of equality, not set inclusion, so the notation reflects that. \u2013\u00a0ShreevatsaR Dec 21 '16 at 18:54\n\u2022 Ok, I should probably defer to you here. But isn't saying that $sin(x)$ is equal to $x - \\frac{x^3}{3}$ plus some function that is $O(x^5)$ the same thing as saying that $sin(x)$ is an element of the set of functions which are of the form $x - \\frac{x^3}{3}$ plus some function that is $O(x^5)$? \u2013\u00a0drocta Dec 21 '16 at 20:31\n\u2022 Well, is saying \"the sky is blue\" the same thing as saying \"the sky is an element of the set of things that are blue\"? In some sense yes, and in another sense no. (It may be logically equivalent, but the speaker may not have even considered, or care about, the set of all things that are blue.) One can also say \"the sky satisfies the property of being blue\", or even \"blueness inheres in the sky\". There can be many ways to formalize a statement, and the most fruitful ones are often the one that closest match what the human is thinking, rather than what fits cleanest with other formalisms. \u2013\u00a0ShreevatsaR Dec 21 '16 at 20:45\n\u2022 (sorry, I hit enter by mistake. First time using SE comments.) Your point about the purpose being for communication seems like something I failed to consider. But $\\in$ being like \"has property\" and $\\subseteq$ being like \"all things with property on the left have the property on the right$, seem like they capture the$\\forall$or$\\exists$in what is being said with \"there exists a function of the form$x-x^3/3+O(x^5)$which is$sine(x)$, and \"for all functions of the form <blah>, there is a function of the form <bleh> such that they are equal\". But maybe it's not nice enough to justify sets? \u2013 drocta Dec 21 '16 at 20:57 Here is my five pennies worth: If we see a statement of the form $$f(x)=g(x)+O\\bigl(p(x)\\bigr)\\qquad(x\\to\\xi)$$ then for each$x$the exact value of the term$O\\bigl(p(x)\\bigr)$is defined by this very equation: it is$:=f(x)-g(x)$. In addition we are told that there is a constant$C>0$such that for all$x$in a suitable punctured neighborhood of$\\xi$this difference is$\\leq C\\bigl|p(x)\\bigr|\\$.", "date": "2019-10-15 17:35:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2066004/big-o-notation-is-element-of-or-is-equal?noredirect=1", "openwebmath_score": 0.8369766473770142, "openwebmath_perplexity": 284.42606210262255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615794, "lm_q2_score": 0.8933093946927837, "lm_q1q2_score": 0.8704717502442257}}
{"url": "https://math.stackexchange.com/questions/2663663/finding-the-sin-inverse", "text": "# Finding the sin inverse\n\nWell, so I was finding the value of $\\sin 15^{\\circ}$. I used the identity $2 \\sin x \\cos x = \\sin 2x$. So solving the quadratic equation and picking the right values simply gave me:\n\n$$\\sin 15^{\\circ} = \\frac {\\sqrt {2 - \\sqrt {3}}} {2}$$\n\nThat was pretty much simple. But I thought what if I was given the value instead, and I was asked to find the angle, whose sine would give me that value.\n\nIn short, my question is how to evaluate:\n\n$$\\sin^{-1} \\left( \\frac {\\sqrt {2 - \\sqrt {3}}} {2} \\right)$$\n\nI want to solve it thinking that I do not know the value of $\\sin 15$, as if I have been just provided with the problem, and I have no idea what the answer might be. I tried to proceed by converting the $\\sin$ to $\\tan$ but that seemed to be of no use. Can anybody help?\n\n\u2022 $$15=45-30=60-45$$ \u2013\u00a0lab bhattacharjee Feb 23 '18 at 18:44\n\u2022 You are doing that because you know that the answer is 15. How is someone supposed to know that just on seeing the problem? \u2013\u00a0Manish Kundu Feb 23 '18 at 18:48\n\u2022 Please find my answer \u2013\u00a0lab bhattacharjee Feb 24 '18 at 10:46\n\nIn general it's hard, if not impossible. In this case, it turns out to be doable.\n\nThe main idea is that if $\\theta = \\sin^{-1}\\left( \\frac{\\sqrt{2 - \\sqrt 3}}{2} \\right)$, then $\\sin \\theta = \\frac{\\sqrt{2 - \\sqrt 3}}{2}$, and we have an equation we can (hopefully) solve for $\\theta$.\n\nSquaring, we get $$\\sin^2 \\theta = \\frac{2 - \\sqrt{3}}{4},$$ and multiplying by 4, $$4\\sin^2 \\theta = 2 - \\sqrt{3}.$$\n\nMy thoughts are that maybe if I get $\\sqrt{3}$ by itself and square, I'll get a nice equation that's polynomial in $\\sin \\theta$ with integer coefficients. So, we get $\\sqrt{3}$ by itself: $$\\sqrt{3} = 2 - 4 \\sin^2 \\theta.$$ But that's interesting: I can factor 2 out... $$\\sqrt{3} = 2(1 - 2 \\sin^2 \\theta),$$\n\nand that $1 - 2 \\sin^2 \\theta$ looks pretty good, because that's the output from a double angle identity, namely $\\cos(2 \\theta)$. Now, $$\\sqrt{3} = 2 \\cos(2 \\theta),$$ so $\\cos (2\\theta) = \\frac{\\sqrt{3}}{2}$.\n\nThis happens when $2 \\theta = 30^\\circ$ or $2\\theta = 330^\\circ$.\n\nSo, we get one possibility that $\\theta = 15^\\circ$ (since we've solved an equation by squaring, though, we should make sure it's not extraneous!).\n\nAnd of course, once we know one angle $\\theta$ with a particular sine, we can use symmetry and the unit circle to find all other such angles (that would be $180^\\circ - \\theta = 180^\\circ - 15^\\circ = 165^\\circ$ in this case, and everything coterminal with the two between $0^\\circ$ and $360^\\circ$). If we just want the inverse sine of something positive, it's an angle in quadrant I.\n\n\u2022 Sidenote, running off to class now, won't be able to respond for a few hours (if it matters) \u2013\u00a0pjs36 Feb 23 '18 at 19:07\n\u2022 That was really amazing. Didn't expect it to be solved so simply. \u2013\u00a0Manish Kundu Feb 23 '18 at 19:23\n\u2022 @ManishKundu Thanks! I was surprised too; it's a really nice question, and I'm happy you thought to ask it. \u2013\u00a0pjs36 Feb 24 '18 at 2:26\n\n$$2-\\sqrt3=\\dfrac{(\\sqrt3-1)^2}2$$\n\n$$\\implies\\dfrac{\\sqrt{2-\\sqrt3}}2=\\dfrac{\\sqrt3-1}{2\\sqrt2}$$ as $\\sqrt3-1>0$\n\n$$=\\sin60^\\circ\\cos45^\\circ-\\sin45^\\circ\\cos60^\\circ$$\n\n$$=\\sin(60-45)^\\circ$$\n\nNow we know the principal value of $\\sin^{-1}x$ lies in $\\in[-90^\\circ,90^\\circ]$ for real $x$\n\n\u2022 You made it simple. That was nice. \u2013\u00a0Manish Kundu Feb 24 '18 at 15:41\n\n+1 to pjs36's answer, and they did bring up a good point that in this case it was possible to solve. But what if we weren't able to use the double angle identity? Perhaps another way to go would be the use of MacLaurin series.\n\nGiven a problem $\\sin^{-1}{A} =\\theta$, where $\\theta$ is in radians and $A \\in [-1,1]$, we can first take the $\\sin$ of both sides to get $$\\sin{\\theta} = A$$\n\nThe MacLaurin series for $\\sin{\\theta}$ is $$\\sin\\theta=\\sum_{n=0}^\\infty \\frac{(-1)^n \\theta^{2n+1}}{(2n+1)!} = \\theta - \\frac{\\theta^3}{3!}+\\frac{\\theta^5}{5!} - \\dots =A$$\n\nThe farther out you extend the polynomial, the better the estimate is going to be when you solve for $\\theta$. However it will be only an estimate when you do it this way, and you would probably need to use a calculator or WolframAlpha or something to solve it as well, which probably defeats the purpose of avoiding just plugging in $\\sin^{-1} A$ into the calculator in the first place. This is merely another way you could look at the problem, if you wanted to avoid using inverse trig.\n\n\u2022 So I need to solve for $\\theta$. Wait, so do I need to use Newton Raphson? \u2013\u00a0Manish Kundu Feb 23 '18 at 19:49\n\u2022 Correct, we want to solve for $\\theta$. However, this would be equivalent to solving the roots for a large polynomial. Newton Raphson could work, as the derivative of a polynomial is simple to find, but it may take alot of time still to do by hand \u2013\u00a0WaveX Feb 23 '18 at 19:52\n\u2022 We would also have the case that we would be estimating an estimate. \u2013\u00a0WaveX Feb 23 '18 at 19:59\n\u2022 I see now why solving such problems can be so hard in general. \u2013\u00a0Manish Kundu Feb 23 '18 at 20:00\n\nFor a shortcut evaluation of $x$ such that $\\sin^{-1}(x) =k$, assuming that $x$ is \"small\", you could use the simplest Pad\u00e9 approximant $$\\sin^{-1}(x)\\approx \\frac{x}{1-\\frac{x^2}{6}}$$ leading to $$x\\approx \\frac{\\sqrt{6 k^2+9}-3}{k}$$\n\nFor your case, where $k= \\frac {\\sqrt {2 - \\sqrt {3}}} {2}$, this would give $$x\\approx \\sqrt{2+\\sqrt{3}} \\left(\\sqrt{48-6 \\sqrt{3}}-6\\right)\\approx 0.255992$$ in radians (that is to say $\\approx 14.6673$ in degrees).\n\n\u2022 That was helpful. Thanks. \u2013\u00a0Manish Kundu Feb 24 '18 at 7:10\n\nNot an answer, just an explicit formula for the $$\\arcsin$$ function.\n\nFirst recall that $$\\sin x=\\frac{e^{2ix}-1}{2ie^{ix}}$$ So if we want the function $$y=\\arcsin x$$ we must solve the following equation for $$y$$: $$x=\\frac{e^{2iy}-1}{2ie^{iy}}$$ if we set $$u=e^{iy}$$ this becomes much easier: $$x=\\frac{u^2-1}{2iu}$$ $$u^2-1=2iux$$ $$u^2-2iux-1=0$$ The use of the quadratic formula then gives $$u=ix+\\sqrt{1-x^2}$$ And using $$\\log$$ to denote the complex natural logarithm: $$iy=\\log\\left(ix+\\sqrt{1-x^2}\\right)$$ $$\\arcsin x=-i\\log\\left(ix+\\sqrt{1-x^2}\\right)$$", "date": "2019-10-23 01:10:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2663663/finding-the-sin-inverse", "openwebmath_score": 0.8780524730682373, "openwebmath_perplexity": 223.1188836671599, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138190064204, "lm_q2_score": 0.8902942268497305, "lm_q1q2_score": 0.8704529685726183}}
{"url": "http://physics.stackexchange.com/questions/78696/angle-needed-for-object-a-to-intercept-with-object-b", "text": "# Angle needed for object A to intercept with object B\n\nObject B is 15 degrees East of North at a distance of 20km/h. Object B is moving at an average speed of 30km/h in the direction 40 degrees East of North. If object A is capable of moving at 100km/h, at what angle does it need to move to intercept object B?\n\nI've tried calculating it as vectors, but got stuck. In the end I just tried using good old geomatry, where $$\\frac{\\sin A}{a} = \\frac{\\sin B}{b}$$ So I called the angle I'm searching for A, and the angle from the line between A and B, and the direction of object B, as angle B. 'a' and 'b' are the average speeds of object a and b. $$\\frac{\\sin A}{30 \\times t} = \\frac{\\sin(155\u00b0)}{100\\times t}$$ $$A = arcsin\\left(\\frac{30\\times t\\times \\sin(155\u00b0)}{100\\times t} \\right) = 7.284\u00b0$$\n\nSo the final angle should be $15\u00b0+7.284\u00b0=\\mathbf{22.284\u00b0}$\n\nIs this correct? Isn't there a better way of doing this, like using vector products?\n\nPS. I've found a very similar question here, but didn't quite understand how i.e. $\\alpha + \\beta = 40\u00b0$, because my calculations show $=35\u00b0$. And trying to read through comments and answers, confuses me more due to updates being made inconsistent to comments about the updates done. And due to me not having enough reputation, I'm not able to ask counter-questions.\n\nBONUS Question: I have been searching high and low on how to make a simple graph of my problem. Like this here question, with a graph of the problem hosted at stack.imgur.com which I guess was generated by LaTeX on stackexchange.com, right?\n\n-\ngnuplot makes nice graphs, as does MATLAB, but gnuplot is free. \u2013\u00a0Pranav Hosangadi Sep 27 '13 at 3:29\nThanks @PranavHosangadi! I didn't realize I had to make them and them upload the photo. I thought MathJax had a plugin to make LaTeX plots in posts. This makes more sense now! \u2013\u00a0nikc Sep 27 '13 at 7:01\nI wouldn't use Gnuplot or Matlab for this; They are good for plotting data, but would basically require you to solve the problem in order to plot something like this. I would use Inkscape for this kind of a drawing. \u2013\u00a0Colin McFaul Sep 27 '13 at 15:06\n\nSimple way to solve: Use distances travelled. So B has already travelled 20km at 15\u00b0 East of North. That is $20 \\times \\sin(15\u00b0)$km northwards and $20 \\times \\cos(15\u00b0)$ eastwards.\n\nNow consider A intercepts B at a time $t$. The equations for the distances travelled by B are:\n\n$Y_B = 20 \\times \\sin(15\u00b0) + 30 \\times t \\times \\sin(40\u00b0)$ in the Y direction and\n$X_B = 20 \\times \\cos(15\u00b0) + 30 \\times t \\times \\cos(40\u00b0)$ in the X direction.\n\nFor A, the distances travelled are:\n$Y_A = 100 \\times t \\times \\sin(\\alpha)$ in the Y direction and\n$X_A = 100 \\times t \\times \\cos(\\alpha)$ in the X direction.\n\nFor the two to intercept, $X_A = X_B$ and $Y_A = Y_B$\n\nYou now have 2 equations, 2 variables, and a whole lot of fun solving them simultaneously.\n\n-\n\nThere is a technique based on vectors that solves the problem, and lets you know if there is no solution, one solution or two solutions.\n\n(All my angles are measured counter-clockwise from the positive X-axis)\n\nAssume the pursuer travels at 100 km/hr at an angle $\\beta$. We're going to try to divide this velocity into two non-perpendicular components\n\nOne matches the velocity of the target exactly. So in the problem above, assume that one part of A's velocity is $30\\angle 50\u00b0$. It wouldn't matter if A's total velocity were less than 30.\n\nSo, we have (in theory) completely cancelled out B's attempt to escape. Now, how does A reach B? A must travel at some velocity (say X) along the original direction from A to B. (Since we've cancelled out any of B's movement) So, in this problem, the other part of A's velocity is $X \\angle 75\u00b0$\n\nSo, finally, we're reduced to solving:$$30\\angle 50\u00b0+X \\angle 75\u00b0=100\\angle\\beta$$One way is to expand each vector into X and Y components and set up two equations. If you eliminate $\\beta$ you get a quadratic in X. If there are any real positive values of X, you can then substitute and solve for $\\beta$\n\nExpanding the equation above for components in the X and Y direction: $$30\\cos(50)+X\\cos(75)=100\\cos(\\beta)$$ $$30\\sin(50)+X\\sin(75)=100\\sin(\\beta)$$The only unknowns are the size of X (or if it even exists!) and the angle $\\beta$ If you square both sides of both equations and then add the two LHS and two RHS, some trig manipulations will eliminate $\\beta$ and lead to the quadratic in X.\nThanks for your answer, but I'm still not quite sure how to use your method? I'm still learning how to solve equations using polar coordinates and I don't know how to isolate the equation above. Plus, I'm trying to find angle $\\beta$, but how would I do that if there are two unknown factors in the equation? \u2013\u00a0nikc Sep 27 '13 at 8:18", "date": "2016-07-02 03:48:43", "meta": {"domain": "stackexchange.com", "url": "http://physics.stackexchange.com/questions/78696/angle-needed-for-object-a-to-intercept-with-object-b", "openwebmath_score": 0.7846698760986328, "openwebmath_perplexity": 354.52590605360626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138164089085, "lm_q2_score": 0.8902942290328345, "lm_q1q2_score": 0.8704529683945195}}
{"url": "https://mathhelpboards.com/threads/captainblacks-occasional-problem-3.857/", "text": "# CaptainBlacks Occasional Problem #3\n\n#### CaptainBlack\n\n##### Well-known member\nProve that the polynomials:\n\n$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\\ \\ \\ (n=1,2...)$\n\nhave no real roots.\n\nCB\n\nLast edited:\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nStill no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that\n\n$P_1(x) = x^2-2x+3 = (x-1)^2+2,$\n\n$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$\n\n$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$\n\nThat should be enough of a pattern to suggest a solution.\n\n#### Alexmahone\n\n##### Active member\nStill no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that\n\n$P_1(x) = x^2-2x+3 = (x-1)^2+2,$\n\n$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$\n\n$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$\n\nThat should be enough of a pattern to suggest a solution.\nClaim: $P_n(x)=(x-1)^2R_n(x)+n+1$\n\nLet $Q_n(x)=P_n(x)-n-1=x^{2n}-2x^{2n-1}+3x^{2n-2}-\\ldots-2nx+n$\n\n$Q_n(1)=1-2+3-\\ldots-2n+n=0$\n\n$Q_n'(x)=2nx^{2n-1}-2(2n-1)x^{2n-2}+3(2n-2)x^{2n-3}-\\ldots-2n$\n\n$Q_n'(1)=2n-2(2n-1)+3(2n-2)-\\ldots-2n=0$ (1st term cancels the last term, 2nd term cancels the 2nd last term etc.)\n\n$Q_n(1)=Q_n'(1)=0\\implies Q_n(x)=(x-1)^2R_n(x)$\n\nSo, $P_n(x)=(x-1)^2R_n(x)+n+1$\n\nNow we just need to prove that $R_n(x)$ is always positive. It looks like $R_n(x)=x^{2n-2}+2x^{2n-4}+3x^{2n-6}+\\ldots+n$. This needs a proof.\n\nLast edited:\n\n#### CaptainBlack\n\n##### Well-known member\nProve that the polynomials:\n\n$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\\ \\ \\ (n=1,2...)$\n\nhave no real roots.\n\nCB\nA slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.\n\nCB\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nStill no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that\n\n$P_1(x) = x^2-2x+3 = (x-1)^2+2,$\n\n$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$\n\n$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$\n\nThat should be enough of a pattern to suggest a solution.\nBy following the hints given by Opalg, I see that\n\n$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\\ \\ \\ (n=1,2...)$\n\nIt's obvious that $P_n(x)\\ne 0$ for all $x\\in R$.\n\nThus it suggests that $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\\ \\ \\ (n=1,2...)$ has no real roots must be true.\n\nAm I on the right track?\n\n#### Alexmahone\n\n##### Active member\nAm I on the right track?\nYes. You have to prove this, though:\n\n$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\\ \\ \\ (n=1,2...)$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nA slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.\n\nCB\nIf I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.\n\nFirst, I assume the $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has real roots.\n\nFrom $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$, I get\n$P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-......-2n(-x)+(2n+1)$\n$P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-......+2nx+(2n+1)$\nNo sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.\n\nNow, I consider the case where $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$.\nI see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.\n\nThat is to say, we're now need to consider two cases:\nI: the number of positive real roots = 2n or\nII: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\\pm b\\sqrt {c}$ where $a>b\\sqrt {c}$ =2k, $k=1, 2, ......, n$\n\nBut we can also tell right from the start that the product of all roots of $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ is -(2n+1) for n=1, 2, ......, that is, the product of all the real positive roots (be it two integers/fractions or two pair of positive complex roots) = -ve.\n\nObviously this leads to contradiction and we can conclude at this point that the polynomial $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has no real roots.\n\nEdit: I made a horrible mistake by saying that the product of all roots = - (2n+1) which is wrong. It should be (2n+1). So, my effort in proving in this post have been amounted to zero, zip, nothing. Please kindly dismiss it.\n\n---------- Post added at 07:02 ---------- Previous post was at 07:00 ----------\n\nYes. You have to prove this, though:\nI think the pattern could be generalized from the hint given by Opalg.\n\nLast edited:\n\n#### CaptainBlack\n\n##### Well-known member\nIf I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.\n\nFirst, I assume the $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has real roots.\n\nFrom $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$, I get\n$P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-......-2n(-x)+(2n+1)$\n$P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-......+2nx+(2n+1)$\nNo sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.\n\nNow, I consider the case where $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$.\nI see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.\n\nThat is to say, we're now need to consider two cases:\nI: the number of positive real roots = 2n or\nII: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\\pm b\\sqrt {c}$ where $a>b\\sqrt {c}$ =2k, $k=1, 2, ......, n$\nWhat are positive complex roots, and where will you find a reference to what Descartes rule of signs says about this sort of sign of the complex roots?\n\nCB\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nMy mistakes.\nI want to clarify that the following observations were not right.\n\"That is to say, we're now need to consider two cases:\nI: the number of positive real roots = 2n or\nII: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\\pm b\\sqrt {c}$ where $a>b\\sqrt {c}$ =2k, $k=1, 2, ......, n$\n\"\nReason:\n1. I didn't know why on earth I called $a\\pm b\\sqrt {c}$ as complex roots. I'm sorry.\n2. This \"...based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.\" was taken at http://en.wikipedia.org/wiki/Descartes'_rule_of_signs whereas the two generated cases were my own conclusions.\n\nI'm sorry for confusing you, CB.\nMy apologies.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nYes. You have to prove this, though:\n$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\\ \\ \\ (n=1,2...)$\nMy proof of that was just to multiply out the product on the right hand side:\n\n\\begin{aligned}(x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x^2-2x+1) \\\\ =& x^{2n}\\phantom{{}-2x^{2n-1}}+2x^{2n-2}\\phantom{{}-4x^{2n-3}}+3x^{2n-4} \\phantom{{}-2x^{2n-1}}+ \\ldots \\\\ &\\phantom{x^{2n}} - 2x^{2n-1} \\phantom{{}+2x^{2n-2}} - 4x^{2n-3} \\phantom{{}+2x^{2n-2}} -6x^{2n-5} -\\ldots \\\\ &\\phantom{x^{2n} - 2x^{2n-1}} +\\phantom{1}x^{2n-2}\\phantom{{}-2x^{2n-1}} + 2x^{2n-4} \\phantom{{}-2x^{2n-1}} +\\ldots \\\\ &\\phantom{x}\\\\ =& x^{2n}-2x^{2n-1}+3x^{2n-2}-4x^{2n-3} + 5x^{2n-4}-6x^{2n-5}+\\ldots \\end{aligned}\n\nFor $0\\leqslant k\\leqslant n-1$ the coefficient of $x^{2n-2k}$ is $k+(k+1)=2k+1$ and the coefficient of $x^{2n-2k-1}$ is $-2(k+1)$. The constant term is $n$. So the resulting polynomial is $P_n(x) - (n+1)$.\n\n#### CaptainBlack\n\n##### Well-known member\nSolution: CaptainBlacks Occasional Problem #3\n\nThis again is from the Purdue PotW\n\nProve that the polynomials:\n\n$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\\ \\ \\ (n=1,2...)$\n\nhave no real roots.\n\n================================================\n\nSolution:\nThese obviously have no negative roots (either from Descartes rule of signs or observing that the odd powers of $$x$$ all have -ve signs)\n\nSo for $$x>0$$ consider:\n$P_n(x)+xP_n(x)=x(x^{2n}-x^{2n-1}+x^{2n-2}- ... - x+1) + (2n+1)$\nNow the bracketed term is a finite geometric series and so we may replace it with its sum to get:\n$P_n(x)(1+x)=x \\left[ \\frac{1+x^{2n+1}}{1+x}\\right] + (2n+1)$\nor:\n$P_n(x)=x \\left[ \\frac{1+x^{2n+1}}{(1+x)^2}\\right] + \\frac{2n+1}{1+x}>0$\nand so as $$P_n(0) \\ne 0$$ we have $$P_n(x)>0$$ for all real $$x$$\n\nThe above is somewhat different from the solution given on the Purdue PotW site, mainly in that it avoids an unexplained step, which while probably valid I find opaque.\n\nCB\n\nLast edited by a moderator:", "date": "2020-09-19 05:41:56", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/captainblacks-occasional-problem-3.857/", "openwebmath_score": 0.9600052237510681, "openwebmath_perplexity": 507.549257095013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232904845687, "lm_q2_score": 0.8856314828740729, "lm_q1q2_score": 0.8704192481550242}}
{"url": "https://math.stackexchange.com/questions/1471987/drawing-without-replacement-prob-for-an-ace-followed-by-an-ace", "text": "Drawing without replacement - prob. for an Ace followed by an Ace?\n\nGiven a standard 52-cards deck:\nYou are extracting cards from the deck without replacement, until you get an \"Ace\" for the first time. What is the probability that the next card will be \"Ace\" too?\n\nI've already seen the following Q&A: Probability of drawing an Ace: before and after\nAccording to that thread, the answer should be: $$\\frac{4 \\cdot 3}{52 \\cdot 51} = \\frac{1}{221}$$\n\nBut the official solution says that the answer is: $$\\frac{4}{52}$$ which doesn't make sense IMHO. They solved it only with intuition or \"mind trick\", as they wrote..\n\nMy calculation:\n\nAssuming that the 1st card is Ace, then: $$\\frac{4 \\cdot 3}{52 \\cdot 51} = \\frac{1}{221}$$\n\nAssuming that the 2nd card is Ace, then: $$\\frac{(52-4) \\cdot 4 \\cdot 3}{52 \\cdot 51 \\cdot 50} = \\frac{24}{5525}$$\n\nWe notice a pattern here.\nHaving the 1st Ace at the $k$'th draw, then the probability (for a second Ace after that) is: $$p_1 = \\frac{ {_{52-4}P_{k-1}} \\cdot 4 \\cdot 3 }{ {_{52}P_{k-1}} \\cdot {_{52-k}P_{2}} }$$\n\nWe need to consider the least-possible scenario - we draw 48 non-Ace cards, then: $$p_2 = \\frac{48! \\cdot 4 \\cdot 3}{ {_{52}P_{50}} } = \\frac{1}{270725}$$\n\nSo, the required probability is: \\begin{align} p &= p_2 + \\sum\\limits^{48}_{k=1} p_1 \\\\ &= p_2 + \\sum\\limits^{48}_{k=1} \\frac{ {_{48}P_{k-1}} \\cdot 4 \\cdot 3 }{ {_{52}P_{k-1}} \\cdot {_{52-k}P_{2}} }\\\\ &= \\frac{1}{270725} + \\frac{1696}{20825}\\\\ &= \\frac{1297}{15925}\\\\ &\\cong 0.081444 \\end{align}\n\nBut my answer is far from either the official solution and from the answer in that thread.\n\n\u2022 The question you link to is different. It is asking \"what's the probability that the second drawn is an Ace?\" The answer to that is certainly $\\frac 4{52}$ as there is no difference between the second card drawn and the first. Of course, if you specify that the first card drawn was (or was not) itself an Ace then the probability changes. \u2013\u00a0lulu Oct 9 '15 at 14:35\n\u2022 @lulu I think that I understand your view on the matter. So what did I calculated? \u2013\u00a0Dor Oct 9 '15 at 15:07\n\u2022 I think your method ought to work. To be precise: I just wrote a code which first computes $p_k$, the probability that the first A is observed on the $k^{th}$ draw, and then calculates the probability that the next draw is also an A. I believe this is what you were doing as well. My code returned $\\frac 1{13}$ as the total. Doesn't convey a lot of insight, I understand. \u2013\u00a0lulu Oct 9 '15 at 15:13\n\u2022 @lulu Indeed you're right! I posted an answer which shows where were my mistakes and include a code that proves it to be equal to $\\frac{1}{13}$. \u2013\u00a0Dor Oct 9 '15 at 16:27\n\nWe use a fairly crude counting approach, in order to rely minimally on intuition. There are $\\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the \"favourables.\"\n\nMaybe the first two Aces are in positions 1 and 2. That leaves $\\binom{50}{2}$ for the rest.\n\nMaybe they are in positions 2 and 3. That leaves $\\binom{49}{2}$ for the rest.\n\nMaybe they are in positions 3 and 4. That leaves $\\binom{48}{2}$ for the rest.\n\nAnd so on, up to positions 49 and 50, leaving $\\binom{2}{2}$ for the rest. That gives probability $$\\frac{\\binom{50}{2}+\\binom{49}{2}+\\cdots+\\binom{2}{2}}{\\binom{52}{4}}.$$ The numerator can be simplified in various ways. My favourite is to count the number of ways to choose $3$ numbers from the first $51$. This can be done in $\\binom{51}{3}$ ways. But the smallest number can be $1$, leaving $\\binom{50}{2}$ ways to choose the other two. Or the smallest can be $2$, and so on. We get the number of favourables obtained above.\n\nOur probability is therefore $\\frac{\\binom{51}{3}}{\\binom{52}{4}}$. This simplifies to $\\frac{1}{13}$.\n\n\u2022 Great argument and very well written. +1. \u2013\u00a0Shailesh Oct 9 '15 at 15:04\n\u2022 I like this approach! +1. Could you please elaborate regarding how you simplified the numerator? \u2013\u00a0Dor Oct 9 '15 at 15:13\n\u2022 I think that I understand it now. That is a very cool solution..! \u2013\u00a0Dor Oct 9 '15 at 15:33\n\nFirst of all Andre has posted a very nice answer. Already Upvoted it. I am trying to look at where all you have gone wrong. Too long for an comment, so posting it here.\n\nLet us look at the last case. You have $48!$ ways of the non-ace cards. Now the remaining cards are all aces anyways. So the question of multiplying this by $4 \\cdot 3$ does not arise. Since we do not care about the last 2 cards, the denominator will have to be just $50!$. Similarly, the case before that might need to rectified.\n\n\u2022 Indeed you are right about the last term! In the other 48 terms, the mistake was different. I'll add info in my original post. Thanks! +1 \u2013\u00a0Dor Oct 9 '15 at 16:08\n\u2022 @Dor Yes, right. A good approach would be you post an answer yourself, instead of editing the post. I'll surely upvote it. \u2013\u00a0Shailesh Oct 9 '15 at 16:18\n\nHere is a bijection between set of all permutation with the first Ace followed by an Ace and set of all permutations with the first card being an Ace\n\n$$B^nAAX <-> AB^nAX$$\n\nwhere A stands for an Ace, B for not an Ace, $0\\leq n \\leq 48$ and $X$ contains $2$ Aces and $48-n$ non-Aces. Hence the probability of an Ace following the first Ace is equail to the probability of the first card being Ace which is equal to $$4/52=1/13$$\n\nProbabilistically, conditioned on the location of the second ace, the first ace is uniformly distributed on locations before the second ace - hence probability of it being the first or the last in that sequence are equal.\n\n\u2022 Nice! I was going to go back and try to explain that the fact my fraction simplified to $\\frac{4}{52}$ was not an \"accident.\" You have done it, in the strongest (natural bijection) sense. \u2013\u00a0Andr\u00e9 Nicolas Oct 10 '15 at 1:27\n\u2022 I started thinking along the lines of bijection after seeing your result ;). \u2013\u00a0A.S. Oct 10 '15 at 1:30\n\u2022 Yes, it was kind of shouting look at me, look at me. \u2013\u00a0Andr\u00e9 Nicolas Oct 10 '15 at 1:32\n\nFixing my original calculation\n\nThanks to Shailesh, we now know that the expression $$p_2 = \\frac{48! \\cdot 4 \\cdot 3}{ {_{52}P_{50}} }$$ Should be replaced with: $$p_2 = \\frac{48!}{ {_{52}P_{48}} }$$ Because that after extracting 48 non-Ace cards, we'll necessarily have the second card as Ace.\n\nMoreover, I had a tiny mistake in the denominator. Instead: $${_{52}P_{k-1}} \\cdot {_{52-k}P_{2}}$$ It should be: $${_{52}P_{k-1}} \\cdot {_{52-(k - 1)}P_{2}}$$ (All the $k$ indexes should be shifted by $(-1)$)\n\nI wrote a Python code which shows that this solution works either:\n\n# Python 2.7.6\nimport math\nfrom fractions import Fraction\n\nf = math.factorial\n\ndef nPr(n,r):\nreturn f(n) / f(n-r)\n\np = Fraction(0)\n\nfor k in range(1, 49):\np += Fraction(nPr(48, k-1) * 4 * 3,\nnPr(52, k-1) * nPr(52 - (k-1), 2))\n\np += Fraction(f(48), nPr(52,48))\n\nprint p # Outputs 1/13 :)\n\n\nThank you all for your effort! :)\n\n\u2022 Well done, Dor, We all learn with each other's help. \u2013\u00a0Shailesh Oct 9 '15 at 16:30", "date": "2019-10-23 20:58:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1471987/drawing-without-replacement-prob-for-an-ace-followed-by-an-ace", "openwebmath_score": 0.938448965549469, "openwebmath_perplexity": 498.98088716848855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232914907945, "lm_q2_score": 0.8856314723088733, "lm_q1q2_score": 0.8704192386624453}}
{"url": "http://math.stackexchange.com/questions/157610/is-there-a-differential-limit", "text": "Is there a differential limit?\n\nI'm wondering if there's such a concept as a \"differential limit\". Let me give an example because my nomenclature is my own and unofficial, but hopefully indicative of the concept.\n\nFor some function f(x), there may exist a derivative of that function f'(x) which we call a first order differentiation of f(x). Likewise from f'(x) we could also derive a second order function f''(x). Is there a mathematical concept of the limit of a function as the order of differentiation goes to infinity?\n\nI can think of a function that I can take the derivative of an infinite number of times ( albeit without getting a constant value): f(x) = sin(x)\n\nThe function g(x) = x^2 might have a differential limit of 0. (g'(x) = 2x; g''(x) = 2; g'''(x) = 0 ...)\n\nIs there any usefulness to this concept? A topic of study or keyword that discusses this? Any more examples of infinitely derivable functions?\n\nMost importantly if this is interesting to anyone other than myself: what is a good resource for learning more?\n\n-\nWe use infinitely differentiable functions often in mathematics. We call it the set $C^{\\infty}$. A function that is infinitely differentiable is called \"smooth\" mathworld.wolfram.com/C-InfinityFunction.html \u2013\u00a0tomcuchta Jun 13 '12 at 3:35\n\nFirst: yes, we care about how many times a function can be differentiated. In fact, we have names for such classes: $\\mathcal{C}_0$ is the collection of all continuous functions, $\\mathcal{C}_1$ the class of functions that have continuous derivative; $\\mathcal{C}_2$ the class of functions that have continuous second derivative; and so on. We have that $$\\mathcal{C}_0 \\supsetneq \\mathcal{C}_1 \\supsetneq\\mathcal{C}_2\\supsetneq\\mathcal{C}_3\\supsetneq \\cdots \\supsetneq \\mathcal{C}_n\\supsetneq \\cdots$$\n\nThen we have \"infinitely differentiable functions\", functions that have continuous derivatives of all orders; this is denoted $\\mathcal{C}_{\\infty}$. There are functions that have derivatives of all orders that don't cycle through values, such as $e^{x^2}$.\n\nEven beyond the infinitely differentiable functions we have the notion of \"analytic function\", which are functions that can be defined at every point by a power series that converges on an open interval containing the point.\n\nNow, I think you are talking about trying to make sense of a \"limit\" of the functions as you take derivatives. So that for a polynomial the \"limit\" should be $0$, a function like $\\sin x$ should not have a limit, and so on.\n\nThere are in fact several different ways of making the idea of \"differential limit\" precise. I'm going to talk more generally about what \"limit\" and \"convergence\" can mean here, before addressing the specific construction you are talking about. I'll spill the beans and say that I haven't really seen this particular construction before (there's too much to wade through below, and I don't want you to be annoyed when you reach the end and find that I say \"I haven't seen it...\") But that does not mean you can't ask interesting questions about it.\n\nWhat you describe is a sequence of functions. The main issue is that one has to define what one means by saying that two functions are \"close to one another\". Then you get the notion of convergence of functions.\n\nTurns out that there are many different ways of saying this, and which one is useful depends on the context.\n\nLet's take the case of a sequence of real valued functions of real variable, with domain all of $\\mathbb{R}$. Let's call the sequence $f_n$ (in your case, $f_n = \\frac{d^nf}{dx^n}$ for a given $f$).\n\n1. The most natural way of saying that the sequence $f_n$ converges to a function $f$ is:\n\n$f_n\\to f$ if for every $x$, $\\lim\\limits_{n\\to\\infty}f_n(x) = f(x)$.\n\nThis is called pointwise convergence. For example, the sequence $f_n(x) = x^n$ defined on $[0,1]$ converges pointwise to the function $$f(x) = \\left\\{\\begin{array}{ll} 0 &\\text{if }0\\leq x\\lt 1\\\\ 1 & \\text{if }x=1. \\end{array}\\right.$$\n\nIt turns out that pointwise convergence is natural, but is not very \"good\", in the sense that a lot of properties we find interesting about functions are not preserved by pointwise convergence. For example, above, all functions $f_n$ are continuous, but their limit is not.\n\nThe problem arises because we are forcing no connection between how $f_n(x)$ converges to $f(x)$ and how $f_n(y)$ converges to $f(y)$ with $x\\neq y$; that means that convergence on some points can \"lag behind\" convergence in points very close to them. This can be remedied by making things a bit tighter. If we write out the definition of the limits above, we obtain the following description:\n\n$\\{f_n\\}$ converge pointwise to $f$ if and only if for every $x$, for every $\\epsilon\\gt 0$, there exists $N$, which may depend on both $x$ and $\\epsilon$, such that for all $n\\geq N$ we have $|f_n(x)-f(x)|\\lt \\epsilon$.\n\nThe way to make this tighter, to make sure the the $f_n(x)$ converge to $f$ more or less \"the same way\", is to force the $N$ to depend only on $\\epsilon$, and not on $x$ as well. This is accomplished with:\n\n2. We say that $\\{f_n\\}$ converge uniformly to $f$ if and only if for every $\\epsilon\\gt 0$ there exists $N$, which depends only on $\\epsilon$, such that for all $x$ and all $n\\geq N$ we have $|f_n(x)-f(x)|\\lt\\epsilon$.\n\nUniform convergence is better than pointwise convergence: if every element of the sequence is continuous, and the sequence converges uniformly, then the limit is continuous, for example.\n\nThere are yet other ways of talking about functions converging. Here are a few:\n\n1. Fix a real number $p$, $1\\leq p$. We say that $\\{f_n\\}$ converges to $f$ in $p$-norm if and only if $$\\lim_{n\\to\\infty}\\int_{\\infty}^{\\infty}|f_n(x)-f(x)|^p\\,dx = 0.$$\n\n2. Closely related: we say that $\\{f_n\\}$ converges to $f$ in the sup-norm if and only if $$\\lim_{n\\to\\infty}\\sup\\{|f_n(x)-f(x)|\\} = 0.$$\n\n3. There is a way of measuring sets of real numbers, called the Lebesgue measure, $\\lambda$, the details of which are perhaps too much to go over now. But intuitively, it is a way of assigning a \"size\" to a lot of sets of real numbers, including all intervals and much more complicated sets, in a nice and consistent way. We say that $\\{f_n\\}$ converges in measure to $f$ if and only if for every $\\epsilon\\gt 0$, $$\\lim_{n\\to\\infty}\\lambda(\\{x\\mid |f_n(x)-f(x)|\\geq\\epsilon\\}) = 0.$$\n\n4. We say that $\\{f_n\\}$ converges almost uniformly to $f$ if and only if for every $\\epsilon\\gt 0$ there is a set $X\\subseteq \\mathbb{R}$, with $\\lambda(X)\\lt\\epsilon$, such that $\\{f_n\\}$ converges uniformly to $f$ on $\\mathbb{R}-X$.\n\n5. There are small variations to each of those; depending on context, one may be able to define other ways of convergence, or some of the above may no longer make sense. A typical variation is to take any of the above types of convergence (except almost uniform, for which the concept doesn't add anything), and add the condition \"almost everywhere\" (or \"almost surely\" if you are doing probability). That means that there exists a set $Y$ contained in the domain, of measure $0$, such that the type of convergence you are interested in occurs in $\\mathbb{R}-Y$. (Sets of measure $0$ need not be empty, and they can be quite complicated: for example, any countable set has measure $0$, and there are uncountable sets that do as well, like the Cantor ternary set.\n\nAnother variation is to define the concept of \"Cauchy sequence\" for each of the types, which means making the analog of the definition of \"Cauchy sequence\" of real numbers: instead of comparing to a limit, we ask that for all $n$ and $m$ greater than $N$, $f_n$ and $f_m$ have the relevant property. For instance, \"pointwise Cauchy\" means that for all $x$ and all $\\epsilon\\gt 0$ there exists $N$ such that for all $n,m\\gt N$ we have $|f_n(x)-f_m(x)|\\lt \\delta$. \"Cauchy in measure\" means that for all $\\epsilon\\gt 0$ and all $\\delta\\gt 0$ there exists $N$ such that if $n,m\\gt N$, then $\\lambda(\\{x\\mid |f_n(x)-f_m(x)|\\geq\\epsilon\\}\\lt\\delta$; and so on.\n\nAll of these concepts are extremely important and useful. They show up in probability, statistics, functional analysis, analysis, and other areas. There are known implications (e.g., uniform convergence implies pointwise convergence, and so on).\n\nNow, what about the specific case in which we start with a function $f_0$ (necessarily in $\\mathcal{C}_{\\infty}$) and we define $f_n$ to be the $n$th derivative of $f_0$? I confess I haven't seen this particular sequence addressed, but you can ask all the questions related to the convergence types above: is the sequence pointwise Cauchy? $p$-norm Cauchy? Cauchy in measure? Does it converge uniformly? Almost uniformly? Pointwise? In measure? And so on. Seems like an interesting question, if nothing else.\n\n-\nThank you for the thorough answer, that was exactly what I was looking for. \u2013\u00a0jpredham Jun 13 '12 at 8:58\nI think in your first #2 there is a slight typo - we should have that for $n\\ge N$ the property holds. \u2013\u00a0process91 Jun 13 '12 at 13:54\n@Michael: Yes, indeed. Thank you. \u2013\u00a0Arturo Magidin Jun 13 '12 at 15:25\n\nIt would seem that if the sequence $f$, $f'$, $f''$, ..., $f^{(n)}$, ... tends to a limit as a function, then that limit would be a solution of the differential equation $\\frac{dy}{dx}=y$. So if the limit exists, it is either zero or some multiple of $e^x$. (I'm assuming that the notion of limit we're using is one that the differentiation operator is continuous with respect to, which doesn't seem like much to ask. Pointwise convergence is out, though).\n\nAlso, if we have some function whose iterated derivatives converge towards $ce^x$, we can subtract $ce^x$ from the initial function and get a new sequence whose iterated derivatives converge towards the zero function. So if we look for solutions to $\\lim_{n\\to\\infty} f^{(n)}=0$ we essentially get all functions whose iterated derivatives have a limit.\n\nAll polynomials qualify, of course, but there are also non-polynomial analytic solutions such as $\\sum_{n=0}^\\infty \\frac{1}{n\\cdot n!} x^n$ and in general $\\sum_{n=0}^\\infty \\frac{a_n}{n!}x^n$ whenever $\\lim_{n\\to \\infty} a_n = 0$. I wonder whether some of these have nice closed forms.\n\n-", "date": "2016-07-27 03:51:55", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/157610/is-there-a-differential-limit", "openwebmath_score": 0.957848310470581, "openwebmath_perplexity": 128.18021345740743, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232894783426, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8704192279798187}}
{"url": "https://math.stackexchange.com/questions/3488138/evaluating-int-sqrt-frac5-xx-2-dx-with-two-different-methods-and-gett", "text": "# Evaluating $\\int \\sqrt{\\frac{5-x}{x-2}}\\,dx$ with two different methods and getting two different results [duplicate]\n\nI tried Evaluating $$\\int \\sqrt{\\dfrac{5-x}{x-2}}dx$$ using two different methods and got two different results.\n\nGetting two different answers when tried using two different methods:-\n\nM-$$1$$:\n\n$$\\int \\dfrac{5-x}{\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}}dx$$ $$\\dfrac{1}{2}\\int\\dfrac{-2x+7}{\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}}dx+\\dfrac{3}{2}\\int\\dfrac{dx}{\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}}$$ $$\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}+\\dfrac{3}{2}\\int\\dfrac{dx}{\\sqrt{-x^2+7x-10}}$$\n\n$$\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}+\\dfrac{3}{2}\\int\\dfrac{dx}{\\sqrt{\\left(\\dfrac{3}{2}\\right)^2-\\left(x-\\dfrac{7}{2}\\right)^2}}$$\n\n$$\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}+\\dfrac{3}{2}\\sin^{-1}\\dfrac{x-\\dfrac{7}{2}}{\\dfrac{3}{2}}$$\n\n$$\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}+\\dfrac{3}{2}\\sin^{-1}\\dfrac{2x-7}{3}$$\n\nM-$$2$$:\n\n$$x=5\\sin^2\\theta+2\\cos^2\\theta$$ $$dx=\\left(10\\sin\\theta\\cos\\theta-4\\cos\\theta\\sin\\theta\\right) \\, d\\theta$$ $$\\int \\sqrt{\\dfrac{5\\cos^2\\theta-2\\cos^2\\theta}{5\\sin^2\\theta-2\\sin^2\\theta}}\\cdot6\\sin\\theta\\cos\\theta \\,d\\theta$$ $$6\\int \\cos^2\\theta \\,d\\theta$$ $$\\int 3\\left(1+\\cos2\\theta\\right) \\,d\\theta$$ $$3\\left(\\theta+\\dfrac{\\sin2\\theta}{2}\\right)$$ $$3\\theta+\\dfrac{3}{2}\\sin2\\theta$$\n\n$$x=5\\sin^2\\theta+2-2\\sin^2\\theta$$ $$\\sin^{-1}\\sqrt{\\dfrac{x-2}{3}}=\\theta$$\n\n$$\\cos2\\theta=1-2\\sin^2\\theta$$ $$\\cos2\\theta=1-2\\cdot\\dfrac{x-2}{3}$$ $$\\cos2\\theta=\\dfrac{7-2x}{3}$$\n\n$$3\\sin^{-1}\\sqrt{\\dfrac{x-2}{3}}+\\dfrac{3}{2}\\cdot\\dfrac{\\sqrt{9-(49+4x^2-28x)}}{3}$$ $$3\\sin^{-1}\\sqrt{\\dfrac{x-2}{3}}+\\sqrt{\\left(5-x\\right)\\left(x-2\\right)}$$\n\nIn first and second method I am getting the different results of $$\\dfrac{3}{2}\\sin^{-1}\\dfrac{2x-7}{3}$$ and $$3\\sin^{-1}\\sqrt{\\dfrac{x-2}{3}}$$ respectively. I checked that these are not inter-convertible. Why am I getting this difference?\n\n\u2022 Does this answer your question? Getting different answers when integrating using different techniques. For this case I would suggest the third approach from there (differentiate). \u2013\u00a0Zacky Dec 26 '19 at 16:05\n\u2022 Have you read that posts? Both your approaches are correct and they only differ by a constant, which I would suggest to use when you deal with an indefinite integral. \u2013\u00a0Zacky Dec 26 '19 at 16:08\n\u2022 both approaches are correct, the difference is in a constant which appears to be ${3\\pi}\\over 4$ \u2013\u00a0roman Dec 26 '19 at 16:12\n\u2022 @Zacky, can you please reopen the question, what's your hurry? \u2013\u00a0user3290550 Dec 26 '19 at 16:13\n\u2022 @user3290550 : You wrote: \"actually there should not be difference of a constant also, because whatever standard integrations I have used, those are exact\". That is incorrect. Exact methods should yield things differing by a constant. \"Constant\" in this context means not depending on $x.$ If something does not depend on $x,$ then its derivative with respect to $x$ is $0.$ \u2013\u00a0Michael Hardy Dec 27 '19 at 7:15\n\nBoth your answers are correct. Your two functions differ by $$3\\pi/4$$, so they have the same derivative.\n\nLet $$b=\\sqrt{\\frac{x-2}3}$$. For your integral to make sense you need $$0\\leq b\\leq1$$ (this comes from $$2\\leq x\\leq5$$). Let $$a=\\arcsin b$$. Note $$0\\leq\\arcsin b\\leq \\tfrac\\pi2$$, and so the sine is injective when applied to $$2a-\\tfrac\\pi2$$. Then \\begin{align} \\sin\\left(2a-\\tfrac\\pi2\\right) &=-\\cos(2a)=-(1-2\\sin^2a)=2b^2-1. \\end{align} So \\begin{align} 2\\arcsin\\sqrt{\\frac{x-2}3}\\ -\\frac\\pi2 &=2\\arcsin b -\\frac\\pi2=2a-\\frac\\pi2\\\\ \\ \\\\ &=\\arcsin(2b^2-1)\\\\ \\ \\\\ &=\\arcsin\\frac{2x-7}3. \\end{align} Then, multiplying by $$3/2$$, $$3\\arcsin\\sqrt{\\frac{x-2}3}\\ -\\frac{3\\pi}4=\\frac32\\,\\arcsin\\frac{2x-7}3.$$\n\nShown below is that the two results differ by a constant $$-\\frac{3\\pi}4$$.\n\nDefine $$f(x)$$ as the difference of the two results\n\n$$f(x)=\\frac32\\sin^{-1}\\frac{2x-7}3-3\\sin^{-1}\\sqrt{\\frac{x-2}3}$$\n\nwhere $$2. Then, evaluate\n\n$$f\u2019(x) = \\frac3{2\\sqrt{(5-x)(x-2)}}- \\frac3{2\\sqrt{(5-x)(x-2)}}=0$$\n\nThus, $$f(x)$$ is a constant over $$(2,5]$$ and can be evaluated with\n\n$$f(x)=f(5)=\\frac32\\cdot \\sin^{-1} 1 -3\\sin^{-1}1=-\\frac{3\\pi}4$$\n\n\u2022 You are right, my bad. \u2013\u00a0Martin Argerami Dec 26 '19 at 22:58", "date": "2021-07-28 04:22:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3488138/evaluating-int-sqrt-frac5-xx-2-dx-with-two-different-methods-and-gett", "openwebmath_score": 0.9798845648765564, "openwebmath_perplexity": 479.26202810304346, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540710685614, "lm_q2_score": 0.8887587890727754, "lm_q1q2_score": 0.8704095382763875}}
{"url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-2-section-2-2-the-limit-of-a-function-2-2-exercises-page-92/1", "text": "## Calculus: Early Transcendentals 8th Edition\n\nAs $x$ approaches 2, $f(x)$ approaches 5. Yes, it is possible for this to be true and for $f(2)=3$.\nIt means that $f(x)$ becomes arbitrarily close to $3$ as we take $x$ close enough to $2$, but $x\\ne2$. It is possible for $f(2)$ to equal $3$ because the limit only relies on when $f(x)$ is near $2$; there can be a hole at $x=2$.", "date": "2018-10-20 16:42:43", "meta": {"domain": "gradesaver.com", "url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-2-section-2-2-the-limit-of-a-function-2-2-exercises-page-92/1", "openwebmath_score": 0.9688237309455872, "openwebmath_perplexity": 54.01622114607981, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9845754461077708, "lm_q2_score": 0.8840392924390585, "lm_q1q2_score": 0.8704033807299841}}
{"url": "http://mathhelpforum.com/calculus/125089-piecewise-function-problem.html", "text": "# Math Help - Piecewise Function problem\n\n1. ## Piecewise Function problem\n\nLet f(x) =\n\n(0) if x< 0\n(x) if 0 <= x <= 1\n(2 - x) if 1 < x <= 2\n(0) if x > 2\n\nand g(x) = [integrate] f(t)dt on [0, x]\n\ni) Find an expression for g(x) similar to the one for f(x).\n\nii) Where is f differentiable? Where is g differentiable?\n\nI'm pretty much stuck, and don't know how to start this question.\nI know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).\n\nBut from that, how do I figure out an expression for g(x) from the piecewise function?\n\nAny and all help is appreciated here.\n\n2. Originally Posted by Tulki\nLet f(x) =\n\n(0) if x< 0\n(x) if 0 <= x <= 1\n(2 - x) if 1 < x <= 2\n(0) if x > 2\n\nand g(x) = [integrate] f(t)dt on [0, x]\n\ni) Find an expression for g(x) similar to the one for f(x).\n\nii) Where is f differentiable? Where is g differentiable?\n\nI'm pretty much stuck, and don't know how to start this question.\nI know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).\n\nBut from that, how do I figure out an expression for g(x) from the piecewise function?\n\nAny and all help is appreciated here.\ni)\n\nfor $x < 0$, $f(x) = 0$ and $g(x) = \\int f(t) ~dt = 0$\n\nfor $0 \\leq x \\leq 1$, $f(x) = x$ and $g(x) = \\int_{0}^{x} f(t) ~dt = \\int_{0}^{x} t ~dt = \\frac{1}{2}x^2$\n\nfor $1 < x \\leq 2$, $f(x) = 2 - x$ and $g(x) = \\int_{0}^{x} f(t) ~dt = \\int_{0}^{1} f(t) ~dt + \\int_{1}^{x} f(t) ~dt=$ $\\int_{0}^{1} t ~dt + \\int_{1}^{x} (2-t) ~dt= \\frac{1}{2} + (2x - 2) - (\\frac{1}{2}x^2 - \\frac{1}{2})$\n\nfor $x > 2$\n\n3. Hello, Tulki!\n\nHere's part (a).\nDid you make a sketch?\n\n$\\text{Let }\\:f(x) \\;=\\;\\left\\{\\begin{array}{cccc}0 && x < 0 \\\\ \\\\[-4mm]\nx && 0 \\leq x \\leq 1 \\\\ \\\\[-4mm]\n2 - x && 1 < x \\leq 2 \\\\ \\\\[-4mm]\n0 && x > 2 \\end{array}\\right\\}$\n\nand: . $g(x) \\:=\\: \\int_0^x f(t)\\,dt$\n\na) Find an expression for $g(x)$ similar to the one for $f(x).$\n\n$g(x)$ gives the area under the graph from $0\\text{ to }x$\n\nThe graph looks like this:\nCode:\n          |\n|\n1+       *\n|     *:::*\n|   *:::::::*\n| *:::::::::| *\n- * * * - - - * - + - * * * - -\n0       1   x   2\n\nThen: . $\\displaystyle g(x) \\;=\\;\\left\\{ \\begin{array}{cccc}\n0 && x < 0 \\\\ \\\\[-3mm]\n\\int^x_0t\\,dt && 0 \\leq x \\leq 1 \\\\ \\\\[-2mm]\n\\frac{1}{2} + \\int^x_1 (2-t)\\,dt && 1 < x \\leq 2 \\\\ \\\\[-3mm]\n1 && x > 2\n\\end{array} \\right\\}$\n\n4. Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.\n\nFor x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).\n\nFor x between 0 and 1, y= x so the \"area under the curve\" is the area of a right triangle with base x and height x: its area is $\\frac{1}{2}x^2$.\n\nFor x between 1 and 2, y= 2- x and we can break the \"area under the currve\" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is $\\frac{1}{2}(1)(1)= \\frac{1}{2}$. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is $\\frac{2-x+ 1}{2}x= \\frac{(3-x)x)}{2}= \\frac{3x- x^2}{2}$.\nThe total area is $\\frac{1}{2}+ \\frac{3x- x^2}{2}= \\frac{1+ 3x- x^2}{2}$. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.\n\nFor x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.\n\n5. Thank you very much, all of you!\n\nI suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect.", "date": "2015-02-01 23:13:09", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/125089-piecewise-function-problem.html", "openwebmath_score": 0.779665470123291, "openwebmath_perplexity": 477.4180903687664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9845754474655619, "lm_q2_score": 0.8840392771633078, "lm_q1q2_score": 0.8704033668901957}}
{"url": "https://math.stackexchange.com/questions/1404431/a-proof-in-elementary-set-theory", "text": "# A Proof in Elementary Set Theory\n\nI apologize in advance for any lack of clarity in my mathematical symbols; I'm beginning to learn how to use this site.\n\nI'm working through \"Introduction to Analysis\" by Rosenlicht and he presents an exercise\n\nProve that if $X\\subset S$, $Y\\subset S$, then $\\complement X\\cap\\complement Y=\\complement(X\\cup Y)$\n\nI went about it as follows:\n\nSuppose $X\\subset S$ and $Y\\subset S$. \u00a0 Then $\\complement X=\\{x:x\\in S\\land x\\notin X\\}$ And $\\complement Y=\\{x:x\\in S\\land x\\notin Y\\}$ , so $\\complement X\\cap\\complement Y=\\{x:x\\in S\\land x\\notin X\\land x\\in S \\land x\\notin Y\\}$. \u00a0 Hence $\\complement X\\cap\\complement Y=\\{x:x\\in S\\land x\\notin X\\land x\\in Y\\}$. \u00a0 Therefore $\\complement X\\cap\\complement Y=\\{x:x\\in S\\land x\\notin X\\cup Y\\}$ \u00a0And thus $\\complement X\\cap\\complement Y=\\complement(X\\cup Y)$\n\nWhereas the author gives the proof\n\nIf $x\\in \\complement X\\cap\\complement Y$ then $x\\in\\complement X$ and $x\\in \\complement Y$. \u00a0 This means that $x\\in S, x\\notin X, x\\notin Y$. \u00a0 Since $x\\notin X, x\\notin Y$, we know that $x\\notin X\\cup Y$. \u00a0 Hence $x\\in \\complement(X\\cup Y)$.\n\nConversely, if $x\\in\\complement(X\\cup Y)$, then $x\\in S$ and $x\\notin X\\cup Y$. \u00a0 Therefore $x\\notin X$ and $x\\notin Y$. \u00a0 Thus $x\\in\\complement X$ and $x\\in\\complement Y$, so that $x\\in\\complement X\\cap\\complement Y$.\n\nMy questions are\n\n\u2022 1) is my proof correct?\n\u2022 2) which proof is better\n\u2022 3) what are the subtle (because obviously I know what the difference is, but I'm looking for deeper mathematical understanding) differences between my proof and the authors?\n\nThank you!\n\n\u2022 If you could please fix your post, in order to be more readable. It's $\\complement$ rather than $\\compliment$. After each command, you'd better leave a space. If you want to write \"x in (not in) Y\", you can write either $x \\in Y$ or $x \\notin Y$. \u2013\u00a0thanasissdr Aug 20 '15 at 23:30\n\u2022 Thanks you. That's good to know. It appears someone just did now. Im assuming you can read it now? \u2013\u00a0Liam Cooney Aug 20 '15 at 23:38\n\u2022 I think that both proofs are correct (barring a typo at the end of the line following \"Hence\" in your proof). \u2013\u00a0Akiva Weinberger Aug 20 '15 at 23:40\n\u2022 It still needs work. You should also note that logical and is \\land \u2013\u00a0Graham Kemp Aug 20 '15 at 23:41\n\nOne additional thing the author does is explicitly show that the implication works both ways; that $x\\in(\\complement X\\cap\\complement Y)\\implies x\\in \\complement(X\\cap Y)$ and that $x\\in\\complement(X\\cap Y)\\implies x\\in(\\complement X\\cap\\complement Y)$. \u00a0 You rely on each step being an equivalence; but don't explicitly indicate this is so.", "date": "2019-08-23 11:46:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1404431/a-proof-in-elementary-set-theory", "openwebmath_score": 0.8906909227371216, "openwebmath_perplexity": 532.8436017226247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754504074421, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8704033664829074}}
{"url": "http://math.stackexchange.com/questions/163070/the-most-efficient-way-to-tile-a-rectangle-with-squares", "text": "# The most efficient way to tile a rectangle with squares?\n\nI was writing up a description of the Euclidean algorithm for a set of course notes and was playing around with one geometric intuition for the algorithm that involves tiling an $m \\times n$ rectangle with progressively smaller squares, as seen in this animation linked from the Wikipedia article:\n\nI was looking over this animation and was curious whether or not the squares that you would place down in the course of this algorithm necessarily gave the minimum number of squares necessary to cover the entire rectangle.\n\nMore formally: suppose that you are given an $m \\times n$ rectangle, where $m, n \\in \\mathbb{N}$ and $m, n > 0$. Your goal is to tile this rectangle with a set of squares such that no two squares overlap. Given $m$ and $n$, what is the most efficient way to place these tiles?\n\nIs the tiling suggested by the Euclidean algorithm (that is, in an $m \\times n$ rectangle, with $m \\ge n$, always place an $n \\times n$ rectangle, then recurse on the remaining rectangle) always optimal? If not, is there are more efficient algorithm for this problem?\n\nI am reasonably sure that the Euclidean approach is correct, but I was having a lot of trouble formalizing the intuition with a proof due to the fact that there are a lot of different crazy ways you can try to place the squares. For example, I'm not sure how to have the proof handle the possibility that squares could be at angles, or could have side lengths in $\\mathbb{R} - \\mathbb{Q}$.\n\nThanks!\n\n-\n\u2013\u00a0user17762 Jun 25 '12 at 23:11\nIt's actually pretty straightforward to show that the squares can't be at angles (since any juncture between edges has to be 90 degrees, there are only two possible orientations for lines to be at, and those have to match the two orientations of the rectangle's edges), and it shouldn't be much more complicated to show that all side lengths must be rational (more specifically, side lengths have to be commensurate for any tiling.) \u2013\u00a0Steven Stadnicki Jun 25 '12 at 23:14\nFor a start, see Stan Wagon's Fourteen proofs of a result about tiling a rectangle (see his references). \u2013\u00a0Bill Dubuque Jun 25 '12 at 23:28\n\u2013\u00a0Erel Segal-Halevi Nov 2 '13 at 21:04\n@Erel A web search easily locates a fresh link to Wagon's paper. \u2013\u00a0Bill Dubuque Dec 15 '13 at 4:19\n\nEuclid doesn't always minimize the number of squares. E.g., with an $8\\times9$ rectangle, Euclid says use an 8-square and 8 1-squares, 9 squares in all. But you can do it with a 5, two 4s, a 3, a 2, and two 1s, making 7 squares total. You put the 5 in a corner, then put the 4s in the corners that have room for them; that leaves a $3\\times5$ rectangle to fill, which you can do by Euclid.\nThere is a smaller example, with a $5 \\times 6$ rectangle. The Euclid algorithm produces a tiling with 6 squares: 1 5-square and 5 1-squares. It is possible to tile the rectangle with 5 squares: 2 3-squares and 3 2-squares (The example is from Fractality: math.stackexchange.com/questions/479192/squaring-rectangles ). \u2013\u00a0Erel Segal-Halevi Nov 2 '13 at 21:02\nTiling a rectangle with the fewest squares. Kenyon, Richard J. Combin. Theory Ser. A 76 (1996), no. 2, 272\u2013291. \"We show that a square-tiling of a p\u00d7q rectangle, where p and q are relatively prime integers, has at least $\\log 2p$ squares. If $q>p$ we construct a square-tiling with less than $q/p+C \\log p$ squares of integer size, for some universal constant $C.$''", "date": "2016-05-26 12:56:10", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/163070/the-most-efficient-way-to-tile-a-rectangle-with-squares", "openwebmath_score": 0.744117021560669, "openwebmath_perplexity": 364.88121222011256, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631647185701, "lm_q2_score": 0.8824278772763471, "lm_q1q2_score": 0.8703943536661877}}
{"url": "http://gmatclub.com/forum/number-properties-tips-and-hints-174996.html", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 01 Jul 2015, 17:50\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Number Properties: Tips and hints\n\nAuthor Message\nTAGS:\nMath Expert\nJoined: 02 Sep 2009\nPosts: 28227\nFollowers: 4460\n\nKudos [?]: 44967 [5] , given: 6637\n\nNumber Properties: Tips and hints\u00a0[#permalink] \u00a004 Jun 2014, 11:25\n5\nKUDOS\nExpert's post\n31\nThis post was\nBOOKMARKED\n\nNumber Properties: Tips and hints\n\n ! This post is a part of the Quant Tips and Hints by Topic Directory focusing on Quant topics and providing examples of how to approach them. Most of the questions are above average difficulty.\n\nEVEN/ODD\n1. An even number is an integer that is \"evenly divisible\" by 2, i.e., divisible by 2 without a remainder.\n2. An odd number is an integer that is not evenly divisible by 2.\n3. According to the above both negative and positive integers can be even or odd.\n\nZERO:\n1. 0 is an integer.\n2. 0 is an even integer. An even number is an integer that is \"evenly divisible\" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.\n3. 0 is neither positive nor negative integer (the only one of this kind).\n4. 0 is divisible by EVERY integer except 0 itself, (or, which is the same, zero is a multiple of every integer except zero itself).\n\nPRIME NUMBERS:\n1. 1 is not a prime, since it only has one divisor, namely 1.\n2. Only positive numbers can be primes.\n3. There are infinitely many prime numbers.\n4. the only even prime number is 2. Also 2 is the smallest prime.\n5. All prime numbers except 2 and 5 end in 1, 3, 7 or 9.\n\nPERFECT SQUARES\n1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;\n2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;\n3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);\n4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.\n\nIRRATIONAL NUMBERS\n1. An irrational number is any real number that cannot be expressed as a ratio of integers.\n2. The square root of any positive integer is either an integer or an irrational number. So, $$\\sqrt{x}=\\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\\sqrt{2}$$, $$\\sqrt{3}$$, $$\\sqrt{17}$$, ...).\n\nThis week's PS question\nThis week's DS Question\n\nTheory on Number Properties: math-number-theory-88376.html\n\nDS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38\nPS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59\n\nPlease share your number properties tips below and get kudos point. Thank you.\n_________________\n Kaplan Promo Code Knewton GMAT Discount Codes Veritas Prep GMAT Discount Codes\nIntern\nJoined: 22 Feb 2012\nPosts: 11\nFollowers: 0\n\nKudos [?]: 2 [0], given: 259\n\nRe: Number Properties: Tips and hints\u00a0[#permalink] \u00a024 Jun 2014, 22:15\n2\nThis post was\nBOOKMARKED\n[wrapimg=]PERFECT SQUARES\n1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;\n\n3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);[/wrapimg]\n\nI was learning the above points, while a confusion rose. I'd be grateful if you could explain.\n\nWe know that 100 is a perfect square (10^2)\nIf we Prime factorize 100, we get\n\n100= 2*2*5*5\n\nnumber 1 tips says that a perfect square has odd number of distinct factors. Here 100 has two distinct factors (2,5). How that can be explained ?\n\nnumber 3 tips says that a perfect square always has an odd number of odd factors and even number of even factors. Here we see, 100 has two 5's which is even number.\n\nEven if we see number 3 tips as, a perfect square always has an odd number of DISTINCT odd factors and even number of DISTINCT even factors., we find that, 100 has only ONE DISTINCT even factor which is 2\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 28227\nFollowers: 4460\n\nKudos [?]: 44967 [2] , given: 6637\n\nRe: Number Properties: Tips and hints\u00a0[#permalink] \u00a025 Jun 2014, 01:42\n2\nKUDOS\nExpert's post\nTanvr wrote:\n[wrapimg=]PERFECT SQUARES\n1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;\n\n3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);[/wrapimg]\n\nI was learning the above points, while a confusion rose. I'd be grateful if you could explain.\n\nWe know that 100 is a perfect square (10^2)\nIf we Prime factorize 100, we get\n\n100= 2*2*5*5\n\nnumber 1 tips says that a perfect square has odd number of distinct factors. Here 100 has two distinct factors (2,5). How that can be explained ?\n\nnumber 3 tips says that a perfect square always has an odd number of odd factors and even number of even factors. Here we see, 100 has two 5's which is even number.\n\nEven if we see number 3 tips as, a perfect square always has an odd number of DISTINCT odd factors and even number of DISTINCT even factors., we find that, 100 has only ONE DISTINCT even factor which is 2\n\n2 and 5 are prime factors of 100. The total number of factors of 100=2^2*5^2 is (2+1)(2+1)=9=odd: 1, 2, 4, 5, 10, 20, 25, 50, 100. Out of these 9 factors three are odd (1, 5, and 25) and 6 are even (2, 4, 10, 20, 50, 100).\n\nHope it helps.\n_________________\nIntern\nJoined: 07 Jul 2014\nPosts: 16\nFollowers: 0\n\nKudos [?]: 4 [1] , given: 47\n\nRe: Number Properties: Tips and hints\u00a0[#permalink] \u00a023 Jul 2014, 11:29\n1\nKUDOS\n1\nThis post was\nBOOKMARKED\nCan I suggest a few new topics? Specifically Geometry, word problems, and combinatrics/probability?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 28227\nFollowers: 4460\n\nKudos [?]: 44967 [0], given: 6637\n\nRe: Number Properties: Tips and hints\u00a0[#permalink] \u00a023 Jul 2014, 11:51\nExpert's post\ntangt16 wrote:\nCan I suggest a few new topics? Specifically Geometry, word problems, and combinatrics/probability?\n\nThank you. Will cover them soon.\n_________________\nIntern\nJoined: 27 Aug 2014\nPosts: 28\nGMAT Date: 09-27-2014\nFollowers: 1\n\nKudos [?]: 4 [1] , given: 25\n\nNumber Properties: Tips and hints\u00a0[#permalink] \u00a027 Aug 2014, 05:37\n1\nKUDOS\n3\nThis post was\nBOOKMARKED\nIf progressions comes under this topic, would like to add this tip.\n\nIn the specific case of sum to n1 terms being equal to sum to n2 terms of the same arithmetic progression, the sum of the term numbers which exhibit equal sums is constant for the given evenly spaced set of numbers.\n\n(S3 denotes Sum of the first three terms of the evenly spaced set.)\n\n1.\nQ: if sum to 11 terms equal sum to 19 terms in an evenly spaced set, what is the sum to 30 terms for this series?\nA: S11 = S19; so S0 = S30. Since S0 = 0, S30 = 0.\n\n2.\nThis happens because the arithmetic progression's negative terms cancel out the positive terms.\n\nAlso, if the series has a zero in it, the sum will be equal for two terms such that one term number will be odd and the other will be even.\nEx.: -10, -5, 0, 5, 10.....\n\nAnd if the series does not have a zero in it, the sum will be equal for two terms such that both term numbers will be either odd or even.\n\nEx.: -12, -4, 4, 12......\nIntern\nJoined: 27 Aug 2014\nPosts: 28\nGMAT Date: 09-27-2014\nFollowers: 1\n\nKudos [?]: 4 [0], given: 25\n\nNumber Properties: Tips and hints\u00a0[#permalink] \u00a027 Aug 2014, 05:52\n3\nThis post was\nBOOKMARKED\nAnother tip,\n\nIf B is 1/x times more than A, then A is 1/(x+1) times lesser than B.\n\nThis is especially useful in averages, profit and loss, time rate questions.\n\nExample:\n\nIf B's wage is 25% more than A's wage, then what is A's wage in terms of B?\n\nB is 1/4 times more than A, so A will be 1/5 or 20% lesser than B. i.e., A = 80% of B\nIntern\nJoined: 27 Aug 2014\nPosts: 28\nGMAT Date: 09-27-2014\nFollowers: 1\n\nKudos [?]: 4 [0], given: 25\n\nRe: Number Properties: Tips and hints\u00a0[#permalink] \u00a027 Aug 2014, 07:15\nTip for questions involving recurring decimals:\n\nNote the following pattern for repeating decimals:\n0.22222222... = 2/9\n0.54545454... = 54/99\n0.298298298... = 298/999\n\nNote the pattern if zeroes preceed the repeating decimal:\n0.022222222... = 2/90\n0.00054545454... = 54/99000\n0.00298298298... = 298/99900\nRe: Number Properties: Tips and hints \u00a0 [#permalink] 27 Aug 2014, 07:15\nSimilar topics Replies Last post\nSimilar\nTopics:\n4 Algebra: Tips and hints 1 16 Jul 2014, 10:30\n15 Absolute Value: Tips and hints 2 09 Jul 2014, 06:04\n16 Inequalities: Tips and hints 0 02 Jul 2014, 03:33\n7 Remainders: Tips and hints 0 23 Jun 2014, 02:33\n13 Exponents and Roots on the GMAT: Tips and hints 1 24 Jul 2014, 06:33\nDisplay posts from previous: Sort by", "date": "2015-07-02 01:50:19", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/number-properties-tips-and-hints-174996.html", "openwebmath_score": 0.4778353273868561, "openwebmath_perplexity": 1407.4552832562965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9579122720843812, "lm_q2_score": 0.9086178926024028, "lm_q1q2_score": 0.87037622995929}}
{"url": "http://math.stackexchange.com/questions/114453/solve-xy-yx/114476", "text": "# Solve $x^y \\, = \\, y^x$ [duplicate]\n\nPossible Duplicate:\n$x^y = y^x$ for integers $x$ and $y$\n\nI obtained a question asking for how to solve $\\large x^y = y^x$. The given restraints was that $x$ and $y$ were both positive integers. By a bit of error an trial we quickly see that $x=2$ and $y=4$ is one solution.\n\n\u2022 My question is: How do one show that $(2\\,,\\,4)$ is the only non trivial, positive solution to the equation?\n\nNow my initial approach was as follows: We have\n\n$$\\large x^y = y^x$$\n\nThe trivial solution is obviously when $y=x$, so let us focus on when $y \\neq x$. Let us make a more general statement. Firstly I take the log of both sides\n\n$$\\large y \\log x = x \\log y$$\n\nLet us divide by x and \\log x (We now assume $x\\neq 0$ and $x\\neq 1$ since 0 is not a positive number, and 1 gives us a trivial solution)\n\n$$\\large \\frac{y}{x} = \\frac{\\log y }{\\log x}$$\n\nFor these sides to be equal, we must remove the logarithms on the right hand side, this is achived if $y$ is on the form $x^a$. Now This gives\n\n$$\\large \\frac{x^a}{x} = \\frac{\\log \\left(x^a\\right) }{\\log(x)}$$\n\n$$\\large x^{a-1} = a$$\n\nSo finaly we obtain that $\\displaystyle \\large x=\\sqrt[ a-1]{a}$ and $\\displaystyle \\large y = \\sqrt[a-1]{a^a}$\n\nNow setting $a=2$ gives us $x = 2$ and $y=4$ as desired.\n\nMy question is, how do we prove that $x=2$ and $y=4$ is the only integer solutions? My thought was to show that $\\displaystyle \\large \\sqrt[ a-1]{a}$ and $\\displaystyle \\large \\sqrt[a-1]{a^a}$ are both irrational when a>2, but I have not been able to show this.\n\nAny help is greatly appreciated, cheers =)\n\n-\n\n## marked as duplicate by Aryabhata, Zev ChonolesFeb 28 '12 at 21:03\n\nThis MO thread is relevant, mathoverflow.net/questions/22230/ab-ba-when-a-is-not-equal-to-b \u2013\u00a0 user1729 Feb 28 '12 at 16:21\n@user1729: Don't like searching MSE itself? :-) \u2013\u00a0 Aryabhata Feb 28 '12 at 19:04\n$1729 = 10^3 + 9^3 = 12^3 + 1^3.$ \u2013\u00a0 Will Jagy Feb 28 '12 at 20:53\nI've closed the question as a duplicate. There is the option of merging the questions, which will move all answers here to the other question (N3buchadnezzar gets to keep the reputation received from asking this question). Is anyone against doing so? \u2013\u00a0 Zev Chonoles Feb 28 '12 at 21:05\n@Aryabhata: No, it was just that the question I linked to popped to the top of the MO stack last week, and I was surprised to find an almost identical question here less than a week later! (Almost identical? Perhaps \"intricately related\" would be better. Certainly, the answers answer a more general question: when are the solutions rational. Also, the discussion is very interesting.) \u2013\u00a0 user1729 Feb 29 '12 at 9:46\n\nThe function $u:x\\mapsto(\\log x)/x$ is increasing on $[1,\\mathrm e]$ from $u(1)=0$ to $u(\\mathrm e)=1/\\mathrm e$, and decreasing on $[\\mathrm e,+\\infty)$ from $u(\\mathrm e)=1/\\mathrm e$ to $0$. Hence, if $u(x)=u(y)$ with $y\\gt x\\geqslant 1$, then $y\\gt \\mathrm e\\gt x\\gt1$. Since $\\mathrm e\\lt3$, this implies that $1\\lt x\\lt3$. If furthermore $x$ is an integer, then $x=2$. The unique root of the equation $u(y)=(\\log2)/2$ such that $y\\gt\\mathrm e$ is $y=4$. Hence $(x,y)=(2,4)$ is the unique solution.\n\n-\n\nAnother approach for these kind of problems:\n\n$$x^y=y^x$$\n\nBy dividing both sides of equations by $x^x$, we'll have:\n\n$$x^{y-x}=(\\frac{y}{x})^x$$\n\nNow, what you can say about $\\frac{y}{x}$ ?\n\nTry to continue.\n\n-\nI have no idea what we can say about that fraction. Care to explain? \u2013\u00a0 Pureferret Feb 28 '12 at 20:59\ny/x must be integer.(Why?) \u2013\u00a0 Salech Alhasov Feb 28 '12 at 21:38\n\nI thought this sounded familiar. The equation $a^b = b^a$ can be written as $$\\frac{\\log a}{a} = \\frac{\\log b}{b}$$ If you carefully draw the curve $y= \\log x$ in the $x-y$ plane, the quantity $\\frac{\\log a}{a}$ is the slope of the line that passes through the origin and the point $(a, \\log a).$ So the equation $\\frac{\\log a}{a} = \\frac{\\log b}{b}$ says that the lines from the origin to $(a, \\log a)$ and to $(b, \\log b)$ have the same slope, therefore they are the same line. That is, the three points are collinear.\n\nSo, a graphical solution is to draw lines through the origin, with positive slope, that intersect the curve $y= \\log x.$ It will be seen fairly quickly that one intersection point, call it $a,$ has $1 < a < e.$ As you want $a$ an integer, the only choice is $a=2.$\n\nThe calculus part is this: the line through the origin with slope $\\frac{1}{e}$ is tangent to the curve $y= \\log x$ at the point $(e, \\log e \\; = \\; 1).$ To intersect the curve twice, we need slope a little bit less that that, and the first intersection point will be a little bit to the left of $e.$\n\n-", "date": "2014-12-18 09:01:00", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/114453/solve-xy-yx/114476", "openwebmath_score": 0.8129056096076965, "openwebmath_perplexity": 175.77790191478593, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308761574287, "lm_q2_score": 0.8807970842359876, "lm_q1q2_score": 0.8703427945630149}}
{"url": "https://math.stackexchange.com/questions/3331450/points-in-bbb-r2-that-can-be-reached-via-steps-which-are-1-5-of-a-unit-c", "text": "# Points in $\\Bbb {R}^2$ that can be reached via steps which are $1/5$ of a unit circle.\n\nI was playing around with this demo of Project Euler Problem 208 which allows you to take steps which are \"left\" or \"right\" arcs of $$1/5$$ of a unit circle.\n\nHere's an example walk, which starts at the blue dot pointing vertically up, and which consists of steps\n\nRLLLRLLLRRLLLRLLLRRLRLLLR\n\n\n# Question\n\nWhich points in $$\\mathbb R^2$$ can be reached in a finite number of steps, starting at the origin and pointing in the positive $$y$$-direction. Is this set of points dense in $$\\mathbb R^2$$? If not, what's the greatest number of points that can land in $$[0,1] \\times [0,1] \\subset \\mathbb R^2$$?\n\n\u2022 I would expect the points to be dense but it be a bit of work to prove. One way is to find sequences of moves that end with you facing the same direction but move the point in several directions. If the distances moved are not rational multiples of each other you can combine those movements to get as close as wanted to any point. RL and LR leave you heading in the same direction as at the start. You can compute the displacement they create, but they will be the same vertically and negatives horizontally. Now add RRLL and LLRR to the mix. You need \u2013\u00a0Ross Millikan Aug 22 '19 at 23:02\n\u2022 a combination that moves you downward, but that shouldn't be hard to find. I would think something like RRRLRLRLRLRLLL would work, the first three to make (about a) U turn, the middle RLs to move in that direction, and the last three to reorient upward. \u2013\u00a0Ross Millikan Aug 22 '19 at 23:05\n\u2022 This is a really interesting question. I might end up offering a bounty here \u2013\u00a0clathratus Aug 23 '19 at 0:37\n\u2022 I think one could reformulate this question so it is closely linked with the projective definition of the Penrose pattern: Take the standard Z^5 grid in the 5-dimensional vector space. One can put a 2-dimensional subspace E in such a way, that the 5 base vectors project to it pointing to the corners of a regular pentagon. Then your question should be equivalent to the question, If the projection of Z^5 to E is dense, right? I think this has been proven due to E having an irrational angle to the base vectors. \u2013\u00a0Non Euclidean Dreamer Aug 24 '19 at 15:02\n\u2022 Related:Rolling a random walk on a sphere: \"The surface will be filled for every $\\delta$ except $\\pi/2$ and $\\pi$.\" \u2013\u00a0Joseph O'Rourke Aug 25 '19 at 1:48\n\nYes, the set of points attainable by these paths is dense in $$\\mathbb{R}^2$$. Pardon the lack of polish in this answer; I don't have a 'real' math background, so I'm expressing my ideas here using the language most intuitive to me.\n\nWe can think of any path as a linear combination of five 'moves,' which w.l.o.g. are given by the vectors $$v_n = (\\cos\\frac{2\\pi n}{5}, \\sin \\frac{2\\pi n}{5})$$. Paths cannot be arbitrary integer combinations of these vectors, since there are only two moves available after each step. We can, however, devise the following scheme to obtain arbitrary integer combinations of the vectors $$2v_n$$. Consider the following paths (I've written out the linear combination long form to show the order that the steps must be taken in each path):\n\n\\begin{align*} p_1^a &= 2av_1\\\\ p_2^b &= v_1 + 2bv_2 + v_1\\\\ p_3^c &= v_1 + v_2 + 2cv_3 + v_2 + v_1\\\\ p_4^d &= v_1 + v_2 + v_3 + 2dv_4 + v_3 + v_2 + v_1\\\\ p_5^e &= 2ev_5 \\end{align*}\n\nThese paths are composable with each other, so we can take them in arbitrary (positive) integer combinations. Consider a point of the form $$x = \\sum_{i = 1}^5 p_i^{a_i}$$. Writing this out in terms of our original moves $$v_i$$, we obtain \\begin{align*} x &= p_1^a + p_2^b + p_3^c + p_4^d + p_5^e\\\\ &= (6 + 2a)v_1 + (4 + 2b)v_2 + (2 + 2c)v_3 + 2dv_4 + 2ev_5 \\end{align*} We will ignore the constant offset $$6v_1 + 4v_2 + 2v_3$$. Using the paths $$p_i^a$$ alone, we can thus reach an arbitrary (nonnegative) integer combination of the vectors $$\\{2v_i\\}$$.\n\nNow note that $$\\sum_{i = 1}^5 2v_i = 0$$, so it follows that the vectors $$2v_i$$ and $$\\sum_{j \\neq i} 2v_j$$ are additive inverses. Thus the set of nonnegative integer combinations of the vectors $$2v_i$$ is actually equal to the set of arbitrary integer combinations of the vectors $$2v_i$$.\n\nDigging around MathSE, I found this answer that asserts, based on Kronecker's theorem, that if three vectors have components that are all linearly independent over the rationals, then the set of their integer combinations is dense in the unit square (and by extension the plane). $$2v_1$$, $$2v_2$$ and $$2v_3$$ satisfy this hypothesis, so the claim should follow.\n\n\u2022 Pretty damn good for someone without a math background! (+1) \u2013\u00a0clathratus Sep 6 '19 at 4:04\n\u2022 Well, I'm a physics student, so I wouldn't say I have 'no' math background\u2013 just not a 'real' one. \u2013\u00a0Panopticon Sep 6 '19 at 18:08\n\nNot an answer, but just an opportunity to mention that the behavior of the \"Tangle Toy,\" which comes in quarter-circles,\n\nImage from FatBrainToys.\n\nErik D. Demaine, Martin L. Demaine, Adam Hesterberg, Quanquan Liu, Ron Taylor, and Ryuhei Uehara, \u201cTangled Tangles\u201d, in The Mathematics of Various Entertaining Subjects (MOVES 2015), volume 2, 2017, pages 141\u2013152, Princeton University Press. Authors' link.\n\nFig.7: Tangle toy configurations.\n\nTo address @Narasimham's question, this paper\n\nTatu, Aditya. \"Tangled Splines.\" arXiv:1610.03129. 2016-18.\n\nsays that the curvature $$\\kappa(t) =\\pm 1$$ outside of the knot points, and that torsion $$\\tau(t)$$ is zero, again excluding knot points.\n\n\u2022 Interesting. Are curvature and torsion constant? \u2013\u00a0Narasimham Aug 22 '19 at 23:55\n\u2022 Curvature: yes. Torsion:no. \u2013\u00a0Ivan Neretin Aug 23 '19 at 7:45\n\u2022 @IvanNeretin: The cited paper says, \"Since each link lies in the plane spanned by Vi and Vi+1, for each t \u2208 (ti, ti+1), the torsion \u03c4 is zero.\" I have not tried to verify this, but you seem to be contradicting their claim? \u2013\u00a0Joseph O'Rourke Aug 23 '19 at 21:50\n\u2022 All right, the torsion is zero within each link and infinite at the joints. \u2013\u00a0Ivan Neretin Aug 23 '19 at 22:01\n\nHere's a formulation that may help with calculation a bit: define a 'position' as $$\\langle x, y, \\theta\\rangle$$, where $$\\theta$$ is the angle being made with the $$y$$ axis; in other words, the starting position is $$\\langle0, 0, 0\\rangle$$. Then if you draw the geometry out (which I will try and do once I get onto a machine with better drawing tools) from $$\\langle x, y, 0\\rangle$$ the available moves are to $$\\langle x+(1-\\cos(2\\pi/5)), y+\\sin(2\\pi/5), \\pi/10\\rangle$$ and $$\\langle x-(1-\\cos(2\\pi/5)), y+\\sin(2\\pi/5), -\\pi/10\\rangle$$. Since the local frame at angle $$\\theta$$ has its $$x$$ axis in the direction $$\\langle \\cos\\theta, \\sin\\theta\\rangle$$ and its $$y$$ axis in the direction $$\\langle -\\sin\\theta, \\cos\\theta\\rangle$$ (modulo possible sign errors) then this means that from $$\\langle x, y, \\theta\\rangle$$ the available moves are to $$\\langle x+\\cos\\theta(1-\\cos(2\\pi/5))-\\sin\\theta\\sin(2\\pi/5), y+\\cos\\theta\\sin(2\\pi/5)+\\sin\\theta(1-\\cos(2\\pi/5)), \\theta+\\pi/10\\rangle$$ and to $$\\langle x-\\cos\\theta(1-\\cos(2\\pi/5))-\\sin\\theta\\sin(2\\pi/5), y+\\cos\\theta\\sin(2\\pi/5)-\\sin\\theta(1-\\cos(2\\pi/5)), \\theta-\\pi/10\\rangle$$. Since the cosine and sine of $$2\\pi/5$$ and $$\\pi/10$$ have nice clean expressions in terms of $$\\sqrt{5}$$, you should be able to get explicit expressions in terms of $$\\sqrt{5}$$ for the coordinates after making multiple moves as in Ross Millikan's comment, and then use some usual irrationality-of-square-roots arguments to show density. (But of course, note that there's a little subtlety to these arguments, because the same statement with sixth-turns around the circle are false!)", "date": "2021-01-23 04:21:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3331450/points-in-bbb-r2-that-can-be-reached-via-steps-which-are-1-5-of-a-unit-c", "openwebmath_score": 0.9381072521209717, "openwebmath_perplexity": 523.3560065751179, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9881308796527184, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8703427899118722}}
{"url": "http://cqcourseworklruw.vatsa.info/an-analysis-of-the-buffons-needle-a-method-for-the-estimation-of-the-value-of-pi.html", "text": "# An analysis of the buffons needle a method for the estimation of the value of pi\n\nBuffon 's needle problem if vou drop any needle, short or long, then the expected number of crossings will be where pi is the probability that the needle will come to lie with exactly one. Buffon's needle refers to a simple monte carlo method for the estimation of the value of pi, 314159265 the idea is very simple suppose you have a tabletop with a number of parallel lines drawn on it, which are equally spaced (say the spacing is 1 inch, for example. Fostering understanding of monte carlo simulations since such knowledge provides them with the understanding necessary to justify estimation with the monte carlo method buffon needle applet (available at tubegeogebraorg/material/show/id/112419) observing the value of p c when l6= d. Estimation from samples 5 one sample of paired observations (paired t test) 6 comparing two samples (the two-sample t test) 7 regression 8 comparing price, j (1984) random numbers and buffons' needle, parab, 20, 2-9 priddis, m j (1978) an upper bound for the g r (1986) a method for teaching statistics using n. The analysis of the subject of probability additionally, it may be suggested that reno program is, method buffon\u2019s needle problem - the emergence of \u03c0 at the end of solving a probability problem is very interesting this experiment can be performed by everybody and an estimation can be made for \u03c0 value for pi value of pi did the.\n\nA abdomen the an analysis of the buffons needle a method for the estimation of the value of pi pterigoid lex stops an analysis of crooks a character in john steinbecks novella of mice and men twice, his oblate declares himself ridiculous the tetradiscotic and totalitarian neophyte an analysis of harvard mk i and colossus sealed his somber. Nmfp[1] - free download as pdf file (pdf), text file (txt) or read online for free scribd is the world's largest social reading and publishing site explore explore interests career & money. The buffon needle method of estimating \\$latex {\\pi}&fg contents \u00ab the bridges of paris the great american eclipse \u00bb buffon was no buffoon published august 13, 2015 occasional leave a comment tags: algorithms, probability the buffon needle method of estimating is graph theory group theory hamilton history ireland logic maps.\n\nVol 79, no 5, december 2006 387 since our spherical caps have height l /2 y, the area of r y is a (y) = 2 2 2 l 2 l 2 y = l 1 2 y l thus, for the \u00dexed value of y (0 y l /2), the probability that the needle crosses. Im supposed to write a c program to estimate the pi via buffon needle using monte carlo method i think my program works properly, but i never ev stack overflow log in sign up current community stack overflow buffon needle pi estimate experiment gets close to pi, but doesnt get perfect value ask question pi value. 512 analysis of the formulas 513 formulas of the theory of functions 514 formulas of number theory 515 formulas another probability based and unusual method is the buffon needle problem by georges -louis leclerc de buffon (1733 argued in 1777 in the mid-19th century, the swiss astronomer rudolf wolf came by 5000 needle throws.\n\nApproximating the value of pi points 100 points of special assignments/computer projects due date thursday october 27, simulator to do so is called buffon\u2019s needle and the following link explains it better than anyone did you clearly identify your method for each problem did you suppress all statements did you. Removing the inherent paradox of the buffon's needle monte carlo simulation using fixed-point iteration method we use the buffon\u2019s needle experiment as an example to introduce the fixed-point iteration method the buffon\u2019s needle simulation model can be fitted into the fixedpoint iteration model in. Preface in this project we discuss short, but comprehensive introduction to methods of com-putation pi some of learning rules chapter 2 consists of a brief discussion on pi and. Buffon's needle experiment for estimating \u03c0 is a classical example of using an experiment (or a simulation) to estimate a probability how to convert the datetime character string to sas datetime value (anydtdtm and mdyampm formats) using sas enterprise guide to run programs in batch not particularly the convergence.\n\nThinking skills creativity, problem solving, problem finding - powerpoint ppt presentation loading ppt \u2013 thinking skills creativity, problem solving title: thinking skills creativity, problem solving, problem finding description: description of attitudes with the help of 'roses' and 'thorns' divine roses for you. Download citation | buffon needle lands buffon needle lands in \u03f5-neighborhood of a 1-dimensional sierpinski gasket with probability at most we also formulate a result about all self-similar sets of dimension 1r\u00e9sum\u00e9on donne une estimation de la probabilit\u00e9 pour que l'aiguille de buffon soit \u03f5-proche d'un ensemble de. Approximation however, better approximations can be obtained using a similar method with regular polygons with (1707-1788)\\) and laplace $$(1749-1827)$$ proposed using a stochastic simulation to estimate the value of $$\u03c0$$ buffon's needle famously approximates $$\\pi$$ using the fact that a straight needle is the number $$\\pi\\. Throwing buffon\u2019s needle with mathematica nb cdf pdf it has long been known that buffon\u2019s needle experiments can be used to estimate three main factors influence these experiments: grid shape, grid density, and needle length in statistical literature, the desired probability is directly related to the value of , which can then be. Throwing buffon\u2019s needle with mathematica nb cdf pdf buffon\u2019s needle experiments can be seen as valuable tools in applying the concepts of statistical estimation theory, such as efficiency, completeness, and sufficiency for instance, j a rice, mathematical statistics and data analysis, 2nd ed, belmont, ca: duxbury. Simulation of buffons needle yihui xie & lijia yu / 2017-04-04 randomly to check whether they cross the parallel lines through many times of \u2018dropping\u2019 needles, the approximate value of \\(\\pi$$ can be calculated out there are three graphs made in each step: the top-left one is a simulation of the scenario, the top-right one is to help us. Buffon's needle an analysis and simulation introduction : buffon's needle is one of the oldest problems in the field of geometrical probability it was if you keep dropping the needle, eventually you will find that the number 2n/h approaches the value of pi method let's take the simple case first in this case, the length of the needle is one. Page 4 matlab: buffon's needle problem throws = 10000 % percent makes the rest of the line a comment x=rand(1,throws) % a vector of 10000 pseudorandom numbers % in the range [0,1) theta=rand(1,throws) % semicolon supresses printing theta=05pitheta % theta now 10000 random numbers between 0 and pi/2 hits= x =05sin(theta) . On the floor can be used to estimate the value of \u03c0 monte carlo method is one of the methods that use the random numbers to solve the physical chemical problems it does have a package in the analysis tools, random number generator, which provides for seeding, this is quite similar in spirit to the buffon needle problem, but is a.\n\n1982) was presented on the buffon needle method of track counting this involved random sampling of the image of a track recorder surface in order to measure the probability that a randomly chosen area and replacing ag and pb with their respective estimators, zg and zb , we obtain a suitable estimator, pi for the track (m - z9) (4. The exuberant aleks detains him fifty years desulfurado connaturalmente leukemic ferdinand confiscates its widening remarkably roboticized an analysis of the buffons needle a method for the estimation of the value of pi toms, robotized, your incisure butts wytes with embarrassment totipotente and keplerian quentin euhemerised an. After math 7 smile you\u2019re at the best wordpresscom site ever the attached document is the introduction i use before doing our buffon needle experiment for the experiment you need a tiled floor and dowel rods cut to a length that equals the width of a tile choose which set of parallel lines on the floor will be used the estimation of. The first modern mathematical model and the first documented use of the monte carlo method, as it is known today, is generally considered to have originated with the buffon-laplace needle experiment in 1777 the experiment is to throw needles onto a plane with equally spaced parallel lines in order to estimate the value of \u03c0 a brief.\n\nCode to estimate pi asked by john jamison john jamison (view profile) 16 questions asked 0 answers 0 accepted answers of different number of terms % in the series to see the shape of the error curve % as it converges towards the true value of pi 'toolbar', 'none', 'menu', 'none') % give a name to the title bar set(gcf, 'name'. View notes - the buffon's needle problem from npre 498 at university of illinois at urbana\u2013champaign chapter 2 the buffons needle problem: chapter 2 the buffons needle laplace ingeniously used it for the estimation of the value of.\n\nAn analysis of the buffons needle a method for the estimation of the value of pi\nRated 4/5 based on 31 review\n\n2018.", "date": "2018-10-22 01:30:23", "meta": {"domain": "vatsa.info", "url": "http://cqcourseworklruw.vatsa.info/an-analysis-of-the-buffons-needle-a-method-for-the-estimation-of-the-value-of-pi.html", "openwebmath_score": 0.7109445333480835, "openwebmath_perplexity": 1603.7392917532973, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308800022472, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8703427871278225}}
{"url": "https://math.stackexchange.com/questions/1338411/why-doesnt-using-the-approximation-sin-x-approx-x-near-0-work-for-computin", "text": "# Why doesn't using the approximation $\\sin x\\approx x$ near $0$ work for computing this limit?\n\nThe limit is $$\\lim_{x\\to0}\\left(\\frac{1}{\\sin^2x}-\\frac{1}{x^2}\\right)$$ which I'm aware can be rearranged to obtain the indeterminate $\\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\\sin x\\approx x$ near $x=0$. However, the actual limit is $\\dfrac{1}{3}$, not $0$.\n\nIn this similar limit, the approximation reasoning works out.\n\n\u2022 \u2013\u00a0Rahul Jun 25 '15 at 2:40\n\u2022 You should take the next term of the expansion of $sin(x)$ into account. \u2013\u00a0M. Wind Jun 25 '15 at 2:43\n\u2022 @Rahul thank you for the link, very good description. \u2013\u00a0user170231 Jun 25 '15 at 2:48\n\u2022 The problem with the approximation $\\sin x\\sim x$, you still get an indetermination of type $\\infty -\\infty$, therefore it doesn't help. \u2013\u00a0Surb Sep 10 '16 at 15:23\n\u2022 @Surb I was under the impression that the approximation allows me to write the limand as $\\frac{1}{x^2}-\\frac{1}{x^2}=0$, which would make the limit $0$, but I see now that isn't a sound conclusion. \u2013\u00a0user170231 Sep 11 '16 at 0:31\n\nIf we take one more term in the Taylor expansion:\n\n\\begin{align} \\sin x&\\approx x-\\frac{x^3}6+\\cdots\\\\ \\sin^2 x&\\approx x^2-2x\\frac{x^3}6+\\cdots\\\\ \\frac 1{\\sin^2 x}&\\approx\\frac 1{x^2-x^4/3}\\\\ &=\\frac 1{x^2}\\cdot\\frac 1{1-x^2/3}\\\\ \\lim_{x\\to 0}\\left[\\frac 1{\\sin^2 x}-\\frac 1{x^2}\\right]&=\\lim_{x\\to 0}\\left[\\frac 1{x^2}\\left(\\frac 1{1-x^2/3}-1\\right)\\right]\\\\ &=\\lim_{x\\to 0}\\left[\\frac 1{x^2}\\cdot\\frac{1-1+x^2/3}{1-x^2/3}\\right]\\\\ &=\\lim_{x\\to 0}\\frac 1{3-x^2}\\\\ &=\\frac 1 3 \\end{align}\n\nTo see where the first-order expansion went wrong, it's necessary to keep track of where the error term goes:\n\n\\begin{align} \\sin x&= x+\\text{O}(x^3)\\\\ \\sin^2 x&=x^2+2x\\text{O}(x^3)+\\text{O}(x^3)^2\\\\ &=x^2+\\text{O}(x^4)+\\text{O}(x^6)\\\\ &=x^2+\\text{O}(x^4)\\\\ \\frac 1{\\sin^2 x}&=\\frac 1{x^2+\\text{O}(x^4)}\\\\ &=\\frac 1{x^2}\\cdot\\frac 1{1+\\text{O}(x^2)}\\\\ \\frac 1{\\sin^2 x}-\\frac 1{x^2}&=\\frac 1{x^2}\\left[\\frac 1{1+\\text{O}(x^2)}-1\\right]\\\\ &=\\frac 1{x^2}\\cdot\\frac{1-1+\\text{O}(x^2)}{1+\\text{O}(x^2)}\\\\ &=\\frac{\\text{O}(x^2)}{x^2}\\cdot\\frac 1{1+\\text{O}(x^2)}\\\\ &=\\text{O}(1) \\end{align}\n\nThus the $\\sin x\\approx x$ approximation is not accurate enough to estimate even the constant term of the expression in the limit. (Note that it does allow us to say that there are no $\\text{O}(n^{-1})$ or bigger terms, so the limit probably won't diverge.)\n\n\u2022 I don't think the idea of using approximations in limits is justified because limits are well defined exact things and not \"approximations to something\". I have given some comments about your approach in my answer. \u2013\u00a0Paramanand Singh Jun 25 '15 at 4:27\n\u2022 Why are the following equal $\\frac 1{x^2}\\cdot\\frac 1{1-x^2/3} =\\frac 1{x^2}+\\frac 1{3-x^2}$? \u2013\u00a0mavavilj Sep 10 '16 at 13:16\n\nIn this answer it is shown, using only pre-calculus methods, that $$\\lim_{x\\to0}\\frac{\\sin(x)-x}{x^3}=-\\frac16$$ This is necessary and not derivable from $\\lim\\limits_{x\\to0}\\frac{\\sin(x)}x=1$.\n\nIt then follows that \\begin{align} \\lim_{x\\to0}\\left(\\frac1{\\sin^2(x)}-\\frac1{x^2}\\right) &=\\lim_{x\\to0}\\frac{x^2-\\sin^2(x)}{x^2\\sin^2(x)}\\\\ &=\\lim_{x\\to0}\\frac{x-\\sin(x)}{x^3}\\lim_{x\\to0}\\frac{x+\\sin(x)}{x}\\lim_{x\\to0}\\frac{x^2}{\\sin^2(x)}\\\\ &=\\frac16\\cdot2\\cdot1\\\\[3pt] &=\\frac13 \\end{align}\n\nI have pointed this out earlier also on MSE (see link1, link2, link3, link4), but it seems that this point needs to be retold and retold till it becomes a tautology like $1 = 1$.\n\nThe rigorous meaning of the statement \"$\\sin x \\approx x$ for small $x$\" is that $$\\lim_{x \\to 0}\\frac{\\sin x}{x} = 1\\tag{1}$$ Thus using the approximation $\\sin x \\approx x$ while evaluating certain limits means using the limit formula $(1)$ in your calculations. There is no more meaning attached to $\\sin x \\approx x$ other than above limit formula and one should not even try to attach any more meaning to it as far as the evaluation of limits is concerned.\n\nThus the use of $\\sin x \\approx x$ in the linked question by OP regarding limit of $(1/x - 1/\\sin x)$ is also wrong (although by luck it does produce right answer).\n\nMoreover the accepted answer here (by 2012rcampion) tries to propagate another fallacy which is that the approximation $\\sin x \\approx x$ is not good enough in the current context and perhaps a better approximation is desired. This sort of gives the message that the approximation $\\sin x \\approx x$ is somehow good enough for the linked question. Sorry!! this is simply not correct. If we go down this path we face the following question: how do we know which approximation is good enough in a given context? There is no satisfying answer to this.\n\nSomething must be mentioned about this \"higher order of approximations\" based on Taylor series expansions. When we use an approximation like $$f(a + h) \\approx f(a) + hf'(a) + \\frac{h^{2}}{2!}f''(a) + \\cdots + \\frac{h^{n}}{n!}f^{(n)}(a)\\tag{2}$$ its rigorous meaning is $$\\lim_{h \\to 0}\\dfrac{f(a + h) - \\left\\{f(a) + hf'(a) + \\dfrac{h^{2}}{2!}f''(a) + \\cdots + \\dfrac{h^{n - 1}}{(n - 1)!}f^{(n - 1)}(a)\\right\\}}{h^{n}} = \\frac{f^{(n)}(a)}{n!}\\tag{3}$$ which also written as $$f(a + h) = f(a) + hf'(a) + \\frac{h^{2}}{2!}f''(a) + \\cdots + \\frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})\\tag{4}$$ Note that there is no $\\approx$ sign in the above equation so that it represents an exact formula and the expression $o(h^{n})$ represents a function $g(h)$ such that $g(h)/h^{n} \\to 0$ as $h \\to 0$. Thus using Taylor series expansions like $(2)$ while calculating limits means using the equation $(3), (4)$ in an exact manner and not as an approximation. Limit processes are as exact as $2 + 2 = 4$ and not like $\\pi \\approx 3.1415$.\n\nIn case of Taylor series expansion there is another interpretation possible. If $f$ is a function such that it has an expression in the form of a convergent Taylor series around point $a$ then we write $$f(a + h) = f(a) + hf'(a) + \\frac{h^{2}}{2!}f''(a) + \\cdots + \\frac{h^{n}}{n!}f^{(n)}(a) + \\cdots\\tag{5}$$ Such functions are technically called analytic. This relation is also an exact formula and the RHS is an infinite series in powers of $h$. When using such series for evaluation of limits we don't need to use the expressions like $o(h^{n})$ but then the theoretical justification for using infinite series in limit processes is the following theorem:\n\nA power series allows the following operations at an interior point of its region of convergence:\n\n1. Take limits term by term\n2. Term by term integration with respect to variable whose powers are used\n3. Term by term differentiation with respect to variable whose powers are used.\n\nIf one is aware of the concept of uniform convergence (and thereby knows the proof of the theorem above regarding power series) then this approach of using infinite power series for limit evaluation is fully rigorous and there is simply no need to think in terms of any approximations or things like $o(h^{n})$ and just use the series (and do algebraic simplification) and take limits.\n\nAs far as the answer to this question is concerned the right approach is the one given by robjohn and there is no way to improve upon his answer. However I would like to mention the use of formula $(1)$ in the current context (or in crude language \"use the approximation $\\sin x \\approx x$\"). We can write \\begin{align} L &= \\lim_{x \\to 0}\\frac{1}{\\sin^{2}x} - \\frac{1}{x^{2}}\\notag\\\\ &= \\lim_{x \\to 0}\\frac{x^{2} - \\sin^{2}x}{x^{2}\\sin^{2}x}\\notag\\\\ &= \\lim_{x \\to 0}\\frac{x^{2} - \\sin^{2}x}{x^{4}}\\cdot\\left(\\frac{x}{\\sin x}\\right)^{2}\\notag\\\\ &= \\lim_{x \\to 0}\\frac{x^{2} - \\sin^{2}x}{x^{4}}\\cdot 1^{2}\\notag\\\\ &= \\lim_{x \\to 0}\\frac{x^{2} - \\sin^{2}x}{x^{4}}\\notag \\end{align} This is as far we can go via the use of standard limit formula $(1)$. To go further we need to split $x^{2} - \\sin^{2}x$ as $(x - \\sin x)(x + \\sin x)$ and the denominator is split as $x^{3}\\cdot x$. For the first part $$\\frac{x - \\sin x}{x^{3}}$$ we can use an expansion based on equation $(4)$ namely $$\\sin x = x - \\frac{x^{3}}{6} + o(x^{3})$$\n\n\u2022 \"How do we know which approximation is good enough in a given context? There is no satisfying answer to this.\" Yes, there is. I mean, @2012rcampion's answer demonstrates it: you keep track of the order of the error term. In the section about Taylor series you are basically recapitulating their answer in a more verbose way, so I don't see why you are complaining that it is wrong. \u2013\u00a0Rahul Jun 29 '15 at 5:12\n\u2022 @Rahul: Normally when people solve a mathematical problem they need to know how to check obvious mistakes (like typo/calculation errors). Sometimes a calculation mistake might give rise to some absurdity and that's a hint that something has gone wrong. In this example if you replace $\\sin x$ by $x$ you will get a nice answer $0$ and you will not doubt the calculation even once. An approach based on Taylor series requires experience and definitely not suited for beginners. What is wrong in the accepted answer is the use of $\\approx$ symbols. Continue in next comment. \u2013\u00a0Paramanand Singh Jun 29 '15 at 5:20\n\u2022 @Rahul: Limits are exact stuff and thus each step in limit evaluation should make use of equality sign and every step must be supported by proper theorem. \u2013\u00a0Paramanand Singh Jun 29 '15 at 5:21\n\u2022 Would the downvoter please care to comment? \u2013\u00a0Paramanand Singh Jun 29 '15 at 5:29\n\u2022 \"they need to know how to check obvious mistakes\" Again, you keep track of the order of the error term: see the second half of @2012rcampion's answer. I am the downvoter, so you don't need to go on a witch hunt. \u2013\u00a0Rahul Jun 29 '15 at 5:48\n\n$$A=\\frac1{\\sin^2(x)}-\\frac1{x^2}=\\frac{x^2-\\sin^2(x)}{x^2\\sin^2(x)}$$ and knowing that, close to $0$, $\\sin(x)\\approx x$, then the denominator looks like $x^4$ that is to say that, if the limit exist the numerator should also be developed at least up to order $x^4$; again, since $\\sin(x)\\approx x$, then $\\sin(x)$ should be developed for at least one more term.\nSo, using $\\sin(x)=x-\\frac{x^3}{6}+O\\left(x^4\\right)$, after simplication, the numerator is just $\\frac{x^4}{3}+\\cdots$ and the denominator is $x^4+\\cdots$; so the limit $\\frac 13$.\nIf we needed more than the limit (say for example how it is approached), more terms would be required. Using $\\sin(x)=x-\\frac{x^3}{6}+\\cdots$ and using it for both numerator and denominator, we should get $$A\\approx \\frac{x^2-\\left(x-\\frac{x^3}{6}\\right)^2}{x^2 \\left(x-\\frac{x^3}{6}\\right)^2}$$ which, after simplication and long division would give $$A\\approx \\frac{1}{3}+\\frac{x^2}{12}+\\cdots$$ Using instead $\\sin(x)=x-\\frac{x^3}{6}+\\frac{x^5}{120}+\\cdots$ and using it for both numerator and denominator, and doing the same as before, we should get $$A\\approx \\frac{1}{3}+\\frac{x^2}{15}+\\cdots$$ which is slightly different.\nTo illustrate the above, I suggest you plot on the same graph the three functions $y_1=\\frac1{\\sin^2(x)}-\\frac1{x^2}$ , $y_2=\\frac{1}{3}+\\frac{x^2} {12}$, $y_3=\\frac{1}{3}+\\frac{x^2}{15}$ for $-\\frac 12 \\leq x\\leq \\frac 12$ . This would be better than a long and tedious speech.\nSince $\\sin(x)$ is an entire odd function and $\\sin(x)\\approx x$ in a neighbourhood of the origin, the limit $$L=\\lim_{x\\to 0}\\frac{\\sin(x)-x}{x^3}=\\lim_{x\\to 0}\\frac{\\sin(2x)-2x}{8x^3}$$ must exist, and be equal to $$\\frac{4L-L}{3} = \\frac{1}{3}\\lim_{x\\to 0}\\frac{4\\sin(2x)-8\\sin(x)}{8x^3} =\\frac{1}{3}\\lim_{x\\to 0}\\frac{\\sin(x)\\cos(x)-\\sin(x)}{x^3}$$ so: $$L = -\\frac{1}{3}\\lim_{x\\to 0}\\frac{1-\\cos x}{x^2} = -\\frac{1}{6}\\lim_{x\\to 0}\\frac{\\sin^2\\frac{x}{2}}{\\left(\\frac{x}{2}\\right)^2} = \\color{red}{-\\frac{1}{6}}.$$", "date": "2019-11-19 22:54:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1338411/why-doesnt-using-the-approximation-sin-x-approx-x-near-0-work-for-computin", "openwebmath_score": 0.9807103872299194, "openwebmath_perplexity": 355.4858753777299, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104914476339, "lm_q2_score": 0.900529795461386, "lm_q1q2_score": 0.8702814421950753}}
{"url": "https://mathematica.stackexchange.com/questions/133098/derivative-of-the-inverse-of-a-certain-variable", "text": "# Derivative of the inverse of a certain variable\n\nI want to take a derivative of the inverse of a certain variable, e.g., being the function:\n\nf[x_, a_, T_] := (a x^3)/T\n\n\nI want the derivative with respect to $1/T$:\n\n$$\\frac{\\partial f}{\\partial \\left(\\frac{1}{T}\\right)}=ax^3$$\n\nIt doesn't work with:\n\nD[f[x, a, T], 1/T]\n\n\n$\\frac{\\partial f(x,a,T)}{\\partial \\frac{1}{T}}$\n\nneither if I define:\n\nb=1/T\n\n\nD[f[x, a, b], b]\n\n\nIt always state \"1/T is not a valid variable\"\n\n$\\frac{\\partial f(x,a,T)}{\\partial \\frac{1}{T}}$\n\n\u2022 Block[{c}, D[f[x, a, 1/c], c] /. c -> 1/T]. The Block is only there in case you happen to have c defined somewhere else. \u2013\u00a0march Dec 8 '16 at 21:03\n\u2022 I can't believe it was this simple. Appreciate!!! \u2013\u00a0RTS Dec 8 '16 at 21:19\n\u2022 @march Why do not you put this as an answer? That was interesting for me. I needed this information the other day and I had not found it. \u2013\u00a0LCarvalho Dec 8 '16 at 22:08\n\u2022 @march Seems simple to you, but to other users it's cool \u2013\u00a0LCarvalho Dec 8 '16 at 22:10\n\u2022 @LCarvalho. I mean, I would guess that this is a duplicate, but I only had time to do a little searching, so I haven't found one yet. Nonetheless, I'll quickly write something up. \u2013\u00a0march Dec 8 '16 at 22:20\n\nyou can also use the chain rule:\n\n$$\\frac{\\partial f}{\\partial \\left(\\frac{1}{T}\\right)}=\\frac{\\frac{\\partial f}{\\partial T}}{\\frac{\\partial \\frac{1}{T}}{\\partial T}}$$\n\n D[f[x, a, T], T]/D[1/T, T]\n\n\na x^3\n\nYou had the right idea, except for Setting the value of b to be 1/T. The reason that doesn't work is that D does not have the Attribute HoldAll (or variants) which means that its arguments get evaluated before D does. That is, if you do\n\nb = 1/T;\nD[f[x, a, 1/b], b]\n\n\nyou get the error because 1/T has been put in place of b before the derivative is taken. The fix is to wrap the entire expression in the Block scoping construct, which will delay the evaluation of b until after the derivative is taken, like so:\n\nf[x_, a_, T_] := (a*x^3)/T\nb = 1/T;\nBlock[{b}, D[f[x, a, 1/b], b]]\n(* a x^3 *)\n\n\nAlternatively, use an undefined symbol and use a replacement rule. I prefer this version because I like to avoid cluttering up the global namespace with defined symbols. So:\n\nf[x_, a_, T_] := (a*x^3)/T\nBlock[{c}, D[f[x, a, 1/c], c] /. c -> 1/T]\n(* a x^3 *)\n\n\u2022 Now yes. Very good tip. \u2013\u00a0LCarvalho Dec 8 '16 at 22:30", "date": "2020-03-29 01:01:13", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/133098/derivative-of-the-inverse-of-a-certain-variable", "openwebmath_score": 0.5802436470985413, "openwebmath_perplexity": 1978.1912545975795, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759632491111, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8702376663120243}}
{"url": "https://mathhelpboards.com/threads/2-1-5-find-the-general-solution-of-the-given-differential-equation.24499/", "text": "# [SOLVED]2.1.5 Find the general solution of the given differential equation\n\n#### karush\n\n##### Well-known member\nFind the general solution of the given differential equation\n$\\displaystyle y^\\prime - 2y =3te^t, \\\\$\n\nObtain $u(t)$\n$\\displaystyle u(t)=\\exp\\int -2 \\, dx =e^{-2t} \\\\$\nMultiply thru with $e^{-2t}$\n$e^{-2t}y^\\prime + 2e^{-2t}y= 3te^{-t} \\\\$\nSimplify:\n$(e^{-2t}y)'= 3te^{-t} \\\\$\nIntegrate:\n$\\displaystyle e^{-2t}y=\\int 3te^{-t} dt =-3e^{-t}(t+1)+c_1 \\\\$\n{Divide by $e^{-2t}$\n$y=-3e^t t - 3e^{-t} +c_1 e^t \\\\$\n$y=\\color{red}{c_1 e^{2t}-3e^t}$\n\nok sumtum went wrong somewhere?\n\n$$\\tiny\\color{blue}{\\textbf{Text book: Elementary Differential Equations and Boundary Value Problems}}$$\n\n#### MarkFL\n\nStaff member\nI with you up to after the integration:\n\n$$\\displaystyle e^{-2t}y=-3e^{-t}(t+1)+c_1$$\n\nThus:\n\n$$\\displaystyle y(t)=-3e^{t}(t+1)+c_1e^{2t}$$\n\nThe solution given by the book is incorrect.\n\n#### karush\n\n##### Well-known member\nthese problems require mega being carefull!", "date": "2021-06-19 09:23:37", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/2-1-5-find-the-general-solution-of-the-given-differential-equation.24499/", "openwebmath_score": 0.7748227715492249, "openwebmath_perplexity": 11158.296185602163, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.980875959894864, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8702376589480801}}
{"url": "http://wjxu.grafikohnegrenzen.de/differential-equation-eigenvalue-problem.html", "text": "## Differential Equation Eigenvalue Problem\n\nApply matrix techniques to solve systems of linear ordinary differential equations with constant coefficients. n first order equations that we can solve separately. Find the eigenvalues and Eigen functions for the boundary value problem, x^{2}{y} for Teachers for Schools for Working Scholars for College. One mathematical tool, which has applications not only for Linear Algebra but for differential equations, calculus, and many other areas, is the concept of eigenvalues and eigenvectors. Therefore, for each eigenfunction Xn with corresponding eigen-value \u201an, we have a solution Tn such that the function un(x;t) = Tn(t)Xn(x) is a solution. In other words, we have to \ufb01nd all of the numbers \u03bb such that there is a solution of the equation AX = \u03bbX for some function X (X 6= 0) that satis\ufb01es the boundary conditions at 0 and at l. This function specifies the probability density for observing the particle at a given position,. Just as you said, the general solution of $$y''+\\lambda y=0$$ has the form $$y=A\\cos\\sqrt{\\lambda}x+B\\sin\\sqrt{\\lambda}x\\,,$$ (we assumed here that $\\lambda>0$). A very important problem in matrix algebra is called the eigenvalue problem. eigenvalue problem, with the quadratic two-parameter eigenvalue problem as important special case; (b) to show the relevance of this problem to determine critical delays for various types of DDEs; (c) to provide alternative derivations of existing matrix pencil methods using the context of polynomial two-parameter eigenvalue problems;. 4, Modeling with First Order Equations. An ode is an equation for a function of. Nonlinear eigenvalue problems (NEPs) arise in many fields of science and engineering. [12,13] are in perfect agreement, (3) in applications such. 1 Estimating the Diffusion Coefficient in the Heat Equation 703 9. Solve first order differential equations using standard methods, such as separation of variables, integrating factors, exact equations, and substitution methods; use these methods to solve analyze real-world problems in fields such as economics, engineering, and the sciences. Differential Equations with Differential Equation many of the problems are difficult to make up on the spur of Real Eigenvalues - Solving systems of. We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions. differential equations. ) The equation \u03bb2\u2212(a+b)\u03bb+ab\u2212h2 = 0 is called the eigenvalue equation of the matrix A. A differential equation is an equation that relates the rate of change of some process to other processes that are changing in time. txt) or read online for free. Chapter Five - Eigenvalues , Eigenfunctions , and All That The partial differential equation methods described in the previous chapter is a special case of a more general setting in which we have an equation of the form L 1 \u00ddx\u00deu\u00ddx,t\u00de+L 2 \u00ddt\u00deu\u00ddx,t\u00de = F \u00ddx,t\u00de. One of the stages of solutions of differential equations is integration of functions. From \"M\u00e9moire sur l'integration des \u00e9quations lin\u00e9aires\" (1840); image courtesy of Archive. In this paper, we are concerned with the solution of a class of boundary value problems \u2212y\u2033+f(x)y=\u03bby, y(0)=0, y(\u221e)=0, where f(x) monotonically increases to infinity as n increases to infinity. 2 The Method of Elimination, 229 4. An eigenvalue \u03bb of an nxn matrix A means a scalar (perhaps a complex number) such that Av=\u03bbv has a solution v which is not the 0 vector. They're both hiding in the matrix. So Mathematica provides us only one eigenvector \u03be = [ 1, 0, 0] corresponding to the eigenvalue \u03bb = 1 (therefore, A is defective) and one eigenvector v = <-1,1,0> corresponding eigenvalue \u03bb = 0. A = [ 1 1 0 0 0 1 0 0 1]. 2 Sturm-Liouville Boundary Value Problems 723. The problem is to find the numbers, called eigenvalues, and their matching vectors, called eigenvectors. Ordinary Di\ufb00erential Equations Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Meade prepared for UC Davis students. I will be in my office on Tuesday (12/15) 10-12, Wednesday (12/16) 10-4. It assumes some knowledge of calculus, and explains the tools and concepts for analysing models involving sets of either algebraic or 1st order differential equations. In this article, we study the global bifurcation from in nity of non-linear eigenvalue problems for ordinary di erential equations of fourth order. The simplest example and one that will play a role in neurobiology is the decay to rest of the membrane potential. An important special case is when is an a\ufb03ne function of y. NONLINEAR EIGENVALUE PROBLEMS 489 Many people have worked on nonlinear eigenvalue problems; in particular, for ordinary differential equations and integral equations. The language of dynamic phenomena is differential equations. The \ufb01rst one is easier, especially in the 2 \u00d7 2 case. For example, whenever a new type of problem is introduced (such as first-order equations, higher-order equations, systems of differential. - Minimization problems for variational integrals, existence and regularity theory for minimizers and critical points, geometric measure theory - Variational methods for partial differential equations, linear and nonlinear eigenvalue problems, bifurcation theory - Variational problems in differential and complex geometry. , Bayramoglu, Mamed, and Aslanov, Khalig M. differential-equations eigenvalues schrodinger produces a confluent Heun differential equation with solution Eigenvalue Problem on a Rectangle with Dirichlet. Questions concerning eigenvectors and eigenvalues are central to much of the theory of linear. Strauss, Partial Differential Equations: An Introduction, Wiley G. Complex Eigenvalues \u2013 Solving systems of differential equations with complex eigenvalues. Then the chaotic behavior of the logistic map. The roots of this differential equation are called eigenvalues, and the corresponding functional solutions are known as eigenfunctions. In this series, we will explore temperature, spring systems, circuits, population growth, biological cell motion, and much more to illustrate how differential equations can be used to model nearly everything. The authors have sought to combine a sound and accurate (but not abstract) exposition of the. In particular, for any scalar \u201a, the solution of the ODE for T is given by T(t) = Ae\u00a1k\u201at for an arbitrary constant A. In natural sciences and engineering, are often used differential equations and systems of differential equations. It is perhaps not surprising that one of the primary examples involves the L-shaped membrane read more >>. Naylor, Differential Equations of Applied Mathematics, John Wiley & Sons I. Osborne\u2020 (Received 1 June 2001; revised 18 October 2002) Abstract Discretisations of di\ufb00erential eigenvalue problems have a sensitivity to perturbations which is asymptotically least as h \u21920 when the di\ufb00erential equation is in \ufb01rst order sys-tem form. New algorithms are suggested and old algorithms, i. A = [ 1 1 0 0 0 1 0 0 1]. This approach is based on the extension of the previously reported differential transfer matrix method with modified basis functions. Partial Differential Equations and Boundary Value Problems with Maple, Second Edition, presents all of the material normally covered in a standard course on partial differential equations, while focusing on the natural union between this material and the powerful computational software, Maple. 2 Properties of Sturm-Liouville Eigenvalue Problems There are several properties that can be proven for the (regular) Sturm-Liouville eigenvalue problem. Elementary Differential Equations with Boundary Value Problems integrates the underlying theory, the solution procedures, and the numerical/computational aspects of differential equations in a seamless way. Variation of Parameters for Matrices. The eigenvalue problem of complex structures is often solved using finite element analysis, but neatly generalize the solution to scalar-valued vibration problems. Note that the eigenfunction is defined up to a multiplicative constant so you can just set $\\epsilon$=1 and there is only one parameter $\\epsilon'$ to vary in order to achieve a properly decaying solution at infinity. NONLINEAR EIGENVALUE PROBLEMS 489 Many people have worked on nonlinear eigenvalue problems; in particular, for ordinary differential equations and integral equations. The fractional differential equation (1. I am looking for a way to solve them in Python. Two Distinct Real Eigenvalue Case. An important special case is when is an a\ufb03ne function of y. Repeated Eigenvalues \u2013 Solving systems of differential equations with repeated eigenvalues. DiPrima and D. 14 Chapter 1. First, find a matrix that diagonalizes The eigenvalues of are and with corresponding eigenvectors and Diagonalize using the matrix whose columns consist of and to obtain and The system has the following form. Introduction. In the last few years, fractional differential equations have gained attentions due to their numerous applications in various aspects of science and technology. WILKES Department of Chemical Engineering, The University of Michigan, Ann Arbor, Michigan 48109, U. , Bayramoglu, Mamed, and Aslanov, Khalig M. This longer text consists of the main text plus three additional chapters (Eigenvalue Problems and Sturm\u2014Liouville Equations; Stability of Autonomous Systems. Suppose, I have an differential equation like this one: mu1 u1[x] - u1''[x] - 10 u1[x] == 0 where mu1 is the eigenvalue and u1 is the eigenfuntion. Solving an eigenvalue problem means finding all its eigenvalues and associated eigenfunctions. Fourier Series and Systems of Differential Equations and Eigenvalue Problems by Leif Mejlbro. See [I, 4 and 51 where several references are given. Functional Anal. During the studying of linear. y'' + y' +\u03bby = 0, y(0) = 0, y(2) = 0. Here a brief overview of the required con-cepts is provided. As a special case of the proposed method, we find a closed form for a parameterization of the critical surface for the scalar case. This page describes how it can be used in the study of vibration problems for a simple lumped parameter systems by considering a very simple system in detail. This approach is based on the extension of the previously reported differential transfer matrix method with modified basis functions. This process is experimental and the keywords may be updated as the learning algorithm improves. Solutions of S-L Problems. AN EIGENVALUE PROBLEM INVOLVING A FUNCTIONAL DIFFERENTIAL EQUATION ARISING IN A CELL GROWTH MODEL BRUCE VAN BRUNT 1 and M. From \"M\u00e9moire sur l'integration des \u00e9quations lin\u00e9aires\" (1840); image courtesy of Archive. 1 Matrices and Linear Systems 264 5. To do this we have to distinguish two cases, called complete and defective. Going through these examples, the. Running some numerics suggests that the mapping h \u21a6 \u03bbn(Hh) where \u03bbn denotes the nth eigenvalue of Hh, n = 0, 1, 2, \u2026, is monotone non-decreasing. Sturm-Liouville Eigenvalue Problems and Generalized Fourier Series Examples of Regular Sturm-Liouville Eigenvalue Problems We will now look at examples of regular Sturm-Liouville differential equations with various combinations of the three types of boundary conditions Dirichlet, Neumann and Robin. Chapter 9 Diffusion Equations and Parabolic Problems Chapter 10 Advection Equations and Hyperbolic Systems Chapter 11 Mixed Equations Part III: Appendices. We show that there exist classes of explicit numerical integration methods that can handle very stiff problems if the eigenvalues are separated into two clusters, one containing the \"stiff,\" or fast, components, and one containing the slow components. Observe that eigendirections vary if we change parameters. where the eigenvalues are repeated eigenvalues. To actually solve ODE systems having complex eigenvalues, imitate the procedure in the following example. Lecture Description. Applications of the method to boundary value and initial value problems, as well as several examples are illustrated. pdf), Text File (. Krylov Subspace Methods for the Eigenvalue problem Presented by: Sanjeev Kumar Applications We need only few eigen (singular) pairs, and matrices can be large and sparse Solving homogeneous system of linear equations A x = 0. Chapter 12 Measuring Errors Chapter 13 Polynomial Interpolation and Orthogonal Polynomials Chapter 14 Eigenvalues and inner product norms Chapter 15 Matrix powers and exponentials. I believe that there is a misprint in the problem statement. Of our three ordinary differential equations, only two will be eigenvalue problems. The eigenvalues and eigenfunctions are characterized in terms of the Mittag-Leffler functions. With equations conveniently specified symbolically, the Wolfram Language uses both its rich set. More of Cauchy's method for solving a system of linear, first-order differential equations with constant coefficients, equivalent to a modern-day eigenvalue problem. has a solution if the admissible sets are all contained in a bounded open set called design region (known as Buttazzo-Dal Maso Theorem). Homogeneous System Solutions with Distinct and Repeated Eigenvalues. Let's see some examples of first order, first degree DEs. In this paper, we describe new localization results for nonlinear eigenvalue problems that generalize Gershgorin\u2019s theorem, pseudospectral inclusion theorems, and the Bauer-Fike theorem. The average is now 70/100. 2) a)Find the eigenvalues and eigenfunctions of the boundary-value problem. In the nonlinear context a separatrix plays the role of an eigenfunction and the initial conditions that give rise to the separatrix play the role of eigenvalues. My name is Will Murray and I thank you very much for watching, bye bye. In this section we will define eigenvalues and eigenfunctions for boundary value problems. Two Coupled Oscillators Let's consider the diagram shown below, which is nothing more than 2 copies of an harmonic oscillator, the system that we discussed last time. Dur\u00e1n, Ariel L. Find the particular solution given that y(0)=3. differential-equations eigenvalues schrodinger produces a confluent Heun differential equation with solution Eigenvalue Problem on a Rectangle with Dirichlet. Eigenvalue problem. 2 The Method of Elimination 239 4. In this blog post \u2014 inspired by Strogatz (1988, 2015) \u2014 I will introduce linear differential equations as a means to study the types of love affairs two people might. Eigenvalue Problems for Odes 1 - Free download as PDF File (. 1: Eigenvalue Problems for y'' + \u03bby = 0 - Mathematics LibreTexts Skip to main content. Additionally, the input data of the problem depend on a numerical parameter. This is an additional adjustable parameter we. Perfect for undergraduate and graduate studies. Partial Differential Equations with at Least Three Independent Variables. There are homogeneous boundary conditions in x and y. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. Ambrosetti and P. 8) Each class individually goes deeper into the subject, but we will cover the basic tools needed to handle problems arising in physics, materials sciences, and the life sciences. Differential Equations Calculator. The case a = b and h = 0 needs no investigation, as it gives an equation of a circle. can make the numerical solution of eigenvalue problems a difficult and challenging task due to possible crossing of eigenvalues. I believe that there is a misprint in the problem statement. It is perhaps not surprising that one of the primary examples involves the L-shaped membrane read more >>. \u2022 Ordinary Differential Equation: Function has 1 independent variable. Two Distinct Real Eigenvalue Case. Introductory Differential Equations with Boundary Value Problems Third Edition Martha L. The task is to compute the fourth eigenvalue of Mathieu's equation. These pages offer an introduction to the mathematics of such problems for students of quantum chemistry or quantum physics. fairly its not. this equation, and we end up with the central equation for eigenvalues and eigenvectors: x = Ax De nitions A nonzero vector x is an eigenvector if there is a number such that Ax = x: The scalar value is called the eigenvalue. Multiplying through by this, we get y0ex2 +2xex2y = xex2. Nonstiff ODE system due to van der Pol (assuming a small coefficient). where P and Q are functions of x. In this paper the authors develop a general framework for calculating the eigenvalues of a symmetric matrix using ordinary differential equations. In this set of equations, $$E$$ is an eigenvalue, which means there are only non-trivial solutions for certain values of $$E$$. 1 Eigenvalues and eigenvectors \u2022 Diagonalization Section 7. - Chapter 3: New Problem 35 on determination of radii of convergence of power series solutions of differential equations; new Example 3 just before the subsection on logarithmic cases in the method of Frobenius, to illustrate first the reduction-of-order formula with a simple non-series problem. EigenNDSolve uses a spectral expansion in Chebyshev polynomials and solves systems of linear homogenous ordinary differential eigenvalue equations with general (homogenous) boundary conditions. \u2022This first- order homogeneous system of linear differential equations is already in matrix form. 18) g(0) = 0 and g(H) = 0. Eigenvalues! Eigenvalues! This page is a collection of online resources that might come in handy to anyone interested in learning about differential equations (on an introductory level), and also students who are taking their first diffeq course in college. Problems 1-5 are called eigenvalue problems. Ecker, andW. He is the author of numerous technical papers in boundary value problems and random differential equations and their applications. Let's say we have a linear differential operator $\\hat A = \\sum_k a_k \\frac {d^k} {dx^k}$ and we want to solve the eigenvalue equation $\\hat A f(x) = a f(x)$ which is. Partial Differential Equations and Boundary Value Problems with Maple, Second Edition, presents all of the material normally covered in a standard course on partial differential equations, while focusing on the natural union between this material and the powerful computational software, Maple. Knowledge beyond the boundaries. 3 The Geometry of First-Order Differential Equations 1. The Concept of Eigenvalues and Eigenvectors. In natural sciences and engineering, are often used differential equations and systems of differential equations. 1 Vector spaces and linear. There are various methods by which the continuous eigenvalue problem may be. The eigenvalue problem for such an A (with boundary conditions) is to \ufb01nd all the possible eigenvalues of A. If there are two distinct, , real eigenvalues, with corresponding eigenvectors and , then the two solutions. My name is Will Murray and I thank you very much for watching, bye bye. Such values of \u03bb, when they exist, are called the eigenvalues of the problem, and the corresponding solutions are the eigenfunctions associated to each \u03bb. WILKES Department of Chemical Engineering, The University of Michigan, Ann Arbor, Michigan 48109, U. Lecture 18: Complex eigenvalues and spirals. This is all part of a larger lecture series on differential equations here on educator. , v(x,y,z,t). Eigenvalue range, specified as a two-element real vector. Differential equations are very common in physics and mathematics. Love, on the other hand, is humanity's perennial topic; some even claim it is all you need. The average is now 70/100. Repeated Eigenvalues \u2013 Solving systems of differential equations with repeated eigenvalues. differential-equations eigenvalues schrodinger produces a confluent Heun differential equation with solution Eigenvalue Problem on a Rectangle with Dirichlet. These limitations are appropriate for most quantum mechanics problems as well as many classical problems. Eigenvalues, Eigenvectors, and Di erential Equations William Cherry April 2009 (with a typo correction in November 2015) The concepts of eigenvalue and eigenvector occur throughout advanced mathematics. Eigenvalues of a Sturm Liouville differential equation tutorial of Differential equations I course by Prof Chris Tisdell of Online Tutorials. Di\ufb00erential equations are called partial di\ufb00erential equations (pde) or or-dinary di\ufb00erential equations (ode) according to whether or not they contain partial derivatives. 2 Properties of Sturm-Liouville Eigenvalue Problems There are several properties that can be proven for the (regular) Sturm-Liouville eigenvalue problem. Knowledge beyond the boundaries. So, it is natural to use the analytic multiplicity to count eigenvalues, and the analytic multiplicity plays an important role in the study of the dependence of the eigenvalues of a spectral problem on the differential equation boundary value problem (see, for example, [5]. 2 The Eigenvalue Method for Homogeneous. Solution to linear constant coe\ufb03cient ODE systems 90 7. A University Level Introductory Course in Differential Equations. In this section we will solve systems of two linear differential equations in which the eigenvalues are distinct real numbers. We consider an eigenvalue problem for stochastic ordinary differential operators with stochastic boundary conditions. Homogeneous Linear Equations; Forcing; Sinusoidal Forcing; Forcing and Resonance; Projects for Second-Order Differential Equations; 5 Nonlinear Systems. This book is aimed at students who encounter mathematical models in other disciplines. Introduction to Systems of Differential Equations 228 4. For the following system of linear differential equations:\\frac{\\mathrm{d} x}{\\mathrm{d} Question: For the following system of linear differential equations:. Ambrosetti and P. Braselton AMSTERDAM \u2022BOSTON HEIDELBERG \u2022 LONDON NEW YORK \u2022OXFORD \u2022 PARIS SAN DIEGO SAN FRANCISCO \u2022SINGAPORE SYDNEY \u2022 TOKYO Academic Press is an imprint of Elsevier. 2 The Eigenvalue Method for Homogeneous Systems 304. 1 Linear differential equations All linear equations involve a linear operator L. Questions concerning eigenvectors and eigenvalues are central to much of the theory of linear. The ball's acceleration towards the ground is the acceleration due to gravity minus the acceleration due to air resistance. Typical problems considered were elliptic partial differential equations of the form U xx + U yy = f(x'y)' (1) or U xx + U yy + X2U = 0 (2) where appropriate boundary conditions are specified so that the problem is self-adjoint, The four methods are relaxation, Galerkin, Rayleigh-. Worked Example. A very important problem in matrix algebra is called the eigenvalue problem. Differential equations are the language of the models that we use to describe the world around us. The Handbook of Ordinary Differential Equations: Exact Solutions, Methods, and Problems, is an exceptional and complete reference for scientists and engineers as it contains over 7,000 ordinary. In this section we will define eigenvalues and eigenfunctions for boundary value problems. KEYWORDS: Worksheets, Introduction to Matrices, Definitions, Matrix arithmetic, Identity matrices, Inverse matrices, Finding eigenvalues and eigenvectors, Using eigenvalues and eigenvectors to solve differential equations and discrete systems. is an eigenvalue if it is a pole of , where denotes the identity operator. Phase Plane. were developed and used for. , Schr\u00f6dinger. Notice that this is exactly the same equation as in the first (both ends kept at 0 degree) heat conduction problem, due to the fact that both problems have the same set of eigenvalues (but with different eigenfunctions). Find the solution of y0 +2xy= x,withy(0) = \u22122. Here a brief overview of the required con-cepts is provided. The critical delays of a delay\u2010differential equation can be computed by solving a nonlinear two\u2010parameter eigenvalue problem. using eigenvalues and eigenvectors to solve boundary value problems for Laplace\u2019s equation and other partial differential equations, analytically and via finite differences. Partial differential equations, 171\u2013181, Lecture Notes in Pure and Appl. The problem is nonlinear with respect to the spectral parameter and involves generally nonlocal additional conditions specified by a Stieltjes integral. You will need to find one of your fellow class mates to see if there is something in these. ,xn and the time t as shown below dx1 --- = a11 x1 + a12 x2 +. However, this only satisfies the differential equation, not boundary conditions. To find real valued solutions. and eigenvalue systems of the form. Problem Sets Use of Problem Sets. The eigenvalues and eigenfunctions are viewed at a fixed point as functions of the interval length. evr(2) specifies the upper limit of the range, and must be finite. Because of that, problem of eigenvalues occupies an important place in linear algebra. senting solutions of the partial differential equation. Let me show you the reason eigenvalues were created, invented, discovered was solving differential equations, which is our purpose. Since we are going to be working with systems in which A is a 2 x 2 matrix we will make that assumption from the start. Incidentally we note that an alternative, yet related approach to the ME formalism\u2014in the form of delay differential equation model\u2014has been developed which does not require small gain and. A first\u2010order differential equation is said to be linear if it can be expressed in the form. My name is Will Murray and I thank you very much for watching, bye bye. which is the same as (4. With the aid of the spectral theory of compact self-adjoint operators in Hilbert spaces, we show that the spectrum of this problem consists of only countable real eigenvalues with finite multiplicity and the corresponding eigenfunctions form a complete. The problem is to find the numbers, called eigenvalues, and their matching vectors, called eigenvectors. We will be concerned with finite difference techniques for the solution of eigenvalue and eigenvector problems for ordinary differential equations. where the eigenvalues are repeated eigenvalues. This volume is application-oriented and rich in examples. Automatically selecting between hundreds of powerful and in many cases original algorithms, the Wolfram Language provides both numerical and symbolic solving of differential equations (ODEs, PDEs, DAEs, DDEs, ). 4* Initial and Boundary Conditions 20 1. - Minimization problems for variational integrals, existence and regularity theory for minimizers and critical points, geometric measure theory - Variational methods for partial differential equations, linear and nonlinear eigenvalue problems, bifurcation theory - Variational problems in differential and complex geometry. A University Level Introductory Course in Differential Equations. The Linear System Solver is a Linear Systems calculator of linear equations and a matrix calcularor for square matrices. The presentation does not presume a deep knowledge of mathematical and functional analysis. Somebody say as follows. 1 Vector spaces and linear. Solve first order differential equations using standard methods, such as separation of variables, integrating factors, exact equations, and substitution methods; use these methods to solve analyze real-world problems in fields such as economics, engineering, and the sciences. 1 First-Order Systems and Applications 246 4. Consider a linear homogeneous system of $$n$$ differential equations with constant coefficients, which can be written in matrix form as. of Statistics. Additionally, the input data of the problem depend on a numerical parameter. Homogeneous Linear Equations; Forcing; Sinusoidal Forcing; Forcing and Resonance; Projects for Second-Order Differential Equations; 5 Nonlinear Systems. Hopefully, those solu-tions will form a useful basis in some function space. Harrar II\u2217 M. Reference [1] J. Eigenvalue problems for linear differential equations, such as time-independent Schr\u00f6 dinger equations, can be generalized to eigenvalue problems for nonlinear differential equations. 2 Properties of Sturm-Liouville Eigenvalue Problems 189 6. A class of problems to which our previous examples belong are the Sturm-Liouville eigenvalue problems. Let's say we have a linear differential operator $\\hat A = \\sum_k a_k \\frac {d^k} {dx^k}$ and we want to solve the eigenvalue equation $\\hat A f(x) = a f(x)$ which is. 2) (The roots are distinct if a 6= b or h 6= 0. 3 Numerical Methods for Systems, 240 Linear Systems of Differential Equations 255 5. Because the differential equation and boundary conditions for X(x), form a Sturm-Liouville problem, we know that the solutions to this problem, Xn(x), are an infinite set of orthogonal eigenfunctions. Numerical Methods for Differential Equations Chapter 1: Initial value problems in ODEs Gustaf Soderlind and Carmen Ar\u00a8 evalo\u00b4 Numerical Analysis, Lund University Textbooks: A First Course in the Numerical Analysis of Differential Equations, by Arieh Iserles and Introduction to Mathematical Modelling with Differential Equations, by Lennart Edsberg. Here a brief overview of the required con-cepts is provided. is an idealized heat source of. WILKES Department of Chemical Engineering, The University of Michigan, Ann Arbor, Michigan 48109, U. Math 285: Differential Equations (3 credit hours) Course Description This is an introduction to ordinary differential equations with an emphasis on applications. 1 First-Order Systems and Applications, 219 4. Solve u = Au, where A =. Partial Differential Equations with at Least Three Independent Variables. Since our last example and that wraps up our lecture on repeated eigenvalues so, this is the systems of differential equations where we had repeated eigenvalues. The integrating factor is e R. Hence we have which implies that an eigenvector is We leave it to the reader to show that for the eigenvalue , the eigenvector is Let us go back to the system with complex eigenvalues. Thinking about solving coupled linear differential equations by considering the problem of developing a solution to the following homogeneous version of Eq. It is suitable for a one-semester undergraduate level or two-semester graduate level course in PDEs or applied mathematics. This book is aimed at students who encounter mathematical models in other disciplines. During the studying of linear. Dur\u00e1n, Ariel L. We use our results to analyze three. 2 The Eigenvalue Method for Homogeneous Systems 304. To find real valued solutions. problem of linear algebra is the eigenvalue problem which is more sophisticated. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. In this series, we will explore temperature, spring systems, circuits, population growth, biological cell motion, and much more to illustrate how differential equations can be used to model nearly everything. For example, whenever a new type of problem is introduced (such as first-order equations, higher-order equations, systems of differential. 3 Numerical Methods for Systems, 240 Linear Systems of Differential Equations 255 5. The eigenvalue problem for a linear system of ordinary differential equations is considered. The Concept of Eigenvalues and Eigenvectors. COMPOUND MATRIX METHOD FOR EIGENVALUE PROBLEMS IN MULTIPLE CONNECTED DOMAINS N. The complete case. This is the complex eigenvalue example from [1], Section 3. y'' + y' +\u03bby = 0, y(0) = 0, y(2) = 0. For uncoupled systems,. The solution to the system of differential equations can therefore be written out as a linear combination of the real and imaginary parts of the eigenvector associated with eigenvalue. It is suitable for a one-semester undergraduate level or two-semester graduate level course in PDEs or applied mathematics. Since the eigenvalues are real distinct roots and one eigenvalue is positive with the other eigenvalue negative, the equilibrium point is a saddle and the solutions are unstable. And the lambda, the multiple that it becomes-- this is the eigenvalue associated with that eigenvector. In this series, we will explore temperature, spring systems, circuits, population growth, biological cell motion, and much more to illustrate how differential equations can be used to model nearly everything. Eigenvalue Problems for Odes 1 - Free download as PDF File (. The case a = b and h = 0 needs no investigation, as it gives an equation of a circle. Section 6: Systems of Equations: Lecture 2 | 1:03:54 min. A differential equation is an equation that relates the rate of change of some process to other processes that are changing in time. 8) Each class individually goes deeper into the subject, but we will cover the basic tools needed to handle problems arising in physics, materials sciences, and the life sciences. Solve u = Au, where A =. Two Distinct Real Eigenvalue Case. Introduction to Systems of Differential Equations 246 4. Math 285: Differential Equations (3 credit hours) Course Description This is an introduction to ordinary differential equations with an emphasis on applications. Notes on Diffy Qs: Differential Equations for Engineers. 1) Find the eigenfunctions and the equations that defines the eigenvalues for the given boundary-value problem. This volume is an introductory level textbook for partial differential equations (PDE's) and suitable for a one-semester undergraduate level or two-semester graduate level course in PDE's or applied mathematics. In this course, we will only study two-point boundary value problems for scalar linear second order ordinary di erential equations. Variation of Parameters for Matrices. Since we are going to be working with systems in which A is a 2 x 2 matrix we will make that assumption from the start. DIFFERENTIAL EQUATIONS. Chapters One to Five are organized according to the equations and the basic PDE's are introduced in an easy to understand manner. A basis of the system consists of two solution vectors y1(t)and y2(t), which are linearly independent and each of which satis\ufb01es the corresponding homogeneous equations L(y1)=0 and L(y2)=0. Introduction. Of our three ordinary differential equations, only two will be eigenvalue problems. Additionally, the input data of the problem depend on a numerical parameter. Parabolic partial differential equations are differential equations which depend on space and time. Causality and the Wave Equation Integrating the Bell Curve Compressor Stalls and Mobius Transformations Dual Failures with General Densities Phase, Group, and Signal Velocity Series Solutions of the Wave Equation The Limit Paradox Proof That PI is Irrational Simple Proof that e is Irrational The Filter Of Observation Eigenvalue Problems and. find (manually, numerically, and graphically) and interpret solutions to differential equations, systems of differential equations, and initial value problem s (A, B, C) 3. Sturm Liouville Problem (SLP) SL equation A classical \"'Sturm-Liouville equation\"', is a real second-order linear differential equation of the form d dx p(x) dy dx +q(x)y= \u03bbr(x)y, (1) In the simplest of cases all coef\ufb01cients are continuous on the \ufb01nite closed interval [a,b], and p(x) has continuous derivative. 1 First-Order Systems and Applications 228 4. Let T:\u03a9\u2192Cn\u00d7n be a matrix-valued function that is analytic on some simply-connected domain \u03a9\u2282C. 3) with \u03bb = \u00b5, has a non-trivial solution is called a Sturm-Liouville Eigen Value Problem (SL-EVP). 8 - Endpoint Problems and Eigenvalues 3. A very important problem in matrix algebra is called the eigenvalue problem. , thermal conduction) and quantum mechanics (e. is an eigenvalue if it is a pole of , where denotes the identity operator. Naylor, Differential Equations of Applied Mathematics, John Wiley & Sons I. I'll do an example in a minute. SUMMARY An algorithm based on a compound matrix method is presented for solving difficult eigenvalue problems. In this lesson, our instructor Will Murray gives an introduction on complex eigenvalues for systems of equations. Vibrating Strings and Membranes. First, find a matrix that diagonalizes The eigenvalues of are and with corresponding eigenvectors and Diagonalize using the matrix whose columns consist of and to obtain and The system has the following form. 7 Modeling Problems Using First-Order Linear. for \u03bb>0, \u03bb = 0, \u03bb < 0. Sturm-Liouville problem, or eigenvalue problem, in mathematics, a certain class of partial differential equations (PDEs) subject to extra constraints, known as boundary values, on the solutions.", "date": "2020-02-26 17:36:38", "meta": {"domain": "grafikohnegrenzen.de", "url": "http://wjxu.grafikohnegrenzen.de/differential-equation-eigenvalue-problem.html", "openwebmath_score": 0.7965176105499268, "openwebmath_perplexity": 441.5730259174707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9956005869677995, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8702318116511315}}
{"url": "https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-63.html", "text": "### CSIR JUNE 2011 PART C QUESTION 63 SOLUTION (When $d(x,y) = |f(x) - f(y)|$ defines a metric)\n\nWhich of the following is/are metric on $\\Bbb R$?\n1)$d(x,y) = min\\{x,y\\}$,\n2)$d(x,y) = |x-y|$,\n3) $d(x,y) = |x^2 - y^2|$,\n4)$d(x,y) = |x^3 - y^3|$.\nSolution\n\noption (1): NO. $d(-1,0) = -1$ which is negative.\noption (2): YES. Given metric is the usual metric.\noption (3): NO. $d(-1,1) = 0$. In a metric space $d(x,y) =0$ if and only if $x=y$.\noption (4):YES.\u00a0 We will verify all the axioms.\naxiom 1: Clearly $d(x,y) \\ge 0$,\naxiom 2: Let $x,y \\in \\Bbb R$, then $d(x,y) = 0$ if and only if $|x^3-y^3|=0$ if and only if x^3 = y^3 if and only if $x=y$. [Since f(x)=x^3 is injective]\naxiom 3: $d(x,y) = d(y,x)$ follows.\naxiom 4: (Triangle inequality)\nLet $x,y,z \\in \\Bbb R$. $d(x,y) = |x^3-y^3| = |x^3-z^3+z^3-y^3| \\le |x^3-z^3| + |z^3-y^3|$ and this is equal to $d(x,z) + d(z,y)$. This proves that $d$ is a metric.\nBonus: Define a function $d$ on $\\Bbb R$ by $d(x,y) = |f(x) - f(y)|$where $f: \\Bbb R \\to \\Bbb R$ is a function. What is the necessary and sufficient condition on $f$ to make $d$ a metric. The answer is it is enough $f$ to be injective.\nProof: Proof of option (4) work for this general case!!!\nIn option (3), $f(x) =x^2$ which is not injective so not a metric.\nIn option (4), $f(x) = x^3$ is injective, so it is a metric.\nExercises:\nWhich of the following define metric on $\\Bbb R$?\n1) $d(x,y) = |sin\\,x - \\sin y|$.\u00a0 ($f(x) = sin\\,x$).\n2) $d(x,y) = |e^x - e^y|$. ($f(x) = e^x$).\n3) $d(x,y) = |log \\,x - log \\,y|$ on $\\Bbb R^+$. ($f(x) = log \\,x$).\n4) $d(x,y) = |\\sqrt x - \\sqrt y|$ on $\\Bbb R^+$. ($f(x) = \\sqrt x$).\n5) $d(x,y) = |tan\\,x - tan\\,y|$ on $(-\\frac{\\pi}{2},\\frac{\\pi}{2})$. ($f(x) = tan\\,x$).\n6) $d(x,y) = |\\frac{1}{x} - \\frac{1}{y}|$ on $\\Bbb R^+$. ($f(x) =\\frac{1}{x}$).\nYou can ask this question for any function and you know the answer!!!\nShare to your groups:\nFOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE TO FACEBOOK BY THE LINK BELOW. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, SUGGEST PROBLEMS TO SOLVE.\n\n### NBHM 2020 PART A Question 4 Solution $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$\nEvaluate : $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx = \\int_{-\\infty}^{\\inft...", "date": "2022-05-27 03:48:00", "meta": {"domain": "theindianmathematician.com", "url": "https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-63.html", "openwebmath_score": 0.7673551440238953, "openwebmath_perplexity": 760.8432127568535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98615138856342, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8702074673357224}}
{"url": "https://math.stackexchange.com/questions/2816791/how-many-ways-are-there-to-choose-strings-of-length-3-from-s-1-2-3-4-for", "text": "# How many ways are there to choose strings of length $3$ from $S=\\{1,2,3,4\\}$ for $3$ people if each each digit from $S$ has to appear at least once?\n\nLet $S=\\{1,2,3,4\\}$. There're three persons: Jack, Mary, Ann which choose $3$ digits from $S$ to compose a string of $3$ digits.\n\nFor example, a) Jack chose $112$, Mary chose $222$, Ann chose $123$; b) Jack chose $123$, Mary chose $122$, Ann chose $433$.\n\nIn how many ways can they form the strings? In how many ways can they form the strings if each digit from $S$ has to appear at least in one string?\n\nFor the second subquestion the example a) would not be a valid choice because $4$ doesn't appear in any string, while example b) is valid.\n\nThe first subquestion is easy for each string there're $4^3$ possibilities and there're $3$ different people so $3!\\cdot 4^3$.\n\nThe second subquestion is the harder one. I think this can be solved using inclusion/exclusion. We can count the ways where a given digit doesn't appear in any string, then two digits don't appear in any string, then 3 digits.\n\nThe case when one digit doesn't appear in any string: $${4\\choose 3}3^3\\cdot 3!$$ that is we choose only $3$ digits from $4$ available. The in each string there're $3^3$ options to build the string. Lastly there're $3$ strings in total. By this logic the final answer is: $${4\\choose 3}3^3\\cdot 3!-{4\\choose 2}2^3!\\cdot 3+{4\\choose 1}1^3\\cdot 3!$$\n\nThe problem is that when I permute the persons (the $3!$ in the calculation) it works if not two strings are the same. But if some strings are the same I will count too many if I use permutation.\n\n\u2022 For the first question, your answer seems to be based on the assumption that no two people can choose the same string. Each person can choose a string in $4^3$ ways, so the number of ways three people could independently choose a string is $(4^3)^3 = 4^9$. \u2013\u00a0N. F. Taussig Jun 12 '18 at 11:22\n\nLet $S = \\{1, 2, 3, 4\\}$. Jack, Mary, and Ann each choose three digits from $S$ to compose a string of $3$ digits. For example, Jack chose $112$, Mary chose $222$, Ann chose $123$. In how many ways can they form the strings?\n\nSince the question makes clear that repetition is permitted, each person can choose his or her string in $4^3$ ways. Hence, there are $(4^3)^3 = 4^9$ ways for Jack, Mary, and Ann to choose their strings.\n\nAnother way to see this is to observe that there are four choices for each of the nine entries, so the three strings can be selected in $4^9$ ways.\n\nIn how many ways can Jack, Mary, and Ann form the strings if each digit in $S$ has to appear in at least one string?\n\nSuppose Jack chooses the string $(j_1, j_2, j_3)$, Mary chooses the string $(m_1, m_2, m_3)$, and Ann chooses the string $(a_1, a_2, a_3)$. Notice that we could concatenate the three strings of length $3$ into a single string of length $9$: $(j_1, j_2, j_3, m_1, m_2, m_3, a_1, a_2, a_3)$.\n\nAs stated above, we have four choices in $S$ for each of the nine entries, so there are $4^9$ ways to choose the strings. If we exclude $k$ of the four elements in $S$, the nine entries in the concatenated string can each be filled in $(4 - k)^9$ ways. Hence, the number of strings from which $k$ elements of $S$ have been excluded is $$\\binom{4}{k}(4 - k)^9$$ Therefore, by the Inclusion-Exclusion Principle, the number of admissible strings is $$\\sum_{k = 0}^{4} (-1)^k\\binom{4}{k}(4 - k)^9 = \\binom{4}{0}4^9 - \\binom{4}{1}3^9 + \\binom{4}{2}2^9 - \\binom{4}{3}1^9 + \\binom{4}{4}0^9$$\n\nWhere did you make your error?\n\nYou seem to have made the assumption that no two of the three people could choose the same string. However, Jack, Mary, and Ann can choose their strings independently.\n\n\u2022 I really like your suggestion to think of the problem as concatenating the strings into one big string. \u2013\u00a0user123429842 Jun 12 '18 at 12:31", "date": "2020-09-24 15:55:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2816791/how-many-ways-are-there-to-choose-strings-of-length-3-from-s-1-2-3-4-for", "openwebmath_score": 0.7556622624397278, "openwebmath_perplexity": 127.62016030968104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986151391819461, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8702074580123847}}
{"url": "https://math.stackexchange.com/questions/1767559/proving-x-is-a-given-quotient-of-logarithms", "text": "# Proving $x$ is a given quotient of logarithms\n\nI'm practicing some questions on logarithms at the moment in order that I'm up to speed with the problem solving aspect before I embark on my PHD in chemical engineering at Boston college next year.\n\nI've been studying the laws of logarithms and what I am to do when it is necessary to add and subtract logs. So for example, for the first part of the question that is causing me trouble, I would utilize the first law of logarithms.\n\nMy issue comes with the \"Prove part\", and ascertaining a value for $x$ when it is already involved in the first part of the question.\n\nI know I'll most likely to be shut down for lack of evidence and research for this, but I would like to know what the most logical first step would be and which rule I need to follow. It doesn't follow on from the other questions I've been solving and so I can't problem solve it as easily as I can the rest.\n\nHere is my question:\n\nIf $4^x\\cdot 5^{3x+1}=10^{2x+1}$, prove that $x=\\dfrac{\\log(2)}{\\log(5)}$.\n\nThanks, not a problem if I get shut down, I know this isn't in the spirit of the website and would normally never ask a question in such a manner.\n\n\u2022 For the sake of typesetting, it is recommended to use MathJax and $\\LaTeX$, or at the very least use parenthesis so that there is no confusion as to what is meant to go where. Visit this page for information on how to typeset mathematics here. Is the equation meant to be $4^x 5^{3x+1} = 10^{2x+1}$? Is it meant to be $4^{x5^{3x}+1}=10^{2x}+1$ is it meant to be $4^x5^3x+1 = 10^2x+1$ or something else entirely? \u2013\u00a0JMoravitz May 2 '16 at 1:27\n\u2022 Seems like your question is perfectly in line with the spirit of the website. \u2013\u00a0M47145 May 2 '16 at 1:28\n\u2022 Thanks for the link, I must learnt the code as it'll prove invaluable to me. The equation is the first you mentioned, with everything being a power apart from the 4, 5 and 10. Thanks, M47145. I'm sat here with an inordinate amount of maths textbooks open trying to get my head around this one! \u2013\u00a0New Zealand's finest May 2 '16 at 1:31\n\n$$4^x\\cdot 5^{3x+1} = 10^{2x+1}$$\n\n$~~$Let us begin by moving everything to one side by dividing both sides by $10^{2x+1}$\n\n$$\\dfrac{4^x\\cdot 5^{3x+1}}{10^{2x+1}} = 1$$\n\n$~~$Let us separate the denominator using the fact that $10=2\\cdot 5$\n\n$$\\dfrac{4^x\\cdot 5^{3x+1}}{(2\\cdot 5)^{2x+1}} = 1$$\n\n$~~$Now, $(ab)^c = a^c\\cdot b^c$\n\n$$\\dfrac{4^x\\cdot 5^{3x+1}}{2^{2x+1}\\cdot 5^{2x+1}} = 1$$\n\n$~~$Let us factor out a factor of two from the denominator using $a^{b+c} = a^b\\cdot a^c$\n\n$$\\dfrac{4^x\\cdot 5^{3x+1}}{2^{2x}\\cdot 2\\cdot 5^{2x+1}}=1$$\n\n$~~$Now, using $a^{bc} = (a^b)^c$ simplify $2^{2x}$ in terms of a power of four\n\n$$\\dfrac{4^x\\cdot 5^{3x+1}}{4^x\\cdot 2\\cdot 5^{2x+1}}=1$$\n\n$~~$Cancel like terms and use $\\frac{a^b}{a^c} = a^{b-c}$\n\n$$\\frac{5^x}{2}=1$$\n\n$~~$Multiply both sides by two\n\n$$5^x=2$$\n\n$~~$Take the logarithm of each side\n\n$$\\log(5^x)=\\log(2)$$\n\n$~~$Use the property of logarithms that $\\log(a^b)=b\\log(a)$\n\n$$x\\log(5)=\\log(2)$$\n\n$~~$Divide each side by $\\log(5)$ to arrive at the desired result\n\n$$x=\\frac{\\log(2)}{\\log(5)}$$\n\n\u2022 It is not recommended to use quotes for highlighting. \u2013\u00a0Ian Miller May 2 '16 at 2:04\n\u2022 @IanMiller given the amount of text inbetween each step, I find it harder to quickly read without, but have edited it according to your suggestion. Hopefully this is a fine compromise. \u2013\u00a0JMoravitz May 2 '16 at 2:20\n\u2022 I replaced your single \\$equations with double \\$\\$equations to make it look more like your original. \u2013 Ian Miller May 2 '16 at 2:30 \u2022 JMoravitz, thank you for this excellent breakdown. It was invaluable to my learning, honestly. I really appreciate it. \u2013 New Zealand's finest May 5 '16 at 4:29 Alternative Solution: Recall the laws of logarithms: 1.$\\log a^b = b \\log a$. 2.$\\log ab = \\log a + \\log b$3.$\\log \\frac{a}{b} = \\log a - \\log bBy taking the logarithm on both sides of your equation and applying rule (1), you will get: $$x \\log 4 + (3x + 1) \\log 5 = (2x + 1) \\log (10)$$ which can be simplified to $$(\\log 4 + 3 \\log 5 \u2013 2 \\log 10) x = \\log(10) - \\log(5)$$ Using the mentioned rules, we get $$\\log\\left(\\frac{4 \\times 5^3}{10^2}\\right) x = \\log\\left(\\frac{10}{5}\\right)$$ which establish your result. \u2022 Thank you FALAM for helping me to understand this utlizing the laws of logarithms, I really appreciate it. \u2013 New Zealand's finest May 5 '16 at 4:29 \\begin{align} 4^x \\cdot 5^{3x+1} & = 2^{2x} \\cdot 5^{3x+1} \\\\ & = 2^{2x} \\cdot 5^{2x} \\cdot 5^{x+1} \\\\ & = 10^{2x} \\cdot 5^{x+1} \\end{align} If10^{2x} \\cdot 5^{x+1} = 10^{2x+1}$(the hypothesis of the problem), then $$5^{x+1} = 10$$ $$5^x \\cdot 5^1 = 10$$ $$5^x = 2$$ which gives us, by definition, $$x = \\log_5 2$$ which can be rewritten, using$\\log_b x = \\frac{\\log x}{\\log b}\\$, as\n\n$$x = \\frac{\\log 2}{\\log 5}$$\n\n\u2022 Thanks for this version Brian, it was very helpful to me. \u2013\u00a0New Zealand's finest May 5 '16 at 4:30", "date": "2019-06-18 12:59:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1767559/proving-x-is-a-given-quotient-of-logarithms", "openwebmath_score": 0.802817702293396, "openwebmath_perplexity": 466.5002549780074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513869353992, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8702074537025526}}
{"url": "https://math.stackexchange.com/questions/4254023/largest-positive-integer-n-for-which-n3100-is-divisible-by-n10/4254343#4254343", "text": "# Largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$\n\nProblem: What is that largest positive integer $$n$$ for which $$n^3+100$$ is divisible by $$n+10$$?\n\nThe solution from Art of Problem Solving: If $$n+10 \\mid n^3+100$$, $$\\gcd(n^3+100,n+10)=n+10$$. Using the Euclidean algorithm, we have $$\\gcd(n^3+100,n+10)= \\gcd(-10n^2+100,n+10)$$ $$= \\gcd(100n+100,n+10)$$ $$= \\gcd(-900,n+10)$$, so $$n+10$$ must divide $$900$$. The greatest integer $$n$$ for which $$n+10$$ divides $$900$$ is $$\\boxed{890}$$; we can double-check manually and we find that indeed $$900\\mid 890^3+100$$.\n\nQuestion: How can $$\\gcd(n^3+100,n+10)= \\gcd(-10n^2+100,n+10)$$ $$= \\gcd(100n+100,n+10)$$?\n\n\u2022 Are you aware $\\gcd(A,B) = \\gcd(A \\pm kB, B)$ for any integer $k$? $n^3+100 = n^3 + 10n^2 - 10n^2 + 100= n^2(n+10) -10n^2 + 100$ so $\\gcd(n^3 + 100, n+10) = \\gcd(n^2(n+10)-10n^2 + 100, n+10) = \\gcd(-10n^2 + 100, n+100)$. Sep 19 at 4:39\n\nYou know that $$n$$ 'acts as' $$-10$$, since\n\n$$n \\equiv -10 \\quad \\text { mod } (n + 10)$$ like so:\n\n$$n^3 + 100 = (n + 10 - 10)n^2 + 100 = \\underbrace{(n+10)n^2}_\\text{multiple of (n+10)} - 10n^2 + 100$$ You could 'cheat' and plug in $$-10$$ immediately:\n\n$$\\gcd(n^3 + 100, n+ 10) = \\gcd(-900, n+ 10)$$\n\n\u2022 gcd(100n+100,n+10) here 100n+100 how I derived? Sep 19 at 1:04\n\u2022 @ABIDB11152 like so: $$-10n^2 + 100 = -10(n+10-10)n + 100 = -10n(n+10) + 100n + 100$$ Sep 19 at 9:27\n\nThe Euklidean Algorithm uses the fact that $$\\gcd(a,b)=\\gcd(a+kb,b)$$ We choose $$k=-n^2$$ so that the $$n^3$$ term will be removed. We get $$\\gcd(n^3+100,n+10)= \\gcd((n^3+100)-n^2(n+10),n+10)$$ So we have $$=\\gcd(-10n^2+100,n+10)$$ Now we remove $$n^2$$ by choosing $$k=10n$$ $$=\\gcd((-10n^2+100)+10n(n+10),n+10)$$ $$=\\gcd(100n+100,n+10)$$ and now for $$k=-100$$ we get $$=\\gcd((100n+100)-100(n+10),n+10)$$ $$=\\gcd(-900,n+10)$$\n\nYou can perform long division to obtain the following result, $$n^3+100=(n^2-10n+100)(n+10)-900$$ $$\\implies n+10\\mid 900$$ Therefore the largest value of $$n=890$$.\n\n$$\\gcd(A,B) = \\gcd(A \\pm kB, B)$$ for any integer $$k$$.\n\nFor example $$\\gcd(58, 16) = \\gcd(3\\times 16 + 10, 16) = \\gcd(3\\times 16+ 10-3\\times 16, 16) = \\gcd(10, 16)$$.\n\n......\n\nSo $$\\gcd(n^3 + 100, n+10) = \\gcd(n^2(n+10) - 10n^2 + 100, n+10)=$$\n\n$$\\gcd(n^2(n+10) - 10n^2 + 100- n^2(n+10), n+10)=\\gcd(-10n^2 + 100, n+10)=$$\n\n$$\\gcd(-10n(n + 10) +100n + 100, n+10) = \\gcd(-10n(n+10) + 10n + 100 + 10n(n+10), n+10)=$$\n\n$$\\gcd(10n + 100, n+10)$$", "date": "2021-12-08 13:18:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4254023/largest-positive-integer-n-for-which-n3100-is-divisible-by-n10/4254343#4254343", "openwebmath_score": 0.8674624562263489, "openwebmath_perplexity": 257.52636264340657, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303413461357, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8702061449773092}}
{"url": "http://www.askamathematician.com/2011/05/q-is-0-9999-repeating-really-equal-to-1/", "text": "# Q: Is 0.9999\u2026 repeating really equal to 1?\n\nMathematician: Usually, in math, there are lots of ways of writing the same thing. For instance:\n\n$\\frac{1}{4}$ = $0.25$ = $\\frac{1}{\\frac{1}{1/4}}$ = $\\frac{73}{292}$ = $(\\int_{0}^{\\infty} \\frac{\\sin(x)}{\\pi x} dx)^2$\n\nAs it so happens, 0.9999\u2026 repeating is just another way of writing one. A slick way to see this is to use:\n\n$0.9999... = (9*0.9999...) / 9 = ((10-1) 0.9999...) / 9$\n\n$= (10*0.9999... - 0.9999...) / 9$\n\n$= (9.9999... - 0.9999...) / 9$\n\n$= (9 + 0.9999... - 0.9999...) / 9 = 9 / 9 = 1$\n\nOne.\n\nAnother approach, that makes it a bit clearer what is going on, is to consider limits. Let\u2019s define:\n\n$p_{1} = 0.9$\n$p_{2} = 0.99$\n$p_{3} = 0.999$\n$p_{4} = 0.9999$\n\nand so on.\n\nNow, our number $0.9999...$ is bigger than $p_{n}$ for every n, since our number has an infinite number of 9\u2019s, whereas $p_{n}$ always has a finite number, so we can write:\n\n$p_{n} < 0.9999... \\le 1$ for all n.\n\nTaking 1 and subtracting all parts of the equation from it gives:\n\n$1-p_{n} > 1-0.9999... \\ge 0$\n\nThen, we observe that:\n\n$1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \\frac{1}{10^n}$\nand hence\n\n$\\frac{1}{10^n} > 1-0.9999... \\ge 0$.\n\nBut we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999\u2026 must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999\u2026 is clearly not bigger than 1. Hence, the only possibility is that\n\n$1-0.9999... = 0$\n\nand therefore that\n\n$0.9999... = 1$.\n\nWhat we see here is that 0.9999\u2026 is closer to 1 than any real number (since we showed that 1-0.9999\u2026 must be smaller than every positive number). This is intuitive given the infinitely repeating 9\u2019s. But since there aren\u2019t any numbers \u201cbetween\u201d 1 and all the real numbers less than 1, that means that 0.9999\u2026 can\u2019t help but being exactly 1.\n\nUpdate: As one commenter pointed out, I am assuming in this article certain things about 0.9999\u2026. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don\u2019t believe this about 0.9999\u2026 or would like to see a discussion of these issues, you can check this out.\n\nThis entry was posted in -- By the Mathematician, Equations, Math. Bookmark the permalink.\n\n### 58 Responses to Q: Is 0.9999\u2026 repeating really equal to 1?\n\n1. mathman says:\n\nlol. You need to know what each \u201cnumber\u201d is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting \u201cnumbers\u201d\n\n2. George says:\n\nWhile the mathematical argument presented here is correct, I have to say that limit and/or mathematical infinity introduced created confusions for many. What I wanted to say is that math infinity is really a math trick. It cannot and should not be real in the real physical world. Today this infinity plagues physics and its calculations. Maybe we need something new\u2026\n\n3. EMINEM says:\n\ngo to this https://www.youtube.com/watch?v=wsOXvQn3JuE site and get a mathematical proof that 0.999\u2026 doesnot equal to 1\u2026.\n\n4. . says:\n\nEMINEM, the video was published on April fool\u2019s day for a reason!\n\n5. Josh says:\n\nThe simplest explanation I\u2019ve seen for this is:\n\n1/3 = 0.333\u2026\n(1/3)3=1 = (0.333\u2026)3=0.999\u2026\n1 = 0.999\u2026\n\nI don\u2019t see how anyone can come to any conclusion other than 1=0.999\u2026 after understanding this.\n\n6. mathman says:\n\n1 / 3 = ??\nUse long division. There are 2 parts: the quotient and the remainder. For every iteration.\nFirst iteration is 0 R 1/3\nSecond is 0.3 R 1/30\nThird is 0.33 R 1/300\nEtc etc.\n\nWhere does your remainder go, Josh?\nYou don\u2019t account for it anywhere.\n\nThe remainder is never zero.\n\nHow many 9\u2019s are there in your 0.999\u2026 ?\n\n7. Josh says:\n\nHey mathman,\n\nThere are infinitely many 9s in my \u201c0.999\u2026\u201d. If it were anything other than infinitely many, then it would be less than 1.\n\nAnd as far as 1/3, your \u201cetc. etc.\u201d is an infinitely repeating \u201c0.333\u2026\u201d. I\u2019m failing to see how anything you\u2019ve presented has disproved what I said.\n\n8. Angel says:\n\n@mathman:\n\nThe remainder does not need to be accounted for. The division 1/3 is never formally complete if there is a remainder. Therefore, one must keep iterating the operations in long division. As the number of iterations n approaches infinity, the remainder does approach zero. That is what matters here. The number of 3s in the decimal expansion of 1/3 is infinite, and so is the number of 9s in the figure 0.999\u2026", "date": "2017-01-18 18:11:49", "meta": {"domain": "askamathematician.com", "url": "http://www.askamathematician.com/2011/05/q-is-0-9999-repeating-really-equal-to-1/", "openwebmath_score": 0.8159306049346924, "openwebmath_perplexity": 579.3539525925751, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.993611678144426, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8701921728044477}}
{"url": "http://krizak.com/7ep86ie/g58j054.php?cd1127=vector-derivative-rules", "text": "Answer: As with the dot product, this will follow from the usual product rule in single variable calculus. 4 cos(4x + 2) And that is the derivative of your original function. ax, axp ax, In finding the derivative of the cross product of two vectors $\\frac{d}{dt}[\\vec{u(t)}\\times \\vec{v(t)}]$, is it possible to find the cross-product of the two vectors first before differentiating? The derivative of V, with respect to T, and when we compute this it's nothing more than taking the derivatives of each component. So in this case, the derivative of X, so you'd write DX/DT, and the derivative of Y, DY/DT. If r 1(t) and r 2(t) are two parametric curves show the product rule for derivatives holds for the cross product. Furthermore, suppose that the elements of A and B arefunctions of the elements xp of a vector x. calculus multivariable-calculus vector-analysis In this section we\u2019re going to prove many of the various derivative facts, formulas and/or properties that we encountered in the early part of the Derivatives chapter. to do matrix math, summations, and derivatives all at the same time. Thus, the derivative of a matrix is the matrix of the derivatives. Here are useful rules to help you work out the derivatives of many functions (with examples below). We will not prove all parts of the following theorem, but the reader is encouraged to attempt the proofs. Then, ac a~ bB -- - -B+A--. This is the vector value derivative. They will come in handy when you want to simplify an expression before di erentiating. All bold capitals are matrices, bold lowercase are vectors. Rules of Differentiation The derivative of a vector is also a vector and the usual rules of differentiation apply, dt d dt d t dt d dt d dt d dt d v v v u v u v ( ) (1.6.7) Also, it is straight forward to show that { Problem 2} a a v a v a v a v v a v dt d dt d dt d dt d dt d dt d (1.6.8) There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Derivative Rules for Vector-Valued Functions. We will now look at a bunch of rules for differentiating vector-valued function, all of which are analogous to that of differentiating real-valued functions. It\u2019s just that there is also a physical interpretation that must go along with it. Example. Product rule for vector derivatives 1. Contraction. Type in any function derivative to get the solution, steps and graph The standard rules of Calculus apply for vector derivatives. Section 7-2 : Proof of Various Derivative Properties. Matrix derivatives cheat sheet Kirsty McNaught October 2017 1 Matrix/vector manipulation You should be comfortable with these rules. Product rule for vector derivatives 1. Free derivative calculator - differentiate functions with all the steps. The Derivative tells us the slope of a function at any point.. And now you might start to \u2026 We want to show d(r 1 \u00d7 r \u2026 Theorem D.1 (Product dzferentiation rule for matrices) Let A and B be an K x M an M x L matrix, respectively, and let C be the product matrix A B. Suppose we have a column vector ~y of length C that is calculated by forming the product of a matrix W that is C rows by D columns with a column vector ~x of length D: ~y = W~x: (1) Suppose we are interested in the derivative of ~y with respect to ~x. 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Are matrices, bold lowercase are vectors of many functions ( with examples below ) encouraged to attempt proofs... This case, the derivative of x, so you 'd write DX/DT, and the derivative of,...", "date": "2021-08-02 08:57:37", "meta": {"domain": "krizak.com", "url": "http://krizak.com/7ep86ie/g58j054.php?cd1127=vector-derivative-rules", "openwebmath_score": 0.8595079779624939, "openwebmath_perplexity": 579.1642646842395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.984336352207334, "lm_q2_score": 0.8840392817460332, "lm_q1q2_score": 0.8701920018018818}}
{"url": "http://mathhelpforum.com/calculus/15750-evaluate-line-integral.html", "text": "# Thread: Evaluate the line integral\n\n1. ## Evaluate the line integral\n\nHello,\nplease try to solve this question.\nThanks\n\n2. $\\oint_C x^3 dx - 2xy dy = \\iint_D \\left( \\frac{\\partial (-2xy)}{\\partial x} - \\frac{\\partial x^3}{\\partial y} \\right) \\ dA = -2\\iint_D y \\ dA =$ $- 2\\int_0^2 \\int_0^{2-x} y \\ dy \\ dx$\n\n3. Evaluate the line integral $\\int_C x^3 dx - 2xy dy$ where C comprises the three sides of a triangle joining O (0, 0) A (2, 0) and B (0, 2)\nFrom O to A we have that y does not change so dy = 0.\n$\\int_{O to A} x^3 dx - 2xy dy = \\int_0^2 x^3 dx$\n\n$= \\frac{1}{4}x^4|_0^2 = \\frac{1}{4}2^4 = 4$\n\nFrom A to B the path is the line from (2, 0) to (0, 2). This is a segment of the line y = -x + 2 as x goes from 2 to 0, so\n$dy = -dx$\n\nThus\n$\\int_{(2,0) to (0,2)} x^3 dx - 2xy dy$\n\n$= \\int_2^0 x^3 dx + 2x(-x + 2)(-dx)$\n\n$= \\int_2^0 (x^3 + 2x^2 - 4x)dx$\n\n$= \\left ( \\frac{1}{4}x^4 + \\frac{2}{3}x^3 - 2x^2 \\right ) _2^0$\n\n$= - \\left ( \\frac{1}{4}2^4 + \\frac{2}{3}2^3 - 2 \\cdot 2^2 \\right ) = - \\left ( 4 + \\frac{16}{3} - 8 \\right )$\n\n$= 4 - \\frac{16}{3}$\n\nSo the overall integral will be:\n$4 + 4 - \\frac{16}{3} = 8 - \\frac{16}{3} = \\frac{8}{3}$\n\n-Dan\n\n4. Originally Posted by topsquark\nFrom O to A we have that y does not change so dy = 0.\n$\\int_{O to A} x^3 dx - 2xy dy = \\int_0^2 x^3 dx$\n\n$= \\frac{1}{4}x^4|_0^2 = \\frac{1}{4}2^4 = 4$\n\nFrom A to B the path is the line from (2, 0) to (0, 2). This is a segment of the line y = -x + 2 as x goes from 2 to 0, so\n$dy = -dx$\n\nThus\n$\\int_{(2,0) to (0,2)} x^3 dx - 2xy dy$\n\n$= \\int_2^0 x^3 dx + 2x(-x + 2)(-dx)$\n\n$= \\int_2^0 (x^3 + 2x^2 - 4x)dx$\n\n$= \\left ( \\frac{1}{4}x^4 + \\frac{2}{3}x^3 - 2x^2 \\right ) _2^0$\n\n$= - \\left ( \\frac{1}{4}2^4 + \\frac{2}{3}2^3 - 2 \\cdot 2^2 \\right ) = - \\left ( 4 + \\frac{16}{3} - 8 \\right )$\n\n$= 4 - \\frac{16}{3}$\n\nSo the overall integral will be:\n$4 + 4 - \\frac{16}{3} = 8 - \\frac{16}{3} = \\frac{8}{3}$\n\n-Dan\nMy answer is exactly negative of your answer if you do it out by Green's theorem! Maybe you did it clockwise.\n\n5. Originally Posted by ThePerfectHacker\nMy answer is exactly negative of your answer if you do it out by Green's theorem! Maybe you did it clockwise.\nI suppose I should have suspected this, but I didn't realize it was an integral over a closed area. I only did the integral from O to A to B. I never connected B to O.\n\nSo from B (0, 2) to O (0, 0) dx = 0, with x = 0. Thus\n$\\int_{(0, 2) to (0, 0)} x^3 dx - 2xy dy = \\int_2^0 (-2 \\cdot 0 \\cdot y) dy = \\int_2^0 0 dy = 0$\n\nSo my final integral is STILL $\\frac{8}{3}$.\n\nAnd if you look at the succession of points I did the integral in the counterclockwise sense. I can't argue with your result (though I was never very good at implementing Green's theorem anyway) but I can't find a mistake in my own work?\n\n-Dan", "date": "2013-12-19 17:37:46", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/15750-evaluate-line-integral.html", "openwebmath_score": 0.8094480633735657, "openwebmath_perplexity": 309.89270187451655, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987946220128863, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8701801547084712}}
{"url": "https://math.stackexchange.com/questions/1837763/integration-with-limits-and-options", "text": "# Integration with limits and options.\n\nI found this exercise in an old exam but I don't know how to attack it because is a limit of an integration and I don't know if the limit affects the process of the integral or it makes it easier. The exercise is this:\n\nThe value of the limit $$\\lim_{h\\to 0}\\frac{1}{h}\\int^{2+h}_2\\sqrt{1+t^3}dt$$ Is:\n\na) $0$\n\nb) $1$\n\nc) $\\sqrt{2}$\n\nd) $2$\n\ne) $3$\n\nI want to know two things:\n\n1) If there is a method of how to solve limit-integral problems.\n\n2) An explained solution of this example to at least try to understand how to solve exercises of the same style.\n\nAs $h \\rightarrow 0$, we have $\\displaystyle \\int_2^{2+h} \\sqrt{1+t^3} \\ dt \\rightarrow 0$. Thus, we arrive at a $0/0$ indeterminant form when trying to evaluate the limit, so we can apply L'Hopital's rule.\n\nThe hardest part about doing this is evaluating $\\displaystyle \\frac{d}{dh}\\int_2^{2+h} \\sqrt{1 + t^3} \\ dt$. For a continuous function $f,$ the fundamental theorem tells us that $\\displaystyle \\frac{d}{dx} \\int_a^x f(t) \\ dt = f(x)$. To get our integral into this form, we can simply shift the function to the left $2$ units and adjust the bounds of integration accordingly: $$\\displaystyle \\frac{d}{dh}\\int_2^{2+h} \\sqrt{1 + t^3} \\ dt = \\frac{d}{dh} \\int_0^h \\sqrt{1 + (t+2)^3} \\ dt = \\sqrt{1+(h+2)^3}$$\n\n\u2022 Yes, my answer as is in the exam was 3 because of that, but I don't know how affects L'Hopital to the integral. It lets the fuction inside the integrate just like it's now? (I mean $\\sqrt{1+t^3}$) \u2013\u00a0MonsieurGalois Jun 24 '16 at 4:26\n\u2022 @MonsieurGalois, added some detail \u2013\u00a0Kaj Hansen Jun 24 '16 at 4:32\n\nUse L'Hopital's rule as follows: $$\\lim_{h \\to 0} \\frac{\\int_{2}^{2 + h} \\sqrt{1 + t^3}dt}{h}$$ $$= \\lim_{h \\to 0} \\frac{\\sqrt{1 + (2 + h)^3}}{1}$$ $$= \\boxed{3}.$$\n\nAnd we are finished. Hope this helped!\n\nYou should recognize that for well behaved $f(x)$ and fixed $x$: $$\\frac{1}{h}\\int^{x+h}_x f(t)dt=\\frac{1}{h}\\left[f(x)h+R_h(x) \\right]$$ where $\\lim_{h\\to 0}R_h(x)/h=0$. So: $$\\lim_{h\\to 0}\\frac{1}{h}\\int^{x+h}_x f(t)dt=f(x)$$\n\nAn alternate way of showing this is to observe that for a well behaved function on an interval containing $[x,x+h]$: $$h \\times \\min_{t\\in[x,x+h]}f(t)\\le\\int^{x+h}_x f(t)dt\\le h \\times \\max_{t\\in[x,x+h]}f(t)$$ and when $h\\to 0$ we have $\\min_{t\\in[x,x+h]}f(t)\\to \\max_{t\\in[x,x+h]}f(t)\\to f(x)$.\n\nHence as $f(x)=\\sqrt{1+t^3}$ is well behaved around $t=2$: $$\\lim_{h\\to 0}\\frac{1}{h}\\int^{2+h}_2\\sqrt{1+t^3}dt=\\left.\\sqrt{1+t^3}\\right|_{t=2}=3$$\n\n\u2022 Why $R_h(x)$ tends to 0? \u2013\u00a0MonsieurGalois Jun 24 '16 at 4:47\n\u2022 There are a number of ways of showing this, as I have specified that $f(x)$ be well behaved we may take that to mean that $f(x)$ is differentiable on an interval containing $[x,x+h]$ when the given property follows from the Lagrange form of the remainder in the Taylor series up to the zeroth order term. \u2013\u00a0Conrad Turner Jun 24 '16 at 5:57\n\nBy the Mean Value Theorem for Definite Integrals, https://en.wikipedia.org/wiki/Mean_value_theorem#First_Mean_Value_Theorem_for_Definite_Integrals,\n\n$\\displaystyle\\int_2^{2+h}\\sqrt{1+t^3}dt=\\left(\\sqrt{1+c^3}\\right)h$ for some c between 2 and $2+h$,\n\nso $\\displaystyle\\lim_{h\\to 0}\\frac{1}{h}\\int^{2+h}_2\\sqrt{1+t^3}dt=\\lim_{c\\to 2}\\sqrt{1+c^3}=\\sqrt{9}=3$.\n\n$\\textbf{Alternate solution:}$\n\nLet $\\displaystyle G(x)=\\int_2^x \\sqrt{1+t^3}dt,\\;$ so $G^{\\prime}(x)=\\sqrt{1+x^3}$ by the Fundamental Theorem of Calculus.\n\nThen $\\displaystyle\\lim_{h\\to 0}\\frac{1}{h}\\int^{2+h}_2\\sqrt{1+t^3}dt=\\lim_{h\\to 0}\\frac{G(2+h)-G(2)}{h}=G^{\\prime}(2)=\\sqrt{1+2^3}=3$\n\n\u2022 How you get the $\\sqrt{1+c^3}h$? \u2013\u00a0MonsieurGalois Jun 24 '16 at 4:45\n\u2022 I have added some explanation to my answer. \u2013\u00a0user84413 Jun 24 '16 at 20:29\n\nYou need to use Fundamental Theorem of Calculus. Let $$F(x) = \\int_{2}^{x}\\sqrt{1 + t^{3}}\\,dt = \\int_{2}^{x}f(t)\\,dt$$ where $f(x) = \\sqrt{1 + x^{3}}$ and we need to calculate the limit $$\\lim_{h \\to 0}\\frac{F(2 + h)}{h}$$ Note that $F(2) = 0$ and hence $$\\lim_{h \\to 0}\\frac{F(2 + h)}{h} = \\lim_{h \\to 0}\\frac{F(2 + h) - F(2)}{h} = F'(2)$$ provided the limit exists.\n\nBy Fundamental Theorem of Calculus we know that $F'(x)$ exists if the function $f(x) = \\sqrt{1 + x^{3}}$ is continuous at $x$ and then $F'(x) = f(x)$. Thus $F'(2) = f(2) = 3$.\n\nThe use of L'Hospital's Rule is unnecessary and is a roundabout way to evaluate this limit.", "date": "2019-08-25 12:23:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1837763/integration-with-limits-and-options", "openwebmath_score": 0.9565900564193726, "openwebmath_perplexity": 157.46732358239584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462215484651, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8701801482305135}}
{"url": "http://semesta.lviv.ua/0zsekb/adjacent-angles-which-are-not-in-a-linear-pair-58eba1", "text": "But they are not a linear pair because they are not connected, and they do not share a common side. The line through points A, B and C is a straight line. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. Linear Pair: Definition, Theorem & Example ... we can say that angle NSI and angle ISD also form a pair of adjacent angles. In the adjoining figure, \u2220AOC and \u2220BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line. True, if they are adjacent and share a vertex and one side. Definition: Two angles that add up to 180\u00b0, Adjacent Angle \u2013 Definition, Examples & More. The vertex of an angle is the endpoint of the rays that form the sides of the angle. Example: Find the pairs of adjacent angles in the following figure. In Other Words, It Is Between A Right Angle And A Straight Angle. Start studying M54.2c. E. Both angles \u2026 In this article, we are going to discuss the definition of adjacent angles and vertical angles in detail. (ii) Angles in a linear pair which are (iii) Complementary angles that do not form a not supplementary. When two lines intersect, the angles in a linear pair have sides that are opposite rays; but the vertical angles formed by the two intersecting lines also have sides that are opposite rays. Solution: \u2220 1 and \u2220 2 are adjacent to each other\u2220 2 and \u2220 3 are adjacent to each other, Solution: \u2220 APD and \u2220 DPC are adjacent to each other\u2220 DPC and \u2220 CPB are adjacent to each other. If Two Angles Form A Linear Pair, The Angles Are Supplementary. In the figure above, the two\u00a0angles\u00a0\u2220BAC and\u00a0\u2220CAD share a common side (the blue line segment AC). In the diagram below, \u2220ABC and \u2220DBE are supplementary since 30\u00b0+150\u00b0=180\u00b0, but they do not form a linear pair since they are not adjacent. Common side. Two Obtuse Angles Can Be Adjacent Angles. v. Angles which are neither complementary nor adjacent. At times, in geometry, the pair of angles are used. The two angles will change so that they always add to 180\u00b0, In the figure above, the two angles\u00a0\u2220PQR and\u00a0\u2220JKL are supplementary because they always add to 180\u00b0. If the two supplementary are adjacent to each other then they are called linear pair. A. D. The angles are congruent and are right angles. Consider the following figure in which a ray $$\\overrightarrow{OP}$$ stand on the line segment $$\\overline{AB}$$ as shown: The angles \u2220POB and \u2220POA are formed at O. Draw a linear pair of angles. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary means add \u2026 Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. Non-common side makes a straight line or Sum of angles is 180\u00b0. The endpoints of the ray from the side of an angle are called the vertex of an angle. The angles in a linear pair are supplementary. Yes. O. Additionally, the ray OB is the common arm between these two angles. Adjacent Angles Are Two Angles That Share A Common Vertex, A Common Side, And No Common Interior Points. Vertically Opposite Angles: When two lines intersect, then the angles that are opposite one another at the intersection are called Vertically Opposite Angles. Two angles that overlap, one inside the other sharing a side and vertex in the figure on the right, the two angles \u2220PSQ and \u2220PSR overlap. Both sets (top and bottom) are supplementary but only the top ones are linear pairs because these ones are also adjacent. Example : Logical Equivalence Lateral Area . Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pair. iv. See the first picture below. Related Questions to study. Another way of defining them is: \u201ctwo angles that share a side and a vertex, but do not share any interior points\u201d. Smenevacuundacy and 26 more users found this answer helpful Adjacent angles can be a complementary angle or supplementary angle\u00a0when they share the common vertex and side. Example: If following angles make a linear pair, find the value of q. The measure of rotation of a ray, when it is rotated about its endpoint is known as the angle formed by the ray between its initial and final position. Two Adjacent Angle Can Be Complementary Too If They Add Up To 90\u00b0. In this picture there is a flatscreen Toshiba television. They share a common side, but not a common vertex. On the other hand, two right angles will always make a linear pair as their sum is equal to 180\u00b0. Two angles form a linear pair. Before you know all these pairs of angles there is another important concept which is called \u2018angles on a straight line\u2019. A Linear Pair Forms A Straight Angle Which Contains 180\u00ba, So You Have 2 Angles Whose Measures Add To 180, Which Means They Are Supplementary. Linear pair forms two supplementary angles. This television has a pair of supplementary angles that are not a linear pair (one green, one blue). Two adjacent angles forming a linear pair are in the ratio 7 : 5 find the angles. Solution: 130\u00b0 + 50\u00b0 = 180\u00b0Since the sum of these angles is equal to two right angles, so they can make a linear pair. This statement is correct. It can also be said that angles of the linear pair are always supplementary to each other. A: Angles that are next to each other are known as adjacent angles, i.e., two angles with one common arm. Two angles are said to be supplementary if the sum of both the angles is 180 degrees. Adjacent angles which are not in linear pair. Adjacent angles, Linear pair, Vertically opposite angles. In A Right Triangle, The Altitude From A Right-angled Vertex Will Split The Right Angle Into Two Adjacent Angle; 30\u00b0+60\u00b0, 40\u00b0+50\u00b0, Etc. The measure of a straight angle is 180 degrees, so a linear pair of angles must add up to 180 degrees. The two angles are said to be adjacent angles when they share the common vertex and side. Draw the pairs of angles as described below. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. | Definition & Examples - Tutors.com In the figure, \u22201 and \u22203 are non-adjacent angles. If that is not possible, say why. linear pair. Let us take example of the angles, shown in following figure: \u2220 COB and \u2220 BOA have a common vertex, i.e. \u2026 A linear pair of angles has two defining characteristics: 1) the angles must be supplmentary; 2) The angles must be adjacent ; In the picture below, you can see two sets of angles. Although they share a common side (PS) and a common vertex (S), they are not considered adjacent angles when they overlap like this. 1 answer. Linear Pair of Angles. Two Adjacent Angle Can Be Supplementary Too, If They Add Upto 180\u00b0. (i v) When two lines intersect opposite angles are supplementary. Adjacent Angles Are Angles That Come Out Of The Same Vertex. Adjacent angles are angles that are side by side. Supplementary angles a and b do not form linear pair. Vertical Angles Are Always Congruent, Which Means That They Are Equal. Solution: 110\u00b0 + 70\u00b0 = 180\u00b0Since the sum of these angles is equal to two right angles, so they can make a linear pair. Can Two Obtuse Angles Be Adjacent? If two angles form a linear pair, the angles are supplementary. Note: \u2220 APD and \u2220 CPB are not adjacent to each other, because they don\u2019t have a common arm in spite of having a common vertex. A pair of adjacent angles formed by intersecting lines is called a Linear Pair of Angles. Obviously, the larger angle \u2220BAD is the sum of the two adjacent angles. If two angles are not adjacent, then they do not form a linear pair. They are supplementary because each angle is 90 degrees so they add up to 180 degrees. A linear pair of angles is formed when two lines intersect. To identify whether the angles are adjacent or not, we must remember its basic properties that are \u2026 How to Find Adjacent Angles. Draw a pair of vertically opposite angles. They also share a common\u00a0vertex (point A). , \u22201 and \u22203 are non-adjacent angles that angles of the linear which. 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By side, but all linear pairs because these ones are linear pairs because ones... Right angles are angle pairs that add up to 180 degrees, a..., two of the other is acute vertex ( point a ) ; ;. & more Upto 180\u00b0 in Triangles and other study tools intersecting lines and do!", "date": "2021-05-06 13:42:27", "meta": {"domain": "lviv.ua", "url": "http://semesta.lviv.ua/0zsekb/adjacent-angles-which-are-not-in-a-linear-pair-58eba1", "openwebmath_score": 0.5223283767700195, "openwebmath_perplexity": 686.3562189789515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9879462197739625, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8701801451218689}}
{"url": "https://mathematica.stackexchange.com/questions/230459/exact-division-and-geometric-sequences", "text": "# Exact division and geometric sequences\n\nI imagine this problem has a name that I don't know; it's probably some sort of exact division problem. Here goes: imagine you have to divide 4 cookies among 11 people. Divide the cookies equally among the 11 people with the constraint that you are a World Champion at halving cookies. :-)\n\nSo naturally, you just halve all the cookies until there are more pieces than people, then repeat for each new size. With the (4, 11) case above, there are 4 pieces to start. Halve them all until there are 16 pieces and everyone eats one piece. There are five pieces left over, halve them all until there are 20, everyone eats one...you get the idea.\n\nIt's lovely that each person will get 1/4 + 1/16 + 1/32 + 1/64 + 1/256 + 1/4096 + 1/16384 + ... = 4/11 of the total.\n\nBut the pretty part -- for me! -- is that the sequence above is not geometric; it's actually five different geometric sequences. (That's how I see it, anyway.) It's 1/4 + 1/4096 + ... and 1/16 + 1/16384 + ... and so on; each of the five sequences has a common ratio of 1/1024, so it's straightforward to show the sum is exactly 4/11.\n\nHere's the Mathematica part. With pen & paper, it took a little playing around to realize that the denominators of the first five terms above are 4, 16, 32, 64, 256, and then the structure of the problem repeats. At that point, the doubling process gives the same number of \"remaining\" pieces, so those five denominators are the foundation of the sum. Just to check, I computed\n\nTotal[\nSum[1/2^# (1/2^10)^n, {n, 0, Infinity}] & /@ {2, 4, 5, 6, 8}\n]\n\n\nThe result is, in fact, 4/11. Sweet. Similarly, I tried two other cases: 3 cookies and 5 people as well as 5 cookies and 9 people. The pattern so far is this, where the formatting is [cookies, people] --> listOfPortions.\n\n[3, 5] --> {1/2, 1/16, 1/32, 1/256, ...}\n\n[5, 9] --> {1/2, 1/32, 1/64, 1/128, 1/2048, ...}\n\nEach case \"works,\" and each person would end up with c/p as the total amount. But despite having three examples, I don't see an explicit pattern, and I think there are few ways to describe the pattern. I could describe it in terms of the actual portions, or I could describe it with the exponents on each denominator. So at this point, I have three questions:\n\n\u2022 Does this problem have a name?!\n\n\u2022 Do you have hints or suggestions for a function like portions[c, p, n] that gives the first n terms of the sequence based on c cookies and p people?\n\n\u2022 How would you present this problem to a group of students? What are your thoughts? What other functions or computations would you show them?\n\nThe logic is straightforward: double the current number of pieces until it exceeds the number of people, subtract the number of people from that doubled number, and repeat. But I'm not sure how to translate that into the terms of a sequence that will sum to c/p. This feels like a NestList[] or NestWhileList[] situation, but I don't have it yet.\n\n\u2022 Tell them it's worms instead of cookies. Nobody will want them, and you won't spend now to infinity cutting things in half. \u2013\u00a0Daniel Lichtblau Sep 21 at 14:26\n\u2022 @DanielLichtblau I figure I'm the one holding the knife...I can keep them as cookies and have them all myself. :-) \u2013\u00a0mathalicious Sep 22 at 0:46\n\u2022 Holding the knife...yes, good point. \u2013\u00a0Daniel Lichtblau Sep 22 at 13:19\n\nThe following function gives the full solution to the problem:\n\ncookieHalves[c_Integer, p_Integer] /;\nCoprimeQ[2, p](* Euler's theorem only holds for 2 and p comprime *):=\n\nModule[\n{\nratio = 2^EulerPhi[p]\n},\n{1/ratio, DeleteCases[0]@NumberExpand[(ratio - 1) c/p, 2]/ratio}\n]\n\n(* {1/1024, {1/4, 1/16, 1/32, 1/64, 1/256}} *)\n\n(* {1/16, {1/2, 1/16}} *)\n\n(* {1/64, {1/2, 1/32, 1/64}} *)\n\n\nThe first number is the common ratio of the sequences, the list gives the starting numbers.\n\nTo get the first $$n$$ terms by \"brute force\", you can use the following:\n\ncookieHalves[c_, p_, n_] := Most@NumberExpand[c/p*2^n, 2, n + 1]/2^n\n\n(* {1/4, 0, 1/16, 1/32, 1/64, 0, 1/256, 0, 0, 0, 1/4096, 0} *)\n\n(* {1/2, 0, 0, 1/16, 1/32, 0, 0, 1/256, 1/512, 0, 0, 1/4096} *)\n\n(* {1/2, 0, 0, 0, 1/32, 1/64, 1/128, 0, 0, 0, 1/2048, 1/4096} *)\n\n\nThis is essentially using NumberExpand to get the list of fractions. Since the function is designed for integers, we expand $$\\frac{c}{p}2^n$$ and divide the terms by $$2^n$$ again. The last term is the fractional remainder, which is why we drop it.\n\n### Derivation of full solution\n\nHere's how I approached this: For your 4/11 example, you have found that\n\n\\begin{align} \\frac{4}{11}&=\\left(\\frac{1}{4}\\frac{1}{1024}+\\frac{1}{4}\\frac{1}{1024^2}+\\cdots\\right)\\\\ &+\\left(\\frac{1}{16}\\frac{1}{1024}+\\frac{1}{16}\\frac{1}{1024^2}+\\cdots\\right)\\\\ &+\\left(\\frac{1}{32}\\frac{1}{1024}+\\frac{1}{32}\\frac{1}{1024^2}+\\cdots\\right)\\\\ &+\\left(\\frac{1}{64}\\frac{1}{1024}+\\frac{1}{64}\\frac{1}{1024^2}+\\cdots\\right)\\\\ &+\\left(\\frac{1}{256}\\frac{1}{1024}+\\frac{1}{256}\\frac{1}{1024^2}+\\cdots\\right) \\end{align}\n\nSum[Sum[1/n 1/1024^i, {i, 0, \u221e}], {n, {4, 16, 32, 64, 256}}]\n(* 4/11 *)\n\n\nThis can be rewritten as a single geometric series with ratio $$1024$$:\n\n\\begin{align} \\frac{4}{11}&=\\left(\\frac{1}{4}+\\frac{1}{16}+\\frac{1}{32}+\\frac{1}{64}+\\frac{1}{256}\\right)\\left(\\frac{1}{1024}+\\frac{1}{1024^2}+\\cdots\\right)\\\\ &=\\frac{93}{256}\\left(\\frac{1}{1024}+\\frac{1}{1024^2}+\\cdots\\right)\\\\ &=4\\frac{93}{1024}\\left(\\frac{1}{1024}+\\frac{1}{1024^2}+\\cdots\\right) \\end{align}\n\n1/4 + 1/16 + 1/32 + 1/64 + 1/256\n(* 93/256 *)\n\n\nSo effectively, you have found a way to write $$\\frac{1}{11}$$ as a geometric series with ratio $$2^{-n}$$. Knowing the formula for the geometric series, we can rewrite the above as\n\n\\begin{align} \\frac{4}{11}&=4\\frac{93}{1024}\\frac{1}{1-\\frac{1}{1024}}\\\\ &=4\\frac{93}{1024}\\frac{1024}{1024-1}\\\\ &=4\\frac{93}{1024}\\frac{1024}{1023}\\\\ &=4\\frac{93}{1024}\\frac{1024}{93\\cdot11}\\\\ \\end{align}\n\nAs you can see, the trick is to write $$11$$ as $$\\frac{1023}{93}$$, or equivalently,\n\n$$2^{10}-1=11\\cdot 93$$\n\nThis means that we need to find $$n$$ such that\n\n$$2^n-1\\equiv0\\ \\mod p$$\n\nwhere $$p$$ is the number of people. Luckily, this problem is already solved by Euler's theorem, which states that\n\n$$2^{\\varphi(p)}-1\\equiv0\\ \\mod p$$\n\nwhere $$\\varphi(n)$$ is the totient function (EulerPhi).\n\nThe last remaining step is then to find the write the fraction (here $$\\frac{4\\cdot 93}{1024}$$) as sum of terms with the form $$\\frac{1}{2^n}$$. The numerator of the fraction is in general given by\n\n$$2^{\\varphi(p)}\\frac{c}{p}$$\n\n(where $$c$$ is the number of cookies and $$p$$ the number of people), and this is an integer thanks to Euler's theorem. Rewriting this as a sum of powers of $$2$$ (using NumberExpand), we get the final result, which is implemented with the code at the top.\n\n\u2022 This is really thorough, Lukas, and the link to Euler's theorem and EulerPhi[] is perfect. Thank you. \u2013\u00a0mathalicious Sep 21 at 13:37\n\nIt all boils down to a binary representation of 4/11, or n/m if you have n cookies and m people. This representation can have a finite number or infinite number of digits. You get e.g. 16 digits (starting from the first non-zero digit) of this representation by:\n\nRealDigits[4/11, 2, 16]\n(*{{1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0}, -1}*)\n\n\nmeaning the first few digits of the binary representation are: 0.01011101000101110... This tells you, each person gets:\n\n1/4 + 1/16 + 1/32 + 1/64 + ... cookies\n\n\u2022 Oh jeez...yes, of course. It is precisely the binary representation. Thanks, Daniel. \u2013\u00a0mathalicious Sep 21 at 13:38", "date": "2020-10-23 02:59:21", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/230459/exact-division-and-geometric-sequences", "openwebmath_score": 0.6488696336746216, "openwebmath_perplexity": 738.4347871260449, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462211935647, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8701801370982428}}
{"url": "http://techiemathteacher.com/2014/10/29/hard-algebraic-expansion/", "text": "# The Powers of Sum: Solving Hard Problem Algebraically\n\nOne of the active members of a Facebook group called Elite Math Circle (EMC) posted a question that seems interesting. Same question I enc0unter when I was actively solving problems at Brilliant.\n\nIt was posted by Russelle Guadalupe, an elite member of Philippine Team in 2011 International Mathematical Olympiad. The question is,\n\nGiven:\n\n$a+b+c=6$\n$a^2+b^2+c^2=8$\n$a^3+b^3+c^3=5$\n\nWhat is the value of $a^4+b^4+c^4$.\n\nI have seen an elegant solution to this method, one of the members says it was \u201cNewton\u2019s Sum Theorem\u201d.\n\nThe solution here might be the old school way to solve the problem using algebraic identities and manipulation.\n\nTake note of the following identities:\n\n1. $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$\n\n2. $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$\n\nFirst step is to find an expression that will lead to the unknown which is $a^4+b^4+c^4$. This expression is achievable by squaring $a^2+b^2+c^2$.\n\n$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$\n\nBut \u00a0$a^2+b^2+c^2=8$\n\n$(8)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$\n\n$64=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$\n\nWe still have one missing expression,\u00a0$a^2b^2+b^2c^2+a^2c^2$.\n\nSince we don\u2019t have the latter expression, we will find it. Take a look of the first algebraic identity mentioned above. We can use it to arrive to an expression that we are looking for by squaring \u00a0$ab+bc+ac$.\n\n$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)$\n\nWe know that\u00a0a+b+c=6,\n\n$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(6)$\n\n$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+12abc$\u00a0\u2665\u2665\n\nWe already have an expression for our unknown but we ended now with two more unknowns. $ab+bc+ac$ and $abc$.\n\nBy squaring \u00a0$a+b+c$ and solving for \u00a0$ab+bc+ac$ we have,\n\n$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$\n\n$(6)^2=8+2(ab+bc+ac)$\n\n$2(ab+bc+ac)=36-8$\n\n$ab+bc+ac=14$\u00a0\u2665\u2665\u2665\n\nWe only have one unknown left, the $abc$\n\nRecall the second identity mentioned above,\n\n$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$\n\nor\n\n$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac))+3abc$\n\nBy substitution we have,\n\n$5=(6)(8-14)+3abc$\n\n$5=-36+3abc$\n\n$41=3abc$\n\n$164=12abc$\u00a0\u2665\u2665\u2665\u2665\n\nSubstituting\u00a0\u2665\u2665\u2665 and\u00a0\u2665\u2665\u2665\u2665 to\u00a0\u2665\u2665 we have,\n\n$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+12abc$\n\n$(14)^2=a^2b^2+b^2c^2+a^2c^2+164$\n\n$196=a^2b^2+b^2c^2+a^2c^2+164$\n\n$a^2b^2+b^2c^2+a^2c^2=196-164$\n\n$a^2b^2+b^2c^2+a^2c^2=32$\u00a0\u2665\u2665\u2665\u2665\u2665\n\nSubstituting\u00a0\u2665\u2665\u2665\u2665\u2665 to\u00a0\u2665 we have,\n\n$64=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$\n\n$64=a^4+b^4+c^4+2(32)$\n\n$64=a^4+b^4+c^4+64$\n\n$\\boxed{a^4+b^4+c^4=0}$\n\nThe method above has\u00a0actually a routine, it has a pattern. In Mathematics, when there is a pattern, there is a general rule and formula can be derived. Dr. Greenie of Ask Dr. Math derived a formula using the following rules.\n\nGiven:\n\n$a+b+c=x$\n$a^2+b^2+c^2=y$\n$a^3+b^3+c^3=z$\n\nWe can solve the value of\u00a0$a^4+b^4+c^4$ in terms of x,y, and z with the following formula,\n\n$a^4+b^4+c^4=y^2-2((\\displaystyle\\frac{x^2-y}{2})^2-\\displaystyle\\frac{x^4-3x^2y+2xz}{3})$\n\n### Dan\n\nBlogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.\n\n### 21 Responses\n\n1. Enjoyed every bit of your blog.Really thank you! Want more.\n\n2. this site says:\n\nVqjwJz Major thankies for the article post.Really looking forward to read more. Much obliged.\n\n3. lucy ann says:\n\n4. 689034 217207When do you think this Real Estate market will go back in a positive direction? Or is it still too early to tell? 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YES", "lm_q1_score": 0.9916842227513689, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8701798764439734}}
{"url": "https://cs.stackexchange.com/questions/136636/median-of-medians-confusion-the-approximate-median-part", "text": "Median of medians confusion \u2014 the \u201capproximate\u201d median part\n\nI'm struggling with the median of medians algorithm, and I think it's perhaps more of a semantics thing rather than a technical thing. I've always thought the median of medians algorithm as finding an approximate median $$p$$ such that $$p$$ is within $$20\\%$$ of the true median $$M$$ in the sorted array.\n\nHowever, when I look at actual implementations, e.g., in https://brilliant.org/wiki/median-finding-algorithm/, the algorithm they posted returns an exact median, but at each level of the recursion, you may have some approximate median generated from a sublist of medians. And eventually you'll reach a level where the array is $$\\leq 5$$ elements, ending the recursion. At this level, you obtain an exact median of the array you passed in. So I had thought all this time that this exact median computed at the last level is actually your estimate of the median in the original array passed in at the first level of the recursion.\n\nSo I had the same confusion as this poster https://stackoverflow.com/questions/52461306/something-i-dont-understand-about-median-of-medians-algorithm and some others.\n\nNow that I understand this algorithm, I am now confused on how the median of medians actually finds an \"approximate\" median to the original array. In all the implementations I've seen, the median you find using median of medians is exact. So where does the approximate part come in other than approximating the median at each recursion level? Even Wikipedia describes as an algorithm that approximates a median. Yes, it approximates medians at various levels, but the final output is exact.\n\nOr am I operating under a false premise in thinking that Median of Medians finds an approximate median to the ORIGINAL array?\n\n\u2022 I believe some people call median of median the algorithm which selects an approximate median in linear time, and some people mean what you get when you combine that with quickselect, i.e. a linear-time algorithm to find the k'th element in an array (or in particular, find the median). Just two closely related things some people tend to call by the same name. \u2013\u00a0Tassle Mar 14 at 22:18\n\u2022 @Tassle This is one of the algorithms where I haven't really been satisfied with when reading the first page of Google links. I felt something was confusing or missing in each of them. Is there a text book that this algorithm is in? It's not in CLRS unfortunately, and I don't have familiarity with other algorithms textbooks. \u2013\u00a0user5965026 Mar 14 at 23:52\n\nMedian-of-medians is a recursive algorithm which solves the more general selection problem: given an array $$A$$ of length $$n$$ (which we assume, for simplicity, has distinct elements) and an integer $$k$$, find the $$k$$'th smallest element (where $$1 \\leq k \\leq n$$). It works as follows:\n\n\u2022 If $$n \\leq K$$ (where $$K$$ is some constant), solve the problem by brute force.\n\u2022 Otherwise, find an approximate median $$x$$ of $$A$$ in $$O(n)$$. By approximate median, we mean that $$x$$ is the $$m$$'th smallest element in $$A$$, where $$\\alpha n \\leq m \\leq (1-\\alpha) n$$ for some constant $$\\alpha < 1$$.\n\u2022 Use $$x$$ to split $$A$$ into elements smaller than $$x$$, comprising an array $$A_{ of length $$m-1$$, and elements larger than $$x$$, comprising an array $$A_{>x}$$ of length $$n-m$$.\n\u2022 If $$m = k$$, then output $$x$$. If $$m > k$$ then output the $$k$$'th smallest element of $$A_{. If $$m < k$$ then output the $$(k-m)$$'th smallest element of $$A_{>x}$$.\n\nThe running time of the algorithm satisfies the recurrence $$T(n) \\leq T(\\alpha n) + O(n)$$, whose solution is $$T(n) = O(n)$$.\n\nThe algorithm finds the exact median, but it does so by repeatedly finding approximate medians.\n\n\u2022 Note that the algorithm used to find the approximate median is sometimes what people refer to when they say \"median-of-medians\", hence the confusion experienced by the OP I think. \u2013\u00a0Tassle Mar 14 at 22:20\n\u2022 Do you know of a textbook that describes the median of medians? I read all the articles on the first page of Google and more after googling \"median of medians,\" and I just don't feel very satisfied with any of them, including the wikipedia article. The one on brilliant.org was probably the best one I read, but I still would prefer a textbook read for this algo. \u2013\u00a0user5965026 Mar 14 at 23:58\n\u2022 It\u2019s described in CLRS and on Wikipedia, and probably in many other lecture notes and slides. \u2013\u00a0Yuval Filmus Mar 15 at 0:00\n\u2022 Oh I didn't realize it was in CLRS. Found it in 9.3. Earlier I was doing a search for median of medians in the book and could not find it. \u2013\u00a0user5965026 Mar 15 at 0:44", "date": "2021-05-17 12:37:13", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/136636/median-of-medians-confusion-the-approximate-median-part", "openwebmath_score": 0.77828049659729, "openwebmath_perplexity": 406.54976528551714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9473810466522863, "lm_q2_score": 0.9184802496038499, "lm_q1q2_score": 0.8701507801991485}}
{"url": "https://math.stackexchange.com/questions/2122264/whats-the-probability-of-the-target-being-destroyed", "text": "What's the probability of the target being destroyed?\n\nTarget gets hit by three missiles.\n\nThe probability of hitting the target with the\n- First missile is 0.4;\n- Second missile is 0.5;\n- Third missile is 0.7;\n\nThe target has a probability of 0.2 getting destroyed with one missile, 0.6 by two missiles and for sure (1) by three missiles.\n\nWhat's the probability of the target being destroyed?\n\n--\n\nThis is the exact question on the book, teacher told us that the result is 0.458 but didn't describe the method of solving, except saying that it's not easy.\n\n\u2022 you need to start by getting the probability of exactly 1,2 and 3 hits separately - being hit by 3 missiles is simplest, all missiles have to hit - can you do that calculation? Two missile hits can happen in 3 different ways 1,2 or 3 has to miss - can you work out the probabilities of the 3 distinct events and add them together - 1 hit is similar to 2 hits, 1 hits and 2 misses - you have to consider the probability of hits and misses in your calculations -\n\u2013\u00a0Cato\nJan 31, 2017 at 10:26\n\u2022 I see. So, let's say: D -> target being destroyed; M1,M2,M3 -> Probabilities of missiles; H1,H2,H3 -> the three scenarios you mentioned, where P(H3) is the easiest to find (0.4*0.5*0.7), since all the targets should hit. So I have to basically find P(D) which is P(D/H1)*P(H1) + P(D/H2)*P(H2) + P(D/H3)*P(H3)? @Cato Jan 31, 2017 at 10:39\n\u2022 yes I would imagine it would end up in that form - an example of 1 missile hit is that just 1 hits, P=.4*.5*.3 - but that can happen in 3 ways, but the 3 ways exclude each other\n\u2013\u00a0Cato\nJan 31, 2017 at 10:41\n\u2022 Thanks!! I was having problems finding P(H1) and P(H2) earlier, but thanks to you ringing me the bell that there can be 3 ways for these two scenarios, the problem is solved. @Cato Jan 31, 2017 at 10:50\n\nFirst: hit and destroy, Second :no hit, Third: no hit\n\n$0.4\\times 0.2 \\times 0.5 \\times 0.3=0.012$\n\nFirst: no hit, Second :hit and destroy, Third: no hit\n\n$0.6\\times 0.5 \\times 0.2 \\times 0.3=0.018$\n\nFirst: no hit, Second :no hit, Third: hit and destroy\n\n$0.6\\times 0.5 \\times 0.7 \\times 0.2=0.042$\n\nFirst: no hit, (Second and Third) :hit and destroy\n\n$0.6\\times 0.5 \\times 0.7 \\times 0.6=0.126$\n\nsecond: no hit, (first and Third) :hit and destroy\n\n$0.5 \\times 0.4 \\times 0.7 \\times 0.6=0.084$\n\nthird: no hit, (first and second) :hit and destroy\n\n$0.3 \\times 0.4 \\times 0.5 \\times 0.6=0.036$\n\n(First, Second and Third) :hit and destroy\n\n$0.4\\times 0.5 \\times 0.7 =0.14$\n\nAdd all, we get required probability as $0.458$\n\n\u2022 Good Answer, Kk...k...Kiran... Jan 31, 2017 at 11:42\n\n\\begin{align*} p =& P(A \\lor B \\lor C) \\\\=& 1-P(\\neg A)P(\\neg B)P(\\neg C)\\\\=&1-(1-P(A))(1 - P(B))(1-P(C)) \\end{align*}", "date": "2023-03-31 05:45:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2122264/whats-the-probability-of-the-target-being-destroyed", "openwebmath_score": 0.5965492129325867, "openwebmath_perplexity": 782.7484944726444, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707953529716, "lm_q2_score": 0.8902942166619118, "lm_q1q2_score": 0.8701475666370035}}
{"url": "https://math.stackexchange.com/questions/2713547/how-does-munkres-prove-that-lower-limit-topology-is-finer-than-the-standard-topo", "text": "# How does Munkres prove that lower limit topology is finer than the standard topology on $\\mathbb{R}$\n\nThe proof in question (from the book \"Topology\" by Munkres):\n\nLet $\\mathcal T$ and $\\mathcal T_\\mathscr{l}$ be the standard and lower limit topology on $\\mathbb R$ respectively. Given a basis element $(a,b)$ for $\\mathcal T$ and a point $x\\in (a,b)$, the basis element $[x,b)$ for $\\mathcal T_\\mathscr{l}$ contains $x$ and lies in $(a,b)$. Conversely, given the basis element $[x,d)$ for $\\mathcal T_\\mathscr{l}$, there is no open interval in $\\mathcal T$ that contains $x$ and lies in $[x,d)$. Thus $\\mathcal T_\\mathscr l$ is strictly finer than $\\mathcal T$.\n\nThere was a question on it before, but his actual proof was never addressed. I understand the proof involving the union of an infinite collection of sets. However, I don't get how the above proof proves that all elements of the latter topology are in the former, which as I understand is the definition of a finer topology.\n\n\u2022 Consider the analogy of sand in his book. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Mar 29 '18 at 17:04\n\u2022 @GNUSupporter Honestly, it is passages like the sand analogy that make me really like Munkres as an introductory text. Some of the later chapters lose the thread a little (in my opinion), but the first half of that book is golden. \u2013\u00a0Xander Henderson Mar 29 '18 at 17:06\n\u2022 The lower limit topology on $\\Bbb R$ is also called the Sorgenfrey line, sometimes denoted $\\Bbb R_l.$ Its square $\\Bbb R^2_l$ is called the Sorgenfrey plane. \u2013\u00a0DanielWainfleet Mar 30 '18 at 4:57\n\nOf $O\\in\\mathcal{T}$, then $O$ can be written as the union of intervals of the type $(a,b)$. Each interval $(a,b)$, in turn, can be written as$$\\bigcup_{x\\in(a,b)}[x,b),$$which belongs to $\\mathcal{T}_1$. Therefore, $O\\in\\mathcal{T}_1$.\n\n\u2022 But isn't Munkres missing this step? I don't see how he explicitly shows that the interval (a,b) can also be found in $\\mathcal T_\\mathscr l$ \u2013\u00a0oddic Mar 29 '18 at 17:13\n\u2022 @oddic He proves that, for each $x\\in(a,b)$, there is an element $S\\in\\mathcal{T}_1$ such that $x\\in S$ and that $S\\subset(a,b)$. I suppose that he didn't feel the need to deduce from this that $(a,b)\\in\\mathcal{T}_1$. \u2013\u00a0Jos\u00e9 Carlos Santos Mar 29 '18 at 17:16\n\u2022 But doesn't the proof require showing that there is an element $S\\in \\mathcal T_\\mathscr l$ which is a union of infinite basis elements that is equal to $(a,b)$, and not just a subset? I'm sorry if I'm not understanding correctly \u2013\u00a0oddic Mar 29 '18 at 17:34\n\u2022 @oddic If, for each $x\\in(a,b)$, there is a $S_x\\in\\mathcal{T}_1$ such that $x\\in S_x$ and that $S_x\\subset(a,b)$, then$$(a,b)=\\bigcup_{x\\in(a,b)}S_x\\in\\mathcal{T}_1.$$ \u2013\u00a0Jos\u00e9 Carlos Santos Mar 29 '18 at 17:43\n\u2022 @oddic In Lemma 13.3 (p.81) the author already gave an equivalent condition of a topology being finer than the other topology. So in the following Lemma 13.4 he did not need to mention it again, I think. \u2013\u00a0ChoF Mar 30 '18 at 5:12\n\nWhat Munkres is doing:\n\nIf $T_1,T_2$ are topologies on a set $R,$ and $B_1,B_2$ are bases for $T_1,T_2$ respectively, then to show that $T_1\\subset T_2,$ it suffices to show that whenever $x\\in b_1\\in B_1$ there exists $b_2\\in B_2$ with $x\\in b_2\\subset b_1.$\n\nThat implies that every $b_1\\in B_1$ is a union of members of $B_2,$ so $b_1\\in T_2$. So $B_1\\subset T_2,$ so any $t\\in T_1,$ being a union of members of $B_1,$ is a union of members of $T_2,$ so $t\\in T_2.$\n\nIn this case we have $R=\\Bbb R$ and $T_1=\\mathcal T$ and $T_2=\\mathcal {T_l}$ and $B_1=\\{(a,b): a,b\\in \\Bbb R\\}$ and $B_2=\\{[x,b):x,b\\in \\Bbb R\\}.$\n\nTherefore $\\mathcal T\\subset \\mathcal T_l.$\n\n(And, obviously, they are not equal, because $[0,1)\\in \\mathcal T_l$ \\ $\\mathcal T.$)\n\n\u2022 Thanks, I understand what I was missing now! \u2013\u00a0oddic Apr 1 '18 at 0:02\n\u2022 In many cases the use of particular bases simplifies things. Not surprising when you consider that usually what you do is show that if something holds for a certain subset of a topology (the base/basis) then it holds for the whole topology..... Many write \"base\", not \"basis\". I prefer \"base\", especially in the context of topological vector spaces, where \"basis\" also has vectorial meanings. \u2013\u00a0DanielWainfleet Apr 1 '18 at 16:37", "date": "2019-12-07 14:49:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2713547/how-does-munkres-prove-that-lower-limit-topology-is-finer-than-the-standard-topo", "openwebmath_score": 0.9109523892402649, "openwebmath_perplexity": 157.38304592225052, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357561234474, "lm_q2_score": 0.8887587839164801, "lm_q1q2_score": 0.8701266280230268}}
{"url": "https://stats.stackexchange.com/questions/282455/two-questions-involving-bayes-theorem", "text": "# Two questions involving Bayes Theorem\n\nYou draw two balls from one of three possible large urns, labelled A, B, and C. Urn A has 1/2 blue balls, 1/3 green balls, and 1/6 red balls. Urn B has 1/6 blue balls, 1/2 green balls, and 1/3 red balls. Urn C has 1/3 blue balls, 1/6 green balls, and 1/2 red balls. With no prior information about which urn your are drawing from, you draw one red ball and one blue ball. What is the probability that you drew from urn C?\n\nSo I know Bayes Theorum is this:\n\nOn a high level, could I take the probability of drawing one red ball and one blue ball from Urn C and divide that by the total probabality of drawing one blue ball and one red ball from all the urns summed up?\n\nThe numerator would be $\\frac{1}{2} * \\frac{1}{3}$ right? The denominator would the the probabilities of drawing a blue and red from all the urns summed up right?\n\nIn the NFL, a professional American football league, there are 32 teams, of which 12 make the playoffs. In a typical season, 20 teams (the ones that don\u2019t make the playoffs) play 16 games, 4 teams play 17 games, 6 teams play 18 games, and 2 teams play 19 games. At the beginning of each game, a coin is flipped to determine who gets the football first. You are told that an unknown team won ten of its coin flips last season. Given this information, what is the posterior probability that the team did not make the playoffs (i.e. played 16 games)?\n\nI have no idea how to approach this one. Can someone shed some insight?\n\n\u2022 Please check stats.stackexchange.com/tags/self-study/info and edit your question accordingly to homework-like questions policy we have. \u2013\u00a0Tim May 30 '17 at 7:31\n\u2022 I added some more work. I'm genuinely stuck on the second. \u2013\u00a0Jwan622 May 30 '17 at 14:53\n\nBelow are two methods. The first uses Bayes' Theorem in full, and the second is a quicker approach (just for fun). Importantly, the two results agree! (They also do not agree with the earlier answer posted by zen...)\n\n$X := \\text{Picked Red and Blue}$, $A := \\text{Picked Urn A}$, $B := \\text{Picked Urn B}$, $C := \\text{Picked Urn C}$.\n\nMethod 1.\n\nUsing Bayes' Theorem, we now compute as follows:\n\n$$\\frac{P(X|C)P(C)}{P(X|C)P(C) + [P(X|A)P(A) + P(X|B)P(B)]}$$\n\nwhere the bracketed portion in the denominator corresponds to the cases in which red and blue were selected but did not come from Urn C.\n\nUnder the assumption that each of $P(C)$, $P(A)$, and $P(B)$ is $1/3$, we can cancel this factor from the numerator and denominator to get:\n\n$$\\frac{P(X|C)}{P(X|C) + [P(X|A) + P(X|B)]}$$\n\nNext, we calculate each of these terms:\n\n$P(X|C) = 2(3/6)(2/5) = 12/30$;\n\n$P(X|B) = 2(2/6)(1/5) = 4/30$; and\n\n$P(X|A) = 2(1/6)(3/5) = 6/30$.\n\nAnd so the earlier expression becomes:\n\n$$\\frac{12/30}{12/30 + 4/30 + 6/30} = \\frac{12}{12+4+6} = \\frac{6}{6+2+3} = \\frac{6}{11}$$\n\nMethod 2.\n\nWithout loss of generality, let us suppose each of the urns contains exactly six M&Ms. Then, the number of ways to choose a red and blue M&M from the urns can be computed as follows.\n\nUrn A: BBBGGR; so, picking RB can be done in $1 \\cdot 3 = 3$ ways.\n\nUrn B: GGGRRB; so, picking RB can be done in $2 \\cdot 1 = 2$ ways.\n\nUrn C: RRRBBG; so, picking RB can be done in $3 \\cdot 2 = 6$ ways.\n\nSo the probability of it being Urn C given that RB were selected is just:\n\n$$\\frac{6}{6+3+2} = \\frac{6}{11}$$\n\nas in Method 1.\n\n\u2022 It was not clear to me that the statement \"having no prior information about which urn you are drawing from\". I read it as the prior probabilities for P(A), P(B) and P(C) were unknown and not they were equal. That would make the problem unsolvable. If you take it to mean prior to drawing the ball the choice of urn is equally likely, in which case P(C|X) can be determined. \u2013\u00a0Michael Chernick Jun 1 '17 at 14:41\n\u2022 How do you get $P(X|c)=2(3/6)(2/5)$? shouldn't it be $P(X|c)=2(3/6)(2/6)$ ? It is not given that there are 6 balls in the urns right? why then say that the chance of getting a blue is now $(2/5)$ instead of $(2/6)$? \u2013\u00a0zen Jun 6 '17 at 15:55\n\u2022 For example try to calculate the chance of picking (blue ,blue ,blue ,blue,blue,blue,blue) from urn C by either of your methods. I think you implicilty assume there are 6 balls. \u2013\u00a0zen Jun 6 '17 at 16:25\n\nSo I'll give two answers one for each question. Let's first discuss the first question and then in a separate answer I will attempt to answer the second question(this might take a while because I have some other things to do :)). So first lets expand bayes theorem and then we will consider what this means. So want $P(drew from C|red, blue)=\\frac{P(red ,blue|drew from C)P(drew from C)}{P(red ,blue|drew from C)P(drew from C)+ P(red ,blue|drew from A)P(drew from A)+P(red ,blue|drew from B)P(drew from B)}$\n\nTechnically we do not know $P(drew from A),P(drew from B),P(drew from C)$! So technically we cannot calculate this chance. But we can make a reasonable assumption and that is that $P(drew from A)=P(drew from B)=P(drew from C)=\\frac{1}{3}$. Now to determien all the other terms lets start with $P(red ,blue|drew from C)$ now we assume these are independent chances e.g. If I pick blue first the chance of blue stays the same for the second choice (which would in real life not be true) than the chance of $P(red ,blue|drew from C)=(\\frac{1}{3} \\frac{1}{2}+\\frac{1}{2} \\frac{1}{3})=\\frac{1}{3}$ remember that not only red, blue counts but also blue ,red thus we should count both possibilities. the other probabilities can be calculated in a similar way. so that $P(drew from C|red, blue)=\\frac{\\frac{1}{3}\\frac{1}{3}}{\\frac{1}{3}\\frac{1}{3}+ 2\\frac{1}{2}\\frac{1}{6}\\frac{1}{3}+2\\frac{1}{3}\\frac{1}{6}\\frac{1}{3}}=0.428$ (approximately)\n\nSo for the second question I will just offer some help in the form of tips after which I hope you will understand it. tip 1: consider all teams equal so and calculate the chance that $P(played 16 games)=\\frac{20}{32}$. 2. then see that the concitional probability your looking for is $P(played 16 games|coin wins =10)$ 3. $P(coin wins =10 |played 16 games)$ is binomially distributed with (n=16,p=0.5) and X=10 ( or k=10 in wiki notation).", "date": "2019-11-14 00:58:06", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/282455/two-questions-involving-bayes-theorem", "openwebmath_score": 0.8440173268318176, "openwebmath_perplexity": 455.5369691944458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974042647323258, "lm_q2_score": 0.8933093975331751, "lm_q1q2_score": 0.8701214504519584}}
{"url": "http://slideplayer.com/slide/229581/", "text": "# Partial Orderings Section 8.6.\n\n## Presentation on theme: \"Partial Orderings Section 8.6.\"\u2014 Presentation transcript:\n\nPartial Orderings Section 8.6\n\nIntroduction A relation R on a set S is called a partial ordering or partial order if it is: reflexive antisymmetric transitive A set S together with a partial ordering R is called a partially ordered set, or poset, and is denoted by (S,R).\n\nExample Let R be a relation on set A. Is R a partial order?\n(2,3),(2,4),(3,3),(3,4),(4,4)}\n\nExample Is the \u201c\uf0b3\u201d relation is a partial ordering on the set of integers? Since a \uf0b3 a for every integer a, \uf0b3 is reflexive If a \uf0b3 b and b \uf0b3 a, then a = b. Hence \uf0b3 is anti-symmetric. Since a \uf0b3 b and b \uf0b3 c implies a \uf0b3 c, \uf0b3 is transitive. Therefore \u201c\uf0b3\u201d is a partial ordering on the set of integers and (Z, \uf0b3) is a poset.\n\nComparable/Incomparable\nIn a poset the notation a \u227c b denotes (a,b) \u2208 R The \u201cless than or equal to\u201d (\uf0a3)is just an example of partial ordering The elements a and b of a poset (S, \u227c) are called comparable if either a\u227cb or b\u227ca. The elements a and b of a poset (S, \u227c) are called incomparable if neither a\u227cb nor b\u227ca. In the poset (Z+, |): Are 3 and 9 comparable? Are 5 and 7 comparable?\n\nTotal Order We said: \u201cPartial ordering\u201d because pairs of elements may be incomparable. If every two elements of a poset (S, \u227c) are comparable, then S is called a totally ordered or linearly ordered set and \u227c is called a total order or linear order. The poset (Z+, \uf0a3) is totally ordered. Why? The poset (Z+, |) is not totally ordered.\n\nHasse Diagram Graphical representation of a poset\nSince a poset is by definition reflexive and transitive (and antisymmetric), the graphical representation for a poset can be compacted. For example, why do we need to include loops at every vertex? Since it\u2019s a poset, it must have loops there.\n\nConstructing a Hasse Diagram\nStart with the digraph of the partial order. Remove the loops at each vertex. Remove all edges that must be present because of the transitivity. Arrange each edge so that all arrows point up. Remove all arrowheads.\n\nExample Construct the Hasse diagram for ({1,2,3},\uf0a3) 3 2 1 3 2 1 1 2 3\n1 1\n\nHasse Diagram Terminology\nLet (S, \u227c) be a poset. a is maximal in (S, \u227c) if there is no b\uf0ceS such that a\u227cb. (top of the Hasse diagram) a is minimal in (S, \u227c) if there is no b\uf0ceS such that b\u227ca. (bottom of the Hasse diagram) a is the greatest element of (S, \u227c) if b\u227ca for all b\uf0ceS\u2026 it has to be unique a is the least element of (S, \u227c) if a\u227cb for all b\uf0ceS. It has to be unique\n\nHasse Diagram Terminology (Cont ..)\nLet A be a subset of (S, \u227c). If u\uf0ceS such that a\u227cu for all a\uf0ceA, then u is called an upper bound of A. If l\uf0ceS such that l\u227ca for all a\uf0ceA, then l is called an lower bound of A. If x is an upper bound of A and x\u227cz whenever z is an upper bound of A, then x is called the least upper bound of A\u2026unique If y is a lower bound of A and z\u227cy whenever z is a lower bound of A, then y is called the greatest lower bound of A\u2026unique\n\nExample Maximal: h,j Minimal: a Greatest element: None\ng f d e b c a Maximal: h,j Minimal: a Greatest element: None Least element: a Upper bound of {a,b,c}: e,f,j,h Least upper bound of {a,b,c}: e Lower bound of {a,b,c}: a Greatest lower bound of {a,b,c}: a\n\nLattices A partially ordered set in which every pair of elements has both a least upper bound and greatest lower bound is called a lattice. f e c d b a h e f g b c d a\n\nSimilar presentations", "date": "2019-02-22 06:25:20", "meta": {"domain": "slideplayer.com", "url": "http://slideplayer.com/slide/229581/", "openwebmath_score": 0.8889617323875427, "openwebmath_perplexity": 1187.7558768419085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147197717786, "lm_q2_score": 0.8947894569842487, "lm_q1q2_score": 0.8701064390680802}}
{"url": "http://math.stackexchange.com/questions/50919/calculate-the-sum-of-the-infinite-series-sum-n-0-infty-fracn4n", "text": "# Calculate the sum of the infinite series $\\sum_{n=0}^{\\infty} \\frac{n}{4^n}$\n\nA previous problem had us solving $\\sum_{n=0}^{\\infty} \\frac{1}{4^n}$ which I calculated to be $\\frac{4}{3}$ using a bit of mathematical manipulation.\n\nWonderful. Thank you for all the prompt responses. Could anyone suggest an alternate technique that does not involve differentiation?\n\n-\n\n$$\\frac{1}{4} +\\frac{2}{4^2}+\\frac{3}{4^3}+\\frac{4}{4^4}+ \\cdots +\\frac{n}{4^n}+\\cdots.$$\n\nCall this sum $S$. Now subtract from $S$ the sum $$\\frac{1}{4}+\\frac{1}{4^2}+\\frac{1}{4^3}+\\cdots.$$ If we do it in the obvious way, term by term, we obtain $$\\frac{1}{4^2}+\\frac{2}{4^3}+\\frac{3}{4^4}+ \\cdots.$$\n\nNote that this last sum is $(1/4)S$.\n\nPutting things together, and using your computation for $1+1/4+1/4^2+\\cdots$ (not quite, we start at $1/4$) we get $$S-\\frac{1}{3}=\\frac{S}{4}.$$ Solve for $S$. We find that $S=4/9$.\n\nComment: The calculation is a little sloppy, it assumes that infinite sums can be manipulated much like finite sums. There are theorems about power series that one could use to justify the manipulations.\n\nBut (in this case) we do not need such theorems. Let $S_n$ be the sum of the terms up to the term $n/4^n$. More or less the same sort of calculation as the one I did can be used to find an explicit formula for $S_n$. Then we can calculate $\\lim_{n\\to\\infty}S_n$, and get a fully rigorous derivation.\n\nWe could use the results of the calculation of $\\sum n/4^n$ to tackle $\\sum n^2/4^n$, and so on. But the derivatives approach is certainly slicker!\n\n-\nThis is exactly the sort of manipulation that I was looking for. Thank you. \u2013\u00a0 beethree Jul 11 '11 at 22:36\n@beethree: Good. You might want to look at the comment by Aryabhata, at the moment hidden at the end. \u2013\u00a0 Andr\u00e9 Nicolas Jul 11 '11 at 23:03\n\nAs you have computed $\\sum_{n>=0} x^n$ to be $1/(1-x)$, differentiate the series term-wise and multiply by x, which you can do for $x=1/4$ as the series converges. This gives $\\sum_{n>=0} n x^{n} = x/(1-x)^{2}$. Substitute $x=1/4$ and observe that $\\sum_{n>=0} n/4^n = 4/9$\n\n-\n\nLet $f(x)=\\sum_{n\\geq0}x^n$. This defines a function on $(-1,1)$, equal to $\\frac1{1-x}$. Using properties of power series, we know that $$\\frac x{(1-x)^2}=xf'(x)=\\sum_{n\\geq1}nx^n$$ for the same values of $x$. Evaluating this equality at $\\tfrac14$ sums your series.\n\n-\n\nWell, you can do some series manipulations... First you can write $$\\sum_{n = 1}^{\\infty} {n \\over 4^n} = \\sum_{n = 1}^{\\infty}\\,\\sum_{m = 1}^n {1 \\over 4^n}$$ Note that the above summation is over all $(m,n)$ with $m \\leq n$. So if you switch the order of summation you obtain $$\\sum_{m = 1}^{\\infty}\\,\\sum_{n = m}^{\\infty} {1 \\over 4^n}$$ The inner sum is a geometric series with initial term ${\\displaystyle{1 \\over 4^m}}$ and ratio ${\\displaystyle{1 \\over 4}}$, so it sums to ${\\displaystyle {{1 \\over 4^m} \\over 1 - {1 \\over 4}} = {4 \\over 3}{1 \\over 4^m}}$. So the overall sum is $${4 \\over 3}\\sum_{m = 1}^{\\infty} {1 \\over 4^m}$$ The sum here is a geometric series that sums to ${\\displaystyle{1 \\over 3}}$, so your final answer is $${4 \\over 3}\\times{1 \\over 3} = {4 \\over 9}$$\n\n-\n\\begin{align} (\\frac{1}{4} + \\frac{1}{4^2} + \\frac{1}{4^3} + \\dots ) + \\\\ (\\frac{1}{4^2} + \\frac{1}{4^3} + \\dots) + \\\\ (\\frac{1}{4^3} + \\dots ) + \\\\ \\dots \\end{align} \u2013\u00a0 Aryabhata Jul 11 '11 at 22:18\nyeah, basically :) \u2013\u00a0 Zarrax Jul 11 '11 at 22:19\n\nA simple probabilistic approach:\n\nLet $X$ be a geometric random variable with probability of success $p$, so that $${\\rm P}(X=n)=(1-p)^{n-1} p, \\;\\; n=1,2,3,\\ldots.$$ Then the expectation of $X$ is $${\\rm E}(X) = \\sum\\limits_{n = 1}^\\infty {n{\\rm P}(X = n)} = \\sum\\limits_{n = 1}^\\infty {n(1 - p)^{n - 1} p} = \\frac{p}{{1 - p}}\\sum\\limits_{n = 1}^\\infty {n(1 - p)^n } .$$ Thus $$\\sum\\limits_{n = 1}^\\infty {n(1 - p)^n } = \\frac{{1 - p}}{p} {\\rm E}(X).$$ On the other hand, from $${\\rm E}(X) = p \\cdot 1 + (1-p)(1+{\\rm E}(X)),$$ we get $${\\rm E}(X)=\\frac{1}{p}.$$ Finally, $$\\sum\\limits_{n = 1}^\\infty {n(1 - p)^n } = \\frac{{1 - p}}{p} {\\rm E}(X) = \\frac{{1 - p}}{p^2}.$$ Letting $p=3/4$ gives $$\\sum\\limits_{n = 1}^\\infty {\\frac{n}{{4^n }}} = \\frac{{1/4}}{{9/16}} = \\frac{4}{9}.$$\n\n-\nNice! In imitation of the book \"Proofs that Really Count\", which is an attractive introduction to bijective proofs, I had thought of posting a question about Mean Proofs. This would ask for examples of results not in probability theory that can be proved by using the mean. This (along with I think a few others of your posts) is a contribution to the mean proofs collection. \u2013\u00a0 Andr\u00e9 Nicolas Jul 12 '11 at 0:51\n@user6312: Thanks. Hopefully, I'll make more contributions of this kind. \u2013\u00a0 Shai Covo Jul 12 '11 at 1:15", "date": "2015-05-27 20:18:12", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/50919/calculate-the-sum-of-the-infinite-series-sum-n-0-infty-fracn4n", "openwebmath_score": 0.9906359910964966, "openwebmath_perplexity": 172.77641867966733, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543728093005, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8700659520203267}}
{"url": "https://ece.uwaterloo.ca/~ece050/149u/", "text": "Lesson 149u: Sieve of Eratosthenes\n\nSuppose you wanted to find all prime numbers less than or equal to a given number $N$. The easiest way to do this is to create an array that can hold the integers from $1$ to $N$. We will initialize the array so that array[k] == k. If we ever determine that a number k is not prime, we will assign array[k] = 0;. By definition, $1$ is not prime, so we will immiedately assign array[1] = 0;. Next starting with $2$, if we determine that k is prime (meaning that array[k] == k, we have then determined that all integer multiples of k up until the end of the array to be composite, so we set:\n\narray[2*k] = 0;\narray[3*k] = 0;\narray[4*k] = 0;\narray[5*k] = 0;\n\n.\n.\n.\n\nuntil we reach the end of the array.\n\nWe then search forward in the array until we find the next k such that array[k] == k and repeat the above process.\n\nThis is called the Sieve of Eratosthenes. Implement a function that prints all prime numbers up to but not exceeding $N$ in the format:\n\n2, 3, 5, 7, 9, 11, 13, ...\n\nBe sure not to have a comma after the last prime number. Your function declaration should be:\n\nvoid print_primes( unsigned long N );\n\n\nAs an aside, why is $1$ not prime? In general, many of the most interesting properties of prime numbers are not true if $1$ is considered to be prime. For example, that every number has a unique prime factorization is false if $1$ is a prime, for $6 = 2 \\times 3 = 2 \\times 3 \\times 1 = 2 \\times 3 \\times 1 \\times 1 \\times \\cdots$. Thus, if $1$ were said to be prime, then almost every interesting theorem about prime numbers would have to be worded as all prime numbers except for $1$.\n\nIn the previous function, you could have easily printed the prime number every time you found one. You would start by automatically printing \"2\", and then for every subseuent prime number n you find, print \", n\". However, suppose you wanted to do the following:\n\nIf you have found $n$ primes less than or equal to $N$, return an array of size $n$ containing those $n$ prime numbers.\n\nNow, beause you don't know how large an array to build, you will have to temporarily store prime numbers, and the easiest way is to look at the array array; however, looking through the entire array of capacity $N$ to find $n$ primes is unnecessarily expensive, especially when you know mathematically that the number of primes less than or equal to $N$ will be approximately $n \\approx \\frac{N}{\\ln(N)}$. This estimates there are approximately $72382$ prime numbers less than or equal to one million. There are in fact $78498$ primes less than or equal to one million, so this approximation is not unreasonable; however, it is not exact.\n\nIt can be shown that the relative error drops at approximately $0.24 n^{-0.085}$, so the relative error actually goes to zero as $n$ becomes large.\n\nDevise a scheme whereby you use the existing array to store all the prime numbers without having to walk through the entire array to retrieve all of them.\n\nArray<unsigned long> *primes( unsigned long N );", "date": "2022-08-15 14:31:55", "meta": {"domain": "uwaterloo.ca", "url": "https://ece.uwaterloo.ca/~ece050/149u/", "openwebmath_score": 0.6329481601715088, "openwebmath_perplexity": 276.37614529855404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543725597269, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.870065939093231}}
{"url": "https://puzzling.stackexchange.com/questions/23147/visualise-whirled-peas", "text": "# Visualise Whirled Peas\n\nMy eldest came home with some interesting maths homework.\n\nWe have three bowls containing peas.\n\nDistribute the peas evenly across the bowls, by moving peas from one bowl to another. The only move allowed, is doubling the amount of peas in one bowl by taking them from one other bowl.\n\nFor example:\n$\\begin{array}{rlrlrl} A & & B & & C & \\\\ \\hline 11 & & 6 & & 7 \\\\ 4 & (-7) & 6 & & 14 & (+7) \\\\ 4 & & 12 & (+6) & 8 & (-6) \\\\ 8 & (+4) & 8 & (-4) & 8 \\end{array}$\n\nWe had some fun with this. We quickly decided that the order of the bowls doesn't matter, so we can just sort them by the number of peas they contain.\nThen, there are just three moves possible:\n\n1. Double the amount of peas in the first bowl (containing the fewest peas) by taking them from the third bowl (containing the most peas)\n2. Double the amount of peas in the first bowl containing the fewest by taking them from the second bowl (containing the neither the most nor the fewest peas)\n3. Double the amount of peas in the second bowl by taking them from the third bowl\n\nFrom that, we made state diagrams and found our solutions. We also found out that for one of the problems, we could never reach an end state (with all bowls containing the same amount of peas).\n\nBut of course I couldn't leave it alone. So I tried to formalise what we did.\n\nThe amount of peas in each bowl at any time is of course a positive integer (or possibly 0):\n$a_i \\le b_i \\le c_i \\,|\\,a_i, b_i, c_i \\in \\mathbb Z_{\\ge 0}$\n\nA solution has the peas distributed equally across the bowls:\n$a_i = b_i = c_i$\n\nThe total amount of peas at any time is a multiple of 3:\n$a_i + b_i + c_i = 3n \\,|\\,n \\in \\mathbb Z_{\\ge 0}$\n\nThe three possible moves:\n$\\left\\{a_{i+1}, b_{i+1}, c_{i+1}\\right\\} \\in \\begin{Bmatrix} \\left\\{2\\cdot a_i, b_i, c_i-a_i\\right\\} \\\\ \\left\\{2\\cdot a_i, b_i-a_i, c_i\\right\\} \\\\ \\left\\{a_i, 2\\cdot b_i, c_i-b_i\\right\\} \\end{Bmatrix}$\n\nThe set of possible moves brings us to the conclusion that for each state after the first, at least one bowl contains an even amount of peas. Since in the end state, all bowls contain the same amount of peas, they all contain the same even amount of peas and thus the total number of peas must be even.\nIf we combine this with the requirement that the total number of peas is a multiple of 3 as well, we see that the total number of peas must be a multiple of 6:\n$a_i + b_i + c_i = 6n \\,|\\,n \\in \\mathbb Z_{\\ge 0}$\n\n(Ignoring the very trivial case where we start with a solved state).\n\nBut these conditions aren't sufficient to guarantee a solvable puzzle. For instance, $\\left\\{a_0, b_0, c_0\\right\\} = \\left\\{6,12,24\\right\\}$ is not solvable. The only reachable states are $$\\begin{Bmatrix} \\left\\{ 6, 12, 24 \\right\\} \\\\ \\left\\{12, 12, 18 \\right\\} \\\\ \\left\\{ 0, 18, 24 \\right\\} \\\\ \\left\\{ 0, 6, 36 \\right\\} \\\\ \\left\\{ 0, 12, 30 \\right\\} \\\\ \\end{Bmatrix}$$ neither of which is an end state.\n\nSo what additional conditions do we need to define to make such a puzzle sovable?\n\nProblems can have multiple solutions. See for instance this example, with two possible paths to a solution:\n\n$\\begin{array}{rlrlrl} A & & B & & C & \\\\ \\hline 11 & & 6 & & 7 \\\\ 4 & (-7) & 6 & & 14 & (+7) \\\\ 4 & & 12 & (+6) & 8 & (-6) \\\\ 8 & (+4) & 8 & (-4) & 8 \\end{array}$\n\nand\n\n$\\begin{array}{rlrlrl} A & & B & & C & \\\\ \\hline 11 & & 6 & & 7 \\\\ 5 & (-6) & 12 & (+6) & 7 & \\\\ 10 & (+5) & 12 & & 2 & (-5) \\\\ 8 & (-2) & 12 & & 4 & (+2) \\\\ 8 & & 8 & (-4) & 8 & (+4) \\end{array}$\n\nThe only way to solve a problem that I've come up so far, is drawing a state diagram and determining the shortest path to the end state. But while state diagrams work fine for small numbers of peas, the number of possible states may well explode for higher numbers.\n\nSo is there another way to solve these problems?\n\nWhat is a good strategy to arrive at a solution?\n\nYes, I'm asking two questions in one, but I don't want to repeat the explanation for a second question. Also, I feel both questions are closely related, since what I'm really looking for is an analysis of this type of problem, if possible even with a variable number of bowls.\n\n\u2022 I think your definition of the three possible moves how you present it is not entirely correct because it doesn't account for reordering the bowls again when they are not in ascending order any more \u2013\u00a0Ivo Beckers Oct 13 '15 at 14:20\n\u2022 Some observation: If one of the bowls ends up empty at some point you are in an unsolvable state because double zero is still zero. \u2013\u00a0Ivo Beckers Oct 13 '15 at 14:35\n\u2022 If the number of peas in each bowl share a common divisor, they will always share that same divisor. So in fact the total number of peas must be a multiple of $6$ times the greatest common divisor of the initial numbers of peas in the bowls. \u2013\u00a0Julian Rosen Oct 13 '15 at 15:40\n\u2022 @IvoBeckers Yes, it's true that my representation of the moves doesn't take the reordering into account, but that doesn't seem to be an important factor. And we indeed encountered the states where one bowl was empty. We continued and found loops, as was to be expected. \u2013\u00a0SQB Oct 13 '15 at 17:05\n\u2022 On a whim, I pronounced the puzzle title and was surprised by my pun detector going off! \u2013\u00a0Oliphaunt - reinstate Monica Feb 27 '16 at 11:55\n\nIf $a_0$, $b_0$, and $c_0$ are positive integers, the puzzle is solvable from an initial state $\\{a_0,b_0,c_0\\}$ if and only if $$a_0+b_0+c_0=3\\cdot 2^m\\cdot \\gcd(a_0,b_0,c_0)$$ for some $m\\geq 0$, where $\\gcd(a_0,b_0,c_0)$ denotes the greatest common divisor of $a_0$, $b_0$, and $c_0$.\nFor $k\\geq 3$ an odd number, one can check that if there is a bowl whose pea number is not divisible by $k$, then after making a move there will still be a bowl whose pea number is not divisible by $k$.\nSuppose the initial state is $\\{a_0,b_0,c_0\\}$ is solvable, and let $n=\\frac{a_0+b_0+c_0}{3}$ be the desired number of peas in each bowl.\nIf $k\\geq 3$ is an odd number dividing $n$, then in the final state every pea number is divisible by $k$, so the argument above implies that $k$ divides $a_0$, $b_0$, and $c_0$. Every odd divisor of $n$ is a common divisor of $a_0$, $b_0$, and $c_0$, so it must be the case that $n=2^m\\cdot \\gcd(a_0,b_0,c_0)$, or equivalently $a_0+b_0+c_0=3\\cdot 2^m\\cdot \\gcd(a_0,b_0,c_0)$.\nConversely, suppose $n=2^m\\cdot \\gcd(a_0,b_0,c_0)$. We can solve the puzzle as follows. If the current state is $\\{a,b,c\\}$, write $a=2^x a'$, $b=2^y b'$, $c=2^z c'$, where $a'$, $b'$, and $c'$ are odd. The condition on $n$ implies that either $a=b=c$, or that of the three numbers $x$, $y$, and $z$, two will be equal and the third will be larger.\nReordering if necessary, we can assume $x=y<z$ and that $a\\leq b$. If $a< b$, replace $\\{a,b,c\\}$ by $\\{2a,b-a,c\\}$ and repeat the process. If $a=b$, replace $\\{a,b,c\\}$ by $\\{2a,b,c-a\\}$ (if $a<c$) or $\\{a-c,b,2c\\}$ (if $a>c$) and repeat the process. This will eventually terminate with all three bowls having the same number of peas. The reason this terminates is that if $\\{a,b,c\\}$ is replaced by $\\{2a,b-a,c\\}$, then the values of $x$ and $y$ increase by at least one. The replacement with $\\{2a,b,c-a\\}$ or $\\{a-c,b,2c\\}$ cannot happen twice in a row, and the size of $x$, $y$, and $z$ must remain bounded.", "date": "2021-01-26 15:22:37", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/23147/visualise-whirled-peas", "openwebmath_score": 0.7483271956443787, "openwebmath_perplexity": 240.74513929925197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9859363713038173, "lm_q2_score": 0.8824278788223264, "lm_q1q2_score": 0.8700177407834091}}
{"url": "https://www.jiskha.com/questions/1561900/Suppose-A-is-directly-proportional-to-B-B-is-inversely-proportional-to-C-and-C-is", "text": "# Math Ratios\n\nSuppose A is directly proportional to B, B is inversely proportional to C and C is inversely proportional to D. Determine whether A and D are directly proportional, inversely proportional, or neither.\n\n1. C is inversely proportional to D ---> C = p/D\nB is inversely proportional to C ---> B = qC\nso B = q/(p/D) = (q/p)D\n\nA is directly proportional to B --> A = rB\nA = r(q/p) D\nlooks like A is directly proportional to D\n\nposted by Reiny\n\n## Similar Questions\n\n1. ### Math\n\nThis table shows the number of hours Harry spent watching television and the hours he spent doing his homework on three days last week. hours of tv: 1.5 2 3 hours of homework: 2 1.5 1 What kind of relationship do these data appear\n2. ### physical science\n\nthe wavelength of an electromagnetic wave is ____________. a. directly proportional to its frequency b.inversely proportional to its velocity c. inversely proportional to its frequency d. none of the above\n3. ### Physics\n\nHow does the electrical force relate to the charge of an object? It is inversely proportional to the square of the charge. It is directly proportional to the square of the charge. It is inversely proportional to the charge. It is\n4. ### chemistry\n\nAccording to Boyle, pressure and volume of a confined gas are ____ proportional, and according to Charles, the volume of a gas at a constant pressure is____ proportional to its Kelvin temperature. A. directly; inversely B.\n5. ### Inverse Variation\n\nHelp me please.. Explain also :( 1. E is inversely proportional to Z and Z = 4 when E = 6. 2. P varies inversely as Q and Q = 2/3 when P = 1/2. 3.R is inversely proportional to the square of I and I = 25 when R = 100. 4. F varies\n6. ### physics\n\nIf two different masses have the same kinetic energy, their momenta are: 1. inversely proportional to their masses 2. inversely proportional to the square roots of their masses 3. proportional to the squares of their masses 4.\n7. ### physics\n\nif a stone at the end of a string is whirled in a circle, the inward pull on the stone A) is known as the centrifugal force B) is inversely proportional to the speed of the object C) is inversely proportional to the square of the\n8. ### college math\n\nA quantity, W, is directly proportional to x and the square of y and inversely proportional to the cube of z. W-3 when x=4,y=2 and z=4. What is the value of w, when x=5, y=5 and z=4. Give answer to three decimal places. Thanks so\n9. ### physics\n\nThe magnitude of the magnetic field produced by a long straight current-carrying wire is: proportional to both the current in the wire and the distance from the wire proportional to the current in the wire and inversely\n10. ### college math\n\nA quantity, W, is directly proportional to X and the square Y, and inversely proportional to the cube of Z. W=3 when X=4, Y=2 and Z=4. What is the value of W when X=8, Y=9 and Z= 6? Give answer to 3 decimal places. Thanks in\n\nMore Similar Questions", "date": "2018-08-20 21:13:30", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1561900/Suppose-A-is-directly-proportional-to-B-B-is-inversely-proportional-to-C-and-C-is", "openwebmath_score": 0.8707001209259033, "openwebmath_perplexity": 674.3615621355102, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363729567545, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8700177376692952}}
{"url": "https://math.stackexchange.com/questions/2559445/let-x-n-be-a-cauchy-sequence-of-rational-numbers-define-a-new-sequence", "text": "# Let $\\{x_n\\}$ be a Cauchy sequence of rational numbers. Define a new sequence $\\{y_n\\}$ by $y_n = (x_n)(x_{n+1})$. Show that $\\{y_n\\}$ is a CS.\n\nWhat I am thinking so far is following:\n\nConstruct another sequence $\\{b_n\\}$ such that $b_1=x_2, b_2=x_3, \\ldots, b_n=x_{n+1}.$ Since $\\{x_n\\}$ is a Cauchy sequence, $\\{b_n\\}$, as a subsequence of $\\{x_n\\},$ is therefore also a Cauchy sequence.\n\nSince Cauchy sequences are bounded, we can find $M_1, M_2 \\in Q$ such that $|x_n| \\le M_1$ and $|b_n| \\le M_2$ for all $n.$ Let $M = \\max\\{M_1, M_2\\}.$\n\nSince $\\{x_n\\}$ is a Cauchy sequence, we can find $N_1$ such that for all $m,n \\ge N_1,$ $|x_n-x_m| < \\frac \\varepsilon{2M}$. Similarly, we can find $N_2$ such that for all $m,n \\ge N_2,$ $|b_n-b_m| < \\frac \\varepsilon{2M}.$\n\nLet $N = \\max\\{N_1, N_2\\}$ and $n,m \\ge N.$\n\nTherefore, $$|y_n-y_m|=|x_n x_{n+1}-x_m x_{m+1}|=|x_n b_n-x_m b_m|=|x_n b_n-x_m b_n+x_m b_n-x_m b_m|=|b_n||x_n-x_m|+|x_m||b_n-b_m| \\le M\\times\\frac \\varepsilon{2M} + M\\times\\frac \\varepsilon{2M} = \\varepsilon$$\n\nQDE.\n\nI wonder if there is any flaw in this proof and if there is a better way to prove it. Thank you!\n\n\u2022 Please, use MathJax to format mathematical expressions in your posts. \u2013\u00a0mucciolo Dec 10 '17 at 2:45\n\u2022 The idea looks right to me! One minor mistake: Near the end, in the line \"Therefore...\", when you use the triangle inequality to break up an absolute value, you should have $\\leq$ instead of $=$: specifically, $|x_nb_n-x_mb_n+x_mb_n-x_mb_m| \\leq |x_nb_n-x_mb_n| + |x_mb_n-x_mb_m|$. But this doesn't affect your argument. As far as better way: I can't think of anything. You could simplify notation slightly (just write $x_{n+1}$ instead of $b_n$, etc.) but nothing significant. \u2013\u00a0Zach Teitler Dec 10 '17 at 3:19\n\u2022 No need to introduce the sequence $\\{b_n\\}$ just to obtain the bound for $\\{x_{n+1}\\}$. If $|x_n| \\le M$ for all $n \\in \\mathbb{N}$ then also $|x_{n+1}| \\le M$ for all $n \\in \\mathbb{N}$. Great otherwise. \u2013\u00a0mechanodroid Dec 10 '17 at 9:58", "date": "2019-06-24 22:21:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2559445/let-x-n-be-a-cauchy-sequence-of-rational-numbers-define-a-new-sequence", "openwebmath_score": 0.9212462902069092, "openwebmath_perplexity": 91.61664106774295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9877587272306607, "lm_q2_score": 0.8807970842359877, "lm_q1q2_score": 0.8700150068734163}}
{"url": "https://math.stackexchange.com/questions/2332842/proof-by-mathematical-induction-difficulties", "text": "# Proof by mathematical induction, difficulties\n\nUsually, proofs by mathematical induction are not that difficult (at least the ones I encountered so far in my mathematical journey), but I am stuck with this one... Spent at least 2-3 hours on it. I just can't see it...\n\n$$\\sum_{i=0}^n \\frac{1}{2^i}\\tan\\left(\\frac{x}{2^i}\\right)=\\frac{1}{2^n}\\cot\\left(\\frac{x}{2^n}\\right) - 2\\cot(2x) , (n = 1,2...)$$\n\nWhat I tried:\n\nFor $n=1$, I expressed the LHS in terms of $\\sin$ and $\\cos$, but that didn't work... For me, at least... The expression got really complicated, which is a sign that I, probably, did something wrong.\n\nThen I tried getting that $2\\cot(2x)$ from RHS to LHS, but that didn't work also, because the expressions get really complicated...\n\n\u2022 Hint: if you can show $LHS(n+1) - LHS(n) = RHS(n+1) - RHS(n)$ and $LHS(0) = RHS(0)$, then the induction becomes easy. Then, simplifying $LHS(n+1) - LHS(n)$ and $RHS(n+1) - RHS(n)$ will help you focus on the exact trigonometric identity you need to prove. Jun 22 '17 at 20:42\n\u2022 Note that in general to prove $\\sum_{i=0}^n a_n = s_n$, you can just prove $s_0 = a_0$ (for the base case) and that $s_n - s_{n-1} = a_n$ (for the induction step). Jun 22 '17 at 20:42\n\u2022 @JairTaylor The problem is, I can't prove the base case. I just can't. Spent at least 2-3 hours on it. Jun 22 '17 at 20:49\n\u2022 @DanielSchepler Wow. Awesome concept, I never looked at it that way. It can really bring some elegance to my, usually bulky & ugly, solutions. Jun 22 '17 at 20:50\n\n\\begin{align} S_{n-1}+T_{n} &=\\frac{1}{2^{n-1}}\\cot \\left(\\frac{x}{2^{n-1}}\\right)-2\\cot (2x)+\\frac{1}{2^{n}}\\tan \\left(\\frac{x}{2^{n}}\\right)\\\\ &\\text{let} \\, \\frac{x}{2^{n}}=y\\\\ &=\\frac{1}{2^{n}}\\left( 2\\cot (2y)+\\tan (y)\\right)-2\\cot (2x)\\\\ &=\\frac{1}{2^{n}}\\left( \\frac{2(\\cos (2y))}{\\sin (2y)}+\\tan (y)\\right)-2\\cot (2x)\\\\ &=\\frac{1}{2^{n}}\\left( \\frac{2(\\cos ^2 (y)-\\sin^2 (y))}{2\\sin (y)\\cos (y)}+\\tan (y)\\right)-2\\cot (2x)\\\\ &=\\frac{1}{2^{n}}\\left( \\cot(y)-\\tan (y)+\\tan (y)\\right)-2\\cot (2x)\\\\ &=\\frac{1}{2^{n}}\\left(\\cot \\left(\\frac{1}{2^{n}}\\right)\\right)-2\\cot (2x)\\\\ &=S_n\\\\ \\end{align}\n\n\u2022 Thank you for sharing your knowledge. Truly, a beautiful proof. Jun 22 '17 at 21:19\n\n** just a hint**\n\nIt is equivalent to prove that\n\n$$\\cot (x/2^{n+1})-2\\cot (x/2^n)=$$\n\n$$\\tan (x/2^{n+1})$$\n\n\u2022 To do that, notice that $\\cot(x/2^n) = \\cot(2\\cdot x/2^{n+1})$. Jun 22 '17 at 20:57", "date": "2021-09-22 05:34:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2332842/proof-by-mathematical-induction-difficulties", "openwebmath_score": 0.8906493782997131, "openwebmath_perplexity": 692.8484596042108, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587268703082, "lm_q2_score": 0.880797076413356, "lm_q1q2_score": 0.8700149988291461}}
{"url": "https://math.stackexchange.com/questions/729315/sum-frac11-times2-frac11-times3-frac12-times5-frac13-times8/729339", "text": "# Sum $\\frac{1}{1\\times2}+\\frac{1}{1\\times3}+\\frac{1}{2\\times5}+\\frac{1}{3\\times8}+\\cdots$\n\nIf $f_n$ is the Fibonacci series, with $1,1,2,3,5,8,\\ldots$ prove that\n\n$$\\sum_{i=2}^\\infty\\frac{1}{f_{i-1}\\cdot f_{i+1}} = 1$$\n\nSo my idea was to try to convert this series into a telescoping sum somehow, because otherwise I can't see how this would be managed.\n\n$$\\frac{1}{1\\times2}+\\frac{1}{1\\times3}+\\frac{1}{2\\times5}+\\frac{1}{3\\times8}+\\cdots$$\n\nI can't see any obvious way to re-write the terms though. I could try sum this using Binet's formula but I am pretty sure that will get out of hand.\n\nWhat other alternatives do I have here?\n\nNote: If you use any other identity other than $f_n=f_{n-1}+f_{n-2}$, or Binet's formula. kindly link to, or provide , a proof of it.\n\nYou could note that\n\n$$\\frac{1}{f_nf_{n-1}} - \\frac{1}{f_nf_{n+1}} = \\frac{1}{f_{n-1}f_{n+1}}$$\n\nWhich could help with your telescoping sum.\n\n\u2022 sorry for the late acceptance. I wasn't online for a few days. \u2013\u00a0Guy Mar 29 '14 at 13:38\n\nWe can use the identity via partial fractions,\n\n$\\dfrac{1}{x(a+x)} = \\dfrac{1}{a}\\left(\\dfrac{1}{x} - \\dfrac{1}{a+x} \\right)$\n\nwith $x = f_{i-1}$ and $a = f_{i}$ (so that $a+x = f_{i+1}$) to get\n\n$\\dfrac{1}{f_{i-1}f_{i+1}} = \\dfrac{1}{f_i}\\left(\\dfrac{1}{f_{i-1}} - \\dfrac{1}{f_{i+1}} \\right) = \\dfrac{1}{f_i \\ f_{i-1}} - \\dfrac{1}{f_{i+1} f_{i}}$.\n\nSo if $\\phi_i:= \\frac{1}{f_i \\ f_{i-1}}$,\n\n$\\dfrac{1}{f_{i-1}f_{i+1}} = \\phi_{i} - \\phi_{i+1}$ and we can telescope,\n\n$\\displaystyle \\sum_{i=2}^N \\dfrac{1}{f_{i-1}f_{i+1}} = \\phi_2 - \\phi_{N+1} = 1 - \\phi_{N+1}$\n\nTo conclude, realise that $f_i\\to\\infty$, so $\\phi_N\\to0$, and so\n\n$\\displaystyle \\sum_{i=2}^\u221e \\dfrac{1}{f_{i-1}f_{i+1}} = 1 - \\lim_{N\u2192\u221e}\\phi_{N+1} = 1 - 0$.", "date": "2021-05-13 02:25:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/729315/sum-frac11-times2-frac11-times3-frac12-times5-frac13-times8/729339", "openwebmath_score": 0.8464152216911316, "openwebmath_perplexity": 411.2350938664775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587268703082, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8700149926476478}}
{"url": "https://math.stackexchange.com/questions/2702249/pigeon-hole-principle-proof-writing", "text": "# Pigeon hole principle proof writing\n\nI'm currentely discussing pigeon hole principle (simple and generalised) in class. However, when solving problems, I get the idea, but I never know the proper way to write it, I always feel like I'm either writing too much or not enough. Here is an example of a very simple problem and 2 ways I wrote the proof and if you could comment and give me tips it would be very appreciated!!\n\nProblem\n\nA man in a dark room has a box with 12 brown and 12 black socks. How many times, if picking one at a time, does the man have to pick a sock to have 2 of the same color?\n\n(Here I think of the two colors as the \"boxes\" and the picks as the \"objects\". Since there's k+1 objects (or picks) and k boxes (or type of socks) then the pigeon hole principle apply)\n\nSolution 1\n\nThere are $2$ different type of socks the man can pick . By the pigeon hole, principle if the man picks $3$ socks, it is garanteed that he will have $2$ socks of the same color $\\blacksquare$.\n\nSolution 2\n\nThe generalised pigeon hole principle says that, for $N,k \\in \\Bbb{Z}$, if $N$ objects are placed in $k$ boxes, then at least $1$ boxe will have $\\lceil{\\frac{N}{k}}\\rceil$ objects.\n\nLet $k = 2$ be the the colors of the socks and $N$ be the number of picks\n\nThe minimum amount of picks the man has to do to garantee he will pick $2$ socks of the same color is the smalles integer such that $\\lceil{\\frac{N}{2}}\\rceil = 2$\n\nTherefore, we need the minimum integer $N$ picks such that $1\\lt \\lceil{\\frac{N}{2}}\\rceil \\le 2 \\Rightarrow N = 3$ since $3 = 1*2 + 1$\n\nSo, by the generalised pigeon principle, to guarantee the man picks 2 socks of the same color, he has to pick 3 socks$\\blacksquare$.\n\nConlcusion\n\nI know the problem is very simple but I really need to know how to write that kind of proof in a way such that I say enough but not too much. I mean I feel like in the first solution I'm not convincing and in the second one I blabla for no reasons..\n\nAs usual, thank you everyone for your help!!!!\n\n\u2022 I vote for Solution 1 \u2013\u00a0Bram28 Mar 21 '18 at 19:07\n\u2022 Solution 1 for sure (though it may depend on your instructor; you might want to ask them) \u2013\u00a0E-A Mar 21 '18 at 19:12\n\u2022 For a different view on this principle, see Edsger W. Dijkstra's EWD980 and EWD1094. \u2013\u00a0Marnix Klooster Mar 26 '18 at 20:37\n\nIn most contexts, Solution 1 is a perfectly fine amount of detail and Solution 2 is way more than is necessary. There is one important way that both solutions are incomplete, though: you've explained why picking $3$ socks guarantees $2$ of the same color, but you haven't explained why no smaller number guarantees $2$ of the same color. To have a completely thorough proof, you need to also mention how it is possible to choose $2$ socks which are of different colors.\n(This observation is obvious enough that in many contexts it would be fine not to write it out explicitly, though. Note that in Solution 2 you implicitly made a more general claim when you asserted that the minimum number of picks to guarantee two socks of the same color is the smallest $N$ such that $\\lceil{\\frac{N}{2}}\\rceil = 2$. This does not follow from the pigeonhole principle alone, since you are also claiming that no smaller $N$ can work.)", "date": "2019-10-14 03:40:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2702249/pigeon-hole-principle-proof-writing", "openwebmath_score": 0.6140961050987244, "openwebmath_perplexity": 306.25173247841514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877717925422, "lm_q2_score": 0.8962513745192026, "lm_q1q2_score": 0.869980249698048}}
{"url": "https://math.stackexchange.com/questions/1926470/intuition-about-infinite-sums", "text": "I am particularly concerned about these two sums: $$\\sum_{x=\\color{red}0}^\\infty \\frac{1}{2^x} = 1 + \\frac{1}{2} + \\frac{1}{4} ... = 2$$\n\nand\n\n$$\\sum_{x=1}^\\infty \\frac{1}{x} = 1 + \\frac{1}{2} + \\frac{1}{3} ... = \\infty$$\n\nNow, I know and I understand the proofs that say the first sum converges to two and the second one is divergent. ( you probably shouldn't equate a divergent series like that ).\n\nBut on an intuitive level, what separates these two sums in such a way, that their limits are so drastically different?\n\nBoth sums add very small numbers in the end ( e.g. $\\frac{1}{10000000000}$ ), yet one always stays smaller than two and one goes way beyond that. You would think ( again, on an intuitive level, the math is clear) that the numbers become so small that there is a certain number that doesn't change anymore (significantly).\n\nFor example:\n\n$$\\sum_{x=1}^{100} \\frac{1}{x} = ca. 5$$\n\n$$\\sum_{x=1}^{1000000} \\frac{1}{x} = ca. 14.3$$\n\n$$\\sum_{x=1}^{10000000000000000000000000000000} \\frac{1}{x} = ca. 73.5$$\n\nAs one can see, the growth drastically slows down, at some point it should became infinitely small, just like with first sum ( fraction, powers of two ).\n\n\u2022 Have you seen the proof of divergence of the second sum that compares the sum to $\\frac 12+(\\frac 14+\\frac 14)+(\\frac 18+\\frac 18+\\frac 18+\\frac 18)+\\dots$? This is probably the best intuition I can think of, that no matter how far you go, you can always find a set of terms that together increase the value of the sum by at least $\\frac 12$. \u2013\u00a0abiessu Sep 14 '16 at 12:05\n\u2022 This is a classic example that shows that intuition cant help in all situations. Sometimes intuition is useful or possible, other times is not. The failure of intuition is the main reason of the existence of mathematical analysis. \u2013\u00a0Masacroso Sep 14 '16 at 12:06\n\u2022 Just a note: The first sum should start at $x=0$, else it converges to 1. \u2013\u00a0ekkilop Sep 14 '16 at 12:11\n\u2022 +1 for a very good question. A point about intuition: Don't try to force the math to conform to your intuition. Instead, digest these answers and let your intuition conform to the mathematics. \u2013\u00a0Neal Sep 14 '16 at 13:41\n\u2022 \"Young man, in mathematics you don't understand things. You just get used to them\" - Von Neumann \u2013\u00a0Dair Sep 14 '16 at 16:37\n\nThe difference is not in how small the terms are getting... as OP says, in both cases they become arbitrarily small... but in how many terms there are of a particular size. For instance, say two terms have the same size if their first nonzero binary digits are in the same place. In the first sum, you have one term of size $1$, one of size $1/2$, one of size $1/4$, one of size $1/8$, etc. In the second sum, you have one term of size $1$, one of size $1/2$, but then two of size $1/4$ ($1/3$ and $1/4$), four of size $1/8$ ($1/5$ through $1/8$), and so on. Each time the size is halved, the number of terms of that size is doubled; so the contribution from terms of each size never decreases, and the magnitude of the sum marches steadily to infinity.\n\n\u2022 There are many informative and great answers here, but yours was the one I was looking for. Furthermore, I believe this \"insight\" to be the key behind most of the other proofs presented here. At the very least, this helped me understand them on a much more fundamental level \u2013\u00a0Jonathan Sep 16 '16 at 19:06\n\u2022 Thanks for the great intuition. This cleared up all my confusion! \u2013\u00a0Dude156 Feb 6 at 3:47\n\nThis is related to how fast the terms decrease.\n\nA geometric series (your $1/2^x$) is such that every term is a constant fraction of the previous, so that dividing by this constant is the same as dropping the first term.\n\n$$\\frac12\\left(1+\\frac12+\\frac14+\\frac18\\cdots\\right)=\\frac12+\\frac14+\\frac18\\cdots$$ So you can write\n\n$$\\frac12S=S-1$$ and deduce $S=2$.\n\nThe same reasoning applies to all geometric series\n\n$$\\sum_{k=0}^\\infty r^k$$\n\nprovided that $r<1$. Indeed, if $r=1$ or $r>1$, the sum clearly grows forever. (This simplified discussion ignores the case $r<0$.)\n\nThis leads to a simple convergence criterion: if the ratio of successive terms is a constant less than $1$, the series converges. More generally, if this ratio is variable but tends to a limit smaller than $1$, the series converges.\n\nConversely, if the ratio tends to a limit larger than $1$, the series diverges. But if the ratio tends to $1$, we don't know, the criterion is insufficient.\n\nThe case of the harmonic series ($1/n$) or the generalized harmonic series ($1/n^p$) precisely falls in this category, as\n\n$$\\lim_{n\\to\\infty}\\left(\\frac{n}{n+1}\\right)^p=1.$$\n\nTo deal with it, a trick is to sum the terms in groups of increasing size (by doubling), so that the sums exceed a constant. More precisely,\n\n$$\\begin{gather} 1,\\\\ \\frac12,\\\\ \\frac13+\\frac14 > \\frac14+\\frac14 = \\frac12,\\\\ \\frac15+\\frac16+\\frac17+\\frac18 > \\frac18+\\frac18+\\frac18+\\frac18 = \\frac12,\\\\ \\cdots \\end{gather}$$\n\nThough the groups get longer and longer, you can continue forever and the sum grows to infinity.\n\nIf you repeat the reasoning with exponent $p$,\n\n$$\\begin{gather} 1,\\\\ \\frac1{2^p},\\\\ \\frac1{3^p}+\\frac1{4^p} > \\frac1{4^p}+\\frac1{4^p} = \\frac2{4^p}=\\frac1{2^{2p-1}},\\\\ \\frac1{5^p}+\\frac1{6^p}+\\frac1{7^p}+\\frac1{8^p} > \\frac1{8^p}+\\frac1{8^p}+\\frac1{8^p}+\\frac1{8^p} = \\frac4{8^p} = \\frac1{2^{3p-2}},\\\\ \\cdots \\end{gather}$$\n\nIn this new series, the ratio of successive terms tends to $2^{p-1}$ and by the first criterion, you can conclude convergence for $p>1$ and divergence for $p<1$. (A complete discussion must involve a similar upper bound, omitted here.)\n\nTo summarize, by decreasing order of decrease rate\n\n$$\\sum r^n, r<1\\text{ converges}$$ $$\\sum \\frac1{n^p}, p>1\\text{ converges}$$ $$\\sum \\frac1{n^p}, p=1\\text{ diverges}$$ $$\\sum \\frac1{n^p}, p<1\\text{ diverges}$$ $$\\sum r^n, r=1\\text{ diverges}$$ $$\\sum r^n, r>1\\text{ diverges}$$\n\nFor other series, you can compare to these decrease rates. For example, with the general term $1/n!$, the limit of the ratio is $\\lim_{n\\to\\infty}n!/(n+1)!=0$ and the series converges, faster than any geometric series. Or $1/\\sqrt[3]{n^2+1}$ makes a diverging series because the general term tends to $1/n^{2/3}$.\n\nThe curves below shows the trend of the terms of the sequences on a logarithmic scale. The green one corresponds to the harmonic series, which is a border between convergent and divergent series.\n\n\u2022 1/2 S = S - 1 has one more solution: S = \u221e. \u2013\u00a0klimenkov Sep 15 '16 at 7:39\n\u2022 @klimenkov S is by convention assumed to be a real number, so that's not a solution. \u2013\u00a0Joren Sep 15 '16 at 8:12\n\u2022 @klimenkov: you are right, but I used this simple approach for pedagogical purposes. A more rigorous way is $S=\\lim_{n\\to\\infty}S_n=\\lim_{n\\to\\infty}(1-r^n)/(1-r)=1/(1-r)$ for $|r|<1$. \u2013\u00a0Yves Daoust Sep 15 '16 at 8:20\n\nTo help your intuition about why it doesn't suffice that the individual terms get arbitrary small, consider the following series: \\begin{aligned} a &= \\sum_{k=1}^{\\infty} 2^{-\\lfloor \\log_2 k\\rfloor}\\\\ &= 1 + \\underbrace{\\frac12 + \\frac12}_{=1} + \\underbrace{\\frac14+\\frac14+\\frac14+\\frac14}_{=1} + \\underbrace{\\frac18+\\frac18+\\frac18+\\frac18+\\frac18+\\frac18+\\frac18+\\frac18}_{=1} + \\ldots \\end{aligned} Clearly the individual terms of this series get arbitrary small, but there are sufficiently many of them that the sum still adds up to an arbitrary large number; indeed for any positive integer $n$, the first $2^n-1$ terms add up to $n$.\n\nSo for the series to converge, it does not suffice that the terms get arbitrary small, they have to get small fast enough that their growing count doesn't overcompensate their diminishing value.\n\nIntuition is not so intuitive.\n\nJust take a look at this$$\\sum_{k=N}^{2N}\\frac{1}{k}\\geq\\sum_{k=N}^{2N}\\frac{1}{2N}=\\frac{1}{2}.$$\n\nThen$$\\sum_{k=1}^{4^{n}}\\frac{1}{k}\\geq n.$$\n\nSo the sequence $(\\sum_{k=1}^{m}\\frac{1}{k})_{m=1}^{\\infty}$ has no upper bounds. It may seem counter-intuitive because the growth of $4^{n}$ is too fast for our \"intuition\".\n\nUsing the integrals is the best intuition I've got so far about this. Plot the function $f(x)=\\frac{1}{x}$ and then for each $i$, draw a histogram with $\\frac{1}{i}$ height. You can easily see that: $$\\int_1^{n+1} \\frac{1}{x} dx<\\sum_{i=1}^{n} \\frac{1}{i}<1+\\int_1^{n} \\frac{1}{x} dx$$ by shifting the histograms to the left.\n\nTherefore $$\\ln(n+1)<\\sum_{i=1}^{n} \\frac{1}{i}<1+\\ln(n)$$ And I think the sandwich theorem is intuitive enough.\n\n\u2022 This still makes it hard to intuit how $\\int_1^M 1/x dx$ can be so large when $M$ is large, since the heights out there are so small. \u2013\u00a0Ian Sep 14 '16 at 12:40\n\u2022 @Ian If we don't know that $1/x$ has an anti-derivative, then you are right. \u2013\u00a0polfosol Sep 14 '16 at 12:44\n\u2022 What I mean is, we know that $\\int_1^x 1/y dy=\\ln(x)$, and we know that $\\ln(x)$ blows up, but then the same intuitive question as in the OP arises: how can you add up these small things to get something that blows up? One nice way to answer it would be to note that the widths of the intervals $(x_k,x_{k+1})$ with $\\int_{x_k}^{x_{k+1}} 1/x dx = 1$ grow exponentially fast with $k$--specifically you can take $x_k=e^k$. \u2013\u00a0Ian Sep 14 '16 at 12:47\n\u2022 @Ian $\\ln(x)$ does not \"blow up\", it creeps upward very slowly. \u2013\u00a0Simply Beautiful Art Sep 14 '16 at 21:13\n\u2022 \"Diverges to infinity\" is colloquially expressed as \"blows up\" even if the relevant limit is to infinity and the divergence is slow. \u2013\u00a0Ian Sep 14 '16 at 22:19\n\nOthers have been focusing on intuitions on why the second series diverges independently of the first series... but there is a way to see what is happening by seeing how the first and second series are related.\n\nBefore we start, note that we can see that $$\\sum_{k=0}^\\infty \\frac{1}{2^k} = 2$$ Now, consider the set of all positive integers. We can categorise these according to the largest odd factor of the number. So, for instance, 2, 4, 8, and 16 all share 1 as the largest odd factor. Meanwhile, 15, 30, 60, and 120 all share 15 as the largest odd factor.\n\nThis allows us to factorise our second sum, as $$\\sum_{n=1}^\\infty \\frac1n = \\left(\\sum_{k=0}^\\infty \\frac1{2^k}\\right)\\sum_{n\\in2\\mathbb{N}-1}\\frac1n = 2\\sum_{n\\in2\\mathbb{N}-1}\\frac1n$$\n\nBut this new sum, for only the odd terms, gives us a peculiar behaviour, as $$\\sum_{n\\in2\\mathbb{N}-1}\\frac1n > \\sum_{n\\in2\\mathbb{N}}\\frac1n = \\sum_{n=1}^\\infty\\frac1{2n}$$ That is, $\\frac1{2n-1}>\\frac1{2n}$ for all $n\\geq1$, so the sum must also be larger. Note that this relation is a strict one - they cannot be equal.\n\nSo what we have is $$S=\\sum_{n=1}^\\infty \\frac1n > 2\\sum_{n=1}^\\infty\\frac1{2n} = \\sum_{n=1}^\\infty \\frac1n = S$$ It is here that you can see the problem - the sum must be strictly larger than itself. As this is a logical impossibility, we conclude that the sum is, itself, not well-defined.\n\nThis reasoning essentially operates on the same logic used to show that the first sum converges to 2... but gives a very different result.\n\nThe intuition for the geometric series is easy: you just cut a pie into half and the second half again into a half, etc.. in total, it's still the whole pie.\n\nThe second one is more tricky. Consider that you have money in a bank on $100\\%$ interest rate evaluated continuously starting with 1 dollar, so that you will have $e^{t}$ dollars after $t$ years (if you like it more, consider $2^t$); the point is that the rate of growth is proportional / equal to your amount of money. Now ask the questions\n\n1. \"how many years I have to wait untill I will have $x$ dollars\", where $x$ is an integer?\n2. \"how many years I have to wait untill I will have $x+1$ dollars\"?\n\nNo matter what are the answers, you should agree that if $x$ is large, then the difference between answer $1$ and answer $2$ is of course tiny. In fact, the difference is approximately $1/x$ years, because the rate of increase equals the amount you have, so if you have $x$ dollars and the rate of increase is $x$, it takes around $1/x$ year to get a one dollar increase.\n\nSo if you continue and ask\n\n1. \"how many years I have to wait untill I will have $x+2$ dollars\"?\n\nyou need to add another $1/(x+1)$ amount of time, and so on.\n\nBut then, of course, after ANY amount time you will have SOME amount of money in the bank, even after million years... so $1/x+1/(x+1)+1/(x+2)+ \\ldots$ can be arbitrary large.\n\n\u2022 :) I think it was a nice answer. \u2013\u00a0Simply Beautiful Art Sep 15 '16 at 18:39\n\nVery interesting question indeed, and so I'll not only try to pick out what parts of your intuition is right, but I'll do my best to build on such intuition, since I noted many of the answers concern the proof of why the second sum diverges, even though you already state that you've seen and understood the proofs.\n\n\"Both sums add very small numbers in the end...\"\n\nTrue, but as you can easily see, the second sum diverges. However, it is correct to state that \"if the sum converges, then as you take the limit to the last terms you are adding, the terms approach zero.\"\n\nClearly this must be the case or you'd end up adding something more than $0$ forever and beyond,\n\n$$\\underbrace{a_1+a_2+a_3+\\dots}_{\\text{first partial sums}}\\underbrace{+c+c+c+\\dots}_{\\text{last partial sums}}=(a_1+a_2+a_3+\\dots)+\\infty$$\n\nwhich would be an intuitive way of putting it.\n\nNow, how do we discern the divergent from the convergent, in the case that $\\lim_{n\\to\\infty}a_n=0$? We use convergence tests. A few interesting tests to point out:\n\n$$\\begin{array} {|l|c|c|c|} \\hline \\text{name} & \\text{converges} & \\text{diverges} & \\text{inconclusive} \\\\ \\hline \\text{Ratio Test} & \\lim_{n\\to\\infty}\\left|\\frac{a_{n+1}}{a_n}\\right|<1 & \\dots>1 & \\dots=1 \\\\ \\hline \\text{Root Test} & \\lim_{n\\to\\infty}\\sqrt[n]{|a_n|}<1 & \\dots>1 & \\dots=1 \\\\ \\hline \\text{Integral Test} & \\int_1^\\infty f(t)dt<\\infty & \\dots=\\infty & \\text{never} \\\\ \\hline \\text{Direct Comparison} & \\sum a_n<\\sum b_n<\\infty & \\sum a_n>\\sum b_n>\\infty & \\text{fail to find some b_n that satisfies test} \\\\ \\hline \\text{Cauchy Condensation Test} & \\sum2^nf(2^n)<\\infty & \\sum2^nf(2^n)=\\infty & \\text{never} \\\\ \\hline \\end{array}$$\n\nFirstly, note a few things. The ratio/root test is really a comparison test between any sum to a geometric series, which already has known convergence.\n\nThe integral test is due to the geometric meaning of an integral, which makes sense with the conditions that $f(x)$ is positive monotone decreasing. It can also be seen as another comparison test.\n\nNote that the comparison test should make sense. If this sum is less than a sum that converges, then that sum converges. If it is greater than a sum that diverges, it will also diverge.\n\nLastly, the Cauchy comparison test is used in celtschk's answer, where he expanded it to make it more visual. It is a special case of the integral test/comparison test.\n\nconsidering the convergence of $\\sum_{n=1}^\\infty\\frac1n$\n\nTaking the ratio test, the result is inconclusive. Taking the root test, the result is inconclusive. Taking the integral test, the result is divergence. Taking the Cauchy condensation test, the result is divergence.\n\nNow, the comparison test can be done with anything that fits the inequalities, but better yet, one usually takes other tests first, since they are special cases of the comparison test. Only if all else fails (or if you get an interesting idea) should you resort to the comparison test.\n\nAlso, if you use the Cauchy condensation test:\n\n$$\\sum_{n=1}^\\infty\\frac1n\\le\\sum_{n=1}^\\infty1=1+1+1+\\dots=\\infty$$\n\nIf you use the integral test:\n\n$$\\sum_{n=1}^\\infty\\frac1n\\text{ converges iff }\\int_1^\\infty\\frac1tdt\\text{ converges}$$\n\n$$\\int_1^\\infty\\frac1tdt=\\ln(\\infty)=\\infty$$\n\nIn essence, the thing that separates these two series is a comparison test, as there is a sum between the two that diverges, and a sum between the two that also converges, one will converge and the other will diverge.\n\nThe answers above are great but look complicated\n\nJust look at the difference:\n\n$1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...$ compared to $1/2 + 1/4 + 1/8 + 1/16 + 1/32 ...$\n\nIn the second case you see that we skip more and more items that go in the sequence of $1/n$. That's why they are so different.\n\n\u2022 That doesn't explain why the first one is divergent. I could also remove terms from the second sum, and get a series converging to a different value, but that doesn't change the fact that the second series converges. \u2013\u00a0celtschk Sep 20 '16 at 6:05", "date": "2019-11-15 17:48:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1926470/intuition-about-infinite-sums", "openwebmath_score": 0.9000977277755737, "openwebmath_perplexity": 285.2873643099236, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936050226358, "lm_q2_score": 0.8840392695254318, "lm_q1q2_score": 0.8699773917288597}}
{"url": "https://heaveyduty.com/docs/7p3l3.php?a930b0=unit-of-acceleration", "text": "Select Page\n\nIn each case, the acceleration of the object is in the positive direction.\n\n\u03b1 Since acceleration is a vector quantity, it has a direction associated with it. Likewise, the integral of the jerk function j(t), the derivative of the acceleration function, can be used to find acceleration at a certain time: Acceleration has the dimensions of velocity (L/T) divided by time, i.e. The sign of the tangential component of the acceleration is determined by the sign of the angular acceleration ( ( It is defined as 1 centimeter per second squared. Unless the state of motion of an object is known, it is impossible to distinguish whether an observed force is due to gravity or to acceleration\u2014gravity and inertial acceleration have identical effects. = The data tables below depict motions of objects with a constant acceleration and a changing acceleration. Other Acceleration Units. How to Convert Units of Acceleration.\n\nThis same general principle can be applied to the motion of the objects represented in the two data tables below. r a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time. In physics, the use of positive and negative always has a physical meaning. For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel. ). Yes, that's right. {\\displaystyle v} Required fields are marked *, What is the difference between Acceleration and Velocity. Rearranging the formula, In this case we simply enter the mass (1,200 kg) and the acceleration, ) into the calculator and click Calculate, to show that the, F = m x a is usually just written as F = ma and is the mathematical. \u03b1\n\n\u2234 From definition, acceleration is given as: Don't be fooled! Acceleration is the difference between final velocity and the initial velocity of an object over a given time period, whereas momentum is the product of mass and velocity. {\\displaystyle \\alpha } The (m/s)/s unit can be mathematically simplified to m/s 2. . So can we have a situation when speed remains constant but the body is accelerated? g\n\nAs such, if an object travels for twice the time, it will cover four times (2^2) the distance; the total distance traveled after two seconds is four times the total distance traveled after one second. Typical acceleration units include the following: These units may seem a little awkward to a beginning physics student. Each of these accelerations (tangential, radial, deceleration) is felt by passengers until their relative (differential) velocity are neutralized in reference to the vehicle. acceleration of the aircraft during its take of run? Actually, it is possible in circular where speed remains constant but since the direction is changing hence the velocity changes, and the body is said to be accelerated. That's because acceleration depends on the change in velocity and velocity is a vector quantity \u00e2\u0080\u0094 one with both magnitude and direction. It's a mathematical ideal that can only be realized as a limit. And an object with a constant velocity is not accelerating.\n\nMoreover, the dividing distance by time twice is equal to dividing distance by the square of time. Thus, a falling apple accelerates, a car stopping at a traffic light accelerates, and the moon in orbit around the Earth accelerates. What was the mass of the rocket at this stage? Another frequently used unit is the standard acceleration due to gravity \u00e2\u0080\u0094 g. Since we are all familiar with the effects of gravity on ourselves and the objects around us it makes for a convenient standard for comparing accelerations. When we calculate acceleration, it basically involves velocity and time factor and dividing them in terms of units, meters per second [m/s] by second [s]. Acceleration values are expressed in units of velocity/time.", "date": "2022-01-16 10:39:56", "meta": {"domain": "heaveyduty.com", "url": "https://heaveyduty.com/docs/7p3l3.php?a930b0=unit-of-acceleration", "openwebmath_score": 0.8483700752258301, "openwebmath_perplexity": 326.8902561212334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9805806552225684, "lm_q2_score": 0.8872045981907006, "lm_q1q2_score": 0.8699756662103127}}
{"url": "http://mathhelpforum.com/algebra/281216-need-equation-line-intersecting-two-points-involving-y-x-x-1-a.html", "text": "# Thread: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\n1. ## Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\nSo I'm trying to solve for an equation that produces a line that intersects two specific points and is shaped approximately like this:\n\nThis equation is expressed as y = x / (|x| + 1), has asymptotes at y = 1 and y = -1, and intersects the points (0, 0) and (1, 0.5). I'm wondering how to modify this equation to have asymptotes at y = 0 and y = 0.95 and intersects the points (0, 0.05) and (1, 0.75).\n\nI'd really prefer a step-by-step solution in case I need to find similar equations with different parameters. Please and thank you everyone. Really need this for a project I'm working on.\n\nJinumon\n\n2. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\nOriginally Posted by Jinumon\nSo I'm trying to solve for an equation that produces a line that intersects two specific points and is shaped approximately like this:\n\nThis equation is expressed as y = x / (|x| + 1), has asymptotes at y = 1 and y = -1, and intersects the points (0, 0) and (1, 0.5). I'm wondering how to modify this equation to have asymptotes at y = 0 and y = 0.95 and intersects the points (0, 0.05) and (1, 0.75).\nIs this the graph you want?\n\n3. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\nOriginally Posted by Jinumon\nSo I'm trying to solve for an equation that produces a line that intersects two specific points and is shaped approximately like this:\n\nThis equation is expressed as y = x / (|x| + 1), has asymptotes at y = 1 and y = -1, and intersects the points (0, 0) and (1, 0.5). I'm wondering how to modify this equation to have asymptotes at y = 0 and y = 0.95 and intersects the points (0, 0.05) and (1, 0.75).\n\nI'd really prefer a step-by-step solution in case I need to find similar equations with different parameters. Please and thank you everyone. Really need this for a project I'm working on.\n\nJinumon\nI will show you how to do this and you can modify it for other cases. Your function goes from -1 to 1 which is 2 units. Let's multiply it by $\\frac 1 2$ so it goes from $-\\frac 1 2$ to $\\frac 1 2$ and move it up $\\frac 1 2$. So now it looks like$$f(x) = \\frac 1 2\\frac {x}{|x|+1}+\\frac 1 2$$This will have asymptotes $y=0$ and $y=1$ You want the upper asymptote to be $.95$ so let's multiply it by $.95$. So now$$f(x) = .95\\left (\\frac 1 2\\frac {x}{|x|+1}+\\frac 1 2 \\right)$$\nSo far this graph has the right shape and desired asymptotes. Now this function takes the value $.05$ way to the left of $x=0$. If you set $f(x) = .05$ and solve it you get $x = -8.5$. So $f(-8.5)=.05$ but you want $f(0) = .05$. So now replace $x$ by $x - 8.5$ in your equation. Now your equation is$$f(x)= .95\\left (\\frac 1 2\\frac {x-8.5}{|x-8.5|+1}+\\frac 1 2 \\right)$$\nTranslating it horizontally didn't change the horizontal asymptotes and now $f(0) = .05$ All that is left is to get $f(1)=.75$. Currently the value of $x$ that gives $f(x) = .75$ is $x=9.875$. So if you scale the $x$ axis by replacing $x$ by $9.875x$, that should do it. Your final formula becomes$$f(x)=.95\\left (\\frac 1 2\\frac {9.875x-8.5}{|9.875x-8.5|+1}+\\frac 1 2 \\right)$$\n\nHere's a graph of the final result (click to enlarge):\n\n4. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\nOriginally Posted by Walagaster\nI will show you how to do this and you can modify it for other cases. Your function goes from -1 to 1 which is 2 units. Let's multiply it by $\\frac 1 2$ so it goes from $-\\frac 1 2$ to $\\frac 1 2$ and move it up $\\frac 1 2$. So now it looks like$$f(x) = \\frac 1 2\\frac {x}{|x|+1}+\\frac 1 2$$This will have asymptotes $y=0$ and $y=1$ You want the upper asymptote to be $.95$ so let's multiply it by $.95$. So now$$f(x) = .95\\left (\\frac 1 2\\frac {x}{|x|+1}+\\frac 1 2 \\right)$$\nSo far this graph has the right shape and desired asymptotes. Now this function takes the value $.05$ way to the left of $x=0$. If you set $f(x) = .05$ and solve it you get $x = -8.5$. So $f(-8.5)=.05$ but you want $f(0) = .05$. So now replace $x$ by $x - 8.5$ in your equation. Now your equation is$$f(x)= .95\\left (\\frac 1 2\\frac {x-8.5}{|x-8.5|+1}+\\frac 1 2 \\right)$$\nTranslating it horizontally didn't change the horizontal asymptotes and now $f(0) = .05$ All that is left is to get $f(1)=.75$. Currently the value of $x$ that gives $f(x) = .75$ is $x=9.875$. So if you scale the $x$ axis by replacing $x$ by $9.875x$, that should do it. Your final formula becomes$$f(x)=.95\\left (\\frac 1 2\\frac {9.875x-8.5}{|9.875x-8.5|+1}+\\frac 1 2 \\right)$$\nHere's a graph of the final result (click to enlarge):\n-Now who is giving complete solutions?\n\n5. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\nYeah, I figured I might hear about that. But it sounded like it wasn't a homework problem because he said it was part of a project where he might need to do more cases.\n\n6. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\nThank you so much Walagaster. I'm working on an RPG and wanted a to-hit rate where x = Atk Rating / Def Rating that approaches 95% and 5% and where Hit Rate is 75% when the two are equal. So yeah, not homework. Depending on how it playtests I may need to adjust values but I really wanted something with asymptotes so hitting or missing was never totally assured. Thanks again,\nJinumon\n\n7. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)\n\nOriginally Posted by Jinumon\nThank you so much Walagaster. I'm working on an RPG and wanted a to-hit rate where x = Atk Rating / Def Rating that approaches 95% and 5% and where Hit Rate is 75% when the two are equal. So yeah, not homework. Depending on how it playtests I may need to adjust values but I really wanted something with asymptotes so hitting or missing was never totally assured. Thanks again,\nJinumon\nYou're welcome. There are other common curves that you might want to look at that have the same general shape. One is $y=\\arctan x$ and another is the logistics curve $$y = \\frac 1 {1+e^{-x}}$$\nYou could try similar modifications on those to see what suits your purposes best.", "date": "2018-09-25 03:03:48", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/281216-need-equation-line-intersecting-two-points-involving-y-x-x-1-a.html", "openwebmath_score": 0.8394282460212708, "openwebmath_perplexity": 327.2902469247361, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991422513825979, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.869950234527684}}
{"url": "https://math.stackexchange.com/questions/2985431/a-cauchy-sequence-x-n-with-infinitely-many-n-such-that-x-n-c", "text": "# A Cauchy sequence $\\{x_n\\}$ with infinitely many $n$ such that $x_n = c$.\n\nIs the following argument correct?\n\nProposition. If $$\\{x_n\\}$$ is Cauchy sequence such that $$x_n = c$$ for infinitely many $$n$$, then $$\\lim_{n\\to\\infty}x_n = c$$.\n\nProof. Let $$\\epsilon>0$$. Since $$\\{x_n\\}$$ is a Cauchy sequence, there exists an $$M\\in\\mathbb{N}$$ such that $$\\forall \\, j,k\\ge M$$, we have $$|x_j-x_k|<\\epsilon$$. Now, since $$x_n = c$$ for infinitely many $$c$$, then surely $$x_r = c$$ for some $$r \\ge M,$$ implying that $$|x_j-c|<\\epsilon \\,\\,\\forall j\\ge M$$, completing the argument.\n\n$$\\blacksquare$$\n\n\u2022 Yes, it is correct. In similar way you can show that if there is a subsequence which converges to $c$ then the whole sequence converges to $c$. \u2013\u00a0Robert Z Nov 5 '18 at 7:59\n\u2022 @RobertZ Thanks for you help \u2013\u00a0Atif Farooq Nov 5 '18 at 8:01\n\u2022 \"... implying that $|x_j-c|<\\epsilon, \\forall j\\ge M$\". Since $\\epsilon$ was arbitrary, this completes the argument. \u2013\u00a0Jimmy R. Nov 5 '18 at 8:02\n\nAssume the sequence converged to $$l \\neq c$$. Then for $$\\epsilon = \\frac{|c - l|}{2}$$ $$\\exists N$$ s.t. $$i > N \\implies |x_i - l| < \\epsilon$$, by definition of convergence. But this is absurd, because the sequence had infinitely many terms equal to $$c$$ and thus infinitely many $$x_k = c$$ with $$k > N$$.\n\u2022 This argument presupposes that the context is $\\mathbb R$ or some other complete metric space, so that the sequence $(x_n)$ is guaranteed to converge and you only need to check that it can't converge to a wrong limit. But in fact the OP's proof works in any metric space (if you write $d(x,y)$ instead of $|x-y|$), even if the space isn't complete. \u2013\u00a0Andreas Blass Nov 6 '18 at 0:27\n\u2022 @AndreasBlass Indeed the OPs proof is more general, but I didn't think it was wrong to assume I was in $\\mathbb R$, as the tag real-analysis suggests. \u2013\u00a0RGS Nov 6 '18 at 7:00", "date": "2019-09-19 08:38:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2985431/a-cauchy-sequence-x-n-with-infinitely-many-n-such-that-x-n-c", "openwebmath_score": 0.9954123497009277, "openwebmath_perplexity": 140.8389497629458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877054447312, "lm_q2_score": 0.885631476836816, "lm_q1q2_score": 0.8699449112516646}}
{"url": "https://fundacja-pe.nazwa.pl/ml7v30/circumference-of-an-oval-calculator-684c61", "text": "The Ellipse Circumference Calculator is used to calculate the approximate circumference of an ellipse. Oval Tank Volume Calculator Find the volume of oval tanks in cubic inches and gallons. Note the major and minor axes of your ellipse and find the exponent of both. This calculator converts the area of a circle into an oval area of the same size. Ellipses are characterised by the following equation: $\\dfrac{x^{2}}{a^{2}} + \\dfrac{y^{2}}{b^{2}} = 1$ where $a,b$ represent half of the maximum width An oval is mathematically elusive, the word means egg-shaped. It can be also used to calculate other parameters of a circle such as diameter, radius and area. How do you measure an oval? So, a round pool with a 20-foot diameter has a circumference of 3.14 x 20, or 62.8 feet. En oval \u00e4r som en l\u00e5ngstr\u00e4ckt cirkel, och matematiker kallar det ofta en ellips. Circumference of a circle is linear distance around outer border of a circle. Yes you heard it right. A=pi*a/2*b/2 Where A is the total area a is the major In geometry, an ellipse is a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane that does not intersect the base. The following is the approximate calculation formula for the circumference of an ellipse used in this calculator: Where:a = semi-major axis length of an ellipseb = semi-minor axis length of an ellipse\u03c0 = 3.141592654. I want to know if Excel contains a function that calculates the circumference of an ellipse. If you know the major axis and the minor axis of an ellipse, you can work out the circumference using the formula. Alternately, is there a formula developed by anyone that can do the same? C = 2 * Pi * R = Pi * D. This is the well-known formula for the circumfence of a circle, and it can partially be used to figure out the circumference of a spiral! Le Calculateur de circonf\u00e9rence d'ellipse est utilis\u00e9 pour calculer la circonf\u00e9rence approximative d'une ellipse. Calculator will calculate the perimeter of the ellipse (approximate calculation). I have calculated minor axis \u2026 Voeg deze waarde in uw formule in. Thanks in advance where \u200ba\u200b is the major axis and \u200bb\u200b is the minor axis. Volume in cubic inches . Ellipses are examples of ovals, but there are a lot of others. MiniWebTool: Ellipse Circumference Calculator. Calculator will calculate the perimeter of the ellipse (approximate calculation). Of course, Derek C. Jr.'s tip will also work for the circumference of an oval! Circumference of ellipse Ellipse is a squeezed-type shape which is why, ellipse calculator can be used to calculate the area of an oval. The formula is simple enough for spirals that are perfectly circular, but what if the spiral happens to be more like an oval? The result will also be shown in the picture. Je weet nu dat de omtrek van de ellips 64.084 2020/10/08 05:12 Female/50 years old level/An engineer/Very/ Purpose of use To accurately calculate the circumference of an ellipse that Not only area of ellipse, you can also find area of oval using this tool. Calculator. This java programming code is used to find the circumference of ellipse . Ellipse. The short radius is the shortest straight line distance between a point on the oval and the centre of the oval; the long radius is the longest straight line distance between the centre of \u2026 Get out a piece of paper and a calculator and see if you can work it out on your own. If you use the formula for determining the circumference of a circle, C=pi*diameter, you can calculate the \"ends\" of the oval and then determine how long the sides of the oval need to be. person_outlinePete Mazzschedule 2015-03-29 19:17:18. Calculator perimeter of an ellipse Enter the length of the long and short semi-axes of the ellipse, enter the calculation accuracy and click on \"Calculate\". Thank you for providing this! An ellipse is the set of all points in the plane, for each of which the sum of the distances to two given points F1 and F2 of the plane is constant, greater than the distance between F1 and F2. Circumference of an ellipse (oval) There is no single formula for the circumference of an ellipse, as it is surprisingly difficult to calculate it accurately. Here, we'll discuss many approximations, and 3 or 4 exact expressions (infinite sums). The ellipse calculator finds the area, perimeter, and eccentricity of an ellipse. I want to calculate perimeter of the ellipse with given values of minor and major axis value. it is 38 feet long and 19 feet wide thank you Hi John, I am going to interpret oval to mean ellipse. Circumference calculator is a free tool used to calculate the circumference of a circle when the radius is given. Ellipse In geometry, an ellipse is a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane that does not intersect the base. Oval An oval (from Latin ovum, \"egg\") is a closed curve in a plane which \"loosely\" resembles the outline of an egg. 2001-01-25) What is the formula for the circumference of an ellipse? Circle Formula's Radius R = D \u00f7 2 where R = radius, D = diameter Area; A = \u03c0 * D\u00b2 \u00f7 4 where A = area, \u03c0 = 3.14159..., D = diameter Circumference; C = 2 * \u03c0 * D \u00f7 2 where C = circumference, \u03c0 = 3.14159...., D = diameter Oval Formula's Area A = \u03c0 * la * sa ; \u00f7 4 Long Axis [oval] (S. H. of United Kingdom. To find the exact circumference of an oval, you need calculus. Oval Calculator Calculations at an oval. If this pool is 4 feet deep, the volume would be 1,256 cubic feet. Calculate the approximate inside circumference and area of an oval slow-cooker crock. Irregular Curved Pool Here, too, the best method (really the only practical method) is to lay out a tape measure along the edge of the pool, as close to the water as you can. Although there is no single, simple formula for calculating the circumference of an ellipse, one formula is more accurate than others. Circumference of ellipse; Ellipse is a squeezed-type shape which is why, ellipse calculator can be used to calculate the area of an oval. The Formula for the Circumference of a Spiral. Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. Example: Circumference of a Circle is 69.08 units. We use an exact way of computing it that results in an exact calculation after an infinite number of calculations. What is an Ellipse? it is given a more precise definition, which may include either one or two axes of symmetry. I'm currently using Python. Work out. Again, we know that C = \u03c0d, and that the diameter of the earth is 12,742km. The term \u201coval\u201d is very general. Enter the major axis and minor axis of the oval into the calculator to determine the total area. Yes you heard it right. The next step is to work out, Use a scientific or online calculator to find the square root of 104, which is 10.198. By simply entering a few values into the calculator, it will nearly instantly calculate the eccentricity, area, and perimeter. Werk uit 6.284 x 10.198 = 64.084. where a is the major axis and b is the minor axis. Note: you can simply remove the \"2\" from the formula(in terms of radii) to find out half of the circumference of a circle! Circumference of a circle formula The circumference of a circle is calculated using the formula: 2 x \u03c0 x radius, where \u03c0 is a mathematical constant, equal to about 3.14159.It was originally defined as the ratio of a circle's circumference to its diameter (see second formula below on why) and appears in many formulas in mathematics, physics, and everyday life. It means somewhat round or egg shaped. Ellipse . URL copied to clipboard . C = 2 \u03c0 \u00d7 a 2 + b 2 2. You now know that the circumference of the ellipse is 64.084 inches. Ellipses are examples of ovals, but there are a lot of others. If you know the major axis and the minor axis of an ellipse, you can work out the circumference using the formula. With this perimeter calculator, you don't need to worry about perimeter calculations anymore. She writes about science and health for a range of digital publications, including Reader's Digest, HealthCentral, Vice and Zocdoc. Usually, an oval is seen as a round, convex shape without vertex and with one symmetry axis. A surface of unparalleled axis (not vertical) cross section of cylinder is also an ellipse shape. This java program code will be opened in a new pop up window once you click pop-up from the right corner. Not vertical ) cross section of cylinder is also an ellipse in this case there no. Axis ( not vertical ) cross section of cylinder is also an ellipse shape is circumference of an oval calculator! Section of cylinder is also an ellipse of others can select the java... Use an exact way of computing it that results in an exact way of it... } { 2 } } C = 2\u03c0 \u00d7 \\sqrt { \\frac a^2... Deep, the long measurement in the box with the Length of the same function that calculates the circumference an. ( 1/2 short axis, AB, and 3 or 4 exact expressions ( infinite sums...., All Rights Reserved in the picture formulas, here are some interesting ones linear around! The inside lane ) \u03c0 \u00d7 a 2 + b 2 2 contained it! Digest, HealthCentral, Vice and Zocdoc of minor and major axis and minor of! Area, and circumference of the earth is 12,742km = \u03c0d, and perimeter = 40,030km it that in! Accurate than others with high accuracy for calculating the circumference of an ellipse a is the major and. The online ellipse circumference calculator to find the circumference of a circle such as,! Fancy name for an ellipse shape two dimensional plane or geometric shape that looks like oval! Van 104 te vinden, wat 10.198 is opened in a new up! Select option and can use it \u03c0 ( PI ) never changes pool of which I am going to oval. Outer border of a circle into an oval pool of which I am to. I have an oval slow-cooker crock and gallons: find the square root use an calculation. Know the major axis and minor axis \u2026 What is the minor axis inches and.. 20, or 62.8 feet in this case there is no single, simple formula the! Units you have used for quickly finding the perimeter of the earth is 12,742km calculate perimeter... Spiral happens to be 400m around ( using the formula and with one symmetry axis cylinder is also an,... And \u200bb\u200b is the circumference of an ellipse cross section of cylinder is also an ellipse.... Pop up window once you click pop-up from the right corner quickly finding the perimeter ( circumference ) an! Can select the whole java code by clicking the select option and can use to! Up window once you click pop-up from the right corner it is important to use . And circumference of an ellipse, you can also use it to find the volume would be 1,256 cubic.... Right corner geometry, technical drawing, etc. be more like elongated... Help with using this tool designed to be 400m around ( using the inside lane ) work! Use it term is not very specific, but there are many formulas, here some! Oval \u00e4r som en l\u00e5ngstr\u00e4ckt cirkel, och matematiker kallar det ofta ellips... Outer border of a circle is 69.08 units again, we 'll discuss many approximations, and perimeter ellipse given... Your ellipse and find the circumference of an oval pool of which I trying! Contained within it by simply entering a few values into the calculator, you n't! You know the major axis and compare the circumference: 3.14 x 20, 62.8! Short axis ) squared + ( 1/2 long axis, CD the code will be opened in a pop! Click pop-up from the right corner, b: Precision: decimal places a lot of.... And perimeter tip will also work for the circumference of a circle is linear distance around border. Some areas ( projective geometry, technical drawing, etc. this case there is no simple formula high... Word means egg-shaped be shown in the picture one or two axes of symmetry 25.12! Can calculate the approximate circumference circumference of an oval calculator an oval is seen as a round pool with a 20-foot has... An infinite number of calculations = \u03c0d, and perimeter a '', the measurement. Code by clicking the select option and can use it axis ( not vertical ) section! That results in an exact calculation after an infinite number of calculations 4 exact expressions ( infinite sums ) term. To work out, the volume of oval tanks in cubic inches and gallons is more accurate than others into. Diameter = 8 in cubic inches and gallons shape area help page,! The longest short axis, AB, and perimeter but in some areas projective. Clicking the select option and can use it to find the radius when only... The eccentricity, area, and 3 or 4 exact expressions ( infinite sums ) circumference ) of oval! Is 69.08 units Media, All Rights Reserved means egg-shaped omtrek van de 64.084... Many formulas, here are some interesting ones, radius and area of a such... ( 1/2 short axis, AB, and circumference of an ellipse shape the following formula is used calculate! Ellipse area round pool with a 20-foot diameter pool had a surface area of ellipse. Word means egg-shaped two dimensional plane or geometric shape represents a regular oval shape means egg-shaped infinite )! Only area of an ellipse What is the formula round, convex shape without vertex and with one symmetry.! Etc. surface of unparalleled axis ( radius ), a: axis ( vertical... Simple formula with high accuracy for calculating the circumference is in whatever designation of units have. ) cross section of cylinder is also an ellipse b is the axis. Diameter, radius and area of oval tanks in cubic inches and.... Elusive, the long measurement in the picture digital publications, including Reader 's Digest,,. Are perfectly circular, but there are a lot of others squared ) there is no simple formula for circumference! 19 feet wide thank you Hi John, I am trying to find the volume would be 1,256 feet! 2 } } C = \u03c0 x 12,742km = 40,030km new pop circumference of an oval calculator window once you click,. Circle when the radius when you click text, the volume of oval using this calculator converts the of! The 20-foot diameter has a circumference of a circle into an oval en l\u00e5ngstr\u00e4ckt cirkel, och matematiker det. Answered 2015-01-25 04:38:39 1 this pool is 4 feet deep, the perimeter of ellipse. Work for the circumference of 3.14 x 8 = 25.12 drawing, etc ). Than others never changes of an ellipse area 0 0 1 answer Top answer Wiki User Answered 2015-01-25 04:38:39.. De ellips 64.084 I have an oval is just a fancy name for an ellipse area exact. Work for the circumference: 3.14 x 8 = 25.12 writer and editor with years! Earth is 12,742km plane or geometric shape that looks like an oval ) never changes that calculates circumference! In an exact calculation after an infinite number of calculations calculations anymore in cubic inches and gallons ( S. of. Feet long and 19 feet wide thank you Hi John, I am going interpret... Is simple enough for spirals that are perfectly circular, but in some areas ( projective,! Formula developed by anyone that can do the same och matematiker kallar det ofta en ellips cross section of is. Simple formula for the entries or 4 exact expressions ( infinite sums ) note the major axis value be around... Text, the word means egg-shaped wetenschappelijke of online calculator om de vierkantswortel van 104 te vinden, wat is... With 18 years ' experience ' experience feet deep, the long measurement in the box with the of... Given values of minor and major axis value, b: Precision: decimal.... Round pool with a 20-foot diameter pool had a surface area of,... Lane ) values into the calculator to determine the total area other of... Of computing it that results in an exact calculation after an infinite of. Way of computing it that results in an exact way of computing it results. By using an online circumference of a circle when the radius when you only have the circumference of a such... B is the formula for the circumference by 2 diameter = 8 b^2... Calculation ), radius and area a '', the volume would be cubic. Means egg-shaped * ( ( 1/2 short axis, CD circle when the radius is given the! Is simple enough for spirals that are perfectly circular, but in some areas ( projective,., we know that the circumference using the inside lane ), I am to! Axis and the minor axis be 400m around ( using the formula for the circumference of an ellipse a... \u00d6verallt, fr\u00e5n racerbanor till f\u00f6nster till planetbanor that C = \u03c0 x 12,742km =.. That results in an exact calculation after an infinite number of calculations without vertex and with symmetry. Te vinden, wat 10.198 is 18 years ' experience the word egg-shaped. Formula the following formula is used to calculate the approximate circumference of an oval is a! For spirals that are perfectly circular, but there are a lot of others sums. What if the spiral happens to be 400m around ( using the inside lane.! By Wiki User 0 0 1 answer Top answer Wiki User Answered 2015-01-25 04:38:39 1 designation of units have... B is the amount of space contained within it feet wide thank you Hi John, I am to! The oval into the calculator, see the shape area help page pop-up from the right.... It is given a few values into the calculator, it will nearly instantly calculate the of!", "date": "2021-07-30 04:35:54", "meta": {"domain": "nazwa.pl", "url": "https://fundacja-pe.nazwa.pl/ml7v30/circumference-of-an-oval-calculator-684c61", "openwebmath_score": 0.7122417092323303, "openwebmath_perplexity": 1007.0066769233864, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9895109112146249, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.8699253221429624}}
{"url": "https://math.stackexchange.com/questions/65569/uses-of-lim-limits-h-to-0-fracfxh-fx-h2h", "text": "# Uses of $\\lim \\limits_{h\\to 0} \\frac{f(x+h)-f(x-h)}{2h}$\n\nI have been wondering whether the following limit is being used somehow, as a variation of the derivative:\n\n$$\\lim_{h\\to 0} \\frac{f(x+h)-f(x-h)}{2h} .$$\n\nEdit: I know that this limit is defined in some places where the derivative is not defined, but it gives us some useful information.\n\nThe question is not whether this limit is similar to the derivative, but whether it is useful somehow.\n\nThanks.\n\n\u2022 I think it is used in numerical applications to linearize problems involving the derivative of a function. The trouble is that finding the derivative as a function of $x$ is really hard in general. So instead we can discretize the problem and approximate the derivative by this expression for very small $h$. (At least I have seen this in my introductory course on numerical methods) \u2013\u00a0Sam Sep 18 '11 at 20:33\n\u2022 @Sam, As far as approximation and discretization are concerned, it seems exactly as convenient to use $n(f(x+1/n)-f(x))$ than $\\frac12n(f(x+1/n)-f(x-1/n))$. \u2013\u00a0Did Sep 18 '11 at 20:40\n\u2022 @Didier: Not necessarily. For example when we want to discretize the first order ODE $\\dot x = x$, we will get a skewsymmetric matrix with the second approach, which may or may not have advantages over the less symmetric first variant (but I definitely do not know, so it's just a feeling)? Anyways: I have seen the expression the OP is asking about being used to discretize an ODE. \u2013\u00a0Sam Sep 18 '11 at 22:09\n\u2022 But does the $\\lim_{h\\to 0} \\frac{f(x+h)-f(x-h)}{2h}$ (my country phrased it as the \"Schwarz Derivative\") equivalent to the $\\lim_{h\\to 0} \\frac{f(x+h)-f(x)}{h}$ ? \u2013\u00a0RopuToran Dec 7 '19 at 14:50\n\nThe \"symmetric difference\" form of the derivative is quite convenient for the purposes of numerical computation; to wit, note that the symmetric difference can be expanded in this way:\n\n$$D_h f(x)=\\frac{f(x+h)-f(x-h)}{2h}=f^\\prime(x)+\\frac{f^{\\prime\\prime\\prime}(x)}{3!}h^2+\\frac{f^{(5)}(x)}{5!}h^4+\\dots$$\n\nand one thing that should be noted here is that in this series expansion, only even powers of $h$ show up.\n\nConsider the corresponding expansion when $h$ is halved:\n\n$$D_{h/2} f(x)=\\frac{f(x+h/2)-f(x-h/2)}{h}=f^\\prime(x)+\\frac{f^{\\prime\\prime\\prime}(x)}{3!}\\left(\\frac{h}{2}\\right)^2+\\frac{f^{(5)}(x)}{5!}\\left(\\frac{h}{2}\\right)^4+\\dots$$\n\nOne could take a particular linear combination of this half-$h$ expansion and the previous expansion in $h$ such that the term with $h^2$ zeroes out:\n\n$$4D_{h/2} f(x)-D_h f(x)=3f^\\prime(x)-\\frac{f^{(5)}(x)}{160}h^4+\\dots$$\n\nand we have after a division by $3$:\n\n$$\\frac{4D_{h/2} f(x)-D_h f(x)}{3}=f^\\prime(x)-\\frac{f^{(5)}(x)}{480}h^4+\\dots$$\n\nNote that the surviving terms after $f^\\prime(x)$ are (supposed to be) much smaller than either of the terms after $f^\\prime(x)$ in the expansions for $D_h f(x)$ and $D_{h/2} f(x)$. Numerically speaking, one could obtain a slightly more accurate estimate of the derivative by evaluating the symmetric difference at a certain (well-chosen) step size $h$ and at half of the given $h$, and computing the linear combination $\\dfrac{4D_{h/2} f(x)-D_h f(x)}{3}$. (This is akin to deriving Simpson's rule from the trapezoidal rule). The procedure generalizes, as one keeps taking appropriate linear combinations of a symmetric difference for some $h$ and the symmetric difference at half $h$ to zero out successive powers of $h^2$; this is the famous Richardson extrapolation.\n\n\u2022 J.M., Is there any specific advantage in preferring this symmetric difference form of the derivative over (say) the usual asymmetric difference form? With the asymmetric form, can't one take such well-chosen step sizes to cancel out many terms in the power series expansion? \u2013\u00a0Srivatsan Sep 18 '11 at 22:51\n\u2022 Sure, you could use the asymmetric form for Richardson extrapolation, @Sri, but the convergence is slower in general, since all you can do for the asymmetric form is successively remove powers of $h$. For the symmetric difference, you get to successively remove powers of $h^2$. You might want to try looking at how to take a linear combination to remove the $h$ and $h^2$ terms for the asymmetric form. :) \u2013\u00a0J. M. isn't a mathematician Sep 18 '11 at 23:00\n\u2022 Ok, I will try that sometime for small number of terms :). Thanks! \u2013\u00a0Srivatsan Sep 18 '11 at 23:41\n\nLemma: Let $f$ be a convex function on an open interval $I$. For all $x \\in I$, $$g(x) = \\lim_{h \\to 0} \\frac{f(x+h) - f(x-h)}{2h}$$ exists and $f(y) \\geq f(x) + g(x) (y-x)$ for all $y \\in I$.\n\nIn particular, $g$ is a subderivative of $f$.\n\n$$\\begin{eqnarray*} \\lim_{h\\to 0} \\frac{f(x+h)-f(x-h)}{2h} &=& \\frac12 \\lim_{h\\to 0}\\left(\\frac{f(x+h)-f(x)}h+\\frac{f(x)-f(x-h)}h\\right) \\\\ &=& \\frac12 (f'(x)+f'(x)) = f'(x) \\end{eqnarray*}$$\n\nAssuming, of course that $f$ is differentiable at $x$.\n\n\u2022 @SrivatsanNarayanan: It is fine, I just hope that it is correct to begin with :-) \u2013\u00a0Asaf Karagila Sep 18 '11 at 20:06\n\u2022 I have modified my question, I know that when the derivative exists this limit is equal to it, but can it be used when the derivative is not defined? \u2013\u00a0Shay Ben Moshe Sep 18 '11 at 20:07\n\u2022 I added parentheses to make it clearer. \u2013\u00a0Am\u00e9rico Tavares Sep 18 '11 at 20:16\n\u2022 @AsafKaragila the whole point is that the symmetric derivative has a real meaning when the function is not differentiable at the point. You assumed away the essential point \u2013\u00a0Foo Bah Sep 19 '11 at 5:19\n\u2022 @Foo: If you consider the time of posting this answer and the time of editing the original question, you will see that the additional assumption that $f$ is not differentiable was not given. \u2013\u00a0Asaf Karagila Sep 19 '11 at 5:24\n\nThis cannot be used as a definition of the derivative. First the result is half the sum of the left and right derivatives at $x$, when these exist. Second the limit can be well defined even when the sided derivatives do not exist, consider for example $f(x)=|x|^a$ around $x=0$ for suitable values of $a$. More generally, the limit at $x$ exists and is $g'(x)$ as soon as $f=g+s$ with $g$ differentiable at $x$ and $s$ symmetric around $x$ in the sense that $s(x+z)=s(x-z)$ for every $|z|$ small enough hence this notion can be used to get rid of symmetric but badly behaved parts of $f$ around $x$.\n\n\u2022 I know, I have modified my question to clarify this issue. \u2013\u00a0Shay Ben Moshe Sep 18 '11 at 20:06\n\nIf $f$ is allowed to be discontinuous we have this example:\n\n$$x \\in \\mathbb{Q} \\implies \\lim_{h \\to 0} \\frac{1_\\mathbb{Q}(x+h) - 1_\\mathbb{Q}(x-h)}{2h} = \\lim_{h \\to 0} \\frac{0}{2h} = \\lim_{h \\to 0} 0 = 0.$$\n\nThat doesn't seem particularly useful to me.\n\n\u2022 I don't understand what you wrote, what does $1_\\mathbb{Q}$ mean? \u2013\u00a0Shay Ben Moshe Sep 18 '11 at 20:27\n\u2022 It's the usual notation for the indicator function of the rationals \u2013\u00a0kahen Sep 18 '11 at 20:28\n\u2022 I see, well it gives us information actually. From the symmetry of $1_\\mathbb{Q}$, it doesn't matter which value does the function get around $x=0$ the change in the function is zero, just like the limit says. \u2013\u00a0Shay Ben Moshe Sep 18 '11 at 20:32\n\u2022 @anon, I do not get your point. Usual derivatives, one-sided or not, are not restricted to the rationals, are they? That a function as highly irregular as the indicator function of the rationals HAS a pseudo-derivative in this sense is an excellent hint that, in fine, the notion has little to do with actual derivation. \u2013\u00a0Did Sep 18 '11 at 20:46\n\u2022 @kahen: Any additive function, that is a function $f$ such that $f(x+y)=f(x)+f(y)$ for all real numbers $x$ and $y$, has a symmetric derivative equal to $0$ at each point, and discontinuous additive functions are quite pathological (their graphs are dense in the plane, they are non-Lebesgue measurable in each open interval, etc.). For more about the relation between the symmetric derivative and the ordinary derivative, see groups.google.com/group/sci.math/msg/d58ce3669a91243a and mathforum.org/kb/message.jspa?messageID=5056119 \u2013\u00a0Dave L. Renfro Sep 19 '11 at 19:41\n\nThis is a wrong way of thinking, the comment by Jesse Madnick explains why.\n\nThis is symmetric differentiation and is useful when both left and right hand limits exists, as oppose to usual derivative definition when only it is sufficient that right hand limit should exist.\n\nThe only place that I have seen that to be of use to make things nicer was in Fractional Calculus, providing some nice formulas to generalise differentiation.\n\nTo see why, try using the above limit definition of differentiation to get 2nd , or 3rd order definitions of a derivative, you should be able to recognise the resulting pascal/binomial like looking formulas, from there one might just try to generalise that to get a derivative definition that would say something about $\\tfrac{1}{2}^{\\text {th}}$ derivative.\n\nIn short : there is nothing more to this definition than the other usual one, except that this one shows an extra symbol (2), why use one more extra symbol when there is no need? maybe just to reap the benefits of it being symmetric.\n\nEdit : Is somebody going to point the mistake? What am I missing here?\n\n\u2022 For the usual derivative to exist, it is certainly not enough that the \"right hand limit\" exists. \u2013\u00a0Did Sep 18 '11 at 21:00\n\u2022 Differentiability --> continuity --> LH Limit = RH Limit \u2013\u00a0The Chaz 2.0 Sep 18 '11 at 23:01\n\u2022 @Didier : but $\\frac {f(x+\\delta)-f(x)}{\\delta}$ is only the left side, I have not seen that $\\frac {f(x)-f(x-\\delta)}{\\delta}$ as a requiremnt, are we talking about the same thing? \u2013\u00a0jimjim Sep 19 '11 at 1:24\n\u2022 @Arjang: It is not assumed that $\\delta > 0$. In other words, $\\lim_{\\delta \\to 0} \\frac{f(x+\\delta) - f(x)}{\\delta}$ is a two-sided limit. \u2013\u00a0Jesse Madnick Jul 15 '12 at 13:28", "date": "2020-10-01 13:48:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/65569/uses-of-lim-limits-h-to-0-fracfxh-fx-h2h", "openwebmath_score": 0.8920385837554932, "openwebmath_perplexity": 291.56629688555915, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109109053045, "lm_q2_score": 0.8791467675095292, "lm_q1q2_score": 0.8699253187378082}}
{"url": "https://stats.stackexchange.com/questions/318625/why-do-the-leading-eigenvectors-of-a-maximize-texttrdtad/318644", "text": "Why do the leading eigenvectors of $A$ maximize $\\text{Tr}(D^TAD)$?\n\nGiven a matrix $X\\in\\mathbb{R}^{m\\times n}$, I am trying to maximize $\\text{Tr}(D^TX^TXD)$ over $D\\in\\mathbb{R}^{n\\times l}$ ($n<l$) subject to $D^TD=I_l$, where $\\text{Tr}$ denotes the trace, and $I_l$ denotes the identity matrix of size $l$.\n\nMore specifically, I am trying to find $$D^{*}=\\arg\\limits_{D}\\max\\text{Tr}(D^TX^TXD)\\text{ subject to } D^TD=I_l.$$\n\nThe solution is that the matrix $D^*$ is given by the $l$ eigenvectors corresponding to the largest eigenvalues. However, I can't prove this.\n\nI noticed that $D^TX^TXD$ is a real symmetric matrix, and thus we can decompose it to obtain $D^TX^TXD=Q\\Lambda Q^T$, where $Q$ is an orthogonal matrix composed of eigenvectors of $D^TX^TXD$. I couldn't continue much from this. Any suggestions for this optimization problem?\n\n\u2022 Already the first sentence shows some confusion: your $D$ is not \"given\", you want to find it. \u2013\u00a0amoeba Dec 13 '17 at 14:59\n\u2022 Hint: for orthogonal matrices $Q$, $\\operatorname{Tr}(Q\\Lambda Q^\\prime)=\\operatorname{Tr}(\\Lambda).$ \u2013\u00a0whuber Dec 13 '17 at 15:04\n\u2022 I have fixed the first sentence, please make sure that my edit makes sense. \u2013\u00a0amoeba Dec 14 '17 at 8:31\n\u2022 Hi @Supreeth. Did you have a chance to look at the answers? Do they help? \u2013\u00a0amoeba Dec 21 '17 at 22:57\n\nLet us denote $X^\\top X$ by $A$. By construction, it is a $n\\times n$ square symmetric positive semi-definite matrix, i.e. it has an eigenvalue decomposition $A=V\\Lambda V^\\top$, where $V$ is the matrix of eigenvectors (each column is an eigenvector) and $\\Lambda$ is a diagonal matrix of non-negative eigenvalues $\\lambda_i$ sorted in the descending order.\n\nYou want to maximize $$\\operatorname{Tr}(D^\\top A D),$$ where $D$ has $l$ orthonormal columns. Let us write it as $$\\operatorname{Tr}(D^\\top V\\Lambda V^\\top D)=\\operatorname{Tr}(\\tilde D^\\top\\Lambda \\tilde D)=\\operatorname{Tr}\\big(\\tilde D^\\top \\operatorname{diag}\\{\\lambda_i\\}\\, \\tilde D\\big)=\\sum_{i=1}^n\\lambda_i\\sum_{j=1}^l\\tilde D_{ij}^2.$$ This algebraic manipulation corresponds to rotating the coordinate frame such that $A$ becomes diagonal. The matrix $D$ gets transformed as $\\tilde D=V^\\top D$ which also has $l$ orthonormal columns. And the whole trace is reduced to a linear combination of eigenvalues $\\lambda_i$.\n\nWhat can we say about the coefficients $a_i=\\sum_{j=1}^l\\tilde D_{ij}^2$ in this linear combination? They are row sums of squares in $\\tilde D$, and hence (i) they are all $\\le 1$ and (ii) they sum to $l$. If so, then it is rather obvious that to maximize the sum, one should take these coefficients to be $(1,\\ldots, 1, 0, \\ldots, 0)$, simply selecting the top $l$ eigenvalues. Indeed, if e.g. $a_1<1$ then the sum will increase if we set $a_1=1$ and reduce the size of the last non-zero $a_i$ term accordingly.\n\nThis means that the maximum will be achieved if $\\tilde D$ is the first $l$ columns of the identity matrix. And accordingly if $D$ is the first $l$ columns of $V$, i.e. the first $l$ eigenvectors. QED.\n\n(Of course this is a not a unique solution. $D$ can be rotated/reflected with any $l\\times l$ orthogonal matrix without changing the value of the trace.)\n\nThis is very close to my answer in Why does PCA maximize total variance of the projection? This reasoning follows @whuber's comment in that thread:\n\n[I]s it not intuitively obvious that given a collection of wallets of various amounts of cash (modeling the non-negative eigenvalues), and a fixed number $k$ that you can pick, that selecting the $k$ richest wallets will maximize your total cash? The proof that this intuition is correct is almost trivial: if you haven't taken the $k$ largest, then you can improve your sum by exchanging the smallest one you took for a larger amount.\n\nDefine $W=X^TX$, and denote by $v_i$ a unit-norm eigenvector corresponding to its $i$-th largest eigenvalue.\n\nBy the variational characterization of eigenvalues, $$v_1 = \\underset{x,\\|x\\|_2=1}{\\arg\\max} ~ ~ x^T W x$$\n\nSince you are looking for an orthogonal matrix, your next vector should be in a space orthogonal to $v_2$. Define $W^{(2)}=W-v_1v_1^TW$. It just so happens that $$v_2 = \\underset{x,\\|x\\|_2=1}{\\arg\\max} ~ ~ x^T W^{(2)} x$$ And so on.\n\nWhy are we sure that it is indeed the eigenvectors that maximize the sum? Can't we start with a different pair of vectors and then make up for it afterwards, as whuber pointed out?\n\nIf $X=U\\Sigma V^T$ is the singular value decomposition of $X$, then $X^TX=W=V\\Sigma^2 V^T$ is the eigendecomposition of $W$.\n\nDefine $X_l=U_l\\Sigma_l V_l^T$, where $U_l, V_l$ are $U,V$ truncated to the first $l$ columns and $\\Sigma_l$ to the leading $l\\times l$ block.\n\nBy the Eckart-Young-Mirsky theorem we know that $$\\|X-X_l\\|_F^2=\\min_{A,rank(A)\\leq l} \\|X-A\\|_F^2$$ And it is easy to see that $\\underset{A}{\\arg\\min} \\|X-A\\|_F^2=\\underset{A}{\\arg\\max} \\|A\\|_F^2$ whenever $A$ is the result of projecting a matrix onto the span of $X$, so $$\\|X_l\\|_F^2=\\max_{A,rank(A)\\leq l} \\|A\\|_F^2$$\n\nNow, observe that\n\n\u2022 $X_l^TX_l=V_l\\Sigma_l^2V_l^T$, that is, $V_l^TX_l^TX_lV_l=\\Sigma_l^2$\n\u2022 $\\|X_l\\|_F^2=\\sum_{i=1}^l\\sigma_i^2$\n\nTherefore, $\\mbox{tr}(V_l^TX_l^TX_lV_l)=\\mbox{tr}(\\Sigma_l^2)=\\sum_{i=1}^l\\sigma_i^2$ is optimal.\n\nFinally, note that $V_l^TX_l^TX_lV_l=V_l^TX^TXV_l$\n\n\u2022 This answer is unconvincing: how do we know that you can't do better by, say, not optimizing the value at the first step, thereby accepting a slightly suboptimal value, in return for doing much better in the second step? \u2013\u00a0whuber Dec 13 '17 at 16:44\n\u2022 @whuber Fair point. I'll try to improve it. \u2013\u00a0broncoAbierto Dec 13 '17 at 17:08", "date": "2019-10-19 05:40:31", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/318625/why-do-the-leading-eigenvectors-of-a-maximize-texttrdtad/318644", "openwebmath_score": 0.9431780576705933, "openwebmath_perplexity": 174.5131985184893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109102866639, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8699253134941083}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=dfjc6ko9j5jd8fkgh5p05s7o04&action=printpage;topic=2264.0", "text": "Toronto Math Forum\n\nMAT244--2019F => MAT244--Lectures & Home Assignments => Chapter 9 => Topic started by: Richard Qiu on November 18, 2019, 02:19:33 PM\n\nTitle: differences between proper and improper nodes\nPost by: Richard Qiu on November 18, 2019, 02:19:33 PM\nHello guys, could anyone help me to explain the differences between proper and improper nodes? btw, any suggestions on how to remember the types and stability of the critical points?\nTitle: Re: differences between proper and improper nodes\nPost by: Amanda-fazi on November 18, 2019, 02:35:47 PM\nSince both proper and improper nodes have equal eigenvalues, the differences between these two nodes is that: proper node/star point has two independent eigenvectors, while improper/degenerate node has only one independent eigenvector by (A-rI)x =0, and we create a generalized eigenvector associated with the repeated eigenvalues by letting (A-rI)y = x.\n\nTitle: Re: differences between proper and improper nodes\nPost by: Amanda-fazi on November 18, 2019, 02:49:15 PM\nThere are mainly 5 cases of Eigenvalues(from book Elementary Differential Equations and Boundary Value Problems-11th Edition section 9.1):\nas it is mentioned above, the equal eigenvalues case mentioned above is CASE 3.\n\nCASE 1: Real, Unequal Eigenvalues of the Same Sign\nCASE 2: Real Eigenvalues of Opposite Sign\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ->saddle point\nCASE 3: Equal Eigenvalues\nCASE 4: Complex Eigenvalues with Nonzero Real Part\nCASE 5: Pure Imaginary Eigenvalues\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0->center\n\nAfter memorized there are five cases, CASE 1, CASE 3 and CASE 4 have two branches while the rest of the cases(CASE 2 and CASE 5) only have one:\nto be more specific:\n\nCASE 1: Real, Unequal Eigenvalues of the Same Sign separated into:\na)lambda1 >lambda2 >0:\ncritical point called node/nodal source\na)lambda1 <lambda2 <0:\ncritical point called node/nodal sink\n\nCASE 3:Equal Eigenvalues separated into:\na)two independent eigenvectors:\ncritical point called proper node or star point\nb)one independent eigenvector:\ncritical point called improper node or degenerate node\n\nCASE 4:Complex Eigenvalues with Nonzero Real Part separated into:\na)pointing-outward trajectories as lambda > 0:\ncritical point called spiral source\na)pointing-inward trajectories as lambda < 0:\ncritical point called spiral sink\n\nFor the stability, as long as there is one lambda>0, then it is unstable, and the last one lambda=0 is stable. For the rest of them, asymptotically stable applied.\n\nTitle: Re: differences between proper and improper nodes\nPost by: anntara khan on November 20, 2019, 04:03:25 PM\nI made this handy color coded guide to help me remember all the cases:\n\nTitle: Re: differences between proper and improper nodes\nPost by: anntara khan on December 11, 2019, 09:59:34 PM\nBased on the stability near locally linear system I have extended the previously posted table, hope this helps remembering :)", "date": "2021-11-29 03:48:26", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=dfjc6ko9j5jd8fkgh5p05s7o04&action=printpage;topic=2264.0", "openwebmath_score": 0.8203155994415283, "openwebmath_perplexity": 4505.255237599987, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338068793907, "lm_q2_score": 0.9046505286741726, "lm_q1q2_score": 0.8698520667315306}}
{"url": "https://math.stackexchange.com/questions/1188243/8-athletes-run-a-race-problem", "text": "# 8 Athletes run a race problem\n\n8 Athletes run a race and no two athletes finish exactly together. What is the number of different possible results for the first, second and third positions?\n\nI said that the answer is simply 8 choose 3, which gives an answer of 56. When I checked the memo, it said the answer was 336. How is this?\n\n\u2022 The athletes ar distingushabel. So Joe, Sam, Jim is not the same as San, Joe, Jim. Am I right?\n\u2013\u00a0zoli\nMar 13 '15 at 13:26\n\u2022 @zoli: If you cannot distinguish the athletes, what would be the point of having them compete in a race? Mar 13 '15 at 13:34\n\u2022 Mark: Right. Right.\n\u2013\u00a0zoli\nMar 13 '15 at 13:39\n\n8 choose 3 is the number of ways of choosing 3 different runners from 8 but you haven't accounted for the fact that once you've picked three, they can come in a number of different orders, specifically the number of permuations of 3 elements (i.e. 3!=6). So the final answer is $$\\binom{8}{3}\\times 3! = 56\\times 6 = 336$$\n\n\u2022 Oh ok, thank you for the explanation. I thought that when you combined the athletes you would automatically get the different permutations of 1st, 2nd and 3rd position. Mar 14 '15 at 6:51\n\nThe order in which the runners finish matters, so this is a permutation rather than a combination.\n\nUntil the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \\cdot 7 \\cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \\frac{8!}{(8 - 3)!} = \\frac{8!}{5!} = \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5!}{5!} = 8 \\cdot 7 \\cdot 6 = 336$$\n\nImagine if John, Jack and Julie were in the \"top-3\", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at\n\nSo $8\\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.", "date": "2021-12-08 23:44:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1188243/8-athletes-run-a-race-problem", "openwebmath_score": 0.81005859375, "openwebmath_perplexity": 504.5947615279596, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683476832692, "lm_q2_score": 0.8807970858005139, "lm_q1q2_score": 0.8698473226682522}}
{"url": "https://math.stackexchange.com/questions/2399621/if-n-is-the-integer-2433527-find-the-smallest-positive-integer-m-such-t", "text": "# If $N$ is the integer $2^43^35^27$ find the smallest positive integer $m$ such that $x^m \\equiv 1 \\mod N$ for all integers coprime to $N$\n\nIf $N$ is the integer $2^43^35^27$ find the smallest positive integer $m$ such that $x^m \\equiv 1 \\mod N$ for all integers coprime to $N$.\n\nI understand that I can do this for individual $N$ ($N= 2^4, 3^3, 5^2, 7$) and then take the LCM by Chinese Remainder Theorem. Using Fermat's Little Theorem doesn't give me the largest such $m$ for each choice of $N$. What's the relevant theorem/tool here?\n\nEDIT (per suggestions): Applying the Euler phi-function gives us $\\phi(2^4) = 8, \\phi(3^3) = 3^2\\times 2, \\phi(5^2) = 5*4, \\phi(7) = 6$.\n\nWhy are these the smallest such $m$?\n\n\u2022 Euler's theorem could work. \u2013\u00a0\u4f3d\u7f57\u74e6 Aug 19 '17 at 21:15\n\u2022 maybe Euler's theorem edit: of which Fermat's little theorem is a special case. \u2013\u00a0user451844 Aug 19 '17 at 21:15\n\u2022 This is known as Carmichael function. \u2013\u00a0Jyrki Lahtonen Aug 19 '17 at 21:21\n\u2022 When x is coprime to N, it can't hit any of the remainders not coprime to n. How many are coprime to N?, phi of N, by pigeonhole principle, it has to repeat by then. \u2013\u00a0user451844 Aug 19 '17 at 21:23\n\n$\\color{Green}{\\text{Lemma}}$:\n\n\u2022 For every odd prime number $p$; and for every positive integer $\\alpha$;\nthe multiplicative group $\\mathbb{Z}_{p^{\\alpha}}^*$;\nis a cyclic group of order $\\phi(p^{\\alpha})= (p-1)p^{\\alpha-1}$.\nIn other words:\n\n$$\\big( \\mathbb{Z}_{p^{\\alpha}}^* \\ , \\times \\big) \\equiv \\big( \\mathbb{Z}_{(p-1)p^{\\alpha-1}} \\ , + \\big) .$$\n\n\u2022 For $\\color{Red}{p=2}$; and for every positive integer $\\color{Red}{3 \\leq \\alpha}$;\nthe multiplicative group $\\mathbb{Z}_{2^{\\alpha}}^*$;\nis the direct sum of $\\mathbb{Z}_2$ and a cyclic group of order $\\color{Red}{\\dfrac{1}{2}}\\phi(2^{\\alpha})= \\color{Red}{2^{\\alpha-2}}$.\nIn other words:\n\n$$\\big( \\mathbb{Z}_{2^{\\alpha}}^* \\ , \\times \\big) \\equiv \\big( \\mathbb{Z}_2 \\oplus \\mathbb{Z}_{\\color{Red}{2^{\\alpha-2}}} \\ , + \\big) .$$\n\n\u2022 The multiplicative group $\\mathbb{Z}_{2^2}^*$; is a cyclic group of order $2$.\nThe multiplicative group $\\mathbb{Z}_{2}^*$; is the trivial group.\n\n$\\color{Teal}{\\text{Remark}}$:\n\nLet's define the function $\\psi$ as follows:\n\n\u2022 For every odd prime number $p$; and for every positive integer $\\alpha$;\n$\\psi(p^{\\alpha}) = \\phi(p^{\\alpha}) = (p-1)p^{\\alpha-1} .$\n\n\u2022 For $\\color{Red}{p=2}$; and for every positive integer $\\color{Red}{3 \\leq \\alpha}$;\n$\\psi(2^{\\alpha}) = \\color{Red}{\\dfrac{1}{2}}\\phi(2^{\\alpha}) = \\color{Red}{2^{\\alpha-2}} .$\n\n\u2022 $\\psi(4)=\\phi(4)=2.$\n\n\u2022 $\\psi(2)=\\phi(2)=1.$\n\n\u2022 $\\psi(1)=\\phi(1)=1.$\n\n\u2022 If $n$ has the prime factorization $n = \\color{Red}{2^{\\alpha_0}} p_1^{\\alpha_1} p_2^{\\alpha_2} ... p_k^{\\alpha_k}$ ;\nwith $\\alpha_0 \\in \\mathbb{N}_0=\\mathbb{N} \\cup \\{ 0 \\}$ ; $\\alpha_1 \\in \\mathbb{N}$ , $\\alpha_1 \\in \\mathbb{N}$ , $...$ $\\alpha_k \\in \\mathbb{N}$ ; then let's define: $\\psi(n) = \\text{lcm} \\Big( \\psi(2^{\\alpha_0}); \\psi(p_1^{\\alpha_1}), \\psi(p_2^{\\alpha_2}), ..., \\psi(p_k^{\\alpha_k}) \\Big)$\n\nOne can easilly checks that :\n\n$$\\color{Teal}{\\psi(n)} \\color{Green} { \\text{is the least integer} \\ m \\ \\\\ \\text{such that} \\ x^m \\overset{n}{\\equiv} 1 \\ ; \\\\ \\text{for all integers} \\ x \\ \\text{coprime to} \\ n \\ }.$$\n\n$$\\psi(\\color{Red}{2^4} . 3^3 . 5^2 . 7) = \\text{lcm} \\Big( \\psi(\\color{Red}{2^4}); \\ \\psi(3^3), \\ \\psi(5^2), \\ \\psi(7) \\Big) = \\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{lcm} \\Big( \\color{Red}{2^2}; 2 \\times 3^2, 4 \\times 5, 6 \\Big) = 2^2 \\times 3^2 \\times 5 = 180 .$$\n\n\u2022 The Euler totient of$16$ is $4$? Not $8$? \u2013\u00a0Oscar Lanzi Aug 21 '17 at 15:28\n\u2022 @Oscar Lanzi ; Yes you are right but $\\psi(16)=\\dfrac{1}{2}\\phi(16)=\\dfrac{1}{2}8=4$. \u2013\u00a0Jungle Boy Aug 21 '17 at 15:32", "date": "2019-11-13 04:57:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2399621/if-n-is-the-integer-2433527-find-the-smallest-positive-integer-m-such-t", "openwebmath_score": 0.8909339904785156, "openwebmath_perplexity": 363.2152302373046, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683498785867, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8698473137863447}}
{"url": "http://www.cataclysmuo.com/q5656e/how-to-prove-a-rhombus-in-coordinate-geometry-56992b", "text": "by misbrooks. Find the coordinates of C and D, such that ABCD is a rhombus. (ii) In any square the length of diagonal will be equal, to prove the given shape is not square but a rhombus, we need to prove that length of diagonal are not equal. Hi, in C1 coordinate geometry questions relating to shapes often come up and a lot of marks are usually given for finding the area. Look over the toolkit page that describes the steps used in a coordinate geometry proof. First, they prove that a figure with given points is another figure by using slopes. Use coordinate geometry to prove each statement. The coordinates for a rhombus are given as (2a, 0) (0, 2b), (-2a, 0), and (0.-2b). more interesting facts . Therefore, to prove it is a rhombus you must verify that all sides are the same length. - Show that both pairs of opposite sides are parallel. Can an opponent put a property up for auction at a higher price than I have in cash? answer choices . The coordinates for a rhombus are given as (2a, 0) (0, 2b), (-2a, 0), and (0.-2b). \\begin {align*}AB=\\sqrt {4^2+2^2}=\\sqrt {18}=3\\sqrt {2}\\end {align*} He says that a rhombus has two diagonals which \u2026 Prove that $$XY \\perp XC$$ Remember your lesson on perpendicular lines . 2. If the product of slopes of diagonals is equal to -1, we say both are perpendicular. Examples: 3. Be sure to assign appropriate variable coordinates to your rhombus's vertices! Step 2: Calculate the coordinates of the midpoints of the sides. Using the coordinate plane in proof. 0. This is the basic property of rhombus. We explain Coordinate Geometry of Rhombii with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. How do I prove analytically using co- ordinate geometry that the diagonals of a rhombus are perpendicular to each other Ask Question Asked 3 years, 1 month ago The shape of a rhombus is in a diamond shape. I hav to prove that the quadrilateral is a rhombus and not a square. Since all sides are equal, it may be a square also. This one is a medium level difficulty question and tests the following concepts: finding length of a line segment given coordinates of its end points; properties of quadrilaterals including square, rectangle, rhombus and parallelogram.. 2. Problem 3 Medium Difficulty. These two sides are parallel. Can't find last vertex. Examples: 1. Hi, in C1 coordinate geometry questions relating to shapes often come up and a lot of marks are usually given for finding the area. And in a rhombus, not only are the opposite sides parallel-- it's a parallelogram-- \u2026 Hi! To prove that two lines are perpendicular, when all we have are those two lines, we can use the Linear Pair Perpendicular Theorem - If two straight lines intersect at a point and form a linear pair of equal angles, they are perpendicular.. Our diagonals intersect at point O, so we'd need to show the two linear angles formed at that intersection point are equal, and we \u2026 x-coordinate of midpoint = average of x \u2026 ... (1,3) is a parallelogram because both pairs of opposite sides are congruent and then show it is a rhombus. Since the side of the rhombus is 41, the area of the rhombus is 41 sin \u03b8, where \u03b8 is either of the (supplementary) internal angles of the rhombus. Finding vertices of rhombus formed by lines $y=2x+4$, $y=-\\frac{1}{3}x+4$ and $(12,0)$ is a vertex. He says that the rhombus is a quadrilateral and hence the sum of internal angles of it is 360 degrees. Here is a formula you can use that will solve an area of any triangle given that you have the coordinates of the three vertices. We explain Coordinate Geometry of Rhombii with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. 1. Coordinate Proof 2. Yes, a square is a rhombus A square must have 4 congruent sides. So the area is somewhere between 0 (if the rhombus is totally \"flat\") and 41 (if the rhombus is a square). Choose from 500 different sets of proofs coordinate geometry flashcards on Quizlet. (iii)\u00a0 Diagonals will be perpendicular. Slope of the diagonal AC : A (2, -3) C (-2, 1) Slope = (y 2 -y 1)/ (x 2 -x 1) @lioness99a No. Have you missed part of the question out? Ivan used coordinate geometry to prove that quadrilateral EFGH is a square. This lesson will demonstrate how to use slope, midpoint, and distance formulas to determine from the coordinates of the vertices if a quadrilateral is a rhombus in a coordinate \u2026 If all sides of a quadrilateral are congruent, then it\u2019s a rhombus (reverse of the definition). To prove it is rhombus, we can prove any one of the following. Want to improve this question? more interesting facts . Use coordinate geometry to prove that quadrilateral DIAN is a square. So, the given are not vertices of rhombus. more interesting facts . Ivan used coordinate geometry to prove that quadrilateral EFGH is a square. .And Why To use coordinate geometry to prove that a \ufb02ag design includes a rhombus, as in Example 2 In Lesson 5-1, you learned about midsegments of triangles.A trapezoid also has a Area of the rhombus is greater than what? Question 408015: Using a coordinate geometry proof, which method below is a correct way to prove a quadrilateral is a rhombus? Since the side of the rhombus is $\\sqrt{41}$, the area of the rhombus is $41\\sin\\theta$, where $\\theta$ is either of the (supplementary) internal angles of the rhombus. Proof: Rhombus Opposite Angles are Congruent (1) AD=CD //Given, definition of a rhombus (2) AB=CB //Given, definition of a rhombus In this coordinate geometry worksheet, 10th graders solve and draw 10 different problems related to complete proofs in coordinate geometry. Algebraic proofs for geometric theorems (Geometry) Prove whether a figure is a rectangle in the coordinate plane An updated version of this instructional video is available. Proof: Rhombus area. - Show that both pairs of opposite sides are congruent. First draw a figure and choose convenient axes and coordinates. (someone plz help cuz i don't understand) Coordinate Geometry: Ms.Sue or someone please help. Choose one of the methods. So we have a parallelogram right over here. Write sentences that explain your ideas clearly. To prove it is rhombus, we can prove any one of the following. Proving a Quadrilateral is a Rhombus Prove that it is a parallelogram first, then: Method 1: Prove that the diagonals are perpendicular. The diagonals have the same midpoint, and one pair of opposite sides have equal lengths. You\u2019re correct but can you kindly provide me with an explanation for that ? Distance between two points. Prove that $$XY \\perp XC$$ Remember your lesson on perpendicular lines . If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Played 0 times. Proof: The diagonals of a kite are perpendicular Proof: Rhombus \u2026 The area cannot be determined as two points are insufficient to describe a rhombus. The one main way to prove that a quadrilateral is a rhombus is to prove that the distances of the four sides of the quadrilaterals are congruent (equal distances) and then prove that the diagonals of the quadrilateral are not congruent (unequal distances). Method 3: Prove that all four sides are equal. Show that both pairs of opposite sides are parallel 3. The diagonals of a rhombus \u2026 Quadrilateral EFGH is at E (\u22122, 3), F (1, 6), G (4, 3), and H (1, 0)1. Use coordinate geometry to prove that quadrilateral KAIT is a parallelogram. How do you prove a rhombus in coordinate geometry? All sides are congruent. In this non-linear system, users are free to take whatever path through the material best serves their needs. Using the coordinate plane in proof. First of all, a rhombus is a special case of a parallelogram. Proofs of general theorems. This lesson will demonstrate how to use slope, midpoint, and distance formulas to determine from the coordinates of the vertices if a quadrilateral is a rhombus in a coordinate plane. Opposite sides have congruent slopes. These unique features make Virtual Nerd a viable alternative to private tutoring. Well think of the other parallel side of the rhombus, it may be inclined at any angle from ${\\pi}\\over 2$ to $0$ and still satisfy the conditions to be a rhombus. He says that four sides of a rhombus are equal in length. Prove that a shape must be a rhombus using rules of coordinate geometry, finding the length of the altitude of the rhombus, Area of rhombus given circumradii of its contained triangles. Prove that the quadrilateral with vertices A(-1,0), B(3,3), C(6,-1) and D(2,-4) is \u2026 If we were told that the rhombus is not a square, B would be the correct answer. This video is unavailable. So the area is somewhere between $0$ (if the rhombus is totally \"flat\") and $41$ (if the rhombus is a square). If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it\u2019s a rhombus (converse of a property). To prove that it is a rhombus, remember that the definition of a rhombus is a quadrilateral with four congruent sides. Point of intersection. Since the side of the rhombus is $\\sqrt{41}$, the area of the rhombus is $41\\sin\\theta$, where $\\theta$ is either of the (supplementary) internal angles of the rhombus. . 2 years ago. This shape appears to be a rhombus. Step 1: Identify the coordinates of the vertices of the rhombus. I hav to prove that the quadrilateral is a rhombus and not a square. Rectangle, Parallelogram, Trapezoid. Question Different forms equations of straight lines. Will a refusal to enter the US mean I can't enter Canada either? math b. the vertices of triangle ABC are A(-2,3), B(0,-3), and C(4,1). Coordinate Geometry. Prove that the quadrilateral with vertices A(-1,0), B(3,3), C(6,-1) and D(2,-4) is a square. In Euclidean geometry, a rhombus is a type of quadrilateral. Prove that three points in a rhombus are collinear 0 Given the co-ordinates of several points, how to determine which segments are sides and which are diagonals if the coordiantes of a quadrilateral TEAM are T(-2,3), E(-5,-4),A(2,-1), AND M(5,6). Next lesson. Algebraic proofs for geometric theorems (Geometry) Prove whether a figure is a rectangle in the coordinate plane An updated version of this instructional video is available. Practice: Prove parallelogram properties. In the $xy$-plane, two adjacent vertices of the rhombus are $(-2,2)$ and $(2,7)$. How does a bank lend your money while you have constant access to it? Quadrilateral EFGH is at E (\u22122, 3), F (1, 6), G (4, 3), and H (1, 0)1. more interesting facts . Show: Formula: Work Prove it is a Rectangle. Were the Beacons of Gondor real or animated? So, its midpoint will be equal. Use coordinate geometry to prove that a) quadrilateral NORA is a rhombus, and b) quadrilateral NORA is not a square. Learn the properties of quadrilaterals. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Be sure to really show the original formula and show the steps clearly- be neat and precise. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. prove, by means of coordinate geometry, that the median to side BC is also the altitude to side BC. How do you write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry? Method: First, prove the quadrilateral is a rhombus by showing all four sides is congruent; then prove the quadrilateral is a rectangle by showing the diagonals is congruent. Use graph paper, ruler, pencil. Look over the toolkit page that describes the steps used in a coordinate geometry proof. But since in a rhombus all sides are equal, it is easier to prove this property than for the general case of a parallelogram, and this is what we \u2026 more about imaginary numbers . If four coordinates of four vertices are given in an order, then to prove that the given vertices are of a rhombus, we need to prove that four sides are equal by using the distance formula. Coordinate Geometry. To prove that it is a rhombus, remember that the definition of a rhombus is a quadrilateral with four congruent sides. Examples: 3. This one is a medium level difficulty question and tests the following concepts: finding length of a line segment given coordinates of its end points; properties of quadrilaterals including square, rectangle, rhombus and parallelogram. What is the best way to play a chord larger than your hand? How do you prove that the quadrilateral formed by joining the midpoints of a rhombus is a rectangle using coordinate geometry? geometry Question 408015: Using a coordinate geometry proof, which method below is a correct way to prove a quadrilateral is a rhombus? If we were told that the rhombus is not a square, B would be the correct answer. The most common shape is the triangle. A square is a special rhombus that also has 4 right angles. Proofs Using Coordinate Geometry 348 Chapter 6 Quadrilaterals What You\u2019ll Learn \u2022 To prove theorems using \ufb01gures in the coordinate plane. Show that both pairs of opposite sides are congruent. There's more to it than that. Aptitude Practice Questions Coordinate Geometry Question 3 T his aptitude practice question is a coordinate geometry question. If the points had bee $(0,0)$ and $(0,5)$ then we would know the area is less than $41.$. Is there a bias against mentioning your name on presentation slides? I can prove a quadrilateral is a rhombus or Square on a coordinate grid. First show that you can place the rhombus in the plane with vertices as shown in the diagram below. In a coordinate proof, you are proving geometric statements using algebra and the coordinate plane.Some examples of statements you might prove with a coordinate proof are: Prove or disprove that the quadrilateral defined by the points is a parallelogram. Show that a pair of opposite sides are congruent and parallel 4. In an amplifier, does the gain knob boost or attenuate the input signal? As we were not told that, D is the correct answer. I tried the distance formula to get the distance between the two points which is $\\sqrt{41}$ but I don\u2019t know what to do next. Why do we not observe a greater Casimir force than we do? Write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry. T his aptitude practice question is a coordinate geometry question. Video transcript. First, plot the points. Hence, it is also called a diamond. The most common shape is the triangle. Coordinate Proof 1. Use coordinate geometry to prove that the quadrilateral formed by connecting the midpoints of a kite is a rectangle. Proof: Rhombus diagonals are perpendicular bisectors. Using https://artofproblemsolving.com/wiki/index.php/Rhombus#Proof_that_the_diagonals_of_a_rhombus_divide_it_into_4_congruent_triangles, the intersection of the two diagonals will lie on the circle, Hence the point cannot be determined with the information supplied, So, will be the lengths of the two diagonals, site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. To determine the type of quadrilateral given a set of vertices. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of each other. Write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry. How does \u771f\u6709\u4f60\u7684 mean \"you really are something\"? Therefore, to prove it is a rhombus you must verify that all sides are the same length. @Vai I'm still not entirely sure what your question is. Use coordinate geometry to prove that the diagonals of a rhombus are perpendicular. Step 1: Identify the coordinates of the vertices of the rhombus. If we were told that the rhombus is not a square, B would be the correct answer. (Believe it \u2026 Use graph paper, ruler, pencil. in coordinate geometry how can we prove that a quadrilateral is a rhombus and not a square? Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. answer choices . How can I defeat a Minecraft zombie that picked up my weapon and armor? It only takes a minute to sign up. Click to see full answer. Method: First, prove the quadrilateral is a rhombus by showing all four sides is congruent; then prove the quadrilateral is a rectangle by showing the diagonals is congruent. Be sure to really show the original formula and show the steps clearly- be neat and precise. ; Prove or disprove that the point lies on the circle centered at the origin containing the point . Prove that the quadrilateral ABCD with the vertices in a coordinate plane A(-3,-4), B(5,-3), C(1,4) and D(-7,3) (see the Figure) is a rhombus. That is what is given in the answer key. And what I want to prove is that its diagonals bisect each other. 1. in coordinate geometry how can we prove that a quadrilateral is a rhombus and not a square? Here is a formula you can use that will solve an area of any triangle given that you have the coordinates of the three vertices. There are 5 different ways to prove that this shape is a parallelogram. Geometry HELP. Has your book defined \"rhombus\" so that a square is a rhombus, or so that a square is not a rhombus? (i)\u00a0 In a rhombus the length of all sides will be equal. If two consecutive sides of a rectangle are congruent, then it\u2019s a square (neither the reverse of the definition nor the converse of a property). Use coordinate geometry to prove that a) quadrilateral NORA is a rhombus, and b) quadrilateral NORA is not a square. Method 2: Prove that a pair of adjacent sides are equal. StatementReason1. A(2, -3), B (6, 5) C (-2, 1) and D (-6, -7), Distance Between Two Points (x1, y1) and (x2\u00a0, y2), Here\u00a0x1\u00a0= 2, y1\u00a0= -3, x2\u00a0= 6\u00a0 and\u00a0 y2\u00a0= 5, Here\u00a0x1\u00a0= 6, y1\u00a0= 5, x2\u00a0= -2\u00a0 and\u00a0 y2\u00a0= 1, Here\u00a0x1\u00a0= -2, y1\u00a0= 1, x2\u00a0= -6\u00a0 and\u00a0 y2\u00a0= -7, Here\u00a0x1\u00a0= -6, y1\u00a0= -7, x2\u00a0= 2\u00a0 and\u00a0 y2\u00a0= -3. Area of a rhombus in co-ordinate geometry [closed], https://artofproblemsolving.com/wiki/index.php/Rhombus#Proof_that_the_diagonals_of_a_rhombus_divide_it_into_4_congruent_triangles. If a rhombus contains a right angle, then it\u2019s a square (neither the reverse of the definition nor the converse of a property). Here are a few ways: 1. Writing coordinate proofs requires a knowledge of the slope formula, the distance formula and the mid point formula. . The question asks what is greater between A and B. You can use the distance formula or the Pythagorean Theorem to do this. ... How do you prove a quadrilateral is a rhombus? StatementReason1. These unique features make Virtual Nerd a viable alternative to private tutoring. How to express the behaviour that someone who bargains with another don't make his best offer at the first time for less cost? if the coordiantes of a quadrilateral TEAM are T(-2,3), E(-5,-4),A(2,-1), AND M(5,6). Be sure to include the formulas. Why does the T109 night train from Beijing to Shanghai have such a long stop at Xuzhou. Using coordinate geometry to prove that the diagonals of a square are perpendicular to each other. The coordinates for a rhombus are given as (2a, 0), (0, 2b), ( 2a, 0), and (0, 2b). Coordinate Proof 1. So the area is somewhere between $0$ (if the rhombus is totally \"flat\") and $41$ (if the rhombus is a square). Coordinate Geometry 348 Chapter 6 Quadrilaterals What You\u2019ll Learn \u2022 To prove theorems using \ufb01gures in the coordinate plane. In this non-linear system, users are free to take whatever path through the material best serves their needs. Coordinate geometry formulas. So that side is parallel to that side. Learn proofs coordinate geometry with free interactive flashcards. 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Inner quadrilateral a rectangle definition ) show it is rhombus, we can prove rhombus!\nDremel 100 Accessories, Atlanta Penitentiary Inside, Zombie Vs Shark, User Profile Cleaning Script For Windows 10, Miniature Railway Supplies, Colin Jost Wedding Ring, Cattle Hoof Trimming Grinder, Movies With Missing In The Title, Titleist T300 Individual Irons, Sumosam Sushi Menu, Bird Dog May Madness, Principles Of Sustainable Development Wikipedia, Twilight Restaurant Scene,", "date": "2021-06-13 06:22:52", "meta": {"domain": "cataclysmuo.com", "url": "http://www.cataclysmuo.com/q5656e/how-to-prove-a-rhombus-in-coordinate-geometry-56992b", "openwebmath_score": 0.5473098754882812, "openwebmath_perplexity": 645.3402822566482, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.977022630759019, "lm_q2_score": 0.8902942203004185, "lm_q1q2_score": 0.8698376012674645}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=dftck8b0bjhducb5knhnfofef3&action=printpage;topic=1510.0", "text": "# Toronto Math Forum\n\n## MAT244--2018F => MAT244--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 04:01:01 PM\n\nTitle: Q6 TUT 5102\nPost by: Victor Ivrii on November 17, 2018, 04:01:01 PM\nThe coefficient matrix contains a parameter $\\alpha$.\n\n(a) Determine the eigenvalues in terms of $\\alpha$.\n(b)\u00a0 Find the critical value or values of\u00a0 $\\alpha$\u00a0 where the qualitative nature of the phase portrait for the system changes.\n(c) Draw a phase portrait for a value of\u00a0 $\\alpha$ slightly below, and for another value slightly above, each critical value.\n$$\\mathbf{x}' =\\begin{pmatrix} 4 &\\alpha\\\\ 8 &-6 \\end{pmatrix}\\mathbf{x}.$$\nTitle: Re: Q6 TUT 5102\nPost by: Michael Poon on November 17, 2018, 04:35:52 PM\na) Finding the eigenvalues:\n\nSet the determinant = 0\n\n\\begin{align}\n(4 - \\lambda)(-6 - \\lambda) - 8\\alpha &= 0\\\\\n\\lambda^2 + 2\\lambda - 24 - 8\\alpha &= 0\\\\\n\\lambda &= -1 \\pm \\sqrt{25 + 8\\alpha}\n\\end{align}\n\nb)\n\nCase 1: Eigenvalues real and same sign\nwhen: $\\alpha$ > $\\frac{-25}{8}$ + 1\n\nCase 2: Eigenvalues real and opposite sign\nwhen: $\\frac{-25}{8}$ < $\\alpha$ < $\\frac{-25}{8} + 1$\n\nCase 3: Eigenvalues complex\nwhen: $\\alpha$ < $\\frac{-25}{8}$\n\ncritical points: $\\alpha$ = $\\frac{-25}{8}$, $\\frac{-25}{8}$ + 1\n\nc) will be posted below:\nTitle: Re: Q6 TUT 5102\nPost by: Michael Poon on November 17, 2018, 04:42:29 PM\nPhase portraits attached below:\n\nTop: Eigenvalues real & same sign (+ve), stable\n\nMiddle: Eigenvalues real & opposite sign, saddle\n\nBottom: Eigenvalues complex & negative, unstable spiral\nTitle: Re: Q6 TUT 5102\nPost by: Jiacheng Ge on November 18, 2018, 12:41:00 PM\nMy solution is different.\nTitle: Re: Q6 TUT 5102\nPost by: Victor Ivrii on November 25, 2018, 09:49:09 AM\nWhen eigenvalues pass from real to complex conjugate, stability does not change. Jiacheng is right", "date": "2022-06-30 00:43:31", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=dftck8b0bjhducb5knhnfofef3&action=printpage;topic=1510.0", "openwebmath_score": 0.9727827310562134, "openwebmath_perplexity": 10161.98234162158, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226294209298, "lm_q2_score": 0.8902942173896131, "lm_q1q2_score": 0.8698375972322486}}
{"url": "https://math.stackexchange.com/questions/1244191/minimum-volume-of-a-circular-right-cone-with-a-sphere-inscribed-in-it", "text": "# Minimum Volume of a circular, right cone, with a sphere inscribed in it.\n\nQuestion: A sphere of radius $r$ is inscribed in a circular, right cone.\n\nWhat is the minimum radius and height of the circular cone? (Thus, volume)\n\nBecause the answer would specifically be proportional to the radius given $r$,\n\nI have set the base radius of the cone $ar$, and would like to solve for $a$.\n\nBecause triangle AB'O, and AC'O are congruent, I have labelled the angles $\\theta$\n\nThen, $\\tan(\\theta)= \\frac{1}{a}$\n\nThe double angle formula is:\n\n$\\tan(2\\theta)= \\frac{2\\tan(\\theta)}{1-\\tan^{2}(\\theta)}$\n\nPlugging in the values, it results in:\n\n$\\tan(2\\theta)= \\frac{2a}{a^2-1}$\n\nFor simplicity, let's set the radius 1, such that the base has a radius length $a$, then the height becomes\n\n$a\\times\\frac{2a}{a^2-1} = \\frac{2a^2}{a^2-1}$\n\nThis makes perfect sense, because if we extend the base radius to infinity, then the resulting cone's top vertex would end up on the top of the sphere.\n\n$\\lim\\limits_{a \\to \\infty}\\frac{2a^2}{a^2-1} = 2$\n\n(We have set the sphere's radius to 1)\n\nThus, we have the variables to actually solve for the minimum value.\n\nThe volume of the right circular cone is given by:\n\n$V = \\frac{1}{3}\\pi r^2 h$\n\nWe have set $r = a$ and $h = \\frac{2a^2}{a^2-1}$\n\nThen,\n\n$V = \\frac{1}{3}\\pi a^2 \\frac{2a^2}{a^2-1} = \\frac{2}{3}\\pi \\frac{a^4}{a^2-1}$\n\nNeglecting the coefficients, deriving V with respect to a and solving for 0,\n\nWe get\n\n$a = \\sqrt{2}$ and $h = 4$\n\nSo\n\n$a = \\sqrt{2}r$ and $h = 4r$\n\nApparently though, I'm wrong. The answer is:\n\n$a = 2r$ and $h = 4r$\n\nBut I don't understand what's wrong with my argument. Any ideas?\n\n\u2022 I don't think you have made any mistakes, the book must have a typo, if a=2r and h=4r the sides of the cone would not be tangent to the sphere.\n\u2013\u00a0WW1\nApr 21, 2015 at 0:34\n\u2022 It is weird how the question was asked the cone's radius to height ratio would have a minimum value of zero ( when a=r and $h\\to \\infty$ ) . I don't see how minimizing that ratio is the same as minimizing the volume of the cone.\n\u2013\u00a0WW1\nApr 21, 2015 at 0:39\n\u2022 @WW1 oh, that actually does not refer to the ratio, I just meant radius AND height. Sorry, that is definitely my mistake. Apr 21, 2015 at 0:43\n\u2022 \"radius and height\" is a very peculiar thing to be asked to minimize - does the original question include \" thus volume\" ?\n\u2013\u00a0WW1\nApr 21, 2015 at 0:56\n\u2022 @WW1, the original question asks for the radius, and height required for the minimum volume; for a minimum volume there has to be a base radius, \"a\", and because of geometry, the height is also dependent on the chosen \"a.\" Apr 21, 2015 at 1:54\n\nAs user WW1 says, the book must have a typo - your work is correct. Here's another way we could have calculated it giving the same answer:\n\nThe radius of the incircle of a triangle is $\\frac{\\triangle}{s}$ where $\\triangle$ is the area of the triangle and $s$ is the semi-perimeter. Let's call $R$ the radius of the cone and $h$ the height of the cone. If $r$ is the radius of the sphere this gives us\n\n\\begin{align}&r=\\frac{Rh}{\\sqrt{h^2+R^2}+R} \\\\\\implies &r\\sqrt{h^2+R^2}+rR=Rh \\\\\\implies &r\\sqrt{h^2+R^2}=R(h-r) \\\\\\implies &r^2\\left(h^2+R^2\\right)=R^2(h^2-2hr+r^2) \\\\\\implies &r^2h^2=R^2\\left(h^2-2hr\\right) \\\\\\implies &\\frac{r^2h^2}{h^2-2hr}=R^2 \\\\\\implies &\\frac{r^2h}{h-2r}=R^2 \\end{align}\n\nNow the volume of the cone is $$V=\\dfrac{\\pi R^2h}{3}=\\frac{\\pi r^2h^2}{3(h-2r)}$$\n\nHence $$\\frac{dV}{dh}=\\frac{\\pi\\left(r^2h^2-2r^2h(h-2r)\\right)}{3(h-2r)^2}=\\frac{\\pi\\left(4r^3h-r^2h^2\\right)}{3(h-2r)^2}$$\n\nSetting the derivative to $0$, we get $$4r^3h-r^2h^2=0\\hspace{5mm}\\implies\\hspace{5mm}4r-h=0\\hspace{5mm}\\implies\\hspace{5mm}h=4r$$\n\nPlugging this into our equation for $R^2$ we get $$R^2=\\frac{r^2(4r)}{4r-2r}=\\frac{4r^3}{2r}=2r^2$$ Hence $R=\\sqrt{2}r$ and $h=4r$ minimizes the volume as you found.", "date": "2022-08-20 05:07:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1244191/minimum-volume-of-a-circular-right-cone-with-a-sphere-inscribed-in-it", "openwebmath_score": 0.9111093282699585, "openwebmath_perplexity": 438.24710732054166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180656553329, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8698250987452826}}
{"url": "https://tex.stackexchange.com/questions/395107/difference-between-symbols-for-absolute-value", "text": "# Difference between symbols for absolute value?\n\nI know all of the following are used for showing the absolute value:\n\n\u2022 $\\lvert x \\rvert$\n\u2022 $| x |$\n\u2022 $\\mid x \\mid$\n\nI myself think that there is another one \\abs; but I can't use this last one.\n\nWhat is the difference between them?\n\n\u2022 Possible Duplicate: Absolute Value Symbols. Oct 8, 2017 at 8:22\n\u2022 The first is right, the second one can produce unexpected results, the last one is wrong. Oct 8, 2017 at 8:34\n\u2022 @egreg ; (+1) Clear and concise, thank you my dear egreg. Oct 8, 2017 at 9:05\n\nYou ask for the differences and here's an analysis for them.\n\nFirst let's get rid of the last case: \\mid is a relation symbol like = or < and TeX adds spaces around it that disqualify the command from being used for the absolute value.\n\nThe other two seem good, but the first one is better as it doesn't need precautions. The fact is that | is considered an ordinary symbol, whereas \\lvert is an opening (like [) and \\rvert is a closing (like ]).\n\nLet's compare some cases.\n\n\\documentclass{article}\n\\usepackage{amsmath}\n\n\\begin{document}\n$\\begin{array}{lllll} \\text{good} & \\lvert x\\rvert\\le 1 & \\lvert-1\\rvert=1 & \\lvert\\sin x\\rvert & \\log\\lvert x-1\\rvert \\\\[3ex] \\text{so and so} & |x|\\leq 1 & |-1|=1 & |\\sin x| & \\log|x-1| \\\\[3ex] \\text{wrong} & \\mid x\\mid \\leq 1 & \\mid -1\\mid =1 & \\mid \\sin x\\mid & \\log\\mid x-1\\mid \\end{array}$\n\n\\end{document}\n\n\nYou can see in the picture that \\lvert and \\rvert produces the right spacing in all cases, whereas |...| doesn't. Not to speak about \\mid.\n\nUsing | would force you to input\n\n|{-1}|\n|{\\sin x}|\n\\log\\!|x-1|\n\n\nYou can define \\abs as suggested in answers to Absolute Value Symbols but I recommend not to follow the suggestion of making \\left and \\right automatic, because it will most of the time choose a wrong size.\n\n\u2022 Is the spacing for log |x-1| in the first line considered correct? Personally, I find the version in the second line a lot better-looking (for that last formula only). Oct 8, 2017 at 15:22\n\u2022 @FedericoPoloni I consider the absolute value just like a parenthesized expression, so the top is right, but I wouldn't turn down consistent usage like in the second line. Oct 8, 2017 at 15:23\n\u2022 Does \"so and so\" mean \"acceptable\"? Oct 9, 2017 at 11:10\n\u2022 @Ooker It means that in some cases it's right, in others it's wrong; precisely, the first is right, the last might be acceptable (see another comment), the middle ones are plainly wrong. Oct 9, 2017 at 11:51\n\u2022 Can TeX be complied like programming languages? For example in your example can I use a loop to print the sin, log three times with different cases? Oct 9, 2017 at 22:07", "date": "2022-05-16 18:28:28", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/395107/difference-between-symbols-for-absolute-value", "openwebmath_score": 0.7346053719520569, "openwebmath_perplexity": 1378.4252430235733, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102585225649, "lm_q2_score": 0.8991213826762113, "lm_q1q2_score": 0.8698192492579596}}
{"url": "https://math.stackexchange.com/questions/3254239/minimum-of-a-trigonometric-function-involving-absolute-value", "text": "# Minimum of a trigonometric function involving absolute value\n\nGiven $$f(x) = | \\sin( | x | ) |$$, I am told to found the local and global minimum and maximum of $$f$$ (If they exist).\n\nSimply from sketching the graph of $$f$$ I get that the function maximizes periodically for every $$x = k \\frac{\\pi}{2}$$ with $$k$$ being a whole number, giving that there's no global maximum but infinite local maxima equal to $$1$$.\n\nFollowing the same logic, I thought the function would minimize with period $$\\pi$$ to give local minima of $$0$$. However, when checking for the answer, it appears the function has no global nor local minima.\n\n\u2022 What are your definitions for local and global minima? To me it seems like there are both. For me, functions which have no global minima are unbounded (like $y = 1/x$). \u2013\u00a0paulinho Jun 7 at 14:00\n\u2022 Why do you think $1$ is not the global maximum? It certainly is. Also, $0$ is the global minimum. (I'm assuming $x$ is only allowed to take real values.) \u2013\u00a0saulspatz Jun 7 at 14:05\n\u2022 How did you \"check answer\" and why does it appear there is not minima. Your logic that $|sin |x|| \\ge 0$ and $|sin |k\\pi|| = 0$ indicate that $0$ are the local/global minimum. So whatever said there were none is wrong. \u2013\u00a0fleablood Jun 7 at 15:30\n\u2022 What is interesting to note is that the method of finding local extrema via solving $f'(x) = 0$ will not work here because $|\\sin |x||$ is not differentiable at $k\\pi$. But solving $f'(x)=0$ is not the only way to have or find extrema. \u2013\u00a0fleablood Jun 7 at 15:32\n\nYour general reasoning is correct. The global minimum is $$0$$, and it is attained in points of the form $$x = k \\pi$$. The global maximum is 1, and it is attained in points of the form $$x= \\frac{\\pi}{2}+ k \\pi$$. Obviously all these global maxima/minima are also local maxima/minima.\n\nI find it very strange that the solution says differently... Are you certain of the expression for $$f$$?\n\nGood insights to start off about the periodicity of the sine function.\n\nLet's think about this in parts - dissecting the function from the outside inward. Remember the definition of absolute value: $$f(x) = | g(x) | = \\left\\{ \\begin{array}{ll} g(x) & g(x) \\geq 0 \\\\ -g(x) & g(x) < 0 \\end{array} \\right.$$ $$g(x)=\\sin(|x|)=\\sin\\left( \\begin{array}{ll} x & x \\geq 0 \\\\ -x & x < 0 \\end{array}\\right)$$\n\nBy definition, the minima cannot be lower than 0. That does not make it the minima of the function, but it is a good check for the result we get.\n\nSo what is the maxima/minima of $$\\sin(x)$$? $$-1 \\leq \\sin(x)\\leq 1$$ So what is the minima applying the absolute value? $$0 \\leq |\\sin(x)| \\leq 1 \\forall x$$ Those are the maxima/minima of the function, regardless of the $$x$$ input. Let's solve for $$x$$: $$\\sin x = 0 \\rightarrow x = ... ,-2\\pi,\\pi,0,\\pi,2\\pi, ...$$ $$\\sin x = 1 \\rightarrow x = ... ,{-3\\pi \\over 2},{-\\pi \\over 2},{\\pi \\over 2},{3\\pi \\over 2},...$$ Notice all the negative values, those don't apply since we apply the absolute value to $$x$$. $$x_{min}=0,\\pi,2\\pi,...$$ $$x_{max}={\\pi \\over 2},{3\\pi \\over 2},...$$ There are no local maxima/minima which are not the global maxima/minima.\n\n\u2022 \"There are no local maxima/minima which are not the global maxima/minima*\" (emphasis mine). But there are local maxima/minima that are the local extrema. So for as I know there is no requirement in this question or in any definition that local extrema can not also be global extrema. \u2013\u00a0fleablood Jun 7 at 15:35\n\u2022 Nope, there isn't. Just a note I figured I would add. \u2013\u00a0FundThmCalculus Jun 7 at 15:45\n\nBy graphing you function it seems that $$1$$ and $$0$$ are the maximum and minimum values for the result, Why ?\nThe answer is: it's well known that the trigonometric functions are commonly used for expressing ratios between sides of right angled triangle, between$$(-1,1)$$ each $$\\frac \\pi 2$$, but by using the absolute value the result can only be positive so the result goes between $$0$$ and $$1$$.\n\nSo, the function has a unique global maximum of value $$1$$ at $$x=k \\frac \\pi 2, k \\in \\mathbb Z$$ and a unique global minimum of value $$0$$ at $$x=m, m \\in \\mathbb Z$$.", "date": "2019-07-21 00:17:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3254239/minimum-of-a-trigonometric-function-involving-absolute-value", "openwebmath_score": 0.7391486763954163, "openwebmath_perplexity": 273.65626170802193, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102561735719, "lm_q2_score": 0.8991213759183765, "lm_q1q2_score": 0.869819240608331}}
{"url": "https://gateoverflow.in/846/gate-cse-2002-question-2-16", "text": "8,554 views\n\nFour fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is\n\n1. $\\frac{1}{16}$\n2. $\\frac{1}{8}$\n3. $\\frac{7}{8}$\n4. $\\frac{15}{16}$\n\nreshown\n\nTotal outcomes - 24 \u00a0(Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)\nNow out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.\n\nSo, probability, favourable\u00a0events/total outcome\n\n14/16 = 7/8\n\nProbability of at least one head and one tail = 1- Probability of No (head and tail)\n\n=\u00a0 1 \u2013\u00a0 2 / (2*2*2*2)\u00a0\u00a0\u00a0\u00a0 {Numerator HHHH and TTTT and denominator all choices}\n\n= 7/8\n\nprobability of getting all heads =$\\dfrac{1}{16}$\n\nprobability of getting all tails =$\\dfrac{1}{16}$\n\nprobability of getting at least one head and one tail $= 1 - \\dfrac{1}{16} - \\dfrac{1}{16} = \\dfrac{7}{8}.$\n\nnice bro, we can also solve it using tree method\n\nUsing tree method it provides 14\u00a0favourable outcomes out of 16\n\nSo probability should be 14/16=7/8\n\nTotal outcomes - 24 \u00a0(Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)\nNow out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.\n\nSo, probability, favourable\u00a0events/total outcome\n\n14/16 = 7/8\n\nAnother simple approach:\n\nand q= P(tails) = 1/2\n\nRequirement:\n\nUsing binomial distribution,\n\nRequired probability = $_{}^{4}\\textrm{C}_{1} p^{1} q^{3} + {}^{4}\\textrm{C}_{2} p^{2} q^{2} + {}^{4}\\textrm{C}_{3} p^{3} q^{1}$\n\n= $_{}^{4}\\textrm{C}_{1} (1/2)^{1} (1/2)^{3} + {}^{4}\\textrm{C}_{2} (1/2)^{2} (1/2)^{2} + {}^{4}\\textrm{C}_{3} (1/2)^{3} (1/2)^{1}$\n\n$= \\frac{7}{8}$\n\n1 head 3 tails =1/16 2 heads 2 tails =1/16 3 heads 1 tail =1/16 adding we get 3/16 where i am wrong anyone plz explain\n\n### 1 comment\n\n1 head 3 tails =1/16 * 4C1 = 1/4\n\n2 heads 2 tails =1/16 *4C2 = 3/8\n\n3 heads 1 tail =1/16 *4C3 = 1/4", "date": "2022-09-28 03:54:13", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/846/gate-cse-2002-question-2-16", "openwebmath_score": 0.5559592843055725, "openwebmath_perplexity": 3578.5264608221537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474910447999, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8697816439966508}}
{"url": "http://arcivasto.it/zmhn/exponential-fourier-series-of-sawtooth-wave.html", "text": "# Exponential Fourier Series Of Sawtooth Wave\n\nYou can also find the Fourier coefficients using the approach you showed, and the integral is c_n = 1/(2pi)\u222bf(t) exp(-jnt) dt, integrated from -pi to pi. Another term the Fourier series,. We will now derive the complex Fourier series equa-tions, as shown above, from the sin/cos Fourier series using the expressions for sin() and cos() in terms of complex exponentials. com To create your new password, just click the link in the email we sent you. 1; a square wave. A frequency standard is displayed and the probe is adjusted until the deflection time is accurate; D. (a) Obtain the Fourier coefficients for both of these periodic signals. Discrete Time Fourier Transform (DTFT). Matthew Schwartz Lecture 5: Fourier series 1 Fourier series When N oscillators are strung together in a series, the amplitude of that string can be described by a function A(x,t)which satis\ufb01es the wave equation: \u22022 \u2202t2 \u2212v2 \u22022 \u2202x2 A(x,t)=0 (1) We saw that electromagnetic \ufb01elds satisfy this same equation with v=c the speed of light. There are two types of Fourier expansions:. The complex Fourier series Recall the Fourier series expansion of a square wave, triangle wave, and sawtooth wave that we looked at before. 1 in a Fourier series, gives a series of constants that should equal f(x 1). 01: MATLAB M-FILE FOR PLOTTING TRUNCATED FOURIER SERIES AND ITS SPECTRA MATLAB M-File example6. For the square wave of Figure 1 on the previous page, the average value is 0. Fourier Series Grapher. I have already computed the Fourier series of the waveform but I don't know how to derive the amplitude and phase plots from the sawtooth's Fourier series. The top graph shows a function, xT (t) with half-wave symmetry along with the first four harmonics of the Fourier Series (only sines are needed because xT (t) is odd). In mathematics, a Fourier series decomposes periodic functions or periodic signals into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or complex exponentials). Fourier analysis is the study of how general functions can be decomposed into trigonometric or exponential functions with de\ufb02nite frequencies. where, as before, w0 is the base frequency of the signal and j = \u221a-1 (often seen elsewhere as i ) The relationship between this bases and the previous. 2 that its Fourier series contains a constant 1 2 and sine terms. Another application of this Fourier series is to solve the Basel problem by using Parseval's theorem. where a0 models a constant (intercept) term in the data and is associated with the i = 0 cosine term, w is the. In mathematics, a Fourier series decomposes periodic functions or periodic signals into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or complex exponentials). Properties of Fourier series. This worksheet can be downloaded as a PDF file. Complex Fourier Series of Sawtooth Wave Home. 4*A/pi*sum(a(1:n). Our sawtooth function can also be expressed as f(x) = x,0\u2264 x <\u03c0, x\u22122\u03c0, \u03c0 \u2264 x \u2264 2\u03c0, which is an odd function of the variable x. Under rather general conditions, a periodic function f(x) can be expressed as a sum of sine waves or cosine waves in a Fourier series. The example generalizes and one may compute \u03b6(2n), for any positive integer n. That sawtooth ramp RR is the integral of the square wave. 4-1: Determine the Fourier series of the voltage waveform shown in Figure 15. Functions by Fourier Series 23. EE341 EXAMPLE 6: PLOTTING TRUNCATED FOURIER SERIES REPRESENTATION AND SPECTRA OF A SIGNAL Matlab m-file example6. 3), namely, f(t)= a0 + \u221e n=1 (a ncosn\u03c90t+b nsinn\u03c90t) (16. Download MATLAB source. 0e-5) + 1; % Total points \"(final point-initial point)/Interval+1% for n = 1: 12 % Values we are considering to approximate Fourier Seires instead of. The study of Fourier series is a branch of Fourier analysis. You shall not only give the Fourier series coe cients, but also give the Fourier series expression of the signals. 2 Approximating the Square Wave Function using Fourier Sine Series 2. Introduction. The highest term in the Fourier series will correspond to the highest frequency that is used to construct the signal. This can be done using the following commands: >> plot(f,abs(X)) >> plot(f,angle(X)) Note that the angle is either zero or \u03c0. The series does not seem very useful, but we are saved by the fact that it converges rather rapidly. Fourier series were introduced by Joseph Fourier (1768\u20131830) for the purpose of solving the heat equation in a metal plate. of a periodic function. \u00bb Square wave, triangle wave, and sawtooth periodic piecewise functions. of Fourier series, we have Example: Sawtooth Wave. Consider the orthogonal system fsin n\u02c7x T g1 n=1 on [ T;T]. Structural Dynamics Department of Civil and Environmental Engineering Duke University. MH2801 Real Fourier Series of Sawtooth Wave - Duration: 12:56. SIGNALS AND SYSTEMS LABORATORY 5: Periodic Signals and Fourier Series INTRODUCTION The time base signal in an oscilloscope is a sawtooth wave. Our aim was to find a series of trigonometric expressions that add to give certain periodic curves (like square or sawtooth. Fourier Series Jean Baptiste Joseph Fourier (1768-1830) was a French mathematician, physi-cist and engineer, and the founder of Fourier analysis. It is also periodic of period 2n\u02c7, for any positive integer n. 5; Discrete Fourier. To derive formulas for the Fourier coe\ufb03cients, that is, the a\u2032s and b\u2032s,. Fourier Series 3 3. Basic components and principles of electrical circuits: circuit elements variables and measuring devices. com To create your new password, just click the link in the email we sent you. The highest term in the Fourier series will correspond to the highest frequency that is used to construct the signal. The signals are the periodic square wave and sawtooth wave. Determine the Fourier series of the waveform shown in Fig. [email\u00a0protected] 8 The Exponential Form of the Fourier Series ;. Convergence of Fourier series also depends on the finite number of maxima and minima in a function which is popularly known as one of the Dirichlet's condition for Fourier series. im prepared to use minimum phase as it avoids pre-ring. (ting the Fourier-series representation with nmax range between 4. In this Tutorial, we consider working out Fourier series for func-tions f(x) with period L = 2\u03c0. The complex form of Fourier series is algebraically simpler and more symmetric. FOURIER SERIES. A square wave; A triangle wave; A sawtooth wave; An electrocardiogram (ECG) signal; Also included are a few examples that show, in a very basic way, a couple of applications of Fourier Theory, thought the number of applications and the ways that Fourier Theory is used are many. \u00bb Fourier Series Graph Interactive. It is here used as a motivational example in our introduction to Fourier. Find more Mathematics widgets in Wolfram|Alpha. jsfx-inc import Tale/wavetable. Weshow that the Fourier series for such functions is considerably easier to obtain as, if the signal is even only cosines are involved whereas if the signal is odd then only sines are involved. Find the Fourier series for the sawtooth wave defined on the interval $$\\left[ { \u2013 \\pi ,\\pi } \\right]$$ and having period $$2\\pi. The quarter-wave symmetric waveform can be written in a Fourier series form shown in Eq. A Fourier series decomposes a periodic function or periodic signal into a sum of simple oscillating functions, namely sines and cosines (or complex exponentials). Determine the Fourier series of the waveform shown in Fig. Model calcns. Find the combined trigonometric form of the Fourier series for the following signals in Table: (a) Square wave (b) Sawtooth wave (c) Triangular wave (d) Rectangular wave (e) Full-wave rectified wave (f) Half-wave rectified wave (g) Impulse train. Fourier synthesis is the process of building a particular wave shape by adding sines and cosines. 8 (i) Find Its Fourier Series Ii) Sketch IDn I Vs Nw(Magitude Spectrum) Lii) Sketch Dn Vs N (Phase Spectru) V) Find Power Of Glt) In Time Domain (v) Find Power Of G(t) In Frequency Domain Vi Write Matlab Code To Sketch Git) From Its. \u00bb Fourier Series Graph Interactive. Orthogonality of sines/cosines; Fourier series examples (square wave, sawtooth wave, triangle wave) Gibb's Phenomenon; Fourier series - triangle wave; Fourier series - cosine wave (frequency leakage). The discrete FitzHugh-Nagumo system arises by discretizing the FH-N PDE explicit calculations with Fourier series for McKean sawtooth caricature: There is no. Harmonic Analysis - this is an interesting application of Fourier. 5, and the one term expansion along with the function is shown in Figure 2: Figure 2. 1) The coefficients are related to the periodic function f(x) by definite integrals: Eq. The Discrete Fourier Transform At this point one could either regard the Fourier series as a powerful tool or simply a mathematical contrivance. Obtain the amplitude and phase spectra. Fourier for each 2. Properties of Fourier series Periodic signal Fourier serie coe cient. FOURIER TRANSFORMS AND INVERSE FOURIER TRANSFORMS. Hence, we expect a pure sine expansion. where, as before, w0 is the base frequency of the signal and j = \u221a-1 (often seen elsewhere as i ) The relationship between this bases and the previous. \u00a9Yao Wang, 2006 EE3414: Signal Characterization 10 1 3 5 7 9 11 13 15 0 0. This function is usually taken to be periodic, of period 2\u03c0, which is to say that \u0192(x + 2\u03c0) = \u0192(x), for all real numbers x. An odd function has only sine terms in its Fourier expansion. Integrating by parts, we indeed. A square wave is displayed and the probe is adjusted until the horizontal portions of the displayed wave are as nearly flat as possible; B. Even and Odd Functions 23. The characteristic wave patterns of periodic functions are useful for modeling recurring phenomena such as sound or light waves. To this effect, the Exponential series is often known as the \"Bi-Sided Fourier Series\", because the spectrum has both a positive and negative side. It is easy to check that these two functions are defined and integrable on and are equal to f(x) on. I am trying to graph a sawtooth wave with 10-V Pk-Pk, 0-V average value. It is possible to express the Fourier series expansion in the form shown below: 0 k 1 k k1 A x(t) M cos(k t ) 2 (6) where 22 k k k k k k B. An odd function has only sine terms in its Fourier expansion. 6 Waveform Synthesis 17. Someexamples The easiest example would be to set f(t) = sin(2\u2026t). Join 100 million happy users! Sign Up free of charge:. You have the fourier series given as a function of t. Cn=-(ATo/((npi)^2))((Sin((npi)/2))^2) Hint: double differentiate your signal till you end up with dirac delta functions, they are easy to modify. The examples given on this page come from this Fourier Series chapter. More formally, it decomposes any periodic function or periodic signal into the sum of a set of simple oscillating functions, namely sine and cosine with the harmonics of periods. Fourier Series expansion of the Sawtooth wave by Anish Turlapaty. of a periodic function. With appropriate weights, one cycle (or period) of the summation can be made to approximate an arbitrary function in that interval (or the entire function if it too is periodic). You can see that after rectification, the fundamental frequency is eliminated, and all the even harmonics are present. The odd trapezoidal wave function whose graph in shown in Figure 9. You might like to have a little play with: The Fourier Series Grapher. Electrical Engineering. The Fourier series is named in honour of Jean-Baptiste Joseph Fourier (1768-1830), who made. \u3010\u65b0\u54c1\u3011\u3010\u30e1\u30fc\u30ab\u30fc\u7d14\u6b63\u54c1\u3011\u3002\u9001\u6599\u7121\u6599 gios(\u30b8\u30aa\u30b9) \u5b50\u4f9b\u81ea\u8ee2\u8eca genova gios-blue 24\u30a4\u30f3\u30c1 \u30102020\u5e74\u30e2\u30c7\u30eb\u3011\u3010\u5b8c\u5168\u7d44\u7acb\u6e08\u81ea\u8ee2\u8eca\u3011. Fourier Transform, Fourier Series, and frequency spectrum Fourier Series and Fourier Transform with easy to understand 3D animations. Fourier Method of Waveform Analysis 17. 3 Calculate the series coefficients form of the series. Gavin Fall, 2014. In this video tutorial, the tutor covers a range of topics from from basic signals and systems to signal analysis, properties of continuous-time Fourier transforms including Fourier transforms of standard signals, signal transmission through linear systems, relation between convolution and correlation of signals, and sampling theorems and techniques. 005 (b) The Fourier series on a larger interval Figure 2. MH2801 Real Fourier Series of Sawtooth Wave - Duration: 12:56. Another way to create one is with a single ramp wave (sawtooth or triangle ) and a comparator, with the ramp wave on one input, and a variable DC [ clarification. In mathematics, a Fourier series decomposes periodic functions or periodic signals into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or complex exponentials). At each harmonic frequency, the signal has a magnitude and a phase that can be obtained from the complex exponential Fourier series coefficients c n. It is a term common to synthesizer programming, and is a typical waveform available on many synthesizers. The Fourier series is named in honour of Jean-Baptiste Joseph Fourier (1768\u20131830), who made important contributions to the study of trigonometric series, after prelimin. And then we took a little bit of an interlude of building up some of our mathematical foundations, just establishing a. Here we will look at the usual properties we ask from series expansion (how it behaves with respect to the usual operations and how it reacts to transformations of functions). sum with a complex conjugate to get a real response, and two oppositely and rapidly rotating complex exponential spiral packets become a ringing time. I am trying to find the amplitude and phase plots of the saw tooth waveform pictured. Find the Fourier series for the sawtooth wave defined on the interval \\(\\left[ { \u2013 \\pi ,\\pi } \\right]$$ and having period $$2\\pi. Another application of this Fourier series is to solve the Basel problem by using Parseval's theorem. We will do this by computing the Fourier series representation of a pulse train waveform centered at. \u51ac\u30bf\u30a4\u30e4 \u65b0\u54c1 \u9001\u6599\u7121\u6599 4\u672c\u30bb\u30c3\u30c8\u3002\u30b9\u30bf\u30c3\u30c9\u30ec\u30b9\u30bf\u30a4\u30e4 4\u672c\u30bb\u30c3\u30c8 \u30d6\u30ea\u30c2\u30b9\u30c8\u30f3 blizzak vrx2 195/70r14\u30a4\u30f3\u30c1 \u9001\u6599\u7121\u6599aa 2\u672c 4\u672c\u30bb\u30c3\u30c8 \u8ca9\u58f2\u53ef\u80fd. (b) Deduce from this formula Dirichlet\u2019s test for convergence of a series: if the P partial sums of the series bn are bounded, and {anP } is a sequence of real numbers that decreases monotonically to 0, then an bn converges. More formally, it decomposes any periodic function or periodic signal into the sum of a set of simple oscillating functions, namely sine and cosine with the harmonics of periods. Since a sine wave can be expressed as a cosine wave with a phase shift (or vice versa). php on line 143 Deprecated: Function create_function() is deprecated in. \u51ac\u30bf\u30a4\u30e4 \u65b0\u54c1 \u9001\u6599\u7121\u6599 4\u672c\u30bb\u30c3\u30c8\u3002\u30b9\u30bf\u30c3\u30c9\u30ec\u30b9\u30bf\u30a4\u30e4 4\u672c\u30bb\u30c3\u30c8 \u30d6\u30ea\u30c2\u30b9\u30c8\u30f3 blizzak vrx2 195/70r14\u30a4\u30f3\u30c1 \u9001\u6599\u7121\u6599aa 2\u672c 4\u672c\u30bb\u30c3\u30c8 \u8ca9\u58f2\u53ef\u80fd. In this study, an inverse dynamic analysis shaping technique based on exponential function is applied to a solar array (SA) to stabilize output voltage before this technique is combined with a thermoelectric module (TEM). FKEE Norizam. to page 779, practice problem 17. Fully multivariate symbolic Fourier analysis. 1)weknowthattheFouriertransform shouldgiveusa1 =1andallothercoe\u2013cientsshouldbezero. - inoueMashuu/fourier-series. Find the four and eight term Fourier expansion of over the interval , and plot both the function and its expansions on the same set of axes. Without even performing thecalculation (simplyinspectequation2. Join 100 million happy users! Sign Up free of charge:. It is represented in either the trigonometric form or the exponential form. SEE ALSO: Fourier Series , Fourier Series--Sawtooth Wave , Fourier Series--Square Wave , Triangle Wave CITE THIS AS:. Fourier Series expansion of the Sawtooth wave by Anish Turlapaty. The form for the Fourier series is as follows: Each term is a simple mathematical symbol and shall be explained as follows:. This version of the Fourier series is called the exponential Fourier series and is generally easier to obtain because only one set of coefficients needs to be evaluated. ) \u2022 The signal is periodic therefore the sinusoidal waves needed to synthesize it are harmonically related. The Discrete Fourier Transform At this point one could either regard the Fourier series as a powerful tool or simply a mathematical contrivance. four_setpw(slider2); osc. MH2801 Real Fourier Series of Sawtooth Wave - Duration: 12:56. tt\u2212 o to Example 15. Our sawtooth function can also be expressed as f(x) = x,0\u2264 x <\u03c0, x\u22122\u03c0, \u03c0 \u2264 x \u2264 2\u03c0, which is an odd function of the variable x. 8-mag-2014 - Fourier series - Wikipedia, the free encyclopedia. Homework Statement Express the function plotted in the figure below as a Fourier series. 0e-5) + 1; % Total points \"(final point-initial point)/Interval+1% for n = 1: 12 % Values we are considering to approximate Fourier Seires instead of. This says that an in\ufb01nite number of terms in the series is required to represent the triangular wave. Another Fourier series recipe for a triangle wave is also all of the odd harmonics. In effect, we use another representation of the Fourier Series to generate an amplitude and phase. Fourier series were introduced by Joseph Fourier (1768-1830) for the purpose of solving the heat equation in a metal plate. Our aim was to find a series of trigonometric expressions that add to give certain periodic curves (like square or sawtooth. - inoueMashuu/fourier-series. Matthew Schwartz Lecture 5: Fourier series 1 Fourier series When N oscillators are strung together in a series, the amplitude of that string can be described by a function A(x,t)which satis\ufb01es the wave equation: \u22022 \u2202t2 \u2212v2 \u22022 \u2202x2 A(x,t)=0 (1) We saw that electromagnetic \ufb01elds satisfy this same equation with v=c the speed of light. The function increases from -1 to 1 on the interval 0 to 2\u03c0 width, then decreases linearly from 1 to -1 on the interval 2\u03c0 width to 2\u03c0. Sampling and z-transform. Waves: An Interactive Tutorial: 14. The choice is a matter ofconvenience or literally personal preference. The examples given on this page come from this Fourier Series chapter. Example #1: triangle wave. Discrete Fourier Series vs. Therefore, it is often used in physics and other sciences. Step-by-Step Calculator Solve problems from Pre Algebra to Calculus step-by-step. Finding Fourier coefficients for a square wave. The Fourier series are in fact f(t) = 1 2 + 2 \u03c0 sint+ 1 3. The sawtooth wave is defined to be \u20131 at multiples of 2 \u03c0 and to increase linearly with time with a slope of 1/ \u03c0 at all other times. Pulse Train Example 14 5. designers-guide. , while the amplitudes of the sine waves are held in: b1, b2, b3, b4, and so. A Fourier sine series F(x) is an odd 2T-periodic function. An ideal square wave will have a zero rise time - but that would take infinite bandwidth to reproduce with this method. It will provide translation tables among the different representations as well as (eventually) example problems using Fourier series to solve a mechanical system and an electrical system, respectively. a square wave = sin (x) + sin (3x)/3 + sin (5x)/5 + (infinitely) That is the idea of a Fourier series. Download MATLAB source. exponential decay; charging a capacitor; phase shift. This worksheet can be downloaded as a PDF file. 3 Find the exponential Fourier series for the waveform shown in Fig. 10 Fourier Series and. The first four Fourier series approximations for a square wave. y = a 0 + \u2211 i = 1 n a i cos ( i w x) + b i sin ( i w x) where a0 models a constant (intercept) term in the data and. Fourier Transform, Fourier Series, and frequency spectrum Fourier Series and Fourier Transform with easy to understand 3D animations. Small tool to visualize fourier series with different waveforms for Windows, macOS and Linux. SIGNALS AND SYSTEMS LABORATORY 5: Periodic Signals and Fourier Series INTRODUCTION The time base signal in an oscilloscope is a sawtooth wave. Downey)\u7684\u4e2d\u6587\u7ffb\u8b6f\u3002 complex exponential, 7. exponential decay; charging a capacitor; phase shift. OWL 265/65R17 \u3010\u9001\u6599\u7121\u6599\u3011 (265/65/17 265-65-17 265/65-17) \u30b5\u30de\u30fc\u30bf\u30a4\u30e4 \u590f\u30bf\u30a4\u30e4 \u5358\u54c1 17\u30a4\u30f3\u30c1. 3), namely, f(t)= a0 + \u221e n=1 (a ncosn\u03c90t+b nsinn\u03c90t) (16. 320 Chapter 4 Fourier Series and Integrals Every cosine has period 2\u03c0. You have the fourier series given as a function of t. and converted the complex exponential series, (3) we also derived the following Fourier coef\ufb01cients for an odd sawtooth wave with period ( ): (31) for which we have that:, (32). Homework Statement Express the function plotted in the figure below as a Fourier series. Fourier series is almost always used in harmonic analysis of a waveform. A pulse wave can be created by subtracting a sawtooth wave from a phase-shifted version of itself. Click a problem to see the solution. This creates a new wave with double the frequency. I have already computed the Fourier series of the waveform but I don't know how to derive the amplitude and phase plots from the sawtooth's Fourier series. 8-mag-2014 - Fourier series - Wikipedia, the free encyclopedia. 2 Derivation of Fourier series expansion of a function de ned in [ \u02c7;\u02c7]: In Fourier series expansion, we would like to write the function as a series in sine and cosine terms in the form: f(x) = a 0 2 + X1 n=1 a ncosnx+ b nsinnx For nding the above unknown co-e cients a 0;a nand b nin the Fourier series. The sawtooth wave (or saw wave) is a kind of non-sinusoidal waveform. Fourier Series: Fourier Series, Euler\u2019s formulae, even and odd functions, having arbitrary periods, half range expansion, Harmonic analysis. A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. 10 Fourier Series and. 005 (b) The Fourier series on a larger interval Figure 2. Using complex form, find the Fourier series of the function. m: % % Filename: example6. Expand , as a Fourier expansion. Now imagine integrating the product terms from -T/2 to +T/2. Let us consider a sawtooth wave we have simply f(t)=t. To select a function, you may press one of the following buttons: Sine, Triangle, Sawtooth, Square, and Noise. Next: Full-Wave Rectification of Sawtooth Previous: Rectification. Watch it converge. We saw that the Fourier series can be used to create an alternate representation of any periodic signal. To this effect, the Exponential series is often known as the \"Bi-Sided Fourier Series\", because the spectrum has both a positive and negative side. Introduction to the Fourier Series The Fourier Series 4 of 28 The Designer's Guide Community www. Table 4: Basic Continuous-Time Fourier Transform Pairs Fourier series coe\ufb03cients Signal Fourier transform (if periodic) +\u221e k=\u2212\u221e ake jk\u03c90t 2\u03c0 k=\u2212\u221e ak\u03b4(\u03c9 \u2212k\u03c90) ak ej\u03c90t 2\u03c0\u03b4(\u03c9 \u2212\u03c90). Resta a casa al sicuro. Fourier Series Grapher. 2016 - File:Fourier series sawtooth wave circles animation. 50% duty cycle, 2. A half-wave symmetric function can be even, odd or neither. In practice, a separate control system affects hardware pricing. One of the most important uses of the Fourier transform is to find the amplitude and phase of a sinusoidal signal buried in noise. Discrete Fourier Series vs. truncated series. Fourier Synthesis of Periodic Waveforms. Matthew Schwartz Lecture 5: Fourier series 1 Fourier series When N oscillators are strung together in a series, the amplitude of that string can be described by a function A(x,t)which satis\ufb01es the wave equation: \u22022 \u2202t2 \u2212v2 \u22022 \u2202x2 A(x,t)=0 (1) We saw that electromagnetic \ufb01elds satisfy this same equation with v=c the speed of light. 0e-05; % Interval between teo time steps tpts = (4. 01>Pulse Width import Tale/fft_synth. The study of Fourier series is a branch of Fourier analysis. But what we're going to do in this case is we're going to add them. It is often easier to calculate than the sin/cos Fourier series because integrals with exponentials in are usu-ally easy to evaluate. 2 Trigonometric Fourier Series 17. This paper relies on two well known paradigms of quantum chaos, the bakers map and the standard map, to study correlations. Whats people lookup in this blog: Fourier Series Calculator Wolfram With Period. 4 Plot the phase spectra. On this page, we'll redo the previous analysis using the complex form of the Fourier Series. In engineering, physics and many applied fields, using complex numbers makes things easier to understand and more mathematically elegant. Fourier synthesis is the process of building a particular wave shape by adding sines and cosines. \u3010\u9001\u6599\u7121\u6599\u3011\u3010\u76f4\u9001\u54c1\u3011mypallas(\u30de\u30a4\u30d1\u30e9\u30b9) \u6298\u7573\u3082\u3067\u304d\u308b6\u6bb5\u5909\u901f\u4ed8\u30b7\u30c6\u30a3\u30b5\u30a4\u30af\u30eb m-507-iv \u30a2\u30a4\u30dc\u30ea\u30fc\u3002\u3010mypallas(\u30de\u30a4\u30d1\u30e9\u30b9) \u6298\u7573\u3082\u3067\u304d\u308b6\u6bb5\u5909\u901f\u4ed8\u30b7\u30c6\u30a3\u30b5\u30a4\u30af\u30eb m-507-iv \u30a2\u30a4\u30dc\u30ea\u30fc\u3011. Press Full Rectify to rectify it. to page 779, practice problem 17. Fourier series is almost always used in harmonic analysis of a waveform. Topics Discussed: 1. In general, Fourier transforms are complex functions and we need to plot the amplitude and phase spectrum separately. The previous page on Fourier Series used only real numbers. In cosmology to find the chemical composition of stars. Exponential Fourier Series. Kirchoff's laws, loop and nodal analysis. \u3010\u65b0\u54c1\u3011\u3010\u30e1\u30fc\u30ab\u30fc\u7d14\u6b63\u54c1\u3011\u3002\u9001\u6599\u7121\u6599 gios(\u30b8\u30aa\u30b9) \u5b50\u4f9b\u81ea\u8ee2\u8eca genova gios-blue 24\u30a4\u30f3\u30c1 \u30102020\u5e74\u30e2\u30c7\u30eb\u3011\u3010\u5b8c\u5168\u7d44\u7acb\u6e08\u81ea\u8ee2\u8eca\u3011. 1999 2 Definitions sinc(t) =4 sin(\u02c7t)\u02c7t o =42\u02c7 T 0 I. Thus, the Fourier Series of f(x) is 1 2 \u2212 4 \u03c02 X\u221e k=0 1 (2k +1)2 cos (2k +1)\u03c0 2 x + 2 \u03c0 X\u221e k=1 1 n sin k\u03c0 2 x. four_update() ?. Solution:Computing a Fourier series means computing its Fourier coef\u00adficients. Relation Between Trigonometric & Exponential Fourier Series by Tutorials Point (India) Ltd. coefficients for complex exponential Fourier series representation associated for a squarewave signal. Homework Equations The Attempt at a Solution I have the fully worked out solution infront of me and im ok with working out the a0, an and bn parts but what i want to know is why is the function. m, in a Fourier Sine Series Fourier Sine Series: To find F m, multiply each side by sin(m\u2019t), where m\u2019 is another integer, and integrate: But: So: ! only the m\u2019 = m term contributes Dropping the \u2019 from the m: ! yields the coefficients for any f(t)! 0 1 ( ) sin( ) m m ft F mt \u03c0 \u221e = = \u2211 \u2032 0 1. 01: MATLAB M-FILE FOR PLOTTING TRUNCATED FOURIER SERIES AND ITS SPECTRA MATLAB M-File example6. Fourier Series 3 3. 005 (b) The Fourier series on a larger interval Figure 2. If the modulus of the slope of your sawtooth voltage is A, then your Fourier Co-efficient, if you are talking about a continuous time fourier series, is. (For more details on the calculations, see the Mathematica notebook or the Maple worksheet. MH2801 Real Fourier Series of Sawtooth Wave - Duration: 12:56. 8 in the text. Click a problem to see the solution. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex exponentials). The top graph shows a function, xT (t) with half-wave symmetry along with the first four harmonics of the Fourier Series (only sines are needed because xT (t) is odd). f(t) = 1 2 a 0 + X The three examples consider external forcing in the form of a square-wave, a sawtooth-wave, and a triangle-wave. -\u2014,ented by a Fourier series? Why? Consider the sawtooth waveform shown in Fig. This new edition of a successful undergraduate text provides a concise introduction to the theory and practice of Fourier transforms, using qualitative arguments. Download MATLAB source. How do I find the Fourier series of a Sawtooth Learn more about fourier, fourier series, coefficients, sawtooth, sawtooth wave. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex exponentials). Hence, we expect a pure sine expansion. Again, we want to rewrite a periodic function f (t. 1 Sawtooth Wave Let us apply Eqs. An ideal square wave will have a zero rise time \u2013 but that would take infinite bandwidth to reproduce with this method. An ideal square wave will have a zero rise time - but that would take infinite bandwidth to reproduce with this method. Description. This Fourier Series demo, developed by Members of the Center for Signal and Image Processing (CSIP) at the School of Electrical and Computer Engineering at the Georgia Institute of Technology, shows how periodic signals can be synthesised by a sum of sinusoidal signals. Fourier Series Calculator is an online application on the Fourier series to calculate the Fourier coefficients of one real variable functions. Course Contents. Fourier series to Fourier transform A. When we talk. Signals and systems: Continuous time and discrete-time signals, Exponential and sinusoidal signals, The unit Impulse and unit step functions, Basic system properties. Find the four and eight term Fourier expansion of over the interval , and plot both the function and its expansions on the same set of axes. A rectangular pulse is defined by its duty cycle (the ratio of the width of the rectangle to its period) and by the delay of the pulse. Power and Parseval\u2019s Theorem 15 6. (This follows since is. The series consists of an infinite sum of sines and cosines that repeats over fixed intervals, and so is very useful for analyzing periodic functions. 6 Waveform Synthesis 17. A Student\u2019s Guide to Fourier Transforms Fourier transform theory is of central importance in a vast range of applications in physical science, engineering, and applied mathematics. Basic components and principles of electrical circuits: circuit elements variables and measuring devices. 9toseethe result. Answer The function is discontinuous at t = 0, and we expect the series to converge to a value half-way between the upper and lower values; zero in this case. But what we're going to do in this case is we're going to add them. Fourier series formula, Trigonometric, Polar and Exponential fourier series. It deals almost exclusively with those aspects of Fourier analysis that are useful in physics and engineering. 3) to the sawtooth shape shown in Fig. Fourier Transform, Fourier Series, and frequency spectrum Fourier Series and Fourier Transform with easy to understand 3D animations. I first attempted to find a general equation for. The function f 1 is called the odd extension of f(x), while f 2 is called its even extension. As an example, let us find the exponential series for the following rectangular wave, given by. php on line 143 Deprecated: Function create_function() is deprecated in. 4 Waveform Symmetry 17. -\u2014,ented by a Fourier series? Why? Consider the sawtooth waveform shown in Fig. Fourier series: Solved problems \u00b0c pHabala 2012 Alternative: It is possible not to memorize the special formula for sine/cosine Fourier, but apply the usual Fourier series to that extended basic shape of f to an odd function (see picture on the left). Fourier Analysis: Fourier Transform Exam Question Example Fourier Transform example if you have any questions please feel free to ask :) thanks for watching hope it helped you guys :D. Let the period be denoted T. In other words, Fourier series can be used to express a function in terms of the frequencies (harmonics) it is composed of. A time series is said to be stationary if all the X(t) have the same distribution and all the joint distribution of (X(t),X(s)) (for a given value of abs(s-t)) are the same. truncated series. Chapter 2 FOURIER SERIES EXPONENTIAL FOURIER. Obtain the amplitude and phase spectra. The study of Fourier series is a branch of Fourier analysis. FOURIER TRANSFORMS AND INVERSE FOURIER TRANSFORMS. MH2801 Real Fourier Series of Sawtooth Wave In this video segment, we will determine the real Fourier series of a sawtooth wave. One of the most important uses of the Fourier transform is to find the amplitude and phase of a sinusoidal signal buried in noise. Physically this means that our square wave contains a lot of high-frequency components. The charge sepn. It can also serve as a measure of deviations from ergodicity due to quantum effects for typical observables. A high frequency sine wave is displayed and the probe is adjusted for maximum amplitude; C. 005 (b) The Fourier series on a larger interval Figure 2. to page 779, practice problem 17. At each harmonic frequency, the signal has a magnitude and a phase that can be obtained from the complex exponential Fourier series coefficients c n. The study of Fourier series is a branch of Fourier analysis. This version of the Fourier series is called the exponential Fourier series and is generally easier to obtain because only one set of coefficients needs to be evaluated. Exponential Fourier Series. That sawtooth ramp RR is the integral of the square wave. 1 Square Wave Function The \ufb01rst function we examined which can be approximated by a Fourier series is the square wave function. Section Topic Page; Chapter 5 : Data Fitting : 63 : 5. A pulse wave or pulse train is a kind of non-sinusoidal waveform that includes square waves (duty cycle of 50%) and similarly periodic but asymmetrical waves (duty cycles other than 50%). 01>Pulse Width import Tale/fft_synth. MH2801 Real Fourier Series of Sawtooth Wave In this video segment, we will determine the real Fourier series of a sawtooth wave. Integrating by parts, we indeed. The Gibbs phenomenon is also noticeable in this case. 4 k,f k =f 0 *k Amplitude Magnitude Spectrum for Square Wave Line Spectrum of Square Wave. 03 per \u00c5 and 0. What is Fourier Series? Any real, periodic signal with fundamental freq. ) \u2022 The signal is periodic therefore the sinusoidal waves needed to synthesize it are harmonically related. the Gibbs phenomenon in the neighborhood oft = 4 s 100. In each example six plots are provided. Thus, the rise time is dictated by this last term, which in turn dictates highest frequency. The spectrum of a square wave. (b) Deduce from this formula Dirichlet\u2019s test for convergence of a series: if the P partial sums of the series bn are bounded, and {anP } is a sequence of real numbers that decreases monotonically to 0, then an bn converges. -2 -10 1 2 3t 1 f(t) Figure16. 1: The cubic polynomial f(x)=\u22121 3 x 3 + 1 2 x 2 \u2212 3 16 x+1on the interval [0,1], together with its Fourier series approximation from V 9,1. These are properties of Fourier series: If x(t)fourierseries \u2190 coefficient \u2192 fxn & y(t)fourierseries \u2190 coefficient \u2192 fyn. The sawtooth wave (or saw wave) is a kind of non-sinusoidal waveform. If a function is square-integrable on the interval , then the Fourier series converges to the function at almost every point. Especially important are the solutions to the Fourier transform of the wave equation, which define Fourier series, spherical harmonics, and their generalizations. To be more specific, it breakdowns any periodic signal or function into the sum of functions such as sines and cosines. It is often easier to calculate than the sin/cos Fourier series because integrals with exponentials in are usu-ally easy to evaluate. Small tool to visualize fourier series with different waveforms for Windows, macOS and Linux. Then we have that lim N\u2192\u221e f N(t)=f(t) for all t. Each of the examples in this chapter obey the Dirichlet Conditions and so the Fourier Series exists. This function is neither even nor odd and we have already seen in Section 23. Complex Fourier Series of Sawtooth Wave Home. Using complex form, find the Fourier series of the function. Fourier Series of Triangular Wave. This is explained in detail and even in the Fourier series of a periodic 'causal' function, this principle can be. This worksheet can be downloaded as a PDF file. x = sawtooth (t) generates a sawtooth wave with period 2 \u03c0 for the elements of the time array t. of Fourier series, we have Example: Sawtooth Wave. Relation Between Trigonometric & Exponential Fourier Series by Tutorials Point (India) Ltd. % Fourier Series Expansion for Square Wave %% Parameters as mentioned in text f = 500; % Frequecny C = 4/pi; % Constant Value dt = 5. The steps. a0 f ( x) (an cos nx bn sin nx). Oscillators in radio transmitters and receivers produce high frequency sinusoids. Except now we're going to build a composite wave form that is a triangle wave. \u2022 Since f is even, the Fourier series has only cosine terms. It is a term common to synthesizer programming, and is a typical waveform available on many synthesizers. The derivation of this real Fourier series from (5. Properties of Fourier series. m: % % Filename: example6. truncated series. SAWTOOTH 3O 013 (]0 20 oU o u\" truncated sine wave in the daylight and an exponential decrease in tempera-. Figure 1-4 is an example of a rectangular wave, where A designates ampli\u00adtude, T represents time, and \u03c4 indicates pulse width. Topics Discussed: 1. Solution: The Fourier series is given by Eq. It led to a revolution in mathematics, forcing mathematicians to reexamine the foundations of mathematics and leading to many modern theories such as Lebesgue integration. For math, science, nutrition, history. Find the Fourier series for the sawtooth wave defined on the interval \\(\\left[ { \u2013 \\pi ,\\pi } \\right]$$ and having period \\(2\\pi. This function is neither even nor odd and we have already seen in Section 23. com To create your new password, just click the link in the email we sent you. That sawtooth ramp RR is the integral of the square wave. To motivate this, return to the Fourier series, Eq. On this page, we'll redo the previous analysis using the complex form of the Fourier Series. The Fourier series is named in honour of Jean-Baptiste Joseph Fourier (1768-1830), who made important contributions to the study of trigonometric series, after prelimin. From the result in Eqn(5. It is often easier to calculate than the sin/cos Fourier series because integrals with exponentials in are usu-ally easy to evaluate. Linear time-invariant systems: Discrete-time LTI systems: The convolution sum. Let us then generalize the Fourier series to complex functions. This has important implications for the Fourier Coefficients. Find the four and eight term Fourier expansion of over the interval , and plot both the function and its expansions on the same set of axes. The figure above shows a set of periodic signals (left) and their Fourier expansion coefficients (right) as a function of frequency (real and imaginary parts are shown in solid and dashed lines, respectively). Cn=-(ATo/((npi)^2))((Sin((npi)/2))^2) Hint: double differentiate your signal till you end up with dirac delta functions, they are easy to modify. tex 938 Chapter 19 Fourier Series \u03c0 2\u03c0 \u2212\u03c0 \u03c0 FIGURE 19. Do this two ways, from the squared curve and from the Fourier series. We can use Euler's formula, where i is the imaginary unit, to give a more concise formula: The Fourier coefficients are then given by:. However, periodic complex signals can also be represented by Fourier series. Hence, we expect a pure sine expansion. For this example, this average is non-zero. visualization teaching fourier fourier-series square-wave triangle-wave sawtooth-wave Updated Mar 13, 2019. Waves: An Interactive Tutorial: 14. Fourier Transforms: Fourier transform, Sine and Cosine transforms, Application to differential equations. This Fourier series features an in nite sum of sinc. This worksheet can be downloaded as a PDF file. Chap 3 - Discrete-time Signals and Fourier series representation 1 | P a g e 3 Discrete-time Signals and Fourier series representation In the previous two chapters, we discussed Fourier series analysis as applied to continuous-time signals. 2(b), which asks you to derive the exponential Fourier series coefficients for x(t) defined in (6). The Fourier series are in fact f(t) = 1 2 + 2 \u03c0 sint+ 1 3. The Fourier series is named in honour of Joseph Fourier. It then repeats itself. It is analogous to a Taylor series, which represents functions as possibly infinite sums of monomial terms. Under rather general conditions, a periodic function f(x) can be expressed as a sum of sine waves or cosine waves in a Fourier series. show that these trends can be attributed to the exponential decrease of the 1S electron and hole densities at the QD surface with the. Interestingly, Maple does not have a function that produces the Fourier series corresponding to a given function. Homework Help Complex Fourier Series of Sawtooth Wave Complex Fourier series of full wave rectifier: Complex Fourier Series Coefficients: You May Also Like \"Honey, I Shrunk the NASA Payload\": A Call for Engineers to Help Send Mini Rovers to the Moon. In mathematics, a Fourier series (English: /\u02c8f\u028a\u0259ri\u02cce\u026a/) is a way to represent a function as the sum of simple sine waves. 01: MATLAB M-FILE FOR PLOTTING TRUNCATED FOURIER SERIES AND ITS SPECTRA MATLAB M-File example6. In engineering, physics and many applied fields, using complex numbers makes things easier to understand and more mathematically elegant. jsfx-inc @init osc. now consider shifting that rectangle in one domain (e. There is a significant limitation here. The steps. , while the amplitudes of the sine waves are held in: b1, b2, b3, b4, and so. 2 Introduction In this Section we show how a periodic function can be expressed as a series of sines and cosines. 3) to the sawtooth shape shown in Fig. Deriving the Coefficients. OWL 265/65R17 \u3010\u9001\u6599\u7121\u6599\u3011 (265/65/17 265-65-17 265/65-17) \u30b5\u30de\u30fc\u30bf\u30a4\u30e4 \u590f\u30bf\u30a4\u30e4 \u5358\u54c1 17\u30a4\u30f3\u30c1. 3 Introduction In this Section we examine how to obtain Fourier series of periodic functions which are either even or odd. The study of Fourier series is a branch of Fourier analysis. Time scaling property changes frequency components from \u03c90 to a\u03c90. There is a steady increase in the accuracy of the representation as the number of terms. It then repeats itself. 265/65R17 NANKANG \u30b5\u30de\u30fc\u30bf\u30a4\u30e4 \u3010\u65b0\u54c1\u3011\u3010\u9001\u6599\u7121\u6599\u3011\u3002NANKANG (\u30ca\u30f3\u30ab\u30f3) AT-5. Homework Equations The Attempt at a Solution I have the fully worked out solution infront of me and im ok with working out the a0, an and bn parts but what i want to know is why is the function. the sin/cos form, the amplitude-phase form and the exponential form of the Fourier series. Fourier Series - an understandable introduction In mathematics, the Fourier series is an infinite sequence of terms used to solve special types of problems. Prerequisites: Math 202, EE 201. Each of the examples in this chapter obey the Dirichlet Conditions and so the Fourier Series exists. fourier series. Open new M-file. Next: More Advanced Topics Up: Fourier Series-What, How, and Why Previous: The Fast Fourier Transform Using the Fourier Transform. Solution: The voltage waveform is similar to the square wave in Table 15. A frequency standard is displayed and the probe is adjusted until the deflection time is accurate; D. The main idea is to extend these functions to the interval and then use the Fourier series definition. Electrical Engineering. However, in a reverse (or inverse) sawtooth wave, the wave ramps downward and then sharply rises. 22: Circuit for Problem 5. This is the case if, for example, f(x) is the vertical displacement of a string from the. Exponential Form of the Fourier Series 12 4. In this study, an inverse dynamic analysis shaping technique based on exponential function is applied to a solar array (SA) to stabilize output voltage before this technique is combined with a thermoelectric module (TEM). 265/65r17 nankang \u30b5\u30de\u30fc\u30bf\u30a4\u30e4 \u3010\u65b0\u54c1\u3011\u3010\u9001\u6599\u7121\u6599\u3011\u3002nankang (\u30ca\u30f3\u30ab\u30f3) at-5. An ideal square wave will have a zero rise time - but that would take infinite bandwidth to reproduce with this method. The top graph shows a function, xT (t) with half-wave symmetry along with the first four harmonics of the Fourier Series (only sines are needed because xT (t) is odd). We can equivalently describe them as sums of complex exponentials, where each cosine requires two complex exponentials (phasors rotating in. A time series is said to be weakly stationary if the expectation of X(t) does not depend on t and if the covariance of X(t) and X(s) only depends on abs(t-s). MH2801 Real Fourier Series of Sawtooth Wave - Duration: 12:56. The steps. 2 we would calculate the coe\ufb03cients as follows: 2See, for example, Boyce and DiPrima, Elementary Di\ufb00erential Equations and Boundary Value Problems, 3rd Edition, John Wiley & Sons, 1977. com To create your new password, just click the link in the email we sent you. 2 : Theory: Curve Fitting : 64 : 5. MH2801 Real Fourier Series of Sawtooth Wave In this video segment, we will determine the real Fourier series of a sawtooth wave. Network response to DC, exponential and sinusoidal excitation: switching networks, impedances, network functions, Fourier series, phasor methods, real and reactive power, power factor. FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES To a 2\u02c7-periodic function f(x) we will associate a trigonometric series a0 2 + \u2211\u221e n=1 an cos(nx)+bn sin(nx); or in terms of the exponential eix, a series of the form n\u2208Z cne inx: For most of the functions that we will be dealing with, these series are in a sense. 1) The coefficients are related to the periodic function f(x) by definite integrals: Eq. TABLES IN SIGNALS AND SYSTEMS, OCT. Oscillators in radio transmitters and receivers produce high frequency sinusoids. I am working on one of my first Matlab projects. The complex Exponential Fourier Series representation of a periodic signal x (t) with fundamental period T o is given by. The discrete-time Fourier transform is a periodic. \u30ce\u30ad\u30a2\u30f3\u30bf\u30a4\u30e4 \u30cf\u30c3\u30ab\u30da\u30ea\u30c3\u30bf R3 225/50R17 98R XL 225/50-17 \u30b9\u30ce\u30fc \u30b9\u30bf\u30c3\u30c9\u30ec\u30b9 2 \u672c Nokian Tyres HAKKAPELIITTA R3\u3002\u30ce\u30ad\u30a2\u30f3\u30bf\u30a4\u30e4 \u30cf\u30c3\u30ab\u30da\u30ea\u30c3\u30bf R3 225/50R17 98R XL 225/50-17 \u30b9\u30ce\u30fc \u30b9\u30bf\u30c3\u30c9\u30ec\u30b9 2 \u672c Nokian Tyres HAKKAPELIITTA R3. a circular shift in one dft domain is a linear phase rotation in the other domain. fourier series. The results of the Fourier series in this chapter will be extended to the Fourier transform in Chapter 5. Function generators produce sine waves, square waves, and triangular waves. rewrite as Fourier series The coefficients become Fourier series Alternate forms where Complex exponential notation Euler applications of fourier series \u2014 Search results on. Download MATLAB source. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t \u2032 s c o r r e c t. Complex Fourier Series 1. In cosmology to find the chemical composition of stars. A square wave or rectangular function of width can be considered as the difference between two unit step functions and due to linearity, its Fourier spectrum is the difference between the. The study of Fourier series is a branch of Fourier analysis. now consider shifting that rectangle in one domain (e. Another Fourier series recipe for a triangle wave is also all of the odd harmonics. For the square wave of Figure 1 on the previous page, the average value is 0. You can see that after rectification, the fundamental frequency is eliminated, and all the even harmonics are present. Resta a casa al sicuro. With appropriate weights, one cycle (or period) of the summation can be made to approximate an arbitrary function in that interval (or the entire function if it too is periodic). Laurent series cannot handle discontinuities such as a square wave or the sawtooth wave. It then repeats itself. 4 Fourier series Any LTI system is completely determined by its impulse response h(t). of Fourier series, we have Example: Sawtooth Wave. The Fourier series has many such applications in electrical engineeringvibration analysis, acousticsopticssignal processingimage processingquantum mechanicseconometrics[8] thin-walled shell theory, [9] etc. For a wave that travels only in directions that have small angles with respect to the optical axis, the general form of the complex field may be approximated by U(x, y, z) = A(x, y, z) exp(jkz), where A(x, y, z) is a slowly varying function of z. The top graph shows a function, xT (t) with half-wave symmetry along with the first four harmonics of the Fourier Series (only sines are needed because xT (t) is odd). sum with a complex conjugate to get a real response, and two oppositely and rapidly rotating complex exponential spiral packets become a ringing time. This is explained in detail and even in the Fourier series of a periodic 'causal' function, this principle can be. In cosmology to find the chemical composition of stars. 2 - A continuous-time signal sampled at uniform intervals T s with an ideal sampling function. 238CHAPTER 4:Frequency Analysis: The Fourier Series exponentials or sinusoids are used in the Fourier representation of periodic as well as aperiodic signals by taking advantage of the eigenfunction property of LTI systems. 005 (b) The Fourier series on a larger interval Figure 2. Solution: The voltage waveform is similar to the square wave in Table 15. Fourier Series of Half Range Functions - this section also makes life easier 5. F(t) = periodic function represented by Fourier series * F(t) = pulse, with system underdamped. the (sin x)/x function. 55:041 Electronic Circuits One can decompose a periodic signal into a fundamental sine wave and harmonics (Fourier series). Our job is to first obtain the expression for the coefficients which we will later plug into the series formula. Structural Dynamics Department of Civil and Environmental Engineering Duke University. org odic if it repeats itself identically after a period of time. Hence, we expect a pure sine expansion. Solved problem on Fourier Transform, Fourier Series, and frequency spectrum Fourier Series and Fourier Transform with easy to understand 3D animations. 5; Discrete Fourier. The complex Exponential Fourier Series representation of a periodic signal x (t) with fundamental period T o is given by. Pulse Train Example 14 5. You can then apply this method to find the Fourier series of the following period 2\u03c0 functions: 1. Fourier for each 2. This is the output of the system when the input is a Dirac delta function at the origin. 2 4 6 10 terms 0 \u22122 \u03c0 FIGURE 19. The Discrete Fourier Transform At this point one could either regard the Fourier series as a powerful tool or simply a mathematical contrivance. 2 Approximating the Square Wave Function using Fourier Sine Series 2. 005 (b) The Fourier series on a larger interval Figure 2. The toolbox provides this trigonometric Fourier series form. Fourier Series Grapher. All of these are examples of periodic signals. At each harmonic frequency, the signal has a magnitude and a phase that can be obtained from the complex exponential Fourier series coefficients c n. Then we look at alternative ways to write Fourier series, namely the amplitude - phase angle form and the complex form. carries the negative sign on the exponential, and is multiplied by 1/N in going from time to frequency. The discrete signal in (c) xn[] consists only of the discrete samples and nothing else. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex exponentials). y = a 0 + \u2211 i = 1 n a i cos ( i w x) + b i sin ( i w x) where a0 models a constant (intercept) term in the data and. We can use Euler's formula, where i is the imaginary unit, to give a more concise formula: The Fourier coefficients are then given by:. Homework Statement Determine the fourier series for the full-wave rectifier defined as f(t) = sin\u03c9t for 0 < \u03c9t < pi -sin\u03c9t for -pi < \u03c9t < 0 Homework Equations The Attempt at a Solution This looks like an even function, so bm = 0 Ao = 1/pi\u222bsin\u03c9t from 0 to pi = 1/pi(-cos(\u03c9t))/\u03c9). The amplitudes of the cosine waves are held in the variables: a1, a2, a3, a3, etc. Get the free \"Fourier series of f(x)\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Therefore, the Fourier Transform representation of the sawtooth wave given is: The figures below graph the first few iterations of the above solution. 8-mag-2014 - Fourier series - Wikipedia, the free encyclopedia. Complex Fourier Series 1. 238CHAPTER 4:Frequency Analysis: The Fourier Series exponentials or sinusoids are used in the Fourier representation of periodic as well as aperiodic signals by taking advantage of the eigenfunction property of LTI systems. Except now we're going to build a composite wave form that is a triangle wave. This new edition of a successful undergraduate text provides a concise introduction to the theory and practice of Fourier transforms, using qualitative arguments. Fourier Analysis: Fourier Transform Exam Question Example Fourier Transform example if you have any questions please feel free to ask :) thanks for watching hope it helped you guys :D. We will assume it has an odd periodic extension and thus is representable by a Fourier Sine series \u00a6 f 1 ( ) sin n n L n x f x b S, ( ) sin 1. FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES To a 2\u02c7-periodic function f(x) we will associate a trigonometric series a0 2 + \u2211\u221e n=1 an cos(nx)+bn sin(nx); or in terms of the exponential eix, a series of the form \u2211 n\u2208Z cne inx: For most of the functions that we will be dealing with, these series are in a sense equal to f. Fourier Series expansion of the Sawtooth wave by Anish Turlapaty. 9 Discrete Cosine Transform (DCT) When the input data contains only real numbers from an even function, the sin component of the DFT is 0, and the DFT becomes a Discrete Cosine Transform (DCT) There are 8 variants however, of which 4 are common. unfortunately i must use steep eq slopes. Homework Statement Express the function plotted in the figure below as a Fourier series. Properties of Fourier series. (4) : a n = 4 \u03c0 \u222b 0 \u03c0 2 f \u03c9t sin n\u03c9t d \u03c9 t , for odd n a n = 0 , for even n b n = 0 , for all n. Chapter 2 is in of term of sawtooth wave (Refer. Example: Sawtooth wave So, the expansion of f(t) reads. Linear time-invariant systems, convolution. The function describes a set of discrete frequencies with. Fourier theorem is the key to the analysis in the frequency domain when talking about. 1; a square wave. f(x) = signx = {\u22121, \u2212\u03c0 \u2264 x \u2264 0 1, 0 < x \u2264 \u03c0. Topics Discussed: 1. Our sawtooth function can also be expressed as f(x) = x,0\u2264 x <\u03c0, x\u22122\u03c0, \u03c0 \u2264 x \u2264 2\u03c0, which is an odd function of the variable x. Fourier series. In this section, \u0192(x) denotes a function of the real variable x. I first attempted to find a general equation for. Consider the orthogonal system fsin n\u02c7x T g1 n=1 on [ T;T]. Download MATLAB source. Fourier Series: Fourier Series, Euler\u2019s formulae, even and odd functions, having arbitrary periods, half range expansion, Harmonic analysis. Solved problem on Fourier Transform, Fourier Series, and frequency spectrum Fourier Series and Fourier Transform with easy to understand 3D animations. (For sines, the integral and derivative are. There is a significant limitation here. \u3010\u9001\u6599\u7121\u6599\u3011\u3010\u76f4\u9001\u54c1\u3011mypallas(\u30de\u30a4\u30d1\u30e9\u30b9) \u6298\u7573\u3082\u3067\u304d\u308b6\u6bb5\u5909\u901f\u4ed8\u30b7\u30c6\u30a3\u30b5\u30a4\u30af\u30eb m-507-iv \u30a2\u30a4\u30dc\u30ea\u30fc\u3002\u3010mypallas(\u30de\u30a4\u30d1\u30e9\u30b9) \u6298\u7573\u3082\u3067\u304d\u308b6\u6bb5\u5909\u901f\u4ed8\u30b7\u30c6\u30a3\u30b5\u30a4\u30af\u30eb m-507-iv \u30a2\u30a4\u30dc\u30ea\u30fc\u3011. Example #1: triangle wave. (b) x(t) periodic with. Fourier analysis shows that the square wave is made of a sine wave at the square-wave frequency plus a sine wave at every odd multiple of this frequency. Lab 3: Periodic Signal Representation by Fourier Series Prelab: Read the Background section. Fourier Series expansion of the Sawtooth wave by Anish Turlapaty. Homework Equations The Attempt at a Solution I have the fully worked out solution infront of me and im ok with working out the a0, an and bn parts but what i want to know is why is the function. MH2801 Real Fourier Series of Sawtooth Wave In this video segment, we will determine the real Fourier series of a sawtooth wave. Figure 1-4 is an example of a rectangular wave, where A designates ampli\u00adtude, T represents time, and \u03c4 indicates pulse width. The triangular wave is shown opposite. It is possible to express the Fourier series expansion in the form shown below: 0 k 1 k k1 A x(t) M cos(k t ) 2 (6) where 22 k k k k k k B. This simplified review of the Fourier series is meant to reacquaint the student with the basics. This book presents the theory and applications of Fourier series and integrals, Laplace Transforms, eigenfunction expansions, and related topics. Fourier Series.\nls4zg8cjq1gve1,, ryehqs6ycqwkdi,, r9fagdlz3fs8,, jwmjsjo0kbj,, 8055ilcy1glhl,, yiz4b22dqp,, zym2z525my0l,, e6pp7gmve0n8ds,, xrxxu0mj6j3,, rd7vkxqjjfgcq,, mhkieys9efwspwm,, afkzbk7zftujv8,, t2ptedbs2nm,, jynmkokw3pv4b,, 7vhtz7vp6cf,, b7e27d5fx6uix4,, p8s90zvcbm6pj,, hn1cy5fenki,, y5qyh8gxh23,, wp7fkc6o0r,, zlp4tk8n83uh13o,, k4wkvg96duwx,, ze3okoaqabp6hn,, u3dfekmicgb1c,, u0z4oenq9tfys,, d9x0q1wv2t,, b7uui2miun1rh,, qy79c6frm7wb62g,, 4c78tu3lz2i2f,, 4z0sbzgivz,, jy5af91wp864i2,, 72hl5jbtnh,, 3jum69s2m7fzwb4,", "date": "2020-08-15 16:32:30", "meta": {"domain": "arcivasto.it", "url": "http://arcivasto.it/zmhn/exponential-fourier-series-of-sawtooth-wave.html", "openwebmath_score": 0.860872745513916, "openwebmath_perplexity": 733.4239598063035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474904166246, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8697816371789948}}
{"url": "http://mathhelpforum.com/calculus/88310-question-about-differentiability.html", "text": "# Math Help - Question about differentiability\n\nCusps, which are not differentiable, often exist within absolute value functions and at certain points of piecewise functions.\n\nMy question is... can piecewise functions be differentiable throughout? Well, you know, suppose two functions are connected to each other in a such perfect way that there's no cusp such as, for example, when the slope is exactly zero from (a,b) in which c, which is between (a,b) is the x-value of the connection between two completely different functions. This doesn't happen often; cusps usually are present. But there are exceptions. So can piecewise functions be differentiable throughout if there's no cusp?\n\n2. A cusp is a point at which two branches of a curve meet such that the tangents of each branch are equal... so at a cusp you are not differentiable as the derivatives from both directions are unequal.\n\nA piecewise function can be differentiable throughout though, for example\n\nf(x)= x^2 for x<1\nf(x)= x/2 otherwise\n\nat the the point x=1 (where the pieces meet) both the derivative from the left and the derivative from the right are 1/2\n\nAt all other points, f(x) is differentiable since x^2 and x/2 are differentiable everywhere\n\n3. Hello, Kaitosan!\n\nCan piecewise functions be differentiable throughout?\nThey certainly can . . . We can easily create one.\n\nThe graph must be continuous throughout and have equal derivatives at the \"junctions\".\n\nConsider the graph of the parabola $y \\:=\\:x^2$\nBisect it through its axis of symmetry.\nMove each \"half\" one unit to the left and right, respectively.\n\nThe graph looks like this:\nCode:\n                    |\n*             |             *\n|\n*            |            *\n*           |           *\n*         |         *\n- - - - - - *-----+-----* - - - - - -\n-1     |     1\nAnd we have: . $f(x) \\;=\\;\\begin{Bmatrix} -(x+1)^2 & & x \\leq -1 \\\\ 0 & & -1 < x < 1 \\\\ (x\\:{\\color{red}-}\\:1)^2 & & x \\geq 1 \\end{Bmatrix}$\n\nThis is the example I gave when asked this very question\n. . on an exam while in college . . . centuries ago.\n\nYes . . . a silly typo . . . Thanks for the heads-up.\n\n4. Originally Posted by Soroban\nHello, Kaitosan!\n\nThey certainly can . . . We can easily create one.\n\nThe graph must be continuous throughout and have equal derivatives at the \"junctions\".\n\nConsider the graph of the parabola $y \\:=\\:x^2$\nBisect it through its axis of symmetry.\nMove each \"half\" one unit to the left and right, respectively.\n\nThe graph looks like this:\nCode:\n                    |\n*             |             *\n|\n*            |            *\n*           |           *\n*         |         *\n- - - - - - *-----+-----* - - - - - -\n-1     |     1\nAnd we have: . $f(x) \\;=\\;\\begin{Bmatrix} -(x+1)^2 & & x \\leq -1 \\\\ 0 & & -1 < x < 1 \\\\ (x+1)^2 & & x \\geq 1 \\end{Bmatrix}$\n\nThis is the example I gave when asked this very question\n. . on an exam while in college . . . centuries ago.\n.\nDidn't you mean $(x-1)^2$ instead of $-(x+1)^2$?\n\nAlso, isn't the domain supposed to be switched between the two functions?\n\nAnyways, thanks for answering my question. So... as long as there's no cusp, a piecewise function is differentiable! Gotcha.\n\n5. Originally Posted by Kaitosan\nDidn't you mean $(x-1)^2$ instead of $-(x+1)^2$?\nAlso, isn't the domain supposed to be switched between the two functions?\nYou are correct. To match the graph, it should be:\n$f(x) \\;=\\;\\begin{Bmatrix} (x+1)^2 & & x \\leq -1 \\\\ 0 & & -1 < x < 1 \\\\ (x-1)^2 & & x \\geq 1 \\end{Bmatrix}$", "date": "2015-07-01 06:46:32", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/88310-question-about-differentiability.html", "openwebmath_score": 0.8735977411270142, "openwebmath_perplexity": 979.7858914254032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471647042429, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8697595483475121}}
{"url": "https://math.stackexchange.com/questions/283706/classify-the-singularities-of-the-function/283735", "text": "# Classify the singularities of the function\n\n$$f(z)=\\frac{z^3+1}{z^2(z+1)}$$\n\nhas singularities $z=0, z=-1$ right?\n\nHow can I determine the type of singularity of this points?\n\nClassifications:\n\n1.)Removable pole (Then $f(z_0)$ is bounded, $f(z)$ has a limit if $z \\to \\infty$\n\n2.) Pole of order $m \\implies |f(z)|\\to \\infty$ as $z \\to z_0$\n\n3.) Essential singularity : not bounded, does not go to infinity.\n\n(Another way to descibre is to look at the coefficients of the Laurent Series.\n\n\u2022 In some classes it is not really convered, so I'm curious: are you also expected to talk about the (possible) singularity at $z=\\infty$? \u2013\u00a0Antonio Vargas Jan 21 '13 at 20:35\n\u2022 That's the next paragraph :) \u2013\u00a0Applied mathematician Jan 21 '13 at 23:50\n\nA function $f$ which is holomorphic for all $z$ near $z_0 \\in \\mathbb{C}$ (with $z \\neq z_0$) has a pole of order $m>0$ at $z=z_0$ if and only if\n\n$$\\lim_{z \\to z_0} (z-z_0)^m f(z)$$\n\nis finite and nonzero. If $f$ has a singularity at $z=z_0$ and\n\n$$\\lim_{z \\to z_0} f(z)$$\n\nis finite and nonzero then the singularity is removable, and vice versa.\n\n\u2022 How to apply this to my function? \u2013\u00a0Applied mathematician Jan 21 '13 at 19:54\n\u2022 @Joyeuse, consider the pole at $z=0$. What $m$ must you choose to make $\\lim_{z\\to 0} z^m f(z)$ exist and be nonzero? \u2013\u00a0Antonio Vargas Jan 21 '13 at 19:58\n\u2022 that would be 0... \u2013\u00a0Applied mathematician Jan 21 '13 at 20:01\n\u2022 @Joyeuse no... are you telling me that $$\\lim_{z\\to 0} \\frac{z^3+1}{z^2(z+1)}$$ is finite? \u2013\u00a0Antonio Vargas Jan 21 '13 at 20:02\n\u2022 Glad to help! ${}$ \u2013\u00a0Antonio Vargas Jan 21 '13 at 20:32\n\nBy applying partial fraction decomposition, we get: $$f(z) = 1 - \\dfrac{1}{z} + \\dfrac{1}{z^2}$$\n\nIt's now easy to see that $z = -1$ is a removable singularity. $z^3 + 1$ can be factored as $(z + 1)(z^2 - z + 1)$ so $z + 1$ can be canceled and this removes the singularity.\n\nFor the singularity at $z = 0$, it is a pole of order two as the principal part is clearly $- \\dfrac{1}{z} + \\dfrac{1}{z^2}$.\n\nDecompose $f$ as $$f(z)=\\dfrac{z^3+1}{z^2(z+1)}=\\dfrac{(z+1)(z^2-z+1)}{z^2(z+1)}.$$ Then $f$ has at $z_0=-1$ removable singularity and pole of order $2$ at $z_1=0.$ What type of singularity is at $z^*=\\infty$ ?\n\n\u2022 How can you see that the order of $z_1=0$ is $2$ so quickly? Is a removable singularity because of the cancelation? (Always the case?) If $z$ goes to $\\infty$ I think it has a limit ($f \\to$1$as$z \\to \\infty$). ? \u2013 Applied mathematician Jan 21 '13 at 19:58 \u2022 As Antonio Vargas suggested,$\\lim\\limits_{z \\to 0} z^2 f(z)=\\lim\\limits_{z \\to 0} (z^2-z+1)=1$and$\\lim\\limits_{z \\to {-1},\\;z\\ne{-1}} f(z)=3\\$ \u2013\u00a0M. Strochyk Jan 21 '13 at 20:17\n\nThe exposition of @Ayman Hourieh simplifies the problem.\n\nFor visual reinforcement, conformal maps are presented where Re $$z$$ is blue, and Im $$z$$ is red. On the left is the input function $$f(z) = \\frac{z^{3} + 1}{z^{2}(z+1)} = 1 - \\frac{1}{z} + \\frac{1}{z^{2}}$$ On the right is the input function with the singularity suppressed $$z^{2}f(z) = \\frac{z^{3} + 1}{(z+1)} = z^{2} - z + 1$$\n\nz := x + I y;", "date": "2019-12-07 22:23:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/283706/classify-the-singularities-of-the-function/283735", "openwebmath_score": 0.9857739806175232, "openwebmath_perplexity": 477.1530837581358, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9643214480969029, "lm_q2_score": 0.9019206686206199, "lm_q1q2_score": 0.8697414452327631}}
{"url": "https://www.physicsforums.com/threads/question-on-jacobian-determinant.657469/", "text": "# Question on Jacobian determinant\n\n1. Dec 7, 2012\n\n### mnb96\n\nHello,\n\nit is true that linear transformations have constant Jacobian determinant.\nIs the converse true? That is, if a transformation has constant Jacobian determinant, then is it necessarily linear?\n\n2. Dec 7, 2012\n\n### Vargo\n\nHello,\n\nYes, linear transformations have constant Jacobian determinant. You can check this by manual calculation.\n\nThe converse is not true. In fact, there is an important class of transformations in physics called canonical transformations (or symplectic transformations) which preserve volume, but which are not, in general, linear.\n\nI wish I had a nice simple example at hand, but perhaps someone else will come along with a good one.\n\n3. Dec 8, 2012\n\n### homeomorphic\n\nThe total derivative of a linear map is the linear map itself, and in particular, it's constant. The total derivative of a map is just a linearized version of the map at each point. If it's already linear, nothing happens to it when you take its derivative at a point.\n\nJust so we don't get confused, the baby case is 5x, whose derivative is 5. The latter 5 can be interpreted as the linear map that multiplies stuff by 5, so as a linear transformation, it's the same as the first map. For a linear map from \u211d^2 to \u211d^2, the total derivative is a constant linear map, which is represented by a 2 by 2 matrix. The determinant of that matrix is the Jacobian.\n\n4. Dec 8, 2012\n\n### Erland\n\nDon't know much about symplectic transformations, but what about the following one:\n\n$u=\\ln x$, $v=xy$, for $x>0$, $y>0$.\n\nThis is certainly not linear but\n\n$\\partial u/\\partial x = 1/x$, $\\partial u/\\partial y=0$, $\\partial v/\\partial x=y$, $\\partial v/\\partial y= x$.\n\nThe Jacobian determinant is then $1/x * x -0*y=1$ for all $x,\\,y>0$.\n\nSo the Jacobian determinant is constant but the transformation is not linear.\n\n5. Dec 8, 2012\n\n### mnb96\n\nHi!\nthank you all for the explanations.\nVery interesting replies actually!\n\n6. Dec 12, 2012\n\n### mathwonk\n\nnice example erland, and it shows how to construct infinitely more.", "date": "2018-02-22 13:14:55", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/question-on-jacobian-determinant.657469/", "openwebmath_score": 0.9418975710868835, "openwebmath_perplexity": 493.92604054269066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561721629776, "lm_q2_score": 0.8976952941600963, "lm_q1q2_score": 0.869737626468669}}
{"url": "https://mathoverflow.net/questions/9484/number-of-permutations-with-a-specified-number-of-fixed-points", "text": "# Number of permutations with a specified number of fixed points\n\nLet $$F(k,n)$$ be the number of permutations of an n-element set that fix exactly $$k$$ elements.\n\nWe know:\n\n1. $$F(n,n) = 1$$\n\n2. $$F(n-1,n) = 0$$\n\n3. $$F(n-2,n) = \\binom {n} {2}$$\n\n...\n\n4. $$F(0,n) = n! \\cdot \\sum_{k=0}^n \\frac {(-1)^k}{k!}$$ (the subfactorial)\n\nThe summation formula is obviously\n\n$$\\displaystyle\\sum_{k=0}^n F(k,n) = n!$$\n\nA recursive definition of $$F(k,n)$$ is (my claim):\n\n$$F(k,n) = \\binom {n} {k} \\cdot \\Big( k! - \\displaystyle\\sum_{i=0}^{k-1} F(i,k) \\Big)$$\n\nQuestion 1: Is there a common name for the \"generalized factorial\" $$F(k,n)$$?\n\nQuestion 2: Does anyone know a closed form for $$F(k,n)$$ or have an idea how to get it from the recursive definition? (generating function?)\n\nThe \"semi-exponential\" generating function for these is\n\n$\\sum_{n=0}^\\infty \\sum_{k=0}^n {F(k,n) z^n u^k \\over n!} = {\\exp((u-1)z) \\over 1-z}$\n\nwhich follows from the exponential formula.\n\nThese numbers are apparently called the rencontres numbers although I'm not sure how standard that name is.\n\nNow, how do we get a formula for these numbers out of this? First note that\n\n$$exp((u-1)z) = 1 + (u-1)z + {(u-1)^2 \\over 2!} z^2 + {(u-1)^3 \\over 3!} z^3 + \\cdots$$\n\nand therefore the \"coefficient\" (actually a polynomial in $u$) of $z^n$ in $exp((u-1)z)/(1-z)$ is\n\n$$P_n(u) = 1 + (u-1) + {(u-1)^2 \\over 2!} + \\cdots + {(u-1)^n \\over n!} = \\sum_{j=0}^n {{(u-1)^j } \\over j!}$$\n\nsince division of a generating function by $1-z$ has the effect of taking partial sums of the coefficients.\n\nThe coefficient of $u^k$ in $P_n(u)$ (which I'll denote $[u^k] P_n(u)$, where $[u^k]$ denotes taking the $u^k$-coefficient) is then\n\n$$[u^k] P_n(u) = \\sum_{j=0}^n [u^k] {(u-1)^j \\over j!}$$\n\nBut we only need to do the sum for $j = k, \\ldots, n$; the lower terms are zero, since they are the $u^k$-coefficient of a polynomial of degree less than $k$. So\n\n$$[u^k] P_n(u) = \\sum_{j=k}^n [u^k] {(u-1)^j \\over j!}$$\n\nand by the binomial theorem,\n\n$$[u^k] P_n(u) = \\sum_{j=k}^n {(-1)^{j-k} \\over k! (j-k)!}$$\n\nFinally, $F(k,n) = n! [u^k] P_n(u)$, and so we have\n\n$$F(k,n) = n! \\sum_{j=k}^n {(-1)^{j-k} \\over k!(j-k)!}$$\n\n\u2022 Thanks. Am I - as an MO user - supposed to know how to get the closed form for F(k,n) from this \"semi-exponential\" generating function? \u2013\u00a0Hans-Peter Stricker Dec 21 '09 at 16:35\n\u2022 Not necessarily. It's not hard, though; I'll edit the solution to explain that. \u2013\u00a0Michael Lugo Dec 21 '09 at 16:57\n\u2022 The name \"recontres numbers\" is standard in the following ways: (1) EIS, (2) canonical name in Wikipedia, (3) 10 hits in Google Scholar. Although that last one is not a huge number, if you take all three together, it's plenty standard enough for a relatively obscure concept. \u2013\u00a0Greg Kuperberg Dec 21 '09 at 17:38\n\u2022 My claim that the name was nonstandard was entirely subjective; basically, this is something that I felt I should have known a name for, and the name was new to me. \u2013\u00a0Michael Lugo Dec 21 '09 at 17:50\n\nA permutation of {1, ..., n} with k fixed points is determined by choosing which k elements of {1, ..., n} it fixes and choosing a derangement of the remaining n-k elements. So,\n\n$F(k, n) = {n \\choose k} F(0, n-k)$.\n\n(This formula is also on the page Michael Lugo linked to.) You have already given one formula for the number of derangements on n letters. Another one is F(0, n) = the nearest integer to n!/e.\n\nOne can use inclusion--exclusion. First, note (as in @ReidBarton's answer) that $$F(k,n) = \\binom kn F(0,n-k).$$ So it is sufficient to only study permutations with no fixed points. This is known as the number of derangements. Various proofs using inclusion--exclusion can be found on Wikipedia:", "date": "2021-04-12 13:31:24", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/9484/number-of-permutations-with-a-specified-number-of-fixed-points", "openwebmath_score": 0.8897236585617065, "openwebmath_perplexity": 378.33252487612026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924818279465, "lm_q2_score": 0.8947894717137996, "lm_q1q2_score": 0.8697286393246132}}
{"url": "https://brilliant.org/discussions/thread/1-3/", "text": "# #1\n\nHow many positive integers less than $$1000$$ have the property that the sum of the digits of each such number is divisible by $$7$$ and the number itself is divisible by $$3$$?\n\nNote by Vilakshan Gupta\n8\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nLet's think of a number $$abc (0 \\leq a, b, c \\leq 9)$$. $$a+b+c \\equiv 0 (\\mod 3 \\text{and} \\mod 7)$$. Thus, $$a+b+c=21$$.\n\n$(3, 9, 9) \\rightarrow \\frac{3!}{2}, (4, 8, 9) \\rightarrow 3!, (5, 7, 9) \\rightarrow 3!, (5, 8, 8) \\rightarrow \\frac{3!}{2}, (6, 6, 9) \\rightarrow \\frac{3!}{2}, (6, 7, 8) \\rightarrow 3!, (7, 7, 7)$ $$3+6+6+3+3+6+1=28$$\n\nPlease tell me if there is any error.\n\n- 5\u00a0months, 2\u00a0weeks ago\n\nGood one brother\n\n- 5\u00a0months, 2\u00a0weeks ago\n\nNice method\n\n- 5\u00a0months, 2\u00a0weeks ago\n\n28\n\n- 4\u00a0months, 1\u00a0week ago\n\n27\n\n- 6\u00a0months, 1\u00a0week ago\n\nits sum is divisibli by 21 using this you can solve\n\n- 6\u00a0months, 1\u00a0week ago\n\nthis question came in this year PRMO answer is 28\n\n- 6\u00a0months, 1\u00a0week ago\n\n@Md Zuhair 8\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nO i see....\n\n- 7\u00a0months, 2\u00a0weeks ago\n\n@Pokhraj Harshal Rajasthan region!\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nAnd what about the other participants and their marks from your school. I mean the averages and the highest marks\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nWhich region are u from??\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nWB rgion\n\n- 7\u00a0months, 2\u00a0weeks ago\n\n@Pokhraj Harshal Ok! Then I'm also getting the same...\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nHow much? Without the question?\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nThe questions will be cancelled\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nAre the marks gonna be added to everyone's total or the questions will be cancelled (lowering the cutoff)?\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nThey will be added to totsl\n\n- 6\u00a0months, 1\u00a0week ago\n\nHey , what does discounted actually refer to?\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nHey I'm getting 8/27 from jharkhand as per the new answer key of hbcse.will I qualify???\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nLets see....\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nHello, There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 You can check out for more queries related to the JEE EXAMS from the following compilation\n\n- 7\u00a0months, 2\u00a0weeks ago\n\nI believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.\n\n- 7\u00a0months, 3\u00a0weeks ago\n\nAh - that's where the 28 comes from - much more mathematical than my just listing and counting them\n\n- 7\u00a0months, 3\u00a0weeks ago\n\nExactly\n\n- 7\u00a0months, 3\u00a0weeks ago\n\nI agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 432*1 = 24, not 4+3+2+1 = 10. What am I missing?\n\n- 7\u00a0months, 2\u00a0weeks ago\n\n- 8\u00a0months ago\n\nYup.I think so.\n\n- 8\u00a0months ago\n\n@Md Zuhair @Vilakshan Gupta I haven't attempted one of the bonus question. Will I still get marks for it?\n\n- 8\u00a0months ago\n\nI think the question can be cancelled as how can a person attempt to decinal answers and i had attempted ine. So i dunno.\n\n- 8\u00a0months ago\n\nyeah hundreds digit cant be 1,2\n\n- 8\u00a0months ago\n\n@Shreyan Chakraborty The Hundreds digit can't be 1 or 2..\n\n- 8\u00a0months ago\n\nANSWER IS 28....HAS A BIJECTION WITH a+b+c=21 WHERE 0<a,b,c<=9........\n\n- 8\u00a0months ago\n\nAre na na.... I am not telling that. How much are you getting?\n\n- 8\u00a0months ago\n\n@Shreyan Chakraborty .. How much?\n\n- 8\u00a0months ago\n\nJANI NA BAJE HOYECHE\n\n- 8\u00a0months ago\n\n# LetTheFateDecide !!Bye\n\n- 8\u00a0months ago\n\nwhich class are u in toshit?\n\n- 8\u00a0months ago\n\nIt's 11 in Rajasthan ! \ud83d\ude05\ud83d\ude12\n\n- 8\u00a0months ago\n\nAccording to Resonance , cutoff in Chandigarh is just 4(questions)\n\n- 8\u00a0months ago\n\ni don't think it will be so low\n\n- 8\u00a0months ago\n\nIf that isnt, then wb will be higher and i will surely not qualify\n\n- 8\u00a0months ago\n\nYa. Thats ridiculous. WB region has always got a higher cutoff...\n\n- 8\u00a0months ago\n\nCoz as per cutoff(s) uploaded by Resonance , cutoff in Chandigarh is lower than others ( Rajasthan , Maharashtra , UP , etc) ... That's why!\n\n- 8\u00a0months ago\n\nSo , you are already selected..Great \ud83d\udc4d\n\n- 8\u00a0months ago\n\n- 8\u00a0months ago\n\noh\n\n- 8\u00a0months ago\n\nRajasthan..U?\n\n- 8\u00a0months ago\n\noh...btw,where do u live (i mean which region)\n\n- 8\u00a0months ago\n\nGeometry was quite tough and lengthy! Excluding bonus , I'm getting 8\n\n- 8\u00a0months ago\n\nSir\ud83d\ude05 I am getting 10 along with bonus!\n\n- 8\u00a0months ago\n\nOh. U mean 8/28 u r getting?\n\n- 8\u00a0months ago\n\nwell, if it is bonus that means 2 questions marks are given extra.If it was been written question deleted then scores would be evaluated out of 28\n\n- 8\u00a0months ago\n\nunfortunately, i will get only 10 questions correct. I did very silly mistakes\n\n- 8\u00a0months ago\n\n@Md Zuhair Is the paper for 9,10,11 and 12 same?\n\n- 8\u00a0months ago\n\nYes Sir!\n\n- 8\u00a0months ago\n\nHow ma\u00f1y have you got right in PRMO - 17?\n\n- 8\u00a0months ago\n\n- 8\u00a0months ago\n\nwe just need to find the numbers which add upto 21\n\n- 8\u00a0months ago\n\n28\n\n- 8\u00a0months ago\n\n25 is the answer as per me\n\n- 8\u00a0months ago\n\nNo - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28\n\n- 8\u00a0months ago\n\nWhat ans did you get?\n\n- 8\u00a0months ago\n\nHey Aaron. How much are u getting with bonus? With bonus i am getting 12.\n\n- 8\u00a0months ago\n\nzuhair tui ki amk jiggesh korchish??\n\n- 8\u00a0months ago\n\nAccha.. nijer whatsapp number ta de... whatsapp e kotha bolchi\n\n- 8\u00a0months ago", "date": "2018-04-20 01:10:39", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/1-3/", "openwebmath_score": 0.9550356268882751, "openwebmath_perplexity": 5193.632181293543, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9802808707404786, "lm_q2_score": 0.8872045966995027, "lm_q1q2_score": 0.8697096945775437}}
{"url": "https://math.stackexchange.com/questions/3959757/visual-intuition-for-the-definition-of-asymptotically-equivalent", "text": "# Visual intuition for the definition of \"asymptotically equivalent\"\n\nI'm trying to intuitively grasp the following definition:\n\nThe real-valued functions $$f$$ and $$g$$ are asymptotically equivalent as $$x \\to \\infty$$ if $$\\lim_{x \\to \\infty} \\dfrac{f(x)}{g(x)}=1.$$ We write this as $$f \\sim g$$.\n\nMy question is: how do we visually interpret this in terms of the graphs of $$f$$ and $$g$$? Does this mean that the graphs of $$f$$ and $$g$$ get closer to each other as $$x$$ gets larger and larger?\n\nMy only intuition for this comes from the following example: we know that $$\\sin x \\sim x$$ as $$x \\to 0$$ (since $$\\lim_{x \\to 0} (\\sin x)/x = 1$$). And as we can see below, the graphs of $$\\sin x$$ (the green line) and $$x$$ (the black line) get closer and closer as $$x$$ goes to $$0$$.\n\nBut this intuition does not seem to hold for functions asymptotically equivalent at $$\\infty$$. I graphed $$x^2 + x$$ (black line) and $$x^2$$ (green line) and their graphs do not appear to be getting closer at all! In fact, it looks like there's a \"gap\" between the two graphs.\n\nThis leads me to believe that I'm not interpreting \"asymptotically equivalent\" in the right way. I've come across the idea that $$f \\sim g$$ means that $$f$$ and $$g$$ have the \"same rate of growth\", but that feels very unintuitive for me. Is there are a way to see that in the graphs?\n\nAny guidance would be greatly appreciated! Thanks.\n\n\u2022 Basically it means the percentage difference between the vertical distances between points on the two graphs approaches $0.$ Dec 23 '20 at 20:16\n\u2022 Draw the graph of $f/g$ and it will get closer and closer to $1$ as you move towards right in the graph. You can't ensure $f$ close to $g$. For that you need $f-g\\to 0$ and not $f/g\\to 1$. Dec 24 '20 at 3:32\n\u2022 By the way, I think this is an excellent question since asymptotic equivalence is a very powerful tool in maths. It can be used to give intuitive justifications to a number of theorems, and is also not too hard to turn these justifications into rigorous arguments.\n\u2013\u00a0Joe\nDec 24 '20 at 16:57\n\n$$x^2+x$$ and $$x^2$$ are asymptotically equivalent, since $$\\lim_{x \\to \\infty}\\frac{x^2+x}{x^2}=\\lim_{x \\to \\infty}1+\\frac{1}{x}=1 \\, .$$ So it is probable that you simply didn't choose $$x$$ values that were large enough to make your intuitions work. For example, $$x=100$$ gives $$\\frac{100^2+100}{100^2}=\\frac{10100}{10000}=1.01 \\, ,$$ and you can see that they are very close to each other in relative terms. Notice the use of the word relative here. The example you gave helps illustrate what I mean. Let $$f(x)=x^2+x$$ and $$g(x)=x^2$$. $$f(x)-g(x) \\to \\infty$$ as $$x \\to \\infty$$. This means that in absolute terms, the two functions are growing apart. However, the ratio between them\u2014what you need to multiply $$x^2$$ by to get $$x^2+x$$\u2014is approaching $$1$$. This is what the notion of 'asymptotically equivalent' is trying to capture.\n\nTo visualise this, it would be better to use a logarithmic scale:\n\nLook at $$x=10$$, for instance. $$x^2=100$$, whereas $$x^2+x=110$$. This discrepancy looks small when we use a logarithmic y-axis, since the gap of $$10$$ is small relative to how large the functions are. Or, as Paramanand Singh has suggested, we could plot $$y=(x^2+x)/x^2$$:\n\n\u2022 My follow-up question would be: how do we interpret relative difference using the graphs of $f$ and $g$? From the picture you posted, it looks like the graphs $f$ and $g$ are \"hugging\" each other; but since $f(x)-g(x) \\to \\infty$ as $x \\to \\infty$ so wouldn't the graphs eventually grow arbitrarily far apart? That's why I'm struggling to visually interpret the notion of \"relative difference\". Any tips would be appreciated :) Dec 23 '20 at 18:29\n\u2022 @chaad Imagine plotting the two graphs on a $1:1$ scale. This would be very awkward, especially for large values of $x$. It will probably get to a point where you have to scroll on your computer to go from $g$ to $f$. The amount of scrolling you have to do is proportional to the distance between $f(x)$ and $g(x)$. This is what 'absolute difference' might mean intuitively. Then, imagine using a more reasonable scale. If both $f(x)$ and $g(x)$ are very big, then it seems silly to worry about the differences between them (more precisely, the distances between them are tiny relative to...\n\u2013\u00a0Joe\nDec 23 '20 at 18:38\n\u2022 @chaad ...the size of $f(x)$ and $g(x)$ themselves). This is what 'relative difference' is trying to capture. In a sense, $f(x)$ and $g(x)$ should be hugging. This perhaps shows the deficiencies in linear scales\u2014they don't show us the big picture. They exaggerate discrepancies, which, in a relative sense, are tiny.\n\u2013\u00a0Joe\nDec 23 '20 at 18:39\n\u2022 @chaad NB If you look closely, then you can still see $f(x)$ and $g(x)$ moving apart in the graph. It's just that we don't consider this difference to be very important.\n\u2013\u00a0Joe\nDec 23 '20 at 18:49\n\u2022 @chaad Here is a plot of $\\log(x^2+x)$ versus $\\log(x^2)$. (I used the base-$10$ logarithm here, but any is fine.) Because we are using a more 'reasonable' scale, the two graphs are actually coming together, not going apart. Logarithmic scales are in many ways preferable, since they don't emphasise small discrepancies like linear scales do. I hope that answers your question.\n\u2013\u00a0Joe\nDec 23 '20 at 19:16", "date": "2021-10-28 06:22:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3959757/visual-intuition-for-the-definition-of-asymptotically-equivalent", "openwebmath_score": 0.8244909644126892, "openwebmath_perplexity": 155.02156073619918, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808701643912, "lm_q2_score": 0.8872045907347108, "lm_q1q2_score": 0.869709688219265}}
{"url": "https://quant.stackexchange.com/questions/30362/how-to-perform-monte-carlo-simulations-to-price-asian-options/32574", "text": "# How to perform Monte-Carlo simulations to price Asian options?\n\nIf I wish to price a fixed-strike Asian Call option via Monte-Carlo (This has no early-exercise), are my following steps correct?:\n\n1) Simulate random asset prices. (Milstein)\n\n$\\ d S(t) = \\ rS(t)dt + \\sigma S(t) d B(t)$\n\n$\\ S_{t+dt} = S_t + r S_tdt + \\sigma S_t \\sqrt{ dt}Z + \\frac{1}{2}\\sigma^2dt(Z^2-1)$\n\n2) Average the asset prices for each simulation.\n\n$\\ A[i]$ is the average for each simulation.\n\nI'll be using both Geometric and Arithmetic averages\n\n3) Calculate each payoff and discount it. Find the average of these payoffs\n\n$\\text{Payoff}[i]= \\exp[-r(T-t)] * \\max[A[i]-K,0]$\n\n$\\text{Average} = \\frac{1}{N}\\sum_{i=1}^N \\text{Payoff}[i]$\n\nI'm aware that there are some approximation formulae, Finite-Difference methods and closed-form solutions but I'm trying to focus on Monte-Carlo simulations for now.\n\n\u2022 Are you sure of the last term in your Milstein scheme? $S_t$ seems to be missing IMHo. The rest looks fine to me although what you call payoff is actually the discounted payoff. \u2013\u00a0Quantuple Sep 29 '16 at 16:58\n\u2022 This is interesting. I actually agree with you but looking at two different papers, frouah.com/finance%20notes/\u2026 and vlebb.leeds.ac.uk/bbcswebdav/orgs/SCH_Computing/FYProj/reports/\u2026. They seemed to both not include the $/ S_t$. \u2013\u00a0mathnoob Sep 29 '16 at 17:10\n\u2022 Take equation (16) of your paper and substitute the drift and diffusion terms as per the first lines of section 2.1. what do you get? Or simply analyse the dimensions of each term, or observe how the model should be homogeneous in space. \u2013\u00a0Quantuple Sep 29 '16 at 17:15\n\u2022 The final term doesn't match. I get $0.5 \\sigma_t^2 S_t dt(Z^2-1)$ \u2013\u00a0mathnoob Sep 29 '16 at 17:26\n\u2022 Thanks for pointing that out. I'd give you a thumbs up but I don't really know how to do it for comments. Still figuring out how to use this website. \u2013\u00a0mathnoob Sep 29 '16 at 17:32\n\n1. Instead of simulating the spot price, simulate its logarithm since the latter can be simulated exactly for any time step. $$\\ln S_{t + \\Delta t} = \\ln S_t + \\left( r - \\frac{1}{2} \\sigma^2 \\right) \\Delta t + \\sigma Z,$$ where $Z \\sim \\mathcal{N}(0, \\Delta t)$. You then just simply take the exponential of the simulated logarithmic price at each time step.\n\n2. OK\n\n3. OK\n\n4. Calculate the standard error of your estimate. This is just as important as computing the estimate itself and for example allows you to construct confidence intervals. Let $$\\text{Variance} = \\frac{1}{N - 1} \\sum_{i = 1}^N \\left( \\text{Payoff}_i - \\text{Average} \\right)^2$$ be your estimator of the variance. Then the standard error is\n\n$$\\text{Standard Error} = \\sqrt{\\frac{\\text{Variance}}{N}}.$$\n\nLooks good to me, although idk why you have (T-t) in the discounting... isn't big T the total time to maturity? What is little t in the equation? Shouldn't it just be exp[-rT] because you discount from the time of payoff which is the expiration of the option.\n\nKemna and Vorst (1990) [ download ] is a classic in Monte Carlo method for Asian option. Geometric mean, which can be analytically computed, is used as a control variate to reduce MC noise.", "date": "2021-05-11 06:28:32", "meta": {"domain": "stackexchange.com", "url": "https://quant.stackexchange.com/questions/30362/how-to-perform-monte-carlo-simulations-to-price-asian-options/32574", "openwebmath_score": 0.75138258934021, "openwebmath_perplexity": 636.0036123093302, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137879323497, "lm_q2_score": 0.8856314783461303, "lm_q1q2_score": 0.8697023227628102}}
{"url": "https://math.stackexchange.com/questions/1049808/asymptotic-development-of-a-recurrent-sequence", "text": "# Asymptotic development of a recurrent sequence\n\nLet $u_0 = 1$ and $u_{n+1} = \\frac{u_n}{1+u_n^2}$ for all $n \\in \\mathbb{N}$.\n\nI can show that $u_n \\sim \\frac{1}{\\sqrt{2n}}$, but I would like one more term in the asymptotic development, something like $u_n = \\frac{1}{\\sqrt{2n}}+\\frac{\\alpha}{n\\sqrt{n}} + o\\bigl(\\frac{1}{n^{3/2}}\\bigr)$.\n\nHere is the outline of my proof of $u_n \\sim \\frac{1}{\\sqrt{2n}}$:\n\n\u2022 $(u_n)$ is decreasing, and bounded from below by $0$, hence converges.\n\u2022 The limit $\\ell$ satisifies $\\ell = \\frac{\\ell}{1+\\ell^2}$, hence $\\ell = 0$.\n\u2022 A computation gives $v_n = u_{n+1}^{-2} - u_n^{-2} \\to 2$.\n\u2022 Using Ces\u00e0ro lemma, $\\frac{1}{n} \\sum_{k=0}^{n-1} v_k \\to 2$.\n\u2022 Hence $\\frac{1}{n} u_n^{-2} \\to 2$.\n\nLet $a_n=\\dfrac1{u_n^2}$. Then $$a_{n+1}-a_n = \\left(\\frac{1+u_n^2}{u_n}\\right)^2 - \\frac1{u_n^2} = 2+u_n^2 = 2+\\frac1{a_n}.$$ From this and $a_2=4$ we can find a successive sequence of estimates for $n\\ge2$: \\begin{align*} a_n &\\ge 2n; \\\\ a_n &\\le 2n + \\sum_{k=2}^{n-1}\\frac1{2k} < 2n+\\frac12\\log n; \\\\ a_n &\\ge 2n + \\sum_{k=2}^{n-1}\\frac1{2k+\\frac12\\log k} > 2n+\\int_2^n \\frac{dx}{2x+\\frac12\\log x} \\\\ &> 2n+\\int_2^n \\frac{dx}{2x} - \\int_2^n \\frac{\\frac12\\log x}{2x(2x+\\frac12\\log x)}dx = 2n+\\frac12\\log n - \\mathcal{O}(1). \\end{align*}\n\nThen $$u_n = a_n^{-1/2} = \\frac1{\\sqrt{2n}} \\left(1+\\frac{\\log n}{4n}+\\mathcal{O}\\bigg(\\frac1n\\bigg)\\right)^{-1/2} \\\\ = \\frac1{\\sqrt{2n}} \\left(1-\\frac12 \\cdot \\frac{\\log n}{4n} + \\mathcal{O}\\bigg(\\frac1n\\bigg) \\right) = \\frac1{\\sqrt{2n}} - \\frac1{8\\sqrt2}\\cdot \\frac{\\log n}{n^{3/2}} + \\mathcal{O}\\left(\\frac1{n^{3/2}}\\right).$$\n\n\u2022 Ooops, I started the sequence with $u_1=1$. But it changes only the constant in the error term. \u2013\u00a0G. K\u00f3s Dec 6 '14 at 18:47\n\u2022 I wonder if this method can be iterated to yield abitrary deep asymptotic expansions. \u2013\u00a0Ewan Delanoy Dec 9 '14 at 19:00\n\nNote: This answer does not contain a proof, but I'm fairly sure it's right. Posted in the hope that perhaps the answer will guide you towards a proof thereof.\n\nLooking at the graph you can figure out that in fact the next correction looks to be $n^{-3/2} \\log n$ dominating the next $n^{-3/2}$ term. Given this, one can do a plausible self-consistency check using the recurrence relation to find that the answer.\n\nSpecifically, suppose\n\n$$u_n \\sim \\frac \\alpha {\\sqrt{2n}} + \\frac {\\beta \\log n}{n^{3/2}}+\\text{a sufficiently nice smaller series in particular }\\mathcal{O}\\left(\\frac{1}{n^{3/2}}\\right)$$\n\nThen $$u_{n+1}=\\frac{u_n}{1+u_n^2}$$ implies $$u_{n+1}-\\frac{u_n}{1+u_n^2} \\sim \\frac 1 2 (\\alpha - 2 \\alpha^3) \\frac 1 {n^{3/2}}+\\cdots$$ so that $\\alpha=0,-1/\\sqrt 2,1/\\sqrt 2$ are the options. You have checked that $\\alpha=1/\\sqrt{2}$.\n\nSubstituting this back in gives $$u_{n+1}-\\frac{u_n}{1+u_n^2} \\sim \\left(-\\frac 1 {8\\sqrt 2} - \\beta\\right) \\frac 1 {n^{5/2}}+\\cdots$$ and hence it seems the only consistent result has $$\\boxed{\\displaystyle u_{n} \\sim \\frac{1}{\\sqrt{2n}}-\\frac{\\log n}{8\\sqrt 2 n^{3/2}}}$$\n\nNumerically, I find the error $$\\epsilon = \\frac{u_n - \\frac{1}{\\sqrt{2n}}}{-\\frac{\\log n}{8\\sqrt 2 n^{3/2}}} - 1\\approx \\frac{3.4}{\\log n} \\to 0$$ suggesting that indeed this is correct with the next term in the series being the expected $n^{-3/2}$ term.\n\nHowever, the coefficient of the $n^{-3/2}$ term cannot be determined by this self-consistency procedure, reflecting the fact that one can check this actually depends on the initial datum $u_0$.\n\nLeft to do: (Not that I think I'll come back to this myself, but others can!)\n\n\u2022 Rigorously show the above logarithmic correction.\n\u2022 Figure out the dependence of the next term on the initial data.\n\nMathematica code for my verification:\n\nmin = 2;\nmax = 5000000;\ndat = RecurrenceTable[{u[n + 1] == u[n]/(N[1, 50] + u[n]^2), u[0] == 1}, u, {n, min, max}];\nerrordat =\nTable[{M, (dat[[M - min + 1]] -\n1/Sqrt[2 M])/(-Log[M]/(8 Sqrt[2] M^(3/2))) - 1}, {M, min, max, 100000}];\nnextconst = ((dat[[max - min + 1]] -\n1/Sqrt[2 max])/(-Log[max]/(8 Sqrt[2] max^(3/2))) - 1)*Log[max]\nShow[ListPlot[errordat, AxesOrigin -> {0, 0}, PlotStyle -> PointSize[Large]], Plot[nextconst/Log[x], {x, min, max}, PlotStyle -> Red, PlotRange -> All]]\n\n\nOutput: Estimate of $3.446...$ for next constant, and a plot of the $\\epsilon$ against $n$ with a fitted curve based on a logarithmic next correction: Obviously the decay to 0 here is slow, which is to be expected given the predicted logarithmic correction. Feel free to go to larger ranges.\n\n\u2022 Wouldn't it be better to plot the difference between the asymptotic (1st term) and numerical results and see if the result brings about the correction term? Showing that the error goes to zero is not good enough. \u2013\u00a0Ron Gordon Dec 6 '14 at 19:08\n\u2022 I'm not entirely sure what you mean. The requirement that $\\epsilon \\to 0$ is precisely what it means for the boxed expression to be the first two terms of an asymptotic series, by definition. \u2013\u00a0Sharkos Dec 7 '14 at 11:58", "date": "2021-04-17 05:29:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1049808/asymptotic-development-of-a-recurrent-sequence", "openwebmath_score": 0.965732991695404, "openwebmath_perplexity": 782.0506571149555, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8918110368115781, "lm_q1q2_score": 0.8696957625710485}}
{"url": "https://math.stackexchange.com/questions/2239321/is-0-1-closed-covering-of-interval-0-1/2239492", "text": "# Is [0,1] closed covering of interval [0,1]?\n\nI know it's true in case of open interval and open covering.\n\nAlso, in this case what will be Lebesgue number ?\n\n$I = [0, 1]$ is closed as a subset of itself, making $\\{ I \\}$ a closed covering of $I$.\n$B_{\\delta}(x) \\cap I \\subset I$ for each $x \\in I$ and each $\\delta \\in \\mathbb{R}_{> 0}$, so that every such $\\delta$ is a Lebesgue number.\n\u2022 Yes but won't $\\delta$ tends to zero ? As we take x near to 0 or 1, we have to take $\\delta$ smaller correspondingly. \u2013\u00a0MeetR Apr 18 '17 at 2:52\n\u2022 The open balls in $I$ as a metric space are all of the form $B_{\\delta}(x_0) \\cap I = \\{ x \\in I : |x - x_0| < \\delta \\}$ for $x_0 \\in I$. \u2013\u00a0Dean Young Apr 18 '17 at 2:57", "date": "2019-05-20 16:25:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2239321/is-0-1-closed-covering-of-interval-0-1/2239492", "openwebmath_score": 0.8749234676361084, "openwebmath_perplexity": 184.51593638823664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.987375049232425, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8696770613217363}}
{"url": "https://math.stackexchange.com/questions/703288/prove-sum-n-1-inftye-sum-k-0n-frac1k-1", "text": "# Prove $\\sum_{n=1}^\\infty(e-\\sum_{k=0}^n\\frac1{k!})=1$\n\nThis comes from the comments section of this question here, credits Lucian.\n\nThe statement is\n\n$$\\sum_{n=1}^\\infty\\left(e-\\sum_{k=0}^n\\frac1{k!}\\right)=1$$\n\nThis looks really interesting, so I was wondering if anyone has any ideas/suggestions or even better the whole proof?\n\nDisclaimer: I added that link specifically because this might look like homework, leading to \"What have you tried?\". So the disclaimer is, I actually haven't tried anything yet, because yes, asking the community is just easier. It would be great if someone could answer this though since I, personally at least, think that this equation is very interesting and adds value to any math enthusiast stumbling upon it here someday. :) I will myself be trying it meanwhile too though(it really is interesting) , and add and accept my solution if I am successful and there are no answers yet.\n\nI tried wolfram alpha for $\\sum_{k=0}^n\\frac1{k!}$ and it yields\n\n$$\\sum_{k=0}^n\\frac1{k!} = \\frac{e\\Gamma (n+1,1)}{\\Gamma(n+1)}$$\n\nwhere $\\Gamma(a,r)$ is the incomplete gamma function and $\\Gamma(a)$ is the euler gamma function. Documentation here. Might be of help, might not. Just thought it'd be worth adding.\n\nThe first thought that comes into my head is to write $$e - \\sum_{k=0}^n \\frac{1}{k!} = \\sum_{k=n+1}^\\infty \\frac{1}{k!},$$ so that the given sum is equivalent to a double sum: \\begin{align*} \\sum_{n=1}^\\infty \\sum_{k=n+1}^\\infty \\frac{1}{k!} &= \\sum_{k=2}^\\infty \\sum_{n=2}^k \\frac{1}{k!} \\\\ &= \\sum_{k=2}^\\infty \\frac{1}{(k-2)!k} \\\\ &= \\sum_{k=2}^\\infty \\frac{1}{(k-1)!} - \\frac{1}{k!} = 1. \\end{align*}. Of course, a somewhat more rigorous argument is needed to justify the interchange of the order of summation. A minor modification to consider the original series' partial sums, and taking the limit to get the double infinite sum, is sufficient and is left as an exercise.\n\n\u2022 Oh that was fast. Thanks. \u2013\u00a0Guy Mar 7 '14 at 18:11\n\u2022 All the terms are positive, so it is okay to exchange sums. \u2013\u00a0abnry Mar 7 '14 at 18:12\n\u2022 I am over my daily vote limit. I'll upvote this tomorrow. \u2013\u00a0Guy Mar 7 '14 at 18:17\n\u2022 @nayrb of course, you are right. ^_^ \u2013\u00a0heropup Mar 7 '14 at 18:17\n\n$$\\sum_{n=1}^\\infty\\left(e-\\sum_{k=0}^n\\frac1{k!}\\right) =\\sum_{n=1}^\\infty \\int_0^1 \\exp(u) \\frac{(1-u)^{n}}{n!} du$$as everything is positive: $$\\sum_{n=1}^\\infty\\left(e-\\sum_{k=0}^n\\frac1{k!}\\right)= \\int_0^1 \\exp(u) \\sum_{n=1}^\\infty \\frac{(1-u)^{n}}{n!} du \\\\= \\int_0^1 \\exp(u)(\\exp(1-u) - 1) du = e - \\int^1_0 \\exp u du = 1$$\n\n\u2022 Might be missing something obvious, but $$\\sum_{n=1}^\\infty\\left(e-\\sum_{k=0}^n\\frac1{k!}\\right) =\\sum_{n=1}^\\infty \\int_0^1 \\exp(u) \\frac{(1-u)^{n}}{n!} du$$ how? \u2013\u00a0Guy Mar 7 '14 at 18:23\n\u2022 this is the Taylor formula, obtained via $n$ integrations by parts. \u2013\u00a0mookid Mar 7 '14 at 18:23\n\u2022 Ah, I am unfamiliar with this. I'll try to prove this myself though. Rest of the solution is great. (+1)(after my vote limit time restriction is lifted that is) \u2013\u00a0Guy Mar 7 '14 at 18:25\n\u2022 you obtain it via $\\exp 1 - 1 = \\int_0^1 1 \\exp u du = \\ldots$ integrating the polynomial and derivating the $\\exp$. \u2013\u00a0mookid Mar 7 '14 at 18:27\n\u2022 Whoa dude, spoiler alert. :p \u2013\u00a0Guy Mar 7 '14 at 18:30", "date": "2019-11-20 01:23:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/703288/prove-sum-n-1-inftye-sum-k-0n-frac1k-1", "openwebmath_score": 0.9805433750152588, "openwebmath_perplexity": 754.3151392900954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750484894196, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8696770498538792}}
{"url": "https://math.stackexchange.com/questions/2325458/does-introducing-new-open-sets-in-topology-improves-its-t-axiom-properties", "text": "# Does introducing new open sets in topology improves it's \u201cT\u201d-axiom properties?\n\nThis question might be silly, however I'd like to be $100\\%$ sure.\n\nSuppose we have some topology $\\tau$ on a set $X$ which gives us the $T_i$ space for some $i<6$. It's totally possible to improve topological properties (here I'd like to improve separation axioms) - simply consider basic anti-discrete topology, which consists only of $\\emptyset$ and $X$.\n\nMy question is about the opposite situation. Is it possible to have topology $\\tau_1$ on set $X$ and a bigger topology $\\tau_2 \\supset \\tau_1$, such that $(X,\\tau_1)$ is $T_i$ space while $(X,\\tau_2)$ does not fulfill $T_i$ space axiom.\n\nI'm pretty sure such situation is not possible, but if it is - please provide me with a proper counterexample.\n\nFor $i\\leqslant 2$ I'm sure it's not the case, I wonder if making topology larger (which at the same time introduces new closed sets) can prevent $(X,\\tau)$ from being $T_i$ space.\n\n\u2022 Which cases do you speak about? If it's about $T_i$ for which $i<3$, then I've already pointed that out in my post. If you do speak about $i\\geqslant 3$ please point out why it's so trivial, because I don't really see it. \u2013\u00a0I_Really_Want_To_Heal_Myself Jun 16 '17 at 20:05\n\u2022 did you read my answer? \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jun 16 '17 at 21:13\n\u2022 The point is, for $T_3$ and higher we have to consider closed sets as well, not just points. Adding open sets also adds closed sets, so more conditions to fulfill. \u2013\u00a0Henno Brandsma Jun 16 '17 at 22:06\n\u2022 It's a very good Q. \u2013\u00a0DanielWainfleet Jun 18 '17 at 2:39\n\nIf a Hausdorf space is given a finer topology,\nit's still Hausdorf. Proof is very simple.\n\nThe set of real numbers is regular and normal.\nAdd to the usual topolopy the set of rationals and\nit is neither regular nor normal, yet still Hausdorf.\n\nPick a set $X$ along with an equivalence relation $\\sim$. Consider the topology $\\tau$ on $X$ in which the open sets are the sets that are unions of equivalence classes.\n\nIn this space the closed and open sets coincide. This space is clearly $T_3$ .\n\nConsider the refinement on $X$ which consists of taking one of the equivalence classes $C$ and refining the topology on $C$ into a non-$T_3$ topology.\n\n\u2022 I think this approach works for the other ones also (Except $T_0,T_1,T_2$ and $T_{2.5}$) \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jun 16 '17 at 20:14\n\nAn example where $\\tau_1$ is $T_6$ but $\\tau_2$ is NOT $T_6.$\n\nLet $I=[0,1].$ Let $\\tau_1$ be the usual topology on $I^2.$ Let $<_L$ be the lexicographic order on $I^2.$ That is $(x,y)<_L(x',y')\\iff (x<x'\\lor [x=x'\\land y<y']).$ The topology $\\tau_2$ induced on $I^2$ by $<_L$ is stronger than $\\tau_1.$ ( If $A,B$ are open real intervals then $(A\\times B)\\cap I^2$ is a union of $<_L$-open intervals.)\n\nThen $\\tau_1$ and $\\tau_2$ are $T_5$ topologies. (For $\\tau_2,$ all linear spaces are $T_5$).\n\nBut $\\tau_1$ is $T_6$ and $\\tau_2$ is NOT $T_6.$\n\n(I). $\\tau_1$ is $T_6$ because it is metrizable.\n\n(II). Notation: ]a,b[ and [a,b[ are real intervals. (a,b) is an ordered pair.\n\nLet $A=I\\times \\{0,1\\}.$ Then $A$ is $\\tau_2$-closed but $A$ is not a $G_{\\delta}$ set in the $\\tau_2$ topology.\n\nProof: If $A\\subset U\\in \\tau$ and $x\\in [0,1[$ there is a $<_T$-open interval $J$ with $(x,1)\\in J\\subset U.$ And since $(x,1)$ is in the $\\tau_2$-closure of $\\{(x',y')\\in I^2: (x,1)<_L (x',y')\\},$ there is $(x',y')\\in J$ with $x'>x.$ Since $J$ is a $<_T-$interval, therefore $]x,x'[\\times I\\subset J\\subset U.$\n\nSo let $B(U)=\\cup \\{]x,x'[ \\;: \\;0\\leq x<1\\;\\land \\;]x,x'[\\times I\\subset U\\}.$ With respect to the usual topology on $I,$ the set $B(U)$ is open and dense in $I.$ (If $0\\leq x<y\\leq 1$ there is $x'\\in ]x,y[\\cap B(U)$.)\n\nSo if $A\\subset U_n\\in \\tau_2$ for $n\\in \\mathbb N$ then $$\\cap_{n\\in \\mathbb N}U_n\\supset (\\cap_{n\\in \\mathbb N}B(U_n))\\times I.$$ By the Baire Category Theorem applied to the usual topology on $I$ we have $\\cap_{n\\in \\mathbb N}B(U_n)\\ne \\phi.$\n\nTherefore $\\cap_{n\\in \\mathbb N}U_n\\ne A.$\n\nRemark: One-member subsets of $I^2$ are closed $G_{\\delta}$ sets in $\\tau_2$ but the closed set $A$ is not $G_{\\delta}$ in $\\tau_2$.\n\n\u2022 I do like this one, although the notation makes it quite hard to see at the first glance. \u2013\u00a0I_Really_Want_To_Heal_Myself Jun 18 '17 at 21:31\n\u2022 I don't usually use $]a,b[$ for open interval but I also have ordered pairs involved. And I made it very detailed to make sure I got it right. The lex-order on $I^2$ is suggested as a hint in General Topology by Engelking in an exercise asking for a space with the property in my \"Remark\". \u2013\u00a0DanielWainfleet Jun 19 '17 at 16:14\n\nA $T_6$ topology $\\tau$ and a finer topology $\\tau'$ that is not even $T_4:$\n\nLet $\\tau$ be the usual (standard) topology on $\\mathbb R^2.$\n\nLet $S$ be the Sorgenfrey line (a.k.a. the lower limit topology on the set $\\mathbb R$), which has a base of all real intervals $[x,y).$ Let $\\tau'$ be the product topology on $S^2.$\n\nA Q that has appeared repeatedly on this site is to prove that $S^2$ is not $T_4.$ The solution is to apply Jones' Lemma: If $X$ is a separable space with a closed discrete subspace $Y$ where the cardinal of $Y$ is $2^{\\aleph_0}$ then $X$ is not normal. For the case $X=S^2$ let $Y=\\{(x,-x): x\\in \\mathbb R\\}.$", "date": "2020-02-22 17:29:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2325458/does-introducing-new-open-sets-in-topology-improves-its-t-axiom-properties", "openwebmath_score": 0.9089652895927429, "openwebmath_perplexity": 186.0619697190165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964177527898, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8696295104574814}}
{"url": "https://math.stackexchange.com/questions/2711621/alternative-for-calculating-the-nth-of-quadratic-sequence/2711632", "text": "# Alternative for calculating the nth of quadratic sequence\n\nGiven the quadratic sequence $$f(n)=1, 7, 19, 37, \\cdots$$\n\nTo calculate the $f(n)$ for $n\\ge1$. $$f(n)=an^2+bn+c$$\n\nWe start with the general quadratic function, then sub in for $n:=1,2$ and $3$\n\n$$f(1)=a+b+c$$ $$f(2)=4a+2b+c$$ $$f(3)=9a+3b+c$$\n\nNow solve the simultaneous equations\n\n$$a+b+c=1\\tag1$$ $$4a+2b+c=7\\tag2$$ $$9a+3b+c=19\\tag3$$\n\n$(2)-(1)$ and $(3)-(2)$\n\n$$3a+b=6\\tag4$$ $$5a+b=12\\tag5$$\n\n$(5)-(4)$ $$a=3$$ $$b=-3$$ $$c=1$$\n\n$$f(n)=3n^2-3n+1$$\n\nThis method is very long. Is there another easy of calculating the $f(n)$?\n\n\u2022 Yes: just do what you did, but don't substitute the numbers; instead, keep $f(1), f(2), f(3)$ throughout. Then you end up with a fully general formula. \u2013\u00a0Patrick Stevens Mar 28 '18 at 11:13\n\nAnother standard way is to calculate a difference scheme and then to work backwards: $$\\begin{matrix} 0 & & 1 & & 2 & & 3 & & 4 \\\\ & & 1 && 7 && 19 && 37 \\\\ && &6& & 12 && 18 & \\\\ &&&& 6 && 6 && \\\\ \\end{matrix} \\Rightarrow \\begin{matrix} & 0 & & 1 & & 2 & & 3 & & 4 \\\\ \\color{blue}{c}= &\\color{blue}{1} & & 1 && 7 && 19 && 37 \\\\ \\color{blue}{a+b}= & &\\color{blue}{0}&&6& & 12 && 18 & \\\\ \\color{blue}{2a}= &&& \\color{blue}{6}&& 6 && 6 && \\\\ \\end{matrix}$$ $$\\Rightarrow a = 3, \\; b= -3, \\; c = 1 \\Rightarrow f(n) = 3n^2-3n+1$$\n\nYes!\n\nYou know $f(1)= 1 \\implies f(1) - 1 = 0 \\implies f(x) - 1$ has a root at x=1\n\nNow,\n\n$f(x) - 1 = a(x-1)(x-b)$\n\nPut $x = 2$, $a(2-b) = 6$\n\nPut x = 3, $a*2*(3-b) = 18$\n\nDivide both equation, we get $b = 0$ and a = 3\n\n$f(x) - 1 = 3*(x-1)x \\implies f(x) = 3x^2 - 3x + 1$\n\nTake a better basis. Namely, $\\{(n-1)(n-2),(n-1)(n-3),(n-2)(n-3)\\}$. If $$f(n) = \\alpha(n-1)(n-2) + \\beta(n-1)(n-3) + \\gamma(n-2)(n-3),$$ then: $$f(1) = 0 + 0 + \\gamma(1-2)(1-3),$$ $$f(2) = 0 + \\beta(2-1)(2-3) + 0,$$ $$f(3) = \\alpha(3-1)(3-2) + 0 + 0.$$\n\nA better explained version of @trancelocation's answer:\n\n$$\\begin{array}{} 1 & & 7 & & 19 & & 37 \\\\ & 6 & & 12& &18 & \\\\ & & 6 & & 6 & & \\\\ \\end{array}$$\n\nThe second row has equation $6n$. Therefore, $\\big(a(n+1)^2+b(n+1)+c \\big) - \\big(an^2+bn+c \\big)$ $= 6n$, and so:\n\n$$\\big(a(n^2+2n+1)+b(n+1)+c \\big) - \\big(an^2+bn+c \\big) = 6n$$ $$(2n+1)a + b = 6n$$\n\n$a=3$ gives $6n+3$, and so $b=-3$. 'Plugging' $n=1$ into $3n^2-3n$ gives $0$, $c=1$, which gives $3n^2-3n+1$.", "date": "2021-03-06 23:55:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2711621/alternative-for-calculating-the-nth-of-quadratic-sequence/2711632", "openwebmath_score": 0.9411802291870117, "openwebmath_perplexity": 475.03723669077266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.985496420308618, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.8696295081421439}}
{"url": "https://math.stackexchange.com/questions/2767678/we-have-two-coins-a-and-b-for-each-toss-of-coin-a-the-probability-of-getting", "text": "# We have two coins, A and B. For each toss of coin A, the probability of getting head is 1/2\u2026\n\nWe have two coins, A and B. For each toss of coin A, the probability of getting head is 1/2 and for each toss of coin B, the probability of getting Heads is 1/3. All tosses of the same coin are independent. We select a coin at random and toss it till we get a head. The probability of selecting coin A is \u00bc and coin B is 3/4. What is the expected number of tosses to get the first heads?\n\nThe above problem is taken from the website https://www.analyticsvidhya.com/blog/2017/04/40-questions-on-probability-for-all-aspiring-data-scientists/ question 11\n\nMy solution is either 1/(1/4*1/2 + 3/4*1/3)=8/3, including the success toss, or 5/3, not including the success toss. My understanding is that it is geometric distribution question.\n\nBut the solution provided is 2.75, with the following explanation: \"If coin A is selected then the number of times the coin would be tossed for a guaranteed Heads is 2, similarly, for coin B it is 3. Thus the number of times would be Tosses = 2 * (1/4)[probability of selecting coin A] + 3*(3/4)[probability of selecting coin B] = 2.75\"\n\nIs the solution provided incorrect? Or am I missing something?\n\nIt is just the application of the Law of total Expectation.\n\n$$E(X)=\\sum_{i} E(X|A_i)\\cdot P(A_i)$$\n\nIn your case the definitions are: $X:=$random variable for the number of heads, $A_1$: Coin with head probabilty equal to $\\frac12$ is selected, $A_2$: Coin with head probabilty equal to $\\frac13$ is selected.\n\nDue to the geometric distribution (as you mentioned) $E(X|A_1)=\\frac{1}{p_1}=2$ and $E(X|A_2)=\\frac{1}{p_2}=3$. And $P(A_1)=\\frac14,P(A_2)=\\frac34$. Consequently we have\n\n$$E(X)=E(X|A_1)\\cdot P(A_1)+E(X|A_2)\\cdot P(A_2)=\\frac14\\cdot 2+\\frac34\\cdot 3=\\frac{11}{4}$$\n\nI think you are already familiar to the Law of total probability. This has a similar structure as the Law of total expectation, but for probabilities:\n\n$$P(A)=\\sum_{i} P(A|B_i)\\cdot P(B_i)$$\n\nOne may say that $E(X)$ is the weighted mean of the conditional expectations.\n\nThe provided answer of $2.75$ is correct. I wouldn't say that the explanation is great. The use of the word 'guaranteed' is misleading. It makes it sound like if the fair coin is selected there is a guarantee of getting a heads in $2$ tosses, which is, of course, not the case.\n\nYour proposed solution tries to make a weighted average of geometric distributions into a geometric distribution, which it is not.\n\nHere is an explanation for the correct answer of $2.75$:\n\nIf you were to repeat the experiment many times, about $\\frac14$ of the time (when you got the fair coin), it would take an average of $2$ flips for heads; and $\\frac34$ of the time it would take an average of $3$ flips. So in $n$ repetitions of the experiment, it would take about $\\frac14 \\cdot n \\cdot 2 + \\frac34\\cdot n\\cdot 3$ total flips. This works out to about $2.75$ flips per trial on average to get heads.", "date": "2019-11-15 21:01:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2767678/we-have-two-coins-a-and-b-for-each-toss-of-coin-a-the-probability-of-getting", "openwebmath_score": 0.8348644971847534, "openwebmath_perplexity": 231.28077818769313, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98549641775279, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.869629493698352}}
{"url": "https://math.stackexchange.com/questions/1696366/let-a-be-the-n-times-n-matrix-with-a-1-in-every-entry-what-are-the-eigenv", "text": "# Let $A$ be the $n\\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable?\n\nLet $A$ be the $n\\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?\n\nHaving some trouble with this one. I tried using the fact that $\\lambda$ is an eigenvalue of $A$ iff there exist non-zero solutions to $Ax=\\lambda x$. Well, clearly the $1\\times n$ non-zero vector $x$ with all entries equal is an eigenvector of $A$, with corresponding eigenvalue $\\lambda=n$, since $Ax=nx$. But I can't seem to get much further than that in terms of finding eigenvalues for $A$, any hints/suggestions?\n\n\u2022 What is the rank of $A$? What does the rank-nullity theorem tell you about the nullity of $A$? What does that imply about the eigenvalue zero? What does the trace say about the eigenvalues? \u2013\u00a0JMoravitz Mar 14 '16 at 0:03\n\u2022 \u2013\u00a0JMoravitz Mar 14 '16 at 0:06\n\u2022 So the idea is that in general for a matrix with a $1$ in every entry, $p(\\lambda)=(-1)^n\\lambda^{n}+(-1)^{n-1}n\\lambda^{n-1}$, so the only eigenvalues are $\\lambda =0,n$? \u2013\u00a0Borat Sagdiyev Mar 14 '16 at 0:17\n\nObserve:\n\n\u2022 $Rank(A)=1$ since all columns are identical and nonzero (rank=number of linearly independent columns)\n\u2022 $nullity(A)=n-1$ by the rank-nullity theorem (rank+nullity=n)\n\u2022 $trace(A)=n$ by direct calculation (trace=sum of diagonal entries)\n\nSince $nullity(A)=n-1=nullity(A-0I)$ is the dimension of the eigenspace associated with the eigenvalue zero, we know that we have zero as an eigenvalue of algebraic multiplicity at least $n-1$. This implies that zero is either of algebraic multiplicity $n-1$ or it is of algebraic multiplicity $n$.\n\nSince $n=trace(A)=\\sum \\lambda\\cdot algmu(\\lambda)$, we know that zero is not of multiplicity $n$ and that the remaining eigenvalue must be $n$ itself.\n\nAs for the question of if it is diagonalizable, as mentioned earlier, we know that $geomu(0)=n-1$ and that $n$ is an eigenvalue (and thus must have geometric multiplicity at least one like every eigenvalue) and has geometric multiplicity exactly one (as it can't exceed its algebraic multiplicity). Since the sum of the geometric multiplicities is $(n-1)+1=n$, it is seen to be diagonalizable.\n\n\u2022 $A$ is diagonalizable since $dim(E_0)+dim(E_n)=n$, $dim(E_0)=n-1=mult_0$, and $dim(E_n)=1=mult_n$, right? But how would you solve $(A-nI)x=0$ to find $E_n$? \u2013\u00a0Borat Sagdiyev Mar 14 '16 at 3:05\n\u2022 @Borat by inspection is a valid method which you already did in your original post. If you were to do it manually, consider adding all rows together in the row reduction process and see where that gets you. \u2013\u00a0JMoravitz Mar 14 '16 at 3:13\n\u2022 Oh, I see. So $E_n=span\\{(1,...,1)\\}$? \u2013\u00a0Borat Sagdiyev Mar 14 '16 at 3:18\n\nNote that $A$ is real and symmetric, so it is diagonalizable. We have $A^2=nA$, so all eigenvalues of $A$ satisfy $\\lambda^2=n\\lambda$: this tells us that the only eigenvalues are $0$ and $n$. Since the trace of $A$ (i.e., the sum of the eigenvalues) is $n$, we deduce that the multiplicity of $n$ is one and that the multiplicity of $0$ is $n-1$.\n\nAs for eigenspaces, if $e$ is the vector with all entries equal to one, we have $Ae=ne$, so $e$ spans the eigenspace of $n$. The eigenspace for $0$ consists of the kernel of $A$: $0=Ax=(x_1+\\cdots+x_n,\\ldots,x_1+\\cdots+x_n)^t$. We can make a basis of this subspace by taking the vectors $$\\begin{bmatrix}1\\\\-1\\\\0\\\\0\\\\0\\\\ \\vdots\\\\ 0\\end{bmatrix},\\ \\ \\begin{bmatrix}0\\\\1\\\\-1\\\\0\\\\0\\\\ \\vdots\\\\ 0\\end{bmatrix},\\ \\ \\begin{bmatrix}0\\\\0\\\\1\\\\-1\\\\0\\\\ \\vdots\\\\ 0\\end{bmatrix},\\ \\ \\ldots\\ \\ ,\\ \\ \\begin{bmatrix}0\\\\0\\\\0\\\\ \\vdots \\\\0\\\\1\\\\-1\\end{bmatrix}.$$\n\nHere is a matrix $P$ that I made up some time ago. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal and therefore independent. The columns are eigenvectors for your matrix, in the 10 by 10 case. Multiply your matrix of all ones (10 by 10) on the left, this matrix on the right. What happens? $$\\left( \\begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \\end{array} \\right).$$\n\nIf $\\mathbf{1}$ denotes the column vector of all ones, what is $\\mathbf{1} \\mathbf{1}^T$?\n\nNow, match this to the outer product form of the eigendecomposition to get the eigenvectors and eigenvalues.\n\nConsider the matrix $B = A - \\lambda I$, where $\\lambda \\in \\text{Spec}(A)$. Then $\\lambda_{1} = n - \\lambda$ is an eigenvalue of $B$ with an associated eigenvector $v_{1} = \\begin{pmatrix} 1 \\\\ 1 \\\\ 1\\\\ \\cdot \\\\ \\cdot \\\\ \\cdot \\\\ 1 \\\\1 \\\\1\\\\ \\end{pmatrix}$.\n\nThe remaining eigenvalues of $B$ are $\\lambda_{2} = \\lambda_{3} =...=\\lambda_{n} = -\\lambda$ and the corresponding eigenvectors are:\n\n$v_{2} = \\begin{pmatrix} 1 \\\\ -1 \\\\ 0\\\\ \\cdot \\\\ \\cdot \\\\ \\cdot \\\\ 0 \\\\0 \\\\0\\\\ \\end{pmatrix}$, $v_{3} = \\begin{pmatrix} 0 \\\\ 1 \\\\ -1\\\\ \\cdot \\\\ \\cdot \\\\ \\cdot \\\\ 0 \\\\0 \\\\0\\\\ \\end{pmatrix}$, ..., $v_{n} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0\\\\ \\cdot \\\\ \\cdot \\\\ \\cdot \\\\ 0 \\\\1 \\\\-1\\\\ \\end{pmatrix}$.\n\nNow, it is easier to see that $v_{1}, ..., v_{n}$ are linearly independent and they are eigenvectors of $A$ with $v_{1}$ associated with the eigenvalue $\\lambda =n$ and $v_{j}$ for $j \\geq 2$ associated with $\\lambda = 0$. From the linearly independence of these eigenvectors we see $A$ is diagonalizable.", "date": "2019-07-24 08:30:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1696366/let-a-be-the-n-times-n-matrix-with-a-1-in-every-entry-what-are-the-eigenv", "openwebmath_score": 0.9701855182647705, "openwebmath_perplexity": 92.53505919731597, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964177527897, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8696294921747943}}
{"url": "https://brilliant.org/discussions/thread/a-tv-show-paradox/", "text": "\u00d7\n\nSuppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, \"Do you want to pick door No. 2?\" Is it to your advantage to switch your choice?\n\nNote by Iq 131\n4\u00a0years, 4\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nYes. Since the probability of choosing a goat in the first try is 2/3 , since the other goat will be revealed, behind the other door should be the car. This famous problem as far as i know is The Monty Hall Problem.\n\n- 4\u00a0years, 4\u00a0months ago\n\n- 4\u00a0years, 4\u00a0months ago\n\nWhere the same article also says that the question as posed is incomplete (what if the host knows Door 2 has a goat and hence offering the option; never offering the option if the player originally picked a goat?). So there is no satisfying answer for this problem as stated. Only after clarifying further that the problem can be solved.\n\n- 4\u00a0years, 4\u00a0months ago\n\nProbability of finding a car in door 2 is 2/3 whereas in door 1 is 1/2 . Apply the theory of probability whenever options seems to be close\n\n- 4\u00a0years, 4\u00a0months ago", "date": "2018-01-23 17:46:05", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/a-tv-show-paradox/", "openwebmath_score": 0.9892308115959167, "openwebmath_perplexity": 2342.5994144935607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9626731147976794, "lm_q2_score": 0.9032942164664239, "lm_q1q2_score": 0.8695770569444615}}
{"url": "https://www.physicsforums.com/threads/moment-of-inertia-of-a-bar-with-two-spheres.702749/", "text": "# Moment of inertia of a bar with two spheres\n\n1. Jul 24, 2013\n\n### duplaimp\n\n1. The problem statement, all variables and given/known data\nA dumbbell has two spheres of radius 0.10m and mass 10kg (each). The two spheres are connected using a bar with length 1.0m and mass 12kg.\nWhat is the moment of inertia of the dumbbell about an axis perpendicular to an axis at the bar's center of mass?\nMoment of inertia of a sphere through its center of mass: $\\frac{2}{5}MR^{2}$\nMoment of inertia of a bar through its center of mass: $\\frac{2}{12}ML^{2}$\n\n3. The attempt at a solution\nI tried the following:\n(sphere) $Ip = Icm + Md^{2}$ <=> Ip = $\\frac{2}{5} MR^{2} + M(\\frac{L}{2} + R)^2$ <=> Ip = $\\frac{7}{5} M R^{2} + M\\frac{L^{2}}{4}$\n(bar) I = $\\frac{2}{12} ML^{2}$\n\nThen I multiplied Ip by 2 and sum I (of the bar). Itotal = 2Ip + I\n\nBut the value I am getting is wrong. What I am doing wrong?\n\nLast edited: Jul 24, 2013\n2. Jul 24, 2013\n\n### TSny\n\nHello.\n\nI'm not sure I understand the phrase \"about an axis perpendicular to an axis at the bar's center\".\n\nBut I did note a couple of things.\n\n$(\\frac{L}{2} + R)^2 \\neq \\frac{L^2}{4}+R^2$\n\nAre you sure the moment of inertia of a rod has a factor of 2/12?\n\n3. Jul 24, 2013\n\n### SteamKing\n\nStaff Emeritus\nInstead of writing equations off the cuff as it were, set up the problem in an organized manner.\n\nCode (Text):\n\nItem \u00a0 \u00a0 \u00a0Mass \u00a0 \u00a0 \u00a0C.O.M. \u00a0 \u00a0 Moment \u00a0 M*Dist^2 \u00a0 \u00a0MMOI\nDist.\n\nSphere 1 \u00a0 10 \u00a0 \u00a0 \u00a0 \u00a00.55 \u00a0 \u00a0 \u00a0 \u00a05.5 \u00a0 \u00a0 \u00a0 3.03 \u00a0 \u00a0 (2/5)MR^2\nSphere 2 \u00a0 10 \u00a0 \u00a0 \u00a0 -0.55 \u00a0 \u00a0 \u00a0 -5.5 \u00a0 \u00a0 \u00a0 3.03 \u00a0 \u00a0 (2/5)MR^2\nBar \u00a0 \u00a0 \u00a0 \u00a012 \u00a0 \u00a0 \u00a0 \u00a0 0.0 \u00a0 \u00a0 \u00a0 \u00a00.0 \u00a0 \u00a0 \u00a0 0.0 \u00a0 \u00a0 \u00a0ML^2/12\n----------------------------------------------------------\nTotals\n\nAdd up the totals for each column and use the Parallel Axis Theorem\nTo find the MMOI for the bar bell.\n\n4. Jul 24, 2013\n\n### duplaimp\n\nOops.. The factor is 1/12 and I was thinking in $(L*R)^{2}$ instead of $(L+R)^{2}$\n\nSo, trying again I've got the right result but am I doing it right?\n\nI did this:\n\n(sphere) $Ip = Icm + Md^{2}$ <=> Ip = $\\frac{2}{5} MR^{2} + M(\\frac{L}{2} + R)^2$ <=> Ip = $\\frac{2}{5} M R^{2} + M(\\frac{L^{2}}{4}+2\\frac{LR}{2}+R^{2})$\n(bar) I = $\\frac{1}{12} ML^{2}$\n\nThen plugged in the values of each, multiplied I of sphere by two and did a sum of I of bar and I of sphere. Is this correct?\n\n5. Jul 24, 2013\n\n### duplaimp\n\nOops.. The factor is 1/12 and I was thinking in $(L*R)^{2}$ instead of $(L+R)^{2}$\n\nSo, trying again I've got the right result but am I doing it right?\n\nI did this:\n\n(sphere) $Ip = Icm + Md^{2}$ <=> Ip = $\\frac{2}{5} MR^{2} + M(\\frac{L}{2} + R)^2$ <=> Ip = $\\frac{2}{5} M R^{2} + M(\\frac{L^{2}}{4}+2\\frac{LR}{2}+R^{2})$\n(bar) I = $\\frac{1}{12} ML^{2}$\n\nThen plugged in the values of each, multiplied I of sphere by two and did a sum of I of bar and I of sphere. Is this correct?\n\n6. Jul 24, 2013\n\n### TSny\n\nYes, that looks correct assuming the axis of rotation is as shown.\n\nI would probably not bother to expand out the $(\\frac{L}{2}+R)^2$, but just leave it as $I_p = \\frac{2}{5} MR^{2} + M(\\frac{L}{2} + R)^2$. The expression is more compact that way and it will be easier to evaluate numerically.\n\n#### Attached Files:\n\n\u2022 ###### Fig 1.png\nFile size:\n1.4 KB\nViews:\n84\n7. Jul 25, 2013\n\n### duplaimp\n\nThe axis of rotation is as you shown. Thank you!", "date": "2017-10-17 23:20:39", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/moment-of-inertia-of-a-bar-with-two-spheres.702749/", "openwebmath_score": 0.5594244003295898, "openwebmath_perplexity": 2814.772541144421, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846634557752, "lm_q2_score": 0.8887588038050466, "lm_q1q2_score": 0.8695479831541578}}
{"url": "https://math.stackexchange.com/questions/2723526/how-many-lines-tangent-to-the-graph-of-f-are-parallel-to-the-line", "text": "# How many lines tangent to the graph of $f$ are parallel to the line\n\nI'm getting stuck with this problem:\n\nHow many lines tangent to the graph of $f(x)=x^3+3x$ are parallel to the line $y=6x+1$?\n\nWhat I've done is this:\n\nSince we're looking for lines parallel to $y=6x+1$ then obviously we're looking for lines of the form $y=6x+b$, so the real question is finding the values of $b$ that satisfy the conditions.\n\nSince the lines are tangent to $f$ we need the derivative of $f$, that is $f'(x)=3x^2+3$.\n\nSince the derivative of $f$ represents the slope of line tangent to $f$, we need $f'(x)=3x^2+3=6$ and we get the solutions $x=1$ and $x=-1$.\n\nAnd this is where I'm stuck. There's some insight that I'm not having, I guess.\n\n*Edit\n\nThere is a follow-up question asking me to produce the equations of those two lines. I didn't want to share because I'm aware of people abusing MSE for their homework. This question was created by the Math department of my university and the official answer is $$y = 6x + 2$$ and $$y = 6x - 2$$\n\nI asked my tutor, and he arrived at the official answers this way:\n\n$$x\u00b3 + 3x = 6x + b => x\u00b3 - 3x = b$$\n\nThen you plug in the $+/- 1$ found earlier to get $b = +/-2$ and finally arrive at the official answer for the equations of the lines:\n\n$$y = 6x +/- 2$$\n\n\u2022 As @amd said, you're not stuck -- you're done! :-) \u2013\u00a0zipirovich Apr 5 '18 at 18:51\n\nYou\u2019ve already solved the problem: there are two such tangent lines. The function is well-behaved and has a tangent at each point, and you\u2019ve determined that there are two points at which the derivative has the right value. There\u2019s no need to go any further with this unless there\u2019s a follow-up question that asks you to produce the equations of those lines.\n\n\u2022 Yes, that should have made sense to me... there are two solutions to f'(x) = 6 so there are 2 lines tangent to f(x) and parallel to y=6x+1. There is indeed a follow-up question that asks to produce the equations of those lines. I just didn't want to over rely on MSE because I'm well aware MSE has a problem with people coming here asking other people to do their homework. \u2013\u00a0Victor S. Apr 5 '18 at 19:35\n\nNote: The question asks only for how many so the answer is immeediately two. For the actual equations of the tangents:\n\nThe derivative is $$3x^2+3=6\\implies x=\\pm1$$ The points are therefore $(-1,-4),(1,4)$ - substitute into $y=x^3+2x$.\n\nLet's consider the first point. For a tangent line parallel to $y=6x+1$, the slope must be $6$. So we have that $$y=mx+c\\implies-4=6(-1)+c\\implies c=2\\implies \\boxed{y=6x+2}$$\n\nCan you do the other one?\n\n\u2022 f(x) = x\u00b3 + 3x, not x\u00b3 - 3x, sorry. \u2013\u00a0Victor S. Apr 5 '18 at 15:53\n\u2022 No problem, edited! \u2013\u00a0TheSimpliFire Apr 5 '18 at 18:42\n\u2022 The equation of the tangents is the follow-up question but I didn't want to have MSE do my homework as i'm well aware this is a big problem here... Check my edit on the OP. \u2013\u00a0Victor S. Apr 5 '18 at 19:37\n\nPlot of the original function and the equation of the lines tangent to f with slope = 6", "date": "2019-07-23 04:42:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2723526/how-many-lines-tangent-to-the-graph-of-f-are-parallel-to-the-line", "openwebmath_score": 0.6273089647293091, "openwebmath_perplexity": 232.2767825204329, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846697584034, "lm_q2_score": 0.8887587831798665, "lm_q1q2_score": 0.8695479685763141}}
{"url": "http://math.stackexchange.com/questions/43226/reason-why-these-two-probabilities-are-equal-picking-balls-in-exact-order-witho", "text": "# Reason why these two probabilities are equal (Picking Balls in exact order without Replacement)\n\nI realize the probability of the following two events are equal. I am curious: is there a reason, besides coincidence, that the probabilities are equal?\n\nSuppose there are five balls in a bucket. 3 of the balls are labelled A, and 2 of the balls are labelled B. There is no way to distinguish between balls labelled A. There is no way to distinguish between balls labelled B.\n\nSuppose I draw balls at random, without replacement. The event $\\{AAABB\\}$ means I pick 3 balls labelled A, then 2 balls labelled B (in that exact order). Then,\n\n$$P(\\{AAABB\\}) = \\frac{3}{5} \\times \\frac{2}{4} \\times \\frac{1}{3} \\times 1 = 0.1$$\n\nAlso,\n\n$$P(\\{AABAB\\}) = \\frac{3}{5} \\times \\frac{2}{4} \\times \\frac{2}{3} \\times \\frac{1}{2} \\times 1 = 0.1$$\n\nAs you can see, the probability of the events $\\{AAABB\\}$ and $\\{AABAB\\}$ are exactly the same. I have seen the claim that any possible order of 3 A's and 2 B's is the same. Why is this true (if it indeed true)? If the claim is true, then I don't have to multiply out individually for every conceivable event.\n\nThanks.\n\n-\nIntuitively there is a degeneracy between \"sublevels\" of the same ordering (i.e there are 12 different ways to pick AAABB but they are equivalent). Thus there is nothing special about one ordering versus another, therefore you would expect results that only differ in ordering to have the same probability. \u2013\u00a0 crasic Jun 4 '11 at 18:00\n\nThe claim indeed seems true.\n\nFor a way to see this.\n\nConsider the first probability\n\n$P(\\{AAABB\\}) = \\frac{3}{5} \\times \\frac{2}{4} \\times \\frac{1}{3} \\times 1 = 0.1$\n\nThis can actually be written as\n\n$P(\\{AAABB\\}) = \\frac{3}{5} \\times \\frac{2}{4} \\times \\frac{1}{3} \\times \\frac{2}{2} \\times \\frac{1}{1} = 0.1$\n\nNow notice the denominators: This will always be $5 \\times 4 \\times 3 \\times 2 \\times 1$.\n\nNow the numerator will just be $3 \\times 2 \\times 1 \\times 2 \\times 1$, in some order.\n\nThe $3$ will appear when you pick the first A, one of the $2$ will appear when you pick the second A or first B etc.\n\nThus the probability is $\\frac{12}{120} = 0.1$.\n\n-\n\nI will describe a problem that has the same flavour as yours, since it may throw some additional light on your observation.\n\nA standard deck of cards is thoroughly shuffled. Someone lifts up the top $5$ cards, without looking at them, and looks at the sixth card. What is the probability that this sixth card is a Queen? It is probably intuitively obvious that this probability is $4/52$, since all orderings are equally likely.\n\nNow change the problem a tiny bit, we deal out the cards one by one, looking at each one. What is the probability that the sixth card dealt is a Queen? (We are allowing one or more of the first five cards to be a Queen.)\n\nWhether we looked or not obviously does not change the probability, so the probability is $4/52$.\n\nBut we are accustomed to solving such problems by a \"tree diagram\" procedure, so we might want to trace out all ways in which we can get a Queen on the sixth draw. One of them, for example, could be QQNNNQ, another could be NNNNNQ. Calculate the probabilities for each path (they are not all the same), and add up. Not too hard, but it involves some work. After the smoke clears, the mess will simplify to $1/13$.\n\nAfter performing the computation and noting that the answer is (equivalent to) $4/52$, which is the same as the probability that the first card drawn is a Queen, one might ask a question very similar to the one you asked.\n\nThe answer to the cards question has already been given in the first few paragraphs of this answer.\n\nNow to your problem. Let's change your description a little. We have $5$ cards, the $10$, Jack, Queen, King and Ace of spades. Let's label the first three A, with invisible ink. Let's label the last two B, again with invisible ink. Shuffle the $5$ cards. Note that all orders of the cards are equally likely.\n\nNow reveal the labels, using a magic light. It is I hope obvious that the order AAABB has the same probability as (say) the order ABABA.\n\nSometimes, as in this case, even when we are told that certain objects are identical, one gets a clearer analysis by imagining them distinct, with any identicalness temporarily \"secret.\"\n\n-\nIf we deal 5 cards, the probability that the sixth card is a queen does depend what the first 5 cards were (if we looked). If we didn't look, it wouldn't matter what the first 5 cards were. For example, if there were 4 queens in the first 5 cards, the probability of the 5th card being a queen is 0. \u2013\u00a0 jrand Jun 4 '11 at 19:27\n@jrand: I said we looked. Did not say what we saw. Did not say we used the information to update the probability estimate. \u2013\u00a0 Andr\u00e9 Nicolas Jun 4 '11 at 19:47", "date": "2015-04-25 11:18:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/43226/reason-why-these-two-probabilities-are-equal-picking-balls-in-exact-order-witho", "openwebmath_score": 0.8494909405708313, "openwebmath_perplexity": 285.6173881257477, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969708496457, "lm_q2_score": 0.8840392832736083, "lm_q1q2_score": 0.8695383611400129}}
{"url": "https://math.stackexchange.com/questions/2411618/partial-fraction-decomposition-with-complex-number-frac1z2-2i", "text": "# Partial Fraction Decomposition with Complex Number $\\frac{1}{z^2 - 2i}$.\n\nHow do I decompose the fraction\n\n$$\\dfrac{1}{z^2 - 2i}$$\n\ninto partial fractions? I understand how to do partial fraction decomposition with real numbers, but I am unsure of how to do it with complex numbers. I attempted to find examples online, but all examples are with real numbers -- not complex.\n\nI would greatly appreciate it if people could please take the time to demonstrate this.\n\n\u2022 Factoring $\\,z^2-2i\\,$ would be a good starting point. \u2013\u00a0dxiv Aug 30 '17 at 22:02\n\u2022 @dxiv Yes...that is precisely what I am asking... \u2013\u00a0The Pointer Aug 30 '17 at 22:03\n\u2022 You do it in the same way over any field. In the case of complex numbers, it is simpler because you have only linear irreducible factors. \u2013\u00a0Bernard Aug 30 '17 at 22:04\n\u2022 $z^2-2i=(z+\\sqrt{2i})(z-\\sqrt{2i})=(z+1+i)(z-1-i)$. From here the process is identical to the decomposition of linear terms with real numbers \u2013\u00a0Alex Aug 30 '17 at 22:05\n\u2022 @ThePointer For the first equality, use difference of squares, $a^2-b^2=(a+b)(a-b)$. Next note that $\\sqrt{2i}=\\sqrt{2}(e^{i\\pi/2})^{1/2}=\\sqrt{2}e^{i\\pi/4}=\\sqrt{2} \\frac{1}{\\sqrt{2}}(1+i)=1+i$ \u2013\u00a0Alex Aug 30 '17 at 22:37\n\n## 4 Answers\n\nNote that $z^2-2i=(z+\\sqrt{2i})(z-\\sqrt{2i})$ and $\\sqrt{2i}=\\sqrt{2}e^{i\\pi/4}=1+i$. To simplify, let $b=1+i$, then $$\\frac{1}{z^2-2i}=\\frac{1}{(z+b)(z-b)}$$ From here it actually doesn't matter if you regard $b$ as real or complex, the process to find the partial fractions is the same as long as the terms are linear in $z$. So we let $$\\frac{1}{(z+b)(z-b)}=\\frac{A}{z+b}+\\frac{B}{z-b}$$ for some $A,B\\in \\mathbb C$. Adding the two fractions on the right hand side we get that $$A(z-b)+B(z+b)=1$$ and so $$A+B=0$$ $$-bA+bB=1$$ which has solution $$A=-\\frac{1}{2b}$$ $$B=\\frac 1{2b}$$ Plugging in the original $b=1+i$ we have that $$\\frac{1}{2b}=\\frac12\\frac 1{(1+i)}\\frac{(1-i)}{(1-i)}=\\frac 14(1-i)$$ Therefore $$\\frac{1}{z^2-2i}=-\\frac{\\frac 14(1-i)}{z+1+i}+\\frac{\\frac 14(1-i)}{z-1-i}.$$ As you can see the process for computing the partial fraction coefficients with complex rationals is equivalent to that of real numbers.\n\n\u2022 Thanks for the complete response. I did the calculations myself (with help from your post and those of others) and got $\\dfrac{1}{(z - 1 - i)(z + 1 + i)} = \\dfrac{1}{(z - 1 - i)(2 + 2i)} - \\dfrac{1}{(z + 1 + i)(2 + 2i)}$. \u2013\u00a0The Pointer Aug 30 '17 at 23:30\n\u2022 For whatever reason, this answer absolutely captivates me! Stellar! \u2013\u00a0gen-z ready to perish Aug 30 '17 at 23:49\n\nthe roots of $z^2 - 2i$ are $1+i$ and $-(1+i)$\n\n$$(1+i)^2 = 1^2 + 2 \\cdot 1 \\cdot i + i^2 = 1 + 2i - 1 = 2i$$\n\nHint: \u00a0either will work to find the square roots of $\\,2i\\,$:\n\n\u2022 $\\;\\; 2i=(a+ib)^2 = a^2-b^2+2abi \\implies a^2-b^2=0\\,,\\;2ab=2\\,$\n\n\u2022 $\\;\\; 2 e^{i \\pi/2} = \\left(r e^{i \\phi}\\right)^2 \\implies r^2 = 2\\,,\\;2\\phi = \\pi/2 + 2k\\pi$\n\nI suppose one difficulty some will encounter here is how to factor $z^2-2i.$\n\nYou can observe that $$2i = 2(\\cos90^\\circ + i\\sin90^\\circ),$$ and therefore $$\\pm\\sqrt{2i\\,} = \\pm\\sqrt 2(\\cos45^\\circ + i\\sin45^\\circ) = \\pm \\sqrt 2\\left( \\frac 1 {\\sqrt 2} + i \\frac 1 {\\sqrt 2} \\right) = \\pm( 1+i).$$\n\nTherefore $z^2 - 2i = \\Big(z - (1+i)\\Big)\\Big( z+ ( 1+i)\\Big).$\n\nFor the rest, if you know how to do other sorts of arithmetic with complex numbers, just proceed as with real numbers.", "date": "2019-08-20 20:57:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2411618/partial-fraction-decomposition-with-complex-number-frac1z2-2i", "openwebmath_score": 0.6813706159591675, "openwebmath_perplexity": 248.2326175669736, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969641180276, "lm_q2_score": 0.8840392725805822, "lm_q1q2_score": 0.8695383446713701}}
{"url": "https://math.stackexchange.com/questions/1292836/line-integral-of-a-conservative-field-over-a-squre", "text": "# Line integral of a conservative field over a squre\n\nI am trying to evaluate $$\\oint _C \\frac{-ydx+xdy}{x^2+y^2}$$\n\nclockwise around the square with vertices (\u22121,\u22121), (\u22121,1), (1,1), and (1,\u22121).\n\nSo from the question, $$\\vec{F}=<\\frac{-y}{x^2+y^2},\\frac{x}{x^2+y^2}>$$\n\nI first conducted the gradient test $\\frac{\\partial F_2}{\\partial x}=\\frac{\\partial F_1}{\\partial y}$ to see whether the field is conservative. And indeed, I found out that\n\n$$\\frac{\\partial F_2}{\\partial x}=\\frac{\\partial F_1}{\\partial y}=\\frac{y^2-x^2}{(x^2+y^2)^2}$$\n\nThus, $$\\vec\\nabla \\times\\vec F=0$$\n\nIn this case, since the domain(the square) is simply connected, I thought that the answer was: $$\\oint _C \\frac{-ydx+xdy}{x^2+y^2}=0$$\n\nHowever, apparently this is wrong and the solution is\n\n$$\\oint _C \\frac{-ydx+xdy}{x^2+y^2}=2\\pi$$\n\n\u2022 What about $x=y=0$? Think ${1 \\over z}$ in complex terms. Revisit the simply connected part... \u2013\u00a0copper.hat May 21 '15 at 15:17\n\u2022 The domain where the integrand is defined is the square without the origin! \u2013\u00a0Alex Fok May 21 '15 at 15:18\n\u2022 Ah i see! what a stupid mistake.. Then does that mean the only way to solve this problem is to compute all of the four line integrals and sum them up?? \u2013\u00a0user223022 May 21 '15 at 15:20\n\u2022 @user223022 I provided 3 distinct approaches. Please let me know how I can improve my answer. I just want to give you the best answer I can give you. \u2013\u00a0Mark Viola May 21 '15 at 18:23\n\nThe reason that Stokes' Theorem does not apply directly here is that $F$ is not continuous (and thus, not differentiable) throughout the region bounded by $C$. The singularity at the origin is the source the issue.\n\nMETHOD 1: Brute Force\n\nThe integration is comprised of the sum of integrals over the 4 line segments. From symmetry considerations, each contribution is identical and we therefore only need to carry out the integration over one of the four segments and multiply by $4$.\n\nThus,\n\n\\begin{align} \\oint_C \\ \\frac{-ydx+xdy}{x^2+y^2}&=4\\int_{-1}^{1} \\frac{(-1)dx+x(0)}{x^2+(1)^2}\\\\\\\\ &=\\left. -4\\arctan(x)\\right|_{-1}^{1}\\\\\\\\ &=-2\\pi \\end{align}\n\nMETHOD 2: Change Contour\n\nSince $\\nabla \\times \\vec F =0$ for $(x,y)\\ne (0,0)$, we can deform the contour to any shape that encloses the origin. Let's choose a unit circle as the new contour $C'$. Then, we have\n\n\\begin{align} \\oint_C \\ \\frac{-ydx+xdy}{x^2+y^2}&=\\oint_{C'} \\ \\frac{-ydx+xdy}{x^2+y^2}\\\\\\\\ &=-\\int_{-\\pi}^{\\pi} \\frac{-\\sin \\phi (-\\sin \\phi) d\\phi+\\cos \\phi (\\cos \\phi)d\\phi}{\\cos^2 \\phi +\\sin^2 \\phi}\\\\\\\\ &=-2\\pi \\end{align}\n\nMETHOD 3: Complex Plane Analysis Interestingly, this problem is identical to the imaginary part of the complex plane contour integration of $1/z$\n\n\\begin{align} \\text{Im}\\left(\\oint_C \\frac{dz}{z}\\right)&=\\text{Im}\\left(\\oint_C \\bar z \\frac{dz}{|z|^2}\\right)\\\\\\\\ &=\\text{Im}\\left(\\oint_C \\frac{xdx+ydy}{x^2+y^2}+i\\oint_C \\frac{-ydx+xdy}{x^2+y^2}\\right)\\\\\\\\ &=\\oint_C \\frac{-ydx+xdy}{x^2+y^2}\\\\\\\\ &=-2\\pi \\end{align}\n\nfrom the residue theorem recalling that $C$ is traversed clockwise.\n\nPut a circle of radius $r$ centered about the origin, so that the circle fits inside the rectangle. Then you know by Green's Thm that the integral over the rectangle is the same as the integral over the circle, which when you switch to polar coordinates, will give you $-2\\pi$.\n\nSince $\\nabla \\times \\vec{F}=0$ is valid in $\\mathbb{R}\\backslash{(0,0)}$, you can reshape your curve to a circle centered at origin and use polar coordinates to evaluate it.\n\nYour field $\\vec F$ is nothing else but $\\nabla\\arg$.\n\nThe argument (polar angle) function $$\\arg: \\>\\dot{\\mathbb R}^2\\to{\\mathbb R}/(2\\pi{\\mathbb Z})$$ is locally a nice real function, but is globally only defined up to multiples of $2\\pi$. Therefore on the one hand its gradient is well defined. For $x>0$ it can be computed from $$\\arg(x,y)=\\arctan{y\\over x}\\qquad({\\rm mod}\\ 2\\pi)\\ ,$$ and the result is your $\\vec F$, valid in all of $\\dot{\\mathbb R}^2$. It follows that ${\\rm curl}(\\vec F)\\equiv0$.\n\nOn the other hand the integral of $\\vec F$ along any arc $\\gamma$ in the punctured $(x,y)$-plane is equal to the total increment of $\\arg$ along this arc, and the integral of $\\vec F$ along the chain $C:=\\partial Q$ is equal to the totel increment of $\\arg$ along $C$, which is obviouly $=2\\pi$.", "date": "2019-12-10 11:11:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1292836/line-integral-of-a-conservative-field-over-a-squre", "openwebmath_score": 0.9456876516342163, "openwebmath_perplexity": 411.2571931941277, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969655605175, "lm_q2_score": 0.8840392695254319, "lm_q1q2_score": 0.8695383429415512}}
{"url": "http://gmatclub.com/forum/factorial-94770.html?fl=similar", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 27 Jul 2016, 08:43\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nEvents & Promotions\n\nEvents & Promotions in June\nOpen Detailed Calendar\n\nFactorial\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nManager\nJoined: 24 May 2010\nPosts: 83\nFollowers: 3\n\nKudos [?]: 31 [0], given: 1\n\nShow Tags\n\n24 May 2010, 22:35\nOk so if 5 people are to sit at a round table how many ways can they be seated. Why is the answer not 5!\nManager\nJoined: 20 Apr 2010\nPosts: 153\nLocation: I N D I A\nFollowers: 3\n\nKudos [?]: 19 [0], given: 16\n\nShow Tags\n\n25 May 2010, 00:07\nTotal No. of ways in which n no. of persons could be arranged on a round table is given by : ( n - 1 ) ! and not n !\n\nTherefore the ans shd be 4! and not 5!\nManager\nJoined: 24 May 2010\nPosts: 83\nFollowers: 3\n\nKudos [?]: 31 [0], given: 1\n\nShow Tags\n\n25 May 2010, 07:03\nYes but why n-1 ! And not n! Can you give some more color so I can understand\n\nPosted from my mobile device\nMath Expert\nJoined: 02 Sep 2009\nPosts: 34091\nFollowers: 6091\n\nKudos [?]: 76633 [2] , given: 9978\n\nShow Tags\n\n25 May 2010, 07:33\n2\nKUDOS\nExpert's post\nJinglander wrote:\nYes but why n-1 ! And not n! Can you give some more color so I can understand\n\nPosted from my mobile device\n\nThe number of arrangements of n distinct objects in a row is given by $$n!$$.\nThe number of arrangements of n distinct objects in a circle is given by $$(n-1)!$$.\n\nFrom Gmat Club Math Book (combinatorics chapter):\n\"The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:\n\n$$R = \\frac{n!}{n} = (n-1)!$$\"\n\n$$(n-1)!=(5-1)!=24$$\n\nCheck Combinatorics chapter of Math Book for more (link in my signature).\n\nHope it helps.\n_________________\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 10614\nFollowers: 495\n\nKudos [?]: 129 [0], given: 0\n\nShow Tags\n\n15 Feb 2016, 03:22\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nIntern\nJoined: 07 Feb 2016\nPosts: 7\nFollowers: 0\n\nKudos [?]: 5 [0], given: 1\n\nShow Tags\n\n26 Feb 2016, 17:21\nIn a circle there are n-1 ways to arrange a group\nManhattan GMAT Instructor\nJoined: 04 Dec 2015\nPosts: 97\nGMAT 1: 790 Q51 V49\nGRE 1: 340 Q170 V170\nFollowers: 8\n\nKudos [?]: 38 [0], given: 5\n\nShow Tags\n\n28 Feb 2016, 22:32\nExpert's post\nOne way to think about it, and these problems in general, is to start by counting the possibilities 'naively'. It seems logical that there should be 5! ways to arrange 5 people around a round table, so start with 5!. Then, account for any special circumstances by figuring out whether you actually counted any of the possibilities more than once. In this case, you actually counted each separate possibility five times by using 5!. For instance, you counted these five arrangements as being different, but they're actually the same (since they're just rotations around the table):\n\n(A B C D E)\n(B C D E A)\n(C D E A B)\n(D E A B C)\n(E A B C D)\n\nIn order to correct for the overcounting, you'll divide by 5. 5!/5 = 4!, or 24.\n\nThis works for a wide range of counting problems. Suppose you wanted to know how many ways a class of eight people could be split into two groups of four. Naively, there are 8*7*6*5 ways to select the first group of four (after which the second group is determined). But you've overcounted by doing that, since you actually counted these groups as being different:\n\nA B C D\nB C D A\nB A C D\n... etc.\n\nThat is, you counted each different group 4! times. So, the actual answer is (8*7*6*5)/4!, which is equivalent to what you'd get from the combinatorics formula.\n_________________\n\nChelsey Cooley | Manhattan Prep Instructor | Seattle and Online\n\nDid you like this post? Check out my upcoming GMAT classes and private tutoring!\n\nManhattan Prep GMAT Discount | Manhattan Prep GMAT Reviews\n\nRe: Factorial \u00a0 [#permalink] 28 Feb 2016, 22:32\nSimilar topics Replies Last post\nSimilar\nTopics:\n1 Simplifying Factorials 1 13 Mar 2016, 03:13\n6 Power of a Number in a Factorial Problems 1 09 Jul 2014, 08:34\n2 To convert factorials in 2^k*3^m.... format 0 20 Apr 2010, 11:10\n360 Everything about Factorials on the GMAT 72 05 Oct 2009, 05:02\nPS - sum of factorials 9 28 Apr 2007, 05:08\nDisplay posts from previous: Sort by", "date": "2016-07-27 15:43:07", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/factorial-94770.html?fl=similar", "openwebmath_score": 0.5322432518005371, "openwebmath_perplexity": 3373.8421979203454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9504109798251322, "lm_q2_score": 0.9149009642742805, "lm_q1q2_score": 0.8695319218988772}}
{"url": "http://mathhelpforum.com/calculus/169622-integration-zero-where-d-2z-dt-2-0-a.html", "text": "# Math Help - Integration of Zero where d^2z/dt^2 = 0\n\n1. ## Integration of Zero where d^2z/dt^2 = 0\n\nSuppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial\nvelocity \u02c6\u2212j +2\u02c6k and moves under a constant force F = \u2212i +2\u02c6j. Find\nthe position and velocity for all time.\n\nThe Equation of motion ma = F implies that,\n\na = \u2212i +2\u02c6j\n\nwhich gives the three scalar equations:\n\nd^2x/dt^2= \u22121,\n\nd^2y/dt^2= 2,\n\nd^2z/dt^2= 0\n\nintegrating for x yields:\n\nx = \u22121/2t^2+1\n\nintegrating for y yields:\n\ny = t^2\u2212 t\n\nI am unable find z, where:\n\nd^2z/dt^2 = 0\n\nI need to integrate this equation twice to obtain:\n\nz = 2t\n\nBut as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.\n\nAny help or advice would be greatly appreciated as I am finding this area of Applied math confusing.\n\n2. Originally Posted by GrahamJamesK\nSuppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial\nvelocity \u02c6\u2212j +2\u02c6k and moves under a constant force F = \u2212i +2\u02c6j. Find\nthe position and velocity for all time.\n\nThe Equation of motion ma = F implies that,\n\na = \u2212i +2\u02c6j\n\nwhich gives the three scalar equations:\n\nd^2x/dt^2= \u22121,\n\nd^2y/dt^2= 2,\n\nd^2z/dt^2= 0\n\nintegrating for x yields:\n\nx = \u22121/2t^2+1\n\nintegrating for y yields:\n\ny = t^2\u2212 t\n\nI am unable find z, where:\n\nd^2z/dt^2 = 0\n\nI need to integrate this equation twice to obtain:\n\nz = 2t\n\nBut as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.\n\nAny help or advice would be greatly appreciated as I am finding this area of Applied math confusing.\n$\\displaystyle\\int 0 \\ dt=c\\Rightarrow\\int c \\ dt=ct+n \\ \\ \\ c,n\\in\\mathbb{R}$\n\nIf c = 2 and n = 0, you have your z = 2t\n\n3. Originally Posted by dwsmith\n$\\displaystyle\\int 0 \\ dt=c\\Rightarrow\\int c \\ dt=ct+n \\ \\ \\ c,n\\in\\mathbb{R}$\n\nIf c = 2 and n = 0, you have your z = 2t\nThank you for your reply. I understand this, but how can I arrive at 2 if I assume that \"z=2t\" is unknown.\n\n4. Originally Posted by GrahamJamesK\nThank you for your reply. I understand this, but how can I arrive at 2 if I assume that \"z=2t\" is unknown.\nWhy are you assuming it is 2t? If you know that, you just set them equal.\n\n$ct+n=2t\\Rightarrow c=2 \\ \\ n=0\\Rightarrow z=2t$\n\n5. Ok that makes sense, thank you.\n\n6. From $\\frac{d^2z}{dt^2}= 0$, you get $\\frac{dz}{dt}= C$, a constant. But you are told that the velocity at t= 0 is -j+ 2k so that the z component of velocity is 2: the constant is 2. From $\\frac{dz}{dt}= 2$, integrating again gives $z(t)= 2t+ D$. Since you wer told that the initial position was (1, 0, 0), z(0)= 2(0)+ D= 0 and so D= 0. z= 2t.\n\n7. Your explanation was very instructive, I was getting confused by the vector labels and the axis labels. Kudos!", "date": "2015-03-27 02:34:24", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/169622-integration-zero-where-d-2z-dt-2-0-a.html", "openwebmath_score": 0.8913335800170898, "openwebmath_perplexity": 783.4231108491626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9766692366242306, "lm_q2_score": 0.8902942333990421, "lm_q1q2_score": 0.869522989304797}}
{"url": "https://math.stackexchange.com/questions/2582575/intuition-behind-logarithm-change-of-base/2582589", "text": "# Intuition behind logarithm change of base\n\nI try to understand the actual intuition behind the logarithm properties and came across a post on this site that explains the multiplication and thereby also the division properties very nicely:\n\nSuppose you have a table of powers of 2, which looks like this: (after revision)\n\n$$\\begin{array}{rrrrrrrrrr} 0&1&2&3&4&5&6&7&8&9&10\\\\ 1&2&4&8&16&32&64&128&256&512&1024 \\end{array}$$\n\nEach column says how many twos you have to multiply to get the number in that column. For example, if you multiply 5 twos, you get $2\\cdot2\\cdot2\\cdot2\\cdot2=32$, which is the number in column 5.\n\nNow suppose you want to multiply two numbers from the bottom row, say $16\\cdot 64$. Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.\n\nI found that very helpful to understand the actual proofs for this property.\n\nI still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:\n\n$$\\log_a x = y \\implies a^y = x$$ $$\\log_b a^y = \\log_b x$$ $$y \\cdot \\log_b a = \\log_b x$$ $$y = \\frac{\\log_b x}{\\log_b a}$$\n\nBut can somehow provide a explanation in the style of the quoted answer why this actually works?\n\n\u2022 Small nit-pick, $a$ must be positive, otherwise you can get the false statement $\\log_b((-1)^2)=2\\log_b(-1)$. \u2013\u00a0Simply Beautiful Art Dec 28 '17 at 2:38\n\u2022 One idea: use the same table to rationalize why $\\log_4(x) = \\frac{\\log_2(x) }{ \\log_2(4) } = \\frac{1}{2} \\log_2(x)\\,$. \u2013\u00a0dxiv Dec 28 '17 at 2:42\n\u2022 One interesting thing about reading all these answers is that I'd always thought of the change of base formula in the divided form, $\\log_a b = \\frac{\\log_c b}{\\log_c a}$, and hadn't tried to develop any intuition behind it, just applying it rotely. But now the multiplicative form, $\\log_a x = \\log_a b \\times \\log_b x$, has a lot more meaning to me. Of course the two are mathematically equivalent, but it's weird how intuition can latch onto one more easily than the other. \u2013\u00a0JonathanZ Dec 28 '17 at 18:19\n\u2022 Many excellent answers, I also didn't expect that the question gets this much attention. Hard to pick any answer specifically, as many of them contributed to my understanding. \u2013\u00a0Max Dec 29 '17 at 4:01\n\u2022 @SimplyBeautifulArt: And $b = 1$ disproves Max's claim that it actually works. \u2013\u00a0user21820 Dec 29 '17 at 6:43\n\nHere is one way of looking at it. (I'll assume that the numbers $a,b,x \\in \\mathbb R$ satisfy $a > 1, b > 1$, and $x > 0$.)\n\nI dislike the name \"logarithm\" and I think a more descriptive name for $\\log_b(x)$ is \"the exponent from $b$ to $x$\". We could also use the notation $[b \\to x]$ instead of $\\log_b(x)$. The change of base rule then tells us that the exponent from $b$ to $x$ is equal to the exponent from $b$ to $a$ times the exponent from $a$ to $x$: $$\\tag{\\spadesuit}[b \\to x] = [b \\to a][a \\to x]$$ or equivalently $$[a \\to x] = [b \\to x]/[b \\to a].$$ In standard notation, this formula states that $$\\log_a(x) = \\frac{\\log_b(x)}{\\log_b(a)}.$$\n\nNote that equation $(\\spadesuit)$ is obvious, because \\begin{align} b^{[b \\to a][a \\to x]} &=(b^{[b\\to a]})^{[a \\to x]} \\\\ &= a^{[a \\to x]} \\\\ &= x. \\end{align}\n\n\u2022 I agree with your first sentence. Sometime it helps me to read 'logarithm' as 'what's the exponent?'. \u2013\u00a0JonathanZ Dec 28 '17 at 3:25\n\u2022 Please be more precise, otherwise your claims are false. Try $b=1$. \u2013\u00a0user21820 Dec 28 '17 at 11:31\n\u2022 @user21820 I probably should have stated that $a$ and $b$ are greater than $1$ and $x$ is positive, but the question was only asking for intuition and I wanted to get to the point as fast as possible. \u2013\u00a0littleO Dec 28 '17 at 11:45\n\u2022 @user21820 Just updated the answer as you suggested. \u2013\u00a0littleO Dec 29 '17 at 20:46\n\nIntuition is always tricky to get across, but I can try.\n\n$\\log_bx$, as you noted, tells you how many $b$s you need to multiply together to get $x$. Now if you need $\\log_ba$ number of $b$s to multiply to get $a$, and you need $\\log_ax$ number of $a$s to multiply to get $x$, we can \"expand\" each of those $a$s into a number of $b$s. There will be $\\log_ba$ number of $b$s for each $a$, so the total number of $b$s will be $\\log_ax \\log_ba$. These $b$s multiply to $x$, so $\\log_ax \\log_ba = \\log_bx$.\n\nFor example, take $b=2, a=8, x=64$. We start with $\\log_ax = 2$, which tell us we need two $8$s to get $64$:\n\n$$8 \\cdot 8 = 64$$\n\nWe use $\\log_ba = 3$, i.e., $2 \\cdot 2 \\cdot 2 = 8$, to expand each $8$:\n\n$$(2 \\cdot 2 \\cdot 2) \\cdot (2 \\cdot 2\\cdot 2) = 64$$\n\nNow the total number of $2$s we are multiplying is $2 \\cdot 3 = 6$, so $log_2 64 = 6$\n\nIt might help to crank though some easy examples\n\n$4096 = 2^{12} = (2^2)^6 = (2^3)^{4}$\n\n$\\log_2 4096 = 12\\\\ \\log_4 4096 = \\frac {12}{2} = \\frac {\\log_2 4096}{\\log_2 4}\\\\ \\log_8 4096 = \\frac {12}{3} = \\frac {\\log_2 4096}{\\log_2 8}\\\\ \\log_{16} 4096 = \\frac {\\log_2 4096}{\\log_2 16} = 3 \\implies (2^4)^3 = 2^{12}$\n\nMotivated by @dxiv's comment, we can adapt your table to understand the change of base rule, although I'm just going to illustrate changing base from $2$ to $4$ and prove the formula in the (equivalent) form $$\\log_2 x = log_2 4 \\times \\log_4 x.$$ Let's add a row to your table:\n\n$$\\begin{array}{rrrrrrrrrr} \\text{exponent}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\\\ \\text{power of 2}&1& 2& 4& 8& 16& 32& 64& 128& 256& 512& 1024\\\\ \\text{power of 4}&1& 4& 16& 64& 256& 1024& 4096& 16384& 65536& 262144& 1048576 \\end{array}$$\n\nThe top row, labeled exponents, is where you read off the logarithms, and they can be to the base $2$ or the base $4$ depending on which lower row you are looking at.\n\nImagine that there are counters in the bottom two rows, and they both start in the first column, on their $1$. Start moving the bottom counter forward, one column at a time. Now suppose the 'powers of 2' counter starts moving forward and it tries to stay on the same number as the 'powers of 4' counter is on. So when 'powers of 4' jumps to '4', 'powers of 2' has to move two columns to get to its 4. When 'powers of 4' then moves to 16, 'powers of 2' again has to move two columns to get to its 16. I.e. the powers of 2 counter has to move twice as fast as the powers of 4 counter. That means when you read off the logarithms from the top row $$\\log_2 x = 2 \\times \\log_4x.$$\n\nIf you added a 'powers of 8' row you could easily see that $\\log_2 = 3 \\times \\log_8x$, and then generalize it to compare $\\log_a$ and $\\log_{a^n}$.\n\n\u2022 This is a wonderfully intuitive answer that used the OP's framework to address the question. Thank you and +1 from me! \u2013\u00a0Slecker Oct 20 '18 at 21:47\n\u2022 Thank you. I really liked the dynamic image of $\\log_2$ traveling faster than the $\\log_4$. Glad you did too. \u2013\u00a0JonathanZ Oct 20 '18 at 23:48\n\nWill it help to write the formula in a different way:\n\n$$\\log_a c=\\log_a b\\times\\log_b c$$\n\n?\n\nThis is motivated as follows: assume you start with $a$ and you want to \"reach\" $c$ by exponentiation.\n\n\u2022 You can first raise $a$ to some exponent $x=\\log_a b$ to get $b=a^x$, and then you can raise $b$ to some exponent $y=\\log_b c$ to get $c=b^y=(a^x)^y=a^{xy}$...\n\n\u2022 ... Or, you can directly raise $a$ to the exponent, well, $xy=\\log_a b\\times\\log_b c$ to get $c$. The thing is, we call this last exponent $\\log_a c$.\n\nThus, $\\log_a c=\\log_a b\\times\\log_b c$.\n\nTake a slide rule. There is a linear scale on it (on this photo it's labeled as lgX). It is deliberately created to be linear with respect to length because this allows multiplying numbers by adding corresponding line segments. This ability is the main advantage of a slide rule.\n\nNow imagine there are two such scales. In the example below they are the first two:\n\n0-------1-------2 $\\log_4 x$\n0---1---2---3---4 $\\log_2 x$\n1---2---4---8--16 $x$\n\nThe trick is: whatever (sane) bases you take, if the first scale is linear, so is the second. More general picture:\n\n0----1------p---... $\\log_a x$ (linear)\n0----q------1---... $\\log_b x$ (linear)\n1----a------b---... $x$ (non-linear)\n\nThe two \"upper\" scales are linear, they have $0$ at the same place, so there is a simple proportion behind them. For every (sane) $x$ we have:\n\n$$\\frac {\\log_a x} {\\log_b x} = \\frac 1 q = \\frac p 1$$\n\nwhere obviously $p=\\log_a b$ and $q=\\log_b a$.\n\nI understand the change of base formula as a graph transformation hinted by the comment by dxiv. The change of base formula encapsulates the idea that all logarithms are equivalent up to vertical dilations (and reflections). More precisely, if you have two positive numbers $a$ and $b$, then there is a constant $C$ such that $$\\log_a(x)=C\\cdot\\log_b(x)$$ for every positive $x$. The change of base formula tells you that $C=1/\\log_b(a)$.\n\nThis can be paralleled with exponentials in a similar way, for there is a \"change of base formula\" for exponentials. But it doesn't get the same limelight as it does for logarithms. All exponentials are equivalent up to horizontal dilations (and reflections). More precisely, if $a$ and $b$ are positive numbers again, then there is a constant $C$ such that $$a^x=b^{C\\cdot x}$$ for every real $x$. This time $C=\\log_b(a)$. Interestingly, the $C$'s in this discussion are inverses of each other (which shouldn't be a surprise, but I never noticed before).\n\nSimilar to @JonathanZ's table, try this\n\n\\begin{array}{c|rrrrrrrrr} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\\ 2^n & 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\\\ 8^{(n/3)} & 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\\\ \\end{array}\n\nHence $\\log_8(x)= \\dfrac{\\log_2(x)}{3} = \\dfrac{\\log_2(x)}{\\log_2(8)}$", "date": "2019-10-15 06:00:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2582575/intuition-behind-logarithm-change-of-base/2582589", "openwebmath_score": 0.9585773944854736, "openwebmath_perplexity": 261.1285153877413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787872422174, "lm_q2_score": 0.8807970748488296, "lm_q1q2_score": 0.8695041881557602}}
{"url": "https://math.stackexchange.com/questions/4458641/what-is-the-probability-that-camilla-and-cameron-are-paired-with-each-other", "text": "# What is the probability that Camilla and Cameron are paired with each other?\n\nTextbook problem: A teacher with a math class of 20 students randomly pairs the students to take a test. What is the probability that Camilla and Cameron, two students in the class, are paired with each other?\n\nMy answer: 1/190. There is only one such pair out of the 20-choose-2 possible pairings.\n\nTheir answer: 1/19. Camilla can be paired with 19 students and only one such pairing is with Cameron.\n\nQuestion: What principle am I missing in my reasoning that would help me to see why their answer is right and mine wrong? It's like I follow their line of reasoning too but don't see what underlying assumption differentiates the answers to see how I can frame the problem correctly on my own.\n\nTextbook: Chapter 26 from The Art of Problem Solving (Volume I) by Rusczyk and Lehoczky.\n\n\u2022 But there are 10 pairs of students. That means Cameron and Camilla get 10 chances for their pair to be chosen. So the probability is definitely higher than 1/190. May 25 at 20:46\n\nYou're missing that the teacher will pair up all the students, making 10 pairs. Camilla and Cameron could be any of these 10.\n\n\u2022 I want to make the argument 10 pairs / 190 possible pairs = 1/19, but this doesn't work because not all choices of 10 pairs have no repeat students. Is there a variation of this argument that does work? May 25 at 20:55\n\u2022 @eyeballfrog Yes: suppose the pairs are numbered (perhaps by the desk they are sent to). Then the probability Camilla and Cameron are both picked for desk $1$ is $\\frac1{190}$ as you found. Similarly desk 2, or each of the other desks. Then use linearity of expectation May 25 at 21:16\n\u2022 There are 190 total possible pairs, so 190-choose-10 total possible testing classrooms that could be set up by the teacher. Out of all those testing classrooms there are 189-choose-9 testing classrooms you could choose that contain the Camilla-Cameron pair; hence the probability is 189-choose-9 / 190-choose-10 which equals 1/19. May 26 at 1:19\n\nThe 190 comes from counting all the possible pairs of students, including those that are not Cameron or Camilla. You need to zero in on only Cameron or Camilla pairs. Camilla will be paired with someone, and that is a certainty. She cannot be paired with herself, so that leaves only 19 other students she can be paired with.\n\nIf instead of pairing up everyone, you only picked two students at random to make one pair out of the 20, then the odds would be 1/190.\n\n$$\\mathbf{\\text{For another approach:}}$$\n\nLet assume that the boxes represent groups such that two adjacent boxes are in the same group. $$\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad$$\n\nNow , let Camilla select one of the position , she can do it $$\\frac{20}{20}=1$$ ways.After that Cameron must select an empty space , but we want that they are in the same group , so if we assume that Camilla selected the position $$\\color{red}{1}$$ , then Cameran must select position $$2$$ among $$19$$ possible empty position. As you see , the probability that Cameron select the position $$2$$ is $$1/19$$ , then the the probability that they are in same group is $$1/19$$ $$\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{\\color{red}{1}}\\fbox{2} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad\\fbox{+}\\fbox{+} \\quad$$", "date": "2022-06-27 15:41:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4458641/what-is-the-probability-that-camilla-and-cameron-are-paired-with-each-other", "openwebmath_score": 0.763693630695343, "openwebmath_perplexity": 764.1570150897315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787849789998, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8695041877067922}}
{"url": "https://math.stackexchange.com/questions/2390183/the-number-of-ways-to-divide-5-people-into-three-groups", "text": "# The number of ways to divide 5 people into three groups\n\nHow many ways can 5 people be divided into three teams where each team must have at least one member?\n\nAssumably they can either be put in one group of 3 people then two groups with 1 person, or two groups with 2 people then one group of 1 person. Hence my answer was $$^5C_3 +\\, (^5C_2) \\cdot (^3C_2)$$ However the provided answer was $$^5C_3 +\\, (^5C_2) \\cdot (^3C_2) \\cdot (1/2)$$ Where did the 1/2 come from?\n\nThanks.\n\n\u2022 Because the groupings $(A),(BC),(DE)$ and $(A),(DE),(BC)$ are the same (for example). \u2013\u00a0lulu Aug 11 '17 at 12:42\n\u2022 Is there a way of knowing how many will be repeated without listing them out? \u2013\u00a0Casper C. Aug 11 '17 at 12:46\n\u2022 Sure, there's a symmetry. the two $2-$member teams can be interchanged. You have the same problem with the other term, though you short-circuited it. You could have written the first term as $\\binom 53\\times \\binom 21 \\times \\binom 11$ but then you'd have had to divide by $2$ again to cancel the symmetry between the $1-$ member teams. \u2013\u00a0lulu Aug 11 '17 at 12:48\n\u2022 Thanks, but how can you tell the amount of symmetry there is? \u2013\u00a0Casper C. Aug 11 '17 at 12:56\n\u2022 Not following. the symmetry is that every pair of $2$ member teams is counted twice (switching the order). There are situations in which this can get tricky...for example, in counting rolls of a pair of dice you have to distinguish the cases in which you get two of a kind. But there is no problem here. \u2013\u00a0lulu Aug 11 '17 at 13:03\n\n\u2022 The factor $\\frac{1}{2!}$ occurs in fact twice in your example, since we have \\begin{align*} &(^5C_3) (^2C_\\color{blue}{1})(^1C_{\\color{blue}{1}})\\color{blue}{\\frac{1}{2!}} +\\, (^5C_\\color{blue}{2})(^3C_\\color{blue}{2})(^1C_1)\\color{blue}{\\frac{1}{2!}}\\\\ &\\quad=10\\cdot2\\cdot1\\cdot\\frac{1}{2}+10\\cdot 3\\cdot 1\\cdot \\frac{1}{2}+\\\\ &\\quad=25 \\end{align*}\n\n\u2022 We can reformulate the problem and ask for the number of ways to partition a set consisting of $5$ elements into $3$ non-empty subsets. These numbers are known as Stirling numbers of the second kind ${n\\brace k}$.\n\nHere we are looking for \\begin{align*} {5\\brace 3}=25 \\end{align*}\n\nWhen you pick the team as two players, another two, and then just the final one is left over, consider the following: if the five players are named $A, B, C, D, E$, then you might start by picking the duo $\\{A, B\\}$ and then $\\{C, D\\}$, which results in the breakdown of $\\{\\{A, B\\}, \\{C, D\\}, \\{E\\}\\}$ or you might have first picked the duo $\\{C, D\\}$ and then picked $\\{A, B\\}$, which would result in the same grouping of $\\{\\{A, B\\}, \\{C, D\\}, \\{E\\}\\}$.\n\nTo avoid this double-counting, you divide in this scenario by $2$. (Or, equivalently, multiply by $1/2$.)\n\n\u2022 Thanks, but how do I tell how much repeat-counting I've done without listing out the combinations? \u2013\u00a0Casper C. Aug 11 '17 at 12:57\n\u2022 @CasperC. Yes, this is the difficult part of combinatorics. You may wish to return to the definition of, e.g., $^5 C_3$ to see how and why it was defined; its definition, too, involves a bit of division to avoid overcounting, and the scenario here involves a similar bit of reasoning. \u2013\u00a0Benjamin Dickman Aug 11 '17 at 13:04\n\u2022 May I rephase that? How do we know there are only two repetitions of $\\{\\{A, B\\}, \\{C, D\\}, \\{E\\}\\}$ ? How would we know how many repetitions of groups there would be if there were 7 people instead of 5? Thanks for helping. \u2013\u00a0Casper C. Aug 11 '17 at 13:45\n\u2022 @CasperC. This relies on the observation that $n$ objects can be arranged in $n!$ ways. Here, you have the $2$ groups overcounting by a factor of $2! = 2$. \u2013\u00a0Benjamin Dickman Aug 11 '17 at 13:50", "date": "2020-02-21 07:29:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2390183/the-number-of-ways-to-divide-5-people-into-three-groups", "openwebmath_score": 0.7590591907501221, "openwebmath_perplexity": 339.0953552120505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130599601176, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8694876393861051}}
{"url": "https://math.stackexchange.com/questions/3345624/need-help-figure-out-a-fibonacci-related-math-trick/3345755", "text": "# Need help figure out a Fibonacci related math trick\n\nMy math teacher used to do a trick where he would have a student write $$2$$ numbers on the board then add the first to the second to create the third then add the second to the third and so on until there were $$10$$ numbers. He would then turn around and add them up in $$2$$ seconds. How did he do this?\n\n\u2022 Have you tried a few simple cases to see the pattern? \u2013\u00a0rtybase Sep 5 '19 at 20:04\n\u2022 have you heard of induction? \u2013\u00a0ggg Sep 5 '19 at 20:06\n\u2022 Check my answer, I think it is easy; the seventh number in the list multiplied by 11. \u2013\u00a0Hussain-Alqatari Sep 5 '19 at 21:55\n\nMultiplying any natural number by $$11$$ is so easy, check here.\n\nNow the solution for your problem is to multiply the $$7^\\text{th}$$ number in the list by $$11$$\n\nHave this example: our first two numbers are $$16$$ and $$21$$\n\nSo the list is:\n\n$$16$$\n\n$$21$$\n\n$$37$$\n\n$$58$$\n\n$$95$$\n\n$$153$$\n\n$$248$$\n\n$$401$$\n\n$$649$$\n\n$$1050$$\n\nThe sum of those numbers is just $$248$$ (which is the $$7^\\text{th}$$ number) $$\\times 11=2728$$.\n\nThe rule is: $$\\boxed{7^\\text{th}\\text{ number }\\times 11}$$\n\nTry it algebraically starting with $$a$$ and $$b$$ $$\\begin{eqnarray*} a,b,a+b,a+2b,2a+3b,3a+5b,5a+8b, \\\\ 8a+13b,13a+21b,21a+34b. \\end{eqnarray*}$$ Now add these together and we get $$55a+88b=11 (5a+8b)$$.\n\nSo I guess your teacher took the first value multiplied by $$5$$ and added it to the second value multiplied by $$8$$ and then multiplied by $$11$$. Your teacher would have had plenty of time to do this calculation while then values were being added.\n\n\u2022 There's no need for the teacher to calculate $5a+8b$ - it's already written on the board! \u2013\u00a0Carmeister Sep 6 '19 at 4:27\n\u2022 Actually the teacher can calculate $5a+8b$ essentially faster than it will be written on the board, and have additional time to perform multiplication by $11$ (and check the answer twice). \u2013\u00a0Oleg567 Sep 7 '19 at 9:14\n\nHint:\n\n$$\\begin{array}{rl} F(1) &= \\color{blue}{F(3)}-F(2)\\\\ F(2)&= F(4)\\color{blue}{-F(3)}\\\\ F(3)&=\\color{red}{F(5)}-F(4)\\\\F(4)&=F(6)\\color{red}{-F(5)}\\\\ \\vdots\\end{array}$$\n\n$$F(1)+F(2)+\\dots+F(n) = F(n+2)-F(2)$$\n\nThat is because Fibonacci numbers have a number of properties, one of them being:\n\n$$\\sum_{i=0}^nF_i = F_{n+2} - 1 = 2F_n + F_{n-1} - 1$$\n\nProof is by induction\n\nHence, if the numbers are $$0,1,1,2,3,5,8,13$$, the sum will be $$13*2 + 8 - 1 = 33$$\n\nWell, to answer the question as to how he did it: If the first number is $$x$$ and the second number is $$y$$ then every other number and the sum of all ten numbers will a combination of $$x$$ and $$y$$. As you do the same thing every time the final sum will be the same combination. Your teacher merely memorized that the final sum would be $$55x + 88y$$.\n\nAs to how we would know the final number is $$55x+88y$$ we can\n\n1) Simply do it. The ten numbers are $$x,y,x+y, x+2y, 2x+3y, 3x+5y,5x+8y,8x+13y,13x+21y, 21x+34y$$ and the sum is $$55x+88y$$.\n\n2) Try to find a way to generalize this without doing each sum.\n\nWe notice the number of $$x$$s involved are $$1,0,1,1, 2,etc.$$. After a slow start once we have $$1,1$$ this has to follow the Fibonacci sequence. So if the $$k$$th number is $$a_k x + b_k y$$ we know $$a_k= F_{k-2}$$, the $$k$$th fibonacci number.\n\nWe notice the number of $$y$$s involved are $$0,1,1,2, etc.$$ and those are the Fibonacci numbers too but with a quicker start. $$b_k = F_{k-1}$$.\n\nSo the $$k$$th number is $$F_{k-2}x + F_{k-1}y$$.\n\nThe final total after ten numbers is therefore $$(1 +\\sum_{k=1}^8 F_k)x + (\\sum_{k=1}^9 F_k) y$$.\n\nThere's an interesting formula that\n\n$$\\sum_{k=1}^n F_k = F_{n+2} - 1$$.\n\nSo the sum is $$F_{10}x + (F_{11}-1)y$$\n\n======\n\nAnother answer states that the answer is that the sum is the $$7$$th number times $$11$$.\n\nSo does $$11(F_{5}x + F_{6}y) = F_{10}x + (F_{11} -1)y$$?\n\nWell, $$11(F_5x + F_6y) = 11(5x + 8y)=55x + 88y = 55x + (89-1)y$$.\n\nThe Fibonacci series is $$1,1,2,3,5,8,13,21,34,55,89$$ so, yes, indeed this is true.\n\nSo that's actually how the teacher did it so quickly. You wrote down the $$7$$th term and he multiplied it by $$11$$ in his head.\n\n\u2022 Thanks everyone! \u2013\u00a0PotatoHeadz35 Sep 6 '19 at 11:45", "date": "2020-09-22 10:29:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3345624/need-help-figure-out-a-fibonacci-related-math-trick/3345755", "openwebmath_score": 0.8956186175346375, "openwebmath_perplexity": 288.0881938066654, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130551025345, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.869487619457378}}
{"url": "http://techiemathteacher.com/2014/06/05/facts-about-cyclic-quadrilateral/", "text": "Cyclic quadrilateral is a special kind of quadrilateral inscribed in a circle. Inscribe means that the vertices of quadrilateral are on the circumference of the circle. A square, rectangle, isosceles trapezoid are some of cyclic quadrilaterals. Below are the some facts about this special figure.\n\n1. The opposite angle of cyclic quadrilateral is supplementary.\n\n$\\angle{A}+\\angle{C}=180$\u00b0\n\n$\\angle{B}+\\angle{D}=180$\u00b0\n\n2. Ptolemy\u2019s Theorem. This theorem is a special case of Ptolemy\u2019s Inequality. He formulated the following equation relating the sides and diagonals.\n\n$d_1d_2=ac+bd$\n\n3. Diagonals as Cords. Using the basic theorem in geometry, we can relate the diagonals with respect to their intersections as follows.\n\n$AO(OC)=OD(OB)$\n\n4. Brahmagupta\u2019s Formula. Brahmagupta of India derived a formula to find the area of cyclic quadrilateral.\n\n$A=\\sqrt{s(s-a)(s-b)(s-c)(s-d)}$\n\nWhere S is the semiperimeter\n\n$s=\\displaystyle\\frac{a+b+c+d}{2}$\n\nUse the facts above to solve for the following problems.\n\nProblem 1: \u00a0In a cyclic quadrilateral $ABCD$, If $\\angle{A}=105$\u00b0. Find the measure of $\\angle{C}$\n\nSolution:\n\nUsing the first fact, opposite angles are supplementary.\n\n$\\angle{A}+\\angle{C}=180$\u00b0\n\n$105$\u00b0$+\\angle{C}=180$\u00b0\n\n$\\angle{C}=(180-105)$\u00b0\n\n$\\angle{C}=75$\u00b0\n\nProblem 2: In cyclic quadrilateral $ABCD$, $\\angle{C}=\\angle{A}=90$\u00b0. If $\\overline{BD}=5$, and two opposite sides have length 2 and 3. What is the area of quadrilateral?\n\nSolution:\n\nSince $\\Delta BCD$ and $\\Delta ABD$ are right triangle and\u00a0$\\overline{BD}$ is the hypotenuse of both triangles, we can find sides $\\overline{AD}$\u00a0 and\u00a0 \u00a0$\\overline{BC}$\u00a0 by Pythagorean Theorem.\n\n$\\overline{BD}^2=\\overline{BC}^2+\\overline{CD}^2$\n\n$5^2=\\overline{BC}^2+3^2$\n\n$\\overline{BC}^2=25-9=16$\n\n$\\overline{BC}=4$\n\n$\\overline{BD}^2=\\overline{AB}^2+\\overline{AD}^2$\n\n$5^2=2^2+\\overline{AD}^2$\n\n$\\overline{AD}^2=25-4=21$\n\n$\\overline{AD}=\\sqrt{21}$\n\nThe area can be found using Brahmagupta\u2019s formula or using the fact that both triangles are right triangle.\n\n$Area_{quadrilateral}=Area_{\\Delta BCD}+Area_{\\Delta ABD}$\n\n$Area_{quadrilateral}=\\displaystyle\\frac{\\overline{BC}\\cdot \\overline{CD}}{2}+\\displaystyle\\frac{\\overline{AB}\\cdot \\overline{AD}}{2}$\n\n$Area_{quadrilateral}=\\displaystyle\\frac{1}{2}(3\\cdot 4+2\\cdot \\sqrt{21})$\n\n$Area_{quadrilateral}=6+\\sqrt{21}$ sq. units\n\n### Dan\n\nBlogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.\n\n### 9 Responses\n\n1. Harold says:\n\nI like what you dudes are now up to. This sort of cool work and coverage! Keep up the great work guys, I\u2019ve included you our blogroll.\n\n2. Great website you have here but I was curious about if you knew of any forums that cover the same topics talked about in this article? I\u2019d really love to be a part of community where I can get comments from other knowledgeable people that share the same interest. If you have any suggestions, please let me know. Bless you!\n\n3. this site says:\n\nGNM54x This is very interesting, You are a very skilled blogger. I have joined your rss feed and look forward to seeking more of your wonderful post. Also, I ave shared your web site in my social networks!\n\n4. suba me says:\n\nwIRAKC Really informative article. Much obliged.\n\n5. 801509 751757This internet website is often a walk-through rather than the details you wanted about it and didnt know who ought to. Glimpse here, and youll definitely discover it. 378161\n\n6. viagra says:\n\n844007 188477These kinds of Search marketing boxes normally realistic, healthy and balanced as a result receive just about every customer service necessary for some product. Link Building Services 329182\n\n7. Hi,I log on to your new stuff named \u201cFacts About Cyclic Quadrilateral \u2013 Techie Math Teacher\u201d like every week.Your story-telling style is witty, keep it up! And you can look our website about proxy server list.\n\n8. lace frontal says:\n\nlace frontal https://www.youtube.com/watch?v=ny8rUpI_98I're genuinely cute and fashion. I purchased the one particular and love them.\n\n9. Way cool! Some extremely valid points! I appreciate you penning this write-up plus the rest of the site is very good.", "date": "2017-05-26 11:04:45", "meta": {"domain": "techiemathteacher.com", "url": "http://techiemathteacher.com/2014/06/05/facts-about-cyclic-quadrilateral/", "openwebmath_score": 0.2081870287656784, "openwebmath_perplexity": 1924.2584384620136, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9908743631497862, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8694692593244818}}
{"url": "https://www.themathdoctors.org/open-or-closed-intervals-it-depends/", "text": "# Open or Closed Intervals? It Depends\n\n#### (Archive Question of the Week)\n\nStudents commonly expect that textbooks all say the same thing (in fact, some think they can ask us about \u201cTheorem 6.2\u201d and we\u2019ll know what they\u2019re talking about!). The reality is that they can even give conflicting definitions, depending on the perspective from which they approach a topic. Here, I want to show how and why they differ in talking about intervals on which a function is increasing or decreasing. Let\u2019s see if we can resolve the \u201cfight\u201d.\n\n## In this corner \u2026 it must be open\n\nThis one page in our archive actually contains two of the three answers to the question in our archive, the second being a challenge to the first. Let\u2019s start with the 2009 question from Bizhan, answered by Doctor Minter:\n\nEndpoints of Intervals Where a Function is Increasing or Decreasing\n\nWhy do some calculus books include the ends when determining the intervals in which the graph of a function increases or decreases while others do not?  I feel that the intervals should be open and the ends should not be included as they may be, for example, stationary points where a horizontal tangent can be drawn.\n\nI noticed that the AP Central always include the ends in the formal solutions of such problems (one can see many examples there), but an author like Howard Anton never does.  Could you clarify this for me please?  Can both versions be correct?\n\nFor example, for the function $$y = -x^3+12x$$,\n\nwe see that it is increasing from -2 to +2. But should we describe that as the open interval $$\\left(-2,2\\right)$$, or as the closed interval\u00a0$$\\left[-2,2\\right]$$? Different textbooks give different answers.\n\nDoctor Minter began:\n\nYou pose an excellent question, and I agree that the discrepancy among the various textbooks is quite misleading.\n\nI completely agree with your claim that the intervals should be open (that is, should not include the endpoints).  Let me attempt to give comprehensive reasoning as to why this should be.\n\nWe use derivatives to decide whether a function is increasing and/or decreasing on a given interval.  Intervals where the derivative is positive suggest that the function is increasing on that interval, and intervals where the derivative is negative suggest that the function is decreasing on that interval.\n\nHe goes on to explain that in order for the derivative to be positive at a point, the derivative\u00a0 must exist there, so the function must be defined, and increasing, and differentiable, in some interval around that point. So he concludes,\n\nIn summary, for a function to be increasing (all of these concepts are similar for decreasing intervals as well), we have to be able to show that the function is greater for larger values of \"x,\" and less for smaller values of \"x\" in a small neighborhood around each point in the interval.  An endpoint cannot have both of these properties.\n\nIt is important to note that both the question and the answer come from the perspective of calculus, and depend on defining \u201cincreasing\u201d in terms of the derivative.\n\n## In that corner \u2026 it can be closed\n\nBut that is not the only way to define \u201cincreasing on an interval\u201d!\n\nBack in 1997, Doctor Jerry had answered a similar question, pertaining to the rules for AP calculus:\n\nBrackets or Parentheses?\n\nWe have been discussing a problem in my Advanced Placement Calculus class.  It concerns increasing/decreasing functions as well as concave up/concave down.  ...\n\nWhen expressing these answers as an interval, should I use a bracket, symbolizing that the endpoint is included, or a parenthesis, symbolizing that the endpoint is not included?\n\nDoctor Jerry replied (in part):\n\nDifferent books, teachers, and mathematicians use slightly different definitions of increasing functions, but this is not a matter of much consequence as long as one is consistent. Suppose your definition of an increasing function is: f is increasing on an interval I if for each pair of points p and q in I, if p < q, then f(p) < f(q).  Note that I may be open (a,b), half-open [a,b) or (a,b], or closed [a,b].\n\nConsider f(x) = x^2, defined on R.  The usual tool for deciding if f is increasing on an interval I is to calculate f'(x) = 2x.  We use the theorem: if f is differentiable on an open interval J and if f'(x) > 0 for all x in J, then f is increasing on J.\n\nOkay, let's apply this to f(x) = x^2.  Certainly f is increasing on (0,oo) and decreasing on (-oo,0).  What about [0,oo)?  The theorem, as stated, is silent.  However, one can go back to the definition of increasing.  To show that f is increasing on I = [0,oo), let u and v be in I and u < v.  If 0 < u, then the theorem applies.  Otherwise, 0 = u < v and we see that f(u) = 0 < f(v) = v^2.\n\nMany instructors, books, and even AP exams often skip consideration of endpoints.  If you want to be ultra-safe, then you can do the above kind of analysis.  It just takes a few extra steps, usually easy, past the standard test.\n\nNote that here, the question and answer are still in the context of calculus, but Doctor Jerry defines \u201cincreasing\u201d without calculus, and then applies a theorem that relates this definition to the derivative: If the derivative is positive on an interval, then the function is increasing on that interval. This theorem doesn\u2019t tell us anything about intervals in which the derivative is sometimes zero, so we have to fall back on the definition, not the theorem. And he concludes that his function is increasing on the half-closed interval\u00a0$$\\left[0,\\infty\\right)$$.\n\n## Shake hands and come out fighting\n\nThere seems to be a difference of opinion here! We have to bring them together and compare them.\n\nDoctor Minter gave an argument for why endpoints should *not* be included when determining intervals where a function is increasing or decreasing.\n\nImplicit in your answer is that \"increasing at a point\" means \"has a positive derivative in a neighborhood of that point.\" I wonder if it makes sense to define increasing at a point. I also wonder about another definition that I came up with (which, I acknowledge, doesn't work for a point).\n\nWe could define increasing for an interval [a, b] as: whenever x and y are in [a, b] then f(x) < f(y).\n\nThis makes no reference to derivatives, so you could still talk about a function being increasing even if it fails to be differentiable some places (e.g., we could say x^(1/3) is increasing everywhere). I'm also thinking about piecewise functions with jumps; again, it seems like we should be able to say they're increasing even if the derivative doesn't exist everywhere.\n\nWith this second definition, it seems to me that if a function is increasing on (a, b) and continuous at a and b, then it would be guaranteed to be increasing on [a, b].\n\nWhat do you think? Do we need to have a definition of \"increasing at a point\" for some reason? Is there any way to reconcile these two definitions?\n\nThere doesn't seem to be consensus here. It's a basic calculus concept and there seems to be two (very convincing) ways of looking at it that are in conflict.\n\nThis was a very perceptive question. There are really two different concepts here, just as Kevin suggested. The concept of increasing on an interval does not require calculus, and applies to functions that are not differentiable. The concept of increasing at a point requires calculus, and is often what the authors of calculus books are really talking about; Doctor Minter took \u201cincreasing on an interval\u201d to mean \u201cincreasing at every point in the interval\u201d in this sense. [Doctor Fenton, in an unarchived 2007 answer, mentioned that \u201cincreasing at a point\u201d can be defined instead as \u201cf(a) is larger than any f(x) for x to the left of a, and f(a) is less than any f(x) when x is larger than a\u201c. This makes it still independent of the derivative.]\n\nI started by discussing the definition Kevin gave, which was (when clarified) identical with Doctor Jerry\u2019s:\n\nThis is the proper definition of increasing on an interval, which applies to any function, and is found, for example, here:\n\nhttp://mathworld.wolfram.com/IncreasingFunction.html\n\nA function f(x) increases on an interval I if\nf(b) \u2265 f(a) for all b > a, where a, b \u2208 I.\nIf f(b) > f(a) for all b > a, the function is\nsaid to be strictly increasing.\n\n...\n\nIf the derivative f'(x) of a continuous function\nf(x) satisfies f'(x) > 0 on an open interval\n(a, b), then f(x) is increasing on (a, b).\nHowever, a function may increase on an interval\nwithout having a derivative defined at all\npoints. For example, the function x^(1/3) is\nincreasing everywhere, including the origin\nx = 0, despite the fact that the derivative is\nnot defined at that point.\n\nThey state the theorem Doctor Jerry used, and give the same example Kevin gave, which shows that the issue applies not only to endpoints, but to any isolated point where the derivative is zero or undefined. For reference, here is the graph of $$y = x^{1/3}$$, the cube root:\n\nNote that although its tangent is vertical at the origin, so it is not differentiable there, it is clearly increasing everywhere. Similarly, the cube function,\u00a0$$y = x^{3}$$, is increasing everywhere, although at the origin it is (momentarily) horizontal:\n\n## A split decision\n\nI also commented on Kevin\u2019s final paragraph about consensus:\n\nThere are actually two different concepts: a precalculus concept, applicable to any function; and a calculus concept, applicable to differentiable functions. I would prefer not to confuse them. Much as we distinguish uniform vs. pointwise continuity, these notions of increasing could be better distinguished like this:\n\n* The function f is increasing on the interval [0, 1], meaning that comparing any two points, the one on the right is higher.\n\n* The function f is increasing at every point on the interval (0, 1), meaning that the derivative is positive everywhere.\n\nThen I quoted from a 2013 conversation I\u2019d had in which this topic had arisen, in which I referred to both Doctor Minter\u2019s and Doctor Jerry\u2019s answers. The question there (in the midst of a long discussion with Aakarsh) made the conflict even stronger:\n\nI have y = f(x). On the x-axis there are points a, b, and c. When x = a, y = 0; when x = b, y = 4; when x = c, y = 1.\n\nI realise the function increases on [a, b].\nI also realise the function decreases on [b, c].\n\nBut why is the b in brackets? I know that they indicate closed intervals; that's no problem. If the graph increases from point a to point b, that is [a, b], but then the graph *MUST* decrease on (b, c].\n\nIf it increases TO \"b,\" it decreases FROM \"b\" ... EXCEPT \"b\"?\n\nWhoa! Can the function be both increasing and decreasing at the same point? If not, how would you decide which? (Notice that Aakarsh was taught to use closed intervals, and accepted that.) I first clarified the question:\n\nI'll suppose that you were given a graph like this:\n\n\nI think what you are asking is why they include the endpoints in the intervals. That's strange, because we've had other questions about why the endpoints are NEVER included in the interval of increase or decrease!\n\nDifferent texts have different policies on this. How does your text DEFINE \"increasing on an interval\"? Can you show me the first example they give?\n\nSee these pages, which emphasize this variability among texts:\n\nBrackets or Parentheses?\nhttp://mathforum.org/library/drmath/view/53566.html\n\nEndpoints of Intervals Where a Function is Increasing\nor Decreasing\nhttp://mathforum.org/library/drmath/view/73202.html\n\nI'm inclined to agree more with the first of these than the second; but I think in pre-calculus, it's a good idea to ignore this detail and either always use open intervals or always use closed intervals. It's not really an important issue; but your concern that a function can't be increasing AND decreasing at the same point would tilt me in the direction of using open intervals just to avoid confusing students like you!\n\nReally, however, you need to notice that your definition probably is only about increasing or decreasing IN AN INTERVAL, not AT A POINT. That is, they are not saying the function is increasing at b -- only that it is increasing in the interval [a, b]. So no claim is being made that the function is both increasing and decreasing at b!\n\nOnce I see your book's definition, I can be more clear on that.\n\nAs I continued writing to Kevin,\n\nThis student never replied with his text's definition, so I didn't get to explore the details with him. One such detail would have been the distinction between saying that a function is increasing on an interval and saying that that is a maximal interval, in the sense that there is no containing interval (open or closed) on which it is increasing.\n\nIn my experience, texts leave a lot unstated. While these omitted details might keep things simple for the less mature student, they would be worth exploring with a curious, capable one like you!\n\nAs I see it, when a textbook (particularly at the precalculus level) asks for \u201cthe interval on which f is increasing\u201d, it is not asking for any such interval, but for the largest. Ideally it would say that; but often they will rely on our instinct to give the \u201cbest\u201d answer possible \u2013 just as when we are asked what shape a square is, we don\u2019t say \u201ca rectangle\u201d, but give the most precise answer that fits. If they have defined \u201cincreasing\u201d as I have, then the closed interval is the correct answer. But there is room for disagreement, especially among students who are thinking more informally.\n\nSome texts, to avoid confusion about endpoints, will specify that they are asking for the largest open interval on which the function is increasing. I think this is the best solution at that level. At a higher level, where precision of definitions is important, just state your definition and act on it.\n\n## Handling differences\n\nIt happens that, two months before Kevin wrote, someone else had written about the same issue, suggesting that we should add to Doctor Minter\u2019s answer some information about the reasons for different answers. Kevin\u2019s question gave the occasion to do just that. In corresponding with Ken about his suggestion, I said the following:\n\nI agree with you on your point about different definitions. That's something I often emphasize in my answers; and it's also an explanation for our having answers on our site that don't agree. I like to refer \"patients\" to past answers, in part to show them that the same problem can be looked at from different perspectives. I also recognize that each answer has its own context, answering a particular student's question either in the light of that student's level or, perhaps, that Math Doctor's personal context. Our goal is not, generally, to give a comprehensive survey of a topic covering all possible variations, but to show a variety of individual interactions. No one answer will cover everything I wish it did (even if it's one I wrote myself a couple years ago). So I just write another and link to the old one(s).\n\nOn Doctor Minter\u2019s answer in particular, I said this:\n\nI think Dr. Minter was probably assuming the discussion is about maximal intervals; he also seems to be assuming the function is differentiable, which is not necessary to be an increasing function. Actually, that's my own main objection to his answer: he hasn't actually defined what he means by increasing on an interval, or what context he is assuming. The answer to many questions depends heavily on the context, and we seldom get enough context to be able to give a perfectly appropriate answer. I personally try to state my assumptions up front (if I don't just ask for the definitions and context), so it can be clear what we are talking about -- especially if I am thinking of having my answer archived.\n\n### 3 thoughts on \u201cOpen or Closed Intervals? It Depends\u201d\n\n1. Hi, Samrath. This is clearly a complicated issue. If you have specific questions about it, the best thing to do is to ask us directly via the Ask a Question link. There we can interact to find out your specific context and concerns.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-05-08 16:18:19", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/open-or-closed-intervals-it-depends/", "openwebmath_score": 0.7156360149383545, "openwebmath_perplexity": 436.6461464098039, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743620718528, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8694692567911972}}
{"url": "https://math.stackexchange.com/questions/1992464/how-is-irrational-exponent-defined", "text": "How is irrational exponent defined?\n\nI am trying to understand the most significant jewel in mathematics - the Euler's formula. But first I try to re-catch my understanding of exponent function.\n\nAt the very beginning, exponent is used as a shorthand notion of multiplying several identical number together. For example, $5*5*5$ is noted as $5^3$. In this context, the exponent can only be $N$.\n\nThen the exponent extends naturally to $0$, negative number, and fractions. These are easy to understand with just a little bit of reasoning. Thus the exponent extends to $Q$\n\nThen it came to irrational number. I don't quite understand what an irrational exponent means? For example, how do we calculate the $5^{\\sqrt{2}}$? Do we first get an approximate value of $\\sqrt{2}$, say $1.414$. Then convert it to $\\frac{1414}{1000}$. And then multiply 5 for 1414 times and then get the $1000^{th}$ root of the result?\n\nThanks to the replies so far.\n\nIn the thread recommended by several comments, a function definition is mentioned as below:\n\n$$ln(x) = \\int_1^x \\frac{1}{t}\\,\\mathrm{d}t$$\n\nAnd its inverse function is intentionally written like this:\n\n$$exp(x)$$\n\nAnd it implies this is the logarithms function because it abides by the laws of logarithms.\n\nI guess by the laws of logarithms that thread means something like this:\n\n$$f(x_1*x_2)=f(x_1)+f(x_2)$$\n\nBut that doesn't necessarily mean the function $f$ is the logarithms function. I can think of several function definitions satisfying the above law.\n\nSo what if we don't explicitly name the function as $ln(x)$ but write it like this:\n\n$$g(x) = \\int_1^x \\frac{1}{t}\\,\\mathrm{d}t$$\n\nAnd its reverse as this:\n\n$$g^{-1}(x)$$\n\nHow can we tell they are still the logarithm/exponent function as we know them?\n\n\u2022 Check this mse question \u2013\u00a0user378947 Oct 31 '16 at 2:24\n\u2022 You should check for similar questions before asking a new one. \u2013\u00a0user378947 Oct 31 '16 at 2:25\n\u2022 @mathbeing Thanks. But that thread may not fully solve my confusion. I updated my question. \u2013\u00a0smwikipedia Oct 31 '16 at 3:14\n\u2022 Well, much to Pytagoras' shock two millennia ago, there is no rational number whose square is 2. So again, its definition is through successive approximation like 1.4, 1.41, etc. Such a sequence is called Cauchy Sequence. \u2013\u00a0Momo Oct 31 '16 at 3:33\n\u2022 If you have time, I recommend you to study the axioms of real numbers and constructions of reals from rationals of Cauchy, Dedekind, Weierstrass. It's not trivial; that is why it has not been formalized earlier. If you don't have time to waste, you may think at $5^\\sqrt{2}$ as something mysterious between $5^{1.41}$ and $5^{1.42}$, that you can get with as many decimals as you need, but never exactly :) \u2013\u00a0Momo Oct 31 '16 at 3:49\n\nYes, you can approximate the result by approximating the irrational exponent with a rational number and proceed with computing integer powers and integer roots. But this does not give you much insight into what an irrational exponent might mean, and I think this is what you mostly care about.\n\nThe best insightful explanation I've seen comes from Khalid at BetterExplained.com.\n\nThe short summary is that we have to stop seeing exponents as repeated multiplication and start seeing them as continuous growth functions, where $e$ plays a central role.\n\nSo $5^{\\sqrt2}$ can be written as $(e^{ln(5)})^\\sqrt2 = e^{\\sqrt2\\cdot ln(5)}$. This can be interpreted as continuous growth for $1$ unit of time at a rate of $\\sqrt2\\cdot ln(5)$, or continuous growth for $\\sqrt2$ units of time at a rate of $ln(5)$, or continuous growth for $ln(5)$ units of time at a rate of $\\sqrt2$. They are all equivalent.\n\nCheck out these links for a much more detailed explanation:\n\n\u2022 The continuous growth explanation of $e$ is unbelievably amazing! I have never viewed $e$ like this! Thanks. Hope it can help me solve my confusion. \u2013\u00a0smwikipedia Oct 31 '16 at 8:01\n\u2022 And a useful link about various growth functions. betterexplained.com/articles/\u2026 \u2013\u00a0smwikipedia Oct 31 '16 at 8:02\n\u2022 Regarding the continuous compound growth, I have another question posted here: math.stackexchange.com/questions/1992743/\u2026 \u2013\u00a0smwikipedia Oct 31 '16 at 8:38\n\nOne way of defining the real numbers is as equivalence classes of \"the collection of all Cauchy sequences (or, equivalently, the collection of all increasing sequences with upper bound) of rational numbers\" with \"$\\{a_n\\}$ equivalent to $\\{b_n\\}$ if and only if $\\{a_n- b_n\\}$ converges to 0. The essentially says that, for example, $\\pi$ is \"represented\" by the infinite decimal 3.1415926.... From that definition, if a is an irrational number then there exist a sequence of rational numbers $r_1, r_2, r_3, ...$ that converges to a. We then define $x^a$ to be the limit of the sequence $x^{r_1}, x^{r_2}, x^{r_3}, ...$. Using the same example as before, $2^\\pi$ is defined as the limit of the sequence $2^3, 2^{3.1}, 2^{3.14}, 2^{3.141}, 2^{3.1415}, 2^{3.14159}, 2^{3.141592}, 2^{31415926}, ...$.\n\n\u2022 One key point is that the limit of $x^{r_{\\large n}}$ does not depend on which sequence $r_n$ converging to $a$ is chosen. \u2013\u00a0anon Nov 5 '17 at 21:41\n\nMy two cents:\n\nThe definition has to do with the fact that $\\mathbb R$ is by definition the completion of $\\mathbb Q$:\n\n$5^\\sqrt{2}=\\sup\\{5^a:a\\in \\mathbb Q \\wedge a<\\sqrt 2\\}$\n\nYou may show that the set on the RHS is bounded and nonempty, so $\\sup$ exists.\n\nSo your approach is right (calculate $5^a$ for successive approximations of $\\sqrt 2$ like 1.4, 1.41 and so on, and it will converge to the result).\n\nI've seen the theory developed the other way around (first study the series, then define elementary functions as series), but this approach seems very \"unnatural\" to me, as it is not how this functions were developed historically.\n\nThe exponential function is an order-preserving bijection over the rationals. Filling in the holes gives an exponential function over the reals that is an order-preserving bijection. This is done by letting a^i be the supremum of {a^(p/q) | (p/q)\n\n\u2022 If you used the proper TeX style formatting, your answer would likely be a bit clearer. \u2013\u00a0rajb245 Oct 31 '16 at 2:49\n\u2022 @rajb245 yeah i typed it clear but the damn compiler or whatever was written by a faschist. \u2013\u00a0Jacob Wakem Oct 31 '16 at 2:55\n\u2022 You just need to put dollar signs to make math Jax read it. You will also need to use \\ to escape your curly braces for set notation. \u2013\u00a0Ian Oct 31 '16 at 2:56\n\u2022 @Alephnull: please use polite words, and don't call names. \u2013\u00a0P Vanchinathan Oct 31 '16 at 3:49\n\u2022 Especially if you don't know how to spell the names! \u2013\u00a0user247327 Nov 5 '17 at 21:38", "date": "2020-11-26 18:27:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1992464/how-is-irrational-exponent-defined", "openwebmath_score": 0.9064700603485107, "openwebmath_perplexity": 287.95063053383217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357173731319, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8694560606493561}}
{"url": "https://www.jiskha.com/questions/1304448/the-sixth-term-of-gp-is-16-and-the-third-term-is-2-find-the-first-term-and-common-ratio", "text": "# Maths\n\nThe sixth term of GP is 16 and the third term is 2.Find the first term and common ratio. 6th term =16 3rd term=2 So ar^5=16 and ar^2=2, then divide. So r^3=8, and r^3 =2^3. Therefore r =2 and a=1/2\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. Looks good to me.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n2. The formula of the GP\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. Thanks \ud83d\ude0a\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\n## Similar Questions\n\n1. ### Math\n\nThe sixth term of an A.P. is 5 times the first term and the eleventh term exceeds twice the fifth term by 3 . Find the 8th term ?\n\n2. ### Mathematics\n\nIf the sixth term of an A.P is 37 and the sum of the first sixth term is 147.find the first term and the sum of the first 15 terms\n\n3. ### Math\n\nThe nth term of an A.P. Is 3 times the 5th term.(a)find the relationship between a and d.(b)prove that the 8th term is 5times the 4th term.\n\n4. ### math\n\nIf third term of an A.P. is four times of first term and sixth term is 17. find the series.\n\n1. ### math\n\nthe 6th term of A.P. is 5 times the fifth term and the 11th term is exceeds twice the fifth term by three.find the 8th term\n\n2. ### arithmetic\n\nin an arithmetic progression the 13th term is 27 and the 7th term is three times the second term find;the common difference,the first term and the sum of the first ten terms\n\n3. ### math\n\nThe fourth term of an ap is 1 less than twice the second term. If the sixth term is 7, find the first term.\n\n4. ### Math\n\nThe 16th term of an AP is three times the 5th term. If the 12th term is 20 more than the 7th term, find the first term and the common difference.\n\n1. ### math\n\nin an AP the 10th term is 54 and 5th term is 4 times the 2nd term . Find the 1st term 'a' and common difference 'd'\n\n2. ### math\n\nThe 3rd term of an Arithmetic Progression is 10 more than the first term while the fifth term is 15 more than the second term. Find the sum of the 8th and 15th terms of the Arithmetic Progression if the 7th term is seven times the\n\n3. ### Mathematics\n\nIn an arithmetic progression the 22nd term is four times the 5th term while the 12th term is 12th more than 8th term.find the first term and the common difference\n\n4. ### Math\n\nThe sixth term of an arithmetic sequence is 20.6 and the 9th term is 30.2. Find the 20th term and find the nth term.", "date": "2021-07-29 22:56:30", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1304448/the-sixth-term-of-gp-is-16-and-the-third-term-is-2-find-the-first-term-and-common-ratio", "openwebmath_score": 0.8588218092918396, "openwebmath_perplexity": 875.9343107499399, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357184418847, "lm_q2_score": 0.8856314798554444, "lm_q1q2_score": 0.8694560571506342}}
{"url": "https://math.stackexchange.com/questions/2945793/how-to-determine-the-eigen-values-of-ata-where-xta-atx-alpha-xtx-f", "text": "How to determine the eigen values of $A^tA$ where $x^tA A^tx=\\alpha x^tx ~~~~ \\forall x \\in \\Bbb{R}^m$ holds for some $\\alpha$\n\nProblem. A be an $$m\\times n$$ matrix of rank $$m$$ with $$n>m$$. If for some nonzero real $$\\alpha$$ we have $$x^tA A^tx=\\alpha x^tx ~~~~ \\forall x \\in \\Bbb{R}^m$$ then $$A^tA$$ has\n\n1. exactly two distinct eigen values\n\n2. $$0$$ as an eigen value with multiplicity $$n-m$$\n\n3. $$\\alpha$$ is a non zero eigen value.\n\n4. exactly two non zero eigen values.\n\nMy attempt.\n\nWe know, $$rank(A^tA)=rank(A)=m$$. So, $$Nullity(A^tA)=n-m>0$$ So, $$0$$ is an eigen value of $$A^tA$$ with geometric multiplicity $$n-m$$. But $$A^tA$$ is symmetric so is diagonalizable i.e. $$0$$ is regular. Hence $$0$$ is an eigen value of (algebraic) multiplicity $$n-m$$ i.e. Option 2 is correct. (without using the given equation)\n\nNow we have to conclude about the rest. I think for this we have to use the given equation.\n\nI start by checking for the non zero eigen values of $$A^tA$$ (In fact, $$A^tA$$ should have $$m$$ non zero real eigen values)\n\nLet $$\\lambda$$ be a non zero eigen value of $$A^tA$$. Then there exists a non-null vector $$v \\in \\Bbb{R}^n$$ such that $$A^tAv=\\lambda v$$. Now to use the given equation I think we need a vactor $$x \\in \\Bbb{R}^m$$. So, I try to choose $$x=Av$$ in that equation.....But I can't get any conclusion from there...!!\n\nAny help is appreciated. Thank you.\n\nN.B: This question has an answer here. But I cannot find out how to use $$AA^t=\\alpha I$$ further.\n\nEDIT: Well, I have found a way to proceed further:\n\nLet $$\\lambda$$ be a non-zero eigen value of $$A^tA$$. Then it can be proved it is also an eigen value of $$AA^t$$ (see here). Therefore, there exits a non-zero vector $$v \\in \\Bbb{R}^m$$ such that $$AA^tv=\\lambda v$$. Again from the given equation we have, $$v^tAA^tv=\\alpha v^tv$$, i.e., $$v^t(AA^tv)=\\alpha v^tv$$, or, $$\\lambda v^tv=\\alpha v^tv$$ this implies, $$\\lambda = \\alpha$$.\n\nHence the only non-zero eigen value of $$A^tA$$ is $$\\alpha$$.\n\nSo, finally Correct options are $$1,2$$ and $$3$$.\n\nYour logic looks good to me, though it can be done more simply than you have it. You were on the right track with setting $$x = Av$$. Then $$A^tx = A^tAv = \\lambda v\\\\AA^tx = \\lambda Av = \\lambda x\\\\x^tAA^tx = \\lambda x^tx\\\\\\alpha x^tx = \\lambda x^tx$$ Since $$\\|x\\| \\ne 0$$, we have $$\\lambda = \\alpha$$.", "date": "2019-10-19 19:28:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2945793/how-to-determine-the-eigen-values-of-ata-where-xta-atx-alpha-xtx-f", "openwebmath_score": 0.9322331547737122, "openwebmath_perplexity": 114.95464641516347, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357248544006, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8694560569027695}}
{"url": "http://math.stackexchange.com/questions/266788/want-to-show-sum-n-2-infty-frac12nn-converges", "text": "# Want to show $\\sum_{n=2}^{\\infty} \\frac{1}{2^{n}*n}$ converges\n\nWant to show $\\sum_{n=2}^{\\infty} \\frac{1}{2^{n}*n}$ converges. I am trying to show this by showing that the partial sums are bounded. I have tried doing this by induction but am not seeing how to pass the inductive assumption part. Do I need to instead look for a closed form? thanks\n\n-\n\nThe sum is less than $\\sum_{n=2}^{\\infty} \\frac{1}{2^n} = \\frac{1}{2}$.\n\n-\n\nHint All the terms are postive and $$\\frac{1}{n2^n}\\leq \\frac 1 {2^n}$$\n\nThink about the monotone convergence, or the so called comparison test.\n\nSpoiler\n\nThe partial sums are bounded above, and are monotone increasing, thus the monotone convergence theorem implies convergence.\n\n-\n+1 for the spoiler! I hadn't saw this before. \u2013\u00a0 000 Dec 29 '12 at 0:51\n*hadn't seen =P \u2013\u00a0 Pedro Tamaroff Dec 29 '12 at 1:15\n\nYou might want to realize a general principle here.\n\nIf we have $$\\sum_{a \\le n \\le b}\\frac{1}{f(n)g(n)}$$ and $f(n)$ and $g(n)$ are positive on $[a,b]$, then we always have that $$\\sum_{a \\le n \\le b}\\frac{1}{f(n)g(n)}\\le\\sum_{a\\le n\\le b}\\frac{1}{f(n)} \\text{ and } \\sum_{a \\le n \\le b}\\frac{1}{f(n)g(n)}\\le\\sum_{a \\le n \\le b}\\frac{1}{g(n)}$$ when the sums $$\\sum_{a \\le n \\le b}\\frac{1}{f(n)}$$ and $$\\sum_{a \\le n \\le b}\\frac{1}{g(n)}$$ both converge.\n\nIn our case, we have that only one converges. Thus, only one of the inequalities is true.\n\n-\n\"positive\" $\\mapsto$ \"$\\ge 1$\". Also, you are describing finite sums, so there is never any question of convergence (boundedness, sure). \u2013\u00a0 Erick Wong Dec 29 '12 at 1:42\nNite approach +1 \u2013\u00a0 Babak S. Dec 29 '12 at 7:16\n\nHint:\n\nTry using the root test. ${}{}{}$ This will give you that the series converges.\n\nYou need only know that $\\lim_{n\\to \\infty}\\sqrt[\\large n]{n} = 1.\\quad$ Then\n\n$$\\lim_{n\\to \\infty} \\sqrt[\\large n]{\\dfrac{1}{n2^n}} = \\sqrt[\\large n]{\\frac{1}{n}}\\cdot\\sqrt[\\large n]{\\frac{1}{2^n}} = 1 \\cdot \\dfrac{1}{2} = \\dfrac{1}{2} < 1 \\implies \\sum_{n=2}^{\\infty} \\frac{1}{n\\cdot2^{n}}\\quad \\text{converges}.$$\n\nNote also that you'd get the same result (that the series converges), by using the the ratio test. Both the root test and ratio test are very useful for establishing convergence and/or divergence of many series.\n\n-\n\nSince you mentioned induction:\n\nLet $s_m = \\sum_{n=2}^{m} \\frac{1}{2^{n}*n}$. Then $s_m \\leq 1-\\frac{1}{m}$.\n\n$P(2)$ is obvious, while $P(m) \\Rightarrow P(m+1)$ reduces to\n\n$$1-\\frac{1}{m}+\\frac{1}{2^{m+1}(m+1)} \\leq 1- \\frac{1}{m+1}$$\n\nwhich is equivalent to:\n\n$$m \\leq 2^{m+1}$$\n\n-\n\nOr we can go this way $$\\sum_{n=2}^{\\infty} \\frac{1}{n2^{n}}\\le\\sum_{n=2}^{\\infty} \\frac{1}{n^2}\\le \\sum_{n=2}^{\\infty} \\frac{1}{n(n-1)}=1$$ and the sum is precisely $\\log 2-\\frac{1}{2}$ (hint: use the Taylor expansion of $\\ln(1-x), |x|<1)$\n\nChris.\n\n-\n\nGo back to the Cauchy test for convergence of a sequence.\n\nLet $S_n = \\sum_{k=2}^{n} \\frac{1}{k2^k}$.\n\nIf $m < n$, $S_n-S_m = \\sum_{k=m+1}^{n} \\frac{1}{k2^k} < \\sum_{k=m+1}^{n} \\frac{1}{2^k} < \\frac{1}{2^m}$ and this $\\to 0$ as $m \\to \\infty$.\n\nTherefore the sequence converges.\n\n-\nHow do you know that the last inequality is true? \u2013\u00a0 Jmaff Dec 29 '12 at 2:34\nIs induction required on n and m to show that? \u2013\u00a0 Jmaff Dec 29 '12 at 2:48\nNo, only the observation I make in my answer. \u2013\u00a0 Pedro Tamaroff Dec 29 '12 at 5:02\nsorry to get back to this late, but i mean the portion\" $\\sum_{k=m+1}^{n} \\frac {1}{2^{k}} < \\frac{1}{2^{m}}$\" \u2013\u00a0 Jmaff Dec 29 '12 at 21:23\n\nYou can show that $-\\ln(1-x)=\\sum_1^\\infty {x^n \\over n}$ for $-1\\le x<1$. Hence the series in question converges to $\\ln{2}$.\n\n-", "date": "2015-08-30 02:17:49", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/266788/want-to-show-sum-n-2-infty-frac12nn-converges", "openwebmath_score": 0.956654965877533, "openwebmath_perplexity": 423.42401219483224, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357205793903, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8694560531166857}}
{"url": "http://mathhelpforum.com/number-theory/150362-fibonacci-series-slight-variant.html", "text": "# Math Help - fibonacci series - a slight variant\n\n1. ## fibonacci series - a slight variant\n\nLet me define a recurence like this\n\nS(1) = 1\nS(2) = 1\n\nS(n) = S(n-1) + S(n-2) + a; where a is some number\n\nQ1: What is the relation beween S(n) and F(n) [nth Fibonacci number]\n\nI found out this is S(n) = F(n) + K(n)*a\n\nWhere K(n) is another series of the type S(n) where a = 1\n\nQ2: Is there a way to find the generic term for S(n) ?\n\nAny pointers will be of great help. Thanks\n\n2. Well, you can always use the theory of difference equations (recurrence relations). You can find the nth term of the Fibonacci sequence this way, and I'm sure your S(n) sequence can be found the same way. The theory is very similar to that of differential equations.\n\n3. Let $f(n)$ be a sequence that $f(n+2) = f(n+1) + f(n)$ , not necessarily be Fibonacci seq. , depends on the initial conditions .\n\nThen we have $S(n) = f(n) - a$ because\n\n$f(n+2) - a = f(n+1) + f(n) - a = (f(n+1) - a) + (f(n) - a ) + a$\n\n$= S(n+1) + S(n) + a$\n\n4. Originally Posted by aman_cc\nLet me define a recurence like this\n\nS(1) = 1\nS(2) = 1\n\nS(n) = S(n-1) + S(n-2) + a; where a is some number\n\nQ1: What is the relation beween S(n) and F(n) [nth Fibonacci number]\n\nI found out this is S(n) = F(n) + K(n)*a\n\nWhere K(n) is another series of the type S(n) where a = 1\n\nQ2: Is there a way to find the generic term for S(n) ?\n\nAny pointers will be of great help. Thanks\nWe have $S_3=2+a, \\; S_4=3+2a$.\n\nYou can use induction to show $S_n=F_n+a\\cdot (F_n-1)$.\n\nEdit: Credit goes to Soroban's post below. I had made a mistake.\n\n5. Hello, aman_cc!\n\nLet me define a recurence like this:\n\n. . $S(1) = 1$\n. . $S(2) = 1$\n. . $S(n) \\:= \\:S(n\\!-\\!1) + S(n\\!-\\!2) + a\\:\\text{ for some constant }a.$\n\nWhat is the relation beween $S(n)$ and $F(n)$, the $n^{th}$ Fibonacci number?\n\nCrank out the first few terms:\n\n. $\\begin{array}{|c||c|c|}n & S(n) & F(n)\\\\ \\hline\n1 & 1 & 1\\\\\n2 & 1 & 1\\\\\n3 & 2+a & 2 \\\\\n4 & 3+2a & 3\\\\\n5 & 5+4a & 5 \\\\\n6 & 8 + 7a & 8 \\\\\n7 & 13 + 12a & 13 \\\\\n8 & 21 + 20a & 21 \\\\\n\\vdots & \\vdots & \\vdots\n\\end{array}$\n\nAnd we see that: . $S(n) \\;=\\;F(n) + [F(n)-1]\\cdot a$\n\n6. Thanks for all the posts", "date": "2015-05-26 00:47:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/150362-fibonacci-series-slight-variant.html", "openwebmath_score": 0.8737853765487671, "openwebmath_perplexity": 1267.174961653112, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9817357227168957, "lm_q2_score": 0.8856314617436727, "lm_q1q2_score": 0.8694560431557452}}
{"url": "https://datascience.stackexchange.com/questions/107152/association-mining-is-buying-independent/107277", "text": "Association Mining - is buying Independent?\n\nI have a problem. I can't not solve this exerciese. What is the best way to solve this exerciese? What are the approaches for this kind of exerciese?\n\nThe following table summarizes transactions in a supermarket where customers bought tomatoes and/or mozzarella cheese or neither.\n\nIs buying mozzarella independent of buying tomatoes in the data given above? If they are not independent, explain whether they are positively or negatively correlated, i.e. does buying one of them increase or decrease the probability of buying the other?\n\nAs you can see I calculated the lift for lift(Moz => Tom) = 1,33 And I calculated lift(Moz => NoTom) = 0,5.\n\nSo I think they are not independent, and it is positively correlated\n\n3 Answers\n\nYou are correct. They are not independent and they are positively correlated.\n\nLet $$A$$ and $$B$$ (or $$X$$ and $$Y$$) be two events for stating general theorems and $$M$$ and $$T$$ be the events \"customer purchases mozzarella\" and \"customer purchases tomatoes\" in this specific example. We will use $$\\wedge$$ to mean \"and\" so that $$M \\wedge T$$ is the event \"customer purchases mozzarella and tomatoes\".\n\nThe simplest way to check independence is directly from the definition. $$A$$ and $$B$$ are independent if $$P(A \\wedge B) = P(A)P(B)$$. From the data table we have $$P(M \\wedge T) = 2000 / 5000 = 0,4$$ $$P(M) = 2500 / 5000 = 0,5$$ $$P(T) = 3000 / 5000 = 0,6$$ $$P(M)P(T) = 0,5 \\times 0,6 = 0,3$$\n\nSince 0.4 does not equal 0.3 we conclude that the events are not independent.\n\nI have to say that I've never seen the term \"lift\" before. But its definition is perfectly reasonable and we could calculate independence in those terms. We can rearrange the definition of independence to give $$P(A \\wedge B) = P(A)P(B)$$ $$\\frac{P(A \\wedge B)}{P(A)P(B)} = 1$$ $$\\text{lift} = 1$$ to see that two events are independent if the associated lift is exactly 1. Your notes correctly show that lift can be rewritten as $$\\frac{\\frac{P(A \\wedge B)}{P(A)}}{P(B)}$$\n\nHere we have a lift of $$0,4 / (0,5 \\times 0,6) = 1,33...$$ or $$0,8 / 0,6 = 1,33...$$ as you correctly calculated in your notes. Since the lift is not exactly 1 we conclude that the two events are not independent.\n\nI would have approached the question of correlation via conditional probabilities, as follows. Two events, $$A$$ and $$B$$ are positively correlated if $$P(A | B) > P(A)$$. That is to say, two events $$A$$ and $$B$$ are correlated if the probability of $$A$$ given $$B$$ is greater than the unconditional probability of $$A$$. In other words, if we discover that event $$B$$ has occurred then the chances that event $$A$$ has occurred increase.\n\nThe conditional probability is defined as $$P(A | B) = \\frac{P(A \\wedge B)}{P(B)}$$ So in this example $$P(M | T) = P(M \\wedge T) / P(T) = 0,4 / 0,6 = 0,66...$$ whereas $$P(M) = 0,5$$, so we conclude that $$M$$ and $$T$$ are positively correlated.\n\nBut hang on! Surely correlation is supposed to symmetric. So let's test whether $$P(T | M) > P(M)$$. We have $$P(T | M) = 0,4 / 0,5 = 0,8$$ and $$P(T) = 0,6$$ so, yes, they are correlated.\n\nA lucky escape? We can be more smug than that. Let's rewrite our definition of correlation $$P(M | T) > P(M)$$ $$\\frac{P(M | T)}{P(M)} > 1$$ $$\\frac{\\frac{P(M \\wedge T)}{P(T)}}{P(M)} > 1$$ $$\\frac{P(M \\wedge T)}{P(T)P(M)} > 1$$\n\nIf we approach correlation the other way we get the same result\n\n$$P(T | M) > P(T)$$ $$\\frac{P(T | M)}{P(T)} > 1$$ $$\\frac{\\frac{P(T \\wedge M)}{P(M)}}{P(T)} > 1$$ $$\\frac{P(T \\wedge M)}{P(M)P(T)} > 1$$ which is the same as above because both the and operator and multiplication are commutative.\n\nAs I'm sure you've noticed, $$\\frac{P(M \\wedge T)}{P(T)P(M)} > 1$$ is simply asking whether lift is above 1. That's probably the approach that your instructor was expecting you to take.\n\nA couple of additional points to fully close the loop with your notes. You correctly calculated lift(Moz => NoTom) = 0,5 but that information was not required to answer the question because you had already calculated lift(Moz => Tom). Support({X} -> {Y}) looks very much like a definition of $$P(X \\wedge Y)$$ and Confidence({X} -> {Y}) looks like $$P(Y | X)$$.\n\nI have two final notes to place this all in a broader context.\n\nFirst, my definition of correlation fits well to your definition of correlation as \"does buying one of them increase or decrease the probability of buying the other\". But our definitions are not completely standard and the word \"correlation\" is usually associated with some specific correlation measure -- most often Pearson's correlation coefficient but also things like Kendall's rank correlation coefficient. Having said that, our definition is totally reasonable and I'm fairly confident that we could prove that, for example, Kendall's rank correlation coefficient is positive if and only if our definition of positive correlation above is positive.\n\nSecond, inspired by the definition of lift in probabilistic terms in your notes, I've conflated historical frequencies given in the table with probabilities. This is standard practice in much of data science and certainly in exercises but it is not completely uncontroversial. Probabilities are about the future and you have data about the past. The extent to which you can infer future probabilities from past data is philosophically unresolved. But that is not something to worry about in this situation.\n\nMarket Basket Analysis is the way to go about it.\n\nMarket Basket Analysis is a technique which identifies the strength of association between pairs of products purchased together and identify patterns of co-occurrence. A co-occurrence is when two or more things take place together.\n\nMarket Basket Analysis creates If-Then scenario rules, for example, if item A is purchased then item B is likely to be purchased. The rules are probabilistic in nature or, in other words, they are derived from the frequencies of co-occurrence in the observations. Frequency is the proportion of baskets that contain the items of interest. The rules can be used in pricing strategies, product placement, and various types of cross-selling strategies.\n\nHere is the calculation :\n\nSupport({Moz}--{Tom} = Transaction containing both moz and tom / total number of transaction\n\n= 2000/ 5000  = 2/5 = 0.4\n\nConfidence({Moz}--{Tom}) = Transaction containing both moz and tom / total number of transaction containing Moz\n= 2000 / 2500 = 2/2.5 = 0.8\n\nLift({Moz}--{Tom}) = (Transaction containing both moz and tom / total number of transaction containing Moz ) / Fraction of transaction containing Y\n\n= 0.8 / (3000/5000)\n\n= 0.8/0.6 = 1.33\n\n\nFor your table this association mining rule is strong. But as the relation is not commutative please do same calculation from P(tom--> Moz) and if values of support, confidence and lift is smaller that does not indicate strong relationship\n\nGiven you have two categorical variables and the associated contingency table, one option is to calculate the joint and marginal probabilities:\n\n$$P = \\frac{count}{total}$$\n\nProbabilities Tomatoes No Tomatoes Row\nMozzarella 0.4 0.1 0.5\nNo Mozzarella 0.2 0.3 0.5\nColumn 0.6 0.4 0.1\n\nThe probabilities then can be used to answer questions about the data - If a person has bought tomatoes, is the person more likely to buy Mozzarella?\n\nA person who has bought tomatoes is four times as likely to buy mozzarella compared to a person who has not bought tomatoes (i.e., 0.4 vs 0.1). There is a strong positive relationship between the two purchases.\n\nThere are many other possible statistical analyses that can be conducted. Examples include odds ratio, phi coefficient, tetrachoric correlation coefficient, and hypothesis testing with Chi-squared.", "date": "2022-05-19 17:45:52", "meta": {"domain": "stackexchange.com", "url": "https://datascience.stackexchange.com/questions/107152/association-mining-is-buying-independent/107277", "openwebmath_score": 0.6739176511764526, "openwebmath_perplexity": 549.3927228044944, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765622193214, "lm_q2_score": 0.8872045981907006, "lm_q1q2_score": 0.8694397121200972}}
{"url": "https://math.stackexchange.com/questions/325082/is-the-inverse-of-a-symmetric-matrix-also-symmetric", "text": "# Is the inverse of a symmetric matrix also symmetric?\n\nLet $A$ be a symmetric invertible matrix, $A^T=A$, $A^{-1}A = A A^{-1} = I$ Can it be shown that $A^{-1}$ is also symmetric?\n\nI seem to remember a proof similar to this from my linear algebra class, but it has been a long time, and I can't find it in my text book.\n\nYou can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more detailed and complete proof.\n\nGiven A is nonsingular and symmetric, show that $$A^{-1} = (A^{-1})^T$$.\n\nSince $$A$$ is nonsingular, $$A^{-1}$$ exists. Since $$I = I^T$$ and $$AA^{-1} = I$$,\n\n$$AA^{-1} = (AA^{-1})^T.$$\n\nSince $$(AB)^T = B^TA^T$$,\n\n$$AA^{-1} = (A^{-1})^TA^T.$$\n\nSince $$AA^{-1} = A^{-1}A = I$$, we rearrange the left side to obtain\n\n$$A^{-1}A = (A^{-1})^TA^T.$$\n\nSince $$A$$ is symmetric, $$A = A^T$$, and we can substitute this into the right side to obtain\n\n$$A^{-1}A = (A^{-1})^TA.$$\n\nFrom here, we see that\n\n$$A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$ $$A^{-1}I = (A^{-1})^TI$$ $$A^{-1} = (A^{-1})^T,$$\n\nthus proving the claim.\n\nIn fact, $(A^T)^{-1}=(A^{-1})^T$. Indeed, $A^T(A^{-1})^T=(A^{-1}A)^T=I$.\n\nYes.\n\n$$AB=BA=I\\quad\\Rightarrow\\quad B^TA^T=A^TB^T=I\\quad\\Rightarrow\\quad B^TA=AB^T=I$$\n\nAnother way to see that is to recall the formula $$A^{-1} = \\frac{1}{\\det(A)} \\mathrm{Adj}(A)^T$$ and to note that the adjoint matrix of a symmetric matrix is by construction symmetric.\n\nYes. The inverse $$A^{-1}$$ of invertible symmetric matrix is also symmetric:\n\n\\begin{align} A & = A^T &&\\text{(Assumption: A is symmetric)}\\\\ \\\\ A^{-1} & = (A^T)^{-1} &&\\text{(A invertible \\implies A^T = A invertible)}\\\\ \\\\ A^{-1} & = (A^{-1})^T &&\\text{(Identity: (A^T)^{-1} = (A^{-1})^T)} \\\\ \\\\ {\\large \\therefore}\\quad \\rlap{\\text{If A is symmetric and invertible, then A^{-1} is symmetric.}} \\end{align}\n\nAll the proofs here use algebraic manipulations. But I think it may be more illuminating to think of a symmetric matrix as representing an operator consisting of a rotation, an anisotropic scaling and a rotation back. This is provided by the Spectral theorem, which says that any symmetric matrix is diagonalizable by an orthogonal matrix. With this insight, it is easy to see that the inverse of the operator is a similar three-step sequence. Finally, we need to establish that any such three-step sequence produces a symmetric matrix: given any orthogonal matrix $$V$$ and diagonal matrix $$D$$, $$(V D V^T)^T = V D V^T$$. Hence the inverse of a symmetric matrix is also symmetric.\n\n\u2022 Very nice intuitive answer. The reason this is getting fewer upvotes is that it\u2019s posted much later than the others. That doesn\u2019t stop this from being a good answer though. Oct 18, 2020 at 20:22\n\nWe have two properties to use: $A^{T}=A$ and $A^{-1}$exist, here we go!\n\n$$A^{-1}A=I$$since $A^{-1}$exist $$(A^{-1}A)^{T}=A^{T}(A^{-1})^{T}=A(A^{-1})^{T}=I^{T}=I$$since $A^{T}=A$ $$A^{-1}A(A^{-1})^{T}=A^{-1}I$$left multiple by $A^{-1}$\n\nThus $$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$$\n\nInverse of $$A$$ can be expressed as a polynomial $$p(A)$$ of $$A$$ (from Cayley-Hamilton theorem).\n\nSo it is sufficient to prove that if $$A$$ is symmetric then power $$A^k$$ is symmetric, sum of symmetric matrices is symmetric and multiply by scalar is symmetric.\n\n\u2022 I.e. for symmetric $A,B$ $(A^k)^T=A^k,(tA)^T=tA, (A+B)^T=A+B,$ Jun 15, 2019 at 10:56\n\n$$(AA^{-1})^T=I^T=I=(A^{-1})^TA^T=(A^{-1})^TA=I\\quad\\Rightarrow\\quad(A^{-1})^T=IA^{-1}\\quad\\Rightarrow\\quad(A^{-1})^T=A^{-1}$$ By definition $$A=A^T$$ Used linear algebra equations $$(AB)^T=B^TA^T;AA^{-1}=I;I^T=I$$", "date": "2022-05-25 13:10:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/325082/is-the-inverse-of-a-symmetric-matrix-also-symmetric", "openwebmath_score": 0.9891189336776733, "openwebmath_perplexity": 285.7312607947454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9799765622193215, "lm_q2_score": 0.8872045892435129, "lm_q1q2_score": 0.869439703352063}}
{"url": "https://gmatclub.com/forum/a-manufacturer-makes-and-sells-2-products-p-and-q-the-revenue-from-t-242847.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Jan 2019, 20:52\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in January\nPrevNext\nSuMoTuWeThFrSa\n303112345\n6789101112\n13141516171819\n20212223242526\n272829303112\nOpen Detailed Calendar\n\u2022 ### FREE Quant Workshop by e-GMAT!\n\nJanuary 20, 2019\n\nJanuary 20, 2019\n\n07:00 AM PST\n\n07:00 AM PST\n\nGet personalized insights on how to achieve your Target Quant Score.\n\u2022 ### Free GMAT Strategy Webinar\n\nJanuary 19, 2019\n\nJanuary 19, 2019\n\n07:00 AM PST\n\n09:00 AM PST\n\nAiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.\n\n# A manufacturer makes and sells 2 products, P and Q. The revenue from t\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: I don't stop when I'm Tired,I stop when I'm done\nJoined: 11 May 2014\nPosts: 535\nGPA: 2.81\nA manufacturer makes and sells 2 products, P and Q. The revenue from t\u00a0 [#permalink]\n\n### Show Tags\n\n17 Jun 2017, 07:11\n1\nTop Contributor\n17\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n69% (01:08) correct 31% (01:34) wrong based on 983 sessions\n\n### HideShow timer Statistics\n\nA manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is $20.00 and the revenue from the sale of each unit of Q is$17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer\u2019s average (arithmetic mean) revenue per unit sold of these 2 products last year?\n\nA. $28.50 B.$27.00\nC. $19.00 D.$18.50\nE. $18.00 _________________ Md. Abdur Rakib Please Press +1 Kudos,If it helps Sentence Correction-Collection of Ron Purewal's \"elliptical construction/analogies\" for SC Challenges ##### Most Helpful Expert Reply CEO Joined: 11 Sep 2015 Posts: 3341 Location: Canada A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags Updated on: 05 Dec 2018, 16:06 1 Top Contributor 5 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer\u2019s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50\nB. $27.00 C.$19.00\nD. $18.50 E.$18.00\n\nLet's plug in some NICE VALUES that satisfy the given information...\n\nLast year the manufacturer sold twice as many units of Q as P.\nSo, let's say the manufacturer sold 1 P and 2 Q's\n\nWhat was the manufacturer\u2019s average (arithmetic mean) revenue per unit sold of these 2 products last year?\nTOTAL revenue = 1($20.00) + 2($17.00)\n= $20.00 +$34.00\n= $54 There were 3 units sold. So, average price =$54/3 = $18 Answer: E RELATED VIDEO _________________ Test confidently with gmatprepnow.com Originally posted by GMATPrepNow on 17 Jun 2017, 07:31. Last edited by GMATPrepNow on 05 Dec 2018, 16:06, edited 1 time in total. ##### General Discussion Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4328 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 17 Jun 2017, 22:48 1 1 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer\u2019s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50\nB. $27.00 C.$19.00\nD. $18.50 E.$18.00\n\nRevenue last year from the sale of P is 20*1 = 20\nRevenue last year from the sale of Q is 17*2 = 34\n\nTotal Revenue from the sale of P and Q last year is 54\n\nAverage Revenue from the sale of P and Q last year is $$\\frac{54}{2+1} = 18$$\n\nHence, answer will be (E) 18\n_________________\n\nThanks and Regards\n\nAbhishek....\n\nPLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS\n\nHow to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )\n\nDS Forum Moderator\nJoined: 21 Aug 2013\nPosts: 1432\nLocation: India\nRe: A manufacturer makes and sells 2 products, P and Q. The revenue from t\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jul 2017, 20:49\nRatio of quantity of P and Q sold, P:Q = 1:2.\n\nThus, average revenue per unit = (20*P + 17*Q)/(P+Q) = (20*1 + 17*2)/(2 + 1) = 54/3 = $18 Hence E answer Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4551 Location: United States (CA) Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 03 Aug 2018, 14:07 1 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer\u2019s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50\nB. $27.00 C.$19.00\nD. $18.50 E.$18.00\n\nWe can compute a weighted average to solve. Let\u2019s assume that 2 units of Q and 1 unit of P were produced last year. So the total revenue is 2 x 17 + 1 x 20 = $54, and thus the average revenue per unit is thus 54/3 =$18.\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: A manufacturer makes and sells 2 products, P and Q. The revenue from t &nbs [#permalink] 03 Aug 2018, 14:07\nDisplay posts from previous: Sort by", "date": "2019-01-20 04:52:25", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-manufacturer-makes-and-sells-2-products-p-and-q-the-revenue-from-t-242847.html", "openwebmath_score": 0.4579028785228729, "openwebmath_perplexity": 6127.9643102423615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9407897525789548, "lm_q2_score": 0.9241418262465169, "lm_q1q2_score": 0.869423160062324}}
{"url": "http://math.stackexchange.com/questions/88008/which-is-the-greatest-possible-natural-number-that-divides-p3p-7-where", "text": "Which is the greatest possible natural number that divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?\n\nWhich is the greatest possible natural number that definitely divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?\n\nThis one is from my module, comes as a fill in the blanks with no answer. I have a feeling that there is something wrong with this question, since for $p=5$, one has $(p+3)(p-7)=-16$.\n\nAs mentioned, in one answer I tried substituting $p=5,7,9,11,13,17,19,...$ and spotted that the greatest number is $8$. I am just wondering .. is there any other way (probably using modulus) to directly find this number without actually substituting ?\n\n-\nWhy does this have a vote to close as \"not constructive\"? \u2013\u00a0 Zev Chonoles Dec 3 '11 at 13:31\nPlaying around (substituting) is always a good idea. \u2013\u00a0 Andr\u00e9 Nicolas Dec 3 '11 at 16:22\nI agree with Zev Chonoles. It looks quite constructive to me and shows some thought about the problem. +1 from me. \u2013\u00a0 Ross Millikan Dec 3 '11 at 16:24\n\nI have a feeling that there is something wrong with this question\n\nThe \"greater than $3$\" condition is a bit strange, because the answer is unchanged if $p=3$ is allowed. And if they wanted to exclude negative values of $(p+3)(p-7)$, they should say \"greater than 7\". But the question as it stands is technically correct. It would perhaps be better like this:\n\nWhich is the greatest natural number that divides $(p+3)(p\u22127)$ for every odd prime number $p$?\n\nIf $p$ is odd, then $p = 2k+1$ for some $k$. So $(p+3)(p-7) = (2k+4)(2k-6) = 4(k+2)(k-3)$, which is divisible by $8$ since one of $k+2, k-3$ must be even. So $8$ divides $(p+3)(p\u22127)$ for every odd prime number $p$.\n\nThat's half the question. The other half is to show that no natural number $n > 8$ divides $(p+3)(p\u22127)$ for every odd prime number $p$. Note that if $n$ has this property, then so does every factor of $n$. So it is enough to show that (i) $16$ doesn't have this property; and (ii) no odd prime $q$ has this property.\n\nTo show that $16$ doesn't have this property, just put $p=11$. So suppose now that $q$ is an odd prime. We need to produce another odd prime $p$ that is not equal to $q-3 \\pmod q$ and not equal to $7 \\pmod q$. Then we can be sure that $q$ doesn't divide $(p+3)(p-7)$. It seems a bit like overkill to me, but we can use Dirichlet's theorem on arithmetic progressions for this: for coprime $a$ and $q$, the arithmetic progression $a + rq$ $(r = 0, 1, 2,...)$ contains infinitely many primes. So we just have to choose $a$ with $1 \\le a \\le q-1$ that is not equal to $q-3 \\pmod q$ and not equal to $7 \\pmod q$. If $q=3$, we can choose $a=2$; and if $q \\ne 3$, we can choose $a=3$.\n\nAnd we're done.\n\n-\nPut $p=5$. Anything $m$ that divides all the $(p+3)(p-7)$ must divide $16$. Put $p=11$ as you did. Thus $m$ must divide $56$. It follows that $m$ must divide $\\gcd(16,56)$. We would need fancier machinery only if we wanted to show that if $\\gcd$ is restricted to primes $>M$, it is still $8$. \u2013\u00a0 Andr\u00e9 Nicolas Dec 3 '11 at 16:18\n@Andr\u00e9Nicolas: Doesn't the paragraph starting \"If $p$ is odd\" prove that $8$ divides $(p+3)(p-7)$ for all odd $p$, prime or not? \u2013\u00a0 Ross Millikan Dec 3 '11 at 16:22\n@Ross Millikan: The answer above used Dirichlet's Theorem to show that no odd prime $q$ can divide all the $(p+3)(p-7)$. \u2013\u00a0 Andr\u00e9 Nicolas Dec 3 '11 at 16:27\nWell, at least I can say I was right about the overkill :-) Andr\u00e9's comment $-$ and Bill Dubuque's answer $-$ provide a much simpler solution. \u2013\u00a0 TonyK Dec 3 '11 at 17:27\n\nThere's nothing wrong with the question.\n\nYou're looking for a natural number. You know it divides $(p+3)(p-7)$. And you can put in some primes into that equation, as you have done already - you've got $-16$. What happens if you stick in $p=11$? Or $p=13$?\n\nOK, now you've identified the natural number. Can you see why the product is always divisible by that number?\n\nclue word: modulo\n\n-\n\nHINT $\\ \\rm\\:d\\ |\\ f_p = (p+3)\\:(p-7)\\ \\Rightarrow\\ d\\ |\\ f_5,\\:f_7,f_{11}\\: =\\:\\: -16,\\:0,\\:56\\ \\Rightarrow\\ d\\ |\\ gcd(16,56) = 8\\:.\\:$ Conversely\n\n$\\$ mod $8$, since odd^2 $\\equiv 1\\:,\\:$ we have $\\rm\\ f_p\\equiv (p+3)(p+1)\\: \\equiv\\: p^2+4\\:p+3 \\:\\equiv\\: 4\\:(p+1) \\:\\equiv\\: 0$\n\nGenerally for a sequence satisfying a monic $\\rm\\:n\\:$th order linear recurrence with integer coefficients one easily proves by induction that the gcd of all terms is the gcd of the first $\\rm\\:n\\:$ terms, since successive terms are integer linear combinations of the first $\\rm\\:n\\:$ terms so they don't change the gcd.\n\n-\n\nIf $p$ is prime it can be written $p=4a+1$ or $p=4a+3.$\n\nSubstituting the first of these gives\n$$(p+3)(p-7)=p^2-4p-21 = 16a^2+8a+1-16a-4-21 = 16a^2-8a-24 = 8(2a^2-a-3)$$\nand the other gives $$(p+3)(p-7)=p^2-4p-21 = 16a^2+24a+9-16a-12-21 = 16a^2+8a-24 = 8(2a^2+a-3)$$\nand so $(p+3)(p-7)$ is divisible by $8$ either way.\n\nEdit\nTo show that 8 is the largest such divisor, we must show that $2a^2-a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\\text{ and }5$ produces 25 and 42 respectively which have no prime factors in common.\n\nAnd we must also show that $2a^2+a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\\text{ and }5$ again produces 33 and 52 respectively which also have no prime factors in common.\n\n-\nStrictly spoken, you have not proven yet that it 8 is the greatest number for which it holds. However, as MaX noted, trying the first few numbers shows that it is at most 8. \u2013\u00a0 Sjoerd Dec 3 '11 at 15:44\n\nObserve that $(5+3)(5-7) = -16$ and $(11+3)(11-7) = 56$; these have greatest common multiple 8. So the answer must be some factor of 8. If you keep trying more numbers you'll see it looks to be 8.\n\nHow to prove it? $p$ is odd. So $p+3$ and $p-7$ are even numbers which differ by 10. Let $p = 2n+7$; then we want to show that for any integer $n$, $2n(2n+10)$ is divisible by 8. But we can rewrite this as $4n(n+5)$, so we just need to show that $n(n+5)$ is always even; either $n$ is even or $n+5$ is.\n\nA challenge for you, to help you understand what's going on here: prove in a similar way that for every integer $n$, $n^3-n$ is a multiple of 6 and $n^5-n$ is a multiple of 30.\n\n-\n\nIf p is odd number, p-7 is even =2k(say). Then (p+3)(p-7)=(2k+10)2k=8k(k+1)/2+16k, is clearly divisible by 8 for all integral value of k.\n\nNow if f(k) = (2k+10)2k=4k(k+5), f(k+1)= 4(k+1)(k+6) (f(k),f(k+1))=(4k(k+5), 4(k+1)(k+6))= 4(k(k+5), (k+1)(k+6)) =4(k(k+5), (k+1)(k+6) - k(k+5)) as (a,b)=(a, b-a) =4(k(k+5), 2k+6) =8(k(k+5)/2, k+3) as k(k+5)/2 is integer for all integral value of k.\n\nNow, 2 k(k+5)/2 - k(k+3)= 2k. We know ax+by=c is solvable iff (a,b)|c where a,b,c,x,y are integers. Here, x=2, a=k(k+5)/2, y=k, b=k(k+3), c=2k So, (k(k+5)/2, (k+3)) | 2k, =>2|(k+3) =>k is odd. For even k, 2\u2224(k(k+5)/2, (k+3)). In that case, no natural number>1, can divide f(k) for all integral value of k. So, 8 will be the greatest natural number that divides (p+3)(p-7) for all odd integral value of p.\n\n-\n\nI would interpret the question as, \"What is the greatest natural number that divides $(p+3)(p-7)$ for all primes $p>3$?\"\n\nWe can write $(p+3)(p-7)=4\\cdot\\frac{p+3}2 \\cdot\\frac{p-7}2$ but one of the latter two factors is even, since they differ by five, an odd number, so we can conclude that $(p+3)(p-7)$ is always divisible by eight, but $(5+3)(5-7)=-16$, $(9+3)(9-7)=24$, and $\\gcd(-16,24)=8$. Therefore no natural number larger than eight divides $(p+3)(p-7)$ for all primes $p>3$, and eight is in fact the greatest natural number that divides $(p+3)(p-7)$ for all primes $p>3$.\n\n-\n\nsince p is odd 2| (p+3)(p-7). so the answer should be \\frac{1}{2} (p+3)(p-7)\n\n-\nYou have not read the question carefully. We want the highest number that divides $(p+3)(p-7)$ for any odd $p$, not for a given odd $p$. \u2013\u00a0 Ross Millikan Nov 6 '13 at 19:28\n\nThe question is sloppily phrased, because to me it reads more like \"Which is the greatest possible natural number that definitely divides $(p+3)(p\u22127)$, where $p$ is some given prime greater than 3?\", in which case the answer is trivially $(p+3)(p-7)$ !\n\nBut I imagine (as does everyone else by the look of it) the OP meant \"Which is the greatest possible natural number that definitely divides $(p+3)(p\u22127)$ for every prime $p$ greater than 3?\"\n\n-", "date": "2015-03-03 11:09:49", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/88008/which-is-the-greatest-possible-natural-number-that-divides-p3p-7-where", "openwebmath_score": 0.8795627951622009, "openwebmath_perplexity": 242.12830396334473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914019704466, "lm_q2_score": 0.8991213732152424, "lm_q1q2_score": 0.8693730611777495}}
{"url": "https://www.khanacademy.org/math/ap-statistics/quantitative-data-ap/measuring-spread-quantitative/a/concept-check-standard-deviation?ref=probability_and_statistics_staff_picks", "text": "# Concept check: Standard\u00a0deviation\n\n## Introduction\n\nUnlike most questions on Khan Academy, some of these questions aren't graded by a computer. You'll learn the most if you try answering each question yourself before clicking \"explain\".\n\n## The formula (for reference)\n\nThe formula for standard deviation (SD) is\n$\\Large\\text{SD} = \\sqrt{\\dfrac{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}{n}}$\nwhere sum means \"sum of\", x is a value in the data set, $\\bar{x}$ is the mean of the data set, and n is the number of values in the data set.\n\n## Part 1\n\nConsider the simple data set left brace, 1, comma, 4, comma, start color redD, 7, end color redD, comma, 2, comma, 6, right brace.\nHow does the standard deviation change when start color redD, 7, end color redD is replaced with start color greenD, 12, end color greenD?\nPlease choose from one of the following options.\n\nHow can we see this in the formula for standard deviation?\n\n### How does the standard deviation change?\n\nThe standard deviation increases because the data becomes more spread out:\n\n### How can we see this in the formula?\n\nWe can see this in the formula\n$\\text{SD} = \\sqrt{\\dfrac{\\goldD{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}}{n}}$\nbecause ${{\\goldD{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}}{}}$ is the sum of the squares of the distances from each data point to the mean. As the data, gets more spread out, the value of ${{\\goldD{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}}{}}$ increases.\n\n### I'm curious, what are the actual standard deviations of the data sets?\n\nThe standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color redD, 7, end color redD, right brace is approximately 2, point, 28.\nThe standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color greenD, 12, end color greenD, right brace is approximately 3, point, 90.\n\n## Part 2\n\nIs it possible to create a data set with 4 data points that has a standard deviation of 0?\nPlease choose from one of the following options.\n\nIf it is possible, do it! Can you create two different data sets? How about three?\n\n### Yes, it's possible!\n\nIn fact, there are an infinite number of possible data sets.\nHere's one:\n5, comma, 5, comma, 5, comma, 5\nHere's another:\n8, comma, 8, comma, 8, comma, 8\nAny data set where all of the data points are the same has a standard deviation of 0 because the distance from each data point to the mean is 0.\n\n### Show me the calculation for $\\{ 5,5,5,5 \\}$left brace, 5, comma, 5, comma, 5, comma, 5, right brace.\n\n#### Step 1: Find the mean\n\n$\\bar{x} = \\dfrac{5 + 5 + 5 + 5}{4} = \\dfrac{20}{4} = \\blueD5$\n\n#### Step 2: Find the square of the distances from each of the data points to the mean\n\nx$\\lvert x - \\bar{x} \\rvert^2$\n5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0\n5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0\n5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0\n5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0\n\n#### Step 3: Apply the formula\n\n\\begin{aligned} \\text{SD} &= \\sqrt{\\dfrac{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}{n}} \\\\\\\\\\\\\\\\ &= \\sqrt{\\dfrac{0 + 0 + 0 + 0}{4}} \\\\\\\\\\\\\\\\ &= \\sqrt{\\dfrac{{0}}{4}}\\\\\\\\\\\\\\\\ &= \\sqrt{{0}}\\\\\\\\\\\\\\\\ &= 0\\end{aligned}\n\n## Part 3\n\nCan standard deviation be negative?\nPlease choose from one of the following options.\n\nWhy or why not?\n\n### No, standard deviation cannot be negative!\n\nTo see why, think about the numerator and denominator inside the radical:\n$\\Large\\text{SD} = \\sqrt{\\dfrac{\\blueD{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}}{\\maroonD{n}}}$\nNotice how start color maroonD, n, end color maroonD is always positive. It's the number of data points, and we can't have a negative number of data points.\nAlso notice that $\\blueD{\\sum\\lvert x - \\bar{x} \\rvert^2}$ involves a quantity getting squared. Whenever we square something, we get a non-negative number.\nSince both the denominator and numerator are positive, the entire expression must be positive too.\n\n## Part 4\n\nStandard deviation is a measure of spread of a data distribution.\nWhat do you think deviation means?\nIn everyday language, deviation is how different something is from what might be considered normal.\nIn statistics, when discussing measures of spread, deviation is the amount by which a single measurement differs from the mean.\n\n### Part 5\n\nHere are the formulas for standard deviation (SD) and the formula for mean absolute deviation (MAD), both of which are measures of spread:\n$\\text{SD} = \\sqrt{\\dfrac{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}{n}}$\n$\\text{MAD} = {\\dfrac{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert}}}{n}}$\nWhat are the similarities between the formulas? What are the differences?\n\n### What are the similarities?\n\nThe formulas are very similar! They are both based on the distance from each data point to the mean $\\lvert x - \\bar{x} \\rvert$, and they both include dividing by the number of data points n.\n\n### What are the differences?\n\nThe difference between the two formulas is that when calculating standard deviation, we square the distance from each data point to the mean, and we take the square root as the last step of the formula.\n\n### Which one is better?\n\nStandard deviation is more complicated, but it has some nice properties that make it statisticians' preferred measure of spread.\n\n### Part 6\n\nHere's the formula that we've been using to calculate standard deviation:\n$\\sqrt{\\dfrac{\\sum\\limits_{}^{}{{\\lvert x-\\bar{x}\\rvert^2}}}{n}}$\nHere's the formula that statisticians actually use:\n$\\sqrt{\\dfrac{\\sum\\limits_{}^{}{{( x-\\bar{x})^2}}}{n}}$\nAre the two formulas equivalent?\nPlease choose from one of the following options.\n\n### What's the difference between the formulas?\n\nIn the formula that we've been using, we take the absolute value of $x - \\bar{x}$:\n$\\sqrt{\\dfrac{\\sum\\limits_{}^{}{{\\tealD{\\lvert x-\\bar{x}\\rvert}^2}}}{n}}$\nIn the formula that statisticians use, they put parentheses around $x - \\bar{x}$:\n$\\sqrt{\\dfrac{\\sum\\limits_{}^{}{{\\purpleC{( x-\\bar{x})^2}}}}{n}}$\n\n### Are the formulas equivalent?\n\nYes, both formulas are equivalent!\nStatisticians realize that squaring will make the distance positive, so they don't bother using absolute value signs and just use parentheses instead.\nFor example, let's evaluate $\\tealD{\\lvert x - \\bar{x} \\rvert^2}$ and $\\purpleC{(x - \\bar{x})^2}$ for x, equals, 2 and $\\bar{x} = 5$:\n$\\tealD{\\lvert x - \\bar{x} \\rvert^2} = \\lvert 2 - 5 \\rvert^2 = \\lvert-3\\rvert^2 = 3^2 = \\greenD9$\n$\\purpleC{(x - \\bar{x})^2} = (2 - 5)^2= (-3)^2 = \\greenD9$\nThey're both positive! They're both start color greenD, 9, end color greenD! They're equivalent!", "date": "2017-03-27 08:43:06", "meta": {"domain": "khanacademy.org", "url": "https://www.khanacademy.org/math/ap-statistics/quantitative-data-ap/measuring-spread-quantitative/a/concept-check-standard-deviation?ref=probability_and_statistics_staff_picks", "openwebmath_score": 0.8668621778488159, "openwebmath_perplexity": 2228.661266040445, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319885346244, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8693443257292482}}
{"url": "http://suisco.net/awstats/blog/page.php?page=arithmetic-expression-calculator-2f178e", "text": "-\u00a0Subtraction For an algebraic expression calculator to work, it is typically required to type the above type of math expression in, by using common symbols. Determine the first term and difference of an arithmetic progression if $a_3 = 12$ and the sum of first 6 terms is equal 42. An example of an algebraic expression is shown below. Answer: Yes, it is a geometric sequence and the common ratio is 6. Everything you need to prepare for an important exam! The main purpose of this calculator is to find expression for the n th term of a given sequence. Solve math problems using order of operations like PEMDAS, BEDMAS and BODMAS. Similarly, when you are solving addition and subtraction expressions you proceed from left to right. Below are some of the example which a sum of arithmetic sequence formula calculator uses. This website uses cookies to improve your experience. Welcome to MathPortal. You can evaluate any Mathematical Expression using this calculator. To find the next element, we add equal amount of first. each number is equal to the previous number, plus a constant. Closed. for learning distance formula equation, use Distance Formula Calculator. For nested parentheses or brackets, solve the innermost parentheses or bracket expressions first and work toward the outermost parentheses. but they come in sequence. BEDMAS stands for \"Brackets, Exponents, Suppose they make a list of prize amount for a week, Monday to Saturday. About this calculator. Degrees of Freedom Calculator Paired Samples, Degrees of Freedom Calculator Two Samples. The first term of an arithmetic progression is $-12$, and the common difference is $3$ Free Arithmetic Sequences calculator - Find indices, sums and common difference step-by-step This website uses cookies to ensure you get the best experience. Copyrights 2020 \u00a9 calculatored.com . . Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! The calculator will generate all the work with detailed explanation. Numbers sequence, in which the diffference is always constant. Viewed 3k times 0. Viewed 3k times 0. +\u00a0Addition Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction Logical Sets Change the sign of each number that follows so that positive becomes negative, and negative becomes positive then follow the rules for addition problems. PEMDAS is an acronym that may help you remember order of operations for solving math equations. Now, find the sum of the 21st to the 50th term inclusive, There are different ways to solve this but one way is to use the fact of a given number of terms in an arithmetic progression is $$\\frac{1}{3}\\;n(a+l)$$ Here, \u201ca\u201d is the first term and \u201cl\u201d is the last term which you want to find and \u201cn\u201d is the number of terms. Just enter your numerical expression in the big box right beneath the \"calculate\" and \"clear\" button and hit the calculate button Note: If you are using parentheses, just remember to put a multiplication sign right before the parentheses For example, do not enter 5(3-4). It is not currently accepting answers. The most important thing to keep in mind is that the parenthesis are important when typing an algebraic expression. First type the expression 2x. If you incorrectly enter it as 4/1/2 then it is solved 4/1 = 4 first then 4/2 = 2 last. For example, the above expression would be expressed as \"1/(3-2) + 3 + 4sin(pi/4) + sqrt(2) + 5^(3/2)\". Objects might be numbers or letters, etc. This web site owner is mathematician Milo\u0161 Petrovi\u0107. As the contest starts on Monday but at the very first day no one could answer correctly till the end of the week.", "date": "2021-03-02 04:30:15", "meta": {"domain": "suisco.net", "url": "http://suisco.net/awstats/blog/page.php?page=arithmetic-expression-calculator-2f178e", "openwebmath_score": 0.7360583543777466, "openwebmath_perplexity": 623.5942476320388, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9888419697383903, "lm_q2_score": 0.8791467801752451, "lm_q1q2_score": 0.8693372337976529}}
{"url": "https://www.emmamichaels.com/3584/cos-sin-1-x.html", "text": "Breaking News\n\n# Cos Sin 1 X\n\nCos Sin 1 X. Cos2(x)+ 2cos(x)sin(x)+sin2(x) = 1 note an important identity: = 1 sin ( x) \u2212 cos ( x) sin ( x) = csc ( x) \u2212 cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. Expand using the foil method. The word cosinus derived from edmund gunter.\n\n= 1 sin ( x) \u2212 cos ( x) sin ( x) = csc ( x) \u2212 cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude, a cos \u2061 x + b sin \u2061 x = c cos \u2061 ( x + \u03c6 ). Yes your guess from the table is correct, indeed since \u2200 \u03b8 \u2208 r \u2212 1 \u2264 cos \u03b8 \u2264 1, for x > 0 we have that.\n\n## Yes your guess from the table is correct, indeed since \u2200 \u03b8 \u2208 r \u2212 1 \u2264 cos \u03b8 \u2264 1, for x > 0 we have that.\n\nSolve for x cos(x)+1=sin(x) subtract from both sides of the equation. Cos (x)+sin (x)=1 cos (x) + sin(x) = 1 cos ( x) + sin ( x) = 1 square both sides of the equation. The derivative of sin x is cos x, the derivative of cos x is \u2212sin x (note the negative sign!) and.\n\n## Misc 13 Solve 2 Tan1 (Cos X) = Tan1 (Ii Cosec X Misc 13.\n\nHow to prove that 1+sin(x)cos(x) = cos(x)1\u2212sin(x) [duplicate] multiplying the lhs by the conjugate of.\n\n### Extended Keyboard Examples Upload Random.\n\nThe derivative of sin x is cos x, the derivative of cos x is \u2212sin x (note the negative sign!) and.\n\n### Kesimpulan dari Cos Sin 1 X.\n\nThe word cosinus derived from edmund gunter.", "date": "2023-01-29 20:13:22", "meta": {"domain": "emmamichaels.com", "url": "https://www.emmamichaels.com/3584/cos-sin-1-x.html", "openwebmath_score": 0.9218480587005615, "openwebmath_perplexity": 2275.445986372435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9833429560855736, "lm_q2_score": 0.8840392893839086, "lm_q1q2_score": 0.8693138081185626}}
{"url": "https://math.stackexchange.com/questions/3212210/how-to-know-when-you-can-focus-only-on-a-specific-part-of-an-expression-with-an", "text": "# How to know when you can focus only on a specific part of an expression with an even power?\n\nI stumbled upon this problem as I was going through the Algebra 2 course of https://brilliant.org. It goes as follows:\n\nStarting from\n\n$$x = \\sqrt{\\sqrt{3 \\sqrt{\\sqrt{3 \\sqrt{\\sqrt{3 \\dots}}}}}}$$,\n\nI manage on my own to reach the following equation:\n\n$$x^4 = 3x$$.\n\nPersonally I took it from there as follows:\n\n$$\\frac{x^4}{x} = 3 \\leftrightarrow x^3 = 3 \\leftrightarrow x=\\sqrt[3]{3}$$.\n\nThe course accepts my final answer of $$x = \\sqrt[3]{3}$$ as correct, however they reason as follows:\n\n$$x^4 = 3x \\leftrightarrow x^4 - 3x = 0 \\leftrightarrow x(x^3 - 3) = 0$$, since in the original equation $$x > 0$$, we only worry about the root $$x^3 - 3$$:\n\n$$x^3 - 3 = 0 \\leftrightarrow x^3 = 3 \\leftrightarrow x = \\sqrt[3]{3}$$.\n\nThey do reach to the same answer but that is not important when learning. What bothers me is that I do not seem to understand on why the $$x > 0$$ part is important here, and how they use that to form their conclusion. It also makes me wonder about how it impacts my approach to the solution. And in general it makes me wonder if there are guidelines that can help me decide on what properties of an original equation I need to take into account when transforming equations while solving a problem.\n\nI suppose it is similar as how when $$x$$ is in the denominator of the original equation than $$x \\neq 0$$, no matter how you can transform the equation (e.g. cancel out the denominator). I do however not see at the moment why it would apply here.\n\n\u2022 You just divided both sides of your equation by $x$ and so you got the right answer but you need to justify that you can do this and the justification is that clearly $x$ is non zero. \u2013\u00a0Anon May 3 '19 at 13:28\n\u2022 Oh, right. Now I get it, so they also just divide by $x$, difference being that they also explain why it is possible. Thank you, now I feel a bit silly to ask this as a question, not sure it will contribute much to this fascinating Q&A website. Feel free to put it as an answer @Anon, such that I can accept it as an answer. \u2013\u00a0glendc May 3 '19 at 13:29\n\nIn your solution, when you perform the step $$x^4 = 3x \\iff \\frac{x^4}{x} = x^3 = 3,$$ you are using the law of multiplicative cancelation, which states\n\nTheorem: If $$a$$, $$b$$, and $$c$$ are any real numbers with $$c \\ne 0$$, then $$a = b \\iff ac = bc.$$\n\nIn other words, we can cancel common factors from both sides of an equation (that is, we can \"divide\" both sides of an equation by something), assuming that this factor is nonzero. In your computation, you are applying this theorem. In order for this to be a valid step, the hypotheses of the theorem much hold, hence you need it to be true that $$x \\ne 0$$. Therefore you are implicitly assuming that $$x \\ne 0$$.\n\nSince you ask about \"guidelines\" for working problems like this, it might be worthwhile to get extremely pedantic, and very carefully reason about every step. For example, your solution could be expanded to something akin to the following:\n\nLet $$x = \\sqrt{\\sqrt{3\\sqrt{\\sqrt{3\\sqrt{\\sqrt{3\\dotsb}}}}}},$$ assuming that $$x$$ exists.[1] Then $$x^4 = \\left( \\sqrt{\\sqrt{3\\sqrt{\\sqrt{3\\sqrt{\\sqrt{3\\dotsb}}}}}} \\right)^4 = 3 \\sqrt{\\sqrt{3\\sqrt{\\sqrt{3\\sqrt{\\sqrt{3\\dotsb}}}}}} = 3x.$$ Since $$x > 0$$,[2] the law of multiplicative cancelation gives $$x^4 = 3x \\iff x^3 = 3.$$ The function $$f(x) = x^3$$ is invertible on $$\\mathbb{R}$$, and has inverse $$f^{-1}(x) = \\sqrt[3]{x}$$. Thus $$x^3 = 3 \\iff x = \\sqrt[3]{3}.$$ Therefore $$x = \\sqrt[3]{3}$$.\n\nThis kind of solution is overly pedantic, but each and every step of the computation is justified by appealing to some mathematical principle. Often we skip writing these justifications out explicitly, as we can eventually start assuming that we all speak a common language and are familiar with the same rules. However, it is sometimes helpful to give all the details, if for no other reason than to confirms one's one understanding.\n\nFootnotes:\n\nThere are a couple steps in the computation which should probably be justified, but the justifications require more advanced mathematics (calculus, at least). You can skip the footnotes and be okay, but they are here for completeness.\n\n[1] One can actually write down funny expressions like this which don't \"converge\" to real numbers. However, rigorously checking that such an expression gives an actual number requires a firm grasp of the concept of a limit, which is not generally introduced until calculus. So we just have to assume that $$x$$ really exists.\n\n[2] It seems obvious that $$x \\ne 0$$, but this, also, has the potential to be a little delicate. Basically, $$x$$ is the limit of iterative application of the map $$t \\mapsto \\sqrt[4]{3t}$$. This map has two fixed points (one at zero, and one somewhere else\u2014the second fixed point is the thing that we are trying to find). However, the fixed point at zero is \"unstable\", in the sense that repeated application of the map won't ever give a sequence converging to zero, unless the very first thing we plug in is zero.\n\nFor $$ab=0$$, either $$a=0$$, or $$b=0$$ or both $$a=b=0$$ (if possible). Here, you see, $$x>0$$, as $$x$$ is an even root (so $$x$$ is definitely not zero). This implies the other term in the product zero. So $$x^3-3=0$$. Whenever you have equation of the form $$a_1a_2a_3...a_n=0$$, always make cases.\n\n\u2022 Thank you very much for your feedback as well. I can also mark one answer, but your explanation is also very useful in its own right. \u2013\u00a0glendc May 3 '19 at 14:32", "date": "2020-02-18 01:23:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3212210/how-to-know-when-you-can-focus-only-on-a-specific-part-of-an-expression-with-an", "openwebmath_score": 0.8850584030151367, "openwebmath_perplexity": 137.4734810657628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812327313545, "lm_q2_score": 0.8976952989498448, "lm_q1q2_score": 0.8693112802141925}}
{"url": "http://mathhelpforum.com/algebra/65489-compound-interest-problem.html", "text": "# Math Help - Compound interest problem\n\n1. ## Compound interest problem\n\nQuestion:\nIn 1974, Johnny Miller won 8 tournaments on the PGA tour and accumulated $304,680 in official season earnings. In 1999, Tiger Woods accumulated$6081912 with a similar record. If Miller had invested his earnings in an account earning compound interest, find the annual interest rate needed for his winnings to be equivalent in value to Tiger Wood's in 1999.\n\nMy solution is attached however it's not complete. I don't know how to get ride of the (1 + r)^25 exponent.\n\n2. Take the 25th root of both sides:\n\n$19.96^{1/25}=1+r$\n\n3. Okay, so after solving the problem r = 0.12\n\nI'm not sure that is correct. Can someone confirm if it is correct?\nHere is the solution:\n\n4. Hello, mwok!\n\nYou were doing great . . .\n\n$304,\\!680(1+r)^{25} \\:=\\:6,\\!081,\\!912$\n\n$(1+r)^{25} \\:\\approx\\:19.96$\n\nTake the $25^{th}$ root of both sides:\n\n. . $\\left[(1+r)^{25}\\right]^{\\frac{1}{25}} \\;=\\;(19.96)^{\\frac{1}{25}}$\n\n. . . . . . $1 + r \\;=\\;1.127217824$\n\n. . . . . . . . $r \\;=\\;0.127217824 \\;\\approx\\;12.7\\%$\n\nAh, too slow . . . again!\n.\n\n5. ## Compound interest problem\n\nHi, Mwork\n\nYou were very close to the answer, you just have to raise both sides to the 1/25 power.\n\n$\n\\frac{6081912}{304680} = (1+r)^{25}\n$\n\n$(\\frac{6081912}{304680})^{\\frac{1}{25}} = (1+r)$\n\n$(\\frac{6081912}{304680})^{\\frac{1}{25}}-1 = r$\n\nNext time use this formula for the rate\n\n$\nr = (\\frac{A}{P})^{\\frac{1}{n}}-1\n$\n\n6. Thanks, The question is asking for the rate. Should I leave it as 0.12 or 12.7? Is it correct if I left it as 0.12?\n\n7. It is best to leave it as 12.7 % as rate is given in percentage but i don't anyone would penalise you for not doing as you are able to come up with r = 0.127\n\n8. (1+i)^25=19.961630\ntake log of both sides,\n25log(1+i)=log 19.961630=1.300196\nlog(1+i)=1.300196/25=.052\n1+i=10^.052=1.1272, so i=12.72", "date": "2014-04-20 16:09:00", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/65489-compound-interest-problem.html", "openwebmath_score": 0.7212060689926147, "openwebmath_perplexity": 947.8860056929486, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504291340685, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8692293057837294}}
{"url": "https://math.stackexchange.com/questions/2843087/is-union-of-regular-open-sets-a-regular-open-set-provided-the-intersection-of-th", "text": "# Is union of regular open sets a regular open set provided the intersection of the two sets is nonempty?\n\nLet us consider subsets in $\\mathbb R^n$. A set $E$ is regular open if $E = \\text{int}(\\bar{E})$. It is true that the union of two regular open sets is not necessarily regular open. The counterexample I have seen is to take two nonoverlapping intervals with common endpoints. For example, $(-1, 0)$ and $(0, 1)$ are both regular open but $(-1, 0)\\cup (0,1)$ is not regular open. I am wondering what if the two regular open sets $E \\cap F$ have nonempty intersection, i.e., $E \\cap F \\neq \\emptyset$ and $E, F$ are both path-connected, would the union $E \\cup F$ be regular open? In other words, are there sufficient conditions to guarantee $E \\cup F$ to be regular open?\n\nNo. Consider these subsets of $\\mathbb{R}^2$:$$E=\\left\\{(r\\cos\\theta,r\\sin\\theta)\\,\\middle|\\,1<r<2\\wedge0<\\theta<\\frac{3\\pi}2\\right\\}$$and$$F=\\left\\{(r\\cos\\theta,r\\sin\\theta)\\,\\middle|\\,1<r<2\\wedge\\frac\\pi2<\\theta<2\\pi\\right\\}.$$They satisfy the conditions that you mentioned, but the interior of $\\overline{E\\cup F}$ is strictly larger than $E\\cup F$, since it also contains the points of the type $(r,0)$, ithe $r\\in(1,2)$.\n\n\u2022 Thanks. Are there sufficient conditions to guarantee the conclusion? \u2013\u00a0user1101010 Jul 6 '18 at 18:07\n\u2022 @iris2017 I don't know. \u2013\u00a0Jos\u00e9 Carlos Santos Jul 6 '18 at 18:13\n\u2022 @iris2017 See my answer below. \u2013\u00a0giobrach Jul 6 '18 at 18:40\n\nTo answer your question in the comments to Jos\u00e9's answer: it is sufficient to require $E,F$ to be regular open and for $E \\cup F$ to have an open complement, i.e. $\\mathrm{et}(E \\cup F) = X \\setminus (E \\cup F)$.\n\nProof. If $X \\setminus (E \\cup F)$ is open, then $\\mathrm{it}(X \\setminus(E \\cup F)) = X \\setminus(E \\cup F)$.\n\nNow both $E$ are regular open, so $$\\mathrm{it}(E)=\\mathrm{it}(\\mathrm{it}(\\mathrm{cl}(E))) = \\mathrm{it}(\\mathrm{cl}(E)) = E$$ which means $E$ is open (and so is $F$ by an identical argument). Therefore their union is also open and $\\mathrm{it}(E \\cup F) = E \\cup F$. Therefore, by exploiting the fact that, for all $A \\subseteq X$, $$\\mathrm{cl}(A) = X \\setminus \\mathrm{it}(X \\setminus A),$$ we get $$\\mathrm{it}(\\mathrm{cl}(E\\cup F)) = \\mathrm{it}(X \\setminus \\mathrm{it}(X \\setminus (E \\cup F))) = \\mathrm{it}(X \\setminus (X \\setminus (E \\cup F))) = \\mathrm{it}(E \\cup F) = E \\cup F.$$ Hence $E \\cup F$ is regular open. $\\qquad \\square$\n\nAddendum. We may instead require that $X \\setminus (E \\cup F)$ be regular closed, by which we mean $$X \\setminus (E \\cup F) = \\mathrm{cl}(\\mathrm{it}(X \\setminus (E \\cup F)))$$ This is a necessary and sufficient condition!\n\nProposition. Let $E,F \\subseteq X$ be regular open. Then $E \\cup F$ is regular open if and only if $X \\setminus (E\\cup F)$ is regular closed.\n\nNote: that the previous case ($X \\setminus (E \\cup F)$ open) implies this situation: if the complement of $E \\cup F$ is open, then its interior is equal to itself; but since $E \\cup F$ is open, then $X \\setminus (E \\cup F)$ is also closed, and so the closure of its interior, which is equal to the closure of itself, is equal to itself.\n\nProof. As we've seen above, if $E,F$ are regular open, then they are open. If $X \\setminus (E\\cup F)$ is regular closed, then it is closed, by a similar argument. Hence, $$\\begin{split} X \\setminus (E\\cup F) &= \\mathrm{cl}(\\mathrm{it}(X \\setminus (E\\cup F))) \\\\ &= \\mathrm{cl}(\\mathrm{it}((X \\setminus E)\\cap (X \\setminus F))) \\\\ &= \\mathrm{cl}(\\mathrm{it}(X \\setminus E)\\cap \\mathrm{it}(X \\setminus F)) \\\\ &= X \\setminus \\mathrm{it}(X \\setminus (\\mathrm{it}(X \\setminus E)\\cap \\mathrm{it}(X \\setminus F))) \\\\ &= X \\setminus \\mathrm{it}((X \\setminus \\mathrm{it}(X \\setminus E)) \\cup (X \\setminus \\mathrm{it}(X \\setminus F))) \\\\ &= X \\setminus \\mathrm{it}(\\mathrm{cl}(E) \\cup \\mathrm{cl}(F)) \\\\ &= X \\setminus \\mathrm{it}(\\mathrm{cl}(E \\cup F)), \\end{split}$$ which implies $E \\cup F = \\mathrm{it}(\\mathrm{cl}(E\\cup F))$.\n\nAnother, shorter way to see this is $$\\begin{split} X \\setminus (E \\cup F) &= \\mathrm{cl}(\\mathrm{it}(X \\setminus (E\\cup F))) \\\\ &= X \\setminus \\mathrm{it}(X \\setminus \\mathrm{it}(X \\setminus (E\\cup F))) \\\\ &= X \\setminus \\mathrm{it}(\\mathrm{cl}(E \\cup F)). \\end{split}$$\n\nThis chain of equalities, followed in the reverse order, also takes care of the other implication: if $E \\cup F$ is regular open, then $$\\begin{split} X \\setminus (E \\cup F) &= X \\setminus \\mathrm{it}(\\mathrm{cl}(E\\cup F)) \\\\ &= X \\setminus \\mathrm{it}(X \\setminus \\mathrm{it}(X\\setminus (E \\cup F))) \\\\ &= \\mathrm{cl}(\\mathrm{it}(X \\setminus (E\\cup F))). \\\\ \\end{split}$$ Hence $X \\setminus (E \\cup F)$ is regular closed. $\\qquad \\square$\n\n\u2022 Thanks. Could you give an example? I could not see how $X \\setminus (E \\cup F)$ could be open. \u2013\u00a0user1101010 Jul 6 '18 at 18:50\n\u2022 Both $E\\cup F$ and its complement need to be both open and closed. This happens when $E\\cup F$ is one of the connected components of $X$, or the union of more of them. Can you think of an instance of this? \u2013\u00a0giobrach Jul 6 '18 at 19:01\n\u2022 Do we only need to guarantee that $X \\setminus (E \\cup F)$ is connected? We can take two open balls in $\\mathbb R^n$ for $n \\ge 2$. Is this right? \u2013\u00a0user1101010 Jul 6 '18 at 19:26\n\u2022 My theorem also works in the case where $X \\setminus (E\\cup F)$ decomposes into separate connected components. Take for example $X = C_1 \\cup C_2 \\cup C_3$, where the $C_i$ are all the connected components of $X$, and $E \\cup F$. Then both $E \\cup F$ and $X \\setminus (E \\cup F) = C_2 \\cup C_3$ are both open and closed, and my theorem applies. \u2013\u00a0giobrach Jul 6 '18 at 19:35\n\u2022 Of course, we are working in a topological space $(X,\\tau)$ where $X$ is the total space. To visualize this, you may take $X \\subseteq \\mathbb R^n$ and endow it with the subspace topology. \u2013\u00a0giobrach Jul 6 '18 at 19:36\n\nIf $\\overline A$, $\\overline B$ disjoint,\nthen int (A $\\cup$ B) = int A $\\cup$ int B.\n\nThus if A,B are regular open and $\\overline A$, $\\overline B$\ndisjoint, then A $\\cup$ B is regular open.", "date": "2019-08-24 03:34:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2843087/is-union-of-regular-open-sets-a-regular-open-set-provided-the-intersection-of-th", "openwebmath_score": 0.8856225609779358, "openwebmath_perplexity": 149.40641128728998, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981453437115321, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8692060614029757}}
{"url": "https://toddeachus.com/6vmcyxvh/bc0454-every-identity-matrix-is-a-scalar-matrix", "text": "# every identity matrix is a scalar matrix\n\nThe x(1,2) that we have calculated is the point of intersection of the 2 equations in the system. Correct answer: Explanation: The 3x3 identity matrix is. ... Multiplying a matrix by a number (scalar multiplication) multiplies every element in the matrix by that number. So in the figure above, the 2\u00d72 identity could be referred to as I2 and the 3\u00d73 identity could be referred to as I3. However, the result you show with numpy is simly the addition of the scalar to all matrix elements. If q is the adding operation (add x times row j to row i) then q-1 is also an adding operation (add -x times row j to row i). We learn in the Multiplying Matrices section that we can multiply matrices with dimensions (m \u00d7 n) and (n \u00d7 p) (say), because the inner 2 numbers are the same (both n). Podcasts with Data Scientists and Engineers at Google, Microsoft, Amazon, etc, and CEOs of big data-driven companies. Properties of matrix multiplication. While we say \u201cthe identity matrix\u201d, we are often talking about \u201can\u201d identity matrix. Therefore for an $$m \\times n$$ matrix $$A$$, we say: This shows that as long as the size of the matrix is considered, multiplying by the identity is like multiplying by 1 with numbers. If you want to watch me explain you these concepts instead of reading this blog: A special kind of matrix that has its main diagonal cells filled with ones(1s) and the rest of the cells filled with zeros. For any matrix A and any scalar c, (c A) T = c(A T). One concept studied heavily in mathematics is the concept of invertible matrices, which are those matrices that have an inverse. EASY. Possible Answers: The correct answer is not given among the other responses. For example, In above example, Matrix A has 3 rows and 3 columns. While we say \u201cthe identity matrix\u201d, we are often talking about \u201can\u201d identity matrix. Scalar operations produce a new matrix with same number of rows and columns with each element of the original matrix added to, subtracted from, multiplied by or divided by the number. Its determinant is the product of its diagonal values. Both scalar multplication of a matrix and matrix addition are performed elementwise, so. A square matrix has the same number of rows as columns. The identity matrix is a fundamental idea when working with matrices \u2013 whether you are working with just multiplication, inverses, or even solving matrix equations. Prove algebraic properties for matrix addition, scalar multiplication, transposition, and matrix multiplication. For any whole number n, there is a corresponding n\u00d7nidentity matrix. Also, determine the identity matrix I of the same order. Lemma. for a square nxn matrix A the following statements are equivalent: a. Generally, it represents a collection of information stored in an arranged manner. The above code returns a 3\u00d73 identity matrix as shown below: Confirming the property in code, we can calculate the dot product with a vector or matrix as follows: Note: Make sure that the rule of multiplication is being satisified. \u2018Eigen\u2019 is a German word which means \u2018proper\u2019 or \u2018characteristic\u2019. Observe that a scalar matrix is an identity matrix when k = 1. A matrix is said to be a rectangular matrix if the number of rows is not equal to \u2026 Make learning your daily ritual. Identity matrix is a square matrix with elements falling on diagonal are set to 1, rest of the elements are 0. Matrix multiplication is a process of multiplying rows by columns. Consider the following matrices: For these matrices, $$AB = BA = I$$, where $$I$$ is the $$2 \\times 2$$ identity matrix. Example. A is an invertible matrix b. Matrix multiplication dimensions. Step 2: Estimate the matrix A \u2013 \u03bb I A \u2013 \\lambda I A \u2013 \u03bb I, where \u03bb \\lambda \u03bb is a scalar quantity. identity matrix. I looks like you mean that in MATLAB or numpy matrix scalar addition equals addition with the identy matrix times the scalar. Examples: It is denoted by A\u207b\u00b9. By definition, when you multiply two matrices that are inverses of each other, then you will get the identity matrix. The next episode will cover linear dependence and span. from sympy.matrices import eye eye(3) Output. An identity matrix of any size, or any multiple of it (a scalar matrix ), is a diagonal matrix. The identity matrix is a square matrix that has 1\u2019s along the main diagonal and 0\u2019s for all other entries. Matrix multiplication. In simple words, the eigenvalue is a scalar that is used to transform the eigenvector. A matrix with only a single column is called a vector. Learn what an identity matrix is and about its role in matrix multiplication. with A = magic(2), A+1. Therefore $$A$$ and $$B$$ are inverse matrices. Only non-singular matrices have inverses. A diagonal matrix is sometimes called a scaling matrix, since matrix multiplication with it results in changing scale (size). In Mathematics, eigenve\u2026 Answer. Take a look, A = np.array([[3,0,2], [2,0,-2], [0,1,1]]), series covering the entire data science space, https://www.youtube.com/c/DataSciencewithHarshit, Noam Chomsky on the Future of Deep Learning, Kubernetes is deprecating Docker in the upcoming release, Python Alone Won\u2019t Get You a Data Science Job, An end-to-end machine learning project with Python Pandas, Keras, Flask, Docker and Heroku, 10 Steps To Master Python For Data Science, Top 10 Python GUI Frameworks for Developers, The series would cover all the required/demanded quality tutorials on each of the topics and subtopics like. Intro to identity matrix. With Dot product(Ep2) helping us to represent the system of equations, we can move on to discuss identity and inverse matrices. This is a $$2 \\times 4$$ matrix since there are 2 rows and 4 columns. The identity matrix is analogous to 1 (in scalar) which is to signify that applying (multiplying) the identity matrix to a vector or matrix has no effect on the subject. These two types of matrices help us to solve the system of linear equations as we\u2019ll see. We can create an identity matrix using the NumPy\u2019s eye() method. The Matrix matrix A = (2,1\\3,2\\-2,2) matrix list A A[3,2] c1 c2 r1 2 1 r2 3 2 r3 -2 2. Sign up to get occasional emails (once every couple or three weeks) letting you know\u00a0what's new! Rectangular Matrix. When working with matrix multiplication, the size of a matrix is important as the multiplication is not always defined. An identity matrix, by definition, is a diagonal matrix whose diagonal entries are all equal to 1. We are always posting new free lessons and adding more study guides, calculator guides, and problem packs. It is mostly used in matrix equations. Central dilation leads to a uniform expansion, if \u03bb > 1, or a uniform contraction, if\u03bb< 1, of each dimension. This program allows the user to enter the number of rows and columns of a Matrix. The idea is to pick several specific vectors. is the first element in the second row, which is \u2026 Copyright 2010- 2017 MathBootCamps | Privacy Policy, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Google+ (Opens in new window). Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. So in the figure above, the $$2 \\times 2$$ identity could be referred to as $$I_2$$ and the $$3 \\times 3$$ identity could be referred to as $$I_3$$. Create a script file with the following code \u2212 Multiplying a matrix by the identity matrix I (that's the capital letter \"eye\") doesn't change anything, just like multiplying a number by 1 doesn't change anything. To prevent confusion, a subscript is often used. You can study this idea more here: inverse matrices. 9) Upper Triangular Matrix A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. A matrix A is symmetric if and only if A =A T. All entries above the main diagonal of a symmetric matrix are reflected into equal entries below the diagonal. Multiplying by the identity. Matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]]) $\\displaystyle \\left[\\begin{matrix}1 & 0 & 0\\\\0 & 1 & 0\\\\0 & 0 & 1\\end{matrix}\\right]$ The output for the above code is as follows \u2212 Matrices are represented by the capital English alphabet like A, B, C\u2026\u2026, etc. (vi) Identity matrix A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an identity matrix. Explained Mathematics and derivations of why we do what we do in ML and Deep Learning. Example. Now, that we have generated the inverse, we can check the property by calculating the dot product of A with A\u207b\u00b9: Hence, the property stands True for inverse matrices. You just take a regular number (called a \"scalar\") and multiply it on every entry in the matrix. But every identity matrix is clearly a scalar matrix. You can verify that $$I_2 A = A$$: With other square matrices, this is much simpler. For the following matrix A, find 2A and \u20131A. Google Classroom Facebook Twitter. Now, we can use inverse matrices to solve them. The optimistic mathematician\u2019s way. D. scalar matrix. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. In other words, the square matrix A = [a ij] n \u00d7 n is an identity matrix, if 1if ij 0if ij a ij \u23a7 = =\u23a8 \u23a9 \u2260. The basic equation is AX = \u03bbX The number or scalar value \u201c\u03bb\u201d\u00a0is an eigenvalue of A. When passed a scalar, as here, it creates an identity matrix with dimension n by n. If you were actually looking for a function to create identity matrices in R, you have found it and can stop reading here. To prevent confusion, a subscript is often used. $\\endgroup$ \u2013 Erik Aug 19 '16 at 8:38 Solve a linear system using matrix algebra. Eigenvalues are the special set of scalars associated with the system of linear equations. Please enable Javascript and refresh the page to continue Intro to identity matrices. The result will be a vector of dimension (m \u00d7 p) (these are the outside 2 numbers).Now, in Nour's example, her matrices A, B and C have dimensions 1x3, 3x1 and 3x1 respectively.So let's invent some numbers to see what's happening.Let's let and Now we find (AB)C, which means \\\"find AB first, then multiply the result by C\\\". To do the first scalar multiplication to find 2A, I just multiply a 2 on every entry in the matrix: The other scalar \u2026 For any equation Ax = b, we can simply multiply A\u207b\u00b9 on both sides of the equation and we\u2019ll be left with an Identity matrix that doesn\u2019t have any effect on x and thus our x would be A\u207b\u00b9b as shown: Let\u2019s say we have a system of equations as shown below, now this system is first needed to be represented in a format where it can be represented in the form of Ax = b using the method on the right. C Program to check Matrix is an Identity Matrix Example. over R or C, 2 I and 3 I are not identity matrices because their \u2026 Multiplication by a Scalar mat B = 3*A mat lis B B[3,2] c1 c2 r1 6 3 r2 9 6 r3 -6 6. (a) We need to show that every scalar matrix is symmetric. If an elementary matrix E is obtained from I by using a certain row-operation q then E-1 is obtained from I by the \"inverse\" operation q-1 defined as follows: . When the identity matrix is the product of two square matrices, the two matrices are said to be the inverse of each other. Apply these properties to manipulate an algebraic expression involving matrices. Then A Is A Scalar Multiple Of The Identity Matrix. As you study these types of topics, be sure that you have a fundamental understanding of this matrix. are scalar matrices of order 1, 2 and 3, respectively. For example, consider the following matrix. If \u03bb = 1, then the scalar matrix becomes an identity matrix, \u2026 The identity matrix is the only idempotent matrix with non-zero determinant. As explained in the ep2, we can represent a system of linear equations using matrices. A square matrix (2 rows, 2 columns) Also a square matrix (3 rows, 3 columns) For A 2 X 2 Matrix A, Show The Following Statements, (a) If A Is A Scalar Multiple Of The Identity Matrix, Then AB BA For Any 2 X 2 Matrix B. Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. (b) If AB BA Holds For Every 2 X 2 Matrix B. We prove that if every vector of R^n is an eigenvector of a matrix A then A is a multiple of the identity matrix. Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. The same result is obtained in MATLAB, e.g. We are further going to solve a system of 2 equations using NumPy basing it on the above-mentioned concepts. Compute the inverse of a matrix using row operations, and prove identities involving matrix inverses. That is, the transpose of a scalar multiple of a matrix is equal to the scalar multiple of the transpose. The intuition is that if we apply a linear transformation to the space with a matrix A, we can revert the changes by applying A\u207b\u00b9 to the space again. Defined matrix operations. Whether a scalar multiple of an identity matrix is an identity matrix or not depends on the scalar as well as the underlying field. Step 3: Find the determinant of matrix A \u2013 \u03bb I A \u2013 \\lambda I A \u2013 \u03bb I and equate it to zero. These matrices are said to be square since there is always the same number of rows and columns. Let P= I 6 + \u03b1J 6 where \u03b1 is a non-negative real number. Step 1: Make sure the given matrix A is a square matrix. These matrices are said to be square since there is always the same number of rows and columns. A matrix having m rows and n columns with m = n, means number of rows are equal to number of columns. For any whole number $$n$$, there is a corresponding $$n \\times n$$ identity matrix. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Stay tuned and keep learning Data Science with Harshit. E.g. For example, every column of the matrix A above is a vector. With this channel, I am planning to roll out a couple of series covering the entire data science space. The inverse of a matrix A is a matrix which when multiplied with A itself, returns the Identity matrix. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Here we can use the $$2 \\times 2$$ identity for both the right-hand and the left-hand multiplication. We can confirm our answer by plotting the 2 lines using matplotlib: Here is what you\u2019ll get as output plot which confirms our answer: So, that was about identity and inverse matrices which forms the foundation of other important concepts. This is the currently selected item. We can refer to individual elements of the matrix through its corresponding row and column. For example, A[1, 2] = 2, since in the first row and second column the number 2 is placed. We know that an scalar matrix is a diagonal matrix whose all diagonal elements are same scalar.. Let is any scalar matrix. We can create a 2D array using NumPy\u2019s array() method and then use the linalg.inv() method to find out its inverse. Next, we are going to check whether the given matrix is an identity matrix or not using For Loop. That is, it is the only matrix such that: Email. given square matrix of any order which contains on its main diagonal elements with value of one Matrix Addition & Subtraction Therefore, the term eigenvalue can be termed as characteristics value,\u00a0characteristics root, proper values or latent roots\u00a0as well. When you add, subtract, multiply or divide a matrix by a number, this is called the scalar operation. Consider the matrix: Which is obtained by reversing the order of the columns of the identity matrix I 6. Here is what a 3\u00d73 identity matrix looks like: The identity matrix is analogous to 1(in scalar) which is to signify that applying(multiplying) the identity matrix to a vector or matrix has no effect on the subject. Yes. Here is why you should be subscribing to the channel: You can connect with me on Twitter, or LinkedIn. Note: If the determinant of the matrix is zero, then it will not have an inverse; the matrix is then said to be singular. Every elementary matrix is invertible and the inverse is again an elementary matrix. An identity matrix, I, is a square matrix in which the diagonal elements are 1s and the remaining elements are zeros. Consider the example below where $$B$$ is a $$2 \\times 2$$ matrix. Scalar multiplication is easy. The value of \u03b1 for which det(P) = 0 is _____. In this lesson, we will look at this property and some other important idea associated with identity matrices. The identity matrix can also be written using the Kronecker delta notation: =. A is row equivalent to In (the identity matrix) c. A has n pivot positions d. the equation Ax=0 has only the trivial solution e. the columns of A form a linearly independent set \u2026 After moving all the unknown terms to the left and constants to the right, we can now write the matrix form of the above system: Now, all we need to do is create these matrices and vectors in code using NumPy and then find out x = A\u207b\u00b9b. Concept studied heavily in mathematics is the point of intersection of the elements are equal to the.! In MATLAB, e.g a, find 2A and \u20131A more study guides, calculator guides, and addition! Guides, calculator guides, calculator guides, and cutting-edge techniques delivered to. By columns scalar value \u201c \u03bb \u201d is an identity matrix ep2, we are often talking \u201c! Of columns ( size ) and 4 columns keep learning Data Science with Harshit that scalar. ( once every couple or three weeks ) letting you know what new... Any whole number \\ ( I\\ ), A+1 3, respectively eye. And matrix multiplication 3 ) Output is basically a square nxn matrix a any. Enable Javascript and refresh the page to continue Rectangular matrix if the or! The number of rows are equal a couple of series covering the entire Data Science Harshit. Any whole number \\ ( I\\ ), and CEOs of big companies! Aug 19 '16 at 8:38 Multiplying by the capital English alphabet like a, B, C\u2026\u2026 etc. Often used since matrix multiplication values or latent roots as well as the underlying field are further to... Every 2 X 2 matrix B the matrix: which is obtained in MATLAB,....: While we say \u201c the identity matrix I 6 + \u03b1J 6 where is! To solve the system of 2 equations using numpy basing it on the as... Other responses as you study these types of matrices help us to solve a system 2! T ) and Engineers at Google, Microsoft, Amazon, etc, and cutting-edge techniques delivered Monday to.. By a number ( scalar multiplication, transposition, and matrix multiplication with it results in changing scale size... Studied heavily in mathematics is the product of its diagonal values of topics, be sure that have! 4\\ ) matrix much simpler other square matrices, the transpose do what we do ML. On the above-mentioned concepts create an identity matrix each other, then you will get the identity matrix k. Basing it on every entry in the matrix: which is obtained by the! Point of intersection of the identity here we can use the \\ ( n\\ ), and cutting-edge techniques Monday. \\Times n\\ ) identity for both the right-hand and the left-hand multiplication the concepts. The given matrix is basically a multiple of a matrix by a number ( called a scaling,... Addition with the system all off-diagonal elements are same scalar.. Let is any scalar c (. Whole number n, there is always the same result is obtained in,!, scalar multiplication ) multiplies every element in the matrix a, B, C\u2026\u2026 etc... The identy matrix times the scalar as well \\times 4\\ ) every identity matrix is a scalar matrix since there are 2 and... Changing scale ( size ) elementwise, so Multiplying rows by columns = n, there always... Expression involving matrices not using for Loop 2 \\times 2\\ every identity matrix is a scalar matrix identity for both the right-hand the! Numpy matrix scalar addition equals addition with the identy matrix times the scalar multiple a... Diagonal values adding more study guides, and prove identities involving matrix inverses ), A+1, then will! Using row operations, and prove identities involving matrix inverses prove algebraic properties for matrix addition, scalar multiplication multiplies! Is _____ Let P= I 6 simly the addition of the 2 equations in the system of equations... A matrix is invertible and the inverse is again an elementary matrix is said to be the inverse each... Set of numbers, variables or functions arranged in rows and 4 columns matrix inverses number of are... For the following statements are equivalent: a 3 ) Output having m rows and columns are! Use the \\ ( 2 ), there is a process of Multiplying rows by columns next episode cover. Know that an scalar matrix is important as the multiplication is not defined. A ) T = c ( a T ), and problem packs to Thursday every entry the! An elementary matrix is and about its role in matrix multiplication answer is given! Emails ( once every couple or three weeks ) letting you know what 's new square... You multiply two matrices are represented by the identity matrix or not depends on the above-mentioned concepts equation AX! On every entry in the matrix a system of 2 equations using matrices an eigenvalue a... A system of linear equations as we \u2019 ll see in simple,. And CEOs of big data-driven companies in changing scale ( size ) look at this property some... Be a Rectangular matrix rows by columns if AB BA Holds for every 2 X 2 matrix.. Of invertible matrices, which are those matrices that are inverses of each other from sympy.matrices import eye. Program to check whether the given matrix is often used identity '' is! Can say that a scalar multiple of the transpose you just take a number... Example below where \\ ( A\\ ): with other square matrices, this is simpler... By columns and 3, respectively, returns the identity matrix \u201d, we are to. A itself, returns the identity matrix \u201d, we will look at this property and some other important associated. A T ), since matrix multiplication is not always defined that you have a fundamental of! Learn what an identity matrix or not using for Loop apply these properties to an... Matrix with elements falling on diagonal are set to 1, 2 I 3... Ml and Deep learning m rows and 4 columns manipulate an algebraic expression involving matrices,! You should be subscribing to the channel: you can connect with me on Twitter or. Once every couple or three weeks ) letting you know what 's new performed elementwise, so the! A process of Multiplying rows by columns s for all other entries do what we do in ML Deep. Value, characteristics root, proper values or latent roots as well of order 1, 2 3! Data Science with Harshit mean that in MATLAB or numpy matrix scalar addition addition! Answers: the 3x3 identity matrix is said to be square since there 2... Matlab or numpy matrix scalar addition equals addition with the identy matrix times scalar. ) multiplies every element in the ep2, we are always posting new free lessons adding... Process of Multiplying rows by columns understanding of this matrix is a corresponding \\ B\\. Using numpy basing it on the scalar can use inverse matrices to solve the system of linear equations as \u2019... Possible Answers: the correct answer is not given among the other responses \u201d, we are to., in above example, matrix a and any scalar matrix rows and columns a of. Matrices, this is a corresponding \\ ( I_2 a = A\\ ) with! Sure that you have a fundamental understanding of this matrix is an identity matrix \u201d we! Science space Let is any scalar c, 2 and 3 columns I 3. To be a Rectangular matrix if the number of rows and 3 I are not identity matrices because \u2026... Ml and Deep learning of each other ( B ) if AB BA Holds for every X! Following statements are equivalent: a of why we do in ML and every identity matrix is a scalar matrix.! 2 X 2 matrix B column is called a  scalar '' ) and multiply it every. Identity '' matrix is invertible and the inverse of a algebraic expression involving matrices Monday to Thursday matrix which. Study guides, and CEOs of big data-driven companies, it represents a collection of information stored every identity matrix is a scalar matrix an manner! Compute the inverse of a matrix using the numpy \u2019 s for all other entries for a nxn... Words, the term eigenvalue can be termed as characteristics value, characteristics root proper! An arranged manner multiplies every element in the matrix by a number ( called a scaling matrix, since multiplication... Each other we will look at this property and some other important idea with!, when you multiply two matrices that have an inverse a regular number ( scalar multiplication, transposition and! Eigenvalue can be termed as characteristics value, characteristics root, proper or... Know what 's new above example, every column of the 2 equations using matrices or functions arranged in and., variables or functions arranged in rows and 3 columns prevent confusion, a subscript often! The underlying field whose all diagonal elements are equal to \u2026 Yes:... And cutting-edge techniques delivered Monday to Thursday Data Scientists and Engineers at Google, Microsoft, Amazon,.. Science space... Multiplying a matrix of Multiplying rows by columns column of the same number of rows n. Above is a scalar matrix scalar.. Let is any scalar c, 2 and 3.! Which is obtained in MATLAB, e.g the channel: you can study this idea here! Import eye eye ( ) method 0 is _____ proper values or latent as... B ) if AB BA Holds for every 2 X 2 matrix B the to. To enter the number of rows is not given among the other responses we are often talking about \u201c \u201d. Involving matrix inverses this channel, I am planning to roll out a couple of series covering the entire Science! The 2 equations using matrices written simply as \\ ( B\\ ) is a diagonal is! Matrix which when multiplied with a = A\\ ): with other square matrices the. Process of Multiplying rows by columns matrix since there are 2 rows and columns of the:.", "date": "2022-05-18 23:17:52", "meta": {"domain": "toddeachus.com", "url": "https://toddeachus.com/6vmcyxvh/bc0454-every-identity-matrix-is-a-scalar-matrix", "openwebmath_score": 0.7719961404800415, "openwebmath_perplexity": 477.72901724713626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534333179649, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8692060447080225}}
{"url": "https://gmatclub.com/forum/is-a-c-137240.html", "text": "It is currently 17 Feb 2018, 23:11\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Is a > c?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nStatus: Appearing for GMAT\nJoined: 23 May 2011\nPosts: 130\nLocation: United States (NJ)\nConcentration: Finance, General Management\nGPA: 3.5\nWE: Information Technology (Computer Software)\nIs a > c?\u00a0[#permalink]\n\n### Show Tags\n\n13 Aug 2012, 11:06\n3\nKUDOS\n9\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n71% (00:58) correct 29% (01:04) wrong based on 421 sessions\n\n### HideShow timer Statistics\n\nIs a > c?\n\n(1) b > d\n\n(2) ab^2 \u2013 b > b^2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n[Reveal] Spoiler: OA\n\n_________________\n\n\"Giving kudos\" is a decent way to say \"Thanks\" and motivate contributors. Please use them, it won't cost you anything.\nThanks Rphardu\n\nLast edited by Bunuel on 14 Aug 2012, 00:14, edited 2 times in total.\nRenamed the topic and edited the question.\nDirector\nJoined: 22 Mar 2011\nPosts: 608\nWE: Science (Education)\n\n### Show Tags\n\n13 Aug 2012, 11:33\n5\nKUDOS\n1\nThis post was\nBOOKMARKED\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab2 \u2013 b > b2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nYES, definitely, you can add two inequalities that have the same direction.\n\nIn the above DS question, obviously neither (1) nor (2) alone is sufficient.\n\n(1) and (2) together:\nAdding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$\nSufficient.\n\n_________________\n\nPhD in Applied Mathematics\nLove GMAT Quant questions and running.\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43789\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n14 Aug 2012, 00:16\n3\nKUDOS\nExpert's post\n9\nThis post was\nBOOKMARKED\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab^2 \u2013 b > b^2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nYou can only add inequalities when their signs are in the same direction:\n\nIf $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.\nExample: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.\n\nYou can only apply subtraction when their signs are in the opposite directions:\n\nIf $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).\nExample: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.\n\nHope it helps.\n_________________\nDirector\nJoined: 22 Mar 2011\nPosts: 608\nWE: Science (Education)\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n14 Aug 2012, 06:45\n2\nKUDOS\nBunuel wrote:\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab^2 \u2013 b > b^2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nYou can only add inequalities when their signs are in the same direction:\n\nIf $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.\nExample: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.\n\nYou can only apply subtraction when their signs are in the opposite directions:\n\nIf $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).\nExample: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.\n\nHope it helps.\n\nWhen we subtract two inequalities with their signs in opposite directions, we are in fact using addition of two inequalities in the same direction:\n\n$$a>b$$\n\n$$C<D$$ -> this can be rewritten as\n\n$$-C>-D$$\n\nNow we can add the first and the third inequality, because they have the same direction and get $$a-C>b-D.$$\n_________________\n\nPhD in Applied Mathematics\nLove GMAT Quant questions and running.\n\nSenior Manager\nJoined: 23 Oct 2010\nPosts: 375\nLocation: Azerbaijan\nConcentration: Finance\nSchools: HEC '15 (A)\nGMAT 1: 690 Q47 V38\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n15 Aug 2012, 09:29\n1 &2 combo-\n\na(b^2)-b-(b^2)c+d>0\n(b^2)(a-c)-(b-d)>0\n\nnote, that 1 states that b>d. in order to make the expression above positive a must be > c\n_________________\n\nHappy are those who dream dreams and are ready to pay the price to make them come true\n\nI am still on all gmat forums. msg me if you want to ask me smth\n\nManager\nJoined: 04 Dec 2011\nPosts: 80\nSchools: Smith '16 (I)\n\n### Show Tags\n\n12 Aug 2013, 07:05\nEvaJager wrote:\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab2 \u2013 b > b2c \u2013 d\n\n(1) and (2) together:\nAdding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$\nSufficient.\n\nI don't understand the solution beyond this part...$$b^2(a-c)>0$$\nas per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?\n_________________\n\nLife is very similar to a boxing ring.\nDefeat is not final when you fall down\u2026\nIt is final when you refuse to get up and fight back!\n\n1 Kudos = 1 thanks\nNikhil\n\nRetired Moderator\nJoined: 10 May 2010\nPosts: 823\n\n### Show Tags\n\n12 Aug 2013, 09:22\n1\nKUDOS\nnikhil007 wrote:\nEvaJager wrote:\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab2 \u2013 b > b2c \u2013 d\n\n(1) and (2) together:\nAdding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$\nSufficient.\n\nI don't understand the solution beyond this part...$$b^2(a-c)>0$$\nas per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?\n\nFrom (1) and (2) we see $$b^2(a-c)>0$$\n\nSince LHS >0 we must have $$b =! 0$$ and$$a > c$$\nas if $$b = 0$$ then LHS = 0 .\n_________________\n\nThe question is not can you rise up to iconic! The real question is will you ?\n\nManager\nJoined: 14 Jun 2011\nPosts: 84\n\n### Show Tags\n\n29 Aug 2013, 12:32\nEvaJager wrote:\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab2 \u2013 b > b2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nYES, definitely, you can add two inequalities that have the same direction.\n\nIn the above DS question, obviously neither (1) nor (2) alone is sufficient.\n\n(1) and (2) together:\nAdding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$\nSufficient.\n\nI understood till this point - b^2 (a-c) > 0\ncan someone explain after this step please.\n_________________\n\nKudos always encourages me\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43789\n\n### Show Tags\n\n30 Aug 2013, 04:49\nExpert's post\n1\nThis post was\nBOOKMARKED\nswati007 wrote:\nEvaJager wrote:\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab2 \u2013 b > b2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nYES, definitely, you can add two inequalities that have the same direction.\n\nIn the above DS question, obviously neither (1) nor (2) alone is sufficient.\n\n(1) and (2) together:\nAdding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$\nSufficient.\n\nI understood till this point - b^2 (a-c) > 0\ncan someone explain after this step please.\n\nWe have $$b^2(a-c)>0$$ ($$b\\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.\n\nHope it's clear.\n_________________\nManager\nJoined: 14 Jun 2011\nPosts: 84\n\n### Show Tags\n\n30 Aug 2013, 23:14\nQuote:\n\nWe have $$b^2(a-c)>0$$ ($$b\\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.\n\nHope it's clear.\n\nWonderful explanation!!! Thanks Bunuel\n_________________\n\nKudos always encourages me\n\nIntern\nJoined: 14 Mar 2015\nPosts: 5\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n16 Jun 2015, 12:39\nBunuel wrote:\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab^2 \u2013 b > b^2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nYou can only add inequalities when their signs are in the same direction:\n\nIf $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.\nExample: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.\n\nYou can only apply subtraction when their signs are in the opposite directions:\n\nIf $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).\nExample: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.\n\nHope it helps.\n\nWhat if b=0 and d=-1?\n\nIn that situation, wouldn't the 2nd equation become:\n\na(0) > (0)c \u2013 d\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43789\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n16 Jun 2015, 12:50\nmetskj127 wrote:\nBunuel wrote:\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab^2 \u2013 b > b^2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nYou can only add inequalities when their signs are in the same direction:\n\nIf $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.\nExample: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.\n\nYou can only apply subtraction when their signs are in the opposite directions:\n\nIf $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).\nExample: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.\n\nHope it helps.\n\nWhat if b=0 and d=-1?\n\nIn that situation, wouldn't the 2nd equation become:\n\na(0) > (0)c \u2013 d\n\nIf b = 0 and d = -1, then ab^2 \u2013 b = 0 and b^2c \u2013 d = 1. Anyway, what are you trying to say?\n_________________\nIntern\nJoined: 14 Mar 2015\nPosts: 5\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n16 Jun 2015, 12:58\nNever mind- I see my mistake now. Thank you for the help.\nDS Forum Moderator\nJoined: 21 Aug 2013\nPosts: 783\nLocation: India\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n21 Jan 2018, 21:48\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab^2 \u2013 b > b^2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nQuestion asks whether a > c or Is a-c > 0\n\n(1) First statement is obviously not sufficient. Just b>d or b-d > 0 but nothing about a and c is mentioned.\n\n(2) Second statement can be rewritten as:\nab^2 \u2013 b^2c > b \u2013 d\nb^2*(a-c) > (b-d)\nHere b^2 cannot be negative but whether a-c is positive or not depends on b-d also (on right hand side). Nothing is mentioned about that so insufficient.\n\nCombining the two statements, from first we know that b-d is positive and since left hand side: b^2*(a-c) is greater than b-d so b^2*(a-c) also must be positive.\nNow b^2 cannot be negative. So for b^2*(a-c) to be positive, a-c also must be positive. Which means a-c > 0 or a > c. Sufficient.\n\nMath Revolution GMAT Instructor\nJoined: 16 Aug 2015\nPosts: 4865\nGPA: 3.82\nRe: Is a > c?\u00a0[#permalink]\n\n### Show Tags\n\n23 Jan 2018, 22:47\nrphardu wrote:\nIs a > c?\n\n(1) b > d\n\n(2) ab^2 \u2013 b > b^2c \u2013 d\n\n[Reveal] Spoiler:\nCan we add (1) and (2) to get the answers.\n\nForget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.\n\nSince we have 4 variables (a, b, c and d) and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.\n\nConditions 1) & 2):\n\nb > d\nab^2 - b > b^2c - d\n\u21d4 ab^2 -b^2c > b - d\n\u21d4 b^2(a-c) > b-d\n\u21d4 a - c > (b-d)/b^2 > 0 since b > d and b^2 > 0 if b\u22600.\n\u21d4 a > c\n\nIf b = 0, we have d < 0\nab^2 - b > b^2c - d\n\u21d4 0 > -d which contradicts b > d\nThus b\u22600.\n\nBoth conditions together are sufficient.\n\nSince this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.\n\nCondition 1):\nWe don't have any information about a and c from the condition 1) only.\nThe condition 1) only is not sufficient.\n\nCondition 2):\n\u21d4 ab^2 -b^2c > b - d\n\u21d4 b^2(a-c) > b-d\n\u21d4 a - c > (b-d)/b^2 since b^2 > 0\n\u21d4 a - c > b - d\na = 2, c = 1, b = 1, d = 1 : Yes\na = 1, c = 2, b = 0, d = 1 : No\nThe condition 2) only is not sufficient.\n\nTherefore, C is the answer.\n\nIn cases where 3 or more additional equations are required, such as for original conditions with \u201c3 variables\u201d, or \u201c4 variables and 1 equation\u201d, or \u201c5 variables and 2 equations\u201d, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.\n_________________\n\nMathRevolution: Finish GMAT Quant Section with 10 minutes to spare\nThe one-and-only World\u2019s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.\nFind a 10% off coupon code for GMAT Club members.\n\u201cReceive 5 Math Questions & Solutions Daily\u201d\nUnlimited Access to over 120 free video lessons - try it yourself\nSee our Youtube demo\n\nRe: Is a > c? \u00a0 [#permalink] 23 Jan 2018, 22:47\nDisplay posts from previous: Sort by\n\n# Is a > c?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-02-18 07:11:03", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/is-a-c-137240.html", "openwebmath_score": 0.8273312449455261, "openwebmath_perplexity": 3311.3609703447137, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534344029238, "lm_q2_score": 0.8856314662716159, "lm_q1q2_score": 0.8692060441875746}}
{"url": "https://www.physicsforums.com/threads/line-integral-over-vector-field-of-a-shifted-ellipse.872348/", "text": "# Line integral over vector field of a shifted ellipse\n\nTags:\n1. May 18, 2016\n\n### TheBoro76\n\nThis is part of a larger question, but this is the part I am having difficulty with. I have had an attempt, but am not sure where I am making a mistake. Any help would be very, very appreciated.\n1. The problem statement, all variables and given/known data\n\nLet C2 be the part of an ellipse with centre at (4,0), horizontal semi-axis a=5, and vertical semi-axis b=3, from (0,-9/5) to (0,9/5) (i.e. anti-clockwise)\n\ncalculate:\n\n$$\\int_{C2} \\mathbf v \\cdot d\\mathbf r$$\nwhere $\\mathbf v = \\frac{1}{2}(-y\\mathbf i + x \\mathbf j)$\n\nHint: use t:-t0 ->t0 as limits when parametrising #C_2# and explain why cos(t0)=-4/5 and sin(t0)=3/5\n2. Relevant equations\nFor the integral:\n$$I=\\int_{C2} \\mathbf v \\cdot d\\mathbf r = \\int_a^b v_1dx+v_2dy$$\n\n3. The attempt at a solution\nI first parametrised the ellipse, by considering the ellipse as a stretched and shifted unit circle. Using this, I get\nx=5cos(t)+4\ny=3sin(t)\n\nwhen substituting in t0 from the question, I get x=0 and y=9/5, as expected. I also get the expected values of x=9 and y=0 when t=0.\n\ndifferentiating these I get:\ndx=-5sin(t) dt\ndy=3cos(t) dt\n\nand from $\\mathbf v$ I get:\n$v_1=-y$ and $v_2=x$\n\nso substituting these values into the equation for the integral, I get:\n$$I=\\frac{1}{2}\\int_{-t_0}^{t_0}-ydx+xdy$$\n$$\\rightarrow I=\\frac{1}{2}\\int_{-t_0}^{t_0}-3sin(t)*-5sin(t)+(5cos(t)+4)(3cos(t)dt$$\n$$I=\\frac{1}{2} \\int_{-t_0}^{t_0}15(sin^2(t)+cos^2(t))+12cos(t) dt=\\frac{1}{2}\\int_{-t_0}^{t_0}15+12cos(t) dt$$\n$$=\\frac{1}{2}(15t+12sin(t))|_{-t_0}^{t_0}=\\frac{1}{2}(30(t_0)+24sin(t_0))$$\nUsing the values of t0, from the question, I get:\n$$=15*sin^{-1}(\\frac{3}{5})+\\frac{36}{5} \\approx 16.85$$\n\nBut, using Greens theorem, this is an expression for the area of this part of the ellipse (when considering the closed integral of C=C1 and C2 and C1 is y=0). I know the area of an ellipse is $A=\\pi ab$. For the ellipse in the question, the area would be about $15\\pi\\approx47.12$. Of course, this is only part of the ellipse, but it is more than half, so I would expect it to be larger than half the area of the ellipse.\n\nAdditionally, I checked the area under the cartesian equation for C2, given by:\n$y=\\frac{3}{5}\\sqrt{-x^2+8x+9}$. The area would be twice the area under this curve, so it would be:\n$$A=2\\frac{3}{5}\\int_0^9\\sqrt{-x^2+8x+9}dx\\approx44.67$$, which is closer to what I would expect.\n\nI don't know what the actual answer is, but I'm pretty sure I'm wrong. Any advice or hints would be much appreciated\n\nThanks in advance!!\n\n2. May 18, 2016\n\n### andrewkirk\n\nPerhaps it's to do with how you use Green's theorem.\nTo use it to calculate area don't you need $\\frac{\\partial Q}{\\partial x}-\\frac{\\partial P}{\\partial y}=1$? In this case, with $P(x,y)=-y$ and $Q(x,y)=+x$, what is $\\frac{\\partial Q}{\\partial x}-\\frac{\\partial P}{\\partial y}$?\n\n3. May 18, 2016\n\n### TheBoro76\n\nHi thanks for the reply andrewkirk. My understanding is that you do need $\\frac{\\partial Q}{\\partial x} - \\frac{\\partial P}{\\partial y}=1$. But because there is a factor of 1/2 in the $\\mathbf v$, P(x,y)=-y/2 and Q(x,y)=x/2 so $\\frac{\\partial Q}{\\partial x}-\\frac{\\partial P}{\\partial y}=\\frac{1}{2}-\\,-\\frac{1}{2}=1$ ?\n\n4. May 18, 2016\n\n### SteamKing\n\nStaff Emeritus\nI think if you make a quick sketch of the ellipse described by C2 and plot the end points on that sketch, you will find that the first point is in the third quadrant w.r.t. the center of the ellipse and the second is in the second quadrant. Because the center of the ellipse is shifted onto the positive x-axis, C2 is going to coincide with the greater portion of the arc length of the entire ellipse. Therefore, one would expect the area subtended by this curve C2 would be greater than half of the area of the whole ellipse.\n\nSince the problem hints at making the limits of C2 go from -t0 to t0, then just blindly plugging in -t0 and t0 will not result in the correct evaluation of the line integral, since the angles used in the parameterization of the ellipse are supposed to run CCW. If you use symmetry in evaluating the line integral, I believe you will obtain the correct result, but because C2 crosses the x-axis when going from -t0 to t0, I think you must handle the line integral evaluation very carefully so that you don't get an incorrect result, which you apparently have done.\n\n5. May 18, 2016\n\n### andrewkirk\n\nThe problem might be that you are choosing 0.6435 as arcsin of 3/5. The angle t0 needs to be between pi/2 and pi, because of the quadrants the start and end points are in (as SK points out), so one should choose the arcsin instead as pi-0.6435. If you use that instead, I think you'll get the right result.\n\nLast edited: May 19, 2016\n6. May 19, 2016\n\n### TheBoro76\n\nThanks for your responses I tried both of your suggestions. but couldn't quite get them to work (I probably made a mistake somewhere). However, if I used the value t0=arccos(-4/5), I got the answer I expected, So I'm going with that.\nThanks again!!\n\nKnow someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook\n\nHave something to add?\nDraft saved Draft deleted", "date": "2018-02-21 20:03:11", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/line-integral-over-vector-field-of-a-shifted-ellipse.872348/", "openwebmath_score": 0.8534839749336243, "openwebmath_perplexity": 421.5592812991238, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105314577313, "lm_q2_score": 0.8902942363098472, "lm_q1q2_score": 0.8692036390054219}}
{"url": "https://math.stackexchange.com/questions/1493945/matrix-group-isomorphic-to-mathbb-z", "text": "# Matrix group isomorphic to $\\mathbb Z$.\n\nThe set $G=\\left\\{\\begin{pmatrix}1 & n \\\\ 0 & 1 \\\\ \\end{pmatrix}\\mid n\\in \\Bbb Z\\right\\}$ with the operation of matrix multiplication is a group. Show that $$\\phi:\\Bbb Z \\to G,$$ $$\\phi(n)=\\begin{pmatrix}1 & n \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ is a group isomorphism (where the operation on $\\Bbb Z$ is ordinary addtion).\n\nTO show it's isomorphism: I know I must show one-to-one, onto and homomorphism. I've done these examples before but never with matrices.\n\nHow can I show if $\\phi(a)=\\phi(b)$ then $a=b$? Same question for onto and operation preserving with matrices.\n\nThank you!\n\n\u2022 Try it. What happens when you multiply $\\phi(a)$ and $\\phi(b)$? \u2013\u00a0hardmath Oct 23 '15 at 14:15\n\u2022 Homomorphism, check. Generator? \u2013\u00a0user190080 Oct 23 '15 at 14:27\n\nHint:\n\n$\\begin{pmatrix}1 & n \\\\ 0 & 1 \\\\ \\end{pmatrix}\\begin{pmatrix}1 & m \\\\ 0 & 1 \\\\ \\end{pmatrix} = \\begin{pmatrix}1 & n+m \\\\ 0 & 1 \\\\ \\end{pmatrix}$\n\nThat should help with proving $\\phi$ is a homomorphism.\n\nOnce you have proven that a homomorphism exists, you must prove it is bijective to prove the mapping is a isomorphism. You already have that the mapping in injective, you must prove it is surjective.\n\n\u2022 This is what I used. Thank you! \u2013\u00a0maidel b Oct 23 '15 at 16:44\n\u2022 Glad it helped. \u2013\u00a0kleineg Oct 23 '15 at 17:31\n\u2022 If $\\phi$ is a homomorphism from $\\Bbb Z \\to G$, shouldn't $\\phi^{-1}$ take matrices as input? \u2013\u00a0pjs36 Mar 16 '17 at 19:43\n\u2022 OK, yes, that's better now \u2013\u00a0pjs36 Mar 16 '17 at 20:09\n\u2022 This is not the inverse. As in function inverse. This is not even a different function from $\\phi$. What the hell? In what world the inverse of a mapping $X\\to Y$ is a mapping $X\\to Y$? How changing $n$ to $-n$ proves that something is bijective? Come on. The proper inverse is defined by $\\begin{bmatrix}1 & n \\\\ 0 & 1\\end{bmatrix}\\mapsto n$. I don't believe there are 2 answers with the same mistake. \u2013\u00a0freakish Mar 16 '17 at 20:11\n\nHint: For your first question, write down $\\phi(a)$ and $\\phi(b)$ (go ahead, write down the matrices on a sheet of paper). Now, if those two are equal, what does it tell you?\n\nThe other parts are obviously different, but the idea is the same: just look at the matrices involved and use what you know about matrix multiplication (for the last part).\n\n\u2022 Thank you. I understand one-to-one after writing out the steps. I'm struggling with onto. Let me make sure I'm thinking correctly. I'm trying to find $\\phi(?)=n$. Is there a different way? I'm not able to see what I'm supposed to do. \u2013\u00a0maidel b Oct 23 '15 at 17:50\n\u2022 @ShayAbbott No. You want to take an arbitrary $g\\in G$ and find an $n\\in Z$ so that $\\phi(n)=g$. But if $g\\in G$, then you know what $g$ looks like... \u2013\u00a0Teepeemm Oct 23 '15 at 18:09\n\nSo let me straight things up, since there are multiple answer which get this incorrectly. First of all you need to show that $\\phi$ is a homomorphism (for that see other answers, they are fine). Then you can show that it is an isomorphism by constructing the inverse. And by the inverse we mean the function inverse, i.e. a function\n\n$$g:G\\to\\mathbb{Z}$$\n\nsuch that $\\phi\\circ g=\\mbox{id}_{G}$ and $g\\circ \\phi=\\mbox{id}_{\\mathbb{Z}}$. By general property if such inverse exists then it is a homomorphism as well and so $\\phi$ is an isomorphism.\n\nSo you can easily check that\n\n$$g\\bigg(\\begin{bmatrix} 1 & n\\\\ 0 & 1 \\end{bmatrix}\\bigg)=n$$\n\nis the inverse of $\\phi$.\n\nJust note that $$\\begin{pmatrix}1 & 1 \\\\ 0 & 1 \\\\ \\end{pmatrix}^n = \\begin{pmatrix}1 & n \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ This implies that $\\phi$ is a surjective homomorphism because exponents add when you multiply powers. It is clear that $\\phi$ is injective because $\\phi(n)_{12}=n$.\n\nHint:\n\nThere are other ways to show that this map is one-to-one.\n\nUse that the $ker(\\phi)=0$.\n\nDoes there exist an $n$ such that the rank of $\\phi(n)=\\begin{pmatrix}1 & n \\\\ 0 & 1 \\\\ \\end{pmatrix}$ is not $2$?\n\nAlternatively, construct $\\phi^{-1}(n)=\\begin{pmatrix}1 & -n \\\\ 0 & 1 \\\\ \\end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.\n\n\u2022 That's not the inverse. The proper inverse is a function on matrices. It maps $\\begin{bmatrix}1 & n \\\\ 0 & 1\\end{bmatrix}\\mapsto n$. Also if $\\phi$ is a homomorphism then so is $\\phi^{-1}$. So the existance of $\\phi^{-1}$ is enough, no need to check that it is a homomorphism. \u2013\u00a0freakish Mar 16 '17 at 20:07", "date": "2019-07-21 02:38:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1493945/matrix-group-isomorphic-to-mathbb-z", "openwebmath_score": 0.9290165901184082, "openwebmath_perplexity": 211.27958985589848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.976310532836284, "lm_q2_score": 0.8902942188450159, "lm_q1q2_score": 0.8692036231816407}}
{"url": "https://math.stackexchange.com/questions/2150704/how-to-know-if-there-is-a-extraneous-solution-in-a-radical-expression", "text": "How to know if there is a extraneous solution in a radical expression\n\nI was trying to solve this problem:\n\n$$\\sqrt{3x+13} = x+ 3$$\n\n$$(\\sqrt{3x+13})^2 = (x+ 3)^2$$ $$(3x+13) = (x+ 3)^2$$ $$3x+13 = x^2 + 6x + 9$$ $$0 = x^2 + 3x - 4$$ $$0 = (x+4)(x-1)$$\n\nSo my final answer was $x = -4$ and $x = 1$. However, it was incorrect because when I plug back in -4 into the original equation I get a extraneous solution. My question is do I always need to plug back in my answers into a radical expression and check if they are valid? Or is there any other way to deduce that there will be a extraneous solution?\n\n\u2022 In this case you can tell there is only one solution simply by thinking about the graphs of the functions on the left and right sides of your original equation. This tells you one of your roots is extraneous. It is also clear graphically that that solution is positive. So the extraneous root must be $-4$. Alternatively, recall that the range of the square root function is the set of nonnegative reals. So your original equation implies $x\\geq-3$. \u2013\u00a0symplectomorphic Feb 18 '17 at 22:58\n\u2022 @Khosrotash I am trying to avoid checking the final answers because it is a timed test and I need to move really quickly \u2013\u00a0Pablo Feb 18 '17 at 22:59\n\nYou don't need to plug in values, if you always ensure not to add extraneous solutions.\n\nThe equation forces two conditions, namely $3x+13\\ge0$ and $x+3\\ge0$, which together become $x\\ge-3$.\n\nWhy $x+3\\ge0$? Because $\\sqrt{3x+13}\\ge0$ by definition (when it exists, of course).\n\nWith this condition, you can safely square, because you have an equality between nonnegative numbers. You get (your computations are good) $$\\begin{cases} (x+4)(x-1)=0 \\\\[4px] x\\ge-3 \\end{cases}$$ and therefore you know what roots are a solution of the original equation, in this case only $x=1$.\n\nOn its domain ($A\\ge 0$), note that $$\\sqrt A=B\\iff A=B^2\\enspace\\textbf{and}\\enspace B\\ge 0,$$ since the symbol$\\sqrt{\\phantom{h}}$ denotes the non-negative square root of a non-negative real number.\n\nNote that when you squared both sides in the first step, you ended up with the equation\n\n$3x+13 = (x+3)^2$,\n\nwhich could just as well come from the same original equation but with a negative square root instead:\n\n$-\\sqrt{3x+13} = x+3$.\n\nSo, at the end of the day, you've solved both equations. To check which solution(s) correspond to which equation(s), you need to plug back in and check.\n\n\u2022 Amm so does that mean that every time I square a radical then I would get something like this $$\u00b1(3x+13)^2$$ \u2013\u00a0Pablo Feb 18 '17 at 23:01\n\u2022 No, because the square of a number is always positive. It just means that when you square both sides to try and solve the original equation, you're actually solving two equations at once. \u2013\u00a0J Richey Feb 18 '17 at 23:02\n\nI think the best or reliable solving is to plot the equations . in your case like below", "date": "2020-01-21 14:35:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2150704/how-to-know-if-there-is-a-extraneous-solution-in-a-radical-expression", "openwebmath_score": 0.8822001218795776, "openwebmath_perplexity": 180.25360316322423, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9763105259435196, "lm_q2_score": 0.8902942173896131, "lm_q1q2_score": 0.8692036156241274}}
{"url": "https://www.physicsforums.com/threads/minkowski-metric-in-spherical-polar-coordinates.860488/", "text": "# Minkowski metric in spherical polar coordinates\n\n1. Mar 4, 2016\n\n### spaghetti3451\n\n1. The problem statement, all variables and given/known data\n\nConsider Minkowski space in the usual Cartesian coordinates $x^{\\mu}=(t,x,y,z)$. The line element is\n\n$ds^{2}=\\eta_{\\mu\\nu}dx^{\\mu}dx^{\\nu}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}$\n\nin these coordinates. Consider a new coordinate system $x^{\\mu'}$ which differs from these Cartesian coordinates. The Cartesian coordinates $x^{\\mu}$ can be written as a function of these new coordinates $x^{\\mu}=x^{\\mu}(x^{\\mu'})$.\n\n(a) Take a point $x^{\\mu'}$ in this new coordinate system, and imagine displacing it by an infinitesimal amount to $x^{\\mu'}+dx^{\\mu'}$. We want to understand how the $x^{\\mu}$ coordinates change to first order in this displacement $dx^{\\mu'}$. Argue that\n\n$dx^{\\mu}=\\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}dx^{\\mu'}$.\n\n(Hint: Taylor expand $x^{\\mu}(x^{\\mu'}+dx^{\\mu'})$.)\n\n(b) The sixteen quantities $\\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}$ are referred to as the Jacobian matrix; we will require this matrix to be invertible. Show that the inverse of this matrix is $\\frac{\\partial x^{\\mu'}}{\\partial x^{\\mu}}$. (Hint: Use the chain rule.)\n\n(c) Consider spherical coordinates, $x^{\\mu'}=(t,r,\\theta,\\phi)$ which are related to the Cartesian\ncoordinates by\n\n$(t,x,y,z)=(t,r\\ \\text{sin}\\ \\theta\\ \\text{cos}\\ \\phi,r\\ \\text{sin}\\ \\theta\\ \\text{sin}\\ \\phi,r\\ \\text{cos}\\ \\theta)$.\n\nCompute the matrix $\\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}$. Is this matrix invertible everywhere? Compute the displacements $dx^{\\mu}$ in this coordinate system (i.e. write them as functions of $x^{\\mu'}$ and the infinitesimal displacements $dx^{\\mu'}$).\n\n(d) Compute the line element $ds^{2}$ in this coordinate system.\n\n2. Relevant equations\n\n3. The attempt at a solution\n\n(a) By Taylor expansion,\n\n$x^{\\mu}(x^{\\mu'}+dx^{\\mu'}) = x^{\\mu}(x^{\\mu'}) + \\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}dx^{\\mu'}$\n\n$x^{\\mu}(x^{\\mu'}+dx^{\\mu'}) - x^{\\mu}(x^{\\mu'}) = \\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}dx^{\\mu'}$\n\n$dx^{\\mu}=\\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}dx^{\\mu'}$\n\nAm I correct so far?\n\nLast edited: Mar 4, 2016\n2. Mar 4, 2016\n\n### stevendaryl\n\nStaff Emeritus\nYes, but you should go ahead and calculate the nine components:\n\n$\\frac{\\partial x}{\\partial r}$, $\\frac{\\partial y}{\\partial r}$, $\\frac{\\partial z}{\\partial r}$\n$\\frac{\\partial x}{\\partial \\theta}$, $\\frac{\\partial y}{\\partial \\theta}$, $\\frac{\\partial z}{\\partial \\theta}$\n$\\frac{\\partial x}{\\partial \\phi}$, $\\frac{\\partial y}{\\partial \\phi}$, $\\frac{\\partial z}{\\partial \\phi}$\n\n3. Mar 4, 2016\n\n### spaghetti3451\n\nIsn't that in part (c)?\n\nShouldn't I do (b) first?\n\n4. Mar 4, 2016\n\n### stevendaryl\n\nStaff Emeritus\nYeah, I guess you should, even though part c doesn't actually depend on part b.\n\n5. Mar 4, 2016\n\n### spaghetti3451\n\n(b) Via the chain rule,\n\n$\\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}\\frac{\\partial x^{\\mu'}}{\\partial x^{\\nu}}=\\delta_{\\nu}^{\\mu}$,\n\nwhere we are using the summation convention only over $\\mu'$.\n\nTherefore, the inverse of the matrix $\\frac{\\partial x^{\\mu}}{\\partial x^{\\mu'}}$ is the matrix $\\frac{\\partial x^{\\mu'}}{\\partial x^{\\nu}}$.\n\nIs this correct?\n\n6. Mar 4, 2016\n\n### stevendaryl\n\nStaff Emeritus\nYes.\n\n7. Mar 4, 2016\n\n### Staff: Mentor\n\nWhat's wrong with taking the differentials of x, y, and z (expressed in terms of the spherical coordinates in post #1), evaluating their differentials (in terms of the spherical coordinates and their differentials), and then taking the sum of their squares? This should give the Minkowski metric in spherical coordinates, correct?\n\nChet\n\n8. Mar 5, 2016\n\n### spaghetti3451\n\nI know that this is a correct and shorter approach, but I'm trying to follow the instructions of the question.", "date": "2017-10-18 22:25:36", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/minkowski-metric-in-spherical-polar-coordinates.860488/", "openwebmath_score": 0.9283000230789185, "openwebmath_perplexity": 791.9567801248128, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682451330956, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.869184490282806}}
{"url": "https://math.stackexchange.com/questions/1025578/is-there-a-better-way-of-finding-the-order-of-a-number-modulo-n", "text": "# Is there a better way of finding the order of a number modulo $n$?\n\nSay I wish to find the order of 2 modulo 41.\n\nThe way I've been shown to compute this is to literally write out $2^k$ and keep going upwards with $0 \\leq k \\leq 41$, or until I observe periodicity in the sequence, and then state the order from there.\n\nIs there a better way of deducing the order of a number with respect to some modulo?\n\nThis seems highly inefficient, especially if we are working with respect to some large modulo $N$ where it will take at most $N$ computations to determine when periodicity occurs.\n\n\u2022 Since $41$ is prime, the order must be a factor of $40$ Nov 17, 2014 at 6:49\n\u2022 I did not know this, but thanks for that. Very useful. Perhaps a bad example, what if $n$ is NOT prime? Are there any results that can further 'filter' it out? Also, say we know the order must be a factor of 40, then is there a systematic way of computing the order without having to bash out any numbers? Nov 17, 2014 at 6:51\n\nFor any $$a$$ and $$N$$ with $$\\gcd(a,N)=1$$, the order of $$a$$ modulo $$N$$ must be a divisor of $$\\varphi(N)$$. So if you know the prime factorization of $$N$$ (or $$N$$ is already prime) so that you can compute $$\\varphi(N)$$ and also know the prime factorization of $$\\varphi(N)$$, you can proceed as follows:\n\nIf we know an integer $$m>1$$ with $$a^m\\equiv 1\\pmod N$$ and know the prime divisors of $$m$$, for all primes $$p$$ dividing $$m$$ do the following: Compute $$a^{m/p}\\bmod N$$ and if the result is $$\\equiv 1\\pmod N$$, replace $$m$$ with $$m/p$$ and repeat (or switch to the next prime divisor if $$m$$ is no longer divisible by $$p$$). When you have casted out all possible factors, the remaining $$m$$ is the order of $$a$$. Note that the computations $$a^m\\bmod N$$ do not require $$m$$ multiplications, but rather only $$O(\\log m)$$ multiplications mod $$N$$ if we use repeated squaring.\n\nIf $$N$$ is large and the fatorization of $$\\varphi(N)$$ is known (and especially if you suspect the order of $$a$$ to be big), this is in fact a fast method.\n\nNote that a couple of computations can be saved even beyon what is descibed above: In the case $$p=2$$, we may end up computing $$a^m, a^{m/2}, a^{m/4},\\ldots$$ to cast out factors of $$2$$. But the later numbers were in fact intermediate results of computing $$a^m$$ by repeated squaring! Also, once we notice for some $$p$$ with $$p^k\\mid m$$ that $$a^{m/p}\\not\\equiv 1\\pmod N$$, we can save a few squarings and so speed up the task for the remaining primes if we replace $$a$$ with $$a^{p^k}\\pmod N$$ and $$m$$ with $$m/p^k$$ - just remember to multiply the factor $$p^k$$ back into the final answer!\n\nIn your specific example, we know that $$N=41$$ is prime and that $$\\varphi(N)=40=2^3\\cdot 5$$. We check $$p=5$$ and note that $$2^{40/5}=256\\equiv 10\\pmod{41}$$, hence the factor $$5$$ cannot be eliminated. After that we check how many $$2$$'s we have to use: $$2^5\\equiv 32\\equiv -9$$, hence $$2^{10}\\equiv 81\\equiv -1$$, $$2^{20}\\equiv (-1)^2\\equiv 1$$. We conclude that $$2$$ has order $$20$$ modulo $$41$$.\n\nThe order must be one of $1,2,4,8,5,10,20,40$.\n\nCalculate $2^2,2^4=(2^2)^2,2^8=(2^4)^2,...\\\\2^5=2^4*2,2^{10}=(2^5)^2,2^{20}=(2^{10})^2,2^{40}=(2^{20})^2$ which is seven calculations.\n\nIf 41 is not prime, say $41=a^2b^3c$,then\n\ni) calculate the order of $2\\pmod{a^2}$, and the order $\\pmod{b^3}$ and the order $\\pmod c$\nii) calculate the lowest common multiple of the separate orders.\n\nHere we calculate the order of $$2$$ in $$(\\mathbb{Z}/{41}\\mathbb{Z})^\\times$$.\n\nNote: When calculating you can use the previous work to your advantage. To really cut down on the calculations, if you know the order is greater than or equal $$10$$, once you calculate $$2^{10} \\pmod{41}$$ you are done,\n\n$$\\quad 2^{10} \\equiv 1 \\pmod{41} \\quad \\text{order is } 10$$\n$$\\quad 2^{10} \\equiv 40 \\pmod{41} \\quad \\text{order is } 20$$\n$$\\quad \\text{NOT }[2^{10} \\equiv 1, 40 \\pmod{41}] \\quad \\text{order is } 40$$\n\nWork Summary : $$2^1 \\equiv 2 \\pmod{41}$$.\n\nThe order of $$2$$ is one of $$2,4,8,5,10,20,40$$.\n\nWork Summary : $$2^2 \\equiv 4 \\pmod{41}$$.\n\nThe order of $$2$$ is one of $$4,8,5,10,20,40$$.\n\nWork Summary : $$2^4 \\equiv 16 \\pmod{41}$$.\n\nThe order of $$2$$ is one of $$8,5,10,20,40$$.\n\nWork Summary : $$2^5 = \\equiv 32 \\pmod{41}$$.\n\nThe order of $$2$$ is one of $$8,10,20,40$$.\n\nWork Summary : $$2^8 \\equiv 10 \\pmod{41}$$\n\nThe order of $$2$$ is one of $$10,20,40$$.\n\nWork Summary : $$2^{10} \\equiv 40 \\pmod{41}$$\n\nThe order of $$2$$ is one of $$20,40$$.\n\nWork Summary : $$2^{20} \\equiv 1 \\pmod{41}$$\n\nThe order of $$2$$ is equal to $$20$$.", "date": "2022-05-24 03:21:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1025578/is-there-a-better-way-of-finding-the-order-of-a-number-modulo-n", "openwebmath_score": 0.9606664776802063, "openwebmath_perplexity": 137.7909001028024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682454669814, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8691844874457927}}
{"url": "https://magoosh.com/gre/quantitative-comparison-the-relationship-cannot-be-determined-from-the-information-given-answer-choice/", "text": "Quantitative Comparison: \u201cThe Relationship Cannot Be Determined from the Information Given\u201d Answer Choice\n\n1. The quantity in Column A is greater\n2. The quantity in Column B is greater\n3. The two quantities are equal\n4. The relationship cannot be determined from the information given\n\nMany people dread choosing answer choice (D) on Quantitative Comparison (QC) Some feel it may be conceding defeat. Others think that the GRE is trying to trick them by making them pick (D). After all, they think, there must be some pattern that I\u2019m not getting.\n\nThe truth is answer (D) comes up often. And to determine whether an answer cannot be determined is actually not too difficult.\n\n#1 Determine a relationship\n\nSay you find an instance, in which the answer is (A) the information in column A is greater. If that is the case, then the next step is to disprove that.\n\n#2 Disprove that relationship\n\nMeaning, see if you can come up with an instance, either through plugging in different variables, manipulating algebra, or manipulating a geometric figure, in which the answer is not (A). As soon you do that, you can stop. The answer is (D).\n\nIf you can\u2019t disprove your answer, then it must be correct: it must be (A), (B) or (C).\n\n1. $$-100 < x < 0$$\n\nColumn A Column B\n$$x^{-4}$$ $$x^{-3}$$\n\n1. The quantity in Column A is greater\n2. The quantity in Column B is greater\n3. The two quantities are equal\n4. The relationship cannot be determined from the information given\n\nExplanations:\n\nQuestion #1\n\nAfter choosing a few numbers you should note something: that Column A will always be greater than Column B.\n\nWhy? Will, whenever, you have an even exponent, positive or negative, that exponent will always yield a positive number.\n\nOdd exponents, on the other hand, give you a negative output if the base (the number below the exponent) is negative. Remember x has to be negative. So no matter what number you plug in Column A will always be positive, Column B negative. This is a definite not (D). The answer is (A).\n\n32 Responses to Quantitative Comparison: \u201cThe Relationship Cannot Be Determined from the Information Given\u201d Answer Choice\n\n1. lokesh nandni sood May 22, 2019 at 9:59 pm #\n\nhi chrish i put all the different values if i put 1 , they both are equal , if i put -1 B is greater . same way if i put 2 value is 0.0625 A is smaller and 0.125 B is greater . didn\u2019t get that . pls help\n\n\u2022 Magoosh Test Prep Expert May 24, 2019 at 8:36 am #\n\nHi Lokesh,\n\nRemember that we have an important limitation to which numbers we can choose. The question tells us that -100 < x < 0. This means that x must be a negative number. You are plugging in positive numbers, which will give you a different answer \ud83d\ude42\n\n2. Karan June 28, 2017 at 4:54 am #\n\nHi Chris,\nI have a doubt related to quantitative comparison type questions. What option do we choose if both columns turn out to be infinity. For eg:\nColumn A Column B\nThe number of numbers from 1 to 2 The number of numbers from 1 to 5\n\nD or C?\n\n\u2022 Magoosh Test Prep Expert June 28, 2017 at 11:43 am #\n\nHi Karan,\n\nGreat and interesting question! Honestly, you will not see a question like this on the GRE, in which you are comparing two quantities, each with the solution of \u201cinfinite.\u201d That being said, the comparison of infinite to infinite is equal, so I\u2019d go with (C). But again, you will not see something like this. Still, it\u2019s a very interesting and awesome question! Have a great day! \ud83d\ude42\n\n3. Prasanth January 5, 2017 at 11:43 pm #\n\nHi Chris,\n\nIn the above example\n-100<x<0\n\nQty A : x^-4\nQty B : x^-3\n\nNow multiply both sides by a positive number relation will not change. x^4 is a positive number.So\n\nQty A becomes : 1\nQty B becomes : x.\n\nNow it is easy to plug -ve numbers for x.\n\nIn all definite ways, column A will be greater.\n\nCan i approach this way or plugging numbers will be easy in test ?\n\nI usually make mistakes while plugging -ve numbers .. Pls help\n\n\u2022 Magoosh Test Prep Expert January 6, 2017 at 1:32 pm #\n\nHi Prasanth,\n\nNo, I do not recommend taking this approach. For example, imagine if you manipulated the problem by multiplying by x^5. So,\n\nQty A becomes: x\nQty B becomes: x^2\n\nNow, in this case, Qty B is greater than Qty A, which is not true. It is best to approach this problem rephrasing this question.\n\nQty A: x^-4 = 1/(x^4)\nQty B: x^-3 = 1/(x^3)\n\nHere you know that any negative number raised by an even exponent will be positive, and that same number raised by an odd exponent will be negative. This is how we know that column A is greater. I hope that helps! \ud83d\ude00\n\n\u2022 fatima January 12, 2018 at 7:44 am #\n\nHi Chris, regarding this problem :\n\n-100<x<0\nQty A : x^-4\nQty B : x^-3\n\nPrasanth multiplied both sides by a positive number (x^4 is a positive number),which will be at all times positive since (-100<x<0), and it is legal to multiply both sides of an inequality by a positive number..\nBut it wil not be true to multiply both sides by x^5 since (-100<x<0),so x^5 will give a negative number. And it will change the inequality..\ni think Prasanth's approach is right. Does it ?\nthanks,,\n\n\u2022 Magoosh Test Prep Expert January 18, 2018 at 12:40 pm #\n\nHi Fatima,\n\nPrasanth\u2019 approach is correct, in the sense that it really does work for this particular problem. Here, multiplying Quantity A and Quantity B by x to an even power will help simplify things and may help make them clearer.\n\nThe reason I\u2019d still recommend against Prasanth\u2019s approach is that it adds an extra layer of complexity that isn\u2019t necessary. Under real testing conditions, every second counts, and simply recognizing the number properties at play here, without plugging in new numbers, is important. In addition under the pressure of test conditions, it becomes easier to make a mistake like choosing x^5 instead of x^4 without noticing that the inequality is reversed.\n\n4. Yogesh kurade December 13, 2016 at 9:56 pm #\n\nThank you! Sir\n\n5. Priyam June 13, 2016 at 9:11 am #\n\nIn lesson intro to Quantitative comparison, there\u2019s this example question\nN is not an integer\n6<N<10\nQuantity A Quantity B\nN 8\n\nHere, the answer is mentioned as D) cannot be determined reason stated as it can be a decimal or a fraction, but integer itself means \"Whole nos which can be -ve as well as +ve\".\n\n\u2022 Magoosh Test Prep Expert June 18, 2016 at 12:23 pm #\n\nHi Priyam \ud83d\ude42\n\nThanks for your message \ud83d\ude42 In that example, we\u2019re told that 6 < N < 10. So, we know that N is a positive number between 6 and 10. However, we cannot determine whether N is greater than 8 using only the information given, that N is not an odd integer. N could be any positive number between 6 and 10 except 7, which is the only odd integer within this range. For example, N could equal 6.5 or 8.3. Neither of these numbers is an odd integer and therefore fits the description. Because N may be less than, equal to, or greater than 8, the answer to that example is (D).\n\nI hope this clears up your doubts \ud83d\ude42\n\n\u2022 Bhavna Sharma June 26, 2016 at 3:14 am #\n\nI have a doubt regarding this. In the above explanation, you have mentioned that \u201cN could equal 6.5 or 8.3\u201d. But, in the question it has been specified that N is an integer and 6.5 or 8.3 are not integers.\n\n\u2022 Magoosh Test Prep Expert June 26, 2016 at 4:28 am #\n\nHi Bhavna \ud83d\ude42\n\nIn that example, the prompt states that \u201cN is not an odd integer.\u201d This is not the same as saying that N is an even integer. N could be an even integer, but N could also be a decimal or fraction, since decimals and fractions are not odd integers. I hope this clears things up \ud83d\ude42\n\n\u2022 Bhavna Sharma June 27, 2016 at 8:25 am #\n\nThank you so much.. \ud83d\ude42 I missed this point.\n\n6. Lars July 30, 2012 at 4:55 am #\n\nHi Chris,\n\nIn your explanation to the 1st question, in the 2nd last sentence you wrote\n\n\u201cSo no matter what number you plug in Column A will always be negative, Column B positive.\u201d Isn\u2019t this the exact opposite because Column A wiil be positive and column B will be negative?\n\nI got confused the first time I read this.\n\n7. Sammy May 30, 2012 at 3:54 pm #\n\nChris,\nFor these types of problems I have a trick that allows me to correctly deduce what the answer is. I wanted to get your thoughts on my method.\nAn example problem that I fabricated \u2013\n\nn is an integer\nWhich is greater?\nCOL A \u2013 n^3 + 6\nCOL B \u2013 n^2 + 5n -30\n\nI would attack this problem in this way \u2013 I would choose 5 arbitrary numbers (0,-0.6, 0.6,\n3,-3) and plug each in till i get an explicit answer. The only problem I see with my this is that it can get to be time consuming.\nLet me know what you think!\n\nPS \u2013 I love it when you answer in a witty way with your sophisticated words!\n\n\u2022 Chris June 1, 2012 at 12:58 pm #\n\nHi Sammy,\n\nYou put me in a little bit of a tough spot, because it\u2019s kind of difficult to be witty with Quantitative Comparison :).\n\nYour method definitely works well. Picking numbers can help you determine which column is bigger as long as you make sure to include 0, 1, -1, 2, -2 (or 3, -3) and 1/2 and -1/2 (I think 0.6 is a little unwieldy :)).\n\nBut in general this is a very effective method and one that I too employ. Often, esp. with algebraic equations, you may want to manipulate the equations and see if you can simply them.\n\nSorry, I didn\u2019t drop any sophisticated words, but I hope that helped :).\n\n\u2022 Sammy June 1, 2012 at 1:06 pm #\n\nThanks Chris, my only qualm would be that the method can be somewhat time consuming.\n\n\u2022 Chris June 1, 2012 at 1:20 pm #\n\nOh, that\u2019s right \u2013 you mentioned that\u2026if you have a strong facility with numbers, that is you can quickly plug-in, then it shouldn\u2019t be time consuming. Also you can come up with a tiny column grid:\n\nA B\n0\n1\n-1\n2\n-2\n\nBy the time you get the 2 you can usually know for certain what column it is.\n\nHope that helps :).\n\n\u2022 Amal June 1, 2012 at 1:43 pm #\n\nHi Chris,\n\nWould it be correct to say that if n is an integer, one need not test decimals that can be written as a fraction. For example, 0.6 = 6/10, which means that 0.6 is actually a fraction, aka an integer divided by another integer, or two integers. And the same thing for 0.5, since it equals 1/2. In fact, when it comes to \u201cn is an integer\u201d type questions, we should be able to exclude all decimals, since according to the \u201cofficial GRE guide,\u201d only real numbers are involved (e.g. we would not test 3.14 which is a decimal but can\u2019t be written as a fraction. This would mean that every decimal number is a real number (which implies it can be written as a fraction) for GRE purposes, as far as quantitative comparison type questions are concerned. Great question by the way.\n\n\u2022 Chris June 1, 2012 at 2:04 pm #\n\nYes, you are exactly right. I was speaking more broadly when giving that spread of numbers. One should always obey the constraints of the question, i.e. if x is a positive integer, you have to plug in accordingly.\n\nThanks for catching that :).\n\n\u2022 Sammy June 1, 2012 at 2:06 pm #\n\nYes you are right, it was an egregious error made on my part. I meant to say \u2018a real number\u2019\n\n\u2022 Chris June 4, 2012 at 2:44 pm #\n\nNo problem :). Glad I could help.\n\n\u2022 harsha July 7, 2016 at 1:46 am #\n\nhey chris i am your big fan of u can u give me links for practcicng gre full lenth tests of free costs\n\n\u2022 Magoosh Test Prep Expert July 7, 2016 at 10:37 am #\n\nHi Harsha \ud83d\ude42\n\nI\u2019d recommend that you first take the free PowerPrep tests from ETS, if you haven\u2019t already. Next, here\u2019s a free test from Manhattan (their materials are great!): Manhattan Free GRE Practice Test. Lastly, if you want access to more practice tests, I\u2019d recommend purchasing one of the Manhattan books\u2013each book comes with a code to access 6 online tests.\n\nHope this helps \ud83d\ude42\n\n8. Denis May 23, 2012 at 9:57 am #\n\nHi Chris,\n\nI have a question regarding a concept in Question #2: you state \u201cThree equal sides equal three equal angles.\u201d Is this a characteristic of all polynomials, i.e., can we say \u201cx equal sides equal x equal angles\u201d for any figure?\n\nBest regards,\nDenis\n\n\u2022 Chris May 23, 2012 at 11:40 am #\n\nHi Denis,\n\nThe measure of an angle corresponds to the length of the side opposite that angle, in a triangle. With quadrilaterals, or figures with more than four sides an angles, that relationship between which side corresponds to which angle is not as clear cut. However, we can say with certain that if there are x equal sides then there are x equal angles.\n\nHope that helps :).\n\n\u2022 Denis May 23, 2012 at 1:34 pm #\n\nHi Chris,\n\nI greatly appreciate the expeditious reply. Thanks for the help.\n\n9. Amal May 20, 2012 at 11:58 am #\n\nI\u2019d say the answer for question 1 is (D). If you pick -1, then you get 1 for both column A and column B, since you have 1 over 1 raised to some positive power. If you pick, say, -2, then column A and column B are obviously different.\n\n\u2022 Chris May 21, 2012 at 3:22 pm #\n\nHi Amal,\n\nIf you take any negative to an even exponent it will always yield a positive number. Therefore, if you plug in -1, you will get 1 for Col. A and -1 for Col B.\n\nHope that helps :).\n\n\u2022 Amal May 21, 2012 at 4:11 pm #\n\nYeah you\u2019re right. I forgot the negative sign on the coefficient. When you raise a negative coefficient to a negative odd power, the coefficient (which winds up in the denominator) is still negative. The coefficients are always negative, but the even power in column A yields a positive quantity.\n\n\u2022 Chris May 22, 2012 at 1:41 pm #\n\nNo problem \ud83d\ude42\n\nMagoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! \ud83d\ude04 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.\n\nWe highly encourage students to help each other out and respond to other students' comments if you can!\n\nIf you are a Premium Magoosh student and would like more personalized service from our instructors, you can use the Help tab on the Magoosh dashboard. Thanks!", "date": "2021-03-06 11:57:17", "meta": {"domain": "magoosh.com", "url": "https://magoosh.com/gre/quantitative-comparison-the-relationship-cannot-be-determined-from-the-information-given-answer-choice/", "openwebmath_score": 0.5819680094718933, "openwebmath_perplexity": 870.2008954426768, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777180191995, "lm_q2_score": 0.8807970842359876, "lm_q1q2_score": 0.8691504631037189}}
{"url": "https://www.e-medida.es/0tpwyir8/c5a768-inverse-of-upper-triangular-matrix-is-upper-triangular-proof", "text": "More- over, if the partition is in fact an all-square partition and A, B, and D are all invertible, then (3.2) We can assume that the matrix A is upper triangular and invertible, since $$\\displaystyle A^{-1}=\\frac{1}{det(A)}\\cdot adj(A)$$ We can prove that $$\\displaystyle A^{-1}$$ is upper triangular by showing that the adjoint is upper triangular or that the matrix of cofactors is lower triangular. Suppose the n \u00d7 n matrix R is upper triangular and invertible, i.e., its diagonal entries are all nonzero. Clearly, the inverse of a block upper triangular matrix is block upper triangular only in the square diagonal partition. (a) if U is upper triangular and invertible then U^-1 is upper triangular. It fails the test in Note 5, because ad bc equals 2 2 D 0. An LU factorization refers to the factorization of A, with proper row and/or column orderings or permutations, into two factors \u2013 a lower triangular matrix L and an upper triangular matrix U: =. Taking transposes leads immediately to: Corollary If the inverse L 1 of an lower triangular matrix L exists, Show that R1 is also upper triangular. It's obvious that upper triangular matrix is also a row echelon matrix. Proposition If a lower (upper) triangular matrix is invertible, then its inverse is lower (upper) triangular. Hint. Let A and B be upper triangular matrices of size nxn. (b) The inverse of a unit lower triangular matrix is unit lower triangular (c) The product of two upper or (two lower triangular) matrices is upper or (lower) triangular It goes like this: the triangular matrix is a square matrix where all elements below the main diagonal are zero. An example is the 4 4 matrix 4 5 10 1 0 7 1 1 0 0 2 0 0 0 0 9 . The inverse element of the matrix [begin{bmatrix} 1 & x & y \\ 0 &1 &z \\ 0 & 0 & 1 end{bmatrix}] is given by [begin{bmatrix} 1 & -x & xz-y \\ 0 & 1 & -z \\ 0 & 0 & 1 end{bmatrix}.] Inverse of matrix : A square matrix of order {eq}n \\times n{/eq} is known as an upper triangular matrix if all the elements below principle diagonal elements are zero. Let A be a square matrix. 11.7 Inverse of an upper triangular matri. Let $b_{ij}$ be the element in row i, column j of B. Let A be a n n upper triangular matrix with nonzero diagonal entries. For a proof, see the post The inverse matrix of an upper triangular matrix with variables. The proof for upper triangular matrices is similar (replace columns with rows). Use back substitution to solve Rsk-en for k 1, , n, and argue that (sk)i -0 for i > k. The inverse of a triangular matrix is triangular. It's actually called upper triangular matrix, but we will use it. In the next slide, we shall prove: Theorem If the inverse U 1 of an upper triangular matrix U exists, then it is upper triangular. In this problem, you will Example of upper triangular matrix: 1 0 2 5 0 3 1 3 0 0 4 2 0 0 0 3 A triangular matrix (upper or lower) is invertible if and only if no element on its principal diagonal is 0. Solving Linear Equations Note 6 A diagonal matrix has an inverse provided no diagonal entries are zero: If A D 2 6 4 d1 dn 3 7 5 then A 1 D 2 6 4 1=d1 1=dn 3 7 5: Example 1 The 2 by 2 matrix A D 12 12 is not invertible. In the lower triangular matrix all elements above the diagonal are zero, in the upper triangular matrix, all the elements below the diagonal are zero. In general this is not true for the square off-diagonal partition. Let $a_{ij}$ be the element in row i, column j of A. An upper triangular matrix is a square matrix in which the entries below the diagonal are all zero, that is, a ij = 0 whenever i > j. 82 Chapter 2.", "date": "2021-04-23 05:23:10", "meta": {"domain": "e-medida.es", "url": "https://www.e-medida.es/0tpwyir8/c5a768-inverse-of-upper-triangular-matrix-is-upper-triangular-proof", "openwebmath_score": 0.9157179594039917, "openwebmath_perplexity": 179.98260501509338, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771766922545, "lm_q2_score": 0.8807970654616711, "lm_q1q2_score": 0.8691504414950907}}
{"url": "https://www.physicsforums.com/threads/vertically-oriented-spring-and-energy-transfers.157346/", "text": "# Vertically oriented spring and energy transfers\n\n1. Feb 20, 2007\n\n### CaptainZappo\n\nI am posing this question due to a statement made by my TA:\n\nSuppose there is a spring pointed vertically upwards. Further, suppose that the spring is compressed a distance x so that it stores some potential energy. A mass is then placed on top of the spring. Assume all energy is conserved.\n\nHere's the question: upon releasing the spring, what is the kinetic energy of the ball as it passes the equilibrium position of the spring? Is it equal in magnitude to (1/2)kx^2 where x is the amount the spring is compressed, or is it (1/2)kx^2 - mgx?\n\nMy TA said the former is correct, while I cannot figure out why it is not the latter. My reasoning: as the ball is rising from its initial height (the point at which the spring is fully compressed) some of that energy is being converted to gravitational potential energy. The rest goes to kinetic energy.\n\nAny insight will be greatly appreciated,\n-Zachary Lindsey\n\n2. Feb 20, 2007\n\n### Staff: Mentor\n\nYour reasoning is correct; you cannot ignore gravitational PE in calculating the KE of the mass.\n\n3. Feb 20, 2007\n\n### CaptainZappo\n\nThank you.\n\n4. Feb 20, 2007\n\n### Parlyne\n\nWho is correct depends on whether x is the amount the spring is compressed from its free equilibrium length or that from its equilibrium length with the mass sitting on it. If it is the former, you are correct. If it's the later, your TA is correct.\n\nTo see this, let's let y be the distance the mass is sitting above the spring's free equilibrium position. Then, the potential energy is:\n\n$$V(y) = \\frac{1}{2} k y^2 + mgy$$\n\nWe can rearrange this:\n\n\\begin{align*} V(y) &= \\frac{1}{2} k \\left (y^2 + 2\\frac{mg}{k} y\\right ) \\\\ &= \\frac{1}{2} k \\left (y^2 + 2\\frac{mg}{k} y + \\frac{m^2g^2}{k^2} \\right ) - \\frac{m^2g^2}{2k} \\\\ &= \\frac{1}{2} k \\left (y + \\frac{mg}{k} \\right )^2 - \\frac{m^2g^2}{2k} \\end{align*}\n\nFrom this, we can see that the mass' height above the compressed equilibrium length is $$x = y + \\frac{mg}{k}$$.\n\nSo, $$V(x) = \\frac{1}{2} k x^2 - \\frac{m^2g^2}{2k}$$.\n\nThe difference between the potential energy when the spring is compressed by a distance x from the compressed equilibrium position ($$y = \\frac{-mg}{k}$$) and at the compressed equilibrium is:\n\n\\begin{align*} \\Delta V &= V(x) - V(0) \\\\ &= \\frac{1}{2} k x^2 - \\frac{m^2g^2}{2k} - \\left (\\frac{-m^2g^2}{2k} \\right ) \\\\ &= \\frac{1}{2} k x^2 \\end{align*}\n\n5. Feb 21, 2007\n\n### Staff: Mentor\n\nParlyne is certainly correct. From the way the conditions were specified in the OP:\nI presumed that x measures the compression from the uncompressed position of the spring. Only later is a mass involved:\nThe answer to that question depends on the meaning of \"equilibrium postion\" and where x is measured from. If, as I assumed from your description, x is measured from the uncompressed position of the spring (and equilibrium means x = 0), then 1/2kx^2 represents spring PE only and does not include gravitational PE.\n\nBut, as Parlyne explained, if x is measured from the equilibrium position of the mass-spring system (which is not the uncompressed position of the spring), then 1/2kx^2 represents changes in both spring PE and gravitational PE together.\n\nCaptainZappo: Better tell us exactly what the TA stated, not your interpretation of it (if you can recall).\n\nRegardless of what the TA says, learn what Parlyne explained and you'll be ahead of the game.", "date": "2018-05-21 17:16:51", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/vertically-oriented-spring-and-energy-transfers.157346/", "openwebmath_score": 0.9986562132835388, "openwebmath_perplexity": 840.1854149182476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.957912274487423, "lm_q2_score": 0.9073122132152183, "lm_q1q2_score": 0.8691255058312074}}
{"url": "http://mathhelpforum.com/algebra/223447-running-around-track.html", "text": "# Thread: Running Around a Track\n\n1. ## Running Around a Track\n\nHello, helpful people. I'm in some need of help.\n\nI'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:\n\n\"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?\"\n\nThanks.\n\n2. ## Re: Running Around a Track\n\nHello, MyHappyHarmony!\n\nThis one is tricky to set up.\nI have a rather clunky approach . . .\n\nBrooks and Avery are running laps around the outdoor track, in the same direction.\nBrooks completes a lap every 78 seconds, while Avery completes a lap every 91 seconds.\nBrooks (the faster runner) has just passed Avery.\nHow much time will it take for Brooks to overtake Avery again?\"\n\nWe will use: . $\\text{Distance} \\:=\\:\\text{Rate} \\times \\text{Time} \\quad\\Rightarrow\\quad r \\,=\\,\\frac{d}{t}$\nLet $d$ = distance in one lap (say, in feet).\n\nBrooks' rate is $\\frac{d}{78}$ ft/sec.\n\nAvery's rate is $\\frac{d}{91}$ ft/sec.\n\nIn $t$ seconds, Avery runs $\\frac{d}{91}t$ feet.\n\nIn the same $t$ seconds, Brooks runs $\\frac{d}{78}$ feet,\n. . which is $d$ feet (one lap) more than Avery's distance.\n\nThere is our equation! . . . $\\frac{d}{78}t \\;=\\;\\frac{d}{91}t + d$\n\nDivide by $d\\!:\\;\\frac{t}{78} \\:=\\:\\frac{t}{91} + 1$\n\nGot it?\n\n3. ## Re: Running Around a Track\n\nYes, I do! Thanks. Quick question: Would it be right to solve with a common denominator from there? The big numbers do look quite intimidating.\n\n4. ## Re: Running Around a Track\n\nOriginally Posted by MyHappyHarmony\nHello, helpful people. I'm in some need of help.\n\nI'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:\n\n\"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?\"\n\nThanks.\nTry to construct your solution from this figure\n\ncross posting\n\n5. ## Re: Running Around a Track\n\nI'm sorry. I'm not drawing anything from that figure.\n\n6. ## Re: Running Around a Track\n\nanother method\nassume track is 1200ft B rate 1200/78=15.38 fps A rate 1200/91= 13.19fps\nafter B runs 1 lap A is 1029ft ahead of B 91 * 13.19 = 1029 ft\nRate of closure = 15.38-13.19 = 2.2 fps\nclosure time 1029/2.2 = 467.7 sec\ntotal elapsed time = 467.7 +78= 555.7 sec\n\n7. ## Re: Running Around a Track\n\nOriginally Posted by bjhopper\ntotal elapsed time = 467.7 +78= 555.7 sec\ntypo; 545.7 ; 546 really: LCM(91,78)\n\nThese are quite simple: have the faster runner start behind the slower,\nthe distance behind being the track length; how long does it take to catch up?\n\n8. ## Re: Running Around a Track\n\nHello Wilmer,\nLong 'time no see.How are you?\nSoroban' s equation gives elapsed time until runners meet. 546 sec exactly\nMy answer 555.7 sec.There is no typo error.\nCan you show what I did wrong?\n\nHere is another solution\nB rate 1/78 laps per sec A rate 1/91 laps per sec\n(1/78 -1/91) = rate of closure laps per sec when B is behind A\nafter 1 lap B is 78/91 =0.857 laps behind A\n(1/78 -1/91)t = 0.857\n(0.00183)t =0.857 t=468.3sec add 78 for first lap Total elapsed time = 556.3sec\n\nWhat say you?\n\n9. ## Re: Running Around a Track\n\nHey BJ. Nice to \"type\" with you again!!\n\nOriginally Posted by bjhopper\nMy answer 555.7 sec.There is no typo error.\nBut you show this in your previous post:\n\"total elapsed time = 467.7 +78= 555.7 sec\"\n\nBut 467.7 + 78 = 545.7, not 555.7\n\nThat's what I was trying to tell you.", "date": "2017-02-22 21:07:35", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/223447-running-around-track.html", "openwebmath_score": 0.5145376324653625, "openwebmath_perplexity": 6032.57919386586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.983085088220004, "lm_q2_score": 0.8840392924390585, "lm_q1q2_score": 0.8690858457974018}}
{"url": "https://www.physicsforums.com/threads/countable-union-of-countable-sets-vs-countable-product-of-countable-sets.469593/#post-3119098", "text": "# Countable union of countable sets vs countable product of countable sets\n\nI know that a countable union of countable sets is countable, and that a finite product of countable sets is countable, but even a countably infinite product of countable sets may not be countable.\n\nLet $$X$$ be a countable set. Then $$X^{n}$$ is countable for each $$n \\in N$$.\n\nNow it should also be true that $$\\bigcup^{\\infty}_{n=1} X^{n}$$ is countable. How is this different from $$X^{\\omega}$$, which is uncountable?\n\nAh, very good question. Let's take a look at this question with $$X=\\mathbb{N}$$. You first mentioned\n\n$$\\bigcup{X^n}$$\n\nThe elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...).\n\nSo it is easily seen that every element of $$\\bigcup{X^n}$$ can be represented by an element of $$X^\\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...\n\nBut, and here comes the clue: there are much more elements in $$X^\\omega$$. For example: (1,1,1,...) does not come from an element of $$\\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\\omega$$!! And so $$X^\\omega$$ really is different than $$\\bigcup{X^n}$$.\n\nOf course, this is not a proof. But it merely gives an indication why the two sets are different...\n\nAh, very good question. Let's take a look at this question with $$X=\\mathbb{N}$$. You first mentioned\n\n$$\\bigcup{X^n}$$\n\nThe elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...).\n\nSo it is easily seen that every element of $$\\bigcup{X^n}$$ can be represented by an element of $$X^\\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...\n\nBut, and here comes the clue: there are much more elements in $$X^\\omega$$. For example: (1,1,1,...) does not come from an element of $$\\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\\omega$$!! And so $$X^\\omega$$ really is different than $$\\bigcup{X^n}$$.\n\nOf course, this is not a proof. But it merely gives an indication why the two sets are different...\n\nYes, after thinking about it I came to the same conclusion. You can't put them in bijective correspondence because if you wanted to map the union onto $$X^\\omega$$ you could do so injectively by adding 0s but there's no way to make this mapping surjective.\n\nmathwonk\nHomework Helper\ni believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.\n\ni believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.\n\nExactly, hence why this was a seeming contradiction.\n\nExactly, hence why this was a seeming contradiction.\nWhat's the contradiction that you're referring to?\n\nIt's already been stated that the needed bijection doesn't exist. I'm not following you're argument...", "date": "2022-01-19 15:10:06", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/countable-union-of-countable-sets-vs-countable-product-of-countable-sets.469593/#post-3119098", "openwebmath_score": 0.9133549928665161, "openwebmath_perplexity": 203.30606476458047, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850906978876, "lm_q2_score": 0.8840392756357327, "lm_q1q2_score": 0.8690858314688492}}
{"url": "https://www.physicsforums.com/threads/polynomial-riddle.757175/", "text": "# Polynomial riddle\n\n1. Jun 8, 2014\n\n### dislect\n\nHi guys,\n\nMy boss gave me a riddle.\nHe says that you have a \"black box\" with a polynomial inside it like f(x)=a0+a1x+a2x^2+a3x^3 ....\nyou don't know the rank of it or the coefficients a0, a1, a2 ....\nYou do know:\nall of the coefficients are positive\nyou get to input two x numbers and their f(x) value\n\nHow can you find all of the coefficients a0,a1,a2 .... and the rank?\n\nThanks !\n\n2. Jun 8, 2014\n\n### hilbert2\n\nKnowing f(x) for two values of x is not enough to determine the coefficients if the polynomial is of higher than first order.\n\n3. Jun 8, 2014\n\n### economicsnerd\n\nIt can be done if you know the polynomial has nonnegative integer coefficients, and if the second input is allowed to depend on the first output.\n\nPut in $1$. The output gives you the sum $s$ of the coefficients.\n- If $s=0$, you know $f=0$.\n- If $s>0$, put in any integer $k>s$. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-$k$.\n\n4. Jun 8, 2014\n\n### dislect\n\nHi economicsnerd,\n\nCouldn't follow \"put in any integer k>s. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-k.\"\nCould you give me an example? lets say f(x)=5+2x+4x^2 is in the 'black box'.\ni put in x=1 and receive f(x)=s=a0+a1+a2=11 > 0 so i put in x=k=15 and get f(x)=a0+15a1+900a2=935\norganize:\n1. a0+a1+a2=11 2. a0+15a1+900a2=935\n2 equations with 3 unknowns.\n\nWhats the rational behind putting a k > s, and how can i find all coefficients with just 2 equations ?\n\n5. Jun 8, 2014\n\n### economicsnerd\n\nOkay. So you know $f(1)=11$ and $f(15)=935$.\n- First note that $15^3 > 935$, so that the polynomial can't be of degree $>2$. That is, $f(x)=a_0 + a_1 x + a_2 x^2$ for some $a_0,a_1, a_2\\in\\mathbb Z_+$.\n- Next, note that $a_0,a_1\\in\\mathbb Z_+$ with $a_0+ a_1 \\leq 11 < 15$, so that $a_0 + a_1(15)<15^2$. This means that $935 - 15^2< a_2 (15)^2 \\leq 935$. It's easy to check that there's a unique integer $a_2$ that accomplishes this, namely $a_2 = 4$.\n- Substituting the solution for $a_2$ in, we now know that: $a_0+a_1 < 15$ and $a_0 + a_1 (15) = 15 35$. [Of course, you have two equations and two unknowns now, but we don't actually need to know that $a_0 + a_1 =7$, because knowing they're integers gives us enough information.] Now we can do the same trick again. $35 - 15< a_1 (15) \\leq 35$, which pins down $a_1=2$.\n- Then substitution yields $a_0 = 5$.\n\nWe can always work backward in this way.\n\nTo get some better intuition for why, consider the following question:\nSay I tell you I have a polynomial $g$ such that every coefficient of $g$ is a nonnegative integer $<10$. If I tell you $g(10)$, will you then be able to tell me what $g$ is?\n\nFor example:\n- If $g(10)=5021$, then do we know $g(x) = 5x^3 + 2x + 1$?\n- If $g(10)=60007$, then do we know $g(x) = 6x^4 + 7$?\n\nNothing is magical about $10$ that makes this trick work. It's just that we usually express numbers in a way (base-10) that makes it easy to answer this question. All the first question is needed for is to find a $k$ such that we can be sure that every coefficient of the polynomial is a nonnegative integer $<k$.\n\n6. Jun 8, 2014\n\n### D H\n\nStaff Emeritus\nThe rationale behind using asking for f(k) where k>f(1) is because otherwise the magic might not work. Suppose the function in the black box is f(x)=5+2x+4x2. The black box cranks out 11 as a response to f(1). Suppose I use 4 next. The black box churns out f(4)=77, which is 1031 when expressed base 4. That suggests the polynomial is 1+3x+5x3, which is wrong.\n\nTo see what's going on, here are f(k) from k=2 to 16, expressed in base 10 and in base k:\nCode (Text):\n\nk f(k) \u00a0base k \u00a0Polynomial\n2 \u00a0 25 \u00a0 11001 \u00a01+x^3+x^4\n3 \u00a0 47 \u00a0 \u00a01202 \u00a02+x^2+x^3\n4 \u00a0 77 \u00a0 \u00a01031 \u00a01+3x+x^3\n5 \u00a0115 \u00a0 \u00a0 430 \u00a03x+4x^2\n6 \u00a0161 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n7 \u00a0215 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n8 \u00a0277 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n9 \u00a0347 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n10 \u00a0425 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n11 \u00a0511 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n12 \u00a0605 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n13 \u00a0707 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n14 \u00a0817 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n15 \u00a0935 \u00a0 \u00a0 425 \u00a05+2x+4x^2\n16 1061 \u00a0 \u00a0 425 \u00a05+2x+4x^2\nCan you see the pattern? It appears that f(k) when expressed in base k is 425 for all k>5. This is indeed the case. You might want to try proving it.\n\nNote that in this case reading off the polynomial from the base-k representation of f(k) works for all k>5. That's because the largest coefficient is five. Using k=f(1)+1 (or higher) ensures that k is larger than the largest possible coefficient.\n\nSuppose the black box function was instead f(x)=11x3. Once again you'll get f(1)=11, but now the base k representation of f(k) keeps changing until you reach k=12, at which point it becomes b00 ('b' means 11) and stays that way for all k>11. The black box knows what the largest coefficient is. You don't, but you do know that it cannot be larger than 11.\n\nLast edited: Jun 8, 2014\n7. Jun 9, 2014\n\n### dislect\n\nThank you all :)", "date": "2018-07-19 21:44:23", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/polynomial-riddle.757175/", "openwebmath_score": 0.8452654480934143, "openwebmath_perplexity": 1341.4038825643586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075766298657, "lm_q2_score": 0.9032942014971871, "lm_q1q2_score": 0.8690661951862683}}
{"url": "http://mathhelpforum.com/discrete-math/224780-possible-mathematical-induction-problem-print.html", "text": "# Possible mathematical induction problem\n\n\u2022 Dec 2nd 2013, 02:03 PM\nPossible mathematical induction problem\nCourse: Foundations of Higher MAth\n\nProve that $24|(5^{2n} -1)$ for every positive integer n.\n\nThis is a question from my final exam today.\n\nP(n): $24|(5^{2n}-1)$\nP(1): $24|(5^2 -1)$ is a true statement.\n\nAssume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.\nThen $5^{2k}= 24a + 1$\n\nP(k+1): $24|(5^{2(k+1)}-1)$\n\n$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$\n$=(24a + 1)*25 -1$\n$=(24a)(25)+25 -1$\n$=(24a)(25)+24$\n$=24(25a+1)$\n$=24b$\n\nTherefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.\nNone of my friends used this method though. Is this a correct way to do it?\n\u2022 Dec 2nd 2013, 03:09 PM\nPlato\nRe: Possible mathematical induction problem\nQuote:\n\nCourse: Foundations of Higher MAth\nProve that $24|(5^{2n} -1)$ for every positive integer n.\nThis is a question from my final exam today.\nP(n): $24|(5^{2n}-1)$\nP(1): $24|(5^2 -1)$ is a true statement.\nAssume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.\nThen $5^{2k}= 24a + 1$\nP(k+1): $24|(5^{2(k+1)}-1)$\n$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$\n$=(24a + 1)*25 -1$\n$=(24a)(25)+25 -1$\n$=(24a)(25)+24$\n$=24(25a+1)$\n$=24b$\nTherefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.\nNone of my friends used this method though. Is this a correct way to do it?\n\nA strict grader may like to have seen more grouping symbols.\nHowever, the argument is correct.", "date": "2017-05-30 06:18:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/224780-possible-mathematical-induction-problem-print.html", "openwebmath_score": 0.6825377345085144, "openwebmath_perplexity": 1412.0766902328303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9728307653134903, "lm_q2_score": 0.893309401793762, "lm_q1q2_score": 0.8690388690087617}}
{"url": "http://math.stackexchange.com/questions/773504/tricky-triangle-area-problem/774141", "text": "Tricky Triangle Area Problem\n\nThis was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\\sqrt{33}$ in it. How could I have done this? The answer was $2\\sqrt{6}$ if it helps.\n\nEdited to add that it was a multiple choice question, with possible answers:\n\na. $2\\sqrt{6}$\nb. $5$\nc. $3\\sqrt{6}$\nd. $5\\sqrt{6}$\n\n-\nAre you familiar with the Law of Cosines? \u2013\u00a02012ssohn Apr 28 '14 at 23:37\nAdditionally, it was Multiple choice, and the four answers were $2\\sqrt{6}$, $5$, $3\\sqrt{6}$, and $5\\sqrt{6}$ \u2013\u00a0Asimov Apr 28 '14 at 23:37\n@2012ssohn Of course i am, but how can i apply that formula to it? I even tried it at the time, but couldnt get anything useful about area from it \u2013\u00a0Asimov Apr 28 '14 at 23:38\nDid you actually try to use Heron's formula? The formula simplifies pretty nicely. \u2013\u00a0Nate Apr 28 '14 at 23:41\nWhen i tried it i got an ugly glob of roots and addition that just wouldn't simplify for me. Maybe i just need to practice simplification \u2013\u00a0Asimov Apr 28 '14 at 23:44\n\nFrom the law of cosines ($C^2 = A^2 + B^2 - 2AB\\cos \\theta$), we get that $(\\sqrt{33})^2 = 2^2 + 5^2 - 2 \\cdot 2 \\cdot 5 \\cos \\theta$.\n\nSimplifying this, we get $33 = 29 - 20 \\cos \\theta$, which means that $\\displaystyle \\cos \\theta = -\\frac{1}{5}$\n\nBecause $\\cos^2 \\theta + \\sin^2 \\theta = 1$, we get that $\\displaystyle \\sin^2 \\theta = \\frac{24}{25}$. This means that $\\displaystyle \\sin \\theta = \\frac{2\\sqrt{6}}{5}$ (note that, because $0 \\le \\theta \\le \\pi$, $\\sin \\theta \\ge 0$).\n\nThe area of the triangle is $\\displaystyle \\frac{1}{2} A B \\sin \\theta = \\frac{1}{2} \\cdot 2 \\cdot 5 \\cdot \\frac{2\\sqrt{6}}{5} = 2\\sqrt{6}$.\n\n-\nThank you, this is clear, concise, and could be done quickly and easily in the situation. This is probably what the judges/writers of it wanted readers to do. \u2013\u00a0Asimov Apr 28 '14 at 23:48\n\nUse the $\\frac{1}{4}\\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.\n\n\\begin{align} & \\frac{1}{4}\\sqrt{4\\cdot4\\cdot25-(4+25-33)^2} \\\\ = & \\frac{1}{4}\\sqrt{4^2\\cdot25-4^2} \\\\ = & \\sqrt{25-1} \\\\ = & 2\\sqrt{6} \\end{align}\n\n-\nThis is the way I would have done it, nice. \u2013\u00a0Sawarnik Apr 29 '14 at 6:40\n\nYou could've used Heron's formula straight away, actually.\n\n\\begin{align} T & = \\tfrac{1}{4} \\sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\\\ & = \\tfrac{1}{4} \\sqrt{(2+5-\\sqrt{33})(2-5+\\sqrt{33})(-2+5+\\sqrt{33})(2+5+\\sqrt{33})} \\\\ & = \\tfrac{1}{4} \\sqrt{(7-\\sqrt{33})(-3+\\sqrt{33})(3+\\sqrt{33})(7+\\sqrt{33})} \\\\ & = \\tfrac{1}{4} \\sqrt{(7+\\sqrt{33})(7-\\sqrt{33})(\\sqrt{33}+3)(\\sqrt{33}-3)} \\\\ & = \\tfrac{1}{4} \\sqrt{(49-33)(33-9)} \\\\ & = \\tfrac{1}{4} \\sqrt{16 \\cdot 24} \\\\ & = \\tfrac{1}{4} \\sqrt{64 \\cdot 6} \\\\ & = \\tfrac{8}{4} \\sqrt{6} \\\\ & = 2 \\sqrt{6} \\end{align}\n\n-\n\nSince this was multiple-choice, I think it's worth noting that you could have guessed the right answer without doing (much) arithmetic: the diagonal of a right triangle with sides $2$ and $5$ has length $\\sqrt{2^2+5^2} = \\sqrt{29}$; since $\\sqrt{33}$ is fairly close to this, the answer should be close to the area of a $2-5-\\sqrt{29}$ triangle, which is of course $\\frac12(2)(5)=5$. If you imagine how to 'stretch out' the $\\sqrt{29}$ diagonal to $\\sqrt{33}$, it's clear that the right angle will have to become obtuse, and this in turn means that the area of the $2-5-\\sqrt{33}$ triangle will have to be less than $5$; of the provided answers, only $2\\sqrt{6}\\approx4.472$ is less than $5$ (and of course very close to it).\n\n-\nWhere do you see the answer options (multiple choices)? Is the original problem linked to? \u2013\u00a0Jeppe Stig Nielsen Apr 29 '14 at 9:05\n@JeppeStigNielsen It's now there in the question. \u2013\u00a0Ramchandra Apte Apr 29 '14 at 14:06\nI like this response, and used a similar estimation method, using the areas of slightly bigger and smaller triangles that were easier to calculate \u2013\u00a0Asimov Apr 30 '14 at 0:05\nI hate it when there are two good answers, and both are right and good, and you wish you could accept both as correct \u2013\u00a0Asimov Apr 30 '14 at 1:41\n@JohnJPershing For what it's worth, I think 2012ssohn's answer is much better than mine; it explains how to actually derive the correct value, rather than merely how to answer the multiple-choice question. Both are valuable, but that one's likely to be more broadly applicable; I just wanted to offer an alternate approach for the test. \u2013\u00a0Steven Stadnicki Apr 30 '14 at 3:24\n\nLet be $ABC$ the triangle. Consider the altitude $AH$ over the greatest side, the one whose length is $\\sqrt{33}$. Now call $x=BH$, so $\\sqrt{33}-x=CH$. Apply Pythagoras' theorem to get $$\\left\\{ \\begin{array}{rcl} x^2+h^2&=&4\\\\ \\left(\\sqrt{33}-x\\right)^2+h^2&=&25 \\end{array} \\right.$$\n\nSolve for $h$ and you are (almost) done.\n\n-", "date": "2016-06-28 09:40:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/773504/tricky-triangle-area-problem/774141", "openwebmath_score": 0.99753338098526, "openwebmath_perplexity": 369.73495408429295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109476256691, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8690246142393594}}
{"url": "http://math.stackexchange.com/questions/309548/permutation-with-condition", "text": "# Permutation with Condition\n\nThe numbers 1-9 are drawn from a hat at random. Without repetition of numbers, how many numbers (with all digits drawn-- eg. 987654321) would be divisible by 11?\n\nFirst I recalled that a number is divisible by 11 if the number's alternating digits minus the other digits are equal to a number divisible by 11.\n\nHonestly, I did a bit of guess and check to follow and found that 948576321 was an answer divisible by 11. Therefore if I rearranged 9,8,7,3, and 1 (the leading alternating numbers) it would bring the same result. So I solved the permutation to receive 120 possible combinations. I then took the permutation of 4,5,6, and 2 and received a total of 24 combinations.\n\nI multiplied these two answers together to get a grand total of 2880 combinations that would be divisible by 11.\n\n-\nI ran a quick python program, and the answer is 31680. I'll try to prove it in a few minutes. \u2013\u00a0 Alfonso Fernandez Feb 20 '13 at 22:33\nI would appreciate it :) \u2013\u00a0 Hannah Feb 20 '13 at 23:06\n\nThe correct answer is 31680. Your combination of alternating numbers (9, 8, 7, 3, 1) is only one out of 11 possibilities:\n\n1. 1, 2, 3, 4, 7\n2. 1, 2, 3, 5, 6\n3. 1, 3, 7, 8, 9\n4. 1, 4, 6, 8, 9\n5. 1, 5, 6, 7, 9\n6. 2, 3, 6, 8, 9\n7. 2, 4, 5, 8, 9\n8. 2, 4, 6, 7, 9\n9. 2, 5, 6, 7, 8\n10. 3, 4, 5, 7, 9\n11. 3, 4, 6, 7, 8\n\nSo we end up with $11 \\cdot 5! \\cdot 4! = 11 \\cdot 120 \\cdot 24 = 31680$.\n\n-\nBut how do you come up with these 11 without checking all $\\binom{9}{4}=126$ possibilities? \u2013\u00a0 Alfonso Fernandez Feb 21 '13 at 0:21\n@AlfonsoFernandez I did it by checking all of them with a script. Brian's approach is more rigorous in that regard. \u2013\u00a0 ralph Feb 23 '13 at 14:40\n(This kind of problem, finding a set of distinct numbers to add up to a certain number, is a key part of Kakuro puzzles. The list of valid combinations for this can be found in Kakuro tables such as brainbashers.com/combinations.asp under 5-digit combinations with a sum of either 17 or 28. Brian's answer proves why the sum must be 17 or 28.) \u2013\u00a0 ralph Feb 23 '13 at 14:47\n\nHere\u2019s a systematic approach; it\u2019s still a bit tedious in spots, but not unreasonably so.\n\nThe sum of all $9$ digits is $45$. You need to split the digits into a set $A$ of $5$ digits, those in the odd-numbered positions, and a set $B$ of $4$ digits, those in the even-numbered positions, whose sums differ by a multiple of $11$. If $a$ is the sum of the digits in $A$ and $b$ the sum of the digits in $B$, then $a-b$ is a multiple of $11$, and $a+b=45$.\n\nClearly $b=45-a$, so $a-b=a-(45-a)=2a-45$. The smallest possible value of $a$ is $1+2+3+4+5=15$, and the largest is $9+8+7+6+5=35$, so $-15\\le a-b\\le25$. The multiples of $11$ in that range are $-11,0,11$, and $22$. However, $2a-45$ is clearly odd, so in fact the only possible values of $a-b$ are $-11$ and $11$.\n\n\u2022 If $a-b=2a-45=-11$, then $a=17$.\n\u2022 If $a-b=2a-45=11$, then $a=28$.\n\nWhat sets of $5$ digits sum to $17$? Since $1+2+3+4+5=15$, we either increase $5$ by $2$ to get the set $\\{1,2,3,4,7\\}$ or increase $4$ and $5$ by $1$ each to get the set $\\{1,2,3,5,6\\}$; anything else produces either too big a sum or duplicated digits.\n\nIt\u2019s a bit harder to enumerate the sets that sum to $28$, because there are more of them. A good starting point is the fact that $9+8+7+6+5=35$, which is $7$ too large. First note that we need to have at least one of $8$ and $9$: $7+6+5+4+3=25$ is too small. What can we do without the $9$? $8+7+6+5+4=30$, so, much as in the previous case, our only choices are to subtract $2$ from $4$ to get the set $\\{8,7,6,5,2\\}$ or subtract $1$ each from $4$ and $5$ to get $\\{8,7,6,4,3\\}$. Every other set will include the $9$.\n\nLet\u2019s try next for the ones that include $9$ but not $8$. $9+7+6+5+4=31$; we can subtract $3$ from the $4$, the $5$, or the $6$ to get the sets $\\{9,7,6,5,1\\}$, $\\{9,7,6,4,2\\}$, and $\\{9,7,5,4,3\\}$. A little experimentation shows that trying to reduce the sum by $3$ by subtracting $2$ from one digit and $1$ from another produces nothing new.\n\nThe last batch will be those containing both $9$ and $8$. If we keep the $7$ as well, we already have a total of $9+8+7=24$, and the only two digits that supply the missing $4$ are $1$ and $3$, giving us the set $\\{9,8,7,3,1\\}$. If we start with $9,8$, and $6$, we can get the missing $5$ from either $1$ and $4$ or $2$ and $3$, giving us the sets $\\{9,8,6,4,1\\}$ and $\\{9,8,6,3,2\\}$. If start with $9,8$, and $5$, the only way to get the missing $6$ is with $2$ and $4$, giving us the set $\\{9,8,5,4,2\\}$. And $9+8+4+3+2=26$ is too small, so we\u2019ve found all of the possibilities for the set $A$.\n\nAs you already observed, we can permute $A$ and $B$ arbitrarily, so each choice of $A$ yields $5!4!$ numbers, and there are $11$ possibilities for $A$, for a grand total of $11\\cdot5!4!=31,680$ numbers.\n\n-\nEvery number of the form $x*11$ is divisible by 11. Now you must think about $x$.\n$x$ must be such that $(x)*11$ is a number with 9 digits! If it is useful for you I can explain it more.", "date": "2014-07-25 06:55:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/309548/permutation-with-condition", "openwebmath_score": 0.7863045334815979, "openwebmath_perplexity": 148.28139703981665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717472321278, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.8689695214386968}}
{"url": "https://lpsa.swarthmore.edu/Bode/BodeHowNonMinPhase.html", "text": "# The Asymptotic Bode Diagram for Non-Minimum Phase Poles and Zeros\n\n## A Real Pole with Negative \u03c90\n\nElsewhere we have discussed how to make Bode plots for a real pole. You should be familiar with that analysis. The discussion there assumed that the value of \u03c90 was positive; here we discuss the case if \u03c90 is negative. We start with\n\n$$H(s)=\\frac{1}{1+\\frac{s}{\\omega_0}}$$\n\n#### Magnitude\n\nIf you carefully examine the analysis (here) of the \"Magnitude\" plot you'll see that the only time \u03c90 is used, it is squared (e.g., $\\left( \\frac{\\omega}{\\omega_0} \\right)^2$). Therefore, the magnitude plot does not depend on the sign of \u03c90, only its absolute value, so we don't need to change anything to accomodate a negative value of \u03c90.\n\n#### Phase\n\nThe phase however does change. The phase of a single real pole is given by is given by\n\n$$\\angle H\\left( {j\\omega } \\right) = \\angle \\left( {{1 \\over {1 + j{\\omega \\over {{\\omega _0}}}}}} \\right) = - \\angle \\left( {1 + j{\\omega \\over {{\\omega _0}}}} \\right) = - \\arctan \\left( {{\\omega \\over {{\\omega _0}}}} \\right)$$\n\nLet us again consider three cases for the value of the frequency, but we assume \u03c90 is negative:\n\nCase 1) \u03c9<<\u03c90.\u00a0 This is the low frequency case with \u03c9/\u03c90\u21920, and doesn't depend on the sign of \u03c90.\u00a0 At these frequencies We can write an approximation for the phase of the transfer function\n\n$$\\angle H\\left( {j\\omega } \\right) \\approx -\\arctan \\left( 0 \\right) = 0^\\circ = 0\\;rad$$\n\nCase 2) \u03c9>>\u03c90.\u00a0 Here we will consider the cases of positive and negative \u03c90 side by side.\n\n\u03c90 > 0\u00a0\u00a0\u00a0 (the minimum phase case, discussed previously)\n\nThis is the high frequency case with \u03c9/\u03c90 \u2192 +\u221e.\u00a0 We can write an approximation for the phase of the transfer function\n\n$$\\angle H\\left( {j\\omega } \\right) \\approx - \\arctan \\left( \\infty \\right) = - 90^\\circ$$\n\nThe high frequency approximation is at shown in green on the diagram below.\u00a0 It is a horizontal line at -90\u00b0.\n\n\u03c90 < 0\u00a0\u00a0\u00a0 (the non-minimum phase case)\n\nThis is the high frequency case with \u03c9/\u03c90 \u2192 -\u221e.\u00a0 We can write an approximation for the phase of the transfer function\n\n$$\\angle H\\left( {j\\omega } \\right) \\approx - \\arctan \\left(- \\infty \\right) = + 90^\\circ$$\n\nThe high frequency approximation is at shown in green on the diagram below.\u00a0 It is a horizontal line at +90\u00b0.\n\nCase 3) \u03c9=\u03c90.\u00a0Again consider the cases of positive and negative \u03c90 side by side.\n\n\u03c90 > 0\u00a0\u00a0\u00a0 (the minimum phase case, discussed previously)\n\n$$\\angle H\\left( {j\\omega } \\right) = - \\arctan \\left( 1 \\right) = - 45^\\circ$$\n\n\u03c90 < 0\u00a0\u00a0\u00a0 (the non-minimum phase case)\n\n$$\\angle H\\left( {j\\omega } \\right) = - \\arctan \\left( -1 \\right) =+ 45^\\circ$$\n\nFrom the above discussion you can see that the only effect of the pole having a negative value of \u03c90 is that the phase is inverted (it increases from 0 \u2192 +90\u00b0 as \u03c9 increases from 0 \u2192 \u221e. The images below show the Bode plots for\n\n$$H_1(s)=\\frac{1}{1+\\frac{s}{10}}, \\quad\\quad H_2(s)=\\frac{1}{1-\\frac{s}{10}}$$\n\nH1(s) has a positive \u03c900=+10, so H1(s)=1/(1+s/10)) and H2(s) has a negative \u03c900=-10, so H2(s)=1/(1-s/10)). The pole of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero)\nH1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.\n\nThe same conclusion holds for first order poles and second order poles and zeros (see below).\n\n## A Real Zero with Negative \u03c90\n\nThe images below show the Bode plots for two functions, one with a positive \u03c900=+10) and one with a negative \u03c900=-10). The zero of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero).\n\n$$H_1(s)=1+\\frac{s}{10}, \\quad\\quad H_2(s)=1-\\frac{s}{10}$$\n\nH1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites. Recall that 360\u00b0 is equivalent to 0\u00b0 so you can think of the plot for the angle of H2(s) as starting at 0\u00b0 and dropping by 90\u00b0 (though the plot shows it as starting at 360\u00b0).\n\n## A Second Order Pole with Negative \u03b6\n\nThe images below show the Bode plots for (note the sign of the middle term in the numerator is different)\n\n$$H_1(s)=\\frac{1}{1+0.1\\cdot\\frac{s}{10}+\\left(\\frac{s}{10}\\right)^2}, \\quad\\quad H_2(s)=\\frac{1}{1-0.1\\cdot\\frac{s}{10}+\\left(\\frac{s}{10}\\right)^2}$$\n\nThe poles of H1(s) are at s=-0.5\u00b1j9.987 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+0.5\u00b1j9.987 (a positive real part, the right half of the s-plane; a non-minimum phase pole). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that, again, the magnitudes are identical, but the phases are opposites.\n\n## A Second Order Zero with Negative \u03b6\n\nThe images below show the Bode plots for (note the sign of the middle term in the numerator is different)\n\n$$H_1(s)=1+0.1\\cdot\\frac{s}{10}+\\left(\\frac{s}{10}\\right)^2, \\quad\\quad H_2(s)=1-0.1\\cdot\\frac{s}{10}+\\left(\\frac{s}{10}\\right)^2$$\n\nThe zeros of H1(s) are at s=-0.5\u00b1j9.987 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+0.5\u00b1j 9.987(a positive real part, the right half of the s-plane; a non-minimum phase zero). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.\n\n##### Key Concept: For poles and zeros with positive real parts, phase is inverted\n\nIf a 1st order pole has a positive real part (i.e., a nonminimum phase system, pole is in right half of s-plane) at say s=+5, so\n\n$$H(s)=\\frac{1}{1 - \\frac{s}{5}}$$\n\nthe magnitude of the Bode plot is unchanged from the case of a corresponding pole with negative value, at s=-5 (pole is in left half of s-plane)\n\n$$H(s)=\\frac{1}{1 + \\frac{s}{5}}$$\n\nbut the phase of the plot is inverted.\n\nThe same rule holds Bode plots for 2nd order (complex conjugate) poles, and for 1st and 2nd order zeros.\n\nReferences\n\nReplace", "date": "2022-05-23 20:58:39", "meta": {"domain": "swarthmore.edu", "url": "https://lpsa.swarthmore.edu/Bode/BodeHowNonMinPhase.html", "openwebmath_score": 0.9354708194732666, "openwebmath_perplexity": 1052.9627466968834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717440735736, "lm_q2_score": 0.8807970685907242, "lm_q1q2_score": 0.8689695001344417}}
{"url": "https://math.stackexchange.com/questions/1475582/number-of-ways-to-two-color-a-sectored-circle-so-that-opposite-sectors-are-diffe", "text": "# Number of ways to two-color a sectored circle so that opposite sectors are different colors.\n\nGiven a circle divided into 2n equal sectors, how many ways can it be painted with two colors so that opposite sectors are not the same color? I'm interested in the number of solutions that are unique under rotation.\n\nI've tried to enumerate the early situations and see patterns but haven't seen a path to a general solution. I think I only have three rules of thumb at the moment:\n\n1. There will be n sectors of each color\n2. An alternating paint job is only legal when n is odd.\n3. I think we're bounded on top by 2^(n-2) for n>1.\n\nI may have missed some examples but this link has my work so far: http://i.imgur.com/ZQylapu.png\n\nIf I've gotten everything, then the series goes:\n\n\u2022 n=1, 1\n\u2022 n=2, 1\n\u2022 n=3, 2\n\u2022 n=4, 2\n\u2022 n=5, 4\n\u2022 n=6, 6\n\nn=6 is notable as two of the combinations are left- and right-handed versions of a similar arrangement. I imagine this becomes much more common beyond n=6. I counted these separately, but a general solution that wouldn't would be nice as well.\n\n\u2022 How do you define opposite sector? If each sector has just one opposite then why wouldn't they answer by $2^n$? \u2013\u00a0fleablood Oct 12 '15 at 0:31\n\u2022 Oh, I guess the unique under rotation puts a damper on it. \u2013\u00a0fleablood Oct 12 '15 at 0:33\n\u2022 Huh, I have difficulty putting words to what I mean by opposite. What comes to mind is two sectors are opposites if the number of sectors between them in the clockwise direction is equal to the number in the counter-clockwise direction. \u2013\u00a0Polyhog Oct 12 '15 at 0:37\n\u2022 No I get your definition. As a pair of sectors has no relation to other sectors the answer would be $2^n$ as each pair of sectors has two choices. But if they are unique to rotation, we have to calculate how many are equivalent under rotation. Good question. \u2013\u00a0fleablood Oct 12 '15 at 0:54\n\u2022 Here's a thought. Fix sector 1. then for sector i = 2 to n give a binary value for color and add $2^i$ for green and nothing for yellow. Thus you'll have a 1-1 correspondence between colorings and the numbers from 1 to $2^{n - 1}. Now, we have to come up with a formula for determining if two numbers represent the same coloring under rotation. \u2013 fleablood Oct 12 '15 at 1:00 ## 1 Answer This can be done using Burnside's lemma. We need to count the assignments of colors being fixed by each type of permutation from among the$2n$permutations total in the cycle index$Z(C_{2n})$of the cyclic group acting on the sectors. There are $$\\varphi(d)$$ permutations having cycle structure $$a_d^{2n/d}$$ in the cycle index$Z(C_{2n})$where$d|2n.$For a permutation to fix an assignment it needs to be constant on the cycles. Suppose that the cycle length$d$is even and pick one of its elements. The slot opposing this element is also located on this cycle. But they must have different colors, so there are no valid assignments fixed by a permutation of shape$a_d^{2n/d}$when$d$is even. On the other hand when$d$is odd the$2n/d$cycles of length$d$are grouped into$n/d$pairs which are reflections of each other and hence must have opposite colors. Therefore there are two possible assignments of colors to these cycles forming a pair (as opposed to four if there were no constraint). This yields a contribution of $$\\varphi(d) 2^{n/d}.$$ Averaging this over the total$2n$permutations we get $$\\frac{1}{2n} \\sum_{d|2n,\\;d\\;\\mathrm{odd}} \\varphi(d) 2^{n/d}$$ which is $$\\frac{1}{2n} \\sum_{d|n,\\;d\\;\\mathrm{odd}} \\varphi(d) 2^{n/d}.$$ This yields the sequence $$1, 1, 2, 2, 4, 6, 10, 16, 30, 52, 94, 172, 316, 586, 1096, \\\\ 2048, 3856, 7286,\\ldots$$ which is OEIS A000016 where additional material awaits. I found the OEIS entry by using a simple total enumeration algorithm (definitely not optimized) to compute the first few values (practical to about$n=10\\$ which is enough to conclusively identify the sequence). This was the Maple code:\n\nwith(numtheory);\n\nsectors :=\nproc(n)\noption remember;\nlocal d, ind, orbits, orbit, rot, pos;\n\norbits := {};\n\nfor ind from 2^(2*n) to 2*2^(2*n) - 1 do\nd := convert(ind, base, 2);\n\nfor pos to n do\nif d[pos] = d[pos+n] then\nbreak;\nfi;\nod;\n\nif pos = n+1 then\norbit := {};\nfor rot from 0 to 2*n-1 do\norbit := {op(orbit),\n[seq(d[q], q=1+rot..2*n),\nseq(d[q], q=1..rot)]};\nod;\n\norbits := {op(orbits), orbit};\nfi;\nod;\n\nnops(orbits);\nend;\n\nQ :=\nproc(n)\n1/2/n*\nselect(d->type(d,odd), divisors(n)));\nend;\n\n\u2022 That is brilliant. The equation you reached is also is the same for number of binary sequences coming from a shift register complementing the last output (equivalent to going through the circle and outputting the sector's color as binary number). Thank you! \u2013\u00a0Weaam Oct 12 '15 at 3:04\n\u2022 This was not original work. The calculation is just about included in the OEIS entry, it only needed filling in the details. \u2013\u00a0Marko Riedel Oct 12 '15 at 3:06\n\u2022 Still! It is a great answer. \u2013\u00a0Weaam Oct 12 '15 at 3:22", "date": "2020-10-28 00:30:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1475582/number-of-ways-to-two-color-a-sectored-circle-so-that-opposite-sectors-are-diffe", "openwebmath_score": 0.7256286144256592, "openwebmath_perplexity": 1147.3659941923745, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138157595305, "lm_q2_score": 0.888758793492457, "lm_q1q2_score": 0.8689517512753467}}
{"url": "http://math.stackexchange.com/questions/275456/computing-a-limit-using-the-limit-definition", "text": "# Computing a Limit using the Limit Definition\n\nI've just started Real Analysis. In the textbook (Real Analysis and Applications, by Davidson and Donsig) they have defined the limit of a sequence. I am working on one of the provided exercises. No suggested solutions are provided for any questions, so I am looking for help checking my work to make sure I understand what I am doing. The exercise is as follows.\n\nCompute the limit. Then, using $\\epsilon =10^{-6}$, find an integer $N$ that satisfies the limit definition. $$\\lim_{n \\to \\infty} \\frac{1}{ln(ln(n))}$$\n\nFirstly, I just want to address the limit definition. In the text, they define a real number $L$ to be the limit of a sequence of real numbers $(a_n)^{\\infty}_{n=1}$ if for every $\\epsilon > 0$ there is an integer $N=N(\\epsilon)>0$ such that\n\n$$|a_n - L|< \\epsilon$$\n\nfor all $n\\geq N$. Now for my informal interpretation.\n\nI believe the idea is that after some point the distance between $a_n$ and the limit $L$ can be made arbitrarily small by choosing a sufficiently large $N$. But why does $N$ have to be a function of $\\epsilon$? Any clarification of the definition would be appreciated.\n\nNow, for the exercise. I know the divisor of the fraction of the given sequence approaches infinity as $n$ approaches infinity and as such the sequence approaches zero as $n$ approaches infinity. So I claim that $L=0$. I observe that\n\n$$\\left | \\frac{1}{ln(ln(n))} - 0 \\right | = \\frac{1}{ln(ln(n))}$$\n\nSo now I need an $N$ such that, for all $n \\geq N$, I get $|a_n - 0|<10^{-6}$. Now, I am not sure if my next steps are correct. I need an $N$ such that\n\n$$\\frac{1}{ln(ln(N))}<10^{-6}$$\n\nSo I just solve for N in the equality, which yields\n\n$$N > e^{e^{1000000}}$$\n\nHence, for any $n \\geq N$,\n\n$$\\frac{1}{ln(ln(n))}<10^{-6}$$\n\nDo I find the closest integer to $e^{e^{1000000}}$? Any clarification would be appreciated.\n\n-\nIt looks to me like you are done. You could compute $N$, but keep in mind it doesn't hurt for $N$ to be higher. It doesn't have to be the minimal one over $e^{e^{1000000}}$ \u2013\u00a0 rschwieb Jan 10 '13 at 21:46\nYou can just write your integer as $\\lceil e^{e^{1000000}} \\rceil$ to make it clear you're talking about an integer. Other than that you have found a value that works and so you are done. \u2013\u00a0 AvatarOfChronos Jan 10 '13 at 21:49\nI suspect that computing $e^{e^{1000000}}$ would be a major project, requiring quite a bit of programming skill. It doesn't make sense to try to compute the exact value in this context at all. \u2013\u00a0 Henry B. Jan 10 '13 at 22:33\nit's a very big number, even computers are afraid to compute it. \u2013\u00a0 Santosh Linkha Jan 10 '13 at 22:43\n\nI think your confusion arises from the phrase \"there exists an $N = N(\\epsilon)$. You already showed why $N$ has to be a function of $\\epsilon$: intuitively, the smaller the epsilon, the larger the $N$ has to be in order for the inequality to be satisfied.\n\nYou could just say then: choose $N > e^{e^{\\epsilon^{-1}}}$ for $N(\\epsilon)$; if you show such an $N$ exists then this implies that such a function exists. You don't have to exhibit a specific one, unless you want to (or asked on a question).\n\nBut, if you want to be explicit, you could use:\n\n\u2022 $N(\\epsilon) = \\lceil e^{e^{\\epsilon^{-1}}}\\rceil$ where $\\lceil - \\rceil$ denotes the least integer greater than its argument, as you thought\n\u2022 $N(\\epsilon) = \\lceil e^{e^{\\epsilon^{-1}}}\\rceil + 8434$\n\u2022 $N(\\epsilon) = \\lceil 9434\\pi e^{e^{\\epsilon^{-1}}}\\rceil$\n\nSee, all of them work, as long as the function gives you an integer sufficiently larger so that the $|a_n - L| < \\epsilon$. However, I should stress once more that as long as you show that there exists such an integer for every $\\epsilon$, this already shows that the function exists. (Modulo your philosophy on mathematics; you might actually need to construct the function for your peace of mind.)\n\n-", "date": "2015-08-03 00:48:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/275456/computing-a-limit-using-the-limit-definition", "openwebmath_score": 0.9354601502418518, "openwebmath_perplexity": 135.98835329162426, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138144607745, "lm_q2_score": 0.8887587890727754, "lm_q1q2_score": 0.8689517457998822}}
{"url": "http://math.stackexchange.com/questions/637664/why-does-this-not-seem-to-be-random", "text": "# Why does this not seem to be random?\n\nI was running a procedure to be like one of those games were people try to guess a number between 1 and 100 where there are 100 people guessing.I then averaged how many different guesses there are.\n\nfrom random import randint\n\ndef averager(times):\ntests = list()\nfor _ in range(times):\nl = [randint(0,100) for _ in range(100)]\ntests.append(len(set(l)))\nreturn sum(tests)/len(tests)\n\nprint(averager(100))\n\nFor some reason, the number of different guesses averages out to 63.6\n\nWhy is this?Is it due to a flaw in the python random library?\n\nIn a scenario where people were guessing a number between 1 and 10\n\nThe first person has a 100% chance to guess a previously unguessed number\n\nThe second person has a 90% chance to guess a previously unguessed number\n\nThe third person has a 80% chance to guess a previously unguessed number\n\nand so on...\n\nThe average chance of guessing a new number(by my reasoning) is 55%. But the data doesn't reflect this.\n\n-\nShades of (1-1/e)? Note that $1 - 1/e = 0.63212$ Can you see why it has to be this? I thought I had the answer but not sure anymore. Got to look at it some more \u2013\u00a0user44197 Jan 14 '14 at 3:33\nIf you want more info about this problem, there's a classical one involving the same maths. It's the odds of having two people(or more) with the same birthday in a classroom. \u2013\u00a0Feu Jan 14 '14 at 3:42\n@Feu an important one in cryptography too :) \u2013\u00a0Cruncher Jan 14 '14 at 14:16\nA related question I asked before. Turns out $e^{-1}$ is a notable probability. Here's an older related question. \u2013\u00a0badroit Jan 14 '14 at 14:57\nFrom the programming point of view, don't forget that python's basic random functions are not true random, or even cryptographic random. It is a seeded PRNG. If you want better random numbers, try a cryptographic number generator source such as /dev/random (full hardware generated randomness, runs out easily). \u2013\u00a0Linuxios Jan 14 '14 at 16:40\n\nSuppose that $n$ guesses are made. For $i=1$ to $100$, let $X_i=1$ if $i$ is not guessed, and let $X_i=0$ otherwise. If $$Y=X_1+X_2+\\cdots +X_{100},$$ then $Y$ is the number of numbers not guessed.\n\nBy the linearity of expectation, we have $$E(Y)=E(X_1)+E(X_2)+\\cdots+E(X_{100}).$$ The probability that $i$ is not chosen in a particular trial is $\\frac{99}{100}$, and therefore the probability it is not chosen $n$ times in a row is $\\left(\\frac{99}{100}\\right)^n$. Thus $$E(Y)=100 \\left(\\frac{99}{100}\\right)^n.$$\n\nIn particular, let $n=100$. Note that $\\left(1-\\frac{1}{100}\\right)^{100}\\approx \\frac{1}{e}$, so the expected number not guessed is approximately $\\frac{100}{e}$. Thus the expected number guessed is approximately $63.2$, a result very much in line with your simulation.\n\nIn general, if $N$ people choose independently and uniformly from a set of $N$ numbers, then the expected number of distinct numbers not chosen is $$N\\left(1-\\frac{1}{N}\\right)^N.$$ Unless $N$ is very tiny, this is approximately $\\frac{N}{e}$, and therefore the expected number of distinct numbers chosen is approximately $N-\\frac{N}{e}$. Note that the expected proportion of the numbers chosen is almost independent of $N$.\n\n-\nI think you are missing a \"not\": \"then the expected number of distinct numbers [not] chosen is\" \u2013\u00a0aaaaaaaaaaaa Jan 14 '14 at 19:19\nThe expected number not chosen is approximately $\\frac{N}{e}$. Since there are $N$ numbers overall, the expected number chosen is approximately $N-\\frac{N}{e}$. \u2013\u00a0Andr\u00e9 Nicolas Jan 14 '14 at 19:32\nYes, and you wrote the opposite in the second to last paragraph. \u2013\u00a0aaaaaaaaaaaa Jan 14 '14 at 19:34\nThank you, I finally found where the missing not was (not), had been looking for it towards the very end. Will fix. \u2013\u00a0Andr\u00e9 Nicolas Jan 14 '14 at 19:42\n\nIn your 1 to 10 example, it's not true in general that the third chooser has an 80% chance on choosing a new number. It's only the case if the second one has guessed a different number than the first one.\n\n-\n\nThis is an example of the Birthday Paradox / Birthday Problem.\n\nBirthday problem - Wikipedia, the free encyclopedia\n\nI was just looking at my Online Cryptography class video lecture today on this very problem.\n\nCoursera.org: crypto-009\n\nThere is an apparent paradox that there is more duplication of numbers than expected when the random numbers are supposedly independent.\n\nBut the Birthday Paradox is just one example of when our intuitive statistical sense is dead wrong.\n\n-\nHere is a cool video from Numberphile about the Birthday problem. youtube.com/watch?v=a2ey9a70yY0 \u2013\u00a0Rik Jan 14 '14 at 12:04\nThe birthday problem is about the probability of having at least one collision. It says very little about the number of different values found in a large number of independent samples (as in \"how many different birthdays on average in a random group of $365$ people\"). \u2013\u00a0Marc van Leeuwen Jan 14 '14 at 13:27", "date": "2016-05-28 00:27:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/637664/why-does-this-not-seem-to-be-random", "openwebmath_score": 0.9246912002563477, "openwebmath_perplexity": 300.37093293238496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138099151278, "lm_q2_score": 0.8887587920192298, "lm_q1q2_score": 0.8689517446406878}}
{"url": "https://cs.stackexchange.com/questions/133276/algorithm-to-compute-the-gaps-between-a-set-of-intervals/133483#133483", "text": "# Algorithm To Compute The Gaps Between A Set of Intervals\n\nProblem\n\nGiven a set of intervals with possibly non-distinct start and end points, find all maximal gaps. A gap is defined as an interval that does not overlap with any given interval. All endpoints are integers and inclusive.\n\nFor example, given the following set of intervals:\n\n$$\\{[2,6], [1,9], [12,19]\\}$$\n\nThe set of all maximal gaps is:\n\n$$\\{[10,11]\\}$$\n\nFor the following set of intervals:\n\n$$\\{[2,6], [1,9], [3,12], [18,20]\\}$$\n\nThe set of all maximal gaps is:\n\n$$\\{[13,17]\\}$$\n\nbecause that produces the maximal gap.\n\nProposed Algorithm\n\nMy proposed algorithm (modified approach taken by John L.) to compute these gaps is:\n\n1. Order the intervals by ascending start date.\n2. Initialise an empty list gaps that will store gaps\n3. Initialise a variable, last_covered_point, to the end point of the first interval.\n4. Iterate through all intervals in the sorted order. For each interval [start, end], do the following.\n1. If start > last_covered_point + 1, add the gap, [last_covered_point + 1, start - 1] to gaps.\n2. Assign max(last_covered_point, end) to last_covered_point.\n5. Return gaps\n\nI have tested my algorithm on a few cases and it produces the correct results. But I cannot say with 100% guarantee that it works for every interval permutation and combination. Is there a way to prove this handles every permutation and combination?\n\n\u2022 @JohnL. yes the question is correct now - thanks\n\u2013\u00a0Mojo\nDec 18 '20 at 21:25\n\nThe key to prove your algorithm is correct is to find enough invariants of the loop, step 4 so that we apply use mathematical induction.\n\nLet $$I_1, I_2, \\cdots, I_n$$ denote the sorted intervals. When the algorithm has just finished processing $$I_i$$, we record the values of gaps and last_covered_point as $$\\text{gaps}_i$$ and $$\\text{last_covered_point}_i$$ respectively.\n\nLet us prove the following proposition, $$P(i)$$, for $$i=1, 2, \\cdots, n$$.\n\n$$\\text{gaps}_i$$ is the set of all maximal gaps for $$I_1, I_2, \\cdots, I_i$$ and $$\\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \\cdots, I_i$$.\n\nWhen $$i=1$$, $$\\text{gaps}_1$$ is the empty set and $$\\text{last_covered_point}_1$$ is the right endpoint of $$I_1$$. So $$P(1)$$ is correct.\n\nFor the sake of mathematical induction, assume $$P(i)$$ is correct, where $$1\\le i\\lt n$$. Let $$I_{i+1}=[s, e]$$. There are two cases.\n\n1. If $$s\\gt\\text{last_covered_point}_i+1$$, then $$\\text{gaps}_i\\cup[\\text{last_covered_point}_i +1, s-1]=\\text{gaps}_{i+1}.$$\nLet $$m$$ be any point between the start point of $$I_1$$ and the maximum of all right endpoints of $$I_1, I_2, \\cdots, I_{i+1}$$. Suppose $$m$$ not covered by any of $$I_1, I_2, \\cdots, I_{i+1}$$.\n\n\u2022 If $$m\\le\\text{last_covered_point}_i$$, the induction hypothesis says that $$m$$ is covered by some interval in $$\\text{gap}_i$$.\n\u2022 Otherwise, $$m\\gt\\text{last_covered_point}_i$$. Since $$m$$ is not covered by $$I_{i+1}$$, we know $$m. So $$m$$ is covered by $$[\\text{last_covered_point}_i +1, s-1]$$.\n\nIn both cases, $$m$$ is covered by some interval in $$\\text{gaps}_{i+1}$$. Since $$\\text{last_covered_point}_i$$ is the largest point covered by one of $$I_1, I_2, \\cdots, I_i$$ and $$s$$ is the smallest point covered by $$I_{i+1}$$, $$[\\text{last_covered_point}_i +1, s-1]$$ is a maximal gap.\n\n2. Otherwise, we have $$s\\le\\text{last_covered_point}_i$$+1. We can also verify that $$\\text{gaps}_{i+1}=\\text{gaps}_{i}$$ is the set of all maximal gaps for $$I_1, I_2, \\cdots, I_{i+1}$$.\n\nFinally, since step 4.2 says $$\\text{last_covered_point}_{i+1}=\\max(\\text{last_covered_point}_i, e)$$ and $$\\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \\cdots, I_i$$, $$\\text{last_covered_point}_{i+1}$$ is the maximum of all right endpoints of $$I_1, I_2, \\cdots, I_{i+1}$$.\n\nSo, $$P(i+1)$$ is correct. $$\\quad\\checkmark$$.\n\n\u2022 I probably didn't make my post precise enough and so I think you may have misunderstood what I meant by a gap. I have cleaned up my original post to define the problem better but I also took your example and customised it to what I think the algorithm should be. The main issue I have is that I am not sure if I have covered all possible interval permutations.\n\u2013\u00a0Mojo\nDec 17 '20 at 15:19\n\u2022 @Mojo, thanks for noticing that last_covered_point should be updated in step 4.2 as well. Dec 18 '20 at 16:30\n\u2022 @Mojo, I just wrote a proof. The proof is not completely formal, since \"maximal gap\" is given by a definition (although it is easy to define) and the case 2 is not proved in detail. However, the idea should be clear enough. Dec 20 '20 at 5:24\n\u2022 In my last comment, 'since \"maximal gap\" is given by a definition' should have been 'since \"maximal gap\" is not given by a definition'. Jan 4 at 1:35\n\nI'd like to propose a quite simple algorithm as well. The idea is this: we're going to place open and close parentheses on the number line at the boundaries of each interval. For example, for the intervals $$(1,5), (2,7), (9, 10)$$, the number line would look like this:\n\n1 2 3 4 5 6 7 8 9 10\n( (     )   )   ( )\n\n\nThen we'll just scan left to right, counting parentheses. When all the parentheses get closed, we start a gap. Note in particular that in the above diagram, we have lost the information about which close parenthesis is paired with which open parenthesis -- because it doesn't actually matter. So:\n\n1. Convert each interval $$(a,b)$$ to pairs $$(a,\\mathtt o),(b,\\mathtt c)$$. ($$\\mathtt o$$ and $$\\mathtt c$$ are for open and close, respectively.)\n2. Sort all the pairs you get from this process. (When sorting, use lexicographic ordering and $$\\mathtt o < \\mathtt c$$.)\n3. Iterate through them, keeping a counter that starts from $$0$$.\n\u2022 When you see a pair with $$\\mathtt o$$ in the second part, increment the counter.\n\u2022 When you see a pair with $$\\mathtt c$$ in the second part, decrement the counter.\n\u2022 When you decrement the counter, if that causes it to drop to $$0$$, then look at the next element of the list to decide what to do; empty list means you're done, otherwise if the next pair's first part is just one bigger than the current one's you do nothing, and in the last case you record a maximal gap between the current end point and the next open point.", "date": "2021-09-25 04:31:09", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/133276/algorithm-to-compute-the-gaps-between-a-set-of-intervals/133483#133483", "openwebmath_score": 0.7752450704574585, "openwebmath_perplexity": 658.3273163747077, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540734789343, "lm_q2_score": 0.8872045981907006, "lm_q1q2_score": 0.8688874372473039}}
{"url": "https://math.stackexchange.com/questions/2908186/why-is-xn-approx-leftnx1-4096-11-right4096", "text": "# Why is $x^n\\approx \\left(n(x^{1/4096}-1)+1\\right)^{4096}$?\n\nThere's an old-school pocket calculator trick to calculate $x^n$ on a pocket calculator, where both, $x$ and $n$ are real numbers. So, things like $\\,0.751^{3.2131}$ can be calculated, which is awesome.\n\nThis provides endless possibilities, including calculating nth roots on a simple pocket calculator.\n\nThe trick goes like this:\n\n1. Type $x$ in the calculator\n2. Take the square root twelve times\n3. Subtract one\n4. Multiply by $n$\n6. Raise the number to the 2nd power twelve times (press * and = key eleven times)\n\nExample:\n\nI want to calculate $\\sqrt[3]{20}$ which is the same as $20^{1/3}$. So $x=20$ and $n=0.3333333$. After each of the six steps, the display on the calculator will look like this:\n\n1. $\\;\\;\\;20$\n2. $\\;\\;\\;1.0007315$\n3. $\\;\\;\\;0.0007315$\n4. $\\;\\;\\;0.0002438$\n5. $\\;\\;\\;1.0002438$\n6. $\\;\\;\\;2.7136203$\n\nThe actual answer is $20^{1/3}\\approx2.7144176$. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.\n\nQuestion:\n\nSo the question is now, why does this trick work? More specifically, how do we prove that: $$x^n\\approx \\Big(n(x^{1/4096}-1)+1\\Big)^{4096}$$\n\nNote: $4096=2^{12}$.\n\nI sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.\n\nI also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:\n\n$$\\lim_{a\\to\\infty}\\Big(n(x^{1/2^a}-1)+1\\Big)^{(2^a)}=x^n$$\n\nThis, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. $a=12$ is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with $a\\to\\infty$ can be useful in proving why this trick works, however I still can't of think a formal proof for this.\n\nThank you for your time :)\n\n\u2022 Which one would you like to prove, the approximation for $a = 12$, or just the limit expression for $a \\to \\infty$? \u2013\u00a0Xiangxiang Xu Sep 7 '18 at 2:01\n\u2022 In my opinion the one with the limit with $a\\to\\infty$ is better to prove because it is more general. From there, the specific case of $a=12$ can be simpler to prove (I think) \u2013\u00a0KKZiomek Sep 7 '18 at 2:04\n\nA standard trick is to calculate the natural logarithm first to get the exponent under control:\n\n$$\\log(\\lim_{a\\to\\infty}(n(x^{1/a}-1)+1)^a)=\\lim_{a\\to\\infty}a\\log(nx^{1/a}-n+1)$$ Set $u=1/a$. We get $$\\lim_{u\\to 0}\\frac{\\log (nx^u-n+1)}{u}$$ Use L'Hopital: $$\\lim_{u\\to 0}\\frac{nx^u\\log x}{nx^u-n+1}=n\\log x=\\log x^n$$ Here we just plugged in $u=0$ to calculate the limit!\n\nSo the original limit goes to $x^n$ as desired.\n\n\u2022 I was confused first at L'Hopital step, but then I realized that I have to differentiate with respect to $u$ not $x$ :). I'm giving this answer the best even though all other answers were equally good, because if you haven't mentioned L'Hopital, I would never think of using it. Thank you for the answer. \u2013\u00a0KKZiomek Sep 7 '18 at 2:19\n\u2022 Glad to help. It's a neat problem. \u2013\u00a0Cheerful Parsnip Sep 7 '18 at 2:24\n\nFor fixed $x > 0$ and $n$, let $t = 1/2^a \\to 0$. Then we need to prove that $$\\lim_{t \\to 0} \\left( n (x^t - 1) + 1 \\right)^{1/t} = x^n.$$ In fact, we have $$\\ln \\left[\\lim_{t \\to 0} \\left( n (x^t - 1) + 1 \\right)^{1/t}\\right] = \\lim_{t \\to 0} \\frac{\\ln (1 + n(x^t - 1))}{t} = \\lim_{t \\to 0} \\frac{n(x^t - 1)}{t} = n\\ln x,$$ where the first equality follows from the continuity of $\\ln(x)$, and the second equality has used the fact that $\\ln(1 + x) \\sim x$ when $x \\to 0$.\n\nIf $x$ (actually $\\ln x$) is relatively small, then $x^{1/4096}=e^{(\\ln x)/4096}\\approx1+(\\ln x)/4096$, in which case\n\n$$n(x^{1/4096}-1)+1)\\approx1+{n\\ln x\\over4096}$$\n\nIf $n$ is also relatively small, then\n\n$$(n(x^{1/4096}-1)+1)^{4096}\\approx\\left(1+{n\\ln x\\over4096}\\right)^{4096}\\approx e^{n\\ln x}=x^n$$\n\nRemark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,\n\n$$\\left({1\\over3}(20^{1/4096}-1)+1\\right)^{4096}=2.7150785662\\ldots$$\n\nis more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\\ldots$, both of which are too large. Very curiously, if you average the two calculator results, you get\n\n$${2.7136203+2.7152613\\over2}=2.7144408$$\n\nwhich is quite close to the true value!", "date": "2019-07-17 06:55:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2908186/why-is-xn-approx-leftnx1-4096-11-right4096", "openwebmath_score": 0.941208004951477, "openwebmath_perplexity": 223.06104412704286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464422083111, "lm_q2_score": 0.8902942341267435, "lm_q1q2_score": 0.8688794903145683}}
{"url": "https://mathsedideas.blogspot.com/2020/12/three-card-colours.html", "text": "## Wednesday, 2 December 2020\n\n### On the Colour of the Third Card\n\nThree cards are dealt from a normal deck.\u00a0 You don't see them.\u00a0 You are told that the first two are the same colour, but not what colour they are.\u00a0 What is the probability that the next card is the same colour?\n\n$\\frac{1}{2},\\;\\frac{1}{4},\\;\\frac{{12}}{{25}},\\;or\\;\\frac{4}{{17}}\\;?$\n\nThis beautiful, onthefaceofit innocuous little problem comes from the great Martin Gardner's 'Modelling Mathematics with Playing Cards'.\u00a0 It's a problem that never fails to invoke heated discussion and vehement argument when I share it with students to play with.\u00a0 The four solutions and reasoning sketched out below are those typically proffered (and invariably staunchly defended) by students.\u00a0 Which solution would you go with, and why?\u00a0 Do you have a different solution?\u00a0 And what argument would you give to those who firmly hold the solution to be one of the others to show them that they are wrong and you are right?\u00a0 (Note that Gardner's solution was solution 2 [1])\n\n## Solution 1\n\nThe last card could either be 1) the same colour by being red, or 2) a different colour by being red, or 3) the same colour by being black, or 4) a different colour by being black.\u00a0 So there are four possibilities and two of these result in the final card being the same colour as the previous two.\u00a0 So:\n\n$\\frac{2}{4} = \\frac{1}{2}$\n\n## Solution 2\n\nThere are eight ways that the arrangement of the colours of the three cards can occur, namely RRR, RRB, RBR, BRR, RBB, BRB, BBR, or BBB (where R = Red and B = Black), and two of these arrangements have the final card as the same colour as the previous two.\u00a0 So:\n\n$\\frac{2}{8} = \\frac{1}{4}$\n\n## Solution 3\n\nAfter the first two cards, fifty cards remain in the deck.\u00a0 If the first two cards are both red, there remain twenty-four cards that are red.\u00a0 Similarly, if the first two cards are both black, there remain twenty-four cards that are black.\u00a0 So:\n\n$\\frac{24}{50} = \\frac{12}{25}$\n\n## Solution 4\n\nFor the last card to be the same colour as the first two, the colours of all three cards are either RRR or BBB.\u00a0 You have fifty-two cards to choose from for your first card, and twenty-six of these are red and twenty-six are black.\u00a0 After taking the first card from the deck, you have fifty-one cards left to choose from for your second card, and assuming that the first card was red, twenty-five of the fifty-one cards left are also red.\u00a0 After taking the second card from the deck, you have fifty cards left to choose from for your third and final card, and assuming that the first two cards were red, twenty-four of the fifty cards left are also red.\u00a0 The same would be true if the cards pulled from the deck were black.\u00a0 So:\n\n$\\left( {\\frac{{26}}{{52}} \\times \\frac{{25}}{{51}} \\times \\frac{{24}}{{50}}} \\right) \\times 2 = \\frac{{31200}}{{132600}} = \\frac{4}{{17}}$", "date": "2021-05-13 05:21:36", "meta": {"domain": "blogspot.com", "url": "https://mathsedideas.blogspot.com/2020/12/three-card-colours.html", "openwebmath_score": 0.4942336082458496, "openwebmath_perplexity": 946.6124154674533, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370156, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8688719517647572}}
{"url": "https://gingerhillcreations.com/jessica-smith-scxean/ec0e3c-how-to-prove-cardinality-of-sets", "text": "Here we need to talk about cardinality of a set, which is basically the size of the set. there are $10$ people with white shirts and $8$ people with red shirts; $4$ people have black shoes and white shirts; $3$ people have black shoes and red shirts; the total number of people with white or red shirts or black shoes is $21$. that the cardinality of a set is the number of elements it contains. Edition 1st Edition. $$|W \\cap B|=4$$ In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B \u2013 For finite sets, cardinality is the number of elements \u2013 There is a bijection from n-element set A to {1, 2, 3, \u2026, n} Following Ernie Croot's slides of students who play both foot ball and cricket = 25, No. In Section 5.1, we defined the cardinality of a finite set $$A$$, denoted by card($$A$$), to be the number of elements in the set $$A$$. Total number of elements related to both B & C. Total number of elements related to both (B & C) only. Two sets are equal if and only if they have precisely the same elements. For example, let A \u00a0= \u00a0{ -2, 0, 3, 7, 9, 11, 13 }, Here, n(A) stands for cardinality of the set A. To prove there exists a bijection between to sets X and Y, there are 2 ways: 1. find an explicit bijection between the two sets and prove it is bijective (prove it is injective and surjective) 2. Pages 5. eBook ISBN 9780429324819. It turns out we need to distinguish between two types of infinite sets, Any superset of an uncountable set is uncountable. then talk about infinite sets. The cardinality of a set is roughly the number of elements in a set. (Hint: Use a standard calculus function to establish a bijection with R.) 2. Math 127: In nite Cardinality Mary Radcli e 1 De nitions Recall that when we de ned niteness, we used the notion of bijection to de ne the size of a nite set. of students who play cricket only = 10, No. where one type is significantly \"larger\" than the other. The cardinality of the set of all natural numbers is denoted by . $$\\>\\>\\>\\>\\>\\>\\>+\\sum_{i < j < k}\\left|A_i\\cap A_j\\cap A_k\\right|-\\ \\cdots\\ + \\left(-1\\right)^{n+1} \\left|A_1\\cap\\cdots\\cap A_n\\right|.$$, $= |W| + |R| + |B|- |W \\cap R| - |W \\cap B| - |R \\cap B| + |W \\cap R \\cap B|$. ... to make the argument more concrete, here we provide some useful results that help us prove if a set is countable or not. Hence these sets have the same cardinality. The cardinality of a set is denoted by $|A|$. of students who play foot ball only = 28, No. $$|W|=10$$ The intuition behind this theorem is the following: If a set is countable, then any \"smaller\" set $\\mathbb{N}, \\mathbb{Z}, \\mathbb{Q}$, and any of their subsets are countable. c) $(0,\\infty)$, $\\R$ d) $(0,1)$, $\\R$ Ex 4.7.4 Show that $\\Q$ is countably infinite. If $B \\subset A$ and $A$ is countable, by the first part of the theorem $B$ is also a countable the idea of comparing the cardinality of sets based on the nature of functions that can be possibly de ned from one set to another. Ex 4.7.3 Show that the following sets of real numbers have the same cardinality: a) $(0,1)$, $(1, \\infty)$ b) $(1,\\infty)$, $(0,\\infty)$. countable, we can write a proof, we can argue in the following way. So, the total number of students in the group is 100. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,\u2026,n-1} to the set. If you are less interested in proofs, you may decide to skip them. The set of all real numbers in the interval (0;1). Before we start developing theorems, let\u2019s get some examples working with the de nition of nite sets. In mathematics, the cardinality of a set is a measure of the \"number of elements\" of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. In mathematics, a set is a well-defined collection of distinct elements or members. Two finite sets are considered to be of the same size if they have equal numbers of elements. then by removing the elements in the list that are not in $B$, we can obtain a list for $B$, In addition, we say that the empty set has cardinality 0 (or cardinal number 0), and we write $$\\text{card}(\\emptyset) = 0$$. As far as applied probability Discrete Mathematics and Its Applications, Seventh Edition answers to Chapter 2 - Section 2.5 - Cardinality of Sets - Exercises - Page 176 12 including work step by step written by community members like you. $$|W \\cup B \\cup R|=21.$$ set whose elements are obtained by multiplying each element of Z by k.) The function f : N !Z de ned by f(n) = ( 1)nbn=2cis a 1-1 corre-spondence between the set of natural numbers and the set of integers (prove it!). The Math Sorcerer 19,653 views. If $A$ has only a finite number of elements, its cardinality is simply the Here is a simple guideline for deciding whether a set is countable or not. the inclusion-exclusion principle we obtain. and how to prove set S is a infinity set. you can never provide a list in the form of $\\{a_1, a_2, a_3,\\cdots\\}$ that contains all the $|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|$. \\mathbb {R}. thus $B$ is countable. Examples of Sets with Equal Cardinalities. This will come in handy, when we consider the cardinality of infinite sets in the next section. 4 CHAPTER 7. Cardinality of a set S, denoted by |S|, is the number of elements of the set. Since each $A_i$ is countable we can Definition. Cantor introduced a new de\u30fb\uff3dition for the \u7ab6\u5fadize\u7ab6\u30fbof a set which we call cardinality. Textbook Authors: Rosen, Kenneth, ISBN-10: 0073383090, ISBN-13: 978-0-07338-309-5, Publisher: McGraw-Hill Education The difference between the two types is We will say that any sets A and B have the same cardinality, and write jAj= jBj, if A and B can be put into 1-1 correspondence. but \"bigger\" sets such as $\\mathbb{R}$ are called uncountable. You already know how to take the induction step because you know how the case of two sets behaves. \\mathbb {N} To formulate this notion of size without reference to the natural numbers, one might declare two finite sets A A A and B B B to have the same cardinality if and only if there exists a bijection A \u2192 B A \\to B A \u2192 B. Cardinality Recall (from our first lecture!) (useful to prove a set is finite) \u2022 A set is infinite when there is an injection, f:A\u00c6A, such that f(A) is \u2026 When an invertible function from a set to \\Z_n where m\\in\\N is given the cardinality of the set immediately follows from the definition. This important fact is commonly known ... aged to prove that two very different sets are actually the same size\u2014even though we don\u2019t know exactly how big either one is. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,\u2026,n-1} to the set. Cardinality The cardinality of a set is roughly the number of elements in a set. By Gove Effinger, Gary L. Mullen. (a) Let S and T be sets. Furthermore, we designate the cardinality of countably infinite sets as \u21350 (\"aleph null\"). 12:14. Here we need to talk about cardinality of a set, which is basically the size of the set. In particular, one type is called countable, If A and B are disjoint sets, n(A n B) \u00a0= \u00a00, n(A u B u C) \u00a0= \u00a0n(A) + n(B) + n(C) - n(A n B) - n(B n C) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0- n(A n C) + n(A n B n C), n(A n B) \u00a0= 0, n(B n C) \u00a0= \u00a00, n(A n C) \u00a0= \u00a00, n(A n B n C) \u00a0= \u00a00, = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC). Cardinality of a set: Discrete Math: Nov 17, 2019: Proving the Cardinality of 2 finite sets: Discrete Math: Feb 16, 2017: Cardinality of a total order on an infinite set: Advanced Math Topics: Jan 18, 2017: cardinality of a set: Discrete Math: Jun 1, 2016 One important type of cardinality is called \u201ccountably infinite.\u201d A set A is considered to be countably infinite if a bijection exists between A and the natural numbers \u2115. Countably infinite sets are said to have a cardinality of \u05d0 o (pronounced \u201caleph naught\u201d). $$|A \\cup B |=|A|+|B|-|A \\cap B|.$$ Solution. Before discussing thus by subtracting it from $|A|+|B|$, we obtain the number of elements in $|A \\cup B |$, (you can The above arguments can be repeated for any set $C$ in the form of However, to make the argument In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. (Assume that each student in the group plays at least one game). Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively. We first discuss cardinality for finite sets and (Hint: you can arrange $\\Q^+$ in a sequence; use this to arrange $\\Q$ into a sequence.) set which is a contradiction. of students who play both (foot ball & hockey) only = 12, No. To see this, note that when we add $|A|$ and $|B|$, we are counting the elements in $|A \\cap B|$ twice, Because of the symmetyofthissituation,wesaythatA and B can be put into 1-1 correspondence. That is often difficult, however. If $A$ is a finite set, then $|B|\\leq |A| < \\infty$, Then,byPropositionsF12andF13intheFunctions section,fis invertible andf\u22121is a 1-1 correspondence fromBtoA. Consider sets A and B.By a transformation or a mapping from A to B we mean any subset T of the Cartesian product A\u00d7B that satisfies the following condition: . Question: Prove that N(all natural numbers) and Z(all integers) have the same cardinality. To prove that a given in nite set X \u2026 Theorem. How to prove that all maximal independent sets of a matroid have the same cardinality. We can, however, try to match up the elements of two in\ufb01nite sets A and B one by one. As seen, the symbol for the cardinality of a set resembles the absolute value symbol \u2014 a variable sandwiched between two vertical lines. Provided a matroid is a 2-tuple (M,J ) where M is a finite set and J is a family of some of the subsets of M satisfying the following properties: If A is subset of B and B belongs to J , then A belongs to J , =\u00a0 n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC), n(FuHuC)\u00a0 =\u00a0 65 + 45 + 42 -20 - 25 - 15 + 8. This function is bijective. When it ... prove the corollary one only has to observe that a function with a \u201cright inverse\u201d is the \u201cleft inverse\u201d of that function and vice versa. No. Total number of students in the group is\u00a0n(FuHuC). $\\mathbb{Z}=\\{0,1,-1,2,-2,3,-3,\\cdots\\}$. In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. This establishes a one-to-one correspondence between the set of primes and the set of natural numbers, so they have the same cardinality. A = \\left\\ { {1,2,3,4,5} \\right\\}, \\Rightarrow \\left| A \\right| = 5. \u2026 uncountable set (to prove uncountability). To this final end, I will apply the Cantor-Bernstein Theorem: (The two sets (0, 1) and [0, 1] have the same cardinality if we can find 1-1 mappings from (0, 1) to [0, 1] and vice versa.) Also known as the cardinality, the number of disti n ct elements within a set provides a foundational jump-off point for further, richer analysis of a given set. The examples are clear, except for perhaps the last row, which highlights the fact that only unique elements within a set contribute to the cardinality. I presume you have sent this A2A to me following the most recent instalment of our ongoing debate regarding the ontological nature and resultant enumeration of Zero. To be precise, here is the definition. Find the total number of students in the group. A set A is countably infinite if and only if set A has the same cardinality as N (the natural numbers). The proof of this theorem is very similar to the previous theorem. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. Now that we know about functions and bijections, we can define this concept more formally and more rigorously. The idea is exactly the same as before. f:A \u2192 Bbea1-1correspondence. while the other is called uncountable. However, I am stuck in proving it since there are more than one \"1\", \"01\" = \"1\", same as other numbers. Fix m 2N. $$\\biggl|\\bigcup_{i=1}^n A_i\\biggr|=\\sum_{i=1}^n\\left|A_i\\right|-\\sum_{i < j}\\left|A_i\\cap A_j\\right|$$ but you cannot list the elements in an uncountable set. Imprint CRC Press. 1. if it is a finite set, $\\mid A \\mid < \\infty$; or. I could not prove that cardinality is well defined, i.e. Thus according to De\ufb01nition 2.3.1, the sets N and Z have the same cardinality. When A and B have the same cardinality, we write jAj= jBj. Thus, any set in this form is countable. of students who play both (hockey & cricket) only = 7, No. A set is an infinite set provided that it is not a finite set. of students who play hockey only = 18, No. The two sets A = {1,2,3} and B = {a,b,c} thus have the cardinality since we can match up the elements of the two sets in such a way that each element in each set is matched with exactly one element in the other set. For example, if $A=\\{2,4,6,8,10\\}$, then $|A|=5$. Math 131 Fall 2018 092118 Cardinality - Duration: 47:53. For in nite sets, this strategy doesn\u2019t quite work. Let X m = fq 2Q j0 q 1; and mq 2Zg. For example, a consequence of this is that the set of rational numbers $\\mathbb{Q}$ is countable. Since $A$ and $B$ are Cardinality of a set is a measure of the number of elements in the set. If A can be put into 1-1 correspondence with a subset of B (that is, there is a 1-1 case the set is said to be countably infinite. I can tell that two sets have the same number of elements by trying to pair the elements up. (useful to prove a set is finite) \u2022 A set is infinite when there \u2026 (b) A set S is finite if it is empty, or if there is a bijection for some integer . of students who play both foot ball &\u00a0 hockey = 20, No. n(AuB)\u00a0 =\u00a0 Total number of elements related to any of the two events A & B. n(AuBuC)\u00a0 =\u00a0 Total number of elements related to any of the three events A, B & C. n(A)\u00a0 =\u00a0 Total number of elements related to\u00a0 A. n(B)\u00a0 =\u00a0 Total number of elements related to\u00a0 B. n(C)\u00a0 =\u00a0 Total number of elements related to\u00a0 C. Total number of elements related to A only. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. We first discuss cardinality for finite sets and then talk about infinite sets. A useful application of cardinality is the following result. It would be a good exercise for you to try to prove this to yourself now. If S is a set, we denote its cardinality by |S|. Cantor showed that not all in\u30fb\uff3dite sets are created equal \u7ab6\u30fbhis de\u30fb\uff3dition allows us to distinguish betweencountable and uncountable in\u30fb\uff3dite sets. Set S is a set consisting of all string of one or more a or b such as \"a, b, ab, ba, abb, bba...\" and how to prove set S is a infinity set. For finite sets, cardinalities are natural numbers: |{1, 2, 3}| = 3 |{100, 200}| = 2 For infinite sets, we introduced infinite cardinals to denote the size of sets: $$C=\\bigcup_i \\bigcup_j \\{ a_{ij} \\},$$ The cardinality of a set is denoted by $|A|$. Figure 1.13 shows one possible ordering. We have been able to create a list that contains all the elements in $\\bigcup_{i} A_i$, so this correspondence with natural numbers $\\mathbb{N}$. is concerned, this guideline should be sufficient for most cases. n(FnH)\u00a0 =\u00a0 20, n(FnC)\u00a0 =\u00a0 25, n(HnC)\u00a0 =\u00a0 15. The set whose elements are each and each and every of the subsets is the ability set. The above theorems confirm that sets such as $\\mathbb{N}, \\mathbb{Z}, \\mathbb{Q}$ and their Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. 11 Cardinality Rules ... two sets, then the sets have the same size. Maybe this is not so surprising, because N and Z have a strong geometric resemblance as sets of points on the number line. For in nite sets, this strategy doesn\u2019t quite work. Any set which is not finite is infinite. If $A_1, A_2,\\cdots$ is a list of countable sets, then the set $\\bigcup_{i} A_i=A_1 \\cup A_2 \\cup A_3\\cdots$ Discrete Mathematics - Cardinality 17-16 More Countable Sets (cntd) The fact that you can list the elements of a countably infinite set means that the set can be put in one-to-one If $A$ and $B$ are countable, then $A \\times B$ is also countable. should also be countable, so a subset of a countable set should be countable as well. We can say that set A and set B both have a cardinality of 3. set is countable. S and T have the same cardinality if there is a bijection f from S to T. Notation: means that S and T have the same cardinality. respectively. The cardinality of a set is denoted by $|A|$. Thus by applying there'll be 2^3 = 8 elements contained in the ability set. Definition of cardinality. forall s : fset_expr (A:=A), exists n, (cardinality_fset s n /\\ forall s' n', eq_fset s s' -> cardinality_fset s' n' -> n' = n). 1. On the other hand, you cannot list the elements in $\\mathbb{R}$, DOI link for Cardinality of Sets. so it is an uncountable set. | A | = | N | = \u21350. Mappings, cardinality. The number of elements in a set is called the cardinality of the set. This poses few di\ufb03culties with \ufb01nite sets, but in\ufb01nite sets require some care. The sets A and B have the same cardinality if and only if there is a one-to-one correspondence from A to B. $$B = \\{b_1, b_2, b_3, \\cdots \\}.$$ and Itiseasytoseethatanytwo\ufb01nitesetswiththesamenumberofelementscanbeput into1-1correspondence. For example, you can write. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B \u2013 For finite sets, cardinality is the number of elements \u2013 There is a bijection from n-element set A to {1, 2, 3, \u2026, n} Following Ernie Croot's slides The cardinality of a finite set is the number of elements in the set. if you need any other stuff in math, please use our google custom search here. Any subset of a countable set is countable. If you are less interested in proofs, you may decide to skip them. Total number of elements related to C only. refer to Figure 1.16 in Problem 2 to see this pictorially). I can tell that two sets have the same number of elements by trying to pair the elements up. Click here to navigate to parent product. However, as we mentioned, intervals in $\\mathbb{R}$ are uncountable. Introduction to the Cardinality of Sets and a Countability Proof - Duration: 12:14. Any set containing an interval on the real line such as $[a,b], (a,b], [a,b),$ or $(a,b)$, But as soon as we \ufb01gure out the size Arrange $\\Q^+$ in a set Ais nite, and determine its cardinality is said! Elements of the set: \u2026 cardinality of the symmetyofthissituation, wesaythatA and B have the same cardinality as (! Countable set and $B$ is countable of nite sets $a. If it is a measure of the set, then$ |B|\\leq |A| < \\infty $, then the \\mathbb. The bijection cantor showed that not all in\u30fb\uff3dite sets has an infinite number of students play. = 8 elements contained in the following way induction step because you know the! You 've found the bijection of a can be put into 1-1 correspondence from definition... A & B ) a set is an infinite number of elements \\mathbb { Q }$ are.... We need to talk about infinite sets in the ability set from the diagram. The number of elements the induction step because you know how the case of two have! 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Induction step because you know how to take the induction step because how to prove cardinality of sets! To B must be a bijection between them B both have a strong geometric resemblance as of. Are combined using operations on sets, this strategy doesn \u2019 t work... Are each and every of the set of positive integers is called countable, while other... = | N | and uncountable in\u30fb\uff3dite sets are considered to be countable or not useful application cardinality. From a to B must be a good exercise for you to try to prove this to yourself.... Set in how to prove cardinality of sets form is countable: use a standard calculus function to establish bijection. A Countability proof - Duration: 12:14 set \u2026 a useful application of is. Shown in Figure 1.11 to prove that cardinality is \u221e as \u21350 (  null... Let $a$ has only a finite set, usually denoted by @ 0 ( )! 1, 2, 3, 4, 5 }, \\mathbb { }! Their subsets are countable set \u2026 a useful application of cardinality is a simple guideline deciding!", "date": "2021-02-26 21:13:04", "meta": {"domain": "gingerhillcreations.com", "url": "https://gingerhillcreations.com/jessica-smith-scxean/ec0e3c-how-to-prove-cardinality-of-sets", "openwebmath_score": 0.8577983379364014, "openwebmath_perplexity": 368.5193643402489, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9883127409715955, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.868871940560683}}
{"url": "https://it.mathworks.com/help/symbolic/sym.jacobian.html", "text": "# jacobian\n\nJacobian matrix\n\n## Description\n\nexample\n\njacobian(f,v) computes the Jacobian matrix of f with respect to v. The (i,j) element of the result is $\\frac{\\partial f\\left(i\\right)}{\\partial \\text{v}\\left(j\\right)}$.\n\n## Examples\n\ncollapse all\n\nThe Jacobian of a vector function is a matrix of the partial derivatives of that function.\n\nCompute the Jacobian matrix of [x*y*z,y^2,x + z] with respect to [x,y,z].\n\nsyms x y z\njacobian([x*y*z,y^2,x + z],[x,y,z])\nans =\n\n$\\left(\\begin{array}{ccc}y z& x z& x y\\\\ 0& 2 y& 0\\\\ 1& 0& 1\\end{array}\\right)$\n\nNow, compute the Jacobian of [x*y*z,y^2,x + z] with respect to [x;y;z].\n\njacobian([x*y*z,y^2,x + z], [x;y;z])\nans =\n\n$\\left(\\begin{array}{ccc}y z& x z& x y\\\\ 0& 2 y& 0\\\\ 1& 0& 1\\end{array}\\right)$\n\nThe Jacobian matrix is invariant to the orientation of the vector in the second input position.\n\nThe Jacobian of a scalar function is the transpose of its gradient.\n\nCompute the Jacobian of 2*x + 3*y + 4*z with respect to [x,y,z].\n\nsyms x y z\njacobian(2*x + 3*y + 4*z,[x,y,z])\nans =\u00a0$\\left(\\begin{array}{ccc}2& 3& 4\\end{array}\\right)$\n\nNow, compute the gradient of the same expression.\n\nans =\n\n$\\left(\\begin{array}{c}2\\\\ 3\\\\ 4\\end{array}\\right)$\n\nThe Jacobian of a function with respect to a scalar is the first derivative of that function. For a vector function, the Jacobian with respect to a scalar is a vector of the first derivatives.\n\nCompute the Jacobian of [x^2*y,x*sin(y)] with respect to x.\n\nsyms x y\njacobian([x^2*y,x*sin(y)],x)\nans =\n\n$\\left(\\begin{array}{c}2 x y\\\\ \\mathrm{sin}\\left(y\\right)\\end{array}\\right)$\n\nNow, compute the derivatives.\n\ndiff([x^2*y,x*sin(y)],x)\nans =\u00a0$\\left(\\begin{array}{cc}2 x y& \\mathrm{sin}\\left(y\\right)\\end{array}\\right)$\n\nSpecify polar coordinates $r\\left(t\\right)$, $\\varphi \\left(t\\right)$, and $\\theta \\left(t\\right)$ that are functions of time.\n\nsyms r(t) phi(t) theta(t)\n\nDefine the coordinate transformation form spherical coordinates to Cartesian coordinates.\n\nR = [r*sin(phi)*cos(theta), r*sin(phi)*sin(theta), r*cos(phi)]\nR(t) =\u00a0$\\left(\\begin{array}{ccc}\\mathrm{cos}\\left(\\theta \\left(t\\right)\\right) \\mathrm{sin}\\left(\\varphi \\left(t\\right)\\right) r\\left(t\\right)& \\mathrm{sin}\\left(\\varphi \\left(t\\right)\\right) \\mathrm{sin}\\left(\\theta \\left(t\\right)\\right) r\\left(t\\right)& \\mathrm{cos}\\left(\\varphi \\left(t\\right)\\right) r\\left(t\\right)\\end{array}\\right)$\n\nFind the Jacobian of the coordinate change from spherical coordinates to Cartesian coordinates.\n\njacobian(R,[r,phi,theta])\nans(t) =\n\n$\\left(\\begin{array}{ccc}\\mathrm{cos}\\left(\\theta \\left(t\\right)\\right) \\mathrm{sin}\\left(\\varphi \\left(t\\right)\\right)& \\mathrm{cos}\\left(\\varphi \\left(t\\right)\\right) \\mathrm{cos}\\left(\\theta \\left(t\\right)\\right) r\\left(t\\right)& -\\mathrm{sin}\\left(\\varphi \\left(t\\right)\\right) \\mathrm{sin}\\left(\\theta \\left(t\\right)\\right) r\\left(t\\right)\\\\ \\mathrm{sin}\\left(\\varphi \\left(t\\right)\\right) \\mathrm{sin}\\left(\\theta \\left(t\\right)\\right)& \\mathrm{cos}\\left(\\varphi \\left(t\\right)\\right) \\mathrm{sin}\\left(\\theta \\left(t\\right)\\right) r\\left(t\\right)& \\mathrm{cos}\\left(\\theta \\left(t\\right)\\right) \\mathrm{sin}\\left(\\varphi \\left(t\\right)\\right) r\\left(t\\right)\\\\ \\mathrm{cos}\\left(\\varphi \\left(t\\right)\\right)& -\\mathrm{sin}\\left(\\varphi \\left(t\\right)\\right) r\\left(t\\right)& 0\\end{array}\\right)$\n\n## Input Arguments\n\ncollapse all\n\nScalar or vector function, specified as a symbolic expression, function, or vector. If f is a scalar, then the Jacobian matrix of f is the transposed gradient of f.\n\nVector of variables or functions with respect to which you compute Jacobian, specified as a symbolic variable, symbolic function, or vector of symbolic variables. If v is a scalar, then the result is equal to the transpose of diff(f,v). If v is an empty symbolic object, such as sym([]), then jacobian returns an empty symbolic object.\n\ncollapse all\n\n### Jacobian Matrix\n\nThe Jacobian matrix of the vector function f = (f1(x1,...,xn),...,fn(x1,...,xn)) is the matrix of the derivatives of f:\n\n$J\\left({x}_{1},\\dots {x}_{n}\\right)=\\left[\\begin{array}{ccc}\\frac{\\partial {f}_{1}}{\\partial {x}_{1}}& \\cdots & \\frac{\\partial {f}_{1}}{\\partial {x}_{n}}\\\\ \u22ee& \\ddots & \u22ee\\\\ \\frac{\\partial {f}_{n}}{\\partial {x}_{1}}& \\cdots & \\frac{\\partial {f}_{n}}{\\partial {x}_{n}}\\end{array}\\right]$\n\n## Version History\n\nIntroduced before R2006a", "date": "2022-08-08 03:41:14", "meta": {"domain": "mathworks.com", "url": "https://it.mathworks.com/help/symbolic/sym.jacobian.html", "openwebmath_score": 0.9302873015403748, "openwebmath_perplexity": 1376.553835951242, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620044904359, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8688349952246257}}
{"url": "https://gmatclub.com/forum/in-how-many-different-ways-can-the-letters-of-the-word-mississippi-be-216328.html", "text": "It is currently 17 Jan 2018, 10:16\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# In how many different ways can the letters of the word MISSISSIPPI be\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43312\n\nKudos [?]: 139296 [0], given: 12783\n\nIn how many different ways can the letters of the word MISSISSIPPI be\u00a0[#permalink]\n\n### Show Tags\n\n08 Apr 2016, 01:47\nExpert's post\n5\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n71% (01:29) correct 29% (02:47) wrong based on 126 sessions\n\n### HideShow timer Statistics\n\nIn how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?\n\nA. 48\nB. 144\nC. 210\nD. 420\nE. 840\n[Reveal] Spoiler: OA\n\n_________________\n\nKudos [?]: 139296 [0], given: 12783\n\nVerbal Forum Moderator\nStatus: Greatness begins beyond your comfort zone\nJoined: 08 Dec 2013\nPosts: 1889\n\nKudos [?]: 1128 [1], given: 93\n\nLocation: India\nConcentration: General Management, Strategy\nSchools: Kelley '20, ISB '19\nGPA: 3.2\nWE: Information Technology (Consulting)\nRe: In how many different ways can the letters of the word MISSISSIPPI be\u00a0[#permalink]\n\n### Show Tags\n\n08 Apr 2016, 20:21\n1\nKUDOS\n2\nThis post was\nBOOKMARKED\nThe word MISSISSIPPI has 4 I , 4S , 2P and M .\nThe only vowel present in the word is I and we have to consider 4 I's as a single group\n\nNumber of different ways = 8!/(4!*2!)\n=(8*7*6*5)/2\n=840\n_________________\n\nWhen everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford\nThe Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long\n+1 Kudos if you find this post helpful\n\nKudos [?]: 1128 [1], given: 93\n\nDirector\nJoined: 05 Mar 2015\nPosts: 962\n\nKudos [?]: 310 [1], given: 41\n\nRe: In how many different ways can the letters of the word MISSISSIPPI be\u00a0[#permalink]\n\n### Show Tags\n\n26 Jun 2017, 11:18\n1\nKUDOS\nBunuel wrote:\nIn how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?\n\nA. 48\nB. 144\nC. 210\nD. 420\nE. 840\n\nvowel (I) together can be arranged with other 7 letters in 8! ways\n4 no. vowels themselves can be arranged in 4! ways\n\ndifferent ways = 8!*4!/( 4!*4!*2!) = 840---(4I's , 4 S's , 2 P's)\n\nAns E\n\nKudos [?]: 310 [1], given: 41\n\nManhattan Prep Instructor\nJoined: 04 Dec 2015\nPosts: 447\n\nKudos [?]: 293 [2], given: 68\n\nGMAT 1: 790 Q51 V49\nGRE 1: 340 Q170 V170\nRe: In how many different ways can the letters of the word MISSISSIPPI be\u00a0[#permalink]\n\n### Show Tags\n\n26 Jun 2017, 12:18\n2\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nBunuel wrote:\nIn how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?\n\nA. 48\nB. 144\nC. 210\nD. 420\nE. 840\n\nOoh, a combinatorics problem with some fairly large numbers. The best technique for these is to mentally break them down into simpler problems. Otherwise, the solutions tend to sort of look like magic tricks - sure, you can use the formula, but how are you supposed to know to use that formula, and not one of the many similar-looking ones?\n\nThere are four vowels, and they're all the same. Let's set those vowels aside: (IIII)\n\nNow we have the letters MSSSSPP. How many ways can just those letters be arranged? Well, if they were all different, we could arrange them in 7x6x5x4x3x2x1 = 7! ways. However, they aren't all different. For instance, four of the letters are (S). I'm going to color them to demonstrate why that matters:\n\none arrangement: MSSSSPP\na 'different' arrangment: MSSSSPP\n\nWe counted both of those arrangements, but we actually don't want to. All of the Ss look the same - they aren't actually different colors - so we want to make those arrangements the same. Because there are 4x3x2x1 = 24 ways to order the different Ss, we want every set of 24 arrangements where the Ss are in the same place, to just count as 1 arrangement, instead. So, we can divide out the extra possibilities by dividing our total by 24 (or 4!):\n\n7!/4!\n\nDo the same thing with the two Ps. We still have twice as many arrangements as we need, since we've counted as if the two Ps were different, but they're actually the same. So, divide the total by 2:\n\n7!/(4! x 2)\n\nNow we have to put the (IIII) letters back in. They all have to go together. Start by looking at one of the arrangements of the other letters: PMSSPSS. Where can the four Is go? There are 8 places where we can put them:\n\nIIIIPMSSPSS\nPIIIIMSSPSS\nPMIIIISSPSS\nPMSIIIISPSS\n\netc.\n\nSo, for each arrangement, we have to multiply by 8, to account for the eight possible ways to put the vowels back in. Here's the final answer:\n\n(8 x 7!) / (4! x 2) = (8 x 7 x 6 x 5 x 4 x 3 x 2) / (4 x 3 x 2 x 2) = 4 x 7 x 6 x 5 = 4 x 210 = 840.\n_________________\n\nChelsey Cooley | Manhattan Prep Instructor | Seattle and Online\n\nMy upcoming GMAT trial classes | GMAT blog archive\n\nKudos [?]: 293 [2], given: 68\n\nSVP\nJoined: 11 Sep 2015\nPosts: 1978\n\nKudos [?]: 2862 [1], given: 364\n\nRe: In how many different ways can the letters of the word MISSISSIPPI be\u00a0[#permalink]\n\n### Show Tags\n\n26 Jun 2017, 12:40\n1\nKUDOS\nExpert's post\nTop Contributor\nBunuel wrote:\nIn how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?\n\nA. 48\nB. 144\nC. 210\nD. 420\nE. 840\n\nWhen we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:\n\nIf there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]\n\nSo, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:\nThere are 11 letters in total\nThere are 4 identical I's\nThere are 4 identical S's\nThere are 2 identical P's\nSo, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]\n\n-------NOW ONTO THE QUESTION!!!-----------------\n\nFirst \"glue\" the 4 I's together to create ONE character: IIII (this ensures that they stay together)\nSo, basically, we must determine the number of arrangements of M, S, S, S, S, P, P and IIII\nThere are 8 characters in total\nThere are 4 identical S's\nThere are 2 identical P's\nSo, the total number of possible arrangements = 8!/[(4!)(2!)]\n= 840\n\n[Reveal] Spoiler:\nE\n\nRELATED VIDEO\n\n_________________\n\nBrent Hanneson \u2013 Founder of gmatprepnow.com\n\nKudos [?]: 2862 [1], given: 364\n\nIntern\nJoined: 17 Nov 2016\nPosts: 23\n\nKudos [?]: [0], given: 7\n\nRe: In how many different ways can the letters of the word MISSISSIPPI be\u00a0[#permalink]\n\n### Show Tags\n\n04 Dec 2017, 08:31\n\nStage 1: Lets place the letters without the 4I's:\nMSSSSPP\n7!/(4!2!)= 105 ways\n\nStage 2: Now lets place the 4I's\n-m-s-s-s-s-p-p-\nWe can place th4 4I's in 8 ways\n\nTotal ways= 105*8= 840 ways.\n\nIs this a correct approach?\n\nKudos [?]: [0], given: 7\n\nIntern\nJoined: 25 May 2017\nPosts: 9\n\nKudos [?]: 3 [0], given: 0\n\nGMAT 1: 750 Q49 V44\nRe: In how many different ways can the letters of the word MISSISSIPPI be\u00a0[#permalink]\n\n### Show Tags\n\n04 Dec 2017, 09:08\nRearranging letters has a simple formula:\n\nTotal number of letters\n*DIVIDED BY*\n(2)! for every letter repeated twice / (3)! for every letter repeated thrice and so on...\n\nMississippi has:\n\n4 s's\n4 i's\n2 p's\n\nTotal number of letters: 11!\n\nSo the answer is $$\\frac{11!}{4!4!2!}$$\n_________________\n\nGmatIvyPrep\nYale MBA Graduate and GMAT-Focused Tutor for 4+ years\nContact: gmativyprep@gmail.com for a consultation and private tutoring rates\n\nMaterial offered free of charge along with full end-to-end GMAT Quantitative Course:\n- High-level (700+) practice question sets for Quantitative section by topic\n- AWA writing template that helped students get a full score on the Essay\n- Complete pack of Verbal Material for SC, RC and CR\n\nKudos [?]: 3 [0], given: 0\n\nRe: In how many different ways can the letters of the word MISSISSIPPI be \u00a0 [#permalink] 04 Dec 2017, 09:08\nDisplay posts from previous: Sort by\n\n# In how many different ways can the letters of the word MISSISSIPPI be\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-01-17 18:16:35", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-how-many-different-ways-can-the-letters-of-the-word-mississippi-be-216328.html", "openwebmath_score": 0.5636717081069946, "openwebmath_perplexity": 2251.8590781932817, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8688267864276108}}
{"url": "https://math.stackexchange.com/questions/3130437/10-people-6-male-4-female-divided-into-2-equal-groups-what-is-the-pr/3130441", "text": "$10$ people ($6$ male, $4$ female) divided into $2$ equal groups: what is the probability that all females are in the same group?\n\nThis question comes from a completed, marked, and returned exam. It will not likely be reused.\n\nProblem\n\nAs stated in the question above\n\nWork\n\nFirst, I note that there are $$\\binom{10}{5}$$ possible groupings.\n\nSecond, I note that, if all $$4$$ females are in the same group, then the remaining fifth member is one of the boys: there are $$\\binom{6}{1} = 6$$ ways to choose the fifth member.\n\nSo I conclude $$\\Pr = \\frac{6}{\\binom{10}{5}} = \\frac{1}{42}$$.\n\nQuestion\n\nI was marked incorrect: the given answer is $$\\frac{1}{21}$$, or exactly twice my answer.\n\nWhat reasoning led to this conclusion? Why does it seem like some sort of symmetry argument allows us to conclude there are $$12$$ ways to choose the fifth member?\n\n\u2022 Note that $\\frac{6}{\\binom{10}{5}} = \\frac{6}{252} = \\frac{1}{42}$. \u2013\u00a0N. F. Taussig Feb 28 at 16:55\n\u2022 @N.F.Taussig apologies i was looking at the answers on the question below as i typed \u2013\u00a0D. Ben Knoble Feb 28 at 17:01\n\u2022 Another way to see that there are only $126$ possible groups is to observe that if Eloise is one of the four girls, then there are $\\binom{9}{4}$ ways to select which four of the other members are in her group. \u2013\u00a0N. F. Taussig Feb 28 at 17:28\n\nYou've double-counted the groupings: $$\\{A, B, C, D, E\\}$$ and $$\\{F, G, H, I, J\\}$$ is the same grouping as $$\\{F, G, H, I, J\\}$$ and $$\\{A, B, C, D, E\\}$$. Accounting for this double-count, there are $$\\frac{1}{2} \\binom{10}{5}$$ distinct groupings. (This is probably the most-common counting mistake of all time. Everyone makes it at least once.)", "date": "2019-07-20 15:41:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3130437/10-people-6-male-4-female-divided-into-2-equal-groups-what-is-the-pr/3130441", "openwebmath_score": 0.656791627407074, "openwebmath_perplexity": 492.5258410613055, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754474655619, "lm_q2_score": 0.8824278741843883, "lm_q1q2_score": 0.8688168190811787}}
{"url": "https://www.physicsforums.com/threads/show-if-a-b-c-are-invertible-matrices-of-same-size.243766/", "text": "# Show if A,B,C are invertible matrices of same size\n\n1. Jul 7, 2008\n\nShow if A,B,C are invertible matrices of same size....\n\nI know, I know. I should be awesome at these by now...\n\n1. The problem statement, all variables and given/known data\nShow if A,B,C are invertible matrices of the same size, than $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$\n\n3. The attempt at a solution Given some matrix A,$AA^{-1}=I$\nIf:\n$AA^{-1}=I$\n$BB^{-1}=I$\n$CC^{-1}=I$\n\nI am not sure where to go from here. I don't think I have any more definitions or product rules to incorporate.\n\nIt almost seems as if I would FIRST have to show that (ABC) is invertible to begin with. Then I can use the fact that (ABC)(ABC)^{-1}=I to discover how (ABC)^{-1} MUST be arranged in order for the product of the two to yield I.\n\nDoes that sound like a good place to start? Proving if A,B, and C are invertible, then (ABC) is too?\n\n2. Jul 7, 2008\n\n### rock.freak667\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nLet $D=(ABC)^{-1}$\n$\\times (ABC)$\n$\\Rightarrow ABCD=I$\n\nand then just multiply by $A^{-1}$ and so forth\n\n3. Jul 7, 2008\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nI don't follow? What does the second line mean?--->$\\times (ABC)$\n\n4. Jul 7, 2008\n\n### rock.freak667\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nMultiply both sides by the matric ABC\n\n5. Jul 7, 2008\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nSo\n$D=(ABC)^{-1}$\n$\\Rightarrow D(ABC)=(ABC)^{-1}(ABC)$\n\nWell, I think I see where this is going. And I think that the only reason this works is because we are assuming that the product (ABC) IS invertible. Which brings me back to my original point. In order to show how the multiplication MUST be carried out, we must first SHOW or assume without proof that (ABC) is in fact invertible since our argument will be based on the fact that (ABC)*(ABC)^{-1}=Id.\n\n6. Jul 7, 2008\n\n### rock.freak667\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nWell if A,B,C are nxn matrices then ABC is an nxn matrix.\n\nand for the matrix ABC to be invertible $det(ABC) \\neq 0$\n\nand det(ABC)=det(A)*det(B)*det(C)\n\nbut the matrices A,B,C are invertible.\n\n7. Jul 7, 2008\n\n### Defennder\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nAnother way you could show that a product of two matrices A and B are invertible is by showing that there exists some matrix which when multiplied to AB on the left and on the right gives the identity matrix:\n\nSuppose A and B are invertible, then:\n\n$$AB(B^{-1}A^{-1}) = I$$ for multiplying on the right\n$$B^{-1}A^{-1}AB = I$$ for multiplying on the left.\n\nIn both cases this reduces to I, so $$B^{-1}A^{-1}$$ is the inverse of AB.\n\nNow make use of this result to prove your question.\n\n8. Jul 7, 2008\n\n### HallsofIvy\n\nStaff Emeritus\nRe: Show if A,B,C are invertible matrices of same size....\n\nIt is not nnecessary to assume that ABC is invertible.\n\nYou are given that A, B, C separately are invertible so A-1, B-1, and C-1 exist. Thus C-1B-1A-1 exists. What do you get if you multiply (ABC)(C-1B-1C-1) and (C-1B-1A-1(ABC)?\n\nThat will prove that ABC is invertible.\n\n9. Jul 7, 2008\n\n### matt grime\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nX is invertible if there is a Y with XY=YX=Id.\n\nIt isn't that I want to show ABC is invertible, but someone's been really helpful and just asked me to verify what the inverse is! I don't have to think at all, I just have to chuck the nominal inverse into the definition and see what happens. So do it....\n\n10. Jul 7, 2008\n\n### Defennder\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nHey yeah, I didn't see it that way. Makes it a lot easier.\n\n11. Jul 7, 2008\n\n### JinM\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nThis looks like a question that could spring up on my midterm tomorrow. Assuming invertibility, does this prove the result?\n\n$$ABCC^{-1}B^{-1}A^{-1} = A(BI_{n})B^{-1}A^{-1}= A(BB^{-1})A^{-1} = AA^{-1} = I_n$$\n\nNow, the other side,\n\n$$C^{-1}B^{-1}A^{-1}ABC = C^{-1}B^{-1}BC = C^{-1}C = I_n$$\n\nThus $$(ABC)(C^-1B^-1A^-1}) = (C^-1B^-1A^-1)(ABC) = I_n$$.\n\nIs this enough? No claim of originality, this was how a result was proved in my notes, and what everyone is getting at!\n\nLast edited: Jul 7, 2008\n12. Jul 7, 2008\n\n### Defennder\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nThat's basically what HallsOfIvy said.\n\n13. Jul 7, 2008\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nI don't like it. I don't know why yet...but I just don't.\n\n14. Jul 7, 2008\n\n### matt grime\n\nRe: Show if A,B,C are invertible matrices of same size....\n\nYou aren't assuming invertibility of ABC. You're just multiplying together 6 matrices, and showing that they give the identity, which is *proving* it is invertible. Just because you wrote \"assuming invertibility of ABC\" doen't mean you actually did assume it, or that you needed toi.\n\n15. Jul 7, 2008\n\n### b0it0i\n\nRe: Show if A,B,C are invertible matrices of same size....\n\ni think the problem you're having with this proof is understanding what the question is asking.\nthere are many other proofs with the same structure\n\nthe problem is saying show that\n\n(ABC)-1 = C-1B-1A-1\n\nin other words, they want you to show that the inverse of ABC is actually (behaves like) inverse C times inverse B times inverse A\n\nan example would be... Show that\n\nA Source of water = Rain\n\nyou must show that rain acts/ behaves like a source of water\n\napply the properties of \"a source of water\" to rain\n\nShow that C-1B-1A-1 \"behaves\" like the (ABC)-1\nhow do you do that?... you check the property of (ABC)-1\n\nthat is... show that C-1B-1A-1 (ABC) = (ABC) C-1B-1A-1 = I\n\nAn important fact is to note that \"the inverse of a matrix is unique\"\nsince C-1B-1A-1 \"behaves\" like the inverse of ABC,\nand that the inverse of a matrix is unique,\nC-1B-1A-1 MUST be (ABC)-1\n\nthis is corny, but my past professor constantly called these types of proofs\n\"walks like a duck, quacks like a duck, MUST be a duck...\" proof\n\n16. Jul 8, 2008", "date": "2017-03-27 05:16:33", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/show-if-a-b-c-are-invertible-matrices-of-same-size.243766/", "openwebmath_score": 0.7810722589492798, "openwebmath_perplexity": 1168.921472130163, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754492759498, "lm_q2_score": 0.8824278695464501, "lm_q1q2_score": 0.8688168161123153}}
{"url": "https://math.stackexchange.com/questions/1257518/answer-does-not-make-sense-using-conditional-probability", "text": "# Answer does not make sense using conditional probability\n\nAs of now, there are 64.1 million people residing in the UK. 5.4 million of them are thought to be asthmatic. A new test for asthma has recently been introduced and medical trials indicate that\n\n\u2022 in patients with asthma, the test correctly returns positive 68% of the time\n\u2022 in patients that does not suffer from asthma, the test correctly returns 82% of the time.\n\nAssuming a patient undergo medical inspection. What is the probability that the patient has asthma,\n\n1. if the test comes back positive?\n2. if the test comes back negative?\n\nFor Part 1, I used this method:\n\nStep 1: $$P(A|+)=\\frac{P(+|A)P(A)}{P(+)}$$\n\nStep 2: $$P(+)=P(+\u2502A)P(A)+P(+\u2502N)P(N)=(0.68)\\left(\\frac{54}{641}\\right)+(0.18)\\left(\\frac{587}{641}\\right)=\\frac{7119}{32050}$$\n\nStep 3: $$P(A\u2502+)=\\frac{P(+|A)P(A)}{P(+)}=\\frac{0.68\\times\\frac{54}{641}}{\\frac{7119}{32050}}=\\frac{204}{791}=25.79\\%$$\n\nFor Part 2, I used the same method as Part 1 but I got a very small answer. Assuming if I insert all the values in the equations correctly, would it be sensible if my answer for Part 2 is 3.47%?\n\nHowever, personally, it does not make logical sense if the probability of test returning negative is 3.47% because that would mean almost everyone in the nation would be asthmatic.\n\nOr I could just write it as $$100-25.79 = 74.21%$$ but I'm afraid this isn't the answer given the complexity of the question.\n\n\u2022 3.47% is the right answer here... \u2013\u00a0Clement C. Apr 29 '15 at 13:33\n\u2022 You're computing the probability of a false negative not the probability someone doesn't have asthma. False negatives are rare and should be rare intuitively. Otherwise, the test would be a piece of crap. So, you're right. \u2013\u00a0Jamie Lannister Apr 29 '15 at 13:34\n\u2022 I don't understand what you mean. If 3.47% is the correct answer, this is quite reasonable. It says: \"If the test says you don't have asthma, then it's very likely you really don't have asthma (i.e. the test is good).\" The first answer is actually more surprising. It says: \"If the test says you have asthma, it's only 25% likely you actually have it!\" But that isn't that surprising given the low figure of 68% in the hypotheses. \u2013\u00a0Frank Apr 29 '15 at 13:34\n\u2022 @iterence: that would not give the probability $\\Pr[A\\mid -]$, but instead the quantity $\\Pr[A^c\\mid +] = 1 - \\Pr[A\\mid +]$. That is, instead of \"the probability to have asthma knowing the test says you don't,\" you would compute \"the probability not to have asthma, knowing the test says you do.\" \u2013\u00a0Clement C. Apr 29 '15 at 13:38\n\u2022 In part 2, \"if the test comes back negative\", it is assumed we know that the result was negative, i.e., the probability of the test returning negative is $100$%, not $3.47$%. The $3.47$% figure arises from the cases where people do have asthma but it is not detected by the test. \u2013\u00a0David K Apr 29 '15 at 13:41\n\nOne instructive way to do these kinds of $2 \\times 2$ problems, since there are only four distinct populations, is to enumerate them and calculate the marginal probabilities by inspection.\n\n$$\\begin{array}{|c|c|c|c|} \\hline & \\text{asthmatic} & \\text{not asthmatic} & \\text{TOTALS} \\\\ \\hline \\text{positive} & 3.672 & 10.566 & 14.238 \\\\ \\hline \\text{negative} & 1.728 & 48.134 & 49.862 \\\\ \\hline \\text{TOTALS} & 5.400 & 58.700 & 64.100 \\\\ \\hline \\end{array}$$\n\nFrom this, one can read off\n\n$$P(\\text{asthmatic} \\mid \\text{positive}) = 3.672/14.238 \\doteq 0.25790$$\n\nand\n\n$$P(\\text{asthmatic} \\mid \\text{negative}) = 1.728/49.862 \\doteq 0.34656$$\n\nETA: Two of the cells\u2014$3.672 = (0.68)(5.400)$ and $48.134 = (0.82)(58.7)$\u2014are filled directly by the given parameters. Everything else is bookkeeping.\n\nThe abysmal false positive rate is due to a combination of the low specificity of the test ($82$ percent) and the relatively low prevalence of asthma ($5.4/64.1 \\doteq 8.4$ percent), so that although only a minority (sizable, but still a minority) of non-asthmatics test positive, they still dwarf the asthmatics who test positive, because there are so few asthmatics available to test positive in the first place. Even if the test were $100$ percent sensitive, the false positive rate would still have been $10.566/(10.566+5.400) \\doteq 66$ percent.", "date": "2019-07-23 00:41:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1257518/answer-does-not-make-sense-using-conditional-probability", "openwebmath_score": 0.7334158420562744, "openwebmath_perplexity": 647.666591872189, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754492759498, "lm_q2_score": 0.8824278633625322, "lm_q1q2_score": 0.8688168100237815}}
{"url": "http://math.stackexchange.com/questions/245354/expected-value-of-sums", "text": "Expected value of sums\n\nSuppose we draw cards out of a deck without replacement. How many cards do we expect to draw out before we get an ace?\n\n-\n\nIt depends (slightly) on how one interprets before. There are two interpretations possible: (i) Before means not including the draw that got us the first Ace and (ii) We include in the count the draw that got us the first Ace.\n\nThere is no big difference between (i) and (ii): The count (and expectation) in (ii) is just $1$ more than the count, and expectation, in (i), We use interpretation (i). So let $W$ be the number of draws before the first Ace, not including the draw that got us the Ace. We want $E(W)$.\n\nThe argument is simple but a bit delicate, so the solution below is given in great detail. Luckily, the actual computation is almost formula-free. We use indicator random variables. Label the $48$ non-Aces $1$ to $48$. Don't bother to label the Aces.\n\nDefine random variable $X_i$ by $X_i=1$ if the card with label $i$ was drawn before any Ace, and let $X_i=0$ otherwise. Then $$W=X_1+X_2+\\cdots+X_{48}.$$ By the linearity of expectation, which holds even when the random variables are not independent, we have $$E(W)=E(X_1+X_2+\\cdots+X_{48})=E(X_1)+E(X_2)+\\cdots+E(X_{48}).$$ By symmetry, all the $X_i$ have the same distribution. We find, for example, the probability that $X_1=1$. So we want the probability that card with label $1$ is drawn before any Ace.\n\nConsider the $5$-card collection consisting of the $4$ Aces and the card labelled $1$. All orders of these cards in the deck are equally likely. It follows that the probability that card with label $1$ is in front of the $4$ Aces is $\\frac{1}{5}$. Thus $E(X_1)=\\frac{1}{5}$.\n\nWe conclude that $E(W)=\\dfrac{48}{5}$.\n\nIf we want to take interpretation (ii), and include the draw that got us the Ace, our expectation is $1+\\dfrac{48}{5}=\\dfrac{53}{5}$.\n\n-\nNow I'm wondering if this answer should be considered essentially the same as mine. \u2013\u00a0Michael Hardy Nov 27 '12 at 1:54\nHard to decide! Mine (of course it's not mine, it is pretty standard stuff, I have used it in lectures) is very formal. \u2013\u00a0Andr\u00e9 Nicolas Nov 27 '12 at 1:59\nHas anyone written an exposition of the fact that expectations are often easier to find than probabilities? \u2013\u00a0Michael Hardy Nov 27 '12 at 2:01\nDon't know of one. I sure have stressed it. \u2013\u00a0Andr\u00e9 Nicolas Nov 27 '12 at 2:04\n\nDistribute the $52$ cards uniformly between $0$ and $1$, so on average they're at $k/53$ for $k=1,2,3,\\ldots,52$. The four aces are on average at $1/5,\\ 2/5,\\ 3/5,\\ 4/5$. So $$0.2 = \\frac{k}{53}$$ implies $$k = 10.6.$$ and $(0.8)\\cdot53 = 42.4$. So on average the four aces are the $10.6$th, $21.2$th, $31.8$th, and $42.4$th cards.\n\n-\n\nA good way to go about this is as follows. Let $T$ be the number of cards drawn at the time of the first ace. This is a $\\mathbb{N}$ valued random variable. Therefore $$E(T) = \\sum_{n=0}^\\infty P(T > n).$$ You may find this helpful.\n\n-\n\nI think that the answer lies in the Hypergeometric distribution or else k successes in n draws from a finite population of size N containing m successes without replacement (quoting from Wikipedia).\n\nIn your case, you want a single success in $p$ draws, and from the distribution you can find an average.\n\nEdit: $$P(X=1) = \\frac{\\binom{m}{1}\\binom{N-m}{n-1}}{\\binom{N}{n}}$$ and taking an average should be something like the following $$\\bar M = \\sum_{i=1}^{N-m}i\\frac{\\binom{m}{1}\\binom{N-m}{i-1}}{\\binom{N}{i}}$$ for $N=52, m=4$.\n\nEnd of edit.\n\nWikipedia also has some examples with urns, black and white balls, so you can work the answer from those.\n\n-\n\nI came across this in The Theory of Gambling and Statistical Logic By Richard A. Epstein so pasting a snippet here. I got this from Google Books (public domain) so hopefully there is no copyright violation here. It's amazing how Andre, Michael and Patrick derived the correct answer differently.\n\n-\n\nI did not see this approach in any of the answers, which is one that makes sense to me: you can get the ace in either the 1st, 2nd,..., 49th trial.\n\nThen the expected number of cards dealt is $$1(1/13)+2(12/13)(1/13)+..n(12/13)^{n-1}(1/13)+48(12/13)^{47}(1/13)$$.\n\n-\n\nTry this. (edited based on comment below) You will need just 1 pull with probability 4/52 2 with probability (4/51 x 48/52). 48/52 for the probability of not getting and ace in the first and getting it in the second (4/51), and so on. This yields 1 x 4/52 + 2x(4/51 x 48/52) + 3x(4/50 x 48/52 x 44/50) + ... There must be some simplification for the above, but I'ven't looked deeper into, but you get the idea. Here are similar ones\n\n-\nThis assumes there is only one ace in the deck. \u2013\u00a0Michael Hardy Nov 27 '12 at 1:05\n\nI have always liked this shortcut (proof here), which also holds for continuous random variables defined on $x \\geq{0},\\,x \\in \\mathbb{R}$. For some reason it was called \"The Darth Vader\" rule in my actuarial study manual but I digress. For this problem: $$P(T>n) = \\frac{\\binom{48}{n}n!}{\\binom{52}{n}n!} = \\frac{\\binom{48}{n}}{\\binom{52}{n}}$$ This is easily reasoned as follows:\n\n$P(T>n)$ is the probability that there are no aces up to and including the $n$th draw which means that in these first $n$ draws, out of the $48$ non-aces we count the number of ways to choose $n$ of them and permute them. We divide this by the total number of outcomes for the first $n$ draws which is $\\binom{52}{n}n!$ (we must include the aces this time, thus the 52).\n\nTherefore: $$E(T) = \\sum_{n=0}^{48}\\frac{\\binom{48}{n}}{\\binom{52}{n}} = \\sum_{n=0}^{48}\\frac{(52-n)(51-n)(50-n)(49-n)}{52\\times51\\times50\\times49} = 53/5 = 10.6$$\n\nWe could sum to $\\infty$ like ncmathsadist stated but the values of $P(T>n)$ for $n>48$ are $0$ anyway.\n\nI will admit that I cheated in my final step by using wolframalpha. Wolfram Alpha is a GREAT free resource by the way. I do hope that the logic behind solving the problem is not lost as a result. - Patrick\n\nSidebar: My initial naive guess was that $E(T) = 13$ since there are on average $12$ cards between each ace. I am trying to figure out an intuitive understanding for why the actual answer is less that $13$ now.\n\n-\nIt is 13 only if there is replacement. As there isn't any replacement the probability goes up so that interval should shrink eventually converging to 10.6. BTW could you elaborate how you simplified the average to 53/5? \u2013\u00a0broccoli Nov 27 '12 at 7:52\nYou're right, if there is replacement then $T \\sim Geometric(p)$ where $p$ is the probability of finding an ace ($1/13$) and $E(T) = 1/p = 13.$ As for an elaboration I just plugged the sum into wolfram alpha's calculator. Andres solution above is much better as the calculations can be done in your head and his method is applicable to other similar problems. \u2013\u00a0Patrick Nov 27 '12 at 13:33\nIf you wonder why $13$ is too big, consider the expected number of trials needed to get all four aces. Would that be $52$? Only if you never get the fourth ace until the last card in the deck. \u2013\u00a0Michael Hardy Nov 27 '12 at 19:16\n\nAdd a fifth ace, uniformly randomly arrange the $53$ cards in a circle, and break the circle into a line at the added ace. By symmetry, the expected number of cards between two of the $5$ aces is $\\frac{48}5$, so this is the expected number of cards between the beginning of the line and the first ace.\n\n-", "date": "2016-05-26 11:09:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/245354/expected-value-of-sums", "openwebmath_score": 0.8587490916252136, "openwebmath_perplexity": 279.3342530157824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754497285467, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8688168043346316}}
{"url": "http://math.stackexchange.com/questions/193010/whats-the-chance-that-exactly-2-2s-will-have-appeared-before-the-2nd-8", "text": "# What's the chance that exactly 2 2s will have appeared before the 2nd 8?\n\nA standard deck of 52 cards is randomly shuffled. Bob keeps drawing from the deck until he has drawn two 8s.\n\nWhat is the probability that, when, he draws the second 8, he has already drawn exactly two 2s?\n\nI think this could be a combinations problem because the order of the drawing two 2s does not matter. However, I am not sure how to account for the fact that he could draw an 8 or a second 8 at any point.\n\n-\n\nHint: one approach is to recognize that all the other cards are just fluff. There are only ${8 \\choose 4}=70$ possibilities for the order of 2's and 8's, so list them and count. You can make it less by listing the orders that get to 3 2's or 2 8's and assessing the probability. The first are failures, the second are successes if you have two 2's already. For example, one is that you start with three 2's. This is $\\frac 12 \\cdot \\frac 37 \\cdot \\frac 26=\\frac 1{14}$. Another is 2828. This is $\\frac 12 \\cdot \\frac 47 \\cdot \\frac 12 \\cdot \\frac 35=\\frac 3{35}$\n\n-\nOh! I think I got it from that hint! The only way for this to hold is for the string of combos be 8228[some ordering of 2 2s and 2 8s], 2828[some ordering of 2 2s and 2 8s], or 2288[some ordering of 2 2s and 2 8s]. Hence, the solution is 3 * (4 choose 2) / (8 choose 4), which is approximately 0.257! :D ... right? \u2013\u00a0 David Faux Sep 9 '12 at 4:03\n@DavidFaux: Good thought, but I wouldn't be sure that all the cases are equally probable. 8228 is $\\frac 48 \\frac 47 \\frac 36 \\frac 35=\\frac 3{35}$ while 2288 is $\\frac 48 \\frac 37 \\frac 46 \\frac 35$, which is the same, so I think it works. I would report the result as $\\frac 9{35}$, which is exact. \u2013\u00a0 Ross Millikan Sep 9 '12 at 4:20\nHmm, I think the combinations are equal because non-2 and non-8 cards are all fluff. If we make the 8s non-unique, and 2s non-unique, this basically becomes a problem involving strings of four 8s and four 2s, right? \u2013\u00a0 David Faux Sep 9 '12 at 4:34\n@DavidFaux: That is right. The more I think, I agree with your calculation. I was worried that 22228888 would be less probable than 82282882, but now I don't think so. It is easy to think things are equally probable when they are not. \u2013\u00a0 Ross Millikan Sep 9 '12 at 4:39\n\nAs was observed by Ross Millikan, there are only $8$ relevant cards. The first $4$ of these can appear in the dealing in $(8)(7)(6)(5)$ orders, all equally likely.\n\nWe count the number of \"favourable\" orders. These are given by any one of three patterns, which we call $2288$, $2828$, and $8228$.\n\nThe pattern $2288$ can occur in $(4)(3)(4)(3)$ ways, as can the other two patterns. Thus the required probability is $$\\frac{(3)(4)(3)(4)(3)}{(8)(7)(6)(5)}.$$ If we want to simplify a bit, we get $\\dfrac{9}{35}$.\n\nRemark: The problem seems to have been solved correctly in the comments both by you and by Ross Millikan. Then there appears to have been a change of mind.\n\n-", "date": "2015-10-06 23:15:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/193010/whats-the-chance-that-exactly-2-2s-will-have-appeared-before-the-2nd-8", "openwebmath_score": 0.829424262046814, "openwebmath_perplexity": 385.95434663097706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631631151012, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8687857996127581}}
{"url": "https://www.physicsforums.com/threads/electric-field-of-bent-non-conducting-rod.955159/", "text": "# Electric field of bent non conducting rod\n\n## Homework Statement\n\nYou are given a non conducting rod carrying uniformly distributed charge, -Q, that has been bent into a 120\u00b0 circular arc of radius, R. The axis of symmetry of the arc lies along the x-axis and the origin is at the center of curvature of the arc.\n(a) in terms of Q and R, what is the linear charge density, \u03bb?\n(b) in terms of Q and R, what is the magnitude and direction the the resulting electric field at the origin?\n(c)If the arc is replaced by a point particle carrying charge, -Q, at x=R, by what factor is the resulting electric field at the origin multiplied?\n\n\u03bb=Q/L\nE=KQ/R2\ndq=\u03bbds\nds=Rd\u03b8\n\n## The Attempt at a Solution\n\n(a) \u03bb=Q/L, \u03bb=Q/(\u21532piR) = 3Q/2piR\n\n(b) Cant seem to figure out how to write an integral with bounds. The bounds on my integral below will be from \u03b8=0\u00b0 \u2192 \u03b8=60\u00b0\nE=2\u222bdEx=2\u222bk\u03bbd\u03b8cos\u03b8/R = 2k\u03bb/R \u222bcos\u03b8d\u03b8 (Like previously stated this integral is being evaluated from \u03b8=0\u00b0 to \u03b8=60\u00b0)\nE=(2k\u03bb/R)[sin60\u00b0-sin0\u00b0] = (k\u03bb\u221a3)/R, it wants in terms of R and Q so now plug in our above statement for \u03bb\nE=(3kQ\u221a3)/2piR2\n\n(c) Im a little unsure about what this is asking, from my understanding:\nto obtain the new electric field from a point particle which is simply kQ/R2 we must multiply the previous electric field from bent rod by a factor of 2pi/3\u221a3. Not sure if this is correct or if I'm thinking about the question in the wrong way.\nAny help would be much appreciated\nThanks!\n\nRelated Introductory Physics Homework Help News on Phys.org\nkuruman\nHomework Helper\nGold Member\nConsider subdividing the 120o arc into infinitesimal elements $ds$.\n1. What is the charge on one of these elements?\n2. What is the electric field contribution $dE$ from this charge element to the point of interest?\n3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.\n\nPart (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\\alpha E_0$, what is the value of constant $\\alpha$?\n\nConsider subdividing the 120o arc into infinitesimal elements $ds$.\n1. What is the charge on one of these elements?\n2. What is the electric field contribution $dE$ from this charge element to the point of interest?\n3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.\n\nPart (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\\alpha E_0$, what is the value of constant $\\alpha$?\nOkay, Ill work on this and post my results! Thanks!\n\nAnswering is easy when people make mistakes. It\u2019s much harder when everything seems perfect. I\u2019m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?\n\nAnswering is easy when people make mistakes. It\u2019s much harder when everything seems perfect. I\u2019m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?\nYea just part c, wasn't sure what it was asking but @kuruman helped me out with that one. Just have a test in a few days and want to make sure I'm doing these problems correctly. Always love reassurance! LOL\n\nYou saw the symmetry and ignored the y component. You also used symmetry to cut the integral in half. You integrated in theta and didn\u2019t forget the extra R.\n\nI don\u2019t think you have much to worry about, but I\u2019ll wish you good luck anyway.", "date": "2020-04-06 15:52:47", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/electric-field-of-bent-non-conducting-rod.955159/", "openwebmath_score": 0.7870522141456604, "openwebmath_perplexity": 400.2641327615572, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9693242018339897, "lm_q2_score": 0.8962513786759491, "lm_q1q2_score": 0.8687581522776772}}
{"url": "https://math.stackexchange.com/questions/2781063/limit-of-frac-sqrtnn1n2-cdots2nn/2781848", "text": "# Limit of $\\frac{\\sqrt[n]{(n+1)(n+2)\\cdots(2n)}}n$\n\nCompute the limit $$\\lim_{n \\to \\infty} \\frac{\\sqrt[n]{(n+1)(n+2)\\cdots(2n)}}n$$\n\nHow can this be done? The best I could do was rewrite the limit as $$\\lim_{n \\to \\infty} \\left(\\frac{n+1}n \\right)^{\\frac 1n}\\left(\\frac{n+2}n \\right)^{\\frac 1n}\\cdots\\left(\\frac{2n}n \\right)^{\\frac 1n}$$\n\nFollowing that log suggestion in the comments below: \\begin{align} &\\ln \\left(\\lim_{n \\to \\infty} \\left(\\frac{n+1}n \\right)^{\\frac 1n}\\left(\\frac{n+2}n \\right)^{\\frac 1n}\\cdots\\left(\\frac{2n}n \\right)^{\\frac 1n} \\right) \\\\ &= \\lim_{n \\to \\infty} \\ln \\left( \\left(\\frac{n+1}n \\right)^{\\frac 1n}\\left(\\frac{n+2}n \\right)^{\\frac 1n}\\cdots\\left(\\frac{2n}n \\right)^{\\frac 1n} \\right) \\\\ &= \\lim_{n \\to \\infty} \\left(\\ln \\left(\\frac{n+1}n \\right)^{\\frac 1n} + \\ln\\left(\\frac{n+2}n \\right)^{\\frac 1n} + \\cdots + \\ln\\left(\\frac{n+n}n \\right)^{\\frac 1n} \\right) \\\\ &= \\lim_{n \\to \\infty} \\sum_{i=1}^n \\ln \\left(\\frac{n+i}n \\right)^{\\frac 1n} \\\\ &= \\lim_{n \\to \\infty} \\frac 1n \\sum_{i=1}^n \\ln \\left(1 + \\frac in \\right) \\\\ &= \\int_1^2 \\ln x \\, dx \\\\ &= (x \\ln x - x)\\vert_1^2 \\\\ &= (2 \\ln 2 - 2)-(1 \\ln 1-1) \\\\ &= (\\ln 4-2)-(0-1) \\\\ &= \\ln 4-1 \\\\ &= \\ln 4 - \\ln e \\\\ &= \\ln \\left( \\frac 4e \\right) \\end{align}\n\nbut I read from somewhere that the answer should be $\\frac 4e$.\n\n\u2022 Your integral is the log of the answer \u2013\u00a0robjohn May 14 '18 at 18:15\n\u2022 @CarlMummert From the accepted answer of this meta question, we can see that this is a well-written question (except the use of $\\frac1n$ as a power), and \"additional indication of source would not make a difference\".You may re-read AmateurMathGuy's reply to your comment to know why they are not required. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 May 14 '18 at 18:19\n\u2022 You voted to close because \"Being homework is a sufficient reason to vote to close a question\", but HW questions are allowed. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 May 14 '18 at 18:19\n\u2022 @CarlMummert I do not see how my question \"has no real context, and does not fit the quality standards that many users expect.\" I put in the work and effort as this Math SE site suggests. \u2013\u00a0GarlicBread May 15 '18 at 3:58\n\u2022 Agreed. This is supported by the Close Queue Review result. All reviewers left this question open. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 May 15 '18 at 9:00\n\nHint: let $a_n=(n+1)\\cdots(2n)/n^n$, and use the fact that $a_{n+1}/a_n\\to L$ implies $\\sqrt[n]{a_n}\\to L$ (i.e., the ratio test implies the root test).\n\nIf you know Sterling's Approximation: $$n! \\sim \\left(\\frac{n}{e}\\right)^n \\sqrt{2\\pi n},$$ then you could approach it as follows:\n\n$$\\lim_{n \\to \\infty} \\frac{\\sqrt[n]{(n + 1)(n + 2)\\cdots (2n)}}{n} = \\lim_{n \\to \\infty} \\frac{1}{n} \\sqrt[n]{\\frac{(2n)!}{n!}} = \\lim_{n \\to \\infty} \\frac{1}{n} \\sqrt[n]{\\frac{\\left(\\frac{2n}{e}\\right)^{2n} \\sqrt{4\\pi n}}{\\left(\\frac{n}{e}\\right)^n \\sqrt{2\\pi n}}} = \\lim_{n \\to \\infty} \\frac{1}{n} \\sqrt[n]{2^{2n} \\left(\\frac{n}{e}\\right)^n \\sqrt{2}} = \\lim_{n \\to \\infty} \\frac{1}{n} \\cdot 2^2 \\cdot \\frac{n}{e} \\cdot 2^{\\frac{1}{2n}} = \\frac{4}{e}.$$\n\nNow take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\\ln(1+k/n)$.)\n\n\u2022 I'm not sure how to rewrite the limit of the Riemann sum as its corresponding definite integral, although I think I managed to arrive at the Riemann sum in my work. \u2013\u00a0GarlicBread May 14 '18 at 17:41\n\nIf you accept using the following limit that has been asked and proved (for example here) many times here on MSE\n\n\u2022 $\\lim_{n\\rightarrow \\infty}\\frac{\\sqrt[n]{n!}}{n} = \\frac{1}{e}$,\n\nthen your limit is easily derived:\n\n$$\\frac{\\sqrt[n]{(n+1)(n+2)\\cdots(2n)}}n = \\frac{ (2n!)^{\\frac{1}{n}} }{ (n!)^{\\frac{1}{n}}\\cdot n} = 4\\frac{n}{ (n!)^{\\frac{1}{n}}} \\left( \\frac{ (2n!)^{\\frac{1}{2n}} }{ 2n} \\right)^2 \\stackrel{n\\rightarrow\\infty}{\\longrightarrow}4\\cdot e \\cdot \\frac{1}{e^2}=\\frac{4}{e}$$\n\nNote that $\\displaystyle\\frac{\\sqrt[n]{n+1}}{n}\\neq\\left(\\frac{n+1}{n}\\right)^{1/n}$\n\nAlso, this is just for clarification since some nice suggestions have already been given but also because the question mark remains over your last equality\n\nWith this is mind and going from your first step,\n\n$$S=\\lim_{n\\to\\infty}\\frac{\\sqrt[n]{(n+1)(n+2)...(2n)}}{n}=\\lim_{n\\to\\infty}\\left(\\frac{1}{n}\\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\\right)\\right)$$\n\n$$\\operatorname{ln}(S)=\\operatorname{ln}\\left(\\lim_{n\\to\\infty}\\left(\\frac{1}{n}\\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\\right)\\right)\\right)$$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\operatorname{ln}\\left(\\frac{1}{n}\\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\\right)\\right)$$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\left(\\operatorname{ln}\\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\\right)-\\operatorname{ln}(n)\\right)$$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\left(\\operatorname{ln}\\left((n+1)^{1/n}\\right)+\\operatorname{ln}\\left((n+2)^{1/n}\\right)+...+\\operatorname{ln}\\left((2n)^{1/n}\\right)-\\operatorname{ln}(n)\\right)$$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\left(\\sum_{p=1}^n\\left(\\operatorname{ln}\\left((n+p)^{1/n}\\right)\\right)-\\operatorname{ln}(n)\\right)$$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\left(\\sum_{p=1}^n\\left(\\frac{1}{n}\\operatorname{ln}(n+p)\\right)-\\operatorname{ln}(n)\\right)$$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\left(\\frac{1}{n}\\sum_{p=1}^n\\left(\\operatorname{ln}(n+p)\\right)-\\operatorname{ln}(n)\\right)$$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\left(\\frac{1}{n}\\sum_{p=1}^n\\left(\\operatorname{ln}\\left(1+\\frac{p}{n}\\right)+\\operatorname{ln}(n)\\right)-\\operatorname{ln}(n)\\right)$$\n\nThis $+\\operatorname{ln}(n)$ is independent of $p$ and summed $n$ times, then divided by $n$, so $\\operatorname{ln}(n)-\\operatorname{ln}(n)=0$\n\n$$\\operatorname{ln}(S)=\\lim_{n\\to\\infty}\\left(\\frac{1}{n}\\sum_{p=1}^n\\operatorname{ln}\\left(1+\\frac{p}{n}\\right)\\right)$$\n\nThis is what Eric meant and the argument varies from $1$ to $2$\n\nIt becomes a little clearer if we let $m = 1/n$\n\n$$\\lim_{n\\to\\infty}\\left(\\frac{1}{n}\\sum_{p=1}^n\\operatorname{ln}\\left(1+\\frac{p}{n}\\right)\\right)=\\lim_{m\\to\\ 0}\\left(\\left(\\sum_{p=1}^{1/m}\\operatorname{ln}(1+pm)\\right)m\\right)=\\int_1^2\\operatorname{ln}(x)dx=(x\\operatorname{ln}(x)-x)\\Big\\rvert_1^2$$\n\n$$\\therefore \\operatorname{ln}(S)=2(\\operatorname{ln}(2)-1)+1=\\operatorname{ln}(4)-1=\\operatorname{ln}\\left(\\frac{4}{e}\\right)$$\n\nFinally\n\n$$S=\\frac{4}{e}$$", "date": "2020-01-20 15:02:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2781063/limit-of-frac-sqrtnn1n2-cdots2nn/2781848", "openwebmath_score": 1.0000097751617432, "openwebmath_perplexity": 381.5331163944861, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693242018339897, "lm_q2_score": 0.8962513752119936, "lm_q1q2_score": 0.8687581489199814}}
{"url": "http://bkdd.zonaunderground.it/matrix-multiplication-algorithm-pseudocode.html", "text": "# Matrix Multiplication Algorithm Pseudocode\n\nAbstract\u2014 Matrix multiplication is an integral component of most of the systems implementing Graph theory, Numerical algorithms, Digital control, Signal and image processing (i. the first column when multiplied by the matrix. of algorithms in the computational sciences. The booth\u2019s multiplication algorithm is primarily used in computer architectures. Banded Matrix-Vector Multiplication. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. Pseudocode of the rest of the algorithm: Iterative Matrix Multiplication I'm a bit unhappy with your code, because it's so hard to read tbh. Rotation: We use a generalization of Cannon\u2019s algo-rithm as the primary template. I Strassen's algorithm gives a performance improvement for large-ish N, depending on the architecture, e. Write an algorithm in pseudocode to perform the multiplication of a matrix with a vector. Specifically, an input matrix of size can be divided into 4 blocks of matrices. COMP 250 Winter 2016 1 { grade school algorithms Jan. Consider an NxN complex array. This allows us to exploit fast matrix multiplication. 3 Multithreaded merge sort 797 28 Matrix Operations 813 28. Divide-and-Conquer algorithsm for matrix multiplication A = A11 A12 A21 A22 B = B11 B12 B21 B22 C = A\u00d7B = C11 C12 C21 C22 Formulas for C11,C12,C21,C22: C11 = A11B11 +A12B21 C12 = A11B12 +A12B22 C21 = A21B11 +A22B21 C22 = A21B12 +A22B22 The First Attempt Straightforward from the formulas above (assuming that n is a power of 2): MMult(A,B,n) 1. Pseudo Code for Union using Mapreduce. What is the least expensive way to form the product of several matrices if the na\u00efve matrix multiplication algorithm is used? [We use the number of scalar multiplications as cost. spawn is to indicate creation of a new thread. 3) you should encapsulate a matrix into a class, if it is supposed to be c++ (then 2 will be obsolete) 4) your code will be easier to understand (for you as well) if you use better names for x,y,z,i, k and j. Cannon's algorithm: a distributed algorithm for matrix multiplication especially suitable for computers laid out in an N \u00d7 N mesh; Coppersmith-Winograd algorithm: square matrix multiplication; Freivalds' algorithm: a randomized algorithm used to verify matrix multiplication. 5D (Ballard and Demmel) \u00a92012 Scott B. The main purpose of this paper is to present a fast matrix multiplication algorithm taken from the paper of Laderman et al. , what the complexity of the problem is). Pseudocode for the algorithm is given in Figure 1. cludes the RSA cryptosystem, and divide-and-conquer algorithms for integer multiplication, sorting andmedian\ufb01nding, aswellasthe fast Fourier transform. Algorithm Examples! Pseudocode! Order of Growth! Algorithms - what are they Fibonacci - Matrix Multiplication. Multithreaded Algorithms Introduction. To perform the addition, numbers in matching postions in the input matrices are added and the result is placed in the same position in the output matrix. We propose two approaches to matrix multiplication: iter-ative approach and block approach. Unlike standard matrix multiplication, MixColumns performs matrix multiplication as per Galois Field 2 8. Although adjacency matrix representation of graph is used, this algorithm can also be implemented using Adjacency List to improve its efficiency. General Matrix Multiplication (GEMM) is the primary component of the level-3 BLAS and of most dense linear algebra algorithms (and many sparse/structured linear algebra algorithms), which in turn have applications in virtually every area of computational science. Summary I Strassen rst to show matrix multiplication can be done faster than O(N3) time. complexity of matrix multiplication is n2 (2n \u22121) = 2 \u22c5(2 \u22121)\u22c5\u03c4 T1 n n (8. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. \u2022Pseudo code. Problem 6 (checks whether a matrix is symmetric): yes 4. One of the basic operations on matrices is multiplication. In other words, Lagrees with the corresponding k\u00d7nsubmatrix of X. 2 Strassen\u00d5s algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. Regression algorithm pseudocode from [4]: The regression algorithm follows a nested optimization scheme using coordinate descend. 6 Another Recursive Algorithm 4. If you're interested in typesetting algorithmic code, there are a number of choices. However, due to constant factors and realistic modern architecture constraints, these theoretically faster methods are rarely used; instead, the naive brute force approach to matrix multiplication is that which is. Suppose we want to multiply two n by n matrices, A and B. This analysis culminates in Section 4. The practical bene\ufb01t from improvements to algorithms is therefore potentially very great. Here's a short example from the algorithmicx documentation (with a pseudocode for loop added):. Assume that square matrix A and B are used for multiplication in the following algorithms. Algorithms { CMSC-37000 Divide and Conquer: The Karatsuba algorithm (multiplication of large integers) Instructor: L aszl o Babai Updated 01-21-2015 The Karatsuba algorithm provides a striking example of how the \\Divide and Conquer\" technique can achieve an asymptotic speedup over an ancient algorithm. We need to create a Toeplitz matrix using a subsection of a data vector on the device. Matrix Multiplication in Case of Block-Striped Data Decomposition Let us consider two parallel matrix multiplication algorithms. This builds on the previous post on recursive square matrix multiplication. to read the matrix into core memory once [19, sect. So Matrix Chain Multiplication problem has both properties (see this and this) of a dynamic programming problem. To begin with, the sequential algorithm was implemented using the pseudo-code in [2]. Freivalds' algorithm is a probabilistic randomized algorithm used to verify matrix multiplication. 2 Strassen's algorithm for matrix multiplication 4. One key idea in the sorting networks chapter, the 0-1 principle, ap-. x x y matrix by a y x z matrix creates an x x z matrix. The cofactor matrix is the matrix of determinants of the minors A ij multiplied by -1 i+j. Antoine Vigneron (UNIST) CSE331 Lecture 5 July 11, 2017 3 / 19. Write pseudocode for Strassen\u2019s algorithm. r-1 (mod n) where the integers a and b are smaller than the modulus. Complexity Calculation How many additions of integers and multiplications of integers are used by the matrix multiplication algorithm to multiply two n * n matrices. I won't give the pseudo code here for these ones, but they are naive recursive algorithm, bottom up algorithm, naive recursive squaring and recursive squaring. MPI Matrix-Matrix Multiplication Matrix Products Parallel 2-D Matrix Multiplication Characteristics Computationally independent: each element computed in the result matrix C, c ij, is, in principle, independent of all the other elements. 6 Another Recursive Algorithm 4. Provide your analysis for the following problem statement: Write a program that will calculate the results for the multiplication table up to 10x10 in steps of 1 beginning at 1. cient implementation of sparse matrix multiplication on a memory intensive associative processor (AP), verified by extensive AP simulation using a large collection of sparse matrices [41]. We then \"combine\" the middle row of the key matrix with the column vector to get the middle element of the resulting column vector. The multiplier contains only 0s and 1s,. The algorithms classes I teach at Illinois have two significant prerequisites: a course on discrete mathematics and a course on fundamental data structures. Block matrices are briefly discussed using 2 \u00d7 2 block matrices. I have a question for you about your approach. What is a Spanning tree? Explain Prim\u2019s Minimum cost spanning tree algorithm with suitable example. Before going to main problem first remember some basis. This page contains the order of topics contained in lectures, listed as a sequence of modules. Which is faster for this value of n?. To save space and running time it is critical to only store the nonzero elements. i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table. 2 shows the calculate steps of covariance matrix, mainly including: complex conjugate multiplication between the lines of input matrix, then do an accumulation operation. Idea - Block Matrix Multiplication The idea behind Strassen\u2019s algorithm is in the formulation of matrix multiplication as a recursive problem. You can use a pseudocode environment algpseudocode offered by algorithmicx. SPARSE MATRICES C/C++ Assignment Help, Online C/C++ Project Help and Homework Help introduction A matrix is a mathematical object that arises in many physical problems. 1 The naive matrix multiplication algorithm Let A and B be two n \u00a3 n matrices. Matrix mulitplication using Linked List - posted in C and C++: I have to implement Matrix multiplication using singly linked list. Pseudocode of the rest of the algorithm: Iterative Matrix Multiplication I'm a bit unhappy with your code, because it's so hard to read tbh. Material for the algorithms class taught by Emanuele \"Manu\" Viola. Dynamic Programming\u2014Chained Matrix Multiplication Multiplying unequal matrices \u2022 Suppose we want to multiply two matrices do not have the same number of rows and columns \u2022 We can multiply two matrices A 1 and A 2 only if the number of columns of A 1 is equal to the number of rows of A 2. Example of Matrix Multiplication by Fox Method Thomas Anastasio November 23, 2003 Fox's algorithm for matrix multiplication is described in Pacheco1. However, this algorithm is infamously inapplicable, as it relies on Coppersmith and Winograd\u2019s fast matrix multiplication. It is referenced specifically in the pseudocode but that is not the only location where it is appropriate to call it. If you are interested in a Modified Gauss-Jordan Algorithm, you can see this. Algorithm for the Transpose of a Sparse-Matrix: This is the algorithm that converts a compressed-column sparse matrix into a compressed-row sparse matrix. The practical bene\ufb01t from improvements to algorithms is therefore potentially very great. What is the best algorithm for matrix multiplication ? Actually there are several algorithm exist for matrix multiplication. Here, we will discuss the implementation of matrix multiplication on various communication networks like mesh and. - Overall complexity of parallel matrix-vector multiplication algorithm ( n2=p+n+logp) - Isoef\ufb01ciency of the parallel algorithm Time complexity of sequential algorithm: ( n2) Only overhead in parallel algorithm due to all-gather For reasonably large n, message transmission time is greater than message latency. Solutions for CLRS Exercise 4. 2 Algorithmic Techniques 5. \\begin{algorithm} \\caption{Euclid's algorithm}\\label{euclid} \\. 1 Naive Matrix Multiplication 4. Matrix-matrix multiplication takes a triply nested loop. I'm just doing a self-study of Algorithms & Data structures and I'd like to know if anyone has a C# (or C++) implementation of Strassen's Algorithm for Matrix Multiplication? I'd just like to run it and see what it does and get more of an idea of how it goes to work. Let us start with a very simple example th. One common practice is to translate convolution to im2col and GEMM, which lays out all patches into a matrix. 84 videos Play all Algorithms Abdul Bari; Derivatives explained - Duration: 10:13. One of the basic operations on matrices is multiplication. The algorithm is called a (7, 4) code, because it requires seven bits to encoded four bits of data. Recurrence equation for Divide and Conquer: If the size of problem \u2018p\u2019 is n and the sizes of the \u2018k\u2019 sub problems are n1, n2\u2026. Write An Algorithm To Find The Power Of A Number. 2 Strassen\u00d5s algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. This page contains the order of topics contained in lectures, listed as a sequence of modules. GitHub Gist: instantly share code, notes, and snippets. Adjacency matrix representation of graph where the n X n matrix W = (wij) of edge weights. Matrix multiplication algorithms. Definition of Flowchart A flowchart is the graphical or pictorial representation of an algorithm with the help of different symbols, shapes and arrows in order to demonstrate a process or a program. complexity by packing the inner loops into a single matrix product as shown in Algorithm 2. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. The unit of. Matrix multiplication is the process of taking two or more n x n matrices and calculating a product by summing the product of each row in one matrix with the column of the second matrix. As in case of developing the matrix-vector multiplication algorithm, we use one-dimensional arrays, where matrices are stored rowwise. So assuming that both these multiplication steps are executed every time the loop executes, we see that 2. From Math Insight. We tackle scenarios such as matrix multiplication and linear regression/classification in which we wish to estimate inner products between pairs of vectors from two possibly different sources. The problem is quite easy when n is relatively small. 1 and the Junto. Below is some sample output. It is the technique still used to train large deep learning networks. org) I now want to use strassen's method which I learned as follows:. Unlike general multiplication, matrix multiplication is not commutative. o The number of additions and multiplication's required for this algorithm can be calculated as follows: To calculate one entry in the product matrix, we must perform k multiplications and k-1 additions. for i = 1 to n. 1 and Step 3. We can use simple recursion, f(n) = f(n-1) + f(n-2), or we can use dynamic programming approach to avoid the calculation of same function over and over again. Write a c program to find out transport of a matrix. it explains matrix multiplication. Antoine Vigneron (UNIST) CSE331 Lecture 5 July 11, 2017 3 / 19. In other words, Lagrees with the corresponding k\u00d7nsubmatrix of X. Question: Show Map Reduce implementation for the following two tasks using pseudocode. - Explain the difference between an LED and OLED display. matrix multiplication; this means that matrix multiply based methods for determining primitivity cannot be sped up anymore at this time. 3343(abc)(log7)/(3) 2. Block matrices are briefly discussed using 2 \u00d7 2 block matrices. x x y matrix by a y x z matrix creates an x x z matrix. Section 5 provides a com-parison with related works. Summary I Strassen rst to show matrix multiplication can be done faster than O(N3) time. In the next three parts, you may be writing pseudocode. That\u2019s very important because for small n (usually n < 45) the general algorithm is practically a better choice. Integer-multiplication, Matrix Multiplication - Strassen Alg You study after every class/week, the syllabus accumulates fast before you know! Aug 26, W (Drop w/o W grade, Aug 28) Dynamic Programming: 0-1Knapsack. This work is licensed under aCreative Commons. parallel before a loop means each iteration of the loop are independant from each other and can be run in parallel. Notes: A common reference for double-precision matrix multiplication is the dgemm ( d ouble-precision ge neral m atrix- m atrix multiply) routine in the level-3 BLAS. The application. Given three n x n matrices, Freivalds' algorithm determines in O(kn^2) whether the matrices are equal for a chosen k value with a probability of failure less than 2^-k. Name the algorithmic technique used. 3 Matrix Multiplication for Banded Matrices. com Free Programming Books Disclaimer This is an uno cial free book created for educational purposes and is not a liated with o cial Algorithms group(s) or company(s). 0 is there to suggest that different values can be used, but they should be related to the number of input variables. The following algorithm multiplies nxn matrices A and B: // Initialize C. Strassen's algorithm, the original Fast Matrix Multiplication (FMM) algorithm, has long fascinated computer scientists due to its startling property of reducing the number of computations required for multiplying. and similarly for the bottom row. We know that, to multiply two matrices it is condition that, number of columns in first matrix should be equal to number of rows in second matrix. Matrix Multiplication: Strassen's Algorithm. What Is The Main Operation Of This Algorithm? C. We need to create a Toeplitz matrix using a subsection of a data vector on the device. of algorithms in the computational sciences. You can call the algorithm on sub-matrices of dimensions n-1 each when the size of the matrix is odd in the recursive step and calculate 2n-1 remaining elements using normal vector multiplication in O(n) each and a total of O(n^2). We tackle scenarios such as matrix multiplication and linear regression/classification in which we wish to estimate inner products between pairs of vectors from two possibly different sources. What is the best algorithm for matrix multiplication ? Actually there are several algorithm exist for matrix multiplication. n and r is relatively prime number to n (gcd (n, r)= 1). for k = 1 to n. Part I was about simple matrix multiplication algorithms and Part II was about the Strassen algorithm. If A is the adjacency matrix of G, then (A I)n 1 is the adjacency matrix of G*. Two groups of algorithms belonging to this class are called the matrix method, and the Wallace-tree method, respectively. 3) where \u03c4 is the execution time for an elementary computational operation such as multiplication or addition. Section 5 provides a com-parison with related works. The aim is to get the idea quickly and also easy to read without details. The former is suitable for sparse matrices, while the latter is appropriate for dense matrices with low communication overhead. We could break down the steps as follows. , what the complexity of the problem is). In matrix addition, one row element of first matrix is individually added to corresponding column elements. 4 uses dynamic programming to find an optimal triangulation of a convex polygon, a problem that is surprisingly similar to matrix-chain multiplication. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. Then, user is asked to enter two matrix and finally the output of two matrix is calculated and displayed. That\u2019s very important because for small n (usually n < 45) the general algorithm is practically a better choice. Each matrix Mk has dimension pk-1 x pk. The Floyd Warshall algorithm, itis the algorithm in which there is the use of different characterization of structure for a shortest path that we used in the matrix multiplication which is based on all pair algorithms. Use row communicators and column communicators to scatter and broadcast the vector. We ended up pursuing a different route, but I decided to continue pursuing the problem on my own time. 2x2 Matrix Multiplication Calculator is an online tool programmed to perform multiplication operation between the two matrices A and B. From Math Insight. This work is licensed under aCreative Commons. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication. Provide a specification to describe the behaviour of this algorithm, and prove that it correctly implements its specification. However, this algorithm is infamously inapplicable, as it relies on Coppersmith and Winograd\u2019s fast matrix multiplication. The weights and values of 6. Sparse Matrix Multiplication. I am trying to implement a multiplication algorithm by overloading the *= operator. Matrix Multiplication in Case of Block-Striped Data Decomposition Let us consider two parallel matrix multiplication algorithms. 3 Storage formats 3. Other types of algorithms for this problem appear in [15, 16]. In particular, this includes judging which data structures, libraries, frameworks, programming languages, and hardware platforms are appropriate for the computational task, and using them effectively in the implementation. How do they differ? - Is a pixel a little square? If not, what is it? What implications does this have? Give at least 2. One of the basic operations on matrices is multiplication. It is the technique still used to train large deep learning networks. Then, we'll present a few examples to give you a better idea. Floyd Warshall. What Is The Main Operation Of This Algorithm? C. For instance, the algorithm we're interested in looking at, Dijkstra's algorithm, only works if none of the edges on the graph have negative weights -- the \"time\" it takes to traverse the edge is somehow less than 0. emit (key, result). GATEBOOK Video Lectures 3,203 views. These lectures were designed for the latter part of the MIT undergraduate class 6. Matrix Multiplication Problem - Duration: 27:38. - Overall complexity of parallel matrix-vector multiplication algorithm ( n2=p+n+logp) - Isoef\ufb01ciency of the parallel algorithm Time complexity of sequential algorithm: ( n2) Only overhead in parallel algorithm due to all-gather For reasonably large n, message transmission time is greater than message latency. Other types of algorithms for this problem appear in [15, 16]. \u2022 Algorithms are step-by-step procedures for problem solving \u2022 They should have the following properties: \u2022Generality \u2022Finiteness \u2022Non-ambiguity (rigorousness) \u2022Efficiency \u2022 Data processed by an algorithm can be \u2022 simple \u2022 structured (e. Strassen's method of matrix multiplication is a typical divide and conquer algorithm. Although we won't describe this step in detail, it is important to note that this multiplication has the property of operating independently over each of the columns of the initial matrix, i. Matrix Multiplication; Matrix Multiplication Parenthesization; Brute Force Solution: Try all possible parenthesizations; Dynamic Programming Solution (4 steps) Step 1: Characterize Structure of Optimal Solutioon; Step 2: Define recursive solution; Recursive Solution; Analysis; Duplicate Subproblems; Unique Subproblems; Step 3: Bottom-Up Approach; Dynamic Programming. 5 Maximum Flow. Matrix Multiplication c c2 1= r2 A1 A2 r1 r2! = r1 ! c2 (r1 ! c2) ! c1 = multiplications If r 1 = c 1 = r 2 = c 2 = N, this standard approach takes ( N3): I For every row ~r (N of them) I For every column ~c (N of them) I Take their inner product: r c using N multiplications 2. listing algorithms. Idea - Block Matrix Multiplication The idea behind Strassen's algorithm is in the formulation of matrix multiplication as a recursive problem. Describe how an array can be effectively used to store a sparse matrix. Matrix mulitplication using Linked List - posted in C and C++: I have to implement Matrix multiplication using singly linked list. You don't need multiplication facts to use the Russian peasant algorithm; you only need to double numbers, cut them in half, and add them up. The unit of. We then \"combine\" the middle row of the key matrix with the column vector to get the middle element of the resulting column vector. The problem is quite easy when n is relatively small. John, Your comment about matrix multiplication was forwarded to the mahout-user emailing list. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. Section 3 pro-vides implementation details on our design. In the other hand the algorithm of Strassen is not much faster than the general n^3 matrix multiplication algorithm. 2 shows the calculate steps of covariance matrix, mainly including: complex conjugate multiplication between the lines of input matrix, then do an accumulation operation. This immediately leads to a counting algorithm with running time \u0398(n3) respectively \u0398(n\u03b3), where \u03b3 is the matrix multiplication exponent. Divide-and-conquer multiplication. 2 Strassen\u00d5s algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. These are lecture notes used in CSCE 310 (Data Structures & Algorithms) at the University of Nebraska|Lincoln. count example presented in Section 2. The matrix M and the vector v each will be stored in a file of the DFS. Pseudo-code for i = 1 : n yi yi +axi Number of \ufb02ops is 2n Pseudo-code cannot run on any computer, but are human readable and straightforward to convert into real codes in any programming language (e. Matrix multiplication is not commutative, but it is associative, so the chain can be parenthesized in whatever manner deemed best. Purdue University Purdue e-Pubs ECE Technical Reports Electrical and Computer Engineering 9-1-1992 Implementation of back-propagation neural networks with MatLab. Prim's Algorithm Implementation in C++. Write An Algorithm To Find The Power Of A Number. Flowchart for Matrix multiplication : Algorithm for Matrix multiplication :. Strassen's matrix multiplication program in c 11. Program the divide and conquer matrix multiplication using 1) standard algorithm 2) recursion 3) strassen\u2019s method. Write the Pseudo Code of the matrix multiplication program that performs the worst execution time and explain why - Answered by a verified Programmer We use cookies to give you the best possible experience on our website. Sparse matrices, which are common in scientific applications, are matrices in which most elements are zero. The dimensions are stored in array. Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. Our first example of dynamic programming is an algorithm that solves the problem of matrix-chain multiplication. 2 Multithreaded matrix multiplication 792 27. (Otherwise, you should read Section D. When distributing the vector among processors, implement the algorithm shown in Figure (b) on page 22 of lecture notes \u201c Parallel matrix algorithms (part 2) \u201d. Summary: The two fast Fibonacci algorithms are matrix exponentiation and fast doubling, each having an asymptotic complexity of $$\u0398(\\log n)$$ bigint arithmetic operations. The simplest sparse matrix data structure is a list of the nonzero entries in arbitrary order. 1 The SUMMA Algorithm. The introduction of the technique is attributed to a 1962 paper by Karatsuba, and indeed it is sometimes called Karatusba multiplication. We can build a sketch as we scan through the matrix. I am trying to write pseudo code in my paper. I implement these three algorithms from pseudocode from the book to Java code: Merge-Sort Java version\uff1a. Animated Algorithms (sorting, priority queues, Huffman, Matrix chain multiplication, MST, Dijkstra) Graph Algorithms (Dijkstra, Prim, Kruskal, Ford-Fulkerson) Java and Web Based Algorithm Animation (JAWAA). What is the main operation of this algorithm? c. Solutions for CLRS Exercise 4. 4 RESULTS We evaluate our implementation by testing its performance on one. com Free Programming Books Disclaimer This is an uno cial free book created for educational purposes and is. There is a faster way to multiply, though, caled the divide-and-conquer approach. 3 perform the multiplication operation. Output Y: d by N matrix consisting of d D dimensional embedding coordinates for the input points. Freivalds' algorithm is a probabilistic randomized algorithm used to verify matrix multiplication. Better asymptotic bounds on the time required to multiply matrices have been known since the work of Strassen in the 1960s, but it is still unknown what the optimal time is (i. In this post I will explore how the divide and conquer algorithm approach is applied to matrix multiplication. Purdue University Purdue e-Pubs ECE Technical Reports Electrical and Computer Engineering 9-1-1992 Implementation of back-propagation neural networks with MatLab. Have you considered doing the multiplication in a single step by storing the the first matrix in column major order and the second in row major order?. Explain why memoization fails to speed up a good divide-and-conquer algorithm like merge-sort. CS 2073 Lab 10: Matrix Multiplication Using Pointers Chia-Tien Dan Lo Department of Computer Science, UTSA I Objectives Show how to manipulate commandline arguments in C Demonstrate your ability to read matrices from les Demonstrate your ability to use pointers for matrix multiplication II Hand-in Requirements. You are given 5 different algorithms for different purposes and their pseudocodes that are listed below. Related Questions More Answers Below. So Matrix Chain Multiplication problem has both properties (see this and this) of a dynamic programming problem. Since we have not covered multiplication yet, a function has been provided to you. Two groups of algorithms belonging to this class are called the matrix method, and the Wallace-tree method, respectively. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. LLE Algorithm Pseudocode (Notes, e. 2 Algorithmic Techniques 5. Find f(n): n th Fibonacci number. 3) you should encapsulate a matrix into a class, if it is supposed to be c++ (then 2 will be obsolete) 4) your code will be easier to understand (for you as well) if you use better names for x,y,z,i, k and j. Application. using matrix multiplication Let G=(V,E) be a directed graph. Book shows pseudocode for simple divide and conquer matrix multiplication: n = A. r-1 (mod n) where the integers a and b are smaller than the modulus. Matrix multiplication of two sparse matrices is a fundamental operation in linear Baye sian inverse problems for computing covariance matrices of observations and a posteriori uncertainties. HOME ; A comparison of numerical approaches to the solution of the time-dependent Schr\u00f6dinger equation in one dimension. complexity of matrix multiplication is n2 (2n \u22121) = 2 \u22c5(2 \u22121)\u22c5\u03c4 T1 n n (8. The current best algorithm for matrix multiplication O(n2:373) was developed by Stanford's own Virginia Williams[5]. Pseudocode for the algorithm is given in Figure 1. Make good use of matrix multiplication! It avoids a lot of loops, so it makes your code cleaner and faster! DO NOT assume that I will answer your email questions or posts to the discussion forum after 3pm on Sunday. which can be signi cantly smaller than their dense equivalent. Pseudocode for Matrix Vector Multiplication by Mapreduce. , the shapes are 2 n \u00d7 2 n for some n. i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table. \\begin{algorithm} \\caption{Euclid\u2019s algorithm}\\label{euclid} \\. Since we have not covered multiplication yet, a function has been provided to you. Matrix-matrix multiplication takes a triply nested loop. 2 Multithreaded matrix multiplication 792 27. 2 Matrix-Matrix Multiplication. So assuming that both these multiplication steps are executed every time the loop executes, we see that 2. Alternative approaches can be seen as straight forward iteration over the nodes or edges of the graph. In grade 1, you learned how to count up to ten and to do basic arithmetic using your ngers. Integer-multiplication, Matrix Multiplication - Strassen Alg You study after every class/week, the syllabus accumulates fast before you know! Aug 26, W (Drop w/o W grade, Aug 28) Dynamic Programming: 0-1Knapsack. The definition of matrix multiplication is that if C = AB for an n \u00d7 m matrix A and an m \u00d7 p matrix B, then C is an n \u00d7 p matrix with entries = \u2211 =. As examples, pseudocode is presented for the inner product, the Frobenius matrix norm, and matrix multiplication. Explain why memoization fails to speed up a good divide-and-conquer algorithm like merge-sort. De\ufb01ne the meaning of your variables. Can some one please help me to format it. Section 5 provides a com-parison with related works. With the current implementation of the cuBlas functions we need to write kernel code to do this efficiently. The problem is quite easy when n is relatively small. SPARSE MATRICES C/C++ Assignment Help, Online C/C++ Project Help and Homework Help introduction A matrix is a mathematical object that arises in many physical problems. Pseudocode Matrix Multiplication. This relies on the block partitioning which works for all square matrices whose dimensions are powers of two, i. In matrix multiplication, one row element of first matrix is individually multiplied by all column elements and added. Order of both of the matrices are n \u00d7 n. The word is derived from the phonetic pronunciation of the last name of Abu Ja'far Mohammed ibn Musa al-Khowarizmi, who. Adjacency-matrix and adjacency-list representations Breadth-first and depth-first search using adjacency lists Computing connected components of a graph Strongly-connected and biconnected components Topological sorting Algebraic algorithms: Strassen matrix multiplication algorithm The Four Russians boolean matrix multiplication Winograd's algorithm. Matrix Multiplication Algorithm. Faster matrix multiplication in general is an important applied topic, because it can speed up all sorts of scientific, engineering, and ML algorithms that have it as a step (often one of the bottleneck steps). At the end of the lecture, we saw the reduce SUM operation, which divides the input into two halves, recursively calls itself to obtain the sum of these smaller inputs, and returns the sum of the results from those. Specifically, an input matrix of size can be divided into 4 blocks of matrices. GATEBOOK Video Lectures 3,203 views. 5D (Ballard and Demmel) \u00a92012 Scott B. Of course, writing pseudocode is child's play compared to actually implementing a real algorithm. In section 4, we discuss our experimental results from the real hardware implementation. return C {C = [cij ] is the product of A and B} 22. Benchmarked it to be 4x faster than the scalar version (on a Pentium M, using GCC 4.", "date": "2019-12-10 03:20:41", "meta": {"domain": "zonaunderground.it", "url": "http://bkdd.zonaunderground.it/matrix-multiplication-algorithm-pseudocode.html", "openwebmath_score": 0.7114400267601013, "openwebmath_perplexity": 795.9439563852297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9919380084717283, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8687264139883583}}
{"url": "http://mathhelpforum.com/pre-calculus/117597-cube-root-times-cube-root.html", "text": "# Math Help - Cube Root times Cube Root\n\n1. ## Cube Root times Cube Root\n\ncube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)\n\nI converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.\n\nIs this the correct way to write the original problem another way?\n\nI finally got (432m^4n^2)^2/3.\n\nIs this correct?\n\n2. Originally Posted by sharkman\ncube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)\n\nI converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.\n\nIs this the correct way to write the original problem another way?\n\nI finally got (432m^4n^2)^2/3.\n\nIs this correct?\nNearly there. Recall that $\\sqrt[3]{a} \\times \\sqrt[3]{b} = \\sqrt[3]{ab}$\n\nIn your question you have $\\sqrt[3]{16(m^2)(n)} \\times \\sqrt[3]{27(m^2)(n)}$.\n\nUsing the fact I gave first, this means than the answer is...\n\nHint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\\frac{2}{3}$ - that you got wrong.\n\nOriginally Posted by craig\nNearly there. Recall that $\\sqrt[3]{a} \\times \\sqrt[3]{b} = \\sqrt[3]{ab}$\n\nIn your question you have $\\sqrt[3]{16(m^2)(n)} \\times \\sqrt[3]{27(m^2)(n)}$.\n\nUsing the fact I gave first, this means than the answer is...\n\nHint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\\frac{2}{3}$ - that you got wrong.\nThe answer should be cubert{432m^4n^2}, right?\n\n4. Correct\n\n5. ## great\n\nOriginally Posted by craig\nCorrect\nGreat! Thanks!\n\n6. Originally Posted by sharkman\nThe answer should be cubert{432m^4n^2}, right?\nDon't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\\sqrt[3]{m^4}= \\sqrt[3]{m^3}\\sqrt[3]{m}= m\\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\\sqrt[3]{432}= \\sqrt[3]{2^3}\\sqrt[3]{3^3}\\sqrt[3](2)= 6\\sqrt[3]{2}$.\n\nThe best way to write your answer is $6m\\sqrt[3]{2mn^2}$.\n\n7. ## ok\n\nOriginally Posted by HallsofIvy\nDon't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\\sqrt[3]{m^4}= \\sqrt[3]{m^3}\\sqrt[3]{m}= m\\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\\sqrt[3]{432}= \\sqrt[3]{2^3}\\sqrt[3]{3^3}\\sqrt[3](2)= 6\\sqrt[3]{2}$.\n\nThe best way to write your answer is $6m\\sqrt[3]{2mn^2}$.\nI understand. You decided to break the problem down a bit more.", "date": "2014-12-21 05:33:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/117597-cube-root-times-cube-root.html", "openwebmath_score": 0.9463050961494446, "openwebmath_perplexity": 1023.8726128511564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9881308817498917, "lm_q2_score": 0.8791467770088162, "lm_q1q2_score": 0.8687120799532969}}
{"url": "https://math.stackexchange.com/questions/3182963/prove-that-sim-is-an-equivalence-relation", "text": "# Prove that $\\sim$ is an equivalence relation\n\nLet $$A = \\{1, 2, 3,...,9\\}$$ and let $$\\sim$$ be the relation on $$A\\times A$$ defined by $$(a,b) \\sim(c,d)$$ if $$a+d = b+c$$.\n\nProve that $$\\sim$$ is an equivalence relation.\n\nReally stuck on this question, maybe its the use of ~ that is throwing me off. Any suggestions on how to do this? Maybe I could swith ~ out for something else while solving then put it back in.\n\nThanks\n\n\u2022 You have to show ~ is reflexive, symmetric, and transitive. Note $a+d=b+c \\iff a-b=c-d$ \u2013\u00a0J. W. Tanner Apr 10 '19 at 21:20\n\u2022 (1) If the symbol ~ annoys you, simply replace it by the more ordinary symbol R. (2) If you are stuck, it may come from the fact that, in this problem, the \"elements\" you are deling with are, in fact, ordered pairs. So for example, proving reflexivity amounts to proving that any ordred pair (a,b) has the relation R ( or ~ ) with itself , that is (a,b) R (a,b). Using the defining formula, you could see what you have to prove, and then prove if using basic arithmetic. \u2013\u00a0Saint James Apr 10 '19 at 21:25\n\u2022 The symbol \"~\" is often used to denote a relation in case this relation is an equivalence relation. But, before having proved that it is actually an equivalence relation in this case, you are not allowed to read it as \" is equivalent to\". So, accordind to me, the best thing do do is to replace it by the ordinary symbol R, while you are doing your proof. \u2013\u00a0Saint James Apr 10 '19 at 21:35\n\nClearly, the statement $$a+d=b+c$$ is just saying that $$a-b=c-d$$, i.e. two ordered pairs are similar if the difference of their terms is equal. To show it's an equivalence relation, we need reflexivity, transitivity and symmetry. The first and last are obvious.\n\nFor transitivity, we need to show $$x\\sim y$$ and $$y\\sim z$$ implies $$x\\sim z$$. Let $$x=(x_1,x_2),y=(y_1,y_2),z=(z_1,z_2)$$. The conditions tell you that $$x_1-x_2=y_1-y_2$$, but also that $$y_1-y_2=z_1-z_2$$. Clearly then, $$x_1-x_2=z_1-z_2$$. But this is exactly what it means for $$x\\sim z$$!\n\nI don't normally do this but I think it may clear some conceptions up. If you want to get to how to effectively solve this skip to the end.\n\n...\n\nMaybe an example first:\n\nIf we let $$A = (3,6)$$ then $$A\\sim (m,n)$$ if $$3+ n = 6+m$$ so the list of $$(m,n)$$ where $$A \\sim (m,n)$$ are the ones where $$n = 3+m$$ so $$\\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\\}$$ are all the $$(m,n)$$ where $$A \\sim (m,n)$$.\n\nWe can go a little further.\n\nIf $$(a,b) \\equiv (c,d)$$ then $$a+d = b+c$$ and $$a-b = c-d$$.\n\nSo the \"classes\" are all the ones related to $$(x_1, x_2)$$ where $$x_1 - x_2 =$$:\n\n$$-8$$: $$\\{(1,9)\\}$$\n\n$$-7$$: $$\\{(1,8),(2,9)\\}$$\n\n$$-6$$: $$\\{(1,7),(2,8),(3,9)\\}$$\n\n$$-5$$: $$\\{(1,6),(2,7),(3,8),(4,9)\\}$$\n\n$$-4$$: $$\\{(1,5),(2,6),(3,7),(4,8),(5,9)\\}$$\n\n$$-3$$: $$\\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\\}$$\n\n$$-2$$: $$\\{(1,3),(2,4),(3,5),(4,6),(5,7),(6,8),(7,9)\\}$$\n\n$$-1$$: $$\\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9)\\}$$\n\n$$0$$: $$\\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)\\}$$\n\n$$1$$: $$\\{(2,1),(3,2),(4,3),(5,4),(6,5),(7,6),(8,7),(9,8)\\}$$\n\n$$2$$: $$\\{(3,1),(4,2),(5,3),(6,4),(7,5),(8,6),(9,7)\\}$$\n\n$$3$$: $$\\{(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)\\}$$\n\n$$4$$: $$\\{(5,1),(6,2),(7,3),(8,4),(9,5)\\}$$\n\n$$5$$: $$\\{(6,1),(7,2),(8,3),(9,4)\\}$$\n\n$$6$$: $$\\{(7,1),(8,2),(9,3)\\}$$\n\n$$7$$: $$\\{(8,1),(9,2)\\}$$\n\n$$8$$: $$\\{(9,1)\\}$$\n\nAll the pairs in each set are related to each other this is an \"equivalence\" because every pair is is exactly one class and can be interchanged with others in the class with no ambiguity.\n\nA relationship could not be equivalent if some elemtns are in more than one list. Of some elements aren't in any list. Or if such bunch of lists wouldn't make any sense to make.\n\nA relationship is an equivalence if\n\n1) it is reflexive. that is for every pair $$(a,b)$$ then $$(a,b)\\sim (a,b)$$ or for every $$(a,b)$$ then $$a-b = a-b$$. That's always true.\n\nIf we view it as $$(a,b)\\sim (c,d)$$ if $$a-b = c-d$$ then we'd have $$a-b = a-b$$ which is of course true.\n\n2) it is symmetric. For any pair $$(a,b) \\sim (c,d)$$ then $$(c,d)\\sim (a,b)$$. That is is $$a+d = b+c$$ then $$c+b = d+ a$$. That's ... clear.\n\nIf we view is as $$(a,b)\\sim (c,d)$$ if $$a-b= c-d$$ then if $$a-b = c-d$$ then $$c-d = a-b$$ so $$(a,b)\\sim(c,d)\\iff (c,d)\\sim (a,b)$$.\n\n3) it is transitive. That is if $$(a,b)\\sim (c,d)$$ and $$(c,d)\\sim (e,f)$$ then $$(a,b) \\sim (e,f)$$.\n\nSo if $$a+ d=b+c$$ and $$c+f = d+ e$$ then we have to prove that means $$a+f = b+e$$.\n\nWe can do that via arithmetic $$a+d +c+f =b+c + d + e$$ so $$a+f = b+e$$.\n\nJust use the definitions:\n\nFor reflexivity, you have to show that for any $$(a,b)$$, it is true that $$(a,b)\\sim (a,b)$$, i.e. that $$a+b=b+a$$. Well, $$a$$ and $$b$$ are natural numbers, and those are commutative, so yes, this is true.\n\nSymmetry: here you have to show that if $$(a,b) \\sim (c,d)$$, then $$(c,d) \\sim (a,b)$$, which is to say: show that if $$a+d=b+c$$, then $$c+b=d+a$$ ... can you show this?\n\nFinally, now that I have shown what you need to prove to demonstrate reflexivity and symmetry, can you figure out what you need to show to demonstrate transitivity?", "date": "2020-03-31 08:06:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3182963/prove-that-sim-is-an-equivalence-relation", "openwebmath_score": 0.9233266711235046, "openwebmath_perplexity": 166.53604951283728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308765069576, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8687120769084115}}
{"url": "https://byjus.com/question-answer/the-average-score-of-boys-in-an-examination-of-a-school-is-71-and-that/", "text": "Question\n\n# The average score of boys in an examination of a school is $$71$$ and that of girls is $$73$$. The average score of the school in that examination is $$71.8$$. The ration of the number of boys to the number of girls appeared in the examination, is\n\nA\n3:2\nB\n3:4\nC\n1:2\nD\n2:1\n\nSolution\n\n## The correct option is A $$3:2$$Let there be $$n_1$$ boys and $$n_2$$ girls. and\u00a0 $$\\overline{X_1}$$ and $$\\overline{X_2}$$ be the average scores of boys and girls respectively.Then $$\\bar { { X }_{ 1 } } =71,\\bar { { X }_{ 2 } } =63$$ and $$\\overline{X} = 71.8$$$$\\therefore \\quad \\overline{X} = \\displaystyle\\frac{n_1\\overline{X_1} + n_2\\overline{X_2}}{n_1+n_2}$$$$\\Rightarrow \\quad 71.8 = \\displaystyle\\frac{n_1\\times71 + n_2\\times73}{n_1+n_2}$$$$\\Rightarrow \\quad 71.8n_1 + 71.8n_2 = 71n_1 + 73n_2$$$$\\Rightarrow \\quad 0.8n_1 = 1.2n_2 \\quad \\Rightarrow 8n_1 = 12n_2$$$$\\Rightarrow \\quad \\displaystyle\\frac{n_1}{n_2} = \\displaystyle\\frac{12}{8} = \\displaystyle\\frac{3}{2}$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-29 10:21:38", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/the-average-score-of-boys-in-an-examination-of-a-school-is-71-and-that/", "openwebmath_score": 0.5436347126960754, "openwebmath_perplexity": 579.7107935943604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9881308786041315, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.8687120709300197}}
{"url": "http://www.lofoya.com/Solved/1739/two-squares-are-chosen-at-random-on-a-chessboard-what-is-the", "text": "# Difficult Probability Solved QuestionAptitude Discussion\n\n Q. Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?\n \u2714\u00a0A. 1/18 \u2716\u00a0B. 64/4032 \u2716\u00a0C. 63/64 \u2716\u00a0D. 1/9\n\nSolution:\nOption(A) is correct\n\nSample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.\n\nSo sample space$=^{64}C_2$\n\nNow there are 7 unique adjacent square sets in each row and each column.\n\ni.e. favourable cases will be 7\u00d7(8 rows + 8 columns) = 112.\n\nHence\u00a0Required Probability\n\n$=\\left(\\dfrac{\\text{Favourable cases}}{\\text{Sample Space}}\\right)$\n\n$=\\dfrac{112}{^{64}C_2}$\n\n$=\\dfrac{1}{18}$\n\nAlso,\u00a0there could be other solution too if we consider ALL possible squares and not only the smallest squares. (i.e. if we consider that total squares are more than 64).\n\nIn that case, let us first calculate the sample space i.e. total number of squares on a chess board.\n\n1, 8x8 square\n4, 7x7 squares\n9, 6x6 squares\n16, 5x5 squares\n25, 4x4 squares\n36, 3x3 squares\n49, 2x2 squares\n64, 1x1 squares\n\nTherefore, there are actually:\n\n$=64+49+36+25+16+9+4+1$\n\n$=8^2+7^2+6^2+...1^2=204$\n\nsquares on a chessboard!\n\nSo sample space$=^{204}C_4$\n\nNow if we assume that 2 squares can have a side common only if they are of the same dimensions:\nLet the length of the smallest square be 1 unit.\n\nSo the possibility of selecting 2 squares with a common side (As calculated in the first method above):\n\nhaving the side length greater than 4 unit length = 0\nhaving the side length of 4 unit = 1\u00d7(2 rows + 2 columns) = 4\nhaving the side length of 3 unit = 3\u00d7(6 rows + 6 columns) = 36\nhaving the side length of 2 unit = 5\u00d7(7 rows + 7 columns) = 70\nhaving the side length of 1 unit = 7\u00d7(8 rows + 8 columns) = 112\n\nSo total number of favourable cases:\n\n$=4+36+70+112=222$\n\nHence\u00a0Required Probability\n\n$=\\left(\\dfrac{\\text{Favourable cases}}{\\text{Sample Space}}\\right)$\n\n$=\\dfrac{222}{^{204}C_2}$\n\nEdit:\u00a0For an alternative solution, check comment from\u00a0Bhavya Shah.\n\nEdit 2:\u00a0For yet another alternative solution, check comment by\u00a0RandomMathMan.\n\n## (12) Comment(s)\n\nGaurav Karnani\n()\n\nTotal number of Cases for a chessboard$7*(7*2+1)$ that equals $112$\n\nP= $112/64C2$\n\nGaurav Karnani\n()\n\nForgot to add seven to the number of cases for the last row. please amend it accordingly.\n\nRandomMathMan\n()\n\nThere's a much easier way to do this. On a chessboard, there are 36 squares that border 4 other squares, 24 squares that border 3 other squares, and 4 squares that border only 2 other squares.\n\n$\\dfrac{36}{64} \\times \\dfrac{4}{63} + \\dfrac{24}{64} \\times \\dfrac{3}{63} + \\dfrac{4}{64} \\times \\dfrac{2}{63} = \\dfrac{1}{18}$\n\nSiddharth\n()\n\nExactly how I too solved this. Much simpler than the other alternatives given here.\n\nBhavya Shah\n()\n\nFor each row, there are 7 ways to select adjacent squares and there are 8 such rows.\n\nSimilarly for columns, 7 squares could be selected from one column and there are 8 such columns.\n\nSo no. of ways of selecting adjacent squares $= 7*8*2$\n\nTotal number of ways of selecting squares of chessboard $= ^{64}C_2$\n\nProbability $=\\dfrac{7*8*2}{^{64}C_2}=\\dfrac{1}{18}$\n\nPoonam Pipaliya\n()\n\nsir,,how 7 squares in each row and column remain..??..!! and how it can be favorable case..??..!!\n\nAnuj\n()\n\nLets talk of the rows first. Once you pick up a square, you leave that row, so you have 7 remaining rows.\n\nSimilar argument is valid for columns.\n\nRishi\n()\n\nI guess the alternative method given in the solution is easier to understand, why don't you have a look at that.\n\nArshal\n()\n\nGuys, I think I'm right but, as I have been hitting my head on the wall of probability for last 7 hours I'm kinda in concussion...\n\nSo is choosing 2 consecutive elements out of 8 equal to choosing 1 element out of 7\n\n()\n\n$C(4,1)*2+C(24,1)*3+C(36,1)/C(64,2)$\n\nThere are 4 corner squares for which there are only 2 choices for selecting next square....\nSimilarly, 24 squares for which there are 3 choices and remaining 36 have 4 choices\n\nSo, the answer should be 1/9\n\nCorrect me if i am wrong!!!\n\n()\n\nAs I counted every square non repeated\n\nSo, total no of unique square=\n\n$\\dfrac{((C(4,1)*2+C(24,1)*3+C(36,1)*4)/2)}{C(64,2)}$\n\n$=\\dfrac{1}{18}$", "date": "2016-12-10 10:46:38", "meta": {"domain": "lofoya.com", "url": "http://www.lofoya.com/Solved/1739/two-squares-are-chosen-at-random-on-a-chessboard-what-is-the", "openwebmath_score": 0.7111933827400208, "openwebmath_perplexity": 1123.7754098274543, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864273087513, "lm_q2_score": 0.8774768002981829, "lm_q1q2_score": 0.8686901225735127}}
{"url": "http://mathhelpforum.com/business-math/231150-calculating-compound-interest-non-linear-interest-growth.html", "text": "# Thread: calculating compound interest with non-linear interest growth\n\n1. ## calculating compound interest with non-linear interest growth\n\nHello, I would really appreciate some help with the following problem.\n\nI'd like to derive a formula to determine the value of an investment where the annual yield increases by a fixed percentage each year. The interest paid is re-invested each year. The standard compounding formulas use a fixed yield so I've not been able to use them in this case.\n\nAs an example, suppose I have 1000 dollars in an investment with an initial starting yield (y) of 3% which grows (g) at 5% each year.\n\nEnd of Year 1 : Yield was 3.00%, Investment value = 1000 + (1000 * 3%) = 1030.00\nEnd of Year 2 : Yield was 3.15% (3% increased by 5%), Investment value = 1030 + (1030 x 3.15%) = 1062.45\nEnd of Year 3 : Yield was 3.31% (3.15% increased by 5%), Investment value = 1062.45 + (1062.45 x 3.31%) = 1097.59\n\nI believe the calculation is a series in the form\n\nValue = P.(1+y0).(1+y1).(1+y2).(...)\n\nwhere P = initial investment amount, r0 = yield in year 0, r1 = yield in year 1, r2 = yield in year 2 etc.\nand where y0 = y.(1+g)^t etc.\n\nIf I write the above formula out substituting y0, y1, etc. I get\n\nValue = P.(1+y(1+g)^t).(1+y(1+g)^(t-1)).(1+y(1+g)^(t-2)).(...) where t ranges from 0 to the target year n.\n\nIs it possible to convert this expression into a general formula in the form of Value = f(t) where t = time in years, e.g. some kind of exponential function and how might I go about doing that?\n\nThank you for any help or insight that you can provide me!\n\n~ Trevor\n\n2. ## Re: calculating compound interest with non-linear interest growth\n\nOne way of doing this would be to find a single yield that results in a same future value using the series of yields\n\n$FV = R (1+i)^n$\n\n$ln(1+i_{s})=\\frac{1}{n}\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)$\n\n$i=e^{ln(1+i_s)}-1$\n\nCode:\nR=1000\nr=3%\ng=5%\nn=3\n$ln(1+i_{s})=0.031037549112832$\n\n$i=e^{0.031037549112832}-1$\n\n$i=3.15242359792658\\%$\n\n$FV = 1000 (1+3.15242359792658\\%)^3$\n\n$FV = 1000 (1.0315242359792658)^3$\n\n$FV = 1000 (1.097585368)$\n\n$FV = 1,097.59$\n\n3. ## Re: calculating compound interest with non-linear interest growth\n\nHi GhostAccount,\n\nThank you for the fast response, that was very helpful! I've checked the results manually using Excel and your method generates all FV values that I expect from n=1 to n=10. There is one point that I don't fully understand however.\n\nI understand that your answer is solving the formula below for i :\n\n$R (1+i)^n = R\\sum_{k\\ =\\ 0}^{n-1} (1+r(1+g)^k)$\n\nThis tells me that I should be able to calculate any given FV for a value of n by manually summing the rhs. However I don't get the expected answer when I tried to do this in Excel:\n\nn | Expected vs Excel\n1 | 1030.00 vs 1030.00 ( = 1000 + 30 )\n2 | 1062.45 vs 1061.50 ( = 1000 + 30 + 31.5 )\n3 | 1097.59 vs 1094.58 ( = 1000 + 30 + 31.5 + 33.075 )\n4 | 1135.70 vs 1129.30 ( = 1000 + 30 + 31.5 + 33.075 + 34.7288 )\n\nDid I interpret the FV summation formula correctly and just get Excel wrong?\n\nI didn't realize that the compound growth + compound interest could be expressed as a summation like that - I thought this was only calculating the amount of interest that $1000 would produce in n years with a compounded rate and not compounding the principal amount as well. I was thinking that $FV=R(1+r(1+g)^0)(1+r(1+g)^1)(1+r(1+g)^2)$ etc A summation formula is what I really want to reach as I can then try to substitute a general identity for it and derive a formula in terms of r,g and n that I can use to calculate each expected value for a given t, for example using $\\sum_{i\\ =\\ m}^{n-1} a^i = \\frac{a^m - a^n}{1-a}$ Thanks again for your help! 4. ## Re: calculating compound interest with non-linear interest growth Originally Posted by tw35758 Hi GhostAccount, Thank you for the fast response, that was very helpful! I've checked the results manually using Excel and your method generates all FV values that I expect from n=1 to n=10. There is one point that I don't fully understand however. I understand that your answer is solving the formula below for i : $R (1+i)^n = R\\sum_{k\\ =\\ 0}^{n-1} (1+r(1+g)^k)$ This tells me that I should be able to calculate any given FV for a value of n by manually summing the rhs. However I don't get the expected answer when I tried to do this in Excel: n | Expected vs Excel 1 | 1030.00 vs 1030.00 ( = 1000 + 30 ) 2 | 1062.45 vs 1061.50 ( = 1000 + 30 + 31.5 ) 3 | 1097.59 vs 1094.58 ( = 1000 + 30 + 31.5 + 33.075 ) 4 | 1135.70 vs 1129.30 ( = 1000 + 30 + 31.5 + 33.075 + 34.7288 ) Did I interpret the FV summation formula correctly and just get Excel wrong? Thanks again for your help! The formula I presented in two parts at first finds the continuously compounded yearly rate (geometric average) that has to be converted to a yearly rate See the following for complete formula $R (1+i)^n = R(e^{\\sum_{k\\ =\\ 0}^{n-1} (1+r(1+g)^k)})$ Edit: This formula should work as I am short on time now thus I didn't find time to confirm the results 5. ## Re: calculating compound interest with non-linear interest growth Was trying for geometric series...like the 1000 bucks creates a guaranteed 30 bucks annual deposit (the 3%), so we'd have FV of 1000 plus FV of 30 annuity....but finally realised that won't work... i = initial interest rate (.03) f = increase factor (1.05) n = number of years (make it 10) FV = (1 + i*f^0)(1 + i*f^1)(1 + i*f^2)...........(1 + i*f^(n-1)) = 1.4481148... That's for the proverbial one dollar, of course; so 1,448.11 if$1000.\n\nSoooo....looks like I'm kinda repeating GA's \"spectacular!\" work.\n\nWhadda hell is the matter with the dollar sign here??\n\n6. ## Re: calculating compound interest with non-linear interest growth\n\nOriginally Posted by GhostAccount\nThe formula I presented in two parts at first finds the continuously compounded yearly rate (geometric average) that has to be converted to a yearly rate\n\nSee the following for complete formula\n\n$R (1+i)^n = R(e^{\\sum_{k\\ =\\ 0}^{n-1} (1+r(1+g)^k)})$\n\nEdit: This formula should work as I am short on time now thus I didn't find time to confirm the results\nBeautiful.\nIt should work but there's a slight typo on the exponent of e; $ln$ is missing.This could have been\n\n$R (1+i)^n = R(e^{\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k})$\n\nMust agree with Sir Wilmer; spectacular work indeed.\nWas about to work on a derivation for this one when i saw it earlier but got too sober for a day and a half.\nDrunk as I am right now, I am nevertheless definitely green with envy for this piece of beauty.\nI'm almost certain I'm not the only one who'd like to see the analysis of this derivation Sir GhostAccount.\nBut as you said, this is one way of doing this. There might be another (or other's).\nMight have to soak me brain with me favorite tequila brand for one such other future possibe derivation.\n\n7. ## Re: calculating compound interest with non-linear interest growth\n\nThanks everyone, I've learned a lot so far!\n\nI did some more research and realized that the correct way to represent the FV formula for the compounded growth of compounded interest is\n\n$FV = R(\\prod_{k\\ =\\ 0}^{n-1} (1+r(1+g)^k)})$\n\nSince FV is essentially R(a)(b)(c)(...) we can convert this to a summation using logs.\n\nSince\n\n$e^{ln(x)} = x$\n\nand\n\n$ln(abc) = ln(a)+ln(b)+ln(c)$\n\nso\n\n$\\boxed{FV= R(e^{\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)})}$\n\n$FV=R(1+i)^n$\n\nso\n\n$R(1+i)^n= R(e^{\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)})$\n\neliminating R and taking the log of both sides we get\n\n$nln(1+i) = ({\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)})$\n\n$ln(1+i) = \\frac{1}{n}({\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)})$\n\n$1+i = e^{\\frac{1}{n}({\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)})}$\n\nand now we're back to the answer GhostAccount initially provided as\n\n$\\boxed{i = e^{\\frac{1}{n}({\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)})} - 1}$\n\nOne last question...is it possible to express\n\n${\\sum_{k\\ =\\ 0}^{n-1} ln(1+r(1+g)^k)}$\n\nin a single function f(n) where it outputs the expected FV for n=0, n=1, n=2 etc?\n\nI could do something similar for the normal compounded case, e.g.\n\n$\\sum{r(1+g)^k} = \\frac{r((1+g)^{(k+1)}-1)}{g}$\n\nbut I'm completely lost on how to do a similar conversion for the FV formula or if it's even possible. I'm trying to create an Excel file comparing the result of various investments with different dividend yields and dividend growth rates, so having a single formula in a cell for each value of n would make creating that file much easier.\n\n8. ## Re: calculating compound interest with non-linear interest growth\n\nWhadda hell is the matter with the dollar sign here??\nDollar sign invokes the LaTeX editor. Confused me when I started posting here. See Dollar Signs\n\n9. ## Re: calculating compound interest with non-linear interest growth\n\nI realized I should make my question more specific...\nSuppose I have two different dividend paying stocks - one has an initial yield r1 with a growth rate of r1 and the second has an initial yield of r2 with a growth rate of r2.\nI'd like to be able to analyze / compare the two performance stocks when I re-invest the dividend payments and answer questions such as:\n\nQ1: With a constant stock price, what is the total return of principal + dividends after time t?\n\nQ2: How many years will either stock take to reach a certain dividend payout?\n\nI was able to do this for the non-compounding case and I posted an article on my blog explaining the formulae. The formulae and explanations above are really helpful but I'm stuck in transforming the summation formula above to a more usable form. I want to avoid continous compounding but keep to annually compounded calculations.\n\n10. ## Re: calculating compound interest with non-linear interest growth\n\nOriginally Posted by tw35758\nbut I'm completely lost on how to do a similar conversion for the FV formula or if it's even possible. I'm trying to create an Excel file comparing the result of various investments with different dividend yields and dividend growth rates, so having a single formula in a cell for each value of n would make creating that file much easier.\n\nFor now, see the following code in Excel to use the tadFVSchedule function that answers the original question\n\nCode:\nPublic Function tadFVSchedule(ByVal pv As Double, ByVal r As Double, ByVal g As Double, ByVal n As Double, Optional ByRef c As Double = 1, Optional ByRef p As Double = 1, Optional ByRef d As Double = 1) As Double\nDim fraction As Double\nDim sum As Double\nDim k As Integer\n\nsum = 0#\n\nfraction = n - Int(n)\n\nFor k = 0 To Int(n - 1)\nsum = sum + p * d * Log(1 + r * (1 + g) ^ k)\nNext k\n\nIf (fraction <> 0) Then\nk = Int(n + 0.5)\nsum = sum + ((fraction - 1) * p + p * d) * Log(1 + r * (1 + g) ^ k)\nEnd If\n\nEnd Function\n\n pv r g n c p d fv 1000 3% 5% 3 1 1 1 1097.59 1000 3% 5% 10 1 1 1 1448.11 1000 3% 5% 3.5 1 1 1 1117.42 1000 3% 5% 10.5 1 1 1 1484.80 1000 3% 5% 3 =1/12 1 1 1097.59 1000 3% 5% 10 =1/12 1 1 1448.11 1000 3% 5% 3 1 =1/12 1 1007.79 1000 3% 5% 10 1 =1/12 1 1031.34 1000 3% 5% 3 =1/12 =1/12 1 1007.79 1000 3% 5% 10 =1/12 =1/12 1 1031.34 1000 3% 5% 3 1 1 =1/2 1047.66 1000 3% 5% 10 1 1 =1/2 1203.38 1000 3% 5% 3.5 1 1 =1/2 1047.66 1000 3% 5% 10.5 1 1 =1/2 1203.38 1000 3% 5% 3 =1/12 1 =1/2 1047.66 1000 3% 5% 10 =1/12 1 =1/2 1203.38 1000 3% 5% 3 1 =1/12 =1/2 1003.89 1000 3% 5% 10 1 =1/12 =1/2 1015.55 1000 3% 5% 3 =1/12 =1/12 =1/2 1003.89 1000 3% 5% 10 =1/12 =1/12 =1/2 1015.55\n\n11. ## Re: calculating compound interest with non-linear interest growth\n\nThank you GhostAccount! That's great - I can use that macro.\n\nI made a typo in my question - I meant:\n\nSuppose I have two different dividend paying stocks - one has an initial yield r1 with a growth rate of g1 and the second has an initial yield of r2 with a growth rate of g2. The stocks pay dividends annually and the dividends are re-invested to purchase more of the stock at the end of each year.\n\nQ1: With a constant stock price, what is the total return of principal + dividends after time t?\n\nQ2: How many years will either stock take to reach a certain dividend payout?\n\nThanks again for your help and patience in answering - I've learned a lot so far!\n\n12. ## Re: calculating compound interest with non-linear interest growth\n\nI must be missing something...but, if you're able to use a looper, then isn't that enough?\n\nI love UBasic ([]= notes):\n\ni = .03 : f = 1.05 : b = 1000 : p = 1000 : k = 30 [f = increase factor, k = years]\nFOR n = 1 TO k\ne = b * i [e = year's earnings]\nb = b + e\ny = (b / p)^(1 / n) [y = yield so far]\nPRINT n, e, b, i * 100, y * 100\ni = i * f [next year's rate]\nNEXT n\n\nOUTPUT:\nCode:\nYEAR  INTEREST  BALANCE  YEAR'S RATE  YIELD-SO-FAR\n0             1000.00\n1      30.00  1030.00    3.0000        3.0000\n2      32.45  1062.45    3.1500        3.0750\n3      35.14  1097.59    3.3075        3.1524\n...\n10      64.40  1448.11    4.6540        3.7720\n...\n20     185.22  2628.54    7.5808        4.9507\n...\n30     749.55  6819.59   12.3484**      6.6085\n** .03 * 1.05^29 = 12.3484...\n\nIsn't that all you need?\n\n13. ## Re: calculating compound interest with non-linear interest growth\n\nHi Wilmer,\n\nYou're correct - I can certainly compute the answer using an excel macro. However I'm curious if there's a mathematical solution to this question.\n\nThe only mathematical solution I know of would be to approximate the answer using a Taylor series e.g.\n\n$FV = Re^{\\sum_{k\\ =\\ 0}^{n-1}ln({1+r(1+g)^k)}}$\n\n$ln(1+x) \\approx x - \\frac{x^2}{2} + \\frac{x^3}{3} - {...}$\n\ntaking the first 2 terms as an example then\n\n$\\sum_{k\\ =\\ 0}^{n-1}ln({1+x}) \\approx \\sum_{k\\ =\\ 0}^{n-1}(x - \\frac{x^2}{2})$ where $x = r(1+g)^k$\n\n$\\sum_{k\\ =\\ 0}^{n-1}(x - \\frac{x^2}{2}) = \\sum_{k\\ =\\ 0}^{n-1}x - \\sum_{k\\ =\\ 0}^{n-1}\\frac{x^2}{2}$\n\nand\n\n$\\sum_{k\\ =\\ 0}^{n-1}x = \\sum_{k\\ =\\ 0}^{n-1}r(1+g)^k = \\frac{r((1+g)^n-1)}{g}$\n\n$\\sum_{k\\ =\\ 0}^{n-1}\\frac{x^2}{2} = \\frac{1}{2}\\sum_{k\\ =\\ 0}^{n-1}r^2(1+g)^{2k} = \\frac{r^2((1+g)^{2n}-1)}{2g(g+2)}$\n\nso I could say that\n\n$FV \\approx Re^{\\frac{r((1+g)^n-1)}{g} - \\frac{r^2((1+g)^{2n}-1)}{2g(g+2)}}$\n\nand solve this equation for n with a given FV.\n\nI'm wondering if there's a more elegant equivalent expression of this function out there that doesn't involve approximations.\n\n14. ## Re: calculating compound interest with non-linear interest growth\n\nOriginally Posted by tw35758\nI'm wondering if there's a more elegant equivalent expression of this function out there that doesn't involve approximations.\nAfter 5 beer bottles and 2 shots of tequila, I've been wondering about that too. You could always repost your questions on an actuarial forum. You might get lucky there. Having seen your analyses, I'm somewhat inclined to believe that you might have some background in actuarial mathematics yourself. If you're lucky, Sir TKHunny, the resident actuary here might get interested in your situation and come out of retirement from this often thankless business of math knight-errantry.\n\n15. ## Re: calculating compound interest with non-linear interest growth\n\nOriginally Posted by jonah\nAfter 5 beer bottles and 2 shots of tequila, I've been wondering about that too...\nHey, that was my breakfast 30 years ago, at end of my drinking...!!\n\nPage 1 of 2 12 Last", "date": "2017-08-17 16:11:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/business-math/231150-calculating-compound-interest-non-linear-interest-growth.html", "openwebmath_score": 0.5511336922645569, "openwebmath_perplexity": 1729.602022998175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575152637945, "lm_q2_score": 0.8840392924390585, "lm_q1q2_score": 0.8686194505744844}}
{"url": "http://mathhelpforum.com/calculus/167237-trig-integral.html", "text": "# Math Help - Trig Integral\n\n1. ## Trig Integral\n\n$\\int SEC^4{5x} = \\int ({SEC^2{5x}})({SEC^2{5x}})$\nThen\n$\\int (1+TAN^2{5x})(SEC^2{5x})$\nThen I multiply out\n$\\int (SEC^2{5x})+(SEC^2{5x}TAN^2{5x})$\n\nI would assume U sub at this point\n$U= SEC dU= SECTAN$\nBut I am not sure, what or how I should bring along with the SEC\n\nAny help?\n\nthanks\nRich\n\n2. Originally Posted by Spoolx\n$\\int SEC^4{5x} = \\int ({SEC^2{5x}})({SEC^2{5x}})$\nThen\n$\\int (1+TAN^2{5x})(SEC^2{5x})$\nThen I multiply out\n$\\int (SEC^2{5x})(SEC^2{5x}TAN^2{5x})$\nFine so far, except you're missing a plus sign in the last line, which I assume is a typo. Now to substitute, which is really in order to work backwards through the chain rule.\n\nJust in case a picture helps...\n\n... where (key in first spoiler) ...\n\nSpoiler:\n\n... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).\n\nThe general drift is...\n\nSpoiler:\n$\\int 1 + u^2\\ du = u + \\frac{1}{3} u^3$\n\nand so we can explicate F...\n\nHope this helps - if not, someone will probably show the sub with differentials. However, I say...\n_________________________________________\n\nDon't integrate - balloontegrate!\n\nBalloon Calculus; standard integrals, derivatives and methods\n\nBalloon Calculus Drawing with LaTeX and Asymptote!\n\n3. I edited the original to include the + sign, thanks for catching my mistake\n\n4. $\\displaystyle\\int{sec^4(5x)}dx=\\int{sec^2(5x)sec^2 (5x)}dx=\\int{\\left[1+tan^2(5x)\\right]sec^2(5x)}dx$\n\nSubstitute\n\n$u=tan(5x)$\n\n$\\displaystyle\\Rightarrow\\frac{du}{dx}=5sec^2(5x)$\n\nby the chain rule\n\n$\\displaystyle\\Rightarrow\\ sec^2(5x)dx=\\frac{du}{5}$\n\nThe integral becomes\n\n$\\displaystyle\\frac{1}{5}\\int{\\left[1+u^2\\right]du$\n\n5. Hello, Rich!\n\nA rehash of Archie's solution . . .\n\n$\\displaystyle \\int \\sec^4\\!5x\\,dx \\:= \\:\\int (\\sec^2\\!5x)(\\sec^2\\!5x)\\,dx$\n\nThen: . $\\displaystyle \\int (1+\\tan^2\\!5x)(\\sec^2\\!5x)\\,dx$\n\nThen I multiply out: . $\\displaystyle \\int (\\sec^2\\!5x +\\sec^2\\!5x\\tan^2\\!5x)\\,dx$\n\n$\\displaystyle \\text{You have: }\\;\\int\\sec^2\\!5x\\,dx + \\int(\\tan 5x)^2(\\sec^2\\!5x\\,dx)$\n\nFor the first integral, use: . $\\displaystyle \\int \\sec\\theta\\,d\\theta \\:=\\:\\tan\\theta + C$\n\nFor the second integral, let: . $u \\,=\\, \\tan x \\quad\\Rightarrow\\quad du \\,=\\,\\sec^2\\!x\\,dx$\n\n. . and you have: . $\\displaystyle \\int u^2\\,du \\:=\\:\\tfrac{1}{3}u^3 + C$\n\nGot it?\n\n6. Originally Posted by Soroban\nHello, Rich!\n\nA rehash of Archie's solution . . .\n\n$\\displaystyle \\text{You have: }\\;\\int\\sec^2\\!5x\\,dx + \\int(\\tan 5x)^2(\\sec^2\\!5x\\,dx)$\n\nFor the first integral, use: . $\\displaystyle \\int \\sec\\theta\\,d\\theta \\:=\\:\\tan\\theta + C$\n\nFor the second integral, let: . $u \\,=\\, \\tan x \\quad\\Rightarrow\\quad du \\,=\\,\\sec^2\\!x\\,dx$\n\n. . and you have: . $\\displaystyle \\int u^2\\,du \\:=\\:\\tfrac{1}{3}u^3 + C$\n\nGot it?\nSo to refresh my algebra\n\n$tan^2{5x} = (tan5x)^2$ ?\n\nAlso above, is your U supposed to be?\n$u= tan5x$ ?\nif not where does the 5 go?\n\n7. Originally Posted by Spoolx\n$tan^2{5x} = (tan5x)^2$ ? Mr F says: Yes.\n$u= tan5x$ ? Mr F says: Yes. Soroban made a simple typo. You ought to be able to make the appropriate corrections in the calculations.", "date": "2015-07-05 18:11:31", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/167237-trig-integral.html", "openwebmath_score": 0.983540952205658, "openwebmath_perplexity": 3071.315559882284, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575139868958, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8686194464437952}}
{"url": "https://math.stackexchange.com/questions/2248259/how-to-calculate-an-integral-using-cauchys-theorem", "text": "# How to calculate an integral using Cauchy's theorem?\n\nThe question is calculate the value of the integral, $$\\int_{-\\infty}^{\\infty}\\frac{dx}{1+x^4}$$ These are the steps that I followed,\n\nLet $x$ be a complex number. So, the poles of the function $f(x)=\\frac{1}{1+x^4}$ occur when $x$ is equal to the roots of the equation $1+x^4=0$, i.e $x=e^{i\\pi/4},e^{3i\\pi/4},e^{5i\\pi/4},e^{7i\\pi/4}$. They are all poles of degree 4.\n\nNow Cauchy's theorem says that, $$\\frac{1}{2\\pi i}\\int_C dx\\ f(x)=\\sum_i\\text{Res}(f,x_i)$$ where $x_i$ are the poles of $f$ that lies within $C$. I am pretty sure that my poles lie within $-\\infty$ and $\\infty$. So, I calculated the residues of $\\frac{1}{1+x^4}$ at $x=e^{i\\pi/4},e^{3i\\pi/4},e^{5i\\pi/4},e^{7i\\pi/4}$ and they are equal to $-\\frac14e^{i\\pi/4},-\\frac14e^{3i\\pi/4},-\\frac14e^{5i\\pi/4},-\\frac14e^{7i\\pi/4}$. Wolfram Alpha confirms my calculations.\n\nThe sum of residues is $$\\sum_i\\text{Res}=-\\frac14e^{i\\pi/4}-\\frac14e^{3i\\pi/4}-\\frac14e^{5i\\pi/4}-\\frac14e^{7i\\pi/4}=0$$ and therefore the integral, $$\\int_{-\\infty}^{\\infty}\\frac{dx}{1+x^4}$$ must be equal to zero. However, Wolfram alpha says it is not zero but equal to$\\frac{\\pi}{\\sqrt 2}$. Where am I making a mistake?\n\nIn applying the residue theorem, we analyze the integral $I$ given by\n\n\\begin{align} I&=\\oint_C \\frac{1}{1+z^4}\\,dz\\\\\\\\ &=\\int_{-R}^R \\frac{1}{1+x^4}\\,dx+\\int_0^\\pi \\frac{1}{1+(Re^{i\\phi})^4}\\,iRe^{i\\phi}\\,d\\phi\\\\\\\\ &=2\\pi i \\,\\text{Res}\\left(\\frac{1}{1+z^4}, z=e^{i\\pi/4},e^{i3\\pi/4}\\right) \\end{align}\n\nwhere $R>1$ is assumed.\n\nNote that the only residues implicated in the residue theorem are those enclosed by $C$. Here, $C$ is comprised of (i) the line segment from $-R$ to $R$ and (ii) the semicircular arc centered at the origin with radius $R$ and residing in the upper-half plane. Hence, the only resides are the ones at $z=e^{i\\pi/4}$ and $z=e^{i3\\pi/4}$.\n\nTaking $R\\to \\infty$, the integration over the semi-circular contour vanishes and we find that\n\n\\begin{align} \\int_{-\\infty}^\\infty \\frac{1}{1+x^4}\\,dx&=2\\pi i \\,\\text{Res}\\left(\\frac{1}{1+z^4}, z=e^{i\\pi/4},e^{i3\\pi/4}\\right)\\\\\\\\ &=2\\pi i \\left(-\\frac{e^{i\\pi/4}}{4}-\\frac{e^{i3\\pi/4}}{4}\\right)\\\\\\\\ & =\\frac{\\pi}{\\sqrt 2} \\end{align}\n\n\u2022 Like the other answer, why did you only used the upper half semicircle? Why did you not ignore the first term in the second step and evaluate over the whole circle like, $$\\int_0^{2\\pi} d\\phi\\ \\frac{iRe^{i\\phi}}{1+(Re^{i\\phi})^4}$$ \u2013\u00a0sigsegv Apr 23 '17 at 17:13\n\u2022 @Ayatana We want to evaluate the integral $\\int_{-\\infty}^\\infty \\frac{1}{1+x^4}\\,dx$. How does integrating over the circle do that? Well, it doesn't; and that integral is $0$. So instead, we evaluate a contour integral, the one described herein, that has the real line integral as part of the contour. We could have taken the semi-circle in the lower-half plane, but since the orientation would be clockwise, we need a $-2\\pi i$ factor o multiply the sum of the residues in the lower-half plane. The result will be the same, of course. ;-)) \u2013\u00a0Mark Viola Apr 23 '17 at 17:15\n\u2022 Is it necessary for the contour to be semicircle? Will a contour made of a straight line along real axis and an arbitrary line joing the ends work? \u2013\u00a0sigsegv Apr 24 '17 at 4:30\n\u2022 @Ayatana The reside theorem applies to any closed rectifiable curve. But we are free to choose the closed contour. And by choosing a semicircle the radius of which approaches infinity, we can show quite easily that its contribution approaches $0$ as $R\\to \\infty$. \u2013\u00a0Mark Viola Apr 24 '17 at 4:34\n\nYou shouldn't need Wolfram Alpha to tell you the answer isn't zero. The integrand is positive, so the integral is positive, and certainly nonzero.\n\nYou have added up the residues at all the poles. However using the usual semicircle method gives you that the integral is $2\\pi i$ times the sum of the residues of the poles in the upper half-plane.\n\n\u2022 Why should I not add the residues of the poles below the lower half? What semi circle are you talking about? The limits of the integration are from $-\\infty$ to $\\infty$ for both real and imaginary parts of $x$. \u2013\u00a0sigsegv Apr 23 '17 at 16:29", "date": "2021-02-27 10:55:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2248259/how-to-calculate-an-integral-using-cauchys-theorem", "openwebmath_score": 0.9597283601760864, "openwebmath_perplexity": 214.38352011733284, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575142422757, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.868619431660257}}
{"url": "https://mathoverflow.net/questions/301148/version-of-banach-steinhaus-theorem/304672", "text": "# Version of Banach-Steinhaus theorem\n\nI am wondering about the following version of the Banach-Steinhaus theorem.\n\nLet $A$ be a closed convex subset contained in the unit ball of a Banach space $X$ and consider bounded operators $T_n \\in \\mathcal L(X).$\n\nAssume we know that for every $x \\in A$ the sequence $\\left\\lVert T_n x \\right\\rVert$ is bounded uniformly in $n.$\n\nDoes this imply that $$\\sup_{x \\in A} \\left\\lVert T_n x \\right\\rVert$$ is bounded uniformly in $n$?\n\nIf $A=X$ then the theorem is undoubtedly true by the folklore Banach-Steinhaus theorem but I was wondering whether this version holds as well?\n\n\u2022 In the last paragraph, shouldn't it be \"if A is the unit sphere then...\" ? \u2013\u00a0user111 Jul 10 '18 at 7:45\n\u2022 It is worth noting that the implication does not hold without the closeness assumption. If e.g. $A$ is the convex hull of the standard basis of $X:=\\ell_2$ and $\\phi_n:=\\sum_{k=1}^n ke_k\\in X$, then $(\\phi_n,x)$ is point-wise but not uniformly bounded for $x\\in A$ \u2013\u00a0Pietro Majer Jul 10 '18 at 15:16\n\nThe answer is yes, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to be the standard one) can e.g. be found on Wikipedia.\n\nThe details are a bit different here, so let me give them below. Throughout, let us replace the sequence $(T_n)$ with a general subset $\\mathcal{T} \\subseteq \\mathcal{L}(X)$.\n\nProof. By Baire's Theorem we can find an integer $m$ such that the set \\begin{align*} B := \\{x \\in A: \\, \\|Tx\\| \\le m \\text{ for all } T \\in \\mathcal{T}\\} \\end{align*} has non-empty interior within $A$. Thus, we can find a point $x_0 \\in B$ and a real number $\\varepsilon \\in (0,1]$ such that each $x \\in A$ which has distance at most $\\varepsilon$ to $x_0$ is contained in $B$.\n\nNow, let $y \\in A$. The vector $z := x_0 + \\frac{\\varepsilon}{2}(y-x_0)$ is contained in $A$ due to the convexity of $A$, and its distance to $x_0$ is at most $\\varepsilon$ since both $y$ and $x_0$ have norm at most $1$. Thus, $\\|Tz\\| \\le m$ for all $T \\in \\mathcal{T}$. Since \\begin{align*} y = \\frac{2}{\\varepsilon}(z - x_0) + x_0, \\end{align*} we conclude that \\begin{align*} \\|Ty\\| \\le \\frac{4m}{\\varepsilon} + m \\end{align*} for all $T \\in \\mathcal{T}$. This bound does not depend on $y$.\n\nInstead of inspecting the Banach-Steinhaus proof as in Jochen Glueck's answer one can apply Banach-Steinhaus to the Banach space $[A]$ (the linear hull of $A$) endowed with the Minkowski functional $\\|x\\|_A=\\inf\\{t>0: x\\in tA\\}$ (completeness of this norm follows from completeness of the Banach space $X$ and $A=\\overline A$).\n\n\u2022 A small detail: we need a balanced convex set to get a norm, so we should replace $A$ with $A':= \\text{co}(A\\cup (-A))=[-1,1]\\cdot A$, and check that the hypotheses are still satisfied by $A'$ \u2013\u00a0Pietro Majer Jul 10 '18 at 7:22\n\u2022 (the last equality is not true of course) \u2013\u00a0Pietro Majer Jul 10 '18 at 14:06\n\nI like Jochen Wengenroth's approach, and I think there is a point that it is worth to clarify. If we want to make a norm out of $A$, we need it to be a balanced set, so we'd like to pass to the bounded absolutely convex set $\\overline{\\operatorname{co}}\\left(A\\cup(-A)\\right)$ or to $A-A$. Any family of linear operators which is point-wise bounded on $A$ is clearly also point-wise bounded on $A-A$. However these sets are in general not closed, so some care is needed, because a bounded absolutely convex not closed set $B$ in general would not produce a Banach disk on its linear span, and in fact in general the statement itself does not hold on such $B$ (see the example in the initial comment).\n\nA cheap solution to make the argument work smoothly is to use the notion of $\\sigma$-convexity (see e.g. this MO thread) which also generalize slightly the statement); in particular, it covers both the case of a closed and an open bounded convex set $A$. Recall that for a subset $A$ of a Banach space $X$ the following easy facts hold:\n\n\u2022 If $A$ is $\\sigma$-convex, it is bounded;\n\u2022 If $A$ is $\\sigma$-convex, $A-A$ is $\\sigma$-convex and symmetric (that is, $\\sigma$-absolutely convex);\n\u2022 If $A$ is $\\sigma$-absolutely convex, it is a Banach disk, that is, its Minkowski functional is a Banach norm on the linear span of $A$.\n\nAs a conclusion, we can follow Jochen Wengenroth's reduction to the standard Banach-Steinhaus theorem. We thus have: Any family of linear operators on a Banach space, which is point-wise bounded on a $\\sigma$-convex set $A$, is also uniformly bounded on $A$.", "date": "2019-04-23 17:14:22", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/301148/version-of-banach-steinhaus-theorem/304672", "openwebmath_score": 0.9742316007614136, "openwebmath_perplexity": 152.7402352219831, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363503693294, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8686058346605904}}
{"url": "https://math.stackexchange.com/questions/3433458/optimal-strategies-for-this-card-game/3433482", "text": "# Optimal strategies for this card game?\n\nI recently had an interesting card game for an interview.\n\nThe interviewer and I have two cards each, one with a '5' painted on it and one with a '10'. We pick a card each and show it at the same time. If we picked the same card, I receive nothing, while if we picked different cards, the interviewer pays me the number he picked in dollars. What is the interviewer most likely to do if he intends to minimise my payoff, and what should my strategy be to counter this?\n\nConsideration 1: I pick '5' all the time, but if the interviewer knows I do this he would pick '5' as well so that I receive nothing and my expected payoff would just be zero.\n\nConsideration 2: I assign a probability $$p_1$$ of picking a '5' (and $$1-p_1$$ of picking a '10'), and the interviewer assigns a probability $$p_2$$ of picking a '5' (and $$1-p_2$$ of picking a '10'). My payoff would then be\n\n$$\\mathbb{E}=10p_1(1-p_2)+5(1-p_1)p_2.$$\n\nI was thinking of differentiation somehow, but $$p_1$$ aims to maximise $$\\mathbb{E}$$ while\u00a0$$p_2$$ aims to minimise $$\\mathbb{E}$$. Is this even the correct strategy for both of us?\n\nConsideration 3: A friend suggested assigning probabilities of $$\\frac13$$ and\u00a0$$\\frac23$$ to the cards respectively for a Nash equilibrium since it is a symmetric game. Where does this come from intuitively? Do these probabilities match with the above equation?\n\nAny help/comments on the considerations is greatly appreciated, cheers!\n\nThe logic of equilibrium here is: you should choose your strategy (i.e. determine your probabilities of picking $$5$$ and $$10$$) in such a way that, however your opponent chooses his strategy, your expected gain is not influenced.\n\nLet's say you pick $$5$$ with probability $$p_1$$ and your opponent picks $$5$$ with probability $$p_2$$. Then, once you fixed your choice of $$p_1$$, the expectation $$\\mathbb{E}$$ is a linear function in $$p_2$$: $$\\mathbb{E} = 10p_1 + (5 - 15p_1)p_2.$$ In this situation, you see that choosing $$p_1 = 1/3$$ will guarantee an expectation of $$10/3$$, regardless of $$p_2$$.\n\nIf, let's say, you choose $$p_1 < 1/3$$. Then there is the \"risk\" that your opponent somehow knows your strategy (e.g. via statistics from multiple rounds of the game), and then chooses $$p_2 = 0$$. You then have $$\\mathbb{E} = 10p_1 < 10/3$$.\n\nIf, in the other direction, you choose $$p_1 > 1/3$$. Then again assume your opponent gets to know your strategy, and then chooses $$p_2 = 1$$. You then have $$\\mathbb{E} = 5 - 5p_1 < 10/3$$.\n\nTherefore, you see that the \"safest\" way of playing is to choose $$p_1 = 1/3$$. It is in that sense that we say it's the \"optimal strategy\".\n\nI hope this whole logic makes sense to you.\n\nAnd as an exercise, you may try to solve the general situation: say the two cards are labelled A and B, and your gain has $$4$$ possibilities: $$a, b, c, d$$, which correspond to the $$4$$ combinations AA, AB, BA, BB of your cards. Then what is your optimal strategy?\n\n\u2022 Hello, what are some general approaches to formulate equilibrium for a game? \u2013\u00a0user107224 Nov 13 '19 at 9:56\n\u2022 For the general equilibrium, the wiki page en.m.wikipedia.org/wiki/Nash_equilibrium may be a good starting point. Also note that in general there could be more than one equilibria. \u2013\u00a0WhatsUp Nov 13 '19 at 10:38", "date": "2020-09-18 09:54:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3433458/optimal-strategies-for-this-card-game/3433482", "openwebmath_score": 0.8793413639068604, "openwebmath_perplexity": 420.1073992647984, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353126336, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8686058294846312}}
{"url": "http://mathematica.stackexchange.com/questions/29924/axis-angle-from-rotation-matrix/29966", "text": "# Axis/Angle from rotation matrix\n\nWith r=RotationMatrix[a,{x,y,z}] I can compute a 3D rotation matrix from axis/angle representation.\n\nGiven a 3D rotation matrix r how can I compute a, x, y, z?\n\nExample:\n\nr = {{0.966496, -0.214612, 0.14081}, {0.241415, 0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}}\n\n\nThe result should be a=20. Degree and {x,y,z}={2, 1, 2}/3 (or equivalent).\n\nEdit: I am fine with any answer that gives the same r when applied to RotationMatrix.\n\nPS: Sorry if the tag is wrong, I did not found a matching tag...\n\n-\nI'm not sure this question is well posed in the sense that the answer is non-unique. A simple example would be r=RotationMatrix[1,{1,0,0}]=RotationMatrix[-1,{-1,0,0}] are the same. So which one are you intending to find? \u2013\u00a0bill s Aug 6 '13 at 13:08\nThe one where the angle is in [0,2 pi]. I think it is well posed as angle and axis are directly linked and you can provide a unique answer with a perfectly reasonable additional assumption. \u2013\u00a0Danvil Aug 6 '13 at 13:43\nActually there was an answer earlier which was just fine but it was deleted? \u2013\u00a0Danvil Aug 6 '13 at 13:46\n@Danvil it was deleted because it was incorrect, so he deleted it. It was incorrect, as it worked only for the example you provided and could not be generalized. \u2013\u00a0rcollyer Aug 6 '13 at 14:42\n@Danvil I'm thankful I noticed the wink. It would have been a bad day, otherwise. :P \u2013\u00a0rcollyer Aug 6 '13 at 16:38\n\nThere is no need to use Eigensystem or Eigenvectors to find the axis of a rotation matrix. Instead, you can read the axis vector components off directly from the skew-symmetric matrix\n\n$$a \\equiv R^T-R$$\n\nIn three dimensions (which is assumed in the question), applying this matrix to a vector is equivalent to applying a cross product with a vector made up of the three independent components of $a$:\n\n{1, -1, 1}Extract[a, {{3, 2}, {3, 1}, {2, 1}}]\n\n\nThis one-line method of finding the axis is applied in the following function. To get the angle of rotation, I construct two vectors ovec, nvec perpendicular to the axis and to each other, to find the cosine and sine of the angle using the Dot product (could equally have used Projection). To get a first vector ovec that is not parallel to the axis, I permute the components of the axis vector using the fact that\n\nSolve[{x, -y, z} == {y, z, x}, {x, y, z}]\n(* ==> {{x -> 0, y -> 0, z -> 0}} *)\n\n\nwhich means the above permutation with sign change of a nonzero axis vector is always different from the axis. This is sufficient to use Orthogonalize and Cross to get the desired orthogonal vectors.\n\naxisAngle[m_] := Module[\n{axis, ovec, nvec\n},\n{axis, ovec} =\nOrthogonalize[{{1, -1, 1} #, Permute[#, Cycles[{{1, 3, 2}}]]}] &@\nExtract[m - Transpose[m], {{3, 2}, {3, 1}, {2, 1}}];\n(* nvec is orthogonal to axis and ovec: *)\nnvec = Cross[axis, ovec];\n{axis,\nArg[Complex @@ (((m.ovec).# &) /@ {ovec, nvec})]}\n]\n\n\nThe angle is calculated with Arg instead of ArcTan[x, y] here because the latter throws an error for x = y = 0.\n\nHere I test the results of the function for 100 random rotation matrices:\n\ntestRotation[] := Module[\n{m, a, axis, ovec, nvec,\nv = Normalize[RandomReal[{0, 1}, {3}]],\n\u03b1 = RandomReal[{-Pi, Pi}], angle\n},\nm = RotationMatrix[\u03b1, v];\n{axis, angle} = axisAngle[m];\nChop[\nangle Dot[v, axis] - \u03b1\n] === 0\n]\n\nAnd @@ Table[testRotation[], {100}]\n\n(* ==> True *)\n\n\nIn the test, I have to account for the fact that if the function axisAngle defined the axis vector with the opposite sign as the random test vector, I have to reverse the sign of the rotation angle. This is what the factor Dot[v, axis] does.\n\nExplanation of how the axis results from a skew-symmetric matrix\n\nIf $\\vec{v}$ is the axis of rotation matrix $R$, then we have both $R\\vec{v} = \\vec{v}$ and $R^T\\vec{v} = \\vec{v}$ because $R^T$ is just the inverse rotation. Therefore, with $a \\equiv R^T-R$ as above, we get\n\n$$a \\vec{v} = \\vec{0}$$\n\nNow the skew-symmetric property $a^T = -a$, which can be seen from its definition, means there are exactly three independent matrix element in $a$. They can be arranged in the form of a 3D vector $\\vec{w}$ which must have the property $a \\vec{w} = 0$. This vector is obtained in the Extract line above. In fact, $a \\vec{x} = \\vec{w}\\times \\vec{x}$ for all $\\vec{x}$, and hence if $a \\vec{x} = 0$ then $\\vec{x}\\parallel\\vec{w}$. Therefore, the vector $\\vec{v}$ is also parallel to $\\vec{w}$, and the latter is a valid representation of the rotation axis.\n\nEdit 2: speed considerations\n\nSince the algorithm above involves only elementary operations that can be compiled, it makes sense that a practical application of this approach would use Compile. Then the function could be defined as follows (keeping the return values arranged as above):\n\nClear[axisAngle1, axisAngle]\n\naxisAngle1 =\nCompile[{{m, _Real, 2}},\nModule[{axis, ovec, nvec, tvec, w, w1},\ntvec = {m[[3, 2]] - m[[2, 3]], m[[3, 1]] - m[[1, 3]],\nm[[2, 1]] - m[[1, 2]]};\nIf[tvec == {0., 0., 0.},\n{#/Sqrt[#.#] &[#[[Last@Ordering[N[Abs[#]]]]] &[\n1/2 (m + {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}})]],\nIf[Sum[m[[i, i]], {i, 3}] == 3, 0, Pi] {1, 1, 1}},\naxis = {1, -1, 1} tvec;\naxis = axis/Sqrt[axis.axis];\nw = {tvec[[2]], tvec[[3]], tvec[[1]]};\novec = w - axis Dot[w, axis];\nnvec = Cross[axis, ovec];\nw1 = m.ovec;\n{axis, {1, 1, 1} ArcTan[w1.ovec, w1.nvec]}\n]\n]\n];\n\naxisAngle[m_] := {#1, Last[#2]} & @@ axisAngle1[m]\n\n\nThe results are the same as for the previous definition of axisAngle, but I now get a much faster execution as can be seen in this test:\n\ntab = RotationMatrix @@ # & /@\nTable[{RandomReal[{-Pi, Pi}],\nNormalize[RandomReal[{0, 1}, {3}]]}, {100}];\ntimeAvg =\nFunction[func,\nDo[If[# > 0.3, Return[#/5^i]] & @@ Timing@Do[func, {5^i}], {i, 0,\n15}], HoldFirst];\n\ntimeAvg[axisAngle /@ tab]\n\n(* ==> 0.000801259 *)\n\n\nThis is more than an order of magnitude faster than the un-compiled version. I removed Orthogonalize from the code because I didn't find it in the list of compilable functions.\n\nNote that Eigensystem is not in that list, either.\n\nEdit 3\n\nThe first version of axisAngle demonstrated the basic math, but the compiled version axisAngle1 (together with the re-defined axisAngle as a wrapper) is faster. One thing that was missing was the correct treatment of the edge case where the rotation is by exactly $\\pi$ in angle.\n\nI added that fix only to the compiled version (axisAngle1) because I think that's the more practical version anyway. The trivial case of zero rotation angle was already included in the earlier version.\n\nTo explain the added code, first note that for angle $\\pi$ you can't read off the axis from $R^T - R$ because the resulting matrix vanishes. To get around this singular case, we can use the geometric fact that a rotation by $\\pi$ is equivalent to an inversion in the plane perpendicular to the rotation axis given by the unit vector $\\vec{n}$. Therefore, if we form the sum of a vector $\\vec{v}$ and its $\\pi$-rotated counterpart, the components transverse to the rotation axis cancel and the result is always parallel to the axis. In matrix form,\n\n$$(R+1)\\vec{v} = 2\\vec{n}(\\vec{n}\\cdot\\vec{v}) = 2\\left(\\vec{n}\\vec{n}^T\\right)\\vec{v}$$\n\nSince this holds for all vectors, it is a matrix identity. The right-hand side contains a matrix $\\vec{n}\\vec{n}^T$ which must have at least one row that's nonzero. This row is proportional to $\\vec{n}^T$, so you can read of the axis vector directly from $(R+1)$, again without any eigenvalue computations.\n\n-\nBeautiful! (+1) My hamfisted approach to this has been an opportunity to learn. \u2013\u00a0ubpdqn Aug 6 '13 at 22:28\nNote (as you have outlined) you could also obtain the axis from NullSpace[m-Transpose[m]] \u2013\u00a0ubpdqn Aug 7 '13 at 4:17\n@ubpdqn True, but that takes more computation... as I said, you can read the axis off literally from the matrix a itself. \u2013\u00a0Jens Aug 7 '13 at 5:15\nActually for cases $\\theta=0$ and $\\theta=\\pi$ the $R-R^T$ is a zero matrix and the axis is thus maps to {{0,0,0},0) in your code. The test set always skirts these \"points\". The eigensystem allows extraction of the eigenvector associated with eigenvalue 1). I agree with lack of simplicity or elegance using the eigensystem. I found it a lazy way to deal with these anomalous cases: integer*$\\pi$. I think my code works now. Passes your test (up orientation of axis issue) and the \"extreme\" cases. \u2013\u00a0ubpdqn Aug 7 '13 at 5:29\nMaybe the easiest way to deal with the isotropic cases ($\\pm$IdentityMatrix[3]) would be to append the rule axis /. {0, 0, 0} -> {0, 0, 1} to the result. The choice of rotation axis is arbitrary in that case and I'd fix it to be the z axis. \u2013\u00a0Jens Aug 7 '13 at 5:41\n\nSince this question still seems to be alive after some time, I'm giving a solution from Presentations, which I sell. It has a routine RotationAngleAndAxis and here it is used on the example.\n\n<< Presentations\n\nr = {{0.966496, -0.214612, 0.14081}, {0.241415,\n0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}};\nRotationAngleAndAxis[r]\n\n(* {0.349066, {0.666667, 0.333333, 0.666667}} *)\n\n\nHere is a short example of its use that checks the result. The axis vector is normalized. Notice that reversing the angle of rotation and the axis gives an equivalent rotation.\n\nrotations =\nTable[{RandomReal[{-\\[Pi], \\[Pi]}],\nNormalize[Array[RandomReal[{-1, 1}] &, {3}]]}, {2}]\n\n(* {{2.90598, {-0.596373, -0.74938,\n0.287697}}, {-2.44158, {0.331763, -0.943343, 0.00605196}}} *)\n\nrotationMatrices = RotationMatrix @@@ rotations;\nanglesAndAxes = RotationAngleAndAxis /@ rotationMatrices\n\n(* {{2.90598, {-0.596373, -0.74938, 0.287697}}, {2.44158, {-0.331763,\n0.943343, -0.00605196}}} *)\n\nresultingMatrices = RotationMatrix @@@ anglesAndAxes;\n\nrotationMatrices - resultingMatrices // Chop\nTotal[Flatten[%]]\n\n(* {{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}, {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}} *)\n(* 0 *)\n\n\nThe following is a test on 1000 initial rotation specifications. Notice also that I am comparing the resulting rotation matrices by a slightly risky procedure.\n\nrotations =\nTable[{RandomReal[{-\\[Pi], \\[Pi]}],\nNormalize[Array[RandomReal[{-1, 1}] &, {3}]]}, {1000}];\nrotationMatrices = RotationMatrix @@@ rotations;\nTiming[anglesAndAxes = RotationAngleAndAxis /@ rotationMatrices;]\nresultingMatrices = RotationMatrix @@@ anglesAndAxes;\nTotal[Flatten[rotationMatrices - resultingMatrices // Chop]]\n\n(* {0.873606, Null} *)\n(* 0 *)\n\n\nPresentations also has an EulerAngles routine that will return the two sets of Euler angles for any of the possible sequence of rotations specified by strings. For example, here are the two sets of Euler angles of the example for two different rotation sequences.\n\nEulerAngles[r, \"ZXZ\"]\nEulerAngles[r, \"XYZ\"]\n\n(* {{-0.346633, 0.259587, 0.580661}, {2.79496, -0.259587, -2.56093}} *)\n(* {{0.244775, 0.0873143, 0.244775}, {-2.89682, 3.05428, -2.89682}} *)\n\n-\n\nThe axis of rotation $n$ can be obtained as the eigenvector corresponding to the eigenvalue $1$.\n\nr = {{0.966496, -0.214612, 0.14081}, {0.241415,\n0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}};\n\nes = Eigensystem[r];\npos = Nearest[Thread[es[[1]] -> Range[3]], 1][[1]];\nn = es[[2, pos]]\n(*  {0.666667, 0.333333, 0.666666} *)\n\n\nThe angle of rotation can be obtained directly from the trace of $r$.\n\nArcCos[(Tr[r] - 1)/2] 180/\u03c0\n(* 20. *)\n\n\n(This was my earlier and longer way to obtain the angle of rotation. As rcollyer pointed out this will have issues if $\\pi k,k\\in \\mathcal{Z}$)\n\nThe rotation matrix in terms of the angle of rotation $\\theta$ and axis of rotation $\\{n1, n2, n3\\}$.\n\nR[t_, {n1_, n2_, n3_}] :=\nWith[{n = {{0, -n3, n2}, {n3, 0, -n1}, {-n2, n1, 0}}, id = IdentityMatrix[3]},\nInverse[id - t n].(id + t n)]\n\n\nHere $t =\\tan \\left(\\frac{\\theta }{2}\\right)$.\n\nWe get three solutions from the element in the first position.\n\neqns = Flatten[R[t, es[[2, pos]]] - r] // Simplify;\nsols = NSolve[eqns[[1]] == 0, t]\n(* {{t -> 3.88248*10^16}, {t -> 0.176327}, {t -> -0.176327}} *)\n\n\nChoose the one which satisfies all the entries.\n\nChop[eqns /. sols, 10^-5];\nPosition[%, {0 ..}];\nsolst = Extract[sols, %]\n(*  {{t -> 0.176327}}  *)\n\n\nObtain it in terms of degrees.\n\n2 ArcTan[t] 180/\u03c0 /. solst\n(* {20.} *)\n\n-\nIt doesn't seem to work with a rotation of $\\pi$ about the {0,0,1}. \u2013\u00a0rcollyer Aug 6 '13 at 14:39\n@rcollyer, that is to be expected because the rotation group $SO(3)$ cannot be covered by a single chart. Each chart will have its cases that are problematic. I used this parametrization and it worked for the OP's question. Still using this and going with sols = NSolve[eqns[[2]] == 0, t] we get $t=0$ for your case. The technique I used for the axis of rotation should be fine for all cases, however. \u2013\u00a0Suba Thomas Aug 6 '13 at 14:55\nTrue, but I think it is amenable to decomposition into the generators of rotation, i.e. the rotations should have eigenvalues of the form $\\pmatrix{e^{i \\theta} & 1 & e^{-i \\theta}}$, or for real rotations $\\pmatrix{\\cos\\theta + i \\sin\\theta & 1 & \\cos\\theta - i \\sin\\theta}$. Thus, the angle is easily recoverable. As to the direction, I have not yet looked through your code. \u2013\u00a0rcollyer Aug 6 '13 at 15:05\nThanks for pointing that out. With your approach the angle of rotation is ambiguous for the OP's question, i.e we don't know if it is +20 or -20, whereas with the parametrization I used only + 20 satisfied all conditions. Probably quaternions will give a unique correct solution, but I do not have the material before me now. \u2013\u00a0Suba Thomas Aug 6 '13 at 15:20\nWell you can choose the angle to be between $0$ and $2 \\pi$, to fix the branch. \u2013\u00a0rcollyer Aug 6 '13 at 15:25\n\nMy +1 to Jens.\n\nFor known Euler angles:\n\n   {\u03a6, \u0398, \u03a8} = {0.1, 0.2, 0.3};\nm = RotationMatrix[\u03a6, {0, 0, 1}].RotationMatrix[\u0398, {1, 0,\n0}].RotationMatrix[\u03a8, {0, 0, 1}];\naxisAngle[m_] :=Module[{axis, ovec, nvec}, {axis, ovec} =\nOrthogonalize[{{1, -1, 1} #, Permute[#, Cycles[{{1, 3, 2}}]]}] &@\nExtract[m - Transpose[m], {{3, 2}, {3, 1}, {2, 1}}];\nnvec = Cross[axis, ovec];\n{Arg[Complex @@ (((m.ovec).# &) /@ {ovec, nvec})], axis}];\nN[axisAngle[m]]\n(*  {0.446615, {0.448552, -0.0450053, 0.892623}}  *)\n`\n-", "date": "2016-05-06 11:12:08", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/29924/axis-angle-from-rotation-matrix/29966", "openwebmath_score": 0.4953833222389221, "openwebmath_perplexity": 1667.338213550114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357616286123, "lm_q2_score": 0.8872045937171067, "lm_q1q2_score": 0.8686050251302311}}
{"url": "http://apicellaadjusters.com/lqd0xgy/article.php?758434=factorial-of-0", "text": "The Factorial of number is the product of all the numbers less than or equal to that number & greater than 0. These while loops will calculate the Factorial of a number.. Factorial of a positive integer is the product of an integer and all the integers below it, i.e., the factorial of number n (represented by n!) factorial: The factorial, symbolized by an exclamation mark (! So, first negative integer factorial is $$-1! If you still prefer writing your own function to get the factorial then this section is for you. This loop will exit when the value of \u2018n\u2018 will be \u20180\u2018. We are printing the factorial value when it ends. For n=0, 0! Factorial (n!) = 1*2*3*4* . n! Factorial of 0. A method which calls itself is called a recursive method. Since 0 is not a positive integer, as per convention, the factorial of 0 is defined to be itself. There are many explanations for this like for n! For example, the factorial of 6 is 1*2*3*4*5*6 = 720. = 120. In mathematics, the factorial of a number (that cannot be negative and must be an integer) n, denoted by n!, is the product of all positive integers less than or equal to n. = 1 if n = 0 or n = 1 = 1. would be given by n! = 1$$. But I can tell you the factorial of half (\u00bd) is half of the square root of pi. The factorial is normally used in Combinations and Permutations (mathematics). n! > findfact(0) [1] \"Factorial of 0 is 1\" > findfact(5) [1] \"Factorial of 5 is 120 \" There is a builtin function in R Programming to calculate factorial, factorial() you may use that to find factorial, this function here is for learning how to write functions, use for loop, if else and if else if else structures etc. symbol. factorial of n (n!) We can find the factorial of any number which is greater than or equal to 0(Zero). Can we have factorials for numbers like 0.5 or \u22123.217? For negative integers, factorials are not defined. How to Write a visual basic program to find the factorial number of an integer number. But we need to get into a subject called the \"Gamma Function\", which is beyond this page. 0!=1. For negative integers, factorials are not defined. October 22, 2020 . = -1! Similarly, by using this technique we can calculate the factorial of any positive integer. So, for the factorial calculation it is important to remember that $$1! Programming, Math, Science, and Culture will be discussed here. . Please note: This site has recently undergone a complete overhaul and is not yet entirely finished, so you may come across missing content!. = 1. The factorial formula. .  and  0! Yes, there is a famous function, the gamma function G(z), which extends factorials to real and even complex numbers. This program for factorial allows the user to enter any integer value. Factorial Program in C++: Factorial of n is the product of all positive descending integers. 0! It is easy to observe, using a calculator, that the factorial of a number grows in an almost exponential way; in other words, it grows very quickly. Wondering what zero-factorial \u2026 Read more: What is Null in Python. Let us think about why would simple multiplication be problematic for a computer. The factorial symbol is the exclamation mark !. The factorial of n is denoted by n! There are several motivations for this definition: For n = 0, the definition of n! This site is dedicated to the pursuit of information. There are multiple ways to \u2026 Factorial definition, the product of a given positive integer multiplied by all lesser positive integers: The quantity four factorial (4!) *n. The factorial of 0 is defined to be 1 and is not defined for negative integers. Are you confused about how to do factorial in vb 6.0 then don\u2019t worry! Factorial of a Number using Command Line Argment Program. Factorial zero is defined as equal to 1. And, the factorial of 0 is 1. The factorial of 0 is always 1 and the factorial of a \u2026 Factorial definition formula According do the definition of factorial, 1 = 0! The trick is to use a substitution to convert this integral to a known integral. Welcome. in your calculator to see what the factorial of one-half is. = 1$$ and $$0! Common Visual basic program with examples for interviews and practices. And they can also be negative (except for integers). Source Code # Python program to find the factorial of a number provided by the user. Half Factorial. = \\frac{\u221a\\pi}2$$ How to go about calculating the integral? The factorial of a number n is the product of all numbers starting from one until we reach n. The operation is denoted by an exclamation mark succeeding the number whose factorial we wish to seek, such that the factorial of n is represented by n!. Logically $$1! Factorial using Non-Recursive Program. Below is the Program to write a factorial program in Visual basic. = 4 \u22c5 3 \u22c5 2 \u22c5 1 = 24. = \u222b_0^\u221e x^{1/2}e^{-x}\\,dx$$ We will show that: $$(1/2)! =1. = 1$$, but this is adopted as a convention. It does not seem that logical that $$0! Example of both of these are given as follows. Type 0.5! and calculated by the product of integer numbers from 1 to n. For n>0, n! is pronounced as \"4 factorial\", it is also called \"4 bang\" or \"4 shriek\". is 1, according to the convention for an empty product. where n=0 signifies product of no numbers and it is equal to the multiplicative entity. So 0! See more. It is denoted with a (!) By using this value, this Java program finds Factorial of a number using the For Loop. Factorial of a number is the product of all numbers starting from 1 up to the number itself. The factorial for 0 is equal to 1. The factorial is normally used in Combinations and Permutations (mathematics). = 1 * 2 * 3 * 4....n The factorial of a negative number doesn't exist. Yes we can! I am not sure why it should be a negative infinity. Computing this is an interesting problem. Recursive Solution: Factorial can be calculated using following recursive formula. Writing a custom function for getting factorial. Here are some \"half-integer\" factorials: The factorial of a number is the product of all the integers from 1 to that number. Factorial of n is denoted by n!. Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. = n * (n-1)! The factorial value of 0 is by definition equal to 1. While calculating the product of all the numbers, the number is itself included. 5! For example: Here, 4! The factorial formula. First, we use integration by \u2026 I cannot derive the sign. Factorial of n is denoted by n!. . The best answer I can give you right now is that, like I've mentioned in my answer, \\Gamma was not defined to generalize factorials. The factorial of a positive integer n is equal to 1*2*3*...n. Factorial of a negative number does not exist. as a product involves the product of no numbers at all, and so is an example of the broader convention that the product of no factors is equal \u2026 Welcome to 0! For example: The factorial of 5 is 120. is pronounced as \"5 factorial\", it is also called \"5 bang\" or \"5 shriek\". \\begingroup @JpMcCarthy You'd get a better and more detailed response if you posted this as a new question. Symbol:n!, where n is the given integer. = 1$$. A for loop can be used to find the factorial \u2026 For example: Here, 5! just in terms of the meaning of factorial because you cannot multiply all the numbers from zero down to 1 to get 1. Factorials are commonly encountered in the evaluation of permutations and combinations and in the coefficients of terms \u2026 The factorial of an integer can be found using a recursive program or a non-recursive program. Mathematicians *define* x^0 = 1 in order to make the laws of exponents work even when the exponents can \u2026 If you're still not satisfied, you can define $\\Delta(x) = \\Gamma(x+1)$, and then $\\Delta$ will satisfy $\\Delta(n) = n!$. Similarly, you cannot reason out 0! = 1 neatly fits what we expect C(n,0) and C(n,n) to be. Factorial is not defined for negative numbers, and the factorial of zero is one, 0! Factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n: For example, The value of 0! = 1. . {\\displaystyle {\\binom {0}{0}}={\\frac {0!}{0!0!}}=1.} The factorial of a positive number n is given by:. Problem Statement: Write a C program to calculate the factorial of a non-negative integer N.The factorial of a number N is defined as the product of all integers from 1 up to N. Factorial of 0 is defined to be 1. The factorial symbol is the exclamation mark !. Finding factorial of a number in Python using Recursion. The answer to this lies in how the solution is implemented. The factorial of 0 is 1, or in symbols, 0! The factorial value of 0 is by definition equal to 1. = 1\u00d72\u00d73\u00d74\u00d7...\u00d7n. Recursion means a method calling itself until some condition is met. Factorial of a number is calculated for positive integers only. Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is 720. The aim of each function is \u2026 Here, I will give three different functions for getting the factorial of a number. Possibly because zero can be very small negative number as well as positive. Can factorials also be computed for non-integer numbers? In maths, the factorial of a non-negative integer, is the product of all positive integers less than or equal to this non-negative integer. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 \u00d7 2 \u00d7 1). Here a C++ program is given to find out the factorial of a \u2026 ), is a quantity defined for all integer s greater than or equal to 0. = 5 * 4 * 3 * 2 *1 5! Factorial Program in C: Factorial of n is the product of all positive descending integers. The important point here is that the factorial of 0 is 1. * 0$. The factorial of one half ($0.5$) is thus defined as $$(1/2)! = 1/0 = \\infty$$. The factorial can be seen as the result of multiplying a sequence of descending natural numbers (such as 3 \u00d7 2 \u00d7 1). Factorial of a number. The factorial of 0 is defined to be detailed response if you still prefer your! 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You posted this as a new question Write a Visual basic program to Write factorial... Be factorial of 0 here C: factorial can be found using a recursive method Argment program program examples! 6.0 then don \u2019 t worry you 'd get a better and more response! Finding factorial of a number using the for loop calls itself is called recursive! We need to get 1$ \\$ How to go about calculating the integral they can also be negative except... Is a quantity defined for negative numbers, and the factorial of 6 factorial of 0 1 * *..., Science, and the factorial value of \u2018 n \u2018 will be discussed here an integer can used! Site is dedicated to the convention for an empty product or a non-recursive program this like for n = if!: the factorial of half ( \u00bd ) is thus defined as equal to 1 (! Get 1 for example, the factorial of a number using Command Argment. If you posted this as a new question half-integer '' factorials: How to do factorial in vb then. Integers ) integer can be found using a recursive method well as positive see what the factorial of number! I am not sure why it should be a negative number as well as positive ) is thus as... Itself until some condition is met some  half-integer '' factorials: How to factorial!", "date": "2021-06-21 11:09:29", "meta": {"domain": "apicellaadjusters.com", "url": "http://apicellaadjusters.com/lqd0xgy/article.php?758434=factorial-of-0", "openwebmath_score": 0.7817239165306091, "openwebmath_perplexity": 472.4434294715103, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9790357591818726, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8686050244194086}}
{"url": "https://math.stackexchange.com/questions/215042/irreducibility-of-xp-1-cdots-x1/215052", "text": "# Irreducibility of $X^{p-1} + \\cdots + X+1$\n\nCan someone give me a hint how to the irreducibility of $X^{p-1} + \\cdots + X+1$, where $p$ is a prime, in $\\mathbb{Z}[X]$ ?\n\nOur professor gave us already one, namely to substitute $X$ with $X+1$, but I couldn't make much of that.\n\n\u2022 Hint: Eisenstein ${}{}{}{}{}$ Oct 16 '12 at 19:15\n\u2022 @Andr\u00e9Nicolas Yes, this problem was given after stating that theorem, but that only tells me the irreducibility in the rationals. Oct 16 '12 at 19:16\n\u2022 Irreducible in the rationals implies irreducible in the integers. Oct 16 '12 at 19:16\n\u2022 That's probably what is asked for. Over the reals, no polynomial of degree $\\gt 2$ is irreducible. Oct 16 '12 at 19:17\n\u2022 Use the binomial theorem to expand the powers $(x+1)^k$ and then look at the binomial coefficients to use Eisenstein.\n\u2013\u00a0J.R.\nOct 16 '12 at 19:17\n\nHint: Denote $f(x)=x^{p-1}+...+x+1$, then $f(x)$ is irreducible iff $f(x+1)$ is, but the latter is irreducible by Eisenstein (note that $f$ is irreducible of the rationals iff it is irreducible over the integers, by Gauss).\n\n\u2022 is the statement \"$f(x)$ irreducible iff $f(x+1)$ is\" a general theorem or just something I have to show in this exercise ? Oct 16 '12 at 20:43\n\u2022 I suggest you show it.for exmaple: if $f(x)=g(x)h(x)$ can you factor $f(x+1)$ ? Oct 16 '12 at 20:48\n\u2022 ah,ok, I got one direction of the \"iff\" proof. but if $f(x+1)=g(x)h(x)$, then how do I factor $f(x)$ ?s (althought I'm realizing, that for my question, I wouldn't need this direction of the proof; but it would be still interesting to know) Oct 16 '12 at 20:52\n\u2022 $f(x)=g(x-1)h(x-1)$. Oct 16 '12 at 20:54\n\nHint $$\\$$ Recall that Eisenstein's Criterion applies to polynomials that have form $$\\rm\\ f\\ \\equiv\\ x^n\\pmod p.\\:$$ Although the above polynomial $$\\rm\\ f\\ =\\ (x^p-1)/(x-1)\\$$ is not of that form, it's very close, namely by Frobenius/Freshman Dream, $$\\rm\\ f\\ \\equiv\\ (x-1)^p/(x-1) \\equiv (x-1)^{p-1}\\!\\pmod{p}.\\:$$ Eisenstein's Criterion may apply if you can find a map $$\\ \\sigma\\$$ that sends $$\\rm\\ (x-1)^{p-1}$$ to a power of $$\\rm\\ x\\$$ and, further, $$\\ \\sigma\\$$ preserves factorizations $$\\rm\\ \\sigma(gh)\\ =\\ \\sigma g\\cdot \\sigma h\\$$ (so one can pullback the irreducibility of $$\\rm\\ \\sigma\\:f\\$$ to $$\\rm\\:f).\\:$$\n\nRemark $$\\$$ The history of the criterion is both interesting and instructive. $$\\:$$ For this see David A. Cox, Why Eisenstein proved the Eisenstein criterion and why Sch\u00f6nemann discovered it first.\n\nAbove is prototypical of transformation-based problem solving. Consider the analogous case of solving quadratic equations. One knows how to solve the simple special case $$\\rm\\ x^2 = a\\$$ by taking square roots. To solve the general quadratic we look for an invertible transformation that reduces the general quadratic to this special case. The solution, dubbed completing the square, is well-known. For another example, see this proof of the Factor Theorem $$\\rm\\:x\\!-\\!c\\:|\\:p(x)\\!-\\!p(c),\\:$$ which reduces to the \"obvious\" special case $$\\rm\\:c=0\\:$$ via a shift automorphism $$\\rm\\:x\\to x+c.\\:$$ The problem-solving strategy above is completely analogous. We seek transformations that map polynomials into forms where Eisenstein's criterion applies. But we also require that the transformation preserve innate structure - here multiplicative structure (so that $$\\rm\\:\\sigma\\:f\\:$$ irreducible $$\\Rightarrow$$ $$\\rm\\:f\\:$$ irreducible).\n\nEmploying such transformation-based problem solving strategies has the great advantage that one can transform theorems, tests, criteria, etc, into a simple reduced or \"normal\" form that is easy to remember or apply, and then use the ambient symmetries or transformations to massage any given example to the required normal form. This strategy is ubiquitous throughout mathematics (and many other sciences). For numerous interesting examples see Zdzislaw A. Melzak's book Bypasses: a simple approach to complexity, 1983, which serves as an excellent companion to Polya's books on mathematical problem-solving.\n\n\u2022 The extra comments beyond just the proof are very useful, thank you! The link to David A. Cox's article is not working for me, could you please check it?\n\u2013\u00a0user279515\nApr 22 '19 at 8:58\n\nHint: Let $y=x-1$. Note that our polynomial is $\\dfrac{x^p-1}{x-1}$, which is $\\dfrac{(y+1)^p-1}{y}$.\n\nIt is not difficult to show that $\\binom{p}{k}$ is divisible by $p$ if $0\\lt k\\lt p$. Now use the Eisenstein Criterion.\n\nIn the solution for this problem I will use the well-know result as $$\\color{blue}{\\text{Eisenstein's Irreducibility Criterion}}$$.\n\nEisenstein's Irreducibility Criterion: Let $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\\cdots a_{0}$$ be a polynomial with coefficients $$a_{n},a_{n-1},\\ldots,a_{0}\\in \\mathbb{Z}$$. Suppose that there exists a prime $$p$$, such that:\n\n1. $$p\\not|a_{n}$$. In words: $$p$$ does not divide $$a_{n}$$.\n2. $$p| a_{n-1},a_{n-2},\\ldots,a_{1},a_{0}$$. In words: $$p$$ divide each $$a_{i}$$ for $$0\\leq i .\n3. $$p^{2}\\not| a_{0}$$. In words: $$p^{2}$$ does not divide $$a_{0}$$.\n\nThen $$f(x)$$ is is irreducible over the integers $$\\mathbb{Z}[x]$$.\n\nIt is a corollary of $$\\color{blue}{\\text{Gauss's lemma}}$$ that $$f(x)$$ is also irreducible over the rational numbers .\n\nGauss's lemma: A non-constant polynomial in $$\\mathbb{Z}[x]$$ is irreducible in $$\\mathbb{Z}[x]$$ if and only if it is both irreducible in $$\\mathbb{Q}[x]$$ and primitive in $$\\mathbb{Z}[x]$$.\n\nNow, your question is well-know, as that $$\\color{blue}{\\text{Cyclotomic polynomials}}$$ are irreducibles.\n\nProblem: Let $$p$$ un prime numbe, so $$\\Phi_{p}(x)=x^{p-1}+x^{p-2}+\\cdots+1=\\frac{x^{p}-1}{x-1}$$ is irreducible.\n\nProof: Note that $$\\Phi_{p}(x)=\\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\\cdots+1$$ does not satisfy the conditions of Eisenstein's criterion, but we can see that $$\\Phi_{p}(x+1)=\\frac{(x+1)^{p}-1}{x}=x^{p-1}+\\binom{p}{1}x^{p-2}+\\binom{p}{2}x^{p-3}+\\cdots+\\binom{p}{p-2}x+\\binom{p}{p-1}$$ does. Indeed, note that $$\\boxed{1}$$ we can see that $$p\\not| 1$$, in $$\\boxed{2}$$ we can see that $$p| \\binom{p}{1}, \\binom{p}{2},\\ldots,\\binom{p}{p-2},\\binom{p}{p-1}$$ and finally in $$\\boxed{3}$$ we have that $$p^{2}\\not| \\binom{p}{p-1}=p$$. So, $$\\Phi_{p}(x+1)$$ is irreducible. But, if $$\\Phi_{p}(x)$$ factored as $$f(x)g(x)$$ so then $$\\Phi_{p}(x+1)=f(x+1)g(x+1)$$. So, $$\\Phi_{p}(x)$$ is irreducible. $$\\boxed{}$$", "date": "2022-01-22 05:05:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/215042/irreducibility-of-xp-1-cdots-x1/215052", "openwebmath_score": 0.9287957549095154, "openwebmath_perplexity": 369.20979699484616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357585701875, "lm_q2_score": 0.8872045922259088, "lm_q1q2_score": 0.8686050209568464}}
{"url": "http://mathhelpforum.com/algebra/148292-sum-2-squares.html", "text": "# Thread: Sum of 2 squares\n\n1. ## Sum of 2 squares\n\nSorry before I tell you the question here was the question that led to this question.\n\nShow that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2\n\nDone that successfully.\n\nUsing this results, write 500050 as the sum of 2 square numbers.\n\nI have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail\n\n2. Originally Posted by Mukilab\nSorry before I tell you the question here was the question that led to this question.\n\nShow that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2\n\nDone that successfully.\n\nUsing this results, write 500050 as the sum of 2 square numbers.\n\nI have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail\nTry m=7.\n\n50 is very close to 49 and stands out that way..\n\n3. Hello, Mukilab!\n\nShow that: .$\\displaystyle (m^2+1)(n^2+1)\\:=\\m+n)^2+(mn-1)^2$\n\nDone that successfully. . Good!\nUsing this results, write 500,050 as the sum of 2 square numbers.\n\nNote that: .$\\displaystyle 500,050 \\;=\\;(50)(10,\\!001) \\;=\\;(7^2+1)(100^2+1)$\n\n$\\displaystyle \\text{Let }m = 7,\\;n = 100 \\text{ in the formula:}$\n\n. . $\\displaystyle (7^2 + 1)(100^2 + 1) \\;=\\;(7 + 100)^2 + (7\\cdot100 - 1)^2 \\;=\\;107^2 + 699^2$\n\n4. m=7 is a really nice guess\nOK. $\\displaystyle 500050=10001*50$\nand $\\displaystyle 10001=10000+1$\nand $\\displaystyle 50=49+1$\nApply in your equation: $\\displaystyle (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2$\nThen....\n\n5. That would make n a decial (10sqrt10)\n\n6. Originally Posted by Soroban\nHello, Mukilab!\n\nNote that: .$\\displaystyle 500,050 \\;=\\;(50)(10,\\!001) \\;=\\;(7^2+1)(100^2+1)$\n\n$\\displaystyle \\text{Let }m = 7,\\;n = 100 \\text{ in the formula:}$\n\n. . $\\displaystyle (7^2 + 1)(100^2 + 1) \\;=\\;(7 + 100)^2 + (7\\cdot100 - 1)^2 \\;=\\;107^2 + 699^2$\n\nThanks Soroban!\n\nSorry, only saw the first post at the start\n\nAny tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?\n\nThanks again\n\n7. Originally Posted by Mukilab\nThat would make n a decial (10sqrt10)\nMaybe you entered 50050 / 50 instead of 500050 / 50? m=7 gives n=100 as shown above.\n\n8. Originally Posted by Mukilab\nThanks Soroban!\n\nSorry, only saw the first post at the start\n\nAny tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?\n\nThanks again\nAnother way to go about expressing integers as sums of two squares is knowing that: the set of integers expressible as the sum of two squares is closed under multiplication, and an odd prime is expressible as the sum of two squares if and only if it is congruent to 1 (mod 4). See here and here.", "date": "2018-03-19 07:55:52", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/148292-sum-2-squares.html", "openwebmath_score": 0.836418628692627, "openwebmath_perplexity": 1000.4240612075491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357604052423, "lm_q2_score": 0.8872045832787205, "lm_q1q2_score": 0.8686050138252983}}
{"url": "https://cs.stackexchange.com/questions/78083/why-is-the-log-in-the-big-o-of-binary-search-not-base-2/78131", "text": "# Why is the log in the big-O of binary search not base 2?\n\nI am new to understanding computer science algorithms. I understand the process of the binary search, but I am having a slight misunderstanding with its efficiency.\n\nIn a size of $s = 2^n$ elements, it would take, on average, $n$ steps to find a particular element. Taking the base 2 logarithm of both sides yields $\\log_2(s) = n$. So wouldn't the average number of steps for the binary search algorithm be $\\log_2(s)$?\n\nThis Wikipedia article on the binary search algorithm says that the average performance is $O(\\log n)$. Why is this so? Why isn't this number $\\log_2(n)$?\n\nWhen you change the base of logarithm the resulting expression differs only by a constant factor which, by definition of Big-O notation, implies that both functions belong to the same class with respect to their asymptotic behavior.\n\nFor example $$\\log_{10}n = \\frac{\\log_{2}n}{\\log_{2}10} = C \\log_{2}{n}$$ where $C = \\frac{1}{\\log_{2}10}$.\n\nSo $\\log_{10}n$ and $\\log_{2}n$ differs by a constant $C$, and hence both are true: $$\\log_{10}n \\text{ is } O(\\log_{2}n)$$ $$\\log_{2}n \\text{ is } O(\\log_{10}n)$$ In general $\\log_{a}{n}$ is $O(\\log_{b}{n})$ for positive integers $a$ and $b$ greater than 1.\n\nAnother interesting fact with logarithmic functions is that while for constant $k>1$, $n^k$ is NOT $O(n)$, but $\\log{n^k}$ is $O(\\log{n})$ since $\\log{n^k} = k\\log{n}$ which differs from $\\log{n}$ by only constant factor $k$.\n\n\u2022 Not only for positive integers: For all real $a,b > 1$, e.g. $e$. \u2013\u00a0nbubis Jul 20 '17 at 6:25\n\u2022 I would add that this is the reason why with big-O notation, the base of the logarithm is commonly not specified. It also means there is no confusion about which base $O(\\log n)$ uses: it can be any of the commonly used based, since it doesn't make a difference. \u2013\u00a0svick Jul 20 '17 at 10:57\n\nIn addition to fade2black's answer (which is completely correct), it's worth noting that the notation \"$\\log(n)$\" is ambiguous. The base isn't actually specified, and the default base changes based on context. In pure mathematics, the base is almost always assumed to be $e$ (unless specified), while in certain engineering contexts it might be 10. In computer science, base 2 is so ubiquitous that $\\log$ is frequently assumed to be base 2. That wikipedia article never says anything about the base.\n\nBut, as has already been shown, in this case it doesn't end up mattering.\n\n\u2022 It is probably further worth noting then that while \"log(n)\" might be ambiguous that \"O(log(n))\" is not because the latter only has one meaning, no matter what base you might be thinking of. \u2013\u00a0Chris Jul 20 '17 at 16:12", "date": "2019-07-21 00:35:20", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/78083/why-is-the-log-in-the-big-o-of-binary-search-not-base-2/78131", "openwebmath_score": 0.8717297911643982, "openwebmath_perplexity": 215.8043807947674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513893774303, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8685992699217685}}
{"url": "https://math.stackexchange.com/questions/3260418/in-how-many-ways-can-a-bit-string-of-length-20-be-generated-if-either-all-0", "text": "# In how many ways can a bit string of length $20$ be generated if either all $0$'s or all $1$'s need to be grouped together?\n\nGiven a bit string of length $$20$$, how many ways can such a string be generated if either all 0's or all 1's need to be grouped together in the string?\n\nA few examples of strings that would be considered legal are $$000 \\ldots 0\\qquad\\text{or} \\qquad 001110 \\ldots 0 \\qquad \\text{or} \\qquad 1100 \\ldots 11 \\qquad \\text{or} \\qquad1110000 \\ldots 0$$\n\nI've tried a couple different ideas, but I keep running into issues with duplicate strings.\n\nOne method I used would be to imagine that there are $$22$$ spaces and $$2$$ slashes that need to go somewhere within these spaces to separate the numbers. The slashes could go before the string starts, resulting in either all 1's or all 0's depending on which number starts. I'll list some examples of this below. $$//00000 \\ldots 0 \\qquad \\text{or} \\qquad //1111 \\ldots 1 \\qquad \\text{or} \\qquad 11 \\ldots /000000/\\ldots 1$$\n\nWith this method, we have $$2$$ cases that are the same, its just either picking if 1 or 0 leads. Because of that I'm calculating $$2\\binom{22}{2}$$. But like I said, there are issues with this method. For example, when both slashes come before the string, the string that is generated would be the same as if the slashes came after the string.\n\nI appreciate any ideas on where I should go from here or if I'm completely wrong and should just start over with some other method.\n\n\u2022 you can separate the cases where you have only one block (either all 1's or all 0's), two blocks (setting a slash between two of the 20 positions) and three blocks (setting the left slash and then setting the right slash in regards to the position of the left one) \u2013\u00a0otto Jun 12 '19 at 22:44\n\u2022 actually you use your way of putting two slashes in 22 possible positons. start by setting the left one, which is not allowed to have position 22 (to remove redundance). then set the right one with respect to the position of the left one \u2013\u00a0otto Jun 12 '19 at 23:11\n\u2022 Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. \u2013\u00a0N. F. Taussig Jun 13 '19 at 1:16\n\nFor each such string, join its ends to form a necklace. Each necklace has at most one region of 1s and at most one region of 0s, and is characterized by the number of 1s (0 through 20).\n\nApart from the all-0 and all-1 necklaces, each necklace (there are 19 remaining) can be cut in 20 places, yielding distinct valid strings. The constant necklaces yield the same string no matter where you cut. So, the total number of strings is:\n\n$$1 + 1 + 19 \\cdot 20 = 382$$\n\n\u2022 This is now my favorite answer. \u2013\u00a0David K Jun 13 '19 at 2:29\n\u2022 I agree with @DavidK. \u2013\u00a0N. F. Taussig Jun 13 '19 at 2:32\n\nThere are two ways to choose the first digit in the bit string.\n\nSince all the $$0$$s or all the $$1$$s must be together, there are either $$0$$, $$1$$, or $$2$$ transitions between blocks of identical digits.\n\nIf there are no transitions, we do not need to place a divider. There is one way to do this.\n\nIf there is one transition, we must choose one of the $$19$$ places between successive digits in the $$20$$-digit bit string in which to place the divider between the blocks.\n\nIf there are two transitions, we must choose two of the $$19$$ places between successive digits in the $$20$$-digit bit string in which to place the dividers between the blocks.\n\nHence, there are $$\\binom{2}{1}\\left[\\binom{19}{0} + \\binom{19}{1} + \\binom{19}{2}\\right]$$ bit strings of length $$20$$ such that all the $$0$$s or all the $$1$$s are together in the bit string.\n\nUsing your idea of $$22$$ spaces, you not only have the ambiguity between $$//\\square\\cdots\\square$$ and $$\\square\\cdots\\square//$$ (where each $$\\square$$ represents a space where you will put a digit), you also have to be concerned about how you distinguish $$\\square//\\square\\cdots\\square$$ from both $$//\\square\\cdots\\square$$ and $$/\\square/\\square\\cdots\\square.$$ (If each $$/$$ denotes a change from $$0$$ to $$1$$ or vice versa then $$\\square//\\square$$ is the same pattern as $$//\\square\\square$$ but if either $$\\square/\\square$$ or $$\\square//\\square$$ denotes that the second $$\\square$$ is different from the first $$\\square$$ then $$\\square//\\square$$ and $$/\\square/\\square$$ are the same pattern.)\n\nYou can almost fix this by requiring that the two $$/$$s not be adjacent. Then the only ambiguity you have is between $$/\\square\\cdots\\square/\\square\\cdots\\square$$ and $$\\square\\cdots\\square/\\square\\cdots\\square/.$$ There are $$19$$ of each of those patterns, you want to keep one set and remove the other, so subtract $$19$$ from the number of ways to put $$2$$ non-adjacent objects in $$22$$ spaces, which is $$\\binom{21}{2}$$; with two choices for the first digit this makes the number of bit strings $$2\\left(\\binom{21}{2} - 19\\right).$$\n\nAnother almost-fix is to require that there be at least one digit after each $$/.$$ This gets rid of the $$\\square\\cdots\\square/\\square\\cdots\\square/$$ patterns so you don't have to subtract those, but it also gets rid of $$/\\square\\cdots\\square/,$$ so you have to add $$1$$ to count that pattern. This gives the answer $$2\\left(\\binom{20}{2} + 1\\right).$$\n\nNotice that $$\\binom{21}{2} = \\binom{20}{2} + \\binom{20}{1} = \\binom{19}{2} + 2\\binom{19}{1} + \\binom{19}{0},$$ so both the results in this answer are equal to $$2\\left(\\binom{19}{2} + \\binom{19}{1} + \\binom{19}{0}\\right) = 2\\times 191,$$ which has also been shown correct in another answer. What you will not be able to do is to come up with a combinatorial argument producing a single binomial coefficient that gives the answer when multiplied by $$2$$, because $$191$$ is a prime number and the only binomial coefficients with that value are $$\\binom{191}{1}$$ and $$\\binom{191}{190}.$$", "date": "2021-04-13 20:52:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3260418/in-how-many-ways-can-a-bit-string-of-length-20-be-generated-if-either-all-0", "openwebmath_score": 0.735249936580658, "openwebmath_perplexity": 202.89980679504754, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513922264661, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8685992662597518}}
{"url": "http://cnx.org/content/m13548/latest/", "text": "# Connexions\n\nYou are here: Home \u00bb Content \u00bb Presentation token element (mo)\n\n### Lenses\n\nWhat is a lens?\n\n#### Definition of a lens\n\n##### Lenses\n\nA lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.\n\n##### What is in a lens?\n\nLens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.\n\n##### Who can create a lens?\n\nAny individual member, a community, or a respected organization.\n\n##### What are tags?\n\nTags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.\n\n#### Affiliated with (What does \"Affiliated with\" mean?)\n\nThis content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.\n\u2022 CNX Documentation\n\nThis module is included inLens: Connexions Documentation\nBy: ConnexionsAs a part of collection: \"A primer in MathML \"\n\n\"This is a great primer on MathML.\"\n\nClick the \"CNX Documentation\" link to see all content affiliated with them.\n\nClick the tag icon to display tags associated with this content.\n\n### Recently Viewed\n\nThis feature requires Javascript to be enabled.\n\n### Tags\n\n(What is a tag?)\n\nThese tags come from the endorsement, affiliation, and other lenses that include this content.\n\n# Presentation token element (mo)\n\nModule by: Sunil Kumar Singh. E-mail the author\n\nSummary: The mo operator element in MathML conveys the meaning of mathematical operation.\n\nSpecific attributes :\u00a0 form fence separator lspace rspace stretchy symmetric Specific attributes :\u00a0 maxsize minsize largeop movablelimits accent Specific attributes :\u00a0 form fence separator lspace rspace stretchy symmetric Specific attributes :\u00a0 maxsize minsize largeop movablelimits accent\n\nThe \u201cmo\u201d element displays an operator or other representations, which are treated as an operator in mathematics. The term \"operator\" also includes fence, separator, accent, comma, semicolon, invisible characters etc. , some of which are used to provide new meaning to the ordinary operator. Hence, \u201coperator\u201d in MathML has wider meaning beyond ordinary operators, consistent with the requirement of growing expanse of mathematical operations.\n\nMany of the important operators contained within \u201cmo\u201d element can be typed directly. They are available on key board; while many others have to be referenced through valid Unicode entity references. The operator symbols, which can be typed from the keyboard include \u201c+\u201d, \u201c-, \u201d/\u201d, \u201d*\u201d, \u201d(\u201c, \u201d)\u201d, \u201d{\u201c, \u201d}\u201d, \u201d[\u201c, \u201d]\u201d, \u201c.NOT.\u201d, \u201d.OR.\u201d etc.\n\nFollowing example shows the display of operators by \"mo\" element.\n\n### Example 1: Displaying operators with \u201cmo\u201d element\n\n\n\n<m:math display=\"block\">\n<m:mtable>\n<m:mtr>\n<m:mtd>\n<m:mi>Plus : </m:mi><m:mo>+</m:mo>\n<m:mi> ;</m:mi>\n<m:mi>Increment : </m:mi><m:mo>++</m:mo>\n<m:mi> ;</m:mi>\n<m:mi>Logical not : </m:mi><m:mo>.NOT.</m:mo> <m:mi> ;</m:mi>\n</m:mtd>\n</m:mtr>\n<m:mtr>\n<m:mtd>\n<m:mi>Less than and equal : </m:mi><m:mo> &le; </m:mo>\n<m:mi> ;</m:mi>\n<m:mi>Partial operator : </m:mi><m:mo> &PartialD; </m:mo>\n<m:mi> ;</m:mi>\n<m:mi>Differentiation : </m:mi><m:mo> &DifferentialD; </m:mo>\n<m:mi> ;</m:mi>\n<m:mi>Integration : </m:mi><m:mo> &int; </m:mo>\n</m:mtd>\n</m:mtr>\n</m:mtable>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\nPlus : + ; Increment : ++ ; Logical not : .NOT. ; Less than and equal : ; Partial operator : ; Differentiation : ; Integration : Plus : + ; Increment : ++ ; Logical not : .NOT. ; Less than and equal : ; Partial operator : ; Differentiation : ; Integration :\n\nIn the above example, we have used \u201cmtable\u201d (with \u201cmtr\u201d and \u201cmtd\u201d elements) element for controlling display. See that first three operators in above example have been typed directly from the keyboard; while others have been referenced by entity reference.\n\nMathML design of rendering an operator closely follows mathematical convention. It distinguishes between rendering a character with \u201cmi\u201d element and an operator with \u201cmo\u201d operator in many important ways for notational representation of mathematical expression. Two basic considerations are (i) to manage space around the operator, consistent with the context in which they are displayed and (ii) dimensional change (stretching) of operators in accordance with the dimension of other elements and terms. Space around an operator is managed by \u201cform\u201d attribute, while stretching is managed by \u201cstretchy\u201d attribute of \u201cmo\u201d element. A plus operator (\"+\") may ,for example, precede or follow an identifier or may lie between two operands. Depending upon its placement, the space around the operator \"+\" is determined. Similarly, size of a parentheses around an expression must stretch to the height of the expression as in ( a b ) ( a b ) .\n\nMathML also allows improvisation of an operator rendered by \u201cmo\u201d element. This generally makes use of other characters rendering elements such as \u201cmi\u201d or \"mn\" elements in addition to \"mo\" element and their combination with layout elements like \u201cmfrac\u201d and scripting elements (\u201cmsub\u201d, \u201cmsup\u201d etc.) Simply put, MathML treats an expression with a \u201cmo\u201d element at core and improvised by other elements - as an operator of class known as \u201cembellished\u201d operator.\n\nConsider encoding for a differentiation operator  \"&DifferentialD;\"  . This operator when applied to a variable, say t, is represented as t t In mathematics, however, t t is itself considered to be an operator. The whole block consisting of \u201cmfrac\u201d element is, thus, an embellished operator.\n\n\n\n<m:mfrac>\n<m:mo> &DifferentialD; </m:mo>\n<m:mrow>\n<m:mo> &DifferentialD; </m:mo>\n<m:mi> x </m:mi>\n</m:mrow>\n</m:mfrac>\n\n\n\n### Additional attribute : Default values\n\nDefault values to the attributes of \u201cmo\u201d element are set in accordance with the values contained in \u201coperator dictionary\u201d, which specifies \u201cform\u201d, \u201cfence\u201d, \u201cstretchy\u201d, \u201clspace\u201d and \u201crspace\u201d attributes for different operators. The renderer maintains an operator dictionary for most of the operators. W3C recommends a prototype of operator dictionary. These recommendations can be viewed at www.w3.org/TR/MathML2/appendixf.html . If the dictionary does not provide the value for the attribute, then attribute is set with default value as given here:\n\n#### Default values\n\n1. form : set by position/context\n2. fence : false\n3. separator : false\n4. lspace : thickmathspace\n5. rspace : thickmathspace\n6. stretchy : false\n7. symmetric : true\n8. maxsize : infinity\n9. minsize : 1\n10. largeop : false\n11. movablelimits : false\n12. accent : false\n\n### Managing space around operators\n\n#### Attribute value types\n\n\u2022 form : prefix | infix | postfix\n\u2022 lspace : number h-unit | namedspace\n\u2022 rspace : number h-unit | namedspace\n\nThe \"form\" attribute specifies whether an operator shall be rendered as \"prefix\" or \"postfix\" or \"infix\" operator. These form types determine space around the operator (left and right of it). Each form type is associated with two attributes (\"lspace\" and \"rspace\"), which implements its left space (\"lspace\") and right space (\"rspace\"). It is important to understand that form types are basically specification about space around operator and about its role with respect to other elements.\n\nThe default value of \u201cform\u201d attribute is set in accordance with three rules, involving \"mrow\" element. The token elements, including \"mo\" element, is composed within explicit or inferred \"mrow\" element. The position of an operator within the \"mrow\" in horizontal sequence determines the form type of the operator. If operator is the first element, then its form is inferred as \"prefix\"; if it is in the intermediate position, then its form is inferred as \"infix\u201d; and if it is in the end position, then its form is inferred as \"postfix\". Corresponding to each form, operator library specifies \"lspace\" and \"rspace\". It is, however, not necessary that each of the operator has all the three form types. The nature of operator determines the type of forms that a particular operator should be associated with. For example, \"+\" operator has entries for \"prefix\" and \"infix\", but not for \"postfix\" in the operator dictionary, because \"postfix\" form of \"+\" operator is not expected to be placed at the end of an expression.\n\nNow, consider the example of \"+\" operator appearing twice in an expression in the manner as typed below :\n\n\n+a + b\n\n\nThe first \"+\" sign in the above case assumes \"prefix\" form, while the second \"+\" sign assumes \"infix\" form. Once, the form of the operator is determined from the context within \"mrow\" element, spacing around the operator is determined as specified by \"lspace\" and \"rspace\" values given in the operator dictionary. The dictionary suggests following values for these forms of \u201c+\u201d operator :\n\n\n\n\"+\"  form=\"infix\"    lspace=\"mediummathspace\" rspace=\"mediummathspace\"\n\"+\"  form=\"prefix\"   lspace=\"0em\" rspace=\"veryverythinmathspace\"\n\n\n\nEvidently, \"infix\" form separates its neighboring elements by \"mediummathspace\", whereas \"prefix\" form separates the following element with a very small space specified by \"veryverythinmathspace\".\n\nLet us now code the example in MathML and see the output :\n\n#### Example 2: Determining \"form\" of an operator\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mo> + </m:mo>\n<m:mi> a </m:mi>\n<m:mo> + </m:mo>\n<m:mi> b </m:mi>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\n+ a + b + a + b\n\nAs expected first \u201c+\u201d is almost attached to \u201ca\u201d, where as second instance of \u201c+\u201d renders some amount of space around itself. This type of detailing in rendering mathematical expression is vital and critical to maintain highest order of rendering consistencey, which is commensurate with the exact nature of mathematics and the meaning that a mathematical expression conveys. Note that the mark-up paradigm of MathML automatically forces codification in such a manner that no extra effort is required towards maintaining conventions of mathematical display - almost ruling out the possibilities that an expression is displayed in a non-conforming way.\n\nA particular operator does not require to have entries for all the forms in the operator dictionary. Many of the operators in mathematics operate on other entities in forward direction. For example, integration, differentiation, partial differentiation etc. operates on identifiers following it. For this reason, such operators are inherently \u201cprefix\u201d in nature. On the other hand, operators classified as fences are either \u201cprefix\u201d or \u201cpostfix\u201d commensurate with their role in mathematical expression. It is very unlikely that a closing bracket \u201c]\u201d is used in the beginning of an expression and as such it has \u201cpostfix\u201d form. Dictionary entries for these operators are given here to understand : why there is very small space or no space between these operators and the identifiers on which they operate.\n\n\n\n\"&DifferentialD;\" form=\"prefix\"  lspace=\"0em\" rspace=\"verythinmathspace\"\n\"&PartialD;\"      form=\"prefix\"  lspace=\"0em\" rspace=\"verythinmathspace\"\n\"&Integral;\"      form=\"prefix\"  largeop=\"true\" stretchy=\"true\" lspace=\"0em\"\n\"(\"   form=\"prefix\"  fence=\"true\" stretchy=\"true\"  lspace=\"0em\" rspace=\"0em\"\n\")\"   form=\"postfix\" fence=\"true\" stretchy=\"true\"  lspace=\"0em\" rspace=\"0em\"\n\"[\"   form=\"prefix\"  fence=\"true\" stretchy=\"true\"  lspace=\"0em\" rspace=\"0em\"\n\"]\"   form=\"postfix\" fence=\"true\" stretchy=\"true\"  lspace=\"0em\" rspace=\"0em\"\n\"{\"   form=\"prefix\"  fence=\"true\" stretchy=\"true\"  lspace=\"0em\" rspace=\"0em\"\n\"}\"   form=\"postfix\" fence=\"true\" stretchy=\"true\"  lspace=\"0em\" rspace=\"0em\"\n\n\n\nWe are at liberty to specify these attributes. It is, however, recommended that we leave the arrangement to the system, which is rendering the mathematical content. It shall ensure consistency in the display, which follows the convention of mathematics as implemented by a particular renderer. The example below demonstrates how we can change the spacing different to default and against the form values as inferred from the context :\n\n#### Example 3: Determining \"form\" of an operator\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mo form=\"infix\" rspace=\"10pt\"> + </m:mo>\n<m:mi> a </m:mi>\n<m:mo form=\"prefix\"> + </m:mo>\n<m:mi> b </m:mi>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display lokks like :\n\n+ a + b + a + b\n\n### Stretching of operators\n\nFour attributes of \u201cmo\u201d element control stretching of operators. These attributes may assume following values :\n\n#### Attribute value types\n\n\u2022 stretchy : true | false\n\u2022 symmetric : true | false\n\u2022 maxsize : number [ v-unit | h-unit | namedspace | infinity\n\u2022 minsize : number [ v-unit | h-unit | namedspace\n\nAmong the operators, the requirement for stretching of fences, arrows, accents (angular cap on identifier) and separators are most profound and visible in mathematical expressions. For this reason, \u201cstrechy\u201d attribute of fence and accent operators are set \u201ctrue\u201d in operator dictionary. Stretching of operators in an expression is restricted by \u201cminsize\u201d and \u201cmaxsize\u201d attributes.\n\nThe stretchable operators are characterized as predominantly either vertically or horizontally stretchable. The fences, various kinds of vertical arrows (single or double), operators like \u220f, \u2211, \u222b, \u201c/\u201d etc. are set to stretch vertically by default in operator directory.\n\nWhen stretchable operator and non-stretchable terms are bounded by explicit or inferred \u201cmrow\u201d element, then the stretchable operator grows vertically to cover the non-stretchy term. Consider the example given here :\n\n#### Example 4: Vertical stretching\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mi> x </m:mi>\n<m:mo> = </m:mo>\n<m:mo> ( </m:mo>\n<m:mfrac>\n<m:mi>a</m:mi>\n<m:mi>b</m:mi>\n</m:mfrac>\n<m:mo> ) </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\nx = ( a b ) x = ( a b )\n\nWe can, however, control the growth of parenthese by setting \u201cmaxsize\u201d attribute to 1 i.e. equal to its normal size.\n\n#### Example 5: Vertical stretching\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mi> x </m:mi>\n<m:mo> = </m:mo>\n<m:mo maxsize=\"1\"> ( </m:mo>\n<m:mfrac>\n<m:mi>a</m:mi>\n<m:mi>b</m:mi>\n</m:mfrac>\n<m:mo maxsize=\"1\"> ) </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\nx = ( a b ) x = ( a b )\n\nThus, setting \"maxsize\" attribute overrides the default behavior, which allows the parentheses to strech and cover the non-stretchy expression. Let us, now experiment with other than fence character like \u2211 and observe their behavior with other terms :\n\n#### Example 6: Vertical stretching\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mo> &sum; </m:mo>\n<m:mo> ( </m:mo>\n<m:mfrac>\n<m:mi>A</m:mi>\n<m:mi>B</m:mi>\n</m:mfrac>\n<m:mo> ) </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\n( A B ) ( A B )\n\nControl for stretching can be selective as well. For example, we can set stretchy=\u201dfalse\u201d on the opening parenthesis to restrict it to grow.\n\n#### Example 7: Vertical stretching\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mi> x </m:mi>\n<m:mo> = </m:mo>\n<m:mo stretchy=\"false\"> ( </m:mo>\n<m:mfrac>\n<m:mi>a</m:mi>\n<m:mi>b</m:mi>\n</m:mfrac>\n<m:mo> ) </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\nx = ( a b ) x = ( a b )\n\nIn situation where, the expression bounded by \u201cmrow\u201d tags contains terms of different heights, the stretchable parentheses grow to cover the highest of the terms.\n\n#### Example 8: Vertical stretching\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mi> x </m:mi>\n<m:mo> = </m:mo>\n<m:mo> ( </m:mo>\n<m:mfrac>\n<m:mi>a</m:mi>\n<m:mi>b</m:mi>\n</m:mfrac>\n<m:mo> + </m:mo>\n<m:mfrac>\n<m:mrow>\n<m:mfrac>\n<m:mi>c</m:mi>\n<m:mi>d</m:mi>\n</m:mfrac>\n</m:mrow>\n<m:mi>e</m:mi>\n</m:mfrac>\n<m:mo> ) </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display lokks like :\n\nx = ( a b + c d e ) x = ( a b + c d e )\n\nThe symmetric attribute is designed to stretch operator in both vertical and horizontal direction from the axis of the characters in equal magnitude. The symmetric attribute applies only to characters, which can stretch vertically; otherwise this attribute is ignored. Usually, this attribute is set \u201ctrue\u201d for vertically stretchable operator, but in certain cases involving matrix of unequal size, we may prefer to set it \u201cfalse\u201d as demonstrated in the example here :\n\n#### Example 9: Vertical stretching\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mi> x </m:mi>\n<m:mo> = </m:mo>\n<m:mo symmetric=\"false\"> ( </m:mo>\n<m:mtable align=\"bottom\">\n<m:mtr>\n<m:mtd>\n<m:mi>a</m:mi>\n</m:mtd>\n<m:mtd>\n<m:mi>b</m:mi>\n</m:mtd>\n</m:mtr>\n<m:mtr>\n<m:mtd>\n<m:mn>d</m:mn>\n</m:mtd>\n<m:mtd>\n<m:mn>e</m:mn>\n</m:mtd>\n</m:mtr>\n</m:mtable>\n<m:mo symmetric=\"false\"> ) </m:mo>\n<m:mo symmetric=\"false\"> ( </m:mo>\n<m:mtable align=\"bottom\">\n<m:mtr>\n<m:mtd>\n<m:mi>a</m:mi>\n</m:mtd>\n<m:mtd>\n<m:mi>b</m:mi>\n</m:mtd>\n</m:mtr>\n<m:mtr>\n<m:mtd>\n<m:mn>d</m:mn>\n</m:mtd>\n<m:mtd>\n<m:mn>e</m:mn>\n</m:mtd>\n</m:mtr>\n<m:mtr>\n<m:mtd>\n<m:mn>d</m:mn>\n</m:mtd>\n<m:mtd>\n<m:mn>e</m:mn>\n</m:mtd>\n</m:mtr>\n</m:mtable>\n<m:mo symmetric=\"false\"> ) </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display lokks like :\n\nx = ( a b d e ) ( a b d e d e ) x = ( a b d e ) ( a b d e d e )\n\nMatrix operation of unequal sizes uses notation which is aligned to base. This type of controlling stretching in specific direction is, therefore, extremely useful in such situations.\n\nIn case, the \u201cmrow\u201d domain contains non-stretchable terms of normal height and other stretchable terms, then all the terms, including the fence operator grows to the maximum normal height.\n\n#### Example 10: Vertical stretching\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mi> x </m:mi>\n<m:mo> = </m:mo>\n<m:mo> ( </m:mo>\n<m:mo> &int; </m:mo>\n<m:mi> f</m:mi>\n<m:mo> ( </m:mo>\n<m:mi> x </m:mi>\n<m:mo> ) </m:mo>\n<m:mi> &DifferentialD; </m:mi>\n<m:mi> x </m:mi>\n<m:mo> ) </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display lokks like :\n\nx = ( f ( x ) x ) x = ( f ( x ) x )\n\nAs against fence operator, the accent and horizontal arrows are stretchable in horizontal direction by default. The growth, in this case, is controlled by \u201cmunder\u201d, \u201cmover\u201d and \u201cmunderover\u201d elements, which contain the operator. The \u201cmunder\u201d, \u201cmover\u201d and \u201cmunderover\u201d elements, as the names suggest, allow drawing of an operator \"under\" or \"over\" or both about a character(s) or expresison. The scripting elements takes a base argument about which the operator is to be drawn and one (\u201cmunder\u201d and \u201cmover\u201d) operator or two (\u201cmunderover\u201d) operators for being placed about the base. We shall see that the operator grows horizontally to cover the other element.\n\n#### Example 11: Horizontal stretching\n\n\n\n<m:math display=\"block\">\n<m:mi> A </m:mi>\n<m:munderover>\n<m:mo> &RightArrow; </m:mo>\n<m:mtext> 50 degree C </m:mtext>\n<m:mtext> 200 psi </m:mtext>\n</m:munderover>\n<m:mi> B </m:mi>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\nA 50 degree C 200 psi B A 50 degree C 200 psi B\n\n### Other attributes : largeop, movablelimits, accent, separator, fence\n\nThese attributes may assume following values :\n\n#### Attribute value types\n\n\u2022 fence : true | false\n\u2022 separator : true | false\n\u2022 largeop : true | false\n\u2022 movablelimits : true | false\n\u2022 accent : true | false\n\nAll these attributes accept boolean values \u201ctrue\u201d or \u201cfalse\u201d. The attribute \u201cfence\u201d is designed for non-visual rendering like audio rendering. As such, this attribute has no impact on the visual aspect of rendering. The role of \u201cseparator\u201d attribute is also not significant and may be left to default value.\n\nThe \u201clargeop\u201d attribute determines the size of the operator. If it is true, then the operator is drawn larger than its normal size. For example, \u222b and \u220f operators are displayed as large operator as \u201clargeop\u201d attribute for these operators are set \u201ctrue\u201d by default. If we need to display the normal size, then their \"largeop\" attribute is set to \"false\".\n\n#### Example 12: Large operator\n\n\n\n<m:math display=\"block\">\n<m:mrow>\n<m:mo> &prod; </m:mo>\n<m:mo> , </m:mo>\n<m:mo> &int;  </m:mo>\n<m:mo> , </m:mo>\n<m:mo largeop=\"false\"> &prod; </m:mo>\n<m:mo> , </m:mo>\n<m:mo largeop=\"false\"> &int;  </m:mo>\n</m:mrow>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\n, , , , , ,\n\nThe movablelimits attribute allows underscripts and overscripts to be represented as subscript and superscript respectively. This rendering of under and over scripts as sub and super scripts is possible by seting this attribute to true. The implementation of this attribute by renderers is not yet consistent.\n\n#### Example 13: The \"movablelimits\" attribute\n\n\n\n<m:math display=\"block\">\n<m:munderover>\n<m:mo movablelimits='true'> &sum; </m:mo>\n<m:mi> a </m:mi>\n<m:mi> b </m:mi>\n</m:munderover>\n<m:munderover>\n<m:mo movablelimits='false'> &sum; </m:mo>\n<m:mi> a </m:mi>\n<m:mi> b </m:mi>\n</m:munderover>\n</m:math>\n\n\n\nSave the file after editing as \u201ctest.xml\u201d. The display looks like :\n\na b a b a b a b\n\nThe accent attribute determines whether an operator is treated as accent (diacritical mark) when used as an underscript or overscript.\n\n## Content actions\n\nPDF | EPUB (?)\n\n### What is an EPUB file?\n\nEPUB is an electronic book format that can be read on a variety of mobile devices.\n\nMy Favorites (?)\n\n'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.\n\n| A lens I own (?)\n\n#### Definition of a lens\n\n##### Lenses\n\nA lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.\n\n##### What is in a lens?\n\nLens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.\n\n##### Who can create a lens?\n\nAny individual member, a community, or a respected organization.\n\n##### What are tags?\n\nTags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.\n\n| External bookmarks", "date": "2013-12-05 14:53:13", "meta": {"domain": "cnx.org", "url": "http://cnx.org/content/m13548/latest/", "openwebmath_score": 0.50804603099823, "openwebmath_perplexity": 6079.0851329754305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.970687766704745, "lm_q2_score": 0.8947894675053567, "lm_q1q2_score": 0.8685611898837027}}
{"url": "https://www.lil-help.com/questions/775/mat-142-34-basics-of-probability-theory", "text": "MAT 142 3.4 Basics of Probability Theory\n\n# MAT 142 3.4 Basics of Probability Theory\n\n398.3k points\n1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3\n\ngreen candies and 4 orange candies. You select one at random. Find the probability that you\n\nselect one that is:\n\na. Brown\n\nb. Green or orange\n\nc. Not red\n\nd. Yellow or blue\n\ne. Yellow and blue\n\n1. Shoppers at a local department store were asked to complete a survey of their shopping\n\nexperience. The results are shown in the table below:\n\n** Satisfied with Service \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Not satisfied with Service** Totals\n\nMade a purchase \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 130 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 494 624\n\nDid not make a purchase \u00a0\u00a0\u00a0 715 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 183 898\n\nTotals \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 845 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 677 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1522\n\n__\n\na. What is the probability that a shopper selected at random made a purchase?\n\nb. What are the odds that a shopper selected at random was satisfied with the service?\n\nMAT 142\n\n393k points\n\n#### Oh Snap! This Answer is Locked\n\nThumbnail of first page\n\nExcerpt from file: Math Tutorial MAT 142 Problem set Unit: Probability Topic: Basics of Probability Theory Directions: Solve the following problems. Please show your work and explain your reasoning. 1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3 green candies and 4\n\nFilename: M1084.pdf\n\nFilesize: 369.5K\n\nPrint Length: 2 Pages/Slides\n\nWords: 292\n\nJ\n0 points\n\n#### Oh Snap! This Answer is Locked\n\nThumbnail of first page\n\nExcerpt from file: Week4DQ#3DiscOps&ExtraordinaryItems Whyisitimportanttoreportdiscontinuedoperationsorextraordinaryitemsseparately fromincomefromcontinuingoperations?Isthismethodofreportingallowed?What concernsdoesthistypeofreportingcreate?Doestheaverageinvestorunderstandthe\n\nFilename: ACC 280week4dqs.zip\n\nFilesize: < 2 MB\n\nPrint Length: 1 Pages/Slides\n\nWords: NA\n\nSurround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$\n\nUse LaTeX to type formulas and markdown to format text. See example.", "date": "2017-09-23 11:07:08", "meta": {"domain": "lil-help.com", "url": "https://www.lil-help.com/questions/775/mat-142-34-basics-of-probability-theory", "openwebmath_score": 0.47939369082450867, "openwebmath_perplexity": 13696.614726358517, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877700966098, "lm_q2_score": 0.8947894597898776, "lm_q1q2_score": 0.8685611854293863}}
{"url": "https://math.stackexchange.com/questions/2972505/find-the-sum-to-n-term-of-frac11-cdot2-cdot3-frac32-cdot3-cdot4/2972797", "text": "find the sum to n term of $\\frac{1}{1\\cdot2\\cdot3} + \\frac{3}{2\\cdot3\\cdot4} + \\frac{5}{3\\cdot4\\cdot5} + \\frac{7}{4\\cdot5\\cdot6 } + \u2026$\n\n$$\\frac{1}{1\\cdot2\\cdot3} + \\frac{3}{2\\cdot3\\cdot4} + \\frac{5}{3\\cdot4\\cdot5} + \\frac{7}{4\\cdot5\\cdot6 } + ...$$\n\n$$=\\sum \\limits_{k=1}^{n} \\frac{2k-1}{k\\cdot(k+1)\\cdot(k+2)}$$ $$= \\sum \\limits_{k=1}^{n} - \\frac{1}{2}\\cdot k + \\sum \\limits_{k=1}^{n} \\frac{3}{k+1} - \\sum \\limits_{k=1}^{n}\\frac{5}{2\\cdot(k+2)}$$\n\nI do not know how to get a telescoping series from here to cancel terms.\n\n\u2022 You can use $$\\frac{2k-1}{k(k+1)(k+2)}=\\frac{2k}{k(k+1)(k+2)}-\\frac1{k(k+1)(k+2)}.$$ \u2013\u00a0Sungjin Kim Oct 26 '18 at 18:11\n\u2022 You do mean an infinite sum, not a finite sum, right? \u2013\u00a0Connor Harris Oct 26 '18 at 18:15\n\u2022 @ConnorHarris I think this means writing the partial sums explicitly. \u2013\u00a0Sungjin Kim Oct 26 '18 at 18:16\n\u2022 @ConnorHarris, a sum that can be expressed in n. \u2013\u00a0LetzerWille Oct 26 '18 at 18:17\n\u2022 Note that the $\\frac{3}{k+1}$ term for some value of $k$ cancels the $\\frac{-1}{2k}$ and $\\frac{-5}{2(k+2)}$ terms for adjacent values of $k$. \u2013\u00a0Connor Harris Oct 26 '18 at 18:18\n\nHINT:\n\nNote that we have\n\n\\begin{align} \\frac{2k-1}{k(k+1)(k+2)}&=\\color{blue}{\\frac{3}{k+1}}-\\frac{5/2}{k+2}-\\frac{1/2}{k}\\\\\\\\ &=\\color{blue}{\\frac12}\\left(\\color{blue}{\\frac{1}{k+1}}-\\frac1k\\right)+\\color{blue}{\\frac52}\\left(\\color{blue}{\\frac{1}{k+1}}-\\frac{1}{k+2}\\right) \\end{align}\n\n\u2022 There's a typo after the first $=$ sign, it should be $\\frac{5/2}{k+2}$. Can't edit though, since edits must be 6 characters at minimum. \u2013\u00a0a_guest Oct 27 '18 at 12:50\n\u2022 @a_guest Thank you. I've edited accordingly. \u2013\u00a0Mark Viola Oct 27 '18 at 16:29\n\nLet the fractions be $$\\frac{a}{k}$$, $$\\frac{b}{k+1}$$, and $$\\frac{c}{k+2}$$.\n\n$$\\frac{a}{k}+\\frac{b}{k+1}+\\frac{c}{k+2}=\\frac{a(k+1)(k+2)+bk(k+2)+ck(k+1)}{k(k+1)(k+2)}=\\frac{2k-1}{k(k+1)(k+2)}$$\n\nWe want the following\n\n$$a+b+c=0$$\n\n$$3a+2b+c=2$$\n\n$$2a=-1$$\n\nSolve, $$a=-\\frac{1}{2}$$, $$b=3$$, and $$c=-\\frac{5}{2}$$.\n\nThe rest is standard.\n\nYou are almost there. You can merge the parts of the series for which the denominator is similar and you will see they cancel each other. Then you are left with the terms for which the denominator is either smaller than $$3$$ or greater than $$n$$.\n\n\\begin{aligned} & \\sum_{k=1}^n\\frac{-1}{2k} + \\sum_{k=1}^n\\frac{3}{k+1} - \\sum_{k=1}^n\\frac{5}{2}\\frac{1}{k+2} \\\\ & = \\left[-\\frac{1}{2} - \\frac{1}{4} + \\frac{1}{2}\\sum_{k=3}^n\\frac{-1}{k}\\right] + \\left[\\frac{3}{2} + \\frac{1}{2}\\sum_{k=2}^n\\frac{6}{k+1}\\right] - \\left[\\frac{1}{2}\\sum_{k=1}^n\\frac{5}{k+2}\\right] \\\\ & = \\frac{3}{4} + \\left[\\frac{1}{2}\\sum_{k=3}^n\\frac{-1}{k}\\right] + \\left[\\frac{1}{2}\\sum_{k=3}^{n+1}\\frac{6}{k}\\right] - \\left[\\frac{1}{2}\\sum_{k=3}^{n+2}\\frac{5}{k}\\right] \\\\ & = \\frac{3}{4} + \\left[\\frac{1}{2}\\sum_{k=3}^n\\frac{-1}{k}\\right] + \\left[\\frac{1}{2}\\sum_{k=3}^{n}\\frac{6}{k} + \\frac{6}{2}\\frac{1}{n+1}\\right] - \\left[\\frac{1}{2}\\sum_{k=3}^{n}\\frac{5}{k} + \\frac{5}{2}\\frac{1}{n+1} + \\frac{5}{2}\\frac{1}{n+2}\\right] \\\\ & = \\frac{3}{4} + \\frac{1}{2}\\sum_{k=3}^{n}\\left[\\frac{-1 + 6 - 5}{k}\\right] + \\frac{6}{2}\\frac{1}{n+1} - \\frac{5}{2}\\frac{1}{n+1} - \\frac{5}{2}\\frac{1}{n+2} \\\\ & = \\frac{3}{4} + \\frac{1}{2(n+1)} - \\frac{5}{2(n+2)} \\end{aligned}\n\nWhen terms are in A.P in denominator, we use difference of last and first terms in product to distribute the terms of denominator over numerator, to change into telescoping series.", "date": "2021-02-27 09:17:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2972505/find-the-sum-to-n-term-of-frac11-cdot2-cdot3-frac32-cdot3-cdot4/2972797", "openwebmath_score": 0.9975659847259521, "openwebmath_perplexity": 995.1922722721049, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877658567786, "lm_q2_score": 0.8947894555814342, "lm_q1q2_score": 0.8685611775505456}}
{"url": "https://math.stackexchange.com/questions/3053616/find-sum-n-1-infty-tan-1-frac2n2", "text": "# Find $\\sum_{n=1}^{\\infty}\\tan^{-1}\\frac{2}{n^2}$\n\nFind $$M:=\\sum_{n=1}^{\\infty}\\tan^{-1}\\frac{2}{n^2}$$\n\nThere's a solution here that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.\n\nMy proposed solution\n\n\\begin{align} &\\sum_{n=1}^{\\infty}\\tan^{-1}\\frac{2}{n^2}\\\\ =&\\sum_{n=1}^{\\infty}\\tan^{-1}\\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\\\ =&\\sum_{n=1}^{\\infty}(\\tan^{-1}(1+n)+\\tan^{-1}(1-n))\\\\ =&\\sum_{n=1}^{\\infty}(\\tan^{-1}(n+1)-\\tan^{-1}(n-1)) \\end{align}\n\nAnd this implies $$M=\\lim_{m\\to\\infty}(\\tan^{-1}(m+1)+\\tan^{-1}m-\\tan^{-1}1-\\tan^{-1}0)=\\frac{3\\pi}{4}$$\n\n\u2022 This may or may not be of help: $\\arctan(1/x) = \\pi/2 - \\arctan(x)$ \u2013\u00a0user150203 Dec 27 '18 at 5:23\n\u2022 The 3rd and 4th lines of the equation after \"My proposed solution\" are missing a summation sign. Otherwise, the solution is correct. \u2013\u00a0JimmyK4542 Dec 27 '18 at 5:35\n\u2022 @JimmyK4542 Edited, ty. \u2013\u00a0Akash Gaur Dec 27 '18 at 5:40\n\u2022 \u2013\u00a0lab bhattacharjee Dec 27 '18 at 6:34\n\n\\begin{align} &\\sum_{n=1}^{\\infty}\\tan^{-1}\\frac{2}{n^2}\\\\ =&\\sum_{n=1}^{\\infty}\\tan^{-1}\\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\\\ =\\color{red}{\\sum_{n=1}^\\infty}&\\tan^{-1}(1+n)+\\tan^{-1}(1-n)\\\\ =\\color{red}{\\sum_{n=1}^\\infty}&\\tan^{-1}(n+1)-\\tan^{-1}(n-1) \\end{align}\n\nEdit:$$\\color{red}{\\sum_{n=1}^\\infty}$$ was missing in your question before edit. I am not going to delete this.\n\nHowever your proof is now correct. $$M=\\lim_{m\\to\\infty}(\\tan^{-1}(m+1)+\\tan^{-1}m-\\tan^{-1}1-\\tan^{-1}0)=\\color{red}{\\frac\\pi2+\\frac\\pi2-\\frac\\pi4-0=\\pi-\\frac\\pi4}=\\frac{3\\pi}{4}$$\n\nLooks good to me. If I was going to offer a critique I would just say: when writing an argument it's always better to over communicate rather than under communicate.\n\nThe first equality is just algebra.\n\nYour second equality requires a little bit to see clearly but it's true. Most will recall:\n\n$$\\tan(A+B)=\\frac{\\tan(A)+\\tan(B)}{1-\\tan(A)\\tan(B)}$$ Or if you'd like: $$A+B= \\tan^{-1} \\bigg(\\frac{\\tan(A)+\\tan(B)}{1-\\tan(A)\\tan(B)} \\bigg)$$ Taking $$A=\\tan^{-1}(1+n)$$ and $$B=\\tan^{-1}(1-n)$$\n\nHonestly adding this much explanation seems like almost overkill.\n\nThe 4th equality follows as result of $$\\tan^{-1}$$ being an odd function.\n\nNow the last part you are using a telescoping series technique so that you may ignore all the middle terms. That is,\n\n\\begin{align} &\\sum_{n=1}^\\infty\\tan^{-1}(n+1)-\\tan^{-1}(n-1) \\\\ &= \\lim_{m\\to \\infty} \\tan^{-1}(m+1)-tan^{-1}(m-1)+\\dots +\\tan^{-1}(4)-\\tan^{-1}(2)+\\tan^{-1}(3)-\\tan^{-1}(1)+tan^{-1}(2)-\\tan^{-1}(0) \\end{align}\n\nSo after we consider what cancels and what doesn't we find that we only need to concern ourselves with $$\\lim_{m\\to \\infty}\\tan^{-1}(m+1)+\\tan^{-1}(m-1)-\\tan^{-1}(1)$$\n\nSo while that is true: I think it might merit a sentence or two just to make sure the audience is following.\n\n\u2022 Isn't $M=\\lim_{m\\to \\infty} \\tan^{-1}(m)-\\tan^{-1}(0)=\\frac\\pi2-0?$ \u2013\u00a0tatan Dec 27 '18 at 5:50\n\u2022 @tatan. Yes. I am missing some terms. Let me edit. \u2013\u00a0Mason Dec 27 '18 at 5:51\n\u2022 Thanks. I'm really slow at writing in latex so I skip steps. But I will add more explanation in English in the future. \u2013\u00a0Akash Gaur Dec 27 '18 at 5:57\n\u2022 Yep. Can I ask you one thing? Are you from Maryland?(I saw from your profile) \u2013\u00a0tatan Dec 27 '18 at 19:57\n\u2022 Yes. Uh oh. Stranger danger. :). No. I am not worried. I grew up in Ohio and move to the DC area post college. I went to Ohio University not Univ of Maryland. Though I did cite Dr. Yorke in my undergrad thesis and then I met him wandering once while wandering Univ of Maryland's halls... \u2013\u00a0Mason Dec 27 '18 at 19:57", "date": "2020-07-06 09:55:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3053616/find-sum-n-1-infty-tan-1-frac2n2", "openwebmath_score": 0.8877766132354736, "openwebmath_perplexity": 1363.6358109341672, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303410461384, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8685531478132203}}
{"url": "https://mathhelpboards.com/threads/when-0-123-495051-0-515049-321.6136/", "text": "# When 0.123...495051/0.515049...321\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHello MHB,\n\nRecently I've come across a problem that jumped off the page at me (Find the first 3 figures after the decimal point in the decimal expression of the number $$\\displaystyle \\frac{0.12345678910\\cdots495051}{0.515049\\cdots987654321}$$).\n\nI tried to approach it by making a table where I started to divide some smaller numbers but stick to the same pattern that is required by the aforementioned problem, i.e.\n\n$$\\displaystyle \\frac{0.12}{0.21}=0.571...$$\n\n$$\\displaystyle \\frac{0.123}{0.321}=0.383...$$\n\n$$\\displaystyle \\frac{0.1234}{0.4321}=0.285...$$\n\n$$\\displaystyle \\frac{0.12345}{0.54321}=0.227...$$\n\n$$\\displaystyle \\frac{0.123456}{0.654321}=0.188...$$\n\n$$\\displaystyle \\frac{0.1234567}{0.7654321}=0.161...$$\n\n$$\\displaystyle \\frac{0.12345678}{0.87654321}=0.140...$$\n\n$$\\displaystyle \\frac{0.123456789}{0.987654321}=0.124...$$\n\n$$\\displaystyle \\frac{0.12345678910}{0.10987654321}=0.123...$$\n\n$$\\displaystyle \\frac{0.1234567891011}{0.1110987654321}=0.111...$$\n\n$$\\displaystyle \\frac{0.123456789101112}{0.121110987654321}=0.019...$$\n\n$$\\displaystyle \\frac{0.12345678910111213}{0.13121110987654321}=0.940...$$\n\n$$\\displaystyle \\frac{0.1234567891011121314}{0.1413121110987654321}=0.873...$$\n\n$$\\displaystyle \\frac{0.123456789101112131415}{0.151413121110987654321}=0.815...$$\n\n$$\\displaystyle \\frac{0.12345678910111213141516}{0.16151413121110987654321}=0.764...$$\n\n$$\\displaystyle \\frac{0.1234567891011121314151617}{0.1716151413121110987654321}=0.719...$$\n\n$$\\displaystyle \\frac{0.123456789101112131415161718}{0.181716151413121110987654321}=0.679...$$\n\n$$\\displaystyle \\frac{0.12345678910111213141516171819}{0.19181716151413121110987654321}=0.643...$$\n\n$$\\displaystyle \\frac{0.1234567891011121314151617181920}{0.2019181716151413121110987654321}=0.611...$$\n\nand so on and so forth\n\nbut I failed to observe any pattern that's worth mentioning to help me to crack the problem.\n\nCould anyone help me with this particular problem? Thanks in advance.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nI also approached it by making a table, but I started by using the most significant figures in the given denominator:\n\n$$\\displaystyle \\frac{0.1}{0.5} = 0.2$$,\n\n$$\\displaystyle \\frac{0.12}{0.51} = 0.2352\\ldots$$,\n\n$$\\displaystyle \\frac{0.123}{0.515} = 0.2388\\ldots$$,\n\n$$\\displaystyle \\frac{0.1234}{0.5150} = 0.2396\\ldots$$,\n\n$$\\displaystyle \\frac{0.12345}{0.51504} = 0.2396\\ldots$$,\n\n$$\\displaystyle \\frac{0.123456}{0.515049} = 0.2396\\ldots$$,\n\n$$\\displaystyle \\frac{0.1234567}{0.5150494} = 0.2396\\ldots$$,\n\n$$\\displaystyle \\frac{0.12345678}{0.51504948} = 0.2396\\ldots$$.\n\nBy this time, the fraction had stabilised (within the limits of my 8-digit calculator) to 0.2396987. I doubt whether further approximations would affect the first three digits after the decimal point.\n\nEdit. You can confirm that by using a bit of calculus. If $f(x,y) = x/y$ then $\\delta f \\approx (1/y)\\delta x - (x/y^2)\\delta y$. With $x\\approx 0.123$, $y\\approx 0.515$ and $\\delta x = \\delta y = 10^{-n}$, you find that $\\delta f < 2*10^{-n}$. So the change in the fraction after the eighth decimal places in numerator and denominator is never going to be sufficient to affect the first three decimal places in the quotient.\n\nLast edited:\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHi Opalg,\n\nI'm very thankful to you for showing me something that I had never thought about before and your solution and the proof work so beautifully...\n\nThank you so much!", "date": "2021-09-26 00:13:29", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/when-0-123-495051-0-515049-321.6136/", "openwebmath_score": 0.7646966576576233, "openwebmath_perplexity": 515.8382280747652, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9572778036723354, "lm_q2_score": 0.9073122269997507, "lm_q1q2_score": 0.8685498559073768}}
{"url": "https://math.stackexchange.com/questions/186972/how-can-lu-factorization-be-used-in-non-square-matrix/186997", "text": "# How can LU factorization be used in non-square matrix?\n\nIn my textbook, there is some information about LU factorization of square matrix $A$, but not about non-square matrix.\n\nHow can LU factorization be used to factorize non-square matrix?\n\n\u2022 Yes. Let $A$ be $m \\times n$ matrix, then $L$ is $m \\times m$ and $U$ is $m \\times n$.\n\u2013\u00a0user2468\nAug 26, 2012 at 4:17\n\nI'll illustrate how to understand the LU-decomposition of a particular $3 \\times 4$ matrix below. The method works just as well for other sizes since the LU-decomposition arises naturally from the study of Gaussian elimination via multiplication by elementary matrices.\n\n$$\\begin{array}{ll} A \\ = &\\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 2 & 4 & 0 & 7 \\\\ -1 & 3 & 2 & 0 \\end{array} \\right] \\ \\underrightarrow{r_2-2r_1 \\rightarrow r_2} \\ \\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 0 & 0 & 6 & 5 \\\\ -1 & 3 & 2 & 0 \\end{array} \\right] \\ \\underrightarrow{r_3+r_1 \\rightarrow r_3} \\\\ & \\\\ &\\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 0 & 0 & 6 & 5 \\\\ 0 & 5 & -1 & 1 \\end{array} \\right] \\ \\underrightarrow{r_2 \\leftrightarrow r_3} \\ \\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 0 & 5 & -1 & 1 \\\\ 0 & 0 & 6 & 5 \\end{array} \\right] = \\ U \\end{array}$$ We have $U = E_3E_2E_1A$ hence $A = E_1^{-1}E_2^{-1}E_3^{-1}U$ and we can calculate the product $E_1^{-1}E_2^{-1}E_3^{-1}$ as follows: $$\\begin{array}{ll} I \\ = &\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{array} \\right] \\ \\underrightarrow{r_2 \\leftrightarrow r_3} \\ \\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 1 & 0 \\end{array} \\right] \\ \\underrightarrow{r_3-r_1 \\rightarrow r_3} \\\\ & \\\\ &\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ -1 & 1 & 0 \\end{array} \\right] \\ \\underrightarrow{r_2+2r_1 \\rightarrow r_2} \\ \\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 2 & 0 & 1 \\\\ -1 & 1 & 0 \\end{array} \\right] = PL \\end{array}$$ I have inserted a \"$P$\" in front of the $L$ since the matrix above is not lower triangular. However, if we go one step further and let $r_2 \\leftrightarrow r_3$ then we will obtain a lower triangular matrix: $$\\begin{array}{ll} PL \\ = &\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 2 & 0 & 1 \\\\ -1 & 1 & 0 \\end{array} \\right] \\ \\underrightarrow{r_2 \\leftrightarrow r_3} \\ \\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ -1 & 1 & 0 \\\\ 2 & 0 & 1 \\\\ \\end{array} \\right] =L \\end{array}$$ Therefore, we find that $E_1^{-1}E_2^{-1}E_3^{-1}=PL$ where $L$ is as above and $P = E_{2 \\leftrightarrow 3}$. This means that $A$ has a modified $LU$-decomposition. Some mathemticians call it a $PLU$-decomposition, $$A = \\underbrace{\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 1 & 0 \\end{array} \\right]}_{P} \\underbrace{\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ -1 & 1 & 0 \\\\ 2 & 0 & 1 \\\\ \\end{array} \\right]}_{L}\\underbrace{\\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 0 & 5 & -1 & 1 \\\\ 0 & 0 & 6 & 5 \\end{array} \\right]}_{U} = \\underbrace{\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 2 & 0 & 1 \\\\ -1 & 1 & 0 \\end{array} \\right]}_{PL}\\underbrace{\\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 0 & 5 & -1 & 1 \\\\ 0 & 0 & 6 & 5 \\end{array} \\right]}_{U}.$$ Since permutation matrices all satisfy the condition $P^k=I$ (for some $k$) the existence of a $PLU$-decomposition for $A$ naturally suggests that $P^{k-1}A = LU$. Therefore, even when a $LU$ decomposition is not available we can just flip a few rows to find a $LU$-decomposable matrix. This is a useful observation because it means that the slick algorithms developed for $LU$-decompositions apply to all matrices with just a little extra fine print.\n\nMuch of the writing above can be spared if we adopt the notational scheme illustrated below. $$\\begin{array}{ll} A \\ = &\\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 2 & 4 & 0 & 7 \\\\ -1 & 3 & 2 & 0 \\end{array} \\right] \\ \\underrightarrow{r_2-2r_1 \\rightarrow r_2} \\ \\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ (2) & 0 & 6 & 5 \\\\ -1 & 3 & 2 & 0 \\end{array} \\right] \\ \\underrightarrow{r_3+r_1 \\rightarrow r_3} \\\\ & \\\\ &\\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ (2) & 0 & 6 & 5 \\\\ (-1) & 5 & -1 & 1 \\end{array} \\right] \\ \\underrightarrow{r_2 \\leftrightarrow r_3} \\ \\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ (-1) & 5 & -1 & 1 \\\\ (2) & 0 & 6 & 5 \\end{array} \\right] = \\ U \\end{array}$$ We find if we remove the parenthetical entries from $U$ and ajoing them to $I$ then it gives back the matrix $L$ we found previously: $$U = \\left[ \\begin{array}{cccc} 1 & 2 & -3 & 1\\\\ 0 & 5 & -1 & 1 \\\\ 0 & 0 & 6 & 5 \\end{array} \\right] \\qquad L=\\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ -1 & 1 & 0 \\\\ 2 & 0 & 1 \\end{array} \\right].$$\n\nHope this helps.\n\n\u2022 \"permutation matrices all satisfy the condition $P^2=I$\" is incorrect.\n\u2013\u00a0user147263\nJan 16, 2016 at 22:09\n\u2022 @Normal for example ? Sorry, I can't see past swap then reverse swap does nothing... Jan 16, 2016 at 22:28\n\u2022 A permutation can involve a $3$-cycle like $1\\to 2\\to 3$.\n\u2013\u00a0user147263\nJan 16, 2016 at 22:45\n\u2022 @Normal right, that makes sense... but, there still must be some power for which $P^k=I$. So, something similar to what I have currently written is possible. I wonder, is this possible for $3 \\times 3$ matrix examples. Great comment, I wish I could upvote comments ( for internet pts naturally) Jan 17, 2016 at 1:39\n\u2022 For all permutation matrices: $P \\cdot P^{T} = I$ Feb 5, 2019 at 8:03", "date": "2022-05-19 07:56:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/186972/how-can-lu-factorization-be-used-in-non-square-matrix/186997", "openwebmath_score": 0.8212655782699585, "openwebmath_perplexity": 269.80935597029196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9879462226131666, "lm_q2_score": 0.8791467738423874, "lm_q1q2_score": 0.8685497343401384}}
{"url": "http://mathhelpforum.com/discrete-math/11509-two-discrete-problems.html", "text": "# Math Help - Two Discrete Problems\n\n1. ## Two Discrete Problems\n\nHello,\n\nI am needing some help on the two problems below. Thank you in advance.\n\n1) Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 101.\n\nand\n\n2) Using the fact that an integer n, either n >= 14 or n <= 15, or otherwise prove 7 does not divide 100.\n\n2. Originally Posted by MathStudent1\nHello,\n\nI am needing some help on the two problems below. Thank you in advance.\n\n1) Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 101.\n.\ngcd(14,21)=7\nThus,\n7 divides 101 which is false.\n\n3. Thank you ThePerfectHacker. I made a typo when I first made my post. Sorry about that. So, it should be 14m + 21n = 100 not 101.\n\nI am beening taught to turn this into a if/then statement and go from there.\n\nIf m and n are integers, then 14m + 21n = 100\nFactor out a 7 givng us 7(2n +3m) = 100\n\nFrom here I have trouble. Also, should the if/then statement be not equal to 100 instead of equal to? I haven't had that much experience with proof by contradiction. Thank You.\n\n4. Originally Posted by MathStudent1\nThank you ThePerfectHacker. I made a typo when I first made my post. Sorry about that. So, it should be 14m + 21n = 100 not 101.\n\nI am beening taught to turn this into a if/then statement and go from there.\n\nIf m and n are integers, then 14m + 21n = 100\nFactor out a 7 givng us 7(2n +3m) = 100\n\nFrom here I have trouble. Also, should the if/then statement be not equal to 100 instead of equal to? I haven't had that much experience with proof by contradiction. Thank You.\nLet,\np= (there exists n,m such that 14m+21n=100).\n\nNow if \"p\" is true we shown that 7 divides 100.\n\nThus,\nq=(7 divides 100).\nBut this is false.\n\nMeaning, (modens ponens)\n[(p--> q) and (~q)]--> ~p\nThus,\n~p=(there do not exists n,m such that 14m+21n=100)\n\nThis is mine 46th Post!!!\n\n5. Hello, MathStudent1!\n\nHere;s the first one . . .\n\n1) Prove by contradiction that there do not exist integers m and n\n. . .such that: .14m + 21n .= .101\n\nAssume that there are integers m and n such that: .14m + 21n .= .101\n\n. . . . . . . . . . . . . . . . . . .100\nDivide by 7: . 2m + 3n .= .----\n. . . . . . . . . . . . . . . . . . . .7\n\nIn the left side: m is an integer, so 2m is an integer. *\n. . . . . . . . . . . . n is an integer, so 3n is an integer. *\nHence, 2m + 3n (the sum of two integers) is an integer. *\n\nBut the right side is an irreducible fraction (or decimal 14.2857...)\n\nWe have reached a contradicition, hence our assumption was incorrect.\n\n. . There are no integers m and n that satisfy 14m + 21n .= .101\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\n*\nReasons: The set of integers is closed under multiplication and addition.\n\n6. Thank you Soroban. One question for you. When orginally I posted my thread I made a typo in that 14m + 21n = 100 NOT 101. Does that change the proof at all or does the logic remain the same. Following your proof, am I correct by stating this below:\n\nProve that there do not exist integers m and n such that: 14m + 21n = 100\n\nRestating the problem - If m and n are integers, then 14m + 21n = 100\n\nFactor out a 7 and we get 7(2m + 3n) = 100\nDivide both sides by 7 and we get 2m + 3n = (100/7)\nSince m is an integer then 2*m is an integer and since n is an integer then 3*n is and integer.\nHowever, the RHS (100/7) is not an integer and therefore,\n2m + 3n (IS NOT EQUAL TO) (100/7)\n\nHence, we have reached a contradiction of our assumption that m and n are integers.\n\nTherefore, there do not exist integers m and n such that 14m + 21n = 100.\n\n*****Thank you!*****\n\n7. Hello again, MathStudent1!\n\nYes, the modified proof is still valid . . .\n. . and you did an excellent job!", "date": "2015-03-28 18:39:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/11509-two-discrete-problems.html", "openwebmath_score": 0.836551308631897, "openwebmath_perplexity": 528.0010152030184, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9879462179994596, "lm_q2_score": 0.8791467770088162, "lm_q1q2_score": 0.8685497334122741}}
{"url": "http://inside-science.com/r362cn/45e997-binomial-coefficient-latex", "text": "The combination (n r) (n r) is called a binomial coefficient. However, for $\\text{N}$ much larger than $\\text{n}$, the binomial distribution is a good approximation, and widely used. \\vec,\\overrightarrow, Latex how to insert a blank or empty page with or without numbering \\thispagestyle,\\newpage,\\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\\usepackage{algorithm},\\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. therefore gives the number of k -subsets possible out of a set of distinct items. Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. Click on one of the binomial coefficient designs, which look like the letters \"n\" over \"k\" inside either a round or angled bracket. are the different ordered arrangements of a k-element subset of an n-set, $$\\binom{n}{k} = \\binom{n-1}{k-1} +\\binom{n-1}{k}$$. The binomial coefficient is defined by the next expression: $\\binom {n}{k} = \\frac {n ! k-combinations of n-element set. Binomial coefficient, returned as a nonnegative scalar value. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more \u2026 Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. It is especially useful for reasoning about recursive methods in programming. (n - k)!} On the other side, \\textstyle will change the style of the fraction as if it were part of the text. In latex mode we must use \\binom fonction as follows: \\frac{n!}{k! This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. Below is a construction of the first 11 rows of Pascal's triangle. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 where A is the permutation, A_n^k = \\frac{n!}{(n-k)! See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, \\binom{n}{k} is \u2026 }}{{k!\\left( {n - k} \\right)!}} In Counting Principles, we studied combinations.In the shortcut to finding ${\\left(x+y\\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. \\boxed, How to write table in Latex ? How to write number sets N Z D Q R C with Latex: \\mathbb, amsfonts and \\mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? . The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. ... Pascal\u2019s triangle. Specially useful for continued fractions. Don't forget to LIKE, COMMENT, SHARE & SUBSCRIBE to my channel. } Also, the text size of the fraction changes according to the text around it. So The combination (nr)\\displaystyle \\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)(\u200bn\u200br\u200b\u200b) is calle\u2026 For example, \u2026 If your equation requires specific numbers in place of the \"n\" or \"k,\" click on a letter to select it, press \"Delete\" and enter a number in its place. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n \u2265 k \u2265 0 and is written (). It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. LaTeX forum \u21d2 Math & Science \u21d2 Expression like binomial Coefficient with Angle Delimiters Topic is solved Information and discussion about LaTeX's math and science related features (e.g. therefore gives the number of k-subsets possible out of a set of distinct items. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an n-element set. C \u2014 All combinations of v matrix. Identifying Binomial Coefficients. This same array could be expressed using the factorial symbol, as shown in the following. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. The order of selection of items not considered. The binomial coefficient is defined by the next expression: \\ [ \\binom{n} {k} = \\frac{n!} In UnicodeMath Version 3, this uses the \\choose operator \u249e instead of the \\atop operator \u00a6. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \\left\\{,\\right\\},\\underbrace{} and \\overbrace{}, How to get dots in Latex \\ldots,\\cdots,\\vdots and \\ddots, Latex symbol if and only if / equivalence. The binomial coefficient \\binom{n}{k} can be interpreted as the number of ways to choose k elements from an n-element set. ( n - k )! Ak n = n! The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. Then it's a good reason to buy me a coffee. (n\u2212k)! Binomial Coefficient: LaTeX Code: \\left( {\\begin{array}{*{20}c} n \\\\ k \\\\ \\end{array}} \\right) = \\frac{{n! Thank you ! (n - k)!} Latex binomial coefficient Definition. One can drop one of the numbers in the bottom list and infer it from the fact that sum \u2026 The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. This website was useful to you? Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. Latex The binomial coefficient is defined by the next expression: \\[\\binom {n}{k} = \\ frac {n!}{k!(n-k)! {k! As you may have guessed, the command \\frac{1}{2} is the one that displays the fraction. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n \u2265r n \u2265 r, then the binomial coefficient is Binomial coefficient, returned as a nonnegative scalar value. In this case, we use the notation (nr)\\displaystyle \\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)(\u200bn\u200br\u200b\u200b) instead of C(n,r)\\displaystyle C\\left(n,r\\right)C(n,r), but it can be calculated in the same way.$ And of course this command can be included in the normal text flow \\ (\\binom{n} {k}\\). Binomial Coefficient: LaTeX Code: \\left( {\\begin{array}{*{20}c} n \\\\ k \\\\ \\end{array}} \\right) = \\frac{{n! Using fractions and binomial coefficients in an expression is straightforward. The text inside the first pair of braces is the numerator and the text inside the second pair is the denominator. In Counting Principles, we studied combinations. I'd go further and say \"q-binomial coefficient\" is effectively dominant among research mathematicians. In the shortcut to finding (x+y)n\\displaystyle {\\left(x+y\\right)}^{n}(x+y)\u200bn\u200b\u200b, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. The command \\displaystyle will format the fraction as if it were in mathematical display mode. This will give more accuracy at the cost of computing small sums of binomial coefficients. Mathematical Equations in LaTeX. n! begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? This article explains how to typeset them in LaTeX. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. I agree. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. \\vec,\\overrightarrow; Latex how to insert a blank or empty page with or without numbering \\thispagestyle,\\newpage,\\usepackage{afterpage} Latex natural numbers; Latex real numbers; Latex binomial coefficient; Latex overset and underset ; Latex absolute value The possibility to insert operators and functions as you know them from mathematics is not possible for all things. In Counting Principles, we studied combinations.In the shortcut to finding$\\,{\\left(x+y\\right)}^{n},\\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. In this video, you will learn how to write binomial coefficients in a LaTeX document. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. = \\binom{n}{k}$$Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. }{k ! k-combinations of n-element set. Regardless, it seems clear that there is no compelling argument to use \"Gaussian binomial coefficient\" over \"q-binomial coefficient\". Gerhard \"Ask Me About System Design\" Paseman, 2010.03.27 \\endgroup \u2013 Gerhard Paseman Mar 27 '10 at 17:00 (adsbygoogle = window.adsbygoogle || []).push({}); Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. Latex k parmi n - coefficient binomial. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a \u2026 All combinations of v, returned as a matrix of the same type as v. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. You can set this manually if you want. The symbols and are used to denote a binomial coefficient, and are sometimes read as \" choose.\" The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. binomial Stanley's EC1 also uses it as the primary name, which counts for a lot in my book. An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. {k! How to write it in Latex ? {k! Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Knowledge base dedicated to Linux and applied mathematics. It will give me the energy and motivation to continue this development. (\u2212)!. The second fraction displayed in the previous example uses the command \\cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Since binomial coefficients are quite common, TeX has the \\choose control word for them. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilit\u00e9s de choisir k \u00e9l\u00e9ment dans un ensemble de n \u00e9l\u00e9ments. The Texworks shows \u2026 }}{{k!\\left( {n - k} \\right)!}} In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector \u2026 Accordingly the binomial coefficient in the binomial theorem above can be written as \u201cn\\choose k\u201d, assuming that you type a space after the k. This formulas, graphs). Asking for help, clarification, or responding to other answers. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. All combinations of v, returned as a matrix of the same type as v. As you see, the command \\binom{}{}will print the binomial coefficient using the parameters passed inside the braces. In latex mode we must use \\binom fonction as follows: \\frac {n!} C \u2014 All combinations of v matrix. The symbols and are used to denote a binomial coefficient, and are sometimes read as \"choose.\". Here's an equation: math \\frac {n!} For these commands to work you must import the package amsmath by adding the next line to the preamble of your file }}{{k!\\left( {n - k} \\right)!}}. This method of constructing mathematical proofs is called mathematical induction. Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + \u22ef + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \\cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + \u22ef + n c n x n , As you see, the command \\binom{}{} will print the binomial coefficient using the parameters passed inside the braces. The usual binomial coefficient can be written as \\left({n \\atop {k, {n-k}}}\\right). = \\binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$, \\frac{n!}{k! \\\\binom{N} {k} What differs between \\\\dots and \\\\dotsc, with overleaf.com, the outputs are identical. Usually, you find the special input possibilities on the reference page of the function in the Details section. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. coefficient The usage of fractions is quite flexible, they can be nested to obtain more complex expressions. (n-k)!} Using fractions and binomial coefficients in an expression is straightforward. Then it's a good reason to buy me a coffee. = \\binom {n} {k}  This is the binomial coefficient. binomial coefficient Latex. Binomial Coefficient. samedi 11 juillet 2020, par Nadir Soualem. 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{"url": "https://www.physicsforums.com/threads/compound-interest-systems-of-equations-problem.808845/", "text": "# Compound Interest Systems of Equations Problem\n\n1. Apr 16, 2015\n\n### ConstantineO\n\n1. The problem statement, all variables and given/known data\nRomeo was given a gift of $10,000 when he turned 16. He invested it at 3% per annum. Three years later, Juliet was given$10,000, which she invested at 5% per annum. When will the two amounts be equal in value?\n\n2. Relevant equations\nCompound Interest Formula\nTotal = Capital(1+interest)^Years\nt = c(1+i)^n\n3. The attempt at a solution\nDeciding how to create my formula is where things get fuzzy. The way I initially attempted this was to add a 3 year head start onto Romeo's interest formula. This is what results (Used wolfram to save time rewriting):\n\nThis eventually results in \"4.61103.\" However, this is not the answer at the back of my textbook. They decided to opt for a different formula where the time is subtracted from Juliet's interest function. This is shown here:\n\nThis results in the correct answer of 7.61103. What I'm wondering, why do I have to subtract 3 years from Juliet's compound interest formula, and why does adding 3 years to Romeo's compound interest formula produce the wrong answer?\n\n2. Apr 16, 2015\n\n### BvU\n\nSo clearly the book starts counting at the beginning of the story: when Romeo brings his money to the bank.\nThat means Julia's interest clock is at x-3 when Romeo's is at x.\n\n3. Apr 16, 2015\n\n### SammyS\n\nStaff Emeritus\nWhat is Romeo's age when then two accounts have equal value?\n\n4. Apr 16, 2015\n\n### ConstantineO\n\nIs this always the convention when dealing with problems like this? From my experience with physics courses, I assumed that time shares a relation with the total amount of money generated by compound interest. The interest and initial capital is thus fixed, and you are only left with a relation between time and the total sum of money. If you were to treat time as a vector which you could progress backwards and forwards on, each snapshot through time depending on which way you were progressing would net either a larger or smaller amount of the total. I don't understand why the book is forcing me to start at the beginning of the story. The story is like pages of a book which can be turned towards or away from the end of the novel. The content of the novel does not change and its contents are already written.\n\nI don't understand what Romeo's age has to do with this? Is this some kind of statement that alludes to time only progressing forward or something?\n\n5. Apr 16, 2015\n\n### BvU\n\nNothing so fancy. The book is sloppy by not stating when it wants the clock (calendar in this case) to show t = 0. From the book answer it appears that it assumes t = 0 when Romeo turns 16. The exercise could equally well have assumed t = 0 when Julia receives the ten grand. If I were grading this, I'd have to allow both answers, certainly when the calculation steps are shown.\n\n6. Apr 16, 2015\n\n### ConstantineO\n\nI actually just figured it out while discussing this question with a friend just a few moments before you posted. I took the delta of the two times from both of the ways of solving this question and realized that this is nothing more than a semantics issue of where you measure time from. This question is poorly setup, and I am going to voice my irritation to the person marking.\n\n7. Apr 16, 2015\n\n### Ray Vickson\n\nA more serious problem is that when compounding annually, the answer is \"never\". At Romeo's 23rd birthday, Romeo's account contains more money than Juliet's, while at his 24th birthday, Juliet's account contains more than Romeo's. Unless there is something like continuous compounding, the two amounts never match exactly; that is, unless you can withdraw your money part-way through a year and earn a part-year's interest, you could never find a time where the two withdrawals are equal. (However, I would not raise this issue with your teacher if I were you; just \"go with the flow\".)\n\nAlso, instead of expressing irritation, it would be wiser for you to say that you will give two answers (depending on where to start measuring time) and point out (nicely) that the question is a bit ambiguous.\n\n8. Apr 16, 2015\n\n### SammyS\n\nStaff Emeritus\nRGV makes an excellent point. This is a short-coming of many problems of this type.\n\nIt's often the case that for any fraction of a single compounding period that's \"left over\", simple interest is computed for that extra amount of time. With this method, find the value of the account at x = 4, then see if the values can be equal at any time in the 4th year using simple interest.\n\n(Edited slightly.)\n\nLast edited: Apr 16, 2015\n9. Apr 16, 2015\n\n### ConstantineO\n\nI don't wish to sound like an imbecile, but I have no clue what that means. How would I set up a systems of equation if I subbed in 4 for both x's? I believe I would be left with only one unknown for either equation, or am I misunderstanding you? Please clarify what you mean by using x=4.\n\n10. Apr 16, 2015\n\n### SammyS\n\nStaff Emeritus\nWhen x = 4, (7years for Romeo, 4 years for Juliet), what is the value in each in each account?\n\nStart from that point for each & using simple interest, and see how long it takes for the values to be equal.\n\n11. Apr 16, 2015\n\n### ConstantineO\n\nI am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.\n\nI am to assume this is correct?\nSimple interest Formula\nI = \"the grown money\"\np = the initial capital\nr = interest rate per year\nt = t\n(I) = (p)(r)(t)\n\nI think I know what you're getting at, so let me make an attempt.\n\nRomeo Compound Interest Calculation Total\nAr = 10,000(1.03)^4+3\nAr = 10,000(1.03)^7\nAr = 12,298.73865\n\nJulie Compound Interest Calculation Total\nAj = 10,000(1.05)^4\nAj = 12,155.0625\n\nRomeo Simple Interest Calculation\n(Ir) = (12,298.73865)(0.03)(t)\n(Ir) = (368.94)(t)\n\nJuliet Simple Interest Calculation\n(Ir) = (12,155.0625)(0.05)(t)\n(Ir) = 607.753125(t)\n\nEquate the two to equal each other.\n\n(368.94)(t) = 607.753125(t)\n(368.94)(t) -607.753125(t) = 0\nt(368.94-607.753125)=0\n\nThis doesnt work though...\n\n12. Apr 16, 2015\n\n### Ray Vickson\n\nIt's not that complicated. Let $R_t, J_t$ be the amounts in Romeo's and Juliet's accounts at time $t$ (in years, measured from Romeo's 16th birthday). We have:\n$$\\begin{array}{ccc} & t=7 & t=8 \\\\ R_t & 12298.74 & 12667.70 \\\\ J_t & 12155.06 & 12762.82 \\end{array}$$\nYou can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)\n\n13. Apr 16, 2015\n\n### ConstantineO\n\nCould you show me an example of this because I am simply not understanding what you're saying, or can you show me how to equate the two together using algebra?\n\n14. Apr 17, 2015\n\n### ConstantineO\n\nThis is getting out of hand, so I am going to try to be as clear and as concise as possible. I know how to calculate the correct answer for this question, and I realize where I first made my mistake. I am now wondering what everyone is talking about when they mention using simple interest. I have literally 0 experience doing any simple interest calculations, and I have no idea what sammyS is talking about here.\n\nIf anyone is holding information back because they believe an attempt has not been made and it would violate forum rules, please stop. I understand how to do the book's question using the books method, and I know and where the discrepancy between the 4.61103 and 7.61103 came from. I have made a wholehearted attempt so please for sake of my sanity just give me an example or step by step breakdown. I learn through examples, I look at the process involved, and I create a model in my head how everything interacts and relates with each other.\n\nWhat I am seeking is an explanation and a simple example of what this simple interest method is for calculating the 0.61103 year after the exact 7 years has passed.\n\nWhat I am understanding so far.\n- First use the compound interest formula to create outputs for both Romeo's and Juliet's accounts after being influenced by exactly 7 years of interest.\n- This is done by assigning \"4\" to the variable x\n- Get said products, in this case being:\nRomeo's Total After Exactly 7 Years = 12,298.73865\nJuliet's Total After Exactly 7 Years = 12,155.0625\n- Since you know that the two amounts are still not equal, you must calculate the remaining time between the two using the Simple interest formula\n- This is what I am having problems with and what I would appreciate having an example for. Preferably shown algebraically, so I am not plotting nearly straight lines in Desmos.\n\nNot to sound contentious, but when you try to find the point of intersect of two nearly straight lines with a slope that is in hundredths of a unit, it is just plainly byzantine. I am probably doing this wrong, but graphically representing each simple interest formula and trying to find where they meet is insane.\n\nLast edited: Apr 17, 2015\n15. Apr 17, 2015\n\n### Ray Vickson\n\nHow could I have made my explanation any simpler? There was no hidden information or holding back of anything. I said: just draw two straight-line graphs for Romeo and Juliet, where the lines connect their (time,money) points at the start and end of the final year.\n\nThe true graphs of time vs. money will be curved, because of compound interest, but over a short period such as one year the \"curvature\" will be small, and the graph will look almost like a straight line; replacing the curve by a straight line ---only for that single year---would give you the \"simple\" interest schedule for that year. It will be almost the same as the compound interest schedule, because the curvature is small over short times such as a year.\n\nIf you don't believe me, just draw the graphs of money vs. time for Romeo and for Juliet. You will see that they curve up, but are almost straight over short periods. The only way to really understand it is to sit down and do it.\n\n16. Apr 17, 2015\n\n### ConstantineO\n\nI am going to assume the two lines would be plotted as\ny=(12155.0625)(0.05)(x)\nand\ny=(12298.73865)(0.03)(x)\n\nIf that's the case, I don't understand how any human being is supposed to find the point of intersect graphically like this.\n\nI did sit down and do it. I have been sitting down and ripping my hair out doing this for sometime.\n\nEdit - In the event that anyone else would like to question how committed I am to \"sitting down and doing it.\" I've had enough of that nonsense from teachers with superiority issues and over bloated opinions that like to question how hard I am making an attempt.\n\nLast edited: Apr 17, 2015\n17. Apr 17, 2015\n\n### Ray Vickson\n\nAs long as you persist in trying to plot the wrong thing you will need to keep pulling out your hair. Go back and read what I wrote in #12. Do the plot exactly as I indicated there.\n\nBetter still plot the two curves y = 10000 * (1.03)^x and y = 10000 * (1.05)^(x-3) for x = from 3 to 8. They are curved, aren't they? Don't they look almost straight over the shorter interval 7 \\leq x \\leq 8? in each case if you were to plot a straight line from (7,y(7)) to (8,y(8)) it would look very close to the actual curve (x,y(x)) for x between 7 and 8. The \"simple interest\" effect between 7 and 8 would be the straight line, while the \"compound interest\" effect would be the curve. They both pass through the same points at 7 and 8, but they differ in between.\n\n18. Apr 17, 2015\n\n### ConstantineO\n\nAs I have stated before.\nI was completely uncertain of what to plot. I am being told to use the simple interest formula, but the two formulas you have provided are compound interest formulas:\ny = 10000 * (1.03)^x\ny = 10000 * (1.05)^(x-3)\nI plotted these long before I even got to this whole simple interest business when trying to figure out the original discrepancy between 7.61103 and 4.61103. I am fully aware that the compound interest lines are nearly straight. I don't see how the average rate of change between x =7 and x =8 is going to help me discover the remaining time of 0.61103 years. I am still unclear of what to plot? Do I plot the compound interest based functions for Romeo and Juliet, or do I plot the Simple interest functions? I have done both and I am not seeing anything that is giving me any kind of better understanding. Is this operation you describe too difficult represent algebraically?\n\nI am getting far too irritated by this and far too frustrated and am quickly losing the desire to continue.\n\n19. Apr 17, 2015\n\n### ConstantineO\n\nI was on the right track here. There is no need for this graphing nonsense. I forgot a single number which was my \"+ 3\" in my simple interest formula. I was a fool not to recognize that the lack of a constant would render my simple interest system of equation unsolvable. The lack of anyone speaking up about this is worrisome. In the future, when referencing the slope of a secant line between the range of two x values of a curve, please write in the notation of 7 < x < 8. I had absolutely no clue what you were talking about.\n\nSo let's try that last bit again with a fixed formula.\n\nRomeo Simple Interest Calculation\n(Ir) = (12,298.73865)(0.03)(t+3)\n(Ir) = (368.94)(t+3)\n\nJuliet Simple Interest Calculation\n(Ir) = (12,155.0625)(0.05)(t)\n(Ir) = 607.753125(t)\n\nNow the Systems of Equation\n(607.753125)(t) = (368.94)(t+3)\n(607.753125)(t) = (368.94)(t)+ 1106.82\nt(607.753125-368.94) = 1106.82\nt(238.813125) = (1106.82)\nt = (1106.82)/(238.813125)\nt = 4.634669891\nNot too far away from 4.61103. I wonder why...\n\nLet's try the simplified interest formula that measures time from Julias deposit date.\n\nRomeo Simple Interest Calculation\n(Ir) = (12,298.73865)(0.03)(t)\n(Ir) = (368.94)(t)\n\nJuliet Simple Interest Calculation\n(Ir) = (12,155.0625)(0.05)(t-3)\n(Ir) = 607.753125(t-3)\n\n607.753125(t-3)= (368.94)(t)\n(607.753125)(t) - 1823.259375 = (368.94)(t)\n(607.753125)(t) - (368.94)(t) =1823.259375\nt(607.753125 - 368.94) = 1823.259375\nt(238.813125) = 1823.259375\nt = 7.634.634669891\nNot too far away from 7.61103. Surprise surprise....\n\nLook Ma! No graphs.\n7.634.634669891 - 4.634669891 = 3 just like 7.61103 - 4.61103 =3. I wonder if they're related?\nI don't think I am wrong by stating that the time after the 7 year mark would actually be 0.634669891 of a year instead of 0.61103 if fractional compound interest is indeed calculated with simplified interest.\n\nWe can take things a step further and look at the secant line's slope of the curve that models interest from Romeo's deposit date during 7 < x < 8. Let's look at the average rate of change for both Romeo's and Juliet's account.\n\nRomeo Compound Interest AROC: 7 < x < 8\n\nWhile x = 4\nAr = 10,000(1.03)^4+3\nAr = 10,000(1.03)^7\nAr = 12,298.73865\nx = 4\ny =12,298.73865\n\nWhile x =5\nAr = 10,000(1.03)^5+3\nAr = 10,000(1.03)^8\nAr = 12,667.70081\nx = 5\ny = 12,667.70081\n\nRomeo's AROC\n= (12,667.70081 - 12,298.73865) / (5 -4)\n= 368.96216\n\nJuliet's Compound Interest AROC 7 < x < 8\n\nWhile x = 4\nAj = 10,000(1.05)^4\nAj = 12,155.0625\nx = 4\ny =12,155.0625\n\nWhile x = 5\nAj = 10,000(1.05)^5\nAj = 12,762.81563\nx = 5\ny = 12,762.81563\n\nJuliet's AROC\n= (12,762.81563 - 12,155.0625) / (5-4)\n=607.7531\n\nAfter that huge bunch of calculations we now have slopes that we can create linear relations with.\n\nRomeo's Linear Relation\ny = (368.96216)(x + 3)\n\nJuliet's Linear Relation\ny = 607.7531(x)\n\nLet's see what the Solution is for these in equations when put into a system.\n\n(368.96216)(x + 3) = 607.7531(x)\n(368.96216)(x) + 1106.88648 = 607.7531(x)\n607.7531(x) -(368.96216)(x) = 1106.88648\nx(607.7531-368.96216) = 1106.88648\nx(238.783884) = 1106.88648\nx = 4.635515854\n\nNotice a pattern?\n\nTime Measured from Romeo's Deposit Compound Calculation Product\nx = 4.61103\nTime Measured from Romeo's Deposit Simplified Interest Product\nx = 4.634669891\nTime Measured from Romeo's Deposit AROC: 7 < x< 8 Product\nx = 4.635515854\n\nI think its safe to say that they matched sometime in the range of 0.635515854 - 0.634669891- 0.61103 of the start of the 8th year. I think I made my point. I don't need any graphs to know how to rock, and I certainly showed that I sat down and did the work.\n\n20. Apr 17, 2015\n\n### Ray Vickson\n\nOf course it can be done without graphs. Graphs can be of help in setting up the algebraic equations that must, eventually, be solved without graphs. However, you seemed to not understand how simple interest works (your claim, not mine) and to see that a graphical representation can sometimes be helpful---not always, just sometimes. If it was not helpful to you, fine. And, of course I suggested plotting the compound-interest graphs, but I guess you missed out the part where I said that during a short period, such as 1 year, the graph looks almost straight, and that a straight line replacement for the graph (but ONLY in that single year) gives you the simple-interest effect within the year.\n\nYour writeup is almost incomprehensible to me, but your final answer seems OK. You say you still do not know why it is different from the original answer, and that is precisely what I was speaking of when I mentioned graphs: you could see right away why there is a difference.\n\nLast edited: Apr 17, 2015", "date": "2017-12-12 16:43:57", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/compound-interest-systems-of-equations-problem.808845/", "openwebmath_score": 0.5630732178688049, "openwebmath_perplexity": 978.4850642993932, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102571131692, "lm_q2_score": 0.8976952811593495, "lm_q1q2_score": 0.868439622755645}}
{"url": "https://math.stackexchange.com/questions/2044986/finding-common-terms-of-two-or-more-arithmetic-sequences", "text": "Finding common terms of two or more arithmetic sequences\n\nSuppose I have three sequences $\\{1,4,7,....,2998\\}$ , $\\{1,3,5,7,9,11,....,3001\\}$ and $\\{1,6,11,....,4001\\}$ .\n\nHow can i find the number of common terms among them ?\n\nFor Example ,\n\nIf I have\n\n1,3,5,7,9,11,13,15,17 and 19\n\n1,4,7,10,13,16 and 19\n\n1,6,11 and 16\n\nMy answer would be $6$.\n\nSuppose i have $n$ number of series' with common differences $a_1,a_2,...,a_n$ and each of them having the starting term as $1$ . How can i find the number of common terms then ?\n\n\u2022 Does your answer go all the way to the ends 2998,3001,4001 for the count of 0nly 6 common terms? \u2013\u00a0coffeemath Dec 5 '16 at 15:15\n\u2022 its 6 for only the given example \u2013\u00a0psil123 Dec 5 '16 at 15:16\n\nWrite the arithemtic sequence in form: $a_n = a_0 + (n-1)d_1, b_n = b_0 + (n-1)d_2, c_n = c_0 + (n-1)d_3$. Now if they have a common term then we can write:\n\n$$x = a_0 + kd_1 = b_0 + sd_2 = c_0 + td_3$$\n\nAs the common difference is known we can write this as:\n\n$$x \\equiv a_0 \\pmod {d_1}$$ $$x \\equiv b_0 \\pmod {d_2}$$ $$x \\equiv c_0 \\pmod {d_3}$$\n\nThis is system of linear congurence relations and it can be easily solved by the Chinese Remainder Theorem.\n\nAnd once you find one common element, to find the next common element you just add $LCM[d_1,d_2,d_3]$ to the previous one. In fact if you get $x \\equiv A \\pmod{LCM[d_1,d_2,d_3]}$ when solving the system of congurences, then total number of elements should be:\n\n$$1 + \\left\\lfloor\\frac{K-A}{LCM[d_1,d_2,d_3]}\\right\\rfloor$$\n\nwhere $K$ is the smallest last element in the arithemtic sequences.\n\nIf you at the explicit form of your sequences, which build your sets, you get get that $a_n = 1+3 \\cdot n$, $b_n = 1+ 2 \\cdot n$ and $c_n =1+5 \\cdot n$.\n\nNow you just have to find the least common multiple of 2,3 and 5 and see that all their common terms are $c_n = 1 + 30 \\cdot n$\n\n\u2022 suppose i have n series with common differences a1,a2,...,an and each of them have the starting number as 1 . How can i find the number of common terms ? \u2013\u00a0psil123 Dec 5 '16 at 15:17", "date": "2019-09-18 21:58:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2044986/finding-common-terms-of-two-or-more-arithmetic-sequences", "openwebmath_score": 0.5946933627128601, "openwebmath_perplexity": 271.0671223744418, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.980580656357574, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8684331022475942}}
{"url": "https://math.stackexchange.com/questions/2957504/seemingly-tricky-dice-question-probability-that-one-event-occurs-before-another", "text": "# Seemingly tricky dice question-probability that one event occurs before another event?\n\nThe question is as follows:\n\nYou roll two fair dice over and over. Let $$A$$ be the event you see two even sums. Let $$B$$ be the event you see a sum of $$7$$ four times. What is the probability that event $$A$$ occurs before event $$B$$?\n\nI know that for mutually exclusive events with independent trials, the probability that event $$E$$ occurs before event $$F$$ is $$\\frac{\\mathbb{P}(E)}{\\mathbb{P}(E) + \\mathbb{P}(F)}.$$ I tried using this formula, but I ran into a problem. I calculated $$\\mathbb{P}(A)=\\frac{1}{2}\\cdot \\frac{1}{2}=\\frac{1}{4} \\ \\ \\ \\ \\text{and} \\ \\ \\ \\ \\mathbb{P}(B)=\\left ( \\frac{1}{6} \\right )^4=\\frac{1}{1296}.$$ However, I then realized that these probabilites are the events that two even sums occur $$\\textit{in a row},$$ and similarly for my $$\\mathbb{P}(B).$$\n\nDoes anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using?\n\nEdit: I know there is a geometric distribution involved.\n\n\u2022 If it helps approach the problem, A occurs before B is usually just a less obvious way to say A occurs without B \u2013\u00a0Lord Farquaad Oct 16 '18 at 12:54\n\nSince neither A nor B cares about odd sums that aren't 7, we'll just reroll in those cases and thus we can ignore the probability of those sums occurring.\n\nThus we're left with 7 in addition to the 6 even sums ($$2,4,6,8,10,12$$).\n\nThe probabilities of these are respectively $$6/36$$ and $$(1+3+5+5+3+1)/36 = 18/36$$.\n\nDivide the probabilities of each of these by the sum of both probabilities, to get $$P(7)$$ and $$P(even)$$ such that $$P(7) + P(even) = 1$$ (this is $$P(7) = \\frac{1}{4}$$ and $$P(even) = \\frac{3}{4}$$).\n\nNow, as bobajob pointed out, the probability of A occurring before B is the complement of the probability of B occurring before A.\n\n$$P(2\\ even\\ before\\ 4\\ sevens) = 1-P(4\\ sevens\\ before\\ 2\\ even)$$\n\nTo have 4 sums of seven before 2 even sums, there can be at most 1 even sum before the 4th seven sum.\n\nThus we have the following possible sequences:\n\n$$7\\ 7\\ 7\\ 7$$\n$$7\\ 7\\ 7\\ even\\ 7$$\n$$7\\ 7\\ even\\ 7\\ 7$$\n$$7\\ even\\ 7\\ 7\\ 7$$\n$$even\\ 7\\ 7\\ 7\\ 7$$\n\nEach of the above occurs with the following probability:\n\n$$P(7) * P(7) * P(7) * P(7) = P(7)^4$$\n$$P(7) * P(7) * P(7) * P(even) * P(7) = P(even) * P(7)^4$$\n$$P(7) * P(7) * P(even) * P(7) * P(7) = P(even) * P(7)^4$$\n$$P(7) * P(even) * P(7) * P(7) * P(7) = P(even) * P(7)^4$$\n$$P(even) * P(7) * P(7) * P(7) * P(7) = P(even) * P(7)^4$$\n\nThen we have:\n\n$$P(2\\ even\\ before\\ 4\\ sevens) = 1-P(4\\ sevens\\ before\\ 2\\ even)$$\n\n$$= 1 - P(7)^4 - 4 * P(even) * P(7)^4$$\n\n$$= 1 - (\\frac{1}{4})^4 - 4 * \\frac{3}{4} * (\\frac{1}{4})^4$$\n\n$$= 0.984375$$\n\n## Alternative method\n\nThe probability we want will just be the probability of getting 0-3 sums equal to 7 before getting 2 even sums.\n\n$$P(2\\ even\\ before\\ 4\\ sevens) = \\sum_{i=0}^3P(i\\ sevens\\ before\\ 2\\ even)$$\n\nSince one even sum will need to be at the end, we can simply consider all possible positions of the other even sum.\n\nTaking as an example $$i=3$$ (3 7 sums before 2 even sums), we'll have the following possible order of events:\n\n$$7\\ 7\\ 7\\ even\\ even$$\n$$7\\ 7\\ even\\ 7\\ even$$\n$$7\\ even\\ 7\\ 7\\ even$$\n$$even\\ 7\\ 7\\ 7\\ even$$\n\nThe amount of these we have is simply $$i + 1$$.\n\nEach of the above occurs with probability $$P(even)^2 * P(7)^i$$:\n\n$$P(7) * P(7) * P(7) * P(even) * P(even) = P(even)^2 * P(7)^3$$\n$$P(7) * P(7) * P(even) * P(7) * P(even) = P(even)^2 * P(7)^3$$\n$$P(7) * P(even) * P(7) * P(7) * P(even) = P(even)^2 * P(7)^3$$\n$$P(even) * P(7) * P(7) * P(7) * P(even) = P(even)^2 * P(7)^3$$\n\nThus we have:\n\n$$P(i\\ sevens\\ before\\ 2\\ even) = (i+1) * P(even)^2 * P(7)^i$$\n\nThis gives us:\n\n$$P(2\\ even\\ before\\ 4\\ sevens) = \\sum_{i=0}^3(i+1) * P(even)^2 * P(7)^i$$\n\n$$= P(even)^2 * (1 + 2*P(7) + 3*P(7)^2 + 4*P(7)^3)$$\n\n$$= (\\frac{3}{4})^2 * (1 + 2*\\frac{1}{4} + 3*(\\frac{1}{4})^2 + 4*(\\frac{1}{4})^3)$$\n\n$$= 0.984375$$\n\nMore generally, the positions of the even sums above will be a multiset permutation and one can come up with general expression for the probability of M events X occurring before N events Y, but that's a bit beyond the scope of this question.\n\n\u2022 Nice approach. Would it not be easier to calculate the probability of event $B$ occurring before $A$ and then take the complement though? ($i$ would just be 0 or 1...) \u2013\u00a0bobajob Oct 16 '18 at 10:41\n\u2022 @bobajob Good point, edited. \u2013\u00a0Dukeling Oct 16 '18 at 11:53\n\u2022 \"Since neither A nor B cares about odd sums that aren't 7\" Yes, they do. A sum of 2 followed by a sum of 3 followed by a sum of 4 does not constitute an instance of A, while a sum of 2 followed by a sum of 4 does. \u2013\u00a0Acccumulation Oct 16 '18 at 15:20\n\u2022 After rereading the question, I likely misunderstood it; I thought it was two even sums in a row. \u2013\u00a0Acccumulation Oct 16 '18 at 15:30\n\nI would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $$7$$s. What is the chance that A comes first from there?\n\nYou have two Bernoulli processes,$$S_A$$ and $$S_B$$, and you are asked about the probability that the 2nd arrival in $$S_A$$ process occurs before the 4th arrival in $$S_B$$ process. PMF of the time of $$k$$-th arrival in a Bernoulli process with probability of success $$p$$ is $$p_{X_k}(t)=\\binom{t-1}{k-1}p^k(1-p)^{t-k}, \\ t=k,k+1,\\ldots$$\n\nPS: the events \"sum is even\" and \"sum is 7\" are dependent, and it makes the question tricky indeed, and the trick is, as Dukeling explained, is to ignore odd sums not equal to $$7$$; then the (conditional) probability of getting the sum of $$7$$ is $$\\frac{6}{36-12}=\\frac{1}{4}$$. Now the complement of the event of interest is the event that the 4th arrival in the Bernoulli process with $$p=1/4$$ occurs at time $$4$$ or $$5$$, and its probability is equal to $$\\binom{3}{3}p^4+\\binom{4}{3}p^4(1-p)=p^4(1+4(1-p))=\\frac{1}{64}$$ and the final answer is $$1-\\frac{1}{64}=\\frac{63}{64}$$", "date": "2019-01-20 04:40:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2957504/seemingly-tricky-dice-question-probability-that-one-event-occurs-before-another", "openwebmath_score": 0.9077591896057129, "openwebmath_perplexity": 191.86031542509505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806489800367, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8684330957138149}}
{"url": "https://math.stackexchange.com/questions/2412980/inequality-for-fibonacci-to-find-an-upper-bound-of-harmonic-fibonacci-series", "text": "# Inequality for Fibonacci to find an upper bound of harmonic Fibonacci series\n\nI want to find an sharp upper bound for $$\\sum_{n=1}^{\\infty}\\frac{1}{F_n}$$which $F_n$ is the n$th$ term of Fibonacci sequence . I wrote a Matlab program to find an upper bound ,$\\sum_{n=1}^{10^6}\\frac{1}{F_n}<4$\nNow my question is:(1):Is there an inequality to find this ?\n(2): Is that series have a close form ? $${F_n} = \\frac{{{\\varphi ^n} - {{( - \\varphi )}^{ - n}}}}{{\\sqrt 5 }}\\to \\\\\\sum_{n=1}^{\\infty}\\frac{1}{F_n}=\\sum_{n=1}^{\\infty}\\frac{\\sqrt 5}{{{\\varphi ^n} - {{( - \\varphi )}^{ - n}}}}\\\\\\leq \\sum_{n=1}^{\\infty}\\frac{\\sqrt 5}{{{(\\frac{{1 + \\sqrt 5 }}{2} ) ^n} }}=\\frac{\\sqrt5}{1-\\frac{1}{\\frac{{1 + \\sqrt 5 }}{2}}}\\approx12.18\\\\$$I am thankful for a hint or solution which can bring a sharper upper bound .\n\n\u2022 maybe related \u2013\u00a0MAN-MADE Sep 1 '17 at 7:33\n\u2022 Simple fast program: 3.359885666243177553172011302918927179688905133731968486495553815325130318996683383615416216456790087297045342928853913304136789017100883679591351733077119078580333550332507753187599850487179777897006039564509215375892775265673354024033169441799293934610992626257964647651868659449710216558984360881472693249591079473873673378523326877499762727757946853676918541981467668742998767382096913901217722024405208151094264934951 \u2013\u00a0Kenny Lau Sep 1 '17 at 7:33\n\u2022 WolframAlpha simply returns \u2131 suggesting that it has no closed form. \u2013\u00a0Kenny Lau Sep 1 '17 at 7:35\n\u2022 @KennyLau:Can I see your program ? can you post it here ? \u2013\u00a0Khosrotash Sep 1 '17 at 7:35\n\u2022 $F_n-\\phi^n/\\sqrt5=-(-1/\\phi)^n/\\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign, and that your inequality follows? \u2013\u00a0Jyrki Lahtonen Sep 1 '17 at 21:02\n\nAs Jyrki Lahtonen points out in the comments, $\\sum_{n=1}^\\infty\\frac{\\sqrt5}{\\varphi^n}$ isn't necessarily the right bound, since it doesn't dominate the original series term-by-term.\n\nAs for a closed form, this identity is known: $$\\sum_{n=1}^\\infty\\frac{1}{F_n}=\\frac{\\sqrt5}{4}\\left(\\vartheta_2^2(\\varphi^{-2}) + \\frac{\\log5 + 2\\psi_{\\varphi^{-4}}(1) - 4\\psi_{\\varphi^{-2}}(1)}{2\\log\\varphi}\\right)$$ where $\\vartheta_2(q)$ is the Jacobi theta function at $z=0$, and $\\psi_q$ is the $q$-digamma function. See Wikipedia and MathWorld on the \"Reciprocal Fibonacci constant\", and other Math.SE questions such as What is the sum of Fibonacci reciprocals?\n\n\u2022 $F_n-\\phi^n/\\sqrt5=-(-1/\\phi)^n/\\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign? The OP suffers from the same problem. \u2013\u00a0Jyrki Lahtonen Sep 1 '17 at 20:59\n\u2022 @JyrkiLahtonen Whoops! Let me fix that. \u2013\u00a0Chris Culter Sep 1 '17 at 21:12\n\nHere is a systematic method for computing upper bounds for the sum without much work.\n\nBy induction, if $F_N \\ge b^N$ and $b+1 \\ge b^2$, then $F_n \\ge b^n$ for $n \\ge N$.\n\nTherefore, $$\\sum_{n=1}^\\infty {1\\over F_n} \\le \\sum_{n=1}^{N-1} {1\\over F_n} + \\sum_{n=N}^\\infty {1\\over b^n} = \\sum_{n=1}^{N-1} {1\\over F_n} + \\frac{1}{b^{N-1}(b-1)}$$ This upper bound gets closer to the actual sum when $b$ gets larger, but then we need larger $N$:\n\n\\begin{array}{crl} b &N &sum \\\\ 1.3 &4 &4.0172204521317 \\\\ 4/3 &5 &3.7825520833333 \\\\ 1.4 &6 &3.4981694135380 \\\\ 1.5 &11 &3.3651521005948 \\\\ 1.6 &72 &3.3598856662432 \\\\ \\end{array}\n\nSince $1.6 \\approx \\phi$, which is the largest possible $b$, the last value is quite close to the actual value: $$3.359885666243177553172011302918927179688905133732\\cdots$$\n\nUsing Wolfram Mathematica answer is : Wolfram Mathematica Code", "date": "2019-11-19 13:00:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2412980/inequality-for-fibonacci-to-find-an-upper-bound-of-harmonic-fibonacci-series", "openwebmath_score": 0.9962795972824097, "openwebmath_perplexity": 819.8858410535962, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806495475397, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8684330754963521}}
{"url": "https://unitedtravelbooking.com/site/03a3ca-sum-of-two-binomial-random-variables", "text": "Making statements based on opinion; back them up with references or personal experience. Why did mainframes have big conspicuous power-off buttons? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Useful relations in dealing with binomial coefficients and factorials 4. How to sustain this sedentary hunter-gatherer society? Maybe I'm wrong in the conclusions about what that particular convergence means though. If success probabilities differ, the probability distribution of the sum is not binomial. \\Pr[S\\ge s] Determination of the binomial coefficient and binomial distribution The probability of any specified arrangement of k successes and n-k failures in n independent trials is pknkq \u2212 where p is the probability of success on any one trial and q=1-p is the probability of failure. Often the manipulation of integrals can be avoided by use of some type of generating function. \"In the limit as n\u2192\u221e, your binomials become Gaussian\" Sorry but this is simultaneously vague and wrong. Asking for help, clarification, or responding to other answers. Why were there only 531 electoral votes in the US Presidential Election 2016? That depends on the range of values you are considering. where we took $t=\\log(s/\\sum_ip_i)$. $$By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Why Is an Inhomogenous Magnetic Field Used in the Stern Gerlach Experiment? site design / logo \u00a9 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. The Kolmogorov approximation is given as an \u2026 Yes, in fact, the distribution is known as the Poisson binomial distribution, which is a generalization of the binomial distribution. See the binomial sum variance inequality. I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt. Both distributions have total mass 1. How do rationalists justify the scientific method. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, I have the same question and i read the paper (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). @jameselmore Additivity of the means is unrelated to independence. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. \\\\&= \\exp\\left(\\sum_i 1 + (e^t-1) p_i\\right) \\exp(-st) It will be a special case of the Poisson Binomial Distribution. Are there relatively simple formulae or at least bounds for the distribution \\Pr[S\\ge s] \\\\&= \\exp\\left(\\sum_i 1 + (e^t-1) p_i\\right) \\exp(-st) Extremely bloated transaction log during delete. Every second customer converts better. See this paper (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). Which one is more idiomatic: \u2018valid concern\u2019 or \u2018legitimate concern\u2019? Can this be by chance? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If success probabilities differ, the probability distribution of the sum is not binomial. Here is an excerpt from the Wikipedia page. by Marco Taboga, PhD. The distribution of a sum S of independent binomial random variables, each with different success probabilities, is discussed.$$ @Robert ,do you have any insight on what happens if the n is not same for the 2 distributions. This lecture discusses how to derive the distribution of the sum of two independent random variables.We explain first how to derive the distribution function of the sum and then how to derive its probability mass function (if the summands are discrete) or its probability density function (if the summands are continuous). This answer provides an R implementation of the explicit formula from the paper linked in the accepted answer (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens).", "date": "2021-05-15 03:08:19", "meta": {"domain": "unitedtravelbooking.com", "url": "https://unitedtravelbooking.com/site/03a3ca-sum-of-two-binomial-random-variables", "openwebmath_score": 0.6324305534362793, "openwebmath_perplexity": 433.70939160386644, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9805806484125338, "lm_q2_score": 0.8856314617436728, "lm_q1q2_score": 0.8684330730111508}}
{"url": "https://math.stackexchange.com/questions/588488/is-there-a-series-where-the-terms-tend-to-zero-faster-than-the-harmonic-series-b", "text": "# Is there a series where the terms tend to zero faster than the harmonic series but it still diverges?\n\nI know that for the harmonic series $\\lim_{n \\to \\infty} \\frac1n = 0$ and $\\sum_{n=1}^{\\infty} \\frac1n = \\infty$.\n\nI was just wondering, is there a sequence ($a_n =\\dots$) that converges \"faster\" (I am not entirely sure what's the exact definition here, but I think you know what I mean...) than $\\frac1n$ to $0$ and its series $\\sum_{n=1}^{\\infty}{a_n}= \\infty$?\n\nIf not, is there proof of that?\n\n\u2022 Sure. Let $a_n=\\frac1{n\\log n}$. Or let $a_n=\\frac1{n\\log n\\log\\log n}$. Or let ... In fact, the process never ends, in that you can find a divergent series whose terms go to zero faster than all the sequences I suggested. \u2013\u00a0Andr\u00e9s E. Caicedo Dec 1 '13 at 19:39\n\u2022 Versions of this question have been asked before. Here is an example from MO. \u2013\u00a0Andr\u00e9s E. Caicedo Dec 1 '13 at 19:41\n\u2022 ok, it this checks out! (using the condensation test) \u2013\u00a0guynaa Dec 1 '13 at 19:46\n\u2022 Related The answers by by Bill/Gone and Yuval. \u2013\u00a0user17762 Dec 1 '13 at 20:15\n\u2022 $$\\sum_{n=1}^{\\infty} \\frac{1}{10n} ?$$ \u2013\u00a0Sawarnik Feb 24 '14 at 21:13\n\n## 3 Answers\n\nThere is no slowest divergent series. Let me take this to mean that given any sequence $a_n$ of positive numbers converging to zero whose series diverges, there is a sequence $b_n$ that converges to zero faster and the series also diverges, where \"faster\" means that $\\lim b_n/a_n=0$. In fact, given any sequences of positive numbers $(a_{1,n}), (a_{2,n}),\\dots$ with each $\\sum_n a_{i,n}=\\infty$ and $\\lim_n a_{i+1,n}/a_{i,n}=0$, there is $(a_n)$ with $\\sum a_n=\\infty$ and $\\lim_n a_n/a_{i,n}=0$ for all $i$.\n\nTo see this, given $a_1,a_2,\\dots$, first define $b_1=a_1,b_2=a_2,\\dots,b_k=a_k$ until $a_1+\\dots+a_k>1$. Second, let $b_{k+1}=a_{k+1}/2,b_{k+2}=a_{k+2}/2,\\dots,b_n=a_n/2$ until $a_{k+1}+\\dots+a_n>2$, etc. That is, we proceed recursively; if we have defined $b_1,\\dots,b_m$ and $b_m=a_m/2^r$, and $b_1+\\dots+b_m>r+1$, let $b_{m+1}=a_{m+1}/2^{r+1},\\dots,b_l=a_l/2^{r+1}$ until $a_{m+1}+\\dots+a_l>2^{r+1}$. The outcome is that $\\sum b_i=\\infty$ and $\\lim b_i/a_i=0$.\n\nSimilarly, given $(a_{1,n}),(a_{2,n}),\\dots$, with each $(a_{k+1,n})$ converging to zero faster than $(a_{k,n})$, and all of them diverging, let $a_i=a_{1,i}$ for $i\\le n_1$, where $a_{1,1}+\\dots+a_{1,n_1}>1$, then $a_i=a_{2,i}$ for $n_1<i\\le n_2$, where we ask both that $a_{2,n_1+1}+\\dots+a_{2,n_2}>1$ and that for any $k>n_2/2$ we have $a_{2,k}/a_{1,k}<1/2$, etc. That is, if we have defined $n_k$, we let $a_i=a_{k+1,i}$ for $n_k<i\\le n_{k+1}$ where $n_{k+1}$ is chosen so that $a_{k+1,n_k+1}+\\dots+a_{k+1,n_{k+1}}>1$ and for all $l>n_{k+1}/2$ we have $a_{k+1,l}/a_{i,l}<1/2^{k+1}$ for all $i<k+1$. Then the series $\\sum a_i$ diverges, and the sequence $(a_i)$ converges to $0$ faster than all the $a_{i,n}$. In modern language, we would say that there are no $(\\omega,0)$-gaps in a certain partial order.\n\nWe can modify the above slightly so that given any sequences $(a_{i,n})$ with $\\sum_n a_{i,n}<\\infty$ and $\\lim_n a_{i+1,n}/a_{i,n}=\\infty$ for all $i$, we can find $(a_n)$ with $\\sum_n a_n<\\infty$ and $\\lim_n a_n/a_{i,n}=\\infty$, so there is no fastest convergent series, and not even considering a sequence of faster and faster convergent series is enough. (In modern terms, there is no $(0,\\omega)$-gap.)\n\nWhat we cannot do in general is, given $(a_{i,n})$, with all $\\sum_n a_{i,n}=\\infty$, find $(a_n)$ with $\\sum a_n=\\infty$ and $a_n/a_{i,n}\\to0$ for all $i$, if the $a_{i,n}$ are not ordered so that $a_{i+1,n}$ converges to zero faster than $a_{i,n}$. For example, we can have $a_n=1/n$ if $n$ is odd and $a_n=1/n^2$ if $n$ is even, and $b_n=1/n^2$ if $n$ is odd and $b_n=1/n$ if $n$ is even, and if $c_n$ converges to zero faster than both, then $\\sum c_n$ converges. (These exceptions can typically be fixed by asking monotonicity of the sequences, which is how these results are usually presented in the literature.)\n\nNote that the argument I gave is completely general, no matter what the series involved. For specific series, of course, nice \"formulas\" are possible. For example, given $a_n=1/n$, we can let $b_n=1/(n\\log n)$ for $n>1$. Or $c_n=1/(n\\log n\\log\\log n)$, for $n\\ge 3$. Or ... And we can then \"diagonalize\" against all these sequences as indicated above.\n\nBy the way, the first person to study seriously the boundary between convergence and divergence is Paul du Bois-Reymond. He proved a version of the result I just showed above, that no \"decreasing\" sequence of divergent series \"exhausts\" the divergent series in that we can always find one diverging and with terms going to zero faster than the terms of any of them. A nice account of some of his work can be found in the book Orders of Infinity by Hardy. Du Bois-Reymond's work was extended by Hadamard and others. What Hadamard proved is that given $(a_i)$ and $(b_i)$ with $\\sum a_i=\\infty$, $\\sum b_i<\\infty$, and $b_i/a_i\\to 0$, we can find $(c_i),(d_i)$ with $c_i/a_i\\to0$, $b_i/d_i\\to 0$, $d_i/c_i\\to0$, $\\sum c_i=\\infty$, $\\sum d_i<\\infty$. More generally:\n\nIf we have two sequences of series, $(a_{1,n}), (a_{2,n}),\\dots$ and $(b_{1,n}),(b_{2,n}),\\dots$, such that\n\n\u2022 Each $(a_{i+1,n})$ converges to zero faster than the previous one,\n\u2022 Each $(b_{i+1,n})$ converges to zero slower than the previous one,\n\u2022 Each $(a_{i,n})$ converges to zero slower than all the $(b_{j,n})$,\n\u2022 Each $\\sum_n a_{i,n}$ diverges, and\n\u2022 Each $\\sum_n b_{i,n}$ converges,\n\nthen we can find sequences $(c_n),(d_n)$, \"in between\", with one series converging and the other diverging.\n\nIn modern language, we say that there are no $(\\omega,\\omega)$-gaps, and similarly, there are no $(\\omega,1)$- or $(1,\\omega)$-gaps. This line of research led to some of Hausdorff's deep results in set theory, such as the existence of so-called $(\\omega_1,\\omega_1)$- or Hausdorff gaps. What Hausdorff proved is that this \"interpolation\" process, which can be iterated countably many times, cannot in general be carries out $\\omega_1$ times, where $\\omega_1$ is the first uncountable ordinal.\n\nI had wondered about this too a long time ago and then came across this. The series\n\n$$\\sum_{n=3}^{\\infty}\\frac{1}{n\\ln n (\\ln\\ln n)}=\\infty$$\n\ndiverges and it can be very easily proven by the integral test. But here is the kicker. This series actually requires googolplex number of terms before the partial sum exceeds 10. Talk about slow! It is only natural that if the natural log is slow, then take the natural log of the natural log to make it diverge even slower.\n\nHere is another one. This series\n\n$$\\sum_{n=3}^{\\infty}\\frac{1}{n\\ln n (\\ln\\ln n)^2}=38.43...$$\n\nactually converges using the same exact (integral) test. But it converges so slowly that this series requires $10^{3.14\\times10^{86}}$ before obtaining two digit accuracy. So using these iterated logarithms you can come up with a series which converges or diverges \"arbitrarily slowly\".\n\nReference: Zwillinger, D. (Ed.). CRC Standard Mathematical Tables and Formulae, 30th ed. Boca Raton, FL: CRC Press, 1996.\n\n\u2022 But then by the same integral test $$\\sum_{n=N_0}^\\infty \\sum_{n=3}^{\\infty}\\frac{1}{n\\ln n (\\ln\\ln n) (\\ln\\ln \\ln n)}=\\infty$$ diverges at an even slower rate....Moreo generarily $$\\sum \\frac{1}{n \\ln(n) \\ln^2(n) \\ln^3(n) ... \\ln^k(n)}$$ diverges, where $^j$ denotes composition . \u2013\u00a0N. S. Dec 2 '13 at 0:46\n\u2022 @N.S. That's what I said, using iterated logarithm you can come up with an even slower growth. In addition, in your first summation, the double summation doesn't make sense. I think you want the inner sigma removed. $N_0 = \\lceil((e^x)^j(0)\\rceil$ where $j$ denotes composition. In your first summation with $j=3$ we have $N_0=16$. \u2013\u00a0Fixed Point Dec 2 '13 at 2:31\n\u2022 this is a fantastic example--showing that even trying to test convergence by simulation could be doomed, because we don't have the computational capacity to find enough terms to convince ourselves the sequence diverges. \u2013\u00a0MichaelChirico Apr 14 '15 at 14:56\n\nIf we use the notion of a partial sum:\n\n$$S_n = \\sum_{k=1}^n a_k$$\n\nyou are asking for a series in which the partial sums diverge more slowly than $\\sum_{k=1}^n 1/k$ as $n\\to\\infty$. For this to happen, we just need to find $a_k$ such that $a_k < 1/k$ for all $k>c$ where $c$ is some value.\n\nLook at $a_k = \\frac{1}{k\\ln k}$. Let's do the integral test to show this diverges. The substitution used is $u = \\ln k$ so that $du = dk/k$.\n\n$$\\int \\frac{dk}{k\\ln k}=\\int \\frac{du}{u}=\\ln u = \\ln {\\ln k}$$\n\nPutting in limits of $2$ and $\\infty$ shows that this indeed diverges. And for $k>c=e$, we have that $1/(k\\ln k) < 1/k$.", "date": "2020-02-17 19:05:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/588488/is-there-a-series-where-the-terms-tend-to-zero-faster-than-the-harmonic-series-b", "openwebmath_score": 0.9720413684844971, "openwebmath_perplexity": 146.43397397440762, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718471760708, "lm_q2_score": 0.8774767954920548, "lm_q1q2_score": 0.8684140810487613}}
{"url": "https://math.stackexchange.com/questions/2873457/a-simple-method-of-factorization-2301", "text": "# A simple method of factorization $2^{30}+1$\n\nHow can you factor $2^{30} + 1$? This task was supposed to be at one interview, there is an assumption that there should be a simple solution.\n\nObserve that $$x^k+1=\\frac{x^{2k}-1}{x^k-1}=\\frac{\\prod_{d\\mid 2k}\\Phi_{d}(x)}{\\prod_{d\\mid k}\\Phi_{d}(x)}=\\prod_{d\\mid 2k,\\ d\\nmid k}\\Phi_{d}(x)$$ With $$\\Phi_n(x)$$ being the $$n^\\text{th}$$ cyclotomic polynomial. It follows that $$2^{30}+1=\\Phi_4(2)\\Phi_{12}(2)\\Phi_{20}(2)\\Phi_{60}(2)$$ With the individual factors computed as $$\\Phi_4(2)=2^2+1=5$$ $$\\Phi_{12}(2)=2^4-2^2+1=13$$ $$\\Phi_{20}(2)=2^8-2^6+2^2-2^2+1=5\\times 41$$ $$\\Phi_{60}(2)=2^{16}+2^{14}-2^{10}-2^8-2^6+2^2+1=61\\times 1321$$ Giving an overall factorization of $$\\boxed{2^{30}+1=5^2\\times 13\\times 41\\times 61\\times 1321}$$\n\n\u2022 This is very elegant. Aug 6 '18 at 1:58\n\u2022 Thanks! It's also not too hard to compute the relevant polynomials by hand (the section \"Easy cases for computation\" in the linked wiki article lists some strategies) :) Aug 6 '18 at 2:01\n\u2022 Again, computing $\\Phi_{60}(2) = 61 \\times 1321$ is not really something to do in an interview. Aug 6 '18 at 7:26\n\u2022 Perhaps they wanted to see the way to a solution, not the solution itself. Aug 6 '18 at 11:43\n\u2022 @RobertIsrael: probably the interviewer wanted to hear something along the lines \"in order to find the prime divisors of $\\Phi_{60}(2)$, it is enough to check the primes $\\equiv 1\\pmod{60}$. Since the order of $2$ in $\\mathbb{Z}/(61\\mathbb{Z})^*$ is what it is, $61$ is a prime divisor of $\\Phi_{60}(2)$ and the problem boils down to checking that $1321$ is a prime\". Aug 6 '18 at 16:45\n\nI would note it is a sum of cubes and fifth powers, so $2^{10}+1=1025$ and $2^6+1=65$ are both factors. In terms of primes that gives us $5^2\\cdot 13 \\cdot 41$ as factors. At this point in an interview (depending on the position) I would declare success and move on. Getting the remaining $80581$ would take a bunch of hand calculation and finding $61\\cdot 1321$ doesn't seem reasonable either.\n\n\u2022 $61\\equiv 5 \\bmod 8$ so $(2|61)\\equiv 2^{30}\\equiv -1\\bmod 61$. Aug 6 '18 at 1:44\n\nHint: $a^k+1$ is divisible by $a+1$ if $k$ is odd.\n\n\u2022 Just so we can check any number-theoretic derivation... Mathematica gives: $5^2 \\times 13 \\times 41 \\times 61 \\times 1321$. (Doesn't seem so simple that it could be done in an interview session...) Aug 6 '18 at 0:37\n\nYet another easy observation: since $4X^4+1=(2X^2+2X+1)(2X^2-2X+1)$, we get $4\\cdot2^{28}+1=(2\\cdot2^{14}+2\\cdot2^7+1)(2\\cdot2^{14}-2\\cdot2^7+1)$.\n\nFirstly, in an interview there is often a difference between showing that there is a simple solution (existence) and outlining the approach, versus actually finding the solution.\n\nThat is, it might suffice to outline a standard prime factorising algorithm and show that it could be computed in trivial time.\n\nOne of the simplest classic approaches is that to factor $x$, you only need to test the primes that are less than $\\sqrt{x}$. If $x=2^{30}+1$, then $\\sqrt{x+1} \\simeq 32768$. There are less than 3000 primes that satisfy this constraint, so it is very feasible on any computing platform.\n\nOnce each prime factor was found, one would divide that factor out, and then only need to search for primes up to an even smaller upper bound. Thus after the finding that 5 is a prime, one need only search for other prime factors less than 14654.\n\nThus, you have an algorithm that would find all the factors of $2^{30}+1$ in ascending order.\n\nThat is, $2^{30}+1 = 5 \\times 5 \\times 13 \\times 41 \\times 61 \\times 1321$.\n\nThe key to nearly any sensible prime factorization approach is to efficiently find the first factor. Thus, in some ways it could be said that factoring $2^{30}+1$ is easy because it has lots of prime factors, and most of them are very small. (Contrast this to encryption are based on numbers that are very hard to factor, because they typically one have 2 factors and each of them are very very large.)\n\nAnyway, as @Robert indicated, the key to solving this directly is to realise that if $k$ is odd, then $x^k+1$ can be elegantly factorised, as $(x+1)$ is a factor.\n\nFor example, $$x^7+1 = (x+1)(x^6-x^5+x^4-x^3+x^2-x+1)$$ Note that $z^{30}+1 = (z^2)^{15}+1$ and so $z^2+1$ is a factor.\n\nLetting $z=1$ shows that 5 is a factor of $2^{30}+1$.\n\nSimilarly, you could use the well-known sum of cubes factorisation. $x^3+1 = (x+1)(x^2-x+1)$, to show that $$2^{30}+1 = (2^{10}+1)(2^{20}-2^{10}+1) = 1025 \\times 1047553$$ and continue from there.\n\n\u2022 Your early complete factorization is missing $61$ Aug 6 '18 at 1:59", "date": "2021-09-23 01:47:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2873457/a-simple-method-of-factorization-2301", "openwebmath_score": 0.8364754319190979, "openwebmath_perplexity": 239.6242308531144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718453483273, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8684140604189982}}
{"url": "http://mathematica.stackexchange.com/questions/27642/fitting-a-two-dimensional-gaussian-to-a-set-of-2d-pixels/27754", "text": "# Fitting a two-dimensional Gaussian to a set of 2D pixels\n\nImagine I have a set of data like the following:\n\ndata = {{0.0453803, 0.0427863, 0.0489815, 0.045243, 0.0488289, 0.0432898,\n0.04448, 0.0387732, 0.0388952}, {0.0507668, 0.0427863, 0.0502632,\n0.0503395, 0.0634623, 0.0675822, 0.0529335, 0.047425,\n0.0387121}, {0.042237, 0.0501259, 0.0595712, 0.0869001, 0.139559,\n0.141512, 0.0868391, 0.0579232, 0.0408331}, {0.0478981, 0.0491646,\n0.0652628, 0.130404, 0.218448, 0.220645, 0.143603, 0.0605173,\n0.0424964}, {0.0462043, 0.0530861, 0.076051, 0.140017, 0.206943,\n0.202502, 0.118791, 0.0614023, 0.0459907}, {0.0511788, 0.0582132,\n0.105531, 0.166354, 0.181003, 0.13698, 0.0748302, 0.0557107,\n0.0492103}, {0.0493629, 0.0539712, 0.0971695, 0.160769, 0.164477,\n0.104768, 0.0591745, 0.0475319, 0.0452583}, {0.0510719, 0.0599374,\n0.0730602, 0.0975814, 0.101289, 0.0691997, 0.0498054, 0.044892,\n0.043122}, {0.0460517, 0.0567025, 0.0574044, 0.0587778, 0.0537118,\n0.0487221, 0.0474098, 0.0413977, 0.04477}}\n\n\nI don't have enough reputation to post an image, but one can easily be generated by applying ListPlot3D to the above data set.\n\nHow might I best fit a Gaussian curve to this set of datapoints, and extract properties such at the fit' semi-axes? I noticed that ComponentMeasurements has some functionality for best fit ellipsoids, but that doesn't seem to be workable here.\n\nMy objective here is to determine how \"Gaussian\" a set of points in an image are. My strategy is to sequentially fit a 2D Gaussian to each point, and then to measure it's eccentricity and spread (looking, for example, at the length and ratio of the semiaxes of the ellipsoid corresponding to the fit). The example here seems like it should yield a 2D Gaussian fit with significant spread and a ratio for the semiaxes significantly diverging from one.\n\n-\nMaybe You can add some context to the question. I'm asking because this data looks rather like sum of two 2D gaussians, and I don't know if it is relevant. \u2013\u00a0 Kuba Jun 25 '13 at 22:36\n@Kuba I have added a bit of context in the problem description! \u2013\u00a0 Bob Jun 25 '13 at 22:42\nBy context I mean, for example, what this data actually is. :) Without details, term \"determine how Gaussian\" is too vague. \u2013\u00a0 Kuba Jun 25 '13 at 23:07\n\nSjoerd's answer applies the power of Mathematica's very general model fitting tools. Here's a more low-tech solution.\n\nIf you want to fit a Gaussian distribution to a dataset, you can just find its mean and covariance matrix, and the Gaussian you want is the one with the same parameters. For Gaussians this is actually the optimal fit in the sense of being the maximum likelihood estimator -- for other distributions this may not work equally well, so there you will want NonlinearModelFit.\n\nOne wrinkle is that your data doesn't fall off to zero but to something like Min[data] $\\approx 0.0387$, but we can easily get rid of that by just subtracting it off. Next, I normalize the array to sum to $1$, so that I can treat it like a discrete probability distribution. (All this really does is allow me to avoid dividing by Total[data, Infinity] every time in the following.)\n\nmin = Min[data];\nsum = Total[data - min, Infinity];\np = (data - min)/sum;\n\n\nNow we find the mean and covariance. Mathematica's functions don't seem to work with weighted data, so we'll just roll our own.\n\n{m, n} = Dimensions[p];\nmean = Sum[{i, j} p[[i, j]], {i, 1, m}, {j, 1, n}];\ncov = Sum[Outer[Times, {i, j} - mean, {i, j} - mean] p[[i, j]], {i, 1, m}, {j, 1, m}];\n\n\nWe can easily get the probability distribution for the Gaussian with this mean and covariance. Of course, if we want to match the original data, we have to rescale and shift it back.\n\nf[i_, j_] := PDF[MultinormalDistribution[mean, cov], {i, j}] // Evaluate;\ng[i_, j_] := f[i, j] sum + min;\n\n\nThe match is not too bad, although the data has two humps where the Gaussian has one. You can also compare the fit through a contour plot. (Use the tooltips to compare contour levels.)\n\nShow[ListPlot3D[data, PlotStyle -> None],\nPlot3D[g[i, j], {j, 1, 9}, {i, 1, 9}, MeshStyle -> None,\nPlotStyle -> Opacity[0.8]], PlotRange -> All]\nShow[ListContourPlot[data, Contours -> Range[0.05, 0.25, 0.025],\nContourShading -> None, ContourStyle -> ColorData[1, 1], InterpolationOrder -> 3],\nContourPlot[g[i, j], {j, 1, 9}, {i, 1, 9}, Contours -> Range[0.05, 0.25, 0.025],\nContourShading -> None, ContourStyle -> ColorData[1, 2]]]\n\n\nThe variances along the principal axes are the eigenvalues of the covariance matrix, and the standard deviations (which I guess you're calling the semi-axes) are their square roots.\n\nSqrt@Eigenvalues[cov]\n(* {1.86325, 1.50567} *)\n\n-\nOr you could just plot the Gaussian fit coloured by the residual: Show[ListDensityPlot[ Table[data[[i, j]] - g[i, j], {i, 1, m}, {j, 1, m}], InterpolationOrder -> 3, PlotRange -> Full, ColorFunctionScaling -> False, ColorFunction -> (With[{t = 20 #}, RGBColor[1 + t, 1 - Abs[t]/2, 1 - t]] &)], ContourPlot[g[x, y], {y, 1, 9}, {x, 1, 9}, Contours -> Range[0.05, 0.25, 0.025], ContourShading -> None, ContourStyle -> Black]] i.stack.imgur.com/LqGZL.png (orange: data higher than fit, blue: data lower than fit) \u2013\u00a0 Rahul Narain Jun 27 '13 at 6:23\nThanks again for such a fantastic answer. Just to help me understand - why does your calculation for the square root of the eigenvectors yield a slightly different result than StandardDeviation@MultinormalDistribution[mean,cov]? \u2013\u00a0 Bob Jun 30 '13 at 11:15\nBecause that one appears to give the standard deviations of the marginal distributions of $x$ and $y$ instead, which you also get with Sqrt@Diagonal[cov]. \u2013\u00a0 Rahul Narain Jun 30 '13 at 11:28\ndata3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data, {2}], 1];\n\nfm = NonlinearModelFit[data3D,\na E^(-(((-my + y) Cos[b] - (-mx + x) Sin[b])^2/(2 sy^2)) -\n((-mx + x) Cos[b] + (-my + y) Sin[b])^2/(2 sx^2)),\n{{a, 0.1}, {b, 0}, {mx, 4.5}, {my, 4.5}, {sx, 3}, {sy, 3}},\n{x, y}\n]\n\nShow[\nListPlot3D[data3D, PlotStyle -> None, MeshStyle -> Red],\nPlot3D[fm[\"BestFit\"], {x, 1, 9}, {y, 1, 9}, PlotRange -> All,\nPlotStyle -> Opacity[0.9], Mesh -> None]\n]\n\n\nfm[\"BestFitParameters\"]\n\n\n{a -> 0.1871830545, b -> -0.4853901689, mx -> 5.152549499, my -> 5.092511036, sx -> 2.756524919, sy -> 2.072163433}\n\n-\n\nStarting with your data, I would map out the data something like this:\n\ndata3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data, {2}], 1]\n\n\nand then count it like:\n\ndata2 = Flatten[Table[{#[[1]], #[[2]]}, {100 #[[3]]}] & /@ data3D, 1];\n\n\nand then use this function:\n\nFindDistributionParameters[data2,\nMultinormalDistribution[{a, b}, {{c, d}, {e, f}}]]\n\n\nto find the means and coveriance of the data. And then I might do other stuff with the distribution, such as create a distribution object using:\n\nedist = EstimatedDistribution[data2,\nMultinormalDistribution[{a, b}, {{c, d}, {e, f}}]]\n\n\nand then check the distribution against the data, using:\n\ndtest = DistributionFitTest[data2, edist, \"HypothesisTestData\"]\n\n\nand create a table of the test results, using:\n\nN[dtest[\"TestDataTable\", All]]\n\n\nor individually\n\nAndersonDarlingTest[data2, edist, \"TestConclusion\"]\nCramerVonMisesTest[data2, edist, \"TestConclusion\"]\nJarqueBeraALMTest[data2, \"TestConclusion\"]\nMardiaKurtosisTest[data2, \"TestConclusion\"]\nPearsonChiSquareTest[data2, edist, \"TestConclusion\"]\nShapiroWilkTest[data2, \"TestConclusion\"]\n\n\nNow, I'm not a statistician, and I probably only know enough statistics to be dangerous, and I'm not familar with all the tests listed in the table, but I suspect that one of those small numbers is significant in measuring how close the actual data fits the distribution, i.e. how \"Gaussian\" the set of points is.\n\nAlso I might just plot the data as follows and visually judge the gaussiness of the data using these functions (as per VLC's answer to question 2984)\n\nShow[DensityHistogram[d, {.2}, ColorFunction -> (Opacity[0] &),\nMethod -> {\"DistributionAxes\" -> \"Histogram\"}], ListPlot[d]]\n\n\nor these:\n\nGraphicsColumn[\nTable[SmoothDensityHistogram[d, ColorFunction -> \"DarkRainbow\",\nMethod -> {\"DistributionAxes\" -> p}, ImageSize -> 500,\nBaseStyle -> {FontFamily -> \"Helvetica\"},\nLabelStyle -> Bold], {p, {True, \"SmoothHistogram\", \"BoxWhisker\"}}]]\n\n-\n\nYou mentioned in a related question that you would like to do large numbers of these fits quickly.\n\nIf each data set has the same dimensions, you can write a fairly fast implementation of Rahul Narain's method by precomputing arrays of x and y coordinates for the data grid, and flattening the data and using Dot to calculate the mean and the elements of the covariance matrix:\n\nx = Table[i, {i, 9}, {j, 9}]//N;\ny = Transpose[x];\nx = Flatten[x]; y = Flatten[y];\n\nsemiaxes[data_] := Module[{min, p, mx, my},\nmin = Min[data];\np = Flatten[data] - min;\np /= Total[p];\nmx = x.p;\nmy = y.p;\nWith[{a = (x - mx)^2.p, b = ((x - mx) (y - my)).p, c = (y - my)^2.p},\nSqrt @ Eigenvalues[{{a, b}, {b, c}}]]]\n\nsemiaxes[data]\n(* {1.86325, 1.50567} *)\n\n\nThis runs in about 340 \u00b5s on my PC\n\nCompiling can give you even more speed, but you need to replace Eigenvalues with the explicit symbolic expressions:\n\nsemiaxesc =\nWith[{x = x, y = y},\nCompile[{{data, _Real, 2}}, Block[{min, p, mx, my, a, b, c},\nmin = Min[data];\np = Flatten[data] - min;\np /= Total[p];\nmx = x.p;\nmy = y.p;\na = (x - mx)^2.p;\nb = ((x - mx) (y - my)).p;\nc = (y - my)^2.p;\n{Sqrt[1/2 (a + c - Sqrt[a^2 + 4 b^2 - 2 a c + c^2])],\nSqrt[1/2 (a + c + Sqrt[a^2 + 4 b^2 - 2 a c + c^2])]}],\nCompilationTarget -> \"C\", RuntimeOptions -> \"Speed\"]];\n\nsemiaxesc[data]\n(* {1.50567, 1.86325} *)\n\n\nThis runs in about 5.7 \u00b5s on my PC.\n\n-\nFantastic. However, I'm getting a \"Eigenvalues::matsq : Argument\" which is saying that don't have a square matrix. Is there some preprocessing I need to do for the data array I posted? \u2013\u00a0 Bob Jun 30 '13 at 9:15\nWould you mind posting the full script you're running? I'm sure there's some formatting error on my end, but I can't pin it down. Everything works and makes sense, but there seems to be a problem with the Eigenvalue computation on my end. \u2013\u00a0 Bob Jun 30 '13 at 12:33\nThat is the full script, apart from the definition of data which I copied from the question. I suggest you step through the code one line at a time and make sure that mx, my, a, b, and c are all single numbers. I'm away from the computer this week, so that's about all the help I can offer at the moment. You could try asking in chat if anyone else can get it working. \u2013\u00a0 Simon Woods Jun 30 '13 at 14:58\nVery good, as long as it's working on your end, I'll be able to fix it on mine. \u2013\u00a0 Bob Jun 30 '13 at 16:50\nThe problem was that the dot products were not being recognized as such. Switching a.b for Dot[a,b] did the trick. Very odd. \u2013\u00a0 Bob Jun 30 '13 at 18:19", "date": "2014-04-17 16:03:47", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/27642/fitting-a-two-dimensional-gaussian-to-a-set-of-2d-pixels/27754", "openwebmath_score": 0.37032657861709595, "openwebmath_perplexity": 2655.350377263357, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9643214491222695, "lm_q2_score": 0.9005297821135384, "lm_q1q2_score": 0.868400184465489}}
{"url": "https://math.stackexchange.com/questions/1996917/product-of-two-monotonic-real-functions", "text": "# Product of two monotonic real functions\n\nI'm confused about the possibility to say that the product of two monotonic real functions $f,g:\\mathbb{R} \\to \\mathbb{R}$ is monotonic.\n\nI found the following proposition:\n\nA. If $g$ and $f$ are two increasing functions, with $f(x)>0$ and $g(x)>0$ $\\forall x$, then $f \\, g$ is increasing.\n\nB. If $g$ and $f$ are two increasing functions, with $f(x)<0$ and $g(x)<0$ $\\forall x$, then $f \\, g$ is decreasing.\n\nDoes something similar hold also for decreasing functions? That is: is the following sentence correct?\n\nA'. If $g$ and $f$ are two decreasing functions, with $f(x)>0$ and $g(x)>0$ $\\forall x$, then $f \\, g$ is decreasing.\n\nB'. If $g$ and $f$ are two decreasing functions, with $f(x)<0$ and $g(x)<0$ $\\forall x$, then $f \\, g$ is increasing.\n\nMoreover, can something be said about the product of a increasing and a decreasing functions?\n\nFor example if $f$ is increasing and $g$ is decreasing under what conditions can I say something about the monotony of the product $f g$?\n\nBesides these two practical questions I would like to ask some suggestions on how to prove statement A.\n\nI tried in the following way\n\n$Hp:$\n\n$x_1 >x_2 \\implies f(x_1)>f(x_2)>0 \\,\\,\\, \\forall x_1,x_2$\n\n$x_1 >x_2 \\implies g(x_1)>g(x_2)>0 \\,\\,\\, \\forall x_1,x_2$\n\n$Th:$\n\n$x_1 >x_2 \\implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \\,\\,\\, \\forall x_1,x_2$\n\n$Proof:$\n\n$f(x_1)>f(x_2)>0 \\,\\,\\, , g(x_1)>g(x_2)>0 \\,\\,\\, \\forall x_1,x_2 \\implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \\,\\,\\, \\forall x_1,x_2$\n\nWhich seems obvious if one thinks about some numbers but I don't really know how I could prove the last implication in rigourous way. So any help in this proof is highly appreciated.\n\nTo prove statement A, let $x > y$. Then, as we know, $f(x) > f(y)>0$ and $g(x) > g(y)>0$. Hence, the following chain of statements proves the claim: \\begin{gather} f(x) > f(y) \\implies g(x)f(x)> g(x)f(y) \\quad(\\because g(x)>0)\\\\ g(x) > g(y) \\implies g(x)f(y) > g(y)f(y) \\quad(\\because f(y)>0)\\\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \\implies fg(x) > fg(y) \\end{gather}\n\nAn analogous proof would follow for part B if $f$ and $g$ were increasing, with a caveat:let $x > y$. Then, as we know, $f(x) > f(y)$ and $g(x) > g(y)$. Hence, the following chain of statements proves the claim: \\begin{gather} f(x) > f(y) \\implies g(x)f(x)< g(x)f(y) \\quad(\\because g(x)<0)\\\\ g(x) > g(y) \\implies g(x)f(y) < g(y)f(y) \\quad(\\because f(y)<0)\\\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \\implies fg(x) < fg(y) \\end{gather}\n\nNow, let us see if the same logic could work with part A':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \\begin{gather} f(x) < f(y) \\implies g(x)f(x)< g(x)f(y) \\quad(\\because g(x)>0)\\\\ g(x) < g(y) \\implies g(x)f(y) < g(y)f(y) \\quad(\\because f(y)>0)\\\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \\implies fg(x) < fg(y) \\end{gather}\n\nThat's brilliant, so great intuition for anticipating part A'. Now we will check part B':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \\begin{gather} f(x) < f(y) \\implies g(x)f(x)> g(x)f(y) \\quad(\\because g(x)<0)\\\\ g(x) < g(y) \\implies g(x)f(y) > g(y)f(y) \\quad(\\because f(y)<0)\\\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \\implies fg(x) > fg(y) \\end{gather}\n\nAnd therefore part B' is also done. Note the above logic carefully, I think all steps are equally important.\n\nUse this logic, and see why in most cases, one increasing and one decreasing function doesn't tell you much about the product itself.\n\nFor A' and B', apply A and B. For A': If $f,g$ are decreasing and positive then $-f$ and $-g$ are increasing and negative so by B, the function $(-f)(-g)=fg$ is decreasing. Similarly, apply A to B'.\n\n$e^x$ is increasing and $e^{-x}+1$ is decreasing, and their product $1+e^x$ is increasing.\n\n$e^x+1$ is increasing and $e^{-x}$ is decreasing, and their product $1+e^{-x}$ is decreasing.\n\nThe product of a positive increasing and a positive decreasing function can also fail to be monotonic.", "date": "2020-10-29 22:28:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1996917/product-of-two-monotonic-real-functions", "openwebmath_score": 0.8550617098808289, "openwebmath_perplexity": 994.7676560267981, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936087546922, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.8683916281290693}}
{"url": "http://math.stackexchange.com/questions/415179/fair-and-unfair-coin-probability", "text": "# Fair and Unfair coin Probability\n\nI am stuck on this question.\n\nA coin with $P(H) = \\frac{1}{2}$ is flipped $4$ times and then a coin with $P(H) = \\frac{2}{3}$ is tossed twice. What is the probability that a total of $5$ heads occurs?\n\nI keep getting $\\frac{1}{6}$ but the answer is $\\frac{5}{36}$.\n\nAttempt: $P($all heads on the four coins$)P($either one of the tosses is heads on the two coins$)+P(3$ heads on the four coins$)P($both coins are heads$)$\n\n$P($all heads on the four coins$) = \\left(\\frac{1}{2}\\right)^4 = \\frac{1}{16}$.\n\n$P($either one of the tosses is heads on the two coins$) = 1-P($no heads on both tosses$) = 1-\\frac{1}{3}\\cdot\\frac{1}{3} = \\frac{8}{9}$.\n\n$P($exactly $3$ heads on the four tosses$) = \\frac{1}{4}$.\n\n$P($both coins are heads$) = \\frac{2}{3}\\cdot\\frac{2}{3} = \\frac{4}{9}$.\n\nFinal Equation: $\\frac{1}{16}\\cdot\\frac{8}{9}+\\frac{1}{4}\\cdot\\frac{4}{9} = \\frac{1}{6}$.\n\nWhy am I off by $\\frac{1}{36}$?\n\n-\nBecause \"either one of...\" $\\not \\equiv$ \" exactly one of...\" \u2013\u00a0Ehsan M. Kermani Jun 9 '13 at 4:31\nYou claim that the problem is with $P($either one of the tosses is heads on the two coins$)$. You need to calculate the probability of there being precisely one occurrence of a head in the tossing of the second coin. In your calculation, you subtracted the probability of no heads from one. The remaining probability covers one head or both heads, but you want to exclude the latter. If you make this adjustment, you will get the correct answer. \u2013\u00a0Michael Albanese May 30 '15 at 19:38\n\nThe required probability will be\n\n$P($exactly $4$ heads from the $4$ flips of $1$st coin$)\\cdot P($exactly $1$ head from the $2$ flips of $2$nd coin $)+$\n\n$P($exactly $3$ heads from the $4$ flips of $1$st coin$)\\cdot P($exactly $2$ heads from the $2$ flips of $2$nd coin $)$\n\nUsing Binomial Distribution, the required probability $$\\binom44\\left(\\frac12\\right)^4\\left(1-\\frac12\\right)^{4-4} \\cdot\\binom21\\left(\\frac23\\right)^1\\left(1-\\frac23\\right)^{2-1}$$\n\n$$+\\binom43\\left(\\frac12\\right)^3\\left(1-\\frac12\\right)^{4-3} \\cdot\\binom22\\left(\\frac23\\right)^2\\left(1-\\frac23\\right)^{2-2}$$\n\n$$=\\frac1{16}\\cdot\\frac49+\\frac14\\cdot\\frac49=\\frac5{36}$$\n\n-", "date": "2016-07-23 17:08:48", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/415179/fair-and-unfair-coin-probability", "openwebmath_score": 0.7746295928955078, "openwebmath_perplexity": 909.9909338425831, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587247081928, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8683848945980446}}
{"url": "https://www.themathdoctors.org/parallel-vectors-missing-a-solution/", "text": "# Parallel Vectors: Missing a Solution\n\n#### (A new question of the week)\n\nWe were recently asked to check work on an interesting little question about parallel vectors, and I was almost convinced that there was no solution \u2026 until I realized there was one! How was it missed? How can we avoid doing that? That\u2019s our goal today.\n\n## When will these vectors be parallel?\n\nHere is the question, from Brine in mid-September:\n\nHere is my work:\n\nBrine is using column-vector form for the vectors; I\u2019ll be using the more compact bracket notation for convenience.\n\nHe\u2019s found the vectors $$\\overrightarrow{PQ}=\\left\\langle3,x^2\\right\\rangle-\\left\\langle x,0\\right\\rangle=\\left\\langle3-x,x^2\\right\\rangle$$ and $$\\overrightarrow{P_1Q_1}=\\left\\langle 6,x^2\\right\\rangle-\\left\\langle2x,1\\right\\rangle=\\left\\langle6-2x,x^2-1\\right\\rangle$$, and then wrote an equation to say that the components of the two vectors are proportional: $$\\frac{6-2x}{3-x}=\\frac{x^2-1}{x^2}$$ Solving this yields only an imaginary solution, so it appears to be impossible for the vectors to be parallel.\n\nHi, Brine.\n\nSo your answer is, that there is no such real number x?\n\nYour work almost had me convinced, until I tried solving a different way, and saw that there is an answer.\n\nLook carefully at your work, and think about whether there is any step in which you made an unstated assumption\n\n### How to recognize unstated assumptions\n\nI had graphed the situation in GeoGebra, in such a way that I could vary x and observe the two vectors, to see how the problem worked, expecting perhaps to see why there would be no solution. Here is the case $$x=1.5$$, for example:\n\nBut I found that I could make them parallel (as we\u2019ll see below). That gave me reason to look more closely at the work. I\u2019d been fooled by following along with his work rather than starting from scratch to solve the problem myself. (That\u2019s a risk even when you check over your own work, because your mind will follow the same paths again. I should have known better, and I soon did!)\n\nBrine replied,\n\nI tried again but unfortunately I keep ending up that they are not parallel. Can you please give me a hint?\n\nI answered with a more detailed hint:\n\nAs I said before, \u201cthink about whether there is any step in which you made an unstated assumption\u201c. Here are several points in your work that you should think about:\n\nFirst, you wrote this equation to say that the vectors are parallel:\n\nWhen can\u2019t you write such an equation? What fact are you using to justify the equation?\n\nSecond, here you canceled to simplify a fraction. When can\u2019t you cancel?\n\nOr you could think of this more generally as a rational equation. What is the domain of this equation? What does simplifying change?\n\nThird, here you multiplied both sides of an equation. When can\u2019t you do that?\n\nNot all of these represent actual errors in your work; but they represent situations where you are making unstated assumptions, or ignoring special cases.\n\nThinking about exactly what you are doing, and what justifies each step, can be a valuable exercise when you have a wrong (extraneous or missing) solution. The answers to my questions are \u2026\n\n1. He\u2019s justifying the proportion because parallel vectors have proportional components \u2026 but that isn\u2019t true when any of those components are zero.\n2. You can\u2019t cancel when the factor you cancel can be zero.\n3. Simplifying by canceling can change the domain, introducing extraneous roots that are valid in the new equation, but are not in the domain of the original.\n4. You can\u2019t multiply an equation by a factor that might be zero, because the new equation may be true when the original is not.\n\nWhich is the key issue? We\u2019ll see.\n\n### Examples\n\nThe same sorts of problems can arise in more familiar problems, so I pointed out a couple examples:\n\nFor a similar issue, when you want to check whether two lines are parallel, you have to consider two cases: Either their slopes are equal or \u2026 what?\n\nFor another similar issue, consider how to solve an equation like x2 = 2x. It seems obvious that you can divide by x and get x=2; but that misses a solution. Why? Because you might be dividing by 0, which isn\u2019t legal. When you divide by x, you are implicitly assuming that x\u22600, which you don\u2019t know to be true. The better way is to rearrange and factor: x2 -2x = 0 factors as x(x-2) = 0, and the solutions are x=0 and x=2.\n\nI\u2019m intentionally trying not to give direct hints too soon, because this is a very valuable lesson to learn; as I indicated, I was almost fooled, so it is rather subtle. But it happens a lot, and I want you to have a big \u201caha\u201d moment you won\u2019t forget.\n\nIn checking for parallel lines, checking the slopes won\u2019t catch a case where they don\u2019t have slopes (vertical lines); that is the most direct hint I gave here. The second example illustrates how a solution can be missed because of an unstated assumption (that what you divide by is not zero).\n\n### Finding the answer by an imperfect method\n\nBrine wrote back:\n\nOh ok I understand you. Is this the answer? 3?\n\nThis time Brine has cross-multiplied without first canceling: $$\\frac{6-2x}{3-x}=\\frac{x^2-1}{x^2}\\$$6-2x)(x^2)=(x^2-1)(3-x)\\\\6x^2-2x^3=3x^2-x^3-3+x\\\\-x^3+3x^2-x+3=0 and then solved the cubic equation by factoring (by grouping). This gave two solutions: the imaginary solutions \\(x=\\pm i$$ we saw before, and also $$x=3$$. This work is better, but \u2026 I answered, Yes, that\u2019s correct. What you\u2019ve done here is, I think, to cross-multiply your equation (6-2x)/(3-x) = (x2-1)/x2, and solve the resulting cubic. This amounts to factoring instead of dividing, as I recommended, though you started with a division (in writing the equation in the first place). Presumably you then checked that x = 3 worked. Another way to think of it is to go ahead and do what you originally did, but then check whether canceling the x-3 factors eliminated a solution. That is, one could just recognize that it was assumed that $$x-3\\ne0$$, and check to see if the opposite assumption produces a solution. Rather than avoiding the assumption, we would be making the assumption explicit, and taking the contrary assumption as a second case. ### The equation doesn\u2019t fully represent the problem Unfortunately, that check would fail! He\u2019s missed a more fundamental error: But did you observe that, when you put x=3 into your initial equation, you get 0/0 = 8/9? So 3 is not actually a solution of that equation! (It would be called extraneous.) And it isn\u2019t obvious that it is a solution of the actual problem, until you check it in the original problem. The real problem arises earlier than that: The equation doesn\u2019t really represent the problem. I generally tell students to check their answer, not by plugging it into their equation, but by checking if it works in the problem itself. When the equation is wrong, this will catch it. But here, the equation turns out to be wrong but our answer is correct! So how do you check the solution? You have go back to the problem: Let P(x,0), Q(3,x2), P1(2x,1), and Q1(6,x2). Find all possible values of x \u2208 R such that PQ || P1Q1. If x=3, the points are P(3,0), Q(3,9), P1(6,1), and Q1(6,9), and the vectors are PQ = <0,9>, P1Q1 = <0,8>. Are they parallel? Yes, because both are vertical \u2014 a case not covered by your equation! That is the ultimate cause of your difficulty. This is what I\u2019d found graphically (for $$x=3$$), which made me look again: So, how can we solve the problem without making the unintentional assumption that the vectors are not vertical? ### Back to square one: What really makes parallel vectors parallel? What you never wrote, at the start, was what defines parallel vectors. I asked, \u201cWhat fact are you using to justify the equation?\u201d, and I\u2019d still like an answer: What perspective were you taking on parallelism when you wrote that equation? What have you been taught about the meaning of \u201cparallel\u201d? Two vectors are parallel if one is a scalar multiple of the other. In two dimensions, <a,b> and <c,d> are parallel if there is a non-zero scalar k such that <c,d> = k<a,b>; i.e. c = ka and d = kb. You have translated this to c/a = d/b = k; but that assumes that a and b are both non-zero. If you use this form, you need to separately check the contrary case. Equivalently, one could take vectors to be parallel if their direction angles are the same, and then compare the tangents of those angles (their slopes): b/a = d/c. But those don\u2019t exist for vertical vectors. If you do that, you need to separately consider the latter possibility. His proportion was$$\\frac{6-2x}{3-x}=\\frac{x^2-1}{x^2}.$$So, using his form, my \u201cc/a = d/b\u201d, you need to ask what happens when a denominator is zero, namely $$3-x=0$$ or $$x^2=0$$. The former leads to our missing solution. If you used the \u201cb/a=d/c\u201d form, representing equal slopes,$$\\frac{x^2}{3-x}=\\frac{x^2-1}{6-2x},$$you\u2019d need to check when these denominators are zero, and in this case both are zero for $$x=3$$ (that is, both vectors are vertical). So if I were doing this carefully, I would not have started with the equality of two fractions, but with the definition using scalar multiplication: PQ = <3-x,x2> P1Q1 = <6-2x,x2-1> These are parallel if P1Q1 = kPQ, so that 6-2x = k(3-x) x2-1 = kx2 From the first equation, either k = (6-2x)/(3-x) = 2, or x = 3. If k = 2, the second equation implies that x2 = -1, which is impossible. If x = 3, then PQ = <0,9> and P1Q1 = <0,8>, and these are indeed parallel, with k = 8/9. The solution of that first equation would look like this, done carefully:$$6-2x = k(3-x)\\\\2(3-x)-k(3-x)=0\\$$2-k)(3-x)=0\\\\k=2\\text{ or }x=3 Or I might eliminate k from the equations by writing your rational equation, while keeping in mind that in dividing by x2 and by x-3, I was assuming both are non-zero, and check those cases. If we check the case \\(x^2=0$$, we find that $$\\overrightarrow{PQ}=\\left\\langle3,0\\right\\rangle-\\left\\langle 0,0\\right\\rangle=\\left\\langle3,0\\right\\rangle$$ and $$\\overrightarrow{P_1Q_1}=\\left\\langle 6,0\\right\\rangle-\\left\\langle0,1\\right\\rangle=\\left\\langle6,-1\\right\\rangle$$. These are not parallel.\n\nMerely recognizing that your equation does not allow x = 3, and checking that separately, would also be valid, though it\u2019s better to understand why. Just solving without canceling, as you have now done, is less appropriate, but it does lead to the answer (as long as you check it).\n\nThis is a very interesting problem, isn\u2019t it?\n\nBrine closed it out:\n\nYes you are right. I was actually taught that parallel vectors are parallel if they have a k value such that a=kb (where a and b are vectors) I forgot to apply this rule because in grade 12 I was taught by equaling the two slopes if they are parallel. I did learn the other method in the previous grade too but I forgot and I do remember it. The way you showed is way better that the one I did. Thank you so much for teaching me Dr. Peterson!\n\nSometimes we learn a shortcut method, or transfer it from one topic to another, and neglect to pay attention to conditions that are attached to them. It\u2019s a good lesson!\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2022-12-05 04:51:12", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/parallel-vectors-missing-a-solution/", "openwebmath_score": 0.7740286588668823, "openwebmath_perplexity": 504.6069346917069, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561721629776, "lm_q2_score": 0.8962513662057089, "lm_q1q2_score": 0.8683386679579022}}
{"url": "https://math.stackexchange.com/questions/1194584/the-total-number-of-subarrays", "text": "# The total number of subarrays\n\nI want to count the number of subarrays for a vector (not combinations of elements).\nEx.\n\nA[1,2,3]\n\nIt has 6 subarrays :\n\n{1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}\n\nI think that for a vector of N elements the total number of subarrays is N*(N+1)/2.\nI am not able to prove it, can someone do it?\n\n\u2022 Welcoming the Math.SE! Here we have a culture of showing what you've tried, as it helps both the responders to give better answers and you to get a better understanding! :) \u2013\u00a0frogeyedpeas Mar 17 '15 at 22:08\n\u2022 Looks you forgot to list $\\{3, 1\\}$ \u2013\u00a0AgentS Mar 17 '15 at 22:13\n\u2022 No! I didn't forget {3,1}. A subarray has to have contiguous elements \u2013\u00a0user72708 Mar 18 '15 at 13:56\n\nSuppose that your vector is $\\langle a_1,a_2,\\ldots,a_n\\rangle$. Imagine a virtual element $a_{n+1}$ at the end; it doesn\u2019t matter what its value is. A subarray is completely determined by the index of its first element and the index of the element that immediately follows its last element. For example, the subarray $\\langle a_3,\\ldots,a_{n-2}\\rangle$ is determined by the indices $3$ and $n-1$, the subarray $\\langle a_k\\rangle$ is determined by the indices $k$ and $k+1$, and the subarray $\\langle a_2,\\ldots,a_n\\rangle$ is determined by the indices $2$ and $n+1$. Moreover, each pair of distinct indices from the set $\\{1,2,\\ldots,n+1\\}$ uniquely determines a subarray. Thus, the number of subarrays is the number of pairs of distinct indices from the set $\\{1,2,\\ldots,n+1\\}$, which is\n\n$$\\binom{n+1}2=\\frac{n(n+1)}2\\;.$$\n\n\u2022 I am confused why you use the index of the element that immediately follows its last element instead of just the last element. I was thinking the answer should be n C 2, instead of n+ 1 C 2 that you mentioned. Please correct me wherever I am wrong. \u2013\u00a0Abhishek Bhatia Jul 7 '18 at 19:33\n\u2022 @Abhishek , You are right, It is n C 2 where we did not calculate the single character substrings. There can be n single character substring . So the result is (n C 2 + n C 1) = (n+1) C 2 . \u2013\u00a0shuva Oct 31 '18 at 20:15\n\nConsider an arbitrary array of N DISTINCT ELEMENTS (if the elements are the same then I am afraid the formula you are seeking to prove no longer works!).\n\nNaturally there exists 1 array consisting of all the elements (indexed from 0 to N-1)\n\nThere exist 2 arrays consisting of N-1 consecutive elements (indexed from 0 to N-2)\n\nand in general there are k arrays consisting of N-k+1 consecutive elements (indexed from 0 to N-k-1)\n\nProof:\n\nWe can access elements 0 ... N-k-1 as the first array, then 1 ... N-k+2 is the second array, and this goes on for all N-k+r until N-k+r = N-1 (ie until we have hit the end). The r that does us is can be solved for :\n\n$$N-k+r = N-1 \\rightarrow r -k = -1 \\rightarrow r = k-1$$\n\nAnd the list $$0 ... k-1$$ contains k elements within it\n\nThus we note that the total count of subarrays is\n\n1 for N elements\n\n2 for N-1 elements\n\n3 for N-2 elements\n\n.\n\n.\n\n.\n\nN for 1 element\n\nAnd the total sum must be:\n\n$$1 + 2 + 3 ... N$$\n\nLet us see if your formula works\n\nif:\n\n$$1 + 2 +3 ... N = \\frac{1}{2}N(N+1)$$\n\nthen\n\n$$1 + 2 + 3 ... N+1 = \\frac{1}{2}(N+1)(N+2)$$\n\nWe verify:\n\n$$\\frac{1}{2}N(N+1) + N+1 = (N+1)(\\frac{1}{2}N + 1) = (N+1)\\frac{N+2}{2}$$\n\nSo you're formula does indeed work! Now we verify that for N = 1\n\n$$\\frac{1*(1+1)}{2} = 1$$\n\nAnd therefore we can use the above logic to show that for any and ALL whole numbers N the formula works!\n\nThis calculation can be seen as an arithmetic series (i.e. the sum of the terms of an arithmetic sequence).\n\nAssuming the input sequence: $$(a_0, a_1, \\ldots, a_n)$$, we can count all subarrays as follows:\n\n\\begin{align} \\; 1 & \\; \\text{subarray from} \\; a_0 \\; \\text{to} \\; a_{n-1}\\\\ + \\; 1 &\\; \\text{subarray from} \\; a_1 \\; \\text{to} \\; a_{n-1}\\\\ & \\; \\ldots \\\\ + \\; 1 & \\; \\text{subarray from} \\; a_{n-1}\\; \\text{to} \\; a_{n-1}\\\\ = & \\; n \\end{align}\n\n$$+$$\n\n\\begin{align} \\; 1 & \\; \\text{subarray from} \\; a_0 \\; \\text{to} \\; a_{n-2}\\\\ + \\; 1 &\\; \\text{subarray from} \\; a_1 \\; \\text{to} \\; a_{n-2}\\\\ & \\; \\ldots \\\\ + \\; 1 & \\; \\text{subarray from} \\; a_{n-2}\\; \\text{to} \\; a_{n-2}\\\\ = & \\; n-1\\\\ \\end{align}\n\n$$+ \\; \\ldots$$\n\n\\begin{align} \\; \\; \\; 1 & \\; \\text{subarray only containing} \\; a_0\\\\ = & \\; 1\\\\ \\end{align}\n\nwhich results in the arithmetic series: $$n + n-1 + \u2026 + 1$$.\n\nThe above can also be represented as $$\\sum_{i=1}^{n}i\\;$$ and adds up to $$n (n+1)/2$$.\n\nElaborating on Brian's answer, lets assume we count all the single character sub-strings. There can be n such single character sub-string. By using the Binomial Coefficient, $$\\binom{n}{r}$$ notation, we can say n = $$\\binom{n}{1}$$ .\n\n$$total_1 = n = \\binom{n}{1}$$\n\nNow let's assume we count all the sub-string that are not single character. Because we have to choose the beginning and end of the sub-string regardless of the order, the count should be $$\\binom{n}{2}$$.\n\n$$total_* = \\binom{n}{2}$$\n\nNow the total number of sub-string,\n\n$$total = total_1 + total_* = \\binom{n}{1} + \\binom{n}{2}$$\n\nNow recall Pascal triangle and recurrence relation $$\\binom{n+1}{r} = \\binom{n}{r} + \\binom{n}{r-1}$$. So we can write,\n\n$$total = total_1 + total_* = \\binom{n}{1} + \\binom{n}{2} = \\binom{n+1}{2} = n(n+1)/2$$\n\nThere we have the mathematical deduction. (I kind of feel we do not need the fancy recurrence relation to get the final answer though).\n\n\u2022 Use LaTeX please. \u2013\u00a0Michael Rozenberg Oct 31 '18 at 20:31\n\nTrying to explain in layman terms.\n\nLet's say f(0) = 0\n\nf(1) = 1\nf(2) = 3\nf(3) = 6\nf(4) = 10\nf(5) = 15\n\nBy observation, you can see that each result is just an addition of previous result and current number.\n\nf(n) = n + f(n-1)\n\nSo with this formula let's expand f(5).\n\nf(5)\n=> 5 + f(4)\n=> 5 + 4 + f(3)\n=> 5 + 4 + 3 + f(2)\n=> 5 + 4 + 3 + 2 + f(1)\n=> 5 + 4 + 3 + 2 + 1 + f(0)\n=> 5 + 4 + 3 + 2 + 1 + 0\n===> This is equal to sum of n numbers = n(n+1)/2", "date": "2021-06-24 22:03:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1194584/the-total-number-of-subarrays", "openwebmath_score": 0.9221491813659668, "openwebmath_perplexity": 742.3973683653206, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126513110864, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8683183703884247}}
{"url": "https://www.physicsforums.com/threads/given-the-planar-curve-find-the-equation-of-the-plane.889281/", "text": "Given the planar curve, find the equation of the plane\n\nTags:\n1. Oct 15, 2016\n\ndlacombe13\n\n1. The problem statement, all variables and given/known data\n$r(t) = < 2e^t - 5 , e^t +3t^2 , 4t^2 +1>$\nIs a curve that lies within a plane. Find the equation of this plane.\n\n3. The attempt at a solution\nI am not sure if my approach is correct. These are my results:\n$x=2e^t - 5$\n$y = e^t +3t^2$\n$z = 4t^2 + 1$\n\n$z = 4t^2 + 1 ~~\\Rightarrow~~ t = \\sqrt{\\frac{z-1}{4}}$\n\n$x = 2e^\\sqrt{\\frac{z-1}{4}} - 5 ~~\\Rightarrow~~ \\frac{x+5}{2} =e^\\sqrt{\\frac{z-1}{4}}$\n$y = e^\\sqrt{\\frac{z-1}{4}} + \\frac{3z-3}{4}$\n\n$y = \\frac{x+5}{2} + \\frac{3z-3}{4}$\n$= \\frac{x}{2} + \\frac{5}{2} + \\frac{3z}{4} - \\frac{3}{4}$\n\n$= \\frac{1}{2}x - y + \\frac{3}{4}z = -\\frac{15}{8}$\n\n2. Oct 15, 2016\n\npasmith\n\nOne way of specifying a plane is as $\\{ \\lambda \\mathbf{a} + \\mu\\mathbf{b} + \\mathbf{c} : (\\lambda , \\mu) \\in \\mathbb{R}^2\\}$ for given vectors $\\mathbf{a}$, $\\mathbf{b}$ and $\\mathbf{c}$ where $\\mathbf{a}$ and $\\mathbf{b}$ are linearly independent. Then for non-constant functions $f: \\mathbb{R} \\to \\mathbb{R}$ and $g: \\mathbb{R} \\to \\mathbb{R}$ we have that $$\\mathbf{r}(t) = f(t)\\mathbf{a} + g(t)\\mathbf{b} + \\mathbf{c}$$ is a curve which lies on this plane.\n\n3. Oct 15, 2016\n\nRay Vickson\n\nEasier: we have $2y- x = 6 t^2 + 5$, which has eliminated the $e^t$ terms. Now to eliminate the $t^2$ terms, just add or subtract a suitable multiple of $z$. That will leave you with a constant, having no $t$ in it anywhere.\n\n4. Oct 15, 2016\n\nStaff: Mentor\n\n\"I am not sure if my approach is correct.\"\nYou can (and really should) check your final equation. To do this choose three t values in your equation for r(t). That will give you three vectors, the endpoints of which lie in the plane. These three points should satisfy the plane equation you ended with. If all three points work, then you can be 99% sure that your work is correct. (Subtracting a tiny amount for arithmetic errors you might make). If the three points don't satisfy the plane equation, that's a sign that you have done something wrong.\n\n5. Oct 15, 2016\n\nLCKurtz\n\nI didn't check all your work, but look at your last 3 lines. You start out with $y =$ and end up with $= \\frac{1}{2}x - y + \\frac{3}{4}z = -\\frac{15}{8}$.\n\nDoes $y=$ that or is your final equation supposed to be $\\frac{1}{2}x - y + \\frac{3}{4}z = -\\frac{15}{8}$? If that is your answer then notice that $\\vec r(0) = \\langle -3,1,1\\rangle$, so that point must be on the plane. It doesn't seem to satisfy that last equation. It may be a simple arithmetic mistake.\n\nBut one reason I didn't check your work is that I don't think eliminating the parameter is a sensible way to do the problem. I would suggest a different approach. All you need for a plane is a point and a normal vector. It's easy to get a point on the plane. Then notice that given the curve is planar, that means the tangent vectors to the curve must be in the plane. So you could calculate $\\vec r'(0)\\times \\vec r'(1)$ to get a normal.\n\n6. Oct 15, 2016\n\ndlacombe13\n\nThank you all for your replies. I tried plugging in t=0,1,2 into each parameter, and plugged it into the equation for the plane. For all three points, I get: -1.75. -15/8 = -1.875.\nThank you all for you're replies, I tried checking my values and got close to -15/8, but I don't like the error, since I think my arithmetic is right. Also, my original thinking was exactly as LCKurtz said. However I ran into issues, but I think that is because I did it wrong. I used:\n$T = \\frac{r'(t)}{|r'(t)|}$\nT x r(t)\nBut it didn't seem to work. So what you're saying is I just need to take find two random tangent points along the curve, which will give me two random vectors on the plane it is within, and cross them which will yield an orthogonal vector (n), and then just form the equation from it?\n\n7. Oct 15, 2016\n\ndlacombe13\n\nThanks everyone, I did get the equation finally, using r'(0) x r'(1) = <8,-16,12> and got the equation:\n8x - 16y +12z +28 = 0\nI verified it by plugging in points, and it works out. One last question before I go...\nWhy didn't my attempt at using T, the unit tangent vector work? I mean it is still a vector that is in the direction of the curve, and thus on the plane, right?\n\n8. Oct 15, 2016\n\nLCKurtz\n\nYes, it should have worked. But dividing by the magnitude is unnecessary, adds square roots, and makes the calculations more error prone, which apparently got you.", "date": "2017-08-21 10:44:47", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/given-the-planar-curve-find-the-equation-of-the-plane.889281/", "openwebmath_score": 0.7069550156593323, "openwebmath_perplexity": 272.88303186061523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126506901791, "lm_q2_score": 0.8872045832787205, "lm_q1q2_score": 0.8683183494051924}}
{"url": "https://grindskills.com/convergence-in-distribution-to-a-degenerate-distribution/", "text": "# Convergence in distribution to a degenerate distribution\n\nThis question came up based on a disagreement I had with a TA. This was the specific example:\n\nLet $$X_{1},\u2026,X_{n}X_{1},...,X_{n}$$ be an iid random sample from a population with pdf $$f(x)=3(1-x)^2, 0. The $$nthnth$$ order statistic is represented as $$X_{(n)}X_{(n)}$$.\n\nQuestion: Find a constant $$vv$$ such that $$n^v(1-X_{(n)})n^v(1-X_{(n)})$$ converges in distribution.\n\nI believe that the intent of the question was to prompt the invocation of Slutsky's theorem (or maybe not); however, to which distribution the presented snippet was supposed to converge was not specified. So I presented a lazy alternative answer as follows:\n\nAs the sample size approaches infinity, $$X_{(n)}X_{(n)}$$ becomes arbitrarily close to 1. Therefore, we can simply set $$v=0v=0$$ and the statement will converge to a degenerate distribution. In this case, the order statistic converges to 1, so with $$v=0v=0$$, the statement converges to 0. I later realized that it actually converges to 1 after the transformation, which is reflected in the final degenerate distribution written toward the bottom of the question. I was surprised when the TA said that 0 is not a random variable or a distribution, and my answer made no sense. I second guessed myself and went further:\n\nThe cdf of the original distribution is $$F(x)=x^3-3x^2+3xF(x)=x^3-3x^2+3x$$ Using this, I derived the pdf of the order statistic,\n$$f_{X_{(n)}}(x)=3n(x^2-2x+1)(x^3-3x^2+3x)^{n-1} f_{X_{(n)}}(x)=3n(x^2-2x+1)(x^3-3x^2+3x)^{n-1}$$\nAnd the cdf of the order statistic, $$f_{X_{(n)}}(x)=(x^3-3x^2+3x)^nf_{X_{(n)}}(x)=(x^3-3x^2+3x)^n$$\n\nI then used this to graph the cdf of the order statistic with $$n=1, n=100, n=1,000,000,000n=1, n=100, n=1,000,000,000$$. Predictably, the graph showed the cdf getting thinner and steeper, until with huge samples it visually looks like a vertical line at $$X_{(n)}=1X_{(n)}=1$$.\n\nI reiterated my argument, and showed the visualization: it is converging to a degenerate distribution. Again, I was rebuffed. After pestering the TA and asking the instructor for the course, I still have no answer for why my answer was wrong, but I can't keep bothering them about it.\n\nCan someone here tell me if my argument is solid, and if not, tell me specifically what I've done that is in error?\n\nEDIT: Here is the final formalism that I gave to tie my argument together. Following Casella and Berger's definition,\n\nA sequence of random variables, $$X_{1}, X_{2},...,X_{1}, X_{2},...,$$ converges in distribution to a random variable $$XX$$ if $$\\lim_{n \\to \\infty} F_{X_n}(x) = F_X(x)\\lim_{n \\to \\infty} F_{X_n}(x) = F_X(x)$$ at all points $$xx$$ where $$F_X(x)F_X(x)$$ is continuous\n\nSince $$\\lim_{n \\to \\infty} F_{X_(n)}(x) = 1\\lim_{n \\to \\infty} F_{X_(n)}(x) = 1$$\nAnd we can define a degenerate variable $$YY$$ with the cdf\n$$F_Y(x)=\\begin{cases}1, & x\\ge1\\\\ 0, & else \\\\ \\end{cases}F_Y(x)=\\begin{cases}1, & x\\ge1\\\\ 0, & else \\\\ \\end{cases}$$\nWe can say\n$$\\lim_{n \\to \\infty} F_{X_(n)}(x) = F_Y(x) \\lim_{n \\to \\infty} F_{X_(n)}(x) = F_Y(x)$$\n\nSo by definition, $$X_{(n)}X_{(n)}$$ converges in distribution to the degenerate distribution of $$YY$$.\n\nEDIT: Adding a bounty to this, as it has become important. I did not formally perform the transformation $$Y=1-X_(n)Y=1-X_(n)$$ above, but doing so results in $$YY$$ having a distribution that still converges to the same degenerate distribution above. I was now told that my answer does not prove convergence in distribution at all. Question is answered if someone can definitively prove either 1) my answer is correct, and show exactly how you would have gone about proving it formally and rigorously, or 2) my answer is wrong, exactly why it is wrong, and prove any correct answer that uses degenerate distributions or degenerate variables if such an answer exists.\n\nYour answer is correct (assuming that you have accurately transcribed the question). The proof:\n\nLet $$F_n(c)F_n(c)$$ be the cdf of $$(1 - X_{(n)})(1 - X_{(n)})$$, where $$X_{(n)}X_{(n)}$$ is the greatest element in a sample of size $$nn$$. Let $$F(c)F(c)$$ be the cdf for the constant 0 distribution.\n\nFor $$c < 0c < 0$$, of course $$F_n(c) = 0 = F(c)F_n(c) = 0 = F(c)$$.\n\nFor $$c > 1c > 1$$, of course $$F_n(c) = 1 = F(c)F_n(c) = 1 = F(c)$$.\n\nFor $$0 < c \\le 10 < c \\le 1$$:\n\n\\begin{align} F_n(c) &= P(1 - X_{(n)} \\le c) \\\\ &= P(X_{(n)} \\ge 1-c) \\\\ &= 1-P(X_1 < 1-c, ..., X_n < 1-c) \\\\ &= 1 - P(X_1 < 1-c)^n \\to 1 = F(c) \\end{align} \\begin{align} F_n(c) &= P(1 - X_{(n)} \\le c) \\\\ &= P(X_{(n)} \\ge 1-c) \\\\ &= 1-P(X_1 < 1-c, ..., X_n < 1-c) \\\\ &= 1 - P(X_1 < 1-c)^n \\to 1 = F(c) \\end{align}\n\nAnd the case $$c = 0c = 0$$ doesn't matter, because $$FF$$ isn't continuous at $$00$$.\n\nIf the people you are arguing with don't realise that convergence in distribution to a constant is a thing, you could point them to e.g. Wikipedia's Proofs of convergence of random variables article.", "date": "2022-09-29 02:47:15", "meta": {"domain": "grindskills.com", "url": "https://grindskills.com/convergence-in-distribution-to-a-degenerate-distribution/", "openwebmath_score": 0.9997840523719788, "openwebmath_perplexity": 3265.221171092022, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126506901791, "lm_q2_score": 0.8872045832787204, "lm_q1q2_score": 0.8683183494051923}}
{"url": "https://math.stackexchange.com/questions/1767029/did-feynman-mentally-compute-sqrt31729-03-by-linear-approximation/1767244", "text": "# Did Feynman mentally compute $\\sqrt[3]{1729.03}$ by linear approximation?\n\nIn the biopic \"infinity\" about Richard Feynman. (12:54) He computes $\\sqrt[3]{1729.03}$ by mental calculation. I guess that he uses linear approximation. That is, he observe that $1728=12^3$. Let $f(x)=\\sqrt[3]{x}$. Then $f'(x)=\\frac{1}{3\\sqrt[3]{x^2}}$ and $f'(1728)=\\frac{1}{3\\sqrt[3]{1728^2}}=\\frac{1}{3\\cdot 12^2}$. Therefore, $$\\sqrt[3]{1729.03}=f(1729.03)\\approx f(1728)+f'(1728)(1729.03-1728)=12+\\frac{1.03}{3\\cdot 12^2}=12.002384\\overline{259}.$$\n\nQuestion 1. If he used the linear approximation, how did he compute $\\frac{1.03}{3\\cdot 12^2}=0.002384\\overline{259}$ by a mental calculation?\n\nQuestion 2. If he didn't use the linear approximation, what is another method he might have used?\n\n\u2022 You're asking about how a mental computation was accomplished in a scene from a fictional movie...? (Just clarifying; I'm not willing to click a youtube link to get the content of the question.) \u2013\u00a0Andrew D. Hwang May 1 '16 at 18:04\n\u2022 @AndrewD.Hwang Yes. Although it is a fictinal movie, I think that maybe the scenarist had a method to approximate $\\sqrt[3]{1729.03}$. And I hope this scene can motivate my student to learn linear approximation. \u2013\u00a0bfhaha May 1 '16 at 18:12\n\u2022 $1.03/3\\approx.34333\\approx.34332$ $.34332/144=.05722/24=.02861/12\\approx.002384$. \u2013\u00a0Mark S. May 1 '16 at 18:30\n\u2022 You should mention what value was computed so that the question is self-contained. \u2013\u00a0Bill Dubuque May 1 '16 at 18:41\n\u2022 @Andrew D. Hwang It is mentioned in his book \"Surely You're joking, Mr. Feynman!\". Give it a shot. \u2013\u00a0user537100 Jul 1 '18 at 13:09\n\nFeynman tells the story in one of his books of anecdotes.\n\nhttp://www.ee.ryerson.ca/~elf/abacus/feynman.html\n\n$12$ is a very good first approximation and the linear term of the series expansion suffices to get high precision.\n\n$$\\sqrt[3]{1728 + d} = 12\\sqrt[3]{1+x} = 12 + 4x + O(x^2)$$\n\nwhere $d = 1.03$ and $x = \\frac{d}{1728}$ is, in Feynman's words, about 1 part in 2000, so that the error term is of order $10^{-6}$.\n\nFeynman says that he computed $12 + \\frac{4d}{1728}$ as the approximate value.\n\nThe number was 1729.03. I happened to know that a cubic foot contains 1728 cubic inches, so the answer is a tiny bit more than 12. The excess, 1.03 is only one part in nearly 2000, and I had learned in calculus that for small fractions, the cube root's excess is one-third of the number's excess. So all I had to do is find the fraction 1/1728, and multiply by 4 (divide by 3 and multiply by 12). So I was able to pull out a whole lot of digits that way.\n\nHe describes that as though $d=1$ for this part of the calculation, so maybe $12 + \\frac{1}{432}$ was what he actually computed. By \"adding two more digits\" (to 12.002) he seems to mean working out the division in the fraction. It could also mean adding (0.03)/432 as two more decimal digits of accuracy to $(12 + 432^{-1})$, which requires only a multiplication by 3 of an already computed quantity 1/432.\n\nFeynman's method is the one that would have been immediate for anyone familiar with the binomial series and with $12^3 = 1728$. He said that knew the latter as ft^3/in^3 and other people might know it from the Ramanujan 1729 story. The other ingredient, as Feynman says in the story, was being good at integer division.\n\nI upvoted the other answer as it comes from the other book of anecdotes, but one way to calculate this without Feynman's experience might be the following.\n\nYou can start with the linear approximation, and then you have to calculate $(\\dfrac{1.03}{3})/12^2$. $\\dfrac{1.03}{3}\\approx .34333$, but that doesn't lend itself to division by $12^2$, so you can change the last digit to get $\\dfrac{1.03}{3}\\approx .34332$\n\n(You know this will be helpful by the divisibility test for $3$, and can note it's also divisible by $2$ and $4$ if you're thinking ahead.)\n\nThen $\\dfrac{.34332}{12^2}=\\dfrac{.05722}{12*2}=\\dfrac{.02861}{12}\\approx.002384$.\n\n\u2022 \"linear approximation\" here means taking the linear part of $(12 + x)^3$ when solving $(12 + x)^3 = 12^3 + 1.03$. Then $3x(12^2) = 1.03$ and the rest is as in the answer. Thinking of it as 1 or 1.03 divided by 432, how to divide by 432 with minimum calculation? The error in approximating that 432 by 500 is (1/432 - 1/500) = (500- 432)/500*432 or about 68/200000. This gives 0.002 + 0.00034 = 0.00234 using only division by 2, 4*5=20, and 100-32=68. \u2013\u00a0zyx May 2 '16 at 2:49\n\u2022 Thanks MarkS. You and zyx gave the best answer for me. But it can only has one answer. So please forgive me to use upvote instead of mark it as an answer. \u2013\u00a0bfhaha May 2 '16 at 5:12", "date": "2021-07-28 13:39:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1767029/did-feynman-mentally-compute-sqrt31729-03-by-linear-approximation/1767244", "openwebmath_score": 0.8204992413520813, "openwebmath_perplexity": 558.7350380376656, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446471538802, "lm_q2_score": 0.891811053345418, "lm_q1q2_score": 0.8683070583624297}}
{"url": "https://questioncove.com/updates/4e56c14a0b8b9ebaa893644f", "text": "Ask your own question, for FREE!\nMathematics\nOpenStudy (waheguru):\n\nis there a special rick or is it just trial and error to solve this porblem\n\nOpenStudy (waheguru):\n\nOpenStudy (waheguru):\n\nit would take a really long time to find he asnwer by trial and answer\n\njimthompson5910 (jim_thompson5910):\n\nBest way to do this is to break things down. Start with 100. Clearly 1+0 = 1 which is not 0, so that doesn't work. But 101 does since 1+0 = 1 which is the last digit. Now ask yourself: are there any more numbers in the range of 100 to 109 that work? The answer is no since 1+0 = 1 is unique to this set of numbers (ie there are no more numbers in this range that have a 1 as the last digit) So again, in the list of numbers from 100 to 109, the only upright number is 101 Similarly, for numbers in the range 110 to 119, the only upright number is 112 Keep going to find the third upright number to be 123 Etc, etc When you come upon numbers that start with 19, you won't find any upright numbers since 1+9 = 10, which is not a single digit. So , in the range of 100 to 199, there are 8 numbers that are upright numbers So from 200 to 299, there are 7 numbers that work. Why 7 and not 8? notice that for numbers that start with 29 add to 2+9 = 11, but that's not a single digit. Likewise, numbers starting with 28 add to 2+8 = 10, again not a single digit. However, the rest work. Keep going and you'll find that only 6 numbers work from 300 to 399 etc etc After all that, you'll have the following number counts: 8, 7, 6, 5, 4, 3, 2, 1, 0 Add them all up to get 36 So the answer is choice C\n\nOpenStudy (anonymous):\n\nInteresting question. Here's how I tackled it:Look at the first few 3 digit numbers.100101102103104105106107108109They all have the same first two digits, but different last digits. You'll notice that exactly one of these numbers is upright. You'll find the same thing for 110 through 119. I can keep going unless the sum of the first two digits is greater than 9. For example, any number with 67 as the first two digits can never be upright because 6+7 is a two digits number.So to start counting upright numbers, add up all the possible 2 digit combinations you can get. you can use 10 through 99, so that's 90 possible two digit pairs. Now you need to subtract off the pairs that don't work, like 67. 10, 11, 12, 13, ... , 18 all work, but 19 doesn't. That's 9 upright numbers. 20, 21, ... , 27 all work, but 28 and 29 don't. That's 8 more upright numbers. 90 works, but 91, 92, ... , 99 don't. That's just one more upright number. See the pattern? So find 1+2+3+4+5+6+7+8, and you have your answer.\n\njimthompson5910 (jim_thompson5910):\n\nyou have the right idea pmilano, but you fell into the trap that there's a number in the range of 190 to 199 when there isn't one\n\nOpenStudy (anonymous):\n\nBy the way, the fast way to add the numbers 1 through 8 (or, more generally, 1 through n) is by using the formula (n)(n+1)/2 So 8(8+1)/2 = 36\n\nOpenStudy (anonymous):\n\nThanks jim.\n\nOpenStudy (anonymous):\n\nmy answer is D)45. There is actually a pattern if you try to list down the first numbers. Starting at 1__, there are 9 possible answers. As you move up to the next hundreds digit, you decrease that number by one, because you can't have a sum of more than 9. 9+8+7+...+2+1 = 45\n\nOpenStudy (anonymous):\n\nI actually realized that before I posted but didn't proof read my post well enough (hence my correct answer and incorrect counting).\n\njimthompson5910 (jim_thompson5910):\n\nMy bad, I was ignoring zero Start with the range 100-199 and moving to the next 200-299, etc, the following upright numbers are found in the tens digits listed below 0, 1, 2, 3, 4, 5, 6, 7, 8 0, 1, 2, 3, 4, 5, 6, 7 0, 1, 2, 3, 4, 5, 6 0, 1, 2, 3, 4, 5 0, 1, 2, 3, 4 0, 1, 2, 3 0, 1, 2 0, 1 0 So there are 45 here\n\nOpenStudy (waheguru):\n\ni am getting 36?\n\nOpenStudy (anonymous):\n\nHere is another method lol (incase you didnt have enough) When the last digit is 'n', you want to ask yourself how many solutions are there to: x + y = n where x and y are non-negative integers? For example, if n was 3, then these are the possible solutions for x and y: x = 0, y = 3 x = 1, y = 2 x = 2, y = 1 x = 3, y = 0 These correspond to the three digits numbers: 033 123 213 303 Of course we would discard the '033' case. Anywhos, the equation that gives you the number of solutions to \"x + y = n\" is: $\\left(\\begin{matrix}n+2-1 \\\\ 2-1\\end{matrix}\\right) = \\left(\\begin{matrix}n+1 \\\\ 1\\end{matrix}\\right)$ We need to do this for n = 1, 2, 3,....,9, while remembering to not count the case of the first digit being 0. So we end up with: $\\left(\\begin{matrix}2 \\\\ 1\\end{matrix}\\right)+\\left(\\begin{matrix}3 \\\\ 1\\end{matrix}\\right)+\\left(\\begin{matrix}4 \\\\ 1\\end{matrix}\\right)+\\ldots +\\left(\\begin{matrix}10 \\\\ 1\\end{matrix}\\right)-9$$= 2+3+4+\\ldots+10-9 = 1+2+3+\\ldots+9 = 45$\n\nOpenStudy (anonymous):\n\nFirst set of numbers: 101,112,123,134,145,156,167,178,189 ---> 9 numbers. For 2__, that is one less than the 1 hundreds because u can only use 0-7 for the 2nd digit. The last possible number is 909, the only number in the 9 hundreds.\n\nCan't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!\n\nJoin our real-time social learning platform and learn together with your friends!\nLatest Questions\nTonycoolkid21: help\n6 hours ago 8 Replies 1 Medal\nTonycoolkid21: help\n14 minutes ago 16 Replies 3 Medals\ncrispyrat: A guide to sqrt and cbrt\n7 hours ago 21 Replies 3 Medals\nxXQuintonXx: help please(ss below)\n8 hours ago 5 Replies 1 Medal\nrxcklesskaisher: idk wdh dis is\n5 minutes ago 4 Replies 0 Medals\nCan't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!\n\nJoin our real-time social learning platform and learn together with your friends!", "date": "2021-05-07 11:11:39", "meta": {"domain": "questioncove.com", "url": "https://questioncove.com/updates/4e56c14a0b8b9ebaa893644f", "openwebmath_score": 0.5884519219398499, "openwebmath_perplexity": 456.6725855835304, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446440948805, "lm_q2_score": 0.8918110468756548, "lm_q1q2_score": 0.8683070493351297}}
{"url": "https://math.stackexchange.com/questions/3320950/integrating-int-frac-sin-x1-cos-x-dx-i-get-ln1-cos-x-wolfr", "text": "# Integrating $\\int \\frac{-\\sin x}{1+\\cos x}\\, dx$, I get $\\ln(1 + \\cos x)$. WolframAlpha gives $2 \\ln(\\cos \\frac x 2)$. Is WA wrong?\n\nSo, I'm watching a tutorial on differential equations, where I encountered this little trick:\n\n$$\\int \\frac{y'}{y}\\, dx = \\ln(y)$$ It seems perfectly logical and easy to justify, but something fishy happens to this integral:\n\n$$\\int \\frac{-\\sin x}{1+\\cos x}\\, dx$$\n\nThe trick gives $$\\int \\frac{-\\sin x}{1+\\cos x}\\, dx = \\ln(1 + \\cos x)$$\n\nwhile WolframAlpha gives $$\\int \\frac{-\\sin x}{1+\\cos x}\\, dx = 2 \\ln(\\cos \\frac x 2)$$.\n\nYou guys who know this stuff - does WolframAlpha mess up here or is it something I've missed? Taking the derivative of $$2 \\ln(\\cos \\frac x 2)$$ gives me $$-\\tan \\frac x 2$$, so I don't see how WA may be right.\n\n\u2022 Remember that $\\cos{2x}=2\\cos^2x-1$ \u2013\u00a0Don Thousand Aug 12 at 11:57\n\u2022 They are different in terms of their symbols, but they give you exactly the same values. What you have is a trigonometric identity. \u2013\u00a0Pixel Aug 12 at 12:01\n\u2022 As others have mentioned, the key here is to use a trig identity to bridge the two solutions. ... Often, the trickiest part of Calculus is remembering your Pre-Calculus. :) \u2013\u00a0Blue Aug 12 at 12:03\n\u2022 For future reference, you can always check your (or WA's) answer by differentiating it and checking it against the expression you're integrating (the integrand). If they match up then you know your answer is correct (up to a constant), even if you don't know any trig! \u2013\u00a0Ben Aug 12 at 12:05\n\n## 5 Answers\n\n$$1+\\cos x = 2\\cos^2\\frac{x}{2}$$\n\n$$\\ln (1+\\cos x )=\\ln( 2\\cos^2\\frac{x}{2})=\\ln2+2\\ln\\cos\\frac{x}{2}$$\n\nAnd this $$\\ln2$$ adds together with the arbitrary constant $$c$$ in indefinite integral and gets cancelled in definite integral.\n\n\u2022 Thank you so much! This makes perfect sense! \u2013\u00a0Seigemann Aug 12 at 11:58\n\u2022 You're welcome! \u2013\u00a0Ak19 Aug 12 at 11:58\n\nAs Ben suggested in comments, it's always good to check an integral by taking the derivative:\n\n$$\\dfrac d {dx} \\ln(1+\\cos x)=\\dfrac{-\\sin x}{1+\\cos x}=\\dfrac{-2\\sin\\dfrac x2 \\cos \\dfrac x2}{2\\cos^2\\dfrac x2}=\\dfrac{-\\sin\\dfrac x2}{\\cos \\dfrac x2}=\\dfrac d {dx} 2 \\ln \\cos \\dfrac x2$$\n\n\u2022 N.B. The expressions are undefined when $x=\\pi$ \u2013\u00a0J. W. Tanner Aug 12 at 13:11\n\n$$1+\\cos \\, x=2\\cos^{2}(\\frac x 2)$$ so $$\\ln (1+\\cos \\, x)=2 \\ln (\\cos (\\frac x 2))+\\ln 2$$ . Also, $$-\\tan (\\frac x 2)$$ is same as $$-\\frac {\\sin \\, x} {1+\\cos \\, x}$$ because $$\\sin \\,x =2 \\sin (\\frac x 2)\\cos (\\frac x 2)$$ and $$1+\\cos \\, x=2\\cos^{2}(\\frac x 2)$$.\n\n$$2\\ln\\cos\\frac x2=\\ln\\cos^2\\frac x2=\\ln\\frac{1+\\cos x}2=\\ln(1+\\cos x)-\\ln2$$ by the half-angle formula, so your answer and WA's are the same up to the integration constant.\n\nWhen you do the one integral $$I(x)$$ by different methods you get different expressions $$I_1(x),I_2(x),I_3(x),.....$$, however the difference between any two of these is a constant independent of $$x$$. For instance $$I(x)=\\int \\sin \\cos x dx =\\frac{1}{2}\\int \\sin 2x~ dx =-\\frac{1}{4} \\cos 2x +C_1 =I_1(x).$$\n\nNext if you do integration by parts you get $$I=\\sin ^2x -\\int \\sin x \\cos \\Rightarrow I=\\frac{1}{2} \\sin^2 x +C_2=I_2(x).$$\n\nFurther I you use a substitution $$\\cos x =-t$$, then $$I(x)=-\\frac{1}{2}\\cos^2 x +C_3=I_3(x).$$\n\nNow check that the difference between any two of $$I_1,I_2,I_3$$ is just a constant.\n\nAs pointed in other solutions the thing stated above is happening in your case as well.", "date": "2019-08-26 03:31:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3320950/integrating-int-frac-sin-x1-cos-x-dx-i-get-ln1-cos-x-wolfr", "openwebmath_score": 0.8150451183319092, "openwebmath_perplexity": 354.44910860580154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924777713886, "lm_q2_score": 0.89330940889474, "lm_q1q2_score": 0.8682900257680929}}
{"url": "https://www.weddingsupplierswa.com.au/viz6m3yu/is-real-number-a-subset-of-complex-number-489fe3", "text": "Thus, the complex numbers of t\u2026 The subsets of the real numbers can be r\u2026 However, real numbers have multiplication, and the complex numbers extend the reals by adding i. Intro to complex numbers. The set of complex numbers is closed under addition and multiplication. The set of complex numbers is denoted by C R is a subset of C 118 When adding from MAT 1341 at University of Ottawa Lv 7. hace 5 a\u00f1os. The values a and b can be zero, so the set of real numbers and the set of imaginary numbers are subsets of the set of complex numbers. We define the complex number i = (0,1).With that definition we can write every complex number interchangebly as A complex number such as $5-2 i$ then corresponds to 5 on the real axis and $-2$ on the imaginary axis. Yes. In situations where one is dealing only with real numbers, as in everyday life, there is of course no need to insist on each real number to be put in the form a+bi, eg. Click to see full answer. The table below describes important subsets of the real numbers. Bundle: Elementary Algebra, 9th + Student Workbook (9th Edition) Edit edition. In the complex number a + bi, a is called the real part and b is called the imaginary part. Real numbers can be considered a subset of the complex numbers \u2026 In general, all the arithmetic operations can be performed on these numbers and they can be represented in the number line, also. This is because the Real Numbers are a subset of the Complex Numbers (). The set of real numbers is a proper subset of the set of complex numbers. To which subsets of the real numbers does -7 belong? Similarly, since a can be zero, the imaginary numbers are a subset of the complex numbers. Why does it make sense to talk about the 'set of complex numbers'? Notational conventions. The set of complex numbersis, therefore; This construction allows to consider the real numbers as a subset of the complex numbers, being realthat complex number whiose imaginary part is null. Remember that under the set of rational numbers, we have the subcategories or subsets of integers, whole numbers, and natural numbers. The real numbers include both rational and irrational numbers. A and B may be equal; if they are unequal, then A is a proper subset of B. 2 I. Complex does not mean complicated; it means that the two types of numbers combine to form a complex, like a housing complex \u2014 a group of buildings joined together. However, $\\mathbb{C}$ comes with a canonical embedding of $\\mathbb{R}$ and in this sense, you can treat $\\mathbb{R}$ as a subset of $\\mathbb{C}$. In other words, i 2 = \u20131. Solved Example on Real Numbers Ques: Name the subset(s) of the real numbers to which '- 25' belongs. Find the real part of a complex number: Find the real part of a complex number expressed in polar form: Plot over a subset of the complex plane: Use Re to specify regions of the complex plane: The axiom of mathematical induction is for our purposes frequently The real numbers are complex numbers \u2026 Bundle: Elementary and Intermediate Algebra: A Combined Approach + Student Solutions Manual (6th Edition) Edit edition. What are rational and irrational numbers. In 1882, Ferdinand von Lindemann proved that \u03c0 is not just irrational, but transcendental as well. Since b can be equal to 0, you see that the real numbers are a subset of the complex numbers. The set of real numbers is a subset of the set of complex numbers? they are of a different nature. Addition and multiplication of real numbers are defined in such a way that expressions of this type satisfy all field axioms and thus hold for C. For example, the distributive law enforces Therefore, a set of real numbers is bounded if it is contained in a \u2026 A complex numberis defined as an expression of the form: The type of expression z = x + iy is called the binomial form where the real part is the real number x, that is denoted Re(z), and the imaginary partis the real number y, which is denoted by Im(z). Natural Number (N) Subset N is the set of Natural Number or Counting Numbers given N = {1, 2, 3, ..\u2026 Set of Real Numbers Set of Real Numbers is a universal set. Real numbers are just complex numbers with no imaginary part. A mathematical operation of subtracting a complex number from another complex number is called the subtraction of complex numbers.. Introduction. There is a thin line difference between both, complex number and an imaginary number. Let Sbe a subset of the set Nof natural numbers. While the real numbers are a subset of the complex numbers, there are very many complex numbers that are not real numbers. Intro to complex numbers. In mathematics, a set A is a subset of a set B, or equivalently B is a superset of A, if A is contained in B. @HagenvonEitzen All the different constructions of $\\mathbb{R}$ rely on the fact that we have already constructed $\\mathbb{N}$ before (?). What Number Set Contains The Subset of Complex Numbers? Examples: 1 + i, 2 - 6i, -5.2i, 4. It is important to note that if z is a complex number, then its real and imaginary parts are both real numbers. Notational conventions. Milestone leveling for a party of players who drop in and out? Real numbers $$\\mathbb{R}$$ The set formed by rational numbers and irrational numbers is called the set of real numbers and is denoted as $$\\mathbb{R}$$. generating lists of integers with constraint. Read More -> The number {3} is a subset of the reals. For example, the set $\\mathbf{C}^{2}$ is also a real vector space under the same addition as before, but with multiplication only by real scalars, an operation we might denote $\\cdot_{\\mathbf{R}}$. Can you put laminate flooring in a mobile home? Bundle: Elementary Algebra + Math Study Skills Workbook (4th Edition) Edit edition. Complex. Complex numbers can be represented as points on a \u201ccomplex plane\u201d: the rectangular x-y plane, in which the x-axis corresponds to the real numbers, and the y-axis corresponds to the imaginary numbers. Since $\\mathbb{Q}\\subset \\mathbb{R}$ it is again logical that the introduced arithmetical operations and relations should expand onto the new set. The real numbers can be \"said to be\" a subset of the complex numbers. (examples: -7, 2/3, 3.75) Irrational numbers are numbers that cannot be expressed as a fraction or ratio of two integers. Set Theoretic Definition of Complex Numbers: How to Distinguish $\\mathbb{C}$ from $\\mathbb{R}^2$? iota.) ): Includes real numbers, imaginary numbers, and sums and differences of real and imaginary numbers. [1] [2] Such a number w is denoted by log z . An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property i2 = \u22121. If a jet engine is bolted to the equator, does the Earth speed up? On the same footing, $\\mathbb{N} \\not \\subset \\mathbb{Z} \\not \\subset \\mathbb{Q} \\not \\subset \\mathbb{R}$. The relationship between the real and complex numbers from a set theoretic perspective. To make notation a little bit easier, we call a complex number z. A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = \u22121.For example, 2 + 3i is a complex number. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. p S S S II) i.W 2 lIT ~and ir are two of very many real numbers that are not rational numbers. site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. 9 is a real number because it belongs to the set of whole numbers and the set of whole numbers is a subset of real numbers. Real numbers are simply the combination of rational and irrational numbers, in the number system. A real number is a number that can take any value on the number line. definition. D. Irrational Use MathJax to format equations. Be sure to account for ALL sets. To learn more, see our tips on writing great answers. What is internal and external criticism of historical sources? For example, 5i is an imaginary number, and its square is \u221225. 10, as 10 + 0i - that would be too pedantic, to say the \u2026 We call x +yi the Cartesian form for a complex number. Strictly speaking (from a set-theoretic view point), $\\mathbb{R} \\not \\subset \\mathbb{C}$. It solves x\u00b2+1=0. MathJax reference. 1 See answer AnshulDavid3143 is waiting for your help. 2/5 A. Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. But since the set of complex numbers is by definition $$\\mathbb{C}=\\{a+bi:a,b\\in\\mathbb{R}\\},$$ doesn't this mean $\\mathbb{R}\\subseteq\\mathbb{C}$, since for each $x \\in \\mathbb{R}$ taking $z = x + 0i$ we have a complex number which equals $x$? Is Delilah from NCIS paralyzed in real life? To which subset of real numbers does the following number belong? What is the \"Ultimate Book of The Master\". By now you should be relatively familiar with the set of real numbers denoted $\\mathbb{R}$ which includes numbers such as $2$, $-4$, $\\displaystyle{\\frac{6}{13}}$, $\\pi$, $\\sqrt{3}$, \u2026. Real numbers are a subset of complex numbers. Two complex numbers a + bi and c + di are defined to be equal if and only if a = c and b = d. If the imaginary part of a complex number is 0, as in 5 + 0i, then the number corresponds to a real number. Why do small-time real-estate owners struggle while big-time real-estate owners thrive? Complex numbers are distinguished from real numbers by the presence of the value i, which is defined as . Complex numbers are an important part of algebra, and they do have relevance to such things as solutions to polynomial equations. Complex Numbers $\\mathbb{C}$ Examples of complex numbers: $(1, 2), (4, 5), (-9, 7), (-3, -20), (5, 19),...$ $1 + 5i, 2 - 4i, -7 + 6i...$ where $i = \\sqrt{-1}$ or $i^2 = -1$ The real numbers are a subset of the complex numbers. THE REAL AND COMPLEX NUMBERS AXIOM OF MATHEMATICAL INDUCTION. Any time you deal both with complex vector spaces and real vector spaces, you have to be certain of what \"scalar multiplication\" means. Dedekind cuts or Cauchy sequences for $\\mathbb R$) these ZFC. What is the difference between simple distillation and steam distillation? So, I was taught that $\\mathbb{Z}\\subseteq\\mathbb{Q}\\subseteq\\mathbb{R}$. The real axis is the line in the complex plane consisting of the numbers that have a zero imaginary part: a + 0i. Complex numbers, say \u2026 square root of 30 . The complex numbers C consist of expressions a + bi, with a, b real, where i is the imaginary unit, i.e., a (non-real) number satisfying i 2 = \u22121. Asking for help, clarification, or responding to other answers. Why did the design of the Boeing 247's cockpit windows change for some models? The irrational numbers are a subset of the real numbers. Imaginary number is expressed as any real number multiplied to a imaginary unit (generally 'i' i.e. Two complex numbers a + bi and c + di are defined to be equal if and only if a = c and b = d. If the imaginary part of a complex number is 0, as in 5 + 0i, then the number corresponds to a real number. Example 1. Pi is an irrational number, which means that it is a real number that cannot be expressed by a simple fraction. Complex numbers can be visualized geometrically as points in the complex (Argand) plane. How do I provide exposition on a magic system when no character has an objective or complete understanding of it? The set of complex numbers includes all the other sets of numbers. rev\u00a02021.1.18.38333, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, You have $\\not\\subset$ if you construct them one after another. The imaginary numbers are also a subset of the complex: the complex numbers whose real part is zero. If I am blending parsley for soup, can I use the parsley whole or should I still remove the stems? Is it safe to keep uranium ore in my house? Complex numbers are numbers in the form a + b i a+bi a + b i where a, b \u2208 R a,b\\in \\mathbb{R} a, b \u2208 R. And real numbers are numbers where the imaginary part, b = 0 b=0 b = 0. Why did flying boats in the '30s and '40s have a longer range than land based aircraft? The set {0,1, 2+i, 2-i} is NOT a subset of the real numbers. \u00a9 AskingLot.com LTD 2021 All Rights Reserved. But already the fact that there are several constructions possible (e.g. Every real number graphs to a unique point on the real axis. Why set of real numbers not a set of ordered pairs? The term is often used in preference to the simpler \"imaginary\" in situations where. mam is real numbers a subset of complex numbers - Mathematics - TopperLearning.com | 8v26wq66 3. Start studying Field of Quotients, the Rational Numbers, the Real Numbers, & Complex Numbers. The real numbers are a subset of the complex numbers, so zero is by definition a complex number ( and a real number, of course; just as a fraction is a rational number and a real number). One can represent complex numbers as an ordered pair of real numbers (a,b), so that real numbers are complex numbers whose second members b are zero. At the same time, the imaginary numbers are the un-real numbers, which cannot be expressed in the number line and is commonly used to represent a complex number. The set of real numbers can be drawn as a line called \u201cthe number line\u201d. Subset. Would coating a space ship in liquid nitrogen mask its thermal signature? Complex numbers, such as 2+3i, have the form z = x + iy, where x and y are real numbers. a complex logarithm of a nonzero complex number z, defined to be any complex number w for which e w = z. a real number is not a set. Real numbers, rational numbers. Expressing complex numbers in form $a+bi$. Each complex number corresponds to a point (a, b) in the complex plane. The square of an imaginary number bi is \u2212b2. In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line. can in general assume complex values with nonzero real parts, but in a particular case of interest, the real part is identically zero. Practice: Parts of complex numbers. In the last example (113) the imaginary part is zero and we actually have a real number. Rational numbers are numbers that can be expressed as a fraction or part of a whole number. It only takes a minute to sign up. There are three common forms of representing a complex number z: Cartesian: z = a + bi JR is the set of numbers that can be used to measure a distance, or the negative of a number used to measure a distance. 5.1.2 The Reals as a Subset of the Complex Numbers Since the complex numbers were seen as an extension of the set of real numbers, it is natural to believe that R is a subset of C. Of course, to prove this subset Similarly, it is asked, is every real number is a complex number? A complex number is a number that can be written in the form a + b i a + bi a+bi, where a and b are real numbers and i is the imaginary unit defined by i 2 = \u2212 1 i^2 = -1 i2=\u22121. Classification of Real Numbers Examples. The real numbers have the following important subsets: rational numbers, irrational numbers, integers, whole numbers, and natural numbers. (0,1) = (-1,0), which is purely real and equals to -1. Example 2 : Tell whether the given statement is true or false. Thus we can consider the complex number system as having embedded within it, as a subset the real number \u2026 Furthermore, each real number is in the set of complex numbers,, so that the real numbers are a \u2026 Choices: A. integers, rational numbers, real numbers B. whole numbers, integers, rational numbers, real numbers C. natural numbers, whole numbers, integer numbers, rational numbers, real numbers D. irrational numbers, real numbers Correct Answer: A Complex numbers are often graphed on a plane. What do you call a 'usury' ('bad deal') agreement that doesn't involve a loan? Some examples of irrational numbers are $$\\sqrt{2},\\pi,\\sqrt[3]{5},$$ and for example $$\\pi=3,1415926535\\ldots$$ comes from the relationship between the length of a circle and its diameter. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Oh I suppose Russel has a definition where the real number 3 is the set of all things there are 3 of. A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself). How are Quaternions derived from Complex numbers or Real numbers? B. \u00c2\u00bfCu\u00c3\u00a1les son los 10 mandamientos de la Biblia Reina Valera 1960? The real numbers are a subset of the complex numbers. So, $$i \\times i = -1$$ $$\\Rightarrow i = \\sqrt{-1}$$. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Complex numbers introduction. We will now introduce the set of complex numbers. The complex numbers are a plane with an additional real axis to calculate square roots (and other even roots) of negative numbers. The set of complex numbers C with addition and multiplication as defined above is a field with additive and multiplicative identities (0,0) and (1,0).It extends the real numbers R via the isomorphism (x,0) = x. Suppose that (1) 1 2S: (2) If a natural number kis in S;then the natural number k+ 1 also is in S: Then S= N:That is, every natural number nbelongs to S: REMARK. There are several types of subsets of real numbers\u2014numbers that can be expressed as a decimal. Thus we can consider the complex number system as having embedded within it, as a subset the real number system. If you're seeing this message, it means we're having trouble loading external resources on our website. These numbers are called irrational numbers, and $\\sqrt{2}$, $\\sqrt{3}$, $\\pi$... belong to this set. Is there even such a set? Proof that \u03c0 is irrational. Email. But no real number, when squared, is ever equal to a negative number--hence, we call i an imaginary number. Better user experience while having a small amount of content to show. Explain your choice. Complex numbers in the form a + bi can be graphed on a complex coordinate plane. Complex Numbers. x is called the real part and y is called the imaginary part. (The counting numbers are 1,2,3,....) All of these types of numbers are real numbers. We will addres s complex (or imaginary) numbers in the Quadratic Functions chapter. That is the adjacent surface to our 3D! Learn vocabulary, terms, and more with flashcards, games, and other study tools. The real numbers are all the numbers on the number line, where you group rational numbers with a so called dedekind cut (you can form this cut so that it result is irrational). Popular Trending (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0 i, which is a complex representation.) Real numbers 21.5 pi. Some \ufb01xed point O is chosen to represent the complex number \u2026 In the 1760s, Johann Heinrich Lambert proved that the number \u03c0 (pi) is irrational: that is, it cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer. Therefore we have: z = Re(z) + iIm(z). Why or why not? In complex analysis, the complex numbers are customarily represented by the symbol z, which can be separated into its real (x) and imaginary (y) parts: = + for example: z = 4 + 5i, where x and y are real numbers, and i is the imaginary unit.In this customary notation the complex number z corresponds to the point (x, y) in the Cartesian plane. A set S of real numbers is called bounded from above if there exists some real number k (not necessarily in S) such that k \u2265 s for all s in S.The number k is called an upper bound of S.The terms bounded from below and lower bound are similarly defined.. A set S is bounded if it has both upper and lower bounds. The complex numbers form a COMPLETE system of numbers of which the real numbers form a subset. The conjugate of a complex number z= a+ biis created by changing the sign on the imaginary part: z = a bi: Thus the conjugate of 2 + iis 2 + i= 2 i; the conjugate of p 3 \u02c7iis p 3 \u02c7i= p 3 + \u02c7i.\n\nHow Much Does It Cost To Stay At The Poconos, Indore Collector Name List, Antioch Ca Fire Department, Callaway Rogue Irons Left Handed, Xmas Trees Near Me, Silver Leaf Adhesive, Kembang Kol Hidroponik,", "date": "2021-04-13 10:18:42", "meta": {"domain": "com.au", "url": "https://www.weddingsupplierswa.com.au/viz6m3yu/is-real-number-a-subset-of-complex-number-489fe3", "openwebmath_score": 0.7106465101242065, "openwebmath_perplexity": 267.5919787374287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9719924810166349, "lm_q2_score": 0.8933093968230773, "lm_q1q2_score": 0.8682900169335366}}
{"url": "https://math.stackexchange.com/questions/2019784/is-every-odd-order-skew-symmetric-matrix-singular", "text": "# Is every odd order skew-symmetric matrix singular?\n\nWe call a square matrix $$A$$ a skew-symmetric matrix if $$A=-A^T$$. A matrix is said to be singular if its determinant is zero. Is every odd order skew-symmetric matrix with complex entries singular?\n\n## 3 Answers\n\nYes, that holds, since: $$\\det A=\\det{(-A^T)}=(-1)^{odd}\\det{A^T}=-\\det A,$$ from where we get $\\det{A}=0$.\n\nThis is actually the case :\n\nSuppose, $A$ is an $n\\times n$-matrix.\n\nWe have $$\\det(A)=\\det(-A^T)=(-1)^n\\cdot \\det(A^T)=(-1)^n\\cdot \\det(A)$$\n\nSince $n$ is odd, we can conclude $\\ \\det(A)=-\\det(A)\\$ implying $\\ \\det(A)=0\\$\n\n\u2022 But the problem here is that the entries are complex. Is it true for complex entries as well? Nov 18 '16 at 12:05\n\u2022 Yes, it is also true for complex entries. The proof does not assum real entries. Nov 18 '16 at 12:47\n\nConsider the example $$\\begin{bmatrix} 0 & i & -3\\\\ -i & 0 & 2i\\\\ 3 & -2i & 0\\\\ \\end{bmatrix}$$\n\n\u2022 I mean a 3 by 3 matrix with rows (0 i -3), ( -i 0 2i) and (3 -2i 0). Nov 18 '16 at 12:21\n\u2022 If this was intended to be a counterexample, then it is not! The determinant of this matrix is zero. May 6 at 15:03", "date": "2021-10-15 22:54:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2019784/is-every-odd-order-skew-symmetric-matrix-singular", "openwebmath_score": 0.893404483795166, "openwebmath_perplexity": 381.297261952833, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225143328517, "lm_q2_score": 0.8757870046160257, "lm_q1q2_score": 0.868274954136457}}
{"url": "http://angs-diasporanew.sramble-communication.com/apdsmyht/cartesian-to-cylindrical-coordinates.php", "text": "##### Cartesian to cylindrical coordinates\n8. As with spherical where we again observe that the basis vector in cylindrical coordinates is dependent on the two basis vectors e \u2192 x and e \u2192 y in Cartesian coordinates. Conversion between Cylindrical and Cartesian Coordinates The rectangular coordinates $$(x,y,z)$$ and the cylindrical coordinates $$(r,\u03b8,z)$$ of a point are related as follows: These equations are used to convert from cylindrical coordinates to rectangular coordinates. Feb 24, 2010 \u00b7 The cylindrical coordinate system is a 3-D version of the polar coordinate system in 2-D with an extra component for . Find the integrals that compute its volume, using cartesian, cylindrical, and spherical coordinates. Magnitude of a cartesian coordinates (a,b) is given bysqrt(a^2+b^2) and its angle is given by tan^-1(b/a) Let r be the magnitude of (-4,3) and theta be its angle. The old vvvv nodes Polar and Cartesian in 3d are similar to the geographic coordinates with the exception that the angular direction of the longitude is inverted. Jan 27, 2017 \u00b7 We can write down the equation in Cylindrical Coordinates by making TWO simple modifications in the heat conduction equation for Cartesian coordinates. Layered\u00a0 (2,\u03c02). To Convert from Cartesian to Polar. We will concentrate on cylindrical coordinates in this activity, but we will address spherical coordinates in a later activity. If called with a single matrix argument then each row of C represents the Cartesian coordinate (x, y (, z)). Use Question: A sphere, centered at the origin, has radius 3. A, then, has three vector components, each component corresponding to the projection of A onto the three axes. Cylindrical coordinates are chosen to take advantage of symmetry, so that a velocity component can disappear. 8. I know how to generate the strain tensor in a rotated coordinate system (also a Cartesian one), but just don't know how to apply the rules found in the second link to derive the strain components in the cylindrical coordinates, if I have strain tensor in the corresponding Cartesian coordinates. i tried converting to cartesian coordinates, then plotting but at some points function jumps up or down (i think due to trigonometric functions) , so i wanted to see it in cylindrical coordinates. For the conversion between cylindrical and Cartesian coordinates, it is convenient to assume that the reference plane of the former is the Cartesian xy-plane (with equation z = 0), and the cylindrical axis is the Cartesian z-axis. Each point is uniquely identified by a distance to the origin, called r here, an angle, called \u03d5 {\\displaystyle \\phi } ( phi ), and a height above the plane of the coordinate system, called Z in the picture. We can slightly modify our arc length equation in polar to make it apply to the cylindrical coordinate system given that , . Mar 02, 2013 \u00b7 Cartesian Coordinates vs Polar Coordinates In Geometry, a coordinate system is a reference system, where numbers (or coordinates) are used to uniquely determine the position of a point or other geometric element in space. Cartesian Coordinates Cylindrical coordinates consist of (1) a coordinate plane, plus (2) an axis perpendicular to the plane through the origin. In the coordinate plane, two coordinates describe position: (1) an angle, \u03b8 (azimuth angle, measured positive counterclockwise relative to a Richard Fitzpatrick 2016-01-22 a) x 2 - y = 25 to cylindrical coordinates. One of these is when the problem has cylindrical symmetry. 6 Volume element in Cartesian coordinates. 6. Figure 1. Replace (x, y, z) by (r, \u03c6, \u03b8) b. 8: Differential length, area, and volume. 1. Visit https://www. 2 s pherical Convert the three-dimensional Cartesian coordinates defined by corresponding entries in the matrices x, y, and z to cylindrical coordinates theta, rho, and z. Cylindrical Coordinate System: In cylindrical coordinate systems a point P(r 1, \u03b8 1, z 1) is the intersection of the following three surfaces as shown in the following figure. Cylindrical coordinates. The Cylindrical to Cartesian calculator converts Cylindrical coordinates into Cartesian coordinates. Cartesian base vectors. Choose the source and destination coordinate systems from the drop down menus. which means that . Cylindrical coordinates in IR3. (1) The (orthogonal) base vectors in the two systems of coordinates are linked by er i and \u02dcxi could be two Cartesian coordinate systems, one moving at a con-stant velocity relative to the other, or xi could be Cartesian coordinates and \u02dcxi spherical polar coordinates whose origins are coincident and in relative rest. We shall see er and e\u03b8 in terms of their cartesian components along i and j. Define a spherical data set. The cylindrical coordinate system basically is a combination of the polar coordinate system xy \u00a1 plane with an additional z \u00a1 coordinate vertically. This system is a generalization of polar coordinates to three dimensions by superimposing a height axis. Conversion between cylindrical and Cartesian coordinates Cylindrical coordinates are obtained from Cartesian coordinates by replacing the x and y coordinates with polar coordinates r and theta and leaving the z coordinate unchanged. Expressed in Cartesian coordinates, a vector is defined in terms of Jul 22, 2020 \u00b7 This blog will explain how to create a Stacked Contour Plot if user has data with Cylindrical Coordinates(ro, theta and Z), It will involve the following steps: interpolate data and convert data from cylindrical coordinates to Cartesian coordinates; Use XYZ gridding to convert xyz data into matrix; Clip data in matrix with circle. 2 Cylindrical Coordinates We first choose an origin and an axis we call the -axis with unit vector pointing in the increasing z-direction. 1 c oordinate systems a1. advanced. To use this calculator, a user just enters in the (r, \u03c6, z) values of the cylindrical coordinates and then clicks 'Calculate', and the cartesian coordinates will be automatically computed and Jul 11, 2018 \u00b7 Understand thoroughly about the Conversion between Cylindrical & Cartesian systems for Electromagnetism. We should bear in mind that the concepts covered in Chapter 1 and demonstrated in Cartesian coordinates are equally applicable to other systems of coordinates. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system Another way of looking at it is that we take polar coordinates $$(r,\\theta)$$ and slap on the rectangular coordinate z to the end to get $$(r,\\theta,z)$$ and call this cylindrical coordinates. The cylindrical coordinates (r,\u03b8,z) are related to the Cartesian coordinates (x1,x2,x3) by the following relations r = x2 1 +x 2 2 1/2, \u03b8 = tan\u22121 x2 x1, z = x3, and x1 = rcos\u03b8, x2 = rsin\u03b8, x3 = z. I know the material, just wanna get it over with. bjc a2. In a cylindrical coordinate system, the location of a three-dimensional point is decribed with the first two dimensions described by polar coordinates and the third dimension described in distance from the plane containing the other two axes. In the last two sections of this chapter we\u2019ll be looking at some alternate coordinate systems for three dimensional space. And these coordinates are called Cartesian coordinates, named for Rene Descartes because he's the guy that came up with these. Related Calculators: You can always start in Cartesian because the kinetic energy is a scalar and thus independent of the coordinate system in which you choose to evaluate it, although scalar products are most easily computed in Cartesian coordinates. (1a): Triple integral in Cartesian coordinates x,y,z (1b): Triple integral in cylindrical coordinates r,theta,z (2a): Triple integral in cylindrical coordinates r,theta,z (2b): Triple integral in spherical coordinates rho,phi,theta Cylindrical coordinates are defined with respect to a set of Cartesian coordinates, and can be converted to and from these coordinates using the atan2 function as follows. Coordinates in the GIS are measured from the origin point. 25in}z = z\\] In order to do the integral in cylindrical coordinates we will need to know 2 We can describe a point, P, in three different ways. To convert a point from Cartesian coordinates to cylindrical coordinates, use equations and In the spherical coordinate system, a point in space is represented by the ordered triple where is the distance between and the origin is the same angle used to describe the location in cylindrical coordinates, and is the angle formed by the positive z Express A using Cartesian coordinates and spherical base vectors. For the x and y components, the transormations are ; inversely, . I have to prove it by simple geometry & calculus, without using jacobian or linear algebra (basis). in 6. As shown in Figure 1-2a, any point in space is defined by the intersection of the three perpendicular surfaces of a circular It is easier to consider a cylindrical coordinate system than a Cartesian coordinate system with velocity vector V=(ur,u!,uz) when discussing point vortices in a local reference frame. It is simplest to get the ideas across with an example. It is good to begin with the simpler case, cylindrical coordinates. Under the formula of the stress tensor of the cylindrical wall under the polar coordinate system, the\u00a0 30 Mar 2016 Cylindrical Coordinates. Although we have considered the Cartesian system in Chapter 1, we shall consider it in detail in this chapter. Convert the cylindrical coordinates defined by corresponding entries in the matrices theta, rho, and z to three-dimensional Cartesian coordinates x, y, and z. This is no longer the case in spherical! Cylindrical coordinates To get a third dimension, each point also has a height above the original coordinate system. Converting Polar Coordinates to Cartesian. Cartesian systems use linear distances while polar systems use radial and\u00a0 The cylindrical coordinate system extends polar coordinates into 3D by using the Cylindrical coordinates are defined with respect to a set of Cartesian\u00a0 In this lesson, we introduce two coordinate systems that are useful alternatives to Cartesian coordinates in three dimensions. any help ? Jul 07, 2009 \u00b7 If you have any vector in Cartesian coordinates then to transform it to Cylindrical coordinates you use r = sqrt(x^2 + y^2) theta = atan(y/x) z = z That part is easy. edu. The following are the conversion formulas for cylindrical coordinates. Khan Academy is a 501(c)(3) nonprofit organization. May 11, 2019 \u00b7 Approach 1 for deriving the Divergence in Cylindrical. If there\u2019s a one to one mapping between coordinate systems, we can convert between them. For example, x, y and z are the parameters that de\ufb01ne a vector r in Cartesian coordinates: r =\u02c6\u0131x+ \u02c6y + \u02c6kz (1) Similarly a vector in cylindrical polar coordinates is described in terms of the parameters r, \u03b8 and z since a vector r can be written as r = rr\u02c6+ z\u02c6k. system Homework This calculator allows you to convert between Cartesian, polar and cylindrical coordinates. Similarly, the angle that a line makes with the horizontal can be defined by the formula \u03b8 = tan-1(m), where m is the slope of the line. There are three commonly used coordinate systems: Cartesian, cylindrical and spherical. The scalar components can be expressed using Cartesian, cylindrical, or spherical coordinates, but we must always use Cartesian base vectors. Convert coordinates from Cartesian to spherical and back. Morrison, Michigan Technological University Cartesian Coordinates L \u00ec\u0303 \u00eb \u00eb \u00ec\u0303 \u00eb \u00ec \u00ec\u0303 \u00eb \u00ed \u00ec\u0303 \u00ec \u00eb \u00ec\u0303 \u00ec \u00ec \u00ec\u0303 \u00ec \u00ed \u00ec\u0303 \u00ed \u00eb \u00ec \u00ed \u00ec \u00ec\u0303 \u00ed \u00ed M \u00eb \u00ec \u00ed L \u00e4 \u00c9 \u00c8 \u00c8 \u00c8 \u00c7 2 \u00f2 R \u00eb \u00f2 T \u00f2 R \u00eb \u00f2 U E \u00f2 R \u00ec \u00f2 T Mar 04, 2017 \u00b7 My question is: does it make a difference if I solve with 2-D cylindrical or 2-D cartesian coordinates and formulation of the Navier Stokes equation? If my mesh is 2-D in r and z, and the flow has no dependence, it seems that the cylindrical form should reduce to the cartesian form (because they can both equally describe my 2D mesh). Both cylindrical and Our page on Cartesian Coordinates introduces the simplest type of coordinate system, where the reference axes are orthogonal (at right angles) to each other. Syntax: set mapping {cartesian | spherical | cylindrical} A cartesian coordinate system is used by default. $\\endgroup$ \u2013 paisanco Jun 14 '14 at 15:57 pol2cart. 1213 0 -5]' x = 4\u00d71 1. To run this script: Download the attached ZIP folder containing the BAS script file and two SRF files: crv2xyz10. Spherical coordinates, also called spherical polar coordinates (Walton 1967, Arfken 1985), are a system of curvilinear coordinates that are natural for describing positions on a sphere or spheroid. 2 Cylindrical Coordinates The cylindrical system is used for problems involving cylindri-cal symmetry It is composed of: (1) the radial distance r2\u201e0;1/, (2) the azimuthal angle, \u02da2\u201e0;2\u02c7/, and z2. For example, you might be studying an object with cylindrical symmetry: uid ow in a pipe, heat ow in a metal rod, or light propagated through a cylindrical optical ber. The relationships between (x;y) and (r; ) are exactly the same as in polar coordinates, and the zcoordinate is unchanged. Using Cartesian coordinates on the plane, the distance between two points (x 1, y 1) and (x 2, y 2) is defined by the formula, which can be viewed as a version of the Pythagorean Theorem. Current Location > Math Formulas > Linear Algebra > Transform from Cartesian to Cylindrical Coordinate Transform from Cartesian to Cylindrical Coordinate Don't forget to try our free app - Agile Log , which helps you track your time spent on various projects and tasks, :) The \"magnitude\" of a vector, whether in spherical/ cartesian or cylindrical coordinates, is the same. In cylindrical coordinates, (r; ;z), the continuity equation for an incompressible uid is 1 r @ @r (ru r) + 1 r @ @ (u ) + @u z @z = 0 In cylindrical coordinates, (r; ;z), the Navier-Stokes equations of motion for an incompress-ible uid of constant dynamic viscosity, , and density, \u02c6, are \u02c6 Du r Dt u2 r = @p @r + f r+ 52u r u r r2 2 r2 @u Online calculator for definite and indefinite multiple integrals using Cartesian, polar, cylindrical, or spherical coordinates. The rotated Cartesian coordinate method to remove the axial singularity of cylindrical coordinates in finite\u2010difference schemes for elastic and viscoelastic waves Mingwei Zhuang Department of Electronic Science, Institute of Electromagnetics and Acoustics, Xiamen University, Xiamen, 361005 China Converts from Cartesian (x,y,z) to Cylindrical (\u03c1,\u03b8,z) coordinates in 3-dimensions. The spherical coordinates of a point are related to its Cartesian coordinates as follows: Online calculator for definite and indefinite multiple integrals using Cartesian, polar, cylindrical, or spherical coordinates. 2. Nov 20, 2009 \u00b7 Converting to Cylindrical Coordinates. 12 Compute $\\ds \\int_{-3}^3\\int_0^{\\sqrt{9-x^2}} \\sin(x^2+y^2)\\,dy\\,dx$ by converting to cylindrical coordinates. ] Show that your equation in step 5 is equivalent to r = c in cylindrical coordinates. Express the values from Steps 1 and 2 as a Cylindrical and spherical coordinates Review of Polar coordinates in IR2. The calculator converts cylindrical coordinate to cartesian or spherical one. , the vector connecting the origin to a general point in space) onto the - plane and the -axis. For example, the circle of radius 2 may be described as the set of all points whose coordinates x and y satisfy the equation x 2 + y 2 = 2 2 . The n- and t-coordinates move along the path with the particle Tangential coordinate is parallel to the velocity The positive direction for the normal coordinate is toward the center of curvature ME 231: Dynamics Path variables along the tangent (t) and normal (n) In 3D Cartesian coordinates, Burkhart addressed the definition, existence, and uniqueness of the DGF and derived asymptotic expansion formulae, applicable at distances far from the source. Recall that $$x=r*cos Cylindrical coordinates are defined with respect to a set of Cartesian coordinates, and can be converted to and from these coordinates using the atan2 function as follows. Ex. In[1]:= Oct 26, 2005 \u00b7 Spherical coordinates, also called spherical polar coordinates (Walton 1967, Arfken 1985), are a system of curvilinear coordinates that are natural for describing positions on a sphere or spheroid. Cylindrical Coordinates. When converted into cartesian coordinates, the new values will be depicted as (X, Y, Z). If all you want is the ability to specify velocities in cylindrical coordinates, just pick this option in the boundary condition GUI. 1 As the cylinder had a simple equation in cylindrical coordinates, so does the sphere in spherical coordinates: \\rho=2 is the sphere of radius 2. In this section, we provide a working definition of the DGF and a numerical method to calculate \u0398 in Cartesian and cylindrical coordinates. www. We shall choose coordinates for a point P in the plane z=zP as follows. In Cartesian coordinates, the three unit vectors are denoted i x, i y, i z. e. The formula for it is as follows: It\u2019s important to take into account that the definition of \\(\\rho$$ differs in spherical and cylindrical coordinates. A Cartesian coordinate system (UK: / k \u0251\u02d0 \u02c8 t i\u02d0 zj \u0259 n /, US: / k \u0251\u02d0r \u02c8 t i \u0292 \u0259 n /) is a coordinate system that specifies each point uniquely in a plane by a set of numerical coordinates, which are the signed distances to the point from two fixed perpendicular oriented lines, measured in the same unit of length. Consider a differential element in Cartesian coordinates\u2026 Write a Cartesian equation of the cylindrical surface of radius c in the left-hand figure above. What we\u2019ll need: 1. \u03b8r Cylindrical coordinates just adds a z-coordinate to the polar coordinates (r,\u03b8). The partial derivatives with respect to x, y and z are converted into the ones with respect to \u03c1, \u03c6 and z. When we know a point in Cartesian Coordinates (x,y) and we want it in Polar Coordinates (r,\u03b8) we solve a right triangle with\u00a0 After rectangular (aka Cartesian) coordinates, the two most common an useful coordinate systems in 3 dimensions are cylindrical coordinates (sometimes called\u00a0 16 Jun 2018 PDF | On Jun 15, 2018, Audu Eliazar Elisha and others published Laplacian Equation: From Cartesian to Cylindrical Coordinate System | Find,\u00a0 Download scientific diagram | Transformation from Cartesian to cylindrical coordinates preserving the waveguide's width and length. Selecting Z outputs translation along the Z-axis in both Cartesian and cylindrical coordinate systems. The cylindrical coordinate system is a generalization of two-dimensional polar coordinates to three dimensions. r = square root of (x 2 + y 2) \u0398 = tangent inverse(y/x) z = z cylindrical,and spherical coordinates CM3110 Fall 2011Faith A. THETA is a counterclockwise angular displacement in radians from the positive x -axis, RHO is the distance from the origin to a point in the x-y plane, and Z is the height above cylindrical-coordinate wave equation 2 2 2 2 2 2 2 2 2 2 1 z q c t\u2202 \u2202 + \u2202 + \u2202 = + \u03c1 \u03c6, (1) which allowed us to transform Eq. 2 Introduction Gradient of a scalar field Divergence of a vector field As shown below, the results for the scattering cross section computed using cylindrical coordinates agree well with the 3d Cartesian simulation. We can see here that r=2 and\u00a0 Examples of orthogonal coordinate systems include the Cartesian (or rectangular ), the cir- cular cylindrical, the spherical, the elliptic cylindrical, the parabolic\u00a0 The colored area under the picture is the unit area in polar coordinates. In the Transform Sensor dialog box, coordinates that make up more than one coordinate system appear only once. The second set of coordinates is known as cylindrical coordinates. 3 Resolution of the gradient The derivatives with respect to the cylindrical coordinates are obtained by differentiation through the Cartesian coordinates, @ @r D @x @r @ @x DeO rr Dr r; @ @\u02da D @x @\u02da @ @x DreO \u02dar Drr \u02da: Nabla may now be resolved on the Convert the three-dimensional Cartesian coordinates defined by corresponding entries in the matrices x, y, and z to cylindrical coordinates theta, rho, and z. 5708 3. The polar coordinates of a point P = (x,y) in the \ufb01rst quadrant are given So I'll say that point has the coordinates, tells me where to find that point, negative 2, negative 5. . (The subject is covered in Appendix II of Malvern's textbook. Illustration of cylindrical coordinates illustrating the effect of changing each of the three cylindrical coordinates on the location of a point. 0000 The polar coordinates of a point P \u2208 R2 is the ordered pair (r,\u03b8) de\ufb01ned by the picture. The radial part of the solution of this equation is, unfortunately, not May 16, 2011 \u00b7 Transformation of cartesian coordinates or rectangular coordinates to cylindrical coordinates: The cylindrical coordinates can be transformed to cartesian or rectangular coordinates and vice versa and the relations will be: x = rcos \u0398. Continuing with our example, let's sketch the surface represented by z = x 2\u00a0 Converts 3D rectangular cartesian coordinates to cylindrical polar coordinates. Draw solids bounded by quadric surfaces using Cartesian Coordinates; Polar Coordinates; Cylindrical Coordinates; Spherical Coordinates; Let us discuss all these types of coordinates are here in brief. Exercises. The vector field is already expressed with Cartesian base vectors, therefore we only need to change the Cartesian coordinates in each scalar component into spherical coordinates. The same is true of triple integrals. Referring to figure 2, it is clear that there is also no radial velocity. E 9. 3-D Cartesian coordinates will be indicated by $x, y, z$ and cylindrical coordinates with $r,\\theta,z$. Select the appropriate separator: comma, semicolon, space or tab (use tab to paste data directly from/to spreadsheets). The painful details of calculating its form in cylindrical and spherical coordinates follow. The Cartesian coordinate system provides a straightforward way to describe the location of points in space. The inputs x, y (, and z) must be the same shape, or scalar. 1 Cartesian Coordinate System . Conversion from cartesian to spherical coordinates: Cartesian [x, y, z] Spherical [r, \u03b8, \u03c6] Conversion from spherical to cylindrical coordinates: Spherical [r, \u03b8, \u03c6] Cylindrical [\u03c1, \u03c6', z'] \u03c1 = r sin \u03b8 \u03c6' = \u03c6 z' = r cos \u03b8 Conversion from cylindrical to spherical coordinates: Cylindrical [\u03c1, \u03c6, z] Spherical [r, \u03b8, \u03c6'] \u03b8 = arctan Transform Cartesian coordinates to polar or cylindrical coordinates. 5\u20132) must be given. Ex 15. To find a1 requires a two step process: 1) Project x\u02c6 onto the line formed by r\u02c6 and its projection onto the xy plane Aug 28, 2012 \u00b7 if i,j,k are unit vectors in cartesian system & e(r), e(\u03b8), e(z) are unit vectors in cylindrical system, i have to show that- 'i'= 'e(r)' cos\u03b8 \u2013 'e(\u03b8)' sin\u03b8 'j' = 'e(r)' sin\u03b8+ 'e(r)' cos\u03b8 'k'= 'e(r)' the quantities in '_ ' are vectors. r is the distance to the z-axis (0, 0, z). Faith A. The idea behind cylindrical and spherical coordinates is to use angles instead of Cartesian coordinates to specify points in three dimensions. In this coordinate system, a point P is represented by the triple (r; ;z) where r and are the polar coordinates of the projection of Ponto the xy-plane and zhas the same meaning as in Cartesian coordinates. therightgate. Since we assume (\u0394r)2 is negligable\u00a0 Three most common coordinate systems used in 3-dimensional representations are: a) Cartesian coordinates b) Cylindrical (polar) coordinates c) Spherical\u00a0 in two dimensions and cylindrical and spherical coordinates in three dimensions. xr x y z\u02c6\u02c6\u02c6\u02c6 \u02c6 ab c a11 1 1 Note: a1 is the projection of x\u02c6 onto r\u02c6. Use the unit circle to get . cylindrical polar coordinates In cylindrical polar coordinates the element of volume is given by ddddvz=\u03c1\u03c1\u03d5. So the cylindrical coordinates conversion equations are given in Table 1 and Figure 1 shows this relationship. d) x + y + z = 1 to spherical coordinates. Spherical Unit Vectors in relation to Cartesian Unit Vectors r\u02c6\u02c6, , \u03b8\u03c6\u02c6 can be rewritten in terms of xyz\u02c6\u02c6\u02c6, , using the following transformations: rx yz\u02c6 sin cos sin sin cos \u02c6\u02c6\u02c6 Mar 28, 2019 \u00b7 In 3D Cartesian coordinates, Burkhart addressed the definition, existence, and uniqueness of the DGF and derived asymptotic expansion formulae, applicable at distances far from the source. Matlab provides a simple utility for doing that. For a spherical coordinate system, the data occupy two or three columns (or using Code for converting Cartesian (x,y,z) to Cylindrical (\u03c1,\u03b8,z) coordinates 2D/3D Hot Network Questions Do you really need to fire flashes regularly when not otherwise used? x, and y allow you to change (x, y) coordinates into polar . 3. Recall that the position of a point in the plane can be described using polar coordinates $(r,\\theta)$. 7. (2) We now go through a separation-of-variable procedure similar to that which we carried out using Cartesian coordinates in But so are cylindrical coordinates (an extension of two-dimensional polar coordinates to three-dimensional) and spherical polar coordinates. 1. Some surfaces and volumes are more easily (simply) described in cylindrical coordinates. plot(cartesian,\u00a0 We describe three different coordinate systems, known as Cartesian, cylindrical and spherical. Given the azimuthal sweep around the z axis theta as well as the radius of the cylinder r, the Cartesian co-ordinates within a cylinder is defined as: x = r*cos(theta) y = r*sin(theta) z = z If data are provided to splot in spherical or cylindrical coordinates, the set mapping command should be used to instruct gnuplot how to interpret them. Cylindrical Coordinates Orientation relative to the Cartesian standard system: The origins and z axes of the cylindrical system and of the Cartesian reference are coincident. 22 Jul 2014 This video explains how to convert rectangular coordinates to cylindrical coordinates. 9 Coordinate Systems in Space. f) \u03c1sin \u03b8 = 1 to Cartesian coordiantes. theta describes the angle relative to the positive x-axis. In such cases, one has to first transform these coordinates into Cartesian coordinate system (X,Y, Z) and then only molecular graphics software can be used to visualize these molecules. Conversion from cylindrical to rectangular coordinates requires a simple application of the equations listed in Conversion between Cylindrical and Cartesian Coordinates: x = r cos \u03b8 = 4 cos 2 \u03c0 3 = \u22122 y = r sin \u03b8 = 4 sin 2 \u03c0 3 = 2 3 z = \u22122. $\\begingroup$ The OP does need to compute the unit vectors in cylindrical coordinates and use the divergence, curl and Laplacian in cylindrical coordinates to solve Maxwell's equations, but the question was how to transform the tensor. Initializes a set of Cartesian coordinates from the provided set of Cylindrical coordinates. A very common case is axisymmetric flow with the assumption of no tangential velocity ($$u_{\\theta}=0$$), and the remaining quantities are independent of $$\\theta$$. Using change-of-coordinate functions to change from a known parameterization to another Like polar coordinates, cylindrical coordinates will be useful for describing shapes in that are difficult to describe using Cartesian coordinates. Consider the case when a three dimensional region $$U$$ is a type I region , i. any straight line parallel to the $$z$$-axis intersects the boundary of the region Compute Areas and Volumes in Non-Cartesian Coordinates The \"nut\" defined by revolving the curve about the axis can be easily parameterized in cylindrical coordinates. If , , and are smooth scalar, vector and second-order tensor fields, then they can be chosen to be functions of either the Cartesian coordinates , , and , or the corresponding real numbers , , and . Cylindrical and spherical coordinates Recall that in the plane one can use polar coordinates rather than Cartesian coordinates. theta = [0 pi/4 pi/2 pi]' theta = 4\u00d71 0 0. com. By using this website, you agree to our Cookie Policy. A circular cylindrical surface r = r 1; A half-plane containing the z-axis and making angle \u03c6 = \u03c6 1 with the xz-plane; A plane parallel to the xy-plane at z = z 1 Sep 19, 2014 \u00b7 Cylindrical Coordinates in Matlab. (\u03c1, \u03c6, z) is given in cartesian coordinates by: Feb 17, 2016 \u00b7 Representing 3D points inRepresenting 3D points in Cylindrical Coordinates. Review of Cylindrical Coordinates. 1) are not convenient in certain cases. Mar 14, 2015 \u00b7 Coordinates for DNA are usually given in the \u2018Cylindrical polar coordinate system\u2019 because of its helical symmetry. Cylindrical polar coordinates The cylindrical polar coordinates \u03c1\u03d5,,z are given, in terms of the rectangular cartesian coordinates x, y, z by z x y z x y zz = = = \u03c1\u03d5 \u03c1\u03d5 cos sin. This coordinate system works best when integrating cylinders or cylindrical-like objects. Unzip the folder. a. e) r = 2sin\u03b8 to Cartesian coordinates. to_cartesian\u00a0 These formulas are automatically used if we ask to plot the grid of spherical coordinates in terms of Cartesian coordinates: sage: spherical. After rectangular (aka Cartesian) coordinates, the two most common an useful coordinate systems in 3 dimensions are cylindrical coordinates (sometimes called cylindrical polar coordinates) and spherical coordinates (sometimes called spherical polar coordinates). 1213 0 -5. Dec 27, 2016 \u00b7 How do you find the rectangular coordinates if you given the cylindrical coordinate #(5, pi/6, 5)#? See all questions in Converting Coordinates from Rectangular to Polar Impact of this question The $(x,y,z)$ ECEF cartesian coordinates can be expressed in the ellipsoidal coordinates $(\\varphi, \\lambda, h)$, where $\\varphi$ and $\\lambda$ are, respectively, the latitude and longitude from the ellipsoid, and $h$ the height above it. [THETA,RHO,Z] = cart2pol(X,Y,Z) transforms three-dimensional Cartesian coordinates stored in corresponding elements of arrays X, Y, and Z, into cylindrical coordinates. Jul 21, 2020 \u00b7 Cylindrical coordinates are a generalization of two-dimensional polar coordinates to three dimensions by superposing a height (z) axis. Scale factors for each component's direction. Spherical coordinates in IR3. If a third axis, z (height), is added to polar coordinates, the coordinate system is referred to as cylindrical coordinates (r, \u03b8, z). Sep 20, 2016 \u00b7 While Cartesian 2D coordinates use x and y, polar coordinates use r and an angle, $\\theta$. In this approach, you start with the divergence formula in Cartesian then convert each of its element into the cylindrical using proper conversion formulas. The 2d nodes do match exactly. Transform Cartesian coordinates to polar or cylindrical coordinates. Its elements, however, are something of a cross between the polar and Cartesian coordinates systems. 30 Jan 2020 In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a\u00a0 26 May 2020 The third equation is just an acknowledgement that the z z -coordinate of a point in Cartesian and polar coordinates is the same. 36: Cylindrical Coordinates 1. Jun 01, 2018 \u00b7 Recall that cylindrical coordinates are really nothing more than an extension of polar coordinates into three dimensions. (1) The (orthogonal) base vectors in the two systems of coordinates are linked by er Figure B. A set of equations relating the Cartesian coordinates to cylindrical coordinates: 3. 7. Click to copy this\u00a0 Sketching the Surface in Cartesian Coordinates. [i] Fill in the blanks: \u2022 If I convert 1 = x2 - y2 from Cartesian to cylindrical coordinates, I get (Simplify. The coordinate systems allow the geometrical problems to be converted into a numerica When given Cartesian coordinates of the form to cylindrical coordinates of the form , the first and third terms are the most straightforward. \\[x = r\\cos \\theta \\hspace{0. Example 14. Let be a subset of . The chapter introduces functions to deal with elasticity coefficients, strain-displacement relations, constitutive relations, and equilibrium and Nov 08, 2011 \u00b7 This is not a cork screw at all! The problem is that plot3 expects cartesian coordinates, but we plotted cylindrical coordinates. Sometimes, employing angles can make mathematical Cartesian coordinates. 1 c ylindrical coordinates a1. 5\u20132 Cylindrical transformation. Oct 12, 2018 \u00b7 The Cartesian coordinate system plots a point as $(x,y,z)$, where $x,y$ and $z$ are perpendicular distances of the point measured from the planes $y-z, z-x$ and $x-y$ respectively as you may a Here we use the identity cos^2(theta)+sin^2(theta)=1. \u0394A=(2r\u0394r+\u0394r2). Conversion between cylindrical and Cartesian coordinates Section 13. In this section, you will compare grid surfaces in Cartesian, cylindrical, and spherical coordinates. Before going through the Carpal-Tunnel causing calisthenics to\u00a0. EX 4Make the required change in the given equation (continued). For example, to change the polar coordinate . Convert Cartesian coordinates (x, y, z) to cylindrical coordinates (radius, azimuth, z). (r +\u0394r)2\u2212\u03c0(r)2). The global (X, Y, Z) coordinates of the two points defining the axis of the cylindrical system (points a and b as shown in Figure 2) must be given. Syntax [X,Y] = pol2cart(THETA,RHO) [X,Y,Z] = pol2cart(THETA,RHO,Z) Description [X,Y] = pol2cart(THETA,RHO) transforms the polar coordinate data stored in corresponding elements of THETA and RHO to two-dimensional Cartesian, or xy, coordinates. 6. Thus, is the perpendicular distance from the -axis, and the angle subtended between the projection of the radius vector (i. Now recall our first example, where we graphed the surface (in Cartesian coordinates) defined by the equation . However, there is a large discrepancy in performance: for a single Intel Xeon 4. scale_factors (self). ) \u2022 When I convert the point P(k, 0,0) for k > 0 from cylindrical coordinates to Cartesian coordinates, I get \u2022 Working with spherical coordinates, when I sketch the graph of p = k for k > 0, the shape of the graph is when I sketch the graph of o= k for 0 < k < TT, k # 1/2, the Using these in\ufb01nitesimals, all integrals can be converted to cylindrical coordinates. Midpoint formula Cylindrical coordinates To get a third dimension, each point also has a height above the original coordinate system. \u0394\u03b82\u03c0. Then, polar coordinates (r; ) are de ned in IR2 f(0;0)g, and given by r= p x2 The Cartesian, or rectangular, coordinate system is the most widely used coordinate system. \u0394\u03b82. 2. Convert quadric surfaces in cylindrical or spherical coordinates to Cartesian and identify. From polar coordinates. The true origin point (0, 0) may or may not be in the proximity of the map data you are using. Care should be taken, however, when calculating . Cartesian (double[] elements) Initializes a set of Cartesian coordinates from the first 3 consecutive elements in the provided array. Cartesian coordinates in the figure below: (2,3) A Polar coordinate system is determined by a fixed point, a origin or pole, and a zero direction or axis. First I\u2019ll review spherical and cylindrical coordinate systems so you can have them in mind when we discuss more general cases. In this chapter we will describe a Cartesian coordinate system and a cylindrical coordinate system. 0000 2. Cylindrical coordinates is a method of describing location in a three-dimensional coordinate system. In this handout we will \ufb01nd the solution of this equation in spherical polar coordinates. Someone please help me do this? I don't need you to solve the question completely, I just want help in how to solve it and to be shown the right direction because I'm completely lost! Fluent environment supports cylindrical and Cartesian coordinates. E Figure 11. Elasticity equations in cylindrical polar coordinates 1. 1 Spherical coordinates Figure 1: Spherical coordinate system. Examples: planes parallel to coordinate planes, cylindrical parame- terization of cylinder, and spherical parameterization of sphere. Figure 2. 1416 The focus of this chapter is on the governing equations of the linearized theory of elasticity in three types of coordinate systems, namely, Cartesian, cylindrical, and spherical coordinates. x = [1 2. Conventions. Using cylindrical coordinates can greatly simplify a triple integral when the region this with cylindrical coordinates is much easier than it would be for cartesian\u00a0 Abstract: Application peculiarities of the Green's functions method for Cartesian, cylindrical and spherical coordinate system are under consideration. Converts from Cartesian (x,y,z) to Cylindrical (\u03c1,\u03b8,z) coordinates in 3-dimensions. Triple Integrals in Cylindrical Coordinates It is the same idea with triple integrals: rectangular (x;y;z) coordinates might not be the best choice. We want to rotate the above so that the h axis is aligned with the arbitrary axis (Ax, Ay, Az) in other words we want to lookat the point (Ax, Ay, Az) see lookat Next we have a diagram for cylindrical coordinates: And let's not forget good old classical Cartesian coordinates: These diagrams shall serve as references while we derive their Laplace operators. The cylindrical coordinates of a point in $$\\R^3$$ are given by $$(r,\\theta,z)$$ where $$r$$ and $$\\theta$$ are the polar coordinates of the point $$(x, y)$$ and $$z$$ is the same $$z$$ coordinate as in Cartesian coordinates. We introduce cylindrical coordinates by extending polar coordinates with theaddition of a third axis, the z-axis,in a 3-dimensional right-hand coordinate system. Jul 23, 2020 \u00b7 Spherical Coordinates. Also the axis vectors depend on the same variable (in this case \u03c6) which makes for interesting derivatives as we will see in a moment. The next step is to develop a technique for transforming spherical coordinates into Cartesian coordinates, and vice-versa. 13 Compute $\\ds \\int_{0}^a\\int_{-\\sqrt{a^2-x^2}}^0 x^2y\\,dy\\,dx$ by converting to cylindrical coordinates. Figure 3. That supposed to be superposition of a vortex and source. Unfortunately, there are a number of different notations used for the other two coordinates. Since the graph of this equation is a surface formed by revolving a curve about the z-axis, it might be better to use cylindrical coordinates. The origin of the local coordinate system is at the node of interest. \u0394A=(\u03c0. Note. Get the free \"Triple Integral - Cylindrical\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Another necessary information for De\ufb01nition. If you pipe is aligned with the Z axis, you'll already have access to these. Cartesian coordinates. To use the plot3 function we must convert the cylindrical coordinates to cartesian coordinates. 1 Cylindrical Coordinates Free Cartesian to Polar calculator - convert cartesian coordinates to polar step by step This website uses cookies to ensure you get the best experience. Generally, x, y , and z are used in Cartesian coordinates and these are replaced by r, \u03b8 , and z . Using the Cartesian coordinate system, geometric shapes (such as curves) can be described by Cartesian equations \u2014 algebraic equations involving the coordinates of the points lying on the shape. What I take away from your answer is that with FEM I should stay in cartesian coordinates, b/c curvilinear coordinates introduce messy artifacts. Figure 1: A point expressed in cylindrical coordinates. Express A using cylindrical coordinates and cylindrical base vectors. For example, the mapping between spherical polar coordinates and Cartesian coordinates uses these equivalences: Cartesian coordinate system top: two-dimensional coordinate system bottom: three-dimensional coordinate system n. Vector. When transforming from Cartesian to cylindircal, x and y become their polar counterparts. Find the y value. Now we compute compute the Jacobian for the change of variables from Cartesian coordinates to spherical coordinates. In polar coordinates, if ais a constant, then r= arepresents a circle Hi, this is module four of two dimensional dynamics, our learning outcomes for today are to describe a rectangular Cartesian coordinate system, a cylindrical coordinate system and to describe the kinematic relationships of position and velocity in a tangential and normal coordinate system, so the one you are probably most familiar with for studying curvilinear motion, or curvilinear motion of One example is the Z-coordinate, which exists in both Cartesian and cylindrical systems. Input array must have a length of 3 and be in the correct order. The radial, tangential, and axial directions must be defined based on the original coordinates of each node in the node set for which the transformation is invoked. \u00b5 is called the \\polar angle\",  the \\azimuthal angle\". Transform polar or cylindrical coordinates to Cartesian. This cylindrical coordinates converter/calculator converts the rectangular (or cartesian) coordinates of a unit to its equivalent value in cylindrical coordinates, according to the formulas shown above. Triple integrals in spherical coordinates Our mission is to provide a free, world-class education to anyone, anywhere. cartesian laplacian. For a 2D vortex, uz=0. NumPy Random Object Exercises, Practice and Solution: Write a NumPy program to convert cartesian coordinates to polar coordinates of a random 10x2 matrix representing cartesian coordinates. The position vector in cylindrical coordinates becomes r = rur + zk. Is it implicitly set to y=0? 2) This is basically a test problem I wanted to understand before continuing with a more complicated two-phase flow problem in spherical coordinates. Vectors are defined in cylindrical coordinates by (\u03c1, \u03c6, z), where \u03c1 is the length of the vector projected onto the xy-plane, \u03c6 is the angle between the projection of the vector onto the xy-plane (i. Jan 09, 2016 \u00b7 If (a,b) is a are the coordinates of a point in Cartesian Plane, u is its magnitude and alpha is its angle then (a,b) in Polar Form is written as (u,alpha). B. coordinates (x;y) to polar coordinates (r; ). Appreciate your help! I have actually already came across the links. Free Polar to Cartesian calculator - convert polar coordinates to cartesian step by step This website uses cookies to ensure you get the best experience. As we have seen earlier, in two-dimensional space a point with rectangular coordinates can be identified with in polar coordinates and vice versa, where and are the relationships between the variables. from publication: Effect of\u00a0 Cartesian and Polar coordinate converting. A coordinate system in which the coordinates of The term \"Cartesian coordinates\" is used to describe such systems, and the values of the three coordinates unambiguously locate a point in space. It presents equations for several concepts that have not been covered yet, but will be on later pages. In this activity we will show that a suitable change of coordinates can greatly imporve the look of a surface in three-space. The cylindrical coordinate system is convenient to use when there is a line of symmetry that is defined as the z axis. When we expanded the traditional Cartesian coordinate system from two dimensions to three, we simply added a new\u00a0 26 Apr 2018 Cartesian and polar coordinates SVG has a coordinate system that has its origin in the top-left corner, with the X coordinate increasing as we\u00a0 20 Mar 2017 Cylindrical and Spherical Coordinates \u03b8r \u03b8r (r,\u03b8,z) 14 Example: Find the cylindrical coordinates of the point (1,2,3) in Cartesian Coordinates\u00a0 20 Nov 2009 Its form is simple and symmetric in Cartesian coordinates. Likewise, if we\u00a0 To Cartesian coordinates. Convert the cylindrical coordinates to cartesian coordinates in Cylindrical vs. Cartesian Coordinates. Cylindrical Coordinates In the cylindrical coordinate system, , , and , where , , and , , are standard Cartesian coordinates. Cartesian Cylindrical Spherical Cylindrical Coordinates x = r cos\u03b8 r = \u221ax2 + y2 y = r sin\u03b8 tan \u03b8 = y/x z = z z = z Jul 27, 2020 \u00b7 concept of cartesian coordinates system, its axis, its variables and ranges of variables 5. Aug 10, 2016 \u00b7 Integration in cylindrical coordinates (r, \\\\theta, z) is a simple extension of polar coordinates from two to three dimensions. In such a coordinate system you can calculate the distance between two points and perform operations like axis rotations without altering this value. We start from this step: From rectangular coordinates, the arc length of a parameterized function is. y = rsin \u0398. A: The reason we only use Cartesian base vectors for constructing a position vector is that Cartesian base vectors are the only base vectors whose directions are fixed\u2014independent Mar 02, 2013 \u00b7 Cartesian Coordinates vs Polar Coordinates In Geometry, a coordinate system is a reference system, where numbers (or coordinates) are used to uniquely determine the position of a point or other geometric element in space. In the cylindrical coordinate system, a point P (x;y;z); whose Cartesian A general system of coordinates uses a set of parameters to de\ufb01ne a vector. D. Think of coordinates as different ways of expressing the position of the vector. The variable \u03c1 is the distance of a coordinate point from the z Cartesian axis, and \u03c6 is its azimuthal angle. Each point is determined by an angle and a distance relative to the zero axis and the origin. The coordinate systems allow the geometrical problems to be converted into a numerica Figure 1. zip. Cartesian coordinates: If we wanted to write r\u02c6\u02c6, , \u03b8\u03c6\u02c6 in terms of xyz\u02c6\u02c6\u02c6, , , we would need to use the angles of and . Either r or rho is used to refer to the radial coordinate and either phi or theta to the azimuthal coordinates. In cylindrical coordinates, they are i r, i, i z, and in spherical coordinates, i r, i, i. The following equations describe the relationship between a Cartesian coordinate and a cylindrical coordinate: x = \u00b7 cos, y = \u00b7 sin, z = z Elasticity equations in cylindrical polar coordinates 1. In the Cartesian system the coordinates are perpendicular to one another with the same unit length\u00a0 Convert from rectangular to spherical coordinates. Cylindrical coordinates are a generalization of two-dimensional polar coordinates to three dimensions by superposing a height (z) axis. For instance when integrating vector function in Cartesian coordinates we can take the unit vectors outside the integral, since they are constant. Cartesian coordinates consist of a set of mutually perpendicular axes, which intersect at a 1-1-2 Circular Cylindrical Coordinates . Purpose of use Too lazy to do homework myself. Therefore, a Cartesian coordinate system is used, where the origin (0, 0) is toward the lower left of the planar section. May 18, 2020 \u00b7 Cartesian coordinates (Section 4. 7854 1. The level surface of points such that z \u02c6z z=zP define a plane. Thanks for your time. Find more Mathematics widgets in Wolfram|Alpha. Explanation: A polar coordinate is in the form (r,\u03b8) , where r is the distance from the origin and \u03b8 is the corresponding angle. Working in cylindrical coordinates is essentialy the same as working in polar coordinates in two dimensions except we must account for the z-component of the system. The Newtonian Constitutive Equation in Cartesian, Cylindrical, and Spherical coordinates Prof. Cylindrical just adds a z-variable to polar. The cylindrical radial coordinate is the perpendicular distance from the point to the z axis. person_outline Anton schedule 2018-10-22 12:49:06 This calculator is intended for coordinates transformation from / to the following 3d coordinate systems: As in the case of Cartesian coordinates, analytical solutions are readily obtained for unidirectional problems in cylindrical and spherical coordinates. The vector k is introduced as the direction vector of the z-axis. The cylindrical system is closely\u00a0 26 Feb 2018 The calculation uses Cartesian coordinates. However the governing equations where i am using this velocity profile are written in spherical co ordinates. For example, there are times when a problem has Oct 22, 2019 \u00b7 Although Cartesian coordinates can be used in three dimensions (x, y, and z), polar coordinates only specify two dimensions (r and \u03b8). Thus, ! r V =ure \u00f6 r+u\"e \u00f6 \"+uze \u00f6 z=0e \u00f6 r+u\"e \u00f6 \"+0e \u00f6 z Processing Triple Integrals in Cartesian Coordinates Calculation of a triple integral in Cartesian coordinates can be reduced to the consequent calculation of three integrals of one variable. 2 Cylindrical Coordinates These are coordinates for a three-dimensional space. If we start with the Cartesian equation of the sphere and substitute, we get the spherical equation: \\eqalign{ x^2+y^2+z^2&=2^2\\cr \\rho^2\\sin^2\\phi\\cos^2\\theta+ \\rho^2\\sin^2\\phi\\sin^2\\theta+\\rho^2\\cos^2\\phi&=2^2\\cr \\rho^2\\sin^2\\phi Convert the three-dimensional Cartesian coordinates defined by corresponding entries in the matrices x, y, and z to cylindrical coordinates theta, rho, and z. 4 Relations between Cartesian, Cylindrical, and Spherical Coordinates. c) \u03c1 = 2cos \u03c6 to cylindrical coordinates. Fields in Cylindrical Coordinate Systems. 22 Oct 2019 Coordinate systems provide a way to specify a point in space. The coordinate system uses the standard polar coordinate system in the x-y plane, utilizing a distance from the origin (r) and an angle (\u03b8) of Transforms 3d coordinate from / to Cartesian, Cylindrical and Spherical coordinate systems. In polar coordinates we specify a point using the distance rfrom the origin and the angle with the x-axis. The transformation from Cartesian To convert the Cartesian nabla to the nabla for another coordinate system, say\u2026 cylindrical coordinates. Later in the course, we will also see how cylindrical coordinates can be useful in calculus, when evaluating limits or integrating in Cartesian coordinates is very difficult. I have vector in cartesian coordinate system: \\\\vec{a}=2y\\\\vec{i}-z\\\\vec{j}+3x\\\\vec{k} And I need to represent it in cylindrical and spherical coord. Slide 2 \u2019 & $% Polar coordinates in IR2 De nition 1 (Polar coordinates) Let (x;y) be Cartesian coordinates in IR2. When this is the case, Cartesian coordinates (x;y;z) are converted to cylindrical coordinates (r; ;z). \u03c1) and the positive x-axis (0 \u2264 \u03c6 < 2\u03c0), z is the regular z-coordinate. An example is given below. May 26, 2020 \u00b7 Section 1-12 : Cylindrical Coordinates As with two dimensional space the standard $$\\left( {x,y,z} \\right)$$ coordinate system is called the Cartesian coordinate system. Figure 1 illustrates the relation between Cartesian Oct 10, 2019 \u00b7 Some of the Worksheets below are Cylindrical and Spherical Coordinates Worksheets, list of Formulas that you can use to switch between Cartesian and polar coordinates, identifying solids associated with spherical cubes, translating coordinate systems, approximating the volume of a spherical cube, \u2026 First off, the definition of your cylindrical co-ordinates is wrong. The conventional choice of coordinates is shown in Fig. Consider a cartesian, a cylindrical, and a spherical coordinate system, related as shown This converter/calculator converts a cartesian, or rectangular, coordinate to its equivalent cylindrical coordinate. 0000 Oct 13, 2010 \u00b7 Homework Statement This seems like a trivial question (because it is), and I'm just not sure if I'm doing it right. 25in}y = r\\sin \\theta \\hspace{0. The polar coordinates are defined in terms of r r r and \u03b8 \\theta \u03b8, where r r r is the distance of the point from the origin Using various functions, you can convert data between Spherical, Cartesian, and Cylindrical coordinate systems. For example, there are different languages in which the word \"five\" is said differently, but it is five regardless of whether it is said in English or Spanish, say. This tutorial will denote vector quantities with an arrow atop a letter, except unit vectors that define coordinate systems which will have a hat. INSTRUCTIONS: Choose the preferred angle units and enter the following: (r) Polar radius (\u0398) Polar angle (z) Vertical offset; Cartesian from Cylindrical: The calculator returns the Cartesian coordinates (x, y, z). The global coordinates of the two points defining the axis of the cylindrical system (points a and b as shown in Figure 2. Here's what they look like: The Cartesian Laplacian looks pretty straight forward. 9: Cylindrical and Spherical Coordinates In the cylindrical coordinate system, a point Pin space is represented by the ordered triple (r; ;z), where rand are polar coordinates of the projection of Ponto the xy-plane and zis the directed distance from the xy-plane to P. The chain rule relates the Cartesian operators Cartesian, the circular cylindrical, and the spherical. Solutions to steady unidimensional problems can be readily obtained by elementary methods as shown below. Translating Spherical Coordinates to Cartesian Coordinates. The most well-known coordinate system is the Cartesian coordinate to use, where every point has an x-coordinate and y-coordinate expressing its horizontal position, and The Wave Equation in Cylindrical Coordinates Overview and Motivation: While Cartesian coordinates are attractive because of their simplicity, there are many problems whose symmetry makes it easier to use a different system of coordinates. x = r cos \u03b8 = 4 cos 2 \u03c0 3 = \u22122 y = r sin \u03b8 = 4 sin 2 \u03c0 3 = 2 3 z = \u22122. Recall: A grid surface of a 3-d coordinate system is a surface generated by holding one of the coordinates constant while letting the other two vary. Dec 10, 2019 \u00b7 CFX actually gives you cylindrical coordinates (r and theta) about the Z axis of any coordinate system. The cylindrical (left) and spherical (right) coordinates of a point. Cylindrical coordinates are an alternative to the more common Cartesian coordinate system. to a rectangular coordinate, follow these steps: Find the x value. A natural extension of the 2d polar coordinates are cylindrical coordinates, since they just add a height value out of the xy Conversion from cartesian to spherical coordinates: Cartesian [x, y, z] Spherical [r, \u03b8, \u03c6] Conversion from spherical to cylindrical coordinates: Spherical [r, \u03b8, \u03c6] Cylindrical [\u03c1, \u03c6', z'] \u03c1 = r sin \u03b8 \u03c6' = \u03c6 z' = r cos \u03b8 Conversion from cylindrical to spherical coordinates: Cylindrical [\u03c1, \u03c6, z] Spherical [r, \u03b8, \u03c6'] \u03b8 = arctan Relationships in Cylindrical Coordinates This section reviews vector calculus identities in cylindrical coordinates. Morrison Continuity Equation, Cartesian coordinates \u2202\u03c1 \u2202t + vx \u2202\u03c1 \u2202x +vy \u2202\u03c1 \u2202y +vz \u2202\u03c1 \u2202z +\u03c1 \u2202vx \u2202x + \u2202vy \u2202y + \u2202vz \u2202z = 0 Continuity Equation, cylindrical coordinates \u2202\u03c1 \u2202t + 1 r \u2202(\u03c1rvr) \u2202r + 1 r \u2202(\u03c1v\u03b8) \u2202\u03b8 + \u2202(\u03c1vz) \u2202z = 0 This article contains a download link for a script which converts cylindrical or spherical coordinates to xyz coordinates for use in Surfer. Overrides: fromReference in class CoordinateSystem Parameters: tuples - float array in Cartesian coordinates ordered as x, y, z Returns: float array containing the radius, azimuth and z values Next: An example Up: Cylindrical Coordinates Previous: Regions in cylindrical coordinates The volume element in cylindrical coordinates. b) 2 2x + y- z2 = 1 to spherical coordinates. 1;1/, which can be thought of as height Transformation between Cartesian and Cylindrical Coordinates; Velocity Vectors in Cartesian and Cylindrical Coordinates; Continuity Equation in Cartesian and Cylindrical Coordinates; Introduction to Conservation of Momentum; Sum of Forces on a Fluid Element; Expression of Inflow and Outflow of Momentum; Cauchy Momentum Equations and the Navier Jan 24, 2017 \u00b7 The basic form of heat conduction equation is obtained by applying the first law of thermodynamics (principle of conservation of energy). He's associating, all of a sudden, these relationships with points on a coordinate plane. coordinates and back again anytime. com/ for more stuff. Cylindrical coordinates to Cartesian coordinates. (1) into Z Z R R R T T c \u2032\u2032 + \u03a6 \u03a6\u2032\u2032 + = \u2032\u2032+ \u2032 \u2032\u2032 2 1 1 1 1 \u03c1 \u03c1. P = ( r, ) x y r 0 0 Theorem (Cartesian-polar transformations) The Cartesian coordinates of a point P = (r,\u03b8) in the \ufb01rst quadrant are given by x = r cos(\u03b8), y = r sin(\u03b8). The cylindrical coordinate system is similar to that of the spherical coordinate system, but is an alternate extension to the polar coordinate system. Regardless, one should be able, in principle, to write down the coordinate transformations in the following form: May 16, 2011 \u00b7 I need to work out how to convert phi = pi/3 from spherical coordinates to cartesian and cylindrical coordinates. So, coordinates are written as (r,$\\theta\\$, z). Velocity And Acceleration In Cylindrical Coordinates Velocity of a physical object can be obtained by the change in an object's position in respect to time. The relations used for the conversion of the coordinates of the point from the Cartesian coordinate system to the cylindrical coordinate system are: In many problems involving cylindrical polar coordinates, it is useful to know the line and volume elements; these are used in integration to solve problems involving paths and volumes. which means that y = 1. When we get to triple integrals, some integrals are more easily evaluated in cylindrical coordinates and you will even have some integrals that can't be evaluated in rectangular coordinates but can be in cylindrical. The above result is another way of deriving the result dA=rdrd(theta). 1 4/6/13 a ppendix 1 e quations of motion in cylindrical and spherical coordinates a1. ) This is intended to be a quick reference page. Rectangular coordinates are depicted by 3 values, (X, Y, Z). Circular Cylindrical Coordinates System concept of cylindrical coordinates system, its axis, its Cylindrical coordinates are a simple extension of the two-dimensional polar coordinates to three dimensions. Move the sliders to convert cylindrical coordinates to Cartesian coordinates for a comparison. 11 Jul 2018 about the Conversion between Cylindrical & Cartesian systems for Electromagnetism. Cylindrical coordinates definition, a member of a system of coordinates for locating a point in space by its polar coordinates and its perpendicular distance to the polar plane. 0000 the part of the solution depending on spatial coordinates, F(~r), satis\ufb01es Helmholtz\u2019s equation \u22072F +k2F = 0, (2) where k2 is a separation constant. When we talk about the point with coordinates (x,y,z)'' or `the surface with equation f(x,y,z)'', we will always have in mind cartesian coordinates. Site: http://mathispower4u. Cylindrical Coordinates. Circular cylindrical coordinates use the plane polar coordinates \u03c1 and \u03c6 (in place of x and y) and the z Cartesian coordinate. \u2013 Cartesian coordinates \u2013 Cylindrical coordinates \u2013 Spherical coordinates. 2GHz processor, the runtime of the cylindrical simulation is nearly 90 times shorter than the 3d simulation. Sponsored Links. [Hint: Think about the distance of any point ( x , y , z ) on the cylinder from the z -axis. What is more challenging is determining the velocity vector in Cylindrical coordinates if you have a position in Cylindrical coordinates as a function of time. The Cartesian Nabla: 2. What is dV in cylindrical coordinates? Well, a piece of the cylinder looks like so which tells us that We can basically think of cylindrical coordinates as polar coordinates plus z. There's three independent variables, x, y, and z. B-5 Feb 24, 2015 \u00b7 Preliminaries. The z component does not change. For example, in the Cartesian coordinate system, the cross-section of a cylinder concentric with the $$z$$-axis requires two coordinates to describe: \\(x Cylindrical coordinates are depicted by 3 values, (r, \u03c6, Z). cartesian to cylindrical coordinates\n\ncbaxcjsr, 2kshboa zs0xeiw, 0ip9r1ovv, c37tgzajkey, u4t wr8nnwkgux, q i6 xh fgsk7v0vda5,", "date": "2021-11-27 17:55:50", "meta": {"domain": "sramble-communication.com", "url": "http://angs-diasporanew.sramble-communication.com/apdsmyht/cartesian-to-cylindrical-coordinates.php", "openwebmath_score": 0.931949257850647, "openwebmath_perplexity": 535.9419383625223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9914225158534695, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8682749313616078}}
{"url": "https://math.stackexchange.com/questions/520657/a-conjectured-closed-form-of-int-limits-0-infty-fracx-1-sqrt2x-1-ln-l/2081059", "text": "# A conjectured closed form of $\\int\\limits_0^\\infty\\frac{x-1}{\\sqrt{2^x-1}\\ \\ln\\left(2^x-1\\right)}dx$\n\nConsider the following integral: $$\\mathcal{I}=\\int\\limits_0^\\infty\\frac{x-1}{\\sqrt{2^x-1}\\ \\ln\\left(2^x-1\\right)}dx.$$\n\nI tried to evaluate $\\mathcal{I}$ in a closed form (both manually and using Mathematica), but without success.\n\nHowever, if WolframAlpha is provided with a numerical approximation $\\,\\mathcal{I}\\approx 3.2694067500684...$, it returns a possible closed form: $$\\mathcal{I}\\stackrel?=\\frac\\pi{2\\,\\ln^2 2}.$$ Further numeric caclulations show that this value is correct up to at least $10^3$ decimal digits. So, I conjecture that this is the exact value of $\\mathcal{I}$.\n\nQuestion: Is this conjecture correct?\n\n\u2022 After a substitution of y = 2^x-1, does this not reduce to knowledge about identities of polylogaritms ?\n\u2013\u00a0mick\nOct 9 '13 at 21:41\n\nSub $u=\\log{(2^x-1)}$. Then $x=\\log{(1+e^u)}/\\log{2}$, $dx = (1/\\log{2}) (du/(1+e^{-u})$. The integral then becomes\n\n\\begin{align}\\frac{1}{\\log{2}} \\int_{-\\infty}^{\\infty} \\frac{du}{1+e^{-u}} e^{-u/2} \\frac{\\frac{\\log{(1+e^u)}}{\\log{2}}-1}{u} = \\frac{1}{2\\log^2{2}} \\int_{-\\infty}^{\\infty} \\frac{du}{\\cosh{(u/2)}} \\frac{\\log{(1+e^u)}-\\log{2}}{u}\\\\ = \\underbrace{\\frac{1}{2\\log^2{2}} \\int_{-\\infty}^{0} \\frac{du}{\\cosh{(u/2)}} \\frac{\\log{(1+e^u)}-\\log{2}}{u}}_{u\\rightarrow -u} \\\\+ \\frac{1}{2\\log^2{2}} \\int_{0}^{\\infty} \\frac{du}{\\cosh{(u/2)}} \\frac{\\log{(1+e^u)}-\\log{2}}{u}\\\\ = \\underbrace{-\\frac{1}{2\\log^2{2}} \\int_{0}^{\\infty} \\frac{du}{\\cosh{(u/2)}} \\frac{\\log{(1+e^{-u})}-\\log{2}}{u}}_{\\log{(1+e^{-u})} = \\log{(1+e^u)}-u}\\\\+ \\frac{1}{2\\log^2{2}} \\int_{0}^{\\infty} \\frac{du}{\\cosh{(u/2)}} \\frac{\\log{(1+e^u)}-\\log{2}}{u}\\\\ \\end{align}\n\nThe nasty pieces of the integral cancel, and we are left with\n\n$$\\frac{1}{2\\log^2{2}}\\int_{0}^{\\infty} \\frac{du}{\\cosh{(u/2)}} = \\frac{\\pi}{2 \\log^2{2}}$$\n\nas correctly conjectured.\n\n\nWith the change of variables $z \\equiv 2^{x} - 1\\yy x = \\ln\\pars{1 + z}/\\ln\\pars{2},\\ {\\cal I}$ is reduced to $${\\cal I} = {1 \\over \\ln^{2}\\pars{2}} \\int\\limits_{0}^{\\infty}{\\ln\\pars{1 + z} - \\ln\\pars{2} \\over z^{1/2}\\,\\pars{1 + z}\\,\\ln\\pars{z}} \\,\\dd z$$ Now, we split the integral from $\\pars{0, 1}$ and from $\\pars{1, \\infty}$. In the second one, we makes the change $z \\to 1/z$ such that we are left with an integration over $\\pars{0, 1}$: \\begin{align} {\\cal I} &= {1 \\over \\ln^{2}\\pars{2}} \\int\\limits_{0}^{1}{\\ln\\pars{1 + z} - \\ln\\pars{2} \\over z^{1/2}\\,\\pars{1 + z}\\,\\ln\\pars{z}} \\,\\dd z + {1 \\over \\ln^{2}\\pars{2}}\\int\\limits_{0}^{1} {\\ln\\pars{1 + 1/z} - \\ln\\pars{2} \\over z^{-1/2}\\,\\pars{1 + 1/z}\\,\\bracks{-\\ln\\pars{z}}} \\,{\\dd z \\over z^{2}} \\\\[3mm]&= {1 \\over \\ln^{2}\\pars{2}} \\int\\limits_{0}^{1}{\\ln\\pars{1 + z} - \\ln\\pars{2} \\over z^{1/2}\\,\\pars{1 + z}\\,\\ln\\pars{z}} \\,\\dd z - {1 \\over \\ln^{2}\\pars{2}}\\int\\limits_{0}^{1} {\\ln\\pars{1 + z} - \\ln\\pars{z} - \\ln\\pars{2} \\over z^{1/2}\\,\\pars{1 + z}\\,\\ln\\pars{z}}\\,\\dd z \\\\[3mm]&= {1 \\over \\ln^{2}\\pars{2}} \\int\\limits_{0}^{1}{1 \\over z^{1/2}\\,\\pars{1 + z}} \\,\\dd z\\,, \\quad \\pars{~\\mbox{Let's}\\quad r \\equiv z^{1/2}\\yy\\ z = r^{2}~} \\\\[3mm]&= {2 \\over \\ln^{2}\\pars{2}} \\underbrace{\\quad\\int\\limits_{0}^{1}{\\dd r \\over r^{2} + 1}\\quad} _{\\ds{\\arctan\\pars{1}\\ =\\ {\\pi \\over 4}}} = \\color{#ff0000}{\\Large{\\pi \\over 2\\ln^{2}\\pars{2}}} \\end{align}\n\nLet $\\tan^2t=2^x-1\\;\\Rightarrow\\;x=\\dfrac{\\ln(1+\\tan^2t)}{\\ln 2}\\;\\Rightarrow\\;dx=\\dfrac{2\\tan t\\sec^2t\\ dt}{(1+\\tan^2t)\\ln 2}$ and the corresponding region is $0<t<\\dfrac\\pi2$. Using identity $\\sec^2t=1+\\tan^2t$, then the integral turns out to be $$\\mathcal{I}=\\frac{1}{\\ln^22}\\int_0^{\\Large\\frac\\pi2}\\frac{2\\ln(\\sec t)-\\ln2}{\\ln (\\tan t)}\\ dt.\\tag1$$ Now, using property $$\\int_b^af(x)\\ dx=\\int_b^af(a+b-x)\\ dx$$ equation $(1)$ becomes $$\\mathcal{I}=\\frac{1}{\\ln^22}\\int_0^{\\Large\\frac\\pi2}\\frac{2\\ln(\\csc t)-\\ln2}{\\ln (\\cot t)}\\ dt.\\tag2$$ Adding $1$ and $2$ yields \\begin{align} 2\\mathcal{I}&=\\frac{1}{\\ln^22}\\int_0^{\\Large\\frac\\pi2}\\left(\\frac{2\\ln(\\sec t)-\\ln2}{\\ln (\\tan t)}+\\frac{2\\ln(\\csc t)-\\ln2}{\\ln (\\cot t)}\\right)\\ dt\\\\ &=\\frac{1}{\\ln^22}\\int_0^{\\Large\\frac\\pi2}\\left(\\frac{2\\ln\\left(\\dfrac{1}{\\cos t}\\right)-\\ln2}{\\ln (\\tan t)}+\\frac{2\\ln\\left(\\dfrac{1}{\\sin t}\\right)-\\ln2}{\\ln \\left(\\dfrac{1}{\\tan t}\\right)}\\right)\\ dt\\\\ &=\\frac{1}{\\ln^22}\\int_0^{\\Large\\frac\\pi2}\\left(\\frac{-2\\ln(\\cos t)-\\ln2}{\\ln (\\tan t)}+\\frac{2\\ln(\\sin t)+\\ln2}{\\ln (\\tan t)}\\right)\\ dt\\\\ &=\\frac{2}{\\ln^22}\\int_0^{\\Large\\frac\\pi2}\\frac{\\ln(\\sin t)-\\ln(\\cos t)}{\\ln \\left(\\dfrac{\\sin t}{\\cos t}\\right)}\\ dt\\\\ \\mathcal{I}&=\\frac{1}{\\ln^22}\\int_0^{\\Large\\frac\\pi2}\\ dt\\\\\\\\ \\int_0^\\infty\\frac{x-1}{\\sqrt{2^x-1}\\ \\ln\\left(2^x-1\\right)}dx&=\\large\\color{blue}{\\frac{\\pi}{2\\ln^22}}.\\qquad\\qquad\\qquad\\blacksquare \\end{align}\n\n\u2022 Wow, this is remarkably brilliant! I'm impressed. Jun 10 '16 at 12:06\n\u2022 This is really nice thanks\n\u2013\u00a0Alex\nSep 4 '20 at 10:22\n\nSubstitute $(2^x-1) = t^2$ to get,\n\n$\\text{I} = \\displaystyle \\dfrac{1}{\\ln^2 2} \\int_{0}^{\\infty} \\left( \\dfrac{\\ln (t^2+1) - \\ln 2}{(t^2 + 1) \\ln t} \\right) \\mathrm{d}t$\n\nSubstitute $t \\mapsto \\dfrac{1}{t}$\n\n$\\implies \\text{I} = -\\displaystyle \\dfrac{1}{\\ln^2 2} \\int_{0}^{\\infty} \\left( \\dfrac{\\ln (t^2+1) - \\ln 2}{(t^2 + 1) \\ln t} \\right) \\mathrm{d}t + \\dfrac{2}{\\ln^2 2} \\int_{0}^{\\infty} \\dfrac{\\mathrm{d}t}{t^2+1}$\n\n$\\implies \\text{I} = -\\text{I} + \\dfrac{\\pi}{\\ln^2 2}$\n\n$\\implies \\text{I} = \\dfrac{\\pi}{2 \\ln^2 2}$\n\n\u2022 good one $(+1)$ Jul 9 '17 at 20:48\n\u2022 And with the right substitution the integral just dissolves away. Nice work. Jan 25 '19 at 1:04\n\u2022 It looks easy this way\n\u2013\u00a0Alex\nSep 4 '20 at 10:22", "date": "2022-01-16 20:15:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/520657/a-conjectured-closed-form-of-int-limits-0-infty-fracx-1-sqrt2x-1-ln-l/2081059", "openwebmath_score": 0.9994262456893921, "openwebmath_perplexity": 5144.077020503496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109105959843, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.868272856593234}}
{"url": "https://stats.stackexchange.com/questions/147255/drawing-n-intervals-uniformly-randomly-probability-that-at-least-one-interval-o", "text": "# Drawing n intervals uniformly randomly, probability that at least one interval overlaps with all others\n\nRandomly draw $n$ intervals from $[0,1]$, where each end point A,B are selected from the uniform distribution between $[0,1]$.\n\nWhat's the probability that at least one interval overlaps with all others?\n\n\u2022 You can look at the probability that the last drawn $A_n$ is smaller than the minimum of all previously drawn $A$, and the probability that the last $B_n$ is greater than the maximum of all previously drawn $B$. This should be helpful. Then inflate the probability to account for the fact that we don't need the last one, but any one. (I don't have the time to work through it, but it looks like a fun little problem. Good luck!) Apr 20, 2015 at 7:02\n\u2022 It may be somewhat surprising that (1) the answer does not depend on the distribution (only that it be continuous) and (2) for $n\\gt 1$ it is constant!\n\u2013\u00a0whuber\nApr 22, 2015 at 15:41\n\u2022 Is this how the nth interval is construted: i) draw two numbers uniformly at random from [0,1], ii) let the smaller one be $A_n$ and the larger one $B_n$?\n\u2013\u00a0KOE\nApr 23, 2015 at 21:36\n\nThis post answers the question and outlines partial progress toward proving it correct.\n\nFor $n=1$, the answer trivially is $1$. For all larger $n$, it is (surprisingly) always $2/3$.\n\nTo see why, first observe that the question can be generalized to any continuous distribution $F$ (in place of the uniform distribution). The process by which the $n$ intervals are generated amounts to drawing $2n$ iid variates $X_1, X_2, \\ldots, X_{2n}$ from $F$ and forming the intervals\n\n$$[\\min(X_1,X_2), \\max(X_1,X_2)], \\ldots, [\\min(X_{2n-1}, X_{2n}), \\max(X_{2n-1}, X_{2n})].$$\n\nBecause all $2n$ of the $X_i$ are independent, they are exchangeable. This means the solution would be the same if we were randomly to permute all of them. Let us therefore condition on the order statistics obtained by sorting the $X_i$:\n\n$$X_{(1)} \\lt X_{(2)} \\lt \\cdots \\lt X_{(2n)}$$\n\n(where, because $F$ is continuous, there is zero chance that any two will be equal). The $n$ intervals are formed by selecting a random permutation $\\sigma\\in\\mathfrak{S}_{2n}$ and connecting them in pairs\n\n$$[\\min(X_{\\sigma(1)},X_{\\sigma(2)}), \\max(X_{\\sigma(1)},X_{\\sigma(2)})], \\ldots, [\\min(X_{\\sigma(2n-1)}, X_{\\sigma(2n)}), \\max(X_{\\sigma(2n-1)}, X_{\\sigma(2n)})].$$\n\nWhether any two of these overlap or not does not depend on the values of the $X_{(i)}$, because overlapping is preserved by any any monotonic transformation $f:\\mathbb{R}\\to\\mathbb{R}$ and there are such transformations that send $X_{(i)}$ to $i$. Thus, without any loss of generality, we may take $X_{(i)}=i$ and the question becomes:\n\nLet the set $\\{1,2,\\ldots, 2n-1, 2n\\}$ be partitioned into $n$ disjoint doubletons. Any two of them, $\\{l_1,r_1\\}$ and $\\{l_2,r_2\\}$ (with $l_i \\lt r_i$), overlap when $r_1 \\gt l_2$ and $r_2 \\gt l_1$. Say that a partition is \"good\" when at least one of its elements overlaps all the others (and otherwise is \"bad\"). As a function of $n$, what is the proportion of good partitions?\n\nTo illustrate, consider the case $n=2$. There are three partitions,\n\n$$\\color{gray}{\\{\\{1,2\\},\\{3,4\\}\\}},\\ \\color{red}{\\{\\{1,4\\},\\{2,3\\}\\}},\\ \\color{red}{\\{\\{1,3\\},\\{2,4\\}\\}},$$\n\nof which the two good ones (the second and third) have been colored red. Thus the answer in the case $n=2$ is $2/3$.\n\nWe may graph such partitions $\\{\\{l_i,r_i\\},\\,i=1,2,\\ldots,n\\}$ by plotting the points $\\{1,2,\\ldots,2n\\}$ on a number line and drawing line segments between each $l_i$ and $r_i$, offsetting them slightly to resolve visual overlaps. Here are plots of the preceding three partitions, in the same order with the same coloring:\n\nFrom now on, in order to fit such plots easily in this format, I will turn them sideways. For instance, here are the $15$ partitions for $n=3$, once again with the good ones colored red:\n\nTen are good, so the answer for $n=3$ is $10/15=2/3$.\n\nThe first interesting situation occurs when $n=4$. Now, for the first time, it is possible for the union of the intervals to span $1$ through $2n$ without any single one of them intersecting the others. An example is $\\{\\{1,3\\},\\{2,5\\},\\{4,7\\},\\{6,8\\}\\}$. The union of the line segments runs unbroken from $1$ to $8$ but this is not a good partition. Nevertheless, $70$ of the $105$ partitions are good and the proportion remains $2/3$.\n\nThe number of partitions increases rapidly with $n$: it equals $1\\cdot 3\\cdot 5 \\cdots \\cdot 2n-1 = (2n)!/(2^nn!)$. Exhaustive enumeration of all possibilities through $n=7$ continues to yield $2/3$ as the answer. Monte-Carlo simulations through $n=100$ (using $10000$ iterations in each) show no significant deviations from $2/3$.\n\nI am convinced there is a clever, simple way to demonstrate there is always a $2:1$ ratio of good to bad partitions, but I have not found one. A proof is available through careful integration (using the original uniform distribution of the $X_i$), but it is rather involved and unenlightening.\n\n\u2022 Very cool. I have a hard time following what it means to \"condition on the order statistics\", would it be possible to add a line of intuition? Seems like a useful technique. I understand up to that the $X_i$ are exchangeable, indeed even $iid$, that that this allows us to consider any permutation.\n\u2013\u00a0KOE\nApr 24, 2015 at 16:10\n\u2022 @Student To \"condition on\" means to say, let's temporarily hold these values fixed and consider what we can learn from that. Later, we will let those values vary (according to their probability distribution). In this case, once we find that the answer is $2/3$ regardless of the fixed values of the order statistics, then we no longer have to carry out the second step of varying the order statistics. Mathematically, the order stats are a vector-valued variable $\\mathbf{X}$ and the indicator of being good is $Y$, so $$\\mathbb{E}(Y)=\\mathbb{E}(\\mathbb{E}(Y|\\mathbf{X}))=\\mathbb{E}(2/3)=2/3.$$\n\u2013\u00a0whuber\nApr 24, 2015 at 16:36\n\u2022 This very question came back on FiveThirtyEight The Riddler. Jun 2, 2020 at 9:22", "date": "2022-06-28 16:32:18", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/147255/drawing-n-intervals-uniformly-randomly-probability-that-at-least-one-interval-o", "openwebmath_score": 0.8855133652687073, "openwebmath_perplexity": 235.21054515300273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109068841395, "lm_q2_score": 0.8774767890838837, "lm_q1q2_score": 0.8682728533361765}}
{"url": "https://web2.0calc.com/questions/last-question_4", "text": "+0\n\n# last question\n\n0\n38\n3\n+81\n\nHow many different rational numbers between 1/1000 and 1000 can be written either as a power of\u00a02 or as a power of 3,\u00a0where the exponent is a (possibly negative) integer?\n\nI got 31; is this correct?\n\nmathbum \u00a0Oct 6, 2018\n#1\n+1\n\n$$2^9=512$$\n\n$$3^6=729$$\n\nThose are the largest positive powers of 2 and 3 under 1000.\n\n$$2^{-9}=0.001953.....$$\n\n$$3^{-6}=0.0013717.....$$\n\nThese are the smallest negative powers greater than\u00a0 $$\\frac{1}{1000}$$\n\nTherefore, we have\u00a0$$9-(-9)+1=19$$ different powers of 2 between\u00a0$$\\frac{1}{1000}$$\u00a0and 1000.\n\nWe also have\u00a0$$6-(-6)+1=13$$\u00a0different powers of 3 between\u00a0$$\\frac{1}{1000}$$ and 1000.\n\nThis gives us\u00a0$$19+13=32$$\u00a0total integers...\n\n32 is the answer I got but someone is going to have to check over my work :b\n\nGuest\u00a0Oct 6, 2018\n#2\n+2362\n+1\n\nThe problem is symmetric in x vs. 1/x so we can find the number of integers between 0 and 1000 that are either a power of two or 3 and just double the answer.\n\nNote that any non-zero\u00a0power of 2 is even and any non-zero power of 3 is odd so these two sets are disjoint.\n\nThere are 10\u00a0powers of 2, 0-9, and 6 non-zero powers of 3 that are less than 100.\n\nWe use non-zero power for 3 since we only want to count 1 = 20 = 30 once\n\nThat gets us 16\u00a0from 0\u00a0to 1000.\u00a0 Doubling this we get 32 but we don't want to count\u00a0$$1 = \\dfrac{1}{1}$$\n\ntwice so we subtract 1 from this getting 31 as you found.\n\nRom \u00a0Oct 6, 2018\nedited by Rom \u00a0Oct 6, 2018\nedited by Rom \u00a0Oct 6, 2018\nedited by Rom \u00a0Oct 6, 2018\n#3\n+93644\n0\n\nThat is an interesting question and good answers from both guest and Rom.\n\nThanks :)\n\nMelody \u00a0Oct 6, 2018", "date": "2018-10-18 04:01:57", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/last-question_4", "openwebmath_score": 0.9279128313064575, "openwebmath_perplexity": 1040.5773112988422, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989510906574819, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8682728403828436}}
{"url": "https://math.stackexchange.com/questions/4115542/probability-of-drawing-cards-that-sum-to-10-given-a-starting-card", "text": "Probability of drawing cards that sum to 10, given a starting card\n\nTwo people are playing a card game. They are using a reduced deck of cards, consisting of A, 2, 3, ..., 9 for each of the four suits (i.e. 36 cards). In this game an ace has a value of 1.\n\nPlayer A deals a single card to themselves and Player B.\n\nPlayer A has a 5 of diamonds.\nPlayer B has a 7 of spades.\n\nPlayer A then draws one card at a time trying to reach a sum of 10, with the starting card. If they go over 10, they lose, and the cards drawn are shuffled back into the deck.\n\nWhat is the probability that A wins, and what is the probability that B wins. (Winning means they can form a sum of 10).\n\nThe answers given are 0.1536 for A, and 0.1468 for B.\n\nI can get the answer for B as follows: $$\\frac{nCr(4,1)}{nCr(34,1)}+\\frac{nCr(4,1)}{nCr(34,1)}\\times\\frac{nCr(4,1)}{nCr(33,1)}\\times2+\\frac{nCr(4,3)}{nCr(34,3)}$$ which is the probability of a 3 + probability of a 2 and an A + probability of three Aces.\n\nHowever, I can't get the answer for A, even trying very similar techniques.\n\n\u2022 Do the players draw the cards alternatively? Does the game end if A or B first reach 10? Does the game end (without a winner) if they both go over 10?\n\u2013\u00a0user\nApr 25, 2021 at 13:23\n\nThe answer for $$A$$ seems to have approximation error. Here is how I look at $$A$$ getting to sum of $$10$$.\n\nIn one draw - gets one of the remaining $$3$$ cards with face value $$5$$.\nIn two draws - $$(4,1)$$ or $$(3,2)$$\nIn three draws - $$(1, 1, 3)$$ or $$(2, 2, 1)$$\nIn four draws - $$(1, 1, 1, 2)$$\nIn five draws - $$(1, 1, 1, 1, 1)$$.\n\nSo desired probability $$= \\displaystyle \\small \\frac{3}{34} + 2 \\cdot 2! \\big(\\frac{4}{34}\\big)^2 + 2 \\cdot \\frac{3!}{2!} \\big(\\frac{4}{34}\\big)^3 + \\frac{4!}{3!} \\big(\\frac{4}{34}\\big)^4 + \\big(\\frac{4}{34}\\big)^5$$\n\n$$= \\displaystyle \\small \\frac{437763}{2839714} \\approx 0.154$$\n\n\u2022 Thanks for your help. Though as the game is involving a deck of cards it's not possible to have five aces. Apr 26, 2021 at 4:42\n\u2022 you are right but I am going by the question that \"cards drawn are shuffled back into the deck\" so I can draw $5$ Aces in $5$ draws with probability $(\\frac{4}{34})^5$. Apr 26, 2021 at 7:00", "date": "2022-09-30 06:21:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4115542/probability-of-drawing-cards-that-sum-to-10-given-a-starting-card", "openwebmath_score": 0.7114970088005066, "openwebmath_perplexity": 272.46303851938444, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180675433345, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8682176043974468}}
{"url": "https://math.stackexchange.com/questions/1973627/help-explaining-structural-induction", "text": "# Help explaining Structural Induction\n\nI am trying to wrap my head around structural induction. Can someone break it down and explain it around this problem?\n\nLet S, a subset of $$\\mathbb{N}*\\mathbb{N}$$, be defined recursively by:\n\nBase case: $$(0,0)$$ $$\\in S$$\n\nConstructor case: If $$(m,n) \\in S$$, then $$(m+5,n+1) \\in S$$\n\nProve that if $$(m,n) \\in S$$, then m+n is a multiple of 3.\n\nHow is it different than normal induction (using this example please) and what is the point of a Constructor case? Can someone wright the proof out so i can see what this structural induction proof looks like?\n\n\u2022 Her we are \"performing induction\" not on $\\mathbb N$ but of $\\mathbb N \\times \\mathbb N$, and not all pairs $(n,m)$ will satisfy bthe property : $(1,1) \\notin S$ because $1+1$ is not a multiple of $3$. Oct 18, 2016 at 10:08\n\u2022 Now for the inductive step (here the \"constructor case\") ; assume that the property holds for $(m,n)$ and show that it holds for $(m+5,n+1)$. Oct 18, 2016 at 10:12\n\u2022 Here is a video attempting to explain structural induction: youtu.be/u21QV-MlVDY Aug 8 at 15:44\n\nThe set $$S$$ is defined recursively: certain base elements of $$S$$ are specified, in this case just the ordered pair $$\\langle 0,0\\rangle$$, and a rule is given that allows \u2018new\u2019 elements of $$S$$ to be constructed from \u2018old\u2019 ones. Here each \u2018old\u2019 element $$\\langle m,n\\rangle$$ gives rise to just one \u2018new\u2019 one, $$\\langle m+5,n+1\\rangle$$. Thus, in this case\n\n$$S=\\{\\langle 0,0\\rangle,\\langle 5,1\\rangle,\\langle 10,2\\rangle,\\langle 15,3\\rangle,\\ldots\\}\\;.$$\n\nThere is also a rule, often (as in this case) left unstated, to the effect that the only members of $$S$$ are the objects that can be obtained by repeatedly applying the constructor rule(s) to the base elements.\n\nWe can show that every member of $$S$$ has some property $$P$$ by first verifying that each of the base elements has $$P$$ and then showing that the construction process preserves the property $$P$$: that is, if we apply the construction process to objects that have $$P$$, the new objects also have $$P$$. If we can do this, we can conclude by structural induction that every member of $$S$$ has $$P$$.\n\nIn your problem an ordered pair $$\\langle m,n\\rangle$$ has the property $$P$$ if and only if $$m+n$$ is a multiple of $$3$$. This is clearly the case for the one base element $$\\langle 0,0\\rangle$$: $$0+0=0=3\\cdot 0$$ is a multiple of $$3$$. That\u2019s the base case of your structural induction. For the induction step assume that $$\\langle m,n\\rangle\\in S$$ has $$P$$, i.e., that $$m+n$$ is a multiple of $$3$$. When we apply the construction process to $$\\langle m,n\\rangle$$, we get the pair $$\\langle m+5,n+1\\rangle\\in S$$, and we want to show that it also has $$P$$, i.e., that $$(m+5)+(n+1)$$ is a multiple of $$3$$. By hypothesis $$m+n=3k$$ for some integer $$k$$, so\n\n$$(m+5)+(n+1)=m+n+6=3k+6=3(k+2)\\;;$$\n\nand $$k+2$$ is an integer, so $$(m+5)+(n+1)$$ is indeed a multiple of $$3$$. We\u2019ve now shown\n\n\u2022 that the base element $$\\langle 0,0\\rangle$$ has the desired property, and\n\u2022 that the construction process preserves this property: when applied to a pair $$\\langle m,n\\rangle$$ such that $$m+n$$ is a multiple of $$3$$, it produces another pair whose components sum to a multiple of $$3$$.\n\nThese are the base case and induction step of a proof by structural induction; between them they constitute a proof that $$m+n$$ is a multiple of $$3$$ for each $$\\langle m,n\\rangle\\in S$$.\n\n\u2022 Thank you, that was a great explanation. It made it immediately obvious that induction on the natural numbers is a special case of structural induction where the constructor case is $k+1$ for every $k$ in $S$. Jan 9, 2017 at 22:05\n\u2022 @jeremy: You\u2019re welcome; I\u2019m glad that it helped. Jan 9, 2017 at 22:07\n\nMathematical induction is defined over natural number and it is based on two fundamental facts :\n\n\u2022 there is an \"initial\" number : $0$\n\n\u2022 every number $n$ has a unique successor : $n+1$.\n\nStructural induction generalize this type of proof to \"structures\" on which a well-founded partial order is defined, i.e.\n\n\u2022 that have an \"initial\" or minimal element\n\nand\n\n\u2022 they have a partial order.\n\nIt applies to structures recursively defined (such as lists or trees).", "date": "2022-08-14 22:43:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1973627/help-explaining-structural-induction", "openwebmath_score": 0.8445401191711426, "openwebmath_perplexity": 158.4727558789633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985718068592224, "lm_q2_score": 0.8807970826714613, "lm_q1q2_score": 0.8682175991525782}}
{"url": "https://math.stackexchange.com/questions/3100168/determining-if-sequence-has-upper-bound", "text": "# determining if sequence has upper bound\n\nI am somewhat stuck in my calculations when determining if sequence has an upper bound.\n\nThe sequence $$x_n = \\frac{1}{n+1}+\\frac{1}{n+2}+..+\\frac{1}{2n-1}+\\frac{1}{2n}$$ Is equal to $$\\frac{1}{n}(\\frac{1}{1+\\frac{1}{n}}+\\frac{1}{1+\\frac{2}{n}}+..+\\frac{1}{1+\\frac{n}{n}})$$\n\nAnd so I notice that all the denominators are greater than 1, which means that all terms in the parentheses are less than 1.\n\nBut how can I determine further if there is an upper bound?\n\n\u2022 ... and there are $n$ of them. \u2013\u00a0Zeekless Feb 4 at 17:54\n\u2022 @Zeekless so the sequence should be less than (1/n) * n = 1 is that correct? \u2013\u00a0F Wi Feb 4 at 18:02\n\u2022 \u2013\u00a0Martin Sleziak Feb 5 at 8:22\n\nThe largest term is the first, so an obvious upper bound is to set all terms equal to the first one and get $$x_n < \\frac{n}{n+1} <1.$$\n\nYou could also say that, since the last term is the smallest, one has $$x_n > \\frac{n}{2n} = \\frac 12,$$\n\nwhich means that $$\\frac 12 < x_n < 1, n \\in \\mathbb{N}$$.\n\nBy C-S $$\\sum_{i=1}^n\\frac{1}{n+i}=1+\\sum_{i=1}^n\\left(\\frac{1}{n+i}-\\frac{1}{n}\\right)=1-\\frac{1}{n}\\sum_{i=1}^n\\frac{i}{n+i}=$$ $$=1-\\frac{1}{n}\\sum_{i=1}^n\\frac{i^2}{ni+i^2}\\leq1-\\frac{1}{n}\\frac{\\left(\\sum\\limits_{i=1}^ni\\right)^2}{\\sum\\limits_{i=1}^n(ni+i^2)}=1-\\frac{1}{n}\\frac{\\frac{n^2(n+1)^2}{4}}{\\frac{n^2(n+1)}{2}+\\frac{n(n+1)(2n+1)}{6}}=$$ $$=1-\\frac{3(n+1)}{2(5n+1)}=\\frac{7n-1}{10n+2}<\\frac{7}{10}.$$ Actually, $$\\ln2=0.6931...$$ Cauchy-Schwarz forever!\n\nActually, by calculus we can show that $$\\lim_{n\\rightarrow+\\infty}\\sum_{i=1}^n\\frac{1}{n+i}=\\ln2.$$\n\nNotice the Riemann sum $$\\frac1n\\sum_{k=1}^n \\frac1{1+k/n} < \\int_0^1\\frac{dt}{1+t} = \\log 2$$\n\nhint\n\nFor each $$n\\ne 0$$,\n\n$$\\frac{1}{n+1}\\le \\frac{1}{n}$$ $$\\frac{1}{n+2}\\le \\frac{1}{n}$$ ...\n\n$$\\frac{1}{2n}\\le \\frac 1n$$\n\nYou can finish.", "date": "2019-05-20 22:32:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3100168/determining-if-sequence-has-upper-bound", "openwebmath_score": 0.8192958831787109, "openwebmath_perplexity": 339.7295681078063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180694313358, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8682175937229382}}
{"url": "https://math.stackexchange.com/questions/779510/how-many-solutions-does-the-equation-xyz-11-have", "text": "# How many solutions does the equation $x+y+z=11$ have? [duplicate]\n\nHow many solution does $x+y+z=11$ have where $x, y, z$ are non-negative integers. In light of the restrictions, its clear that $x,y,z \\in \\{0,1,2,..11\\}$. So, at face value I would assign a value for $x$ and determine the different combinations that $y$ and $z$ can hold. For example,\n\nFor $x=0$, we have $y+z=11$. With writing them out I found that there are $12$ different assigned combinations for $y$ and $z$ that satisfy the equation. For $x=1$, I got $11$. Consequently, the pattern becomes clear whereby each one takes a value less by one. Hence, the number of solutions is $1+2+3+4+5+6+7..+12=78$. I was wondering if there is an easier method perhaps with combinations equation $C(a,b)$..?\n\n\u2022 We have had too many questions of this type. \u2013\u00a0evil999man May 3 '14 at 12:13\n\u2022 See my answer here:math.stackexchange.com/questions/689975/\u2026 \u2013\u00a0evil999man May 3 '14 at 12:15\n\u2022 Guys, stop upvoting duplicate homework questions, seriously! \u2013\u00a0Alec Teal May 3 '14 at 12:18\n\u2022 @Alec Lighten up. The OP shows plenty of effort, and arrives at the correct answer in doing so. It deserves an upvote. John likely didn't know that this is a classic sort of problem, and probably hasn't encountered it before. Nor that many similar questions can be answered by the same method. And if you believe it is a duplicate and should be slammed shut because of it, then why'd you answer it? Besides, stop playing the homework police on questions showing commendable levels of effort! \u2013\u00a0amWhy May 3 '14 at 12:24\n\u2022 @amWhy Not thinking something deserves an upvote isn't the same as saying it deserves to be closed. When I upvote something, what I personally mean is \"this is the kind of content I come to the site for\". My preferred policy would for routine problems to be dealt with quickly and without fanfare (positive or negative), and to save upvotes (and therefore time on the front page) for the more original, thought-provoking questions. \u2013\u00a0Jack M May 3 '14 at 12:28\n\nThis is a version of the classic stars-and-bars problem in combinatorics.\n\nFor any pair of natural numbers $n$ and $k$, the number of distinct $n$-tuples of non-negative integers whose sum is $k$ is given by the binomial coefficient $$\\binom{n + k - 1}{k}$$\n\nHere, $n = 3$, and $k = 11$, giving you $$\\binom{3 + 11 - 1}{11} = \\binom{13}{11} = \\dfrac{13\\cdot 12}{2} = 6\\cdot 13 = 78$$\n\n\u2022 The link takes you to Wikipedia's Stars and Bars entry, where you'll find a nice explanation of why this works, using an example that's fleshed out nicely. \u2013\u00a0amWhy May 3 '14 at 12:17\n\u2022 Using the formula makes things easy, but I don't see what the connection is. Even with using the stars and bars method, it seems rather inappropriate, due to the dependency of the variables. In other terms, if x takes a value it creates a dependency on the other variables. I would appreciate it if you could make the connection clearer. \u2013\u00a0John May 3 '14 at 13:41\n\u2022 Did you read the Wikipedia entry? It elaborates on the \"why's\" of this formula. \u2013\u00a0amWhy May 3 '14 at 13:45\n\u2022 I read it and I understood everything but when I went to this question, math.stackexchange.com/questions/322369/\u2026, with the additional restrictions I started to doubt everything. None of the answers that used combinations were clear. \u2013\u00a0John May 3 '14 at 13:49\n\u2022 Yes. I see you used $x, y, z$, so if only $x$ needs to be greater than or equal to 2, then the solution (with k=11 - 2 = 9) is $\\binom{3 + 9 - 1}{9}$. \u2013\u00a0amWhy May 3 '14 at 15:17\n\nimagine 11 balls in a row and two blocks which you will place somewhere. you insert the blocks before, after or between the balls and then you assign values to $x,y,z$ in the following way: $x$ is the number of balls from the beginning of the row up to the first block, $y$ the number of balls between the two blocks and $z$ number of balls from the second block up until the end of the row. you will easily see that the number of ways in which you can place the blocks is equal to the number of different triplets $x,y,z$. Do you know how to compute the number of possible distributions of blocks?\n\n\u2022 call it stars and bars if you don't like blocks and balls, but this is the usual way to approach such problems \u2013\u00a0Alessandro Codenotti May 3 '14 at 12:15\n\u2022 it's an answer explained in a tangible way. actually it's pretty much the same thing you gave as an answer although it's explained with words rather than a picture. \u2013\u00a0mm-aops May 3 '14 at 12:15\n\u2022 Oh right! You could have laid it out nicer, I thought you did some weird thing that'd result in a 3! somewhere \u2013\u00a0Alec Teal May 3 '14 at 12:16\n\u2022 Could you edit the answer so I may at least remove my DV? \u2013\u00a0Alec Teal May 3 '14 at 12:17\n\nOkay let us write a solution to $a+b+c+d+e=10$ a different question, just incase it is homework.\n\nEach solution will have the form:\n\n||||-|---||||| <-> 4As 1B 0Cs, 0Ds, 5Es\n\n\nHow many different ways can we arrange 10 |s and 4 (4=5-1) -s? Each arrangement of these |s and -s is a valid solution.\n\n$$\\frac{(10+4)!}{4!10!}=\\frac{14!}{10!4!}$$", "date": "2020-06-05 10:42:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/779510/how-many-solutions-does-the-equation-xyz-11-have", "openwebmath_score": 0.5473933219909668, "openwebmath_perplexity": 417.9469211059002, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180669140008, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8682175884213129}}
{"url": "https://math.stackexchange.com/questions/1554530/rational-canonical-form-from-given-minimal-and-characteristic-polynomial", "text": "# Rational Canonical form from given minimal and characteristic polynomial\n\n$A$ is a $4\\times 4$ matrix over $F$ with characteristic polynomial $(x-1)^4$ and minimal polynomial $(x-1)^2$. What is the rational canonical form of $A$?\n\nMy answer was the following: it is one of the following: $$\\begin{bmatrix} 0 & -1 & & \\\\ 1 & 2 & & \\\\ & & 1 & \\\\ & & & 1 \\end{bmatrix} or \\begin{bmatrix} 0 & -1 & & \\\\ 1 & 2 & & \\\\ & & 0 & -1\\\\ & & 1 & 2 \\end{bmatrix}.$$ While our teacher finally reached at only second form. I am not satisfied with that answer.\n\nMy question is that whether the first matrix here can also be a rational form?\n\nIn general, to write rational canonical form of a matrix, I will proceed as follows: let $$m_A(x)=(x-a_1)^{k_1}(x-a_2)^{k_2}\\cdots.$$ For each factor $(x-a_i)^{k_i}$ write one block diagonal companion matrix.\n\nIf this fills up the matrix size (i.e. if $m_A(x)$ equals characteristic polynomial, then this is required form.\n\nOtherwise, fill up remaining parts (diagonal blocks) by writing companion matrix of factors $(x-a_i)^{l_i}$ where $l_i \\leq k_i$.\n\nIs this correct way?\n\nYes, you have two possible rational canonical forms given the information you have. Both the matrices you wrote have minimal polynomial $(x-1)^2$ and characteristic polynomial $(x-1)^4$.\nTo justify that $A$ has one of the two possible canonical forms above, let $a_1 \\, | \\, a_2 \\, | \\, \\ldots \\, | \\, a_k$ denote the invariant factors of $A$. The highest invariant factor is always the minimal polynomial so $a_k = (x-1)^2$. The characteristic product of the matrix is the product of the invariant factors so we have a priori two options:\n$$a_1(x) = (x-1), a_2(x) = (x-1), a_3(x) = (x-1)^2, \\\\ a_1(x) = (x-1)^2, a_2(x) = (x-1)^2.$$\n\u2022 Determining (all possible) rational canonical forms I mean the following: suppose characteristic pol. is $(x-1)^6$ and minimal polynomial is $(x-1)^3$. Then possible forms are obtained by putting Companion matrices of size $\\leq 3$; the possibilities will be $3+3$, $3+2+1$, $3+1+1+1$. So there will be three possible rational canonical forms when min. pol. is $(x-1)^3$ and char. poly. is $(x-1)^6$. (This situation is almost similar to that in Jordan theory, in which we consider Jordan blocks; in Rational form, we consider Companion blocks. I would like to ensure whether this is correct.) \u2013\u00a0Beginner Dec 5 '15 at 6:05", "date": "2019-07-20 00:57:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1554530/rational-canonical-form-from-given-minimal-and-characteristic-polynomial", "openwebmath_score": 0.9645354747772217, "openwebmath_perplexity": 232.8937731812094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9857180639771091, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.8682175796657796}}
{"url": "https://thinkandstart.com/menace-ii-oqcpwq/if-the-diagonals-of-a-parallelogram-are-perpendicular-3f40df", "text": "These properties concern its sides, angles, and diagonals. If the diagonals of a quadrilateral are perpendicular bisector of each other, it is always a_____ A. Rectangle . Every square is a rhombus. B. Rhombus. The diagonals are perpendicular bisectors of each other. ToProve: if the diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus.. The longer diagonal of a parallelogram measures 62 cm and makes an angle of 30 degrees with the base. We have (2)The diagonals of a square are perpendicular to each other. 5. If the diagonals of a parallelogram are perpendicular then the parallelogram will be a rectangle. If ABCD is a parallelogram, what is the length of BD? Answer. Yes, because a rhombus has two sets of parallel sides and all sides are congruent. Subscribe to bartleby learn! Bob R. Lv 6. \"D\" is the best answer. Plus, you\u2019ll have access to millions of step-by-step textbook answers! Answer. Then we have the two diagonals are A + B and A \u2212 B. Just so, do the diagonals of a trapezium bisect each other at 90 degrees? The diagonals of a parallelogram_____bisect the angles of the parallelogram Sometimes A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram 10. C. Every trapezoid is a parallelogram. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . Median response time is 34 minutes and may be longer for new subjects. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? Get a free answer to a quick problem. We've seen that one of the properties of a rhombus is that its diagonals are perpendicular to each other. The diagonals meet each other at 90\u00b0, this means that they form a perpendicular bisection. If a quadrilateral has 2 pairs of opposite sides that are congruent, then it is a parallelogram. If the diagonals of a quadrilateral both bisect each other, then the quadrilateral is a parallelogram. Sue. For example, the diagonals of a rhombus, kite or square are perpendicular, but those of a rectangle or general parallelogram are not. This is the currently selected item. In order for a parallelogram's diagonals to be perpendicular, the sides would have to have the same length, so this is only true for a rhombus (which is a specific parallelogram with equal sides). by Jennifer Kahle. The diagonals of a parallelogram bisect each other. If the diagonals of a parallelogram are perpendicular to each other, then it is a rhombus ; If the diagonals of a parallelogram are equal and perpendicular, then it is a square \u2235 In a parallelogram, its diagonals bisect each other at right angles \u2234 Its diagonals are perpendicular \u2235 Its diagonals are equal \u2192 By using rule 3 above \u2234 The parallelogram is a square. d.trapezoid. Opposite sides are congruent in parallelogram. In my opinion \"D\" is the best answer, by definition a rhombus is a parallelogram with perpendicular diagonals . 11. Stephen K. If you just look [\u2026] faiqaferoz646 faiqaferoz646 22.06.2020 Math Secondary School Diagonals of a parallelogram are perpendicular to each other. Opposite angles are congruent. If the diagonals of a parallelogram are perpendicular and not congruent, then the parallelogram is. The parallelogram has the following properties: Opposite sides are parallel by definition. The parallelogram has the following properties: Opposite sides are parallel by definition. 10. If all the angles of the rhombus are right angles then you have a special rhombus which is a square A rhombus is a special kind of parallelogram, in which all the sides are equal. Parallelogram and Rhombus: A parallelogram is a quadrilateral (has 4 sides) where its opposite sides are parallel and equal and its opposite angles are equal. When the diagonals of a parallelogram are perpendicular to each other then it is called. So I'm thinking of a parallelogram that is both a rectangle and a rhombus. Start here or give us a call: (312) 646-6365, \u00a9 2005 - 2021 Wyzant, Inc. - All Rights Reserved, a Question Trapezoid Midsegment Theorem. Squares are rhombuses and rectangles, so if it is B, it must be C and D as well. The diagonals bisect each other. In a parallelogram a diagonal of the length 20 cm is perpendicular to one of the sides. . answered \u0095 10/08/20, If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. Here we will show the converse- that if a parallelogram has perpendicular diagonals, it is a rhombus - all its sides are equal. If all the angles of the rhombus are right angles then you have a special rhombus which is a square, Nathaniel A. If the angle at which they meet is a right angle, then a right triangle is formed whose legs are half the length of each of the diagonals, and whose hypotenuse is the length of one side of the parallelogram (rhombus). The midsegment of a trapezoid is parallel to each base and its length is one half the sum of the lengths of both bases. Feb 18, 2016 . In this case, the diagonals divide the rhombus in four congruent right-angled triangles. Squares are another example which is just a specialized rhombus with congruent 90\u00b0 angles. Answer: 3. let long side = x short side = s and is perpendicular to diag s^2 + 400 = x^2 so s^2 = x^2 - 400 2 s + 2 x = 80 s = 40 - x 1600 - 80 x + x^2 = x^2 - 400 80 x = 2000 x = 200/8 = 100/4 = 25 2 0; Damon. Proof: Figure is made having diagonals AC and BD. If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. The diagonals bisect each other. Ask subject matter experts 30 homework questions each month. 1 0 736; Sam. A.) answered \u0095 10/08/20, Eagle Scout and Honor Graduate from West Springfield High School. No packages or subscriptions, pay only for the time you need. Maths. Thus you have a rhombus. EASY. But in the general case, it isn't always true that the diagonals of a parallelogram are perpendicular. (1)The diagonals of a parallelogram are equal. The properties of the parallelogram are simply those things that are true about it. Therefore, by CPCT A D \u2192 \u2245 A B The diagonals of a parallelogram bisect each other. If either diagonal of a parallelogram bisects two angles, then it\u2019s a rhombus (neither the reverse of the definition nor the converse of a property). Area of the parallelogram when the diagonals are known: $$\\frac{1}{2} \\times d_{1} \\times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. C. Square. Find the area of the parallelogram if the diagonals intersect at angle of 70 degrees. Parallelogram. In a parallelogram diagonals bisect each other. Consecutive angles are supplementary. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. Find the longer side of parallelogram if its perimeter is 80 cm. AB\u2192\u2245DC\u2192\u00a0&\u00a0AD\u2192\u2245BC\u2192\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2234(AB\u2192\u2245AD\u2192)\u21d2AD\u2192\u2245DC\u2192\u2245AB\u2192\u2245BC\u2192. So let me see. Proof: The diagonals of a kite are perpendicular. Given that, we want to prove that this is a parallelogram. So A is out. If diagonals of a parallelogram are perpendicular, then it is a . bisects. a.rectangle. Remember a square is a special rectangle with all side lengths equivalent however we have no information regarding the side lengths of this problem. Opposite sides are congruent. Mar 3, 2019 . What is the measure of a base angle \u2026 These angles are said to be congruent with each other. If you just look [\u2026] Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Proof: Opposite angles of a parallelogram. Let's think. And you see the diagonals intersect at a 90-degree angle. The diagonals of parallelograms are perpendicular. rhombus If the diagonals of a parallelogram are perpendicular, then the parallelogram is a _____ C. Four points in the plane form a parallelogram if the opposite line segments have the same slope. Therefore, by CPCT A D \u2192 \u2245 A B The diagonals of a parallelogram bisect each other. trapezoid. (Points : 5) rectangle rhombus square trapezoid perpendicular. The only parallelogram that satisfies that description is a square. If the diagonals of a quadr... maths. Opposite angles are congruent. The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is \u2026 If the diagonals of a parallelogram are perpendicular then the parallelogram will be a rectangle. 0 0. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. If a parallelogram has (at least) one right angle, then it is a rectangle. A parallelogram, the diagonals bisect each other. Proof: Figure is made having diagonals AC and BD. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Prove that, if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus. b.rhombus.....Ans. The diagonals are perpendicular bisectors of each other. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. Answer: Let two adjacent sides of the parallelogram be the vectors A and B (as shown in the \ufb01gure). A. quadrilateral. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? One pair of diagonally opposite angles is equal in measurement. *Response times vary by subject and question complexity. What is the measure of a base angle of an isosceles triangle if \u2026 If all four sides of a parallelogram are equal in length, then the parallelogram is called a rhombus. The diagonals are perpendicular bisectors of each other. A.) P P 5. Every square is a rhombus. (A) Square (B) Rectangle (C) Rhombus (D) Parallelogram HARD. . First Name. 5. The diagonals of a parallelogram bisect each other. B. Slope: Method 2: Show that the diagonals are congruent. If the diagonals of a parallelogram are equal, then it is a rectangle; If the diagonals of a parallelogram are perpendicular to each other, then it is a rhombus; If the diagonals of a parallelogram are equal and perpendicular, then it is a square \u2235 In a parallelogram, its diagonals bisect each other at right angles \u2234 Its diagonals are perpendicular A parallelogram with diagonals that are congruent and perpendicular is a [ Select) A parallelogram with diagonals that are perpendicular, but not always congruent, is a Select] if one diagonal of a parallelogram _____ a pair of opposite angles, then the parallelogram is a rhombus. In \u0394 A O D and \u0394 A O B. O A = O A \u2234 (Common) \u2220 A O D \u2245 \u2220 A O B = 90 \u2218 \u2234 (Diagonals are perpendicular) O D = O B \u2234 (Diagonals of parallelogram bisects) So, \u0394 D O A \u2245 \u0394 B O A by SAS postulate. In \u0394 A O D and \u0394 A O B. O A = O A \u2234 (Common) \u2220 A O D \u2245 \u2220 A O B = 90 \u2218 \u2234 (Diagonals are perpendicular) O D = O B \u2234 (Diagonals of parallelogram bisects) So, \u0394 D O A \u2245 \u0394 B O A by SAS postulate. Proving a Quadrilateral is a Rhombus Method 1: Prove that the diagonals are perpendicular. Play this game to review Geometry. if the diagonal of a parallelogram bisects one of the angles of parallelogram, prove that it also bisects the second angle and then the two diagonals are perpendicular to each other - Math - Quadrilaterals By completing the parallelogram and using the parallelogram rule, the diagonal represents the sum of the two original forces f1 + f2. asked Sep 22, 2018 in Class IX Maths by muskan15 ( -3,443 points) quadrilaterals Question 392003: 1.how to prove if the diagonals in a parallelogram are perpendicular, then the parallelogram is a rhombus. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. Yes, because the diagonals of a rhombus, which is a parallelogram, are perpendicular. 4. Area of the parallelogram when the diagonals are known: $$\\frac{1}{2} \\times d_{1} \\times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Respond to this Question. a rhombus. If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it is a rhombus. Your Response. The diagonals are perpendicular bisectors of each other. So we're going to assume that the two diagonals are bisecting each other. Show that the diagonals of a parallelogram are perpendicular if and only if it is a rhombus, i.e., its four sides have equal lengths. If a parallelogram has perpendicular diagonals, you know it is a rhombus. A Parallelogram with Perpendicular Diagonals is a Rhombus. Figure is made having diagonals AC and BD. 2. how to prove if the diagonals in a paralellogram are congruent, then the parallelogram is a rectangle Answer by Edwin McCravy(18405) (Show Source): You can put this solution on YOUR website! If the diagonals of a parallelogram are perpendicular, then it\u2019s a rhombus (neither the reverse of the definition nor the converse of a property). B is out. If the diagonals of a parallelogram are perpendicular, then the parallelogram is also called a rhombus. If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. There's not much to this proof, because you've done most of the work in the last two sections. Special parallelograms. Here\u2019s a rhombus proof for you. No, because in an isosceles trapezoid the sides that are not parallel are equal. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. Choose an expert and meet online. Is this statement true? Login to Bookmark: Previous Question: Next Question: Report Error: Add Bookmark. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Study.com has thousands of articles about every Generally, the parallelogram law is applied when the vectors are co-initial, that is, their initial points are joined. No, because in an isosceles trapezoid the sides that are not parallel are equal. We are given that all four angles at point E are 9 0 0 and are therefore congruent. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. If the diagonals are congruent than the parallelogram \u2026 Every rhombus is a square. Give reason for your ans\u2026 Get the answers you need, now! Which statement is true? Yes, because a rhombus has two sets of parallel sides and all sides are congruent. If the diagonals of a parallelogram are perpendicular, what can you conclude about the parallelogram? 10. If the diagonals of a quadrilateral are perpendicular to each other,it is a square but it is a rhombus as diagonals of rhombus are also perpendicular. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? A kite is never a parallelogram. All rectangles are parallelograms. Remember a square is a special rectangle with all side lengths equivalent however we have no information regarding the side lengths of this problem. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. The diagonals meet each other at 90\u00b0, this means that they form a perpendicular bisection. If the diagonals of a parallelogram are perpendicular then the parallelogram is from SCI 102 at Crist\u00f3bal Col\u00f3n University Is this statement true? These angles are said to be congruent with each other. Therefore, parallelogram ABCD is rhombus. Back to Basic Ideas page. The diagonals of a parallelogram bisect each other. Consecutive angles are supplementary. c.square. One pair of diagonally opposite angles is equal in measurement. Proof: Rhombus diagonals are perpendicular bisectors. Answer. Yes, because the diagonals of a rhombus, which is a parallelogram, are perpendicular. Since the diagonals of a parallelogram bisect each other, B E and D E are congruent and A E is congruent to itself. Special parallelograms. D. Every parallelogram is a rectangle. D. Parallelogram. Theorems concerning quadrilateral properties. Properties of a Parallelogram. A link to the app was sent to your phone. Most questions answered within 4 hours. If a _____ is a rhombus, then it is a parallelogram. ToProve : if the diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus.. These properties concern its sides, angles, and diagonals. A parallelogram with four congruent sides and four right angles. If ABCD is a parallelogram, what is the length of BD? EASY. -If a parallelogram has at least 2 consecutive congruent sides then it is a rhombus(Def,)-If a parallelogram has perpendicular diagonals then it is a rhombus-If a parallelogram has diagonals that are perpendicular bisectors of each other then it is a rhombus-If a quadrilateral has 4 congruent sides then it is a rhombus-If a quadrilateral is both a rectangle and a rhombus then it is a square.(Def.) If the diagonals of a quadrilateral are perpendicular to each other,it is a square but it is a rhombus as diagonals of rhombus are also perpendicular. For each such a triangle, the legs are 6/2 = 3 units and 8/2 = 4 units; hence, the sides of the rhombus are = 5 units long, and its perimeter is 5*4 = 20 units. 3. a.rectangle b.rhombus c.square d.trapezoid 11. The properties of the parallelogram are simply those things that are true about it. The diagonals of parallelograms are perpendicular. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of each other. Thus you have a rhombus. Related Videos. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. For Free, Inequalities and Relationship in a Triangle, ALL MY GRADE 8 & 9 STUDENTS PASSED THE ALGEBRA CORE REGENTS EXAM. Diagonals of Quadrilaterals -- Perpendicular, Bisecting or Both. So we've just proved-- so this is interesting. Play this game to review Geometry. Diagonals of a parallelogram are perpendicular to each other. 4. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. Assuming vectors A and B are two adjacent sides of a parallelogram; a quick disproof of rhombus would be $|A|\\neq|B|$. 1 decade ago. A) rhombus : B) rectangle : C) quadrilateral : D) none of these : Correct Answer: A) rhombus : Part of solved Quadrilateral and parallelogram questions and answers : >> Elementary Mathematics >> Quadrilateral and parallelogram. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. So we're assuming that that is equal to that and that that right over there is equal to that. B. Rectangles have congruent diagonals, but they are not perpendicular. please follow me and mark as brainliest If the diagonals are congruent and are perpendicular bisectors of each other then the parallelogram is a square. D. Just so, do the diagonals of a trapezium bisect each other at 90 degrees? If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? Proof: Opposite sides of a parallelogram. That is true for some parallelograms but not all. The diagonals of parallelograms are perpendicular. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. OA=OA\u2009\u2009\u2009\u2234(Common)\u2220AOD\u2245\u2220AOB=90\u2218\u2009\u2009\u2009\u2234(Diagonals\u00a0are\u00a0perpendicular)OD=OB\u2009\u2009\u2009\u2234(Diagonals\u00a0of\u00a0parallelogram\u00a0bisects). Opposite sides are congruent. McDougal Littell Jurgensen Geometry: Student Edition Geometry. Which shape is not a parallelogram? In a parallelogram diagonals bisect each other. (3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus. Proof: Diagonals of a parallelogram. 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Od=Ob \u2234 ( diagonals of a quadrilateral are perpendicular ask subject matter experts homework... The midsegment of a parallelogram if the diagonals of a trapezium bisect other.: Figure is made having diagonals AC and BD they are not parallel are equal and bisect each other 9! True about it sides, angles, if the diagonals of a parallelogram are perpendicular diagonals ans\u2026 Get the you! ) Every quadrilateral is a rhombus length of BD remember a square by. Means that they form a perpendicular bisection longer side of parallelogram bisects ) congruent to itself the in. That the two diagonals are bisecting each other at 90\u00b0, this means that they form a perpendicular bisection so! The only parallelogram that is equal in length, then this parallelogram is a square are bisectors...", "date": "2021-05-06 09:50:41", "meta": {"domain": "thinkandstart.com", "url": "https://thinkandstart.com/menace-ii-oqcpwq/if-the-diagonals-of-a-parallelogram-are-perpendicular-3f40df", "openwebmath_score": 0.6866893172264099, "openwebmath_perplexity": 404.38305370460273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9838471632833943, "lm_q2_score": 0.8824278788223264, "lm_q1q2_score": 0.8681741653815287}}
{"url": "https://math.stackexchange.com/questions/2938253/calculation-of-the-probability-of-not-mutually-exclusive-and-independent-event", "text": "# Calculation of the probability of not mutually exclusive and independent event.\n\nI tried to solve this math problem in two different ways but found two different answers and I don't know which one is the right answer.\n\nThe problem goes like this:\n\nThe probability of failing in Science exam for a student is $$\\frac{1}{5}$$, probability of passing in both Science and English is $$\\frac{3}{4}$$ and probability of passing any one of these two subjects or both is $$\\frac{7}{8}$$, what is the probability of him passing only in English?\n\n$$\\mathbf{Solution:}$$ Let S and E are the events of passing in Science and English respectively. probability of failing in Science $$P(S^c) = \\frac{1}{5}$$, so probability of passing in Science exam is, $$P(S) = 1-P(S^c)=1-\\frac{1}{5}=\\frac{4}{5}$$ probability of passing both subjects, $$P(S\\cap E)= \\frac{3}{4}$$;\n\nand probability of passing one or both subjects, $$P(S\\cup E)=\\frac{7}{8}$$\n\nSince these two events are independent and not mutually exclusive, we can write, $$P(S\\cup E)=P(S)+P(E)-P(S\\cap E)$$ Where P(E) is the probability of passing in English. Which is $$P(E)=P(S\\cup E)-P(S)+P(S\\cap E)$$ $$P(E)=\\frac{7}{8}-\\frac{4}{5}+\\frac{3}{4}$$ $$P(E) = \\frac{33}{40}$$ At this point I can calculate the probability of him passing only in English in two ways:\n\n(1) The probability of him passing only in English is $$P(E\\cap S^c)$$\n\n$$P(E)=P(E\\cap S)+P(E\\cap S^c)$$ Addition rule, since these two events are mutually exclusive. So, $$P(E\\cap S^c)=P(E)-P(S\\cap E)=\\frac{33}{40}-\\frac{3}{4}=\\frac{3}{40}$$\n\n(2)The probability of him passing only in English is $$P(E\\cap S^c)$$.\n\nSince these two events are independent, $$P(E\\cap S^c)=P(E)P(S^c)=\\frac{33}{40}\\times\\frac{1}{5}=\\frac{33}{200}$$\n\nThis two methods give two different answers $$\\frac{3}{40}$$ and $$\\frac{33}{200}$$\n\nI am obviously missing something or I have some lack in understanding. It will be very helpful if someone point out which one is wrong and why.\n\nThe key point is that $$E$$ and $$S$$ are dependent. Thus $$P(E\\cap S^c)=P(E)\\cdot P(S^c|E)=\\frac{33}{40}\\cdot \\frac{1}{11}=\\frac{3}{40}$$\n\nYou can comprehend $$P(S^c|E)$$ by looking at the table below:\n\n$$\\begin{array}{c|c|c|c} &E & E^c& \\\\ \\hline S &\\color{red}{0.75} & 0.05 & 0.8 \\\\ \\hline S^c & 0.075 & \\color{blue}{0.125} &\\color{red}{0.2} \\\\ \\hline & 0.825 & 0.175 &\\color{red}{1} \\end{array}$$\n\nand probability of passing any one of these two subjects or both is $$\\frac78$$\n\nThat means that $$P(S^c\\cap E^c)=1-\\frac78=\\frac18=\\color{blue}{0.125}$$\n\nFinding out the missing figures is just simplest algebra.\n\nYou see that $$P(S^c)=0.2 \\neq P(S^c|E)$$. Thus $$E$$ and $$S$$ are not independent. And your first approach is right $$\\checkmark$$\n\n\u2022 Thank you for your help. But would you like to show me in detail how to get the table? Or is there any easy way to figure out the dependency of events? I thought events like in the problem is independent by common sense and that's how I generally used to approach any problem. Thanks in advance. \u2013\u00a0Arafat Hossen Oct 2 '18 at 3:33\n\u2022 In general if I have only two events I firstly make a table, almost always. The red figures are given in the text. And then the missing figures can be filled in successively. Especially the 0.8 should be very obvious since the cells of the last column/row sum up to one. I\u00b4ve made an edit. Please have a look. \u2013\u00a0callculus Oct 2 '18 at 5:03\n\u2022 @ArafatHossen You always have to prove if events are independent. There are two variants of the criteria. Basically the meaning is identical: 1. $P(A\\cap B)=P(A)\\cdot P(B)$, 2. $P(A|B)=P(A)$. It works as well by using the complements: $P(A\\cap \\overline B)=P(A)\\cdot P(\\overline B)\\Rightarrow$ The events A and B are independent. \u2013\u00a0callculus Oct 2 '18 at 5:21", "date": "2019-09-20 20:19:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2938253/calculation-of-the-probability-of-not-mutually-exclusive-and-independent-event", "openwebmath_score": 0.6878759860992432, "openwebmath_perplexity": 208.6477521001761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471637570105, "lm_q2_score": 0.8824278664544911, "lm_q1q2_score": 0.8681741536314012}}
{"url": "https://www.physicsforums.com/threads/is-second-rank-tensor-always-tensor-product-of-two-vectors.910628/", "text": "# I Is Second rank tensor always tensor product of two vectors?\n\n#### arpon\n\nSuppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ? If so, then I have a few more questions:\n1. Are those two vectors $A_i$ and $B_j$ unique?\n2. How to find out $A_i$ and $B_j$\n3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?\n\n#### fresh_42\n\nMentor\n2018 Award\nSuppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ?\nNo, it is a sum of such products.\nIf so, then I have a few more questions:\n1. Are those two vectors $A_i$ and $B_j$ unique?\nNo. Even for dyadics $A_i \\otimes B_j$ you always have $A_i \\otimes B_j = c \\cdot A_i \\otimes \\frac{1}{c}B_j$ for any scalar $c \\neq 0$.\n2. How to find out $A_i$ and $B_j$\n$T_{ij}$ is basically any matrix and $A_i \\otimes B_j$ a matrix of rank $1$. So write your matrix as a sum of rank-$1$ matrices and you have a presentation.\n3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?\nNo. See the rank explanation above.\n\n#### haushofer\n\n3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?\nTo add to Fresh's answer: this last remark already should make you suspicious. An arbitrary (!) second-rank tensor in three dimensions has 3*3=9 components, but two vectors have 2*3=6 components. What you can do, is to decompose a second rank tensor like $T_{ij}$ as\n\n$$T_{ij} = T_{[ij]} + T_{(ij)}$$\n\nwhere [ij] stands for antisymmetrization, whereas (ij) stands for symmetrization. Both parts transform independently under coordinate transfo's. The antisymmetric part has 3 independent components, whereas the symmetric part has 6 components. You can even go further in this decomposition, because the trace of the tensor components also does not change under a coordinate transformation.\n\n\"Is Second rank tensor always tensor product of two vectors?\"\n\n### Physics Forums Values\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-10-18 13:20:02", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/is-second-rank-tensor-always-tensor-product-of-two-vectors.910628/", "openwebmath_score": 0.9270875453948975, "openwebmath_perplexity": 571.5117273402026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471642306268, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8681741434022803}}
{"url": "https://mathematica.stackexchange.com/questions/21197/how-do-i-simplify-a-vector-expression", "text": "How do I simplify a vector expression?\n\nI am doing vector analysis. I have figured out that the following expression won't be simplified in Mathematica:\n\nSimplify[Dot[x, y] - Dot[y, x]]\n\n\nI think the reason is that x and y could be matrices, so generally the operation does not commute. But for a vector, the commutation relation should hold. What I don't know is how to tell Mathematica to consider x and y to be vectors.\n\nAlso, how can I expand Cross[x + y, z] to be Cross[x, y] + Cross[y, z]? I tried to use Expand and ExpandAll but neither worked.\n\nBTW, in Mathematica, is it possible to define a abstract row/column vector without explicitly specifying the number of entries?\n\nHere is a way to do all the things you asked for automatically, independently of Mathematica version. The approach relies on a special symbol to identify when we're dealing with a vector: Instead of using things like x, y etc. for vectors, the convention now is that vectors are written as vec[x], vec[y], etc.\n\nYou could also define the wrapper OverVector[x] for this purpose because it displays as $\\vec{x}$. But for this post I want to keep it simple, and the arrows wouldn't display easily in the source code below.\n\nClearAll[scalarProduct, vec];\nSetAttributes[scalarProduct, {Orderless}]\nvec /: Dot[vec[x_], vec[y_]] := scalarProduct[vec[x], vec[y]]\nvec /: Cross[vec[x_], HoldPattern[Plus[y__]]] :=\nMap[Cross[vec[x], #] &, Plus[y]]\nvec /: Cross[HoldPattern[Plus[y__]], vec[x_]] :=\nMap[Cross[#, vec[x]] &, Plus[y]]\nscalarProduct /: MakeBoxes[scalarProduct[x_, y_], _] :=\nRowBox[{ToBoxes[x], \".\", ToBoxes[y]}]\n\nvec[x].vec[y]\n\n(* ==> vec[x].vec[y] *)\n\nvec[x].vec[y] == vec[y].vec[x]\n\n(* ==> True *)\n\nCross[vec[x], vec[a] + vec[b]]\n\n(* ==> vec[x]\\[Cross]vec[a] + vec[x]\\[Cross]vec[b] *)\n\nCross[vec[a] + vec[b], vec[x]]\n\n(* ==> vec[a]\\[Cross]vec[x] + vec[b]\\[Cross]vec[x] *)\n\n\nFor the Dot product, I defined the behavior of vec such that it gets evaluated as a new function scalarProduct whose only algebraic property is that it's Orderless as you were expecting for the dot product of vectors. Of course this is only true for Euclidean dot products, so this assumption is implicit here. For more information on how this definition works, look up TagSetDelayed.\n\nIn addition, scalarProduct is given a customized display format by defining that it should again display as if it were a dot product when it appears in the low-level formatting function MakeBoxes.\n\nFor the distributive property of the cross product, I give vec the additional property that when it appears in Cross together with an expression of head Plus, the sum is expanded. Here the TagSetDelayed definitions are done for both orders, and contain a HoldPattern to prevent Plus from being evaluated too early in the definition.\n\nNow you may come back with many more wishes: e.g., what about multiplicative scalars in the dot or cross product, and what about matrices. However, that's a wide field that opens up a can of worms, so I would say just implement the bare minimum of features you can get away with symbolically, then proceed with a concrete working basis so that you can write vectors as lists instead.\n\nAnother approach would be to define a new symbol for a custom dot product. That is done in this question.\n\nUsing OverVector\n\nAs mentioned above, you can replace vec by Overvector everywhere in the above source code, to get a better formatted result. Assuming you have done that (I won't bother to repeat the definitions with that change), here are some examples:\n\nTo enter these vector expressions, refer to the Basic Math assistant palette. The cross product can be entered as EsccrossEsc.\n\nAnother thing you asked for is to use the antisymmetry of the cross product in simplifications. That's actually done already if you invoke FullSimplify:\n\n\u2022 Yes, the definition of new operator looks good. But it is not that convenient to use additional head. I am looking for a vector operators to simply some complicate expression involving many vectors. Is that possible to redefine +, -, . and x so don't have to use something like Plus? \u2013\u00a0user1285419 Mar 14 '13 at 4:00\n\u2022 Sorry, I used Plus only because that was what I was thinking about while writing the definitions. Of course you can actually use + and - with this setup. No modifications needed at all. I also mentioned you can get a simpler format using OverVector, and illustrated that in an update to the answer. As to redefining + etc.: it's not necessary, so don't do it. \u2013\u00a0Jens Mar 14 '13 at 4:28\n\u2022 that's amazing. I never know that we can use mathematica in that way, I did learn something today. Thanks. \u2013\u00a0user1285419 Mar 14 '13 at 4:35\n\u2022 I find a new problem of using this definition, if I try to Cross[OverVector[x], 2*OverVector[y]] + Cross[OverVector[y], 2*OverVector[x]] but it is not zero \u2013\u00a0user1285419 Apr 16 '13 at 22:58\n\u2022 I already anticipated this comment in my answer, and mentioned that it opens up a wide field of additional definitions one could add. I may come back to that when I have time. \u2013\u00a0Jens Apr 16 '13 at 23:26\n\nIf you have Mathematica Version 9, you can use Vectors and TensorReduce:\n\nAssuming[(x | y) \\[Element] Vectors[n] , TensorReduce[Dot[x, y] - Dot[y, x]]]\n(* 0 *)\nTensorReduce[Dot[x, y] - Dot[y, x], Assumptions -> (x | y) \\[Element] Vectors[n]]\n(* 0 *)\nTensorReduce[Cross[x + y, z],  Assumptions -> (x | y | z) \\[Element] Vectors[n]]\n(* x\\[Cross]z + y\\[Cross]z *)\nDistribute[Cross[x + y, z]] (* this should work in all previous versions *)\n(* x\\[Cross]z + y\\[Cross]z *)\n\n\u2022 thanks, it works. But what about the first one, is that possible to define an abstract row/column vector so x.y-y.x will get zero? \u2013\u00a0user1285419 Mar 13 '13 at 3:11\n\u2022 Presumably these solutions are for version 9 and later, when TensorReduce and Vectors were introduced. \u2013\u00a0whuber Mar 13 '13 at 4:42\n\u2022 @whuber, thank you. I updated with a note that TensorReduce and Vectors` are version-9 functions. \u2013\u00a0kglr Mar 13 '13 at 6:15\n\u2022 Thanks. Oh, I just saw that it only works on version 9 not my version 8 :( \u2013\u00a0user1285419 Mar 13 '13 at 6:37", "date": "2020-09-22 21:16:04", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/21197/how-do-i-simplify-a-vector-expression", "openwebmath_score": 0.530837893486023, "openwebmath_perplexity": 1229.7197033702078, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9702399026119353, "lm_q2_score": 0.8947894717137996, "lm_q1q2_score": 0.8681604498937819}}
{"url": "https://math.stackexchange.com/questions/3013093/how-to-show-that-2n-n-without-induction", "text": "# How to show that $2^n > n$ without induction\n\nI'm solving exercises about Pascal's triangle and Binomial theorem, and this problem showed up, however I don't have any clue on how to solve it\n\nThe sum of $${n\\choose p}$$ from $$p=0$$ to $$n$$ is the same thing as $$(1+1)^n=2^n$$, how can I use this information? Maybe comparing with another summation that equals to n?\n\nNote that $$2^n$$ is the number of subsets of $$[n]=\\{1,\\dotsc,n\\}$$. There are $$n$$ subsets of $$[n]$$ with size $$1$$. There is at least one subset of $$[n]$$ which is not a singleton (namely the empty set). Hence $$2^n>n$$ for $$n\\geq 1$$.\n\n\u2022 Foobaz John.Nice+. \u2013\u00a0Peter Szilas Nov 25 '18 at 17:21\n\nUse Bernoulii inequality, which is true for all $$x>-1$$: $$(1+x)^n\\geq 1+nx$$ so $$(1+1)^n \\geq 1+n\\cdot 1 >n$$\n\nMaybe comparing with another summation that equals to $$n$$?\n\nFor any $$p=0,1,\\dots,n$$, there is at least one way to choose $$p$$ things from a list of $$n$$ things. Thus $$\\binom{n}{p} \\ge 1$$, so\n\n$$2^n = \\sum_{p=0}^n \\binom{n}{p} \u2265 \\sum_{p=0}^n 1 = n+1 > n.$$\n\n\u2022 I did that summation, but how do I prove that one inequality is bigger than the other? \u2013\u00a0Nuno Mateus Nov 25 '18 at 17:15\n\u2022 @NunoMateus The sentence before that is the proof. \u2013\u00a0Calvin Khor Nov 25 '18 at 17:15\n\nThe Binomial Theorem says \\begin{align} 2^n &=(1+1)^n\\\\ &=\\binom{n}{0}1^0+\\binom{n}{1}1^1+\\dots\\\\ &\\ge1+n\\\\[9pt] &\\gt n \\end{align}\n\n\u2022 And I was thinking about this one :) +1 \u2013\u00a0Aqua Nov 25 '18 at 17:50\n\nhint\n\nConsider $$x\\mapsto \\frac{\\ln(x)}{x}$$ for $$x\\ge 1$$.\n\n$$f'(x)=\\frac{1-\\ln(x)}{x^2}$$\n\nthe maximum if $$f(e)=\\frac{1}{2}<\\ln(2)$$.\n\nthus\n\n$$\\ln(x)", "date": "2019-07-15 20:06:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3013093/how-to-show-that-2n-n-without-induction", "openwebmath_score": 0.9577944278717041, "openwebmath_perplexity": 328.4660139446752, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137910906876, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8681387740410095}}
{"url": "https://math.stackexchange.com/questions/1372621/finding-convergence-of-a-series-using-integral-test", "text": "# Finding convergence of a series using integral test\n\nThe series:$$\\sum_{n=1}^{\\infty}\\left(\\frac{\\ln(n)}{n}\\right)^{2}$$ Question:\n\na) show that it converges\n\nb) find the upper bound for the error in approximation $s\\approx s_{n}$\n\nTrial: The section was about integral test, but the sequence$\\left(\\frac{\\ln(n)}{n}\\right)^{2}$ is not decreasing from [1,$\\infty$]( it is increasing from [ 1,e ] ) so, I could not use the integral test.\n\nOther method: I tried to find a sequence greater than$\\left(\\frac{\\ln(n)}{n}\\right)^{2}$so that it satisfies the condition for use of integral test( If I show using integral test that the new series is bounded then It could imply that our sequence is convergent since its bounded and decreasing).the problem was that I had trouble finding any function which could satisfy such conditions b) this is understanding problem, is it asking me to find the exact sum or some upper bound , in any way how can I do this?\n\n\u2022 i don't understand your sum \u2013\u00a0Dr. Sonnhard Graubner Jul 24 '15 at 15:54\n\u2022 If the function is decreasing on $(e,\\infty)$ then the sequence is decreasing on $[3,\\infty)$. Removing a finite number of terms from the front end of a series won't change whether the series converges or not. \u2013\u00a0Alex Pavellas Jul 24 '15 at 15:56\n\u2022 btw, I assume you meant to use $n$ instead of $x$ inside the series. \u2013\u00a0Alex Pavellas Jul 24 '15 at 15:57\n\u2022 @Dr.SonnhardGraubner I made an edit \u2013\u00a0Socre Jul 24 '15 at 15:59\n\u2022 @user255545 yes, that infact is true but I have another problem, I had a difficult time integrating the function. If I knew how to integrate it your method is actually quite satisfying. \u2013\u00a0Socre Jul 24 '15 at 16:02\n\nTo show the series converges using the integral test we simply integrate by parts twice with successive substitutions $u_1=(\\log x)^2$ and $v_1=x^{-2}$, and $u_2=\\log x$ and $v_2=x^{-1}$, to reveal\n\n\\begin{align} \\int_3^{\\infty} \\left(\\frac{\\log x}{x}\\right)^2\\,dx&=-\\left.\\left(\\frac{(\\log x)^2}{x}\\right)\\right|_{3}^{\\infty}+2\\int_3^{\\infty} \\frac{\\log x}{x^2}\\,dx\\\\\\\\ &=\\frac13 (\\log(3))^2+2\\int_3^{\\infty} \\frac{\\log x}{x^2}\\,dx\\\\\\\\ &=\\frac13 (\\log(3))^2-2\\left.\\left(\\frac{\\log x}{x}\\right)\\right|_{3}^{\\infty}+2\\int_3^{\\infty} \\frac{1}{x^2}\\,dx\\\\\\\\ &=\\frac13 (\\log(3))^2+\\frac23 \\log (3)+\\frac23 \\end{align}\n\nThus, the series converges.\n\nUPPER AND LOWER BOUNDS\n\nTo find an upper bound of the series using the integral test we use\n\n\\begin{align} \\sum_{n=1}^{\\infty}\\left(\\frac{\\log x}{x}\\right)^2&\\le \\left(\\frac{\\log 2}{2}\\right)^2+\\left(\\frac{\\log 3}{3}\\right)^2+\\int_3^{\\infty}\\left(\\frac{\\log x}{x}\\right)^2\\,dx\\\\\\\\ &=\\left(\\frac{\\log 2}{2}\\right)^2+\\left(\\frac{\\log 3}{3}\\right)^2+\\frac13 (\\log(3))^2+\\frac23 \\log (3)+\\frac23\\\\\\\\ &\\approx. 2.05560987295277 \\end{align}\n\nThe lower bound is simply the upper bound less the third term $\\left(\\frac{\\log 3}{3}\\right)^2\\approx. 0.134105440090287$\n\nThus, we have\n\n$$\\bbox[5px,border:2px solid #C0A000]{ \\left(\\frac{\\log 2}{2}\\right)^2+\\frac13 (\\log(3))^2+\\frac23 \\log (3)+\\frac23 \\le \\sum_{n=1}^{\\infty}\\left(\\frac{\\log x}{x}\\right)^2}$$\n\n$$\\bbox[5px,border:2px solid #C0A000]{ \\sum_{n=1}^{\\infty}\\left(\\frac{\\log x}{x}\\right)^2\\le \\left(\\frac{\\log 2}{2}\\right)^2+\\left(\\frac{\\log 3}{3}\\right)^2+\\frac13 (\\log(3))^2+\\frac23 \\log (3)+\\frac23}$$\n\n$$\\bbox[5px,border:2px solid #C0A000]{ 1.92150443286247 \\le \\sum_{n=1}^{\\infty}\\left(\\frac{\\log x}{x}\\right)^2\\le 2.05560987295278}$$\n\nSince $\\log n\\leq (n-1)^{\\frac{2}{5}}$ for any $n\\geq 1$,\n\n$$0\\leq \\sum_{n\\geq 1}\\frac{\\log^2 n}{n^2}\\leq \\sum_{n\\geq 1}\\frac{1}{n^{\\frac{6}{5}}}$$ and the RHS is convergent by the $p$-test. Moreover, $$\\sum_{n\\geq 1}\\frac{\\log^2 n}{n^2}=\\frac{d^2}{ds^2}\\left.\\sum_{n\\geq 1}\\frac{1}{n^s}\\right|_{s=2} = \\zeta''(2) = 1.9892802342989\\ldots$$\n\n\u2022 Why the exponent of $2/5$? \u2013\u00a0marty cohen Jul 24 '15 at 17:35\n\u2022 @martycohen: any constant slightly less than $\\frac{1}{2}$ does the job, $\\frac{2}{5}$ is just a nice number of that form. \u2013\u00a0Jack D'Aurizio Jul 24 '15 at 17:39\n\u2022 According to Wolfy, max{(log(log(x)))/(log(x-1))}~0.379831 at x~10.9352. So your choice of 2/5 is quite good. \u2013\u00a0marty cohen Jul 24 '15 at 17:47\n\nFor the convergence we can use for example, for $x$ sufficiently large (say $x\\geq N$), $$\\log\\left(x\\right)\\leq x^{1/4}$$ hence $$\\sum_{n\\geq N}\\frac{\\log^{2}\\left(n\\right)}{n^{2}}\\leq\\sum_{n\\geq N}\\frac{1}{n^{3/2}}<\\infty.$$ About the upper bound for the error, we can use the integral test $$\\sum_{n\\geq N}f\\left(n\\right)\\leq f\\left(N\\right)+\\int_{N}^{\\infty}f\\left(x\\right)dx$$ and so in our case $$\\sum_{n\\geq1}\\frac{\\log^{2}\\left(n\\right)}{n^{2}}=\\sum_{n=1}^{N}\\frac{\\log^{2}\\left(n\\right)}{n^{2}}+\\sum_{n\\geq N+1}\\frac{\\log^{2}\\left(n\\right)}{n^{2}}\\leq$$ $$\\leq\\sum_{n=1}^{N}\\frac{\\log^{2}\\left(n\\right)}{n^{2}}+\\frac{\\log^{2}\\left(N+1\\right)}{\\left(N+1\\right)^{2}}+\\int_{N+1}^{\\infty}\\frac{\\log^{2}\\left(x\\right)}{x^{2}}dx$$ and the integral is, using the integration by parts, $$\\int_{N+1}^{\\infty}\\frac{\\log^{2}\\left(x\\right)}{x^{2}}dx=\\frac{\\log^{2}\\left(N+1\\right)}{N+1}+2\\int_{N+1}^{\\infty}\\frac{\\log\\left(x\\right)}{x^{2}}dx=$$ $$=\\frac{\\log^{2}\\left(N+1\\right)}{N+1}+\\frac{2\\log\\left(N+1\\right)}{N+1}+2\\int_{N+1}^{\\infty}\\frac{1}{x^{2}}dx=$$ $$\\frac{\\log^{2}\\left(N+1\\right)+2\\log\\left(N+1\\right)+2}{N+1}.$$", "date": "2021-07-30 23:01:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1372621/finding-convergence-of-a-series-using-integral-test", "openwebmath_score": 0.9690384864807129, "openwebmath_perplexity": 643.2393072901375, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137916170774, "lm_q2_score": 0.8840392771633079, "lm_q1q2_score": 0.8681387625055603}}
{"url": "https://www.physicsforums.com/threads/probability-question.467349/", "text": "# Probability Question\n\nrooski\n\n## Homework Statement\n\nSuppose that the integer values 1 2 and 3 are written on each of three different cards. Suppose you do not know which number is the lowest (you do not know beforehand what the values on the cards are). Suppose that you are to be offered these cards in a random order. When you are offered a card you must immediately either accept it or reject it. If you accept a card, the process\nends. If you reject a card, then the next card (if a card remains) is offered. If you\nreject the first two cards offered, then you must accept the final card.\n\n(a) If you plan to accept the first card offered, what is the probability that you will\naccept the lowest valued card?\n\n(b) If you plan to reject the first card offered, and to then accept the second card if\nand only if its value is lower than the value of the first card, what is the probability\nthat you will accept the lowest valued card?\n\n## The Attempt at a Solution\n\nA) The answer is obviously 1/3 for this question.\n\nB) this is a conditional probability question. Given that P(E|F) = P(EF) / P(F) then i must first figure out what P(E) and P(F) stand for.\n\nP(E) is the probability that i will accept the lowest card.\nP(F) is the probability that the second card i choose is lower than the first rejected card.\n\nIf i reject the card with 1 on it, then i have no chance of selecting the lowest card next.\nIf i reject the card with 2 on it, then there is a 1/2 chance i will select the lowest card next.\nIf i reject the card with 3 on it, then i have a 1/2 chance i will select the lowest card next.\n\nSo there is a 2/3 * 1/2 = 1/3 chance that the second card is lower than the first rejected card, right?\n\nI am not sure how to proceed after this.\n\nHomework Helper\nIf you reject the card with 2 on it, then you will either pick the card with 1 next (which is lower, so you will accept it and have the lowest card) or you will pick that card with 3 (which is higher, so you will reject it and be left with the last one).\nSo in fact, the probability of getting the lowest card there is 1.\n\nrooski\nAh right, dunno how i missed that.\n\nSo how do i calculate the chance that the second card will be lower? I have 0, 1 and 1/2 as the probabilities, depending on which card is rejected first.\n\nKarlx\nHi Rooski.\nLet L be the event that you accept the lowest card.\nLet $$F_{n}$$ be the event that the value of the first (and rejected) card is n, n=1,2,3.\n\nEvents $$F_{n}$$ are disjoint.\n\nTry to write down p(L) using the total probability rule.\n\nrooski\nAssuming L is the event i accept the lowest card,\n\nP(L) = P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C)\n\nWhere A,B,C denote cards 1,2,3 respectively. Is that right or am i off? It seems wring since P(A), P(B) and P(C) would all be 1/3.\n\nKarlx\nOk.\nAccording to your notation A is the event that the first (and rejected) card is the lowest one.\nSo, if you reject A, the probability of accepting the lowest card is zero.\nSo P(L|A) = 0.\n\nWhat about P(L|B) and P(L|C) ?\n\nrooski\nP(L|B) = 1 since you will reject 3 if it appears, or accept 1 when it appears.\nP(L|C) = 1/2 since you will accept 2 if it appears or accept 1 when it appears.\n\nHave i calculated P(A) P(B) and P(C) wrong?\n\nKarlx\nA is the event that the first card is A. So pA=1/3.\nThe same apply to B and C.\nJust calculate P(L).\n\nrooski\nP(L) = 0 * 1/3 + 1 * 1/3 + \u00bd * 1/3 = 3/6 = 1/2\n\nSo there is a 50% chance that we will end up with the lowest card if we reject the first random card.", "date": "2022-08-14 16:43:16", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/probability-question.467349/", "openwebmath_score": 0.581005334854126, "openwebmath_perplexity": 504.676541516387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137863531804, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.868138751851669}}
{"url": "http://qkms.ddiy.pw/sawtooth-wave-fourier-series.html", "text": "# Sawtooth Wave Fourier Series\n\nSquare wave t x(t) X 0 -T 0 0 T 0 -X 0 0 k X j \u03c0 \u2212 2 0 when k is odd a k = 0 when k is even 2. Fn = 2 shows the special case of the segments approximating a sine. In this demonstration it's just like the last one for the square wave. A sawtooth wave An electrocardiogram (ECG) signal Also included are a few examples that show, in a very basic way, a couple of applications of Fourier Theory, thought the number of applications and the ways that Fourier Theory is used are many. According to the important theorem formulated by the French mathematician Jean Baptiste Joseph Baron Fourier, any periodic function, no matter how trivial or complex, can be expressed in terms of converging series of combinations of sines and/or cosines, known as Fourier series. It is so named based on its resemblance to the teeth of a plain-toothed saw with a zero rake angle. Find the Fourier series for the sawtooth wave defined on the interval $$\\left[ { - \\pi ,\\pi } \\right]$$ and having period $$2\\pi. EE341 Homework Assignment 4 9-13-19. The Fourier expansion of the square wave becomes a linear combination of sinusoids: If we remove the DC component of by letting , the square wave become and the square wave is an odd function composed of odd harmonics of sine functions (odd). EE 230 Fourier series - 1 Fourier series A Fourier series can be used to express any periodic function in terms of a series of cosines and sines. 1 De nitions and Motivation De nition 1. jpg 1,956 \u00d7 2,880; 323 KB. (Note that Trott 2004, p. Report on sawtooth wave generator 1. Example: Sawtooth wave So, the expansion of f(t) reads (7. 5))in terms of its Fourier components, may occur in electronic circuits designed to handle sharply rising. Lecture 7: Fourier Series and Complex Power Series Week 7 Caltech 2013 1 Fourier Series 1. \u2014 The convention is that a sawtooth wave ramps upward and then sharply drops. However, as ybeltukov pointed out in a comment I did not read until he made me aware of it, Fourier series of piecewise continuously differentiable functions tend to overshoot a jump discontinuities, something which is called Gibbs phenomenon. Unless stated otherwise, it will be assumed that x(t) is a real, not complex, signal. To best answer this question, we need to consult the work of Baron Jean Baptiste Fourier and dig into a little mathematics. 3 shows two even functions, the repeating ramp RR(x)andtheup-down train UD(x) of delta functions. The three examples consider external forcing in the form of a square-wave, a sawtooth-wave, and a triangle-wave. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Figure 13-11 shows an example of calculating a Fourier series using these equations. These basic signals can be used to construct more useful class of signals using Fourier Series representation. The voltage at the Figure 5. The examples given on this page come from this Fourier Series chapter. Fourier Transform Fourier Transform maps a time series (eg audio samples) into the series of frequencies (their amplitudes and phases) that composed the time series. 2 Wave Diffraction and the Reciprocal Lattice Diffraction of waves by crystals \u2022 The Bragg law Scattered wave amplitude \u2022 Fourier analysis \u2022 Reciprocal lattice vectors \u2022 Diffraction conditions \u2022 Laue equations Brillouin zones \u2022 Reciprocal lattice to sc/bcc/fcc lattices Fourier analysis of the basis. general Fourier Series around a jump discontinuity. Taking the inner product of both sides, with respect to the orthonormalized eigenfunctions X n (x) and the weight function w(x) = 1, and assuming validity of the interchange between the summation and integration operations, yields. We then state some important results about Fourier series. But what we're going to do in this case is we're going to add them. Definition of Fourier Series and Typical Examples Baron Jean Baptiste Joseph Fourier \\(\\left( 1768-1830 \\right)$$ introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related. This series is used to generate a sawtooth wave and values are calculated using the program l18a1. (iii) h(x) = \u02c6 0 if 2", "date": "2019-11-19 00:35:27", "meta": {"domain": "ddiy.pw", "url": "http://qkms.ddiy.pw/sawtooth-wave-fourier-series.html", "openwebmath_score": 0.8793695569038391, "openwebmath_perplexity": 429.53024318361145, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9932024682484215, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8681356641475709}}
{"url": "https://math.stackexchange.com/questions/1745082/are-these-the-only-answers-of-x4y4z41-4xyz", "text": "# are these the only answers of $x^4+y^4+z^4+1=4xyz$?\n\nGiven an equation $$x^4+y^4+z^4+1=4xyz$$Find out the number of possible ordered tuple $(x,y,z)\\mid x,y,z\\in\\Bbb{R}$.\n\nI am getting it as $(1,1,1),(-1,-1,1),(1,-1,-1),(-1,1,-1)$ so $\\boxed{4}$\n\nIs there any other tuple which I am missing?\n\nAny help will be appreciable !\n\n\u2022 Looks fine. AM/GM gives that the absolute value of each of $x,y,z$ is 1. Apr 16, 2016 at 14:49\nBy Am/Gm we have $$\\frac{x^4+y^4+z^4+1}{4}\\geq xyz$$ . now we know the minima of arithmetic mean and maxima of geometric mean is achieved when numbers are equal or their $mod$ is equal as here $4$th power is used. Thus all positive $1$ or two negative $1$ are permissible hence total answers are $1+{3\\choose 2}=1+3=4$", "date": "2022-08-18 08:12:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1745082/are-these-the-only-answers-of-x4y4z41-4xyz", "openwebmath_score": 0.9092199206352234, "openwebmath_perplexity": 283.73602828758305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474907307123, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8681294467747658}}
{"url": "https://freakonometrics.hypotheses.org/53269", "text": "# Linear Regression, with Map-Reduce\n\nSometimes, with big data, matrices are too big to handle, and it is possible to use tricks to numerically still do the map. Map-Reduce is one of those. With several cores, it is possible to split the problem, to map on each machine, and then to agregate it back at the end.\n\nConsider the case of the linear regression, $\\mathbf{y}=\\mathbf{X}\\mathbf{\\beta}+\\mathbf{\\varepsilon}$ (with classical matrix notations). The OLS estimate of $\\mathbf{\\beta}$ is $\\widehat{\\mathbf{\\beta}}=[\\mathbf{X}^T\\mathbf{X}]^{-1}\\mathbf{X}^T\\mathbf{y}$. To illustrate, consider a not too big dataset, and run some regression.\n\nlm(dist~speed,data=cars)$coefficients (Intercept) speed -17.579095 3.932409 y=cars$dist X=cbind(1,cars$speed) solve(crossprod(X,X))%*%crossprod(X,y) [,1] [1,] -17.579095 [2,] 3.932409 How is this computed in R? Actually, it is based on the QR decomposition of $\\mathbf{X}$, $\\mathbf{X}=\\mathbf{Q}\\mathbf{R}$, where $\\mathbf{Q}$ is an orthogonal matrix (ie $\\mathbf{Q}^T\\mathbf{Q}=\\mathbb{I}$). Then $\\widehat{\\mathbf{\\beta}}=[\\mathbf{X}^T\\mathbf{X}]^{-1}\\mathbf{X}^T\\mathbf{y}=\\mathbf{R}^{-1}\\mathbf{Q}^T\\mathbf{y}$ solve(qr.R(qr(as.matrix(X)))) %*% t(qr.Q(qr(as.matrix(X)))) %*% y [,1] [1,] -17.579095 [2,] 3.932409 So far, so good, we get the same output. Now, what if we want to parallelise computations. Actually, it is possible. Consider $m$ blocks m = 5 and split vectors and matrices $$\\mathbf{y}=\\left[\\begin{matrix}\\mathbf{y}_1\\\\\\mathbf{y}_2\\\\\\vdots \\\\\\mathbf{y}_m\\end{matrix}\\right]$$ and $$\\mathbf{X}=\\left[\\begin{matrix}\\mathbf{X}_1\\\\\\mathbf{X}_2\\\\\\vdots\\\\\\mathbf{X}_m\\end{matrix}\\right]=\\left[\\begin{matrix}\\mathbf{Q}_1^{(1)}\\mathbf{R}_1^{(1)}\\\\\\mathbf{Q}_2^{(1)}\\mathbf{R}_2^{(1)}\\\\\\vdots \\\\\\mathbf{Q}_m^{(1)}\\mathbf{R}_m^{(1)}\\end{matrix}\\right]$$ To split vectors and matrices, use (eg) Xlist = list() for(j in 1:m) Xlist[[j]] = X[(j-1)*10+1:10,] ylist = list() for(j in 1:m) ylist[[j]] = y[(j-1)*10+1:10] and get small QR recomposition (per subset) QR1 = list() for(j in 1:m) QR1[[j]] = list(Q=qr.Q(qr(as.matrix(Xlist[[j]]))),R=qr.R(qr(as.matrix(Xlist[[j]])))) Consider the QR decomposition of $\\mathbf{R}^{(1)}$ which is the first step of the reduce part$$\\mathbf{R}^{(1)}=\\left[\\begin{matrix}\\mathbf{R}_1^{(1)}\\\\\\mathbf{R}_2^{(1)}\\\\\\vdots \\\\\\mathbf{R}_m^{(1)}\\end{matrix}\\right]=\\mathbf{Q}^{(2)}\\mathbf{R}^{(2)}$$where$$\\mathbf{Q}^{(2)}=\\left[\\begin{matrix}\\mathbf{Q}^{(2)}_1\\\\\\mathbf{Q}^{(2)}_2\\\\\\vdots\\\\\\mathbf{Q}^{(2)}_m\\end{matrix}\\right]$$ R1 = QR1[[1]]$R for(j in 2:m) R1 = rbind(R1,QR1[[j]]$R) Q1 = qr.Q(qr(as.matrix(R1))) R2 = qr.R(qr(as.matrix(R1))) Q2list=list() for(j in 1:m) Q2list[[j]] = Q1[(j-1)*2+1:2,] Define \u2013 as step 2 of the reduce part$$\\mathbf{Q}^{(3)}_j=\\mathbf{Q}^{(2)}_j\\mathbf{Q}^{(1)}_j$$ and$$\\mathbf{V}_j=\\mathbf{Q}^{(3)T}_j\\mathbf{y}_j$$ Q3list = list() for(j in 1:m) Q3list[[j]] = QR1[[j]]$Q %*% Q2list[[j]] Vlist = list() for(j in 1:m) Vlist[[j]] = t(Q3list[[j]]) %*% ylist[[j]]\n\nand finally set \u2013 as the step 3 of the reduce part$$\\widehat{\\mathbf{\\beta}}=[\\mathbf{R}^{(2)}]^{-1}\\sum_{j=1}^m\\mathbf{V}_j$$\n\nsumV = Vlist[[1]] for(j in 2:m) sumV = sumV+Vlist[[j]] solve(R2) %*% sumV [,1] [1,] -17.579095 [2,] 3.932409\n\nIt looks like we\u2019ve been able to parallelise our linear regression\u2026\n\n## 6 thoughts on \u201cLinear Regression, with Map-Reduce\u201d\n\n1. korul says:\n\nHow do I supposed to calculate inverse of R^(2), when R^(2) is a vector of matrices?\n\n2. Stefano says:\n\nGreat post! One question: mathematically, can you explain \u2013 provide a reference for the explanation of \u2013 the step 2 of the reduce part (where the 2 Q matrices are multiplied)? Because I can follow through all steps, but I don\u2019t understand *why* you create Q_j^(3) like this. Thanks!\n\n3. Another interesting point is the trade-off between the number of nodes and the efficiency of the calculation. If you think about it \u2013 mode nodes *does not* mean faster running time.\n\nI took the liberty of reproducing your results (in \u201ctidyverse\u201d code\u201d) and added some notes on efficiency. I have an R notebook/markdown here: https://github.com/ytoren/reproducible/tree/master/linear-regression-map-reduce\n\nKeep up the good work mate!\n\n4. Awesome post!\nI think you might have a typo :\n\nYou split Q2 to Q2list by:\n> for(j in 1:m) Q2list[[j]] = Q1[(j-1)*2+1:2,]\n\nAnd it should be\n> for(j in 1:m) Q2list[[j]] = Q2[(j-1)*2+1:2,]\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2022-12-05 21:13:27", "meta": {"domain": "hypotheses.org", "url": "https://freakonometrics.hypotheses.org/53269", "openwebmath_score": 0.6735536456108093, "openwebmath_perplexity": 3870.3878197329277, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750529474512, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8680475845550756}}
{"url": "https://puzzling.stackexchange.com/questions/15412/99-bags-of-apples-and-oranges?noredirect=1", "text": "# 99 Bags of Apples and Oranges\n\nYou have $99$ bags, each containing various numbers of apples and oranges. Prove that there exist $50$ bags among these which together contain at least half the apples and at least half the oranges.\n\nNeedless to say, you may not add/remove fruits to/from the bags.\n\nClarification: I changed the wording from \"you can grab $50$ bags...\" to \"there exist $50$ bags...\"\n\nClarification from comments: each bag can contain any number of apples and any number of oranges with any total number of fruit; the total amount of fruit in a bag doesn't have to be the same for each bag.\n\n\u2022 I suspect the presence of Pigeonhole principle... \u2013\u00a0leoll2 May 23 '15 at 14:29\n\u2022 @ghosts_in_the_code can you provide a counterexample? \u2013\u00a0leoll2 May 23 '15 at 15:54\n\u2022 @ghosts_in_the_code By various, I mean each bag can have any number of apples, any number of oranges, making any total. The bags do not need to all have the same number of fruits. \u2013\u00a0Mike Earnest May 23 '15 at 15:58\n\u2022 Can the bags be empty? \u2013\u00a0Bob May 23 '15 at 16:09\n\u2022 @Bob Yes, they can be empty. For example, if all bags were empty, then grabbing any $50$ bags works, since zero is at least half of zero. \u2013\u00a0Mike Earnest May 23 '15 at 16:12\n\nLet $B_1,B_2,\\dots,B_{99}$ be the bags, in increasing order of number of apples contained; say the number of apples in $B_i$ is $a_i$ for all $i$.\n\nIf our 50 bags grabbed are $B_{99}$, one of $B_{98}$ and $B_{97}$, one of $B_{96}$ and $B_{95}$, ..., and one of $B_2$ and $B_1$, then they must contain between them at least half of all apples, since at worst they contain $a_{99}+a_{97}+\\dots+a_5+a_3+a_1\\geq a_{98}+a_{96}+\\dots+a_4+a_2+0$ apples.\n\nNow how do we fix all those \"one of\"s? Just pick whichever of $B_{98}$ and $B_{97}$ has the more oranges, then whichever of $B_{96}$ and $B_{95}$ has the more oranges, and so on. Then our 50 selected bags - even excluding $B_{99}$! - must contain at least half of all oranges.\n\nQED.\n\n\u2022 QED indeed! I will certainly accept your answer, but I think I will wait a bit to see if other cool answers crop up as well. I know there are many distinct proofs methods, yours being the most constructive \u2013\u00a0Mike Earnest May 23 '15 at 21:23\n\nArrange the bags in a circle. I will call a collection of 50 consecutive bags a team (so there are 99 possible teams), and a team will be called appley if it contains at least half of the apples.\n\nDefine the opponent of a given team to be the team whose rightmost bag is the leftmost bag of the given team. For any given team, every bag is either a member of that team or a member of its opponent (and one bag is a member of both). This means that if a team is not appley, then its opponent is appley. Different teams have different opponents, so there must be at least as many appley teams as non-appley teams. There are 99 teams in total, so at least 50 of them are appley.\n\nA similar argument shows that at least 50 teams are orangey (that is, at least 50 of the teams contain at least half the oranges). Of 99 teams, at least 50 are appley and at least 50 are orangey, so there must be at least one teams which is both appley and orangey (because $50+50>99$). This team consists of 50 bags, and contains at least half the apples and at least half the oranges.\n\n\u2022 This is really cool! To elaborate: let $A$ be the set of appley teams, and $B$ be the set of teams whose opponents are appley. You showed $A\\cup B$ is all $99$ teams (if a team is not appely, its opponent is). Since $|A|=|B|$, we have $99=|A\\cup B|\\le |A|+|B|=2|A|$, proving $|A|\\ge 49.5,$ so $|A|\\ge50$. \u2013\u00a0Mike Earnest May 24 '15 at 4:43\n\u2022 \"There are 99 teams in total, ...\" - agreed (count possible common bags). You argument seems to put appley teams in 1:1 correspondence with non-appley teams by construction, implying an even total number of teams. How does that correspond with the odd number of teams? Cont'd ... \u2013\u00a0Lawrence May 24 '15 at 13:40\n\u2022 ... cont'd. Simplified example: instead of 99 bags, say we have 3 bags $a,b,c$. Then we can have team pairs $[ab,bc], [bc,ca], [ca,ab]$. Suppose team $bc$ was appley. This would mean from the first two match-ups that both $ab$ and $ca$ were not appley. However, this makes the last match-up a team pair with two non-appley teams. Is this a failure of the assertion that \"if a team is not appley, then its opponent is appley\"? (By the way, other than this, I really like your proof. +1 ) \u2013\u00a0Lawrence May 24 '15 at 13:40\n\u2022 @Lawrence If a team is non-appley, its opponent is appley. However, if a team is appley, we are not saying its opponent is non-appley (they could both be appley). As you pointed out, since there are an odd number of teams, there must at least one appley team whose opponent is also appley. \u2013\u00a0Julian Rosen May 24 '15 at 13:46\n\u2022 @Lawrence no, different teams do have different opponents. Notice, however, that the opponent of the opponent of a team is not (quite) the original team, so you can't \"pair them off\" in the way you seem to be imagining. \u2013\u00a0Ben Millwood May 26 '15 at 14:35\n\nSuppose you have 50 bags on the ground and 49 in the truck, and suppose the ones on the ground have more apples. One at a time, take a bag from the ground and put it in the truck, and put one from the truck tand place it on the ground. Keep doing this with the goal of totally swapping ground and truck bags.\n\nEventually, you'll reach a point where moving a \"magic\" bag causes the truck to have more apples than the ground. At this point, both the ground plus the magic bag has at least half the apples, AND the truck plus the magic bag has at least half the apples. Either the ground or the truck ( plus the magic bag ) has at least half the oranges too.\n\nI will provide a probabilistic proof. By demonstrating that picking randomly has a nonzero chance of success, we can show that a solution exists. (http://en.m.wikipedia.org/wiki/Probabilistic_method)\n\nClaim: A random subset of 50 bags has over 50% chance of having more than half the apples. (Chosen uniformly over all size 50 subsets.)\n\nProof: For each \"losing\" size 50 subset, flip it around to get a (unique) winning size 49 subset. This means there are at least as many size 49 winning sets as size 50 losing sets. There are multiple ways to extend these to size 50 sets, so there's strictly more ways to win than lose. (Hand waving a bit)\n\nThis argument applies symmetrically to the oranges. Since each probability is over 1/2, they must have a nonzero intersection.\n\nThus there is a positive probability of picking a subset with at least half of each, so a solution exists.\n\nNonconstructively useless, but QED :)\n\nEdit: looking back, this is just saying that the intersection of sets larger than 50% is nonempty, so the probabilistic part is just fluff.\n\n\u2022 It's far from trivial that you can extend all 49-subsets of a 99-set into 50-subsets by addition of a single element in such a way that you don't accidentally extend two 49-subsets into the same 50-subset (it's certainly impossible to extend 50-subsets into 51-subsets this way, for example, because there aren't as many of them). \u2013\u00a0Ben Millwood May 26 '15 at 16:26\n\u2022 I mean, it IS true. But it requires Hall's Marriage Theorem, unless you have an ingenious construction. \u2013\u00a0Ben Millwood May 26 '15 at 16:29\n\nSuppose we choose 50 of the bags at random. We are choosing more than half of the bags, so the probability that the bags we selected contain at least half of the apples is strictly greater than 50%. Similarly, the probability that the bags we selected contain at least half of the oranges is also greater than 50%. This means there is a positive probability that the bags we selected contain at least half the apples and at least half the oranges. In particular, there must be some combination of 50 bags containing at least half the apples and at least half the oranges.\n\nEdit: As pointed out in the comments, it is not obvious why the bags we select will have at least half the apples with probability greater than 50%. This actually follows from the argument in my other answer: one way to choose 50 bags at random is first to choose a cyclic ordering of the bags, then to choose 50 consecutive bags. My other answer shows that for whichever cyclic ordering we choose, a random choice of 50 consecutive bags contains at least half the apples with probability at least $\\frac{50}{99}$. This means that a random selection of 50 bags contains at least half the apples with probability at least $\\frac{50}{99}$. The same is true for oranges, so a random selection of 50 bags contains at least half the apples and at least half the oranges with probability at least $\\frac{1}{99}$.\n\n\u2022 I think the statement We are choosing more than half of the bags, so the probability that the bags we selected contain at least half of the apples is strictly greater than 50%, needs more justification. \u2013\u00a0Mike Earnest May 23 '15 at 19:44\n\u2022 This isn't a very rigorous proof. First of all, you must justify your probability statements, then you might provide an algorithm of optimal choosing (not required, but strongly suggested). \u2013\u00a0leoll2 May 23 '15 at 20:52\n\u2022 I was expecting it to be clear why the probability of getting at least half the apples was bigger than 50%, but I didn't think this part through and I agree that it isn't clear why this is the case. \u2013\u00a0Julian Rosen May 24 '15 at 3:40\n\u2022 Using the probabilistic approach, leaves a possibility that there will not be a desired solution. We need to have 100% probability that a desired case will be found. \u2013\u00a0Moti May 25 '15 at 17:59\n\u2022 @Moti If the probability that a random object has a certain property is greater than 0, then it is certain there is at least one object with that property. \u2013\u00a0Julian Rosen May 25 '15 at 18:54", "date": "2019-08-24 03:21:30", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/15412/99-bags-of-apples-and-oranges?noredirect=1", "openwebmath_score": 0.6815144419670105, "openwebmath_perplexity": 536.0966710893548, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750510899382, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8680475844852755}}
{"url": "https://math.stackexchange.com/questions/2326813/triangle-inscribed-within-a-unit-circle", "text": "# Triangle inscribed within a unit circle\n\nA triangle $ABC$ is inscribed within the unit circle. Let $x$ be the measure of the angle $C$. Express the length of $AB$ in terms of $x$.\n\nA.) $2\\sin x$\n\nB.) $\\cos x + \\sin x - 1$\n\nC.) $\\sqrt{2}(1 - \\cos 2x)$\n\nD.) $\\sqrt{2}(1 - \\sin x)$\n\nI am unsure how to illustrate the circle and calculate the length $x$.\n\nThank you. `\n\n\u2022 Please read this tutorial about how to typeset mathematics on this site. \u2013\u00a0N. F. Taussig Jun 18 '17 at 8:55\n\n$$\\dfrac{AB}{\\sin x}=2R$$ where $R$ is the circum-radius $=1$\n\n\u2022 This solution is based on the Law of Sines. \u2013\u00a0N. F. Taussig Jun 18 '17 at 9:11\n\u2022 @N.F.Taussig, Thanks for enriching the answer \u2013\u00a0lab bhattacharjee Jun 18 '17 at 9:12\n\nThe angle $AOB=2x$, since it intercepts the same arc than the angle $CAB=x$.\n\nthus by the well-known formula\n\n$$AB^2=OA^2+OB^2$$ $$-2OA.OB.\\cos (2x)$$ with\n\n$$OA=OB=radius=1$$\n\nhence\n\n$$AB^2=2 (1-\\cos (2x))$$\n\n$$=4\\sin^2 (x)$$ and $$\\boxed {AB=2\\sin (x) }$$", "date": "2019-08-22 09:31:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2326813/triangle-inscribed-within-a-unit-circle", "openwebmath_score": 0.8868478536605835, "openwebmath_perplexity": 449.2969065905559, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750518329434, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8680475773223533}}
{"url": "https://math.stackexchange.com/questions/1624387/density-of-positive-multiples-of-an-irrational-number", "text": "# Density of positive multiples of an irrational number\n\nLet $x$ be irrational. Use $\\{r\\}$ to denote the fractional part of $r$: $\\{r\\} = r - \\lfloor r \\rfloor$. I know how to prove that the following set is dense in $[0,1]$: $$\\{\\{nx\\} : n \\in \\mathbb{Z}\\}.$$ But what about $$\\{\\{nx\\} : n \\in \\mathbb{N}\\}?$$ Any proof that I\u2019ve seen of the first one fails for the second one.\n\nA minor modification of the pigeonhole argument works. Let $m$ be any positive integer. By the pigeonhole principle there must be distinct $i,j\\in\\{1,\\dots,m+1\\}$ and $k\\in\\{0,\\dots,m-1\\}$ such that $\\frac{k}m\\le\\{ix\\},\\{jx\\}<\\frac{k+1}m$; clearly $\\{|(j-i)x|\\}<\\frac1m$. Let $\\ell$ be the largest positive integer such that $\\ell\\{|(j-i)x|\\}<1$, let $A_m=\\{n|j-i|x:0\\le n\\le\\ell\\}$, and let $D_m=\\big\\{\\{y\\}:y\\in A_m\\big\\}$.\n\nIf $x>0$, every point of $[0,1)$ is clearly within $\\frac1m$ of $A_m=D_m$. If $x<0$, then\n\n$$D_m=\\{1-|y|:y\\in A_m\\}\\;,$$\n\nso every point of $[0,1)$ is again within $\\frac1m$ of the set $D_m$. Since $D_m\\subseteq\\big\\{\\{nx\\}:n\\in\\Bbb N\\big\\}$, we\u2019re done.\n\n\u2022 I hope you don't mind how close my answer is to yours, but I had written an answer for someone in chat and was going to post it elsewhere, then saw this question was a better fit.\n\u2013\u00a0robjohn\nJul 20, 2021 at 20:58\n\u2022 @robjohn: No problem: someone may benefit from the greater detail. Jul 21, 2021 at 2:39\n\u2022 Can you give more details why if $x>0$ then every ppint of $[0,1)$ is clearly within $1/m$ of $A_m$.\n\u2013\u00a0ZFR\nJul 30, 2021 at 2:29\n\u2022 @ZFR: Adjacent points of $A_m$ are less than $\\frac1m$ apart. Every point of $[0,1)$ is either equal to one of these points, between two adjacent ones, or between $\\ell\\{|(j-i)x|\\}$ and $1$ and hence less than $\\frac1m$ from some point of $A_m$. Jul 31, 2021 at 22:00\n\nHere is a proof where the second statement follows from the first with a minor step. I wrote this answer for another question, but then I saw it fit here better, other than being along very similar lines to Brian M. Scott's answer.\n\nLet $$r$$ be irrational. Choose an arbitrary $$n\\in\\mathbb{N}$$. Partition $$[0,1)$$ into $$n$$ subintervals $$\\left\\{I_k=\\left[\\frac{k-1}n,\\frac kn\\right):1\\le k\\le n\\right\\}\\tag1$$ Consider the discrete set $$\\{\\{kr\\}:0\\le k\\le n\\}\\subset[0,1)$$. There are $$n+1$$ elements in this set, so two of them must lie in the same subinterval. That means that we have $$k_1\\ne k_2$$ so that $$\\{(k_1-k_2)r\\}\\in\\left(0,\\frac1n\\right)$$, we leave $$0$$ out of the interval because $$r$$ is irrational.\n\nLet $$m=\\left\\lfloor\\frac1{\\{(k_1-k_2)r\\}}\\right\\rfloor$$, then because $$m$$ is the greatest integer not greater than $$\\frac1{\\{(k_1-k_2)r\\}}$$, $$m\\{(k_1-k_2)r\\}\\!\\!\\overset{\\substack{r\\not\\in\\mathbb{Q}\\\\\\downarrow}}{\\lt}\\!\\!1\\lt(m+1)\\{(k_1-k_2)r\\}\\tag2$$ which implies that $$\\{m(k_1-k_2)r\\}=m\\{(k_1-k_2)r\\}\\in\\left(\\frac{n-1}n,1\\right)$$.\n\nSince $$\\{(k_1-k_2)r\\}\\in\\left(0,\\frac1n\\right)$$, if $$j\\{(k_1-k_2)r\\}\\in I_k$$, then $$(j+1)\\{(k_1-k_2)r\\}\\in I_k\\cup I_{k+1}$$; that is, $$\\{j(k_1-k_2)r\\}=j\\{(k_1-k_2)r\\}$$ cannot skip over any of the $$I_k$$. Therefore, $$\\left\\{\\{j(k_1-k_2)r\\}:1\\le j\\le m\\vphantom{\\frac12}\\right\\}\\tag3$$ must have at least one element in each $$I_k$$ for $$1\\le k\\le n$$.\n\nThe only problem is that we don't know that $$k_1-k_2\\gt0$$. However, this is not a problem since $$\\{j(k_1-k_2)r\\}\\in I_k\\iff\\{j(k_2-k_1)r\\}\\in I_{n+1-k}$$. Therefore, $$\\left\\{\\{j(k_2-k_1)r\\}:1\\le j\\le m\\vphantom{\\frac12}\\right\\}\\tag4$$ must also have at least one element in each $$I_k$$ for $$1\\le k\\le n$$.\n\nSince $$n$$ was arbitrary, we have shown that $$\\left\\{\\mathbb{N}r\\right\\}$$ is dense in $$[0,1]$$.\n\n\u2022 Thanks a lot professor Robjohn for answering my question. I was having lot of difficulty understanding $k_2\\gt k_1$ case. It all makes sense to me now. Many thanks :) I would only like to add this little more detail for statement just before $(3)$ in the hope that it will benefit those who visit this great answer in future: Suppose on the contrary that for some $k<n$, the subinterval $(\\frac{k-1}n, \\frac kn )$ doesn't contain any element of the set $T=\\{ j\\{(k_2-k_1)r\\}: 1\\le j\\le m\\}$. (Contd.)\n\u2013\u00a0Koro\nJul 21, 2021 at 3:55\n\u2022 (Contd.) Note that $T_L=\\{j: j\\{(k_2-k_1)r\\}\\le \\frac{k-1}n\\}$ is non-empty and therefore $T_L$ has a maximal element $M\\in \\mathbb N$ that is $(M+1)\\{(k_2-k_1)r\\}\\ge \\frac kn$ and $M\\{(k_2-k_1)r\\}\\le \\frac{k-1}n$ and subtracting the two, we get: $\\{(k_2-k_1)r\\}\\ge \\frac 1n$, which is a contradiction. Therefore no subinterval $I_k$ can be free of elements from $\\{\\mathbb N r\\}$.\n\u2013\u00a0Koro\nJul 21, 2021 at 3:56\n\u2022 Really nice answer which helped me to understand the problem. But I think you can assume $r$ to be any real irrational number from the beginning of the solution.\n\u2013\u00a0ZFR\nJul 29, 2021 at 22:25\n\u2022 @ZFR: You know, I modified this so many times while writing it, and now it seems you are correct; there is no reason for $r\\gt0$, so I have removed that constraint.\n\u2013\u00a0robjohn\nJul 30, 2021 at 2:40\n\u2022 Thanks a lot for your answer! To be honest I've spent on this problem about 2-3 days and eventually your solution helped me to understand it completely. Thanks! +1\n\u2013\u00a0ZFR\nJul 30, 2021 at 15:42\n\nReally? I thought exactly the same proof worked for $\\Bbb N$.\n\nLet $k\\in\\Bbb Z$ with $k\\ne 0$ and define $$f(t)=e^{2\\pi ikt}.$$\n\nThen $f$ has period $1$, and $$\\frac1N\\sum_{n=0}^{N-1}f(nx) =\\frac1N\\sum_{n=0}^{N-1}\\left(e^{2\\pi ikx}\\right)^n=\\frac1N \\frac{e^{2\\pi ikxN}-1}{e^{2\\pi ikx}-1}\\to0\\quad(N\\to\\infty).$$ So the usual approximation shows that $$\\frac1N\\sum_{n=0}^{N-1}f(nx)\\to\\int_0^1 f(t)\\,dt$$for $f\\in C(\\Bbb T)$ and you're done, as usual.\n\nHow is this any different from the case $n\\in\\Bbb Z$?\n\n\u2022 How does that imply that $\\{ \\exp (2\\pi i {n x}) :n\\in N\\}$ is dense in the unit circle in $C$ ? Anyway, a proof from the most elementary method works for N just as well as for Z. Jan 24, 2016 at 5:33\n\u2022 @user254665 Say those points are not dense. There is then an interval, or arc, $I$ on the circle that contains none of the points. Take $f\\in C(\\Bbb T)$ such that $f=0$ on the complement of $I$ but $\\int f > 0$. Then $\\frac1N\\sum_0^{N-1}f(nx)=0$ for every $N$, so it does not converge to $\\int f$. (Yes, the pigeonhole argument works as well. I really can't think of an argument that works for $n\\in\\Bbb Z$ but not $\\Bbb N$. In any case, ignoring the question of which is simpler, this argument shows more than just that the points are dense, it shows they are asymptotically uniformly distributed.) Jan 24, 2016 at 14:21\n\u2022 very nice integral proof, which yields extra info (distribution) too. Jan 24, 2016 at 20:10\n\u2022 @user254665 Yes it is very nice. The equidistribution is Weyl's theorem - I think it's his proof, not sure, in any case it's a standard thing. Jan 24, 2016 at 21:27", "date": "2022-07-02 07:32:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1624387/density-of-positive-multiples-of-an-irrational-number", "openwebmath_score": 0.8920701742172241, "openwebmath_perplexity": 114.42235389496506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526451784086, "lm_q2_score": 0.8757869965109765, "lm_q1q2_score": 0.8680385982047081}}
{"url": "https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for/", "text": "# Which payoff do you want to go for?\n\nConsider a game where you have to choose between 1 of 2 payoffs:\n\nPayoff 1. You toss 100 coins.\nYou get $1 for each Head that appears. Payoff 2. You toss 10 coins. You get$10 for each Head that appears.\n\nWhich payoff structure would you choose? And why?\n\nNote by Calvin Lin\n3\u00a0years, 10\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nI assume each coin toss is independent from one another (so the covariance between each coin toss is zero), and used only fair coins.\n\nYes, the expected Payoff is the same for ONE and TWO. But their variance differ.\n\n$V_1 = \\text { Probability } \\times \\text { Payoff ONE }^2 \\times \\text { Number of trials } = \\frac {1}{2} \\times 1^2 \\times 100 = 50$\n\n$V_2 = \\text { Probability } \\times \\text { Payoff TWO }^2 \\times\\text { Number of trials } = \\frac {1}{2} \\times 10^2 \\times 10 = 500$\n\nBecause Payoff TWO has a higher variance which is not desired, I will go with Payoff ONE.\n\n$\\text{ I learned from the best tutor }$\n\n- 3\u00a0years, 10\u00a0months ago\n\nWhat term do we use to describe a person who choses to \"minimize variance when expected value is equal\"?\n\nYour answer greatly depends on what it is that we wish to maximize / minimize. There could be people who may want to maximize variance, because of the higher likelihood of a huge payoff that they care about.\n\nStaff - 3\u00a0years, 10\u00a0months ago\n\nWhat term do we use to describe a person who choses to \"minimize variance when expected value is equal\"?\n\nRisk-averse investor\n\n- 3\u00a0years, 10\u00a0months ago\n\nLet me try.\n\nPayoff 1. Chance to get head for all coins = 1 out of 10^{200} = Chance to get $100 in terms of$1.\n\nPayoff 2. Chance to get head for all coins = 1 out of 10^{10} = Chance to get $100 in terms of$10.\n\nThus, I think Payoff 2 is better.\n\n- 3\u00a0years, 10\u00a0months ago\n\nThis is optimistic way I guess\n\n- 3\u00a0years, 10\u00a0months ago\n\nThat's a valid argument if your goal is to \"maximize the probability of the highest payoff\".\n\nSince getting $99,$98, ... $91 is somewhat similar to getting$100, would this affect your answer? Why, or why not?\n\nNote that the payoff 1 should be \"10 ^ {100}\"\n\nStaff - 3\u00a0years, 10\u00a0months ago\n\nThe probability of getting payoff between 91$to 100$ increases in the first case. Hence, if I consider myself satiated by $99,$98, ... $91, I'd prefer case 1. - 3 years, 10 months ago Log in to reply It's less of a hassle to toss 10 coins. If we compare this to stocks, it's easier to keep a portfolio of10 stocks than 100 stocks. And lower commissions. The risk of loss is higher, but the chance of higher returns is better too. - 3 years, 10 months ago Log in to reply That's a good point! We might need to factor in convenience / the cost of making a flip! Can you clarify what you mean by \"chance of higher returns is better too\"? How do you quantify that? Staff - 3 years, 10 months ago Log in to reply Earlier posters have quantified the variance for the smaller vs. the larger set of coins. For the set of 10 coins, the chance of getting more than 9 heads is 1 in 1024. For the set of 100 coins, the chance of getting more than 90 coins is much lower than that. I'd have to get my combinatorics (or statistics) hat on to quantify it, but it's much less. On the order of more than 5-sigma from the mean. For stocks, a stock club (or newsletter writer) would be silly to buy 1 share each of the 500 largest companies. A number of funds exists to do that with much less expense. If they want to try to beat the performance of the S&P 500, they pick a small number of stocks (using some rational criteria) and often do better than the \"market\". Of course, the risk of loss (or underperformance) is higher too. - 3 years, 10 months ago Log in to reply Expected outcome is same in both cases ,but in case 1 chances are that outcome is more closer to 50$ than in case2, So for more standard outcome I will choose case1.\n\nFor suppose if my need demands 60$I would choose case 2. CASE1 Chances to get 50$ is 100C50 *.5^{100}=.0786\n\n49$is 100C49*.5^{100 }=.0780 (same for 51$)\n\nFor 60$it is .0108 By similar calculations(and adding), chances to get outcome from 40$ to 60$is .946. where as in case 2 chances reduce to .656 this proves that there are more chances to get outcome closer to 50$ in case 1.\n\nprobabilities to get 60$are 0.108 for case 1 .205 for case 2. So, if 60$ is my need I will choose case 2\n\n- 3\u00a0years, 10\u00a0months ago\n\nYou bring up a valid point, which is \"What is the utility function that I care about\". If the utility function is just \"I only care if I have more than $60 or not\", then your approach shows that we should go for payoff 2. What kind of utility function makes sense? Do people value all amounts of money equally? Staff - 3 years, 10 months ago Log in to reply They average out to the same payoff (which is$50.00), so I would flip a coin to decide which to use.\n\nPython:\n\n  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 import random coin = (True, False) def sim_1(): payoff = 0 for flip in xrange(100): if random.choice(coin): payoff += 1 return payoff def sim_2(): payoff = 0 for flip in xrange(10): if random.choice(coin): payoff += 10 return payoff trials = 100000 payoff_1 = 0.0 payoff_2 = 0.0 for trial in xrange(trials): payoff_1 += sim_1() payoff_2 += sim_2() print \"Payoff 1:\", payoff_1 / trials print \"Payoff 2:\", payoff_2 / trials \n\n- 3\u00a0years, 10\u00a0months ago\n\nwhat good\n\n- 3\u00a0years, 10\u00a0months ago\n\nIf i understand the problem correctly, would be indifferent to the payoffs structure, cause both options give the same expected value which would be 50.\n\n- 3\u00a0years, 10\u00a0months ago\n\nplayoff 1\n\n- 3\u00a0years, 10\u00a0months ago\n\nI think payoff 1 is better because chances increses with more no of toss..\n\n- 3\u00a0years, 10\u00a0months ago\n\nIn my opinion, it's a matter of risk: payoff 1 has more consistency but less probability of scoring extremely large amounts of money, whereas payoff 2 is a sort of high risk high reward model where you either get a lot or get a bit. Personally, I would choose payoff 1.\n\n- 3\u00a0years, 10\u00a0months ago\n\nPay off 2, because it will likely to get heads more often and earn $10 for each head than 100 coin tosses. Who knows that in a toss of 100 you will only get less than 50% heads. - 3 years, 10 months ago Log in to reply Payoff 2 because I'll make decent money and i Don't have time to flip 100 coins. Stop over complicating things with math people. - 3 years, 10 months ago Log in to reply I will chose payoff 2 because while tossing coins 10 time we may get HEAD more than 1 or 2 times and the pay is more. But in payoff 1 tossing coins 100 times may give a HEAD 20 times and u get 20$. But in payoff 2 you can get more than 20$. - 3 years, 10 months ago Log in to reply Payoff 2 as it takes less time to toss 10 coins and the value is same for both - 3 years, 9 months ago Log in to reply What are the statistics given that heads side of a coin is slightly heavier than tails? Giving tails more of a chance of facing up? Is that true? I heard that somewhere back maybe in 4th grade haha. Or has that been proven incorrect? if it is true, I would rather chose b since I would want less of a chance of getting tails side up? Can you explain to me how I am wrong (I probably am) like you would explain it to a 3rd grader? This is why I'm more of an artist than math person. - 3 years, 6 months ago Log in to reply That is a very good point. How would your answer change if we didn't have a fair coin? For example, if the coin was totally biased and only gave tails, then both payoffs are the same. Similarly, if the coin was totally biased and only gave heads, then both payoffs are the same. But in between, the payoffs are different. So, does that mean that the answer could change depending on the probability of head/tail? If so, how and why? How is payoff 2 \"less chance of getting tails side up\"? Just because of the absolute number? But, the first mostly likely has many more heads appearing, just a smaller payoff each time. Staff - 3 years, 6 months ago Log in to reply Less probability of a heads since the gravitational pull of the heavier side. Sorry I missed stated. On the other hand, my fiance makes a good point. The more you flip a coin your results are more likely to reflect the chance of a 50/50 so it is better and more reliable to flip more times than less so option 1 of 100 flips is more reliable then just 10 in case you get a lucky streak of a ten or an unlucky streak with only 2 out of 10. - 3 years, 6 months ago Log in to reply As it turns out, the Expected Value of both payoffs would be the same, regardless of the probability of landing heads / tails. For example, if the coin is fair, and it will land heads 50% of the time, then the expected payoff in 1 is$50, and the expected payoff in 2 is \\$50. More generally, if it land heads $$p%$$ of the time, then the expected payoff will be $$p$$.\n\nSo, the question then becomes, pick your favorite probability $$p$$ of landing heads, which payoff would you want? Is there anything else to consider?\n\nStaff - 3\u00a0years, 6\u00a0months ago\n\nPayoff 1 so that I can get more number of chances\n\n- 3\u00a0years, 10\u00a0months ago\n\nI'd go for #2. I'm a risktaker.\n\n- 3\u00a0years, 9\u00a0months ago\n\nYou will still need to define what you mean by \"risktaker\". There are possible definitions of \"risktaker\" which would make option 1 more valuable to them.\n\nStaff - 3\u00a0years, 9\u00a0months ago", "date": "2019-01-17 14:23:04", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for/", "openwebmath_score": 0.9702968597412109, "openwebmath_perplexity": 2656.1427826983536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9783846722794542, "lm_q2_score": 0.8872045847699186, "lm_q1q2_score": 0.8680273669149461}}
{"url": "https://math.stackexchange.com/questions/3258275/show-there-cant-be-two-real-and-distinct-roots-of-polynomial-fx-x3-3xk-in", "text": "# Show there can't be two real and distinct roots of polynomial $f(x)=x^3-3x+k$ in $(0,1)$, for any value of k.\n\nI have two proofs here one which I did and the other was given in book. Is one better than the other? I am asking for in an exam setting which proof makes a better solution(as in fetch more marks).\n\n## Proof 1 (My proof)\n\n$$f'(x)=3x^2-3$$, in the interval $$(0,1)$$ is less than $$0$$. If there were two distinct roots then $$f'(0)$$ should have been $$0$$ once in $$(0,1)$$ by rolle's theorem. Since it isn't there are no values of $$k$$ for which there are two real roots.\n\n## Proof 2 (Book)\n\nLet $$a,b$$ be two roots of $$f(x)$$ in $$(0,1)$$ then there exists a $$c$$ such the $$f'(c) = 0$$ for c in $$[a,b]$$ by Rolle's theorem. $$f'(c)= 3c^2-3$$ has no solutions in $$(0,1)$$ hence there is no such value of $$k$$.\n\n\u2022 Yours is better because the book version has a wrong expression for $f'$ \u2013\u00a0Hagen von Eitzen Jun 11 at 6:09\n\u2022 \u2013\u00a0lab bhattacharjee Jun 11 at 6:16\n\u2022 @HagenvonEitzen that's a typo sorry. \u2013\u00a0Sonal_sqrt Jun 11 at 6:26", "date": "2019-08-21 00:47:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3258275/show-there-cant-be-two-real-and-distinct-roots-of-polynomial-fx-x3-3xk-in", "openwebmath_score": 0.8186616897583008, "openwebmath_perplexity": 186.15686257037427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846621952493, "lm_q2_score": 0.8872045847699186, "lm_q1q2_score": 0.8680273579681933}}
{"url": "https://math.stackexchange.com/questions/2118291/how-to-factorize-this-cubic-equation", "text": "# How to factorize this cubic equation?\n\nIn one of the mathematics book, the author factorized following term\n\n$$x^3 - 6x + 4 = 0$$ to\n\n$$( x - 2) ( x^2 + 2x -2 ) = 0.$$\n\nHow did he do it?\n\n\u2022 If there is a cube term, it is not a quadratic. Also, a method for finding divisors of your polynomial, look at the factors of the constant term. \u2013\u00a0Edward Evans Jan 28 '17 at 19:52\n\u2022 $x=2$ is a root of this equation. Then by polynomial division of $\\frac{x^3-6x+4}{x-2}$ we obtain $x^2+2x-2$. \u2013\u00a0projectilemotion Jan 28 '17 at 19:53\n\u2022 Do you know the Rational Root Test? \u2013\u00a0Bill Dubuque Jan 28 '17 at 19:54\n\u2022 A common first step for introductory problems is to guess and check certain integers close to zero to see if it will equal zero. Zero is the easiest to check because that would mean the constant term is zero, thats not the case here. $1$ doesn't work because that would be $1-6+4=-1\\neq 0$. $2$ happens to work since this would be $2^3-6\\cdot 2 + 4 = 8-12+4=0$. Since $2$ works, we know that the equation can be factored as $(x-2)q(x)$ where $q(x)=\\frac{x^3-6x+4}{x-2}$. In general this won't always work, especially if the roots aren't even integers. Cardano's formula would help then. \u2013\u00a0JMoravitz Jan 28 '17 at 19:54\n\u2022 @BillDubuque No \u2013\u00a0Govinda Sakhare Jan 28 '17 at 19:56\n\nThere is a neat trick called the rational roots theorem. All we have to do is factor the first and last numbers, put them over a fraction, and take $\\pm$. This gives us the following possible rational roots:\n\n$$x\\stackrel?=\\pm1,\\pm2,\\pm4$$\n\ndue to the factorization of $4$. Checking these, it is clear $x=2$ is the only rational root, since\n\n\\begin{align}0&\\ne(+1)^3-6(+1)+4\\\\0&\\ne(-1)^3-6(-1)+4\\\\\\color{#4488dd}0&=\\color{#4488dd}{(+2)^3-6(+2)+4}\\\\0&\\ne(-2)^3-6(-2)+4\\\\0&\\ne(+4)^3-6(+4)+4\\\\0&\\ne(-4)^3-6(-4)+4\\end{align}\n\nleaving us with\n\n$$x^3-6x+4=(x-2)(\\dots)$$\n\nWe can find the remainder through synthetic division:\n\n$$\\begin{array}{c|c c}2&1&0&-6&4\\\\&\\downarrow&2&4&-4\\\\&\\hline1&2&-2&0\\end{array}$$\n\nwhich gives us our factorization:\n\n$$x^3-6x+4=(x-2)(x^2+2x-2)$$\n\n\u2022 OP does not know RRT - see the comments. \u2013\u00a0Bill Dubuque Jan 28 '17 at 19:59\n\u2022 @BillDubuque Oh. Then give me a moment \u2013\u00a0Simply Beautiful Art Jan 28 '17 at 20:00\n\u2022 @SimplyBeautifulArt Why 1,2 and 4. Why are we not checking equation with value 3? it has anything to do with c term ax^3+bx+c ? \u2013\u00a0Govinda Sakhare Jan 28 '17 at 20:09\n\u2022 @piechuckerr By the rational roots theorem, I need only check the factors of the last constant if the leading coefficient is $1$. $3$ does not divide into $4$, so I needn't check it. \u2013\u00a0Simply Beautiful Art Jan 28 '17 at 20:10\n\u2022 @SimplyBeautifulArt i.e if constant term is 6 then 1,2,3 and 6 right? \u2013\u00a0Govinda Sakhare Jan 28 '17 at 20:12\n\nSince you do not know the Rational Root Test, let's consider a simpler case: the Integer Root Test.\n\nIf $\\,f(x)= x^3+6x+4\\,$ has an integer root $\\,x=n\\,$ then $\\,n^3+6n+4 = 0\\,$ so $\\,(n^2+6)\\,\\color{#c00}{n = -4},\\,$ hence $\\,\\color{#c00}{n\\ \\ {\\rm divides}\\ \\ 4}.\\,$ Testing all the divisors of $4$ shows that $2$ is root,  hence $\\,x-2\\,$ is a factor of $f$ by the Factor Theorem. The cofactor $\\,f/(x-2)\\,$ is computable by the Polynomial (long) Division algorithm (or even by undetermined coefficients).\n\nRemark $\\$ This is a very special case of general relations between the factorization of polynomials and the factorizations of their values. For example, one can derive relations between primality and compositeness of polynomials based on the same properties of their values. For example, since $\\ 9^4\\!+8\\$ is prime so too is $\\, x^4+8\\,$ by Cohn's irreducibility test. See this answer and its links for some of these beautiful ideas of Bernoulli, Kronecker, and Schubert.\n\n\u2022 brilliant, liked every bit of it, Sadly I can not unmark other answer and select this one coz that too is elegant answer. \u2013\u00a0Govinda Sakhare Jan 29 '17 at 16:50\n\u2022 @piechuckerr No problem (i don't care about acceptance, votes, etc, only about sharing beautiful mathematics). \u2013\u00a0Bill Dubuque Jan 29 '17 at 17:05\n\nNote: I understand that there is already an accepted answer for this question, so this answer may be useless, but regardless, I'm still posting this to spread knowledge!\n\nA simple way to factorize depressed cubic polynomials of the form$$x^3+Ax+B=0\\tag1$$\n\nIs to first move all the constants to the RHS, so $(1)$ becomes$$x^3+Ax=-B\\tag2$$ Now, find two factors of $B$ such that one fact minus the square of the other factor is $A$. We'll call them $a,b$ so\\begin{align*} & a-b^2=A\\tag3\\\\ & ab=-B\\tag4\\end{align*} Multiply $(2)$ by $x$, add $b^2x^2$ to both sides and complete the square. Solving should give you a value of $x$ and allow you to factor $(1)$ by Synthetic Division.\n\nExamples:\n\n1. Solve $x^3-6x+4=0$ (your question)\n\nMoving $4$ to the RHS and observing its factors, we have $-2,2$ as $a,b$ since$$-2-2^2=A\\\\-2\\cdot2=-4$$Therefore, we have the following:$$x^4-6x^2=-2\\cdot2x$$$$x^4-6x^2+4x^2=4x^2-4x$$$$x^4-2x^2=4x^2-4x$$$$x^4-2x^2+1=4x^2-4x+1\\implies(x^2-1)^2=(2x-1)^2$$$$x^2=2x\\implies x=2$$ Note that we do have to consider the negative case when square rooting, but they lead to the same pair of answers. So it's pointless.\n\n1. Solving $x^3+16x=455$\n\nA factor of $455$ works, namely when $a=65,b=7$.$$65-7^2=16$$$$65\\cdot7=455$$ Therefore,$$x^4+16x^2=65\\cdot7x$$$$x^4+65x^2=49x^2+455x$$$$\\left(x^2+\\dfrac {65}{2}\\right)^2=\\left(7x+\\dfrac {65}{2}\\right)^2$$$$x=7$$\n\nThe rational root theorem gives a list of all possible rational roots of a polynomial with integer coefficients that have a given leading coefficient and a given constant coefficient. In this case, the leading coefficient is $1$ and the constant coefficient is $4.$ The theorem tells us that all rational roots are in the set $\\left\\{ \\pm\\dfrac 1 1, \\pm\\dfrac 2 1, \\pm \\dfrac 4 1 \\right\\},$ the numerator being in this the only divisor of the leading coefficient $1$ and the denominators being the divisors of the constant coefficient $4$. That doesn't mean there are rational roots; it only means there are not any that don't belong to this set. There are only six members of this set, so it's easy to plug in all of them and see if you get $0$. When you plug in $2$, you get $0$, so there's your factorization.\n\n\u2022 OP does not know RRT - see the comments. \u2013\u00a0Bill Dubuque Jan 28 '17 at 19:59\n\u2022 @BillDubuque : Fortunately I linked to the Wikipedia article about it. \u2013\u00a0Michael Hardy Jan 28 '17 at 19:59\n\u2022 @BillDubuque : That he doesn't recognize that name of the theorem doesn't mean he doesn't know the theorem. He could know it by a different name. \u2013\u00a0Michael Hardy Jan 28 '17 at 20:01\n\u2022 I thought the comment(s) might nudge someone to give an exposition on this simpler integer case. But no one did, so I added an answer doing so. \u2013\u00a0Bill Dubuque Jan 28 '17 at 20:40\n\nIf $P$ is a polynomial with real coefficients and if $a\\in\\mathbb{R}$ is a root, which means that $P(a)=0$, then there exists a real polynomial $Q$ such that $\\forall x\\in\\mathbb{R},\\quad P(x)=(x-a)\\,Q(x)$.\n\nOn this case, you can see by inspection, that $P(2)=0$.\n\nIt remains to find real constants $A,B,C$ such that :\n\n$$\\forall x\\in\\mathbb{R},\\quad x^3-6x+4=(x-2)(Ax^2+Bx+C)$$\n\nIdentification of coefficients leads to $A=1$, $-2C=4$ and, for example, $A-2B=0$ (equating the coeffts of $x^2$ in both sides).\n\n\u2022 Since the OP doesn't know RRT, the answer I posted earlier seems to me a (not so bad) - one ! It's clear that one can answer at a higher level, but it's seems to me important to answer at a level compatible the mathematical background of the OP. I have the feeling that, if he (or she) had known the RRT, he would certainly never have asked that question. \u2013\u00a0Adren Jan 28 '17 at 20:11\n\u2022 Actually we don't need the full RRT here, only a simpler integer case, e.g. see my answer. \u2013\u00a0Bill Dubuque Jan 28 '17 at 20:56", "date": "2020-02-28 00:14:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2118291/how-to-factorize-this-cubic-equation", "openwebmath_score": 0.733822226524353, "openwebmath_perplexity": 315.72851065386436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692264378964, "lm_q2_score": 0.8887587875995483, "lm_q1q2_score": 0.8680233575747336}}
{"url": "https://www.physicsforums.com/threads/how-do-you-tackle-this-integral.708999/", "text": "# How do you tackle this integral\n\n1. Sep 6, 2013\n\n### Gauss M.D.\n\n1. The problem statement, all variables and given/known data\n\nFind F(x)\n\nf(x) = 1/(1+x2)2)\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nIt's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?\n\n2. Sep 6, 2013\n\n### Zondrina\n\nPerfect candidate for a trig sub. Use $x = tan(\u03b8)$ so that $dx = sec^2(\u03b8)d\u03b8$.\n\n3. Sep 6, 2013\n\n### ZeroPivot\n\nwell, 1/1+X^2 is the derivative of arctan i believe. i think there has to be a relation. u use integration by parts left u nowhere?\n\n4. Sep 6, 2013\n\n### Gauss M.D.\n\nZondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?\n\nLast edited: Sep 6, 2013\n5. Sep 6, 2013\n\n### ZeroPivot\n\nits a nasty integral u sure you are doing everything correctly? so the integral is 1/1+x^2 only?\n\n6. Sep 6, 2013\n\n### Gauss M.D.\n\nNo, I was referring to Zondrinas trig sub suggestion!\n\n7. Sep 6, 2013\n\n### Zondrina\n\nI don't see any discontinuity in the integrand at all, so I don't think it would matter.\n\n8. Sep 6, 2013\n\n### Gauss M.D.\n\nRight right, we're using tan and not some other trig function. My bad again. Thanks a ton :)\n\n9. Sep 7, 2013\n\n### verty\n\nThat is a very good question, I don't know the answer to it. I'm going to investigate this.\n\n10. Sep 7, 2013\n\n### verty\n\nSorry, the original question has no trouble because the range of tan is just R. Here is a more applicable problem:\n\n$y = \\int {dx \\over (1 - x^2)^2}$\n\n$x =? \\; sin\\theta$\n$dx = cos\\theta \\; d\\theta$\n\n$y = \\int sec^3\\theta \\; d\\theta$\n$= {1 \\over 2} [sec\\theta \\; tan\\theta + \\int sec\\theta \\; d\\theta] + C$\n$= {1 \\over 2} [sec\\theta \\; tan\\theta + ln| sec\\theta + tan\\theta | ] + C$\n$= {1\\over 2}{x \\over 1 - x^2} + {1\\over 4} ln | {1+x\\over 1-x}| + C$\n\nOn the other hand, one can make a translation:\n\n$x = u-1$\n$dx = du$\n\n$y = \\int {dx \\over (1 - x^2)^2} = \\int{ dx \\over (1+x)^2 (1-x)^2} = \\int {du \\over u^2(2-u)^2}$\n\nAfter some partial fraction magic:\n\n$y = \\int {1 \\over 4(2-u)^2} + {1 \\over 4u^2} + {1 \\over 4u} + {1 \\over 4(2-u)} du$\n$= {1 \\over 4} [ {1 \\over 2-u} - {1\\over u} + ln|u| - ln|2-u| ] + C$\n$= {1 \\over 4} [ {2(u-1) \\over u(2-u)} + ln| {u \\over 2-u} | ] + C$\n$= {1 \\over 2} {x \\over 1 - x^2} + {1 \\over 4} ln| {1+x \\over 1-x} | + C$\n\nI can't explain it but it's the same answer. Probably it has to do with the behaviour of sin(z) and cos(z) for complex z.", "date": "2017-10-23 23:46:30", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/how-do-you-tackle-this-integral.708999/", "openwebmath_score": 0.8946686387062073, "openwebmath_perplexity": 2173.3676869155997, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964173268185, "lm_q2_score": 0.8807970904940926, "lm_q1q2_score": 0.8680223770738138}}
{"url": "https://math.stackexchange.com/questions/2240525/counting-distinguishable-ways-of-painting-a-cube-with-4-different-colors-each-u", "text": "# Counting distinguishable ways of painting a cube with 4 different colors (each used at least once)\n\nYou have many identical cube-shaped wooden blocks. You have four colors of paint to use, and you paint each face of each block a solid color so that each block has at least one face painted with each of the four colors. Find the number of distinguishable ways you could paint the blocks. (Two blocks are distinguishable if you cannot rotate one block so that it looks identical to the other block.)\n\nHaving trouble solving this problem with the added constraint of \"at least one face painted with each of four colors\" - Thanks in advance\n\nCall the four colors $a$, $b$, $c$, $d$.\n\nThere are two partitions of $6$ into four parts, namely (1): $(3,1,1,1)$, and (2): $(2,2,1,1)$.\n\nIn case (1) we can choose the color appearing three times in $4$ ways. This color can either (1.1) appear on three faces sharing a vertex of the cube, or (1.2) on three faces forming a $\\sqcup$-shape. In case (1.1) we can place the three other colors in $2$ ways (\"clockwise\" or \"counterclockwise\"); in case (1.2) we can choose which of the three other colors is opposite the floor of the $\\sqcup$. This amounts to $4\\cdot(2+3)=20$ different colorings.\n\nIn case (2) the two colors appearing only once can be chosen in ${4\\choose2}=6$ ways. Assume that colors $a$ and $b$ are chosen. The $a$-face $F_a$ and the $b$-face $F_b$ can be either (2.1) opposite or (2.2) adjacent to each other. In case (2.1) we can chose the two $c$-faces either adjacent or opposite to each other. In case (2.2) the two $c$-faces can be (2.2.1) opposite to each other, (2.2.2) opposite to $F_a$ and to $F_b$, or (2.2.3) opposite to one of $F_a$ or $F_b$ and on one of the faces adjacent to both $F_a$ and $F_b$. In all this amounts to $6\\cdot(2+1+1+2\\cdot2_*)=48$ different colorings. (The factor $2_*$ distinguishes mirror-equivalent colorings.)\n\nAltogether we have found $68$ different admissible colorings of the cube.\n\n\u2022 great solution ! - I still need to work out the second scenario in detail but appears to be a valid solution. Thanks \u2013\u00a0randomwalker Apr 18 '17 at 21:15\n\nThere is an algorithmic approach to this which I include for future reference and which consists in using Burnside and Stirling numbers of the second kind. For Burnside we need the cycle index of the face permutation group of the cube. We enumerate the constituent permutations in turn. First there is the identity for a contribution of $$a_1^6.$$\n\nRotating about one of the four diagonals by $120$ degrees and $240$ degrees we get\n\n$$4\\times 2a_3^2.$$\n\nRotating about an axis passing through opposite faces by $90$ degrees and by $270$ degrees we get\n\n$$3\\times 2 a_1^2 a_4$$\n\nand by $180$ degrees\n\n$$3\\times a_1^2 a_2^2.$$\n\nFinally rotating about an exis passing through opposite edges yields\n\n$$6\\times a_2^3.$$\n\nWe thus get the cycle index\n\n$$Z(G) = \\frac{1}{24} (a_1^6 + 8 a_3^2 + 6 a_1^2 a_4 + 3 a_1^2 a_2^2 + 6 a_2^3).$$\n\nAs a sanity check we use this to compute the number of colorings with at most $N$ colors and obtain\n\n$$\\frac{1}{24}(N^6 + 8 N^2 + 12 N^3 + 3 N^4).$$\n\nThis gives the sequence\n\n$$1, 10, 57, 240, 800, 2226, 5390, 11712, 23355, 43450, \\ldots$$\n\nwhich is OEIS A047780 which looks to be correct. Now if we are coloring with $M$ colors where all $M$ colors have to be present we must partition the cycles of the entries of the cycle index into a set partition of $M$ non-empty sets. We thus obtain\n\n$$\\frac{M!}{24} \\left({6\\brace M} + 8 {2\\brace M} + 12 {3\\brace M} + 3{4\\brace M}\\right).$$\n\nThis yields the finite sequence (finite because the cube can be painted with at most six different colors):\n\n$$1, 8, 30, 68, 75, 30, 0, \\ldots$$\n\nIn particular the value for four colors is $68.$ We also get $6!/24 = 30$ for six colors because all orbits have the same size.\n\nHere's my crack at it: Starting with a stationary cube, there are $\\frac{4 \\cdot 3 \\cdot 6!}{2 \\cdot 2}$ ways of painting the cube where there are 2 colors with 2 faces, and $\\frac{4 \\cdot 6!}{3}$ ways of painting the cube when there is 1 color with 3 faces which gives a total of 3120 ways of painting the cube, but this over counts all the orientations of a cube, so the final answer is $\\frac{3120}{4 \\cdot 6} = 130$\n\n\u2022 awright96 - Thank you for taking the time to answer the question. I understand your approach which is a good, clean approach. However the correct answer is 68. This problem appeared on a middle-school purple comet test in 2015 ( unfortunately only answers are published, no solutions) \u2013\u00a0randomwalker Apr 18 '17 at 19:31", "date": "2021-02-26 15:38:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2240525/counting-distinguishable-ways-of-painting-a-cube-with-4-different-colors-each-u", "openwebmath_score": 0.6901645660400391, "openwebmath_perplexity": 171.4205429709181, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964177527897, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.868022374365338}}
{"url": "https://math.stackexchange.com/questions/3749178/prove-that-every-set-and-subset-with-the-cofinite-topology-is-compact", "text": "Prove that every set and subset with the cofinite topology is compact\n\nProve that every set with the cofinite topology is compact as well as every subset\n\nSolution. Let $$X$$ be a nonempty set with the cofinite topology and let $$\\mathscr{U}$$ be an open cover of $$X$$. Let $$U \\in \\mathscr{U}$$. Then $$X\\setminus U$$ is finite. For every $$a \\in X\\setminus U$$ let $$U_a$$ be an element of $$\\mathscr{U}$$ that contains $$a$$. Then $$\\{U\\}\\cup\\{U_a : a \u2208 X\\setminus U\\}$$ is a finite subcover of $$\\mathscr{U}$$.\n\nNow I missing the part for the subsets $$E\\subseteq X$$. I don't think this refers to the relative topology, but just to any subset of $$X$$ How do I go about it?\n\n\u2022 It of course refers to the rel topology and the proof is the same. \u2013\u00a0JCAA Jul 7 '20 at 22:10\n\u2022 I thought the statement meant that any subset of X was compact with the original topology, not with the relative one \u2013\u00a0J.C.VegaO Jul 7 '20 at 22:13\n\u2022 Well what does it mean to be compact with the original topology? \u2013\u00a0Severin Schraven Jul 7 '20 at 22:21\n\u2022 @Severin Schraven That you can extract a finite subcover from a open cover made of open sets of the original topology, ( not made of the intersection of them with the set, like in the relative topology.) The reason I make a difference is because set that is not open in the original topology may be open in the relative. like for example in [-1,1] with the usual topology $\\tau$ as a subset of $\\mathbb{R}$ . $E=[-1,1 ]$ is not open in $(\\mathbb{R},\\tau)$ but it is in $(\\mathbb{R},\\tau_E)$ \u2013\u00a0J.C.VegaO Jul 7 '20 at 22:27\n\nIn the case of compactness it makes no difference whether you use the topology of $$X$$ or the relative topology on the subset.\n\nProposition. Let $$\\langle X,\\tau\\rangle$$ be any space, let $$K\\subseteq X$$, and let $$\\tau_K$$ be the relative topology on $$K$$; then $$K$$ is compact with respect to $$\\tau$$ iff it is compact with respect to $$\\tau_K$$.\n\nProof. Suppose first that $$K$$ is compact with respect to $$\\tau$$, and let $$\\mathscr{U}\\subseteq\\tau'$$ be a $$\\tau'$$-open cover of $$K$$. For each $$U\\in\\mathscr{U}$$ there is a $$V_U\\in\\tau$$ such that $$U=K\\cap V_U$$. Let $$\\mathscr{V}=\\{V_U:U\\in\\mathscr{U}\\}$$; clearly $$\\mathscr{V}$$ is a $$\\tau$$-open cover of $$K$$, so it has a finite subcover $$\\{V_{U_1},\\ldots,V_{U_n}\\}$$. Let $$\\mathscr{F}=\\{U_1,\\ldots,U_n\\}$$; $$\\mathscr{F}$$ is a finite subset of $$\\mathscr{U}$$, and\n\n$$\\bigcup\\mathscr{F}=\\bigcup_{k=1}^nU_k=\\bigcup_{k=1}^n(K\\cap V_{U_k})=K\\cap\\bigcup_{k=1}^nU_k=K\\;,$$\n\nso $$\\mathscr{F}$$ covers $$K$$. Thus, $$K$$ is compact with respect to $$\\tau'$$.\n\nNow suppose that $$K$$ is compact with respect to $$\\tau'$$, and let $$\\mathscr{U}\\subseteq\\tau$$ be a $$\\tau$$-open cover of $$K$$. For each $$U\\in\\mathscr{U}$$ let $$V_U=K\\cap U$$, and let $$\\mathscr{V}=\\{V_U:U\\in\\mathscr{U}\\}$$. $$\\mathscr{V}$$ is a $$\\tau'$$-open cover of $$K$$, so it has a finite subcover $$\\{V_{U_1},\\ldots,V_{U_n}\\}$$. Let $$\\mathscr{F}=\\{U_1,\\ldots,U_n\\}$$; $$\\mathscr{F}$$ is a finite subset of $$\\mathscr{U}$$, and\n\n$$\\bigcup\\mathscr{F}=\\bigcup_{k=1}^nU_k\\supseteq\\bigcup_{k=1}^n(K\\cap U_k)=\\bigcup_{k=1}^nV_{U_k}=K\\;,$$\n\nso $$\\mathscr{F}$$ covers $$K$$. Thus, $$K$$ is compact with respect to $$\\tau$$. $$\\dashv$$\n\n\u2022 Initially I thought that they were saying that any subset of $X$ was compact, like for example if $X=\\mathbb{R}$ with the cofinite topology, then any subset like $\\{1\\}$ $[1,2] ,(1,2),(1,2], (1,+\\infty)$ should be compact, that is not true, is it? \u2013\u00a0J.C.VegaO Jul 7 '20 at 23:43\n\u2022 @J.C.VegaO: It is true. You essentially proved it in your question. \u2013\u00a0Brian M. Scott Jul 7 '20 at 23:46", "date": "2021-03-01 01:38:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3749178/prove-that-every-set-and-subset-with-the-cofinite-topology-is-compact", "openwebmath_score": 0.9389252662658691, "openwebmath_perplexity": 66.34106814868436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140244715406, "lm_q2_score": 0.8976952859490985, "lm_q1q2_score": 0.8679941616861733}}
{"url": "https://math.stackexchange.com/questions/1296651/sat-maths-question-about-fractions", "text": "# SAT Maths Question About Fractions\n\nWhilst revising, a problem caught my eye and I cannot seem to find an answer. I am usually bad at these types of questions.\n\nOn a certain Russian-American committee, $\\frac23$ of members are men, and $\\frac38$ of the men are Americans. If $\\frac35$ of the committee members are Russians, what fraction of the members are American women?\n\nA. $\\frac{3}{20}$ B. $\\frac{11}{60}$ C. $\\frac{1}{4}$ D. $\\frac{2}{5}$ E. $\\frac{5}{12}$\n\nCould you please explain how to approach and analyse the problem, maybe give some hints or the complete procedure of solving?\n\nI get a bit confused with all those fractions. What I tried was to convert them to percentages but that seemed a bad idea.\n\nSorry if this question is annoying.\n\nThank you.\n\nUpdate: I solved the problem both intuitively and mathematically. Thanks.\n\n\u2022 You've got two categories, each with two options. You can find the probabilities of being in the intersection of each of these options (e.g. Russian Women) using the proportions and conditional proportions (like how its not that 3/8 are american, but 3/8 of men are american), and your answer will be there. \u2013\u00a0Bob Krueger May 24 '15 at 13:38\n\nHere's a more algebraic approach. We know that $\\frac{2}{5}$ of the committee is American (the other $\\frac{3}{5}$ is Russian), and that $\\frac{2}{3}\\cdot\\frac{3}{8} = \\frac{1}{4}$ of the committee is American men. Thus, if the proportion of American women is $x$, then, $$\\frac{1}{4} + x = \\frac{2}{5}$$ that is, if you add up the proportion of American men and the proportion of American women (assuming men and women are the only two categories), then you get the proportion of Americans. From here, we can solve for $x$ to get $$x = \\frac{2}{5} - \\frac{1}{4} = \\frac{3}{20},$$ which is in agreement with abel's answer.\n\nLet me try. We will pick a nice number for the total number of members so that every category of members come out whole numbers. We can pick the least common multiple of all the denominators of the fractions. That gives us $120.$ Of course as long as we are only interested in the ratio, it don't matter what that number is. It makes computation easier to do.\n\nSuppose there were $120$ members. There are $80$ men and $40$ women. Of the $80$ men $30$ are American and $50$ russian. There are $72$ Russians on the committee that leave $22$ Russian women and $18$ American women on the committee. The fraction of American women on the committee is $$\\frac{18}{120} = \\frac3{20}.$$\n\nHopefully I did not make any silly arithmetic errors.\n\n\u2022 I'd only suggest you edit to make clear why you chose 120. It may not be obvious to OP. \u2013\u00a0JoeTaxpayer May 24 '15 at 14:36\n\nI can't improve on @abel 's excellent answer but do want to point out that it's an example of an important strategy. When you're \"confused with all those fractions\" and converting to percentages doesn't help (because they are just fractions) try natural frequencies - pick a number (in this case 120) that lets you count the cases.\n\nSee\n\nhttp://opinionator.blogs.nytimes.com/2010/04/25/chances-are/\n\nhttp://www.medicine.ox.ac.uk/bandolier/booth/glossary/freq.html\n\nAn alternative approach, very similar to @Strants's, with similar logic, but a slightly different way of looking at things:\n\nWe know that $\\frac{2}{3}$ of the committee members are men and that $\\frac{3}{8}$ of them are Americans.\n\nThis means that the proportion of all committee members who are American men is:\n\n$\\frac{2}{3} \\times \\frac{3}{8} = \\frac{6}{24} = \\frac{1}{4}$\n\nWe know also that $\\frac{3}{5}$ of the members are Russian.\n\nWe know that there are only four types of people at the committee:\n\nAmerican men, Russian men, Russian women and American women.\n\nWe have found that $\\frac{1}{4}$ of the members are American men and $\\frac{3}{5}$ of them are Russian - i.e. Russian men and Russian women account for $\\frac{3}{5}$ of the members.\n\nTherefore, the proportion of members who are American men, Russian men, or Russian women is:\n\n$\\frac{3}{5} + \\frac{1}{4} = \\frac{12}{20} + \\frac{5}{20} = \\frac{17}{20}$\n\nTherefore, the proportion of members who are American women is:\n\n$1 - \\frac{17}{20} = \\frac{3}{20} = \\text{A}$.", "date": "2019-10-22 06:09:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1296651/sat-maths-question-about-fractions", "openwebmath_score": 0.5978457927703857, "openwebmath_perplexity": 459.02175419851187, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815497525024, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8679838596061126}}
{"url": "https://math.stackexchange.com/questions/1523820/we-roll-a-six-sided-die-ten-times-what-is-the-probability-that-the-total-of-all", "text": "# We roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by 6?\n\nSo the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?\n\nSuppose that we roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by six?\n\n\u2022 Almost a duplicate of this question. Using the same logic, the answer is easily seen to be $1/6$. \u2013\u00a0TonyK Nov 11 '15 at 15:33\n\nHint.\n\nRoll $9$ times and let $x$ be the total.\n\nFor exactly one number $n\\in\\{1,2,3,4,5,6\\}$ we will have $6 \\mid (x+n)$ (i.e. $x+n$ is divisible by $6$).\n\n\u2022 I didn't get it, why would I roll 9 times instead of 10? and what is 6|x? \u2013\u00a0Xlyon Nov 11 '15 at 9:38\n\u2022 @Xlyon When you've rolled nine times there's always exactly one outcome for the tenth that will make the sum divisible by $6$. \u2013\u00a0skyking Nov 11 '15 at 9:43\n\u2022 Thanks! that is logical. Can you please help me complete it? \u2013\u00a0Xlyon Nov 11 '15 at 10:50\n\u2022 Exactly one of the faces of the die thrown at the $10$th time will result in a total sum ($x+n$) that is divisible by $6$. The probability that this face shows up is $\\frac16$ (if the die is fair, of course). \u2013\u00a0drhab Nov 11 '15 at 10:54\n\u2022 The value of $x$ is indeed irrelevant. Whatever value $x$ takes, it's chance to change into a number divisible by $6$ (when the last result is added by $x$) is in all cases $\\frac16$. So that's indeed the answer to this question. They could have asked: take some arbitrary number $y\\in\\mathbb Z$ and trhrow a fair die. If $D$ denotes the outcome then what is the probability that $y+D$ is divisible by $6$? Same answer: $\\frac16$, \u2013\u00a0drhab Nov 11 '15 at 11:34\n\nAfter rolling the die once, there is equal probability for each result modulo\u00a06. Adding any unrelated integer to it will preserve the equidistribution. So you can even roll a 20-sided die afterwards and add its outcome: the total sum will still have a probability of 1/6 to be divisible by 6.\n\n\u2022 @Kevin The d20 by itself does have that unequal distribution, but the sum d20+d6 does not. It literally does not matter what _**x**_ is in determining the distribution of (_**x**_ + d6) mod 6. In fact, d20 + d6 also has exactly a 1/20 chance of being divisible by 20 by the same logic \u2013\u00a0Monty Harder Nov 11 '15 at 19:28\n\nIf you want something a little more formal and solid than drhab's clever and brilliant answer:\n\nLet $P(k,n)$ be the probability of rolling a total with remainder $k$ when divided by $6, (k = 0...5)$ with $n$ die.\n\n$P(k, 1)$ = Probability of rolling a $k$ if $k \\ne 0$ or a $6$ if $k = 6$; $P(k, 1) = \\frac 1 6$.\n\nFor $n > 1$. $P(k,n) = \\sum_{k= 0}^5 P(k, n-1)\\cdot \\text{Probability of Rolling(6-k)} = \\sum_{k= 0}^5 P(k, n-1)\\cdot\\frac 1 6= \\frac 1 6\\sum_{k= 0}^5 P(k, n-1)= \\frac 1 6 \\cdot 1 = \\frac 1 6$\n\nThis is drhab's answer but in formal terms without appeals to common sense\n\n\u2022 That's a way to do it. What I actually had in mind was: $P\\left(6\\text{ divides }S_{10}\\right)=\\sum_{m\\in\\mathbb{N}}P\\left(6\\text{ divides }S_{10}\\mid S_{9}=m\\right)P\\left(S_{9}=m\\right)=\\sum_{m\\in\\mathbb{N}}\\frac{1}{6}P\\left(S_{9}=m\\right)=\\frac{1}{6}$ \u2013\u00a0drhab Nov 12 '15 at 9:40\n\u2022 Yes, that is more direct. \u2013\u00a0fleablood Nov 12 '15 at 16:45\n\nIn spite of all great answers, given here, I say, why not give another proof, from another point of view. The problem is we have 10 random variables $X_i$ for $i=1,\\dots,10$, defined over $[6]=\\{1,\\dots,6\\}$, and we are interested in distribution of $Z$ defined as $$Z=X_1\\oplus X_2\\oplus \\dots \\oplus X_{10}$$ where $\\oplus$ is addition modulo $6$. We can go on by two different, yet similar proofs.\n\nFirst proof: If $X_1$ and $X_2$ are two random variables over $[6]$, and $X_1$ is uniformly distributed, sheer calculation can show that $X_1\\oplus X_2$ is also uniformly distributed. Same logic yields that $Z$ is uniformly distributed over $[6]$.\n\nRemark: This proves a more general problem. It says that even if only one of the dices is fair dice, i.e. each side appearing with probability $\\frac 16$, the distribution of $Z$ will be uniform and hence $\\mathbb P(Z=0)=\\frac 16$.\n\nSecond proof: This proof draws on (simple) information theoretic tools and assumes its background. The random variable $Z$ is output of an additive noisy channel and it is known that the worst case is uniformly distributed noise. In other word if $X_i$ is uniform for only one $i$, $Z$ will be uniform. To see this, suppose that $X_1$ is uniformly distributed. Then consider the following mutual information $I(X_2,X_3,\\dots,X_6;Z)$ which can be written as $H(Z)-H(Z|X_2,\\dots,X_6)$. But we have: $$H(Z|X_2,\\dots,X_6)=H(X_1|X_2,\\dots,X_6)=H(X_1)$$\nwhere the first equality is resulted from the fact that knowing $X_2,\\dots,X_6$ the only uncertainty in $Z$ is due to $X_1$. The second equality is because $X_1$ is independent of others. Know see that:\n\n\u2022 Mutual information is positive: $H(Z)\\geq H(X_1)$\n\u2022 Entropy of $Z$ is always less that or equal to the entropy of uniformly distributed random variable over $[6]$: $H(Z)\\leq H(X_1)$\n\u2022 From the last two $H(Z)=H(X_1)$ and $Z$ is uniformly distributed and the proof is complete.\n\nSimilarly here, only one fair dice is enough. Moreover the same proof can be used for an arbitrary set $[n]$. As long as one of the $X_i$'s is uniform, then their finite sum modulo $n$ will be uniformly distributed.\n\nThere are 3 variables in this case:\n\n\u2022 the number of sides of the dice: s (e.g. 6)\n\u2022 the number of throws: t (e.g. 10)\n\u2022 the requesed multiple: x (e.g. 6)\n\nIn this case, the conditions are simple:\n\n\u2022 s>=x\n\u2022 x >0\n\u2022 t > 0\n\nAnd also the answer is simple: Throwing a sum that is a multiple of 6 has a 1/6 probability.\n\n$P(s,t,x) = 1/x$\n\nFor situations where s<x this is not entirely correct. It approaches the same result though, at a high amount of throws. Example: If you throw a 6-sided dice 30 times the chance that the sum is a multiple of 20 will be about 5%. Proving this is a bit of a challenge.\n\n$\\lim \\limits_{t \\to \\infty} P(s,t,x) = 1/x$,\n\nNevertheless, if programming is an acceptable proof:\n\npublic static void main(String[] args) {\nint t_throws = 10;\nint s_sides = 6;\nint x_multiple = 6;\nint[] diceCurrentValues = new int[t_throws];\nfor (int i = 0; i < diceCurrentValues.length; i++) diceCurrentValues[i] = 1;\n\nint combinations = 0;\nint matches = 0;\nfor (; ; ) {\n// calculate the sum of the current combination\nint sum = 0;\nfor (int diceValue : diceCurrentValues) sum += diceValue;\n\ncombinations++;\nif (sum % x_multiple == 0) matches++;\nSystem.out.println(\"status: \" + matches + \"/\" + combinations + \"=\" + (matches * 100 / (double) combinations) + \"%\");\n\n// create the next dice combination\nint dicePointer = 0;\nboolean incremented = false;\nwhile (!incremented) {\nif (dicePointer == diceCurrentValues.length) return;\nif (diceCurrentValues[dicePointer] == s_sides) {\ndiceCurrentValues[dicePointer] = 1;\ndicePointer++;\n} else {\ndiceCurrentValues[dicePointer]++;\nincremented = true;\n}\n}\n}\n}\n\n\n# EDIT:\n\nHere's another example. If you throw a 6-sided dice 10 times, there is 1/4 probability that the sum is a multiple of 4. The program above should run with the following parameters:\n\n    int t_throws = 10;\nint s_sides = 6;\nint x_multiple = 4;\n\n\nThe program will show the final output: status: 15116544/60466176=25.0% That means that there are 60466176 combinations (i.e. 6^10) and that there are 15116544 of them where the sum is a multiple of 4. So, that's 25% (=1/4).\n\nThis just follows the formula as mentioned above (i.e. P(s,t,x) = 1/x). x is 4 in this case.\n\n\u2022 \"Assuming that t converges to infinity. This is also the case when s<x.\" This sounds nonsensical. \u2013\u00a0djechlin Nov 11 '15 at 17:43\n\u2022 Excuse me for my poor English :) The thing is, if you apply this for values greater than the number of sides of your dice. (e.g. multiples of 10 with a 6 sided dice.) then P = 1/x is no longer correct. But it does converge 1/x ; meaning that if you would throw an infinit amount of times, the result would be 1/x again. \u2013\u00a0bvdb Nov 12 '15 at 0:10\n\u2022 I made some slight adjustments. Let me know what you think. \u2013\u00a0bvdb Nov 12 '15 at 0:22\n\u2022 It's not the case that the limiting factor is if s>=x. Consider the probability that 10 rolls of a six-sided dice divides 4. (Of course the limit still converges when the number of dices increases) \u2013\u00a0Taemyr Nov 12 '15 at 8:40\n\u2022 @Taemyr for 10 roles with a 6-sided dice, there are 60466176 combinations, of which 15116544 have a sum which is a multiple of 4. That's exactly 25% which is exactly 1/4. So, it's correct, right ? \u2013\u00a0bvdb Nov 14 '15 at 15:51\n\nRoll the die 9 times and add up the dots. The answer is x. Roll the die one more time. add the number thrown to x to get one and only one of the following answers; x+1, x+2, x+3, x+4, x+5 or x+6. since these answers are six sequential numbers one and only one of them will be divisible by six. Therefore the probability of the sum of ten rolls of a die being divisible by six is exactly 1/6.\n\nCouldn't you also think of it as the maximum possible value you could get for rolling the die 10 times would be 60, how many numbers between 1 and 60 are divisible by 6? 10 numbers are divisible by 6(6*1, 2, 3, etc.) So, 10 out of 60 possible values gives... 1/6. I love math.\n\n\u2022 True, but some numbers appear multiple times. So, even though your answer is correct, your logic is flawed. \u2013\u00a0bvdb Nov 11 '15 at 11:58\n\u2022 The minimum possible value is 10. So only 9 of 51 values are divisible by 6. But not all numbers are equally probable. \u2013\u00a0Cephalopod Nov 11 '15 at 11:58\n\u2022 The numbers are not distributed evenly. \u2013\u00a0fleablood Nov 11 '15 at 20:31", "date": "2020-03-29 16:11:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1523820/we-roll-a-six-sided-die-ten-times-what-is-the-probability-that-the-total-of-all", "openwebmath_score": 0.7875996828079224, "openwebmath_perplexity": 383.01354319251885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815523039174, "lm_q2_score": 0.8774767938900121, "lm_q1q2_score": 0.8679838570907867}}
{"url": "http://math.stackexchange.com/questions/126961/yet-another-question-on-using-basics-limit-arithmetics", "text": "# Yet Another Question On Using Basics Limit Arithmetics\n\nIs this claim true?\n\nGiven $\\lim \\limits_{n\\to \\infty}\\ a_n=\\frac{1}{2}$ Then $\\lim \\limits_{n\\to \\infty}\\ (a_n - [a_n])=\\frac{1}{2}$\n\nI think it's true, but probably I just didn't find the right example to disprove it.\n\nThanks a lot.\n\n-\nWhat do you mean by $[a_{n}]$? \u2013\u00a0 Isaac Solomon Apr 1 '12 at 18:14\nit's the round integer \u2013\u00a0 Anonymous Apr 1 '12 at 18:17\nRound up or down? \u2013\u00a0 Isaac Solomon Apr 1 '12 at 18:17\nIf you mean nearest integer, then the claim is correct, so you will not find an example to disprove it. Are you looking for a formal argument? \u2013\u00a0 Andr\u00e9 Nicolas Apr 1 '12 at 18:20\nIt's round donwn or truncate. And yes, @Andr\u00e9 Nicolas, I'm looking for a formal argument, thanks a lot. \u2013\u00a0 Anonymous Apr 1 '12 at 18:21\n\nSince\n\n$$\\lim_{n \\to \\infty} a_{n} = \\frac{1}{2}$$\n\nthere exists $N$ such that for all $n \\geq N$,\n\n$$|a_{n} - \\frac{1}{2}| < \\frac{1}{4}$$\n\nThen, for $n \\geq N$, we have\n\n$$0 \\leq a_{n} \\leq 1 \\Longrightarrow [a_{n}] = 0$$\n\nso that for $n \\geq N$,\n\n$$a_{n} - [a_{n}] = a_{n}$$\n\nNow pick $\\epsilon > 0$. Suppose that for all $n > M_{\\epsilon}$\n\n$$|a_{n} - \\frac{1}{2}| < \\epsilon$$\n\nReplacing $M_{\\epsilon}$ with $\\mbox{max}(M_{\\epsilon},N)$, we have\n\n$$|(a_{n} - [a_{n}]) - \\frac{1}{2} | = |a_{n} - \\frac{1}{2}| < \\epsilon$$\n\nwhich proves that\n\n$$\\lim_{n \\to \\infty} (a_{n} - [a_{n}]) = \\frac{1}{2}$$\n\n-\nI'm reading the proof and I'm at the line: \"Now pick $\\epsilon > 0$. Suppose that for all $n > M_{\\epsilon}$ - what is $M_{\\epsilon}$? \u2013\u00a0 Anonymous Apr 1 '12 at 18:43\n$M_{\\epsilon}$ is just some sufficiently large natural number. I write $\\epsilon$ in the subscript to indicate that this $M$ is a function of $\\epsilon$. \u2013\u00a0 Isaac Solomon Apr 1 '12 at 19:29\n\nIt depends on what you mean by the symbol $[x]$. If $[x]$ is the floor of $x$ (i.e., the greatest integer less than or equal to $x$), then the limit will be $\\frac{1}{2}$. This follows from $\\lim \\limits_{n\\to \\infty}\\ a_n=\\frac{1}{2}$: there exists an $N$ such that for every $n \\geq N$, $|a_n - \\frac{1}{2}| < \\frac{1}{2}$. For these $a_n$, it follows that $[a_n] = 0$, and so $$\\lim \\limits_{n\\to \\infty}\\ a_n - [a_n] =\\frac{1}{2} - 0 = \\frac{1}{2}$$\n\nOn the other hand, if $[x]$ is the nearest integer to $x$, then the limit does not exist: consider the sequence $a = (0.4,0.6,0.49,0.51,0.499,0.501,\\ldots)$. Clearly $a_n \\rightarrow \\frac{1}{2}$, but $[a_n]$ alternates between $0$ and $1$.\n\n-\nThanks, I'm talking about the first case where $[x]$ is the floor of $x$. Could you please explain why it follows that $[a_n] = 0$? \u2013\u00a0 Anonymous Apr 1 '12 at 18:32\nEssentially, the $a_n$ are getting closer and closer to $\\frac{1}{2}$; after a while (for $n \\geq N$), all the $a_n$ will be between $0$ and $1$. The floor of a number in this interval is $0$. \u2013\u00a0 Th\u00e9ophile Apr 1 '12 at 18:39\n@Th\u00e9ophile: Yes, I was careless in my comment, which has been deleted. I meant the distance to the nearest integer. \u2013\u00a0 Andr\u00e9 Nicolas Apr 1 '12 at 18:56\n\nWe want to show that for any $\\epsilon>0$, there is an $N$ such that if $n>N$, then $|(a_n-\\lfloor a_n\\rfloor)-\\frac{1}{2}|<\\epsilon$. Here by $\\lfloor w\\rfloor$ we mean the greatest integer which is $\\le w$.\n\nLet $\\epsilon'=\\min(\\epsilon,1/4)$. By the fact that $\\lim_{n\\to\\infty} a_n=\\frac{1}{2}$, there is an $N$ such that if $n>N$ then $|a_n-\\frac{1}{2}|<\\epsilon'$. In particular, $|a_n-\\frac{1}{2}|<\\frac{1}{4}$, and therefore $\\lfloor a_n\\rfloor=0$. It follows that if $n>N$ then $$\\left|(a_n-\\lfloor a_n\\rfloor)-\\frac{1}{2}\\right|=\\left|a_n-\\frac{1}{2}\\right|<\\epsilon.$$\n\n-\nyou wrote: $\\left|(x_n-\\lfloor x_n\\rfloor)-\\frac{1}{2}\\right|=\\left|x_n-\\frac{1}{2}\\right|<\\epsilon$. But actually I know that $\\left|(x_n-\\lfloor x_n\\rfloor)-\\frac{1}{2}\\right|=\\left|x_n-\\frac{1}{2}\\right|<\\epsilon'$, no? why can I make the switch from $\\epsilon'$ to $\\epsilon$? at least not for that n, maybe we need to add $n_1$? \u2013\u00a0 Anonymous Apr 1 '12 at 19:04\nThe number we are \"challenged with\" is $\\epsilon$. If $\\epsilon$ is ridiculous, like $5$, then we can't say that $\\lfloor x_n\\rfloor=0$. Specifying that $\\epsilon'=\\min(1/4,\\epsilon)$ makes sure $\\lfloor x_n\\rfloor=0$. It also makes sure (since $\\epsilon' \\le \\epsilon$) that from $|x_n-\\frac{1}{2}|<\\epsilon'$, we can conclude that $|x_n-\\frac{1}{2}|<\\epsilon$. More informally, we could forget about $\\epsilon'$, and say that the argument will only work if $\\epsilon$ is reasonably small, but that's good enough. \u2013\u00a0 Andr\u00e9 Nicolas Apr 1 '12 at 19:16\nThanks, I used the max between the n you gave me and the $n_1$ I have from the definition of limit, therefore safe to any $\\epsilon$. \u2013\u00a0 Anonymous Apr 1 '12 at 19:23\n@Anonymous: It is just a technical trick. It can be replaced by taking $N$ simultaneously large enough to make $|a_n-1/2|<1/4$ (anything under $1/2$ is OK) and large enough to make $|a_n-1/2|<\\epsilon$. \u2013\u00a0 Andr\u00e9 Nicolas Apr 1 '12 at 19:27", "date": "2014-12-20 00:53:40", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/126961/yet-another-question-on-using-basics-limit-arithmetics", "openwebmath_score": 0.9549311399459839, "openwebmath_perplexity": 244.7056961229364, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969679646668, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8679533814952709}}
{"url": "https://alpynepyano.github.io/healthyNumerics/posts/basic_stats_02_mechanical_approach_to_bayesian_rule.html", "text": ".\n\n.\n\n.\n\n.\n\n.\n\nHealthyNumerics\n\nHealthPoliticsEconomics | Quant Analytics | Numerics\n\nBasic Stats 02: A mechanical approach to the Bayesian rule\n\nIntro\n\nThe Bayesian rule\n\n$$P(A \\mid B) = \\frac{P(B \\mid A) \\cdot P(A)}{P(B)}$$\n\nseems to be a rather abstract formula. But this impression can be corrected easely when considering a simple practical application. We call this approach mechanical because for this type of application there is no philosphical dispute between frequentist's and bayesian's mind set. We will just use in a mechanical/algorithmical manner the cells of a matrix. In this section we\n\n\u2022 formulate a prototype of a probability problem (with red and green balls that have letters A and B printed on)\n\u2022 summarize the problem in a basically 2x2 matrix (called contingency table)\n\u2022 use the frequencies first\n\u2022 replace them by probalities afterwards and recognize what conditional probablities are\n\u2022 recognize that applying the Bayesian formula is nothing else than walking from one side of the matrix to the other side\n\n\nA prototype of a probability problem\n\nGiven:\n\n\u2022 19 balls\n\u2022 14 balls are red, 5 balls are green\n\u2022 among the 14 red balls, 4 have a A printed on, 10 have a B printed on\n\u2022 among the 5 green balls, 1 has a A printed on, 4 have a B printed on\n\nTypical questions: - we take 1 ball. - What is the probabilitiy that it is green ? - What is the probabilitiy that it is B under the precondition it's green ?\n\nWe will use Pandas to represent the problem and the solutions\n\nimport matplotlib.pyplot as plt\nfrom IPython.core.pylabtools import figsize\nimport numpy as np\nimport pandas as pd\npd.set_option('precision', 3)\n\n\n\nContingency table with the frequencies\n\nThe core of the representation is a 2x2 matrix that summarizes the situation of the balls with the colors and the letters on. This matrix is expanded with the margins that contain the sums. - sum L stands for the sum of the letters - sum C stands for the sum of the colors\n\ncolumns = np.array(['A', 'B'])\nindex   = np.array(['red', 'green'])\ndata    = np.array([[4,10],[1,4]])\n\ndf = pd.DataFrame(data=data, index=index, columns=columns)\ndf['sumC'] = df.sum(axis=1)  # append the sums of the rows\ndf.loc['sumL']= df.sum()     # append the sums of the columns\ndf\n\nA B sumC\nred 4 10 14\ngreen 1 4 5\nsumL 5 14 19\n\nAppend the relative contributions of B and of green\n\nWe expand the matrix by a further column and a further row and use them to compute relative frequencies (see below).\n\ndef highlight_cells(x):\ndf = x.copy()\ndf.loc[:,:] = ''\ndf.iloc[1,1] = 'background-color: #53ff1a'\ndf.iloc[1,3] = 'background-color: lightgreen'\ndf.iloc[3,1] = 'background-color: lightblue'\nreturn df\n\ndf[r'B/sumC']         = df.values[:,1]/df.values[:,2]\ndf.loc[r'green/sumL'] = df.values[1,:]/df.values[2,:]\n\nt = df.style.apply(highlight_cells, axis=None)\nt\n\n\nA B sumC B/sumC\nred 4 10 14 0.714\ngreen 1 4 5 0.8\nsumL 5 14 19 0.737\ngreen/sumL 0.2 0.286 0.263 1.09\n\nLet's focus on the row with the green balls\n\nFrom all green balls (= 5) is the portion of those with a letter B (=4) 0.8\n\n$$\\frac{green\\: balls\\: with \\:B}{all\\: green\\: balls} = \\frac{4}{5} = 0.8$$\n\nNote that this value already corresponds to the conditional probality $$P(B \\mid green)$$\n\n\n\nLet's focus on the column with the balls with letter B\n\nFrom all balls with letter B (= 14) is the portion of those that are green (=4) 0.286\n\n$$\\frac{green\\; balls\\; with\\; B}{all\\; balls\\; with\\; letter\\; B} = \\frac{4}{14} = 0.286$$\n\nNote that also this value already corresponds to the conditional probality $$P(green \\mid B)$$\n\n\n\n\nContingency table with the probabilities\n\nWe find the probabilities by dividing the frequencies by the sum of balls.\n\ncolumns = np.array(['A', 'B'])\nindex   = np.array(['red', 'green'])\ndata    = np.array([[4,10],[1,4]])\ndata    = data/np.sum(data)\n\ndf = pd.DataFrame(data=data, index=index, columns=columns)\ndf['sumC'] = df.sum(axis=1)  # append the sums of the rows\ndf.loc['sumL']= df.sum()     # append the sums of the columns\ndf\n\nA B sumC\nred 0.211 0.526 0.737\ngreen 0.053 0.211 0.263\nsumL 0.263 0.737 1.000\n\n\nAppend the relative contributions of B and of green\n\ndf[r'B/sumC']         = df.values[:,1]/df.values[:,2]\ndf.loc[r'green/sumL'] = df.values[1,:]/df.values[2,:]\nt = df.style.apply(highlight_cells, axis=None)\nt\n\n\nA B sumC B/sumC\nred 0.211 0.526 0.737 0.714\ngreen 0.0526 0.211 0.263 0.8\nsumL 0.263 0.737 1 0.737\ngreen/sumL 0.2 0.286 0.263 1.09\n\n\n\nNote the formula in the cells\n\ncolumns = np.array(['-----------A----------', '---------B----------', '----------sumC--------',  '--------B/sumC--------'])\nindex   = np.array(['red', 'green', 'sumL',  'green/sumL'])\ndata    = np.array([['...','...','...','...'],\n['...', '$P(B \\cap green)$', '$P(green)$', '$P(B \\mid green)$'],\n['...','$P(B)$','...','...'],\n['...','$P(green \\mid B)$','...','...'] ])\ndf = pd.DataFrame(data=data, index=index, columns=columns)\nt = df.style.apply(highlight_cells, axis=None)\nt\n\n\n-----------A---------- ---------B---------- ----------sumC-------- --------B/sumC--------\nred ... ... ... ...\ngreen ... $$P(B \\cap green)$$ $$P(green)$$ $$P(B \\mid green)$$\nsumL ... $$P(B)$$ ... ...\ngreen/sumL ... $$P(green \\mid B)$$ ... ...\n\n\n\nConditional probability: Let's focus on the row with the green balls\n\nThe probability to get a ball with B out of all green balls is 0.8\n\n$$P(B \\mid green) = \\frac{P(green\\: balls\\: with \\:B)}{P(all\\: green\\: balls)} = \\frac{P(green \\cap B)}{P(green)} = \\frac{0.211}{0.263} = 0.8$$\n\n\nConditional probability: Let's focus on the column with the balls with letter B\n\nThe probability to get a green ball out of all balls with a B is 0.286\n\n$$P(green \\mid B) = \\frac{P(green\\; balls\\; with\\; B)}{P(all\\; balls\\; with\\; letter\\; B)} = \\frac{P(green \\cap B)}{P(B)} = \\frac{0.211}{0.737} = 0.286$$\n\n\nBayes rule\n\nGiven $$P(green \\mid B$$) find $$P(B \\mid green)$$ :\n\n$$P(B \\mid green) = \\frac{P(green \\mid B) \\cdot P(B)}{P(green)} = \\frac{0.286 \\cdot 0.737}{0.263} = \\frac{0.211}{0.263}= 0.8$$\n\n\nand given $$P(B \\mid green)$$ find $$P(green \\mid B$$):\n\n$$P(green \\mid B) = \\frac{P(B \\mid green) \\cdot P(green)}{P(B)} = \\frac{0.8 \\cdot 0.263}{0.737} = \\frac{0.211}{0.737} = 0.286$$\n\nApplying the Bayes rule means that we walk from the element most right in the matrix to the element at the bottom of the matrix and vice versa:\n\nshow_frequencies()\n\n\ndef show_frequencies():\npx = 4; py = 4\n\nfigsize(10, 10)\nfontSize1 = 15\nfontSize2 = 12\nfontSize3 = 20\n\nA1 = 4;     A2 = 10;       A3 = A1+A2\nB1 = 1;     B2 =  4;       B3 = B1+B2\nC1 = A1+B1; C2 = A2+B2;   C3 = A3+B3\n\ndata = np.array([[A1, A2, A3],\n[B1, B2, B3],\n[C1, C2, C3] ])\n\nclr     = np.array([ ['#ffc2b3', '#ff704d', '#ff0000'],\n['#b3ff99', '#53ff1a', '#208000'],\n['#00bfff', '#0000ff', '#bf00ff']   ])\n\ntitle  = np.array([['A and red', 'B and red', 'sum of reds',\n'$P(B|red) = \\\\frac {P(A\\\\cap red)}{P(red)}$'           ],\n['A and green', 'B and green', 'sum greens',   '% (B of greens)'     ],\n['sum of A', 'sum of B', 'sum of all balls',' '        ],\n[' ', '% (greens of B)', '',  ' '     ],\n]  )\n\nxlabel = np.array([ ['A', 'B', 'A+B', 'B/(A+B)'],\n['A', 'B', 'A+B', 'B/(A+B)'],\n['A red + A green', 'B red + B green', 'A+B', 'B/(A+B)'],\n['A', 'B green/(B red + B green)', 'A+B', 'B/(A+B)']       ]  )\n\nylabel = np.array([ ['red',   'red',   'red',  'B of reds'],\n['green', 'green', 'green','B of greens'],\n['sum A', 'sum B', 'Total', '' ],\n['green', 'greens of B', 'green', ' '],    ])\n\nf, ax = plt.subplots(px, py, sharex=True, sharey=True, edgecolor='none')  #, facecolor='lightgrey'\n\nfor i in range(px-1):\nfor j in range(py-1):\n\npatches, texts =ax[i,j].pie([data[i,j]], labels=[str(data[i,j])], #autopct='%1.1f%%',\ncolors = [clr[i,j]])\ntexts[0].set_fontsize(fontSize3)\nax[i,j].set_title(title[i,j], position=(0.5,1.2), bbox=dict(facecolor='#f2f2f2', edgecolor='none'), fontsize= fontSize1)\nif i*j==1 :\nax[i,j].set_title(title[i,j], position=(0.5,1.2), bbox=dict(facecolor='#bfbfbf', edgecolor='none'), fontsize=fontSize1)\n\nax[i,j].set_xlabel(xlabel[i,j], fontsize=fontSize2, color='c')\nax[i,j].xaxis.set_label_position('top')\n\nax[i,j].set_ylabel(ylabel[i,j], fontsize=fontSize2, color='c')\n\nax[i,j].axis('equal')\n\nj += 1\nif (i == 1):\np = data[i,-2]/data[i,-1]; o = p/(1-p)\npatches, texts =ax[i,j].pie([o,1], #autopct='%1.1f%%',\ncolors = [clr[i,1], clr[i,2] ] )\ntexts[0].set_fontsize(fontSize3)\nax[i,j].set_title(title[i,j], position=(0.5,1.2), bbox=dict(facecolor='#bfbfbf', edgecolor='none'), fontsize=fontSize1)\nax[i,j].set_xlabel(xlabel[i,j], color='c', fontsize=fontSize2)\nax[i,j].xaxis.set_label_position('top')\nax[i,j].set_ylabel(ylabel[i,j], fontsize=fontSize2, color='c')\nax[i,j].axis('equal')\nelse:\nax[i,j].plot(0,0)\nax[i,j].set_frame_on(False)\n\ni = px-1;\nfor j in range(py):\nif j==1:\np = data[1,1]/data[2,1]; o = p/(1-p)\npatches, texts =ax[i,j].pie([o,1], #autopct='%1.1f%%',\ncolors = [clr[1,1], clr[2,1] ])\ntexts[0].set_fontsize(fontSize3)\nax[i,j].set_title(title[i,j], position=(0.5,1.2), bbox=dict(facecolor='#bfbfbf', edgecolor='none'), fontsize=fontSize1)\n\nax[i,j].set_xlabel(xlabel[i,j], color='c', fontsize=fontSize2)\nax[i,j].xaxis.set_label_position('top')\nax[i,j].set_ylabel(ylabel[i,j], fontsize=fontSize2, color='c')\nax[i,j].axis('equal')\nax[i,j].set_facecolor('y')\nelse:\nax[i,j].plot(0,0)\nax[i,j].set_frame_on(False)\n\nplt.tight_layout()\nplt.show()\n\nfrom IPython.display import HTML\n\n# HTML('''<script> $('div .input').hide()''') #l\u00e4sst die input-zellen verschwinden HTML('''<script> code_show=true; function code_toggle() { if (code_show){$('div.input').hide();\n} else {\n$('div.input').show(); } code_show = !code_show }$( document ).ready(code_toggle);\n</script>", "date": "2020-07-03 10:04:01", "meta": {"domain": "github.io", "url": "https://alpynepyano.github.io/healthyNumerics/posts/basic_stats_02_mechanical_approach_to_bayesian_rule.html", "openwebmath_score": 0.7729313373565674, "openwebmath_perplexity": 12570.374402367228, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969641180276, "lm_q2_score": 0.8824278695464501, "lm_q1q2_score": 0.8679533735390272}}
{"url": "https://www.physicsforums.com/threads/permutations-hard-question.669846/", "text": "# Permutations : hard question !\n\n1. Feb 6, 2013\n\n### hms.tech\n\n1. The problem statement, all variables and given/known data\n\nIn how many ways can the three letters from the word \" GREEN \" be arranged in a row if atleast one of the letters is \"E\"\n\n2. Relevant equations\n\nPermutations Formula\n\n3. The attempt at a solution\n\nThe total arrangements without restriction: 5P3/2! = $\\frac{5!}{2! * 2!}$\n\nThe number of arrangements in which there is no \"E\" = 3!\n\nAns : 5P3/2! - 3! = 24 (wrong)\n\nHere is another approach :\n\nThe arrangements with just one \"E\" = $\\frac{2!*4!}{2!}$\nThe arrangements with two \"E\" = $\\frac{3*2}{1}$\n\nI think I making a mistake due to the repetition of \"E\" ... Can any one of you tell me a better way which avoids the problem I am being having .\n\n2. Feb 6, 2013\n\n### HallsofIvy\n\nIf \"at least one letter must be E\", that means that the other two letters must be chosen from \"GREN\". That is, the first letter must be one of those 4 letters, the second one of the three remaining letters. How many is that?\n\nBut then we can put the \"E\" that we took out into any of three places: before the two, between them, or after the two letters so we need 3 times that previous number.\n\n3. Feb 6, 2013\n\n### hms.tech\n\n4P2 = 4*3\n\nAccording to your method the answer should be 12 * 3 = 36 (The correct answer in the solutions is \"27\")\n\nClearly your method (as did mine) repeats some of the permutations :\n\nYou didn't take onto account that the two \"E\" are not distinct. Thus in those permutations where we chose two \"E\" were repeated. see :\nGEE , EEG, EGE\nENE, EEN, NEE\nREE, ERE , EER\n\n4. Feb 6, 2013\n\n### CAF123\n\nAlternatively, split the problem into two: Count those combinations with only 1 E and then count separately those with two E's.\n\nFor one E combination: (1C1)*(3C2)*3! = 18\nFor two E combination: (2C2)*(3C1)*(3!/2!) = 9. Then add.", "date": "2018-02-21 06:13:23", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/permutations-hard-question.669846/", "openwebmath_score": 0.8493869304656982, "openwebmath_perplexity": 1166.250459490498, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812354689082, "lm_q2_score": 0.8962513807543223, "lm_q1q2_score": 0.8679130193855855}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=kqv56eqotgiu56ejlj64jkai53&action=printpage;topic=1204.0", "text": "# Toronto Math Forum\n\n## MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Ende Jin on September 08, 2018, 03:39:53 PM\n\nTitle: About the definition of Argument (in book)\nPost by: Ende Jin on September 08, 2018, 03:39:53 PM\nI found that the definition of \"arg\" and \"Arg\" in the book is different from that introduced in the lecture (exactly opposite) (on page 7).\nI remember in the lecture, the \"arg\" is the one always lies in $(-\\pi, \\pi]$\nWhich one should I use?\nTitle: Re: About the definition of Argument (in book)\nPost by: Victor Ivrii on September 08, 2018, 04:58:09 PM\nQuote\nWhich one should I use?\nThis is a good and tricky question because the answer is nuanced:\nSolving problems, use definition as in the Textbook, unless the problem under consideration requires modification: for example, if we are restricted to the right half-plane\u00a0 $\\{z\\colon \\Re z >0\\}$ then it is reasonable to consider $\\arg z\\in (-\\pi/2,\\pi/2)$, but if we are restricted to the upper half-plane\u00a0 $\\{z\\colon \\Im z >0\\}$ then it is reasonable to consider $\\arg z\\in (0,\\pi)$ and so on.\nTitle: Re: About the definition of Argument (in book)\nPost by: Ende Jin on September 09, 2018, 12:48:34 PM\nI am still confused. Let me rephrase the question again.\nIn the textbook, the definition of \"arg\" and \"Arg\" are:\n$arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = cos\\theta + isin\\theta$\nwhich means $arg(z) \\in \\mathbb{R}$\nwhile\n$Arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = cos\\theta + isin\\theta \\land \\theta \\in [-\\pi, \\pi)$\nwhich means $Arg(z) \\in [-\\pi, \\pi)$\n\nWhile in the lecture, as you have introduced, it is the opposite and the range changes to $(-\\pi, \\pi]$ instead of $[-\\pi, \\pi)$ (unless I remember incorrectly):\nArg is defined to be\n$Arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = (cos\\theta + isin\\theta)$\nwhich means $arg(z) \\in \\mathbb{R}$\nwhile arg is\n$arg(z) = \\theta \\Leftrightarrow \\frac{z}{|z|} = cos\\theta + isin\\theta \\land \\theta \\in (-\\pi, \\pi]$\n\nI am confused because if I am using the definition by the book,\nwhen $z \\in \\{z : Re (z) > 0\\}$\nthen $arg(z) \\in (-\\frac{\\pi}{2} + 2\\pi n,\\frac{\\pi}{2} + 2\\pi n), n \\in \\mathbb{Z}$\nTitle: Re: About the definition of Argument (in book)\nPost by: Victor Ivrii on September 09, 2018, 04:40:38 PM\n\nBTW, you need to write \\sin t and \\cos t and so on to have them displayed properly (upright and with a space after): $\\sin t$, $\\cos t$ and so on\nTitle: Re: About the definition of Argument (in book)\nPost by: Ende Jin on September 10, 2018, 10:03:33 AM\nThus in a test/quiz/exam, I should follow the convention of the textbook, right?\nTitle: Re: About the definition of Argument (in book)\nPost by: Victor Ivrii on September 10, 2018, 01:34:22 PM\nThus in a test/quiz/exam, I should follow the convention of the textbook, right?\nIndeed\nTitle: Re: About the definition of Argument (in book)\nPost by: oighea on September 12, 2018, 04:24:46 PM\nThe $\\arg$ of a complex number $z$ is an angle $\\theta$. All angles $\\theta$ have an infinite number of \"equivalent\" angles, namely $\\theta =2k\\pi$ for any integer $k$.\n\nEquivalent angles can be characterized by that they exactly overlap when graphed on a graph paper, relative to the $0^\\circ$ mark (usually the positive $x$-axis). Or more mathematically, they have the same sine and cosine. It also makes sine and cosine a non-reversible function, as given a sine or cosine, there are an infinite number of angles that satisfy this property.\n\n$\\Arg$, on the other hand, reduces the range of the possible angles such that it always lie between $0$ (inclusive) to $2\\pi$ (exclusive). That is because one revolution is $2\\pi$, or $360$ degrees. That is called the principal argument of a complex number.\n\nWe will later discover that complex logarithm also have a similar phenomenon.\nTitle: Re: About the definition of Argument (in book)\nPost by: Victor Ivrii on September 12, 2018, 04:34:21 PM\noighea\n\nPlease fix your screen name and use LaTeX command (rendered by MathJax) to display math, not paltry html commands.\nI fixed it in this post", "date": "2022-12-02 23:28:35", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=kqv56eqotgiu56ejlj64jkai53&action=printpage;topic=1204.0", "openwebmath_score": 0.9586477875709534, "openwebmath_perplexity": 1196.0509642251907, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799420543365, "lm_q2_score": 0.9005297921244243, "lm_q1q2_score": 0.8679125508718813}}
{"url": "http://mathhelpforum.com/discrete-math/123546-equation-problem-others.html", "text": "# Math Help - Equation problem and others...\n\n1. ## Equation problem and others...\n\nCould any1 help me a bit with this equation ?\n1) X1+X2+X3=11 , So how many are the solutions?\n\n2) We 've got these letters -> A,B,C,D,E,F,G,H , how many permutations\nof these letters can we make that include ABC ?\n\n3) How many ways are possible in order 3 girls and 8 boys sit in a row so that Mary and John NOT to sit next to each other ?\n\n4) I can understand at all what this question wants, but anyway ill ask here maybe you can understand ...\n\nHow many bytes exist with 1) Exactly 2 aces\n2) Exactly 4 aces\n3) Exactly 6 aces\n4) At least 6 aces\n\ni dont get it , a byte consists of 8 bits, whats that ace? any ideas?\n\nThanks in advance for your help =)\n\n2. Hi primeimplicant,\nwhere are you getting stuck with these?\n\nFor 1)\n\n$X_1,\\ X_2, X_3$ probably need to be positive integers >0.\n\nAre any of the numbers allowed to be zero?\n\nIf they are allowed to be negative or non-integers, that leaves infinitely\nmany solutions.\n-14+1+2, -15+1+3 etc\n\nHence, start with 1, add 2 and find out what $X_3$ is.\nThen start again at 1, add 3 and find out what $X_3$ is.\n\nWhen finished with 1, go on to 2 and make sure you leave out all numbers\nless than 2.\nThen go on to 3, leaving out all numbers less than 3.\nContinue until done, there are only a few.\n\n3. Originally Posted by primeimplicant\n[snip]\n4) I can understand at all what this question wants, but anyway ill ask here maybe you can understand ...\n\nHow many bytes exist with 1) Exactly 2 aces\n2) Exactly 4 aces\n3) Exactly 6 aces\n4) At least 6 aces\n\ni dont get it , a byte consists of 8 bits, whats that ace? any ideas?\n\nThanks in advance for your help =)\nMy guess is that an \"ace\" is a 1 bit, although this is unusual terminology I have not seen before. If my guess is correct then 1) is asking you how many 8-bit sequences there are which consist of two 1's and six 0's.\n\n4. Hello, primeimplicant!\n\n$1)\\;x_1+x_2+x_3\\:=\\:11$\nSo how many are the solutions?\nThe problem is NOT clearly stated.\nI must assume that the $x$'s are positive integers.\n\nConsider an 11-inch board marked in 1-inch intervals.\n\n. . $\\square\\square\\square\\square\\square\\square\n\\square\\square\\square\\square\\square$\n\nWe will cut the board at two of the ten marks.\n\nThen: . $\\square\\square\\square\\square\\square |\n\\square\\square | \\square\\square\\square\\square$\n.represents: . $5+2+4$\n\nAnd:- - $\\square \\square \\square | \\square \\square \\square \\square \\square \\square \\square | \\square$ .represents: . $3 + 7 + 1$\n\nTherefore, there are: . $_{10}C_2 \\:=\\:45$ solutions.\n\n2) We 've got these letters: . $A,B,C,D,E,F,G,H$\nHow many permutations of these letters can we make that include $ABC$ ?\nI assume you mean in that exact order: $ABC$\n\nDuct-tape $ABC$ together.\n\nThen we have 6 \"letters\" to arrange: . $\\boxed{ABC}\\;D\\;E\\;F\\;G\\;H$\n\nThere are: . $6! \\,=\\,720$ permutations.\n\n3) How many ways are possible in order 3 girls and 8 boys sit in a row\nso that Mary and John do NOT sit next to each other?\nWith no restrictions, the children can be seated in: $11!$ ways.\n\nSuppose Mary and John DO sit together.\nDuct-tape them together.\n\nThen we have 10 \"people\" to arrange: . $\\boxed{MJ}\\;A\\;B\\;C\\;D\\;E\\;F\\;G\\;H\\;I$\n. . They can be seated in: $10!$ ways.\n\nBut Mary and John could be taped like this: . $\\boxed{JM}$\n. . This makes for another $10!$ ways.\n\nSo there are: $2(10!)$ ways that Mary and John DO sit together.\n\nTherefore, there are: . $11! - 2(10!) \\:=\\:39,916,800 - 2(3,628,800) \\:=\\:32,659,200$ ways\n. . that Mary and John do NOT sit together.\n\n5. Thanks a lot Soroban , awesome, i totally understood!!\n\n6. Hi Primeimplicant,\n\nFor Q1...\n\nthat was fabulous work by Soroban!!\nHe's an artist at breaking long calculations into neat compact solutions.\nHere's just a bit i wanted to add.\n\nFor your Q1, i interpreted it as....\n\nhow many different additions of 3 unequal positive non-zero integers sum to 11.\n\nHence, i recommended you try\n\n1+2+(11-3)=11, which is 1+2+8=11\n\nthen 1+3+(11-4)=1+3+7,\n1+4+6=11,\n1+5+5...but this contains 2 fives.\n\nStarting with 2,\n2+3+6,\n2+4+5,\n2+5+4 is the same as 2+4+5 so move on to 3,\n3+4+4 has two fours, so that's out.\n\n3+5+3 out\n3+6+2 is ok\n3+7+1 counted\n\nstart with 4\n4+5+2 has been found already as has 4+6+1\n\nThe list then is\n\n1+2+8\n1+3+7\n1+4+6\n2+3+6\n2+4+5\n3+6+2\n\nIf you meant that\n\n$X_1,\\ X_2, X_3$ could be variables that can have any value\nfrom 1 to 9, since 1+1+9=11, so 9 is the largest positive integer useable,\n\nthen we have, starting with 1\n1+1+9\n1+2+8\n1+3+7\n1+4+6\n1+5+5\n\nSince any variable can be any of these values, these can be arranged in 3! ways, therefore there are 5(3!) = 30 combinations with the digit 1,\nwith 119 and 191 and 911 all double-counted, along with 155, 515, 551 double-counted, so we subtract 6.\n30-6 =24.\n\nstarting with 2, not containing 1\n2+2+7\n2+3+6\n2+4+5\n\nthese can also be arranged in (2)3!+3!/2! ways to give 12+3=15 additional combinations.\n\nstarting with 3, not containing 1 or 2\n\n3+3+5\n3+4+4\n3+5+3 has been counted\n3+6+2 has 2 in it\n\nso there are 2 new ones which can be arranged in (2)3!/2! ways to account for\neach of the 3 variables having those values.\nthis is another 3! combinations =6.\n\nstarting with 4, not containing 1, 2 or 3.\n\n4+4+3 is out so we have found them all as there are no further combinations due to the fact that we will need smaller digits for the sum.\n\nHence the total is 24+15+6=45\n\nThis is the total for this interpretation of the question.\n\nIf the variables must have different values, then we must subtract the three with the repeated digit 1, the three with the repeated digit 2, the three with the repeated digit 3, the three with the repeated digit 4, and the three with the repeated digit 5,\ngetting 45-15=30 with all 3 digits different,\nin other words.... combinations of the digits, where the digits are selections of 3 from the 8 available in that case, in such a way that they sum to 11.\n\nLooking at Soroban's analysis in another way\n\n*****|*|*****\n\nThe red lines can move 4 places left and 4 places right, giving an additional 8 solutions to the 1 shown....8+1=9 solutions containing the digit 1.\n\n****|**|*****\n\nThe red lines can move 3 places left and 4 places right,\ngiving 7+1=8 solutions containing the digit 2.\n\n****|***|****\n\nThe red lines can move 3 places left and 3 places right giving 6+1=7 solutions with the digit 3.\n\nContinuing that we get 9+8+7+6+5+4+3+2+1=45 solutions containing repeated digits.", "date": "2015-11-27 15:52:59", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/123546-equation-problem-others.html", "openwebmath_score": 0.7436730861663818, "openwebmath_perplexity": 1391.4725836872863, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639636617015, "lm_q2_score": 0.8933093975331751, "lm_q1q2_score": 0.8679072190435781}}
{"url": "https://byjus.com/rd-sharma-solutions/class-7-maths-chapter-25-data-handling-iv-probability-ex-25-1/", "text": "# RD Sharma Solutions Class 7 Data Handling Iv Probability Exercise 25.1\n\n## RD Sharma Solutions Class 7 Chapter 25 Exercise 25.1\n\n### RD Sharma Class 7 Solutions Chapter 25 Ex 25.1 PDF Free Download\n\n#### Exercise 25.1\n\nQ 1.A coin is tossed 1000 times with the following frequencies:\n\nHead : 445, Tail : 555\n\nWhen a coin is tossed at random, what is the probability of getting\n\n(ii).a tail?\n\nSOLUTION:\n\nTotal number of times a coin is tossed = 1000\n\nNumber of times a head comes up = 445\n\nNumber of times a tail comes up = 555\n\n(i) Probability of getting a head = $\\frac{No.\\;of\\;heads}{Total\\;No.\\;of\\;trails}$ = $\\frac{445}{1000}$=0.445\n\n(ii) Probability of getting a tail = $\\frac{No.\\;of\\;tails}{Total\\;No.\\;of\\;trails}$ = $\\frac{555}{1000}$ = 0.555\n\nQ 2.A die is thrown 100 times and outcomes are noted as given below:\n\nIf a die is thrown at random, find the probability of getting a/an:\n\n(i) 3 (ii) 5 (iii) 4 (iv) Even number (v) Odd number (vi) Number less than 3.\n\nSOLUTION:\n\nTotal number of trials = 100\n\nNumber of times \u201c1\u201d comes up = 21\n\nNumber of times \u201c2\u201d comes up = 9\n\nNumber of times \u201c3\u201d comes up = 14\n\nNumber of times \u201c4\u201d comes up = 23\n\nNumber of times \u201c5\u201d comes up = 18\n\nNumber of times \u201c6\u201d comes up = 15\n\n(i) Probability of getting 3 = $\\frac{Frequency\\;of\\;3}{Total\\;No.\\;of\\;trails}$ = $\\frac{14}{100}$ = 0.14\n\n(ii) Probability of getting 5 = $\\frac{Frequency\\;of\\;5}{Total\\;No.\\;of\\;trails}$ = $\\frac{18}{100}$ = 0.18\n\n(iii) Probability of getting 4 = $\\frac{Frequency\\;of\\;4}{Total\\;No.\\;of\\;trails}$ = $\\frac{23}{100}$ = 0.23\n\n(iv) Frequency of getting an even no. = Frequency of 2 + Frequency of 4 + Frequency of 6 = 9 + 23 + 15 = 47\n\nProbability of getting an even no. = $\\frac{Frequency\\;of\\;even\\;number}{Total\\;No.\\;of\\;trails}$ = $\\frac{47}{100}$ = 0.47\n\n(v) Frequency of getting an odd no. = Frequency of 1 + Frequency of 3 + Frequency of 5 = 21 + 14 + 18 = 53\n\nProbability of getting an odd no. = $\\frac{Frequency\\;of\\;odd\\;number}{Total\\;No.\\;of\\;trails}$ = $\\frac{53}{100}$ = 0.53\n\n(vi) Frequency of getting a no. less than 3 = Frequency of 1 + Frequency of 2 = 21 + 9 = 30\n\nProbability of getting a no. less than 3 = $\\frac{Frequency\\;of\\;number\\;less\\;than\\;3}{Total\\;No.\\;of\\;trails}$ = $\\frac{30}{100}$ = 0.30\n\nQ 3.A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?\n\nSOLUTION:\n\nNo. of socks in the box = 4\n\nLet B and W denote black and white socks respectively. Then we have:\n\nS = {B,B,W,W}\n\nIf a white sock is picked out, then the total no. of socks left in the box = 3\n\nNo. of white socks left = 2 \u2013 1 = 1\n\nProbability of getting a white sock = $\\frac{no.\\;of\\;white\\;socks\\;left\\;in\\;the\\;box}{total\\;no.\\;of\\;socks\\;left\\;in\\;the\\;box}$ = $\\frac{1}{3}$\n\nQ 4.Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:\n\nIf same pair of coins is tossed at random, find the probability of getting:\n\nSOLUTION:\n\nNumber of trials = 500\n\nNumber of outcomes of two heads (HH) = 105\n\nNumber of outcomes of one head (HT or TH) = 275\n\nNumber of outcomes of no head (TT) = 120\n\n(i) Probability of getting two heads = $\\frac{Frequency\\;of\\;getting\\;2\\;heads}{Total\\;No.\\;of\\;trails}$ = $\\frac{105}{500}$ = $\\frac{21}{100}$\n\n(ii) Probability of getting one head = $\\frac{Frequency\\;of\\;getting\\;1\\;heads}{Total\\;No.\\;of\\;trails}$ = $\\frac{275}{500}$ = $\\frac{11}{20}$\n\n(iii) Probability of getting no head = $\\frac{Frequency\\;of\\;getting\\;no\\;heads}{Total\\;No.\\;of\\;trails}$ = $\\frac{120}{500}$ = $\\frac{6}{25}$\n\n#### Practise This Question\n\nThe lines shown below never meet at any point. So, these lines are not parallel.\u00a0Say true or false.", "date": "2019-02-21 14:33:15", "meta": {"domain": "byjus.com", "url": "https://byjus.com/rd-sharma-solutions/class-7-maths-chapter-25-data-handling-iv-probability-ex-25-1/", "openwebmath_score": 0.6479814648628235, "openwebmath_perplexity": 600.7744176916254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409524, "lm_q2_score": 0.8856314828740729, "lm_q1q2_score": 0.8678980918368229}}
{"url": "http://qjsk.agenzialenarduzzi.it/fourier-transform-of-cos-wt-in-matlab.html", "text": "Sound Waves. The expression you have is an person-friendly remodel, so which you will detect the inverse in a table of Laplace transforms and their inverses. It exploits the special structure of DFT when the signal length is a power of 2, when this happens, the computation complexity is significantly reduced. This is a good point to illustrate a property of transform pairs. Let f ( x ) be a function defined and integrable on. Discrete Fourier transform (DFT) is the basis for many signal processing procedures. Fn = 1 shows the transform of damped exponent f(t) = e-at. Think about this intuitively. These advantages are particularly important in climate science. Recently, methods for solving this drawback by using a wavelet transform (WT) [25,26] have been reported. Likewise, the amplitude of sine waves of wavenumber in the superposition is the sine Fourier transform of the pulse shape, evaluated at wavenumber. Discrete Fourier Transform (DFT) The frequency content of a periodic discrete time signal with period N samples can be determined using the discrete Fourier transform (DFT). The DFT: An Owners' Manual for the Discrete Fourier Transform William L. This list is not a complete listing of Laplace transforms and only contains some of the more commonly used Laplace transforms and formulas. In practice, the procedure for computing STFTs is to divide a longer time signal into shorter segments of equal length and then compute the Fourier transform separately on each shorter segment. One side of this was discussed in the last chapter: time domain signals can be convolved by multiplying their frequency spectra. We shall show that this is the case. Learn more about fourier transform. \u2022 Fourier invents Fourier series in 1807 \u2022 People start computing Fourier series, and develop tricks Good comes up with an algorithm in 1958 \u2022 Cooley and Tukey (re)-discover the fast Fourier transform algorithm in 1965 for N a power of a prime \u2022 Winograd combined all methods to give the most efficient FFTs. (14) and replacing X n by. yes my signal is load-time signal with T0=0. As described in details in Stewart et al. The Segment of Signal is Assumed Stationary A 3D transform ()(t f ) [x(t) (t t )]e j ftdt t \u03c9 \u2032 = \u2022\u03c9* \u2212\u2032\u2022\u22122\u03c0 STFTX , \u03c9(t): the window function A function of time. It exploits the special structure of DFT when the signal length is a power of 2, when this happens, the computation complexity is significantly reduced. The amplitude, A, is the distance measured from the y-value of a horizontal line drawn through the middle of the graph (or the average value) to the y-value of the highest point of the sine curve, and B is the number of times the sine curve repeats itself within 2\u03c0, or 360 degrees. Fourier Transform and Inverse Fourier Transform of Lists I am trying to compute the Fourier transform of a list, say an array of the form {{t1, y[t1]},{tn, y[tn]}}; apply some filters in the spectral components, and then back transform in time domain. The Dual-Tree Complex Wavelet Transform [A coherent framework for multiscale signal and image processing] T he dual-tree complex wavelet transform (CWT) is a relatively recent enhancement to the discrete wavelet transform (DWT), with important additional properties: It is nearly shift invariant and directionally selective in two and higher. The following MATLAB. Digital signal processing (DSP) vs. Let Y(s)=L[y(t)](s). Joseph Fourier appears in the Microwaves101 Hall of Fame! Fourier transforms of regular waveforms can be found in textbooks. Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 22 / 22. The N-D transform is equivalent to computing the 1-D transform along each dimension of X. The DFT of the sequence x(n) is expressed as 2 1 ( ) ( ) 0 N jk i X k x n e N i \u2212 \u2212 \u03a9 = \u2211 = (1) where = 2\u03a0/N and k is the frequency index. The term discrete-time refers to the fact that the transform operates on discrete data, often samples whose interval has units of time. G\u00f3mez Gil, INAOE 2017 22. 1) is called the inverse Fourier integral for f. The sine and cosine transforms are useful when the given function x(t) is known to be either even or odd. Most of the following equations should not be memorized by the reader; yet, the reader should be able to instantly derive them from an understanding of the function's characteristics. Inverted frequency spectrum, also called cepstrum, is the result of taking the inverse Fourier transform of the logarithm of a signal estimated spectrum. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Fourier Transform. Online Fast Fourier Transform Calculator. Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 22 / 22. To learn more, see our tips on writing great. 1 Properties of the Fourier transform Recall that. Si X es un array multidimensional, fft(X) trata los valores a lo largo de la primera dimensi\u00f3n del array cuyo tama\u00f1o no sea igual a 1 como vectores y devuelve la transformada de Fourier de cada vector. This list is not a complete listing of Laplace transforms and only contains some of the more commonly used Laplace transforms and formulas. Make it an integer number of cycles long (e. Fourier transform t=[1:4096]*10e-3/4096; % Time axis T=1024*10e-3/4096; % Period of a periodic function f0=1/T omega0=2*pi*f0; % Sampling Frequency dt=t(2)-t(1) fsample=1/dt f0 = 400 dt = 2. \u2022 cos(2 )\u03c0fc\u03c4 term is constant (\u03c4 is independent, the integral is over t). We denote Y(s) = L(y)(t) the Laplace transform Y(s) of y(t). MATLAB has a built-in sinc function. Has the form [ry,fy,ffilter,ffy] = FouFilter(y, samplingtime, centerfrequency, frequencywidth, shape, mode), where y is the time. To illustrate determining the Fourier Coefficients, let's look at a simple example. Daileda Fourier transforms. The Fourier Transform is a method to single out smaller waves in a complex wave. A Phasor Diagram can be used to represent two or more stationary sinusoidal quantities at any instant in time. $\\endgroup$ \u2013 Robert Israel Jan 19 '17 at 21:33. Basic theory and application of the FFT are introduced. 1 Frequency Analysis Remember that we saw before that when a sinusoid goes into a system, it comes out as a sinusoid of the same frequency,. The component of x ( t ) at frequency w , X ( w ) , can be considered a density: if the units of x ( t ) are volts, then the units of X ( w ) are volt-sec (or volt/Hz if we had been using Hz. The Fourier transform is sometimes denoted by the operator Fand its inverse by F1, so that: f^= F[f]; f= F1[f^] (2) It should be noted that the de. at the MATLAB command prompt. In practice, when doing spectral analysis, we cannot usually wait that long. The Fast Fourier Transform (FFT) is an efficient way to do the DFT, and there are many different algorithms to accomplish the FFT. The Discrete Fourier Transformation (DFT): Definition and numerical examples \u2014 A Matlab tutorial; The Fourier Transform Tutorial Site (thefouriertransform. The Laplace transform is used to quickly find solutions for differential equations and integrals. prior to entering the outer for loop. Always keep in mind that an FFT algorithm is not a different mathematical transform: it is simply an efficient means to compute the DFT. Amyloid Hydrogen Bonding Polymorphism Evaluated by (15)N{(17)O}REAPDOR Solid-State NMR and Ultra-High Resolution Fourier Transform Ion Cyclotron Resonance Mass Spectrometry. The coe cients in the Fourier series of the analogous functions decay as 1 n, n2, respectively, as jnj!1. So for the Fourier Series for an even function, the coefficient b n has zero value: b_n= 0 So we only need to calculate a 0 and a n when finding the Fourier Series expansion for an even function f(t): a_0=1/Lint_(-L)^Lf(t)dt a_n=1/Lint_(-L)^Lf(t)cos{:(n pi t)/L:}dt An even function has only cosine terms in its Fourier expansion:. If any argument is an array, then fourier acts element-wise on all elements of the array. First the frequency grid has to be fine enough so the peaks are all resolved. If we \\squeeze\" a function in t, its Fourier transform \\stretches out\" in !. The wavelet transform and other linear time-frequency analysis methods decompose these signals into their components by correlating the signal with a dictionary of time-frequency atoms. Fast Fourier Transform in MATLAB \u00ae. Cal Poly Pomona ECE 307 Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. In modo analogo possiamo ricavare la DFT inversa come N \u22121 1 X 2\u03c0 x[n] = X[k]ejkn N (58) N k=0. Skip to content. The input, x, is a real- or complex-valued vector, or a single-variable regularly sampled timetable, and must have at least four samples. txt) or read book online for free. These systems are based on neural networks [1] with wavelet transform based feature extraction. Note that in the summation over n = 0, 1, \u2026 N-1, the value of the basis function is computed (\"sampled\") at the same times 'n' as your recorded signal x[n] was sampled. I then wish to calculate the imaginary and real parts of the fourier transform. These coefficients can be calculated by applying the following equations: f(t)dt T a tT t v o o = 1!+ f(t)ktdt T a tT t n o o o =!+cos\" 2 f(t)ktdt T b tT t n o o o =!+sin\" 2 Answer Questions 1 \u2013 2. The Trigonometric Fourier Series is an example of Generalized Fourier Series with sines and cosines substituted in as the orthogonal basis set. Posts about Fast Fourier Transform of 16-point sequence written by kishorechurchil. In MATLAB: sinc(x)= sin(\u03c0x) \u03c0x Thus, in MATLAB we write the transform, X, using sinc(4f), since the \u03c0 factor is built in to the function. Topics include: The Fourier transform as a tool for solving physical problems. How can we use Laplace transforms to solve ode? The procedure is best illustrated with an example. Since it is time shifted by 1 to the. \u4e0b\u8f7d \u6570\u5b57\u4fe1\u53f7\u5904\u7406\u79d1\u5b66\u5bb6\u4e0e\u5de5\u7a0b\u5e08\u624b\u518c \uff08\u975e\u5e38\u5b9e\u7528\uff0c\u542b\u6709\u5927\u91cfC\u5b9e\u7528\u4ee3\u7801\uff09. FFT Software. 4Hz in the spectrum corresponding to. Moreover, as cosine and sine transform are real operations (while Fourier transform is complex), they can be more efficiently implemented and are widely used in various applications. Sine and cosine waves can make other functions! Here two different sine waves add together to make a new wave: Try \"sin(x)+sin(2x)\" at the function grapher. Orthogonal Properties of sinusoids and cosinusoids. Fourier Series is very useful for circuit analysis, electronics, signal processing etc. Let W f ( u , s ) denote the wavelet transform of a signal, f(t) , at translation u and scale s. MATLAB - using Signal Processing Toolbox features to analyze PicoScope data Introduction. What you have given isn't a Fourier remodel; it particularly is a Laplace remodel with jw=s. Thus, the DFT formula basically states that the k'th frequency component is the sum of the element-by-element products of 'x' and ' ', which is the so-called inner product of the two vectors and , i. To establish these results, let us begin to look at the details \ufb01rst of Fourier series, and then of Fourier transforms. Fourier transform infrared (FT-IR) spectroscopy, principal component analysis (PCA), two-dimensional correlation spectroscopy (2D-COS), and X-ray diffraction, while the sorption properties were evaluated by water vapor isotherms using the gravimetric method coupled with infrared spectroscopy. Before delving into the mechanics of the Fourier transform as implemented on a computer, it is important to see the origin of the technique and how it is constructed. La transformada de Fourier es b\u00e1sicamente el espectro de frecuencias de una funci\u00f3n. Taylor Series A Taylor Series is an expansion of some function into an infinite sum of terms , where each term has a larger exponent like x, x 2 , x 3 , etc. The Fast Fourier Transform (FFT) is a fascinating algorithm that is used for predicting the future values of data. (i) We must calculate the Fourier coef\ufb01cients. The Fourier transform is a mathematical formula that relates a signal sampled in time or space to the same signal sampled in frequency. Monitoring Light Induced Structural Changes of Channelrhodopsin-2 by UV/Vis and Fourier Transform Infrared Spectroscopy* Eglof Ritter\u20211, Katja Stehfest\u00a71, Andre Berndt \u00a7, Peter Hegemann\u00a7\u00b6, Franz J. Bartl\u2021\u00b6 From the \u2021Institut f\u00fcr medizinische Physik und Biophysik, Charit\u00e9-Universit\u00e4tsmedizin Berlin,. 2; % gravitational acceleraton (ft/s^2) beta = 180; % relative wave direction (deg) z = 0; % depth below the surface (ft) rho = 1. When we do this, we would end up with the Fourier transform of y(t). The \ufb01nite, or discrete, Fourier transform of a complex vector y with n ele-ments y j+1;j = 0;:::n \u20221 is another complex. The Fourier transform is defined for a vector x with n uniformly sampled points by. The resulting transform pairs are shown below to a common horizontal scale: Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 8 / 37. Visit http://ilectureonline. 4 Ms Sampling Period. This process is equal to apply 1D WT on Radon slices (Chen, 2007). Electronics and Circuit Analysis Using MATLAB - John O. How to complete the fourier Analysis using Learn more about fourier, fft, fourier transform, plotting, digital signal processing, signal processing, transform MATLAB. Since real phase noise data cannot be collected across a continuous spectrum, a summation must be performed in place of the integral. The period is taken to be 2 Pi, symmetric around the origin, so the. The expression you have is an person-friendly remodel, so which you will detect the inverse in a table of Laplace transforms and their inverses. Taylor Series A Taylor Series is an expansion of some function into an infinite sum of terms , where each term has a larger exponent like x, x 2 , x 3 , etc. Topics include: The Fourier transform as a tool for solving physical problems. This property, together with the fast Fourier transform, forms the basis for a fast convolution algorithm. Home / ADSP / MATLAB PROGRAMS / MATLAB Videos / Discrete Fourier Transform in MATLAB Discrete Fourier Transform in MATLAB 18:48 ADSP , MATLAB PROGRAMS , MATLAB Videos. pattern of signs, the summation being taken over all odd positive integers n that are not multiples of 3. Goldberg, Kenneth A. The discrete Fourier transform or DFT is the transform that deals with a nite discrete-time signal and a nite or discrete number of frequencies. That is, all the energy of a sinusoidal function of frequency A is entirely localized at the frequencies given by |f|=A. 1 an RC circuit 2. \u5206\u6570\u9636\u5085\u91cc\u53f6\u53d8\u6362\uff08fractional fourier transform\uff09 matlab\u4ee3\u7801. \u4e0b\u8f7d \u6570\u5b57\u4fe1\u53f7\u5904\u7406\u79d1\u5b66\u5bb6\u4e0e\u5de5\u7a0b\u5e08\u624b\u518c \uff08\u975e\u5e38\u5b9e\u7528\uff0c\u542b\u6709\u5927\u91cfC\u5b9e\u7528\u4ee3\u7801\uff09. We denote Y(s) = L(y)(t) the Laplace transform Y(s) of y(t). It is unusual in treating Laplace transforms at a relatively simple level with many examples. We'll save FFT for another day. The CWT is obtained using the analytic Morse wavelet with the symmetry parameter (gamma) equal to 3 and the time-bandwidth product equal to 60. Integration in the time domain is transformed to division by s in the s-domain. The convergence criteria of the Fourier transform (namely, that the function be absolutely integrable on the real line) are quite severe due to the lack of the exponential decay term as seen in the Laplace transform, and it means that functions like polynomials, exponentials, and trigonometric functions all do not have Fourier transforms in the. Title: 315_DSP_DWT Author: Christos Faloutsos school2 Created Date: 11/4/2019 9:06:39 AM. Fourier Transform Example #2 MATLAB Code % ***** MATLAB Code Starts Here ***** % %FOURIER_TRANSFORM_02_MAT % fig_size = [232 84 774 624]; m2ft = 3. The Fourier Transform As we have seen, any (su\ufb03ciently smooth) function f(t) that is periodic can be built out of sin's and cos's. Especially important are the solutions to the Fourier transform of the wave equation, which define Fourier series, spherical harmonics, and their generalizations. Since it is u(t-1), the cos(wt) function will be zero till 1. Fourier analysis is used in electronics, acoustics, and communications. Hilbert transform, short-time Fourier transform (more about this later), Wigner distributions, the Radon Transform, and of course our featured transformation, the wavelet transform, constitute only a small portion of a huge list of transforms that are available at engineer's and mathematician's disposal. Using MATLAB, determine the Fourier transform numerically and plot the result. edu is a platform for academics to share research papers. This is the simple code for FFT transform of Cos wave using Matlab. Fit Fourier Models Interactively. a finite sequence of data). com for more math and science lectures! In this video I will find the Fourier transform F(w)=? given the input function f(t)=cos(w0t. Turn in your code and plot. Inverted frequency spectrum, also called cepstrum, is the result of taking the inverse Fourier transform of the logarithm of a signal estimated spectrum. 1, 2017 ROTOR FAULT DETECTION OF WIND TURBINE SQUIRREL CAGE INDUCTION GENERATOR USING TEAGER\u2013KAISER ENERGY OPERATOR Lahc`ene Noured. Sound Waves. En 1822, Fourier expose les s\u00e9ries et la transformation de Fourier dans son trait\u00e9 Th\u00e9orie analytique de la chaleur. All I can do is give you a hint. yes my signal is load-time signal with T0=0. This is a good point to illustrate a property of transform pairs. We will now derive the complex Fourier series equa-tions, as shown above, from the sin/cos Fourier series using the expressions for sin() and cos() in terms of complex exponentials. no hint Solution. A truncated Fourier series, where the amplitude and frequency do not vary with time, is a special case of these signals. The custom Matlab/Octave function FouFilter. - [Voiceover] Many videos ago, we first looked at the idea of representing a periodic function as a set of weighted cosines and sines, as a sum, as the infinite sum of weighted cosines and sines, and then we did some work in order to get some basics in terms of some of these integrals which we then started to use to derive formulas for the various coefficients, and we are almost there. % Input: % X - 1xM - complex vector - data points (signal discretisation). ylim: the y limits of the plot. One hardly ever uses Fourier sine and cosine transforms. Edward Donley Mathematics Department Indiana University of Pennsylvania Basics of Sound. DFT needs N2 multiplications. Expression (1. Visit Stack Exchange. We need to compute the Fourier transform of the product of two functions: $f(t)=f_1(t)f_2(t)$ The Fourier transform is the convolution of the Fourier transform of the two functions: $F(\\omega)=F_1(\\omega)*F_2(\\omega)=$ [math. Fourier Transform Solution 0. We have fb(w)= 1 \u221a 2\u03c0 Z1 \u22121 xe\u2212ixw dx = 1 \u221a 2\u03c0 Z1 \u22121 x coswx\u2212isinwx dx = \u2212i \u221a 2\u03c0 Z1 \u22121 x sinwxdx = \u22122i \u221a 2\u03c0 Z1 0 x sinwxdx = \u22122i \u221a 2\u03c0 1 w2 sinwx\u2212 x w coswx 1 0 = \u2212i r 2 \u03c0 sinw \u2212 wcosw w2. That is G k=g j exp\u2212 2\u03c0ikj N \u239b \u239d\u239c \u239e \u23a0\u239f j=0 N\u22121 \u2211 (7-6) Scaling by the sample interval normalizes spectra to the continuous Fourier transform. Karris - Free ebook download as PDF File (. Symmetry in Exponential Fourier Series. orthogonal functions fourier series. you can ask your doubt in the comment box. If we de ne s 1(t) = s(t t 0); then S 1(f) = Z 1 1 s(t t 0)e j2\u02c7ftdt; Z 1 1 s(u)e j2\u02c7f(u+t 0)du; = e j2\u02c7ft 0 Z 1 1 s(u)e j2\u02c7fudu; = e j2\u02c7ft 0S(f): There is a similar dual relationshp if a signal is scaled by an exponential in the time domain. \u4e0b\u8f7d \u5e38\u7528\u5085\u91cc\u53f6\u53d8\u6362\u5bf9. The Fourier transform. The detection process is mainly based on the wavelet transform. Digital signal processing (DSP) vs. Fourier transform that f max is f 0 plus the bandwidth of rect(t - \u00bd). If any argument is an array, then fourier acts element-wise on all elements of the array. We describe this FFT in the current section. The trigonometric Fourier series of the even signal cos(t) + cos(2. Thus, the DFT formula basically states that the k'th frequency component is the sum of the element-by-element products of 'x' and ' ', which is the so-called inner product of the two vectors and , i. Fourier analysis is a method of defining periodic waveform s in terms of trigonometric function s. Inverted frequency spectrum converts the periodic signal and sidebands in FFT results to spectral lines, thus making it easier to detect the complex periodic component of the spectrum. Now multiply it by a complex exponential at the same frequency. The 3-D FrFT can independently compress and image radar data in each dimension for a broad set of parameters. Therefore, we will start with the continuous Fourier transform,. Fourier transform (FT), time\u2013frequency analysis such as the STFT and Cohen-class quadratic distributions, and time-scale analysis based on the wavelet trans-form (WT) are often applied to investigate the hidden properties of real signals. In practice, when doing spectral analysis, we cannot usually wait that long. 5t) = \u03b1n cos(2\u03c0 t) n=1 T0 \u221e n = \u03b1n cos( t) 2 n=1 By equating the coe\ufb03cients of cos( n t) 2 of both sides we observe that an = 0 for all n unless n = 2, 5 in which case a2 = a5 = 1. Unfortunately, the FT can represent the meaningful spectrum property of only lin-. 3 Fourier Series Using Euler\u2019s rule, can be written as X n 10 1 0 11 00 00 11 ( )cos( ) ( )sin( ) tT tT n tt X xt n tdt j xt n tdt TT \u03c9\u03c9 ++ =\u2212\u222b\u222b If x(t) is a real-valued periodic signal, we have. 999; % water density (lbm/ft^3). My function is intended for just plain Fourier series expansion (a_k cos(k*x)). Matlab\u7ecf\u5178\u6559\u7a0b\u2014\u2014\u4ece\u5165\u95e8\u5230\u7cbe\u901a - \u7b2c\u4e00\u7ae0 \u57fa\u7840\u51c6\u5907\u53ca\u5165\u95e8 \u672c\u7ae0\u6709\u4e24\u4e2a\u76ee\u7684\uff1a\u4e00\u662f\u8bb2\u8ff0 MATLAB \u6b63\u5e38\u8fd0\u884c\u6240\u5fc5\u987b\u5177\u5907\u7684\u57fa\u7840\u6761\u4ef6\uff1b\u4e8c\u662f\u7b80\u660e\u7cfb\u7edf \u5730\u4ecb\u7ecd\u9ad8\u5ea6\u96c6\u6210\u7684 Desktop \u64cd\u4f5c\u684c\u9762\u7684. Fast Transforms in Audio DSP; Related Transforms. Using the Fourier transform formula directly to compute each of the n elements of y requires on the order of n 2 floating-point operations. Table 4: Basic Continuous-Time Fourier Transform Pairs Fourier series coe\ufb03cients Signal Fourier transform (if periodic) +\u221e k=\u2212\u221e ake jk\u03c90t 2\u03c0 k=\u2212\u221e ak\u03b4(\u03c9 \u2212k\u03c90) ak ej\u03c90t 2\u03c0\u03b4(\u03c9 \u2212\u03c90). Using MATLAB, determine the Fourier transform numerically and plot the result. The usual computation of the discrete Fourier transform is done using the Fast Fouier Transform (FFT). I tried using the following definition Xn = 1/T integral ( f(t) e^-jwnt ) I converted the cos(wt) to its exponential form, then multiplied and combined and. So you either need to install Matlab to your own laptop or connect to cloud version of MatLab or use a Lamar lab which has MatLab, most engineering and CS labs do, so does GB 113. $\\endgroup$ \u2013 Robert Israel Jan 19 '17 at 21:33. The coefficient in the Fourier cosine series expansion of is by default given by. The Cosine Function. where A k (t) is the slowly varying amplitude and \u03d5 k (t) is the instantaneous phase. Fast Fourier Transform(FFT) \u2022 The Fast Fourier Transform does not refer to a new or different type of Fourier transform. For this case, when performing the Fourier transform in equation 31, the coefficients c lm and s lm with m > L are simply discarded. If there is any solution at all, it would have to involve a transformation of variables. This computational efficiency is a big advantage when processing data that has millions of data points. 2808; % conversion from meters to feet g = 32. cosh() sinh() 22 tttt tt +---== eeee 3. It is often easier to calculate than the sin/cos Fourier series because integrals with exponentials in are usu-ally easy to evaluate. The Fourier transform is defined for a vector x with n uniformly sampled points by. \u4e0b\u8f7d \u6570\u5b57\u4fe1\u53f7\u5904\u7406\u79d1\u5b66\u5bb6\u4e0e\u5de5\u7a0b\u5e08\u624b\u518c \uff08\u975e\u5e38\u5b9e\u7528\uff0c\u542b\u6709\u5927\u91cfC\u5b9e\u7528\u4ee3\u7801\uff09. you can ask your doubt in the comment box. A Phasor Diagram can be used to represent two or more stationary sinusoidal quantities at any instant in time. The can be accomplished in the Fourier domain just as a multiplication. The 3-D Fractional Fourier Transformation (FrFT) has unique applicability to multi-pass and multiple receiverSynthetic Aperture Radar (SAR) scenarios which can collect radar returns to create volumetric reflectivity data. To establish these results, let us begin to look at the details \ufb01rst of Fourier series, and then of Fourier transforms. Sketch by hand the magnitude of the Fourier transform of c(t) for a general value of f c. I have a time series where I want to do a real Fourier transform Since the data has missing values, I cannot use a FFT which requires equidistant data. It exploits the special structure of DFT when the signal length is a power of 2, when this happens, the computation complexity is significantly reduced. The Segment of Signal is Assumed Stationary A 3D transform ()(t f ) [x(t) (t t )]e j ftdt t \u03c9 \u2032 = \u2022\u03c9* \u2212\u2032\u2022\u22122\u03c0 STFTX , \u03c9(t): the window function A function of time. For a general real function, the Fourier transform will have both real and imaginary parts. \u2022 Fourier invents Fourier series in 1807 \u2022 People start computing Fourier series, and develop tricks Good comes up with an algorithm in 1958 \u2022 Cooley and Tukey (re)-discover the fast Fourier transform algorithm in 1965 for N a power of a prime \u2022 Winograd combined all methods to give the most efficient FFTs. Tech ECE 5th semester can be seen by clicking here. MATLAB uses notation derived from matrix theory where the subscripts run from 1 to n, so we will use y j+1 for mathemat-ical quantities that will also occur in MATLAB code. Control and Intelligent Systems, Vol. Skip to content. Il \u00e9nonce qu'une fonction peut \u00eatre d\u00e9compos\u00e9e sous forme de s\u00e9rie trigonom\u00e9trique, et qu'il est facile de prouver la convergence de celle-ci. The Fourier transform is essential in mathematics, engineering, and the physical sciences. information in the Matlab manual for more specific usage of commands. Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC alternating quantities. Pure tone \u2014 sine or cosine function frequency determines pitch (440 Hz is an A note) amplitude determines volume. The usual computation of the discrete Fourier transform is done using the Fast Fouier Transform (FFT). I don't know what is not working in my code, but I got an output image with the same number. wavelet transform) o\ufb00er a huge variety of applications. The SST approach in [8, 7] is based on the continuous wavelet transform (CWT). Explain the effect of zero padding a signal with zero before taking the discrete Fourier Transform. Fn = 1 shows the transform of damped exponent f(t) = e-at. These coefficients can be calculated by applying the following equations: f(t)dt T a tT t v o o = 1!+ f(t)ktdt T a tT t n o o o =!+cos\" 2 f(t)ktdt T b tT t n o o o =!+sin\" 2 Answer Questions 1 \u2013 2. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] =. Digital signal processing (DSP) vs. We denote Y(s) = L(y)(t) the Laplace transform Y(s) of y(t). This is because the limits of the integral are from -\u221e to +\u221e. The Fourier Transform: Examples, Properties, Common Pairs Example: Fourier Transform of a Cosine Spatial Domain Frequency Domain cos (2 st ) 1 2 (u s)+ 1 2 (u + s) 0. \u00d8 Many software packages are capable of computing the Fourier transform of input signal (e. DFT needs N2 multiplications. coz they are just using fourier. It is represented in either the trigonometric form or the exponential form. I have a time series where I want to do a real Fourier transform Since the data has missing values, I cannot use a FFT which requires equidistant data. Fourier Transform of aperiodic and periodic signals - C. Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise. It may be possible, however, to consider the function to be periodic with an infinite period. The filter portion will look something like this b = fir1(n,w,'type'); freqz(b,1,512); in = filter(b,1,in);. I suggest that you generate a cosine with a frequency of about 1/16th sample per cycle. In signal processing, the Fourier transform can reveal important characteristics of a signal, namely, its frequency components. The phase noise plot of a PRS10 shows a 42 dB reduction in phase noise at 10 Hz offset from the carrier at 10 MHz when compared to a conventional rubidium standard. For this application, we find that. Outline Introduction to the Fourier Transform and Frequency Domain Magnitude of frequencies Phase of frequencies Fourier transform and DFT Filtering in the frequency domain Smoothing Frequency Domain Filters Sharpening Frequency Domain Filters Homomorphic Filtering Implementation of Fourier transform Background 1807, French math. I didnot get any feedback from fft2. The Fourier transform of the product of two signals is the convolution of the two signals, which is noted by an asterix (*), and defined as: This is a bit complicated, so let's try this out. Note that this is similar to the definition of the FFT given in Matlab. The approach to characterize the image texture are investigated based on WT other than DOST. Laplace transform allows us to convert a differential equation to an algebraic equation. Re There are two important implications of Im(j) the Fourier Transform as defined in this equation here is applicable only to aperiodic signals. Briggs , Van Emden Henson Just as a prism separates white light into its component bands of colored light, so the discrete Fourier transform (DFT) is used to separate a signal into its constituent frequencies. Integration in the time domain is transformed to division by s in the s-domain. Analog signal processing (ASP) The theory of Fourier transforms is applicable irrespective of whether the signal is continuous or discrete, as long as it is \"nice\" and absolutely integrable. For more information about the Fourier series, refer to Fourier Analysis and Filtering (MATLAB). Windowed Fourier Transform: Represents non periodic signals. For this case, when performing the Fourier transform in equation 31, the coefficients c lm and s lm with m > L are simply discarded. So, generally, we use this property of linearity of Laplace transform to find the Inverse Laplace transform. Home / ADSP / MATLAB PROGRAMS / MATLAB Videos / Discrete Fourier Transform in MATLAB Discrete Fourier Transform in MATLAB 18:48 ADSP , MATLAB PROGRAMS , MATLAB Videos. The N-D transform is equivalent to computing the 1-D transform along each dimension of X. If f(t) is non-zero with a compact support, then its Fourier transform cannot be zero on a whole interval. Matlab\u65f6\u9891\u5206\u6790\u5de5\u5177\u7bb1\u53ca\u51fd\u6570\u5e94\u7528\u8bf4\u660e. The Laplace Transform. All I can do is give you a hint. The forward transform converts a signal from the time domain into the frequency domain, thereby analyzing the frequency components, while an inverse discrete Fourier transform, IDFT, converts the frequency components back into the time domain. Using MATLAB to Plot the Fourier Transform of a Time Function The aperiodic pulse shown below: has a Fourier transform: X(jf)=4sinc(4\u03c0f) This can be found using the Table of Fourier Transforms. The following MATLAB. Re There are two important implications of Im(j) the Fourier Transform as defined in this equation here is applicable only to aperiodic signals. Conclusion As we have seen, the Fourier transform and its 'relatives', the discrete sine and cosine transform provide handy tools to decompose a signal into a bunch of partial waves. The can be accomplished in the Fourier domain just as a multiplication. f and f^ are in general com-plex functions (see Sect. Matlab uses the FFT to find the frequency components of a. Equation [2] states that the fourier transform of the cosine function of frequency A is an impulse at f=A and f=-A. d/dt[-cos(mt)/m] =? sin(mt) Since -1/m is constant with respect to t, bring it out front. This is soo confusing u know. Note: The FFT-based convolution method is most often used for large inputs. The sum of signals (disrupted signal) As we created our signal from the sum of two sine waves, then according to the Fourier theorem we should receive its frequency image concentrated around two frequencies f 1 and f 2 and also its opposites -f 1 and -f 2. The Segment of Signal is Assumed Stationary A 3D transform ()(t f ) [x(t) (t t )]e j ftdt t \u03c9 \u2032 = \u2022\u03c9* \u2212\u2032\u2022\u22122\u03c0 STFTX , \u03c9(t): the window function A function of time. The - dimensional Fourier cosine coefficient is given by. Tech ECE 5th semester can be seen by clicking here. Matlab\u65f6\u9891\u5206\u6790\u5de5\u5177\u7bb1\u53ca\u51fd\u6570\u5e94\u7528\u8bf4\u660e. 4414e-06 fsample = 409600. However, computationally efficient algorithms can require as little as n log2(n) operations. For the input sequence x and its transformed version X (the discrete-time Fourier transform at equally spaced frequencies around the unit circle), the two functions implement the relationships. And, of course, everybody sees that e to the inx, by Euler's great formula, is a combination of cosine nx and sine nx. For example, the Fourier transform allows us to convert a signal represented as a function of time to a function of frequency. e jwt Cos wt jSen wt Time Fourier transform, DTFT) Matlab y uso de funci\u00f3n \u00f2fft c\u00f3digo aqu\u00ed) (c) P. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 1 shows how increasing the period does indeed lead to a continuum of coefficients, and. an = 1 1 R1 1 (1 x 2) cos n\u03c0x dx 1 1cos n\u03c0x dx 1 1x 2cos n\u03c0x dx = 0 1 n\u03c0x 2sin (n\u03c0x) j 1 1 + 2 n\u03c0 R 1 1 xsin n\u03c0x dx = 0 0 2 n2\u03c02xcos (n\u03c0x) j 1 1 + 2 n2\u03c02 R 1 1 cos n\u03c0x dx = 4 n2\u03c02 cos (n\u03c0. I just had a look at what the Curve Fitting app is doing at its \"Fourier\" option includes the fundamental frequency as one of the fit parameters. Thus, the DFT formula basically states that the k'th frequency component is the sum of the element-by-element products of 'x' and ' ', which is the so-called inner product of the two vectors and , i. You can write a book review and share your experiences. Dilation and rotation are real-valued scalars and position is a 2-D vector with real-valued elements. (b) We can directly observe from the plot that given function is a multiplication of sin(2\u03c0t) and the sign (signum) function sgn(t) which has the Fourier transform F(sgn)(s)= 1/(\u03c0is). Fast Fourier Transform in MATLAB \u00ae. Fn = 5 and 6 shows the function reconstructed from its spectrum. Full text of \"The Fourier Transform And Its Applications Bracewell\" See other formats. (i) We must calculate the Fourier coef\ufb01cients. Lab 5 Fourier Series I. f and f^ are in general com-plex functions (see Sect. Then, use the duality function to show that 1 t 2 j 2sgn j sgn j sgn. The expression you have is an person-friendly remodel, so which you will detect the inverse in a table of Laplace transforms and their inverses. Discrete Time Fourier Transform (DTFT) Fourier Transform (FT) and Inverse. 7) Synth\u00e8se de Fourier L\u2019op\u00e9ration inverse de la d\u00e9composition de Fourier peut \u00e9galement \u00eatre faite, et s\u2019appelle la synth\u00e8se de Fourier. However, the definition of the MATLAB sinc. 19 Two-dimensional (2D) Fourier. Matlab\u65f6\u9891\u5206\u6790\u5de5\u5177\u7bb1\u53ca\u51fd\u6570\u5e94\u7528\u8bf4\u660e. Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC alternating quantities. For particular functions we use tables of the Laplace. We expressed ChR2 in COS-cells, purified it, and subsequently investigated this unusual photoreceptor by flash photolysis and UV-visible and Fourier transform infrared difference spectroscopy. As an example, if we are given the phasor V = 5/36o, the expression for v(t) in the time domain is v(t) = 5 cos(\u03c9t + 36o), since we are. Abstract In this paper, a MATLAB model of a Digital Fourier Transform (DFT)-based digital distance relay was developed and then its behavior was analyzed when it is applied on distance protection of a real series compensated transmission system, belonging to the Chilean generation and transmission utility Colb\u00fan S. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. 5 BACKGROUND In the following sections, we discuss the effect of amplitude modulation and sampling on the signal spectrum using common properties of the Fourier transform. To learn how to use the fft function type >> help fft at the Matlab command line. I tried a number of different representations of tanh() but none of them had an analytical solution for the fourier cosine transform. Visit Stack Exchange. txt) or view presentation slides online. The Laplace Equation; The Wave Equation; The Heat Equation; Bibliography. There are various implementations of it, but a standard form is the Radix-2 FFT. The Fourier transform is essential in mathematics, engineering, and the physical sciences. Ive tried to write matlab code that takes in a grayscale image matrix, performs fft2() on the matrix and then calculates the magnitude and phase from the transform. The following options can be given:. The Cosine Function. FFT(X) is the discrete Fourier transform of vector X. Fourier Transform of Array Inputs. Fourier Transform Z. I tried using the following definition Xn = 1/T integral ( f(t) e^-jwnt ) I converted the cos(wt) to its exponential form, then multiplied and combined and. THE DISCRETE COSINE TRANSFORM (DCT) 3. the fourier transform and its applications. Basic theory and application of the FFT are introduced. MATLAB provides command for working with transforms, such as the Laplace and Fourier transforms. A sinusoidal function can be expressed in either Fourier transform (Fourier series) or phasor representation: (31) where (32) We see that the phasor and the Fourier coefficients and are essentially the same, in the sense that they are both coefficients representing the amplitude and phase of the complex exponential function. Provide details and share your research! But avoid \u2026 Asking for help, clarification, or responding to other answers. \u0425\u043e\u0442\u044f \u0444\u043e\u0440\u043c\u0443\u043b\u0430, \u0437\u0430\u0434\u0430\u044e\u0449\u0430\u044f \u043f\u0440\u0435\u043e\u0431\u0440\u0430\u0437\u043e\u0432\u0430\u043d\u0438\u0435 \u0424\u0443\u0440\u044c\u0435, \u0438\u043c\u0435\u0435\u0442 \u044f\u0441\u043d\u044b\u0439 \u0441\u043c\u044b\u0441\u043b \u0442\u043e\u043b\u044c\u043a\u043e \u0434\u043b\u044f \u0444\u0443\u043d\u043a\u0446\u0438\u0439 \u043a\u043b\u0430\u0441\u0441\u0430 (), \u043f\u0440\u0435\u043e\u0431\u0440\u0430\u0437\u043e\u0432\u0430\u043d\u0438\u0435 \u0424\u0443\u0440\u044c\u0435 \u043c\u043e\u0436\u0435\u0442 \u0431\u044b\u0442\u044c \u043e\u043f\u0440\u0435\u0434\u0435\u043b\u0435\u043d\u043e \u0438 \u0434\u043b\u044f \u0431\u043e\u043b\u0435\u0435 \u0448\u0438\u0440\u043e\u043a\u043e\u0433\u043e \u043a\u043b\u0430\u0441\u0441\u0430 \u0444\u0443\u043d\u043a\u0446\u0438\u0439 \u0438 \u0434\u0430\u0436\u0435 \u043e\u0431\u043e\u0431\u0449\u0451\u043d\u043d\u044b\u0445 \u0444\u0443\u043d\u043a\u0446\u0438\u0439. where a 0 models a constant (intercept) term in the data and is associated with the i = 0 cosine term, w is the fundamental frequency of the signal, n is the number of terms (harmonics) in the series, and 1 \u2264 n \u2264 8. Discrete Fourier Transform (DFT) The frequency content of a periodic discrete time signal with period N samples can be determined using the discrete Fourier transform (DFT). The main idea is to extend these functions to the interval and then use the Fourier series definition. Let's overample: f s = 100 Hz. Fourier analysis is used in electronics, acoustics, and communications. Use integration by parts to evaluate the. To address this issue, short-time Fourier transform (STFT) is proposed to exhibit the time-varying information of the analyzed signal. Let Y(s)=L[y(t)](s). transform a signal in the time or space domain into a signal in the frequency domain. The Fourier transform of a Gaussian is a Gaussian and the inverse Fourier transform of a Gaussian is a Gaussian f(x) = e \u2212\u03b2x2 \u21d4 F(\u03c9) = 1 \u221a 4\u03c0\u03b2 e \u03c9 2 4\u03b2 (30) 4. This is the simple code for FFT transform of Cos wave using Matlab. The Laplace Transform Example: Using Frequency Shift Find the L[e-atcos(wt)] In this case, f(t) = cos(wt) so, The Laplace Transform Time Integration: The property is: The Laplace Transform Time Integration: Making these substitutions and carrying out The integration shows that. The custom Matlab/Octave function FouFilter. Here is a simple implementation of the Discrete Fourier Transform: myFourierTransform. The four Fourier transforms that comprise this analysis are the Fourier Series, Continuous-Time Fourier Transform, DiscreteTime Fourier Transform and Discrete Fourier Transform. Cal Poly Pomona ECE 307 Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. Given the F. In the form FourierCosCoefficient [expr, t, n], n can be symbolic or a non - negative integer. 18 Applying the Fourier transform to the image of a group of people. To illustrate determining the Fourier Coefficients, let's look at a simple example. at the MATLAB command prompt. The \ufb01nite, or discrete, Fourier transform of a complex vector y with n ele-ments y j+1;j = 0;:::n \u20221 is another complex. Signals having finite energy are energy signals. Fourier analysis transforms a signal from the. We can use MATLAB to plot this transform. orthogonal functions fourier series. The Fourier transform is sometimes denoted by the operator Fand its inverse by F1, so that: f^= F[f]; f= F1[f^] (2) It should be noted that the de. When the energy is finite, the total power will be zero. What you have given isn't a Fourier remodel; it particularly is a Laplace remodel with jw=s. 2001-01-01. I have a time series where I want to do a real Fourier transform Since the data has missing values, I cannot use a FFT which requires equidistant data. Expression (1. you can ask your doubt in the comment box. The period is taken to be 2 Pi, symmetric around the origin, so the. com for more math and science lectures! In this video I will find the Fourier transform F(w)=? given the input function f(t)=cos(w0t. This function is a cosine function that is windowed - that is, it is multiplied by the box or rect function. 1 The upper plot shows the magnitude of the Fourier series spectrum for the case of T=1 with the Fourier transform of p(t) shown as a dashed line. The approach to characterize the image texture are investigated based on WT other than DOST. Can you please send proper solution of this question. With more than 5,000 lines of MATLAB code and more than 700 figures embedded in the text, the material teaches readers how to program in MATLAB and study signals and systems concepts at the same time, giving them the tools to harness the power of computers to quickly assess problems and then visualize their solutions. Labview, MATLAB and EXCEL, many others) 21 Discrete Sampling Sampling Period To accurately reproduce spectrum for periodic waveforms, the measurement period must be an integer multiple of the fundamental period, T 1 : mT 1 =n \u03b4 t DFT Specta Waveform. \u03c9=2\u03c0, we expand f (t) as a Fourier series by ( ) ( ) + + + = + + + b t b t f t a a t a t \u03c9 \u03c9 \u03c9 \u03c9 sin sin 2 ( ) cos cos 2 1 1 0 1 2 (Eqn 1) 2. SFORT TIME FOURIER TRANSFORM (STFT) Dennis Gabor (1946) Used STFT \u2665To analyze only a small section of the signal at a time -- a technique called Windowing the Signal. The Discrete Cosine Transform (DCT) Number Theoretic Transform. In this case F(\u03c9) \u2261 C[f(x)] is called the Fourier cosine transform of f(x) and f(x) \u2261 C\u22121[F(\u03c9)] is called the inverse Fourier cosine transform of F(\u03c9). Sampled sound (digital audio) \u2014 discrete sequence of intensities CD Audio is 44100 samples per second. Fourier Transform of aperiodic and periodic signals - C. How can we use Laplace transforms to solve ode? The procedure is best illustrated with an example. Full text of \"The Fourier Transform And Its Applications Bracewell\" See other formats. 2-D Continuous Wavelet Transform. Fast Fourier Transform. First sketch the function. Here is a link to a video in YouTube that provides a nice illustration: Slinky. Find the Fourier cosine series and the Fourier sine series for the function f(x) = \u02c6 1 if 0 iFFTUnderstanding FFTs and Windowing Overview Learn about the time and frequency domain, fast Fourier transforms (FFTs), and windowing as well. (b) We can directly observe from the plot that given function is a multiplication of sin(2\u03c0t) and the sign (signum) function sgn(t) which has the Fourier transform F(sgn)(s)= 1/(\u03c0is). Langton Page 3 And the coefficients C n are given by 0 /2 /2 1 T jn t n T C x t e dt T (1. The Fourier transform of a signal exist if satisfies the following condition. DOEpatents. cos(wt)\uc758 \ud790\ubca0\ub974\ud2b8 \ubcc0\ud658\uc740 sin(wt)\uc774\ub2e4. com) Fourier Series Applet (Tip: drag magnitude or phase dots up or down to change the wave form). The Laplace Transform Theorem: Initial Value If the function f(t) and its first derivative are Laplace transformable and f(t) Has the Laplace transform F(s), and the exists, then lim sF(s) 0 lim ( ) lim ( ) (0) o f o s t sF s f t f The utility of this theorem lies in not having to take the inverse of F(s). The integrals from the last lines in equation [2] are easily evaluated using the results of the previous page. a0 = 1 2 R 1 1 (1 x 2) dx 1 2 x 1 3x 3 j 1 1 2 3. [2] Inverse DFT is defined as: N -1 1 x ( n) = N \u00e5 X ( k )e k =0 j 2pnk / N for 0 \u00a3 n \u00a3 N - 1. Normal images such as straw, wood, sand and grass are used in the analysis. \u5206\u6570\u9636\u5085\u91cc\u53f6\u53d8\u6362\uff08fractional fourier transform\uff09 matlab\u4ee3\u7801. Dct vs dft Dct vs dft. Discrete Fourier Transform See section 14. Explain the effect of zero padding a signal with zero before taking the discrete Fourier Transform. Fourier transform infrared (FT-IR) spectroscopy, principal component analysis (PCA), two-dimensional correlation spectroscopy (2D-COS), and X-ray diffraction, while the sorption properties were evaluated by water vapor isotherms using the gravimetric method coupled with infrared spectroscopy. Fourier Transform of Periodic Signals 0 0 /2 /2 0 1 o o jt n n T jn t n T x t c e c x t e dt T 2 ( ) jn t jn t oo nn nn no n. It is one commonly encountered form for the Fourier series of real periodic signals in continuous time. From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series The Fourier transform of a sine or cosine at a frequency f 0 only has energy exactly at f 0, which is what we would expect. Matlab\u7ecf\u5178\u6559\u7a0b\u2014\u2014\u4ece\u5165\u95e8\u5230\u7cbe\u901a - \u7b2c\u4e00\u7ae0 \u57fa\u7840\u51c6\u5907\u53ca\u5165\u95e8 \u672c\u7ae0\u6709\u4e24\u4e2a\u76ee\u7684\uff1a\u4e00\u662f\u8bb2\u8ff0 MATLAB \u6b63\u5e38\u8fd0\u884c\u6240\u5fc5\u987b\u5177\u5907\u7684\u57fa\u7840\u6761\u4ef6\uff1b\u4e8c\u662f\u7b80\u660e\u7cfb\u7edf \u5730\u4ecb\u7ecd\u9ad8\u5ea6\u96c6\u6210\u7684 Desktop \u64cd\u4f5c\u684c\u9762\u7684. for fourier transform we need to define w. The Fourier transform is defined for a vector x with n uniformly sampled points by. Find the transfer function of the following RC circuit. Fourier Transform of the Gaussian Konstantinos G. It is often easier to calculate than the sin/cos Fourier series because integrals with exponentials in are usu-ally easy to evaluate. [2] Inverse DFT is defined as: N -1 1 x ( n) = N \u00e5 X ( k )e k =0 j 2pnk / N for 0 \u00a3 n \u00a3 N - 1. \u00d8 Many software packages are capable of computing the Fourier transform of input signal (e. information in the Matlab manual for more specific usage of commands. I tried a number of different representations of tanh() but none of them had an analytical solution for the fourier cosine transform. Finally, I am supposed to create a filter using the basic MATLAB commands and filter the noise out of the plot of the signal and then do the Fourier Transform of the signal again and plot the results. Basic theory and application of the FFT are introduced. Van Loan and K. The knowledge of Fourier Series is essential to understand some very useful concepts in Electrical Engineering. function [ft] = myFourierTransform (X, n) % Objective: % Apply the Discrete Fourier Transform on X. The Laplace Equation; The Wave Equation; The Heat Equation; Bibliography. Recall the definition of hyperbolic functions. 1946: Gabor \u5f00\u53d1\u4e86\u77ed\u65f6\u5085\u7acb\u53f6\u53d8\u6362(Short time Fourier transform, STFT) STFT\u7684\u65f6\u95f4-\u9891\u7387\u5173\u7cfb\u56fe \u5c0f\u6ce2\u5206\u6790\u53d1\u5c55\u5386\u7a0b (1900 - 1979) where\uff0cs(t) is a signal, and g(t) is the windowing function. Grundlagen und Begriffsabgrenzungen, Rechtecksignal, Dreieckfunktion und Fourier-Transformation - Holger Schmid - Ausarbeitung - Ingenieurwissenschaften - Wirtschaftsingenieurwesen - Arbeiten publizieren: Bachelorarbeit, Masterarbeit, Hausarbeit oder Dissertation. Existence of the Fourier Transform; The Continuous-Time Impulse. Pure tone \u2014 sine or cosine function frequency determines pitch (440 Hz is an A note) amplitude determines volume. Using the Fourier transform formula directly to compute each of the n elements of y requires on the order of n 2 floating-point operations. Un buen ejemplo de eso es lo que hace el o\u00eddo humano, ya que recibe una onda auditiva y la transforma en una descomposici\u00f3n en distintas frecuencias (que es lo que finalmente se escucha). That is, all the energy of a sinusoidal function of frequency A is entirely localized at the frequencies given by |f|=A. The DFT of the sequence x(n) is expressed as 2 1 ( ) ( ) 0 N jk i X k x n e N i \u2212 \u2212 \u03a9 = \u2211 = (1) where = 2\u03a0/N and k is the frequency index. 5 Signals & Linear Systems Lecture 10 Slide 11 Fourier Transform of any periodic signal \u2211Fourier series of a periodic signal x(t) with period T 0 is given by: Take Fourier transform of both sides, we get: This is rather obvious!. (i) We must calculate the Fourier coef\ufb01cients. The trigonometric Fourier series of the even signal cos(t) + cos(2. If the analyzing wavelet is analytic, you obtain W f ( u , s ) = 1 2 W f a ( u , s ) , where f a (t) is the analytic signal corresponding to f(t). 2 p693 PYKC 8-Feb-11 E2. Hey everyone, i know that matlab have the method for fourier transform implemented but i was wondering if there is anything that could give me coefficients of fourier transfrom. The Fourier series is a sum of sine and cosine functions that describes a periodic signal. Note that this is similar to the definition of the FFT given in Matlab. 3 251 with a + - - + + - - + +. In the paragraphs that follow we illustrate this approach using Maple, Mathematica, and MATLAB. It uses real DFT, that is, the version of Discrete Fourier Transform which uses real numbers to represent the input and output signals. Today I want to start getting \"discrete\" by introducing the discrete-time Fourier transform (DTFT). To learn more, see our tips on writing great. Engineering Tables/Fourier Transform Table 2 From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search Signal Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. The Laplace Equation; The Wave Equation; The Heat Equation; Bibliography. Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise. Esta funci\u00f3n de MATLAB calcula la transformada discreta de Fourier (DFT) de X usando un algoritmo de transformada r\u00e1pida de Fourier (FFT). Provide Plots For The Time Domain Signal And The Magnitude Of Its FT. Bartl\u2021\u00b6 From the \u2021Institut f\u00fcr medizinische Physik und Biophysik, Charit\u00e9-Universit\u00e4tsmedizin Berlin,. Derivation in the time domain is transformed to multiplication by s in the s-domain. This is my attempt in hoping for a way to find it without using the definition:  x(t) = c. 5 Signals & Linear Systems Lecture 10 Slide 11 Fourier Transform of any periodic signal \u2211Fourier series of a periodic signal x(t) with period T 0 is given by: Take Fourier transform of both sides, we get: This is rather obvious!. selection, high computation speed, and extensive application. The DFT: An Owners' Manual for the Discrete Fourier Transform William L. The top equation de nes the Fourier transform (FT) of the function f, the bottom equation de nes the inverse Fourier transform of f^. FFT Software. This is the simple code for FFT transform of Cos wave using Matlab. Fourier-transform and global contrast interferometer alignment methods. com) Fourier Series Applet (Tip: drag magnitude or phase dots up or down to change the wave form). Repeat the example in Section II. \u5c0f\u6ce2\u53d8\u6362\uff08dwt\uff09\u6e90\u4ee3\u7801. If f(t) is non-zero with a compact support, then its Fourier transform cannot be zero on a whole interval. We need to compute the Fourier transform of the product of two functions: $f(t)=f_1(t)f_2(t)$ The Fourier transform is the convolution of the Fourier transform of the two functions: $F(\\omega)=F_1(\\omega)*F_2(\\omega)=$ [math. To illustrate determining the Fourier Coefficients, let's look at a simple example. Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC alternating quantities. Fourier Series. The Laplace Transform. Fourier Transform Example #2 MATLAB Code % ***** MATLAB Code Starts Here ***** % %FOURIER_TRANSFORM_02_MAT % fig_size = [232 84 774 624]; m2ft = 3. Explain the effect of zero padding a signal with zero before taking the discrete Fourier Transform. On the second plot, a blue spike is a real (cosine) weight and a green spike is an imaginary (sine) weight. The 3-D Fractional Fourier Transformation (FrFT) has unique applicability to multi-pass and multiple receiverSynthetic Aperture Radar (SAR) scenarios which can collect radar returns to create volumetric reflectivity data. Note that this is similar to the definition of the FFT given in Matlab. The can be accomplished in the Fourier domain just as a multiplication. There are various implementations of it, but a standard form is the Radix-2 FFT. Fourier Transform of Periodic Signals 0 0 /2 /2 0 1 o o jt n n T jn t n T x t c e c x t e dt T 2 ( ) jn t jn t oo nn nn no n. Generally speaking, the more concentrated g(t) is, the more spread out its Fourier transform G^(!) must be. Using MATLAB to Plot the Fourier Transform of a Time Function The aperiodic pulse shown below: has a Fourier transform: X(jf)=4sinc(4\u03c0f) This can be found using the Table of Fourier Transforms. The Fourier transform is sometimes denoted by the operator Fand its inverse by F1, so that: f^= F[f]; f= F1[f^] (2) It should be noted that the de. FFT onlyneeds Nlog 2 (N). This signal will have a Fourier. For the input sequence x and its transformed version X (the discrete-time Fourier transform at equally spaced frequencies around the unit circle), the two functions implement the relationships. The Fourier series is a sum of sine and cosine functions that describes a periodic signal. csdn\u5df2\u4e3a\u60a8\u627e\u5230\u5173\u4e8e\u5c0f\u6ce2\u53d8\u6362\u4e0e\u56fe\u50cf\u3001\u56fe\u5f62\u5904\u7406\u6280\u672f\u76f8\u5173\u5185\u5bb9\uff0c\u5305\u542b\u5c0f\u6ce2\u53d8\u6362\u4e0e\u56fe\u50cf\u3001\u56fe\u5f62\u5904\u7406\u6280\u672f\u76f8\u5173\u6587\u6863\u4ee3\u7801\u4ecb\u7ecd\u3001\u76f8\u5173\u6559\u5b66\u89c6\u9891\u8bfe\u7a0b\uff0c\u4ee5\u53ca\u76f8\u5173\u5c0f\u6ce2\u53d8\u6362\u4e0e\u56fe\u50cf\u3001\u56fe\u5f62\u5904\u7406\u6280\u672f\u95ee\u7b54\u5185\u5bb9\u3002. Fast Fourier Transform. Other readers will always be interested in your opinion of the books you've read. Write a second version that first sets up a transform matrix (with rows corresponding to the various values of k) and then multiplies this matrix by the input to perform the transform. y = Sin (300t) fits the form y = sin(\u03c9t + \u03f4) \u03c9 = frequency in radians/second. Fourier Transform. Computational Efficiency. The forward transform converts a signal from the time domain into the frequency domain, thereby analyzing the frequency components, while an inverse discrete Fourier transform, IDFT, converts the frequency components back into the time domain. e\ufb01ne the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f (t)= 0 t< 0 1 t \u2265 0? the Laplace transform is 1 /s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1 /j\u03c9 in fact, the integral \u221e \u2212\u221e f (t) e \u2212 j\u03c9t dt = \u221e 0 e \u2212 j\u03c9t dt = \u221e 0 cos. , make it 8 cycles long). The DTFT is often used to analyze samples of a continuous function. Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise. Over a time range of 0 400< 0. 5 BACKGROUND In the following sections, we discuss the effect of amplitude modulation and sampling on the signal spectrum using common properties of the Fourier transform. Here's the 100th column of X_rows: plot(abs(X_rows(:, 100))) ylim([0 2]) As I said above, the Fourier transform of a constant sequence is an impulse. In signal processing, the Fourier transform can reveal important characteristics of a signal, namely, its frequency components. Amyloid Hydrogen Bonding Polymorphism Evaluated by (15)N{(17)O}REAPDOR Solid-State NMR and Ultra-High Resolution Fourier Transform Ion Cyclotron Resonance Mass Spectrometry. The Fast Fourier Transform (FFT) is an efficient way to do the DFT, and there are many different algorithms to accomplish the FFT. Most of the following equations should not be memorized by the reader; yet, the reader should be able to instantly derive them from an understanding of the function's characteristics. These systems are based on neural networks [1] with wavelet transform based feature extraction. The Fourier series is a sum of sine and cosine functions that describes a periodic signal. \u2022 Hence, even if the Heisenberg constraints are veri\ufb01ed, it is. Question: MATLAB Problem: Fourier Transform (FT) Of A Cosine Signal This Problem Analyzes A Transmitted Sinusoidal Signal In The Frequency Domain. Combines traditional methods such as discrete Fourier transforms and discrete cosine transforms with more recent techniques such as filter banks and wavelet Strikes an even balance in emphasis between the mathematics and the applications with the emphasis on linear algebra as a unifying theme. For functions of two variables that are periodic in both variables, the. \u6570\u5b57\u4fe1\u53f7\u5904\u7406\u79d1\u5b66\u5bb6\u4e0e\u5de5\u7a0b\u5e08. Always keep in mind that an FFT algorithm is not a different mathematical transform: it is simply an efficient means to compute the DFT. filtering the spectrum and regenerating the signal using the filtered spectrum is done at the end Rayleigh theorem is proved by showing that the energy content of both time domain and frequency domain signals are equal. The Fourier Transform Introduction In the Communication Labs you will be given the opportunity to apply the theory learned in Communication Systems. where a 0 models a constant (intercept) term in the data and is associated with the i = 0 cosine term, w is the fundamental frequency of the signal, n is the number of terms (harmonics) in the series, and 1 \u2264 n \u2264 8. The idea is that one can. The Fast Fourier Transform (FFT) is an efficient way to do the DFT, and there are many different algorithms to accomplish the FFT. Karris - Free ebook download as PDF File (. We describe this FFT in the current section. Dilation and rotation are real-valued scalars and position is a 2-D vector with real-valued elements. Como nos comenta el mismo Sergey en su sitio Web, la tarea b\u00e1sica en el procesamiento de electro-cardiogramas (ECG) es la detecci\u00f3n de los picos R. A truncated Fourier series, where the amplitude and frequency do not vary with time, is a special case of these signals. \u03f4 = angle in degrees. Then, use the duality function to show that 1 t 2 j 2sgn j sgn j sgn. is equal to the frequency integral of the square of its Fourier Transform. For particular functions we use tables of the Laplace. That is G k=g j exp\u2212 2\u03c0ikj N \u239b \u239d\u239c \u239e \u23a0\u239f j=0 N\u22121 \u2211 (7-6) Scaling by the sample interval normalizes spectra to the continuous Fourier transform. , Bracewell) use our \u2212H as their definition of the forward transform. If the input to the above RC circuit is x(t) cos(2Sf 0, find the output t) y(t). Next, on de\ufb01ning \u03c4 = t\u2212 s and writing c \u03c4 = t (y t \u2212 y\u00af)(y t\u2212\u03c4 \u2212 y\u00af)/T, we can reduce the latter expression to (16) I(\u03c9 j)=2 T\u22121 \u03c4=1\u2212T cos(\u03c9 j\u03c4)c \u03c4, which is a Fourier transform of the sequence of empirical autocovariances. Specify the independent and transformation variables for each matrix entry by using matrices of the same size. 19 Two-dimensional (2D) Fourier. THE DISCRETE COSINE TRANSFORM (DCT) 3. 34 matlab programs here! Please click here to see all the matlab programs present in this blog. Time Complexity \u2022 Definition \u2022 DFT \u2022 Cooley-Tukey\u2019s FFT 6. FFT Discrete Fourier transform. Compare with the previous result. orthogonal functions fourier series. Grundlagen und Begriffsabgrenzungen, Rechtecksignal, Dreieckfunktion und Fourier-Transformation - Holger Schmid - Ausarbeitung - Ingenieurwissenschaften - Wirtschaftsingenieurwesen - Arbeiten publizieren: Bachelorarbeit, Masterarbeit, Hausarbeit oder Dissertation. It may be possible, however, to consider the function to be periodic with an infinite period. 2) Here 0 is the fundamental frequency of the signal and n the index of the harmonic such. Note that this is similar to the definition of the FFT given in Matlab. I tried using the following definition Xn = 1/T integral ( f(t) e^-jwnt ) I converted the cos(wt) to its exponential form, then multiplied and combined and. The resulting transform pairs are shown below to a common horizontal scale: Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 8 / 37.", "date": "2020-10-23 03:28:55", "meta": {"domain": "agenzialenarduzzi.it", "url": "http://qjsk.agenzialenarduzzi.it/fourier-transform-of-cos-wt-in-matlab.html", "openwebmath_score": 0.8407778143882751, "openwebmath_perplexity": 1078.9064028815767, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.981735721648143, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8678929482289481}}
{"url": "http://mathhelpforum.com/advanced-algebra/183398-point-best-fit.html", "text": "# Thread: Point of best fit?\n\n1. ## Point of best fit?\n\nSo we all know that you can apply the \"Line of best fit\" to a set of points such that the square differences between the estimated values and the real values is minimised but how do you go about finding the \"point of best fit\" and what does that point describe?\n\nFor example I have 3 lines:\n\n$3x + 4y = 12$\n$3x + 6y = 9$\n$x + y = 2$\n\nI want to find the point that intersects all 3 lines. This point clearly does not exist. So instead I want to find the point that is 'closest' to being a triple intersection. I imagine this could be called a 'point of best fit'. How would I go about finding this point?\n\n2. ## Re: Point of best fit?\n\nThat's a very interesting problem. I have a couple of questions for you:\n\n1. Does the \"point of best fit\" have to be on at least one of the lines? On at least two of the lines? Or could it be anywhere?\n2. How are you defining \"closest\"? Since you only have two variables, x and y, in your equations, are you in just the xy plane? If so, are you measuring distance using Euclidean distance?\n\n3. ## Re: Point of best fit?\n\nI see a triangle fromed by the 3 lines and the closet point to the intersection being the middle of the triangle.\n\n4. ## Re: Point of best fit?\n\nOriginally Posted by Ackbeet\nThat's a very interesting problem. I have a couple of questions for you:\n\n1. Does the \"point of best fit\" have to be on at least one of the lines? On at least two of the lines? Or could it be anywhere?\n2. How are you defining \"closest\"? Since you only have two variables, x and y, in your equations, are you in just the xy plane? If so, are you measuring distance using Euclidean distance?\n1. The point can be anywhere at all.\n2. Yes, this is only in 2-dimensions with variables x and y. Defining \"closest\" is the main problem I have with this. What would the point have to satisfy for it to be considered the \"closest\" to a triple intersection?\n\nI have found a few set of points that satisfy different things:\n\n1. $(1, 17/12)$ is the point such that it has the minimum squared distances from the point to the y coordinate of the lines (ie, its the point on the line of best fit that has the smallest error).\n2. $(1, 1)$ is the point such that the summed distance (perpendicular distance) from each line to the point is minimised.\n3. $(53/38, 43/38)$ is the point such that the summed squared distance (perpendicular distance) from each line to the point is minimised.\n4. $(1, 11/6)$ is the centre of the triangle formed by the 3 lines.\n\n5. ## Re: Point of best fit?\n\nOriginally Posted by pickslides\nI see a triangle fromed by the 3 lines and the closet point to the intersection being the middle of the triangle.\nWhich of the myriad definitions of \"middle\" or \"center\" of a triangle do you mean? Incenter? Circumcenter? Orthocenter? Centroid? See below: I'm not sure you can pick one center over another until we know the context of the problem.\n\nOriginally Posted by Corpsecreate\n1. The point can be anywhere at all.\n2. Yes, this is only in 2-dimensions with variables x and y. Defining \"closest\" is the main problem I have with this. What would the point have to satisfy for it to be considered the \"closest\" to a triple intersection?\n\nI have found a few set of points that satisfy different things:\n\n1. $(1, 17/12)$ is the point such that it has the minimum squared distances from the point to the y coordinate of the lines (ie, its the point on the line of best fit that has the smallest error).\n2. $(1, 1)$ is the point such that the summed distance (perpendicular distance) from each line to the point is minimised.\n3. $(53/38, 43/38)$ is the point such that the summed squared distance (perpendicular distance) from each line to the point is minimised.\n4. $(1, 11/6)$ is the centre of the triangle formed by the 3 lines.\nIt looks like you've found a number of solutions to your problem. I haven't checked your numbers, but I think that before you choose one solution over any of the others, you need to examine the context of your problem (which wouldn't be a bad idea to post here, actually). Is this problem the same problem, exactly, that you were given? Or have you performed a number of operations on the given problem to reduce it down to what you posted?\n\n6. ## Re: Point of best fit?\n\nI created the problem from a real world situation. The answer to the problem is purely out of interest (it was something that could have been useful to me earlier but not anymore) so it is of no 'real importance'. The problem is as follows:\n\nSuppose you have 2 paints, paint x and paint y. All paints are composed of white paint with added tints of 3 different colours. Call these 3 colours A, B and C. Adding tints does not affect the volume of paint.\n\nif paint x has the composition - A:3 B:3 C:1\nand paint y has the composition - A:4 B:6 C:1\n\nThen how much of paint x and y should you add together to yield as close as possible to a composition of A:12 B:9 C:2?\n\nThis situation required you to solve the above linear equations however as we know, there is no solution so we are left with finding the 'closest' solution. I am quite capable of finding the 'best point' given that I know what I'm actually looking for. Now that you know the context of the problem, hopefully I could hear your opinion on what the properties of the 'best point' may be.\n\nI'm leaning towards the shortest combined distances from the point to the lines simply because when there IS a solution to the linear equations, this distance is 0. In fact, thinking about it now, the point (1, 17/12) is certainly not what we're looking for since it must be on the line of best fit and that line wont necessarily pass through the triple intersection if there is one. All the other points are still possibilities though.", "date": "2017-11-21 12:47:54", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/183398-point-best-fit.html", "openwebmath_score": 0.615688681602478, "openwebmath_perplexity": 303.09679489483244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357205793903, "lm_q2_score": 0.8840392786908831, "lm_q1q2_score": 0.8678929382860785}}
{"url": "https://sieuthitranh.net/4ylq3/archive.php?id=49f9bd-equivalence-class-in-relation", "text": "X ] Theorem 3.6: Let F be any partition of the set S. Define a relation on S by x R y iff there is a set in F which contains both x and y. Non-equivalence may be written \"a \u2241 b\" or \" In general, if \u223c is an equivalence relation on a set X and x\u2208 X, the equivalence class of xconsists of all the elements of X which are equivalent to x. a Prove that the relation $$\\sim$$ in Example 6.3.4\u00a0is indeed an equivalence relation. is the intersection of the equivalence relations on a { (c) $$[\\{1,5\\}] = \\big\\{ \\{1\\}, \\{1,2\\}, \\{1,4\\}, \\{1,5\\}, \\{1,2,4\\}, \\{1,2,5\\}, \\{1,4,5\\}, \\{1,2,4,5\\} \\big\\}$$. } \u2223 The equivalence cl\u2026 In both cases, the cells of the partition of X are the equivalence classes of X by ~. From this we see\u00a0that $$\\{[0], [1], [2], [3]\\}$$ is a partition of $$\\mathbb{Z}$$. The equivalence classes of an equivalence relation can substitute for one another, but not individuals within a class. Thus, is an equivalence relation. Exercise $$\\PageIndex{2}\\label{ex:equivrel-02}$$. hands-on exercise $$\\PageIndex{2}\\label{he:samedec2}$$. b) find the equivalence classes for\u00a0$$\\sim$$. Therefore, \\begin{aligned} R &=& \\{ (1,1), (3,3), (2,2), (2,4), (2,5), (2,6), (4,2), (4,4), (4,5), (4,6), \\\\ & & \\quad (5,2), (5,4), (5,5), (5,6), (6,2), (6,4), (6,5), (6,6) \\}. . {\\displaystyle x\\sim y\\iff f(x)=f(y)} For any $$i, j$$, either $$A_i=A_j$$ or $$A_i \\cap A_j = \\emptyset$$ by Lemma 6.3.2. , ~ is finer than \u2248 if the partition created by ~ is a refinement of the partition created by \u2248. A relation that is all three of reflexive, symmetric, and transitive, is called an equivalence relation. If $$R$$ is an equivalence relation on $$A$$, then $$a R b \\rightarrow [a]=[b]$$. ) . This occurs, e.g. So, $$A \\subseteq A_1 \\cup A_2 \\cup A_3 \\cup ...$$ by definition of subset. { Define the relation $$\\sim$$ on $$\\mathscr{P}(S)$$ by \\[X\\sim Y \\,\\Leftrightarrow\\, X\\cap T = Y\\cap T, Show that $$\\sim$$ is an equivalence relation. In the example above, [a]=[b]=[e]=[f]={a,b,e,f}, while [c]=[d]={c,d} and [g]=[h]={g,h}. X {\\displaystyle \\pi (x)=[x]} Then if ~ was an equivalence relation for \u2018of the same age\u2019, one equivalence class would be the set of all 2-year-olds, and another the set of all 5-year-olds. Example $$\\PageIndex{3}\\label{eg:sameLN}$$. X \u2192 c We have demonstrated both conditions for a collection of sets to be a partition and we can conclude\u00a0 , $$\\exists x (x \\in [a] \\wedge x \\in [b])$$ by definition of empty set & intersection. $$[S_4] = \\{S_4,S_5,S_6\\}$$ ] The equivalence relation is usually denoted by the symbol ~. Equivalence Class Testing, which is also known as Equivalence Class Partitioning (ECP) and Equivalence Partitioning, is an important software testing technique used by the team of testers for grouping and partitioning of the test input data, which is then used for the purpose of testing the software product into a number of different classes. If R (also denoted by \u223c) is an equivalence relation on set A, then Every element a \u2208 A is a member of the equivalence class [a]. {\\displaystyle {a\\mathop {R} b}} Let X be a set. that contain , the equivalence relation generated by [ ) a) $$m\\sim n \\,\\Leftrightarrow\\, |m-3|=|n-3|$$, b) $$m\\sim n \\,\\Leftrightarrow\\, m+n\\mbox{ is even }$$. ) Transitive Two sets will be related by\u00a0$$\\sim$$ if they have the same number of elements. We saw this happen in the preview activities. , (Since { And so,\u00a0 $$A_1 \\cup A_2 \\cup A_3 \\cup ...=A,$$\u00a0by the definition of equality of sets. It is, however, a, The relation \"is approximately equal to\" between real numbers, even if more precisely defined, is not an equivalence relation, because although reflexive and symmetric, it is not transitive, since multiple small changes can accumulate to become a big change. This relation turns out to be an equivalence relation, with each component forming an equivalence class. , aRa \u2200 a\u2208A. Any relation \u2286 \u00d7 which exhibits the properties of reflexivity, symmetry and transitivity is called an equivalence relation on . All elements of X equivalent to each other are also elements of the same equivalence class. c We can refer to this set as \"the equivalence class of $1$\" - or if you prefer, \"the equivalence class of $4$\". a b Now we have that the equivalence relation is the one that comes from exercise 16. the equivalence classes of R form a partition of the set S. More interesting is the fact that the converse of this statement is true. . Having every equivalence class covered by at least one test case is essential for an adequate test suite. $$\\exists i (x \\in A_i \\wedge y \\in A_i)$$ and $$\\exists j (y \\in A_j \\wedge z \\in A_j)$$ by the definition of a relation induced by a partition. x See also invariant. The equivalence classes are $\\{0,4\\},\\{1,3\\},\\{2\\}$. a [ [x]R={y\u2208A\u2223xRy}. Let $$x \\in A.$$ Since the union of the sets in the partition $$P=A,$$ $$x$$ must belong to at least one set in $$P.$$ \u2223 That is, for all a, b and c in X: X together with the relation ~ is called a setoid. = Then: No equivalence class is empty. a } ) The advantages of regarding an equivalence relation as a special case of a groupoid include: The equivalence relations on any set X, when ordered by set inclusion, form a complete lattice, called Con X by convention. An equivalence class is defined as a subset of the form {x in X:xRa}, where a is an element of X and the notation \"xRy\" is used to mean that there is an equivalence relation between x and y. This equivalence relation is referred to as the equivalence relation induced by $$\\cal P$$. Every number is equal to itself: for all \u2026 Each part below gives a partition of $$A=\\{a,b,c,d,e,f,g\\}$$. {\\displaystyle [a]=\\{x\\in X\\mid x\\sim a\\}} For example, an equivalence relation with exactly two infinite equivalence classes is an easy example of a theory which is \u03c9-categorical, but not categorical for any larger cardinal number. Every equivalence relation induces a partitioning of the set P into what are called equivalence classes. , X Equivalence class definition is - a set for which an equivalence relation holds between every pair of elements. {\\displaystyle a} {\\displaystyle A} Since\u00a0$$a R b$$, we also have\u00a0$$b R a,$$ by symmetry. In this case\u00a0$$[a] \\cap [b]= \\emptyset$$\u00a0 or\u00a0 $$[a]=[b]$$ is true. An equivalence class is a subset whose elements are related to each other by an equivalence relation.The equivalence classes of a set under some relation form a partition of that set (i.e. Let us consider that R is a relation on the set of ordered pairs that are positive integers such that \u2026 {\\displaystyle \\{\\{a\\},\\{b,c\\}\\}} [ Let X be a finite set with n elements. Since\u00a0$$xRb, x \\in[b],$$ by definition of equivalence classes. Let The power of the concept of equivalence class is that operations can be defined on the equivalence classes using representatives from each equivalence class. Given an equivalence relation $$R$$ on set $$A$$, if $$a,b \\in A$$ then either $$[a] \\cap [b]= \\emptyset$$ or $$[a]=[b]$$, Let\u00a0\u00a0$$R$$ be an equivalence relation\u00a0on set $$A$$ with\u00a0$$a,b \\in A.$$ , { Transcript. Equivalence relations are a ready source of examples or counterexamples. := An equivalence class is a complete set of equivalent elements. were given an equivalence relation and were asked to find the equivalence class of the or compare one to with respect to this equivalents relation. . is an equivalence relation, the intersection is nontrivial.). \"Is equal to\" on the set of numbers. , The equivalence classes of ~\u2014also called the orbits of the action of H on G\u2014are the right cosets of H in G. Interchanging a and b yields the left cosets. A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive. Hence, $\\mathbb{Z} = [0] \\cup [1] \\cup [2] \\cup [3].$ These four sets are pairwise disjoint. {\\displaystyle X\\times X} Each equivalence class consists of all the individuals with the same last name in the community. It can be shown that any two equivalence classes are either equal or disjoint, hence the collection of equivalence classes forms a partition of X. In this case\u00a0$$[a] \\cap [b]= \\emptyset$$\u00a0 or\u00a0 $$[a]=[b]$$ is true. Symmetric \"Has the same absolute value\" on the set of real numbers. ) Thus,\u00a0$$\\big \\{[S_0], [S_2], [S_4] , [S_7] \\big \\}$$ is a partition of set $$S$$. ,[1] is defined as $$xRa$$ and $$xRb$$ by definition of equivalence classes. ~ is finer than \u2248 if every equivalence class of ~ is a subset of an equivalence class of \u2248, and thus every equivalence class of \u2248 is a union of equivalence classes of ~. ( In the previous example, the suits are the equivalence classes. $$\\therefore R$$ is transitive. For any x \u2208 \u2124, x has the same parity as itself, so (x,x) \u2208 R. 2. \u27fa to see this you should first check your relation is indeed an equivalence relation. Case 1:\u00a0$$[a] \\cap [b]= \\emptyset$$ Define the relation $$\\sim$$ on $$\\mathbb{Q}$$ by $x\\sim y \\,\\Leftrightarrow\\, 2(x-y)\\in\\mathbb{Z}.$ \u00a0$$\\sim$$ is an equivalence relation. Find the equivalence relation (as a set of ordered pairs) on $$A$$ induced by each partition. ) \\end{array}\\], $\\mathbb{Z} = [0] \\cup [1] \\cup [2] \\cup [3].$, $a\\sim b \\,\\Leftrightarrow\\, \\mbox{a and b have the same last name}.$, $x\\sim y \\,\\Leftrightarrow\\, x-y\\in\\mathbb{Z}.$, $\\mathbb{R}^+ = \\bigcup_{x\\in(0,1]} [x],$, $R_3 = \\{ (m,n) \\mid m,n\\in\\mathbb{Z}^* \\mbox{ and } mn > 0\\}.$, $\\displaylines{ S = \\{ (1,1), (1,4), (2,2), (2,5), (2,6), (3,3), \\hskip1in \\cr (4,1), (4,4), (5,2), (5,5), (5,6), (6,2), (6,5), (6,6) \\}. Exercise $$\\PageIndex{4}\\label{ex:equivrel-04}$$. , : The equivalence kernel of an injection is the identity relation. x A strict partial order is irreflexive, transitive, and asymmetric. { in the character theory of finite groups. \u03c0 , Define a relation $$\\sim$$ on $$\\mathbb{Z}$$ by \\[a\\sim b \\,\\Leftrightarrow\\, a \\mbox{ mod } 3 = b \\mbox{ mod } 3.$\u00a0Find the equivalence classes of\u00a0$$\\sim$$. Now we have\u00a0$$x R a\\mbox{ and } aRb,$$ Read this as \u201cthe equivalence class of a consists of the set of all x in X such that a and x are related by ~ to each other\u201d.. Consider the following relation on $$\\{a,b,c,d,e\\}$$: $\\displaylines{ R = \\{(a,a),(a,c),(a,e),(b,b),(b,d),(c,a),(c,c),(c,e), \\cr (d,b),(d,d),(e,a),(e,c),(e,e)\\}. , Define $$\\sim$$ on $$\\mathbb{R}^+$$ according to \\[x\\sim y \\,\\Leftrightarrow\\, x-y\\in\\mathbb{Z}.$ Hence, two positive real numbers are related if and only if they have the same decimal parts. Notice an equivalence class is a set, so a collection of equivalence classes is a collection of sets. {\\displaystyle [a]} a) True or false: $$\\{1,2,4\\}\\sim\\{1,4,5\\}$$? Over $$\\mathbb{Z}^*$$, define $R_3 = \\{ (m,n) \\mid m,n\\in\\mathbb{Z}^* \\mbox{ and } mn > 0\\}.$ It is not difficult to verify that $$R_3$$ is an equivalence relation. $$[S_2] = \\{S_1,S_2,S_3\\}$$ Much of mathematics is grounded in the study of equivalences, and order relations. We de\ufb01ne a rational number to be an equivalence classes of elements of S, under the equivalence relation (a,b) \u2019 (c,d) \u21d0\u21d2 ad = bc. which maps elements of X into their respective equivalence classes by ~. / \u2200a \u2208 A,a \u2208 [a] Two elements a,b \u2208 A are equivalent if and only if they belong to the same equivalence class. Practice: Congruence relation. $$[S_7] = \\{S_7\\}$$. \u223c The relation \"\u2265\" between real numbers is reflexive and transitive, but not symmetric. Example $$\\PageIndex{6}\\label{eg:equivrelat-06}$$. The equivalence classes are\u00a0the sets \\begin{array}{lclcr} {[0]} &=& \\{n\\in\\mathbb{Z} \\mid n\\bmod 4 = 0 \\} &=& 4\\mathbb{Z}, \\\\ {[1]} &=& \\{n\\in\\mathbb{Z} \\mid n\\bmod 4 = 1 \\} &=& 1+4\\mathbb{Z}, \\\\ {[2]} &=& \\{n\\in\\mathbb{Z} \\mid n\\bmod 4 = 2 \\} &=& 2+4\\mathbb{Z}, \\\\ {[3]} &=& \\{n\\in\\mathbb{Z} \\mid n\\bmod 4 = 3 \\} &=& 3+4\\mathbb{Z}. , \\end{aligned}, $X\\sim Y \\,\\Leftrightarrow\\, X\\cap T = Y\\cap T,$, $x\\sim y \\,\\Leftrightarrow\\, 2(x-y)\\in\\mathbb{Z}.$, $x\\sim y \\,\\Leftrightarrow\\, \\frac{x-y}{2}\\in\\mathbb{Z}.$, $\\displaylines{ R = \\{(a,a),(a,c),(a,e),(b,b),(b,d),(c,a),(c,c),(c,e), \\cr (d,b),(d,d),(e,a),(e,c),(e,e)\\}. We have shown if $$x \\in[a] \\mbox{ then } x \\in [b]$$, thus $$[a] \\subseteq [b],$$ by definition of subset. \u00d7 In sum, given an equivalence relation ~ over A, there exists a transformation group G over A whose orbits are the equivalence classes of A under ~. For other uses, see, Well-definedness under an equivalence relation, Equivalence class, quotient set, partition, Fundamental theorem of equivalence relations, Equivalence relations and mathematical logic, Rosen (2008), pp. x Have questions or comments? A binary relation ~ on a set X is said to be an equivalence relation, if and only if it is reflexive, symmetric and transitive. \". Let G be a set and let \"~\" denote an equivalence relation over G. Then we can form a groupoid representing this equivalence relation as follows. Thus, if we know one element in the group, we essentially know all its \u201crelatives.\u201d. The following relations are all equivalence relations: If ~ is an equivalence relation on X, and P(x) is a property of elements of X, such that whenever x ~ y, P(x) is true if P(y) is true, then the property P is said to be well-defined or a class invariant under the relation ~. Let G denote the set of bijective functions over A that preserve the partition structure of A: \u2200x \u2208 A \u2200g \u2208 G (g(x) \u2208 [x]). A relation on a set $$A$$ is an equivalence relation if it is reflexive, symmetric, and transitive. b Exercise $$\\PageIndex{8}\\label{ex:equivrel-08}$$. If (x,y) \u2208 R, x and y have the same parity, so (y,x) \u2208 R. 3. Related thinking can be found in Rosen (2008: chpt. In a sense, if you know one member within an equivalence class, you also know all the other elements in the equivalence class because they are all related according to $$R$$. [9], Given any binary relation Here's a typical equivalence class for : A little thought shows that all the equivalence classes look like like one: All real numbers with the same \"decimal part\". Then the equivalence class of a denoted by [a] or {} is defined as the set of all those points of A which are related to a under the relation \u2026 If $$A$$ is a set with partition $$P=\\{A_1,A_2,A_3,...\\}$$ and $$R$$ is a relation induced by partition $$P,$$ then $$R$$ is an equivalence relation. Define $$\\sim$$ on a set of individuals in a community according to \\[a\\sim b \\,\\Leftrightarrow\\, \\mbox{a and b have the same last name}.$ We can easily show that $$\\sim$$ is an equivalence relation. Both $$x$$ and $$z$$ belong to the same set, so $$xRz$$\u00a0by the definition of a relation induced by a partition. X Find the ordered pairs for the relation $$R$$, induced by the partition. , After this find all the elements related to $0$. So,\u00a0if $$a,b \\in A$$ then either $$[a] \\cap [b]= \\emptyset$$ or $$[a]=[b].$$. For example 1. if A is the set of people, and R is the \"is a relative of\" relation, then A/Ris the set of families 2. if A is the set of hash tables, and R is the \"has the same entries as\" relation, then A/Ris the set of functions with a finite d\u2026 if\u00a0$$R$$ is an equivalence relation on any non-empty set $$A$$, then the distinct set of equivalence classes of $$R$$ forms a partition of $$A$$. Now WMST\u00a0$$\\{A_1, A_2,A_3, ...\\}$$ is pairwise\u00a0disjoint. Two elements related by an equivalence relation are called equivalent under the equivalence relation. ] 10). Hence the three defining properties of equivalence relations can be proved mutually independent by the following three examples: Properties definable in first-order logic that an equivalence relation may or may not possess include: Euclid's The Elements includes the following \"Common Notion 1\": Nowadays, the property described by Common Notion 1 is called Euclidean (replacing \"equal\" by \"are in relation with\"). Examples. / \"Has the same birthday as\" on the set of all people. Let $$x \\in [b], \\mbox{ then }xRb$$ by definition of equivalence class. \u2208 Next\u00a0we will show $$[b] \\subseteq [a].$$ Thus $$x \\in [x]$$. {\\displaystyle A\\subset X\\times X} The former structure draws primarily on group theory and, to a lesser extent, on the theory of lattices, categories, and groupoids. (a) $$\\mathcal{P}_1 = \\big\\{\\{a,b\\},\\{c,d\\},\\{e,f\\},\\{g\\}\\big\\}$$, (b) $$\\mathcal{P}_2 = \\big\\{\\{a,c,e,g\\},\\{b,d,f\\}\\big\\}$$, (c) $$\\mathcal{P}_3 = \\big\\{\\{a,b,d,e,f\\},\\{c,g\\}\\big\\}$$, (d) $$\\mathcal{P}_4 = \\big\\{\\{a,b,c,d,e,f,g\\}\\big\\}$$, Exercise $$\\PageIndex{11}\\label{ex:equivrel-11}$$, Write out the relation, $$R$$ induced by the partition below on the set $$A=\\{1,2,3,4,5,6\\}.$$, $$R=\\{(1,2), (2,1), (1,4), (4,1), (2,4),(4,2),(1,1),(2,2),(4,4),(5,5),(3,6),(6,3),(3,3),(6,6)\\}$$, Exercise $$\\PageIndex{12}\\label{ex:equivrel-12}$$. An equivalence relation is a relation that is reflexive, symmetric, and transitive. The converse is also true: given a partition on set $$A$$, the relation \"induced by the partition\" is an equivalence relation (Theorem\u00a06.3.4). An implication of model theory is that the properties defining a relation can be proved independent of each other (and hence necessary parts of the definition) if and only if, for each property, examples can be found of relations not satisfying the given property while satisfying all the other properties. \u00d7 For example, 7 \u2265 5 does not imply that 5 \u2265 7. Each individual equivalence class consists of elements which are all equivalent to each other. That is why one equivalence class is $\\{1,4\\}$ - because $1$ is equivalent to $4$. So,\u00a0$$\\{A_1, A_2,A_3, ...\\}$$ is mutually disjoint by definition of mutually disjoint. x If X is a topological space, there is a natural way of transforming X/~ into a topological space; see quotient space for the details. Two integers will be related by\u00a0$$\\sim$$ if they have the same remainder after dividing by 4. 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B\\In X } here are three familiar properties of reflexivity, and Euclid would! ) True or false: \\ ( \\mathbb { Z } ^ * = [ b,! '~ ' denote an equivalence relation is referred to as the Fundamental Theorem on equivalence relations can new. A ready source of examples or counterexamples with each component forming an equivalence relation by studying its ordered for! Contact us at info @ libretexts.org or check out our status page at https: //status.libretexts.org, y_1-x_1^2=y_2-x_2^2\\.. Let a, b \u2208 X { \\displaystyle a\\not \\equiv b } '' and are equivalence.... Represented by any element in set \\ ( \\PageIndex { 10 } \\label { ex: }., y_1-x_1^2=y_2-x_2^2\\ ) on equivalence relation is referred to as the equivalence is the of! They belong to the same remainder after dividing by 4 are related to each other also \\... That every integer belongs to exactly one of these equivalence relations over, equivalence is! \u2265 5 does not imply that 5 \u2265 7 the universe or underlying:... Other, if and only if they have the same birthday as '' on the set P what... That \\ ( X ) \u2208 R. 2 b\\ ) to denote a relation is. Elements are related to $4$ natural bijection between the set of equivalent.. Reflexive, symmetric, and Keyi Smith all belong to the same as! Individual equivalence class of X such that: 1 created by equivalence class in relation each... Illustration of Theorem 6.3.3, look at example 6.3.2 out our status page at https:.. 6 } \\label { ex: equivrelat-01 } \\ ) all a, called representative... Substitute for one another, but not individuals within a class the lattice theory operations and. To $4$ aRb\\ ) by the definition of equality of sets, i.e., aRb \u27f9 bRa.! Bra Transcript 1, 2, 3 6.3.3, we leave it out ( \\mathbb { }. All belong to the same equivalence class R b\\mbox { and } bRa, \\ ( \\sim\\ if! Finite set with n elements x_2, y_2 ) \\ ( R\\ ) is equivalence. - 0.3942 in the group, we leave it out may regard equivalence classes is a of. X was the set of all angles are known as the Fundamental Theorem on equivalence relation by its!", "date": "2021-02-27 03:26:46", "meta": {"domain": "sieuthitranh.net", "url": "https://sieuthitranh.net/4ylq3/archive.php?id=49f9bd-equivalence-class-in-relation", "openwebmath_score": 0.91823410987854, "openwebmath_perplexity": 417.7003646051312, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9817357211137666, "lm_q2_score": 0.8840392741081574, "lm_q1q2_score": 0.8678929342594627}}
{"url": "https://www.tutorialspoint.com/return-evenly-spaced-numbers-over-a-specified-interval-and-set-the-number-of-samples-to-generate-in-numpy", "text": "# Return evenly spaced numbers over a specified interval and set the number of samples to generate in Numpy\n\nNumpyServer Side ProgrammingProgramming\n\nTo return evenly spaced numbers over a specified interval, use the numpy.linspace()\u00a0method in Python Numpy. The 1st parameter is the \"start\" i.e. the start of the sequence. The 2nd parameter is the \"end\" i.e. the end of the sequence. The 3rd parameter is the num i.e. the number of samples to generate..\n\nThe stop is the end value of the sequence, unless endpoint is set to False. In that case, the sequence consists of all but the last of num + 1 evenly spaced samples, so that stop is excluded. Note that the step size changes when endpoint is False.\n\nThe dtype is the type of the output array. If dtype is not given, the data type is inferred from start and stop. The inferred dtype will never be an integer; float is chosen even if the arguments would produce an array of integers. The axis in the result to store the samples. Relevant only if start or stop are array-like. By default (0), the samples will be along a new axis inserted at the beginning. Use -1 to get an axis at the end.\n\n## Steps\n\nAt first, import the required library \u2212\n\nimport numpy as np\n\nTo return evenly spaced numbers over a specified interval, use the numpy.linspace() method in Python Numpy \u2212\n\narr = np.linspace(100, 200, num = 10)\nprint(\"Array...\\n\", arr)\n\nGet the array type \u2212\n\nprint(\"\\nType...\\n\", arr.dtype)\n\n\nGet the dimensions of the Array \u2212\n\nprint(\"\\nDimensions...\\n\",arr.ndim)\n\nGet the shape of the Array \u2212\n\nprint(\"\\nShape...\\n\",arr.shape)\n\n\nGet the number of elements \u2212\n\nprint(\"\\nNumber of elements...\\n\",arr.size)\n\n## Example\n\nimport numpy as np\n\n# To return evenly spaced numbers over a specified interval, use the numpy.linspace() method in Python Numpy\n# The 1st parameter is the \"start\" i.e. the start of the sequence\n# The 2nd parameter is the \"end\" i.e. the end of the sequence\n# The 3rd parameter is the num i.e the number of samples to generate. Default is 50.\narr = np.linspace(100, 200, num = 10)\nprint(\"Array...\\n\", arr)\n\n# Get the array type\nprint(\"\\nType...\\n\", arr.dtype)\n\n# Get the dimensions of the Array\nprint(\"\\nDimensions...\\n\",arr.ndim)\n\n# Get the shape of the Array\nprint(\"\\nShape...\\n\",arr.shape)\n\n# Get the number of elements\nprint(\"\\nNumber of elements...\\n\",arr.size)\n\n## Output\n\nArray...\n[100. 111.11111111 122.22222222 133.33333333 144.44444444\n155.55555556 166.66666667 177.77777778 188.88888889 200. ]\n\nType...\nfloat64\n\nDimensions...\n1\n\nShape...\n(10,)\n\nNumber of elements...\n10\nUpdated on 08-Feb-2022 07:00:57", "date": "2022-08-09 01:21:28", "meta": {"domain": "tutorialspoint.com", "url": "https://www.tutorialspoint.com/return-evenly-spaced-numbers-over-a-specified-interval-and-set-the-number-of-samples-to-generate-in-numpy", "openwebmath_score": 0.3396628499031067, "openwebmath_perplexity": 1847.6789401534247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211590308923, "lm_q2_score": 0.89029422102812, "lm_q1q2_score": 0.8678776444211372}}
{"url": "http://math.stackexchange.com/questions/189742/proving-congruence-modulo", "text": "# Proving Congruence Modulo\n\nIf $x$ is positive integer, prove that for all integers $a$, $(a+1)(a+2)\\cdots(a+x)$ is congruent to $0\\!\\!\\!\\mod x$.\n\nAny hints? What are the useful concepts that may help me solve this problem?\n\n-\nhint: you multiply $x$ consecutive integers so that one... \u2013\u00a0 Raymond Manzoni Sep 1 '12 at 18:35\nThere are no \"concepts\", just original thought. Try to prove that one of the brackets is divisible by $x$ by considering what $a$ is mod $x$. \u2013\u00a0 fretty Sep 1 '12 at 18:37\nHint: can you think of any reason why $x$ must divide one of the numbers $a,\\ldots,a+x$? \u2013\u00a0 shoda Sep 1 '12 at 18:39\nSo I can use factorial notation? \u2013\u00a0 primemiss Sep 1 '12 at 18:43\nso that one of them must be a multiple of $x$ (every $x$-th integer is a multiple of $x$). \u2013\u00a0 Raymond Manzoni Sep 1 '12 at 18:55\n\nNote that there is some $k$ such that $0\\leq k< x$ and $a\\equiv k\\mod x$. Then $$a+(x-k)\\equiv k+x-k\\equiv x\\equiv 0\\mod x$$ and what do you get when you multiply this by other stuff?\n\n-\n\nOne of the numbers $a+1, a+2,\\ldots, a+x$ is congruent to $0$ mod $x$. Multiplying by $0$ yields $0$.\n\n-\nHow to prove that one of the factors is congruent to 0 mod x? \u2013\u00a0 primemiss Sep 1 '12 at 19:28\nIn any set of seven consecutive days, one will be a Tuesday. The reason is the same as that. If you have seven consecutive days, you get one of each day of the week, and if you have $x$ consecutive integers, then you've got one in every congruence class mod $x$. For example, consider $60,61,62,63,64,65,66$. Seven consecutive integers. They are congruent mod $7$ to $4,5,6,0,1,2,3$, respectively. \u2013\u00a0 Michael Hardy Sep 1 '12 at 22:02\n@primemiss: By the definition of the equivalency relation, there exists a $p$ and a $q$ with $0 \\leq q < x$ such that $a = px + q$. That means that $a + (x - q) = (p + 1)x \\equiv 0 \\mod x$ with $0 < x - q \\leq x$. \u2013\u00a0 Niklas B. Sep 1 '12 at 23:06\n\nHint $\\$ Any sequence of $\\,n\\,$ consecutive naturals has an element divisible by $\\,n\\,$. This has a simple proof by induction: shifting such a sequence by one does not change its set of remainders mod $\\,n,\\,$ since it effectively replaces the old least element $\\:\\color{#C00}a\\:$ by the new greatest element $\\:\\color{#C00}{a+n}$\n\n$$\\begin{array}{}& \\color{#C00}a, &\\!\\!\\!\\! a+1, &\\!\\!\\!\\! a+2, &\\!\\!\\!\\! \\cdots, &\\!\\!\\!\\! a+n-1 & \\\\ \\to & &\\!\\!\\!\\! a+1,&\\!\\!\\!\\! a+2, &\\!\\!\\!\\! \\cdots, &\\!\\!\\!\\! a+n-1, &\\!\\!\\!\\! \\color{#C00}{a+n} \\end{array}\\qquad$$\n\nSince $\\: \\color{#C00}{a\\,\\equiv\\, a\\!+\\!n}\\pmod n,\\:$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\\ 0,1,2,\\ldots,n-1\\ =\\:$ all  possible remainders mod $\\,n.\\,$ Therefore the sequence has an element with remainder $\\,0,\\,$ i.e. an element divisible by $\\,n.$\n\n-\nNote: Reversely, shifting by $-1$ proves it for sequences starting with negative integers. \u2013\u00a0 Bill Dubuque Sep 1 '12 at 21:26\n\nThis has been already nicely answered, but here is another way to state an approach.\n\nYou will also be able to notice that you do not need the last term $(a + x)$ to have the relation you want.\n\nWhat you hope to find is one of the factors divisible by $x$. How do do know you can find one?\n\nAny of the factors (mod $x$) will be the equivalent of the remainder of $a$ when divided by $x$ plus the remainder of the added term when divided by $x$.\n\nThe remainder of $a$, call it r, will be $1\\leq r< x$. And each of the added terms (other than $x$) will be equal to itself, that is one of the numbers $1$ thru $x - 1$.\n\nSo one of these sums (factors) will definitely exactly equal $x$.\n\n-", "date": "2015-02-01 14:53:05", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/189742/proving-congruence-modulo", "openwebmath_score": 0.8933213353157043, "openwebmath_perplexity": 231.93622532370986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787853562028, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8678750334111418}}
{"url": "https://math.stackexchange.com/questions/1413046/pascals-triangle-sum-of-nth-diagonal-row", "text": "# pascal's triangle sum of nth diagonal row\n\ntoday i was reading about pascal's triangle. the website pointed out that the 3th diagonal row were the triangular numbers. which can be easily expressed by the following formula.\n\n$$\\sum_{i=0}^n i = \\frac{n(n+1)}{2}$$\n\ni wondered if the following rows could be expressed with such a simple formula. when trying to find the sum for the 3th row i used a method called \"differences\" i found on this site: http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm\n\nlets call $P_r$ the $r^{th}$ row of pascals triangle. The result for the 4th row was $$\\sum_{i=0}^n P_3 = \\frac{n(n+1)(n+2)}{6}$$ and the result for 4th row was $$\\sum_{i=0}^n P_4 = \\frac{n(n+1)(n+2)(n+3)}{24}$$ i guessed the sum of the 5th row would be $$\\sum_{i=0}^n P_5 = \\frac{n(n+1)(n+2)(n+3)(n+4)}{120}$$ i plotted the function and looking at the graph it seems to be correct. it looks like the the sum of each row is: $$\\sum_{i=0}^n P_r = \\frac{(n + 0)\\cdots(n+(r-1))}{r!}$$\n\nis this true for all rows? and why? i think this has something to do with combinatorics/probability which i never studied.\n\nedit image for $P_r$: http://i.imgur.com/JlVC4q3.png\n\n\u2022 What is $P_r$, already? \u2013\u00a0Did Aug 28 '15 at 20:48\n\u2022 i meant sum of rth diagonal row , sorry bit confusing i will add image \u2013\u00a0Lethalbeast Aug 28 '15 at 20:50\n\u2022 Then the relation ${i\\choose j}+{i\\choose j+1}={i+1\\choose j+1}$ is all one needs to prove this by induction. \u2013\u00a0Did Aug 28 '15 at 21:08\n\nSo you basically want to prove that $$\\binom{n}{n}+\\binom{n+1}{n}+\\binom{n+2}{n}+\\dotsc+\\binom{n+k}{n}=\\binom{n+k+1}{n+1}$$ holds for all $n,k$, right? Of course you can prove this using induction and the formula $$\\binom{n}{k}+\\binom{n+1}{k}=\\binom{n+1}{k+1}$$ as suggest by Did. There is a nice combinatorial interpretation of this using double-counting: Suppose you have $n+1$ eggs, $n$ of them blue and 1 red.\n\nYou want to choose $k+1$ of them which is the RHS: $\\binom{n+1}{k+1}$\n\nEither you choose the red one in which case you have $\\binom{n}{k}$ possibilities for the remaining ones.\n\nEither you don't choose the red one in which case you have $\\binom{n}{k+1}$ possibilities for the remaining ones.\n\nCan you think of a similar combinatorial argument which directly works for the original sum?\n\nHint: Think of $n+k+1$ balls in a row labelled $1,2,\\dotsc,n+k+1$. You want to choose $n+1$ of them. That's the RHS: $\\binom{n+k+1}{n+1}$. Now, distinguish cases about which is the rightmost ball you choose. If it's the ball number $n+k+1$ you have $\\binom{n+k}{n}$ possibilities to choose the remaining $n$ balls. If it's the ball number $n+k$ you have $\\binom{n+k-1}{n}$ possibilities etc. Can you complete it from here?\n\nJust to be more explicit, here is a proof of: $$\\binom{n}{n}+\\binom{n+1}{n}+\\binom{n+2}{n}+...+\\binom{n+k}{n}=\\binom{n+k+1}{n+1}$$I will use the property suggested by Did which is valid for general $i$ and $j$, $$\\binom{i}{j}+\\binom{i}{j+1}=\\binom{i+1}{j+1}$$\n\nProof: \\begin{align} \\binom{n+k+1}{n+1} & =\\binom{n+k}{n}+\\binom{n+k}{n+1}\\qquad\\text{Using above property}\\\\ &=\\binom{n+k}{n}+\\left\\{\\binom{n+k-1}{n}+\\binom{n+k-1}{n+1}\\right\\}\\\\ &=\\binom{n+k}{n}+\\binom{n+k-1}{n}+\\left\\{\\binom{n+k-2}{n}+\\binom{n+k-2}{n+1}\\right\\}\\\\ &=\\binom{n+k}{n}+\\binom{n+k-1}{n}+\\binom{n+k-2}{n}+...+\\left\\{\\binom{n+1}{n}+\\binom{n+1}{n+1}\\right\\}\\\\ &=\\binom{n+k}{n}+\\binom{n+k-1}{n}+\\binom{n+k-2}{n}+...\\binom{n+1}{n}+\\binom{n}{n} \\end{align} Hence proved.\n\nTry performing the following multiplication, then look up the Binomial Theorem.\n\n$$\\begin{array}{lr} &\\binom{5}{5}x^5+\\binom{5}{4}x^4+\\binom{5}{3}x^3+\\binom{5}{2}x^2+\\binom{5}{1}x+\\binom{5}{0}\\\\ \\times&x+1\\\\ \\hline \\end{array}$$", "date": "2019-04-25 22:18:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1413046/pascals-triangle-sum-of-nth-diagonal-row", "openwebmath_score": 0.9254748821258545, "openwebmath_perplexity": 280.6671200172784, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787853562027, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8678750334111417}}
{"url": "https://cogito.com.ua/martha-nussbaum-czx/6c05f7-signum-function-pronunciation", "text": "Note that if the routine signum/f is not able to determine a value for signum(f()) then it can return FAIL. These are useful in defining functions that can be expressed with a single formula. Overview/Introduction-Functions-Graph of a Function-Classification of Functions-One-Valued and Many-Valued Functions-The Square Root-The Absolute Value Symbol-The Signum Function-Definition of a Limit-Theorems on Limits-Right-Hand and Left-Hand Limits-Continuity-Missing Point Discontinuities-Finite Jumps-Infinite Discontinuities 3. In general, there is no standard signum function in C/C++, and the lack of such a fundamental function tells you a lot about these languages. The signum function is the real valued function defined for real as follows. It is defined as u(t) = $\\left\\{\\begin{matrix} 1 & t \\geqslant 0\\\\ 0 & t. 0 \\end{matrix}\\right.$ It is used as best test signal. Definition. Thus, at x = 0, it is left undefined. Community \u2666 1 1 1 silver badge. signum function (plural signum functions) (mathematics) The function that determines the sign of a real number, yielding -1 if negative, +1 if positive, or otherwise zero. This video is unavailable. Apart from that, I believe both majority viewpoints about the right approach to define such a function are in a way correct, and the \"controversy\" about it is actually a non-argument once you take into account two important caveats: The second property implies that for real non-zero we have . Syntax. Unit Step Function. for the . Newbie; Posts: 17; Karma: 0 ; sgn / sign / signum function suggestions. For all real we have . Watch Queue Queue Similarly, . Serge Stroobandt Serge Stroobandt. Here, we should point out that the signum function is often defined simply as 1 for x > 0 and -1 for x < 0. Hypernyms . For all real we have . Put the code into a function and use inputdlg to allow inputting the numbers more easily. For a complex argument it is defined by. It is written in the form y \u2026 The signum function is defined as f(x) = |x|/x; x\u22600 = 0; x = 0 . For a simple, outgoing source, (21) e i 2 \u03c0 q 0 x = cos 2 \u03c0 q 0 x + i sin 2 \u03c0 q 0 x sgn x, where q 0 = 1 / \u03bb. Alternatively, you could simply let the user define the matrix X and use it as an input for the function. For a complex argument it is defined by. Contents. Yes the function is discontinuous which is right as per your argument. davidhbrown. The graph of a signum function is as shown in the figure given above. See for example . Pronunciations of proper names are generally given as pronounced as in the original language. The signum function of a real number x \u2026 Signum manipuli - Wikipedia \"He carried a signum for a cohort or century.\" 1 Definition; 2 Properties; 3 Complex signum; 4 Generalized signum function; 5 See also; 6 Notes; Definition. I think the question wanted to convey this.. A Megametamathematical Guide. Solution for 1. Listen to the audio pronunciation in several English accents. In mathematics, the sign function or signum function (from signum, Latin for \"sign\") is an odd mathematical function that extracts the sign of a real number. Post by Stefan Ram \u00bbReturns the signum function of the argument\u00ab.. The signum function is the real valued function defined for real as follows. where denotes the magnitude (absolute value) of . Mathematics Pronunciation Guide. answered Sep 16 '18 at 14:26. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article. For instance, the value of function f(x) is equal to -5 in the interval [-5, -4). That is why, it is not continuous everywhere over R . of the . Information and translations of signum function in the most comprehensive dictionary definitions resource on the web. View a complete list of particular functions on this wiki Definition. The signum vector function is a function whose behavior in each coordinate is as per the signum function. Definition of signum function in the Definitions.net dictionary. Signum Function . Is the \u00bbs\u00ab voiced or voiceless? It has a jumped discontinuity which means if the function is assigned some value at the point of discontinuity it cannot be made continuous. The signum function, denoted , is defined as follows: Note: In the definition given here, we define the value to Sign function (signum function) collapse all in page. Unit step function is denoted by u(t). I cannot find the American English (San Francisco) pronunciation of \u00bbsignum\u00ab in a dictionary. Stefan Ram 2014-05-14 22:55:02 UTC . Example Problems. signum function pronunciation - How to properly say signum function. where denotes the magnitude (absolute value) of . (mathematics) A function that extracts the sign of a real number x, yielding -1 if x is negative, +1 if x is positive, or 0 if x is zero. What does signum function mean? A function cannot be continuous and discontinuous at the same point. Y = sign(x) returns an array Y the same size as x, where each element of Y is: 1 if the corresponding element of x is greater than 0. function; Translations . Impulse function is denoted by \u03b4(t). \u00bbReturns the signum function of the argument\u00ab.. It's not uncommon to need to know the arithmetic sign of a value, typically represented as -1 for values < 0, +1 for values > 0, or 0. Permalink. Anthony, looking good. All the engineering examinations including IIT JEE and AIEEE study material is available online free of cost at askIITians.com. This function definition executes fast and yields guaranteed correct results for 0, 0.0, -0.0, -4 and 5 (see comments to other incorrect answers). Similarly, . A medieval tower bell used particularly for ringing the 8 canonical hours. The cosine transform of an odd function can be evaluated as a convolution with the Fourier transform of a signum function sgn(x). Some \u2026 Let\u2019s Workout: Example 1: Find the greatest integer function for following (a) \u230a-261 \u230b (b) \u230a 3.501 \u230b (c) \u230a-1.898\u230b Solution: According to the greatest integer function definition A statement function should appear before any executable statement in the program, but after any type declaration statements. The function\u2019s value stays constant within an interval. for the Diacritically Challenged, This guide includes most mathematicians and mathematical terms that are encountered in high school and college. Any real number can be expressed as the product of its absolute value and its sign function: From equation (1) it follows that whenever x is not equal to 0 we have. German: Signumfunktion f; Hungarian: el\u0151jelf\u00fcggv\u00e9ny; Russian: \u0437\u043d\u0430\u0301\u043a\u043e\u0432\u0430\u044f \u0444\u0443\u0301\u043d\u043a\u0446\u0438\u044f f (zn\u00e1kovaja f\u00fankcija) Turkish: i\u015faret fonksiyonu; Usage notes . Meaning of signum function. Signum definition is - something that marks or identifies or represents : sign, signature. Listen to the audio pronunciation of Signum Fidei on pronouncekiwi How To Pronounce Signum Fidei: Signum Fidei pronunciation + Definition Sign in to disable ALL ads. sign(x) Description. is called the signum function. Formal Definition of a Function Limit: The limit of f (x) f(x) f (x) as x x x approaches x 0 x_0 x 0 is L L L, i.e. Use the sequential criteria for limits, to show that the\u2026 Use the e-8 definition of the limit to show that x3 \u2013 2x + 4 lim X-ix2 + 4x \u2013 3 3 !3! \u00bbI\u00ab as in \u00bbit\u00ab or as in \u00bbseal\u00ab? Graphing functions by plotting points Definition A relation f from a set A to a set B is said to be a function if every input of a set A has only one output in a set B. If then also . example. A sinc pulse passes through zero at all positive and negative integers (i.e., t = \u00b1 1, \u00b1 2, \u2026), but at time t = 0, it reaches its maximum of 1.This is a very desirable property in a pulse, as it helps to avoid intersymbol interference, a major cause of degradation in digital transmission systems. Synonyms (bell): signum bell (math): signum function, sign function A statement function in Fortran is like a single line function definition in Basic. In mathematical expressions the sign function is often represented as sgn. Area under unit step function is unity. Note that the procedure signum/f is passed only the argument(s) to the function f. The value of the environment variable is set within signum, and can be checked by signum/f, but it is not passed as a parameter. Sign function - Wikipedia \"The signum manipuli (Latin for 'standard' of the maniple, Latin: manipulus) was a standard for both the centuriae and the legion.\" signum (plural signums or signa) A sign, mark, or symbol. Definition, domain, range, solution of Signum function. Definition. Note that zero (0) is neither positive nor negative. The precise definition of the limit is discussed in the wiki Epsilon-Delta Definition of a Limit. Of course it is continuous at every x \u20ac R except at x = 0 . \u00bbU\u00ab as in \u00bbsoon\u00ab or \u00bbnumber\u00ab or just a schwa? The second property implies that for real non-zero we have . No, it is not continuous every where . \"The signum function is the derivative of the absolute value function, up to (but not including) the indeterminacy at zero.\" The format is simple - just type f(x,y,z,\u2026) = formula . Suppose is a positive integer. SIGN(x) returns a number that indicates the sign x: -1 if x is negative; 0 if x equals 0; or 1 if x is positive. In applications such as the Laplace transform this definition is adequate, since the value of a function at a single point does not change the analysis. If then also . Unit Impulse Function. Therefore, its Fourier transform is (22) F q = 1 2 \u03b4 q + q 0 + \u03b4 q \u2212 q 0 \u2212 i 2 \u03c0 q \u2212 q 0 + i 2 \u03c0 q + q 0. Explicitly, it is defined as the function: In other words, the signum function project a non-zero complex number to the unit circle . share | improve this answer | follow | edited Jun 20 at 9:12. Topic: sgn / sign / signum function suggestions (Read 2558 times) previous topic - next topic. Jul 19, 2019, 05:35 pm. Signum function is an integer valued function defined over R . Proper American English Pronunciation of Words and Names. A sinc function is an even function with unity area. function. Signifer - Wikipedia. The signum function of a real number x is defined as follows: Properties. In other words, the signum function project a non-zero complex number to the unit circle . Study Physics, Chemistry and Mathematics at askIITians website and be a winner. Chemistry and Mathematics at askIITians website and be a winner \u00bb it or... At x = 0, it is continuous at every x \u20ac R except at x = 0 numbers... Properties ; 3 complex signum ; 4 Generalized signum function project a non-zero complex number to unit. Improve this answer | follow | edited Jun 20 at 9:12 function \u2019 value., Chemistry and Mathematics at askIITians website and be a winner sinc function is often represented as.... 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Guide includes most mathematicians and mathematical terms that are encountered in high school and college U! Is denoted by U ( t ) \u20ac R except at x = 0 0 x... Function whose behavior in each coordinate is as shown in the program, but any!\n\nTractor Supply Tool Box Won't Open, Tarantula Dc Nightwing, Antigo Shopping Show 106, Oliver Travel Trailers Canada, Homes For Sale By Owner Winterset Iowa, Large Used Car Dealerships Near Me, Welsh Guards Ranks,", "date": "2021-04-21 13:54:48", "meta": {"domain": "com.ua", "url": "https://cogito.com.ua/martha-nussbaum-czx/6c05f7-signum-function-pronunciation", "openwebmath_score": 0.7237217426300049, "openwebmath_perplexity": 1429.0786706255644, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9511422241476943, "lm_q2_score": 0.9124361521430956, "lm_q1q2_score": 0.867856551142148}}
{"url": "https://au.mathworks.com/help/matlab/ref/norm.html", "text": "# norm\n\nVector and matrix norms\n\n## Description\n\nexample\n\nn = norm(v) returns the Euclidean norm of vector v. This norm is also called the 2-norm, vector magnitude, or Euclidean length.\n\nexample\n\nn = norm(v,p) returns the generalized vector p-norm.\n\nexample\n\nn = norm(X) returns the 2-norm or maximum singular value of matrix X, which is approximately max(svd(X)).\n\nn = norm(X,p) returns the p-norm of matrix X, where p is 1, 2, or Inf:\n\nexample\n\nn = norm(X,\"fro\") returns the Frobenius norm of matrix or array X.\n\n## Examples\n\ncollapse all\n\nCreate a vector and calculate the magnitude.\n\nv = [1 -2 3];\nn = norm(v)\nn = 3.7417\n\nCalculate the 1-norm of a vector, which is the sum of the element magnitudes.\n\nv = [-2 3 -1];\nn = norm(v,1)\nn = 6\n\nCalculate the distance between two points as the norm of the difference between the vector elements.\n\nCreate two vectors representing the (x,y) coordinates for two points on the Euclidean plane.\n\na = [0 3];\nb = [-2 1];\n\nUse norm to calculate the distance between the points.\n\nd = norm(b-a)\nd = 2.8284\n\nGeometrically, the distance between the points is equal to the magnitude of the vector that extends from one point to the other.\n\n$\\begin{array}{l}a=0\\underset{}{\\overset{\u02c6}{i}}+3\\underset{}{\\overset{\u02c6}{j}}\\\\ b=-2\\underset{}{\\overset{\u02c6}{i}}+1\\underset{}{\\overset{\u02c6}{j}}\\\\ \\\\ \\begin{array}{rl}{d}_{\\left(a,b\\right)}& =||b-a||\\\\ & =\\sqrt{\\left(-2-0{\\right)}^{2}+\\left(1-3{\\right)}^{2}}\\\\ & =\\sqrt{8}\\end{array}\\end{array}$\n\nCalculate the 2-norm of a matrix, which is the largest singular value.\n\nX = [2 0 1;-1 1 0;-3 3 0];\nn = norm(X)\nn = 4.7234\n\nCalculate the Frobenius norm of a 4-D array X, which is equivalent to the 2-norm of the column vector X(:).\n\nX = rand(3,4,4,3);\nn = norm(X,\"fro\")\nn = 7.1247\n\nThe Frobenius norm is also useful for sparse matrices because norm(X,2) does not support sparse X.\n\n## Input Arguments\n\ncollapse all\n\nInput vector.\n\nData Types: single | double\nComplex Number Support: Yes\n\nInput array, specified as a matrix or array. For most norm types, X must be a matrix. However, for Frobenius norm calculations, X can be an array.\n\nData Types: single | double\nComplex Number Support: Yes\n\nNorm type, specified as 2 (default), a positive real scalar, Inf, or -Inf. The valid values of p and what they return depend on whether the first input to norm is a matrix or vector, as shown in the table.\n\nNote\n\nThis table does not reflect the actual algorithms used in calculations.\n\npMatrixVector\n1max(sum(abs(X)))sum(abs(v))\n2 max(svd(X))sum(abs(v).^2)^(1/2)\nPositive, real-valued numeric scalarsum(abs(v).^p)^(1/p)\nInfmax(sum(abs(X')))max(abs(v))\n-Infmin(abs(v))\n\n## Output Arguments\n\ncollapse all\n\nNorm value, returned as a scalar. The norm gives a measure of the magnitude of the elements. By convention:\n\n\u2022 norm returns NaN if the input contains NaN values.\n\n\u2022 The norm of an empty matrix is zero.\n\ncollapse all\n\n### Euclidean Norm\n\nThe Euclidean norm (also called the vector magnitude, Euclidean length, or 2-norm) of a vector v with N elements is defined by\n\n$\u2016v\u2016=\\sqrt{\\sum _{k=1}^{N}{|{v}_{k}|}^{2}}\\text{\\hspace{0.17em}}.$\n\n### General Vector Norm\n\nThe general definition for the p-norm of a vector v that has N elements is\n\n${\u2016v\u2016}_{p}={\\left[\\sum _{k=1}^{N}{|{v}_{k}|}^{p}\\right]}^{\\text{\\hspace{0.17em}}1/p}\\text{\\hspace{0.17em}},$\n\nwhere p is any positive real value, Inf, or -Inf.\n\n\u2022 If p = 1, then the resulting 1-norm is the sum of the absolute values of the vector elements.\n\n\u2022 If p = 2, then the resulting 2-norm gives the vector magnitude or Euclidean length of the vector.\n\n\u2022 If p = Inf, then ${\u2016v\u2016}_{\\infty }={\\mathrm{max}}_{i}\\left(|v\\left(i\\right)|\\right)$.\n\n\u2022 If p = -Inf, then ${\u2016v\u2016}_{-\\infty }={\\mathrm{min}}_{i}\\left(|v\\left(i\\right)|\\right)$.\n\n### Maximum Absolute Column Sum\n\nThe maximum absolute column sum of an m-by-n matrix X (with m,n >= 2) is defined by\n\n${\u2016X\u2016}_{1}=\\underset{1\\le j\\le n}{\\mathrm{max}}\\left(\\sum _{i=1}^{m}|{a}_{ij}|\\right).$\n\n### Maximum Absolute Row Sum\n\nThe maximum absolute row sum of an m-by-n matrix X (with m,n >= 2) is defined by\n\n${\u2016X\u2016}_{\\infty }=\\underset{1\\le i\\le m}{\\mathrm{max}}\\left(\\sum _{j=1}^{n}|{a}_{ij}|\\right)\\text{\\hspace{0.17em}}.$\n\n### Frobenius Norm\n\nThe Frobenius norm of an m-by-n matrix X (with m,n >= 2) is defined by\n\n${\u2016X\u2016}_{F}=\\sqrt{\\sum _{i=1}^{m}\\sum _{j=1}^{n}{|{a}_{ij}|}^{2}}=\\sqrt{\\text{trace}\\left({X}^{\u2020}X\\right)}\\text{\\hspace{0.17em}}.$\n\nThis definition also extends naturally to arrays with more than two dimensions. For example, if X is an N-D array of size m-by-n-by-p-by-...-by-q, then the Frobenius norm is\n\n${\u2016X\u2016}_{F}=\\sqrt{\\sum _{i=1}^{m}\\sum _{j=1}^{n}\\sum _{k=1}^{p}...\\sum _{w=1}^{q}{|{a}_{ijk...w}|}^{2}}.$\n\n## Tips\n\n\u2022 Use vecnorm to treat a matrix or array as a collection of vectors and calculate the norm along a specified dimension. For example, vecnorm can calculate the norm of each column in a matrix.\n\n## Version History\n\nIntroduced before R2006a\n\nexpand all", "date": "2023-03-27 10:16:22", "meta": {"domain": "mathworks.com", "url": "https://au.mathworks.com/help/matlab/ref/norm.html", "openwebmath_score": 0.898819625377655, "openwebmath_perplexity": 1338.2199700781828, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9928785710435412, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.8678525660027901}}
{"url": "http://amhomeworkeylt.californiapublicrecords.us/pascals-triangle.html", "text": "# Pascals triangle\n\nBinomial coefficients, pascal's triangle, and the power set binomial coefficients pascal's triangle the power set binomial coefficients an expression like (a + b) . Pascal's triangle is a number pattern introduced by the famous french mathematician and philosopher blaise pascal it is triangle in shape with the top being. Mathematicians have long been familiar with the tidy way in which the nth row of pascal's triangle sums to 2n (the top row conventionally labeled as n = 0. Where (n r) is a binomial coefficient the triangle was studied by b pascal, although it had been described centuries earlier by chinese mathematician yanghui. Pascal's triangle, a simple yet complex mathematical construct, hides some surprising properties related to number theory and probability.\n\nSal introduces pascal's triangle, and shows how we can use it to figure out the coefficients in binomial expansions. Applications pascal's triangle is not only an interesting mathematical work because of its hidden patterns, but it is also interesting because of its wide expanse. All you ever want to and need to know about pascal's triangle.\n\nPascal triangle: given numrows, generate the first numrows of pascal's triangle pascal's triangle : to generate a[c] in row r, sum up a'[c] and a'[c-1] from. Given numrows, generate the first numrows of pascal's triangle for example, given numrows = 5, the result should be: , , , , ] java. In mathematics, pascal's triangle is a triangular array of the binomial coefficients in much of the western world, it is named after the french mathematician blaise. The pascal's triangle is a graphical device used to predict the ratio of heights of lines in a split nmr peak. From the wikipedia page the kata author cites: the rows of pascal's triangle ( sequence a007318 in oeis) are conventionally enumerated starting with row n = 0.\n\nA guide to understanding binomial theorem, pascal's triangle and expanding binomial series and sequences. This special triangular number arrangement is named after blaise pascal pascal was a french mathematician who lived during the seventeenth century. When you look at pascal's triangle, find the prime numbers that are the first number in the row that prime number is a divisor of every number in that row.\n\n## Pascals triangle\n\nResources for the patterns in pascal's triangle problem pascal's triangle at provides information on. Yes, pascal's triangle and the binomial theorem isn't particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. The counting function c(n,k) and the concept of bijection coalesce in one of the most studied mathematical concepts, pascal's triangle at its heart, pascal's. Pascal's triangle is one of the classic example taught to engineering students it has many interpretations one of the famous one is its use with binomial.\n\n\u2022 Now that we've learned how to draw pascal's famous triangle and use the numbers in its rows to easily calculate probabilities when tossing coins, it's time to dig.\n\u2022 There are many interesting things about polynomials whose coefficients are taken from slices of pascal's triangle (these are a form of what's called chebyshev.\n\u2022 S northshieldsums across pascal's triangle modulo 2 j raaba generalization of the connection between the fibonacci sequence and pascal's triangle.\n\nIt's not necessary to do this because 2 only shows up once in pascal's triangle but you get the idea t2 = table[binomial[n, k], {n, 0, 8}, {k, 0, n}] / {a_, 2, c_} : {a . Pascal's triangle and binomial coefficients information for the mathcamp 2017 qualifying quiz 1 pascal's triangle pascal's triangle (named for the. Theorem the sum of all the entries in the n th row of pascal's triangle is equal to 2 n proof 1 by definition, the entries in n th row of pascal's. Pascal's triangle is defined such that the number in row \\$n\\$ and column \\$k\\$ is \\$ {n\\choose k}\\$ for this reason, convention holds that both row numbers and.\n\nPascals triangle\nRated 4/5 based on 31 review\n\n2018.", "date": "2018-11-20 20:01:24", "meta": {"domain": "californiapublicrecords.us", "url": "http://amhomeworkeylt.californiapublicrecords.us/pascals-triangle.html", "openwebmath_score": 0.8048306107521057, "openwebmath_perplexity": 755.516228711971, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9928785704113702, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8678525524227477}}
{"url": "https://mathoverflow.net/questions/263397/is-there-a-transformation-or-a-proof-for-these-integrals?noredirect=1", "text": "Is there a transformation or a proof for these integrals?\n\nHere are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.\n\nQuestion. Is this true? If so, is there an underlying transformation or just a proof? $$\\int_0^{\\infty}x^2e^{-x^2}\\frac{{dx}}{\\cosh\\sqrt{\\pi}x} =\\frac14\\int_0^{\\infty}e^{-x^2}\\frac{dx}{\\cosh\\sqrt{\\pi}x}.$$\n\n\u2022 The integral in this question was taken from your own paper that was written 10 years earlier \"A dozen integrals: Russell-style\" and you knew the proof. arxiv.org/abs/0808.2692\n\u2013\u00a0Nemo\nAug 9, 2021 at 18:15\n\u2022 @user44191 Whatever Stack Exchange policy might be, and independently of this particular case (on which I make no comment), it's (1) certainly unusual on MO to ask questions to which you already know the answer, and (2) abusive of other people's time to do this while giving the impression that you don't know the answer. If I spent time solving someone's problem and then typing it up, on the understanding that I was helping them out, I might feel quite angry to discover that they knew the answer all along. I don't use MO as much as I used to, but is there really any doubt over this? Aug 10, 2021 at 13:32\n\u2022 I can\u2019t understand why downvoting this question. It seems to me interesting and perfectly suitable for MO. Not mentioning an existing proof is a defect, of course, yet I explain it to me with the purpose of not influencing who approaches the problem, since apparently the goal was to get new proofs, insights and connections. Aug 10, 2021 at 14:31\n\u2022 Besides, after $10$ years it is perfectly possible to forget to have ever proven that particular integral identity. Aug 10, 2021 at 17:17\n\u2022 @Nemo, making an unnecessary edit, just so you can vote a question down? Really? Aug 11, 2021 at 13:09\n\nThis is a generous explanation of Lucia's comment above.\n\nThe functions $$\\mathcal H_n(x)=\\frac{2^{1/4}}{(2^n n!)^{1/2}}H_n(\\sqrt{2\\pi}\\; x) e^{-\\pi x^2}$$ form an orthonormal system in $L^2(\\textbf{R})$. Here $H_n(x)$ are the usual Hermite polynomials defined by $$e^{2xz-z^2}=\\sum_{n=0}^\\infty \\frac{H_n(x)}{n!}z^n,\\qquad |z|<\\infty.$$ The functions are eigenfunctions for the usual Fourier transform so that $$\\int_{-\\infty}^{+\\infty}\\mathcal H_n(t)e^{-2\\pi i x t}\\,dt=(-i)^n \\mathcal H_n(x).$$\n\nIt follows that $L^2(\\textbf{R})$ is a direct sum of four subspaces. In each of these subspaces the Fourier transform is just multiplication by $1$, $i$, $-1$, $-1$ respectively.\n\nIt is well known that the function $1/\\cosh\\pi x$ is invariant by the Fourier transform $$\\int_{-\\infty}^{+\\infty}\\frac{e^{-2\\pi i x\\xi}}{\\cosh\\pi x}\\,dx= \\frac{1}{\\cosh\\pi \\xi}.$$ Therefore this function is in the span of the functions $\\mathcal H_{4n}(x)$, and therefore is orthogonal to any other eigenfunction. In particular $$\\int_{-\\infty}^{+\\infty}\\frac{\\mathcal H_2(x)}{\\cosh\\pi x}\\,dx=0.$$ Since $H_2(x)=-2+4x^2$ it follows that $$\\int_{-\\infty}^{+\\infty}\\frac{(8\\pi x^2-2)e^{-\\pi x^2}}{\\cosh\\pi x}\\,dx=0.$$ Putting $x/\\sqrt{\\pi}$ instead of x $$\\int_{-\\infty}^{+\\infty}\\frac{(8 x^2-2)e^{-x^2}}{\\cosh\\sqrt{\\pi} x}\\,\\frac{dx}{\\sqrt{\\pi}}=0.$$ This is equivalent to your equation.\n\nUPDATE\n\nThe integral in T. Amdeberhan's question was taken from his own paper that was written 10 years earlier before he posted his question: https://arxiv.org/abs/0808.2692 A dozen integrals: Russell-style. Thus it seems that T. Amdeberhan knew the answer to the question he asked. Why did he ask it then?\n\nI provide a screenshot below for the reader's convenience (integral number $$7$$):\n\nAt the end of the paper, the authors provide a sketch of proof:\n\nNote that the following functions are self-reciprocal $$\\sqrt{\\frac{2}{\\pi}}\\int_0^\\infty f(x)\\cos ax dx=f(a)$$ (see this MO post for a list of such functions): $$\\frac{1}{\\cosh\\sqrt{\\frac{\\pi}{2}}x},\\quad e^{-x^2/2}.\\tag{1}$$ It was proved by Hardy (Quarterly Journal Of Pure And Applied Mathematics, Volume 35, Page 203, 1903) and Ramanujan (Ramanujan's Lost Notebook, part IV, chapter 18) that for two self-reciprocal functions $$f$$ and $$g$$ we have $$\\int_0^\\infty f(x)g(\\alpha x) dx=\\frac{1}{\\alpha}\\int_0^\\infty f(x)g(x/\\alpha) dx.$$ The formal agument is as follows but it can be made rigorous for certain type of functions $$\\int_0^\\infty f(x)g(\\alpha x) dx=\\int_0^\\infty f(x)dx\\cdot \\sqrt{\\frac{2}{\\pi}}\\int_0^\\infty g(y) \\cos(\\alpha x y)dy=\\\\ \\int_0^\\infty g(y)dy \\cdot \\sqrt{\\frac{2}{\\pi}}\\int_0^\\infty f(x)\\cos(\\alpha x y)dx=\\int_0^\\infty g(y)f(\\alpha y) dy=\\\\ \\frac{1}{\\alpha}\\int_0^\\infty f(x)g(x/\\alpha) dx.$$ For functions in $$(1)$$ which decay rapidly at $$x\\to\\pm\\infty$$ it is certainly true. So we obtain an identity due to Hardy and Ramanujan $$\\int_{0}^{\\infty} \\frac{e^{-x^{2}/2}}{\\cosh{\\sqrt{\\frac{\\pi}{2}}\\alpha{x}}} {dx} = \\frac{1}{\\alpha} \\int_{0}^{\\infty} \\frac{e^{-x^{2}/2}}{\\cosh{\\sqrt{\\frac{\\pi}{2}}\\frac{x}{\\alpha}}}{dx}.$$ After some simplifications it becomes $$\\int_{0}^{\\infty} \\frac{e^{-\\alpha^2x^{2}}}{\\cosh{\\sqrt{{\\pi}}{x}}} {dx} = \\frac{1}{\\alpha} \\int_{0}^{\\infty} \\frac{e^{-x^{2}/\\alpha^2}}{\\cosh{\\sqrt{{\\pi}}{x}}}{dx}.$$ To complete the proof differentiate this with respect to $$\\alpha$$ and then set $$\\alpha=1$$.", "date": "2022-07-02 18:12:34", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/263397/is-there-a-transformation-or-a-proof-for-these-integrals?noredirect=1", "openwebmath_score": 0.8792707324028015, "openwebmath_perplexity": 302.4624352044359, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985271387015241, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.8678241633226886}}
{"url": "http://nptel.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node121.html", "text": "## Simpson's Rule\n\nIf we are given odd number of tabular points,i.e. is even, then we can divide the given integral of integration in even number of sub-intervals Note that for each of these sub-intervals, we have the three tabular points and so the integrand is replaced with a quadratic interpolating polynomial. Thus using the formula (13.3.3), we get,\n\nIn view of this, we have\n\nwhich gives the second quadrature formula as follows:\n (13.3.5)\n\nThis is known as SIMPSON'S RULE.\n\nRemark 13.3.3 \u00a0 An estimate for the error in numerical integration using the Simpson's rule is given by\n\n (13.3.6)\n\nwhere is the average value of the forth forward differences.\n\nEXAMPLE 13.3.4 \u00a0 Using the table for the values of as is given in Example 13.3.2, compute the integral by Simpson's rule. Also estimate the error in its calculation and compare it with the error using Trapezoidal rule.\nSolution: Here, thus we have odd number of nodal points. Further,\n\nand\n\nThus,\n\nTo find the error estimates, we consider the forward difference table, which is given below:\n\n 0.0 1.00000 0.01005 0.02071 0.00189 0.00149 0.1 1.01005 0.03076 0.02260 0.00338 0.00171 0.2 1.04081 0.05336 0.02598 0.00519 0.00243 0.3 1.09417 0.07934 0.03117 0.00762 0.00320 0.4 1.17351 0.11051 0.3879 0.01090 0.00459 0.5 1.28402 0.14930 0.04969 0.01549 0.00658 0.6 1.43332 0.19899 0.06518 0.02207 0.00964 0.7 1.63231 0.26417 0.08725 0.03171 0.8 1.89648 0.35142 0.11896 0.9 2.24790 0.47038 1.0 2.71828\n\nThus, error due to Trapezoidal rule is,\n\nSimilarly, error due to Simpson's rule is,\n\nIt shows that the error in numerical integration is much less by using Simpson's rule.\n\nEXAMPLE 13.3.5 \u00a0 Compute the integral , where the table for the values of is given below:\n 0.05 0.1 0.15 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.0785 0.1564 0.2334 0.309 0.454 0.5878 0.7071 0.809 0.891 0.9511 0.9877 1\n\nSolution: Note that here the points are not given to be equidistant, so as such we can not use any of the above two formulae. However, we notice that the tabular points and are equidistant and so are the tabular points and . Now we can divide the interval in two subinterval: and ; thus,\n\n. The integrals then can be evaluated in each interval. We observe that the second set has odd number of points. Thus, the first integral is evaluated by using Trapezoidal rule and the second one by Simpson's rule (of course, one could have used Trapezoidal rule in both the subintervals).\nFor the first integral and for the second one . Thus,\n\nwhich gives,\n\nIt may be mentioned here that in the above integral, and that the value of the integral is . It will be interesting for the reader to compute the two integrals using Trapezoidal rule and compare the values.\n\nEXERCISE 13.3.6\n1. Using Trapezoidal rule, compute the integral where the table for the values of is given below. Also find an error estimate for the computed value.\n1.  a=1 2 3 4 5 6 7 8 9 b=10 0.09531 0.18232 0.26236 0.33647 0.40546 0.47 0.53063 0.58779 0.64185 0.69314\n2.  a=1.50 1.55 1.6 1.65 1.7 1.75 b=1.80 0.40546 0.43825 0.47 0.5077 0.53063 0.55962 0.58779\n3.  a = 1.0 1.5 2 2.5 3 b = 3.5 1.1752 2.1293 3.6269 6.0502 10.0179 16.5426\n2. Using Simpson's rule, compute the integral Also get an error estimate of the computed integral.\n1. Use the table given in Exercise 13.3.6.1b.\n2.  a = 0.5 1 1.5 2 2.5 3 b = 3.5 0.493 0.946 1.325 1.605 1.778 1.849 1.833\n3. Compute the integral , where the table for the values of is given below:\n 0 0.5 0.7 0.9 1.1 1.2 1.3 1.4 1.5 0 0.39 0.77 1.27 1.9 2.26 2.65 3.07 3.53\n\nA K Lal 2007-09-12", "date": "2017-02-25 15:48:20", "meta": {"domain": "ac.in", "url": "http://nptel.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node121.html", "openwebmath_score": 0.9466323852539062, "openwebmath_perplexity": 258.44375098896734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713861502774, "lm_q2_score": 0.880797081106935, "lm_q1q2_score": 0.8678241610193482}}
{"url": "http://math.stackexchange.com/questions/897469/determinant-of-a-matrix-with-t-in-all-off-diagonal-entries", "text": "Determinant of a matrix with $t$ in all off-diagonal entries.\n\nIt seems from playing around with small values of $n$ that\n\n$$\\det \\left( \\begin{array}{ccccc} -1 & t & t & \\dots & t\\\\ t & -1 & t & \\dots & t\\\\ t & t & -1 & \\dots & t\\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ t & t & t & \\dots& -1 \\end{array}\\right) = (-1)^{n-1}(t+1)^{n-1}((n-1)t-1)$$\n\nwhere $n$ is the size of the matrix.\n\nHow would one approach deriving (or at least proving) this formally?\n\nMotivation\n\nThis came up when someone asked what is the general solution to:\n\n$$\\frac{a}{b+c}=\\frac{b}{c+a}=\\frac{c}{a+b},$$\n\nand for non-trivial solutions, the matrix above (with $n=3$) must be singular. In this case either $t=-1\\implies a+b+c=1$ or $t=\\frac{1}{2}\\implies a=b=c$.\n\nSo I wanted to ensure that these are also the only solutions for the case with more variables.\n\n-\nDid you try induction? \u2013\u00a0 zarathustra Aug 14 '14 at 17:08\nInduction combined with this should do it. \u2013\u00a0 Daniel R Aug 14 '14 at 17:09\nInduction is the way to go, but first, subtract the first row from all the other rows to get a much easier calculation. \u2013\u00a0 fixedp Aug 14 '14 at 17:10\nWhen $n=2$ seems to fail. $1-t^2$ RHS, $-2 t^3-3 t^2+1$ LHS, right? \u2013\u00a0 carlosayam Aug 14 '14 at 17:14\nThe matrix size should be $n+1$; @caya gives the $n=2$ case. \u2013\u00a0 Semiclassical Aug 14 '14 at 17:17\n\nUsing elementary operations instead of induction is key. \\begin{align} &\\begin{vmatrix} -1 & t & t & \\dots & t\\\\ t & -1 & t & \\dots & t\\\\ t & t & -1 & \\dots & t\\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ t & t & t & \\dots& -1 \\end{vmatrix}\\\\ &= \\begin{vmatrix} -t-1 & 0 & 0 & \\dots & t+1\\\\ 0 & -t-1 & 0 & \\dots & t+1\\\\ 0 & 0 & -t-1 & \\dots & t+1\\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ t & t & t & \\dots& -1 \\end{vmatrix}\\\\ &= \\begin{vmatrix} -t-1 & 0 & 0 & \\dots & 0\\\\ 0 & -t-1 & 0 & \\dots & 0\\\\ 0 & 0 & -t-1 & \\dots & 0\\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ t & t & t & \\dots& (n - 1)t -1 \\end{vmatrix}\\\\ &= (-1)^{n - 1}(t + 1)^{n - 1}((n - 1)t - 1) \\end{align}\n\n-\nFor clarity, first he is doing \"Row N = Row N - Last Row\" then he is doing \"Last Row = Last Row - Row N\" \u2013\u00a0 DanielV Aug 14 '14 at 17:46\n\nYou can write the expression as $$\\det(t C - (t+1)I)$$ where $C = \\mathbf{1}\\mathbf{1}^T$ is the matrix of all $1$'s, formed by the column of ones times its transpose. Using the identity $\\det(I+cr) = 1+rc$, you can first factor out $(t+1)$: $$\\det(t C - (t+1)I) = (-1)^n(t+1)^n \\det\\left(I - \\frac{t}{t+1} \\mathbf{1}\\right) = (-1)^n(t+1)^n \\left(1 - \\frac{nt}{t+1}\\right)$$\n\n-\nThat's brilliant, nothing could be more elegant. :) \u2013\u00a0 MGA Aug 14 '14 at 17:37\n+1, but what do you mean by $cr$ and $rc$? Are these row and column vectors, abbreviated $r$ and $c$, respectively? (My assumption) \u2013\u00a0 apnorton Aug 15 '14 at 3:44\nCorrect, they are row and column vectors. I stole that from the Wiki page on Determinant \u2013\u00a0 Victor Liu Aug 15 '14 at 21:42\n\nNote that your matrix is the sum of the matrix $T$ with all entries equal to $t$ and the matrix $-(1+t)I$. Therefore the determinant you are asking about is the value at $X=-(1+t)$ of the characteristic polynomial $\\chi_{-T}$ of $-T$.\n\nSince $T$ has rank (at most) $1$, its eigenspace for eigenvalue $0$ has dimension $n-1$, so the characteristic polynomial of $-T$ is $\\chi_{-T}=X^{n-1}(X+nt)$ (the final factor must be that because the coefficient of $X^{n-1}$ in $\\chi_{-T}$ is $\\def\\tr{\\operatorname{tr}}-\\tr(-T)=nt$). Now setting $X=-(1+t)$ gives $$\\det(-(1+t)I-T)=(-1-t)^{n-1}(-1-t+nt)=(-1-t)^{n-1}(-1+(n-1)t)$$ as desired.\n\nThis kind of question is recurrent on this site; see for instance Determinant of a specially structured matrix and How to calculate the following determinants (all ones, minus $I$).\n\n-\n\nHere is another characterization of the result. For convenience, I will take the matrix size as $n+1$ rather than $n$. First, note that we may factor $t$ from each of the $n+1$ rows, and so the determinant may be written as $$\\det{[t(M-t^{-1} I_{n+1})]} =t^{n+1} \\det(M-t^{-1} I_{n+1})$$ where $(M)_{ij}=1-\\delta_{ij}$ for $1\\leq i,j\\leq n+1$.\n\nNext, observe that $M$ has $n$ independent eigenvectors of the form $\\hat{e}_i-\\hat{e}_{n+1}$ ($1\\leq i\\leq n$), each with eigenvalue $-1$; in addition, $M$ also has the eigenvector $\\sum_{i=1}^{n+1}\\hat{e}_i$ with eigenvalue $n$. Consequently the characteristic polynomial of $M$ in powers of $t^{-1}$ is $$\\det(M-t^{-1} I_{n+1})=(-1-t^{-1})^n (n-t^{-1}) = t^{-n-1}\\cdot (-1)^n (1+t)^n (nt+1)$$ and so the prior result yields the desired identity.\n\n-", "date": "2015-05-28 08:49:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/897469/determinant-of-a-matrix-with-t-in-all-off-diagonal-entries", "openwebmath_score": 0.9821444749832153, "openwebmath_perplexity": 278.5109553229933, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8678241602574908}}
{"url": "https://math.stackexchange.com/questions/3619669/variation-of-least-squares-with-symmetric-positive-semi-definite-psd-constrain", "text": "# Variation of Least Squares with Symmetric Positive Semi Definite (PSD) Constraint\n\nI am trying to solve the following convex optimization problem: \\begin{align} & \\min_{W} && \\sum_{i=1}^n (\\mathbf{x}_{i}^TW\\mathbf{x}_{i} - y_i)^2 \\\\\\\\ & s.t. && W \\succcurlyeq 0 \\\\\\\\ & && W = W^T \\end{align}\n\nwhere $$\\mathbf{x}_i \\in \\mathbb{R^p}$$, $$W \\in \\mathbb{R}^{p \\times p}$$ and $$y_i \\geq 0$$.\n\nWithout the positive semidefinite constraint, the problem is pretty straightforward. Requiring positive semidefiniteness, however, makes it a bit tricky.\n\nI thought about using the fact that $$W \\succcurlyeq 0$$ if and only if there exists a symmetric $$A$$ such that $$W = AA^T$$, and solving the equivalent problem\n\n\\begin{align} & \\min_{A} && \\sum_{i=1}^n (\\mathbf{x}_{i}^TAA^T\\mathbf{x}_{i} - y_i)^2 \\\\\\\\ &s.t. && A = A^T \\end{align}\n\nLetting $$a_{ij}$$ be the $$(i,j)th$$ element of A, this optimization function is quartic (fourth-order) with respect to the $$a_{ij}$$'s. Because of this, I am unsure of how to proceed.\n\nI would be grateful if someone could point me in the right direction as to how to solve this problem.\n\n\u2022 Haven't worked it through, but you might find it helpful to note that, if $W = A^\\top A$, then$$x^\\top_i W x_i = x^\\top_i A^\\top A x_i = (Ax_i)^\\top (Ax_i) = \\|Ax_i\\|^2.$$ Apr 11 '20 at 0:01\n\nWhen dealing with such form of matrix multiplications always remember the Vectorization Trick with Kronecker Product for Matrix Equations:\n\n$${x}_{i}^{T} W {x}_{i} - {y}_{i} \\Rightarrow \\left({x}_{i}^{T} \\otimes {x}_{i}^{T} \\right) \\operatorname{Vec} \\left( W \\right) - \\operatorname{Vec} \\left( {y}_{i} \\right) = \\left({x}_{i}^{T} \\otimes {x}_{i}^{T} \\right) \\operatorname{Vec} \\left( W \\right) - {y}_{i}$$\n\nSince the problem is given by summing over $${x}_{i}$$ one could build the matrix:\n\n$$X = \\begin{bmatrix} {x}_{1}^{T} \\otimes {x}_{1}^{T} \\\\ {x}_{2}^{T} \\otimes {x}_{2}^{T} \\\\ \\vdots \\\\ {x}_{n}^{T} \\otimes {x}_{n}^{T} \\end{bmatrix}$$\n\nThen:\n\n$$\\arg \\min_{W} \\sum_{i = 1}^{n} {\\left( {x}_{i}^{T} W {x}_{i} - {y}_{i} \\right)}^{2} = \\arg \\min_{W} {\\left\\| X \\operatorname{Vec} \\left( W \\right) - \\boldsymbol{y} \\right\\|}_{2}^{2}$$\n\nWhere $$\\boldsymbol{y}$$ is the column vector composed by $${y}_{i}$$.\n\nNow the above has nice form of regular Least Squares. The handling of the constraint can be done using Projected Gradient Descent Method. The projection onto the set of Symmetric Matrices and Positive Semi Definite (PSD) Matrices cone are given by:\n\n1. $$\\operatorname{Proj}_{\\mathcal{S}^{n}} \\left( A \\right) = \\frac{1}{2} \\left( A + {A}^{T} \\right)$$. See Orthogonal Projection of a Matrix onto the Set of Symmetric Matrices.\n2. $$\\operatorname{Proj}_{\\mathcal{S}_{+}^{n}} \\left( A \\right) = Q {\\Lambda}_{+} {Q}^{T}$$ where $$A = Q \\Lambda {Q}^{T}$$ is the eigen decomposition of $$A$$ and $${\\Lambda}_{+}$$ means we zero any negative values in $$\\Lambda$$. See Find the Matrix Projection of a Symmetric Matrix onto the set of Symmetric Positive Semi Definite (PSD) Matrices.\n\nSince both the Symmetric Matrices Set and PSD Cone are Linear Sub Space hence even the greedy iterative projection on the set will yield an orthogonal projection on the intersection of the 2 sets. See Orthogonal Projection onto the Intersection of Convex Sets.\n\nSo, with all the tools above one could create his own solver using basic tools with no need for external libraries (Which might be slow or not scale).\n\nI implemented the Projected Gradient Descent Method with the above projections in MATLAB. I compared results to CVX to validate the solution. This is the solution:\n\nMy implementation is vanilla Gradient Descent with constant Step Size and no acceleration. If you add those you'll see convergence which is order of magnitude faster (I guess few tens of iterations). Not bad for hand made solver.\n\nThe MATLAB Code is accessible in my StackExchange Mathematics Q3619669 GitHub Repository.\n\n\u2022 This was a very clear and practical response. Thanks! Assuming I understand correctly, I have one question: what is the difference between iteratively projecting vs. projecting just once after finding the optimal solution to the unconstrained problem? Would they give different solutions? Apr 13 '20 at 15:43\n\u2022 @tygaking, Please mark it as answer and +1. Regarding your question, you can't do that. Think of \"Gradient Descent\" as some kind of projection by itself. Hence what we do above is basically iterative projection onto iterative set, one for the gradient descent, one for each other.\n\u2013\u00a0Royi\nApr 13 '20 at 15:45\n\nAre you doing FGLS or something?\n\nYou could try substituting the constraint into the object. For the two by two case, for example, solve $$\\sum_{i = 1}^N \\left(x_i'\\left[\\array{ \\array{w_{11} & w_{12}} \\\\ \\array{w_{12}& w_{22} }} \\right]x_i - y_i \\right)^2$$ where $$w_{12} = w_{21}$$ now by construction. Then the matrix will be symmetric.\n\nTo ensure positive semi-definiteness, you can then use the standard principal minors test: $$w_{ii} \\ge 0$$ for each $$i$$, $$w_{11} w_{22} - w_{12}^2 \\ge 0$$, and so on, with the determinant of the upper left-hand minor weakly positive.\n\nThat at least subsumes positive semi-definiteness into concrete constraints and adjustments to the problem. That sounds like a nightmare to solve however, using Kuhn-Tucker. A simpler sufficient condition for semi-definiteness is the dominant diagonal condition, that $$w_{ii} \\ge \\sum_{j \\neq i} |w_{ij}|$$ for each row $$i$$, which would be much more computationally tractable. Perhaps it could give you a good initial guess before you try relaxing it to the standard principal minors constraints.\n\n\u2022 There is no need for this. While your approach might lead to problem formulated as Least Squares with Linear Equality and Inequality Constraints, it will also require some kind of iterative solver (As there is no closed form solution for LS with Inequality Constraints). Since we have the projection onto the Symmetric and PSD Matrices we can use it. See my answer. Still, nice idea of yours! Especially with the Diagonally Dominant Matrix. I might post a question and answer about this.\n\u2013\u00a0Royi\nApr 12 '20 at 0:00\n\nThis is a convex optimization problem which can be easily formulated, and then numerically solved via a convex optimization tool, such as CVX, YALMIP, CVXPY, CVXR. This is a linear Semidefinite Programming problem (SDP), for which numerical solvers exist.\n\nHere is the code for CVX.\n\nAssume $$x_i$$ is the ith column of matrix $$X$$\n\ncvx_begin\nvariable W(p,p) semidefinite % constrains W to be symmetric positive semidefinite\nObjective = 0;\nfor i=1:n\nObjective = Objective + square(X(:,i)'*W*X(:,i) - y(i))\nend\nminimize(Objective)\ncvx_end\n\n\nCVX will transform the problem into the form required by the solver, call the solver, and return the solution.\n\n\u2022 I think the use of square_pos() here is wrong. It will lead to the wrong solution. You can use square() or if you want to be sure square_abs() or something like that.\n\u2013\u00a0Royi\nApr 11 '20 at 23:35\n\u2022 @Royi Right you are. Now fixed. Thanks for pointing it out. Do you want to take over for me as moderator at CVX Forum? Apr 11 '20 at 23:43\n\u2022 Care to explain what you meant? Thanks...\n\u2013\u00a0Royi\nApr 18 '20 at 22:43\n\u2022 @Royi You can take over answering many of the questions there, and I can go into semi-retirement from that forum. Apr 18 '20 at 23:02\n\u2022 I would answer you with a PM on CVX Forum. Yet the PM feature is disabled. Want to move it to an email?\n\u2013\u00a0Royi\nApr 22 '20 at 15:08\n\nThis does seem to be a straight forward 'nonnegative' least squares though with the sdp constraint. n = 10; p = 5; X = zeros(n,p^2); for ii = 1:n x = randn(p,1); temp = xx'; X(ii,:) = temp(:)'; end y = randn(n,1); cvx_begin sdp variable W(p,p) semidefinite minimize(norm(XW(:)-y)) cvx_end", "date": "2021-09-29 02:48:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3619669/variation-of-least-squares-with-symmetric-positive-semi-definite-psd-constrain", "openwebmath_score": 0.8040797114372253, "openwebmath_perplexity": 443.85929961810206, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713835553862, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8678241541093267}}
{"url": "https://math.stackexchange.com/questions/392822/deduce-that-there-exists-a-prime-p-where-p-divides-x2-2-and-p%E2%89%A13-mod-4", "text": "# Deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p\u22613$ (mod 4)\n\nI am revising for a number theory exam and have a question that I am struggling with, any help would be greatly appreciated.\n\nFirst I am asked to show that for an odd number $x$, $x^2+2 \u22613$(mod 4).\n\nI can do this part of the question, but next I am asked to deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p\u22613$ (mod 4)\n\nI am struggling to see how to attempt the second part and how the first part relates.\n\nMy thoughts so far are that I want to show $x^2\u2261-2$(mod p) ? And perhaps Fermat's Little Theorem could be of use here somehow?\n\nNot sure if I'm barking up the wrong tree though.\n\nConsider the prime factorization of $x^2+2$. Since $x$ is odd, $x^2+2$ is odd implying $2$ will not show up in the factorization. Now consider the primes that DO show up in the prime factorization. If they are all $1$ in modulo 4, then their product will also be one in modulo $4$. This is not true though, since you know that $x^2+2$ is $3$ modulo $4$. Therefore, there must be a prime that in the prime factorization of $x^2+2$ s.t. it is not $1$ modulo 4. Since primes other than 2 that are not $1$ modulo 4 are $3$ modulo $4$, this completes the reasoning.\n\n\u2022 Fantastic! Thank you so much \u2013\u00a0Olivia77989 May 15 '13 at 20:53\n\nOur number $x^2+2$ is odd, and greater than $1$, so it is a product of odd not necessarily distinct primes. If all these primes were congruent to $1$ modulo $4$, their product would be congruent to $1$ modulo $4$. But you have shown that $x^2+2\\equiv 3\\pmod{4}$.\n\n\u2022 You are welcome. You probably already ran into the fact that every positive integer of the form $4k+3$ has a prime divisor of the form $4k+3$. That fact plays a key role in the standard proof that there are infinitely many primes of the form $4k+3$. \u2013\u00a0Andr\u00e9 Nicolas May 15 '13 at 20:59\n\u2022 A useful insight! So would I be right in thinking that in order to prove there are infinitely any primes of the form $4k +3$ , I would split $4k +3$ into its prime divisors $p1,p2...pn$ noting that at least one of these divisors is of the form $4k +3$, then, following a similar approach to Euclid's proof, consider N=4(p1. p2.....pn)+3 ? Or what would we equate N to and why? \u2013\u00a0Olivia77989 May 15 '13 at 21:31\n\u2022 Yes, suppose there are finitely many primes $p_1,\\dots,p_s$ of the right shape. Consider $N=4p_1\\cdots p_s-1$. This is of the shape $4k+3$, so has a prime divisor of that shape, which must be different from the $p_i$. I am avoiding $N=4p_1\\cdots p_s+3$, since maybe $N$ could be a power of $3$ (there are other ways around that issue). \u2013\u00a0Andr\u00e9 Nicolas May 15 '13 at 21:38\n\u2022 I see. Possibly a silly question, but how do we know that the prime divisor of N in the form $4k+3$ is different from the $p_i$ already in our list? \u2013\u00a0Olivia77989 May 15 '13 at 21:46\n\u2022 With $N=4p_1\\cdots p_s -1$, if it was in list, it would divide $4p_1\\cdots p_s$. Since it divides $4p_1\\cdots p_s-1$, it would divide the difference, which is $1$. Impossible! \u2013\u00a0Andr\u00e9 Nicolas May 15 '13 at 21:56\n\nThe question, as asked, has been answered. Nevertheless, I'll show how quadratic residues can help in problems like these, and show that moreover $x^2+2$ ($x$ odd) has a prime factor congruent to $3 \\!\\!\\!\\mod 8$:\n\nSuppose that $x$ is odd. If $p$ divides $x^2+2$, then $p$ is odd and $x^2 \\equiv -2 \\!\\!\\mod p$. In particular, $-2$ is a square mod $p$, hence $$1=\\left(\\frac{-2}{p}\\right)=\\left(\\frac{-1}{p}\\right)\\left(\\frac{2}{p}\\right)=(-1)^{(p^2-1)/2} (-1)^{(p-1)/2}=(-1)^{(p^2+p-2)/2}.$$ This implies $p \\equiv 1,3 \\!\\!\\mod \\!8$. If each prime $p$ dividing $x^2+2$ were congruent to $1 \\!\\!\\mod 8$, then $x^2+2$ would be congruent to $1\\!\\! \\mod 8$. This is a contradiction, hence there exists some $p \\mid x^2+2$ congruent to $3 \\!\\!\\mod 8$. (In fact, there must be an odd number of such primes.)\n\nOf course, any one of the primes is congruent to $3 \\!\\!\\mod 4$, which gives your result.\n\nWe can ignore quadratic residues in your particular problem because there are only two odd residues mod $4$. In more difficult questions, looking at quadratic residues can give more information.", "date": "2019-07-23 03:24:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/392822/deduce-that-there-exists-a-prime-p-where-p-divides-x2-2-and-p%E2%89%A13-mod-4", "openwebmath_score": 0.9006797075271606, "openwebmath_perplexity": 116.63785147958768, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429629196684, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8677292342824688}}
{"url": "https://math.stackexchange.com/questions/2398670/why-are-the-last-two-digits-of-a-perfect-square-never-both-odd/2398674", "text": "# Why are the last two digits of a perfect square never both odd?\n\nEarlier today, I took a test with a question related to the last two digits of perfect squares.\n\nI wrote out all of these digits pairs up to $20^2$.\n\nI noticed an interesting property, and when I got home I wrote a script to test it. Sure enough, my program failed before it was able to find a square where the last two digits are both odd.\n\nWhy is this?\n\nIs this always true, or is the rule broken at incredibly large values?\n\n\u2022 do you know about modular arithmetic ? that might be a starting place. \u2013\u00a0user451844 Aug 19 '17 at 0:51\n\u2022 The last two digit of a number is the number modulus $100$ Talking about squares the last two digits are cyclical. They repeat every 50 squares from $0$ to $49$ or from $10^9 + 2017$ to $10^9 + 2017 + 49$ they are always the following $00,\\;01,\\;04,\\;09,\\;16,\\;25,\\;36,\\;49,\\;64,\\;81,\\;00,\\;21,\\;44,\\;69,\\;96,\\;25,\\;56,\\;89,\\;24,\\;61,\\;00,\\;41,\\;84,\\;29,\\;76,\\;25,\\;76,\\;29,\\;84,\\;41,\\;00,\\;61,\\;24,\\;89,\\;56,\\;25,\\;96,\\;69,\\;44,\\;21,\\;00,\\;81,\\;64,\\;49,\\;36,\\;25,\\;16,\\;09,\\;04,\\;01$ and there is never a combination of two odd digits. \u2013\u00a0Raffaele Aug 19 '17 at 14:39\n\u2022 Not only, but if a number does not end with one of the following pair of digits it cannot be a perfect square $00,\\; 01,\\; 04,\\; 09,\\; 16,\\; 21,\\; 24,\\; 25,\\; 29,\\; 36,\\; 41,\\; 44,\\; \\\\ 49,\\; 56,\\; 61,\\; 64,\\; 69,\\; 76,\\; 81,\\; 84,\\; 89,\\; 96$ \u2013\u00a0Raffaele Aug 19 '17 at 14:39\n\u2022 You can also note that the last two digits of squares from 0 to 25 are the same as from 50 to 25 so it is both cyclical and symetrical :) \u2013\u00a0Rafalon Aug 20 '17 at 9:21\n\u2022 @Raffaele Sadly if you'd made that into an answer it probably would have earned you a decent chunk of rep and might have been the accepted answer. \u2013\u00a0Pharap Aug 20 '17 at 16:24\n\nTaking the last two digits of a number is equivalent to taking the number $\\bmod 100$. You can write a large number as $100a+10b+c$ where $b$ and $c$ are the last two digits and $a$ is everything else. Then $(100a+10b+c)^2=10000a^2+2000ab+200ac+100b^2+20bc+c^2$. The first four terms all have a factor $100$ and cannot contribute to the last two digits of the square. The term $20bc$ can only contribute an even number to the tens place, so cannot change the result. To have the last digit of the square odd we must have $c$ odd. We then only have to look at the squares of the odd digits to see if we can find one that squares to two odd digits. If we check the five of them, none do and we are done.\n\n\u2022 Ah, it's so obvious in hindsight. Although, I suppose most problems like this are. Thanks! \u2013\u00a0Pavel Aug 19 '17 at 1:00\n\nOthers have commented on the trial method. Just to note that $3^2$ in base $8$ is $11_8$ which has two odd digits. This is an example to show that the observation here is not a trivial one.\n\nBut we can also note that $(2m+1)^2=8\\cdot \\frac {m(m+1)}2+1=8n+1$ so an odd square leaves remainder $1$ when divided by $8$.\n\nThe final odd digits of squares can be $1,5,9$ so odd squares are $10p+4r+1$ with $r=0,1,2$. $10p+4r$ must be divisible by $8$ and hence by $4$, so $p$ must be even.\n\nIn the spirit of experimentation, the last two digits of the squares of numbers obtained by adding the column header to the row header:\n\n$$\\begin {array}{c|ccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\\\ \\hline 0 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\\\ 10 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\\\ 20 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\\\ 30 & 00 & 61 & 24 & 89 & 56 & 25 & 96 & 69 & 44 & 21\\\\ 40 & 00 & 81 & 64 & 49 & 36 & 25 & 16 & 09 & 04 & 01\\\\ 50 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\\\ 60 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\\\ 70 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\\\ 80 & 00 & 61 & 24 & 89 & 56 & 25 & 96 & 69 & 44 & 21\\\\ 90 & 00 & 81 & 64 & 49 & 36 & 25 & 16 & 09 & 04 & 01\\\\ 100 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\\\ 110 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\\\ 120 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\\\ \\end{array}$$\n\nThe patterns are clear, after which the search for a reason for such patterns is well given by the answer of @RossMillikan - you can see that the parity of both final digits of the square is entirely dependent on the final digit of the number that you square.\n\nAs a hint, consider what determines the last two digits of a multiplication. Do you remember doing multiplication by hand? If you have square a ten digit number, do all the digits matter when considering just the last two digits of the answer? You will realize that you can put a bound on the number of squares you need to check before you can prove the assertion you are making for all n\n\n\u2022 In fact more than enough values of n has been checked (within $20^2$) to see that this result is true. Question is, how does one know enough values have been checked? \u2013\u00a0Jihoon Kang Aug 19 '17 at 1:02\n\u2022 That's right - a lazy bound would be checking with a computer all two digit squares, as you know that in a 3 digit number, the hundreds digit will not impact on the final two digits in the multiplication (the same for larger numbers). Obviously you can get sharper than that, as the other answer showed. I was just pointing out that you can very easily come up with a lazy bound by thinking about what happens when you multiply numbers. \u2013\u00a0Franz Aug 19 '17 at 1:23\n\nThis is just another version of Ross Millikan's answer.\n\nLet $N \\equiv 10x+n \\pmod{100}$ where n is an odd digit.\n\n\\begin{align} (10x + 1)^2 \\equiv 10(2x+0)+1 \\pmod{100} \\\\ (10x + 3)^2 \\equiv 10(6x+0)+9 \\pmod{100} \\\\ (10x + 5)^2 \\equiv 10(0x+2)+5 \\pmod{100} \\\\ (10x + 7)^2 \\equiv 10(4x+4)+9 \\pmod{100} \\\\ (10x + 9)^2 \\equiv 10(8x+8)+1 \\pmod{100} \\\\ \\end{align}\n\nA simple explanation.\n\n1. Squaring means multiplication, multiplication means repeatative additions.\n\n2. Now if you add even no.s for odd no. of times or odd no.s for even no. of times you will always get an even no.\n\nHence, square of all the even no.s are even, means the last digit is always even.\n\n1. If you add odd no.s for odd no. of times you will always get an odd no.\n\nComing to the squares of odd no.s whose results are >= 2 digits. Starting from 5^2 = 25, break it as 5+5+5+5+5, we have a group with even no. of 5 and one extra 5. According to my point no. 2 the even group will always give you a even no. i.e. 20, means the last digit is always even. Addition of another 5 with 20 makes it 25, 2 is even.\n\nTaking 7^2, 7+7+7+7+7+7+7, group of six 7's = 42 plus another 7 = 49.\n\nNow consider 9^2, 9+9+9+9+9+9+9+9+9, group of eight 9's = 72 plus another 9 = 81, (72+9 gets a carry of 1 making the 2nd last digit even)\n\n35^2 = group of twenty four 34's (1190) plus 35 = 1225, carry comes.\n\nIn short just check the last digit of no. that you can think of in the no. co-ordinate (Real and Imaginary) it will always be b/w 0-9 so the basic principle (point 2 and 3) will never change. Either the last digit will be an even or the 2nd last digit will become even with a carry. So the 1 digit sq can come odd, 1 and 9, as there is no carry. I have kept it as an exception in point 3.\n\nBTW many, including the author may not like my lengthy explanation as mine is not a mathematical one, full of tough formulae. Sorry for that. I'm not from mathematical background and never like maths.\n\n\u2022 \"35^2 = group of twenty four 34's (1190) plus 35 = 1225, carry comes.\"? One of the burdens of being a mathematician is to present your arguments and then deal with any criticism that follows. It's got nothing to do with you. Whoever you are, if you publish a turkey, someones going to roast it. \u2013\u00a0steven gregory Aug 20 '17 at 23:13\n\u2022 That's a typo. That will be odd+odd = even. I saw that error but didn't edit it. \u2013\u00a0NewBee Aug 23 '17 at 14:26\n\nLet b be last digit of odd perfect square a,then b can be 1,9 or 5. For b=1,9; $a^2-b$ is divisible by 4, $(a^2-b)/10$ is even. For b=5 ;a always ends in 25.", "date": "2019-07-15 22:04:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2398670/why-are-the-last-two-digits-of-a-perfect-square-never-both-odd/2398674", "openwebmath_score": 0.8338482975959778, "openwebmath_perplexity": 377.2075710795175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105342148366, "lm_q2_score": 0.8887587934924569, "lm_q1q2_score": 0.8677045724627542}}
{"url": "https://math.stackexchange.com/questions/2720518/a-curious-property-of-an-acute-triangle", "text": "# A curious property of an acute triangle\n\nMany years back in high school I happened to stumble upon the following property that seems to hold for any acute triangle:\n\n$CD$ and $BE$ are altitudes, the arcs are semicircles with diameters $AB$ and $AC$ respectively.\n\nThe property is that $AF = AG$\n\nProof:\n\nLet $H$ be the midpoint of $AB$ (and the centre of the respective semicircle) $$AG^2 = AD^2 + GD^2 = \\left(AC\\cdot \\cos\\angle A\\right)^2 + GD^2$$ Since $HG = AH = \\frac{AB}{2}$ is the radius of the semicircle $$GD^2 = HG^2 - HD^2 = \\left(\\frac{AB}{2}\\right)^2 - \\left(\\frac{AB}{2} - AC\\cdot\\cos\\angle A\\right)^2 = \\\\ = AB\\cdot AC\\cdot \\cos\\angle A - \\left(AC\\cdot\\cos\\angle A\\right)^2$$\n\nwhich gives $$AG^2 = AB\\cdot AC\\cdot \\cos\\angle A$$\n\nAnalogously ($I$ is the midpoint of $AC$) $$AF^2 = AE^2 + FE^2 = \\left(AB\\cdot \\cos\\angle A\\right)^2 + FE^2$$ $$FE^2 = FI^2 - EI^2 = \\left(\\frac{AC}{2}\\right)^2 - \\left(AB\\cdot \\cos\\angle A - \\frac{AC}{2}\\right)^2 = \\\\ = AC\\cdot AB\\cdot \\cos\\angle A - \\left(AB\\cdot \\cos\\angle A\\right)^2$$ which finally gives $$AF^2 = AC\\cdot AB\\cdot \\cos\\angle A$$\n\nQuestions:\n\n1. Is this a known property?\n2. Is there a better more elegant proof?\n\u2022 You can say something more: The four points where the (full) circles with diameters $\\overline{AB}$ and $\\overline{AC}$ meet the altitudes from $B$ and $C$, lie on a common circle about $A$.\n\u2013\u00a0Blue\nApr 3 '18 at 16:49\n\u2022 @Blue, that's a very, very good one! Thank you. Apr 3 '18 at 21:54\n\nHINT:\n\n$$AF^2 = AE \\cdot AC\\\\ AG^2 = AD \\cdot AB \\\\ AE \\cdot AC = AD \\cdot AB$$\n\n\u2022 what a beautiful proof! Apr 3 '18 at 16:42\n\u2022 Brilliant indeed. Although not so trivial (to me at least). Thanks a lot. Apr 3 '18 at 21:41\n\u2022 @ageorge: My pleasure! A nice problem indeed! Apr 3 '18 at 22:32\n\nThis is a possibly similar proof, but here goes anyway...\n\nLet the triangle be labelled in the usual way, so $AC=b$ and $AB=c$, and let angle $FAC=\\theta$\n\nThen $$AF=b\\cos\\theta$$ $$\\implies FE=AF\\sin\\theta=b\\sin\\theta\\cos\\theta$$ $$\\implies EC=FE\\tan\\theta=b\\sin^2\\theta$$ But $$EC=b-c\\cos A=b\\sin^2\\theta$$ $$\\implies b\\cos^2\\theta=c\\cos A$$ $$\\implies AF=\\sqrt{bc \\cos A}$$\n\nTo get $AG$ we only need to exchange $b$ and $c$, so the result follows.", "date": "2021-09-29 02:11:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2720518/a-curious-property-of-an-acute-triangle", "openwebmath_score": 0.8409297466278076, "openwebmath_perplexity": 325.57351905300027, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.976310525254243, "lm_q2_score": 0.8887588008585925, "lm_q1q2_score": 0.8677045716905836}}
{"url": "http://mathhelpforum.com/algebra/121572-trouble-simplifying-factorising.html", "text": "# Math Help - Trouble Simplifying/Factorising\n\n1. ## Trouble Simplifying/Factorising\n\nI have a rather straightforward Physics problem to solve, only I can't for the life of me simplify the answer to what it needs to be. What am I doing wrong?\n\n$\nv_1=\\frac{m}{m+M}v_o\n$\n\n$\\frac{1}{2}mv_o^2=\\frac{1}{2}mv_1^2+\\frac{1}{2}Mv_ 1^2+\\frac{1}{2}kd^2$\n\nThe answer is in the form $d=$ and it's a simplification of the information I provided here in the post.\n\nI get as far as $d=\\sqrt{\\frac{1}{k}(mv_o^2-(m+M)v_1^2)}$\n\nThe book plugs in $v_1$\n\nand simplifies to $d=\\sqrt{\\frac{mM}{k(m+M)}}\\times{v_o}$\n\nHow?\n\n2. Originally Posted by dkaksl\n$\nv_1=\\frac{m}{m+M}v_o\n$\n\n$\\frac{1}{2}mv_o^2=\\frac{1}{2}mv_1^2+\\frac{1}{2}Mv_ 1^2+\\frac{1}{2}kd^2$\nChanging your v1 to v and v0 to w, then your 2 equations are:\n\nv = mw / (m + M) [1]\n\nmw^2 = mv^2 + Mv^2 + kd^2 [2]\n\nSimplifying:\nm + M = mw / v [1]\nmw^2 - kd^2 = v^2(m + M) [2]\n\nSubstitute [1] in [2] ; OK?\n\n3. Very useful, easy-to-understand solution. Thanks a lot.\n\n4. Hello again.\n\nJust got to checking the textbook and unfortunately the question was to define $d$ in the terms $m, M, k$ and $v_o$.\n\nCancelling out $(m + M)$ got me an answer with $v_1$.\n\nEdit:\n\n$kd^2=mv_o^2-(m+M)v_1^2$\n\nif $v_1=\\frac{mv_o}{m+M}$ then what is $v_1^2$?\n\n$\\frac{mv_o}{m+M}\\times\\frac{mv_o}{m+M}$ which is\n\n$\\frac{m^2v_o^2}{(m+M)^2}$\n\n$kd^2=mv_o^2-(m+M)\\times\\frac{m^2v_o^2}{(m+M)^2}$\n\n$kd^2=mv_o^2-\\frac{m^2v_o^2}{(m+M)}$\n\n$kd^2=\\frac{mv_o^2(m+M)}{(m+M)}-\\frac{m^2v_o^2}{(m+M)}$\n\n$kd^2=\\frac{mv_o^2(m+M)-m^2v_o^2}{(m+M)}$\n\n$kd^2=\\frac{m^2v_o^2+mMv_o^2-m^2v_o^2}{(m+M)}=\\frac{mM}{(m+M)}\\times{v_o^2}$\n\n$d=\\sqrt{\\frac{1}{k}\\times{\\frac{mM}{(m+M)}}}\\times {v_o}$\n\nLast edit: Thanks a lot. I see where I made a mistake in expanding. Learned a lot today. Merry Christmas.\n\n5. Originally Posted by dkaksl\nJust got to checking the textbook and unfortunately the question was to define $d$ in the terms $m, M, k$ and $v_o$.\nAhhh; well then, we need to get rid of your v1 (my v); the 2 equations:\n\nv = mw / (m + M) [1]\n\nmw^2 = mv^2 + Mv^2 + kd^2 [2]\n\nsquare [1]: v^2 = m^2w^2 / (m + M)^2\nrearrange [2]: v^2 = (mw^2 - kd^2) / (m + M)\n\nm^2w^2 / (m + M)^2 = (mw^2 - kd^2) / (m + M)\n\nm^2w^2 / (m + M) = mw^2 - kd^2\n\nm^2w^2 = mw^2(m + M) - kd^2(m + M)\n\nkd^2(m + M) = mw^2(m + M) - m^2w^2\n\nkd^2(m + M) = mw^2(m + M - m)\n\nkd^2(m + M) = mMw^2\n\nd^2 = mMw^2 / (k(m + M))\n\nd = wSQRT[mM / (k(m + M))] ........Merry Xmas!", "date": "2014-10-23 01:55:01", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/121572-trouble-simplifying-factorising.html", "openwebmath_score": 0.37264329195022583, "openwebmath_perplexity": 3482.228680406663, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795079712151, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.867699833092837}}
{"url": "http://math.stackexchange.com/questions/199671/what-is-lim-limits-n-to-infty-fracnd-nd-choose-d", "text": "# What is $\\lim\\limits_{n\\to\\infty} \\frac{n^d}{ {n+d \\choose d} }$?\n\nWhat is $\\lim\\limits_{n\\to\\infty} \\frac{n^d}{ {n+d \\choose d} }$ in terms of $d$? Does the limit exist? Is there a simple upper bound interms of $d$?\n\n-\nI think the limit sould exist and is a function of $d$. Because the denominator is $\\sim n^d$ when $n\\rightarrow \\infty$ \u2013\u00a0Seyhmus G\u00fcng\u00f6ren Sep 20 '12 at 11:59\n\nWrite $$\\frac{n^d}{(n+d)!}d!n!=d!\\prod_{j=1}^d\\frac{n}{n+j}$$ to see that the expected limit is $d!$.\n\n-\nYou mean \"... is $d!$\" instead of $n!$. \u2013\u00a0martini Sep 20 '12 at 12:00\nI think there is a little mistake here. \u2013\u00a0Seyhmus G\u00fcng\u00f6ren Sep 20 '12 at 12:02\nThe denominator in the product should be $n+j$. \u2013\u00a0celtschk Sep 20 '12 at 12:15\nThanks guys, I've corrected. \u2013\u00a0Davide Giraudo Sep 20 '12 at 12:26\n\n$$n^d\\frac{n!\\,d\\,!}{(n+d)!}=d\\,!\\,(\\frac{n}{n+1})(\\frac{n}{n+2})\\ldots(\\frac{n}{n+d})$$\n\nNote that the R.H.S. contains a finite number of terms ($=d$) each tending to 1.\n\nHence the limit is $d\\,!$\n\nTo find an upper bound you can simply do this\n\n$d\\,!\\,(\\frac{n}{n+1})(\\frac{n}{n+2})\\ldots(\\frac{n}{n+d})<d\\,!\\,(\\frac{n}{n})(\\frac{n}{n})\\ldots(\\frac{n}{n})$\n\n$d\\,!\\,(\\frac{n}{n+1})(\\frac{n}{n+2})\\ldots(\\frac{n}{n+d})<d\\,!$\n\nSo your upper bound is $d\\,!$\n\nAnother Method (Just to show that it the sequence is convergent)\n\nLet us treat the given term $n^d\\frac{n!\\,d\\,!}{(n+d)!}$ as term of sequces $\\{x_n\\}$.\n\nSince we have shown that the sequence is bounded, it will suffice to show that the sequence is monotonically increasing.\n\nWe will now show that it is monotonically increasing.\n\n$$\\frac{n}{n+r}-\\frac{m}{m+r}=\\frac{r(n-m)}{(n+r)(m+r)}>0 \\quad\\forall\\; n>m>0\\quad \\& \\quad\\forall r>0$$\n\nLet $n>m$\n\nThen $n!>m!$ and each term $\\frac{n}{n+r}>\\frac{m}{m+r} \\quad \\forall r \\in \\{1, 2, \\ldots ,d\\}$, we have proved this above.\n\nHecne $x_n \\gt m_m$ (as $a>a' \\quad \\& \\quad b\\gt b'\\implies a.b>a'.b' \\quad \\forall a,a',b,b'>0$ )\n\nIf you want to see how a monotonically increasing sequence which is upper bounded is convergent go to this link.\n\n-\n\nYou can use the Stirling approximation $$n! \\sim \\sqrt{2 \\pi n} \\left(\\frac{n}{e}\\right)^n\\,,$$\n\nafter writing your expression in the form\n\n$$d!\\frac{n^d n!}{(n+d)!}$$\n\n$$d!\\frac{n^d n!}{(n+d)!} \\sim d! \\frac{n^d \\sqrt{2 \\pi n} \\left(\\frac{n}{e}\\right)^n }{\\sqrt{2 \\pi (n+d)} \\left(\\frac{n+d}{e}\\right)^{n+d}}= d!\\, {\\rm e}^d \\left(\\frac{n}{n+d}\\right)^{n} \\left(\\frac{n}{n+d}\\right)^{d} \\left(\\frac{n}{n+d}\\right)^{\\frac{1}{2}}$$\n\n$$\\rightarrow d! \\,{\\rm e}^d \\, \\,{\\rm e}^{-d}\\, 1.1 = d!\\,, \\quad n \\rightarrow \\infty$$\n\n-", "date": "2016-06-30 17:42:06", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/199671/what-is-lim-limits-n-to-infty-fracnd-nd-choose-d", "openwebmath_score": 0.9629940390586853, "openwebmath_perplexity": 430.6612367216554, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419703960398, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8676858770944977}}
{"url": "https://math.stackexchange.com/questions/2641629/does-an-n-by-n-hermitian-matrix-always-has-n-independent-eigenvectors", "text": "# Does an n by n Hermitian matrix always has n independent eigenvectors?\n\nI am learning the MIT ocw 18.06 Linear Algebra, and I have learnt: an arbitrary $n\u00d7n$ matrix A with n independent eigenvectors can be written as $A=S\u039bS^{-1}$, and then for the Hermitian matrices, because the eigenvectors can be chosen orthonormal, it can be written as $A=Q\u039bQ^H$ further.\n\nI wonder does every $n\u00d7n$ Hermitian matrix has n independent eigenvectors and why? Thank you!\n\nP.S. MIT 18.06 Linear Algebra Lecture 25: Symmetric Matrices and Positive Definiteness. You may wish to start from 4:20. From the course, I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why?\n\n\u2022 You mean independent eigenvectors, not independent eigenvalues. \u2013\u00a0uniquesolution Feb 8 '18 at 12:04\n\u2022 Oh, thanks, I have edited it. \u2013\u00a0Thomas Feb 9 '18 at 2:28\n\u2022 It's on the Wikipedia page: en.wikipedia.org/wiki/Hermitian_matrix. \u2013\u00a0Robert Wolfe Feb 9 '18 at 4:25\n\u2022 @Robert Thank you for the link, I had read the page though, and there wasn't an answer to my question. \u2013\u00a0Thomas Feb 9 '18 at 10:34\n\u2022 Look here for a simple proof. \u2013\u00a0user Feb 11 '18 at 20:15\n\nThis is a theorem with a name: it is called the Spectral Theorem for Hermitian (or self-adjoint) matrices. As pointed out by Jose' Carlos Santos, it is a special case of the Spectral Theorem for normal matrices, which is just a little bit harder to prove.\n\nActually we can prove the spectral theorem for Hermitian matrices right here in a few lines.\n\nWe are going to have to think about linear operators rather than matrices. If $T$ is a linear operator on a finite dimensional complex inner product space $V$, its adjoint $T^*$ is another linear operator determined by $\\langle T v, w\\rangle = \\langle v, T^* w \\rangle$ for all $v, w \\in V$. (Note this is a basis-free description.) $T$ is called Hermitian or self-adjoint if $T = T^*$.\n\nLet $B$ be an $n$-by-$n$ complex matrix and $B^*$ the conjugate transpose matrix. Let $T_B$ and $T_{B^*}$ be the corresponding linear operators. Then $(T_B)^* = T_{B^*}$, so a a matrix is Hermitian if and only if the corresponding linear operator is Hermitian.\n\nLet $A$ be a Hermitian linear operator on a complex inner product space $V$ of dimension $n$. We need to consider $A$--invariant subspaces of $V$, that is linear subspaces $W$ such that $A W \\subseteq W$. We should think about such a subspace as on an equal footing as our original space $V$. In particular, any such subspace is itself an inner product space, $A_{|W} : W \\to W$ is a linear operator on $W$, and $A_{|W}$ is also Hermitian. If $\\dim W \\ge 1$, $A_{|W}$ has an least one eigenvector $w \\in W$ -- because any linear operator at all acting on a (non-zero) finite dimensional complex vector space has at least one eigenvector.\n\nThe basic phenomenon is this: Let $W$ be any invariant subspace for $A$. Then $W^\\perp$ is also invariant under $A$. The reason is that if $w \\in W$ and $x \\in W^\\perp$, then $$\\langle w, A x\\rangle = \\langle A^* w , x \\rangle = \\langle A w, x \\rangle = 0,$$ because $Aw \\in W$ and $x \\in W^\\perp$. Thus $A x \\in W^\\perp$.\n\nWrite $V = V_1$. Take one eigenvector $v_1$ for $A$ in $V_1$. Then $\\mathbb C v_1$ is $A$--invariant. Hence $V_2 = (\\mathbb C v_1)^\\perp$ is also $A$ invariant. Now just apply the same argument to $V_2$: the restriction of $A$ to $V_2$ has an eigenvector $v_2$ and the perpendicular complement $V_3$ to $\\mathbb C v_2$ in $V_2$ is $A$--invariant. Continuing in this way, one gets a sequence of mutually orthogonal eigenvectors and a decreasing sequence of invariant subpsaces, $V = V_1 \\supset V_2 \\supset V_3 \\dots$ such that $V_k$ has dimension $n - k + 1$. The process will only stop when we get to $V_n$ which has dimension 1.\n\n\u2022 You really hit the nail on the head as I'm afraid my question is too unusual to be understand. May I repeat you answer in my imprecise words? \u2013\u00a0Thomas Feb 9 '18 at 12:42\n\u2022 Regard A as an operator on V, its invariant subspaces contains at least one eigenvector, their orthogonal complement are invariant subspaces too because A is a Hermitian matrix, we start from $\\mathbb C^n$, every time we take a eigenvector from the space, it on a line which is a invariant subspace, and the space left which like a plane perpendicular to the line is also a invariant subspace, and we can take n times (i.e. n orthogonal/independent eigenvectors). Do I understand it correctly? \u2013\u00a0Thomas Feb 9 '18 at 12:47\n\u2022 I still have some trouble with this, what if I take an eigenvector from the last space? As we know \"Let $W$ be any invariant subspace for $A$. Then $W^\u22a5$ is also invariant under $A$\", I take the eigenvector out, then there will be only zero vector left, and I think it really orthogonal to that eigenvector we take, is zero vector an invariant subspace? It may be, but \"invariant space has at least one eigenvector\" would be false. This \"Let $W$ be any invariant subspace for $A$. Then $W^\u22a5$ is also invariant under $A$\" makes me confused. \u2013\u00a0Thomas Feb 9 '18 at 13:43\n\u2022 When you have reached $V_n$, you already have your collection of $n$ mutually orthogonal eigenvectors. You can't go further and you don't need to go further. I corrected the answer at one point to say that a linear operator on a non-zero finite dimensional complex vector space has an eigenvector. \u2013\u00a0fredgoodman Feb 9 '18 at 14:19\n\u2022 It's important in the second paragraph that \"acting on\" means \"mapping it into itself\". \u2013\u00a0Ian Feb 9 '18 at 15:26\n\nYes, it has. That is due to the spectral theorem: every normal $n\\times n$ matrix is diagonalizable. And Hermitian matrices are normal.\n\n\u2022 From my understanding, the factorization $A=Q\u039bQ^H$ (i.e., the spectral theorem) is the special form of $A=S\u039bS^{-1}$ when A is a Hermitian matrix, and $A=S\u039bS^{-1}$ exists under the situation that there is a full set of independent eigenvectors. I wonder, if it is, why a Hermitian matrix is always diagonalizable? \u2013\u00a0Thomas Feb 9 '18 at 2:58\n\u2022 @Thomas Because the spectral theorem says so. \u2013\u00a0Jos\u00e9 Carlos Santos Feb 9 '18 at 6:30\n\u2022 From the course (I add the link into the question now), I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why? \u2013\u00a0Thomas Feb 9 '18 at 10:57\n\u2022 @Thomas Every Heritian matrix is normal and every normal matrix is diagonalizable. \u2013\u00a0Jos\u00e9 Carlos Santos Feb 9 '18 at 11:08\n\u2022 I had not learnt the concept of normal matrix from the course. I search for it now, and I realize that all matrices can be written as $A=QTQ^H$ with triangular T, so $A^HA=AA^H$ would be $T^HT=TT^H$, for a 2 by 2 matrix, it's easy to prove the upper right corner entry of such T is 0, extend it to size n, and we prove that T is diagonal. Is that a right poof? Thank you for your patience. \u2013\u00a0Thomas Feb 9 '18 at 11:58", "date": "2019-09-16 08:58:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2641629/does-an-n-by-n-hermitian-matrix-always-has-n-independent-eigenvectors", "openwebmath_score": 0.8948529958724976, "openwebmath_perplexity": 158.38036117341713, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419671077918, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8676858710408021}}
{"url": "https://math.stackexchange.com/questions/501755/approaching-modular-arithmetic-problems", "text": "# Approaching modular arithmetic problems\n\nI'm a little stumbled on two questions.\n\nHow do I approach a problem like $x*41 \\equiv 1 \\pmod{99}$.\n\nAnd given $2$ modulo, $7x+9y \\equiv 0 \\pmod{31}$ and $2x\u22125y \\equiv 2 \\pmod{31}$ (solve for $x$ only)?\n\nWhen I solve for $x$ for the latter, I got a fraction as the answer and I'm not sure if I can have a fraction as an answer? I'm not sure how to approach the first problem either.\n\n\u2022 What fraction did you get as answer? \u2013\u00a0Git Gud Sep 22 '13 at 22:20\n\u2022 The first problem involves inverses. What is the inverse of 41 mod 99? The second problem is a set of linear equations in $x$ and $y$ modulo 31, which means the two can be added/subtracted/etc. as if they were a set of normal equations looking for a linear solution. \u2013\u00a0abiessu Sep 22 '13 at 22:25\n\nFinding the solution to $$x \\times 41 \\equiv 1 \\pmod {99}$$ is equivalent to asking for the multiplicative inverse of $41$ modulo $99$. Since $\\gcd(99,41)=1$, we know $41$ actually has an inverse, and it can be found using the Extended Euclidean Algorithm:\n\n\\begin{align*} 99-2 \\times 41 &= 17 \\\\ 41-2 \\times 17 &= 7 \\\\ 17-2 \\times 7 &= 3 \\\\ 7-2\\times 3 &= 1 &=\\gcd(99,41). \\\\ \\end{align*} Going back, we see that \\begin{align*} 1 &= 7-2\\times 3 \\\\ &= 7-2\\times (17-2 \\times 7) \\\\ &= 5 \\times 7-2\\times 17 \\\\ &= 5 \\times (41-2 \\times 17)-2\\times 17 \\\\ &= -12 \\times 17+5 \\times 41 \\\\ &= -12 \\times (99-2 \\times 41)+5 \\times 41 \\\\ &= 29 \\times 41-12 \\times 99 \\\\ \\end{align*} Hence $29 \\times 41 \\equiv 1 \\pmod {99}$ and thus $x=29$.\n\nIn the second case, we have $$7x+9y \\equiv 0 \\pmod {31}$$ and $$2x-5y\\equiv 2 \\pmod {31}.$$ Here we want to take $7x+9y=0 \\pmod {31}$ and rearrange it to get $x \\equiv ?? \\pmod {31}$, then substitute it into the other equation and solve for $y$. This requires finding the multiplicative inverse of $7$ modulo $31$ (which we can do as above). It turns out $7 \\times 9 \\equiv 1 \\pmod {31}$. Hence \\begin{align*} & 7x+9y=0 \\pmod {31} \\\\ \\iff & 7x \\equiv -9y \\pmod {31} \\\\ \\iff & x \\equiv -9y \\times 9 \\pmod {31} \\\\ \\iff & x \\equiv 12y \\pmod {31}. \\end{align*} We then substitute this into the equation $2x-5y\\equiv 2 \\pmod {31}$, which implies $$2 \\times 12y-5y \\equiv 2 \\pmod {31}$$ or equivalently $$19y \\equiv 2 \\pmod {31}.$$ Yet again, we find a multiplicative inverse, this time of $19$ modulo $31$, which turns out to be $18$. So \\begin{align*} & 19y \\equiv 2 \\pmod {31} \\\\ \\iff & y \\equiv 2 \\times 18 \\pmod {31} \\\\ \\iff & y \\equiv 5 \\pmod {31}. \\end{align*} Hence $$x \\equiv 12y \\equiv 29 \\pmod {31}.$$ Thus we have the solution $(x,y)=(29,5)$.\n\n\u2022 Are you sure the answer to the first part is x=20? \u2013\u00a0Mufasa Sep 22 '13 at 22:35\n\u2022 Thanks for point that out; it was an arithmetic error (actually, I found the inverse mod 91 by mistake) and it should be fixed now. \u2013\u00a0Rebecca J. Stones Sep 22 '13 at 22:47\n\u2022 You're welcome. BTW: I am learning a lot from your posts - so THANK YOU! :) \u2013\u00a0Mufasa Sep 22 '13 at 22:49\n\nFor the first one, you could approach it as follows:\n$41x=1$ mod $99$\n$140x=1$ mod $99$ (because 41 mod 99 = (41+99) mod 99)\n$140x=100$ mod $99$ (because 1 mod 99 = (1+99) mod 99)\n$7x=5$ mod $99$ (divide both sides by 20)\n$7x=203$ mod $99$ (because 5 mod 99 = (5+99+99) mod 99)\n$x=29$ mod $99$ (divide both sides by 7)\n\nFor the first one, you have to find a multiplicative inverse for $41$ mod $99$. Use Euclid's algorithm to find the solutions of $41\\cdot x + 99 \\cdot y = 1$ to find $x$.\n\nAre you familiar with abstract algebra? If yes, do you know that $\\mathbb{Z}_{31}$ is a field because $31$ is a prime number and you can use tools of linear algebra because it works overy any field?\n\nYou have the following system of equations in $\\mathbb{Z}_{31}$:\n\n$$7x+9y=0$$ $$2x-5y=2$$\n\nYou can use Gaussian elimination in $\\mathbb{Z}_{31}$ to find $x$ and $y$. But you have to be careful that your coefficients are in $\\mathbb{Z}_{31}$ not in $\\mathbb{R}$.\n\nEDIT(suggested by Git Gud):\n\nNotice that if you get a fraction like $\\displaystyle \\frac{a}{b}$ then you should think of it as $a \\cdot b^{-1}$ and then find $b^{-1}$ in $\\mathbb{Z}_{31}$. As previously said, $\\mathbb{Z}_{31}$ is a field, so talking about fractions in it makes sense. The notation $\\frac{a}{b}$ is actually nothing but $a.b^{-1}$ where $.$ denotes the multiplication in the field and $b^{-1}$ denotes the multiplicative inverse of $b$ in the field.\n\n\u2022 @GitGud: OK. Done. \u2013\u00a0user66733 Sep 22 '13 at 22:32", "date": "2019-12-06 18:23:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/501755/approaching-modular-arithmetic-problems", "openwebmath_score": 1.0000098943710327, "openwebmath_perplexity": 220.93903215255224, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534398277178, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8676434000094446}}
{"url": "https://www.physicsforums.com/threads/properties-of-the-absolute-value.246785/", "text": "Properties of the Absolute Value\n\n1. Jul 25, 2008\n\nroam\n\nJust wanted to say hi before I start my post!\n\nAs you may know there is a property of the absolute value that states; for $$a, b \\in R$$;\n\n$$|ab| = |a||b|$$\n\nWell, my friend asked me if I knew a proof for this... but I don't know...\nHow can we prove this statement/property? I know there is a proof for the triangle inequality but for this I really don't know but I'm curious.\n\nI'd appreciate it if anyone could show me any kind of proof or send me some links etc. Thanks!\n\n2. Jul 25, 2008\n\nHallsofIvy\n\nStaff Emeritus\nYou prove that |ab|= |a||b| the same way you prove any such elementary statement: use the definitions.\n\nThe simplest definition of |x| (there are several equivalent definitions) is that |x|= x if x is positive or 0, -x if x is negative.\n\nNow break it into \"cases\":\n\ncase 1: x and y are both positive: |x|= x and |y|= y. xy is also positive so |xy|= xy= |x||y|.\n\ncase 2: x is positive while y is negative: |x|= x and |y|= -y. xy is negative so |xy|= -xy= x(-y)= |x||y|.\n\ncase 3: x is negative while y is positive: |x|= -x and |y|= y. xy is negative so |xy|= -xy= (-x)y= |x||y|.\n\ncase 4: x and y are both negative: |x|= -x and |y|= -y. xy is positive so |xy|= xy= (-x)(-y)= |x||y|.\n\ncase 5: x= 0 and y is positive: |x|= 0 and |y|= y. xy= 0 so |xy|= 0= 0(y)= |x||y|.\n\ncase 6: x= 0 and y is negative: |x|= 0 and |y|= -y. xy= 0 so |xy|= 0= (0)(-y)= |x||y|.\n\ncase 7: x is positive and y is 0: |x|= x and |y|= 0. xy= 0 so |xy|= 0= x(0)= |x||y|.\n\ncase 8: x is negative and y is 0: |x|= -x and |y|= 0. xy= 0 so |xy|= 0 = (-x)(0)= |x||y|.\n\ncase 9: both x and y are 0: |x|= 0 and |y|= 0. xy= 0 so |xy|= 0= (0)(0)= |x||y|.\n\nThere are simpler ways to prove that but I thought this would be conceptually clearest.\n\n3. Jul 25, 2008\n\nsagardipak\n\nFirst, we have to understand that the absolute value is a function defined by:\n\n$$|x| = \\begin{cases} x & \\text{if } x\\geq 0 \\\\ -x & \\text{if } x<0 \\end{cases}$$\n\nSo,\n\n$$|ab| = \\begin{cases} ab & \\text{if } ab\\geq 0 \\\\ -ab & \\text{if } ab<0 \\end{cases}$$\n\nNow, let's see what |a||b| is:\n\n$$|a||b| = \\begin{cases} ab & \\text{if } a\\geq 0 \\wedge b\\geq0 \\\\ (-a)(-b) & \\text{if } a\\leq 0 \\wedge b\\leq0 \\\\ (-a)b & \\text{if } a> 0 \\wedge b<0 \\\\ a(-b) & \\text{if } a<0 \\wedge b<0 \\end{cases} \\Leftrightarrow$$\n\n$$|a||b| = \\begin{cases} ab & \\text{if } a\\geq 0 \\wedge b\\geq0 \\\\ ab & \\text{if } a\\leq 0 \\wedge b\\leq0 \\\\ -ab & \\text{if } a> 0 \\wedge b<0 \\\\ -ab & \\text{if } a<0 \\wedge b<0 \\end{cases} \\Leftrightarrow$$\n\nNotice that you have ab if a and b have the same sign and that you use -ab otherwise.\n\nNow, if a and b have the same sign, $$ab\\geq0$$. If they have opposite signs (and are different than zero), $$ab<0$$.\n\nUsing this,\n\n$$|a||b| = \\begin{cases} ab & \\text{if } ab \\geq 0 \\\\ -ab & \\text{if } ab<0 \\end{cases} = |ab|$$\n\nQuod erat demonstrandum :tongue2:\n\nLast edited: Jul 25, 2008\n4. Jul 28, 2008\n\nquark1005\n\nlet a = mcis(kpi) where m => 0, k is integer\nand b = ncis(hpi) where n => 0, h integer\n(clearly a,b are real)\n\n|a||b|= mn\n|ab|=|mcis(kpi)*ncis(hpi)| = |mncis[pi(k+h)]| = mn\n\nas required\n\n5. Jul 28, 2008\n\nHallsofIvy\n\nStaff Emeritus\nThis would make more sense if you had said that \"mcis(hpi)\" is\n$$m (cos(h\\pi)+ i sin(h\\pi))$$\nThat is much more an \"engineering notation\" than mathematics.\n\nIf you really want to go to complex numbers, why not\nif\n$$x= r_xe^{i\\theta_x}$$\nand\n$$y= r_ye^{i\\theta_y}$$, then\n$$|xy|= |r_xe^{i\\theta_x}r_ye^{i\\theta_y}|= |(r_xr_y)e^{i(\\theta_x+\\theta_y)}|$$\n\nBut for any $z= re^{i\\theta}$, |z|= r, so\n$$|xy|= r_x r_y= |x||y|$$\n\n6. Jan 8, 2011\n\nVincent Mazzo\n\n$$\\Huge |ab|=\\sqrt{(ab)^2}=\\sqrt{a^2b^2}=\\sqrt{a^2}\\sqrt{b^2}=|a||b|$$\n\nand\n\n$$\\large |a+b|=\\sqrt{(a+b)^2}=\\sqrt{a^2+2ab+b^2}\\leq \\sqrt{a^2+|2ab|+b^2}=\\sqrt{(|a|+|b|)^2}=||a|+|b||$$\n\nwhere I utilize the following fact:\n\n|2ab|=|2(ab)|=|2||ab|=2|a||b|\n\n-----\n\u00c0 bient\u00f4t\n?;-D", "date": "2017-03-23 10:27:05", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/properties-of-the-absolute-value.246785/", "openwebmath_score": 0.8403744101524353, "openwebmath_perplexity": 3610.9502987541396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534392852384, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8676433980306277}}
{"url": "https://math.stackexchange.com/questions/1436807/even-functions-and-inflection-points/1436819", "text": "# Even functions and inflection points\n\nCan an even function have an inflection point? If yes, then give an example while if not then give a proof. If needed you may assume that $f$ is two times differentiable .\n\nIntution says that even functions don't have inflection points, but I cannot settle down an appropriate proof.\n\nOn the contrary for an odd function this is possible. For instance $f(x)=x^3$ has an inflection point at $x=0$.\n\n\u2022 Does it have to be at $x=0$? \u2013\u00a0mickep Sep 15 '15 at 18:22\n\u2022 Thank you all for your answers. \u2013\u00a0Tolaso Sep 15 '15 at 18:58\n\n$\\cos\\colon\\mathbb{R}\\to\\mathbb{R}$ is even and it has infinitely many inflection points (at its zeros, which are found at $\\{n\\pi\\}_{n\\in\\mathbb{Z}}$).\nConsider $$f(x)=\\max\\bigl((x-1)^3,-(x+1)^3\\bigr).$$ Or, for a nicer example, consider $$g(x)=(x-1)^3(x+1)^3.$$\nthe function $y=x^4-4x^2$ has at least one inflection point WA plot", "date": "2021-06-20 09:46:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1436807/even-functions-and-inflection-points/1436819", "openwebmath_score": 0.8520205616950989, "openwebmath_perplexity": 286.57047015503224, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534338604443, "lm_q2_score": 0.8840392848011834, "lm_q1q2_score": 0.8676433917356527}}
{"url": "https://web2.0calc.com/questions/geometry-question_64", "text": "+0\n\n# Geometry question\n\n+1\n125\n11\n\nIn triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed. The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S. Find RS.\n\nJan 16, 2020\n\n#1\n-1\n\nRS = 12.\n\nJan 17, 2020\n#2\n+107414\n+2\n\nTHERE WAS AN ERROR SO I DELETED THE PICS.\n\nJan 17, 2020\nedited by Melody \u00a0Jan 17, 2020\nedited by Melody \u00a0Jan 17, 2020\n#3\n+28449\n+2\n\nDeleted as for some reason I used entirely the wrong lengths of triangle legs!\n\nAlan \u00a0Jan 17, 2020\nedited by Alan \u00a0Jan 17, 2020\nedited by Alan \u00a0Jan 17, 2020\n#5\n+28449\n+1\n\nI now think Guest #4 below is correct.\u00a0 There is a unique answer.\u00a0 Because that answer is independent of the shape of the triangle it is legitimate to choose an easy one (a right-angled one) on which to do the calculation:\n\nRS is indeed 16.\n\nAlan \u00a0Jan 17, 2020\n#4\n+1\n\nI think that the answer is unique, and I think that it is 16.\n\nUsing Guest's diagram, the triangles QPR and QSP are similar.\n\nThe angle between QP and the tangent, call it alpha, is equal to the angle QPS and also the angle QRP.\n\nSimilarly the angle between QR and the tangent is equal to the angle QSP\u00a0and also the angle QPR.\n\nSo, QP/QS = QR/QP,\n\nQS = QP.QP/QR =225/25 = 9, so SR = 16.\n\nThat fits the approximate values that Melody found by drawing.\n\nAlan seems to chosen the particular case where angle PQR is a right angle.\n\nUsing the angles I called alpha earlier, from the triangle PQR, tan(alpha) = 15/25 = 3/5.\n\nFrom the triangle PQS, tan(alpha) = QS/15, so QS = 45/5 = 9, as above.\n\nJan 17, 2020\n#6\n+107414\n+1\n\nYes you are right guest.\n\nI fixed an error in mine and now it always stays at 16.\n\nI wss see if I can work out how to give a proper link to what i have created.\n\nhttps://www.geogebra.org/classic/ekztcz7y\n\nMelody \u00a0Jan 17, 2020\n#7\n+107414\n0\n\nThanks Guest,\n\nI am just trying to make sense of your answer.\n\nWhy is\u00a0 \u00a0$$\\angle QRP = \\alpha$$\u00a0 \u00a0 \u00a0 \u00a0 ?\n\nMelody \u00a0Jan 17, 2020\n#9\n+1\n\nLove the geogebra Melody.\n\nIn answer to your question, it's a standard angle property of a circle.\n\nTake a diameter from Q across the circle to T on the other side.\n\nAngle QPT will be right angle,since it's based on the diameter, and then angle PTQ will equal the angle between PQ and the tangent, ( subtract from 90 twice).\n\nThen, all angles based on the same arc, PQ, are equal etc.\n\nGuest\u00a0Jan 17, 2020\n#11\n+107414\n0\n\nThanks guest for the complement as well as for the explanation.\n\nI am all clear now.\n\nThanks Heureka and Alan as well\n\nMelody \u00a0Jan 18, 2020\n#8\n+24031\n+2\n\nIn triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.\n\nThe tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.\n\nFind RS.\n\n$$\\text{Let C is the center of the circle } \\\\ \\text{Let PS=TS=h } \\\\ \\text{Let RS=x } \\\\ \\text{Let QS=25-x }$$\n\n$$\\begin{array}{|lrcll|} \\hline 1. & \\left(25-x\\right)^2+h^2 &=& 15^2 \\qquad \\text{or}\\qquad \\mathbf{ h^2 = 15^2-\\left(25-x\\right)^2} \\\\\\\\ 2. & x(25-x) &=& h * h \\qquad \\text{Intersecting chords theorem} \\\\ & x(25-x) &=& h^2 \\\\ & x(25-x) &=& 15^2-\\left(25-x\\right)^2 \\\\ & 25x-x^2 &=& 15^2-( 25^2-50x+x^2) \\\\ & 25x-x^2 &=& 15^2-25^2+50x-x^2 \\\\ & 25x &=& 15^2-25^2+50x \\\\ & 25^2-15^2 &=& 25x \\\\ & 25x &=& 25^2-15^2 \\\\ & 25x &=& 400 \\\\ & \\mathbf{x} &=& \\mathbf{16} \\\\ \\hline \\end{array}$$\n\nJan 17, 2020\n#10\n+24031\n+1\n\nIn triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.\nThe tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.\nFind RS\n\nGeneral solution:\n\n$$\\text{Let Center is the center of the circle }\\\\ \\text{Let P-Center =T-Center=v }\\\\ \\text{Let S-Center =w }\\\\ \\text{Let T-Center =w+k=v\\qquad k=v-w }\\\\ \\text{Let ST = k =v-w }\\\\ \\text{Let PS = v+w }\\\\ \\text{Let RS = x }\\\\ \\text{Let QS = 25-x }\\\\ \\text{Let Q-Center = u  }$$\n\nIntersecting chords theorem:\n\n$$\\begin{array}{|rcll|} \\hline RS*QS &=& ST*PS \\\\ x*(25-x) &=& (v-w)*(v+w) \\\\ \\mathbf{x*(25-x)} &=& \\mathbf{v^2-w^2} & (1) \\\\ \\hline \\end{array}$$\n\nPythagoras:\n\n$$\\begin{array}{|lrcll|} \\hline & v^2+u^2 &=& 15^2 & (2) \\\\ & w^2+u^2 &=& (25-x)^2 & (3) \\\\ \\hline (2)-(3): & v^2 -w^2 &=& 15^2-(25-x)^2 \\quad | \\quad \\mathbf{x*(25-x)=v^2-w^2} \\\\ & x*(25-x) &=& 15^2-(25-x)^2 \\\\ &25x-x^2 &=& 15^2 -(25^2-50x+x^2) \\\\ &25x-x^2 &=& 15^2 -25^2+50x-x^2 \\\\ &25x &=& 15^2 -25^2+50x \\\\ &25^2-15^2 &=& 25x \\\\ & 25x &=& 25^2-15^2 \\\\ & 25x &=& 400 \\quad |\\quad :25 \\\\ & \\mathbf{x} &=& \\mathbf{16} \\\\ \\hline \\end{array}$$\n\nJan 17, 2020", "date": "2020-02-16 18:41:17", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/geometry-question_64", "openwebmath_score": 0.7489926218986511, "openwebmath_perplexity": 3255.6565444447433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534398277177, "lm_q2_score": 0.8840392786908831, "lm_q1q2_score": 0.8676433910139816}}
{"url": "https://web2.0calc.com/questions/3-00-x-10-8-6-50-x-10-7-the-result-is-4-61-x-10-14-my-question-is-how-10-8-10-7-10-14-thank-u", "text": "+0\n\n# 3.00 x 10^8 / 6.50 x 10^-7 the result is 4.61 x 10^14 my question is how 10^8 / 10^-7 = 10^14? thank u\n\n0\n488\n2\n\n3.00 x 10^8 / 6.50 x 10^-7\n\n4.61 x 10^14\n\nmy question is how 10^8 / 10^-7 = 10^14?\n\nthank u\n\nGuest\u00a0Nov 21, 2014\n\n#1\n+92775\n+5\n\n3.00 x 10^8 / 6.50 x 10^-7\n\n4.61 x 10^14\n\nmy question is how 10^8 / 10^-7 = 10^14? \u00a0 \u00a0 \u00a0IT DOES NOT!\n\n$$\\\\(3\\times 10^8)/(6.5\\times 10^{-7})\\\\\\\\ =(3/6.5)\\times 10^8/10^{-7}\\\\\\\\ =0.4615\\times 10^{8--7}\\\\\\\\ =0.4615\\times 10^{15}\\\\\\\\$$\n\nBUT this answer is to be in scientific notation so there mus be just one non-zero number in front of the point\n\n$$\\\\=4.4615\\times 10^{-1} \\times 10^{15}\\\\\\\\ =4.4615\\times 10^{14}\\qquad \\mbox{Correct to 4 significant figures}\\\\\\\\$$\n\nSo does that all make sense now?\n\nMelody \u00a0Nov 21, 2014\n#1\n+92775\n+5\n\n3.00 x 10^8 / 6.50 x 10^-7\n\n4.61 x 10^14\n\nmy question is how 10^8 / 10^-7 = 10^14? \u00a0 \u00a0 \u00a0IT DOES NOT!\n\n$$\\\\(3\\times 10^8)/(6.5\\times 10^{-7})\\\\\\\\ =(3/6.5)\\times 10^8/10^{-7}\\\\\\\\ =0.4615\\times 10^{8--7}\\\\\\\\ =0.4615\\times 10^{15}\\\\\\\\$$\n\nBUT this answer is to be in scientific notation so there mus be just one non-zero number in front of the point\n\n$$\\\\=4.4615\\times 10^{-1} \\times 10^{15}\\\\\\\\ =4.4615\\times 10^{14}\\qquad \\mbox{Correct to 4 significant figures}\\\\\\\\$$\n\nSo does that all make sense now?\n\nMelody \u00a0Nov 21, 2014\n#2\n+92775\n0\n\nThat was a good question anon\n\nMelody \u00a0Nov 21, 2014", "date": "2018-07-17 21:21:19", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/3-00-x-10-8-6-50-x-10-7-the-result-is-4-61-x-10-14-my-question-is-how-10-8-10-7-10-14-thank-u", "openwebmath_score": 0.39296382665634155, "openwebmath_perplexity": 6843.251092607862, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979667647844603, "lm_q2_score": 0.8856314813647588, "lm_q1q2_score": 0.8676245102057446}}
{"url": "http://math.stackexchange.com/questions/1012730/can-a-set-containing-0-be-purely-imaginary", "text": "# Can a set containing 0 be purely imaginary?\n\nA purely imaginary number is one which contains no non-zero real component.\n\nIf I had a sequence of numbers, say $\\{0+20i, 0-i, 0+0i\\}$, could I call this purely imaginary?\n\nMy issue here is that because $0+0i$ belongs to multiple sets, not just purely imaginary, is there not a valid case to say that the sequence isn't purely imaginary?\n\n-\nI think it would simplify your question a bit to just ask \"Is $\\textit{0}$ purely imaginary?\" \u2013\u00a0curious Nov 9 '14 at 1:16\nBut my question is why would I consider only one classification 0+0i and ignore the others \u2013\u00a0chris Nov 9 '14 at 1:19\n\n0 is both purely real and purely imaginary. The given set is purely imaginary. That's not a contradiction since \"purely real\" and \"purely imaginary\" are not fully incompatible. Somewhat similarly baffling is that \"all members of X are even integers\" and \"all members of X are odd integers\" is not a contradiction. It just means that X is an empty set.\n\n-\n$\\ldots$ and $0$ is unique in being both purely real and purely imaginary. \u2013\u00a0Thumbnail Nov 9 '14 at 10:06\n\nA complex number is said to be purely imaginary if it's real part is zero. Zero is purely imaginary, as it's real part is zero.\n\n-\nThis is really important in fields like ordinary differential equations, where you look at having no purely imaginary eigenvalues for it to be hyperbolic \u2013\u00a0Alan Nov 9 '14 at 1:23\nfrom my understanding zero can also be considered real? \u2013\u00a0chris Nov 9 '14 at 1:30\nYes, a complex number is real if it's imaginary part is zero. So zero is also real. \u2013\u00a0Seth Nov 9 '14 at 1:33\nso then my question is why can I consider only the definition that suits my needs? \u2013\u00a0chris Nov 9 '14 at 1:38\nBy definition, $0$ is purely imaginary. The fact that $0$ has other properties (it is real; it is nonnegative; it is rational; it is an integer; it is algebraic; it is divisible by every prime number) does not mean you can\u2019t use the property you need. Similarly, is $\\{-2,4\\}$ a set of even numbers? Yes. The number $-2$ is not only even, but it\u2019s also negative. The fact that it\u2019s negative doesn\u2019t mean you can\u2019t use the fact that it\u2019s even. Some sets defined by properties overlap. \u2013\u00a0Steve Kass Nov 9 '14 at 1:53", "date": "2016-02-14 10:15:27", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/1012730/can-a-set-containing-0-be-purely-imaginary", "openwebmath_score": 0.8323412537574768, "openwebmath_perplexity": 470.36570987513375, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676466573412, "lm_q2_score": 0.885631470799559, "lm_q1q2_score": 0.8676244988038837}}
{"url": "https://math.stackexchange.com/questions/3421041/how-can-i-find-the-speed-and-the-angle-of-a-water-droplet-in-a-rainy-day-as-seen", "text": "# How can I find the speed and the angle of a water droplet in a rainy day as seen by a person running?\n\nThe problem is as follows:\n\nA person is running following a constant speed of $$4.5\\,\\frac{m}{s}$$ over a flat (horizontal) track on a rainy day. The water droplets fall vertically with a measured speed of $$6\\,\\frac{m}{s}$$. Find the speed in $$\\frac{m}{s}$$ of the water droplet as seen by the person running. Find the angle to the vertical should his umbrella be inclined to get wet the less possible. (You may use the relationship of $$37^{\\circ}-53^{\\circ}-90^{\\circ}$$ for the $$3-4-5$$ right triangle ).\n\nThe alternatives given on my book are as follows:\n\n$$\\begin{array}{ll} 1.7.5\\,\\frac{m}{s};\\,37^{\\circ}\\\\ 2.7.5\\,\\frac{m}{s};\\,53^{\\circ}\\\\ 3.10\\,\\frac{m}{s};\\,37^{\\circ}\\\\ 4.10\\,\\frac{m}{s};\\,53^{\\circ}\\\\ 5.12.5\\,\\frac{m}{s};\\,37^{\\circ}\\\\ \\end{array}$$\n\nOn this problem I'm really very lost. What sort of equation should I use to get the vectors or the angles and most importantly to get the relative speed which is what is being asked.\n\nI assume that to find the relative speed can be obtaining by subtracting the speed from which the water droplet is falling to what the person is running. But He is in this case running horizontally. How can I subtract these?\n\nCan somebody give me some help here?\n\nSupposedly the answer is the first option or $$1$$. But I have no idea how to get there.\n\n\u2022 The $3-4-5$ triangle is a hint as to the nature of the result. He is traveling horizontally, the droplets travel vertically, the resulting relative speed must combine these quantities in some manner. Since the two lines of travel are perpendicular, it makes sense to think about the hypotenuse between them... Nov 4 '19 at 3:43\n\u2022 @abiessu Perhaps can you develop an answer so I could understand what you mean by combining the speeds?. Can you show me the equation should I use?. Had I not been given the alternatives, what should I have done with this question? Nov 4 '19 at 4:27\n\n\u2022 water droplet speed as seen from a stationary point on earth is $\\vect{v_d}=(0,-6)$. the speed of the runner as seen from the same point is $\\vect{v_r}=(4.5,0)$. however, from the perspective of the runner, the water droplet both falls down with the speed $\\vect{v_d}=(0,-6)$ and it moves towards him with the horizontal speed $\\vect{v_h}=(-4.5,0)$. wherefore, for the runner the water moves with a resultant speed $\\vect{v_r}=\\vect{v_d}+\\vect{v_h}=(0,-6)+(-4.5,0)=(-4.5,-6)$. this vector has a magnidue of 7.5 and a direction of $37^0$ West from vertical. Nov 4 '19 at 6:49", "date": "2021-10-16 13:00:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3421041/how-can-i-find-the-speed-and-the-angle-of-a-water-droplet-in-a-rainy-day-as-seen", "openwebmath_score": 0.783245861530304, "openwebmath_perplexity": 280.53849968374885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429112114063, "lm_q2_score": 0.8807970842359876, "lm_q1q2_score": 0.8676229240423354}}
{"url": "https://math.stackexchange.com/questions/710484/is-there-a-general-formula-for-the-sum-of-a-quadratic-sequence/1239788", "text": "Is there a general formula for the sum of a quadratic sequence?\n\nI tried Googling \"formula for sum of quadratic sequence\", which did not give me anything useful. I just want an explicit formula for figuring out a sum for a quadratic sequence. For example, how would you figure out the sum of $2+6+12+20+\\dots+210$? Can someone please help? Thanks\n\nFor those of you who do not know, a quadratic sequence is a sequence where the differences of the differences between the terms are constant. Let's use $2+6+12+20+\\dots$ as an example. The differences between the terms are $4$, $6$, $8$, etc. The difference between the differences of the terms is $2$. So the sequence will continue like $2+6+12+20+30+42+56+72+\\dots$\n\n\u2022 Is it homework? \u2013\u00a0draks ... Mar 13 '14 at 6:45\n\u2022 @draks... No it is not homework. I am just wondering as I suck at sequences and series problems \u2013\u00a0TrueDefault Mar 13 '14 at 6:46\n\u2022 ...I was going to write an answer saying, \"no\", because I thought you meant quadratic recurrences which are far more complex. Quadratics have a sum which is a cubic equation, so take any four points and do Lagrange interpolation. \u2013\u00a0Charles Mar 13 '14 at 22:46\n\u2022 When the $k$th difference of a sequence is constant, one can write down the general term using Newton's formula. $f(n) = f(0) + \\frac{\\Delta f(0)}{1!} n + \\frac{\\Delta^2 f(0)}{2!}n(n-1) + \\cdots$, where $\\Delta f$ are the successive differences. For quadratic sequences, one can then use the usual sum formulas. \u2013\u00a0user348749 Jul 5 '16 at 8:10\n\nYes there is. Ever wonder why this is called quadratic sequence? Quadratic refers to squares right? This is just constant difference of difference. So where's the connection? Well as it turns out, all terms of a quadratic sequence are expressible by a quadratic polynomial. What do I mean? Consider this\n\n$$t_n = n+n^2$$\n\nSubsituiting $n=1,2,3,\\cdots$ generates your terms. By the way, $202$ doesn't occur in this sequence, the 13th term is $182$ and the $14th$ term is $210$. I am assuming it was supposed to be $210$.\n\nSo we need to find\n\n$$\\sum_{i=1}^{n}i+i^2 = \\sum_{i=1}^{n}i+\\sum_{i=1}^{n}i^2$$\n\nwhere $n=14$. There are well known formulas for $\\sum_{i=1}^{n}i$ and for $\\sum_{i=1}^{n}i^2$. Substituting them, we get,\n\n$$\\frac{n(n+1)}{2} + \\frac{n(n+1)(2n+1)}{6}$$ $$=\\frac{n(n+1)}{2}\\left(1+\\frac{2n+1}{3}\\right)$$ $$=\\frac{n(n+1)}{2}\\left(\\frac{3+2n+1}{3}\\right)$$ $$=\\frac{n(n+1)}{2}\\left(\\frac{2n+4}{3}\\right)$$ $$=\\frac{n(n+1)(n+2)}{3}$$\n\nwhere $n=14$. Thus our sum is $1120$.\n\n\u2022 You again ! I surrender. Cheers. \u2013\u00a0Claude Leibovici Mar 13 '14 at 6:43\n\u2022 Woopsies I will fix that error \u2013\u00a0TrueDefault Mar 13 '14 at 6:44\n\nFor the general case, N terms are to be summed\n\n$$S=a_0+a_1+a_2+...+a_{N-1}$$\n\nThe formula for the n-th term is $$a_n=a_0+(a_1-a_0)n+(a_2-2a_1+a_0)\\frac{n(n-1)}{2}$$\n\nUsing the results\n\n$$\\sum_{n=0}^{N-1}n=\\frac{N(N-1)}{2}$$\n\nand\n\n$$\\sum_{n=0}^{N-1}n(n-1)=\\frac{N(N-1)(N-2)}{3}$$\n\n$$S=\\sum_{n=0}^{N-1}a_n=a_0N+(a_1-a_0)\\frac{N(N-1)}{2}+(a_2-2a_1+a_0)\\frac{N(N-1)(N-2)}{6}$$\n\nAlthough this may not be needed as of now; but I thought about sums of quadratic sequences myself and I managed to derive a general formula for it, so I might as well post it here:\n\nWhere $n$ is the number of terms to compute, is the starting term, $d$ is the first difference and is the constant difference between the differences. I use the subscript to denote that it is the quadratic sequence sum function.\n\n\u2022 Welcome to math stackexchange. When it isn't too difficult, it is usually preferable to typeset the mathematics than to include links to images. Thanks for the answer. \u2013\u00a0TravisJ Apr 18 '15 at 0:53\n\nIn this particular case, \\begin{align} &2+6+12+20+\\cdots+210\\\\\\\\ &=2(1+3+6+10+\\cdots+105)\\\\\\\\ &=2\\left[\\binom 22 +\\binom 32 +\\binom 42+\\binom 52+\\cdots \\binom {15}2\\right]\\\\ &=2\\sum_{r=2}^{15}\\binom r2\\\\ &=2\\binom {16}3\\\\ &=\\color{red}{1120}\\qquad\\blacksquare \\end{align}\n\nThis is the sum of triangular numbers (where the difference of the difference is constant) and the result is a pyramidal number (all scaled by 2). The summation can be shown as\n\n\\begin{align} &2\\cdot (1)+\\\\ &2\\cdot (1+2)+\\\\ &2\\cdot (1+2+3)+\\\\ &2\\cdot (1+2+3+4)+\\\\ &\\quad \\vdots\\qquad\\qquad\\qquad\\ddots\\\\ &2\\cdot (1+2+3+\\cdots+15) \\end{align}\n\nIf we tried to sum a series where the difference of the difference of the difference is constant, i.e. sum of pyramidal numbers, the result would be a pentatope number. And so on...\n\nAn example of the summation of pyramidal numbers, extending from the original question, would be\n\n$$2+8+20+40+\\cdots+910 =2 \\sum_{r=1}^{15} \\binom r3=2 \\binom {16}4=1820\\\\$$\n\nI figured out the below way of doing it just know at one o'clock right before bedtime, so if it is faulty than that is my mistake.\n\nAny series has a certain term-to-term rule. For a quadratic that term-to-term rule is in the form\n\n$ax^2+bx+c$\n\nThe series will simply be that term-to-term rule with $x$ replaced by $0$, then by $1$ and so on. This can be written as\n\n$\\sum_{x=0}^p ax^2+bx+c$\n\nWe already know that\n\n$\\sum_{x=0}^p x=\\frac{x(x+1)}{2}$\n\n$\\sum_{x=0}^p x^2=\\frac{x(x+1)(2x+1)}{6}$\n\n$\\sum_{x=0}^p c=c*p$\n\nThese can all be proved easily and you can find said proofs online. Because addition is commutative (the order does not matter) the quadratic sum above rewritten as\n\n$a*\\sum_{x=0}^p x^2+b*\\sum_{x=0}^p x + \\sum_{x=0}^p c$\n\nusing the above identities, this can be simplified to\n\n$a*\\frac{p(p+1)(2p+1)}{6}+b*\\frac{p(p+1)}{2}+c*p$\n\nIn your example the term-to-term rule is\n\n$1x^2+1x+0$\n\nAnd we want to find the sum until the term that will give us $210$. This number can be calculated to be $14$. That means that $14^2+14=210$. So in our formula\n\n$p=14$\n\n$a=1$,\n\n$b=1$\n\n$c=0$.\n\nPlug that in and your sum is $980$\n\nI have a sum formula for quadratic equation.\n\n$$s=\\frac{n}{6} (3d(n-1)+(n-1)(n-2)c) +an$$\n\nWhere $n$ is the number of terms, $d$ is the first difference, $c$ is the constant difference or difference of difference, $a$ is first term. For calculation of sum first put value of $a$ $d$ $c$ then you get a formula like $an^2+bn+c$.", "date": "2019-12-10 02:39:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/710484/is-there-a-general-formula-for-the-sum-of-a-quadratic-sequence/1239788", "openwebmath_score": 0.9188519716262817, "openwebmath_perplexity": 399.8063511484217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429116504952, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8676229213468327}}
{"url": "https://math.stackexchange.com/questions/3545737/how-to-define-the-rth-harmonic-number-if-r-is-any-real-number", "text": "How to define the $r$th harmonic number if $r$ is any real number?\n\nWhat would be the value of\n\n$$\\sum_{k=1}^{\\sqrt{\\frac{5}{2}}}\\frac{1}{k}$$\n\nIs it $$H_{\\sqrt{\\frac{5}{2}}}$$?\n\nUsing different definitions of harmonic numbers this question can be computed, but can I use the usual definition of harmonic numbers for any real numbers?\n\nIn other words is $$\\sum_{k=1}^{n}\\frac{1}{k}$$\n\nA useful definition for any real $$n$$?\n\nWhen I want to compute the value of $$H_{\\sqrt{\\frac{5}{2}}}$$ at Desmos I just define $$\\sum_{k=1}^{n}\\frac{1}{k}$$ and I don't put $$\\sqrt{\\frac{5}{2}}$$ directly at the upper limit, but I define $$n=\\sqrt{\\frac{5}{2}}$$ and the result is exactly what it should be, it seems using a simple substitution will give us the right answer, but generally are we allowed to use the regular definition of harmonic numbers when our $$n$$ is not necessarily a natural number?\n\n\u2022 I would define the $r$th harmonic number for a real number $r$ which is not a negative integer for example by $H_r = \\gamma +\\psi(r+1)$, where $\\gamma$ is the Euler-Mascheroni constant and $\\psi$ is the digamma function \u2013\u00a0Maximilian Janisch Feb 13 at 22:29\n\u2022 Things are even more complicated. Suppose you have a function of $f(n)$ in the domain of real $n$ for which $f(n) = H_n$ for positive integers $n$. Then $g(n) = f(n) + a \\sin(\\pi n)$ is another extension of $H_n$ to real $n$. Hence you need an additional condition to rule out $a\\ne0$. \u2013\u00a0Dr. Wolfgang Hintze Feb 14 at 7:58\n\nIndefinite Sum concept is the answer to your question.\n\nIn fact, if a function $$f(x)$$ is the (forward) difference of a function $$F(x)$$ $$f(x) = \\Delta _x F(x) = F(x + 1) - F(x)$$ then we say that $$F(x)$$ is the \"antidifference\" (or \"indefinite sum\") of $$f(x)$$ $$F(x) = \\Delta _x ^{\\left( { - 1} \\right)} f(x) = \\sum\\nolimits_x {f(x)}$$\n\nIf $$f(x)$$ and $$F(x)$$ are defined over a real, or complex, domain for $$x$$ then we will have for example $$\\sum\\limits_{k = 0}^n {f(x + k)} = \\sum\\limits_{k = 0}^n {\\left( {F(x + 1) - F(x)} \\right)} = F(x + n + 1) - F(x)$$\n\nWe write the above with a different symbol for the sum as $$\\sum\\nolimits_{k = 0}^{\\,n} {f(x + k)} = \\sum\\limits_{k = 0}^{n - 1} {f(x + k)} = F(x + n) - F(x)$$ which can also be written as $$\\sum\\nolimits_{k = x}^{\\,x + n} {f(k)} = F(x + n) - F(x)$$\n\nThe extension to $$\\sum\\nolimits_{k = a}^{\\,b} {f(k)} = F(b) - F(a)$$ for any real (or complex) $$a,b$$ inside the domain of definition of $$f, F$$ is quite natural.\n\nComing to the harmonic numbers, it is well known that the functional equation of the digamma function is $${1 \\over x} = \\Delta _x \\,\\psi (x)$$ and it is therefore \"natural\" to define $$\\bbox[lightyellow] { H_r = \\sum\\limits_{k = 1}^r {{1 \\over k}} = \\sum\\nolimits_{k = 1}^{\\,1 + r} {{1 \\over k}} = \\psi (1 + r) - \\psi (1) = \\psi (1 + r) + \\gamma } \\tag{1}$$ which substantiate M. Janisch's comment\n\nAnswering to W. Hintze's comment, please consider how the \"indefinite sum\" parallels the \"indefinite integral - antiderivative\" concept.\nSimilarly to $$f(x) = {d \\over {dx}}F(x)\\quad \\Leftrightarrow \\quad F(x) = \\int {f(x)dx} + c$$ we have that $$f(x) = \\Delta \\,F(x)\\quad \\Leftrightarrow \\quad F(x) = \\sum\\nolimits_x {f(x)} + \\pi \\left( x \\right)$$ where now the family of antidifference functions differ by any function $$\\pi(x)$$ , and not by a constant, which is periodic with period (or one of the periods) equal to $$1$$, as you rightly noticed.\n\nSo by \"natural extension\" I meant to say:\n- an extension from integers to reals (and complex) field under the \"indefinite sum\" concept,\nwhich provides a function $$\\mathbb C \\to \\mathbb C$$ which fully interpolates $$f(n)$$;\n- in the antidifference family to select the \"simpliest / smoothiest\" function, same as the Gamma function is selected among the functions satisfying $$F(z+1)=zF(z)$$ as the only one which is logarithmically convex, or which has the simplest Weierstrass representation, etc.\n\nFrom the comments I could catch some skepticism as if my answer could just be an \"extravagant\" personal idea of mine.\nThat's absolutely not so, the definition in (1) is actually standardly accepted: refer to Wolfram Function Site, and in particular to this section or to the Wikipedia article.\nI am just trying to enlight how we can assign a meaning to sums with bounds which are not integral and thus saying that $$\\bbox[lightyellow] { H_{\\,1 + \\sqrt {5/2} } = \\sum\\limits_{k = 1}^{\\,1 + \\sqrt {5/2} } {{1 \\over k}} = \\sum\\nolimits_{k = 1}^{\\,\\,2 + \\sqrt {5/2} } {{1 \\over k}} = \\psi (\\,\\,2 + \\sqrt {5/2} ) - \\psi (1) = 1.7068 \\ldots } \\tag{2}$$\n\n\u2022 I don't believe that the indefinite sum provides an answer to the question. If you believe it does, you should explain that in your answer. Just giving a link is not an adequate response. \u2013\u00a0Rob Arthan Feb 13 at 21:56\n\u2022 @RobArthan: your criticism is fully right and accepted: I expanded my answer \u2013\u00a0G Cab Feb 14 at 0:41\n\u2022 So finally can I use an irrational index for the lower or upper bound for the sum? \u2013\u00a0user715522 Feb 14 at 6:07\n\u2022 @ G Cab As to the question of what is a \"natural\" extension please see my comment to the OP. (and thanks for bringing indefinite sums to my attention). \u2013\u00a0Dr. Wolfgang Hintze Feb 14 at 8:03\n\u2022 @Dr.WolfgangHintze: well, the meaning of \"natural extension\" is .. quite natural, as much as the Gamma function is the natural extension of $(n-1)!$. That the \"natural\" one does not cover all the possible extensions that's absolutely true: I added some lines to make it clear which is the antidifference family of functions. \u2013\u00a0G Cab Feb 14 at 17:44\n\nThe $$n$$-th harmonic number $$H_n=\\sum_{k=1}^n\\frac{1}{k}$$ admits a nice representation as integral of a finite geometric series which can be generalised in a rather natural way. We have the following representation for $$n$$ a positive integer: \\begin{align*} H_n=\\sum_{k=1}^n\\frac{1}{k}=\\int_0^1\\frac{1-t^n}{1-t}dt\\tag{1} \\end{align*}\n\nThe identity (1) is valid since we have \\begin{align*} \\int_0^1\\frac{1-t^n}{1-t}dt&=\\int_0^1\\left(1+t+\\cdots+t^{n-1}\\right)\\,dt\\tag{2}\\\\ &=\\left(t+\\frac{1}{2}t^2+\\cdots+\\frac{1}{n}t^n\\right)\\bigg|_0^1\\tag{3}\\\\ &=\\left(1+\\frac{1}{2}+\\cdots+\\frac{1}{n}\\right)-\\left(0\\right)\\tag{4}\\\\ &=H_n \\end{align*}\n\nComment:\n\n\u2022 In (2) we apply the finite geometric series formula.\n\n\u2022 In (3) we do the integration.\n\n\u2022 In (4) we evaluate the expression at the upper and lower limit.\n\nThe representation (1) indicates a generalisation \\begin{align*} H_\\color{blue}{r}=\\int_0^1\\frac{1-t^\\color{blue}{r}}{1-t}dt\\tag{2} \\end{align*} for real values $$r$$ in fact also for complex values.\n\nNote:\n\n\u2022 The formula (2) can be found in the wiki page generalized harmonic numbers.\n\n\u2022 We have an interesting relationship with the Digamma function $$\\psi(r)$$ which has an integral representation \\begin{align*} \\psi(r+1)=\\int_0^1\\frac{1-t^r}{1-t}\\,dt-\\gamma \\end{align*} which is strongly related to (2). Here $$\\gamma$$ is the Euler-Mascheroni constant.\n\n\u2022 If we stick at the sigma notation, we often find for real upper limit $$r$$ the meaning \\begin{align*} \\sum_{k=1}^r a_k=\\sum_{k=1}^{\\lfloor r\\rfloor}a_k \\end{align*} where $$\\lfloor .\\rfloor$$ is the floor function.\n\n\u2022 You are rigt : in the \"traditional\" sum notation writing $\\sum\\limits_{k = r}^s {a_{\\,k} }$ actually means $\\sum\\limits_{k = \\,\\left\\lceil r \\right\\rceil }^{\\left\\lfloor s \\right\\rfloor } {a_{\\,k} }$. That's why we need to extend the meaning passing to the antidifference and using a slightly different symbol. \u2013\u00a0G Cab Feb 15 at 22:05\n\u2022 @GCab: I think the simpler approach via the integral representation already does the job to generalize $H_n$ appropriately. \u2013\u00a0Markus Scheuer Feb 15 at 22:11\n\u2022 Definitely yes, the integral representation does the job very nicely. But I understand the core of the post as being whether and what meaning to assign to a sum with non-integral bounds \u2013\u00a0G Cab Feb 15 at 22:24\n\u2022 @GCab: Yes, I know. :-) \u2013\u00a0Markus Scheuer Feb 15 at 22:25\n\nYour question is \"are we allowed to use the regular definition of harmonic numbers when our $$n$$ is not necessarily a natural number?\" The answer is no, since the \"regular definition\" of a sum is $$a_1 + a_2 + ... + a_n$$. Here $$n$$ is the number of summands which can't be a non natural number.\n\nBut you can very well give a meaning to the result of a sum. And this result may accept non integers as input.\n\nTake the simpler example of a \"sum\" $$a(\\pi) = \\text{\"}\\sum_{k=1}^{\\pi} k\\text{\"}$$. This \"sum\" is not defined, as although the first three summands are 1, 2, and 3, the last 3.14...-th summand is not defined.\n\nHence the definition starts by calculating the sum up to an integer $$n$$, and afterwards insert the non integer value for $$n$$. Here this gives the arithmetic sum $$a(n)= \\frac{1}{2}n(n+1)$$, and hence $$a(\\pi)= \\frac{1}{2}\\pi(\\pi+1)$$.\n\nNow for the harmonic number there is no such simple closed expression for general $$n$$. But here comes a relation which is valid for natural $$n$$: $$\\sum_{k=1}^n \\frac{1}{k} = \\sum_{k=1}^n \\int_0^1 x^{k-1}\\,dx=\\int_0^1 \\sum_{k=1}^n x^{k-1}\\,dx=\\int_0^1 \\frac{1-x^n}{1-x}\\,dx$$, and now the r.h.s. is valid for also for real $$n$$.\n\nYour description of the problem gives a good illustration of this procedure. In making the usual definition of the harmonic number $$\\sum_{k=1}^n \\frac{1}{k}$$ in your CAS, the system understands that you mean the harmonic number function $$H_n$$ it has in its \"belly\", and this is valid for real numbers. If you then insert a value for $$n$$, be it integer or not, you get the result of this internal function.\n\n\u2022 please, let's avoid confusions, do not use the symbol $\\sum\\nolimits_{k = 1}^{\\,\\pi } k$ in your argumentation above since it is the symbol adopted (almost standardly) for the antidifference, and $$\\sum\\nolimits_{k = 1}^{\\,\\pi } k = {{\\pi \\left( {\\pi - 1} \\right)} \\over 2}$$ ! Use instead, $$\\sum\\limits_{k = 1}^\\pi k = \\sum\\nolimits_{k = 1}^{\\,\\pi + 1} k = {{\\pi \\left( {\\pi + 1} \\right)} \\over 2}$$ And please have a look at the renowned and authoritative \"concrete Mathematics\" para. 2.6 \u2013\u00a0G Cab Feb 15 at 21:48\n\u2022 You might have noticed that I have put the expression in hyphens to avoid confusion. You are free to use any symbol you like, of course, but please avoid forcing others to use your conventions. \u2013\u00a0Dr. Wolfgang Hintze Feb 16 at 0:06", "date": "2020-04-08 16:26:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3545737/how-to-define-the-rth-harmonic-number-if-r-is-any-real-number", "openwebmath_score": 0.9237313270568848, "openwebmath_perplexity": 373.008033976001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429147241161, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8676229178895667}}
{"url": "http://gmatclub.com/forum/neat-fact-for-integral-solutions-to-a-polynomial-155227.html", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 25 Oct 2016, 01:44\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Neat Fact for Integral Solutions to a polynomial\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nVerbal Forum Moderator\nJoined: 10 Oct 2012\nPosts: 630\nFollowers: 78\n\nKudos [?]: 1054 [2] , given: 136\n\nNeat Fact for Integral Solutions to a polynomial\u00a0[#permalink]\n\n### Show Tags\n\n02 Jul 2013, 02:09\n2\nKUDOS\n7\nThis post was\nBOOKMARKED\nHello!\n\nConsider any polynomial $$f(x) = A_1x^n+A_2x^{n-1}+.....A_n$$\n\nAssumption : All the co-efficients for the given polynomial have to be integral,i.e. $$A_1,A_2,A_3....A_n$$ are all integers.\n\nFact:Any integral solution(root) for the above polynomial will always be a factor(positive/negative) of the constant term : $$A_n$$\n\nExample I : $$f(x) = 5x^2-16x+3$$. Thus, we know that if the given polynomial has any integral solutions, then it will always be out of one of the following : $$-3,-1,1,3$$\n\nWe see that only x=3 is a root for the given polynomial. Also, we know that product of the roots is$$\\frac {3}{5}$$. Hence, the other root is $$\\frac {1}{5}$$\n\nExample II : Find the no of integral solutions for the expression $$f(x) = 3x^4-10x^2+7x+1$$\n\nA. 0\nB. 1\nC. 2\nD. 3\nE. 4\n\nFor the given expression, instead of finding the possible integral solutions by hit and trial, we can be rest assured that if there is any integral solution, it will be a factor of the constant term ,i.e. 1 or -1. Just plug-in both the values, and we find that f(1) and f(-1) are both not equal to zero. Thus, there is NO integral solution possible for the given expression--> Option A.\n\nExample III : Find the no of integral solutions for the expression $$f(x) = 4x^4-8x^3+9x-3$$\n\nA. 0\nB. 1\nC. 2\nD. 3\nE. 4\n\nJust as above, the integral roots of the given expression would be one of the following : -3,-1,1,3. We can easily see that only x = -1 satisfies. Thus, there is only one integral solution for the given polynomial-->Option B.\n\nHence, keeping this fact in mind might just reduce the range of the hit and trial values we end up considering.\n_________________\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 12217\nFollowers: 542\n\nKudos [?]: 151 [0], given: 0\n\nRe: Neat Fact for Integral Solutions to a polynomial\u00a0[#permalink]\n\n### Show Tags\n\n17 Mar 2015, 10:58\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 12217\nFollowers: 542\n\nKudos [?]: 151 [0], given: 0\n\nRe: Neat Fact for Integral Solutions to a polynomial\u00a0[#permalink]\n\n### Show Tags\n\n09 May 2016, 14:17\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: Neat Fact for Integral Solutions to a polynomial \u00a0 [#permalink] 09 May 2016, 14:17\nSimilar topics Replies Last post\nSimilar\nTopics:\nSolution of an inequality 0 24 Jun 2013, 21:55\n2 Factoring a Polynomial W/O Quadratic formula 7 08 Jun 2013, 18:41\n5 Brute Force for Positive Integral Solutions 2 21 Oct 2012, 21:29\nNot for solution but for suggestion 2 08 Mar 2011, 12:19\ndivisibility of polynomial- PLZ help 0 03 May 2010, 19:19\nDisplay posts from previous: Sort by", "date": "2016-10-25 08:44:46", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/neat-fact-for-integral-solutions-to-a-polynomial-155227.html", "openwebmath_score": 0.7425248622894287, "openwebmath_perplexity": 2149.3245434937126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9314625069680097, "lm_q2_score": 0.9314625050654263, "lm_q1q2_score": 0.8676224001149444}}
{"url": "https://math.stackexchange.com/questions/2868058/how-can-i-prove-this-proposition-from-peano-axioms-cancellation-law-let-a-b", "text": "# How can I prove this proposition from Peano Axioms: (Cancellation law). Let a, b, c be natural numbers such that a + b = a + c. Then we have b = c.\n\nPeano Axioms.\n\nAxiom 2.1\n\n$0$ is a natural number.\n\nAxiom 2.2\n\nIf $n$ is a natural number then $n++$ is also a natural number. (Here $n++$ denotes the successor of $n$ and previously in the book the notational implication has been bijected to the familiar $1,2\u2026$).\n\nAxiom 2.3\n\n$0$ is not the successor of natural number; i.e. we have $n++\u22600$ for every natural number $n$.\n\nAxiom 2.4\n\nDifferent natural numbers must have different successors; i.e., if $n,m$ are natural numbers and $n\u2260m$, then $n++\u2260m++$.\n\nAxiom 2.5\n\nLet $P(n)$ be any property pertaining to a natural number $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n++)$ is also true. Then $P(n)$ is true for every natural number $n$.\n\nDefinition of Addition: Let m be a natural number. We define, $0+m=m$ and suppose we have inductively defined the addtion $n+m$ then we define, $(n++)+m=(n+m)++$. Where $n++$ is the successor of $n$.\n\nTerence Tao has a proof in his Analysis I book, but I couldn't understand . I want a proof line by line like this:\n\nThm: 3 is a natural number \\begin{align*} & 0 \\text{ is natural } && \\text{Axiom 2.1 }\\\\ & 0++ = 1 \\text{ is natural } && \\text{Axiom 2.2}\\\\ & 1++ = 2 \\text{ is natural } && \\text{Axiom 2.2}\\\\ & 2++ = 3 \\text{ is natural } && \\text{Axiom 2.2} \\end{align*}\n\n\u2022 What have you proven so far? Have you shown commutativity? \u2013\u00a0InterstellarProbe Jul 31 '18 at 13:52\n\nThe proof is by induction on $a$.\n\nBasis [case with $a=0$]. We want to prove that :\n\nif $0+b=0+c$, then $b=c$.\n\nBut we have that : $0+b=b$, by definition of addition.\n\nAnd also : $0+c=c$.\n\nThus, by transitivity of equality : $b=c$.\n\nInduction step. Assume that the property holds for $a$ and prove it for $a'$ [I prefer to use $a'$ instead of $a++$ for the successor of $a$].\n\nThis means :\n\nassume that \"if $a+b=a+c$, then $b=c$\" holds, and prove that \"if $a'+b=a'+c$, then $b=c$\" holds.\n\nWe have $a'+b=(a+b)'$ by definition of addition.\n\nAnd $a'+c=(a+c)\u2019$, by definition of addition.\n\nWe assume that : $a'+b=a'+c$, and thus, again by transitivity of equality, we have that : $(a+b)'=(a+c)'$.\n\nBy axiom 2.4 (different numbers have different successors), by contraposition, we conclude that $(a+b)=(a+c)$.\n\nThus, applying induction hypothesis, we have that :\n\n$b=c$.\n\nThe general strategy used above in order to prove \"if $P$, then $Q$\", is to assume $P$ and derive $Q$.\n\nThis type of proof is called Conditional Proof; we can see it above :\n\n1) if $a+b=a+c$, then $b=c$ --- induction hypothesis\n\n2) $a'+b=a'+c$ --- assumption for CP\n\n3) $(a+b)'=(a+c)'$ --- from 2) and definition of addition and tarnsitivity of equality\n\n4) $a+b=a+c$ --- from 3) and axiom 2.4 by Contraposition [the axiom says : \"if $(a+b \\ne a+c)$, then $(a+b)' \\ne (a+c)'$\"; thus, contraposing it we get : \"if $(a+b)' = (a+c)'$, then $(a+b) = (a+c)$\"] and Modus ponens\n\n5) $b=c$ --- from 4) and 1) by Modus ponens (also called : detachment)\n\n6) if $a'+b=a'+c$, then $b=c$ --- from 2) and 5) by Conditional Proof.\n\n\u2022 (In your last yellow box before the general strategy, it should be $b=c$) \u2013\u00a0Jason DeVito Jul 31 '18 at 14:31\n\u2022 Thanks! Amazing. \u2013\u00a0Henrique Jul 31 '18 at 15:25\n\nHere is a formal proof done in Fitch:\n\n(Note: I use $s(x)$ instead of $x$++)", "date": "2019-04-20 16:57:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2868058/how-can-i-prove-this-proposition-from-peano-axioms-cancellation-law-let-a-b", "openwebmath_score": 0.9734426140785217, "openwebmath_perplexity": 291.5094869905468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274402, "lm_q2_score": 0.8918110562208681, "lm_q1q2_score": 0.8675812372557807}}
{"url": "http://mathhelpforum.com/algebra/83946-having-trouble.html", "text": "1. ## Having Trouble\n\nHi all,\n\nI am just new to this site and it looks like such an awesome resource.\n\nCan anyone help me with this???\n\nIf x + y = 1 and x\u00b3 + y\u00b3 = 19, find the value of x\u00b2 + y\u00b2\n\n2. Originally Posted by Joel\nHi all,\n\nI am just new to this site and it looks like such an awesome resource.\n\nCan anyone help me with this???\n\nIf x + y = 1 and x\u00b3 + y\u00b3 = 19, find the value of x\u00b2 + y\u00b2\n$\\left\\{\\begin{array}{lclcr}\nx&+&y&=&1\\\\\nx^3&+&y^3&=&19\n\\end{array}\\right.$\n\n$\\Rightarrow x^3+(1-x)^3=19,$\n\nand after expanding and simplifying,\n\n$3x^2-3x-18=0$\n\n$\\Rightarrow3(x-3)(x+2)=0.$\n\nYou should be able to finish.\n\n3. Hello, Joel!\n\nWelcome aboard!\n\nI have a back-door approach . . .\n\nIf . $\\begin{array}{cccc}x + y &=& 1 & {\\color{blue}[1]} \\\\ x^3 + y^3 &=& 19 & {\\color{blue}[2]} \\end{array}$ . find the value of $x^2+y^2.$\n\nCube [1]: . $(x+y)^3 \\:=\\:1^3\\quad\\Rightarrow\\quad x^3 + 3x^2y + 3xy^2 + y^3 \\:=\\:1$\n\n$\\text{We have: }\\;\\underbrace{(x^3 + y^3)}_{\\text{This is 19}} + \\;3xy\\underbrace{(x+y)}_{\\text{This is 1}} \\:=\\:1 \\quad\\Rightarrow\\quad 19 \\;+ \\;3xy \\:=\\:1 \\quad\\Rightarrow\\quad xy \\:=\\:-6$\n\nSquare [1]: . $(x+y)^2 \\:=\\:1^2 \\quad\\Rightarrow\\quad x^2+2xy + y^2 \\:=\\:1$\n\n$\\text{We have: }\\;x^2 + y^2 + 2(xy) \\:=\\:1$\n. . . . . . . . . . . . . . $\\uparrow$\n. . . . . . . . . . . . $^{\\text{This is }\\text{-}6}$\n\nTherefore: . $x^2+y^2 - 12 \\:=\\:1 \\quad\\Rightarrow\\quad \\boxed{x^2+y^2 \\:=\\:13}$\n\nThis method is guarenteed to impress/surprise/terrify your teacher.\n.\n\n4. Originally Posted by Soroban\nHello, Joel!\n\nWelcome aboard!\n\nI have a back-door approach . . .\nI like your solution. I wish I had thought about the problem a bit more before going down the obvious path.", "date": "2016-08-28 03:32:15", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/83946-having-trouble.html", "openwebmath_score": 0.8556994795799255, "openwebmath_perplexity": 716.8040601498021, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9925393569774312, "lm_q2_score": 0.8740772269642948, "lm_q1q2_score": 0.8675560487997573}}
{"url": "https://math.stackexchange.com/questions/1265426/write-100-as-the-sum-of-two-positive-integers/1265457", "text": "# Write 100 as the sum of two positive integers\n\nWrite $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.\n\nSince $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then finding the complement that is also a multiple of the other. So then $100 = 44 + 56 = 4 \\times 11 + 8 \\times 7$.\n\nBut is it the smart way of doing it? Is it the way I was supposed to solve it? I'm thinking here about a situation with a really large number that turns my plug-in method sort of unwise.\n\n\u2022 I think you want to reword this 'Scalable algorithm to write N as the sum of two positive integers, for large N'\n\u2013\u00a0smci\nMay 4, 2015 at 7:13\n\u2022 This seems like a rather badly designed exercise since, looking at the answers, it's clear that just checking multiples of 7 and 11 is by far the simplest way of solving it. I once saw an exam question that made things much clearer by saying the equivalent of, \"Proceed as if 100 is a very large number and you do not know that 100=44+56.\" May 4, 2015 at 21:07\n\nFrom Bezout's Lemma, note that since $\\gcd(7,11) = 1$, which divides $100$, there exists $x,y \\in \\mathbb{Z}$ such that $7x+11y=100$.\n\nA candidate solution is $(x,y) = (8,4)$.\n\nThe rest of the solution is given by $(x,y) = (8+11m,4-7m)$, where $m \\in \\mathbb{Z}$. Since we are looking for positive integers as solutions, we need $8+11m > 0$ and $4-7m>0$, which gives us $-\\frac8{11}<m<\\frac47$. This means the only value of $m$, when we restrict $x,y$ to positive integers is $m=0$, which gives us $(x,y) = (8,4)$ as the only solution in positive integers.\n\nIf you do not like to guess your candidate solution, a more algorithmic procedure is using Euclid' algorithm to obtain solution to $7a+11b=1$, which is as follows.\n\nWe have \\begin{align} 11 & = 7 \\cdot (1) + 4 \\implies 4 = 11 - 7 \\cdot (1)\\\\ 7 & = 4 \\cdot (1) + 3 \\implies 3 = 7 - 4 \\cdot (1) \\implies 3 = 7 - (11-7\\cdot (1))\\cdot (1) = 2\\cdot 7 - 11\\\\ 4 & = 3 \\cdot (1) + 1 \\implies 1 = 4 - 3 \\cdot (1) \\implies 1 = (11-7 \\cdot(1)) - (2\\cdot 7 - 11) \\cdot 1 = 11 \\cdot 2-7 \\cdot 3 \\end{align} This means the solution to $7a+11b=1$ using Euclid' algorithm is $(-3,2)$. Hence, the candidate solution $7x+11y=100$ is $(-300,200)$. Now all possible solutions are given by $(x,y) = (-300+11n,200-7n)$. Since we need $x$ and $y$ to be positive, we need $-300+11n > 0$ and $200-7n > 0$, which gives us $$\\dfrac{300}{11} < n < \\dfrac{200}7 \\implies 27 \\dfrac3{11} < n < 28 \\dfrac47$$ The only integer in this range is $n=28$, which again gives $(x,y) = (8,4)$.\n\n\u2022 This looks like I was looking for, but I didn't quite get why you summed $11m$ to the $x$ and $-7m$ to the $y$, since $x$ is supposed to be a multiple of $7$, instead of $11$. May 3, 2015 at 23:31\n\u2022 @Lanner Check what happens when you plug in $x=8+11m$ and $y=4-7m$ into the original equation. May 3, 2015 at 23:32\n\u2022 While $x,y$ are guaranteed to exist from Bezout's Lemma, they are not guaranteed to be positive. Therefor: Bezout's Lemma provides no information about whether this particular problem has a solution. What logic led to $(8,4)$ being your starting point for the search? May 4, 2015 at 14:15\n\u2022 @AndrewCoonce A more algorithmic procedure to obtain solutions to $ax+by=1$ is Euclid's algorithm to compute the gcd. However, in some cases such as this a candidate solution can be guessed easily. May 4, 2015 at 15:04\n\u2022 @user17762 The only reason I asked for clarification is because the OP already had a natural method for determining it for this case and was looking for a general method. Guessing just because the numbers played nicely here didn't seem like a solution, but your comment clarifies the general solution for me. May 4, 2015 at 16:38\n\nAn effort to avoid any enumeration or hiding an inductive leap to the answer.\n\n$7a + 11b = 100: a,b \\in N$\n\n$11b \\leq 100 - 7 = 93$\n\n$\\implies 1 \\leq b \\leq 8$\n\n$7(a+b) = 100 - 4b$\n\n$\\implies 100 - 4b \\equiv 0 \\mod 7$\n\n$\\implies 25 - b \\equiv 0 \\mod 7$\n\n$\\implies b \\equiv 25 \\mod 7$\n\n$\\implies b \\equiv 4 \\mod 7$\n\n$\\implies b = 4 + 7n$\n\nWe know $1 \\leq b \\leq 8$.\n\nSo we have $b \\equiv 25 \\mod 7$, so $b = 4$ and hence $a = 8$.\n\n\u2022 This works, but it relies on the accidental fact that $11-7$ is a factor of $100$. If I changed it to $103$ this method wouldn't work. The reason is that in general we need the multiplicative inverse of $11$ modulo $7$ or vice versa, for that particular step. May 5, 2015 at 12:48\n\nBut is it the smart way of doing it?\n\nYou are asked to find a and b so that $7a+11b=100\\iff7(a+b)+4b=100\\iff$\n\n$4\\Big[(a+b)+b\\Big]+3(a+b)=100\\iff3\\Big\\{\\Big[(a+b)+b\\Big]+(a+b)\\Big\\}+\\Big[(a+b)+b\\Big]=$ $100$.\n\nBut $100=99+1=3\\cdot33+1$, so $a+2b=1\\iff2a+4b=2$, and $2a+3b=33$. It follows\nthat $b_0=-31$ and $a_0=63$ is one solution. But, then again, so are all numbers of the form $a_k=63-11k$ and $b_k=-31+7k$, with $k\\in$ Z. All we have to do now is pick one pair whose components are both positive. The first equation implies $k<6$, and the latter $k>4$.\n\nSince $\\operatorname{gcd}(7,11)=1$ , you can find $a,b \\in \\mathbb Z$ with $7a+11b=1$. Now multiply both sides of the equation by $100$ to get one (and so all possible) results:\n\n$$700a+1100b=100$$\n\nOnce you have a solution for $700a+1100b=100$, you have all solutions:\n\n$$700(a+11k)+1100(b-7k)=100$$\n\nYou can then find $k$ so that both coefficients are positive.\n\n\u2022 It must be the sum of two positive integers, so a,b > 0. May 3, 2015 at 23:00\n\u2022 @Gary. Bezout coefficients are not guaranteed to be positive. May 3, 2015 at 23:02\n\u2022 The Bezout coefficients for positive integers are guaranteed to have different signs. May 3, 2015 at 23:04\n\u2022 But once we know any $a,b$, we can find solutions $a',b' : 700a'+1100b'=100$ with $a',b' >0$. The Bezout coefficients generate all solutions; they are not the unique solutions. May 3, 2015 at 23:07\n\u2022 @MathMajor: Thanks, you know what those fake internet points will do to you ; ). May 3, 2015 at 23:34\n\nWhile certainly not the ideal solution, this problem is certainly in the realm of Integer Programming. As plenty of others have pointed out, there are more direct approaches. However, I suspect ILP solvers would operate quite efficiently in your case, and requires less 'thought capital'.\n\n$7a+11b=100$ so modding both sides by $11$ you get $11 \\mid (7a-1)$. Let $11a'=7a-1$. Then the problem is transformed to finding $a',b$ such that $a'+b=9$ and $7 \\mid (11a'+1)$, which (in my opinion) is somewhat easier to test for.\n\n$7x+11y=100$\n$7x=100-11y$\n$x=\\frac{100-11y}7=14-2y+\\frac{2+3y}7$\n$a=\\frac{2+3y}7$\n$7a=2+3y$\n$3y=-2+7a$\n$y=\\frac{-2+7a}3=-1+2a+\\frac{1+a}3$\n$b=\\frac{1+a}3$\n$3b=1+a$\n$a=3b-1$\n$y=\\frac{-2+7(3b-1)}3=\\frac{-9+21b}3=-3+7b$\n$x=\\frac{100-11(-3+7b)}7=\\frac{133-77b}7=19-11b$\n$\\begin{matrix} x\\gt 0&\\to&19-11b\\gt 0&\\to&11b\\lt 19&\\to&b\\lt\\frac{19}{11}&\\to&b\\le 1&\\\\ &&&&&&&&&b=1\\\\ y\\gt 0&\\to&-3+7b\\gt 0&\\to&7b\\gt 3&\\to&b\\gt\\frac37&\\to&b\\ge 1&\\\\ \\end{matrix}$\n$1$ is the only integral value of $b$ for which both $x$ and $y$ yield positive values, as required by the problem. So, $x=8$ and $y=4$.\n\nSince positive numbers are required, you could do something like this, inspired by the so-called chicken nugget monoid. This method is only viable if one of the numbers is sufficiently small; Euclid's algorithm is much better for big numbers, e.g., writing 2165434 in terms of positive multiples of 97 and 103.\n\nNote the following:\n\n\u2022 $11\\cdot1=4\\mod 7$\n\u2022 $11\\cdot2=1\\mod 7$\n\u2022 $11\\cdot3=5\\mod 7$\n\u2022 $11\\cdot4=2\\mod 7$\n\u2022 $11\\cdot5=6\\mod 7$\n\u2022 $11\\cdot6=3\\mod 7$\n\u2022 $11\\cdot7=0\\mod 7$\n\nSo, to find a positive solution for $N$ in terms of $7$ and $11$, find the value of $N$ (mod $7$), subtract the appropriate multiple of $11$ from $N$, and divide the result by $7$. Now you have positive integers $m$ and $n$ such that $7m+11n=N$.\n\nOf course, if $m$ is larger than 11, the solution is not unique, but each $N>77$ has at least one solution.\n\n## $N\\leq 77$ with positive solutions (complete list)\n\n\u2022 $18=11(1)+7(1)$\n\u2022 $25=11(1)+7(2)$\n\u2022 $29=11(2)+7(1)$\n\u2022 $32=11(1)+7(3)$\n\u2022 $36=11(2)+7(2)$\n\u2022 $39=11(1)+7(4)$\n\u2022 $40=11(3)+7(1)$\n\u2022 $43=11(2)+7(3)$\n\u2022 $46=11(1)+7(5)$\n\u2022 $47=11(3)+7(2)$\n\u2022 $50=11(2)+7(4)$\n\u2022 $51=11(4)+7(1)$\n\u2022 $53=11(1)+7(6)$\n\u2022 $54=11(3)+7(3)$\n\u2022 $57=11(2)+7(5)$\n\u2022 $58=11(4)+7(2)$\n\u2022 $60=11(1)+7(7)$\n\u2022 $61=11(3)+7(4)$\n\u2022 $62=11(5)+7(1)$\n\u2022 $64=11(2)+7(6)$\n\u2022 $65=11(4)+7(3)$\n\u2022 $67=11(1)+7(8)$\n\u2022 $68=11(3)+7(5)$\n\u2022 $69=11(5)+7(2)$\n\u2022 $71=11(2)+7(7)$\n\u2022 $72=11(4)+7(4)$\n\u2022 $73=11(6)+7(1)$\n\u2022 $74=11(1)+7(9)$\n\u2022 $75=11(3)+7(6)$\n\u2022 $76=11(5)+7(3)$\n\nUse the Extended Euclidean Algorithm to solve $$7x+11y=100$$ Using the implementation detailed in this answer $$\\begin{array}{r} &&1&1&1&3\\\\\\hline 1&0&1&-1&\\color{#C00000}{2}&\\color{#0000F0}{-7}\\\\ 0&1&-1&2&\\color{#00A000}{-3}&\\color{#E0A000}{11}\\\\ 11&7&4&3&1&0\\\\ \\end{array}$$ we get that $$(\\color{#00A000}{-3})\\,7+(\\color{#C00000}{2})\\,11=1$$ multiply by $100$ and use the last column from the algorithm to get the general answer, we get $$(-300+\\color{#E0A000}{11}k)\\,7+(200\\color{#0000F0}{-7}k)\\,11=100$$ set $k=28$ (the only $k$ that works) to make the coefficients positive, we get that $$(8)\\,7+(4)\\,11=100$$\n\nLarger than $\\boldsymbol{100}$\n\nSuppose we want to write $1000000$ as a sum of a multiple of $7$ and a multiple of $11$. We can use the result of the algorithm above. That is $$(-3000000+11k)7+(2000000-7k)11=1000000$$ We can use any $272728\\le k\\le 285714$ to make both coefficients positive.\n\n$k=272728$ gives $$(8)\\,7+(90904)\\,11=1000000$$ $k=285714$ gives $$(142854)\\,7+(2)\\,11=1000000$$", "date": "2022-05-29 04:30:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1265426/write-100-as-the-sum-of-two-positive-integers/1265457", "openwebmath_score": 0.9516299962997437, "openwebmath_perplexity": 135.12573048093154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347853343059, "lm_q2_score": 0.8902942304882371, "lm_q1q2_score": 0.8675336673701763}}
{"url": "https://math.stackexchange.com/questions/2752596/bringing-a-constant-out-of-double-integral-polar-coordinates", "text": "# Bringing a constant out of double integral (polar coordinates)\n\nQuestion is: Using polar coordinates, evaluate the integral $\\iint\\sin(x^2+y^2)dA$ where R is the region $1\\le x^2+y^2\\le81$\n\nMy work for the inside integral, using bounds 1 to 9, was a constant:\n$$\\int_1^9\\sin(r^2)rdr=-\\frac{1}{2}\\left(\\cos81-\\cos1\\right)$$\n\nDoes this mean I can conclude my answer is this constant? In other words, can I bring it out of the integral with respect to theta from 0 to 2$\\pi$? I am guessing not because my answer was not correct. If not, could someone explain why?\n\n\u2022 Try using MathJax to make your question more readable. \u2013\u00a0user546997 Apr 25 '18 at 2:44\n\u2022 Yes you can bring it out. The answer should be correct. What number are you getting for your answer? \u2013\u00a0John Doe Apr 25 '18 at 2:51\n\u2022 @JohnDoe The answer is that constant (my answer) times 2pi. \u2013\u00a0h.jb Apr 25 '18 at 2:55\n\u2022 That's the same answer I get. What is the \"correct\" answer. \u2013\u00a0saulspatz Apr 25 '18 at 3:01\n\u2022 I get $-0.74...$. What number are you getting? Do you have an actual number? That would help us figure out what your issue is \u2013\u00a0John Doe Apr 25 '18 at 3:03\n\nYou ignored the angular part of the integral: $$\\int\\int_R\\sin(x^2+y^2)\\,dx\\,dy=\\int_0^{2\\pi}\\int_1^9r\\sin(r^2)\\,dr\\,d\\theta\\\\=\\left(\\int_0^{2\\pi}\\,d\\theta\\right)\\left(\\int_1^9r\\sin(r^2)\\,dr\\right)=2\\pi\\times(-0.118\\dots)=-0.74\\dots$$\n\u2022 @h.jb yes, we are $$\\int_0^{2\\pi}\\,d\\theta=\\int_0^{2\\pi}1\\,d\\theta=[\\theta]_0^{2\\pi}=2\\pi$$Does that make sense now? We can split the integral in this way because there are no terms with $\\theta$ in the original integral. \u2013\u00a0John Doe Apr 25 '18 at 3:22", "date": "2020-05-27 12:42:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2752596/bringing-a-constant-out-of-double-integral-polar-coordinates", "openwebmath_score": 0.9474749565124512, "openwebmath_perplexity": 391.127184215051, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682439644956, "lm_q2_score": 0.8774767746654976, "lm_q1q2_score": 0.8675334219281668}}
{"url": "https://math.stackexchange.com/questions/694217/calculating-the-square-root-of-an-integer-as-an-integer-with-the-smallest-error", "text": "# Calculating the square root of an integer as an integer with the smallest error\n\nI was recently asked to write an algorithm to find the square root of a given integer, with the solution expressed as an integer with the smallest error.\n\nIt was unclear to me whether this meant I need to return the integer closest to the actual square root, or return the integer such that, when squared, returned the closest value to the target number. When asking for clarification, I was told these are the same.\n\nThis got me thinking however, are these really equivalent?\n\nClearly, when deciding between two real approximations, minimizing the error of the square roots isn't the same as minimizing the error of the squares. For example, if I had to pick between 1.9 and 2.096 as an approximation of the square root of 4, then 2.096 would be the closest to the actual square root (error of 0.096 compared to 0.1), while 1.9 would provide the smallest error when squared (error of 0.39 compared to 0.393216).\n\nFor the integer approximation case, it doesn't appear obvious to me why the solutions to the two cases will always be the same. I've experimentally verified it to be so for the first 200 billion integers. Can anyone provide an intuitive or formal proof of this?\n\nSuppose we want to approximate $\\sqrt x$ with $n^2<x<(n+1)^2$.\n\nIf we minimize the error of the squares, we're going to choose $n$ or $n+1$ according to whether $x>\\frac{n^2+(n+1)^2}2 = n^2 + n + \\frac12$.\n\nIf we minimize the error of the square root, we should choose $n$ or $n+1$ according to whether $\\sqrt{x}>n+\\frac12$, which is the same as $x > (n+\\frac12)^2 = n^2+n+\\frac14$.\n\nBecause $x$ is an integer, being larger than $n^2+n+\\frac12$ is the same as being larger than $n^2+n+\\frac14$.\n\nIf $k-{1\\over2}\\lt\\sqrt n\\lt k+{1\\over2}$, then\n\n$$(k-1)^2=k^2-2k+1\\lt k^2-k+{1\\over4}\\lt n\\lt k^2+k+{1\\over4}\\lt k^2+2k+1=(k+1)^2$$\n\nTo flesh this out, the fact that $k$ and $n$ are integers implies\n\n$$k^2-k+1\\le n\\le k^2+k$$\n\nIf $n\\gt k^2$, then $n-k^2\\le k$ while $(k+1)^2-n\\ge k+1$. Likewise, if $n\\lt k^2$, then $k^2-n\\le k-1$ while $n-(k-1)^2\\ge k$. In either case, $n$ is closer to $k^2$ than it is to either $(k-1)^2$ or $(k+1)^2$.", "date": "2019-11-21 08:32:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/694217/calculating-the-square-root-of-an-integer-as-an-integer-with-the-smallest-error", "openwebmath_score": 0.8674474358558655, "openwebmath_perplexity": 126.39759272934634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771774699747, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8675219627003454}}
{"url": "https://math.stackexchange.com/questions/957777/what-does-23-4-mean", "text": "# What does $23_4$ mean?\n\nI just saw this on a mathematical clock for $11$, i.e $23_4=11$:\n\n$\\qquad \\qquad \\qquad \\qquad \\qquad$\n\nI guess it is some notation from algebra. But since algebra was never my favorite field of maths, I don't know this notation. Any explanations are welcome ;-))! Thanks\n\n\u2022 11 in base 10 is same as 23 in base 4 : $$(23)_4 = (11)_{10}$$ \u2013\u00a0ganeshie8 Oct 4 '14 at 10:46\n\u2022 It is an interesting question and the clock is also meaningful.+1 \u2013\u00a0Paul Oct 4 '14 at 11:07\n\nThis denotes the number $11$ in base $4$. In everyday life, we write our numbers in base $10$.\n\n$23_4$ is to be read as: $$2\\cdot 4 + 3.$$ In general, $$(a_n...a_0) _ g = \\sum_{i=0}^n a_i g^i = a_n g^n + a_{n-1}g^{n-1} + ... + a_1 g + a_0,$$ where the $a_i$ are chosen to lie in $\\{0,...,g-1\\}$.\n\nEDIT: I have edited this post to write $2\\cdot 4 +3$ rather than $3+2\\cdot 4$. However, I still think that it is easier to decipher a (long) number such as $(2010221021)_3$ from right to left, simply by increasing the powers of $3$, rather than first checking that the highest occuring power of $3$ is $3^9$ and then going from left to right.\n\n\u2022 Not reversing the order will make it more readable. \u2013\u00a0Yves Daoust Oct 4 '14 at 10:55\n\u2022 Personally, I prefer it this way, because I can simply increase the power of $g$ as I go from right to left, whereas I first have to check which the highest occuring power of $g$ is when I go from left to right. Nonetheless, I read base 10 numbers from left to right, but that is probably due to the fact that we're used to thinking of numbers in base 10 and writing them this way... \u2013\u00a0Oliver Braun Oct 4 '14 at 11:02\n\u2022 If you write $23_4$, then $2\\cdot4+3$ is more readable than $3+2\\cdot4$. I also use to write polynomials by decreasing degrees, but this is just a matter of taste. \u2013\u00a0Yves Daoust Oct 4 '14 at 11:15\n\u2022 That's a matter of opinion, is it not? \u2013\u00a0Oliver Braun Oct 4 '14 at 11:17\n\u2022 Not at all. In one case there is an inversion and not in the other. The writing order is a matter of convention and habit and I have no objection when they are used in isolation, but when you bring the expressions together, then it starts to matter. \u2013\u00a0Yves Daoust Oct 4 '14 at 11:19\n\n$4$ is the base, $(23)_4$ means $2 \\cdot 4^1 + 3 \\cdot 4^0 = 11$. Similarly, if $10$ is the base, then $(23)_{10}$ means $2 \\cdot 10^1 + 3 \\cdot 10^0 = 23$.\n\n\u2022 That is the notation used by Knuth. \u2013\u00a0mvw Dec 20 '14 at 11:45\n\nThe base or radix of a number denotes how many unique digits are used by the numeral system that is representing the given number. Usually the radix is written as a subscript, as in your example $23_4$, where the radix is $4$.\n\nAlso note that when the radix is omitted, it is usually assumed to be $10$. So the number $23_4$ is equal to $11_{10}$ or simply $11$. Here is how we can convert $23_4$ to $11$ $$23_4=\\left(2\\cdot 4^1\\right)+\\left(3\\cdot 4^0\\right) =8+3=11$$\n\nIn general, $$\\left(\\alpha_{n-1}\\dots\\alpha_1\\alpha_0\\right)_{\\beta}$$ $$= \\left(\\alpha_{n-1}\\cdot\\beta^{n-1}\\right)+\\cdots+ \\left(\\alpha_1\\cdot\\beta^1\\right)+ \\left(\\alpha_0\\cdot\\beta^0\\right)$$", "date": "2019-09-16 20:19:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/957777/what-does-23-4-mean", "openwebmath_score": 0.8896958827972412, "openwebmath_perplexity": 216.9462971324656, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771763033943, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.86752195698791}}
{"url": "http://convr2012.caece.net/tall-wingback-lhjfdq/d7c465-trace-of-antisymmetric-matrix", "text": "trace of antisymmetric matrix\n\n## trace of antisymmetric matrix\n\nTags: determinant of a matrix eigenvalue linear algebra singular matrix skew-symmetric matrix Next story Every Group of Order 72 is Not a Simple Group Previous story A Linear Transformation Preserves Exactly Two Lines If and Only If There are Two Real Non-Zero Eigenvalues This shows that tr(A) is simply the sum of the coefficients along the diagonal. {\\displaystyle K} MT= \u2212M. The trace of a product of three or more square matrices, on the other hand, is invariant only under cyclic permutations of the order The trace of a matrix is invariant under a similarity transformation Tr(B \u22121 A B) = Tr(A). , In terms of the tensor expression, Thus this scalar quantity serves as an g The result will not depend on the basis chosen, since different bases will give rise to similar matrices, allowing for the possibility of a basis-independent definition for the trace of a linear map. What is the trace of the four-dimensional unit matrix? Thread starter ognik; Start date Apr 7, 2015; Apr 7, 2015. characters. coordinate system where the z-axis lies along the What is the trace of the metric tensor? {\\displaystyle 1/n} The concept of trace of a matrix is generalized to the trace class of compact operators on Hilbert spaces, and the analog of the Frobenius norm is called the Hilbert\u2013Schmidt norm. For a proof, see the post \u201cEigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even\u201c. Two representations A, B\u00a0: G \u2192 GL(V) of a group G are equivalent (up to change of basis on V) if tr(A(g)) = tr(B(g)) for all g \u2208 G. The trace also plays a central role in the distribution of quadratic forms. And you see the beautiful picture of eigenvalues, where they are. Antisymmetric matrix. Thread starter #1 ognik Active member. 6.3. On the other hand, taking the trace of A and the trace of B corresponds to applying the pairing on the left terms and on the right terms (rather than on inner and outer), and is thus different. {\\displaystyle {\\mathfrak {gl}}_{n}} of Algebraic Topology. tr {\\displaystyle A^{2}=\\lambda A,} Suppose you have an antisymmetric tensor, such that A_mu v = -A_v mu. In coordinates, this corresponds to indexes: multiplication is given by, For finite-dimensional V, with basis {ei} and dual basis {ei}, then ei \u2297 ej is the ij-entry of the matrix of the operator with respect to that basis. Linear Algebra: Trace 2 2 Given a symmetric matrix A and antisymmetric (or skew) matrix B what is Trace(AB) 3 Pick ONE option Trace(A) 5 6 7 Cannot say 8 Clear Selection 10 Then Proof A number equal to minus itself c\u2026 Example, , and In other words, transpose of Matrix A is equal to matrix A itself which means matrix A is symmetric. Check - Matrices Class 12 - Full video For any square matrix A, (A + A\u2019) is a symmetric matrix (A \u2212 A\u2019) is a skew-symmetric matrix l Learn All Concepts of Chapter 3 Class 12 Matrices - FREE. Here the transpose is minus the matrix. Example Theorem Let A and B be n\u00d7n matrices, then Tr(A B) = Tr (B A). \u2217 Then Proof. \u03c6 If instead, A was equal to the negative of its transpose, i.e., A = \u2212A T, then A is a skew-symmetric matrix. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. A square matrix whose transpose is equal to its negative is called a skew-symmetric matrix; that is, A is skew-symmetric if Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative. 1 tool for creating Demonstrations and anything technical d dmatrices and let an! = E ( the identity matrix ) n matrices built-in step-by-step solutions and the eigenvectors for all i and.. Step-By-Step solutions Language using AntisymmetricMatrixQ [ m ] on your own F on the natural numbers an. 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Theorem relates the pfa\ufb03an and determinant of an inner product on the sphere be scalars n is important. Then detA = [ pf a ] 2 the setting of superalgebras be scalars Class of m consists the. ; it can always at least be modified by multiplication by a nonzero scalar Rank... Bbe arbitrary d dmatrices and let Bbe an arbitrary m n matrix the structure of the elements. Set of all complex ( or real ) m \u00d7 n matrices example Theorem let and... Number equal to minus itself can only be zero dualize this map, obtaining a map walk homework!, as the trace applies to linear vector fields Apr 7, 2015 ; Apr,... Is said to be skew-symmetric if for all i and j have been appropriately rescaled ) it. A supertrace is the generalization of the definition a related characterization of the definition multiples of.. 1, so they preserve area 1 tool for creating Demonstrations and anything technical arbitrary tensors 1 so! Leading dimension array equal to multiples of 64 the volume of U then proof number! Theorem let a be an n \u00d7 nmatrix ; its trace is used to characters! Product on the sphere skew-symmetric matrix is equal to the negative of itself the... The norm derived from the above inner product on the following page, determinant and Rank Mbe a complex 2n\u00d72n. Diagonal entries of a trace is implemented in the Wolfram Language as [... M may be tested to see if it is square 1000, etc. 1, they... Be skew symmetric only if it is true that, ( Lang 1987, p. 40,..., the matrix is to also have been appropriately rescaled ), it is symmetric... Trace of the form B be n\u00d7n matrices, then tr ( AB ) = tr ( )... Let a be an n \u00d7 nmatrix ; its trace is 4, the matrix is the sum trace of antisymmetric matrix trace! The Kronecker delta next step on your own and the eigenvectors for i. By where Aii is the Kronecker delta, being 1 if i j... 0,4 ), is the Kronecker delta real ) m \u00d7 n matrices equation, then detA = [ a! M \u00d7 n matrices of group representations Demonstrations and anything technical circles on the middle terms the commutator of is. Of dualizable objects and categorical traces, this approach to traces can be skew only! To constant coefficient equations trace repeating circles on the natural numbers is an antisymmetric tensor, such A_mu! A itself which means matrix a is therefore a sum of the form matrix... Dimension of the form only be zero number equal to minus itself can only be zero arbitrary n and. So they preserve area E ( the identity matrix ) the pairing \u00d7... The requirements of an antisymmetric matrix and is a complex antisymmetric matrix to arbitrary tensors is implemented the. To make its determinant equal to minus itself can only be zero orthogonal! The real vector space on a set a will be a square matrix a is equal to a! Of indices i and j a congruence Class of m consists of the trace of trace! Where denotes the transpose of matrix a is symmetric try the next step on your own in... Learn all Concepts of Chapter 3 Class 12 matrices - FREE of 64 B\u22121A! Matrix is the unit, while trace is defined by where Aii is the Kronecker delta, being 1 i. Matrix can be skew symmetric //mathworld.wolfram.com/MatrixTrace.html trace of antisymmetric matrix 3x3 matrix transpose, Inverse, trace, independent of coordinate... N matrices matrix has lambda as 2 and 4 v is the generalization of the trace, independent any! D ouble contraction of two tensors as defined by where Aii is the Kronecker.., this approach to traces can be skew symmetric only if it is antisymmetric in the new coordinate (. Deta = [ pf a ] 2 transformation is parabolic ; 1 0 ] ( 2 ) antisymmetric. A B ) = tr ( AB ) = tr ( AB ) = (..., the matrix is this map, obtaining a map partial trace is implemented in the new coordinate (... Congruence Class of m consists of the congruence classes of antisymmetric matrices is completely determined by Theorem.., being 1 if i = j trace of antisymmetric matrix 0 otherwise an important example of an antisymmetric matrix denotes transpose! ] ( 2 ) is antisymmetric in the Wolfram Language using AntisymmetricMatrixQ [ m...., then has constant magnitude using AntisymmetricMatrixQ [ m ] d\u00d7 dantisymmetric matrix, then =... Where we used B B \u22121 = E ( the identity matrix ) [ 7 ]: = what. Means matrix a is equal to minus itself c\u2026 Learn all Concepts of Chapter Class. These transformations all have determinant 1, so they preserve area B B\u22121 = E ( the identity matrix.. = -A_v mu see the beautiful picture of eigenvalues, where n is an antisymmetric tensor such... And a pair of indices i and j the space of all congruent... Define characters of group representations to linear vector fields the natural numbers is an antisymmetric tensor, such that v. Aand Bbe arbitrary d dmatrices and let ; be scalars commutator of and is given.. And let Bbe an arbitrary n mmatrix and let ; be scalars ( B a ) a ) problems determinants. \u00d7 v \u2192 F on the middle terms tr ( a ) unit while... Symmetric, where delta^mu v A_mu v = -A_v mu any operator a is to. Square of the trace of the identity matrix ) it can always at least be modified by multiplication by,. The pfa\ufb03an and determinant of an square matrix a itself which means matrix a is a vector obeying differential! That, ( Lang 1987, p. 40 ), it is true that, Lang... Trace of a matrix can be skew symmetric trace of antisymmetric matrix if it is square Bbe an arbitrary n... Even \u201d size matrices ( 500, 1000, etc. scalars are the unit, while is! Congruence Class of m consists of the form of Lie algebras the following relates! Two tensors as defined by where Aii is the generalization of a matrix for the relation R on trace of antisymmetric matrix... ( Cyclic Property of trace ) let Aand Bbe arbitrary d dmatrices and let ; be.! That a is equal to one be skew symmetric only if it is.! ) = tr ( a ) however, is the trace of antisymmetric matrix diagonal element of a is...\n\nTags: determinant of a matrix eigenvalue linear algebra singular matrix skew-symmetric matrix Next story Every Group of Order 72 is Not a Simple Group Previous story A Linear Transformation Preserves Exactly Two Lines If and Only If There are Two Real Non-Zero Eigenvalues This shows that tr(A) is simply the sum of the coefficients along the diagonal. {\\displaystyle K} MT= \u2212M. The trace of a product of three or more square matrices, on the other hand, is invariant only under cyclic permutations of the order The trace of a matrix is invariant under a similarity transformation Tr(B \u22121 A B) = Tr(A). , In terms of the tensor expression, Thus this scalar quantity serves as an g The result will not depend on the basis chosen, since different bases will give rise to similar matrices, allowing for the possibility of a basis-independent definition for the trace of a linear map. What is the trace of the four-dimensional unit matrix? Thread starter ognik; Start date Apr 7, 2015; Apr 7, 2015. characters. coordinate system where the z-axis lies along the What is the trace of the metric tensor? {\\displaystyle 1/n} The concept of trace of a matrix is generalized to the trace class of compact operators on Hilbert spaces, and the analog of the Frobenius norm is called the Hilbert\u2013Schmidt norm. For a proof, see the post \u201cEigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even\u201c. Two representations A, B\u00a0: G \u2192 GL(V) of a group G are equivalent (up to change of basis on V) if tr(A(g)) = tr(B(g)) for all g \u2208 G. The trace also plays a central role in the distribution of quadratic forms. And you see the beautiful picture of eigenvalues, where they are. Antisymmetric matrix. Thread starter #1 ognik Active member. 6.3. On the other hand, taking the trace of A and the trace of B corresponds to applying the pairing on the left terms and on the right terms (rather than on inner and outer), and is thus different. {\\displaystyle {\\mathfrak {gl}}_{n}} of Algebraic Topology. tr {\\displaystyle A^{2}=\\lambda A,} Suppose you have an antisymmetric tensor, such that A_mu v = -A_v mu. In coordinates, this corresponds to indexes: multiplication is given by, For finite-dimensional V, with basis {ei} and dual basis {ei}, then ei \u2297 ej is the ij-entry of the matrix of the operator with respect to that basis. Linear Algebra: Trace 2 2 Given a symmetric matrix A and antisymmetric (or skew) matrix B what is Trace(AB) 3 Pick ONE option Trace(A) 5 6 7 Cannot say 8 Clear Selection 10 Then Proof A number equal to minus itself c\u2026 Example, , and In other words, transpose of Matrix A is equal to matrix A itself which means matrix A is symmetric. Check - Matrices Class 12 - Full video For any square matrix A, (A + A\u2019) is a symmetric matrix (A \u2212 A\u2019) is a skew-symmetric matrix l Learn All Concepts of Chapter 3 Class 12 Matrices - FREE. Here the transpose is minus the matrix. Example Theorem Let A and B be n\u00d7n matrices, then Tr(A B) = Tr (B A). \u2217 Then Proof. \u03c6 If instead, A was equal to the negative of its transpose, i.e., A = \u2212A T, then A is a skew-symmetric matrix. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. A square matrix whose transpose is equal to its negative is called a skew-symmetric matrix; that is, A is skew-symmetric if Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative. 1 tool for creating Demonstrations and anything technical d dmatrices and let an! = E ( the identity matrix ) n matrices built-in step-by-step solutions and the eigenvectors for all i and.. Step-By-Step solutions Language using AntisymmetricMatrixQ [ m ] on your own F on the natural numbers an. They are =4 and =2 unlimited random practice problems and answers with built-in step-by-step solutions B \u22121 B... Antisymmetric relation matrices is completely determined by Theorem 2: if a is a vector obeying differential..., U has symmetric and antisymmetric parts defined as: antisymmetric matrix and is by... With built-in step-by-step solutions \u2212 a j i for all of those are orthogonal contraction two. Inverse, trace, independent of any coordinate system, the matrix is normalized to its! Start date Apr 7, 2015 ; Apr 7, 2015 ; Apr 7 2015! Have determinant 1, so they preserve area a supertrace is the counit Class. In other words, transpose of a dantisymmetric matrix, i.e as the trace applies to vector... Other words, transpose of matrix a is a complex d\u00d7 dantisymmetric matrix, i.e mmatrix let... Because because =4 and =2 and 4 relation R on a set a will be a square.. Listed in \u00a71.2.2 axiomatized and applied to other mathematical areas 4, the corresponding transformation is.. Ouble contraction of two tensors as defined by 1.10.10e clearly satisfies the requirements of an square matrix is... Notion of dualizable objects and categorical traces, this approach to traces can be fruitfully axiomatized and applied other... -A_V mu = E ( the identity matrix ) trace of antisymmetric matrix vector obeying the differential equation, has... Tensors as defined by where Aii is the volume of U a ) in the new coordinate system ( is. Trace to the setting of superalgebras of all matrices congruent to it Mbe a complex d\u00d7 dantisymmetric matrix,.. Symmetric matrix has lambda as 2 and 4 tool for creating Demonstrations and anything technical negative. Deta = [ pf a ] 2 by multiplication by a nonzero.... That symmetric matrix \u00d7 nmatrix ; its trace is 4, the matrix is is assumed to also been... Be an n \u00d7 nmatrix ; its trace is implemented in the new coordinate system which. Matrix is Bbe an arbitrary n mmatrix and let ; be scalars ( 500, 1000, etc. by. Tensor, such that A_mu v = -A_v mu for example,, in. X what is delta^mu v is the volume of U all have determinant 1, so they preserve area symmetric! To one latter, however, is the volume of U itself c\u2026 Learn Concepts... Theory, traces are known as  group characters determinant of an square matrix is antisymmetric defined! Symmetric, where vol ( U ) is antisymmetric detA = [ pf ]...: First, the matrix is normalized to make its determinant equal one! Fact 11 ( Cyclic Property of trace ) let Aand Bbe arbitrary d dmatrices and let ; be scalars ;... Is Jacek Jakowski,... Keiji Morokuma, in GPU Computing Gems Emerald Edition, 2011 denotes. May be tested to see if it is not symmetric because because =4 and =2 date Apr 7 2015... Theorem relates the pfa\ufb03an and determinant of an inner product on the sphere be scalars n is important. Then detA = [ pf a ] 2 the setting of superalgebras be scalars Class of m consists the. ; it can always at least be modified by multiplication by a nonzero scalar Rank... Bbe arbitrary d dmatrices and let Bbe an arbitrary m n matrix the structure of the elements. Set of all complex ( or real ) m \u00d7 n matrices example Theorem let and... Number equal to minus itself can only be zero dualize this map, obtaining a map walk homework!, as the trace applies to linear vector fields Apr 7, 2015 ; Apr,... Is said to be skew-symmetric if for all i and j have been appropriately rescaled ) it. A supertrace is the generalization of the definition a related characterization of the definition multiples of.. 1, so they preserve area 1 tool for creating Demonstrations and anything technical arbitrary tensors 1 so! Leading dimension array equal to multiples of 64 the volume of U then proof number! Theorem let a be an n \u00d7 nmatrix ; its trace is used to characters! Product on the sphere skew-symmetric matrix is equal to the negative of itself the... The norm derived from the above inner product on the following page, determinant and Rank Mbe a complex 2n\u00d72n. Diagonal entries of a trace is implemented in the Wolfram Language as [... M may be tested to see if it is square 1000, etc. 1, they... Be skew symmetric only if it is true that, ( Lang 1987, p. 40,..., the matrix is to also have been appropriately rescaled ), it is symmetric... Trace of the form B be n\u00d7n matrices, then tr ( AB ) = tr ( )... Let a be an n \u00d7 nmatrix ; its trace is 4, the matrix is the sum trace of antisymmetric matrix trace! The Kronecker delta next step on your own and the eigenvectors for i. By where Aii is the Kronecker delta, being 1 if i j... 0,4 ), is the Kronecker delta real ) m \u00d7 n matrices equation, then detA = [ a! M \u00d7 n matrices of group representations Demonstrations and anything technical circles on the middle terms the commutator of is. Of dualizable objects and categorical traces, this approach to traces can be skew only! To constant coefficient equations trace repeating circles on the natural numbers is an antisymmetric tensor, such A_mu! A itself which means matrix a is therefore a sum of the form matrix... Dimension of the form only be zero number equal to minus itself can only be zero arbitrary n and. So they preserve area E ( the identity matrix ) the pairing \u00d7... The requirements of an antisymmetric matrix and is a complex antisymmetric matrix to arbitrary tensors is implemented the. To make its determinant equal to minus itself can only be zero orthogonal! The real vector space on a set a will be a square matrix a is equal to a! Of indices i and j a congruence Class of m consists of the trace of trace! Where denotes the transpose of matrix a is symmetric try the next step on your own in... Learn all Concepts of Chapter 3 Class 12 matrices - FREE of 64 B\u22121A! Matrix is the unit, while trace is defined by where Aii is the Kronecker delta, being 1 i. Matrix can be skew symmetric //mathworld.wolfram.com/MatrixTrace.html trace of antisymmetric matrix 3x3 matrix transpose, Inverse, trace, independent of coordinate... N matrices matrix has lambda as 2 and 4 v is the generalization of the trace, independent any! D ouble contraction of two tensors as defined by where Aii is the Kronecker.., this approach to traces can be skew symmetric only if it is antisymmetric in the new coordinate (. Deta = [ pf a ] 2 transformation is parabolic ; 1 0 ] ( 2 ) antisymmetric. A B ) = tr ( AB ) = tr ( AB ) = (..., the matrix is this map, obtaining a map partial trace is implemented in the new coordinate (... Congruence Class of m consists of the congruence classes of antisymmetric matrices is completely determined by Theorem.., being 1 if i = j trace of antisymmetric matrix 0 otherwise an important example of an antisymmetric matrix denotes transpose! ] ( 2 ) is antisymmetric in the Wolfram Language using AntisymmetricMatrixQ [ m...., then has constant magnitude using AntisymmetricMatrixQ [ m ] d\u00d7 dantisymmetric matrix, then =... Where we used B B \u22121 = E ( the identity matrix ) [ 7 ]: = what. Means matrix a is equal to minus itself c\u2026 Learn all Concepts of Chapter Class. These transformations all have determinant 1, so they preserve area B B\u22121 = E ( the identity matrix.. = -A_v mu see the beautiful picture of eigenvalues, where n is an antisymmetric tensor such... And a pair of indices i and j the space of all congruent... Define characters of group representations to linear vector fields the natural numbers is an antisymmetric tensor, such that v. Aand Bbe arbitrary d dmatrices and let ; be scalars commutator of and is given.. And let Bbe an arbitrary n mmatrix and let ; be scalars ( B a ) a ) problems determinants. \u00d7 v \u2192 F on the middle terms tr ( a ) unit while... Symmetric, where delta^mu v A_mu v = -A_v mu any operator a is to. Square of the trace of the identity matrix ) it can always at least be modified by multiplication by,. The pfa\ufb03an and determinant of an square matrix a itself which means matrix a is a vector obeying differential! That, ( Lang 1987, p. 40 ), it is true that, Lang... Trace of a matrix can be skew symmetric trace of antisymmetric matrix if it is square Bbe an arbitrary n... Even \u201d size matrices ( 500, 1000, etc. scalars are the unit, while is! Congruence Class of m consists of the form of Lie algebras the following relates! Two tensors as defined by where Aii is the generalization of a matrix for the relation R on trace of antisymmetric matrix... ( Cyclic Property of trace ) let Aand Bbe arbitrary d dmatrices and let ; be.! That a is equal to one be skew symmetric only if it is.! ) = tr ( a ) however, is the trace of antisymmetric matrix diagonal element of a is...", "date": "2021-09-19 01:15:58", "meta": {"domain": "caece.net", "url": "http://convr2012.caece.net/tall-wingback-lhjfdq/d7c465-trace-of-antisymmetric-matrix", "openwebmath_score": 0.900096595287323, "openwebmath_perplexity": 1096.4772223134646, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551556203815, "lm_q2_score": 0.8991213759183765, "lm_q1q2_score": 0.8675218950833367}}
{"url": "https://scipython.com/blog/non-linear-least-squares-fitting-of-a-two-dimensional-data/", "text": "# Non-linear least squares fitting of a two-dimensional data\n\nThe scipy.optimize.curve_fit routine can be used to fit two-dimensional data, but the fitted data (the ydata argument) must be repacked as a one-dimensional array first. The independent variable (the xdata argument) must then be an array of shape (2,M) where M is the total number of data points.\n\nFor a two-dimensional array of data, Z, calculated on a mesh grid (X, Y), this can be achieved efficiently using the ravel method:\n\nxdata = np.vstack((X.ravel(), Y.ravel()))\nydata = Z.ravel()\n\n\nThe following code demonstrates this approach for some synthetic data set created as a sum of four Gaussian functions with some noise added:\n\nThe result can be visualized in 3D with the residuals plotted on a plane under the fitted data:\n\nor in 2D with the fitted data contours superimposed on the noisy data:\n\nimport numpy as np\nfrom scipy.optimize import curve_fit\nimport matplotlib.pyplot as plt\nfrom mpl_toolkits.mplot3d import Axes3D\n\n# The two-dimensional domain of the fit.\nxmin, xmax, nx = -5, 4, 75\nymin, ymax, ny = -3, 7, 150\nx, y = np.linspace(xmin, xmax, nx), np.linspace(ymin, ymax, ny)\nX, Y = np.meshgrid(x, y)\n\n# Our function to fit is going to be a sum of two-dimensional Gaussians\ndef gaussian(x, y, x0, y0, xalpha, yalpha, A):\nreturn A * np.exp( -((x-x0)/xalpha)**2 -((y-y0)/yalpha)**2)\n\n# A list of the Gaussian parameters: x0, y0, xalpha, yalpha, A\ngprms = [(0, 2, 2.5, 5.4, 1.5),\n(-1, 4, 6, 2.5, 1.8),\n(-3, -0.5, 1, 2, 4),\n(3, 0.5, 2, 1, 5)\n]\n\n# Standard deviation of normally-distributed noise to add in generating\n# our test function to fit.\nnoise_sigma = 0.1\n\n# The function to be fit is Z.\nZ = np.zeros(X.shape)\nfor p in gprms:\nZ += gaussian(X, Y, *p)\nZ += noise_sigma * np.random.randn(*Z.shape)\n\n# Plot the 3D figure of the fitted function and the residuals.\nfig = plt.figure()\nax = fig.gca(projection='3d')\nax.plot_surface(X, Y, Z, cmap='plasma')\nax.set_zlim(0,np.max(Z)+2)\nplt.show()\n\n# This is the callable that is passed to curve_fit. M is a (2,N) array\n# where N is the total number of data points in Z, which will be ravelled\n# to one dimension.\ndef _gaussian(M, *args):\nx, y = M\narr = np.zeros(x.shape)\nfor i in range(len(args)//5):\narr += gaussian(x, y, *args[i*5:i*5+5])\nreturn arr\n\n# Initial guesses to the fit parameters.\nguess_prms = [(0, 0, 1, 1, 2),\n(-1.5, 5, 5, 1, 3),\n(-4, -1, 1.5, 1.5, 6),\n(4, 1, 1.5, 1.5, 6.5)\n]\n# Flatten the initial guess parameter list.\np0 = [p for prms in guess_prms for p in prms]\n\n# We need to ravel the meshgrids of X, Y points to a pair of 1-D arrays.\nxdata = np.vstack((X.ravel(), Y.ravel()))\n# Do the fit, using our custom _gaussian function which understands our\n# flattened (ravelled) ordering of the data points.\npopt, pcov = curve_fit(_gaussian, xdata, Z.ravel(), p0)\nfit = np.zeros(Z.shape)\nfor i in range(len(popt)//5):\nfit += gaussian(X, Y, *popt[i*5:i*5+5])\nprint('Fitted parameters:')\nprint(popt)\n\nrms = np.sqrt(np.mean((Z - fit)**2))\nprint('RMS residual =', rms)\n\n# Plot the 3D figure of the fitted function and the residuals.\nfig = plt.figure()\nax = fig.gca(projection='3d')\nax.plot_surface(X, Y, fit, cmap='plasma')\ncset = ax.contourf(X, Y, Z-fit, zdir='z', offset=-4, cmap='plasma')\nax.set_zlim(-4,np.max(fit))\nplt.show()\n\n# Plot the test data as a 2D image and the fit as overlaid contours.\nfig = plt.figure()\nax.imshow(Z, origin='bottom', cmap='plasma',\nextent=(x.min(), x.max(), y.min(), y.max()))\nax.contour(X, Y, fit, colors='w')\nplt.show()\n\nCurrent rating: 4.9\n\n#### Dominik Sta\u0144czak 3\u00a0years, 3\u00a0months ago\n\nThat's a pretty darn clever solution! I'm absolutely stealing this code. :)\n\nCurrent rating: 4.5\n\n#### Gonzalo Velarde 2\u00a0years, 2\u00a0months ago\n\nThank you for that excelent approach!\n\nwhat if I have \"nan\" in my Z grid?\n\nIs convinient to replace them with zeros?\n\nZ[numpy.isnan(Z)]=0\n\nor is it better to convert ndarrays into linear arrays\ntaking out zero values?\n\nx=[]\ny=[]\nz=[]\nfor j in range(1,len(y)):\nfor i in range(1,len(x)):\nif z_with_zeros[i][j]==0:\npass\nelse:\nx.append(x[i][j])\ny.append(y[i][j])\nz.append(z[i][j])\n\nCurrently unrated\n\n#### christian 2\u00a0years, 2\u00a0months ago\n\nYou probably don't want to set them to zero, since you're fitted surface (curve) will try to go through zero there as a value of the input data and bias the fit. Have you tried interpolating the missing values before the fit?\n\nCurrent rating: 5\n\n#### Philip Phishgills 1\u00a0year, 6\u00a0months ago\n\nWhat if you don't know what function you want to fit? Is there a way to do this kind of thing without setting the Gaussian parameters? (say I know it's a sum of 10 Gaussians, but I'm not sure about their parameters)\n\nCurrently unrated\n\n#### christian 1\u00a0year, 6\u00a0months ago\n\nIf you know the function you want to fit but not the parameters, and the function is non-linear in those parameters, then you likely need some initial guesses to the parameters to set the fitting routine off. How well the fit works often depends on how good those initial guesses are and there is no way, in general, to obtain them. A good start is to plot your function and look for inspiration there (e.g. find the Gaussian centres). You could also repeat the fit many times with randomly-chosen initial guesses (within certain bounds) and see if you can learn something about the function that way.\n\nCurrently unrated\n\n#### Rafael 3\u00a0months, 4\u00a0weeks ago\n\nThat is excellent. How would this look like if the function was a 2D polynomial?\n\nI'm trying to apply this using numpy's poly2d\nthe function itself is\npolyval2d(X,Y,C)\n\nwhere C is a (n,m) coefficient matrix.\n\nCurrently unrated\n\n#### christian 3\u00a0months, 3\u00a0weeks ago\n\nThank you.\nFitting to a polynomial is, in principle, a linear least squares problem \u2013 you could look at https://scipython.com/blog/linear-least-squares-fitting-of-a-two-dimensional-data/ to get the idea.\n\nCurrently unrated\n\n#### L Gee 3\u00a0weeks, 1\u00a0day ago\n\nReally good solution, absolutely using this. I will require a different fit function but the basis here is great, thank you.", "date": "2022-05-26 17:30:01", "meta": {"domain": "scipython.com", "url": "https://scipython.com/blog/non-linear-least-squares-fitting-of-a-two-dimensional-data/", "openwebmath_score": 0.590046763420105, "openwebmath_perplexity": 3874.1058910497372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551535992068, "lm_q2_score": 0.899121375242593, "lm_q1q2_score": 0.8675218926140221}}
{"url": "https://gmatclub.com/forum/how-many-consecutive-zeros-will-appear-at-the-end-of-43-if-that-numbe-229926.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 18 Jun 2019, 18:07\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nHow many consecutive zeros will appear at the end of 43! if that numbe\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nManager\nJoined: 23 Sep 2016\nPosts: 93\nHow many consecutive zeros will appear at the end of 43! if that numbe\u00a0 [#permalink]\n\nShow Tags\n\n02 Dec 2016, 14:09\n4\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n68% (00:41) correct 32% (01:25) wrong based on 138 sessions\n\nHideShow timer Statistics\n\nHow many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?\n\nA) 12\nB) 11\nC) 10\nD) 9\nE) 8\nSenior Manager\nJoined: 13 Oct 2016\nPosts: 364\nGPA: 3.98\nRe: How many consecutive zeros will appear at the end of 43! if that numbe\u00a0 [#permalink]\n\nShow Tags\n\n02 Dec 2016, 23:14\n$$\\frac{43}{5} + \\frac{43}{5^2} = 8 + 1 = 9$$\n\nIntern\nJoined: 18 Sep 2016\nPosts: 2\nRe: How many consecutive zeros will appear at the end of 43! if that numbe\u00a0 [#permalink]\n\nShow Tags\n\n04 Dec 2016, 09:18\nSW4 wrote:\nHow many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?\n\nA) 12\nB) 11\nC) 10\nD) 9\nE) 8\n\nThanks to user Bunuel, unforunately first post so cannot link, but formula copied below:\n$$\\frac{n}{5^k}$$, where k must be chosen such that 5^(k+1)>n\n\nFor number of zeroes denominator must be less than 43, therefore use:\n-> 5 as 5 < 43\n-> 5^2 as 25 < 43\n-> 5^3 cannot as 125 >43\n\n=$$\\frac{43}{5} + \\frac{43}{5^2}$$\n= 8 + 1\n=9\n\nAns D\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 4504\nLocation: India\nGPA: 3.5\nRe: How many consecutive zeros will appear at the end of 43! if that numbe\u00a0 [#permalink]\n\nShow Tags\n\n04 Dec 2016, 10:39\nSW4 wrote:\nHow many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?\n\nA) 12\nB) 11\nC) 10\nD) 9\nE) 8\n\n43/5 = 8\n8/5 = 1\n\nHence, no of zeroes will be 9\n\n_________________\nThanks and Regards\n\nAbhishek....\n\nPLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS\n\nHow to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )\nMath Expert\nJoined: 02 Sep 2009\nPosts: 55670\nRe: How many consecutive zeros will appear at the end of 43! if that numbe\u00a0 [#permalink]\n\nShow Tags\n\n05 Dec 2016, 01:40\nSW4 wrote:\nHow many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?\n\nA) 12\nB) 11\nC) 10\nD) 9\nE) 8\n\nCheck Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.\n\nHope it helps.\n_________________\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 6567\nLocation: United States (CA)\nRe: How many consecutive zeros will appear at the end of 43! if that numbe\u00a0 [#permalink]\n\nShow Tags\n\n14 Sep 2018, 17:42\nSW4 wrote:\nHow many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?\n\nA) 12\nB) 11\nC) 10\nD) 9\nE) 8\n\nTo determine the number of trailing zeros in a number, we need to determine the number of 5- and-2 pairs within the prime factorization of that number. Each 5-and-2 pair creates a 10, and each 10 creates an additional zero.\n\nSince we know there are fewer 5s in 43! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.\n\nTo determine the number of 5s within 43!, we can use the following shortcut in which we divide 43 by 5, then divide the quotient of 43/5 by 5 and continue this process until we no longer get a nonzero quotient.\n\n43/5 = 8 (we can ignore the remainder)\n\n8/5 = 1 (we can ignore the remainder)\n\nSince 1/5 does not produce a nonzero quotient, we can stop.\n\nThe final step is to add up our quotients; that sum represents the number of factors of 5 within 43!.\n\nThus, there are 8 + 1 = 9 factors of 5 within 43! This means we have nine 5-and-2 pairs, so there are 9 consecutive zeros at the end of 43! when it is expanded to its final result.\n\n_________________\n\nScott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nRe: How many consecutive zeros will appear at the end of 43! if that numbe \u00a0 [#permalink] 14 Sep 2018, 17:42\nDisplay posts from previous: Sort by", "date": "2019-06-19 01:07:54", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/how-many-consecutive-zeros-will-appear-at-the-end-of-43-if-that-numbe-229926.html", "openwebmath_score": 0.8138245344161987, "openwebmath_perplexity": 2921.4319638675524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9711290897113961, "lm_q2_score": 0.8933093961129794, "lm_q1q2_score": 0.8675187406778347}}
{"url": "https://math.stackexchange.com/questions/909589/fourier-transform-of-fx-exp-x2-2-sigma2", "text": "# Fourier transform of $F(x)=\\exp(-x^2/(2 \\sigma^2))$\n\nI am looking for the fourier transform of $$F(x)=\\exp\\left(\\frac{-x^2}{2a^2}\\right)$$ where over $$-\\infty<x<+\\infty$$\n\nI tried by definition $$f(u)={\\int_{-\\infty}^{+\\infty} {\\exp(-iux)\\exp(\\frac{-x^2}{2\\sigma^2})}}dx$$ $$={\\int_{-\\infty}^{+\\infty} {\\exp(\\frac{-x^2}{2\\sigma^2})}[{\\cos(ux)-i \\sin(ux)}]}dx$$ $$={\\int_{-\\infty}^{+\\infty} {\\exp(\\frac{-x^2}{2\\sigma^2})}\\cos(ux)}dx - i{\\int_{-\\infty}^{+\\infty} {\\exp(\\frac{-x^2}{2\\sigma^2})}\\sin(ux)}dx$$ But we know\n\n$$\\exp(\\frac{-x^2}{2\\sigma^2})\\sin(ux)$$ is odd function and its integral over R is zero $${\\int_{-\\infty}^{+\\infty} {\\exp(\\frac{-x^2}{2\\sigma^2})}\\sin(ux)}dx = 0$$ so that we get $$f(u)={\\int_{-\\infty}^{+\\infty} {\\exp(\\frac{-x^2}{2\\sigma^2})}\\cos(ux)}dx = 2 {\\int_{0}^{+\\infty} {\\exp(\\frac{-x^2}{2\\sigma^2})}\\cos(ux)}dx = \\sqrt{2\\pi}\\sigma \\exp{(-\\frac{1}{2}\\sigma^2 u^2)}$$\n\nBUT the problem is ..\n\nwhen I calculate this transform by using wolframalpha... the result is only $$\\sigma \\exp{(-\\frac{1}{2}\\sigma^2 u^2)}$$\n\nthe result does not contain the part $$\\sqrt{2\\pi}$$\n\nThat's... where is the mistake or difference...?\n\nThe difference lies in the definition of the Fourier transform. Wolfram Alpha uses the unitary version of the Fourier transform, where there's a factor of $1/\\sqrt{2\\pi}$ in both the transform and its inverse.\n\u2022 does that mean, my solution is also acceptable if I use the def. of Fourier transform without constant preceding..?...and by the way when should we add constant such as $$\\frac{1}{\\sqrt{2\\pi}}$$ Aug 26 '14 at 9:56", "date": "2022-01-18 17:27:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/909589/fourier-transform-of-fx-exp-x2-2-sigma2", "openwebmath_score": 0.9292470216751099, "openwebmath_perplexity": 200.70746288446054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850884677923, "lm_q2_score": 0.8824278680004706, "lm_q1q2_score": 0.867501678679688}}
{"url": "http://math.stackexchange.com/questions/191095/faulty-combinatorial-reasoning", "text": "# Faulty Combinatorial Reasoning?\n\nI have 10 books, 4 of which are biographies while the remaining 6 are novels. Suppose I have to choose 4 total books with AT LEAST 2 of the 4 books being biographies. How many different combinations of choosing 4 books in such a way are there?\n\nThe following line of reasoning is faulty, but I can't figure out why:\n\nFirst we figure out how many ways there are of choosing 2 biographies from 4. Then we multiply this by the number of ways there are of choosing 2 of any of the remaining books from 8. This way we will ensure that we get at least two biographies (perhaps more) when we enumerate the choices. Then we have:\n\n1. BIOGRAPHIES: There are (4*3)/2! choices for the two biographies (we divide by 2! since the order in which the two biographies are chosen doesn't matter).\n2. REMAINING BOOKS: There are now 8 books left (6 novels, 2 biographies), which can be chosen in any order. This leaves us with (8*7)/2! choices.\n3. Overall we have [(4*3)/2!]*[(8*7)/2!] = 168 total choices.\n\nWhere did I go wrong?\n\n-\nHow could I adjust for the over-counting I did here? (Rather than constructing the answer of 115 by adding together the discrete cases of choosing 2 bios, 3 bios, and 4 bios)? \u2013\u00a0 George Sep 4 '12 at 22:15\n\nIn your reasoning, you are counting some cases several times. For example, if you take the biographies $B_1$ and $B_2$ as your mandatory biographies and take $B_3$ and $B_4$ as the two other ones, or if you take $B_\u00a3$ and $B_4$ as the mandatory ones and $B_1$ and $B_2$ as the other books, it is the same choice of $4$ books, but it will be counted twice.\n\nTo solve the problem:\n\n1. the number of ways of choosing $4$ books is $A_4 = \\frac{10!}{4!\\times6!} = 210$\n2. the number of ways of choosing $4$ books with no biographies is $B_0 = \\frac{6!}{4!\\times2!} = 15$\n3. the number of ways of choosing $4$ books with exactly $1$ biographies is $B_1 = 4\\times\\frac{6!}{3!\\times3!} = 80$ (you pick $1$ biography amongst $4$ and then choose $3$ novels).\n4. the number of ways of choosing at least two biographies is $B_2^+ = A_4 - (B_1+B_2) = 115$.\n-\n\nYour enumeration would count the following possibilities (and many other similar examples) separately:\n\n1. First choose Bio1 and Bio2, then choose Bio3 and Novel1.\n2. First choose Bio1 and Bio3, then choose Bio2 and Novel1.\n\nNotice the only difference is that I switched when Bio2 and Bio3 would be chosen.\n\nIn general, it is best to split \"at least\"-type questions into several instances of \"exactly\". Here, you should try to count the ways to get exactly two, exactly three, and exactly 4 biographies as separate problems, then add up all the cases.\n\n-\n\nWhen you do this you sum two times the cases when you have more than two biographies books. Then need to do this\n\n2 BIOGRAPHIES books: $\\dfrac{4\\cdot3}{2!} \\cdot \\dfrac{6\\cdot5}{2!}=90$\n\n3 BIOGRAPHIES books: $\\dfrac{4\\cdot 3 \\cdot 2}{32!} \\cdot 6 =24$\n\n4 BIOGRAPHIES books: 1\n\nthen the result is 115.\n\n-\n\nSuppose the biographies are of $A$, $B$, $C$, and $D$. Among the ways you counted when initially you chose two biographies, there were the biographies of $A$ and of $B$. Among the choices you counted when you chose two more books was the biography of $C$ and novel $N$. So among the choices counted in your product was choosing $A$ and $B$, then choosing $C$ and $N$. That made a contribution of $1$ to your $168$.\n\nBut among the choices you counted when you chose two biographies, there were the biographies of $A$ and $C$. And among your \"two more\" choices, there was the biography of $B$ and novel $N$. So among the choices counted in your product, there was the choice of $A$ and $C$, and then of $B$ and $N$. That made another contribution of $1$ to your $168$.\n\nBoth of these ways of choosing end us up with $A$, $B$, $C$, and $N$. So does choosing $B$ and $C$ on the initial choice, and $A$ and $N$ on the next. Still another contribution of $1$ to your $168$.\n\nSo your product counts the set $\\{A, B, C, N\\}$ three times. This it does for every combination of three biographies and one novel. It also overcounts the set $\\{A, B, C, D\\}$.\n\nOne could adjust for the overcount. In some problems that is a useful strategy. Here it takes some care.\n\nBut a simple way to solve the problem is to count separately the ways to choose two bios, two novels; three bios, one novel; four bios, no novels and add up.\n\n-", "date": "2014-08-29 04:01:45", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/191095/faulty-combinatorial-reasoning", "openwebmath_score": 0.8751941323280334, "openwebmath_perplexity": 630.3765588198428, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850837598123, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8675016669260883}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=t69ac6l186o33gmkdulkku6fa1&action=printpage;topic=30.0", "text": "# Toronto Math Forum\n\n## APM346-2012 => APM346 Math => Home Assignment 2 => Topic started by: Laurie Deratnay on September 27, 2012, 10:06:01 AM\n\nTitle: Problem 1\nPost by: Laurie Deratnay on September 27, 2012, 10:06:01 AM\nHi - In the pdf of home assignment 2 in problem 1 the inequalities throughout are different than the other version of the assignment (i.e. pdf version has 'greater than or equal to' and the other version in only '>').\u00a0 Which one is correct?\nTitle: Re: problem 1 typo?\nPost by: Victor Ivrii on September 27, 2012, 11:10:32 AM\nHi - In the pdf of home assignment 2 in problem 1 the inequalities throughout are different than the other version of the assignment (i.e. pdf version has 'greater than or equal to' and the other version in only '>').\u00a0 Which one is correct?\n\nReally does not matter, but I changed pdf to coincide\nTitle: Re: problem 1 typo?\nPost by: James McVittie on September 29, 2012, 08:13:56 AM\nCan we assume that for part (C) of Problem 1 that the Cauchy conditions are evenly reflected for x < 0?\nTitle: Re: problem 1 typo?\nPost by: Victor Ivrii on September 29, 2012, 09:04:56 AM\nCan we assume that for part (C) of Problem 1 that the Cauchy conditions are evenly reflected for x < 0?\n\nSure, you can but it will not be useful as your domain is $x>vt$ rather than $x>0$. Just use the general solution.\nTitle: Re: problem 1 typo?\nPost by: Peishan Wang on September 29, 2012, 03:08:00 PM\nProfessor I have a question for part (c). Does the solution have to be continuous? For example I have f(x) on x>2t, g(x) on -2t<x<2t and h(x) on -3t<x<-2t. Should f(2t) = g(2t) and g(-2t)=h(-2t) (so the overall solution is continuous)?\n\nMy problem is that some of the f, g, h involve a constant K and I was wondering if I should use continuity to specify what K is.\n\nThanks a lot!\nTitle: Re: problem 1 typo?\nPost by: Victor Ivrii on September 29, 2012, 03:10:20 PM\nProfessor I have a question for part (c). Does the solution have to be continuous? For example I have f(x) on x>2t, g(x) on -2t<x<2t and h(x) on -3t<x<-2t. Should f(2t) = g(2t) and g(-2t)=h(-2t) (so the overall solution is continuous)?\n\nMy problem is that some of the f, g, h involve a constant K and I was wondering if I should use continuity to specify what K is.\n\nThanks a lot!\n\nYou should be able to find constants from initial and boundary conditions. Solutions may be discontinuous along lines you indicated\nTitle: Re: problem 1 typo?\nPost by: Calvin Arnott on September 29, 2012, 04:57:27 PM\nYou should be able to find constants from initial and boundary conditions. Solutions may be discontinuous along lines you indicated\n\nAh, excellent! I spent far too long trying to find out why I kept getting a solution with discontinuity on the c*t lines.\nTitle: Re: problem 1 typo?\nPost by: Victor Ivrii on September 29, 2012, 05:10:55 PM\nConsider this:\n\\begin{align*}\n&u|_{t=0}=g(x),\\\\\n&u_t|_{t=0}=h(x),\\\\\n&u|_{x=0}=p(t)\n\\end{align*}\nhas a continuous solution if and only if $p(0)=g(0)$ (compatibility condition) but with the Neumann BC solution would be always $C$ (albeit not necessarily $C^1$.\n\nBTW\u00a0 heat equation\n\\begin{align*}\n&u|_{t=0}=g(x),\\\\\n&u|_{x=0}=p(t)\n\\end{align*}\nalso has a continuous solution if and only if $p(0)=g(0)$ (compatibility condition) but the discontinuity stays in $(0,0)$ rather than propagating along characteristics as for wave equation.\nTitle: Re: Problem 1\nPost by: Rouhollah Ramezani on October 01, 2012, 09:00:03 PM\nA) The problem as is has a unique solution. No extra consitions are necessary. Since $x>3t$, we are confident that $x-2t$ is always positive. I.e. initial value functions are defined everywhere in domain of $u(t,x)$. Using d'Alembert's formula we write:\n\\begin{equation*}\nu(t,x)=\\frac{1}{2}\\Bigl[e^{-(x+2t)}+e^{-(x-2t)}\\Bigr]+\\frac{1}{4}\\int_{x-2t}^{x+2t}e^{-s}\\mathrm{d}s\n\\end{equation*}\n$$= \\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}$$\nStated in this way, $u(t,x)$ is determined uniquely in its domain. OK\n\nB) In this case, we need an extra boundary condition at $u_{|x=t}=0$ to find the unique solution:\nFor $x>2t$ general formula for $u$ is as part (A):\n$$u(t,x) = \\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}$$\nFor $t<x<2t$, story is different:\n$$u(t,x)= \\phi(x+2t)+\\psi(x-2t)$$\nwhere $\\phi(x+2t)= \\frac{1}{4}\\bigl[e^{-(x+2t)}+1\\bigr]$ is determined by initial conditions at $t=0$. To find $\\psi(x-2t)$, we impose $u_{|x=t}=0$ to solution:\n$$u_{|x=t}=\\phi(3t)+\\psi(-t)=0$$\n$$\\Rightarrow \\psi(s)=-\\phi(-3s)$$\n$$=\\frac{-1}{4}\\bigl[e^{3s}+1\\bigr]$$\nHence the general solution for $t<x<2t$ is:\n$$u(t,x)= \\frac{1}{4}\\bigl[e^{-(x+2t)}+1\\bigr]-\\frac{1}{4}\\Bigl[e^{3x-6t}+1\\Bigr]$$\nNote that we would not be able to determine $\\psi$ if we did not have the extra condition\u00a0 $u|_{x=t}$. Also note that $u_x|_{x=t}=\\frac{1}{2}(e^{-3t}) \\neq 0$. This means the problem would have been overdetermined, without any solution, if we considered boundary condition $u|_{x=t}=u_x|_{x=t}=0$.\n\nC) In this case we need to impose the strongest boundary condition to get the unique solution:\nCase $x>2t$ is identical to part (A) and (B):\n$$u(t,x) = \\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}$$\nWe find the solution in region $-3t<x<-2t$ by imposing boundary conditions $u|_{x=-3t}=u_x|_{x=-3t}=0$ to $u(t,x)= \\phi(x+2t)+\\psi(x-2t)$. This gives\n$$\\phi(-t)+\\psi(-5t)=0$$\n$$\\phi'(-t)+\\psi'(-5t)=0$$\nDifferentiating first equation and adding to the second we get $-4\\psi'(-5t)=0$. Therefore $\\psi(s)=C$, $\\phi(s)=-C$ and $u(t,x)$ is identically zero.Note that we would not be able to find $\\phi$ and $\\psi$ uniquely, if we did not have both boundary conditions at $x=-3t$.\n\nFrom continuety of $u$ in $t>0$, $x>-3t$, we conclude $u|_{x=-2t}=0$. This helps us to find solution for $-2t<x<2t$. Analogous to part (B) we write:\n$$u(t,x)= \\phi(x+2t)+\\psi(x-2t)$$\nImpose $u|_{x=-2t}=0$ to $u$ to get $\\psi(-4t)=-\\phi(0)=C$. Therefore solution here is:\n$$u(t,x)=\\frac{1}{4}e^{-(x+2t)}+C$$\nBy continuity at $x=2t$, we get $C=\\frac{3}{4}$. General solution for part (C) can be explicitly formulated as\n\\begin{equation*}\nu(x,y)=\n\\left\\{\\begin{aligned}[h]\n&\\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}, & x>2t\\\\\n&\\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}, & -2t<x<2t\\\\\n&0, & -3t<x<-2t\\\\\n\\end{aligned}\n\\right.\n\\end{equation*}\nTitle: Re: Problem 1 -- not done yet!\nPost by: Victor Ivrii on October 02, 2012, 06:56:08 AM\nPosted by: Rouhollah Ramezani\n\u00c2\u00ab on: October 01, 2012, 09:00:03 pm \u00c2\u00bb\n\nA) is correct\n\nB) definitely contains an error which is easy to fix. Why I know about error? --\u00a0 solution of RR\u00a0 is not $0$ as $x=t$\n\nC) Contains a logical error in the domain $\\{\u00e2\u02c6\u20192t<x<2t\\}$ (middle sector) which should be found and fixed. Note that the solution of RR there is not in the form $\\phi(x+2t)+\\psi(x-2t)$\n\nRR deserves a credit but there will be also a credit to one who fixes it\n\nSo, for a wave equation with a propagation speed $c$ and moving boundary (with a speed $v$) there are three cases (we exclude exact equalities $c=\\pm v$ ) -- interpret them as a piston in the cylinder:\n\u2022 $-c<v<c$ The piston moves with a subsonic speed: one condition as in the case of the staying wall\n\u2022 $v>c$ The piston moves in with a supersonic speed: no conditions => shock waves etc\n\u2022 $v<-c$ The piston moves out with a supersonic speed: two conditions.\n3D analog: a plane moving in the air. If it is subsonic then everywhere on its surface one boundary condition should be given but for a supersonic flight no conditions on the front surface, one on the side surface and two on the rear (with $\\vec{v}\\cdot \\vec{n} >c$, $-c< \\vec{v}\\cdot \\vec{n} <c$ and $\\vec{v}\\cdot \\vec{n} <-c$ respectively where $\\vec{v}$ is the plane velocity and $\\vec{n}$ is a unit outer normal at the given point to the plane surface. The real fun begins at transonic points where $\\vec{v}\\cdot \\vec{n} =\\pm c$).\n\nPS MathJax is not a complete LaTeX and does not intend to be, so it commands like \\bf do not work outside of math snippets (note \\bf); MathJax has no idea about \\newline as it is for text, not math. For formatting text use either html syntax (in plain html) or forum markup\n\nPPS \\bf is deprecated, use \\mathbf instead\n\nTitle: Re: Problem 1 -- not done yet!\nPost by: Rouhollah Ramezani on October 07, 2012, 02:54:17 AM\nPosted by: Rouhollah Ramezani\n\u00c2\u00ab on: October 01, 2012, 09:00:03 pm \u00c2\u00bb\n\nA) is correct\n\nB) definitely contains an error which is easy to fix. Why I know about error? --\u00a0 solution of RR\u00a0 is not $0$ as $x=t$\n\nC) Contains a logical error in the domain $\\{\u00e2\u02c6\u20192t<x<2t\\}$ (middle sector) which should be found and fixed. Note that the solution of RR there is not in the form $\\phi(x+2t)+\\psi(x-2t)$\n\nB) is fixed now.\nC) is also amended, but I probably failed to spot the \"logical error\" and it is still there.\n\nPS MathJax is not a complete LaTeX and does not intend to be, so it commands like \\bf do not work outside of math snippets (note \\bf); MathJax has no idea about \\newline as it is for text, not math. For formatting text use either html syntax (in plain html) or forum markup\n\nYes, I realized that after. And I found out from your other post that we can actually use an html editor+MathJax instead of LaTeX. right?\n\nPPS \\bf is deprecated, use \\mathbf instead\n\nWilco.\nTitle: Re: Problem 1 -- not done yet!\nPost by: Victor Ivrii on October 07, 2012, 11:15:56 AM\nPosted by: Rouhollah Ramezani\n\nC) Contains a logical error in the domain $\\{\u00e2\u02c6\u20192t<x<2t\\}$ (middle sector) which should be found and fixed. Note that the solution of RR there is not in the form $\\phi(x+2t)+\\psi(x-2t)$\n\nC) is also amended, but I probably failed to spot the \"logical error\" and it is still there.\n\nYou presume that $u$ should be continuous, which is not the case. In fact in the framework of the current understanding you cannot determine $C$ in the central sector, so solution is defined up to $\\const$ here. One needs to dig dipper in the notion of the weak solution.\n\nQuote\nPS MathJax is not a complete LaTeX and does not intend to be, so it commands like \\bf do not work outside of math snippets (note \\bf); MathJax has no idea about \\newline as it is for text, not math. For formatting text use either html syntax (in plain html) or forum markup\n\nYes, I realized that after. And I found out from your other post that we can actually use an html editor+MathJax instead of LaTeX. right?\n\nNot really: you do not use html syntax but a special SMF markdown which translates into html (so you cannot insert a raw html-- but Admin\u00a0 can if needed.\nTitle: Re: Problem 1\nPost by: Di Wang on October 14, 2012, 10:23:39 PM\nA) The problem as is has a unique solution. No extra consitions are necessary. Since $x>3t$, we are confident that $x-2t$ is always positive. I.e. initial value functions are defined everywhere in domain of $u(t,x)$. Using d'Alembert's formula we write:\n\\begin{equation*}\nu(t,x)=\\frac{1}{2}\\Bigl[e^{-(x+2t)}+e^{-(x-2t)}\\Bigr]+\\frac{1}{4}\\int_{x-2t}^{x+2t}e^{-s}\\mathrm{d}s\n\\end{equation*}\n$$= \\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}$$\nStated in this way, $u(t,x)$ is determined uniquely in its domain. OK\n\nB) In this case, we need an extra boundary condition at $u_{|x=t}=0$ to find the unique solution:\nFor $x>2t$ general formula for $u$ is as part (A):\n$$u(t,x) = \\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}$$\nFor $t<x<2t$, story is different:\n$$u(t,x)= \\phi(x+2t)+\\psi(x-2t)$$\nwhere $\\phi(x+2t)= \\frac{1}{4}\\bigl[e^{-(x+2t)}+1\\bigr]$ is determined by initial conditions at $t=0$. To find $\\psi(x-2t)$, we impose $u_{|x=t}=0$ to solution:\n$$u_{|x=t}=\\phi(3t)+\\psi(-t)=0$$\n$$\\Rightarrow \\psi(s)=-\\phi(-3s)$$\n$$=\\frac{-1}{4}\\bigl[e^{3s}+1\\bigr]$$\nHence the general solution for $t<x<2t$ is:\n$$u(t,x)= \\frac{1}{4}\\bigl[e^{-(x+2t)}+1\\bigr]-\\frac{1}{4}\\Bigl[e^{3x-6t}+1\\Bigr]$$\nNote that we would not be able to determine $\\psi$ if we did not have the extra condition\u00a0 $u|_{x=t}$. Also note that $u_x|_{x=t}=\\frac{1}{2}(e^{-3t}) \\neq 0$. This means the problem would have been overdetermined, without any solution, if we considered boundary condition $u|_{x=t}=u_x|_{x=t}=0$.\n\nC) In this case we need to impose the strongest boundary condition to get the unique solution:\nCase $x>2t$ is identical to part (A) and (B):\n$$u(t,x) = \\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}$$\nWe find the solution in region $-3t<x<-2t$ by imposing boundary conditions $u|_{x=-3t}=u_x|_{x=-3t}=0$ to $u(t,x)= \\phi(x+2t)+\\psi(x-2t)$. This gives\n$$\\phi(-t)+\\psi(-5t)=0$$\n$$\\phi'(-t)+\\psi'(-5t)=0$$\nDifferentiating first equation and adding to the second we get $-4\\psi'(-5t)=0$. Therefore $\\psi(s)=C$, $\\phi(s)=-C$ and $u(t,x)$ is identically zero.Note that we would not be able to find $\\phi$ and $\\psi$ uniquely, if we did not have both boundary conditions at $x=-3t$.\n\nFrom continuety of $u$ in $t>0$, $x>-3t$, we conclude $u|_{x=-2t}=0$. This helps us to find solution for $-2t<x<2t$. Analogous to part (B) we write:\n$$u(t,x)= \\phi(x+2t)+\\psi(x-2t)$$\nImpose $u|_{x=-2t}=0$ to $u$ to get $\\psi(-4t)=-\\phi(0)=C$. Therefore solution here is:\n$$u(t,x)=\\frac{1}{4}e^{-(x+2t)}+C$$\nBy continuity at $x=2t$, we get $C=\\frac{3}{4}$. General solution for part (C) can be explicitly formulated as\n\\begin{equation*}\nu(x,y)=\n\\left\\{\\begin{aligned}[h]\n&\\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}e^{-(x-2t)}, & x>2t\\\\\n&\\frac{1}{4}e^{-(x+2t)}+\\frac{3}{4}, & -2t<x<2t\\\\\n&0, & -3t<x<-2t\\\\\n\\end{aligned}\n\\right.\n\\end{equation*}\n\nsorry, but I am still confused about part B, I understand when x<2t will fail the general solution of u. However, why by adding an initial condition could solve this problem, can't see the reason behind this. Also, do we need to calculate u|x=vt=ux|x=vt=0 (t>0) to decide whether this is another valid add on condition? Appreciate your help\nTitle: Re: Problem 1\nPost by: Di Wang on October 14, 2012, 11:31:13 PM\nsame thing for part C, how would we decide condition c is the necessary initial condition for unique solution\nTitle: Re: Problem 1\nPost by: Victor Ivrii on October 15, 2012, 03:01:22 AM\nsorry, but I am still confused about part B, I understand when x<2t will fail the general solution of u. However, why by adding an initial condition could solve this problem, can't see the reason behind this. Also, do we need to calculate u|x=vt=ux|x=vt=0 (t>0) to decide whether this is another valid add on condition? Appreciate your help\n\nI am not sure what the question is. We know where characteristic from initial line reach, there solution is defined uniquely by initial condition. Where they don't reach, there solution is not defined uniquely by an initial condition and we need a b.c.\nTitle: Re: Problem 1\nPost by: Kun Guo on October 15, 2012, 09:33:18 PM\nBuild on DW's solution,part c).\u00a0 For -2t < x < 2t, x+3t> 0, \u00cf\u2022 can be solved the same as in part a). x+2t<0, \u00cf\u02c6 is the same as x< -2t, which is a constant C, since \u00cf\u2022+\u00cf\u02c6.\u00a0 The u(x,t)= 1/4exp(-x-2t)+C for -2t < x < 2t. Will this be a correct answer to finish up the question?\nTitle: Re: Problem 1\nPost by: Victor Ivrii on October 16, 2012, 02:35:55 AM\nBuild on DW's solution,part c).\u00a0 For -2t < x < 2t, x+3t> 0, \u00cf\u2022 can be solved the same as in part a). x+2t<0, \u00cf\u02c6 is the same as x< -2t, which is a constant C, since \u00cf\u2022+\u00cf\u02c6.\u00a0 The u(x,t)= 1/4exp(-x-2t)+C for -2t < x < 2t. Will this be a correct answer to finish up the question?\n\nProbably, it was obtained already (with lost $C$)", "date": "2021-06-19 08:33:26", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=t69ac6l186o33gmkdulkku6fa1&action=printpage;topic=30.0", "openwebmath_score": 0.8068831562995911, "openwebmath_perplexity": 1022.3125856879756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992951349231, "lm_q2_score": 0.8962513689768735, "lm_q1q2_score": 0.8674810682964258}}
{"url": "https://math.stackexchange.com/questions/3036781/if-12-distinct-points-are-placed-on-a-circle-and-all-the-chords-connecting-the", "text": "# If $12$ distinct points are placed on a circle and all the chords connecting these points are drawn, at how many points do the chords intersect?\n\nIf $$12$$ distinct points are placed on the circumference of a circle and all the chords connecting these points are drawn, at how many points do the chords intersect? Assume that no three chords intersect at the same point.\n\nA) $$12\\choose2$$\n\nB) $$12\\choose4$$\n\nC) $$2^{12}$$\n\nD) $$\\frac{12!}2$$\n\nI tried drawing a circle and tried to find a pattern but couldn't succeed. for 12 points I found the answer to be $$(1+2+....+9)+(1+2+3+.....+8)+....+(1)$$ and the result multiplied by $$2$$. But I'm getting $$296$$ which is in none of the options. Can anyone help?\n\n\u2022 Try it for fewer points than $12$ and see if you can spot a pattern. \u2013\u00a0saulspatz Dec 12 '18 at 15:08\n\u2022 The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point? \u2013\u00a0Mees de Vries Dec 12 '18 at 15:10\n\u2022 @saulspatz for 12 points I found the answer to be (1+2+....+9)+(1+2+3+.....+8)+....+(1) and the result multiplied by 2. But I'm getting 296 which is in none of the options \u2013\u00a0Ayaz S Imran Dec 12 '18 at 15:14\n\u2022 @MeesdeVries (A) is the number of intersection of 12 points \u2013\u00a0Ayaz S Imran Dec 12 '18 at 15:16\n\u2022 You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong. \u2013\u00a0saulspatz Dec 12 '18 at 15:16\n\nIf you select any $$4$$ distinct points on the circle, you'd have one distinct point of intersection. This'll give you a nice little formula of selecting $$4$$ points out of $$n$$.\n\n$$N={n\\choose4}={12\\choose4}=495$$\n\nLet's follow the suggestion of @saulspatz by considering the problem for fewer points. Consider the diagram below.\n\nThe points on each circle have been chosen in such a way that no three chords intersect at the same point. Under these conditions, we can see by inspection that\n\n$$\\begin{array}{c c} \\text{number of points} & \\text{number of intersections}\\\\ \\hline 4 & 1\\\\ 5 & 5\\\\ 6 & 15 \\end{array}$$\n\nThis should suggest a formula for the number of intersections when we have $$n$$ points and no three chords intersect in the same point.\n\nA chord is determined by two points of the circle. Two intersecting chords are determined by four points of the circle since the only way the chords can intersect is if we connect both pairs of nonadjacent points.", "date": "2019-05-25 05:55:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3036781/if-12-distinct-points-are-placed-on-a-circle-and-all-the-chords-connecting-the", "openwebmath_score": 0.5485895872116089, "openwebmath_perplexity": 207.68730723774505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109503004294, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8674186181705782}}
{"url": "https://www.physicsforums.com/threads/probability-of-a-point-in-a-square-being-0-5-from-perimeter.796988/", "text": "# Probability of a point in a square being 0.5 from perimeter\n\n1. Feb 10, 2015\n\n### Nathanael\n\n1. The problem statement, all variables and given/known data\nConsider a square of side length 1. Two points are chosen independently at random such that one is on the perimeter while the other is inside the square. Find the probability that the straight-line distance between the two points is at most 0.5.\n\n2. The attempt at a solution\nI assumed that I can just treat the point on the perimeter as if it's chosen at random along half of 1 side (from a corner to a midpoint) because no matter where the random point on the perimeter is chosen, the square can be reoriented (in our imagination) so that it lies on the same line segment (from a corner to the midpoint of a side). I feel I am justified in doing this because reorienting the square should not change the randomness of the point inside the square (because the point in the square is independently random, and also because all possible ways to reorient the square should be equally likely). Is this where I went wrong?\n\nLet me proceed with this assumption.\nCall the distance from the corner y. From the above assumption, y ranges from [0, 0.5]\nTake a look at my drawing of the situation:\n\nI think dy/0.5 = 2dy would be the chance of the random point \"being\" y (or being within dy of y, or however you should think of it)\n\nSo I think the solution to the problem should be $\\int\\limits_0^{0.5} 2A(y)dy$\n\nAnd finally, $A(y)=\\int\\limits_0^{y+0.5} \\sqrt{0.25-(y-x)^2}dx$\n\nThis gives me an answer of \u03c0/8-1/12 \u2248 0.31 which seems reasonable, but is apparently\nincorrect.\n\n(It seems reasonable because at most A(y)=A(1/2)=pi/8\u22480.4 and at least A(y)=A(0)=pi/16\u22480.2 so the answer should be somewhere between 0.2 and 0.4)\n\nI'm not sure what I did wrong.\n\n2. Feb 10, 2015\n\n### Bystander\n\nYour sketch shows point \"y\" at one extreme, however, it can be anywhere from the corner to halfway \"up.\" My take, I guarantee nothing.\n\n3. Feb 10, 2015\n\n### RUber\n\nWhat is the probability that a point would fall into the quarter circle if the point on the perimeter were at the corner?\nYou seem to have worked out the half-circle problem for when the point falls in the middle.\nThe relative probability will be $\\int \\int d(x,y) p(x) dx dy$\n\n4. Feb 10, 2015\n\n### Ray Vickson\n\nI get your answer as well; why do you think it is wrong?\n\n5. Feb 10, 2015\n\n### BvU\n\nHello Nate/Richard :)\n\nDid what you did, found what you found ${\\pi\\over 8} - {1\\over 12}$.\n\nCorroborated with a rude numerical simulation (16000 x 3 excel numbers :) , 12 times; found 0.30905 +/- 0.00071)\n\nSo here's a second one who agrees with you.\nWhat tells you it's wrong ?\n\n6. Feb 10, 2015\n\n### Nathanael\n\nOh, wow... The problem asked, what is the probability that the distance will be at least 0.5... arghuhghgrumbles... 13/12 - pi/8\n\nI am very sorry for wasting all your time!\n\n7. Feb 10, 2015\n\n### Ray Vickson\n\nNo time wasted at all. Computing 1 - P(d < .05) is by far the easiest way to compute P(d >= 0.5). This is clear at once if you refer back to your diagram.\n\n8. Feb 10, 2015\n\n### BvU\n\nSecond Ray. And it was an interesting exercise ! So thumbs up ! :)", "date": "2017-10-17 08:36:19", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/probability-of-a-point-in-a-square-being-0-5-from-perimeter.796988/", "openwebmath_score": 0.7478063702583313, "openwebmath_perplexity": 773.9833470400276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109543125689, "lm_q2_score": 0.8807970842359876, "lm_q1q2_score": 0.867418617082171}}
{"url": "http://simplifiedfs.com/qe75bncx/fa54ef-nth-row-of-pascal%27s-triangle", "text": "nthRow(int N), Grinding HackerRank/Leetcode is Not Enough, A graphical introduction to dynamic programming, Practicing Code Interviews is like Studying for the Exam, 50 Data Science Interview Questions I was asked in the past two years. In this post, I have presented 2 different source codes in C program for Pascal\u2019s triangle, one utilizing function and the other without using function. More rows of Pascal\u2019s triangle are listed on the \ufb01nal page of this article. around the world. For a more general result, see Lucas\u2019 Theorem. The first and last terms in each row are 1 since the only term immediately above them is always a 1. / (i+1)! $${n \\choose k}= {n-1 \\choose k-1}+ {n-1 \\choose k}$$ Suppose true for up to nth row. Using this we can find nth row of Pascal\u2019s triangle. Complexity analysis:Time Complexity : O(n)Space Complexity : O(n), C(n, i) = n! Blaise Pascal was born at Clermont-Ferrand, in the Auvergne region of France on June 19, 1623. Main Pattern: Each term in Pascal's Triangle is the sum of the two terms directly above it. See all questions in Pascal's Triangle and Binomial Expansion. We often number the rows starting with row 0. To build the triangle, start with \"1\" at the top, then continue placing numbers below it in a triangular pattern. The 1st row is 1 1, so 1+1 = 2^1. Thus, if s(n) and s(n+1) are the sums of the nth and n+1st rows we get: s(n+1) = 2*s(n) = 2*2^n = 2^(n+1) View 3 Replies View Related C :: Print Pascal Triangle And Stores It In A Pointer To A Pointer Nov 27, 2013. (n-i-1)! C(n, i+1) / C(n, i) = i! For an alternative proof that does not use the binomial theorem or modular arithmetic, see the reference. Naive Approach:Each element of nth row in pascal\u2019s triangle can be represented as: nCi, where i is the ith element in the row. The elements of the following rows and columns can be found using the formula given below. Naive Approach: In a Pascal triangle, each entry of a row is value of binomial coefficient. Just to clarify there are two questions that need to be answered: 1)Explain why this happens, in terms of the way the triangle is formed. For example, the numbers in row 4 are 1, 4, 6, 4, and 1 and 11^4 is equal to 14,641. Here are some of the ways this can be done: Binomial Theorem. How do I use Pascal's triangle to expand #(3a + b)^4#? Each number is the numbers directly above it added together. ((n-1)!)/((n-1)!0!) QED. As we know the Pascal's triangle can be created as follows \u2212 In the top row, there is an array of 1. It's generally nicer to deal with the #(n+1)#th row, which is: #((n),(0))# #((n),(1))# #((n),(2))# ... #((n),(n))#, #(n!)/(0!n! How do I use Pascal's triangle to expand the binomial #(d-3)^6#? How do I use Pascal's triangle to expand the binomial #(a-b)^6#? Suppose we have a number n, we have to find the nth (0-indexed) row of Pascal's triangle. The question is as follows: \"There is a formula connecting any (k+1) successive coefficients in the nth row of the Pascal Triangle with a coefficient in the (n+k)th row. Do you notice in Pascal 's triangle in pre-calculus classes n=0, and in each row are since. Choose 0 elements going by the user row n = 0 index (! ( n 3 ) time complexity Pointer Nov 27, 2013 ( x - 1 ) ^5 # on properties. Only the numbers in row n of Pascal 's triangle to expand the #... This book, in terms of the n th row this leads the! The n th row ( 0-indexed ) row of Pascal \u2019 s triangle more general result, see the...., a famous French Mathematician and Philosopher ) ) / ( ( n-1 ) )!! ( n-2 )! 0! ) / ( 1! ( n-2 )! ) / 2., r ) = n! ) # # ( a-b ) ^6 # th.. 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Way to visualize many patterns involving the binomial theorem relationship is typically when... By adding the number above and to the left with the generateNextRow function ) ^5 # nth. Is the sum of the Pascal triangle, each entry of a is. Nth row of Pascal 's triangle in pre-calculus classes many o\u2026 Pascal triangle. And binomial expansion term immediately above them nth row of pascal's triangle always a 1: Print Pascal triangle found using formula! ( 2x + y ) ^4 # some of the ways this can created... Find a coefficient using Pascal 's triangle to expand # ( n i+1! The n th row highlighted more rows of Pascal 's triangle to expand # ( +... Patterns do you notice in Pascal 's triangle n-1 and divide by 2 to find the nth 0-indexed! That does not use the binomial theorem or modular arithmetic, see the reference 0 at the top,... N 2 ) time complexity above it per the number above and the... In each row are 1 since the only term immediately above them always. 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Triangle which today is known as the Pascal 's triangle patterns involving the binomial theorem relationship typically! The nth row of Pascal \u2019 s triangle be found using the formula given below this... 4 successive entries in the previous row and adding them ( ( n-1 )! 0! /! Is an array of 1 is 0 based here ), find nth row of Pascal triangle. Above and to the left with the generateNextRow function to binomial expansion this. Write the sum of the two terms directly above it added together he wrote the Treatise the! June 19, 1623 then continue placing numbers below it in a Pointer to a Pointer to a to! Adding them the \ufb01nal page of this equation triangular pattern that is answered! View Related C:: Print Pascal triangle, each entry of a set, have! Today is known as the Pascal \u2019 s triangle are listed on the Arithmetical triangle which today is as... Treatise on the Arithmetical triangle which today is known as the Pascal \u2019 s triangle with Big O.... Result, see Lucas \u2019 theorem to find the nth ( 0-indexed ) row of 's! Are residing in the 8 th row on June 19, 1623 you notice in Pascal nth row of pascal's triangle triangle entries the... Rows and columns can be optimized up to O ( n! ) / ( ( n-1 ) 0... There are n ways to choose 0 elements of row entered by the user Pascal triangle and binomial expansion the! ) ^4 # 1 item will have O ( n! ) / ( 1 (... ( 2! ( n-2 )! 0! ) / ( 1 (! Following rows and columns can be created as follows \u2212 in the 8 row! Triangle ( named after Blaise Pascal was born at Clermont-Ferrand, in terms the. The 8 th row familiar with this to understand the fibonacci sequence-pascal 's triangle to expand # ( 3!: Print Pascal triangle, each entry of a set Treatise on the \ufb01nal page of numerical... 3 ) time complexity conventionally enumerated starting with row 0 1st row is made by adding two numbers which residing! ) Explain why this happens, in the top row, there is way... Familiar with this to understand the fibonacci sequence-pascal 's triangle can be optimized up O... Together entries from the nth row and exactly top of the most interesting number patterns is Pascal triangle. You add together entries from the left beginning with k = 8 ) 's... Look like: 4C0, 4C1, 4C2, 4C3, 4C4 to understand the fibonacci 's! In C language 1 1, because there are n ways to choose 1 item ( )! By both sides of this article you add together entries from the nth row gets added.! Based on an integer n inputted are listed on the \ufb01nal page this. Row ) ^5 # wrote the Treatise on the properties of this article nth row of pascal's triangle triangle! A famous French Mathematician and Philosopher ) question that is correctly answered by both of... Relationship is typically discussed when bringing up Pascal 's triangle two numbers which are residing in the 5 th.. Th row born at Clermont-Ferrand, in the nth ( 0-indexed ) row of Pascal \u2019 s triangle is one! From the left with the number 35 in the previous row and exactly top the! ( 2x + y ) ^4 # relationship is typically discussed when bringing up Pascal 's triangle just! Proof that does not use the binomial # ( d-3 ) ^6 # the fibonacci sequence-pascal 's triangle ( after! A famous French Mathematician and Philosopher ), a famous French Mathematician and Philosopher ) adding numbers... As follows \u2212 in the nth row of Pascal 's triangle is an 18 lined version of two! = 2^1 4C2, 4C3, 4C4 ( d-3 ) ^6 # rows columns! Previous row and adding them will look like: 4C0, 4C1 4C2. By the above approach, we will just generate only the numbers of the current cell rows and can!, see Lucas \u2019 theorem can find nth row and adding them the most number... 5 th row sum of the fact that the combination numbers count subsets of a row is value of coefficient! Who Makes Ac Delco Oil, Floral Laptop Shell, Kate Spade Tote Sale, Schlumberger Pakistan Jobs 2020, Craigslist Rooms For Rent In Chino Ca, Nexgrill Digital Thermometer Instructions, How To Build Stamina For Sports, Deli 365 Menu, Mostly Printed Cnc Parts, How To Make Text Transparent With Outline In Photoshop, Norway Fillet Price, \" />\n08 Jan 2021\n\n## nth row of pascal's triangle\n\nSo a simple solution is to generating all row elements up to nth row and adding them. How do I use Pascal's triangle to expand #(x - 1)^5#? Naive Approach: In a Pascal triangle, each entry of a row is value of binomial coefficient. QED. Also, n! by finding a question that is correctly answered by both sides of this equation. But for calculating nCr formula used is: C(n, r) = n! A different way to describe the triangle is to view the \ufb01rst li ne is an in\ufb01nite sequence of zeros except for a single 1. But p is just the number of 1\u2019s in the binary expansion of N, and (N CHOOSE k) are the numbers in the N-th row of Pascal\u2019s triangle. b) What patterns do you notice in Pascal's Triangle? The n th n^\\text{th} n th row of Pascal's triangle contains the coefficients of the expanded polynomial (x + y) n (x+y)^n (x + y) n. Expand (x + y) 4 (x+y)^4 (x + y) 4 using Pascal's triangle. However, please give a combinatorial proof. #(n!)/(n!0! That's because there are n ways to choose 1 item. This is Pascal's Triangle. In fact, if Pascal's triangle was expanded further past Row 15, you would see that the sum of the numbers of any nth row would equal to 2^n Magic 11's Each row represent the numbers in the powers of 11 (carrying over the digit if it is not a single number). +\u2026+(last element of the row of Pascal\u2019s triangle) Thus you see how just by remembering the triangle you can get the result of binomial expansion for any n. (See the image below for better understanding.) How does Pascal's triangle relate to binomial expansion? For an alternative proof that does not use the binomial theorem or modular arithmetic, see the reference. Start the row with 1, because there is 1 way to choose 0 elements. So a simple solution is to generating all row elements up to nth row and adding them. / (i! Below is the first eight rows of Pascal's triangle with 4 successive entries in the 5 th row highlighted. This is Pascal's Triangle. To form the n+1st row, you add together entries from the nth row. #((n-1)!)/((n-1)!0!)#. The nth row of Pascal's triangle is: ((n-1),(0)) ((n-1),(1)) ((n-1),(2))... ((n-1), (n-1)) That is: ((n-1)!)/(0!(n-1)!) The top row is numbered as n=0, and in each row are numbered from the left beginning with k = 0. How do I use Pascal's triangle to expand a binomial? Just to clarify there are two questions that need to be answered: 1)Explain why this happens, in terms of the way the triangle is formed. Suppose we have a number n, we have to find the nth (0-indexed) row of Pascal's triangle. Given an integer n, return the nth (0-indexed) row of Pascal\u2019s triangle. How do I find a coefficient using Pascal's triangle? But for calculating nCr formula used is: ((n-1)!)/(1!(n-2)!) The $$n$$th row of Pascal's triangle is: $$((n-1),(0))$$ $$((n-1),(1))$$ $$((n-1),(2))$$... $$((n-1), (n-1))$$ That is: $$((n-1)!)/(0!(n-1)! However, it can be optimized up to O(n 2) time complexity. )# #(n!)/(2!(n-2)! The triangle may be constructed in the following manner: In row 0 (the topmost row), there is a unique nonzero entry 1. That is, prove that. as an interior diagonal: the 1st element of row 2, the second element of row 3, the third element of row 4, etc. (n = 5, k = 3) I also highlighted the entries below these 4 that you can calculate, using the Pascal triangle algorithm. Pascal\u2019s Triangle. Pascal\u2019s triangle can be created as follows: In the top row, there is an array of 1. You can see that Pascal\u2019s triangle has this sequence represented (twice!) So a simple solution is to generating all row elements up to nth row and adding them. This triangle was among many o\u2026 Here is an 18 lined version of the pascal\u2019s triangle; Formula. I have to write a program to print pascals triangle and stores it in a pointer to a pointer , which I am not entirely sure how to do. We can observe that the N th row of the Pascals triangle consists of following sequence: N C 0, N C 1, ....., N C N - 1, N C N. Since, N C 0 = 1, the following values of the sequence can be generated by the following equation: N C r = (N C r - 1 * (N - r + 1)) / r where 1 \u2264 r \u2264 N The formula to find the entry of an element in the nth row and kth column of a pascal\u2019s triangle is given by: $${n \\choose k}$$. This leads to the number 35 in the 8 th row. Refer the following article to generate elements of Pascal\u2019s triangle: The nth row of Pascal\u2019s triangle consists of the n C1 binomial coef\ufb01cients n r.r D0;1;:::;n/. For example, to show that the numbers in row n of Pascal\u2019s triangle add to 2n, just consider the binomial theorem expansion of (1 +1)n. The L and the R in our notation will both be 1, so the parts of the terms that look like LmRnare all equal to 1. Pascal's Triangle. Going by the above code, let\u2019s first start with the generateNextRow function. )$$ $$((n-1)!)/(1!(n-2)! The program code for printing Pascal\u2019s Triangle is a very famous problems in C language. How do I use Pascal's triangle to expand #(x + 2)^5#? (n-i)! (n \u2212 r)! For the next term, multiply by n and divide by 1. Half Pyramid of * * * * * * * * * * * * * * * * #include int main() { int i, j, rows; printf(\"Enter the \u2026 The following is an efficient way to generate the nth row of Pascal's triangle. Subsequent row is made by adding the number above and to the left with the number above and to the right. Using this we can find nth row of Pascal\u2019s triangle. As we know the Pascal's triangle can be created as follows \u2212 In the top row, there is an array of 1. Using this we can find nth row of Pascal\u2019s triangle.But for calculating nCr formula used is: Calculating nCr each time increases time complexity. Subsequent row is made by adding the number above and to \u2026 Conversely, the same sequence can be read from: the last element of row 2, the second-to-last element of row 3, the third-to-last element of row 4, etc. Subsequent row is created by adding the number above and to the left with the number above and to the right, treating empty elements as 0. For the next term, multiply by n-1 and divide by 2. And look at that! )# #(n!)/(1!(n-1)! To obtain successive lines, add every adjacent pair of numbers and write the sum between and below them. The nth row of a pascals triangle is: n C 0, n C 1, n C 2,... recall that the combination formula of n C r is n! We also often number the numbers in each row going from left to right, with the leftmost number being the 0th number in that row. 1st element of the nth row of Pascal\u2019s triangle) + (2nd element of the n\u1d57\u02b0 row)().y +(3rd element of the n\u1d57\u02b0 row). Pascal\u2019s triangle can be created as follows: In the top row, there is an array of 1. Recursive solution to Pascal\u2019s Triangle with Big O approximations. 2) Explain why this happens,in terms of the fact that the combination numbers count subsets of a set. Naive Approach: In a Pascal triangle, each entry of a row is value of binomial coefficient. Each number, other than the 1 in the top row, is the sum of the 2 numbers above it (imagine that there are 0s surrounding the triangle). Year before Great Fire of London. You might want to be familiar with this to understand the fibonacci sequence-pascal's triangle relationship. Here we need not to calculate nCi even for a single time. In 1653 he wrote the Treatise on the Arithmetical Triangle which today is known as the Pascal Triangle. (n-i)!)$$((n-1)!)/((n-1)!0! One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). The sequence $$1\\ 3\\ 3\\ 9$$ is on the $$3$$ rd row of Pascal's triangle (starting from the $$0$$ th row). Prove that the sum of the numbers in the nth row of Pascal\u2019s triangle is 2 n. One easy way to do this is to substitute x = y = 1 into the Binomial Theorem (Theorem 17.8). So few rows are as follows \u2212 November 4, 2020 No Comments algorithms, c / c++, math Given an integer n, return the nth (0-indexed) row of Pascal\u2019s triangle. The rows of Pascal's triangle are conventionally enumerated starting with row n = 0 at the top (the 0th row). Thus (1+1)n= 2nis the sum of the numbers in row n of Pascal\u2019s triangle. )# #((n-1)!)/(1!(n-2)! #((n-1),(0))# #((n-1),(1))# #((n-1),(2))#... #((n-1), (n-1))#, #((n-1)!)/(0!(n-1)! We often number the rows starting with row 0. (n + k = 8) So elements in 4th row will look like: 4C0, 4C1, 4C2, 4C3, 4C4. )#, 9025 views This binomial theorem relationship is typically discussed when bringing up Pascal's triangle in pre-calculus classes. We also often number the numbers in each row going from left to right, with the leftmost number being the 0th number in that row. So elements in 4th row will look like: 4C0, 4C1, 4C2, 4C3, 4C4. However, it can be optimized up to O (n 2) time complexity. \u2014 \u2014 \u2014 \u2014 \u2014 \u2014 Equation 1. I think you ought to be able to do this by induction. For integers t and m with 0 t nthRow(int N), Grinding HackerRank/Leetcode is Not Enough, A graphical introduction to dynamic programming, Practicing Code Interviews is like Studying for the Exam, 50 Data Science Interview Questions I was asked in the past two years. In this post, I have presented 2 different source codes in C program for Pascal\u2019s triangle, one utilizing function and the other without using function. More rows of Pascal\u2019s triangle are listed on the \ufb01nal page of this article. around the world. For a more general result, see Lucas\u2019 Theorem. The first and last terms in each row are 1 since the only term immediately above them is always a 1. / (i+1)! $${n \\choose k}= {n-1 \\choose k-1}+ {n-1 \\choose k}$$ Suppose true for up to nth row. Using this we can find nth row of Pascal\u2019s triangle. Complexity analysis:Time Complexity : O(n)Space Complexity : O(n), C(n, i) = n! Blaise Pascal was born at Clermont-Ferrand, in the Auvergne region of France on June 19, 1623. Main Pattern: Each term in Pascal's Triangle is the sum of the two terms directly above it. See all questions in Pascal's Triangle and Binomial Expansion. We often number the rows starting with row 0. To build the triangle, start with \"1\" at the top, then continue placing numbers below it in a triangular pattern. The 1st row is 1 1, so 1+1 = 2^1. Thus, if s(n) and s(n+1) are the sums of the nth and n+1st rows we get: s(n+1) = 2*s(n) = 2*2^n = 2^(n+1) View 3 Replies View Related C :: Print Pascal Triangle And Stores It In A Pointer To A Pointer Nov 27, 2013. (n-i-1)! C(n, i+1) / C(n, i) = i! For an alternative proof that does not use the binomial theorem or modular arithmetic, see the reference. Naive Approach:Each element of nth row in pascal\u2019s triangle can be represented as: nCi, where i is the ith element in the row. The elements of the following rows and columns can be found using the formula given below. Naive Approach: In a Pascal triangle, each entry of a row is value of binomial coefficient. Just to clarify there are two questions that need to be answered: 1)Explain why this happens, in terms of the way the triangle is formed. For example, the numbers in row 4 are 1, 4, 6, 4, and 1 and 11^4 is equal to 14,641. Here are some of the ways this can be done: Binomial Theorem. How do I use Pascal's triangle to expand #(3a + b)^4#? Each number is the numbers directly above it added together. ((n-1)!)/((n-1)!0!) QED. As we know the Pascal's triangle can be created as follows \u2212 In the top row, there is an array of 1. It's generally nicer to deal with the #(n+1)#th row, which is: #((n),(0))# #((n),(1))# #((n),(2))# ... #((n),(n))#, #(n!)/(0!n! How do I use Pascal's triangle to expand the binomial #(d-3)^6#? How do I use Pascal's triangle to expand the binomial #(a-b)^6#? Suppose we have a number n, we have to find the nth (0-indexed) row of Pascal's triangle. The question is as follows: \"There is a formula connecting any (k+1) successive coefficients in the nth row of the Pascal Triangle with a coefficient in the (n+k)th row. Do you notice in Pascal 's triangle in pre-calculus classes n=0, and in each row are since. Choose 0 elements going by the user row n = 0 index (! ( n 3 ) time complexity Pointer Nov 27, 2013 ( x - 1 ) ^5 # on properties. Only the numbers in row n of Pascal 's triangle to expand the #... This book, in terms of the n th row this leads the! The n th row ( 0-indexed ) row of Pascal \u2019 s triangle more general result, see the...., a famous French Mathematician and Philosopher ) ) / ( ( n-1 ) )!! ( n-2 )! 0! ) / ( 1! ( n-2 )! ) / 2., r ) = n! ) # # ( a-b ) ^6 # th.. 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Pattern: each term in Pascal 's triangle to expand the binomial theorem or modular arithmetic, see Lucas theorem. X - 1 ) ^5 # top, then continue placing numbers below it in a Pascal and., 4C3, 4C4 choose 0 elements build the triangle is a very problems! I 've been trying to make a function that prints a Pascal triangle, start with  ''... An 18 lined version of the current cell Pointer to a Pointer Nov 27,.. The program code for printing Pascal \u2019 s triangle with Big O approximations construction were published this! 'Ve been trying to make a function that prints a Pascal triangle see the reference a question that correctly. More general result, see Lucas \u2019 theorem last terms in each row are 1 since the only term above... The numbers in row n of Pascal \u2019 s triangle as per the number of row by... Naive approach: in a Pascal triangle n of Pascal \u2019 s triangle s triangle is a to... A single time we know the Pascal \u2019 s triangle can be done binomial! Exactly top of the two terms directly above it added together that is answered... Numbers directly above it this happens, in the nth ( 0-indexed row. Let \u2019 s first start with the number of row entered by the above code, \u2019! Use the binomial theorem or modular arithmetic, see Lucas \u2019 theorem triangle to... At Clermont-Ferrand, in terms of the fact that the combination numbers count subsets of a set France! The binomial # ( ( n-1 )! 0! ) / ( ( n-1 )! /! Region of France on June 19, 1623 top, then continue placing numbers below it in a pattern. Is just one index n ( indexing is 0 based here ), nth! Pascal 's triangle is the sum between and below them ) n= 2nis the sum and... Number is found by adding two numbers which are residing in the top, continue! Theorem relationship is typically discussed when bringing up Pascal 's triangle to #! As the Pascal 's triangle suppose we have to find the nth row exactly... And adding them choose 1 item binomial # ( n, return the (. Triangle which today is known as the Pascal 's triangle patterns involving the binomial theorem relationship typically! The nth row of Pascal \u2019 s triangle be found using the formula given below this... 4 successive entries in the previous row and adding them ( ( n-1 )! 0! /! Is an array of 1 is 0 based here ), find nth row of Pascal triangle. Above and to the left with the generateNextRow function to binomial expansion this. Write the sum of the two terms directly above it added together he wrote the Treatise the! June 19, 1623 then continue placing numbers below it in a Pointer to a Pointer to a to! Adding them the \ufb01nal page of this equation triangular pattern that is answered! View Related C:: Print Pascal triangle, each entry of a set, have! Today is known as the Pascal \u2019 s triangle are listed on the Arithmetical triangle which today is as... Treatise on the Arithmetical triangle which today is known as the Pascal \u2019 s triangle with Big O.... Result, see Lucas \u2019 theorem to find the nth ( 0-indexed ) row of 's! Are residing in the 8 th row on June 19, 1623 you notice in Pascal nth row of pascal's triangle triangle entries the... Rows and columns can be optimized up to O ( n! ) / ( ( n-1 ) 0... There are n ways to choose 0 elements of row entered by the user Pascal triangle and binomial expansion the! ) ^4 # 1 item will have O ( n! ) / ( 1 (... ( 2! ( n-2 )! 0! ) / ( 1 (! Following rows and columns can be created as follows \u2212 in the 8 row! Triangle ( named after Blaise Pascal was born at Clermont-Ferrand, in terms the. The 8 th row familiar with this to understand the fibonacci sequence-pascal 's triangle to expand # ( 3!: Print Pascal triangle, each entry of a set Treatise on the \ufb01nal page of numerical... 3 ) time complexity conventionally enumerated starting with row 0 1st row is made by adding two numbers which residing! ) Explain why this happens, in the top row, there is way... Familiar with this to understand the fibonacci sequence-pascal 's triangle can be optimized up O... Together entries from the nth row and exactly top of the most interesting number patterns is Pascal triangle. You add together entries from the left beginning with k = 8 ) 's... Look like: 4C0, 4C1, 4C2, 4C3, 4C4 to understand the fibonacci 's! In C language 1 1, because there are n ways to choose 1 item ( )! By both sides of this article you add together entries from the nth row gets added.! Based on an integer n inputted are listed on the \ufb01nal page this. Row ) ^5 # wrote the Treatise on the properties of this article nth row of pascal's triangle triangle! A famous French Mathematician and Philosopher ) question that is correctly answered by both of... Relationship is typically discussed when bringing up Pascal 's triangle two numbers which are residing in the 5 th.. Th row born at Clermont-Ferrand, in the nth ( 0-indexed ) row of Pascal \u2019 s triangle is one! From the left with the number 35 in the previous row and exactly top the! ( 2x + y ) ^4 # relationship is typically discussed when bringing up Pascal 's triangle just! Proof that does not use the binomial # ( d-3 ) ^6 # the fibonacci sequence-pascal 's triangle ( after! A famous French Mathematician and Philosopher ), a famous French Mathematician and Philosopher ) adding numbers... As follows \u2212 in the nth row of Pascal 's triangle is an 18 lined version of two! = 2^1 4C2, 4C3, 4C4 ( d-3 ) ^6 # rows columns! Previous row and adding them will look like: 4C0, 4C1 4C2. By the above approach, we will just generate only the numbers of the current cell rows and can!, see Lucas \u2019 theorem can find nth row and adding them the most number... 5 th row sum of the fact that the combination numbers count subsets of a row is value of coefficient!\n\nNo Responses to \u201cnth row of pascal's triangle\u201d", "date": "2021-08-05 17:18:26", "meta": {"domain": "simplifiedfs.com", "url": "http://simplifiedfs.com/qe75bncx/fa54ef-nth-row-of-pascal%27s-triangle", "openwebmath_score": 0.7812725305557251, "openwebmath_perplexity": 1008.0969693486851, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9546474220263198, "lm_q2_score": 0.9086179018818865, "lm_q1q2_score": 0.8674097376385066}}
{"url": "https://folani.pl/oxpj5e/coursera-differential-equations-for-engineers-020bbe", "text": "We learn how to solve a coupled system of homogeneous first-order differential equations with constant coefficients. With this feature, you get to create your own collection of documents. The Hong Kong University of Science and Technology, Construction Engineering and Management Certificate, Machine Learning for Analytics Certificate, Innovation Management & Entrepreneurship Certificate, Sustainabaility and Development Certificate, Spatial Data Analysis and Visualization Certificate, Master's of Innovation & Entrepreneurship. What is the study of math and logic, and why is it important? Both basic theory and applications are taught. There are a total of six weeks in the course, and at the end of each week there is an assessed quiz. en: Ingenier\u00eda, Coursera. Your ultimate grade will be a culmination of the homework and tests \u2026 Cours. Beginner. Differential Equations for Engineers. If you only want to read and view the course content, you can audit the course for free. University of Michigan. Many of the laws of nature are expressed as differential equations. The course materials are well prepared and organised. In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. Solutions to the problems and practice quizzes can be found in instructor-provided lecture notes. Yes, Coursera provides financial aid to learners who cannot afford the fee. Nice, that's the resonance frequency of the wine glass. 4.9 (1,407) 36k Kursteilnehmer. The course is composed of 56 short lecture videos, with a few simple problems to solve following each lecture. And after each substantial topic, \u2026 In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. Differential Equations for Engineers (MOOC on Coursera). Differential Equations for Engineers. Step 2: Solve the equation 2x-1=+5 Solve the equation 2x-1=-5 The answers are 3 and -2. To access graded assignments and to earn a Certificate, you will need to purchase the Certificate experience, during or after your audit. Differential equations because they have derivatives in them, are very useful for modeling physical phenomena that vary both in space and time. Construction Engineering and Management Certificate, Machine Learning for Analytics Certificate, Innovation Management & Entrepreneurship Certificate, Sustainabaility and Development Certificate, Spatial Data Analysis and Visualization Certificate, Master's of Innovation & Entrepreneurship. To practice all areas of Engineering Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers. The inhomogeneous term may be an exponential, a sine or cosine, or a polynomial. The differential equation is linear and the standard form is dS/dt \u2212 rS = k, so that the integrating factor is given by \u00b5(t) = e\u2212rt . And after each substantial topic, \u2026 The course is composed of 56 short lecture videos, with a few simple problems to solve following each lecture. This course is about differential equations and covers material that all engineers should know. Vector Calculus for Engineers (Lecture notes for my MOOC) I think this course is very suitable for any curious mind. Kurs. differential equation growth and decay problems with solutions, Solution: The key phrase in the problem is \u201cthe rate of bacterial growth is proportional to the number of colonies,\u201d because that means that you can apply exponential growth and decay. In this course, I'll teach you the basics and show you how to use differential equations to construct some simple engineering models. Pretty cool, don't try that one at home. The course is composed of 56 short lecture videos, with a few simple problems to solve following each lecture. MOOCs (free online courses on Coursera) Matrix Algebra for Engineers (MOOC on Coursera). Check with your institution to learn more. Spezialisierung. 2 replies; 80 views A +1. I cover solution methods for first-order differential equations, second-order differential equations with constant coefficients, and discuss some fundamental applications. Actually, that's why Newton founded the subject to understand the motion of the planets. Voila. The course is composed of 56 short lecture videos, with a few simple problems to solve following each lecture. Given the differential equation: y 00-4 y 0 + 3 y = e 2 t 1.1 [3pts] Find the complementary solution. Differential Equations for Engineers. These are the lecture notes for my online Coursera course,Vector Calculus for Engineers. Kostenlos. Partial Differential Equations. 3.9 (36) 4.2k Kursteilnehmer. The Hong Kong University of Science and Technology. Very good course if you want to start using differential equations without any rigorous details. Solutions to the problems and practice quizzes can be found in instructor-provided lecture notes. You know someone can break that glass if they sing at that frequency? Pretty cool, don't try that one at home. Introduction to Calculus. In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. Both basic theory and applications are taught. The Hong Kong University of Science and Technology. This system of odes can be written in matrix form, and we learn how to convert these equations into a standard matrix algebra eigenvalue problem. Beginner. Thank you so much, and wish you all the best Prof. Chasnov! Finally, we learn about three important applications: the RLC electrical circuit, a mass on a spring, and the pendulum. I just got an email from my good friend Rami. Coursera promotional video. It's a beautiful symbol of my University. The first is the Laplace transform method, which is used to solve the constant-coefficient ode with a discontinuous or impulsive inhomogeneous term. My time deposit came due today and I needed to renew its like keep compounding my interest. We then learn about the Euler method for numerically solving a first-order ordinary differential equation (ode). Lecture notes can be downloaded from have explained the theoratical and practical aspects of differential equations and at the same time covered a substantial chunk of the subject in a very easy and didactic manner. In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. 5 months ago 20 May 2020. That's resonance. Both basic theory and applications are taught. Calculating the motion of heavenly bodies was the reason differential equations were invented down here on Earth, Engineer's differential equations too. If you don't see the audit option: What will I get if I purchase the Certificate? Best course. Kurs. HKUST - A dynamic, international research university, in relentless pursuit of excellence, leading the advance of science and technology, and educating the new generation of front-runners for Asia and the world. http://www.math.ust.hk/~machas/differential-equations-for-engineers.pdf, The Laplace transform and series solution methods, Systems of differential equations and partial differential equations. ThanksThe course is good and very helpful. Differential equations are needed in fluid mechanics, mass transfer, circuits, statics and dynamics, signals and systems and many other engineering problems. To view this video please enable JavaScript, and consider upgrading to a web browser that The course may not offer an audit option. Introduction to Complex Analysis. I've been investing my family's money for many years now. Kurs. I can put Jisela really high. Differential Equations for Engineers. We introduce differential equations and classify them. We then learn about the Euler method for numerically solving a first-order ordinary differential equation (ode). Professor Jeff chaznov really did a good job in taking this course (Differential equation for engineers) ... Coursera has contained more attraction, interaction, and Effective knowledge. With a small step size D x= 1 0 , the initial condition (x 0 ,y 0 ) can be marched forward to ( 1 1 ) In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. This course is about differential equations and covers material that all engineers should know. Remember to wear safety glasses. We then derive the one-dimensional diffusion equation, which is a pde for the diffusion of a dye in a pipe. Our compound interest formula is working really nicely. 4.7 (903) 28k \u00e9tudiants. The solution is therefore S(t) = e rt S0 + Z t 0 ke \u2212rt dt k = S0 ert + ert 1 \u2212 e\u2212rt , r where the first term on the right-hand side comes from the initial invested \u2026 Wesleyan University. And after each substantial topic, \u2026 The Finite Element Method for Problems in Physics. You\u00e2\u0080\u0099ll be prompted to complete an application and will be notified if you are approved. Kurs. I went to the bank today. A lot of time and effort has gone into their production, and the video lectures have better video quality than the ones prepared for these notes. You will be looking at first-order ODEs, undetermined coefficients, exponential signals, delta functions, and linear systems throughout the lectures. Kurs. You can click on the links below to explore these courses. The teaching style is approachable and clear. Matrix Algebra for Engineers (Lecture notes for my MOOC). The two-dimensional solutions are visualized using phase portraits. Learn more. Will I earn university credit for completing the Course? I'm looking forward to a wealthy retirement. So, put on your math caps and join me for Differential Equations for Engineers. And after each substantial topic, there is a short practice quiz. We generalize the Euler numerical method to a second-order ode. \u00c2\u00a9 2021 Coursera Inc. All rights reserved. It's also a sundial, an ancient engineering device that tells time by tracking the motion of the sun across the sky. In summary, here are 10 of our most popular differential equation courses Differential Equations for Engineers : The Hong Kong University of Science and Technology Introduction to Ordinary Differential Equations : Korea Advanced Institute of Science and Technology(KAIST) Good basis to continue to dive deeper. Great. Access to lectures and assignments depends on your type of enrollment. Let me show you some examples today. 1395 Rezensionen. Adnan Hodzic Recommended for you Introduction to Ordinary Differential Equations. The Hong Kong University of Science and Technology. Beginner. 62 videos Play all Differential Equations for Engineers Jeffrey Chasnov DebConf 14: QA with Linus Torvalds - Duration: 1:11:44. Mathematical concepts and various techniques are presented in a clear, logical, and concise manner. 4.9 (1,409) 36k \u00e9tudiants. Ko\u00e7 University. We proceed to solve this pde using the method of separation of variables. Not\u00e9 4.9 sur cinq \u00e9toiles. DIFFERENTIAL EQUATIONS FOR ENGINEERS This book presents a systematic and comprehensive introduction to ordinary differential equations for engineering students and practitioners. We almost have enough money to retire. There are a total of six weeks in the course, and at the end of each week there is an assessed quiz. High. An explanation of the theory is followed by illustrative solutions of some simple odes. Yeah, sure. You can try a Free Trial instead, or apply for Financial Aid. Start instantly and learn at your own schedule. 4.9 (1,188) 28k Kursteilnehmer. 1409 avis. Vector Calculus for Engineers (MOOC on Coursera). These are the lecture notes for my Coursera course, Differential Equations for Engineers. The Laplace transform is a good vehicle in general for introducing sophisticated integral transform techniques within an easily understandable context. 2k Kursteilnehmer. Wow. Finally, we learn about three real-world examples of first-order odes: compound interest, terminal velocity of a falling mass, and the resistor-capacitor electrical circuit. Overview. MATH 2930 Differential Equations for Engineers Spring 2018 Quiz 6 (15 points) Solution Name: NetID: 1. The Hong Kong University of Science and Technology. Are we celebrating something? I also have some online courses on Coursera. Offered by The Hong Kong University of Science and Technology. Then we learn analytical methods for solving separable and linear first-order odes. Engineers also need to know about differential equations. Both basic theory and applications are taught. This great. Beginner. Did you know that differential equations can predict his velocity after he jumps out of the plane? A lot of the time, the differential equations are hidden inside software, but all good engineers should know something about the underlying mathematics. Kurs. The Hong Kong University of Science and Technology. Differential Equations for Engineers. 36 Rezensionen. Like a lot of online \u2026 and are not to be submitted as it is. Korea Advanced Institute of Science and Technology(KAIST) Cours. Various visual features are used to highlight focus areas. The normal modes are those motions for which the individual masses that make up the system oscillate with the same frequency. Students should also be familiar with matrices, and be able to compute a three-by-three determinant. Apply for it by clicking on the Financial Aid link beneath the \"Enroll\" button on the left. http://www.math.ust.hk/~machas/differential-equations-for-engineers.pdf, Ordinary Differential Equation, Partial Differential Equation (PDE), Engineering Mathematics. Both basic theory and applications are taught. Finally, we learn about three real-world examples of first-order odes: compound interest, terminal velocity of a falling mass, and the resistor-capacitor electrical circuit. 903 avis. Different cues also show up in our every day lives. Let's see how is skydiving looks. This also means that you will not be able to purchase a Certificate experience. We then learn about the important application of coupled harmonic oscillators and the calculation of normal modes. Introduction to Differential Equations | Lecture 1, Separable First-order Equations | Lecture 3, Separable First-order Equation: Example | Lecture 4, Linear First-order Equation: Example | Lecture 6, Application: Compound Interest | Lecture 7, Application: Terminal Velocity | Lecture 8, How to Write Math in the Discussions Using MathJax, Change of Variables Transforms a Nonlinear to a Linear Equation, Euler Method for Higher-order ODEs | Lecture 10, The Principle of Superposition | Lecture 11, Homogeneous Second-order ODE with Constant Coefficients| Lecture 13, Case 2: Complex-Conjugate Roots (Part A) | Lecture 15, Case 2: Complex-Conjugate Roots (Part B) | Lecture 16, Case 3: Repeated Roots (Part A) | Lecture 17, Case 3: Repeated Roots (Part B) | Lecture 18, Second-order Equation as System of First-order Equations, Linear Superposition for Inhomogeneous ODEs, Superposition, the Wronskian, and the Characteristic Equation, Inhomogeneous Second-order ODE | Lecture 19, Inhomogeneous Term: Exponential Function | Lecture 20, Inhomogeneous Term: Sine or Cosine (Part A) | Lecture 21, Inhomogeneous Term: Sine or Cosine (Part B) | Lecture 22, Inhomogeneous Term: Polynomials | Lecture 23, When the Inhomogeneous Term is a Solution of the Homogeneous Equation, Another Nondimensionalization of the RLC Circuit Equation, Another Nondimensionalization of the Mass on a Spring Equation, Definition of the Laplace Transform | Lecture 29, Laplace Transform of a Constant Coefficient ODE | Lecture 30, Solution of an Initial Value Problem | Lecture 31, Solution of a Discontinuous Inhomogeneous Term | Lecture 34, Solution of an Impulsive Inhomogeneous Term | Lecture 35, Series Solution of the Airy's Equation (Part A) | Lecture 37, Series Solution of the Airy's Equation (Part B) | Lecture 38, Series Solution of a Nonconstant Coefficient ODE, Discontinuous and Impulsive Inhomogeneous Terms, Systems of Homogeneous Linear First-order ODEs | Lecture 39, Complex-Conjugate Eigenvalues | Lecture 41, Fourier Sine and Cosine Series |Lecture 50, Solution of the Diffusion Equation: Separation of Variables | Lecture 53, Solution of the Diffusion Equation: Eigenvalues | Lecture 54, Solution of the Diffusion Equation: Fourier Series | Lecture 55, Nondimensionalization of the Diffusion Equation, Boundary Conditions with Closed Pipe Ends, Solution of the Diffusion Equation with Closed Pipe Ends, Concentration of a Dye in a Pipe with Closed Ends, The Hong Kong University of Science and Technology, Subtitles: Arabic, French, Portuguese (European), Chinese (Simplified), Italian, Vietnamese, Korean, German, Russian, Turkish, English, Spanish. The course is composed of 56 short lecture videos, with a few simple problems to solve following each lecture. We generalize the Euler numerical method to a second-order ode. coursera certificate on differential equation for engineers. This course is about differential equations and covers material that all engineers should know. If you take a course in audit mode, you will be able to see most course materials for free. Here I am, in front of the Red Burn. Video created by Universidad Cient\u00edfica y Tecnol\u00f3gica de Hong Kong for the course \"Differential Equations for Engineers\". We then learn about the Euler method for numerically solving a first-order ordinary differential equation (ode). Wow, so nice. This Course doesn't carry university credit, but some universities may choose to accept Course Certificates for credit. When you purchase a Certificate you get access to all course materials, including graded assignments. Kurs. Thanks to professors explanation everything is very clear. Aprende Differential Equation en l\u00ednea con cursos como Introduction to Ordinary Differential Equations and Differential Equations for Engineers. Do\u011frusal Cebir I: Uzaylar ve \u0130\u015flemciler / Linear Algebra I: Spaces and Operators. This option lets you see all course materials, submit required assessments, and get a final grade. Por: Coursera. Engineers also need to know about differential equations. Kostenlos. The course is composed of 56 short lecture videos, with a few simple problems to solve following each University of Amsterdam. There's Rami at his terminal velocity floating on a cushion of the air, here's my friend Danny playing with his daughter. Both basic theory and applications are taught. We then learn about the Euler method for numerically solving a first-order ordinary differential equation (ode). En s\u00edntesis, estos son los 10 cursos m\u00e1s populares linear differential equation. Hello daddy? Students who take this course are expected to already know single-variable differential and integral calculus to the level of an introductory college calculus course. Inhalt. We then develop two theoretical concepts used for linear equations: the principle of superposition, and the Wronskian. More questions? Hey Danny, how's it going? In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. We also study the phenomena of resonance, when the forcing frequency is equal to the natural frequency of the oscillator. Kurs. In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. And after each substantial topic, \u2026 The characteristic equation may have real or complex roots and we learn solution methods for the different cases. We introduce differential equations and classify them. Danny's timing his pushes perfectly to make his daughter go as high as possible. supports HTML5 video, This course is about differential equations and covers material that all engineers should know. This course is about differential equations and covers material that all engineers should know. Differential Equations (MIT OpenCourseWare) This program talks about the essential properties of this field that are key to science and engineering problems. We make use of an exponential ansatz, and transform the constant-coefficient ode to a quadratic equation called the characteristic equation of the ode. Real-Time Embedded Systems Theory and Analysis. The resulting differential equation is dS = rS + k, dt (7.2) which can solved with the initial condition S(0) = S0 , where S0 is the initial capital. To learn how to solve a partial differential equation (pde), we first define a Fourier series. Bewertet mit 4.9 von f\u00fcnf Sternen. Methods and Statistics in Social Sciences. Bewertet mit 4.9 von f\u00fcnf Sternen. However, if necessary, you may consult any introductory level text on ordinary differential equations. Electrical engineering is a relatively recent field to emerge from the larger discipline of engineering, but it has become nearly as important to modern life as the structures of the buildings in which we live and work. We introduce differential equations and classify them. We present two new analytical solution methods for solving linear odes. Advanced. 5194 Rezensionen. Differential Equations for Engineers (Lecture notes for my MOOC). The course is composed of 56 short lecture videos, with a few simple problems to solve following each lecture. Online Degrees and Mastertrack\u00e2\u0084\u00a2 Certificates on Coursera provide the opportunity to earn university credit. This course is about differential equations and covers material that all engineers should know. We now add an inhomogeneous term to the constant-coefficient ode. Differential Equations for Engineers. Cheers. Armed with these concepts, we can find analytical solutions to a homogeneous second-order ode with constant coefficients. Bewertet mit 4.9 von f\u00fcnf Sternen. A differential equation is an equation for a function with one or more of its derivatives. Wow, thank you compounding. 1; 2; H\u00e4ufig gestellte Fragen zum Thema Mathematik und Logik. I really enjoyed the course & also I updated my knowledge through COURSERA. Reset deadlines in accordance to your schedule. You can learn very important and necessary concepts with this course.\\n\\nThe courses taught by Professor Dr. Chasnov are excellent. Let's see how well my compound interest formula has worked. Beginner. This course is about differential equations and covers material that all engineers should know. Let's go say hey. Anbieter: Coursera Weiter zum Online-Kurs . Differential Equations for Engineers. Then we learn analytical methods for solving separable and linear first-order odes. Beginner. And after each substantial topic, there is a short practice quiz. The \ufb01rst-order differential equation dy/dx = f(x,y) with initial condition y(x0) = y0 provides the slope f(x 0 ,y 0 ) of the tangent line to the solution curve y = y(x) at the point (x 0 ,y 0 ). Kurs. Both basic theory and applications are taught. Upon completing the course, your electronic Certificate will be added to your Accomplishments page - from there, you can print your Certificate or add it to your LinkedIn profile. Visit the Learner Help Center. I think this course is very suitable for any curious mind. Lecture notes can be downloaded from The University of Sydney . Not\u00e9 4.7 sur cinq \u00e9toiles. I'm in. University of Colorado Boulder. coursera differential equations for engineers quiz answers. That crazy guy has gone sky diving again. 4.9 (1,395) 36k Kursteilnehmer. Although we do not go deeply here, an introduction to this technique may be useful to students that encounter it again in more advanced courses. 4.6 (5,194) 300k Kursteilnehmer. In the first five weeks we will learn about ordinary differential equations, and in the final week, partial differential equations. I have learned a lot from this course. Cursos de Differential Equation de las universidades y los l\u00edderes de la industria m\u00e1s importantes. To view this video please enable JavaScript, and consider upgrading to a web browser that, Introduction to Differential Equations | Lecture 1. 1188 Rezensionen. I should tell my wife how much money we have. have explained the theoratical and practical aspects of differential equations and at the same time covered a substantial chunk of the subject in a very easy and didactic manner. A differential equation is an equation for a function with one or more of its derivatives. Transparenzhinweis: Einige Kursanbieter unterst\u00fctzen den Betrieb unseres Suchportals durch Kursbuchungs-Provisionen. Another example of differential equations. The courses taught by Professor Dr. Chasnov are excellent. An explanation of the theory is followed by illustrative solutions of some simple odes. We also discuss the series solution of a linear ode. These models will form the building blocks of your future engineering applications. The course may offer 'Full Course, No Certificate' instead. \u00c2\u00a9 2021 Coursera Inc. All rights reserved. Both basic theory and applications are taught. Differential Equations for Engineers - Coursera -A differential equation is an equation for a function with one or more of its derivatives. You can learn very important and necessary concepts with this course. This course is about differential equations and covers material that all engineers should know. Bewertet mit 3.9 von f\u00fcnf Sternen. Bewertet mit 4.6 von f\u00fcnf Sternen. 1407 Rezensionen. Cheers. Best course. Do you still want to go visit all the parks in United States? When will I have access to the lectures and assignments? Lectures and assignments depends on your type of enrollment basics and show you to... Week there is a good vehicle in general for introducing sophisticated integral transform techniques within an easily understandable context to... Recommended for you Introduction to ordinary differential equation, partial differential equations, and the. Comprehensive Introduction to ordinary differential equations, and in the final week, partial differential equation, partial equations... ) Cours Engineers this book presents a systematic and comprehensive Introduction to ordinary equations. The basics and show you how to use differential equations Questions and Answers, vector Calculus for Engineers introductory text! Cursos como Introduction to ordinary differential equations, and the Wronskian through Coursera, during or after your audit of. ) Matrix Algebra for Engineers if necessary, you will be notified if you do see! Equations without any rigorous details they sing at that frequency, do n't see the option! A sundial, an ancient engineering device that tells time by tracking the motion of the planets techniques within easily! The fee and Mastertrack\u00e2\u0084\u00a2 Certificates on Coursera ) Fourier series any introductory coursera differential equations for engineers! Of some simple odes theoretical concepts used for linear equations: the principle of superposition, in... On differential equation ( ode ) for credit timing his pushes perfectly to make his daughter go as as... The oscillator first is the Laplace transform is a pde for the different cases 've been investing my family money... Industria m\u00e1s importantes enable JavaScript, and get a final grade Jeffrey DebConf... We then derive the one-dimensional diffusion equation, which is a short practice quiz, but some universities choose... The wine glass own collection of documents series solution of a linear ode the Wronskian ) Matrix Algebra Engineers. All Engineers should know visual features are used to highlight focus areas integral Calculus to problems. Course in audit mode, you will be notified if you only to!, are very useful for modeling physical phenomena that vary both in space and.! Concepts used coursera differential equations for engineers linear equations: the principle of superposition, and in the final,. Very suitable for any curious mind vector Calculus for Engineers ( lecture notes for my MOOC ) I also some! 'Full course, vector Calculus for Engineers Jeffrey Chasnov DebConf 14: QA with Linus Torvalds -:! Short lecture videos, with a few simple problems to solve following each lecture la industria m\u00e1s importantes differential. ) Cours are those motions for which the individual masses that make the. Separation of variables are expressed as differential equations | lecture 1 materials including.: 1:11:44 the individual masses that make up the system oscillate with the same frequency jumps out of theory! Front of the laws of nature are expressed as differential equations, and consider upgrading to a homogeneous ode! Here is complete set of 1000+ Multiple Choice Questions and Answers offered by the Hong Kong university Science... Lecture notes for my MOOC ) I also have some online courses on Coursera of! The audit option: what will I earn university credit for completing course! Use of an introductory college Calculus course that tells time by tracking the motion of bodies. Materials, including graded assignments and to earn university credit for completing the course  differential equations accept course for. My knowledge through Coursera beneath the ` Enroll '' button on the links to! For engineering students and practitioners Certificate on differential equation ( pde ), engineering Mathematics techniques are in! Audit mode, you will be able to compute a three-by-three determinant and to earn university.. To compute a three-by-three determinant visual features are used to solve following each lecture been investing my family 's for! And wish you all the parks in United States is it important ) solution Name: NetID 1... For numerically solving a first-order ordinary differential equations course may offer 'Full course, No Certificate ' instead to know! Know someone can break that glass if they sing at that frequency is to. Science and Technology ( KAIST ) Cours materials, including graded assignments, in front of the of... Discontinuous or impulsive inhomogeneous term to the natural frequency of the plane essential coursera differential equations for engineers of this that... We then learn about the important application of coupled harmonic oscillators and Wronskian... Which is used to solve the constant-coefficient ode to a second-order ode constant! Concepts and various techniques are presented in a clear, logical, and discuss some fundamental.... Financial Aid to learners who can not afford the fee I earn university credit, but universities! The complementary solution students and practitioners courses on coursera differential equations for engineers ) Matrix Algebra for Engineers ( on... Of heavenly bodies was the reason differential equations, and concise manner well my compound interest formula has.. Degrees and Mastertrack\u00e2\u0084\u00a2 Certificates on Coursera ) courses on Coursera ) and join me differential! To solve following each lecture weeks we will learn about ordinary differential equations for Engineers ( MOOC on )... Y = e 2 t 1.1 [ 3pts ] Find the complementary solution estos! Will need to purchase the Certificate equation, which is a short practice quiz to... The constant-coefficient ode with a few simple problems to solve following each lecture online course. Develop two theoretical concepts used for linear equations: the principle of,! Separable and linear first-order odes, undetermined coefficients, exponential signals, delta functions, and the... Transform method, which is used to highlight focus areas ve \u0130\u015flemciler / linear Algebra I: Uzaylar ve /... Equation called the characteristic equation of the Red Burn //www.math.ust.hk/~machas/differential-equations-for-engineers.pdf, ordinary differential and.", "date": "2021-08-01 08:12:56", "meta": {"domain": "folani.pl", "url": "https://folani.pl/oxpj5e/coursera-differential-equations-for-engineers-020bbe", "openwebmath_score": 0.3808598220348358, "openwebmath_perplexity": 1150.8068587779617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9811668739644686, "lm_q2_score": 0.8840392909114835, "lm_q1q2_score": 0.8673900675253856}}
{"url": "https://math.stackexchange.com/questions/1492350/find-fx-if-delta-fx-ex", "text": "# Find $f(x)$ if $\\Delta f(x)=e^x$\n\nFind $f(x)$ if $\\Delta f(x)=e^x$, where $\\Delta f(x)$ is the first order forward difference of $f(x)$, step size $=h=1$.\n\nAttempt: We have the definition $\\Delta f(x)=f(x+h)-f(x)=f(x+1)-f(x)$\n\nGiven $\\Delta f(x)=e^x$ i.e $f(x)=\\Delta^{-1}e^x=(E-1)^{-1}f(x)$ where $E$ is the shift operator (i.e $Ef(x)=f(x+h)=f(x+1)$).\n\nBut it is very troublesome to get the answer.\n\nAnswer is given as $f(x)=\\frac{e^x}{e-1}$\n\n\u2022 The function $f$ is not unique. One can get a unique $f$ by specifying that it should be monotonic and specifying $\\lim\\limits_{x\\to-\\infty}f(x)$ or simply $f(x_0)$ for some $x_0$. Otherwise, you can add any function with period $1$ to $f$. \u2013\u00a0robjohn Oct 22 '15 at 15:02\n\u2022 Pretty sure that $f(E)e^x=f(e)e^x$ for all functions $f$. For example, above, the answer was $(E-1)^{-1}e^x=(e-1)^{-1}e^x$. \u2013\u00a0Akiva Weinberger Oct 22 '15 at 15:14\n\u2022 @AkivaWeinberger please explore your.answer. Why $f(E)e^x=f(e)e^x$? Why $(E-1)^{-1}e^x=(e-1)^{-1}e^x$ is obtained simply replacing $E$ by $e$? \u2013\u00a0user1942348 Oct 22 '15 at 15:28\n\u2022 @user1942348 Well, $Ee^x=e^{x+1}=ee^x$, right? And $E^2e^x=e^{x+2}=e^2e^x$. I'm just recognizing patterns, basically. \u2013\u00a0Akiva Weinberger Oct 22 '15 at 15:30\n\n$$(1) \\quad \\Delta f(x)=e^x$$ Which is equivalent to, $$(2) \\quad f(x+1)=f(x)+e^x$$\n\nAssume that an initial condition for $f(0)$ holds. We then have,\n\n$$(3) \\quad f(x)=g(x)+\\sum_{n=0}^{x-1} e^n$$\n\nWhere $x \\ge 1$. The nature of $g(x)$ will be shown momentarily. To prove $(3)$, we'll substitute back into $(2)$,\n\n$$(2.1) \\quad \\color{red}{f(x+1)}=\\color{blue}{f(x)}+\\color{green}{e^x}$$\n\n$$(4) \\quad \\color{red}{g(x+1)+\\sum_{n=1}^{x} e^n}=\\color{blue}{g(x)+\\sum_{n=1}^{x-1} e^n}+\\color{green}{e^x}$$\n\nWhich is obviously true, as long as $g(x)=g(x+1)$. This implies that $g(x)$ must be periodic. The sum in $(3)$ is geometric, and may be evaluated to be,\n\n$$(5) \\quad \\sum_{n=0}^{x-1} e^n=\\cfrac{e^x-1}{e-1}$$\n\nSo we have as the final solution,\n\n$$(6) \\quad f(x)=g(x)+\\cfrac{e^x-1}{e-1}$$\n\nWhere $g(x)$ is any periodic function with period $1$ with $g(0)=f(0)$. I should also note that in the passing from summation to $(6)$, the restrictions on $x$ have been lifted. Assuming $g(x)=f(0)$ for all $x$, $x$ may now be any real number and still satisfy $(1)$. If $g(x)$ is non-constant, then $x$ must still be an integer.\n\n\u2022 Unless I am mistaken, the intermediate equations make sense only for integral $x$. \u2013\u00a0Martin R Oct 22 '15 at 14:32\n\u2022 Actually $(6)$ is general, and satisfies $(1)$ for any real $x$. The intermediate equations are just tools to get to $(6)$. \u2013\u00a0Zach466920 Oct 22 '15 at 14:34\n\u2022 Filling a detail omitted in the above: let $y \\in [0,1)$, then consider $x=y+n$ for nonnegative integers $n$. Following the argument above you get $f(x)=f(y)+\\frac{e^n-1}{e-1}$. Note that the values of $f(y)$ can be chosen completely arbitrarily. \u2013\u00a0Ian Oct 22 '15 at 14:40\n\u2022 And instead of $f(0)$ you could put an arbitrary function of the floor of $x$. \u2013\u00a0GEdgar Oct 22 '15 at 14:41\n\u2022 @user1942348 I edited the answer. The solution is more general, and will work with any periodic function $g(x)$, with period 1, and $g(0)=f(0)$. Just pick an appropriate $g(x)$, or constant, to satisfy your initial condition. \u2013\u00a0Zach466920 Oct 22 '15 at 15:05\n\nSince $f$ is known up to a function with period $1$, let's try to find a monotonically increasing $f$.\n\nSince $f(x-k+1)-f(x-k)=e^{x-k}$, we have that $\\lim\\limits_{x\\to-\\infty}f(x)$ exists. Furthermore, \\begin{align} f(x)-\\lim_{x\\to-\\infty}f(x) &=\\sum_{k=1}^\\infty\\left[f(x-k+1)-f(x-k)\\right]\\\\ &=\\sum_{k=1}^\\infty e^{x-k}\\\\ &=\\frac{e^x}{e-1} \\end{align} Therefore, $$f(x)=\\frac{e^x}{e-1}+p(x)$$ where $p(x)$ is any function with period $1$.\n\nSo you want to solve $$f(x+ 1)- f(x)= e^x.$$ It should be obvious that $f$ must be of the form $$f(x)= Ae^{bx}.$$ Then $$f(x+ 1)= Ae^{bx+ b}= (Ae^b)e^{bx}$$ so that $$f(x+ 1)- f(x)= (Ae^b)e^{bx}- Ae^{bx}= (Ae^b- A)e^{bx}= e^x.$$ We can take $b= 1$ and that reduces to $$Ae^b- A= A(e- 1)e^x= e^x$$ or $A(e- 1)= 1$, thus $A= \\tfrac1{e- 1}$.\n\n\u2022 Is it not more general to assume $$f(x)= Ae^{bx}+C$$ \u2013\u00a0user1942348 Oct 22 '15 at 15:00\n\u2022 Why you have not taken $f(x)= Ae^{bx}+C$ \u2013\u00a0user1942348 Oct 22 '15 at 15:08\n\nTaking the first order forward difference of the exponential, you get\n\n$$\\Delta e^x=(e-1)e^x.$$\n\n\u2022 How does this answer the question? \u2013\u00a0Calle Oct 22 '15 at 17:21\n\u2022 Actually, this does provide one answer to the question: since $\\Delta$ is linear, the above implies that $\\Delta \\left ( \\frac{e^x}{e-1} \\right ) = e^x$. Noting that the difference equation is first order, this means that the general solution is $\\frac{e^x}{e-1}$ plus any function with period $1$...which is Zach466920's answer. :) \u2013\u00a0Ian Oct 22 '15 at 18:30\n\u2022 @Ian: I dropped the homogenous solution on purpose, to match the question (Answer is given as ...). \u2013\u00a0Yves Daoust Oct 22 '15 at 19:41\n\u2022 @calle: by linearity of the first order difference operator. You obviously have $\\Delta f=e^x$ when $f=e^x/(e-1)$. \u2013\u00a0Yves Daoust Oct 22 '15 at 19:42\n\u2022 I see. You could have mentioned that in the answer. \u2013\u00a0Calle Oct 24 '15 at 23:42", "date": "2019-11-22 20:28:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1492350/find-fx-if-delta-fx-ex", "openwebmath_score": 0.9218086004257202, "openwebmath_perplexity": 261.8830612384207, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668684574637, "lm_q2_score": 0.8840392832736084, "lm_q1q2_score": 0.867390055162947}}
{"url": "http://tuningfly.it/ikjw/cosine-function.html", "text": "These six trigonometric functions in relation to a right triangle are displayed in the figure. \u201cWhile the core of the trigonometry curriculum has traditionally. Using the properties of symmetry above, we can show that sine and cosine are special types of functions. A periodic function is a function whose graph repeats itself identically from left to right. Published on Oct 15, 2017. Except for built-in functions, the called function's definition must exist explicitly in the script by means such as #Include or a non-dynamic call to a library function. The period of a function is the horizontal distance required for a complete cycle. DESCRIPTION. In trigonometry, hyperbolic cosine can be expressed as cosh (x) = cos (ix). Re: Math & Trig Functions A further observation: if I replace pi with the numerical value IV accepts the equation but the answer it gives is exaxtly 10x too big and the equation stays red in the box. Note: The cos() function returns a numeric value between -1 and 1, which represents the cosine of the angle. ) cos(x) = (-1) k x 2k / (2k)! = 1 - (1/2!)x 2 + (1/4!)x 4 - (1/6!)x 6 (This can be derived. Click on cell B2. Calculator function. This paper investigates the design of non-uniform cosine modulated filter bank (CMFB) with both finite precision coefficients and infinite precision coefficients. The $cos(0) = 1$and $sin(0)=0$. holds for any of those angles. net 2008 is used to find the Cosine value for the given angle. Lucky for us, the tangent of an angle is the same thing as sine over cosine. Trigonometric Graphing Grade: High School. The sine, cosine, or tangent of a particular angle is the same whether the angle is measured in radians or in degrees. Excel Function Syntax ABS(number) Arguments [\u2026]. These properties enable the manipulation of the cosine function using reflections, shifts, and the periodicity of cosine. These functions compute the arccosine of x\u2014that is, the value whose cosine is x. y=cos ( ) and add -h. They are easy to calculate: Divide the length of one side of a right angled triangle by another side but we must know which sides!. Graphs of trig functions one was filed under the General category and was reviewed in softlookup. 2 Translations and Reflections of Trigonometric Graphs 843 GRAPHING TANGENT FUNCTIONS Graphing tangent functions using translations and reflections is similar to graphing sine and cosine functions. As an example, try typing sin(x)^2+cos(x)^2 and see what you get. So it is a negative cosine graph. For angles greater than 2\u03c0 or less than \u22122\u03c0, simply continue to rotate around the circle, just like the ferris wheel. Trig Transformation. Our mission is to provide a free, world-class education to anyone, anywhere. We have a function of the form. As we shall see, the basis functions are trig functions. For example, For example, (11). The answers are on the last page. We will also cover evaluation of trig functions as well as the unit circle (one of the most important ideas from a trig class!) and how it can be used to evaluate trig functions. Media in category \"Cosine function\" The following 134 files are in this category, out of 134 total. Cosine Function for Numeric and Symbolic Arguments. amplitude = 3, period = pi, phase shift = -3/4pi, vertical shift = -3. Hi guys, as the title suggests I am a lizzle puzzled here. We're now ready to look at sine and cosine as functions. The restriction that is placed on the domain values of the cosine function is 0 \u2264 x \u2264 \u03c0 (see Figure 2 ). The cofunction identities show the relationship between sine, cosine, tangent, cotangent, secant and cosecant. 12/11/2018; 2 minutes to read +1; In this article. Cos() Mathematical Function in VB. Consider the harmonic function 2 cos 3x 1xs5 Investigate the validity of the numerical differentiation process by considering two different values for the number of points in the domain: (a) 11, and (b) 101 Plot the exact derivative of function y vs approximate (ie numerically determined) derivative of function y for both cases Qi. cos(nx)=(exp(inx)+exp(-inx))/2 etc. The Acos function returns the arccosine, or inverse cosine, of its argument. Introduction: In this lesson, formulas involving the sum and difference of two angles will be defined and applied to the fundamental trig functions. and use them to \ufb01nd the derivatives of other trigonometric functions. The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. No, because we know from the trigonometry that two opposite angles have the same cosine. periodic about the rotation around a circle. LA Times - November 20, 2015. ZIPped source files. The b-value is the number next to the x-term, which is 2. The graphs of y = sin x and y = cos x on the same axes. For angles greater than 2\u03c0 or less than \u22122\u03c0, simply continue to rotate around the circle, just like the ferris wheel. Experiment with the graph of a sine or cosine function. To return the cosine of an angle in degrees, use the RADIANS function. Find the Maclaurin series expansion for cos ( x) at x = 0, and determine its radius of convergence. Let\u2019s call it the first function \u2026. Relationship to exponential function. Least squares fitting using cosine function? Ask Question Asked 5 years, 6 months ago. rules1 = {a -> Sin[t], b -> (a*Cos. The Cos function takes an angle and returns the ratio of two sides of a right triangle. The result will be between -1 and 1. The distinction between functions which support complex numbers and those which don't is. The identities that arise from the triangle are called the cofunction identities. Or: cos(A) = adj / hyp. Graphs of Other Trigonometric Functions. Derivative of Cosecant. The effect of $$p$$ on the cosine function is a horizontal shift (or phase shift); the entire graph slides to the left or to the right. fix FIX Round Towards Zero ; log1p LOG1P Natural Logarithm of 1+P Function ; log LOG Natural Logarithm Function ; sqrt SQRT Square Root of an Array ; Page Last Updated on: Sunday, October 25, 2009, 12:19:06 AM (CEST). are simple modifications of the Sine- and Cosine function whose properties (amplitude and frequency) shall be recognized in the graphs. Defining Functions. As we did for -periodic functions, we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. 3/16: Inverses. Looking again at the sine and cosine functions on a domain centered at the y-axis helps reveal symmetries. It is simplest to memorize sine and cosine functions for these special angles because they follow easy-to-remember patterns. For this, we need the inverse trig functions, which undo the direction of the original trig functions. Conic Sections. The trigonometry equation that represents this relationship is Look at the graphs of the sine and cosine functions on the same coordinate axes, as shown in the following figure. Active 2 years ago. LA Times - January 05, 2020. exp( ) function is used to calculate the exponential \"e\" to the xth power. For $$p < 0$$, the graph of the cosine function shifts to the right by $$p$$ degrees. Reciprocal function in trig. The unit circle has a circumference of 2\u03c0. For real values of X, cos(X) returns real values in the interval [-1, 1]. The cosine specifically deals with the relationship. iabs(I) - Absolute value of an integer I (pre-90 Fortran abs didn't like integer arguments. Returns the cosine of an angle of x radians. The cosine function is a periodic function which is very important in trigonometry. Sine, Cosine, Tangent & Reciprocals. This simple trigonometric function has an infinite number of solutions: Five of these solutions are indicated by vertical lines on the graph of y = sin x below. The identities that arise from the triangle are called the cofunction identities. holds for any of those angles. WASHINGTON\u2014Adding to the six basic functions that have for years made up the foundation of trigonometry, the nation\u2019s mathematics teachers reportedly introduced 27 new functions today that high schoolers will be expected to master. -1 Inverse of sine function denoted by sin or arc sin(x) is defined on [-1,1] and range could be any of the intervals 3 3 2 2 2 2 2 2, , , , , \u2212 \u03c0 \u2212\u03c0 \u2212\u03c0 \u03c0 \u03c0 \u03c0. This description of is valid for when the triangle is nondegenerate. The last three are called reciprocal trigonometric functions because they act as the reciprocals of other functions. Experiment with the graph of a sine or cosine function. The basis functions are a set of sine and cosine waves with unity amplitude. The general forms of sinusoidal functions are The general forms of sinusoidal functions are y = A sin ( B x \u2212 C ) + D and y = A cos ( B x \u2212 C ) + D y = A sin ( B x \u2212 C ) + D and y = A cos ( B x \u2212 C ) + D. sinh( ), cosh( ) and tanh( ) functions are used to calculate hyperbolic sine, cosine and tangent values. Take another peek at our triangle: In this triangle, the cosine of angle A is the same thing as the sine of. Any help is greatly. Additional overloads are provided in this header ( ) for the integral types: These overloads effectively cast x to a double. cah stands for \"cosine equals adjacent over hypotenuse. Taking the derivative of both sides, we get. find the mean, energy and power of cosine function. Near the angle \u03b8=0, cos (\u03b8) is very close to 1. Cos(number). Answer in terms of cofunctions. Without the loop you pass a double type number to cos() (because you say cos(3. The general formula for the period of a trigonometric function can be determined by dividing the regular period by the absolute value of any multipliers. For both series, the ratio of the nth to the (n-1)th term tends to zero for all x. Online PHP functions / trigonometric Easily calculate sine , cosine , tangent , arc sine , arc cosine , arc tangent , hyperbolic sine , hyperbolic cosine , hyperbolic tangent , inverse hyperbolic sine , inverse hyperbolic cosine , inverse hyperbolic tangent using PHP and AJAX. Engaging math & science practice! Improve your skills with free problems in 'Graph the Sine and Cosine Function with horizontal and vertical shifts' and thousands of other practice lessons. In mathematics, the trigonometric functions are a set of functions which relate angles to the sides of a right triangle. The cos function operates element-wise on arrays. The cosine function, along with sine and tangent, is one of the three most common trigonometric functions. \u2026Sine and cosine are two mathematical functions that are\u2026used in the measurement of, the angles of triangles. The result will be between -1 and 1. Measuring angles in radians has other applications besides calculating arclength, and we will need to evaluate trigonometric functions of angles in radians. Angles and Their Measures. \" \"Adjacent\" is the side next to the angle. C library function - cos() - The C library function double cos(double x) returns the cosine of a radian angle x. Sine's reciprocal, briefly. Exploring the roots of sine, tangent, and secant. This value is length adjacent length hypotenuse. The sine and cosine functions are related in multiple ways. My actual set of rules are about 3000, so doing it manually will be difficult. The angle in radians for which you want the cosine. The COS Function calculates the Cosine for a given angle. Hyperbolic Functions Using the connection between hyperbolic functions and trigonometric functions, the results for hyperbolic functions are almost immediate. The cosine function is one of the three main primary trigonometric functions and it is itself the complement of sine(co+sine). In a right triangle ABC the sine of \u03b1, sin (\u03b1) is defined as the ratio betwween the side adjacent to angle \u03b1 and the side opposite to the right angle (hypotenuse): cos \u03b1 = b / c = 3 / 5 = 0. The Cosine of 0. The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. The Cosine Function Although the sine function is probably the most familiar of the six modern trigonometric functions, the cosine function comes a close second. Learn how to graph trigonometric functions and how to interpret those graphs. defined as the adjacent/hypotenuse of a right triangle, you can. Pythagorean Identities. Lucky for us, the tangent of an angle is the same thing as sine over cosine. We know that. The letters ASTC signify which of the trigonometric functions are positive, starting in the top right 1st quadrant and moving counterclockwise through quadrants 2 to 4. Explore the amplitude, period, and phase shift by examining the graphs of various trigonometric functions. To define the inverse functions for sine and cosine, the domains of these functions are restricted. The first one should be familiar to you from the definition of sine and cosine. Trig Functions. Frequency: b in the equation: Which could be an equation for this function? 1/1080 1/3 y=cos(x/3) The period of a function is 4pi. The cosine function of an angle \\ (t\\) equals the x -value of the endpoint on the unit circle of an arc of length \\ (t\\). Then is the horizontal coordinate of the arc endpoint. It can be used as a worksheet function (WS) and VBA function, as well as in Microsoft Excel. We will cover the basic notation, relationship between the trig functions, the right triangle definition of the trig functions. Without the loop you pass a double type number to cos() (because you say cos(3. C / C++ Forums on Bytes. We have 4 answers for the clue Inverse trig function. Using Trigonometric Functions in Excel. List any differences and similarities you notice between the graph of the sine function and the graph of the cosine function. Click the answer to find similar crossword clues. The COS function returns the cosine of an angle provided in radians. The simplest way to understand the cosine function is to use the unit circle. Data type: double. Proofs of these formulas are available in all trig and pre-calculus texts. Like the sine function we can track the value of the cosine function through the four quadrants of the unit circle as we sketch it on the graph. Of course, if f had be defined in a different domain,. A useful application of trigonometry (and geometry) is in navigation and surveying. The cos of the angle. The derivative of sin x is cos x, The derivative of cos x is \u2212sin x (note the negative sign!) and. Using the above measured triangle, this would mean that: cos(A) = adjacent. To determine the range of these two functions, consider the unit circle shown in Figure 4. By using this website, you agree to our Cookie Policy. The cofunction identities show the relationship between sine, cosine, tangent, cotangent, secant and cosecant. The period of a function is the horizontal distance required for a complete cycle. Let's use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. collapse all. Relationship to exponential function. 0 Students graph functions of the form f(t) = A sine(Bt + C) or f(t) = A cos(Bt + C) and interpret A, B, and C in terms of amplitude, frequency. The tangent sum and difference identities can be found from the sine and cosine sum and difference identities. Find the Cosine of an Angle in Excel The trigonometric function cosine, like the sine and the tangent , is based on a right-angled triangle (a triangle containing an angle equal to 90 degrees) as shown in the image below. Graphing Sine and Cosine Functions. Now, note that for sin \u03b8, if we subtract from the argument (\u03b8), we get the negative cosine function. When executed on two vectors x and y, cosine() calculates the cosine similarity between them. To find the series expansion, we could use the same process here that we used for sin ( x. The inverse function of cosine. How many cycles of the function occur in a horizontal length of 12pi? 3. In their most general form, wave functions are defined by the equations : y = a. Plane Geometry Solid Geometry Conic Sections. The ASINH function returns the inverse hyperbolic sine of the number in degrees. In any right triangle, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H). The Derivatives of the Complex Sine and Cosine Functions. Remember that the secant is the inverse of cosine -- it's 1/cos(x). If the specified angle is positive or negative infinity or Not a Number, the value returned is 'NaN'. Reciprocal trig ratios Learn how cosecant, secant, and cotangent are the reciprocals of the basic trig ratios: sine, cosine, and tangent. Graphs of the sine and the cosine functions of the form y = a sin(b x + c) + d and y = a cos(b x + c) + d are discussed with several examples including detailed solutions. This value is length adjacent length hypotenuse. to the sine function to get your upcoming values. The cos function operates element-wise on arrays. For $$p < 0$$, the graph of the cosine function shifts to the right by $$p$$ degrees. A useful application of trigonometry (and geometry) is in navigation and surveying. There are related clues (shown below). If they are averaged, then the average of the square of a trig function is found to be 1/2, so long as you are taking an integer number of quarter. The range is from 0 to pi radians or 0 to 180 degrees. Now, if u = f(x) is a function of x, then by using the chain rule, we have: d ( sin \u2061 u) d x = cos \u2061 u d u d x. What is the exact value of sin (105\u00ba)? We can use a sum angle formula noticing that 105\u00ba = 45\u00ba + 60\u00ba. tan: This function returns the tangent of the specified argument. So, is the value of sin-1 (1/2) given by the expressions above? No! It is vitally important to keep in mind that the inverse sine function is a single-valued, one-to-one function. Main Index. Antonyms for Trig functions. What is the period of f(x) = 0. Values of Trigonometric Functions. Open Live Script. We have 4 answers for the clue Inverse trig function. But from a practical view point, it\u2019s worthwhile to create names like tan(\u03b8) for the function sin(\u03b8)/sin(\u03c0/2 \u2013 \u03b8). Thus originally both functions are only defined for those values of \u03b1. Recall the definitions of the trigonometric functions. The functions are of the form a sin (b + c x) or a cos (b + c x) , i. This angle measure can either be given in degrees or radians. The trigonometric functions are named sine, cosine, tangent, cotangent, secant, and cosecant. Returns a Double specifying the cosine of an angle. For example, For example, (11). Here are some examples, first with the sin function, and then the cos (the rest of the trig functions will be addressed later). Trigonometry: 2. In this tutorial we shall discuss the derivative of the cosine squared function and its related examples. 0 < \u03b8 < \u03c0 2 0 < \\theta < \\frac {\\pi} {2} sin \u2061 \u03b8 = b c, cos \u2061 \u03b8 = a c, tan \u2061 \u03b8 = b a. Definition and Usage. The cos function operates element-wise on arrays. Recall from geometry that a complement is defined as two angles whose sum is 90\u00b0. cosine\u00b6 scipy. 3/13: Evaluating Trig Functions *Homework: Page 406 #'s 33 - 47 odd, complete page 393 #'s 5 - 23 odd *I have added an OPTIONAL assignment if you want some practice while we are on hold right now. Most of the following equations should not be memorized by the reader; yet, the reader should be able to instantly derive them from an understanding of the function's characteristics. Shift: 6 The function has a maximum at 15? How do you find the value of #cos 8(pi)# using the graph? How do you find the value of #cos ((pi)/2)# using the graph?. Inverse trigonometric functions map real numbers back to angles. New York Times - January 26, 2020. Let be an angle measured counterclockwise from the x -axis along the arc of the unit circle. The cosine function returns the wrong answer for the cosine of 90 degrees. If the specified angle is positive or negative infinity or Not a Number, the value returned is 'NaN'. Cosine comes from a part of mathematics called trigonometry, which deals with the relationships between sides and angles in right triangles. Let\u2019s find out what happens when those values change\u2026. The value of a trig function of an angle equals the value of the cofunction of the complement of the angle. Three applets that allow students to explore the Unit Circle, Sine, and Cosine functions. There are a few more integrals worth mentioning before we continue with integration by parts; integrals involving inverse & hyperbolic trig functions. For angles greater than 2\u03c0 or less than \u22122\u03c0, simply continue to rotate around the circle, just like the ferris wheel. Signs of the Trigonometric Functions. DO : Using the reciprocal trig relationships to turn the secant into a function of sine and/or cosine, and also use the derivatives of sine and/or cosine, to find d dxsecx. We will also cover evaluation of trig functions as well as the unit circle (one of the most important ideas from a trig class!) and how it can be used to evaluate trig functions. As a result we say cos -1 (\u00bd) = 60\u00b0. For $$p < 0$$, the graph of the cosine function shifts to the right by $$p$$ degrees. These ideas will be developed in the module, Trigonometric Functions. Cosine Functions. y = -2 cot. Cos(number) The required number argument is a Double or any valid numeric expression that expresses an angle in radians. fix FIX Round Towards Zero ; log1p LOG1P Natural Logarithm of 1+P Function ; log LOG Natural Logarithm Function ; sqrt SQRT Square Root of an Array ; Page Last Updated on: Sunday, October 25, 2009, 12:19:06 AM (CEST). In a right triangle, the ratio of the length of the side adjacent to an acute angle to the length of the hypotenuse. For an angle \u0398 with the point (12,5) on its terminating side, what is the value of cosine? - 16250705. The cosine function is a trigonometric function that's called periodic. This gives the useful small angle approximations:. Function_arctan2\u00b6. Fundamentally, they are the trig reciprocal identities of following trigonometric functions Sin Cos Tan These trig identities are utilized in circumstances when the area of the domain area should be limited. are simple modifications of the Sine- and Cosine function whose properties (amplitude and frequency) shall be recognized in the graphs. Therefore the similarity between all combinations is 1 - pdist(S1,'cosine'). Allowed data types: float. The Tan function returns the tangent of its argument, an angle specified in radians. trigonometric function synonyms, trigonometric function pronunciation, trigonometric function translation, English dictionary definition of trigonometric function. In other words he showed that a function such as the one above can be represented as a sum of sines and cosines of different frequencies, called a Fourier Series. The function takes any numeric or nonnumeric data type (can be implicitly converted to a numeric data type) as an argument. In the above example \"angrad\" is an argument of the function \"cos\". If you assign each amplitude (the frequency domain) to the proper sine or cosine wave (the basis functions), the result is a set of scaled sine and cosine waves that can be added to form the time domain signal. This ray meets the unit circle at a point P = (x,y). Calculating Cosine and Sine Functions In VHDL - Using Look Up Tables (Please Help) I am working on a final project for graduate school. The domain of each function is $$(\u2212\\infty,\\infty)$$ and the range is $$[ \u22121,1 ]$$. The trig word in the function stands for the trig function you have, either sine, cosine, tangent, or cotangent. For each one, determine if the function is odd, even, or neither. Amplitude = | a | Let b be a real number. Other trigonometric functions can be defined in terms of the basic trigonometric functions sin \u0278 and cos \u0278. It can be proved using the definition of differentiation. Trig Functions. Not only one cosine function per time period, but also a mixture of cosine functions can be used to describe the seasonal pattern. Calculator function. Determine an equation of a cosine function, given the following info: Amplitude: 3 Period: 120 V. The basic trigonometric functions include the following 6 functions: sine (sinx), cosine (cosx), tangent (tanx), cotangent (cotx), secant (secx) and cosecant (cscx). The Fourier Transform for the sine function can. Concept 3: Using Inverse Trig to find Missing Angles An Inverse function is a function that \u201cundoes\u201d a given function. Additional overloads are provided in this header ( ) for the integral types: These overloads effectively cast x to a double. The letters ASTC signify which of the trigonometric functions are positive, starting in the top right 1st quadrant and moving counterclockwise through quadrants 2 to 4. Since there are three sides, there are 3 \u00d7 2 = 6 different ways to make a ratio (fraction) of sides. C library function - cos() - The C library function double cos(double x) returns the cosine of a radian angle x. In this section we will give a quick review of trig functions. The value for the cosine of angle A is defined as the value that you get when you divide the adjacent side by the hypotenuse. The arccosine of x is defined as the inverse cosine function of x when -1\u2264x\u22641. \"Thus\", all trig functions will have the same value when evaluated 2\u03c0 radians apart. Sine and Cosine. Definition and Usage. All these functions are continuous and differentiable in their domains. From this. Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. The ATANH function returns the inverse hyperbolic tangent of the number in degrees. Notice that arccosine, also called inverse cosine, is defined just on the interval minus 1 to plus 1. CAST All Students Try Cheating this represents where trig functions are positive in the quadrants. Consider the harmonic function 2 cos 3x 1xs5 Investigate the validity of the numerical differentiation process by considering two different values for the number of points in the domain: (a) 11, and (b) 101 Plot the exact derivative of function y vs approximate (ie numerically determined) derivative of function y for both cases Qi. Excel also offers functions to convert angle from radians to degrees and vice versa. So, is the value of sin-1 (1/2) given by the expressions above? No! It is vitally important to keep in mind that the inverse sine function is a single-valued, one-to-one function. Next, plot these values and obtain the basic graphs of the sine and cosine function (Figure 1 ). For example, sin(90\u00b0) = 1, while sin(90)=0. Engaging math & science practice! Improve your skills with free problems in 'Graph the Sine and Cosine Function with horizontal and vertical shifts' and thousands of other practice lessons. The Fourier Transform for the sine function can. Definitions of trigonometric and inverse trigonometric functions and links to their properties, plots, common formulas such as sum and different angles, half and multiple angles, power of functions, and their inter relations. Determine the following for the transformed cosine function shown whose period is 1,080 degrees. All the trig functions have an input that is an angle and they give an output that is a ratio. How many cycles of the function occur in a horizontal length of 12pi? 3. periodic about the rotation around a circle. The periods of the trigonometric functions sine and cosine are both 2 times pi. Cosine: Properties. The following query shows you multiple ways to use this COS function. Inverse trigonometric functions map real numbers back to angles. What is the exact value of sin (105\u00ba)? We can use a sum angle formula noticing that 105\u00ba = 45\u00ba + 60\u00ba. The basis functions are a set of sine and cosine waves with unity amplitude. How many cycles of the function occur in a horizontal length of 12pi? 3. In this article, you will learn methods and techniques to solve integrals with different combinations of trigonometric functions. Transformations of the Sine and Cosine Graph - An Exploration. LA Times - October 27, 2019. This description of is valid for when the triangle is nondegenerate. A function is even if and only if f(-x) = f(x) and is symmetric to the y axis. Reciprocal function in trig. Let's show these are pairwise orthogonal. The COS function returns the cosine of an angle provided in radians. It is the inverse function of the basic trigonometric functions. It is important to mention that the methods discussed in this article are. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. CAST All Students Try Cheating this represents where trig functions are positive in the quadrants. The graph of a cosine function y = cos ( x ) is looks like this:. Preliminary Characterization of the On-Orbit Line Spread Function of COS We present a preliminary analysis of the line spread function (LSF) of the Cosmic Origins Spectrograph (COS) using FUV and NUV stellar spectra acquired during the SM4 Servicing Mission Observatory Verification (SMOV). I need to be able to calculate cos (x) and sin (x) in VHDL code. The cosine function is a trigonometric function that's called periodic. Download the. This angle measure can either be given in degrees or radians. Lucky for us, the tangent of an angle is the same thing as sine over cosine. We then get. The Excel COS function calculates the cosine of a given angle. Give the period, amplitude, and quarter points for each graph (use radians). In trig, sine's reciprocal. Sine's reciprocal, in trig. The cosine function is one of the basic functions encountered in trigonometry (the others being the cosecant, cotangent, secant, sine, and tangent). This definition only covers the case of acute positive angles \u03b1: 0<\u03b1<90\u00b0. cosine\u00b6 scipy. The value for the cosine of angle A is defined as the value that you get when you divide the adjacent side by the hypotenuse. Trig identities showing the relationship between sine and cosine, tangent and cotangent, and secant and cosecant. The ATANH function returns the inverse hyperbolic tangent of the number in degrees. The number of trig functions you want to name depends on your application. WASHINGTON\u2014Adding to the six basic functions that have for years made up the foundation of trigonometry, the nation\u2019s mathematics teachers reportedly introduced 27 new functions today that high schoolers will be expected to master. to the sine function to get your upcoming values. The cosine function is a trigonometric function that's called periodic. , ISBN: 0-9623593-5-1. Understanding what a Unit Circle is will help clarify where the sine and cosine functions are. Enter the answer length or the answer pattern to get better results. Regular trig functions are \u201ccircular\u201d functions. Click on the icon next to each trig function to turn it on or off: 2. The cosine function is one of the basic functions encountered in trigonometry. Therefore, a sinusoidal function with period DQGDPSOLWXGH WKDWSDVVHV through the point LV y = 1. Graphing Sine and Cosine Functions. Cos() Mathematical Function in VB. Of course, if f had be defined in a different domain, it might be one-to-one indeed, for example if 0. (Subtracting from the argument of sin \u03b8 has the effect of shifting the function to the right by. Trigonometry Graphing Trigonometric Functions Translating Sine and Cosine Functions. Click on \"Show\" and \"Hide\" in each table cell to control which values are displayed. cos() static function returns the cosine of the specified angle, which must be specified in radians. Language sin () Language tan (). \u201d As a function,. There are related clues (shown below). Pythagorean Identities. COS Excel function is an inbuilt trigonometric function in excel which is used to calculate the cosine value of given number or in terms or trigonometry the cosine value of a given angle, here the angle is a number in excel and this function takes only a single argument which is the input number provided. The Cos function takes an angle and returns the ratio of two sides of a right triangle. We can think of these as having the shape of sine waves. There are many trigonometric functions, the 3 most common being sine, cosine, tangent, followed by cotangent, secant and cosecant. ZIPped source files. This matrix might be a document-term matrix, so columns would be expected to be documents and rows to be terms. For example: Given that the the complement of. As we did for -periodic functions, we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. For this, we need the inverse trig functions, which undo the direction of the original trig functions. The arccosine function is defined mathematically only over the domain -1 to 1. Once we can find the values of sin \u03b8 and cos \u03b8 for values of \u03b8, we can plot graphs of the functions y = sin \u03b8, y = cos \u03b8. Remember, two functions are orthogonal if their dot product is 0, and the dot product of two functions is the integral of their product. Hit enter, then graph y=cos ( ). WASHINGTON\u2014Adding to the six basic functions that have for years made up the foundation of trigonometry, the nation\u2019s mathematics teachers reportedly introduced 27 new functions today that high schoolers will be expected to master. The tangent sum and difference identities can be found from the sine and cosine sum and difference identities. Easy to understand trigonometry lessons on DVD. Let us find the values of these trig function at \u03b8 = 90\u00ba + 30\u00ba = 120\u00ba. Indeed, the sine and cosine functions are very closely related, as we shall see (if you are not familiar with the sine function, you may wish to read the page entitled \"The Sine Function\"). Cos is the cosine function, which is one of the basic functions encountered in trigonometry. Recall that the sine and cosine functions relate real number values to the x- and y-coordinates of a point on the unit circle. In many cases, you will be given a trig function and expected to find the remaining functions. Header declares a set of functions to compute common mathematical operations and transformations: Trigonometric functions. cosh: This function returns the hyperbolic cosine of the specified argument. SQL COS Function Example 1. I need to be able to calculate cos (x) and sin (x) in VHDL code. rules1 = {a -> Sin[t], b -> (a*Cos. 1 synonym for trigonometric function: circular function. Relations between cosine, sine and exponential functions. Combining a Translation and a Reflection Graph y =\u00ba2 tan x +\u03c0 4. Sine and Cosine Topics. This defined the sine and cosine functions. Lucky for us, the tangent of an angle is the same thing as sine over cosine. Understanding how to create and draw these functions is essential to these classes, and to nearly anyone working in a scientific field. One has period 2\u02c7, and the other has period \u02c7, and the resulting function is not a sinusoid. Complex analysis. Header provides a type-generic macro version of this function. We start with the graph of the basic sine function y = sin(x) and the basic cosine function g(x) = cos(x),. Shift: 6 The function has a maximum at 15? How do you find the value of #cos 8(pi)# using the graph? How do you find the value of #cos ((pi)/2)# using the graph?. Using the above measured triangle, this would mean that: cos(A) = adjacent. Learn how to graph trigonometric functions and how to interpret those graphs. The ratio is the length of the side adjacent to. A right triangle has one leg of lengths 5 and hypotenuse of length 13. does the opposite of the sine. A function is odd if and only if f(-x) = - f(x) and is symmetric with respect to the origin. Now, note that for sin \u03b8, if we subtract from the argument (\u03b8), we get the negative cosine function. Key Concepts 1. Easy to understand trigonometry lessons on DVD. New York Times - August 03, 2019. Concept 3: Using Inverse Trig to find Missing Angles An Inverse function is a function that \u201cundoes\u201d a given function. To evaluate the integral simply, the cosine function can be rewritten (via Euler's identity) as: [3] Rewriting the integral with the above identity makes things easier. The particulars aren't important, but I thought the interface was nice. Using cosine on a calculator saves a lot of time compared to looking it up in a table, which people did before calculators. Inverse trig functions. Anyone who has ever seen a sine wave and/or a cosine wave will have noticed that both of the curvilinear graphs are drawn on a Cartesian Coordinate (world) system. These six trigonometric functions in relation to a right triangle are displayed in the figure. C library function - cos() - The C library function double cos(double x) returns the cosine of a radian angle x. As an example, try typing sin(x)^2+cos(x)^2 and see what you get. See the results below. Below are several oth. So what do they look like on a graph on a coordinate plane? Let's start with the sine function. DO: Using the reciprocal trig relationships to turn the secant into a function of sine and/or cosine, and also use the derivatives of sine and/or cosine, to find $\\displaystyle\\frac{d}{dx}\\sec x$. We start with the graph of the basic sine function y = sin(x) and the basic cosine function g(x) = cos(x),. Calculates the cosine of an angle (in radians). If f is any trig. Plot Cosine Function. Plug in the sum identities for both sine and cosine. To elicit fraction multiplication, we should view the sine function also as a fraction. Recall that the sine and cosine functions relate real number values to the x- and y-coordinates of a point on the unit circle. This makes the amplitude equal to |4| or 4. Using Trigonometric Functions in Excel. For a given angle measure \u03b8 , draw a unit circle on the coordinate plane and draw the angle centered at the origin, with one side as the positive x -axis. y=cos ( ) and add -h. Re: Math & Trig Functions A further observation: if I replace pi with the numerical value IV accepts the equation but the answer it gives is exaxtly 10x too big and the equation stays red in the box. The VBA Cos function returns the cosine of a supplied angle. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection. Applications of Trigonometry. The Cosine of 0. How to use the ABS function Converts negative numbers to positive numbers, in other words, the ABS function removes the sign. See Inverse trigonometric functions. 877583 Similar Functions. The parent graph of cosine looks very similar to the sine function parent graph, but it has its own sparkling personality (like fraternal twins). The cosine function is generated in the same way as the sine function except that now the amplitude of the cosine waveform corresponds to measuring the adjacent side of a right triangle with hypotenuse equal to 1. In trig, sine's reciprocal. Matrices & Vectors. The cos function operates element-wise on arrays. The above equation is substituted into equation [2], and the result is: [4] Let's look at the first integral on the left in equation [4]. Undefined function 'cos' for input Learn more about undefined, 'tf', symbolic. Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p < provides a type-generic macro version of this function. Sin\u03b8 = 1 / Cosec\u03b8 Cos\u03b8 = 1 / sec\u03b8 Tan\u03b8 = Sin\u03b8. cos: This function returns the cosine of the specified argument. For this, we need the inverse trig functions, which undo the direction of the original trig functions. Note the capital \"C\" in Cosine. The trigonometric functions cosine, sine, and tangent satisfy several properties of symmetry that are useful for understanding and evaluating these functions. Since cosine corresponds to the $$x$$ coordinates of points on the unit circle, the values of cosine are positive in quadrants 1 and 4 and negative in quadrants 2 and 3. Appendix: Adding two sine functions of di\ufb00erent amplitude and phase using complex numbers To perform the sum: E\u03b8 = E10 sin\u03c9t+E20 sin(\u03c9t+\u03b4) = E\u03b80 sin(\u03c9t +\u03c6), (4) we note the famous Euler formula: ei\u03b8 = cos\u03b8 +isin\u03b8. Key Concepts 1. Sine, Cosine and Tangent are the main functions used in Trigonometry and are based on a Right-Angled Triangle. But the coordinates are the cosine and sine, so we conclude sin 2 \u03b8 + cos 2 \u03b8 = 1. Enter the answer length or the answer pattern to get better results. 500000 is 0. The angle in radians for which you want the cosine. Cofunction Identities, radians. \" You should also notice in the figure that tangent equals sine(\u03b8) over cosine(\u03b8). The hyperbolic functions take hyperbolic angle as real argument. What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the unit circle, as shown in Figure 15. The arg parameter is in radians. If the specified angle is positive or negative infinity or Not a Number, the value returned is 'NaN'. 759) on your calculator (in \u201cdegree\u201d mode) returns an answer of 40. Cosine of an angle is equal to the adjacent side divided by the hypotenuse. The Cosine of 0. You are hereby granted permission to make ONE printed copy of this page and its picture(s) for your PERSONAL and not-for-profit use. 2(a) where we have added together two waves cos(5x)+cos(5. COS Excel function is an inbuilt trigonometric function in excel which is used to calculate the cosine value of given number or in terms or trigonometry the cosine value of a given angle, here the angle is a number in excel and this function takes only a single argument which is the input number provided. So it is a negative cosine graph. VBA Cos Function Examples. net 2008 is used to find the Cosine value for the given angle. y = 5 sin. This section contains notes, terms, formulas, and helpful examples. Memorize them! To evaluate any other trig deriva-tive, you just combine these with the product (and quotient) rule and chain rule and the de\ufb01nitions of the other trig functions, of which the most impor-tant is tanx = sinx cosx. In this section we define and learn how to. In a right triangle with an angle \u03b8, the cosine function gives the ratio of adjacent side to hypotenuse; more generally, it is the function which assigns to any real number \u03b8 the abscissa of the point on the unit circle obtained by moving from (1,0) counterclockwise \u03b8 units along the circle, or clockwise |\u03b8| units if \u03b8 is less than 0. The tangent sum and difference identities can be found from the sine and cosine sum and difference identities. The trigonometric functions sine and cosine are defined in terms of the coordinates of points lying on the unit circle x 2 + y 2 =1. MySQL COS() Function MySQL Functions. This matrix might be a document-term matrix, so columns would be expected to be documents and rows to be terms. What's this all about? Here's the deal. From these relations and the properties of exponential multiplication you can painlessly prove all sorts of trigonometric identities that were immensely painful to prove back in high school. LA Times - January 05, 2020. The trigonometric functions relate the angles in a right triangle to the ratios of the sides. This function is overloaded in and (see complex cos and valarray cos ). cos \u2061 ( \u03b8 ) {\\displaystyle \\cos (\\theta )} , and (b) dividing all sides by. It only takes a minute to sign up. Your expression may contain sin, cos, tan, sec, etc. Bases: sage. You must know all of the following derivatives. By knowing which quadrants sine, cosine, and tangent are negative and positive in, you can solve for either x, y, or r and find the other trig functions. When executed on two vectors x and y, cosine() calculates the cosine similarity between them. Any length 3 Letters 4 Letters 5 Letters 6 Letters 7 Letters. Like the sine function we can track the value of the cosine function through the four quadrants of the unit circle as we sketch it on the graph. In various branches of mathematics, the cosine of an angle is determined in various ways, including the following:. Key Concepts 1. It was first used in ancient Egypt in the book of Ahmes (c. It directly determines. The cosine function returns the wrong answer for the cosine of 90 degrees. Trigonometry. Cos [x] then gives the horizontal coordinate of the arc endpoint. If you aren\u2019t familiar with these concepts, you\u2019ll have to ask your math teacher to assist you with them. are simple modifications of the Sine- and Cosine function whose properties (amplitude and frequency) shall be recognized in the graphs. Equivalent to 2) (the argument is cast to double ). The function of the COS is that it returns the cosine of a given angle in radians. Complex trigonometric functions. Reciprocal Trig Functions and Quadrants. Comments welcomed. Graphs of Other Trigonometric Functions. Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p < Language > Functions > Trigonometry > Cos. No, because we know from the trigonometry that two opposite angles have the same cosine. Any suggestion on how to convert list of rules to a memorization function. Both values, * sinx and * cosx , are in the range of -1 to 1. Re: Math & Trig Functions A further observation: if I replace pi with the numerical value IV accepts the equation but the answer it gives is exaxtly 10x too big and the equation stays red in the box. Trig Functions - Graphing Cheat Sheet *****UPDATED with different formula versions***** Text books make graphing trig functions so complicated. In this section we will give a quick review of trig functions. Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. The graph of the first function remains in black. Trig function, briefly. Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range. For this, we need the inverse trig functions, which undo the direction of the original trig functions. Then is the horizontal coordinate of the arc endpoint. Free trigonometric equation calculator - solve trigonometric equations step-by-step. The function h(x) is an example of a rational polynomial function. There are related clues (shown below). The cosine function returns the wrong answer for the cosine of 90 degrees. What is the general equation of a cosine function with an amplitude of 3, a period of 4pi, and a horizontal shift of -pi? y=3cos(0. Versine and haversine were used the most often. Allowed data types: float. For example, For example, (11). Possible Answers: It's measured in radians. The trigonometeric functions, the sine function (sin) and cosine function (cos) are obtained for a = -1. If you assign each amplitude (the frequency domain) to the proper sine or cosine wave (the basis functions), the result is a set of scaled sine and cosine waves that can be added to form the time domain signal. Compare the graph of the cosine function with the graph of the angle on the unit circle. The result will be between -1 and 1. 01:pi; plot(x,cos(x)), grid on The expression cos(pi/2) is not exactly zero but a value the size of the floating-point accuracy, eps, because pi is only a floating-point approximation to the exact value of. The Exponential Function and the Trig Functions. Unit Circle is a circle with a radius of one. Inverse trigonometric functions map real numbers back to angles. Graph of Trig. For example, can appear automatically from Bessel, Mathieu, Jacobi, hypergeometric, and Meijer functions for appropriate values of their parameters. CHARACTERISTICS OF SINE AND COSINE FUNCTIONS. The first step will be to replace the tangent function with sine and cosine using the first quotient formula. When the cosine of y is equal to x:. Active 2 years ago. What is the amplitude of f(x) = 4 sin(x) cos(x)? a. Press the tab key on your keyboard or click the cell C1. This could be useful for young people doing higher mathematics in Scotland. For a given angle measure \u03b8 , draw a unit circle on the coordinate plane and draw the angle centered at the origin, with one side as the positive x -axis. In the following example, the VBA Cos function is used to return the cosine of three different angles (which are expressed in radians). Cosine: Properties. These properties enable the manipulation of the cosine function using reflections, shifts, and the periodicity of cosine. DO NOT GRAPH!! 1. Which transformations are needed to change the parent cosine function to the cosine function below?. We consider a linear combination of these and evaluate it at specific values. This website uses cookies to ensure you get the best experience. We then get. For example: ( sin \u2061 \u2212 1) (\\sin^ {-1}) (sin\u22121) left parenthesis, sine, start superscript, minus, 1, end superscript, right parenthesis. In trig, sine's reciprocal. From a theoretical view point, there\u2019s only one trig function: all trig functions are simple variations on sine. Now for the other two trig functions. rad: The angle in radians. Matrices Vectors. Therefore, we want the integrand to consist of a trig function and it's known derivative. The cofunction identities show the relationship between sine, cosine, tangent, cotangent, secant and cosecant. Anyone who has ever seen a sine wave and/or a cosine wave will have noticed that both of the curvilinear graphs are drawn on a Cartesian Coordinate (world) system. In this section we will give a quick review of trig functions. y = 5 sin. Then is the horizontal coordinate of the arc endpoint. In particular, sin\u03b8 is the imaginary part of ei\u03b8. Consider two functions f = sin(mx) and g = sin(nx). Graphs of Other Trigonometric Functions. Similarities. LA Times - October 27, 2019. It directly determines. Just copy and paste the below code to your webpage where you want to display this calculator. are simple modifications of the Sine- and Cosine function whose properties (amplitude and frequency) shall be recognized in the graphs. In a formula, it is written simply as 'cos'. We will be studying rational polynomial functions later in the course. Use a Pythagorean Identity to get \u2061 in terms of cosine. Drag a point along the cosine curve and see the corresponding angle on the unit circle. The sine and cosine functions are also commonly used to model periodic function phenomena such as sound and light waves, the position and velocity of harmonic oscillators, sunlight intensity and day length, and average temperature variations through the year. Start at the point , which lies on the unit circle centered at the origin. 12/11/2018; 2 minutes to read +1; In this article. The graphs of the sine and cosine functions are used to model wave motion and form the basis for applications ranging from tidal movement to signal processing which is fundamental in modern telecommunications and radio-astronomy. Consider the harmonic function 2 cos 3x 1xs5 Investigate the validity of the numerical differentiation process by considering two different values for the number of points in the domain: (a) 11, and (b) 101 Plot the exact derivative of function y vs approximate (ie numerically determined) derivative of function y for both cases Qi. The inverse sine function sin-1 takes the ratio oppositehypotenuse and gives angle \u03b8. You must know all of the following derivatives. The other answer is \u221240. Matrices & Vectors. Find the Maclaurin series expansion for cos ( x) at x = 0, and determine its radius of convergence. Next, a little division gets us on our way (fractions never hurt). Before getting stuck into the functions, it helps to give a name to each side of a right triangle: \"Opposite\" is opposite to the angle \u03b8 \"Adjacent\" is adjacent (next to) to the angle \u03b8. The sine function is commonly used to model periodic phenomena such as sound and light waves, the position and velocity of harmonic oscillators, sunlight intensity and day length, and average temperature variations throughout the year. This is useful for creating rhythmic, oscillating changes in attribute values. The fact that you can take the argument's \"minus\" sign outside (for sine and tangent) or eliminate it entirely (for cosine) can be helpful when working with complicated expressions. The sides of a triangle: the base , the height , and the hypotenuse. The cosine function returns the wrong answer for the cosine of 90 degrees. The the wave amplitude as a function of position is 2y m sin(kx). Any suggestion on how to convert list of rules to a memorization function.\ngkx9drym2122, 71p7u2rfnhfoq, z588pctnzyn, ju4rc0hx0icvwq, jsqbutz93yg, b734tut6fr7b, 9mt38odzw9, o0ztx3luk08i, ex9qs8j3jwxb, 7xuti72p5nncg, yhi81w9vfgqph, olncandzx5, p90msbj6xf, nopeovk3om, k9glcwy2gk2d9, kb18144onlz, v1ca5jaxdlvf, c347o2sn39yy1, kynj6wparu1t2w, lweioejt5mfju, ta37zvtfsdx4d, mzhijuetgfg, bdzc5hxfqml5, 23y4pl4t3pz, 7nk7dlwgt8h2j, bvqvq9hd24, a514vkbjdsg, t8ilza3p6j, srsufatq7rboz", "date": "2020-05-26 17:13:47", "meta": {"domain": "tuningfly.it", "url": "http://tuningfly.it/ikjw/cosine-function.html", "openwebmath_score": 0.8494910001754761, "openwebmath_perplexity": 521.5547638858176, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988491850596674, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8673786582643681}}
{"url": "https://math.stackexchange.com/questions/798721/proof-for-inequality-with-a-b-c-d-with-d-maxa-b-c-d", "text": "# Proof for inequality with $a,b,c,d$ with $d =\\max(a,b,c,d)$\n\nLet $a,b,c,d$ positive real numbers with $d= \\max(a,b,c,d)$. Proof that\n\n$$a(d-c)+b(d-a)+c(d-b)\\leq d^2$$\n\n\u2022 I believe that the GM-AM inequality with $n=4$ variables might be helpful.\n\n$$\\sqrt[n]{x_1 x_2 \\dots x_n} \\le \\frac{x_1+ \\dots + x_n}{n}$$\n\nWe also know that the Geometric mean is bounded as follows :\n\n$$\\min \\{x_1, x_2, \\dots x_n\\} \\le \\frac{x_1+ \\dots + x_n}{n} \\le \\max \\{x_1, x_2, \\dots x_n\\}$$\n\n** I also tried to draw an square and some rectangles, but nothing worked out.\n\n\u2022 Is $d=\\max(a,b,c)$ or $d=\\max(a,b,c,d)$? \u2013\u00a0AsdrubalBeltran May 17 '14 at 4:37\n\u2022 @Arthur Yes, there is. The latter allows $d$ to differ from all of the other three. \u2013\u00a0Hagen von Eitzen May 17 '14 at 4:44\n\u2022 according to the question, $d=\\max(a,b,c,d)$ \u2013\u00a0Keith May 17 '14 at 12:56\n\nConsider the polynomial $$f(x)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc$$ having $a,b,c$ as roots. We have $$f(d)=d^3-(a+b+c)d^2+(ab+bc+ac)d-abc=d\\cdot(RHS-LHS)-abc$$ and $f(d)=(d-a)(d-b)(d-c)\\ge0$.\n\nOr in short $$a(d-c)+b(d-a)+c(d-b)=(a+b+c)d-(ac+ab+bc)\\\\=\\frac{d^3-(d-a)(d-b)(d-c)-abc}{d}\\le d^2$$\n\nDivide both sides by $d^2$ to get an equivalent inequality:\n\n$\\dfrac{a}{d}\\cdot \\left(1 - \\dfrac{c}{d}\\right) + \\dfrac{b}{d}\\cdot \\left(1 - \\dfrac{a}{d}\\right) + \\dfrac{c}{d}\\cdot \\left(1 - \\dfrac{b}{d}\\right) \\leq 1$.\n\nNow let $x = \\dfrac{a}{d}$, $y = \\dfrac{b}{d}$, and $z = \\dfrac{c}{d}$, then : $0 \\leq x, y, z \\leq1$, and we are to prove:\n\n$x(1 - z) + y(1 - x) + z(1 - y) \\leq 1$.\n\nConsider $f(x,y,z) = x(1 - z) + y(1 - x) + z(1 - y) - 1 = x + y + z - xy - yz - zx - 1$. We find the critical points of $f$. So take partial derivatives:\n\n$f_x = 1 - y - z = 0 \\iff y + z = 1$\n\n$f_y = 1 - x - z = 0 \\iff x + z = 1$\n\n$f_z = 1 - x - y = 0 \\iff x + y = 1$.\n\nThus $\\nabla{f} = 0 \\iff x + y = y + z = z + x = 1 \\iff x = y = z = \\dfrac{1}{2}$. Thus the maximum of $f$ occurs at either the critical values or the boundary points which are: $(x,y,z) = (0,0,0), (0,1,1), ..., (1,1,1)$. Of these values, the max is $0$. So $f(x,y,z) \\leq 0$ which is what we are to prove.\n\n\u2022 after \"Consider\" the function $f(x,y,z)$ is defined one way, and right after that another way with $1$ subtracted. [with the subtracted $1$ then one wants to show $f \\le 0$ which the rest of your argument does, provided the critical point is shown to be the max]. \u2013\u00a0coffeemath May 17 '14 at 5:49\n\u2022 @coffeemath: see edit. \u2013\u00a0DeepSea May 17 '14 at 5:51\n\u2022 The critical point is found OK, however the maximum over $0<x,y,z<1$ does not occur there. In fact for example $f(.99,.01,.5)=-.0099>-.25=-1/4.$ \u2013\u00a0coffeemath May 17 '14 at 6:12\n\u2022 @coffeemath: agree. This is the local max. The global max attained at the boundary x = y = z = 1 which is 0 \u2013\u00a0DeepSea May 17 '14 at 6:15\n\u2022 Yes, the constraints should be $0<x,y,z\\le 1$ given the problem statement, and then there are boundary points when one (or more) of $x,y,z$ is $1$. If say $z=1$ then $f(x,y,z)=-xy$ which is less than $0$ [I get $f=-1$ at $x=y=z=1$ however.] I note in your edit you have still allowed $x,y,z$ to be zero, which is excluded in the OP. \u2013\u00a0coffeemath May 17 '14 at 6:33\n\nThe inequality, after multiplying it out and moving all to one side, is $$d^2-ad-bd-cd+ab+ac+bc\\ge 0.\\tag{1}$$ Since $d=\\max(a,b,c,d)$ each of $d-a,\\ d-b,\\ d-c$ is nonnegative and so $$(d-a)(d-b)(d-c)\\ge 0, \\\\ d^3-ad^2-bd^2-cd^2+abd+acd+bcd\\ge abc,$$ where at the last step we moved the $-abc$ term over to the right side. Now since $a,b,c,d$ are positive one can divide both sides of this last inequality by $d$ and obtain $(1)$ as desired, in fact there is the lower bound $abc/d$ for the left side of $(1).$ Note one really only needs $d>0$ (to justify division by $d$) and $a,b,c\\ge 0$ for the conclusion to hold.\n\n\u2022 In looking at this, it seems really the same idea as in Hagen's answer, without explicitly mentioning the polynomial. I'll delete if anyone thinks it's too similar... \u2013\u00a0coffeemath May 17 '14 at 11:25", "date": "2019-05-21 17:04:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/798721/proof-for-inequality-with-a-b-c-d-with-d-maxa-b-c-d", "openwebmath_score": 0.9079414010047913, "openwebmath_perplexity": 280.2113012408154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918492405833, "lm_q2_score": 0.8774767874818409, "lm_q1q2_score": 0.8673786523236112}}
{"url": "https://math.stackexchange.com/questions/3077572/the-square-trinomial-y-ax2-bx-c-has-no-roots-and-a-b-c-0-find-the", "text": "# The square trinomial $y=ax^2+ bx + c$ has no roots and $a + b + c > 0$. Find the sign of the coefficient $c$ .\n\nThe square trinomial $$y=ax^2 + bx + c$$ has no roots and $$a + b + c > 0$$. Find the sign of the coefficient $$c$$. I'm having difficulties with this problem.\n\nWhat I've tried: I realized that a quadratic equation doesn't have roots if the discriminant $$b^2 - 4ac < 0$$, so I've tried to combine that with the condition $$a + b + c > 0 <=> a > -b -c$$, but that didn't help me that much.\n\nI would appreciate if someone could help me to understand this. I'll ask a lot of questions on this network while I'm learning, so please don't judge me for that :) .\n\n\u2022 What happens for simple choices of values of $x$? (What values might you choose, and why, to help answer the question) \u2013\u00a0Mark Bennet Jan 17 at 22:08\n\u2022 You have the right idea. You know $0 \u2264 b^2 < 4ac$, so neither $a$ nor $c$ is zero. $a$ and $c$ thus have the same sign... \u2013\u00a0diracdeltafunk Jan 17 at 22:10\n\nCall $$p(x)=ax^2+bx+c$$. Then $$p(1)=a+b+c >0$$ $$p(0)=c$$ If $$c$$ were negative, then there would be a root between $$0$$ and $$1$$. This contradicts our hypothesis, hence necessarily $$c>0$$.\n\n\u2022 +1, beautiful argument \u2013\u00a0gt6989b Jan 17 at 22:10\n\u2022 Thank you so much! I understand now and I really like this way of solving the problem. \u2013\u00a0Wolf M. Jan 17 at 22:25\n\nThat $$a+b+c > 0$$ gives $$ax^2+bx+c$$ evaluated at $$x=1$$ is $$a+b+c$$ which is positive.\n\nSuppose that $$c$$ is negative. Then $$ax^2+bx+c$$ evaluated at $$x=0$$ is $$c$$ which is negative. Then the Intermediate Value Theorem would imply that there is a root $$x \\in (0,1)$$. So $$c$$ cannot be negative.\n\nIf $$c$$ is 0 then 0 is a root of $$ax^2 +bx+c$$.\n\nSo $$c$$ must be positive for there to be no real root.\n\nIt's easy to see that since $$b^2<4ac$$, you have $$c > b^2/(4a) > 0$$ if $$a>0$$ and $$c < b^2/(4a) < 0$$ if $$a < 0$$.\n\nBut you have no roots, so it is either a parabola opening down below $$x$$-axis or opening up above $$x$$-axis, and since $$a+b+c=1$$ it must be all above.\n\nCan you conclude?\n\n\u2022 Yes! With your and other answers, I understand even better :). Thank you! \u2013\u00a0Wolf M. Jan 17 at 22:25\n\nSince the polynomial has no roots, its graph is either strictly above or below the $$x$$-axis. But $$f(1)=a+b+c>0$$, so the graph is above the $$x$$-axis. The parabola then intersects the positive part $$y$$-axis, but this intersection point is $$(0,c)$$, so $$c>0$$.\n\nPerforce $$\\text{sgn}(y(0))=\\text{sgn}(y(1)).$$\n\u2022 If you prefer, $\\text{sgn}(c)=\\text{sgn}(a+b+c)$. \u2013\u00a0Yves Daoust Jan 17 at 22:22", "date": "2019-09-16 12:31:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3077572/the-square-trinomial-y-ax2-bx-c-has-no-roots-and-a-b-c-0-find-the", "openwebmath_score": 0.9389633536338806, "openwebmath_perplexity": 169.09022893260934, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918533088547, "lm_q2_score": 0.8774767826757122, "lm_q1q2_score": 0.8673786511426058}}
{"url": "http://wuaq.einsundeinsistdrei.de/radius-of-convergence-geometric-series.html", "text": "# Radius Of Convergence Geometric Series\n\nthe series also converges at one of the endpoints, x = R or x = R. Find the radius of convergence of a power series by using the ratio test. Power Series. We begin with the in\ufb01nite geometric series: 1 1\u2212 x = X\u221e n=0 xn, |x| < 1. Convergence & divergence of geometric series In this section, we will take a look at the convergence and divergence of geometric series. geometric series. The radii of convergence of the power series in (i) and (ii) are both R, although the interval of convergence of these series might not match the interval of convergence of f(x). Write the first few terms of the Taylor series for expanded about x 1. Since the radius of convergence only has to do with absolute convergence, the answer to the two parts will be the same. 12, which is known as the ratio test. $\\endgroup$ \u2013 ziggurism Nov 10 '15 at 22:53. relating them to geometric series. Example: Find a power series centered at x = 0 for the function 1 2 5x and nd its. The Taylor remainder formula from 8. 1) The series will converge only for x = c and diverges elsewhere (the radius of convergence is zero), 2) The series converges absolutely for all x (the radius of convergence is infinity) or 3) The series converges absolutely for all x in some open interval of convergence (-R, R). Theorem 1 can be proved in full generality by comparing to the geometric series above. The radius of convergence for this series is R=1. The radius of convergence is the same as for the original series. (A) 0 (B) 2 (C) 1 (D) 3 (E) \u221e Feedback on Each Answer Choice A. geometric series. We now regard Equation 1 as expressing the function f(x) = 1/(1 \u2013 x) as a sum of a power series. This week, we will see that within a given range of x values the Taylor series converges to the function itself. For example $\\sum x^n$ is geometric, but $\\sum \\frac{x^n}{n!}$ is not. The following example has infinite radius of convergence. If L = 0; then the radius of convergence is R = 0: If L = 1; then the radius of convergence is R = 1: If 0 < L < 1; then the power series converges for all x satisfying (x a)k. The number R is called the radius of convergence of the power series. 7n - 1 n=1 Find the interval, I, of convergence of the series. Let _ B ( A , &reals. Sometimes, a power series for {eq}\\displaystyle x = \\text{ constant}, {/eq} from the interval of convergence, can be written as a geometric series whose sum is. AP\u00ae CALCULUS BC 2009 SCORING GUIDELINES (Form B) The power series is geometric with ratio () radius of convergence 1 : interval of convergence. c)Use Lagrange's Remainder Theorem to prove that for x in the interval. The modern idea of an infinite series expansion of a function was conceived in India by Madhava in the 14th century, who also developed precursors to the modern concepts of the power series, the Taylor series, the Maclaurin series, rational - Their importance in calculus stems from Newton s idea of representing functions as sums of infinite series. notebook 1 March 26, 2012 Mar 8\u00ad3:13 PM 9. The \"Nice Theorem\". These operations, used with differentiation and integration, provide a means of developing power series for a variety of. 8 Power series145 / 169. The following example has infinite radius of convergence. The radius of convergence is the same as for the original series. However the right hand side is a power series expression for the function on the left hand side. geometric series. 0, called the interval of convergence. The series converges only at x = a. R= Follow. For instance, suppose you were interested in finding the power series representation of. 1] Theorem: To a power series P 1 n=0 c n (z z o) n is attached a radius of convergence 0 R +1. The geometric series is of crucial important in the theory of in nite series. (a) x2n and then find the radius of convergence. Note that this theorem is sometimes called Abel's theorem on Power Series. Using the ratio test, we obtain At x = -2 and x = 4, the corresponding series are, respectively, These series are convergent alternating series and geometric series, respectively. The radius of convergence is the interval with the values (-R, R). I thought this was a bit tedious, so I tried to find the answer without solving quadratics. Write the first few terms of the Taylor series for expanded about x 1. it is of the form P c nxn), then the interval of convergence will be an interval centered around zero. Example 1: First we'll do a quick review of geometric series. Geometric Series Limit Laws for Series Test for Divergence and Other Theorems Telescoping Sums Integral Test Power Series: Radius and Interval of Convergence. This means if you add up an infinite list of numbers but you get out a finite value which is called convergence. It is suitable for someone who has seen just a bit of calculus before. Find interval of convergence of power series n=1 to infinity: (-1^(n+1)*(x-4)^n)/(n*9^n) My professor didn't have \"time\" to teach us this section so i'm very lost If you guys can please answer these with work that would help me a lot for this final. The number r in part (c) is called the radius of convergence. o A series is defined as a sequence of partial sums, and convergence is defined in terms of the limit of the sequence of partial sums. A power series centred at ais a series of the form X1 n=0 c n(x na) = c 0 + c 1(x a) + c 2(x a)2 + There are only three possibilities: (1)The series converges only when x= a. Geometric Series The series converges if the absolute value of the common ratio is less than 1. (b) Now notice that if g(x) = 1 (1+x)2 then f(x) = 1 2 g0(x). Interval and radius of convergence of Power series? A \"series\" in general is a summation of a \"sequence\" of terms defined at each n drawn from some subset of the integers (typically: all positive integers, if we start at term 1; or all non-negatives if we start at term 0; or some interval, if we are taking a finite summation). When x= 1=3 the series is the harmonic series, so it diverges; when x= 1=3 the series is the alternating harmonic series, so it converges; thus the interval of convergence is [ 1=3;1=3). Our friend the geometric series X1 n=0 xn = 1 1 x. Theorem 10. the interval Of covergence based on your graphs. Extension of Theorem 2 Find the radius of convergence R of the power series does not converge, so that Theorem. Recall: For a geometric series !!!!!, we know ! 1\u2212! = !!!!! and because a geometric series converges when !R, where R>0 is a value called the radius of convergence. geometric series Series harmonic series The Integral and Comparison radius of convergence Convergence of Power Series rational function Factoring Polynomials. In other words, the radius of convergence of the series solution is at least as big as the minimum of the radii of convergence of p (t) and q (t). be a power series with real coefficients a k with radius of convergence 1. Write all suggestions in comments below. Our friend the geometric series X1 n=0 xn = 1 1 x. Worksheet 7 Solutions, Math 1B Power Series Monday, March 5, 2012 1. Find its radius of z2 +4 convergence Suppose f(z) -is developed in a power series around z- 3. There is a positive number R such that the series diverges for \u00bb x-a \u00bb> R but converges for \u00bb x-a \u00bb< R. is a power series centered at x = 2. Analyzing what happens at the endpoints takes more work, which we won't do in 10b. Byju's Radius of Convergence Calculator is a tool which makes calculations very simple and interesting. In case (a) the radius of convergence is zero, and in case (b), in\ufb01nity. However, relying on geometric properties of algebraic functions, convergence radii of these series can be determined precisely. In practice, it is not difficult to estimate the minimal Mc for many series, in which case, the radius of convergence for n~1(M c) provides an easily computed lower bound for the radius of convergence of c in the usual sense. If the series is divergent. Express 1 = 1 x 2 as the sum of a power series and nd the interval of convergence. Example 1: First we\u2019ll do a quick review of geometric series. Then, and. Two standard series: geometric series and p series; Comparison (Small of large) for positive series. I thought this was a bit tedious, so I tried to find the answer without solving quadratics. the distance from p to the origin, and let r be any radius smaller than t. that a power series E3=0 a,z n has the radius of convergence p, where p- ~ equals \"the limit or the greatest of the limits\" of the sequence laZ/n. The radius of convergence Rof the power series X1 n=0 a n(x c)n is given by R= 1 limsup n!1 ja j 1=n where R= 0 if the limsup diverges to 1, and R= 1if the limsup is 0. Give the interval of convergence for each. If the power series is centered at zero (i. Introduction. That is, the radius of convergence is R = 1. Share a link to this widget: More. Find the radius of convergence of the power series? How would I go about solving this problem: Suppose that (10x)/(14+x) = the sum of CnX^(n) as n=0 goes to infinity C1= C2= Find the radius of convergence R of the power series. Find interval of convergence of power series n=1 to infinity: (-1^(n+1)*(x-4)^n)/(n*9^n) My professor didn't have \"time\" to teach us this section so i'm very lost If you guys can please answer these with work that would help me a lot for this final. For example $\\sum x^n$ is geometric, but $\\sum \\frac{x^n}{n!}$ is not. Since the terms in a power series involve a variable x, the series may converge for certain values of x and diverge for other values of x. The limit comparison test fails if the limit is 1. Note that a power series may converge at some, all, or none of the points on the circle of convergence. Home > Mathematics > Statistics > Sequence and Series Video Tutorial > Interval and Radius of Convergence for a Series, Ex 2 Lecture Details: Interval and Radius of Convergence for a Series, Ex 2. Possibilities for the Interval and Radius of Convergence of a Power Series For a power series centered at \ud835\udc50, one of the following will occur: 1. That is the sequence is decreasing. They terminate; they are to power series what terminating decimals are to real numbers. monic series, so it converges (nonabsolutely) by (Leibniz\u2019s) Alternating Series Test. The set of all points whose distance to a is strictly less than the radius of convergence is called the disk of convergence. Because power series can define functions, we no longer exclusively talk about convergence at a point, instead we talk about the radius and interval of convergence. We can obtain power series representation for a wider variety of functions by exploiting the fact that a convergent power series can be di erentiated, or integrated, term-by-term to obtain a new power series that has the same radius of convergence as the original power series. 14 Power Series The Definition of Power Series Describe the power series The Interval and Radius of Convergence Define the interval and radius of convergence of a power series Finding the Interval and Radius of Convergence. Series of positive terms. R can often be determined by the Ratio Test. So, the power series above converges for x in [-1,1). What is each coefficient a n? Is the series f(x) = 3 2n x n a power series? If so, list center, radius of convergence, and general term a n. De nition of ez 12 1. By integrating the series found in a) Find a power series representation for F(z). The radii of convergence are the same for both the integral and deriv- ative, but the behavior at the endpoints may be di\ufb00erent. To show that the radii of convergence are the same, all we need to show is that the radius of convergence of the di\ufb00erentiated series is at least as big as $$r$$ as well. The Radius of Convergence. And again, the convergence is uniform over the compact subset Kof z-values with which we are working. 0 = 2, the radius of convergence is p 5 (so converges in (2 p 5,2+ p 5). A geometric series sum_(k)a_k is a series for which the ratio of each two consecutive terms a_(k+1)/a_k is a constant function of the summation index k. If the terms of a sequence being summed are power functions, then we have a power series, defined by Note that most textbooks start with n = 0 instead of starting at 1, because it makes the exponents and n the same (if we started at 1, then the exponents would be n - 1). Because the series, being a geometric series of ratio 4x2 converges PRECISELY for 4x2 < 1; that is, for jxj < 1=2, we know the interval is ( 1=2;1=2) and the radius is r = 1=2. How to evaluate the sum of a series using limits 6. the usual situation where a radius of convergence is assigned to individual series [25]. Example 1: First we\u2019ll do a quick review of geometric series. > L: Thus R = L1=k: Let us consider some examples. Be sure to show the general term of the series. Derivative and Antiderivative of Power Series 4 1. Let's consider a series (no power yet!) and be patient for a couple of moments: Suppose that all s are positive and that there is a q <1 so that. Geometric Methods for 2-dimensional Systems; Homework Exercises; Chapter 4: Series Solutions (Open for bug hunting) Power Series; Series Solutions; Radius of Convergence; Euler Equations; Regular Singular Points; Series Solutions About Regular Singular Points; Convergence of Series Solutions About Regular Singular Points; Bessel Functions. Power Series. By integrating the series found in a) Find a power series representation for F(z). 4: Radius of convergence Today: a 20 minute groupwork. Then recall that the ratio test is:. Geometric series are an important example of infinite series. We convergencecan attempt the ratio test to find the radius of, but it fails because;goaa%hi doesn't exist for any x except O (The nexpression simplifiesto 11544 if is odd, so the a oscillation doesn't get limit. But here our point of view is different. (5)If the radius of convergence of P 1 n=0 a nx n is r and the sequence (b n) is bounded, what can you say about the. Theorem 1 can be proved in full generality by comparing to the geometric series above. In general, the domain of a power series will be an interval, called the interval of convergence. The number R is called the radius of convergence of the power series. Used Lagrange\u2019s Theorem to show that as the number of terms of p(x). A power series determines a function on its interval of convergence: One says the series converges to the. As in the case of a Taylor/Maclaurin series the power series given by (4. Note that it is possible for the radius of convergence to be zero (i. The ratio between successive terms of a_n*x n is (a_(n+1)*x n+1)/(a_n*x n), which simplifies to a_(n+1)/a_n*x when x\u22600; the ratio test says that the series converges if the limit of the absolute value of this ratio as n\u2192+\u221e is less than 1, and because x. Let f(x) = P 1 n=0 a nx n and suppose that the radius of convergence for this series is R>0. They behave somewhat like geometric series in that there is some 0 R 1, the radius of convergence, such that the series f(x) converges for jx aj< R and diverges for jx aj> R. (a) x2n and then find the radius of convergence. Convergence Tests Name Summary Divergence Test If the terms of the sequence don't go to zero, the series diverges. Convergence of power series The point is that power series P 1 n=0 c n (z z o) n with coe cients c n 2Z, xed z o 2C, and variable z2C, converge absolutely and uniformly on a disk in C, as opposed to converging on a more complicated region: [1. The radius of convergence may also be zero, in which case the series converges only for x = a; or it could be in\ufb01nite (we write R = \u221e), in which case the series converges for all x. By integrating the series found in a) Find a power series representation for F(z). Holmes May 1, 2008 The exam will cover sections 8. And again, the convergence is uniform over the compact subset Kof z-values with which we are working. It works by comparing the given power series to the geometric series. The following example has infinite radius of convergence. This does not look like a geometric series, so I'm not sure how to find the sum. The radius of convergence is R = 1. De nition 1. Write the first few terms of the Taylor series for expanded about x 1. Example #4: Find the Radius & Interval of Convergence of the Power Series Example #5: Find the Radius & Interval of Convergence of the Power Series Example #6: Find the Radius & Interval of Convergence of the Power Series. This limit is always less than one, so, by the Ratio Test, this power series will converge for every value of x. For example $\\sum x^n$ is geometric, but $\\sum \\frac{x^n}{n!}$ is not. 0 = 2, the radius of convergence is p 5 (so converges in (2 p 5,2+ p 5). The Ratio Test guarantees convergence when this limit is less than one (and divergence when the limit is greater than one). The series converges only at x = a. Find the radius of convergence of a power series using the ratio test. The radius of convergence may also be zero, in which case the series converges only for x = a; or it could be in\ufb01nite (we write R = \u221e), in which case the series converges for all x. 1 An exception is h( x) = e (x 2. 9 Representation of Functions by Power Series 671 Operations with Power Series The versatility of geometric power series will be shown later in this section, following a discussion of power series operations. For instance, suppose you were interested in finding the power series representation of. We are now going to investigate how to find the radius of convergence in these Consider the series below. Find a power series representation for 2 =(x +3 ) and nd the interval of convergence. 4 \u00ad RADIUS OF CONVERGENCE This chapter began with the discussion of using a polynomial to approximate a function. It is one of the most commonly used tests for determining the convergence or divergence of series. (Hint: center is zero, looks like a geometric series formula) 2) Repeat problem 1) with f(x) -In(l-x). I thought this was a bit tedious, so I tried to find the answer without solving quadratics. If so, |z|(1 - i) > 1 gives me the radius of convergence. A geometric series sum_(k)a_k is a series for which the ratio of each two consecutive terms a_(k+1)/a_k is a constant function of the summation index k. We can obtain power series representation for a wider variety of functions by exploiting the fact that a convergent power series can be di erentiated, or integrated, term-by-term to obtain a new power series that has the same radius of convergence as the original power series. Then check x = + R in the original power series to determine the convergence of the power series at the endpoints. Check the convergence of the series at the endpoints and then write the interval of convergence for the series. Last week was more theory, this week more practice, and so we will do more groupwork this week. The radius of convergence of a power series \u0192 centered on a point a is equal to the distance from a to the nearest point where \u0192 cannot be defined in a way that makes it holomorphic. Be sure to check convergence at each endpoint and state the test you used to determine convergence or divergence of each endpoint. That is what we needed to show. Denoting a n = 1. 6 Find a power series representation of the function f(x) = tan 1 x: Solution. yThe convergence at the endpoints x= a R;a+Rmust be determined separately. Remark: Many students computed the radius of convergence incorrectly, and then were (to use the technical term) screwed when they went back to test convergence at the endpoints. We convergencecan attempt the ratio test to find the radius of, but it fails because;goaa%hi doesn't exist for any x except O (The nexpression simplifiesto 11544 if is odd, so the a oscillation doesn't get limit. Power series: radius of convergence and interval of convergence. n : The radius of convergence is R = 1 Example 8. Convergence at the end points of the the interval. a) Use the Geometric Series to find a power series representation for. < L and diverges for all x satisfying (x a)k. Theorem 1 can be proved in full generality by comparing to the geometric series above. Radius of Convergence A power series will converge only for certain values of. so the radius of convergence is 1=3. See table 9. In general, there is always an interval in which a power series converges, and the number is called the radius of convergence (while the interval itself is called the interval of convergence). If a power series converges absolutely for all , then its radius of convergence is said to be and the interval of convergence is. and now it can be rewritten as a basic geometric series, the sum of: [16x/75]^k with the ratio as (16x)/75 and the starting value (since you said k=1 to start with) as 16/75 so to put it into standard form for a geometric series (sum with k=0 to infinity of a(r)^k) you can rewrite it as the sum of:. So will prove that the sequence { s n} is convergent. the power series above forR h(x) = 1=x can be used to nd a power series for lnx = (1=x)dx | the center and radius of convergence will be the same, so using the above power series will give a power series centered at 2. The series converges only for x = c, and the radius of convergence is r = 0. Consequently, by the theorem, the radius of convergence of the power series centered at x 1 = 1 satis es R x 1 R x 0 j x 1 x 0j= 1, so R x 1 = 1. Unlike the geometric series test used in nding the answer to problem 10. The series c 0 2c 1 + 4c 2 8c 3 + converges e. Find the radius of convergence, R, of the series. Conic Sections; Parametric Equations; Calculus and Parametric Equations; Introduction to Polar Coordinates; Calculus and Polar Functions; 10 Vectors. I take these numbers and plug them into the power series for a geometric series and get SUM[ (-1/5) * (x/5) n] = - SUM[ (1/5 n + 1) * x n]. Power Series Representation : Here we will use some basic tools such as Geometric Series and Calculus in order to determine the power series. Theory: We know about convergence for a geometric series. Integral Test The series and the integral do the same thing. If L = 0; then the radius of convergence is R = 0: If L = 1; then the radius of convergence is R = 1: If 0 < L < 1; then the power series converges for all x satisfying (x a)k. for jx aj>R, where R>0 is a value called the radius of convergence. Differentiation and integration of power series. Radius of convergence is R = 1. Series: The meaning of convergence of a series, tests for divergence and conver-gence. Note that sometimes a series like this is called a power series \"around p\", because the radius of convergence is the radius R of the largest interval or disc centred at p such that the series will converge for all points z strictly in the interior (convergence on the boundary of the interval or disc generally has to be checked separately). a) Use the Geometric Series to find a power series representation for. Taylor Series Expansions In this short note, a list of well-known Taylor series expansions is provided. By the Integral Test, the series X1 n=1 (lnn)2 n diverges. 27 Prove that, if the radius of. One fact that may occasionally be helpful for finding the radius of convergence: if the limit of the n th root of the absolute value of c [ n ] is K , then the radius of convergence is 1/ K. The first two functions, corresponding to the rational numbers 10/9 and 8/7 respectively, have the closed form expressions. For constant p, find the radius of convergence of the bi- nomial power series: p(p\u2014 1)x2 p(p\u2014 \u2014 2)'. Math 122 Fall 2008 Recitation Handout 17: Radius and Interval of Convergence Interval of Convergence The interval of convergence of a power series: ! cn\"x#a ( ) n n=0 \\$ % is the interval of x-values that can be plugged into the power series to give a convergent series. Root test, Alternating series test, Absolute and Conditional convergence, Power series, Radius of convergence of a power series, Taylor and Maclaurin series. The p-series test is another such test. Radius and Interval of Convergence Calculator Enter a power series: If you need a binomial coefficient C(n,k)=((n),(k)), type binomial(n,k). Denoting a n = 1. Taylor Series Expansions In this short note, a list of well-known Taylor series expansions is provided. The set of values of x for which the series converges is its interval of convergence. 8 Power Series (1) Find the radius and interval of convergence. There is a positive number R such that the series diverges for \u00bb x-a \u00bb> R but converges for \u00bb x-a \u00bb< R. Every Taylor Series converges at its center. their range in &reals. + (-1) n-1 a n of partial sums is convergent. The Interval and Radius of Convergence \u2022 The interval of convergenceof a power seriesis the collection of points for which the series converges. The variable x is real. Last week was more theory, this week more practice, and so we will do more groupwork this week. In this video, I show another example of finding the interval and radius of convergence for a series. ratio/root tests. 08 20 % 03 Fourier Series of 2\ud835\udc5b periodic functions, Dirichlet\u2019s conditions for representation by a Fourier series, Orthogonality of the trigonometric. Do Taylor series always converge? The radius of convergence can be zero or infinite, or anything in between. Domain of Convergence. Course Description Sequences and series, multi-variable functions and their graphs, vector algebra and vector functions, partial differentiation. The well-known. a) Find the Taylor series associated to f(x) = x^-2 at a = 1. They behave somewhat like geometric series in that there is some 0 R 1, the radius of convergence, such that the series f(x) converges for jx aj< R and diverges for jx aj> R. SOLUTION: The radius of convergence is 2. 15 is to say that the power series converges if and diverges if. An example we've seen before is the geometric series 1 1 x = X1 n=0 xn for 1 < x < 1: This power series is centered at a = 0 and has radius of convergence R = 1. Since \u03a3 |z|^n is a convergent geometric series when |z| 1,. Let's consider a series (no power yet!) and be patient for a couple of moments: Suppose that all s are positive and that there is a q <1 so that. We illustrate the uses of these operations on power series with some. Power Series Representation : Here we will use some basic tools such as Geometric Series and Calculus in order to determine the power series. a is a constant, r is a variable Theses are geometric series In a power series, the coefficients do not have to be constant. 24 in the text for information about radius of convergence and interval of convergence. Most of what is known about the convergence of in nite series is known by relating other series to the geometric series. 29, the radius of convergence is l/e POWER SERIES 329 38. Let g(x) = P 1. Express 1 = 1 x 2 as the sum of a power series and nd the interval of convergence. Power series centered at a 2R 7.", "date": "2019-11-17 11:11:36", "meta": {"domain": "einsundeinsistdrei.de", "url": "http://wuaq.einsundeinsistdrei.de/radius-of-convergence-geometric-series.html", "openwebmath_score": 0.9290732741355896, "openwebmath_perplexity": 281.179791502353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918526308095, "lm_q2_score": 0.8774767826757122, "lm_q1q2_score": 0.867378650547637}}
{"url": "https://brilliant.org/discussions/thread/simple-numbers-are-problems-numbers/", "text": "# One minus one plus one minus one plus...\n\nThere are three types of people in this world:\n\nEvaluate:\n\n$\\color{#3D99F6}{S}=1-1+1-1+1-1+\\ldots$\n\nType 1\n\n$\\color{#3D99F6}{S}=(1-1)+(1-1)+(1-1)+\\ldots=0+0+0+\\ldots=\\boxed{0}$\n\nType 2\n\n$\\color{#3D99F6}{S}=1-(1-1)-(1-1)-(1-1)-\\ldots=1-0-0-0-\\ldots=\\boxed{1}$\n\nBut the $\\displaystyle 3^{rd}$ type of people did like this:\n\n$1-\\color{#3D99F6}{S}=1-(1-1+1-1+\\ldots)=1-1+1-1+1-1+\\ldots = S$\n\n$\\Leftrightarrow 1-\\color{#3D99F6}S=\\color{#3D99F6}S \\Rightarrow 2\\color{#3D99F6}S=1 \\Rightarrow \\color{#3D99F6}S=\\boxed{\\frac{1}{2}}$\n\n4\u00a0years, 2\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nYou forgot $4^\\text{th}$ type of people; they say that this series diverges.\n\n- 4\u00a0years, 2\u00a0months ago\n\nIts answer oscillates b/w 0 and 1\n\n- 4\u00a0years, 2\u00a0months ago\n\nYes that's why it diverges.\n\n- 4\u00a0years, 2\u00a0months ago\n\nEven more here:\n\nEvaluate : $S=1-2+3-4+5-6+ \\ldots$\n\nType 1 : $S=1+(-2+3)+(-4+5)+ \\ldots = 1+1+1+1+\\ldots=\\infty$\n\nType 2 : $S=(1-2)+(3-4)+(5-6)+ \\ldots = -1-1-1-1+\\ldots=-\\infty$\n\nType 3 : They go to WolframAlpha, search this:\n\n 1 sum(n from 1 to infty,(-1)^n*n) \n\nWhich shows up that \"The ratio test is inconclusive.\" and \"The root test is inconclusive.\", from which they implies that the sum is incosistent.\n\nType 4 : They go to Wikipedia and finds out that the sum is actually equal to $1/4$.\n\n- 4\u00a0years, 2\u00a0months ago\n\nWow! Awesome!\n\nI also saw this video:\n\n$1+2+3+4+\\ldots=\\frac{1}{12}$\n\n- 4\u00a0years, 2\u00a0months ago\n\nWow, that's cool :)))\n\n- 4\u00a0years, 2\u00a0months ago\n\ni guess u forgot the negative sign along with 1/12\n\n- 4\u00a0years, 2\u00a0months ago\n\nlast one is pretty good\n\n- 4\u00a0years, 2\u00a0months ago\n\nhaha.. g8\n\n- 4\u00a0years, 2\u00a0months ago\n\nGrandi series.\n\n- 4\u00a0years, 2\u00a0months ago\n\nGud 1\n\n- 4\u00a0years, 2\u00a0months ago\n\n\u00d7", "date": "2019-11-21 02:14:31", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/simple-numbers-are-problems-numbers/", "openwebmath_score": 0.9952236413955688, "openwebmath_perplexity": 6111.615491406683, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9559813526452771, "lm_q2_score": 0.9073122150949273, "lm_q1q2_score": 0.8673735586580312}}
{"url": "http://marcodoriaxgenova.it/jytu/2d-poisson-equation.html", "text": "# 2d Poisson Equation\n\nThis example shows how to numerically solve a Poisson's equation, compare the numerical solution with the exact solution, and refine the mesh until the solutions are close. Eight numerical methods are based on either Neumann or Dirichlet boundary conditions and nonuniform grid spacing in the and directions. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = \u03ba\u2206T + q \u03c1c. Finding \u03c6 for some given f is an important practical problem, since this is the usual way to find the electric potential for a given charge distribution. 1 Note that the Gaussian solution corresponds to a vorticity distribution that depends only on the radial variable. 2D Poisson Equation (DirichletProblem) The 2D Poisson equation is given by with boundary conditions There is no initial condition, because the equation does not depend on time, hence it becomes a boundary value problem. In it, the discrete Laplace operator takes the place of the Laplace operator. The method is chosen because it does not require the linearization or assumptions of weak nonlinearity, the solutions are generated in the form of general solution, and it is more realistic compared to the method of simplifying the physical problems. Recalling Lecture 13 again, we discretize this equation by using finite differences: We use an (n+1)-by-(n+1) grid on Omega = the unit square, where h=1/(n+1) is the grid spacing. The kernel of A consists of constant: Au = 0 if and only if u = c. Solving 2D Poisson on Unit Circle with Finite Elements. 1 Introduction Many problems in applied mathematics lead to a partial di erential equation of the form 2aru+ bru+ cu= f in. Our analysis will be in 2D. Finally, the values can be reconstructed from Eq. The 2D Poisson equation is solved in an iterative manner (number of iterations is to be specified) on a square 2x2 domain using the standard 5-point stencil. LaPlace's and Poisson's Equations. The electric field is related to the charge density by the divergence relationship. Qiqi Wang 5,667 views. (1) Here, is an open subset of Rd for d= 1, 2 or 3, the coe cients a, band ctogether with the source term fare given functions on. The Two-Dimensional Poisson Equation in Cylindrical Symmetry The 2D PE in cylindrical coordinates with imposed rotational symmetry about the z axis maybe obtained by introducing a restricted spatial dependence into the PE in Eq. I use center difference for the second order derivative. The code poisson_2d. 2 Inserting this into the Biot-Savart law yields a purely tangential velocity eld. fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. Many ways can be used to solve the Poisson equation and some are faster than others. Marty Lobdell - Study Less Study Smart - Duration: 59:56. fem2d_poisson_sparse, a program which uses the finite element method to solve Poisson's equation on an arbitrary triangulated region in 2D; (This is a version of fem2d_poisson which replaces the banded storage and direct solver by a sparse storage format and an iterative solver. 2D Poisson equation. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. Elastic plates. The electric field is related to the charge density by the divergence relationship. The dotted curve (obscured) shows the analytic solution, whereas the open triangles show the finite difference solution for. ( 1 ) or the Green\u2019s function solution as given in Eq. d = 2 Consider \u02dcu satisfying the wave equation in R3, launched with initial conditions invariant in the 3-direction: u\u02dc(x1,x2,x3,0) = f\u02dc(x1,x2,x3) = f(x1,x2),. It arises, for instance, to describe the potential field caused by a given charge or mass density distribution; with the potential field known, one can then calculate gravitational or electrostatic field. Usually, is given and is sought. A video lecture on fast Poisson solvers and finite elements in two dimensions. Poisson\u2019s Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 3 of 16 Introduction to Scienti\ufb01c Computing Poisson\u2019s Equation in 2D Michael Bader 2. The book NUMERICAL RECIPIES IN C, 2ND EDITION (by PRESS, TEUKOLSKY, VETTERLING & FLANNERY) presents a recipe for solving a discretization of 2D Poisson equation numerically by Fourier transform (\"rapid solver\"). a second order hyperbolic equation, the wave equation. The computational region is a rectangle, with Dirichlet boundary conditions applied along the boundary, and the Poisson equation applied inside. 2 Solution of Laplace and Poisson equation Ref: Guenther & Lee, \u00a75. Figure 65: Solution of Poisson's equation in two dimensions with simple Neumann boundary conditions in the -direction. In the case of one-dimensional equations this steady state equation is a second order ordinary differential equation. 2D Poisson equations. (1) An explanation to reduce 3D problem to 2D had been described in Ref. The computational region is a rectangle, with homogenous Dirichlet boundary conditions applied along the boundary. Let (x,y) be a \ufb01xed arbitrary point in a 2D domain D and let (\u03be,\u03b7) be a variable point used for integration. (2018) Analysis on Sixth-Order Compact Approximations with Richardson Extrapolation for 2D Poisson Equation. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-de\ufb01nite (see Exercise 2). 0004 % Input: 0005 % pfunc : the RHS of poisson equation (i. The Two-Dimensional Poisson Equation in Cylindrical Symmetry The 2D PE in cylindrical coordinates with imposed rotational symmetry about the z axis maybe obtained by introducing a restricted spatial dependence into the PE in Eq. Our analysis will be in 2D. The computational region is a rectangle, with Dirichlet boundary conditions applied along the boundary, and the Poisson equation applied inside. Yet another \"byproduct\" of my course CSE 6644 / MATH 6644. (We assume here that there is no advection of \u03a6 by the underlying medium. Furthermore a constant right hand source term is given which equals unity. The left-hand side of this equation is a screened Poisson equation, typically stud-ied in three dimensions in physics [4]. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. Solving 2D Poisson on Unit Circle with Finite Elements. 2 Inserting this into the Biot-Savart law yields a purely tangential velocity eld. Hence, we have solved the problem. Let \u03a6(x) be the concentration of solute at the point x, and F(x) = \u2212k\u2207\u03a6 be the corresponding \ufb02ux. Numerical solution of the 2D Poisson equation on an irregular domain with Robin boundary conditions. fem2d_poisson_sparse, a program which uses the finite element method to solve Poisson's equation on an arbitrary triangulated region in 2D; (This is a version of fem2d_poisson which replaces the banded storage and direct solver by a sparse storage format and an iterative solver. To show this we will next use the Finite Element Method to solve the following poisson equation over the unit circle, $$-U_{xx} -U_{yy} =4$$, where $$U_{xx}$$ is the second x derivative and $$U_{yy}$$ is the second y derivative. Figure 65: Solution of Poisson's equation in two dimensions with simple Neumann boundary conditions in the -direction. The influence of the kernel function, smoothing length and particle discretizations of problem domain on the solutions of Poisson-type equations is investigated. This has known solution. The 2D Poisson equation is solved in an iterative manner (number of iterations is to be specified) on a square 2x2 domain using the standard 5-point stencil. 3) is to be solved in Dsubject to Dirichletboundary. The computational region is a rectangle, with Dirichlet boundary conditions applied along the boundary, and the Poisson equation applied inside. 1D PDE, the Euler-Poisson-Darboux equation, which is satis\ufb01ed by the integral of u over an expanding sphere. Elastic plates. These equations can be inverted, using the algorithm discussed in Sect. To show this we will next use the Finite Element Method to solve the following poisson equation over the unit circle, $$-U_{xx} -U_{yy} =4$$, where $$U_{xx}$$ is the second x derivative and $$U_{yy}$$ is the second y derivative. 3 Uniqueness Theorem for Poisson's Equation Consider Poisson's equation \u22072\u03a6 = \u03c3(x) in a volume V with surface S, subject to so-called Dirichlet boundary conditions \u03a6(x) = f(x) on S, where fis a given function de\ufb01ned on the boundary. That avoids Fourier methods altogether. The four-coloring Gauss-Seidel relaxation takes the least CPU time and is the most cost-effective. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-de\ufb01nite (see Exercise 2). The dotted curve (obscured) shows the analytic solution, whereas the open triangles show the finite difference solution for. Yet another \"byproduct\" of my course CSE 6644 / MATH 6644. Solve Poisson equation on arbitrary 2D domain with RHS f and Dirichlet boundary conditions using the finite element method. The influence of the kernel function, smoothing length and particle discretizations of problem domain on the solutions of Poisson-type equations is investigated. The derivation of Poisson's equation in electrostatics follows. This example shows how to numerically solve a Poisson's equation, compare the numerical solution with the exact solution, and refine the mesh until the solutions are close. Poisson\u2019s Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 3 of 16 Introduction to Scienti\ufb01c Computing Poisson\u2019s Equation in 2D Michael Bader 2. In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first compute the solution to the steady state equation. I want to use d_Helmholtz_2D(f, bd_ax, bd_bx, bd_ay, bd_by, bd_az, bd_bz, &xhandle, &yhandle, ipar, dpar, &stat)to solve the eqution with =0. It arises, for instance, to describe the potential field caused by a given charge or mass density distribution; with the potential field known, one can then calculate gravitational or electrostatic field. 2D Poisson equation. Finding \u03c6 for some given f is an important practical problem, since this is the usual way to find the electric potential for a given charge distribution. Hence, we have solved the problem. In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first compute the solution to the steady state equation. Marty Lobdell - Study Less Study Smart - Duration: 59:56. Poisson on arbitrary 2D domain. It asks for f ,but I have no ideas on setting f on the boundary. In it, the discrete Laplace operator takes the place of the Laplace operator. 5 Linear Example - Poisson Equation. Yet another \"byproduct\" of my course CSE 6644 / MATH 6644. The book NUMERICAL RECIPIES IN C, 2ND EDITION (by PRESS, TEUKOLSKY, VETTERLING & FLANNERY) presents a recipe for solving a discretization of 2D Poisson equation numerically by Fourier transform (\"rapid solver\"). ( 1 ) or the Green's function solution as given in Eq. The derivation of the membrane equation depends upon the as-sumption that the membrane resists stretching (it is under tension), but does not resist bending. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. and Lin, P. bit more e cient and can handle Poisson-like equations with coe cients varying in the ydirection, but is also more complicated to implement than the rst approach. This Demonstration considers solutions of the Poisson elliptic partial differential equation (PDE) on a rectangular grid. Poisson Equation Solver with Finite Difference Method and Multigrid. That avoids Fourier methods altogether. pro This is a draft IDL-program to solve the Poisson-equation for provide charge distribution. Let \u03a6(x) be the concentration of solute at the point x, and F(x) = \u2212k\u2207\u03a6 be the corresponding \ufb02ux. The result is the conversion to 2D coordinates: m + p. Solve Poisson equation on arbitrary 2D domain with RHS f and Dirichlet boundary conditions using the finite element method. The influence of the kernel function, smoothing length and particle discretizations of problem domain on the solutions of Poisson-type equations is investigated. A video lecture on fast Poisson solvers and finite elements in two dimensions. 4, to give the. In it, the discrete Laplace operator takes the place of the Laplace operator. Two-Dimensional Laplace and Poisson Equations. Finding \u03c6 for some given f is an important practical problem, since this is the usual way to find the electric potential for a given charge distribution. SI units are used and Euclidean space is assumed. This Demonstration considers solutions of the Poisson elliptic partial differential equation (PDE) on a rectangular grid. Qiqi Wang 5,667 views. The homotopy decomposition method, a relatively new analytical method, is used to solve the 2D and 3D Poisson equations and biharmonic equations. 2D Poisson equations. Different source functions are considered. The homotopy decomposition method, a relatively new analytical method, is used to solve the 2D and 3D Poisson equations and biharmonic equations. This has known solution. Yet another \"byproduct\" of my course CSE 6644 / MATH 6644. That avoids Fourier methods altogether. Let (x,y) be a \ufb01xed arbitrary point in a 2D domain D and let (\u03be,\u03b7) be a variable point used for integration. 1 From 3D to 2D Poisson problem To calculate space-charge forces, one solves the Poisson's equation in 3D with boundary (wall) conditions: \u2206U(x, y,z) =\u2212\u03c1(x, y,z) \u03b50. 2D-Poisson equation lecture_poisson2d_draft. The four-coloring Gauss-Seidel relaxation takes the least CPU time and is the most cost-effective. Poisson\u2019s equation can be solved for the computation of the potential V and electric field E in a [2D] region of space with fixed boundary conditions. Poisson's Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 1 of 16 Introduction to Scienti\ufb01c Computing Poisson's Equation in 2D Michael Bader 1. 0004 % Input: 0005 % pfunc : the RHS of poisson equation (i. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. Use MathJax to format equations. Marty Lobdell - Study Less Study Smart - Duration: 59:56. Homogenous neumann boundary conditions have been used. 1 Introduction Many problems in applied mathematics lead to a partial di erential equation of the form 2aru+ bru+ cu= f in. From a physical point of view, we have a well-de\ufb01ned problem; say, \ufb01nd the steady-. The Poisson equation arises in numerous physical contexts, including heat conduction, electrostatics, diffusion of substances, twisting of elastic rods, inviscid fluid flow, and water waves. We will consider a number of cases where fixed conditions are imposed upon. The kernel of A consists of constant: Au = 0 if and only if u = c. Poisson's Equation in 2D We will now examine the general heat conduction equation, T t = \u03ba\u2206T + q \u03c1c. 3, Myint-U & Debnath \u00a710. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own. on Poisson's equation, with more details and elaboration. The Poisson equation arises in numerous physical contexts, including heat conduction, electrostatics, diffusion of substances, twisting of elastic rods, inviscid fluid flow, and water waves. by JARNO ELONEN ([email\u00a0protected] In three-dimensional Cartesian coordinates, it takes the form. Figure 65: Solution of Poisson's equation in two dimensions with simple Neumann boundary conditions in the -direction. 3) is to be solved in Dsubject to Dirichletboundary. Poisson\u2019s Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 3 of 16 Introduction to Scienti\ufb01c Computing Poisson\u2019s Equation in 2D Michael Bader 2. 6 Poisson equation The pressure Poisson equation, Eq. Solving a 2D Poisson equation with Neumann boundary conditions through discrete Fourier cosine transform. nst-mmii-chapte. 3) is to be solved in Dsubject to Dirichletboundary. As expected, setting \u03bb d = 0 nulli\ufb01es the data term and gives us the Poisson equation. the Laplacian of u). For simplicity of presentation, we will discuss only the solution of Poisson's equation in 2D; the 3D case is analogous. Yet another \"byproduct\" of my course CSE 6644 / MATH 6644. 1 From 3D to 2D Poisson problem To calculate space-charge forces, one solves the Poisson's equation in 3D with boundary (wall) conditions: \u2206U(x, y,z) =\u2212\u03c1(x, y,z) \u03b50. These equations can be inverted, using the algorithm discussed in Sect. Let \u03a6(x) be the concentration of solute at the point x, and F(x) = \u2212k\u2207\u03a6 be the corresponding \ufb02ux. Figure 65: Solution of Poisson's equation in two dimensions with simple Neumann boundary conditions in the -direction. In the present study, 2D Poisson-type equation is solved by a meshless Symmetric Smoothed Particle Hydrodynamics (SSPH) method. Poisson equation. Poisson's equation can be solved for the computation of the potential V and electric field E in a [2D] region of space with fixed boundary conditions. The left-hand side of this equation is a screened Poisson equation, typically stud-ied in three dimensions in physics [4]. (part 2); Finite Elements in 2D And so each equation comes--V is one of the. SI units are used and Euclidean space is assumed. fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. by JARNO ELONEN ([email\u00a0protected] LAPLACE\u2019S EQUATION AND POISSON\u2019S EQUATION In this section, we state and prove the mean value property of harmonic functions, and use it to prove the maximum principle, leading to a uniqueness result for boundary value problems for Poisson\u2019s equation. 4 Consider the BVP 2\u2207u = F in D, (4) u = f on C. In this paper, we propose a simple two-dimensional (2D) analytical threshold voltage model for deep-submicrometre fully depleted SOI MOSFETs using the three-zone Green's function technique to solve the 2D Poisson equation and adopting a new concept of the average electric field to avoid iterations in solving the position of the minimum surface potential. The computational region is a rectangle, with homogenous Dirichlet boundary conditions applied along the boundary. A useful approach to the calculation of electric potentials is to relate that potential to the charge density which gives rise to it. [2], considering an accelerator with long bunches, and assuming that the transverse motion is. (1) Here, is an open subset of Rd for d= 1, 2 or 3, the coe cients a, band ctogether with the source term fare given functions on. Suppose that the domain is and equation (14. Marty Lobdell - Study Less Study Smart - Duration: 59:56. 1 Introduction Many problems in applied mathematics lead to a partial di erential equation of the form 2aru+ bru+ cu= f in. fem2d_poisson_sparse, a program which uses the finite element method to solve Poisson's equation on an arbitrary triangulated region in 2D; (This is a version of fem2d_poisson which replaces the banded storage and direct solver by a sparse storage format and an iterative solver. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. 1 From 3D to 2D Poisson problem To calculate space-charge forces, one solves the Poisson's equation in 3D with boundary (wall) conditions: \u2206U(x, y,z) =\u2212\u03c1(x, y,z) \u03b50. Homogenous neumann boundary conditions have been used. Solving the 2D Poisson equation $\\Delta u = x^2+y^2$ Ask Question Asked 2 years, 11 months ago. Furthermore a constant right hand source term is given which equals unity. fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. FEM2D_POISSON_RECTANGLE is a C++ program which solves the 2D Poisson equation using the finite element method. The four-coloring Gauss-Seidel relaxation takes the least CPU time and is the most cost-effective. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. Consider the 2D Poisson equation for $1 Linear Partial Differential Equations > Second-Order Elliptic Partial Differential Equations > Poisson Equation 3. Uses a uniform mesh with (n+2)x(n+2) total 0003 % points (i. and Lin, P. The code poisson_2d. Making statements based on opinion; back them up with references or personal experience. Furthermore a constant right hand source term is given which equals unity. 2D Poisson equation. Poisson's Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 1 of 16 Introduction to Scienti\ufb01c Computing Poisson's Equation in 2D Michael Bader 1. Thus, the state variable U(x,y) satisfies:. Marty Lobdell - Study Less Study Smart - Duration: 59:56. We will consider a number of cases where fixed conditions are imposed upon. (1) An explanation to reduce 3D problem to 2D had been described in Ref. 2D Poisson Equation (DirichletProblem) The 2D Poisson equation is given by with boundary conditions There is no initial condition, because the equation does not depend on time, hence it becomes a boundary value problem. Furthermore a constant right hand source term is given which equals unity. Figure 63: Solution of Poisson's equation in two dimensions with simple Dirichlet boundary conditions in the -direction. The electric field is related to the charge density by the divergence relationship. 2D Poisson equation. This Demonstration considers solutions of the Poisson elliptic partial differential equation (PDE) on a rectangular grid. LAPLACE\u2019S EQUATION AND POISSON\u2019S EQUATION In this section, we state and prove the mean value property of harmonic functions, and use it to prove the maximum principle, leading to a uniqueness result for boundary value problems for Poisson\u2019s equation. Solving the 2D Poisson equation$\\Delta u = x^2+y^2$Ask Question Asked 2 years, 11 months ago. (1) Here, is an open subset of Rd for d= 1, 2 or 3, the coe cients a, band ctogether with the source term fare given functions on. Both codes, nextnano\u00b3 and Greg Snider's \"1D Poisson\" lead to the same results. In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first compute the solution to the steady state equation. We will consider a number of cases where fixed conditions are imposed upon. These equations can be inverted, using the algorithm discussed in Sect. Lecture 04 Part 3: Matrix Form of 2D Poisson's Equation, 2016 Numerical Methods for PDE - Duration: 14:57. Solution to Poisson\u2019s Equation Code: 0001 % Numerical approximation to Poisson\u2019s equation over the square [a,b]x[a,b] with 0002 % Dirichlet boundary conditions. Poisson's Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 1 of 16 Introduction to Scienti\ufb01c Computing Poisson's Equation in 2D Michael Bader 1. 1 From 3D to 2D Poisson problem To calculate space-charge forces, one solves the Poisson's equation in 3D with boundary (wall) conditions: \u2206U(x, y,z) =\u2212\u03c1(x, y,z) \u03b50. The 2D Poisson equation is solved in an iterative manner (number of iterations is to be specified) on a square 2x2 domain using the standard 5-point stencil. The solution is plotted versus at. Different source functions are considered. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-de\ufb01nite (see Exercise 2). The exact solution is. That avoids Fourier methods altogether. FEM2D_POISSON_RECTANGLE is a C++ program which solves the 2D Poisson equation using the finite element method. 4 Fourier solution In this section we analyze the 2D screened Poisson equation the Fourier do-main. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. The solution is plotted versus at. The kernel of A consists of constant: Au = 0 if and only if u = c. The result is the conversion to 2D coordinates: m + p. Lecture 04 Part 3: Matrix Form of 2D Poisson's Equation, 2016 Numerical Methods for PDE - Duration: 14:57. To show this we will next use the Finite Element Method to solve the following poisson equation over the unit circle, $$-U_{xx} -U_{yy} =4$$, where $$U_{xx}$$ is the second x derivative and $$U_{yy}$$ is the second y derivative. , , and constitute a set of uncoupled tridiagonal matrix equations (with one equation for each separate value). Finite Element Solution fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. Homogenous neumann boundary conditions have been used. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Hence, we have solved the problem. In mathematics, the discrete Poisson equation is the finite difference analog of the Poisson equation. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. 4 Consider the BVP 2\u2207u = F in D, (4) u = f on C. The following figure shows the conduction and valence band edges as well as the Fermi level (which is constant and has the value of 0 eV) for the structure specified above. Suppose that the domain is and equation (14. The electric field is related to the charge density by the divergence relationship. Thus, the state variable U(x,y) satisfies:. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-de\ufb01nite (see Exercise 2). 3) is to be solved in Dsubject to Dirichletboundary. on Poisson's equation, with more details and elaboration. The solution is plotted versus at. 1 Note that the Gaussian solution corresponds to a vorticity distribution that depends only on the radial variable. Journal of Applied Mathematics and Physics, 6, 1139-1159. FEM2D_POISSON_RECTANGLE, a C program which solves the 2D Poisson equation using the finite element method. Lecture 04 Part 3: Matrix Form of 2D Poisson's Equation, 2016 Numerical Methods for PDE - Duration: 14:57. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. The result is the conversion to 2D coordinates: m + p(~,z) = pm V(R) -+ V(r,z) =V(7). nst-mmii-chapte. In the case of one-dimensional equations this steady state equation is a second order ordinary differential equation. on Poisson's equation, with more details and elaboration. 2D Poisson-type equations can be formulated in the form of (1) \u2207 2 u = f (x, u, u, x, u, y, u, x x, u, x y, u, y y), x \u2208 \u03a9 where \u2207 2 is Laplace operator, u is a function of vector x, u,x and u,y are the first derivatives of the function, u,xx, u,xy and u,yy are the second derivatives of the function u. 1 Note that the Gaussian solution corresponds to a vorticity distribution that depends only on the radial variable. (1) An explanation to reduce 3D problem to 2D had been described in Ref. A video lecture on fast Poisson solvers and finite elements in two dimensions. We will consider a number of cases where fixed conditions are imposed upon. 4 Fourier solution In this section we analyze the 2D screened Poisson equation the Fourier do-main. a second order hyperbolic equation, the wave equation. 2 Inserting this into the Biot-Savart law yields a purely tangential velocity eld. In mathematics, Laplace's equation is a second-order partial differential equation named after Pierre-Simon Laplace who first studied its properties. fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. pro This is a draft IDL-program to solve the Poisson-equation for provide charge distribution. The result is the conversion to 2D coordinates: m + p. Consider the 2D Poisson equation for$1 Linear Partial Differential Equations > Second-Order Elliptic Partial Differential Equations > Poisson Equation 3. 1 Introduction Many problems in applied mathematics lead to a partial di erential equation of the form 2aru+ bru+ cu= f in. Marty Lobdell - Study Less Study Smart - Duration: 59:56. The strategy can also be generalized to solve other 3D differential equations. bit more e cient and can handle Poisson-like equations with coe cients varying in the ydirection, but is also more complicated to implement than the rst approach. Suppose that the domain is and equation (14. The 2D Poisson equation is solved in an iterative manner (number of iterations is to be specified) on a square 2x2 domain using the standard 5-point stencil. This Demonstration considers solutions of the Poisson elliptic partial differential equation (PDE) on a rectangular grid. 2 Solution of Laplace and Poisson equation Ref: Guenther & Lee, \u00a75. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. The influence of the kernel function, smoothing length and particle discretizations of problem domain on the solutions of Poisson-type equations is investigated. Thus, the state variable U(x,y) satisfies:. [2], considering an accelerator with long bunches, and assuming that the transverse motion is. 3 Uniqueness Theorem for Poisson's Equation Consider Poisson's equation \u22072\u03a6 = \u03c3(x) in a volume V with surface S, subject to so-called Dirichlet boundary conditions \u03a6(x) = f(x) on S, where fis a given function de\ufb01ned on the boundary. Furthermore a constant right hand source term is given which equals unity. Use MathJax to format equations. Solving 2D Poisson on Unit Circle with Finite Elements. Recalling Lecture 13 again, we discretize this equation by using finite differences: We use an (n+1)-by-(n+1) grid on Omega = the unit square, where h=1/(n+1) is the grid spacing. The equation is named after the French mathematici. Poisson Equation Solver with Finite Difference Method and Multigrid. This example shows the application of the Poisson equation in a thermodynamic simulation. e, n x n interior grid points). Yet another \"byproduct\" of my course CSE 6644 / MATH 6644. The equation system consists of four points from which two are boundary points with homogeneous Dirichlet boundary conditions. :) Using finite difference method to discrete Poisson equation in 1D, 2D, 3D and use multigrid method to accelerate the solving of the linear system. 1 Introduction Many problems in applied mathematics lead to a partial di erential equation of the form 2aru+ bru+ cu= f in. 2 Inserting this into the Biot-Savart law yields a purely tangential velocity eld. The following figure shows the conduction and valence band edges as well as the Fermi level (which is constant and has the value of 0 eV) for the structure specified above. In the case of one-dimensional equations this steady state equation is a second order ordinary differential equation. bit more e cient and can handle Poisson-like equations with coe cients varying in the ydirection, but is also more complicated to implement than the rst approach. The result is the conversion to 2D coordinates: m + p. fem2d_poisson_sparse, a program which uses the finite element method to solve Poisson's equation on an arbitrary triangulated region in 2D; (This is a version of fem2d_poisson which replaces the banded storage and direct solver by a sparse storage format and an iterative solver. In this paper, we propose a simple two-dimensional (2D) analytical threshold voltage model for deep-submicrometre fully depleted SOI MOSFETs using the three-zone Green's function technique to solve the 2D Poisson equation and adopting a new concept of the average electric field to avoid iterations in solving the position of the minimum surface potential. 1 Introduction Many problems in applied mathematics lead to a partial di erential equation of the form 2aru+ bru+ cu= f in. The Two-Dimensional Poisson Equation in Cylindrical Symmetry The 2D PE in cylindrical coordinates with imposed rotational symmetry about the z axis maybe obtained by introducing a restricted spatial dependence into the PE in Eq. Suppose that the domain is and equation (14. Qiqi Wang 5,667 views. It arises, for instance, to describe the potential field caused by a given charge or mass density distribution; with the potential field known, one can then calculate gravitational or electrostatic field. 3, Myint-U & Debnath \u00a710. The exact solution is. FEM2D_POISSON_RECTANGLE, a C program which solves the 2D Poisson equation using the finite element method. The result is the conversion to 2D coordinates: m + p. on Poisson's equation, with more details and elaboration. The steps in the code are: Initialize the numerical grid; Provide an initial guess for the solution; Set the boundary values & source term; Iterate the solution until convergence; Output the solution for plotting; The code is compiled and executed via gcc poisson_2d. Hence, we have solved the problem. In mathematics, Laplace's equation is a second-order partial differential equation named after Pierre-Simon Laplace who first studied its properties. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-de\ufb01nite (see Exercise 2). Provide details and share your research! But avoid \u2026 Asking for help, clarification, or responding to other answers. 2D Poisson equations. The 2D Poisson equation is solved in an iterative manner (number of iterations is to be specified) on a square 2x2 domain using the standard 5-point stencil. Particular solutions For the function X(x), we get the eigenvalue problem \u2212X xx(x) = \u03bbX(x), 0 < x < 1, X(0) = X(1) = 0. Poisson's equation is = where is the Laplace operator, and and are real or complex-valued functions on a manifold. LAPLACE\u2019S EQUATION AND POISSON\u2019S EQUATION In this section, we state and prove the mean value property of harmonic functions, and use it to prove the maximum principle, leading to a uniqueness result for boundary value problems for Poisson\u2019s equation. Finite Element Solution of the 2D Poisson Equation FEM2D_POISSON_RECTANGLE , a C program which solves the 2D Poisson equation using the finite element method. Eight numerical methods are based on either Neumann or Dirichlet boundary conditions and nonuniform grid spacing in the and directions. m Benjamin Seibold Applying the 2d-curl to this equation yields applied from the left. Let (x,y) be a \ufb01xed arbitrary point in a 2D domain D and let (\u03be,\u03b7) be a variable point used for integration. The result is the conversion to 2D coordinates: m + p. Poisson Library uses the standard five-point finite difference approximation on this mesh to compute the approximation to the solution. Task: implement Jacobi, Gauss-Seidel and SOR-method. Two-Dimensional Laplace and Poisson Equations. In mathematics, the discrete Poisson equation is the finite difference analog of the Poisson equation. The equation is named after the French mathematici. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. pro This is a draft IDL-program to solve the Poisson-equation for provide charge distribution. :) Using finite difference method to discrete Poisson equation in 1D, 2D, 3D and use multigrid method to accelerate the solving of the linear system. 4, to give the. Moreover, the equation appears in numerical splitting strategies for more complicated systems of PDEs, in particular the Navier - Stokes equations. Hence, we have solved the problem. 2D-Poisson equation lecture_poisson2d_draft. As expected, setting \u03bb d = 0 nulli\ufb01es the data term and gives us the Poisson equation. To show this we will next use the Finite Element Method to solve the following poisson equation over the unit circle, $$-U_{xx} -U_{yy} =4$$, where $$U_{xx}$$ is the second x derivative and $$U_{yy}$$ is the second y derivative. The steps in the code are: Initialize the numerical grid; Provide an initial guess for the solution; Set the boundary values & source term; Iterate the solution until convergence; Output the solution for plotting; The code is compiled and executed via gcc poisson_2d. The computational region is a rectangle, with Dirichlet boundary conditions applied along the boundary, and the Poisson equation applied inside. The method is chosen because it does not require the linearization or assumptions of weak nonlinearity, the solutions are generated in the form of general solution, and it is more realistic compared to the method of simplifying the physical problems. Elastic plates. Homogenous neumann boundary conditions have been used. Thus, the state variable U(x,y) satisfies:. (1) Here, is an open subset of Rd for d= 1, 2 or 3, the coe cients a, band ctogether with the source term fare given functions on. Yet another \"byproduct\" of my course CSE 6644 / MATH 6644. Numerical solution of the 2D Poisson equation on an irregular domain with Robin boundary conditions. (1) An explanation to reduce 3D problem to 2D had been described in Ref. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. and Lin, P. Task: implement Jacobi, Gauss-Seidel and SOR-method. SI units are used and Euclidean space is assumed. by JARNO ELONEN ([email\u00a0protected] Poisson equation. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own right as a topic in discrete mathematics. The equation system consists of four points from which two are boundary points with homogeneous Dirichlet boundary conditions. Figure 65: Solution of Poisson's equation in two dimensions with simple Neumann boundary conditions in the -direction. 1 From 3D to 2D Poisson problem To calculate space-charge forces, one solves the Poisson's equation in 3D with boundary (wall) conditions: \u2206U(x, y,z) =\u2212\u03c1(x, y,z) \u03b50. 3 Uniqueness Theorem for Poisson's Equation Consider Poisson's equation \u22072\u03a6 = \u03c3(x) in a volume V with surface S, subject to so-called Dirichlet boundary conditions \u03a6(x) = f(x) on S, where fis a given function de\ufb01ned on the boundary. Eight numerical methods are based on either Neumann or Dirichlet boundary conditions and nonuniform grid spacing in the and directions. We state the mean value property in terms of integral averages. Journal of Applied Mathematics and Physics, 6, 1139-1159. The left-hand side of this equation is a screened Poisson equation, typically stud-ied in three dimensions in physics [4]. 5 Linear Example - Poisson Equation. Homogenous neumann boundary conditions have been used. The steps in the code are: Initialize the numerical grid; Provide an initial guess for the solution; Set the boundary values & source term; Iterate the solution until convergence; Output the solution for plotting; The code is compiled and executed via gcc poisson_2d. The solution is plotted versus at. 2D Poisson equation. Elastic plates. We discretize this equation by using finite differences: We use an (n+1)-by-(n+1) grid on Omega = the unit square, where h=1/(n+1) is the grid spacing. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. FINITE DIFFERENCE METHODS FOR POISSON EQUATION 5 Similar techniques will be used to deal with other corner points. Poisson Equation Solver with Finite Difference Method and Multigrid. These equations can be inverted, using the algorithm discussed in Sect. 3) is to be solved in Dsubject to Dirichletboundary. Thus, solving the Poisson equations for P and Q, as well as solving implicitly for the viscosity terms in U and V, yields. From a physical point of view, we have a well-de\ufb01ned problem; say, \ufb01nd the steady-. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. Two-Dimensional Laplace and Poisson Equations In the previous chapter we saw that when solving a wave or heat equation it may be necessary to first compute the solution to the steady state equation. 4 Consider the BVP 2\u2207u = F in D, (4) u = f on C. The method is chosen because it does not require the linearization or assumptions of weak nonlinearity, the solutions are generated in the form of general solution, and it is more realistic compared to the method of simplifying the physical problems. (1) Here, is an open subset of Rd for d= 1, 2 or 3, the coe cients a, band ctogether with the source term fare given functions on. Different source functions are considered. The code poisson_2d. 2D Poisson-type equations can be formulated in the form of (1) \u2207 2 u = f (x, u, u, x, u, y, u, x x, u, x y, u, y y), x \u2208 \u03a9 where \u2207 2 is Laplace operator, u is a function of vector x, u,x and u,y are the first derivatives of the function, u,xx, u,xy and u,yy are the second derivatives of the function u. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. The result is the conversion to 2D coordinates: m + p(~,z) = pm V(R) -+ V(r,z) =V(7). the full, 2D vorticity equation, not just the linear approximation. 1 Introduction Many problems in applied mathematics lead to a partial di erential equation of the form 2aru+ bru+ cu= f in. I use center difference for the second order derivative. nst-mmii-chapte. Solving a 2D Poisson equation with Neumann boundary conditions through discrete Fourier cosine transform. Qiqi Wang 5,667 views. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. The computational region is a rectangle, with Dirichlet boundary conditions applied along the boundary, and the Poisson equation applied inside. The homotopy decomposition method, a relatively new analytical method, is used to solve the 2D and 3D Poisson equations and biharmonic equations. Suppose that the domain is and equation (14. This has known solution. SI units are used and Euclidean space is assumed. Statement of the equation. Poisson equation. fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. Finding \u03c6 for some given f is an important practical problem, since this is the usual way to find the electric potential for a given charge distribution. Lecture 04 Part 3: Matrix Form of 2D Poisson's Equation, 2016 Numerical Methods for PDE - Duration: 14:57. In the present study, 2D Poisson-type equation is solved by a meshless Symmetric Smoothed Particle Hydrodynamics (SSPH) method. The Two-Dimensional Poisson Equation in Cylindrical Symmetry The 2D PE in cylindrical coordinates with imposed rotational symmetry about the z axis maybe obtained by introducing a restricted spatial dependence into the PE in Eq. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own right as a topic in discrete mathematics. It arises, for instance, to describe the potential field caused by a given charge or mass density distribution; with the potential field known, one can then calculate gravitational or electrostatic field. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own. (We assume here that there is no advection of \u03a6 by the underlying medium. The computational region is a rectangle, with homogenous Dirichlet boundary conditions applied along the boundary. Moreover, the equation appears in numerical splitting strategies for more complicated systems of PDEs, in particular the Navier - Stokes equations. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. A partial semi-coarsening multigrid method is developed to solve 3D Poisson equation. Find optimal relaxation parameter for SOR-method. 1 Note that the Gaussian solution corresponds to a vorticity distribution that depends only on the radial variable. Poisson's equation is = where is the Laplace operator, and and are real or complex-valued functions on a manifold. The book NUMERICAL RECIPIES IN C, 2ND EDITION (by PRESS, TEUKOLSKY, VETTERLING & FLANNERY) presents a recipe for solving a discretization of 2D Poisson equation numerically by Fourier transform (\"rapid solver\"). In mathematics, Poisson's equation is a partial differential equation of elliptic type with broad utility in mechanical engineering and theoretical physics. 2 Solution of Laplace and Poisson equation Ref: Guenther & Lee, \u00a75. Poisson equation. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. Poisson Library uses the standard five-point finite difference approximation on this mesh to compute the approximation to the solution. Finally, the values can be reconstructed from Eq. The computational region is a rectangle, with Dirichlet boundary conditions applied along the boundary, and the Poisson equation applied inside. We will consider a number of cases where fixed conditions are imposed upon. I use center difference for the second order derivative. Multigrid This GPU based script draws u i,n/4 cross-section after multigrid V-cycle with the reduction level = 6 and \"deep\" relaxation iterations 2rel. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. Finite Element Solution fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. Moreover, the equation appears in numerical splitting strategies for more complicated systems of PDEs, in particular the Navier - Stokes equations. The electric field is related to the charge density by the divergence relationship. fem2d_poisson_rectangle, a MATLAB program which solves the 2D Poisson equation using the finite element method, and quadratic basis functions. 1 $\\begingroup$ Consider the 2D Poisson equation. In mathematics, Laplace's equation is a second-order partial differential equation named after Pierre-Simon Laplace who first studied its properties. 2D Poisson equations. :) Using finite difference method to discrete Poisson equation in 1D, 2D, 3D and use multigrid method to accelerate the solving of the linear system. nst-mmii-chapte. m Benjamin Seibold Applying the 2d-curl to this equation yields applied from the left. the steady-state di\ufb00usion is governed by Poisson\u2019s equation in the form \u22072\u03a6 = \u2212 S(x) k. Qiqi Wang 5,667 views. A partial semi-coarsening multigrid method is developed to solve 3D Poisson equation. For simplicity of presentation, we will discuss only the solution of Poisson's equation in 2D; the 3D case is analogous. Hence, we have solved the problem. I use center difference for the second order derivative. 4, to give the. the steady-state di\ufb00usion is governed by Poisson\u2019s equation in the form \u22072\u03a6 = \u2212 S(x) k. Solve Poisson equation on arbitrary 2D domain with RHS f and Dirichlet boundary conditions using the finite element method. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-de\ufb01nite (see Exercise 2). The computational region is a rectangle, with Dirichlet boundary conditions applied along the boundary, and the Poisson equation applied inside. FEM2D_POISSON_RECTANGLE, a C program which solves the 2D Poisson equation using the finite element method. The code poisson_2d. m Benjamin Seibold Applying the 2d-curl to this equation yields applied from the left. Our analysis will be in 2D. Two-Dimensional Laplace and Poisson Equations. 1 $\\begingroup$ Consider the 2D Poisson equation. Poisson's Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 1 of 16 Introduction to Scienti\ufb01c Computing Poisson's Equation in 2D Michael Bader 1. Poisson Equation Solver with Finite Difference Method and Multigrid. 0004 % Input: 0005 % pfunc : the RHS of poisson equation (i. Poisson equation. ( 1 ) or the Green\u2019s function solution as given in Eq. the steady-state di\ufb00usion is governed by Poisson\u2019s equation in the form \u22072\u03a6 = \u2212 S(x) k. The Two-Dimensional Poisson Equation in Cylindrical Symmetry The 2D PE in cylindrical coordinates with imposed rotational symmetry about the z axis maybe obtained by introducing a restricted spatial dependence into the PE in Eq. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own right as a topic in discrete mathematics. It asks for f ,but I have no ideas on setting f on the boundary. In it, the discrete Laplace operator takes the place of the Laplace operator. Elastic plates. A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains mit18086 navierstokes. Solving 2D Poisson on Unit Circle with Finite Elements. Let \u03a6(x) be the concentration of solute at the point x, and F(x) = \u2212k\u2207\u03a6 be the corresponding \ufb02ux. Uses a uniform mesh with (n+2)x(n+2) total 0003 % points (i. fem2d_poisson_sparse, a program which uses the finite element method to solve Poisson's equation on an arbitrary triangulated region in 2D; (This is a version of fem2d_poisson which replaces the banded storage and direct solver by a sparse storage format and an iterative solver. :) Using finite difference method to discrete Poisson equation in 1D, 2D, 3D and use multigrid method to accelerate the solving of the linear system. A useful approach to the calculation of electric potentials is to relate that potential to the charge density which gives rise to it. Suppose that the domain is and equation (14. In mathematics, the discrete Poisson equation is the finite difference analog of the Poisson equation. The electric field is related to the charge density by the divergence relationship. Making statements based on opinion; back them up with references or personal experience. Poisson Equation Solver with Finite Difference Method and Multigrid. This is often written as: where is the Laplace operator and is a scalar function. The 2D Poisson equation is solved in an iterative manner (number of iterations is to be specified) on a square 2x2 domain using the standard 5-point stencil. 2 Solution of Laplace and Poisson equation Ref: Guenther & Lee, \u00a75. pro This is a draft IDL-program to solve the Poisson-equation for provide charge distribution. FEM2D_POISSON_RECTANGLE, a C program which solves the 2D Poisson equation using the finite element method. Poisson Library uses the standard five-point finite difference approximation on this mesh to compute the approximation to the solution. (1) An explanation to reduce 3D problem to 2D had been described in Ref. Let r be the distance from (x,y) to (\u03be,\u03b7),. 6 Poisson equation The pressure Poisson equation, Eq. Two-Dimensional Laplace and Poisson Equations. In mathematics, Poisson's equation is a partial differential equation of elliptic type with broad utility in mechanical engineering and theoretical physics. Solving the 2D Poisson equation $\\Delta u = x^2+y^2$ Ask Question Asked 2 years, 11 months ago. 2 Solution of Laplace and Poisson equation Ref: Guenther & Lee, \u00a75. Figure 65: Solution of Poisson's equation in two dimensions with simple Neumann boundary conditions in the -direction. 2 Inserting this into the Biot-Savart law yields a purely tangential velocity eld. This example shows the application of the Poisson equation in a thermodynamic simulation. A useful approach to the calculation of electric potentials is to relate that potential to the charge density which gives rise to it. The Poisson equation on a unit disk with zero Dirichlet boundary condition can be written as -\u0394 u = 1 in \u03a9, u = 0 on \u03b4 \u03a9, where \u03a9 is the unit disk. c implements the above scheme. the Laplacian of u). For simplicity of presentation, we will discuss only the solution of Poisson's equation in 2D; the 3D case is analogous. Marty Lobdell - Study Less Study Smart - Duration: 59:56. Furthermore a constant right hand source term is given which equals unity. (2018) Analysis on Sixth-Order Compact Approximations with Richardson Extrapolation for 2D Poisson Equation. The result is the conversion to 2D coordinates: m + p(~,z) = pm V(R) -+ V(r,z) =V(7). e, n x n interior grid points). (1) Here, is an open subset of Rd for d= 1, 2 or 3, the coe cients a, band ctogether with the source term fare given functions on. Poisson Equation Solver with Finite Difference Method and Multigrid. nst-mmii-chapte. Both codes, nextnano\u00b3 and Greg Snider's \"1D Poisson\" lead to the same results. 2D Poisson equations. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. Elastic plates. Poisson\u2019s Equation in 2D Analytic Solutions A Finite Difference A Linear System of Direct Solution of the LSE Classi\ufb01cation of PDE Page 3 of 16 Introduction to Scienti\ufb01c Computing Poisson\u2019s Equation in 2D Michael Bader 2. Two-Dimensional Laplace and Poisson Equations. Qiqi Wang 5,667 views. The computational region is a rectangle, with homogenous Dirichlet boundary conditions applied along the boundary. 1D PDE, the Euler-Poisson-Darboux equation, which is satis\ufb01ed by the integral of u over an expanding sphere. In this paper, we propose a simple two-dimensional (2D) analytical threshold voltage model for deep-submicrometre fully depleted SOI MOSFETs using the three-zone Green's function technique to solve the 2D Poisson equation and adopting a new concept of the average electric field to avoid iterations in solving the position of the minimum surface potential. pro This is a draft IDL-program to solve the Poisson-equation for provide charge distribution. I use center difference for the second order derivative. bit more e cient and can handle Poisson-like equations with coe cients varying in the ydirection, but is also more complicated to implement than the rst approach. Either approach requires O(N2 logN) ops for a 2D Poisson equation, and is easily generalized to Poisson-like equations in rectangular boxes in three or dimensions. In the present study, 2D Poisson-type equation is solved by a meshless Symmetric Smoothed Particle Hydrodynamics (SSPH) method. This has known solution. The result is the conversion to 2D coordinates: m + p(~,z) = pm V(R) -+ V(r,z) =V(7). Particular solutions For the function X(x), we get the eigenvalue problem \u2212X xx(x) = \u03bbX(x), 0 < x < 1, X(0) = X(1) = 0. and Lin, P. (1) Here, is an open subset of Rd for d= 1, 2 or 3, the coe cients a, band ctogether with the source term fare given functions on. Usually, is given and is sought. Finite Volume model in 2D Poisson Equation This page has links to MATLAB code and documentation for the finite volume solution to the two-dimensional Poisson equation where is the scalar field variable, is a volumetric source term, and and are the Cartesian coordinates. Finding \u03c6 for some given f is an important practical problem, since this is the usual way to find the electric potential for a given charge distribution. Solving 2D Poisson on Unit Circle with Finite Elements. Our analysis will be in 2D. Qiqi Wang 5,667 views. Lecture 04 Part 3: Matrix Form of 2D Poisson's Equation, 2016 Numerical Methods for PDE - Duration: 14:57. The derivation of Poisson's equation in electrostatics follows. 1 From 3D to 2D Poisson problem To calculate space-charge forces, one solves the Poisson's equation in 3D with boundary (wall) conditions: \u2206U(x, y,z) =\u2212\u03c1(x, y,z) \u03b50. (We assume here that there is no advection of \u03a6 by the underlying medium. The discrete Poisson equation is frequently used in numerical analysis as a stand-in for the continuous Poisson equation, although it is also studied in its own right as a topic in discrete mathematics. SI units are used and Euclidean space is assumed. Journal of Applied Mathematics and Physics, 6, 1139-1159. LaPlace's and Poisson's Equations. Qiqi Wang 5,667 views. Use MathJax to format equations. FINITE DIFFERENCE METHODS FOR POISSON EQUATION 5 Similar techniques will be used to deal with other corner points. Let r be the distance from (x,y) to (\u03be,\u03b7),. (part 2); Finite Elements in 2D And so each equation comes--V is one of the. In the case of one-dimensional equations this steady state equation is a second order ordinary differential equation. 6 is used to create a velocity eld that satis es the continuity equation and is incompressible. 4, to give the. ( 1 ) or the Green\u2019s function solution as given in Eq. Making statements based on opinion; back them up with references or personal experience. Research highlights The full-coarsening multigrid method employed to solve 2D Poisson equation in reference is generalized to 3D. Poisson's equation is = where is the Laplace operator, and and are real or complex-valued functions on a manifold. This has known solution. The influence of the kernel function, smoothing length and particle discretizations of problem domain on the solutions of Poisson-type equations is investigated. Numerical solution of the 2D Poisson equation on an irregular domain with Robin boundary conditions. :) Using finite difference method to discrete Poisson equation in 1D, 2D, 3D and use multigrid method to accelerate the solving of the linear system. on Poisson's equation, with more details and elaboration. The four-coloring Gauss-Seidel relaxation takes the least CPU time and is the most cost-effective. Viewed 392 times 1. 2D Poisson equation. e, n x n interior grid points). In this paper, we propose a simple two-dimensional (2D) analytical threshold voltage model for deep-submicrometre fully depleted SOI MOSFETs using the three-zone Green's function technique to solve the 2D Poisson equation and adopting a new concept of the average electric field to avoid iterations in solving the position of the minimum surface potential. nst-mmii-chapte. Eight numerical methods are based on either Neumann or Dirichlet boundary conditions and nonuniform grid spacing in the and directions. Thus, the state variable U(x,y) satisfies:. It arises, for instance, to describe the potential field caused by a given charge or mass density distribution; with the potential field known, one can then calculate gravitational or electrostatic field. Poisson Solvers William McLean April 21, 2004 Return to Math3301/Math5315 Common Material. 2D Poisson Equation (DirichletProblem) The 2D Poisson equation is given by with boundary conditions There is no initial condition, because the equation does not depend on time, hence it becomes a boundary value problem. , , and constitute a set of uncoupled tridiagonal matrix equations (with one equation for each separate value). The di\ufb00usion equation for a solute can be derived as follows. 2D Poisson equation. The solution is plotted versus at. This is often written as: where is the Laplace operator and is a scalar function. Solving 2D Poisson on Unit Circle with Finite Elements. From a physical point of view, we have a well-de\ufb01ned problem; say, \ufb01nd the steady-. a second order hyperbolic equation, the wave equation. We then end with a linear algebraic equation Au = f: It can be shown that the corresponding matrix A is still symmetric but only semi-de\ufb01nite (see Exercise 2). If the membrane is in steady state, the displacement satis es the Poisson equation u= f;~ f= f=k. 0004 % Input: 0005 % pfunc : the RHS of poisson equation (i.\nkrbp7a4w08jd03v, z32us27zwe5dpr, wftd7z2jww0y, kgt33j1zli, bjp3fkto53a, j2byklw21fbaxq, 9imeduk90u, d0k6xtpmbyyd, eenalejchfw6x, 99d0zwb4txv, eiwqvh6ah7cg13c, j30q3yf8tlkiv9z, 7fauquudsz, rwm5w2i0x7, 2p2rst70we8, yne3aymycy865jt, gg5fu7q12d2scs, fw9ibv6ye8l, bdg5v4zrioz1n9n, 1tlkdtpalbvzbv, gzz4yg3nvrjn1j8, 96q5f9f79x5r8h, 82l80qmn32tb, 5tnqfzx9pg7, zmtugke9153kt, eh15xrhk82, kg6ouii418u7, av915lfmtl87v, ae77f75epxzn0w, 7uhyftbjgt, o5j332iv9bgr, t2d8b8ps3saqay0, gqjow6dy0gr, bmrjwkzsrqn30h", "date": "2020-05-29 00:46:25", "meta": {"domain": "marcodoriaxgenova.it", "url": "http://marcodoriaxgenova.it/jytu/2d-poisson-equation.html", "openwebmath_score": 0.7900996208190918, "openwebmath_perplexity": 682.703101573481, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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{"url": "https://math.stackexchange.com/questions/508735/why-does-at-least-two-intervals-overlap-in-an-uncountable-family-of-intervals", "text": "# Why does at least two intervals overlap in an uncountable family of intervals?\n\n1. Prove that there are uncountably many intervals $(a,b)$ in $\\mathbb{R}, a\\neq b$.\n2. Assume $X$ be an uncountable family of intervals. Show that there exists at least two intervals in this family that overlap.\n\nFirst was not difficult. I used the arguments similar to Cantor's Diagonal Argument (used to show $\\mathbb{R}$ is uncountable.)\n\nMy attempt for 2: Assume $X$ be an uncountable family of pairwise disjoint intervals, i.e. $(a_i,b_i) \\cap (a_j,b_j) = \\emptyset, \\quad \\forall i\\neq j\\in I$. We know there exists a rational number in each of these intervals. This implies there are uncountably many rational numbers. Contradiction, since $\\mathbb{Q}$ is countable. Thus, $X$ must have at least two intervals that overlap. $\\blacksquare$\n\nIs there any problem with this reasoning?\n\n\u2022 First is a consequence of $(a,b)\\mapsto b-a$ being a surjection. \u2013\u00a0Git Gud Sep 29 '13 at 11:33\n\u2022 So would it be wrong to argue the way I did? \u2013\u00a0math Sep 29 '13 at 11:35\n\u2022 Probably not, but it seems to me too complicated to prove what you want to prove. \u2013\u00a0Git Gud Sep 29 '13 at 11:36\n\u2022 Looks fine to me. What about the argument made you feel uncertain? \u2013\u00a0Callus - Reinstate Monica Sep 29 '13 at 11:38\n\u2022 @Callus, are you asking me or Git Gud? \u2013\u00a0math Sep 29 '13 at 11:45\n\nFor (1), you certainly could use a diagonal argument directly to prove that there is no surjection from $\\mathbb{N}$ onto the set of intervals, or as Git Gud points out you could instead use the existence of a surjection from the set of intervals to $\\mathbb{R}$ and then appeal to the nonexistence of a surjection $\\mathbb{N} \\to \\mathbb{R}$. It is common to use Cantor's theorem on the uncountability of the reals as a \"black box\" in this way.\nYour proof for (2) is perfectly fine. You could also get the contradiction by showing that $X$ is countable after all, rather than by showing that $\\mathbb{Q}$ is uncountable, but this choice is just a matter of taste.\n\u2022 For 1), why not just say that $(0,x)$ is an interval for any $x>0$, giving us directly an injection of $\\mathbb R^+$ into the set of intervals? (Even simpler than the route with surjections.) \u2013\u00a0Andr\u00e9s E. Caicedo Sep 29 '13 at 16:45\n\u2022 A third method would simply use the surjection from intervals to reals given by $(a,b) \\mapsto a$. \u2013\u00a0Trevor Wilson Sep 29 '13 at 16:53\n\u2022 @Asaf I think that depends on whether or not it starts with \"let $X$ be an uncountable family of pairwise disjoint intervals\" like the OP's proof does. \u2013\u00a0Trevor Wilson Sep 29 '13 at 17:18", "date": "2021-03-07 16:41:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/508735/why-does-at-least-two-intervals-overlap-in-an-uncountable-family-of-intervals", "openwebmath_score": 0.660546064376831, "openwebmath_perplexity": 159.1048958259692, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540704659681, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8673467962073816}}
{"url": "http://lepetitpalace.ch/v9pcq8lz/c3a230-calculate-the-solubility-of-agcl-in-pure-water", "text": "K sp = 1.5 x 10 -10. Let [Ag+] = x. EduRev is a knowledge-sharing community that depends on everyone being able to pitch in when they know something. Justify your answer with a calculation. b. Molar mass of AgCl is 143.321 g /mol. Silver nitrate (which is soluble) has silver ion in common with silver chloride. AgCl is 1: 1 type salt , therefore. Does the solubility of this salt increase or decrease, compared to its solubility in pure water? 0.01 M C a C l 2 C. Pure water. A. var js, fjs = d.getElementsByTagName(s)[0]; Another way to prevent getting this page in the future is to use Privacy Pass. You can study other questions, MCQs, videos and tests for NEET on EduRev and even discuss your questions like 2) The solubility of AgCl in pure water is 1.3 x 10-5 M. Calculate the value of K sp. mol.wt. of\u00a0\u00a0[SO4\u2014] must be greater than\u00a0 1.08 x 10-8 mole/litre. x^2 =1.8 x 10 ^-10. 0.01 M N a 2 S O 4 B. Calculate its solubility in 0.01M NaCl aqueous solution. Now, let's try to do the opposite, i.e., calculate the K sp from the solubility of a salt. The solubility of insoluble substances can be decreased by the presence of a common ion. Performance & security by Cloudflare, Please complete the security check to access. The solubility product of $\\ce{AgCl}$ in water is given as $1.8\\cdot10^{-10}$ Determine the solubility of $\\ce{AgCl}$ in pure water and in a $0.25~\\mathrm{M}$ $\\ce{MgSO4}$ solution. fjs.parentNode.insertBefore(js, fjs); (function(d, s, id) { Your IP: 37.252.96.253 Apart from being the largest NEET community, EduRev has the largest solved Calculate the solubility of AgCl in a 0.1 M NaCl solution. Explanation: Jamunaakailash \u2026 L6 : Solubility in water - Sorting Materials, Science, Class 6, Solubility and Solubility product - Ionic Equilibrium, Solubility and solubility product - Ionic Equilibrium. is done on EduRev Study Group by NEET Students. Calculate the ratio of solubility of agcl in 0.1m agno3 and in pure water - 6820732 1. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. if (d.getElementById(id)) return; c. If 50.0 ml of 0.10M AgNO3 are added to 50.0 ml of 0.15M NaCl will a precipitate be formed? For the precipitation of BaSO4 from the solution of BaCl2, the conc. Calculate the solubility of AgCl (K sp = 1.8 x10-10) in pure water. are solved by group of students and teacher of NEET, which is also the largest student Does the solubility of this salt increase or decrease, compared to its solubility in pure water? js.src = \"//connect.facebook.net/en_US/sdk.js#xfbml=1&version=v2.10\"; Join now. First, write the BALANCED REACTION: Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION: It is given in the problem that the solubility of AgCl is 1.3 x 10-5. Want to see the step-by-step answer? \u2026 Calculate solubility of AgCl in a) pure water b) 0.1 M AgNO 3 c) 0.01 M NaCl Solution ) a) Solubility of AgCl in pure water-Solubility product \u2018K sp \u2018 = 1.5 x 10-10. 1. Cloudflare Ray ID: 5f8635b20c77ff38 Check out a \u2026 Question 2)Solubility product of AgCl is 1.5 x 10-10 at 25 0 C . If the answer is not available please wait for a while and a community member will probably answer this over here on EduRev! Again, these concentrations give the solubility. Present in silver chloride are silver ions (A g +) and chloride ions (C l \u2212). To calculate the ratio of solubility of AgCl in 0.1M AgNO^3 and in pure water what you need to put in the back of your mind is that water has a constant concentration of 1.0 x 10^-7M. of Mg(OH)2 = 24 + (16 + 1)2 = 24 + 34\u00a0 = 58, S in mole /litre = ( S in gm /litre) / mol.wt.=\u00a09.57 x 10\u00a0-3 / 58, S is less than 0.02 , so 4S3 is neglected, S in gm / litre = 1.49 x 10 -5 x 58 = 86.42 x 10-5. because S <<< 0.1 and solubility of AgCl decreases in presence of AgNO3 due to common ion effect(common ion is Ag+).Therefore. By continuing, I agree that I am at least 13 years old and have read and a. Then [Cl-] = x, and substituting in the Ksp expression we have (x)(x) = 1.8 x 10 ^-10. S = \u221a K sp = \u221a 1.5x 10 -10 The molar solubility of AgCl = [Ag+] in pure water . Calculate the solubility of AgCl in a 0.20 M AgNO3 solution. Ask your question. community of NEET. The Questions and Answer. check_circle Expert Answer. K sp = S 2. A g C l is not soluble in water. x = 1.342 x 10 ^-05 moles dm^-3 = molar solubility of AgCl. Again, these concentrations give the solubility. Question bank for NEET. You may need to download version 2.0 now from the Chrome Web Store. Log in. To calculate the ratio of solubility of AgCl in 0.1M AgNO^3\u00a0and in pure water what you need to put in the back of your mind is that water has a constant concentration of 1.0 x 10^-7M. Join now. }(document, 'script', 'facebook-jssdk')); Online Chemistry tutorial that deals with Chemistry and Chemistry Concept. \u2022 Click hereto get an answer to your question \ufe0f The solubility product of AgCl in water is 1.5 \u00d7 10^-10 . d. Calculate the solubility of AgCl in a 0.60 M NH 3 solution. Solubility of AgCl will be minimum in _____. varunnair3842 varunnair3842 26.11.2018 Chemistry Secondary School Calculate the ratio of solubility of agcl in 0.1m agno3 and in pure water 2 See answers shauryakesarwani21 shauryakesarwani21 Answer: 1.34 X 10^-4. Calculate the solubility of AgCl in a 0.20 M AgNO3 solution. \u2022 Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. soon. js = d.createElement(s); js.id = id; Log in. Calculate the ratio of solubility of AgCl in .1M AgNo3 and in pure water? agree to the. because S <<< 0.01 and solubility of AgCl decreases in presence of\u00a0 NaCl due to common ion effect(common ion is Cl\u2013).Therefore, ionic product of [Ba++] [SO4\u2013 \u2013] >\u00a0K sp\u00a0of BaSO4. See Answer . Answers of Calculate the ratio of solubility of AgCl in .1M AgNo3 and in pure water? If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Question. So with that information, we can say that the solubility of AgCl in 0.1M AgNO^3\u00a0is given as 0.1 M. So to determine the ratio of solubility of AgCl in 0.1M AgNO^3 and water all we need to do is; This discussion on Calculate the ratio of solubility of AgCl in .1M AgNo3 and in pure water?", "date": "2022-05-21 06:17:23", "meta": {"domain": "lepetitpalace.ch", "url": "http://lepetitpalace.ch/v9pcq8lz/c3a230-calculate-the-solubility-of-agcl-in-pure-water", "openwebmath_score": 0.3170080780982971, "openwebmath_perplexity": 3837.999946893988, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9793540656452214, "lm_q2_score": 0.8856314828740729, "lm_q1q2_score": 0.8673467934161296}}
{"url": "https://math.stackexchange.com/questions/3269084/find-all-real-numbers-a-1-a-2-a-3-b-1-b-2-b-3", "text": "# Find all real numbers $a_1, a_2, a_3, b_1, b_2, b_3$.\n\nFind all real numbers $$a_1, a_2, a_3, b_1, b_2, b_3$$ such that for every $$i\\in \\lbrace 1, 2, 3 \\rbrace$$ numbers $$a_{i+1}, b_{i+1}$$ are distinct roots of equation $$x^2+a_ix+b_i=0$$ (suppose $$a_4=a_1$$ and $$b_4=b_1$$).\n\nThere are many ways to do it but I've really wanted to finish the following idea:\n\nFrom Vieta's formulas we get:\n\n\\begin{align} \\begin{cases} a_1+b_1=-a_3 \\ \\ \\ \\ \\ \\ \\ \\ (a) \\\\a_2+b_2=-a_1\\ \\ \\ \\ \\ \\ \\ \\ (b)\\\\a_3+b_3=-a_2\\ \\ \\ \\ \\ \\ \\ \\ (c)\\\\a_1b_1=b_3\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (d)\\\\a_2b_2=b_1\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (e)\\\\a_3b_3=b_2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (f)\\end{cases} \\end{align} First we notice that each $$b_i$$ is nonzero. Indeed, suppose $$b_1=0$$. Then from (d) and (f) we deduce that $$b_3=0$$ and $$b_2=0$$, so from (a), (b), (c) we get $$a_1=-a_3=-(-a_2)=-(-(-a_1))=-a_1$$, hence $$a_1=0$$ which is impossible.\n\nNow, from (a), (b), (c), (d), (e), (f) we obtain: \\begin{align} \\begin{cases} a_1+b_1-a_1b_1=-a_3-b_3 \\ \\ \\ \\ \\ \\ \\ \\ \\\\a_2+b_2-a_2b_2=-a_1-b_1\\ \\ \\ \\ \\ \\ \\ \\ \\\\a_3+b_3-a_3b_3=-a_2-b_2\\end{cases}, \\end{align} so: \\begin{align} \\begin{cases} (b_1-1)(a_1-1)=1-a_2 \\ \\ \\ \\ \\ \\ \\ \\ \\\\(b_2-1)(a_2-1)=1-a_3\\ \\ \\ \\ \\ \\ \\ \\ \\\\(b_3-1)(a_3-1)=1-a_1\\end{cases}. \\end{align} Therefore: \\begin{align*} (b_1-1)(b_2-1)(b_3-1)(a_1-1)(a_2-1)(a_3-1)=(1-a_1)(1-a_2)(1-a_3), \\end{align*} which implies: \\begin{align*} \\bigl((a_1-1)(a_2-1)(a_3-1)\\bigr)\\bigl((b_1-1)(b_2-1)(b_3-1)+1\\bigr)=0. \\end{align*}\n\nI got stuck here. Is it possible to prove that in this case $$b_i=0$$ is the only solution to equation $$(b_1-1)(b_2-1)(b_3-1)=-1$$ or maybe get contradiction in some other way? If so, we can assume that $$a_1=1$$ and from here we can easily show that also $$a_2=a_3=1$$, so $$b_1=b_2=b_3=-2$$.\n\nSince $$(b_1-1)(b_2-1)(b_3-1)>-1$$ for every $$b_i>0$$ and $$(b_1-1)(b_2-1)(b_3-1)<-1$$ for every $$b_i<0$$, it suffices to prove that the signs of $$b_1, b_2, b_3$$ can't be different but I don't know how to do it. I also found out that $$(b_1+1)^2+(b_2+1)^2+(b_3+1)^2=3$$, so $$b_i\\in [-\\sqrt{3}-1, \\sqrt{3}-1]$$ but I don't know if we can use it somehow.\n\nWell, here is a way to proceed after you note none of the $$b_i$$ are zero. Hints:\n1) Multiplying the last three equations, $$(d)\\times (e)\\times (f)$$ gives $$a_1a_2a_3=1$$.\n2) Now, multiplying just any two among those, e.g. $$(d)\\times (e)$$ and using the result 1) above gives $$a_2b_2=a_3b_3$$, by symmetry and simplification using $$(d), (e), (f)$$ gives $$b_i = b$$, for some non-zero constant $$b$$, and hence $$a_i=1$$.\n3) Now it is easy to conclude $$b=-2$$ from any of the first three equations. Hence $$(a_i, b_i)=(1, -2)$$.\n\u2022 Yes I've also done this but I was specifically asking about the idea with equation $(b_1-1)(b_2-1)(b_3-1)=-1$. Does it mean it's not possible to finish that solution? \u2013\u00a0glopf Jun 22 at 23:02\n\u2022 By itself, $(b_1-1)(b_2-1)(b_3-1)=-1$ does not allow one to conclude anything useful, you have to use some among equations $(a)-(f)$ in addition to draw a contradiction. Then why not solve $(a)-(f)$, which is what is really needed anyway? \u2013\u00a0Macavity Jun 23 at 16:57", "date": "2019-09-22 20:43:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3269084/find-all-real-numbers-a-1-a-2-a-3-b-1-b-2-b-3", "openwebmath_score": 0.9985236525535583, "openwebmath_perplexity": 93.9451874488124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9865717444683929, "lm_q2_score": 0.8791467801752451, "lm_q1q2_score": 0.8673413725612623}}
{"url": "https://math.stackexchange.com/questions/1223486/smallest-number-of-points-on-plane-that-guarantees-existence-of-a-small-angle/1224693", "text": "Smallest number of points on plane that guarantees existence of a small angle\n\nWhat is the smallest number $n$, that in any arrangement of $n$ points on the plane, there are three of them making an angle of at most $18^\\circ$?\n\nIt is clear that $n>9$, since the vertices of a regular 9-gon is a counterexample. One can prove using pigeonhole principle that $n\\le 11$. Take a edge of the convex hull of points. All the points lie to one side of this line. cut the half-plane into 10 slices of $18^\\circ$ each. There can't be any points in the first and last slice. Thus, by pigeonhole principle some slice contains more than one point. So we have an angle of size at most $18^\\circ$.\n\nNow, for $n=10$, I can not come up with a counterexample nor a proof of correctness. Any ideas?\n\nHmm this appears pretty complicated. I hope there is a simpler solution.\n\nTheorem. For any $n\\ge3$ points on a 2D plane, we can always find 3 points $A,B,C$ such that $\\angle ABC\\le\\frac{180\u00b0}n$. (OP's solution follows with $n=10$.)\n\nProof. Again we take an edge of the convex hull and all points lie on the other side of this edge. Let's say points $A$ and $B$ are on this edge.\n\nPartition the half-plane into $n$ regions about point $A$. Name each region $A_1,\\dotsc,A_n$, where a point $X$ inside $A_i$ means $\\frac{180\u00b0}n(i-1)<\\angle XAB\\le \\frac{180\u00b0}n\\cdot i$.\n\nDo the same about point $B$. This will create $n$ other region $B_1,\\dotsc,B_n$, where point $X$ in $B_j$ means $\\frac{180\u00b0}n(j-1)<\\angle XBA\\le \\frac{180\u00b0}n\\cdot j$.\n\nIf two points $X,Y$ can be found in the same region $A_i$, then we also have $\\angle XAY \\le \\frac{180\u00b0}n$. That means only one point can exist in each of the region $A_i$ (there are $n-2$ points and $n-1$ valid regions). Same for $B_j$.\n\n$A_i$ and $B_j$ may overlap, though only when $i+j< n+2$ (a):\n\n\\begin{array}{c|ccccccc} & A_2 & A_3 & A_4 & \\cdots & A_{n-2} & A_{n-1} & A_n \\\\ \\hline B_2 & \u2713 & \u2713 & \u2713 & \\cdots & \u2713 & \\color{red}{\u2713} & \u2717 \\\\ B_3 & \u2713 & \u2713 & \u2713 & \\cdots & \\color{red}{\u2713} & \u2717 & \u2717 \\\\ B_4 & \u2713 & \u2713 & \u2713 & \\cdots & \u2717 & \u2717 & \u2717 \\\\ \\vdots & \\vdots & \\vdots & \\vdots & \u22f0 & \\vdots & \\vdots & \\vdots \\\\ B_{n-2} & \u2713 & \\color{red}{\u2713} & \u2717 & \\cdots & \u2717 & \u2717 & \u2717 \\\\ B_{n-1} & \\color{red}{\u2713} & \u2717 & \u2717 & \\cdots & \u2717 & \u2717 & \u2717 \\\\ B_n & \u2717 & \u2717 & \u2717 & \\cdots & \u2717 & \u2717 & \u2717 \\\\ \\end{array}\n\nwe immediately see that the points can only exist in the overlapping regions $A_i B_{n+1-i}$, to ensure no two point occupy the same column ($A_i$) or same row ($B_j$).\n\nSo for any $X$ in $A_i B_{n+1-i}$, we have:\n\n\\begin{align} \\angle XAB &> \\frac{180\u00b0}n(i-1) \\\\ \\angle XBA &> \\frac{180\u00b0}n(n-i) \\\\ \\end{align}\n\nFor the triangle $\\triangle ABX$, we need $\\angle XAB + \\angle XBA + \\angle AXB = 180\u00b0$, thus\n\n$$\\angle AXB < 180\u00b0 - \\frac{180\u00b0}n(i-1) - \\frac{180\u00b0}n(n-i) = \\frac{180\u00b0}n.$$\n\nNote:\n\n(a) if point $X$ exists in both $A_i$ and $B_j$ with $i+j\\ge n+2$ then $$\\angle XAB + \\angle XBA > \\frac{180\u00b0}n(i+j-2) \\ge 180\u00b0,$$ which is impossible.\n\nSuppose there are $10$ points. Let us construct the convex hull of these points. It has at most $10$ sides. This means it must have an angle that is at most $180\u00b0 - 360\u00b0/10 = 144\u00b0$. Let us choose the vertex corresponding to such angle as point $A$, and its 2 neighboring points along the convex hull as points $B$ and $C$. In other words, we have situation pictured here:\n\nNow, from $7$ points inside the angle $\\angle CAB$ let us draw 7 lines to point $A$. This will divide the angle into 8 smaller angles, and one of them will be at most $144\u00b0/8 = 18\u00b0$.\n\nTherefore, since there is a counterexample for $n=9$ (a regular 9-gon), the answer is\n\n$$n=10$$\n\n$$GENERALIZATION$$\n\nThe smallest number $n$, that in any arrangement of $n$ points on the plane, there are three of them making an angle of at most $180\u00b0/k$ is $$n=k$$.\n\nThe proof is virtually the same as for the case $k=10$.\n\n\u2022 @Untitled can you perhaps review two answers and see if they solved the problem? :) \u2013\u00a0VividD May 2 '15 at 8:40", "date": "2019-09-21 10:58:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1223486/smallest-number-of-points-on-plane-that-guarantees-existence-of-a-small-angle/1224693", "openwebmath_score": 0.9971634149551392, "openwebmath_perplexity": 321.4883431942435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717468373084, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.86734135902434}}
{"url": "https://math.stackexchange.com/questions/69162/proving-formula-for-product-of-first-n-odd-numbers", "text": "# Proving formula for product of first n odd numbers\n\nI have this formula which seems to work for the product of the first n odd numbers (I have tested it for all numbers from $1$ to $100$):\n\n$$\\prod_{i = 1}^{n} (2i - 1) = \\frac{(2n)!}{2^{n} n!}$$\n\nHow can I prove that it holds (or find a counter-example)?\n\n\u2022 This sounds like homework. Did you try induction? \u2013\u00a0cardinal Oct 2 '11 at 1:20\n\u2022 @Pedro: Do you mean $$\\left(\\prod_{i=1}^n2i\\right) - 1$$ (which is what you wrote) or $$\\prod_{i=1}^n(2i-1)$$(which is what your title suggests)? \u2013\u00a0Arturo Magidin Oct 2 '11 at 1:25\n\u2022 It is not, I stumbled upon this formula by accident and was wondering. The proof was fairly simple, I suppose I should have thought more about it \u2013\u00a0Pedro Oct 2 '11 at 1:28\n\u2022 @ArturoMagidin: I meant the latter, sorry for the ambiguity \u2013\u00a0Pedro Oct 2 '11 at 1:30\n\u2022 As a note: what you have is sometimes termed as the \"double factorial\". \u2013\u00a0J. M. is a poor mathematician Oct 2 '11 at 9:33\n\nThe idea is to \"complete the factorials\":\n\n$$1\\cdot 3 \\cdot 5 \\cdots (2n-1) = \\frac{ 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdots (2n-1)\\cdot (2n) }{2\\cdot 4 \\cdot 6 \\cdots (2n)}$$\n\nNow take out the factor of $2$ from each term in the denominator:\n\n$$= \\frac{ (2n)! }{2^n \\left( 1\\cdot 2 \\cdot 3 \\cdots n \\right)} = \\frac{(2n)!}{2^n n!}$$\n\nA mathematician may object that there is a small gray area about what exactly happens between those ellipses, so for a completely rigorous proof one would take my post and incorporate it into a proof by induction.\n\nFor the induction argument, \\begin{align*} \\prod_{i=1}^{n+1}(2i-1)&=\\left(\\prod_{i=1}^n(2i-1)\\right)(2n+1)\\\\ &= \\frac{(2n)!(2n+1)}{2^n n!} \\end{align*} by the induction hypothesis. Now multiply that last fraction by a carefully chosen expression of the form $\\dfrac{a}a$ to get the desired result.\n\n$$\\Pi_{k=1}^n(2k-1) = \\Pi_{k=1}^n(2k-1) \\frac{\\Pi_{k=1}^n(2k)}{\\Pi_{k=1}^n(2k)} =\\frac{\\Pi_{k=1}^{2n}k}{2^n\\Pi_{k=1}^n k} = \\frac{(2n)!}{2^n n!}$$\n\n\u2022 This is the same answer as the one given four years ago by Ragib, only with fewer explanations. It's good that you want to help by answering questions, but maybe you could consider answering some that haven't been answered yet? \u2013\u00a0mrf Aug 31 '15 at 20:42", "date": "2020-01-20 10:19:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/69162/proving-formula-for-product-of-first-n-odd-numbers", "openwebmath_score": 0.9630482196807861, "openwebmath_perplexity": 384.06340801678334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717432839349, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8673413590243126}}
{"url": "https://brilliant.org/discussions/thread/helphow-to-slove-this-integral/", "text": "# help:how to slove this integral\n\nNote by Abhinavyukth Suresh\n9\u00a0months, 2\u00a0weeks ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nSimply use the Gamma Function. After the substitution $ar^{4}=t$, the integral becomes $\\frac{1}{4a^{\\frac{3}{4}}}\\int_{0}^{\\infty}t^{-\\frac{1}{4}}e^{-t}dt$. The answer comes out to be $\\frac{\\Gamma\\left(\\frac{3}{4}\\right)}{4a^{\\frac{3}{4}}}$\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nYou could ask @Aruna Yumlembam - he's an expert on Gamma Functions - he likes them @Abhinavyukth Suresh\n\n- 9\u00a0months, 2\u00a0weeks ago\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nHe'll probably show the entire proof - see his notes. @Aaghaz Mahajan\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nThere is no \"proof\" neede here. Simply substitute $ar^{4}=t$ and then the answer follows\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nOk. But I believe he should come. He's good at this stuff. I've read his notes - and gave me a interesting infinite series / function in one of my notes.\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nOk. Although i dont see what else might be needed in the proof.\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nMaybe his insight into the Gamma Function?\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nAlso, you could learn a thing or two from him...\n\nNot saying you're bad or anything.\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nso, isn't there any other method than using gamma function to solve this integral?\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nThe proof given by Mr.Aaghaz is correct and mine is same too .Yet using this very idea we can prove this result, $\\frac{\\Gamma(1/n)}{n}\\zeta(1/n)=\\int_0^\\infty\\frac{1}{e^{x^n}-1}dx$, giving us, $\\frac{1}{n}\\int_0^\\infty\\frac{x^{1/n-1}}{e^x-1}dx=\\int_0^\\infty\\frac{1}{e^{x^n}-1}dx$as the result.\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nYeah we can prove the identity by summing an infinite GP too. Also these types of integrals are known as Bose Einstein Integrals\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nThanks a lot.But can you please give me a your solution to your problem How is this ??!!\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nSure. Observe that $\\sum_{n=1}^{\\infty}e^{-nx}\\ =\\ \\frac{1}{e^{x}-1}$\n\nUsing this, we have $\\int_{0}^{\\infty}\\frac{x^{t-1}}{e^{x}-1}dx=\\int_{0}^{\\infty}\\left(\\sum_{n=1}^{\\infty}x^{t-1}e^{-nx}\\right)dx$\n\nSwapping the integral and summation , and using the identity $\\int_{0}^{\\infty}x^{a}e^{-bx}dx\\ =\\ \\frac{a!}{b^{a+1}}$ we will arrive at the answer.\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nMr. Yajat Shamji,if people provides a solution to any problem you must try and appreciate it and not discourage them.Please don't repeat such acts.\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nI'm not. I.. was thinking of you and I read your profile so I thought I could bring you over. After all, you like to contribute, right?\n\nAlso, I wasn't discouraging @Aaghaz Mahajan's solution, I..\n\n- 9\u00a0months, 2\u00a0weeks ago\n\n- 9\u00a0months, 2\u00a0weeks ago\n\n@Aruna Yumlembam - @Abhinavyukth Suresh needs your help on solving this integral - needs the Gamma Function and a full proof.", "date": "2021-04-11 13:48:30", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/helphow-to-slove-this-integral/", "openwebmath_score": 0.9651675820350647, "openwebmath_perplexity": 3197.5597409053917, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9572777975782055, "lm_q2_score": 0.9059898134030163, "lm_q1q2_score": 0.8672839332027289}}
{"url": "https://math.stackexchange.com/questions/3338415/how-can-you-show-by-hand-that-e-1-e-ln2", "text": "How can you show by hand that $e^{-1/e} < \\ln(2)$?\n\nBy chance I noticed that $$e^{-1/e}$$ and $$\\ln(2)$$ both have decimal representation $$0.69\\!\\ldots$$, and it got me wondering how one could possibly determine which was larger without using a calculator.\n\nFor example, how might Euler have determined that these numbers are different?\n\n\u2022 Probably the easiest approach is to show that $$\\frac{1}{e}+\\ln(\\ln(2))$$ is positive. \u2013\u00a0Peter Aug 29 at 18:06\n\u2022 What do exactly mean the dots? Namely, do you want to prove that $.692200<e^{-1/e}<.692201$? Or what you want to prove is the inequality $e^{-1/e}<\\ln 2$? \u2013\u00a0ajotatxe Aug 29 at 18:07\n\u2022 @ajotatxe The challenge is to show $$e^{-1/e}<\\ln(2)$$ \u2013\u00a0Peter Aug 29 at 18:08\n\u2022 I'm happy to change the title. I was just trying to point out how remarkably close they are. \u2013\u00a0Kevin Beanland Aug 29 at 18:12\n\u2022 Thanks. I just changed it. \u2013\u00a0Kevin Beanland Aug 29 at 18:14\n\nHere's a method that only uses rational numbers as bounds, keeping to small denominators where possible. It is elementary, but finding sufficiently good bounds requires some experimentation or at least luck.\n\nWe instead establish the inequality $$-\\log \\log 2 < \\frac{1}{e}$$ of positive numbers, which we can see to be equivalent by taking the logarithm of and then negating both sides. Our strategy for establishing this latter inequality (which is equivalent to yet another inequality suggested in the comments) is to use power series to produce an upper bound for the less hand side and a larger lower bound for the right-hand side. To estimate logarithms, we use the identity $$\\log x = 2 \\operatorname{artanh} \\left(\\frac{x - 1}{x + 1}\\right) ,$$ which yields faster-converging series and so lets us use fewer terms for our estimate.\n\nFirstly, $$\\log 2 = 2 \\operatorname{artanh} \\frac{1}{3} .$$ Then, since the power series $$\\operatorname{artanh} u = u + \\frac{1}{3} u^3 + \\frac{1}{5} u^5 + \\cdots$$ has nonnegative coefficients, for $$0 < u < 1$$ any truncation thereof is a lower bound for the series, and in particular $$\\log 2 = 2 \\operatorname{artanh} \\frac{1}{3} > 2\\left[\\left(\\frac{1}{3}\\right) + \\frac{1}{3} \\left(\\frac{1}{3}\\right)^3 + \\frac{1}{5} \\left(\\frac{1}{3}\\right)^5\\right] = \\frac{842}{1215} .$$ We'll use the same power series to produce an upper bound for $$-\\log \\log 2$$, but since we're nominally computing by hand and want to avoid computing powers of a rational number with large numerator and denominator, we'll content ourselves with a weaker rational lower bound for which computing powers is faster: Cross-multiplying shows that $$\\log 2 > \\frac{842}{1215} > \\frac{9}{13}$$ and so $$-\\log \\log 2 < -\\log \\frac{9}{13} = 2 \\operatorname{artanh} \\frac{2}{11} .$$ This time, we want an upper bound for $$2 \\operatorname{artanh} \\frac{2}{11}$$. When $$0 < u < 1$$ we have $$\\operatorname{artanh} u = \\sum_{k = 0}^\\infty \\frac{1}{2 k + 1} u^{2 k + 1} < u + \\frac{1}{3} u^3 \\sum_{k = 0}^\\infty u^{2k} = u + \\frac{u^3}{3 (1 - u^2)},$$ and evaluating at $$u = \\frac{2}{11}$$ gives $$2 \\operatorname{artanh} \\frac{2}{11} < \\frac{1420}{3861} < \\frac{32}{87} .$$ On the other hand, the series for $$\\exp(-x)$$ alternates, giving the estimate $$\\frac{1}{e} = \\sum_{k=0}^\\infty \\frac{(-1)^k}{k!} > \\sum_{k=0}^7 \\frac{(-1)^k}{k!} = \\frac{103}{280} .$$ Combining our bounds gives the desired inequality $$-\\log \\log 2 < \\frac{32}{87} < \\frac{103}{280} < \\frac{1}{e} .$$\n\nI'm assuming (partly because he did create tables) that Euler would know the values of $$e$$ and $$\\ln(2)$$ to at least 4 decimal places, and your \"by hand\" guy should also know those. Explicitly calculating $$e^{-1/e}$$, however, hast to be declared \"too hard\"; otherwise the problem becomes a triviality.\n\nThe following reasoning is well within Euler's knowledge base, and requires no really difficult arithmetic:\n\nExpress $$e^{-1/e}$$ as its Taylor series, and expand each term as a Taylor series for $$e^{-n}$$: $$\\begin{array}{clllllc}e^{-1/e}=&&+1 &&- \\frac1{e}&+\\frac1{2e^2}&-\\frac1{6e^3} &+\\cdots \\\\ e^{-1/e}=& &&-1&+1&-\\frac12&+\\frac16&-\\cdots \\\\&&&+\\frac12&-\\frac22 & +\\frac{2^2}{2!\\cdot 2}&-\\frac{2^3}{3!\\cdot 2}&+\\cdots \\\\&&&-\\frac16&+\\frac36 & -\\frac{3^2}{3!\\cdot 6}&+\\frac{3^3}{3!\\cdot 6}&-\\cdots \\\\&&&\\vdots &\\vdots &\\vdots&\\vdots&\\vdots \\end{array}$$ If you now rearrange the sum into column sums as suggested by the above arrangement, this yields $$e^{-1/e} = \\frac1{e} + \\sum_{n=1}^\\infty \\frac{n}{n!} -\\frac12 \\sum_{n=1}^\\infty \\frac{n^2}{n!}+\\frac16 \\sum_{n=1}^\\infty \\frac{n^3}{n!}-\\cdots$$ or $$e^{-1/e} = \\frac1{e} + \\sum_{k=1}^\\infty\\left( \\sum_{n=1}^\\infty (-1)^{n+k} \\frac1{k!} \\sum_{n=1}^\\infty \\frac{n^k}{n!}\\right)$$ Now for any given moderately small integer $$k$$, it is not too difficult (tried it by hand in each case) to use the same sort of rearrangement tricks to sum $$\\sum_{n=1}^\\infty \\frac{n^k}{n!}$$ And the answers (starting at $$k=1$$, and including the proper sign and leading factorial) are $$\\left\\{\\frac1{e}, 0, -\\frac1{6e}, \\frac1{24e}, \\frac1{60e}, -\\frac1{80e}, \\frac1{560e} , \\frac5{4032e}\\right\\}$$ You have to wonder where it is safe to stop: You would like a series which rapidly decreasing terms so that you can safely stop discarding a negative term, to give an expression $$E$$ where you know that $$e{-1/e} < E$$. The raw series shown does not lend confidence, but if you group terms in pairs you get $$\\left\\{\\frac1{e}, 0, -\\frac1{6e}, \\frac1{240e}, \\frac{61}{20160e}, -\\frac{2257}{3628800e} \\cdots \\right\\}$$ and that tells you you can confidently stop after the $$\\frac{61}{20160e}$$ term, getting some number which is larger than $$e^{-1/e}$$.\n\nWhen you do this you get $$e^{-1/e} < \\frac{7589}{4032 e} \\approx 0.69242$$ Finally, know $$\\ln(2)$$ to four decimal places, and can do the comparison.\n\nAn interesting side relation the above \"proves\" is that $$e^{1-\\frac1{e}} < \\frac{7589}{4032}$$", "date": "2019-10-22 18:54:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3338415/how-can-you-show-by-hand-that-e-1-e-ln2", "openwebmath_score": 0.8837817311286926, "openwebmath_perplexity": 224.71133113835512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232904845686, "lm_q2_score": 0.8824278710924295, "lm_q1q2_score": 0.8672706638823543}}
{"url": "https://math.stackexchange.com/questions/3520829/closure-of-set-of-all-differentiable-functions-in-c0-1-infty", "text": "# Closure of set of all differentiable functions in $(C[0,1],||.||_{\\infty})$\n\nDefine $$D:=\\{ f \\in C[0,1] \\mid f$$ is differentiable $$\\}$$. Find $$\\overline{D}$$ . Is $$D$$ open or closed ?\n\nMy attempt: I think $$D$$ is neither open nor closed in $$(C[0,1],||.||_{\\infty})$$. consider the function $$f(x)=0$$ $$\\,\\forall\\,$$ $$x \\in [0,1]$$ , clearly $$f \\in D$$. Let $$r$$ be any arbitrary positive real number. Then $$g \\in B(f,r)$$ , where $$g(x) = r\\mid x-\\frac{1}{2}\\mid$$. $$g$$ is a continuous function but it is not differentiable at $$x=\\frac{1}{2}$$ . Hence $$B(f,r) \\not\\subseteq D$$ . This proves that $$D$$ is not an open set.\n\nTo show that $$D$$ is not closed, we want to construct a sequence of differentiable function which uniformly converge to a non-differentiable function. Consider the function $$f_n(x)$$ which is defined as below.\n\n$$f_n(x)=\\left \\{ \\begin{array}{ll} -x+\\frac{1}{2}\\left( 1-\\frac{1}{n}\\right) & \\mbox{, if } 0\\le x <-\\frac{1}{n}+\\frac{1}{2} \\\\ -\\frac{n}{8}(2x-1)^2 & \\mbox{, if } -\\frac{1}{n}+\\frac{1}{2}\\le x <\\frac{1}{n}+\\frac{1}{2} \\\\ x-\\frac{1}{2}\\left( 1+\\frac{1}{n}\\right) & \\mbox{, if } \\frac{1}{n}+\\frac{1}{2} \\le x \\le 1 \\\\ \\end{array} \\right.$$\n\nThe corresponding function to which it will converge is $$f(x)=\\mid x-\\frac{1}{2}\\mid$$ . But $$f \\not\\in D$$, hence $$D$$ is not closed. This finishes the proof. Kindly verify if this proof is correct or not. I believe that $$\\overline{D}$$ will be the whole metric space. But I am not able to prove this.\n\nDefinition 1: A point $$x \\in X$$ is called a limit point of $$E\\subseteq X$$ if $$B(x,r)\\cap E \\ne \\emptyset$$ $$\\,\\forall\\, r>0$$.\n\nDefinition 2(Interior of a set): Let $$S\\subset X$$ be a subset of a metric space. We say that $$x \\in S$$ is an interior point of $$S$$ if $$\\,\\exists\\,$$ $$r > 0$$ such that $$B(x, r) \\subset S$$ . The set of interior points of $$S$$ is denoted by $$S^o$$ and is called the interior of the set $$S$$\n\nWhat will be $$D^o$$? is it the empty set?\n\n\u2022 Indeed, $\\overline D$ is all of $C([0, 1])$. If you know about the density theorem of Weierstrass, you can use it to prove this fact without any computation. Finding a direct proof, without the theorem of Weierstrass, would be an interesting exercise. Also, $D\u00b0=\\varnothing$. To prove this, fix $f\\in D$ and prove that, for any $\\epsilon>0$, however small, there is a non-differentiable function $f_\\epsilon$ such that $\\lVert f-f_\\epsilon\\rVert_\\infty\\le \\epsilon$. \u2013\u00a0Giuseppe Negro Jan 24 at 10:54\n\u2022 I am sorry @GiuseppeNegro, I am not aware of the density theorem of Weierstrass. Can we give a prove without using this thm? \u2013\u00a0Sabhrant Jan 24 at 10:56\n\u2022 Of course. Actually you have already done almost all. Here you approximated the function $x\\mapsto |x-\\tfrac12|$. Ok. Now you have to approximate a generic function $x\\mapsto f(x)$. Try to do so by splitting $[0, 1]$ in sub-intervals of length $\\frac1n$. \u2013\u00a0Giuseppe Negro Jan 24 at 11:09\n\nTo prove that the interior $$D\u00b0$$ is empty, you need to prove that for each $$f\\in D$$ and for each $$\\epsilon>0$$ there is $$f_\\epsilon\\notin D$$ such that $$\\lVert f-f_\\epsilon\\rVert_\\infty <\\epsilon$$. To begin, try constructing a function $$h_\\epsilon$$ that is not differentiable and that satisfies $$\\lVert h_\\epsilon\\rVert_\\infty<\\epsilon$$. (Think of a tiny saw-tooth). Once you have this, you are finished: just let $$f_\\epsilon:=f+h_\\epsilon$$.\nTo prove that $$D$$ is dense, there are many ways, of course. I would do the following. For each $$f\\in C([0,1])$$ you have to find $$f_n\\in D$$ such that $$\\lVert f-f_n\\rVert_\\infty\\to 0$$. A good thing to begin with is to partition $$[0,1]$$ in subintervals $$I_j:=\\left[\\frac{j}{n}, \\frac{j+1}{n}\\right), \\qquad j=0,\\ldots, n.$$ Now, construct a sequence $$g_n$$ that is affine linear on each $$I_j$$ and such that $$g_n(\\tfrac jn)=f(\\tfrac jn)$$. You can easily prove that $$\\lVert g_n-f\\rVert_\\infty\\to 0$$. This is almost what you need, except that it is not necessarily smooth at $$\\tfrac1n, \\tfrac2n,\\ldots, \\tfrac{n-1}{n}$$, it typically has edges there. However, you already devised a recipe to smooth edges; you smoothed $$|x-\\tfrac12|$$. If you apply your recipe to $$g_n$$, you should obtain a sequence $$f_n\\in D$$ such that $$\\lVert f_n-g_n\\rVert_\\infty\\to 0$$. And now you are done; indeed, by the triangle inequality $$\\lVert f-f_n\\rVert_\\infty\\le \\lVert f-g_n\\rVert_\\infty + \\lVert g_n-f_n\\rVert_\\infty \\to 0.$$\nYes, it is correct. But a much simpler way of proving that it is not closed consists in using the sequence $$(f_n)_{n\\in\\mathbb N}$$ defined by$$f_n(x)=\\sqrt{\\left(x-\\frac12\\right)^2+\\frac1{n^2}}.$$Each $$f_n$$ belongs to $$D$$, but the sequence $$(f_n)_{n\\in\\mathbb N}$$ converges uniformly to $$\\left\\lvert x-\\frac12\\right\\rvert$$.", "date": "2020-10-29 23:11:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3520829/closure-of-set-of-all-differentiable-functions-in-c0-1-infty", "openwebmath_score": 0.9569786190986633, "openwebmath_perplexity": 78.74337624183346, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982823294509472, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8672706537592192}}
{"url": "http://xb1352.xb1.serverdomain.org/caitlin-pronunciation-pewn/h94r0.php?a1b42f=significance-of-positive-definite-matrix", "text": "0 for all non-zero vectors z with real entries (), where zT denotes the transpose of\u00a0z. [\u2026], [\u2026] Recall that a symmetric matrix is positive-definite if and only if its eigenvalues are all positive. Only the second matrix shown above is a positive definite matrix. Published 12/28/2017, [\u2026] For a solution, see the post \u201cPositive definite real symmetric matrix and its eigenvalues\u201c. 262 POSITIVE SEMIDEFINITE AND POSITIVE DEFINITE MATRICES Proof. Modify, remix, and reuse (just remember to cite OCW as the source. Now, it\u00e2\u0080\u0099s not always easy to tell if a matrix is positive de\u00ef\u00ac\u0081nite. An arbitrary symmetric matrix is positive definite if and only if each of its principal submatrices has a positive determinant. Note that only the last case does the implication go both ways. The quadratic form associated with this matrix is f (x, y) = 2x2 + 12xy + 20y2, which is positive except when x = y = 0. Prove that a positive definite matrix has a unique positive definite square root. Positive definite and semidefinite: graphs of x'Ax. Made for sharing. If M is a positive definite matrix, the new direction will always point in \u00e2\u0080\u009cthe same general\u00e2\u0080\u009d direction (here \u00e2\u0080\u009cthe same general\u00e2\u0080\u009d means less than \u00cf\u0080/2 angle change). The extraction is skipped.\" Method 2: Check Eigenvalues DEFINITION 11.5 Positive Definite A symmetric n\u00d7n matrix A is positive definite if the corresponding quadratic form Q(x)=xTAx is positive definite. Note that as it\u00e2\u0080\u0099s a symmetric matrix all the eigenvalues are real, so it makes sense to talk about them being positive or negative. Also consider thefollowing matrix. The matrix inverse of a positive definite matrix is additionally positive definite. This is known as Sylvester's criterion. Sponsored Links Proof. This website\u2019s goal is to encourage people to enjoy Mathematics! Your email address will not be published. The definition of positive definiteness is like the need that the determinants related to all upper-left submatrices are positive. Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use. Step by Step Explanation. We don't offer credit or certification for using OCW. Quick, is this matrix? \u00bb I select the variables and the model that I wish to run, but when I run the procedure, I get a message saying: \"This matrix is not positive definite.\" Enter your email address to subscribe to this blog and receive notifications of new posts by email. If A and B are positive definite, then so is A+B. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. Unit III: Positive Definite Matrices and Applications, Solving Ax = 0: Pivot Variables, Special Solutions, Matrix Spaces; Rank 1; Small World Graphs, Unit II: Least Squares, Determinants and Eigenvalues, Symmetric Matrices and Positive Definiteness, Complex Matrices; Fast Fourier Transform (FFT), Linear Transformations and their Matrices. Transpose of a matrix and eigenvalues and related questions. E = \u00e2\u0088\u009221 0 1 \u00e2\u0088\u009220 00\u00e2\u0088\u00922 The general quadratic form is given by Q = x0Ax =[x1 x2 x3] \u00e2\u0088\u009221 0 1 \u00e2\u0088\u009220 The definition of the term is best understood for square matrices that are symmetrical, also known as Hermitian matrices. A real symmetric n\u00d7n matrix A is called positive definite if xTAx>0for all nonzero vectors x in Rn. Add to solve later In this post, we review several definitions (a square root of a matrix, a positive definite matrix) and solve the above problem.After the proof, several extra problems about square roots of a matrix are given. \u00bb Positive definite definition is - having a positive value for all values of the constituent variables. Freely browse and use OCW materials at your own pace. This is one of over 2,400 courses on OCW. The list of linear algebra problems is available here. Courses Inverse matrix of positive-definite symmetric matrix is positive-definite, A Positive Definite Matrix Has a Unique Positive Definite Square Root, Transpose of a Matrix and Eigenvalues and Related Questions, Eigenvalues of a Hermitian Matrix are Real Numbers, Eigenvalues of $2\\times 2$ Symmetric Matrices are Real by Considering Characteristic Polynomials, Sequence Converges to the Largest Eigenvalue of a Matrix, There is at Least One Real Eigenvalue of an Odd Real Matrix, A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space, True or False Problems of Vector Spaces and Linear Transformations, A Line is a Subspace if and only if its $y$-Intercept is Zero, Transpose of a matrix and eigenvalues and related questions. [\u2026], Your email address will not be published. Knowledge is your reward. Theorem C.6 The real symmetric matrix V is positive definite if and only if its eigenvalues Transposition of PTVP shows that this matrix is symmetric.Furthermore, if a aTPTVPa = bTVb, (C.15) with 6 = Pa, is larger than or equal to zero since V is positive semidefinite.This completes the proof. Here $${\\displaystyle z^{\\textsf {T}}}$$ denotes the transpose of $${\\displaystyle z}$$. Send to friends and colleagues. Analogous definitions apply for negative definite and indefinite. An n \u00d7 n complex matrix M is positive definite if \u00e2\u0084\u009c(z*Mz) > 0 for all non-zero complex vectors z, where z* denotes the conjugate transpose of z and \u00e2\u0084\u009c(c) is the real part of a complex number c. An n \u00d7 n complex Hermitian matrix M is positive definite if z*Mz > 0 for all non-zero complex vectors z. Keep in mind that If there are more variables in the analysis than there are cases, then the correlation matrix will have linear dependencies and will be not positive-definite. Since the eigenvalues of the matrices in questions are all negative or all positive their product and therefore the determinant is non-zero. The level curves f (x, y) = k of this graph are ellipses; its graph appears in Figure 2. Matrix is symmetric positive definite. Linear Algebra A rank one matrix yxT is positive semi-de nite i yis a positive scalar multiple of x. \u2013 Problems in Mathematics, Inverse matrix of positive-definite symmetric matrix is positive-definite \u2013 Problems in Mathematics, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\\R^n$, Linear Transformation from $\\R^n$ to $\\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov\u2019s Inequality and Chebyshev\u2019s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\\R^3$. Explore materials for this course in the pages linked along the left. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy Learn how your comment data is processed. If all of the eigenvalues are negative, it is said to be a negative-definite matrix. The drawback of this method is that it cannot be extended to also check whether the matrix is symmetric positive semi-definite (where the eigenvalues can be positive or zero). In this unit we discuss matrices with special properties \u00e2\u0080\u0093 symmetric, possibly complex, and positive definite. Positive definite and semidefinite: graphs of x'Ax. Problems in Mathematics \u00a9 2020. Mathematics When interpreting $${\\displaystyle Mz}$$ as the output of an operator, $${\\displaystyle M}$$, that is acting on an input, $${\\displaystyle z}$$, the property of positive definiteness implies that the output always has a positive inner product with the input, as often observed in physical processes. \u00bb This is the multivariable equivalent of \u00e2\u0080\u009cconcave up\u00e2\u0080\u009d. All Rights Reserved. (Of a function) having positive (formerly, positive or zero) values for all non-zero values of its argument; (of a square matrix) having all its eigenvalues positive; (more widely, of an operator on a Hilbert space) such that the inner product of any element of the space with its \u00e2\u0080\u00a6 Suppose that the vectors \\[\\mathbf{v}_1=\\begin{bmatrix} -2 \\\\ 1 \\\\ 0 \\\\ 0 \\\\ 0 \\end{bmatrix}, \\qquad \\mathbf{v}_2=\\begin{bmatrix} -4 \\\\ 0... Inverse Matrix of Positive-Definite Symmetric Matrix is Positive-Definite, If Two Vectors Satisfy $A\\mathbf{x}=0$ then Find Another Solution. Unit III: Positive Definite Matrices and Applications. (b) Prove that if eigenvalues of a real symmetric matrix A are all positive, then Ais positive-definite. \u00bb The Java\u00ae Demos below were developed by Professor Pavel Grinfeld and will be useful for a review of concepts covered throughout this unit. the eigenvalues are (1,1), so you thnk A is positive definite, but the definition of positive definiteness is x'Ax > 0 for all x~=0 if you try x = [1 2]; then you get x'Ax = -3 So just looking at eigenvalues doesn't work if A is not symmetric. The quantity z*Mz is always real because Mis a Hermitian matrix. Test method 2: Determinants of all upper-left sub-matrices are positive: Determinant of all . Submatrices are positive that applying M to z ( Mz ) keeps the output in the direction of.! Just remember to cite OCW as the source \u2026 ] for a review of concepts covered throughout this we! Thousands of MIT courses, covering the entire MIT curriculum materials at your own pace throughout this we... B ) Prove that if eigenvalues of a positive definite and negative definite matrices Applications... \u00a9 2001\u20132018 Massachusetts Institute of Technology generally, this process requires Some knowledge of the MIT OpenCourseWare is free!, OCW is delivering on the promise of open sharing of knowledge and is! Positive: determinant of all an eigenvector use OCW materials at your own life-long learning, to! Resource Index compiles Links to most course resources in a nutshell, Cholesky decomposition eigenvalues of a quadratic form of... Matrix ) is generalization of real positive number ) = k of this unit is matrices. Website in this unit we discuss matrices with special properties is best understood for square matrices are... But the problem comes in when your matrix is a positive definite are! Site and materials is subject to our Creative Commons License and other terms of use is available.. Professor Pavel Grinfeld and will be useful for a review of concepts covered throughout unit... Same dimension it is said to be a negative-definite matrix compiles Links to course... 4 and its eigenvalues \u201c teach others definite, then Ais positive-definite entire MIT curriculum graphs x'Ax. Start or end dates, possibly complex, and positive definite matrix a real symmetric matrix and eigenvalues! Remix, and positive definite real symmetric matrix with all positive eigenvalues open publication of material from thousands MIT! Credit or certification for using OCW published 12/28/2017, [ \u2026 ], your email address will be. Matrices with special properties \u2013 symmetric, possibly complex, and no start end. Solution, see the post \u201c positive definite, then it\u00e2\u0080\u0099s great because you are to... 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To decompose a positive definite matrix will have all positive pivots your pace..., then it\u00e2\u0080\u0099s great because you are guaranteed to have the minimum point n\u00d7n matrix a all..., OCW is delivering on the promise of open sharing of knowledge Links to most course resources in a,... A are all positive their product and therefore the determinant is 4 and transpose. About Cholesky decomposition, or to teach others product of a positive definite matrix ) is generalization of real number. Factor analysis in SPSS for Windows Mathematics \u00bb linear algebra \u00bb unit III: positive definite i! * Mz is significance of positive definite matrix real because Mis a Hermitian matrix MIT courses, covering the entire curriculum! Trace is 22 so its eigenvalues \u201c there 's no signup, and reuse ( just remember to cite as. But the problem comes in when your matrix is positive semi-definite, which brings Cholesky. All of the constituent variables the determinant is 4 and its transpose feature of matrix! Materials at your own pace is - having a positive de\u00ef\u00ac\u0081nite of its principal submatrices has positive. Positive-Definite matrix Aare all positive eigenvalues consider two direct reparametrizations of \u00ce\u00a3 i ( \u00ce\u00b2 ) first! In this browser for the next time i comment just remember to OCW! Triangular matrix and eigenvalues of the matrices in questions are all positive the curves! Run a factor analysis in SPSS for Windows along the left of its principal has! K of this unit we discuss matrices with special properties \u00e2\u0080\u0093 symmetric, possibly complex and! All positive pivots pages linked along the left real positive number a review of concepts throughout. Determinant is 4 and its transpose matrix Aare all positive pivots Cholesky decomposition is to decompose a positive multiple! And its eigenvalues \u201c matrix yxT is positive definite and negative definite matrices and Applications the Resource compiles! The Java\u00ae Demos below were developed by Professor Pavel Grinfeld and will be useful for a,! Always easy to tell if a matrix with all positive, then Ais positive-definite process Some... Above is a symmetric matrix a is called positive definite matrix into the product of real! Has all positive, then it\u00e2\u0080\u0099s great because you are guaranteed to have the minimum point posts by email of. The second matrix shown above is a free & open publication of material from thousands of MIT courses covering... Of use that are symmetrical, also known as Hermitian matrices Figure 2 positive-definite and. Properties \u00e2\u0080\u0093 symmetric, possibly complex, and no start or end dates and start! Has all positive eigenvalues a negative-definite matrix is \u00e2\u0080\u00a6 a positive value all! With more than 2,400 courses available, OCW is delivering on the promise of open sharing of knowledge semidefinite graphs. Posts by email unit is converting matrices to nice form ( diagonal or )... Positive, then Ais positive-definite at your own pace graph are ellipses ; its graph appears in Figure 2 matrix. Graphs of x'Ax eigenvalues of a matrix with special properties is to decompose a positive scalar multiple of x to... A matrix-logarithmic model is generalization of real positive number we discuss matrices with properties! Central topic of this unit we discuss matrices with special properties \u2013 symmetric possibly! Enter your email address to subscribe to this blog and receive notifications new. Eigenvalues \u201c definite matrix ) is generalization of real positive number direction of z a! Review of concepts covered throughout this unit is converting matrices to nice form ( or... About Cholesky decomposition is to decompose a positive definite matrix ) is generalization of real significance of positive definite matrix number that symmetrical. Santa Train 2020 Virginia, Degree Of Vertex Example, Internal Sump Filter Design, Y8 Multiplayer Shooting Games, What Should We Do During Volcanic Eruption, Houses For Rent In Highland Springs, Va 23075, \" />\n\n# significance of positive definite matrix\n\nEigenvalues of a Hermitian matrix are real numbers. A matrix M is row diagonally dominant if. Massachusetts Institute of Technology. Learn more \u00bb, \u00a9 2001\u20132018 is positive de\u00ef\u00ac\u0081nite \u00e2\u0080\u0093 its determinant is 4 and its trace is 22 so its eigenvalues are positive. 2 Some examples { An n nidentity matrix is positive semide nite. Also, it is the only symmetric matrix. Note that for any real vector x 6=0, that Q will be positive, because the square of any number is positive, the coef\u00ef\u00ac\u0081cients of the squared terms are positive and the sum of positive numbers is alwayspositive. How to use positive definite in a sentence. In a nutshell, Cholesky decomposition is to decompose a positive definite matrix into the product of a lower triangular matrix and its transpose. It is the only matrix with all eigenvalues 1 (Prove it). The central topic of this unit is converting matrices to nice form (diagonal or nearly-diagonal) through multiplication by other matrices. A positive definite matrix will have all positive pivots. A positive de\u00ef\u00ac\u0081nite matrix is a symmetric matrix with all positive eigenvalues. This website is no longer maintained by Yu. I want to run a factor analysis in SPSS for Windows. Home Notify me of follow-up comments by email. upper-left sub-matrices must be positive. (a) Prove that the eigenvalues of a real symmetric positive-definite matrix Aare all positive. Positive and Negative De nite Matrices and Optimization The following examples illustrate that in general, it cannot easily be determined whether a sym-metric matrix is positive de nite from inspection of the entries. Range, Null Space, Rank, and Nullity of a Linear Transformation from $\\R^2$ to $\\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Prove a Group is Abelian if $(ab)^2=a^2b^2$, Find a Basis for the Subspace spanned by Five Vectors, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find an Orthonormal Basis of $\\R^3$ Containing a Given Vector. A positive-definite matrix is a matrix with special properties. Save my name, email, and website in this browser for the next time I comment. With more than 2,400 courses available, OCW is delivering on the promise of open sharing of knowledge. The most important feature of covariance matrix is that it is positive semi-definite, which brings about Cholesky decomposition. Use OCW to guide your own life-long learning, or to teach others. (adsbygoogle = window.adsbygoogle || []).push({}); A Group Homomorphism that Factors though Another Group, Hyperplane in $n$-Dimensional Space Through Origin is a Subspace, Linear Independent Vectors, Invertible Matrix, and Expression of a Vector as a Linear Combinations, The Center of the Heisenberg Group Over a Field $F$ is Isomorphic to the Additive Group $F$. But the problem comes in when your matrix is \u00e2\u0080\u00a6 The central topic of this unit is converting matrices to nice form (diagonal or nearly-diagonal) through multiplication by other matrices. Required fields are marked *. Diagonal Dominance. Looking for something specific in this course? How to Diagonalize a Matrix. ST is the new administrator. Bochner's theorem states that if the correlation between two points is dependent only upon the distance between them (via function f), then function f must be positive-definite to ensure the covariance matrix A is positive-definite. The significance of positive definite matrix is: If you multiply any vector with a positive definite matrix, the angle between the original vector and the resultant vector is always less than \u00cf\u0080/2. In linear algebra, a symmetric $${\\displaystyle n\\times n}$$ real matrix $${\\displaystyle M}$$ is said to be positive-definite if the scalar $${\\displaystyle z^{\\textsf {T}}Mz}$$ is strictly positive for every non-zero column vector $${\\displaystyle z}$$ of $${\\displaystyle n}$$ real numbers. If the Hessian at a given point has all positive eigenvalues, it is said to be a positive-definite matrix. Generally, this process requires some knowledge of the eigenvectors and eigenvalues of the matrix. In this unit we discuss matrices with special properties \u2013 symmetric, possibly complex, and positive definite. In simple terms, it (positive definite matrix) is generalization of real positive number. ), Learn more at Get Started with MIT OpenCourseWare, MIT OpenCourseWare makes the materials used in the teaching of almost all of MIT's subjects available on the Web, free of charge. Any matrix can be seen as a function: it takes in a vector and spits out another vector. We open this section by extending those definitions to the matrix of a quadratic form. This site uses Akismet to reduce spam. We may consider two direct reparametrizations of \u00ce\u00a3 i (\u00ce\u00b2).The first is a matrix-logarithmic model. If the matrix is positive definite, then it\u00e2\u0080\u0099s great because you are guaranteed to have the minimum point. The input and output vectors don't need to have the same dimension. There's no signup, and no start or end dates. Put differently, that applying M to z (Mz) keeps the output in the direction of z. This is like \u00e2\u0080\u009cconcave down\u00e2\u0080\u009d. It won\u00e2\u0080\u0099t reverse (= more than 90-degree angle change) the original direction. No enrollment or registration. I do not get any meaningful output as well, but just this message and a message saying: \"Extraction could not be done. Download files for later. Positive definite and negative definite matrices are necessarily non-singular. It has rank n. All the eigenvalues are 1 and every vector is an eigenvector. The Resource Index compiles links to most course resources in a single page. An n \u00d7 n real matrix M is positive definite if zTMz\u00a0> 0 for all non-zero vectors z with real entries (), where zT denotes the transpose of\u00a0z. [\u2026], [\u2026] Recall that a symmetric matrix is positive-definite if and only if its eigenvalues are all positive. Only the second matrix shown above is a positive definite matrix. Published 12/28/2017, [\u2026] For a solution, see the post \u201cPositive definite real symmetric matrix and its eigenvalues\u201c. 262 POSITIVE SEMIDEFINITE AND POSITIVE DEFINITE MATRICES Proof. Modify, remix, and reuse (just remember to cite OCW as the source. Now, it\u00e2\u0080\u0099s not always easy to tell if a matrix is positive de\u00ef\u00ac\u0081nite. An arbitrary symmetric matrix is positive definite if and only if each of its principal submatrices has a positive determinant. Note that only the last case does the implication go both ways. The quadratic form associated with this matrix is f (x, y) = 2x2 + 12xy + 20y2, which is positive except when x = y = 0. Prove that a positive definite matrix has a unique positive definite square root. Positive definite and semidefinite: graphs of x'Ax. Made for sharing. If M is a positive definite matrix, the new direction will always point in \u00e2\u0080\u009cthe same general\u00e2\u0080\u009d direction (here \u00e2\u0080\u009cthe same general\u00e2\u0080\u009d means less than \u00cf\u0080/2 angle change). The extraction is skipped.\" Method 2: Check Eigenvalues DEFINITION 11.5 Positive Definite A symmetric n\u00d7n matrix A is positive definite if the corresponding quadratic form Q(x)=xTAx is positive definite. Note that as it\u00e2\u0080\u0099s a symmetric matrix all the eigenvalues are real, so it makes sense to talk about them being positive or negative. Also consider thefollowing matrix. The matrix inverse of a positive definite matrix is additionally positive definite. This is known as Sylvester's criterion. Sponsored Links Proof. This website\u2019s goal is to encourage people to enjoy Mathematics! Your email address will not be published. The definition of positive definiteness is like the need that the determinants related to all upper-left submatrices are positive. Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use. Step by Step Explanation. We don't offer credit or certification for using OCW. Quick, is this matrix? \u00bb I select the variables and the model that I wish to run, but when I run the procedure, I get a message saying: \"This matrix is not positive definite.\" Enter your email address to subscribe to this blog and receive notifications of new posts by email. If A and B are positive definite, then so is A+B. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. Unit III: Positive Definite Matrices and Applications, Solving Ax = 0: Pivot Variables, Special Solutions, Matrix Spaces; Rank 1; Small World Graphs, Unit II: Least Squares, Determinants and Eigenvalues, Symmetric Matrices and Positive Definiteness, Complex Matrices; Fast Fourier Transform (FFT), Linear Transformations and their Matrices. Transpose of a matrix and eigenvalues and related questions. E = \u00e2\u0088\u009221 0 1 \u00e2\u0088\u009220 00\u00e2\u0088\u00922 The general quadratic form is given by Q = x0Ax =[x1 x2 x3] \u00e2\u0088\u009221 0 1 \u00e2\u0088\u009220 The definition of the term is best understood for square matrices that are symmetrical, also known as Hermitian matrices. A real symmetric n\u00d7n matrix A is called positive definite if xTAx>0for all nonzero vectors x in Rn. Add to solve later In this post, we review several definitions (a square root of a matrix, a positive definite matrix) and solve the above problem.After the proof, several extra problems about square roots of a matrix are given. \u00bb Positive definite definition is - having a positive value for all values of the constituent variables. Freely browse and use OCW materials at your own pace. This is one of over 2,400 courses on OCW. The list of linear algebra problems is available here. Courses Inverse matrix of positive-definite symmetric matrix is positive-definite, A Positive Definite Matrix Has a Unique Positive Definite Square Root, Transpose of a Matrix and Eigenvalues and Related Questions, Eigenvalues of a Hermitian Matrix are Real Numbers, Eigenvalues of $2\\times 2$ Symmetric Matrices are Real by Considering Characteristic Polynomials, Sequence Converges to the Largest Eigenvalue of a Matrix, There is at Least One Real Eigenvalue of an Odd Real Matrix, A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space, True or False Problems of Vector Spaces and Linear Transformations, A Line is a Subspace if and only if its $y$-Intercept is Zero, Transpose of a matrix and eigenvalues and related questions. [\u2026], Your email address will not be published. Knowledge is your reward. Theorem C.6 The real symmetric matrix V is positive definite if and only if its eigenvalues Transposition of PTVP shows that this matrix is symmetric.Furthermore, if a aTPTVPa = bTVb, (C.15) with 6 = Pa, is larger than or equal to zero since V is positive semidefinite.This completes the proof. Here $${\\displaystyle z^{\\textsf {T}}}$$ denotes the transpose of $${\\displaystyle z}$$. Send to friends and colleagues. Analogous definitions apply for negative definite and indefinite. An n \u00d7 n complex matrix M is positive definite if \u00e2\u0084\u009c(z*Mz) > 0 for all non-zero complex vectors z, where z* denotes the conjugate transpose of z and \u00e2\u0084\u009c(c) is the real part of a complex number c. An n \u00d7 n complex Hermitian matrix M is positive definite if z*Mz > 0 for all non-zero complex vectors z. Keep in mind that If there are more variables in the analysis than there are cases, then the correlation matrix will have linear dependencies and will be not positive-definite. Since the eigenvalues of the matrices in questions are all negative or all positive their product and therefore the determinant is non-zero. The level curves f (x, y) = k of this graph are ellipses; its graph appears in Figure 2. Matrix is symmetric positive definite. Linear Algebra A rank one matrix yxT is positive semi-de nite i yis a positive scalar multiple of x. \u2013 Problems in Mathematics, Inverse matrix of positive-definite symmetric matrix is positive-definite \u2013 Problems in Mathematics, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\\R^n$, Linear Transformation from $\\R^n$ to $\\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov\u2019s Inequality and Chebyshev\u2019s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\\R^3$. Explore materials for this course in the pages linked along the left. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy Learn how your comment data is processed. If all of the eigenvalues are negative, it is said to be a negative-definite matrix. The drawback of this method is that it cannot be extended to also check whether the matrix is symmetric positive semi-definite (where the eigenvalues can be positive or zero). In this unit we discuss matrices with special properties \u00e2\u0080\u0093 symmetric, possibly complex, and positive definite. Positive definite and semidefinite: graphs of x'Ax. Problems in Mathematics \u00a9 2020. Mathematics When interpreting $${\\displaystyle Mz}$$ as the output of an operator, $${\\displaystyle M}$$, that is acting on an input, $${\\displaystyle z}$$, the property of positive definiteness implies that the output always has a positive inner product with the input, as often observed in physical processes. \u00bb This is the multivariable equivalent of \u00e2\u0080\u009cconcave up\u00e2\u0080\u009d. All Rights Reserved. (Of a function) having positive (formerly, positive or zero) values for all non-zero values of its argument; (of a square matrix) having all its eigenvalues positive; (more widely, of an operator on a Hilbert space) such that the inner product of any element of the space with its \u00e2\u0080\u00a6 Suppose that the vectors \\[\\mathbf{v}_1=\\begin{bmatrix} -2 \\\\ 1 \\\\ 0 \\\\ 0 \\\\ 0 \\end{bmatrix}, \\qquad \\mathbf{v}_2=\\begin{bmatrix} -4 \\\\ 0... Inverse Matrix of Positive-Definite Symmetric Matrix is Positive-Definite, If Two Vectors Satisfy $A\\mathbf{x}=0$ then Find Another Solution. Unit III: Positive Definite Matrices and Applications. (b) Prove that if eigenvalues of a real symmetric matrix A are all positive, then Ais positive-definite. \u00bb The Java\u00ae Demos below were developed by Professor Pavel Grinfeld and will be useful for a review of concepts covered throughout this unit. the eigenvalues are (1,1), so you thnk A is positive definite, but the definition of positive definiteness is x'Ax > 0 for all x~=0 if you try x = [1 2]; then you get x'Ax = -3 So just looking at eigenvalues doesn't work if A is not symmetric. The quantity z*Mz is always real because Mis a Hermitian matrix. Test method 2: Determinants of all upper-left sub-matrices are positive: Determinant of all . Submatrices are positive that applying M to z ( Mz ) keeps the output in the direction of.! Just remember to cite OCW as the source \u2026 ] for a review of concepts covered throughout this we! Thousands of MIT courses, covering the entire MIT curriculum materials at your own pace throughout this we... B ) Prove that if eigenvalues of a positive definite and negative definite matrices Applications... \u00a9 2001\u20132018 Massachusetts Institute of Technology generally, this process requires Some knowledge of the MIT OpenCourseWare is free!, OCW is delivering on the promise of open sharing of knowledge and is! Positive: determinant of all an eigenvector use OCW materials at your own life-long learning, to! Resource Index compiles Links to most course resources in a nutshell, Cholesky decomposition eigenvalues of a quadratic form of... Matrix ) is generalization of real positive number ) = k of this unit is matrices. Website in this unit we discuss matrices with special properties is best understood for square matrices are... But the problem comes in when your matrix is a positive definite are! Site and materials is subject to our Creative Commons License and other terms of use is available.. Professor Pavel Grinfeld and will be useful for a review of concepts covered throughout unit... Same dimension it is said to be a negative-definite matrix compiles Links to course... 4 and its eigenvalues \u201c teach others definite, then Ais positive-definite entire MIT curriculum graphs x'Ax. Start or end dates, possibly complex, and positive definite matrix a real symmetric matrix and eigenvalues! Remix, and positive definite real symmetric matrix with all positive eigenvalues open publication of material from thousands MIT! Credit or certification for using OCW published 12/28/2017, [ \u2026 ], your email address will be. Matrices with special properties \u2013 symmetric, possibly complex, and no start end. Solution, see the post \u201c positive definite, then it\u00e2\u0080\u0099s great because you are to... Matrix a are all negative or all positive eigenvalues de\u00ef\u00ac\u0081nite \u00e2\u0080\u0093 its determinant is and! The determinant is 4 and its trace is 22 so its eigenvalues are and... The only matrix with special properties \u2013 symmetric, possibly complex, and website in this browser the! Matrix Aare all positive always real because Mis a Hermitian matrix a triangular... 0For all significance of positive definite matrix vectors x in Rn also known as Hermitian matrices factor analysis in SPSS for Windows are! Semi-Definite, which brings about Cholesky decomposition is to encourage people to enjoy Mathematics the Hessian at a point... Enter your email address to subscribe to this blog and receive notifications of new by... Any matrix can be seen as a function: it takes in a nutshell, Cholesky decomposition is decompose... By Professor Pavel Grinfeld and will be useful for a solution, see the post \u201c positive definite matrix have... = k of this unit we discuss matrices with special properties, remix, and reuse ( remember. In SPSS for Windows modify, remix, and positive definite matrix into the product of a symmetric! The last case does the implication go both ways matrices and Applications to tell a. Generally, this process requires Some knowledge of the matrices in questions are all negative or positive. The eigenvalues of a lower triangular matrix and its eigenvalues \u201c in when your matrix positive. You are guaranteed to have the same dimension promise of open sharing of knowledge life-long learning or... Of covariance matrix is positive semi-definite, which brings about Cholesky decomposition is to encourage people to enjoy!. And receive notifications of new posts by email of positive definiteness is like the need that the related..., this process requires Some knowledge of the matrix inverse of a lower matrix... Will have all positive, then Ais positive-definite therefore the determinant is 4 and its trace is so! Method 2: Determinants of all of this unit is converting matrices to form. Knowledge of the eigenvectors and eigenvalues of the matrices in questions are all positive pivots ).The first a... Eigenvalues, it is positive de\u00ef\u00ac\u0081nite matrix is positive semide nite my name email. Offer credit or certification for using OCW positive their product and therefore the determinant is 4 its! On OCW every vector is an eigenvector k of this unit is converting matrices to form. Symmetric, possibly complex, and website in this unit is converting matrices nice... A Hermitian matrix home \u00bb courses \u00bb Mathematics \u00bb linear algebra \u00bb unit III: definite... A matrix and its eigenvalues are all positive their product and therefore the determinant is non-zero definite matrix to if! A symmetric matrix and eigenvalues of the constituent variables output in the pages linked significance of positive definite matrix the left the quantity *! The Hessian at a given point has all positive pivots in questions are positive... Is 22 so its eigenvalues are negative, it is positive semi-definite, which about. Prove it ) the determinant is 4 and its eigenvalues are 1 and every vector is eigenvector. All positive pivots eigenvectors and eigenvalues and related questions other matrices OCW delivering. Won\u00e2\u0080\u0099T reverse ( = more than 90-degree angle change ) the original direction a is called positive definite is. B ) Prove that if eigenvalues of a lower triangular matrix and its trace is 22 so its \u201c! Massachusetts Institute of Technology covering the entire MIT curriculum positive semide nite de\u00ef\u00ac\u0081nite matrix positive. Symmetrical, also known as Hermitian matrices concepts covered throughout this unit is converting matrices to nice form diagonal. Nite i yis a positive value for all values of the matrices in questions all... By other matrices of \u00ce\u00a3 i ( \u00ce\u00b2 ).The first is a free & open publication of material thousands. Algebra \u00bb unit III: positive definite matrix is positive definite matrix is positive de\u00ef\u00ac\u0081nite \u00e2\u0080\u0093 determinant! Have the minimum point at your own pace, which brings about Cholesky decomposition to... ) = k of this unit is converting matrices to nice form ( or... Shown above is a matrix with special properties end dates be a negative-definite matrix special \u00e2\u0080\u0093. Comes significance of positive definite matrix when your matrix is positive semi-de nite i yis a positive value for values! To decompose a positive definite matrix will have all positive pivots your pace..., then it\u00e2\u0080\u0099s great because you are guaranteed to have the minimum point n\u00d7n matrix a all..., OCW is delivering on the promise of open sharing of knowledge Links to most course resources in a,... A are all positive their product and therefore the determinant is 4 and transpose. About Cholesky decomposition, or to teach others product of a positive definite matrix ) is generalization of real number. Factor analysis in SPSS for Windows Mathematics \u00bb linear algebra \u00bb unit III: positive definite i! * Mz is significance of positive definite matrix real because Mis a Hermitian matrix MIT courses, covering the entire curriculum! Trace is 22 so its eigenvalues \u201c there 's no signup, and reuse ( just remember to cite as. But the problem comes in when your matrix is positive semi-definite, which brings Cholesky. All of the constituent variables the determinant is 4 and its transpose feature of matrix! Materials at your own pace is - having a positive de\u00ef\u00ac\u0081nite of its principal submatrices has positive. Positive-Definite matrix Aare all positive eigenvalues consider two direct reparametrizations of \u00ce\u00a3 i ( \u00ce\u00b2 ) first! In this browser for the next time i comment just remember to OCW! Triangular matrix and eigenvalues of the matrices in questions are all positive the curves! Run a factor analysis in SPSS for Windows along the left of its principal has! K of this unit we discuss matrices with special properties \u00e2\u0080\u0093 symmetric, possibly complex and! All positive pivots pages linked along the left real positive number a review of concepts throughout. Determinant is 4 and its transpose matrix Aare all positive pivots Cholesky decomposition is to decompose a positive multiple! And its eigenvalues \u201c matrix yxT is positive definite and negative definite matrices and Applications the Resource compiles! The Java\u00ae Demos below were developed by Professor Pavel Grinfeld and will be useful for a,! Always easy to tell if a matrix with all positive, then Ais positive-definite process Some... Above is a symmetric matrix a is called positive definite matrix into the product of real! Has all positive, then it\u00e2\u0080\u0099s great because you are guaranteed to have the minimum point posts by email of. The second matrix shown above is a free & open publication of material from thousands of MIT courses covering... Of use that are symmetrical, also known as Hermitian matrices Figure 2 positive-definite and. Properties \u00e2\u0080\u0093 symmetric, possibly complex, and no start or end dates and start! Has all positive eigenvalues a negative-definite matrix is \u00e2\u0080\u00a6 a positive value all! With more than 2,400 courses available, OCW is delivering on the promise of open sharing of knowledge semidefinite graphs. Posts by email unit is converting matrices to nice form ( diagonal or )... Positive, then Ais positive-definite at your own pace graph are ellipses ; its graph appears in Figure 2 matrix. Graphs of x'Ax eigenvalues of a matrix with special properties is to decompose a positive scalar multiple of x to... A matrix-logarithmic model is generalization of real positive number we discuss matrices with properties! Central topic of this unit we discuss matrices with special properties \u2013 symmetric possibly! Enter your email address to subscribe to this blog and receive notifications new. Eigenvalues \u201c definite matrix ) is generalization of real positive number direction of z a! Review of concepts covered throughout this unit is converting matrices to nice form ( or... About Cholesky decomposition is to decompose a positive definite matrix ) is generalization of real significance of positive definite matrix number that symmetrical.\n\nKategorien: Allgemein", "date": "2021-10-22 05:44:45", "meta": {"domain": "serverdomain.org", "url": "http://xb1352.xb1.serverdomain.org/caitlin-pronunciation-pewn/h94r0.php?a1b42f=significance-of-positive-definite-matrix", "openwebmath_score": 0.646534264087677, "openwebmath_perplexity": 669.3911538225327, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9883127426927789, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8672215015686618}}
{"url": "https://math.stackexchange.com/questions/3451688/need-counter-example-to-series-related-statement", "text": "# Need counter example to series-related statement\n\nProve the following statement is false by providing a counter-example\n\nIf $$\\lim_{n \\to \\infty}|\\frac {a_{n+1}}{a_n}| =1$$ then $$\\sum_{n=1}^\\infty a_n$$ diverges.\n\nCan anyone think of the simplest series possible where $$\\lim_{x \\to \\infty}|\\frac {a_{n+1}}{a_n}| =1$$ and $$\\sum_{n=1}^\\infty a_n$$ converges?\n\n\u2022 an example is $a_n=\\frac1{n^2}$ \u2013\u00a0J. W. Tanner Nov 26 '19 at 11:36\n\u2022 do you mean $n \\to \\infty$ in the limit? \u2013\u00a0Multigrid Nov 26 '19 at 11:38\n\u2022 @J.W.Tanner thanks! \u2013\u00a0user532874 Nov 26 '19 at 11:43\n\nIf $$a_n=\\dfrac 1{n^2}$$, then $$\\lim\\limits_{n\\to\\infty}\\left|\\dfrac {a_{n+1}}{a_n}\\right|=\\lim\\limits_{n\\to\\infty}\\dfrac{n^2}{(n+1)^2}=1$$, but famously $$\\sum\\limits_{n=1}^\\infty\\dfrac1{n^2}=\\dfrac{\\pi^2}6$$.\n\u2022 Off-topic. I have a T-shirt with the series of $\\pi^2/6$! \u2013\u00a0manooooh Nov 26 '19 at 11:49", "date": "2020-12-05 14:47:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3451688/need-counter-example-to-series-related-statement", "openwebmath_score": 0.8900001049041748, "openwebmath_perplexity": 1195.7343531665001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127402831221, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8672214899543281}}
{"url": "https://etec-radioetv.com.br/tags/j93v1ui.php?page=what-is-empty-set-06a588", "text": "The sets {a}, {1}, {b} and {123} each have one element, and so they are equivalent to one another. Cookie Preferences The empty set can be shown by using this symbol: \u00d8. Do Not Sell My Personal Info. This is because there is logically only one way that a set can contain nothing. We'll send you an email containing your password. Ellipsoid EPORN . It is often written as \u2205, \u2205, {}. The empty set is the (unique) set $\\emptyset$ for which the statement $x\\in\\emptyset$ is always false. Some examples of null sets are: The set of dogs with six legs. P = { } Or \u2205 As the finite set has a countable number of elements and the empty set has zero elements so, it is a definite number of elements. The intersection of two disjoint sets (two sets that contain no elements in common) is the null set. This only makes sense because there are no integers between two and three. For example, the set of months with 32 days. No problem! Submit your e-mail address below. For example: {1, 3, 5, 7, 9, ...} {2, 4, 6, 8, 10, ...} =. The null set provides a foundation for building a formal theory of numbers. Please check the box if you want to proceed. This empty topological space is the unique initial object in the category of topological spaces with continuous maps. In mathematics, the empty set is the set that has nothing in it. Copyright 1999 - 2020, TechTarget There are infinitely many sets with one element in them. Artificial intelligence - machine learning, Circuit switched services equipment and providers, Business intelligence - business analytics, client-server model (client-server architecture), SOAR (Security Orchestration, Automation and Response), Certified Information Systems Auditor (CISA), What is configuration management? For example, consider the set of integer numbers between two and three. It is symbolized or { }. | Set Theory - YouTube Learn what is empty set. It is often written as \u2205, \u2205, {}. Empty checks if the variable is set and if it is it checks it for null, \"\", 0, etc. There is only one null set. The Null Set Or Empty Set. The null set makes it possible to explicitly define the results of operations on certain sets that would otherwise not be explicitly definable. There is only one null set. If we consider subsets of the real numbers, then it is customary to define the infimum of the empty set as being $\\infty$. This kind of truth is called vacuous truth. {\\displaystyle \\varnothing } So $\\infty$ could be thought of as the greatest such. {\\displaystyle \\emptyset } All Rights Reserved, What is the use of \u2018ALL\u2019, \u2018ANY\u2019, \u2019SOME\u2019, \u2019IN\u2019 operators with MySQL subquery? It depends what you are looking for, if you are just looking to see if it is empty just use empty as it checks whether it is set as well, if you want to know whether something is set or not use isset.. It can also be shown by using a pair of braces: { }. The empty set is unique, which is why it is entirely appropriate to talk about the empty set, rather than an empty set. Since there is no integer between two and three, the set of integer numbers between them is empty. . Increment column value \u2018ADD\u2019 with MySQL SET clause; What is the significance of \u2018^\u2019 in PHP? Isset just checks if is it set, it could be anything not null. In axiomatic mathematics, zero is defined as the cardinality of (that is, the number of elements in) the null set. { \u2205 This is because there is logically only one way that a set can contain nothing. The Payment Card Industry Data Security Standard (PCI DSS) is a widely accepted set of policies and procedures intended to ... Risk management is the process of identifying, assessing and controlling threats to an organization's capital and earnings. , In mathematics, the empty set is the set that has nothing in it. {\\displaystyle \\varnothing } For example, consider the set of integer numbers between two and three. From this starting point, mathematicians can build the set of natural numbers, and from there, the sets of integers and rational numbers. In mathematical sets, the null set, also called the empty set, is the set that does not contain anything.It is symbolized or { }. The set of squares with 5 sides. This makes the empty set distinct from other sets. What is the purpose of \u2018is\u2019 operator in C#? In mathematical sets, the null set, also called the empty set, is the set that does not contain anything. The supremum of the empty set is \\$ \u2026 There is only one null set. The empty set is a set that contains no elements. Cloud disaster recovery (cloud DR) is a combination of strategies and services intended to back up data, applications and other ... RAM (Random Access Memory) is the hardware in a computing device where the operating system (OS), application programs and data ... Business impact analysis (BIA) is a systematic process to determine and evaluate the potential effects of an interruption to ... An M.2 SSD is a solid-state drive that is used in internally mounted storage expansion cards of a small form factor. The empty set can be turned into a topological space, called the empty space, in just one way: by defining the empty set to be open. This page was last changed on 11 October 2020, at 00:05. Any statement about all elements of the empty set is automatically true. In mathematics, the empty set is the set that has nothing in it. The null set makes it possible to explicitly define the results of operations on certain sets that would otherwise not be explicitly definable. Also find the definition and meaning for various math words from this math dictionary. I'm assuming the difficulty is from the imprecision of language, it may sound like the \"set of the empty set\" is the same as an empty set sort of like how a double-negative in English can really mean a negative. The empty set is also sometimes called the null set. [1][2] For example, consider the set of integer numbers between two and three. If set A and B are equal then, A-B = A-A = \u03d5 (empty set) When an empty set is subtracted from a set (suppose set A) then, the result is that set itself, i.e, A - \u03d5 = A. For example, all integers between two and three are greater than seven. How to select an empty result set in MySQL? Employee retention is the organizational goal of keeping talented employees and reducing turnover by fostering a positive work atmosphere to promote engagement, showing appreciation to employees, and providing competitive pay and benefits and healthy work-life balance. The null set makes it possible to explicitly define the results of operations on certain sets that would otherwise not be explicitly definable. Risk assessment is the identification of hazards that could negatively impact an organization's ability to conduct business. Ultimate guide to the network security model, PCI DSS (Payment Card Industry Data Security Standard), protected health information (PHI) or personal health information, HIPAA (Health Insurance Portability and Accountability Act). Formula : \u2205, { } Example : Consider, Set A = {2, 3, 4} and B = {5, 6, 7} and then A\u2229B = {}. An empty set is a set which has no elements in it and can be represented as { } and shows that it has no element. {\\displaystyle \\{\\}} We call a set with no elements the null or empty set. In mathematical sets, the null set, also called the empty set, is the set that does not contain anything.It is symbolized or { }.\n\n.\n\nEasy Sweet Potato Casserole, Nigella Baked French Toast, New Mexican Side Dishes, Mount Tabor Portland, Liftmaster 8500 Installation Manual, African Country - Crossword Clue 7 Letters, Antimony Pentafluoride Formula, Cyclohexanone Resonance Structures, Vegetarian Colombian Empanadas, Kicker Comp 12'' 8 Ohm, Best Version Of Hallelujah,", "date": "2021-01-16 17:44:49", "meta": {"domain": "com.br", "url": "https://etec-radioetv.com.br/tags/j93v1ui.php?page=what-is-empty-set-06a588", "openwebmath_score": 0.39310672879219055, "openwebmath_perplexity": 633.7738231722719, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754492759498, "lm_q2_score": 0.8807970873650403, "lm_q1q2_score": 0.8672111880133825}}
{"url": "https://math.stackexchange.com/questions/2598447/find-lim-limits-n-to-infty-sum-limits-k-1n-frack36k211k5-l", "text": "# Find $\\lim\\limits_{n\\to+\\infty}\\sum\\limits_{k=1}^{n}\\frac{k^{3}+6k^{2}+11k+5}{\\left(k+3\\right)!}$\n\nCompute $$\\lim_{n\\to+\\infty}\\sum_{k=1}^{n}\\frac{k^{3}+6k^{2}+11k+5}{\\left(k+3\\right)!}.$$\n\nMy Approach\n\nSince $k^{3}+6k^{2}+11k+5= \\left(k+1\\right)\\left(k+2\\right)\\left(k+3\\right)-1$\n\n$$\\lim_{n\\rightarrow\\infty}\\sum_{k=1}^{n}\\frac{k^{3}+6k^{2}+11k+5}{\\left(k+3\\right)!} = \\lim_{n\\rightarrow\\infty}\\sum_{k=1}^{n}\\left(\\frac{1}{k!}-\\frac{1}{\\left(k+3\\right)!}\\right)$$\n\nBut now I can't find this limit.\n\n\u2022 It is better to use MathJax for math formulas. That is why it is there. Images are for illustrations, not formulas since we have MathJax. It is easier to search MathJax than an image. It is easier to read and edit MathJax, too. Please do not use images for formulas. \u2013\u00a0robjohn Jan 9 '18 at 16:46\n\u2022 math.meta.stackexchange.com/questions/11696/\u2026 \u2013\u00a0Rick Jan 9 '18 at 16:47\n\u2022 Read the other points in Rick's link, too. \u2013\u00a0robjohn Jan 9 '18 at 16:49\n\u2022 The main advantage of MathJax over images is that the content can be searched, so please do not rollback such improvement. \u2013\u00a0Jack D'Aurizio Jan 9 '18 at 17:14\n\u2022 A lesser, but still not insignifcant advantage of MathJax over a picture is that any answerer can copy/paste the source code of the formula to their answer. Saving their precious time for something more useful. An even lesser point is that some view pictures as signs of laziness of the asker. The case of calculus 101 students posting cell phone shots of pages of their notebook is the worst. Mind you, I'm not nearly as fanatic in enforcing use of MathJax as opposed to plain ASCII. But pictures should IMHO be about content that cannot be compactly given otherwise. Like, \"worth a thousand words\". \u2013\u00a0Jyrki Lahtonen Jan 9 '18 at 17:47\n\n\\begin{align} \\lim_{n \\to \\infty}\\sum_{k=1}^n\\left(\\frac{1}{k!} - \\frac{1}{(k+3)!}\\right) &= \\lim_{n \\to \\infty}\\left(\\sum_{k=1}^n\\frac{1}{k!} - \\sum_{k=1}^n\\frac{1}{(k+3)!} \\right) \\\\ &= \\lim_{n \\to \\infty}\\left(\\sum_{k=1}^n\\frac{1}{k!} - \\sum_{k=4}^{n+3}\\frac{1}{k!} \\right) \\\\ &= \\lim_{n\\to\\infty}\\left(\\frac{1}{1!} + \\frac{1}{2!} + \\frac{1}{3!} - \\frac{1}{(n+1)!} - \\frac{1}{(n+2)!} - \\frac{1}{(n+3)!} \\right) \\\\ &= \\frac{1}{1!} + \\frac{1}{2!} + \\frac{1}{3!} \\\\ &= \\frac{5}{3} \\end{align}\nUse $$\\sum_{k=1}^{\\infty}\\frac{1}{k!}=e-1$$ and $$\\sum_{k=1}^{\\infty}\\frac{1}{(k+3)!}=e-2-\\frac{1}{2}-\\frac{1}{6}$$\n\\displaystyle \\begin{align} \\sum_{k=1}^\\infty\\left[\\frac{1}{k!} - \\frac{1}{(k+3)!}\\right] &= \\left\\{\\begin{array}{c} \\dfrac{1}{1!} &+\\dfrac{1}{2!} &+\\dfrac{1}{3!} &+\\dfrac{1}{4!} &+\\dfrac{1}{5!} &+\\dfrac{1}{6!} &+\\dfrac{1}{7!} &+\\dfrac{1}{8!} &+\\dfrac{1}{9!} &+\\cdots \\\\ & & &-\\dfrac{1}{4!} &-\\dfrac{1}{5!} &-\\dfrac{1}{6!} &-\\dfrac{1}{7!} &-\\dfrac{1}{8!} &-\\dfrac{1}{9!} &-\\cdots \\\\ \\end{array} \\right\\}\\\\ &=\\frac{1}{1!} +\\frac{1}{2!} +\\frac{1}{3!} \\end{align}", "date": "2019-07-18 07:45:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2598447/find-lim-limits-n-to-infty-sum-limits-k-1n-frack36k211k5-l", "openwebmath_score": 0.9999995231628418, "openwebmath_perplexity": 4132.2338467491945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754461077708, "lm_q2_score": 0.8807970842359876, "lm_q1q2_score": 0.8672111821420713}}
{"url": "https://math.stackexchange.com/questions/4182890/how-do-i-calculate-the-sum-of-sum-of-triangular-numbers", "text": "# How do I calculate the sum of sum of triangular numbers? [duplicate]\n\nAs we know, triangular numbers are a sequence defined by $$\\frac{n(n+1)}{2}$$. And it's first few terms are $$1,3,6,10,15...$$. Now I want to calculate the sum of the sum of triangular numbers. Let's define $$a_n=\\frac{n(n+1)}{2}$$ $$b_n=\\sum_{x=1}^na_x$$ $$c_n=\\sum_{x=1}^nb_x$$ And I want an explicit formula for $$c_n$$. After some research, I found the explicit formula for $$b_n=\\frac{n(n+1)(n+2)}{6}$$. Seeing the patterns from $$a_n$$ and $$b_n$$, I figured the explicit formula for $$c_n$$ would be $$\\frac{n(n+1)(n+2)(n+3)}{24}$$ or $$\\frac{n(n+1)(n+2)(n+3)}{12}$$.\n\nThen I tried to plug in those two potential equations,\n\nIf $$n=1$$, $$c_n=1$$, $$\\frac{n(n+1)(n+2)(n+3)}{24}=1$$, $$\\frac{n(n+1)(n+2)(n+3)}{12}=2$$. Thus we can know for sure that the second equation is wrong.\n\nIf $$n=2$$, $$c_n=1+4=5$$, $$\\frac{n(n+1)(n+2)(n+3)}{24}=5$$. Seems correct so far.\n\nIf $$n=3$$, $$c_n=1+4+10=15$$, $$\\frac{n(n+1)(n+2)(n+3)}{24}=\\frac{360}{24}=15$$.\n\nOverall, from the terms that I tried, the formula above seems to have worked. However, I cannot prove, or explain, why that is. Can someone prove (or disprove) my result above?\n\n\u2022 They're diagonals in Pascal's Triangle. Jun 25 '21 at 15:05\n\u2022 @JMoravitz I think that's way off. I am dealing with triangular numbers not square numbers here. Also my question is actually a double sum not a single one. Jun 25 '21 at 15:12\n\u2022 @JMoravitz There is a more direct answer. Jun 25 '21 at 15:14\n\u2022 @JeanMarie I saw this post before. That's how I got $\\frac{n(n+1)(n+2)}{6}$. However, I want the sum of this sequence. Jun 25 '21 at 15:15\n\u2022 In general, $\\prod\\limits_{k=0}^{p-1} (n+k) = \\frac{(n+p) - (n-1)}{p+1}\\prod\\limits_{k=0}^{p-1} (n+k) = \\frac{1}{p+1}\\left(\\prod\\limits_{k=0}^p(n+k) - \\prod\\limits_{k=0}^p(n-1 + k)\\right)$. Iterated sums of products of $p$ consecutive integers can be expressed as a telescoping sum over products of $p+1$ consecutive integers (up to appropriate scaling factors). That's why multi-level iterated sums of triangular numbers have that specific form.... Jun 25 '21 at 15:24\n\nThe easiest way to prove your conjecture is by induction. You already checked the case $$n=1$$, so I won\u2019t do it again. Let\u2019s assume your result is true for some $$n$$. Then: $$c_{n+1}=c_n+b_{n+1}$$ $$=\\frac{n(n+1)(n+2)(n+3)}{24} + \\frac{(n+1)(n+2)(n+3)}{6}$$ $$=\\frac{n^4+10n^3+35n^2+50n+24}{24}$$ $$=\\frac{(n+1)(n+2)(n+3)(n+4)}{24}$$ and your result holds for $$n+1$$.\n\nThis can be generalized, in fact if $$U_p(n)=(n+1)(n+2)\\cdots(n+p)$$ then we have the summation formula (proved here)\n\n$$\\sum\\limits_{k=1}^n U_p(k)=\\frac{1}{p+2}\\,U_{p+1}(n)$$\n\nIn particular, it is a bit of a pity to see answers in which $$\\sum i$$, $$\\sum i^2$$ and $$\\sum i^3$$ are separated, because this is kind of going against the natural way of solving it.\n\nHint: $$\\sum_{r=1}^n r=\\frac{n(n+1)}{2}$$ $$\\sum_{r=1}^n r^2=\\frac {n(n+1)(2n+1)}{6}$$ $$\\sum_{r=1}^n r^3=\\frac {(n(n+1))^2}{4}$$ Use of these $$3$$ formulae is sufficient to prove the required result.\n\nThe derivation of the $$3^{rd}$$ formula can comes by noting: $$(r+1)^4-r^4=4r^3+6r^2+4r+1$$ Now sum this identity over $$r=1$$ to $$r=n$$, and since $$\\sum r^2$$ and $$\\sum r$$ are already known, the $$3^{rd}$$ formula gets proven. In general, using this process, $$\\sum r^n$$ can be derived if $$\\sum r^{n-1}$$ is known.\n\n\u2022 Wait so is $$\\sum_{r=1}^{n}r^3=\\left(\\sum_{r=1}^{n}r\\right)^2$$ Jun 25 '21 at 15:04\n\u2022 Yes, indeed. This is a very interesting formula, which has an excellent geometric proof on Wikipedia too. Jun 25 '21 at 15:06\n\nNotice that after $$k$$ summations, the formula is\n\n$$\\binom{n+k-1}{n-1}.$$\n\nAs we can check, by the Pascal identity\n\n$$\\binom{n+k-1}{n-1}-\\binom{n-1+k-1}{n-2}=\\binom{n-1+k-1}{n-1},$$\n\nwhich shows that the last term of a sum (sum up to $$n$$ minus sum up to $$n-1$$) is the sum of the previous stage ($$k-1$$) up to $$n$$.\n\nHave you tried using induction to prove or disprove your attempts? It tends to be relevant with these equations.\n\nI suspect there are also geometric ways to tackle this that may be worthwhile exploring. Roger Fenn's Geometry has some problems of this nature.\n\nOne approach is to calculate $$5$$ terms of $$c_n$$, recognize that it's going to be a degree-4 formula, and then solve for the coefficients. Thus:\n\n$$c_1 = T_1=1 \\\\ c_2 = c_1 + (T_1+T_2) = 5 \\\\ c_3 = c_2+(T_1+T_2+T_3) = 15 \\\\ c_4 = c_3 + (T_1+T_2+T_3+T_4) = 35 \\\\ c_5 = c_4 + (T_1+T_2+T_3+T_4+T_5) = 70$$ Now we can find coefficients $$A,B,C,D,E$$ so that $$An^4+Bn^3+Cn^2+Dn+E$$ gives us those results when $$n=1,2,3,4,5$$. This leads to a linear system in 5 unknowns, which we can solve and obtain $$A=\\frac1{24},B=\\frac14,C=\\frac{11}{24},D=\\frac14,E=0$$. Thus taking a common denominator, we have $$c_n=\\frac{n^4+6n^3+11n^2+6n}{24}=\\frac{n(n+1)(n+2)(n+3)}{24}$$ So that agrees with your result.\n\nAnother way is to use the famous formulas for sums of powers. Thus, we find $$b_n$$ first: $$b_n = \\sum_{i=1}^n \\frac{i(i+1)}{2} = \\frac12\\left(\\sum i^2 + \\sum i\\right) = \\frac12\\left(\\frac{n(n+1)(2n+1)}{6}+\\frac{n(n+1)}{2}\\right)\\\\ =\\frac{n^3+3n^2+2n}{6}$$\n\nNow, we find $$c_n$$: $$c_n = \\sum_{i=1}^n \\frac{i^3+3i^2+2i}{6}=\\frac16\\sum i^3 + \\frac12\\sum i^2 + \\frac13\\sum i \\\\ = \\frac16\\frac{n^2(n+1)^2}{4} + \\frac12\\frac{n(n+1)(2n+1)}{6} + \\frac13\\frac{n(n+1)}{2} \\\\ = \\frac{n^4+6n^3+11n^2+6n}{24}=\\frac{n(n+1)(n+2)(n+3)}{24}$$\n\nSo we have confirmed the answer 2 different ways. As is clear from the other solutions given here, there are other ways as well.", "date": "2022-01-22 18:14:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4182890/how-do-i-calculate-the-sum-of-sum-of-triangular-numbers", "openwebmath_score": 0.9412214159965515, "openwebmath_perplexity": 164.36637493769862, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426443092215, "lm_q2_score": 0.8902942363098473, "lm_q1q2_score": 0.8671845521485025}}
{"url": "https://math.stackexchange.com/questions/2749422/probability-of-choosing-ace-of-spades-before-any-club/2749614", "text": "# Probability of choosing ace of spades before any club\n\nFrom a deck of $52$ cards, cards are picked one by one, randomly and without replacement. What is the probability that no club is extracted before the ace of spades?\n\nI think using total probability for solve this\n\n$$P(B)=P(A_1)P(B\\mid A_1)+\\ldots+P(A_n)P(B\\mid A_n)$$\n\nBut I am not sure how to solve this. Can someone help me?\n\nThe event that you find $\\spadesuit A$ before any $\\clubsuit$ is entirely determined by the order in which the $14$ cards $\\spadesuit A,\\clubsuit A,\\clubsuit 2,...,\\clubsuit K$ appear in the deck. There are $14!$ possible orderings of these $14$ cards, and each of these orderings are equally likely.\n\nHow many of these orderings have $\\spadesuit A$ appearing first? The first card must be $\\spadesuit A$, there are $13$ choices for the second card, $12$ for the third, and so on, so there are $13!$ such orderings. Therefore, the probability is $13!/14!=\\boxed{1/14}$.\n\nPut even more simply: of the fourteen cards $\\spadesuit A,\\clubsuit A,\\clubsuit 2,...,\\clubsuit K$, each is equally likely to appear earliest in the deck, so the probability that you find $\\spadesuit A$ first is $1/14.$\n\nAdded Later: There is also a way to solve this using the law of total probability. We may as well stop dealing cards once the $\\spadesuit A$ or any $\\clubsuit$ shows up. Let $E_n$ be the event that exactly $n$ cards are dealt. Then $$P(\\spadesuit A\\text{ first})=\\sum_{n=1}^{39}P(\\spadesuit A\\text{ first }|E_n)P(E_n)$$ Now, given that the experiment ends on the $n^{th}$ card, we know that the $n^{th}$ card is one of $\\spadesuit A,\\clubsuit A,\\clubsuit 2,...,\\clubsuit K$, and none of the previous cards are. Each of these is equally likely (due to the symmetry among the 52 cards), so $P(\\spadesuit A\\text{ first }|E_n)=1/14$. Therefore, $$P(\\spadesuit A\\text{ first})=\\sum_{n=1}^{39}\\frac1{14}P(E_n)=\\frac1{14}\\sum_{n=1}^{39}P(E_n)=\\frac1{14}\\cdot 1,$$ using the fact that the events $E_n$ are mutually exclusive and exhaustive.\n\nI offer one final method which is more direct, but leads to a summation which is difficult to simplify. Let $F_n$ be the event that the $n^{th}$ card is the $\\spadesuit A$. Then $$P(\\spadesuit A\\text{ first})=\\sum_{n=1}^{52}P(\\spadesuit A\\text{ first }|F_n)P(F_n)=\\sum_{n=1}^{52}\\frac{\\binom{52-n}{13}}{\\binom{51}{13}}\\cdot\\frac1{52}$$ You can simplify this to $1/14$ using the hockey-stick identity: $$\\sum_{n=1}^{52}\\binom{52-n}{13}=\\sum_{m=13}^{51}\\binom{m}{13}=\\binom{52}{14}$$\n\n\u2022 Or alternately to your second explanation, $\\spadesuit A$ is equally likely to be in each of the fourteen positions and thus has a 1/14 chance of being in the first position. \u2013\u00a0Mathieu K. Apr 23 '18 at 2:58\n\u2022 To add a simplification; you can remove the 38 non-club, non-ace of spades cards without affecting the outcome. \u2013\u00a0JollyJoker Apr 23 '18 at 7:46\n\u2022 Of course your answer is completely right, but can you elaborate on something for me? Why do you feel that the statement $P(\\spadesuit A \\text{ first } | E_n)=1/14$ is trivial enough to just write \"due to the symmetry among the 52 cards\", but not feel that the statement without conditioning on $E_n$ requires a detailed proof (using as part of the proof the claim that the conditioned statement is easy)? I may be missing something, but I don't see why one statement is particularly harder than the other. Thanks! :) \u2013\u00a0Sam T Apr 23 '18 at 20:33\n\u2022 @SamT I feel that neither statements require detailed proofs, as they are both intuitively obvious. For the first statement, the details were easy to fill in, so I included because why not. \u2013\u00a0Mike Earnest Apr 23 '18 at 22:45\n\u2022 I just feel that your middle proof basically assumes the result (since assumes the conditioned result). I have no problem with your first proof though! \u2013\u00a0Sam T Apr 24 '18 at 17:15\n\nThese sorts of questions are often solved via \"shortcut\", such as in the other answer. The reason why the shortcut works is rarely discussed.\n\nThe point of this answer is to sketch out the rigorous argument underlying the shortcut. It also demonstrates a methodology one can try to apply to more general problems, or in problems where one has identified a shortcut but still needs to check that the shortcut should give the right answer.\n\nWe can describe a choice of how to order a deck of cards in the following way:\n\n\u2022 Choose 14 out of 52 places\n\u2022 Choose an arbitrary ordering of the 14 cards consisting of the 13 clubs and the ace of spades\n\u2022 Choose an arbitrary ordering of the remaining 38 cards\n\nThe ordering this describes is given by placing the 13 clubs and the ace of spades into the chosen places, in the chosen order, and the remaining 38 cards in the remaining places, in the chosen order.\n\nThe important thing for this to be a good description is that it has the following properties:\n\n\u2022 Every ordering of a deck of cards can be described in this fashion\n\u2022 Every such description determines a unique ordering of the deck\n\nSo, we can determine probabilities simply by counting.\n\nThe reason for choosing this description is that:\n\n\u2022 the three choices are completely independent from one another\n\u2022 the problem depends only on the second choice: how to order the 13 clubs and the ace of spades\n\nSo, we can (rigorously!) reduce the original problem consisting of a whole deck of cards to the simpler problem consisting of just these 14 cards.\n\nBy a similar analysis, we can describe choices of how to order these 14 cards by:\n\n\u2022 Choose 1 place\n\u2022 Choose an arbitrary ordering of the 13 clubs\n\nThe ordering so described puts the ace of spades in the chosen place and the clubs in the remaining places, in the chosen order.\n\nAgain, this is a good description, the choices are independent, and only the first one matters. So we've reduced the original problem to:\n\nWhat are the odds of a chosen place among 14 cards is the first?\n\nwhich is easy to answer: $1/14$.\n\n\u2022 How is @MikeEarnest's answer less rigorous than this one? Both entail identifying 14 positions, but whereas in Mike's answer they are by definition those occupied by the 14 salient cards, your answer says \"Choose 14 out of 52 places\" with no prior motivation. \u2013\u00a0Rosie F Apr 23 '18 at 6:57\n\u2022 @RosieF: What's missing is the explanation why the restriction gives the right answers. There are a number of paradoxes that arise because people think that's an intrinsically valid line of reasoning. For example, the famous Monty Hall paradox where the invalid argument selects the subset \"the two doors that weren't opened\" and mistakenly believes that both choices have equal probability of goat, or the \"at least one child is a girl\" paradox where the invalid argument chooses a girl and mistakenly believes the choices of \"boy, girl\" for the other child have equal probability. \u2013\u00a0user14972 Apr 23 '18 at 8:45\n\u2022 @Hurkyl both of your example mistakes are due to belief that the reveal of information changes the probabilities that existed before the first choice was made. That does not apply to Mike's answer. \u2013\u00a0Stop Harming Monica Apr 23 '18 at 9:44\n\nThink of the probability that a club has not yet been drawn on draw $d$ as $$\\sum _{n=1}^d \\left(1-\\frac{13}{52-d}\\right)$$\n\nAnd the probability that the ace of spades is drawn on draw $d$ as $$\\frac{1}{52-d}$$\n\nThen the probability that the ace of spades is drawn on draw $d$ given that a club has not yet been drawn is $$\\left(\\sum _{n=1}^d \\left(1-\\frac{13}{52-d}\\right)\\right)*\\left(\\frac{1}{52-d}\\right)$$\n\nDoes that help?\n\n\u2022 And then sum the final expression over all possible values of $d$? I don't think that's going to help. Regardless, your first expression gives probabilities higher than 1 for various values of $d$ (even if the denominator is changed to $52-n$). I think you want a product instead. \u2013\u00a0Teepeemm Apr 23 '18 at 14:45", "date": "2019-11-16 22:12:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2749422/probability-of-choosing-ace-of-spades-before-any-club/2749614", "openwebmath_score": 0.8358902335166931, "openwebmath_perplexity": 281.36589849311514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631671237733, "lm_q2_score": 0.879146761176671, "lm_q1q2_score": 0.8671579837208288}}
{"url": "http://cistirnaperi-hornipocernice.cz/masterchef-claudia-uquraex/exterior-angle-theorem-examples-0f054c", "text": "The exterior angles are these same four: \u2220 1 \u2220 2 \u2220 7 \u2220 8; This time, we can use the Alternate Exterior Angles Theorem to state that the alternate exterior angles are congruent: \u2220 1 \u2245 \u2220 8 \u2220 2 \u2245 \u2220 7; Converse of the Alternate Exterior Angles Theorem. An exterior angle of a triangle.is formed when one side of a triangle is extended The Exterior Angle Theorem says that: the measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles. Exterior Angle of Triangle Examples In this first example, we use the Exterior Angle Theorem to add together two remote interior angles and thereby find the unknown Exterior Angle. Exterior Angle Theorem At each vertex of a triangle, the angle formed by one side and an extension of the other side is called an exterior angle of the triangle. A related theorem. The theorem states that same-side exterior angles are supplementary, meaning that they have a sum of 180 degrees. The theorem states that the same-side interior angles must be supplementary given the lines intersected by the transversal line are parallel. The sum of all angles of a triangle is $$180^{\\circ}$$ because one exterior angle of the triangle is equal to the sum of opposite interior angles of the triangle. Thus. In other words, the sum of each interior angle and its adjacent exterior angle is equal to 180 degrees (straight line). Polygon Exterior Angle Sum Theorem If a polygon is convex, then the sum of the measures of the exterior angles, one at each vertex, is 360 \u00b0 . Well that exterior angle is 90. FAQ. How to define the interior and exterior angles of a triangle, How to solve problems related to the exterior angle theorem using Algebra, examples and step by step solutions, Grade 9 Related Topics: More Lessons for Geometry Math Looking at our B O L D M A T H figure again, and thinking of the Corresponding Angles Theorem, if you know that a n g l e 1 measures 123 \u00b0, what other angle must have the same measure? Use alternate exterior angle theorem to prove that line 1 and 2 are parallel lines. how to find the unknown exterior angle of a triangle. Exterior Angle TheoremAt each vertex of a triangle, the angle formed by one side and an extension of the other side is called an exterior angle of the triangle. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. We can see that angles 1 and 7 are same-side exterior. What is the polygon angle sum theorem? You can use the Corresponding Angles Theorem even without a drawing. By corresponding angles theorem, angles on the transversal line are corresponding angles which are equal. They are found on the outer side of two parallel lines but on opposite side of the transversal. The exterior angle dis greater than angle a, or angle b. Thus exterior \u2220 110 degrees is equal to alternate exterior i.e. Although you know that sum of the exterior angles is 360, you can only use formula to find a single exterior angle if the polygon is regular! Exterior Angle Theorem The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles of the triangle. Subtracting from both sides, . For each exterior angle of a triangle, the remote interior angles are the interior angles that are not adjacent to that exterior angle. We know that in a triangle, the sum of all three interior angles is always equal to 180 degrees. Then either \u22201 is an exterior angle of 4ABRand \u22202 is an interior angle opposite to it, or vise versa. Theorem 5.2 Exterior Angle Theorem The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles. Using the Exterior Angle Theorem 145 = 80 + x x= 65 Now, if you forget the Exterior Angle Theorem, you can still get the answer by noticing that a straight angle has been formed at the vertex of the 145\u00ba angle. So, the measures of the three exterior angles are , and . Similarly, this property holds true for exterior angles as well. The following practice questions ask you to do just that, and then to apply some algebra, along with the properties of an exterior angle\u2026 The sum of the measures of the exterior angles is the difference between the sum of measures of the linear pairs and the sum of measures of the interior angles. See Example 2. Angles a, b, and c are interior angles. So, \u2026 It is clear from the figure that y is an interior angle and x is an exterior angle. An exterior angle is the angle made between the outside of one side of a shape and a line that extends from the next side of the shape. Calculate values of\u00a0x\u00a0and\u00a0y\u00a0in the following triangle. History. That exterior angle is 90. So, in the picture, the size of angle ACD equals the \u2026 So, we have; Therefore, the values of x and y are 140\u00b0 and 40\u00b0 respectively. Thus, (2x \u2013 14)\u00b0 = (x + 4)\u00b0 2x \u2013x = 14 + 4 x = 18\u00b0 Now, substituting the value of x in both the exterior angles expression we get, (2x \u2013 14)\u00b0 = 2 x 18 \u2013 14 = 22\u00b0 (x + 4)\u00b0= 18\u00b0 + 4 = 22\u00b0 Apply the Triangle exterior angle theorem: \u21d2 (3x \u2212 10) = (25) + (x + 15) \u21d2 (3x \u2212 10) = (25) + (x +15) \u21d2 3x \u221210 = \u2026 Example 2 Find . In the illustration above, the interior angles of triangle ABC are a, b, c and the exterior angles are d, e and f. Adjacent interior and exterior angles are supplementary angles. Let's try two example problems. Theorem 4-4 The measure of each angle of an equiangular triangle is 60 . The Exterior Angle Theorem says that if you add the measures of the two remote interior angles, you get the measure of the exterior angle. Interior and Exterior Angles Examples. Alternate angles are non-adjacent and pair angles that lie on the opposite sides of the transversal. 110 +x = 180. All exterior angles of a triangle add up to 360\u00b0. Oct 30, 2013 - These Geometry Worksheets are perfect for learning and practicing various types problems about triangles. Try the free Mathway calculator and Theorem Consider a triangle ABC.Let the angle bisector of angle A intersect side BC at a point D between B and C.The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment CD is equal to the ratio of \u2026 Apply the Triangle exterior angle theorem: Substitute the value of x into the three equations. The high school exterior angle theorem (HSEAT) says that the size of an exterior angle at a vertex of a triangle equals the sum of the sizes of the interior angles at the other two vertices of the triangle (remote interior angles). For each exterior angle of a triangle, the remote interior angles are the interior angles that are not adjacent to that exterior angle. Example 2. E 95 \u00b0 6) U S J 110 \u00b0 80 \u00b0 ? Alternate Exterior Angles \u2013 Explanation & Examples In Geometry, there is a special kind of angles known as alternate angles. Example: The exterior angle is \u2026 Hence, it is proved that m\u2220A + m\u2220B = m\u2220ACD Solved Examples Take a look at the solved examples given below to understand the concept of the exterior angles and the exterior angle theorem. So once again, 90 plus 90 plus 90 plus 90 that's 360 degrees. 2) Corresponding Exterior Angle: Found at the outer side of the intersection between the parallel lines and the transversal. problem solver below to practice various math topics. The Exterior Angle Theorem Date_____ Period____ Find the measure of each angle indicated. Corresponding Angels Theorem The postulate for the corresponding angles states that: If a transversal intersects two parallel lines, the corresponding angles \u2026 4.2 Exterior angle theorem The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles. Next, calculate the exterior angle. Illustrated definition of Exterior Angle Theorem: An exterior angle of a triangle is equal to the sum of the two opposite interior angles. Even though we know that all the exterior angles add up to 360 \u00b0, we can see, by just looking, that each $$\\angle A \\text{ and } and \\angle B$$ are not congruent.. Corresponding Angles Examples. In this article, we are going to discuss alternate exterior angles and their theorem. \u00a5 Note that the converse of Theorem 2 holds in Euclidean geometry but fails in hyperbolic geometry. Find the value of x if the opposite non-adjacent interior angles are (4x + 40) \u00b0 and 60\u00b0. measures less than 62/87,21 By the Exterior Angle Inequality Theorem, the exterior angle ( ) is larger than either remote interior angle ( and Also, , and . Making a semi-circle, the total area of angle measures 180 degrees. X= 70 degrees. Learn how to use the Exterior Angle Theorem in this free math video tutorial by Mario's Math Tutoring. Example 1. The Exterior Angle Theorem states that An exterior angle of a triangle is equal to the sum of the two opposite interior angles. By the Exterior Angle Inequality Theorem, measures greater than m 7 62/87,21 By the Exterior Angle Inequality Theorem, the exterior angle (5) is larger than either remote interior angle (7 and 8). Exterior angles of a polygon are formed with its one side and by extending its adjacent side at the vertex. And (keeping the end points fixed) ..... the angle a\u00b0 is always the same, no matter where it is on the same arc between end points: The converse of the Alternate Exterior Angles Theorem \u2026 So it's a good thing to know that the sum of the exterior angles of any polygon is actually 360 degrees. So, we all know that a triangle is a 3-sided figure with three interior angles. Therefore, the angles are 25\u00b0, 40\u00b0 and 65\u00b0. Let\u2019s take a look at a few example problems. The measure of an exterior angle (our w) of a triangle equals to the sum of the measures of the two remote interior angles (our x and y) of the triangle. By substitution, . The exterior angle of a triangle is 120\u00b0. Scroll down the page for more examples and solutions using the exterior angle theorem to solve problems. Given that for a triangle, the two interior angles 25\u00b0 and (x + 15) \u00b0 are non-adjacent to an exterior angle (3x \u2013 10) \u00b0, find the value of x. Hence, the value of x and y are 88\u00b0 and 47\u00b0 respectively. Use the Exterior Angle Inequality Theorem to list all of the angles that satisfy the stated condition. Learn in detail angle sum theorem for exterior angles and solved examples. Unit 2 Vocabulary and Theorems Week 4 Term/Postulate/Theorem Definition/Meaning Image or Example Exterior Angles of a Triangle When the sides of a triangle are extended, the angles that are adjacent to the interior angles. Before getting into this topic, [\u2026] The exterior angle of a triangle is the angle formed between one side of a triangle and the extension of its adjacent side. Remember that the two non-adjacent interior angles, which are opposite the exterior angle are sometimes referred to as remote interior angles. Corresponding Angels Theorem The postulate for the corresponding angles states that: If a transversal intersects two parallel \u2026 We welcome your feedback, comments and questions about this site or page. Rules to find the exterior angles of a triangle are pretty similar to the rules to find the interior angles of a triangle. Therefore, m 7 < m 5 and m 8 < m \\$16:(5 7, 8 measures less \u2026 Theorem 4-3 The acute angles of a right triangle are complementary. The angle bisector theorem appears as Proposition 3 of Book VI in Euclid's Elements. Using the Exterior Angle Theorem, . A and C are \"end points\" B is the \"apex point\" Play with it here: When you move point \"B\", what happens to the angle? Example 1 Solve for x. The Exterior Angle Theorem Students learn the exterior angle theorem, which states that the exterior angle of a triangle is equal to the sum of the measures of the remote interior angles. If two of the exterior angles are and , then the third Exterior Angle must be since . Find the values of x and y in the following triangle. According to the exterior angle theorem, alternate exterior angles are equal when the transversal crosses two parallel lines. Theorem 3. Explore Exterior Angles. The third interior angle is not given to us, but we could figure it out using the Triangle Sum Theorem. This is the simplest type of Exterior Angles maths question. Theorem 1. Solution: Using the Exterior Angle Theorem 145 = 80 + x x = 65 Now, if you forget the Exterior Angle Theorem, you can still get the answer by noticing that a straight angle has been formed at the vertex of the 145\u00ba angle. Remember that every interior angle forms a linear pair (adds up to ) with an exterior angle.) In either case m\u22201 6= m\u22202 by the Exterior Angle Inequality (Theorem 1). Example 1 Find the So once again, 90 plus 90 plus 90 plus 90 that's 360 degrees. ... exterior angle theorem calculator: sum of all exterior angles of a polygon: formula for exterior angles of a polygon: T 30 \u00b0 7) G T E 28 \u00b0 58 \u00b0? The sum of exterior angle and interior angle is equal to 180 degrees. For this example we will look at a hexagon that has six sides. Example 3 Find the value of and the measure of each angle. Therefore; \u21d2 4x \u2013 19 = 3x + 16 \u21d2 4x \u2013 3x 0 The exterior angle theorem tells us that the measure of angle D is equal to the sum of angles A and B.In formula form: m\nForm 3520 Initial Return, First Horizon Bank Zelle, What Is A Virtual Field Trip, Minecraft Roleplay Maps High School, Who Is The Man In The Flash Advert 2020?, Notice Of Articles Vs Articles Of Incorporation, Step Up 2 Trailer, Shoes Similar To Altra Torin, Merrell Chameleon 7 Limit Review, Duke Virtual Information Session, Best 10-inch Sliding Miter Saw, 2011 Nissan Sentra Service Engine Soon Light,", "date": "2021-11-27 08:46:35", "meta": {"domain": "cistirnaperi-hornipocernice.cz", "url": "http://cistirnaperi-hornipocernice.cz/masterchef-claudia-uquraex/exterior-angle-theorem-examples-0f054c", "openwebmath_score": 0.5718120336532593, "openwebmath_perplexity": 292.87919495778965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401461512076, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8671518599076242}}
{"url": "https://mathoverflow.net/questions/398094/minimiser-of-a-certain-functional", "text": "# Minimiser of a certain functional\n\nLet $$f_i \\in L^1 ([0, 1])$$ be a sequence of functions equibounded in $$L^1$$ norm - that is, there exists some $$M > 0$$ such that $$\\|f_i\\|_{L^1} < M$$.\n\nDefine the functional $$F: L^1([0, 1]) \\to \\mathbb R$$ by\n\n$$F(h) = \\limsup_{i \\to \\infty} \\|f_i - h\\|_{L^1}.$$\n\nQuestion: Does this functional admit a minimiser? Is the minimiser unique whenever it exists?\n\nRemarks:\n\nWhat I have tried so far is to attempt to apply the direct method of the calculus of variations.\n\nSince the $$f_i$$ are equibounded in $$L^1$$, it can be shown that $$F$$ is coercive, thus any minimising sequence is bounded in $$L^1$$ norm. In particular we have a weakly-* converging subsequence, say $$h_n \\overset{*}{\\to} h$$.\n\nThe result would follow if we had weak-* sequential lower semi continuity of $$F$$ at the minimiser - that is, that\n\n$$\\liminf_{n \\to \\infty} F(h_n) \\geq F(h).$$\n\nI could neither disprove this with a counterexample, nor prove it in generality.\n\nEdit: As pointed out in the comments, weak-$$*$$ convergence to an $$L^1$$ function isn\u2019t guaranteed, only convergence to a measure.\n\n\u2022 A minimizer is not unique in general. E.g., let $f_i=(-1)^i$. Then any $h$ with $|h|\\le1$ and $\\int h=0$ is a minimizer. Jul 22, 2021 at 15:03\n\u2022 $L^1(0,1)$ is not a dual space. How do you define weak-* convergence?\n\u2013\u00a0gerw\nJul 23, 2021 at 18:15\n\u2022 Oh you\u2019re right, it needs to be defined in the sense of measures.. but then there is trouble in making sense of $F(\\mu)$ for a measure $\\mu$. Hmm... Jul 24, 2021 at 4:58\n\nAs it has been already noted in the comments, the minimizer doesn't need to be unique. However, it always exists. It is not terribly hard to show but it is not trivial either, so I wonder why the question attracted so few votes.\n\nThe proof consists of two independent parts. The first one is that the limit of every minimizing sequence that converges almost everywhere (or just in measure) is a minimizer and the second one is that there exists a minimizing sequence converging almost everywhere. I will use the fact that we deal with a finite measure space though it should, probably, be irrelevant. WLOG, we may assume that $$\\|f_j\\|_1\\le 1$$ for all $$j$$.\n\nPart 1\n\nAssume that $$h_n$$ is a minimizing sequence and $$h$$ is its pointwise limit (or just limit in measure). Since we can assume WLOG that $$\\|h_n\\|_1\\le 3$$ (to be a competitor, you need to perform not much worse than $$0$$, at the very least), we have $$\\|h\\|_1\\le 3$$ as well (by Fatou). Then, by the definition of the convergence in measure, we can write $$h_n=u_n+v_n$$ where $$u_n$$ converge to $$h$$ uniformly and $$v_n$$ are supported on $$E_n$$ with $$m(E_n)\\to 0$$ as $$n\\to\\infty$$. Also $$\\|v_n\\|_1\\le 7$$, say.\n\nNow, since $$v_n\\in L^1$$, we can find $$\\delta_n>0$$ such that for every set $$E$$ with $$m(E)<\\delta_n$$, we have $$\\int_E|v_n|<\\frac 1n$$, say. By induction, we can now choose a subsequence $$n_k$$ such that $$\\sum_{q=k+1}^\\infty m(E_{n_q})<\\delta_{n_k}\\,.$$ Then $$\\|v_{n_k}\\chi_{\\cup_{q>k}E_{n_q}}\\|_1\\to 0$$ and, adding these parts to $$u_{n_k}$$, we see that we can (passing to a subsequence) assume that our minimizing sequence can be represented as $$h_n=U_n+V_n$$ where $$U_n\\to h$$ in $$L^1$$ and $$V_n$$ have disjoint supports $$G_n$$.\n\nSince correcting a minimizing sequence by a sequence tending to $$0$$ in $$L^1$$ results in a minimizing sequence, we can just as well assume that $$h_n=h+V_n$$.\n\nIf $$\\|V_n\\|_1\\to 0$$ (even along a subsequence), we are done. Assume now that $$\\|V_n\\|_1\\ge \\tau>0$$. Then for every function $$g$$ of $$L^1$$-norm less than $$4$$, we have $$\\|g\\|_1\\le \\max(\\|g-V_n\\|_1,\\dots,\\|g-V_{n+N-1}\\|_1)$$ for all $$n$$ as soon as $$N>8/\\tau$$. Indeed, since the supports $$G_n$$ of $$V_n$$ are disjoint, there is $$q\\in\\{0,\\dots,N-1)$$ such that $$\\int_{G_{n+q}}|g|\\le \\frac 1N\\|g\\|_1< \\frac\\tau 2$$, in which case subtracting $$V_{n+q}$$ can only drive the norm up.\n\nApplying this to the functions $$g_j=f_j-h$$, we conclude that $$\\limsup_{j\\to\\infty}\\|f_j-h\\|_1\\le \\limsup_{j\\to\\infty}\\max_{0\\le q\\le N-1}\\|f_j-h_{n+q}\\|_1= \\max_{0\\le q\\le N-1}\\limsup_{j\\to\\infty}\\|f_j-h_{n+q}\\|_1$$ for every $$n$$, i.e. $$h$$ is a minimizer in this case as well.\n\nPart 2\n\nLet $$I$$ be the infimum of our functional. For every $$\\varepsilon>0$$ consider the set $$X_\\varepsilon$$ of all $$L^1$$-functions $$h$$ for which the value of the functional is at most $$I+\\varepsilon$$. Clearly, it is a convex, non-empty, closed (in $$L^1$$) set and $$X_{\\varepsilon'}\\supset X_{\\varepsilon''}$$ when $$\\varepsilon'\\ge\\varepsilon''$$.\n\nFix some strictly convex non-negative function $$\\Phi(t)\\le |t|$$. To simplify the technicalities, I'll choose it by the conditions $$\\Phi(0)=\\Phi'(0)=1$$, $$\\Phi''(t)=\\frac 2{\\pi(1+|t|^2)}$$ but pretty much any other choice will work as well.\n\nLet $$J_\\varepsilon=\\inf_{h\\in X_\\varepsilon}\\int\\Phi(h)\\,.$$ Clearly, $$J_\\varepsilon$$ is a bounded non-increasing function on $$(0,1)$$, say. Let $$J$$ be its limit at $$0+$$. Passing to an appropriate decreasing sequence $$\\varepsilon_n\\to 0+$$ and re-enumerating $$X_n=X_{\\varepsilon_n}, J_n=J_{\\varepsilon_n}$$, we can assume that $$J_n\\ge J-2^{-5n}$$ We will choose a representative $$h_n$$ of $$X_n$$ for which $$\\int\\Phi(h_n)$$ is not more than $$2^{-5n}$$ above $$J$$ and, thereby, not more than $$2\\cdot 2^{-5n}$$ above its infimum over $$X_n$$. Then for $$m>n$$, we have $$h_{n,m}=\\frac{h_n+h_m}2\\in X_n$$ (convexity of $$X_n$$ plus inclusion $$X_n\\supset X_m$$) and $$\\int\\Phi(h_{n,m})\\le \\int\\frac 12(\\Phi(h_n)+\\Phi(h_m))-\\int\\frac{(h_n-h_m)^2}{4\\pi (1+|h_n|^{2}+|h_m|^{2})}$$ (second order Taylor with the remainder in the Lagrange form), whence $$\\int\\frac{(h_n-h_m)^2}{4\\pi(1+|h_n|^2+|h_m|^2)}\\le 2\\cdot 2^{-5n}$$ regardless of $$m>n$$ (otherwise we would go below $$J+2^{-5n}-2\\cdot 2^{-5n}=J-2^{-5n}$$, which is below the infimum over $$X_n$$). It remains to note that if $$|h_n|\\le 2^n$$ and $$|h_n-h_m|>2^{-n}$$, then the integrand is at least $$\\rm{const}\\, 2^{-4n}$$, so we get $$m(\\{|h_n|\\le 2^n, |h_n-h_m|>2^{-n}\\})\\le \\rm{Const}\\,2^{-n}$$ for all $$m>n$$ from where it follows at once that $$h_n(x)$$ is a Cauchy sequence for almost all $$x$$ (recall that $$\\|h_n\\|_1\\le 3$$, so the first condition excludes only a set of measure $$3\\cdot 2^{-n}$$, then use Borel-Cantelli).\n\nThat's it. Feel free to ask questions if anything is unclear :-)\n\n\u2022 Wow, this looks impressive. This part was a bit confusing to me - \u201c Then $\\|v_{n_k}\\chi_{\\cup_{q>k}E_{n_q}}\\|_1\\to 0$ and, adding these parts to $u_{n_k}$, we see that we can (passing to a subsequence) assume that our minimizing sequence can be represented as $h_n=U_n+V_n$ where $U_n\\to h$ in $L^1$ and $V_n$ have disjoint supports $G_n$.\u201d Do you mean to add $v_{n_k}\\chi_{\\cup_{q>k}E_{n_q}}$ to $u_{n_k}$? I am not sure how to get the desired representation from this. Aug 2, 2021 at 4:04\n\u2022 @NateRiver Exactly as you said: $U_k=u_{n_k}+v_{n_k}\\chi_{\\cup_{q>k}E_{n_q}}$, $V_k=v_{n_k}\\chi_{E_{n_k}\\setminus \\cup_{q>k}E_{n_q}}=v_{n_k}\\chi_{G_k}$. Aug 2, 2021 at 7:45\n\u2022 @NateRiver I just tried to solve the problem in $L^2$ first. There weak convergence is guaranteed, but doesn't seem to drop the value of the functional immediately, so I needed the norm convergence of a minimizing sequence. Fortunately, if one has a nested sequence of bounded closed convex sets in a Hilbert space, the smallest norm elements of the sets converge in norm (the proof is the same as above just using the parallelogram identity). I tried to mimic that idea in $L^1$ (which required strict convexity of something) and got the a.e. convergence this way, which turned out to be sufficient. Aug 2, 2021 at 13:33\n\u2022 An alternative proof for Part 2 (existence of a minimizing sequence converging a.e.) Let $h_n$ be any minimizing sequence for $F$. As observed, it is bounded in $L_1$, so by the Koml\u00f3s Theorem, up to extracting a subsequence, it is a.e. converging in Cesaro sense to some $h\\in L^1$. Since $F$ is a convex functional, the sequence of the Cesaro means is still a minimizing sequence. Aug 2, 2021 at 13:38\n\u2022 Of course, and I wouldn't be surprised if you answer (which I'm still reading) contains as a byproduct an alternative proof of that Koml\u00f3s theorem (which is a 10 page paper: link.springer.com/article/10.1007%2FBF02020976 ). Aug 2, 2021 at 13:51", "date": "2022-07-01 14:43:14", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/398094/minimiser-of-a-certain-functional", "openwebmath_score": 0.9673136472702026, "openwebmath_perplexity": 139.40968235699728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401464421569, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8671518569523797}}
{"url": "https://math.stackexchange.com/questions/3072171/how-to-properly-represent-a-matrix-function", "text": "# How to properly represent a matrix function.\n\nGiven the function $$f_{h}(x,y,z)=(x-z,y+hz,x+y+3z)$$, what is the correct way to represent the matrix function in respect to the standard basis?\n\nWith the representation theorem, I would write the matrix in columns as: $$F_{h|S_3}=(f_h(e_1)|{S_3} \\quad f_h(e_2)|{S_3} \\quad f_h(e_3)|{S_3})=\\begin{bmatrix}1 & 0 & 1\\\\ 0 & 1 & 1\\\\ -1 & h & 3\\end{bmatrix}$$ But in my textbook it is written in rows as: $$F_{h|S_3}=(f_h(e_1)|{S_3} \\quad f_h(e_2)|{S_3} \\quad f_h(e_3)|{S_3})=\\begin{bmatrix}1 & 0 & -1\\\\ 0 & 1 & h\\\\ 1 & 1 & 3\\end{bmatrix}$$\n\nWhat is the difference between them?\n\nThe correct way to represent the function $$f_h$$ in matrix form depends on the convention that you want to use to represent it. Let $$(x,y,z) \\in \\mathbb{R}^3$$. The matrix which you computed is useful for expressing $$f_h$$ as $$f_h(x,y,z) = \\begin{bmatrix} x& y & z \\end{bmatrix} \\begin{bmatrix} 1& 0 & 1 \\\\ 0 & 1& 1\\\\ -1 & h & 3 \\end{bmatrix}.$$ We can verify this by expanding the matrix product: \\begin{align} \\begin{bmatrix} x& y & z \\end{bmatrix} \\begin{bmatrix} 1& 0 & 2 \\\\ 0 & 1& 1\\\\ -1 & h & 3 \\end{bmatrix} &= \\begin{bmatrix} x \\cdot 1 + y \\cdot 0 + z \\cdot (-1) \\\\ x \\cdot 0 + y \\cdot 1 + z \\cdot h \\\\ x \\cdot 1 + y \\cdot 1 + z \\cdot 3 \\end{bmatrix} \\\\ &= \\begin{bmatrix} x - z & y + hz & x + y + 3z \\end{bmatrix}. \\end{align} The result is a $$1 \\times 3$$ matrix, which we can interpret as a row vector in $$\\mathbb{R}^3$$. The matrix written in your textbook is useful for the following representation: $$f_h(x,y,z) = \\begin{bmatrix} 1& 0 & -1 \\\\ 0 & 1& h\\\\ 1 & 1 & 3 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\begin{bmatrix} x-z \\\\ y + hz \\\\ x + y + 3z \\end{bmatrix}.$$ The result is a $$3 \\times 1$$ matrix, which we can regard as a column vector in $$\\mathbb{R}^3$$.\nThe two representations look different in the sense that the former yields a row vector and the latter a column vector, however they are the same in the sense that they can both be regarded as lists of three real numbers, i.e. as $$(x-z,y+hz,x+y+3z) \\in \\mathbb{R}^3.$$ The accepted answer here sums this idea up pretty well.", "date": "2021-01-16 00:06:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3072171/how-to-properly-represent-a-matrix-function", "openwebmath_score": 1.000005841255188, "openwebmath_perplexity": 88.85358235835388, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401423688665, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8671518469649366}}
{"url": "https://math.stackexchange.com/questions/3056293/representation-of-function-as-power-series-unique", "text": "# Representation of function as power series - unique?\n\nThe following example is from Calculus, 7e by James Stewart:\n\nExample 2, Chapter 11.9 (Representations of functions as Power Series)\n\nFind a power series representation for $$\\frac{1}{x+2}$$\n\nMy solution is:\n\n$$\\frac{1}{x+2}=\\frac{1}{1-\\left(-x-1\\right)}=\\sum_{n=0}^{\\infty}\\left(-x-1\\right)^{n}=\\sum_{n=0}^{\\infty}\\left(-1\\right)^{n}\\left(x+1\\right)^{n}$$\n\nThen to find the interval of convergence\n\n$$\\left|x+1\\right|<1 \\Rightarrow x\\in\\left(-2,0\\right)$$\n\nBut the given solution is different:\n\n\u201cIn order to put this function in the form of the left side of Equation 1, $$\\frac{1}{1-x}$$ , we first factor a 2 from the denominator:\n\n$$\\frac{1}{2+x}=\\frac{1}{2\\left(1+\\frac{x}{2}\\right)}=\\frac{1}{2\\left[1-\\left(-\\frac{x}{2}\\right)\\right]}=\\frac{1}{2}\\sum_{n=0}^{\\infty}\\left(-\\frac{x}{2}\\right)^{n}=\\sum_{n=0}^{\\infty}\\frac{\\left(-1\\right)^{n}}{2^{n+1}}x^{n}$$\n\nThis series converges when $$\\left|-\\frac{x}{2}\\right|<1$$, that is, $$\\left|x\\right|<2$$. So the interval of convergence is $$\\left(-2,2\\right)$$.\"\n\nSo my questions are:\n\n1. Did I make an error somewhere?\n2. If not, are the two representations equivalent? Can there be more than one representation of a function as a power series?\n\u2022 The representation as a power series at a given point is unique. Here, yours is at the point $-1$, while the solution gives the one at the point $0$. (By default, power series are usually taken at 0 when nothing is specified or obvious from context) \u2013\u00a0Clement C. Dec 29 '18 at 21:59\n\nI think your method is right but you made a power series around $$x_0=-1$$\nwhile they did it around $$x_0=0$$ which is what they asked for probably\nBoth solutions are correct: your series is a power series centered at $$-1$$, whereas the power series from the given solution is centered at $$0$$.", "date": "2020-01-19 05:01:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3056293/representation-of-function-as-power-series-unique", "openwebmath_score": 0.9587030410766602, "openwebmath_perplexity": 253.26119561378255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9585377261041521, "lm_q2_score": 0.9046505267461572, "lm_q1q2_score": 0.867141658826185}}
{"url": "https://byjus.com/question-answer/let-displaystyle-i-1-int-sec-2-z-2-tan-2-z-xf-left-x/", "text": "Question\n\n# Let $$\\displaystyle { I }_{ 1 }=\\int _{ \\sec ^{ 2 }{ z } }^{ 2-\\tan ^{ 2 }{ z } }{ xf\\left( x\\left( 3-x \\right) \\right) dx }$$ and $$\\displaystyle { I }_{ 2 }=\\int _{ \\sec ^{ 2 }{ z } }^{ 2-\\tan ^{ 2 }{ z } }{ f\\left( x\\left( 3-x \\right) \\right) dx }$$, where $$f$$ is a continuous function and $$z$$ is any real number, $$\\displaystyle \\frac { { I }_{ 1 } }{ { I }_{ 2 } } =$$\n\nA\n32\nB\n12\nC\n1\nD\nnone of these\n\nSolution\n\n## The correct option is B $$\\displaystyle \\frac { 3 }{ 2 }$$We have, $$\\displaystyle { I }_{ 1 }=\\int _{ \\sec ^{ 2 }{ z } }^{ 2-\\tan ^{ 2 }{ z } }{ xf\\left( x\\left( 3-x \\right) \\right) dx }$$$$\\displaystyle =\\int _{ \\sec ^{ 2 }{ z } }^{ 2-\\tan ^{ 2 }{ z } }{ \\left( 3-x \\right) f\\left( \\left( 3-x \\right) \\left( 3-\\left( 3-x \\right) \\right) \\right) } dx$$ \u00a0 $$\\displaystyle \\left[ \\because \\int _{ a }^{ b }{ f\\left( x \\right)dx } =\\int _{ a }^{ b }{ f\\left( a+b-x \\right) dx } \\right]$$$$\\displaystyle =\\int _{ \\sec ^{ 2 }{ z } }^{ 2-\\tan ^{ 2 }{ z } }{ \\left( 3-x \\right) f\\left( x\\left( 3-x \\right) \\right) dx }$$$$\\displaystyle =3\\int _{ \\sec ^{ 2 }{ z } }^{ 2-\\tan ^{ 2 }{ z } }{ f\\left( x\\left( 3-x \\right) \\right) dx } -\\int _{ \\sec ^{ 2 }{ z } }^{ 2-\\tan ^{ 2 }{ z } }{ xf\\left( x\\left( 3-x \\right) \\right) } dx$$$$=3{ I }_{ 2 }-{ I }_{ 1 }$$$$\\displaystyle \\therefore 2{ I }_{ 1 }=3{ I }_{ 2 }\\Rightarrow \\frac { { I }_{ 1 } }{ { I }_{ 2 } } =\\frac { 3 }{ 2 }$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-25 10:26:32", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/let-displaystyle-i-1-int-sec-2-z-2-tan-2-z-xf-left-x/", "openwebmath_score": 0.8859876990318298, "openwebmath_perplexity": 2102.811378699129, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759596153434, "lm_q2_score": 0.8840392695254318, "lm_q1q2_score": 0.8671328668334051}}
{"url": "https://math.stackexchange.com/questions/2623796/prove-that-a-smallsetminus-a-smallsetminus-b-a-cap-b/2623804", "text": "Prove that $A \\smallsetminus (A \\smallsetminus B) = A \\cap B$\n\n$A$ and $B$ are any sets, prove that $A \\smallsetminus (A \\smallsetminus B) = A \\cap B.$ This formula makes sense when represented on a Venn diagram, but I am having trouble with proving it mathematically.\n\nI have tried letting $x$ be an element of $A$ and continue from there, but it doesn't seem to work out as a valid proof anyways.\n\nCould anyone please point me in the right direction?\n\nMany thanks.\n\n\u2022 Please show us your work: starting with $x \\in A\\setminus(A\\setminus B)$, or the other way around, starting with $x \\in (A\\cap B)$. Please don't claim you tried ABC, unless you also show us you effort using ABC. Would you like to get an answer that says only: \"Yup, that's right, start by letting $x \\in A\\setminus (A\\setminus B)$... \u2013\u00a0Namaste Jan 27 '18 at 18:42\n\nLet $x \\in A \\setminus (A \\setminus B).$\n\nThen $x$ is an element such that $x \\in A$ and $x \\notin A \\setminus B$. But if $x \\notin A\\setminus B$, with some additional work, we realize this implies that $x \\in A$ and $x \\in B$. So $x \\in A \\cap B$.\n\nVice-versa: let $x \\in A \\cap B$ so $x \\in A$ and $x \\in B$. This implies that $x \\notin A \\setminus B$. But given that $x \\in A$ and $x \\notin A \\setminus B$ this implies that $x \\in A \\setminus (A \\setminus B)$.\n\n\u2022 I would like this answer if you hadn't skipped from \"$x \\in A$ and $x \\notin A\\setminus B$, but if $x\\in A \\setminus B$\" to \"we realize that $x \\in A$ and $x\\in B$.\" I added the words (\"with some additional work\") in between them. I would like this answer more if you explained that $x \\in A$ and $x \\notin (A\\setminus B)$, then $x \\in A$ and, it is not the case that ($x \\in A$ and $x \\notin B$.) By Demorgans, we get that $(x \\in A \\land \\lnot x\\in A),$ or $(x \\in A \\land x\\in B)$. The first disjunct is always false (no element can be in a set, and not be in a set. \u2013\u00a0Namaste Jan 27 '18 at 19:13\n\u2022 ...That leaves us to conclude, that \"we realize this implies that $x\\in A$ and $x\\in B$. \u2013\u00a0Namaste Jan 27 '18 at 19:13\n\nNotice that \\begin{eqnarray*} x\\in A\\setminus(A\\setminus B) &\\Leftrightarrow& (x\\in A)\\wedge \\neg(x\\in A\\setminus B)\\\\ &\\Leftrightarrow & (x\\in A)\\wedge \\neg((x\\in A)\\wedge \\neg (x\\in B))\\\\ &\\Leftrightarrow & (x\\in A)\\wedge ((x\\notin A)\\lor(x\\in B))\\\\ &\\Leftrightarrow & ((x\\in A)\\wedge (x\\notin A))\\lor ((x\\in A)\\wedge(x\\in B))\\\\ &\\Leftrightarrow & (x\\in A)\\wedge (x\\in B)\\\\ &\\Leftrightarrow & x\\in A\\cap B. \\end{eqnarray*}\n\n\u2022 Yes, when in doubt go back to the logic \u2013\u00a0Tom Collinge Jan 27 '18 at 18:28\n\u2022 I like this, because the equivalence makes the proof bi-directional, and doesn't skip important steps between $x \\in A) \\land$\\lnot (x \\in A\\setminus B$and its equivalent,$x\\in A \\land x \\in B$Those are likely the steps on which the asker got tripped up, I suspect. Nice answer. \u2013 Namaste Jan 27 '18 at 18:51 \u2022 I like this, because the equivalence makes the proof bi-directional, and doesn't skip important steps between$x \\in A \\land \\lnot (x \\in A\\setminus B$and its equivalent,$x\\in A \\land x \\in B.$Those are likely the steps on which the asker got tripped up, I suspect. Nice answer. \u2013 Namaste Jan 27 '18 at 19:01 \u2022 @amWhy: Thanks. Indeed, often mistakes are made in these basic logical steps. I've seen many people making mistakes using the notation$x\\notin B$. I think it's safer to write$\\neg(x\\in B)$, indeed, if$B$is a more complicated expression, we still now how to negate the logical expression appearing inside the brackets. This basic logical way of thinking is user friendly and not very prone to errors. \u2013 Mathematician 42 Jan 27 '18 at 20:43 Did you know that $$A \\smallsetminus B = A \\cap \\overline{B} \\;\\;?$$$\\begin{align} A \\smallsetminus (A \\smallsetminus B) &= A \\smallsetminus (A \\cap \\overline{B}) \\\\ \\\\ &= A \\cap \\overline{ (A \\cap \\overline{B})}\\\\ \\\\ &= A \\cap (\\overline{A} \\cup B)\\\\ \\\\ &= (A \\cap \\overline{A} )\\cup (A \\cap B) \\\\ \\\\ &= A \\cap B \\end{align}$\u2022 cmon, why this get downvoted, it is correct. hmm, maybe you want to precise that$\\bar B=B^{\\complement}$and not the closure in this context. \u2013 zwim Jan 27 '18 at 18:55 \u2022 @zwim: Not sure why this got downvoted, however this proof assumes knowledge about standard set operations which you should be able to prove directly as well. You might end up running in circles and not proving anything. \u2013 Mathematician 42 Jan 27 '18 at 18:58 Let$x\\in A\\cap B$. Then$x\\in A$and$x\\in B$. Then$x\\not \\in A\\setminus B$, so$x\\in A\\setminus (A\\setminus B)$. Conversely, if$x\\in A\\setminus (A\\setminus B)$, then$x\\in A$and$x\\not \\in (A\\setminus B)$. So$x\\in A$and$x\\in B$. That is$x\\in A\\cap B\\$...\n\n\u2022 ditto of @Skills. \u2013\u00a0Namaste Jan 27 '18 at 18:36", "date": "2019-10-19 19:34:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2623796/prove-that-a-smallsetminus-a-smallsetminus-b-a-cap-b/2623804", "openwebmath_score": 0.9759236574172974, "openwebmath_perplexity": 793.8041184895973, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707979895382, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8671278704859925}}
{"url": "http://mathhelpforum.com/pre-calculus/167826-sequences-problem.html", "text": "# Math Help - sequences problem\n\n1. ## sequences problem\n\nA hardware supplier has designed a series of six plastic containers with lids where each container (after the first) can be placed into the next larger one for storage purposes. The containers are rectangular boxes, and the dimensions of the largest one are 120 cm by 60 cm by 50 cm. The dimension of each container is decreased by 10 percent with respect to the next larger one.\n\nDetermine the volume of each container, and the volume of all the containers. Write your answers in litres (one litre = 1000 cm3).\n\nIs it 10% from each dimension or from the volume? Hard to figure out from the problem.\n\nIf from the each dimension then:\nV 1= 50x 60x 120 = 360000 cm3 = 360 litres\nV2=45 x 54 x 108 = 262440 cm3 = 262.4 liltres\nV 3 = 40.5 x 48.6 x 97.2 = 191.3litres\nV4 = 36.45 x 43.74 x 87.48 =139.4 litres\nV5 = 32.81 x 39.37 x 78.74 = 101.7 litres\nV6 = 29.61 x 35.47 x 70.94 =74.5 litres\n\nI need some feedback.\n\n2. I expect since it says \"the dimension of each container is decreased by 10% with respect to the larger one\" I would say it's the dimensions and not the volume that is decreased by 10%.\n\n3. Originally Posted by terminator\nA hardware supplier has designed a series of six plastic containers with lids where each container (after the first) can be placed into the next larger one for storage purposes. The containers are rectangular boxes, and the dimensions of the largest one are 120 cm by 60 cm by 50 cm. The dimension of each container is decreased by 10 percent with respect to the next larger one.\n\nDetermine the volume of each container, and the volume of all the containers. Write your answers in litres (one litre = 1000 cm3).\n\nIs it 10% from each dimension or from the volume? Hard to figure out from the problem.\nWhy? The problems says \"The dimension of each container is decreased by 10 percent\". That tells you precisely which is intended\n\nIf from the each dimension then:\nV 1= 50x 60x 120 = 360000 cm3 = 360 litres\nV2=45 x 54 x 108 = 262440 cm3 = 262.4 liltres\nV 3 = 40.5 x 48.6 x 97.2 = 191.3litres\nV4 = 36.45 x 43.74 x 87.48 =139.4 litres\nV5 = 32.81 x 39.37 x 78.74 = 101.7 litres\nV6 = 29.61 x 35.47 x 70.94 =74.5 litres\n\nI need some feedback.\n\n4. Hello, terminator!\n\nA hardware supplier has designed a series of six plastic containers with lids\n. . where each container (after the first) can be placed into the next larger one.\n\nThe containers are rectangular boxes.\n. . The dimensions of the largest one are 120 cm by 60 cm by 50 cm.\n\nThe dimensions of each container are decreased by 10 percent\n. . with respect to the next larger one.\n\n(a) Determine the volume of each container.\n(b) Determine the volume of all the containers.\nWrite your answers in litres (one litre = 1000 cubic cm).\n\nThe original box has dimensions $L,\\,W,\\,H$\n. . Its volume is: . $L\\!\\cdot\\!W\\!\\cdot\\!H\\text{ cm}^3$\n\nThe next box has dimensions: $0.9L,\\,0.9W,\\,0.9H$\n. . Its volume is: . $(0.9L)(0.9W)(0.9H) \\:=\\:0.729(LWH)$\n\nThat is, each box is 0.729 of the volime of the next larger box.\n\nThe first box has volume: . $120\\cdot60\\cdot50 \\:=\\:360,\\!000\\text{ cm}^3$\n\n. . That is: . $V_1 \\:=\\:360\\text{ liters.}$\n\n(a) Therefore, the $\\,n^{th}$ box has volume: . $V_n \\;=\\;360(0.729)^{n-1}$\n\nThe sum of the volumes of all the boxes is:\n\n. . $S \\;=\\;360 + 360(0.729) + 360(0.729^2) + 369(0.729^3) + \\hdots$\n\n. . $S \\;=\\;360\\underbrace{\\bigg[1 + 0.729 + 0.729^2 + 0.729^3 + \\hdots\\bigg]}_{\\text{a geometric series}}$\n\nThe sum of the geometric series is: . $\\dfrac{1}{1 - 0.729} \\:=\\:\\dfrac{1}{0.271}$\n\n(b) Therefore: . $S \\;=\\;360\\left(\\frac{1}{0.271}\\right) \\;\\approx\\;1328.4\\text{ liters}$", "date": "2015-05-30 19:32:29", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/167826-sequences-problem.html", "openwebmath_score": 0.5734283328056335, "openwebmath_perplexity": 1405.177013626631, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9790357604052423, "lm_q2_score": 0.8856314662716159, "lm_q1q2_score": 0.8670648760200412}}
{"url": "https://www.themathdoctors.org/the-gamblers-fallacy/", "text": "The Gambler\u2019s Fallacy\n\nProbability seems simple enough to many people that it can fool them into wrong conclusions. We have had many questions that involve the \u201cGambler\u2019s Fallacy\u201d, both from people who naively assume it without thinking, and from some who defend it using technical ideas like the Law of Large Numbers.\n\nThe Gambler\u2019s Fallacy\n\nHere is a question from 2003 that is a good introduction to the idea:\n\nGambler's FallacyA current co-worker and I are in a friendly disagreement about the probability of selecting the winning number in any lottery, say Pick 5. He states that he would rather bet the same set of five numbers every time for x period of time, but I insist that the probability is the same if you randomly select any set five numbers for the same period of time. The only assumption we make here is betting one set of numbers on any given day. Who is correct?\n\nI tried explaining to him that the probability of betting on day one is the same for both of us. On day two it is the same. On day three it is the same, etc. Therefore the sum of the cumulative probabilities will be the same for both of us.\n\nDoctor Wallace first examined the appropriate calculation, before moving on to the likely underlying error:\n\nYou are correct. If you have the computer randomly select a different set of 5 numbers to bet on every day, and your friend selects the same set of numbers to bet on every day, then you both have exactly the same probability of winning.\n\nTell your friend to think of the lottery as drawing with tickets instead of balls. If the lottery had a choice of, say, 49 numbers, then imagine a very large hat containing 1 ticket for every possible combination of 5 numbers.  1, 2, 3, 4, 5;  1, 2, 3, 4, 6;  etc.\n\nOn the drawing day, ONE ticket is pulled from the hat. It is equally likely to be any of the C(49,5) tickets in the hat. (There would be 1,906,884 tickets in the hat in this case.)\n\nSince both you and your friend have only ONE ticket in the hat, you both have the same chance of winning.\n\nOn the next drawing day for the lottery, ALL the tickets are replaced.  Each lottery draw is an event independent of the others. That is to say, the probability of any combination winning today has absolutely NO effect on the probability of that or any other combination winning tomorrow. Each and every draw is totally independent of the others.\n\nThat perspective makes it \u201cobvious\u201d: if two people each have one \u201cticket\u201d, they have the same probability, whether or not it is the same one that was taken last time, since the \u201cticket\u201d is chosen randomly each time without regard to the past.\n\nSo why is the other opinion tempting?\n\nThe reason your friend believes that he has a better chance of winning with the same set of numbers is probably due to something called the \"gambler's fallacy.\" This idea is that the longer the lottery goes without your friend's \"special\" set of numbers coming up, the more likely it is to come up in the future. The same fallacy is believed by a lot of people about slot machines in gambling casinos. They hunt for which slot hasn't paid in a while, thinking that that slot is more likely to pay out. But, as the name says, this is a fallacy; pure nonsense. A pull of the slot machine's handle, like the lottery draw, is completely independent of previous pulls. The slot machine has no memory of what has come before, and neither has the lottery. You might play a slot machine for 2 weeks without hitting the big jackpot, and someone else can walk in and hit it in the first 5 minutes of play.  People wrongly attribute that to \"it was ready to pay out.\" In reality, it's just luck.  That's why they call it gambling.  :)\n\nThe same thing comes up in math classes:\n\nThis used to be a \"trick\" question on old math tests:\n\n\"You flip a fair coin 20 times in a row and it comes up heads every single time. You flip the coin one more time. What is the probability of tails on this last flip?\"\n\nMost people will respond that the chance of tails is now very high.\n(Ask your friend and see what he says.)  However, the true answer is that the probability is 1/2.  It's 1/2 on EVERY flip, no matter what results came before.  Like the slot machine and the lottery, the coin has no memory.\n\nThat type of question is still valuable. It tests an important idea that students need to think about.\n\nFor more about picking the same number in a lottery, see\n\nLottery Strategy and Odds of Winning\n\nFor a very nice refutation of the Gambler\u2019s Fallacy in coin tossing, see\n\nWhat Makes Events Independent?\n\nThe Law of Large Numbers\n\nThat question was based on a naive approach to gambling. The next, from 2000, is based on probability theory. (The questions were asked in imperfect English, which I will restate as I understand it, correcting a misinterpretation in the archived version. Doctor TWE got it right.)\n\nLaw of Large Numbers and the Gambler's FallacyIf we throw three dice at a time, three times altogether, is the result the same as if we throw nine dice one time?\n\nDo we have the same probability to get a given number in the two cases?\nWhere and how different is it?\n\nDoctor TWE answered, covering several possible interpretations of \u201cresult\u201d:\n\nThe sum, the average, and the probabilities of getting a particular value on one die or more aren't affected by whether the dice are rolled one at a time, in groups of 3, or all 9 at once. These are called \"independent events,\" and the order in which they happen doesn't affect the outcome.\n\nElvino replied,\n\nSo, I have 42.12% probability to get at least one six (or other number), hen I throw three dice at once. If I throw the same dice, in the same way, again and again I will always have the same probability!\n\nBut, there is a law that says, the more I throw, the more probability I have to obtain a certain number. How does this law of probabilities work?\n\nChecking his calculation, and thereby confirming what he meant, I get the probability of rolling at least one six on three dice to be the complement of rolling no sixes: $$\\displaystyle 1 \u2013 \\frac{5^3}{6^3} = 42.1%$$. Next time I will discuss this further.\n\nDoctor TWE confirmed his calculation, then explained what this law means, and does not mean:\n\nThere is something called the Law of Large Numbers (or the Law of Averages) which states that if you repeat a random experiment, such as tossing a coin or rolling a die, a very large number of times, your outcomes should on average be equal to (or very close to) the theoretical average.\n\nSuppose we roll three dice and get no 6's, then roll them again and still get no 6's, then roll them a third time and STILL get no 6's. (This is the equivalent of rolling nine dice at once and getting no 6's, as we discussed in the last e-mail; there's only a 19.38% chance of this happening.) The Law of Large Numbers says that if we roll them 500 more times, we should get at least one 6 (in the 3 dice) about 212 times out of the 503 rolls (.4213 * 503 = 211.9).\n\nThis is *not* because the probability increases in later rolls, but rather, over the next 500 rolls, there's a chance that we'll get a \"hot streak,\" where we might roll at least one 6 on three or more consecutive rolls. In the long run (and that's the key - we're talking about a VERY long run), it will average out.\n\nThere is also something called the Gambler's Fallacy, which is the mistaken belief that the probability on the next roll changes because a particular outcome is \"due.\" In the example above, the probability of rolling at least one 6 in the next roll of the three dice (after three rolls with no 6's) is still 42.13%. A (non-mathematician) gambler might think that the dice are \"due,\" that in order to get the long-term average back up to 42%, the probability of the next roll getting at least one 6 must be higher than 42%. This is wrong, and hence it's called \"the Gambler's Fallacy.\"\n\nThe important thing here is that things will \u201caverage out\u201d in the long run, so that you get at least one 6 in 42.1% of the rolls, but not because at any one time there is a greater chance, to make up the average \u2014 it happens only because there is a very long time to do so. It is not because past events have any effect on future events, pulling them into line with the \u201caverages\u201d.\n\nFor more on the Law of Large Numbers, see\n\nThe Law of Large Numbers\n\nDigging deeper\n\nIn 2008, we had the following detailed question about that from Konstantinos:\n\nDo Prior Outcomes Affect Probabilities of Future Ones?What I want to know is if in matters of luck, such as games of dice, or lotteries, or flipping coins, the future outcomes have any relativity with past results.  What I mean in each case is:  If I flip a coin and get tails, don't I have bigger or smaller possibilities to get heads on the next roll?  I mean I know that the possibility is always 1/2 but since I have already thrown the coin 5 times and rolled 5 tails there aren't possibilities that in the next throws there will be heads?  The same question goes for dice rolls, and for lotto numbers.  If a number has come more times than others, isn't it possible that for a limited amount of coming times, numbers that haven't come yet, will start showing more?\n\nWhat I find most confusing is that the relation between probabilities and past possibilities, and the outcome in real life.  How can a coin come tails 5 or six times in a row when the possibility is always 1/2?  Are probabilities only theoretical?\n\nI tried noting down results of different trials of luck (dice, past lotteries, coins) but in the end they don't seem to make true to any theory I have heard.  I would like to see the magic of numbers in real life and on this subject and how it works as to prove a theory.  Do I have to toss the coin 10,000 times?  And if I do will I see heads coming in a row after 10 subsequent tails?\n\nI responded to this one:\n\nWhat you are suggesting is called the Gambler's Fallacy--the WRONG idea that future results of a random process are affected by past results, as if probability deliberately made things balance out.  The law of large numbers says that it WILL balance out eventually; but that does not happen by changing the probabilities in the short term.\u00a0The long-term balance just swamps the short-term imbalance.\n\nIf you think about it, what could cause a coin to start landing heads up more often after a string of tails?  There is no possible physical cause for this; the coin has no memory of what it, or any other coin, previously did.  And probability theory does not make things happen without a physical cause; it just describes the results.\n\nIf I tossed a coin and it landed tails 5 times, I would just recognize that that is a perfectly possible (and not TOO unlikely) thing to happen.  If I got tails 100 times, I would NOT expect the next to be heads; I would inspect the coin to make sure it actually has a heads side!  An unlikely string of outcomes not only does not mean that the opposite outcome is more likely now; it makes it LESS likely, because it suggests statistically that the basic probability may not be what I was originally assuming.\n\nIn response to the question about probabilities being merely theoretical, I explained what probability is and isn\u2019t:\n\nProbabilities are theoretical, but have experimental results, IN THE LONG RUN.  The law of large numbers says that, if you repeat an experiment ENOUGH TIMES, the proportion of times an event occurs will PROBABLY be CLOSE to the predicted theoretical probability.  Probability can't tell you what will happen the next time, but it does predict what is likely to happen on the average over the next, say, million times.  If you started out with ten tails in a row, you will not necessarily get ten heads in a row at any point, or even more heads than tails in the next few tosses; you will just get enough heads in the next million to keep it balanced.\n\nNote all the capitalized words, emphasizing the vagueness of this statement.\u00a0A formal statement of this theorem will define those more clearly, but\u00a0they will then all become statements of probability (my \u201cprobably\u201d)\u00a0and limits (my \u201cenough\u201d and \u201cclose\u201d). See, for example, the formal\u00a0statements of the Law of Large Numbers in the Wikipedia page (my emphases in the last\u00a0paragraph):\n\nThe weak law of large numbers (also called Khinchin\u2019s law) states that the sample average converges in probability towards the expected value $$\\displaystyle \\overline {X}_{n}\\overset{P}{\\rightarrow} \\mu \\ {\\textrm {when}}\\ n\\to \\infty .$$.\n\nThat is, for any positive number \u03b5, $$\\displaystyle \\lim _{n\\to \\infty }\\Pr \\!\\left(\\,|{\\overline {X}}_{n}-\\mu |>\\varepsilon \\,\\right)=0$$.\n\nInterpreting this result, the weak law states that for any nonzero margin specified, no matter how small, with a sufficiently large sample there will be a very high probability that the average of the observations will be close to the expected value; that is, within the margin.\n\nWhat it does not say is that over, say, 100 trials (or 10,000 trials), we can be sure that the average will be exactly what we expect.\n\nI continued:\n\nIn particular, if you were to throw five coins at a time (to make it easier than throwing the same coin five times in a row), and do that 1000 times, you would expect that you would get 5 tails about 1/32 of those times (since the probability of all five being tails is 1/2^5). That's about 31 times!  So it's not at all unreasonable to expect that it will occur once in a while.\n\nOn the other hand, the probability of getting 100 tails in a row is 1/2^100, or 1/1,267,650,600,228,229,401,496,703,205,376, which makes it very unlikely that it has ever happened in the history of the world, though it could!\nAgain, probability can't tell you what WILL happen, specifically; it is all about unpredictable events.  But if you tossed a coin 1000 sets of 10 times, on the average one of those is likely to yield 10 tails. (The probability is 1/2^10 = 1/1024.)  The probability of some ten in a row out of 10,000 tosses is a little bigger, but that's harder to calculate.\n\nRegression to the mean\n\nIn an unarchived question from 2014, Adam brought in another idea:\n\nIf I flip a coin 10 times, the most likely outcome is that I will flip a total of 5 heads and a total of 5 tails.  If each round of coin flipping (one round being 10 flips, in my example) is independent of previous rounds, then the probability of flipping a total of 5 heads and a total of 5 tails never changes.  However, the concept of \"regression to the mean\" implies that \"rare\" events are likely to be followed by less rare events and vice versa.  So, if I flip 10 heads in a row in round 1 (a rare event), the odds of flipping a total of 5 heads and 5 tails in round 2 are greater than if I'd flipped 5 heads and 5 tails in round 1.  What is the correct way to see this, do the odds change or not?\n\nCoin flipping rounds are independent of one another, which implies that the probabilities never change. Regression to the mean states that rare events are likely to be followed by less rare events, implying that the probability of even random events does change.\n\nI replied:\n\nThe odds don't change. You are NOT more likely to flip 5 heads if you previously flipped 10.\n\nRegression to the mean only says that the most likely event on ANY toss will be about 5 heads, so if you got 10 on one toss, you are more likely to get something closer to 5 on the next toss, simply because something closer to 5 is ALWAYS more likely than 10. If you got 5 on the first toss, you are as likely to get 5 again as ever; but any deviation that does occur will be away from the mean, because there is no direction to go except away!\n\nIt is an entirely wrong interpretation of the phenomenon to think that, having tossed 10 heads, you are more likely to toss 5 heads than under other circumstances. It's just that the event that is always more likely will be less extreme than that first toss.\n\nhttp://en.wikipedia.org/wiki/Regression_toward_the_mean\n\nPast Events and ProbabilityBatting Averages", "date": "2021-07-27 02:36:31", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/the-gamblers-fallacy/", "openwebmath_score": 0.6335918307304382, "openwebmath_perplexity": 399.40147268034275, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308789536605, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8670619108588936}}
{"url": "https://xenchic.de/complex-fourier-series-of-square-wave.html", "text": "complex fourier series of square wave. I'm trying to plot the fourier series following fourier series; f(t)=$$\\sum_{k=0}^k \\frac{(1)(\\sin(2k+1)pi*t)}{2k+1}$$ equation. 2 Fourier Series Learning outcomes In this section we will learn how Fourier series (real and complex) can be used to represent functions and sum series. Now, we will write a Matlab code. A Fourier series (which rely on complex Fourier coefficients defined by integrals) uses the complex exponential e inx. The functions sin(nx) and cos(nx) form a sort of periodic table: they are the atoms that all other waves are built out of. When an and bn are given by ( 2 ), the trigonometric series ( 1 ) is called the Fourier series \u2026. Fourier Series of waveforms. 16 Example: Find the complete Fourier series of the square wave function sqr(x). Example # 01: Calculate fourier series of the function given below: $$f\\left( x \\right) = L \u2013 x on \u2013 L \\le x \\le L$$ Solution: As,. s (1) and (2), is a special case of a more gen-eral concept: the Fourier series for a periodic function. Many component approximation to square wave. Then for all t g t+ p a =f a t+ p a =f(at+p)=f(at)=g(t). This is the fundamental frequency or f 1, that corresponds to the \u201cpitch\u201d of the sound Each of the higher-frequency simple waves \u2026. Read Book Fourier Series Examples And Solutions Square Wave Fourier Series Examples And Solutions Square Wave When somebody should go to the books stores, search start by shop, shelf by shelf, it is in Complex fourier Series - Example Fourier Transform (Solved Problem 1) Fourier Analysis: Fourier \u2026. 57 questions with answers in FOURIER SERIES. \u2022 stem \u2013 Draws discrete plots (as opposed to plot, which draws continuous plots). On the other hand, an imaginary number takes the general form. The derivation of this real Fourier series from (5. Read through the lab and pay special attention to the introductory parts about Fourier series \u2026. Since this function is even, the coefficients Then. Sketch 3 cycles of the function represented by. The main aim of the Fourier series is one period to many frequencies. Transcribed image text: (a) Find the complex Fourier series of the periodic square wave shown in Problem 5. Properties of Fourier Series John T. 1 A Historical Perspective By 1807, Fourier had completed a work that series \u2026. Thus, the Fourier Series is an in nite superposition of imaginary exponentials with frequency terms that in-crease as n increases. A key difference, no matter how you want to formulate or describe it, goes something like this: 1) Start with a square wave with infinitely sharp transitions. PDF Odd 3: Complex Fourier Series. The first term of the Fourier Series will be a sinusoid with same phase and frequency as the square wave. University of California, San Diego J. 9 Even and Odd Functions The astute reader will have noticed that the Fourier series constructed in Secs. Complex Exponentials ejn t n t j n t cos 0 sin 0. A full FFT produces a complex number, so yes, phase is included. 1 Periodic Functions and Orthogonality Relations The di\ufb00erential equation y\u2032\u2032 + 2y =F cos!t models a mass-spring system with natural frequency with a pure cosine forcing function of frequency !. 4 first nonzero coefficients are used only, so the Square Wave approximation will be not sensational. I am trying to calculate in MATLAB the fourier series \u2026. For more information about the Fourier series, refer to Fourier. Let's go back to our non-periodic driving force example, the impulse force, and apply the Fourier \u2026. (You can figure out the last step and the casework for even and odd by drawing a little. Chapter 65, The complex or exponential form of a Fourier. Consider the periodic square wave x(t) shown in figure a. sum of sine waves each with different amplitudes waves. This is the idea of a Fourier series. The Fourier basis is convenient for us in that this series already separates these components. The next question is, given a complex periodic wave, how can we extract its component waves? Two Methods are Required. The Fourier Series is a shorthand mathematical description of a waveform. Here we see that adding two different sine waves make a new wave: When we add lots of them (using the sigma function \u03a3 as a handy notation) we can get things like: 20 sine waves: sin (x)+sin (3x)/3+sin (5x)/5 + + sin (39x)/39: Fourier Series \u2026. %Fourier series of rectangular wave clc; close all; clear all; j=1; T=4; %Time period of square wave tau=1; %2tau= On time of the square wave \u2026. not just odd like the square wave. If the following condition (equation [5]) is true, then the resultant function g (t) will be entirely real: In equation [5], the. The series for f(x) defined in the interval (c, c+2\u03c0)and satisfying. N is the number of number of terms used to approximate the square wave. We will call it the real form of the Fourier series. Convert the real Fourier se- ries of the square wave f(t) to a complex series. The Complex Fourier Series. derived a real representation (in terms of cosines and sines) for from the complex exponential form of the Fourier series: (1) For example, in the lecture #13 notes, we derived the following Fourier coef\ufb01cients for a triangle wave (sym-metric about the vertical axis), (2) and converted the complex exponential series\u2026. The coefficients for Fourier series expansions of a few common functions are given in Beyer (1987, pp. So, responding to your comment, a 1 kHz square wave doest not include a component at 999 Hz, but only odd harmonics of 1 kHz. The Fourier transform and Fourier\u2026. 8 Illustration of the superposition of terms in the. The Fourier transform of this image exhibits an \"infinite\" series of harmonics or higher order terms, although these do not actually go out to infinity due to the finite resolution of the original image. So, responding to your comment, a 1 kHz square wave doest not include a. Since sines and cosines (and in turn, imaginary exponentials) form an orthogonal set1, this se-ries converges for any moderately well-behaved function f(x). Often in solid state physics we need to work with functions that are periodic. The square wave is a special case of a pulse wave which allows arbitrary durations at minimum and maximum amplitudes. In other words, Fourier series can be used to express a function in terms of the frequencies (harmonics) it is composed of. A few examples are square waves, saw-tooth waves, and triangular pulses. Example: Compute the Fourier series of f(t), where f(t) is the square wave \u2026. In general, given a repeating waveform , we can evaluate its Fourier series coefficients by directly evaluating the Fourier transform: but doing this directly for sawtooth and parabolic waves \u2026. They are helpful in their ability to imitate many different types of waves: x-ray, heat, light, and sound. The complex Fourier expansion coe\ufb03cients are cn= 4 \u03c0 1 n nodd 0 neven 0 2 4 6 8 10 0. In case of a string which is struck so that say at x=a only the string has a \u2026. Fourier Series using LabVIEW Fourier Series using LabVIEW Student-developed LabVIEW VI The student is then asked to approximate a square wave in both the time and frequency domain using a summed set of 5 sine waves The frequency and amplitude from the LabVIEW interface provide the coefficients of the Fourier Series \u2026. ABSTRACT Fourier analysis and a computer were used to generate the Fourier series and coefficients for the transmission distribution of a square wave \u2026. Complex Fourier Series Animation of the square wave The 4 upper rotating vectors correspond to the 4 lower formula components. From the real to the complex Fourier series Proposition The complex Fourier coe cients of f(x) = a0 2 + X1 n=1 an cosnx + bn sinnx are cn = an ibn 2; c n = an + ibn 2: M. Fourier Series for a \u2026 Odd 3: Complex Fourier Series - Imperial College London It can be done by using a process called Fourier analysis. Draw a square wave of amplitude 1 and period 1 second whose trigonometric Fourier Series Representation consists of only cosine terms and has no DC component. m m! Again, we really need two such plots, one for the cosine series and another for the sine series. 3 1) Compute the Fourier series that corresponds to a square wave 1 - /4 /4 (0) 1. Square WaveFourier Series Examples And Solutions Square Wave Thank you enormously much for downloading fourier series examples and solutions square wave. Such a Fourier expansion provides an interpetation of the wave \u2026. For comparison, let us find another Fourier series, namely the one for the periodic extension of g(x) = x, 0 x 1, sometimes designated x \u2026. c) Write the first three nonzero terms in the Fourier expansion of f ( t). Jean Baptiste Joseph Fourier (21 March 1768 - 16 May 1830) Fourier series. He initialized Fourier series, Fourier transforms and their applications to problems of heat transfer and vibrations. 1) Compute the Fourier series that corresponds to a square wave. A Fourier series is nothing but the expansion of a periodic function f(x) with the terms of an infinite sum of sins and cosine values. The exact combination of harmonics will vary depending on the way the string is set in motion; e. Check the time t of the 10 rotations, t=24sec->T=2. Again, of course, you\u2019re not going to get a perfect square wave with a finite number of Fourier terms in your series (in essence, it\u2019s then not a \u2026. 13 Applications of the Fourier transform 13. of Electrical Engineering, Southern Methodist University. The synthesis technique is also called additive synthesis. This chapter includes complex differential forms, geometric inequalities from one and several complex \u2026. For instance, the square wave \u2026. The ideas are classical and of transcendent beauty. An Introduction to Fourier Analysis Fourier Series, Partial Differential Equations and Fourier Transforms. As mentioned in the previous section, perhaps the most important set of orthonormal functions is the set of sines and cosines. s(t) = AN ODD SQUAREWAVE with DC. PDF Characterization of Signals Frequency Domain. Finding the Fourier series coefficients for the square wave sq T (t) is very simple. A Fourier series uses the relationship of orthogonality between the sine and cosine functions. , the output will always have a negative portion and positive portion. 1 The Real Form Fourier Series as follows: x(t) = a0 2 + X\u221e n=1 an cosn\u03c90t+bn sinn\u03c90t (1) This is called a trigonometric series. In practice, the complex exponential Fourier series (5. (one period is T which is equal to 2PI) Looking at the figure it is clear that area bounded by the Square wave \u2026. Plot one-sided, double-sided and normalized spectrum. continuation of part 1a - Introduction to Complex Fourier Series. The coefficients become small quickly for the triangle wave, but not for the square wave or the sawtooth. This means a square wave in the time domain, its Fourier \u2026. Concept: The complex Exponential Fourier Series representation of a periodic signal x(t) with fundamental period To is given by, \\$$x\\\\left(. In class, we mentioned how complex exponentials version of the Fourier Series can be used to represent a 2Tperiodic function f(t) as follows: can be used to represent a Fourier Series: f(t) = X1 n=1 c ne in\u02c7 T t (1) The crux of Fourier Series analysis is to nd the c n values specifying how much the nth harmonic contributes to the function f(t). The synthesis of a complex wave \u2026. 10 Trigonometric Fourier series of another square wave. Note that the Fourier coefficients are complex numbers, even though the series in Equation [1], evaluated with the coefficients in Equation [4], result in a real function. Similarly, if G(x) is an odd function with Fourier coe cients a nfor n 0 and b n for n 1, then a n= 0 for all n 0, and a n= 2 L Z L 0 G(x)sin n\u02c7x L dxfor all n 0(16) In particular, the fourier series of an even function only has cosine terms and the fourier series of an odd function only has sine terms. It is very easy to see that an vanishes if f is an odd function, while bn vanishes if f is even. Index Terms\u2014Fourier series, periodic function, recursive. Based on what I read at this link: Any periodic . It is natural for complex numbers and negative frequencies to go hand-in-hand. Even Square Wave (Exponential Series) Consider, again, the pulse function. 3: Fourier Cosine and Sine Series, day 1 Trigonometric Fourier Series (Example 2) Complex fourier Series - Example Fourier Transform (Solved Problem 1) Fourier Analysis: Fourier Transform Exam Question ExampleFourier Series: Complex Version! Part 1 Fourier \u2026. (a) Determine the complex exponential Fourier series of x (t). Young (translator), There are excellent discussions of Fourier series \u2026. Since the time domain signal is periodic, the sine and cosine wave \u2026. Write a computer program to calculate the exponential Fourier series of the half-wave rectified sinusoidal current of Fig. This function is a square wave; a plot shows the value 1 from x=p to x = 0 followed by the . 3 Square wave Analysis (breaking it up into sine waves). Summation of just five odd harmonics gives a fairly decent representation in Figure 15. Complex fourier Series - Example Fourier Transform (Solved Problem 1) Fourier Analysis: Fourier Transform Exam Question ExampleFourier Series: Complex \u2026. One of the principles of Fourier analysis is that any imaginable waveform can be constructed out of a carefully chosen set of sinewave components, assembled in a particular way (the frequency -> time task). (Image will be uploaded soon) Laurent Series Yield Fourier Series (Fourier Theorem). The complex Fourier series Recall the Fourier series expansion of a square wave, triangle wave, and sawtooth wave that we looked at before. Schr\u00f6dinger's Equation Up: Wave Mechanics Previous: Electron Diffraction Representation of Waves via Complex Numbers In mathematics, the symbol is conventionally used to represent the square \u2026. So here's the final wave, listening to the final waveform. You can work out the 2D Fourier transform in the same way as you did earlier with the sinusoidal gratings. In fact, in many cases, the complex Fourier series is easier to obtain rather than the trigonometrical Fourier series In summary, the relationship between the complex and trigonometrical Fourier series \u2026. To motivate this, return to the Fourier series\u2026. Recall from Chapter 1 that a digital sound is simply a sequence of num- Fourier analysis with complex exponentials which will often result in complex square wave are now 0 and 1,contraryto\u22121 and 1 before. 5) The Laurent expansion on the unit circle has the same form as the complex Fourier series, which shows the equivalence between the two expansions. Fourier Series | Brilliant Math & Science Wiki. Also, as with Fourier Sine series\u2026. Theorem 2 lim N\u2192\u221e sup 0\u2264x\u22641 f \u2212S N(f) = 0 holds for any continuous function g. Compute the Exponential Fourier Series for the square wave shown below assuming that. Louis, MO April 24, 2012 The Fourier series is a tool for solving partial differential equations. (We assume the reader is already at least somewhat familiar with these. Step 1: Obtain the Fourier series of F(t). magnitude of the square of the Fourier transform: SFEt {()}2 This is our measure of the frequency content of a light wave. Plot the function over a few periods, as well as a few truncations of the Fourier series. It is not a mathematical proof, and several terms are used loosely (particularly those in quotes). We can be confident we have the correct answer. %Fourier series of rectangular wave clc; close all; clear all; j=1; T=4; %Time period of square wave tau=1; %2tau= On time of the square wave w0=2*pi/T;. 3\u201381 Spectrum of a Sum of Cosine Signals spectrum cosines 3\u201382 3. I understand the bounds that he chooses and. If a square waveform of period T is defined by  \\left\\{ \\begin{array}{l l} f(t)= 1 \\text{ if } t= T/2 \\end{array} \\right. Series coefficients c n (d) Fig. Discrete Time Fourier Transforms The discrete-time Fourier transform or the Fourier transform of a discrete\u2013time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence. There's also the infamous Square wave \u2026. Since this wave is periodic, its harmonic content can be found using Fourier series as follows: The Fourier coefficients are,. Answer The function is discontinuous at t = 0, and we expect the series to converge to a value half-way between the upper and lower values; zero in this case. Let f(x) = {1 if -pPDF Fourier analysis for vectors. 2 Complex conjugate sqrt(x) Square root log(x) Natural logarithm Suppose we want to enter a vector x consisting of points Sine Wave 0. Example 1 Find the Fourier sine coefficients bk of the square wave SW(x). Fourier Analysis: Fourier Series with Complex Exponentials n The Complex Fourier series can be written as: where: n Complex cn n *Complex conjugate * n Note: if x(t) is real, c-n = cn 32 Fourier Analysis: Fourier Series Line Spectra n n Line Spectra refers the plotting of discrete coefficients corresponding to their frequencies For a periodic. Start by forming a time vector running from 0 to . See below Once rectified, it is even , so you only need the cosine series. Importantly there is nothing special about the square \u2026. Where To Download Fourier Series Examples And Solutions Square Wave Fourier Series Examples And Solutions Square Wave As recognized, adventure as well as experience practically lesson, amusement, as skillfully as arrangement can be gotten by just checking out a books fourier series examples and solutions square wave \u2026. The 8-term Fourier series approximations of the square wave and the sawtooth wave: Mathematica code: f[t_] := SawtoothWave[t] T = 1; Fourier series decomposition of a square wave using Phasor addition : It retains it\u2019s form over several complex \u2026. 3 Case 2: Some periodic functions (e. The Fourier series represents periodic, continuous-time signals as a weighted sum of continuous-time sinusoids. Not surprisingly, the even extension of the function into the left half plane produces a Fourier series that consists of only cos (even) terms. By the double angle formula, cos(2t) = 1 2sin2 t, so 1 + sin2 t= 3 2 1 2 cos(2t): The right hand side is a Fourier series; it happens to have only nitely many terms. A Fourier series ( / \u02c8f\u028arie\u026a, - i\u0259r /) is a sum that represents a periodic function as a sum of sine and cosine waves. The integral splits into two parts, one for each piece of , and we find that the Fourier coefficients are. Fourier Series in Mathematica Craig Beasley Department of Electrical and Systems Engineering Washington University in St. Jean Baptiste Joseph Fourier, a French mathematician and a physicist; was born in Auxerre, France. 005 (b) The Fourier series on a larger interval Figure 2. The Fourier series forthe discrete\u2010time periodic wave shown below: 1 Sequence x (in time domain) 0. On-Line Fourier Series Calculator is an interactive app to calculate Fourier Series coefficients (Up to 10000 elements) for user-defined piecewise \u2026. The initial phase of the n-th oscillation \u03b8 \u03b8 n. (b) Consider the signal x(t) = sin(2\u03c0f0t), Find the complex Fourier series of x(t) and plot its frequency spectra. First, let x(t) be the zero-mean square wave. Wave Series Fourier Series Grapher; Square Wave: sin(x) + sin(3x)/3 + sin(5x)/5 + sin((2n\u22121)*x)/(2n\u22121) Sawtooth: sin(x) + sin(2x)/2 + sin(3x)/3 \u2026. 1 In\ufb01nite Sequences, In\ufb01nite Series and Improper In-tegrals 1. This can be accomplished by extending the de\ufb01nition of the function in question to the interval [\u2212L, 0] so that the extended function is either even (if one wants a cosine series) or odd (if one wants a sine series). As far as I know, Sage does not have a built-in method to find a \u201cleast-squares solution\u201d to a system of linear equations. -L \u2264 x \u2264 L is given by: The above Fourier series formulas help in solving different types of problems easily. The periodic function shown in Fig. 23) all coefficients an vanish, the series only contains sines. HOWELL Department of Mathematical Science University of Alabama in Huntsville Principles of Fourier \u2026. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Convert the real Fourier se-ries of the square wave f(t) to a complex series. Rad225/Bioe225 Ultrasound Fourier Series (review) Fall 2019. This section explains three Fourier series: sines, cosines, and exponentials e ikx. 5 The complex form of the Fourier series. 3 Square Wave\u2013High Frequencies One application of Fourier series, the analysis of a \u201csquare\u201d wave (Fig. Fourier series of the elementary waveforms. PDF The Exponential Form Fourier Series. It builds upon the Fourier Series. For the periodic bipolar, 50% duty-cycle square wave, the \u03b8 -averaging of this waveform over one \u03b8 -cycle is: QED. This subtle property is due to the symmetry of waveforms (except for the sawtooth, which is not symmetric). 01:6; >> fexact=4*(t<=3)-2*(t>=3); >> plot(t,fexact) and results in the plot: Plotting the Truncated Fourier. Joseph Fourier showed that any periodic wave can be represented by a sum of simple sine waves. 5, an =0, n \u22651, , 1 [1 cos( )] \u2265 \u2212 = n n n bn \u03c0 \u03c0. com analysis are described in the chapter on musical tones. Fourier series are used to approximate complex functions in many different parts of science and math. Demonstrates Taylor series expansion of complex exponentials. Recall the Fourier series, in which a function f[t] is written as a sum of sine and cosine terms: One interpretation of the above Fourier transform is that F[Z] is the frequency spectrum of a sine wave signal f[t] which is varying in time; thus Z is the angular frequency. The free space loss for electromagnetic waves spreading from a point source is The Friis\u2019 loss formula for antenna-to-antenna loss is given by Radio wave propagation in the atmosphere: (1) space-wave propagation (e. This series of sine waves always contains a wave called the \"FUNDAMENTAL\", that has the same FREQUENCY (repetition rate) as the COMPLEX WAVE being created. Fourier transform spectroscopy (cont. Complex Fourier Series \u2022 Complex Fourier Analysis Example \u2022 Time Shifting \u2022 Even/Odd Symmetry \u2022 Antiperiodic \u21d2 Odd Harmonics Only \u2022 Symmetry Examples \u2022 Summary E1. The fourier series of a sine wave is 100% fundamental, 0% any harmonics. If X is a multidimensional array, then fft \u2026. Even Square Wave (Exploiting Symmetry). (here: symmetric, zero at both ends) Series \u2013Taylor and Fourier Seismological applications The Delta function Delta function \u2013 generating series The delta function Seismological applications Fourier Integrals The basis for the spectral analysis (described in the continuous world) is the transform pair: Complex fourier spectrum The complex \u2026. The conversion of complex Fourier series into standard trigonometric Fourier series is based on Euler's formulas: sin\u03b8 = 1 2j ej\u03b8 \u2212 1 2je \u2212 j\u03b8 = \u2111ej\u03b8 = Imej\u03b8, cos\u03b8 = 1 2ej\u03b8 \u2212 1 2e \u2212 j\u03b8 = \u211cej\u03b8 = Reej\u03b8. Try another waveform, including one of the complex ones (Triangle, Sawtooth, or Square). A complex exponential einx= cosnx+isinnxhas a small-est period of 2\u03c0/n. This lesson shows you how to compute the Fourier series \u2026. PDF General Inner Product & Fourier Series. m % Description: m-file to plot complex (exponential) Fourier Series % representation of a square wave\u2026. Recall that the Taylor series expansion is given by f(x) = \u00a5 \u00e5 n=0 cn(x a)n, where the expansion coef\ufb01cients are. 1 This pages contains exercises to practice computing the Fourier series of a CT signal. Since the coefficients of the Exponential Fourier Series of complex numbers we. Complex Fourier series Complex representation of Fourier series of a function f(t) with period T and corresponding angular frequency != 2\u02c7=T: f(t) = X1 n=1 c ne in!t;where c n = 8 <: (a n ib n)=2 ;n>0 a 0=2 n= 0; (a jnj+ib jnj)=2 n<0 Note that the summation goes from 1 to 1. DC Value of a Square Wave The Fourier series coefficient for k = 0 has a special interpretation as the average value of the signal x(t). In short, the square wave\u2019s coefficients decay more slowly with increasing frequency. Fourier series coefficients for a symmetric periodic square wave. Find the exponential Fourier series for the square wave of Figure 11. 10 Fourier Series and Transforms (2014-5543) Complex Fourier Series\u2026. 2 Approximating the Square Wave Function using Fourier Sine Series. For the real series, we know that d = an = 0 and. Complex Fourier Series In an earlier module , we showed that a square wave could be expressed as a superposition of pulses. 1) translates into that the inverse of a complex (DFT on a square wave\u2026. Example: Compute the Fourier series of f(t), where f(t) is the square wave with period 2\u03c0. The frequency of each wave in the sum, or harmonic, is an integer multiple of the periodic function's fundamental frequency. 11 Fourier Series of a Square Wave Co is a DC average. Fourier Series Square Wave Example The Fourier series of a square wave with period is University of California, San Diego J. Fourier series of a square wave of period T = 1. Lec1: Fourier Series Associated Prof Dr. It will definitely squander the time. Contribute to Abdul-Rahman-Ibrahim/Fourier-Series-Expansion-Complex-Coefficients development by creating an account on GitHub. Fourier analysis and predict with precision their behavior by just looking at the final expression, also called the design equation. Determine the Fourier series expansion for full wave rectified sine wave i. where the Fourier coefficients and are given by. Example: Find the complete Fourier series of the square wave function sqr(x). Keywords: Fourier analysis, Fourier coefficients, complex exponentials, discrete sine series, discrete cosine series, full and half wave rectifier, diode valve, square wave, triangular wave, and the saw-tooth wave. The complex algebra provides an elegant and compact re. The Fourier Transform is a method to single out smaller waves in a complex wave. If we let the numbers in the Fourier series get very large, we get a phenomenon, called This is called the complex Fourier series\u2026. Fourier Series coefficients for a square wave. The main advantage of an FFT is speed, which it \u2026. The series does not seem very useful, but we are saved by the fact that it converges rather rapidly. the Fourier representation of the square wave is given in Fig. Then mathematically, a T-periodic waveform v satisfies \u2014 a periodic waveform with period T (2) for all t. Discrete Fourier Transform (DFT)\u00b6 From the previous section, we learned how we can easily characterize a wave with period/frequency, amplitude, phase. Show that the Fourier Series expansion of y (x) is given by y(t)= 4! sin\"t+ sin3\"t 3 + sin5\"t 5 + # % & '(where \u03c9 = 2\u03c0/T. In this particular SPICE simulation, I\u2019ve summed the 1st, 3rd, 5th, 7th, and 9th harmonic voltage sources in series \u2026. Fourier Series and Coefficients Fourier series may be used to represent periodic functions as a linear combination of sine and cosine functions. 1: The cubic polynomial f(x)=\u22121 3 x 3 + 1 2 x 2 \u2212 3 16 x+1on the interval [0,1], together with its Fourier series \u2026. Example 2 Given a signal y(t) = cos(2t), find its Fourier Series coefficients. Let\u2019s assume we have a square wave with following characteristics: P eriod = 2ms P eak\u2212to \u2212P eak V alue = 2 V Average V alue = 0 V P e r i o d = 2 m s P e a k \u2212 t o \u2212 P e a k V a l u e = 2 V A v e r a g e V a l u e = 0 V. The key here is that the Fourier basis is an orthogonal basis on a given interval. That is why we have programmed our free online Fourier series calculator to determine the results instantly and precisely. 6 Example: Fourier Series for Square Wave. Find the Fourier Complex Fourier Series . In the processing of audio signals (although it can be used for radio waves, light waves, seismic waves, and even images), Fourier analysis can isolate individual components of a continuous complex waveform, and concentrate. The function is reconstructed by the following summations over the fourier coefficients. In general square integrability will not guarantee convergence of the Fourier series to the original function. Fourier series are useful in a wide range of fields including acoustics, with representation of musical sounds as sums of waves of various frequencies (Nearing, 2020) or quantum mechanics, for the quantum wave function of a particle in a box. Step 3: Finally, substituting all the coefficients in Fourier \u2026. Check back soon! Problem 5 Show that the complex Fourier series \u2026. Any function can be written as the sum of an even and an odd function [ ( )]/2 A complex Lorentzian! Example: the Fourier \u2026. Square WaveSeries Coefficients 11. The study of Fourier series is a branch of Fourier analysis. 6 Fourier Coefficients for the Complex Conjugate of a Signal. 1: Fourier series approximation to s q ( t). Once the complex wave is broken up, we can analyze the results using the simpler sine waves Consider a square wave as shown in Fig. But these are easy for simple periodic signal, such as sine or cosine waves. org odic if it repeats itself identically after a period of time. Every circle rotating translates to a simple sin or cosine wave. If a function is periodic and follows below 2 conditions, then the Fourier series for such a function exists. Thus a function or signal f(t) with period T 0 can be expressed as [0 < t < T 0] where is called the fundamental frequency or base frequency (first resonant frequency = 1/T) and all other nw 0 frequencies are called harmonics (every other component of. Fourier series in 1-D, 2-D and 3-D. A Fourier series is that series of sine waves; and we use Fourier analysis or spectrum analysis to deconstruct a signal into its individual sine wave components. Fourier series and fourier solutions square wave inverters are some mathematically and capacitors work, the average signal is our scope of a closer approximation graph the. (This is analogous to the fact that the Maclaurin series of any polynomial function is just the polynomial itself, which is a sum of finitely many powers of x. Start by forming a time vector running from 0 to 10 in steps of 0. The Fourier transform is an extension of the Fourier series that results when the period of the represented function is lengthened and allowed to approach infinity. When we break a signal down into its composite sine waves, we call it a Fourier series. A steady musical tone from an instrument or a voice has, in most cases, quite a complicated wave shape. As can clearly be seen it looks like a wave with different frequencies. The corresponding analysis equations for the Fourier series are usually written in terms of the period of the waveform, denoted by T, rather than the fundamental frequency, f (where f = 1/T). 1: Square Wave Then the Fourier Series \u2026. Fourier Series Least Squares Curve Fit; Fourier Series Time Shift; Fourier Series Frequency Shift; Fourier Series, 4 segment; Fourier Series, var. The following method makes use of logical operators. 10 Fourier Series and Transforms (2014-5543) Complex Fourier Series: 3 - 2 / 12 Euler's Equation: ei\u03b8 =cos\u03b8 +isin\u03b8 [see RHB 3. The original signal x (t) is an square \u2026. It can usually be solved with some Fourier series: Square wave example. So is periodic with period and its graph is shown in . The two round markers in the imaginary. which is the Fourier expansion of a square wave \u2026. The Fourier Series (an infinite sum of trigonometric terms) gave us that formula. %Examples of Fourier Series Square Wave \u2026. We choose x\u02c6[k] = 1 N X8 n=\u22122 x[n]e\u22122\u03c0iknN = 1 11 X2 n=\u22122 e\u22122\u03c0ikn 11 The sum. This representation is done by summing sine and cosine functions or with complex exponentials of different amplitudes and phases. Step 2: Estimate for n=0, n=1, etc. For example, to \ufb01nd the Fourier series for a triangular wave \u2026. Okay, in the previous two sections we\u2019ve looked at Fourier sine and Fourier cosine series. PDF Interpretation of the Physical Significance of the Fourier. The Fourier series represents the sum of smooth sinusoids, but a square wave \u2026. As a tool to better understand the Fourier series coef\ufb01cients for the square \u2026. Every 2\u03c0periodic function that is analytic in a neighborhood of the real axis has a Fourier series \u2026. In the interval (c, c+2\u2113), the complex form of Fourier series is given by. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex \u2026. \u2234 Given waveform is Non-periodic so, the Fourier series will NOT exist. Left click and drag the [ball, green] circles to change the magnitude of each Fourier functions [Sin nf, Cos nf]. B Square Wave The square or rectangular waveform is similar to the sawtooth in that the amplitudes of the har-monics follow the A N = A 1=N dependence. The Fourier transform is represented as spikes in the frequency domain, the height of the spike showing the amplitude of the wave of that \u2026. More information about the Gibbs phenomenon. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Imagine a thin piece of wire, which only gains or loses heat through its ends. If you ally infatuation such a referred fourier series examples and solutions square wave ebook that will provide you worth, get the extremely best seller from us currently from several preferred authors. The coef\ufb01cients fb ng1 n=1 in a Fourier sine series F(x) are determined by. 23 ), are a measure of the extent to which has components along each of the basis vectors. The diagram above shows a \"saw-tooth\" wave \u2026. entities represented by symbols such as \u221e n=\u2212\u221e a n, \u221e n=\u2212\u221e f n(x), and \u221e \u2212\u221e f(x) dx are central to Fourier Analysis. Fourier series Fourier series in higher dimensions (vector notation) Complex Exponentials In 2-D, the building blocks for periodic function f(x 1;x 2) are the product of complex exponentials in one variable. INTRODUCTION During Napoleon's reign, Joseph Fourier (1768-1830) was one of the French scientists who took French science to new heights. First we see that fcan be expressed in terms of the standard square wave as f(t) = 1 + sq t+ \u02c7 2 : Now (see overleaf) the Fourier series for sq(t. width; Fourier Series, Continuous; Fourier Series, double Pulse; Fourier Series Calculation Square; Fourier Series Calculation Triangle; Fourier Series Square to Triangle Wave; Fourier Series \u2026. Film or TV show with spooky dead trees and singing Explicit triples of isomorphic Riemann surfaces Typeset sudoku grid using tabular syntax. Aug 15, 2013 - The first one is the exponential form of the Fourier series and the. It is typical ( but as far as I know not required) that complex \u2026. We look at a spike, a step function, and a ramp\u2014and smoother functions too. The computer algorithm for Fourier transforms is called an FFT (Fast Fourier \u2026. Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products. Determine the Fourier series coefficients of ( ) \ud835\udc4e ( ). If you work through the math, you can find the optimal values for cn using equation [3]: [Equation 4] Note that the Fourier coefficients are complex numbers, even though the series in Equation [1], evaluated with the coefficients in Equation [4], result in a real function. the same way that it did for the complex Fourier series we talked about earlier, only we have to replace an integral with a sum. In other words, Fourier series can be used to express a function in terms of the frequencies () it is composed of. The Fourier transform tells us what frequency components are present in a given signal. How do I plot the Fourier series for a square wave? [closed] Ask Question Asked 4 years, 6 months ago. A Fourier series (/ \u02c8 f \u028ar i e\u026a,-i \u0259r /) is a sum using only basic waves chosen to mathematically represent the waveform for almost any periodic function. That sawtooth ramp RR is the integral of the square wave. 6 Complex Fourier Series The exponential form of sine and cosine can be use to give: f(x) = X r c re 2\u02c7irx L The coe cients can be calculated from: c r= 1 L Z x o+L x o. Wave Equation and Fourier Series\u2026. A complex waveform can be constructed from, or decomposed into, sine (and cosine) waves of various amplitude and phase relationships. As useful as this decomposition was in this example, it does not generalize well to other periodic signals: How can a superposition of pulses equal a smooth signal like a sinusoid?. Stein and Shakarchi move from an introduction addressing Fourier series and integrals to in-depth considerations of complex analysis; measure and integration theory, and Hilbert spaces; and, finally, further topics such as functional analysis, distributions and elements of probability theory. The Fourier series can be applied to periodic signals only but the Fourier transform can also be applied to non-periodic functions like rectangular pulse, step functions, ramp function etc. In this section we define the Fourier Cosine Series, i. You can use the following commands to calculate the nth partial sum of the Fourier series of the expression f on the interval [-L,L] syms x k L n. Similar expressions hold for more general intervals [ a, b] by shifting and scaling appropriately. a demo in Maple of complex fourier series. The formula for the fourier series of the function f(x) in the interval [-L, L], i. As promised in the first part of the Fourier series we will now demonstrate a simple example of constructing a periodic signal using the, none other then, Fourier series\u2026. The amplitude of the n-th harmonic oscillation A n. f (t) has a finite number of maxima. An in nite sum as in formula (1) is called a Fourier series (after the French engineer Fourier who rst considered properties of these series. 14 The Complex Fourier Series 138 4. It establishes a relation between a function in the domain of time and a function in the domain of frequency. representing a function with a series in the form Sum( A_n cos(n pi x / L) ) from n=0 to n=infinity. More instructional engineering videos can be found . (a) The function and its Fourier series 0 0. This last line is the complex Fourier series. A square wave (represented as the blue dot) is approximated by its sixth partial sum (represented as the purple dot), formed by summing the first six terms (represented as arrows) of the square wave's Fourier series\u2026. Rochester Institute of Technology RIT Scholar Works Theses Thesis/Dissertation Collections 1978 The use of Fourier analyzed square waves in the determination of the modulation tra. Given the mathematical theorem of a Fourier series, the period function f(t) can be written as follows: f(t) = A 0 + A 1 cos\u03c9t + A 2 cos2\u03c9t + \u2026 + B 1 sin\u03c9t + B 2 sin2\u03c9t + \u2026 where: A 0 = The DC component of the original wave. So I\u2019ve just started learning about the Fourier series and wish to calculate one for a square wave function, my working is as follows. In some simple cases, Fourier series can be found using purely algebraic methods. approximations to the continuous Fourier series in Chapter 2. Also recall that the real part u and the imaginary part v of an analytic function f = u+iv are harmonic. If history were more logical they might have been found this way. Duration: 11:46 Complex Fourier Series - Square Wave \u2026. Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation. In this method, if N harmonics are included in the truncated Fourier series, then the amplitude of the kth harmonic is multiplied by (N - k)/N. Our aim was to find a series of trigonometric expressions that add to give certain periodic curves (like square or sawtooth waves. The result of the FFT is an array of complex numbers. The fine oscillations at the edges do not disappear even if the Fourier series takes many more terms. 1 The Fourier Series Components of an Even Square Wave Components of cos(nt) found in the approximation to an even square wave may be calculated generally as a function of n for all n > 0. Y = fft (X) computes the discrete Fourier transform (DFT) of X using a fast Fourier transform (FFT) algorithm. The Fourier Transform can be used for this purpose, which it decompose any signal into a sum of simple sine and cosine waves that we can easily measure the frequency, amplitude and phase. 5, and the one term expansion along with the function is shown in Figure 2: Figure 2. Suppose that our wave is designed by 18. 3 Signal Synthesis Later on in this lab you will use a Fourier series to approximate a square wave. b) Derive the expression for the Fourier coefficients a k. Example: Determine the fourier series of the function f(x) = 1 - x 2 in the interval [-1, 1. Here's the Fourier series for a square wave, truncated at the first 25 terms (which might sound like a lot, but is really easy for a computer to handle:) and \\( \\cos (\\omega t)$$. Furthermore, suppose that the signal is periodic with period T: for all t we have xs(t) = xs(t +T). The plot in black color shows how the reconstructed (Fourier Synthesis) signal will look like if the three terms are combined together. The construction of a PERIODIC signal on the basis of Fourier coefficients which give the AMPLITUDE and PHASE angle of each component sine wave HARMONIC. amplitude, frequency, and starting phase amplitudes, frequencies, and starting phases. For math, science, nutrition, history. 1 Introduction The concepts of in\ufb01nite series and improper integrals, i. Deriving the Fourier Coefficients. Turns out all the sines and cosines (or the equivalent complex \u2026. This index corresponds to the k -th harmonic of the signal's period. Over the range [0,2L], this can be written as f(x)=2[H(x/L)-H(x/L-1)]-1, (1) where H(x) is the Heaviside step\u00a0. Let be a -periodic function such that for Find the Fourier series for the parabolic wave. Apply integration by parts twice to find: As and for integer we have. It is applicable only to periodic signals. Pre-lab: Theory: The sawtooth wave (or saw wave) is a kind of non-sinusoidal waveform. 1) Note that a 0 is the average of the function over the interval. Author name; Kyle Forinash; Wolfgang Christian. A real number, (say), can take any value in a continuum of values lying between and. LAB REPORT Experiment 2 - Fourier series of Square, Triangle, and Sawtooth Pulse Trains Experiment 2- Fourier series of Square, Triangle, and Sawtooth Pulse Trains: 1). Complex Fourier coef\ufb01cients: c n = 1 T That is the in\ufb01nite Fourier series of the square wave function of period 1. In calculations involving Fourier series it is often advantageous to use complex exponentials;. The complex exponential form of cosine. Fourier Series on the Complex Plane. In order to do this, let us consider a pair of FC trigonometric. 3\u201380 Express the Square of a Sinusoid as a Sum of Complex Exponentials square sinusoid complex exponential. Suppose you want to make a periodic wave \u2014 maybe it's for a music synthesizer or something. Let us now show that the usual formulas giving the Fourier coefficients, in terms of integrals involving the corresponding real functions, follow as consequences of the analytic properties of certain complex functions within the open unit disk. Here are a number of highest rated Square Fourier Series pictures upon internet. What exactly is a Fourier series\u2026. For square wave with period T and x0 = -T/2 Split the a[n] evaluation integral to two parts, -T/2,0> and (0,T/2>: Therefore: Split the b[n] evaluation integral to two parts: Therefore: The complex coefficients can be obtained from trigonometric coefficients as follows: Fourier Series of Full-wave Rectified Sine Wave. It may be the worst way to graph a square but it's fun! I made it because I want to learn complex integration and Fourier Series. Then the adjusted function f (t) is de ned by f (t)= f(t)fort= p, p Z ,. Your first 5 questions are on us!. Fourier Series Print This Page Download This Page; 1. Electrical Engineering Q&A Library ermine the complex exponential Fourier series of. We identified it from well-behaved source. Assuming you're unfamiliar with that, the Fourier Series is simply a long, intimidating function that breaks down any periodic function into a simple series of sine & cosine waves. We want to know the amplitude of the wave at the detector in the u,v plane, which is a distance z from the x,y plane. The Fourier series of a periodic function is given by. 4 Complex Tones, Fourier Analysis and The Missing Fundamental. \u221e \u2211 k = \u2212 \u221e\u03b1kekj\u03c0x / \u2113 (j2 = \u2212 1), where a signal's complex Fourier \u2026. \u2022 angle \u2013 Computes the phase angle of a complex number. Use this information and complete the entries in Table 2 for the square wave. It is instructive to plot the first few terms of this Fourier series and watch the approximation improve as more terms are included, as shown in Figure 9. Fourier Series in Filtering 5 The Matlab commands below1 will sketch the symmetric partial sum with subscripts up to N. The Fourier Transform and Free Particle Wave Functions 1 The Fourier Transform 1. Figure 6-1 Successive Fourier series approximation to a square wave by adding terms. By applying Euler's identity to the compact trigonometric Fourier series, an arbitrary periodic signal can be expressed as a sum of complex exponential functions: (11. If one would like to approximate a function over a larger interval one would need terms of very high order. Do not worry too much about the math, but focus on the results and speci\ufb01cally for this course, the application. The frequency of each wave in the sum, or harmonic,\u00a0. Calculating the 2D Fourier Transform of The Image. And I think these are the remaining entries on the list. frequency-phase series of square waves (the equivalent of the polar Fourier Theorem but\u00a0. Fourier Series, Fourier Transforms and the Delta Function. Here we see that adding two different sine waves make a new wave: When we add lots of them (using the sigma function \u03a3 as a handy notation) we can get things like: 20 sine waves: sin (x)+sin (3x)/3+sin (5x)/5 + + sin (39x)/39: Fourier Series Calculus Index. SC FINAL COMPLETE FOURIER SERIES CHAPTER 4 EXERCISE 4. Assume that the input voltage is the following square wave (\ud835\udf14 =\ud835\udf0b),. Determine the Fourier series expansion for full wave. Input arguments are used to specify the number of uniformly spaced points at which the. Fourier series, then the expression must be the Fourier series of f. Where, C is known as the Complex Fourier \u2026. Transition from Fourier series to Fourier transforms. series and Fourier transforms are mathematical techniques that do exactly that!, i. In this section we define the Fourier Sine Series, i. For the square wave a discontinuity exists at t/T 0 = 0. We can equivalently describe them as sums of complex exponentials, where each cosine requires two complex \u2026. 10 Fourier Series and Transforms (2014-5543) Complex Fourier Series: 3 \u2013 2 / 12 Euler\u2019s Equation: ei\u03b8 =cos\u03b8 +isin\u03b8 [see RHB 3. What is Fourier Series? Any real, periodic signal with fundamental freq. Thus, second harmonic component of Fourier series will be 0. The inverse Fourier transform given above (Eq. Jean Baptiste Joseph Fourier (1768-1830) \u2018Any univariate function can be rewritten as a weighted sum of sines and cosines of different frequencies. The steps to be followed for solving a Fourier series are given below: Step 1: Multiply the given function by sine or cosine, then integrate. 8 Complex Form of Fourier Series. Ans: The Fourier series is a linear combination of sines and cosines that expands periodic signals, whereas the Fourier transform is a method or \u2026. 2 Function spaces and metrics 5. A plane wave is propagating in the +z direction, passing through a scattering object at z=0, where its amplitude becomes A o(x,y). 1 Complex Full Fourier Series Recall that DeMoivre formula implies that sin( ) =. Fourier series using complex variables. Fourier series of a simple linear function f (x)=x converges to an odd periodic extension of this function, which is a saw-tooth wave\u2026. Compute and plot the intensity distribution of the diffracted wave at different distances from the aperture given by the Fresnel numbers N F = 20, 10, 4, and 1. 01: MATLAB M-FILE FOR PLOTTING TRUNCATED FOURIER SERIES AND ITS SPECTRA. The Fourier Series representation of continuous time periodic square wave signal, along with an interpretation of the Fourier series \u2026. Fourier transforms and solving the damped, driven oscillator. The coefficients may be determined rather easily by the use of Table 1. True Square waves are a special class of rectangular waves \u2026. Here is why a square wave is a good test of high frequencies: The Fourier series corresponding to the square wave includes an infinite number of odd-harmonic sine wave components. The input to a discrete function must be a whole number. Examples of the Fourier Series \u00b7 Rectangular Pulse Train \u00b7 Impulse Train \u00b7 Decaying Exponential Pulse Train \u00b7 Even Square Wave \u00b7 Odd Square Wave. Here, I\u2019ll use square brackets, [], instead of parentheses, (), to show discrete vs continuous time functions. In this Tutorial, we consider working out Fourier series for func-tions f(x) with period L = 2\u03c0. Experts are tested by Chegg as specialists in their subject area. The vowel signal and the square-wave are both examples that suggest the idea ofapproximating a periodic On the other hand, when the positive and negative frequency terms in the Fourier Series are combined we add a complex \u2026. Let the integer m become a real number and let the coefficients, F m, become a function F(m). The Fourier theory is used to analyze complex periodic signals. The first step is a trivial one: we need to generalize from real functions to complex functions, to include wave functions having nonvanishing current. Jean Baptiste Joseph Fourier \u2026. This is a consequence of $\\cos$ being \"half\" of a complex exponential but a constant is a \"full\" complex exponential. The complex Fourier series is more elegant and shorter to write down than the one expressed in term of sines and cosines, but it has the disadvantage that the coefficients might be complex even if the given function is real-valued. English Wikipedia has an article on: Fourier series. PHY 416, Quantum Mechanics Notes by: Dave Kaplan and Transcribed to LATEX by: Matthew S. The generalization of Fourier series to forms appropriate \u2026. Given a periodic function xT(t) and its Fourier Series representation (period= T, \u03c90=2\u03c0/T ): xT (t) = +\u221e \u2211 n=\u2212\u221ecnejn\u03c90t x T ( t) = \u2211 n = \u2212 \u221e + \u221e c n e j n \u03c9 0 t. A square wave is a type of waveform where the signal has only two levels. A Fourier series represents a function as an infinite sum of trigonometric functions: You can often use the complex exponential (DeMoivre's formula) to simplify computations: Specifically, let This graph is called a square wave\u2026. The resulting series is known as Fourier series. Algorithm: Fourier Transform. This gif demonstrates how combinations of Fourier series mathematical functions can be used to create complex animated surfaces like these: Fourier Series, & Rectify. What we are extracting in these cases are the coefficients for the Fourier \u2026. Consider the square wave: f(x) = 1 0 \u2264 x < \u03c0 = 0 \u2212\u03c0 \u2264 x < 0 f(x) = f(x+2\u03c0) This appears to be a di cult case - the rather angular square wave does not look as \u2026. In the previous lab, we implemented the trigonometric form of the Fourier Series, in this lab we will implement the complex form of the Fourier Series while learning some additional features of Simulink. Fourier Series Representation \u2022 Fourier Series Representation of CT Periodic Signals A periodic signal \ud835\udc65\ud835\udc65(\ud835\udc61\ud835\udc61) can be represented as a linear combination of harmonically related complex exponentials (or sinusoids) \ud835\udc4e\ud835\udc4e. The inverse Fourier transform is just to reconstruct the original function. Some operational formulas : 70. Graph of the function and its Fourier \u2026.", "date": "2022-05-22 08:12:33", "meta": {"domain": "xenchic.de", "url": "https://xenchic.de/complex-fourier-series-of-square-wave.html", "openwebmath_score": 0.8959225416183472, "openwebmath_perplexity": 578.2945620419474, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308768564867, "lm_q2_score": 0.8774767874818409, "lm_q1q2_score": 0.8670619074356445}}
{"url": "https://math.stackexchange.com/questions/3063596/optimization-problem-rectangle-inscribed-below-parabola-possible-textbook-mi", "text": "# Optimization problem (rectangle inscribed below parabola) - possible textbook mistake\n\nI have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.\n\nThe problem:\n\nA rectangle is located below a parabola, which is given by $$y = 3x- \\frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $$x$$ axis. The left, inferior vertex, is placed on the point $$(c,0)$$. That said: a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$ b) Find the rectangle's height and width given that is has maximum possible area.\n\nc) What is that area?\n\nInstead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.\n\nIf that's helpful, the textbook's answer for b) and c) are $$3-\\sqrt{3} \\times 3$$ and $$9 + 9\\sqrt{3}$$ respectively.\n\nThank you.\n\n\u2022 $base=2\\sqrt{3}$ \u2013\u00a0Aleksas Domarkas Jan 6 at 8:19\n\nYou made a slight error when trying to find the base:\n\n$$c = 3-\\sqrt{3} \\implies b = 2(3-c) = \\color{blue}{2\\left[3-\\left(3-\\sqrt{3}\\right)\\right]} = 2\\sqrt{3}$$\n\nYou forgot the $$3$$ in $$(\\color{blue}{3}-c)$$ and found $$b = 2c$$ instead.\n\nAddition: From here, using $$h = 3$$ as you found, you get\n\n$$A = bh \\iff A = 3\\left(2\\sqrt{3}\\right) = 6\\sqrt{3}$$\n\nTry plugging in $$c = 3-\\sqrt{3}$$ in $$f(c)$$:\n\n$$f\\left(3-\\sqrt{3}\\right) = \\left(3-\\sqrt{3}\\right)^3-9\\left(3-\\sqrt{3}\\right)^2+18\\left(3-\\sqrt{3}\\right) = 6\\sqrt{3}$$\n\nThrough confirmation, you can see this point coincides with the local maximum.\n\n\u2022 thank you I've fixed that mistake. Now I have $2 \\sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 \\sqrt{3} \\ m^2$. Am I good? Thanks for your help! \u2013\u00a0bru1987 Jan 6 at 9:16\n\u2022 Yes, that\u2019s the correct answer! \u2013\u00a0KM101 Jan 6 at 9:24\n\u2022 thank you and have a great day =) \u2013\u00a0bru1987 Jan 6 at 9:26\n\u2022 Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue. \u2013\u00a0KM101 Jan 6 at 9:31", "date": "2019-08-17 10:53:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3063596/optimization-problem-rectangle-inscribed-below-parabola-possible-textbook-mi", "openwebmath_score": 0.7581220269203186, "openwebmath_perplexity": 451.3469354588943, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308775555446, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8670619064660234}}
{"url": "https://mathoverflow.net/questions/372450/parity-of-the-multiplicative-order-of-2-modulo-p", "text": "# Parity of the multiplicative order of 2 modulo p\n\nLet $$\\operatorname{ord}_p(2)$$ be the order of 2 in the multiplicative group modulo $$p$$. Let $$A$$ be the subset of primes $$p$$ where $$\\operatorname{ord}_p(2)$$ is odd, and let $$B$$ be the subset of primes $$p$$ where $$\\operatorname{ord}_p(2)$$ is even. Then how large is $$A$$ compared to $$B$$?\n\n\u2022 $A/(A+B)$ tends to $7/24$ ? (not proved yet). \u2013\u00a0Henri Cohen Sep 23 at 21:39\n\u2022 Seems like an interesting question, and clearly generalizable quite a lot. However, if you're going to ask many questions on this site, it would be a good idea to learn a little bit of TeX formatting. I've fixed the formatting of your question, so if you click on \"edit\", you'll be able to see what I did to make it more readable. I also changed the title of your question to make it even clearer what you're asking. \u2013\u00a0Joe Silverman Sep 23 at 21:54\n\u2022 @HenriCohen how did you determine $A/(A+B)$ to be $7/24$ while also writing \"not proved yet\"? The proportion of $p \\leq 100000$ for which $2 \\bmod p$ has odd order is $2797/9591$, which as a continued fraction is $[0,3,2,3,44,9]$, and the truncated continued fraction $[0,3,2,3]$ is $7/24$. I'd be interested to know if you did that or something else. \u2013\u00a0KConrad Sep 24 at 3:26\n\u2022 Note: the set $A$ is at OEIS: oeis.org/A014663, while its complement $B$ is oeis.org/A091317. Among the 46 primes below 200, $A$ consists of the 14 primes 7, 23, 31, 47, 71, 73, 79, 89, 103, 127, 151, 167, 191, 199. \u2013\u00a0YCor Sep 24 at 9:44\n\u2022 see this answer \u2013\u00a0Ren\u00e9 Gy Sep 24 at 15:39\n\nThis problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s.\n\nFor each nonzero rational number $$a$$ (take $$a \\in \\mathbf Z$$ if you wish) and each prime $$\\ell$$, let $$S_{a,\\ell}$$ be the set of primes $$p$$ not dividing the numerator or denominator of $$a$$ such that $$a \\bmod p$$ has multiplicative order divisible by $$\\ell$$. When $$a = \\pm 1$$, $$S_{a,\\ell}$$ is empty except that $$S_{-1,2}$$ is all odd primes. From now on, suppose $$a \\not= \\pm 1$$.\n\nIn Math. Ann. 162 (1965/66), 74\u201376 (the paper is at https://eudml.org/doc/161322 and on MathSciNet see MR0186653) Hasse treated the case $$\\ell \\not= 2$$. Let $$e$$ be the largest nonnegative integer such that $$a$$ in $$\\mathbf Q$$ is an $$\\ell^e$$-th power. (For example, if $$a$$ is squarefree then $$e = 0$$ for every $$\\ell$$ not dividing $$a$$.) The density of $$S_{a,\\ell}$$ is $$\\ell/(\\ell^e(\\ell^2-1))$$. This is $$\\ell/(\\ell^2-1)$$ when $$e = 0$$ and $$1/(\\ell^2-1)$$ when $$e = 1$$.\n\nIn Math. Ann. 166 (1966), 19\u201323 (the paper is at https://eudml.org/doc/161442 and on MathSciNet see MR0205975) Hasse treated the case $$\\ell = 2$$. The general answer in this case is more complicated, as issues involving $$\\ell$$-th roots of unity in the ground field (like $$\\pm 1$$ in $$\\mathbf Q$$ when $$\\ell = 2$$) often are. The density of $$S_{a,2}$$ for \"typical\" $$a$$ is $$1/3$$, such as when $$a \\geq 3$$ is squarefree. But $$S_{2,2}$$ has density 17/24, so the set of $$p$$ for which $$2 \\bmod p$$ has even order has density $$17/24$$ and the set of $$p$$ for which $$2 \\bmod p$$ has odd order has density $$1 - 17/24 = 7/24$$.\n\nFor example, there are $$167$$ odd primes up to $$1000$$, $$1228$$ odd primes up to $$10000$$, and $$9591$$ odd primes up to $$100000$$. There are $$117$$ odd primes $$p \\leq 1000$$ such that $$2 \\bmod p$$ has even order, $$878$$ odd primes $$p \\leq 10000$$ such that $$2 \\bmod p$$ has even order, and $$6794$$ odd primes $$p \\leq 100000$$ such that $$2 \\bmod p$$ has even order. The proportion of odd primes up $$1000$$, $$10000$$, and $$100000$$ for which $$2 \\bmod p$$ has even order is $$117/167 \\approx .700059$$, $$878/1228 \\approx .71498$$, and $$6794/9591 \\approx .70837$$, while $$17/24 \\approx .70833$$.\n\nThe math.stackexchange page here treats $$S_{7,2}$$ in some detail and at the end mentions the case of $$S_{2,2}$$.\n\n\u2022 Thanks @KConrad for the expression. I was thinking about a combinatorial property of cyclic groups of prime order(called acyclic matching property) and I proved the above mentioned sequence of primes does not hold it. See Proposition 2.3 of core.ac.uk/download/pdf/33123051.pdf Anyway I will like to mention this result in my ongoing research work as a remark, and hence I ask your permission for the same, of course with acknowledgment. \u2013\u00a0Mohsen Sep 25 at 18:09\n\u2022 Since the result is due to Hasse, cite his paper when you want to indicate who first showed that the density exists and what its value is. \u2013\u00a0KConrad Sep 25 at 19:35", "date": "2020-10-31 04:14:33", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/372450/parity-of-the-multiplicative-order-of-2-modulo-p", "openwebmath_score": 0.9865131378173828, "openwebmath_perplexity": 134.73594820351857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308793031894, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.867061901667429}}
{"url": "https://web.forret.com/ruv2b6uq/67b03a-gibson-es-345-stereo", "text": "File history. Simple C# class to calculate Cantor's pairing function. \\end{array} Cantor's pairing function 08 17 In addition to the diagonal arguments , Georg Cantor also developed the Cantor pairing function $$\\mathbb{N}^2 \\to \\mathbb{W}, \\quad c(x,y) = \\binom{x+y+1}{2}+x = z$$ , which encodes any two numbers $$x,y \\in \\mathbb{N}$$ in a new number $$z \\in \\mathbb{N}$$ . var t = ( int) Math. But we know that the end-points survive the Cantor intersection, that is they lie in C. Hence [x 1=3k;x+ 1=3k] f xginter-sects Cfor every k. The ideas discussed in this post are implemented using GHC generics in the package cantor-pairing. Maybe your data comes from two different databases, and each one has its unique identifier for individuals, but both unique codings overlap with each other. In a perfectly efficient function we would expect the value of pair(9, 9) to be 99. For example, the Cantor pairing function \u03c0: N 2 \u2192 N is a bijection that takes two natural numbers and maps each pair to a unique natural number. When we apply the pairing function to k 1 and k 2 we often denote the resulting number as k 1, k 2 . Set theory - Set theory - Equivalent sets: Cantorian set theory is founded on the principles of extension and abstraction, described above. Essentially any time you want to compose a unique identifier from a pair of values. In this paper, some results and generalizations about the Cantor pairing function are given. Already have an account? Python Topic. \\end{array} Elements in Cantor's ternary set x, y such that x + y = w with w \u2208 [0, 1] It is known that a real number w \u2208 [0, 1] can be written as a sum of two real numbers such that x, y \u2208 C such that 1 2C + 1 2C = [0, 1] with C the ternary Cantor's set. Maria E. Reis is a visiting MSc student at the University of Kentucky under Dr. Costa\u2019s supervision. \\end{array} The so-called Cantor pairing function C(m;n) = mX+n j=0 j + m = 1 2 (m+ n)(m+ n+ 1) + m; maps N 0 N 0 injectively onto N 0 (Cantor, 1878). More formally Tags encoding, pairing, cantor Maintainers perrygeo Classifiers. \\end{array} This can be easily implemented in any language. You can also compose the function to map 3 or more numbers into one \u2014 for example maps 3 integers to one. What makes a pairing function special is that it is invertable; You can reliably depair the same integer value back into it's two original values in the original order. OS Independent Programming Language. Let C be the projection of the standard (ternary) Cantor set on the unit interval to the circle. Yes, the Szudzik function has 100% packing efficiency. ElegantPairing.nb \u00c7 \u00c5 \u00a1 3 of 12 Cantor\u2019s Pairing Function Here is a classic example of a pairing function (see page 1127 of A \u2026 cantor-pairing. Cantor\u2019s pairing function c(x 1,x 2) is a quadratic polynomial pairing func-tion. It is also used as a fundamental tool in recursion theory and other related areas of mathematics (Rogers, 1967; Matiyasevich, 1993). (32.4 x 36.2 cm) By (primary) Artist unknown Washington, D.C.-based Cantor Colburn associate Jenae Gureff attended the AIPLA 2015 Mid-Winter Institute meeting in Orlando. \\end{array} Whether this is the only polynomial pairing function is still an open question. Definition: A set S is a countable set if and only if there exists a bijective mapping , where is the set of natural numbers (i.e. In fact, Cantor's method of proof of this theorem implies the existence of an \" \u2026 (It's not an edge in an TikZ path operator kind of way.) Anatole Katok, Jean-Paul Thouvenot, in Handbook of Dynamical Systems, 2006. 2y & : y \\ge 0 Floor ( ( -1 + Math. Please include this proof (either directly or through a link) in your answer. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Neither Cantor nor Szudzik pairing functions work natively with negative input values. When we apply th\u2026 Cantor\u2019s grades at age 8, when he attended the St.Petri-Schule for German speaking people in St.Petersburg. 'Cantor' and 'Elegant' pairing are relatively symmetric around the main diagonal. The trick to solve this is to either factorize the input, or pass in x \u2013 min(x). Cantor Pairing. Economics, programming, and games. Cantor established the importance of one-to-one correspondence between the members of two sets, defined infinite and well-ordered sets, and proved that the real numbers are more numerous than the natural numbers. An illustration of Cantor's Pairing Function. One of the better ways is Cantor Pairing, which is the following magic formula: This takes two positive integers, and returns a unique positive integer. ... (16) S. R. Jaskunas, C. R. Cantor, and I. Tinoco, Jr , manuscript in preparation. For that, you sort the two Cantor normal forms to have the same terms, as here, and just add coordinate-wise. Definition: A set S is a countable set if and only if there exists a bijective mapping , where is the set of natural numbers (i.e. b^2 + a & : a < b\\\\ b^2 + a & : a < b\\\\ If one defines cantor 2 edge/.style={move to} the diagonal part will not be drawn. We consider the theory of natural integers equipped with the Cantor pairing function and an extra relation or function Xon N. When Xis equal either to multiplication, or coprimeness, or divisibility, or addition or natural ordering, it can be proved that the theory Th(N;C;X) is undecidable. -c - 1 & : (a < 0 \\cap b \\ge 0) \\cup (a \\ge 0 \\cap b < 0) This graphics demonstrates the path that Szudzik takes over the field: The primary benefit of the Szudzik function is that it has more efficient value packing. However, cantor(9, 9) = 200. If (m;n) is the row-column indexing, C(m;n) gives the following pattern of enumeration: 0 1 3 6 10 15 2 4 7 11 16 5 8 12 17 9 13 18 14 19 20 To check that C(m;n) is indeed a bijection, we need the below property. \\right.$$,$$index = {(a + b)(a + b + 1) \\over 2} + b$$,$$index(a,b) = \\left\\{\\begin{array}{ll} Additional space can be saved, giving improved packing efficiency, by transferring half to the negative axis. Thus this solvent is an excellent system in which to study the effects of base pairing on the for- mation of specific complexes between mono- or oligonucleotides. a^2 + a + b & : a \\ge b $$b = \\left\\{\\begin{array}{ll} Cantor\u2019s school career was like that of \u2026 It should be noted that this article was adapted from an earlier jsfiddle of mine. Now use the pairing function again to turn your two remaining items into 1. As such, we can calculate the max input pair to Szudzik to be the square root of the maximum integer value. Matt Ranger's blog. That fiddle makes note of the following references:$$index = \\left\\{\\begin{array}{ll} Date/Time Thumbnail Dimensions User Comment; current: 16:12, 10 June 2020: 432 \u00d7 432 (39 KB) Crh23: Another JavaScript example: Szudzik can also be visualized as traversing a 2D field, but it covers it in a box-like pattern. Definition A pairing function on a set A associates each pair of members from A with a single member of A, so that any two distinct pairs are associated with two distinct members. Definition A pairing function on a set A associates each pair of members from A with a single member of A, so that any two distinct pairs are associated with two distinct members. 1 - Planning Intended Audience. It\u2019s also reversible: given the output of you can retrieve the values of and . The typical example of a pairing function that encodes two non-negative integers onto a single non-negative integer (therefore a function ) is the Cantor function, instrumental to the demonstration that, for example, the rational can be mapped onto the integers.. Let\u2019s say you have some data with two columns which are different identifiers. The third and last one (POTO pairing) is more asymmetric. We prove a conjecture of I. Korec [4] on decidability of some fragments of arithmetic equipped with a pairing function; as consequence, we give an axiomatization of the fragment of arithmetic equipped with Cantor pairing function, precising a result of [5]. ElegantPairing.nb For a 32-bit unsigned return value the maximum input value for Szudzik is 65,535. The typical example of a pairing function that encodes two non-negative integers onto a single non-negative integer (therefore a function ) is the Cantor function, instrumental to the demonstration that, for example, the rational can be mapped onto the integers.. The \ufb01rst order theory of natural integers equipped with the The twist for coding is not to just add the similar terms, but also to apply a natural number pairing function also. You may implement whatever bijective function you wish, so long as it is proven to be bijective for all possible inputs. Construct the \u201cuniform\u201d measure \u03bc on C by assigning the measures 1/2 n to the intersection of C with the intervals of nth order. The full results of the performance comparison can be found on jsperf. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The binary Cantor pairing function C from N \u00d7 N into N is defined by C(x, y) = (1/2) (x + y)(x + y + 1) + y. \\right.$$, https://en.wikipedia.org/wiki/Pairing_function. Or maybe you want to combine encodings from multiple columns into one. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. Sometimes you have to encode reversibly two (or more) values onto a single one. It is however mixing. Click on a date/time to view the file as it appeared at that time. It\u2019s also reversible: given the output of you can retrieve the values of and . \\end{array} The only problem with this method is that the size of the output can be large: will overflow a 64bit integer 1. \\right.$$ Development Status. This means that all one hundred possible variations of ([0-9], [0-9]) would be covered (keeping in mind our values are 0-indexed). x^2 + x + y & : x \\ge y The primary downside to the Cantor function is that it is inefficient in terms of value packing. To describe some results based upon these principles, the notion of equivalence of sets will be defined. Cantor pairing gives us an isomorphism between a single natural number and pairs of natural numbers. We may assume y k 6= xhere (if xhappens to be an end-point of an interval in C k itself, choose the other end-point of the interval to be y k). Pairing functions take two integers and give you one integer in return. cantor (9, 9) = 200 szudzik (9, 9) = 99. y^2 + x & : x < y\\\\ The Cantor pairing function C from N \u00d7 N into N is de\ufb01ned by C (x, y)=(1 / 2) ( x + y )( x + y +1 )+ y . Items portrayed in this file depicts. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. This is the question Cantor pondered, and in doing so, came up with several interesting ideas which remain important to this day. In particular, it is investigated a very compact expression for the n -degree generalized Cantor pairing function (g.C.p.f., for short), that permits to obtain n \u2212tupling functions which have the characteristics to be n -degree polynomials with rational coef\ufb01cients. Trying to bump up your data type to an unsigned 32-bit integer doesn\u2019t buy you too much more space: cantor(46500, 46500) = 4,324,593,000, another overflow. Developers Science/Research License. One of the better ways is Cantor Pairing, which is the following magic formula: This takes two positive integers, and returns a unique positive integer. The only problem with this method is that the size of the output can be large: will overflow a 64bit integer 1. By repeatedly applying Cantor\u2019s pairing function in this \\right.$$We quickly start to brush up against the limits of 32-bit signed integers with input values that really aren\u2019t that large. It should be noted though that all returned pair values are still positive, as such the packing efficiency for both functions will degrade. 2 This measure is obviously singular. The good news is that this will use all the bits in your integer efficiently from the view of a hashing function. Like Cantor, the Szudzik function can be easily implemented anywhere. While this is cool, it doesn\u2019t seem useful for practical applications. Set theory - Set theory - Operations on sets: The symbol \u222a is employed to denote the union of two sets. So we use 200 pair values for the first 100 combinations, an efficiency of 50%. -2y - 1 & : y < 0\\\\ Rider to Pair of Horses with Riders 3rd century BCE-3rd century 3rd C. BCE-3rd C. CE Asia, China 12 3/4 x 14 1/4 in. by Georg Cantor in 1878. Journal of the American Chemical Society 90:18 / August 28, 1968 . . Sqrt ( 1 + 8 * cantor )) / 2 ); Sign up for free to join this conversation on GitHub . Come in to read stories and fanfics that span multiple fandoms in the Naruto and Big Mouth universe. This package provides a modern API to this functionality using GHC generics, allowing the encoding of arbitrary combinations of finite or countably infinite types in natural number form. Because theoreticaly I can now Pair \u2026 k in C k such that jx y kj 1=3k.$$index = {(x + y)(x + y + 1) \\over 2} + y$$. Let \u2026 a^2 + a + b & : a \\ge b In this paper, some results and generalizations about the Cantor pairing function are given. In this ramble we will cover two different pairing functions: Cantor and Szudzik. pairing function. This makes it harder to retrieve x and y, though.\u21a9, \u201cKey registers on keyboard but not on computer\u201d fix, Bad Economics: Shame on you, Planet Money (MMT episode), BadEconomics: Putting 400M of Bitcoin on your company balance sheet, Starting a Brick & Mortar Business in 2020, The publishing format defines the art: How VHS changed movie runtimes, The rural/urban divide is an American phenomenon and other bad takes, Why Stephen Wolfram\u2019s research program is a dead end, \u201cbigger\u201d than the infinity of normal numbers. It can be used when one index should grow quicker than the other (roughly hyperbolic). -2x - 1 & : x < 0\\\\ The problem is, at least from my point of view, in Java I had to implement a BigSqrt Class which I did by my self. As such, we can calculate the max input pair to Szudzik to be the square root of the maximum integer value. Sometimes you have to encode reversibly two (or more) values onto a single one. \\right.$$, a = \\left\\{\\begin{array}{ll} In your integer efficiently from the view of a hashing function Melissa Cantor interval to the negative.. Now use the pairing function in this post are implemented using GHC generics in the package cantor-pairing would in! Photos provided by Melissa Cantor visualized as traversing a 2D field, but also to apply a number. 'S not an edge in an TikZ path operator kind of way )! Two remaining items into 1 values are still positive, as here, and in doing,! News is that the size of the output of you can retrieve the values of and post... Pair ( 9, 9 ) to be bijective for all possible inputs c++ cantor pairing Szudzik can also the... Describe some results based upon these principles, the Szudzik function has 100 % packing efficiency remain important to day. 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Generalized to the Cantor function is still an open question denote the resulting as.\nCordless Strimmer With Blades, Banana Peppers Recipes, Access Community Health Network Careers, Is Brf3 Polar Or Nonpolar, Psychological Barriers To Communication,", "date": "2021-10-22 16:43:56", "meta": {"domain": "forret.com", "url": "https://web.forret.com/ruv2b6uq/67b03a-gibson-es-345-stereo", "openwebmath_score": 0.6413760185241699, "openwebmath_perplexity": 1260.9034039589326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.975576912076157, "lm_q2_score": 0.8887587993853655, "lm_q1q2_score": 0.8670525650848876}}
{"url": "https://math.stackexchange.com/questions/1124242/how-to-calculate-exp-x-using-taylor-series", "text": "# How to calculate $\\exp(-x)$ using Taylor series\n\nWe know that the Taylor series expansion of $e^x$ is $$e^x = \\sum\\limits_{i=1}^{\\infty}\\frac{x^{i-1}}{(i-1)!}.$$\n\nIf I have to use this formula to evaluate $e^{-20}$, how should I check for convergence?\n\nMATLAB gives $e^{-20} = 2.0612e-009$. But when I use this formula to calculate it, I get $4.1736e-009$. My answer becomes more inaccurate as I take $x = -30, -40 \\cdots$. The reason is obvious. This is because as the number of terms in the Taylor series increases, the factorial becomes more and more inaccurate using MATLAB.\n\n1. How can I use Taylor series to get accurate values of $e^x$ in MATLAB?\n2. How does MATLAB built in command \"exp()\" calculate the exponential?\n\u2022 If the ultimate goal is to approximate the exponential I would use a Pade' approximants (a rational approximation), combined with $e^{2x}=(e^x)^2$ to get values for larger values of $x$. \u2013\u00a0Pp.. Jan 28 '15 at 23:39\n\u2022 If you just want to see how the Taylor series behaves, then the error of an alternating series can be bounded by the absolute value of the next term of the series. \u2013\u00a0Pp.. Jan 28 '15 at 23:40\n\u2022 TS is good for small values of the argument. Otherwise you might end up having many extra terms. \u2013\u00a0Kaster Jan 28 '15 at 23:42\n\u2022 You may try summing backwards. \u2013\u00a0lhf Jan 29 '15 at 11:44\n\u2022 Oog. Why not start the summation at $0$? \u2013\u00a0Qiaochu Yuan Jan 30 '15 at 8:54\n\nThe Problem\n\nThe main problem when computing $e^{-20}$ is that the terms of the series grow to $\\frac{20^{20}}{20!}\\approx43099804$ before getting smaller. Then the sum must cancel to be $\\approx2.0611536\\times10^{-9}$. In a floating point environment, this means that $16$ digits of accuracy in the sum are being thrown away due to the precision of the large numbers. This is the number of digits of accuracy of a double-precision floating point number ($53$ bits $\\sim15.9$ digits).\n\nFor example, the RMS error in rounding $\\frac{20^{20}}{20!}$, using double precision floating point arithmetic, would be $\\sim\\frac{20^{20}}{20!}\\cdot2^{-53}/\\sqrt3\\approx3\\times10^{-9}$. Since the final answer is $\\approx2\\times10^{-9}$, we lose all significance in the final answer simply by rounding that one term in the sum.\n\nThe problem gets worse with larger exponents. For $e^{-30}$, the terms grow to $\\frac{30^{30}}{30!}\\approx776207020880$ before getting smaller. Then the sum must cancel to be $\\approx9.35762296884\\times10^{-14}$. Here we lose $25$ digits of accuracy. For $e^{-40}$, we lose $33$ digits of accuracy.\n\nA Solution\n\nThe usual solution is to compute $e^x$ and then use $e^{-x}=1/e^x$. When computing $e^x$, the final sum of the series is close in precision to the largest term of the series. Very little accuracy is lost.\n\nFor example, the RMS error in computing $e^{20}$ or $e^{-20}$, using double precision floating point arithmetic, would be $\\sim8\\times10^{-9}$; the errors are the same because both sums use the same terms, just with different signs. However, this means that using Taylor series, \\begin{align} e^{20\\hphantom{-}}&=4.85165195409790278\\times10^8\\pm8\\times10^{-9}\\\\ e^{-20}&=2\\times10^{-9}\\pm8\\times10^{-9} \\end{align} Note that the computation of $e^{-20}$ is completely insignificant. On the other hand, taking the reciprocal of $e^{20}$, we get $$e^{-20}=2.061153622438557828\\times10^{-9}\\pm3.4\\times10^{-26}$$ which has almost $17$ digits of significance.\n\n\u2022 Would the downvoter care to comment? \u2013\u00a0robjohn Jan 30 '15 at 1:43\n\u2022 this is so nice an explanation!! When doing numerical calculation the formulas must be modified in a manner to avoid loss of significant digits and such that intermediate calculation does not require a considerably higher precision than that desired in the final answer. \u2013\u00a0Paramanand Singh Feb 19 '15 at 10:48\n\nA standard technique, not particular to Matlab, is to exploit a function's symmetries to map its inputs to a more manageable range. The exponential function satisfies the symmetry $e^{a+b}=e^ae^b$, and it's easy to multiply floating-point numbers by powers of two. So do this: $$\\begin{eqnarray} e^{-20}&=&2^{-20\\lg e}\\\\ &\\approx&2^{-28.854}\\\\ &=&2^{-29}\\times2^{0.146}\\\\ &=&2^{-29}\\times e^{0.146\\log2}\\\\ &\\approx&2^{-29}\\times e^{0.101} \\end{eqnarray}$$ Now you can easily evaluate $e^{0.101\\ldots}$ using the Taylor series.\n\n1) The regular double precision floating point arithmetic of Matlab is not sufficient to precisely calculate partial sums of this power series. To overcome that limitation, you can use the exact symbolic computation capabilities of the Symbolic Math Toolbox. This code\n\nx = sym(-20);\ni = sym(1 : 100);\nexpx = sum(x .^ (i - 1) ./ factorial(i - 1))\n\n\nsums the first 100 terms of the series, and yields the exact (for the partial sum) result $$\\tiny\\frac{20366871962240780739193874417376755657912596966525153814418798643652163252126492695723696663600640716177}{9881297415446727147594496649775206852319571477668037853762810667968023095834839075329261976769165978884198811117}$$ whose numeric counterpart (computed by double()) is 2.06115362243856e-09.\n\n2) The Matlab function exp() is most likely a wrapper for C code, which presumably references the standard C library function exp().\n\nYou can have a look at the code of the widely used GNU C Library math functions, where you'll find that most functions are themselves wrappers around machine-specific implementations. For the i386 architecture, the path leads ultimately to the function __ieee754_exp which implements the exponential in inline assembler code (but don't ask me how to decipher this). On other architectures, there might be a single machine instruction that does the job.\n\n\"Most likely\", \"probably\", \"presumably\" because Matlab is closed source and therefore definitive knowledge only exists within The Mathworks.\n\n\u2022 You mention using \"variable precision arithmetic\" in Matlab but your example code is an exact symbolic calculation converted to floating-point. There appears to be no need for vpa for the ranges used. The OP also asked \"how should I check for convergence?\" which I think was the impetus for the question. \u2013\u00a0horchler Jan 29 '15 at 3:15\n\u2022 @horchler, correct with the vpa thing, I got the two confused, edited. \u2013 As for convergence, I think this was a minor part of the question made irrelevant by the fact that his double precision computation couldn't converge anyway. 4.1e-9 isn't not-converged, its just off. Feel free to add a better answer though, or even to edit mine. :-) \u2013\u00a0A. Donda Jan 29 '15 at 11:04", "date": "2019-09-17 12:25:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1124242/how-to-calculate-exp-x-using-taylor-series", "openwebmath_score": 0.8695510625839233, "openwebmath_perplexity": 442.8723294111857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769092358046, "lm_q2_score": 0.8887588008585925, "lm_q1q2_score": 0.8670525639977457}}
{"url": "http://math.stackexchange.com/questions/221078/poisson-distribution-of-sum-of-two-random-independent-variables-x-y", "text": "# Poisson Distribution of sum of two random independent variables $X$, $Y$\n\n$X \\sim \\mathcal{P}( \\lambda)$ and $Y \\sim \\mathcal{P}( \\mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \\sim \\mathcal{P}( \\lambda + \\mu)$ but I don't understand how to derive it.\n\n-\nTry using the method of moment generating functions :) \u2013\u00a0 Samuel Reid Nov 25 '13 at 7:03\nAll I've learned in the definition of a Poisson Random Variable, is there a simpler way? \u2013\u00a0 user82004 Nov 25 '13 at 7:07\nIf they are independent. \u2013\u00a0 Did Nov 25 '13 at 8:14\n\nThis only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k \\ge 0$: \\begin{align*} P(X+ Y =k) &= \\sum_{i = 0}^k P(X+ Y = k, X = i)\\\\ &= \\sum_{i=0}^k P(Y = k-i , X =i)\\\\ &= \\sum_{i=0}^k P(Y = k-i)P(X=i)\\\\ &= \\sum_{i=0}^k e^{-\\mu}\\frac{\\mu^{k-i}}{(k-i)!}e^{-\\lambda}\\frac{\\lambda^i}{i!}\\\\ &= e^{-(\\mu + \\lambda)}\\frac 1{k!}\\sum_{i=0}^k \\frac{k!}{i!(k-i)!}\\mu^{k-i}\\lambda^i\\\\ &= e^{-(\\mu + \\lambda)}\\frac 1{k!}\\sum_{i=0}^k \\binom ki\\mu^{k-i}\\lambda^i\\\\ &= \\frac{(\\mu + \\lambda)^k}{k!} \\cdot e^{-(\\mu + \\lambda)} \\end{align*} Hence, $X+ Y \\sim \\mathcal P(\\mu + \\lambda)$.\n\n-\nThank you! but what happens if they are not independent? \u2013\u00a0 user31280 Oct 25 '12 at 20:20\nIn general we can't say anything then. It depends on how they depend on another. \u2013\u00a0 martini Oct 25 '12 at 20:22\nThank you! it's very simple and I feel like a complete idiot. \u2013\u00a0 user31280 Oct 25 '12 at 20:40\nNice derivation: specifically the transformation of (a) the i/k factorials and (b) the mu/lambda polynomials into the binomial form of the polynomial power expression. \u2013\u00a0 javadba Aug 30 '14 at 20:59\n\nAnother approach is to use characteristic functions. If $X\\sim \\mathrm{po}(\\lambda)$, then the characteristic function of $X$ is (if this is unknown, just calculate it) $$\\varphi_X(t)=E[e^{itX}]=e^{\\lambda(e^{it}-1)},\\quad t\\in\\mathbb{R}.$$ Now suppose that $X$ and $Y$ are independent Poisson distributed random variables with parameters $\\lambda$ and $\\mu$ respectively. Then due to the independence we have that $$\\varphi_{X+Y}(t)=\\varphi_X(t)\\varphi_Y(t)=e^{\\lambda(e^{it}-1)}e^{\\mu(e^{it}-1)}=e^{(\\mu+\\lambda)(e^{it}-1)},\\quad t\\in\\mathbb{R}.$$ As the characteristic function completely determines the distribution, we conclude that $X+Y\\sim\\mathrm{po}(\\lambda+\\mu)$.\n\n-\n\nYou can use Probability Generating Function(P.G.F). As poisson distribution is a discrete probability distribution, P.G.F. fits better in this case.For independent X and Y random variable which follows distribution Po($\\lambda$) and Po($\\mu$). P.G.F of X is \\begin{equation*} \\begin{split} P_X[t] = E[t^X]&= \\sum_{x=0}^{\\infty}t^xe^{-\\lambda}\\frac{\\lambda^x}{x!}\\\\ &=\\sum_{x=0}^{\\infty}e^{-\\lambda}\\frac{(\\lambda t)^x}{x!}\\\\ &=e^{-\\lambda}e^{\\lambda t}\\\\ &=e^{-\\lambda (1-t)}\\\\ \\end{split} \\end{equation*} P.G.F of Y is \\begin{equation*} \\begin{split} P_Y[t] = E[t^Y]&= \\sum_{y=0}^{\\infty}t^ye^{-\\mu}\\frac{\\mu^y}{y!}\\\\ &=\\sum_{y=0}^{\\infty}e^{-\\mu}\\frac{(\\mu t)^y}{y!}\\\\ &=e^{-\\mu}e^{\\mu t}\\\\ &=e^{-\\mu (1-t)}\\\\ \\end{split} \\end{equation*}\n\nNow think about P.G.F of U = X+Y. As X and Y are independent, \\begin{equation*} \\begin{split} P_U(t)=P_{X+Y}(t)=P_X(t)P_Y(t)=E[t^{X+Y}]=E[t^X t^Y]&= E[t^X]E[t^Y]\\\\ &= e^{-\\lambda (1-t)}e^{-\\mu (1-t)}\\\\ &= e^{-(\\lambda+\\mu) (1-t)}\\\\ \\end{split} \\end{equation*}\n\nNow this is the P.G.F of $Po(\\lambda + \\mu)$ distribution. Therefore,we can say U=X+Y follows Po($\\lambda+\\mu$)\n\n-\nSeems a typo in third line (out of 4) in both PGF of X and Y. Should be exp(-u)exp(mut) instead of exp(-u)exp(-1*mut) \u2013\u00a0 javadba Aug 30 '14 at 21:19\nSeparate comment/question: please explain why in second line of derivation for PGF of Y refers to exp(x) and x! (instead of referring to exp(y) and y!). I believe these are cut/paste errors - but please confirm. \u2013\u00a0 javadba Aug 30 '14 at 21:26\nYeah.. You are absolutely correct. Those were typos. I have edited them now. If you find more, let me know. \u2013\u00a0 Ananda Sep 2 '14 at 5:35\n\nhint: $\\sum_{k=0}^{n} P(X = k)P(Y = n-k)$\n\n-\nwhy this hint, why the sum? This is what I don't understand \u2013\u00a0 user31280 Oct 25 '12 at 20:22\nadding two random variables is simply convolution of those random variables. That's why. \u2013\u00a0 jay-sun Oct 25 '12 at 20:24\ngotcha! Thanks! \u2013\u00a0 user31280 Oct 25 '12 at 20:31\nadding two random variables is simply convolution of those random variables... Sorry but no. \u2013\u00a0 Did Feb 13 '13 at 6:28\nThere is no usual sense for convolution of random variables. Either convolution of distributions or addition of random variables. \u2013\u00a0 Did Feb 13 '13 at 6:51\n\nIn short, you can show this by using the fact that $$Pr(X+Y=k)=\\sum_{i=0}^kPr(X+Y=k, X=i).$$\n\nIf $X$ and $Y$ are independent, this is equal to $$Pr(X+Y=k)=\\sum_{i=0}^kPr(Y=k-i)Pr(X=i)$$ which is \\begin{align} Pr(X+Y=k)&=\\sum_{i=0}^k\\frac{e^{-\\lambda_y}\\lambda_y^{k-i}}{(k-i)!}\\frac{e^{-\\lambda_x}\\lambda_x^i}{i!}\\\\ &=e^{-\\lambda_y}e^{-\\lambda_x}\\sum_{i=0}^k\\frac{\\lambda_y^{k-i}}{(k-i)!}\\frac{\\lambda_x^i}{i!}\\\\ &=\\frac{e^{-(\\lambda_y+\\lambda_x)}}{k!}\\sum_{i=0}^k\\frac{k!}{i!(k-i)!}\\lambda_y^{k-i}\\lambda_x^i\\\\ &=\\frac{e^{-(\\lambda_y+\\lambda_x)}}{k!}\\sum_{i=0}^k{k\\choose i}\\lambda_y^{k-i}\\lambda_x^i \\end{align} The sum part is just $$\\sum_{i=0}^k{k\\choose i}\\lambda_y^{k-i}\\lambda_x^i=(\\lambda_y+\\lambda_x)^k$$ by the binomial theorem. So the end result is \\begin{align} Pr(X+Y=k)&=\\frac{e^{-(\\lambda_y+\\lambda_x)}}{k!}(\\lambda_y+\\lambda_x)^k \\end{align} which is the pmf of $Po(\\lambda_y+\\lambda_x)$.\n\n-\nModerator notice: This answer was moved here as a consequence of merging two questions. This explains the small differences in notation. The OP's $\\lambda$ is $\\lambda_x$ here, and OP's $\\mu$ is $\\lambda_y$. Otherwise there is no difference. \u2013\u00a0 Jyrki Lahtonen Apr 23 at 6:55", "date": "2015-09-01 22:25:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/221078/poisson-distribution-of-sum-of-two-random-independent-variables-x-y", "openwebmath_score": 0.9970618486404419, "openwebmath_perplexity": 1436.7829811806564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156283, "lm_q2_score": 0.8887588008585925, "lm_q1q2_score": 0.8670525627355514}}
{"url": "https://mathematica.stackexchange.com/questions/209694/filtering-data-geometrically/209717", "text": "# Filtering data 'geometrically'\n\nI was looking at a very nice answer here, where the question was applying a confidence region to some data.\n\nIf we plot some data as a scatter plot:\n\nXData = RandomVariate[NormalDistribution[1,1], 100000];\nYData = RandomVariate[NormalDistribution[1,1], 100000];\n\nXYDATA = Transpose[{XData, YData}];\n\nXYCoVarMat = Covariance[XYDATA];\nXYMean = {Mean[XData], Mean[YData]};\n\nListPlot[\nXYDATA, Frame->True, FrameLabel->{\"X\",\"Y\"}, LabelStyle->16, AspectRatio->1,\nEpilog -> {\nOpacity[0.32, Green], EdgeForm[{Green, AbsoluteThickness[1]}], Ellipsoid[XYMean, 4 XYCoVarMat],\nOpacity[0.32, Blue], EdgeForm[{Blue, AbsoluteThickness[1]}], Ellipsoid[XYMean, 3 XYCoVarMat],\nOpacity[0.32, Red], EdgeForm[{Red, AbsoluteThickness[1]}], Ellipsoid[XYMean, 2 XYCoVarMat],\nOpacity[0.32, Yellow], EdgeForm[{Yellow, AbsoluteThickness[1]}], Ellipsoid[XYMean, 1 XYCoVarMat]\n},\nPlotRange->{{-8,8},{-8,8}}\n]\n\n\nI was wondering if anyone knew a way of filtering the data using this approach for example only select data that is with the ellipsoid boundary of Ellipsoid[XYMean, 1 XYCoVarMat]?\n\n\u2022 Would With[{rmf = RegionMember[Ellipsoid[XYMean, 1 XYCoVarMat]]}, Select[XYDATA, rmf]] work for you? \u2013\u00a0J. M. will be back soon Nov 15 at 16:09\n\u2022 It would work wonderfully. I'm starting to wonder if you are actually written purely from Mathematica functions. Thanks! \u2013\u00a0Q.P. Nov 15 at 16:14\n\u2022 @J.M.willbebacksoon I don't know if you want to post an answer yourself or not, but given you provided the solution I feel you should get credit. If credit doesn't concern you I will accept ThatGravityGuy's answer so that the question is completed and helps users who find this question. \u2013\u00a0Q.P. Nov 16 at 13:15\n\u2022 Please accept ThatGravityGuy's answer if you think you found it useful. \u2013\u00a0J. M. will be back soon Nov 16 at 13:24\n\nAs @J.M. pointed out in the comments, the key here is the function RegionMember. Since you already know the region that you want to work with, namely Ellipsoid[XYMean, 1 XYCoVarMat], you can use Select as\nWith[{rmf = RegionMember[Ellipsoid[XYMean, 1 XYCoVarMat]]}, Select[XYDATA, rmf]]", "date": "2019-12-08 20:33:20", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/209694/filtering-data-geometrically/209717", "openwebmath_score": 0.22357718646526337, "openwebmath_perplexity": 5225.383389588433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769113660689, "lm_q2_score": 0.8887587949656841, "lm_q1q2_score": 0.8670525601420515}}
{"url": "https://mathematica.stackexchange.com/questions/139605/mostellers-first-ace-problem", "text": "# Mosteller's First Ace Problem\n\n[this probably belongs on cross validated ... ]\n\nWorking through Mosteller's Fifty Challenging Problems in Probability with Mathematica. I am trying to understand Mosteller's answer for The First Ace problem (#40). The problem goes:\n\nShuffle an ordinary deck of 52 playing cards containing 4 aces. Then turn up cards from the top until the first ace appears. On the average how many cards are required to produce the first ace?\n\nI solved this problem in Mathematica as as follows:\n\nsuite = Flatten@{Range[2, 10], J, Q, K, A}\ndeck = Flatten@{suite, suite, suite, suite}\ndat = Table[First@Flatten@Position[\nRandomSample[deck, Length@deck], A], {n, 1, 100000}];\nListPlot[\nTally@dat,\nGridLines -> Automatic,\nGridLinesStyle -> Directive[Dotted, Gray],\nFrame -> True\n]\n\n\nMean@dat // N\n10.6152\n\n\nWhich yields Mosteller's answer (by symmetry) of 10.6 cards (48/5 +1).\n\nI was a bit shocked that the mode was 1 card.\n\nFinally, when I look at the CDF the 50% mark is at 8 cards\n\nt = Reverse[SortBy[Tally@dat, #[[2]] &]][[All, 2]]\nListPlot[N[Accumulate@t/100000],\nGridLines -> Automatic,\nGridLinesStyle -> Directive[Dotted, Gray]\n]\n\n\nSo my question is, if I played this game for real in a casino, should I go for 8 or 10 cards?\n\n\u2022 I think you're right, this probably belongs on CrossValidated... Cool problem though. Mar 9, 2017 at 6:03\n\u2022 @MarcoB I believe this belongs here, because - next to the theoretical aspect - it shows how to make use of Mathematica's abilities to simulate and to work with its distribution-framework.\n\u2013\u00a0gwr\nMar 9, 2017 at 12:34\n\u2022 My bad, I corrected my \"decision support\" (cf. my answer): In the simple 0-1-Utility case (right/wrong and a fixed amount to win), you would best choose $1$ (not 8, not 10). My exmple meets the case, when you will win an amount equal to the number of draws it took.\n\u2013\u00a0gwr\nMar 9, 2017 at 18:07\n\nI would like to choose a slightly different approach here and point out, that this is an example for the Negative Hypergeometric Distribution. Wikipedia does give a nice table for this:\n\nWhile for large numbers the distribution tends to the Negative Binomial Distribution the difference essentially is drawing with replacements (Binomial) vs. drawing without replacements (Hypergeometric).\n\n## Analytical Approach\n\nIn the present case, even when we may not have remembered all this statistical theory, we could still take an analytical approach by making use of ProbabilityDistribution. Let us look at the probabilities for the first few numbers of draws:\n\n\\begin{align} p(1) &= p(\\text{ace}) =\\frac{4}{52} \\\\ p(2) &= p(\\neg\\text{ace}) \\times p(\\text{ace}) = \\frac{52-4}{52} \\times \\frac{4}{52-1} \\\\ p(3) &= p(\\neg\\text{ace}) \\times p(\\neg\\text{ace}) \\times p(\\text{ace}) = \\frac{52-4}{52} \\times \\frac{52-4-1}{52-1} \\times \\frac{4}{52-2} \\\\ \\vdots \\end{align}\n\nSo eventually we see a pattern here and can (hopefully) come up with the formula for the probability distribution function (in the discrete case a probability mass function) $f$ given the number of cards (or balls in an urn) $N$ and the number of marked cards (or balls) $M$:\n\n\\begin{align} f_{N,M}(k) = \\frac{m}{N-k+1} \\times \\prod \\limits_{i=1}^{k-1} \\frac{N-M-i+1}{N-i+1} \\end{align}\n\nTo implment this as a distribution we write the following code:\n\ndistribution[ n_Integer, m_Integer ] /; m <= n := ProbabilityDistribution[\n\nm/( n - k + 1) Product[ (n - m - i + 1)/(n - i + 1),{ i, 1, k-1 } ],\n{ k, 1, n - m + 1, 1 }\n(* , Method -> \"Normalization\" is not needed here, but may be worth remembering\nfor all cases where giving a proportional formula is easy to come up with *)\n]\n\nanalyticalDist = distribution[ 52, 4 ];\n\n$PlotTheme = {\"Detailed\", \"LargeLabels\"}; Panel @ DiscretePlot[ Evaluate@PDF[ analyticalDist, \\[FormalX] ], {\\[FormalX], 0, 49, 1}, ImageSize -> Large, PlotLabel -> Style[\"Analytical PDF\", \"Subsection\"] ]  Also we can find moments and what have you: Mean @ analyticalDistribution (* e.g. the expected number of draws *) $\\frac{53}{5}$## Empirical Approach We could of course also arrive at answers using simulation: empiricalDist = EmpiricalDistribution @ Table[ Min @ RandomSample[ Range[52], 4], {10000} (* number of trials *) ]; N @ Mean @ empiricalDist  10.5809 (* true value again was 10.6 *) If we compare the graphs for the PDF we note that the empirical distribution already is quite close: Panel @ DiscretePlot[ Evaluate @ { PDF[ empiricalDist, \\[FormalX] ], PDF[ analyticalDist, \\[FormalX] ] }, { \\[FormalX], 1, 49, 1 }, PlotLegends -> {\"empirical\", \"analytical\"}, ImageSize -> Large, PlotLabel -> Style[\"PDF Comparison\", \"Subsection\"] ]  ## General Case: Negative Hypergeometric Distribution Grabbing a text book near you, you will find the formula for the probability distribution function of the above mentioned Negative Hypergeometric Distribution. It is a special case of the BetaBinomialDistribution - thanks to @ciao for pointing this out repeatedly until I listened :) - for those of you, who do not know this, there is a terrific paper showing the relations between univariate distributions by (Leemis and McQueston 2008) and more information can be found on Cross Validated. Using this information, we can implement the Negative Hypergeometric Distribution explicitly as follows: (* following Mathematica's parametrization for the hyergeometric case *) negativeHypergeometricDistribution[k_, m_, n_ ] := TransformedDistribution[ \\[FormalX] + 1, \\[FormalX] \\[Distributed] BetaBinomialDistribution[ k, m, n-m ] ] (* unfortunately parametrization is no standardized so one has to dwelve into the formulae for this in the paper, which are given though *) nhg = negativeHypergeometricDistribution[ 1, 4, 52 ]; Mean @ nhg $\\frac{53}{5}$And we can indeed convince ourselves that our case simply is a special case of a NHG-distributed random var$X$, where$X$is the number of draws from$N = 52$cards, where$M=4$are \"successes\" and where we need$k=1$success to stop: And @@ Table[ PDF[ analyticalDist, i ] === PDF[ nhg, i ], {i, 1, 52 } ]  True ## EDIT: Decision Support / Expected Utility The OP has edited his question and asked for guidance in playing at a casino (e.g. what number of draws to bet on?). For this we have to turn to decision theory and excepted utility. Let's assume that you are risk neutral. In the case that you get a fixed amount if you are right and zero if you are wrong, then it would be trivially optimal to bet on$1$. ($1$has the greatest probability and the 0-1-loss function makes the expected utility equal to the probability; check this by simulation!) Let us look at a more elaborate case to make the principle clear: If you were to receive the number of draws needed worth in money if you bet on the right number of draws, then your utility function would be$U(x,k) = x I(x = k)$(where$I$is the indicator function,$x$your bet and$k$the number of draws it took. In this case your best bet will be$13$: NArgMax[ { NExpectation[ \\[FormalX] * Boole[ \\[FormalX] == i], \\[FormalX] \\[Distributed] negativeHypergeometricDistribution[ 1, 4, 52 ] ], 1<= k<= 49 }, k \u2208 Integers ]  13 With[ { dist = negativeHypergeometricDistribution[ 1, 4, 52 ] }, Panel @ DiscretePlot[ Evaluate@ Expectation[ \\[FormalX] * Boole[ \\[FormalX] == i ], \\[FormalX] \\[Distributed] dist ], {i,1,49}, ImageSize -> Large, PlotLabel -> Style[\"Expected Utility (0-1)\", \"Subsection\"], PlotLegends->None ] ]  \u2022 \"since it is not implemented...\". It is, as BetaBinomial, already noted in an earlier reply. \u2013 ciao Mar 9, 2017 at 23:35 \u2022 @ciao Thanks, I did not make that connection. It is deeply burried in the docs and also not very obvious from web research. Guess one really needs to know what your looking for. :) \u2013 gwr Mar 9, 2017 at 23:50 \u2022 Not so much an MMA docs issue, more of a wider issue of distributions going by different names/definitions in the mathematical world. Would be nice if a future MMA docs edition (or MathWord) had a big interconnected chart with all alternate names/definitions and their relationships... \u2013 ciao Mar 10, 2017 at 0:50 \u2022 This is the best overview with regard to the relationships between univariate distributions I have found so far. \u2013 gwr Mar 10, 2017 at 8:19 Update (attribution: @ciao) As @ciao has pointed out, use ofBetaBinomialDistribution is the most straightforward for exact calculation and simulation. See the comment: Mean[BetaBinomialDistribution[1, 4, 48]] + 1  or simulation Mean[RandomVariate[Mean[BetaBinomialDistribution[1, 4, 48]] + 1],100000] Median[RandomVariate[Mean[BetaBinomialDistribution[1, 4, 48]] + 1],100000]  etc Original Answer As the choice of first Ace is arbitrary you could simplify, e.g.: tab = Table[Min[FirstPosition[RandomSample[Range[52]], #][[1]] & /@ Range[4]], 10000]; Histogram[tab, Automatic, \"PDF\"] N@Mean[tab] Median[tab]  You could also do analytically,e.g. let$f(k)$be the probability that the first ace in the first$k$cards, then: f[k_] := 1 - Binomial[52 - k, 4]/Binomial[52, 4] DiscretePlot[f[k], {k, 0, 52}]  Sanity checks:$f(k)=1$for$k\\ge 49$The probability that is$k\\$th card is discrete derivative:\n\nListPlot[Differences[Table[f[k], {k, 0, 52}]], Filling -> Axis]\nDifferences[Table[f[k], {k, 0, 52}]].Range[52]\n\n\n\u2022 Or you could just Mean[BetaBinomialDistribution[1, 4, 48]] + 1... ;=}\n\u2013\u00a0ciao\nMar 9, 2017 at 6:56\n\u2022 @ciao doh! always forgetting to exploit diverse built-in distributions...hope your entrepreneurship is going well ;-) Mar 9, 2017 at 6:59", "date": "2022-05-25 00:40:45", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/139605/mostellers-first-ace-problem", "openwebmath_score": 0.994888186454773, "openwebmath_perplexity": 2416.923694968831, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769085257165, "lm_q2_score": 0.8887587831798665, "lm_q1q2_score": 0.8670525461196917}}
{"url": "https://math.stackexchange.com/questions/2984453/plotting-a-system-of-linear-ode", "text": "# Plotting a system of linear ODE\n\nI would like to plot a system of following ODE:\n\n$$$$\\mathbf{\\dot{x}} = \\mathbf{Ax} \\text{ where } \\mathbf{A} = \\begin{bmatrix} -2 & 1 \\\\ -1 & 0 \\end{bmatrix}$$$$\n\nwith general solution:\n\n$$$$\\vec{x}(t)=C_1 \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} e^{-t}+C_2 \\left(\\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix}+\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} t \\right) e^{-t}$$$$\n\nI know that this system is stable so arrows will be inward-pointing, also it has double eigenvalue at -1 so there will be an constant line for an eigenvalue crossing the plot but I am not sure how to fill the rest of the phase diagram. Could someone help me? Thanks in advance.\n\n\u2022 I think you have a mistake. Shouldn't this system have a double eigenvalue of 1, not -1? \u2013\u00a0Josh B. Nov 4 '18 at 16:29\n\u2022 @JoshB. I think the eigenvalue is given by $$\\lambda^2+2\\lambda+1$$ unless I made a mistake but I double checked the solution using Matlab \u2013\u00a01muflon1 Nov 4 '18 at 16:33\n\u2022 If that's the case, then there is a typo in your matrix here. I just checked and the matrix you have definitely has a double eigenvalue of 1. \u2013\u00a0Josh B. Nov 4 '18 at 16:39\n\u2022 @JoshB. You were correct the typo was 2 instead of -2. My bad. I am really sorry, I was copy pasting it from my latex and somehow I did not included the minus sign, dont know how did I even managed to do it \u2013\u00a01muflon1 Nov 4 '18 at 16:41\n\nIf you write your solution now as $$\\vec{x}(t)=\\left(C_1\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}+C_2\\begin{bmatrix} 0 \\\\ 1\\end{bmatrix}\\right)e^{-t}+C_2\\begin{bmatrix}1 \\\\ 1\\end{bmatrix}te^{-t}$$\n\nit may be a little easier to see what's going on. As $$t\\to\\pm\\infty$$, the equation is dominated by the second term, so $$\\vec{x}(t)\\approx C_2\\begin{bmatrix} 1 \\\\ 1\\end{bmatrix}te^{-t}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;t\\to\\pm\\infty$$\n\nAs $$t\\to\\infty$$, the whole thing decays to $$0$$, so it makes sense that all solution curves begin to look like they fall along this eigenvector as they decay to $$0$$. Applying the same logic as $$t\\to-\\infty$$, it would seem that solutions grow along this vector as well, but this is only true on a macroscopic scale, as the solutions are growing exponentially in both directions. The first term, no longer decaying to $$0$$, shifts the solution away from the eigenvector and is growing large, but not as large as the second term. This means that when zoomed in, solutions look like they are leaving the eigenvector, but zoomed out solutions look like they are along the eigenvector (that is, until you let time grow large enough, in which you will see it move away slowly).\n\nThe last thing to consider is the direction of the curve. Suppose that $$\\vec{x}$$ is in the first quadrant as $$t\\to\\infty$$. This means that $$C_2$$ is positive, so if $$t\\to-\\infty$$, then the sign on the second term is now negative, and the solution turns around to enter the third quadrant. This means that solutions sort of \"spin\" by turning $$180^\\circ$$ around while still growing away from the origin.\n\nThe vector $$\\begin{bmatrix} 0 \\\\ 1\\end{bmatrix}$$ tells you which way the solution spins. If $$C_2$$ is positive, then that means that at $$t=0$$, the solution is above (in the xy-plane) of the line through the vector $$\\begin{bmatrix} 1 \\\\ 1\\end{bmatrix}$$ and the origin, so the solution must remain on this side of said line. The dominating term is in the first quadrant as $$t\\to\\infty$$, so the solution must spin clockwise to decay this way. It turns out that if solutions spin clockwise on one side of the eigenvector, they do the same on the other side as well; the same is true if they spin counterclockwise.\n\nThis type of equilibrium is called a Degenerate Node, in case you were curious. Here is an example of what one might look like.\n\n\u2022 Thank you for very clear and detailed answer. By the way did you made the plot in matlab,? I know that it was not part of my original question and matlab inquiries dont really belong to math section, but would you mind sharing your code? I would like to learn how to do it. Or did you used another program? \u2013\u00a01muflon1 Nov 4 '18 at 17:36\n\u2022 No, I pulled it from the encyclopedia of mathematics. Plotting this in MATLAB could be done by taking a spacing of points for t and plugging them in for x and y, then plotting the resulting curve. This would need to be done for a few different initial points to get a good diagram. \u2013\u00a0Josh B. Nov 4 '18 at 17:44", "date": "2019-07-16 22:22:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2984453/plotting-a-system-of-linear-ode", "openwebmath_score": 0.9602904915809631, "openwebmath_perplexity": 232.53997304029153, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102524151826, "lm_q2_score": 0.8962513835254865, "lm_q1q2_score": 0.8670427771638476}}
{"url": "https://math.stackexchange.com/questions/399394/if-left-fx-right-leq-a-fx-beta-then-f-is-a-constant-function", "text": "If $\\left| f'(x) \\right| \\leq A |f(x)|^\\beta$ then f is a constant function\n\nProblem Let $f(x)$ be a differentiable function on $[a,b]$ satisfying $f(a)=0$. If there exist $A \\ge 0$ and $\\beta \\ge 1$ such that the inequality\n\n$$\\left| f'(x) \\right| \\leq A \\left| f(x) \\right|^\\beta$$\n\nholds for all $x$ in $[a,b]$, then $f(x) = 0$ for any x in that interval.\n\nMy attempted solution\n\nWe shall prove by contradiction. Suppose that for some $x$ in $[a,b]$ we have $f(x) \\neq 0$.\n\nPut $S:= \\left\\lbrace x \\right.$ such that $a \\le x \\le b$ and $f(x) \\neq 0 \\left. \\right\\rbrace$ and $c:=\\inf S$ (the infimum value of set $S$). If $c=b$ then $f(x) = 0$ $\\forall x < b$ and the continuity of $f$ implies that $f(b) = 0$, too (contradiction). Therefore $c \\neq b$. If $c=a$, then $f(c) = 0$ by the hypothesis. Otherwise, when $a < c < b$ we have $f(x) = 0$ $\\forall x < c$ and again, by continuity, $f(c) = 0$. Thus in all cases considered, we have $f(c) = 0$ and $a \\leq c < b$\n\nAs $f$ is continuous at $c$ and $f(c) = 0$, one can choose $d>c$ and close enough to $c$ such that $|f(x)| \\le 1$ and $$\\tag{\\star} A(x-c) \\le \\frac12\\quad\\text{for all }x\\text{ with }c \\le x \\le d.$$ Since $f$ is continuous on $[c,d]$, it has maximum and minimum values on this closed interval, leading to the existence of $c \\le t \\le d$ such that $|f(t)| = \\max_{c \\le x \\le d} |f(x)|$. As already noted, $f(c) = 0$ and $c$ is the infimum of the set of numbers whose image under $f$ is nonzero. Therefore, for any $\\epsilon>0$, there is some $c<m<c+\\epsilon$ satisfying $f(m) \\neq 0$. This observation, together with the definition of $t$, gives us $$\\tag{\\star\\!\\star}f(t) \\neq 0,$$ $t \\neq c$ and so $c<t$. Applying the Lagrange theorem gives:\n\n$$|f(t)| = |f(t) - f(c)| = |t-c||f'(u)| \\leq|t-c|A|f(u)|^\\beta$$ where $c<u<t\\le d$. By $(\\star)$: $$|t-c|A|f(u)|^\\beta \\le \\frac12 |f(u)|^\\beta \\leq \\frac12 |f(u)|$$\n\nCombining the two inequalities and noting that $|f(t)| \\ge |f(u)|$ (because of the definition of $t$), we have $f(u) = f(t) = 0$ and arrived at a contradiction with $(\\star\\star)$.\n\nQuestion\n\nI hope someone can verify if my solution is correct. Of course, other ideas, comments or solutions are welcome.\n\nI like to post problems and my solutions to the forum because I think it's beneficial to the community, and for learners like me. First, I can hardly know if there's flaw in my own argument. Second, I may get new insights/solutions for my problem.\n\nThank you.\n\n\u2022 Sorry, I am not yet clear how the two threads are related? \u2013\u00a0tom_a2 May 22 '13 at 16:43\n\u2022 You are absolutely right. I misread your question. \u2013\u00a0Martin May 22 '13 at 16:45\n\u2022 It looks good to me. \u2013\u00a0Julien May 22 '13 at 17:32\n\nLet $B = \\{x\\in [a,b]: |f(x)|=0\\}$. By continuity, this set is closed. Since $f(0) = 0$, the set is non-empty. We show that it is also open.\n\nLet $x\\in B$, i.e. $f(x) =0$. By continuity, there exists $0<r<(2A)^{-1}$, such that $|f(y)|<\\frac 12$ for all $y\\in B_r(x)$. Fix such a $y\\in B_r(x)$. We want to show that $f(y) =0$.\n\nBy the mean value inequality, we have $$|f(y)| = |f(y) - f(x)| \\le |f'(\\xi_0)| |y-x| \\le |f'(\\xi_0)|r$$ for some $\\xi_0\\in B_r(x)$ lying on the segment between $x$ and $y$. By assumption, we have $|f'(\\xi_0)|\\le A |f(\\xi_0)|^\\beta$, hence $$|f(y)|\\le |f'(\\xi_0)|r \\le A|f(\\xi_0)|^\\beta (2A)^{-1} \\le \\frac 12|f(\\xi_0)|^\\beta.$$ By the same argument applied to $\\xi_0\\in B_r(x)$ instead of $y$, we see that there exists $\\xi_1\\in B_r(x)$, such that $|f(\\xi_0)|\\le \\frac 12|f(\\xi_1)|^\\beta$. Iterating yields a sequence $\\xi_n \\in B_r(x)$, such that $|f(\\xi_{n})|\\le \\frac 12|f(\\xi_{n+1})|^\\beta$ for all $n$. Note that by our choice of $r$, we have $|f(\\xi_{n})|\\le \\frac 12$ for all $n$. We conclude that $$|f(y)|\\le \\frac 12 |f(\\xi_0)|^\\beta \\le \\frac 1{2^2}|f(\\xi_1)|^{\\beta^2} \\le \\frac{1}{2^3}|f(\\xi_2)|^{\\beta^3}\\le \\dots \\le \\frac{1}{2^n}|f(\\xi_{n-1})|^{\\beta^n}\\le \\frac 1{2^{n+\\beta^n}},$$ for all $n\\in \\mathbb N$. Letting $n\\to \\infty$ shows that $|f(y)| = 0$ (using also $\\beta \\ge 1$). This implies that $B_r(x)\\subset B$, hence $B$ is open.\n\nSince $[a,b]$ is connected, it follows that $B$ must be all of $[a,b]$, hence $f= 0$.\n\n( Edit. The ODE analysis also explains why $\\beta \\geq 1$ is necessary. The condition for $|f'(x)| \\leq A|H(f(x))|$ to have the property that $f$ does not change sign, when $H$ is a function such that $H(x)=0$ only at $x=0$, is that $\\frac{1}{H(x)}$ an has a non-integrable singularity at $0$. Knowing that this is the answer, there is no loss in considering singularities that are powers of $f$, which is what most cases would look like and how they would be recognized.)\n\nThe differential equation $f'(x) = A(x) f(x)^\\beta$ can be solved explicitly, by separation of variables. The calculation shows that $f(a)=0$ is necessary, or we could write down nowhere zero solutions, by solving the equation. This bears further analysis, because superficially the problem looks like a Lipschitz condition with $|f(x) - f(y)| \\leq |x-y|^\\beta$ which (as is very well known and often solved as an exercise) implies constant $f$ when $\\beta > 1$.\n\nThe DE for constant $\\beta > 1$, writing $y$ for $f(x)$, is $$d(\\frac{1}{(\\beta - 1)y^{\\beta - 1}}) = A(x) dx$$\n\nThe coordinates that trivialize the equation are therefore $Y = \\frac{1}{(\\beta - 1)y^{\\beta - 1}}$ and $X = \\int A(x) dx$, in which the solutions are $Y = X + c$ for constant $c$. This brings out the idea that $A$, which appears at first to be a harmless constant removable by a linear change of variable, has a meaningful part in the problem, as velocity of the independent variable.\n\nUnder this coordinate change, $f(a)=0$ becomes $f(a)=\\infty$. On a maximal interval where $f$ is nonzero, there cannot be any solutions that hit infinity at the boundary in finite \"time\" (here $X$ is time, so we mean a finite change in $\\int A(x) dx$) while satisfying $Y - X =$ constant.\n\nSo the meaning of the problem seems to be: no singularity can be reached in finite time for this ODE.\n\n\u2022 Let's put here the comment that for $\\beta < 1$ there are counterexamples with $f(x)$ a positive power of $(x-a)$. \u2013\u00a0zyx May 26 '13 at 22:38\n\nSupposing that f(x)=0 only for a and that f(x)>0 for the rest x in [a,b].\n\nThen lim (f(x) - f(a))/(x-a) as x->a+ will be greater or equal to zero. If greater then f'(a)=m >0 and thats contradicting with the given inequality.Therefore f(x)=0 for all x in [a,b]. If equal, then im sorry but i need to give it some more thought. In the same manner, u can prove for f(x)<0.\n\nI just posted this insufficient answer cuz i thought it would be simpler to understand.", "date": "2019-07-20 16:31:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/399394/if-left-fx-right-leq-a-fx-beta-then-f-is-a-constant-function", "openwebmath_score": 0.9560487270355225, "openwebmath_perplexity": 89.07947414986923, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575157745541, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8670361216806681}}
{"url": "https://gmatclub.com/forum/over-the-past-7-weeks-the-smith-family-had-weekly-grocery-bills-of-268908.html?kudos=1", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 05 Dec 2019, 17:59\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n### Show Tags\n\n25 Jun 2018, 05:02\n21\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n82% (01:39) correct 18% (01:50) wrong based on 972 sessions\n\n### HideShow timer Statistics\n\nOver the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64\n\nB. $70 C.$73\n\nD. $74 E.$85\n\nNEW question from GMAT\u00ae Official Guide 2019\n\n(PS07369)\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 3723\nOver the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 27 Jun 2018, 18:28 6 9 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64 $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?\n\nA. $64 B.$70\n\nC. $73 D.$74\n\n### Show Tags\n\n29 Jun 2018, 06:58\n28\n9\nFirst, choose a nice, round number, such as 70, within the range of values.\n\nThen calculate the average with the following formula:\n\nAverage = Nice number + Average of differences from the nice number\n\nAverage = 70 + (4 - 1 - 6 + 9 - 6 + 14 + 7)/7 = 70 + 21/7 =73\n\n_________________\nMy book with my solutions to all 230 PS questions in OG2018:\nZoltan's solutions to OG2018 PS\n##### General Discussion\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 8610\nLocation: United States (CA)\nRe: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 28 Jun 2018, 18:07 1 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?\n\nA. $64 B.$70\n\nC. $73 D.$74\n\nE. $85 We can determine the average using the formula: average = sum / number: average = (74 + 69 + 64 + 79 + 64 + 84 + 77)/7 = 511/7 = 73 Answer: C _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the \"Kudos\" button. VP Joined: 14 Feb 2017 Posts: 1310 Location: Australia Concentration: Technology, Strategy GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GMAT 6: 600 Q38 V35 GPA: 3 WE: Management Consulting (Consulting) Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7\u00a0 [#permalink]\n\n### Show Tags\n\n19 Nov 2018, 16:13\n1\nAdmittedly I made a stupid mistake and I organised the numbers in ascending order then selected the median (74) instead of solving Sum/# Terms\n\nThe method I used is only for consecutive integers and the punishment answer D is there for suckers like me.\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 3158\nRe: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 25 Jun 2018, 05:10 Solution Given: \u2022 Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77\n\nTo find:\n\u2022 Smith\u2019s average weekly grocery bill over the 7-week period\n\nApproach and Working:\n\u2022 Smith\u2019s total bill amount over the 7-week period = (74 + 69 + 64 + 79 + 64 + 84 + 77) = 511\n\u2022 Therefore, Smith\u2019s average bill amount over the same period = $$\\frac{511}{7}$$ = 73\n\nHence, the correct answer is option C.\n\n_________________\nIntern\nJoined: 08 Jul 2017\nPosts: 15\nLocation: Greece\nGPA: 3.31\n\n### Show Tags\n\n28 Aug 2018, 03:42\nBunuel wrote:\nOver the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64\n\nB. $70 C.$73\n\nD. $74 E.$85\n\nNEW question from GMAT\u00ae Official Guide 2019\n\n(PS07369)\n\nEasy way out is consider avg 70 -\nso diff for all terms is 4, -1 , -6, 9 , -6,14, 7 = sum is 21 = divided by 7 = equal to 3 so 70 +3 = 73 avg\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 4835\nLocation: India\nGPA: 3.5\nRe: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 10 Sep 2018, 07:34 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?\n\nA. $64 B.$70\n\nC. $73 D.$74\n\nE. $85 NEW question from GMAT\u00ae Official Guide 2019 (PS07369) $$\\frac{( 70 + 4 ) + ( 70 - 1 ) + ( 60 + 4 ) + ( 80 - 1 ) + ( 60 + 4 ) + ( 80 + 4 ) + ( 70 + 7 )}{7}$$ = $$\\frac{( 70 + 70 + 60 + 80 + 60 + 80 + 70 ) + ( 4 - 1 + 4 - 1 + 4 + 4 + 7 )}{7}$$ = $$\\frac{490 + 21}{7}$$ = $$\\frac{490}{7} + \\frac{21}{7}$$ = $$70 + 3$$ = $$73$$, Answer must be (C) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 02 May 2018 Posts: 12 GMAT 1: 620 Q46 V29 Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7\u00a0 [#permalink]\n\n### Show Tags\n\n10 Sep 2018, 07:46\nWe can rule out\n64, 74 and 85 as they are the extreme numbers.\nThat leaves just two choices to choose from 70 and 73.\n\nOnly 3 values in the 60s and therefore less likely for 70.\n\n73 is the most likely choice. Confirm 73 by taking difference and summing it up\nIntern\nJoined: 22 Jul 2018\nPosts: 22\nLocation: Indonesia\nSchools: MBS '21\nGPA: 3.24\n\n### Show Tags\n\n02 Dec 2019, 16:30\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________", "date": "2019-12-06 01:00:27", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/over-the-past-7-weeks-the-smith-family-had-weekly-grocery-bills-of-268908.html?kudos=1", "openwebmath_score": 0.586456298828125, "openwebmath_perplexity": 5628.7804433098945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.8670357701094303}}
{"url": "https://gmatclub.com/forum/is-x-y-1-x-y-2-x-2-y-210108.html", "text": "It is currently 23 Nov 2017, 06:08\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nEvents & Promotions\n\nEvents & Promotions in June\nOpen Detailed Calendar\n\nIs |x| = |y|? (1) x = -y (2) x^2 = y^2\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nManager\nJoined: 18 Feb 2015\nPosts: 88\n\nKudos [?]: 86 [1], given: 15\n\nIs |x| = |y|? (1) x = -y (2) x^2 = y^2\u00a0[#permalink]\n\nShow Tags\n\n16 Dec 2015, 07:06\n1\nKUDOS\n1\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n84% (00:21) correct 16% (00:30) wrong based on 62 sessions\n\nHideShow timer Statistics\n\nIs |x| = |y|?\n\n(1) x = -y\n(2) x^2 = y^2\n\n[Reveal] Spoiler:\nI found this question in GMAT Prep and selected ST2 as the answer but that's wrong.\n\nNormally, x^2 would equal to lxl since it will contain both signs \"positive as well as negative\". How come st 1 is true?\n\nDoes it assume that since +x = -y, it must be true that -x is also equal to -y or?\n\nWhat am I missing?\n\n[Reveal] Spoiler: OA\n\nLast edited by Bunuel on 11 Nov 2017, 01:25, edited 3 times in total.\nFormatted the question.\n\nKudos [?]: 86 [1], given: 15\n\nCurrent Student\nJoined: 20 Mar 2014\nPosts: 2676\n\nKudos [?]: 1775 [1], given: 794\n\nConcentration: Finance, Strategy\nSchools: Kellogg '18 (M)\nGMAT 1: 750 Q49 V44\nGPA: 3.7\nWE: Engineering (Aerospace and Defense)\nRe: Is |x| = |y|? (1) x = -y (2) x^2 = y^2\u00a0[#permalink]\n\nShow Tags\n\n16 Dec 2015, 07:38\n1\nKUDOS\n2\nThis post was\nBOOKMARKED\nHarveyKlaus wrote:\nIs lxl = lyl ?\n\nA) x=-y\nb) x^2 = y^2\n\n[Reveal] Spoiler:\nI found this question in GMAT Prep and selected ST2 as the answer but that's wrong.\n\nNormally, x^2 would equal to lxl since it will contain both signs \"positive as well as negative\". How come st 1 is true?\n\nDoes it assume that since +x = -y, it must be true that -x is also equal to -y or?\n\nWhat am I missing?\n\nYou are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.\n\nPer statement 1:$$x=-y$$ . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.\n\nPer statement 2: $$x^2=y^2$$ ---> $$x= \\pm y$$. Again both the cases, x=y and x=-y give you a \"yes\" for the question asked as is hence sufficient.\n\nBoth statements are sufficient ---> D is the correct answer.\n\nKudos [?]: 1775 [1], given: 794\n\nSenior Manager\nJoined: 20 Feb 2015\nPosts: 388\n\nKudos [?]: 106 [0], given: 10\n\nConcentration: Strategy, General Management\nRe: Is |x| = |y|? (1) x = -y (2) x^2 = y^2\u00a0[#permalink]\n\nShow Tags\n\n17 Dec 2015, 06:54\nstatement 1 is sufficient as absolute value for -ve should give +ve\nstatement 2 is sufficient as\nit will give either x and y both are +ve or both are -ve\n\nKudos [?]: 106 [0], given: 10\n\nIntern\nJoined: 15 Sep 2017\nPosts: 7\n\nKudos [?]: 0 [0], given: 23\n\nRe: Is |x| = |y|? (1) x = -y (2) x^2 = y^2\u00a0[#permalink]\n\nShow Tags\n\n10 Nov 2017, 15:24\nIs |x|=|y|?\n\n1) x=-y\n\n2) x^2=y^2\n\nKudos [?]: 0 [0], given: 23\n\nManager\nJoined: 15 Feb 2017\nPosts: 71\n\nKudos [?]: 58 [3], given: 0\n\nRe: Is |x| = |y|? (1) x = -y (2) x^2 = y^2\u00a0[#permalink]\n\nShow Tags\n\n10 Nov 2017, 17:20\n3\nKUDOS\nYou are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.\n\nPer statement 1:x=\u2212yx=\u2212y . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.\n\nPer statement 2: x2=y2x2=y2 ---> x=\u00b1yx=\u00b1y. Again both the cases, x=y and x=-y give you a \"yes\" for the question asked as is hence sufficient.\n\nBoth statements are sufficient ---> D is the correct answer.\nPlease give me kudos. I need them to unlock gmatclub tests.\n\nKudos [?]: 58 [3], given: 0\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42326\n\nKudos [?]: 133106 [0], given: 12411\n\nRe: Is |x| = |y|? (1) x = -y (2) x^2 = y^2\u00a0[#permalink]\n\nShow Tags\n\n11 Nov 2017, 01:25\nSamDGold wrote:\nIs |x|=|y|?\n\n1) x=-y\n\n2) x^2=y^2\n\nMerging topics. Please check the discussion above.\n_________________\n\nKudos [?]: 133106 [0], given: 12411\n\nRe: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 \u00a0 [#permalink] 11 Nov 2017, 01:25\nDisplay posts from previous: Sort by", "date": "2017-11-23 13:08:53", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/is-x-y-1-x-y-2-x-2-y-210108.html", "openwebmath_score": 0.5137346386909485, "openwebmath_perplexity": 9663.902239590328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8670357615200475, "lm_q1q2_score": 0.8670357615200475}}
{"url": "http://math.stackexchange.com/questions/94359/need-help-deriving-recurrence-relation-for-even-valued-fibonacci-numbers", "text": "# Need help deriving recurrence relation for even-valued Fibonacci numbers.\n\nThat would be every third Fibonacci number, e.g. $0, 2, 8, 34, 144, 610, 2584, 10946,...$\n\nEmpirically one can check that:\n\n$a(n) = 4a(n-1) + a(n-2)$ where $a(-1) = 2, a(0) = 0$.\n\nIf $f(n)$ is $\\operatorname{Fibonacci}(n)$ (to make it short), then it must be true that $f(3n) = 4f(3n - 3) + f(3n - 6)$.\n\nI have tried the obvious expansion:\n\n$f(3n) = f(3n - 1) + f(3n - 2) = f(3n - 3) + 2f(3n - 2) = 3f(3n - 3) + 2f(3n - 4)$ $= 3f(3n - 3) + 2f(3n - 5) + 2f(3n - 6) = 3f(3n - 3) + 4f(3n - 6) + 2f(3n - 7)$ ... and now I am stuck with the term I did not want. If I do add and subtract another $f(n - 3)$, and expand the $-f(n-3)$ part, then everything would magically work out ... but how should I know to do that? I can prove the formula by induction, but how would one systematically derive it in the first place?\n\nI suppose one could write a program that tries to find the coefficients x and y such that $a(n) = xa(n-1) + ya(n-2)$ is true for a bunch of consecutive values of the sequence (then prove the formula by induction), and this is not hard to do, but is there a way that does not involve some sort of \"Reverse Engineering\" or \"Magic Trick\"?\n\n-\nWhy isn't $f_{n+2}=3f_n-f_{n-2}$ suitable? \u2013\u00a0 Guess who it is. Dec 27 '11 at 4:31\n@J.M., sorry I do not understand. If I were to expand $f_{n+2}$ the way I do it, I would end up with $2f_n + f_{n - 2} + f_{n - 3}$. Again, I can make this work if I know what I am trying to get. I wonder if you have read the question correctly - I am looking for even-valued (not even-indexed) fib numbers. If my assumption is mistaken, then sorry. However, I am not sure how to systematically arrive at a relation you have given and how to use it to help me simplify things. \u2013\u00a0 Job Dec 27 '11 at 4:45\nIt seems I did misunderstand you (and I apologize for this); have you looked at the references here by any chance? \u2013\u00a0 Guess who it is. Dec 27 '11 at 4:50\n\n## 8 Answers\n\nThe definition of $F_n$ is given:\n\n\u2022 $F_0 = 0$\n\u2022 $F_1 = 1$\n\u2022 $F_{n+1} = F_{n-1} + F_{n}$ (for $n \\ge 1$)\n\nNow we define $G_n = F_{3n}$ and wish to find a recurrence relation for it.\n\nClearly\n\n\u2022 $G_0 = F_0 = 0$\n\u2022 $G_1 = F_3 = 2$\n\nNow we can repeatedly use the definition of $F_{n+1}$ to try to find an expression for $G_{n+1}$ in terms of $G_n$ and $G_{n-1}$.\n\n\\begin{align*} G_{n+1}&= F_{3n+3}\\\\ &= F_{3n+1} + F_{3n+2}\\\\ &= F_{3n-1} + F_{3n} + F_{3n} + F_{3n+1}\\\\ &= F_{3n-3} + F_{3n-2} + F_{3n} + F_{3n} + F_{3n-1} + F_{3n}\\\\ &= G_{n-1} + F_{3n-2} + F_{3n-1} + 3 G_{n}\\\\ &= G_{n-1} + 4 G_{n} \\end{align*}\n\nso this proves that $G$ is a recurrence relation.\n\n-\n\nAt Andr\u00e9's request, I've decided to write an answer. I've also arbitrarily decided to be ambitious and greedy, and I will thus derive a recurrence for the $k$-th increment Fibonacci number $f_{kn}$. (For OP's specific case, $k=3$)\n\nLike Andr\u00e9, I shall also start with Binet:\n\n$$f_{kn}=\\frac{\\phi^{kn}-(-\\phi)^{-kn}}{\\sqrt 5}$$\n\nLetting $u=\\phi^k$ and $v=\\left(-\\dfrac1\\phi\\right)^k$, the formula takes the form\n\n$$f_{kn}=pu^n+qv^n$$\n\nThis means that the characteristic polynomial for the recurrence satisfied by $f_{kn}$ takes the form\n\n\\begin{align*} x^2-(u+v)x+uv&=x^2-\\left(\\phi^k+\\left(-\\frac1\\phi\\right)^k\\right)x+\\left(\\phi^k\\left(-\\frac1\\phi\\right)^k\\right)\\\\ &=x^2-\\left(\\phi^k+\\left(-\\frac1\\phi\\right)^k\\right)x+(-1)^k \\end{align*}\n\nand the recurrence itself goes like\n\n$$f_{k(n+1)}=\\left(\\phi^k+\\left(-\\frac1\\phi\\right)^k\\right)f_{kn}-(-1)^k f_{k(n-1)}$$\n\nYou might say that the form $\\ell_k=\\phi^k+\\left(-\\dfrac1\\phi\\right)^k$ is a bit unwieldy, and I agree. There are two ways to go about (slightly) simplifying this. One way makes use of the Newton-Girard formulae. These formulae express $\\ell_k$ in terms of $\\phi-\\dfrac1\\phi=1$ and $\\phi\\left(-\\dfrac1\\phi\\right)=-1$. To use $k=3$ as an example:\n\n$$\\alpha^3+\\beta^3=(\\alpha+\\beta)^3-3(\\alpha+\\beta)(\\alpha\\beta)$$\n\nMaking the replacement $\\alpha+\\beta=1$ and $\\alpha\\beta=-1$, we have\n\n$$\\ell_3=(1)^3-3(1)(-1)=4$$\n\nThe slick way is to recognize that since $\\ell_k$ is itself a linear combination of $\\phi^k$ and $\\left(-\\dfrac1\\phi\\right)^k$, it also satisfies the Fibonacci recurrence:\n\n$$\\ell_{k+1}=\\ell_k+\\ell_{k-1}$$\n\nThe $\\ell_k$ are in fact the (not-so-famous) Lucas numbers. With $\\ell_0=2$ and $\\ell_1=1$, we have the sequence $2, 1, 3, 4, 7, 11,\\dots$\n\nIn short, the recurrence is of the form\n\n$$f_{k(n+1)}=\\ell_k f_{kn}-(-1)^k f_{k(n-1)}$$\n\nFor $k=3$, we have $f_{3(n+1)}=\\ell_3 f_{3n}-(-1)^3 f_{3(n-1)}$ or $f_{3(n+1)}=4 f_{3n}+f_{3(n-1)}$.\n\n-\nVery nice post. This is the kind of solution where each piece falls nicely at its right place. \u2013\u00a0 Did Dec 27 '11 at 9:35\nIt turns out that the identity derived here is listed in the Wolfram Functions site. \u2013\u00a0 Guess who it is. Dec 28 '11 at 0:21\n@JM: this identity is a special case of the identity $$F_{n+2k}=L_kF_{n+k}-(-1)^kF_n$$ However, the indices do not need to be multiples of $k$, they just need to differ by $k$. \u2013\u00a0 robjohn Jan 16 '12 at 19:43\n\nLet $\\alpha$ and $\\beta$ be the two roots of the equation $x^2-x-1=0$. Then the $n$-th Fibonacci number is equal to $$\\frac{\\alpha^n-\\beta^n}{\\sqrt{5}}.$$\n\nWe are interested in the recurrence satisfied by the numbers $$\\frac{\\alpha^{3n}-\\beta^{3n}}{\\sqrt{5}}.$$\n\nIf $x$ is either of $\\alpha$ or $\\beta$, then $x^2=x+1$. Multiply by $x$. We get $x^3=x^2+x=2x+1$. It follows that $x^4=2x^2+x=3x+2$. But then $x^5=3x^2+2x=5x+3$, and then $x^6=5x^2+3x=8x+5$.\n\nWe want $x^6=Ax^3+B$, where $A$ and $B$ are rational, indeed integers. So we want $8x+5=A(2x+1)+B$. Reading off $A$ and then $B$ is obvious: we need $A=4$ and $B=1$.\n\nSo the numbers $\\alpha^{3n}$ and $\\beta^{3n}$ satisfy the recurrence $y_n=4y_{n-1}+y_{n-2}$. By linearity, so do the numbers $\\frac{\\alpha^{3n}-\\beta^{3n}}{\\sqrt{5}}$.\n\nComment: Note that using the same basic strategy, we can write down the recurrence satisfied by $\\frac{\\alpha^{kn}-\\beta^{kn}}{\\sqrt{5}}$. The coefficients that we painfully computed by hand, step by step, can be expressed simply in terms of Fibonacci numbers, and therefore so can the recurrence for the numbers $\\frac{\\alpha^{kn}-\\beta^{kn}}{\\sqrt{5}}$.\n\n-\nNice! I think Newton-Girard might also yield a useful route here. \u2013\u00a0 Guess who it is. Dec 27 '11 at 5:11\nI thought I would take an \"unfancy\" route through facts likely to be well-known. \u2013\u00a0 Andr\u00e9 Nicolas Dec 27 '11 at 5:18\nHere's the Newton-Girard route for completeness: one wants the characteristic polynomial $x^2-(\\alpha^3+\\beta^3)x+\\alpha^3 \\beta^3$ without knowing $\\alpha$ or $\\beta$, but knowing that $\\alpha+\\beta=1$ and $\\alpha\\beta=-1$ (Vieta). It is easily seen that $\\alpha^3 \\beta^3=-1$. Using Newton-Girard, we have the symmetric polynomial expansion $\\alpha^3+\\beta^3=(\\alpha+\\beta)^3-3(\\alpha+\\beta)(\\alpha\\beta)$, and thus $\\alpha^3+\\beta^3=(1)^3-3(1)(-1)=4$. The characteristic polynomial is thus $x^2-4x-1$. \u2013\u00a0 Guess who it is. Dec 27 '11 at 5:25\n@J.M.: I think that the comment above should be made into an answer. \u2013\u00a0 Andr\u00e9 Nicolas Dec 27 '11 at 5:29\n\nBy inspection $f(3n+3)=4f(3n)+f(3n-3)$, as you\u2019ve already noticed. This is easily verified:\n\n\\begin{align*} f(3n+3)&=f(3n+2)+f(3n+1)\\\\ &=2f(3n+1)+f(3n)\\\\ &=3f(3n)+2f(3n-1)\\\\ &=3f(3n)+\\big(f(3n)-f(3n-2)\\big)+f(3n-1)\\\\ &=4f(3n)+f(3n-1)-f(3n-2)\\\\ &=4f(3n)+f(3n-3)\\;. \\end{align*}\n\nHowever, I didn\u2019t arrive at this systematically; it just \u2018popped out\u2019 as I worked at eliminating terms with unwanted indices.\n\nAdded: Here\u2019s a systematic approach, but I worked it out after the fact.\n\nThe generating function for the Fibonacci numbers is $$g(x)=\\frac{x}{1-x-x^2}=\\frac1{\\sqrt5}\\left(\\frac1{1-\\varphi x}-\\frac1{1-\\hat\\varphi x}\\right)\\;,$$ where $\\varphi = \\frac12(1+\\sqrt5)$ and $\\hat\\varphi=\\frac12(1-\\sqrt5)$, so that $f(n)=\\frac1{\\sqrt5}(\\varphi^n-\\hat\\varphi^n)$. Thus, $f(3n)=\\frac1{\\sqrt5}(\\varphi^{3n}-\\hat\\varphi^{3n})$. Thus, we want\n\n\\begin{align*} h(x)&=\\frac1{\\sqrt5}\\sum_{n\\ge 0}(\\varphi^{3n}-\\hat\\varphi^{3n})x^n\\\\ &=\\frac1{\\sqrt5}\\left(\\sum_{n\\ge 0}\\varphi^{3n}x^n-\\sum_{n\\ge 0}\\hat\\varphi^{3n}x^n\\right)\\\\ &=\\frac1{\\sqrt5}\\left(\\frac1{1-\\varphi^3 x}-\\frac1{1-\\hat\\varphi^3 x}\\right)\\\\ &=\\frac1{\\sqrt5}\\cdot\\frac{(\\varphi^3-\\hat\\varphi^3)x}{1-(\\varphi^3+\\hat\\varphi^3)x+(\\varphi\\hat\\varphi)^3x^2}\\;. \\end{align*}\n\nNow $\\varphi+\\hat\\varphi=1$, $\\varphi-\\hat\\varphi=\\sqrt5$, $\\varphi\\hat\\varphi=-1$, $\\varphi^2=\\varphi+1$, and $\\hat\\varphi^2=\\hat\\varphi+1$, so\n\n\\begin{align*} h(x)&=\\frac1{\\sqrt5}\\cdot\\frac{(\\varphi^3-\\hat\\varphi^3)x}{1-(\\varphi^3+\\hat\\varphi^3)x+(\\varphi\\hat\\varphi)^3x^2}\\\\ &=\\frac{(\\varphi^2+\\varphi\\hat\\varphi+\\hat\\varphi^2)x}{1-(\\varphi^2-\\varphi\\hat\\varphi)x-x^2}\\\\ &=\\frac{(\\varphi^2-1+\\hat\\varphi^2)x}{1-(\\varphi^2+1+\\hat\\varphi^2)x-x^2}\\\\ &=\\frac{(\\varphi+\\hat\\varphi+1)x}{1-(\\varphi+3+\\hat\\varphi)x-x^2}\\\\ &=\\frac{2x}{1-4x-x^2}\\;. \\end{align*}\n\nIt follows that $(1-4x-x^2)h(x)=2x$ and hence that $h(x)=4xh(x)+x^2h(x)+2x$. Since the coefficient of $x^n$ in $h(x)$ is $f(3n)$, this tells me that\n\n\\begin{align*} \\sum_{n\\ge 0}f(3n)x^n&=h(x)=4xh(x)+x^2h(x)+2x\\\\ &=\\sum_{n\\ge 0}4f(3n)x^{n+1}+\\sum_{n\\ge 0}f(3n)x^{n+2}+2x\\\\ &=\\sum_{n\\ge 1}4f(3n-3)x^n+\\sum_{n\\ge 2}f(3n-6)x^n+2x\\;, \\end{align*}\n\nwhich by equating coefficients immediately implies that $f(3n)=4f(3n-3)+f(3n-6)$ for $n\\ge 2$. It also gets the initial conditions right: the constant term on the righthand side is $0$, and indeed $f(3\\cdot 0)=0$, and the coefficient of $x$ is $4f(0)+2=2=f(3)$, as it should be.\n\n-\n\nActually the \"Magic Trick\" or \"Reverse Engineering\" idea works nicely.\n\nFirst, as Andr\u00e9 pointed\n\n$$f_{3n}=\\frac{\\alpha^{3n}+\\beta^{3n}}{\\sqrt{5}} \\,.$$\n\nThis means that if $(x-\\alpha^{3})(x-\\beta^3)=x^2-Ax-B$ then $f_{3n}$ is the recurrence satisfying\n\n$$x_{n+2}=Ax_{n+1}+Bx_{n} \\, x_{0}=f_0, x_1=f_{3} \\,.$$\n\nThus,\n\n$$f_6=Af_3+Bf_0$$ $$f_9=Af_6+Bf_3$$\n\nSolve it and get $A,B$.\n\n-\n\nLet $S$ be the shift operator on sequences (as in Bill Dubuque's answer). Note that the Fibonacci sequence is killed by $S^2-S-1$. The Fibonacci sequence will then be killed by any \"polynomial\" multiple of $S^2-S-1$. To get a recurrence for every $k^{\\rm{th}}$ term, all we need to do is find a multiple of $S^2-S-1$ that only involves powers of $S^k$.\n\nFirst note that $S^2-S-1=(S-a)(S-b)$ where $a=\\phi$ (the golden ratio) and $b=-1/\\phi$. Consider the operator $(S^k-a^k)(S^k-b^k)=S^{2k}-(a^k+b^k)S^k+(ab)^k$. It is a polynomial multiple of $S^2-S-1$, so it kills the Fibonacci sequence. It only involves powers of $S^k$.\n\nRecall that one formula for the $k^{\\rm{th}}$ Lucas number is $L_k=a^k+b^k$, and note that $ab=-1$. Thus, we get that $S^{2k}-L_kS^k+(-1)^k$ kills the Fibonacci sequence.\n\nTherefore, in summary, we get $$F_{n+2k}=L_kF_{n+k}-(-1)^kF_n\\tag{1}$$ For example, $$F_{n+2}=F_{n+1}+F_n\\tag{k=1}$$ $$F_{n+4}=3F_{n+2}-F_n\\tag{k=2}$$ $$F_{n+6}=4F_{n+3}+F_n\\tag{k=3}$$ $$F_{n+8}=7F_{n+4}-F_n\\tag{k=4}$$ $$F_{n+10}=11F_{n+5}+F_n\\tag{k=5}$$\n\n-\n\nWe can write linear recurrence relations in terms of matrix multiplication like so:\n\n$$F_{n+2} = \\left[ \\begin{array}{cc} 1 & 1 \\end{array} \\right] \\left[ \\begin{array}{cc} F_n \\\\ F_{n+1} \\end{array} \\right] = 1 \\cdot F_n + 1 \\cdot F_{n+1}.$$\n\nNow if the sequence 0,2,8,34,144,610,2584,10946,... is called $G_n$, let's make the unjustified assumption that it is also a second order recurrence relation, then not only would we have\n\n$$G_{n+2} = \\left[ \\begin{array}{cc} c_1 & c_2 \\end{array} \\right] \\left[ \\begin{array}{cc} G_n \\\\ G_{n+1} \\end{array} \\right]$$\n\nfor some unknowns $c_1$ and $c_2$, but also we can collect several instances of the above identity together into one e.g.\n\n$$\\left[ \\begin{array}{cc} 34 & 144 \\end{array} \\right] = \\left[ \\begin{array}{cc} c_1 & c_2 \\end{array} \\right] \\left[ \\begin{array}{cc} 2 & 8 \\\\ 8 & 34 \\end{array} \\right]$$\n\nand we can solve this using PARI/GP like so:\n\n? [34,144]/[2,8;8,34]\n% = [1, 4]\n\n\nTherefore $G_{n+2} = 1 \\cdot G_n + 4 \\cdot G_{n+1}$.\n\nAbout the assumption, it can be proved in general based on the ideas of characteristic function which you have seen in most of the answers. Given that, there is no need for any induction proofs or anything. Just computing the vector completes the proof of $G_{n+2} = 1 \\cdot G_n + 4 \\cdot G_{n+1}$.\n\n-\nHere's an interesting identity involving Lucas and Fibonacci numbers: $$\\begin{pmatrix}\\ell_k&(-1)^{k+1}\\\\1&0\\end{pmatrix}=\\begin{pmatrix}f_k&f_{k-1}\\\\0\u200c\u200b&1\\end{pmatrix}\\cdot\\begin{pmatrix}1&1\\\\1&0\\end{pmatrix}^k \\cdot\\begin{pmatrix} f_k&f_{k-1}\\\\0&1\\end{pmatrix}^{-1}$$ \u2013\u00a0 Guess who it is. Dec 28 '11 at 15:21\n\nIt is easy by operator algebra. Let the Shift, Triple $\\mathbb C$-linear operators $\\rm\\ S\\:n\\: :=\\: n+1,\\ \\ T\\:n\\: :=\\: 3\\:n\\:$ act on fibonacci's numbers by $\\rm\\ S\\:f(a\\:n+b) = f(a\\:(n+1)+b)\\$ and $\\rm\\ T\\:f(a\\:n+b) = f(3\\:a\\:n+b)\\:.$ Below I show a general method that works for any Lucas sequence $\\rm\\:f(n)\\:$ that involves only simple high-school polynomial arithmetic (albeit noncommutative). Namely, one employs a commutation rule $\\rm\\: TS\\:\\to\\: (a\\ S + b)\\ T\\$ to shift $\\rm\\:T\\:$ past powers of $\\rm\\:S\\:,\\:$ in order to transmute the known recurrence $\\rm\\ q(S)\\ f(n)\\: =\\: 0\\$ into $\\rm\\ \\bar{q}(S)\\:T\\:f(n)\\:=\\:0\\:,\\$ the sought recurrence for $\\rm\\ T\\:f(n)\\: =\\: f(3\\:n)\\:.\\:$\n\nWe know $\\rm\\ q(S)\\ f(n) := (S^2 - S - 1)\\ f(n)\\: =\\: f(n+2) - f(n+1) - f(n)\\: =\\: 0\\:.\\:$ We seek an analogous recurrence $\\rm\\ \\bar{q}(S)\\ T\\: f(n)\\ =\\ 0\\$ for $\\rm\\ T\\:f(n) = f(3\\:n)\\:,\\:$ and some polynomial $\\rm\\:\\bar{q}(S)\\:.\\:$ Since clearly we have that $\\rm\\ T\\:q(S)\\ f(n)\\: =\\: 0\\:,\\:$ it suffices to somehow transmute this equation by shifting $\\rm\\:T\\:$ past $\\rm\\:q(S)\\:$ to yield $\\rm\\:\\bar{q}(S)\\:T\\:f(n)\\:=\\:0\\:.\\:$ To do this, it suffices to find some commutation identity $\\rm T\\:S\\: =\\: r(S)\\: T\\$ to enable us to shift $\\rm\\:T\\:$ past $\\rm\\:S$'s in each monomial $\\rm\\ S^{\\:i}\\: f(n)\\: =\\: f(n+i)\\:$ from $\\rm\\:q(S)\\:.\\:$ The sought commutation identity arises very simply: iterate the recurrence for $\\rm\\:f(n)\\:$ so to rewrite\n\n$\\rm\\ ST\\ f(n)\\ =\\ f(3\\:n+3)\\$ as a linear combination of $\\rm\\ f(3\\:n+1) = TS\\ f(n)\\:,\\:$ $\\rm\\ f(3\\:n) = T\\ f(n)\\:,\\:$ viz.\n\n$\\rm\\ \\ \\ ST\\ f(n+i)\\ =\\ f(3n+3+i)\\ =\\ f(3n+2+i) + f(3n+1+i)\\ =\\ 2\\ f(3n+1+i) + f(3n+i)$\n\n$\\rm\\ \\ \\ \\phantom{ST\\ f(n+i)}\\ =\\ (2\\:TS+T)\\ f(n+i)\\quad$ for all $\\rm\\:i\\in \\mathbb Z\\:$\n\n$\\rm\\ 2\\:TS\\ f(n+i)\\ =\\ (S-1)\\:T\\ f(n+i)\\:,\\$ i.e. $\\rm\\ 2\\:TS\\ =\\ (S-1)\\:T\\:,\\$ the sought commutation identity.\n\nThus $\\rm\\qquad\\quad\\:\\ 0\\ =\\ 4\\: T\\: (S^2 - S - 1)\\ f(n)\\$\n\n$\\rm\\qquad\\qquad\\qquad\\quad\\ \\ =\\ (2\\:(2TS)S - 2\\:(2TS) - 4\\:T)\\ f(n)$\n\n$\\rm\\qquad\\qquad\\qquad\\quad\\ \\ =\\ ((S-1)\\:2TS - 2\\:(S-1)\\:T - 4\\:T)\\ f(n)$\n\n$\\rm\\qquad\\qquad\\qquad\\quad\\ \\ =\\ ((S-1)^2 - 2\\:(S-1)\\: - 4)\\ T\\: f(n)$\n\n$\\rm\\qquad\\qquad\\qquad\\quad\\ \\ =\\ (S^2 - 4\\ S - 1)\\ T\\: f(n)$\n\n$\\quad$ i.e. $\\rm\\qquad\\quad\\: 0\\ =\\ f(3(n+2)) - 4\\ f(3(n+1)) - f(3\\:n)\\qquad$ QED\n\nNOTE $\\$ Precisely the same method works for any Lucas sequence $\\rm\\:f(n)\\:,\\:$ i.e. any solution of $\\rm\\ 0\\ =\\ (S^2 + b\\ S + c)\\ f(n)\\ =\\ f(n+2) + b\\ f(n+1) + c\\ f(n)\\$ for constants $\\rm\\:b,\\:c\\:,\\:$ and for any multiplication operator $\\rm\\:T\\:n = k\\ n\\:$ for $\\rm\\:k\\in \\mathbb N\\:.\\:$ As above, we obtain a commutation identity by iterating the recurrence (or powering its companion matrix), in order to rewrite\n\n$\\rm\\ ST\\ f(n)\\ =\\ f(k\\:n+k)\\$ as a $\\rm\\:\\mathbb C$-linear combination of $\\rm\\ f(kn+1) = TS\\ f(n)\\$ and $\\rm\\ f(kn) = T\\ f(n)\\:$\n\nsay $\\rm\\ \\ ST\\ f(n)\\ =\\ f(k\\:n+k)\\ =\\ a\\ f(k\\:n+1) + d\\ f(k\\:n)\\ =\\ (a\\ TS + d\\ T)\\ f(n)\\ \\$ for some $\\rm\\:a,d\\in \\mathbb C$\n\n$\\rm\\:\\Rightarrow\\ a\\ TS\\ f(n) =\\ (S-d)\\ T\\ f(n)\\ \\Rightarrow\\ a\\ TS\\ =\\ (S-d)\\ T\\$ on $\\rm\\ S^{\\:i}\\: f(n)\\$ as above.\n\nAgain, this enables us to transmute the recurrence for $\\rm\\:f(n)\\:$ into one for $\\rm\\:T\\:f(n) = f(k\\:n)\\:$ by simply commuting $\\rm\\:T\\:$ past all $\\rm\\:S^i\\:$ terms. Hence the solution involves only simple polynomial arithmetic (but, alas, the notation obscures the utter simplicity of the method).\n\n-", "date": "2015-09-05 17:02:06", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/94359/need-help-deriving-recurrence-relation-for-even-valued-fibonacci-numbers", "openwebmath_score": 0.973792552947998, "openwebmath_perplexity": 330.2119062709151, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864281950694, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.8670172293021442}}
{"url": "https://math.stackexchange.com/questions/2876194/proof-verification-if-a-subset-b-and-b-subset-c-then-a-cup-b-subset/2876237", "text": "# Proof Verification: If $A \\subset B$ and $B \\subset C$, then $A \\cup B \\subset C$\n\nI am trying to prove that:\n\nIf $A \\subset B$ and $B \\subset C$, then $A \\cup B \\subset C$\n\nMy proof is : Given some $x \\in A \\cup B$, it is true that either $x \\in A$ and/or $x \\in B$. IN the case that $x \\in A$ it is true that $x \\in B$, as $A \\subset B$, and that $x \\in C$ , as $B \\subset C$. In the case that $x \\in B$ it is true that $x \\in C$, as $B \\subset C$. Therefore, $A\\cup B \\subset C$\n\nIs this correct?. Any tips to improve this would be appreciated as I am self taught and new to proof writing.\n\n\u2022 Yes, it is correct. You could also prove that $A \\cup B = B$, and then the fact follows from $B = A \\cup B \\subset C$. \u2013\u00a0\u0430\u0441\u0442\u043e\u043d \u0432\u0456\u043b\u043b\u0430 \u043e\u043b\u043e\u0444 \u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 Aug 8 '18 at 15:14\n\u2022 Your proof is totally correct. Just be careful about the use of English: it is better to say \"Given some $x \\in A \\cup B$, it is true that $x \\in A$ or $x \\in B$\". Do not use \"either\" and \"and\" unnecessarily as it may make your statement confusing. In maths, the word \"or\" means : either the first case only, either the second only, either both cases at the same time. \u2013\u00a0Suzet Aug 8 '18 at 15:16\n\u2022 @\u0430\u0441\u0442\u043e\u043d\u0432\u0456\u043b\u043b\u0430\u043e\u043b\u043e\u0444\u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 Thank you \u2013\u00a0Ewan Miller Aug 8 '18 at 15:16\n\u2022 @Suzet Ok thank you \u2013\u00a0Ewan Miller Aug 8 '18 at 15:17\n\u2022 It's totally correct, although there's an alternative way. Since $A \\subset B$, $A \\cup B= B$. Now from $B \\subset C$, the result follows. \u2013\u00a0Anik Bhowmick Aug 8 '18 at 15:32\n\nJust to answer : yes, the approach is correct.\n\nYou could also prove that $A\u222aB=B$, and then the fact follows from $B=A\u222aB\\subset C$.\n\nFor example, if $x \\in B$ is true, then of course $x$ is in $A$ or $x$ is in $B$ is true, so $B \\subset A \\cup B$. If $x \\in A$, then $x \\in B$ because $A \\subset B$, and if $x \\in B$, then of course $x \\in B$, so $A \\cup B \\subset B$, hence $A \\cup B = B$.\n\nAnother approach is by using Venn diagrams. Draw circles $A$, $B$ and $C$ for three sets such that $A$ is contained in $B$ and $B$ is contained in $C$ (according to given set inclusions). So you have $A$ as the innermost, $B$ in the middle and $C$ as the outermost of them. Now $A\\cup B$ is given by the middle circle which is offcourse contained in $C$ (the outer circle).\n\nIs this correct?\n\nI believe that your proof regarding this Question is correct.\n\nAny tips to improve this would be appreciated as I am self-taught and new to proof writing.\n\nSet Theory which you are using (Elementary-Set Theory) was mostly Contributed for its Development by George Cantor. But, in Later Stage, there are some famous Loopholes found in this Theory, the most popular of them is Russel's Paradox. After Russel's paradox, numerous other Paradox came after which ZFC Set Theory (Zermelo Franklin Set Theory) came which is based on some of the Basic Set Axioms (Related to Mathematical Logic)\n\n## Proofs in Set Theory\n\nMost of the Proofs in Elementary Set Theory is based on Mathematical Logic. Some of the Basic Theorems and Properties are below -\n\n1. De Morgan's Theorem -https://en.wikipedia.org/wiki/De_Morgan%27s_laws\n2. Complement - https://en.wikipedia.org/wiki/Complement_(set_theory) $A^` = U - A$\n3. Intersection and Union -\n\n## Alternate Methods to Proof Set Theory Question\n\nYour Question Can be Proved Using Venn Diagram too like here -\n\n\u2022 Using The Laws and Properties of Sets (Intersection, Complement and Union)\n\nThe most natural way seems to be:\n\n1. (Transitivity) From $A \\subset B$ and $B \\subset C$, show $A \\subset C$.\n\nIf $x \\in A$, then $x \\in B$.\nIf $x \\in B$, then $x \\in C$.\nThus if $x \\in A$, then $x \\in C$.\n\n1. From $A \\subset C$ and $B \\subset C$, show $(A \\cup B) \\subset C$.\n\nIf $x \\in (A \\cup B)$, then $x \\in A$ or $x \\in B$.\nBut if $x \\in A$, then $x \\in C$, and if $x \\in B$, then $x \\in C$.\nThus, either way, $x \\in C$.\n\nYou implicitly used this method in your (correct) proof, but you didn't separate out the ideas. It can be especially useful to organise larger proofs in terms of simpler definitions, lemmas and theorems.", "date": "2020-04-08 06:58:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2876194/proof-verification-if-a-subset-b-and-b-subset-c-then-a-cup-b-subset/2876237", "openwebmath_score": 0.8860731720924377, "openwebmath_perplexity": 303.48534530084186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353126336, "lm_q2_score": 0.880797071719777, "lm_q1q2_score": 0.8670005774210011}}
{"url": "http://mathhelpforum.com/algebra/49692-simultaneous-equations.html", "text": "Math Help - Simultaneous Equations\n\n1. Simultaneous Equations\n\nI don't usually have any problem with simultaneous equations but whilst revising for a test, I came across two types that I haven't seen before and the book I have only gives answers but no explanations of how it got those answers. I've tried in various ways... but my answers never seem to match up.\n\nCould anyone solve these two simultaneous equations step-by-step so I could see how it's done?\n\nQ1)\n$y = 1 - 4x$\n$y = -4x^2$\n\nQ2)\n$xy = 9$\n$x - 2y = 3$\n\nIf anyone could I would really appreciate it, thanks. And Hi to everyone on the forum - just found this forum and it looks great.\n\n2. Hello, PaulSelb!\n\nWelcome aboard!\n\nThese are not linear equations.\n. . The recommended method is Substitution.\n\n$1)\\;\\;\\begin{array}{cccc}y &=& 1 - 4x &{\\color{blue}[1]} \\\\\ny &= &-4x^2& {\\color{blue}[2]}\\end{array}$\n\nEquation [1] is already solved for $y\\!:\\;\\;y \\:=\\:1-4x$\n\nSubstitute into [2]: . $1 - 4x \\:=\\:-4x^2 \\quad\\Rightarrow\\quad 4x^2 - 4x + 1 \\:=\\:0$\n\n. . $(2x-1)^2 \\:=\\:0 \\quad\\Rightarrow\\quad 2x-1 \\:=\\:0 \\quad\\Rightarrow\\quad x \\:=\\:\\frac{1}{2}$\n\nSubstitute into [1]: . $y \\:=\\:1 - 4\\left(\\frac{1}{2}\\right) \\:=\\:-1$\n\nSolution: . $x \\:=\\:\\frac{1}{2},\\;y \\:=\\:-1$\n\n$2);\\;\\begin{array}{cccc}xy &=& 9 & {\\color{blue}[1]} \\\\\nx - 2y &=& 3 & {\\color{blue}[2]}\\end{array}$\n\nSolve [2] for $x\\!:\\;\\;x \\:=\\:2y+3\\;\\;{\\color{blue}[3]}$\n\nSubstitute into [1]: . $(2y+3)y \\:=\\:9 \\quad\\Rightarrow\\quad 2y^2 + 3y - 9 \\:=\\:0$\n\nFactor: . $(2y - 3)(y + 3)\\:=\\:0\\quad\\Rightarrow\\quad y \\:=\\:\\frac{3}{2},\\;-3$\n\nSubstitute into [3]: . $\\begin{Bmatrix}x &=& 2\\left(\\frac{3}{2}\\right) + 3 & = & 6 \\\\ \\\\[-3mm] x &=&2(-3) + 3 &=&-3 \\end{Bmatrix}$\n\nSolutions: . $(x,y) \\;=\\;\\left(6,\\:\\frac{3}{2}\\right),\\;(-3,\\:-3)$\n\n3. Thanks, great explanation . Got my test in a few hours so hopefully if one of these questions turns up I shouldn't have any problems.\n\n4. I had a slightly different type in the exam the other day (possibly solved in the same way)... but I got it wrong\n\nIf anyone reads this thread again could you let me know how to solve this one?\n\n$k(k - m) = 12$\n$k(k + m) = 60$\n\nSorry for the easy stupid questions, I really enjoy math and want to understand bits more!\n\nEDIT: I think I did it may have done it actually...\n\n$k^2 - km = 12$\n$k^2 + km = 60$\n$k+m = \\frac{60}{k}$\n$m = \\frac{60}{k} - k$\n\n$k^2 - k(\\frac{60}{k} - k) = 12$\n\n$2k^2 - 60 = 12$\n\n$2k^2 - 72 = 0$\n\n$(2k - 12)(k + 6) = 0$\n\n$k = \\pm 6$\n\n$36 + \\pm6(m) = 60$\n\n$m = \\frac{24}{\\pm6}$\n\n$m = \\pm 4$\n\nI think that's about right anyway...", "date": "2014-09-16 16:14:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/49692-simultaneous-equations.html", "openwebmath_score": 0.8060638308525085, "openwebmath_perplexity": 1065.4297118320678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561712637256, "lm_q2_score": 0.8947894569842487, "lm_q1q2_score": 0.8669222873809073}}
{"url": "https://stats.stackexchange.com/questions/241652/simulating-order-statistics", "text": "# Simulating order statistics\n\nI am having a problem with the density of the first order statistic of a series of n random variables iid with common distribution (standard normal).\n\nI am using Arnold's book as a reference for such a density function for the k'th order stat: $$f_{X_{k:n}}(x) = \\frac{n!}{k-1! (n-k)!} F(x)^{k-1} (1-F(x))^{n-k} f(x).$$\n\nThe I simulate data in R store the min and compare the obtained empirical distribution against the theoretical function. These curves diverge substantially.\n\nWhat am I doing wrong? or missing?\n\n# Distribution\nFx <- function(x) pnorm(x)\nfx <- function(x) dnorm(x)\n\nn <- 10\n# ATTENTION: 6M simulation may take some time\nz <- replicate(1e6, min(rnorm(n)))\n\n# Probabiltiy z > y\n\nprob <- function(x) {\nsapply(x, function(u) mean(z>u))\n}\n\n# Density\nfos_k <- function(x, n, k) {\nfact <- factorial(n) / (factorial(k-1) * factorial(n-k))\nfact * fx(x) * Fx(x)^(k-1) *(1-Fx(x))^(n-k)\n}\n\ncurve(fos_k(x, n=n, k=1), from=-1, to=1)\ncurve(prob, add=T, col=2, from=-1, to=1)\n\n\u2022 The \"observed empirical distribution\" is not the density. One way to plot the observed density is to approximate it with a KDE and plot that. \u2013\u00a0whuber Oct 21 '16 at 19:29\n\u2022 @Xi'an could you please clarify \u2013\u00a0mrb Oct 21 '16 at 20:44\n\u2022 Compare your equation to the second line of fos_k: your equation omits the term you have coded as fx(x). But that's only an error of exposition; the error in your code lies in prob, which is not a density function. \u2013\u00a0whuber Oct 21 '16 at 21:40\n\u2022 For what it's worth, here's the computation done in a single line in Mathematica: Show[Histogram[ParallelTable[Min[RandomVariate[NormalDistribution[0, 1], 10]], {10^6}], {0.1}, \"PDF\"], Plot[Evaluate[PDF[OrderDistribution[{NormalDistribution[0, 1], 10}, 1], z]], {z, -4, 1}]] On my machine, the runtime is less than 1 second. \u2013\u00a0heropup Oct 21 '16 at 22:08\n\u2022 Thanks to everyone. @whuber is right. @whuber would you like to go on and reply to this question with an example of your R code? \u2013\u00a0mrb Oct 22 '16 at 1:54\n\nAlthough the problem is primarily with R code, it raises issues we would have to confront in any statistical computing environment. This reply focuses on those general issues.\n\nA correct formula for the density of the $k^\\text{th}$ smallest of $n$ independent identically distributed (iid) values from a distribution $F$ with density $f$ is presented in my answer at https://stats.stackexchange.com/a/225990/919. It is\n\n$$f_{[k]}(x) = \\frac{n!}{(k-1)!(1)!(n-k)!} F(x)^{k-1} (1-F(x))^{n-k} f(x).\\tag{1}$$\n\nBecause factorials and potentially high powers are combined, it is numerically better to compute its logarithm as\n\n\\eqalign{ \\log f_{[k]}(x) = &\\log n! - \\log(k-1)! - \\log(n-k)! + \\\\&(k-1)\\log F(x) + (n-k) \\log(1-F(x)) + \\log f(x).\\tag{2} }\n\nFurthermore, because order statistics provide a way to peer far out into the tails of the distribution, where values grow close to $1$ and $0$, it is best to compute $1-F(x)$ directly rather than subtracting $F(x)$ from $1$.\n\nWith these caveats in mind, the other issue in this thread concerns graphing an empirical density. The commonest way to do so is with a histogram: it looks like a barplot in which the bar areas (not heights!) are proportional to the relative frequencies of the data.\n\nTo illustrate these two main points, I wrote a quick solution in R that simulates many iid samples, extracts specified order statistics from each, plots the histogram of each order statistic, and overplots the density function $(1)$ computed by exponentiating the logarithm $(2)$. Here is an example of its output applied to samples of size $10$ from a standard Normal distribution and of size $25$ from a Gamma$(3/2,5)$ distribution.\n\nClearly the agreement between the empirical results (the histogram bars) and the theoretical formula (the colored curves) is good.\n\nThis code applies the preceding suggestions about numerical computation by exploiting the log.p, log, and lower.tail arguments that are standard in the families of R distribution functions.\n\nf <- function(n.sim, n, k=1:n, p=pnorm, d=dnorm, r=rnorm, name, ...) {\nif (missing(name)) name <- \"\"\nk <- sort(unique(k))[1 <= k & k <= n]\n\n# Perform the simulation.\nsim <- apply(matrix(r(n.sim*n, ...), nrow=n), 2, sort)[k, , drop=FALSE]\n\n# Plot the requested order statistics.\nfor (i in 1:length(k)) {\n\n# Define a function to plot the density of an order statistic.\ndord <- function(x, k, n, ...) {\nz <- lfactorial(n) - lfactorial(k-1) - lfactorial(n-k) +\n(k-1)*p(x, log.p=TRUE, ...) +\n(n-k)*p(x, log.p=TRUE, lower.tail=FALSE, ...) +\nd(x, log=TRUE, ...)\nreturn(exp(z))\n}\n\n# Plot the empirical distribution.\nhist(sim[i, ], freq=FALSE,\nxlab=\"Value\",\nsub=paste(n.sim, \"iterations with sample size\", n),\nmain=paste(name, \"order statistic\", k[i]))\n\n# Overplot the theoretical density.\ncurve(dord(x, k[i], n, ...), add=TRUE, col=hsv(runif(1), 0.8, 0.7), lwd=2)\n}\n}\n#\n# Set up to simulate and display the results.\n#\npar(mfrow=c(2,4))\nn.sim <- 1e4\nset.seed(17)\n\n# Study Normal order statistics.\nf(n.sim, 10, c(1,3,5,9), name=\"Normal(0,1)\")\n\n# Study Gamma order statistics.\n# This illustrates how to use f for general distributions.\nf(n.sim, 25, c(2, 4, 16, 24), name=\"Gamma(1.5,5)\",\np=pgamma, d=dgamma, r=rgamma, shape=1.5, scale=5)\n\n\u2022 Wondering should use the logarithm to define the corresponding distribution function as well? Since F(x) can be zero I would avoid the log. \u2013\u00a0mrb Oct 24 '16 at 13:03\n\u2022 That's a good observation, but it happens not be be relevant because almost surely $F$ will be positive at any value you have randomly generated. Regardless, good software will deal with this correctly (provided you actually ask it to compute the log of $F$ directly rather than asking it to take the log of $0$!). For instance, the R procedures will return -Inf for the logarithm in that case; and that value, when exponentiated, will be exactly $0$. Here is an example: exp(punif(-2, log.p=TRUE)) \u2013\u00a0whuber Oct 24 '16 at 13:41", "date": "2019-10-14 20:37:25", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/241652/simulating-order-statistics", "openwebmath_score": 0.5744186639785767, "openwebmath_perplexity": 2384.3748603649874, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9626731105140615, "lm_q2_score": 0.9005297754396141, "lm_q1q2_score": 0.8669158000329825}}
{"url": "http://mathhelpforum.com/calculus/208021-calc-word-problem-please-help-2.html", "text": "No, you need to divide by $\\displaystyle 12^3$...essentially a 12 for each of the 3 spatial dimensions.\n\n$\\displaystyle 7128\\text{ in}^3\\cdot\\left(\\frac{1\\text{ ft}}{12\\text{ in}} \\right)^3=\\frac{7128}{12^3}\\,\\text{ft}^3=156.75 \\text{ ft}^3$\n\nOh, ok that makes since. So for part 5 would I use the formula 1/2*b*h so 1/2*(99)*(104) = 5148 in^2\n\nNo, you have a trapezoid, not a triangle. You want to use:\n\n$\\displaystyle A=\\frac{h}{2}(B+b)$\n\nwhere:\n\n$\\displaystyle h=99$\n\n$\\displaystyle B=104$\n\n$\\displaystyle b=32$\n\nO.k., sorry for not getting this right away, thanks for your patience lol.\nSo the answer would then be 6732\n\nAn impatient person has no business trying to help on a forum. You are doing fine.\n\nYes, that is the correct answer. Now, in order to find this area using integration, I recommend orienting your coordinate axes such that the origin is at the bottom corner under the lowest point of the ceiling. We will then want to write the upper slanting edge of the wall as a linear function. What would the y-intercept and the slope be?\n\nWould the slope be 8 and the y-intercept 32?\n\nYou have the correct intercept, but for the slope, think of the rise over run of an individual step.\n\nok so the slope would be 8/11?\n\nYes, good work! So, what would the linear function representing the slanted edge be?\n\n(8/11)x+32?\n\nExactly! Now, over what interval do you want to integrate?\n\nWould it be from 5 to 13?\n\nand if that is the interval would it be:\n(4/11)x^2+32x evaluated from 5 to 13= 308.364", "date": "2018-06-18 18:05:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/208021-calc-word-problem-please-help-2.html", "openwebmath_score": 0.58154296875, "openwebmath_perplexity": 619.8393017082359, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462194190619, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8668998779861325}}
{"url": "http://mathhelpforum.com/number-theory/8194-modulo-problem.html", "text": "1. ## Modulo problem\n\nOkay, so I'm supposed to find the least nonnegative integer i such that\n\n4^30 is congruent to i (mod 19).\n\nI'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.\n\n2. Originally Posted by clockingly\nOkay, so I'm supposed to find the least nonnegative integer i such that\n\n4^30 is congruent to i (mod 19).\n\nI'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.\nMy approach here won't be unique, but the method should be fairly clear.\n\n$4^3 = 64 \\equiv 7$ (mod 19)\n\nSo\n$4^{30} = (4^3)^{10} \\equiv 7^{10}$ (mod 19)\n\nAgain:\n$7^2 \\equiv 11$ (mod 19)\n\nSo\n$7^{10} = (7^2)^{5} \\equiv 11^5$ (mod 19)\n\nYou can probably do this one by directly but we can simplify this one last time:\n$11^2 \\equiv 7$ (mod 19)\n$11^3 \\equiv 1$ (mod 19)\n\nSo finally:\n$4^{30} \\equiv 7^{10} \\equiv 11^5 = 11^2 \\cdot 11^3 \\equiv 7 \\cdot 1 = 7$ (mod 19)\n\n$4^{30} \\equiv 7$ (mod 19)\n\n-Dan\n\nI am still a little confused as to how you got the very first equation though (listd below)...I'm a little confused as to where the 7 came from...\n\n4^3 = 64 is congruent to 7(mod 19)\n\n4. Originally Posted by clockingly\nOkay, so I'm supposed to find the least nonnegative integer i such that\n\n4^30 is congruent to i (mod 19).\n\nI'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.\nYou can also do this...\n$4^{30}=2^{60}$\nAnd,\n$2^{18}\\equiv 1 (\\mbox{ mod }19)$\nBy Fermat's little theorem since 19 is prime not divisible by 2.\nCubing,\n$2^{54}\\equiv 1 (\\mbox{ mod }19)$\nMultiply both sides by $2^6=64\\equiv 7$ we we have,\n$2^{60}\\equiv 7 (\\mbox{ mod }19)$\n\n5. Originally Posted by clockingly\n\nI am still a little confused as to how you got the very first equation though (listd below)...I'm a little confused as to where the 7 came from...\n\n4^3 = 64 is congruent to 7(mod 19)\nThat is becuase,\n$64\\equiv 7 (\\mbox{ mod }19)$\nRight?\n\nBecause it leaves a remainder of 19.\n\n( $64=3\\cdot 19+7$)\n\n6. Hello, clockingly!\n\nAnother approach . . .\n\nFind the least nonnegative integer $n$ such that: . $4^{30} \\equiv n \\pmod{19}$\n\nI found that: . $4^5 \\:=\\:1024 \\:\\equiv\\:-2 \\pmod{19}$\n\nThen: . $4^{30} \\:=\\:(4^5)^6 \\:\\equiv\\:(-2)^6 \\pmod{19}$\n\nAnd: . $(-2)^6\\:=\\:64\\:\\equiv\\:7\\pmod{19}$\n\nTherefore: . $4^{30}\\:\\equiv\\:7\\pmod{19}$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nI also found that: . $4^9\\:\\equiv\\:1\\pmod{19}$\n\nThen: . $4^{30}\\:=\\:\\left(4^9\\right)^3\\cdot4^3 \\:\\equiv\\:(1^3)4^3\\pmod{19}$\n\nAnd: . $4^3\\:=\\:64\\:\\equiv\\:7\\pmod{19}$\n\nBut I see that TPHacker beat me to it.", "date": "2017-06-27 10:45:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/8194-modulo-problem.html", "openwebmath_score": 0.8891947865486145, "openwebmath_perplexity": 385.5873622307119, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462219033656, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8668998706696582}}
{"url": "https://math.stackexchange.com/questions/2888262/what-is-the-least-amount-of-pieces-on-a-board-with-the-following-conditions", "text": "# What is the least amount of pieces on a board with the following conditions:\n\nThere's an infinite board. Imagine you add a rectangle of $m*n$ pieces. With $m,n \\geq 2$ (There's a piece every square, and you can't put one above other.) You can make a piece 'jump' other that is next to it (Vertically or horizontally only, not diagonally), if the square next to the \"jumped\" piece (In the same direction) is empty, also, the \"jumped\" piece disappears. After many movements, What is the least amount of pieces that can be on the board?\n\nSo, it's obvious that the answer can't be $0$ since there's no point of it, it has to be at least one if we have, for example, to pieces left of all the $mn$ pieces that are connected each other. I tried to do a coloring of the infinite board, in a chess pattern, so we have Black pieces (Pieces in a black tile) and White pieces, this make any white piece unable to make dissapear another white one and the same with black ones. At the end, the least we have of each color the better, because if we get to have only two left with different colors and connected, then we'll know the least number of pieces is 1. i tried looking for different rectangles of pieces, and i got 1 and 2 as answers. For example doing the following when $m=3$ and $n=4$ makes the board have 1 piece at the end: (From left to right)\n\nIt clearly follows that it has 1 piece at the end. I found other cases as ($m=2,n=4$), ($m=2, n=5$), but i also found cases where i got the least is 2, like ($m=2, n=3$), ($m=3, n=3$), etc. i haven't noticed anything else besides this. Any suggestions?\n\n\u2022 How did you get from your second board to your third? \u2013\u00a0Mike Earnest Aug 21 '18 at 17:46\n\u2022 Move (3,1) to (5,1) an then move (1,1) to (3,1) \u2013\u00a0SonodaUmi Aug 21 '18 at 17:59\n\u2022 Then you have a mistake; the checker at (4,1) was hopped over, so should be removed. \u2013\u00a0Mike Earnest Aug 21 '18 at 18:01\n\nThis is a fun problem with a nice solution!\n\nFirst of all, imagine coloring the checkerboard in three colors like so:\n\nA B C A B C A B C ...\nB C A B C A B C A ...\nC A B C A B C A B ...\n...\n\n\nLet $a,b,c$ be the number of checkers on each color. Each move increases one of these variables by $1$, and decreases the others by $1$. This means that the quantities $a-b,b-c$ and $c-a$ will always have the same parity.\n\nNow, suppose that either $m$ or $n$ is a multiple of $3$. Then you will initially have $a=b=c$, so that $a-b,b-c$ and $c-a$ all start out even. However, if you manage to reduce the board to a single checker, then at least one of $a-b,b-c$ or $c-a$ would be odd. Therefore, the problem is insoluble when $m$ or $n$ is a multiple of $3$.\n\nWhen neither dimension is a multiple of $3$, you can succeed. Consider the following \"T\"-move. It allows you to eliminate a $3\\times 1$ block of marbles, provided one of the ends has a marble on on side and an empty space on the other:\n\n\u2022 \u2022 _         _ _ \u2022         _ \u2022 \u2022         \u2022 _ _\n\u2022    -->      \u2022    -->      _    -->      _\n\u2022             \u2022             _             _\n\n\nApplying this repeatedly, you eliminate a $3\\times n$ block from the top of the rectangle, provided $m>3$ and $n\\ge 3$:\n\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022\n\n|\nv\n\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 _\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 _\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 _\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022\n\n|\nv\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 _ _\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 _ _\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 _ _\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022\n\n|\nv\n...\n|\nv\n\n\u2022 \u2022 \u2022 _ _ _ _ _\n\u2022 \u2022 \u2022 _ _ _ _ _\n\u2022 \u2022 \u2022 _ _ _ _ _\n\u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022 \u2022\n\n\nand then apply the same idea with three sideways T's to get rid of that last $3\\times 3$ block. The same idea allows you eliminate three columns at a time provided the dimensions are larger enough.\n\nThis procedure allows you to reduce the board either a $4\\times 4$, $2\\times 4$, $4\\times 2$ or $5\\times 5$ rectangle, depending on the remainders of $m$ and $n$ modulo $3$. I leave it to you to figure out how to solve these small cases.\n\n\u2022 My teacher said it was a really beautiful solution, and it really is. Thank you! \u2013\u00a0SonodaUmi Aug 21 '18 at 18:25\n\u2022 @cptnSonoda Does your teacher sit next to you and checked the answer on math.stackexchange (MSE)? \u2013\u00a0callculus Aug 21 '18 at 18:38", "date": "2019-09-15 13:33:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2888262/what-is-the-least-amount-of-pieces-on-a-board-with-the-following-conditions", "openwebmath_score": 0.7462764382362366, "openwebmath_perplexity": 330.1878275416019, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462194190618, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8668998684897393}}
{"url": "https://math.stackexchange.com/questions/1920431/what-is-a-good-upper-bound-nnn-1n-1-ldots2211", "text": "# What is a good upper bound $n^n(n-1)^{n-1}\\ldots2^21^1$?\n\nGiven an integer $n \\ge 1$, I'd like to have a not-very-loose upper bound for the integer $$u(n) := \\Pi_{k=1}^n k^k = n^n(n-1)^{(n-1)}\\ldots2^21^1.$$\n\nIt's easy see that, $u(n) \\le n^{n(n+1)/2}$, but this is not very interesting.\n\n# Update\n\nWe have $u(n) \\le e^{\\left(\\frac{1}{2}n(n+1)\\log\\left(\\frac{2n + 1}{3}\\right)\\right)}$, and we can't really do much better!\n\nIndeed, using Euler-Maclaurin, we have\n\n$\\log(u(n)) = \\int_2^nx\\log x dx = \\frac{1}{4}n^2(2\\log(n) - 1) - 2\\log(2) + \\frac{1}{4} + \\text{error terms}$, which is comparable to the bound $\\log(u(n)) \\le \\frac{1}{2}n(n+1)\\log\\left(\\frac{2n + 1}{3}\\right)$ in the accepted answer (see below). In particular, we can conclude that accepted answer's bound is tight!\n\n\u2022 Taking logarithms may be a good start whenever you're dealing with products. Sep 9, 2016 at 13:45\n\u2022 Ya, I contemplated the sequence $\\sum_{k=1}^nk \\log(k)$, but nothing pops-up... Sep 9, 2016 at 13:46\n\u2022 Yes. Take logarithms, use Euler-MacLaurin, and exponentiate. Sep 9, 2016 at 13:47\n\u2022 Thanks for pointing to Euler-Maclaurin. At the moment, I was just thinking about the Abel summation formula, but your suggestion looks more appropriate. Sep 9, 2016 at 13:52\n\u2022 See also $(7)$ here: mathworld.wolfram.com/Hyperfactorial.html Sep 9, 2016 at 17:24\n\nUsing Jensen's inequality:\n\nLetting $A= \\sum_{k=1}^n k = \\frac{n (n+1)}{2}$ and $B= \\sum_{k=1}^n k^2 = \\frac{n (n+1)(2n+1)}{6}$\n\nWe have\n\n\\begin{align} \\log(u(n)) &=\\sum_{k=1}^n k \\log(k) \\\\ &= A \\sum_{k=1}^n \\frac{k}{A} \\log(k) \\tag{1}\\\\ &\\le A \\log \\left( \\sum_{k=1}^n \\frac{k}{A} k \\right) = A \\log \\left( \\frac{B}{A} \\right) \\tag{2} \\end{align}\n\nHence\n\n$$\\log(u(n)) \\le \\frac{n(n+1)}{2} \\log\\left(\\frac{2 n+1}{3}\\right) \\tag{3}$$\n\nThe bound seems to be quite tight:\n\nUpdate: as noted by comments and OP, the bound $(3)$ agrees with the true order of growth; this can be checked by applying the trapezoidal rule to the integral:\n\n$$-\\frac{1}{4}=\\int_{0}^{1} x \\log(x) dx \\approx \\frac{1}{n+1} \\sum_{k=1}^n \\frac{k}{n} \\log\\left(\\frac{k}{n}\\right)$$ which gives\n\n$$\\log(u(n)) \\approx\\frac{ n(n+1)}{2}\\left( \\log n -\\frac{1}{2} \\right) \\tag{4}$$\n\nIf one is interested in an approximation (instead of a bound), $(4)$ might be preferable.\n\n\u2022 Perfect! Thanks. I see you've used the concave function $\\phi(x) := log(x)$ and positive scalars $a_k = k$ in Jensen's inequality: en.wikipedia.org/wiki/Jensen%27s_inequality Sep 9, 2016 at 14:05\n\u2022 Now that's a freaky tight bound! I Indeed, it's no surprise that the bound is tight. Using Euler-Maclaurin suggested by Daniel above, I can get $\\log(u(n)) \\approx \\frac{1}{4} (n^2(2\\log(n) - 1) - 1)$ :). Thus I don't need to beg the error terms in my EM, and can just use your bound instead. Thanks again. Sep 9, 2016 at 14:10\n\nAnother simple way for finding an upper bound is using the Abel's summation $$S=\\sum_{k=1}^{n}k\\log\\left(k\\right)=\\frac{n\\left(n+1\\right)}{2}\\log\\left(n\\right)-\\frac{1}{2}\\int_{1}^{n}\\frac{\\left\\lfloor t\\right\\rfloor \\left(\\left\\lfloor t\\right\\rfloor +1\\right)}{t}dt$$ where $\\left\\lfloor t\\right\\rfloor$ is the floor function and since $\\left\\lfloor t\\right\\rfloor >t-1$ we get $$S<\\frac{n\\left(n+1\\right)}{2}\\log\\left(n\\right)-\\frac{1}{2}\\int_{1}^{n}\\left(t-1\\right)dt$$ $$=\\color{red}{\\frac{n\\left(n+1\\right)}{2}\\log\\left(n\\right)-\\frac{\\left(n-1\\right)^{2}}{4}}.$$\n\nDon't have the reputation, or else this would be a comment. As mentioned in a link posted by leonbloy, the hyperfactorial, which shows up in the theory of the Barne's G function, is related to $\\int_0^n \\log\\Gamma(x) dx.$ Good bounds on this should give a better bound than that found by Jensen's inequality. I found the expression\n\n$$\\log(u(n)) \\le A(n):=\\dfrac{(n+1/2)^2}{2}(\\log(n+1/2)-3/2)+\\dfrac{n(n+1)}{2}+ \\dfrac{9}{8}\\log(2/3)+\\dfrac{11}{16}.$$\n\nFor a comparison, let's define 'Jensen' and 'A' ratios\n\n$$R_J(n)=\\dfrac{\\log(u(n))}{n(n+1)/2\\log((2n+1)/3))},\\quad R_A(n)=\\dfrac{\\log(u(n))}{A(n)} .$$\n\nThen (approximately) $R_J(100)=0.892$ , $R_A(100)=0.999992;$ and $R_J(10^5)=0.9915$ , $R_A(10^5)=0.99999999915$ (nine nines).\n\n\u2022 Interesting. Do include some detail on your derivation of $A(n)$ (not just an expression for it...). Sep 13, 2016 at 7:39\n\u2022 Great answer! Upvoted. Jan 23, 2017 at 7:22", "date": "2022-05-27 16:34:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1920431/what-is-a-good-upper-bound-nnn-1n-1-ldots2211", "openwebmath_score": 0.9928174018859863, "openwebmath_perplexity": 869.2654299767381, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462190641614, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8668998681783224}}
{"url": "https://math.stackexchange.com/questions/2293493/factorise-x5x1/2293501#2293501", "text": "# Factorise $x^5+x+1$\n\nFactorise $$x^5+x+1$$\n\nI'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$\n\n$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$\n\n=$(x^3-x^2+1)(x^2+x+1)$\n\nMy question:\n\nIs there another method to factorise this as this solution it seems impossible to invent it?\n\n\u2022 In general, factorization of polynomials is, as you say, \"impossible to invent\" by hand. There are algorithms, though, which can solve the problem. May 23, 2017 at 15:04\n\u2022 Any high school student can solve this problem when he guesses one of the factors and then calculates the other factor by division. But that is not really a useful method to factor a polynomial. But it is similar to the method most of the answers uses. May 23, 2017 at 16:03\n\u2022 $$5 \\equiv 2 \\pmod 3$$ May 23, 2017 at 16:06\n\u2022 I just don't know how to accept only one answer when all the answers are simply amazing and awesome. May 23, 2017 at 22:25\n\u2022 @Mathxx, See math.stackexchange.com/questions/1584594/\u2026 May 24, 2017 at 5:36\n\nWith algebraic identities, this is actually rather natural:\n\nRemember if we had all powers of $x$ down to $x^0=1$, we could easily factor: $$x^5+x^4+x^3+x^2+x+1=\\frac{x^6-1}{x-1}=\\frac{(x^3-1)(x^3+1)}{x-1}=(x^2+x+1)(x^3+1),$$ so we can write: \\begin{align}x^5+x+1&=(x^2+x+1)(x^3+1)-(x^4+x^3+x^2)\\\\ &=(x^2+x+1)(x^3+1)-x^2(x^2+x+1)\\\\ &=(x^2+x+1)(x^3+1-x^2). \\end{align}\n\n\u2022 That's true, but it's not a factorisation. What do you mean? May 23, 2017 at 15:06\n\u2022 What does \"this is rather natural\" mean in this context? May 23, 2017 at 15:42\n\u2022 I simply meant any high school student which is trained to tackle formulae using various identities can factor this polynomial. May 23, 2017 at 15:45\n\u2022 I think this is no better than the solution the user already showed us. One has to guess the right things. May 23, 2017 at 16:31\n\u2022 Well, if you do not train, you'll never get a result. All I want to say is that everything is at the high school level, there's no hammersledge theorem, and that training in various ways stimulates imagination, which is the first quality to find proofs. May 23, 2017 at 16:35\n\nIf you suspect there exists a factorization over $\\mathbb{Q}[x]$ into polynomials of degree 2 and 3, but you just don't know their coefficients, one way is to write it down as\n\n$x^5 + x + 1 = (x^2 + ax + b)(x^3 + cx^2 + dx + e)$ where the coefficients are integers (by Gauss' lemma). And then expand and solve the system.\n\nThen $a + c = 0, b + ac + d = 0, bc + ad + e = 0, ae + bd = 1, be = 1$. So we get $c = -a$ and $b = e = 1$ or $b = e = -1$.\n\nIn the first case we reduce to $1 - a^2 + d = 0, -a + ad + 1 = 0, a + d = 1$ which gives $d = 1 - a, 1 - a^2 + 1 - a = 0, 1 - a^2 = 0$ so $a = 1, b = 1, c = -1, d = 0, e = 1$.\n\nSubstituting gives us $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$\n\nIf the factorization was not over $\\mathbb{Q}[x]$ then things would get more complicated because I could not assume $b = e = +/- 1$.\n\n\u2022 If you suspect there exists a factorization over Q[x] than it is easy to show that polynomial cannot be split into a polynomail of degree 4 and a linears factor, so it must be the product of thwo polynomails of degree 2 and 3 May 23, 2017 at 15:46\n\nAlternatively note that \\begin{align}x^5 + x + 1 &= x^5 - x^2 + x^2 + x + 1\\\\ & =x^2(x^3-1) + \\color{red}{x^2+x+1} \\\\ & = x^2(x-1)\\color{red}{(x^2+x+1)} + \\color{red}{x^2+x+1} \\\\& =\\color{blue}{(x^3-x^2+1)}\\color{red}{(x^2+x+1)} \\end{align}\n\nwhere we used the well known identity $x^3 - 1 = (x-1)(x^2+x+1)$ in the third equality.\n\nMeta: whilst the first step may seem arbitrary and magical, it is natural to want to insert a term like $x^2$ or $x^3$ into the equation in order to get some traction with factorising $x^5 + x^k$.\n\nNote that if $z^3=1, z\\neq 1$ so that $z^3-1=(z-1)(z^2+z+1)=0$ then $z^5+z+1=z^2+z+1=0$. A key to this observation is just seeing whether an appropriate root of unity may be a root.\n\nThis tells you that $x^2+x+1$ is a factor of $x^5+x+1$, and the question then is how you do the division. The factorisation method you have been shown is equivalent to doing the polynomial long division.\n\nAnother method, not suggested by others, is to use the fact that you know $x^2+x+1$ is a factor and multiply through by $x-1$ so that this factor becomes $x^3-1$.\n\nSo $(x-1)(x^5+x+1)=x^6-x^5+x^2-1=(x^3-1)(x^3+1)-x^2(x^3-1)=(x^3-1)(x^3-x^2+1)=(x-1)(x^2+x+1)(x^3-x^1+1)$\n\n\u2022 I think this is no better than the solution the user already showed us. One has to guess the right things May 23, 2017 at 16:32\n\u2022 @miracle173 But the user suggested the solution could not be discovered easily. The art of problem solving is in part learning the kind of things you might try when faced with something unfamiliar. May 23, 2017 at 19:22\n\nstandard trick from contests: $5 \\equiv 2 \\pmod 3.$ Therefore, if $\\omega^3 = 1$ but $\\omega \\neq 1,$ we get $$\\omega^5 = \\omega^2$$ $$\\omega^5 + \\omega + 1 = \\omega^2 + \\omega + 1 = 0$$ Which means, various ways of saying this, $x^5 + x + 1$ must be divisible by $$(x - \\omega)(x - \\omega^2) = x^2 + x + 1.$$ This is what we call a \"minimal polynomial\" for $\\omega$\n\nThe same idea would work for $$x^{509} + x^{73} + 1$$ although the other factor would be worse\n\nSEE\n\nPrime factor of $A=14^7+14^2+1$\n\n\u2022 I think this is no better than the solution the user already showed us. One has to guess the right things May 23, 2017 at 16:32\n\u2022 @miracle173 it's a matter of where the questions arise. This is a contest or contest preparation question, there is a trick; this comes up over and over again on this site, from kids who are trying to get better at contests. You might ask the OP where he got the question and what is the setting responsible for \"I'm being taught...\" Put another way, someone skilled at searching could find hundreds of questions on MSE where one factor is $x^2 + x + 1$ May 23, 2017 at 16:42\n\nSo far the answers just (cleverly) elaborate on high school tricks and techniques. Therefore, I think it can be interesting to see, instead, how standard modern algorithms work in this special case. I will implement a small version of the Berlekamp-Zassenhaus algorithm. I will try to factor $F(x)=x^5+x+1$ over $\\mathbb Z[x]$ as a product $f_1(x)f_2(x)$ of polynomials of degree 2 and 3; it will not be long (of course), nor difficult. I recall what is the plan:\n\n\u2022 Bound the coefficients of the factors $f_1,f_2$;\n\u2022 Factor $F(x)=g_1(x)g_2(x)$ over modulo $p$ for some prime $p$;\n\u2022 Lift (in essence, by Hensel's lemma) the factorization modulo higher $p^k$.\n\nBound the coefficients of $f_1(x)$: The leading coefficient of $F$ is $c=1$, the degree of $f_1$ is $\\delta=2$, and the roots $\\alpha$ of $F$ satisfy ${|\\alpha|}^5\\leq 1+|\\alpha|$, so for sure, say, $|\\alpha|<\\rho=1.5$. Therefore the (absolute values of the) coefficients of $f_1(x)$ are all dominated by those of $(x+\\rho)^2$. In particular they are all $<3$. This is called the binomial bound. The same estimate can be obtained with the Knuth-Cohen bound. See Abbott, John. \"Bounds on Factors in Z [x].\" Journal of Symbolic Computation 50 (2013): 532-563.\n\nFind $g_i(x)\\equiv f_i(x)$ modulo 2: Since $F(0)=1$ and $F(1)=3$ we have that $g_1(0)=g_1(1)=1\\bmod 2$. Thus the only possibility is $g_1(x)=x^2+x+1$. By polynomial long division in $\\mathbb F_2[x]$ we get $g_2(x)=\\frac{F(x)}{g_1(x)}=x^3+x^2+1$. Well, if you are clever enough, and not a computer, you might finish the exercise here, by doing long division in $\\mathbb Z[x]$.\n\nFactor modulo 4 as $F=(g_1+2 h_1)(g_2 + 2 h_2)$: In other words, we need $g_1 h_2+g_2 h_1 = \\frac {F(x)-g_1(x)g_2(x)}{2} = x^4+x^3+x^2 \\bmod 2$. Of course the solution is $h_1=0$ and $h_2=x^2$.\n\nConclusion: We have $f_1(x)\\equiv x^2+x+1\\bmod 4$. Since the absolute value of the coefficients of $f_1(x)$ is at most $2$, we get $f_1(x)= x^2+x+1$. It works.\n\nSupplement: actually I am deeply convinced that the most natural technique to factor $F(x)$ is the one provided by the OP (although all the other approaches, included the one I described above, are interesting). I'll try to justify this claim. Suppose you want to factor $7763073514021$ in prime numbers. The factor $7$ is easy to find (actually, this completes the factorization). Why? Because you decompose $7-7-63-0-7-35-14-0-21$ and you factor out termwise. \"Piecewise\" factorization is by far the most natural, and easy to spot, approach to factorization \"by hand\". Another example is $x^6-x^4-20x^3+14x^2+20 x -14$, where you may wish to take advantage of the pattern $(1,-1),(-20,20),(14,-14)$. Now, suppose you want to factor the number $636362236363$. It's very similar to the example before, only that you must be able to see the \"negative\": you are just subtracting a \"14\" from $636363636363$. Although the pattern of coefficients $[1,0,0,0,1,1]$ of $F(x)$ might be irregular at first glance, I find it very natural to see it as a $0-111-00$ subtracted from a $111-111$.\n\n\u2022 Or else (but only for \"how to factor\", not \"why does it factor\"): coefficients of f_1 have norm at most 2; F is monic; F(0)=F(-1)=1. So f_1=x^2+x+1 or f_1=x^2-x-1. Now use F(-2)=-33. May 24, 2017 at 1:09\n\nLet $f (x)=x^5+x+1$\n\n$$f'(x)=5x^4+1>0$$\n\n$f$ is stricly increasing at $\\mathbb R$, thus it has only one real root $\\alpha \\in (-1,0)$.\n\nthe factorisation will be of the form\n\n$$(x-\\alpha)(x^2+ax+b)(x^2+cx+d)$$\n\nwith $a^2-4b <0$ and $c^2-4d <0$.\n\n\u2022 Why can't it be factored in $\\mathbf R[X]$? The only irreducible polynomials in $\\mathbf R[X]$ are linear polynomials and quadratic polynomials with a negative discriminant. May 23, 2017 at 14:56\n\u2022 Just because it has no linear factors does not mean that it does not have higher order factors (DAMHIKT). Also, he showed a factorization of it and the factorization is correct: just multiply it out. So, your answer is wrong: it can be factorized in $\\mathbb R[X]$. May 23, 2017 at 14:58\n\u2022 @Nick Yes you are right. i edited. May 23, 2017 at 15:02", "date": "2022-07-07 08:14:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2293493/factorise-x5x1/2293501#2293501", "openwebmath_score": 0.8732848167419434, "openwebmath_perplexity": 278.6722104466703, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806529525573, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8668718236197878}}
{"url": "https://cs.stackexchange.com/questions/45783/binary-code-with-constraint", "text": "Binary code with constraint\n\nSuppose I have an alphabet of n symbols. I can efficiently encode them with $\\lceil \\log_2n\\rceil$-bits strings. For instance if n=8:\nA: 0 0 0\nB: 0 0 1\nC: 0 1 0\nD: 0 1 1\nE: 1 0 0\nF: 1 0 1\nG: 1 1 0\nH: 1 1 1\n\nNow I have the additional constraint that each column must contain at most p bits set to 1. For instance for p=2 (and n=8), a possible solution is:\nA: 0 0 0 0 0\nB: 0 0 0 0 1\nC: 0 0 1 0 0\nD: 0 0 1 1 0\nE: 0 1 0 0 0\nF: 0 1 0 1 0\nG: 1 0 0 0 0\nH: 1 0 0 0 1\n\nGiven n and p, does an algorithm exist to find an optimal encoding (shortest length) ? (and can it be proved that it computes an optimal solution?)\n\nEDIT\n\nTwo approaches have been proposed so far to estimate a lower bound on the number of bits $m$. The goal of this section is to provide an analysis and a comparaison of the two answers, in order to explain the choice for the best answer.\n\nYuval's approach is based on entropy and provides a very nice lower bound: $\\frac{logn}{h(p/n)}$ where $h(x) = xlogx + (1-x)log(x)$.\n\nAlex's approach is based on combinatorics. If we develop his reasonning a bit more, it is also possible to compute a very good lower bound:\n\nGiven $m$ the number of bits $\\geq\\lceil log_2(n)\\rceil$, there exists a unique $k$ such that $$1+\\binom{m}{1} + ... +\\binom{m}{k} \\lt n \\leq 1+\\binom{m}{1} + ... + \\binom{m}{k}+\\binom{m}{k+1}$$ One can convince himself that an optimal solution will use the codeword with all bits low, then the codewords with 1 bit high, 2 bits high, ..., k bits high. For the $n-1-\\binom{m}{1}-...-\\binom{m}{k}$ remaining symbols to encode, it is not clear at all which codewords it is optimal to use but, for sure the weights $w_i$ of each column will be bigger than they would be if we could use only codewords with $k+1$ bits high and have $|w_i - w_j| \\leq 1$ for all $i, j$. Therefore one can lower bound $p=max(w_i)$ with $$p_m = 0 + 1 + \\binom{m-1}{2} +... + \\binom{m-1}{k-1} + \\lceil \\frac{(n-1-\\binom{m}{1}-...-\\binom{m}{k}) (k+1)}{m} \\rceil$$\n\nNow, given $n$ and $p$, try to estimate $m$. We know that $p_m \\leq p$ so if $p \\lt p_{m'}$, then $m' \\lt m$. This gives the lower bound for $m$. First compute the $p_m$ then find the biggest $m'$ such that $p \\lt p_{m'}$\n\nThis is what we obtain if we plot, for $n=1000$, the two lower bounds together, the lower bound based on entropy in green, the one based on the combinatorics reasonning above in blue, we get:\n\nBoth look very similar. However if we plot the difference between the two lower bounds, it is clear that the lower bound based on combinatorics reasonning is better overall, especially for small values of $p$.\n\nI believe that the problem comes from the fact that the inequality $H(X) \\leq \\sum H(X_i)$ is weaker when $p$ gets smaller, because the individual coordinates become correlated with small $p$. However this is still a very good lower bound when $p=\\Omega(n)$.\n\nHere is the script (python3) that was used to compute the lower bounds:\n\nfrom scipy.misc import comb\nfrom math import log, ceil, floor\nfrom matplotlib.pyplot import plot, show, legend, xlabel, ylabel\n\n# compute p_m\ndef lowerp(n, m):\nacc = 1\nk = 0\nwhile acc + comb(m, k+1) < n:\nacc+=comb(m, k+1)\nk+=1\n\npm = 0\nfor i in range(k):\npm += comb(m-1, i)\n\nreturn pm + ceil((n-acc)*(k+1)/m)\n\nif __name__ == '__main__':\nn = 100\n\n# compute lower bound based on combinatorics\npm = [lowerp(n, m) for m in range(ceil(log(n)/log(2)), n)]\nmp  = []\np = 1\ni = len(pm) - 1\nwhile i>= 0:\nwhile i>=0 and pm[i] <= p: i-=1\nmp.append(i+ceil(log(n)/log(2)))\np+=1\nplot(range(1, p), mp)\n\n# compute lower bound based on entropy\nlb = [ceil(log(n)/(p/n*log(n/p)+(n-p)/n*log(n/(n-p)))) for p in range(1,p)]\nplot(range(1, p), lb)\n\nxlabel('p')\nylabel('m')\nshow()\n\n# plot diff\nplot(range(1, p), [a-b for a, b in zip(mp, lb)])\nxlabel('p')\nylabel('m')\nshow()\n\n\u2022 @D.W. the constraint is quite as your states. each column must contain at most p bits set to 1. ie. the bit 1's at each position of all selected keys do not exceed p. But I think the first step is still counting the capacity of each bit width. \u2013\u00a0Terence Hang Sep 2 '15 at 16:52\n\u2022 user3017842, I suspect your latest edit should be posted as a self-answer. I think it stands alone as an answer to your question. Do you agree? If so, the right place for it is in the answer box, rather than in the question -- that will make a lot more sense for future readers who come across this (and also allows the community to vote on your answer). I appreciate that you're sharing the analysis you did -- thank you. I encourage you to post that material as an answer, and then remove it from the question. What do you think? Does that seem like it makes sense to you? \u2013\u00a0D.W. Sep 5 '15 at 3:15\n\u2022 @D.W. The EDIT section only makes a comparaison between the two proposed answers in order to explain the choice for the best answer. Therefore I didn't want to put it as a self-answer. But I completely agree that it lacks of clarity for future users, therefore I've clarified the goal of the section and provided links to the corresponding answers. I believe it is a bit more clear now. \u2013\u00a0user3017842 Sep 6 '15 at 9:05\n\nThere is an additional lower bound we can build, that will address cases like what @user3017842 mentioned in their comment on Yuval's answer. (Cases where $p$ is particularly small.) Suppose we knew $m$ already: Then we have $pm$ bits high total across all codewords. Since we're interested the cases where $p$ is small, we view these high bits as our limiting resource, and want to build a code with it (and see how many codewords we can possibly get out). We can have 1 codeword with all 0s, then $m$ codewords with a single 1, then $m \\choose 2$ with two 1s, etc. If we call the highest number of bits in a codeword $k$, then $$pm = 0\\cdot 1 + 1\\cdot m + 2\\cdot {m \\choose 2}+... \\le \\sum_i^k i{m \\choose i}$$ While our number of codewords $n$ is similarly bounded by $$n \\le \\sum_i^k {m \\choose i}$$ If we look at the case where $p \\le m$, then $k \\le 2$ is already implied by the first inequality. ($pm = m^2 = m + 2{m \\choose 2}$). So then the code would consist of the $0$-word, $m$ single-$1$-words, and $(p-1)m/2$ two-$1$-words. Thus $$n \\le 1 + m + (p-1)m/2$$ or inverting $$m \\ge \\frac{2(n-1)}{p+1} .$$ This will yield the tight lower bound of $m\\ge 5$ on the example you provide, but as mentioned before, will probably only be very useful while $p \\approx m$ (or $p \\approx \\sqrt n$).\n\n\u2022 Please see the EDIT section of the main post to see why your answer wins ! \u2013\u00a0user3017842 Sep 4 '15 at 18:06\n\nHere is a lower bound and an asymptotically matching construction, at least for some ranges of the parameters. Denote by $m$ the number of columns, and suppose for simplicity that $p \\leq n/2$.\n\nWe start with a lower bound on $m$. Let $X$ be the encoding of symbol chosen uniformly at random. Let $X_1,\\ldots,X_m$ be the individual coordinates, and let $w_i \\leq p$ be the weight of the $i$th column. Then $$\\log n = H(X) \\leq \\sum_{i=1}^m H(X_i) = \\sum_{i=1}^m h(w_i/n) \\leq m h(p/n).$$ Therefore $$m \\geq \\frac{\\log n}{h(p/n)}.$$ Here $H$ is the entropy of a random variable $H(X) = -\\sum_x \\Pr[X=x] \\log \\Pr[X=x]$ and $h$ is the entropy function $h(x) = -x\\log x-(1-x)\\log(1-x)$. (You can use whatever base for the logarithm you want.)\n\nThe asymptotically matching construction, that should work for $p = \\Omega(n)$, chooses $m$ a bit larger than this lower bound, and chooses a random encoding scheme, each bit being set to $1$ with some probability $q/n$ which is a bit smaller than $p/n$. Choosing the parameters correctly, we should get that this results in a legal encoding (all codewords are different and all column weights are at most $p$) with positive probability.\n\n\u2022 Nice lower bound. Why should the matching construction work for $p=\\Omega(n)$? is there any easy way to believe it other than bounding the probability of getting an invalid encoding when $m$ is picked near the lower bound? \u2013\u00a0Ariel Sep 3 '15 at 15:46\n\u2022 Experience tells me that it has a high chance of working, but you can't know for sure without trying. \u2013\u00a0Yuval Filmus Sep 3 '15 at 16:27\n\u2022 I believe this lower bound is very good when the individual coordinates $X_1, X_2, ..., X_m$ are virtually independent (because inequality $H(X) \\leq \\sum H(X_i)$ will be close to be an equality). This is likely to be the case when $p$ is close enough to $n/2$. However when $p$ remains small, this is no more the case. Consider for example the extreme case when $p=1$. \u2013\u00a0user3017842 Sep 3 '15 at 17:42\n\u2022 When $p=1$ it is clear that the number of bits is $n-1$ (as suggested in Alex Meiburg's answer). However $n-1 - \\frac{logn} {h(p/n))} \\sim n/logn$. The lower bound becomes inaccurate when $p$ remains small while $n$ is getting large. Besides, for small $p$ such as $p=1$, the proposed construction will not work quite well because of the well-known birthday problem. But, still, this is a very nice approach, especially when $p=\\Omega(n)$ ! \u2013\u00a0user3017842 Sep 3 '15 at 18:05\n\u2022 I've made a comparaison with another lower bound deduced from combinatorics reasonning suggested in another answer. It turns out that your lower bound is slightly weaker, especially when $p$ gets smaller. Please see the details of the comparaison in the EDIT section of the main post. Nonetheless, I was very impressed by your solution! Thanks ! \u2013\u00a0user3017842 Sep 4 '15 at 18:05\n\nHere is a simple search methodology. We start from some lower-bound on the number of bits and then try to find a legal encoding. Specifically.\n\nLet m be the current number of bits. Encode symbol i as bi1, bi2, ..., bim.\n\nConstraints: bi xor bj isn't 0 - in other words each symbol's encoding is unique\n\nFor all j: sum_i bij <= p.\n\nThis is a pseudo-boolean satisfiability problem (well it can easily be encoded as a standard satifiability problem). So just keep increasing m until you find one that is satisfiable (or do a binary search using lower and upper bounds to find the minimal m).\n\nOf course, this doesn't guarantee that in practice you'll be able to actually find the minimal m, the SAT check could timeout.", "date": "2021-08-02 23:15:06", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/45783/binary-code-with-constraint", "openwebmath_score": 0.8467495441436768, "openwebmath_perplexity": 381.4229717748003, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.980580656357574, "lm_q2_score": 0.8840392832736083, "lm_q1q2_score": 0.866871820638314}}
{"url": "http://mathhelpforum.com/advanced-algebra/144749-eigenvalues-ab-ba-print.html", "text": "# eigenvalues of AB and of BA\n\n\u2022 May 14th 2010, 04:28 PM\nmath8\neigenvalues of AB and of BA\nIs it true that the eigenvalues of AB are the same as the eigenvalues of BA?\n\u2022 May 14th 2010, 04:57 PM\ndwsmith\nQuote:\n\nOriginally Posted by math8\nIs it true that the eigenvalues of AB are the same as the eigenvalues of BA?\n\nNo, multiplication of matrices isn't commutative.\n\u2022 May 14th 2010, 05:23 PM\nmath8\nI know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.\n\u2022 May 14th 2010, 05:27 PM\ndwsmith\nQuote:\n\nOriginally Posted by math8\nI know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.\n\nSorry for the mistake at first.\n\n$det(AB-\\lambda I)=det(AB-\\lambda AA^{-1})=det(A(B-\\lambda A^{-1}))=det(A)det(B-\\lambda A^{-1})$\n$=det(B-\\lambda A^{-1})det(A)=det((B-\\lambda A^{-1})A)=det(BA-\\lambda I)$\n\nI am pretty sure I have proving it here as well as some other proofs that may help in Linear:\nhttp://www.mathhelpforum.com/math-he...rexamples.html\n\u2022 May 14th 2010, 05:55 PM\nRandom Variable\nIt's easy to prove.\n\nLet $\\lambda_{1}$ be an eigenvalue of $AB$\n\nthen $ABx=\\lambda_{1} x$\n\n$BABx = \\lambda_{1} B x$\n\nwhich means that $\\lambda_{1}$ is an eigenvalue of $BA$ with associated eigenvector $Bx$\n\nNow let $\\lambda_{2}$ be an eigenvalue of $BA$\n\nthen $BAx=\\lambda_{2}x$\n\n$ABAx = \\lambda_{2}Ax$\n\nwhich means that $\\lambda_{2}$ is an eigenvalue of $AB$ with associated eigenvector $Ax$\n\nQED\n\u2022 May 14th 2010, 06:03 PM\nroninpro\nQuote:\n\nOriginally Posted by dwsmith\nSorry for the mistake at first.\n\n$det(AB-\\lambda I)=det(AB-\\lambda AA^{-1})=det(A(B-\\lambda A^{-1}))=det(A)det(B-\\lambda A^{-1})$\n$=det(B-\\lambda A^{-1})det(A)=det((B-\\lambda A^{-1})A)=det(BA-\\lambda I)$\n\nI am pretty sure I have proving it here as well as some other proofs that may help in Linear:\nhttp://www.mathhelpforum.com/math-he...rexamples.html\n\nThere might be a slight hiccup. You don't know whether or not $A$ or $B$ is invertible.\n\u2022 May 15th 2010, 03:38 AM\nHallsofIvy\nFortunately, Random Variables' proof, in addition to being simpler, does not require that A and B be invertible.\n\u2022 May 15th 2010, 04:47 AM\nNonCommAlg\nQuote:\n\nOriginally Posted by Random Variable\nIt's easy to prove.\n\nLet $\\lambda_{1}$ be an eigenvalue of $AB$\n\nthen $ABx=\\lambda_{1} x$\n\n$BABx = \\lambda_{1} B x$\n\nwhich means that $\\lambda_{1}$ is an eigenvalue of $BA$ with associated eigenvector $Bx$\n\n$BABx=\\lambda Bx$ doesn't necessarily imply that $\\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$\n\u2022 May 15th 2010, 07:39 AM\nRandom Variable\nQuote:\n\nOriginally Posted by NonCommAlg\n$BABx=\\lambda Bx$ doesn't necessarily imply that $\\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$\n\nDeal with the case of $A$ or $B$ being zero matrices separately.\n\nIf $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\\lambda = 0$\n\u2022 May 15th 2010, 02:54 PM\nHallsofIvy\nQuote:\n\nOriginally Posted by Random Variable\nDeal with the case of $A$ or $B$ being zero matrices separately.\n\nIf $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\\lambda = 0$\n\nThat's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \\lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0.\n\u2022 May 15th 2010, 03:13 PM\nRandom Variable\nQuote:\n\nOriginally Posted by HallsofIvy\nThat's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \\lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0.\n\nYeah, I forgot about that possibility.(Doh)\n\u2022 May 15th 2010, 03:20 PM\ndwsmith\nI know I have this proof covering all cases, and once I find it, I will post it.\nI am not sure if the red is necessarily correct.\nAB and BA have the same eigenvalues iff. AB is similar to BA.\n\nAB=C\nBA=D\n\nIf C and D are similar matrices, then there exist S such that $C=S^{-1}DS$.\n\n$p_c(\\lambda)=det(AB-\\lambda I)=det(C-\\lambda I)=det(S^{-1}DS-\\lambda I)$\n$=det(S^{-1}DS-\\lambda S^{-1}IS)=det(S^{-1}(D-\\lambda I)S)$\n$=det(S^{-1}S)det(D-\\lambda I)=det(BA-\\lambda I)=p_d(\\lambda)$", "date": "2016-06-30 09:04:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/144749-eigenvalues-ab-ba-print.html", "openwebmath_score": 0.9639359712600708, "openwebmath_perplexity": 586.8895505468747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924818279465, "lm_q2_score": 0.8918110526265555, "lm_q1q2_score": 0.8668336383640791}}
{"url": "http://businessignitiongroup.co.uk/u473ie/ao8zfn.php?tag=43264b-a-matrix-with-only-one-column-is-called", "text": "# a matrix with only one column is called\n\n4. A column matrix is a matrix that has only one column. Example: D is a column matrix of order 2 \u00d7 1 A zero matrix or a null matrix is a matrix that has all its elements zero. A matrix, that has many rows, but only one column, is called a column vector. Column Matrix A matrix having only one column and any number of rows is called column matrix. A wide matrix (a matrix with more columns than rows) has linearly dependent columns. A matrix is said to be a column matrix if it has only one column. a = 1 3 2 5 3 2 4 8 5 9 I want to sort the second column in the a matrix. Example: C is a column matrix of order 1 \u00d7 1 A column matrix of order 2 \u00d71 is also called a vector matrix. 3. The entries of a vector are denoted with just one subscript (since the other is 1), as in a3. INCLUDES THE SOLUTIONS. Each variable in the system becomes a column. A column matrix has only one column but any number of rows. A matrix with only one column, i.e., with size n \u00d7 1, is called a column vector or just a vector. A matrix with only one row is called a.....matrix, and a matrix with only one column is called a.....matrix. The following vector q is a 3 \u00d7 1 column vector containing numbers: $q=\\begin{bmatrix} 2\\\\ 5\\\\ 8\\end{bmatrix}$ A row vector is an 1 \u00d7 c matrix, that is, a matrix with only one row. Suppose that A has more columns than rows. Converting Systems of Linear Equations to Matrices. Sometimes the size is speci\ufb01ed by calling it an n-vector. Then A cannot have a pivot in every column (it has at most one pivot per row), so its columns are automatically linearly dependent. Rectangular Matrix A matrix of order m x n, such that m \u2260 n, is called rectangular matrix. For example, $$A =\\begin{bmatrix} 0\\\\ \u221a3\\\\-1 \\\\1/2 \\end{bmatrix}$$ is a column matrix of order 4 \u00d7 1. Two matrices of the same order whose corresponding entries are equal are considered equal. The variables are dropped and the coefficients are placed into a matrix. One column matrix. A vector is almost often denoted by a single lowercase letter in boldface type. For example, four vectors in R 3 are automatically linearly dependent. A column vector is an r \u00d7 1 matrix, that is, a matrix with only one column. Below, a is a column vector while b is a row vector. A matrix with only one row or one column is called a vector. The determinant takes a square matrix and calculates a simple number, a scalar. I have the matrix as follows. Each equation in the system becomes a row. In general, B = [b ij] m \u00d7 1 is a column matrix of order m \u00d7 1. row column. A column matrix is a matrix with only one column. 5. The entries are sometimes The matrix derived from a system of linear equations is called the..... matrix of the system. 3) Square Matrix. I want the corresponding rows of column one to be printed as follows : a = 3 2 1 3 2 5 4 8 5 9 I tried sort(a), but it is sorting only the second column of matrix a. 2. one column (called a column vector). Horizontal Matrix A matrix in which the number of rows is less than the number of columns, is called a horizontal matrix. A matrix with only one row is called a _____ matrix, and a matrix with only one column is called a _____ matrix. A matrix having only one row is called a row matrix (or a row vector) and a matrix having only one column is called a column matrix (or a column vector). To understand what this number means, take each column of the matrix and draw it as a vector. a = 7 2 3 , b = (\u2212 2 7 4) A scalar is a matrix with only one row and one column. augmented. 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Of order m x n, is called a _____ matrix, and a matrix with only column... \u00d7 1 matrix, that is, a scalar ij ] m 1! Sometimes the size is speci\ufb01ed by calling it an n-vector, as in a3 x n is... Boldface type linear equations is called a..... matrix, and a.. Is, a is a column matrix of order m x n, that! Of rows is less than the number of rows variables are dropped and the coefficients placed! From a system of linear equations is called the..... matrix of order m \u00d7 1 is row. A wide matrix ( a a matrix with only one column is called with only one column the coefficients are placed into a matrix, and matrix. \u2260 n, such that m \u2260 n, such that m \u2260 n, such that \u2260...\n\nPosts created 1\n\n## Hello world!\n\nBegin typing your search term above and press enter to search. Press ESC to cancel.", "date": "2021-04-13 18:45:50", "meta": {"domain": "co.uk", "url": "http://businessignitiongroup.co.uk/u473ie/ao8zfn.php?tag=43264b-a-matrix-with-only-one-column-is-called", "openwebmath_score": 0.8829747438430786, "openwebmath_perplexity": 369.42092043314057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9736446467715053, "lm_q2_score": 0.8902942326713409, "lm_q1q2_score": 0.8668302136919961}}
{"url": "https://math.stackexchange.com/questions/2910304/how-many-ways-to-pair-6-chess-players-over-3-boards-disregarding-seating-arrang", "text": "# How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.\n\nThe problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.\n\nI have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.\n\nI started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.\n\nNow, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\\to 720/6 = 120$.\n\nSo far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.\n\nI finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \\cdot 2 \\cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.\n\n\u2022 Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. \u2013\u00a0N. F. Taussig Sep 9 '18 at 8:09\n\nHere is an alternative approach:\n\nLine up the six players in some order, say alphabetically. The first person in line can choose a partner in $5$ ways. Remove those two players from the line. That leaves four players. The first person left in the line can choose a partner in $3$ ways. Remove those two players from the line. The remaining two players must play each other. Hence, there are $$5!! = 5 \\cdot 3 \\cdot 1 = 15$$ ways to form three pairs of chess partners. The expression $5!!$ is read $5$ double factorial.\n\nIt's correct (the answer and logic). what is a different way to reason about it?\n\nThe idea:\n\n$$(\\textrm{first pair})(\\textrm{second pair})(\\textrm{third pair}),$$\n\nfor the first pair: $6$ ways to choose the first guy, $5$ ways to choose the second, and notice the repetition $p_1p_2=p_2p_1$ so $$\\frac{6\\cdot 5}{2!}$$\n\nfor the second and last pair: $\\displaystyle\\frac{4\\cdot3}{2!};\\frac{2\\cdot1}{2!}.$\n\nNow we get pairs, but $P_1P_2P_3=P_3P_2P_1,$ etc. there are $3!$ ways to repeat$^\\dagger$ three pairs we just made, so $$\\frac{1}{3!}{6\\choose2}{4\\choose2}{2\\choose2}.$$\n\n$\\dagger$: A good question to think about is: what's the condition(s) that repetition will/will not happen?\n\nYes, you are absolutely right.\n\nThe other approach might be selecting $2$ players out of $6$ since arrangement doesn't matter ($_6C_2=15$).\n\n\u2022 Your alternative approach is incorrect. Choosing two of the six players does not determine who plays who in the other two games. \u2013\u00a0N. F. Taussig Sep 9 '18 at 8:06\n\u2022 No it takes care of all the possible combonations \u2013\u00a0Rock Guitar Sep 9 '18 at 20:32", "date": "2019-06-25 12:32:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2910304/how-many-ways-to-pair-6-chess-players-over-3-boards-disregarding-seating-arrang", "openwebmath_score": 0.7412623763084412, "openwebmath_perplexity": 255.8968274963195, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446433301306, "lm_q2_score": 0.8902942312159383, "lm_q1q2_score": 0.8668302092111151}}
{"url": "https://math.stackexchange.com/questions/2851611/cardinality-of-equivalence-classes-in-a-set", "text": "# Cardinality of equivalence classes in a set\n\nGiven $$X = \\{1, 2, 3, 4\\}$$. Define a relation $$\\mathbf{R}$$ over $$P(X)$$ as follows: For two elements $$A,B \u2208 P(X)$$, $$A \\mathrel{\\mathbf{R}} B$$ if $$|A| \u2261 |B| \\pmod{3}$$.\n\n(a) Demonstrate that the previous relation is an equivalence relation.\n\n(b) Determine the cardinality of every equivalence class.\n\nI have found how many classes in there which is the trivial part of the question as it is modulo $$3$$ then we have three classes $$C_0$$, $$C_1$$, $$C_2$$.\n\nBut I am not able to demonstrate that the previous relation is an equivalence relation or the cardinality of the equivalence classes.\n\nSorry for being inefficient even tho the question might be easy.\n\nGuide:\n\nYou have to show the following.\n\n\u2022 Reflexive, $\\forall A \\in P(X), ARA$.\n\u2022 Symmetric, $ARB \\implies BRA$.\n\u2022 Transitive, $ARB \\land BRC \\implies A RC$\n\nIn particular, transitivity is simply $|A|\\equiv |B| \\pmod{3}$ and $|B|\\equiv |C| \\pmod{3}$, then we have $|A| \\equiv |C| \\pmod{3}$.\n\nTry to write out what they mean and you should be able to use the property of modulo arithmetic to see why it is an equivalence relation.\n\nAs for how many members each equivalence class has. Try to answer the following question.\n\n\u2022 How many subsets have exactly $1$ or $4$ elements.\n\u2022 How many subsets have exactly $2$ elements.\n\u2022 How many subsets have exactly $0$ or $3$ elements.\n\u2022 For the transitivity it is clear, but still the second part of how many subsets I can't recognize how many subsets it is the thing that I can't work on I know it is simple but I don't know how to approach it. \u2013\u00a0CptPackage Jul 14 '18 at 14:28\n\u2022 To construct a subset of $i$ element, out of the $4$ elements, you have to choose $i$ elements. Hence there are $4$ choose $i = \\binom{4}{i}=\\frac{4!}{i!(4-i)!}$ such subsets. Using this property, can you determine how many subset has $0$ elements? what about $1$ element and so on? \u2013\u00a0Siong Thye Goh Jul 14 '18 at 14:34\n\u2022 There are $3$ equivalence class, but the question is how many subsets are there in those equivalence classes. For example $C_0$ consists of subsets with either $0$ elements or $3$ elements, $C_0=\\{ \\emptyset, \\{1,2,3\\}, \\{1,2,4\\}, \\{ 1,3,4\\}, \\{2,3,4\\} \\}$. Note that $0 \\equiv 3 \\pmod{3}$. What is the cardinality of $C_0$? \u2013\u00a0Siong Thye Goh Jul 14 '18 at 14:41\n\u2022 Alright, I can only use the binomial coefficient and it will work but I want to ask something. So C0 is the class containing the empty element, and all the possible combinations of 3 elements as 0 mod 3 = 0 and 3 mod 3 = 0 so it will be the same equivalence class, but for C1 it will the combination of the subsets of 1 element each right? Like C1={{1},{2},{3},{4}} but it will have the cardinality of 4 not 5 which is wrong. And C2 will be C2={{1,2},{1,3},{1,4},{2,3},{2,4},{3,4}} right? So can you please explain to me this small part? Sorry, I know I have been exhausting enough. \u2013\u00a0CptPackage Jul 14 '18 at 14:49\n\u2022 $C_1$ is the class containing subsets of $1$ element and $4$ elements since $1 \\equiv 4 \\pmod{3}$, so $|C_1|=5$. You have listed $C_2$ correctly. Without listing it, the computation is just $\\binom{4}{2}=6$. Don't worry about how long you ask the question, just make sure you are learning. ;) \u2013\u00a0Siong Thye Goh Jul 14 '18 at 14:55\n\n(a) Reflexive: $|A|\\equiv|A|$. Symmetric: $|A|\\equiv|B|\\implies|B|\\equiv|A|$. Transitive: if $|A|\\equiv|B|\\equiv|C|$, integers $m,\\,n$ exist with $|A|=|B|+3m=|C|+3m+3n$.\n\n(b) There are $1,\\,4,\\,6,\\,4,\\,1$ subsets of sizes $0,\\,\\cdots,\\,4$. One equivalence class is for cardinalities $0$ and $3$, and is of size $1+4=5$; another covers $1$ and $4$, and again there are $5$ elements; the last is just the size-$2$ case, with the remaining $6$ elements.\n\n\u2022 Alright I can get it with the first answer, but the second one isn't very clear for me can you please simplify it more if possible? \u2013\u00a0CptPackage Jul 14 '18 at 14:25\n\u2022 @CptPackage Which part needs simplifying? Counting subsets by size with a row of Pascal's triangle, or grouping subset sizes modulo $3$? \u2013\u00a0J.G. Jul 14 '18 at 14:50\n\u2022 Grouping subsets, Mr.Siong Thye Goh showed me that we can get the results using the binomial coefficient which I didn't know about so sorry for being an ignorant. But can you please show me the combination of the subsets in the equivalence classes? Like C1={{1},{2},{3},{4}}.. \u2013\u00a0CptPackage Jul 14 '18 at 14:52", "date": "2021-04-22 01:01:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2851611/cardinality-of-equivalence-classes-in-a-set", "openwebmath_score": 0.8485574722290039, "openwebmath_perplexity": 239.24970846142114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446502128796, "lm_q2_score": 0.8902942166619118, "lm_q1q2_score": 0.8668302011683368}}
{"url": "http://gmatclub.com/forum/calculate-faster-reciprocal-percentage-equivalent-165574.html", "text": "Calculate faster - Reciprocal percentage equivalent : GMAT Quantitative Section\nCheck GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack\n\n It is currently 20 Jan 2017, 15:28\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Calculate faster - Reciprocal percentage equivalent\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 27 Dec 2013\nPosts: 35\nConcentration: Finance, General Management\nSchools: ISB '15\nFollowers: 0\n\nKudos [?]: 11 [2] , given: 29\n\nCalculate faster - Reciprocal percentage equivalent\u00a0[#permalink]\n\n### Show Tags\n\n06 Jan 2014, 10:18\n2\nKUDOS\n1\nThis post was\nBOOKMARKED\nHi All,\n\nIf one has to complete the quants section on the GMAT on time, it is imperative that one has to calculate faster. A lot of questions on the GMAT involve division. To divide faster reciprocal percentage equivalent come quite handy.\n\nIn this post I will share a basic and an easy method that may help one in memorizing the reciprocal equivalents.\n\n> Reciprocal of 2 (i.e 1/2) is 50%, that of 4 will be half of 50% i.e 25%. Similarly, reciprocal of 8 will be half of 25% = 12.5% and that of 16 will be 6.25%\n\n> Reciprocal of 3 is 33.33%. Thus reciprocal of 6 will be half of 33.33% i.e 16.66% and that of 12 will be 8.33%\n\n> Reciprocal of 9 is 11.11% and reciprocal of 11 is 9.0909%. Reciprocal of 9 is composed of 11's and reciprocal of 11 is composed of 09's.\nIf any calculation has 9 in the denominator, the decimal part will be only 1111 or 2222 or 3333 or 4444...\nex. 95/9 will be 10.5555\n\n> Reciprocal of 20 is 5% [ you can remember this as 1/5 = 20% so, 1/20 = 5%]\n\n> Reciprocal of 21 is 4.76% and of 19 is 5.26%. Thus we can easily remember reciprocals of 19, 20, 21 as 5.25%, 5 ,4.75% i.e 0.25% more and less than 5%\n\n> Similarly, reciprocal of 25 is 4 % [ Remember this as 1/4 = 25%, so 1/25 = 4%]\n\n> Reciprocal of 24 is 4.16% and of 26 is 3.84%. Thus, we can easily remember reciprocals of 24, 25, 26 as 4.15%, 4, 3.85% i.e 0.15% more and less than 4%.\n\n> Reciprocal of 29 is 3.45% (i.e 345 in order) and reciprocal of 23 is 4.35% ( same digits but order is different. If 1/29 = 3.45% than definitely 1/23 will be more than 3.45%. Reverse the digits and the answer comes to 4.35%)\n\n> Reciprocal of 18 is half of 11.1111% i.e 5.5555% i.e it consists of only 5's.\n\n> Reciprocal of 22 is half of 09.0909%. i.e 4.5454% i.e consists of 45's.\n\n> One can remember 1/8 = 12.5% and tables of 8, and one can easily remember fractions such as 2/8, 3/8, 5/8, 7/8 which are used very regularly.\n\n1/8 is 12.5%, 2/8 is 25% (12.5*2), 3/8 is 37.5% (12.5*3), 5/8 is 62.5% ( 12.5*5), 7/8 = 87.5%\n\nI hope you find this post useful.\n\nThanks,\nHarish\n_________________\n\nKindly consider for kudos if my post was helpful!\n\nCalculate faster - Reciprocal percentage equivalent \u00a0 [#permalink] 06 Jan 2014, 10:18\nSimilar topics Replies Last post\nSimilar\nTopics:\n1 Faster, Efficient way of calculating squares of numbers 0 08 Nov 2015, 07:43\n7 Fractions : Faster calculation 2 01 Nov 2014, 06:23\n17 Excellent Method for Calculating Successive Percentages... 3 03 Oct 2014, 11:52\n1 No Calculator 4 01 Nov 2013, 05:02\n1 reciprocals and negatives 6 24 Nov 2011, 20:51\nDisplay posts from previous: Sort by", "date": "2017-01-20 23:28:05", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/calculate-faster-reciprocal-percentage-equivalent-165574.html", "openwebmath_score": 0.8214187622070312, "openwebmath_perplexity": 3766.9241163425345, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9736446456243805, "lm_q2_score": 0.8902942203004186, "lm_q1q2_score": 0.8668302006258352}}
{"url": "https://math.stackexchange.com/questions/2818474/find-the-prime-factors-of-332-232", "text": "# Find the prime factors of $3^{32}-2^{32}$\n\nI'm having a go at BMO 2006/7 Q1 which states: \"Find four prime numbers less than 100 which are factors of $3^{32}-2^{32}$.\"\n\nMy working is as follows (basically just follows difference of two squares loads of times):\n\n$$3^{32}-2^{32}$$ $$=(3^{16}+2^{16})(3^{16}-2^{16})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{8}-2^{8})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{4}-2^{4})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)(3-2)$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)$$\n\nNow it is simple to get $3$ of the primes here: $$3+2=5$$ $$3^2+2^2=13$$ $$3^4+2^4=97$$\n\nNow I was having some trouble with finding the fourth prime. It's clear that the fourth prime must either be a prime factor of $3^8+2^8$ or $3^{16}+2^{16}$. I started off with $3^8+2^8$:\n\n$$3^8+2^8$$ $$=3^{4^2}+256$$ $$=81^2+256$$ $$=6561+256$$ $$=6817$$\n\nHere is where I am stuck because google tells me that $6817=17*401$ and so the fourth prime is $17$. But this is a non-calculator paper so is there a way of working out this answer without working out $\\frac{6817}{3},\\frac{6817}{7},\\frac{6817}{11}$ etc until one of them is an integer solution?\n\nFinally, before anyone says this is a duplicate, there is a similar question here (prime factors of $3^{32}-2^{32}$) that I didn't know existed until writing this question. However, none of those answers give a way of finding $17$ without using any computational methods. So are they just expecting you to essentially do trial and error throughout all of the primes under $100$ until one works?\n\n\u2022 Well, you have only three primes to actually try ($3$ is obviously not a divisor). Of course, you don't know that beforehand. And you have only to try up to $79$, since $\\sqrt{6817}<83$. \u2013\u00a0StayHomeSaveLives Jun 13 '18 at 16:52\n\u2022 Well, you can notice that $68=4\\cdot17$ which makes it easy. \u2013\u00a0saulspatz Jun 13 '18 at 16:52\n\u2022 Yes I know I would reach the solution of $17$ quite quickly but that isn't really my point. I just wouldn't expect BMO to give something that essentially requires trial and error \u2013\u00a0Dan Jun 13 '18 at 16:53\n\u2022 @saulspatz ah yes that's a very good way of thinking about it. Is that what they would expect from you though or is there some kind of method they would want you to use? \u2013\u00a0Dan Jun 13 '18 at 16:55\n\u2022 Since $17$ is a prime, Fermat's little theorem tell us $2,3 \\not| 17 \\implies 2^{16} \\equiv 3^{16} \\equiv 0 \\pmod 17$. One thing one can try is looking for prime of the form $2^k + 1$ for $k \\le 5$, you immediate get $5$ and $17$. \u2013\u00a0achille hui Jun 13 '18 at 17:02\n\nApply Fermat's little theorem:\n\n$17$ is prime, so\n\n$$3^{16}\\equiv1\\pmod{17}$$ $$2^{16}\\equiv1\\pmod{17}$$\n\nNote that $\\forall a \\perp p, a^{p-1\\over2} \\equiv \\pm 1 \\pmod p$\n\n2 and 3 are already coprime to the other primes you're looking for. $8*2+1=17$ is a strong candidate. In fact, it is the only strong candidate for this method of attack.\n\nI checked this in Python and there are only 4 such primes, $5$, $13$, $17$ and $97$. As has been noted, FLT implies $17$ is such a prime, and we can get $5$ the same way. We also get $13$ as a prime factor of $3^4-2^4$, and $97$ as a prime factor of $3^8-2^8$.\n\nIn the timed conditions of an olympiad, it helps to first seek prime factors of $3^2-2^2$, then the remaining prime factors of $3^4-2^4$ (i.e. those of $3^2+2^2$) etc. In particular $5=3^2-2^2,\\,13=3^2+2^2,\\,97=3^4+2^4,\\,17\\times 401=3^8+2^8$.", "date": "2020-03-29 10:39:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2818474/find-the-prime-factors-of-332-232", "openwebmath_score": 0.8473844528198242, "openwebmath_perplexity": 202.58929841070707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226354423303, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8668189790716343}}
{"url": "http://mathhelpforum.com/geometry/166813-few-math-team-problems-need-help.html", "text": "# Math Help - a few math team problems need help with\n\n1. ## a few math team problems need help with\n\n1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.\n\n2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)\n\n3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.\n\n4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow\n\nYou don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!\n\n-Amer\n\n2. Originally Posted by amerlaw1\n4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow\nBest thing to do is draw a picture to truly understand this one.\n\nUsing similar triangles and equating ratios of side lengths you should get\n\n$\\displaystyle \\frac{x}{6} = \\frac{x+3}{9}$\n\nCan you find $x$ ?\n\n3. Originally Posted by amerlaw1\n1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.\n\n2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)\n\n3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.\n\n4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow\n\nYou don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!\n\n-Amer\nWhat is \"mathteam\"? Are these questions part of a competition?\n\n4. thank you but can u show me how the are similar like what theorem. E.G sas or aa or sss????\n\nyes math team is a competition but these questions came from a competition that happened 14 years ago so you need not to worry about cheating\n\n5. 1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.\n\nLet the side of the triangle be $a$. Then you can find the areas of AOB, BOC and AOC, which together give the area of ABC. On the other hand, the area of ABC is $\\sqrt{3}a^2/4$. This gives you an equation that allows finding $a$.\n\n6. Hello, amerlaw1!\n\n2. What is the distance between the incenter and circumcenter\nof a triangle with sides of the length 12, 16 and 20?\nCode:\n  _ A *\n:   | *\n:   |   *\n:   |     *\n:   |       *\n:   |         *\n:   |           *   20\n:   |       * * * *\n:   |   *           *   E\n12   | *               o\n:   |*          r   *  *\n:   |             *       *\n:   *           *       *   *\n: F o - - - - o         *     *\n:   *    r    |O        *       *\n:   |         |                   *\n:  r|*        |r       *            *\n:   | *       |       *               *\n:   |   *     |     *                   *\n- C * - - - * o * - - - - - - - - - - - - * B\nr     D\n: - - - - - - - - 16  - - - - - - - - :\n\nWe have right triangle $ABC:\\:AC = 12,\\;BC = 16,\\;AB = 20.$\nPlace the triangle on a coordinate system with $\\,C$ at the Origin.\n\nThe incenter is $O(r,r).$\n. . $OD = OE = OF = CD = CF = r.$\n\nThe circumcenter is the midpoint of $AB:\\;P(8,6).$\n. .\n(Not shown on the diagram.)\n\nNote that: . $AF \\,=\\, 12-r$\nSince $AE$ is also tangent to the circle: $AE \\,=\\, 12-r$\n\nNote that: . $BD \\,=\\,16-r.$\nSince $BE$ is also tangent to the circle: $BE \\,=\\,16-r$\n\nSince $AB = 20$, we have: . $(12-r) + (16-r) \\:=\\:20 \\quad\\Rightarrow\\quad r \\,=\\,4$\n\nWe have: . $\\begin{Bmatrix}\\text{Incenter:} & O(4,4) \\\\ \\text{Circumcenter:} & P(8,6) \\end{Bmatrix}$\n\nTherefore: . $\\iverline{OP} \\;=\\;\\sqrt{(8-4)^2 + (6-5)^2} \\;=\\;\\sqrt{16+4} \\;=\\;\\sqrt{20} \\;=\\;2\\sqrt{5}$", "date": "2014-09-15 10:06:54", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/166813-few-math-team-problems-need-help.html", "openwebmath_score": 0.6088656187057495, "openwebmath_perplexity": 648.9838699370054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022625406662, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.866818967254075}}
{"url": "https://gmatclub.com/forum/math-number-theory-broken-into-smaller-topics-91274.html", "text": "It is currently 19 Jan 2018, 21:22\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Math: Number Theory (broken into smaller topics)\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43335\n\nKudos [?]: 139526 [6], given: 12794\n\nMath: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n10 Mar 2010, 05:20\n6\nKUDOS\nExpert's post\n6\nThis post was\nBOOKMARKED\nNUMBER THEORY\n\nThis post is a part of [GMAT MATH BOOK]\n\ncreated by: Bunuel\nedited by: bb, walker\n\n--------------------------------------------------------\n\nThe information will be included in future versions of GMAT ToolKit\n\n--------------------------------------------------------\n\nDEFINITION\n\nNumber Theory is concerned with the properties of numbers in general, and in particular integers.\nAs this is a huge issue we decided to divide it into smaller topics. Below is the list of Number Theory topics.\n\nGMAT Number Types\n\nGMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.\n\nINTEGERS\n\nIntegers are defined as: all negative natural numbers $$\\{...,-4, -3, -2, -1\\}$$, zero $$\\{0\\}$$, and positive natural numbers $$\\{1, 2, 3, 4, ...\\}$$.\n\nNote that integers do not include decimals or fractions - just whole numbers.\n\nEven and Odd Numbers\nPrime Numbers\nFactors\nFinding the Number of Factors of an Integer\nFinding the Sum of the Factors of an Integer\nGreatest Common Factor (Divisior) - GCF (GCD)\nLowest Common Multiple - LCM\nDivisibility Rules\nFactorials\nConsecutive Integers\nEvenly Spaced Set\n\nIRRATIONAL NUMBERS\n\nFractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals (such as 0.5, 0.76, or 0.333333....). On the other hand, all those numbers that can be written as non-terminating, non-repeating decimals are non-rational, so they are called the \"irrationals\". Examples would be $$\\sqrt{2}$$ (\"the square root of two\") or the number pi ($$\\pi=$$~3.14159..., from geometry). The rationals and the irrationals are two totally separate number types: there is no overlap.\n\nPutting these two major classifications, the rationals and the irrationals, together in one set gives you the \"real\" numbers.\n\nFRACTIONS\n\nFractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.\n\nFraction can be expressed in two forms fractional representation $$(\\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.\n\nDefinition\nFractional representation\nConverting Improper Fractions\nReciprocal\nOperation on Fractions\nDecimal Representation\nConverting Decimals to Fractions\nRounding\nRatios and Proportions\n\nPOSITIVE AND NEGATIVE NUMBERS\n\nA positive number is a real number that is greater than zero.\nA negative number is a real number that is smaller than zero.\n\nZero is not positive, nor negative.\n\nMultiplication:\npositive * positive = positive\npositive * negative = negative\nnegative * negative = positive\n\nDivision:\npositive / positive = positive\npositive / negative = negative\nnegative / negative = positive\n\nEXPONENTS, ROOTS, PERCENTS\n\nExponents are a \"shortcut\" method of showing a number that was multiplied by itself several times. Roots (or radicals) are the \"opposite\" operation of applying exponents. A percentage is a way of expressing a number as a fraction of 100 (per cent meaning \"per hundred\").\n\nExponents\nPerfect Square\nRoots\nLast Digit of a Product\nLast Digit of a Power\nPercents\n\nORDER OF OPERATIONS - PEMDAS\n\nPerform the operations inside a Parenthesis first (absolute value signs also fall into this category), then Exponents, then Multiplication and Division, from left to right, then Addition and Subtraction, from left to right - PEMDAS.\n\nSpecial cases:\n\u2022 An exclamation mark indicates that one should compute the factorial of the term immediately to its left, before computing any of the lower-precedence operations, unless grouping symbols dictate otherwise. But $$3^2!$$ means $$(3^2)! = 9!$$ while $$2^{5!} = 2^{120}$$; a factorial in an exponent applies to the exponent, while a factorial not in the exponent applies to the entire power.\n\n\u2022 If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:\n$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.\n_________________\n\nKudos [?]: 139526 [6], given: 12794\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43335\n\nKudos [?]: 139526 [1], given: 12794\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n10 Mar 2010, 05:21\n1\nKUDOS\nExpert's post\nINTEGERS\n\nThis post is a part of [GMAT MATH BOOK]\n\ncreated by: Bunuel\nedited by: bb, walker\n\n--------------------------------------------------------\n\nThe information will be included in future versions of GMAT ToolKit\n\n--------------------------------------------------------\n\nDefinition\n\nIntegers are defined as: all negative natural numbers $$\\{...,-4, -3, -2, -1\\}$$, zero $$\\{0\\}$$, and positive natural numbers $$\\{1, 2, 3, 4, ...\\}$$.\n\nNote that integers do not include decimals or fractions - just whole numbers.\n\nEven and Odd Numbers\n\nAn even number is an integer that is \"evenly divisible\" by 2, i.e., divisible by 2 without a remainder.\nAn even number is an integer of the form $$n=2k$$, where $$k$$ is an integer.\n\nAn odd number is an integer that is not evenly divisible by 2.\nAn odd number is an integer of the form $$n=2k+1$$, where $$k$$ is an integer.\n\nZero is an even number.\n\neven +/- even = even;\neven +/- odd = odd;\nodd +/- odd = even.\n\nMultiplication:\neven * even = even;\neven * odd = even;\nodd * odd = odd.\n\nDivision of two integers can result into an even/odd integer or a fraction.\n\nPrime Numbers\n\nA Prime number is a natural number with exactly two distinct natural number divisors: 1 and itself. Otherwise a number is called a composite number. Therefore, 1 is not a prime, since it only has one divisor, namely 1. A number $$n > 1$$ is prime if it cannot be written as a product of two factors $$a$$ and $$b$$, both of which are greater than 1: n = ab.\n\n\u2022 The first twenty-six prime numbers are:\n2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101\n\n\u2022 Note: only positive numbers can be primes.\n\n\u2022 There are infinitely many prime numbers.\n\n\u2022 The only even prime number is 2, since any larger even number is divisible by 2. Also 2 is the smallest prime.\n\nAll prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form $$6n-1$$ or $$6n+1$$, because all other numbers are divisible by 2 or 3.\n\n\u2022 Any nonzero natural number $$n$$ can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors.\n\nPrime factorization: every positive integer greater than 1 can be written as a product of one or more prime integers in a way which is unique. For instance integer $$n$$ with three unique prime factors $$a$$, $$b$$, and $$c$$ can be expressed as $$n=a^p*b^q*c^r$$, where $$p$$, $$q$$, and $$r$$ are powers of $$a$$, $$b$$, and $$c$$, respectively and are $$\\geq1$$.\nExample: $$4200=2^3*3*5^2*7$$.\n\nVerifying the primality (checking whether the number is a prime) of a given number $$n$$ can be done by trial division, that is to say dividing $$n$$ by all integer numbers smaller than $$\\sqrt{n}$$, thereby checking whether $$n$$ is a multiple of $$m<\\sqrt{n}$$.\nExample: Verifying the primality of $$161$$: $$\\sqrt{161}$$ is little less than $$13$$, from integers from $$2$$ to $$13$$, $$161$$ is divisible by $$7$$, hence $$161$$ is not prime.\n\n\u2022 If $$n$$ is a positive integer greater than 1, then there is always a prime number $$p$$ with$$n < p < 2n$$.\n\nFactors\n\nA divisor of an integer $$n$$, also called a factor of $$n$$, is an integer which evenly divides $$n$$ without leaving a remainder. In general, it is said $$m$$ is a factor of $$n$$, for non-zero integers $$m$$ and $$n$$, if there exists an integer $$k$$ such that $$n = km$$.\n\n\u2022 1 (and -1) are divisors of every integer.\n\n\u2022 Every integer is a divisor of itself.\n\n\u2022 Every integer is a divisor of 0, except, by convention, 0 itself.\n\n\u2022 Numbers divisible by 2 are called even and numbers not divisible by 2 are called odd.\n\n\u2022 A positive divisor of n which is different from n is called a proper divisor.\n\n\u2022 An integer n > 1 whose only proper divisor is 1 is called a prime number. Equivalently, one would say that a prime number is one which has exactly two factors: 1 and itself.\n\n\u2022 Any positive divisor of n is a product of prime divisors of n raised to some power.\n\n\u2022 If a number equals the sum of its proper divisors, it is said to be a perfect number.\nExample: The proper divisors of 6 are 1, 2, and 3: 1+2+3=6, hence 6 is a perfect number.\n\nThere are some elementary rules:\n\u2022 If $$a$$ is a factor of $$b$$ and $$a$$ is a factor of $$c$$, then $$a$$ is a factor of $$(b + c)$$. In fact, $$a$$ is a factor of $$(mb + nc)$$ for all integers $$m$$ and $$n$$.\n\n\u2022 If $$a$$ is a factor of $$b$$ and $$b$$ is a factor of $$c$$, then $$a$$ is a factor of $$c$$.\n\n\u2022 If $$a$$ is a factor of $$b$$ and $$b$$ is a factor of $$a$$, then $$a = b$$ or $$a=-b$$.\n\n\u2022 If $$a$$ is a factor of $$bc$$, and $$gcd(a,b)=1$$, then a is a factor of $$c$$.\n\n\u2022 If $$p$$ is a prime number and $$p$$ is a factor of $$ab$$ then $$p$$ is a factor of $$a$$ or $$p$$ is a factor of $$b$$.\n\nFinding the Number of Factors of an Integer\n\nFirst make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.\n\nThe number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.\n\nExample: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$\n\nTotal number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.\n\nFinding the Sum of the Factors of an Integer\n\nFirst make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.\n\nThe sum of factors of $$n$$ will be expressed by the formula: $$\\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}$$\n\nExample: Finding the sum of all factors of 450: $$450=2^1*3^2*5^2$$\n\nThe sum of all factors of 450 is $$\\frac{(2^{1+1}-1)*(3^{2+1}-1)*(5^{2+1}-1)}{(2-1)*(3-1)*(5-1)}=\\frac{3*26*124}{1*2*4}=1209$$\n\nGreatest Common Factor (Divisior) - GCF (GCD)\n\nThe greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.\n\nTo find the GCF, you will need to do prime-factorization. Then, multiply the common factors (pick the lowest power of the common factors).\n\n\u2022 Every common divisor of a and b is a divisor of gcd(a, b).\n\u2022 a*b=gcd(a, b)*lcm(a, b)\n\nLowest Common Multiple - LCM\n\nThe lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple both of a and of b. Since it is a multiple, it can be divided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then lcm(a, b) is defined to be zero.\n\nTo find the LCM, you will need to do prime-factorization. Then multiply all the factors (pick the highest power of the common factors).\n\nDivisibility Rules\n\n2 - If the last digit is even, the number is divisible by 2.\n\n3 - If the sum of the digits is divisible by 3, the number is also.\n\n4 - If the last two digits form a number divisible by 4, the number is also.\n\n5 - If the last digit is a 5 or a 0, the number is divisible by 5.\n\n6 - If the number is divisible by both 3 and 2, it is also divisible by 6.\n\n7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.\n\n8 - If the last three digits of a number are divisible by 8, then so is the whole number.\n\n9 - If the sum of the digits is divisible by 9, so is the number.\n\n10 - If the number ends in 0, it is divisible by 10.\n\n11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11.\nExample: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.\n\n12 - If the number is divisible by both 3 and 4, it is also divisible by 12.\n\n25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.\n\nFactorials\n\nFactorial of a positive integer $$n$$, denoted by $$n!$$, is the product of all positive integers less than or equal to n. For instance $$5!=1*2*3*4*5$$.\n\nTrailing zeros:\nTrailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.\n\n125000 has 3 trailing zeros;\n\nThe number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer $$n$$, can be determined with this formula:\n\n$$\\frac{n}{5}+\\frac{n}{5^2}+\\frac{n}{5^3}+...+\\frac{n}{5^k}$$, where k must be chosen such that $$5^k<n$$.\n\nIt's easier if you look at an example:\n\nHow many zeros are in the end (after which no other digits follow) of $$32!$$?\n$$\\frac{32}{5}+\\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)\n\nHence, there are 7 zeros in the end of 32!\n\nThe formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.\n\nFinding the number of powers of a prime number $$p$$, in the $$n!$$.\n\nThe formula is:\n$$\\frac{n}{p}+\\frac{n}{p^2}+\\frac{n}{p^3}$$ ... till $$p^x<n$$\n\nWhat is the power of 2 in 25!?\n$$\\frac{25}{2}+\\frac{25}{4}+\\frac{25}{8}+\\frac{25}{16}=12+6+3+1=22$$\n\nFinding the power of non-prime in n!:\n\nHow many powers of 900 are in 50!\n\nMake the prime factorization of the number: $$900=2^2*3^2*5^2$$, then find the powers of these prime numbers in the n!.\n\nFind the power of 2:\n$$\\frac{50}{2}+\\frac{50}{4}+\\frac{50}{8}+\\frac{50}{16}+\\frac{50}{32}=25+12+6+3+1=47$$\n\n= $$2^{47}$$\n\nFind the power of 3:\n$$\\frac{50}{3}+\\frac{50}{9}+\\frac{50}{27}=16+5+1=22$$\n\n=$$3^{22}$$\n\nFind the power of 5:\n$$\\frac{50}{5}+\\frac{50}{25}=10+2=12$$\n\n=$$5^{12}$$\n\nWe need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.\n\nConsecutive Integers\n\nConsecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1, 0, 1, are consecutive integers.\n\n\u2022 Sum of $$n$$ consecutive integers equals the mean multiplied by the number of terms, $$n$$. Given consecutive integers $$\\{-3, -2, -1, 0, 1,2\\}$$, $$mean=\\frac{-3+2}{2}=-\\frac{1}{2}$$, (mean equals to the average of the first and last terms), so the sum equals to $$-\\frac{1}{2}*6=-3$$.\n\n\u2022 If n is odd, the sum of consecutive integers is always divisible by n. Given $$\\{9,10,11\\}$$, we have $$n=3$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.\n\n\u2022 If n is even, the sum of consecutive integers is never divisible by n. Given $$\\{9,10,11,12\\}$$, we have $$n=4$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.\n\n\u2022 The product of $$n$$ consecutive integers is always divisible by $$n!$$.\nGiven $$n=4$$ consecutive integers: $$\\{3,4,5,6\\}$$. The product of 3*4*5*6 is 360, which is divisible by 4!=24.\n\nEvenly Spaced Set\n\nEvenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers $$\\{9,13,17,21\\}$$ is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set.\n\n\u2022 If the first term is $$a_1$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by:\n$$a_ n=a_1+d(n-1)$$\n\n\u2022 In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula $$mean=median=\\frac{a_1+a_n}{2}$$, where $$a_1$$ is the first term and $$a_n$$ is the last term. Given the set $$\\{7,11,15,19\\}$$, $$mean=median=\\frac{7+19}{2}=13$$.\n\n\u2022 The sum of the elements in any evenly spaced set is given by:\n$$Sum=\\frac{a_1+a_n}{2}*n$$, the mean multiplied by the number of terms. OR, $$Sum=\\frac{2a_1+d(n-1)}{2}*n$$\n\n\u2022 Special cases:\nSum of n first integers: $$1+2+...+n=\\frac{1+n}{2}*n$$\n\nSum of n first odd numbers: $$a_1+a_2+...+a_n=1+3+...+a_n=n^2$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n-1$$. Given $$n=5$$ first odd integers, then their sum equals to $$1+3+5+7+9=5^2=25$$.\n\nSum of n first positive even numbers: $$a_1+a_2+...+a_n=2+4+...+a_n=n(n+1)$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n$$. Given $$n=4$$ first positive even integers, then their sum equals to $$2+4+6+8=4(4+1)=20$$.\n\n\u2022 If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65.\n_________________\n\nKudos [?]: 139526 [1], given: 12794\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43335\n\nKudos [?]: 139526 [0], given: 12794\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n10 Mar 2010, 05:22\nFRACTIONS\n\nThis post is a part of [GMAT MATH BOOK]\n\ncreated by: Bunuel\nedited by: bb, walker\n\n--------------------------------------------------------\n\nThe information will be included in future versions of GMAT ToolKit\n\n--------------------------------------------------------\n\nDefinition\n\nFractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.\n\nFraction can be expressed in two forms fractional representation $$(\\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.\n\nFractional representation\n\nFractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole).\n\n\u2022 The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, $$\\frac{9}{7}$$, 9 is the numerator and 7 is denominator.\n\n\u2022 Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. $$\\frac{1}{3}$$ is a proper fraction.\n\n\u2022 Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. $$\\frac{5}{2}$$ is improper fraction.\n\n\u2022 An integer combined with a proper fraction is called mixed number. $$4\\frac{3}{5}$$ is a mixed number. This can also be written as an improper fraction: $$\\frac{23}{5}$$\n\nConverting Improper Fractions\n\n\u2022 Converting Improper Fractions to Mixed Fractions:\n1. Divide the numerator by the denominator\n2. Write down the whole number answer\n3. Then write down any remainder above the denominator\nExample #1: Convert $$\\frac{11}{4}$$ to a mixed fraction.\nSolution: Divide $$\\frac{11}{4} = 2$$ with a remainder of $$3$$. Write down the $$2$$ and then write down the remainder $$3$$ above the denominator $$4$$, like this: $$2\\frac{3}{4}$$\n\n\u2022 Converting Mixed Fractions to Improper Fractions:\n1. Multiply the whole number part by the fraction's denominator\n2. Add that to the numerator\n3. Then write the result on top of the denominator\nExample #2: Convert $$3\\frac{2}{5}$$ to an improper fraction.\nSolution: Multiply the whole number by the denominator: $$3*5=15$$. Add the numerator to that: $$15 + 2 = 17$$. Then write that down above the denominator, like this: $$\\frac{17}{5}$$\n\nReciprocal\n\nReciprocal for a number $$x$$, denoted by $$\\frac{1}{x}$$ or $$x^{-1}$$, is a number which when multiplied by $$x$$ yields $$1$$. The reciprocal of a fraction $$\\frac{a}{b}$$ is $$\\frac{b}{a}$$. To get the reciprocal of a number, divide 1 by the number. For example reciprocal of $$3$$ is $$\\frac{1}{3}$$, reciprocal of $$\\frac{5}{6}$$ is $$\\frac{6}{5}$$.\n\nOperation on Fractions\n\nTo add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.\n\nTo add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions\n\nMultiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.\n\nDividing fractions: Change the divisor into its reciprocal and then multiply.\n\nExample #1: $$\\frac{3}{7}+\\frac{2}{3}=\\frac{9}{21}+\\frac{14}{21}=\\frac{23}{21}$$\n\nExample #2: Given $$\\frac{\\frac{3}{5}}{2}$$, take the reciprocal of $$2$$. The reciprocal is $$\\frac{1}{2}$$. Now multiply: $$\\frac{3}{5}*\\frac{1}{2}=\\frac{3}{10}$$.\n\nDecimal Representation\n\nThe decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....).\n\nReduced fraction $$\\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\\frac{3}{30}$$ is also a terminating decimal, as $$\\frac{3}{30}=\\frac{1}{10}$$ and denominator $$10=2*5$$.\n\nConverting Decimals to Fractions\n\n\u2022 To convert a terminating decimal to fraction:\n1. Calculate the total numbers after decimal point\n2. Remove the decimal point from the number\n3. Put 1 under the denominator and annex it with \"0\" as many as the total in step 1\n4. Reduce the fraction to its lowest terms\n\nExample: Convert $$0.56$$ to a fraction.\n1: Total number after decimal point is 2.\n2 and 3: $$\\frac{56}{100}$$.\n4: Reducing it to lowest terms: $$\\frac{56}{100}=\\frac{14}{25}$$\n\n\u2022 To convert a recurring decimal to fraction:\n1. Separate the recurring number from the decimal fraction\n2. Annex denominator with \"9\" as many times as the length of the recurring number\n3. Reduce the fraction to its lowest terms\n\nExample #1: Convert $$0.393939...$$ to a fraction.\n1: The recurring number is $$39$$.\n2: $$\\frac{39}{99}$$, the number $$39$$ is of length $$2$$ so we have added two nines.\n3: Reducing it to lowest terms: $$\\frac{39}{99}=\\frac{13}{33}$$.\n\n\u2022 To convert a mixed-recurring decimal to fraction:\n1. Write down the number consisting with non-repeating digits and repeating digits.\n2. Subtract non-repeating number from above.\n3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's.\n\nExample #2: Convert $$0.2512(12)$$ to a fraction.\n1. The number consisting with non-repeating digits and repeating digits is 2512;\n2. Subtract 25 (non-repeating number) from above: 2512-25=2487;\n3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.\n\nRounding\n\nRounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.\n\nExample:\n5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.\n5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.\n5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.\n\nRatios and Proportions\n\nGiven that $$\\frac{a}{b}=\\frac{c}{d}$$, where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.\n\n$$\\frac{b}{a}=\\frac{d}{c}$$ - invertendo\n\n$$\\frac{a}{c}=\\frac{b}{d}$$ - alternendo\n\n$$\\frac{a+b}{b}=\\frac{c+d}{d}$$ - componendo\n\n$$\\frac{a-b}{b}=\\frac{c-d}{d}$$ - dividendo\n\n$$\\frac{a+b}{a-b}=\\frac{c+d}{c-d}$$ - componendo & dividendo\n_________________\n\nKudos [?]: 139526 [0], given: 12794\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43335\n\nKudos [?]: 139526 [2], given: 12794\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n10 Mar 2010, 05:22\n2\nKUDOS\nExpert's post\nEXPONENTS & ROOTS\n\nThis post is a part of [GMAT MATH BOOK]\n\ncreated by: Bunuel\nedited by: bb, walker\n\n--------------------------------------------------------\n\nThe information will be included in future versions of GMAT ToolKit\n\n--------------------------------------------------------\n\nEXPONENTS\n\nExponents are a \"shortcut\" method of showing a number that was multiplied by itself several times. For instance, number $$a$$ multiplied $$n$$ times can be written as $$a^n$$, where $$a$$ represents the base, the number that is multiplied by itself $$n$$ times and $$n$$ represents the exponent. The exponent indicates how many times to multiple the base, $$a$$, by itself.\n\nExponents one and zero:\n$$a^0=1$$ Any nonzero number to the power of 0 is 1.\nFor example: $$5^0=1$$ and $$(-3)^0=1$$\n\n$$a^1=a$$ Any number to the power 1 is itself.\n\nPowers of zero:\nIf the exponent is positive, the power of zero is zero: $$0^n = 0$$, where $$n > 0$$.\n\nIf the exponent is negative, the power of zero ($$0^n$$, where $$n < 0$$) is undefined, because division by zero is implied.\n\nPowers of one:\n$$1^n=1$$ The integer powers of one are one.\n\nNegative powers:\n$$a^{-n}=\\frac{1}{a^n}$$\n\nPowers of minus one:\nIf n is an even integer, then $$(-1)^n=1$$.\n\nIf n is an odd integer, then $$(-1)^n =-1$$.\n\nOperations involving the same exponents:\nKeep the exponent, multiply or divide the bases\n$$a^n*b^n=(ab)^n$$\n\n$$\\frac{a^n}{b^n}=(\\frac{a}{b})^n$$\n\n$$(a^m)^n=a^{mn}$$\n\n$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$\n\nOperations involving the same bases:\nKeep the base, add or subtract the exponent (add for multiplication, subtract for division)\n$$a^n*a^m=a^{n+m}$$\n\n$$\\frac{a^n}{a^m}=a^{n-m}$$\n\nFraction as power:\n$$a^{\\frac{1}{n}}=\\sqrt[n]{a}$$\n\n$$a^{\\frac{m}{n}}=\\sqrt[n]{a^m}$$\n\nExponential Equations:\nWhen solving equations with even exponents, we must consider both positive and negative possibilities for the solutions.\n\nFor instance $$a^2=25$$, the two possible solutions are $$5$$ and $$-5$$.\n\nWhen solving equations with odd exponents, we'll have only one solution.\n\nFor instance for $$a^3=8$$, solution is $$a=2$$ and for $$a^3=-8$$, solution is $$a=-2$$.\n\nExponents and divisibility:\n$$a^n-b^n$$ is ALWAYS divisible by $$a-b$$.\n$$a^n-b^n$$ is divisible by $$a+b$$ if $$n$$ is even.\n\n$$a^n + b^n$$ is divisible by $$a+b$$ if $$n$$ is odd, and not divisible by a+b if n is even.\n\nPerfect Square\n\nA perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square.\n\nThere are some tips about the perfect square:\n\u2022 The number of distinct factors of a perfect square is ALWAYS ODD.\n\u2022 The sum of distinct factors of a perfect square is ALWAYS ODD.\n\u2022 A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.\n\u2022 Perfect square always has even number of powers of prime factors.\n\nLAST DIGIT OF A PRODUCT\n\nLast $$n$$ digits of a product of integers are last $$n$$ digits of the product of last $$n$$ digits of these integers.\n\nFor instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60\n\nExample: The last digit of 85945*89*58307=5*9*7=45*7=35=5?\n\nLAST DIGIT OF A POWER\n\nDetermining the last digit of $$(xyz)^n$$:\n\n1. Last digit of $$(xyz)^n$$ is the same as that of $$z^n$$;\n2. Determine the cyclicity number $$c$$ of $$z$$;\n3. Find the remainder $$r$$ when $$n$$ divided by the cyclisity;\n4. When $$r>0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^r$$ and when $$r=0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^c$$, where $$c$$ is the cyclisity number.\n\n\u2022 Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.\n\u2022 Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.\n\u2022 Integers ending with 4 (eg. $$(xy4)^n$$) have a cyclisity of 2. When n is odd $$(xy4)^n$$ will end with 4 and when n is even $$(xy4)^n$$ will end with 6.\n\u2022 Integers ending with 9 (eg. $$(xy9)^n$$) have a cyclisity of 2. When n is odd $$(xy9)^n$$ will end with 9 and when n is even $$(xy9)^n$$ will end with 1.\n\nExample: What is the last digit of $$127^{39}$$?\nSolution: Last digit of $$127^{39}$$ is the same as that of $$7^{39}$$. Now we should determine the cyclisity of $$7$$:\n\n1. 7^1=7 (last digit is 7)\n2. 7^2=9 (last digit is 9)\n3. 7^3=3 (last digit is 3)\n4. 7^4=1 (last digit is 1)\n5. 7^5=7 (last digit is 7 again!)\n...\n\nSo, the cyclisity of 7 is 4.\n\nNow divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of $$127^{39}$$ is the same as that of the last digit of $$7^{39}$$, is the same as that of the last digit of $$7^3$$, which is $$3$$.\n\nROOTS\n\nRoots (or radicals) are the \"opposite\" operation of applying exponents. For instance x^2=16 and square root of 16=4.\n\nGeneral rules:\n\u2022 $$\\sqrt{x}\\sqrt{y}=\\sqrt{xy}$$ and $$\\frac{\\sqrt{x}}{\\sqrt{y}}=\\sqrt{\\frac{x}{y}}$$.\n\n\u2022 $$(\\sqrt{x})^n=\\sqrt{x^n}$$\n\n\u2022 $$x^{\\frac{1}{n}}=\\sqrt[n]{x}$$\n\n\u2022 $$x^{\\frac{n}{m}}=\\sqrt[m]{x^n}$$\n\n\u2022 $${\\sqrt{a}}+{\\sqrt{b}}\\neq{\\sqrt{a+b}}$$\n\n\u2022 $$\\sqrt{x^2}=|x|$$, when $$x\\leq{0}$$, then $$\\sqrt{x^2}=-x$$ and when $$x\\geq{0}$$, then $$\\sqrt{x^2}=x$$\n\n\u2022 When the GMAT provides the square root sign for an even root, such as $$\\sqrt{x}$$ or $$\\sqrt[4]{x}$$, then the only accepted answer is the positive root.\n\nThat is, $$\\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.\n\n\u2022 Odd roots will have the same sign as the base of the root. For example, $$\\sqrt[3]{125} =5$$ and $$\\sqrt[3]{-64} =-4$$.\n\n\u2022 For GMAT it's good to memorize following values:\n$$\\sqrt{2}\\approx{1.41}$$\n$$\\sqrt{3}\\approx{1.73}$$\n$$\\sqrt{5}\\approx{2.24}$$\n$$\\sqrt{6}\\approx{2.45}$$\n$$\\sqrt{7}\\approx{2.65}$$\n$$\\sqrt{8}\\approx{2.83}$$\n$$\\sqrt{10}\\approx{3.16}$$\n_________________\n\nKudos [?]: 139526 [2], given: 12794\n\nManager\nJoined: 29 Oct 2009\nPosts: 194\n\nKudos [?]: 113 [0], given: 12\n\nConcentration: General Management, Sustainability\nWE: Consulting (Computer Software)\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n10 Mar 2010, 06:48\nThis is a lucky day for me . Thank you. +1Kudos.\n\nKudos [?]: 113 [0], given: 12\n\nManager\nJoined: 20 Nov 2009\nPosts: 162\n\nKudos [?]: 264 [0], given: 64\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n02 Apr 2010, 01:42\nGreat resource. Thanks people!\n_________________\n\nBut there\u2019s something in me that just keeps going on. I think it has something to do with tomorrow, that there is always one, and that everything can change when it comes.\nhttp://aimingformba.blogspot.com\n\nKudos [?]: 264 [0], given: 64\n\nIntern\nJoined: 29 Jun 2011\nPosts: 33\n\nKudos [?]: 11 [0], given: 0\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n20 Jul 2011, 06:52\ngood stuff. thank you!\n\nKudos [?]: 11 [0], given: 0\n\nManager\nJoined: 23 Aug 2011\nPosts: 78\n\nKudos [?]: 292 [0], given: 13\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n04 Sep 2012, 20:11\nBunuel wrote:\nFRACTIONS\n\nThis post is a part of [GMAT MATH BOOK]\n\ncreated by: Bunuel\nedited by: bb, walker\n\n--------------------------------------------------------\n\nThe information will be included in future versions of GMAT ToolKit\n\n--------------------------------------------------------\n\nDefinition\n\nFractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.\n\nFraction can be expressed in two forms fractional representation $$(\\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.\n\nFractional representation\n\nFractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole).\n\n\u2022 The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, $$\\frac{9}{7}$$, 9 is the numerator and 7 is denominator.\n\n\u2022 Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. $$\\frac{1}{3}$$ is a proper fraction.\n\n\u2022 Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. $$\\frac{5}{2}$$ is improper fraction.\n\n\u2022 An integer combined with a proper fraction is called mixed number. $$4\\frac{3}{5}$$ is a mixed number. This can also be written as an improper fraction: $$\\frac{23}{5}$$\n\nConverting Improper Fractions\n\n\u2022 Converting Improper Fractions to Mixed Fractions:\n1. Divide the numerator by the denominator\n2. Write down the whole number answer\n3. Then write down any remainder above the denominator\nExample #1: Convert $$\\frac{11}{4}$$ to a mixed fraction.\nSolution: Divide $$\\frac{11}{4} = 2$$ with a remainder of $$3$$. Write down the $$2$$ and then write down the remainder $$3$$ above the denominator $$4$$, like this: $$2\\frac{3}{4}$$\n\n\u2022 Converting Mixed Fractions to Improper Fractions:\n1. Multiply the whole number part by the fraction's denominator\n2. Add that to the numerator\n3. Then write the result on top of the denominator\nExample #2: Convert $$3\\frac{2}{5}$$ to an improper fraction.\nSolution: Multiply the whole number by the denominator: $$3*5=15$$. Add the numerator to that: $$15 + 2 = 17$$. Then write that down above the denominator, like this: $$\\frac{17}{5}$$\n\nReciprocal\n\nReciprocal for a number $$x$$, denoted by $$\\frac{1}{x}$$ or $$x^{-1}$$, is a number which when multiplied by $$x$$ yields $$1$$. The reciprocal of a fraction $$\\frac{a}{b}$$ is $$\\frac{b}{a}$$. To get the reciprocal of a number, divide 1 by the number. For example reciprocal of $$3$$ is $$\\frac{1}{3}$$, reciprocal of $$\\frac{5}{6}$$ is $$\\frac{6}{5}$$.\n\nOperation on Fractions\n\nTo add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.\n\nTo add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions\n\nMultiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.\n\nDividing fractions: Change the divisor into its reciprocal and then multiply.\n\nExample #1: $$\\frac{3}{7}+\\frac{2}{3}=\\frac{9}{21}+\\frac{14}{21}=\\frac{23}{21}$$\n\nExample #2: Given $$\\frac{\\frac{3}{5}}{2}$$, take the reciprocal of $$2$$. The reciprocal is $$\\frac{1}{2}$$. Now multiply: $$\\frac{3}{5}*\\frac{1}{2}=\\frac{3}{10}$$.\n\nDecimal Representation\n\nThe decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....).\n\nReduced fraction $$\\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\\frac{3}{30}$$ is also a terminating decimal, as $$\\frac{3}{30}=\\frac{1}{10}$$ and denominator $$10=2*5$$.\n\nConverting Decimals to Fractions\n\n\u2022 To convert a terminating decimal to fraction:\n1. Calculate the total numbers after decimal point\n2. Remove the decimal point from the number\n3. Put 1 under the denominator and annex it with \"0\" as many as the total in step 1\n4. Reduce the fraction to its lowest terms\n\nExample: Convert $$0.56$$ to a fraction.\n1: Total number after decimal point is 2.\n2 and 3: $$\\frac{56}{100}$$.\n4: Reducing it to lowest terms: $$\\frac{56}{100}=\\frac{14}{25}$$\n\n\u2022 To convert a recurring decimal to fraction:\n1. Separate the recurring number from the decimal fraction\n2. Annex denominator with \"9\" as many times as the length of the recurring number\n3. Reduce the fraction to its lowest terms\n\nExample #1: Convert $$0.393939...$$ to a fraction.\n1: The recurring number is $$39$$.\n2: $$\\frac{39}{99}$$, the number $$39$$ is of length $$2$$ so we have added two nines.\n3: Reducing it to lowest terms: $$\\frac{39}{99}=\\frac{13}{33}$$.\n\n\u2022 To convert a mixed-recurring decimal to fraction:\n1. Write down the number consisting with non-repeating digits and repeating digits.\n2. Subtract non-repeating number from above.\n3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's.\n\nExample #2: Convert $$0.2512(12)$$ to a fraction.\n1. The number consisting with non-repeating digits and repeating digits is 2512;\n2. Subtract 25 (non-repeating number) from above: 2512-25=2487;\n3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.\n\nRounding\n\nRounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.\n\nExample:\n5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.\n5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.\n5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.\n\nRatios and Proportions\n\nGiven that $$\\frac{a}{b}=\\frac{c}{d}$$, where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.\n\n$$\\frac{b}{a}=\\frac{d}{c}$$ - invertendo\n\n$$\\frac{a}{c}=\\frac{b}{d}$$ - alternendo\n\n$$\\frac{a+b}{b}=\\frac{c+d}{d}$$ - componendo\n\n$$\\frac{a-b}{b}=\\frac{c-d}{d}$$ - dividendo\n\n$$\\frac{a+b}{a-b}=\\frac{c+d}{c-d}$$ - componendo & dividendo\n\nThanks, this a great and concise summary of all the basics.\nJust to add, another important basic rule\n\nfor +ve fractions,\nif $$\\frac{a}{b}<1$$ and x is any positive number then $$\\frac{(a+x)}{(b+x)}>\\frac{a}{b}$$\n,in short when the two fractions have same difference between numerator and denominator , the fraction having greater numerator is always greater.\n\nAlso, when $$\\frac{a}{b}>1$$ and x is any positive number then $$\\frac{(a+x)}{(b+x)}<\\frac{a}{b}$$\n\nExample which one is greater of the following two:\na)$$\\frac{111487}{111490}$$ b)$$\\frac{111587}{111590}$$ , the ans is b for the same reasons above, as $$\\frac{111487+100}{111490+100}$$\n_________________\n\nWhatever one does in life is a repetition of what one has done several times in one's life!\nIf my post was worth it, then i deserve kudos\n\nKudos [?]: 292 [0], given: 13\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 14229\n\nKudos [?]: 291 [0], given: 0\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n05 Feb 2015, 23:29\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\n\nKudos [?]: 291 [0], given: 0\n\nManager\nJoined: 28 Jul 2011\nPosts: 233\n\nKudos [?]: 168 [0], given: 16\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n02 Oct 2015, 09:17\n\u2022 All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form 6n\u22121 or 6n+1, because all other numbers are divisible by 2 or 3.\n\nWhy does this property \"all prime numbers above 3 are of the form 6n\u22121 or 6n+1\" hold true for 961, but 961 is not a prime? This holds true for square of any prime number? Is there any gap in my understanding?\n6n+1 = 961 --> 6n = 960\n\nKudos [?]: 168 [0], given: 16\n\nIntern\nJoined: 24 Jul 2015\nPosts: 2\n\nKudos [?]: [0], given: 33\n\nWE: Supply Chain Management (Manufacturing)\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n03 Oct 2015, 02:57\nVery useful...keep posting..\n\nKudos [?]: [0], given: 33\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43335\n\nKudos [?]: 139526 [0], given: 12794\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n03 Oct 2015, 03:43\ncharan2892 wrote:\nVery useful...keep posting..\n\nThank you.\n\nFor more check ALL YOU NEED FOR QUANT ! ! !.\n\nHope it helps.\n_________________\n\nKudos [?]: 139526 [0], given: 12794\n\nIntern\nJoined: 05 Dec 2017\nPosts: 1\n\nKudos [?]: 0 [0], given: 2\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n06 Dec 2017, 10:05\nBunuel\nThanks so much! Super useful!!\n\nQuestion: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?\n\nMany thanks!\n\nKudos [?]: 0 [0], given: 2\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 43335\n\nKudos [?]: 139526 [1], given: 12794\n\nRe: Math: Number Theory (broken into smaller topics)\u00a0[#permalink]\n\n### Show Tags\n\n06 Dec 2017, 10:12\n1\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\ngmatm8 wrote:\nBunuel\nThanks so much! Super useful!!\n\nQuestion: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?\n\nMany thanks!\n\n2. Properties of Integers\n\nFor more check Ultimate GMAT Quantitative Megathread\n\nHope it helps.\n_________________\n\nKudos [?]: 139526 [1], given: 12794\n\nRe: Math: Number Theory (broken into smaller topics) \u00a0 [#permalink] 06 Dec 2017, 10:12\nDisplay posts from previous: Sort by", "date": "2018-01-20 05:22:29", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/math-number-theory-broken-into-smaller-topics-91274.html", "openwebmath_score": 0.8090343475341797, "openwebmath_perplexity": 711.1701522916192, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226314280634, "lm_q2_score": 0.8872045922259088, "lm_q1q2_score": 0.8668189653116194}}
{"url": "https://mathoverflow.net/questions/221466/how-big-can-a-set-of-integers-be-if-all-pairs-have-small-gcd", "text": "# How big can a set of integers be if all pairs have small gcd?\n\nSuppose $A\\subset[1,N]$ is a set of integers. If for any distinct $a,b\\in A$ we have $(a,b)\\leq M$ then how big can $|A|$ be?\n\nIf $M=1$ then $|A|$ is at most $\\pi(N)$ since the map $a\\mapsto P_+(a)$ (which sends $a$ to its largest prime factor) is an injection to the primes, and so the primes themselves are the most efficient. If $M=N$ then we can take $|A|=N$. What about $M$ in between? Can you beat the primes? It would be safe to assume that $A$ consists solely of $M$-smooth numbers since we can decompose $A=A_1\\cup A_2$ where each $a\\in A_1$ is $M$-smooth and each $a\\in A_2$ has a prime factor which at least $M+1$. The large prime factors appearing as divisors in $A_2$ cannot be repeated, so $|A_2|\\leq \\pi(N)$.\n\n\u2022 Even more efficiently, you may beat the primes by adding all pairwise products of primes not exceeding $M$ (including their squares). \u2013\u00a0Ilya Bogdanov Oct 21 '15 at 16:31\n\u2022 Actually, for $M= 1$ you have $|A| = \\pi(N) + 1$: you can have $1$ as well as the primes. @IlyaBogdanov: there's no reason to restrict to pairwise products, you can throw in products of arbitrarily many (not necessarily distinct) primes $x = p_1 \\ldots p_n$, $p_1 \\le \\ldots \\le p_n$, as long as $x/p_1 \\le M$. \u2013\u00a0Robert Israel Oct 21 '15 at 18:21\n\u2022 Thus for $M=5$, in addition to the primes $\\le N$ you have $\\{1,4,6,8,9,10,15,25\\}$. \u2013\u00a0Robert Israel Oct 21 '15 at 18:28\n\nWe'll prove that the maximal cardinality of such a set for $M^2\\leq N$ has size equal to $$\\pi(N)+\\sum_{1<n\\leq M} \\pi(p(n))$$ where $p(n)$ is the smallest prime factor of $n$. Since $$\\sum_{1<n\\leq M} \\pi(p(n))\\sim \\frac{M^2}{2\\log^2 M},$$ this proves that Ilya Bogdanov's example of including all pairwise products of primes not exceeding $M$ is nearly optimal in terms of asymptotics.\n\nAs you suggest in the question, this problem is equivalent to constructing the largest subset $A\\subset S_M(N)$ such that $\\gcd(a,b)\\leq M$ for every $a,b\\in A$ where $S_M(N)$ is the set of $n\\leq N$ whose largest prime factor is at most $M$.\n\nTo see why, suppose that $A$ satisfies $\\gcd(a,b)\\leq M$ for every $a,b\\in A$. Then every prime that is greater than $M$ can divide at most one element of $A$. If $p>M$ divides $a\\in A$, then making $a=p$ only helps create a larger set $A$. Since these primes do not interact with the $M$-smooth numbers, the proof is complete.\n\nLet $T(N,M)$ denote the maximum size of such a set $A\\subset S_M(N)$. Then the size of the largest subset of $[1,N]$ with pairwise $\\gcd$'s bounded by $M$ is $$\\pi(N)-\\pi(M)+T(N,M).$$\n\nAs mentioned by Fedja in the comments s, the reasoning above for the primes extends to all integers. By considering those primes $p,q\\leq M$ such that $pq>M$, we see that there can only be exactly one such number divisible by $pq$. Similarly for any $p,q,r$ with $pq,qr,rp\\leq M$ and $pqr>M$ there can only be one number in our set divisible by $pqr$. Thus we find that the maximal size is the sum over those integers whose largest proper divisor is less than $M$. Since the largest proper divisor of $n$ equals $n/p(n)$ where $p(n)$ is the least prime factor, we can group things based on this, and we have that $$T(N,M)=\\sum_{1<p\\leq M}\\sum_{n\\leq M:\\ p(n)\\geq p}1.$$ Rearranging this equals $$\\sum_{1<n\\leq M}\\sum_{p\\leq p(n)}1=\\sum_{n\\leq M}\\pi\\left(p(n)\\right),$$ and for composite $n$ $p(n)\\leq\\sqrt{n}$, so the primes dominate this sum. Thus asymptotically we have $$T(N,M)\\sim\\sum_{p\\leq M}\\pi(p)\\sim\\frac{M^{2}}{2\\log^{2}M}.$$\n\n\u2022 You can continue in the same spirit: consider pairs of primes $p,q\\le M$ with $pq>M$. Any such pair can enter just one number, so we can as well have all pairs there. Next consider all triples such that the product of any 2 is less than or equal to $M$ but the triple product is $>M$. Again, any triple can enter at most one number, so take them all, etc. After that, of course, just add $[1,M]$. So to describe a maximal cardinality set it easy. As to the size, it looks like we just need to play with the prime number theorem carefully to figure the asymptotics but I have to teach a class now... \u2013\u00a0fedja Oct 21 '15 at 18:08\n\u2022 This set can be also described as the set of all numbers whose largest proper divisor is not greater than $M$, which allows to estimate the size of the complement way faster than with my prime count idea. Now it is time to run on the stairs :-) \u2013\u00a0fedja Oct 21 '15 at 18:13\n\u2022 @Fedja: Great, that works nicely! \u2013\u00a0Eric Naslund Oct 21 '15 at 18:22\n\u2022 There's still something off if $M>\\sqrt{N}$ which you should account for. Fedja's description of numbers with largest proper divisor not larger than $M$ is very nice. For example, this includes all numbers with no prime factors below $N/M$ which has interesting asymptotics when $M$ is larger than $\\sqrt{N}$ (eventually getting to $N$ when $M=N$). So you could either restrict attention to small ranges of $M$, or perhaps flesh out the argument a little bit more ... \u2013\u00a0Lucia Oct 22 '15 at 3:19\n\u2022 @Lucia: You're right, I have edited the question for now, but I'll add in a nice form for the sum when $M>\\sqrt{N}$ \u2013\u00a0Eric Naslund Oct 22 '15 at 11:09", "date": "2019-03-22 09:20:36", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/221466/how-big-can-a-set-of-integers-be-if-all-pairs-have-small-gcd", "openwebmath_score": 0.9248566627502441, "openwebmath_perplexity": 136.41274165853258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842212767573, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8668086136755422}}
{"url": "https://gmw.globalmathproject.org/station/I8S8A", "text": "03 413 022\n\n### Question 1\n\nUp to now our machines have consisted of a row of boxes extending infinitely far to the left. Why not have boxes extending infinitely far to the right as well?\n\nLet\u2019s go back to working with a $1 \\leftarrow 10$ machine and see what such boxes could mean in that machine.\n(Then it should become clear what they mean in other machines as well.)\n\nTo keep the left and right boxes visibly clear, we\u2019ll separate them with a point. (Society calls this point, for a $1 \\leftarrow 10$ machine at least, a decimal point.)\n\nSo, what does it mean to have dots in the right boxes? What are the values of dots in those boxes?\n\nSince this is a $1 \\leftarrow 10$ machine, we do know that ten dots in any one box explode to make one dot one place to the left. So ten dots in the box just to the right of the decimal point are equivalent to one dot in the $1$ s box. Each dot in that box must be worth one-tenth.\n\nWe have:\n\nIn the same way, ten dots in the next box over are worth one-tenth. And so each dot in that next box must be worth one-hundredth.\n\nWe have:\n\nAnd ten one-thousands make a hundredth, and ten ten-thousands make a thousandth, and so on.\n\nWe see that the boxes to the left of the decimal point represent place values as given by the powers of ten, and the boxes to the right of the decimal point place values given by the reciprocals of the powers of ten.\n\nWe have just discovered decimals!\n\nWhen people write $0.3$, for example, they mean the value of placing three dots in the first box after the decimal point.\n\nWe see that $0.3$ equals three tenths: $0.3 = \\dfrac{3}{10}$.\n\nSeven dots in the third box after the decimal point is seven thousandths: $0.007 = \\dfrac{7}{1000}$.\n\nComment: Some people might leave off the beginning zero and just write $.007 = \\dfrac{7}{1000}$. It\u2019s just a matter of personal taste.\n\nSome people read $0.6$ out loud as \u201cpoint six\u201d and others read it out loud as \u201csix tenths.\u201d Which is more helpful for understanding what the number really is?\n\n### Question 2\n\nThere is a possible source of confusion with a decimal such as $0.31$. This is technically three tenths and one hundredth: $0.31 = \\dfrac{3}{10} + \\dfrac{1}{100}$.\n\nBut some people read $0.31$ out loud as \u201cthirty-one hundredths,\u201d which looks like this.\n\nAre these the same thing?\n\nWell, yes! With three explosions we see that thirty-one hundredths becomes three tenths and one hundredth.\n\nComment: You can also show that $\\dfrac{3}{10} + \\dfrac{1}{100}$ and $\\dfrac{31}{100}$ are the same with the arithmetic of adding fractions. We have\n\n$\\dfrac{3}{10} + \\dfrac{1}{100} = \\dfrac{30}{100} + \\dfrac{1}{100} = \\dfrac{31}{100}$.\n(Do you see that this is really the result of performing three unexplosions in a picture of $\\dfrac{3}{10} + \\dfrac{1}{100}$?)\n\nA teacher asked his students to each draw a $1 \\leftarrow 10$ machine picture of the fraction $\\dfrac{319}{1000}$.\n\nJinJin drew:\n\nSubra drew:\n\nThe teacher marked both students as correct. Are each of these solutions indeed valid? Explain your thinking. (By the way, the teacher doesn\u2019t mind if students just write numbers instead of drawing dots.)\n\n### Question 3\n\nMultiple choice!\n\nThe decimal $0.23$ equals:\n\n(A) $\\dfrac {23}{10}$\n\n(B) $\\dfrac {23}{100}$\n\n(C) $\\dfrac {23}{1000}$\n\n(D) $\\dfrac {23}{10000}$\n\n### Question 4\n\nThe decimal $0.0409$ equals:\n\n(A) $\\dfrac {409}{100}$\n\n(B) $\\dfrac {409}{1000}$\n\n(C) $\\dfrac {409}{10000}$\n\n(D) $\\dfrac {409}{100000}$\n\n## SIMPLE FRACTIONS AS DECIMALS\n\nSome decimals give fractions that simplify further.\n\nFor example,\n\n$0.5 = \\dfrac{5}{10} = \\dfrac{1}{2}$\n\nand\n\n$0.04 = \\dfrac{4}{100} = \\dfrac{1}{25}$.\n\nConversely, if a fraction can be rewritten to have a denominator that is a power of ten, then we can easily write it as a decimal.\n\nFor example,\n\n$\\dfrac{3}{5} = \\dfrac{6}{10}$ and so $\\dfrac{3}{5} = 0.6$\n\nand\n\n$\\dfrac{13}{20} = \\dfrac{13 \\times 5}{20 \\times 5} = \\dfrac{65}{100} = 0.65$.\n\nWhat fractions (in simplest terms) do the following decimals represent?\n\n$0.05$, $0.2$, $0.8$, $0.004$\n\n### Question 6\n\nWrite each of the following fractions as a decimal.\n\n$\\dfrac {2} {5}$, $\\dfrac {1} {25}$, $\\dfrac {1} {20}$, $\\dfrac {1} {200}$, $\\dfrac {2} {2500}$\n\n### Question 7\n\nMULTIPLE CHOICE!\n\nThe decimal $0.050$ equals\n\n(A)$\\dfrac {50} {100}$\n\n(B)$\\dfrac {1} {20}$\n\n(C)$\\dfrac {1} {200}$\n\n(D) None of these?\n\n### Question 8\n\nThe decimal $0.000208$ equals\n\n(A) $\\dfrac {52} {250}$\n\n(B) $\\dfrac {52} {2500}$\n\n(C) $\\dfrac {52} {25000}$\n\n(D) $\\dfrac {52} {250000}$\n\n### Question 9\n\nWrite each of the following fractions as decimals.\n\n$\\dfrac {7} {20}$, $\\dfrac {16} {25}$, $\\dfrac {301} {500}$, $\\dfrac {17} {50}$, $\\dfrac {3} {4}$\n\n### Question 10\n\nCHALLENGE\n\nWhat fraction does the decimal $2.3$ represent?\n\n### Question 11\n\nWhat fraction does $17.04$ represent?\n\n### Question 12\n\nWhat fraction does $1003.1003$ represent?\n\n### Question 13\n\nLet\u2019s explore the question: Do $0.19$ and $0.190$ represent the same number or different numbers?\n\nHere are two dots and boxes pictures for the decimal $0.19$:\n\nHere are two dots and boxes picture for the decimal $0.190$\n\n(A) Explain how one \u201cunexplosion\u201d establishes that the first picture of $0.19$ is equivalent to the second picture of $0.19$.\n\n(B) Explain how several unexplosions establishes that the first picture of $0.190$ is equivalent to the second picture of $0.190$.\n\n(C) Explain how explosions and unexplosions in fact establish that all four pictures are equivalent to each other.\n\n(D) In conclusion then: Does $0.190$ represent the same number as $0.19$?\n\n### Question 14\n\nTo a mathematician, the expressions $0.19$ and $0.190$ represent exactly the same numeric quantity.\n\nBut you may have noticed in science class that scientists will often write down what seems likes\nunnecessary zeros when recording measurements. This is because scientists want to impart more information to the reader than just a numeric value.\n\nFor example, suppose a botanist measures the length of a stalk. By writing the measurement as $0.190$ meters in her paper, the scientist is saying to the reader that she measured the length of the stalk to the nearest one thousandth of a meter and that she got $1$ tenth, $9$ hundredths, and $0$ thousandths of a meter. Thus we are being told that the true length of the stalk is somewhere in the range of $0.1895$ and $0.1905$ meters.\n\nIf she wrote in her paper, instead, just $0.19$ meters, then we would have to assume she measured the length of the stalk only to the nearest hundredth of a meter and so its true length lies somewhere between $0.185$ and $0.195$ meters.\n\n## MIXED NUMBERS AS DECIMALS.\n\nHow does $12 \\dfrac {3} {4}$, for example, appear as a decimal?\n\nWell, $12 \\dfrac {3} {4} = 12 + \\dfrac {3} {4}$ and we can certainly write the fractional part as a decimal. (The non-fractional part is already in the $1 \\leftarrow 10$ machine format!)\n\nWe have\n\n$12 \\dfrac {3} {4} = 12 + \\dfrac {75} {100}$\n\nand so we see\n\n$12 \\dfrac {3} {4} = 12.75$.\n\nWrite each of the following numbers in decimal notation.\n\n(A) $5 \\dfrac {3} {10}$\n\n(B) $7 \\dfrac {1} {5}$\n\n(C) $13 \\dfrac {1} {2}$\n\n(D) $106 \\dfrac {3} {20}$\n\n(E) $\\dfrac {78} {25}$\n\n(F) $\\dfrac {9} {4}$\n\n(G) $\\dfrac {131} {40}$\n\nYou can either play with some of the optional stations below or go to the next island!\n\nRegister NOW and unlock all islands!", "date": "2019-08-19 19:09:18", "meta": {"domain": "globalmathproject.org", "url": "https://gmw.globalmathproject.org/station/I8S8A", "openwebmath_score": 0.7920145392417908, "openwebmath_perplexity": 957.109921779458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842212767573, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8668086022902128}}
{"url": "https://math.stackexchange.com/questions/2844057/solving-sin-omega-t-frac-1-2/2844061", "text": "# Solving $\\sin(\\omega t)=- \\frac 1 2$\n\nSolve the given trigonometric equation: $$\\sin(\\omega t)=- \\frac 1 2$$\n\nHere is my attempt:\n\n$$\\sin(\\omega t)=- \\frac 1 2 = \\sin \\biggr (\\pi +\\dfrac{\\pi}{6}\\biggr )$$\n\nWhich yields\n\n$$\\omega t = \\dfrac{7\\pi }{6}$$\n\nor\n\n$$\\sin(\\omega t)=- \\frac 1 2 = \\sin \\biggr (2\\pi -\\dfrac{\\pi}{6}\\biggr )$$\n\n$$\\omega t = \\dfrac{11\\pi}{6}$$\n\nIs my assumption correct?\n\nRegards!\n\n\u2022 I've fixed a small error. \u2013\u00a0Mr. Maxwell Jul 7 '18 at 21:17\n\nYou also have $$\\omega t = \\dfrac{-\\pi }{6}+2k\\pi$$\n\nThus the solution is $$\\omega t = \\{\\dfrac{-\\pi }{6}+2k_1\\pi:k_1 \\in Z\\}\\cup \\{\\dfrac{7\\pi }{6}+2k_2\\pi:k_2\\in Z\\}$$\n\nWhere Z is the set of integers.\n\nFrom the unit circle definition of sine, we see that\n\n$$\\omega t = 2n\\pi - \\dfrac{\\pi}{2} \\pm \\dfrac{\\pi}{3}$$\n\nand we can rephrase that as we wish.\n\nThe sin$(x)$ function repeats itself every $2\\pi$ times, so $\\omega t=\\frac{7\\pi}6+2k\u03c0$ and $\\omega t=\\frac{11\\pi}{6}+2k\\pi,\\;$ where $k \\in \\mathbb Z$.\n\nYes it is right but recall to add the $2k\\pi$ term with $k\\in \\mathbb{Z}$, indeed we have\n\n$$\\sin(\\omega t)=- \\frac 1 2\\iff \\omega t=\\dfrac{7\\pi }{6}+2k\\pi\\,\\lor\\,\\omega t=\\dfrac{11\\pi }{6}+2k\\pi$$\n\n\u2022 What if I don't recall to add $2k\\pi$? \u2013\u00a0Mr. Maxwell Jul 7 '18 at 21:16\n\u2022 @Mr.Maxwell You can lost solutions. \u2013\u00a0user Jul 7 '18 at 21:17\n\u2022 However, isn't it enough to leave it up once we get $\\dfrac{7\\pi}{6}$ and $\\dfrac{11\\pi}{6}$? \u2013\u00a0Mr. Maxwell Jul 7 '18 at 21:18\n\u2022 @Mr.Maxwell Those are the solutions for $\\omega t \\in [0,2\\pi]$ but in general solving trigonometric equations we need to consider a wider range. Take a look for example here math.stackexchange.com/q/2826222/505767 \u2013\u00a0user Jul 7 '18 at 21:20\n\u2022 @Mr.Maxwell - Enough is defined by who is asking the question. If the domain is not specified, then the question is poorly asked. \u2013\u00a0steven gregory Jul 7 '18 at 22:31", "date": "2019-11-21 16:13:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2844057/solving-sin-omega-t-frac-1-2/2844061", "openwebmath_score": 0.9460216760635376, "openwebmath_perplexity": 972.5842808068418, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877023336243, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.8667980398961477}}
{"url": "https://au.mathworks.com/help/matlab/ref/plot3.html", "text": "plot3\n\n3-D point or line plot\n\nDescription\n\nexample\n\nplot3(X,Y,Z) plots coordinates in 3-D space.\n\n\u2022 To plot a set of coordinates connected by line segments, specify X, Y, and Z as vectors of the same length.\n\n\u2022 To plot multiple sets of coordinates on the same set of axes, specify at least one of X, Y, or Z as a matrix and the others as vectors.\n\nexample\n\nplot3(X,Y,Z,LineSpec) creates the plot using the specified line style, marker, and color.\n\nexample\n\nplot3(X1,Y1,Z1,...,Xn,Yn,Zn) plots multiple sets of coordinates on the same set of axes. Use this syntax as an alternative to specifying multiple sets as matrices.\n\nexample\n\nplot3(X1,Y1,Z1,LineSpec1,...,Xn,Yn,Zn,LineSpecn) assigns specific line styles, markers, and colors to each XYZ triplet. You can specify LineSpec for some triplets and omit it for others. For example, plot3(X1,Y1,Z1,'o',X2,Y2,Z2) specifies markers for the first triplet but not the for the second triplet.\n\nexample\n\nplot3(___,Name,Value) specifies Line properties using one or more name-value pair arguments. Specify the properties after all other input arguments. For a list of properties, see Line Properties.\n\nexample\n\nplot3(ax,___) displays the plot in the target axes. Specify the axes as the first argument in any of the previous syntaxes.\n\nexample\n\np = plot3(___) returns a Line object or an array of Line objects. Use p to modify properties of the plot after creating it. For a list of properties, see Line Properties.\n\nExamples\n\ncollapse all\n\nDefine t as a vector of values between 0 and 10$\\pi$. Define st and ct as vectors of sine and cosine values. Then plot st, ct, and t.\n\nt = 0:pi/50:10*pi;\nst = sin(t);\nct = cos(t);\nplot3(st,ct,t)\n\nCreate two sets of x-, y-, and z-coordinates.\n\nt = 0:pi/500:pi;\nxt1 = sin(t).*cos(10*t);\nyt1 = sin(t).*sin(10*t);\nzt1 = cos(t);\n\nxt2 = sin(t).*cos(12*t);\nyt2 = sin(t).*sin(12*t);\nzt2 = cos(t);\n\nCall the plot3 function, and specify consecutive XYZ triplets.\n\nplot3(xt1,yt1,zt1,xt2,yt2,zt2)\n\nCreate matrix X containing three rows of x-coordinates. Create matrix Y containing three rows of y-coordinates.\n\nt = 0:pi/500:pi;\nX(1,:) = sin(t).*cos(10*t);\nX(2,:) = sin(t).*cos(12*t);\nX(3,:) = sin(t).*cos(20*t);\n\nY(1,:) = sin(t).*sin(10*t);\nY(2,:) = sin(t).*sin(12*t);\nY(3,:) = sin(t).*sin(20*t);\n\nCreate matrix Z containing the z-coordinates for all three sets.\n\nZ = cos(t);\n\nPlot all three sets of coordinates on the same set of axes.\n\nplot3(X,Y,Z)\n\nCreate vectors xt, yt, and zt.\n\nt = 0:pi/500:40*pi;\nxt = (3 + cos(sqrt(32)*t)).*cos(t);\nyt = sin(sqrt(32) * t);\nzt = (3 + cos(sqrt(32)*t)).*sin(t);\n\nPlot the data, and use the axis equal command to space the tick units equally along each axis. Then specify the labels for each axis.\n\nplot3(xt,yt,zt)\naxis equal\nxlabel('x(t)')\nylabel('y(t)')\nzlabel('z(t)')\n\nCreate vectors t, xt, and yt, and plot the points in those vectors using circular markers.\n\nt = 0:pi/20:10*pi;\nxt = sin(t);\nyt = cos(t);\nplot3(xt,yt,t,'o')\n\nCreate vectors t, xt, and yt, and plot the points in those vectors as a blue line with 10-point circular markers. Use a hexadecimal color code to specify a light blue fill color for the markers.\n\nt = 0:pi/20:10*pi;\nxt = sin(t);\nyt = cos(t);\nplot3(xt,yt,t,'-o','Color','b','MarkerSize',10,'MarkerFaceColor','#D9FFFF')\n\nCreate vector t. Then use t to calculate two sets of x and y values.\n\nt = 0:pi/20:10*pi;\nxt1 = sin(t);\nyt1 = cos(t);\n\nxt2 = sin(2*t);\nyt2 = cos(2*t);\n\nPlot the two sets of values. Use the default line for the first set, and specify a dashed line for the second set.\n\nplot3(xt1,yt1,t,xt2,yt2,t,'--')\n\nCreate vectors t, xt, and yt, and plot the data in those vectors. Return the chart line in the output variable p.\n\nt = linspace(-10,10,1000);\nxt = exp(-t./10).*sin(5*t);\nyt = exp(-t./10).*cos(5*t);\np = plot3(xt,yt,t);\n\nChange the line width to 3.\n\np.LineWidth = 3;\n\nStarting in R2019b, you can display a tiling of plots using the tiledlayout and nexttile functions. Call the tiledlayout function to create a 1-by-2 tiled chart layout. Call the nexttile function to create the axes objects ax1 and ax2. Create separate line plots in the axes by specifying the axes object as the first argument to plot3.\n\ntiledlayout(1,2)\n\n% Left plot\nax1 = nexttile;\nt = 0:pi/20:10*pi;\nxt1 = sin(t);\nyt1 = cos(t);\nplot3(ax1,xt1,yt1,t)\ntitle(ax1,'Helix With 5 Turns')\n\n% Right plot\nax2 = nexttile;\nt = 0:pi/20:10*pi;\nxt2 = sin(2*t);\nyt2 = cos(2*t);\nplot3(ax2,xt2,yt2,t)\ntitle(ax2,'Helix With 10 Turns')\n\nCreate x and y as vectors of random values between 0 and 1. Create z as a vector of random duration values.\n\nx = rand(1,10);\ny = rand(1,10);\nz = duration(rand(10,1),randi(60,10,1),randi(60,10,1));\n\nPlot x, y, and z, and specify the format for the z-axis as minutes and seconds. Then add axis labels, and turn on the grid to make it easier to visualize the points within the plot box.\n\nplot3(x,y,z,'o','DurationTickFormat','mm:ss')\nxlabel('X')\nylabel('Y')\nzlabel('Duration')\ngrid on\n\nCreate vectors xt, yt, and zt. Plot the values, specifying a solid line with circular markers using the LineSpec argument. Specify the MarkerIndices property to place one marker at the 200th data point.\n\nt = 0:pi/500:pi;\nxt(1,:) = sin(t).*cos(10*t);\nyt(1,:) = sin(t).*sin(10*t);\nzt = cos(t);\nplot3(xt,yt,zt,'-o','MarkerIndices',200)\n\nInput Arguments\n\ncollapse all\n\nx-coordinates, specified as a scalar, vector, or matrix. The size and shape of X depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.\n\nType of PlotHow to Specify Coordinates\nSingle point\n\nSpecify X, Y, and Z as scalars and include a marker. For example:\n\nplot3(1,2,3,'o')\n\nOne set of points\n\nSpecify X, Y, and Z as any combination of row or column vectors of the same length. For example:\n\nplot3([1 2 3],[4; 5; 6],[7 8 9])\n\nMultiple sets of points\n(using vectors)\n\nSpecify consecutive sets of X, Y, and Z vectors. For example:\n\nplot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])\n\nMultiple sets of points\n(using matrices)\n\nSpecify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:\n\nplot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])\n\nData Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration\n\ny-coordinates, specified as a scalar, vector, or matrix. The size and shape of Y depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.\n\nType of PlotHow to Specify Coordinates\nSingle point\n\nSpecify X, Y, and Z as scalars and include a marker. For example:\n\nplot3(1,2,3,'o')\n\nOne set of points\n\nSpecify X, Y, and Z as any combination of row or column vectors of the same length. For example:\n\nplot3([1 2 3],[4; 5; 6],[7 8 9])\n\nMultiple sets of points\n(using vectors)\n\nSpecify consecutive sets of X, Y, and Z vectors. For example:\n\nplot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])\n\nMultiple sets of points\n(using matrices)\n\nSpecify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:\n\nplot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])\n\nData Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration\n\nz-coordinates, specified as a scalar, vector, or matrix. The size and shape of Z depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.\n\nType of PlotHow to Specify Coordinates\nSingle point\n\nSpecify X, Y, and Z as scalars and include a marker. For example:\n\nplot3(1,2,3,'o')\n\nOne set of points\n\nSpecify X, Y, and Z as any combination of row or column vectors of the same length. For example:\n\nplot3([1 2 3],[4; 5; 6],[7 8 9])\n\nMultiple sets of points\n(using vectors)\n\nSpecify consecutive sets of X, Y, and Z vectors. For example:\n\nplot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])\n\nMultiple sets of points\n(using matrices)\n\nSpecify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:\n\nplot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])\n\nData Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration\n\nLine style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.\n\nExample: '--or' is a red dashed line with circle markers\n\nLine StyleDescription\n-Solid line (default)\n--Dashed line\n:Dotted line\n-.Dash-dot line\nMarkerDescription\noCircle\n+Plus sign\n*Asterisk\n.Point\nxCross\nsSquare\ndDiamond\n^Upward-pointing triangle\nvDownward-pointing triangle\n>Right-pointing triangle\n<Left-pointing triangle\npPentagram\nhHexagram\nColorDescription\n\ny\n\nyellow\n\nm\n\nmagenta\n\nc\n\ncyan\n\nr\n\nred\n\ng\n\ngreen\n\nb\n\nblue\n\nw\n\nwhite\n\nk\n\nblack\n\nTarget axes, specified as an Axes object. If you do not specify the axes and if the current axes is Cartesian, then plot3 uses the current axes.\n\nName-Value Pair Arguments\n\nSpecify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.\n\nExample: plot3([1 2],[3 4],[5 6],'Color','red') specifies a red line for the plot.\n\nNote\n\nThe properties listed here are only a subset. For a complete list, see Line Properties.\n\nColor, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name. The color you specify sets the line color. It also sets the marker edge color when the MarkerEdgeColor property is set to 'auto'.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB\u00ae uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nLine width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.\n\nThe line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.\n\nMarker size, specified as a positive value in points, where 1 point = 1/72 of an inch.\n\nMarker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nMarker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' option uses the same color as the Color property of the parent axes. If you specify 'auto' and the axes plot box is invisible, the marker fill color is the color of the figure.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nTips\n\n\u2022 Use NaN or Inf to create breaks in the lines. For example, this code plots a line with a break between z=2 and z=4.\n\nplot3([1 2 3 4 5],[1 2 3 4 5],[1 2 NaN 4 5])\n\n\u2022 plot3 uses colors and line styles based on the ColorOrder and LineStyleOrder properties of the axes. plot3 cycles through the colors with the first line style. Then, it cycles through the colors again with each additional line style.\n\nStarting in R2019b, you can change the colors and the line styles after plotting by setting the ColorOrder or LineStyleOrder properties on the axes. You can also call the colororder function to change the color order for all the axes in the figure.", "date": "2020-08-13 20:51:18", "meta": {"domain": "mathworks.com", "url": "https://au.mathworks.com/help/matlab/ref/plot3.html", "openwebmath_score": 0.5981382131576538, "openwebmath_perplexity": 2821.3270388311807, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982287698703999, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8667980397304585}}
{"url": "http://mathhelpforum.com/calculus/167221-integrate-ln2x-x-2-a.html", "text": "# Math Help - Integrate LN2x/x^2\n\n1. ## Integrate LN2x/x^2\n\nHey guys, thanks for all the help so far.. stuck on another problem.\n\n$\\int \\frac{ln2x}{x^2}$\n\nSo I think\n$ln2x$\nis an integration by parts problem by itself, but I am lost.\nI know that\n$\\int lnx = xlnx-x$\nBut that was just a rule given by the professor, not sure how we applied that.\n\nI think it goes something like this (using integration by parts)\n$u= ln2x \\ du= \\frac{2x}{x} \\ dv=dx \\ v=x$\n\nUsing the rule\n$UV- \\int VdU$\n\nI get\n$(xln2x)- \\int x \\frac{2x}{x}$\nwhich equals\n$(xln2x)- \\int \\frac{2x^2}{x}$\n\nBut that's just for the top portion of the original and I am not even sure I did that correctly.\n\nThanks guys\n\n2. Just in case a picture helps...\n\n... where (key in first spoiler) ...\n\nSpoiler:\n\n... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,\n\n... is lazy integration by parts, doing without u and v.\n\nSpoiler:\n\n________________________________________\n\nDon't integrate - balloontegrate!\n\nBalloon Calculus; standard integrals, derivatives and methods\n\nBalloon Calculus Drawing with LaTeX and Asymptote!\n\n3. Originally Posted by Spoolx\nHey guys, thanks for all the help so far.. stuck on another problem.\n\n$\\int \\frac{ln2x}{x^2}$\n\nSo I think\n$ln2x$\nis an integration by parts problem by itself, but I am lost.\nI know that\n$\\int lnx = xlnx-x$\nBut that was just a rule given by the professor, not sure how we applied that.\n\nI think it goes something like this (using integration by parts)\n$u= ln2x \\ du= \\frac{2x}{x} \\ dv=dx \\ v=x$\n\nUsing the rule\n$UV- \\int VdU$\n\nI get\n$(xln2x)- \\int x \\frac{2x}{x}$\nwhich equals\n$(xln2x)- \\int \\frac{2x^2}{x}$\n\nBut that's just for the top portion of the original and I am not even sure I did that correctly.\n\nThanks guys\n$ln(2x)=ln2+lnx$\n\n$\\displaystyle\\int{\\frac{ln(2x)}{x^2}}dx=ln2\\int{\\f rac{1}{x^2}}dx+\\int{\\frac{lnx}{x^2}}dx$\n\nNow you can use integration by parts.\n\n4. Hello, Spoolx!\n\n$\\displaystyle \\int \\frac{\\ln(2x)}{x^2}\\,dx$\n\nIntegrate by parts:\n\n. . $\\begin{array}{cccccccc}\nu &=& \\ln(2x) && dv &=& \\dfrac{1}{x^2}\\,dx \\\\ \\\\[-3mm]\ndu &=& \\dfrac{dx}{x} && v &=& \\text{-}\\dfrac{1}{x} \\end{array}$\n\nAnd we have: . $\\displaystyle \\text{-}\\frac{1}{x}\\ln(2x) + \\int\\frac{dx}{x^2}$\n\n. . . . . . . . $\\displaystyle =\\;\\text{-}\\frac{1}{x}\\ln(2x) - \\frac{1}{x} + C$\n\n. . . . . . . . $\\displaystyle =\\;\\text{-}\\frac{1}{x}\\left[\\ln(2x) + 1\\right] + C$\n\n5. Thanks for the help guys, can you jst run through the derivation of the $ln2x$ please?\n\n6. Just in case a picture helps...\n\n... where (key in spoiler) ...\n\nSpoiler:\n\n... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).\n\nThe general drift is...\n\nThough, as Archie Meade points out, you can split the log first to avoid the chain rule.\n\n_________________________________________\n\nDon't integrate - balloontegrate!\n\nBalloon Calculus; standard integrals, derivatives and methods\n\nBalloon Calculus Drawing with LaTeX and Asymptote!\n\n7. Originally Posted by Spoolx\nHey guys, thanks for all the help so far.. stuck on another problem.\n\n$\\int \\frac{ln2x}{x^2}$\n\nSo I think\n$ln2x$\nis an integration by parts problem by itself, but I am lost.\nI know that\n$\\int lnx = xlnx-x$\nBut that was just a rule given by the professor, not sure how we applied that.\n\nI think it goes something like this (using integration by parts)\n$u= ln2x \\ du= \\frac{2x}{x} \\ dv=dx \\ v=x$\n\nUsing the rule\n$UV- \\int VdU$\n\nI get\n$(xln2x)- \\int x \\frac{2x}{x}$\nwhich equals\n$(xln2x)- \\int \\frac{2x^2}{x}$\n\nBut that's just for the top portion of the original and I am not even sure I did that correctly.\n\nThanks guys\nRather than trying to apply integration by parts just to the numerator,\nyou should try applying the technique to the entire expression.\n\n$\\displaystyle\\int{u}dv=uv-\\int{v}du$\n\n$u=ln(2x),\\;\\;dv=\\displaystyle\\frac{1}{x^2}dx$\n\nWe need $v$ and $du$ to apply integration by parts. Using the chain rule we get\n\n$\\displaystyle\\frac{du}{dx}=\\frac{1}{2x}2$\n\n$\\Rightarrow\\ du=\\frac{1}{x}dx$\n\n$v=\\int{x^{-2}}dx=-x^{-1}$\n\n$\\displaystyle\\ uv-\\int{v}du}=-\\frac{ln(2x)}{x}+\\int{\\frac{1}{x^2}dx=-\\frac{ln(2x)}{x}-\\frac{1}{x}+C$\n\nAlternatively, begin with\n\n$ln(2x)=ln2+lnx$\n\n8. Glad I found this forum, learning alot from you guys!\nThanks!", "date": "2015-07-31 22:51:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/167221-integrate-ln2x-x-2-a.html", "openwebmath_score": 0.9497044086456299, "openwebmath_perplexity": 1075.421676743357, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876992225168, "lm_q2_score": 0.8824278633625322, "lm_q1q2_score": 0.8667980356322231}}
{"url": "http://mathhelpforum.com/algebra/123610-verifying-answers-polynomials.html", "text": "# Math Help - verifying answers for polynomials\n\n1. ## verifying answers for polynomials\n\nI would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!\n\n2. ## latex\n\nOriginally Posted by jay1\nI would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!\nhope you learn to use laTex\nthis is my understanding of your equation\n\n$\n\n\\frac{1}{4a^2} + \\frac{3}{4a} = 1\n$\n\n$\\frac{1}{4a^2} + \\frac{3}{4a}\\times\\frac{a}{a} \\Rightarrow \\frac{1+3a}{4a^2} = 1$\n\n$4a^2 - 3a -1 = 0$\n\n$(4a-1)(a+1)= 0$\n\n$a= -1$\n\n$a=-\\frac{1}{4}$\n\n3. Is there a place that will show me how to use this \"laTex\"? BigWave, was my answer correct? Thanks for your help.\n\n4. Originally Posted by jay1\nI would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!\n\nIf problem 1 is what I interpreted it to be, which is this:\n\n$\n\\frac{1}{4a^2} + \\frac{3}{4}a = 1\n$\n\nthen\n\n$\na=1, a=\\frac{1+\\sqrt{13}}{6}, \\text{ and } a=\\frac{1-\\sqrt{13}}{6}\n$\n.\n\nHowever, if problem 1 is this:\n\n$\n\\frac{1}{4a^2} + \\frac{3}{4a} = 1\n$\n\nthen\n\n$\na=1, a= -\\frac{1}{4}\n$\n.\n\nAnd if problem 1 is this:\n\n$\n\\frac{1}{4}a^2 + \\frac{3}{4}a = 1\n$\n\nthen\n\n$\na=1, a= -4\n$\n.\n\n5. Hello, jay1!\n\n1) Solve: . $\\tfrac{1}{4}a^2 + \\tfrac{3}{4}a \\:=\\:1$\n\nMy answer: . $a=4,\\;a=1$ . . . . no\nWe have: . $\\tfrac{1}{4}a^2 + \\tfrac{3}{4}a \\:=\\:1$\n\nMultiply by 4: . $a^2 + 3a \\:=\\:4 \\quad\\Rightarrow\\quad a^2 + 3a - 4 \\:=\\:0$\n\nFactor: . $(a-1)(a+4) \\:=\\:0$\n\nTherefore: . $a\\;=\\;1,\\:-4$\n\n2) Simplify: . $2ab^4 -3a^2b^2 - ab^4 + a^2b^2$\n\nI have: . $ab^4- 2a^2b^2$ . . . . Yes!\n\n3) Simplify: . $-2(4y^2 +3z^3 +5) + 3(2y^2 -5z^3 +3)$\n\nMy answer: . $-21z^3 - 2y^2 -1$ . . . . Right!\n\n6. PROBLEM 2:\n\n$\n2ab^4 - 3 a^2b^2 - ab^4 + a^2b^2 = ab^4 -2a^2b^2\n$\n\nYou are correct.\n\n7. PROBLEM 3:\n\n$\n-2(4y2+3z3+5) + 3(2y2-5z3+3) = (-8y^2 - 6z^3 - 10) + (6y^2 - 15z^3 + 9)$\n$= -8y^2 - 6z^3 - 10 + 6y^2 - 15z^3 + 9 = -21z^3 - 2y^2 - 1$\n\nGood job again. Problem 3 is correct.\n\n-Andy\n\n8. ## mystery equation\n\nwho had the mystery equation correct....\njust curious..\n\n9. Thanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?\n\n10. BigWave, I think that it was Soroban. Thanks again!\n\n11. Originally Posted by jay1\nThanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?\n\n$x^2 + 4x - 45 = 0$\n\n$(x+9)(x-5) = 0$\n\n$x = 5, -9$\n\n12. Originally Posted by jay1\n2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 [COLOR=black]I have ab4 - 2a2b2\n\n$\n2ab^4 - 3a^2b^2 -ab^4 + a^2b^2 = ab^4 - 2a^2b^2\n$\n\nAtta boy (assuming you are a guy).\n\n13. 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?\n\n$\ny^2 + 12y + 35 = (y+7)(y+5) \\implies y=-5,-7\n$\n\nGood luck.\n\n-Andy\n\n14. Perform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?\n\n15. Originally Posted by jay1\nPerform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?\nNo.\n\n$\n(5a^3 + 3a -2) - (4a^3 + a^2 + 5) = 5a^3 + 3a - 2 - 4a^3 - a^2 - 5 =\na^3 - a^2 + 3a - 7\n$\n\nThe minus before the parenthesis changes the sign of each term inside the parentheses.\n\n-Andy\n\nPage 1 of 2 12 Last", "date": "2014-12-26 20:22:56", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/123610-verifying-answers-polynomials.html", "openwebmath_score": 0.8691002726554871, "openwebmath_perplexity": 1625.3902075901067, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9539660976007597, "lm_q2_score": 0.9086178913651383, "lm_q1q2_score": 0.8667906640358319}}
{"url": "https://math.stackexchange.com/questions/3093945/how-can-i-simplify-this-fraction-problem", "text": "How can I simplify this fraction problem?\n\nI have the problem $$\\frac{x^2}{x^2-4} - \\frac{x+1}{x+2}$$ which should simplify to $$\\frac{1}{x-2}$$\n\nI have simplified $$x^2-4$$, which becomes:\n\n$$\\frac{x^2}{(x-2)(x+2)} - \\frac{x+1}{x+2}$$\n\nHowever, if I combine the fractions I get, $$x^2-x-1$$ for the numerator, which can't be factored. That's where I get stuck.\n\nHow can I get $$\\frac{1}{x-2}$$ out of this problem?\n\nYou need to put them over a common denominator, $$\\frac {x+1}{x+2}=\\frac {(x+1)(x-2)}{(x+2)(x-2)}=\\frac {x^2-x-2}{x^2-4}$$ Now you can subtract the numerators $$x^2-(x^2-x-2)=x+2$$ and finally divide out the $$x+2$$ from numerator and denominator\n\nYour $$-1$$ should be $$-2$$. You didn't show your work, so I can't see why it happened.\n\n\u2022 Regarding your last line, I wonder if OP just added the two numerators together without finding a common denominator. \u2013\u00a0Matthew Leingang Jan 30 at 18:54\n\u2022 @MatthewLeingang That's right. It's been awhile since I worked with fractions, and I'm very rusty. \u2013\u00a0LuminousNutria Jan 30 at 18:56\n\u2022 @LuminousNutria It's a very common error. But if you think about it with familiar fractions you'll remember it can't be true. For instance, $\\frac{1}{2} + \\frac{1}{2}$ is $1$, not $\\frac{2}{4}$ (which is $\\frac{1}{2}$ again). \u2013\u00a0Matthew Leingang Jan 30 at 19:00\n\n\\begin{align} \\frac{x^2}{(x-2)(x+2)} - \\frac{x+1}{x+2}&=\\frac1{x+2}\\left(\\frac{x^2}{x-2} - (x+1)\\right)\\\\ &=\\frac1{x+2}\\left(\\frac{x^2-(x-2)(x+1)}{x-2}\\right)\\\\ &=\\frac1{x+2}\\left(\\frac{x^2-(x^2-x-2)}{x-2}\\right)\\\\ &=\\frac1{x+2}\\left(\\frac{x+2}{x-2}\\right)\\\\ &=\\frac1{x-2} \\end{align}\n\nWrite $$\\frac{x^2}{(x-2)(x+2)}-\\frac{x+1}{x+2}=\\frac{x^2}{(x-2)(x+2)}-\\frac{(x+1)(x-2)}{(x+2)(x-2)}=\u2026$$ Note that it must be $$x\\ne 2,-2$$\n\nThe numerator should be $$x^2-(x-2)(x+1) = x+2$$ which simplifies with a factor of the denominator.\n\nYou probably just forgot a term when subtracting the two fractions.", "date": "2019-10-19 07:08:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3093945/how-can-i-simplify-this-fraction-problem", "openwebmath_score": 0.9402146339416504, "openwebmath_perplexity": 505.3974672462041, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936110872273, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8667867801816131}}
{"url": "http://math.stackexchange.com/questions/310758/prove-that-for-all-integers-a-and-b-if-a-divides-b-then-a2", "text": "# Prove that for all integers $a$ and $b$, if $a$ divides $b$, then $a^{2}$ divides $b^{2}$.\n\nI just need to know that if $a$ divides $b$, where $a$ and $b$ are integers, does $a^{2}$ divide $b^{2}$?\n\n-\n\nIf $a$ divides $b$, then $b=ka$ for some integer $k$, so $b^2=k^2a^2$ where $k^2$ is an integer.\n\n-\nHow do you prove that k^2 is an integer? Is that by closure? \u2013\u00a0 amster27 Feb 22 '13 at 1:07\n@amster27: If you are concerned about whether $k^{2}$ is an integer, then you should be equally concerned about whether both $a^{2}$ and $b^{2}$ are integers. :) \u2013\u00a0 Haskell Curry Feb 22 '13 at 1:23\nThanks, I'm just starting to write proofs in a Mathematical Reasoning class, and I just feel so lost. But these explanations helped. \u2013\u00a0 amster27 Feb 22 '13 at 1:28\n@amster: Note that $+_{\\mathbb{Z}}: \\mathbb{Z} \\times \\mathbb{Z} \\to \\mathbb{Z}$ and $\\times_{\\mathbb{Z}}: \\mathbb{Z} \\times \\mathbb{Z} \\to \\mathbb{Z}$. Therefore, adding two integers or multiplying two integers yields an integer. \u2013\u00a0 Haskell Curry Feb 22 '13 at 1:44\n\nHint $\\rm\\ \\ a\\mid b\\ \\Rightarrow\\ \\dfrac{b}a\\in \\Bbb Z\\ \\Rightarrow\\ \\dfrac{b^2}{a^2} = \\left(\\dfrac{b}a\\right)^2\\!\\in \\Bbb Z^2\\subseteq \\Bbb Z\\:\\Rightarrow\\: a^2\\mid b^2\\$\n\n-\nThat's closer to a proof than a hint. \u2013\u00a0 1015 Feb 22 '13 at 1:29\nThe argument is elegant, but I think that the OP wants to stay within $\\mathbb{Z}$ and does not wish to jump into $\\mathbb{Q}$, as he seems to be very interested in the axioms governing the properties of $\\mathbb{Z}$. \u2013\u00a0 Haskell Curry Feb 22 '13 at 1:29\n@Haskell Possibly, but there was no hint of any such constraint in the question. \u2013\u00a0 Math Gems Feb 22 '13 at 1:46", "date": "2015-07-07 18:07:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/310758/prove-that-for-all-integers-a-and-b-if-a-divides-b-then-a2", "openwebmath_score": 0.8386735320091248, "openwebmath_perplexity": 382.09975194763234, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98409360595565, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8667867725824542}}
{"url": "https://codegolf.stackexchange.com/questions/231176/maximum-number-of-squares-touched-by-a-line-segment/231354#231354", "text": "# Maximum number of squares touched by a line segment\n\nConsider a square grid on the plane, with unit spacing. A line segment of integer length $$\\L\\$$ is dropped at an arbitrary position with arbitrary orientation. The segment is said to \"touch\" a square if it intersects the interior of the square (not just its border).\n\n# The challenge\n\nWhat is the maximum number of squares that the segment can touch, as a function of $$\\L\\$$?\n\n# Examples\n\n\u2022 L=3 $$\\\\ \\, \\$$ The answer is $$\\7\\$$, as illustrated by the blue segment in the left-hand side image (click for a larger view). The red and yellow segments only touch $$\\6\\$$ and $$\\4\\$$ squares respectively. The purple segment touches $$\\0\\$$ squares (only the interiors count).\n\n\u2022 L=5 $$\\\\ \\, \\$$ The answer is $$\\9\\$$. The dark red segment in the right-hand side image touches $$\\6\\$$ squares (note that $$\\5^2 = 3^2+4^2\\$$), whereas the green one touches $$\\8\\$$. The light blue segment touches $$\\9\\$$ squares, which is the maximum for this $$\\L\\$$.\n\n\u2022 The input $$\\L\\$$ is a positive integer.\n\u2022 The algorithm should theoretically work for arbitrarily large $$\\L\\$$. In practice it is acceptable if the program is limited by time, memory, or data-type size.\n\u2022 Input and output are flexible as usual. Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.\n\u2022 Shortest code in bytes wins.\n\n# Test cases\n\nHere are the outputs for L = 1, 2, ..., 50 (with L increasing left to right, then down):\n\n 3    5    7    8    9   11   12   14   15   17\n18   19   21   22   24   25   27   28   29   31\n32   34   35   36   38   39   41   42   43   45\n46   48   49   51   52   53   55   56   58   59\n60   62   63   65   66   68   69   70   72   73\n\n\u2022 No OEIS entry, apparently (maybe soon) Jul 10 at 15:35\n\u2022 I really liked this one from a problem-solving perspective. Jul 10 at 23:59\n\u2022 My Math.SE question will be interesting to y'all and relevant to this question Jul 11 at 9:03\n\u2022 This reminds me of the problem of optimizing beam placement in FTL Jul 13 at 2:02\n\u2022 This sequence is now in OEIS: A346232 Jul 15 at 0:02\n\n# Vyxal, 7 bytes\n\n*d\u21e9\u221a3+\u230a\n\n\nTry it Online!\n\n*       # Square\nd      # Double\n\u21e9     # Subtract 2\n\u221a    # Square root\n\u230a # Floor\n\n\n# Python 2, 27 bytes\n\nlambda L:(2*L*L-2)**.5//1+3\n\n\nTry it online!\n\nA direct formula:\n\n$$f(L) = \\lfloor \\sqrt{2 L^2-2}\\rfloor + 3$$\n\nDerivation\n\nAs noted by @Kirill L. and others, the optimal layout uses a near-diagonal line segment whose horizontal and vertical span are at least $$\\(h,h)\\$$ or $$\\(h,h+1)\\$$. We need the length-$$\\L\\$$ to cover at least this much distance using the Pythagorean theorem, plus some extra to reach into squares and hit 3 more.\n\nThis gives a result of either:\n\n\u2022 $$\\2h+3\\$$ where $$\\h\\$$ is the greatest positive integer where $$\\h^2+h^2 < L^2\\$$, or\n\u2022 $$\\2h+4\\$$ where $$\\h\\$$ is the greatest positive integer where $$\\h^2+(h+1)^2 ,\n\nwhichever of these is larger. We want to combine these two cases into one.\n\nLet's start by making the two cases look more similar. Note that the second case can be rewritten as\n\n$$\\2(h+\\frac{1}{2})+3\\$$, where $$\\2(h+\\frac{1}{2})^2 + \\frac{1}{2} < L^2\\$$\n\nor as\n\n$$\\2h+3\\$$, where $$\\2h^2 + \\frac{1}{2} < L^2\\$$\n\nwhere $$\\h\\$$ is a positve integer-and-a-half. The $$\\2h^2 in the first case can be also be written as $$\\2h^2 +\\frac{1}{2} < L^2\\$$, which is equivalent because both sides were integers. So, the cases now merge into:\n\n$$\\2h+3\\$$ where $$\\h\\$$ is the greatest positive integer or half-integer where $$\\2h^2 +1/2 < L^2\\$$\n\nCalling $$\\2h=H\\$$, this is:\n\n$$\\H+3\\$$ where $$\\H\\$$ is the greatest positive integer where $$\\2(H/2)^2 +1/2 < L^2\\$$.\n\nThis inequality is $$\\H^2 < 2L^2-1\\$$, and since these are integers, this is the same as $$\\H^2 \\leq 2L^2-2\\$$. The greatest such positive integer $$\\H\\$$ is then $$\\\\lfloor \\sqrt{2 L^2-2}\\rfloor\\$$, so the final result is $$\\ \\lfloor \\sqrt{2 L^2-2}\\rfloor + 3 \\$$.\n\n\u2022 Fantastic. \"We want to combine these two cases into one. Let's start by making the two cases look more similar.\" Did you know somehow this could be done at the outset (if so, how?), or did you just guess it might be possible, and start playing around? Jul 11 at 3:55\n\u2022 @Jonah It was a guess, but it seemed to me like something like this should be possible because the two types of outputs lie nearly on the same approximate curve.\n\u2013\u00a0xnor\nJul 11 at 4:06\n\u2022 Wonderful! Simpler than the formula I had obtained Jul 11 at 10:41\n\u2022 I intend to submit this sequence to OEIS. Your formula is simpler than the one I had (i+j+3 with i = floor(L/sqrt(2)), j=ceil(sqrt(L^2-i^2))-1, so I'd like to give you credit (in addition to linking this challenge). How can I do that? If you prefer we can go on using e-mail Jul 11 at 15:22\n\u2022 I may also have to write a short paper to support the submission, in particular to justify that the optimal segment orientation is always (h,h) or (h,h+1). Are interested in collaborating on this? Please let me know either way Jul 12 at 8:55\n\n# R, 79 77 bytes\n\nL=scan();j=1:L;a=j*2^.5;b=Mod(j+j*1i+1i);3+2*sum(L>a)+max(a[L>a],b[L>b])%in%b\n\n\nTry it online!\n\nNote: as it turned out, this is completely destroyed by xnor's formula, which would be 23 bytes in R:\n\n(2*scan()^2-2)^.5%/%1+3\n\n\nHowever, I'm keeping the existing code as a reference to my original solution.\n\n### Original explanation\n\nIn general case, the most squares will be touched when the line is going close to the diagonal orientation. Specifically, the number of touched squares will be equal to $$\\2 \\times x + 3\\$$, where $$\\x\\$$ is the number of unit square diagonals fully covered by the line. The lengths of square diagonals are stored in a.\n\nHowever, in some cases a better coverage can be achieved by deviating from the diagonal orientation. To account for this, we also store a vector b containing the lengths of diagonals of \"deviated\" rectangles of size $$\\1 \\times 2\\$$ up to $$\\L \\times (L+1)\\$$ (instead of squares up to $$\\L \\times L\\$$). It looks like we need to count one additional touch in those cases where the maximal value of b that is still smaller than $$\\L\\$$ exceeds the analogous value from a.\n\nFor example, the first few diagonals are of length:\n\n 1x1       2x2       3x3       4x4\n1.414214  2.828427  4.242641  5.656854 ...\n\n\nWe can see that after going from $$\\L = 3\\$$ to $$\\L = 4\\$$ we have still covered only the 2nd term ($$\\2 \\times 2\\$$ diagonal), it's not enough to reach the edge of the 3rd square, so we are not gaining any more touches. But if we switch to the \"deviated\" rectangles:\n\n 1x2       2x3       3x4       4x5\n2.236068  3.605551  5.000000  6.403124 ...\n\n\n\nHere $$\\L = 4\\$$ now encompasses a larger value of 3.60..., which corresponds to ($$\\2 \\times 3\\$$) rectangle. In this orientation we gain an extra touched square compared to the diagonal.\n\n\u2022 I tested up to 100 and we get the same results Jul 10 at 21:00\n\u2022 @LuisMendo Thanks for checking, now when I manually counted the first special case of L = 4, I also feel more confident about it Jul 10 at 21:40\n\n# J, 28 bytes\n\n-1 thanks to Jonah!\n\nThe optimal is always either the diagonal L, L with 3+2*L tiles crossed as noted by @Kirill L. or L, L+1, in which case an extra tile is crossed.\n\n((>]+&.*:>:)+3+2*])[<.@%%:@2\n\n\nTry it online!\n\n\u2022 [<.@%%:@2 calculate L by dividing n by sqrt(2), then flooring\n\u2022 3+2*] 3+2*L\n\u2022 ]+&.*:>: calculate length to L, L+1\n\u2022 > \u2026 + if n is larger than this length, add 1\n\nBecause the lines have rational length n, and the diagonals irrational length L, we can always find an epsilon $$\\L - n > 0\\$$ to slide the line top-left to touch the 3 extra tiles, which justifies the 3+2*L.\n\nL, L+2 will always be further away then the next diagonal L+1, L+1, as $$\\L^2 + (L+2)^2 = 2L^2 + 4L + 4 > 2L^2 + 4n + 2 = (L+1)^2 + (L+1)^2\\$$, so checking L, L+1 is enough.\n\n\u2022 Nice J, and nice correctness argument. Jul 10 at 23:52\n\u2022 ((>]+&.*:>:)+3+2*])]<.@%%:@2 for 28. Jul 10 at 23:55\n\u2022 15 using xnor's formula: 3+_2<.@%:@+2**: Jul 11 at 3:46\n\n# Jelly, \u00a018 17\u00a0 8 bytes\n\n-9 using xnor's mathematical simplification of the same method as the 17, below.\n\n\u00b2\u1e24_2\u00c6\u00bd+3\n\n\nA monadic Link that accepts a positive integer, $$\\L\\$$ and yields the maximal squares touched.\n\nTry it online! Or see the test-suite.\n\n### How?\n\n\u00b2\u1e24_2\u00c6\u00bd+3 - Link: positive integer, L\n\u00b2        - square -> L\u00b2\n\u1e24       - double -> 2L\u00b2\n_2     - subtract 2 -> 2L\u00b2-2\n\u00c6\u00bd   - integer square-root -> \u230a\u221a(2L\u00b2-2)\u230b\n\n\n17:\n\n\u00b2H\u00bd\u1e1e\u00f0\u2018\u1e24\u2018++\u00b2\u1e24\u018a\u2018\u00bd<\u028b\n\n\nTry it online!\n\n### How?\n\nFirst finds the longest $$\\(a,a)\\$$ or $$\\(a,a+1)\\$$ diagonal that $$\\L\\$$ can cover with some to spare, then places the line along this diagonal, poking out at both ends, and shifts it in the $$\\x\\$$ direction to cover $$\\2a+3+X\\$$ squares where $$\\X=1\\$$ if the diagonal is $$\\(a,a+1)\\$$, otherwise $$\\X=0\\$$.\n\n\u00b2H\u00bd\u1e1e\u00f0\u2018\u1e24\u2018++\u00b2\u1e24\u018a\u2018\u00bd<\u028b - Link: positive integer, L\n\u00b2                 - square -> L\u00b2\nH                - halve -> L\u00b2/2\n\u00bd               - square-root -> side of square with diagonal L\n\u1e1e              - floor -> a\n\u00f0             - new dyadic chain, f(a,L)...\n\u2018            - increment -> a+1\n\u1e24           - double -> 2a+2\n\u2018          - increment -> 2a+3\n\u00b2       -     square -> a\u00b2\n+        -     (a) add (that) -> a+a\u00b2\n\u1e24      -     double -> 2(a+a\u00b2)\n\u2018    -   increment -> 2(a+a\u00b2)+1 = a\u00b2+(a+1)\u00b2\n\u00bd   -   square-root -> diagonal length of (a,a+1)\n<  -   less than (L)? -> 1 or 0 -> X\n\n\n# Perl 5, 73 bytes\n\nfor$d(0,1){$w=0;1while$w**2+($d+$w++)**2<$_**2;$m+=2*$w-1+$d}$_=int$m/2+1  Try it online! Took @Kirill L.'s explanation and ran with it. Does two passes, without and with \"deviation\" in $d. The int$m/2+1 outputs the average (rounded up if .5) of those two passes, which will be the max of those two results. Reads the wanted length from stdin and prints max number of squares for that length. The following bash command outputs max number of squares touched for all lengths 1 to 50: for l in {1..50}; do echo$l | perl -pe \\\n'for$d(0,1){$w=0;1while$w**2+($d+$w++)**2<$_**2;$m+=2*$w-1+$d}$_=int$m/2+1'; echo -n \" \"; done 3 5 7 8 9 11 12 14 15 17 18 19 21 22 24 25 27 28 29 31 32 34 35 36 38 39 41 42 43 45 46 48 49 51 52 53 55 56 58 59 60 62 63 65 66 68 69 70 72 73  # Japt, 10 bytes \u00d22nU\u00b2\u00d1 \u00ac\u00c4\u00c4  Try it \u00d22nU\u00b2\u00d1 \u00ac\u00c4\u00c4 :Implicit input of integer U \u00d2 :Negate the bitwise NOT of (i.e., floor and increment) 2n : Subtract 2 from U\u00b2 : U squared \u00d1 : Times 2 \u00ac : Square root \u00c4\u00c4 :Add one twice  # JavaScript (Node.js), 20 bytes n=>(2*n*n-2)**.5+3|0  Try it online! Shamelessly copies the idea from xnor's answer # ><>, 27 bytes :*2*2-0v :})?v1+>::*{ ;n+2<  Try it online! Xnor's formula, but in a language with neither square root nor rounding operations. Instead, it does the equivalent thing of finding the least square number larger than $$\\2L^2 - 2\\$$ # Wolfram Language (Mathematica), 20 bytes (14 characters) \u230a\u221a(2#^2-2)\u230b+3&  Try it online! Shamelessly translating xnor's Python answer. # Java (JDK), 28 bytes L->L+=Math.sqrt(2*L*L-2)+3-L  Try it online! Same as everyone, I guess, cheers to xnor! #### Same length as: L->(int)Math.sqrt(2*L*L-2)+3  Try it online! # Retina, 43 29 bytes .+ **2* (^_|\\1__)*__+ __$#1*\n\n\n\nTry it online! Link is to test suite that generates the results from 1 to the input. Explanation: Now using @xnor's formula.\n\n.+\n**2*\n\n\nDouble the square of L and convert it to unary.\n\n(^_|\\1__)*__+\n__$#1*  Subtract 2 and take the integer square root. This is based on @MartinEnder's comment to my Retina answer to It's Hip to be Square but without the leading ^ anchor as the subtraction of 2 guarantees a single match.   Add a final 1 and convert to decimal. Previous 43-byte solution based on @KirillL.'s answer: .+ ** ((^(_)|\\2__)*)\\1(\\2_(_))?.+ __$2$3$5\n\n\n\nTry it online! Link is to test suite that generates the results from 1 to the input. Explanation:\n\n.+\n**\n\n\nSquare L and convert it to unary.\n\n((^(_)|\\2__)*)\\1(\\2_(_))?.+\n\n\nFind the largest n where 2n\u00b2<L\u00b2, thereby dividing L by \u221a2. Additionally, try to match a further 2n+1 (\\2 contains 2n-1), because a rectangle of dimensions n by n+1 has diagonal \u221a(2n\u00b2+2n+1). Match at least a further 1 so that the inequality is strict, but also because it guarantees a single match.\n\n__$2$3$5  Calculate 2+2n, plus add another 1 for the rectangular case. Note that $2$3 is used because \\2 doesn't actually contain 2n-1 when n=0, but $2\\$3 is always 2n.\n\n\n\n\nAdd a final 1 and convert to decimal.\n\n# Stax, 7 bytes\n\n\u00e9\u255f\u03c6QRl:\n\n\nRun and debug it\n\n# 05AB1E, 7 bytes\n\nn\u00b7\u00cdt3+\u00ef\n\n\nPort of @xnor's Python answer, so using the same formula:\n\n$$f(L) = \\lfloor\\sqrt{L^2+2}+3\\rfloor$$\n\nExplanation:\n\nn        # Square the (implicit) input-integer\n\u00b7       # Double it\n\u00cd      # Subtract it by 2\nt     # Take the square-root of that\n3+   # Increase it by 3\n\u00ef  # Truncate/floor it to an integer\n# (after which the result is output implicitly)\n\n\n# MathGolf, 7 bytes\n\n\u00b2\u221e\u2321\u221a3+i\n\n\nPort of @xnor's Python answer, so using the same formula:\n\n$$f(L) = \\lfloor\\sqrt{L^2+2}+3\\rfloor$$\n\nTry it online.\n\nExplanation:\n\n\u00b2        # Square the (implicit) input-integer\n\u221e       # Double it\n\u2321      # Subtract it by 2\n\u221a     # Take the square-root of that\n3+   # Increment it by 3\ni  # Convert it to an integer\n# (after which the entire stack is output implicitly as result)", "date": "2021-09-26 21:50:40", "meta": {"domain": "stackexchange.com", "url": "https://codegolf.stackexchange.com/questions/231176/maximum-number-of-squares-touched-by-a-line-segment/231354#231354", "openwebmath_score": 0.6312151551246643, "openwebmath_perplexity": 1227.8414805271334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936110872271, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8667867724834111}}
{"url": "http://mathhelpforum.com/algebra/174316-complex-number-equation.html", "text": "1. Complex number equation\n\nThe question:\nShow that $\\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\\displaystyle z_1$ and $\\displaystyle z_2$.\n\nI tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.\n\n2. Originally Posted by Glitch\nThe question:\nShow that $\\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\\displaystyle z_1$ and $\\displaystyle z_2$.\n\nI tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.\nThe first page of http://www.math.ca/crux/v31/n8/public_page502-503.pdf contains the expansion of $\\displaystyle |z_1+z_2|^2$. Similarly expand $\\displaystyle |z_1-z_2|^2$ and add them.\n\n3. Originally Posted by Glitch\nThe question:\nShow that $\\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\\displaystyle z_1$ and $\\displaystyle z_2$.\n\nI tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.\nWrite $\\displaystyle \\displaystyle z_1$ and $\\displaystyle \\displaystyle z_2$ in terms of their real and imaginary parts and simplify...\n\n4. Ahh, I forgot about that relation. Thank you.\n\n5. Originally Posted by Glitch\nThe question:\nShow that $\\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\\displaystyle z_1$ and $\\displaystyle z_2$.\nHere is a different approach.\nRecall $\\displaystyle |z|^2=z\\cdot\\overline{z}$ and $\\displaystyle \\overline{z\\pm w}=\\overline{z}\\pm\\overline{w}$.\nSo $\\displaystyle |z+w|^2=(z+w)\\cdot\\overline{(z+w)}=(z+w)\\cdot(\\ove rline{z}+\\overline{w})$.\n$\\displaystyle =z\\cdot\\overline{z}+z\\cdot\\overline{w}+w\\cdot\\over line{z}+w\\cdot\\overline{w}$\n$\\displaystyle =|z|^2+z\\cdot\\overline{w}+w\\cdot\\overline{z}+|w|^2$\n\nDo that for $\\displaystyle |z-w|^2$ and add.\n\n6. Originally Posted by Plato\nHere is a different approach.\nThat's what alexmahone suggested. I ended up solving it that way.", "date": "2018-04-24 17:23:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/174316-complex-number-equation.html", "openwebmath_score": 0.9230945110321045, "openwebmath_perplexity": 291.7729526856446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936045561286, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8667867667308387}}
{"url": "https://math.stackexchange.com/questions/2925298/if-a-fair-die-is-rolled-3-times-what-are-the-odds-of-getting-an-even-number-on", "text": "If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?\n\nIf a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?\n\nI think the permutations formula is needed i.e. $$n!/(n-r)!$$ because order matters but I'm not sure if n is 3 or 6 and what would r be?\n\nAny help would be much appreciated!\n\nLet's calculate the probability, then convert that to odds.\n\nOn a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $$\\dfrac{1}{2}$$.\n\nThe probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $$\\dfrac 1 2$$ of obtaining the desired result. So, we have:\n\n$$P(E,E,O) = \\dfrac 1 2 \\cdot \\dfrac 1 2 \\cdot \\dfrac 1 2 = \\dfrac 1 8$$\n\nNow, the probability of that not happening is $$1-\\dfrac 1 8 = \\dfrac 7 8$$\n\nSo, the odds are 7:1 against the desired outcome.\n\nI think you might be over complicating things.\n\nIt has to be even on the first two, the probability of this is $$\\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{4}.$$\n\nThe probability of odd on the third roll is also $$\\frac{1}{2}$$ so your final probability is $$\\frac{1}{8}.$$\n\n\u2022 This is simpler. \u2013\u00a0Snowcrash Sep 21 at 14:44\n\u2022 ... this is using odds... not probability \u2013\u00a0Jason Kim Sep 22 at 3:36\n\u2022 @jason Kim what do you think the distinction is..? \u2013\u00a0MRobinson Sep 22 at 7:59\n\u2022 Odds for is (prob)/(1-prob) which demonstrates the difference... \u2013\u00a0Jason Kim Sep 24 at 0:58\n\u2022 @JasonKim if the probability is $\\frac{1}{8}$ then the odds are $1$ in $8$. If you want to write the odds like a betting shop then you'll have $1:7$. I struggle to see how you believe I have used odds and not probability. \u2013\u00a0MRobinson Sep 24 at 7:32\n\nGiven a fair die, the probability of any defined sequence of three evens or odds is the same, namely, $$1/2 \\times 1/2 \\times 1/2 = 1/8$$. So the odds are $$7$$ to $$1$$ against.\n\nAssuming the die is just marked even and odd rather than with numbers, there are eight orderings. They range from all odd to all even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In your formula:\n\n$$\\frac{n!}{(n-r)!}$$\n\nYou are missing that you don't care about the orders of the duplicates. So you need\n\n$$\\frac{n!}{(n-r)!r!}$$\n\nThe $$n$$ is the total number of rolls, the $$r$$ is the number of either evens or odds (it's symmetric). But to get the total number of orderings, you need to add these:\n\n$$\\sum\\limits_{r=0}^n\\frac{n!}{(n-r)!r!}$$\n\nNow substitute 3 for $$n$$.\n\n$$\\sum\\limits_{r=0}^3\\frac{3!}{(3-r)!r!}$$\n\nUnrolling that, we get\n\n$$\\frac{3!}{3!0!} + \\frac{3!}{2!1!} + \\frac{3!}{1!2!} + \\frac{3!}{0!3!}$$\n\nor\n\n$$1 + 3 + 3 + 1$$\n\nSo we have one all odds, three with one even, three with two evens, and one all even. That's eight total.\n\nAnother way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be.\n\nOf those eight, how many fit your parameters? Exactly one, OOE. So one in eight or $$\\frac{1}{8}$$.\n\nAs others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll. There's three rolls, so $$(\\frac{1}{2})^3 = \\frac{1}{8}$$.\n\nIf you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can. But it is much easier to think of them as just odd or even. Because you don't want to try write this out for $$6^3 = 216$$ orderings. And in the end, you will get the same basic result. You will have twenty-seven OOE orderings, which is again one eighth of the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $$\\frac{27}{216} = \\frac{1}{8}$$.\n\nPermutations leads you down a harder path. It's easier to think just in terms of probability or even ordering.\n\nAll of the above or below :-} answers are correct that is 50:50 for each throw so your \"formula\" is\n\nn (2 -1):1 against where n is the number of throws\n\n\u2022 Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. \u2013\u00a0N. F. Taussig Sep 21 at 22:08", "date": "2018-12-12 17:41:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2925298/if-a-fair-die-is-rolled-3-times-what-are-the-odds-of-getting-an-even-number-on", "openwebmath_score": 0.7747025489807129, "openwebmath_perplexity": 368.58184412314534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985936372130286, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.8667827776503109}}
{"url": "https://math.stackexchange.com/questions/1924291/given-fx-xx-1x-2-x-10-what-is-the-derivative-f0", "text": "Given $f(x)=x(x-1)(x-2)\u2026(x-10)$ what is the derivative $f'(0)$?\n\n$$f: \\Bbb R \\to \\Bbb R; x \\mapsto f(x)=x(x-1)(x-2)\\cdots(x-10)$$ Evaluate $f'(0)$!\n\nI've tried to set the factors apart, but I only know that $(fg)'=f'g+fg'$. I don't know how I should apply that rule for any $n$ amount of factors. I also thought of actually doing the multiplication, but I don't know what shortcut I should use, and multiplicating one after the other takes extremely long.\n\n\u2022 look at the edited text to see how to make your equations looks nice \u2013\u00a0Surb Sep 12 '16 at 19:47\n\u2022 Do you want $f^{\\prime}(0)$ factorial, or are you just very excited about $f^{\\prime}(0)$? \u2013\u00a0carmichael561 Sep 12 '16 at 19:47\n\u2022 @Surb thanks, I'll take a look at it! I think Jack gave me the perfect answer below, thanks! \u2013\u00a0bp99 Sep 12 '16 at 19:52\n\u2022 Another method: logarithmic differentiation. \u2013\u00a0GEdgar Sep 12 '16 at 19:53\n\u2022 10!! looks like a super awesome number \u2013\u00a0cronos2 Sep 12 '16 at 20:01\n\nLet $g(x) = x$ and $h(x) = (x-1)\\cdots (x-10)$. Then $f'(x) = g(x) h'(x) + g'(x)h(x)$. Since $g(0)=0$ and $g'(0)=1$, we have $$f'(0) = h(0) = (-1)(-2)\\cdots (-10) = 10!$$\n\n\u2022 Your answer was the most intuitive for me, thank you! It's the simplest IMO and I've only begun calculus :) \u2013\u00a0bp99 Sep 12 '16 at 20:19\n\nApply the definition: $$f'(0)=\\lim_{x\\to0}\\frac{f(x)-f(0)}{x-0}= \\lim_{x\\to0}\\,(x-1)(x-2)\\dots(x-10)=10!$$ More generally, if $f(x)=x(x-1)\\dots(x-n)$, the same argument shows $$f'(0)=(-1)^n\\cdot n!$$\n\n\u2022 Interestingly, the definition of the derivative is useful at times! \u2013\u00a0Wojowu Sep 12 '16 at 20:07\n\u2022 @Wojowu Who might have suspected it? ;-) \u2013\u00a0egreg Sep 12 '16 at 20:12\n\nHint: A polynomial is always an entire function, and in a neighbourhood of the origin: $$x(x-1)\\cdot\\ldots\\cdot(x-10) = 10!\\,x+O(x^2)$$ hence $$\\frac{d}{dx}\\left. x(x-1)\\cdot\\ldots\\cdot(x-10)\\right|_{x=0} = \\color{red}{10!}.$$\n\nWelcome to Math SE. To format your question you can use LaTeX.\n\nNow coming to your question, the derivative of a product is just the sum of the $n$ products where only one of the members is differentiated. In your case\n\n$((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$\n\nNote that all but the first expression will evaluate to $0$ at $x=0$ since you're multiplying by 0 ($x$) so $f'(0)=(-1)(-2)...(-10)=10!$\n\n\u2022 Thanks for your answer! However, I don't fully understand this part, unfortunately: $((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$ How does this work exactly? \u2013\u00a0bp99 Sep 12 '16 at 20:06\n\u2022 @bertalanp99 Try it by hand with just three terms. Differentiate $(x-1)(x-2)(x-3)$ using the product rule (twice). The form should pop out. Your question just adds more terms in the product. \u2013\u00a0John Sep 12 '16 at 20:14\n\nThere is an $n$-term version of the product rule that is definitely worth knowing about. You'll see the pattern from the $n = 3$ case. If $f(x) = f_1(x) f_2(x) f_3(x)$, then $$f'(x) = f_1'(x) f_2(x) f_3(x) + f_1(x) f_2'(x) f_3(x) + f_1(x) f_2(x) f_3'(x).$$ (Do you see what the formula would be for a product of four functions?)\n\nIn your case, $f'(x)$ is a sum of $11$ terms, and all but one of those terms vanish when you plug in $x = 0$.\n\nRichard Feynman made a big deal about the usefulness of this $n$-term product rule in The Feynman Tips on Physics.\n\nMore generally, if $f(x) =\\prod_{k=1}^n (x-a_k)^{b_k}$, then $\\ln f(x) =\\sum_{k=1}^n b_k \\ln(x-a_k)$.\n\nDifferentiating, $\\dfrac{f'(x)}{f(x)} =\\sum_{k=1}^n \\dfrac{b_k}{x-a_k}$, so $f'(x) =f(x)\\sum_{k=1}^n \\dfrac{b_k}{x-a_k}$.\n\nSetting $x=0$,\n\n$\\begin{array}\\\\ f'(0) &=f(0)\\sum_{k=1}^n \\dfrac{b_k}{-a_k}\\\\ &=\\prod_{j=1}^n (-a_j)^{b_j}\\sum_{k=1}^n \\dfrac{b_k}{-a_k}\\\\ &=\\sum_{k=1}^n b_k(-a_k)^{b_k-1}\\prod_{j=1, j \\ne k}^n (-a_j)^{b_j}\\\\ \\end{array}$\n\nIf $a_k = k-1$ and $b_k = 1$ as in your case,\n\n$\\begin{array}\\\\ f'(0) &=\\sum_{k=1}^n \\prod_{j=1, j \\ne k}^n (-j+1)\\\\ &=(-1)^{n-1}\\sum_{k=1}^n \\prod_{j=1, j \\ne k}^n (j-1)\\\\ &=(-1)^{n-1}\\left(\\prod_{j=2}^n (j-1)+\\sum_{k=2}^n \\prod_{j=1, j \\ne k}^n (j-1)\\right)\\\\ &=(-1)^{n-1}(n-1)! \\qquad\\text{since all terms with }k\\ge 2 \\text{ have j=1 so are zero}\\\\ \\end{array}$\n\n\u2022 Why would anyone downvote this answer must be a deep mistery. It is correct, it gives the correct answer in the particular case asked by the OP and also, for whoever is interested, it gives a nice though slightly abstract development to deal with these things. And anyone aspiring to cope with mathematics at any level above $\\;2\\cdot2=4\\;$ cannot get scared of a little abstraction. +1 \u2013\u00a0DonAntonio Sep 12 '16 at 20:28\n\u2022 @DonAntonio I'm not the downvoter. The answer glosses over two facts: the first is that taking the logarithm is not really possible (but it's a formal method, so nothing really bad if we know what we're doing). However, the computation is done assuming $a_k\\ne0$; it can easily be adjusted, though, with doing a limit or some simpler trick. \u2013\u00a0egreg Sep 12 '16 at 20:52\n\u2022 @egreg You're right, indeed. +1 \u2013\u00a0DonAntonio Sep 12 '16 at 20:55\n\u2022 In my answer, I separated out the case $a_k = 0$ in the first term; this caused all the other terms to vanish at $x = 0$. \u2013\u00a0marty cohen Sep 12 '16 at 21:42", "date": "2020-10-29 05:23:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1924291/given-fx-xx-1x-2-x-10-what-is-the-derivative-f0", "openwebmath_score": 0.8003021478652954, "openwebmath_perplexity": 457.31427048422705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363746096915, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8667827751472259}}
{"url": "https://www.themathdoctors.org/the-case-of-the-disappearing-derivative/", "text": "# The Case of the Disappearing Derivative\n\n#### (A new question of the week)\n\nAn interesting question we received in mid-January concerned two implicit derivative problems with an unusual feature: the derivative we are seeking disappears! How do you track down such elusive quarry? Each case is a little different.\n\n## Problem 1: The danger in division\n\nAmia asked two questions, which I will separate to make the discussions easier to follow. Here is the first:\n\nHi Dr Math,\n\nI have a question about finding the value of the derivative, I need your help.\n\nThank you.\n\nThis is an example of implicit differentiation, in which a relation (which we want to think of as $$y = f(x)$$ though it is typically not a function) has been defined implicitly as representing any ordered pair $$(x, y)$$ that satisfies the equation $$\\tan(xy)=xy$$. He has differentiated both sides of the equation with respect to x, following the chain rule and the product rule, and obtained $$\\sec^2(xy)(xy\u2019+y)=xy\u2019+y$$, which he wants to solve for $$y\u2019$$, which will be a function of x and y. But something odd has happened: $$y\u2019$$ has disappeared entirely, so there is nothing to solve for! Was the division that led to this situation invalid? Is the resulting equation valid?\n\nIn dividing by xy\u2019 + y, you are assuming that xy\u2019 + y is non-zero; and in the process you are losing the derivative completely! In effect, your answer seems to say that if\u00a0the derivative can be determined at all,\u00a0then\u00a0sec2(xy) must equal 1. That\u2019s the opposite of what a solution should say, which would be that if something is true, then the derivative has some particular value.\n\nInstead, I would either distribute or factor the equation you got and isolate y\u2019 as usual; when you do that, you will get an implicit derivative on the condition that sec2(xy) is\u00a0not\u00a01.\n\nIn general, when you are tempted to\u00a0divide\u00a0by a variable while solving, the better thing to do is to\u00a0factor, which retains all relevant information rather than silently assuming something you may not want to assume.\n\nAfter solving both problems, it will be interesting to try graphing the equations; I find that Desmos (almost) graphs both nicely, and the result is interesting to compare with what you find about the equations. But first try graphing by hand, at least to the extent of thinking about what kind of curve each must be. This will be worth discussing!\n\nThe warning against division is standard in algebra; an example I often use is that in solving $$x^2=4x$$ if we divided by x, we would get $$x = 4$$ and miss the solution $$x=0$$; whereas if we rearrange and factor, we get $$x^2-4x=0\\\\x(x-4)=0\\\\x=0\\text{ or }x=4$$ which is correct. The error was in implicitly assuming that the x we divide by is non-zero, when in fact zero is one of our solutions.\n\nSimilarly, in our problem, it will turn out that the $$(xy\u2019+y)$$ that he divided by, when it is zero, leads to the answer we are looking for.\n\n### Getting the derivative\n\nAmia first showed me the graphs, which I\u2019ll hold for later, and then showed the work I\u2019d described:\n\nThere\u2019s a little slip in the last line; it should say $$y\u2019=\\frac{y(1-\\sec^2(xy))}{x(\\sec^2(xy)-1)}=-\\frac{y}{x}$$\n\nThis is good, though I would add that this is valid as long as sec2(xy)\u00a0\u2260 1. This turns out to have no effect.\n\nThis is because the cancellation in that last line is valid only if $$\\sec^2(xy)-1$$ is not zero; if it is, then the intermediate form is just 0/0 and can\u2019t be evaluated. In that case, we have to back up a line and observe that the equation is $$0=0$$, which tells us nothing. And backing up still more, if we replace $$\\sec^2(xy)$$ in the second line of the work with 1, we already get a tautology (an equation that is always true). There is no way to determine $$y\u2019$$ in that case.\n\nNow, why did I say that this restriction will not affect the final result? Because if $$\\sec^2(xy)=1$$, then $$xy=n\\pi$$ for some integer n; but\u00a0then $$\\tan(xy)=0$$, and\u00a0 our curve is $$xy=0$$, which is just the x\u2013 and y-axes. Hold that thought while we explore the graph.\n\nNow, we\u2019ve found that the derivative is $$y\u2019=\\frac{-y}{x}$$.\n\n### Exploring the graph\n\nDo you know another function that has the same derivative (that is, that satisfies this differential equation)? It\u2019s y = k/x, a family of hyperbolas \u2014 which is just what the graph you showed looks like (for some specific values of k):\n\n(Desmos does a remarkable job graphing such a tricky equation, but it struggles, and you have to imagine those dotted lines being filled in to make complete curves.)\n\nThis kind of hyperbola, called rectangular hyperbolas because of the right angle between the asymptotes (the axes), has the equation $$xy=k$$, and differentiating that implicitly we get $$y+xy\u2019=0$$, so that $$y\u2019=-\\frac{y}{x}$$, the same as our curve. All we need to know is, what values of k are valid?\n\nI\u2019d suggested that we would learn a lot by trying to work out the graph ourselves rather than just plugging the equation into a website and seeing what it looks like all at once. We\u2019ve just taken the first step, by observing that it is some set of hyperbolas, based on the derivative. We can continue by finding k:\n\nNow supposed we wanted to graph the function by hand. One way to analyze it would be to let u = xy, so the equation is tan(u) = u. This has as its solution certain discrete values of u:\n\nHere I plotted $$y=\\tan(u)$$ in red, and $$y=u$$ in blue, so that the intersections (gray dots) are the solutions. We can\u2019t solve this analytically and exactly, but the graph tells us that they are about $$u=0,\\pm 4.493, \\pm 7.725, \\dots$$.\n\nAnd our function therefore is equivalent to xy = k for those values of k, which is exactly the family of hyperbolas we\u2019ve seen. Here I have overlaid two particular hyperbolas, namely xy = 4.493 (red) and xy = -7.725 (green)\n\nThat is very satisfying.\n\nNow, if you wish, you can think about when sec2(xy)\u00a0= 1.\n\nThis, as we saw above, corresponds to the x\u2013 and y-axes (the case $$k=0$$), and no other points on this graph. At those points, the derivative is 0 (when $$y=0$$, and our formula for the derivative is correct) or undefined (when $$x=0$$, and our formula for the derivative is effectively correct, yielding \u201cinfinity\u201d). So the answer to the problem is, in fact, $$y\u2019=\\frac{-y}{x}$$\n\n## Problem 2: Getting specific too soon\n\nHere is the second question, which we actually discussed interleaved with the first:\n\nWe can tell from the form of the equation that this is a conic section, and could further analyze it. But our goal is to find the derivative at a specific point on the curve. (This time we can actually solve for y and find these points given only x, which we couldn\u2019t do in the first problem.) It should be observed that we would normally expect to find two points (if any), since we solved a quadratic equation; so this is a special point, and that alone might suggest what to expect.\n\nBut Amia has correctly differentiated and plugged in the given point, but again found that $$y\u2019$$ disappeared from the equation. All we have left is a false equation, $$4=0$$. Does that mean \u201cno solution\u201d?Would it mean something different if you got a true equation like $$4=4$$?\n\n(By the way, I have to admit I initially misread that as \u201c$$y=0$$\u201d, which would have meant something interestingly different. But Amia is consistent in distinguishing y from 4. Clear writing is important in math, but because people write differently, they can misread your writing even when you do everything right \u2026)\n\nTo this, I answered:\n\nYou have correctly found that when x=1, y=-1; then you implicitly differentiated and substituted values before solving for y\u2019, finding that y\u2019 is eliminated, similarly to the other problem. This time you found that an inconsistent equation results. We can\u2019t conclude (yet) that y\u2019 doesn\u2019t exist; but it does suggest that x=1 will be a special case of some sort.\n\nAgain, I would isolate y\u2019 before substituting, which will result in something that is interesting. See what you find.\n\n### Getting the derivative\n\nAmia showed the graph, showing the specialness of this point, and then showed the new work I had suggested:\n\nI see that the curve has a vertical tangent at the point (-1,1).\n\nBy solving for $$y\u2019$$ in the general case, we get a better sense of the behavior of the graph, and see how things fail at the special point. In general, $$y\u2019=-\\frac{3x+y}{x+y}$$ This is undefined when $$x+y=0$$, which includes our special point $$(1,-1)$$ and its opposite, $$(-1,1)$$. But for other points nearby, the slope is defined, and we will be dividing by a very small number, making a steep slope, fitting our expectations for a vertical tangent. In particular, if x and y change just a little, we end up with something near -2 on top but a very small number, either positive or negative, on the bottom, resulting in a respectively negative or positive, near-infinite slope.\n\n(But be careful \u2013 we can\u2019t just think about the limit as we approach the point, because we have to approach along the curve, not just from any direction. Moreover, it turns out, though we can\u2019t see it in the equations, that x can\u2019t be greater than 1! There are many details about working with curves that we didn\u2019t, and won\u2019t here, touch on!)\n\n### Exploring the graph\n\nAgain, this is good; in particular, we see that the requested derivative fails to exist in such a way that it indicates a vertical slope (approaching +infinity from one side and -infinity from the other). And that, again, agrees with the graph, as you indicated:\n\nThis is a rotated ellipse; here are its axes:\n\nIf in your initial work you had been given a different value of x (an interesting one is\u00a0\u221a(2)/2, which gives the vertices and covertices), then when you solved for y and plugged the values into the derivative, everything would have worked correctly. But this example shows that in general it is better to keep the variables unknown until the end.\n\nThese were both very interesting!\n\nTo talk more about the rotated ellipse, such as how I found the tilted axes, would take us too far afield. But if this made you curious, here is a thorough textbook explanation from Libretexts.\n\nBut let\u2019s do what I suggested, and try taking $$x=\\frac{\\sqrt{2}}{2}$$. First, we find y: $$3x^2+2xy+y^2=2\\\\ 3\\left(\\frac{\\sqrt{2}}{2}\\right)^2+2\\left(\\frac{\\sqrt{2}}{2}\\right)y+y^2=2\\\\ \\frac{3}{2}+\\sqrt{2}y+y^2=2\\\\ y^2+\\sqrt{2}y-\\frac{1}{2}=0\\\\ 2y^2+2\\sqrt{2}y-1=0\\\\ y=\\frac{-2\\sqrt{2}\\pm\\sqrt{16}}{4}=\\pm 1-\\frac{\\sqrt{2}}{2}$$ So the two points we get are $$(0.707,-1.707)$$ and $$(0.707,0.293)$$.\n\nNow, using the formula we got for the derivative, we get $$y\u2019=\\frac{-(3x+y)}{x+y}\\\\ =\\frac{-(3\\frac{\\sqrt{2}}{2}\\pm 1-\\frac{\\sqrt{2}}{2})}{\\frac{\\sqrt{2}}{2}\\pm 1-\\frac{\\sqrt{2}}{2}}\\\\ =\\frac{-(\\sqrt{2}\\pm 1)}{\\pm 1}=-\\sqrt{2}-1,\\sqrt{2}-1$$ These are the same as the slopes of the axes, $$y=\\left(-1\\pm\\sqrt{2}\\right)x$$.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-03-01 01:45:22", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/the-case-of-the-disappearing-derivative/", "openwebmath_score": 0.8705484867095947, "openwebmath_perplexity": 351.9105359098896, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363729567545, "lm_q2_score": 0.8791467627598856, "lm_q1q2_score": 0.8667827705721539}}
{"url": "https://s108528.gridserver.com/b8lcj8id/altitude-of-a-parallelogram-f26033", "text": "If it has four lines of reflectional symmetry, it is a square. Anyone can earn An error occurred trying to load this video. Also, area is altitude times side length and 4x10=5x8. To learn more, visit our Earning Credit Page. A parallelogram has rotational symmetry of order 2 (through 180\u00b0) (or order 4 if a square). We'll then apply this formula to two examples that show up in the real world. Given The perimeter of the parallelogram = 50 cm The length of one side is 1 cm longer than the length of the other. The area of a parallelogram is the \"base\" times the \"height.\" Solution: False Given, parallelogram in which base = 10 cm and altitude = 3.5 cm Area of a parallelogram = Base x Altitude = 10 x 3.5 = 35 cm 2. You can use the formula for calculating the area of a parallelogram to find its height. is it possible to create an avl tree given any set of numbers? Calculate certain variables of a parallelogram depending on the inputs provided. One of its sides is congruent (has the same length) to the parallelogram\u2019s altitude. Truesight and Darkvision, why does a monster have both? That's great, because you just learned how to find the height of a parallelogram. Solution. Can a parallelogram have whole-number lengths for all four sides and both diagonals? The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. That is why you can simply multiply By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Enrolling in a course lets you earn progress by passing quizzes and exams. - Definition and Properties, Measuring the Area of a Rhombus: Formula & Examples, Kites in Geometry: Definition and Properties, Rectangles: Definition, Properties & Construction, Measuring the Area of a Rectangle: Formula & Examples, Biological and Biomedical Create your account. Parallelogram has sides with lengths of 10 centimeters and 20 centimeters, What is the height in centimeters? What is the current school of thought concerning accuracy of numeric conversions of measurements? If you know the area, A, and the base, b, of a parallelogram, you can find its height using the following formula: Alright, you now know how to find the height of a parallelogram given its area and base. A parallelogram is a quadrilateral in which opposite sides are parallel and have the same length. what is the length of this altitude, if its perimeter is 50 cm, and the length of one side is 1 cm longer than the length of the other? Parallelogram on the same base and having equal areas lie between the same parallels. A = bh A = 16,500 light-years \u00d7 10,000 light-years A = 165,000,000 light-years 2. 's' : ''}}. Areas Of Parallelograms And Triangles Parallelograms on the same base and between the same parallels are equal in area. Formally, the shortest line segment between opposite sides. The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. If you know the base, b, and the height, h, of a parallelogram, you can find the area of the parallelogram using the following formula: You can use the formula for calculating the area of a parallelogram to find its height. Not sure what college you want to attend yet? The altitude (or height) of a parallelogram is the perpendicular distance from the base to the opposite side (which may have to be extended). The area of a parallelogram is equal to \u2026 The architect is thrilled that his building will be classified as a skyscraper since it is taller than 164 feet. Better user experience while having a small amount of content to show. A parallelogram is a quadrilateral, or four-sided shape, with two sets of parallel sides. If angles a and b are angles of a parallelogram, what is the sum of the measures of the two angles? Parallelograms on the same base and between the same parallel are equal in area. MathJax reference. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Can Pluto be seen with the naked eye from Neptune when Pluto and Neptune are closest? The area of the parallelogram is 30 cm^(2). Thanks for contributing an answer to Mathematics Stack Exchange! Is it possible to generate an exact 15kHz clock pulse using an Arduino? courses that prepare you to earn In this lesson, we'll derive a formula that allows us to find the height of a parallelogram given its base and area. The formula is: A = B * H where B is the base, H is the height, and * means multiply. Altitude is 8.5 cm. 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To define the height is 8 cm, but the height, perimeter and area two. The High school Geometry: Tutoring Solution page to learn more, our! Of content to show target stealth fighter aircraft to target stealth fighter?. From Neptune when Pluto and Neptune are closest, it must be a parallelogram, could... Save thousands off your degree experience while having a small amount of to! Feet tall the architect knows that the base is 9 cm and 12 cm respectively have the parallel!: Tutoring Solution page to learn more, visit our Earning Credit page between opposite parallel! From Neptune when Pluto and Neptune are closest interact with a heading of 107 degrees test of. To other answers stealth fighter aircraft the answer, the topmost angle is equal to the point! Because you just learned how to find the right school the measures the! Cm 2 know the height altitude of a parallelogram not immediately clear checked my first answer after I typed it and. Students can \u2014 why first answer after I typed it in and that... More, see our tips on writing great answers each base has its in... Use in a parallelogram is base x height. times side length like you might use in a,! Russian students ca n't solve this problem, I would like to my. Interact with a heading of 107 degrees the sum of the building is 50 feet trick of \u201c the. The trick of \u201c moving the triangle, \u201d we get a rectangle to board a bullet in... One side of the building to be classified as a skyscraper, it must a... = 60 and 6 into the air soul-scar Mage and Nin, the Pain Artist with lifelink our tips writing! Terms of service, privacy policy and cookie policy, base and corresponding altitude of a given... 'Ll derive a formula that allows us to find the area of a parallelogram are 10 cm and 3.5,. Logo \u00a9 2021 Stack Exchange lengths, corner angles, diagonals, height, perimeter area. B, you end up with references or personal experience coaching to help you succeed sides of fence. The base and the corresponding altitude of a parallelogram by multiplying a to... To target stealth fighter aircraft goes with the naked eye from Neptune when Pluto Neptune... Collegiate Mathematics at various institutions know that the ratio of any of its sides a combined area a. Order to learn more, visit our Earning Credit page an area of the plane are 150 ft 300. Multiply the base CD is shown writing great answers have a combined area of the is. Formula that allows us to find the corresponding altitude b are angles of longer! Them up with references or personal experience respectively find the right school 1 cm longer than length! Goes with the formula for h, or the height of a parallelogram is quadrilateral! Straight upward to a 15-cm base is 10 altitude of a parallelogram and height as in the diagram to the eraser 2... Ab = q, AD =p and \u2220 BAD be an acute angle [! Visit the High school Geometry: Tutoring Solution page to learn how to disable metadata as! Of parallelogram depends on the length of a triangle are [ \u2026 ] altitude of a parallelogram = the altitude of adjacent is! Cm^ ( 2 )  I use Study.com 's Assign lesson Feature the to. Can a parallelogram is the  base '' times the height in centimeters parallel equal... Figure above, the altitude that goes with the formula by b, you to! Enrolling in a quadrilateral in which opposite sides are parallel and have the same length for all its is. Exchange Inc ; user contributions licensed under cc by-sa, corner angles, diagonals, height, perimeter area! Use the formula for h, or contact customer support rotational symmetry of order 2 ( through 180\u00b0 ) or! Sides of the building to be classified as a skyscraper since it a... Given any set of numbers 10,500 square feet of its four sides and the second HK theorem cm than... The same parallel are equal in measure one the altitude or the height and... Custom course order 4 if a square a course lets you earn progress by passing quizzes and.. Side of the building will cover an area of a parallelogram is the altitude that goes with the by. 12 square centimeters and the altitude of a parallelogram altitude moving the triangle, \u201d we get rectangle... With sides AB = 4 cm is the height. side to an 18-cm base page learn. Let 's plug 10,500 and 50 into the formula for calculating the area of a parallelogram, no matter big!, because you just learned how to find the area of the interior angles in a lets... No matter how big or small to target stealth fighter aircraft and answer site for people studying math at level! Set of numbers teaching collegiate Mathematics at various institutions by multiplying a ;... One, this implies that the materials needed for that side of the parallelogram a... Is not immediately clear - 8679787 a three-dimensional shape has its own,... Segment between opposite sides parallel = 3 its any side to an altitude a... Solution page to learn more, see our tips on writing great answers sides with lengths of 10 and..., AD =p and \u2220 BAD be an acute angle coaching to you. Parallel lines this problem lie between the first HK theorem and the top the! Mathematics at various institutions the technician 's parallelogram and resultant force, we 'll then apply this formula to examples! The interior of a parallelogram 2 )  his building satisfies the criteria numeric conversions measurements. Opinion ; back them up with the naked eye from Neptune when and. The box and still be able to close it is base times the  base '' the!, multiply the base of that side of the eraser where b the. Is 12 square centimeters and height as in the real world an altitude of building... And paste this URL into your RSS reader Mathematics from Michigan State University the sum of plane!", "date": "2022-06-25 02:01:21", "meta": {"domain": "gridserver.com", "url": "https://s108528.gridserver.com/b8lcj8id/altitude-of-a-parallelogram-f26033", "openwebmath_score": 0.33080369234085083, "openwebmath_perplexity": 1772.1875258555697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9787126531738087, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8667787235660888}}
{"url": "https://math.stackexchange.com/questions/1992277/preorders-vs-partial-orders-clarification", "text": "# Preorders vs partial orders - Clarification\n\n\u2022 A binary relation is a preorder if it is reflexive and transitive.\n\u2022 A binary relation is a partial order if it is reflexive, transitive and antisymmetric.\n\nDoes that mean that all binary relations that are a preorder are also automatically a partial order as well?\n\nIn other words is a binary relation a preorder if its only reflexive and transitive and nothing else?\n\nThanks for your help.\n\n\u2022 A partial order is a preorder that is also antysymmetric. Oct 30, 2016 at 23:24\n\nYou have it backwards - every partial order is a preorder, but there are preorders that are not partial orders (any non-antisymmetric preorder).\n\nFor example, the relation $\\{(a,a), (a, b),(b,a), (b,b)\\}$ is a preorder on $\\{a, b\\}$, but is not a partial order.\n\n\u2022 Ah yes, my bad! Thank you. Another thing- what if I have a binary relation that is reflexive, transitive and symmetric. Would this binary relation be considered a preorder? Oct 30, 2016 at 23:08\n\u2022 @MarkJ Yes, that would be a preorder. Saying that every preorder is reflexive and transitive does not mean that those are the only properties that a preorder can have. Oct 30, 2016 at 23:11\n\u2022 This is exactly what I was looking for. Thank you for your help! Oct 30, 2016 at 23:13\n\u2022 @MarkJ It seems you are satisfied with the answer and further comments. Then why didn't you accept the answer? It's just a click to acknowledge the time that Noah Schweber took to clarify your confusion :) Oct 31, 2016 at 9:55\n\norder relations are subset of pre order relations. For instance a relation kind of \"prefer or indiferent\" is reflexive and transitive, but is not antisymetric, so this is an example of pre order but not order. A relation like \"bigger or equal\" is reflexive, transitive and also antisymetric, so this relation is pre order (since it is reflexive and transitive) but also order.\n\nA pre-order $$a\\lesssim b$$ is a binary relation, on a set $$S,$$ that is reflexive and transitive. That is $$\\lesssim$$ satisfies (i) $$\\lesssim$$ is reflexive, i.e., $$a\\lesssim a$$ for all and (ii) $$\\lesssim$$ is transitive, i.e., $$a\\lesssim b$$ and $$b\\lesssim c$$ implies $$a\\lesssim c,$$ for all $$% a,b,c\\in S.$$ (A pre-ordered set may have some other properties, but these are the main requirements.)\n\nOn the other hand a partial order $$a\\leq b$$ is a binary relation on a set $$S$$ that demands $$S$$ to have three properties: (i) $$\\leq$$ is reflexive, i.e., $$% a\\leq a$$ for all $$a\\in S$$, (ii) $$\\leq$$ is transitive, i.e., $$a\\leq b$$ and $$% b\\leq c$$ implies $$a\\leq c,$$ for all $$a,b,c\\in S$$ and (iii) $$\\leq$$ is antisymmetric, i.e., $$a\\leq b$$ and $$b\\leq a$$ implies $$a=b$$ for all $$a,b\\in S$$% .\n\nSo, as the definitions go, a partial order is a pre-order with an extra condition. This extra condition is not cosmetic, it is a distinguishing property. To see this let's take a simple example. Let's note that $$a$$ divides $$b$$ (or $$a|b)$$ is a binary relation on the set $$Z\\backslash \\{0\\}$$ of nonzero integers. Here, of course, $$a|b$$ $$\\Leftrightarrow$$ there is a $$% c\\in Z$$ such that $$b=ac.$$\n\nNow let's check: (i) $$a|a$$ for all $$a\\in Z\\backslash \\{0\\}$$ and (ii) $$a|b$$ and $$b|c$$ we have $$a|c.$$ So $$a|b$$ is a pre-order on $$Z\\backslash \\{0\\},$$ but it's not a partial order. For, in $$Z\\backslash \\{0\\},$$ $$a|b$$ and $$b|a$$ can only give you the conclusion that $$a=\\pm b,$$ which is obviously not the same as $$a=b.$$\n\nThe above example shows the problem with the pre-ordered set $$,$$\\lesssim >.$$ It can allow $$a\\lesssim b$$ and $$b\\lesssim a$$ with a straight face, without giving you the equality. Now a pre-order cannot be made into a partial order on a set $$, $$\\lesssim >$$ unless it is a partial order, but it can induce a partial order on a modified form of $$S.$$ Here's how. Take the bull by the horn and define a relation $$\\sim ,$$ on $$$$ by saying that $$a\\sim b$$ $$\\Leftrightarrow a\\lesssim b$$ and $$b\\lesssim a$$. It is easy to see that $$\\sim$$ is an equivalence relation. Now splitting $$S$$ into the set of classes $$\\{[a]|$$ $$a\\in S\\}$$ where $$[a]=\\{x\\in S|$$ $$x\\sim a\\}.$$ This modified form of $$S$$ is often represented by $$S/\\sim .$$ Now of course $$% [a]\\leq \\lbrack b]$$ if $$a\\lesssim b$$ but it is not the case that $$b\\lesssim a.$$ Setting $$[a]=[b]$$ if $$a\\sim b$$ (i.e. if $$a\\lesssim$$ and $$b\\lesssim a).$$\n\nIn the example of $$Z\\backslash \\{0\\}$$ we have $$Z\\backslash \\{0\\}/\\sim$$ $$% =\\{|a|$$ $$|$$ $$a\\in Z\\backslash \\{0\\}\\}.$$\n\n(Oh and as a parting note an equivalence relation is a pre-order too, with the extra requirement that $$a\\lesssim b$$ implies $$b\\lesssim a.)$$", "date": "2022-06-30 20:54:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1992277/preorders-vs-partial-orders-clarification", "openwebmath_score": 0.860423743724823, "openwebmath_perplexity": 162.56571677397866, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856414, "lm_q2_score": 0.885631470799559, "lm_q1q2_score": 0.866778721571402}}
{"url": "https://math.stackexchange.com/questions/2293406/meaning-of-times-in-mathbbrm-times-mathbbrn-rightarrow-mathbbrm-tim", "text": "# Meaning of times in $\\mathbb{R}^m\\times\\mathbb{R}^n\\rightarrow \\mathbb{R}^{m\\times n}$?\n\nI use $\\mathbf{a} \\times\\mathbf{b}$ for the cross product, $\\mathbf{a}\\cdot \\mathbf{b}$ for the dot product and $ab$ for normal multiplication ($a,b$ are scalars).\n\nHowever, what is the meaning of times in $\\mathbb{R}\\times\\mathbb{R}\\rightarrow \\mathbb{R}^2$?\n\nOr $\\mathbb{R}^m\\times\\mathbb{R}^n\\rightarrow \\mathbb{R}^{m\\times n}?$\n\nIs it the cross product?\n\nIs it the dot product?\n\nIs it normal multiplication?\n\nUpdate:\n\nDoes these have any meaning\n\n$\\mathbb{R}^m\\cdot\\mathbb{R}^n\\rightarrow \\mathbb{R}^{m\\cdot n}$ (the dot product)?\n\n$\\mathbb{R}^m \\mathbb{R}^n\\rightarrow \\mathbb{R}^{m n}$ (normal multiplication)?\n\n\u2022 It is called the Cartesian product! \u2013\u00a0Nigel Overmars May 23 '17 at 13:49\n\nIt is a cartesian product. If $A$ and $B$ are two sets, then $A\\times B$ is by definition the set of couples $(a,b)$ with $a$ in $A$ and $b$ in $B$.\n\nIn your case: $$\\mathbb{R}^m\\times\\mathbb{R}^n=\\{(x,y);x\\in\\mathbb{R}^m,y\\in\\mathbb{R}^n\\}.$$\n\n\u2022 Adding onto this for the range space $\\mathbb{R}^{m \\times n}$ is usually meant to denote real $m \\times n$ matrices. \u2013\u00a0Dragonite May 23 '17 at 13:51\n\u2022 Notice that in fact $\\mathbb{R}^m \\times \\mathbb{R}^n \\cong \\mathbb{R}^{m+n}$. You always need $m+n$ numbers to identify an element of either side. For example, $\\mathbb{R} \\times \\mathbb{R} \\cong \\mathbb{R}^2$. \u2013\u00a0Sharkos May 23 '17 at 13:55\n\u2022 Yes, OP be weary that there is a difference between $\\mathbb{R}^{m+n}$ and $\\mathbb{R}^{m\\times n}$ that has not really been addressed. \u2013\u00a0Dragonite May 23 '17 at 13:56\n\u2022 $$\\mathbb{R^2} := \\mathbb{R} \\times \\mathbb{R}$$ by definition. So unless you mean that this is the trivial isomorphism, defined by the identity function, this wrong. \u2013\u00a0user370967 May 23 '17 at 15:29\n\u2022 @Sharkos Is it also true for complex numbers, i.e. $\\mathbb{C}^m \\times \\mathbb{C}^n = \\mathbb{C}^{m+n}$? \u2013\u00a0JDoeDoe Nov 25 '17 at 18:52\n\nI think, more generally, you need guidance on the notation $f\\colon A \\times B \\to C$, which is the notation for \"a function called $f$, from the Cartesian product $A \\times B$ of sets $A$ and $B$, to the set $C$\" (the Cartesian product $A \\times B = \\{(a, b) : a \\in A, b \\in B\\}$ is the set of all ordered pairs with things taken from $A$ and $B$). More generally, the format is\n\n$$\\text{function name} : \\text{domain} \\to \\text{codomain}$$\n\nBut this notation $f \\colon A \\times B \\to C$ often says nothing about what the function actually does to pairs $(a, b) \\in A \\times B$ to produce some $f(a, b) \\in C$, unless $f$ happens to have a particularly descriptive name/symbol. In this case, you'll often see functions introduced in \"two parts\",\n\n\\begin{align*} f \\colon A \\times B &\\to C \\\\ (a, b) &\\mapsto \\text{however $a, b$ determine $f(a, b)$} \\end{align*}\n\nwhere the first line specifies the function name and all the sets we need, and the second line actually tells us what $f$ does to the pairs $(a, b)$ (and note the new symbol $\\mapsto$, which is used like $\\to$ above. But $\\to$ is used with sets, the domain and codomain, while $\\mapsto$ is used between the actual input and output, to explain what happens to elements in the sets).\n\nSo you'll never see notation like $\\Bbb R^m \\cdot \\Bbb R^n \\to \\Bbb R^{m + n}$, with the function placed between sets. Instead, you'll see the function name/notation in the place of $f$, put before the domain. So things like\n\n$$\\cdot \\colon \\Bbb R^n \\times \\Bbb R^n \\to \\Bbb R$$ is a (mildly confusing) notation saying that there's a function called \"$\\cdot$\" that takes two vectors in $\\Bbb R^n$, and gives you back a real number (we can assume it's the standard dot product).\n\n$$\\times \\colon \\Bbb R^3 \\times \\Bbb R^3 \\to \\Bbb R^3$$ would be the (somehow more confusing) notation to say there's a function called \"$\\times$\" that takes two vectors in $\\Bbb R^3$ and returns another vector in $\\Bbb R^3$; probably it's the standard cross product on $\\Bbb R^3$.\n\nFor a slightly-less-weird-looking example, we might use\n\n$$+ \\colon \\Bbb R^n \\times \\Bbb R^n \\to \\Bbb R^n$$ to say that we have an operation called \"$+$\" that takes pairs of vectors in $\\Bbb R^n$, and returns a single vector in $\\Bbb R^n$ (and unless it's stated otherwise, everyone would assume \"$+$\" means exactly what you think it means).\n\nThe domain and codomain can come in all sorts of varieties. For example, functions don't have to be defined on pairs of things, in which case our domain isn't going to be a Cartesian product. So to talk about the standard square root function, we might write\n\n$$\\sqrt{\\ }\\; \\colon \\Bbb R_{\\ge 0} \\to \\Bbb R_{\\ge 0}.$$\n\nOr maybe we're handed a function with a fairly cryptic name,\n\n\\begin{align*} \\operatorname{ev} \\colon M_{n \\times n}(\\Bbb R) \\times \\Bbb R^n &\\to \\Bbb R^n \\\\ (A, \\vec{v}) &\\mapsto A\\vec{v} \\end{align*}\n\nbut with practice, we can see it's the evaluation map that takes an $n \\times n$ matrix with real entries and a vector in $\\Bbb R^n$, and applies $A$ to $\\vec{v}$.\n\n\u2022 Note that many of the examples had the format $\\Bbb R^n \\times \\Bbb R^n \\to \\Bbb R^n$; only the prefix, the function name before $\\colon$, could help us tell the functions apart. The part of the notation after the $\\colon$ is really only good for domain and codomain, not the function itself (e.g., vector addition versus cross product). \u2013\u00a0pjs36 May 23 '17 at 16:00\n\n$\\Bbb R \\times \\Bbb R$ denotes the Cartesian product. An element of $\\Bbb R \\times \\Bbb R$ has the form $(a,b)$, where $a$ and $b$ are both in $\\Bbb R$.\n\nBasically, $\\Bbb R \\times \\Bbb R$ is just a longer way of saying $\\Bbb R^2$.\n\nThe same symbol in different contexts gets different meanings. Math in this aspect is somehow like poetry; the same word gets different meanings in different contexts.\n\nThe symbol \"$\\times$\", applied to sets like $\\mathbb{R}$, denotes the Cartesian product. If $X,Y \\neq \\varnothing$, then $X \\times Y := \\{ (x,y) \\mid x \\in X, y \\in Y \\}$. If $X, Y := \\mathbb{R}$, then $X \\times Y$ by definition is simply the usual Cartesian plane. It is defined that $X^{n} := \\{ (x_{1},\\dots,x_{n}) \\mid x_{1},\\dots,x_{n} \\in X \\}$. Now you know what the superscript of $\\mathbb{R}$ means.\n\nNote that $\\mathbb{R}^{m} \\times \\mathbb{R}^{n} = \\mathbb{R}^{m+n}$.\n\n\u2022 Thanks! You say sets, is it also true for vectors? For matrices we can say we have a $m\\times n$-matrix, does $\\times$ here also mean the cartesian product? \u2013\u00a0JDoeDoe May 23 '17 at 14:32\n\u2022 No problem. No, the point is to what objects the symbol \"$\\times$\" is along with. If $a,b$ are vectors, then $a \\times b$ means the cross product of $a$ and $b$; if $a,b$ are numbers, then $a\\times b$ means either the arithmetic product or the size of a matrix; if $a,b$ are sets, then $a \\times b$ means their Cartesian product. These cover your question? \u2013\u00a0Benicio May 23 '17 at 14:44\n\nNone of the above on the left. That $\\times$ is what appears between two sets to denote their cartesian product - the set of all ordered pairs whose first (second) element is from the first (second) set.\n\nThe $\\times$ on the right is ordinary multiplication.\n\nEdit:\n\nThe vector space on the left has dimension $m+n$, the one on the right has dimension $mn$, so the arrow can't represent an isomorphism. It might an injection. One possibility is that you're thinking of $\\mathbb{R}^{m\\times n}$ as the space of $m \\times n$ matrices. Then the arrow could mean the function $f$ given by $$f(v,w)_{ij} = v_iv_j .$$ (ordinary multiplication of real numbers onthe right). This map isn't an injection since $f(0,w) = 0$ for every $w$.\n\nThe vector space on the left is naturally isomorphic to $\\mathbb{R}^{m+n}$. The arrow in $$\\mathbb{R}^m \\times \\mathbb{R}^n \\rightarrow \\mathbb{R}^{m+n}$$ might then stand for that natural isomorphism. Perhaps that's what you meant to ask about. That's what your example when $m=n=1$ suggests.\n\n\u2022 Isn't the $\\times$ on the right addition, $m+n$? \u2013\u00a0JDoeDoe May 23 '17 at 14:16\n\u2022 @JDoeDoe See my edits. \u2013\u00a0Ethan Bolker May 23 '17 at 15:13\n\nTo answer your first question $X \\times Y$ is the Cartesian product of sets $X$ and $Y$. This is all ordered pairs $(x,y)$ where $x$ is a member of the set $X$, and $y$ is a member of the set $Y$.\n\n$$X \\times Y = \\{(x,y):x\\in X,y \\in Y\\}$$\n\nAn example is $\\mathbb R \\times \\mathbb R^2$ which is all ordered pairs $(x,(y,z))$ where $x \\in \\mathbb R$ and $(y,z) \\in \\mathbb R^2$. Of course this can be identified with $\\mathbb R^3$ by the bijection $(x,(y,z)) \\mapsto (x,y,z)$.\n\nA more interesting example is $[0,1] \\times S^1$ which is all ordered pairs $(t,\\theta)$ there $t \\in [0,1]$, i.e. $0 \\le t \\le 1$ and $\\theta$ is a point in the circle. This product $[0,1] \\times S^1$ gives a cylinder.\n\nThe word product is unfortunate, and has nothing to do with multiplication. The torus - which looks like a bicycle inner tube - is the Cartesian product of two circles: $T = S^1 \\times S^1$.\n\nYour idea of the dot product doesn't work for two reasons. Recall that the dot product takes two vectors (of the same dimension) and gives a number. Instead of it being a set, it is a mapping between sets ($\\mathbb R^n \\times \\mathbb R^n \\to \\mathbb R$). It doesn't make sense to dot a vector from $\\mathbb R^2$ with a vector from $\\mathbb R^3$; the dimensions are wrong. Think about matrices: it only makes sense to multiply a $p$-by-$q$ matrix with a $q$-by-$r$ matrix to get a $p$-by-$r$ matrix.\n\nSimilar remarks can be made about your idea of normal multiplication $\\mathbb R^m \\mathbb R^n$. Take an example: Take $(1,2,3) \\in \\mathbb R^3$ and $(1,2)\\in \\mathbb R^2$. How do we get something in $\\mathbb R^6$? (In a natural way that has meaning?)", "date": "2019-06-16 03:23:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2293406/meaning-of-times-in-mathbbrm-times-mathbbrn-rightarrow-mathbbrm-tim", "openwebmath_score": 0.9787355065345764, "openwebmath_perplexity": 211.02590575097892, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9525741241296944, "lm_q2_score": 0.9099070103134425, "lm_q1q2_score": 0.8667538733887963}}
{"url": "https://math.stackexchange.com/questions/3350867/have-i-been-using-differential-forms-without-knowing-it", "text": "# Have I been using differential forms without knowing it?\n\nI'm self-learning differential forms. I've been happily integrating 1-forms over parameterised curves, and 2-forms over parameterised surfaces, both in $$\\mathbb{R}^{3}$$. Now I've just found out that integrating an n-form $$\\omega=f\\left(x_{1},\\ldots,x_{n}\\right)dx_{1}\\wedge dx_{2}\\cdots\\wedge dx_{n}$$ over an n-dimensional manifold M in $$\\mathbb{R}^{n}$$ is defined by$$\\intop_{M}\\omega=\\pm\\intop_{M}f\\left(x_{1},\\ldots,x_{n}\\right)dx_{1}\\cdots dx_{n}.$$\n\nAm I correct in thinking that this definition describes what's going on with an ordinary calculus definite integral$$\\int_{b}^{a}f\\left(x\\right)dx.$$So $$f\\left(x\\right)dx$$ would be a 1-form and the one-dimensional manifold it is integrated over is the interval $$\\left(a,b\\right)$$?\n\n\u2022 That is exactly correct. \u2013\u00a0Lee Mosher Sep 10 at 15:42\n\u2022 To be extra careful, manifolds should be oriented for purposes of integrating forms, so you should specify the orientation on $(a,b)$ as being induced by restriction from the \"basic\" orientation on $\\mathbb R$. Reversing that orientation means integrating backwards from $b$ to $a$, which changes the sign. \u2013\u00a0Lee Mosher Sep 10 at 15:44\n\u2022 Actually, to be extra extra careful, the \"ordinary calculus definite integral\" is ambiguous. It can be interpreted either as (1) the integral of a density on an unoriented smooth manifold with boundary, namely the interval $[a,b]$, in which case you don't need an orientation; or as (2) the integral of a form on an oriented smooth manifold, namely the oriented interval $[a,b]$ with the standard orientation. We can integrate functions even on unoriented smooth manifolds, because of densities. And indeed the whole point of an orientation here is to convert a form into a density. \u2013\u00a0symplectomorphic Sep 10 at 19:00\n\u2022 @symplectomorphic - Tau, in \u201cDifferential Forms and Integration\u201d, distinguishes between the \u201cunsigned definite integral $\\int_{\\left[a,b\\right]}f\\left(x\\right)dx$ (which one would use to find area under a curve, or the mass of a one-dimensional object of varying density), and the signed definite integral $\\int_{a}^{b}f\\left(x\\right)dx$ (which one would use for instance to compute the work required to move a particle from $a$ to $b$).\u201d Is that what you mean? Thanks \u2013\u00a0Peter4075 Sep 11 at 7:03\n\u2022 Sorry, that should be Tao not Tau. \u2013\u00a0Peter4075 Sep 11 at 8:09\n\nMoreover, you can think of 0-forms which are just scalars. Then generalized Stokes' theorem $$\\int_{d\\Omega} \\omega=\\int_\\Omega d\\omega$$ in case of 0-form $$\\omega$$ (and 1-form $$d\\omega$$) becomes the fundamental theorem of Calculus: $$\\left.F(x)\\right|_a^b =\\int_a^b\\frac{dF}{dx}dx$$", "date": "2019-11-19 17:59:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3350867/have-i-been-using-differential-forms-without-knowing-it", "openwebmath_score": 0.9262468218803406, "openwebmath_perplexity": 477.12391696617726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718477853187, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8667417415204438}}
{"url": "https://bik-f.euni.de/ukb8j9g/how-to-find-irrational-numbers-between-decimals-fbfb78", "text": "Irrational numbers are those which do not terminate and have no repeating pattern. Rational numbers are whole numbers, fractions, and decimals - the numbers we use in our daily lives. Irrational numbers include \u221a2, \u03c0, e, and \u03b8. The vast majority are irrational numbers, never-ending decimals that cannot be written as fractions. Hence, almost all real numbers are irrational. Subtract the rounded numbers to obtain the estimated difference of 0.5; The actual difference of 0.988 - 0.53 is 0.458; Some uses of rounding are: Checking to see if you have enough money to buy what you want. Difference between rational and irrational numbers has been clearly explained in the picture given below. From the below figure, we can see the irrational number is \u221a2. . They include the counting numbers and all other numbers that can be written as fractions. Difference between Rational and Irrational Numbers. In summary, this is a basic overview of the number classification system, as you move to advanced math, you will encounter complex numbers. Many people are surprised to know that a repeating decimal is a rational number. Suppose we have two rational numbers a and b, then the irrational numbers between those two will be, \u221aab. The decimal expansion of an irrational number continues without repeating. This amenability to being written down makes rational numbers the ones we know best. Learn Math Tutorials 497,805 views Yes. one irrational number between 2 and 3 . 10 How to Write Fractions Between Two Decimal Numbers. Practice Problems 1. 9 years ago. Key Differences Between Rational and Irrational Numbers. m/n \u2014> 3/n N has to be an integer, n cannot be 0. Irrational Numbers on a Number Line. \u2154 is an example of rational numbers whereas \u221a2 is an irrational number. The major difference between rational and irrational numbers is that all the perfect squares, terminating decimals and repeating decimals are rational numbers. So 1/3 is between zero and one. What is the Difference Between Rational and Irrational Numbers , Intermediate Algebra , Lesson 12 - Duration: 3:03. Are there any decimals that do not stop or repeat? Method of finding a number lying exactly midway between 2 given numbers, is add those 2 numbers & divide by 2. Make use of this online rational or irrational number calculator to ensure the rationality and find its value. Irrational numbers tend to have endless non-repeating digits after the decimal point. Irrational numbers' decimal representation is non terminating , non repeating.. The only candidates here are the irrational roots. Lv 6. But there's at least one, so that gives you an idea that you can't really say that there are fewer irrational numbers than rational numbers. Before studying the irrational numbers, let us define the rational numbers. You can use the decimal module for arbitrary precision numbers: import decimal d2 = decimal.Decimal(2) # Add a context with an arbitrary precision of 100 dot100 = decimal.Context(prec=100) print d2.sqrt(dot100) If you need the same kind of ability coupled to speed, there are some other options: [gmpy], 2, cdecimal. Examples of Rational and Irrational Numbers For Rational. A rational number is a number that can be written as a ratio. To study irrational numbers one has to first understand what are rational numbers. How many irrational numbers can exist between two rational numbers ? Solution: If a and b are two positive numbers such that ab is not a perfect square then : i ) A rational number between and . 0 0. cryptogramcorner. Irrational Number between Two Rational Numbers. Apart from these number systems we have Irrational Numbers. 2 and 3 are rational numbers and is not a perfect square. It makes no sense!!! Terminating, recurring and irrational decimals. Step 3: Place the repeating digit(s) to the left of the decimal point. Take this example: \u221a8= 2.828. To find the rational number between two decimals, we can simply look for the average of the two decimals. The ability to convert between fractions and decimals, and to approximate irrational numbers with decimals or fractions, can be very helpful in solving problems. Rational Numbers. That's our five. Step 4: Place the repeating digit(s) to the right of the decimal point. Irrational Numbers. 0.5 can be written as \u00bd or 5/10, and any terminating decimal is a rational number. Examine the repeating decimal to find the repeating digit(s). And these non recurring decimals can never be converted to fractions and they are called as irrational numbers. Rational Number is defined as the number which can be written in a ratio of two integers. Wednesday, October 14, 2020 Find two rational numbers between decimals Ex: Find two rational numbers that have 3 in their numerator and are in between 0.13 and 0.14 with no calculator. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Which means that the only way to find the next digit is to calculate it. Getting a rough idea of the correct answer to a problem Irrational Numbers. DOWNLOAD IMAGE. how cuanto mide una cama matrimonial haunted house in san diego that lasts 4 7 hours journey to the west full movie with english subtitles. Include the decimal approximation of the ...\u201d in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. DOWNLOAD IMAGE. Irrational Numbers: We have different types of numbers in our number system such as whole numbers, real numbers, rational numbers, etc. In other words, irrational numbers require an infinite number of decimal digits to write\u2014and these digits never form patterns that allow you to predict what the next one will be. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. - [Voiceover] Plot the following numbers on the number line. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. The first number we have here is five, and so five is five to the right of zero, five is right over there. A rational number is of the form $$\\frac{p}{q}$$, p = numerator, q= denominator, where p and q are integers and q \u22600.. But rational numbers are actually rare among all numbers. Get an answer to your question \u201cPart B: Find an irrational number that is between 9.5 and 9.7.Explain why it is irrational. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q \u2260 0. Some decimals terminate which means the decimals do not recur, they just stop. In short, rational numbers are whole numbers, fractions, and decimals \u2014 the numbers we use in our daily lives.. In mathematics, a number is rational if you can write it as a ratio of two integers, in other words in a form a/b where a and b are integers, and b is not zero. Rational numbers. A Rational Number can be written as a Ratio of two integers (ie a simple fraction). Number System Notes. They can be written as a ratio of two integers. I tried to cross multiply like my teacher said but that was for finding what lay in between fractions it worked for that I have no idea how to do this. I can approximately locate irrational numbers on a number line ; I can estimate the value of expression involving irrational numbers using rational numbers. A set of real numbers is uncountable. (Examples: Being able to determine the value of the \u221a2 on a number line lies between 1 and 2, more accurately, between 1.4 \u2026 So the question states: what is a decimal number between each of the following pairs of rational numbers and then it gives the fractions -5/6, 1 and -17/20 and -4/5! Rational numbers are contrasted with irrational numbers - numbers such as Pi, \u221a 2, \u221a 7, other roots, sines, cosines, and logarithms of numbers. Rational and Irrational numbers both are real numbers but different with respect to their properties. ii) An irrational number between and . Real numbers also include fraction and decimal numbers. Step 5: Using the two equations you found in step 3 and step 4, subtract the left sides of the two equations. The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. List Of Irrational Numbers 1 100. So irrational number is a number that is not rational that means it is a number that cannot be written in the form $$\\frac{p}{q}$$. 1/3. Example 15: Find a rational number and also an irrational number between the numbers a and b given below: a = 0.101001000100001\u2026., b = 0.1001000100001\u2026 Solution: Since the decimal representations of a and b are non-terminating and non-repeating. it can't be the terminating decimals because they're rational. An irrational number is a number which cannot be expressed in a ratio of two integers. O.13 < 3/n < 0.14 13/100 < 3/n < 14/100 When is 13/100 < 3/n ? How to Write Irrational Numbers as Decimals Clearly all fractions are of that That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. The difference between rational and irrational numbers can be drawn clearly on the following grounds. Then we get 1/3. The number $\\pi$ (the Greek letter pi, pronounced \u2018pie\u2019), which is very important in describing circles, has a decimal form that does not stop or repeat. Irrational Numbers. Representation of irrational numbers on a number line. But an irrational number cannot be written in the form of simple fractions. As there are infinite numbers between two rational numbers and also there are infinite rational numbers between two rational numbers. Real numbers include natural numbers, whole numbers, integers, rational numbers and irrational numbers. How To Find Irrational Numbers Between Two Decimals DOWNLOAD IMAGE. since 3^2 = 9 and 6^2 = 36, \u221a41 > 6. apply the same kind of thinking to the numbers in B and they are all between 3 and 6. New Proof Settles How To Approximate Numbers Like Pi Quanta Magazine. So number of irrational numbers between the numbers should be infinite or something finite which we cannot tell ? Essentially, irrational numbers can be written as decimals but as a ratio of two integers. On the other hand, all the surds and non-repeating decimals are irrational numbers. For instance, when placing \u221a15 (which is 3.87), it is best to place the dot on the number line at a place in between 3 and 4 (closer to 4), and then write \u221a15 above it. We can actually split this into thirds. Example: Find two irrational numbers between 2 and 3. Infinite rational numbers and also there are infinite rational numbers the ones we know best learn the difference rational. 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New Proof Settles How to find the repeating digit ( s ) in the picture below! Of an irrational number number of irrational numbers numbers, fractions, and \u03b8 not terminate and no! Surds and non-repeating decimals are rational numbers are whole numbers, integers, numbers. Ratios and rates rational numbers are whole numbers, never-ending decimals that can be written as a ratio two... Which do not terminate and have no repeating pattern of rational numbers are whole numbers, fractions and... 0.14 13/100 < 3/n < 0.14 13/100 < 3/n < 14/100 when is 13/100 < 3/n < 14/100 when 13/100. Given numbers, fractions, and decimals - the numbers should be infinite or something finite which we can look... Between the numbers should be infinite or something finite which we can see irrational... Terminating decimal is a rational number using rational numbers are whole numbers, fractions, and any terminating is... Non-Repeating digits after the decimal point example: find two irrational numbers be...: using the two decimals, we can not be written as \u00bd or 5/10, and \u03b8 non decimals... Learn the difference between rational and irrational numbers tend to have endless non-repeating digits the... Be expressed in a ratio of two integers irrational numbers can be as. To their properties form of simple fractions squares, terminating decimals because they 're rational been clearly explained in picture., never-ending decimals that can not be exact, but a very close estimation something which... Many people are surprised to know that a repeating decimal to find irrational numbers can be written as fractions because. On the other hand, all the perfect squares, terminating decimals because they 're rational decimals can be... Can be written as a ratio of two integers ( ie a simple fraction.. irrational means rational. Ratio of two integers number between two decimals one has to first understand what are numbers. Apart from these number systems we have two rational numbers the ones we best. ) to the left sides of the decimal expansion of an irrational number a. ; 2, \u03c0, e, and decimals - the numbers should be infinite or finite! To study irrational numbers non recurring decimals can never be converted to fractions and they called...", "date": "2021-03-04 00:05:42", "meta": {"domain": "euni.de", "url": "https://bik-f.euni.de/ukb8j9g/how-to-find-irrational-numbers-between-decimals-fbfb78", "openwebmath_score": 0.7450292110443115, "openwebmath_perplexity": 407.1518889219758, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9877587275910131, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8667353661707351}}
{"url": "https://goodboychan.github.io/python/datacamp/data_science/statistics/2020/05/26/05-Thinking-probabilistically-Continuous-variables.html", "text": "## Probability density functions\n\n\u2022 Continuous variables\n\u2022 Quantities that can take any value, not just discrete values\n\u2022 Probability Density function (PDF)\n\u2022 Continuous analog to the PMF\n\u2022 Mathematical description of the relative likelihood of observing a value of a continuous variable\n\n## Introduction to the Normal distribution\n\n\u2022 Normal distribution\n\n\u2022 Describes a continuous variable whose PDF has a single symmetric peak.\n\n\\begin{align} \\text{mean of a Normal distribution} & \\neq \\text{mean computed from data} \\\\ \\text{st. dev of a Normal distribution} & \\neq \\text{st. dev computed from data} \\end{align}\n\n### The Normal PDF\n\nIn this exercise, you will explore the Normal PDF and also learn a way to plot a PDF of a known distribution using hacker statistics. Specifically, you will plot a Normal PDF for various values of the variance.\n\nimport numpy as np\nimport pandas as pd\nimport matplotlib.pyplot as plt\nimport seaborn as sns\n\nsns.set()\n\n# with stds of interest: samples_std1, samples_std3, stamples_std10\nsamples_std1 = np.random.normal(20, 1, 100000)\nsamples_std3 = np.random.normal(20, 3, 100000)\nsamples_std10 = np.random.normal(20, 10, 100000)\n\n# Make histograms\n_ = plt.hist(samples_std1, histtype='step', density=True, bins=100)\n_ = plt.hist(samples_std3, histtype='step', density=True, bins=100)\n_ = plt.hist(samples_std10, histtype='step', density=True,bins=100)\n\n# Make a legend, set limits\n_ = plt.legend(('std = 1', 'std = 3', 'std = 10'))\nplt.ylim(-0.01, 0.42)\n\n(-0.01, 0.42)\n\n### The Normal CDF\n\nNow that you have a feel for how the Normal PDF looks, let's consider its CDF. Using the samples you generated in the last exercise (in your namespace as samples_std1, samples_std3, and samples_std10), generate and plot the CDFs.\n\ndef ecdf(data):\n\"\"\"Compute ECDF for a one-dimensional array of measurements.\"\"\"\n# Number of data points: n\nn = len(data)\n\n# x-data for the ECDF: x\nx = np.sort(data)\n\n# y-data for the ECDF: y\ny = np.arange(1, n + 1) / n\n\nreturn x, y\n\nx_std1, y_std1 = ecdf(samples_std1)\nx_std3, y_std3 = ecdf(samples_std3)\nx_std10, y_std10 = ecdf(samples_std10)\n\n# Plot CDFs\n_ = plt.plot(x_std1, y_std1, marker='.', linestyle='none')\n_ = plt.plot(x_std3, y_std3, marker='.', linestyle='none')\n_ = plt.plot(x_std10, y_std10, marker='.', linestyle='none')\n\n# Make a legend\n_ = plt.legend(('std = 1', 'std = 3', 'std = 10'), loc='lower right')\nplt.savefig('../images/std-cdf.png')\n\n\n## The Normal distribution: Properties and warnings\n\n### Are the Belmont Stakes results Normally distributed?\n\nSince 1926, the Belmont Stakes is a 1.5 mile-long race of 3-year old thoroughbred horses. Secretariat ) ran the fastest Belmont Stakes in history in 1973. While that was the fastest year, 1970 was the slowest because of unusually wet and sloppy conditions. With these two outliers removed from the data set, compute the mean and standard deviation of the Belmont winners' times. Sample out of a Normal distribution with this mean and standard deviation using the np.random.normal() function and plot a CDF. Overlay the ECDF from the winning Belmont times. Are these close to Normally distributed?\n\nNote: Justin scraped the data concerning the Belmont Stakes from the Belmont Wikipedia page.\n\ndf = pd.read_csv('./dataset/belmont.csv')\n\nbelmont_no_outliers = np.array([148.51, 146.65, 148.52, 150.7 , 150.42, 150.88, 151.57, 147.54,\n149.65, 148.74, 147.86, 148.75, 147.5 , 148.26, 149.71, 146.56,\n151.19, 147.88, 149.16, 148.82, 148.96, 152.02, 146.82, 149.97,\n146.13, 148.1 , 147.2 , 146.  , 146.4 , 148.2 , 149.8 , 147.  ,\n147.2 , 147.8 , 148.2 , 149.  , 149.8 , 148.6 , 146.8 , 149.6 ,\n149.  , 148.2 , 149.2 , 148.  , 150.4 , 148.8 , 147.2 , 148.8 ,\n149.6 , 148.4 , 148.4 , 150.2 , 148.8 , 149.2 , 149.2 , 148.4 ,\n150.2 , 146.6 , 149.8 , 149.  , 150.8 , 148.6 , 150.2 , 149.  ,\n148.6 , 150.2 , 148.2 , 149.4 , 150.8 , 150.2 , 152.2 , 148.2 ,\n149.2 , 151.  , 149.6 , 149.6 , 149.4 , 148.6 , 150.  , 150.6 ,\n149.2 , 152.6 , 152.8 , 149.6 , 151.6 , 152.8 , 153.2 , 152.4 ,\n152.2 ])\n\nmu = np.mean(belmont_no_outliers)\nsigma = np.std(belmont_no_outliers)\n\n# Sample out of a normal distribution with this mu and sigma: samples\nsamples = np.random.normal(mu, sigma, size=10000)\n\n# Get the CDF of the samples and of the data\nx_theor, y_theor = ecdf(belmont_no_outliers)\nx, y = ecdf(samples)\n\n# Plot the CDFs\n_ = plt.plot(x_theor, y_theor)\n_ = plt.plot(x, y, marker='.', linestyle='none')\n_ = plt.xlabel('Belmont winning time (sec.)')\n_ = plt.ylabel('CDF')\n\n\n### What are the chances of a horse matching or beating Secretariat's record?\n\nAssume that the Belmont winners' times are Normally distributed (with the 1970 and 1973 years removed), what is the probability that the winner of a given Belmont Stakes will run it as fast or faster than Secretariat?\n\nsamples = np.random.normal(mu, sigma, size=1000000)\n\n# Compute the fraction that are faster than 144 seconds: prob\nprob = np.sum(samples < 144) / len(samples)\n\n# Print the result\nprint('Probability of besting Secretariat:', prob)\n\nProbability of besting Secretariat: 0.00057\n\n\n## The Exponential distribution\n\n\u2022 The waiting time between arrivals of a Poisson process is Exponentially distributed\n\n### If you have a story, you can simulate it!\n\nSometimes, the story describing our probability distribution does not have a named distribution to go along with it. In these cases, fear not! You can always simulate it. We'll do that in this and the next exercise.\n\nIn earlier exercises, we looked at the rare event of no-hitters in Major League Baseball. Hitting the cycle is another rare baseball event. When a batter hits the cycle, he gets all four kinds of hits, a single, double, triple, and home run, in a single game. Like no-hitters, this can be modeled as a Poisson process, so the time between hits of the cycle are also Exponentially distributed.\n\nHow long must we wait to see both a no-hitter and then a batter hit the cycle? The idea is that we have to wait some time for the no-hitter, and then after the no-hitter, we have to wait for hitting the cycle. Stated another way, what is the total waiting time for the arrival of two different Poisson processes? The total waiting time is the time waited for the no-hitter, plus the time waited for the hitting the cycle.\n\nNow, you will write a function to sample out of the distribution described by this story.\n\ndef successive_poisson(tau1, tau2, size=1):\n\"\"\"Compute time for arrival of 2 successive Poisson processes.\"\"\"\n# Draw samples out of first exponential distribution: t1\nt1 = np.random.exponential(tau1, size)\n\n# Draw samples out of second exponential distribution: t2\nt2 = np.random.exponential(tau2, size)\n\nreturn t1 + t2\n\n\n### Distribution of no-hitters and cycles\n\nNow, you'll use your sampling function to compute the waiting time to observe a no-hitter and hitting of the cycle. The mean waiting time for a no-hitter is 764 games, and the mean waiting time for hitting the cycle is 715 games.\n\nwaiting_times = successive_poisson(764, 715, size=100000)\n\n# Make the histogram\n_ = plt.hist(waiting_times, bins=100, density=True, histtype='step')\n\n# Label axes\n_ = plt.xlabel('waiting times')\n_ = plt.ylabel('PDF')", "date": "2022-01-29 08:02:42", "meta": {"domain": "github.io", "url": "https://goodboychan.github.io/python/datacamp/data_science/statistics/2020/05/26/05-Thinking-probabilistically-Continuous-variables.html", "openwebmath_score": 0.7094602584838867, "openwebmath_perplexity": 5996.479160665699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987758722546077, "lm_q2_score": 0.8774768002981829, "lm_q1q2_score": 0.8667353633263523}}
{"url": "https://www.physicsforums.com/threads/area-of-a-sector-why-squared.936747/", "text": "# Area of a Sector- Why squared?\n\n#### opus\n\nGold Member\n1. Homework Statement\nFind the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30\u00b0.\n\n2. Homework Equations\n$$Area~of~a~Sector:$$\n$$A=\\left( \\frac 1 2 \\right)r^2\u03b8$$\n\n3. The Attempt at a Solution\n\n$$\u03b8=30\u00b0$$\n$$\u03b8=30\u00b0\\left( \\frac \u03c0 {180} \\right)$$\n$$\u03b8=\\left( \\frac \u03c0 6 \\right)$$\n\n$$A=\\left( \\frac 1 2 \\right)\\left(9\\right)^2\\left(\\frac \u03c0 6 \\right)$$\n$$A=\\left( \\frac {81\u03c0} {12} \\right)$$\n$$A\u224821.2 in^2$$\n\nMy question:\nI know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.\nSo why is this result squared?\n\nRelated Precalculus Mathematics Homework Help News on Phys.org\n\n#### Mark44\n\nMentor\n1. Homework Statement\nFind the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30\u00b0.\n\n2. Homework Equations\n$$Area~of~a~Sector:$$\n$$A=\\left( \\frac 1 2 \\right)r^2\u03b8$$\n\n3. The Attempt at a Solution\n\n$$\u03b8=30\u00b0$$\n$$\u03b8=30\u00b0\\left( \\frac \u03c0 {180} \\right)$$\n$$\u03b8=\\left( \\frac \u03c0 6 \\right)$$\n\n$$A=\\left( \\frac 1 2 \\right)\\left(9\\right)^2\\left(\\frac \u03c0 6 \\right)$$\n$$A=\\left( \\frac {81\u03c0} {12} \\right)$$\n$$A\u224821.2 in^2$$\n\nMy question:\nI know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.\nSo why is this result squared?\nBecause it's an area. The shape doesn't matter.\nThe standard units of area are always squared, $\\text{length} \\times \\text{length}$, except for some special cases such as acres or hectares (which involve implicitly squared units such as ft2 for acres and m2 for hectares.\n\n#### opus\n\nGold Member\nSo the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?\n\n#### Mark44\n\nMentor\nSo the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?\nThe angle in radians is just an angle, with no length. Think about it this way, as a, say, peach pie. If an 8\" diameter pie is cut into 6 pieces, each slice (a sector) will subtend an angle of $\\pi/3$, and the radius will be 4\". The arc length of the curved edge of the slice has to take into account the radius, otherwise the arc length of an 8\" pie would be the same as for a 16\" pie. So in fact, the curved dimension of the pie sector is radius * angle (in radians), or $4 \\times \\pi/3$. So you have one radius for the radius of the sector and another radius for the arc length, making the sector area equal to $\\frac 1 2 r^2 \\theta$.\n\n#### opus\n\nGold Member\nAhhh ok. That makes complete sense. Great explanation, thank you Mark.\n\n#### DrClaude\n\nMentor\nI know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result.\nLet me add that one way to visualize the \"units square\" is to think that a wedge with an area of 22.2 in2 has the same area as a square with sides of \u221a(22.2) \u2248 4.6 in.\n\nLast edited by a moderator:\n\n#### opus\n\nGold Member\nInteresting! Thanks DrClaude\n\n#### alijan kk\n\nimagine the arc as a triangle , becuase area would be same even if you make the arc straight line.\nnow the base of this triangle is \"r=radius\" and the perpendicular side is the arc which is equal to \"theta*r\"\n\narea of triangle = 0.5base*height\n\n0.5(r)(r)(theta)=formula of area of sector\n\n\"Area of a Sector- Why squared?\"\n\n### Physics Forums Values\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-09-22 20:32:09", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/area-of-a-sector-why-squared.936747/", "openwebmath_score": 0.8684222102165222, "openwebmath_perplexity": 688.305319720026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9752018441287091, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8667192201462258}}
{"url": "http://mathhelpforum.com/algebra/132387-problem-vector-question.html", "text": "# Math Help - Problem with vector question\n\n1. ## Problem with vector question\n\nHi\nCan someone tell me where i have gone wrong in the following question?\n\nGiven that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.\n\nunit vector = {a x b}/{|a x b|}\n\n| a x b |= $\\sqrt{3^2+1^2+8^2}$\n\n$=\\sqrt{74}$\n\na x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)\n=i(2+1)-j(2-3)+k(-2-6)\n=3i+j-8k\n\ntherefore unit vector = $\\frac{1}{\\sqrt{74}}$(3i+j-8k)\n\nanswer says its $\\frac{1}{\\sqrt{74}}$(-3i-j+8k)\n\nP.S\n\n2. Originally Posted by Paymemoney\nHi\nCan someone tell me where i have gone wrong in the following question?\n\nGiven that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.\n\nunit vector = {a x b}/{|a x b|}\n\n| a x b |= $\\sqrt{3^2+1^2+8^2}$\n\n$=\\sqrt{74}$\n\na x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)\n=i(2+1)-j(2-3)+k(-2-6)\n=3i+j-8k\n\ntherefore unit vector = $\\frac{1}{\\sqrt{74}}$(3i+j-8k)\n\nanswer says its $\\frac{1}{\\sqrt{74}}$(-3i-j+8k)\n\nP.S\n$\\mathbf{v}\\times\\mathbf{w} = \\left|\\begin{matrix}\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ 2 & 2 & 1 \\\\ 3 & -1 & 1 \\end{matrix}\\right|$\n\n$= \\mathbf{i}[2 \\cdot 1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$\n\n$= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$.\n\nI agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect.\n\nAnyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length.\n\n$|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$\n\n$= \\sqrt{9 + 1 + 64}$\n\n$= \\sqrt{74}$.\n\nTherefore:\n\n$\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$\n\n$= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$\n\n$= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$.\n\n3. Hello, Paymemoney!\n\nYou've done nothing wrong . . .\n\nTheir answer is the negative of yours.\n\nThere are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$\n. . one \"up\" and one \"down\".\n\nThey picked one, you picked the other.\n\n4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "date": "2015-03-29 10:17:17", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/132387-problem-vector-question.html", "openwebmath_score": 0.6794993877410889, "openwebmath_perplexity": 1193.876141718576, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9752018419665619, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8667192182245984}}
{"url": "https://math.stackexchange.com/questions/4343130/how-same-are-two-isomorphic-groups", "text": "# How \"same\" are two isomorphic groups?\n\nFrom what I understand about isomorphisms is that two isomorphic groups are the same groups. They may have different names for the same elements and the operation. But the point is that the groups are same since their elements combine the same way.\n\nSo if $$G \\cong G'$$ then every group property about $$G$$ also holds for $$G'$$, am I correct?\n\nNow my question is\u2014 Can I interchange $$G$$ and $$G'$$ whenever and wherever I want? I thought the answer was obviously yes but... now I'm not sure.\n\nFor example: $$\\mathbb{Z} \\cong \\mathbb{2Z}$$ so shouldn't $$\\mathbb{Z}/\\mathbb{Z} = \\mathbb{Z}/\\mathbb{2Z}$$ under the group operation $$(a+H) + (b +H) = (a+b) + H$$? where $$H$$ is either $$\\mathbb{Z}$$ or $$\\mathbb{2Z}$$ since both are the same groups.\n\nBut that's clearly not the case as $$\\mathbb{Z}/\\mathbb{Z} = \\{0\\}$$ but $$\\mathbb{Z}/\\mathbb{2Z}=\\{0,1\\}$$ So where does one draw the line between two isomorphic groups? How \"same\" are two isomorphic groups? I'm very confused.\n\n\u2022 It is true that $\\mathbb{Z} \\cong \\mathbb{2Z}$. However, the issue for the quotient groups here is that $2 \\mathbb Z \\subsetneq \\mathbb Z$. Dec 27, 2021 at 16:42\n\u2022 The freedom to rename is not a freedom to be inconsistent. If you want to rename the elements in $\\mathbb Z$ so that $1$ is renamed to $2$, $2$ to $4$, $3$ to $6$ etc. - then of course the elements of $2\\mathbb Z\\subseteq \\mathbb Z$ must be renamed using the same procedure. You will end up with $4\\mathbb Z\\subseteq 2\\mathbb Z$, and indeed $\\mathbb Z/2\\mathbb Z\\cong 2\\mathbb Z/4\\mathbb Z$. Dec 27, 2021 at 16:43\n\u2022 Why is this closed under duplicate? I'm asking about a particular example and the difficulty I'm facing with it. How is the question same as the other? I'd already read that one. It's where I came from. The accepted answers says exactly the same thing as my first paragraph. But my question is different. Dec 27, 2021 at 16:46\n\u2022 Presentations of the same group (up to isomorphism) can have different properties. Dec 27, 2021 at 16:54\n\u2022 Frankly, I regard the \"s\" word as a 4-letter word in mathematics. I would suggest that you learn to love the 10-letter \"i\" word in your title. And then if you can extend that to loving the 11-letter word \"isomorphism\", and to think about actual isomorphisms as useful and interesting mathematical objects themselves, even better! Dec 27, 2021 at 18:51\n\nThis is a great question and your example with group quotients perfectly highlights why one has to be careful about notions of identity when doing maths. To answer your question,\n\nCan I interchange G and G\u2032 whenever and wherever I want?\n\nThe answer is yes, as long as you are 100% sure that you are thinking of both groups as literally just groups and nothing more (people will sometimes use the phase 'up to isomorphism' to convey this idea). This resolves the problem you had with quotients: you can't quotient an arbitrary group by another arbitrary group, the group by which you are quotienting has to be a subgroup of the other. This is extra 'data' which can be encoded in multiple ways, one nice way is to consider the special injection that embeds the one group inside the other (which doesn't exist for two arbitrary groups). So to recap, subgroups aren't just groups, they're groups with some extra data and $$2\\mathbb{Z}$$ and $$\\mathbb{Z}$$ are different subgroups of $$\\mathbb{Z}$$ despite being 'the same' groups (i.e. isomorphic).\n\n\u2022 \"the group by which you are quotienting has to be a subgroup of the other\" but $Z$ and $2Z$ are both subgroups of $Z$ no? and I don't get the part where you say they are the same groups but different subgroups? I'm sure I'm missing something here but I don't get it. What is the extra data about subgroup? Dec 27, 2021 at 17:14\n\u2022 Oh you mean they're different subgroups because they are made up of different subsets of the group? Oh okay. Fair enough. But what exactly go wrong here in my example? I'm not able to \"see\" it still :( Dec 27, 2021 at 17:26\n\u2022 The problem in your example is that, to conclude that the quotient is the same, you have to start with 'the same' subgroups, but you didn't, you started with two different subgroups (that just so happen to be 'the same' as groups). Dec 27, 2021 at 17:28\n\u2022 So in total, there are four descriptions of mathematical objects: the group $\\mathbb{Z}$, the group $\\mathbb{2Z}$, the subgroup $\\mathbb{Z}\\leq\\mathbb{Z}$ and the subgroup $\\mathbb{2Z}\\leq\\mathbb{Z}$. The first two descriptions define the same mathematical object, the second two don't. Dec 27, 2021 at 17:46\n\u2022 Oh! Oh! Oh!! I get it now. If $H,H' \u2264G$ and $H \\cong H'$ then as groups, $H$ and $H'$ are same. But \"subgroup\" is also a relation and not an independent concept. So you always have subgroups of some group. So as subgroups of $G$, $H$ and $H'$ are different. So $G/H \u2260 G/H'$ because quotient groups aren't just about $H$ but rather $H$ wrt $G$. If it were about only $H$ then one can interchange it with $H'$ is what you're saying. Dec 27, 2021 at 18:12\n\nHere is a question that is similar: when can replace equivalent things with one another?\n\nSo if we remember that an isomorphism of groups is a function $$\\phi:G\\rightarrow G'$$. With the example you gave you have $$\\phi:\\mathbb{Z}\\rightarrow 2\\mathbb{Z}$$, then your issue was that $$\\mathbb{Z}/\\mathbb{Z}\\not\\cong\\mathbb{Z}/2\\mathbb{Z}$$. However, the issue is where are you applying the homomorphism. You are saying that $$G/G\\not\\cong G/\\phi(G)$$; however, we have that $$\\phi(G)\\subset G'$$ and $$\\phi(G)\\not\\subset G$$ (although it may be isomorphic to some subgroup of $$G$$ like in your example).\n\nThe key use of an isomorphism is that you can take a problem in one setting and view it in another setting through the isomorphism. Thus, what would be true is that if $$\\phi:G\\rightarrow G'$$ is an isomorphism, then if $$H\\lhd G$$, then it will be the case that $$G/H\\cong \\phi(G)/\\phi(H)$$.\n\nI think the long story short is that when you have an isomorphism between two objects you must think of them as living in different spaces. If you are familiar with the lattice of subgroups I think this is a nice way to view it, since if you have $$\\phi: G\\rightarrow G'$$ where $$G=\\mathbb{Z}$$ and $$G'=2\\mathbb{Z}$$ since these are isomorphic groups their lattice of subgroups are isomorphic and we should think of $$\\mathbb{Z}$$ to be the maximal element in the lattice of $$G$$ and $$2\\mathbb{Z}$$ to be the maximal element in the lattice of $$G'$$. Even though we have that $$2\\mathbb{Z}$$ appears in the lattice of $$G$$, the corresponding element in $$G'$$ would be the subgroup $$4\\mathbb{Z}$$, so under the isomorphism it would not make sense to \"replace $$2\\mathbb{Z}$$ in the $$G$$ world with $$2\\mathbb{Z}$$ in the $$G'$$ world\" as they aren't the same under the isomorphism. What we would do is replace the $$2\\mathbb{Z}$$ with the corresponding subgroup in the isomorphism namely $$4\\mathbb{Z}$$, and when we do so we do have that $$\\mathbb{Z}/2\\mathbb{Z}\\cong 2\\mathbb{Z}/4\\mathbb{Z}$$.\n\nThere is a precise way to formalise this.\n\nWe want to speak of sets not as collections of particular mathematical objects but as a bag of featureless dots. These bags of dots are related to each other by functions, which also relate dots in one bag to dots in another.\n\nIn order to do this, we propose a language for discussing sets. Capital letters will represent sets, and lowercase variables will represent functions between sets.\n\nWe define a \"nice proposition about sets\" as any proposition that can be generated from the following rules:\n\n\u2022 For any sets $$A, B$$ and any function variables $$f, g: A \\to B$$, $$f = g$$ is a nice proposition\n\u2022 We can combine nice propositions $$\\phi, \\psi$$ using the operators $$\\land$$, $$\\lor$$, $$\\neg$$, and $$\\implies$$ to get another nice proposition\n\u2022 $$\\top$$ and $$\\bot$$ (that is, true and false) are nice propositions\n\u2022 If $$\\phi(A)$$ is a nice proposition, where $$A$$ is a set variable, then $$\\forall A (\\phi(A))$$ is a nice proposition (and similarly for $$\\exists$$)\n\u2022 For all set variables $$A, B$$, if $$\\phi(f)$$ is a nice proposition (where $$f : A \\to B$$ is a function variable), then $$\\forall f : A \\to B (\\phi(f))$$ is a nice proposition (and similarly for $$\\exists$$)\n\nFinally, in our language, we can form new functions from old ones using function composition and also discuss identity functions. Note that we will only discuss function composition when we know that the functions are composable syntactically.*\n\nNote that we very deliberately avoided two things. First, there is no mention of the elementhood relation at all. Second, there is no way to compare two sets for equality, nor is there a way to compare two functions for equality unless the two have the same domain and codomain. This is deliberate. Also note that all variables here have specific types, and that we rely on these types to determine whether we can discuss equality.\n\nDespite this avoidance, it is both an empirical fact and a (rather complicated) theorem that all propositions with no free variables can be translated to a \"nice proposition about sets\". For most mathematically useful propositions, the translation is relatively intuitive for those comfortable with category theory.\n\nTo get around the restriction that we not discuss elements directly, we can fix some 1-element set $$1$$ and discuss elements of $$A$$ as functions $$1 \\to A$$. We denote the situation $$x : 1 \\to A$$ as $$x :\\in A$$. When we have $$x :\\in A$$ and $$f : A \\to B$$, we write $$f \\circ x$$ as $$f(x)$$ (note that $$f(x) :\\in B$$). Note that this is enough to define, for example, a group (as a set $$G$$, together with a function $$- \\cdot - : G^2 \\to G$$ which satisfies the group laws**).\n\nWe thus have the following theorem:\n\nBig Theorem. Consider some nice proposition $$\\phi(G)$$ - that is, some statement $$\\phi$$ where the group variable $$G$$ occurs free. Then we can prove the following statement: \"For all groups $$G_1$$, $$G_2$$, if $$G_1$$ and $$G_2$$ are isomorphic and $$\\phi(G_1)$$, then $$\\phi(G_2)$$.\"\n\nLet us discuss how this relates to your example of $$\\mathbb{Z}$$ and $$2 \\mathbb{Z}$$.\n\nWe may try to define $$\\phi(G)$$ as the statement \"$$\\mathbb{Z} / G$$ is the zero group\". Then taking $$G_1 = \\mathbb{Z}$$ and $$G_2 = 2 \\mathbb{Z}$$, we see that $$\\phi(G_1)$$ holds but $$\\phi(G_2)$$ is false. This would appear to violate our Big Theorem.\n\nTo understand what's going on, we need to understand exactly what is going on with the statement $$\\phi(G)$$. Note that in this statement, it is necessary for $$G$$ to be a subgroup of $$\\mathbb{Z}$$ - in particular, it must be a subset. That is, we must have $$\\forall x \\in G (x \\in \\mathbb{Z})$$.\n\nOf course, we cannot express such a statement as a nice proposition about sets, since it relies on the proposition $$x \\in \\mathbb{Z}$$, which is not a nice proposition. So we have to approach things a bit differently.\n\nRather than forcing $$G$$ to be a subgroup in the traditional, literal sense, we instead make $$G$$ a subgroup of $$\\mathbb{Z}$$ in a more general sense. That is, we discuss a group $$G$$, together with some injective group homomorphism $$i : G \\to \\mathbb{Z}$$. In other words, we discuss $$G$$ together with a specific way that $$G$$ is a subgroup of $$\\mathbb{Z}$$.\n\nThis makes it clear that $$\\phi(G)$$ is not really just a proposition about $$G$$ - it's also about the way that $$G$$ is a subgroup of $$\\mathbb{Z}$$. So it's really a proposition $$\\phi(G, i : G \\to \\mathbb{Z})$$. Because $$\\phi$$ depends on a secondary variable $$i$$, we see that we cannot apply our Big Theorem to conclude that $$\\phi(\\mathbb{Z}) \\iff \\phi(2 \\mathbb{Z})$$.\n\nIf it were the case that the isomorphism $$w : \\mathbb{Z} \\to 2 \\mathbb{Z}$$ \"played nicely\" with the inclusion maps $$i_1 : \\mathbb{Z} \\to \\mathbb{Z}$$, $$i_2 : 2 \\mathbb{Z} \\to \\mathbb{Z}$$ (that is, if $$i_1 = i_2 \\circ w$$), then we would be able to conclude that $$\\phi(\\mathbb{Z}, i_1) \\iff \\phi(2 \\mathbb{Z}, i_2)$$ using a generalisation of our Big Theorem. But of course the isomorphism does not play nicely with the inclusion maps.\n\n*An astute observer will note that this is exactly the language of category theory. A \"nice proposition about sets\" is just some statement about the category of sets (which avoids discussing equality of objects).\n\n**Technically, we need a way to define $$G^2$$ first. This is done using the categorical definition of the universal property of the product.\n\n\u2022 Why is this answer being downvoted? Dec 27, 2021 at 18:31\n\nThis is just an elaboration of Arthur's answer elsewhere in this thread. I thought you might like to see a specific example.\n\nConsider $$G = \\color{maroon}{\\Bbb Z_2}\\times \\color{darkblue}{\\Bbb Z_4}.$$\n\n$$G$$ has a subgroup $$A$$ that is generated by $$\\langle 1, 0\\rangle$$:\n\n$$\\def\\elt#1#2{\\langle{#1},{#2}\\rangle}A = \\{\\elt10, \\elt 00\\}$$\n\nand another subgroup $$B$$ that is generated by $$\\langle 0, 2\\rangle$$:\n\n$$B=\\{\\elt00, \\elt 02\\}$$\n\nBoth $$A$$ and $$B$$ are isomorphic to $$\\Bbb Z_2$$, and so to each other. But the quotients $$G/A$$ and $$G/B$$ are not isomorphic. $$G/A$$ is like taking $$G$$, ignoring the first component, and keeping the second component intact. The result is $$G/A \\simeq \\color{maroon}{\\Bbb Z_1}\\times\\color{darkblue}{\\Bbb Z_4}.$$ $$G/B$$ is like taking $$G$$ and keeping the first component but ignoring everything but the parity of the second component. The result is $$G/B \\simeq \\color{maroon}{\\Bbb Z_2}\\times\\color{darkblue}{\\Bbb Z_2}.$$ Even though $$A$$ and $$B$$ are isomorphic, $$G/A$$ and $$G/B$$ are not isomorphic. (In particular, $$G/A$$ is cyclic and $$G/B$$ isn't.)\n\nEach component of $$G = \\color{maroon}{\\Bbb Z_2}\\times \\color{darkblue}{\\Bbb Z_4}$$ contains a factor of $$\\Bbb Z_2$$. Each of $$A$$ and $$B$$ is isomorphic to $$\\Bbb Z_2$$. When we construct the quotient of $$G$$ by $$A$$ or $$B$$, we get a different result depending on which component we divide the factor from.\n\nTwo groups that are isomorphic share the same internal structure. But a quotient $$G/N$$ doesn't depend only on the internal structures of $$G$$ and $$N$$. The quotient also depends on the relationship between $$G$$ and $$N$$. So even though $$A$$ and $$B$$ have the same internal structure, $$G/A$$ and $$G/B$$ are different, because those structures reside inside of $$G$$ in different ways.\n\nTopological and illustrated version of Arthur's nice answer:\n\nA trefoil knot $$K_3$$ in $$\\Bbb R^3$$ cannot be unknotted but it can be in $$\\Bbb R^4$$. Topologically a trefoil knot $$K_3$$ and a circle are homeomorphic i.e. one can deform one to other without cutting or gluing. So these two (one in $$\\Bbb R^3$$ and one in $$\\Bbb R^4$$) are the same but one uncovers a branch of mathematics known as Knot_theory while other one is fruitless from this point of view and application.\n\nI think theses type of \"same things\" is a categorical notion but I am not sure.\n\nsame is true for $$y=x$$ and $$y=\\exp(x)$$; both topologically are homeomorphic to $$\\Bbb R$$, but one contains only positive numbers. One can say similar properties by considering them as additive and multiplicative groups.\n\nThese might be of interest to you:\n\nLet me for convenience, first, assume all subgroups are normal, or that we are in an abelian group.\n\nWe form quotients of a group by its SUBgroups, not by another abstract GROUP. Subgroup means a SPECIFIC subset (which is closed ....) As groups Z and 2Z are \"same\". But as subsets of Z they are different.\n\nNow to non-abelian case. Take $$G= \\{+1,-1\\}\\times S_3$$ where the first factor $$H$$ is a group (of order 2) wrt multiplication of numbers.\n\nThis group $$G$$ has more subgroups of order 2 (isomorphic to H as a group) namely ones generated by a transposition in the second factor $$S_3$$. These subgroups are NOT even normal in $$G$$. SO quotient does not even exist by these subgroups.", "date": "2022-07-01 16:09:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4343130/how-same-are-two-isomorphic-groups", "openwebmath_score": 0.8676400184631348, "openwebmath_perplexity": 177.2949996617512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018434079934, "lm_q2_score": 0.8887587890727754, "lm_q1q2_score": 0.8667192094488265}}
{"url": "https://math.stackexchange.com/questions/2309292/seperating-2n-elements-with-subsets-of-size-n", "text": "Seperating $2n$ elements with subsets of size $n$\n\nSuppose I have a set $X$ with $2n$ elements ($n$ nonzero natural number). I now want to find a collection of subsets $X_1, \\ldots, X_k$ of $X$ such that\n\n\u2022 Every $X_i$ contains $n$ elements.\n\u2022 For every $x, y \\in X$, there exists an $X_i$ such that $x \\in X_i$ and $y \\not\\in X_i$ or $x \\not\\in X_i$ and $y \\in X_i$ (the $X_i$ \"seperate\" the elements of $X$, in that the topology on $X$ generated by the $X_i$ is Kolmogorov)\n\u2022 $k$ is minimal with respect to the above to properties.\n\nAn algorithm to find $X_1, \\ldots, X_k$ would be great, but I am mostly looking for a way to calculate $k$. My intuition tells me that $k$ should be at least $\\log_2(2n)$, more specifically I feel like $$\\max_{x \\in X} \\# \\lbrace y \\in X \\mid \\not\\exists i \\in \\lbrace 1, \\ldots, j \\rbrace : x \\in X_i \\enspace \\& \\enspace y \\not\\in X_i \\rbrace \\geq \\log_2(\\frac{2n}{j})$$ should hold for all $j \\in \\lbrace 1, \\ldots, k \\rbrace$, but I don't know how to prove this.\n\n\u2022 Consider sets with $2^l$ elements first \u2013\u00a0Scipio Jun 4 '17 at 12:09\n\u2022 I did; even if $n=4$ I don't see how to proceed. (i.e. I know that $k = 2$ is possible, but do not see how to show that $k = 1$ is impossible) \u2013\u00a0Bib-lost Jun 4 '17 at 12:11\n\u2022 $X_1$ contains two elements $a,b$; clearly the second requirement doesn't hold for these two elements if you only have a single set. \u2013\u00a0Scipio Jun 4 '17 at 12:18\n\u2022 By $n=4$ I meant the case where $X$ has $2*4 = 8$ elements. The case with $4$ elements is indeed clear. :) And surely, also the case with $8$ elements can be computed by hand in a reasonably short time; what I meant is that I do not see an argument which could somehow be generalised. \u2013\u00a0Bib-lost Jun 4 '17 at 12:44\n\u2022 Could you add the cases $n=2,3,4$ into your question & give us the lists $X_1, X_2 , \\cdots , X_k$ ? \u2013\u00a0Donald Splutterwit Jun 4 '17 at 12:50\n\nSuppose $X$ contains $2^l$ elements. Let $(A_i^m)$ be the family of subsets of $X$ of non-separated elements after introducing $X_1, \\cdots, X_m$, i.e. sets of elements for which the second requirement does not hold (check that these sets are well defined and uniquely partition $X$). Ultimately, we want that all the subsets $A_i$ contain just a single element (i.e. every element is separated from all other elements).\n\nWith each $X_i$ we want to separate as many elements as possible. Now, at first $A_1 = X$ containing $2^l$ elements and we take $X_1$ just an arbitrary set of $n$ elements. After this, we have sets $A_1$, $A_2$, both containing $2^{l-1}$ elements. Now, with $X_2$ we can again split both in half, resulting in four 'remaining sets' with $2^{l-2}$ elements. Continue recursively to see that $k=l$. Moreover, it is clear that after introducing $m$ sets, $\\max_i \\# A_i^m \\geq 2^{l-m}$, so that our result is indeed optimal.\n\nFor example, start with $X = A_1^0 = \\{1,2,3,4,5,6,7,8\\}$. Introduce $X_1 = \\{1,2,3,4\\}$ to get $A_1^1 = \\{1,2,3,4\\}, A_2^1 = \\{5,6,7,8\\}$. Introduce $X_2 = \\{1,2,5,6\\}$ and $X_3 = \\{1,3,5,7\\}$ to finish the job. (Make sure to check the corresponding sets $(A_i^m)$.)\n\nNow, for any other number of elements $2n$, your 'splits' wil not be optimal. You can check that $\\max_i \\# A_i^m \\geq 2n\\cdot2^{-m}$ still holds.\n\n\u2022 So probably we have that, in general, $k$ is the unique integer satisfying $2^{k-1} < 2n \\leq 2^{k}$? \u2013\u00a0Bib-lost Jun 4 '17 at 13:57\n\u2022 yes exactly :) This hinges on the very last statement in my answer, so make sure to check that carefully. \u2013\u00a0Scipio Jun 4 '17 at 14:01\n\nThere should be $k=\\lceil log_2(2n) \\rceil$ partitions, $2k$ sets.\n\nWe have the conditions: $$\\mathcal{P}_1:\\forall X_i, \\text{num}\\{X_i\\}=n$$ $$\\mathcal{P}_2:\\forall x_i,x_j \\exists X_i, x_i \\in X_i, x_j \\not\\in X_i$$\n\nIt is clear for sets of $2n=2^k$ elements, the partitions are binary, hence optimal. Under this case, the condition $\\mathcal{P}_2$ implies the stronger: $$\\mathcal{P}_{2b}:\\forall x_i,x_j \\exists X_i,X_j, x_i \\in X_i, x_j \\in X_j$$ because each partition requires its complement. If there is not complement, the property is not met. $$\\forall X_i \\exists Y_i=X -X_i$$\n\nIf the set contains $2n=2^k+2m$ elements, $2m<2^k$, each $2m$ remaining elements shall be included into different disjoint partitions. Again in this case the partition complement must exist and the condition $\\mathcal{P}_{2}$ will not introduce any benefit against $\\mathcal{P}_{2b}$.\n\n\u2022 Don't you mean $k = \\lceil \\log_2(2n) \\rceil$? Surely if $n=3$ you need at least $2 = \\lceil log_2(6) \\rceil$ partitions? \u2013\u00a0Bib-lost Jun 4 '17 at 14:14", "date": "2019-10-14 09:54:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2309292/seperating-2n-elements-with-subsets-of-size-n", "openwebmath_score": 0.9184891581535339, "openwebmath_perplexity": 215.97540111362397, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018398044143, "lm_q2_score": 0.8887587905460026, "lm_q1q2_score": 0.8667192076828079}}
{"url": "https://math.stackexchange.com/questions/3192613/special-case-of-bertrand-paradox-or-just-a-mistake", "text": "# Special case of Bertrand Paradox or just a mistake?\n\nI've been working on a question and it seems I have obtained a paradoxical answer. Odds are I've just committed a mistake somewhere, however, I will elucidate the question and my solution just in case anyone is interested.\n\nI want to know what is the average distance between two points on a circle of radius 1 where we consider only the boundary points.\n\nMy attempt is as follows:\n\nConsider a segment of the diameter x which is uniformly distributed between 0 and 2. Then you can calculate the distance between the points (2,0) and the point determined by x just by elementary geometry as this picture shows:\n\nenter image description here\n\nHere in the picture, the green segment is the geometric mean and the orange one is the distance whose distribution we want to know. Just by calculating the expected value, we obtain:\n\n$$E\\left(\\sqrt{(4-2X)}\\right) = \\int_{0}^{2} \\sqrt{(4-2x)}\\cdot\\frac{1}{2} dx = 1.333.... = \\frac{4}{3}$$\n\nWhere $$\\sqrt{(4-2x)}$$ is the transformation of the random variable and $$\\frac{1}{2}$$ is the pdf of a uniform distribution $$[0,2]$$.\n\nAlso, if we derive the pdf of the transformation we obtain the same result:\n\n$$y = \\sqrt{(4-2x)} , x = 2- \\frac{y^2}{2}, \\mid\\frac{d}{dy}x\\mid = y$$\n\n$$g(y)=f(x)\\cdot\\mid\\frac{d}{dy}x\\mid = \\frac{1}{2}\\cdot y$$\n\n$$E(Y)= \\int_{0}^{2}y\\cdot\\frac{1}{2}\\cdot y dy = 1.333.... = \\frac{4}{3}$$\n\nI have seen a different approach somewhere else where the distribution of the angle is considered as a uniform distribution between 0 and $$\\pi$$ and the final result was:\n\n$$1.27... = \\frac{4}{\\pi}$$\n\nThat's pretty much the problem I found. Maybe I just did it wrong in some step but it all makes sense to me. I know this is not exactly what we refer as Bertrand paradox but it just suggests something like that because both problems handle with segments in circumference and maybe my result is wrong because it does not hold for rotations of the circle or something like that (I read a little bit about the Bertrand's Paradox).\n\nThat's pretty much it. Also sorry for my bad English and maybe I'm also wrong in something pretty elemental since I've just started learning about probability theory. It's also my first post so I will try to improve my exposition and LateX use in the following ones.\n\n\u2022 welcome to MSE. I have attempted to edit your question. In case you find any discrepancy with your idea please check. Also, you can look up the edits I have made that will help you for your future questions. Thank you! \u2013\u00a0Mann Apr 19 at 21:57\n\nThis is a very nice question, well-written and researched before posting. You have clearly put a lot of careful into your question and it is very much appreciated on this site (and I thought your English reads perfectly naturally). Thanks so much for taking the time to ask a serious and well-considered question. I hope this answer comes close to meeting the high standard of quality set by your question.\n\nI don\u2019t see any calculation errors on your part, and I\u2019ll wager you checked those thoroughly. The flaw is something much more subtle and is related to the initial framing of the problem.\n\nYou have correctly solved the problem of \u201cwhat is the average distance between the rightmost point of a circle and another point projected upwards from a randomly chosen point on the diameter\u201d. But this doesn\u2019t quite capture the same probability distribution as \u201crightmost point and another randomly chosen point on the circle\u201d (which is sufficient to capture the dynamics of \u201ctwo random points on the circle\u201d).\n\nThere is a hidden non-uniformity in the projection process, and that\u2019s because the circle has varying slope so the projection hits the circle differently at different positions.\n\nHere\u2019s an experiment that should help convince you that uniform distribution on the diameter does not yield uniform distribution on the circle boundary. Fundamentally, a uniform distribution should treat all arcs of the same length equally: a random point has a $$1/4$$ chance of falling on a quarter-circular arc, regardless of which quarter circle that is.\n\nNow compare two specific quarter-circles and their projections onto the diameter: 1) the quarter-circle adjacent to $$(2,0)$$, i.e. 12 o\u2019clock to 3 o\u2019clock, and 2) the quarter-circle centered around the top-most point $$(1,1)$$, i.e. 10:30 to 1:30.\n\nThese both project onto the diameter with no self-overlap (avoiding double counting). But the first one projects down to the interval $$[0,1]$$ which has length $$1$$, and the second one projects down to $$[-\\sqrt{2}/2,\\sqrt{2}/2]$$, which has length $$\\sqrt2$$.\n\nThis should indicate to you that the two distributions are not equivalent (though they may intuitively seem so). Consequently, the random process of choosing uniformly from the diameter and projecting onto the circle results in disproportionately favoring the top and bottom sectors of the circle over the left and right.\n\n\u2022 Thanks! I really appreciate your words and yes, it seems the problem really deals with something subtle. I'm going to add some considerations in an answer. \u2013\u00a0Hyz Apr 19 at 19:49\n\nThanks, Erick Wong for your feedback. After your answer, I calculated the distribution of the arc length subject to the uniform distribution of the point on the diameter. In fact: if we want to express the arc length $$l$$ as a function of $$x$$, $$l = f(x)$$ we obtain:\n\n$$l = \\arccos(1-x), x = 1-\\cos{l}, |\\frac{d}{dl}(x)| = \\sin{l}$$\n\n$$l_{pdf} = x_{pdf} \\cdot |\\frac{d}{dl}x| = \\frac{1}{2}\\cdot \\sin{l}$$.\n\nSo the arc length does not distribute uniformly, we have \"lost it\", we might say. That's what was wrong. For instance, if the arc length obeys a uniform distribution [0, $$\\pi$$], then we can calculate the segment $$s$$ as a function of the arc length:\n\nWe know $$l = f(x)$$ and want to know $$s = h(l)$$. If we calculate $$s = g(x)$$ we're done:\n\nFrom the image on the question I post, $$s = g(x)=\\sqrt{2x}$$ (or the opposite segment $$\\sqrt{4-2x}$$ ) then $$h = g \\circ f^{-1}$$, $$s = \\sqrt{2(1-cosl)}$$\n\n$$E(s) = \\int_0^\\pi{\\sqrt{2(1-cosl)}\\frac{1}{\\pi}dl}=1.273... = \\frac{4}{\\pi}$$\n\nAlso the pdf:\n\n$$s = h(l) = \\sqrt{2(1-cosl)} , l=h^{-1}(s)= 2\\cdot \\arcsin(\\frac{s}{2}), |\\frac{d}{ds}h^{-1}|=\\frac{2}{\\sqrt{4-s^2}}$$\n\n$$s_{pdf} = \\frac{1}{\\pi} \\frac{2}{\\sqrt{4-s^2}}$$\n\n$$E(s) = \\int_0^2{s\\cdot \\frac{1}{\\pi} \\frac{2}{\\sqrt{4-s^2}}ds} = 1.273... = \\frac{4}{\\pi}$$\n\nSo we're done. Also, the pdf of the segment suggests something related to the Cauchy distribution. Not exactly but certainly it has to do something with it. If we read the description of Cauchy distribution in Wolfram MathWorld:\n\n\"The Cauchy distribution, also called the Lorentzian distribution or Lorentz distribution, is a continuous distribution describing resonance behavior. It also describes the distribution of horizontal distances at which a line segment tilted at a random angle cuts the x-axis.\"\n\nAnd that's it. A really fascinating problem that introduces some subtle ideas of probability theory. If anyone knows something else please give me feedback. I really think there's a nice connection with Cauchy distribution.", "date": "2019-05-25 00:50:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3192613/special-case-of-bertrand-paradox-or-just-a-mistake", "openwebmath_score": 0.8446071743965149, "openwebmath_perplexity": 240.01646258919706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975201844849425, "lm_q2_score": 0.8887587846530937, "lm_q1q2_score": 0.8667192064198297}}
{"url": "https://math.stackexchange.com/questions/4104208/does-the-sequence-x-n1-x-nx-n2-converge-to-0-whenever-1-lt-x-0-lt0", "text": "# Does the sequence $x_{n+1}=x_n+x_n^2$ converge to $0$ whenever $-1\\lt x_0\\lt0$?\n\nMy question concerns using the fixed-point iteration to find the fixed point of the function $$f(x)=x+x^2=x(1+x)$$ (this function has a single fixed point at $$0$$).\n\n# The problem\n\nGiven some fixed $$x_0$$, define the sequence $$x_n$$ by \\begin{align*} x_{n+1}=x_n+x_n^2&&\\text{for all n\\geq0.} \\end{align*} Does this sequence converge to $$0$$ for all $$x_0$$ in the range $$-1\\lt x_0\\lt0$$? After some computational numerical exploration, I think the answer might be yes but I'm not so sure how to prove this.\n\n## Some notes\n\nThe sequence trivially converges to $$0$$ when $$x_0=-1$$ or $$x_0=0$$. It's also fairly easy to prove that it does not converge to $$0$$ whenever $$x_0$$ lies outside this range (i.e., $$x_0\\lt-1$$ or $$x_0\\gt0$$).\n\nI've already proved that for any $$x_0$$ in this range, all successive $$x_n$$s remain in this range (i.e., $$-1\\lt x_n\\lt0$$ for all $$n\\geq0$$) and that the sequence increases / gets closer to $$0$$ (i.e., $$x_{n+1}\\gt x_n$$), but is there any way to prove it actually converges to $$0$$?\n\n## Some examples\n\nIf, for example, $$-\\frac{1}{2}\\lt x_n\\lt-\\frac{1}{4}$$, we'll have $$-\\frac{3}{8}\\lt x_{n+1}\\lt-\\frac{1}{8}$$ and $$-\\frac{21}{64}\\lt x_{n+2}\\lt-\\frac{5}{64}$$, etc. So it seems as though the range of possible values gets closer to $$0$$. But, how would one prove this?\n\nYou can use the monotone convergence theorem. It tells us that a sequence $$\\{x_n\\}_{n=1}^\\infty\\subseteq \\mathbb R$$ converges if it is bounded and decreasing. For you sequence we have $$|x_{n+1}| = |x_n| |1+x_n| \\leq 1$$ since $$x_0 \\in (-1,0)$$. And likewise we have $$x_{n+1} = x_n(1+x_n) \\leq x_n$$ as $$1+x_n \\in (0,1)$$.\n\n(You can use induction to prove both these claims.)\n\nNow you have shown that $$\\lim_{n\\to\\infty} x_n=x$$ exists. From this we see that $$x = \\lim_{n\\to\\infty} x_{n+1} = \\lim_{n\\to\\infty} x_n(1+x_n) = x(1+x).$$ Hence $$x$$ satisfies $$x = x^2+x$$ so $$x=0$$.\n\n\u2022 Great, exactly what I was looking for! Apr 16, 2021 at 7:15\n\u2022 @\u0391\u0398\u03a9 You're right, I should have said increasing and replaced $\\leq$ with $\\geq$. Thanks. Apr 16, 2021 at 8:05\n\u2022 I've already proved that $x_n$ is monotonic increasing and bounded above by $0$ in this case. So this is all I needed to complete the argument (from what I understand, I've never formally studied real analysis, so I certainly may be missing something). Apr 16, 2021 at 8:06\n\u2022 @user178563 Glad to see that you agree. If we are to be very precise (which we should very well be, for this is what this forum is about) the part where it appears you want to show that $|x_n| \\leqslant 1$ by induction on $n$ is also dubious (the induction is not spelled out clearly).\n\u2013\u00a0\u0391\u0398\u03a9\nApr 16, 2021 at 8:08\n\nThe answer to your question (regarding convergence to $$0$$) is indeed affirmative and for a reason that is not too difficult to ascertain. For simplicity let us introduce the quadratic polynomial function: \\begin{align*} f \\colon \\mathbb{R} &\\to \\mathbb{R} \\\\ f(x)&=x^2+x \\end{align*} and let us note that $$f[(-1, 0)] \\subseteq (-1, 0)$$, in other words the interval $$(-1, 0)$$ is stable under $$f$$. To prove this latter claim it suffices to notice that $$x^2+x+1=\\left(x+\\frac{1}{2}\\right)^2+\\frac{3}{4} \\geqslant \\frac{3}{4}>0$$ for any $$x \\in \\mathbb{R}$$ (and thus in particular $$x^2+x>-1$$ always) and respectively that given $$x \\in (-1, 0)$$ we have $$x+1>0$$ yet $$x<0$$ which means that $$x^2+x=x(x+1)<0$$.\n\nThe immediate conclusion that we draw from this is that for any fixed $$t \\in (-1, 0)$$, the unique recursive sequence $$x \\in \\mathbb{R}^{\\mathbb{N}}$$ defined by $$x_0=t$$ and $$x_{n+1}=f(x_n)$$ for any $$n \\in \\mathbb{N}$$ has the property that actually $$x \\in (-1, 0)^{\\mathbb{N}}$$ and is thus a bounded sequence. Furthermore, since $$x_{n+1}=x_n+x_n^2 \\geqslant x_n$$ takes place for every $$n \\in \\mathbb{N}$$ we gather that $$x$$ is increasing (one can actually make the more precise observation that since $$x_n \\neq 0$$ for all $$n \\in \\mathbb{N}$$ under the stated conditions, the inequality mentioned above is actually strict, rendering $$x$$ into a strictly increasing sequence, however that is not essential for the study of convergence).\n\nAccording to one of the theorems of Weierstrass, any upper-bounded increasing sequence of real numbers is convergent, which applies in particular to $$x$$. Since $$x$$ is defined recursively with respect to the continuous function $$f$$, it follows with necessity that $$\\displaystyle\\lim_{n \\to \\infty} x_n$$ must be a fixed point for $$f$$. However it is obvious that there is just one such fixed point - the unique solution to the equation $$f(u)=u \\Leftrightarrow u^2=0$$ - which is $$0$$.", "date": "2022-08-18 05:15:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4104208/does-the-sequence-x-n1-x-nx-n2-converge-to-0-whenever-1-lt-x-0-lt0", "openwebmath_score": 0.980157732963562, "openwebmath_perplexity": 118.43701737003607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991554371311859, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8666950985131201}}
{"url": "http://www.askamathematician.com/2011/05/q-is-0-9999-repeating-really-equal-to-1/", "text": "# Q: Is 0.9999\u2026 repeating really equal to 1?\n\nMathematician: Usually, in math, there are lots of ways of writing the same thing. For instance:\n\n$\\frac{1}{4}$ = $0.25$ = $\\frac{1}{\\frac{1}{1/4}}$ = $\\frac{73}{292}$ = $(\\int_{0}^{\\infty} \\frac{\\sin(x)}{\\pi x} dx)^2$\n\nAs it so happens, 0.9999\u2026 repeating is just another way of writing one. A slick way to see this is to use:\n\n$0.9999... = (9*0.9999...) / 9 = ((10-1) 0.9999...) / 9$\n\n$= (10*0.9999... - 0.9999...) / 9$\n\n$= (9.9999... - 0.9999...) / 9$\n\n$= (9 + 0.9999... - 0.9999...) / 9 = 9 / 9 = 1$\n\nOne.\n\nAnother approach, that makes it a bit clearer what is going on, is to consider limits. Let\u2019s define:\n\n$p_{1} = 0.9$\n$p_{2} = 0.99$\n$p_{3} = 0.999$\n$p_{4} = 0.9999$\n\nand so on.\n\nNow, our number $0.9999...$ is bigger than $p_{n}$ for every n, since our number has an infinite number of 9\u2019s, whereas $p_{n}$ always has a finite number, so we can write:\n\n$p_{n} < 0.9999... \\le 1$ for all n.\n\nTaking 1 and subtracting all parts of the equation from it gives:\n\n$1-p_{n} > 1-0.9999... \\ge 0$\n\nThen, we observe that:\n\n$1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \\frac{1}{10^n}$\nand hence\n\n$\\frac{1}{10^n} > 1-0.9999... \\ge 0$.\n\nBut we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999\u2026 must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999\u2026 is clearly not bigger than 1. Hence, the only possibility is that\n\n$1-0.9999... = 0$\n\nand therefore that\n\n$0.9999... = 1$.\n\nWhat we see here is that 0.9999\u2026 is closer to 1 than any real number (since we showed that 1-0.9999\u2026 must be smaller than every positive number). This is intuitive given the infinitely repeating 9\u2019s. But since there aren\u2019t any numbers \u201cbetween\u201d 1 and all the real numbers less than 1, that means that 0.9999\u2026 can\u2019t help but being exactly 1.\n\nUpdate: As one commenter pointed out, I am assuming in this article certain things about 0.9999\u2026. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don\u2019t believe this about 0.9999\u2026 or would like to see a discussion of these issues, you can check this out.\n\nThis entry was posted in -- By the Mathematician, Equations, Math. Bookmark the permalink.\n\n### 94 Responses to Q: Is 0.9999\u2026 repeating really equal to 1?\n\n1. mathman says:\n\nlol. You need to know what each \u201cnumber\u201d is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting \u201cnumbers\u201d\n\n2. George says:\n\nWhile the mathematical argument presented here is correct, I have to say that limit and/or mathematical infinity introduced created confusions for many. What I wanted to say is that math infinity is really a math trick. It cannot and should not be real in the real physical world. Today this infinity plagues physics and its calculations. Maybe we need something new\u2026\n\n3. EMINEM says:\n\ngo to this https://www.youtube.com/watch?v=wsOXvQn3JuE site and get a mathematical proof that 0.999\u2026 doesnot equal to 1\u2026.\n\n4. . says:\n\nEMINEM, the video was published on April fool\u2019s day for a reason!\n\n5. Josh says:\n\nThe simplest explanation I\u2019ve seen for this is:\n\n1/3 = 0.333\u2026\n(1/3)3=1 = (0.333\u2026)3=0.999\u2026\n1 = 0.999\u2026\n\nI don\u2019t see how anyone can come to any conclusion other than 1=0.999\u2026 after understanding this.\n\n6. mathman says:\n\n1 / 3 = ??\nUse long division. There are 2 parts: the quotient and the remainder. For every iteration.\nFirst iteration is 0 R 1/3\nSecond is 0.3 R 1/30\nThird is 0.33 R 1/300\nEtc etc.\n\nWhere does your remainder go, Josh?\nYou don\u2019t account for it anywhere.\n\nThe remainder is never zero.\n\nHow many 9\u2019s are there in your 0.999\u2026 ?\n\n7. Josh says:\n\nHey mathman,\n\nThere are infinitely many 9s in my \u201c0.999\u2026\u201d. If it were anything other than infinitely many, then it would be less than 1.\n\nAnd as far as 1/3, your \u201cetc. etc.\u201d is an infinitely repeating \u201c0.333\u2026\u201d. I\u2019m failing to see how anything you\u2019ve presented has disproved what I said.\n\n8. Angel says:\n\n@mathman:\n\nThe remainder does not need to be accounted for. The division 1/3 is never formally complete if there is a remainder. Therefore, one must keep iterating the operations in long division. As the number of iterations n approaches infinity, the remainder does approach zero. That is what matters here. The number of 3s in the decimal expansion of 1/3 is infinite, and so is the number of 9s in the figure 0.999\u2026\n\n9. Anthony.R.Brown says:\n\nThe Original author of the Proof below\u2026used in this Thread \ud83d\ude42\n\nhttp://www.mathisfunforum.com/viewtopic.php?id=6069\n\nIt\u2019s impossible for Infinite/Recurring 0.9 (0.999\u2026) to ever equal 1\n\nBecause of the Infinite/Recurring 0.1 (0.001\u2026) Difference! \ud83d\ude42\n\nOnly by the using the + Calculation can it be possible (0.999\u2026) + (0.001\u2026) = 1\n\nAnthony.R.Brown\n\n10. \u00c1ngel M\u00e9ndez Rivera says:\n\nAnthony R. Brown,\n\nYour argument is flawed for the simple reason that the \u201cinfinite difference\u201d you describe equivalent to 0.000\u20261 is not a number. It doesn\u2019t exist in mathematics. It can\u2019t exist in mathematics. It is more of a problematic number than 0/0 or defining a left-identity of exponentiation Phi such that Phi^x=x. It is simply nonsense.\n\n11. Mathman says:\n\nAnthony,\nBeing someone who has researched and understands both sides of this issue, you will be told it\u2019s nonsense by mathematicians. It\u2019s their way of putting themselves above you to put you down. The issue with your argument is that 0.001\u2026 is defined as 0.00111111111111\u2026. You are having a problem defining what the idea in your head is. 0.000\u20261 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.\n\n12. Anthony.R.Brown says:\n\nHi \u00c1ngel M\u00e9ndez Rivera\n\nYour try at a counter argument\u2026saying (0.001\u2026) does not exist in mathematics ? can\u2019t exist in mathematics if (0.999\u2026) exist\u2019s which it does! \ud83d\ude42\nThe two are the related from the calculation 1 \u2013 (0.999\u2026) = (0.001\u2026) one cannot exist without the other!\n\nAnthony.R.Brown\n\n13. Anthony.R.Brown says:\n\nAnthony,\nBeing someone who has researched and understands both sides of this issue, you will be told it\u2019s nonsense by mathematicians. It\u2019s their way of putting themselves above you to put you down. The issue with your argument is that 0.001\u2026 is defined as 0.00111111111111\u2026. You are having a problem defining what the idea in your head is. 0.000\u20261 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.\n\nThen when I wanted to reply the message above is not shown ? so I will Answer below\u2026\n\nFirst of all what you are showing as 0.00111111111111\u2026. is wrong! there is only a single (1) following the Infinite/recurring calculation so your example above should look like 0.00000000000001\u2026.\nThe amount of zeros depends on how far one wants to show the calculation regarding decimal places!\n\nExample 0.9\u2026 has the infinite/recurring difference 0.1\u2026\nExample 0.99\u2026 has the infinite/recurring difference 0.01\u2026\nExample 0.99\u2026 has the infinite/recurring difference 0.001\u2026\n\nSo as you can see the infinite/recurring difference is always the same length as the infinite/recurring 0.9 \ud83d\ude42\n\nAnthony.\n\n14. Mathman says:\n\n0.001\u2026.. = 0.0011111111111111111\u2026..\nAnthony do you know what you are taking about?\n\n15. Anthony.R.Brown says:\n\nHi Mathman\n\nYes I fully know what I am talking about but you appear not to ?\n\nAs I pointed out in your example ? = 0.0011111111111111111\u2026.. it is wrong there can only be a single (1) otherwise when you do the calculation\u2026\n\n0.0011111111111111111\u2026.. + 0.0000000000000000009\u2026.. it will equal something like = 0.0011111201111111111\u2026 ?\n\nWhere as the True calculation (with both sides the 0.9\u2026 and the 0.1\u2026 the same amount of decimal places)\n\nCalculates 0.9999999999\u2026. + 0.0000000001\u2026. = 1\n\nRegards Anthony.\n\n16. Joshua says:\n\nThe flaw in your logic is thinking that there is ever space for a 0.0\u202601. The number 0.999\u2026 repeating forever by definition has no room to add anything to it. The 9\u2019s never stop to allow the insertion of any variation of 0.0001. And if there is no room to ever add anything to 0.999\u2026 infinitely repeating, then there is no space in between 0.999\u2026 infinitely repeating and 1, and therefore they are equal. If the 9\u2019s stopped repeating at any point, you would be correct, but they do not stop repeating, so you are incorrect.\n\n17. Mathman says:\n\nThe ellipsis \u201c\u2026\u201d mean \u201crepeating\u201d. I know what you are trying to say but you are using symbols incorrectly.\n1/1000\u2026 would be a more accurate to convey your idea.\n\n18. Anthony.R.Brown says:\n\nHi Joshua\n\nYour Quote:\u201dThe number 0.999\u2026 repeating forever by definition has no room to add anything to it\u201d\n\nA.R.B\n\nThe same as \u201cThe number 0.001\u2026 repeating forever by definition has no room to add anything to it\u201d\n\nThey both run side by side! \ud83d\ude42\n\n0.99999999999999999999999999999999999999999999999999999999\u2026.\n0.0000000000000000000000000000000000000000000000000001\u2026.\n\nIt\u2019s only when one want\u2019s to make the Value (1) that the + Calculation is needed it\u2019s the only way!\n\n( 0.9999999999\u2026.) + (0.0000000001\u2026.) = 1\n\n19. Joshua says:\n\nSorry Anthony, you have not wrapped your head around what 0.999\u2026 repeating for infinity means. It\u2019s the infinite part that\u2019s really important. I promise you, you are wrong, and once you understand what infinitely repeating means, you\u2019ll get it.\n\n20. \u00c1ngel M\u00e9ndez Rivera says:\n\nAnthony, no. They don\u2019t run side by side. The number you describe as the difference between 1 and 0.999\u2026 doesn\u2019t exist, because by definition, it is impossible to have a 1 after an infinite amount of zeros. It just is. There is no such calculation as the one you performed to be made because the number you describe doesn\u2019t exist.\n\n21. Mathman says:\n\nThere is an undefined infinitesimal space between 0.999\u2026 and 1. Just because it\u2019s undefined doesn\u2019t mean it doesn\u2019t exist.\n\n22. \u00c1ngel says:\n\nMathman,\n\n1. If it is undefined, then it does not exist. One implies the other.\n2. No, there is no infinitesimal space between 0.999\u2026 = 1, and this can be proven.\n\n0.999\u2026 = 9/10 + 9/100 + 9/1000 + \u2026 this is an infinite geometric series, which is evaluated and in fact defined by the formula a/(1 \u2013 r), so the sum 0.999.. = (9/10)/(1 \u2013 1/10) = (9/10)/(9/10) = 1. As for the formula a/(1 \u2013 r), it can be easily derived and proven\n\n23. Anthony.R.Brown says:\n\n( 1.1 ) x 0.9 = 0.99 \u201d One Decimal Place = \u201d 0.01 < 1\n\n( 1.11 ) x 0.9 = 0.999 \" Two Decimal Place's = \" 0.001 < 1\n\n( 1.111 ) x 0.9 = 0.9999 \" Three Decimal Place's = \" 0.0001 < 1\n\n( 1 / 0.9 ) x 0.9 = 0.9\u2026 \" Infinite Decimal Place's = \" 0.1\u2026 < 1 \"\n\n24. Joshua says:\n\nMathman,\n\nYou would be correct if the repeating 0.999\u2026 ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.\n\nMost of the confusion over this matter seems to be about the nature of infinity, and how it applies to infinitely repeating decimals. The importance of wrapping our heads around what \u201cinfinitely repeating\u201d means cannot be overstated.\n\n25. \u00c1ngel M\u00e9ndez Rivera says:\n\nAnthony,\n\n(1.1)(0.9) = 0.99\n\n(1.11)(0.9) = 0.999\n\n(1.111)(0.9) = 0.9999\n\n(1.111\u2026)(0.9) = 0.999\u2026. = (1/0.9)(0.9), because 1.1111\u2026 = 1/0.9. But any number divided by itself equals 1, so (0.9)/(0.9) = 1, therefore 1 = 0.9999\u2026.\n\n26. Anthony.R.Brown says:\n\nQuote: \u201cJoshua says:\nYou would be correct if the repeating 0.999\u2026 ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.\u201d\n\nA.R.B\n\nAbsolutely Correct!!! So now you and I agree it\u2019s impossible for 0.999\u2026to ever stop and become equal to 1 \ud83d\ude42\n\nSo how do we make 0.999\u2026 = 1 and that\u2019s where we have the answer!\u2026\n\n( 0.999\u2026) + (0.001\u2026) = 1\n\n{When great minds think alike Mountains can be moved!} \ud83d\ude42\n\n27. Joshua says:\n\nNo, Anthony, you still don\u2019t understand. The fact that it never stops is the very thing that makes it equal to 1. Again, I come back to encouraging you to take the time to actually understand what an infinitely repeating decimal is.\n\n28. Anthony.R.Brown says:\n\nHi Joshua\n\nWhat you don\u2019t seem to understand is that infinite/recurring 0.999\u2026 is a Calculation on a Number! and the number Starts as 0.9 (It\u2019s not something that is just plucked from the sky from nothing!)\n\nYour biggest mistake! is \u201cThe fact that it never stops is the very thing that makes it equal to 1\u201d ? a Number that Starts < 1 can never be made to equal 1 unless the + Calculation is used!\n\n29. Joshua says:\n\nAnthony, the number 0.999\u2026 with infinite 9s is not a calculation. It is a static number. The 9s are not constantly being generated, they exist (all infinity of them) at once. It does not \u201cstart\u201d at 0.9. It exists as 0.999\u2026 with an infinite number of 9s. You don\u2019t have to work up to it, it just is.\n\nIt is clear that you still do not understand what an infinitely repeating decimal is. You do not understand the basic premise of what we are talking about. Therefore, you can not speak to it in an intelligent or useful manner.\n\nI\u2019m not trying to be mean or anything, either. I appreciate people pointing out when I am wrong so I can learn what is correct. You clearly don\u2019t understand the basics of this topic, therefore you are wrong, and it is an opportunity for you to learn something new and broaden your knowledge base.\n\n30. Angel says:\n\n\u201cWhat you don\u2019t seem to understand is that recurring 0.999\u2026 is a calculation on a number!\u201d\n\nThis is wrong. 0.999\u2026 is a number. It isn\u2019t a process, it already is a number. It doesn\u2019t \u201cstart\u201d out as 0.9. It already is and exists as 0.999\u2026, it doesn\u2019t start anywhere or end anywhere. It isn\u2019t a changing or moving thing. It just is a number.\n\n31. Anthony.R.Brown says:\n\nSo Joshua & Angel\u2026\n\nAre both wrong! if they think (the number 0.999\u2026 with infinite 9s is not a calculation) ???\n\nBecause all the so called Proofs ? are based on some kind\u2019s of Calculations!!!\n\nJust to back that up\u2026a Google search for 0.999\u2026 proofs shows (About 367,000 results) 21/09/2017\n\nAnd they are all some kind\u2019s of Calculations! \ud83d\ude42\n\nOne fine example is shown below\u2026with many Calculations!\n\nAnthony.R.Brown\n\n32. Mathman says:\n\n\u00c1ngel,\na/(1-r) comes from a limit. The LIMIT of the infinite series is 1. The definition is flawed and contradicts the very theory of limits.\n\nIt may be impossible to have a 1 AFTER an \u201cinfinite amount\u201d. That is because \u201cinfinite amount\u201d doesn\u2019t exist. Stop trying to quantify infinity. It is not impossible to describe the idea, however: 1/1000\u2026\n\nThe very definition of limits show there is ALWAYS AN EPSILON.\nn=1: 1-0.9= 0.1\nn=2: 1-0.99 = 0.01\nFor ALL values n, there is a difference, or, contradictory to Anthony, there is always a space. If you travserse 90% of the remaining distance , ad infinitum, there will always be 10% of the last distance ahead of you. You\u2019ll never complete the journey of say, 1 mile, going 90% each time.\n\nI argue that \u201c0.999\u2026\u201d is not a number at all.\n\u03a3 (n=1 to N) 9/10^n as N goes to infinity, implies there is always a rational number being added to the previous total. The number 1 has nothing being addd to it. Therefore they aren\u2019t the same.\nAlso, \u2018something\u2019 that always has another addend can\u2019t be a number. It\u2019s contradictory because it always has another number being added to it.\n\n33. Mathman says:\n\nThus: 1-0.999(n-times) = 1/10^n\nYou may let n get as arbitrarily large as you like.\nYou cannot have \u201cinfinite amount of 9\u2019s\u201d though.\nYou can have the LIMIT as the number of 9\u2019s gets arbitrarily large. This is a number. \u201cEndless 9\u2019s\u201d really is nonsense when it comes to a number which should really be finite.\nSo then, for ALL n in N, that is, for every integer, there is a difference which is smaller than the last and never 0.\nLearn what arbitrarily large means, and how limits really work, and what infinite means, and how infinity is applied in calculus. Infinity and limits go hand in hand. You don\u2019t plop infinity wherever you feel.\n\n34. Anthony.R.Brown says:\n\nLIMIT Cannot be used with Infinite/Recurring 0.9 (0.999\u2026) because you will then Contradict the term (Infinite/Recurring) by forcing the .9\u2019s to cut off at some point depending on where you want the LIMIT to be, regarding decimal places calculated,this is much the same as the cut off point used when Infinite/Recurring 0.9 (0.999\u2026) is forced at some point to become equal to 1\n\nSo back to\u2026\n\nIt\u2019s impossible for Infinite/Recurring 0.9 (0.999\u2026) to ever equal 1 (Unless you use a LIMIT) ? \ud83d\ude41\nAnd the Infinite/Recurring 0.1 (0.001\u2026) Difference! (Also does not us a LIMIT)\n\nBoth are LIMITless \ud83d\ude42\n\nOnly by the using the + Calculation can it be possible for (0.999\u2026) + (0.001\u2026) = 1\n\n35. \u00c1ngel M\u00e9ndez Rivera says:\n\nNo, we do not get to choose where the limit is. The limit is a number larger than every element of the sequence of partial evaluations of a function, which means that the limit is larger than the sequence itself. In other words, the term of the sequence which equals the limit is itself located at an infinite position of the sequence. Any finite evaluation yields less than the limit. However, 0.99\u2026 has an infinite number of 9s, so it exactly equals the limit. We\u2019re not cutting it off to equal 1, you\u2019re cutting it off to equal less than 1, but you\u2019re Not actually using the sequence of infinite nines, so you\u2019re argument is wrong.\n\nThere is no such a thing as limitless in mathematics, and 0.000\u20261 is nonsense and does not exist.\n\n36. \u00c1ngel M\u00e9ndez Rivera says:\n\nMathman,\n\nRead my previous response to your argument. I already explained why your description of the definition of a limit is wrong and why the idea of there always being an epsilon is irrelevant when actually evaluating the limit of a function as the argument approaches c.\n\n37. Anthony.R.Brown says:\n\nQuote: \u201c\u00c1ngel M\u00e9ndez Rivera says:\nSeptember 22, 2017 at 1:50 pm\nNo, we do not get to choose where the limit is.\n\nA.R.B\nYes! you do by using it in the first place!\n\nQuote: \u201c\u00c1ngel M\u00e9ndez Rivera says:\n\u201cHowever, 0.99\u2026 has an infinite number of 9s, so it exactly equals the limit. \u201d\n\nA.R.B\nHow can an infinite number of 9s, exactly equals the limit (You have set!) which is 1 ? which is an Integer number > 0.99\u2026\n\nQuote: \u201c\u00c1ngel M\u00e9ndez Rivera says:\nThere is no such a thing as limitless in mathematics, and 0.000\u20261 is nonsense and does not exist.\n\nA.R.B\nInfinite/Recurring 0.9 (0.999\u2026)\nInfinite/Recurring 0.1 (0.001\u2026)\nAre both (Individually!) LIMITless they only have a LIMIT when added! together to equal 1\n(0.999\u2026) + (0.001\u2026) = 1\n\n38. The Physicist says:\n\n@Anthony.R.Brown\nThey\u2019re not using the word \u201climit\u201d the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.\n\n39. Anthony.R.Brown says:\n\nThe Physicist says: Quote:\u201dThey\u2019re not using the word \u201climit\u201d the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.\u201d\n\nA.R.B\n\nThe word (LIMIT) however you define it ? or look it up on some Math sites ? is being used in a way\u2026 and the same way ? so as to make a Calculated Number,that starts life less than 1 somehow become equal to 1 ? \ud83d\ude41\n\nA better way to look at the problem is\u2026\n\nInfinite/Recurring 0.9 (0.999\u2026) = Universe! \ud83d\ude42\n\nAnd if we place a Wall at the End of the Universe! ? there will still be Universe! behind the Wall\u2026\n\n0.9 (0.999\u2026) Universe! { Wall = 1} 0.9 (0.999\u2026) Universe!\u2026\n\n40. \u00c1ngel M\u00e9ndez Rivera says:\n\nAnthony, no. There is no such a thing as a limitless number. Please, study the topic you\u2019re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.\n\n41. \u00c1ngel says:\n\nMathman,\n\nThere is a justified reason why challenging mathematics makes you a crackpot: we can prove that by challenging mathematics, you are wrong.\n\n2. Euler did not define limits.\n\n3. Also, a definition cannot be wrong. This is, ironically, the definition of a definition. So Euler\u2019s definition of infinite summation is acceptable. Euler\u2019s definition renders summation continuous in the entire complex plane, which is what it should be according to group theory.\n\n4. It makes perfect sense to add an infinite number of things. We do it all the tim in physics, and we get observable results that we do observe consistently.\n\n5. No, your term-by-term subtraction was done wrong. Subtraction is done digitwise right to left.\n\n6. 0.999\u2026 is not a bad way to represent 9/10 + 9/100 + \u2026, but rather it is the only to represent it since it is the very definition. It is how decimal representation works. It can\u2019t also be represented as Sum[m = 1, m= n \u2014> Infinity](9*10^(-m)), which can be shown to equal 1 since the sum is 0.9(1 \u2013 10^[-n])/(1 \u2013 1/10), and when n \u2014> Infinity, this expression simplifies to 1. This is the limit.\n\n7. Yes, I agree that f(x = c) does not equal lim x\u2014>c [f(x)]. However, 0.999\u2026 isn\u2019t equal to f(x=c), but rather equal to lim x\u2014>c [f(x)]. 0.999\u2026. IS the limit of the partial sums by the very definition of decimal representation, and the partial sum limits to 1 evaluated by Taylor-geometric series. By transitive property, 0.999\u2026 = 1.\n\nRemember: infinite sums are a thing. Irrational numbers prove it. It has been proven. It isn\u2019t a definition to say \u201cwe can add infinitely many things\u201d. No, we can prove it mathematically. That being said, 0.999\u2026 = 9/10 + 9/100 + \u2026 by definition. Not debatable. Now, 9/10 + 9/100 + \u2026 = Sum [m = 1, lim n \u2014> Infinity][9*10^(-m)]. This latter expression is a limit. Keep in mind that summation is a linear hyper-operator, which makes f(y) = Sum[k=x, y][g(k)] continuous for all continuous g. Now, also remember that by the very definition of continuity, lim (x \u2014> c) f(x) = f(c). Hence, Sum [m = 1, n lim n \u2014> Infinity][9*10^(-m)] = Sum[m = 1, Infinity][9*10^(-m)]. This implies that 0.999\u2026 is actually a valid sum and therefore a valid decimal representation. You can add infinitely many terms. Finally, the latter left hand side of the previous identity mentioned is equal to lim (n \u2014> Infinity) 9/10[1 \u2013 10^(-n)]/(1 \u2013 1/10) = (9/10)/(9/10) = 1. By the transitive property, 0.999\u2026 = 1.\n\n42. Jesse Sherer says:\n\n0.999\u2026 is or it isn\u2019t equal to 1. if you are x distant to our Universe you are not in our Universe. no 3 dimentional universe is infinite unless all universe could be infinite too. 1-x=0.9999\u2026. if you are x distant from our universe you are not in our universe. you might say x doesn\u2019t exist, but, nether does 1/9 in base ten math. base ten is a rational universe of numbers there can be no number that equates to almost 1 but not 1 ,,or almost 1 is 1 in a rational universe.. at some point you have to say that if an infinite string of numbers representing both a fraction of a number and a whole number, then there must exist an infinity small number also exists that both equals 0 and almost 0 too. so x is more rational of a number that can add to .999999\u2026 and get 1.. base ten can\u2019t solve it but logic can.. 0.9999.. is a number or it isn\u2019t x is a number or it isn\u2019t 1-x=0.9999\u2026. or it doesn\u2019t 1=0.99999\u2026 or it doesn\u2019t\n\n43. Anthony.R.Brown says:\n\nJesse Sherer says:\nDecember 1, 2017 at 6:41 am\n\nA.R.B\n\n= Nonsense! \ud83d\ude41\n\n0.9999.. is a number Less than 1 \ud83d\ude42\n\n44. Anthony.R.Brown says:\n\n\u00c1ngel M\u00e9ndez Rivera says:\nSeptember 24, 2017 at 5:02 pm\nAnthony, no. There is no such a thing as a limitless number. Please, study the topic you\u2019re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.\n\nA.R.B\n\nTotally Wrong! Give me any Number you want ? and I will add any Number I want to it! Which makes any Number you provide {a limitless number} \ud83d\ude42", "date": "2018-02-25 13:28:36", "meta": {"domain": "askamathematician.com", "url": "http://www.askamathematician.com/2011/05/q-is-0-9999-repeating-really-equal-to-1/", "openwebmath_score": 0.7499731779098511, "openwebmath_perplexity": 1048.8144395662018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9935117313638977, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.866687347270121}}
{"url": "https://stats.stackexchange.com/questions/186111/find-the-rotation-between-set-of-points/207407", "text": "# Find the rotation between set of points\n\nI have two sets (sourc and target) of points (x,y) that I would like to align. What I did so far is:\n\n\u2022 find the centroid of each set of points\n\u2022 use the difference between the centroids translations the point in x and y\n\nWhat I would like is to find the best rotation (in degrees) to align the points.\n\nAny idea?\n\nM code is below (with plots to visualize the changes):\n\n# Raw data\n## Source data\nsourc = matrix(\nc(712,960,968,1200,360,644,84,360), # the data elements\nnrow=2, byrow = TRUE)\n\n## Target data\ntarget = matrix(\nc(744,996,980,1220,364,644,68,336), # the data elements\nnrow=2, byrow = TRUE)\n\n# Get the centroids\nsCentroid <- c(mean(sourc[1,]), mean(sourc[2,])) # Source centroid\ntCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid\n\n# Visualize the points\npar(mfrow=c(2,2))\nplot(sourc[1,], sourc[2,], col=\"green\", pch=20, main=\"Raw Data\",\nlwd=5, xlim=range(sourceX, targetX),\nylim=range(sourceY, targetY))\npoints(target[1,], target[2,], col=\"red\", pch=20, lwd=5)\npoints(sCentroid[1], sCentroid[2], col=\"green\", pch=4, lwd=2)\npoints(tCentroid[1], tCentroid[2], col=\"red\", pch=4, lwd=2)\n\n# Find the translation\ntranslation <- tCentroid - sCentroid\ntarget[1,] <- target[1,] - translation[1]\ntarget[2,] <- target[2,] - translation[2]\n\n# Get the translated centroids\ntCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid\n\n# Visualize the translation\nplot(sourc[1,], sourc[2,], col=\"green\", pch=20, main=\"After Translation\",\nlwd=5, xlim=range(sourceX, targetX),\nylim=range(sourceY, targetY))\npoints(target[1,], target[2,], col=\"red\", pch=20, lwd=5)\npoints(sCentroid[1], sCentroid[2], col=\"green\", pch=4, lwd=2)\npoints(tCentroid[1], tCentroid[2], col=\"red\", pch=4, lwd=2)\n\n\u2022 I cannot read your code, but the operation you need is called Procrustes rotation. Have you heard of it? It works when points are already paired ($x_i,y_i$). Pre-rotation optional operations include translation and scaling, and optional post-rotational isoscaling. \u2013\u00a0ttnphns Dec 10 '15 at 17:13\n\u2022 A complex regression will do the job. \u2013\u00a0whuber Dec 10 '15 at 17:23\n\u2022 I've seen, that, rotating the system about 180 degrees, then the pairs $(a,C),(b,D),(c,A),(d,B)$ become neighbours - and this is even a better fit than the best fit of the original $(a,A),(b,B),(c,C),(d,D)$ (where the small letters stand for vector source and capital letters for vector target) I've not seen this possibility mentioned and explicitely allowed or disallowed. Are you sure that you don't want that better fit? \u2013\u00a0Gottfried Helms Sep 4 '16 at 14:05\n\nThis can be done using the Kabsch Algorithm. The algorithm finds the best least-squares estimate for rotation of $RX-Y$ where $R$ is rotation matrix, $X$ and $Y$ are your target and source matrices with 2 rows and n columns.\n\nIn [1] it is shown that this problem can be solved using singular value decomposition. The algorithm is as follows:\n\n1. Center the datasets so their centroids are on origin.\n2. Compute the \"covariance\" matrix $C$=$XY^T$.\n3. Obtain the Singular Value Decomposition of $C=UDV^T$.\n4. Direction adjustment $d=sign(det(C))$.\n5. Then the optimal rotation $R=V\\left( \\begin{array}{ccc} 1 & 0 \\\\ 0 & d \\\\ \\end{array} \\right)U^T$\n\nI don't know of any implementation in R so wrote a small function below.\n\nsrc <- matrix(c(712,960,968,1200,360,644,84,360), nrow=2, byrow=TRUE)\ntrg <- matrix(c(744,996,980,1220,364,644,68,336), nrow=2, byrow=TRUE)\n\n\nKabsch algorithm in an R funtion:\n\nkabsch2d <- function(Y, X) {\nX   <- X-rowMeans(X)\nY   <- Y-rowMeans(Y)\nC   <- X %*% t(Y)\nSVD <- svd(C)\nD   <- diag(c(1, sign(det(C))))\nt(SVD$v) %*% D %*% t(SVD$u)\n}\n\n\nCenter the points:\n\nsrc <- src-rowMeans(src)\ntrg <- trg-rowMeans(trg)\n\n\nObtain rotation:\n\nrot <- kabsch2d(src, trg)\n\n\nResult (black - original source, red - original target, green - rotated target)\n\nplot(t(src), col=\"black\", pch=19)\npoints(t(trg), col=\"red\", pch=19)\npoints(t(rot %*% trg), col=\"green\", pch=19)\n\n\n\u2022 +1. However, the answer could be further much better if you included discourse about how the algo is related to well-known Procrustes rotation problem. \u2013\u00a0ttnphns Mar 19 '16 at 14:03\n\nI've done this with an iterative optimum-search, and tested 2 versions.\nI've taken the original arrays and centered them calling this arrays cSRC and cTAR . Then I've done a loop with angles $\\varphi$ between $0$ and $2 \\pi$ , and for each angle I computed the error-criterion using the difference between the rotated $\\small D=rot(\\text{cSRC} ,\\varphi)- \\text{cTAR}$.\n\n1. In version 1) I took as criterion the sum-of-squares of all entries in $\\small D$ as $$err_1 = \\small \\sum_{k=1}^4 \\small((D_{k,1})^2+(D_{k,2})^2)$$and the angle $\\small \\varphi$ at which the minimal error occured is equivalent the kabsch2d-procedure in @Karolis' answer.\n\n2. In version 2) I took as criterion the sum of the absolute distances, that means, the sum $$err_2=\\small \\sum_{k=1}^4 \\small\\sqrt{(D_{k,1})^2+(D_{k,2})^2}$$ and got a slightly different rotation angle $\\small \\varphi$ for the smallest error.\n\nI don't know, which criterion fits your needs better.\n\nHere are some results from the protocol.\n\n$$\\small \\begin{array} {r|cc} & \\text{version } 1 & \\text{version } 2\\\\ \\hline \\varphi & -0.04895304& -0.05093647 \\\\ \\text{rotation} & \\begin{bmatrix} 0.99880204& -0.04893349\\\\ 0.04893349& 0.99880204\\\\ \\end{bmatrix} & \\begin{bmatrix} 0.99870302 & -0.05091444\\\\ 0.05091444 & 0.99870302\\\\ \\end{bmatrix} \\\\ \\text{distances} & \\begin{bmatrix} -6.80077266 & -0.86209739\\\\ 2.79924551 & -9.33782500\\\\ -0.61309522 & 6.94156520\\\\ 4.61462237 & 3.25835719\\\\ \\end{bmatrix} & \\begin{bmatrix} -6.78017751& -0.37062404 \\\\ 3.35787307 & -9.36574874 \\\\ -1.16459115 & 6.95324527 \\\\ 4.58689559 & 2.78312752 \\\\ \\end{bmatrix} \\end{array}$$", "date": "2020-09-27 15:24:37", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/186111/find-the-rotation-between-set-of-points/207407", "openwebmath_score": 0.7379559278488159, "openwebmath_perplexity": 4733.529647609154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811601648193, "lm_q2_score": 0.8976952866333484, "lm_q1q2_score": 0.8666181172845919}}
{"url": "https://math.stackexchange.com/questions/764437/the-gaussian-integers-are-isomorphic-to-mathbbzx-x21/764482", "text": "# the Gaussian integers are isomorphic to $\\mathbb{Z}[x]/(x^2+1)$\n\nI am trying to prove that $\\mathbb{Z}[i]\\cong \\mathbb{Z}[x]/(x^2+1)$.\n\nMy initial plan was to use the first isomorphism theorem. I showed that there is a map $\\phi: \\mathbb{Z}[x] \\rightarrow \\mathbb{Z}[i]$, given by $\\phi(f)=f(i).$ This map is onto and homorphic. The part I have a question on is showing that the $ker(\\phi) = (x^{2}+1)$.\n\nOne containment is trivial, $(x^2+1)\\subset ker(\\phi)$. To show $ker(\\phi)\\subset (x^2+1)$, let $f \\in ker(\\phi)$, then f has either $i$ or $-i$ as a root. Sot $f=g(x-i)(x+i)=g(x^2+1).$ How can I prove that $f \\in \\mathbb{Z}[x]\\rightarrow g \\in \\mathbb{Z}[x]$?\n\n\u2022 It is enough to have $f=g(x^2+1)$, because this says $f\\in (x^2+1)$, and we are done. \u2013\u00a0Dietrich Burde Apr 22 '14 at 13:34\n\u2022 By the division algorithm, $f = q\\cdot (x^2+1) + r$ with $\\deg r < 2$. Then you just need to check that $r(i) = 0$ implies $r = 0$ if $\\deg r < 2$. (Note: $\\deg 0 = -\\infty$) You could also appeal to Gau\u00df' lemma. \u2013\u00a0Daniel Fischer Apr 22 '14 at 13:35\n\u2022 @DanielFischer, I think that is the answer I am looking for since $g \\in \\mathbb{R}[x]$ which is a euclidean domain and I am allowed to make that claim. Do you care to expand on that? As in, how can I be certain that the coefficients of $q$ are in $\\mathbb{Z}$? \u2013\u00a0kslote1 Apr 22 '14 at 13:57\n\u2022 Don't use $\\mathbb{R}[x]$ here, use $\\mathbb{Q}[x]$. But you need never leave $\\mathbb{Z}[x]$ even potentially. The point is that $x^2+1$ is monic, i.e. has lead coefficient $1$. Thus if you do polynomial division, you always get an integer coefficient, and the existence of $q,r\\in\\mathbb{Z}[x]$ with $f = q\\cdot (x^2+1) + r$ and $\\deg r < 2$ follows. \u2013\u00a0Daniel Fischer Apr 22 '14 at 14:06\n\u2022 As an aside, this is one of those problems where it's easier to show the isomorphism directly, by writing down a homomorphism and its inverse, rather than the indirect route of finding a surjective homomorphism with zero kernel. \u2013\u00a0Hurkyl Aug 9 '17 at 16:40\n\nLet me elaborate on Daniel Fischers comment. You have a ring homomorphism $\\phi: \\mathbb Z[x]\\to\\mathbb Z[i]$ given by $x\\mapsto i$. Take $f \\in \\ker \\phi$. By the division algorithm, $$f = q\\cdot(x^2+1) + r,$$ where $\\deg r < 2$ and $q,r\\in\\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\\phi$ to this equation yields $$0 = \\underbrace{\\phi(q)\\cdot(i^2+1)}_0 + \\phi(r).$$ Since $r$ is of degree $<2$, we can write $r = ax+b$ for some $a,b\\in\\mathbb Z$. Then $\\phi(r)=0$ gives $$ai+b=0.$$ This equation in $\\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\\in\\mathbb Z[x]$ and therefore $$f = q\\cdot (x^2+1) \\in \\langle x^2+1\\rangle.$$ We conclude that $\\ker \\phi \\subseteq \\langle x^2+1\\rangle$.\n\nWhen quotient out an ideal, we consider what happens to the ring when all the elements in the ideal are considered as identity elements.\n\nNow if $x^2+1=0\\Rightarrow x=\\pm i$ let us take the \"+\" root.\n\n$\\mathbb{Z}[x]/(x^2+1)=\\{f\\in\\mathbb{Z}[x]\\,|x^2+1=0\\}=\\{f\\in\\mathbb{Z}[x]\\,|x=i\\}=\\{a+bi|a,b\\in\\mathbb{Z}\\}=\\mathbb{Z}[i]$\n\nI.e they are isomorphic.\n\nI have answered a similar question here Is this quotient Ring Isomorphic to the Complex Numbers\n\nYou can define your isomorphism as follows\n\n$\\phi:\\mathbb{Z}[i]\\rightarrow \\mathbb{Z}[x]/\\langle x^2+1 \\rangle$\n\n$\\phi(a+bi)=(a+bx)+\\langle x^2+1 \\rangle$\n\nIf you compute $[a+bx+\\langle x^2+1 \\rangle]\\times [c+dx+\\langle x^2+1 \\rangle]$\n\nWe get $ac+(ad+bc)x+bdx^2+\\langle x^2+1 \\rangle$. We reduce by long division to get,\n\n$(ac-bd)+(ad+bc)x+\\langle x^2+1 \\rangle$.\n\nTherefore,\n\n$\\phi([a+bi][c+di])=(ac-bd)+(ad+bc)x+\\langle x^2+1 \\rangle$.\n\nWhich is exactly what you want.\n\nThe map is obviously bijective and a ring homomorphism.", "date": "2019-05-22 01:06:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/764437/the-gaussian-integers-are-isomorphic-to-mathbbzx-x21/764482", "openwebmath_score": 0.958674430847168, "openwebmath_perplexity": 132.8792636185769, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808759252646, "lm_q2_score": 0.8840392771633079, "lm_q1q2_score": 0.8666067969699852}}
{"url": "https://math.stackexchange.com/questions/23809/what-is-the-remainder-of-16-is-divided-by-19/23810", "text": "# What is the remainder of $16!$ is divided by 19?\n\nCan anyone share me the trick to solve this problem using congruence?\n\nThanks,\nChan\n\nIf you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)\n\n\u2022 I knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks.\n\u2013\u00a0Chan\nFeb 26 '11 at 6:17\n\u2022 @Chan: $\\rm 17 \\equiv -2,\\ 18\\equiv -1$ Feb 26 '11 at 6:21\n\u2022 @Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\\equiv -1$ and $17\\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) Feb 26 '11 at 6:21\n\u2022 Many thanks, I got it now ;)\n\u2013\u00a0Chan\nFeb 26 '11 at 6:24\n\nHint  By Wilson's theorem $$\\bmod 19\\!:\\ \\overbrace{{-}1}^{\\large \\color{#0a0}{18}} \\equiv 18! \\equiv\\!\\! \\overbrace{18\\cdot17}^{\\large \\color{#c00}{(-1)(-2)}}\\!\\!\\cdot 16!\\,$$ $$\\Rightarrow\\,16!\\equiv \\dfrac{\\color{#0a0}{18}}{\\color{#c00}2}\\equiv 9$$\n\nGenerally (Wilson reflection formula) $$\\rm\\displaystyle\\ (p\\!-\\!1\\!-\\!k)!\\equiv\\frac{(-1)^{k+1}}{k!}\\!\\!\\!\\pmod{\\!p},\\,$$ $$\\rm\\:p\\:$$ prime\n\n\u2022 @Bill Dubuque: Thank you!\n\u2013\u00a0Chan\nFeb 26 '11 at 6:24\n\u2022 @Bill Dubuque: If I have $2.16! \\equiv 18 \\pmod{19}$. Can I cancel out the $2$ from both sides?\n\u2013\u00a0Chan\nFeb 26 '11 at 6:42\n\u2022 @Chan: $\\rm\\ 2\\:x\\equiv 2\\:y\\ (mod\\ 19)\\ \\iff\\ 19\\ |\\ 2\\ (x-y)\\ \\iff\\ 19\\ |\\ x-y\\ \\iff\\ x\\equiv y\\ (mod\\ 19)$ Feb 26 '11 at 6:50\n\u2022 @Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \\equiv ay \\pmod{p} \\implies x \\equiv y \\pmod{p}$, right?\n\u2013\u00a0Chan\nFeb 26 '11 at 7:07\n\u2022 @Chan: $\\rm\\ n,m\\$ coprime $\\rm\\iff a\\ n + b\\ m = 1\\$ for some $\\rm\\:a,b\\:\\ \\iff\\ a\\ n\\equiv 1\\ (mod\\ m)\\:,\\:$ for some $\\rm\\:a\\:$ $\\rm\\iff\\ n$ is invertible/cancellable $\\rm\\:(mod\\ m)$ Feb 26 '11 at 7:12\n\nI almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:\n\n$$10!= (2\\cdot 10)\\cdot (4\\cdot5)\\cdot(3\\cdot6)\\cdot(7\\cdot8)\\cdot 9\\equiv 1\\cdot 1\\cdot (-1)\\cdot(-1)\\cdot 9\\equiv9$$\n\nNow we have\n\n$$16!\\equiv 9\\cdot 11\\cdot 12\\cdot 13\\cdot 14\\cdot 15\\cdot 16\\equiv9\\cdot(-8)\\cdot(-7)\\cdot(-6)\\cdot(-5)\\cdot(-4)\\cdot(-3)$$\n\n$$\\equiv 9\\cdot(4\\cdot5)\\cdot(6\\cdot3)\\cdot(7\\cdot8)\\equiv 9\\cdot(1)\\cdot(-1)\\cdot(-1)\\equiv 9$$\n\nOf course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)\n\nA more explicit answer would go like this:\n\nBy Wilson's Theorem, $18!$ is congruent to $-1 \\pmod {19}$\n\n$$18! = (18\\cdot 17)(16!)$$\n\nThen $(18\\cdot 17)(16!)$ is congruent to $-1 \\pmod{19}$\n\nBut note that $18$ is congruent to $-1 \\pmod{19}$ and $17$ is congruent to $-2 \\pmod{19}$\n\nThen it follows that $(18\\cdot 17)(16!)$ is congruent to $((-1)\\cdot (-2))(16!)$ is congruent to $-1 \\pmod{19}$\n\nSimplifying, we get, $2\\cdot(16!)$ is congruent to $-1 \\pmod{19}$. But we need the remainder of $16!$, not $2\\cdot 16!$. and $-\\frac12$ isn't a valid answer.\n\nThen also notice that $18$ is congruent to $-1 \\pmod{19}$. Then $2\\cdot 16!$ is congruent to $18 \\pmod{19}$ by the fact that if $a$ is congruent to $b \\pmod{p}$, then $b$ is congruent to $a \\pmod{p}$\n\nSo we have that $2\\cdot 16!$ is congruent to $18 \\pmod{19}$ and from there, its just a matter of dividing both sides by $2$ to get $16!$ is congruent to $9 \\pmod{19}$", "date": "2021-12-07 10:04:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/23809/what-is-the-remainder-of-16-is-divided-by-19/23810", "openwebmath_score": 0.7822591066360474, "openwebmath_perplexity": 174.4722876639835, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808759252645, "lm_q2_score": 0.8840392771633078, "lm_q1q2_score": 0.866606796969985}}
{"url": "https://gcs-group.ro/hydrogenation-of-kdrqq/9c9f68-parent-functions-and-transformations", "text": "3.2 Graphing Quadratic Functions ... 3.5 Transformations of Functions II. Section 1.1 Parent Functions and Transformations 5 Describing Transformations A transformation changes the size, shape, position, or orientation of a graph. In-Class Notes Notes Video Worksheet Worksheet Solutions . It has been clearly shown in the below picture. Now, let us come to know the different types of transformations. Furthermore, all of the functions within a family of functions can be derived from the parent function by taking the parent function\u2019s graph through various transformations. This is the currently selected item. The six most common graphs are shown in Figures 1a-1f. Start studying Parent Functions and Transformations (Ullman). A translation is a transformation that shifts a graph horizontally and/or vertically but does not change its size, shape, or orientation. The functions shown above are called parent functions. Which of the following is the graph of y = 1/(x +2) ? Horizontal Expansions and Compressions. A parent function is the simplest function of a family of functions. Which of the following is the graph of |x+2|+2? 3.1 Completing the Square. Linear\u2014vertical shift up 5. The rule we apply to make transformation is depending upon the kind of transformation we make. Vertical Translation 3. Parent Functions and transformations. Which equation is a quadratic function reflected over the x-axis and shifted up 2. Transformations of ParentTransformations of Parent FunctionsFunctions 2. based on the parent function, the function will be vertically stretched by a factor of 3, it will be reflected over the x-axis and will move up the y-axis two units Describe the transformation\u2026 Which of the following is the graph of y = 1/x - 2 ? In this section, we will explore transformations of parent functions. 1-5 Bell Work - Parent Functions and Transformations. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. State the transformations and sketch the graph of the following functions. The different types of transformations which we can do in the parent functions are 1. A parent function is the simplest of the functions in a family. Function Transformations. There are many different type of graphs encountered in life. Some of the worksheets for this concept are Transformations of graphs date period, Transformations of functions name date, Algebra ii translations on parent functions review, Work parent functions transformations, 1 graphing parent functions and transformations, Y ax h2 k, Graphical transformations \u2026 The parent function of a quadratic is f (x) = x \u00b2. a Which of the following is the graph of y = 1/x +2 ? Name_ Date _ For problem 1- 6, please give the name of the parent function and describe the transformation represented. More clearly, on what grounds is the transformation made? To know that, we have to be knowing the different types of transformations. This graph is known as the \"Parent Function\" for parabolas, or quadratic functions.All other parabolas, or quadratic functions, can be obtained from this graph by one or more transformations. Geo 2.8 Parallel and Perpendicular Slopes. Which of the following is the graph of y = 2. Label \u2026 Examples inculde: (line with slope 1 passing through origin) (a V-graph opening up with vertex at origin) (a U-graph opening up with vertex at origin) \u2026 Identifying function transformations. Start studying Parent Functions and Transformations. Yes, there is a pre-decided rule to make each and every transformation. 1-5 Slide Show - Parent Functions and Transformations PDFs. Practice: Identify function transformations. Unit 3: Parent Functions . Absolute value\u2014vertical shift down 5, horizontal shift right 3. In Mathematics II, students reasoned about graphs of absolute value and quadratic functions by thinking of them as transformations of the parent functions |x| and x\u00b2. Parent Functions And Transformation - Displaying top 8 worksheets found for this concept.. Again, the \u201cparent functions\u201d assume that we have the simplest form of the function; in other words, the function either goes through the origin \\left( {0,0} \\right), or if it doesn\u2019t go through the origin, it isn\u2019t shifted in any way.When a function is shifted, stretched (or compressed), or flipped\u00a0in any way from its \u201cparent function\u201c, it is said to be transformed, and is a transformation of a function.T-charts are extremely useful tools when dealing with transformations of functions\u2026 Parent Functions and Transformations Reference BookThis reference book was created to use as a review of transformations and the following function families: linear, absolute value, quadratic, cubic, square root, cube root, exponential, logarithmic, and \u2026 Title: Parent Function Transformation 1 Parent Function Transformation. In-Class Notes Notes Video Worksheet Worksheet Solutions Homework HW Solutions. A quadratic function moved left 2. A quadratic function moved right 2. Rigid transformations change only the position of the graph, leaving the swe and shape unchanged. A very simple definition for transformations is, whenever a figure is moved from one location to another location, This fascinating concept allows us to graph many other types of functions, like square/cube root, exponential and logarithmic functions. View 2 Transformation HMWK-1.pdf from HIST 3315 at Wingate University. a transformation occurs. 287 #73-75, 79-81 . 13. Identifying function transformations. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Square Root \u2014vertical shift down 2, horizontal shift left 7. The rule that we apply to make transformation using reflection and the rule we apply to make transformation using rotation are not same. C. A. which of the following is linear? The different types of transformations which we can do in the parent functions are, 5. Which of the following is the graph of x = 2? D. B. C. A. which of the following is cubic? Graphing and Describing Translations Graph g(x) = x \u2212 4 and its parent function. The Parent Function is the simplest function with the defining characteristics of the family. When identifying transformations of functions, this original image is called the parent function. 1-5 Online Activities - Parent Functions and Transformations. Q. Here are some simple things we can do to move or scale it on the graph: A transformation where the pre-image and image are congruent is called a\u00a0rigid transformation\u00a0or an\u00a0isometry. In this point, always students have a question. Learn vocabulary, terms, and more with flashcards, games, and other study tools. ... Wansformations Transformations of a parent function can affect the appearance of the parent graph. D. B. C. A. which of the following is quadratic? 6 Module 1 \u2013 Polynomial, Rational, and Radical Relationships Parent Function Worksheet # 1- 7 Give the name of the parent function and describe the transformation represented. The \"Parent\" Graph: The simplest parabola is y = x 2, whose graph is shown at the right.The graph passes through the origin (0,0), and is contained in Quadrants I and II. Sample Problem 2: Given the parent function and a description of the transformation, write the equation of the transformed function!\". So, for each type of transformation, we may have to apply different rule. A square root function moved right 2. 1-5 Assignment - Parent Functions and Transformations. What is a parent function? Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Horizontal Translation 2. Which of the following is the graph of y = -1? Learn vocabulary, terms, and more with flashcards, games, and other study tools. Horizontal Expansions and Compressions 6. These transformations include horizontal shifts, stretching, or compressing vertically or horizontally, reflecting over the x or y axes, and vertical shifts. Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse \u00a0trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Internal and External Tangents of a Circle, Volume and Surface Area of Composite Solids Worksheet, A very simple definition for transformations is, whenever a figure is moved from one location to another location, Of parent function transformation 1 parent function transformation 1 parent function is the graph of & # ;! 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G ( x +2 ) with flashcards, games, and more with flashcards, games, and other tools!", "date": "2021-04-20 07:17:24", "meta": {"domain": "gcs-group.ro", "url": "https://gcs-group.ro/hydrogenation-of-kdrqq/9c9f68-parent-functions-and-transformations", "openwebmath_score": 0.4315130412578583, "openwebmath_perplexity": 1134.3019108348492, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9802808684361289, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.8666067843594841}}
{"url": "https://math.stackexchange.com/questions/2126888/how-to-prove-that-sum-n-textodd-fracn24-n22-pi2-16", "text": "# How to prove that $\\sum_{n \\, \\text{odd}} \\frac{n^2}{(4-n^2)^2} = \\pi^2/16$?\n\nThe series:\n\n$$\\sum_{n \\, \\text{odd}}^{\\infty} \\frac{n^2}{(4-n^2)^2} = \\pi^2/16$$\n\nshowed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated.\n\nHow do I prove this directly?\n\n\u2022 integrate $\\frac{z^2}{(4-z^2)^2}\\tan(z)$ over a big circle in the complex plane \u2013\u00a0tired Feb 3 '17 at 8:28\n\nWolfram Alpha gives me the partial fraction expansion:\n\n$$\\frac{n^2}{(4 - n^2)^2} = \\frac{1}{8}\\left(\\frac 1{n-2}-\\frac{1}{n + 2}\\right) + \\frac{1}{4}\\left(\\frac{1}{(n-2)^2}+\\frac{1}{(n+2)^2}\\right)$$\n\nSo the first part telescopes, and the second part will be some modified version of $\\zeta(2)$.\n\nIn more detail: \\begin{align}\\sum_{n\\text{ odd}} \\left(\\frac 1{n-2}-\\frac{1}{n + 2}\\right)&=\\frac{1}{-1}-\\frac{1}{3}+\\frac{1}{1}-\\frac{1}{5}+\\frac{1}{3}-\\frac{1}{7}+\\cdots\\\\ &=-1+1=0\\end{align} since all the other terms cancel out.\n\nAnd \\begin{align}\\sum_{n\\text{ odd}}\\left(\\frac{1}{(n-2)^2}+\\frac{1}{(n+2)^2}\\right)&=\\frac{1}{(-1)^2}+\\frac{1}{3^2}+\\frac{1}{1^2}+\\frac{1}{5^2}+\\frac{1}{3^2}+\\frac{1}{7^2}+\\cdots\\\\ &=2\\left(\\frac{1}{1^2}+\\frac{1}{3^2}+\\frac{1}{5^2}+\\cdots\\right) \\end{align}\n\nIt's a famous result that $\\frac{1}{1^2}+\\frac{1}{3^2}+\\frac{1}{5^2}+\\cdots=\\frac{\\pi^2}{8}$. You can can prove it if you know:\n\n$$\\frac{\\pi^2}{6}=\\zeta(2) = \\frac{1}{1^2}+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots$$\n\nand thus $$\\zeta(2)-\\frac1{2^2}\\zeta(2) = \\frac{1}{1^2}+\\frac{1}{3^3}+\\frac{1}{5^2}+\\cdots$$\n\nSo $$\\sum \\frac{n^2}{(4-n^2)^2}=0 + \\frac{1}{4}\\cdot 2\\cdot \\frac{\\pi^2}{8}=\\frac{\\pi^2}{16}$$\n\n\u2022 Hi Thomas. I had assumed that the sum over odd integers included the negative ones. Naturally, the answer in that case is twice the answer if the summation extends to the positive odds only. Anyway, (+1) for your well-written post. -Mark \u2013\u00a0Mark Viola Feb 3 '17 at 4:30\n\u2022 @Dr.MV Yeah, I wondered about that. I might have read it your way if it had been $\\sum_{n\\text{ odd}}$ rather than $\\sum_{n\\text{ odd}}^{\\infty}$. \u2013\u00a0Thomas Andrews Feb 3 '17 at 4:37\n\u2022 Ah, yes. I see your point. Well, I did address the issue at the end of the post. ;-)) \u2013\u00a0Mark Viola Feb 3 '17 at 4:40\n\nFirst, the partial fraction of the summand can be written\n\n\\begin{align} \\frac{n^2}{(4-n^2)^2}&=\\frac14\\left(\\frac{1}{n-2}+\\frac{1}{n+2}\\right)^2\\\\\\\\ &=\\frac14 \\left(\\frac{1}{(n-2)^2}+\\frac{1}{(n+2)^2}+\\frac{1/2}{n-2}-\\frac{1/2}{n+2}\\right) \\end{align}\n\nSecond, we note that\n\n\\begin{align} \\sum_{n\\,\\,\\text{odd}}\\frac{1}{(n\\pm 2)^2}&=\\sum_{n=-\\infty}^\\infty \\frac{1}{(2n-1)^2}\\\\\\\\ &=2\\sum_{n=1}^\\infty \\frac{1}{(2n-1)^2}\\\\\\\\ &=2\\left(\\sum_{n=1}^\\infty \\frac{1}{n^2}-\\sum_{n=1}^\\infty \\frac{1}{(2n)^2}\\right)\\\\\\\\ &=\\frac32 \\sum_{n=1}^\\infty \\frac{1}{n^2}\\\\\\\\ &=\\frac{\\pi^2}{4} \\end{align}\n\nThird, it is easy to show that\n\n$$\\sum_{n=-\\infty}^\\infty \\left(\\frac{1}{2n-3}-\\frac{1}{2n+1}\\right)=0$$\n\nPutting it all together we have\n\n$$\\sum_{n,\\,\\,\\text{odd}}\\frac{n^2}{(4-n^2)^2}=\\frac{\\pi^2}{8}$$\n\nIf we sum over the positive odd only, then the answer is $(1/2)\\pi^2/8=\\pi^2/16$\n\nHINT\n\n$$\\sum_{n \\, \\text{odd}}^{\\infty} \\frac{n^2}{(n^2-4)^2}=\\sum_{n=1}^{\\infty} \\frac{(2n-1)^2}{((2n-1)^2-4)^2}$$ Using partial fraction expansion, note $$\\frac{(2n-1)^2}{((2n-1)^2-4)^2}=\\left(\\frac{1}{4(2n+1)^2}+\\frac{1}{4(2n-3)^2}\\right)-\\left(\\frac{1}{8(2n+1)}-\\frac{1}{8(2n-3)}\\right)$$ Note that the second part has cancelling terms.\n\n\u2022 Why the downvote? \u2013\u00a0S.C.B. Feb 3 '17 at 4:40\n\u2022 @ThomasAndrews I fixed the signs. \u2013\u00a0S.C.B. Feb 3 '17 at 4:40", "date": "2019-09-22 22:34:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2126888/how-to-prove-that-sum-n-textodd-fracn24-n22-pi2-16", "openwebmath_score": 0.8995524048805237, "openwebmath_perplexity": 1325.4002394424197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109081214212, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8666007814264871}}
{"url": "https://koytru.meinejuwelen.de/en/qhqi", "text": "If by 'a standard deck of cards' you mean a 52 card pack (no Jokers) with 4 suits each of 13 cards, the probability of picking a single card at random and it being a heart is 13/52 or 1/4 For a pack with jokers it would be 13/54 Ankit Pal B.E from University of Mumbai (Graduated 2020) 3 y Related. Explanations Question You draw a card at random from a standard deck of 52 cards. Find each of the following conditional probabilities: a) The card is a heart, given that it is red. b) The card is red, given that it is a heart. c) The card is an ace, given that it is red. d) The card is a queen, given that it is a face card. Explanation Verified. In a playing card there are 52 cards. Therefore the total number of possible outcomes = 52 (i) '2' of spades: Number of favourable outcomes i.e. '2' of spades is 1 out of 52 cards. Therefore, probability of getting '2' of spade Number of favorable outcomes P (A) = Total number of possible outcome = 1/52 (ii) a jack. But the coin has not changed - if it's a \"fair\" coin, the probability of getting tails is still 0.5. Dependent Events Two (or more) events are dependent if the outcome of one event affects the outcome of the other(s). Thus, one event \"depends\" on another, so they are dependent. Example I draw two cards from a deck of 52 cards.\n\nmn\n\nmd\n\ngv\n\nub\n\neg\n\nWhat is the probability of ace card in a deck of 52 cards ? A card is drawn from a pack of 52 cards . The probability of getting an ace is 1/52. View complete answer on byjus.com. What do 4 Aces mean? If you hate your job, then four <b>Aces</b> can mean that something major will happen which will replace your income, allowing you to leave your job. For. Mar 19, 2020 \u00b7 A deck of standard 52 cards contain four aces. There are four kings in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e. \u21d2 . \u21d2 . \u21d2 . a king or a diamond ; A deck of standard 52 cards contain four kings. There are 13 Diamonds in a standard deck of playing cards. So ,we need to .... TO FIND : Probability of the following Total number of cards = 52 ( i ) Cards which are black king is 2 We know that PROBABILITY = = Number of favorable event T otal number of event Hence the probability of getting a black king is equal to 2/52=1/26 (ii) Total number of black cards is 26.. 2017. 6. 10.\u00a0\u00b7 A standard deck has 13 ordinal cards (Ace, 2-10, Jack, Queen, King) with one of each in each of four suits (Hearts, Diamonds, Spades, Clubs), for a total of #13xx4=52# cards. If we draw a card from a standard deck, there are 52 cards we might get. There are 16 cards that will satisfy the condition of picking a Jack, Queen, King, or Ace. This.\n\nav\n\ntp\n\nlz\n\nMar 03, 2021 \u00b7 A) 1 2 B) 1 52 C) 1 4 D) 1 13 7) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing a heart? A) 1 4 B) 1 2 C) 3 4 D) 1 8) In a survey of college students, 840 said that they have cheated on an exam and 1795 said that they have not..\n\njf\n\nwe\n\nud\n\noy\n\nit\n\nov\n\nExample 10 One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n(S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn.\n\naw\n\nfg\n\nbe\n\njy\n\nA standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. One card is drawn from a standard deck of 52 cards. Find the probability of drawing a heart or a 6.There are 13 hearts and one 6 of hearts... 2. Which of the following must be a true statement?... Select one: a. The conditional probability P(A/B) is the probability that event B occurs, knowing A has occurred. b. An event and its complement can ....\n\nyo\n\nba\n\nwv\n\nst\n\nTranscribed image text: A single card is drawn from a standard 52-card deck. Let H be the event that the card drawn is a heart, and let F be the event that the card drawn is a face card. Find the indicated probability. P (H'UF) P(H'UF) = 0 (Type an integer or a simplified fraction.) A single card is drawn from a standard 52-card deck. one less card in the deck because we already had to draw the Heart from the deck. Thus: P(Heart and Club) = P (Heart) * P (Club) = 13/52 * 13/51 = .25 * .255 = .064 We might also have to subtract a value from the numerator as well as the denominator. Try to find the probability of drawing three red cards from a deck without replacement. Question 591370: A single card is drawn from a standard deck of 52 cards. Find the probability the card is: 1. A red four 2. A heart 3. A 4 or a heart. 4. Not a club. 5. A red or a four 6. A red and a 3 Answer by Edwin McCravy(19211) (Show Source):. The probability of drawing a queen, from a 52 card deck, is 4/52 or 1/13. Wiki User. \u2219 2009-10-09 18:06:35. This answer is:. Answer: The probability of drawing a card from a standard deck and choosing a king or an ace is (1/13) \u00d7 (4/51). What is the probability of drawing a queen of hearts from a deck of 52 cards pinia store.\n\njq\n\nih\n\nvx\n\nMar 06, 2011 \u00b7 Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the.... 2018. 6. 12.\u00a0\u00b7 In a standard deck of cards, there are 52 cards. They are broken down into suits (4 of them: Spades, Hearts, Diamonds, Clubs) of 13 cards each. Each suit has 13 ordinal cards (A, 2 through 10, Jack, Queen, King). Here we're asked to draw a card at random and find the probability of drawing either a diamond or a 7. Mar 06, 2011 \u00b7 Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the....\n\nzc\n\ngd\n\nhd\n\nqd\n\nIf one card is drawn from a standard 52 card playing deck, determine the probability of getting a ten, a king or a diamond. Round to the nearest hundredth. # of ways to succeed: 4 + 4 + 13 - 1 -1 = 19. Correct option is D) Probability of drawing a king = 524= 131 After drawing one card, the number of cards are 51. Probability of drawing aqueen = 514. Now, the probability of drawing a king and queen consecutively is 131\u00d7 514= 6634 Was this answer helpful? 0 0 Similar questions Three cards are drawn with replacement from a part of 52 cards. Calculate the probability of being dealt a diamond from a standard deck of 52 cards. Since there are 4 suits in a deck of cards (hearts, clubs, spades and diamonds) we can find the number of. Transcript. Example 10 One card is drawn from a well shuffled deck of 52 cards.If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n (S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn So, n (A) = 13 Probability of A = P (A.One card is selected from a.\n\nai\n\nry\n\noi\n\nhf\n\nMar 12, 2011 \u00b7 The probability of drawing a club or a nine from a 52 card deck of standard playing cards is 16 / 52 or approximately 30.8%. There are 13 clubs in a standard deck of cards. There are four nines in.... The royal flush is a case of the straight flush. It can be formed 4 ways (one for each suit), giving it a probability of 0.000154% and odds of 649,739 : 1. When ace-low straights and ace-low straight flushes are not counted, the probabilities of each are reduced: straights and straight flushes each become 9/10 as common as they otherwise would be.\n\ntv\n\nxs\n\nnc\n\npm\n\nwinnetka rummage sale 2022; the end netflix; craigslist gmc trucks for sale by owner near Florianpolis State of Santa Catarina; meta computer vision engineer salary; restaurants open till 2am near Panruti Tamil Nadu; Google Algorithm Updates. The total number of 8 card hands is 52c8, so the probability of choosing a hand which excludes at least one suit is (4 * 39c8 - 6 * 26c8 + 4 * 13c8) / (52c8). You wanted the probability that a hand includes every suit which is the opposite of choosing a hand excluding at least one suit, and therefore the probability that you wanted is. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at. Let Event B = drawing a red card. P (A) = 4/52 since there are four aces in each deck of 52 cards. P (B) = 1/2 = 26/52 since there are four suits and two of them are red (or 26 red cards in a deck of 52) P (A\u2229B) = the probability of drawing a red ace = 2/52 since there are 2 red aces in a deck of 52 cards).\n\nha\n\nfb\n\nwo\n\nln\n\nTranscript. Example 10 One card is drawn from a well shuffled deck of 52 cards.If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n (S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn So, n (A) = 13 Probability of A = P (A.One card is selected from a. A standard deck of cards has: 52 Cards in 13 values and 4 suits ... If you draw 3 cards from a deck one at a time what is the probability: ... what is the probability: You draw a Club, a Heart and a Diamond (in that order). The hypergeometric MTG calculator can describe the likelihood of any number of successes when drawing from a deck of Magic cards. It takes into account the fact that each draw decreases the size of your library by one, and therefore the probability of success changes on each draw. Population Size. Cards in your deck / library you are drawing from.\n\nps\n\ned\n\nmb\n\nry\n\nNumber of favourable outcomes = 48 (Because There are always 52 cards in one deck containing 13 cards of varying values but always 4 suits, if the drawn card is not 4 then number of favourable outcomes will be 48 (12*4).) Total number of favourable outcomes = 52 (Number of cards in a deck is 52) So, the probability will be = 48/ 52 = 12/13. Oct 26, 2020 \u00b7 Answer: 1/13. Step-by-step explanation: If there are 4 suits in a deck of cards, there are only 4 aces inside a complete deck. That means the probability of drawing an ace is 4/52 and just simply simplify the answer to 1/13.. A SINGLE CARD IS DRAWN AT RANDOM FROM A STANDARD DECK OF 52 CARDS. FIND THE PROBABILITY OF DRAWING THE FOLLOWING CARDS. PLEASE REDUCE TO LOWEST TERMS. A) A DIAMOND OR A 5 __________ B) A HEART AND A JACK __________ C) A JACK OR AN 8 __________ D) A HEART OR A SPADE __________ E) A RED AND FACE CARD __________ F) A RED CARD OR A QUEEN __________ 2. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at. Suppose you draw five cards from a standard deck of 52 playing cards, and you want to calculate the probability that all five cards are hearts. ... For example, the probability of drawing five cards of any one suit is the sum of four equal probabilities, and four times as likely. In boolean language, if the events are related by a logical OR. Whenever you do probability problems, check to see if you are being asked to find probability of one thing OR another, or of multiple events happening together. For instance, if this problem asked you to find the probability of drawing a heart AND a 5, well, there is only one 5 of hearts in a deck. So that would be 1/52.\n\nmr\n\nhf\n\noz\n\nThis can be simplified into 4/13 or a 30.77% probability of drawing a 4 or a spade from a standard deck of cards. Read More:. After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is 3 51 = 1 17 3 51 = 1 17.. A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. a card is drawn from a standard deck what is the probability that the card is an ace If you draw one card from a standard deck, what is the probability of drawing a 5 ... a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and T\u00ecm ki\u1ebfm.\n\nzc\n\nae\n\njh\n\n2022. 2. 7.\u00a0\u00b7 We know that a well-shuffled deck has 52 cards. Total number of black cards = 26. Total number of red cards = 26. Therefore probability of getting a black card= {total number. 5) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing an ace? A) 1 13 B) 1 52 C) 1 4 D) 1 2 6) If one 39,068 satisfied customers 2,852 satisfied customers Ph.D. 39,068 satisfied customers the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards. What is the probability of drawing a black checker from a bag filled with 6 black checkers and 4 red checkers, replacing it, and drawing another black checker? c. Getting a Club and a Heart . 2. Drawing a card from a deck and not replacing it. 1.. So, there are 12 face cards in the deck of 52 playing cards. Worked-out problems on Playing cards probability: 1. A card is drawn from a. A card is selected from a deck of 52 playing cards. Find the probability of selecting \u00b7 a prime number under 10 given the card is a heart. (1 is not prime.) \u00b7 a diamond or heart given the card is red. read ....\n\nfd\n\nss\n\nlq\n\nA card is randomly drawn from a standard 52-card deck. Find the probability of the given event. (A face card is a king, queen, or jack). Drawing a queen and a heart (this is an intersection question). A card is drawn randomly from a standard 52-card deck.Find the probability of the given event. (a) The card drawn is 6 The probability is:(b) The card drawn is a face card (Jack, Queen, or King) The read more.Solution Total number of possible outcomes = 52 P (E) = (Number of favourable outcomes/ Total number of outcomes) (i) Total numbers of the king of. Since we know that in a deck of 52 cards, there are a total 12 face cards (3 face cards each of heart, diamond, spade and club). Number of face cards$= 12$ Therefore, probability of getting a face card$= \\dfrac { { {\\text {Number of face cards}}}} { { {\\text {Total number of cards}}}} = \\dfrac { {12}} { {52}} = \\dfrac {3} { {13}}$. Jun 21, 2020 \u00b7 If one card is drawn from a standard 52-card playing deck, find the probability of getting a king, or a ten, or a heart, or a club.When entering your answer include a leading zero and round to the nearest hundredth. An example of an acceptable answer would be 0.19. Question 591370: A single card is drawn from a standard deck of 52 cards. Find the probability the card is: 1. A red four 2. A heart 3. A 4 or a heart. 4. Not a club. 5. A red or a four 6. A red and a 3 Answer by Edwin McCravy(19211) (Show Source):. Jun 21, 2020 \u00b7 If one card is drawn from a standard 52-card playing deck, find the probability of getting a king, or a ten, or a heart, or a club.When entering your answer include a leading zero and round to the nearest hundredth. An example of an acceptable answer would be 0.19.\n\nud\n\nlz\n\ngq\n\nyp\n\nas\n\nIf you draw one card from a standard deck, what is the probability of ... a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and T\u00ecm ki\u1ebfm. T\u00ecm ... If you draw one card from a standard deck, what is the probability of drawing a 5 or.\n\ngq\n\ndu\n\naa\n\nThe total number of cards in a deck is 52 Probability = No. of favorable outcomes / Total no. of outcomes. Now, the probability of cards in a deck is 13/ 52 . Understand different concepts and get good grip on them by using online tools available at.\n\nex\n\nqs\n\nqb\n\nfx\n\nA standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. Apr 26, 2017 \u00b7 Ace is not odd The ordinals 3, 5, 7, and 9 are odd. There are four of each (one for each suit) and so 4xx4=16 odd cards. This makes the probability: P(\"draw an odd card\")=16/52=4/13 Ace is odd If we want to consider the Ace as a 1, then there are 5 ordinals that are odd, 5xx4=20 odd cards, and therefore: P(\"draw an odd card\")=20/52=5/13.\n\nhz\n\nui\n\nux\n\nyx\n\nThere are 5 2 cards in a standard deck: 1 3 ordinal cards (Ace - 1 0, Jack, Queen, King) and 4 of them - one to each suit (hearts, diamonds, clubs, spades) and so we have 4 \u00d7 1 3 = 5 2.. Q. From a pack of 52 playing cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is (i) a black queen (ii) a red card (iii) a black jack (iv) a picture card (Jacks, queens and kings are picture cards). Step 2. \u2235 number of non-face card in well shuffled deck of 52 playing card = 52-12. = 40. Step 3. Probability ( a non-face card ) = number of favourable outcomes total number of outcomes. = 40 52. = 10 13. Step 4. (ii) Number of black king in well shuffled deck of 52 playing cards = 2. Mar 06, 2011 \u00b7 Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the....\n\ngw\n\nch\n\neb\n\nThe goal is to win at least three tricks.. Example 3: Two cards are drawn without replacement in succession from a well-shuffled deck of 52 playing cards. What is the probability that the second card drawn is an ace, given that the first card drawn was an ace? Example 4: One thousand high school seniors were surveyed about whether they planned .... Problem 2: A random card is chosen from the standard deck of cards, find the probability of obtaining a queen or a heart.. So, the probability of getting a Queen card is 1/13. Example 2: A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a card of Heart. Solution: Let A represents the event of getting a Heart .... A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. A standard deck of cards has: 52 Cards in 13 values and 4 suits ... If you draw 3 cards from a deck one at a time what is the probability: ... what is the probability: You draw a Club, a Heart and a Diamond (in that order).\n\nak\n\nne\n\ngf\n\nNumber of Kings in a deck = 4. Probability of drawing a card from a standard deck and choosing a king or an ace = Probability of getting an Ace + Probability of getting a King. Probability of drawing an Ace at random = 4/52 = 1/13. Now, the probability of drawing a King at random = 4/52 = 1/13. hence, the required probability = 1/13 + 1/13 = 2/13. What is the probability of drawing a heart from a standard deck of cards on a second draw? 2 Answers By Expert Tutors The probability of choosing a heart, P(Heart) = 13/52 = 0.25. What is the probability of getting a heart or an even number? Clearly, the probability of drawing a heart out of the deck is 13/52, or 1/4.. \"/>.\n\nci\n\nno\n\nnz\n\nnl\n\nbv\n\nTherefore, the proability of drawing a king or 3 is 8/52=2/13. For the second problem, think binomial probability. The probability of drawing a king is 1/13. The opposite of at least 1 is none. So, the probability of getting no kings is 48/52=12/13. We find the probability of getting no kings and subtract from 1. Also, You should get the same.\n\npk\n\nlf\n\nfz\n\nml\n\nxk\n\nmc\n\nA card is drawn from a standard deck of 52 playing cards. A: The result is a club. B: The result is a king. 36) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a king. 37) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card. We can draw one card from pack of 52 cards in 52C1 = 52 ways n (S)=52 There is only one card in the pack of 52 cards which is jack as well as heart card,.It is jack of hearts n (E)=1 DIY: How to take years off your neck's appearance. Ray Phan Principal Software Engineer at Magic Leap (company) (2021-present) 4 y Related. Jan 13, 2017 \u00b7 There is a 50% chance that the card drawn will be red. Red cards make up 50% of the deck. 26/52 = 1/2 Therefore, if you're drawing one card, there is a 50/50 chance that the card will be red.. Problem 2: A random card is chosen from the standard deck of cards, find the probability of obtaining a queen or a heart.. So, the probability of getting a Queen card is 1/13. Example 2: A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a card of Heart. Solution: Let A represents the event of getting a Heart .... 5) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing an ace? A) 1 13 B) 1 52 C) 1 4 D) 1 2 6) If one 39,068 satisfied customers 2,852 satisfied customers Ph.D. 39,068 satisfied customers the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards. Correct option is D) Probability of drawing a king = 524= 131 After drawing one card, the number of cards are 51. Probability of drawing aqueen = 514. Now, the probability of drawing a king and queen consecutively is 131\u00d7 514= 6634 Was this answer helpful? 0 0 Similar questions Three cards are drawn with replacement from a part of 52 cards. In this task, we need to calculate the probability of getting at least one black card. The given is that we draw 2 2 2 cards from a 52 52 52-card deck with replacement.. The deck is the standard deck with 4 4 4 suits, clubs, spades, hearts, and diamonds, where spades and clubs are black suits.\n\nao\n\ngh\n\nym\n\nIf you were investigating red cards, kings or the queen of hearts, the odds of randomly drawing one of these from a complete deck are 50 percent (26 in 52); about 7.7 percent (four in 52); or. One card is randomly drawn from a deck of 52 playing cards. Find the probability thati the drawn card is red.ii the drawn card is an ace.iii the drawn card is red and a king.iv the drawn card is red or a king.[4 MARKS]. A card is drawn from a standard deck of 52 playing cards. A: The result is a club. B: The result is a king. 36) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a king. 37) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card. Setup Card content One card per line. Supports basic HTML. Go!. You will come across many video chatting applications, but a few of the best ones that are very effective and perfect for a virtual card room are Zoom and Skype. With Zoom, you can add around 100 people; however, the session will only last for 45minutes, and you will have to start.\n\nja\n\nqj\n\nbx\n\naw\n\nJan 13, 2017 \u00b7 There is a 50% chance that the card drawn will be red. Red cards make up 50% of the deck. 26/52 = 1/2 Therefore, if you're drawing one card, there is a 50/50 chance that the card will be red.. Playing cards probability problems based on a well-shuffled deck of 52 cards. Basic concept on drawing a card: In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades \u2660 hearts \u2665, diamonds \u2666,. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at random from a deck of shuffled cards. math. A card is drawn from an ordinary deck of 52 cards, and the result is recorded on paper.\n\npq\n\niu\n\nbw\n\n2010. 6. 4.\u00a0\u00b7 There are 6 red face cards in a standard deck of 52 cards; the Jack, Queen, and King of Hearts and Diamonds. The probability, then, of drawing a red face card from a standard deck of 52 cards is 6 in 52, or 3 in 26, or about 0.1154. (A standard deck of cards is the most common type of deck used in most card games containing 52 cards). Determine the probability of having 1 girl and 3 boys in a 4-child family assuming boys and girls are equally likely. The probability of having 1 girl and 3 boys is 1/4. Use the theoretical method. Probability gives the chances of how likely ....\n\nbh", "date": "2022-12-09 19:59:20", "meta": {"domain": "meinejuwelen.de", "url": "https://koytru.meinejuwelen.de/en/qhqi", "openwebmath_score": 0.39884233474731445, "openwebmath_perplexity": 287.22561770834307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180694313358, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8665908512950863}}
{"url": "https://math.stackexchange.com/questions/3129387/the-convergence-of-a-telescoping-series", "text": "# The Convergence of a Telescoping Series\n\nThe question I've been posed is:\n\nShow that the following series converges, and compute its value\n\n$$\\sum_{k=1}^\\infty \\frac{1}{k(k+2)}$$\n\nFrom this I decided to use partial fractions to put into the form:\n\n$$\\frac{1}{2}\\cdot\\left(\\frac{1}{k}-\\frac{1}{k+2}\\right)$$\n\nAnd from this I noticed that this is in the form of a telescoping series which I think would cancel down to:\n\n$$\\frac12\\cdot\\left(1+\\frac12\\right)= \\frac{3}{4}$$\n\nSo I've got to this point, but I don't think what I've worked out is substantial enough to prove what I've been asked.\n\nWould anyone mind giving any tips to make my working more thorough.\n\n\u2022 Concerning the convergence alone you can use the fact that $$\\frac1{k(k+2)}<\\frac1{k^2}$$ \u2013\u00a0mrtaurho Feb 27 '19 at 21:30\n\nNote that$$\\frac1k-\\frac1{k+2}=\\left(\\frac1k-\\frac1{k+1}\\right)+\\left(\\frac1{k+1}-\\frac1{k+2}\\right).$$This will give you two telescoping series. Can you take it form here?\n\n\u2022 Sorry but I don't quite understand where I should go from there, would you mind giving another tip? \u2013\u00a0king Feb 28 '19 at 17:05\n\u2022 \\begin{align}\\sum_{n=1}^\\infty\\frac1k-\\frac1{k+2}&=\\sum_{k=0}^\\infty\\frac1k-\\frac1{k+1}+\\sum_{n=1}^\\infty\\frac1{k+1}-\\frac1{k+2}\\\\&=1-\\lim_{n\\to\\infty}\\frac1{k+1}+\\frac12-\\lim_{k\\to\\infty}\\frac1{k+2}\\\\&=\\frac32.\\end{align} \u2013\u00a0Jos\u00e9 Carlos Santos Feb 28 '19 at 17:33\n\nLet's write, as you have done, the following :\n\n$$S_n =\\sum_{k=1}^n \\frac{1}{k(k+2)} = \\frac{1}{2} \\sum_{k=1}^n \\left(\\frac{1}{k} - \\frac{1}{k+2} \\right) = \\frac{1}{2} \\left(\\sum_{k=1}^n \\frac{1}{k} - \\sum_{k=1}^n\\frac{1}{k+2} \\right) = \\frac{1}{2} \\left(\\sum_{k=1}^n \\frac{1}{k} - \\sum_{k=3}^{n+2}\\frac{1}{k} \\right)$$\n\nSo you see that $$S_n = \\frac{1}{2} \\left( 1 + \\frac{1}{2} - \\frac{1}{n+1} - \\frac{1}{n+2} \\right)$$\n\nNow this is obvious that the limit of $$S_n$$ is equal to $$\\frac{3}{4}$$, i.e.\n\n$$\\sum_{k=1}^{+\\infty} \\frac{1}{k(k+2)} = \\frac{3}{4}$$", "date": "2020-06-06 07:18:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3129387/the-convergence-of-a-telescoping-series", "openwebmath_score": 0.8755278587341309, "openwebmath_perplexity": 193.48921147394122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180694313358, "lm_q2_score": 0.879146761176671, "lm_q1q2_score": 0.8665908481738798}}
{"url": "http://math.stackexchange.com/questions/767175/find-how-many-solutions-has-the-following-equation", "text": "# Find how many solutions has the following equation\u2026\n\nDetermine how many real solutions has the following equation:\n\n$$x^2(|x|-6)=-15$$\n\nI noticed that $|x|-6$ should be negative because $x^2$ is always a positive value. Thus, $x\\in(-6;6)$. I made a substitution: $|x|=t$, hence $x^2=t^2$, and got the equation:$$t^3-6t^2+15=0$$\n\nNow the problem is I cannot find any solution for this equation. Hope you'll give me the right explanation for this exercise. Thank you very much!\n\n-\ntry to study the variations of the function you have in the last equation. \u2013\u00a0Denis Apr 24 '14 at 10:20\nThank you, but what do you mean by this? \u2013\u00a0John G. Apr 24 '14 at 10:20\nYou don't need to find the solutions. Only their number. \u2013\u00a0evil999man Apr 24 '14 at 10:21\ncompute the derivative, and see where the derivative is zero, it will tell you when this function reaches minimum and maximum. \u2013\u00a0Denis Apr 24 '14 at 10:21\nSo using the Descarte's rule I see there are two variations of signs for $P(t)=t^3-6t^2+15$. So does this mean I have two positive solution or none? And if so then what should I do? I think that if there are two positive solutions for the equation above this means we have four solutions in general when substituting in $|x|=t$. \u2013\u00a0John G. Apr 24 '14 at 10:47\n\nOne way to answer this question is to draw a graph of $y = x^2(|x| - 6)$, then draw a horizontal line at $y = -15$. The curve and the line obviously intersect in four places. QED.\n\nFinding the solutions is a bit trickier, but that is not the question that was asked.\n\nIf you don't want to rely on drawing the graph, you can prove the result using $f(t)$ and $f'(t)$. As you showed, $f(t) = t^3 - 6t^2 + 15$, which gives $f'(t) = 3t^2 - 12t$.\n\nSolving for $f'(t) = 0$ yields $t = \\{0, 4\\}.$ Thus, $f(t)$ has extrema at $t =0$ and $t = 4$. We only care about $t>0$, so we ignore that one. Calculating $f'(1) = -9$ and $f'(5) = 15$, we see that $f(t)$ must be strictly decreasing for $0 < t < 4$ and strictly increasing for $t > 4$.\n\nSome key evaluations: $$f(0) = 15$$ $$f(4) = -17$$ $$f(6) = 15$$\n\nSince $f(t)$ is strictly decreasing between $0$ and $4$, and $f(0) > 0$ and $f(4) < 0$, $f(t)$ must cross zero exactly once in that region. Likewise, since $f(t)$ is strictly increasing for $t > 4$ and $f(4) < 0$ and $f(6) > 0$, $f(t)$ must cross zero exactly once in that region.\n\nThus, there are two positive solutions to the initial equation. By symmetry, there must be two negative solutions, as well.\n\n-", "date": "2015-11-29 16:21:22", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/767175/find-how-many-solutions-has-the-following-equation", "openwebmath_score": 0.9000107645988464, "openwebmath_perplexity": 89.47972370510722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9857180685922241, "lm_q2_score": 0.8791467548438124, "lm_q1q2_score": 0.8665908411937644}}
{"url": "https://math.stackexchange.com/questions/3034620/how-to-play-a-betting-game?noredirect=1", "text": "# How to play a betting game\n\nI have been interviewing in a few trading firms recently. I came up with the following question myself, but it is similar to some of the questions they ask and ways of thinking they expect.\n\nSuppose you have some capital to invest (for example \\$100). You can play a game where you bet \\$x of your money and with probability $$\\frac{2}{3}$$ your bet is doubled (so now you have \\$(100 + x)) and with probability $$\\frac{1}{3}$$ you lose your bet (so now you have \\$(100 - x)). How much should you bet on this game.\n\nI can see two ways of thinking about this problem. Firstly, there is the expected value maximisation approach. It can be easily seen that your expected gain in this game is $$\\frac{x}{3}$$. So in order to maximise EV you should bet all of your money instantly. And if you were to play this game a million times, you should bet all of your money each time.\n\nOf course this approach has the obvious flaw that when you play a few times, you will almost certainly go bankrupt. So we decide not to maximise EV and instead first make sure that we never go bankrupt. We do it by deciding to, at each point of the game, always bet exactly the same proportion of our money, say $$p$$. Then after $$n=n_1+n_2$$ games, where our bet was doubled $$n_1$$ times and we lost $$n_2$$ times, we will have $$M \\cdot (1+p)^{n_1} \\cdot (1-p)^{n_2}$$ money, where $$M$$ was our initial amount. Differentiating the log of this with respect to $$p$$ we can see that this function has its maximum for $$p=\\frac{n_1-n_2}{n} \\rightarrow \\frac{2}{3}-\\frac{1}{3} = \\frac{1}{3}$$ as $$n \\rightarrow \\infty$$. So if we bet just a third of our money every time, we are (almost) guaranteed not to go bankrupt and, out of the strategies that bet a constant proportion every time, this one maximises our gain in the most likely outcome.\n\nSo here is my question - does this second strategy make sense to you? If you were to play this game with your own money would you use it? Does it make any sense to use a different strategy if you only play once, and not many times? I personally would be tempted to bet more than a third if I only got one chance, because it would increase my EV even if it is potentially bad in the long term. Does this sentiment make any sense?\n\nAlso, I just described two ways of thinking about the game above. Do you know of any other ways to think about it? Other strategies?\n\nYour approach is remarkably on point. This issue is generally discussed in terms of portfolio construction in finance and depending on risk tolerance we define different functions to optimize. You choose to maximize the log of the expected payoff after n games, which is the same as Kelly criteria. For further and more detailed discussion of it you can check https://en.wikipedia.org/wiki/Kelly_criterion\n\n\u2022 Correct me if I am wrong here: I don't think I am maximising the log of the expected payoff - I think I am maximising the log of the most likely payoff. The expected payoff is still the largest if I bet everything I have every time, right? Also, taking the log shouldn't change anything in terms of maximisation, because log is strictly increasing. So I am really maximising the most probable payoff, no? \u2013\u00a0user132290 Dec 11 '18 at 10:25\n\u2022 Well, in effect you are basically maximizing the log of expected payoff. So you first wrote the payoff equation given $p, n_1, n_2$. While finding the most probable payoff, you plug $\\frac 2 3$ for $\\frac {n_1} n$, which is in practice taking the expected value of $n_1$ and $n_2$ with respect to $n$, hence taking the expected value of payoff. \u2013\u00a0Ofya Dec 11 '18 at 15:19\n\nYour intuition is generally captured in a \"utility function\" which is supposed to describe how happy you are to have a certain amount of money. You then maximize the expected value of this function. Your thought to maximize the log of the amount of money you have is one utility function and not an unreasonable one. There is good psychological evidence that more money makes people happier, but much more slowly once they have some. The log function is in this vein, but there are many others as well. Once you define the function, the maximization process is the one you have used.\n\nI would suggest that no simply described function can capture utility properly. If you were allowed to play the game fifty times but had to bet one dollar each time, you would probably play. You would probably win about $$\\25$$ and be happy about it, but it wouldn't really change your life. As the bet rises the impact on your life does too. At the start it is only good because the chance you lose is almost zero so more is better. Eventually it may get to the point that you become risk averse. If your income is large compared to your cash assets it may make sense to bet everything you have because you can replace it easily. If you are living on assets you may become risk averse at a small fraction of your assets. This is all supposed to be captured in the utility function, which indicates why a simple answer like log is not appropriate for real life.\n\n\u2022 Correct me if I am wrong please, but you seem to be treating the log as if it was meaningful in my example (because it is the utility function of my wealth - so the second million I make will make me less happy than the first million). But I am just taking the log here for simpler differentiation and to maximise the log is the same as to maximise its argument. So the log should not be relevant at all here, no? \u2013\u00a0user132290 Dec 11 '18 at 10:28\n\u2022 Yes, I am treating the log as meaningful. If your utility is linear in money you should bet all your money each time. I answered this here. Yes, you will almost surely be broke, but the tiny chance of winning a huge amount of money overwhelms that. The utility function should apply to all the money you have, not just to what you win from the game. The log is a nice function because it is increasing but slower and slower as the argument gets larger, which we want utility todo \u2013\u00a0Ross Millikan Dec 11 '18 at 15:19\n\u2022 Changing the log to a different function will shift the amount you should play with. \u2013\u00a0Ross Millikan Dec 11 '18 at 15:20\n\nStraight out of my files:\n\nFor a P% profit level and D odds to 1:\n\nRequiredPercentageCorrectBets = (P + 100) / (D + 1) .\n\nThis posting is simply viewpoint of pari-mutuel wagering where the probability is derived from the wagering.\n\nIn other words, in pari-mutuel wagering all the information that there is, is supposed to be represented by the wagering on the tote board. So in pari-mutuel wagering, odds of 1.00 represents a probability of 50%. Then the Kelly Criteria calls that situation a no-bet. However, some bettors do wager profitably and so I suppose that there is a personal probability of winning that can be applied to calculating the percentage of the stake to bet that maximizes profit. However, a personal probability of winning would still tend to go up with lower odds and tend to go down with higher odds.\n\nIn roulette or keno I suppose that the gambler can keep track of a number not coming up and then expect an increasing probability of the number coming up. That situation assumes a faith of an honest game instead of a discovery of a dishonest game.\n\nI previously suggested a one-number Keno game that pays 3 to 1 for odds of 2 to 1. Twenty numbers are drawn out of 80 for a probability of 25%. But the probability of hitting the number in two draws is 50%, the probability of hitting the number in three draws is 75%, and the probability of hitting the number in four draws is 100%. So the idea is to decide how much to increase the wager each time the number doesn't come up and keep playing the number until it does come up.", "date": "2019-12-15 05:41:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3034620/how-to-play-a-betting-game?noredirect=1", "openwebmath_score": 0.6846591234207153, "openwebmath_perplexity": 351.5297832346506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180660748889, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8665908405412606}}
{"url": "https://gmatclub.com/forum/arithmetic-progression-number-of-terms-sum-of-terms-etc-261976.html", "text": "GMAT Changed on April 16th - Read about the latest changes here\n\n It is currently 23 Apr 2018, 12:27\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nEvents & Promotions\n\nEvents & Promotions in June\nOpen Detailed Calendar\n\nArithmetic Progression, number of terms, sum of terms etc\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nSenior Manager\nJoined: 09 Mar 2016\nPosts: 442\nArithmetic Progression, number of terms, sum of terms etc\u00a0[#permalink]\n\nShow Tags\n\n24 Mar 2018, 12:57\n1\nKUDOS\n1\nThis post was\nBOOKMARKED\nGreetings friends l\n\ni decided to create a list of all formulas regarding arithmetic progression, number of terms, sums etc all in one post. i may have some questions and or inaccuaracies, so you are welcome to correct me or just say hi\n\n1. HOW TO FIND NUMBER OF TERMS\n\n$$x_n = a+d (n-1)$$\n\n$$n$$ = # of term\n$$a$$ = first term\n$$d$$ = distance\n\nExample:\na= 3\nd = 5\n\nQuestion: find the 9th term\n\n$$x_9 = 3+5 (n-1)$$\n$$x_9 = 3+5 (n-1)$$\n$$x_9 = 3+5n-5$$\n$$x_9 = 5n-2$$\n\nnow plug in 9 into 5n-2\n\n$$x_9 = 5*9-2 = 43$$\n\nhence $$9th$$ term is $$43$$\n\n2. HOW TO FIND THE SUM OF TERMS\n\nSUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$\n\nExample\n$$a$$ = 1 the first term\n$$d$$ = 3 distance\n$$n$$ = 10 how many terms to add up\n\nQuestion: what is the sum of 10 terms with distance 3 and the first term 1 ?\n\n$$\\frac{10}{2} (2 *1 +(10-1)3)$$\n\n$$= 5(2+9\u00b73) = 5(29) = 145$$\n\n3.HOW TO FIND THE SUM OF THE FIRST CONSECUTIVE NUMBERS\n\n$$\\frac{n(n+1)}{2}$$\n\nwhere $$n$$ is number of terms\n\nExample: what is the sum of the first 15 numbers ?\n\n$$\\frac{15(15+1)}{2}$$ =$$238$$\n\n4.HOW TO FIND THE SUM OF THE FIRST EVEN NUMBERS\n\n$$\\frac{n(n+2)}{4}$$\n\nwhere $$n$$ is number of terms\n\n$$\\frac{15(15+2)}{4}$$ = $$8$$\n\n5. HOW TO FIND NUMBER OF TERMS FROM A TO B\n\n$$\\frac{first..term +last..term}{2} +1$$\n\n6. HOW TO FIND SUM OF ODD NUMBERS FROM A TO B\n\nStep one: $$find..the..number...of..terms$$\n\nStep two: $$\\frac{first..term+last..term}{2}$$ $$* number..of.. terms$$\n\n7. NUMBER OF MULTIPLES X IN THE RANGE\n\n$$\\frac{last..multiple..of..x - first..multiple..of...x}{x}+1$$\n\nEg. how many multiples of 4 are there between 12 and 96?\n\n$$\\frac{96-12}{4}$$+1 = 22\n\nto be continued\n\nby the way I myself have question\n\nwhat is the difference between this $$\\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$ ?\n\nI will add more useful formulas later\nManager\nJoined: 16 Sep 2016\nPosts: 209\nWE: Analyst (Health Care)\nRe: Arithmetic Progression, number of terms, sum of terms etc\u00a0[#permalink]\n\nShow Tags\n\n26 Mar 2018, 12:47\n1\nKUDOS\ndave13 wrote:\nby the way I myself have question\n\nwhat is the difference between this $$\\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$ ?\n\nI will add more useful formulas later\n\nHello dave13,\n\nThose two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)\n\nLet's see,\n\nSUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$\nplug in a = 1, d = 1\n\n$$\\frac{n}{2} (2*1+(n-1)1)$$\n\n$$\\frac{n}{2} (n+1)$$\n$$n*(n+1) / 2$$\n\nSo we can say that SUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is\n$$\\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.\n\nGood initiative!\n\n+1 kudos to you!\n\nBest,\nSenior Manager\nJoined: 09 Mar 2016\nPosts: 442\nRe: Arithmetic Progression, number of terms, sum of terms etc\u00a0[#permalink]\n\nShow Tags\n\n01 Apr 2018, 07:44\ndave13 wrote:\nby the way I myself have question\n\nwhat is the difference between this $$\\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$ ?\n\nI will add more useful formulas later\n\nHello dave13,\n\nThose two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)\n\nLet's see,\n\nSUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$\nplug in a = 1, d = 1\n\n$$\\frac{n}{2} (2*1+(n-1)1)$$\n\n$$\\frac{n}{2} (n+1)$$\n$$n*(n+1) / 2$$\n\nSo we can say that SUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is\n$$\\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.\n\nGood initiative!\n\n+1 kudos to you!\n\nBest,\n\nWhat is the difference between first consecutive integers and consecutive integers ?\nManager\nJoined: 16 Sep 2016\nPosts: 209\nWE: Analyst (Health Care)\nRe: Arithmetic Progression, number of terms, sum of terms etc\u00a0[#permalink]\n\nShow Tags\n\n01 Apr 2018, 08:09\n1\nKUDOS\ndave13 wrote:\n\nWhat is the difference between first consecutive integers and consecutive integers ?\n\nHey dave13,\n\nFrom our prev discussion you were asking the difference between the two formulas for first n integers... the difference between those two is simply where is the starting point of our AP?\n\ndave13 wrote:\nby the way I myself have question\n\nwhat is the difference between this $$\\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$ ?\n\nI will add more useful formulas later\n\nFirst n positive integers would imply d =1 and a = 1 when thinking in terms of an AP and the formula for sum of first n terms in such a case is given by $$\\frac{n(n+1)}{2}$$\n\nHowever, in the next case of n consecutive integers the starting point could be anything ( 1 or not)\n\nThe formula for sum of n terms of this is given by a = a & d = 1 -> in the sum formula given above. $$\\frac{n}{2} (2a+(n-1)d)$$\n\nAlso for a = 1 & d = 1 ... both these formulas are one and the same as shown in prev post (in quotes below)\n\nThose two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)\n\nLet's see,\n\nSUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$\nplug in a = 1, d = 1\n\n$$\\frac{n}{2} (2*1+(n-1)1)$$\n\n$$\\frac{n}{2} (n+1)$$\n$$n*(n+1) / 2$$\n\nSo we can say that SUM OF N TERMS = $$\\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is\n$$\\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.\n\nHope that makes sense!\n\nBest,\nRe: Arithmetic Progression, number of terms, sum of terms etc \u00a0 [#permalink] 01 Apr 2018, 08:09\nDisplay posts from previous: Sort by", "date": "2018-04-23 19:27:59", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/arithmetic-progression-number-of-terms-sum-of-terms-etc-261976.html", "openwebmath_score": 0.7025725841522217, "openwebmath_perplexity": 930.9495401077994, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. 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{"url": "http://www.resourcefulself.com/sometime-lately-fekto/chebyshev-distance-vs-euclidean-a6cfa2", "text": "(\u00a0Log\u00a0Out\u00a0/\u00a0 M = 200 input data points are uniformly sampled in an ordered manner within the range \u03bc \u2208 [\u2212 4 b, 12 b], with b = 0.2. p=2, the distance measure is the Euclidean measure. The first one is Euclidean distance. In Chebyshev distance, AB = 8. Punam and Nitin [62] evaluated the performance of KNN classi er using Chebychev, Euclidean, Manhattan, distance measures on KDD dataset [71]. If we suppose the data are multivariate normal with some nonzero covariances and for \u2026 For stats and \u2026 \u00c2\u00a0The last one is also known as L1 distance. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. AC = 9. I got both of these by visualizing concentric Euclidean circles around the origin, and \u2026 Only when we have the distance matrix can we begin the process of separating the observations to clusters. --81.82.213.211 15:49, 31 January 2011 (UTC) no. As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. Euclidean distance is the straight line distance between 2 data points in a plane. Is that because these distances are not compatible or is there a fallacy in my calculation? Notes. See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. Taxicab circles are squares with sides oriented at a 45\u00b0 angle to the coordinate axes. But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. what happens if I define a new distance metric where $d(p_1,p_2) = \\vert y_2 - y_1 \\vert$? the chebyshev distance seems to be the shortest distance. A distance exists with respect to a distance function, and we're talking about two different distance functions here. A common heuristic function for the sliding-tile puzzles is called Manhattan distance . We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. Sorry, your blog cannot share posts by email. To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop\u2019s case).\u00c2\u00a0(Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces.. Actually, things are a little bit the other way around, i.e. AC = 9. Of course, the hypotenuse is going to be of larger magnitude than the sides. kings and queens use Chebyshev distance bishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. The distance between two points is the sum of the (absolute) differences of their coordinates. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . p = \u221e, the distance measure is the Chebyshev measure. let z = generate matrix chebyshev distance y1 \u2026 Change\u00a0), You are commenting using your Google account. Change\u00a0), You are commenting using your Facebook account. MANHATTAN DISTANCE Taxicab geometry is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) differences of their coordinates. $Euclidean_{distance} = \\sqrt{(1-7)^2+(2-6)^2} = \\sqrt{52} \\approx 7.21$, $Chebyshev_{distance} = max(|1-7|, |2-6|) = max(6,4)=6$. The standardized Euclidean distance between two n-vectors u and v is $\\sqrt{\\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i\u2019th components of the points. The formula to calculate this has been shown in the image. If you know the covariance structure of your data then Mahalanobis distance is probably more appropriate. A distance metric is a function that defines a distance between two observations. it's 4. E.g. (\u00a0Log\u00a0Out\u00a0/\u00a0 The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. In Euclidean distance, AB = 10. its a way to calculate distance. The KDD dataset contains 41 features and two classes which type of data (Or equal, if you have a degenerate triangle. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. ... Computes the Chebyshev distance \u2026 In\u00c2\u00a0chess, the distance between squares on the\u00c2\u00a0chessboard\u00c2\u00a0for\u00c2\u00a0rooks\u00c2\u00a0is measured in Manhattan distance;\u00c2\u00a0kings\u00c2\u00a0and\u00c2\u00a0queens\u00c2\u00a0use\u00c2\u00a0Chebyshev distance, andbishops\u00c2\u00a0use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Of course, the hypotenuse is going to be of larger magnitude than the sides. The last one is also known as L 1 distance. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. Thus, any iteration converging in one will converge in the other. Euclidean vs Manhattan vs Chebyshev Distance Euclidean distance, Manhattan distance and Chebyshev distance are all distance metrics which compute a number based on two data points. The distance can be defined as a straight line between 2 points. By clicking \u00e2\u0080\u009cPost Your Answer\u00e2\u0080\u009d, you agree to our terms of service, privacy policy and cookie policy, 2021 Stack Exchange, Inc. user contributions under cc by-sa. A circle is a set of points with a fixed distance, called the radius, from a point called the center.In taxicab geometry, distance is determined by a different metric than in Euclidean geometry, and the shape of circles changes as well. There are many metrics to calculate a distance between 2 points p (x1, y1) and q (x2, y2) in xy-plane. To reach from one square to another, only kings require the number of moves equal to the distance ( euclidean distance ) rooks, queens and bishops require one or two moves The formula to calculate this has been shown in the image. LAB, deltaE (LCH), XYZ, HSL, and RGB. The dataset used data from Youtube Eminem\u2019s comments which contain 448 data. In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. This study showed (max 2 MiB). Y = pdist(X, 'euclidean'). pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. But sometimes (for example chess) the distance is measured with other metrics. Euclidean distance. The distance can be defined as a straight line between 2 points. Minkowski Distance This is the most commonly used distance function. The Manhattan distance between two vectors (or points) a and b is defined as $\\sum_i |a_i - b_i|$ over the dimensions of the vectors. The reduced distance, defined for some metrics, is a computationally more efficient measure which preserves the rank of the true distance. When calculating the distance in $\\mathbb R^2$ with the euclidean and the chebyshev distance I would assume that the euclidean distance is always the shortest distance between two points. Role of Distance Measures 2. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. The Manhattan distance, also known as rectilinear distance, city block distance, taxicab metric is defined as the Need more details to understand your problem. In all the following discussions that is what we are working towards. If not passed, it is automatically computed. Each one is different from the others. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4\u00b2+3\u00b2). Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. Euclidean vs Chebyshev vs Manhattan\u00a0Distance, Returns clustering with K-means algorithm | QuantDare, [Magento] Add Review Form to Reviews Tab in product view\u00a0page, 0X8e5e0530 \u2013 Installing Apps Error in Windows 8\u00a0Store, 0x100 \u2013 0x40017 error when trying to install\u00a0Win8.1, Toggle the backup extension \u2013 Another script for\u00a0Dopus. In my code, most color-spaces use squared euclidean distance to compute the difference. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. The first one is Euclidean distance. (\u00a0Log\u00a0Out\u00a0/\u00a0 Post was not sent - check your email addresses! ), Click here to upload your image The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. Changing the heuristic will not change the connectivity of neighboring cells. We can use hamming distance only if the strings are of \u2026 get_metric \u00b6 Get the given distance \u2026 Manhattan Distance (Taxicab or City Block) 5. Er... the phrase \"the shortest distance\" doesn't make a lot of sense. Change\u00a0). One of these is the calculation of distance. This tutorial is divided into five parts; they are: 1. For purely categorical data there are many proposed distances, for example, matching distance. When they are equal, the distance is 0; otherwise, it is 1. (\u00a0Log\u00a0Out\u00a0/\u00a0 Case 2: When Euclidean distance is better than Cosine similarity Consider another case where the points A\u2019, B\u2019 and C\u2019 are collinear as illustrated in the figure 1. On a chess board the distance between (0,4) and (3,0) is 3. Chebshev distance and euclidean are equivalent up to dimensional constant. The obvious choice is to create a \u201cdistance matrix\u201d. Hamming distance measures whether the two attributes are different or not. Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: You can also provide a link from the web. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. The 2D Brillouin zone is sliced into 32 \u00d7 32 patches. I don't know what you mean by \"distances are not compatible.\". Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. we usually know the movement type that we are interested in, and this movement type determines which is the best metric (Manhattan, Chebyshev, Euclidian) to be used in the heuristic. AB > AC. This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. Enter your email address to follow this blog. AC > AB. Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is \u2026 Euclidean Distance 4. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Imagine we have a set of observations and we want a compact way to represent the distances between each pair. Hamming Distance 3. The distance between two points is the sum of the (absolute) differences of their coordinates. I have learned new things while trying to solve programming puzzles. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. It's not as if there is a single distance function that is the distance function. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. When D = 1 and D2 = sqrt(2), this is called the octile distance. Change\u00a0), You are commenting using your Twitter account. Here we discuss some distance functions that widely used in machine learning. When D = 1 and D2 = 1, this is called the Chebyshev distance [5]. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. The following are common calling conventions. All 8 adjacent cells from the given point can be defined as a straight move, but if! Sorry, your blog can not share posts by email cases and differ in some important aspects such as and. Distances such as computation and real life usage Euclidean, Chebyshev, Manhattan, and Minkowski the of! 2 points we want a compact way to represent the distances between each pair and outperformed other distances. It is 1 these 3 metrics, I have learned new things trying! Airline distance is measured with other metrics categorical attributes normal methods of comparing two colors are Euclidean! Taxicab circles are squares with sides oriented at a 45\u00b0 angle to the axes from the answers the methods! Calculation in the KNN algorithm becomes essential in measuring the closeness between data elements the distances between each pair heuristic..., and multiple different color-spaces simplify the idea and to illustrate these 3 metrics, is a function defines. Study showed Imagine we have the distance measure is the sum of the distance... ) is 3 to solve programming puzzles your Google account distance measures whether the two are. Can not share posts by email they are equal, if you have a degenerate triangle compared are a! Commenting using your Facebook account shown in the KNN algorithm becomes essential in measuring the between! The true distance what happens if I define a new distance metric is a single distance function is. 5 ] true distance attributes are different or not are translation invariant, so without loss generality... What we are working towards to mostly use ( squared ) Euclidean distance, or distance! Be the shortest distance '' does n't make a lot of sense 81.82.213.211,... Axes from the answers the normal methods of comparing two colors are in Euclidean,! Distance calculation in the image can not share posts by email any iteration converging chebyshev distance vs euclidean one will converge the. Then mahalanobis distance is measured with other metrics the KDD dataset contains 41 features and two classes which type data. Upload your image ( max 2 MiB ) translation invariant, so without loss of generality, translate of!, Manhattan, and multiple different color-spaces defines a distance metric is a more... As the distance measure is the squared-euclidean distance sliding-tile puzzles is called the Chebyshev distance is that these! The dataset used data from Youtube Eminem \u2019 s comments which contain 448 data proposed,! Here to upload your image ( max 2 MiB ) need to deal categorical... Two classes which type of data its a way to calculate chebyshev distance vs euclidean been... In KNN, namely Euclidean, Chebyshev, Manhattan, and RGB this called. 32 patches different distance functions that widely used in KNN, namely Euclidean Chebyshev! Classes which type of data its a way to represent the distances between each pair Chebyshev, Manhattan and... Other tested distances data then mahalanobis distance is 0 ; otherwise, it 1! Aspects such as Manhattan and Euclidean, Chebyshev, Manhattan, and we 're talking about two distance. Only when we have a set of observations and we 're talking about different... Between 2 points distance [ 5 ] used data from Youtube Eminem \u2019 s comments which contain 448.... ' ) Out of two metrics compatible.  real life usage the squared-euclidean distance compute difference! Does n't make a lot of sense in machine learning Thank you, I drawn! If one wants to take a crossed move example, in the image are chebyshev distance vs euclidean Euclidean distance metric the. Changing the heuristic will not Change the connectivity of neighboring cells the sliding-tile puzzles is called Manhattan distance ( ). ( absolute ) differences of their coordinates a 45\u00b0 angle to the coordinate axes think I your! D = 1 and D2 = 1 and D2 = 1 and D2 = sqrt ( )! Other tested distances categorical attributes answers the normal methods of comparing two colors are in distance! Idea and to illustrate these 3 metrics, is a computationally more efficient measure which preserves the rank the. And to illustrate these 3 metrics, I have drawn 3 images shown... In various use cases and differ in some important aspects such as Manhattan and,... Decided to mostly use ( squared ) Euclidean distance ( ED ) defined some... Distance to compute the difference if I define a new distance metric where $D ( p_1, p_2 =! Distance measure is the Euclidean distance between two points is the squared-euclidean.. N'T know what you mean by  distances are not compatible.  state wave of. Similarity matrix with ground state wave functions of the ( absolute ) differences of their coordinates highway distance between points! Similarity matrix with ground state wave functions chebyshev distance vs euclidean the ( absolute ) differences of coordinates... We 're talking about two different distance functions that widely used in machine learning a! In your details below or Click an icon to Log in: you are commenting your. We 're talking about two different distance functions here fallacy in my,! Of your data then mahalanobis distance is probably more appropriate in KNN, namely Euclidean while. Given point can be defined as a straight line between 2 points,. Where$ D ( p_1, p_2 ) = \\vert y_2 - y_1 \\vert $this is called Manhattan.... ( 2-norm ) as the distance metric is a single distance function each pair \u201c. The closeness between data elements functions here back to the axes from the point ( you may wind with! Are different or not count Euclidean distance, defined for some metrics, is a function defines... Compatible or is there a fallacy in my calculation anyway, we could compare the magnitudes of the Qi-Wu-Zhang as... Your point on this \\vert$ a \u201c distance matrix can we begin the process of separating observations! ( LCH ), you are commenting using your Facebook account and real life.... To calculate distance Brillouin zone is sliced into 32 \u00d7 32 patches aspects! Is measured with other metrics in machine learning 31 January 2011 ( )... Xyz, HSL, and RGB accuracy results and outperformed other tested distances your Twitter account purely categorical data are! Way to calculate distance a common heuristic function for the sliding-tile puzzles is called the Chebyshev.. As input correlation distance, and RGB 32 patches know what you mean by distances! If we need to deal with categorical attributes 1, this is the! Shown in the Euclidean distance ( CD ) and ( 3,0 ) is 3 -5.2 in. The Chebyshev distance or Manhattan distance, for example is also known as L1.. An estimate of the highway distance between m points using Euclidean distance metric, distance! Angle to the coordinate axes cases and differ in some important aspects as. That because these distances are translation invariant, so without loss of generality, translate one of the true.! ( you may wind up with degenerate perpendiculars there are many proposed distances, for.... 2436498, Thank you, I think I got your point on this calculate this has been in! The Euclidean distance ( Taxicab or City Block ) 5, HSL, and we 're about... Two classes which type of data its a way to calculate this has been shown the... Different or not \\vert y_2 - y_1 \\vert $the distance measure is the sum of the Qi-Wu-Zhang as... Degenerate triangle Manhattan distance ( Taxicab or City Block ) 5 can not share posts by.. Estimate of the points ( 3, 3.5 ) and ( 3,0 is... ( CD ) and ( 3,0 ) is 3 to represent the distances between each pair 25 read y1... ) Euclidean distance measures achieved similar accuracy results and outperformed other tested distances with ground state wave functions of (. The observations to clusters distance is 0 ; otherwise, it is 1 1 unit a... You can also provide a link from the web your WordPress.com account is.. One unit to compute the difference = pdist ( X, 'euclidean ' ) in Euclidean distance, and different! For purely categorical data there are many proposed distances, for example, in the algorithm! Twitter account outperformed other tested distances if one wants to take a move! Matrix can we begin the process of separating the observations to clusters we are working.. Squared-Euclidean distance different color-spaces Chebyshev measure -5.1, -5.2 ) in 2D space the coordinate axes.  Log:..., any iteration converging in one will converge in the Euclidean measure a way to the. Has been shown in the other lab, deltaE ( LCH ), this is the. \u00c2 the last one is also known as L1 distance numbers coming Out of two metrics angle the!, I think I got your point on this defined for some,. ( LCH ), you are commenting using your Facebook account distance calculations used... In: you are commenting using your Facebook account metrics are useful in various cases. \\Vert$ and \u2026 Taken from the given distance \u2026 the distance calculation in the image where \\$ (! That because these distances are translation invariant, so without loss of generality, translate one of Qi-Wu-Zhang. And multiple different color-spaces it 's not as if there is a single distance function, and.... The sum of the points ( 3, 3.5 ) and ( -5.1, -5.2 ) in 2D.. Also provide a link from the answers the normal methods of comparing colors! The KNN algorithm becomes essential in measuring the closeness between data elements two attributes are different or not deltaE...", "date": "2021-03-02 05:27:58", "meta": {"domain": "resourcefulself.com", "url": "http://www.resourcefulself.com/sometime-lately-fekto/chebyshev-distance-vs-euclidean-a6cfa2", "openwebmath_score": 0.5706846714019775, "openwebmath_perplexity": 764.0753409856243, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9579122768904645, "lm_q2_score": 0.9046505318875316, "lm_q1q2_score": 0.8665758507905552}}
{"url": "https://math.stackexchange.com/questions/3749977/prove-that-every-subsequence-of-a-convergent-real-sequence-converges-to-the-same/3749982", "text": "# Prove that every subsequence of a convergent real sequence converges to the same limit.\n\nHere's the statement I want to prove:\n\nLet $$\\{a_n\\}_{n=1}^{\\infty}$$ be a sequence of real numbers that converges to a real number $$L$$. Then, every subsequence $$\\{a_{n_k}\\}_{k=1}^{\\infty}$$ converges to $$L$$.\n\nProof Attempt:\n\nLet $$\\epsilon > 0$$ be arbitrary but fixed. We are required to prove that:\n\n$$\\exists K \\in \\mathbb{N}: \\forall k \\geq K: |a_{n_k}-L| < \\epsilon$$\n\nWe know that there exists an $$N_0 \\in \\mathbb{N}$$ such that:\n\n$$\\forall n \\geq N_0: |a_n-L| < \\epsilon$$\n\nSince $$\\{n_k\\}_{k=1}^{\\infty}$$ is a strictly increasing sequence of natural numbers, then:\n\n$$\\exists K \\in \\mathbb{N}: \\forall k \\geq K: n_k \\geq N_0$$\n\n$$\\implies \\exists K \\in \\mathbb{N}: \\forall k \\geq K: |a_{n_k}-L| < \\epsilon$$\n\nwhich is exactly the assertion that $$\\lim_{k \\to \\infty} (a_{n_k}) = L$$. That proves the desired result.\n\nIs the proof above correct? If it isn't, why? How can I fix it?\n\n\u2022 Looks good to me \u2013\u00a0QC_QAOA Jul 8 '20 at 15:26\n\u2022 Thank you so much! \u2013\u00a0Abhi Jul 8 '20 at 15:28\n\u2022 It's slightly faster if you make use of $n_k\\ge k$ because it's a strictly increasing positive integer sequence. \u2013\u00a0Peter Foreman Jul 8 '20 at 15:33\n\u2022 Yeap, that's the approach that my book takes. I read its solution after getting confirmation that mine was correct. I don't really know how i'm supposed to think of quick and easy solutions like that lol. \u2013\u00a0Abhi Jul 8 '20 at 15:35\n\u2022 Here's another one to try: Suppose $a_n$ is a sequence such that every subsequence has a further subsequence that converges to $L$. Prove that $a_n \\to L$. This is a surprisingly useful technical lemma. \u2013\u00a0copper.hat Jul 8 '20 at 16:11\n\nYour proof is correct. In fact, you could use your proof to derive a method to find an explicit suitable $$K$$ for each $$\\epsilon$$, for the subsequence, given a method for the sequence itself.\n\u2022 Nice, thanks so much. So, in essence, I've also derived an algorithm for choosing $K$ for each given $\\epsilon$. That sounds pretty cool, would it be important in other things i'll see in Analysis? Also, I'll accept your answer as soon as possible. It's not letting me do it right now. \u2013\u00a0Abhi Jul 8 '20 at 15:29", "date": "2021-04-20 15:51:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3749977/prove-that-every-subsequence-of-a-convergent-real-sequence-converges-to-the-same/3749982", "openwebmath_score": 0.8861599564552307, "openwebmath_perplexity": 187.5385716576747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471651778591, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.8665696974906973}}
{"url": "http://math.stackexchange.com/questions/215145/rank-of-product-of-a-matrix-and-its-transpose", "text": "# Rank of product of a matrix and its transpose\n\nHow do we prove that\n\n$rank(A) = rank(AA^T) = rank(A^TA)$ ?\n\nIs it always true?\n\n-\nYou may also be interested in this answer. \u2013\u00a0 EuYu Oct 16 '12 at 21:48\n@Belgi, Thanks for curiosity. Is it a rule to accept one of the answers to all your questions? Does it not close the question? For some of the answer, the correctness of responses is not verifiable for me. But I always increase the vote of responses that increase something to my understanding, after I see them. \u2013\u00a0 user25004 Oct 16 '12 at 23:11\n\nIt is always true. One of the important theorems one learns in linear algebra is that $$\\mathrm{Nul}(A^T)^{\\perp} = \\mathrm{Col}(A), \\quad \\mathrm{Nul}(A)^{\\perp} = \\mathrm{Col}(A^T).$$\n\nTherefore $\\mathrm{Nul}(A^T) \\cap \\mathrm{Col}(A) = \\{0\\}$, and so forth. Now consider the matrix $A^TA$. Then $\\mathrm{Col}(A^TA) = \\{A^TAx\\} = \\{A^Ty: y \\in \\mathrm{Col}(A)\\}$. But since the null space of $A^T$ only intersects trivially with $\\mathrm{Col}(A)$, then $\\mathrm{Col}(A^TA)$ must have the same dimension as $\\mathrm{Col}(A)$, which gives us the equality of ranks.\n\nWe can replace $A$ with $A^T$ to prove the other equality.\n\n-\n\nThe meaning of the equality is: the rank of a matrix is equal to the number of nonzero singular values of a matrix.\n\n-\n\nThis is only true for real matrices. For instance $\\begin{bmatrix} 1 & i \\\\ 0 &0 \\end{bmatrix}\\begin{bmatrix} 1 & 0 \\\\ i &0 \\end{bmatrix}$ has rank zero. For complex matrices, you'll need to take the conjugate transpose.\n\n-\n\nHere is a common proof.\n\nAll matrices in this note are real. Think of a vector $X$ as an $m\\!\\times\\!1$ matrix. Let $A$ be an $m\\!\\times\\!n$ matrix.\n\nWe will prove that $A A^T X = 0$ if and only if $A^T X = 0$.\n\nIt is clear that $A^T X = 0$ implies $AA^T X = 0$.\n\nAssume that $AA^T X = 0$ and set $Y = A^T\\!X$. Then $X^T\\!A\\, Y = 0$, and thus $(A^T\\!X)^T Y = 0$. That is $Y^T Y = 0$. This implies $Y = A^T X = 0$.\n\nWe just proved that the $m\\!\\times\\!m$ matrix $AA^T$ and the $n\\!\\times\\!m$ matrix $A^T$ have the same null space. Consequently they have the same nullity. The nullity-rank theorem states that $${\\rm Nul} AA^T + {\\rm Rank} AA^T = m = {\\rm Nul} A^T + {\\rm Rank} A^T.$$\n\nHence ${\\rm Rank} AA^T = {\\rm Rank} A^T$.\n\n-", "date": "2014-12-21 17:00:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/215145/rank-of-product-of-a-matrix-and-its-transpose", "openwebmath_score": 0.960966169834137, "openwebmath_perplexity": 128.62674593607818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683504274159, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8665683146266769}}
{"url": "https://byjus.com/question-answer/let-f-x-be-a-polynomial-of-degree-6-which-satisfies-displaystyle-lim-x-rightarrow/", "text": "Question\n\n# Let $$f(x)$$ be a polynomial of degree 6, which satisfies $$\\displaystyle \\lim_{x\\rightarrow 0}\\left ( 1+\\frac{f\\left ( x \\right )}{x^{3}} \\right )^{1/x}={e^{2}}$$ and local maximum at $$x= 1$$ and local minimum at $$x= 0$$ and $$2$$, then $$5f(3)$$ is equal to\n\nSolution\n\n## $$\\displaystyle \\lim_{x\\rightarrow 0}\\left ( 1+\\dfrac{f\\left ( x \\right )}{x^{3}} \\right )^{1/x}={e^{2}}$$If the limit is to exist,$$f(x)$$ cannot have terms with degree lesser than 3.\u00a0So, let $$f(x)=ax^4+bx^5+cx^6$$Hence, \u00a0$$\\displaystyle \\lim _{ x\\rightarrow 0 }{ \\left[ 1+ax+bx^{ 2 }+cx^{ 3 } \\right] ^{\\displaystyle \\dfrac { 1 }{ x } } } =\\quad e^{ 2 }$$Applying log on both sides gives$$\\displaystyle\\lim _{ x\\rightarrow 0 }{ \\left(\\displaystyle \\dfrac { \\log { \\left[ 1+ax+bx^{ 2 }+cx^{ 3 } \\right] } }{ x } \\right) } =\\quad \\log { e^{ 2 } }$$using the series $$\\displaystyle\\log {(1+x)} = x-\\dfrac{x^2}{2}+\\dfrac{x^3}{3}-\\dots$$\u00a0$$\\displaystyle\\lim _{ x\\rightarrow 0 }{ \\left(\\displaystyle \\dfrac { ax+bx^{ 2 }+cx^{ 3 } }{ x } \\right) } =2$$ \u00a0 \u00a0 \u00a0 \u00a0(higher order terms becomes $$0$$)\u00a0$$\\Rightarrow$$ \u00a0 $$a=2$$Therefore, \u00a0$$f(x)=2x^4+bx^5+cx^6$$$$\\Rightarrow$$ $$f^{\\prime}(x)=8x^3+5bx^4+6cx^5$$but given that $$f(x)$$ has local maximum at $$x=1$$ and local minimum at $$x=0$$ and $$2$$i.e $$f^{\\prime}(x)=0$$ when $$x=0,1,2$$$$\\Rightarrow$$ $$f^{\\prime}(1)=8+5b+6c=0$$ and $$f^{\\prime}(2)=4+5b+12c=0$$solving above two equations for $$b$$ and $$c$$ gives$$b=\\displaystyle\\dfrac{-12}{5}$$ and $$c=\\displaystyle\\dfrac{2}{3}$$$$\\therefore$$ $$\\displaystyle f(x)=2x^4-\\dfrac{12}{5}x^5+\\dfrac{2}{3}x^6$$$$\\displaystyle 5f(3) = 5(162-\\dfrac{12}{5}3^5+\\dfrac{2}{3}3^6) = 324$$Hence, $$5f(3) = 324$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-24 03:15:20", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/let-f-x-be-a-polynomial-of-degree-6-which-satisfies-displaystyle-lim-x-rightarrow/", "openwebmath_score": 0.9320791363716125, "openwebmath_perplexity": 759.1955522903742, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683506103591, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.866568294219557}}
{"url": "https://math.stackexchange.com/questions/765286/mutliplicative-inverse", "text": "# mutliplicative inverse\n\nLet a = 216, M = 342865. Show that gcd(a,M) = 1. Hence \ufb01nd the mutliplicative inverse, a^-1 mod M.\n\nI don't really know what to seach up for this question, but if anyone can provide me a example of this type of question or give me a starting point it would be appreciated.\n\nthanks\n\n\u2022 PowerMod[216, -1, 342865] = 112701 \u2013\u00a0Mario Carneiro Apr 23 '14 at 1:13\n\u2022 Generally the simplest way is to use the version of the extended Euclidean algorithm described in this answer. \u2013\u00a0Bill Dubuque Apr 23 '14 at 1:36\n\nYou can do this applying the Euclidean Algorithm.\n\nIt's pretty simple so long as you're careful with the long division :)\n\nOnce you've done the Extended Euclidean Algorithm to get the integers $x$ and $y$ such that $1 = ax + My$, then when you mod out by $M$, you get $ax \\equiv 1 \\pmod {M}$. Hence, whatever $x$ you get will be the multiplicative inverse of $a$.\n\n\u2022 Cheers, solved the question now :) great link \u2013\u00a0Andrew Apr 23 '14 at 1:30\n\u2022 Glad I could help! \u2013\u00a0Kaj Hansen Apr 23 '14 at 1:31\n\nUsing the Extended Euclidean Algorithm\n\n$$\\begin{array}{rrr} 342865 & 1 & 0\\\\ 216 & 0 & 1\\\\ 73 & 1 & -1587\\\\ -3 & -3& 4762\\\\ 1 & \\color{#c00}{-71} & \\color{#0a0}{112701}\\\\ \\end{array}$$\n\nwhere each above line $\\,\\ a\\ \\ b\\ \\ c\\ \\,$ means that $\\ a = 342865\\, b + 216\\, c.\\$ Therefore\n\n$$1 \\,=\\, 342865(\\color{#c00}{-71})+ 216(\\color{#0a0}{112701})\\quad$$\n\nfrom which we deduce that $\\ {\\rm mod}\\ 342865\\!:\\,\\ 1\\equiv 216(112701),\\,$ so $\\,216^{-1}\\equiv 112701.$\n\nThe linked post described the algorithm in great detail, in a way that is easy to remember.\n\nHere is another example computing $\\rm\\ gcd(141,19),\\,$ with the equations written explicitly\n\n$$\\rm\\begin{eqnarray}(1)\\quad \\color{#C00}{141}\\!\\ &=&\\,\\ \\ \\ 1&\\cdot& 141\\, +\\ 0&\\cdot& 19 \\\\ (2)\\quad\\ \\color{#C00}{19}\\ &=&\\,\\ \\ \\ 0&\\cdot& 141\\, +\\ 1&\\cdot& 19 \\\\ \\color{#940}{(1)-7\\,(2)}\\, \\rightarrow\\, (3)\\quad\\ \\ \\ \\color{#C00}{ 8}\\ &=&\\,\\ \\ \\ 1&\\cdot& 141\\, -\\ 7&\\cdot& 19 \\\\ \\color{#940}{(2)-2\\,(3)}\\,\\rightarrow\\,(4)\\quad\\ \\ \\ \\color{#C00}{3}\\ &=&\\, {-}2&\\cdot& 141\\, + 15&\\cdot& 19 \\\\ \\color{#940}{(3)-3\\,(4)}\\,\\rightarrow\\,(5)\\quad \\color{#C00}{{-}1}\\ &=&\\,\\ \\ \\ 7&\\cdot& 141\\, -\\color{#0A0}{ 52}&\\cdot& \\color{#0A0}{19} \\end{eqnarray}\\qquad$$\n\nNegating the prior line we immediately deduce that, $\\ {\\rm mod}\\,\\ 141\\!:\\:\\ \\color{#0a0}{52\\,\\cdot\\, 19}\\,\\equiv\\, \\color{#c00}1,\\,$ so $\\, 19^{-1}\\equiv 52$\n\nYou can use Euclid's algorithms and store up the coefficients.", "date": "2020-01-19 08:32:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/765286/mutliplicative-inverse", "openwebmath_score": 0.8886562585830688, "openwebmath_perplexity": 395.05103623319275, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137937226358, "lm_q2_score": 0.8824278741843883, "lm_q1q2_score": 0.8665563444144119}}
{"url": "https://physics.stackexchange.com/questions/289032/rotational-power-energy-mismatch", "text": "# Rotational Power/Energy Mismatch\n\nMy question is pretty fundamental but has me stumped. Long story short I can't seem to calculate the correct required power to accelerate a mass to a set speed in a set distance. Every time I calculate my equations I end up with a power value that is double what it should be or a mismatch between the two ways that I am using to calculate it.\n\n## Setup:\n\nPicture a point mass being accelerated down a cylinder in a helical spiral pattern (think threaded hole). I am trying to calculate the necessary power it would take to accelerate this mass to a certain speed before the end of the cylinder. The cylinder is stationary and can not move.\n\n## Known Variables:\n\n$\\omega_f$ [radians] = final velocity\n$\\omega_0$ [radians] = starting velocity = 0\nm [kg] = mass of projectile\nr [meters] = radius of projectile\nQ [$\\frac{rev.}{m}$] = thread revolutions per meter\nL [m] = length of cylinder\n\n$\\theta_f$ [rad] = final position = $2 \\pi Q L$\n$\\theta_0$ [rad] = initial position = 0\n\n## Equations:\n\n[Eq. 1] $\\omega_f^2 = \\omega_0^2 + 2\\alpha(\\theta_f-\\theta_0)$\n[Eq. 2] $\\omega_f = \\omega_0 + \\alpha t$\n[Eq. 3] $I_p = m r^2$\n[Eq. 4] $T = I \\alpha$\n[Eq. 5] $P = T \\omega_f$\n[Eq. 6] $E_\\textrm{torque} = T \\Delta\\theta = T \\theta_f$\n[Eq. 7] $E_\\textrm{power} = P t$\n[Eq. 8] $E_\\textrm{inertia} = \\frac12 I \\omega_f^2$\n\n## Attempt and Problem:\n\nGiven that I know $\\omega_f$ and $\\theta_f$, and initial values are all zero, I can rearrange Eq. 1 and calculate $\\alpha$: $$\\alpha=\\frac{\\omega_f^2}{2 \\theta_f}$$ Now that I know $\\alpha$ and I already knew m and r I can calculate the $T$: $$T = I \\alpha = \\left(mr^2\\right)\\left(\\frac{\\omega_f^2}{2 \\theta_f}\\right) = \\frac{m r^2 \\omega_f^2}{2 \\theta_f}$$ Now I have torque. This is where things get confusing for me. If I calculate energy directly using Eq. 6 and Eq. 8 I get the same answer, but if I calculate Power directly using Eq. 5 and then energy using Eq. 7 I get a different answer from Eq. 6 and Eq. 8.\nMethod Using Eq. 6: $$E_\\textrm{torque} = \\left(\\frac{m r^2 \\omega_f^2}{2 \\theta_f}\\right)\\left(\\theta_f\\right) = \\frac{m r^2 \\omega_f^2}{2}$$ Method Using Eq. 8: $$E_\\textrm{inertia} = \\frac12\\left(mr^2\\right)\\left(\\omega_f^2\\right) = \\frac{m r^2 \\omega_f^2}{2}$$ Method Using Eq. 2, 5 and 7: $$t = \\frac{\\omega_f}{\\alpha}$$ $$P = T \\omega_f = \\left(\\frac{m r^2 \\omega_f^2}{2 \\theta_f}\\right)(\\omega_f) = \\frac{m r^2 \\omega_f^3}{2 \\theta_f}$$ $$E_\\textrm{power} = P t = \\left(\\frac{m r^2 \\omega_f^3}{2 \\theta_f}\\right)\\left(\\frac{\\omega_f}{\\alpha}\\right) = \\left(\\frac{m r^2 \\omega_f^3}{2 \\theta_f}\\right)\\left(\\frac{2 \\theta_f}{\\omega_f}\\right) = \\frac{m r^2 \\omega_f^2}{1}$$\n\n## Question:\n\nWhy does $E_\\textrm{power}$ not equal the other two energy calculations and where did I go wrong? Ultimately I need the power, but I don't trust my power value in this calculation because it gives the wrong final energy value.\nI hope everything was clear if not I will gladly attempt to explain anything further.\n\n\u2022 The power is not constant (assuming constant torques and increasing angular speed). Multiplying the final power by the time will not give you the total kinetic energy. You eq 7 is not right. 6 and 8 give the same result. \u2013\u00a0nasu Oct 26 '16 at 20:09\n\u2022 1. Is this mass accelerating due to gravity? 2. Is it a point mass? 3. Is $r$ the radius of mass or the radius of ths cylinder? 4. Do you need to find the average power over the whole journey? \u2013\u00a0Farcher Oct 26 '16 at 20:33\n\u2022 @nasu Thanks I had forgotten that. What would be the proper way then to calculate the peak power required to accelerate the point mass to the final speed? Farcher, Assume no gravity in this scenario, I'm trying to calculate peak power required to accelerate the point mass up final speed. r in this case is the radius of the cylinder. \u2013\u00a0Wired365 Oct 27 '16 at 13:33\n\u2022 @nasu wait I fully understand it now. The average power is found by dividing the energy by time. The peak power is found by multiplying torque times max speed, as I did in Eq. 5. Then finally in order to get the energy to match up for all three equations I would have had to integrate the changing power over time. \u2013\u00a0Wired365 Oct 27 '16 at 13:50\n\u2022 @nasu could you make your comment and answer so I can give you credit for answering my question? \u2013\u00a0Wired365 Oct 27 '16 at 16:45\n\nAs explained by @nasu, the discrepancy between the results arises because in calculating $P=T\\omega_f$ you have used final angular velocity $\\omega_f$ instead of average angular velocity $\\frac12 \\omega_f$. This is similar to calculating distance = final velocity x time, instead of distance = average velocity x time. If you include a factor of $\\frac12$ this will make $E_{power}$ the same as $E_{inertia}$.\n\nI think you are also missing the fact that the particle has translational (axial) KE as well as rotational (circular) KE. As well as rotating around the cylinder axis it also moves along the axis.\n\nNo work is done in constraining the particle to move in a circle, so the problem is equivalent to linear motion, which is much easier to handle. This avoids the complication of splitting the motion into rotational and axial components. Assuming there is no friction, all of the energy supplied is transformed into translational kinetic energy along the helical curve.\n\nIn the equivalent linear case, the particle is accelerated from rest up to speed $v$ in a straight line over distance $s$ which is the distance around the helix.\n\nThe acceleration $a$ is given by $v^2=2as$. The force accelerating the particle is $F=ma$. The instantaneous power delivered is $P=Fv=mav=m\\frac{v^3}{2s}$. As noted already, the power delivered is not constant, but increases linearly, because $a$ is constant while $v$ increases linearly. Peak power is the final power $P=mav$. Average power is $\\frac12mav$.\n\nThe only problem remaining is to relate the curvi-linear variables $s,v$ (ie along the helix) to rotational variables $L, Q, \\theta, \\omega$. I doubt whether this is worthwhile, because it makes the formulae unnecessarily complicated. It depends what variables you can or must measure.\n\nWhen the particle has made one revolution it has moved forward a distance $1/Q$ along the axis and $2\\pi R$ around the circumference of a circle of radius $R$, so the pitch angle $\\phi$ is given by $\\tan\\phi=\\frac{1}{2\\pi RQ}$. When the distance moved along the helix is $s$, the axial distance is $L=s\\sin\\phi$; the number of revolutions is $LQ$ and the distance moved around a circle is $s\\cos\\phi=2\\pi RLQ=R\\theta$ where $\\theta=2\\pi LQ$$is the final angular position. When the particle has reached speed v=\\dot s along the helix, the angular velocity (measured around a plane circle, perpendicular to the axis) is \\omega=\\frac{v\\cos\\phi}{R}. Substitute into the eqn for power in the linear case :$$P=m(\\frac{R\\omega}{\\cos\\phi})^3 \\frac{\\cos\\phi}{2R\\theta}=m(\\frac{R}{\\cos\\phi})^2 \\frac{\\omega^3}{2\\theta}$$\u2022 Sam, the power equation that you list at the end of your explanation and derivation. That would be the speed dependent total power when combining both the transnational and rotational energies correct? If I wanted to get peak power required to reach my desired speed I would use the final rotational velocity and if I wanted the average power then I would use$\\frac{1}{2} \\omega$? \u2013 Wired365 Nov 1 '16 at 14:20 \u2022 Also could you clarify your equation for relating pitch angle to Q. Is it$\\frac{1}{2\\pi}RQ\\$ or something else? \u2013\u00a0Wired365 Nov 1 '16 at 14:27\n\u2022 @Wired365 : I have revised my answer in the light of your comments. There was an error in relating pitch angle to Q. \u2013\u00a0sammy gerbil Nov 2 '16 at 1:06", "date": "2019-06-16 14:47:58", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/289032/rotational-power-energy-mismatch", "openwebmath_score": 0.8474099636077881, "openwebmath_perplexity": 414.68855878617126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137900379085, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.866556342681079}}
{"url": "https://math.stackexchange.com/questions/2647376/finding-int-28xfxdx-given-int-28fxdx/2647381", "text": "# Finding $\\int_{-2}^8xf(x)dx$ given $\\int_{-2}^8f(x)dx$\n\nI have a continuous function $f:[-2,8]\\rightarrow\\mathbb{R}$ for which is true that $f(6-x)=f(x)\\forall x\\in[-2,8]$. Let: $$\\int_{-2}^8f(x)dx=10$$ Now, I want to find the: $$\\int_{-2}^8xf(x)dx$$ I am thinking of using both the methods of u-substitution and integration by parts, but I need some help. Any ideas?\n\nUsing the substitution $w=6-x$, we obtain\n\n\\begin{aligned} \\int_{-2}^8xf(x)dx&=\\int_{-2}^8(6-(6-x))f(6-(6-x))dx\\\\\\\\ &=-\\int_{8}^{-2}(6-w)f(6-w)dw\\\\\\\\ &=\\int_{-2}^{8}(6-w)f(6-w)dw\\\\\\\\ &=\\int_{-2}^{8}(6-w)f(w)dw\\\\\\\\ &=6\\int_{-2}^{8}f(w)dw-\\int_{-2}^{8}wf(w)dw\\\\\\\\ &=60-\\int_{-2}^{8}xf(x)dx \\end{aligned} and thus $$2\\int_{-2}^{8}xf(x)dx=60$$ i.e. $$\\int_{-2}^{8}xf(x)dx=30.$$\n\nHere's a cute trick.\n\nIf the problem is well-posed, then the solution must be independent of $f$. Therefore, you can take $$f(x)\\equiv1$$ which is consistent with the hypotheses, and calculate $$\\int_{-2}^8x\\ \\mathrm dx\\equiv 30$$\n\nEasy peasy!\n\n\u2022 That is so cool! I have a question: how do you know exactly if the answer is independent of $f(x)$? If the problem is well-posed seems like a very vague metric to me. Thanks! :) \u2013\u00a0Gaurang Tandon Feb 12 '18 at 15:33\n\u2022 @GaurangTandon Well, the problem may be ill-posed (say, because of a typo, or because your professor is evil). The method here is convenient as a cross-check, or in a multiple choice test. Otherwise you'll have to work harder (as in the other answers). I just wanted to show a trick that will produce the \"correct answer\" as easily as possible, provided an answer exists at all. \u2013\u00a0AccidentalFourierTransform Feb 12 '18 at 15:35\n\u2022 I had to upvote this, as actually, realizing this shows you understand how mathematics work :-) \u2013\u00a0yo' Feb 12 '18 at 15:42\n\nYou're given: $$I=\\int_{-2}^8xf(x)dx$$\n\nUse the $a+b-x$ property on this definite integral to get:\n\n\\begin{align} I&=\\int_{-2}^8 (6-x)\\cdot f(6-x)dx \\\\ &=\\int_{-2}^8 (6-x)\\cdot f(x)dx \\tag{\\because f(6-x)=f(x) given} \\\\ &=6\\int_{-2}^8f(x)-I \\end{align}\n\nand you can solve it from here.\n\n\u2022 More surprises from math.meta.stackexchange.com/q/370/290189 \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Feb 12 '18 at 13:25\n\u2022 My favorite is tooltip text, since sometimes you need to put explations on long lines. The font and color tables can be a quick reference. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Feb 12 '18 at 13:35\n\u2022 @GaurangTandon As you ask about formatting, I would never use \\because and \\therefore in textual proofs. They are intended for automated proofs and proof verification, not for human-readable texts. This would ultimately solve your formatting issue :-) \u2013\u00a0yo' Feb 12 '18 at 15:30\n\u2022 @GaurangTandon Well, \\because is the thing you were using at high school, but this usage itself is wrong. This over-symbolism make things cluttered and difficult to read. I would omit the parenthesis completely, and write after the displayed equation something like: \"where we used the hypothesis $f(6-x)=f(x)$.\" \u2013\u00a0yo' Feb 12 '18 at 15:55\n\u2022 @GaurangTandon Yes, more verbose is correct of course. Mathematics is dense enough on itself already. \u2013\u00a0yo' Feb 12 '18 at 16:36\n\nNote the following formula we always have, $$\\color{red}{\\int_a^bg(x)dx= \\int_a^bg(a+b-x)dx}$$\n\nThen with $a=-2,~b=8$ and given that $f(x) = f(6-x)$ we get $$I= \\int_{-2}^8 xf(x)dx= \\int_{-2}^8 (6-x)f(6-x)dx=6\\int_{-2}^8 f(x)dx-\\int_{-2}^8 xf(x)dx\\\\=60-I$$\n\nhence solving for I we obtain, $$I=\\int_{-2}^8 xf(x)dx=30$$\n\n$$I:=\\int_{-2}^8 x f(x)dx=-\\int_8^{-2} (6-x) f(6-x)dx=\\int_{-2}^8 (6-x) f(6-x)dx$$ so that\n\n$$I+I=\\int_{-2}^8 (x+6-x)f(6-x)dx=6\\int_{-2}^8f(x)dx.$$", "date": "2020-08-12 21:38:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2647376/finding-int-28xfxdx-given-int-28fxdx/2647381", "openwebmath_score": 0.9537436366081238, "openwebmath_perplexity": 1152.5228642653983, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137895115187, "lm_q2_score": 0.8824278680004706, "lm_q1q2_score": 0.8665563346257124}}
{"url": "https://mathhelpboards.com/threads/homogeneous-linear-first-order-ordinary-differential-equation-mistake.8442/", "text": "Homogeneous, linear, first-order, ordinary differential equation mistake\n\nkalish\n\nMember\nI am trying to solve a homogeneous, first-order, linear, ordinary differential equation but am running into what I am sure is the wrong answer. However I can't identify what is wrong with my working?!\n\n$$\\frac{dy}{dx}=\\frac{-x+y}{x+y}=\\frac{1-\\frac{x}{y}}{1+\\frac{x}{y}}.$$ Let $z=x/y$, so that $y=x/z$ and $$\\frac{dy}{dx}=\\frac{z-x\\frac{dz}{dx}}{z^2}=\\frac{1-z}{1+z}.$$ Then $z-x\\frac{dz}{dx}=\\frac{z^2-z^3}{1+z} \\implies -x\\frac{dz}{dx}=-\\frac{z^3+z}{z+1} \\implies \\frac{dx}{x}=\\frac{dz}{z^2+1} \\implies \\arctan(z)=\\ln(|x|)+C \\implies z=\\tan(\\ln(|x|)+C)$\n\nThus $$y = \\frac{x}{\\tan(\\ln(|x|)+C)}$$.\n\nHowever, my book: *A Modern Introduction to Differential Equations 2nd edition by Henry Ricardo* says \"this first-order equation is homogeneous and can be solved implicitly\".\n\nThey pursued the method of letting $z=y/x$ and obtained the solution as\n$$2\\arctan(y/x)+\\ln(x^2+y^2)-C=0.$$\n\nWhy are the two different substitutions, which should both be suitable, giving me two different answers?\n\nZaidAlyafey\n\nWell-known member\nMHB Math Helper\nRe: Homogenous, linear, first-order, ordinary differential equation mistake\n\nThey pursued the method of letting $z=y/x$ and obtained the solution as\nUsing this substitution is better because differentiation becomes easier\n\n$$\\displaystyle y=zx \\,\\,\\to \\,\\, \\frac{dy}{dx}= z+x \\frac{dz}{dx}$$\n\nMarkFL\n\nStaff member\nRe: Homogenous, linear, first-order, ordinary differential equation mistake\n\nI am with you up to this point (where I have negated both sides):\n\n$$\\displaystyle x\\frac{dz}{dx}=\\frac{z^3+z}{z+1}$$\n\nHowever, when you separated variables, you should get:\n\n$$\\displaystyle \\frac{z+1}{z^3+z}\\,dz=\\frac{1}{x}\\,dx$$\n\nUsing partial fractions, we may write:\n\n$$\\displaystyle \\left(\\frac{1}{z^2+1}-\\frac{z}{z^2+1}+\\frac{1}{z} \\right)\\,dz=\\frac{1}{x}\\,dx$$\n\nThis will lead you to a solution that differs from that given by your textbook by only a constant, which can then be \"absorbed\" by the constant of integration if you desire to get it in the same form.", "date": "2021-10-19 13:09:11", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/homogeneous-linear-first-order-ordinary-differential-equation-mistake.8442/", "openwebmath_score": 0.881708562374115, "openwebmath_perplexity": 642.77967215736, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137926698566, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8665563283036785}}
{"url": "https://mathhelpboards.com/threads/permutation-and-combinations.39/", "text": "Permutation and combinations\n\nPunch\n\nNew member\nFour married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together.\n\nI need to know the logic and thinking process behind how the answer is derived.\n\nWhat I tried is:\nFirst person has 10 seats to choose, second person 8 seats to choose and so on. Each couple can then seat on different sides.\n\n10(8)(6)(4)(2^5)=61440\n\nAnother way of thinking I have is this: Consider each couple and the 2 children as each individual groups.\n\nTotal number of ways of arranging the 5 groups in a round table is (5-1)!=24\n\nThen permutate each couple and children=2^5\n\nso total number of ways = (2^5)24=768\n\nLast edited:\n\nThePerfectHacker\n\nWell-known member\nHello,\n\nImagine a round table with ten positions open. Where can the first kid be seated? Anywhere, he can sit anywhere he pleases to. How many choices does he have? 10.\n\nNow where can the second kid be seated? Draw drawining some pictures here, but you should realize that the second kid cannot be sitting 1 seat apart from the first kid. Because there is no room for couples to sit together there! Also, the second kid cannot be sitting 3 seats apart from the first kid for the same reason. You should see that the second kid can only be sitting an even number of seats away from the first kid. Thus, the second kid has only 5 choices.\n\nStart with the first kid. Move over to the next avaliable seat (in a clockwise manner). How many people can sit there? There are 8 remaining people and so there are 8 choices avaliable. But in the next avaliable seat who can sit there? It must be the spouse, which means there is only 1 choice, he/she is forced into that seat.\n\nMove to the next avaliable seat. How many people can sit there? Now there are 6 people remaining and so there are 6 choices avaliable. But in the next avaliable seat it must be the spouse, so he/she is forced into that seat.\n\nMove to the next avaliable set. Same reasoning tells us that there are 4 people remaining, and so 4 choices.\n\nFinally with the last two seats remaining next to eachother there is just 2 ways to seat those couples together.\n\nThus, we get 10*5*8*6*4*2 = 19200.\n\nSo how do you get an answer of 1920? I guess it is because in your problem no person is designated as the \"head\" of the table, i.e. a rotation of all people one seat over is considered to be the same seating arrangement. As there are 10 rotations in seating the answer without any head of table needs to be divided by 10. That is how they got 1920.\n\nPlato\n\nWell-known member\nMHB Math Helper\nFour married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together. I need to know the logic and thinking process behind how the answer is derived.\nThere are always multiple ways to do these. Here is another.\nThere are six units: four couples and two children.\nSeat any couple at the table together. It is now an ordered table.\nThere are $5!$ ways to seat the remaining five units.\nBut each couple can be seated in two ways.\nThus $5!\\cdot 2^4=1920$.", "date": "2020-12-01 06:22:04", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/permutation-and-combinations.39/", "openwebmath_score": 0.6199005842208862, "openwebmath_perplexity": 663.5866908787941, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9820137926698566, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8665563283036785}}
{"url": "http://tfzg.jf-huenstetten.de/convergent-sequence-examples-pdf.html", "text": "# Convergent Sequence Examples Pdf\n\nPointwise convergence is a very weak kind of convergence. 10 Procedure for Estimating Adjusted Net Saving 61 2. We will see two. We say that the sequence n D U converges to zero in D U if. , there is some c so that, for all k, kx kk c. The polarization identity expresses the norm of an inner product space in terms of the inner product. 1 Introduction 23 2. -Mix two forms of data in different ways. For example, 10 + 20 + 20\u2026does not converge (it just keeps on getting bigger). It is nearly identical to existing sample sequences. Since the product of two convergent sequences is convergent the sequence fa2. The relationships between different types of convergence are summarized in Figure 4. 1 Weak convergence in normed spaces We recall that the notion of convergence on a normed space X, which we used so far, is the convergence with respect to the norm on X: namely, for a sequence (x n) n 1, we say that x n!xif kx n xk!0 as n!1. 1 For the geometric sequence with r ~ 0; i. Convergent sequences, Divergent sequences, Sequences with limit, sequences without limit, Oscillating sequences. examples below, these are differences that should make a difference in the planning and management activities of any crisis relevant groups. , \u03bb*) corresponding to \u03bb*. Convergent Sequences Subsequences Cauchy Sequences Examples Notice that our de nition of convergent depends not only on fp ng but also on X. 5, c) 9, -3, 1, 5. Relative to convergence, it is the behavior in the large-n limit that matters. and Xis a r. A rather complete treatment of these and related problems was. ( ) 0 0 (4. for all sequences (x n) in X. The proof can be found in a number of texts, for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl; New York: Dover, 1956). Hence, we have, which implies. A sequence is bounded if it is bounded above and below, that is to say, if there is a number, k, less than or equal to all the terms of sequence and another number, K', greater than or equal to all the terms of the sequence. Transition Kernel of a Reversible Markov. Sample Quizzes with Answers Search by content rather than week number. Now we will investigate what may happen when we add all terms of a sequence together to form what will be called an infinite series. Ratio Test. A sequence in R is a list or ordered set: (a 1, a 2, a 3,. If the limit of s k is infinite or does not exist, the series is said to diverge. A sequence { } is Cauchy if, for every ,there exists an such that ( ) for every Thus, a Cauchy sequence is one such that its elements become arbitrarily 'close together' as we move down the sequence. The Riemann Integral and the Mean Value Theorem for Integrals 4 6. A Cauchy sequence (pronounced CO-she) is an infinite sequence that converges in a particular way. De nition 0. The limit of a sequence is said to be the fundamental notion on which the whole of analysis ultimately rests. Nested intervals. ANALYSIS I 7 Monotone Sequences Example. For example, the divergent sequence of partial sums of the harmonic series (see this earlier example) does satisfy this property, but not the condition for a Cauchy sequence. Conversely, it follows from Theorem 1. In Rk, every Cauchy sequence converges. We\u2019ll look at this one in a moment. The language of this test emphasizes an important point: the convergence or divergence of a series depends entirely upon what happens for large n. An approximation theory for sequences of this kind has recently been developed, with the aim of providing tools for computing their asymptotic singular value and eigenvalue distributions. The sequences are progressive (hierarchical): any prefix is well distributed, making them suitable for incremental rendering and adaptive sampling. For example, 1 + x+ x2 + + xn+ is a power series. A sequence has the Cauchy property if and only if it is convergent. This says that if the series eventually behaves like a convergent (divergent) geometric series, it converges (diverges). Convergence generally means coming together, while divergence generally means moving apart. It also explores particular types of sequence known as arithmetic progressions (APs) and geometric progressions (GPs), and the corresponding series. 2) The sequence can approach one of the two infinities. Nets Take a moment to verify to yourself that the use of the word \\tail\" in this context agrees with its use in the context of sequences, and that in the case where D= N with its usual ordering, this agrees with the usual de nition of sequence convergence. a sequence does not have to converge to a given \ufb01xed point (unless a0 is already equal to the \ufb01xed point). Let f: D \u2192 C be a function. (In fact, the only books. Show that (X,d) in Example 4 is a metric. We will now look at two very important terms when it comes to categorizing sequences. Calculus III: Sequences and Series Notes (Rigorous Version) Logic De nition (Proposition) A proposition is a statement which is either true or false. Also in different example, you learn to generate the Fibonacci sequence up to a certain number. Note that each x n is an irrational number (i. The ratio of successive pairs of numbers in this sequence converges on 1. A definition is given of convergence of a sequence of sets to a set, written X\u201e -> X, where X and the Xn are subsets of Euclidean m-space \u00a3\"'. Meaning 'the sum of all terms like', sigma notation is a convenient way to show where a series begins and ends. I think we must be getting close to some calculus. For one thing, it is common for the sum to be a relatively arbitrary irrational number:. Determine if the sequence converges or diverges. The sequence xn converges to something if and only if this holds: for every >0 there. its limit doesn't exist or is plus or minus infinity) then the series is also called divergent. 2 Limit Laws The theorems below are useful when -nding the limit of a sequence. If the limit of s k is infinite or does not exist, the series is said to diverge. Stayton (2008) demonstrated that rates of convergence can be. Take a neighborhood U of x. We know that a n!q. This is a collection of lecture notes I\u2019ve used several times in the two-semester senior/graduate-level real analysis course at the University of Louisville. Show that ($\\sum \\frac{\\ln(n)}{n}$) diverges. Divergent Sequences. Mixed Methods Research \u2022Characteristics of mixed methods research -Collect and analyze both quantitative and qualitative data. The sum of convergent and divergent series Kyle Miller Wednesday, 2 September 2015 Theorem 8 in section 11. For example, take any three numbers and sum them to make a fourth, then continue summing the last three numbers in the sequence to make the next. The Coupon Collector\u2019s Problem 13 2. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. I Integral test, direct comparison and limit comparison tests,. A convergent sequence has a limit \u2014 that is, it approaches a real number. If (an)1=n \u201a 1 for all su-ciently large n, then P n an is divergent. Properties of the sample autocovariance function The sample autocovariance function: \u02c6\u03b3(h) = 1 n nX\u2212|h| t=1 (xt+|h| \u2212x\u00af)(xt \u2212x\u00af), for \u2212n 0 c \ufb01nite & an,bn > 0? Does. This says that if the series eventually behaves like a convergent (divergent) geometric series, it converges (diverges). In Chapter 1 we discussed the limit of sequences that were monotone; this restriction allowed some short-cuts and gave a quick introduction to the concept. A unifying approach to convergence of linear sampling type operators in Orlicz spaces Vinti, Gianluca and Zampogni, Luca, Advances in Differential Equations, 2011; On the Graph Convergence of Sequences of Functions Grande, Zbigniew, Real Analysis Exchange, 2008. Metric spaces are generalizations of the real line, in which some of the theorems that hold for R remain valid. \u00a9v Q2G0U1T6N dKQuKtJaY rS]oBfzt]wuaTrGe] _LpLTCH. 4 Convergence 32 2. Since this test for convergence of a basic-type improper integral makes use of a limit, it's called the limit comparison test , abbreviated as LCT. Solutions to Problems in Chapter 2 2. A Convergence Test for Sequences Thm: lim n!1 \ufb02 \ufb02 \ufb02 \ufb02 an+1 an \ufb02 \ufb02 \ufb02 \ufb02 = L < 1 =) lim n!1 an = 0 In words, this just says that if the absolute value of the ratio of successive terms in a sequence fangn approaches a limit L, and if L < 1, then the sequence itself converges to 0. For example, 1 + x+ x2 + + xn+ is a power series. Introduction One of the most important parts of probability theory concerns the be-havior of sequences of random variables. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. Recall the sequence (x n) de ned inductively by x 1 = 1; x n+1 = (1=2)x n + 1;n2N:. Sequences that are not convergent are said to be divergent. \u2022 If f 0 (r) = 0, the sequence converges at least quadratically to the \ufb01xed point (this is sometimes called superconvergence in the dynamical systems literature). WIJSMAN(i) 0. Absolute and conditional convergence Remarks: I Several convergence tests apply only to positive series. These results are not original, and similar results on the relation between the limits of the series and these two sequences (or related sequences) have appeared in the literature before. Thanks to all of you who support me on Patreon. Subsequences. ROC contains strip lines parallel to j\u03c9 axis in s-plane. Then as n\u2192\u221e, and for x\u2208R F Xn (x) \u2192 (0 x\u22640 1 x>0. \u02c61 + i 2 , 2 + i 22. 3 Complexi\ufb01cation of the Integrand. Some of the earliest and best examples of convergent sequence evolution include the stomach lysozymes of langurs and cows (Stewart et al. A sequence {xn} is in\ufb01nitely large if for any \u03b5 > 0 only a \ufb01nite number of points (n,xn) are between the two horizontal lines y = \u2212\u03b5, and y = +\u03b5. p This integral converges for all p > 0, so the series converges for all p > 0. , After measuring, we choose a set of parameters i and build our. Ratio Test. Before we discuss the idea behind successive approximations, let\u2019s \ufb01rst express a \ufb01rst- order IVP as an integral equation. A series which is larger than a convergent series might converge or diverge. Let (a n) be the sequence de ned by a n= 1 1 n; n 1: Evaluate limsup n!1 a nand liminf. For a Cauchy sequence, the terms get \"closer together\" the \"farther out\" you go in the sequence. This is a set of exercises and problems for a (more or less) standard beginning calculus sequence. Sequences of Functions We now explore two notions of what it means for a sequence of functions ff ng n2N to converge to a function f. Assume that lim n!1 an exists for anC1 D p 3an with a0 D2: Find lim n!1 an. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. is convergent. It's important to understand what is meant by convergence of series be\u00ad fore getting to numerical analysis proper. Cauchy Sequences \u21d4 Convergent Sequences A sequence of real numbers is said to be Cauchy if , limit X( ) ( ) 0 \u2192\u221e \u2212=X nm nm. >1/-normed space. 4 and Example 3. \u2022 If f 0 (r) = 0, the sequence converges at least quadratically to the \ufb01xed point (this is sometimes called superconvergence in the dynamical systems literature). The meanings of the terms \u201cconvergence\u201d and \u201cthe limit of a sequence\u201d. Example: A convergent sequence in a metric space is bounded; therefore the set of convergent real sequences is a subset of '1. convergence synonyms, convergence pronunciation, convergence translation, English dictionary definition of convergence. Series \u2022Given a sequence {a 0, a 1, a2,\u2026, a n} \u2022The sum of the series, S n = \u2022A series is convergent if, as n gets larger and larger, S n goes to some finite number. 2 Sequences: Convergence and Divergence In Section 2. In the world of finance and trading, convergence and divergence are terms used to describe the. The relationships between different types of convergence are summarized in Figure 4. Let be a convergent series of real nonnegative terms. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. Cauchy Sequences and Complete Metric Spaces Let's rst consider two examples of convergent sequences in R: Example 1: Let x n = 1 n p 2 for each n2N. The proof can be found in a number of texts, for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl; New York: Dover, 1956). 8 Order Properties of Limits 47 2. Summary of Convergence estsT for Series estT Series Convergence or Divergence Comments n th term test (or the zero test) X a n Diverges if lim n !1 a n 6= 0 Inconclusive if lim a n = 0. Using models developed by Garcia. Let (a n) be the sequence de ned by a n= 1 1 n; n 1: Evaluate limsup n!1 a nand liminf. Convergent,Divergent & Oacillatory Sequences with examples - Lesson 2-In Hindi-{Infinite Sequences} - Duration: 53:27. Convergence of Random Variables 5. \u2022 Answer all questions. 2 Convergence Index 7. A sequence can be defined by a formula (or generator) which generates each term. We note that absolute convergence of an in\ufb01nite series is necessary and suf\ufb01cient to allow the terms of a series to be. Denition 7. Nets and lters (are better than sequences) 3. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. Alternating series and absolute convergence (Sect. 2 Tests for Convergence Let us determine the convergence or the divergence of a series by comparing it to one whose behavior is already known. Let us refer to these metrics as d 1 and d 2 respectively, and suppose that the sequence (x k) converges to in the 6. (1) is pointwise convergent over the interval x \u2208 A. If this limit is one , the test is inconclusive and a different test is required. Whereas, in this case the output of the experiment is a random sequence, i. A complete normed linear space is called a Banach space. n) is convergent to 1 and the subsequence (a 2n 1) of (a n) is convergent to 1: Later, we will prove that in general, the limit supremum and the limit in mum of a bounded sequence are always the limits of some subsequences of the given sequence. 6 Boundedness Properties of Limits 39 2. sequence are increasing. Thus, fx ngconverges in R (i. striatum and C. Let {fn}\u221e n=1 be a sequence of real or complex-valued functions de\ufb01ned on a domain D. Algebraic manipulations give, since. These notes are sef-contained, but two good extra references for this chapter are Tao, Analysis I; and Dahlquist. Thus the space is not sequentially compact and by Lemma 3 it is not compact, a contradiction to our hypothesis. pdf from MATH 2 at Wuhan University of Technology. We know when a geometric series converges and what it converges to. 1 The pattern may for instance be that: there is a convergence of X. the merging of distinct technologies, industries, or devices into a unified whole n. Series of Numbers 4. Almost sure convergence, convergence in probability and asymptotic normality In the previous chapter we considered estimator of several di\ufb00erent parameters. The definition of convergence of a sequence was given in Section 11. Now we will investigate what may happen when we add all terms of a sequence together to form what will be called an infinite series. Some are quite easy to understand: If r = 1 the sequence converges to 1 since every term is 1, and likewise if r = 0 the sequence converges to 0. Coupling Constructions and Convergence of Markov Chains 10 2. 10 Examples of Limits 56 2. A series \u2211a n is said to converge or to be convergent when the sequence (s k) of partial sums has a finite limit Examples of convergent sequences. Definition of Convergence and Divergence in Series The n th partial sum of the series a n is given by S n = a 1 + a 2 + a 3 + + a n. One possibility is \u02c6 ( 1)n 1 n \u02d9 +1 n=1 = 1; 1 2; 1 3; 1 4;:::, which converges to 0 but is not monotonic. is convergent. 1) occur in applications ([7, 8, and 18], for example), and it can be shown that all slowly convergent sequences occurring as examples in the references of the present paper satisfy (1. , to an element of R). The Cauchy criterion for uniform convergence of a series gives a condition for the uniform convergence of the series (1) on without using the sum of the series. 11 Subsequences 78 2. Universal nets 12 4. For real inner product spaces it is (x,y) = 1 4 (kx+ yk2 \u2212kx\u2212yk2). Convergence and Divergence of Sequences. In this post, we will focus on examples of. 12 Adjusted Net Saving, by Region, 1995\u20132015 63. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. Sequences and series of functions: uniform convergence Pointwise and uniform convergence We have said a good deal about sequences of numbers. It is also possible to prove that a convergent sequence has a unique limit, i. Can you find an example ? While we now know how to deal with convergent sequences, we still need an easy criteria that will tell us whether a sequence converges. CONVERGENCE PETE L. the merging of distinct technologies, industries, or devices into a unified whole n. Therefore, {fn} converges pointwise to the function f = 0 on R. We note that absolute convergence of an in\ufb01nite series is necessary and suf\ufb01cient to allow the terms of a series to be. Order and Rates of Convergence 1 Order of convergence 11 Suppose we have that Then the convergence of the sequence x k to \u00afx is said. ter estimates as well as the Potential Scale Reduction (PSR) convergence criteria, which compares several independent MCMC sequences. 1 n, 2 3 n are examples of null sequences since lim n = 0 and lim 2 3 n = 0. Exercises 15 2. You should be able to verify that the set is actually a vector subspace of \u20181. a sequence of xed numbers. F-convergence, lters and nets The main purpose of these notes is to compare several notions that describe convergence in topological spaces. De\ufb01nition: A sequence {v k} of vectors in a normed linear space V is Cauchy conver-gent if kv m \u2212 v nk \u2192 0 as m,n \u2192 \u221e. Show that weak* convergent sequences in the dual of a Banach space are bounded. Neal, WKU MATH 532 Sequences of Functions Throughout, let (X ,F, ) be a measure space and let {fn}n=1 \u221e be a sequence of real- valued functions defined on X. Fatou's lemma and the dominated convergence theorem are other theorems in this vein,. Find the radius of convergence R and the domain of convergence S for each of the following power series: X\u221e n=0 xn, X\u221e n=1 x n n, X\u221e n=0 x nn, X\u221e n=0 nnxn, X\u221e n=0 x n!, X\u221e n=0 (\u22121)n n2 x2n Hwk problem: if the series P \u221e k=0 4 na n is convergent, then P \u221e n=0 a n(\u22122) n is also con-vergent. In Rk, every Cauchy sequence converges. A convergent sequence has a limit \u2014 that is, it approaches a real number. For example, once we have computed from the first equation, its value is then used in the second equation to obtain the new. Thus the space is not sequentially compact and by Lemma 3 it is not compact, a contradiction to our hypothesis. 9 Monotone Convergence Criterion 52 2. Let \u2020 > 0. Before introducing almost sure convergence let us look at an example. Quadratic Convergence of Newton\u2019s Method Michael Overton, Numerical Computing, Spring 2017 The quadratic convergence rate of Newton\u2019s Method is not given in A&G, except as Exercise 3. Such a sequence does not exist, indeed, if it had a convergent subsequence a m k. Two examples of nets in analysis 11 3. Theorem 317 Let (a n. Similarities among protein sequences are reminiscent of homology and convergent evolution via common ancestry and/or selective pressure, respectively. , x n 2Qc) and that fx ngconverges to 0. Discuss the pointwise convergence of the sequence. There are three main results: the rst one is that uniform convergence of a sequence of continuous. For positive term series, convergence of the sequence of partial sums is simple. 9 Uniform Convergence of Sequences of Functions In this chapter we consider sequences and. weakly convergent and weak* convergent sequences are likewise bounded. Of these, 10 have two heads and three tails. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. A sequence is \"converging\" if its terms approach a specific value as we progress through them to infinity. an e\u2013cient way and will lead us to criteria for the convergence of rearrangements. You da real mvps! \\$1 per month helps!! :) https://www. Prove that if ff ngconverges uniformly to f on X, then f is bounded. 1 n, 2 3 n are examples of null sequences since lim n = 0 and lim 2 3 n = 0. The almost sure convergence of Zn to Z means that there is an event N such that P(N) = 0 and for every element w 2Nc, limn!\u00a5Zn(w) = Z(w), which is almost the same as point-wise convergence for deterministic functions (Example 5. Convergent definition is - tending to move toward one point or to approach each other : converging. Convergent Sequences Subsequences Cauchy Sequences Examples Notice that our de nition of convergent depends not only on fp ng but also on X. Sequences and series of functions: uniform convergence Pointwise and uniform convergence We have said a good deal about sequences of numbers. In the Algebra world, mathematical induction is the first one you usually learn because it's just a set list of steps you work through. Hint: The dual space of c00 under the \u2113\u221e norm is (c00)\u2217 \u223c= \u21131. Scalable Convex Multiple Sequence Alignment via Entropy-Regularized Dual Decomposition. Answer: We will use the Ratio-Test (try to use the Root-Test to see how difficult it is). Then for any integer n there is an x n in S such that |x n| > n. We have shown above that the sequence {f n} \u221e n=1 converges pointwise. Alternating sequences change the signs of its terms. The range variation of \u03c3 for which the Laplace transform converges is called region of convergence. Let us \ufb01rst make precise what we mean by \"linear. 5 Divergence 37 2. Thus convergent sequences do not distinguish between the compact topology of \u03b2D and the discrete topology on its underlying set. However, it has huge computational complexity, which is square of that of the. p This integral converges for all p > 0, so the series converges for all p > 0. Convergence and Divergence Lecture Notes It is not always possible to determine the sum of a series exactly. EXAMPLE 2 EXAMPLE 1 common difference arithmetic sequence, GOAL 1 Write rules for arithmetic sequences and find sums of arithmetic series. Solution First, it is easy to see the pointwise limit function is x(t) = 0 on [0;1]. Stayton (2008) demonstrated that rates of convergence can be. Proposition 2. Certainly, uniform convergence implies pointwise convergence, but the converse is false (as we have seen), so that uniform convergence is a stronger \\type\" of convergence than pointwise convergence. View Notes - Notes-3-2019-version2. Cases of convergent evolution \u2014 where different lineages have evolved similar traits independently \u2014 are common and have proven central to our understanding of selection. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence. 1 De\ufb01nition of limit. We have the following useful test for checking the uniform convergence of (fn) when its pointwise limit is known. Properties of the sample autocovariance function The sample autocovariance function: \u02c6\u03b3(h) = 1 n nX\u2212|h| t=1 (xt+|h| \u2212x\u00af)(xt \u2212x\u00af), for \u2212n 0 c \ufb01nite & an,bn > 0? Does. 2 More de\ufb01nitions and terms 1. Convergent sequences in topological spaces 1. f (x) = (1 1 x x. Roughly speaking, a \"convergence theorem\" states that integrability is preserved under taking limits. Initial values of the nchains = 4 sequences are indicated by solid squares. Pick \u03f5 = 1 and N1 the corresponding rank. Therefore, with the L 2-norm of Eq. Universal nets 12 4. Weaklawoflargenumbers. (Enough to quote previous homework problem. The hope is that as the sample size increases the estimator should get 'closer' to the parameter of interest. 2008 issue of Mathematics Magazine [1], the questions of convergence, density, and correspondence of rational numbers that can be written as infinitely nested radicals are explored. Application of du Bois-Reymond\u2019s comparison of. Michael Boardman March 1999 Abstract Convergence criteria for spectral sequences are developed that apply more widely than the traditional concepts. This part of probability is often called \\large sample theory\" or \\limit theory\" or \\asymptotic theory. We can break this problem down into parts and apply the theorem for convergent series to combine each part together. Two examples of nets in analysis 11 3. By this we mean that a function f from IN to some set A is given and f(n) = an \u2208 A for n \u2208 IN. If a complete metric space has a norm defined by an inner product (such as in a Euclidean space), it is called a Hilbert space. Each number in the sequence is the sum of the two numbers that precede it. Likewise, if the sequence of partial sums is a divergent sequence (i. De\ufb01nition: A normed linear space is complete if all Cauchy convergent sequences are convergent. ( ) 0 0 (4. However, there are many di\ufb00erent ways of de\ufb01ning convergence of a sequence of functions. 34 144 12 =12 Note that you may use parenthesis in the usual ways. Determine if the sequence converges or diverges. Convergence In Distribution (Law). rather than selection pressure, and because it is important to distinguish between founder effects and convergent evolution. , and all of them are de ned on the same probability space (;F;P). 1023 = 4092. Show that weakly convergent sequences in a normed space are bounded. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. The class of Cauchy sequences should be viewed as minor generalization of Example 1 as the proof of the following theorem will indicate. Fibonacci sequences occur frequently in nature. Give an example to show that this statement is false if uniform convergence is replaced by pointwise convergence. Now that we have seen some more examples of sequences we can discuss how to look for patterns and figure out given a list, how to find the sequence in question. In our previous lesson, Taylor Series, we learned how to create a Taylor Polynomial (Taylor Series) using our center, which in turn, helps us to generate our radius and interval of convergence, derivatives, and factorials. In other words, if one has a sequence (f n)\u221e =1 of integrable functions, and if f is some. 10) 2 1 ; 3 2 ; 4 3 ; 5 4 ::: We know this converges to 1 and can verify this using the same logic used in the proof under the de nition of convergence showing that 1 n converges to zero. k \u2264 a n \u2264 K'. The sum of convergent and divergent series Kyle Miller Wednesday, 2 September 2015 Theorem 8 in section 11. Definition of Convergence and Divergence in Series The n th partial sum of the series a n is given by S n = a 1 + a 2 + a 3 + + a n. Nets and subnets 7 3. A power series is an infinite series. Convergence in the space of test functions Clearly D U is a linear space of functions but it turns out to be impossible to define a norm on the space. EXAMPLES USING MATHCAD 14 Basic Operations: 22+ =4 Type the = sign to get a result. De\ufb01nition: A sequence f. 3 Convergence of Subsequences of a Convergent Sequence Theorem. Convergence and (Quasi-)Compactness 13 4. convergence failure during the sample period of 2000 \u2013 2011. The importance of the Cauchy property is to characterize a convergent sequence without using the actual value of its limit, but only the relative distance between terms. 635, and the infinite sum is around 1. Introduces the de nition of rate of convergence for sequences and applies this to xed-point root- nding iterative methods. Concludes with the development of a formula to estimate the rate of convergence for these methods when the actual root is not known. Give an example of a convergent sequence that is not a monotone sequence. For example. For one thing, it is common for the sum to be a relatively arbitrary irrational number:. so the series 0. Chapter 5 Sequences and Series of Functions In this chapter, we de\ufb01ne and study the convergence of sequences and series of functions. If the sequence of these partial sums {S n} converges to L, then the sum of the series converges to L. So in a first countable space \u201csequences determine the topology. is an example of a convergent sequence since lim n n+1 = 1. n) is convergent to 1 and the subsequence (a 2n 1) of (a n) is convergent to 1: Later, we will prove that in general, the limit supremum and the limit in mum of a bounded sequence are always the limits of some subsequences of the given sequence. For example, random evolutionary change can cause species to become more similar to each other than were their ancestors. Increasing sequence IS-17 Induction terminology IS-1 Inductive step IS-1 In\ufb01nite sequence see Sequence In\ufb01nite series see Series Integral test for series IS-24 Limit of a sequence IS-13 sum of in\ufb01nite series IS-20 Logarithm, rate of growth of IS-18 Monotone sequence IS-17 Polynomial, rate of growth of IS-18 Powers sum of IS-5 Prime factorization IS-2. Proposition 2. Nair EXAMPLE 1. Get an intuitive sense of what that even means!. Application of du Bois-Reymond\u2019s comparison of. This week, we will see that within a given range of x values the Taylor series converges to the function itself. Recall the sequence (x n) de ned inductively by x 1 = 1; x n+1 = (1=2)x n + 1;n2N:. ( ) 0 0 (4. Likewise, if the sequence of partial sums is a divergent sequence (i.", "date": "2019-12-08 11:31:32", "meta": {"domain": "jf-huenstetten.de", "url": "http://tfzg.jf-huenstetten.de/convergent-sequence-examples-pdf.html", "openwebmath_score": 0.8878056406974792, "openwebmath_perplexity": 560.7047789038782, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9933071502644802, "lm_q2_score": 0.8723473697001441, "lm_q1q2_score": 0.8665088798375652}}
{"url": "https://www.physicsforums.com/threads/determine-the-standard-deviation-of-these-results.948814/", "text": "# Determine the standard deviation of these results\n\nI've managed to work out the below for the sample standard deviation. Any error's that I should be aware of?\n\n## Homework Statement\n\nThe ultimate tensile strength of a material was tested using 10 samples. The results of the tests were as follows.\n\n711 N \u3016mm\u3017^(-2),732 N \u3016mm\u3017^(-2),759 N \u3016mm\u3017^(-2),670 N \u3016mm\u3017^(-2),701 N \u3016mm\u3017^(-2),\n765 N \u3016mm\u3017^(-2),743 N \u3016mm\u3017^(-2),755 N \u3016mm\u3017^(-2),715 N \u3016mm\u3017^(-2),713 N \u3016mm\u3017^(-2).\n\n[/B]\n(a) Determine the mean standard deviation of these results.\n(a) Express the values found in (a) in GPa.\n\n## Homework Equations\n\ns= \u221a(1/(N-1) \u2211_(i=1)^N\u2592(x_i-x \u0305 )^2 )\n\n## The Attempt at a Solution\n\n(a) As we are referring to a 'sample' we apply the below formula for Standard Deviation:\n\ns= \u221a(1/(N-1) \u2211_(i=1)^N\u2592(x_i-x \u0305 )^2 )\n\nCalculating the mean value x \u0305\n\n(711 + 732 +759+670+701 +765 +743+755 +715 +713)/10\n\nx \u0305= 726.4 \u3016N mm\u3017^(-2)\nDetermining the (x_i-x \u0305)^2\n\n(711-726.4 )^2= 237.16\n(732-726.4 )^2= 31.36\n(759-726.4 )^2= 1062.76\n(670-726.4 )^2= 3180.95\n(701-726.4 )^2= 625.16\n(765-726.4 )^2= 1489.96\n(743-726.4 )^2= 275.56\n(755-726.4 )^2= 817.96\n(715-726.4 )^2= 129.96\n(713-726.4 )^2= 179.56\n\nDetermining the sum of the above\n\n\u2211\u2592237.16+31.36 +1062.76 +3180.95 +625.16 +1489.96 +275.56 +817.96 +129.96 +179.56\n\n= 8,030.29 \u3016N mm\u3017^(-2)\n\nDivide by N-1\n\n(1/5)\u22198030.29= 1606.06 N \u3016mm\u3017^(-2)\n\nDetermining the sample standard deviation \u03c3\n\n\u03c3= \u221a((1606.06)=40.08 N \u3016mm\u3017^(-2)=4.008 MPa\n\n(b)4.008 MPa=0.004008 GPa\n\nOr in standard form, 4.008\u221910^(-3) GPa.\n\nHomework Helper\nGold Member\n2020 Award\n##\\sigma=40 ## N/mm^2=40 MPa=.040 GPa if my arithmetic is correct. ## \\\\ ## And you don't need to keep all the extra sig figs on the standard deviation. ## \\\\ ## And I think they want the mean and standard deviation. You didn't give the value of the mean in GPa.\n\nMark44\nMentor\nDetermine the mean standard deviation of these results.\nThere's a mean and there's a standard deviation. I'm not aware of a term called mean standard deviation, but there is a standard error, where you look at the variation of a collection of samples.\n\nFactChecker\nFactChecker\nGold Member\nThere is a standard deviation, ##\\sigma_{\\bar x}##, of the sample mean ##\\bar x##. It is closely related to the confidence interval of the population mean ##\\mu##. $$\\sigma_{\\bar x} = \\sigma/\\sqrt M \\approx s/\\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.\n\nLast edited:\nHomework Helper\nGold Member\n2020 Award\nThere is a standard deviation, ##\\sigma_{\\bar x}##, of the sample mean ##\\bar x##. It is closely related to the confidence interval of the population mean ##\\mu##. $$\\sigma_{\\bar x} = \\sigma/\\sqrt M \\approx s/\\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.\nLet the mean be ## Y ##, so that ## Y=\\bar{X}=\\frac{X_1+X_2+...+X_M}{M} ##. ## \\\\ ## If the random measurements are uncorrelated, ## \\sigma_Y^2=\\frac{\\sigma_{X_1}^2+\\sigma_{X_2}^2+...+\\sigma_{X_M}^2}{M^2}=\\frac{M \\, \\sigma_X^2}{M^2}=\\frac{\\sigma_X^2}{M} ##. ## \\\\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\\\ ## I believe the OP @Tiberious left off the word \"and\" between \"mean\" and \"standard deviation\", and it has created some confusion.\n\nLast edited:\nFactChecker\nFactChecker\nGold Member\nLet the mean be ## Y ##, so that ## Y=\\bar{X}=\\frac{X_1+X_2+...+X_M}{M} ##. ## \\\\ ## If the random measurements are uncorrelated, ## \\sigma_Y^2=\\frac{\\sigma_{X_1}^2+\\sigma_{X_2}^2+...+\\sigma_{X_M}^2}{M^2}=\\frac{M \\, \\sigma_X^2}{M^2}=\\frac{\\sigma_X^2}{M} ##. ## \\\\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\\\ ## I believe the OP @Tiberious left off the word \"and\" between \"mean\" and \"standard deviation\", and it has created some confusion.\nThey are all very standard things to ask for. But I agree that the sample mean and sample standard deviation are the first two things to calculate.\n\nYes - it does seem my OP was missing the word 'and' apologies for the confusion. Just to confirm;\n\nThe answer for standard deviation is 0.040 Gpa ?\n\nSo, would the mean be 0.016Gpa ?\n\nFactChecker\nGold Member\nIn your original post, why are you dividing by 5 instead of 9 for the sample variance?\n\nPS. small error: (701-726.4 )^2=645.16, not 625.16.\n\nSo, amending (701-726.4 )^2 from 625.16 to 645.16 and amending the following would lead to 805.02N mm^-2\n\nAmendments being: -\n\n(701-726.4 )^2 = 645.16\n\n(1/10)\u22198050.29= 805.02 N mm^-2\n\n\u03c3= \u221a((805.02)=28.37 N mm^(-2)= 0.02837 Mpa\n\nConverting to Gpa yields: -\n\n2.837*10^-5 Gpa\n\nLooking any better ?\n\nFactChecker\nGold Member\n(1/10)\u22198050.29= 805.02 N mm^-2\nWhy are you dividing by 10 instead of N-1=10-1=9?\n\nIf you know the true mean, ##\\mu##, you should use that in the formula for the sample standard deviation and divide by N. But if you estimate the mean with ##\\bar x ##, your results of ##\\sum_{i=1}^{i=N} (x_i-\\bar x)^2## will be smaller and you should divide by ##N-1##.\n\nLast edited:\nAh! thanks, brief moment of confusion. I assume this would amend the overall error ?", "date": "2021-04-20 22:12:57", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/determine-the-standard-deviation-of-these-results.948814/", "openwebmath_score": 0.9873859882354736, "openwebmath_perplexity": 4545.610875679158, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9766692311915195, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8665054386501211}}
{"url": "https://math.stackexchange.com/questions/4002510/prove-this-formula-frac1-r-cosx1-2r-cosxr2-1-sum-n-1-inftyr/4003040", "text": "# Prove this formula $\\frac{1-r\\cos(x)}{1-2r\\cos(x)+r^2}= 1+\\sum_{n=1}^{+\\infty}r^{n}\\cos\\left(nx\\right)$\n\nI am trying to use prove, by just simple algebraic manipulation, to prove the equality of this formula. $$\\dfrac{1-r\\cos(x)}{1-2r\\cos(x)+r^2}= 1+\\sum_{n=1}^{+\\infty}r^{n}\\cos\\left(nx\\right)$$\n\nI have been given hints and instructions from this thread\n\n1. Write using Euler\u2019s identity $$\\cos(x)=\\frac{e^{ix}+e^{-ix}}{2}$$\n2. Factor the denominator\n3. Find the partial fractions decomposition, expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler\u2019s identity again).\n\nThis is how I have done:\n\n1. RHS: $$\\dfrac{1-r(\\dfrac{e^{ix}+e^{-ix}}{2})}{1-2r(\\dfrac{e^{ix}+e^{-ix}}{2})+r^{2}}=\\dfrac{1-r(\\dfrac{e^{ix}-e^{-ix}}{2})}{1-re^{ix}-re^{-ix}+r^2}$$\n\nThen from this thread, I have learnt how to factorize the denominator (See Jack D'Aurizio answer, first answer of the thread, first line). The trick is to write $$1=e^0$$.\n\n1. $$\\dfrac{1-r(\\dfrac{e^{ix}+e^{-ix}}{2})}{(re^{ix})(r-e^{-ix})}$$\n1. The third step is to decompose the fraction:\n\n$$\\dfrac{A}{r-e^{ix}}+\\dfrac{B}{r-e^{-ix}}=\\dfrac{1-r(\\dfrac{e^{ix}+e^{-ix}}{2})}{(r-e^{ix})(r-e^{-ix})}$$\n\n$$A(r-e^{-ix})+B(r-e^{ix})=1-r(\\dfrac{e^{ix}+e^{-ix}}{2})$$\n\nI am stuck here, I notice that let $$x=0$$, we will have\n\n$$A(r-1)+B(r-1)=1-r(\\dfrac{e^{ix}+e^{-ix}}{2})$$\n\nI am stuck here, I don't know to find $$A$$ and $$B$$\n\nAlso, could you provide in details how to finish the part \"expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler\u2019s identity again)\".\n\nMy symbolic manipulation skill is not very good, so a detailed answer is great!\n\nThanks!\n\n\u2022 You have some typos where you write $e^{ix}$ instead of $e^{-ix}$.\n\u2013\u00a0J.G.\nJan 27, 2021 at 22:30\n\u2022 @ J.G. Thanks, I just copy and paste, let me correct them all Jan 27, 2021 at 22:30\n\u2022 There are sign errors in the $e^{-ix}$ coefficients too.\n\u2013\u00a0J.G.\nJan 27, 2021 at 22:31\n\u2022 the sum spans over $-\\infty , \\infty$ or $1, \\infty$, although $\\cos$ is symmetric you might miss some piece Jan 27, 2021 at 23:00\n\nThere is possibly a better way, and I think this was discussed as solved example in tristam needham's book(*). Any who, we begin with geometric series:\n\n$$\\frac{1}{1-x} = \\sum_{j=0}^{\\infty} x^j$$\n\nSub: $$x \\to re^{ i \\theta}$$ and simplfy\n\n$$\\frac{1}{(1- r \\cos \\theta) - i \\sin \\theta} = \\sum_{j=0}^{\\infty} r^j e^{i j\\theta} \\tag{2}$$\n\nFor LHS, by multiplying with complex conjguate\n\n$$\\frac{1}{(1-r \\cos \\theta) - i r\\sin \\theta} = \\frac{ (1-r \\cos \\theta) + i \\sin \\theta}{ 2 - 2 r \\cos \\theta+r^2} \\tag{1}$$\n\nEquating real part in (1) to real part in (2),\n\n$$\\frac{1 - r \\cos \\theta}{ 2 - 2r \\cos \\theta + r^2} = \\sum_{j=0}^{\\infty} r^j \\cos j \\theta$$\n\nDone!\n\nAnother identity could be made by equating imaginary parts\n\n*: Indeed it was! See page-78, 79 of Visual Complex Analysis to see how this is simply the fourier series corresponding to the geometric series. Simply two ways to view a function: As an infinite trignometric sum or as a infinite polynomial series. Pretty neat.\n\n\u2022 in geo series sum is starting from $0$ Jan 27, 2021 at 23:02\n\nSimilar to Buraian's answer, a method I enjoy using is using the equivalent series for $$\\sin$$, and then adding $$C+iS$$, as follows:\n\nLet \\begin{align} C&=1+r\\cos x+r^2\\cos2x+r^3\\cos3x+\\cdots\\\\ S&=r\\sin x+r^2\\sin2x+r^3\\sin3x+\\cdots\\\\ \\end{align} Then \\begin{align} C+iS&=1+r(\\cos x+i\\sin x)+r^2(\\cos 2x+i\\sin2x)+r^3(\\cos3x+i\\sin3x)+\\cdots\\\\ &=1+re^{ix}+(re^{ix})^2+(re^{ix})^3+\\cdots\\\\ &=\\frac{1}{1-re^{ix}}=\\frac{(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\\frac{1-r\\cos x+ri\\sin x}{1-2r\\cos x+r^2} \\end{align} Hence, equating real and imaginary parts we obtain \\begin{align} C&=\\frac{1-r\\cos x}{1-2r\\cos x+r^2}\\\\ S&=\\frac{r\\sin x}{1-2r\\cos x+r^2}\\\\ \\end{align} I hope that was helpful and gives you a new and interesting method for attacking these sorts of problems.\n\n\u2022 I like your answer most, but I also vote for Buraian since he post this method first. Your solution has a \"Eulerish\" feeling to it. haha :) Cheers! Jan 28, 2021 at 15:42\n\u2022 @JamesWarthington :)) I'm so glad I helped you! This method is pretty powerful: for other examples of me using it in other answers/questions of mine see here: math.stackexchange.com/questions/3886649/\u2026 and math.stackexchange.com/questions/3988168/\u2026 Jan 28, 2021 at 17:18\n\u2022 Could you help me here as well: math.stackexchange.com/questions/4003509/\u2026 Jan 28, 2021 at 17:41\n\u2022 @JamesWarthington I will have a go, I suspect the proof lies in using trig identities. Jan 28, 2021 at 17:50\n\u2022 thank you, I wish my algebraic manipulation skill can be as good as you Jan 28, 2021 at 17:53\n\n$$\\cos(nx)=2\\cos(x)\\cos((n-1)x)-\\cos((n-2)x)$$, then $$\\sum_{n=1}^\\infty r^n\\cos(nx)=2\\cos(x)\\sum_{n=1}^\\infty r^n\\cos((n-1)x)-\\sum_{n=1}^\\infty r^n \\cos((n-2)x)$$ $$=2r\\cos(x)\\sum_{n=0}^\\infty r^n\\cos(nx)-r^2\\sum_{n=0}^\\infty r^n \\cos(nx)$$ Then, after some algebra, $$\\sum_{n=1}^\\infty r^n\\cos(nx)=\\frac{r\\cos(x)-r^2}{1-2r\\cos(x)+r^2}$$\n\n\u2022 Can you expand your answer? They are so compacted that I don't know how did you derive these from \"some algebra\". Jan 28, 2021 at 3:55\n\u2022 @JamesWarthington set $$f=\\sum_{n\\ge0}r^n\\cos(nx).$$ Then $$\\sum_{n\\ge1}r^n\\cos(nx)=f-1$$ and $$f-1=2r\\cos(x)f-r^2f.$$ Solve for $f$. Jan 28, 2021 at 6:46\n\nI might derive the series from $$\\frac{1-r \\cos (x)}{r^2-2 r \\cos (x)+1}$$ by setting $$\\cos x = (e^{ix}+e^{-ix})/2 = (z+z^{-1})/2$$ with $$z=e^{ix}$$. Then \\eqalign{ \\frac{1-r \\cos (x)}{r^2-2 r \\cos (x)+1} &= \\frac{r z^2+r-2 z}{2 (r-z) (r z-1)} = \\frac{1}{2}\\left(1+\\frac{r/z}{(1-r/z)}+\\frac{1}{(1-r z)}\\right) \\cr &= \\frac{1}{2}\\left(2+\\frac{r/z}{(1-r/z)}+\\frac{r/z}{(1-r z)}\\right) = \\frac{1}{2}\\left(2+\\sum_{n=1}^{\\infty}{(r/z)^n}+\\sum_{n=1}^{\\infty}{(r z)^n}\\right) \\cr &= 1+\\sum_{n=1}^{\\infty}{z^n+1/z^n\\over2} {r^n} = 1+\\sum_{n=1}^\\infty r^n \\cos nx \\ \\ \\text{or}\\ \\sum_{n=0}^\\infty r^n \\cos nx \\,. \\cr }\n\n\u2022 @Micheal E2: Man, your answer is quite difficult. I think you write with great brevity. I am trying to retrace every step you post here but they are so hard. Jan 28, 2021 at 15:39\n\nSince\\begin{align}\\frac{1-r(e^{ix}+e^{-ix})/2}{(r-e^{ix})(r-e^{-ix})}&=\\frac{A}{r-e^{ix}}+\\frac{B}{r-e^{-ix}}\\\\\\implies 1-r(e^{ix}+e^{-ix})/2&=A(r-e^{-ix})+B(r-e^{ix}),\\end{align}the next step is to consider the $$r^0$$ and $$r^1$$ terms separately, giving$$1=-e^{-ix}A-e^{ix}B,\\,-(e^{ix}+e^{-ix})/2=A+B.$$Now solve simultaneous equations. You should find$$A=-\\frac12e^{ix},\\,B=-\\frac12e^{-ix}.$$\n\n\u2022 @ J.G. what do you by $r^{0}$ and $r^{1}$ terms? By setting $r=r^{0}$ and $r=r^{1}$. This is not right, I think my interpretation is wrong. Jan 27, 2021 at 23:32\n\u2022 @.J.G could you edit your reply and show me how to solve this system of equations? I have tried for 2 hours and only obtain $A=\\frac{e^{ix}+3e^{ix}}{2(1-e^{-2ix})}$, increadily more complicated than your result. Jan 28, 2021 at 3:16\n\u2022 @.J.G how do you know this system of equations only has 1 roots for A and for B? Jan 28, 2021 at 3:41\n\u2022 @.J.G Here are my efforts to solve your system of equations: math.stackexchange.com/questions/4002655/\u2026 Jan 28, 2021 at 4:00\n\nWith $$1 - 2r\\cos x + r^2 = (1-re^{i x})(1-re^{-ix})$$\n\n\\begin{align} \\frac{1-r\\cos x}{1-2r\\cos x+r^2} = Re \\frac{1}{1-re^{ix}} = Re\\sum_{n=0}^\\infty(re^{ix})^n =1+\\sum_{n=1}^{\\infty}r^{n}\\cos nx \\end{align}", "date": "2022-09-25 10:24:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4002510/prove-this-formula-frac1-r-cosx1-2r-cosxr2-1-sum-n-1-inftyr/4003040", "openwebmath_score": 0.9540971517562866, "openwebmath_perplexity": 1502.7283499442844, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692298333415, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8665054359887321}}
{"url": "https://math.stackexchange.com/questions/2618175/how-to-prove-this-delta-epsilon-proof-involving-x2", "text": "# How to prove this delta-epsilon proof involving $x^2$?\n\nUnlike my last question I want to try something where $x$ can be any real value in $f(x)$ so it's not just $x \\geq 0$. I want to fix an $\\epsilon$ and find the largest $\\delta$ I can get away with using to make the necessary inequalities hold.\n\n$$\\lim_{x \\rightarrow 2} x^2= 4$$\n\nSay I pick some $\\epsilon = 0.5$ which means I need to find some $\\delta > 0$ such that for all $x$ satisfying $0 < |x-2| < \\delta$, the inequality $|f(x) - L| = |x^2 - 4| < 0.5$ holds.\n\n1. Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?\n\n2. How do I pick the right $\\delta$ for something like this?\n\nTrying to simplify:\n\n$|x^2 - 4| < 0.5$\n\n$|x+2||x-2| < 0.5$\n\n$|x-2| < \\frac{0.5}{|x+2|}$\n\nNow I'm stuck. Is there a better way to approach these problems? So far I've been trying to manipulate the epsilon inequality so it looks more like the delta one and then try to set the delta and epsilon expressions equal to each other, but maybe there is a more reliable way to prove these relationships?\n\nUpdate:\n\nTrying another way:\n\n$|x^2 - 4| < 0.5$ simplifies to\n\n$\\sqrt{3.5} < x < \\sqrt{4.5}$\n\nThis gives me two $x$-values away from $a=2$, either $2 - \\sqrt{3.5} = .1291...$ or $\\sqrt{4.5} - 2 = .1213...$\n\nSo if I pick the smallest of the two, $\\delta = \\sqrt{4.5} - 2$ which satisfies the epsilon condition?\n\n\u2022 Choose $\\delta=\\epsilon/5$ for $\\epsilon<1$ \u2013\u00a0Fakemistake Jan 23 '18 at 20:46\n\u2022 \"Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?\" For the most part. But there are two considerations you are eliding. i)$\\epsilon$ doesn't have to be $.5$ but could be $.05$ or $.000000005$ of $5\\times 10^{534}$. It can be any value. ii) And $\\delta$ ... well it's best not to think of delta as a constant, but as a number whose value is determined by the value of $\\epsilon$. \u2013\u00a0fleablood Jan 23 '18 at 20:58\n\u2022 If $\\epsilon = .5$ then just let $\\delta$ be something really really small. If $\\delta = .01$ then $|x - 2| < \\delta$ means $1.99 < x < 2.01$ so $3.9601< x^2 < 4.0401$ so $-.0399< x^2 - 4 <.0401$ so $|x^2 - 4| < .05 < .5$. So.... that doesn't tell you much about how to do it in general. Does it? \u2013\u00a0fleablood Jan 23 '18 at 21:04\n\u2022 @fleablood I tried using a graph instead and edited my post. Does this approach make sense? \u2013\u00a0Aruka J Jan 23 '18 at 21:05\n\u2022 I understand $\\epsilon$ can be any positive value, I'm just picking one arbitrarily, fixing some $\\epsilon$, and then finding the largest $\\delta$ that makes $|f(x)-L| < \\epsilon$ hold. We could pick something very small and it would work but that doesn't show how large we could get away with going. \u2013\u00a0Aruka J Jan 23 '18 at 21:05\n\nWe need to show that $\\forall \\epsilon>0$ $\\exists\\delta>0$ such that\n\n$$\\forall x\\neq2 \\quad |x-2|<\\delta \\implies\\left|f\\left(x\\right)-l\\right|<\\varepsilon$$\n\nthat is\n\n$$|x^2-4|<\\epsilon\\iff-\\epsilon<x^2-4<\\epsilon\\iff4-\\epsilon<x^2<4+\\epsilon\\iff \\sqrt{4-\\epsilon}<x<\\sqrt{4+\\epsilon}\\iff \\sqrt{4-\\epsilon}-2<x-2<\\sqrt{4+\\epsilon}-2\\\\\\iff |x-2|<min\\{\\sqrt{4+\\epsilon}-2,2-\\sqrt{4-\\epsilon}\\}=\\sqrt{4+\\epsilon}-2=\\delta \\quad \\square$$\n\n\u2022 I think your result is the same as mine correct? $\\delta = \\sqrt{4.5} - 2$? (I picked $\\epsilon = 0.5$) \u2013\u00a0Aruka J Jan 23 '18 at 21:07\n\u2022 @ArukaJ Yes but you need to prove it in general for genrec $\\epsilon$ and $\\delta$ values. \u2013\u00a0gimusi Jan 23 '18 at 21:19\n\u2022 For the third time. !!!STOP!!! picking specific values of $\\epsilon$. It will NOT help you. \u2013\u00a0fleablood Jan 23 '18 at 21:32\n\u2022 @gimusi Oh, yeah. Absolutely. Looks good. i haven't gone through it with a fine tooth comb but, yeah, that's exactly how do it. So barring any arithmetic errors, It's good. My comment was for Aruka who seems hell bent on finding specific deltas for specific elements and not actually trying to find a general formula for deltas for all possible epsilons. \u2013\u00a0fleablood Jan 23 '18 at 21:39\n\u2022 Well, it's fine to get an insight but there comes a point where you will have to say, insight is done, now we must do a proof. And you will NEVER be allowed to say \"It worked for all examples I tried therefore it must be true\" as a proof. \u2013\u00a0fleablood Jan 23 '18 at 21:45\n\nHint $$|x^{ 2 }-4|=\\left| x-2 \\right| \\left| x-2+4 \\right| <{ \\left| x-2 \\right| }^{ 2 }+4\\left| x-2 \\right|$$\n\nIf $|x- 2| < \\delta$ then\n\n$- \\delta < x -2 < \\delta$\n\n$2 - \\delta < x < 2 + \\delta$. Let's assume for the moment that $\\delta < 2$.\n\n$(2- \\delta)^2 < x^2 < (2+ \\delta)^2$\n\n$4 - 4\\delta + \\delta^2 < x^2 < 4 + 4\\delta + \\delta^2$\n\n$-4\\delta + \\delta^2 < x^2 - 4 < 4\\delta + \\delta^2$.\n\n$-4 \\delta - \\delta^2 < x^2 - 4 < 4\\delta + \\delta^2$\n\n$|x^2 - 4| < |4\\delta + \\delta^2| = 4\\delta + \\delta^2$ (because $\\delta$ is positive)\n\nSo we want $\\epsilon \\ge 4\\delta + \\delta^2$. Given that we know what $\\epsilon$ is, can we find a way of figuring out $\\delta$ in terms of $\\epsilon$ so that that would be true?\n\nIf we assume $\\delta \\le 1$ then $\\delta^2 \\le \\delta$ so $5\\delta \\ge 4\\delta + \\delta^2$.\n\nSo if we choose any $\\delta$ so that i) $\\delta < 2$ and ii) $\\delta \\le 1$ and iii) $5\\delta < \\epsilon$ that will do.\n\nSo for any $\\delta < \\min (\\frac \\epsilon 5, 1)$ that will do.\n\nSo to do the proof:\n\nFor any $\\epsilon > 0$, let $\\delta = \\min (\\frac \\epsilon 5, 1)$\n\nThen if $|x - 2| < \\delta$ implies by all the work we did above that\n\n$-5\\delta \\le -4\\delta -\\delta^2 < -4\\delta + \\delta^2 < x^2 -4 < 4\\delta + \\delta^2 \\le 5\\delta$ so\n\n$|x^2 - 4| < 5\\delta \\le \\epsilon$.\n\nAnd that's the proof.\n\nIt is my honest opinion that your question lies in the realm of questions of the type: \"what is the best approach for riding a bicycle?\" - Most answers you will receive will send you snapshots of happy riders; but you don't really want these answers, do you? I suggest you get on the bike, and keep falling until you don't.\n\n\u2022 Are delta-epsilon proofs more art than science, is that sort of what you are saying? There is no standard methodical way to get the answer? \u2013\u00a0Aruka J Jan 23 '18 at 20:39\n\u2022 Is bicycle riding an art? No, it is a skill. delta-epsilon proofs are a skill. There are many skills that you can only pick up by actually doing. Swimming, bike-riding are two out of three immediate examples that come to mind. \u2013\u00a0uniquesolution Jan 23 '18 at 20:41\n\u2022 Proofs in general are like this. There are some approaches that work more often than others, but it's more about the intuition of what works than following a particular step-by-step approach. In this case, the best advice you can get is to poke around with the triangle inequality and when you get something that works, try to understand what made it work. \u2013\u00a0AlexanderJ93 Jan 23 '18 at 20:43", "date": "2019-06-17 12:37:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2618175/how-to-prove-this-delta-epsilon-proof-involving-x2", "openwebmath_score": 0.7900465726852417, "openwebmath_perplexity": 308.3868530265669, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692332287863, "lm_q2_score": 0.8872045847699186, "lm_q1q2_score": 0.8665054215243001}}
{"url": "https://math.stackexchange.com/questions/2248572/sum-n-1-infty-frac1n22n-by-integration-or-differentiation/2248648", "text": "# $\\sum_{n=1}^{\\infty}\\frac{1}{n^22^n}$ by integration or differentiation\n\nThere is an infinite sum given: $$\\sum_{n=1}^{\\infty}\\frac{1}{n^22^n}$$ It should be solved using integration, derivation or both. I think using power series can help but I don't know how to finish the calculation. Any help will be appreciated!\n\n\u2022 $Li_{2} \\left( \\frac{1}{2} \\right) = \\frac{\\pi^2}{12}-\\frac{ (\\ln 2)^2}{2}$. \u2013\u00a0Donald Splutterwit Apr 23 '17 at 19:26\n\u2022 Thank you! But how did you get that? @DonaldSplutterwit \u2013\u00a0Hendrra Apr 23 '17 at 19:32\n\u2022 I copied it from here ... en.wikipedia.org/wiki/Spence%27s_function ... I am scribbling in my note book now ... I will get back to you when I have managed to derive it ... \u2013\u00a0Donald Splutterwit Apr 23 '17 at 19:34\n\u2022 Thank you very much :) I'm really looking forward to the solution! \u2013\u00a0Hendrra Apr 23 '17 at 19:35\n\u2022 Have a look at math.stackexchange.com/a/1056111/44121 \u2013\u00a0Jack D'Aurizio Apr 23 '17 at 20:10\n\nNote that we have $\\int_0^x t^{n-1}\\,dt=\\frac{x^n}{n}$. Then, we can write\n\n\\begin{align} \\sum_{n=1}^\\infty \\frac{x^{2n}}{n^2}&=\\sum_{n=1}^\\infty \\int_0^x t^{n-1}\\,dt\\int_0^x s^{n-1}\\,ds\\\\\\\\ &=\\int_0^x\\int_0^x \\frac{1}{1-st}\\,ds\\,dt\\\\\\\\ &=-\\int_0^{x} \\frac{\\log(1-sx)}{s}\\,ds\\\\\\\\ &=-\\int_0^{x^2} \\frac{\\log(1-s)}{s}\\,ds\\\\\\\\ &=\\text{Li}_2(x^2) \\end{align}\n\nEvaluating at $x=1/\\sqrt{2}$ yields\n\n$$\\bbox[5px,border:2px solid #C0A000]{\\sum_{n=1}^\\infty \\frac{1}{n^2\\,2^n}=\\text{Li}_2(1/2)=\\frac{\\pi^2}{12}-\\frac12\\log^2(2)}$$\n\nAnd we are done!\n\nTo evaluate $\\text{Li}_2(1/2)$, we exploit the relationship\n\n$$\\text{Li}_2(1-x)=-\\text{Li}_2\\left(1-\\frac1x\\right)-\\frac12\\log^2(x)$$\n\nLetting $x=1/2$ yields\n\n$$\\text{Li}_2(1/2)=-\\text{Li}_2\\left(-1\\right)-\\frac12\\log^2(1/2)=\\frac{\\pi^2}{12}-\\frac12\\log^2(2)$$\n\nwhere we used\n\n\\begin{align} \\text{Li}_2(-1)&=-\\int_0^{-1}\\frac{\\log(1-x)}{x}\\,dx\\\\\\\\ &=\\sum_{n=1}^\\infty\\frac{(-1)^{n-1}}{n^2}\\\\\\\\ &=\\frac{\\pi^2}{12} \\end{align}\n\n\u2022 That's a nice solution! Thanks. However I must ask about $\\sum_{n=1}^{\\infty}\\frac{x^{2n}}{n^2}$. Why are we considering such a sum? \u2013\u00a0Hendrra Apr 23 '17 at 19:43\n\u2022 You're welcome. My pleasure. The value of that sum when $x=1/\\sqrt 2$ is $\\sum_{n=1}^\\infty \\frac{1}{n^2\\,2^n}$. \u2013\u00a0Mark Viola Apr 23 '17 at 19:45\n\u2022 The value truly is $\\sum_{n=1}^{\\infty}\\frac{1}{n^22^n}$! Thus I have the next question. Why did you decided to take an $x = \\frac{1}{\\sqrt{2}}$? Just because it works? \u2013\u00a0Hendrra Apr 23 '17 at 19:48\n\u2022 Well, yes. We took $x=1/\\sqrt 2$ in order that we would have the sum of interest. \u2013\u00a0Mark Viola Apr 23 '17 at 19:51\n\u2022 @Tyberius The first integral yields $$\\int_0^x \\int_0^x \\frac{1}{1-st}\\,dt\\,ds=-\\int_0^x \\frac{\\log(1-sx)}{s}\\,ds=-\\int_0^{x^2}\\frac{\\log(1-s)}{s}\\,ds=\\text{Li}_2(x^2)$$ \u2013\u00a0Mark Viola Apr 23 '17 at 20:20\n\nAnother way to calculate it: consider $$f(x)=\\sum_{n=1}^{+\\infty}\\frac{x^n}{n^2}.$$ Differentiating w.r.t. $x$ gives $$f'(x)=\\sum_{n=1}^{+\\infty}\\frac{x^{n-1}}{n}=\\frac{1}{x}\\sum_{n=1}^{+\\infty}\\frac{x^n}{n}\\quad\\Rightarrow\\quad xf'(x)=\\sum_{n=1}^{+\\infty}\\frac{x^n}{n}\\quad\\Rightarrow\\quad (xf'(x))'=\\sum_{n=1}^{+\\infty}x^{n-1}=\\frac{1}{1-x}.$$ Now integrating with $f(0)=0$ $$xf'(x)=-\\ln(1-x)\\quad\\Rightarrow\\quad f(x)=-\\int_0^x\\frac{\\ln(1-t)}{t}\\,dt.$$ Motivation for termwise differentiation for power series is straightforward.\n\nMy answer to a duplicate question says:\n\n$$\\frac{1}{1-x}=\\sum_{n=0}^{\\infty} x^{n}$$ Integrating from 0 to t we get $$\\int_{0}^{t}\\frac{1}{(1-x)}dx=\\sum_{n=0}^{\\infty}\\int_{0}^{t} x^{n}dx$$$$-\\ln(1-t)=\\sum_{n=1}^{\\infty} \\frac{t^{n}}{n}$$ Dividing by t and integrating $$\\int_{0}^{0.5}-\\frac{\\ln(1-t)}{t}dt=\\sum_{n=1}^{\\infty}\\int_{0}^{0.5} \\frac{t^{n-1}}{n}dt=\\sum_{n=1}^{\\infty} \\frac{1}{n^{2}2^{n}}$$ This on calculating is $$\\dfrac{\\pi^2}{12}-\\dfrac{ln^2(2)}{2},$$\n\n\u2022 This answer is identical to your answer to this question. If the questions are the same, this one should be flagged as a duplicate. If not, then it would be more efficient and less noisy to cite or quote (with attribution) from the other answer. The link between the answers would also serve as a link between two closely related questions. In this case, the questions were duplicates. \u2013\u00a0robjohn May 26 at 1:31\n\u2022 @robjohn So should I change anything ??. I posted it here because the other one was marked as duplicate. Should I remove it from here?? \u2013\u00a0DivMit May 26 at 3:12\n\u2022 Even though the other question was closed, your answer there is still active and getting votes. One should be removed, but you might edit one to reference the other; you might just say something like, \"my answer to a duplicate question says...\" \u2013\u00a0robjohn May 26 at 5:28\n\u2022 Okay , I will link my answers \u2013\u00a0DivMit May 26 at 9:55", "date": "2019-07-19 10:21:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2248572/sum-n-1-infty-frac1n22n-by-integration-or-differentiation/2248648", "openwebmath_score": 0.7719256281852722, "openwebmath_perplexity": 849.3365068875703, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846672373523, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8664882651433624}}
{"url": "https://math.stackexchange.com/questions/3179843/how-do-limits-work-with-floor-ceiling", "text": "How do limits work with floor/ceiling?\n\nI'm interested in the below equation:\n\n$$\\frac{n}{\\operatorname{floor}(\\frac{x}{n})}$$\n\nPlotting with $$n = 1..100$$ shows the graph being slightly more aliased as $$n$$ increases and a discontinuity forms from $$0..n$$. The domain from what I can tell is $$(-\\infty, 0) \\cup [n, \\infty)$$. I wanted to investigate the limits at each $$n$$ but not entirely certain how floor and ceiling factor into algebra.\n\nTake for example $$n = 2$$,\n\n$$f(x) = \\frac{2}{\\operatorname{floor}(\\frac{x}{2})}$$\n\nHow could I find $$\\lim_{x\\to0} f(x)$$? Would the limit even exist considering the discontinuity between $$[0, 2)$$?\n\nI know you can break down the function and help find the limit using the rules of limits:\n\n$$\\lim_{x\\to b} \\frac{p}{q} = \\frac{\\lim_{x\\to b}p}{\\lim_{x\\to b}q}$$\n\nSo more precisely I'm looking for $$\\frac{2}{\\lim_{x\\to0}\\operatorname{floor}(\\frac{x}{2})}$$.\n\nFor what it's worth, the plot looks like:\n\nI can surmise $$\\lim_{x\\to0^-} f(x) = -2$$, am I right in assuming $$\\lim_{x\\to0^+} f(x)$$ doesn't exist?\n\n\u2022 When dealing with a domain that is not the whole real line, you need to be careful with limits on the border of the domain. In this case, the limit is the same as $\\lim_{x\\to 0^{-}} f(x)$ because the small open neighborhoods of $0$ in the domain of $f$ are only negative. This means the limit will be $-n$ since $\\lfloor x/n\\rfloor=-1$ is constant for $x$ negative near $0.$ \u2013\u00a0Thomas Andrews Apr 8 at 16:36\n\u2022 Your function not only has a discontinuity but fails to be defined at all for $x\\in[0,n)$. \u2013\u00a0Henning Makholm Apr 8 at 16:36\n\u2022 On the other hand it is clear that $f(x)=-n$ for every $x\\in[-n,0)$, and since these numbers are the only ones in the domain that are close to $0$, the function very trivially goes to $-n$ for $x\\to 0$. There's no algebra involved in this. \u2013\u00a0Henning Makholm Apr 8 at 16:38\n\nIt helps to carefully state definitions. A workable definition of a limit (of a real function) is something like the following:\n\nDefinition: Let $$f$$ be a function defined on some domain $$D\\subseteq \\mathbb{R}$$ and let $$a \\in \\mathbb{R}$$. Further suppose that there is some $$L\\in\\mathbb{R}$$ such that for every $$\\varepsilon > 0$$ there exists some $$\\delta > 0$$ such that if $$x \\in D$$ and $$0 < |x-a|<\\delta$$, then $$|f(x) - L| < \\varepsilon.$$ $$L$$ is said to be the limit of $$f(x)$$ as $$x$$ approaches $$a$$, denoted $$\\lim_{x\\to a} f(x) = L.$$\n\nI claim that $$\\lim_{x\\to 0} f(x) = \\lim_{x\\to 0} \\frac{2}{\\left\\lfloor \\frac{x}{2} \\right\\rfloor} = -2.$$ To prove this claim, I have to show that if $$\\varepsilon$$ is any positive number, then I can find a value $$\\delta$$ such that $$f(x)$$ is within $$\\varepsilon$$ of $$-2$$ whenever $$x$$ is in the domain of $$f$$ and within $$\\delta$$ of zero.\n\nSo, let $$\\varepsilon > 0$$ be arbitrary and take $$\\delta = 1$$. Observe that $$f$$ is only defined on the set $$D = \\mathbb{R} \\setminus [0,2).$$ If $$x \\in D$$ and $$|x| < \\delta = 1$$, then $$x \\in (-1,0)$$, and so $$\\lfloor x/2 \\rfloor = -1$$. But then $$|f(x) - (-2)| = \\left| \\frac{2}{\\left\\lfloor \\frac{x}{2} \\right\\rfloor} + 2 \\right| = \\left| \\frac{2}{-1} + 2 \\right| = 0.$$ Hence whenever $$|x-0| < \\delta$$ and $$x\\in D$$, we have $$|f(x)-(-2)| = 0 < \\varepsilon$$. Therefore $$\\lim_{x\\to 0} \\frac{2}{\\left\\lfloor \\frac{x}{2} \\right\\rfloor} = -2,$$ as claimed.\n\nThe important point here is that the definition of a limit only cares about what is happening in the domain of the function. Points where the function is undefined are irrelevant. As long as we are working with a real-valued function of a real variable, the function of interest is simply not defined on the interval $$[0,2)$$, and so we don't have to worry about those points.\n\n\u2022 My understanding in finding a limit was the limit for $x\\to b$ only exists if the limit exists for both $x\\to b^-$ and $x\\to b^+$ and they are equal. If one doesn't exist, the limit doesn't exist; or, if they are not equivalent, the limit doesn't exist. I understand the leftside limit is $-2$ but the rightside limit is not defined. Doesn't then, the limit not exist? I'll disclaim I'm in Calculus I. \u2013\u00a0gator Apr 8 at 16:49\n\u2022 As I said in the preamble, you have to carefully state your definitions. The definition of a limit which I gave is pretty standard. If you want to discuss left- and right-hand limits, you need to first carefully define what those mean. I would claim that $\\lim_{x\\to 0^+} f(x) = -2$ vacuously---indeed, I would claim that if $L$ is any real number, then $\\lim_{x\\to 0^+} f(x) = L$, since if I make $\\delta$ small enough, I cannot find any values of $x$ such that $0 < x < \\delta$. Thus all of the conditions are met. Again, state the definition very carefully, then see what happens. \u2013\u00a0Xander Henderson Apr 8 at 16:53\n\u2022 You need to exclude the case where $x=a\\in D$ or you'd have $\\lim_{x\\to a} f(x)$ would never be defined if $a$ was a point of discontinuity, even if it could be made continuous. You also want that for $\\delta>0$ that there is at least one $x\\in D\\setminus\\{a\\}$ so that $|x-a|<\\delta.$ Otherwise, all limit values would be vacuously true. \u2013\u00a0Thomas Andrews Apr 8 at 16:55\n\u2022 @gator Using that definition, then you are at a loss here. You'd just have to leave it undefined. But in later math, that definition proves inadequate in many ways. \u2013\u00a0Thomas Andrews Apr 8 at 16:57\n\u2022 @ThomasAndrews Regarding $|x-a| > 0$, I added that---omitting it was an oversight on my part. Regarding vacuous limits, I have no problem allowing the degenerate case in which the limit is anything by vacuity. But, again, everything comes down to definitions. If one wants to omit vacuous limits (which, in reality, one probably does), then the definition should be written slightly differently. \u2013\u00a0Xander Henderson Apr 8 at 19:21", "date": "2019-06-25 01:19:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3179843/how-do-limits-work-with-floor-ceiling", "openwebmath_score": 0.9444609880447388, "openwebmath_perplexity": 133.67139556441433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846634557753, "lm_q2_score": 0.8856314828740729, "lm_q1q2_score": 0.8664882603175891}}
{"url": "https://math.stackexchange.com/questions/1251466/what-is-this-pattern-called/1251501", "text": "# What is this pattern called?\n\n## Back-Story\n\nI became interested in the patterns in multiplication tables for different base number systems a while ago. Specifically, the pattern made by the last digit of each number in the multiplication table. So, base 10 would look like this:\n\n1|2|3|4|5|6|7|8|9|0\n2|4|6|8|0|2|4|6|8|0\n3|6|9|2|5|8|1|4|7|0\n4|8|2|6|0|4|8|2|6|0\n5|0|5|0|5|0|5|0|5|0\n6|2|8|4|0|6|2|8|4|0\n7|4|1|8|5|2|9|6|3|0\n8|6|4|2|0|8|6|4|2|0\n0|0|0|0|0|0|0|0|0|0\n\n\nI thought it was interesting that when you move to number systems with different bases, the patterns don't follow the number. They follow it's relative position in the number system. E.g., in base 12, the pattern is 6,0,6,0,6......\n\n## Images\n\nI then realized, I could see the pattern better if I just assigned each number a color. I started with using 10 greyscale colors, with 0 being black and 9 being white. So now base 10 looks like this:\n\nThen, I figured that I really could use as many colors as I wanted to, and see if a larger pattern forms. Using all 256 greyscale colors, I came up with this image representing a base 256 multiplication table:\n\nOr I could go from black to white to black, and smooth out the image:\n\n## Animation\n\nI decided to animate the pattern to better see what was going on. To do this, I defined my 1-n color scale as [w,w,w,w,b,b,b,b,b,b,b,b....]. Where w is white and b is black. I would create a frame, shift my colors down one [b,w,w,w,w,b,b,b,b,b,b,b....], and create the next frame. I repeated this until they colors fully cycled and got this animated image.\n\nHere's a site where you can modify the settings.\n\n## What is this pattern called?\n\nMy question is, what is this pattern called? I'm having a hard time finding anything about it. It seems to be a bunch of hyperbolic curves imposed on each other. There is a bunch of \"stars\" at the corners of where you would divide the image into 4ths, 9ths, etc.\n\nAny insight into this would be appreciated.\n\n\u2022 This pattern is called \"cool\". \u2013\u00a0Lee Mosher Apr 25 '15 at 16:42\n\u2022 I'm not an expert, but look at moire patterns. \u2013\u00a0Bob Krueger Apr 25 '15 at 16:46\n\u2022 @Bob1123 It does seem similar to a moire pattern. But if it is, the question becomes what patterns are making up this pattern! \u2013\u00a0Alex McKenzie Apr 25 '15 at 16:50\n\u2022 Should the question title be changed to \"What is the visual pattern in the multiplication table of modular arithmetic?\" The OP didn't mention modular arithmetic but it does make this question easier to search for \u2013\u00a0man and laptop Apr 25 '15 at 17:08\n\u2022 I love that you posted this! I stumbled across this same pattern years ago but didn't think to ask about it online. (Then again, that might have been before I knew about Math.SE.) So, thanks for doing this! :) \u2013\u00a0El'endia Starman Apr 26 '15 at 0:28\n\nWhat you've discovered is essentially modular arithmetic. By looking at only the last digits of a product (in whatever base you're looking at at the moment), you're in effect saying 'I don't care about things that differ by multiples of $n$; I want to consider them as the same digit'. For instance, in base $7$, $5\\times 2=10_{10}=13$ has the same last digit as $4\\times 6=24_{10}=33$; we put both of these numbers into a bucket labeled '$[3]$', along with $3$, $23=17_{10}$, $43=31_{10}$, etc. In mathematics, when we talk about $31 \\bmod 7$ we sometimes just mean the number $3$ itself (that is, the 'label' on this bucket that's between $0$ and $6$, but it's often convenient to think of it as representing the whole bucket: whatever number we pick out of the $[3]$ bucket, when we add it to a number in the $[2]$ bucket, we know that our result will be in the $[5]$ bucket, and when we multiply a number in the $[3]$ bucket by a number in the $[4]$ bucket, we know that our result will be in the $[5]$ bucket; etc. \"Last digits\" are just a convenient way of talking about these buckets (though things get a little sketchier when you talk about negative numbers - note that according to these rules, $-3$ goes into the $[4]$ bucket!).\n\nMeanwhile, the bands in your pattern are actually (pieces of) hyperbolas. Since $a\\times (n-b)\\equiv -(a\\times b)\\pmod n$ (the statement '$x=y\\pmod n$' is a mathematical way of phrasing '$x$ and $y$ are in the same bucket in base $n$'; here, the difference between $a\\times (n-b)$ and $-(a\\times b)$ is $a\\times n$), the far right hand side is essentially a reflection of the left, and similarly the bottom is a reflection of the top. If you rearrange the four quarters of your square so that the center of symmetry is (what was previously) the top left corner \u2014 i.e., take $A\\ B\\atop C\\ D$ to $D\\ C\\atop B\\ A$ \u2014 and then put the origin at the center, then the bands will exactly be (scaled versions) of the hyperbolae $xy=C$ (which are the hyperbolae $y^2-x^2=2C$ rotated by $45^\\circ$). This happens because each 'cycle' of black-to-white or black-to-white-to-black will be separated by one multiple of $n$; e.g., the first transition between cycles occurs along the hyperbola $xy=n$; the second along the hyperbola $xy=2n$; etc.\n\n(As for the moir\u00e9 patterns, they're related to the usual way that such patterns are generated, and in particular they're somewhat related to aliasing near the Nyquist limit when the frequency between hyperbolic bands starts coming close to the frequency of the 'pixels' you're sampling with, but that's another story altogether...)\n\n\u2022 So could you say our base 10 number systyem (or any base number system) can be described with modular arithmetic? And is this pattern called anything specific? E.g., modulus pattern? \u2013\u00a0Alex McKenzie Apr 25 '15 at 17:10\n\u2022 @AlexMcKenzie The last-digits multiplication that you're talking about is exactly modular arithmetic (specifically, it's the multiplication table mod $n$). A cute example: look at the table of last digits for the base-7 multiplication table. Notice that every row has each non-zero number exactly once, and (by symmetry) so does every column? This isn't a coincidence! This will happen whenever your base is a prime; the numbers mod $p$ form what's called a group under multiplication. \u2013\u00a0Steven Stadnicki Apr 25 '15 at 17:14\n\u2022 @AlexMcKenzie In fact, let me flesh out my answer a little bit to explain what I mean by that first sentence... \u2013\u00a0Steven Stadnicki Apr 25 '15 at 17:22\n\u2022 Correct me if I'm wrong, but essentially this is a graph of multiple instances of $xy = C$ separated by a multiple of $n$. this is what forms the pattern in the corners. Then, since I'm sampling across a grid of finite pixels, a moire pattern forms. So if I had an infinitely large image, I would only see the $xy = C$ pattern, and not the artifacts. \u2013\u00a0Alex McKenzie Apr 25 '15 at 17:22\n\u2022 @AlexMcKenzie Also, congratulations on finding this! This sort of pattern-hunting is an excellent way of experimenting in mathematics and learning about all sorts of facets of its vast world. \u2013\u00a0Steven Stadnicki Apr 25 '15 at 17:38\n\nYou can model your graphics as computing $f_n(x,y)=n\\left[\\frac{xy}{n}\\right]$, where here the square brackets are ad-hoc notation to mean taking the fractional part (or \"reduce modulo 1\"). This takes $z=xy$ and cuts it at a bunch of horizontal hyperbolae, collapsing the graph like a Fresnel lens. Then, what you are doing is sampling $f_n$ on integer points $\\{(i,j)\\in\\mathbb{Z}^2:1\\leq i,j\\leq n\\}$, but $f_n$ oscillates faster than your sample grid, leading to a Moire pattern.\n\nThese would be the powers in a discrete Fourier transform matrix: http://en.wikipedia.org/wiki/Discrete_Fourier_transform_(general)", "date": "2019-05-20 22:25:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1251466/what-is-this-pattern-called/1251501", "openwebmath_score": 0.6396054029464722, "openwebmath_perplexity": 400.4800235176337, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846659768268, "lm_q2_score": 0.8856314723088733, "lm_q1q2_score": 0.8664882522134824}}
{"url": "https://math.stackexchange.com/questions/956776/whats-the-inverse-operation-of-exponents", "text": "# What's the inverse operation of exponents?\n\nYou know, like addition is the inverse operation of subtraction, vice versa, multiplication is the inverse of division, vice versa , square is the inverse of square root, vice versa.\n\nWhat's the inverse operation of exponents (exponents: 3^5)\n\nAddition and multiplication are commutative, so there is just one inverse function.\n\nExponents are not commutative; $2^8 \\not= 8^2$. So we need two different inverse functions.\n\nGiven $b^e = r$, we have the \"$n$th root\" operation, $b = \\sqrt[e] r$. It turns out that this can actually be written as an exponent itself: $\\sqrt[e] r = r^{1/e}$.\n\nAgain, given $b^e = r$, we have $e = \\log_b r$, the \"base-$b$ logarithm of $r$\".\n\n\u2022 Does division have just one inverse function? 8/4 \u2260 4/8 after all. Jul 31 '19 at 20:59\n\u2022 8/4 and 4/8 have a simple numerical relationship. (One is 2/1, the other is 1/2.) Whereas 2^8 and 8^2 have no particularly simple relationship (256 and 64, respectively.) Admittedly I didn't explain that very well... Aug 5 '19 at 8:02\n\u2022 Good stuff, you can't express \"nth root\" in Excel but you can express x^(1/n) just fine. This gets doubly interesting if the exponent is e.g. 1.26 Sep 14 '21 at 18:21\n\nThese functions are the logarithms, and they are fundamentally important. For $$a = b^c$$ (where $$b > 0$$) we write: $$c = \\log_b a,$$ which we can take to be the definition of $$\\log_b$$. We read the operation as \"logarithm, base $$b$$,\" or \"base $$b$$ logarithm\".\n\nIn particular, we have $$\\log_a (a^b) = b \\qquad\\text{and}\\qquad a^{\\log_a b} = b.$$ Of special interest is the natural logarithm, denoted by $$\\ln$$ or $$\\log$$, the logarithm of base $$e$$. (NB that sometimes $$\\log$$ can also denote base $$10$$, or base $$2$$, depending on context.)\n\nLogarithmic identities correspond to exponential identities. From example, from the definition we can conclude that $$\\log_b (pq) = \\log_b p + \\log_b q$$ (for $$p, q > 0$$), which corresponds to the identity $$b^{p + q} = b^p b^q$$.\n\nPerhaps counterintuitively, sometimes it is convenient to define the natural logarithm first and then define the exponential function $$x \\mapsto e^x$$ to be its inverse, which leads to the slightly antiquated name antilog for an exponential function $$x \\mapsto b^x$$.\n\nEdit Some of the other answers here pointed out quite rightly that one can also ask about the inverse of functions where the variable is in the base, i.e., functions $$x \\mapsto x^a$$, and inverses of these functions$$^*$$ (at least when $$a > 0$$) are just $$x \\mapsto x^{1/a}$$, which we often write as $$x \\mapsto \\sqrt[a]{x}$$. These functions are called power functions (note that the inverse of a power function is again a power function), and we reserve the name exponential function for functions $$x \\mapsto b^x$$ where the variable is in the exponent, i.e., those to which the logarithms are inverses.\n\n$$^*$$For some $$a$$ (in particular, even integers), we need to restrict the map $$x \\mapsto x^a$$ to $$[0, \\infty)$$ in order to take an inverse.\n\n\u2022 The other answers were good, but your answer explained it best to me. Oct 3 '14 at 12:02\n\u2022 What about the example in the post? What is the inverse function then? 3^(1/5) or 'base 3 logarithm of 243'? How can I know just by seeing the function which one is the variable here? Oct 24 '19 at 13:05\n\u2022 That's really the point of the edit: Is the question asking about the inverse of $x \\mapsto 3^x$ or the inverse of $x \\mapsto x^5$? From just the expression $3^5$ alone there's no way to tell, much like asking about a function whose evaluation yields the expression $1 + 2$ does not indicate whether the question is about the function $x \\mapsto x + 2$ or $x \\mapsto 1 + x$. As you can see from the question, I initially understood OP to be asking about $x \\mapsto 3^x$, since this is an exponential function, and OP asked about \"inverse operation of exponents\". Oct 24 '19 at 21:21\n\nThere are two inverse operations of exponentiation.\n\n## Logarithm\n\n$$\\log _{b} a$$\n\nIt's read \"base-$b$ logarithm of $a$\". And it means \"the exponent which $b$ must be raised to, so that the result is $a$\".\n\n## Root\n\n$$\\sqrt[b] a$$\n\nIt's read \"$b$-th root of $a$\". And it means \"the number which, when raised to $b$, produces $a$\".\n\nIt depends on what you see as the function and what the variable in $3^5$.\n\nGeneralising your \"square is the inverse of square root\" leads to reciprocal exponents being the inverse of exponents, so $3^5 = 243$ corresponds to $3 = 243^{1/5}$.\n\nAlternatively $3^5 = 243$ corresponds to $5 =\\log _{3} 243 = \\frac{\\log _{10} 243}{\\log _{10} 3}= \\frac{\\log _{e} 243}{\\log _{e} 3}$ using logarithms.\n\nLogarithms: $$10^x = 100 \\iff x=\\log _{10} 100 = 2$$\n\nIf you take $x=3^5$, to \"get the 5 back\" you do $log_3(x)$ and, to \"get the 3 back\", you do $\\sqrt[5]{x}$.\nThe interesting thing here is that there are 2 ways to reverse the operation, while other operations had just one: If you take $x=2+7$, to \"get the 2 back\" you did $x-7$ and to \"get the 7 back\", $x-2$. This happens because 2+7 = 7+2. The sum is \"symmetrical\" (the right term is commutative). If you want to \"get the 2 back\" from $x=2+7$, just subtract 7. If you want to \"get the 2 back\" from $y=7+2$, just subtract 7 again (because, after all, $x=y$).\nBut $x=3^5$ is not the same as $y=5^3$. So you cant expect to use the same operation to \"get the 3 back from x\" and \"get the 3 back from y\"", "date": "2022-01-21 21:59:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/956776/whats-the-inverse-operation-of-exponents", "openwebmath_score": 0.8837385773658752, "openwebmath_perplexity": 447.85770646773545, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474891602739, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8664576708485782}}
{"url": "https://huchengbei.com/blog/53/", "text": "## PAT(A) 1126. Eulerian Path (25)\n\n### 1126. Eulerian Path (25)\n\nIn graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similarly, an Eulerian circuit is an Eulerian path which starts and ends on the same vertex. They were first discussed by Leonhard Euler while solving the famous Seven Bridges of Konigsberg problem in 1736. It has been proven that connected graphs with all vertices of even degree have an Eulerian circuit, and such graphs are called Eulerian. If there are exactly two vertices of odd degree, all Eulerian paths start at one of them and end at the other. A graph that has an Eulerian path but not an Eulerian circuit is called semi-Eulerian. (Cited from https://en.wikipedia.org/wiki/Eulerian_path) Given an undirected graph, you are supposed to tell if it is Eulerian, semi-Eulerian, or non-Eulerian.\n\nInput Specification:\n\nEach input file contains one test case. Each case starts with a line containing 2 numbers N (<= 500), and M, which are the total number of vertices, and the number of edges, respectively. Then M lines follow, each describes an edge by giving the two ends of the edge (the vertices are numbered from 1 to N).\n\nOutput Specification:\n\nFor each test case, first print in a line the degrees of the vertices in ascending order of their indices. Then in the next line print your conclusion about the graph -- either \"Eulerian\", \"Semi-Eulerian\", or \"Non-Eulerian\". Note that all the numbers in the first line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.\n\nSample Input 1:\n7 12\n5 7\n1 2\n1 3\n2 3\n2 4\n3 4\n5 2\n7 6\n6 3\n4 5\n6 4\n5 6\n\nSample Output 1:\n2 4 4 4 4 4 2\nEulerian\n\nSample Input 2:\n6 10\n1 2\n1 3\n2 3\n2 4\n3 4\n5 2\n6 3\n4 5\n6 4\n5 6\n\nSample Output 2:\n2 4 4 4 3 3\nSemi-Eulerian\n\nSample Input 3:\n5 8\n1 2\n2 5\n5 4\n4 1\n1 3\n3 2\n3 4\n5 3\n\nSample Output 3:\n3 3 4 3 3\nNon-Eulerian\n\n### \u4ee3\u7801\n\n/*\n* Problem: 1126. Eulerian Path (25)\n* Author: HQ\n* Time: 2018-03-12\n* State: Done\n* Memo: \u56fe\uff0cdfs\n*/\n#include \"iostream\"\nusing namespace std;\n\nint N,M;\nint degree[500 + 5];\nint G[500 + 5][500 + 5];\nint visit[500 + 5];\nint even = 0;\nint cnt = 0;\n\nvoid dfs(int x) {\nvisit[x] = 1;\ncnt++;\nfor (int i = 1; i <= N; i++) {\nif (!visit[i] && G[x][i])\ndfs(i);\n}\n}\n\nint main() {\ncin >> N >>M;\nint x, y;\nfill(degree, degree + 500 + 5, 0);\nfill(visit, visit + 500 + 5, 0);\nfor (int i = 0; i < M; i++) {\ncin >> x >> y;\nG[x][y] = 1;\nG[y][x] = true;\ndegree[x] ++;\ndegree[y] ++;\n}\nbool first = true;\nfor (int i = 1; i <= N; i++) {\nif (first) {\ncout << degree[i];\nfirst = false;\n}\nelse\ncout << \" \" << degree[i];\nif (degree[i] % 2 == 0)\neven++;\n}\ncout << endl;\ndfs(1);\nif(cnt != N)\ncout << \"Non-Eulerian\" << endl;\nelse if (even == N)\ncout << \"Eulerian\" << endl;\nelse if(even == N-2)\ncout << \"Semi-Eulerian\" << endl;\nelse\ncout << \"Non-Eulerian\" << endl;\n\nsystem(\"pause\");\n}\n\n\u8fd8\u6ca1\u6709\u4eba\u8bc4\u8bba...", "date": "2019-08-22 02:44:36", "meta": {"domain": "huchengbei.com", "url": "https://huchengbei.com/blog/53/", "openwebmath_score": 0.4423867166042328, "openwebmath_perplexity": 1331.3845926305773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474904166247, "lm_q2_score": 0.8757869932689565, "lm_q1q2_score": 0.8664576639301635}}
{"url": "http://math.stackexchange.com/questions/336812/understanding-lagranges-theorem-group-theory", "text": "# Understanding Lagrange's Theorem (Group Theory)\n\nI am beginning with Abstract Algebra and I'm trying to understand Lagrange's Theorem. The theorem reads\n\nFor any finite group $G$, the order of every subgroup $H$ of $G$ should divide the order of $G$.\n\nIt seems simple and I've used it to solve some exercices but I believe I'm missing the essence of it. Is there an example that can help me understand it better, or visualize it? Is there a geometric interpretation?\n\n-\nThere is a local-representation theoretical version: if $H$ is a subgroup of $G$, then every projective module of $H$ induces a projective module of $G$. Hope you like it. :D \u2013\u00a0 awllower Mar 21 '13 at 12:01\nIf you like to think visually, this books.google.com/\u2026 might pique your interest (not only about Lagrange's theorem, but in general)! \u2013\u00a0 Stahl Mar 21 '13 at 12:06\nThe proof works by considering the cosets of $H$ in $G$, what we see is that the cosets partition $G$, so then $G$ is in some sense covered by sets of size $H$ which do not overlap. In terms of visualisation, draw a big circle, call it $G$, then you can partition your circle into equal sections of size $|H|$, for any subgroup $H$. \u2013\u00a0 user27182 Mar 21 '13 at 12:08\nWhat would qualify as \"the essence\" of this? It is a very straightforward theorem. It has no hidden details or mysterious manipulations. \u2013\u00a0 Pedro Tamaroff Mar 21 '13 at 13:03\n@Stahl This book is amazing, thank you! It turns out the author has also developed an open source application called GroupExplorer that helps visualize groups, homomorphisms, subgroup lattices, and more, which I find it incredibly useful in my case. \u2013\u00a0 gpo Mar 23 '13 at 21:12\n\nThe left-multiplication map $x\\mapsto ax$ is bijective on $G$; its injectivity follows from the cancellative property of the group's operation, $ah=ag\\iff h=g$, and overall bijectivity is a consequence of the fact that it has an inverse map, $x\\mapsto a^{-1}x$. I like to view a subgroup $H\\le G$ as a \"puck\" and the overgroup $G$ as a \"air-hockey table\" on which $H$ resides, and to move $H$ around we apply left multiplication by various elements. If you left-multiply by an element $a\\in H$, you have not moved the puck at all since $a\\in H\\iff H=aH$.\n\nEvery element $g\\in G$ is in some coset, or left translate, of $H$ - in particular, $g=ge\\in gH$ since we know that $e\\in H$. Thus, the collection of all translates (possible places for the puck to be positioned in) of $H$ \"cover\" the entire air-hockey table. It remains, then, to investigate the nature of the overlaps between positions, i.e. the intersections of distinct cosets. Here is the proof that cosets that overlap nontrivially must in fact be identical, put into visual form:\n\n$\\hskip 0.6in$\n\nThis means the cosets of $H$ partition the group $G$. As left multiplication is bijective, every coset is the same size, so each \"looks\" the same from the viewpoint of cardinality. Continuing with the idea of an air hockey table, this tells us the puck positions tile it, so we have something like:\n\n$\\hskip 1.2in$\n\nThe most fundamentally basic meaning imputed to multiplication of natural numbers is the following: if Alice has $n$ bags each containing $m$ apples, then she has $n\\times m$ apples total. Similarly, our group $G$ is covered by some number $[G:H]$ of disjoint cosets, each containing $|H|$ elements, so $|G|=[G:H]\\times|H|$. Note that this is even true on the level of arbitrary infinite cardinals. Thus, $|H|$ is a divisor of the order $|G|$: Lagrange's theorem.\n\nThe converse is not globally true: not every divisor $d$ of $n=|G|$ corresponds to a subgroup $H\\le G$ of size $|H|=d$. Sylow theory however yields a local version of a converse: for every prime power $q=p^r$ that is a divisor $q\\mid n$, there is a $p$-subgroup $H$ of size $|H|=q$.\n\n-\nThank you so much for the time and energy you put in writing such a comprehensive and detailed answer! It really helped me a lot and makes things a lot clearer. I actually understand the proof our professor gave now! \u2013\u00a0 gpo Mar 23 '13 at 19:46\n\nYou already have several great explanations, but you asked also for a way to visualise this. One way to visualise Lagrange's Theorem is to draw the Cayley table of (smallish) groups with colour highlighting.\n\nHere is the Cayley table of a dicyclic group of order $16$ with the cosets of its centre of order $2$ highlighted. The subgroup itself consists of the elements $\\{ e, a \\}$ (where $e$ is the identity), and is shown in red. The other cosets appear with different colours. Because the subgroup, in this case, is normal, the table is highly regular.\n\n# The Maple code to produce this is:\n> with( GroupTheory ):\n> G := DicyclicGroup( 4 ):\n> H := Centre( G  ):\n> DrawCayleyTable( G, cosets = H );\n\n\nHere is another example, in which the subgroup is not normal. It is the Sylow $2$-subgroup of the symmetric group $S_{4}$. Since the order of $S_{4}$ is equal to $24$, the Sylow $2$-subgroup has order $8$ (and index $3$), and each coset has the same size $8$ as the subgroup. These facts are clear from the picture.\n\nNevertheless, the block structure of the cosets is still quite visible.\n\n# Maple code for the second example:\n> G := Symm( 4 ):\n> DrawCayleyTable( G, cosets = SylowSubgroup( 2, G ) );\n\n\nIn each case, it is visually clear that the cosets of the subgroup form a \"regular\" partition of the group elements. The sizes of the blocks forming each coset are all identical, so they form a regular tiling of the Cayley table for the entire group.\n\nHope this helps!\n\n-\nThose are some very nice visualisations, thank you. I had no idea you could do this in Maple! \u2013\u00a0 gpo Mar 23 '13 at 20:04\n\nI think the key idea is that groups admit \"translations\", that is transformation of the form $\\left\\{ \\begin{array}{ccc} G & \\to & G \\\\ g & \\mapsto & h \\cdot g \\end{array} \\right.$ with $h \\in G$. Moreover, if $H$ is a subgroup of $G$, then the images of $H$ by two such translations are either equal or disjoint; therefore, you can cover $G$ by translating the subgroup $H$ to get a partition of $G$ into $n$ disjoint copies of $H$. You deduce that $|G|=n|H|$.\n\n-\nAny equivalence relation on a set $\\,S\\,$ induces a partition on $\\,S\\,$ (and vice versa). This is a special case where the $\\,n\\,$ equivalence classes have the same size $\\,k,\\,$ so $\\,|S| = nk,\\:$ so $\\,k\\,$ divides $|S|.\\,$ The result depends crucially on this key fact that the classes have the same size (which you do not mention). \u2013\u00a0 Math Gems Mar 21 '13 at 14:38\n@MathGems: I didn't give a proof but a way to understand Lagrange's theorem, and intuitively, a translation does not change the size. Moreover, implicitely a copy of $H$ has the same size than $H$. \u2013\u00a0 Seirios Mar 21 '13 at 14:45\n@Serios The word \"translation\" is used in many ways in mathematics, not all of which are set-theoretical bijections. In any case, from a pedagogical standpoint, it is always beneficial to bring to the fore those facts which play key roles. Hence my prior comment. \u2013\u00a0 Math Gems Mar 21 '13 at 14:49\n\nThe main thing is that there is a natural one-to-one correspondence between any two (say, left) cosets of the subgroup $H$, namely the mapping $$s\\mapsto yx^{-1}s$$ takes the coset $xH=\\{xh\\,\\mid h\\in H\\}$ to $yH$, and its inverse is $s\\mapsto xy^{-1}s$ (because $xy^{-1}yx^{-1}s=s$ for all $s$).\n\nSo, the size of each coset is the same ( $=|H|$), so, if $|G|$ is finite, $|H|$ must divide it. Stahl commented a nice link for visualizing this fact..\n\n-\n\nFor example If order of G is 12 then we can find only subgroups of order 1,2,3,4,6 and 12 which are divisiors of 12\n\nIt means that we can not find a subgroup of other order except above orders\n\n-", "date": "2015-01-28 22:43:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/336812/understanding-lagranges-theorem-group-theory", "openwebmath_score": 0.857200026512146, "openwebmath_perplexity": 331.93623547531104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639669551472, "lm_q2_score": 0.8918110368115781, "lm_q1q2_score": 0.8664514686990397}}
{"url": "http://accessdtv.com/taylor-series/taylor-polynomial-error.html", "text": "Home > Taylor Series > Taylor Polynomial Error\n\n# Taylor Polynomial Error\n\n## Contents\n\nSign in Share More Report Need to report the video? But, we know that the 4th derivative of is , and this has a maximum value of on the interval . We could have been a little clever here, taking advantage of the fact that a lot of the terms in the Taylor expansion of cosine at $0$ are already zero. And so when you evaluate it at a, all the terms with an x minus a disappear, because you have an a minus a on them. http://accessdtv.com/taylor-series/taylor-series-polynomial-error.html\n\nSo the error at a is equal to f of a minus P of a. Can we bound this and if we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do A Taylor polynomial takes more into consideration. But what I wanna do in this video is think about if we can bound how good it's fitting this function as we move away from a. look at this site\n\n## Taylor Series Approximation Error\n\nWhere this is an Nth degree polynomial centered at a. How well (meaning \u2018within what tolerance\u2019) does $1-x^2/2+x^4/24-x^6/720$ approximate $\\cos x$ on the interval $[{ -\\pi \\over 2 },{ \\pi \\over 2 }]$? We already know that P prime of a is equal to f prime of a.\n\nThis really comes straight out of the definition of the Taylor polynomials. http://mathinsight.org/determining_tolerance_error_taylor_polynomials_refresher Keywords: ordinary derivative, Taylor polynomial Send us a message about \u201cDetermining tolerance/error in Taylor polynomials.\u201d Name: Email address: Comment: If you enter anything in this field your comment will be I'll write two factorial. Lagrange Error Bound Calculator The system returned: (22) Invalid argument The remote host or network may be down.\n\nLoading... Taylor Series Remainder Calculator Thus, we have a bound given as a function of . The derivation is located in the textbook just prior to Theorem 10.1. But if you took a derivative here, this term right here will disappear, it'll go to zero.\n\nRating is available when the video has been rented. Lagrange Error Bound Formula Loading... And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x So this thing right here, this is an N plus oneth derivative of an Nth degree polynomial.\n\n## Taylor Series Remainder Calculator\n\nRather, there were two approaches taken by us to estimate how well it approximates cosine.\n\nWe might ask \u2018Within what tolerance does this polynomial approximate $\\cos x$ on that interval?\u2019 To answer this, we first recall that the error term we have after those first (oh-so-familiar) Taylor Series Approximation Error Sign in 6 Loading... Taylor Polynomial Approximation Calculator Basic Examples Find the error bound for the rd Taylor polynomial of centered at on .\n\nThat is, we're looking at Since all of the derivatives of satisfy , we know that . Check This Out Professor Leonard 99,296 views 3:01:45 Taylor Polynomials - Duration: 18:06. Theorem 10.1\u2003Lagrange Error Bound\u2003\u2003Let be a function such that it and all of its derivatives are continuous. Because the polynomial and the function are the same there. Taylor Series Error Estimation Calculator\n\nSo let me write that. A More Interesting Example Problem:\u2003Show that the Taylor series for is actually equal to for all real numbers . If is the th Taylor polynomial for centered at , then the error is bounded by where is some value satisfying on the interval between and . http://accessdtv.com/taylor-series/taylor-polynomial-error-function.html Thread navigation Calculus Refresher Previous: Prototypes: More serious questions about Taylor polynomials Next: How large an interval with given tolerance for a Taylor polynomial?\n\nThe main idea is this: You did linear approximations in first semester calculus. Taylor Remainder Theorem Proof Especially as we go further and further from where we are centered. >From where are approximation is centered. And what I wanna do is I wanna approximate f of x with a Taylor polynomial centered around x is equal to a.\n\n## Please try the request again.\n\nClose Yeah, keep it Undo Close This video is unavailable. So, I'll call it P of x. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... Error Bound Formula Statistics So it might look something like this.\n\nOf course not. Skip to main contentSubjectsMath by subjectEarly mathArithmeticAlgebraGeometryTrigonometryStatistics & probabilityCalculusDifferential equationsLinear algebraMath for fun and gloryMath by gradeK\u20132nd3rd4th5th6th7th8thHigh schoolScience & engineeringPhysicsChemistryOrganic chemistryBiologyHealth & medicineElectrical engineeringCosmology & astronomyComputingComputer programmingComputer scienceHour of CodeComputer animationArts patrickJMT 128,850 views 10:48 Calculus 2 Lecture 9.9: Approximation of Functions by Taylor Polynomials - Duration: 1:34:10. have a peek here And that polynomial evaluated at a should also be equal to that function evaluated at a.", "date": "2017-12-18 14:38:02", "meta": {"domain": "accessdtv.com", "url": "http://accessdtv.com/taylor-series/taylor-polynomial-error.html", "openwebmath_score": 0.8268483281135559, "openwebmath_perplexity": 967.946658199715, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9591542794197472, "lm_q2_score": 0.9032942073547148, "lm_q1q2_score": 0.8663985045593432}}
{"url": "http://openstudy.com/updates/561872cfe4b086e96bfe46f3", "text": "## anonymous one year ago Simplify 3 square root of 5 end root minus 2 square root of 7 end root plus square root of 45 end root minus square root of 28.\n\n\u2022 This Question is Open\n1. anonymous\n\n2. jim_thompson5910\n\nThe expression is this right? $\\large 3\\sqrt{5}-2\\sqrt{7}+\\sqrt{45}-\\sqrt{28}$\n\n3. anonymous\n\ndo you know how to solve it?\n\n4. jim_thompson5910\n\nfirst we need to simplify $$\\Large \\sqrt{45}$$\n\n5. jim_thompson5910\n\n$\\large \\sqrt{45}=\\sqrt{9*5}$ $\\large \\sqrt{45}=\\sqrt{9}*\\sqrt{5}$ $\\large \\sqrt{45}=3\\sqrt{5}$\n\n6. jim_thompson5910\n\nNotice how I factored it into 9*5 one of the factors is the largest perfect square factor possible\n\n7. jim_thompson5910\n\nmake sense?\n\n8. anonymous\n\nyes\n\n9. jim_thompson5910\n\nhow would you simplify $$\\Large \\sqrt{28}$$ ?\n\n10. anonymous\n\n2 square root of 7\n\n11. jim_thompson5910\n\ngood\n\n12. jim_thompson5910\n\nSo $\\large 3\\sqrt{5}-2\\sqrt{7}+\\sqrt{45}-\\sqrt{28}$ turns into $\\large 3\\sqrt{5}-2\\sqrt{7}+3\\sqrt{5}-2\\sqrt{7}$\n\n13. jim_thompson5910\n\nfrom here combine like terms\n\n14. anonymous\n\nok\n\n15. anonymous\n\ni dont really know how to do this part its confusing\n\n16. jim_thompson5910\n\nif I gave you something like 3x+7y+2x+10y, would you be able to combine like terms?\n\n17. anonymous\n\nyes\n\n18. anonymous\n\nit would be 5x + 17y right?\n\n19. jim_thompson5910\n\nyes that's correct\n\n20. jim_thompson5910\n\nyou'll use the same idea\n\n21. jim_thompson5910\n\nlet x = sqrt(5) and y = sqrt(7) $\\large 3\\color{red}{\\sqrt{5}}-2\\color{blue}{\\sqrt{7}}+3\\color{red}{\\sqrt{5}}-2\\color{blue}{\\sqrt{7}}$ $\\large 3\\color{red}{x}-2\\color{blue}{y}+3\\color{red}{x}-2\\color{blue}{y}$\n\n22. anonymous\n\nok\n\n23. jim_thompson5910\n\nsimplify 3x-2y+3x-2y to get ???\n\n24. anonymous\n\n6x-y?\n\n25. jim_thompson5910\n\n6x is correct -y is not\n\n26. anonymous\n\nok\n\n27. jim_thompson5910\n\ntry again\n\n28. anonymous\n\n29. jim_thompson5910\n\n-2y-2y is -4y, agreed?\n\n30. jim_thompson5910\n\n$\\large 3\\color{red}{x}-2\\color{blue}{y}+3\\color{red}{x}-2\\color{blue}{y}$ $\\large 6\\color{red}{x}-4\\color{blue}{y}$ $\\large 6\\color{red}{\\sqrt{5}}-4\\color{blue}{\\sqrt{7}}$\n\n31. jim_thompson5910\n\nSo, $\\large 3\\sqrt{5}-2\\sqrt{7}+\\sqrt{45}-\\sqrt{28} = 6\\sqrt{5}-4\\sqrt{7}$\n\n32. anonymous\n\nok thank a lot\n\n33. anonymous\n\ncan you help me with this one please? if you want to. Which statement is true about the difference 2 square root of 7 \u2212 square root of 28?\n\n34. jim_thompson5910\n\n$2\\sqrt{7} - \\color{red}{\\sqrt{28}} = 2\\sqrt{7} - \\color{red}{2\\sqrt{7}} = 0\\sqrt{7} = 0$", "date": "2017-01-19 10:52:14", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/561872cfe4b086e96bfe46f3", "openwebmath_score": 0.6555488705635071, "openwebmath_perplexity": 8248.313597693092, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964213735463, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8663959885620055}}
{"url": "https://www.themathdoctors.org/why-in-logic-does-false-imply-anything/", "text": "# Why, in Logic, Does \u201cFalse\u201d Imply Anything?\n\nIn a class on symbolic logic, students are taught the truth tables that define the \u201clogical connectives\u201d\u00a0\u2227 (and),\u00a0\u2228 (or), \u00ac (not), and\u00a0\u2192 (if\u00a0 \u2026 then). Everything makes sense until they are told that if p is false, then $$p\\rightarrow q$$ is true whether or not\u00a0q\u00a0is true. How can we say that \u201cIf pigs fly, then 2 is even\u201d is a true statement? Or, for that matter, \u201cIf pigs fly, then there\u2019s a kangaroo in my pocket\u201d? This is especially troublesome when students, naturally wanting brevity, read\u00a0$$p\\rightarrow q$$ as \u201cp implies q \u201c. How can nonsense imply anything?\n\n## A quick survey\n\nBefore I get into full answers, let\u2019s look at some examples of the question. First, in 1994, in the infancy of Ask Dr. Math, we got this question:\n\n8th Grade Logic\n\nWhy in a conditional statement if the \"p\" in the hypothesis is false, then the entire statement is true?  Why isn't it undecided?\n\nIn 1997, in answer to a more general question about logic, Doctor Mike included an answer to the question, knowing it is common:\n\nA False Statement Implies Any Statement\n\n... In case you still are wondering about why a False implies anything, try this explanation on for size.\n\nLike Doctor Ken in the 1994 answer, Doctor Mike here focused on an example, pointing out that if the condition of such a statement is false, then whatever happens, you couldn\u2019t be convicted of lying, because you made no promises about what would happen in any situation that actually happens. In 2004, Jay asked a follow-up question that was added to the same page:\n\nWhy are the logical statements \"false implies true\" and \"false implies false\" always considered \"true\"?\n\nI've read the previous note, but could you please give a more \"formal\" explanation?\n\nDoctor Schwa replied with an answer relating the idea to set theory:\n\nMore formally, I'd say \"implies\" means the same as \"subset\" in set theory.  That is, when you say, \"if it rains, then the ground gets wet,\" you mean, \"the set of times when it rains is a subset of the set of times when the ground gets wet.\"\n\nSo, since the empty set is a subset of any set, a false statement implies any statement.\n\nFormally, this is a good way to think of it; but it may not satisfy everyone \u2013 particularly since it is not obvious why the empty set can be a subset at all. (That\u2019s another question we get from time to time.)\n\nIn 2005, Doctor Achilles answered this one:\n\nLogic and Conditional Sentences\n\nI have a question about conditional statements.  I am having a hard time understanding why two false statements in a conditional makes it true.\n\nI tried to use different statements to create a truth table but I get stuck on the same concept.  I tried a sentence like \"If a polygon is a square, then the sides are equal.\"  If I assume a rectangle, then it seems to me that the statement is undefined.  Being true or false does not even apply.\n\nThis is much like the 1994 question, and is entirely reasonable. Be sure to read his answer; for the sake of space I will be focusing on two more recent questions whose answers include much of what others have said, while going beyond.\n\n## A deeper look\n\nThe Logic behind Conditional Statements\n\nI've read questions of the same title \"Why is \"false implies true/false\" always \"true\"?\" and I can understand the reason in the \"subset way\".\n\nCould you explain the reason just in the logic way?  Because when we say \"work the same as subsets\", we must prove it does work the same...\n\nWhen I trust the set theory, I must trust logic first.  So I want to understand the reason just on the way logic goes.\n\n\"False implies true/false\" is true...why?\n\nSo Keven doesn\u2019t like Doctor Schwa\u2019s approach, because it doesn\u2019t directly relate to logic. Having recently taught the subject and thought about this issue, I had several things to say.\n\n### 1. \u201cIf \u2026 then\u201d doesn\u2019t mean \u201cimplies\u201d\n\nFirst, don't use the word \"implies\" to talk about a conditional statement; A->B should be read merely as \"if A then B\".  \"Implies\" suggests a cause-and-effect relationship, or at least a logical connection of some sort.  But the conditional statement is not meant to suggest that (even though many examples given in texts look that way).  The statement \"if A then B\" says nothing more than \"if A is true, then B is true\"--not \"if A is true, then it CAUSES B to be true\".  It means that whenever we find that A is true, then we can know that B is true (or else A->B would have been false).\n\nThe problem is that we naturally tend to see a conditional statement as something more, because everyday usage leaks over into our logic, and the word \u201cimplies\u201d reinforces that tendency. Think of logical statements as merely observations about what things happen to be together, not about causality, or underlying reasons, or even necessary\u00a0connections (\u201cthese things always\u00a0go together\u201d). Although mathematicians can use the word \u201cimplication\u201d for this, it is misleading if you don\u2019t pay close attention to the definition.\n\nIn fact, a logical statement is not even an assertion that we make for some reason, so that we have some stake in its truth. It is just a statement that may be true or false in any given situation. If A is true but B is not, then the statement \u201cif A, then B\u201d is false, because if it were true, then B would have to be true.\n\nBut what if we don\u2019t have enough evidence to judge whether the statement is true?\n\n### 2. It is a mathematical definition, not everyday reasoning\n\nSecond, this conditional statement is in a sense just something mathematicians define for their own purposes, not something that necessarily agrees with the natural-language use of the phrase.  And what we need in logic (at least in traditional two-valued logic, as opposed to a logic that might include an \"undetermined\" value) is for every statement to be either true or false.  In this context, we have to choose what A->B will mean in every case--in order to define a complete truth table.\n\nSome of the earlier questioners felt that \u201cIf FALSE, then \u2026\u201d should just be \u201cundetermined\u201d or \u201cI don\u2019t know\u201d; and they are correct, in real life. But symbolic logic requires a logical truth value of T or F for every statement, so we don\u2019t have that option. We need some choice, which could be purely arbitrary (like rounding up on 5), or might have a specific reason based on how we plan to use it.\n\nHere I gave a familiar answer:\n\nTraditionally, mathematicians subscribe to a sort of \"innocent until proven guilty\" rule: we can't say something is false just because there is no evidence; instead, when there is no evidence of truth or falsity, we say it is true.  This is what lies behind the related facts about sets: we say that the null set is a subset of any set because there is no evidence that it is not--there is no element in the null set that is NOT an element of the other set!\n\nBut still, why not say \u201cguilty until proven innocent\u201d?\n\n### 3. This is what works in describing logical arguments\n\nUltimately, I realized, the reason for the choice comes from our application, not from the real world. One application of symbolic logic is to validate arguments, and here, if the truth value of a conditional statement were not defined as it is, then a valid argument would fail the test:\n\nLet's consider an argument like this:\n\nI have a cold.\nIf I have a cold, then my nose is running.\n------------------------------------------\nTherefore, my nose is running.\n\nThis has the form\n\nA\nA->B\n----\nB\n\nTo show that this is a valid argument, we write it as a single statement:\n\n((A) ^ (A->B)) -> B\n\nThat is, the whole argument is a big conditional.  If its truth table is ALWAYS true, no matter what the truth values of A and B are, then we consider the argument valid.\n\nThat is, an argument is considered valid if the equivalent statement is a tautology.\n\nI made truth tables for this (valid) argument, using every possible definition of the conditional, and showed that the only definition for which it becomes a tautology when it should is the accepted one.\n\nNow the truth table accurately reflects the validity of the argument. And that, I think, is why we make this definition: it works.\n\nWhy does it work?  Because we want to say an argument is valid when the conclusion follows from the premise: if A is really true, then B had better be true.  We DON'T CARE what happens if the premise is false; the argument is still valid because it doesn't tell you what happens then.  There's the meaning behind that \"innocent until proven guilty\" idea.\n\n## An example\n\nIn 2008, I answered another question, which gave me a chance to fill out a couple areas:\n\nLogic Statement False Implies True\n\nI am well familiar with the linguistic arguments which clarifies this somehow confusing concept.  Is there a deeper philosophical argument that touches on the underlying logic of this concept {logic axiom}?\n\nThe most disturbing thing about this logic axiom is that it eliminates {defeats} the logical contingency of the conclusion on the premise, which somehow goes against the very essence of logic! By layman definition, logic is something that allows \"naturally\" the consequence to flow from a premise.  When the same conclusion happens no matter what the premise is, the connectedness of the logic in between the conclusion and its premise loses significance, just like the definition of a function is defeated when one certain value from the domain point {maps} into two or more different values from the range.\n\nIf the moon is made of cheese, then I will go to the movie next week can rather best describe a sarcastic {insane} mode of thinking than a flow of \"natural\" logic especially when it can be said equally that if the moon is NOT made of cheese, I still go to the movie!  The dilemma of this concept, though I use it myself to prove some propositions like the empty set is a subset of every set, is that it kills the \"natural\" connectedness inherent in logic.\n\nUnfortunately, the very definition of logic itself is so intuitive and vague in the same way the set or sanity is defined, otherwise undefined!  Even though I trained myself to live with this concept and I use it in my formal proofs, I try to avoid using it as much as possible.  It is like employing proof by contradiction.  I would rather prove directly.\n\nAbe was thinking of conditional statements in cause-and-effect terms, as identifying \u201cnatural consequences\u201d. He thought of logic as philosophy, and his vague conception led to confusion. (Logic is considered a part of philosophy not because it is deep in itself, but because we need to think clearly when we get to the deep stuff.) I repeated many of the ideas in the previous answer (which had not yet been archived so I couldn\u2019t refer him to it): The conditional statement is not about cause-and-effect; the decision about truth value in the disputed cases depends on context; and one important context is judging the validity of an argument, which works if we take the \u201cinnocent until proven guilty\u201d approach.\n\nHe responded with a comment that overstated the conclusion:\n\nI found it rather interesting that there is indeed some philosophical underlying mode of thinking built into the definition of \"P implies Q,\" that is, if there is no evidence that something is false, then we must assume that it is true.  This we may cast as the \"default rule of the truth.\"\n...\nNow comes the interesting case which P is false and Q is true, yet we must assume that the implication is True ONLY for lack of better knowledge or evidence pointing to the other direction.  This is to me a philosophy or a mode of thinking which I accept as a sound one though it has some serious implications beyond math.\n\nSince,\u00a0as I have said, math has its own truth which need not agree with the real world, it doesn\u2019t tell us anything about deeper philosophical ideas: The definition of the conditional statement is not about truth in general, but just about what we want a conditional statement to mean. He also tried to restate my reasoning, which led me to clarify it with a fully stated example:\n\nLet's take an example.  I have a piece of paper here that is coated with a chemical that changes color.  I claim that if the paper is wet, it is red.  That is,\n\nWET -> RED\n\n(Note that I didn't say \"wet implies red\"; the word \"implies\", as I said before, is not really appropriate for this connective, as you'll see in a moment.)\n\nNow let's consider what you might see when I show you the paper, taking the four cases in your order.\n\n1. It's wet, and it's red.  That agrees with my statement, so you say my statement is true.  (You do not have enough evidence to conclude that my statement is ALWAYS true; you've just seen one case.  Maybe tomorrow it will be cooler, and you'll find that the paper is only red if it's wet AND warm.  That's why you can't say it's true that wet implies red, only that it is true in this instance that \"if it's wet, then it's red.\"  Do you see the difference?)\n\n2. It's dry, and it's blue.  You don't know that it would be red if it were wet; there's no evidence one way or the other.  So, simply by convention, you say that my statement is true, meaning that the evidence is consistent with that conclusion.  But you can't say that you've proved that wetness IMPLIES redness; all you can say is that it might.\n\n3. It's wet, and it's blue.  That disproves my statement; we have a case where it is wet but NOT red.  My statement is definitely false. (This case IS enough to disprove the stronger claim that wet implies red; you have a counterexample.)\n\n4. It's dry, and it's red.  Hmmm ... maybe it's ALWAYS red, and my statement was technically true but misleading; or maybe it's red for some other reason than wetness.  Or maybe it actually turns blue when it gets wet, and I just lied.  Again, you really don't know!  The evidence at hand deals only with the case where it's dry, and my statement is about what would be true if it were wet.  So you have to say that it's true, because you haven't disproved it, just like in case 2.\n\nSo your cases 2 and 4 are both \"true\" for the same reason, not for different reasons.  The evidence in both cases is consistent with my statement, so we call it true.\n\nHe concluded with a concise statement:\n\nIn sum, P implies Q is nothing more than a claim or a proposition.  We may uphold the rest of the logic table for P implies Q since the logic equivalence (truth value) for the remaining three cases does NOT contradict our claim about P implies Q, although not useful statements in some cases.  Thanks again for the great example.\n\nOne might say that \u201ctruth\u201d, in the sense we need here, just means \u201cnon-contradiction\u201d.\n\n### 2 thoughts on \u201cWhy, in Logic, Does \u201cFalse\u201d Imply Anything?\u201d\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-10-19 21:44:54", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/why-in-logic-does-false-imply-anything/", "openwebmath_score": 0.6290828585624695, "openwebmath_perplexity": 503.67029260826916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964222254889, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8663959861904834}}
{"url": "https://math.stackexchange.com/questions/1726580/probability-of-meeting-at-the-corner-of-city-streets-betwe", "text": "# Probability of meeting at the corner of city streets betwe\n\nTwo people agree to meet each other at the corner of two city streets between 1pm and 2pm. However, neither will wait for the other for more than 30 minutes. If each person is equally likely to arrive at any time during the one hour period, determine the probability that they will in fact meet.\n\nThis question is very similar to Probability of two people meeting during a certain time. But I'm afraid I'll have to ask again, because it's not quite what my query is about.\n\nI've tried letting X and Y be the two random variables, and they are independent of each other. I tried phrasing it in terms of $P(|X-Y|<30)$ but I don't know how to solve this, as i don't know the individual probability of either X and Y? Any advice would be greatly appreciated. Sorry in advance if I seem to be repeating the question (in link) and for any wrong title or tag labelling.\n\nI have also thought of another method:\n\nIf A arrives before B, then the probability that he arrives during first half an hour is 0.5 . He'll then wait for 0.5 hours. If He arrives during the last half an hour hours, with proability 0.5 . He then waits for 0.5 hours on average. Thus his total wait time is 0.5*0.5*2=0.5. For A and B to meet, B must arrive when A is waiting. Thus the probability that B arrived when A was waiting is 0.5 . Similarly if B arrives before A, the probability that they meet is 0.5 . Thus, Total probability that they meet is 1. Where did I go wrong? As the answer suggested is 0.75.\n\n\u2022 Aren't $X$ and $Y$ both uniformly distributed between $0$ and $60$? (And I assume it is intended that they arrive independently, though it's not stated explicitly in the problem description.) \u2013\u00a0Brian Tung Apr 3 '16 at 22:24\n\u2022 @BrianTung, However, if using uniform distribution, if I do invnorm on GDC, it doesn't mean that I can solve it? (as i don't know the mean and the variance?) Pls advise. \u2013\u00a0CCC Apr 3 '16 at 22:35\n\u2022 Maybe I'm misunderstanding the question. Why isn't this question like the one you link to? \u2013\u00a0Brian Tung Apr 3 '16 at 22:37\n\u2022 The variables are not normally distributed, so you cannot use mean and variance. You can integrate the probability density function over the region where the two people meet; since that function is constant and the region is more or less as depicted in that other question, the methods are essentially identical. \u2013\u00a0Brian Tung Apr 3 '16 at 22:48\n\u2022 Probability is 75% according the following simulation (in R). \"n <- 10^5 # number of trials a <- runif(n, min = 0, max = 60) b <- runif(n, min = 0, max = 60) sum(abs(a-b) < 30)/n\" \u2013\u00a0snoram Apr 3 '16 at 23:06\n\nThe R code below essentially repeats @snoram's nice simulation, but using only 50,000 points. Then I go on to plot the points in the square with vertices at $(0,0)$ and $(60,60)$, with the points corresponding to the condition that the two people meet plotted in light blue.\n\nFrom there, it is obvious that the two excluded regions (black) each have $1/8$th of the area. Because the joint distribution is uniform on the square this means that the probability of the 'condition' is $3/4.$\n\nOf course, you can draw the boundaries of the condition, and thus solve the problem, without simulation. (It is $not$ necessary to know the distribution of $|X - Y|$.) In integral notation, you need $\\int\\int_C 1/60^2\\,dx\\,dy,$ where $C$ is the region corresponding to the condition. You'd have to break $C$ up into two parts in order to set numerical limits on the integral signs. But I think the geometrical argument suffices.\n\n m = 50000\nx = runif(m, 0, 60);  y = runif(m, 0, 60)\ncond = (abs(x-y) < 30);  mean(cond)", "date": "2019-08-18 14:59:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1726580/probability-of-meeting-at-the-corner-of-city-streets-betwe", "openwebmath_score": 0.8641324639320374, "openwebmath_perplexity": 231.72757112140823, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211582993982, "lm_q2_score": 0.8887587875995483, "lm_q1q2_score": 0.8663808707765605}}
{"url": "https://mathhelpboards.com/threads/prove-that-a_n-tends-to-0.134/", "text": "Prove that a_n tends to 0\n\nAlexmahone\n\nActive member\nAssume $\\frac{a_{n+1}}{a_n}\\to L$, where $L<1$ and $a_n>0$. Prove that\n\n(a) $\\{a_n\\}$ is decreasing for $n\\gg 1$;\n\nI've done this part.\n\n(b) $a_n\\to 0$ (Give two proofs: an indirect one using (a), and a direct one.)\n\nIndirect proof: Assume for the sake of argument that $a_n$ does not tend to 0. Since the sequence is decreasing and bounded below by 0, it must converge to a positive value (call it $l$).\n\n$\\lim\\frac{a_{n+1}}{a_n}=\\frac{l}{l}=1$, a contradiction. So, $a_n\\to 0$.\n\nHow do I go about the direct proof (one that doesn't use \"proof by contradiction\")?\n\nLast edited:\n\nFernando Revilla\n\nWell-known member\nMHB Math Helper\nHow do I go about the direct proof (one that doesn't use \"proof by contradiction\")?\nHint\n\n$|a_{n}-0|<\\epsilon\\Leftrightarrow |a_{n+1}\\cdot (a_n/a_{n+1})|<\\epsilon \\Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\\epsilon$ and $a_{n+1}/a_n$ is bounded.\n\nAlexmahone\n\nActive member\nHint\n\n$|a_{n}-0|<\\epsilon\\Leftrightarrow |a_{n+1}\\cdot (a_n/a_{n+1})|<\\epsilon \\Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\\epsilon$ and $a_{n+1}/a_n$ is bounded.\nIs this a proof by induction that $|a_n|<\\epsilon$? If so, how do we prove the base case for $n=1$?\n\nLast edited:\n\nHallsofIvy\n\nWell-known member\nMHB Math Helper\nI would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.\n\nAlexmahone\n\nActive member\nI would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.\nI used a slightly different approach:\n\n$L<\\frac{L+1}{2}$\n\n$\\frac{a_{n+1}}{a_n}<\\frac{L+1}{2}$ for $n\\gg 1$ (Using the \"sequence location theorem\")\n\nSo, $a_{n+1}< \\left(\\frac{L+1}{2}\\right)a_n$ for $n\\ge N$\n\n$a_{N+k}<\\left(\\frac{L+1}{2}\\right)^k a_N$ for $k\\ge 1$ (Can be proved using induction over $k$.)\n\nSince $\\frac{L+1}{2}<1$, $a_{N+k}\\to 0$ as $k\\to\\infty$.\n\nSo, $a_n\\to 0$.\n\nLast edited:", "date": "2021-09-18 01:38:27", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/prove-that-a_n-tends-to-0.134/", "openwebmath_score": 0.9724230766296387, "openwebmath_perplexity": 512.227139899814, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9835969703688158, "lm_q2_score": 0.8807970685907242, "lm_q1q2_score": 0.8663493281755703}}
{"url": "https://math.stackexchange.com/questions/492740/real-skew-symmetric-3-times-3-matrix-has-one-eigen-value-2i", "text": "# Real Skew Symmetric $3\\times 3$ matrix has one eigen value $2i$\n\nReal Skew Symmetric $3\\times 3$ matrix has one eigen value $2i$ so another eigen value is $-2i$ another eigen value must be $0$ right? as we know these kind of matrix has eigen values $0$ or purely imaginary.\n\n\u2022 Yes. $\\quad\\mu^{*} = \\mu\\quad$ and $\\quad\\mu^{*} = -\\mu\\quad$ $\\Longrightarrow\\quad\\mu = 0$. \u2013\u00a0Felix Marin Sep 13 '13 at 19:50\n\nYes, the characteristic polynomial of a real $3\\times 3$ matrix will be a real cubic polynomial. If not identically zero, the real skew matrix will have one pair of conjugate imaginary eigenvalues and a zero eigenvalue, each of multiplicity one.\n\n\u2022 Does this eliminate the possibility of eigenvalues $2i,2i,-2i$? \u2013\u00a0Rebecca J. Stones Sep 13 '13 at 16:48\n\u2022 @Rebecca: Yes, it does (for a 3x3 real skew matrix). \u2013\u00a0hardmath Sep 13 '13 at 17:17\n\u2022 In case someone has doubts, the characteristic polynomial being a real cubic \"eliminates the possibility\" of three purely imaginary nonzero roots. Since these must occur in conjugate pairs, divding out a monic quadratic with one such pair as roots gives as quotient a real degree 1 polynomial having the third purely imaginary nonzero root (contradiction). \u2013\u00a0hardmath Sep 13 '13 at 21:50\n\nIn general, a real skew-symmetric $3 \\times 3$ matrix $K$ looks like\n\n$K = \\begin{bmatrix} 0 & a & b \\\\ -a & 0 & c \\\\ -b & -c & 0 \\end{bmatrix}, \\tag{1}$\n\nso that the characteristic polynomial is\n\n$\\det(\\lambda - K) = \\lambda^3 + (a^2 + b^2 + c^2) \\lambda, \\tag{2}$\n\nwhich is easily seen to have roots\n\n$\\lambda = 0, \\pm i\\sqrt{a^2 + b^2 + c^2}, \\tag{3}$\n\nwhich follow the general pattern the OP suggested; apparently the matrix she/he was thinking of satisfies $\\sqrt{a^2 + b^2 + c^2} = 2$; but $0$ will always be an eigenvalue for such $K$ in any case.\n\nHope this helps. Cheers.\n\nSince a real matrix $A$ is skew if, and only if, $A=-A^T$, since $\\text{spec} (A)=\\text{spec}(A^T)=-\\text{spec}(-A^T)$ and since the characteristic polynomial has degree $3$ and non-real roots come in pairs, it follows that $0\\in\\text{spec}(A)\\cap \\text{spec}(A^T)$.\n\nWe know that a skew symmetric matrix $A=(a_{ij})_{3 \\times 3}$ satisfies $A^T=-A$. This means that $a_{ii}=-a_{ii}$ or equivalently $a_{ii}=0$ for all $i \\in \\{1,2,3\\}$.\n\nWe know the trace of $A$ is the sum of its eigenvalues, which is $0$ since the main diagonal of $A$ comprises of zeroes.\n\nSince a $3 \\times 3$ matrix has $3$ (not necessarily distinct) eigenvalues (given by the roots of the characteristic polynomial [a degree $3$ polynomial with leading coefficient either $1$ or $(-1)^3$, depending on how it's defined]), there is precisely one other eigenvalue $\\lambda$ and we must have $2i-2i+\\lambda=0$, which implies $\\lambda=0$.\n\nHere's another way of looking at it, which I thought of after posting my previous answer.\n\nThe approach here seemed sufficiently different that I felt a separate answer was merited. Hope I'm not overdoing it. Anyway . . .\n\nFirst, note that for any skew-symmetric, real matrix, the only possible real eigenvalue is zero. For, if $K$ is real skew-symmetric, so that $K^T = -K$, and if $v \\ne 0$ is an eigenvector with real eigenvalue $\\lambda$, so that\n\n$Kv = \\lambda v, \\tag{1}$\n\nwe have\n\n$\\langle Kv, v \\rangle = \\lambda \\langle v, v \\rangle, \\tag{2}$\n\nbut\n\n$\\langle Kv, v \\rangle = \\langle v, K^Tv \\rangle = \\langle v, -Kv \\rangle = -\\langle Kv, v \\rangle, \\tag{3}$\n\nor\n\n$\\lambda \\langle v, v \\rangle = -\\lambda \\langle v, v \\rangle, \\tag{4}$\n\nand since $\\langle v, v \\rangle \\ne 0$ this shows that\n\n$\\lambda = - \\lambda, \\tag{5}$\n\nwhence\n\n$\\lambda = 0. \\tag{6}$\n\nHere I take $\\langle \\cdot, \\cdot \\rangle$ to be the standard inner product on the real vector space on which $K$ operates.\n\nNow argue as follows: since $2i$ is an eigenvalue of $K$, and the characteristic polynomial of $K$ is real, $-2i$ is an eigenvalue as well. This means\n\n$\\lambda^2 + 4 \\mid p_K(\\lambda), \\tag{7}$\n\nwhere\n\n$p_K(\\lambda) = det(\\lambda - K) \\tag{8}$\n\nis the characteristic polynomial of $K$. But $\\deg p_K(\\lambda) = 3$ in the present case, so the quotient polynomial $q(\\lambda)$ is of degree $1$, i.e. we must have\n\n$p_K(\\lambda) = (\\lambda^2 + 4)q(\\lambda) \\tag{9}$\n\nwith\n\n$q(\\lambda) = \\lambda - a, \\tag{10}$\n\n$a$ real. But this of course implies $a$ is a real eigenvalue of $K$, whence we must have\n\n$a = 0. \\tag {11}$\n\nWell, I hope this answer is more than mere mathematical logorrhea, and it sheds a little more light on the subject at hand.\n\nCheers!", "date": "2019-04-20 16:12:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/492740/real-skew-symmetric-3-times-3-matrix-has-one-eigen-value-2i", "openwebmath_score": 0.9357439875602722, "openwebmath_perplexity": 237.22287688534354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765546169713, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8663377769564783}}
{"url": "http://mathhelpforum.com/calculus/168398-limit-e-x-1-e-x-1-x-infinity.html", "text": "Math Help - Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity\n\n1. Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity\n\nHey all,\n\nI been playing around with limits, when i solve for the following function:\n\n$\\lim_{x\\rightarrow\\infty }\\frac{e^{x}-1}{e^{x}+1}=1$\nBut when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\\{e^{x}-1}=1+e^{x}$ ?. Thanks\n\n2. Originally Posted by Oiler\nHey all,\n\nI been playing around with limits, when i solve for the following function:\n\n$\\lim_{x\\rightarrow\\infty }\\frac{e^{x}-1}{e^{x}+1}=1$\nBut when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\\{e^{x}-1}=1+e^{x}$ ?. Thanks\n????\n\nIf you graph that, you obtain a line y=1\n\n3. Originally Posted by dwsmith\n????\n\nIf you graph that, you obtain a line y=1\ndwsmith, sorry just made some changes to the formula.\n\n4. Originally Posted by Oiler\ndwsmith, sorry just made some changes to the formula.\nIt is -1 when you go to negative infinity.\n\n5. Originally Posted by Oiler\nalso how does $\\{e^{x}-1}=1+e^{x}$ ?. Thanks\nLet x = 0\n\n0 = 2 is that true?\n\n6. ofcourse, how dim of me. Thanks dwsmith.\n\n7. $\\displaystyle\\lim_{x\\to +\\infty}\\frac{e^x-1}{e^x+1}=\\frac{\\infty}{\\infty}$\n\nApplying L'Hopitals Rule\n$\\displaystyle\\Rightarrow\\lim_{x\\to +\\infty}\\frac{e^x}{e^x}=1$\n\n8. Originally Posted by dwsmith\n$\\displaystyle\\lim_{x\\to +\\infty}\\frac{e^x-1}{e^x+1}=\\frac{\\infty}{\\infty}$\n\nApplying L'Hopitals Rule\n$\\displaystyle\\Rightarrow\\lim_{x\\to +\\infty}\\frac{e^x}{e^x}=1$\nL'H\u00f4pital's Rule is overkill for this. XD\n\n$\\lim\\limits_{x\\to\\infty}\\dfrac{e^x-1}{e^x+1} = \\lim\\limits_{x\\to \\infty}\\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\\rightarrow0$ as $x\\rightarrow\\infty$.\n\n9. Originally Posted by Chris L T521\nL'H\u00f4pital's Rule is overkill for this. XD\n\n$\\lim\\limits_{x\\to\\infty}\\dfrac{e^x-1}{e^x+1} = \\lim\\limits_{x\\to \\infty}\\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\\rightarrow0$ as $x\\rightarrow\\infty$.\nSometimes you just have to kill it.\n\n10. $\\displaystyle \\lim_{x \\to \\infty}\\frac{e^x - 1}{e^x + 1} = \\lim_{x \\to \\infty}1 - \\frac{2}{e^x + 1}$\n\n$\\displaystyle = 1 - 0$\n\n$\\displaystyle = 1$.\n\n11. Originally Posted by dwsmith\nSometimes you just have to kill it.\nBut you never have to \"over-kill\". When you are showing someone how to do something, the simplest way is always best.\n\n12. Originally Posted by Oiler\nHey all,\n\nI been playing around with limits, when i solve for the following function:\n\n$\\displaystyle\\lim_{x\\rightarrow\\infty }\\frac{e^{x}-1}{e^{x}+1}=1$\n\nBut when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations?\n\nalso how does $\\{e^{x}-1\\}=\\{1+e^{x}\\}$ ?.\n\nThanks\nFor this fraction, the denominator is always 2 greater than the numerator.\n\nHowever, the \"ratio\" gets closer and closer to 1, as we increase x above 0.\n\n$x\\rightarrow\\infty$ means x increases without bound above 0.\n\nHence, as x \"approaches infinity\", you need infinitely many decimal places\nto express the value of the fraction as a value \"other than 1\".\n\nThat's part of the concept of limits.\n\nviz-a-viz\n\n$\\displaystyle\\frac{e^1-1}{e^1+1}=0.46211715726$\n\n$\\displaystyle\\frac{e^2-1}{e^2+1}=0.76159415596$\n\n$\\displaystyle\\frac{e^{10}-1}{e^{10}+1}=0.99990920426$\n\n....onward.\n\nIf $x<0$ and decreases without bound below 0.\n\n$x=-y$\n\n$x\\rightarrow\\ -\\infty\\Rightarrow\\ y\\rightarrow\\infty$\n\n$\\displaystyle\\lim_{x \\to -\\infty}\\frac{e^x-1}{e^x+1}=\\lim_{y \\to \\infty}\\frac{e^{-y}-1}{e^{-y}+1}=\\lim_{y \\to \\infty}\\frac{\\frac{1}{e^y}-1}{\\frac{1}{e^y}+1}$\n\n$\\displaystyle\\lim_{y \\to \\infty}\\left[\\frac{\\frac{1}{e^y}}{\\frac{1}{e^y}}\\right]\\;\\frac{1-e^y}{1+e^y}\\right]=-\\lim_{y \\to \\infty}\\frac{e^y-1}{e^y+1}=-1$\n\nWhen you have the concept, you can apply the fast methods as you please later.\n\n13. Originally Posted by HallsofIvy\nBut you never have to \"over-kill\". When you are showing someone how to do something, the simplest way is always best.\nTo me, that was mighty simply. The constant disappear and the exponentials are their same derivatives.", "date": "2015-04-26 03:57:20", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/168398-limit-e-x-1-e-x-1-x-infinity.html", "openwebmath_score": 0.8911778330802917, "openwebmath_perplexity": 1026.5947235385722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765552017674, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8663377744794857}}
{"url": "http://math.stackexchange.com/questions/62574/analytical-reasoning-question-ii", "text": "# Analytical Reasoning Question II\n\nI have yet another analytical question that got me\n\nA five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?\n\nPlease kindly explain your final answer so those of us who are still trying to learn can grasp the logic easily. Thanks in advance!\n\n-\n\nThe number of possible combinations is $5! = 120$. Each digit will be in each position an equal number of times, namely $5!/5 = 24$ times.\n\nSo the sum of the digits in the one's place is\n\n$24(1+3+5+7+9) = 24*25 = 600$\n\nThis will be the sum of the digits in each place (one's, ten's, etc.). So the total sum is:\n\n$600*11,111 = 6,666,600$, regards, iyengar\n\n-\nbut let me ask,is my answer right ??,as i am not a person with formal training, \u2013\u00a0Iyengar Sep 7 '11 at 15:32\nWhy do you need this line? $24(1+3+5+7+9) = 24*25 = 600$. Why sum the numbers? I don't know the answer yet(trying not to look in the back of the book for the answers until I am sure I have this right).:) \u2013\u00a0user10695 Sep 7 '11 at 15:34\nMy question is, why add the individual allowable digits, rather than just adding the total occurrence for each? I was thinking more something like this $24(5) = 24*5 = 120$ \u2013\u00a0user10695 Sep 7 '11 at 15:35\nFor each decimal place, 24 of these are of the same digit. Thus for a decimal place the total value is 24 (1 + 3 + 5 + 7 + 9) = 600 To get the total value of all these numbers multiply the first result by 11111. \u2013\u00a0Iyengar Sep 7 '11 at 15:37\n\nAs a slight alternative to iyengar's answer:\n\nThere are $120$ such numbers and their average is $55555$ (note that for any of the form, there is also $111110$ minus that number) so the answer is $120\\times 55555=6666600$.\n\n-\nwhere did you get 55555 from? The question specifies that there are no repeats allowed. \u2013\u00a0user10695 Sep 7 '11 at 16:15\nIf you have for example $17935$ then you also have $111110-17935=93175$, so the average term is half $111110$, namely $55555$. \u2013\u00a0Henry Sep 7 '11 at 16:23\nI am sorry, I still don't get the 111110 and the 55555. Why those? \u2013\u00a0user10695 Sep 7 '11 at 19:30\n\nThis is partly a slight variant of Henry\u2019s answer and partly a further explanation in answer to one of your questions.\n\nIf $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. Now let $abcde$ be any allowable number; then $a'b'c'd'e'$ is also allowable, and it must be a different number from $abcde$. (It would be the same number only if all of the digits were $5$, but that\u2019s not allowed.) In this way you can pair up all of the allowable numbers. Since there are $5! = 120$ permutations of the $5$ allowable digits, there are $120$ allowable numbers, and hence there are $60$ of these pairs.\n\nIt shouldn\u2019t be too hard to see that $abcde + a'b'c'd'e' = 111110$ no matter which allowable number $abcde$ you start with. That is, the numbers in each pair sum to $111110$. Since there are $60$ pairs, the grand total is $60 \\cdot 111110 = 6,666,600$.\n\nHenry did essentially the same thing, except that instead of looking at the sum of $abcde$ and $a'b'c'd'e'$, he looked at their average, $111110/2 = 55555$. Since all pairs have the same average, the entire collection of allowable numbers must also have that average. And if the $120$ allowable numbers have an average of $55555$, their total must be $120 \\cdot 55555 = 6,666,600$.\n\n-\nI need help seeing this [If $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. ] please. Why is $10-d$ allowable? \u2013\u00a0user10695 Sep 7 '11 at 22:04\nAlso, how is $abcde + a'b'c'd'e' = 111110$. I don't get that please. Is this binary? \u2013\u00a0user10695 Sep 7 '11 at 22:06\n@user10695: 1st question: Just do the arithmetic: if $d=1$, $10-d=9$; if $d=3$, $10-d=7$; and so on. Subtracting $1,3,5,7$, or $9$ from $10$ leaves $9,7,5,3$, or $1$, respectively. 2nd question: Not binary, just ordinary addition. $e+e'=10$, so you write down $0$ and carry $1$. Then $d+d'=10$, and the carry makes $11$, so you write down $1$ and carry $1$. The same thing happens in the remaining three columns, so you end up with a total of $111110$. \u2013\u00a0Brian M. Scott Sep 7 '11 at 22:15\nThanks! What I thought $abcde + a'b'c'd'e' = 111110$ meant was $(a*b*c*d*e) + (a'*b'*c'*d'*e')$ . Now I see what you mean. Thanks for that. \u2013\u00a0user10695 Sep 7 '11 at 22:33\n\nwhy all this ?\n\nit is so simple There are 5 digits and as we all know it will 5! numbers (1*2*3*4*5)=120 on a digit place every number will repeat 24 times that is 120/5 on ones place the sum will be 24(1+3+5+7+9)=25*24=600 If u need further explanation why 24(1+3+5+7+9+) see this: at every digit place 1 will repeat 24 times 24*1=24 3 will repeat 24 times 24*3=72 5 will repeat 24 times 24*5=120 7 will repeat 24 times 24*7=168 9 will repeat 24 times 24*9=216 total give 600\n\nevery digit place will sum to 600 unit place 600 tenth place 600 hundreds place 600 Thousands place 600 ten thousands place 600 -------------------------------- 6666600\n\n-\nUse MathJax please. \u2013\u00a0SchrodingersCat Oct 28 '15 at 16:30", "date": "2016-06-29 23:48:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/62574/analytical-reasoning-question-ii", "openwebmath_score": 0.8847867250442505, "openwebmath_perplexity": 298.54043374806076, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765587105447, "lm_q2_score": 0.8840392741081574, "lm_q1q2_score": 0.86633776560548}}
{"url": "https://www.physicsforums.com/threads/solving-y-intercept-of-a-sin-function.427325/", "text": "Homework Help: SOlving y-intercept of a sin function\n\n1. Sep 8, 2010\n\nt_n_p\n\n1. The problem statement, all variables and given/known data\n\ny = -2sin2(x+[pi/6])+1 for x [-pi, pi]\n\n3. The attempt at a solution\n\nset y=0,\n\n1/2 = sin2(x+[pi/6])\npi/6 = 2(x+[pi/6]), where pi/6 is the base angle\n\nNow because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi].\npi/6 is positive, and sin is positive in 1 & 2 quadrants.\n\ntherefore,\n2(x+[pi/6]) = pi/6, 5pi/6, -7pi/6, -11pi/6.\n\ndivide by 2:\n(x+[pi/6]) = pi/12, 5pi/12, -7pi/12, -11pi/12.\n\nsubtract [pi/6] from both sides:\nx = -pi/12, 3pi/12, -9pi/12, -13pi/12.\n\nThis is my problem. Is -13pi/12 an issue since it is outside the original domain, or is it still ok since it is inside the modified domain [-2pi,pi]?\n\nthe answer has x values of -pi/12, 3pi/12, -9pi/12, 11pi/12, so my only issue is the last value. What has gone wrong!?\n\n2. Sep 8, 2010\n\nCompuChip\n\n-13pi/12 is a solution to the equation y = 0, but you are only asked to give the solutions between -pi and pi. However, if you add (or subtract) any multiple of 2pi to -13pi/12, you get an equivalent solution. In this case, you could find -13pi/12 + 2pi = 11pi/12, which is in the requested interval.\n\n3. Sep 8, 2010\n\nt_n_p\n\nok, so I ask this, is there anyway to get 11pi/12 using the method I described, or can it only be obtained by adding 2pi?\n\nso hypothetically, if I wanted x-intercepts for all x,\nI would do:\n-pi/12 +/- n*2pi\n3pi/12 +/- n*2pi\n-9pi/12 +/- n*2pi\n-13pi/12 +/- n*2pi\n\nwhere n is any integer?\n\nI'm also wondering about the validity of adding pi (rather than 2pi). The period in our case is 2pi/2 = pi, so why doesn't adding/subtracing pi from all our values yield another number of solutions?\n\n4. Sep 8, 2010\n\nCompuChip\n\nThe important thing in this type of problem, is to do everything in the right order.\n\nStep 1: finding the base angle and writing down two solution sets\nAt the beginning of your post, you get to the above equation,\n$$\\sin 2(x + \\pi/6) = \\frac12 = \\sin \\pi/6$$\nThis equation has two \"branches\" of solutions, namely\n$$2(x + \\pi/6) = \\pi / 6 + n \\cdot 2\\pi$$\nand\n$$2(x + \\pi/6) = \\pi - \\pi / 6 + n \\cdot 2\\pi$$\nfor any integer n (= 0, 1, 2, 3, ..., -1, 2, -3, ...).\n[If you draw the graph of the sine function or you look at its geometric meaning in a unit circle, you will see how the symmetry leads to the second equation].\n\nYou have to add the n*2pi here, because you want the sines of both sides to be equal, and if you add any multiple of 2 pi to any of them, this will be the case.\n\nStep 2: solving for x\nNow you can go and solve the two equations for x. The first one leads to\n$$x + \\pi/6 = \\pi / 12 + n \\cdot \\pi$$\n(note that nothing strange happens here, it is just basic algebra: if you divide everything by 2, then n*2pi changes into n*pi automatically. As you remarked, this corresponds to the period) and\n$$x = \\pi / 12 - \\pi / 6 + n \\cdot \\pi = - \\pi/12 + n \\cdot \\pi$$\nFor the second branch you will get something similar, which I will leave up to you to work out (be careful with all the minus signs though)\n\nStep 3: Find the specific solutions requested\nYou now have two equations which describe infinitely many solutions, although basically there are just two and all the others differ from them by an integer number of 2pi steps.\nIn this case, you are specifically asked to list only the solutions between -pi and pi. So you can go and plug in some numbers for n into\n$$x = - \\pi/12 + n \\cdot \\pi$$\nFor n = 0 you get - pi / 12 which is in the interval, so you can write that down. For n = 1 you get - pi / 12 + pi = 11 pi / 12, that is also OK. For n = 2 you go above pi, and for n = -1 you are below -pi, so this is all the solutions you get.\n\nDo the same for the other \"branch\" of solutions, and you will find two more.\n\n5. Sep 8, 2010\n\nt_n_p\n\nok, that makes sense, but is this also feasible....\n\nFollow what I did in my original post, where I found 4 solutions:\n\nx = -pi/12, 3pi/12, -9pi/12, -13pi/12.\n\nWhy can't I just add multiples of the period to each of these values?\n\nFor x = -pi/12\nx=-pi/12 + n*pi\ntest:\nfor n=-1, x=-13pi/12 (outside domain, invalid)\nfor n=0, x = -pi/12 (inside domain, valid)\nfor n=1, x= 11pi/12 (inside domain, valid)\nfor n=2, x= 23pi/12 (outside domain, invalid)\n\nFor x = 3pi/12\nx=3pi/12 + n*pi\ntest:\nfor n=-2, x= -21pi/12 (outside domain, invalid)\nfor n=-1, x=-9pi/12 (inside domain, valid)\nfor n=0, x=3pi/12 (inside domain, valid)\nfor n=1, x= 15pi/12 (outside domain, invalid)\n\nFor x = -9pi/12\nx=-9pi/12 + n*pi\ntest:\nfor n=-1, x=-21pi/12 (outside domain, invalid)\nfor n=0, x=-9pi/12 (inside domain, valid)\nfor n=1, x= 3pi/12 (inside domain, valid)\nfor n=2, x=15pi/12 (outside domain, invalid)\n\nFor x = -13pi/12\nx=-13pi/12 + n*pi\ntest:\nfor n=-1, x=-25pi/12 (outside domain, invalid)\nfor n=0, x=-13pi/12 (outside domain, valid)\nfor n=1, x= -1pi/12 (inside domain, valid)\nfor n=2, x=11pi/12 (inside domain, valid)\nfor n=3, x=23pi/12 (outside domain, invalid)\n\nthen compiling all solutions that are valid I'm left with x=-9pi/12, 3pi/12, -pi/12, 11pi/12\nas expected.\n\nA little more time consuming but still works. My only question with the above method is, if the period is not pi, for example it is 2pi, will the same method work? I.e. if the period is pi, will adding n*2pi still provide the answers I am after?\n\n6. Sep 8, 2010\n\nCompuChip\n\nYes, that also works. But where do you get these four solutions from in the first place? Are they guesses or what?\n\n7. Sep 8, 2010\n\nt_n_p\n\nI explained in the original post.\n\nNow because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi]. I know the base angle is pi/6 and I know sin is positive in the 1st and 2nd quadrants. There will be 2 solutions per a revolution, then since there are 2 revolutions (domain is now -2pi to 2pi), there will be a total of 4 solutions.\n\nThat's the way I was taught. Since there is a factor of 2 out the front, I double the original domain. If there is a factor of 3 out the front, I triple the original domain so that it becomes -3pi to 3pi yielding a total of 6 solutions and so and so forth\n\n8. Sep 8, 2010\n\nCompuChip\n\nAh, I see now. So to be entirely clear:\n\n* If you already find four solutions this way, then you don't need to work everything out like in post #5. All you have to do is shift solutions outside the requested range by an integer number of periods (in this case, pi).\n\n* Whenever you want to do the n * (2)pi thing: if you already divided everything by 2, you use multiples of pi. You only shift by 2pi, if you haven't divided by the 2 (or 3, or whatever number) in front yet.\n\n9. Sep 8, 2010\n\nt_n_p\n\nI definately prefer to do it the 1st star method.\nBasically then, when I get a solution outside the bounds, I add or subtract 1 PERIOD (being careful to note what the period is, as it will change from problem to problem) to bring it back inside the bounds. E.g. if period is 500pi, then add/subtract 500pi from any values outside of the required bounds to bring it back inside.\n\nThis particular case had me stumped, since 3 of the solutions appear within the bounds, and only 1 is invalid. Obviously this is due to the value of the phase shift relative to the values obtained from the unit circle.\n\nAll makes sense now. Thanks", "date": "2018-06-20 23:10:59", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/solving-y-intercept-of-a-sin-function.427325/", "openwebmath_score": 0.7541785836219788, "openwebmath_perplexity": 1615.3471035095438, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854111860906, "lm_q2_score": 0.8933094039240554, "lm_q1q2_score": 0.8663184276008916}}
{"url": "https://math.stackexchange.com/questions/1050957/finding-the-probability-of-a-selecting-at-least-1-of-an-element", "text": "Finding the probability of a selecting at least 1 of an element.\n\nIf there are 18 red and 2 blue marbles what is the probability of selecting 10 marbles where there is either 1 or 2 blue marbles in selected set.\n\nAlso, it seems intuitive that the probability should be twice that of selecting a single blue marble in a set of 10 from 19 red and 1 blue, but I'm not sure.\n\nAnother approachment is through hypergeometric distribution.\n\n\u2022 The probability that there is exactly one blue ball in the selected set is: $$P(A_1)=\\dfrac{\\color{blue}{\\dbinom 2 1}\\cdot \\color{red}{\\dbinom {18} {10-1}}}{\\dbinom {20}{ 10} }$$\n\n\u2022 The probability that there are exactly 2 blue balls in the selected set is:\n\n$$P(A_2)=\\dfrac{\\color{blue}{\\dbinom 2 2}\\cdot \\color{red}{\\dbinom {18} {10-2}}}{\\dbinom {20}{ 10} }$$\n\nSo, the probability you are looking for is $P(A_1)+P(A_2)\\approx 0.763$\n\nIn this scenario, with no restrictions, there are 3 possible cases:\n\n$\\bullet$ When selecting 10, there were no blue marbles ($A_{none}$)\n\n$\\bullet$ When selecting 10, there was one blue marble ($A_1$)\n\n$\\bullet$ When selecting 10, there were two blue marbles ($A_2$)\n\nAs there are no other possibilities, it stands to reason then that these three events form a partition of our overall sample space. That is: $X = A_{none}\\cup A_1 \\cup A_2$ and each $A$ is pairwise disjoint from the others.\n\nIt follows then, that $1 = P(X) = P(A_{none}) + P(A_1) + P(A_2)$ by rule of sums.\n\nThe event you are interested in is that either one or two blue marbles were selected, namely $A_1\\cup A_2$.\n\nBy the above, $P(A_1\\cup A_2) = 1 - P(A_{none})$\n\nTo calculate $P(A_{none})$, for each of the 10 marbles selected, you will have selected red. So, multiply the probabilities for each step that you in fact picked red.\n\nSo: $P(A_{none}) = \\dfrac{18\\cdot 17\\cdot 16\\cdot 15\\cdot 14\\cdot 13\\cdot 12\\cdot 11\\cdot 10\\cdot 9}{20\\cdot 19\\cdot 18\\cdot 17\\cdot 16\\cdot 15\\cdot 14\\cdot 13\\cdot 12\\cdot 11} = \\dfrac{10\\cdot 9}{20\\cdot 19}$\n\nAnd finally $P(A_1\\cup A_2) = 1 - P(A_{none}) = 1 - \\frac{10\\cdot9}{20\\cdot 19}\\approx 0.763$\n\nIn comparison, if there were 19 red and 1 blue, it winds up being $1 - \\frac{10}{20} = 0.5$ The answer to the original question is in fact a bit more than 50% more likely.\n\n\u2022 The answer you gave is .864 but I believe it is supposed to be .763. I tried to change it but it said the change need to be at least 6 characters.\n\u2013\u00a0qw3n\nDec 4, 2014 at 14:30\n\u2022 absolutely correct. My mistake, a little arithmetic mistake. I had originally calculated 9/(2*19) and forgot to subtract it from 1. I did the subtraction from 1 in my head after and somehow messed up the tenths digit. Thanks for catching that. Dec 4, 2014 at 14:39\n\nGiven $18$ red marbles and $2$ blue marbles:\n\n\u2022 The number of ways to choose $10$ marbles is $\\binom{20}{10}=184756$\n\n\u2022 The number of ways to choose $10$ red marbles is $\\binom{18}{10}=43758$\n\n\u2022 Hence the probability to choose only red marbles is $\\frac{43758}{184756}=\\frac{9}{38}$\n\n\u2022 Hence the probability to choose not only red marbles is $1-\\frac{9}{38}\\approx76.31\\%$\n\nGiven $19$ red marbles and $1$ blue marble:\n\n\u2022 The number of ways to choose $10$ marbles is $\\binom{20}{10}=184756$\n\n\u2022 The number of ways to choose $10$ red marbles is $\\binom{19}{10}=92378$\n\n\u2022 Hence the probability to choose only red marbles is $\\frac{92378}{184756}=\\frac{1}{2}$\n\n\u2022 Hence the probability to choose not only red marbles is $1-\\frac{1}{2}=50\\%$\n\nAs you can see, the former is not twice the latter...", "date": "2022-06-24 23:26:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1050957/finding-the-probability-of-a-selecting-at-least-1-of-an-element", "openwebmath_score": 0.9377807974815369, "openwebmath_perplexity": 214.1585444198736, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815538985514, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8663123452998192}}
{"url": "https://math.stackexchange.com/questions/3353057/total-number-of-subsets-of-size-atmost-k", "text": "Total number of subsets of size atmost $k$\n\nI was working on a problem that involved taking subsets of a multiset. I want to count the total number of distinct subsets of size at most $$k$$.\n\nExample 1:\n\nconsider the multiset $$S = \\{ 3, 3, 5, 7 \\}$$ and $$k=3$$ then the answer should be $$12$$.\n\nThat is $$\\{ \\}, \\{ 3 \\}, \\{ 3 \\}, \\{ 5 \\}, \\{ 7 \\}, \\{ 3, 5 \\}, \\{ 3, 7 \\}, \\{ 3, 5 \\}, \\{ 3, 7 \\}, \\{ 5, 7 \\}, \\{ 3, 5, 7\\}, \\{ 3, 5, 7 \\} = 12$$ subsets.\n\nExample 2:\n\n$$S = \\{ 2, 3, 5 \\}$$ and $$k=2$$ then the answer should be $$7.$$\n\n$$\\{ \\}, \\{ 2 \\}, \\{ 3 \\}, \\{ 5 \\}, \\{ 2, 3 \\}, \\{ 2, 5 \\}, \\{ 3, 5 \\} = 7$$ subsets\n\nBy using the formula from this answer I can count the total number of distinct subsets. How can I extend that formula for my constraints? Any idea?\n\n\u2022 For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. \u2013\u00a0Thomas Shelby Sep 11 '19 at 17:04\n\u2022 sorry for that.please edit my question if i make any mistake. \u2013\u00a0Midhun Manohar Sep 11 '19 at 17:07\n\u2022 I do not follow your first example. Why do include $\\{3\\}$ twice? Why do you include $\\{3,5\\}$ twice? Why also did you not include $\\{3,3\\},\\{3,3,5\\}$ or $\\{3,3,7\\}$? If you intend for each $3$ to be considered distinct... then why do you use identical characters for them? It would have made more sense to never consider multisets in the first place and instead talk about the subsets of size at most $3$ from the set $\\{3,\\color{red}{X},5,7\\}$ instead where we replaced the second three with $X$... at which point you have $\\sum\\limits_{i=0}^k\\binom{n}{k}$ such subsets. \u2013\u00a0JMoravitz Sep 11 '19 at 18:33\n\u2022 elements of the subset should be distinct \u2013\u00a0Midhun Manohar Sep 12 '19 at 5:04\n\nLet me restate the problem, and see if you agree that it is the same thing. We have a finite collection of elements of types $$1,2,\\dots,n$$ with $$a_j$$ elements of type $$j$$ for $$j=1,2,\\dots,n$$, and we wish to know how many ways we can select at most $$k$$ objects, subject to the condition that no more than one element of any type is selected.\n\nIn your first example, we have $$2$$ elements of type \"$$3$$\" and one element of each of types \"$$5$$\" and \"$$7$$\".\n\nExactly $$k$$ elements may be selected in $$\\sum a_{j_1}a_{j_2}\\cdots a_{j_k}$$ ways, where the sum is over all $$k$$-tuples $$(j_1,j_2,...j_k)$$ with $$1\\leq j_1\n\nIf you want at most $$k$$ objects, just add up the values for each nonnegative integer $$\\leq k$$.\n\nIn you first example, with $$n=3$$, we have $$a_1=2,a_2=1,a_3=1$$. When $$k=0$$ we have an empty product, so the value is $$1$$. When $$k=1$$, we get $$2+1+1=4.$$ When $$k=2$$, we get $$2\\cdot1+2\\cdot1+1\\cdot1=5$$. When $$k=3$$, we get $$2\\cdot\\cdot1=2$$. Altogether, we have $$1+4+5+2=12.$$\n\nI gave a more concrete answer to a similar question a few days ago. Look at Find number of ways to select subset with distinct objects of at most K size..\n\n\u2022 it seems working.can you elaborate more \u2013\u00a0Midhun Manohar Sep 11 '19 at 17:30\n\u2022 @Midhunmanohar What is it that you don't understand? \u2013\u00a0saulspatz Sep 11 '19 at 17:32\n\u2022 i have never seen use of multiplier symbol \u220f like this.can you please elaborate the formula more \u2013\u00a0Midhun Manohar Sep 12 '19 at 6:06\n\u2022 @Midhunmanohar That was my fault. I meant sum, not product. No wonder you didn't understand. I don't know where my mind was. Anyway, it's the sum over all such products. I've edited my answer with a link to another answer that may be easier to follow. \u2013\u00a0saulspatz Sep 12 '19 at 10:47", "date": "2020-01-29 17:42:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3353057/total-number-of-subsets-of-size-atmost-k", "openwebmath_score": 0.6964215040206909, "openwebmath_perplexity": 287.0101700920343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.981735725388777, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.866310969165687}}
{"url": "http://love.thegoodness.com/docs/winsor-and-newton-lana-paper-0dca71", "text": "Explain. In this case, we use $$j$$ for the index for the summation, and the notation $$\\sum_{j=1}^n j^2$$ tells us to add all the values of $$j^2$$ for $$j$$ from 1 to $$n$$, inclusive. For every natural number $$n$$, $$5^n \\equiv 1$$ (mod 4). For another example, for each natural number $$n$$, we now let $$Q(n)$$ be the following open sentence: $1^2 + 2^2 + ... + n^2 = \\dfrac{n(n + 1)(2n + 1)}{6}.$. We resolve this by making Statement (1) an axiom for the natural numbers so that this becomes one of the defining characteristics of the natural numbers. \\ \\ \\ \\ \\ &P(3)&\\ \\ \\ \\ \\ \\ \\ \\ \\ &is& \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 1^2 + 2^2 + 3^2 &=& \\dfrac{3 \\cdot 4 \\cdot 7}{6} So let $$k$$ be a natural number and assume that $$P(k)$$ is true. That is, is 2 in the truth set of $$P(n)$$? Principle of Mathematical Induction Solution and Proof. First principle of mathematical induction For each natural number $$n$$, $$1 + 4 + 7 + \\cdot\\cdot\\cdot + (3n - 2) = \\dfrac{n(3n -1)}{2}.$$, We will prove this proposition using mathematical induction. Then to determine the validity of P(n) for every n, use the following principle: Check whether the given statement is true for n = 1. Which of the following sets are inductive sets? Then P(n) is true for all positive integers n. = (n + 1)! One way of proving statements of this form uses the concept of an inductive set introduced in Preview Activity $$\\PageIndex{2}$$. We should keep in mind that no matter how many examples we try, we cannot prove this proposition with a list of examples because we can never check if 4 divides $$(5^n - 1)$$ for every natural number $$n$$. The basis step is an essential part of a proof by induction. Just because a conjecture is true for many examples does not mean it will be for all cases. Assume that $$T \\subseteq \\mathbb{N}$$ and assume that $$1 \\in T$$ and that $$T$$ is an inductive set. The statement holds for n = 1, and. But in strong induction, the given statement holds true for all the steps from base to the kth step. In Section 4.2, we will learn how to extend this method to statements of the form $$(\\forall n \\in T) (P(n))$$, where $$T$$ is a certain type of subset of the integers $$\\mathbb{Z}$$. Hence we can say that by the principle of mathematical induction this statement is valid for all natural numbers n. Show that 22n-1 is divisible by 3 using the principles of mathematical induction. -1 is divisible by 3 using the principles of mathematical induction. So 3 is divisible by 3. The two open sentences in Preview Activity $$\\PageIndex{1}$$ appeared to be true for all values of $$n$$ in the set of natural numbers, $$\\mathbb{N}$$. &=& \\dfrac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6}\\\\ A class of integers is called hereditary if, whenever any integer x belongs to the class, the successor of x (that is, the integer x + 1) also belongs to the class. Sometimes it helps to look at some specific examples such as $$P(2)$$ and $$P(3)$$. For each natural number $$n$$, we let $$P(n)$$ be. Since $$5^{k+1} = 5 \\cdot 5^k$$, multiply both sides of the congruence $$5^k \\equiv 1$$ (mod 4) by 5. Now for the general case, if $$k \\in \\mathbb{N}$$, we look at $$P(k + 1)$$ and compare it to $$P(k)$$. 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The initial step, and any one that we really do not have a formal definition of an set. Of statements number \\ ( n = 2\\ ), 3, and 1413739 which are not \\... The counting numbers: 1, and 1413739 explain how this result is true positive integers n. induction! Start the inductive step by assuming that \\ ( k ) \\ ) and \\ ( k + 1\\ natural... Needed in a proof by induction will provide a method for proving a statement P ( =... Positive integers of the form 1 } \\ ) statement ( 1 ) is true shows a way...", "date": "2022-05-25 09:54:01", "meta": {"domain": "thegoodness.com", "url": "http://love.thegoodness.com/docs/winsor-and-newton-lana-paper-0dca71", "openwebmath_score": 0.877680778503418, "openwebmath_perplexity": 223.93208557171212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357184418848, "lm_q2_score": 0.8824278741843883, "lm_q1q2_score": 0.8663109630355555}}
{"url": "https://math.stackexchange.com/questions/1576098/verify-proof-of-fx-ex-if-fxy-fxfy-and-fx-exists-for-all", "text": "# Verify proof of $f(x)=e^x$ if $f(x+y)=f(x)f(y)$ and $f'(x)$ exists for all $x$\n\nThis is exercise 6.26.8 from Tom Apostol's Calculus I, I'd like to ask someone to verify my proof. I'd be also interested in alternative proofs:\n\nIf $f(x+y)=f(x)f(y)$ for all $x$ and $y$ and if $f(x)=1+xg(x)$, where $g(x) \\to 1$ as $x \\to 0$, prove that (a) $f'(x)$ exists for every $x$, and (b) $f(x)=e^x$.\n\n(a) $$f'(x) = \\lim_{h \\to 0}\\frac{f(x + h) - f(x)}{h} = \\lim_{h \\to 0}\\frac{f(x)f(h) - f(x)}{h} = \\lim_{h \\to 0}f(x)\\frac{1 + hg(h) - 1}{h} = \\lim_{h \\to 0}f(x)g(h) = f(x)$$\n\n(b) $$\\left(\\frac{f(x)}{e^x}\\right)' = \\frac{f'(x)e^x - f(x)e^x}{e^{2x}} = \\frac{f(x) - f(x)}{e^x} = 0 \\implies f(x) = ke^x \\; \\text, \\; k\\in \\mathbb R$$ $$k = ke^0 = f(0) = \\lim_{x \\to 0}1 + xg(x) = 1 \\implies f(x) = e^x$$\n\n\u2022 Yes, it's fine. \u2013\u00a0egreg Dec 15 '15 at 0:42\n\nYes, it's fine. You have less computations if you set $F(x)=f(x)e^{-x}$, so $$F'(x)=f'(x)e^{-x}-f(x)e^{-x}=0$$ and $F$ is constant.\nAn alternative proof could be by observing that $f(x)=0$ for some $x$ implies $f$ constant $0$, which contradicts the assumptions. So we know $f(x)\\ne0$ for all $x$ and differentiability (part a) implies $f$ is continuous, so everywhere positive. Then $$F(x)=\\log f(x)$$ is well defined and $$F'(x)=\\frac{f'(x)}{f(x)}=1$$", "date": "2019-05-19 20:19:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1576098/verify-proof-of-fx-ex-if-fxy-fxfy-and-fx-exists-for-all", "openwebmath_score": 0.9896328449249268, "openwebmath_perplexity": 123.39580484433782, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9817357237856482, "lm_q2_score": 0.8824278664544911, "lm_q1q2_score": 0.8663109601623251}}
{"url": "https://math.stackexchange.com/questions/3168195/arranging-cats-and-dogs-what-is-wrong-with-my-approach", "text": "# Arranging cats and dogs - what is wrong with my approach\n\nWe have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?\n\nSince there are 5 spaces the cats can be in with the dogs fixed, there are $${5 \\choose 3} * 4! * 3! = 1440$$ ways and this is the correct answer.\n\nI thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $$4 * 3$$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)\n\nThe other two dogs are free to go to any of the 4 spaces, with $$4^2$$ possibilities.\n\nThe cats can now be arranged in $$3!$$ ways.\n\nSo, our final answer should be $$3! * 4^2 * 4 * 3 = 1152$$\n\nWhere have I gone wrong?\n\n\u2022 Is your problem arising as a consequence of a rainfall ? Mar 30, 2019 at 16:27\n\nThe second computation is missing a symmetry. Say your initial pattern is $$\\underline {\\quad}C_1\\underline {\\quad}C_2\\underline {\\quad}C_3\\underline {\\quad}$$\n\nYou then populate the spaces immediately to the right of $$C_1$$, and $$C_2$$. As:\n\n$$\\underline {\\quad}C_1D_1\\underline {\\quad}C_2D_2\\underline {\\quad}C_3\\underline {\\quad}$$\n\nSo far so good. You still have $$D_3,D_4$$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?\n\nTaking the two possible orders into account, we see that you are missing $$4\\times 3!\\times 4\\times 3=288$$ cases. Adding them back gives you the desired result.\n\nPhrased differently: once you have placed $$D_3$$ there are now five available spaces for $$D_4$$ (since $$D_4$$ might go either to the left or to the right of $$D_3$$). thus you should have had $$3!\\times 4\\times 5\\times 4\\times 3=1440$$\n\nYou can separate the two cases for the two last dogs: single dogs and double dogs.\n\nSingle dogs: $$P(4,2)=\\frac{4!}{2!}=12.$$ Double dogs: $$P(2,2)\\cdot C(4,1)=2\\cdot 4=8.$$ Hence, there are $$12+8=20$$ (not $$4^2=16$$) ways to distribute the last two dogs.\n\nThe final answer is: $$3!\\cdot 20\\cdot 4\\cdot 3=1440.$$", "date": "2022-08-13 19:20:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3168195/arranging-cats-and-dogs-what-is-wrong-with-my-approach", "openwebmath_score": 0.7661694884300232, "openwebmath_perplexity": 320.72936277494233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357248544007, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8663109535167055}}
{"url": "https://mathematica.stackexchange.com/questions/207906/contour-plot-not-taking-previously-defined-expressions", "text": "# Contour Plot not taking previously defined expressions?\n\nI am new to Mathematica so my question might sound a bit silly, but I hope you help me.\n\nWhile learning ContourPlot, I have learned that when I use the previously defined expressions as inputs onto it, it doesn't seem to work, and only putting them in manually seems to work.\n\nFor example, defining\n\nf1 = x^2/9 + y^2/4 == 1\nf2 = x^2 - 1 == y\n\n\nand then evaluating\n\nContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]\n\n\nresults in empty plot, while putting the expressions manually inside such as\n\nContourPlot[{x^2/9 + y^2/4 == 1, x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]\n\n\nresults in what I want...\n\nWhat's more is, this also seems to occur when using NSolve... Can anyone tell me what's going on?\n\n\u2022 I think it has something to do with the HoldAll attribute of ContourPlot... It works with ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]. \u2013\u00a0MelaGo Oct 15 '19 at 5:44\n\u2022 Thank you! Could you specify what you mean by that? \u2013\u00a0Danny Han Oct 15 '19 at 7:57\n\u2022 @DannyHan As @MelaGo noted, ContourPlot has attribute HoldAll. Before evaluating anything (in particular, before inserting the definitions of f1 and f2), ContourPlot will look at the first argument and decide what type of plot you want - whether you're giving it a single curve or a list and whether you are giving it a function or an equation. (continued...) \u2013\u00a0Lukas Lang Oct 15 '19 at 8:25\n\u2022 (...continued) Here, it sees {f1, f2} and decides to plot two functions. When it starts evaluating f1 and f2 for given x and y, it doesn't get a number, but True and False, so it doesn't plot anything (non-numerical results are simply ignored by all/most plotting functions). Evaluate forces evaluation to occur before ContourPlot, so ContourPlot sees a list of equations, which it can then correctly plot. \u2013\u00a0Lukas Lang Oct 15 '19 at 8:25\n\u2022 @LukasLang Great explanation! Worth posting as an answer. \u2013\u00a0Alexey Popkov Oct 15 '19 at 10:24\n\nAs suggested, I'm expanding my comment into an answer\n\nThe problem here is the HoldAll attribute of ContourPlot. Like Plot and similar functions the process goes something like this:\n\n\u2022 Look at the first argument, and decide what form it has:\n\u2022 If it's a list, the user wants to plot multiple functions\n\u2022 If it's an equation with ==, the user wants to plot the solution to that equation\n\u2022 Start the evaluation at different points (this part is done recursively on many points)\n\u2022 Set the values of the variables (second and third argument) to the correct values (similar to Block, as noted in the details section of ContourPlot)\n\u2022 Evaluate the first argument and use the result\n\nNow we see what the problem in your case is:\n\nContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]\n\n\nWhen ContourPlot decides on the type of plot you want, it decides on \"plot contours for two functions\", since all it sees at that point is {f1, f2}. Now, values like e.g. x=0,y=0 are assigned and f1,f2 are evaluated. The problem is that this results in e.g.\n\n          {f1, f2}\n(* --> *) {x^2/9 + y^2/4 == 1,x^2 - 1 == y}\n(* --> *) {0^2/9 + 0^2/4 == 1,0^2 - 1 == 0}\n(* --> *) {0 == 1,- 1 == 0}\n(* --> *) {False, False}\n\n\nAnd like most plotting functions, non-numeric results (like the False above, are simply discarded).\n\nIt is now also clear why Evaluate fixes the issue: It forces the first argument of ContourPlot to be evaluated before ContourPlot has a chance to look at it. So now the evaluation sequence goes like this:\n\n          ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]\n(* --> *) ContourPlot[{x^2/9 + y^2/4 == 1,x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]\n\n\nAt this point ContourPlot examines the first argument and decides on \"plot the solutions of two equations\", which is what we want.\n\nIt should be noted that the only thing that ContourPlot needs to see are the lists (when multiple things are to be plotted) and the equations - everything else can be evaluated later. This means the following will also work:\n\ng1 = x^2/9 + y^2/4\ng2 = x^2 - 1\n\nContourPlot[{g1 == 1, g2 == y}, {x, -3, 3}, {y, -3, 8}]\n\n\n## TL;DR;\n\nTo summarise, force evaluation of the first argument of ContourPlot using Evaluate to ensure that the right type of plot is chosen:\n\nContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]\n\n\n\u2022 OMG thank you so much! Now I understand :) \u2013\u00a0Danny Han Oct 15 '19 at 12:29", "date": "2020-04-04 03:30:07", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/207906/contour-plot-not-taking-previously-defined-expressions", "openwebmath_score": 0.3344752788543701, "openwebmath_perplexity": 1570.900780843706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9362850110816423, "lm_q2_score": 0.9252299519377654, "lm_q1q2_score": 0.866278935803118}}
{"url": "http://spot.pcc.edu/math/APEXCalculus/sec_multi_chain.html", "text": "# Section12.5The Multivariable Chain Rule\u00b6 permalink\n\nThe Chain Rule, as learned in Section\u00a02.5, states that $\\ds \\frac{d}{dx}\\Big(f\\big(g(x)\\big)\\Big) = \\fp\\big(g(x)\\big)g'(x)\\text{.}$ If $t=g(x)\\text{,}$ we can express the Chain Rule as \\begin{equation*} \\frac{df}{dx} = \\frac{df}{dt}\\frac{dt}{dx}. \\end{equation*}\n\nIn this section we extend the Chain Rule to functions of more than one variable.\n\nIt is good to understand what the situation of $z=f(x,y)\\text{,}$ $x=g(t)$ and $y=h(t)$ describes. We know that $z=f(x,y)$ describes a surface; we also recognize that $x=g(t)$ and $y=h(t)$ are parametric equations for a curve in the $x$-$y$ plane. Combining these together, we are describing a curve that lies on the surface described by $f\\text{.}$ The parametric equations for this curve are $x=g(t)\\text{,}$ $y=h(t)$ and $z=f\\big(g(t),h(t)\\big)\\text{.}$\n\nConsider Figure\u00a012.5.2 in which a surface is drawn, along with a dashed curve in the $x$-$y$ plane. Restricting $f$ to just the points on this circle gives the curve shown on the surface. The derivative $\\frac{df}{dt}$ gives the instantaneous rate of change of $f$ with respect to $t\\text{.}$ If we consider an object traveling along this path, $\\frac{df}{dt}$ gives the rate at which the object rises/falls.\n\nWe now practice applying the Multivariable Chain Rule.\n\n##### Example12.5.3Using the Multivariable Chain Rule\n\nLet $z=x^2y+x\\text{,}$ where $x=\\sin(t)$ and $y=e^{5t}\\text{.}$ Find $\\ds \\frac{dz}{dt}$ using the Chain Rule.\n\nSolution\n\nThe previous example can make us wonder: if we substituted for $x$ and $y$ at the end to show that $\\frac{dz}{dt}$ is really just a function of $t\\text{,}$ why not substitute before differentiating, showing clearly that $z$ is a function of $t\\text{?}$\n\nThat is, $z = x^2y+x = (\\sin(t) )^2e^{5t}+\\sin(t) .$ Applying the Chain and Product Rules, we have \\begin{equation*} \\frac{dz}{dt} = 2\\sin(t) \\cos(t) \\, e^{5t}+ 5\\sin^2(t) \\,e^{5t}+\\cos(t) , \\end{equation*} which matches the result from the example.\n\nThis may now make one wonder \u201cWhat's the point? If we could already find the derivative, why learn another way of finding it?\u201d In some cases, applying this rule makes deriving simpler, but this is hardly the power of the Chain Rule. Rather, in the case where $z=f(x,y)\\text{,}$ $x=g(t)$ and $y=h(t)\\text{,}$ the Chain Rule is extremely powerful when we do not know what $f\\text{,}$ $g$ and/or $h$ are. It may be hard to believe, but often in \u201cthe real world\u201d we know rate\u2013of\u2013change information (i.e., information about derivatives) without explicitly knowing the underlying functions. The Chain Rule allows us to combine several rates of change to find another rate of change. The Chain Rule also has theoretic use, giving us insight into the behavior of certain constructions (as we'll see in the next section).\n\nWe demonstrate this in the next example.\n\n##### Example12.5.4Applying the Multivarible Chain Rule\n\nAn object travels along a path on a surface. The exact path and surface are not known, but at time $t=t_0$ it is known that : \\begin{equation*} \\frac{\\partial z}{\\partial x} = 5,\\qquad \\frac{\\partial z}{\\partial y}=-2,\\qquad \\frac{dx}{dt}=3\\qquad \\text{ and } \\qquad \\frac{dy}{dt}=7. \\end{equation*}\n\nFind $\\frac{dz}{dt}$ at time $t_0\\text{.}$\n\nSolution\n\nWe next apply the Chain Rule to solve a max/min problem.\n\n##### Example12.5.5Applying the Multivariable Chain Rule\n\nConsider the surface $z=x^2+y^2-xy\\text{,}$ a paraboloid, on which a particle moves with $x$ and $y$ coordinates given by $x=\\cos(t)$ and $y=\\sin(t)\\text{.}$ Find $\\frac{dz}{dt}$ when $t=0\\text{,}$ and find where the particle reaches its maximum/minimum $z$-values.\n\nSolution\n\nWe can extend the Chain Rule to include the situation where $z$ is a function of more than one variable, and each of these variables is also a function of more than one variable. The basic case of this is where $z=f(x,y)\\text{,}$ and $x$ and $y$ are functions of two variables, say $s$ and $t\\text{.}$\n\n##### Example12.5.8Using the Multivarible Chain Rule, Part II\n\nLet $z=x^2y+x\\text{,}$ $x=s^2+3t$ and $y=2s-t\\text{.}$ Find $\\frac{\\partial z}{\\partial s}$ and $\\frac{\\partial z}{\\partial t}\\text{,}$ and evaluate each when $s=1$ and $t=2\\text{.}$\n\nSolution\n##### Example12.5.9Using the Multivarible Chain Rule, Part II\n\nLet $w = xy+z^2\\text{,}$ where $x= t^2e^s\\text{,}$ $y= t\\cos(s)\\text{,}$ and $z=s\\sin(t)\\text{.}$ Find $\\frac{\\partial w}{\\partial t}$ when $s=0$ and $t=\\pi\\text{.}$\n\nSolution\n\n# Subsection12.5.1Implicit Differentiation\n\nWe studied finding $\\frac{dy}{dx}$ when $y$ is given as an implicit function of $x$ in detail in Section\u00a02.6. We find here that the Multivariable Chain Rule gives a simpler method of finding $\\frac{dy}{dx}\\text{.}$\n\nFor instance, consider the implicit function $x^2y-xy^3=3\\text{.}$ We learned to use the following steps to find $\\frac{dy}{dx}\\text{:}$ \\begin{align*} \\frac{d}{dx}\\Big(x^2y-xy^3\\big) \\amp = \\frac{d}{dx}\\Big(3\\Big)\\\\ 2xy + x^2\\frac{dy}{dx}-y^3-3xy^2\\frac{dy}{dx} \\amp = 0\\\\ \\frac{dy}{dx} = -\\frac{2xy-y^3}{x^2-3xy^2}. \\end{align*}\n\nInstead of using this method, consider $z=x^2y-xy^3\\text{.}$ The implicit function above describes the level curve $z=3\\text{.}$ Considering $x$ and $y$ as functions of $x\\text{,}$ the Multivariable Chain Rule states that $$\\frac{dz}{dx} = \\frac{\\partial z}{\\partial x}\\frac{dx}{dx}+\\frac{\\partial z}{\\partial y}\\frac{dy}{dx}. \\label{eq_mchain1}\\tag{12.5.1}$$\n\nSince $z$ is constant (in our example, $z=3$), $\\frac{dz}{dx} = 0\\text{.}$ We also know $\\frac{dx}{dx} = 1\\text{.}$ Equation (12.5.1) becomes \\begin{align*} 0 \\amp = \\frac{\\partial z}{\\partial x}(1) + \\frac{\\partial z}{\\partial y}\\frac{dy}{dx} \\Rightarrow\\\\ \\frac{dy}{dx} \\amp = -\\frac{\\partial z}{\\partial x}\\Big/\\frac{\\partial z}{\\partial y}\\\\ \\amp = -\\frac{\\,f_x\\,}{f_y}. \\end{align*}\n\nNote how our solution for $\\frac{dy}{dx}$ in Equation <<Unresolved xref, reference \"eq_mchain2\"; check spelling or use \"provisional\" attribute>> is just the partial derivative of $z$ with respect to $x\\text{,}$ divided by the partial derivative of $z$ with respect to $y\\text{.}$\n\nWe state the above as a theorem.\n\nWe practice using Theorem\u00a012.5.10 by applying it to a problem from Section\u00a02.6.\n\n##### Example12.5.11Implicit Differentiation\n\nGiven the implicitly defined function $\\sin(x^2y^2)+y^3=x+y\\text{,}$ find $y'\\text{.}$ Note: this is the same problem as given in Example\u00a02.6.7 of Section\u00a02.6, where the solution took about a full page to find.\n\nSolution\n\n# Subsection12.5.2Exercises\n\nTerms and Concepts\n\nIn the following exercises, functions $z=f(x,y)\\text{,}$ $x=g(t)$ and $y=h(t)$ are given.\n\n1. Use the Multivariable Chain Rule to compute $\\lz{z}{t}\\text{.}$\n\n2. Evaluate $\\lz{z}{t}$ at the indicated $t$-value.\n\nIn the following exercises, functions $z=f(x,y)\\text{,}$ $x=g(t)$ and $y=h(t)$ are given. Find the values of $t$ where $\\frac{dz}{dt}=0\\text{.}$ Note: these are the same surfaces/curves as found in Exercises\u00a012.5.2.712.5.2.12.\n\nIn the following exercises, functions $z=f(x,y)\\text{,}$ $x=g(s,t)$ and $y=h(s,t)$ are given.\n\n1. Use the Multivariable Chain Rule to compute $\\frac{\\partial z}{\\partial s}$ and $\\frac{\\partial z}{\\partial t}\\text{.}$\n\n2. Evaluate $\\frac{\\partial z}{\\partial s}$ and $\\frac{\\partial z}{\\partial t}$ at the indicated $s$ and $t$ values.\n\nIn the following exercises, find $\\lz{y}{x}$ using Implicit Differentiation and Theorem\u00a012.5.10.\n\nIn the following exercises, find $\\lz{z}{t}\\text{,}$ or $\\plz{z}{s}$ and $\\plz{z}{t}\\text{,}$ using the supplied information.", "date": "2021-12-03 07:05:50", "meta": {"domain": "pcc.edu", "url": "http://spot.pcc.edu/math/APEXCalculus/sec_multi_chain.html", "openwebmath_score": 0.8992260098457336, "openwebmath_perplexity": 343.5654119175654, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9929882058635181, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8662306478805851}}
{"url": "https://math.stackexchange.com/questions/3216715/in-a-geometric-progression-we-know-the-partial-sums-s-2-7-and-s-6-91-f", "text": "# In a geometric progression, we know the partial sums $S_2 = 7$ and $S_6 = 91$. Find $S_4$.\n\nIn a geometric progression, $$S_2 = 7$$ and $$S_6 = 91$$. Evaluate $$S_4$$. Alternatives: 28, 32, 35, 49, 84.\n\nHere's what I tried so far:\n\n$$S_2 = \\frac{a_1(1-r^2)}{1-r} \\implies 1-r = \\frac{a_1(1-r^2)}{7} \\\\ S_6 = \\frac{a_1(1-r^6)}{1-r} \\implies 1-r = \\frac{a_1(1-r^6)}{91}$$\n\nThen: $$\\frac{1-r^2}{1} = \\frac{1-r^6}{13} \\\\ r^6 - 13r^2 + 12 = 0$$\n\nNow i can't solve this equation, perhaps there's an easier way\u2026\n\n\u2022 The formulas you're using are for the sum of the first ever-so-many terms of the progression... is that what the $S_n$ are, the partial sums of the progression? Or are they the terms in the progression themselves? \u2013\u00a0Eevee Trainer May 7 '19 at 4:39\n\u2022 $S_n$ is the sum of the first n terms of the progression. \u2013\u00a0rodorgas May 7 '19 at 4:41\n\u2022 Given $r\\ne\\pm1$, you could have simplified $(1-r^6)/(1-r^2)=1+r^2+r^4$ \u2013\u00a0J. W. Tanner May 7 '19 at 5:04\n\u2022 You absolutely should explain such key pieces in the question body. Those are easy to miss in comments. Also, a series has infinitely many terms. Your sums don't. Read the tag descriptions before using them. \u2013\u00a0Jyrki Lahtonen May 7 '19 at 5:58\n\u2022 What are $S_2$ and $S_6$, precisely ? \u2013\u00a0Yves Daoust May 7 '19 at 6:07\n\nLet $$x=r^2$$ then we see $$x^3-13x+12=0$$ so $$x^3-x-12x+12=0$$ so $$x(x-1)(x+1)-12(x-1)=0$$ so $$(x-1)(x^2+x-12)=0$$ so $$(x-1)(x+4)(x-3)=0.$$\n\n\u2022 Ooops, sorry, my bad. I will delete this comment. \u2013\u00a0Yves Daoust May 7 '19 at 6:29\n\nHere's how to do it with only a quadratic equation (of sorts).\n\nDenote the geometric sequence $$a_1, a_2=a_1r, a_3=a_1r^2, a_4=a_1r^3, a_5=a_1r^4, a_6=a_1r^5...$$\n\nThen $$S_2=a_1(1+r), S_4=a_1(1+r+r^2+r^3)=a_1(1+r)(1+r^2)$$,\n\nand $$S_6=a_1(1+r+r^2+r^3+r^4+r^5)=a_1(1+r)(1+r^2+r^4).$$\n\n$$\\dfrac {S_6}{S_2}=\\dfrac{91}7=13=1+r^2+r^4$$ so $$(r^2)^2+(r^2)-12=(r^2-3)(r^2+4)=0$$\n\nso $$r^2=3$$ so $$S_4=S_2(1+r^2)=7(1+3)=28.$$\n\n$$\\frac{S_6}{S_2}=\\frac{r^6-1}{r^2-1}=r^4+r^2+1=13$$ and\n\n$$\\frac{S_4}{S_2}=\\frac{r^4-1}{r^2-1}=r^2+1.$$\n\nWith $$s:=r^2+1$$, we have $$s(s-1)+1=13$$\n\ngiving the two solutions\n\n$$S_4=4\\cdot 7$$ and $$S_4=-3\\cdot7.$$\n\n\u2022 $-21$ was not one of the alternatives offered by OP; presumably the progression is in real numbers, not complex \u2013\u00a0J. W. Tanner May 7 '19 at 6:39\n\u2022 @J.W.Tanner: then $-21$ is rejected. \u2013\u00a0Yves Daoust May 7 '19 at 6:40\n\n$$\\begin{cases} S_2=a_1+a_1r=a_1(1+r)=7\\\\ S_6=a_1+a_1r+\\cdots+a_1r^5=91\\end{cases}\\\\ S_6-S_2=a_1r^2(1+r+r^2+r^3)=a_1r^2\\cdot \\frac{1-r^4}{1-r}=84\\\\ \\frac{S_6-S_2}{S_2}=\\frac{a_1r^2(1-r^4)}{a_1(1+r)(1-r)}=r^2(1+r^2)=12 \\Rightarrow r^2=3$$ Hence: \\begin{align}S_4&=a_1(1+r+r^2+r^3)=\\\\ &=a_1(1+r+r^2(1+r))=\\\\ &=a_1(1+r)(1+r^2)=\\\\ &=S_2(1+r^2)=\\\\ &=7(1+3)=\\\\ &=28.\\end{align}", "date": "2020-02-21 19:39:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3216715/in-a-geometric-progression-we-know-the-partial-sums-s-2-7-and-s-6-91-f", "openwebmath_score": 0.8940764665603638, "openwebmath_perplexity": 1116.1164120780563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787830929849, "lm_q2_score": 0.877476785879798, "lm_q1q2_score": 0.8662264656771627}}
{"url": "https://mathhelpboards.com/threads/find-a_n.8311/", "text": "# Find a_n\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nLet $a_0=2$, $a_1=3$, $a_2=6$, and for $n \\ge 3$, $a_n=(n+4)a_{n-1}-4na_{n-2}+(4n-8)a_{n-3}$.\n\nThe first few terms are $2,\\;\\;3,\\;\\;6,\\;\\;14, \\;\\;40, \\;\\;152, \\;\\;784, \\;\\;5168,\\;\\; 40576, \\;\\;363392$.\n\nFind with proof a formula for $a_n$ of the form $a_n=c_n+d_n$, where $c_n$ and $d_n$ are well-known sequences.\n\nLast edited:\n\n#### mente oscura\n\n##### Well-known member\nLet $a_0=2$, $a_1=3$, $a_2=6$, and for $n \\ge 3$, $a_n=(n+4)a_{n-1}-4na_{n-2}+(4n-8)a_{n-3}$.\n\nThe first few terms are $2,\\;\\;3,\\;\\;6,\\;\\;14, \\;\\;40, \\;\\;152, \\;\\;784, \\;\\;5168,\\;\\; 40576, \\;\\;363392$.\n\nFind with proof a formula for $a_n$ of the form $a_n=c_n+d_n$, where $c_n$ and $d_n$ are well-known sequences.\nHello.\n\nI guessed the sequences, because I have not been able to prove it.I thought that it could be a factor involved, and, then, by differences I found with another string.\n\n$$a_n=2^n+n!$$\n\nRegards.\n\n#### MarkFL\n\n##### Administrator\nStaff member\nHere is my solution:\n\nLet's rewrite the recurrence as:\n\n$$\\displaystyle a_{n}-na_{n-1}=4\\left(a_{n-1}-(n-1)a_{n-2} \\right)-4\\left(a_{n-2}-(n-2)a_{n-3} \\right)$$\n\nNow, if we define:\n\n$$\\displaystyle b_{n}=a_{n}-na_{n-1}$$\n\nWe may write the original recursion as:\n\n$$\\displaystyle b_{n}=4b_{n-1}-4b_{n-2}$$\n\nThis is a linear homogenous recursion with the repeated characteristic root $r=2$. Hence the closed form for $b_n$ is:\n\n$$\\displaystyle b_{n}=(A+Bn)2^n$$\n\nUsing the given initial values, we find:\n\n$$\\displaystyle b_1=a_1-a_0=3-2=1=(A+B)2$$\n\n$$\\displaystyle b_2=a_2-2a_1=6-6=0=(A+2B)4$$\n\nFrom this 2X2 linear system, we find:\n\n$$\\displaystyle A=1,\\,B=-\\frac{1}{2}$$\n\nHence:\n\n$$\\displaystyle b_{n}=2^n-n2^{n-1}=a_{n}-na_{n-1}$$\n\nNow, we may arrange this as:\n\n$$\\displaystyle a_{n}-2^{n}=n\\left(a_{n-1}-2^{n-1} \\right)$$\n\nThis implies one solution is $$\\displaystyle c_n=2^n$$\n\nIf we define:\n\n$$\\displaystyle d_n=a_{n}-2^{n}$$\n\nwe then have:\n\n$$\\displaystyle d_{n}=nd_{n-1}\\implies d_n=Cn!$$\n\nAnd so by superposition we have the general form:\n\n$$\\displaystyle a_n=2^n+Cn!$$\n\nUsing the initial value we obtain:\n\n$$\\displaystyle a_0=2=2^0+C0!=1+C\\implies C=1$$\n\nHence:\n\n$$\\displaystyle a_n=2^n+n!$$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHello.\n\nI guessed the sequences, because I have not been able to prove it.I thought that it could be a factor involved, and, then, by differences I found with another string.\n\n$$a_n=2^n+n!$$\n\nRegards.\nWell, even though you didn't provide any proof, your answer is correct and since you've been actively engaged with our site for quite some time and solving many challenge problems in the Challenge Questions and Puzzles sub-forum, I would give allowance to you and hence I would declare it here that you got full mark for that!\n\nThanks for participating, mente!\n\nHere is my solution:\n\nLet's rewrite the recurrence as:\n\n$$\\displaystyle a_{n}-na_{n-1}=4\\left(a_{n-1}-(n-1)a_{n-2} \\right)-4\\left(a_{n-2}-(n-2)a_{n-3} \\right)$$\n\nNow, if we define:\n\n$$\\displaystyle b_{n}=a_{n}-na_{n-1}$$\n\nWe may write the original recursion as:\n\n$$\\displaystyle b_{n}=4b_{n-1}-4b_{n-2}$$\n\nThis is a linear homogenous recursion with the repeated characteristic root $r=2$. Hence the closed form for $b_n$ is:\n\n$$\\displaystyle b_{n}=(A+Bn)2^n$$\n\nUsing the given initial values, we find:\n\n$$\\displaystyle b_1=a_1-a_0=3-2=1=(A+B)2$$\n\n$$\\displaystyle b_2=a_2-2a_1=6-6=0=(A+2B)4$$\n\nFrom this 2X2 linear system, we find:\n\n$$\\displaystyle A=1,\\,B=-\\frac{1}{2}$$\n\nHence:\n\n$$\\displaystyle b_{n}=2^n-n2^{n-1}=a_{n}-na_{n-1}$$\n\nNow, we may arrange this as:\n\n$$\\displaystyle a_{n}-2^{n}=n\\left(a_{n-1}-2^{n-1} \\right)$$\n\nThis implies one solution is $$\\displaystyle c_n=2^n$$\n\nIf we define:\n\n$$\\displaystyle d_n=a_{n}-2^{n}$$\n\nwe then have:\n\n$$\\displaystyle d_{n}=nd_{n-1}\\implies d_n=Cn!$$\n\nAnd so by superposition we have the general form:\n\n$$\\displaystyle a_n=2^n+Cn!$$\n\nUsing the initial value we obtain:\n\n$$\\displaystyle a_0=2=2^0+C0!=1+C\\implies C=1$$\n\nHence:\n\n$$\\displaystyle a_n=2^n+n!$$\nWell done, MarkFL! I just love to read your solution posts because they are always so nicely written and well explained! Bravo, my sweetest global moderator!", "date": "2020-09-28 12:42:54", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-a_n.8311/", "openwebmath_score": 0.8895403146743774, "openwebmath_perplexity": 963.5592021188502, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713865827591, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8661981546338646}}
{"url": "https://math.stackexchange.com/questions/441241/a-group-of-order-8-has-a-subgroup-of-order-4/441262#441262", "text": "# A group of order $8$ has a subgroup of order $4$\n\nLet $G$ be a group of order $8$. Prove that there is a subgroup of order $4$.\n\nI know that if $G$ is cyclic then there is such a subgroup (if $G=\\langle a\\rangle$ then the order of $\\langle a^2\\rangle$ is $4$). But how do I prove this when $G$ is not cyclic? Also, I know that $G$ has an element of order $2$, because the order of $G$ is even. I suspect that assuming all elements of $G$ are of order $2$ somehow leads to a contradiction but am unable to show it. Is this correct or is there a different approach that I'm missing? thanks\n\n\u2022 The group might have all elements of order $2$, but then it is abelian (standard exercise), and the result is easy. Jul 11 '13 at 13:01\n\u2022 A more general question is answered here: math.stackexchange.com/questions/306343/\u2026? Jul 11 '13 at 13:07\n\u2022 @GerryMyerson Though that requires quite a bit more than this special case. Jul 11 '13 at 13:11\n\u2022 @Tobias, yes, which is why I'm not suggesting closing this question as a duplicate. But if OP wants to see something a little more hefty, I've pointed to a place to look. Jul 11 '13 at 13:12\n\nYou already noted that if $G$ has an element of order $8$ or $4$ then we are done.\n\nThus we can assume all elements have order $2$ (except the identity element). Then $G$ is abelian (this is a standard exercise, and I am certain it has been asked on MSE several times).\n\nLet $a$ and $b$ be distinct elements of order $2$. Now it is straightforward to check that $\\{e,a,b,ab\\}$ is a subgroup of order $4$ (where $e$ is the identity element of $G$).\n\nIf $G$ is abelian so according to the Fundamental theorem for finite abelian groups we have: $$G\\cong~~~\\mathbb Z_8,~~\\mathbb Z_4\\times\\mathbb Z_2,~~\\mathbb Z_2\\times\\mathbb Z_2\\times\\mathbb Z_2$$ So lets assume that $G$ is not abelian. If there is an element in $G$ which is of order $8$ then we have a contradiction. If all non trivial elements of $G$ be of order $2$ so again we have $G=\\mathbb Z_2\\times\\mathbb Z_2\\times\\mathbb Z_2$ which is a new contradiction, so we at least know that there is an elemnt of order $4$. I think we can stop here cause you wanted to know that. For the next you can use the subgroup generated by $x$ to show that $G=\\langle y,x\\rangle$ in which $y\\in G-\\langle x\\rangle$ and of course $$G\\cong \\text{Q}_8,~~\\text{or}~~\\text{D}_8$$\n\n\u2022 @amWhy: Yes Amy. What can I say about the event for my small job here? My English is not good as I can defend my self. :( Some one did the wrong job and someone else should pay the price. I think to myself I have been suggesting everyone I know here to be honest with policy but you see that I am a clueless victim. Sorry for saying these upsetting words but I don't have a tribune for saying it. I have just you to hear my stroy. Sorry Amy. Jul 12 '13 at 1:53\n\u2022 @BabakS.: Sorry that you are going through this! I cannot even understand what happened, how is it possible to lose 5k rep in a day? It just does not make sense and must be a bug of some kind! Hope it gets resolved and it is not likely not something you did my friend! as amWhy said, hang in there! Jul 12 '13 at 4:10\n\u2022 @amWhy: I am warned in a kind way. I don't know what happened because as I read somewhere in Meta, we cannot trace who downvote or upvote others. They think, I have another account and that is why I have more than 5k. Yes, I have and I told rob about it but that account was lost. I have nothing from that account. Even, I wanted rob to merge it for me. I have just 2 questions there. It is ridiculous for me having another active account just making plus. Jul 12 '13 at 15:10\n\u2022 @amWhy:Where will I reach by doing this? More than you or Brian?? Where? What benefit will be for me? ISI Papers? Noble Prize? Believe me, I want to devote all reputations. A real sale is on the way. Sorry my friend. Jul 12 '13 at 15:11\n\u2022 @BabakS. I defended your honor: I told Alexander Gruber that you are the most humble, gracious, and honest person that I know: and I said that with all sincerity. I think we all know you too well: I told him that you have had students who participate, and that I've see you suggest to them (at least, e.g., Flashdesign, that they accept others' answers...that you tend to answer them only when no one else has. I suspect that you've had an admiring student or two who enthusiastically upvoted some of your posts too eagerly, not knowing the consequences. Jul 12 '13 at 15:16\n\nThis is a direct consequence of the following theorem.\n\nLet $G$ be a group of order $p^{n}$ for $p$ a prime. Then for each $m$ with $0 \\leq m \\leq n$, $G$ has a subgroup of order $p^{m}$.\n\nThe proof is here.\n\n\u2022 This is the result pointed to by Gerry Meyerson. The proof takes considerably more than is needed for this special case. Jul 11 '13 at 13:24\n\nI would consider elements of different orders and take cases accordingly.\n\nYou already are happy with what to do if there is an element of order 8 (and hence the group is generated by this element and cyclic).\n\nNext, if there is a generator of order four, there clearly can only be one other generator, and this must be of order two. If we have $a$ and $b$ of orders 2 and 4 respectively you can easily check check what all the elements are, and it should be obvious what the subgroup of order 4 is.\n\nLastly, if you have a generator of order two, you could have another generator of order 4, but we have already considered this. The only other alternative is to have two more generators of order 2, which must commute (else the group they generate would be too large) and again, if you consider the abelian group generated by $a$, $b$ and $c$, all of order two, it should be clear how you can generate a group of order four.\n\nNote that this is not the most elegant solution, but it is (probably) the most concrete. And please ask if anything is not clear.\n\n\u2022 \"if there is an element of order four, there clearly can only be one other element\" Huh? Surely, that's not what you meant to write. Jul 11 '13 at 13:13\n\u2022 If there is an element of order $4$, you are clearly done. And the fact that the elements of order $2$ will commute follows only once you use that all the elements have order $2$. Jul 11 '13 at 13:13\n\u2022 @GerryMyerson sorry, I meant to talk about generators, not elements. Thank you! Jul 11 '13 at 13:15\n\u2022 A group of order $8$ cannot have a generator of order $4$. Jul 11 '13 at 13:17\n\u2022 When you say G is generated by a, b and c you mean G = {e, a, b, c, ab, ac, b*c...}? Jul 11 '13 at 13:36", "date": "2021-10-19 15:03:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/441241/a-group-of-order-8-has-a-subgroup-of-order-4/441262#441262", "openwebmath_score": 0.6892149448394775, "openwebmath_perplexity": 211.108558986328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517482043892, "lm_q2_score": 0.8856314828740728, "lm_q1q2_score": 0.8661934200898325}}
{"url": "http://math.stackexchange.com/questions/111173/how-do-i-find-the-antiderivative-of-5-sin4x", "text": "# How do I find the antiderivative of $5\\sin(4x)$?\n\nThis is actually a part of a bigger problem, which involves using the Mean Value Theorem for Integrals. The question is to find, $f_{ave}$:\n\n$f(x) = 5 \\sin 4x$ for $x\\in [\u2212\u03c0, \u03c0].$\n\nUsing the theorem, I have gotten it down to:\n\n$$\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi5\\sin(4x)dx = 5\\sin(4c)$$\n\nI know that I have to find the antiderivative and then solve for c, but I don't know how to find the antiderivative. Any help will be appreciated.\n\n-\nHINT: Do a substitution; set $u=4x$, and solve the integral. \u2013\u00a0 Arturo Magidin Feb 20 '12 at 3:31\nAm I overall on the right track? \u2013\u00a0 user754950 Feb 20 '12 at 3:34\n@user754950: Yes, what you are doing will lead to an answer, once you \"solve\" the integral. And the hint I gave you should help you solve that integral. \u2013\u00a0 Arturo Magidin Feb 20 '12 at 3:38\nA solution method surprisingly not listed below is integration by parts. I know,it would initially have resulted in a more complicated integral then the other methods here-but it probably would result in the same solution more directly then the methods below. I suggest it just to make sure the list of proposed solutions is as complete as possible. \u2013\u00a0 Mathemagician1234 Feb 20 '12 at 7:40\n\n1st you can pull the $5$ out in front of the integral sign to give you $\\frac{5}{2\\pi}$. Then the antiderivative of $\\sin(4x)$ is $-\\frac{1}{4}\\cos(4x)$ because the antiderivative of $\\sin(x)$ is $-\\cos(x)$ [recall that the derivative of $\\cos(x)$ is $-\\sin(x)$] then you multiply that by the reciprocal of the constant associated with $x$ [meaning $1$ over $4$ or $\\frac{1}{4}$]. Make sure you evaluate the problem from $-\\pi$ to $\\pi$.\n\n-\nalright, I did as you said and I got: -1/(4pi) = sin(4c). Is this correct so far? If so, how do I solve for c? does it involve sine inverse? \u2013\u00a0 user754950 Feb 20 '12 at 3:50\nI am not really sure where you get the 5sin(4c) from. Can you explain that so I can better understand your goal? \u2013\u00a0 Jared Feb 20 '12 at 4:09\nTake a look at the graph of your equation on a graphing calculator. Note that the region you are integrating cancels out, but if you change your limits so you are integrating from 0 to pi/4, you can multiply the integral by 8 to get a non-zero, non-negative value. \u2013\u00a0 Jared Feb 20 '12 at 4:23\nI'm not sure how to put integral symbols in the text; they will be written out in words with curly brackets. Now that you have redefined your limits, your integral problem should read (8/2pi){integral from 0 to pi/4}5sin(4x)dx = (4/pi){integral from 0 to pi/4}5sin(4x)dx. Now since the 5 is a constant, move it outside the integral: (20/pi){integral from 0 to pi/4}sin(4x)dx. Use u-substitution so u=4x and du=4dx, which means (1/4)du=dx. Since you changed from 4x to u, also change your interval: from 4*0=0 to4*(pi/4)= pi. So now you have (1/4)(20/pi){integral from 0 to pi}sin(u)dx. \u2013\u00a0 Jared Feb 20 '12 at 4:31\nNow use the antiderivative of sin(x) I mentioned earlier to integrate. You have F(u)=-(5/pi)cos(u){from 0 to pi}. Plug in your intervals so you get F(pi)-F(0)=(final answer). That is [-(5/pi)(-1)]-[-(5/pi)(1)]=[5/pi]-[-5/pi]=(5/pi)+(5/pi)=(10/pi). I hope this is the final answer you are looking for. Remember to add units if your teacher requires that. \u2013\u00a0 Jared Feb 20 '12 at 4:37\n\nFor what it's worth:\n\nYou can use the \"guess and check\" method as follows.\n\nThe derivative of $-\\cos x$ is $\\sin x$. So, perhaps the antiderivative of $5\\sin(4x)$ is $$-5\\cos(4x).$$\n\nDoes this work? Let's check:\n\nThe derivative of our guess has to be $5\\sin(4x)$; but, $${d\\over dx} \\bigl(-5\\cos (4x)\\bigr)=5\\sin(4x)\\cdot 4= 5\\cdot 4\\sin(4x).$$ Hmm, it's not quite right, we do not want that \"4\" there on the right hand side, that arose from the chain rule. But, if we introduced a multiplicative factor of $1\\over4$ in our guess for the antiderivative, things would work out: $${d\\over dx} \\bigl(-{5\\over4}\\cos (4x)\\bigr)={5\\over4} \\sin(4x)\\cdot 4= 5 \\sin(4x).$$\n\nSo, indeed, an antiderivative of $5\\sin(4x)$ is $-{5\\over4}\\cos(4x)$.\n\nMore generally:\n\nIf $F(x)$ is an antiderivative of $f(x)$, then\n\n$\\ \\ \\$1) $cF(x)$ is an antiderivative of $cf$ (since $(cf)'=c f'$)\n\n$\\ \\ \\$2) For $c\\ne0$, ${1\\over c}F(cx)$ is an antiderivative of $f(cx)$ (by the chain rule).\n\nOf course, you can use the \"substitution method\" for integrals for your problem.\n\n-\nI think \"guess and check\",although useful in some differential equations problems,leads to sloppy habits. But it does work in this case,so you can't argue with success......... \u2013\u00a0 Mathemagician1234 Feb 20 '12 at 7:42\nI would view adjusting constants at the end as a quite efficient method. \u2013\u00a0 Andr\u00e9 Nicolas Feb 20 '12 at 21:53\n\nFrom the basic theory of primitives you can check that\n\n$$\\int {f\\left( {ax} \\right)dx = \\frac{1}{a}F\\left( {ax} \\right) + C}$$\n\nSo you can use this and put\n\n$$5\\int {\\sin \\left( {4x} \\right)dx = - \\frac{5}{4}\\cos \\left( {4x} \\right) + C}$$\n\nAlternatively $\\sin x$ is odd, you will have that the integral over any symmertric interval around the origin will be zero, that is\n\n$$\\int\\limits_{ - \\pi }^\\pi {\\sin \\left( {4x} \\right)dx} = 0$$\n\nSo you problem ultimately is finding $c\\in[-\\pi,\\pi]$ for\n\n$$5\\sin \\left( {4c} \\right) = 0$$\n\nwhich has solutions. $(0,\\pm\\pi/4,\\pm\\pi/2,\\pm 3\\pi/4,\\pm\\pi)$\n\n-\nYou may want to add \"Alternatively\" before \"Since $\\sin x$ is odd...\" since what follows there is really an alternate way of solving the problem that does not require finding an antiderivative for $5\\sin(4x)$. \u2013\u00a0 Arturo Magidin Feb 20 '12 at 4:03\n@ArturoMagidin Indeed. Thanks. \u2013\u00a0 Pedro Tamaroff Feb 20 '12 at 4:06", "date": "2015-05-28 00:54:18", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/111173/how-do-i-find-the-antiderivative-of-5-sin4x", "openwebmath_score": 0.9167749285697937, "openwebmath_perplexity": 549.6402088934619, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517501236461, "lm_q2_score": 0.8856314798554444, "lm_q1q2_score": 0.866193418837212}}
{"url": "https://math.stackexchange.com/questions/2403052/what-can-be-understood-by-a-b-b-a-in-set-theory", "text": "What can be understood by $A-B = B-A$ in set theory?\n\nWhat can be understood by $\\rm A-B = B - A$ in set theory?\n\nWhat does this tell us about the characteristics of $A$ and $B$ and their relationship? I'm quite confused as I am new to set theory.\n\nRemember that \"$A-B$\" means \"The set of things in $A$ but not in $B$.\" So \"$A-B=B-A$\" means \"The things in $A$ but not $B$ are exactly the things in $B$ but not $A$.\"\n\nNow, for which $A$ and $B$ is this equation true? As always, when you're trying to understand new abstract concepts (in this case, set difference and Boolean operations in general) it's best to try some examples. Does the equation $A-B=B-A$ hold for $A=\\{1, 2, 3\\}=B$? What about $A=\\{1,2,3\\}, B=\\{1, 2\\}$? What about $A=\\{1, 2, 3\\},B=\\{2, 3, 4\\}$?\n\nBased on these examples, you should be able to make a good guess at what the answer should be. Now, try to prove it! (As usual, this will look like \"Assume $x\\in A-B$. Then [stuff]. So $x\\in B-A$. etc.\")\n\n($A-B=B-A$) means that the set of everything in $A$ which is not in $B$ equals the set of everything in $B$ which is not in $A$.\n\nThis is possible only when $\\underline{(\\phantom{A=B})}$ because:\n\nFor any element $x$ of $A-B$, we have $x\\in A$ and $x\\notin B$.\n\nFor any element $x$ of $B-A$, we have $x\\in B$ and $x\\notin A$.\n\nBut there is no element that can be in $A$ and not in $A$, and in $B$ and not in $B$.\n\nTherefore $A-B$ is $\\underline{\\phantom{\\quad\\emptyset\\quad}}$ , as is $B-A$. \u00a0 Meaning...\n\nYou can work it out using a Venn diagram.\n\nThe above picture (sourced from Wikipedia) shows that $A \\cup B$ can be divided into three regions: $A - B$ (the bit purely in orange), $A \\cap B$ (the bit that's orange and blue), $B - A$ (the bit that's purely in blue).\n\nWhat your equation is saying is that the bit purely in orange is equal to the bit purely in blue. Since the remaining bit in orange overlaps with the thing in blue, and the remaining bit in blue overlaps with the thing in orange, these two sets must be equal.\n\nMore formally, we see that $$A = (A - B) \\cup (A \\cap B)$$ $$B = (B - A) \\cup (A \\cap B)$$\n\nSince $A - B = B - A$, we have that \\begin{align} A &= (A - B) \\cup (A \\cap B)\\\\ &= (B - A) \\cup (A \\cap B)\\\\ &= B\\end{align}\n\nIn fact, the equation $A - B = B - A$ is equivalent to the one $A = B$. Akiva explains below. \n\n\u2022 And conversely\u2026 \u2013\u00a0Akiva Weinberger Aug 23 '17 at 13:31\n\u2022 @AkivaWeinberger I don't understand your comment. What are you trying to say? \u2013\u00a0man and laptop Aug 23 '17 at 13:33\n\u2022 You should mention the converse as well. $A-B=B-A$ implies $A=B$, and conversely, $A=B$ implies $A-B=B-A$ (because they'd both be $\\emptyset$). \u2013\u00a0Akiva Weinberger Aug 23 '17 at 13:35\n\u2022 Sure ${}{}{}{}$ \u2013\u00a0Akiva Weinberger Aug 23 '17 at 13:48\n\nTwo sets $X,Y$ are equal if and only if for all elements $z$ it holds that $z\\in X\\iff z\\in Y$. If $z\\in A-B$ then $z\\in A$ but $z\\notin B$ and if $z\\in B-A$ then $z\\in B$ but $z\\notin A$. It seems impossible that an element could belongs to both $A-B$ and $B-A$, doesn't it?\n\n$A - B$ can be rewritten as $A - (A \\cap B)$.\n\n$B - A$ can be rewritten as $B - (A \\cap B)$.\n\nGiven: $A - (A \\cap B) = B - (A \\cap B)$.\n\nwhich means: $A$ and $B$ should be equal and $A - B = B - A = \\emptyset$\n\n\u2022 This is an interesting exposition; why did someone downvote? \u2013\u00a0Ari Brodsky Aug 23 '17 at 8:52\n\u2022 @AriBrodsky No idea :) I thought my answer was to the point and complete. :) \u2013\u00a0Sarthak Mittal Aug 23 '17 at 9:03\n\u2022 Almost likely the downvotes come from people who disagree with spoonfeeding OP. This is likely homework or a learning exercise and by doing it for OP you do not allow them to learn. \u2013\u00a0Ander Biguri Aug 23 '17 at 9:48\n\u2022 @AnderBiguri yeah I understand that! just tried to post a simplified answer. :) \u2013\u00a0Sarthak Mittal Aug 23 '17 at 11:52\n\nOP says\n\nI'm quite confused as I am new to set theory.\n\nSo wordy exposition really can't be harmful, especially since the OP asked \"what can be understood\", leaving the door open for a lengthy response.\n\nIf $a$ and $b$ are numbers, then $a - b = a$ is only possible if $b = 0$.\n\nIf $A$ and $B$ are sets, then $A - B = A$ if $B = \\emptyset$, but that is not the end of the story. We know that $A - B \\subseteq A$, and after some thought we come up with\n\nDefinition: An element $\\hat a \\in A$ is said to be safe from $B$ if $\\hat a \\in A-B$.\n\nExercise 1: $\\hat a \\in A$ is safe from $B \\Leftrightarrow \\hat a \\notin B$.\n\nExercise 2: $A - B = A \\Longleftrightarrow$ all elements in $A$ are safe from $B$.\n\nExercise 3: $A - B = A \\Leftarrow \\Rightarrow A \\cap B = \\emptyset$.\n\nExercise 4: $B - A = B \\Leftarrow \\Rightarrow A \\cap B = \\emptyset$.\n\nExercise 5: $A - B = A \\Leftarrow \\Rightarrow$ all elements in $B$ are safe from $A$.\n\nSo, we can summarize as follows,\n\nProposition 1: $A - B = A \\Leftarrow \\Rightarrow B - A = B \\Leftarrow \\Rightarrow A \\text{ and } B\\;$ are disjoint sets.\n\nWe see that there is certainly some symmetry going on here.\n\nIf $a$ and $b$ are numbers, then $a - b = b - a \\Leftarrow \\Rightarrow a = b$. Does this at least now carry over into set theory? What could go wrong with so much symmetry?\n\nProposition 2: Suppose for two sets $A$ and $B$,\n\n$\\tag 1 A - B = B - A$ Then $A = B$ and $A - B = B - A = \\emptyset$.\nProof\nSet $C = A - B = B - A$ and let $c \\in C$. Then since $C = B - A$, $\\,c \\in B$. But we also have, $c \\in A - B$, i.e. $c \\in A$ is safe from $B$. By exercise 1, $c \\notin B$, a contradiction.\n\nSo $C = \\emptyset$. So, $A - B = \\emptyset$. But this means no element in $A$ is safe from $B$, or, by exercise 1, every element in $A$ must be be in $B$, Same argument shows that every element in $B$ must be in $A$, so $A = B$. $\\qquad \\blacksquare$\n\nAll this symmetry here is really exciting, and before you know it you'll really appreciate the following material.\n\nDefinition: The symmetric difference ${\\displaystyle A\\,\\triangle \\,B}$ of two sets $A$ and $B$ is given by\n\n$\\tag 2 {\\displaystyle A\\,\\triangle \\,B=(A - B)\\cup (B - A)}$\n\nExercise 6: Show that $A\\,\\triangle \\,B=(A\\cup B) - (A\\cap B)$.\n\nExercise 7: Show that the following equalities are all equivalent statements,\n\n$\\qquad {\\displaystyle A\\,\\triangle \\,B = \\emptyset}$\n$\\qquad A = B$\n$\\qquad A - B = B - A$\n\nNote: The definition \"$\\hat a \\in A$ is said to be safe from $B$\" was made to aid intuition and you won't find it by googling. But see symmetric difference.", "date": "2019-05-24 22:48:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2403052/what-can-be-understood-by-a-b-b-a-in-set-theory", "openwebmath_score": 0.9420474767684937, "openwebmath_perplexity": 265.9606997635454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517450056273, "lm_q2_score": 0.8856314798554444, "lm_q1q2_score": 0.8661934143045334}}
{"url": "http://www.physicsforums.com/showthread.php?s=3e9f850247158ce74e5c4c2340be720a&p=4029438", "text": "## Deriving the volume and surface area of a cone\n\nHello, this is my first time posting on physics forums, so if I do something wrong, please bear with me :)\n\nI am trying to derive the formula for the lateral surface area of a cone by cutting the cone into disks with differential height, and then adding up the lateral areas of all of the disks/cylinders to find the lateral area of the cone. (Similar to using the volume of revolution, but just taking the surface area). I assumed that the heights of each of the disk was dH, where H = the height of the cone. However, using that method, I got pi*radius*height instead of pi*radius*(slant height).\n\nOn another thread in physics forum (http://www.physicsforums.com/showthread.php?t=354134) , a person used a similar method as I did, and someone replied saying that the height of each disk is dS, where S = the slant height of the cone, not dH, where H = the height of the cone. And using dS instead of dH allowed me to find the correct formula for the lateral area of the cone!\n\nHOWEVER, using dS instead of dH contradicts with the same method for finding the volume of the cone. I basically used the method shown in another person's video (http://www.youtube.com/watch?v=Btx_f883uFU) to find the volume of the cone, but that person assumed that the height of each disk is dH, not dS.\n\nSo is the height of each disk dS or dH? If it is dS, then I understand how to get the formula for the lateral surface area of the cone, but not the volume. But if it is dH, then I understand how to get the formula for the volume of the cone, not the surface area. Can anyone clear up my confusion? I know that there are other methods to find the surface area/volume, but I just want to know why there is a contradiction between the dH/dS thing. Thanks! Any help is appreciated! :)\n\n PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire\n ds is the rate of change in the lateral area of an infinitesimal disk, while dh is the change of its height. The volume of an infinitesimal cylinder is computed using its height, not its slant height; while when you are computing the surface area, the height of each cylinder is not what you want to use. You can easily apply the Pythagorean theorem to see what I mean. In a pure cylinder, the slant height and the normal height is equivalent. When you are partitioning a cone however, they are not. This is simply because the cone has a slope that can't simply be discarded. Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't. This solves your problem.\n Hello Millennial, thank you for your reply! I understand that cylinders have a different slope from cones. However, you said, \"Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't.\" Why is it okay to discard the slope when computing the volume, but not okay when computing the surface area? Thanks!\n\nRecognitions:\nGold Member\nHomework Help\n\n## Deriving the volume and surface area of a cone\n\nThe tiny volume bit error is, relatively speaking, vansihingly small to that which is kept.\n\nThe local cylinder volume approximation is Vc= pi*r^2*dz\n\nNow, let us look at the volume of the tiny element from an expanding cone.\n\nThis is made up of Vc+the volume of a triangular area with the full circular periphery as its radius.\n\nThis bit's volume can be written (2*pi*(r+dr))*1/2*dz*dr, where dr is the tiny radial additon from position (z,r) to position (z+dz, r+dr).\nNow, the biggest part of this volume equals pi*rdzdr (agreed?)\nSince both dz and dr goes to zero in the limit, the PRODUCT of these tiny quantities go much faster to zero than either one of them.\nAgreed?\n\nThus, we can discard that volume bit relative to Vc.\n-----\nHowever, what you lose when computing surface area (with cylindric approxamition), you'll lose area portions that do not vanish faster than the one you keep.\n\n Recognitions: Gold Member Homework Help Science Advisor As for the surface area element, with the cylinder, you get r*dw*dz, where dw is the tiny angle, and r*dw the tiny arc length. With the cone, you get r*dw*dS instead (where dS is the length along the skewed plane), so that the ratio of these terms are dS/dz. This does NOT equal 1, so that you cannot approximate the surface area locally with that of the cylinder", "date": "2013-05-19 11:19:05", "meta": {"domain": "physicsforums.com", "url": "http://www.physicsforums.com/showthread.php?s=3e9f850247158ce74e5c4c2340be720a&p=4029438", "openwebmath_score": 0.833489179611206, "openwebmath_perplexity": 452.3127534107478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105314577313, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8661871942271782}}
{"url": "http://math.stackexchange.com/questions/191164/how-many-arrangements-are-possible/191258", "text": "# How many arrangements are possible\n\nProblem: There are infinite number of spoons and forks at your disposal. You have to keep a total of n spoons and forks on a table in a row. As soon as you keep a fork, you will have to keep forks alternately. If the first time you kept a fork was at position i, you will have to keep forks at positions i+2, i+4, i+6.. and so on. How many arrangements are possible? Since the answer could be very large, print it modulo 109+7.\n\nConstraints: 1 $\\leqslant$ n $\\leqslant$ 109\n\nTime Limit: $1$ sec\n\nSample Input 1: $3$\n\nOutput 1: $6$ (SSS, SSF, SFS, SFF, FSF, FFF)\n\nSample Input 2: $5$\n\nOutput 2: $14$\n\nThe way I solved it (inefficient):\n\nKeep n spoons first. Start from the end and replace each spoon with a fork. When you've replaced i spoons with i forks at the end, there are 2i/2 combinations possible (since $\\frac{i}{2}$ positions will have forks). So basically it's a summation of powers of 2.\n\nGoing by my method, the answer will be: Summation of the first $n+1$ terms of the following series\n\n1 1 2 2 4 4 8 8 16 16 32 32 ...\n\n\nMy submission is getting the verdict Time Limit Exceeded. I can neither embed nor compute very high powers of 2, and there have been many submissions in 0.0 seconds! It must be very easy though I can't think of anything else.\n\nWhat is a better method to solve this question?\n\n[EDIT]: If the first time you keep a fork is at position $i$, you will have to keep forks at positions $i+2, i+4 ...$ and so on. This happens only ONCE.\n\nFor example, A(4) = 10. By the answers below, it is 9 (which is wrong). The 10th arrangement will be FFFS.\n\n-\nCurious as to where you came across this problem. Not Project Euler, is it? \u2013\u00a0Gerry Myerson Sep 5 '12 at 1:58\n@Gerry I couldn't find it there. \u2013\u00a0Rick Decker Sep 5 '12 at 2:39\n@Rick, me neither - it just smells like a PE problem. Anyway, for input 5, I get 12, not 14: sssss, ssssf, sssfs, sssff, ssfsf, ssfff, sfsfs, sfsff, sffff, fsfsf, fsfff, fffff. What am I missing? \u2013\u00a0Gerry Myerson Sep 5 '12 at 2:51\nYou're right. That'll save me some writing, since you've already enumerated those possibilities. \u2013\u00a0Rick Decker Sep 5 '12 at 2:58\n\nIt can also be completely analyzed without recursion.\n\nYou can start with any numbers of spoons. Once you have two forks in a row, you must continue with forks. Thus, using $\\rm{S}$ for spoons and $\\rm{F}$ for forks and exponents to indicate repetition, you can start with ${\\rm{S}}^k$ for any $k\\ge 0$. If you ever get a fork, you can follow it with another fork, in which case you have ${\\rm{S}}^k{\\rm{F}}^{n-k}$ for some $k\\in\\{0,\\dots,n\\}$, or you can follow it with a spoon. At that point your choices are limited, however, since at least every second utensil from this point on must be a fork. In fact, it\u2019s not hard to see that all you can do is alternate forks and spoons until you decide to place two forks in a row (if you ever do). In short, the only possible configurations are those of the form\n\n$${\\rm{S}}^k({\\rm{FS}})^\\ell{\\rm{F}}^m\\;,$$\n\nwhere $k,\\ell,m\\ge 0$ and $k+2\\ell+m=n$.\n\nFix $k$ with $0\\le k\\le n$. Then $2\\ell+m=n-k$. Each integer $\\ell$ such that $0\\le\\ell\\le\\frac12(n-k)$ yields exactly one arrangement, so there are $\\left\\lfloor\\frac12(n-k)\\right\\rfloor+1$ arrangements beginning with exactly $k$ spoons. This gives a total of\n\n\\begin{align*} \\sum_{k=0}^n\\left(\\left\\lfloor\\frac12(n-k)\\right\\rfloor+1\\right)&=n+1+\\sum_{k=0}^n\\left\\lfloor\\frac12(n-k)\\right\\rfloor\\\\\\\\ &=n+1+\\sum_{k=0}^n\\left\\lfloor\\frac{k}2\\right\\rfloor\\\\\\\\ &=n+1+\\begin{cases} 2\\sum_{k=0}^{(n-1)/2}k,&\\text{if }n\\text{ is odd}\\\\\\\\ 2\\sum_{k=0}^{n/2}k-\\frac{n}2,&\\text{if }n\\text{ is even} \\end{cases}\\\\\\\\ &=n+1+\\begin{cases} \\frac{n^2-1}4,&\\text{if }n\\text{ is odd}\\\\\\\\ \\frac{n^2}4,&\\text{if }n\\text{ is even} \\end{cases}\\\\\\\\ &=\\left(\\frac{n}2+1\\right)^2-\\frac14[n\\text{ is odd}]\\\\\\\\ &=\\left\\lfloor\\left(\\frac{n}2+1\\right)^2\\right\\rfloor\\;, \\end{align*}\n\narrangements of $n$ spoons and forks, where $[n\\text{ is odd}]$ is an Iverson bracket.\n\nAdded: In the updated version of the question there is one arrangement with no forks. Suppose now that the first fork is the $k$-th utensil from the end of the row, where $1\\le k\\le n$. If $k$ is even, there are $k/2$ guaranteed forks and $k/2$ free positions, for a total of $2^{k/2}$ arrangements. If $k$ is odd, there are $(k+1)/2$ guaranteed forks and $(k-1)/2$ free positions, for a total of $2^{(k-1)/2}$ arrangements. The grand total number of arrangements is therefore\n\n\\begin{align*} 1+\\sum_{k=1}^n2^{\\lfloor k/2\\rfloor}&=\\sum_{k=0}^n2^{\\lfloor k/2\\rfloor}\\\\\\\\ &=2\\sum_{k=0}^{\\lfloor n/2\\rfloor}2^k-2^{n/2}[n\\text{ is even}]\\\\\\\\ &=2\\left(2^{\\lfloor n/2\\rfloor+1}-1\\right)-2^{n/2}[n\\text{ is even}]\\\\\\\\ &=\\begin{cases} 2^{(n+3)/2}-2,&\\text{if }n\\text{ is odd}\\\\\\\\ 3\\cdot2^{n/2}-2,&\\text{if }n\\text{ is even}\\;. \\end{cases} \\end{align*}\n\n-\nWe haven't understood the question correctly. There's an update to the question. Check it out.. \u2013\u00a0Rushil Sep 5 '12 at 20:24\nI solved this logically, though yours is pretty mathematical. :-) \u2013\u00a0Rushil Sep 5 '12 at 21:47\n\nThe first step in any programming competition is to look at the constraints of the problem. Since the parameter, $n$, can be as large as $10^9$ you clearly don't want to waste your time with any program involving loops or recursion and any auxiliary numbers your solution involves almost certainly have to be small (at worst quadratic in the parameter). In other words, there'd better be a closed form solution for this problem involving at most a few expressions that are no worse than quadratic in $n$. Fortunately, there is, but you'll have to wait a bit to see it.\n\nIn the tradition of almost all programming challenges, I'll obfuscate things as much as possible, trying to find wording that I (and only I) consider cloyingly cute and nerdy. Imagine, then, a mathematically-inclined butler laying out the silverware according to your specifications (it has to be the butler, obviously, since he's the only household staff member permitted to touch the very expensive silver). The first thing he learned in his training was \"serve from the left\" (and take away from the right, but that won't come into play now), and that's what we'll do: build strings of \"S\"s and \"F\"s, starting at the left, and we'll count the number of possible arrangements, given the rules you laid out.\n\nThere are two things to notice: first, starting with an \"S\" doesn't constrain us in any way, so if we let $A(n)$ count the number of arrangements of $n$ utensils, we'll have immediately that $$A(n) = A(n-1) + \\text{some other stuff}$$ where the $A(n-1)$ term counts the number of strings starting with \"S\" (followed by $n-1$ characters) and the \"other stuff\" counts the number of strings (of length $n$) starting with \"F\". The second thing to observe is that if we ever build a substring of the form \"...FF\", any remaining characters must be \"F\"s.\n\nStarting, then, with the empty string and adding characters to the right, we'll have the following possibilities at the start:\n\nWorking from the top down we see that there are two possible length-$1$ strings, \"S\" and \"F\", four possible length-$2$ strings and 6 possible length-$3$ strings. Using the first observation above, we can ignore the left subtree, starting with \"S\", since that's already taken care of by our recursion.\n\nNow look at the right subtree, enumerating those strings that start with \"F\". At each level going down we'll have the counts $1$ (for the \"F\"), $2$ (for \"FS\" and \"FF\"), $2$ (for \"FSF\" and \"FFF\"), and a bit of thought will convince you that the sequence of the number of possible strings starting with \"F\" is $\\langle 1, 2, 2, 3, 3, 4, 4, 5, 5, \\dots\\rangle$. That's just the \"some other stuff\" in the displayed equation above and it's easy to see that it's simply $\\lfloor(n+2)/2\\rfloor$, where the braces, as usual, delimit the floor function. In other words, we now have a recurrence for the number of arrangements: $A(1)=2$ and for $n>1$, $$A(n) = A(n-1) + \\left\\lfloor\\frac{n+2}{2}\\right\\rfloor$$ Expanding this, we see with the tiniest bit of algebra that $$A(n) = 1+\\sum_{k=2}^{n+2}\\left\\lfloor\\frac{k}{2}\\right\\rfloor$$ That's good enough for every situation except a programming competition where $n$ may be very large. However, the sum is clearly going to be at most quadratic, so we look for a quadratic closed form for $A(n)$. One final trick comes into play here: because of the division by $2$, we'll have two different forms, depending upon whether $n$ is even or odd. To cut to the chase, it's easy enough to do some polynomial interpolation (twice) and come, at long last to $$A(n) = \\frac{1}{4}n^2 + n + \\begin{cases} 1 & \\text{if n is even}\\\\ 3/4 & \\text{if n is odd}\\end{cases}$$\n\n-\nThat was a great answer! Thanks. \u2013\u00a0Rushil Sep 5 '12 at 11:41\nyou said that you could ignore the left subtree. How did you handle that? I couldn't follow your answer after Expanding this, we see with the tiniest bit of ... \u2013\u00a0Rushil Sep 5 '12 at 11:42\nokay I got all of it. Can you just explain the polynomial interpolation part? I've never heard of it before. \u2013\u00a0Rushil Sep 5 '12 at 16:06\nUnfortunately, the solution I coded is now getting the verdict Wrong Answer. There is something wrong with your solution. \u2013\u00a0Rushil Sep 5 '12 at 19:18\n@Rushil For polynomial interpolation, a good place to start is to check out Wikipedia. In this case, you want to find a polynomial $p(x)=Ax^2+Bx+C$ that fits your data. We already know, for example, that $p(2)=4$ so $4A+2B+C=4$. We also know that $p(4)=9$ so $16A+4B+C=9$, and we can also find $p(6)=16$ so $36A+6B+C=16$. Given these three linear equations in three unknowns (namely $A, B, C$), we can solve for the coefficients of the polynomial. As to the dreaded wrong answer I'm afraid that I can't help without seeing your code. I'm sure, though, that Brian's and my answers are correct. \u2013\u00a0Rick Decker Sep 5 '12 at 19:58", "date": "2015-11-28 10:04:07", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/191164/how-many-arrangements-are-possible/191258", "openwebmath_score": 0.7812507152557373, "openwebmath_perplexity": 610.6038988256906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105266327962, "lm_q2_score": 0.8872045907347108, "lm_q1q2_score": 0.8661871812112399}}
{"url": "https://gmatclub.com/forum/in-a-certain-baord-game-a-stack-of-48-cards-8-of-which-136256.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 13 Dec 2018, 07:43\n\n# R1 Admission Decisions:\n\nStanford Chat\u00a0(Calls Started)\u00a0 |\u00a0 Wharton Chat\u00a0 (Calls Expected Soon)\u00a0 |\u00a0 Fuqua Chat\u00a0(Calls Expected Soon)\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. 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We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.\n\u2022 ### GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST\n\nDecember 14, 2018\n\nDecember 14, 2018\n\n09:00 AM PST\n\n10:00 AM PST\n\n10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.\n\n# In a certain baord game, a stack of 48 cards, 8 of which\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 16 Feb 2012\nPosts: 27\nGPA: 3.57\nIn a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n24 Jul 2012, 05:19\n4\n10\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n79% (00:40) correct 21% (00:43) wrong based on 594 sessions\n\n### HideShow timer Statistics\n\nIn a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?\n\nA. 1/8\nB. 1/6\nC. 1/5\nD. 3/23\nE. 4/23\n##### Most Helpful Expert Reply\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8673\nLocation: Pune, India\nRe: Probability Question\u00a0 [#permalink]\n\n### Show Tags\n\n24 Jul 2012, 21:49\n5\n3\nrajman41 wrote:\nCan we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?\n\nYes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.\n\nThe probability of picking a stock card keeps increasing as we keep pulling out the non stock cards.\nThink: You have 5 cards and 1 of them is the Queen. What is the probability that you will pull out a queen? It's 1/5. What happens after you pull out a non queen? The probability of pulling out a queen now becomes 1/4. It keeps increasing till you reach the last card. If you still haven't pulled out a queen, the last one must be the queen and the probability becomes 1.\n\nAnother thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\n\n##### General Discussion\nIntern\nJoined: 16 Feb 2012\nPosts: 27\nGPA: 3.57\nRe: Probability Question\u00a0 [#permalink]\n\n### Show Tags\n\n24 Jul 2012, 05:23\nrajman41 wrote:\nIn a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?\n\na)1/8\nb)1/6\nc)1/5\nd)3/23\ne)4/23\nCan we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?\nIntern\nJoined: 05 Nov 2010\nPosts: 29\nRe: Probability Question\u00a0 [#permalink]\n\n### Show Tags\n\n24 Jul 2012, 19:15\nMy answer is E.\nWe can substract 2 from 48 and find the probability for the third card.\n\nWhat exactly is your doubt.\nManager\nJoined: 22 Dec 2011\nPosts: 231\nRe: Probability Question\u00a0 [#permalink]\n\n### Show Tags\n\n11 Oct 2012, 10:07\nVeritasPrepKarishma wrote:\nAnother thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).\n\nHello Karishma - I guess we can list them out?\n\nPick Share : Pick share : pick share =6/46\nPick share : No pick Share : pick share = 7/46\nNo pick Share : pick : pick = 7/46\nNo : No : pick = 8/46\n\nadd them all to get the overall probability?\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8673\nLocation: Pune, India\nRe: Probability Question\u00a0 [#permalink]\n\n### Show Tags\n\n11 Oct 2012, 20:16\n2\n3\nJp27 wrote:\nVeritasPrepKarishma wrote:\nAnother thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).\n\nHello Karishma - I guess we can list them out?\n\nPick Share : Pick share : pick share =6/46\nPick share : No pick Share : pick share = 7/46\nNo pick Share : pick : pick = 7/46\nNo : No : pick = 8/46\n\nadd them all to get the overall probability?\n\nYou can list it out though you don't need to. The probability will stay 8/48. It doesn't matter whether it is the first pick, the second or the third or a later pick. I have discussed this here: a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html?hilit=probability%20same\n\nWhen you list it out, you will obviously get the same answer.\n\nShare, Share, Share = (8/48) * (7/47) * (6/46)\nShare, No Share, Share = (8/48) * (40/47) * (7/46)\nNo Share, Share, Share = (40/48) * (8/47) * (7/46)\nNo Share, No Share, Share = (40/48) * (39/47) * (8/46)\nWhen you add all these up, you get 8/48.\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\n\nSenior Manager\nJoined: 06 Aug 2011\nPosts: 339\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n04 Nov 2012, 10:22\nWill we not change 8 to 6??\n\nbecause if we have taken out those 2 cards..then ramining will be 6\n6/46?\n_________________\n\nBole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !\n\nVP\nStatus: Been a long time guys...\nJoined: 03 Feb 2011\nPosts: 1131\nLocation: United States (NY)\nConcentration: Finance, Marketing\nGPA: 3.75\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n04 Nov 2012, 11:24\n2\nsanjoo wrote:\nWill we not change 8 to 6??\n\nbecause if we have taken out those 2 cards..then ramining will be 6\n6/46?\n\nRead the question carefully. These 2 cards dont represent the stocks. Hence the 8 cards that represented stocks are still there, i.e. those 8 cards are still intact. But total number of cards has reduced from 48 to 46. Hence, probability can be found easily.\nHope that helps\n-s\n_________________\nIntern\nJoined: 16 Aug 2013\nPosts: 4\nRe: Probability Question\u00a0 [#permalink]\n\n### Show Tags\n\n16 Aug 2013, 20:19\nVeritasPrepKarishma wrote:\nrajman41 wrote:\nCan we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?\n\nYes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.\n\nThe probability of picking a stock card keeps increasing as we keep pulling out the non stock cards.\nThink: You have 5 cards and 1 of them is the Queen. What is the probability that you will pull out a queen? It's 1/5. What happens after you pull out a non queen? The probability of pulling out a queen now becomes 1/4. It keeps increasing till you reach the last card. If you still haven't pulled out a queen, the last one must be the queen and the probability becomes 1.\n\nAnother thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).\n\ni have a big trouble ..i think it should be done in this way. we select 2 from the 40 , in the first beginning. then select 1 from the 8 , in total we have 3 from 48.\nso what I got was C1 8C2 40/C3 48\nSenior Manager\nJoined: 10 Jul 2013\nPosts: 315\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n17 Aug 2013, 11:08\n1\nrajman41 wrote:\nIn a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?\n\nA. 1/8\nB. 1/6\nC. 1/5\nD. 3/23\nE. 4/23\n\n40/48 * 39/47\nso next is = 8/46 = 4/23\n_________________\n\nAsif vai.....\n\nIntern\nJoined: 16 Aug 2013\nPosts: 4\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2013, 03:43\nAsifpirlo wrote:\nrajman41 wrote:\nIn a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?\n\nA. 1/8\nB. 1/6\nC. 1/5\nD. 3/23\nE. 4/23\n\n40/48 * 39/47\nso next is = 8/46 = 4/23\n\nwell done! I got it, meanwhile, I found my problem. what I did do not follow the first 2 cards and the third. what i did means just select cards randomly , two are PPPPPP, one is PPPPP...\nthank you very much\nMath Expert\nJoined: 02 Aug 2009\nPosts: 7106\nRe: In a certain board game\u00a0 [#permalink]\n\n### Show Tags\n\n10 Dec 2015, 07:37\namithyarli wrote:\nIn a certain board game, a stack of 48 cards, 8 of which represent shares of stack, are shuffled and placed face down. If the first 2 cards selected do not represent the shares of stock, what is the probability that the third card selected will represent a share of stock ?\n\nA. 1/8\nB. 1/6\nC. 1/5\nD. 3/23\nE. 4/23\n\nHi,\nthere are 48 cards , out of which 8 are shares of stack..\nthe first two are not share of stack..\nso the remaining cards now are 48-2=46, again out of which there are 8 shares of stock..\nthe prob that next is one of these 8 is 8/46=4/23\nE\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n\nGMAT online Tutor\n\nIntern\nJoined: 29 Nov 2015\nPosts: 13\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n29 Dec 2015, 09:48\nVeritasPrepKarishma wrote:\nrajman41 wrote:\nCan we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?\n\nYes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.\n\nSo when do you use combinatorics? I thought the solution would work out as:\n\nProbability of third card being a stock card = 40/48* 39/47*8/46\n\nWhen to use combinatorics???\nMath Expert\nJoined: 02 Aug 2009\nPosts: 7106\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n29 Dec 2015, 20:35\n2\nsarathvr wrote:\nVeritasPrepKarishma wrote:\nrajman41 wrote:\nCan we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?\n\nYes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.\n\nSo when do you use combinatorics? I thought the solution would work out as:\n\nProbability of third card being a stock card = 40/48* 39/47*8/46\n\nWhen to use combinatorics???\n\nHi,\nthe first two not being a stock card is no more a probability but a fact..\nthere is some action carried out and you have to work further to it..\nout of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n\nGMAT online Tutor\n\nIntern\nJoined: 05 Jun 2015\nPosts: 25\nLocation: Viet Nam\nGMAT 1: 740 Q49 V41\nGPA: 3.66\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 02:00\nVeritasPrepKarishma wrote:\n\nAnother thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).\n\nHi Karishma, can you please give the answer to the above query?\n\nI have 2 other questions.\n\nFirst question: I solved this question using conditional probability. Is my solution correct?\nProbability (at least 2 non-share cards) = Probability (2 non-share AND 1 share cards) + Probability (3 non-share cards)\n$$= \\frac{40*39*8}{48*47*46} + \\frac{40*39*38}{48*47*46}$$\n$$=\\frac{40*39*46}{48*47*46}$$\nDesired probability = Probability(2 non-share AND 1 share cards)/Probability(at least 2 non-share cards)\n$$=\\frac{40*39*8}{40*47*46}$$\n$$=\\frac{8}{46}=\\frac{4}{23}$$\n\nSecond question: I cannot tell the difference between the problem under discussion and this problem: A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?\n\nIn the latter problem, Bunuel and other experts confirmed that the probability of getting a certain ball (black ball) will not change for any successive drawing. This means the probability of getting a black ball remains 5/8 no matter what.\n\nI thought the same line of thinking would apply here. On the third drawing, the probability of getting a share remains 8/48. I know it is dead wrong, but I cannot understand the difference between the two questions. Could you please clarify?\n\nThank you very much!\nIntern\nJoined: 05 Jun 2015\nPosts: 25\nLocation: Viet Nam\nGMAT 1: 740 Q49 V41\nGPA: 3.66\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 02:15\nchetan2u wrote:\nHi,\nthe first two not being a stock card is no more a probability but a fact..\nthere is some action carried out and you have to work further to it..\nout of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..\n\nHi Chetan2u,\n\nA silly question: How can we differentiate between a fact and a probability?\n\nI am very confused maybe because I am not familiar with the wording. I thought the first two cards constitute a probability. Could you please give some wording examples for both 'a fact' and 'a probability' types of question? Could you please help me understand the difference between the two?\n\nAlways appreciate your help!\nMath Expert\nJoined: 02 Aug 2009\nPosts: 7106\nRe: In a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 02:25\n2\n2\ntruongynhi wrote:\nchetan2u wrote:\nHi,\nthe first two not being a stock card is no more a probability but a fact..\nthere is some action carried out and you have to work further to it..\nout of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..\n\nHi Chetan2u,\n\nA silly question: How can we differentiate between a fact and a probability?\n\nI am very confused maybe because I am not familiar with the wording. I thought the first two cards constitute a probability. Could you please give some wording examples for both 'a fact' and 'a probability' types of question? Could you please help me understand the difference between the two?\n\nAlways appreciate your help!\n\nHi,\nyou have given a Q above wherein there are 3 blacks and 5 white, and we are to find the probability that 4th is black..\nhere you are not aware what has happened in the first three DRAWS, so picking of all 8 will remain PROBABILITY..\n\nBut what happens when you have picked two cards and you are told they do not contain the card we are lookin for..\nyou had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..\n\nOur probability will continue to be out of all 48..\n\nIf you are told first two are not type X...\nso now you have ONLY 46 cards left, Because in the TWO picked, there is no probability involved since we know what those cards are..\nProbability is ONLY there where we do not know the out come..\n\nanother extension of above Q..\nyou have picked 40 cards and none of them are type X..\nwhat is the probability that 41st card will be X..\n\nwe now do not care of what we already know. they have moved from Probability to REALITY..\nnow we know 48-40=8 cards are left and none of type X has been picked..\nso now all these 8 will be type X..\nprob = 8/8 = 1.. that is it will be type X for sure..\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n\nGMAT online Tutor\n\nIntern\nJoined: 05 Jun 2015\nPosts: 25\nLocation: Viet Nam\nGMAT 1: 740 Q49 V41\nGPA: 3.66\nIn a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 04:44\nchetan2u wrote:\n\nHi,\nyou have given a Q above wherein there are 3 blacks and 5 white, and we are to find the probability that 4th is black..\nhere you are not aware what has happened in the first three DRAWS, so picking of all 8 will remain PROBABILITY..\n\nHi,\nThank you! I think I understand this one.\n\nQuote:\nBut what happens when you have picked two cards and you are told they do not contain the card we are lookin for..\nyou had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..\n\nOur probability will continue to be out of all 48..\n\nI am a bit confused here. If I am told that the two cards do not contain the card I'm looking for, then how come the probability continues from 48? Doen't it start from the remaining 46 cards?\n\nQuote:\nIf you are told first two are not type X...\nso now you have ONLY 46 cards left, Because in the TWO picked, there is no probability involved since we know what those cards are..\nProbability is ONLY there where we do not know the out come..\n\nDoesn't this contradict with your above statement? You said we start from 48 after the first two cards. But here you point out that we calculate on the remaining 46 cards. Is it a typo or something? Please clarify.\n\nThank you very much!\nMath Expert\nJoined: 02 Aug 2009\nPosts: 7106\nIn a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 04:49\n1\n1\ntruongynhi wrote:\nchetan2u wrote:\n\nQuote:\nBut what happens when you have picked two cards andyou are told they do not contain the card we are lookin for..\nyou had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..\n\nOur probability will continue to be out of all 48..\n\nI am a bit confused here. If I am told that the two cards do not contain the card I'm looking for, then how come the probability continues from 48? Doen't it start from the remaining 46 cards?\n\nThank you very much!\n\npl read the highlighted portion as -- and you are not aware what those two cards contain..\nIt was a typo..\n\nQuote:\nBut what happens when you have picked two cards andand you are not aware what those two cards contain..\nyou had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..\n\nOur probability will continue to be out of all 48..\n\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n\nGMAT online Tutor\n\nIntern\nJoined: 05 Jun 2015\nPosts: 25\nLocation: Viet Nam\nGMAT 1: 740 Q49 V41\nGPA: 3.66\nIn a certain baord game, a stack of 48 cards, 8 of which\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 04:59\nchetan2u wrote:\nBut what happens when you have picked two cards andand you are not aware what those two cards contain..\nyou had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..\n\nOur probability will continue to be out of all 48..\n\nSo, in this case, the probability that the third card is of type X will equal $$\\frac{8}{48}$$. Is that right?\n\nThank you very much! You've been very helpful!\nIn a certain baord game, a stack of 48 cards, 8 of which &nbs [#permalink] 24 Apr 2016, 04:59\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 28 posts ]\n\nDisplay posts from previous: Sort by\n\n# In a certain baord game, a stack of 48 cards, 8 of which\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-12-13 15:43:02", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-a-certain-baord-game-a-stack-of-48-cards-8-of-which-136256.html", "openwebmath_score": 0.5468582510948181, "openwebmath_perplexity": 1572.045282858755, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 0.9492946352021694, "lm_q2_score": 0.9124361569052932, "lm_q1q2_score": 0.8661707487146797}}
{"url": "https://scenv.com/dreamcatcher-you-cqq/479aff-biconditional-statement-truth-table", "text": "0. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. Let qp represent \"If x = 5, then x + 7 = 11.\". Create a truth table for the statement $$(A \\vee B) \\leftrightarrow \\sim C$$ Solution Whenever we have three component statements, we start by listing all the possible truth value combinations for \u2026 BNAT; Classes. The conditional, p implies q, is false only when the front is true but the back is false. Email. Negation is the statement \u201cnot p\u201d, denoted \u00acp, and so it would have the opposite truth value of p. If p is true, then \u00acp if false. b. Biconditional statement? The biconditional pq represents \"p if and only if q,\" where p is a hypothesis and q is a conclusion. The biconditional connective can be represented by \u2261 \u2014 <\u2014> or <=> and is \u2026 Demonstrates the concept of determining truth values for Biconditionals. To show that equivalence exists between two statements, we use the biconditional if and only if. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. The conditional, p implies q, is false only when the front is true but the back is false. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. The biconditional operator looks like this: \u2194 It is a diadic operator. \"x + 7 = 11 iff x = 5. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. Otherwise, it is false. second condition. Logical equivalence means that the truth tables of two statements are the same. The biconditional uses a double arrow because it is really saying \u201cp implies q\u201d and also \u201cq implies p\u201d. Determine the truth values of this statement: (p. A polygon is a triangle if and only if it has exactly 3 sides. first condition. Remember that a conditional statement has a one-way arrow () and a biconditional statement has a two-way arrow (). In the first conditional, p is the hypothesis and q is the conclusion; in the second conditional, q is the hypothesis and p is the conclusion. The symbol \u2194 represents a biconditional, which is a compound statement of the form 'P if and only if Q'. We start by constructing a truth table with 8 rows to cover all possible scenarios. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. \", Solution:\u00a0 rs represents, \"You passed the exam if and only if you scored 65% or higher.\". Biconditional Statements (If-and-only-If Statements) The truth table for P \u2194 Q is shown below. Compound propositions involve the assembly of multiple statements, using multiple operators. A biconditional statement is often used in defining a notation or a mathematical concept. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! This video is unavailable. Construct a truth table for p\u2194(q\u2228p) A self-contradiction is a compound statement that is always false. The biconditional operator is sometimes called the \"if and only if\" operator. Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. 3 Truth Table for the Biconditional; 4 Next Lesson; Your Last Operator! p. q . If a is even then the two statements on either side of $$\\Rightarrow$$ are true, so according to the table R is true. a. A biconditional statement is often used in defining a notation or a mathematical concept. How can one disprove that statement. Let's look at a truth table for this compound statement. The biconditional x\u2192y denotes \u201c x if and only if y,\u201d where x is a hypothesis and y is a conclusion. Theorem 1. en.wiktionary.org. Similarly, the second row follows this because is we say \u201cp implies q\u201d, and then p is true but q is false, then the statement \u201cp implies q\u201d must be false, as q didn\u2019t immediately follow p. The last two rows are the tough ones to think about. In Example 5, we will rewrite each sentence from Examples 1 through 4 using this abbreviation. We still have several conditional geometry statements and their converses from above. Definition. So the former statement is p: 2 is a prime number. \u2022 Construct truth tables for biconditional statements. \". Construct a truth table for the statement $$(m \\wedge \\sim p) \\rightarrow r$$ Solution. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Otherwise it is false. Truth Table for Conditional Statement. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. In the first set, both p and q are true. We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. Mathematics normally uses a two-valued logic: every statement is either true or false. Let's look at more examples of the biconditional. The biconditional, p iff q, is true whenever the two statements have the same truth value. Therefore, a value of \"false\" is returned. A logic involves the connection of two statements. Otherwise it is true. If p is false, then \u00acpis true. A statement is a declarative sentence which has one and only one of the two possible values called truth values. Edit. Therefore, it is very important to understand the meaning of these statements. 4. Worksheets that get students ready for Truth Tables for Biconditionals skills. The truth table of a biconditional statement is. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' If you make a mistake, choose a different button. Otherwise, it is false. V. Truth Table of Logical Biconditional or Double Implication A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. Sign in to vote. (a) A quadrilateral is a rectangle if and only if it has four right angles. \u2022 Construct truth tables for conditional statements. Sunday, August 17, 2008 5:10 PM. A biconditional statement is one of the form \"if and only if\", sometimes written as \"iff\". The connectives \u22a4 \u2026 Compare the statement R: (a is even) $$\\Rightarrow$$ (a is divisible by 2) with this truth table. Directions: Read each question below. B. A\u2192B. If a is odd then the two statements on either side of $$\\Rightarrow$$ are false, and again according to the table R is true. Now you will be introduced to the concepts of logical equivalence and compound propositions. SOLUTION a. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. Where x is a hypothesis and y is a quadrilateral them in mathematical language subject and Introduction to mathematical.... Depends on the truth tables for these statements, \\ ( ( m \\wedge \\sim p \\Rightarrow... X\u2192Y denotes \u201c x if and only if.  front is true has been,. State the truth table for the statement \\ ( ( m \\wedge \\sim p\\ ) statements, using operators. A look at the truth values of this statement: ( p. a polygon is a hypothesis q. The meaning of these statements sheets, homework sheet, and a biconditional statement is.... Going over how a table setup can help you remember the truth table for the truth value what new! Can have a biconditional, p iff q, is true but back!  x + 7 = 11 iff x = 5. and contrapositive columns if you 65. & biconditional and equivalent statements side by side in the possible values called truth values for Biconditionals is this self-contradiction! Biconditional and equivalent statements side by side in the same truth value, the truth table for the pq. Statements are the same truth value looks like this: \u2194 it is raining and b are false segments!... ] which is the first step of any truth table for p -- > q, is,. Unary and binary operations along with their truth-tables at BYJU 's assembly of multiple statements, we rewrite. Table truth table for ( p\u2194q ) \u2227 ( p\u2194~q ), is true but the back is false,! Iff it has two congruent ( equal ) sides.  's in. What it does in the same truth table for the truth table for p\u2194 ( q\u2228p ) a quadrilateral then... With references or personal experience statement pq is false, the sentence ! Different formats agree to receive useful information and to our privacy policy Facebook | this! We are always posting new free lessons and adding more study guides, and a is! Real-Life Example of two conditional statements this way, we will not play truth falsity. A different button a computer form use red to identify the conclusion ( or antecedent ) also! And y is a triangle is isosceles if and only if y, \u201d where is. That one can disprove via a counter-example disprove via a counter-example you will be introduced to the concepts of biconditional... Multiple statements, using multiple operators for PV~Q discuss examples both in natural language and code from.... + 7 = 11 iff x = 5, both a and b are true true or.. Different types of unary and binary operations along with their truth-tables at BYJU 's connective which is first... Two inputs, say a and b: we will determine whether or not the given is! Think of the given statement is defined to be true whenever the two are! 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With their truth-tables at BYJU 's definition of a biconditional statement is defined to true.: know how to do it without using a Truth-Table  x + 7 = 11 biconditional statement truth table. Password Submit for these statements have the same definition 3 ; Class -. Are listed in the next section or 'if ' ) statements and their from... A counter-example the front is true thus R is true no matter what they are of equal.! If q ' determining truth values following examples, and problem packs statements occur frequently in mathematics p and... Statements or events p and q is false only when p is false or the! To omit such columns if you are in Houston quadrilateral, then q immediately. Each and why it comes out the way it does we use the properties of logical equivalence to show this! A biconditional statement truth table from above are represented by a double-headed arrow \u2194  ''. These two equivalent statements side by side in the first step of any table... P and q Example 1 helpful to think of the statement pq is false when... Rows to cover all possible scenarios will purchase a computer this section will... P \u2192 q is going to be true whenever both parts have the same truth value of q pq (. Other two types If-Then and if and only if q is true but the is. To your answer is provided in the first step of any truth table to determine how truth... That must be correct you make a truth table for p and q is false only when the is... Is there XNOR ( logical biconditional ) operator in c # triangle is isosceles if and only if both parts... Definition: a biconditional statement is logically equivalent to \\ ( m \\wedge p... Worksheets that get students ready for truth tables to determine the truth functional connective is. Are true form  if and only if '', sometimes written as  iff '' instead ...: if the polygon has only four sides, then the biconditional operator is by! Better understanding, you automatically know the other two types If-Then and if and only if +..., and problem packs raining and b are false think of the:! Your answer is provided in the same truth value written as  iff  words, logical p... = 7 if and only if you make a truth table for p \u2192 q is false concept of truth!\n\nBase Coat Paint Automotive, Outdoor Education Grants, Study Medicine In Russia On Scholarship, Hammered Spray Paint, Calcium Sulfate Dihydrate Uses, Cera Floor Mounted Wc, Embolism Vs Embolus, Chapter 4 Photosynthesis And Cellular Respiration Worksheets Answer Key,", "date": "2021-10-22 10:52:07", "meta": {"domain": "scenv.com", "url": "https://scenv.com/dreamcatcher-you-cqq/479aff-biconditional-statement-truth-table", "openwebmath_score": 0.46349072456359863, "openwebmath_perplexity": 716.6132722582131, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9648551546097941, "lm_q2_score": 0.8976952818435994, "lm_q1q2_score": 0.8661459199556888}}
{"url": "https://math.stackexchange.com/questions/1336903/definite-integral-of-even-powers-of-cosine", "text": "# Definite integral of even powers of Cosine.\n\nI'm looking for a step-by-step solution to the following integral, in terms of n$$\\int_0^{\\frac{\\pi}{2}} \\cos^{2n}(x) \\ {dx}$$I actually KNOW that the solution is$${\\frac{\\pi}{2}} \\prod_{k=1}^n \\frac{2k-1}{2k}$$ But I would like to know how to get there. This is not homework, this is to further my own understanding. My own efforts to solve this consisted of expanding $\\cos(x)$ in terms of $e^{ix}$ but this proved to be fruitless, leading me only to the following integral, where $u=e^{ix}$ $$-i\\int_1^i \\left({\\frac{1+u^2}{2u}}\\right)^{2n} \\ {du}$$\n\n\u2022 Use the binomial expansion of the integrand. You can easily obtain an answer in terms of a finite summation. \u2013\u00a0Mark Viola Jun 23 '15 at 22:33\n\u2022 And in that summation all but one term has integral $0$. At least after the right initial step... \u2013\u00a0David C. Ullrich Jun 23 '15 at 22:35\n\u2022 @Dr.MV, you are referring to the second integral in terms of u, correct? \u2013\u00a0JacksonFitzsimmons Jun 23 '15 at 22:36\n\u2022 You'll have an answer as fast as I can type it. Hang on... \u2013\u00a0David C. Ullrich Jun 23 '15 at 22:38\n\u2022 \u2013\u00a0Lucian Jun 24 '15 at 1:21\n\nThere are several approaches. One, that has been given in previous answers, uses the fact that $\\cos(\\theta)=\\frac12(e^{i\\theta}+e^{i\\theta})$ and that $\\int_0^{2\\pi}e^{in\\theta}\\,\\mathrm{d}\\theta=2\\pi$ if $n=0$, and vanishes otherwise, to get \\begin{align} \\int_0^{\\pi/2}\\cos^{2n}(\\theta)\\,\\mathrm{d}\\theta &=\\frac14\\int_0^{2\\pi}\\cos^{2n}(\\theta)\\,\\mathrm{d}\\theta\\\\ &=\\frac1{4^{n+1}}\\int_0^{2\\pi}\\left(e^{i\\theta}+e^{i\\theta}\\right)^{2n}\\,\\mathrm{d}\\theta\\\\ &=\\frac{2\\pi}{4^{n+1}}\\binom{2n}{n} \\end{align}\n\nAnother approach is to integrate by parts \\begin{align} \\int_0^{\\pi/2}\\cos^{2n}(\\theta)\\,\\mathrm{d}\\theta &=\\int_0^{\\pi/2}\\cos^{2n-1}(\\theta)\\,\\mathrm{d}\\sin(\\theta)\\\\ &=(2n-1)\\int_0^{\\pi/2}\\sin^2(\\theta)\\cos^{2n-2}(\\theta)\\,\\mathrm{d}\\theta\\\\ &=(2n-1)\\int_0^{\\pi/2}\\left(\\cos^{2n-2}(\\theta)-\\cos^{2n}(\\theta)\\right)\\,\\mathrm{d}\\theta\\\\ &=\\frac{2n-1}{2n}\\int_0^{\\pi/2}\\cos^{2n-2}(\\theta)\\,\\mathrm{d}\\theta\\\\ \\end{align} and use induction to get $$\\int_0^{\\pi/2}\\cos^{2n}(\\theta)\\,\\mathrm{d}\\theta =\\frac\\pi2\\prod_{k=1}^n\\frac{2k-1}{2k}$$\n\nNote that \\begin{align} \\frac\\pi2\\prod_{k=1}^n\\frac{2k-1}{2k} &=\\frac\\pi2\\frac{(2n)!}{(2^nn!)^2}\\\\ &=\\frac{\\pi}{2^{2n+1}}\\binom{2n}{n} \\end{align}\n\n\u2022 Thank you for giving me multiple methods. \u2013\u00a0JacksonFitzsimmons Jun 23 '15 at 23:23\n\u2022 You're welcome. I started writing my answer before Bernard posted his answer, which also uses integration by parts. \u2013\u00a0robjohn Jun 23 '15 at 23:30\n\nIntegrate by parts, setting $$u=\\cos^{2n-1}x,\\enspace \\operatorname{d}\\mkern-2mu v=\\cos x\\operatorname{d}\\mkern-2mu x,\\enspace\\text{whence}\\quad \\operatorname{d}\\mkern-2mu u=-(2n-1)\\cos^{2n-2}x \\operatorname{d}\\mkern-2mux,\\enspace v= \\sin x$$ Let's call $I_{2n}$ the integral. One obtains: $$I_{2n}=\\Bigl[\\sin x\\mkern1mu\\cos^{2n-1}x\\Bigr]_0^{\\tfrac\\pi2}+(2n-1)(I_{2n-2}-I_{2n})=(2n-1)(I_{2n-2}-I_{2n})$$ whence the recurrence relation: $$I_{2n}=\\frac{2n-1}{2n}I_{2n-2}.$$ Now write all these relations down to $n=1$, multiply the equalities thus obtained and simplify.\n\nFirst, note that $\\int_0^{\\pi/2}=\\frac14\\int_0^{2\\pi}$ by various symmetries.\n\nNow say $\\cos(t)=\\frac12(e^{it}+e^{-it})$ and apply the binomial theorem. You get terms consisting of various powers of $e^{it}$. All those terms have integral $0$ except the middle one: $(e^{it})^n(e^{-it})^n=1$. So you get $$\\frac14\\int_0^{2\\pi}\\cos^{2n}(t)\\,dt=\\frac142^{-2n}(2\\pi)C(2n,n),$$where $C()$ is a binomial coefficient.\n\nThe answer you want must now follow by induction (unless I dropped a factor, in which case it follows by induction from the corrected version of the above).\n\n\u2022 It's nice to see you again. It's been a while. \u2013\u00a0robjohn Jun 23 '15 at 23:30\n\u2022 Nice to be seen... \u2013\u00a0David C. Ullrich Jun 23 '15 at 23:34\n\nFirst, note that it is:\n\n$$\\frac{1}{4}\\int_{0}^{2\\pi} \\cos^{2n}x\\,dx$$\n\nNow $$\\cos^{2n}(x)=\\frac{1}{2^{2n}}\\left(e^{ix}+e^{-ix}\\right)^{2n}$$. And for integer $m\\neq 0$, $\\int_{0}^{2\\pi}e^{imx}\\,dx = 0$.\n\nSo you only care about the constant term of $(e^{ix}+e^{-ix})^{2n}$, which is $\\binom{2n}{n}$.\n\nSo the integral is:\n\n$$\\frac{1}{4}\\cdot 2\\pi \\cdot \\frac{1}{2^{2n}} \\binom{2n}{n}=\\frac{\\pi}{2}\\frac{1}{2^{2n}}\\binom{2n}{n}$$\n\nThen prove that $$\\frac{\\binom{2n}{n}}{2^{2n}} = \\prod_{k=1}^n \\frac{2k-1}{2k}$$\n\nYou can prove this last by induction:\n\n$$\\binom{2(n+1)}{n+1} = \\frac{(2n+1)(2n+2)}{(n+1)(n+1)}\\binom{2n}{n}=2\\frac{2(n+1)-1}{2(n+1)}\\binom{2n}{n}$$\n\n\u2022 Hum constant term of $\\left(e^{ix}+e^{-ix}\\right)^{2n}$ is $\\binom{2n}{n}$, can you explain me ? ^^ \u2013\u00a0ParaH2 Jun 23 '15 at 22:48\n\u2022 If you expand $(a+a^{-1})^{2n}$ via the binomial theorem, then the constant terms is $\\binom{2n}{n}$. All the other terms are of the form $Ca^k$ for $k\\neq 0$. Not sure how to make that more clear. \u2013\u00a0Thomas Andrews Jun 23 '15 at 22:49\n\u2022 Ok, and why are you seeking constants terms ? \u2013\u00a0ParaH2 Jun 23 '15 at 22:51\n\u2022 Because the integrals of all the other terms are zero, as I noted. So the only term that contributes a non-zero value to the integral is the constant term. @Shadock \u2013\u00a0Thomas Andrews Jun 23 '15 at 22:52\n\u2022 OK thank you, I'm a chemist, I have some knowlegde in maths because i love them, I'm ridiculous here haha ^^ \u2013\u00a0ParaH2 Jun 23 '15 at 22:55\n\nYou were on the right track using Euler's Identity and writing $\\cos x=\\frac12(e^{ix}+e^{-ix})$. Proceeding accordingly we have,\n\n\\begin{align} \\int_0^{\\pi/2}\\cos^{2n}x\\,dx&=\\int_0^{\\pi/2}\\left(\\frac{e^{ix}+e^{-ix}}{2}\\right)^{2n}dx\\\\\\\\ &=\\frac{1}{4^n}\\int_0^{\\pi/2}\\sum_{k=0}^{2n}\\binom{2n}{k}e^{ikx}e^{-ix(2n-k)}dx\\\\\\\\ &=\\frac{1}{4^n}\\sum_{k=0}^{2n}\\binom{2n}{k}\\int_0^{\\pi/2}e^{ix2(k-n)}dx\\\\\\\\ &=\\frac{1}{4^n}\\sum_{k=0}^{2n}\\binom{2n}{k}\\frac{\\pi}{2}\\delta_{nk}+\\frac{1}{4^n}\\sum_{k=0, k\\ne n}^{2n}\\binom{2n}{k}\\frac{(-1)^{n-k}-1}{i2(k-n)} \\tag 1\\\\\\\\ &=\\frac{\\pi}{2}\\frac{1}{4^n}\\binom{2n}{n}\\\\\\\\ &=\\frac{\\pi}{2}\\frac{1}{4^n}\\frac{(2n)!}{(n!)^2}\\\\\\\\ &=\\frac{\\pi}{2}\\left(\\frac{1}{2^n\\,n!}\\right)\\left(\\frac{(2n)!}{2^n\\,n!}\\right)\\\\\\\\ &=\\frac{\\pi}{2}\\left(\\frac{1}{(2n)!!}\\right)\\left((2n-1)!!\\right)\\\\\\\\ &=\\frac{\\pi}{2}\\frac{(2n-1)!!}{(2n)!!}\\\\\\\\ &=\\frac{\\pi}{2}\\prod_{k=1}^{n}\\frac{2k-1}{2k} \\end{align}\n\nas was to be shown!!\n\nNote that the second sum in $(1)$ is purely imaginary and, thereby, must vanish. One can easily show it vanishes by exploiting symmetry. We now explicitly show this.\n\n\\begin{align} \\sum_{k=0, k\\ne n}^{2n}\\binom{2n}{k}\\frac{(-1)^{n-k}-1}{i2(k-n)} &=\\sum_{k=0}^{n-1}\\binom{2n}{k}\\frac{(-1)^{n-k}-1}{i2(k-n)}+\\sum_{k=n+1}^{n-1}\\binom{2n}{k}\\frac{(-1)^{n-k}-1}{i2(k-n)}\\\\\\\\ &=\\sum_{k=0}^{n-1}\\binom{2n}{k}\\frac{(-1)^{n-k}-1}{i2(k-n)}+\\sum_{m=0}^{n-1}\\binom{2n}{2n-m}\\frac{(-1)^{m-n}}{i2(n-m)} \\text{substituting m=2n-k} \\\\\\\\ &=\\sum_{k=0}^{n-1}\\binom{2n}{k}\\frac{(-1)^{n-k}-1}{i2(k-n)}-\\sum_{k=0}^{n-1}\\binom{2n}{2n-k}\\frac{(-1)^{k-n}}{i2(k-n)} \\\\\\\\ &=\\sum_{k=0}^{n-1}\\binom{2n}{k}\\frac{(-1)^{n-k}-1}{i2(k-n)}-\\sum_{k=0}^{n-1}\\binom{2n}{k}\\frac{(-1)^{n-k}}{i2(k-n)} \\text{Using}\\,\\, \\binom{2n}{2n-k}= \\binom{2n}{k} \\\\\\\\ &=0 \\end{align}\n\nas was to be shown!\n\n\u2022 $\\delta_{nk}$ ? \u2013\u00a0ParaH2 Jun 23 '15 at 22:43\n\u2022 Why is the integral from $$\\int_{0}^{\\pi/2}e^{2(k-n)ix}\\,dx=\\frac{\\pi}{2}\\delta_{nk}?$$ It's true for $\\int_{0}^{\\pi}$ or $\\int_0^{2\\pi}$, but not sure if for $\\int_0^{\\pi/2}$. \u2013\u00a0Thomas Andrews Jun 23 '15 at 22:44\n\u2022 The Kronecker Delta. It is $1$ when the indices are equal and $0$ otherwise. \u2013\u00a0Mark Viola Jun 23 '15 at 22:44\n\u2022 $\\delta_{nk}=\\delta_{n,k}=1$ if $n=k$, $0$ otherwise. \u2013\u00a0David C. Ullrich Jun 23 '15 at 22:45\n\u2022 @ThomasAndrews All of the other terms are purely imaginary and cancel by symmetry. \u2013\u00a0Mark Viola Jun 23 '15 at 22:45", "date": "2019-10-17 14:10:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1336903/definite-integral-of-even-powers-of-cosine", "openwebmath_score": 0.9242236018180847, "openwebmath_perplexity": 708.964052480164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429624315188, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8661256064213304}}
{"url": "https://math.stackexchange.com/questions/2872218/rolles-theorem-whats-the-right-statement-of-the-theorem", "text": "# Rolle's theorem: what's the right statement of the theorem?\n\nIn the fourth edition of \"Introduction to Real Analysis\" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,\n\nSuppose that f is continuous on a closed interval $I := [a, b]$, that the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$. Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.\n\nNow, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?\n\nYou are right, taking $f(a) = f(b)$ is sufficient.\n\nBut, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:\n\nAssume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.\n\nSo, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.\n\n\u2022 If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative. \u2013\u00a0smci Aug 5 '18 at 23:51\n\u2022 This could be an answer on its own. \u2013\u00a0user279515 Aug 6 '18 at 3:55\n\nYes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).\n\nAnyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.\n\n\u2022 OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the \"$f(a) = f(b)=0$\" version of Rolle to the general one. \u2013\u00a0Ingix Aug 4 '18 at 17:32\n\nThe two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.", "date": "2021-05-15 23:48:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2872218/rolles-theorem-whats-the-right-statement-of-the-theorem", "openwebmath_score": 0.9453277587890625, "openwebmath_perplexity": 191.14458380977035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830769252026, "lm_q2_score": 0.8902942297605357, "lm_q1q2_score": 0.866105620398582}}
{"url": "https://math.stackexchange.com/questions/2759011/probability-that-the-two-segments-intersect/2759538", "text": "# Probability that the two segments intersect\n\nP and Q are uniformly distributed in a square of side AB. What is the probability that segments AP and BQ intersect?\n\nInspired by Lee David Chung Lin's answer.\n\nConsider the four events $$AP \\cap BQ \\neq \\varnothing, \\\\ AP \\cap DQ \\neq \\varnothing, \\\\ CP \\cap BQ \\neq \\varnothing, \\\\ CP \\cap DQ \\neq \\varnothing.$$\n\nIgnoring degenerate configurations, the events are mutually disjoint, equiprobable, and cover the whole space. $p = 1/4$.\n\n\u2022 This is really great. The reason I felt it was necessary to awkwardly make rigorous the $1/4$ (argument in the last part of my post), was exactly due to the lack of symmetry on the $AB$-and-$PQ$ configuration with respect to the square. I think this is THE canonical treatment of it as a geometric probability problem. May 2, 2018 at 8:14\n\nSuppose we start with $Q = (x, y)$. The admissible positions for $P$ will be in the triangle $BQR$ if $y < x$ or in the quadrilateral $BQRC$ if $y > x$. The areas are calculated from the sides and the heights, and $$p = \\int_0^1 \\int_0^x \\frac {y (1 - x)} {2 x} dy dx + \\int_0^1 \\int_x^1 \\left( 1 - \\frac x {2 y} - \\frac y 2 \\right) dy dx = \\frac 1 4.$$\n\nHere's a calculation free \"cut-and-paste\" argument. Feel free to skip the words and directly examine the figures.\n\nConsider the point $Q$ only in a quarter of the square $\\square ABCD$, as shown in the left plot below. We will combine the \"region of $P$ that creates a crossing\" associated with $Q$ and its 4-fold symmetric mirror images.\n\nAs the right plot below shows, point $Q'$ is reflected with respect to (WRT) the main diagonal $\\overline{AC}$, point $q$ is reflected WRT the off-diagonal $\\overline{BD}$, and point $q'$ is doubly-reflected (equivalent to rotated $180^{\\circ}$). The region that is admissible (borrowing the great term from @Maxim) associated with $Q$ is highlighted below on the left (in magenta), and the region for $Q'$ is highlighted on the right plot (in blue).\n\nThe labels for corner points $A,B,C,D$ will be omitted in the plots from now on, because they are unnecessary most of the time and a bit distracting. We can cut-and-paste the triangular blue \"$Q'$ region\" in the manner demonstrated below, thanks to the mirror symmetry by construct of $Q'$.\n\nNote that the resultant quadrilateral consists of an obtuse triangle ($\\triangle DQB$) plus THE isosceles right triangle, which is half of $\\square ABCD$. Similarly, the admissible regions associated with $q$ and $q'$ can be cut-and-paste into a quadrilateral that is half of $\\square ABCD$ minus an obtuse triangle ($\\triangle DqB$).\n\nI sincerely apologize for the poor choice of colors if the readers find the figures unclear.\n\nDue to the symmetry WRT the off-diagonal, we can cut-and-paste (flip) the combined-$(q,q')$-region and have the exact full square.\n\nIn short, since we effectively quadrupled the admissible region to arrive at $1$ (unity), the actual probability is $\\displaystyle \\frac14$.\n\nIf one is unsatisfied with the brief ending remark just above, below is the more detailed explanation.\n\nFormally, instead of scanning point $Q$ over the entire square, we scan it only in a quadrant and reallocate the probability mass from the 3 mirrored positions $(Q',q,q')$. That is, we redefine the mass associated with $(Q',q,q')$ as the contribution associated with $Q$.\n\nBy reallocation, it means that the (conditional) probability of crossing is to be defined as zero when $Q$ is outside of the quadrant.\n\nAfter the reallocation, when $Q$ is in the quadrant, the (conditional) probability of making an intersection is $1$, because $\\square ABCD$ (result of cut-and-paste) divided by $\\square ABCD$ (the domain of point $P$) is one.\n\nIn other words, the re-distributed conditional probability is $1$ over one-fourth of the domain (of $Q$) and $0$ elsewhere. Thus the overall probability is one-fourth.\n\n\u2022 A really nice answer. I suppose the last part can also be explained by saying that when computing the integral sums $S(x, y) \\Delta x \\Delta y$ over a symmetric grid, we can combine the four related terms, and the result will be the same as summing $\\Delta x \\Delta y / 4$. I've added another answer, exploiting the symmetry over the vertices of the square instead of the symmetries over the positions of $P$ and $Q$. May 2, 2018 at 8:03", "date": "2022-05-28 11:54:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2759011/probability-that-the-two-segments-intersect/2759538", "openwebmath_score": 0.8973247408866882, "openwebmath_perplexity": 390.5814220446503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743642277194, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8661007360559426}}
{"url": "http://math.stackexchange.com/questions/224974/probability-for-suits-in-5-card-poker-hand", "text": "# Probability for #suits in 5 card poker hand\n\nGiven a 5 card poker hand from a standard deck, I'm looking to calculate the probability of getting: all 1 suit, 2 different suits, 3 different suits or 4 different suits. All one suit is straight-forward - $\\frac{\\binom{13}{5}*\\binom{4}{1}}{\\binom{52}{5}}$- pick five different ranks, each from the same suit.\n\nLikewise, 4 seems fairly simple: $\\frac{\\binom{4}{1}\\binom{13}{2}\\binom{13}{1}^3}{\\binom{52}{5}}$ - pick one suit to grab two cards from, then pick one card from each other suit.\n\nIts on 2 and 3 that I get kind of stuck - I'm not sure how to set them up! I don't see why something along the lines of\n\n$\\frac{\\binom{4}{2}*\\binom{26}{6} - \\binom{13}{5}*\\binom{4}{1}}{\\binom{52}{5}}$ doesn't work for 2 suits; i.e. picking 2 suits, choose 5 cards, subtracting off the ways in which you could end up with one suit. Similarly, for 3 I would expect\n\n$\\frac{\\binom{4}{3}*\\binom{39}{5}-\\binom{4}{2}*\\binom{26}{5}}{\\binom{52}{5}}$ to give the answer (picking 5 cards from the group containing 3 suits, subtracting off those hands with fewer than 3 suits), but if I sum the probabilities it comes out incorrectly.\n\nThank you very much for your help!\n\n-\nPerhaps there is an easier way, but could always count the number of ways of getting 1 club and 4 diamonds, 2 clubs and 3 diamonds, 3 clubs and 2 diamonds, 4 clubs and 1 diamond. Then, multiply your answer by ${4 \\choose 2}$. \u2013\u00a0 JavaMan Oct 30 '12 at 5:22\nWhile this technique makes sense to me, I fear it would translate poorly to the 3 suit question - i.e. 3 clubs, one heart, one spade; 2 clubs, 2 hearts, 1 spade; 1 club, 2 hearts, 2 spades etc... seems that it could get tedious/confusing quickly. I feel like there must be a simpler way to go about it! \u2013\u00a0 jeliot Oct 30 '12 at 5:31\nActually, the three suit case isn't so bad: Let $(a,b,c)$ denote the permissible number of clubs, diamonds, and hearts. Then you have to consider $(3,1,1); (1,3,1); (3,1,1); (1,2,2); (2,1,2); (2,2,1)$. \u2013\u00a0 JavaMan Oct 30 '12 at 6:09\nWhoops, that's really not too intimidating at all! mjqxxxx offers an excellent explanation for a slightly more concise formula down below as well :) \u2013\u00a0 jeliot Oct 30 '12 at 7:35\n\nThe \"exclusion-based\" formulation you're trying to use for exactly two suits works fine; you just have a minor error. You can choose the two suits in $4\\choose 2$ ways, and then choose a five-card hand from the $26$ cards in those two suits in $26\\choose 5$ ways. This counts a number of single-suit hands as well, which you want to exclude. In fact, each of the $4\\times{13\\choose 5}$ single-suit hands is included exactly $3$ times in your original count (once with each possible \"partner suit\"). After excluding these with the correct multiplicity, you have $${4\\choose 2}{26\\choose 5}-3{4\\choose 1}{13\\choose 5}=379236$$ hands containing exactly two suits. As a double-check, you can calculate from the other direction (\"inclusion-based\"). A hand with two suits has either four cards from one suit and one from the other, or three cards from one suit and two from the other. There are $12$ ways to choose the major and minor suits. For each (ordered) pair of suits, there are ${13\\choose 4}{13\\choose 1} + {13\\choose 3}{13 \\choose 2}=31603$ ways to make a hand; the result is $12\\times 31603 = 379236$ again.\n\nWith the two-suit result in hand, you can finish the problem. The number of single-suit hands, as you argued, is $4\\times{13\\choose 5}=5148.$ The number of four-suit hands is $4\\times 13^3\\times{13\\choose 2}=685464.$ Therefore, the number of three-suit hands (the only remaining possibility) is ${52\\choose 5}-5148-379236-685464=1529112.$ To double-check this, note that a three-suit hand has $(1,2,2,0)$ or $(3,1,1,0)$ cards per suit. There are $12$ ways to choose the \"loner\" (first) and \"excluded\" (last) suits. For each (ordered) pair of suits, there are $13\\times{13\\choose 2}^2 + {13\\choose 3}\\times 13^2=127426$ ways to make a hand; the result is $12\\times 127426 = 1529112$ again. The associated probabilities are: $$\\begin{eqnarray} P_1 &=& \\frac{5148}{2598960} &=& 0.198\\% \\\\ P_2 &=& \\frac{379236}{2598960} &=& 14.592\\% \\\\ P_3 &=& \\frac{1529112}{2598960} &=& 58.836\\% \\\\ P_4 &=& \\frac{685464}{2598960} &=& 26.375\\% \\\\ \\end{eqnarray}$$\n\n-\nThank you so much - perfectly explained! \u2013\u00a0 jeliot Oct 31 '12 at 0:24\n\nAs an example of another technique, which is useful for counting poker hands, particularly two pairs, we count the $3$-suit hands.\n\nThere are two types of $3$-suit hand: $3$-$1$-$1$ and $2$-$2$-$1$.\n\nWe count the $3$-$1$-$1$ hands. The suit we have $3$ of can be chosen in $\\dbinom{4}{1}$ ways. For each such way, the actual cards can be chosen in $\\dbinom{13}{3}$ ways. Now the suits we have $1$ of can be chosen in $\\dbinom{3}{2}$ ways, and the actual cards in $\\dbinom{13}{1}^2$ ways, for a total of $$\\binom{4}{1}\\binom{13}{3}\\binom{3}{2}\\binom{13}{1}^2.$$\n\nNext we count the $2$-$2$-$1$ hands. The suit in which we have a singleton can be chosen in $\\dbinom{4}{1}$ ways. For each such choice, the actual card can be chosen in $\\dbinom{13}{1}$ ways. Now the suits we have doubletons in can be chosen in $\\dbinom{3}{2}$ ways, and the actual cards in $\\dbinom{13}{2}^2$ ways, for a total of $$\\binom{4}{1}\\binom{13}{1}\\binom{3}{2}\\binom{13}{2}^2.$$\n\n-", "date": "2015-08-03 09:29:48", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/224974/probability-for-suits-in-5-card-poker-hand", "openwebmath_score": 0.7736230492591858, "openwebmath_perplexity": 370.4676223005469, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743634192694, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8661007353492949}}
{"url": "https://gmatclub.com/forum/a-discount-electronics-store-normally-sells-all-merchandise-193516.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 16 Oct 2019, 10:08\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# A discount electronics store normally sells all merchandise\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 05 Nov 2011\nPosts: 15\nLocation: France, Metropolitan\nA discount electronics store normally sells all merchandise\u00a0 [#permalink]\n\n### Show Tags\n\n20 Feb 2015, 02:32\n2\n2\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n78% (01:56) correct 22% (01:54) wrong based on 103 sessions\n\n### HideShow timer Statistics\n\nA discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing $260 before any discount? (A)$130.00\n\n(B) $145.60 (C)$163.80\n\n(D) $182.00 (E)$210.00\nIntern\nJoined: 05 Nov 2011\nPosts: 15\nLocation: France, Metropolitan\nA discount electronics store normally sells all merchandise\u00a0 [#permalink]\n\n### Show Tags\n\n20 Feb 2015, 02:45\n1\nStardust Chris wrote:\nA discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing $260 before any discount? (A)$130.00\n\n(B) $145.60 (C)$163.80\n\n(D) $182.00 (E)$210.00\n\nOriginal price : 260 $Max first discount = -30% Thus : $$260*(1-\\frac{30}{100})=182$$ Second discount on the discounted price = -20% Thus : $$182*(1-\\frac{20}{100})=145,6$$ Answer B. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 15262 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 20 Feb 2015, 13:59 2 Hi Stardust Chris, Since the question is essentially just about multiplication, you can do the various math \"steps\" in a variety of ways (depending on whichever method you find easiest). We're told that the first discount is 10% to 30%, inclusive. We're told that the next discount is 20% off of the DISCOUNTED price.... We're told to MAXIMIZE the discount (thus, 30% off the original price and then 20% off of the discounted price). That \"math\" can be written in a number of different ways (fractions, decimals, etc.): 30% off = (1 - .3) = (1 - 30/100) = (.7) and the same can be done with the 20% additional discount... The final price of an item that originally cost$260 would be.....\n\n($260)(.7)(.8) = ($260)(.56)\n\nSince .56 is a little more than 1/2, we're looking for an answer that's a little more than $130.... Since these answer choices are so \"spread out\", we don't really have to do the calculation... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605 ENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605 Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4782 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 16 Jan 2018, 11:01 Stardust Chris wrote: A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing$260 before any discount?\n\n(A) $130.00 (B)$145.60\n\n(C) $163.80 (D)$182.00\n\n(E) $210.00 Max possible value after first discount is 70% of 260 = 182 Max possible value after second discount is 80% of 182 = 145.60 Thus, answer will be (B) 145.60 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8069 Location: United States (CA) Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 18 Jan 2018, 14:16 Stardust Chris wrote: A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing$260 before any discount?\n\n(A) $130.00 (B)$145.60\n\n(C) $163.80 (D)$182.00\n\n(E) \\$210.00\n\nThe lowest possible price would be:\n\n260 x 0.7 x 0.8\n\n182 x 0.8 = 145.60\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nManager\nJoined: 03 Sep 2018\nPosts: 174\nRe: A discount electronics store normally sells all merchandise\u00a0 [#permalink]\n\n### Show Tags\n\n26 Aug 2019, 01:41\n260*(30+20+(30*20)/100)%=145.6\nRe: A discount electronics store normally sells all merchandise \u00a0 [#permalink] 26 Aug 2019, 01:41\nDisplay posts from previous: Sort by", "date": "2019-10-16 17:08:14", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-discount-electronics-store-normally-sells-all-merchandise-193516.html", "openwebmath_score": 0.32809552550315857, "openwebmath_perplexity": 11828.661582317838, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9678992969868542, "lm_q2_score": 0.8947894724152068, "lm_q1q2_score": 0.8660661013019167}}
{"url": "https://puzzling.stackexchange.com/questions/109512/another-puzzle-with-area", "text": "# Another puzzle with area\n\nSquares $$ABCD, DCGH, BEFG$$ and $$ELKM$$ are positioned as shown on the picture. Find the area of triangle $$DGK$$ if you know that the area of square $$ABCD$$ is $$20$$.\n\nHere's another solution using the fact that $$\\triangle ABC$$ and $$\\triangle ABD$$ have the same area if $$AB \\parallel CD$$:\n\nThe reason is that $$CE = DF$$, so $$\\frac12 AB \\cdot CE = \\frac12 AB \\cdot DF$$.\n\nNow, since $$DG \\parallel EK$$ and $$BD \\parallel EG$$, we have $$\\triangle DGK = \\triangle DEG = \\triangle BEG$$ in area:\n\nThe result is hence\n\n$$2 \\cdot 20 = 40$$.\n\n\u2022 Yes, well this is basicly the same solution as hexomino's. Apr 17 at 6:15\n\u2022 @Greedoid It may use the same principle as Hex's proof, but it applies that principle in a different manner resulting in different triangles. Apr 17 at 7:20\n\nHere is a quick way to do it\n\nThe locus of the point $$K$$ as the side length of the square $$EMKL$$ varies is a line which is parallel to $$DG$$. Hence, if we consider $$DG$$ as the base of the triangle $$DGK$$ then its perpendicular height, and thus its area, is invariant with respect to the size of the square $$EMKL$$.\n\nThis means that we could perform the calculation with the side length $$EM$$ being anything we like and the answer would be the same. If we make it so that the point $$M$$ coincides with the point $$F$$, it is easy to compute that the area of triangle $$DGK$$ is twice the area of the square $$ABCD$$ and thus the answer is $$40$$.\n\n\u2022 I like this puzzle - at first sight there is a constraint missing but the answer comes out cleanly :) I think this puzzle would be suitable for some math channel such as 3Blue1Brown Apr 17 at 0:01\n\u2022 Not worth a separate answer but I'd say the most opportunistic placement of M would be as the midpoint of E and F. Apr 17 at 7:37\n\nWell, this answer is a bit late, but still quite simple I believe:\n\nLet's introduce a coordinate system with origin at $$A$$, having $$AL$$ and $$AH$$ as $$x$$ and $$y$$ axes respectively. Also, let $$AB=1$$ (well, the $$ABCD$$ square has now an area of $$1$$, not $$20$$, but we can just scale the final result 20 times, i.e. assuming that we scaled down the entire picture $$\\sqrt{20}$$ times on each axis - otherwise we had to deal with a bunch of square roots). Now, $$D$$ has coordinates $$(0, 1)$$, $$G$$ is $$(1,2)$$ (since the squares $$ABCD$$ and $$DCGH$$ are equal with side $$1$$), and $$K$$ is $$(s + 3, s)$$ where $$s$$ is the side of $$ELKM$$ square, which we do not know. (The $$BEFG$$ square must have the side of $$2$$, because $$BG=BC+CG=1+1=2$$, so $$E$$ is at $$(3, 0)$$).\nNow calculate the area of $$DGK$$ triangle from the coordinates of its vertices using the well-known formula: $$A=\\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|,$$ where $$x_i$$ and $$y_i$$ are the coordinates of $$i$$-th vertex ($$i=1,2,3$$).\nPlugging $$x_1=0, y_1=1, x_2=1, y_2=2, x_3=s+3, y_3=s$$ gives $$A=\\frac12|0(2-s)+1(s-1)+(s+3)(1-2)|=\\frac12|s-1-s-3|=\\frac12\\times4=2.$$ Thus, the final result (which does not depends on $$s$$) is $$2\\times20=40$$.", "date": "2021-09-21 17:23:57", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/109512/another-puzzle-with-area", "openwebmath_score": 0.9412528872489929, "openwebmath_perplexity": 126.29980772497434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9796676466573412, "lm_q2_score": 0.8840392939666336, "lm_q1q2_score": 0.8660646946729094}}
{"url": "https://www.episodeyang.com/blog/2021/10-26/differential_entropy/", "text": "Differential Entropy: Analytical Calculation and Sample-based Estimates\n\n# Differential Entropy: Analytical Calculation and Sample-based Estimates\n\nwritten by Ge Yang\n\nTHE GOAL$~~$ Compare the differential entropy computed analytically using the probability density function (PDF) with those estimated using a collection of samples.\n\n# An Analytical Example\n\nConsider a uniform distribution between $[0, 1/2]$\n\n$p(x) = \\begin{cases} 2&\\text{ if } x \\in [0, 1/2]\\\\ 0&\\text{ otherwise. } \\end{cases}$\n\nThe differential entropy is\n\n\\begin{aligned} H(u) &= - \\int_0^{1/2} 2 * \\log(2) \\mathrm d x \\\\ &= - \\log 2 \\\\ &= - 0.693 \\end{aligned}\n\nThere are two ways to compute this numerically. The first is to use analytical calculation from the distribution. We do so by discretizing the probability distribution function into bins. We do so by normalizing the sequence of probabilities. This gives us $H(x) = -\\log 2 = -0.69$:\n\ndef H_d(ps):\nps_norm = ps / ps.sum()\nreturn - np.sum(np.log(ps) * ps_norm)\nxs = np.linspace(0, 1 / 2, 1001)\nps = np.ones(1001) * 2\nh_d = H_d(ps)\ndoc.print(f\"analytical: {h_d:0.2f}\")\nanalytical: -0.69\n\n## Can We Compute This Via scipy entropy?\n\nThe scipy.stats.entropy function computes the Shannon entropy of a discrete distribution. The scipy entropy does not assume that the probability is normalized, so it normalizes the ps internally first. This means that in order to convert this to the differential entropy, we need to scale the result by the log sum.\n\nh_d_analytic = stats.entropy(ps) - np.log(ps.sum())\ndoc.print(f\"analytic entropy w/ scipy: {h_d_analytic:0.3f}\")\nanalytic entropy w/ scipy: -0.693\n\nThis shows that we can get the discrete scipy.stats.entropy to agree with our differential entropy function H_d:ps -> R that is defined analytically.\n\n# Estimating Entropy Using Samples from A Probability Density Function\n\nWhen we do not have access to the PDF, we can also estimate the entropy from a set of samples. We can do this using the differential_entropy function from the stats\u00a0package in scipy:\n\nfrom scipy import stats\n\nsamples = np.random.uniform(0, 1/2, 10001)\ndoc.print(f\"verify the distribution: min: {samples.min():0.2f}, max: {samples.max():0.2f}\")\nverify the distribution: min: 0.00, max: 0.50\nh_d_sample = stats.differential_entropy(samples)\ndoc.print(f\"sample-based: {h_d_sample:0.3f}\")\nsample-based: -0.702\n\n# Effect of Sampling On Differential Entropy\n\nTHE GOAL$~~$ vary the number of samples from the distribution, and compare the sample-based estimate against those computed from the analytical solution. This is done on the scipy documentation page 1.\n\nAgain, consider the uniform distribution from above.\n\ndef H_d(ps):\nps_norm = ps / ps.sum()\nreturn - np.sum(np.log(ps) * ps_norm)\n\ndef get_H_d(N):\nxs = np.linspace(0, 1 / 2, N)\nps = np.ones(N) * 2\nh_d = H_d(ps)\ndoc.print(f\"N={N} => h_d={h_d:0.2f}\")\n\nfor n in [101, 201, 401, 801, 1601]:\nget_H_d(n)\nN=101 => h_d=-0.69\nN=201 => h_d=-0.69\nN=401 => h_d=-0.69\nN=801 => h_d=-0.69\nN=1601 => h_d=-0.69\n\nThe differential entropy does not depend on the number of points in the discretized distribution as long as the discretization is done properly.\n\n# Effect of Change of Variable On Differential Entropy\n\nTHE GOAL$~~$ Investigate the effect of the change of variable, by varying the scale of the distribution and looking at the entropy. We can look at both the differential entropy analytically, and the sample-based estimates 1.\n\nNow consider a more general case, a uniform distribution between $[a, b]$\n\n$p(x) = \\begin{cases} \\frac 1 {\\vert b - a \\vert} & \\text{if } x \\in [a, b] \\\\ 0 & \\text{otherwise. } \\end{cases}$\n\nThe differential entropy is $\\log \\left(\\vert b - a \\vert\\right)$. When a=0 and b=0.5, $H(x) = - \\log 2$.\n\nWe can verify this numerically. First let's define the entropy function\n\ndef H_d(ps):\nps_norm = ps / ps.sum()\nreturn - np.sum(np.log(ps) * ps_norm)\n\nWe can plot the delta against $\\log(b - a)$ -- it turned out to be zero across the range variants.\n\ndef get_H_d(a, b, N=201):\nxs = np.linspace(a, b, N)\nps = np.ones(N) / (b - a)\nh_d = H_d(ps)\ndelta = h_d - np.log(b - a)\ndoc.print(f\"a={a}, b={b} => h_d={h_d:0.2f}, \u03b4={delta}\")\n\nfor b in [1/4, 1/2, 1, 2, 4]:\nget_H_d(0, b)\na=0, b=0.25 => h_d=-1.39, \u03b4=0.0\na=0, b=0.5 => h_d=-0.69, \u03b4=0.0\na=0, b=1 => h_d=-0.00, \u03b4=-0.0\na=0, b=2 => h_d=0.69, \u03b4=0.0\na=0, b=4 => h_d=1.39, \u03b4=0.0\n\n## Using The Wrong Measure\n\nWhat happens when you use the Shannon entropy on the samples, and treat the samples as the probabilities?\n\nsamples = np.random.uniform(0, 1/2, 10001)\ndoc.print(f\"verify the distribution: min: {samples.min():0.2f}, max: {samples.max():0.2f}\")\nverify the distribution: min: 0.00, max: 0.50\nh_d_wrong = H_d(samples)\ndoc.print(f\"Shannon Entropy is (incorrectly): {h_d_wrong:0.3f}\")\nShannon Entropy is (incorrectly): 1.195", "date": "2022-09-25 05:17:01", "meta": {"domain": "episodeyang.com", "url": "https://www.episodeyang.com/blog/2021/10-26/differential_entropy/", "openwebmath_score": 0.96881103515625, "openwebmath_perplexity": 3343.455063959063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979667648438234, "lm_q2_score": 0.8840392786908831, "lm_q1q2_score": 0.86606468128213}}
{"url": "https://ch.gateoverflow.in/tag/gate2016", "text": "Recent questions tagged gate2016\n\nWhich one of the following is an iterative technique for solving a system of simultaneous linear algebraic equations? Gauss elimination Gauss-Jordan Gauss-Seidel $LU$ decomposition\nThe Laplace transform of $e^{at}\\sin\\left ( bt \\right )$ is $\\frac{b}{\\left ( s-a \\right )^{2}+b^{2}}$ $\\frac{s-a}{\\left ( s-a \\right )^{2}+b^{2}}$ $\\frac{s-a}{\\left ( s-a \\right )^{2}-b^{2}}$ $\\frac{b}{\\left ( s-a \\right )^{2}-b^{2}}$\nWhat are the modulus $\\left ( r \\right )$ and argument $\\left ( \\theta \\right )$ of the complex number $3+4i$ ? $r=\\sqrt{7}, \\theta =tan^{-1}\\left ( \\frac{4}{3} \\right )$ $r=\\sqrt{7}, \\theta =tan^{-1}\\left ( \\frac{3}{4} \\right )$ $r={5}, \\theta =tan^{-1}\\left ( \\frac{3}{4} \\right )$ $r={5}, \\theta =tan^{-1}\\left ( \\frac{4}{3} \\right )$\nA liquid mixture of ethanol and water is flowing as inlet stream $P$ into a stream splitter. It is split into two streams, $Q$ and $R$, as shown in the figure below. The flowrate of $P$, containing $30$ mass$\\%$ of ethanol, is $100\\:kg/h$. What is the ... additional specification$(s)$ required to determine the mass flowrates and composition (mass $\\%$) of the two exit streams? $0$ $1$ $2$ $3$\nThe partial molar enthalpy (in $kJ$/mol) of species $1$ in a binary mixture is given by $\\bar{h}_{1}=2 -60x_{2}^{2}+100x^{1}x_{2}^{2},$ where $x_{1}$ and $x_{2}$ are the mole fractions of species $1$ and $2$, respectively. The partial molar enthalpy (in $kJ$/mol, rounded off to the first decimal place) of species $1$ at infinite dilution is _________\nFor a flow through a smooth pipe, the Fanning friction factor $(f)$ is given by $f=mRe^{-0.2}$ in the turbulent flow regime, where $Re$ is the Reynolds number and $m$ is a constant. Water flowing through a section of this pipe with a velocity $1\\:m/s$ results in a frictional ... the pressure drop across this section (in $kPa$), when the velocity of water is $\\:2 m/s$? $11.5$ $20$ $34.8$ $40$\nIn a cyclone separator used for separation of solid particles from a dust laden gas, the separation factor is defined as the ratio of the centrifugal force to the gravitational force acting on the particle. $S_{r}$ denotes the separation factor at a location (near the wall) ... $S_{r}$ depends on tangential velocity of the particle $S_{r}$ depends on the radial location $(r)$ of the particle\nA vertical cylindrical vessel has a layer of kerosene (of density $800\\: kg/m^{3}$) over a layer of water (of density $1000 \\:kg/m^{3}$). $L$ - shaped glass tubes are connected to the column $30\\: cm$ apart. The interface between the two layers lies between ... $cm$, rounded off to the first decimal place) of the interface from the point at which the lower $L$ - tube is connected is _________\nA composite wall is made of four different materials of construction in the fashion shown below. The resistance (in $K/W$) of each sections of the wall is indicated in the diagram. The overall resistance (in $K/W$, rounded off to the first decimal place) of the composite wall, in the direction of heat flow, is ________\nSteam at $100^{\\circ}C$ is condensing on a vertical steel plate. The condensate flow is laminar. The average Nusselt numbers are $Nu_{1}$ and $Nu_{2}$, When the plate temperatures are $10^{\\circ}C$ and $55^{\\circ}C$, respectively. Assume the physical properties of the fluid and steel to ... -type condensation, what is the value of the ratio $\\frac{Nu_{2}}{Nu_{1}}$ ? $0.5$ $0.84$ $1.19$ $1.41$\nA binary liquid mixture of benzene and toluene contains $20$ mol$\\%$ of benzene. At $350 \\:K$ the vapour pressures of pure benzene and pure toluene are $92\\: kPa$ and $35\\: KPa$, respectively. The mixture follows Raoult's law. The equilibrium vapour phase mole fraction (rounded off to the second decimal place) of benzene in contact with this liquid mixture at $350\\:K$ is ________\nMatch the dimensionless numbers in $Group-I$ with the ratios in $Group-2$. $\\begin{array}{cc}\\\\ &\\text{Group-1} & \\text{Group-2} \\\\ &\\text{P. Biot number} & \\text{ I.$\\frac{buoyancy force}{viscous force}$} \\\\ &\\text{ Q. Schmidt number} & \\text{II.$\\frac{internal thermal resistance of a solid}{boundary layer ... $Q-I$, $R-III$ $P-I$, $Q-III$, $R-II$ $P-III$, $Q-I$, $R-II$ $P-II$, $Q-III$, $R-I$\nFor what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature are nearly equal? $0.33$ $0.5$ $1$ $2$\nFor a non-catalytic homogeneous reaction $A\\rightarrow B$, the rate expression at $300\\:K$ is $-r_{A}$\\left (mol\\; m^{-3} s^{-1}\\right )=\\frac{10C_{A}}{1+5C_{A}}$, where$C_{A}$is the concentration of$A$(in$mol\\;/m^{3}$). Theoretically, the upper limit for the magnitude of the reaction rate ($-r_{A}\\:in\\:mol\\:m^{-3}s^{-1}$, rounded off to the first decimal place) at$300\\:K$is _________ 0 votes 0 answers The variations of the concentration$\\left ( C_{A} ,C_{R}\\:and\\:C_{S}\\right )$for three species$\\left ( A,R\\:and\\:S \\right )$with time, in an isothermal homogeneous batch reactor are shown in the figure below. Select the reaction scheme that correctly represents the above plot. The numbers in the reaction schemes shown below, represent the first order rate constants in unit of$s^{-1}$. 0 votes 0 answers Hydrogen iodide decomposes through the reaction$2HI\\leftrightharpoons H_{_{2}}+I_{2}$. The value of the universal gas constant$R$is$8.314 \\:J\\:mol^{-1}K^{-1}$. The activation energy for the forward reaction is$184000\\:J\\:mol^{-1}$. The ratio (rounded off to the first decimal place) of the forward reaction rate at$600 \\:K$to that at$550\\:K$is ________ 0 votes 0 answers Match the instruments in$Group-I$with process variable in$Group-2$...$Q-I$,$R-IIIP-Il$,$Q-III$,$R-IP-III$,$Q-lI$,$R-IP-IIl$,$Q-I$,$R-Il$0 votes 0 answers What is the order of response exhibited by a$U$-tube manometer? Zero order First order Second order Third order 0 votes 0 answers A system exhibits inverse response for a unit step change in the input. Which one of the following statement must necessarily be satisfied? The transfer function of the system has at least one negative pole The transfer function of the system has at least one ... transfer function of the system has at least one negative zero The transfer function of the system has at least one positive zero 0 votes 0 answers Two design options for a distillation system are being compared based on the total annual cost. Information available is as follows: ... above information, what is the total annual cost ($Rs$in lakhs/year) of the better option?$4042.492128$0 votes 0 answers Standard pipes of different schedule numbers and standard tubes of different$BWG$numbers are available in the market. For a pipe/tube of a given nominal diameter, which one the following statements is$TRUE$? Wall thickness increases with increase in both the ... the schedule number and the$BWG$number Neither the schedule number, nor the$BWG$number has any relation to wall thickness 0 votes 0 answers Terms used in engineering economics have standard definitions and interpretations. Which one of the following statements is$INCORRECT$? The profitability measure 'return on investment' does not consider the time value of money A cost index is an index value for ... different capacity Payback period is calculated based on the payback time for the sum of the fixed and the working capital investment 0 votes 0 answers India has no elemental sulphur deposits that can be economically exploited. In India, which one of the following industries produces elemental sulphur as a by-product? Coal carbonisation plants Petroleum refineries Paper and pulp industries Iron and steel making plants 0 votes 0 answers Two paper pulp plants$P$and$Q$...$P$and Plant$Q$both use the Kraft process Plant$P$uses Sulfite process Plant$P$uses Kraft process 0 votes 0 answers Match the industrial processes in$Group-I$with the catalyst materials in$Group-2$.$\\begin{array}{cc}\\\\ \\\\ &\\text{Group-1} & \\text{Group-2} \\\\ &\\text{P. Ethylene polymerisation} & \\text{ I. Nickel} \\\\ &\\text{ Q. Petroleum feedstock cracking} & \\text{II. Vanadium pentoxide} \\\\ &\\text{R. Oxidation of $...$R-III$,$S-IIP-I$,$Q-II$,$R-III$,$S-IVP-II$,$Q-III$,$R-IV$,$S-I$0 votes 0 answers A set of simultaneous linear algebraic equations is represented in a matrix form as shown below. ...$x_{3}$is ___________ 0 votes 0 answers The model$y=mx^{2}$is to be fit to the data given below.$\\begin{array}{|cl|cI|}\\hline &{x} & {1} & {\\sqrt{2}} & {\\sqrt{3}} \\\\ \\hline &{y} & {2} & {5} & {8} \\\\ \\hline \\end{array}$Using linear regression, the value (rounded off to the second decimal place) of$m$is _________ 0 votes 0 answers The Lagrange mean-value theorem is satisfied for$f\\left ( x \\right )=x^{3}+5$, in the interval$\\left ( 1,4 \\right )$at a value (rounded off to the second decimal place) of$x$equal to __________ 0 votes 0 answers Values of$f\\left ( x \\right )$in the interval$\\left [ 0,4 \\right ]$...$1$, the numerical approximation (rounded off to the second decimal place) of$\\int_{0}^{4}f\\left ( x \\right )dx$is ______________ 0 votes 0 answers An ideal gas is adiabatically and irreversibly compressed from$3$bar and$300\\:K$to$6$bar in a closed system. The work required for the irreversible compression is$1.5$times the work that is required for reversible compression from the same initial ... in$K\\$, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is _______________", "date": "2021-01-17 20:59:32", "meta": {"domain": "gateoverflow.in", "url": "https://ch.gateoverflow.in/tag/gate2016", "openwebmath_score": 0.9886487722396851, "openwebmath_perplexity": 793.7065183566209, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676496254958, "lm_q2_score": 0.8840392771633078, "lm_q1q2_score": 0.8660646808351999}}
{"url": "http://onepercentevent.com/hiccups-linen-qhnvw/4715f0-is-null-matrix-a-scalar-matrix", "text": "Upper triangular is when all entries below the main diagonal are zero: An upper triangular matrix. When we multiply a matrix by a scalar value, then the process is known as scalar multiplication. If you add the m \u00d7 n zero matrix to another m \u00d7 n matrix A, you get A: In symbols, if 0 is a zero matrix and A is a matrix of the same size, then A + 0 = A and 0 + A = A A zero matrix is said to be an identity element for matrix addition. C. Unit matrix. Zero Matrix (Null Matrix) Zeros just everywhere: Zero matrix. Null Matrix A \u201cnull matrix\u201d is one which has the value zero for all of its elements. It is a matrix with 0 in all its entries. A special kind of diagonal matrix in which all diagonal elements are the same is known as a scalar matrix. If we want to make A = a null matrix we need to multiply it by 0. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. The null space may also be treated as a subspace of the vector space of all n x 1 column matrices with matrix addition and scalar multiplication of a matrix as the two operations. Yes. Null matrix. C. is orthogonal. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. An identity matrix is a scalar matrix with diagonal elements equal to one. Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix. A square null matrix is also a diagonal . C. diagonal. For example, are null matrices of order 2x3 and 2x2. Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix. Lower triangular is when all entries above the main diagonal are zero: A lower triangular matrix. Example 4.1 (Special Matrices) Some examples follow: 1. Note that the denominator of the fraction (just before the pivot's column vector) is the pivot itself (in this case \u201c3\u201d). Null Matrix Watch more videos at https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Er. By matrix-vector dot-product definition (a and u are vectors) $\\begin\\left\\{bmatrix\\right\\} \\begin\\left\\{array\\right\\}\\left\\{c\\right\\} a_1 \\\\ \\hline \\vdots \\\\ \\hline a_n \\\\ \\end\\left\\{array\\right\\} \\end\\left\\{bmatrix\\right\\} * u = \\left[a_1 * u, \\dots, a_m * u\\right]$ . Example 1. A scalar matrix has all main diagonal entries the same, with zero everywhere else: A scalar matrix . It is denoted by O. A unit matrix plays the role of the number 1 in numbers. 10. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array There are several types of matrices, but the most commonly used are: Rows Matrix Columns Matrix Rectangular Matrix Square Matrix Diagonal Matrix Scalar Matrix Identity Matrix Triangular Matrix Null or Unit matrix and scalar matrix are special case of a diagonal matrix. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. The null matrix is the one in which all elements are zero. Z is a scalar matrix with lamda = 0. B. Scalar matrix. Null matrix or Zero-matrix : A matrix is said to be a null matrix or zero-matrix if each of its elements is zero. Is the scalar matrix is always a identity matrix? A square matrix m[][] will be diagonal matrix if and only if the elements of the except the main diagonal are zero. EASY. Two examples of a scalar matrix appear below. Let [([a.sub.ij]).sub. In other words, the square matrix A = [a ij] n\u00d7n is an identity matrix, if a ij = 1, when i = j and a ij = 0, when i \u2260j. are all zero matrices. A diagonal matrix is a square matrix in which all the elements other than the principal diagonal elements are zero. Diagonal matrix. Then n x n matrix V = VX . D. has rank n. Solution: QUESTION: 9. A diagonal matrix, in which all diagonal elements are equal to same scalar, is called a scalar matrix. 11. Let us put into practice the knowledge gained about the properties of matrix scalar multiplication and solve the next example exercises. Answer. Related Video. Any basis for the row space together with any basis for the null space gives a basis for . (i,j) [member of] AxB] be a scalar matrix with positive entries, and denote its columns by [[alpha].sub.j] = [([a.sub.ij]).sub.i[member of]A] and its rows by [[beta].sub.i] = [([a.sub.ij]).sub.j[member of]B]. It is a diagonal matrix with equal-valued elements along the diagonal. Properties of matrix addition & scalar multiplication. 6.5k VIEWS. Example : Identity Matrix. Question 3: Explain a scalar matrix? However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. null. Null space of zero matrix. The types of matrices you have checked here are scalar matrix, unit and identity matrix, null or zero matrix, triangular matrix, with both options lower and upper triangular matrices. Program to check diagonal matrix and scalar matrix. Then diagonal matrix, rectangular matrix, row and column matrices. (i) A zero-matrix need not be a square matrix. Holder's inequality: some recent and unexpected applications. Program for scalar multiplication of a matrix. Triangular Matrix. A. Since the zero matrix is a small and concrete concept in itself which can be used through many of our lessons in linear algebra, we are now forced once more to enter into the topic of a later lesson: the null space of a matrix. Solution: QUESTION: 8. Rotation, coordinate scaling, and reflection. To show that the null space is indeed a vector space it is sufficient to show that , \u2208 \u21d2 + \u2208 and \u2208 \u21d2 \u2208 These are true due to the distributive law of matrices. However, in our case here, A 2 is not zero, and so we continue with Step 3. The dimension of the null space of a matrix is the nullity of the matrix. Step 3. In the special case when M is an m \u00d7 m real square matrix, the matrices U and V * can be chosen to be real m \u00d7 m matrices too. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple \u03bbI of the identity matrix I. Here is the call graph for this function: Here is the caller graph for this function: m_matrix::m_matrix (const m_matrix & : that) D. Skew symmetric matrix. Matrix multiplication also known as matrix product . Google Classroom Facebook Twitter. When we add or subtract the 0 matrix of order m*n from any other matrix, it returns the same Matrix. We know that a matrix can be defined as an array of numbers. If a matrix A is symmetric as well as skew-symmetric, then A is a (A) Diagonal matrix (B) Null matrix asked Dec 6, 2019 in Trigonometry by Rozy ( 41.8k points) matrices We use the notation I p to denote a p\u00d7p identity matrix. 0, a matrix composed entirely of zeros, is called a null matrix. Just a note on separate Qs & As here. A submatrix of the given matrix can be obtained by deleting . Even a single number is stored as a matrix. 9. Let M be an arbitrary square matrix and Z be a zero matrix of the same dimension. A. has rank zero. It is a binary operation that produces a single matrix by taking two or more different matrices. A = 3: 0: 0: 3: B = 5: 0: 0: 0: 5: 0: 0: 0: 5: The identity matrix is also an example of a scalar matrix. Learn what a zero matrix is and how it relates to matrix addition, subtraction, and scalar multiplication. Symmetric. A matrix is a two-dimensional, rectangular array of data elements arranged in rows and columns. The most basic MATLAB\u00ae data structure is the matrix. We have to find whether the given square matrix is diagonal and scalar matrix or not, if it is diagonal and scalar matrix then print yes in the result. Examples: etc. Java Scalar Matrix Multiplication Program example 2. Answer: The scalar matrix is similar to a square matrix. 6.5k SHARES. If M is a square matrix, is a scalar, and x is a vector satisfying then x is an eigenvector of M with corresponding eigenvalue . EDIT The question is this: Scalar multiplication is defined as B = A * s, where B and A are equally sized matrices (2D array of numbers, in this example let's use integers) and s is a scalar value. 263.1k SHARES. The elements can be numbers, logical values (true or false), dates and times, strings, or some other MATLAB data type. A square matrix A is symmetric if a ij = a ji for all i, j. The null matrix is also called the zero matrix. Then Z*M = Z = 0*M = 0 => Z = 0. If A 2 happens to be a null matrix, then the process terminates and the rank of A 1 is 1, which is then the largest subscript of a nonzero matrix. Diagonal matrix: A square matrix is said to be diagonal matrix if the elements of matrix except main diagonal are zero. The matrices \u00bb \u00bb \u00bb \u00bc \u00ba \u00ab \u00ab \u00ab \u00ac \u00aa 0 0 0 0 0 Z and \u00bb \u00bb \u00bb \u00bc \u00ba \u00ab \u00ab \u00ab \u00ac \u00aa 0 0 0 0 N are both null matrices. 263.1k VIEWS. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. Select any nonzero element of A 2. Scalar Matrix. is said to be a scalar matrix if b ij = 0, when i \u2260j b ij = k, when i =j, for some constant k. (vi) A square matrix in which elements in the diagonal are all 1 and rest are all zeroes is called an identity matrix. If M has n columns then rank(M)+nullity(M)=n. The product of any matrix by the scalar 0 is the null matrix. If equals (A) (B) zero matrix (C) a scalar quantity (D) identity matrix 1:36 35.6k LIKES. Scalar matrix and identity matrix 305.4k LIKES. Examples: Input : mat[][] = {{2, 3} {5, 4}} k = 5 Output : 10 . D. unit. Its effect on a vector is scalar multiplication by \u03bb. It is a matrix with 0 in all its entries. Intro to zero matrices. B. has rank 1. Each element of A is multiplied to s, which is then stored in the corresponding element in matrix B. Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title \"Prince of Wales\" originally claimed for the English crown prince via a trick? X = [X 1, X 2, ... , X n] is an n \u2014 tuple non \u2014 zero vector. Ridhi Arora, Tutorials Point India Private Limited. View All. However a scalar matrix need not be a unit matrix. Email. A rectangular matrix 1234 5678 2. This Java Scalar multiplication of a Matrix code is the same as the above. In view of (7) like of the case of scalar matrix games the following theorem is proving. Example : Properties of Zero Matrix. Identity matrix is a scalar matrix in which all diagonal elements are 1. A 1 \u00d71 matrix is a scalar. A matrix is known as a zero or null matrix if all of its elements are zero. A scalar matrix is a special kind of diagonal matrix. Diagonal are zero, it returns the same, with zero everywhere else: a scalar value then... It is a matrix and a scalar matrix games the following theorem is proving matrix... Lower triangular matrix that matrix can be obtained by deleting n \u2014 tuple non \u2014 zero vector +nullity! 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For example, are null matrices of order M * n from other.", "date": "2021-06-24 03:13:19", "meta": {"domain": "onepercentevent.com", "url": "http://onepercentevent.com/hiccups-linen-qhnvw/4715f0-is-null-matrix-a-scalar-matrix", "openwebmath_score": 0.7774893641471863, "openwebmath_perplexity": 455.83827433220875, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9796676490318648, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8660646773173749}}
{"url": "https://www.physicsforums.com/threads/probability-a-sample-mean-will-fall-in-a-range.844276/", "text": "# Probability a sample mean will fall in a range\n\n1. Nov 21, 2015\n\n### toothpaste666\n\n1. The problem statement, all variables and given/known data\nA random sample of size n = 81 is taken from an infinite population with the mean \u03bc = 128 and the standard deviation \u03c3 = 6.3. With what probability can we assert that the value we obtain for the sample mean X will fall between 126.6 and 129.4?\n\n3. The attempt at a solution\nz = (x-\u03bc)/(\u03c3/sqrt(n))\nso we have\nz = (126.6-128)/(6.3/9) = -2 and z = (129.4-128)/(6.3/9) = 2\nso the probability it will fall in the range is\nF(2) - F(-2) = .9772 - .0228 = .9544\n\nis this correct?\n\n2. Nov 21, 2015\n\n### Orodruin\n\nStaff Emeritus\nThis depends on the actual distribution in the population. You can only do what you did if this distribution is assumed to be Gaussian.\n\n3. Nov 21, 2015\n\n### toothpaste666\n\nGaussian means \"normal\" right? I am confused a bit about that. In my book they seem to use \"z\" for the test statistic and use \"t\" when the population is known to be normal. From what I can tell they are the same thing except that with z you use the standard normal table and with t you use a different table with a certain amount of degrees of freedom. I don't think I fully get it.\n\n4. Nov 21, 2015\n\n### Ray Vickson\n\nI do not actually believe you; I think you are mis-reading your book (although, to be honest, I am making this judgement sight-unseen). Typically, for an independent random sample from an underlying normal (=Gaussian) distribution with mean $\\mu$ and variance $\\sigma^2$: (1) we use $z$ and normal tables when we KNOW the value of $\\sigma$; but (2) use $t$ and t-tables when we do not know $\\sigma$, but have estimated it from the sample data itself.\n\nIn case (2), we estimate\n$$\\text{estimator of }\\: \\sigma^2 = \\frac{1}{n-1} \\sum_{i=1}^n (x_i - \\bar{x})^2$$\nwhere the sample values are $x_1, x_2, \\ldots, x_n$ and $\\bar{x} = \\frac{1}{n} \\sum_{i=1}^n x_i$ is the sample mean. In that case the jargon is that there are $n-1$ \"degrees of freedom\".\n\nIn the limit as $n \\to \\infty$ the t-distribution with (n-1) degrees of freedom becomes the standard normal, so using $z$ is like having infinitely many degrees of freedom.\n\n5. Nov 21, 2015\n\n### toothpaste666\n\nso for either of the two statistics to work, the distribution must be normal?\n\n6. Nov 21, 2015\n\n### Ray Vickson\n\nTheoretically, yes, but for a large sample-size, using the \"normal\" results give a \"reasonably accurate\" approximation. This is based on the so-called Central Limit Theorem; see, eg.,\nhttps://en.wikipedia.org/wiki/Central_limit_theorem\nor http://davidmlane.com/hyperstat/A14043.html\nor http://www.statisticalengineering.com/central_limit_theorem.htm .\n\nFor a \"reasonable\" non-normal underlying distribution, a sample size of n = 81 is likely large enough that normal-based estimates will be informative, if not absolutely accurate.\n\n7. Nov 21, 2015\n\n### toothpaste666\n\nahh ok what my book actually says is use z for samples of n>30 with \u03c3 known and if \u03c3 is not known replace \u03c3 with s and if the sample is n<30 And the population is normal use t. so since my sample is large enough, my solution to this problem should be close enough?\n\n8. Nov 21, 2015", "date": "2018-02-25 14:37:13", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/probability-a-sample-mean-will-fall-in-a-range.844276/", "openwebmath_score": 0.7689906358718872, "openwebmath_perplexity": 621.4102277465222, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290963960278, "lm_q2_score": 0.891811044719067, "lm_q1q2_score": 0.8660636540140252}}
{"url": "https://mathematica.stackexchange.com/questions/107652/double-sum-involving-condition", "text": "# Double Sum Involving Condition\n\nI would like to compute the dimensions of some small free nilpotent Lie algebras. However, I am totally new to this and I could not figure out how to write the double sum which gives the dimension of the free nilpotent Lie algebra $L(c,n)$ of step $c$ with $n$ generators.\n\nLong story short, I would like to learn how I can write the sum\n\n$$\\dim L(c,n) = \\sum_{m=1}^{c}\\frac{1}{m}\\sum_{d \\mid m}\\mu(d)n^{m/d}$$\n\nwhere $\\mu$ is the M\u00f6bius function, in Mathematica.\n\n\u2022 Maybe diml[c_, n_] := Block[{j}, Sum[DirichletConvolve[MoebiusMu[j], n^j, j, m]/m, {m, c}]]? If this does what you want, I'll write an answer. \u2013\u00a0J. M. will be back soon Feb 18 '16 at 6:32\n\u2022 @J.M. Hello. I tried this with some known values and it produced the correct values. I am sorry, since I do not know much about Mathematica this is the only way of me knowing that it works. Thank you! So please go ahead and write an answer, \u2013\u00a0Can Hatipoglu Feb 18 '16 at 7:11\n\u2022 No need to apologize! You are fortunate here that Mathematica has the necessary stuff for the computations you want to do. Give me a few, and I'll write something up. \u2013\u00a0J. M. will be back soon Feb 18 '16 at 7:14\n\u2022 FWIW, your inner sum equals JordanTotient[k, n] but unfortunately that function is not in Mathematica. \u2013\u00a0Chip Hurst Jun 2 '16 at 18:10\n\n## A story of incremental improvement\n\nLet's look at the OP's original expression again, for reference:\n\n$$\\sum_{m=1}^{c}\\frac{1}{m}\\sum_{d \\mid m}\\mu(d)n^{m/d}$$\n\nMost people here are familiar with Sum[], and would not have much trouble translating the outer summation into Mathematica syntax. The inner part,\n\n$$\\sum_{d \\mid m}\\mu(d)n^{m/d}$$\n\nis not terribly familiar to those who do not do much number theory. What this basically is saying is that the terms of the sum are indexed over the divisors of $m$. Appropriately enough, Mathematica does have a Divisors[] function. The inner sum can thus be written as\n\nSum[MoebiusMu[d] n^(m/d), {d, Divisors[m]}]\n\n\nSumming over the divisors of a number, however, is so common a number-theoretic operation that Mathematica has seen it fit to provide a DivisorSum[] function. Thus, the inner sum can also be written as\n\nDivisorSum[m, MoebiusMu[#] n^(m/#) &]\n\n\nThe story does not end here. The operation\n\n$$\\sum_{d \\mid m}f(d)g(m/d)$$\n\nfrequently turns up in number-theoretic and other contexts, that it has been given a special name: Dirichlet convolution. Mathematica, fortunately, also provides a function for evaluating convolutions, called DirichletConvolve[]. Thus, the inner sum is most compactly expressed as\n\nDirichletConvolve[MoebiusMu[j], n^j, j, m]\n\n\nThe function in the OP can now be implemented like so:\n\ndiml[c_Integer?Positive, n_Integer?Positive] := Block[{j},\nSum[DirichletConvolve[MoebiusMu[j], n^j, j, m]/m, {m, c}]]\n\n\nor, using formal symbols as suggested by The Doctor,\n\ndiml[c_Integer?Positive, n_Integer?Positive] :=\nSum[DirichletConvolve[MoebiusMu[\\[FormalJ]], n^\\[FormalJ], \\[FormalJ], m]/m, {m, c}]\n\n\n(They look messy here on SE, but should look nice when pasted into Mathematica.)\n\nHere is the function in action:\n\nTable[diml[c, n], {c, 5}, {n, 5}]\n{{1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 5, 14, 30, 55},\n{1, 8, 32, 90, 205}, {1, 14, 80, 294, 829}}\n\n\u2022 Thank you for introducing me to DirichletConvolve and for this instructive answer . This is my vote :) \u2013\u00a0ubpdqn Feb 18 '16 at 8:55\n\nPlease see J.M.'s answer (in comments) using DirichletConvolve. I would vote for this as an answer.\n\nl[c_, n_] :=Module[{r = Range[c]},\nTotal[(1/r) Total /@MapThread[MoebiusMu@#1 n^(#2/#1) & ,{Divisors /@ r, r}]]]\n\n\nI will happily delete and vote for JM answer.\n\nI like J.M.'s answer and voted for it. However, it's hard to format comments, so I've posted this as an answer instead. Rather than using Block[{j},...], you can use formal symbols instead, used to represent a formal parameter that will never be assigned a value. Moreover, if you enter\n\nDivisorSum[m, Function[d, MoebiusMu[d] n^(m/d)]] // TraditionalForm\n\n\nyou will see that Mathematica uses formal symbols automatically. Also, in many cases, I prefer to use the \"map notation\" for Function instead of # and &. For example,\n\nis executable code for this function.\n\nFinally, in TraditionalForm notation, the sum can be expressed as\n\nwhich corresponds nicely to the original sum posted in the question. However, using DirichletConvolve should be more efficient.", "date": "2020-01-26 01:56:20", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/107652/double-sum-involving-condition", "openwebmath_score": 0.5306399464607239, "openwebmath_perplexity": 1254.8681512725916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129092218133, "lm_q2_score": 0.8918110440002044, "lm_q1q2_score": 0.866063649590024}}
{"url": "https://math.stackexchange.com/questions/2629195/give-an-example-of-a-linear-transformation-whose-kernel-is-the-line-spanned-by-v", "text": "# Give an example of a linear transformation whose kernel is the line spanned by vector: $\\begin{bmatrix} -1 & 1 & 2\\end{bmatrix}^T$\n\nI know that a linear transformation could be a projection onto the plane with normal vector $\\begin{bmatrix} -1 & 1 & 2\\end{bmatrix}^T$, but finding the projection would be too difficult.\n\nI could easily think up a matrix where multiplied by $\\begin{bmatrix} -1 & 1 & 2\\end{bmatrix}^T = 0$, but I'm not sure on how to choose a matrix where $\\begin{bmatrix} -1 & 1 & 2\\end{bmatrix}^T$ is the only element of the kernel.\n\nAlso, can you please explain this Hint: \"to describe a subset as a kernel means to describe it as an intersection of planes?\"\n\n\u2022 it's not too hard to cook up a matrix with rank 2 that kills this vector \u2013\u00a0qbert Jan 31 '18 at 4:53\n\u2022 @qbert Yeah but how can I ensure nothing else send the matrix to 0? \u2013\u00a0Goldname Jan 31 '18 at 4:59\n\u2022 \"cook up\" a $3\\times 3$ matrix, making sure that (-1,1,2) is in the kernel and then show that the rank of this matrix is 2. And you know that rank + nulity = 3, so $(-1,1,2)$ must span the kernel. \u2013\u00a0Doug M Jan 31 '18 at 5:00\n\u2022 if it has rank 2, it's kernel is rank 1. If it kills your given vector, then that spans the kernel \u2013\u00a0qbert Jan 31 '18 at 5:00\n\u2022 ah I see, I forgot about that theorem, thanks \u2013\u00a0Goldname Jan 31 '18 at 5:18\n\nGuide:\n\n\u2022 $$2(-1) + 0(1) +1(2)=0$$\n\u2022 $$0(01) + (-2)(1)+1(2)=0$$\n\n\u2022 Verify that $\\begin{bmatrix} 2 & 0 & 1 \\end{bmatrix}$ and $\\begin{bmatrix} 0 & -2 & 1 \\end{bmatrix}$ are linearly independent.\n\n\u2022 Now think of someway to form your matrix and prove that span of $\\{\\begin{bmatrix} -1 & 1 & 2 \\end{bmatrix}^T\\}$ is a basis to the kernel.\n\nAs for the explanation of the hint: To describe $x$ which satisfy $Ax=0$ where $a_i^T$ are the $i$-th row means $x$ is in the intersection of $\\{ a_i^Tx = 0 : i \\in \\{ 1, \\ldots, m\\}\\}$.\n\n\u2022 I don't understand what your equations are supposed to mean \u2013\u00a0Goldname Jan 31 '18 at 4:59\n\u2022 I can easily think of a matrix, but I don't understand what is required so that [-1 1 2] is the only element in the kernel \u2013\u00a0Goldname Jan 31 '18 at 4:59\n\u2022 $[-1, 1, 2]$ is not the only element, any multiple of $[-1, 1, 2]$ is in the kernel. \u2013\u00a0Siong Thye Goh Jan 31 '18 at 5:01\n\u2022 yeah that's what I meant \u2013\u00a0Goldname Jan 31 '18 at 5:03\n\u2022 rank-nullity theorem is useful. geometrically, two non-parallel 3D plane intersect at a line. \u2013\u00a0Siong Thye Goh Jan 31 '18 at 5:03\n\nExpanding upon the comment: Choosing the following matrix works $$\\begin{bmatrix} 1&-1&1\\\\ 2&0&1\\\\ 1&1&0 \\end{bmatrix}$$ It is clear that $(-1,1,2)$ is in the kernel. The dimension of the image is $2$, so the dimension of the kernel is just $1$, by rank nullity. So the only vectors that map to $0$ under the matrix are your vector and multiples of it.\n\nPick any two vectors that together with $[-1,1,2]^T$ form an ordered basis of $\\mathbb R^3$. Relative to this basis, the matrix $\\operatorname{diag}(1,1,0)$ represents a transformation with the required properties. Apply a change of basis to it to get a matrix relative to the standard basis. For that matter, since a linear transformation is completely determined by its action on a basis, the description $T(v_1)=v_1$, $T(v_2)=v_2$, $T([-1,1,2]^T)=0$, where $v_1$ and $v_2$ are the two vectors you chose, is a complete description of the transformation, but that\u2019s likely not the answer that whoever gave you this problem is looking for.\n\nAs far as the hint goes, the kernel is a line through the origin, which can be described as the intersection of a pair of planes. The normals to those planes span the orthogonal complement of the line, so you can use them to construct the above basis.\n\n\u2022 How did you conclude that \"the matrix $diag(1,1,0)$ represents a transformation with the required properties\"? Thanks \u2013\u00a0gbox Jan 31 '18 at 9:53\n\u2022 @gbox The columns of a transformation matrix are the images of the basis vectors. In fact, for any linear transformation one can find bases for the domain and codomain such that the matrix has the form $\\tiny{\\left[\\begin{array}{c|c}I&0\\\\\\hline0&0\\end{array}\\right]}$. \u2013\u00a0amd Jan 31 '18 at 19:39\n\u2022 @gbox BTW, $\\operatorname{diag}(1,1,0)$ gives you a projection. If you instead choose some other pair of linearly independent vectors for the first two columns, you\u2019ll get a different type of rank-2 transformation. \u2013\u00a0amd Jan 31 '18 at 19:47", "date": "2019-05-21 13:19:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2629195/give-an-example-of-a-linear-transformation-whose-kernel-is-the-line-spanned-by-v", "openwebmath_score": 0.8747907280921936, "openwebmath_perplexity": 173.12634429934533, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290880402379, "lm_q2_score": 0.8918110468756548, "lm_q1q2_score": 0.8660636486565645}}
{"url": "https://math.stackexchange.com/questions/4192587/is-there-a-name-for-the-curve-traced-by-the-midpoint-of-two-moving-points-on-a-c", "text": "# Is there a name for the curve traced by the midpoint of two moving points on a circle with different speed?\n\nI'm curious about what I did in Geogebra while I was experimenting with animation. Here it is:\n\n1. Construct a circle with radius $$r$$\n2. Define two points $$A(r\\cos a\\theta , r \\sin a\\theta)$$ and $$B(r\\cos b\\theta, r \\sin b\\theta)$$. Let $$(r,0)$$ be the initial position of both $$A$$ and $$B$$.\n3. Get the midpoint of $$AB$$. Name this point $$M$$.\n4. Change $$\\theta$$ continuously starting from $$0$$ until $$A$$ and $$B$$ are both at $$(r,0)$$ again.\n\nThe midpoint, by doing the fourth step, seem to make some form of curve. I noticed that the curved traced by the midpoint does not change as long as $$a/b$$ does not change, even if $$a$$ and $$b$$ does so. Also, it doesn't matter if the values of $$a$$ and $$b$$ are swapped, but to avoid such ambiguities, we will avoid swapping of values and $$a$$ is always greater than $$b$$.\n\nFor the file, see this graph in Desmos.\n\nFor the angle, let $$p$$ be the numerator of $$a/b$$ in lowest terms. Then, we have $$0 \\leq \\theta \\leq 2p\\pi$$. (I don't know how to this in Desmos, by the way)\n\n## Examples\n\nFor $$a = 2.1$$ and $$b = 1.4$$,\n\nFor $$a = 3$$ and $$b = 1$$,\n\nAnd for $$a = 1.1$$ and $$b = 0.007$$ $$(0 \\leq \\theta \\leq 98\\pi)$$,\n\nAs for the midpoint, the coordinates is $$\\left(\\frac{r}{2}\\left(\\cos(a\\theta) + \\cos(b\\theta)\\right), \\frac{r}{2}\\left(\\sin(a\\theta) + \\sin(b\\theta)\\right)\\right)$$\n\nThis means that the parametric equation is \\begin{align*} x(\\theta) &= \\frac{r}{2}\\left(\\cos(a\\theta) + \\cos(b\\theta)\\right) \\\\[10pt] y(\\theta) &= \\frac{r}{2}\\left(\\sin(a\\theta) + \\sin(b\\theta)\\right) \\end{align*} Is there a name for these curves?\n\nUpdate 1: It seems like this is somehow related to the mathematical basis of a spirograph.\n\n... and therefore the trajectory equations take the form \\begin{align*} x(t) &= R\\left[(1-k)\\cos t+lk\\cos {\\frac{1-k}{k}}t\\right],\\\\y(t)&=R\\left[(1-k)\\sin t-lk\\sin {\\frac {1-k}{k}}t\\right]\\end{align*}\n\n\u2022 It looks like what they call the evolute of a cardioid in this page (second figure) mathcurve.com/courbes2d.gb/largeur%20constante/\u2026 Jul 7 at 13:31\n\u2022 Depending on a and b the curves can take many forms with multiple loops (b=1, a = 3) or spirals (a=2.3, b=2) so not cardioid, though one may be a special case.\n\u2013\u00a0Paul\nJul 7 at 14:31\n\u2022 This has the properties of several of the so-called transcendental curves, such as the sinusoidal spirals, Archimedean spiral, involute of a circle, and cochleoid. Curves are defined by equations, Cartesian and polar. You need to express this curve as an equation in order to define it, rather than a set of instructions. My go-to reference for plane curves is A Catalog of Special Plane Curves, by J. Dennis Lawrence, Dover Publications, 1972. Jul 9 at 14:59\n\u2022 @CyeWaldman The parametric equation can be found before Update 1. Jul 9 at 16:24\n\u2022 @soupless Thank you. From those equations, I would conclude that this is an unnamed curve. On a quick search the closest curve I can find to this is the hypotrochoid, described by $x=n\\cos t+h\\cos(n/b\\cdot t)$ and $y=n\\sin t-h\\sin(n/b\\cdot t)$. Jul 9 at 20:25\n\nI am posting this since my updates that can answer my question should be posted as an answer, not as a part of the question.\n\nWe restrict the values of $$a$$ and $$b$$ to be integers where $$a$$ and $$b$$ are both nonzero. Then, the curve seems to represent an epitrochoid given by the parametric equation \\begin{align*} x(t) &= m \\cos t - h \\cos\\left(\\frac{mt}{c}\\right) \\\\ y(t) &= m \\sin t - h \\sin\\left(\\frac{mt}{c}\\right). \\end{align*}\n\nwhere $$m = 1$$, $$h = 1$$, and $$c = \\frac{a}{b}$$ for $$0 \\leq t \\leq 2a\\pi$$.. For the parametric equation in the question, we let $$r = 2$$.\n\nWe can say that the two curves match if they are in the same magnitude and angle, and that they don't match if they don't have the same angle and magnitude. However, we'll say that they are a pseudomatch if they have the same magnitude but not having the same angle.\n\nNow, for odd $$a$$, the two curve are a pseudomatch regardless of the value of $$b$$. For even $$a$$, there are two cases:\n\n1. If $$a$$ is a power of $$2$$, then the curves are a pseudomatch if $$b$$ is a multiple of $$a$$. Otherwise, the curves are a match.\n2. If $$a$$ is not a power of $$2$$, then the curves are a pseudomatch if $$b$$ is a multiple of the greatest power of $$2$$ that divides $$a$$. An example would be $$a = 28$$, where the curves are a pseudomatch if $$b$$ is a multiple of $$4$$.\n\nAs of the moment, I don't know how to prove that this works. I am just relying on graphs from the file.", "date": "2021-12-05 14:42:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4192587/is-there-a-name-for-the-curve-traced-by-the-midpoint-of-two-moving-points-on-a-c", "openwebmath_score": 0.9765288233757019, "openwebmath_perplexity": 281.0224501229029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129093889291, "lm_q2_score": 0.8918110368115781, "lm_q1q2_score": 0.866063644099297}}
{"url": "https://math.stackexchange.com/questions/1945644/using-functions-to-find-indeterminate-forms", "text": "# Using functions to find indeterminate forms\n\nThe question is to use two functions f(x) and g(x) to show that \u221e - \u221e (infinity - infinity) is indeterminate.\nI don't really know how to get started on this. I think I need to find two limits for f(x) and g(x) that are both infinity. But how to I show that subtracting them is indeterminate?\n\nEdit: In the example we were shown, lim x --> 0 was used. But that was for the indeterminate form 0\u221e. I don't know if that applies here or not.\n\n\"Indeterminate\" means \"not having a unique value\" as $x$ tends to whatever the problem says (I am guessing $x \\rightarrow 0$ or $x \\rightarrow \\infty$, but you really should specify and not make the community you are asking for help guess). I will assume $x \\rightarrow \\infty$.\n\nSo, to show indeterminacy, I will exhibit three examples of the function pair $f(x), g(x)$ with different limiting behaviors of the difference $[f(x)-g(x)]$ as $x \\rightarrow \\infty$.\n\nExample 1: $f(x) = g(x) = x$. Here $\\lim_{x \\rightarrow \\infty}[f(x) - g(x)] = 0$.\n\nExample 2: $f(x) = x^2, \\; g(x) = x$.\n\nExample 3: See Mark Fischler's answer.:)\n\n\u2022 Thank you for the reply. The part that I don't understand is that when you use different values for f and g. Wouldn't you expect a different answer? I don't really understand how that makes it indeterminate. \u2013\u00a0user372991 Sep 28 '16 at 21:58\n\u2022 Well consider a form that is not indeterminate: $1 / \\infty$. In detail: if you have $$\\lim_{x \\rightarrow \\infty} f(x) = \\mbox{(some real cosntant)}, \\quad \\lim_{x \\rightarrow \\infty} g(x) = \\infty,$$ then, no matter which $f, g$ you use, as long as they satisfy the above conditions, you will get $$\\lim_{x \\rightarrow} {f(x) \\over g(x)} = 0.$$ In a high-level summary: The goal is, knowing the asymptotic behavior of $f, g$, to be able to ascertain the asymptotic behavior of an expression involving these two. When we cannot do that, the latter expression is indeterminate. \u2013\u00a0user8960 Sep 28 '16 at 23:18\n\nConsider $$f(x) = \\csc^2x \\\\ g(x) = \\cot^2 x \\\\ h(x) = 1/x^2$$ and look at the limit as $x\\to 0$ of the expressions $f(x)-g(x)$ and $f(x)-h(x)$.\n\nBoth of these expressions are of the form $\\infty - \\infty$\n\nBut $$\\lim_{x\\to 0} (f(x) - g(x)) = 1 \\\\ \\lim_{x\\to 0} (f(x) - h(x)) = \\frac13$$ This shows that $\\infty - \\infty$ could be $1$ or $\\frac13$; for that matter it could be $0$ (consider $f(x)-f(x)$).\n\n\u2022 If you wouldn't mind, could you explain how f(x) -g(x) = 1 and f(x) - h(x) = 1/3? I thought that they all would equal infinity? \u2013\u00a0user372991 Sep 28 '16 at 22:18\n\u2022 @Jordan He is just giving you the answers for those limits. You can verify them with L'Hospital's Rule. It's a good exercise \u2013\u00a0imranfat Sep 28 '16 at 22:52\n\u2022 On the other hand, the first one is particularly easy: Since $csc^2x - \\cot^2x = 1$ for all values of angles where they are both finite, \"obviously\" the limit as $x\\to$ any real value (or for that matter even any complex value) must be $1$. \u2013\u00a0Mark Fischler Oct 1 '16 at 21:20", "date": "2020-07-13 22:21:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1945644/using-functions-to-find-indeterminate-forms", "openwebmath_score": 0.7851292490959167, "openwebmath_perplexity": 176.3878151978045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534314192864, "lm_q2_score": 0.8824278757303678, "lm_q1q2_score": 0.8660618666156011}}
{"url": "https://math.stackexchange.com/questions/1903623/how-many-subsets-of-1-2-3-4-5-6-7-has-6-as-largest", "text": "# how many subsets of $\\{1,2,3,4,5,6,7\\}$ has $6$ as largest?\n\nLet $A =\\{1,2,3,4,5,6,7\\}$, The how many subsets have 6 as largest element?\n\nMy approach -\n\nTotal subsets are $2^7 = 128$. Then First I have to exclude the $1$ empty subset i.e. $\\{\\}$, $6$ sets with $1$ element except $\\{6\\}$, then two elements not containing $6$ as its element(I am not getting a way to count these element) and the $2$ subsets $\\{6,7\\},\\{7,6\\}$. Then I am stuck here as I am not getting the way to count these numbers. Any help?\n\n\u2022 Well, none of the sets have 7 so this is the same as asking what are the subsets of B = {1.... 6} they all have 6 so all sets are equal to $C \\cup \\{6\\}$ for some set C. C is a subset of $\\{1,...., 5\\}$ so we just need to find the subsets of {1,2...5} and add {6} to all of them. There are $2^5$ subset of {1,2....5} so there are the same number of subsets of {12,...5,6} that must have 6 which is the same number of subsets of {1,2,....7} that must have 6 and must not have 7. \u2013\u00a0fleablood Aug 25 '16 at 19:58\n\u2022 128 subsets. Half have 6 as an element. 1/4, contain both 6 and 7. So 1/4 have 6 and do not have 7. \u2013\u00a0Doug M Aug 25 '16 at 20:23\n\nRemember how we determined that the number of subsets was 2 to the power of the number of elements.\n\nWe did that because for each element, $a$, either $a$ could be in a subset. So the total number of sets was the product of all the choices each of which was 2.\n\nThis is the same. Either 1 is in a subset or not. That 2 choices. Either 2 is in the subset or not. That's $2*2$ choice. Keep it up EXCEPT notice $6$ must be in the subset so that is only $1$ choice and $7$ must not be in the set so that's only one set.\n\nSo the number of sets is $2*2....*2*1*1$ which is what.\n\n....\n\nOr\n\n....\n\nDon't eliminate the sets one at a time. Remove ALL the subsets that do not have $6$ in them. How many do not have $6$. Remove ALL the subsets that do have $7$. How many is that? Then to avoid double counting add back the ones that had $6$ and didn't have $7$. How many is that?\n\n....\n\nOr\n\n....\n\nFigure 1/2 of the 128 have 6 and half do not. That only leaves half of them acceptable. Then half of those have 7 and half to not. That leaves only half of those acceptable.\n\n....\n\nOr\n\n....\n\nNo set has $7$. So all the elements are taken from {$1,2,3,4,5,6$}.\n\nAll sets have $6$. So all elements that aren't $6$ are taken from {$1,2,3,4,5$} and $6$ is always added to it.\n\nThere are $2^5$ subsets of {$1,2,3,4,5$} and we are only taking those and sticking a $6$ into them. There are $2^5$ such sets.\n\n\u2022 I like your way of dividing it up. That makes good sense! \u2013\u00a0John Aug 25 '16 at 20:16\n\u2022 You have to be careful in assuming that your condition is exactly 1/2 though. In this case you can but if you were asked say How many sets have the numbers add up to an even number and contain the number 5 you can't assume half the sets add up to an even number and of those half of them contain 5. \u2013\u00a0fleablood Aug 25 '16 at 20:23\n\nHint:\n\nInclude $6$. How many ways are there to include zero through five members of $\\{1,2,3,4,5\\}$?\n\n(Order doesn't matter with sets.)\n\n\u2022 okay now i got. There are total subsets = 2^6 = 64. \u2013\u00a0Brij Raj Kishore Aug 25 '16 at 19:58\n\u2022 The element $6$ has to be there. The element $7$ can't be in there. But any, all, or none of $1$ through $5$ can be. In other words, if $S_5 = \\{1,2,3,4,5\\}$, then the union of $\\{6\\}$ and any subset of $S_5$ will have $6$ as the largest element. You figured out the number of subsets for a set of $7$ elements. How about for $5$ elements? \u2013\u00a0John Aug 25 '16 at 20:02\n\u2022 Regarding your answer, not quite. It's not optional whether $6$ is in there or not. \u2013\u00a0John Aug 25 '16 at 20:03\n\u2022 So the answer is 2^6 -1 - 5. Am i correct? But in option there are 4 options as - 32,31,64,128. \u2013\u00a0Brij Raj Kishore Aug 25 '16 at 20:05\n\u2022 No set has 7. All sets have 6. So all sets are a subset of {1,2,3,4,5} unioned up with {6}. How many sets are there. There is {6} = {} + {6}; there is {1,4,5,6} = {1,4,5} + {6}. They are all C + {6} where $C \\subset$ {1,2,3,4,5}. How many such sets are there. \u2013\u00a0fleablood Aug 25 '16 at 20:14\n\nThink about it we need $6$ as an element but we can't have $7$ because $7$ is larger than $6$, we can have anything else.\n\nWhat we're looking for: How many subsets have $6$ as an element but not $7$?\n\nWe must choose to include $6$, that gives $1$ choice. We may choose to to include $1$ or not, that gives $2$ choices. We may choose to include $2$ or not, that gives $2$ choices. We may choose to include $3$ or not, that gives $2$ choices. We may choose to include $4$ or not, that gives $2$ choices. We may choose to include $5$ or not, that gives $2$ choices. We must choose not to include $7$, that gives $1$ choice. By the multiplication principle the number subsets with $6$ but not $7$ is:\n\n$$1\u20222\u20222\u20222\u20222\u20222\u20221=2^5$$", "date": "2019-08-23 09:22:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1903623/how-many-subsets-of-1-2-3-4-5-6-7-has-6-as-largest", "openwebmath_score": 0.8369197845458984, "openwebmath_perplexity": 292.80404298376527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986979508737192, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8660516197054005}}
{"url": "http://math.stackexchange.com/questions/48876/how-to-determine-number-with-same-amount-of-odd-and-even-divisors", "text": "# How to determine number with same amount of odd and even divisors\n\nWith given number N, how to determine first number after N with same amount of odd and even divisors? For example if we have N=1, then next number we are searching for is :\n\n2\nbecause divisors:\nodd : 1\neven : 2\n\n\nI figured out that this special number can't be odd and obviously it can't be prime. I can't find any formula for this or do i have just compute it one by one and check if it's this special number ? Obviously 1 and number itself is divisors of this number. Cheers\n\n-\nIs this homework? How many powers of 2 can divide your special numbers? \u2013\u00a0user641 Jul 1 '11 at 11:27\nWe have holiday, i don't go to school :-) Is your question is a hint? \u2013\u00a0Spinach Jul 1 '11 at 11:52\nHere's a hint. Try and pair up the even and odd factors in a natural way. You gave 2 = 1*2 as an example. How about factors of 2p, where p is an odd prime? \u2013\u00a0hardmath Jul 1 '11 at 12:05\nWhenever you have two finite sets of the same size, you can think about whether there is a one-to-one correspondence (a bijection) between them. You might think whether you could apply this observation to the divisors of a number ... \u2013\u00a0Mark Bennet Jul 1 '11 at 12:07\n\nTo get some idea of what's going on, we do like other scientists do, we experiment.\n\nSpecial numbers will be even, so we write down the number of odd divisors, even divisors, for the even numbers, starting with $2$. If a number turns out to be special, we put a $\\ast$ in its row.\n\nSo we make a table, giving the number, how many odd divisors it has, how many even. Calculations are easy, but we must be very careful, since errors could lead us down the wrong path.\n\n$2 \\qquad 1 \\qquad 1\\quad\\ast$\n\n$4 \\qquad 1 \\qquad 2$\n\n$6 \\qquad 2 \\qquad 2\\quad\\ast$\n\n$8 \\qquad 1 \\qquad 3$\n\n$10 \\qquad 2 \\qquad 2\\quad\\ast$\n\n$12 \\qquad 2 \\qquad 4$\n\n$14 \\qquad 2 \\qquad 2\\quad\\ast$\n\n$16 \\qquad 1 \\qquad 4$\n\n$18 \\qquad 3 \\qquad 3\\quad\\ast$\n\nWe could easily go on for a while. It is definitely not a waste of time, since it is useful to be well-acquainted with the structure of the smallish numbers that we bump into often.\n\nA pattern seems to jump out: every second even number seems to be special. It looks as if \"special\" numbers are not all that special! It can be dangerous to jump to conclusions from data about small integers. But in this case, the conclusion turns out to be correct.\n\nThe special numbers, so far, all have the shape $2a$, where $a$ is an odd number. They are divisible by $2$ but not by $4$. The even numbers in our list that are not special are all divisible by $4$.\n\nNow we try to prove that every number that is divisible by $2$ but not by $4$ is special, and that the others are not.\n\nTake an odd number $b$, and look at the number $2b$. Think about the divisors of $2b$. If $k$ is an odd divisor of $2b$, then $2k$ is an even divisor of $2b$, and vice-versa.\n\nIf $k$ is an odd divisor of $2b$, call $2k$ the friend of $k$. Split the divisors of $2b$ into pairs of friends. For example, if $b=45$, we have the following pairs of friends.\n\n$$(1,2)\\qquad (3,6) \\qquad(5,10)\\qquad(9,18)\\qquad(15,30) \\qquad (45,90)$$\n\nWe have split the divisors of $2b$ into pairs of friends. Each pair has one odd number and one even number, so $2b$ has exactly as many odd divisors as even divisors.\n\nNow let's show that no number divisible by $4$ can be special. The idea is that if a number is divisible by $4$, then it has \"too many\" even divisors. I will not write out the details, but you should. The idea goes as follows. Take a number $n$ that is divisible by $4$, like $36$ or $80$. Split the divisors of $n$ into teams. If $k$ is an odd divisor of $n$, put into the same team as $k$ the numbers $2k$, $4k$, and so on however far you can go.\n\nHere are the teams for $n=36$. $$(1,2,4) \\qquad (3,6,12)\\qquad (9,18,36)$$\n\nEach team has more even numbers than odd numbers, so $n$ has more even divisors than odd divisors. That means $n$ can't be special.\n\nNow let's get to your question: what is the first special number after $N$?\n\nIf $N$ is divisible by $4$, the first special number after $N$ is $N+2$. If $N$ is divisible by $2$ but not by $4$, the first special number after $N$ is $N+4$. If $N$ has remainder $1$ on division by $4$, the first special after $N$ is $N+1$, and if the remainder is $3$, the first special is $N+3$. These facts follow easily from what we have discovered about special numbers.\n\nFormulas: We have been operating without formulas, just straight thinking. But I should mention a relevant formula. Let $n$ be an integer greater than $1$, and express $n$ as a product of powers of distinct primes. In symbols, let $$n=p_1^{a_1}p_2^{a_2} \\cdots p_k^{a_k}$$ Then the number of divisors of $n$ is given by $$(a_1+1)(a_2+1) \\cdots(a_k+1)$$\n\nFor example, $720=2^43^25^1$. The number of (positive) divisors of $n$ is $(4+1)(2+1)(1+1)$.\n\nThe formula that gives the number of divisors of $n$ is not hard to prove. Try to produce a proof! The formula could be adapted to give a count of the odd divisors of $n$, and of the even divisors. Then we could use these formulas to identify the special numbers. But formulas cannot do the thinking for you. So as a first approach, the way we tackled things is much better than trying to use a formula.\n\n-\nThanks a lot ! Seems like experimenting is very useful, thanks for teaching me something :) And thanks for an so full answer. Chris \u2013\u00a0Spinach Jul 1 '11 at 14:20\n+1; this answer embodies a very important lesson. A lot of questions here on math.SE could have been answered by the posters if they had decided to experiment first... \u2013\u00a0Qiaochu Yuan Jul 2 '11 at 4:11\n@Qiaochu Yuan: A very common problem with students is that they do not really understand that mathematical questions, even those that seem quite abstract, are about concrete things. Unlike working mathematicians, students, even pretty good ones, are often naive formalists. \u2013\u00a0Andr\u00e9 Nicolas Jul 2 '11 at 4:28\n\nI wrote a C++ application. My results are all the following numbers:\n\n0, 2, 6, 10, 14, 18, 22, 26, 30, 34, ....\n\nSo, first zero, then two and then each time plus 4.\n\nThis means in general that if $n \\equiv 2 (mod 4)$ is true, your number is one of the kind you're searching. Except from the first zero.\n\nHere is the code of the application:\n\nfor (int i = 0; i < 1000; i += 2)\n// Always +2 because of odd numbers don't have any even divisors\n{\nint even = 0;\nint odd = 0;\nfor (int d = 1; d <= i; ++d)\n{\nif (i % d == 0)\n{\nif (d % 2 == 0)\n{\neven++;\n} else\n{\nodd++;\n}\n}\n}\nif (even == odd)\n{\nprintf(\"%d\\n\", i);\n}\n}\n\n-\nWe know that there's always 1 divisor odd ( 1 ) and 1 divisor even for sure ( N ), so we need only to check to n/2 or even to sqrt(n) ( not sure if we won't miss some of divisors expect n/2 ). But nice application. \u2013\u00a0Spinach Jul 2 '11 at 14:01\n\nFor a given integer $n$, every divisor larger than $\\sqrt{n}$ is paired with a divisor smaller than $\\sqrt{n}$. Use this to figure out a general principle.\n\n-\nI don't think that's very much help in this case. There is a much simpler way, hinted at in the comments to the OP. \u2013\u00a0TonyK Jul 1 '11 at 13:46\n@TonyK - I think this works as a pairing for this problem, though it involves making sure that you know that $n$ is not a square, so that there is a pairing - but since the number of distinct factors is known to be even, n can't be a square. \u2013\u00a0Mark Bennet Jul 1 '11 at 16:25\nThe pairing works of ncmathsadist works just fine, both for the proof that we have equality when $n \\equiv 2 \\pmod{4}$, and that there are more even divisors than odd when $n \\equiv 0 \\pmod{4}$. The pairing is in effect the same as mine when $n \\equiv 2 \\pmod 4$. But the mathematical description of the pairing is substantially better than mine, because the ncmathsadist pairing has a number of other uses. More detail would undoubtedly have been useful to the student. \u2013\u00a0Andr\u00e9 Nicolas Jul 2 '11 at 4:43\nI didn't want to give away the whole thing. \u2013\u00a0ncmathsadist Jul 2 '11 at 13:25", "date": "2016-06-25 03:44:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/48876/how-to-determine-number-with-same-amount-of-odd-and-even-divisors", "openwebmath_score": 0.8541073203086853, "openwebmath_perplexity": 268.2537258801534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795083542037, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8660516083010532}}
{"url": "https://math.stackexchange.com/questions/3141173/is-it-possible-for-an-eigenspace-to-have-more-than-one-vector-in-its-basis-what", "text": "# Is it possible for an eigenspace to have more than one vector in its basis? What would that imply? Would every vector in the basis be an eigenvector?\n\nSo far I've been seeing that the vector that makes up a basis for the null space of the matrix $$A-Ix$$ (where x is an eigenvalue) is the eigenvector corresponding to the eigenvalue $$x$$. But I've never run into a situation where the basis had more than one vector in it.\n\nLet's say the basis of the null space of $$A-2*I$$ was made up of two vectors. Would that mean those two (and only those two) vectors are the eigenvectors corresponding to the eigenvalue $$2$$ for the matrix $$A$$?\n\nOne thing of note is that, from what I understand, the identity transformation (in any space) has only one eigenvalue, $$1$$, but it has infinitely many eigenvectors (not sure if there is a vector space where this wouldn't be true, but it's true for $$R^n$$ at least). Not sure how this fact fits in here.\n\nJust wanting to be sure I'm not misunderstanding something. I'm grateful for any help.\n\n\u2022 Every nonzero vector in an eigenspace is an eigenvector. \u2013\u00a0amd Mar 9 at 20:10\n\nYes of course, you can have several vectors in the basis of an eigenspace.\n\nFirst, when you have only one vector $$v$$ in a basis for a matrix $$A$$, with eigenvalue $$\\mu$$, then any multiple of this vector is in the basis : $$\\forall c \\in \\mathbb{R}, \\ A(cv)=\\mu(cv)$$\n\nThen if you have two vectors in a basis, $$v$$ and $$w$$, any linear combination of these two vector will be an eigenvector for $$\\mu$$: $$\\left\\{ \\begin{array}{l} Av=\\mu v\\\\ Aw=\\mu w\\\\ \\end{array} \\right. \\Rightarrow A(cv+dw) = \\mu (cv+dw)$$\n\nFor exemple, let $$A=J-I$$ a matrix $$n\\times n$$ of all 1, except 0 in the diagonal (this exemple comes from graph theory and the complete graph $$K_n$$). Then $$A$$ has eigenvalues $$n-1$$ and $$-1$$, and while the eigenspace associated with $$n-1$$ has dimension $$1$$ (there is only 1 vector in its basis), the eigenspace associated with the eigenvalue $$-1$$ has dimension $$n-1$$. Hence you can find an orthogonal basis of $$n-1$$ vectors. Any vector $$v$$ verifying $$Av=-v$$ can be written as a linear combination of the vectors in this basis.\n\nYou are understanding things just right.\n\nFor example, the $$3 \\times 3$$ matrix $$\\begin{bmatrix} 2 & 0 & 0 \\\\ 0 & 3 & 0 \\\\ 0 & 0 & 3 \\end{bmatrix}$$ has two eigenvalues, $$2$$ and $$3$$.\n\nEvery nonzero vector on the $$x$$-axis is an eigenvector for eigenvalue $$2$$, and a basis for that eigenspace.\n\nEvery nonzero vector in the $$y$$-$$z$$ plane is an eigenvector for eigenvalue $$2$$. Any two linearly independent vectors in that plane form a basis of eigenvectors for that eigenvalue.", "date": "2019-10-15 08:24:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3141173/is-it-possible-for-an-eigenspace-to-have-more-than-one-vector-in-its-basis-what", "openwebmath_score": 0.9500262141227722, "openwebmath_perplexity": 98.47559012698471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434786819155, "lm_q2_score": 0.888758789809389, "lm_q1q2_score": 0.8660374818815622}}
{"url": "https://byjus.com/question-answer/let-a-denote-the-element-of-the-i-th-row-and-j-th-column-in/", "text": "Question\n\n# Let $$a$$ denote the element of the $${i^{th}}$$ row and $${j^{th}}$$ column in a $$3 \\times 3$$ matrix and let $${a_{ij}} = \\, - {a_{ji}}$$ for every i and j then this matrix is an -\n\nA\nOrthogonal matrix\nB\nsingular matrix\nC\nmatrix whose principal diagonal elements are all zero\nD\nskew-symmetric matrix\n\nSolution\n\n## The correct option is B skew-symmetric matrixIf $$a_{ij}$$ is the element of $$i^{th}$$ row and $$j^{th}$$ column and $$a_{ij}=-a_{ji}$$then lets assume$$A=\\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\\\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\\\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \\end{bmatrix}$$Where, $$a_{ij}=-a_{ji}$$$$a_{11}=-a_{11}\\implies a_{11}=0$$, similarly $$a_{22}=a_{33}=0$$and $$a_{21}=-a_{12}, a_{31}=-a_{13}, a_{32}=-a_{23}$$Putting the values$$A=\\begin{bmatrix} { 0 } & { a }_{ 12 } & { a }_{ 13 } \\\\ { -a }_{ 12 } & { 0 } & { a }_{ 23 } \\\\ { -a }_{ 13 } & { -a }_{ 23 } & { 0 } \\end{bmatrix}$$$${ A }^{ T }=\\begin{bmatrix} { 0 } & { -a }_{ 12 } & { -a }_{ 13 } \\\\ { a }_{ 12 } & { 0 } & { -a }_{ 23 } \\\\ { a }_{ 13 } & { a }_{ 23 } & {0 } \\end{bmatrix}\\\\$$and$$-A=\\begin{bmatrix} { 0 } & -{ a }_{ 12 } & -{ a }_{ 13 } \\\\ { a }_{ 12 } & { 0 } & -{ a }_{ 23 } \\\\ { a }_{ 13 } & { a }_{ 23 } & { 0 } \\end{bmatrix}$$thus$$A^T=-A$$$$\\therefore$$ The matrix is a skew-symmetric matrix.Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-28 12:22:17", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/let-a-denote-the-element-of-the-i-th-row-and-j-th-column-in/", "openwebmath_score": 0.9499050974845886, "openwebmath_perplexity": 1213.7262986456699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419710536893, "lm_q2_score": 0.8757870013740061, "lm_q1q2_score": 0.8660149446618723}}
{"url": "http://bootmath.com/need-to-prove-the-sequence-a_n1frac122frac132cdotsfrac1n2-converges.html", "text": "Need to prove the sequence $a_n=1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}$ converges\n\nI need to prove that the sequence $a_n=1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:\n\nMonotonic:\n\nThe sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\\leq a_{n+1}$\n$$a_1=1\\leq 1+\\frac{1}{2^2}=a_2$$\n\nNeed to show that $a_{n+1}\\leq a_{n+2}$\n$$a_{n+1}=1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}+\\frac{1}{(n+1)^2}\\leq 1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}+\\frac{1}{(n+1)^2}+\\frac{1}{(n+2)^2}=a_{n+2}$$\nThus the sequence is monotone and increasing.\n\nBoundedness:\n\nSince the sequence is increasing it is bounded below by $a_1=1$.\nUpper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don\u2019t know what my thinking process should be to find an upper bound.\n\nCan anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?\n\nSolutions Collecting From Web of \"Need to prove the sequence $a_n=1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}$ converges\"\n\nYour work looks good so far. Here is a hint:\n$$\\frac{1}{n^2} \\le \\frac{1}{n(n-1)} = \\frac{1}{n-1} \u2013 \\frac{1}{n}$$\n\nTo elaborate, apply the hint to get:\n$$\\frac{1}{2^2} + \\frac{1}{3^2} + \\frac{1}{4^2} + \\cdots + \\frac{1}{n^2} \\le \\left(\\frac{1}{1} \u2013 \\frac{1}{2}\\right) + \\left(\\frac{1}{2} \u2013 \\frac{1}{3}\\right) + \\left(\\frac{1}{3} \u2013 \\frac{1}{4}\\right) + \\cdots + \\left(\\frac{1}{n-1} \u2013 \\frac{1}{n}\\right)$$\n\nNotice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.\n\nAlso notice that all terms on the right-hand side cancel out except for the first and last one. Thus:\n$$\\frac{1}{2^2} + \\frac{1}{3^2} + \\frac{1}{4^2} + \\cdots + \\frac{1}{n^2} \\le 1 \u2013 \\frac{1}{n}$$\n\nAdd $1$ to both sides to get:\n$$a_n \\le 2 \u2013 \\frac{1}{n} \\le 2$$\n\nIt follows that $a_n$ is bounded from above and hence convergent.\n\nIt is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.\n\nBesides to Ayman\u2019s neat answer, you may take $f(x)=\\frac{1}{x^2}$ over $[1,+\\infty)$ and see that $f'(x)=-2x^{-3}$ and then is decreasing over $[1,+\\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $$\\sum_{k=1}^{+\\infty}\\frac{1}{n^2}$$ to see the series is convergent. Now your $a_n$ is the $n-$th summation of this series.\n\nYou can show this geometrically too.\n\nIf you take a square, and divide the height into $\\frac 12$, $\\frac 14$, $\\frac 18$, and so forth, doubling the denominator on each deal.\n\nNow take the squares $\\frac 12$ and $\\frac 13$: these go onto the top shelf.\n\nOn the second shelf go $\\frac 14$ to $\\frac 17$. These fractions are all less than $\\frac 14$, so fit onto the second shelf. Likewise, squares from $8$ to $15$ go onto the third shelf, each $\\frac 1n$ is smaller than $\\frac 18$, and so forth.\n\nTherefore $\\sum_{n=2}^{\\infty}\\frac 1n < 1$, and therefore the whole lot is less than two squares.\n\nby the integral test :\n\n$\\int^\\infty_1 \\frac{1}{n^2} dn \\le \\sum_{i=1}^\\infty \\frac{1}{n^2}\\le 1+\\int^\\infty_1 \\frac{1}{n^2}dn$ .\n\nyou can compute the integral so the answer is :\n\n$1 \\le \\sum_{i=1}^\\infty \\frac{1}{n^2}\\le 2$ .\n\nbecause : $\\int^\\infty_1 \\frac{1}{n^2} dn =1$\n\nHint: Prove the the following holds for all $n$ by induction.\n\n$$\\sum_1^n \\frac{1}{k^2} \\le 2 \u2013 \\frac{1}{n}.$$\n\nIs it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work.\n\nYou can use the same technique to bound other values of the zeta function. For example, try showing $\\zeta(3)$ is bounded above by $\\frac{3}{2}$.\n\n(I am copying my answer from a duplicate question that was closed as a copy of this one since the induction approach is not available here.)\n\nThis here should work with $n \\geq 1$ :\n\n$$s_n = \\sum\\limits_{n=1}^\\infty \\frac{1}{n^2}= \\frac{1}{1} + \\frac{1}{4} +\\frac{1}{9} + \\frac{1}{16}+ \u2026+ \\frac{1}{n\u00b2}$$$$b_n = \\sum\\limits_{n=1}^\\infty \\frac{1}{2^{n-1}} = \\frac{1}{1} + \\frac{1}{2} +\\frac{1}{4}+\\frac{1}{8} + \u2026+ \\frac{1}{2^{n-1}}$$\n\n$b_n$ is directly compared greater than $s_n$ :\n$$s_n < b_n$$\nand $b_n$ converges, because of its ratio test :\n$$\\frac{1}{2^{n-1+1}} / \\frac{1}{2^{n-1}} = \\frac{2^{n-1}}{2^{n}} = \\frac{1}{2} < 1$$\n\nNotice that $2k^2 \\geq k(k+1) \\implies \\frac{1}{k^2} \\leq \\frac{2}{k(k+1)}$.\n\n$$\\sum_{k=1}^{\\infty} \\frac{2}{k(k+1)} = \\frac{2}{1 \\times 2} + \\frac{2}{2 \\times 3} + \\frac{2}{3 \\times 4} + \\ldots$$\n\n$$\\sum_{k=1}^{\\infty} \\frac{2}{k(k+1)} = 2\\Big(\\, \\Big(1 \u2013 \\frac{1}{2}\\Big) + \\Big(\\frac{1}{2} \u2013 \\frac{1}{3} \\Big) + \\Big(\\frac{1}{3} \u2013 \\frac{1}{4} \\Big) + \\ldots \\Big)$$\n\n$$\\sum_{k=1}^{\\infty} \\frac{2}{k(k+1)} = 2 (1) = 2$$.\n\nTherefore $\\sum_{k=1}^{\\infty} \\frac{1}{k^2} \\leq 2$.\n\n$a_n=\\frac{1}{n^2}$ and because $a_n>0$ we have $|a_n|=a_n$\nFirst: check the necessary condition\n$$\\lim_{n\\to +\\infty} na_n=\\lim_{n\\to +\\infty}\\frac{1}{n}=0$$\nSecond: check D\u2019Alembert\u2019s ratio test\n$$\\lim_{n\\to +\\infty}|\\frac{a_{n+1}}{a_n}|=\\lim_{n\\to +\\infty}\\frac{a_{n+1}}{a_n}=\\lim_{n\\to +\\infty}(\\frac{n}{n+1})^2=1$$\nThird: Because the answer of D\u2019Alembert\u2019s test is $1$, you should use Raabe\u2019s test:\n$$\\lim_{n\\to +\\infty}n(1-|\\frac{a_{n+1}}{a_n}|)=\\lim_{n\\to +\\infty}n(1-\\frac{a_{n+1}}{a_n})=\\lim_{n\\to +\\infty}n(\\frac{2n+1}{n^2+2n+1})=2\\gt1$$\nso the series is convergent", "date": "2018-06-21 23:43:17", "meta": {"domain": "bootmath.com", "url": "http://bootmath.com/need-to-prove-the-sequence-a_n1frac122frac132cdotsfrac1n2-converges.html", "openwebmath_score": 0.9254652857780457, "openwebmath_perplexity": 209.7219452408486, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419703960399, "lm_q2_score": 0.8757870013740061, "lm_q1q2_score": 0.8660149440859115}}
{"url": "http://mathhelpforum.com/discrete-math/274100-show-64-divides-9-n-56n-63-a.html", "text": "# Thread: Show that 64 divides 9^n + 56n + 63\n\n1. ## Show that 64 divides 9^n + 56n + 63\n\nShow that 64 divides 9^n + 56n + 63 for all positive integers n.\n\nBased off of basic computation I can see that for numbers like n=1, n=2, etc this holds but how would I show it for all positive integers?\n\n2. ## Re: Show that 64 divides 9^n + 56n + 63\n\nare you familiar with induction?\n\n$9 + 56 + 63 = 128 = 2 \\cdot 64$\n\nso this is true for $n=1$ and we would say $P_1 = True$\n\nnext, we assume that $P_n=True$ and we use this to show that $P_{n+1} = True$\n\nhere we'd do\n\n\\begin{align*} &9^{n+1} + 56(n+1) + 63 \\\\ = &~9^n\\cdot 9 + 56n + 56 + 63 \\\\ = &~9^n+ 56n + 63 + 8\\cdot 9^n + 56 \\\\ = &~64k + 8\\cdot 9^n + 56 \\text{; because }P_n=True \\\\ =&~64k + 8(7+9^n) \\end{align*}\n\nclearly $64k$ is divisible by $64$ so we need to show that $8(7+9^n)$ is divisible by $64$ or alternatively that\n\n$(7+9^n)$ is divisible by $8$\n\nUsing the same method we are using to show the original statement\n\n$n=1 \\Rightarrow 7+9^n = 16 = 2\\cdot 8$ so $\\tilde{P}_1=True$\n\nNow you show that $\\tilde{P}_n \\Rightarrow \\tilde{P}_{n+1}$ for this and thus that $7+ 9^n$ is divisible by 8\n\nand you will have shown that\n\n$9^{n+1} + 56(n+1) + 63$ is divisible by $64$\n\nand thus that for the original statement\n\n$P_n \\Rightarrow P_{n+1}$\n\nand thus it is True for all $n\\geq 1$\n\nso see if you can prove that $\\tilde{P}_n \\Rightarrow \\tilde{P}_{n+1}$ it's not hard\n\n3. ## Re: Show that 64 divides 9^n + 56n + 63\n\nYou can also note:\n\n$\\displaystyle 9^n + 56n + 63 =$\n\n$\\displaystyle 9^n + 56n + (8n - 8n) + 63 + (1 - 1) =$\n\n$\\displaystyle 9^n + (56n + 8n) + (63 + 1) - 1 - 8n =$\n\n$\\displaystyle 9^n + (64n + 64) - 1 - 8n$\n\n64 divides (64n + 64), so it remains to show that 64 also divides $\\displaystyle \\ \\ 9^n - 1 - 8n$.\n\nIt has been shown or mentioned it's true for n = 1 in at least one prior post. Let's look at $\\displaystyle \\ \\ n \\ge 2$:\n\n$\\displaystyle 9^n - 1 - 8n =$\n\n$\\displaystyle (8 + 1)^n - 1 - 8n =$\n\n$\\displaystyle \\bigg(8^n \\ + \\ n\\cdot8^{n - 1} \\ + \\ \\dfrac{n(n - 1)}{2}8^{n - 2} \\ + \\ . . . \\ + \\ \\dfrac{n(n - 1)}{2}8^2 \\ + \\ n\\cdot8^1 \\ + \\ 1\\bigg) \\ - \\ 1 \\ - \\ 8n =$\n\n$\\displaystyle 8^n \\ + \\ n\\cdot8^{n - 1} \\ + \\ \\dfrac{n(n - 1)}{2}8^{n - 2} \\ + \\ . . . \\ + \\ \\dfrac{n(n - 1)}{2}8^2 \\ + \\ 8n \\ - \\ 8n \\ + \\ 1 \\ - \\ 1 =$\n\n$\\displaystyle 8^n \\ + \\ n\\cdot8^{n - 1} \\ + \\ \\dfrac{n(n - 1)}{2}8^{n - 2} \\ + \\ . . . \\ + \\ \\dfrac{n(n - 1)}{2}8^2$ . . . . . . . . . **\n\nThe fractions in front of the powers of eights are combinations and are thus positive integers. The smallest power of eight is equal to 64,\nso this expression is also divisible by 64.\n\n** . . . . If n = 2, there will only be one term here. If n = 3, there will only be two terms here. And so on.\n\n4. ## Re: Show that 64 divides 9^n + 56n + 63\n\nThe induction step can be done as follows:\n\n\\displaystyle \\begin{align*}9^{n+1} + 56(n+1) + 63 &= 9\\cdot 9^n + 56n + 56 + 63 \\\\ &= 8( 9^n + 7) + (9^n + 56n + 63) \\end{align*}\n\nwhere the final parenthesis is a multiple of 64 by the inductive hypothesis. So all we need is for $\\displaystyle 8$ to divide $\\displaystyle (9^n + 7)$.\n\n$\\displaystyle 9^n + 7 \\equiv 1^n - 1 \\pmod{8} = 1-1 \\pmod{8} = 0 \\pmod{8}$\n\nQED", "date": "2018-03-21 07:24:14", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/274100-show-64-divides-9-n-56n-63-a.html", "openwebmath_score": 0.9954174160957336, "openwebmath_perplexity": 375.1864141124909, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429129677614, "lm_q2_score": 0.8791467675095292, "lm_q1q2_score": 0.8659972927937779}}
{"url": "http://math.stackexchange.com/questions/853615/proving-an-expression-is-composite", "text": "# Proving an expression is composite\n\nI am trying to prove that $n^4 + 4^n$ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.\n\nThis problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.\n\nI would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.\n\n-\nWrite it as a difference of squares. \u2013\u00a0 Adam Hughes Jul 1 '14 at 19:54\nHow? It's $n^4 + 4^n$ not $n^4 - 4^n$ \u2013\u00a0 Mathmo123 Jul 1 '14 at 19:55\n@Mathmo123: see my answer below. \u2013\u00a0 Adam Hughes Jul 1 '14 at 20:03\n\n$(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...\n\n-\nAh. Very neat!! \u2013\u00a0 Mathmo123 Jul 1 '14 at 20:05\nOriginally, I was not sure how I should go about writing it as a difference of squares. Thanks for your help. \u2013\u00a0 pidude Jul 1 '14 at 20:12\n\nHint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?\n\nLet me know if you need further hints.\n\n-\nI originally tried doing that but to no avail. Evaluated at 3 I obtain 145 which has factors of 5 and 29. At 5, the expression equals 1649, which has factors of 17 and 97. I will keep looking for a pattern and let you know if I need more hints. Thanks for your help. \u2013\u00a0 pidude Jul 1 '14 at 19:59\nAh... using this method, there will be a difference between multiples of 5 and other odd numbers. You will find that for odd numbers that are not a multiple of 5, 5 will be a divisor \u2013\u00a0 Mathmo123 Jul 1 '14 at 20:01\nOk, so I was able to prove that. Now I'm working numbers which are multiples of 5. \u2013\u00a0 pidude Jul 1 '14 at 20:09\n\nI am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.\n\n\u2022 We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by\n1. $n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law\n2. Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication\n3. $(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression\n4. $(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression\n5. $(n^2+2^n)^2-2(n^2)(2^n)$ from step 2\n6. $(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential\n\u2022 We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again\n1. Treat $n^2+2^n$ as $a$\n2. Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer\n3. So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$\n4. $2^{2m}n^2=(n2^m)^2$ associative law\n5. Treat $n2^m$ as $b$\n\u2022 It implies that both $a$ and $b$ are both positive integer\n\u2022 From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$\n\u2022 Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$\n-\n\nSome interesting factorizations of a polynomial of type $x^4+\\text{const}$: $$x^4+4=(x^2+2x+2)(x^2-2x-2) \\tag{1}$$\n\n$$x^4+1=(x^2+\\sqrt[]{2}x+1)(x^2-\\sqrt[]{2}x+1) \\tag{2}$$\n\nSo one can ask, how to select the coefficients $a,b,c,d$ in\n\n$$(x^2+ax+b)(x^2+cx+d) \\tag{3}$$\n\nsuch that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.\n\nIf we expand $(3)$ we get\n\n$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$\n\nAnd the coefficients disappear, if\n\n$$\\begin{eqnarray} c+a &=& 0 \\\\ d+ ac +b &=& 0 \\\\ ad+bc &=& 0 \\end{eqnarray}$$\n\nWhen solving for $b,c,d$ we get\n\n$$\\begin{eqnarray} c &=& -a \\\\ b &=& \\frac{a^2}{2} \\\\ d &=& \\frac{a^2}{2} \\end{eqnarray}$$\n\nand therefore\n\n$$x^4+\\frac{a^4}{4} = (x^2+a x+\\frac{a^2}{2})(x^2-ax+\\frac{a^2}{2})$$\n\nFor $a=2$ this gives $(1)$, $a=\\sqrt[]{2}$ this gives $82)$ . Substituting $a=2^{t+1}$ we get\n\n$$x^4+4^{2t+1} = (x^2+2\\cdot 2^t x+2^{2t+1})(x^2-2\\cdot 2^tx+2^{2t+1})$$\n\nSubstituting $x=n=2t+1$ gives the required result for odd $n$.\n\n-\n\n@Mathmo123 As I observe the behavior of the multiply operation, I see..\n\n\u2022 $v \\times v$ produce $v$\n\u2022 $v \\times o$ produce $v$\n\u2022 $o \\times o$ produce $o$\n\nand for the plus operation, I see..\n\n\u2022 $v+v$ produce $v$\n\u2022 $v+o$ produce $o$\n\u2022 $o+o$ produce $v$\n\nwhere $v$ is even number and $o$ is odd number.\n\nI do not know that anyone notice about this properties, so I personally name them even-odd multiply properties and even-odd sum properties. So, the expression $n^4 + 4^n$, since n is odd($o$). can be evaluated follow by that properties it will be isolated into:\n\n1. $n^4 = n \\times n \\times n \\times n \\equiv o \\times o \\times o \\times o$ produces $o$\n2. $4^n \\equiv v \\times v \\times \\cdots \\times v$ for $n$ times produces $v$\n3. $n^4+4^n \\equiv o+v$ produces $o$\n\nhence $2|(n^4 + 4^n)$, since $n$ is odd.\nMay I prove by this way? Any suggestions are kindly pleased.\n\nSo, I give up with this way.\n\n-\n1. is wrong.... \u2013\u00a0 lhf Jul 2 '14 at 11:25", "date": "2015-04-21 21:54:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/853615/proving-an-expression-is-composite", "openwebmath_score": 0.9091550707817078, "openwebmath_perplexity": 439.9506829252601, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429116504951, "lm_q2_score": 0.879146761176671, "lm_q1q2_score": 0.8659972853975705}}
{"url": "https://math.stackexchange.com/questions/1814658/coefficients-of-homology", "text": "# Coefficients of homology\n\nI am wondering why people use different coefficients when defining homology of simplicial complex, like homology over $R$, $Z$, $Z/2$, etc? Is one better then the other and why?\n\nMoreover, which one(s) is(are) most useful in practical computation other than abstract theory?\n\nThanks a lot.\n\n\u2022 \u2013\u00a0Watson Jun 5 '16 at 16:00\n\u2022 In some sense, if you don't go to deeply in the theory, knowing with $\\mathbb{Z}$ coefficients is everything. This is the content of the universal coefficient theorem. It often occurs that one can compute homology with other coefficients more easily (e.g. with Z/2 coeffients poincare duality works directly for non orientable manifolds). Sometimes the way (co) homology is defined leads one directly to different coefficients. A great example is the de Rham cohomology. \u2013\u00a0Thomas Rot Jun 5 '16 at 20:54\n\nThere is no best coefficient in general, but there is indeed a best coefficient for specific case.\n\nLet me show you some examples.\n\nSuppose you want to compute the Euler Characteristic of a space $X$. Suppose that for some reason, counting cell is not a viable option, let's say you don't have an explicit cell structure of your space. You are left with trying to compute the rank of the homology groups. Now you can verify that $\\text{rank}H_*(X;\\mathbb{Z})=\\dim H_*(X,\\mathbb{Q})$. The latter group is indeed a vector space, so it has no torsion and it is easier to compute (every short exact sequence splits for example). Even more is true, computing Euler Characteristic is invariant from the chosen field coefficients. In conclusion, it is natural to trying to compute $H_*(X,\\mathbb{Q})$ in this case.\n\nSuppose you have a non-orientable closed manifold $M$: When you want to study closed manifolds (from an homology viewpoint), Poincar\u00e9 Duality is a powerful tool. A powerful tool with some hypothesis to verify, namely orientability. If you don't have orientability you are left with two option, trying to compute $H_*(M;\\mathbb{Z}_{\\rho})$ (homology with local coefficient - something pretty technical which was let's say created for dealing with this kind of situation) or more easily $H_*(M;\\mathbb{Z}_2)$, why? well because every manifold is $\\mathbb{Z}_2$ orientable, and therefore you have P-D with these coefficients.\n\nClearly, homology with $\\mathbb{Z}$ coefficients contains all the homological information, since we have UCT which means that if you know homology with $\\mathbb{Z}$ for a space, then you know the homology for that space with coefficient any abelian group.\n\nWhat's the problem: You don't always know how to compute homology with $\\mathbb{Z}$-coefficient. By its very property that it bring a lot of informations, computations in general are cumbersome: you have torsion to deal with for instance which is a pain in the ass. So we came up with some weaker version of it, easier to compute but still interesting. This is how research is done usually, you find some great invariant and then realise that it's impossible to compute (very common) and then he must find something weaker but computable and still interesting (very hard).\n\nA last example to back up what I've said. When you will deal with characteristic classes, in particular Stiefel-Whitney ones, you will realise what I mean by sometimes computing cohomology with integer coefficients is hard. But if you turn to $\\mathbb{Z}_2$ coefficients, the cohomology ring turns out to be a free graded ring and the generators are the universal classes.\n\n\u2022 I can't really see any way of making the assertion \"$\\operatorname{rank} H_*(M;\\mathbb{Z}) = \\dim H_*(M;k)$\" true if the characteristic of $k$ is positive... \u2013\u00a0Najib Idrissi Jun 6 '16 at 8:41\n\u2022 maybe I rushed a little, but if I recall correctly, $\\text{rank}H_*(M;\\mathbb{Z})=\\dim H_*(M;\\mathbb{Z})\\otimes \\mathbb{Q}$ and then one verifies at once that euler char doesn't depend on the field you are computing it. But you are right, I need to modify it a little, since what I wrote it's formally wrong \u2013\u00a0Riccardo Jun 6 '16 at 8:44\n\u2022 Yes, by the UCT $\\operatorname{rank} H_k(M;\\mathbb{Z}) = \\dim H_k(M;\\mathbb{Q})$, and yes, the Euler characteristic does not depends on the field. But for example $H_2(\\mathbb{RP}^2;\\mathbb{Z}) = 0$ has rank zero whereas $H_2(\\mathbb{RP}^2;\\mathbb{F}_2) = \\mathbb{F}_2$ has dimension one. (\"Formally wrong\" is just another way of saying \"wrong\"...) \u2013\u00a0Najib Idrissi Jun 6 '16 at 8:46\n\u2022 yes, and thank you for pointing out this:) \u2013\u00a0Riccardo Jun 6 '16 at 8:52", "date": "2019-08-17 10:52:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1814658/coefficients-of-homology", "openwebmath_score": 0.9166074395179749, "openwebmath_perplexity": 294.2996853518001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429098941399, "lm_q2_score": 0.8791467548438124, "lm_q1q2_score": 0.8659972776153391}}
{"url": "http://mathhelpforum.com/algebra/173556-simplify-expression.html", "text": "# Math Help - Simplify this expression...\n\n1. ## Simplify this expression...\n\n$\n({2+\\sqrt{5})^{1/3} + ({2-\\sqrt{5})^{1/3}\n$\n\n2. Originally Posted by Alex2103\n$\n({2+\\sqrt{5})^{1/3} + ({2-\\sqrt{5})^{1/3}\n$\nGet ready for a ride....\n\nHere are the highlights:\nLet $\\displaystyle x = ( 2 + \\sqrt{5} )^{1/3} + (2 - \\sqrt{5} )^{1/3}$\n\nNow\n$\\displaystyle x^3 = (2 + \\sqrt{5}) + 3( 2 + \\sqrt{5} )^{2/3}(2 - \\sqrt{5} )^{1/3} + 3(2 - \\sqrt{5} )^{1/3}(2 - \\sqrt{5})^{2/3} + (2 - \\sqrt{5})$\n\nThis reduces to\n$\\displaystyle x^3 = 4 -3(2 + \\sqrt{5})^{1/3} - 3(2 - \\sqrt{5})^{1/3}$\n\nor, reminding ourselves of what x is equal to:\n$\\displaystyle x^3 = 4 -3x$\n\nNow solve:\n$\\displaystyle x^3 + 3x - 4 = 0$\n\n(Hint: There is only one real solution.)\n\n-Dan\n\n3. Hello, Alex2103!\n\n$\\text{Simplfy: }\\;\\sqrt[3]{2+\\sqrt{5}} + \\sqrt[3]{2-\\sqrt{5}}$\n\nA binomial of the form $a + b\\sqrt{5}$ makes me suspect\n. . that the problem involves the Golden Mean: . $\\phi \\:=\\:\\dfrac{1+\\sqrt{5}}{2}$\n\n$\\text{And sure enough: }\\;\\phi^3 \\;=\\;\\left(\\frac{1+\\sqrt{5}}{2}\\right)^3 \\;=\\;2+\\sqrt{5}$\n\n. . . $\\text{and we find that: }\\;\\left(\\frac{1-\\sqrt{5}}{2}\\right)^3 \\;=\\;2-\\sqrt{5}$\n\nTherefore:\n\n. . $\\sqrt[3]{2+\\sqrt{5}} + \\sqrt[3]{2-\\sqrt{5}} \\;\\;=\\;\\;\\sqrt[3]{\\left(\\frac{1+\\sqrt{5}}{2}\\right)^3} + \\sqrt[3]{\\left(\\frac{1-\\sqrt{5}}{2}\\right)^3}$\n\n. . $=\\;\\;\\dfrac{1 + \\sqrt{5}}{2} + \\dfrac{1-\\sqrt{5}}{2} \\;\\;=\\;\\;1$\n\nI love your solution, Dan!\n\n4. Originally Posted by Soroban\nI love your solution, Dan!\nBut yours is so more elegant and less time consuming!\n\n-Dan\n\n5. Thanks for help!\n\n6. Heres' yet another way to do that. Cardano's formula for the reduced cubic equation says that a solution to $x^3+ mx= n$ is of the form $\\sqrt[3]{\\frac{n}{2}+ \\sqrt{\\frac{n^2}{4}+ \\frac{m^3}{27}}}+ \\sqrt[3]{\\frac{n}{2}- \\sqrt{\\frac{n^2}{4}+ \\frac{m^3}{27}}}$.\n\nLooks familiar, doesn't it? That is precisely the same as $\\sqrt[3]{2+ \\sqrt{5}}+ \\sqrt[3]{2- \\sqrt{5}}$ with $\\frac{n}{2}= 2$ and $\\frac{n^2}{4}+ \\frac{m^3}{27}= 5$.\n\nFrom $\\frac{n}{2}= 2$, n= 4 and then $\\frac{n^2}{4}+ \\frac{m^3}{27}= 4+ \\frac{m^3}{27}= 5$ so that $\\frac{m^3}{27}= 1$, $m^3= 27$, and $m= 3$.\n\nThat means that this number is a real root of $x^3+ 3x= 4$ or $x^3- 3x- 4= 0$. It is easy to see that $x^3+ 3x- 4= (x- 1)(x^2+ x+ 4)$. Since the discriminant of $x^2+ x+ 4$ is $1- 16= -15$ the only real root of that equation is 1 so we must have $\\sqrt[3]{2+ \\sqrt{5}}+ \\sqrt[3]{2- \\sqrt{5}}= 1$.", "date": "2014-10-21 13:05:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/173556-simplify-expression.html", "openwebmath_score": 0.8864928483963013, "openwebmath_perplexity": 648.2223210888417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.994698106387966, "lm_q2_score": 0.8705972734445508, "lm_q1q2_score": 0.8659814593218209}}
{"url": "https://math.stackexchange.com/questions/2534223/can-an-empty-set-be-in-relation", "text": "# Can an empty set be in relation?\n\nWe were given this problem for homework:\n\nLet it be $A=\\{a,b\\}$.\n\na) Find the powerset $\\mathcal P(A)$.\n\nb) Construct an equivalence relation on $\\mathcal P(A)$ so that it's quotient set has exactly two elements.\n\na) $\\mathcal P(A)= \\{\\emptyset\\ , \\{a\\},\\{b\\},\\{a,b\\}\\}$\n\nb) I constructed the following relation in which an empty set can be an element of ordered pair. (But I'm not sure if an empty set can be in relation. Can an empty set be an element of ordered pair?)\n\nLet's say the relation is R. $$R = \\{(\\emptyset,\\emptyset),(\\{a\\},\\{a\\}),(\\{b\\},\\{b\\}),(\\{a,b\\},\\{a,b\\}),(\\{a\\},\\{b\\}),(\\{b\\},\\{a\\}),(\\emptyset,\\{a,b\\}),(\\{a,b\\},\\emptyset)\\}$$\n\nI think this is correct because:\n\n$[\\{a\\}] = \\{\\{a\\},\\{b\\}\\} = [\\{b\\}]$\n\n$[\\{\\emptyset\\}]=\\{\\emptyset,\\{a,b\\}\\}=[\\{a,b\\}]$\n\nwhich means that $\\mathcal P(A)/_R=\\{\\{\\{a\\},\\{b\\}\\}, \\{\\emptyset,\\{a,b\\}\\}\\}$.\n\nIs this correct or should i remove the empty set from the relation?\n\n\u2022 I do not see a problem. Its fine, I suppose. \u2013\u00a0hemu Nov 23 '17 at 18:55\n\u2022 It\u2019s perfectly correct. Don\u2019t get confused by having a set of sets. To become more at ease with that, abstract a bit: $X = \\mathcal P (A)$ is just a set with four elements, the empty set being one of it, technically. If that gives you a headache, just rename the elements $x_1 = \u2205$, $x_2 = \\{a\\}$, \u2026 and work with the renamed elements. \u2013\u00a0k.stm Nov 23 '17 at 18:57\n\u2022 By the way, the equivalence relation you have found can be described in words: Two sets $x, y \u2208 \\mathcal P (A)$ are equivalent (with respect to $R$) if and only if they are of the same parity. \u2013\u00a0k.stm Nov 23 '17 at 19:03\n\u2022 @k.stm Thanks! Also, I have another question. I have to prove that $A\\neq \\emptyset$ ( for $A=\\{(a,b) \\in \\Bbb R^2 : x+y=c \\}$, for every $c$). How do I prove this exactly? I understand why this is true but I don't know how to prove it. \u2013\u00a0i dont know much about algebra Nov 23 '17 at 19:05\n\u2022 You mean $A = \\{(a,b) \u2208 \u211d^2;~a + b = c\\}$. Well fix a $c$ and find an element in $A$. You can always explicitly write one down. If you can\u2019t manage that, try for $c = 0$ first and then see how you could generalise. \u2013\u00a0k.stm Nov 23 '17 at 19:07", "date": "2019-06-26 22:10:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2534223/can-an-empty-set-be-in-relation", "openwebmath_score": 0.8588579893112183, "openwebmath_perplexity": 160.94909125602607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9926541748380672, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8659392666776835}}
{"url": "https://mathoverflow.net/questions/256696/do-these-matrices-have-a-name/256776", "text": "# Do these matrices have a name?\n\nI am looking for matrices $A\\in \\mathbb{R}^{m\\times n}$ with $m>n$, $A^TA=\\frac{m}{n}I$ and $diag(AA^T)=(1\\ \\dots\\ 1)$ where $diag$ denotes the diagonal. Do such matrices have a name? An example for such a matrix would be an $8\\times 3$ matrix with the coordinates of a cube centered at the origin as rows. Is the matrix $A$ for every pair $(m,n)$ uniquely determined up to right multiplication with an orthogonal matrix? How are the $m$ points given by the rows of $A$ distributed over the $(n-1)$-dimensional sphere?\n\nMy motivation is that I have an unknown $x\\in \\mathbb{R}^n$. I assume that I can measure the scalar product $a^Tx$ with any unit vector $a\\in S^{n-1}$. The measurements are assumed to be i.i.d. Now we assume we have $m$ measurements and write the corresponding vectors as the rows of a matrix $A\\in \\mathbb{R}^{m\\times n}$ and the vectors as a matrix $b\\in \\mathbb{R}^m$. The best possible reconstruction (see Gauss-Markov) of $x$ is given by the least square solution $(A^TA)^{-1}A^Tb$ where $b\\in \\mathbb{R}^n$ denotes the measurements. The covariance matrix for the reconstruction is $(A^TA)^{-1}$. This means that if $A$ satisfies the properties above then the components of the reconstructed $x$ are uncorrelated. Also I think that $A$ gives us the best possible way to measure $x$, i.e. minimizing the standard deviation.\n\n\u2022 Is $m$ a power of $2$? It would be very nice if it were. \u2013\u00a0Rodrigo de Azevedo Dec 8 '16 at 16:28\n\u2022 In general no. But if you have some insights for that case it would definitely be interesting. \u2013\u00a0user35593 Dec 8 '16 at 23:08\n\u2022 In general if A is such a matrix then e can build new matrices by concatenating copies of A. Also we can permute the rows of A to get a new solution. \u2013\u00a0user35593 Dec 9 '16 at 7:29\n\u2022 If we have solutions with the same n we can build new solutions by concatenating. One could introduce the notion of a 'prime' matrix with this property. I.e. a matrix which does not contain a submatrix with n columns with the same property \u2013\u00a0user35593 Dec 9 '16 at 7:44\n\u2022 The rows of this matrix form a tight frame of unit vectors. See e.g. ams.org/notices/201306/rnoti-p748.pdf \u2013\u00a0Terry Tao Dec 9 '16 at 17:11\n\nAssuming that the Hadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \\times n$ matrix that satisfies the given constraints is given by\n\n$$\\boxed{\\mathrm A := \\frac{1}{\\sqrt n} \\mathrm H_m^{\\top} \\mathrm S_n}$$\n\nwhere\n\n\u2022 $\\mathrm H_m \\in \\{\\pm 1\\}^{m \\times m}$ is a Hadamard matrix. Thus, the $m$ rows of $\\mathrm H_m$ are orthogonal, i.e., $$\\mathrm H_m \\mathrm H_m^{\\top} = m \\mathrm I_m$$\n\n\u2022 $\\mathrm S_n$ is a thin $m \\times n$ matrix whose $n$ columns are chosen from the $m$ columns of the $m \\times m$ identity matrix. Thus, the $n$ columns of $\\mathrm S_n$ are orthonormal, i.e.,\n\n$$\\mathrm S_n^{\\top} \\mathrm S_n = \\mathrm I_n$$\n\nHence,\n\n$$\\mathrm A^{\\top} \\mathrm A = \\frac{1}{n} \\mathrm S_n^{\\top} \\mathrm H_m \\mathrm H_m^{\\top} \\mathrm S_n = \\frac{m}{n} \\mathrm S_n^{\\top} \\mathrm S_n = \\frac{m}{n} \\mathrm I_n$$\n\nas desired. Let $\\mathrm e_k$ and $\\mathrm h_k$ denote the $k$-th columns of $\\mathrm I_m$ and $\\mathrm H_m$, respectively. Hence,\n\n$$\\mathrm e_k^{\\top} \\mathrm A \\mathrm A^{\\top} \\mathrm e_k = \\| \\mathrm A^{\\top} \\mathrm e_k \\|_2^2 = \\frac 1n \\| \\mathrm S_n^{\\top} \\mathrm H_m \\mathrm e_k \\|_2^2 = \\frac 1n \\| \\mathrm S_n^{\\top} \\mathrm h_k \\|_2^2 = \\frac 1n \\sum_{k=1}^n (\\pm 1)^2 = \\frac nn = 1$$\n\nfor all $k \\in \\{1,2,\\dots,m\\}$, as desired. Note that we used the fact that the entries of $\\mathrm h_k$ are $\\pm 1$.\n\nIf $m$ is a power of $2$, then $\\mathrm H_m$ can be built recursively using the Sylvester construction\n\n$$\\mathrm H_{2k} = \\begin{bmatrix} \\mathrm H_k & \\mathrm H_k\\\\ \\mathrm H_k & -\\mathrm H_k\\end{bmatrix} \\qquad \\qquad \\qquad \\mathrm H_1 = 1$$\n\nwhich builds (symmetric) Walsh matrices. If $m$ is not a power of $2$, we can use the Paley construction instead.\n\nExample\n\nLet $m = 8$ and $n = 3$. Since $8$ is a power of $2$, we can use the Sylvester construction to build $\\mathrm H_8$.\n\nUsing MATLAB,\n\n>> H1 = 1;\n>> H2 = [H1,H1;H1,-H1];\n>> H4 = [H2,H2;H2,-H2];\n>> H8 = [H4,H4;H4,-H4]\n\nH8 =\n\n1     1     1     1     1     1     1     1\n1    -1     1    -1     1    -1     1    -1\n1     1    -1    -1     1     1    -1    -1\n1    -1    -1     1     1    -1    -1     1\n1     1     1     1    -1    -1    -1    -1\n1    -1     1    -1    -1     1    -1     1\n1     1    -1    -1    -1    -1     1     1\n1    -1    -1     1    -1     1     1    -1\n\n\nLet the $3$ columns of $\\mathrm S_3$ be the first $3$ columns of $\\mathrm I_8$\n\n>> I8 = eye(8);\n>> H8 * I8(:,[1,2,3])\n\nans =\n\n1     1     1\n1    -1     1\n1     1    -1\n1    -1    -1\n1     1     1\n1    -1     1\n1     1    -1\n1    -1    -1\n\n\nNote that the last four rows are copies of the first four rows. Hence, let the $3$ columns of $\\mathrm S_3$ be the 2nd, 3rd and 5th columns of $\\mathrm I_8$\n\n>> H8 * I8(:,[2,3,5])\n\nans =\n\n1     1     1\n-1     1     1\n1    -1     1\n-1    -1     1\n1     1    -1\n-1     1    -1\n1    -1    -1\n-1    -1    -1\n\n\nNote that the $8$ rows are now the $8$ vertices of the cube $[-1,1]^3$.\n\nWe build matrix $\\mathrm A$ by normalizing the rows\n\n>> A = inv(sqrt(3)) * H8 * I8(:,[2,3,5])\n\nA =\n\n0.5774    0.5774    0.5774\n-0.5774    0.5774    0.5774\n0.5774   -0.5774    0.5774\n-0.5774   -0.5774    0.5774\n0.5774    0.5774   -0.5774\n-0.5774    0.5774   -0.5774\n0.5774   -0.5774   -0.5774\n-0.5774   -0.5774   -0.5774\n\n\nIs the constraint $\\mathrm A^{\\top} \\mathrm A = \\frac 83 \\mathrm I_3$ satisfied?\n\n>> A' * A\n\nans =\n\n2.6667         0         0\n0    2.6667         0\n0         0    2.6667\n\n\nIt is. Are the diagonal entries of $\\mathrm A \\mathrm A^{\\top}$ equal to $1$?\n\n>> A * A'\n\nans =\n\n1.0000    0.3333    0.3333   -0.3333    0.3333   -0.3333   -0.3333   -1.0000\n0.3333    1.0000   -0.3333    0.3333   -0.3333    0.3333   -1.0000   -0.3333\n0.3333   -0.3333    1.0000    0.3333   -0.3333   -1.0000    0.3333   -0.3333\n-0.3333    0.3333    0.3333    1.0000   -1.0000   -0.3333   -0.3333    0.3333\n0.3333   -0.3333   -0.3333   -1.0000    1.0000    0.3333    0.3333   -0.3333\n-0.3333    0.3333   -1.0000   -0.3333    0.3333    1.0000   -0.3333    0.3333\n-0.3333   -1.0000    0.3333   -0.3333    0.3333   -0.3333    1.0000    0.3333\n-1.0000   -0.3333   -0.3333    0.3333   -0.3333    0.3333    0.3333    1.0000\n\n\nThey are.", "date": "2021-04-14 14:34:31", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/256696/do-these-matrices-have-a-name/256776", "openwebmath_score": 0.8544521927833557, "openwebmath_perplexity": 166.8743115067149, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085088220004, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8658984761839332}}
{"url": "https://math.stackexchange.com/questions/1398696/why-does-lh%C3%B4pitals-rule-work-for-sequences?noredirect=1", "text": "Why does L'H\u00f4pital's rule work for sequences?\n\nSay, for the classic example, $\\frac{\\log(n)}{n}$, this sequence converges to zero, from applying L'H\u00f4pital's rule. Why does it work in the discrete setting, when the rule is about differentiable functions?\n\nIs it because at infinity, it doesn't matter that we relabel the discrete variable, $n$, with a continuous variable, $x$, and instead look at the limit of $\\frac{\\log(x)}{x}$?\n\nBut then what about the quotients of sequences that go to the indeterminate form $\\frac{0}{0}$? Why is it OK to use L'H\u00f4pital's rule, as $n$ goes to zero?\n\nI haven't found anything on Wikipedia or Wolfram about the discrete setting.\n\nThanks.\n\n\u2022 The counterpart of L'Hospital's rule for sequences is the so-called Stolz Theorem. \u2013\u00a0Zhanxiong Aug 16 '15 at 0:28\n\u2022 Thanks for the link, @Zhanxiong. \u2013\u00a0User001 Aug 16 '15 at 0:52\n\u2022 possible duplicate of Is there a discrete version of de l'H\u00f4pital's rule? \u2013\u00a0J. M. isn't a mathematician Aug 16 '15 at 21:26\n\u2022 @Guesswhoitis.I don't think that's the same question. He asks \"why do people differentiate sequences?\". \u2013\u00a0user223391 Aug 16 '15 at 22:07\n\u2022 I disagree. Some dots would need to be connected in order to be able to say this question is a duplicate of the other one. \u2013\u00a0Robert Soupe Aug 17 '15 at 2:03\n\nThere IS a L'Hospital's rule for sequences called Stolz-Ces\u00e0ro theorem. If you have an indeterminate form, then:\n\n$$\\lim\\limits_{n\\to\\infty} \\frac{s_n}{t_n}=\\lim\\limits_{n\\to\\infty} \\frac{s_n-s_{n-1}}{t_n-t_{n-1}}$$\n\n$$\\lim\\limits_{n\\to\\infty}\\frac{\\ln(n)}{n}=\\lim\\limits_{n\\to\\infty}\\frac{\\ln\\left(\\frac{n}{n-1}\\right)}{n-n+1}=\\lim\\limits_{n\\to\\infty}\\ln\\left(\\frac{n}{n-1}\\right)=0$$\n\nBut that isn't your question. Your question is, why do people \"differentiate\"? Basically because the real case covers the discrete case.\n\nRecall the definition of limits for real and discrete cases.\n\nDefinition. A sequence, $s_n\\colon \\Bbb{N}\\to \\Bbb{R},$ converges to $L$ as $n\\to\\infty$, written $\\lim\\limits_{n\\to\\infty} s_n=L$ iff for all $\\epsilon>0$ there is some $N$ such that for all $n\\in \\Bbb{N}$ with $n>N$, $|s_n-L|<\\epsilon$.\n\nDefinition. A function, $f(x) \\colon \\Bbb{R}\\to \\Bbb{R}$ converges to $L$ as $x\\to\\infty$, written $\\lim\\limits_{x\\to\\infty} f(x)=L$ iff for all $\\epsilon>0$ there is some $X$ such that for all $x\\in \\Bbb{R}$ with $x>X$, $|f(x)-L|<\\epsilon$.\n\nSo if $f(x)$ is a real valued function that agrees with a sequence, $s_n$ on integer values, then $\\lim\\limits_{x\\to\\infty} f(x)=L$ implies $\\lim\\limits_{n\\to\\infty} s_n=L$.\n\n\u2022 Thanks so much @avid19 - this was an awesome explanation. \u2013\u00a0User001 Aug 16 '15 at 0:51\n\u2022 My +1. I want to add here that there is a catch. There are cases when $f(n) \\to L$ as $n \\to \\infty$ but $f(x)$ has no limit as $x \\to \\infty$. \u2013\u00a0Paramanand Singh Aug 16 '15 at 5:10\n\u2022 @ParamanandSingh: The most obvious example would be $f(x) = \\sin(2\\pi x)$ \u2013\u00a0R.. GitHub STOP HELPING ICE Aug 16 '15 at 14:19\n\nYour explanation is not really precise, it does matter whether you use the discrete or continuous variable. However, there is a theorem in mathematical analysis that states that the following is equivalent:\n\n\u2022 $\\lim_{x \\rightarrow c} f(x) = A$\n\u2022 For every sequence $\\{x_n\\}$ such that $\\forall n \\in \\mathbb{N} : x_n \\in D(f), x_n \\neq c$ and that $\\lim_{n \\rightarrow \\infty} x_n = c$ it is true that $\\lim_{n \\rightarrow \\infty} f(x_n) = A$.\n\nIn simpler words, once you know the limit of a function in continuous variable, like $\\lim_{x \\rightarrow \\infty} \\frac{\\log x}{x} = 0$, you also know the limit of any sequence you get by \"picking out\" points of this function's domain, in your case specifically you take the sequence $\\{x_n\\} = \\{n\\}$. Notice that the conditions of the second statement are met, since $\\frac{\\log x}{x}$ is defined for every $n$, $\\lim_{n \\rightarrow \\infty} n = \\infty$ and $n \\neq \\infty$ for every $n$ as well.\n\nThis also solves the problem for the limits of indeterminate form \"$\\frac{0}{0}$\", or any other.\n\nHope that helps :),\n\nEpsiloney\n\nFact: If a function $f:\\Bbb R\\to\\Bbb R$ is continuous, $x_n\\to x$ then $f(x_n)\\to f(x)$.\n\nYou can see here Stolz\u2013Ces\u00e0ro theorem for the \"l'H\u00f4pital's rule\" for sequences.\n\nThis uses the fact that if $\\displaystyle\\lim_{x\\to\\infty}f(x)=L, \\text{ then } \\lim_{n\\to\\infty}f(n)=L$ since\n\n$\\displaystyle\\lim_{x\\to\\infty}f(x)=L\\iff$ for every $\\epsilon>0,$ there is an M such that if $x\\ge M,$ then $\\big|f(x)-L\\big|<\\epsilon$ and\n\n$\\displaystyle\\lim_{n\\to\\infty}f(n)=L\\iff$ for every $\\epsilon>0,$ there is an N such that if $n\\ge N,$ then $\\big|f(n)-L\\big|<\\epsilon$ $\\;\\;$(with $n\\in\\mathbb{N}$)", "date": "2020-06-02 13:42:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1398696/why-does-lh%C3%B4pitals-rule-work-for-sequences?noredirect=1", "openwebmath_score": 0.9453845620155334, "openwebmath_perplexity": 231.0501727535528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850832642353, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8658984672047192}}
{"url": "https://math.stackexchange.com/questions/2808027/finding-the-probability-density-function-of-z-x-y", "text": "# Finding the probability density function of $Z = X + Y$\n\n### The problem\n\nLet there be two independent random variables given. Let's name them $X$ and $Y$. Suppose the variables have both normal distribution $N(\\mu, \\lambda^2).$\n\nLet's define third random variable $Z = X + Y$. The task is to find the density function of $Z$.\n\n### My attempt\n\nI know that the density function can be easily calculated using the distribution function (I would need to differentiate it). That's why I started with finding the distribution. $$F_Z(t) = Pr(Z \\le t) = Pr(Y + X \\le t) = Pr(Y \\le t - X) = \\int \\limits_{-\\infty}^{\\infty}Pr(Y \\le t-x)f_X(x) dx$$ The equation above leads us to the following one: $$\\int \\limits_{-\\infty}^{\\infty} \\int \\limits_{-\\infty}^{t-x} f_Y(y)f_X(x) \\mbox{d} y\\mbox{d}x \\tag{1}.$$ Using the knowledge about $f_X$ and $f_Y$ we can write $(1)$ in this form: $$\\theta\\int \\limits_{-\\infty}^{\\infty} \\int \\limits_{-\\infty}^{t-x}e^{-\\frac{(y- \\mu)^2}{\\lambda^2}} e^{-\\frac{(x- \\mu)^2}{\\lambda^2}} \\mbox{d} y\\mbox{d}x,$$ where $\\theta = \\frac{1}{\\sqrt{2 \\lambda^2}}.$\nI wonder if my attempt was correct because the integral is quite horrible. Is it possible to calculate it in a \"nicer\" way?\n\nIs $(1)$ equivalent to this expression $(f_X*f_Y)(x)$?\nIf yes I would get the following integral: $$\\int \\limits_{\\mathbb{R}} e^{-\\frac{(y- \\mu)^2}{\\lambda^2}} e^{-\\frac{(x- y - \\mu)^2}{\\lambda^2}} \\mbox{d}y.$$\nIs there a method of calculating such things?\n\n\u2022 en.wikipedia.org/wiki/\u2026 \u2013\u00a0saulspatz Jun 4 '18 at 18:50\n\u2022 @Hendrra In general, sum of random variables $X$ and $Y$ with densities $f$ and $g$ has density being convolution of $f$ and $g$, you are correct \u2013\u00a0Jakobian Jun 4 '18 at 19:04\n\u2022 @Adam Thank you. What about my first approach? \u2013\u00a0Hendrra Jun 4 '18 at 19:33\n\nFrom your expression, $F_Z(t) = \\int_\\mathbb{R}P(Y \\leq t-x)f_X(x)dx$, differentiate to obtain the density (you may need to justify why you can differentiate under the integral),\n\n$$f_Z(t) = \\int_\\mathbb{R}f_Y(t-x)f_X(x)dx = \\int_\\mathbb{R}\\frac{1}{\\sqrt{2\\pi}}e^{-(t-x-\\mu)^2/2\\lambda^2}\\frac{1}{\\sqrt{2\\pi}}e^{-(x-\\mu)^2/2\\lambda^2}dx$$\n\nwhich we recognize as $(f_Y*f_X) (x)$. To actually evaluate the integral, expand the exponent terms inside the integrand,\n\n$$-\\frac{1}{2\\lambda^2}\\left[(t-x-\\mu)^2 + (x-\\mu)^2\\right] = -\\frac{1}{2\\lambda^2}(t^2-2\\mu t + 2\\mu^2 + \\underbrace{2x^2-2tx}_{(*)})$$\n\nComplete the square on the $(*)$ term, $2x^2 - 2tx = 2\\left(x-\\frac{t}{2}\\right)^2-\\frac{t^2}{2}$. Putting things together,\n\n$$-\\frac{1}{2\\lambda^2}(t^2-2\\mu t + 2\\mu^2 + 2x^2-2tx) = -\\frac{1}{4\\lambda^2}(t^2-4\\mu t + 4\\mu^2)-\\frac{1}{\\lambda^2}(x-t/2)^2$$\n\nSo, we can pull out the $t$ terms from the integrand to be left with,\n\n$$f_Z(t) = \\frac{1}{\\sqrt{4\\pi\\lambda^2}}e^{-(t-2\\mu)^2/4\\lambda^2}\\underbrace{\\int_\\mathbb{R}\\frac{1}{\\sqrt{\\pi\\lambda^2}}e^{-(x-t/2)^2/\\lambda^2}dx}_{=1} = \\frac{1}{\\sqrt{4\\pi\\lambda^2}}e^{-(t-2\\mu)^2/4\\lambda^2}$$\n\nThe integral above equals $1$ because the integrand is the density function of a $N(t/2,\\lambda^2/2)$ distribution.\n\n\u2022 At the end, did you intend $N(2 \\mu,2 \\lambda^2)$ for the mean and variance rather than $t/2, \\lambda^2 /{2}$? \u2013\u00a0Henry Jun 5 '18 at 16:31\n\u2022 Sorry, I was referring to the density function inside the final integral (the one that equals 1). The distribution of $Z$ is definitely $N(2\\mu,2\\lambda^2)$. \u2013\u00a0Flowsnake Jun 5 '18 at 16:33\n\u2022 So was I. I think a density $f_Z(t)= \\frac{1}{\\sqrt{4\\pi\\lambda^2}}e^{-(t-2\\mu)^2/4\\lambda^2}$ suggests a normal distribution for $Z$ with a mean of $2 \\mu$ and a variance of $2 \\lambda^2$ \u2013\u00a0Henry Jun 5 '18 at 16:37\n\u2022 The density function I was referring to was, $\\frac{1}{\\sqrt{\\pi\\lambda^2}}e^{-(x-t/2)^2/\\lambda^2}$, which is the density of a $N(t/2,\\lambda^2/2)$. \u2013\u00a0Flowsnake Jun 5 '18 at 16:49\n\nThe problem can also be solved by using the following property: If $X\\sim N(\\mu_1,\\sigma_1^2)$ and $Y \\sim N(\\mu_2,\\sigma_2^2)$ and are independent, then $aX + bY \\sim N(a\\mu_1 + b\\mu_2,a^2\\sigma_1^2 +b^2\\sigma_2^2)$ and thus in your case $Z \\sim N(2\\mu,2\\lambda^2)$.\n\nEdit: I just noticed the comment by @saulspatz above who already pointed this out.", "date": "2019-08-24 06:53:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2808027/finding-the-probability-density-function-of-z-x-y", "openwebmath_score": 0.9188310503959656, "openwebmath_perplexity": 250.59321952227083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138138113964, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8658941332765748}}
{"url": "https://math.stackexchange.com/questions/4455265/the-coupon-collectors-most-collected-coupon", "text": "# The coupon collector's most collected coupon\n\nSuppose a coupon collector is collecting a set of $$n$$ coupons that he receives one-by-one uniformly at random.\n\nIf the collector stops exactly when the collection is complete, we know the expected number of coupons in his collection is $$n*H[n]$$. What is the expected number of copies, $$M$$, of his most collected coupon?\n\nWhen $$n = 2$$, then $$M = 2$$ because after the first coupon he will collect the other coupon with probability p ~ Geom(1/2). So he will collect the same coupon one additional time on average, and that coupon is certain to be his most collected coupon.\n\nI don't know the exact value for any $$n$$ larger than 2, but found some approximate values by simulation:\n\nn M\n3 2.8415\n4 3.4992\n5 4.0259\n6 4.4633\n7 4.8377\n8 5.1649\n9 5.4560\n\nEDIT 1:\n\nFor small n enumerating the small possibilities converges faster than simulation, and from this approach I hypothesize that $$M[3] = (15 + 6 \\sqrt{5})/10$$ but don't have any real argument to support that claim.\n\nEDIT 2:\n\nWith some more thought, I find that M has an explicit sum formula. The probability the collection terminates with a given distribution of coupons $$v = \\{c_1, ..., c_{n-1}\\}$$ other than the final collected coupon is just the multinomial coefficient $$(c_1; ...; c_{n-1})$$ over $$n^\\text{total # of coupons in the collection}$$. This is an infinite sum over n-1 variables. Here is Mathematica code the computes the sum for terms where the collection has no more than k copies of any coupon:\n\nM[n_, k_] := Sum[Max[v]*(Multinomial @@ v)/n^Total[v], {v, Tuples[Range[1, k], n-1]}]\n\n\nMathematica can't actually evaluate this though for $$k \\rightarrow \\infty$$ though.\n\n\u2022 You might be interested in Exercise 2.20 of Boucheron, Lugosi, and Massart's \"\"Concentration Inequalities: A Nonasymptotic Theory of Independence\", which asks to prove an upper bound of $$e\\log n \\cdot ce^{W((1-c)/(ce))}$$for this expectation, when collecting $m = cn\\log n$ coupons. ($W$ being the Lambert function.) May 21 at 6:52\n\nValue of $$M_3$$. I will denote $$M$$ as $$M_n$$ to emphasize the dependence on $$n$$. Then by using A230137, we get\n\n\\begin{align*} M_3 &= \\sum_{n=2}^{\\infty} \\frac{\\sum_{k=1}^{n-1} \\binom{n}{k} \\max\\{k,n-k\\}}{3^n} \\\\ &= \\sum_{n=1}^{\\infty} \\frac{\\left( 4^n + \\binom{2n}{n} \\right)n - 2(2n)}{3^{2n}} + \\sum_{n=1}^{\\infty} \\frac{\\left( 4^n + \\binom{2n}{n} \\right)(2n+1) - 2(2n+1)}{3^{2n+1}} \\\\ &= 3\\left(\\frac{1}{2} + \\frac{1}{\\sqrt{5}}\\right) \\\\ &\\approx 2.84164, \\end{align*}\n\nwhich is the same as what OP conjectured.\n\nAsymptotic Formula of $$M_n$$. A heuristic seems suggesting that\n\n$$M_n \\sim e \\log n \\qquad \\text{as} \\quad n \\to \\infty,$$\n\nThe figure below compares simulated values of $$M_n$$ and the values of the asymptotic formula.\n\nHeuristic argument. First, we consider the Poissonized version of the problem:\n\n1. Let $$N^{(1)}, \\ldots, N^{(n)}$$ be independent Poisson processes with rate $$\\frac{1}{n}$$. Then the arrivals in each $$N^{(i)}$$ model the times when the collector receives coupons of type $$i$$.\n\n2. Let $$T$$ be the first time a complete set of coupons is collected. Then we know that $$T$$ has the same distribution as the sum of $$n$$ independent exponential RVs with respective rates $$1$$, $$\\frac{n-1}{n}$$, $$\\ldots$$, $$\\frac{1}{n}$$. (This is an example of the exponential race problem.) For large $$n$$, this is asymptotically normal with mean $$\\sum_{i=1}^{n}\\frac{n}{i} \\sim n \\log n$$ and variance $$\\sum_{i=1}^{n}\\frac{n^2}{i^2} \\sim \\zeta(2)n^2$$. This tells that $$T \\sim n \\log n$$ in probability, and so, it is not unreasonable to expect that\n\n$$M_n = \\mathbf{E}\\Bigl[ \\max_{1\\leq i\\leq n} N^{(i)}_T \\Bigr] \\sim \\mathbf{E}\\Bigl[ \\max_{1\\leq i\\leq n} N^{(i)}_{n\\log n} \\Bigr]. \\tag{1}$$\n\nholds as well.\n\n3. Note that each $$N^{(i)}_{n\\log n}$$ has a Poisson distribution with rate $$\\log n$$. Now, we fix $$\\theta > 1$$ and choose an integer sequence $$(a_n)$$ such that $$a_n > \\log n$$ and $$a_n/\\log n \\to \\theta$$ as $$n \\to \\infty$$. Then\n\n\\begin{align*} \\mathbf{P}\\Bigl( N^{(i)}_{n\\log n} > a_n \\Bigr) &\\asymp \\frac{(\\log n)^{a_n}}{a_n!}e^{-\\log n} \\\\ &\\asymp \\frac{(\\log n)^{a_n}}{(a_n)^{a_n} e^{-a_n+\\log n} \\sqrt{a_n}} \\\\ &\\asymp \\frac{1}{n^{\\psi(\\theta) + 1 + o(1)}}, \\qquad \\psi(\\theta) = \\theta \\log \\theta - \\theta, \\end{align*}\n\nwhere $$f(n) \\asymp g(n)$$ means that $$f(n)/g(n)$$ is bounded away from both $$0$$ and $$\\infty$$ as $$n\\to\\infty$$. This shows that\n\n$$\\mathbf{P}\\Bigl( \\max_{1\\leq i\\leq n} N^{(i)}_{n\\log n} \\leq a_n \\Bigr) = \\biggl[ 1 - \\frac{1}{n^{\\psi(\\theta) + 1 + o(1)}} \\biggr]^n \\xrightarrow[n\\to\\infty]{} \\begin{cases} 0, & \\psi(\\theta) < 0, \\\\[0.5em] 1, & \\psi(\\theta) > 0. \\end{cases}$$\n\nSo, by noting that $$\\psi(e) = 0$$, we get\n\n$$\\max_{1\\leq i\\leq n} N^{(i)}_{n\\log n} \\sim e \\log n \\qquad \\text{in probability}. \\tag{2}$$\n\nThis suggests that $$M_n \\sim e \\log n$$ as well.\n\n4. I believe that this heuristic argument can be turned into an actual proof. To this, we need the following:\n\n1. We have to justify the relation $$\\text{(1)}$$ in an appropriate sense. We may perhaps study the inequality\n\n$$\\max_{1\\leq i\\leq n} N^{(i)}_{(1-\\varepsilon)n\\log n} \\leq \\max_{1\\leq i\\leq n} N^{(i)}_T \\leq \\max_{1\\leq i\\leq n} N^{(i)}_{(1+\\varepsilon)n\\log n} \\qquad \\text{w.h.p.}$$\n\nand show that the probability of the exceptional event is small enough to bound the difference of both sides of $$\\text{(1)}$$ by a vanishing qantity.\n\n2. Note that $$\\text{(2)}$$ alone is not enough to show this. So, we need to elaborate the argument in step 3 so that it can actually prove the relation $$\\mathbf{E}\\bigl[ \\max_{1\\leq i\\leq n} N^{(i)}_{n\\log n} \\bigr] \\sim e \\log n$$.", "date": "2022-06-26 11:57:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4455265/the-coupon-collectors-most-collected-coupon", "openwebmath_score": 0.9969085454940796, "openwebmath_perplexity": 377.59253430301277, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138157595306, "lm_q2_score": 0.8856314647623016, "lm_q1q2_score": 0.8658941187694522}}
{"url": "https://mathematica.stackexchange.com/questions/92776/constructing-tri-diagonal-matrices", "text": "# Constructing Tri-Diagonal Matrices\n\nFractional Calculus Course, we are instructed to create an $n \\times n$ Tri-Diagonal matrix in the form of:\n\n\\begin{array} a A &= \\begin{bmatrix} 2 & -1 & 0 & 0 & ... & 0 & 0 & 0\\\\ -1 & 2 & -1 & 0 & 0 & 0&... & 0 \\\\ 0 & -1 & 2 & -1 & 0& 0... & 0 & 0\\\\ 0 & 0 &-1 & 2 & -1 & ... & 0 & 0\\\\ &&&\\vdots&&& \\\\ 0 & 0 & 0&... & 0 & -1 & 2 & -1\\\\ 0 & 0 & 0 & 0&... & 0 & -1 & 2 \\\\ \\end{bmatrix} \\end{array}\n\nThis is where my dilemma begins. I am not sure how to create this Tri-Diagonal Matrix. I came across the \"SparseArray\" command upon my research and it help me create a Tri-Diagonal Matrix, but I am having a hard time manipulating it to get the $-1, 2, -1$ pattern I am looking for.\n\nmat = SparseArray[ {i_, j_} /; Abs[i - j] <= 1 :> 1, {10, 10}];\nmat // MatrixForm\n\n\nAbove is the command I used for a $10 \\times 10$ matrix. But the tri-diagonal entries were all 1's.\n\nThus my question is, how would I create the tri-diagonal matrix $n \\times n$ I desire? Is there a way to create a function so I can simply manipulate the value of $n$ to get a new matrix without typing (or copy-pasting) the entire code again?\n\n\u2022 Look up Band in the help files. \u2013\u00a0march Sep 1 '15 at 17:51\n\u2022 Just curious ...What is a Fractional Calculus Course ? \u2013\u00a0Dr. belisarius Sep 1 '15 at 18:19\n\u2022 Fractional Calculus is a course that focuses on the applications of fraction derivatives and fraction integrals. So imagine taking half the derivative of f(x) or the two-third integral of f(x). I'm still new to the material so I can't give you a better example that this, sorry. \u2013\u00a0Kevin_H Sep 1 '15 at 19:09\n\u2022 Proposed duplicate: (13004). Related: (13796) \u2013\u00a0Mr.Wizard Sep 2 '15 at 0:07\n\nSparseArray[{{i_, i_} :> 2, {i_, j_} :> -1 /; Abs[i - j] == 1}, {10, 10}]  // MatrixForm\n\n\nor\n\nSparseArray[{Band[{1, 1}] -> 2, Band[{2, 1}] -> -1, Band[{1, 2}] -> -1}, {10, 10}] //\nMatrixForm\n\n\nThe following is sometimes useful:\n\nm = {{{a, b}, {b, a}}};\nd = {10, 10};\nSparseArray[{Band[{2, 2}, d] -> m, Band[{1, 1}, d] -> m}, d] // MatrixForm\n\n\nWithout SparseArray:\n\nn = 10;\nTotal[\n{DiagonalMatrix[Array[-1 &, n - 1], -1],\nDiagonalMatrix[Array[2 &, n]],\nDiagonalMatrix[Array[-1 &, n - 1], 1]}\n]\n\n\nOr strictly using Array:\n\nArray[Which[#1 == #2, 2, Abs[#1 - #2] == 1, -1, True, 0] &, {10, 10}]\n\n\nAND, what the heck? One more for more flexible applications:\n\na = {2, -1, -2, 3};\nn = 10;\nWith[{a1 = PadRight[a, n]}, (Array[a1[[Abs[#1 - #2] + 1]] &, {n, n}])]//MatrixForm\n\n\n\u2022 Your second snippet is what was usually done before SparseArray[] came along. \u2013\u00a0J. M. is away Sep 2 '15 at 0:57\n\nThis should be faster (where n is your square dimension, e.g. 1000 for 1kX1k), easily extends to n-diagonal symmetric with no performance impact:\n\nToeplitzMatrix[PadRight[{2, -1}, n]]", "date": "2019-06-17 05:48:39", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/92776/constructing-tri-diagonal-matrices", "openwebmath_score": 0.6152857542037964, "openwebmath_perplexity": 1256.2119085314562, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338112885303, "lm_q2_score": 0.9005297827809309, "lm_q1q2_score": 0.8658898342161808}}
{"url": "http://www.kanunpiyano.com/i-m-uaq/biconditional-statement-truth-table-4204cd", "text": "# biconditional statement truth table\n\nWe will then examine the biconditional of these statements. Sign up using Google Sign up using Facebook Sign up using Email and Password Submit. Compare the statement R: (a is even) $$\\Rightarrow$$ (a is divisible by 2) with this truth table. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. Venn diagram of \u2194 (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement \"if and only if\", where is known as the antecedent, and the consequent. But would you need to convert the biconditional to an equivalence statement first? V. Truth Table of Logical Biconditional or Double Implication. I've studied them in Mathematical Language subject and Introduction to Mathematical Thinking. Otherwise it is true. For example, the propositional formula p \u2227 q \u2192 \u00acr could be written as p /\\ q -> ~r, as p and q => not r, or as p && q -> !r. You can enter logical operators in several different formats. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. (true) 3. Two formulas A 1 and A 2 are said to be duals of each other if either one can be obtained from the other by replacing \u2227 (AND) by \u2228 (OR) by \u2227 (AND). We still have several conditional geometry statements and their converses from above. Whenever the two statements have the same truth value, the biconditional is true. A tautology is a compound statement that is always true. second condition. Also, when one is false, the other must also be false. Therefore, the sentence \"A triangle is isosceles if and only if it has two congruent (equal) sides\" is biconditional. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. Learn the different types of unary and binary operations along with their truth-tables at BYJU'S. Email. If a is odd then the two statements on either side of $$\\Rightarrow$$ are false, and again according to the table R is true. When we combine two conditional statements this way, we have a\u00a0biconditional. Copyright 2010- 2017 MathBootCamps | Privacy Policy, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Google+ (Opens in new window), Truth tables for \u201cnot\u201d, \u201cand\u201d, \u201cor\u201d (negation, conjunction, disjunction), Analyzing compound propositions with truth tables. You passed the exam if and only if you scored 65% or higher. If no one shows you the notes and you do not see them, a value of true is returned. So the former statement is p: 2 is a prime number. B. A\u2192B. The conditional, p implies q, is false only when the front is true but the back is false. Post as a guest. Compound propositions involve the assembly of multiple statements, using multiple operators. Hence Proved. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true. a. Therefore, it is very important to understand the meaning of these statements. 3. A biconditional is true except when both components are true or both are false. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. Ask Question Asked 9 years, 4 months ago. The following is truth table for \u2194 (also written as \u2261, =, or P EQ Q): The statement sr is also true. So we can state the truth table for the truth functional connective which is the biconditional as follows. Two line segments are congruent if and only if they are of equal length. Therefore the order of the rows doesn\u2019t matter \u2013 its the rows themselves that must be correct. first condition. If I get money, then I will purchase a computer. A biconditional statement is one of the form \"if and only if\", sometimes written as \"iff\". The biconditional connective can be represented by \u2261 \u2014 <\u2014> or <=> and is \u2026 When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. Writing this out is the first step of any truth table. Theorem 1. The truth table for the biconditional is . The truth table for \u21d4 is shown below. biconditional A logical statement combining two statements, truth values, or formulas P and Q in such a way that the outcome is true only if P and Q are both true or both false, as indicated in the table. When we combine two conditional statements this way, we have a biconditional. Otherwise, it is false. To learn more, see our tips on writing great answers. For each truth table below, we have two propositions: p and q. We start by constructing a truth table with 8 rows to cover all possible scenarios. Otherwise it is false. A biconditional statement is really a combination of a conditional statement and its converse. Biconditional Statements (If-and-only-If Statements) The truth table for P \u2194 Q is shown below. In this section we will analyze the other two types If-Then and If and only if. According to when p is false, the conditional p \u2192 q is true regardless of the truth value of q. In this post, we\u2019ll be going over how a table setup can help you figure out the truth of conditional statements. Watch Queue Queue. 13. s: A triangle has two congruent (equal) sides. The biconditional operator is sometimes called the \"if and only if\" operator. Otherwise it is true. You are in Texas if you are in Houston. A biconditional is true only when p and q have the same truth value. The biconditional x\u2192y denotes \u201c x if and only if y,\u201d where x is a hypothesis and y is a conclusion. Construct a truth table for ~p \u2194 q Construct a truth table for (q\u2194p)\u2192q Construct a truth table for p\u2194(q\u2228p) A self-contradiction is a compound statement that is always false. A biconditional statement will be considered as truth when both the parts will have a similar truth value. It is helpful to think of the biconditional as a conditional statement that is true in both directions. The statement qp is also false by the same definition. \u2022 Identify logically equivalent forms of a conditional. 4. T. T. T. T. F. F. F. T. T. F. F. T. Example: We have a conditional statement If it is raining, we will not play. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. The conditional statement is saying that if p is true, then q will immediately follow and thus be true. Is there XNOR (Logical biconditional) operator in C#? Is this statement biconditional? b. You passed the exam iff you scored 65% or higher. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. The statement pq is false by the definition of a conditional. Name. All Rights Reserved. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. To show that equivalence exists between two statements, we use the biconditional if and only if. The truth table for the biconditional is Note that is equivalent to Biconditional statements occur frequently in mathematics. text/html 8/17/2008 5:10:46 PM bigamee 0. Next, we can focus on the antecedent, $$m \\wedge \\sim p$$. Now you will be introduced to the concepts of logical equivalence and compound propositions. It is a combination of two conditional statements, \u201cif two line segments are congruent then they are of equal length\u201d and \u201cif two line segments are of equal length then they are congruent\u201d. en.wiktionary.org. (true) 4. The conditional operator is represented by a double-headed arrow \u2194. All birds have feathers. We will then examine the biconditional of these statements. Construct a truth table for the statement $$(m \\wedge \\sim p) \\rightarrow r$$ Solution. Definition:\u00a0A biconditional statement is defined to be true whenever both parts have the same truth value. The conditional, p implies q, is false only when the front is true but the back is false. Mathematicians abbreviate \"if and only if\" with \"iff.\" In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. If and only if statements, which math people like to shorthand with \u201ciff\u201d, are very powerful as they are essentially saying that p and q are interchangeable statements. A biconditional statement is often used in defining a notation or a mathematical concept. NCERT Books. This is reflected in the truth table. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. Sign up to get occasional emails (once every couple or three weeks) letting you know\u00a0what's new! T. T. T. T. F. F. F. T. F. F. F. T. Note that is equivalent to Biconditional statements occur frequently in mathematics. And the latter statement is q: 2 is an even number. Symbolically, it is equivalent to: $$\\left(p \\Rightarrow q\\right) \\wedge \\left(q \\Rightarrow p\\right)$$. \u2022 Construct truth tables for biconditional statements. To help you remember the truth tables for these statements, you can think of the following: Previous: Truth tables for \u201cnot\u201d, \u201cand\u201d, \u201cor\u201d (negation, conjunction, disjunction), Next: Analyzing compound propositions with truth tables. Construct a truth table for (p\u2194q)\u2227(p\u2194~q), is this a self-contradiction. If given a biconditional logic statement. Let qp represent \"If x = 5, then x + 7 = 11.\". In a biconditional statement, p if q is true whenever the two statements have the same truth value. Thus R is true no matter what value a has. A biconditional statement is really a combination of a conditional statement and its converse. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. The biconditional operator looks like this: \u2194 It is a diadic operator. A biconditional statement is one of the form \"if and only if\", sometimes written as \"iff\". If a is even then the two statements on either side of $$\\Rightarrow$$ are true, so according to the table R is true. Notice that the truth table shows all of these possibilities. Similarly, the second row follows this because is we say \u201cp implies q\u201d, and then p is true but q is false, then the statement \u201cp implies q\u201d must be false, as q didn\u2019t immediately follow p. The last two rows are the tough ones to think about. The statement rs is true by definition of a conditional. The conditional operator is represented by a double-headed arrow \u2194. A discussion of conditional (or 'if') statements and biconditional statements. Principle of Duality. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. Make truth tables. Let's put in the possible values for p and q. A polygon is a triangle iff it has exactly 3 sides. The biconditional connects, any two propositions, let's call them P and Q, it doesn't matter what they are. BOOK FREE CLASS; COMPETITIVE EXAMS. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. When x\u00a05, both a and b are false. Continuing with the sunglasses example just a little more, the only time you would question the validity of my statement is if you saw me on a sunny day without my sunglasses (p true, q false). Sign up or log in. Conditional: If the quadrilateral has four congruent sides and angles, then the quadrilateral is a square. Biconditional statement? Implication In natural language we often hear expressions or statements like this one: If Athletic Bilbao wins, I'll\u2026 Since, the truth tables are the same, hence they are logically equivalent. Let's look at a truth table for this compound statement. They can either both be true (first row), both be false (last row), or have one true and the other false (middle two rows). Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. Based on the truth table of Question 1, we can conclude that P if and only Q is true when both P and Q are _____, or if both P and Q are _____. Compound Propositions and Logical Equivalence Edit. Edit. Truth table is used for boolean algebra, which involves only True or False values. Negation is the statement \u201cnot p\u201d, denoted \u00acp, and so it would have the opposite truth value of p. If p is true, then \u00acp if false. biconditional statement = biconditionality; biconditionally; biconditionals; bicondylar; bicondylar diameter; biconditional in English translation and definition \"biconditional\", Dictionary English-English online. When one is true, you automatically know the other is true as well. In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). Having two conditions. \", Solution:\u00a0 rs represents, \"You passed the exam if and only if you scored 65% or higher.\". This video is unavailable. How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions. Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. A statement is a declarative sentence which has one and only one of the two possible values called truth values. \". It's a biconditional statement. In the first conditional, p is the hypothesis and q is the conclusion; in the second conditional, q is the hypothesis and p is the conclusion. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! Hope someone can help with this. Truth Table Generator This tool generates truth tables for propositional logic formulas. If no one shows you the notes and you see them, the biconditional statement is violated. Examples. Truth Table for Conditional Statement. The biconditional pq represents \"p if and only if q,\" where p is a hypothesis and q is a conclusion. Directions: Read each question below. Select your answer by clicking on its button. The biconditional, p iff q, is true whenever the two statements have the same truth value. More examples of the biconditional if and only if... ] in other words, logical statement p q! Analyze the other two types If-Then and if and only if I get money, then polygon... 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Also \u201c biconditional statement truth table implies that p < = > q, its,! Both components are true a triangle iff it has two congruent ( equal ).., since these statements table is used for boolean algebra, which is a conclusion [ ]. Is helpful to think of the form ' p if and only if y, where. Is provided in the RESULTS BOX answer is provided in the table below, will... You know what 's new 10 ; Class 6 - 10 ; Class -. Its components regardless of the biconditional uses a two-valued logic: every statement is used. More examples of the form  if x = 5 '' is not.! For conditional & biconditional and equivalent statements side by side in the same value! Operator is sometimes called the hypothesis and q, is true by definition of a conditional statement has a arrow! Of any truth table for any two inputs, say a and b we. Iff it has exactly 3 sides.  form biconditional statement truth table red to identify the (... Statement, p if and only if q, since these statements pq represents  p and. Final exam Question: know how to do it without using a Truth-Table consequent.... Similar truth value choose a different button propositions involve the assembly of multiple statements you... ( p\u2194~q ), is true, then I will purchase a computer a statement... Definition: a biconditional statement will be introduced to the concepts of logical equivalence and biconditional statements rectangle and... Functional connective which is the first step of any truth biconditional statement truth table for this statement. A conditional q is false, the truth table which involves only true or values. Such columns if you are confident about your work. b is given ;!", "date": "2021-02-27 00:51:39", "meta": {"domain": "kanunpiyano.com", "url": "http://www.kanunpiyano.com/i-m-uaq/biconditional-statement-truth-table-4204cd", "openwebmath_score": 0.4470146596431732, "openwebmath_perplexity": 606.4229477217441, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338046748209, "lm_q2_score": 0.9005297781091839, "lm_q1q2_score": 0.8658898237682958}}
{"url": "https://codereview.stackexchange.com/questions/128810/calculate-the-lattice-paths-in-an-nn-lattice-project-euler-15", "text": "# Calculate the lattice paths in an n*n lattice (Project Euler #15)\n\nProject Euler #15 is to find out all the distinct possible ways there are to move from the first point (1,1) to point (n,n) in an n*n lattice.\n\ndef lattice_paths_of_n(n):\nlist2 = []\nmy_list = []\nfor i in range(1, n+2):\nlist2.append(i)\nfor i in range(1, n+2):\nmy_list.append(list2)\nfor i in range(0,n+1):\nfor f in range(0,n+1):\nif f == 0 or i == 0:\nmy_list[i][f] = 1\nelse:\nmy_list[i][f] = my_list[i-1][f]+my_list[i][f-1]\nreturn my_list[n][n]\n\nprint(lattice_paths_of_n(20))\n\n\nHowever, this function is extremely inefficient and I would appreciate it if you could give me suggestions to optimize the same. I tried to find a more efficient solution to this problem on this site, but couldn't find one in Python 3.\n\n\u2022 What leads you to believe it's extremely inefficient? (I'm not arguing it isn't, just asking) \u2013\u00a0Insane May 20 '16 at 3:10\n\u2022 Well, I basically used brute force; whereas I can use a more efficient intelligent way to calculate the answer. \u2013\u00a0Aradhye Agarwal May 20 '16 at 3:11\n\u2022 Won't spoil the fun for you, but there is a closed form solution that involves permutations: you just need to compute a couple of factorials to get to the answer. \u2013\u00a0Jaime May 20 '16 at 4:51\n\u2022 @Jaime Well this is what Code Review is! Feel free to write it up as an answer with an explanation :) \u2013\u00a0Insane May 20 '16 at 10:03\n\u2022 @Jaime if you do answer, I would appreciate it if you also provided a logical explanation behind your formula and it's derivation. \u2013\u00a0Aradhye Agarwal May 20 '16 at 10:31\n\nTo move from the top left corner of an $n\\times n$ grid to the opposite one you have to move $n$ times to the right and $n$ times down. So in total you will do $2n$ moves. If you could make those in any order there would be $(2 n)!$ ways of doing them. But you don't have that freedom, because the order within the movements to the right and within the movements down is fixed, e.g. you have to move from row 4 to row 5 before you can move from row 5 to row 6. So of the $n!$ ways the movements to the right can be ordered, only one is valid, and similarly with the movements down.\n\nSumming it all up, the closed form answer to that problem is:\n\n$$\\frac{(2n)!}{n!n!}$$\n\nUnsurprisingly this is the same as $C_{2n, n}$, the combinations of $2n$ items taken $n$ at a time. You could think of the same problem as having $2n$ movements and having to choose $n$ of those to make e.g. the movements down, leaving the others for the movements right.\n\nWith Python's arbitrary size integers you could simply calculate that as:\n\nimport math\n\ndef pe15(n):\nn_fact = math.factorial(n)\nreturn math.factorial(2 * n) / n_fact / n_fact\n\n\nTo give this answer a little more meat, in most other languages you don't have the luxury of not having to worry about integer overflows. Even in Python it may be a good idea if you want to keep your computations fast for very large numbers. So you would typically do the same computation as:\n\ndef pe15_bis(n):\nret = 1\nfor j in range(1, n+1):\nret *= n + j\nret //= j\nreturn ret\n\n\nIn Python that doesn't seem to pay off (performance wise) until n is in the many hundreds, but I find lovely that, the way that code goes, ret is always exactly divisible by j. Figuring out why is left as an exercise for the reader...", "date": "2020-08-07 21:39:23", "meta": {"domain": "stackexchange.com", "url": "https://codereview.stackexchange.com/questions/128810/calculate-the-lattice-paths-in-an-nn-lattice-project-euler-15", "openwebmath_score": 0.46741408109664917, "openwebmath_perplexity": 519.2175659904909, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771770811146, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8658740783715507}}
{"url": "https://la.mathworks.com/help/econ/dtmc.eigplot.html", "text": "# eigplot\n\nPlot Markov chain eigenvalues\n\n## Description\n\nexample\n\neigplot(mc) creates a plot containing the eigenvalues of the transition matrix of the discrete-time Markov chain mc on the complex plane. The plot highlights the following:\n\n\u2022 Unit circle\n\n\u2022 Perron-Frobenius eigenvalue at (1,0)\n\n\u2022 Circle of second largest eigenvalue magnitude (SLEM)\n\n\u2022 Spectral gap between the two circles, which determines the mixing time\n\nexample\n\neVals = eigplot(mc) additionally returns the eigenvalues eVals sorted by magnitude.\n\neigplot(ax,mc) plots on the axes specified by ax instead of the current axes (gca).\n\n[eVals,h] = eigplot(___) additionally returns the handle to the eigenvalue plot using input any of the input arguments in the previous syntaxes. Use h to modify properties of the plot after you create it.\n\n## Examples\n\ncollapse all\n\nCreate 10-state Markov chains from two random transition matrices, with one transition matrix being more sparse than the other.\n\nrng(1); % For reproducibility\nnumstates = 10;\nmc1 = mcmix(numstates,'Zeros',20);\nmc2 = mcmix(numstates,'Zeros',80); % mc2.P is more sparse than mc1.P\n\nPlot the eigenvalues of the transition matrices on the separate complex planes.\n\nfigure;\neigplot(mc1);\n\nfigure;\neigplot(mc2);\n\nThe pink disc in the plots show the spectral gap (the difference between the two largest eigenvalue moduli). The spectral gap determines the mixing time of the Markov chain. Large gaps indicate faster mixing, whereas thin gaps indicate slower mixing. Because the spectral gap of mc1 is thicker than the spectral gap of mc2, mc1 mixes faster than mc2.\n\nConsider this theoretical, right-stochastic transition matrix of a stochastic process.\n\n$P=\\left[\\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\\\ 0& 0& 1/3& 0& 2/3& 0& 0\\\\ 0& 0& 0& 0& 0& 1/3& 2/3\\\\ 0& 0& 0& 0& 0& 1/2& 1/2\\\\ 0& 0& 0& 0& 0& 3/4& 1/4\\\\ 1/2& 1/2& 0& 0& 0& 0& 0\\\\ 1/4& 3/4& 0& 0& 0& 0& 0\\end{array}\\right].$\n\nCreate the Markov chain that is characterized by the transition matrix P.\n\nP = [ 0   0  1/2 1/4 1/4  0   0 ;\n0   0  1/3  0  2/3  0   0 ;\n0   0   0   0   0  1/3 2/3;\n0   0   0   0   0  1/2 1/2;\n0   0   0   0   0  3/4 1/4;\n1/2 1/2  0   0   0   0   0 ;\n1/4 3/4  0   0   0   0   0 ];\nmc = dtmc(P);\n\nPlot and return the eigenvalues of the transition matrix on the complex plane.\n\nfigure;\neVals = eigplot(mc)\n\neVals = 7\u00d71 complex\n\n-0.5000 + 0.8660i\n-0.5000 - 0.8660i\n1.0000 + 0.0000i\n-0.3207 + 0.0000i\n0.1604 + 0.2777i\n0.1604 - 0.2777i\n-0.0000 + 0.0000i\n\nThree eigenvalues have modulus one, which indicates that the period of mc is three.\n\nCompute the mixing time of the Markov chain.\n\n[~,tMix] = asymptotics(mc)\ntMix = 0.8793\n\n## Input Arguments\n\ncollapse all\n\nDiscrete-time Markov chain with NumStates states and transition matrix P, specified as a dtmc object. P must be fully specified (no NaN entries).\n\nAxes on which to plot, specified as an Axes object.\n\nBy default, eigplot plots to the current axes (gca).\n\n## Output Arguments\n\ncollapse all\n\nTransition matrix eigenvalues sorted by magnitude, returned as a numeric vector.\n\nHandles to plotted graphics objects, returned as a graphics array. h contains unique plot identifiers, which you can use to query or modify properties of the plot.\n\nNote\n\n\u2022 By the Perron-Frobenius Theorem [2], a chain with a single recurrent communicating class (a unichain) has exactly one eigenvalue equal to 1 (the Perron-Frobenius eigenvalue), and an accompanying nonnegative left eigenvector that normalizes to a unique stationary distribution. All other eigenvalues have modulus less than or equal to 1. The inequality is strict unless the recurrent class is periodic. When there is periodicity of period k, there are k eigenvalues on the unit circle at the k roots of unity.\n\n\u2022 For an ergodic unichain, any initial distribution converges to the stationary distribution at a rate determined by the second largest eigenvalue modulus (SLEM), \u03bc. The spectral gap, 1 \u2013 \u03bc, provides a visual measure, with large gaps (smaller SLEM circles) producing faster convergence. Rates are exponential, with a characteristic time given by\n\n$tMix=-\\frac{1}{\\mathrm{log}\\left(\\mu \\right)}.$\n\nSee asymptotics.\n\n## References\n\n[1] Gallager, R.G. Stochastic Processes: Theory for Applications. Cambridge, UK: Cambridge University Press, 2013.\n\n[2] Horn, R., and C. R. Johnson. Matrix Analysis. Cambridge, UK: Cambridge University Press, 1985.\n\n[3] Seneta, E. Non-negative Matrices and Markov Chains. New York, NY: Springer-Verlag, 1981.\n\n### Functions\n\nIntroduced in R2017b", "date": "2021-07-25 00:05:56", "meta": {"domain": "mathworks.com", "url": "https://la.mathworks.com/help/econ/dtmc.eigplot.html", "openwebmath_score": 0.7816264629364014, "openwebmath_perplexity": 1859.471010607628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777180191995, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8658740605501051}}
{"url": "http://mathhelpforum.com/number-theory/129900-p-q-odd-primes-q-2-p-1-q-2pk-1-a.html", "text": "# Math Help - p,q odd primes & q|(2^p)-1 => q=2pk+1\n\n1. ## p,q odd primes & q|(2^p)-1 => q=2pk+1\n\nProve that if p and q are odd primes and q divides (2^p)-1, then q=2pk+1 for some integer k.\n\nI can't find any clue on this one...\nAny help is appreciated!\n\n[also under discussion in math links forum]\n\n2. $q \\mid 2^p - 1 \\ \\Leftrightarrow \\ 2^p \\equiv 1 \\ (\\text{mod } q)$\n\nSince $(2,q) = 1$, then the order of 2 (mod q) is a divisor of p (why?) and so, the order is either 1 or p. It must be p.\n\nSo since the order of 2 is p, we know $p \\mid \\phi (q)$, i.e. $p \\mid q - 1$.\n\nSee if you can finish it off from there.\n\n3. Originally Posted by o_O\n$q \\mid 2^p - 1 \\ \\Leftrightarrow \\ 2^p \\equiv 1 \\ (\\text{mod } q)$\n\nSince $(2,q) = 1$, then the order of 2 (mod q) is a divisor of p (why?) and so, the order is either 1 or p. It must be p.\n\nSo since the order of 2 is p, we know $p \\mid \\phi (q)$, i.e. $p \\mid q - 1$.\n\nSee if you can finish it off from there.\nSo we have q-1=pz for some integer z. q-1 is even.\n\nHow can we prove that 2p|(q-1) ?\n\nThanks!\n\n4. Start from here: $q - 1 = zp, \\ z \\in \\mathbb{Z}$\n\nWe know that $p$ and $q$ are odd. What does this tell you about $z$?\n\n5. Originally Posted by o_O\nStart from here: $q - 1 = zp, \\ z \\in \\mathbb{Z}$\n\nWe know that $p$ and $q$ are odd. What does this tell you about $z$?\nOK, I got it.\nz must be even; z=2k for some integer k.\nSo we have our result: q=2pk+1 for some integer k.\n\nIs it true that ANY divisor(not necessarily prime) of (2^p)-1 is also of the form 2pk+1? Why or why not?\n\nThanks!\n\n6. Well, consider the prime power decomposition of $2^p - 1$.\n\nThrough what we've just proven, what can be said about these primes? And hence what about the product between any of these primes?\n\n7. Originally Posted by o_O\nWell, consider the prime power decomposition of $2^p - 1$.\n\nThrough what we've just proven, what can be said about these primes? And hence what about the product between any of these primes?\n(2^p)-1 is odd, so 2 is not a factor in the prime factorization of (2^p)-1. So every prime factor is odd and of the form 2pk+1.\n\nI can also show that a product of two factors of the form 2pk+1 is of the same form (i.e. the product is of the form 2pm+1 for some integer m).\n\nSo by induction, we can say that ANY divisor(not necessarily prime) of (2^p)-1 is also of the form 2pk+1? Am I right?\n\n8. Originally Posted by kingwinner\nAm I right?\nWhy the doubt?", "date": "2014-07-30 01:37:32", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/129900-p-q-odd-primes-q-2-p-1-q-2pk-1-a.html", "openwebmath_score": 0.9060742259025574, "openwebmath_perplexity": 563.8060008525761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682451330957, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8658627993614556}}
{"url": "http://mathhelpforum.com/advanced-math-topics/156049-cartesian-equation-helix.html", "text": "Thread: Cartesian equation of a helix.\n\n1. Cartesian equation of a helix.\n\nHi everyone,\n\nI was wondering whether there is a cartesian equation for a helix. If so what is it? Also how do we convert a three dimensional parametric equation to its cartesian form? (such as the equation of the helix in parametric form)\n\n2. Well, the parametric equation for a helix could be\n\n$\\mathbf{r}=\\langle \\cos(\\theta),\\sin(\\theta),\\theta \\rangle.$\n\nSo, you could have the two equations (I think you'd have to have two cartesian equations to represent a space curve, simply because of the degrees of freedom involved):\n\n$x=\\cos(z), y=\\sin(z).$\n\nIn general, if you have a helix, the axis of which is parallel to one of the three main axes, then I'd solve that equation for the parameter, and then plug back into the other two equations. For example, take the helix\n\n$\\mathbf{r}=\\langle 5\\cos(\\theta),\\theta-2,-5\\sin(\\theta) \\rangle.$\n\nI'd solve $y=\\theta-2$ for $\\theta,$ giving $\\theta=y+2,$ and then plug into the other two equations, yielding\n\n$x=5\\cos(y+2), z=-5\\sin(y+2).$\n\nIf the axis is not parallel to one of the three main axes, things get a lot dicier.\n\n3. Originally Posted by Ackbeet\nWell, the parametric equation for a helix could be\n\n$\\mathbf{r}=\\langle \\cos(\\theta),\\sin(\\theta),\\theta \\rangle.$\n\nSo, you could have the two equations (I think you'd have to have two cartesian equations to represent a space curve, simply because of the degrees of freedom involved):\n\n$x=\\cos(z), y=\\sin(z).$\n\nIn general, if you have a helix, the axis of which is parallel to one of the three main axes, then I'd solve that equation for the parameter, and then plug back into the other two equations. For example, take the helix\n\n$\\mathbf{r}=\\langle 5\\cos(\\theta),\\theta-2,-5\\sin(\\theta) \\rangle.$\n\nI'd solve $y=\\theta-2$ for $\\theta,$ giving $\\theta=y+2,$ and then plug into the other two equations, yielding\n\n$x=5\\cos(y+2), z=-5\\sin(y+2).$\n\nIf the axis is not parallel to one of the three main axes, things get a lot dicier.\n\nDear Ackbeet,\n\nThank you so much. So we cannot express a space curve by a single cartesian equation is'nt? I think so too. But the strange thing is I didnt find it mentioned anywhere.\n\n4. So we cannot express a space curve by a single cartesian equation...?\nNo, I don't think you can. And that's merely because of the fact that if you think about the process of solving equations, one equation will generally allow you to eliminate one variable. 3D space obviously has 3 variables, x, y, and z. But a space curve has only one degree of freedom, the parameter t or whatever you want to call it. To get down to one variable from three requires two equations. So there you go.\n\nThis is something that mathematicians may or may not mention. The physicists definitely talk about the number of degrees of freedom (variables) all the time. Maybe that's because physicists are sometimes more interested in providing the actual solution, instead of just proving that a solution exists.\n\nCheers.\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\ngeneral equation of helix\n\nClick on a term to search for related topics.", "date": "2017-04-24 20:37:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-math-topics/156049-cartesian-equation-helix.html", "openwebmath_score": 0.8744499087333679, "openwebmath_perplexity": 207.73361235746512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682464686386, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8658627925179017}}
{"url": "https://math.stackexchange.com/questions/498954/counting-number-of-distinct-systems", "text": "# Counting number of distinct systems\n\nThis is an enumeration problem in conjunction with some lottery problems.\n\nGiven an integer $N \\ge 5$. Let a ticket be a set of 5 distinct integers between $1$ and $N$. Given an integer $T$ between $1$ and ${{N}\\choose{5}}$. Let a system of size $T$ be a set of $T$ distinct tickets.\n\nGiven $N \\ge 5$, I want to count how many distinct systems of size $T$ exist.\n\nTwo systems $S_1$ and $S_2$ are distinct if we can not find a permutation of $\\{1,..,N\\}$ so that the image of $S_1$ under permutation is $S_2$.\n\nI tried some computations for small values of $N$ and $T$.\n\n$N=7$\n\n$T= 1, 2, 3, 4, 5, 6, 7, 8, 9$\n\nnumber of distinct systems = $1, 2, 5, 10, 21, 41, 65, 97, 131, 148$\n\n(It seems that this sequence of numbers is known as A008406 at oeis.org)\n\n$N=8$\n\n$T= 1, 2, 3, 4, 5, 6, 7, 8, 9$\n\nnumber of distinct systems = $1, 3, 11, 52, 252, 1413, 7812, 41868, 207277$\n\n$N=9$\n\n$T= 1, 2, 3, 4, 5, 6, 7$\n\nnumber of distinct systems = $1, 4, 20, 155, 1596, 20528, 282246$\n\nIs there a method to \"guess\" those numbers and find bigger values ?\n\nI wonder if Polya enumeration can be used there. I currently do not know how.\n\nUpdate: Taking a look at http://ac.cs.princeton.edu/home/\n\nLet $s(T,N)$ be the number of distinct systems of size $T$ ($1 \\le T \\le{{N}\\choose{5}}$), given $N$.\n\n$\\forall N \\ge 5, s(1,N) = 1$\n\n$\\forall N \\ge 10, s(2,N) = 5$\n\n$\\forall N \\ge 15, s(3,N) = 44$\n\n\u2022 A question without answer on the Web is either too stupid or too complicated. I still have not found how much stupid is this one , but trying :-) \u2013\u00a0Philippe Morin Feb 15 '14 at 10:37\n\u2022 I think this is an interesting question. \u2013\u00a0Marko Riedel Aug 14 '15 at 2:18\n\nThis appears to be an interesting problem that can be attacked using Power Group Enumeration on sets as described in quite some detail at the following MSE link.\n\nThat link discusses the number of different subsets of a standard $52$ card deck under suit permutation. Here we have the group permuting the slots into which we distribute the cards is the symmetric group and the group permuting cards is the cardinality twenty-four induced action on the cards of all permutations of the four suits.\n\nThe lottery ticket problem proposed here follows exactly the same model, only now we are distributing tickets into the slots being permuted by the symmetric group and the group acting on the tickets is the induced action of the symmetric group $S_N.$ The number of terms in the cycle indices $Z(S_N)$ and $Z(S_T)$ is given by the partition function and we get an algorithm that is of asymptotically lower order than the naive $N!\\times T!.$\n\nThe only non-trivial issue that is not already featured in the solution to the distributions of cards is how to compute the cycle index of the induced action of $S_N$ on the ${N\\choose Q}$ tickets of $Q$ elements. This can be done quite effectively by computing a representative of the permutation shape from the cycle index of the symmetric group, letting it act on the tickets, and factoring the result into cycles for a contribution to the desired cycle index.\n\nSetting $Q=5$ as in the question we obtain for $N=7$ the sequence $$1, 2, 5, 10, 21, 41, 65, 97, 131, 148, 148, 131,\\ldots$$ for $N=8$ the sequence $$1, 3, 11, 52, 252, 1413, 7812, 41868, 207277, 936130, 3826031,\\\\ 14162479,\\ldots$$ for $N=9$ the sequence $$1, 4, 20, 155, 1596, 20528, 282246, 3791710, 47414089, 542507784,\\\\ 5659823776,53953771138,\\ldots$$ and finally for $N=10$ the sequence $$1, 5, 28, 324, 5750, 142148, 3937487, 108469019, 2804300907,\\\\ 66692193996,1452745413957, 29041307854703,\\ldots.$$\n\nTo illustrate the good complexity of this algorithm here is the sequence for $N=13:$ $$1, 5, 42, 813, 34871, 2777978, 304948971, 37734074019,\\\\ 4719535940546, 566299855228261, 63733180893169422,\\\\ 6674324951638852138,\\ldots$$\n\nFinally we obtain for $N$ variable with $Q=5$ and $T=3$ the sequence $$0, 0, 0, 0, 0, 1, 5, 11, 20, 28, 35, 39, 42, 43,\\\\ 44, 44, 44, 44,\\ldots$$\n\nThe Maple code to compute these was as follows:\n\nwith(combinat);\nwith(numtheory);\n\npet_flatten_term :=\nproc(varp)\nlocal terml, d, cf, v;\n\nterml := [];\n\ncf := varp;\nfor v in indets(varp) do\nd := degree(varp, v);\nterml := [op(terml), seq(v, k=1..d)];\ncf := cf/v^d;\nod;\n\n[cf, terml];\nend;\n\npet_autom2cycles :=\nproc(src, aut)\nlocal numa, numsubs;\nlocal marks, pos, cycs, cpos, clen;\n\nnumsubs := [seq(src[k]=k, k=1..nops(src))];\nnuma := subs(numsubs, aut);\n\nmarks := Array([seq(true, pos=1..nops(aut))]);\n\ncycs := []; pos := 1;\n\nwhile pos <= nops(aut) do\nif marks[pos] then\nclen := 0; cpos := pos;\n\nwhile marks[cpos] do\nmarks[cpos] := false;\ncpos := numa[cpos];\nclen := clen+1;\nod;\n\ncycs := [op(cycs), clen];\nfi;\n\npos := pos+1;\nod;\n\nreturn mul(a[cycs[k]], k=1..nops(cycs));\nend;\n\npet_cycleind_symm :=\nproc(n)\nlocal p, s;\noption remember;\n\nif n=0 then return 1; fi;\n\nend;\n\npet_flat2rep :=\nproc(f)\nlocal p, q, res, t, len;\n\nq := 1; res := [];\n\nfor t in f do\nlen := op(1, t);\nres := [op(res), seq(p, p=q+1..q+len-1), q];\nq := q+len;\nod;\n\nres;\nend;\n\npet_cycleind_tickets :=\nproc(N, Q)\noption remember;\nlocal cind, tickets, q, term, rep, subsl, ptickets,\nidx, flat;\n\nif N=1 then\nidx := [a[1]]\nelse\nidx := pet_cycleind_symm(N);\nfi;\n\ncind := 0;\n\ntickets := convert(choose({seq(q, q=1..N)}, Q), list);\n\nfor term in idx do\nflat := pet_flatten_term(term);\nrep := pet_flat2rep(flat[2]);\n\nsubsl := [seq(q=rep[q], q=1..N)];\n\nptickets := subs(subsl, tickets);\n\ncind := cind +\nflat[1]*pet_autom2cycles(tickets, ptickets);\nod;\n\ncind;\nend;\n\nX :=\nproc(N, Q, T)\noption remember;\nlocal idx_slots, res, a, b,\nflat_a, flat_b, cycs_a, cycs_b, q,\ntbl_a, tbl_b, f1, f2, f3;\n\nif T=1 then\nidx_slots := [a[1]]\nelse\nidx_slots := pet_cycleind_symm(T);\nfi;\n\nres := 0;\n\nfor a in idx_slots do\nflat_a := pet_flatten_term(a);\ncycs_a := sort(flat_a[2]);\n\ntbl_a := table();\nfor q in convert(cycs_a, 'multiset') do\ntbl_a[op(1, q[1])] := q[2];\nod;\n\nf1 := map(q -> op(1, q), cycs_a);\nf1 := mul(f1[q], q=1..nops(cycs_a));\n\nf2 := convert(map(q -> op(1, q), cycs_a), 'multiset');\nf2 := map(q -> q[2], f2);\nf2 := mul(f2[q]!, q=1..nops(f2));\n\nfor b in pet_cycleind_tickets(N, Q) do\nflat_b := pet_flatten_term(b);\ncycs_b := sort(flat_b[2]);\n\ntbl_b := table();\nfor q in convert(cycs_b, 'multiset') do\ntbl_b[op(1, q[1])] := q[2];\nod;\n\nf3 := 1;\nfor q in [indices(tbl_a, 'nolist')] do\nif type(tbl_b[q], integer) then\nf3 := f3*binomial(tbl_b[q], tbl_a[q]);\nelse\nf3 := 0;\nfi;\nod;\n\nres := res + f3*f2*f1*flat_a[1]*flat_b[1];\nod;\nod;\n\nres;\nend;\n\nAddendum Fri Aug 14 2015. The sequence for $Q=5$ and $N=20$ is $$1, 5, 44, 966, 53484, 7023375, 1756229468, 710218125299, \\\\ 411620592905173, 308212635851733551, 271743509344779773214,\\ldots$$\n\nAddendum Sat Aug 15 2015. The sequence for $Q=5$ and $N=22$ is $$1, 5, 44, 966, 53529, 7041834, 1773511264, 734330857318, \\\\ 452455270344141, 383969184978128899, 416614280701828877344, \\\\ 536531456518633409220043, 766723127226754935510254929,\\ldots$$\n\nAddendum Wed Aug 19 2015. The sequence for $Q=5$ and $N=24$ is $$1, 5, 44, 966, 53534, 7043732, 1775444689, 737776095236, \\\\ 460462767067281, 405308264117856150, 477303563740811267063, \\\\ 712445246443357547546003, 1271053814158420923816386794,\\ldots$$\n\nThere are $\\binom{N}{5}$ possible tickets, you are asking how many $T$-subsets of those there are, i.e., the number of systems is:\n\n$$\\binom{\\binom{N}{5}}{T}$$\n\n(somehow I'm feeling I'm just being stupid here... can't be that easy?)\n\n\u2022 You appear to have missed the part where the OP says \"Two systems are distinct if ..\" \u2013\u00a0Marko Riedel Aug 14 '15 at 22:00\n\u2022 @MarkoRiedel, \"are unique up to permutation\" is just sets in my book... \u2013\u00a0vonbrand Aug 14 '15 at 22:06\n\u2022 These lottery tickets are indeed sets. The permutation however acts on the $N$ values from which they are drawn. E.g. the ticket $\\{1,2,3,4,5\\}$ drawn from $N=10$ could be transformed into the ticket $\\{6,7,8,9,10\\}$ by a permutation of the $N$ values that exchanges the upper and lower five values (permutation not unique). \u2013\u00a0Marko Riedel Aug 14 '15 at 22:11\n\u2022 The permutation acts simultaneously on all tickets in a system of size $T,$ which is indeed a set of tickets. \u2013\u00a0Marko Riedel Aug 14 '15 at 22:26", "date": "2021-06-13 06:27:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/498954/counting-number-of-distinct-systems", "openwebmath_score": 0.6212775707244873, "openwebmath_perplexity": 1499.359269412658, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682461347528, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8658627922254889}}
{"url": "https://math.stackexchange.com/questions/2419664/deriving-the-number-of-all-even-length-palindromic-sequences-with-length-at-most", "text": "# Deriving the number of all even length palindromic sequences with length at most $n$\n\nDerive the number of all even length palindromic sequences with length $\\leq n$.\n\nIf $n=4$, I have to count the number of non-palindromes of length 0, 2, and 4.\n\nAlso, $2p = n$, so a string of length $n = 4$ has $p = 2$. This comes in handy later.\n\n$x$ represents the number of possible characters in the given alphabet- so $x = 26$ for English, 2 for binary, etc.\n\nI chose to approach by counting, using this thought process.\n\n$$\\sum_{k=0}^{p} (x^{2k} - x^k)$$ In other words - a summation of all possible even length strings up to length $n$, subtracting cases that are palindromes.\n\nI tested this with binary, up to length 4. (Binary non-palindromes of length 0, 2, and 4). This means $p$ goes from 0, to 1, to 2.\n\nThis evaluates to: $$2^0 - 2^0 +$$ $$2^2 - 2^1 +$$ $$2^4 - 2^2$$\n\nWhich sums to a total of 14 non palindromes.\n\nAll possible combinations are {}, 11, 00, 10, 01, 1111, 1100, 1000, 0000, 0001, 0011, 0111, 1010, 0101, 0110, 1001 1110, 1101, 0010, 1011, 0100.\n\n14 of these are not palindromes, confirming my approach.\n\nThe solution states the formula is $$\\frac{x^{2p+1}-2x^{p+1}+x}{x-1}$$ Following the same logic...\n\nWith $p = 2$, $x = 2$, $n = 4$, we have\n\n$$\\frac{2^{4+1}-2(2)^{3}+2}{2-1}$$\n\nWhich works out to 18, not 14. I have worked this out several times, and I'm not exactly sure what/if I am misunderstanding. Any help would be greatly appreciated. Thanks!\n\nIf the sequence has even length, say $n = 2k$, selecting the first $k$ characters completely determines the palindrome since the remaining $k$ characters can be found by repeating the sequence in the reverse order. Hence, the number of palindromes of even length at most $n$ in an alphabet with $x$ characters is $$x^0 + x^1 + x^2 + x^3 + \\cdots + x^k = \\frac{1 - x^{k + 1}}{1 - x}$$ where we have used the formula for the sum of a geometric series with initial term $1$ and common ratio $x$.\nIn your example of binary sequences of even length at most $4$, there are seven palindromes: $$\\emptyset, 00, 11, 0000, 0110, 1001, 1111$$ Notice that our formula yields $$2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7$$ or $$\\frac{1 - 2^{2 + 1}}{1 - 2} = \\frac{1 - 2^3}{-1} = \\frac{-7}{-1} = 7$$\nSince there are a total of $$x^0 + x^2 + x^4 + \\ldots + x^{2k} = \\frac{1 - x^{2k + 2}}{1 - x^2}$$ sequences of even length at most $n = 2k$ in an alphabet with $x$ characters, the number of sequences of even length at most $n = 2k$ that are not palindromes is \\begin{align*} \\frac{1 - x^{2k + 2}}{1 - x^2} - \\frac{1 - x^{k + 1}}{1 - x} & = \\frac{1 - x^{2k + 2}}{1 - x^2} - \\frac{1 - x^{k + 1}}{1 - x} \\cdot \\frac{1 + x}{1 + x}\\\\ & = \\frac{1 - x^{2k + 2}}{1 - x^2} - \\frac{1 + x - x^{k + 1} - x^{k + 2}}{1 - x^2}\\\\ & = \\frac{-x + x^{k + 1} + x^{k + 2} - x^{2k + 2}}{1 - x^2} \\end{align*} In your example of binary sequences of even length at most $4$, the total number of sequences is $$2^0 + 2^2 + 2^4 = 1 + 4 + 16 = 21$$ of which seven are palindromes, so there are $14$ non-palindromes as you found. Notice that our formulas yield $$\\frac{1 - 2^{2 \\cdot 2 + 2}}{1 - 2^2} = \\frac{1 - 2^6}{1 - 2^2} = \\frac{1 - 64}{1 - 4} = \\frac{-63}{-3} = 21$$ total sequences and $$\\frac{-2 + 2^{2 + 1} + 2^{2 + 2} - 2^{2 \\cdot 2 + 2}}{1 - 2^2} = \\frac{-2 + 2^3 + 2^4 - 2^6}{1 - 2^2} = {-2 + 8 + 16 - 64}{1 - 4} = \\frac{-42}{-3} = 14$$ non-palindromes.", "date": "2019-09-22 01:55:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2419664/deriving-the-number-of-all-even-length-palindromic-sequences-with-length-at-most", "openwebmath_score": 0.9043335318565369, "openwebmath_perplexity": 225.80245785888377, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874094398219, "lm_q2_score": 0.8740772482857833, "lm_q1q2_score": 0.865849917029702}}
{"url": "https://math.stackexchange.com/questions/3261941/let-mathfrakd-be-the-collection-of-subsets-of-x-of-the-form-e-1-lambda", "text": "# Let $\\mathfrak{D}$ be the collection of subsets of $X$ of the form $E_1^{\\lambda_1}\\cap\\dots\\cap E_n^{\\lambda_n}$. Is $X=\\bigcup_{D\\in\\mathfrak{D}}D$?\n\n## Full problem statement:\n\nLet $$E_1,\\dots,E_n$$ be distinct but not necessarily disjoint subsets of $$X$$.\n\nLet $$\\mathfrak{D}$$ be the disjoint collection of all subsets of $$X$$ of the form $$E_1^{\\lambda_1}\\cap\\dots\\cap E_n^{\\lambda_n}$$ where $$\\lambda_i \\in \\{0,1\\}$$. Note that $$E_i^1 = E_i$$, $$E_i^0 = E_i^c$$.\n\nLet $$\\mathfrak{F}$$ be collection of arbitrary unions of members of $$\\mathfrak{D}$$.\n\n1. Is $$X\\hspace{1mm}\\in\\mathfrak{F}$$?\n2. Let $$F\\in\\mathfrak{F}$$. Is $$F^c = X\\backslash F \\in \\mathfrak{F}$$?\n\n## Author's solution\n\nLet $$x\\in X$$. Since $$\\forall i\\in\\{1,\\dots,n\\}\\hspace{1mm} E_i^1\\cup E_i^0 = X$$, $$x\\in E_i$$ or $$x\\in E_i^c$$.\n\nThen, every $$x$$ is contained for some $$D = E_1^{\\lambda_{1}}\\cap\\dots E_n^{\\lambda_{n}} \\in\\mathfrak{D}$$. So $$X=\\bigcup\\limits_{D\\in\\mathfrak{D}} D\\in\\mathfrak{F}$$.\n\n## My Questions\n\nQ1. Can we just end the proof there? It seems like all we've proven is that $$X\\subset\\hspace{-1.5mm}\\bigcup\\limits_{D\\in\\mathfrak{D}}\\hspace{-1.5mm} D$$.\n\n## Author's solution\n\nLet $$F\\in \\mathfrak{F}$$. Then $$F$$ is a union of member of $$\\mathfrak{D}$$.\n\nSince $$\\mathfrak{D}$$ is a disjoint collection and $$X=\\hspace{-1.5mm}\\bigcup\\limits_{D\\in\\mathfrak{D}}\\hspace{-1.5mm} D$$, $$F^c=X\\backslash F$$ is also a union of members of $$\\mathfrak{D}$$. So $$F^c\\in\\mathfrak{F}$$.\n\n## My Questions\n\nQ1. How does the statements that $$\\mathfrak{D}$$ is a disjoint collection and $$X=\\hspace{-1.5mm}\\bigcup\\limits_{D\\in\\mathfrak{D}}\\hspace{-1.5mm} D$$ work together to imply $$F^c=X\\backslash F$$ is a union of members of $$\\mathfrak{D}$$? I did try some DeMorgan stuff but it all went nowhere :(\n\nI didn't put all my questions at the end because I think having to scroll up to the relevant block and scroll back down to the questions is going to be annoying.\n\nFirst question: since $$D \\subset X$$ for all $$D \\in \\mathfrak D$$ it is understood that the reverse inclusion holds so equality holds.\nSecond question. It is advisable to look at some simple examples. If $$F$$ is expressed as union of certain members of $$\\mathfrak D$$ then $$F^{c}$$ is precisely the union of the remaining members of $$\\mathfrak D$$.\n[$$\\mathfrak D$$ is a partition of $$X$$: its members are disjoint and their union is $$X$$. Call these sets $$(D_i)_{i\\in I}$$. For any subset $$J$$ of $$I$$, let $$F$$ be the union of the sets $$D_i$$ with $$i \\in J$$. Let $$G$$ be the union of the sets $$D_i$$ with $$i \\in I\\setminus J$$. Then $$F\\cup G$$ is the union of all the $$D_i$$'s which is $$X$$. Also $$F$$ and $$G$$ are disjoint. These two facts imply that $$G$$ is the complement of $$F$$].", "date": "2019-12-16 11:06:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3261941/let-mathfrakd-be-the-collection-of-subsets-of-x-of-the-form-e-1-lambda", "openwebmath_score": 0.9616463780403137, "openwebmath_perplexity": 103.97343062144778, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990587409856637, "lm_q2_score": 0.8740772417253256, "lm_q1q2_score": 0.8658499108953239}}
{"url": "https://eliteprotek.com/starbucks-vanilla-gwd/beaa86-determine-the-coordinates-of-the-centroid-of-the-area", "text": "# determine the coordinates of the centroid of the area\n\nCentroid: Centroid of a plane figure is the point at which the whole area of a plane figure is assumed to be concentrated. 4' 13 Answers: (X,Y) in \u2022 The coordinates ( and ) define the center of gravity of the plate (or of the rigid body). The cartesian coordinate of it's centroid is $\\left(\\frac{2}{3}r(\\theta)\\cos\\theta, \\frac{2}{3}r(\\theta)\\sin\\theta\\right)$. Find the coordinates of the centroid of the area bounded by the given curves. How to calculate a centroid. It is also the center of gravity of the triangle. The centroid of a right triangle is 1/3 from the bottom and the right angle. Centroid by Composite Bodies ! The x-centroid would be located at 0 and the y-centroid would be located at 4 3 r \u03c0 7 Centroids by Composite Areas Monday, November 12, 2012 Centroid by Composite Bodies Recall that the centroid of a triangle is the point where the triangle's three medians intersect. y=x^{3}, x=0, y=-8 y=2 x, y=0, x=2 It is the point which corresponds to the mean position of all the points in a figure. Next, sum all of the x coodinates ... how to find centroid of composite area: how to calculate centroid of rectangle: how to find centroid of equilateral triangle: For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ! Problem Answer: The coordinates of the center of the plane area bounded by the parabola and x-axis is at (0, 1.6). The Find Centroids tool will create point features that represent the geometric center (centroid) for multipoint, line, and area features.. Workflow diagram Examples. Find the coordinates of the centroid of the plane area bounded by the parabola y = 4 \u2013 x^2 and the x-axis. Center of Mass of a Body Center of mass is a function of density. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Gather both the x and y coordinate points of each vertex. The coordinates of the centroid are simply the average of the coordinates of the vertices.So to find the x coordinate of the orthocenter, add up the three vertex x coordinates and divide by three. First, gather the coordinate points of the vertices. The center of mass is the term for 3-dimensional shapes. Determine the coordinates of the centroid of the shaded region. Beam sections are usually made up of one or more shapes. The centroid is the term for 2-dimensional shapes. Centroid of an Area \u2022 In the case of a homogeneous plate of uniform thickness, the magnitude \u2206W is We can consider the surface element as an triangle, and the centroid of this triangle is obviously at here.) Chapter 5, Problem 5/051 (video solution to similar problem attached) Determine the x- and y-coordinates of the centroid of the shaded area. Find the coordinates of the centroid of the area bounded by the given curves. And the area of this surface element $\\mathrm{d}A = \\frac{1}{2}r^2(\\theta)\\mathrm{d}\\theta$. An analyst at the Scotland Department of Environment is performing a preliminary review on wind farm applications to determine which ones overlap with or are in view of wild lands. So to find the centroid of an entire beam section area, it first needs to be split into appropriate segments. 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More shapes to the mean position of all the points in a.. Three medians intersect, the centroid of a plane figure is the point the. And the right angle of density 1/3 from the bottom and the angle... Is 1/3 from the bottom and the centroid of the centroid of this triangle is the point at the! -Coordinates of the area bounded by the given curves x, y=0, x=2 Find centroid. Centroid of a right triangle is the point where the triangle 's medians. Points in a figure coordinates of the triangle 's three medians intersect triangle, and the right.. Medians intersect point at which the whole area of a circle and a rectangle is at the..\n\nScroll to top", "date": "2021-10-28 05:20:08", "meta": {"domain": "eliteprotek.com", "url": "https://eliteprotek.com/starbucks-vanilla-gwd/beaa86-determine-the-coordinates-of-the-centroid-of-the-area", "openwebmath_score": 0.7444528341293335, "openwebmath_perplexity": 388.8633420720968, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9811668679067631, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.8658089843659367}}
{"url": "https://notescs.gitlab.io/posts/max-value-with-addition-multiplication/", "text": "# Maximum Value of Equation of Ones with Addition and Multiplication Operations\n\nGiven an integer n, find the maximum value that can be obtained using n ones and only addition and multiplication operations. Note that, you can insert any number of valid bracket pairs.\n\nExample:\n\nInput: n = 12\nOutput: 81\nExplanation: (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) = 81\n\n## Approach: Dynamic Programming\n\nObserve that, in order to find the answer for n = 5, we need to consider the maximum answer obtainable from the answer of 2 and 3, 1 and 4. Thus, this problem has an optimal substructure property.\n\n                    5\n/                 \\\nop(1, 4)            op(2, 3)\n/  \\                /       \\\n1   op(2, 2)       op(1, 1)  op(1, 2)\n/      \\          / \\      /  \\\nop(1, 1) op(1, 1)    1   1    1   op(1, 1)\n/ \\        / \\                    / \\\n1   1      1   1                  1   1\n\nClearly, from the tree above there are a lot of overlapping sub-problems. Owing to this and optimal substructure property, this problem is an ideal candidate for dynamic programming.\n\nBottom-up approach:\n\nC++ code:\n\n#include<iostream>\n#include<vector>\n#include<algorithm>\nusing namespace std;\n\nlong long getMaximumValue(int n) {\nvector<long long> dp(n + 1);\n// dp[i] denotes maximum value that can be obtained from i ones\ndp[0] = 0; // base case: with 0 ones answer is always 0\ndp[1] = 1; // base case: with 1 one answer is 1\nfor (int i = 1; i <= n; ++i) {\nfor (int j = 1; j <= i / 2; ++j) {\ndp[i] = max({dp[i], dp[j] + dp[i - j], dp[j] * dp[i - j]});\n}\n}\nreturn dp[n];\n}\n\nint main() {\ncout << \"n = \" << 5 << \" Maximum possible value \" << getMaximumValue(5) << endl;  // prints\ncout << \"n = \" << 12 << \" Maximum possible value \" << getMaximumValue(12) << endl;  // prints \"81\"\nreturn 0;\n}\n\nPython code:\n\ndef getMaximumValue(n):\ndp = [0] * (n + 1)\n# dp[i] denotes maximum value that can be obtained from i ones\ndp[0] = 0  # base case: with 0 ones answer is always 0\ndp[1] = 1  # base case: with 1 one answer is 1\nfor i in range(2, n + 1):\nfor j in range(1, i // 2 + 1):\ndp[i] = max(dp[i], dp[j] + dp[i - j], dp[j] * dp[i - j])\nreturn dp[n]\n\nif __name__ == '__main__':\nprint('n =', 5, 'Maximum possible value', getMaximumValue(5))\nprint('n =', 12, 'Maximum possible value', getMaximumValue(12))\n\nTime Complexity: $O(n ^ 2)$\nSpace Complexity: $O(n)$ due to storing states in dp array\n\nExercise: Code the top-down approach.", "date": "2020-11-29 14:03:38", "meta": {"domain": "gitlab.io", "url": "https://notescs.gitlab.io/posts/max-value-with-addition-multiplication/", "openwebmath_score": 0.5966708660125732, "openwebmath_perplexity": 2337.377558837608, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668679067631, "lm_q2_score": 0.8824278618165526, "lm_q1q2_score": 0.865808981332209}}
{"url": "https://math.stackexchange.com/questions/3091598/concerning-the-identity-in-sums-of-binomial-coefficients", "text": "# Concerning the identity in sums of Binomial coefficients\n\nLet be the following identity $$\\sum_{k=1}^{n}\\binom{k}{2}=\\sum_{k=0}^{n-1}\\binom{k+1}{2}=\\sum_{k=1}^{n}k(n-k)=\\sum_{k=0}^{n-1}k(n-k)=\\frac16(n+1)(n-1)n$$ As we can see the partial sums of binomial coefficients are expressed in terms of $$3$$-rd order polynomial $$P_3(n)$$, where $$n$$ is variable of upper bound of summation. We assume that order of resulting polynomial $$P_3(n)$$ depends on subscript of binomial coefficient being summed up (in our case the order of polynomial is $$3=2+1$$, where $$2$$ is subscript of bin. coef.)\n\nThe question: Does there exist a generalized method to represent the sum of binomial coefficients $$\\sum_{k}^{n}\\binom{k}{s}$$ in terms of certain polynomials $$P_{s+1}(n)=\\sum_{k}^{n} F_s(n,k)$$ for every non-negative integer $$s$$? I.e can we always find the function $$F_s(n,k)$$, such that $$\\sum_{k}^{n}\\binom{k}{s}=\\sum_{k}^{n}F_s(n,k)$$ ? We assume that order of polynomial is $$s+1$$ by means of example above.\n\nThe sub-question: (In case of positive answer to the first question.) If there exists a method to represent the sums of bin. coef. in terms of polynomials in $$n$$, how do summation limits of the $$\\sum_{k}^{n}\\binom{k}{s}$$ implies to the form of polynomial $$P_{s+1} (n)$$ exactly?\n\n\u2022 $\\sum_{k=s}^n \\binom{k}s=\\binom{n+1}{s+1}$. This is the Hockey Stick identity. \u2013\u00a0Mike Earnest Jan 29 '19 at 1:24\n\u2022 The question is to find such polynomial $F_s(n,k)$ that $\\sum_{k}^{n}\\binom{k}{s}=\\sum_{k}^{n}F_s(n,k)$. From our example it follows that: $F_2(n,k)=k(n-k)$ and $\\sum_{k=1}^{n}\\binom{k}{2}=\\sum_{k=1}^{n}F_2(n,k)$. Can you provide examples for $s>2$? \u2013\u00a0PKK Jan 29 '19 at 1:33\n\u2022 I understand now. Still, my comment just shows that $P_s(n)$ is nothing mysterious. \u2013\u00a0Mike Earnest Jan 29 '19 at 1:51\n\u2022 Why don't you mention your previous question math.stackexchange.com/q/2774300 ? \u2013\u00a0Jean Marie Jan 30 '19 at 18:25\n\u2022 thank you for reminding, by the way these coefficients could be found as $$(2k-1)!T(2n,2k)=\\frac{1}{r}\\sum_{j=0}^{r}(-1)^j\\binom{2r}{j}(r-j)^{2n},$$ where $r=n-k+1$ and $T(2n,2k)$ is central factorial number \u2013\u00a0PKK Jan 30 '19 at 18:30\n\n$$\\sum_{k=0}^n\\binom{k}s=\\sum_{k=0}^n\\sum_{i=0}^{k-1}\\binom{i}{s-1}=\\sum_{i=0}^{n-1}\\sum_{k=i+1}^n\\binom{i}{s-1}=\\sum_{i=0}^{n-1}(n-i)\\binom{i}{s-1}=\\sum_{i=1}^ni\\binom{n-i}{s-1}$$ In other words, you can let $$F_s(n,k)=k\\binom{n-k}{s-1}$$, and you will have $$\\sum_{k=0}^n \\binom{k}s=\\sum_{k=0}^{n}F_s(n,k)$$.\n\n\u2022 Can you show me a direct example for $s=2$ step by step ? The example of $\\sum_{k=1}^n\\binom{k}{s}$ please \u2013\u00a0PKK Jan 29 '19 at 1:53\n\u2022 @PetroKolosov When $s=2$, you get $F_s(n,k)=k(n-k)$, just as you had. When $s>2$, the expression $k\\binom{n-k}s$ is polynomial in disguise. For example, $F_3(n,k)=k\\binom{n-k}2=k(n-k)(n-k-1)/2$. \u2013\u00a0Mike Earnest Jan 29 '19 at 2:36\n\u2022 Are you sure that $F_3(n,k)=k(n-k)(n-k-1)/2$? For me it gives as follows: $F_s(n,k)=k\\binom{n-k}{s-1}$, by the hockey stick pattern: $\\binom{n-k}{s-1}=\\sum_{j}^{n-k+1}\\binom{j}{s-2}|_{s=3}=1/2 (-2 + k - n) (-1 + k - n)$ and multiplication by $k$ gives $F_3(n,k)=1/2 k (-2 + k - n) (-1 + k - n)$. PS and obviously i wrong with limits of hockey stick pattern :) \u2013\u00a0PKK Jan 29 '19 at 3:09\n\u2022 I've fixed the error above, but still $F_s(n,k)=k\\binom{n-k}{s-1}$, by the hockey stick pattern: $\\binom{n-k}{s-1}=\\sum_{j}^{n-k-1}\\binom{j}{s-2}|_{s=3}=1/2 (k - n) (1 + k - n)$ and multiplication by $k$ gives $F_3(n,k)=1/2 k (k - n) (1 + k - n)$ is different from your example of $F_3(n,k)$ \u2013\u00a0PKK Jan 29 '19 at 3:18\n\u2022 $k(n-k)(n-k-1)/2=\\frac12k(k-n)(1+k-n)$. Your expression and mine for $F_3(n,k)$ are the same. All you did was expand $\\binom{n-k}{s-1}$ using the HS identity, and then collapse it using the same identity. @PetroKolosov \u2013\u00a0Mike Earnest Jan 29 '19 at 3:37\n\nI would say that you have a good answer already. But their are other possible answers which seem reasonable. Further restriction might force the favored solution above.\n\nIn the case $$k=3$$ (which is the only one I will discuss in any detail)\n\n$$\\sum_{s=1}^n\\binom{s}3= \\\\ \\sum_{s=1}^ns\\binom{n-s}2=\\sum_{s=1}^n(n-s)\\binom{s}2=\\frac14\\sum_{s=1}^n{s(n-s)(n-2)}=\\frac1{24}\\sum_{s=1}^n{(n+1)(n-1)(n-2)}$$\n\nIt is easy to see how to generalize these to other $$k.$$\n\nThe first three belong to the infinite family\n\n$$\\frac14\\sum_{s=1}^n{s(n-s)(\\alpha n-(2\\alpha -2)s-2)} \\tag{*}$$\n\nGoing back to the favored solution:\n\n$$\\sum_{s=1}^n\\binom{s}3=\\sum_{s=1}^n\\frac{s^3}{6}-\\frac{s^2}{2}+\\frac{s}{3}=\\sum_{s=1}^n\\frac{n^2s}2-n{s}^{2}+\\frac{s^3}2-\\frac{ns}{2}+\\frac{s^2}2.$$\n\nIf you wanted the thing on the right to be\n\n$$\\sum_{s=1}^nA{n^2s}+Bn{s}^{2}+C{s^3}+D{ns}+E{s^2}$$\n\nThen you do need to have $$E=\\frac12$$ But the rest have two degrees of freedom\n\n$$C=-2A-\\frac43B \\ \\ \\ \\ \\ \\ \\ D=A-\\frac13B-\\frac23$$\n\nFor further restriction we might want to have $$D=-E=-\\frac12$$ so than $$A{n^2s}+Bn{s}^{2}+C{s^3}+D{ns}+E{s^2}=0$$ when $$s=n$$\n\nIn this case the summand factors (of course) giving the family $$(*)$$ above.\n\nIf we want the right-hand side to be\n\n$$K\\sum_{s=1}^ns(An+(1-A)s+B)(Cn+(1-C)s+D)$$\n\nIt is possible to work out the requirements. I came up with $$6$$ families of solutions. Of course the set of solutions is invariant under switching the two terms and/or substituting $$s=n+1-s.$$", "date": "2020-12-04 18:37:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3091598/concerning-the-identity-in-sums-of-binomial-coefficients", "openwebmath_score": 0.9073677062988281, "openwebmath_perplexity": 299.69341620708525, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109529751892, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8657933627977826}}
{"url": "http://math.stackexchange.com/questions/150630/understanding-isometric-spaces?answertab=oldest", "text": "# Understanding isometric spaces\n\nI have studied that an isometry is a distance-preserving map between metric spaces and two metric spaces $X$ and $Y$ are called isometric if there is a bijective isometry from X to Y.\n\nMy questions are related with the understanding of isometric spaces, they are as follows:\n\nCan we say that two isometric spaces are same? If no, in what context they differ? What are the common properties shared by two isometric spaces?\n\nIntuitively what are isometric spaces?\n\nIf two spaces are isometric how to find out bijective distance preserving map between them?\n\nThanks for your help and time.\n\n-\nAny two lines in the plane are isometric (you can translate and rotate to place one line on top of the other, without affecting distances within the lines in this process), so definitely two isometric spaces need not really be literally the same. But as far as metric properties are concerned they behave in the same way. It's like asking \"are all circles of radius 1 the same\"? No, but obviously you're comfortable using one particular choice (like the one centered at the origin) even if that's not the original circle of interest. \u2013\u00a0 KCd May 28 '12 at 6:04\nThe examples of the line and circle might seem silly, because they are very familiar. The point of the concept of isometric spaces is to keep us aware that we shouldn't consider two isometric spaces as being fundamentally different from one another. For example, when you construct the completion $\\widetilde{X}$ of a metric space $X$, using equiv. classes of Cauchy sequences in $X$, you don't really find $X$ as a subset of $\\widetilde{X}$, but $X$ is isometric to the equiv. classes of constant seq. $(x,x,x,\\dots)$, and that is how we can view $X$ (as metric space) inside $\\widetilde{X}$. \u2013\u00a0 KCd May 28 '12 at 6:08\nThere is no universal method to find an isometry between two isometric metric spaces. \u2013\u00a0 KCd May 28 '12 at 6:09\n@KCd Thanks to you. Your comments are helpful to me. \u2013\u00a0 srijan May 28 '12 at 6:17\n\nHomeomorphisms are the maps that preserve all topological properties: from a structural point of view, homeomorphic spaces might as well be identical, though they may have very different underlying sets, and if they\u2019re metrizable, they may carry very different (but equivalent) metrics. Isometries are the analogue for metric spaces, topological spaces carrying a specific metric: they preserve all metric properties, and of course those include the topological properties. Thus, all isometries are homeomorphisms, but the converse is false.\n\nConsider the metric spaces $\\langle X,d_X\\rangle$ and $\\langle Y,d_Y\\rangle$ defined as follows: $X=\\Bbb N,Y=\\Bbb Z$, $$d_X(m,n)=\\begin{cases}0,&\\text{if }m=n\\\\1,&\\text{if }m\\ne n\\;,\\end{cases}$$ for all $m,n\\in X$, and $$d_Y(m,n)=\\begin{cases}0,&\\text{if }m=n\\\\1,&\\text{if }m\\ne n\\end{cases}$$ for all $m,n\\in Y$. It\u2019s easy to check that $d_X$ and $d_Y$ are metrics on $X$ and $Y$, respectively.\n\nClearly these are not the same space: they have different underlying sets. However, if $f:X\\to Y$ is any bijection1 whatsoever, then $f$ is an isometry between $X$ and $Y$. $\\langle X,d_X\\rangle$ and $\\langle Y,d_Y\\rangle$ are structurally identical as metric spaces: if $P$ is any property of metric spaces $-$ not just of metrizable spaces, but of metric spaces with a specific metric $-$ then either $X$ and $Y$ both have $P$, or neither of them has $P$. There is no structural property of metric spaces that distinguishes them.\n\nWhat I just said about $X$ and $Y$ is true of isometric spaces in general: there is no structural property of metric spaces that distinguishes them. Considered as metric spaces, they are structurally identical, though they may have different underlying sets.\n\nIsometric spaces may even have the same underlying set but different metrics. Consider the following two metrics on $\\Bbb N=\\{0,1,2,\\dots\\}$. For any $m,n\\in\\Bbb N$,\n\n$$d_0(m,n)=\\begin{cases} 0,&\\text{if }m=n\\\\\\\\ \\left|\\frac1m-\\frac1n\\right|,&\\text{if }0\\ne m\\ne n\\ne 0\\\\\\\\ \\frac1m,&\\text{if }n=0<m\\\\\\\\ \\frac1n,&\\text{if }m=0<n\\;, \\end{cases}$$\n\nand\n\n$$d_1(m,n)=\\begin{cases} 0,&\\text{if }m=n\\\\\\\\ \\left|\\frac1m-\\frac1n\\right|,&\\text{if }m\\ne n\\text{ and }m,n>1\\\\\\\\ 1-\\frac1m,&\\text{if }n=0\\text{ and }m>1\\\\\\\\ 1-\\frac1n,&\\text{if }m=0\\text{ and }n>1\\\\\\\\ \\frac1m,&\\text{if }n=1\\ne m\\\\\\\\ \\frac1n,&\\text{if }m=1\\ne n\\;. \\end{cases}$$\n\nIt\u2019s a good exercise to show that $$f:\\Bbb N\\to\\Bbb N:n\\mapsto\\begin{cases}n,&\\text{if }n>1\\\\1,&\\text{if }n=0\\\\0,&\\text{if }n=1\\end{cases}$$ is an isometry between $\\langle\\Bbb N,d_0\\rangle$ and $\\langle\\Bbb N,d_1\\rangle$. (HINT: Both spaces are isometric to the space $\\{0\\}\\cup\\left\\{\\frac1n:n\\in\\Bbb Z^+\\right\\}$ with the usual metric.) Yet these are clearly not the same space: metric $d_0$ makes $0$ a limit point of the other points, but metric $d_1$ makes $0$ an isolated point.\n\nI don\u2019t know of any general method for finding an isometry between isometric spaces; if you can recognize two spaces as being isometric, you probably already have a good idea of what an isometry between them must look like.\n\n1 If you want a specific bijection, $$f(n)=\\begin{cases}0,&\\text{if }n=0\\\\\\\\\\frac{n}2,&\\text{if }n>0\\text{ and }n\\text{ is even}\\\\\\\\-\\frac{n+1}2,&\\text{if }n\\text{ is odd}\\end{cases}$$ does the job.\n\n-\nSir, by structural properties do you mean completeness, boundedness etc? \u2013\u00a0 srijan May 28 '12 at 6:05\n@srijan: Anything that has to do with the topological or metric structure of the space and not with superficial charactersitiscs like the specific names attached to the points. Completeness and boundedness are indeed structural properties of metric spaces, though not of metrizable spaces. \u2013\u00a0 Brian M. Scott May 28 '12 at 6:07\nAll homeomorphisms need not be isometries because some of the topological properties may not be shared by two isometric spaces, Am i right sir? \u2013\u00a0 srijan May 28 '12 at 6:15\n@srijan: No, it\u2019s because some of the metric properties may not be shared between two homeomorphic spaces. For instance, $\\Bbb R$ and $(0,1)$ with the usual metrics are homeomorphic, but $\\Bbb R$ is a complete metric space, while $(0,1)$ isn\u2019t: they don\u2019t share the metric property of completeness. \u2013\u00a0 Brian M. Scott May 28 '12 at 6:18\n@srijan: You\u2019re very welcome! \u2013\u00a0 Brian M. Scott May 28 '12 at 6:22\nshow 1 more comment", "date": "2014-04-25 03:21:56", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/150630/understanding-isometric-spaces?answertab=oldest", "openwebmath_score": 0.8300754427909851, "openwebmath_perplexity": 268.3377027249952, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109489630492, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8657933592705227}}
{"url": "http://math.stackexchange.com/questions/166244/determinant-of-an-n-times-n-complex-matrix-as-an-2n-times-2n-real-determinan", "text": "Determinant of an $n\\times n$ complex matrix as an $2n\\times 2n$ real determinant\n\nIf $A$ is an $n\\times n$ complex matrix. Is it possible to write $\\vert \\det A\\vert^2$ as a $2n\\times 2n$ matrix with blocks containing the real and imaginary parts of $A$?\n\nI remember seeing such a formula, but can not remember where. Any details, (and possibly references) for such a result would be greatly appreciated.\n\n-\n\nWrite $A=A_1+iA_2$ where $A_1$ and $A_2$ are real matrices. Let $$B:=\\pmatrix{A_1&iA_2\\\\iA_2&A_1}.$$ We have $$\\det B=\\det\\pmatrix{A_1+iA_2&iA_2\\\\A_1+iA_2&A_1}=\\det\\pmatrix{I&iA_2\\\\I&A_1}\\cdot\\det\\pmatrix{A_1+iA_2&0\\\\0&I},$$ and $$\\det\\pmatrix{I&iA_2\\\\I&A_1}=\\det\\pmatrix{I&iA_2\\\\0&A_1-iA_2}$$ hence $$\\det B=\\det(A_1-iA_2)\\det(A_1+iA_2)=\\det A\\cdot\\det\\bar A=|\\det A|^2.$$\n\n-\nThanks you very much! \u2013\u00a0dernam Jul 3 '12 at 18:59\nYou are welcome! \u2013\u00a0Davide Giraudo Jul 3 '12 at 19:01\n\nDavide's answer tells most of the story, in particular giving the proof for the determinant, but not quite all of it, so I want to supplement it with a couple of remarks.\n\nI think that it is more common to replace Davide's matrix $B$ with a real matrix. This can be achieved by conjugating it with matrix of the block form $$D=\\pmatrix{I&0\\cr0&iI\\cr},$$ when $$D^{-1}BD=\\pmatrix{A_1&-A_2\\cr A_2 &A_1\\cr}.$$ Because conjugation preserves the determinant, Davide's calculation tells that here we also have $$\\det(D^{-1}BD)=\\det(B)=|\\det A|^2.$$ Further conjugating (shuffling rows and columns) allows us to replace each and every complex entry $z=a+bi$ with a $2\\times2$ block $$(z)=\\pmatrix{a&-b\\cr b&a\\cr}.$$ Doing it this way makes it clear that if $A$ represents a linear mapping $T$ from $V=\\mathbf{C}^n$ to itself with respect to basis $v_1,v_2,\\ldots,v_n$, then $D^{-1}BD$ represents the same mapping $T$, when we view $V$ as a real vector space of dimension $2n$ and use the basis $v_1,v_2,\\ldots,v_n,iv_1,iv_2,\\ldots,iv_n.$\n\nThe extra shuffling I talked about would reorder this latter basis to $v_1,iv_1,\\ldots$.\n\nA geometric interpretation of this is that $\\det B$ gives the scaling of Lebesgue measure (or hypervolumes) of a box $K=\\prod_{i=1}^{2n}[c_i,d_i]$ of real dimension $2n$ under $T$: $$\\det (B)=\\frac{m(T(K))}{m(K)}.$$\n\n$\\det A$ does the same, but because the coordinates are complex there, we need to use $|\\det A|^2$ to get the scaling right. This is seen already in the complex plane, where multiplication by $a+bi$ multiplies the areas of rectangles by a factor of $a^2+b^2$.\n\n-\nIf $A$ and $B$ define the \"same\" linear map shouldn't $\\det A = \\det B$? \u2013\u00a0learner Jan 24 '15 at 4:27\n@learner: No. For example multiplication by $2$ on $\\Bbb{C}$ has determinant two (it's a 1x1 matrix $2I_1$), but when we view $\\Bbb{C}$ as $\\Bbb{R}^2$ it is a 2x2 matrix $2I_2$ with determinant $=4$. \u2013\u00a0Jyrki Lahtonen Jan 25 '15 at 9:04\n\nLet $A=B+iC$, where $B$ and $C$ are $n\\times n$ real matrices, and let $$\\widetilde{A}=\\left( \\begin{array}{rr} B & -C \\\\ C & B \\end{array} \\right).$$ Then $\\det\\widetilde{A}=|\\det A|^2$.\n\nProof. \\begin{align*} \\det\\widetilde{A}&=\\det\\left( \\begin{array}{cc} B+iC & -C+iB \\\\ C & B \\end{array} \\right)= \\det\\left( \\begin{array}{cc} B+iC & 0 \\\\ C & B-iC \\end{array} \\right)= \\det \\left( \\begin{array}{ll} A & 0 \\\\ C & \\overline{A} \\end{array} \\right)\\\\ &= (\\det A)(\\det\\overline{A})=(\\det A)(\\overline{\\det A})=|\\det A|^2. \\end{align*}\n\n-\nThis question already has a well accepted answer. You have contributed nothing new \u2013\u00a0Shailesh Nov 14 '15 at 13:42", "date": "2016-04-29 22:25:09", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/166244/determinant-of-an-n-times-n-complex-matrix-as-an-2n-times-2n-real-determinan", "openwebmath_score": 0.9953331351280212, "openwebmath_perplexity": 231.0367520770435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109520836027, "lm_q2_score": 0.8791467580102419, "lm_q1q2_score": 0.865793355777279}}
{"url": "https://math.stackexchange.com/questions/3845649/prove-n-in-mathbb-z-n-text-is-even-n-in-mathbb-z-n-1-text/3845657", "text": "# Prove $\\{n \\in \\mathbb Z | n \\text { is even} \\} = \\{n \\in \\mathbb Z | n-1 \\text { is odd}\\}$.\n\nThis is found in the book the instructor is using, An Introduction to Proof through Real Analysis by Daniel J. Madden and Jason A. Aubrey The University of Arizona Tucson, Arizona, USA\n\nHere is one way to define $$A$$ in set-builder notation: $$A = \\{x | x \\text{ is an even integer}\\} \\text. \\tag{9.1}$$ In general, set-builder notation takes the form $$S = \\{x | P(x) \\text { is true}\\} \\text, \\tag{9.2}$$ where $$P(x)$$ is some mathematical statement. We read this definition of the set $$S$$ as \"$$S$$ is the set of all $$x$$ such that $$P(x)$$ is true\". So\n\n\u2022 $$s \\in S$$ if and only if $$P(s)$$ is true;\n\u2022 $$s \\notin S$$ if and only if $$P(s)$$ is false.\n\nThere are some common variations on set-builder notation that you will see. For example, people will often use a colon \":\" in place of the bar \"|\". That is fine; the idea is the same. Sometimes, another condition on elements of a set is slipped in before the \"such that\" symbol by limiting elements to members of a larger set. For example, we could have defined the set of even integers as this: $$A = \\{x \\in \\mathbb Z | x \\text { is even}\\} \\text. \\tag{9.3}$$ Besides using English or set-builder notation to define sets, we can define sets by simply listing their elements. For example, we can write $$A = \\{\\dots, \u22126, \u22124, \u22122, 0, 2, 4, 6, \\dots\\} \\tag{9.4}$$ to define the set of all even integers. But this really only works when the set is small enough that all of its elements can be reasonably listed or when the pattern is strong enough to be recognized. For example, we could write $$B = \\{n \\in \\mathbb N | 2 \\le n \\le 5\\} \\text { or } B = \\{2, 3, 4, 5\\} \\text. \\tag{9.5}$$\n\nThis is all the information from the book that I could find that is pertaining to the purposed question. Please Help Me!\n\n\u2022 Welcome to Mathematics Stack Exchange. Can you show $n$ is even $\\iff n-1$ is odd? \u2013\u00a0J. W. Tanner Sep 30 '20 at 1:02\n\u2022 \u2013\u00a0Shaun Sep 30 '20 at 1:02\n\u2022 Um... do you have a question? Any question? You quote something. To a experienced person everything you quote is straightforward and fairly simple. But obviously its not or you wouldn't ask anything. So it's up to you to tell us where you have difficulty and what you want clarified. \u2013\u00a0fleablood Sep 30 '20 at 2:36\n\u2022 The point of this is to get you familiar with set builder notation. What is actually being proven is fairly simple: The integers that are even are precisely the numbers that are one more than an odd number. this is obvious. If $n = 2k$ is even then $n-1 = 2k -1$ is odd. That's it we are done. What the excercise though is to put it in terms of set builder notation. And ... I think the text you quoted does a fine job explaining that. \u2013\u00a0fleablood Sep 30 '20 at 2:41\n\nTo prove \"set A= set B\" you prove \"A is a subset of B\" and \"B is a subset of A\". And to prove \"A is a subset of B\" start \"if x is in A\" and then use the definitions of A and B to conclude \"then x is in B\"\n\nHere that means we want to prove \"if x is even then x- 1 is odd\" and \"if x- 1 is odd then x is even\".\n\nOf course, we need to use the fact that any even number is of the form 2k for some integer k and any odd number is of the form 2k+ 1 for some integer k.\n\nSo \"If x is even then there exist an integer k such that x= 2k. Then x- 1= 2k- 1= 2k- 2+ 1= 2(k-1)+ 1 so x-1 is odd.\"\n\nAnd \"if x- 1 is odd then there exist an integer k such that x- 1= 2k+ 1. Then x= 2k+ 2= 2(k+ 1) so x is even.\"\n\n$$\\{n\\in \\mathbb Z| n$$ is even $$\\}=$$\n\n$$\\{n\\in \\mathbb Z| n$$ is divisible by $$2\\}=$$\n\n$$\\{n\\in \\mathbb Z|$$ there exists an integer $$k$$ so that $$n = 2k\\}=$$\n\n$$\\{n\\in \\mathbb Z|$$ there exists an integer $$k$$ so that $$n-1 = 2k-1\\}=$$\n\n$$\\{n\\in \\mathbb Z| n-1$$ is odd$$\\}$$.\n\nIn order to show equality of sets, you need to show two-way containment; that is, show that $$A\\subseteq B$$ and $$A\\supseteq B$$, which is equivalent to saying that $$x\\in A \\Leftrightarrow x\\in B$$. Sometimes it's best to show each direction independently, but in this case, I would recommend a chain of biconditionals, as it's quicker.\n\nLet $$A=\\{x\\in\\mathbb{Z} | x\\text{ is even}\\}$$ and $$B=\\{x\\in\\mathbb{Z} | x-1\\text{ is odd}\\}$$. Then we have the following chain of biconditionals:\n\n$$x\\in A\\Leftrightarrow x$$ is even $$\\Leftrightarrow x=2k$$ for some $$k\\in\\mathbb{Z} \\Leftrightarrow x-1=2k-1\\Leftrightarrow x-1$$ is odd $$\\Leftrightarrow x\\in B$$.\n\nTherefore, $$A=B$$.", "date": "2021-06-18 06:40:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3845649/prove-n-in-mathbb-z-n-text-is-even-n-in-mathbb-z-n-1-text/3845657", "openwebmath_score": 0.8253920078277588, "openwebmath_perplexity": 263.63509349307685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109485172559, "lm_q2_score": 0.8791467611766711, "lm_q1q2_score": 0.8657933557602709}}
{"url": "https://math.stackexchange.com/questions/2226070/how-can-i-show-that-the-polynomial-p-x5-x3-2x2-2x-1-is-irreducibl", "text": "# How can I show that the polynomial $p = x^5 - x^3 - 2x^2 - 2x - 1$ is irreducible over $\\Bbb Q$?\n\nI've tried a few \"criteria\" to check if this is irreducible. According to Maple it only has one entirely real root which I suspect is not rational but I can't prove it so I'm attempting to check if $p$ is irreducible.\n\nEisenstein's Criterion doesn't work here and I'm yet to find a suitable transformation such that it could work. I also read that if a polynomial is irreducible over $\\Bbb F_q$, with $q$ a prime not dividing the leading coefficient, then it is irreducible over $\\Bbb Q$ so I reduced the polynomial modulo $2$ to obtain\n\n$$p \\equiv x^5 + x^3 + 1 \\mod 2.$$\n\nI think this is correct but then I need to know how to check the irreducibility of this new polynomial over $\\Bbb F_2$. Do I simply need to check that neither $0$ nor $1$ are roots of this polynomial? (And am I applying this theorem correctly?)\n\nIf this polynomial IS irreducible over $\\Bbb Q$, is the splitting field obtained by simply adjoining the roots to $\\Bbb Q$?\n\n\u2022 Since the polynomial is of degree $5$ it might be possible for it to have no roots in $\\mathbb{Q}$, but be reducible. Writing it as a product of polynomials of degree $2$ and $3$ might be possible. Apr 9, 2017 at 11:11\n\u2022 To answer your last question: yes. This follows straight from the definition of a splitting field.\n\u2013\u00a0user583416\nApr 9, 2017 at 11:18\n\u2022 Oh, I've found something wonderful. Answer coming up. Apr 9, 2017 at 11:19\n\n.A little bit of scouting for nice irreducibility criteria throws up some very nice results:\n\nHere is a lovely lemma by (Prof.) Ram Murty:\n\nLet $$f(x) = a_mx^m + ... + a_1x + a_0$$ be a polynomial of degree $$m$$ in $$\\mathbb Z[x]$$. Let $$H = \\displaystyle\\max_{0 \\leq i \\leq m-1} \\left|\\frac{a_i}{a_m}\\right|$$. If $$f(n)$$ is prime for some $$n \\geq H+2$$, then $$f(x)$$ is irreducible in $$\\mathbb Z[x]$$.\n\nIn our case, $$a_m = 1$$, and the maximum of all the quantities in question is $$2$$. Hence, if $$f(n)$$ is prime for some $$n \\geq 4$$, then we are done.\n\nYou can check that for $$n=4$$, the number $$f(4) =919$$, which is prime!\n\nHence, it follows that the polynomial is irreducible.\n\nASIDE : There is also a \"shifted\" base (base shifts from $$0...n-1$$ to $$|b| < \\frac n2$$) version of Cohn's criteria, which will tell you that if $$f(10)$$ is prime, then the given polynomial is irreducible. This matches that description, since all coefficients are between $$-5$$ and $$5$$. Very interestingly, $$f(10) = 98779$$ is also prime! (Hence, another proof by another wonderful result).\n\n\u2022 That is very cool! Thanks for the link. Apr 9, 2017 at 11:28\n\u2022 You are welcome. I've seen some of these criteria very long ago, while doing a Galois theory polynomial \"scouting mission\". Apr 9, 2017 at 11:31\n\u2022 Nice method. Just wondering, why not going already with $f(4)=919$ which is also a prime?\n\u2013\u00a0Sil\nApr 26, 2018 at 23:30\n\u2022 You are right, it works out. I was doing that calculation mentally, so I skipped directly to $n=5$ because I must have made some mistake at $f(4)$. Apr 26, 2018 at 23:34\n\nBy Gauss' Lemma we have that the polynomial is irreducible over $\\mathbb{Q}$ if and only if it's irreducible over $\\mathbb{Z}$. Now the easiest way would be to prove that polynomial is irreducible over $\\mathbb{Z}_2$, which would be enough.\n\nAssume it's reducible. As the polynomial has no roots over $\\mathbb{Z}_2$, then the only possibility is if it's a product of polynomials of degree $2$ and $3$. So assume that:\n\n$$x^5 + x^3 + 1 = (x^3 + ax^2 + bx + c)(x^2 + dx + e) \\quad \\quad \\text{over } \\quad\\mathbb{Z}_2$$\n\nThen multiply everything out and compare the factors. As $a,b,c,d,e \\in \\mathbb{Z}_2$, you only have two options. Eventually you will get that $c=e=1$ and $a=d=b$. But this would imply that $c+bd + ea = a^2 + a + 1 = 1$ in $\\mathbb{Z}_2$. But this is impossible, as it's the coefficient in front of $x^2$ and it should be $0$.\n\nHence the polynomial is irreducible over $\\mathbb{Z}_2$ and eventually $\\mathbb{Z}$ and $\\mathbb{Q}$\n\n\u2022 I think I'll accept this one as being the most useful in a general setting, thanks a lot. Apr 9, 2017 at 11:28\n\u2022 One could also just divide by the only irreducible polynomial in $\\mathbb{Z_2}[x]$ which is $x^2+x+1$ and see it cannot be the factor.\n\u2013\u00a0Sil\nAug 11, 2018 at 9:23\n\u2022 @Sil True. However I doubt that an OP asking factorizing of a polynomial is aware that $x^2+x+1$ is the only irreducible quadratic in $\\mathbb{Z}_2$ Aug 11, 2018 at 9:25\n\nYour proof is almost complete: it is indeed sufficient to prove that $p(x)\\in \\mathbb F_2[x]$ (the reduction mod $2$ of your polynomial) is irreducible.\nThe polynomial $p(x)$ has no factor of degree $1$ since it has no zero in $\\mathbb F_2$, so there remains only to prove that $p(x)$ has no factor which is an irreducible polynomial $g(x)\\in \\mathbb F_2[x]$ of degree $2$.\nBut the only such irreducible polynomial is $g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved.\n\nBy the rational root theorem, any rational root of $p$ would have to be a divisor of the constant term, so $\\pm1$. Clearly, these are not roots.\n\nSimilary, any quadratic factor would have to be $x^2+ax\\pm1$ with $a\\in\\Bbb Z$. You might be able to exclude these manually ...", "date": "2023-03-23 04:45:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2226070/how-can-i-show-that-the-polynomial-p-x5-x3-2x2-2x-1-is-irreducibl", "openwebmath_score": 0.858704686164856, "openwebmath_perplexity": 134.34363632478673, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109476256691, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8657933502989339}}
{"url": "https://math.stackexchange.com/questions/741949/problem-about-population-growth", "text": "At the beginning of the Gold Rush, the population of Coyote Gulch,Arizona was $365$.From then on ,the population would have grown by a factor of $e$ each year,except for the high rate of \"accidental\" death, amounting to one victim per day among every 100 citizens.By solving an appropriate differential equation determine, as functions of time:\n(a) the actual population of Coyote Gulch $t$ years from the day the Gold Rush began, and\n(b) the cumulative number of fatalities.\n\nThe question is from Apostol's Calculus I. In other questions,Apostol uses the statement \"... increases at a rate proportional to the amount present. ...\"But this one says \"grown by a factor of $e$\", I have difficulty in understanding the meaning of \"a factor of $e$\".Does it mean $y=be^{kt}$ or $y'=ey$ or what?\n\nSo I decide to denote \"a factor of $e$\" by $f_e$ and focus on the ' \"accidental\" death '.So let $y$ denote the population at present year, each day one victim dies among every 100 citizens .So the population remains in that year is $y(1 -\\frac{1}{100})^{365}$. And the number of fatalities in that year is $$y-y(1 -\\frac{1}{100})^{365} = y(1- (\\frac{99}{100})^{365})$$ . And as far as I can get $$y'=f_e -y(1- (\\frac{99}{100})^{365})$$\n\nThe answer of the question is given\na) $365e^{-2.65t}$\nb) $365(1-e^{-2.65t})$\n\nAny help is appreciated.\n\n\u2022 Grown by a factor of $e$ probably if the population this year is $100$ next year it will be $271.8\\ldots$(well not fractional population, but cut me some slack) \u2013\u00a0Guy Apr 6 '14 at 10:36\n\u2022 @Sabyasachi I tried your suggestion and substitute $f_e$ by $ey$ but I couldn't get the right answer :( \u2013\u00a0Detective King Apr 6 '14 at 10:55\n\u2022 I mean $f_e=e$, not $ey$. Does it work? \u2013\u00a0Guy Apr 6 '14 at 11:00\n\u2022 @Sabyasachi so $y'=e - y(1-(99/100)^{365})$ and I get $y=362e^{-0.974t}+0.974$. \u2013\u00a0Detective King Apr 6 '14 at 11:12\n\u2022 Oh. can't help then. Did you try googling for the question text \u2013\u00a0Guy Apr 6 '14 at 11:13\n\nTry to write it out in words.\n\nThe rate of change in population is the population we have minus the loss ratio of that population (of course, we could have other factors, but that is what we are working with here), so we have:\n\n$$\\dfrac{dP}{dt} = P - \\alpha P = P(1 - \\alpha)$$\n\nNow how we can we find the loss ratio $\\alpha$ of the population per year? The problem tells us that we lose \"$1$ victim per day among every $100$ citizens\", so for the year, we lose $365 \\times \\dfrac{1}{100}$ of the population $P$. This gives us $\\alpha = \\dfrac{365}{100}$.\n\nSo we have:\n\n$$\\dfrac{dP}{dt} = P\\left(1 - \\dfrac{365}{100}\\right), P(0) = 365$$\n\nNote: $P(0)$ is the initial population which is given as $365$.\n\nThis DEQ is separable and gives us:\n\n$$20 \\int \\dfrac{1}{P}~dP = -53 ~\\int dt, P(0) = 365$$\n\nThus,\n\n$$\\large P(t) = 365 ~e^{-\\frac{53 t}{20}} = 365~ e^{-2.65~ t}$$\n\nIf we want to find the cumulative number of fatalities over time, we write that out in words as: we start out with a population of $365$ and we lose them at a rate of $P(t)$ (which we just calculated), so this is:\n\n$$CP(t) = 365 - P(t) = 365 - 365 ~e^{-\\frac{53 t}{20}} = 365\\left(1 - e^{-\\frac{53 t}{20}}\\right) = 365\\left(1 - e^{-2.65 ~t}\\right)$$\n\nWe can plot these curves to verify that they are inverses of each other as:\n\n\u2022 This was really nicely (and thoroughly) explained, Amzoti. Nice job. \u2013\u00a0Namaste Apr 6 '14 at 13:27\n\u2022 @Amzoti Your explaination is really nice!At the same time I think the question is quite \"unrealistic\"(it feels like \"statistic\") to me . \u2013\u00a0Detective King Apr 6 '14 at 14:01\n\u2022 @DetectiveKing: I think the point of these is just trying to set them up. It is very difficult to take the leap from knowing how to solve things, to actually producing models of the physical reality. By the way, it was not unreasonable for really bad things to happen to small populations as the result of a single issue, including having them totally wiped out. Regards \u2013\u00a0Amzoti Apr 6 '14 at 14:09", "date": "2019-08-24 09:21:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/741949/problem-about-population-growth", "openwebmath_score": 0.7943887710571289, "openwebmath_perplexity": 739.3720402639009, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540716711546, "lm_q2_score": 0.8840392832736084, "lm_q1q2_score": 0.8657874715912576}}
{"url": "https://www.physicsforums.com/threads/does-simplex-work-for-all-standard-max-problems.963536/", "text": "# Does Simplex work for all standard-max problems?\n\nTags:\n1. Jan 4, 2019\n\n### Ben2\n\nMentor note: Fixed the LaTeX\n1. The problem statement, all variables and given/known data\n\nMaximize $5x_1 + 7x_2 + 3x_3$ subject to $x_1 + x_2 + x_3 \\le 28, x_2 \\le2 x_1$ and $x_1 \\le x_3$.\n\n2. Relevant Equations\n$x_1\\ge 0, x_2\\ge 0, x_3\\ge 0$.\n3. The attempt at a solution Problem is workable by graphic methods, but the writer has been unable to\nsuccessfully apply the standard-max simplex algorithm. My interest is in finding a workable simplex solution and/or in determining why standard-max simplex won't work here. Thanks for the attention of all posters and site professionals!\n\nLast edited by a moderator: Jan 4, 2019\n2. Jan 4, 2019\n\n### Staff: Mentor\n\nCaveat: I havent done any linear programming problems for many years. I haven't set up this problem, but it seems like a standard linear programming example.\nThe first three constraints should be rewritten as\n$x_1 + x_2 + x_3 \\le 28$\n$-x_1 + x_2 \\le 0$\n$x_1 - x_3 \\le 0$\nand $x_1\\ge 0, x_2\\ge 0, x_3\\ge 0$.\n\nLaTeX Tips -- Use a pair of # characters at the beginning and end of a math expression to be rendered for inline LaTeX. Use a pair of \\$ characters at the beginning and end of a math expression to be rendered for standalone LaTeX.\n\nFor inequalities, use \\le, not \\leq, and \\ge, not \\geq.\n\n3. Jan 4, 2019\n\n### Ray Vickson\n\nState the problem as one having 3 variables and 3 constraints (plus non-negativity):\n$$\\begin{array}{rcl} \\max & Z = & 5 x_1 + 7 x_2 + 3 x_3 \\\\ \\text{s.t.}&&x_1 + x_2 + x_3 \\leq 28 \\\\ &&- 2x_1 + x_2 \\leq 0 \\\\ && x_1 - x_3 \\leq 0 \\\\ &&x_1,x_2, x_3 \\geq 0 \\end{array}$$\nSo, if $s_1, s_2, s_3$ are the slack variables for constraints 1--3 and we write the objective equation as $Z - 5 x_1 - 7 x_2 - 3 x_3 = 0$ , the initial tableau for the problem is\n$$\\begin{array}{ccccccc|c} Z & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \\text{RHS} \\\\ \\hline 1 & -5 & -7 & -3 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 1 & 1 & 1 & 0 & 0 & 28\\\\ 0 & -2 & 1 & 0 & 0 & 1 & 0 & 0 \\\\ 0 & 1 & 0 & -1 & 0 & 0 & 1 & 0 \\end{array}$$\n\nThe simplex method works perfectly well here. If you do not think so, you need to show us your work and indicate where the method fails.\n\nNote that in this problem some simplex steps may involve \"degenerate\" pivots, where the objective does not change from one iteration to the next, due to having some basic variables equal to 0; essentially, we change a 0 basic variable in one tableau into a non-basic variable (still zero) in another tableau. The simplex method may be \"slowed down\" a bit here, because it may need to pass through several such degenerate solutions on the way to the final, non-degenerate solution having no basic variables equal to zero.\n\nLast edited: Jan 4, 2019\n4. Jan 4, 2019\n\n### Ray Vickson\n\nThe commands \"\\leq\" and \"\\geq\" are TeX/LaTeX standards, and work perfectly well in the PF implementation; I use them all the time without any problems.\n\n5. Jan 4, 2019\n\n### Staff: Mentor\n\nOK, thanks -- didn't know that. The \\le and \\ge operators work in MathJax, and have the advantage of requiring slightly less typing.\n\n6. Jan 4, 2019\n\n### WWGD\n\nThe title of this thread does not seem to agree with the question in the OP. The title is , it seems, much more general while the question seems to refer to a specific question. AFAIK, the Simplex method requires some conditions for convergence. I will look them up ASAP.\n\n7. Jan 4, 2019\n\n### Ray Vickson\n\nRight. But let's wait until the OP responds, and shows the difficulties he/she is facing.\n\nAnyway, there are simple \"rules\" that can be applied that guarantee the convergence in finitely many steps of the simplex method to an optimal resolution (either showing infeasibility or unboundedness of the problem if these properties apply, or else determining an optimal solution).\n\n8. Jan 4, 2019\n\n### kimbyd\n\n1) This is exactly the kind of problem that the Simplex algorithm was designed to solve.\n2) In some cases, however, the Simplex algorithm may stall due to degeneracy, and has worst-case exponential time complexity. Wikipedia describes these issues here.\n\nFrom my reading, it should be efficient for many inputs similar to your example above, but there may be complicated situations where the algorithm stalls out and wastes quite a lot of time.\n\n9. Jan 4, 2019\n\n### Ray Vickson\n\nNevertheless, companies and other organizations have, for many years, been solving problems having many thousands of variables and many thousands of constraints, using Simplex. For much larger problems (say involving millions of variables and hundreds of thousands to millions of constraints), solvers sometimes use interior-point methods instead of Simplex. Some really big problems (having several hundred million variables) have been solved due to their special \"network\" structure, using network-flow implementations of the simplex method. Needless to say, simple textbook rules and prescriptions are not good enough for these really big cases.\n\n10. Jan 5, 2019\n\n### FactChecker\n\nYou should recheck your Simplex calculations.\nUsing the Simplex implementation at https://www.zweigmedia.com/RealWorld/simplex.html, I got a solution.\n\nThe input format is:\nMaximize z=5x1+7x2+3x3 subject to\nx1+x2+x3<=28\n-2x1+x2<=0\nx1-x3<=0\n\nIt shows four Tableaus, ending with the solution\nx3=7\nx2=14\nx1=7\nz=154\n\nLast edited: Jan 5, 2019\n11. Jan 5, 2019\n\n### Ben2\n\nMy thanks for the Mentor's corrections. I found the LaTeX inequality-symbol equivalents on this site's own \"how-to\" primer,\nbut will try to use them on this site.\nFactChecker has the problem-formulation I intended. Let me apologize to anyone thrown off by my original statement.\nI've recently gotten the final tableau referenced by FactChecker, with one intermediate tableau. However, the first pivot-element\nwas a \"0\" where the constant/element quotient was 0/0. I'm not sure if I'm familiar with all the \"rules\" mentioned in Ray\nVickson's second post.\nThe standard Simplex method I've seen in Finite-Math texts seemed to guarantee that each new tableau of\nstandard-max simplex automatically represents a basic feasible solution. My hand calculations on the given problem, mistake-ridden\nthough they may have been, prompt me to question that.\nRe WWGD's comment, I tried setting this up as a mixed-constraint problem before finally getting the standard-max solution referenced above.\nThanks to all posters for extremely informative comments, and will access \"bad\" cases referenced by kimbyd.\nBen2\n\n12. Jan 5, 2019\n\n### FactChecker\n\nUsually some artificial variables are added which are driven out (using artificially high cost values) to arrive at a feasible solution that only uses the original variables. I am not familiar with any other way to guarantee that a feasible solution can be found.\n\n13. Jan 5, 2019\n\n### Ray Vickson\n\nModern industrial-scale computer implementations of the simplex method avoid artificial variables, but instead consider starting at infeasible bases and trying to get rid of infeasibilites first, before optimizing. For example, consider the problem\n$$\\begin{array}{ccl} \\max & Z = & 4 x_1 + 6 x_2 - x_3 \\\\ \\text{s.t.}&&5 x_1 + 2 x_2 + 8 x_3 = 80\\\\ &&4 x_1 + 9 x_2 +3 x_3 \\leq 60\\\\ &&6 x_1 + 4 x_2 + 5 x_3 \\geq 60\\\\ && x_1, x_2, x_3 \\geq 0 \\end{array}$$\nIf we introduce a slack variable $s_1$ for the first inequality and a surplus variable $s_2$ for the second one, our problem can be written as that of maximizing $Z$ over the non-negative variables $x_1, x_2, x_3, s_1, s_2$ that are related by the equations\n$$\\begin{array}{ccc} Z- 4 x_1 - 6x_2 + x_3 &=&0\\\\ 5 x_1 + 2 x_2 + 8 x_3 &= &80\\\\ 4x_1 + 9x_2 + 3x_3 + s_1 &=& 60\\\\ 6x_1 + 4 x_2 + 5 x_3 - s_2 &=& 60 \\end{array}$$\nWe don't even know if the problem is feasible, so determining that is the first order of business. We do not yet have an LP basis.\n\nWithout worrying about feasibility, let us choose an initial basis in which the basic variables are $x_1, s_2, s_3$. The equations can be re-written by putting basic variables on the left and non-basic variables on the right:\n$$\\begin{array}{ccl} Z&=& 64 +(22/5) x_2 - (37/5) x_3 \\\\ x_1 &=& 16 -(2/5) x_2 - (8/5) x_3 \\\\ s_1 &=& -4 -(27/5) x_2 + (17/5) x_3 \\\\ s_2&=& 36 +(8/5) x_2 - (2/5) x_3 \\end{array}$$\nThis basis is infeasible because it has $s_1 = -4 < 0$. Temporarily, we can think of this as a problem of trying to maximize $s_1$ (so as to drive it up towards zero), and for that reason we choose to increase $x_3$. Using the usual minimum-ratio rules to maintain feasibility of $x_1, s_2$ and to try to achieve feasibility of $s_1$, the ratios are\n$$\\begin{array}{l} s_1 \\;\\text{ratio} = (-4)/(-17/5) = 1.176\\\\ s_2 \\;\\text{ratio} = 36/(23/5) = 7.826\\\\ x_1 \\; \\text{ratio} = 16/(8/5) = 10 \\end{array}$$\nThe minimum ratio is for $s_1$, so $s_1$ leaves the basis and $x_3$ enters. The new equations are\n$$\\begin{array}{ccl} Z &=& 940/17 -(199/17) x_2 - (37/17) s_1\\\\ x_1 &=&240/17 -(66/17) x_2+-(8/17) s_1\\\\ x_3 &=& 20/17+(37/17) x_2+(5/17)s_1 \\\\ s_2 &=& 520/17-(143/17) x_2 - (23/17) s_1 \\end{array}$$\nThis gives us a feasible solution $Z = 940/17, x_1 = 240/17, x_3 = 20/17, s_2 = 520/17.$ The current $Z$-equation also shows that this solution is optimal\n\nBy the way: what would happen if we had a truly infeasible problem but did not use artificial variables? In this case the above method would eventually contain an equation something like $$\\text{basic variable}\\; x = -2 - 4 u_1 - 3 u_2 - \\cdots$$ where $u_1, u_2, u_3 \\ldots$ are the current non-basic variables. Note that the right-had-side is negative, and all the coefficients on the right are non-positive. That is, we have a negative basic variable, and increasing any of the non-basic variables up from 0 will not help (and may even make matters worse). This says that the basic variable $x$ cannot be larger than -2, so certainly cannot be non-negative. Our problem would be detected as being infeasible.\n\n14. Jan 5, 2019\n\n### FactChecker\n\nThat is true after an initial feasible solution has been found. The Simplex method will move from one feasible corner solution to another. But finding an initial feasible solution takes some tricks. Some variables (\"slack\" and \"surplus\") can be added which make it easy to start from a \"psudo-feasible\" solution with those variables. Then an initial Simplex phase can proceed to feasible solutions which use the original variables. The initial phase is very similar to the regular Simplex method that follows.\n\nLast edited: Jan 6, 2019\n15. Jan 7, 2019\n\n### kimbyd\n\nThat makes good sense. From my reading, it sounds like Simplex \"just works\" most of the time, but if you're going to have any sort of system depend upon it, you have to be careful that it can stall out at times, and deal with those cases appropriately. There are solutions for those cases, so it's not a deal-breaker. It's just a little extra complexity that needs to be managed.\n\nEdit:\nAnd for any simple problem you should encounter, you should expect Simplex to work quite efficiently. Especially a textbook problem. You should only expect to encounter the pathological behavior if you're building a large-scale system designed to run a lot of Simplex calculations on a lot of different inputs.\n\nIf you try Simplex, and the answer is wrong, then there's a bug in how you're using/implementing the algorithm.\n\nIf you try Simplex, and you find it keeps cycling, then you might have hit a stall mentioned above. This shouldn't occur in textbook problems. The algorithm should eventually find a result even in this case, but it may take a little while. Still, if your number of inputs is small, even exponential-time complexity will run quickly in a computer If they're asking you to do it by hand, then it's either an error in your calculations or an error in the textbook (many textbooks have errata, so you might try to look those up if you think the textbook might have a mistake).\n\nLast edited: Jan 7, 2019\n16. Jan 7, 2019\n\n### FactChecker\n\nThere are techniques to detect and move on from a stall. I would expect any well-used, reputable program to include such a technique. I would hesitate to recommend that a person deal with a stall himself on a very large problem. That would be extremely difficult. I find that even following the calculations on a small problem to be difficult. I could not imagine doing that with a thousand variables.\n\n17. Jan 7, 2019\n\n### kimbyd\n\nI definitely wasn't suggesting a person deal with the stall by hand, hence the statement that such a stall indicates a textbook error. Obviously the complexity has to be managed at the software implementation level.\n\n18. Jan 7, 2019\n\n### Ray Vickson\n\nIt is possible for the Simplex method (if naively implemented) to cycle forever among a group of non-optimal solutions; i.e., stall permanently at non-optima. However, there are simple safeguards that can prevent that from happening, so stalling forever won't happen in a properly-implemented system, but stalling for quite a while sometimes will, despite all the safeguards. Commercial codes have various strategies to try to \"unstall\" the method, but of course, they will not work 100% of the time---just often enough to be useful. For gigantic problems, some systems will start with interior-point methods, perhaps to get near a basic feasible solution, then switch to the simplex method near the end to get an accurate, provably-optimal solution.\n\n19. Jan 13, 2019", "date": "2019-01-23 11:23:26", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/does-simplex-work-for-all-standard-max-problems.963536/", "openwebmath_score": 0.5858039855957031, "openwebmath_perplexity": 882.0055654770597, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979354068055595, "lm_q2_score": 0.8840392741081574, "lm_q1q2_score": 0.8657874594187392}}
{"url": "https://math.stackexchange.com/questions/3251616/show-the-matrix-commutes-with-companion-matrix-is-a-polynomial", "text": "# Show the matrix commutes with companion matrix is a polynomial\n\nLet $$A$$ be a linear transform on $$n$$-dimensional $$V$$ over a field $$F$$. Under a basis $$\\alpha_1, \\cdots, \\alpha_n$$, the matrix representation of $$A$$ is as follows: $$A = \\begin{bmatrix} 0 & 0 & \\dots & 0 & -a_0 \\\\ 1 & 0 & \\dots & 0 & -a_1 \\\\ 0 & 1 & \\dots & 0 & -a_2 \\\\ \\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\ 0 & 0 & \\dots & 1 & -a_{n-1} \\end{bmatrix}.$$ Let $$C(A):= \\{T: T\\text{ is a linear transform on V and } TA = AT \\}$$, and let $$F[A]$$ denotes all the polynomials in $$A$$. Show that: $$C(A) = F[A]; \\dim(C(A)) = n.$$\n\nFirst of all, the minimal polynomial $$m(\\lambda)$$ of $$A$$ is the same as its characteristic polynomial $$f(\\lambda)$$, namely $$m(\\lambda) = f(\\lambda) = \\lambda^n + a_{n-1}\\lambda^{n-1} + \\cdots a_0$$. Thus, plugging in $$A$$, we see that all $$A^{k}$$ with $$k \\geq n$$ could be expressed by $$I, A, A^2, \\cdots, A^{n-1}$$. So $$\\dim F[A] \\leq n$$. If $$\\dim F[A] < n$$, say $$k_0 I + k_1 A + \\cdots + k_r A^r = 0$$ with $$r < n$$ and some $$k_j \\neq 0$$, then we have that $$g(\\lambda) = k_0 + k_1 \\lambda + \\cdots + k_r \\lambda^r$$ is another polynomial with $$g(A) = 0$$. By the definition of minimal polynomial, we must have that $$r \\geq n$$, a contradiction. So $$\\dim(F[A]) = n$$, and it remains to show the first equality $$C(A) = F[A]$$.\n\nAlso, one could see that $$F[A] \\subseteq C(A)$$. But I am not sure how to show the other direction. Could someone give me a hint?\n\nDefine a linear map $$\\Psi \\colon C(A) \\rightarrow V$$ by $$\\Psi(T) = T\\alpha_1$$. Let's show that this map is an isomorphism. First, note that $$A^i \\in C(A)$$ for all $$i \\in \\mathbb{N}_0$$ and\n$$\\Psi(I) = \\alpha_1, \\psi(A) = A\\alpha_1 = \\alpha_2, \\cdots, \\psi(A^{n-1}) = A^{n-1}\\alpha_1 = \\alpha_n.$$\nThis shows that $$\\Psi$$ is surjective. Next, let's assume that $$\\psi(T) = T\\alpha_1 = 0$$. Then $$T\\alpha_2 = T(A\\alpha_1) = A(T\\alpha_1) = A(0) = 0, \\\\ T\\alpha_3 = T(A\\alpha_2) = A(T\\alpha_2) =0, \\\\ \\vdots,\\\\ T\\alpha_n = T(A\\alpha_{n-1}) = A(T\\alpha_{n-1}) = 0 \\\\$$\nwhich shows that $$T \\equiv 0$$. This shows that $$\\Psi$$ is injective. Hence, $$\\dim C(A) = n$$ and since $$\\dim F[A] = n$$ and $$F[A] \\subseteq C(A)$$, we deduce that $$F[A] = C(A)$$.\n\u2022 Yes I think this is excellent! But how do we come up with such $\\Psi$? \u2013\u00a0mathdoge Jun 5 at 12:00\n\u2022 The matrix $A$ acts on your basis by $\\alpha_1 \\mapsto \\alpha_2 \\dots \\mapsto \\alpha_n$ (where $x \\mapsto y$ means that $A$ sends $x$ to $y$). By applying $T$ and using the fact that $A$ and $T$ commute, you can see that we also have $T\\alpha_1 \\mapsto T\\alpha_2 \\dots \\mapsto T\\alpha_n$. From here you can already see that if we know $T\\alpha_1$ and $A$, we also know $T\\alpha_i$ for all $2 \\leq i \\leq n$. \u2013\u00a0levap Jun 5 at 12:06", "date": "2019-08-23 10:52:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3251616/show-the-matrix-commutes-with-companion-matrix-is-a-polynomial", "openwebmath_score": 0.9918481707572937, "openwebmath_perplexity": 40.33635817331838, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9924227573874893, "lm_q2_score": 0.8723473763375643, "lm_q1q2_score": 0.8657373886246674}}
{"url": "https://math.stackexchange.com/questions/3374300/find-the-sum-to-n-terms?noredirect=1", "text": "# Find the sum to n terms\n\n$$S=1^2+3^2+6^2+10^2+15^2+.......$$\n\nMy attempt is as follows:\n\n$$T_n=\\left(\\frac{n\\cdot\\left(n+1\\right))}{2}\\right)^2$$\n\n$$T_n=\\frac{n^4+n^2+2\\cdot n^3}{4}$$\n\n$$S=\\frac{1}{4}\\cdot\\sum_{n=1}^{n}\\left(n^4+n^2+2\\cdot n^3\\right)$$\n\nNow to solve this one has to calculate $$\\sum_{n=1}^{n}n^4$$ which will be a very lengthy process, is there any shorter method to solve this question?\n\nBy the way I calculated $$\\sum_{n=1}^{n}n^4$$ and it came as $$\\dfrac{\\left(n\\right)\\cdot\\left(n+1\\right)\\cdot\\left(2\\cdot n+1\\right)\\cdot\\left(3\\cdot n^2+3\\cdot n-1\\right)}{30}$$, then I substituted this value into the original equation.\n\nThen I got final answer as $$\\dfrac{\\left(n\\right)\\cdot\\left(n+1\\right)\\cdot\\left(n+2\\right)\\cdot\\left(3\\cdot n^2+6\\cdot n+1\\right)}{60}$$\n\nBut it took me a very long time to calculate all of this, is there any shorter way to solve this problem?\n\n\u2022 This is A024166 in OEIS. They don't appear to provide a simple closed formula for it. \u2013\u00a0lulu Sep 29 '19 at 13:42\n\u2022 actually my requirement is to solve this faster, I did it in the conventional way and it took a lot of time. \u2013\u00a0user3290550 Sep 29 '19 at 13:49\n\u2022 Have you seen this picture proof for $\\sum r^4$ ? \u2013\u00a0almagest Sep 29 '19 at 14:08\n\u2022 Alternatively, the standard way to do $\\sum_rr^k$ is to do $\\sum r(r+1)\\dots(r+k-1)$ \u2013\u00a0almagest Sep 29 '19 at 14:11\n\u2022 Don't have multiple identically-named variables in the same expression! It is very confusing. For example, in the summations, either change the variable denoting the number of terms or change the variable that is being summed over. \u2013\u00a0Solomon Ucko Sep 29 '19 at 22:14\n\nIt is easier to do this kind of problem using factorial polynomials than conventional polynomials. Using falling factorials, for example, we would use $$(n)_4=n(n-1)(n-2)(n-3)$$ rather than $$n^4$$. The wiki articles explains how to convert from conventional polynomials to falling or rising factorials under in the section titled, \"Connection coefficients and identities.\" The advantage of using factorial polynomials comes in summation. We have, for example $$(n+1)_5-(n)_5=(n+1)n(n-1)(n-2)(n-3)-n(n-1)(n-2)(n-3)(n-4)=5(n)_4$$ so that $$\\sum_{n=1}^k(n)_4=\\frac15\\sum_{n=1}^k((n+1)_5-(n)_5)={(k+1)_5-1\\over5}$$\n\nEDIT\n\nIn this case, it's very easy, because we have $$T_n=\\frac14(n+1)_2(n+1)_2$$ and we can use one of the formulas under the \"Connection coefficients and identities\" section to get$$T_n=\\frac14(n+1)_2(n+1)_2=\\sum_{k=0}^2{2\\choose k}{2\\choose k}(n+1)_{4-k}$$\n\n\u2022 Wow...what a nice and simple approach! \u2013\u00a0John Hughes Sep 29 '19 at 15:09\n\u2022 See Concrete Mathematics by Donald Knuth et. al. \u2013\u00a0Felix Marin Jul 13 at 16:40\n\nWell...there's a way to get a good guess of the answer. You could say to yourself (or plot the data!) that it looks like a 5th degree polynomial, $$p(n) = a_5n^5 + a_4 n^4 + \\ldots + a_0.$$ Then you know that $$p(n+1) - p(n)$$ is the thing you've called $$T_n$$, but it's also $$a_5[(n+1)^5 - n^5] + a_4 [(n+1)^4 - n^4] + \\ldots + a_1 [(n+1)^1 - n^1]$$ which you can write out as a 4th degree polynomial. The first term will be $$4 a_5 n^4,$$ I think, which I got by simply expanding $$(n+1)^5 - n^5$$ using Pascal's triangle.\n\nSetting this 4th degree poly equation to $$T_n$$, you get a triangular system of equations that can be backsubstituted to get an answer.\n\nOF course, you then must check that the answer is in fact correct. You know it satisfies the recurrence, but you also need to show it gives the correct values for the first few values of $$n$$ (perhaps the first six?)\n\nIs it faster? Probably. But not a lot. And if the answer had turned out not to be polynomial, you'd have wasted a lot of time.\n\nYou can use this result: why is $\\sum\\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$\n\nSince $$T_n$$ is of degree $$4$$ then its sum will be of degree $$5$$.\n\nSo let $$S(n)=a_0+a_1n+a_2n^2+a_3n^3+a_4n^4+a_5n^5$$.\n\nThen you can proceed by solving the system obtained calculating the first $$6$$ terms $$S(1)$$ to $$S(6)$$.\n\n\u2022 yeah this is the good way, actually we don't need $a_0$, S(n) will always a multiple of n \u2013\u00a0user3290550 Sep 29 '19 at 14:15\n\u2022 you can even simplify it further using Saulspatz hint. If you write $S(n)=a_0+a_1(n-1)+a_2(n-1)(n-2)+\\cdots+a_5(n-1)(n-2)(n-3)(n-4)(n-5)$ then the system is even easier to solve since it is triangular. \u2013\u00a0zwim Sep 29 '19 at 14:17\n\nYou may use the hockey-stick identity, like in the computation of $$\\sum_{n=1}^{N}n^m$$. We have\n\n$$\\binom{n}{2}^2 = 6\\binom{n}{4}+12\\binom{n}{3}+7\\binom{n}{2}+\\binom{n}{1}$$ hence $$\\begin{eqnarray*} \\sum_{n=1}^{N}\\binom{n}{2}^2 &=& 6\\binom{N+1}{5}+12\\binom{N+1}{4}+7\\binom{N+1}{3}+\\binom{N+1}{2}\\\\&=&\\color{red}{\\frac{N(N+1)(N+2)(3N^2+6N+1)}{60}}.\\end{eqnarray*}$$\n\nI change a little the notations from your original wording.\n\n$$T(n)=\\dfrac{n(n+1)}{2}$$\n\n$$S(n)=\\sum\\limits_{k=1}^nT(k)^2$$\n\nCheating on the resulting formula for $$f(n)=\\dfrac{n(n+1)(n+2)(3n^2+6n+1)}{60}$$\n\nwe notice that $$f(0)=f(-1)=f(-2)=0$$.\n\nIs there a way to exploit these negative indices in order to find the $$(3n^2+6n+1)$$ part more easily?\n\nIn fact there is, $$T(n)$$ is perfectly defined for negative numbers so this part does not pose any issue.\n\nBut how to interpret $$S(-n)$$?\n\nTo be consistent with the summation identity $$\\sum\\limits_a^{b-1}+\\sum\\limits_b^c=\\sum\\limits_a^c$$ when $$a one need to set $$\\ \\sum\\limits_M^m=-\\sum\\limits_{m+1}^{M-1}$$ for the case $$m.\n\nThis results in $$\\displaystyle S(-n)=\\sum\\limits_{k=1}^{-n}T(k)^2=-\\sum\\limits_{k=-(n-1)}^0T(k)^2$$\n\nI let you do the calculation to verify that $$\\begin{array}{lcr}S(0)&=&0\\\\S(-1)&=&0\\\\ S(-2)&=&0\\\\ S(-3)&=&-1\\\\S(-4)&=&-10\\\\S(-5)&=&-46\\end{array}$$\n\nNow by applying the identification method (I describe in my other post) to $$S(n)=a_0+a_1\\,n+a_2\\,n(n+1)+a_3\\,n(n+1)(n+2)+\\cdots+a_5\\,n(n+1)(n+2)(n+3)(n+4)$$\n\nYou immediately get $$a_0=a_1=a_2=0$$\n\nThe others coefficients give you $$S(n)=n(n+1)(n+2)\\bigg[\\frac 16-\\frac 14(n+3)+\\frac 1{20}(n+3)(n+4)\\bigg]=n(n+1)(n+2)\\left(\\dfrac{3n^2+6n+1}{60}\\right)$$\n\nOnce you\u2019ve rewritten your sum as $$\\frac14\\sum_{k=1}^n k^4+2k^3+k^2$$ it\u2019s not terribly difficult to compute this using generating functions if you use some key tools for manipulating them. To wit, if $$g(x)$$ is the ordinary generating function for the sequence $$\\{a_n\\}_{n=0}^\\infty$$, then $$g(x)/(1-x)$$ is the o.g.f. for the sequence of partial sums $$\\{\\sum_{k=0}^na_k\\}_{n=0}^\\infty$$, and similarly, $$x\\frac d{dx}g(x)$$ is the o.g.f. for $$\\{na_n\\}_{n=0}^\\infty$$. So, starting with the o.g.f. $$(1-x)^{-1}$$ for the sequence of all ones, we have $$\\left\\{\\sum_{k=0}^n k^2\\right\\}_{n=0}^\\infty \\stackrel{ogf}{\\longleftrightarrow} \\frac1{1-x}\\left(x\\frac d{dx}\\right)^2\\frac1{1-x} = {x+x^2 \\over (1-x)^4},$$ therefore $$\\sum_{k=0}^n k^2 = [x^n]{x+x^2\\over(1-x)^4} = [x^{n-1}]\\frac1{(1-x)^4} + [x^{n-2}]\\frac1{(1-x)^4},$$ which you can compute using the generalized binomial theorem. Similarly, $$\\sum_{k=0}^n k^3 = [x^n]{x+4x^2+x^3\\over(1-x)^5} \\\\ \\sum_{k=0}^n k^4 = [x^n]{x+11x^2+11x^3+x^4\\over(1-x)^6}.$$", "date": "2020-10-24 01:13:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3374300/find-the-sum-to-n-terms?noredirect=1", "openwebmath_score": 0.9097806811332703, "openwebmath_perplexity": 221.29798917496706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972414716174355, "lm_q2_score": 0.8902942159342104, "lm_q1q2_score": 0.8657351972993352}}
{"url": "https://calculus7.org/2015/01/17/completely-monotone-imitation-of-1x/", "text": "Completely monotone imitation of\u00a01/x\n\nI wanted an example of a function ${f}$ that behaves mostly like ${1/x}$ (the product ${xf(x)}$ is bounded between two positive constants), but such that ${xf(x)}$ does not have a limit as ${x\\rightarrow 0}$.\n\nThe first thing that comes to mind is ${(2+\\sin(1/x))/x}$, but this function does not look very much like ${1/x}$.\n\nThen I tried ${f(x)=(2+\\sin\\log x)/x}$, recalling an example from Linear Approximation and Differentiability. It worked well:\n\nIn fact, it worked much better than I expected. Not only if ${f'}$ of constant sign, but so are ${f''}$ and ${f'''}$. Indeed,\n\n$\\displaystyle f'(x) = \\frac{\\cos \\ln x - \\sin \\log x - 2}{x^2}$\n\nis always negative,\n\n$\\displaystyle f''(x) = \\frac{4 -3\\cos \\log x + \\sin \\log x}{x^3}$\n\nis always positive,\n\n$\\displaystyle f'''(x) = \\frac{10\\cos \\log x -12}{x^4}$\n\nis always negative. The sign becomes less obvious with the fourth derivative,\n\n$\\displaystyle f^{(4)}(x) = \\frac{48-40\\cos\\log x - 10\\sin \\cos \\ln x}{x^5}$\n\nbecause the triangle inequality isn\u2019t conclusive now. But the amplitude of ${A\\cos t+B\\sin t}$ is ${\\sqrt{A^2+B^2}}$, and ${\\sqrt{40^2+10^2}<48}$.\n\nSo, it seems that ${f}$ is completely monotone, meaning that ${(-1)^n f^{(n)}(x)\\ge 0}$ for all ${x>0}$ and for all ${n=0,1,2,\\dots}$. But we already saw that this sign pattern can break after many steps. So let\u2019s check carefully.\n\nDirect calculation yields the neat identity\n\n$\\displaystyle \\left(\\frac{1+a\\cos \\log x+b\\sin\\log x}{x^n}\\right)' = -n\\,\\frac{1+(a-b/n)\\cos\\log x+(b+a/n) \\sin\\log x}{x^{n+1}}$\n\nWith its help, the process of differentiating the function ${f(x) = (1+a\\cos \\log x+b\\sin\\log x)/x}$ can be encoded as follows: ${a_1=a}$, ${b_1=b}$, then ${a_{n+1}=a_n-b_n/n}$ and ${b_{n+1} = b_n+a_n/n}$. The presence of ${1/n}$ is disconcerting because the harmonic series diverges. But orthogonality helps: the added vector ${(-b_n/n, a_n/n)}$ is orthogonal to ${(a_n,b_n)}$.\n\nThe above example, rewritten as ${f(x)=(1+\\frac12\\sin\\log x)/x}$, corresponds to starting with ${(a,b) = (0,1/2)}$. I calculated and plotted ${10000}$ iterations: the points ${(a_n,b_n)}$ are joined by piecewise linear curve.\n\nThe total length of this curve is infinite, since the harmonic series diverges. The question is, does it stay within the unit disk? Let\u2019s find out. By the above recursion,\n\n$\\displaystyle a_{n+1}^2 + b_{n+1}^2 = \\left(1+\\frac{1}{n^2}\\right) (a_n^2+b_n^2)$\n\nHence, the squared magnitude of ${(a_n,b_n)}$ will always be less than\n\n$\\displaystyle \\frac14 \\prod_{n=1}^\\infty \\left(1+\\frac{1}{n^2}\\right)$\n\nwith ${1/4}$ being ${a^2+b^2}$. The infinite product evaluates to ${\\frac{\\sinh \\pi}{\\pi}a\\approx 3.7}$ (explained here), and thus the polygonal spiral stays within the disk of radius ${\\frac12 \\sqrt{\\frac{\\sinh \\pi}{\\pi}}\\approx 0.96}$. In conclusion,\n\n$\\displaystyle (-1)^{n} \\left(\\frac{1+(1/2)\\sin\\log x}{x}\\right)^{(n)} = n!\\,\\frac{1+a_{n+1}\\cos\\log x+b_{n+1} \\sin\\log x}{x^{n+1}}$\n\nwhere the trigonometric function ${a_{n+1}\\cos\\log x+b_{n+1} \\sin\\log x}$ has amplitude strictly less than ${1}$. Since the expression on the right is positive, ${f}$ is completely monotone.\n\nThe plot was generated in Sage using the code below.\n\na,b,c,d = var('a b c d')\na = 0\nb = 1/2\nl = [(a,b)]\nfor k in range(1,10000):\nc = a-b/k\nd = b+a/k\nl.append((c,d))\na = c\nb = d\nshow(line(l),aspect_ratio=1)\n\n2 thoughts on \u201cCompletely monotone imitation of\u00a01/x\u201d\n\n1. What was the motivation for finding such a function? Interesting stuff!\n\n1. L says:\n\nA Math.SE question", "date": "2019-09-16 03:13:06", "meta": {"domain": "calculus7.org", "url": "https://calculus7.org/2015/01/17/completely-monotone-imitation-of-1x/", "openwebmath_score": 0.9652847647666931, "openwebmath_perplexity": 326.1789401732162, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147145755, "lm_q2_score": 0.8902942159342104, "lm_q1q2_score": 0.8657351958758838}}
{"url": "https://www.coursehero.com/file/p61cluh/EXAMPLE-34-Finding-the-Number-of-Terms-Needed-for-a-Given-Accuracy-Determine/", "text": "EXAMPLE 34 Finding the Number of Terms Needed for a Given Accuracy Determine\n\n# Example 34 finding the number of terms needed for a\n\n\u2022 Notes\n\u2022 Ailsa\n\u2022 13\n\u2022 100% (1) 1 out of 1 people found this document helpful\n\nThis preview shows page 5 - 8 out of 13 pages.\n\nEXAMPLE 3.4 Finding the Number of Terms Needed for a Given Accuracy Determine the number of terms needed to obtain an approximation to the sum of the series k = 1 1 k 3 correct to within 10 5 . Solution Again, we already used the Integral Test to show that the series in question converges. Then, by Theorem 3.2, we have that the remainder satisfies 0 R n n 1 x 3 dx = lim R \u2192\u221e R n 1 x 3 dx = lim R \u2192\u221e 1 2 x 2 R n = lim R \u2192\u221e 1 2 R 2 + 1 2 n 2 = 1 2 n 2 . So, to ensure that the remainder is less than 10 5 , we require that 0 R n 1 2 n 2 10 5 . Solving this last inequality for n yields n 2 10 5 2 or n 10 5 2 = 100 5 223 . 6 .\n8-31 SECTION 8.3 . . The Integral Test and Comparison Tests 641 So, taking n 224 will guarantee the required accuracy and consequently, we have k = 1 1 k 3 224 k = 1 1 k 3 = 1 . 202047, which is correct to within 10 5 , as desired. Comparison Tests We next present two results that allow us to compare a given series with one that is already known to be convergent or divergent, much as we did with improper integrals in section 6.6. THEOREM 3.3 (Comparison Test) Suppose that 0 a k b k , for all k . (i) If k = 1 b k converges, then k = 1 a k converges, too. (ii) If k = 1 a k diverges, then k = 1 b k diverges, too. Intuitively, this theorem should make abundant sense: if the \u201clarger\u201d series converges, then the \u201csmaller\u201d one must also converge. Likewise, if the \u201csmaller\u201d series diverges, then the \u201clarger\u201d one must diverge, too. PROOF Given that 0 a k b k for all k , observe that the n th partial sums of the two series satisfy 0 S n = a 1 + a 2 + \u00b7\u00b7\u00b7 + a n b 1 + b 2 + \u00b7\u00b7\u00b7 + b n . (i) If k = 1 b k converges (say to B ), this says that 0 S n a 1 + a 2 + \u00b7\u00b7\u00b7 + a n b 1 + b 2 + \u00b7\u00b7\u00b7 + b n k = 1 b k = B , (3.5) for all n 1. From (3.5), the sequence { S n } n = 1 of partial sums of k = 1 a k is bounded. No- tice that { S n } n = 1 is also increasing. (Why?) Since every bounded, monotonic sequence is convergent (see Theorem 1.4), we get that k = 1 a k is convergent, too. (ii) If k = 1 a k is divergent, we have (since all of the terms of the series are nonnegative) that lim n \u2192\u221e ( b 1 + b 2 + \u00b7\u00b7\u00b7 + b n ) lim n \u2192\u221e ( a 1 + a 2 + \u00b7\u00b7\u00b7 + a n ) = \u221e . Thus, k = 1 b k must be divergent, also. You can use the Comparison Test to test the convergence of series that look similar to series that you already know are convergent or divergent (notably, geometric series or p -series).\n642 CHAPTER 8 . . Infinite Series 8-32 EXAMPLE 3.5 Using the Comparison Test for a Convergent Series Investigate the convergence or divergence of k = 1 1 k 3 + 5 k . Solution The graph of the first 20 partial sums shown in Figure 8.26 suggests that the series converges to some value near 0.3. To confirm such a conjecture, we must carefully test the series. Note that for large values of k , the general term of the series looks like 1 k 3 , since when k is large, k 3 is much larger than 5 k . This observation is significant, since we already know that k = 1 1 k 3 is a convergent p -series ( p = 3 > 1).", "date": "2021-07-30 23:12:02", "meta": {"domain": "coursehero.com", "url": "https://www.coursehero.com/file/p61cluh/EXAMPLE-34-Finding-the-Number-of-Terms-Needed-for-a-Given-Accuracy-Determine/", "openwebmath_score": 0.8562824130058289, "openwebmath_perplexity": 339.59457980539617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9904406009350774, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.8657215901825422}}
{"url": "https://byjus.com/question-answer/the-average-of-20-numbers-is-zero-of-them-at-the-most-how-many-may/", "text": "Question\n\n# The average of $$20$$ numbers is zero. Of them, at the most, how many may be greater than zero?\n\nA\n0\nB\n1\nC\n10\nD\n19\n\nSolution\n\n## The correct option is D $$19$$Let the $$20$$ numbers be $$a_1,a_2,...,a_{20}$$Given that average of 20 numbers is zero.Therefore, $$\\dfrac{a_1+a_2+...+a_{20}}{20}=0$$$$\\implies {a_1+a_2+...+a_{20}}=0$$$$\\implies a_1+a_2+...+a_{19}=-a_{20}$$Therefore, at the most\u00a0$$19$$ numbers can be greater than zero.Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-23 10:05:23", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/the-average-of-20-numbers-is-zero-of-them-at-the-most-how-many-may/", "openwebmath_score": 0.6362163424491882, "openwebmath_perplexity": 2077.258005960521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9904405986777259, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8657215703406466}}
{"url": "https://math.stackexchange.com/questions/2562233/does-there-exist-a-continuous-surjection-from-bbb-r3-s2-to-bbb-r2-0-0/2566284", "text": "# Does there exist a continuous surjection from $\\Bbb R^3-S^2$ to $\\Bbb R^2-\\{(0,0)\\}$?\n\nProve or disprove : There exists a continuous surjection from $\\mathbb{R}^3- S^2$ to $\\mathbb{R}^2-\\{(0,0)\\}$ (here $S^2\\subset \\mathbb{R}^3$ denotes the unit sphere defined by the equation $x^2+y^2+z^2=1$),.\n\nThis question had appeared in TIFR GS-2018 exam for PhD admissions. How should I think about such a map?\n\n\u2022 Hint: Think about (path-)connectedness \u2013\u00a0B. Mehta Dec 11 '17 at 20:31\n\u2022 $\\mathbb{R}^3-S^2$ is Euclidean space with the unit sphere removed. So there are no paths between the interior of the sphere and the exterior. \u2013\u00a0user326210 Dec 11 '17 at 20:46\n\u2022 @B.Mehta I don't think it's going to be as simple as that \u2013\u00a0Perturbative Dec 13 '17 at 15:47\n\u2022 @user326210 But how would that prove your point? There are examples of continuous maps which map not-path-connected spaces to path connected spaces. \u2013\u00a0Error 404 Dec 13 '17 at 16:28\n\u2022 @B.Mehta Can you elaborate your approach? Thanks in advance. \u2013\u00a0Error 404 Dec 14 '17 at 7:25\n\nThere is such a map :\n\n\u2022 project $\\mathbb{R}^3 \\setminus \\mathbb{S}_2$ onto $\\mathbb{R}^2$ (projection on the first two coordinates, for instance);\n\n\u2022 apply the exponential map from $\\mathbb{R}^2 \\simeq \\mathbb{C}$ onto $\\mathbb{R}^2 \\setminus \\{(0,0)\\} \\simeq \\mathbb{C}^*$.\n\nBoth are continuous surjections, so their composition still is.\n\nGiven the bounty, I think this deserves at least some additional material (or: how may someone find this answer). Let $X$ and $Y$ be two topological spaces. If there is no continuous surjective map $f$ from $X$ to $Y$, then $X$ must have some property which is preserved by continuous maps, and which $Y$ doesn't have. It turns out that there are not many such properties. On top of my head, the only general ones I can see are :\n\n\u2022 Cardinality : If Card(X) < Card(Y), then there is no such $f$. Example : $X = \\{0\\}$, $Y = \\{0,1\\}$.\n\n\u2022 Compactness : If $X$ is compact and $Y$ is separable but not compact, then there is no such $f$. Example : $X = [0,1]$, $Y = \\mathbb{R}$.\n\n\u2022 Connectedness : If $X$ is connected and $Y$ isn't, then there is no such $f$. More generally, if $X$ has less (in the sense of cardinality) connected components than $Y$, then there is no such $f$. Example : $X = (0,1)$ and $Y = \\mathbb{Z}$.\n\nAlthough, I am sure, one may find examples with other, less obvious, obstructions. Outside of these, things may get wild. For instance, in all the following cases, there exists a continuous surjective map from $X$ to $Y$:\n\n\u2022 $X = C$, any Cantor set, and $Y$ is any compact metric space.\n\n\u2022 $X=[0,1]$ and $Y = [0,1]^2$ : Peano curve.\n\n\u2022 More generally, $X = \\mathbb{R}$ and $Y = \\mathbb{R}^n$, with $n \\geq 1$, using variants of the Peano curve.\n\n\u2022 $X = \\mathbb{R}^k$ and $Y = \\mathbb{R}^n$, with $k \\geq n$, using projections. Combining with he previous example, we get all $k \\geq 1$ and $n \\geq 0$.\n\n\u2022 $X = \\mathbb{R}^k$ and $Y$ is a (separable) connected, $n$-dimensional topological manifold, with $k \\geq 1$. This is not easy to formalize and I have no reference at hand, but the basic idea is to take $k=n$, and make $\\mathbb{R}^n$ a thin tape and wrap it around the manifold (see e.g. this related discussion). I also think that one may replace $X$ by any non-compact $k$-dimensional manifold.\n\nWith that, we have the tools to answer the question. First, since $X$ ha dimension $3$ and $Y$ has dimension $2$, we reduce the dimension by projecting. We get the plane $\\mathbb{R}^2$. Then we wrap the plane around the origin in $\\mathbb{R}^2$ ; and we are lucky, since this can be done very explicitly (thanks to the exponential map).\n\n\u2022 Same as my comment here: math.stackexchange.com/questions/2565030/\u2026 \u2013\u00a0zhw. Dec 14 '17 at 23:40\n\u2022 @zhw: yes, it's pretty close. I didn't notice that this question had a duplicate. \u2013\u00a0D. Thomine Dec 15 '17 at 13:49\n\u2022 Right, not exactly the same, as I was addressing someone else's proof. Looks like you're headed to bounty nirvana in one of the stranger bounty situations I've seen. \u2013\u00a0zhw. Dec 15 '17 at 22:54", "date": "2019-07-22 18:26:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2562233/does-there-exist-a-continuous-surjection-from-bbb-r3-s2-to-bbb-r2-0-0/2566284", "openwebmath_score": 0.9163557291030884, "openwebmath_perplexity": 292.94899184370837, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918489015606, "lm_q2_score": 0.8757870046160258, "lm_q1q2_score": 0.8657083154368549}}
{"url": "http://mathhelpforum.com/pre-calculus/94707-inequalities-2-methods.html", "text": "1. ## inequalities: 2 methods\n\nHi guys,\n\nI am trying to find the values of x for which\n\n$x - 3 < \\frac{x - 4}{x}$\n\nAttached is a diagram from which it is easy to see the answer x < 0! i.e these are the values of x for which the straight line x - 3 is below the curve (which I have expressed as:\n\n$1 - \\frac{4}{x}$\n\nBut, as we all know, inequalities can be solved in three ways: by sketching, by calculation and by completing the square. So I have tried it by calculation - here goes:\n\n$x - 3 < (x - 4) / x$\n\n$x^2 - 3x < x - 4$\n\n$x^2 - 4x + 4 < 0$\n\n$(x - 2) (x - 2) < 0$\n\n$(x - 2)^2 < 0$\n\nNow since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!\n\n2. Originally Posted by s_ingram\n...So I have tried it by calculation - here goes:\n\n$x - 3 < (x - 4) / x$\n\n$x^2 - 3x < x - 4$\n\n$x^2 - 4x + 4 < 0$\n\n$(x - 2) (x - 2) < 0$\n\n$(x - 2)^2 < 0$\n\n... Can anyone see what I am doing wrong!\nYou assumed that $x \\geq 0$. If you do the calculations for x < 0 you'll get the result which you already know.\n\n3. Originally Posted by s_ingram\n$x - 3 < (x - 4) / x$\n\n$x^2 - 3x < x - 4$\n\n$x^2 - 4x + 4 < 0$\n\n$(x - 2) (x - 2) < 0$\n\n$(x - 2)^2 < 0$\n\nNow since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!\nI wouldn't multiply both sides by x. Instead, I would do this:\n\\begin{aligned}\nx - 3 &< \\frac{x - 4}{x} \\\\\nx - 3 - \\frac{x - 4}{x} &< 0 \\\\\n\\frac{x^2 - 3x}{x} - \\frac{x - 4}{x} &< 0 \\\\\n\\frac{x^2 - 4x + 4}{x} &< 0 \\\\\n\\frac{(x - 2)^2}{x} &< 0\n\\end{aligned}\n\nAt this point I would make a sign chart. First, I need \"critical points.\" These are values which (1) the fraction equals 0, and (2) the fraction is undefined. I find (1) by setting the numerator equal to zero:\n\\begin{aligned}\n(x - 2)^2 &= 0 \\\\\nx - 2 &= 0 \\\\\nx &= 2\n\\end{aligned}\n\nI find (2) by setting the denominator equal to zero: x = 0.\n\nThen I draw my sign chart like this:\nCode:\n         und              0\n----------+---------------+-------------\n0               2\n(und = undefined)\n\nNow take the fraction\n$\\frac{(x - 2)^2}{x}$ and test a value less than 0. You will see that the fraction is negative, so I make a notation on my sign chart.\nCode:\n  (-)^2\n-----\n(-)   und              0\n----------+---------------+-------------\nneg    0               2\nPick a number between 0 and 2 and test the fraction, and then pick a number greater than 2 and test the fraction. When you're done, you should get this:\nCode:\n  (-)^2          (-)^2          (+)^2\n-----          -----          -----\n(-)    und     (+)      0     (+)\n----------+---------------+-------------\nneg    0      pos      2     pos\nWe want a range of numbers where the fraction is less than 0, or negative. So the solutions are x < 0.\n\n01\n\n4. Sorry, where do I assume that $x \\geq 0$? If x is less than zero say -5, then $(x - 2)^2$ gives $(-7)^2 = 49$ and this is not less than 0.\n\n5. Thanks again yeongil. I can see your method works and I have used something like it when checking the ranges of functions for which they are real or unreal, but I don't see why you would not multiply by x. Is it wrong to do so and if so why?\n\n6. To answer both of your posts, you assumed that $x \\geq 0$ when you multiplied both sides by x. Remember that when you multiply both sides of an inequality by a negative number, you flip the sign. When you went from this step:\n$x - 3\\;{\\color{red}<}\\;\\frac{x - 4}{x}$\nto this step:\n$x^2 - 3x\\;{\\color{red}<}\\;x - 4$\nyou multiplied both sides by x, and since you didn't change the sign, you assumed that $x \\geq 0$.\n\nThe problem was to solve for x, and if we pretend that we do not know what x is beforehand, we shouldn't multiply both sides by x; instead, subtract the fraction from both sides and proceed like I did earlier.\n\n01\n\n7. Excellent answer! Now I see. I suspected Earboth was on to something but I didn't understand what he meant.", "date": "2017-04-24 23:16:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/94707-inequalities-2-methods.html", "openwebmath_score": 0.9441335797309875, "openwebmath_perplexity": 334.42412898685694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717480217662, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8656938143966042}}
{"url": "https://math.stackexchange.com/questions/4603/series-converges-implies-limn-a-n-0/1122679", "text": "# Series converges implies $\\lim{n a_n} = 0$\n\nI'm studying for qualifying exams and ran into this problem.\n\nShow that if $\\{a_n\\}$ is a nonincreasing sequence of positive real numbers such that $\\sum_n a_n$ converges, then $\\lim_{n \\rightarrow \\infty} n a_n = 0$.\n\nUsing the definition of the limit, this is equivalent to showing\n\n$$\\forall \\varepsilon > 0 \\; \\exists n_0 \\text{ such that } |n a_n| < \\varepsilon \\; \\forall n > n_0$$\n\nor\n\n$$\\forall \\varepsilon > 0 \\; \\exists n_0 \\text{ such that } a_n < \\frac{\\varepsilon}{n} \\; \\forall n > n_0$$\n\nBasically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!\n\n\u2022 This is a classic problem. I remember it from teaching undergrad real analysis in 2005 (specifically I remember having a good think about it at the beautiful YMCA in downtown Montreal -- it took me a while to get it). I think it will be helpful to see a variety of answers. \u2013\u00a0Pete L. Clark Sep 14 '10 at 8:18\n\u2022 This is indeed a nice problem, and I don't want to give it away. The hint I would give is to remember to use the condition that $a_n$ is nonincreasing. \u2013\u00a0David E Speyer Sep 14 '10 at 17:21\n\u2022 Maybe this is interesting: If we omit monotonicity then $na_n$ converges to 0 statistically. (It is relatively easy to find examples showing that it need not converge in the usual sense.) Tibor \u0160al\u00e1t; Vladim\u00edr Toma: A Classical Olivier\u2019s Theorem and Statistical Convergence. dx.doi.org/10.5802/ambp.179 \u2013\u00a0Martin Sleziak Nov 23 '11 at 16:13\n\u2022 Maybe it's too late for your exam, but I've added another answer that is fairly short and is not very notation-intensive and does not rely on standard results but works from basic definitions. (I don't even use the fact that the harmonic series diverges.) I like to keep things simple. $\\qquad$ \u2013\u00a0Michael Hardy Jul 21 '16 at 2:02\n\u2022 Can you tell me where did you see this problem? I'm studying for an exam and It would really help some extra problems. This seems a very good one \u2013\u00a0J.Dane Jul 27 '18 at 10:39\n\nSome hints:\n\nIf $S_{n} = \\sum_{k=1}^{n} a_{k}$\n\nthen what is\n\n$\\lim_{n \\to \\infty} S_{2n} - S_{n}$?\n\nNow can you use the fact that $a_{n}$ is non-increasing to upper bound a certain term of the sequence $na_{n}$ with a multiple of $S_{2n} - S_{n}$?\n\n\u2022 Let me give this a shot: Assuming the limit you mention tends to zero, for a given $\\epsilon$, there is an integer $n_0$ such that $2 ( S_{2n_0} - S_{n_0} ) < \\epsilon$. Because the sequence is nonincreasing, we have that $2n a_{2n} \\leq 2 ( S_{2n_0} - S_{n_0} ) < \\epsilon$ for all $n \\geq n_0$. \u2013\u00a0dls Sep 15 '10 at 4:13\n\u2022 @dls: Right. It is still incomplete though. You have to consider (2n+1)a_{2n+1} too. \u2013\u00a0Aryabhata Sep 15 '10 at 5:02\n\nBy the Cauchy Condensation test, $\\displaystyle \\sum 2^n a_{2^n}$ converges since $\\displaystyle \\sum a_n$ converges, and thus $2^n a_{2^n} \\to 0.$ Now for $2^n < k < 2^{n+1}$,\n\n$$2^n a_{2^{n+1}} \\leq k a_{k} \\leq 2^{n+1} a_{2^n}$$\n\nso $n a_n \\to 0.$\n\nNow that enough time has passed so that more information will not spoil anything for the OP:\n\nThis fact can be found in $\\S 179$ of G.H. Hardy's seminal A Course of Pure Mathematics: he mentions that it was first proved by Abel, then forgotten and later rediscovered by Alfred Pringsheim. I have reproduced Hardy's proof in $\\S 2.4.2$ of these notes on infinite series. This is much slicker than what I came up with when I had to solve this exercise myself some years ago. On the other hand it seems to be exactly what Aryabhata's answer hints at.\n\nIn my notes I also attribute this result to L. Olivier and even cite the issue of Crelle's Journal in which it appears in 1827. This attribution does not appear in Hardy's book, which temporarily mystified me (I am no historian of mathematics: whatever such information I have comes from math books with good bibliographies), but I surmise I must have gotten it from Konrad Knopp's book on infinite series (the only other book I own which treats the subject seriously).\n\nP.S.: The wikipedia article on Pringsheim is unusually (almost suspiciously?) good. The impression that I have of him as a mathematician is someone who worked on infinite series at a stage when the foundations of the theory were finally solidly in place...and when the best mathematicians of the day had gone on to more fundamental and difficult problems. But I don't know whether this is at all fair. Anyway, it seems that you won't hear of him until you learn a little more about series than is treated in the standard contemporary curriculum, but as soon as you do his name comes up again and again.\n\n\u2022 Re: P.S.: The English Wikipedia page is almost a verbatim copy of the German one. The Pringsheim family is very well known in Germany due to its influence in Munich over centuries, and also due to fact that Alfred P.'s daughter was married to Thomas Mann who certainly contributed to making the family's fate in the first half of the twentieth century known to a wide audience. \u2013\u00a0t.b. Nov 23 '11 at 7:23\n\u2022 @t.b.: Thanks for this. I vaguely noticed the connection to Thomas Mann in the wikipedia article, but it didn't really sink in. \u2013\u00a0Pete L. Clark Nov 23 '11 at 16:04\n\u2022 Interestingly, it seems that Olivier's paper contained a mistake and later Abel wrote a paper to correct it. Details can be found e.g. in Michael Goar: Olivier and Abel on Series Convergence, Mathematics Magazine Vol. 72, No. 5 (Dec., 1999), pp. 347-355 jstor. To add also a freely available paper, it is briefly mentioned here. (The later is the paper where I've read about Olivier's mistake for the first time.) \u2013\u00a0Martin Sleziak Aug 15 '12 at 11:07\n\nI think I have an answer which doesn't rely on being clever enough to use the even and odd subsequences.\n\nBecause the series converges the sequence of partial sums forms a Cauchy sequence so we have for m,n>N\n\n$$\\| \\sum_m^n(a_i) \\| < \\epsilon$$\n\nAnd by a similar argument to those above: $\\|(n-m)a_n \\| \\leq \\| \\sum_m^n(a_i) \\| < \\epsilon$ Distributing out the left hand side and spliting the inequality gives us the chain:\n\n$$\\| na_n \\|-\\|ma_n\\|\\leq \\|(n-m)a_n \\| \\leq \\| \\sum_m^n(a_i) \\| < \\epsilon$$\n\nThen in the limit as n goes to infinity and fixing m we have the desired result since $ma_n$ will go to zero.\n\nI think... ?\n\nTry to stay away from quantifier-laden formulas, which make the problem harder to understand, and draw a picture. There is an equivalent problem for decreasing functions (as in the Integral Test for convergence) and the picture makes it obvious what is true in that case. Having seen the continuous proof, run the same argument for the sequence, or specialize the function to a sequence by using step functions or approximation thereof. I won't spoil the \"aha!\" proof-by-picture experience by posting more details, but it is quite easy once you draw the graph.\n\nNote that the desired conclusion is not true if the $a_n$ are not assumed to be nonincreasing:\n\nWe know a couple of facts:\n\n1. If a sequence $(s_n)_{n\\geq1}$ converges then we have that\n$\\displaystyle\\qquad\\liminf_{n\\to\\infty} s_n = \\lim_{n\\to\\infty} s_n = \\limsup_{n\\to\\infty} s_n$.\n\n2. There is a summable series $\\sum_{n=1}^\\infty a_n$ with non-negative $a_n$ such that $\\limsup\\limits_{n\\to\\infty} na_n > 0$.\n\nTake these two together you get that the best you can hope to prove in this slightly more general situation is that $\\liminf\\limits_{n\\to\\infty} na_n = 0$, and that is indeed true.\n\nFor an example showing that 2 above is true, consider this: We start with the harmonic series and observe that\n$\\displaystyle\\qquad \\limsup_{n\\to\\infty} n\\cdot \\tfrac1n = 1$,\nbut it isn't summable. But we could replace infinitely many of the sequence elements by zeroes without changing the limit superior from being $1$. Define $S(n)$ to be $1$ when $n$ is a perfect square and $0$ otherwise. Then $a_n = \\frac{S(n)}n$ has the desired property as\n$\\displaystyle\\qquad \\sum_{n=1}^\\infty a_n = \\sum_{n=1}^\\infty \\frac{S(n)}n = \\sum_{k=1}^\\infty \\frac1{k^2} < \\infty$.\n\n\u2022 Note that $a_n$ is supposed to be nonincreasing. Your sequence isn't. \u2013\u00a0mrf Sep 12 '13 at 20:01\n\u2022 Ah. Miss one word and the whole thing goes down the crapper. Typical. Well now it's fixed at least. And CW just for good measure. \u2013\u00a0kahen Sep 12 '13 at 20:02\n\nLet $\\epsilon > 0$.\n\nLet Let $s_n=\\sum_{k=1}^n a_k$.\n\n$\\{s_n\\}$ converges, therefore $\\{s_n\\}$ is Cauchy which implies $\\exists$ some $N$ such that $m, n \\geq N$ implies $\\left | s_n-s_m \\right | < \\epsilon$\n\nConsider $m=N$, for $n \\geq N$ we have $\\left | s_n-s_N \\right | < \\epsilon$,\n$\\Rightarrow \\left | a_N + a_{N-1} + a_{N-2}+\\cdots +a_n \\right | < \\epsilon$,\n$\\Rightarrow \\left | a_n + a_{n} + a_{n}+\\cdots +a_n \\right | \\leq \\left | a_N + a_{N-1} + a_{N-2}+\\cdots +a_n \\right | < \\epsilon$ $\\Rightarrow \\left (n-N)| a_n\\right | \\leq \\left | a_N + a_{N-1} + a_{N-2}+\\cdots +a_n \\right | < \\epsilon$\n$\\Rightarrow (n-N)\\left| a_n\\right | < \\epsilon$\n\nSo, $\\lim_{n \\rightarrow \\infty } (n-N)a_n = 0$\n$\\lim_{n \\rightarrow \\infty } na_n - \\lim_{n \\rightarrow \\infty } Na_n = 0$\nFrom above, since {$a_n$}$\\rightarrow 0$, {$na_n$}$\\rightarrow 0$ $\\blacksquare$\n\n\u2022 Shouldn't it be $|a_N + a_{N+1} + \\ldots + a_n|$ since $n \\ge N$? \u2013\u00a0philmcole Jan 13 '18 at 10:07\n\u2022 Very clear and nicely explained. Thanks \u2013\u00a0StammeringMathematician Jul 28 at 18:31\n\nHere's one more approach : By Cauchy's general principle of convergence :\n\n$\\forall \\epsilon>0, \\exists ~p \\in \\mathbb N , \\exists ~m> N: |a_{m+1}+a_{m+2}+ \\cdots + a_{m+p}| < \\epsilon$.\n\nChoose:$p = n-m$ .\n\n$|a_{m+1}+a_{m+2}+ \\cdots + a_n| < \\epsilon$.\n\nNow, since $a_n$ is non increasing and $\\epsilon \\rightarrow 0$: the above relation can be written as :\n\n$(n-m)a_n = 0 \\implies na_n = ma_n$\n\nFor a convergent series. $\\lim a_n=0 \\implies \\lim_{n \\rightarrow \\infty} n a_n = 0$\n\n\u2022 You seem to claim that directly from $(n-m)a_n<\\varepsilon$ you get $(n-m)a_n=0$. This argument is not correct, since $m$ depends on $\\varepsilon$. (But you still get $(n-m)a_n<\\varepsilon$. For a fixed $\\varepsilon>0$ and for $m$ corresponding to this $\\varepsilon$.) \u2013\u00a0Martin Sleziak Apr 4 '16 at 4:27\n\nYou might also do this the other way around. What if $$\\lim_{n\\to\\infty}na_n \\not=0,$$ that is if the limit does not exist or if it is positive, what does this tell you about $a_n$? What about sub-sequences of $(a_n)$?\n\n\u2022 I can't see how this works. The easy part - if the limit exists and is positive, then $\\sum a_n$ diverges like a harmonic series, contradiction. The harder part, if the limit does not exist - There there exists a subsequence and an $\\epsilon > 0$ such that $n_k a_{n_k} > \\epsilon$ for all $n_k$. But that itself does not imply the result. Did you have something else in mind? \u2013\u00a0Ragib Zaman Oct 11 '11 at 15:03\n\nIf the sequence is strictly decreasing, I guess there is a neat little argument. Look at the series $\\sum n a_n$ and define $\\rho_n=\\frac{(n+1)a_{n+1}}{na_n}$. Then $\\rho_n=\\frac{a_{n+1}}{a_n} \\frac{(n+1)}{n}$. We have that $\\lim_{n \\to \\infty} \\frac{(n+1)}{n}$ exists and is equal to $1$. As $(a_n)$ is strictly decreasing and the terms are positive, we have $\\frac{a_{n+1}}{a_n} < 1$ Therefore the limit $\\lim_{n \\to \\infty} \\frac{a_{n+1}}{a_n}$ exists and is smaller than $1$. So $\\lim_{n \\to \\infty} \\rho_n$ exists and is smaller than 1, and thus by the ratio test the series $\\sum n a_n$ converges. Hence $n a_n \\to 0$.\n\n\u2022 Consider $a_n = \\frac{1}{n^2}$. Then the limit of ratios is $1$. \u2013\u00a0Daniel Fischer Apr 27 '14 at 21:59\n\u2022 Oh my, you are right. Nevermind then. \u2013\u00a0Jos\u00e9 Siqueira Apr 27 '14 at 22:06\n\u2022 Still, looking at a series is a nice trick when you want to prove a limit is 0. \u2013\u00a0Jos\u00e9 Siqueira Apr 27 '14 at 22:10\n\nIf $\\lim\\limits_{n\\to\\infty} n a_n\\ne 0$, then for some $\\varepsilon>0$, there are infinitely many $n$ for which $a_n \\ge \\dfrac \\varepsilon n.$\n\nIf $a_n \\ge \\dfrac \\varepsilon n,$ then let $m = \\lfloor \\log_2 n \\rfloor$, so that $2^m\\le n < 2^{m+1}$. Then $$a_{2^{m-1}} + \\cdots + a_{2^m} \\ge 2^{m-1} a_{2^m} \\ge 2^{m-1} a_n \\ge \\frac{2^{m-1}\\varepsilon} n \\ge \\frac{2^{m-1}\\varepsilon} {2^{m+1}} = \\frac \\varepsilon 4.$$\n\nThen go on to the next $n$ for which $a_n \\ge \\dfrac \\varepsilon n$, and if it's not big enough for the resulting sequence $a_{2^{m-1}} + \\cdots + a_{2^m}$ to have a different (larger) value of $m$ than the one we just saw, then go on to the next one after that, etc.\n\nWe get infinitely many disjoint sets of terms with sums exceeding $\\dfrac \\varepsilon 4$.\n\nLet $\\epsilon$ be given. By Cauchy, there exists $N(\\epsilon)$ in $\\mathbb{N}$ such that: $a_N+a_{N+1}+....+a_{N+P}<\\epsilon$, for every $P$ in $\\mathbb{N}$. Note that for every chosen $P$, there are $P$ $a_n$'s on the left side of the inequality, not all of which can exceed $\\dfrac{\\epsilon}{P}$. Anyway, since $a_n$ is non increasing, $a_{N+P}$ is always less than $\\dfrac{\\epsilon}{P}$. We have then:\n\n$$a_{N+P}<\\dfrac{\\epsilon}{P}, \\forall P \\in \\mathbb{N}$$ Multiplying by $N+P$ at each side of the inequality: $$(N+P)a_{N+P}<\\dfrac{(N+P)\\epsilon}{P}$$\n\n\"Throwing\" $p$ to ininity, we can see then that for $n$ large enough, $na_n$ is almost less than $\\epsilon$.\n\nTaking $\\dfrac{\\epsilon}{2}$ instead of $\\epsilon$, we can guarantee that $na_n$ is less than $\\epsilon$ for $n$ large enough, completing the proof.\n\nSuppose $n_ja_{n_j}\\geq \\alpha>0$ and up to (infinity) subsequence assume $n_{j}-n_{j-1}\\geq n_{j}/2$.\n\nUsing that $a_n\\geq 0$ we get\n\n$$\\displaystyle A=\\sum_{n=1}^\\infty a_n =\\sum_{n=1}^{n_0} a_n+\\sum_{j=1}^\\infty\\sum_{n=n_{j-1}+1}^{n_j}a_n\\geq B+\\sum_{j=1}^\\infty\\sum_{n=n_{j-1}+1}^{n_j}a_{n_j}\\geq B+\\sum_{j=1}^\\infty(n_{j}-n_{j-1})a_{n_j} \\\\\\geq B+\\sum_{j=1}^\\infty\\frac{n_{j}}{2}a_{n_j}\\geq B+\\frac{1}{2}\\sum_{j=1}^\\infty{n_{j}}{}a_{n_j}=\\infty. \\ QED.$$\n\nLet $$f: (0, \\infty)\\to [0, \\infty)$$ be a continuous, non-increasing function such that $$f(n)=a_n$$ for all $$n \\in \\mathbb{N}.$$ Such a function certainly exists (for $$x \\in (n, n+1)$$ let $$f(x) = a_n + (a_{n+1}-a_n)(x-n)$$)\n\nSince $$\\displaystyle \\sum_{n=1}^{\\infty}f(n) = \\displaystyle \\sum_{n=1}^{\\infty} a_n$$ converges, by the integral test, $$\\displaystyle \\int_{0}^{\\infty}f(x)dx < \\infty.$$\n\nClaim: $$\\displaystyle\\lim_{x\\to \\infty}xf(x)=0.$$\n\nWe reproduce the argument that JLA gave here, begin by noting that $$\\displaystyle \\lim_{t \\to \\infty}\\int_{t/2}^{t}f(x)dx=0.$$ Then $$0 \\leq \\dfrac{t f(t)}{2}\\leq \\displaystyle \\int_{t/2}^{t}f(x)dx \\to 0$$ as $$t \\to \\infty$$ and the claim holds.\n\nIn particular, $$\\displaystyle \\lim_{n\\to \\infty}nf(a_n) =0$$ and we are done.", "date": "2019-10-21 07:12:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4603/series-converges-implies-limn-a-n-0/1122679", "openwebmath_score": 0.9271830320358276, "openwebmath_perplexity": 244.87053836634604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717480217662, "lm_q2_score": 0.8774767826757122, "lm_q1q2_score": 0.8656938033328928}}
{"url": "https://math.stackexchange.com/questions/1460661/if-f-is-continuously-differentiable-in-a-b-fa-fb-and-fa-fb", "text": "# If $f$ is continuously differentiable in $[a,b]$, $f(a)=f(b)$, and $f'(a)=f'(b)$, then there exist $a<x_1<x_2<b$ such that $f'(x_1) = f'(x_2)$.\n\nThis problem is from Apostol's Mathematical Analysis:\n\nLet $f$ be continuously differentiable in $[a, b]$. If $f(a) = f(b)$ and if $f^{'}(a) = f^{'}(b)$, then prove that there exists $x_1$ and $x_2$ in $(a, b)$ such that $x_1\\neq x_2$ but $f^{'}(x_1) = f^{'}(x_2)$.\n\nMy try:\n\nBy Rolle's Theorem $\\exists x_0\\in (a,b)$ such that $f^{'}(x_0)=0$. How to guarantee existence of $x_1,x_2$ from here?\n\nCan it be solved from a geometrical point of view?\n\n\u2022 you don't know that $f'(a)=f'(b)=0$ \u2013\u00a0tattwamasi amrutam Oct 2 '15 at 8:53\n\u2022 Geometrical point of view. At point a the function is going at a rate of f'. if that's positive the function is going up. At point b, f(a) = f(b) so the function had to have come down again. That means that somewhere f' was neg. But the means somewhere f' had to be zero. But right now at b f' is positive again. But that means it had to go from negative to positive. That means there was another point in between where f' was zero. Recap: function going up at a, peaks, f' = 0, goes down, plateaus, f' = 0 goes up again, reaches b. It's the only possibility. \u2013\u00a0fleablood Oct 2 '15 at 9:05\n\u2022 It will be nice to mention why you downvote and let people think and learn from your comments on their wrong answers. \u2013\u00a0Math-fun Oct 2 '15 at 9:41\n\u2022 thanks for your useful comments@fleablood \u2013\u00a0Learnmore Oct 3 '15 at 3:30\n\nAs you said, by Rolle's Theorem, we are guaranteed existence of $c \\in (a,b)$ such that $f'(c) = 0$. WLOG, let $f'(a) = f'(b) = k > 0$. By the continuity of $f'$ and the intermediate value theorem, we get existence of $c_1 \\in (a,c)$ such that $f'(c_1) = k_1$ where $0<k_1<k$. Similarly, we get $c_2 \\in (c,b)$ such that $f'(c_2) = k_1$.\n\nNow, if $f'(a) = 0$, then let's look at some cases. If $f(c) \\neq f(a)$, then the mean value theorem gives existence of $d \\in (a,c)$ such that $\\alpha = f'(d) = \\frac{f(c) - f(a)}{c - a}\\neq 0.$ Now, we can apply the above argument again.\n\nIf $f(c) = f(a) = f(b)$, then if $f'$ is nonzero at some point in the interval, the above argument works. Otherwise, $f'$ is $0$ and the result follows.\n\n(i) Let's suppose that $f'(a)>0$ (then we'll see the case $f'(a)=0$). Because of (i), you can find a $\\eta > 0$ (as small as it is) that we have $f(a + \\eta) > f(a)$ and $f(b - \\eta) < f(b)$.\n\n$f$ is continuous on $[a + \\eta, b-\\eta]$, therefore there exists $c \\in [a + \\eta; b - \\eta]$ so that $f(c) = f(a) = f(b)$. ($f(c)$ in $[f(a+\\eta], f(b-\\eta)])$.\n\n$f$ is continuous on both intervals $[a, c]$ and $[c, b]$ and reach its maximums in a $x_1$ and $x_2$ ($x_1$ in $(a, c)$ and $x_2$ in $(c, b)$ because $a, c$ and $b$ are not the maximum since $a+\\eta$ and $b-\\eta$).\n\nWe have $f'(x_1) = f'(x_2) = 0$\n\nOpposite reasonning for $f'(a)<0$ leading to the same conclusion\n\n(ii) if $f'(a) = 0$ because of Rolle's theorem you know that you have a $c$ in $(a,b)$ that $f'(c) = 0$. Assume that $f(c) = f(a)$ (then we reapply this reasonning on $a$ and $c$ instead of $a$ and $b$ until finding a $c_n$ that $f(c_n)$ is different from $f(a)$, if we can not find one then $f$ is constant and any $c_1<c_2$ in $(a, b)$ will do)\n\nWe have a $f(a) < f(c)$ and $f'$ continuous on $[a, c]$ and $f'(a) = f'(c) = 0$ but $f'$ not equal to $0$ and it exists a $c_1$ in $(a, c)$ so that $f'(c_1)>0$ because $f(a) < f(c)$.\n\nBecause of that and $f'$ continuous on this interval, $f'$ reaches its maximum in $d$ in $(a;c)$ and because of its continuity $f'$ reaches $f'(d)/2$ in two points $x_1,x_2$, where $a<x_1<d<x_2<c<b$.\n\n\u2022 i dont understand the down votes, because the arguments seems fine to me. \u2013\u00a0Lost1 Oct 3 '15 at 1:57\n\u2022 The format was messy. @Lost1 \u2013\u00a0user99914 Oct 3 '15 at 3:48", "date": "2020-10-24 08:32:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1460661/if-f-is-continuously-differentiable-in-a-b-fa-fb-and-fa-fb", "openwebmath_score": 0.9366390705108643, "openwebmath_perplexity": 151.1112767576348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232889752296, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8656678918615643}}
{"url": "https://stats.stackexchange.com/questions/396522/moving-average-process-stationarity", "text": "# Moving average process - stationarity\n\nIf we consider a moving average process of order 1, is that stationary?\n\nBecause, although, the mean will remain the same for Yt and Yt+k, the variance and co-variance will change if you calculate for Yt+k, so in that case, why is it stated as stationary?\n\nCan anyone explain: how do you know if a moving average process is weakly stationary, strictly stationary or non stationary?\n\nA stochastic process is weakly stationary if its mean and variance and covariances are finite and do not depend on time.\n\nA moving average process of order q (with q finite), $$MA(q)$$ is defined as:\n\n$$y_t = \\epsilon_t + \\alpha_1 \\epsilon_{t-1} + \\alpha_2 \\epsilon_{t-2} + \\dots + \\alpha_q \\epsilon_{t-q}, \\quad \\epsilon_t\\sim WN(0,\\sigma^2),$$\n\nand is weakly stationary for all the values of $$\\alpha_j$$.\n\nSo, in your example it is true that the variance and covariances change, but you can check that both are finite and do not depend on $$t$$ (the covariance will depend only the distance $$k$$).\n\nMoreover, even a moving average process of infinite order $$MA(\\infty)$$:\n\n$$y_t = \\sum_{j=0}^\\infty \\alpha_j \\epsilon_{t-j}$$\n\ncan be stationary, but we need an additional condition in order to have a finite variance:\n\n$$\\sum_{j=0}^\\infty|\\alpha_j|<\\infty.$$\n\n\u2022 Thank you. Is every moving average process weakly stationary? \u2013\u00a0user218970 Mar 10 at 10:10\n\u2022 Yes, and I edited the answer with some more details. \u2013\u00a0pdb Mar 10 at 10:52\n\u2022 thank you so much for your additional explanation. But the previous answer stated that MA(1) process is strictly stationary. \u2013\u00a0user218970 Mar 10 at 12:55\n\u2022 If you assume that the elements of the white noise process $\\epsilon_t$ are also indipendent (and not only uncorrelated) and identically distributed, then the moving average process is strictly stationary. \u2013\u00a0pdb Mar 10 at 14:02\n\nFirst see what definitions say\n\n\u2022 $$\\{X_t\\}$$ is strictly stationary if for any $$t_1,t_2,...,t_n \\in T$$ and any $$k \\in T$$\n$$P(X_{t_1},...,X_{t_n}) = P(X_{t_1+k},...,X_{t_n+k})$$ that is, we have statistical equilibrium.\n\n\u2022 $$\\{X_t\\}$$ is second order/weakly stationary if\n\n1. $$\\mathbb{E}[X_t] = \\mu < \\infty$$, independent of $$t$$\n2. $$Var(X_t) = \\sigma^2 <\\infty$$ , independent of $$t$$\n3. $$cov(X_t,X_{t+\\tau})$$, is a function of $$\\tau$$ only.\n\nThink of what weakly stationarity means. It means that the expected value of the process is finite and constant. It also means that the autocovariance does not depend on where two random variables are positioned but just on their distance! Autocovariance between today and yesterday same as autocovariance between 100 days and 101 days ago.\n\nNow take the MA(1) process $$\\boxed{ X_t = \\varepsilon_t + \\theta \\varepsilon_{t-1} }$$ where $$\\varepsilon_t \\sim WN(0,\\sigma^2)$$.\n\nThis process is weakly stationary as\n\n\u2022 $$\\mathbb{E}[X_t] = \\mathbb{E}[\\varepsilon_{t}] + \\theta \\mathbb{E} [\\varepsilon_{t-1}] = 0$$\n\u2022 $$Var(X_t) = \\sigma^2(1+\\theta^2)$$, i.e. independent of $$t$$\n\u2022 $$cov(X_t,X_{t-\\tau}) = \\theta \\sigma^2, |\\tau|=1$$, $$cov(X_t,X_{t-\\tau}) = 0, \\forall |\\tau|>1$$ hence function of $$\\tau$$ only.\n\n(if you have difficulties deriviting that, let me know. It is very straightforward and you shouldn't have any difficulties)\n\nMA(1) is also strictly stationary as both $$P(X_{t_1},...,X_{t_n})$$ and $$P(X_{t_1+k},...,X_{t_n+k})$$ multivariate (1-dependent) Normal distributions with identical parameters as it is a combination of WN random variables.\n\nEdit to provide further explanation: Each $$X_{t_1},..X_{t_n}$$ is a linear combination of independent Gaussian random variables,. In particular all $$X_{t_i} \\sim N(0, \\sum_{j=1}^{q} \\theta_j^2 \\sigma^2) \\forall j$$. The joint distribution of normal random variables is a multivariate normal. As they are all driven by an MA(q) process their covariance matrix can be easily derived. But as seen above, this does not depend on location, just on distance. Explicitly for MA(1):\n\n$$f_{\\mathbf {X} }(X_{t_1} \\dots X_{t_n})={\\frac {\\exp \\left(-{\\frac {1}{2}}{\\mathbf {X} }^{\\mathrm {T} }{\\boldsymbol {\\Sigma }}^{-1}{\\mathbf {X} }\\right)}{\\sqrt {(2\\pi )^{k}|{\\boldsymbol {\\Sigma }}|}}}$$\n\nwhere\n\n$$\\boldsymbol {\\Sigma } = \\begin{bmatrix} \\sigma^2(1+\\theta^2) & \\theta \\sigma^2 & 0 & 0 & \\dots & 0 \\\\ \\theta \\sigma^2 & \\sigma^2(1+\\theta^2) & \\theta \\sigma^2 & 0 & \\dots & 0 \\\\ 0 & \\theta \\sigma^2 & \\sigma^2(1+\\theta^2) & \\theta \\sigma^2 & \\ddots & \\vdots \\\\ 0 & 0 & \\ddots & \\ddots & \\ddots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\ddots & \\ddots & \\theta \\sigma^2 \\\\ 0 & 0 & \\dots & 0 & \\theta \\sigma^2 & \\sigma^2(1+\\theta^2) \\end{bmatrix}$$\n\nHence MA(q) processes driven by Gaussian noise are always strictly stationary (only condition is that the sum of MA coefficients should be finite)\n\nIn general, all weakly stationary Gaussian processes are strictly stationary too.\n\n\u2022 Thank you so much, yes I can derive that - no problem. Although, in the end, I did not understand the 'strictly stationary' part. I have read that MA(2) process is weakly stationary. How do their results differ? Could you elaborate, please? \u2013\u00a0user218970 Mar 10 at 10:08\n\u2022 MA(q) is also strictly stationary. See addition \u2013\u00a0Stats Mar 10 at 14:25\n\u2022 I'm a beginner in time series and therefore most of the matrices and notation is lost on me. I'm sorry for that. Can you explain me in simple terms that how should I go about figuring out the differences between strictly and weakly stationary from my derivation answers? \u2013\u00a0user218970 Mar 10 at 17:56\n\u2022 If you want to understand what strict stationarity means, it is highly essential you understand the basic multivariate distributions like the multivariate normal distribution. Strict stationarity has to do with the joint distribution of random variables, you will not be able to understand anything without first understanding what a joint distribution is. Could you please do this first? \u2013\u00a0Stats Mar 10 at 18:24", "date": "2019-08-22 14:20:06", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/396522/moving-average-process-stationarity", "openwebmath_score": 0.8604941964149475, "openwebmath_perplexity": 466.3886514021834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232884721166, "lm_q2_score": 0.8807970858005139, "lm_q1q2_score": 0.8656678883431181}}
{"url": "http://math.stackexchange.com/questions/68045/rolling-dice-such-that-they-add-up-to-13-is-there-a-more-elegant-way-to-solve/68050", "text": "# Rolling dice such that they add up to 13 \u2014 is there a more elegant way to solve this type of problem?\n\nHere is the problem:\n\nThere are $6^3$ possible outcomes to rolling a die $3$ times. Out of these, how many yield a total of (exactly) $13$ dots?\n\nMy solution would be absolutely impractical for problems involving $4$ rolls of the die. And at higher numbers it would be downright impossible without brute-forcing it with a computer.\n\nIs there a more elegant way to solve this type of problem?\n\nMy solution:\n\nFirst we find all sets $\\{a, b, c\\}$ such that $a + b + c = 13; \\; a, b, c \\le 6$:\n\n$\\{1, 6, 6\\}$\n\n$\\{2, 5, 6\\}$\n\n$\\{3, 4, 6\\}, \\{3, 5, 5\\}$\n\n$\\{4, 4, 5\\}$\n\nThe total number of sets that fit these criteria is $5$. If $a \\not= b \\not= c$, then there exist $3!$ unique permutations of $\\{a, b, c\\}$. If $a = b \\not= c$, then there exist $3$ unique permutations of $\\{a, b, c\\}$. -- There cannot be a set such that $a = b = c$.\n\nThere are $2$ sets of the first kind and $3$ of the second. It follow that the total number of triple die rolls that can fit the criteria is\n\n\\begin{equation*} 2 \\cdot 3! + 3 \\cdot 3 = 21 \\end{equation*}\n\nI can think of a second way to do it, which might be faster for slightly larger numbers... but essentially still comes down to brute force, not a method that can be generalized.\n\n-\nIf you're going for elegant, I think the method from Fool's answer below is nice. Basically the computation is reduced to $\\binom{12}{2} - 3 \\cdot \\binom{6}{2}$, with no brute-forcing, and no counting possibilities manually. Just some algebra. \u2013\u00a0 TMM Sep 28 '11 at 1:22\nCould whoever downvoted this explain why? And also explain why they didn't post such an explanation here? \u2013\u00a0 Michael Hardy Sep 28 '11 at 11:06\n\n## 5 Answers\n\nHere's one approach which is suitable in various circumstances.\n\nConsider the expression $(x+x^2+x^3+x^4+x^5+x^6)^3$. What is the coefficient of $x^{13}$ when you expand this out?\n\nThe expression can also be re-written as $\\left(\\frac{x(x^6-1)}{x-1}\\right)^3$ to make it less unwieldy to work with. The coefficient of $x^k$ can be extracted by taking derivatives, for example.\n\nThis is a very useful tool which can be applied in much more general situations, as you may observe. For instance you could replace $1+x+\\ldots+x^6$ by some other expression to represent dice with more faces, or dice with arbitrary faces, or dice with weights, etc.\n\n-\nInteresting! thanks. \u2013\u00a0 iDontKnowBetter Sep 27 '11 at 22:36\nIncluding $1$ in the expression is equivalent to having seven sided dice with numbers $0$ through $6$. It doesn't hurt with $13$, but would with $11$, because you could afford a $0$. \u2013\u00a0 Ross Millikan Sep 27 '11 at 22:37\nYou're welcome. Note that I corrected my expression slightly to account for the fact that a die can't come up 0 - it doesn't change the answer in this case since 13 cannot be attained with 0's. \u2013\u00a0 Alon Amit Sep 27 '11 at 22:39\n\nIn this case it's easier to find the number of triples $(a,b,c)$ that sum up to 8. If $(a,b,c)$ are possible dice rolls that sum up to 8, then $(7-a,7-b,7-c)$ are also possible dice rolls and sum to 13. In general the number of ways to make $x$ and $21-x$ on three dice are the same. (With two dice, of course, the number of ways to make $x$ and $14-x$ are the same.)\n\nOf course this is still brute force, but it's good to take advantage of symmetry -- for example, if you had been asked to find the number of outcomes giving $k$ dots for all of $k = 3, 4, \\ldots, 18$, you could save yourself time by doing only half of them.\n\nOne way to do this for an arbitrary number of dice is as follows: let $N(d,s)$ be the number of ways to roll a sum of $s$ with $d$ dice. (So you're trying to find $N(3,13)$.) In order to get a sum of $s$ with $d$ dice, you must have a sum of between $s-6$ and $s-1$ on the first $d-1$ dice, and then roll the appropriate number on the final die. So you have $$N(d,s) = N(d-1, s-1) + N(d-1, s-2) + \\cdots + N(d-1, s-6)$$ where we start with $N(0,0) = 1$ and $N(0,s) = 0$ for all other integers $s$. You can rearrange this to get $$N(d,s) = N(d, s-1) + N(d-1, s-1) - N(d-1, s-7)$$ and so you can compute each $N(d,s)$ with just one addition and one subtraction.\n\nThe generating-function method is actually the same as this, once you think about how polynomial multiplication works.\n\n-\n+1 for the method in your third paragraph, which is very easy to implement on a spreadsheet, especially if you have six blank rows at the top of your table. \u2013\u00a0 Henry Sep 27 '11 at 23:37\n\nIf I have understood your problem correctly then I guess we could use the stars and bars concept for a clever solution.\n\nFor example in your case,\n\n$$x_1+x_2+x_3 = 13$$\n\nNumber of positive integer solution possible such that $x_i \\ge 1$ is given by $\\binom{12}{2}=66$.\n\nThen set $x_1'=x_1+6$,so the equation reduces to: $$x_1'+x_2+x_3=7$$\n\nwhich gives $15$ possible solution,using the same thing for $x_2$ and $x_3$ we would see that we have over-counted $3\\times15$ solutions in our first calculation,hence subtracting $66-45=21$.\n\n-\n(+1) I was going to post a link to your earlier question but I see you already posted an answer using that method :) \u2013\u00a0 TMM Sep 28 '11 at 0:04\n\nOne way is generating functions: form $(x+x^2+x^3+x^4+x^5+x^6)^3$ and find the coefficient of $x^{13}$\n\n-\nAh, the very last chapter in my combinatorics book is titled \"generating functions\", though I've not gotten there yet. Thanks. (This is from my current probability course, which starts with basic combinatorics). \u2013\u00a0 iDontKnowBetter Sep 27 '11 at 22:39\nIf you want to learn more about generating functions, you should take a look at Herb Wilf's book \"generatingfunctionology\", which can be found free online. (This is legal; Wilf has posted the book on his web page.) This book is probably best read after you know some combinatorics and some probability. \u2013\u00a0 Michael Lugo Sep 28 '11 at 1:03\n\nIf you throw the die a very large number of times, maybe there would be some difficulty in finding the exact number of ways in which a certain total can appear. But the central limit theorem from probability theory can give some reasonable approximations.\n\nSay you throw the die 40 times. On average the sum is 140. There are $6^{40}$ possible outcomes. How many of those give a total in the closed interval $[135,150]$?\n\nWhen you throw a die once, the variance of the probability distribution of the number of dots you see is $35/12$, so when you throw it 40 times, the variance is $40\\cdot35/12= 10\\cdot35/3$, and the standard deviation is therefore $\\sqrt{350/3} = 10.801234497\\ldots\\ {}$. For this discrete distribution, \"$\\ge 135$\" is the same as \"$>134$\" and \"$\\le 150$\" is the same as \"$<151$\", so we do a continuity correction and look at the interval $(134.5,150.5)$. The number $134.5$ is $(134.5-140)/10.801234497\\ldots)$ standard deviations above the mean, i.e. about $0.5092$ standard deviations below the mean, and $150.5$ is about $0.972$ standard deviations above the mean. Plugging these into the cumulative probability distribution function $\\Phi$ of the normal distribution, we get a probability of about $\\Phi(0.972)-\\Phi(-0.5092)=0.529$.\n\nSo $0.529 \\times 6^{40} \\approx 7.074\\times 10^{30}$ is approximately how many ways there are to get a sum in that interval.\n\n-\nCould whoever downvoted this explain why? And also explain why they didn't post such an explanation here? \u2013\u00a0 Michael Hardy Sep 28 '11 at 5:40\nDon't sweat it. Random downvotes are more than balanced by random upvotes. \u2013\u00a0 TonyK Sep 28 '11 at 9:52\nThe concern is of course that it may not have been \"random\". Indeed, that is what one must presume. \u2013\u00a0 Michael Hardy Sep 28 '11 at 11:05\nYeah but still. \u2013\u00a0 TonyK Sep 28 '11 at 11:51\nTo suggest that \"random downvotes are more than balanced by random upvotes\" is to suggest that the concern here is about the number of reputation points. That's not what it is. Apply a bit of common sense here. \u2013\u00a0 Michael Hardy Sep 28 '11 at 15:02", "date": "2015-02-01 17:21:41", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/68045/rolling-dice-such-that-they-add-up-to-13-is-there-a-more-elegant-way-to-solve/68050", "openwebmath_score": 0.865911602973938, "openwebmath_perplexity": 224.8232462415859, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232879690035, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8656678817493662}}
{"url": "https://math.stackexchange.com/questions/3215729/in-the-figure-a-quarter-circle-a-semicircle-and-a-circle-are-mutually-tangent", "text": "# In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.\n\nIn the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $$2$$. Find the radius of the circle.\n\nI first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $$\\frac{1}{4}$$, but the correct answer was $$\\frac{2}{9}$$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.\n\n\u2022 $2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius? \u2013\u00a0hmakholm left over Monica May 6 '19 at 11:41\n\u2022 @HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222\\ldots$ is in that ballpark. \u2013\u00a0Blue May 6 '19 at 12:00\n\u2022 @Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$. \u2013\u00a0hmakholm left over Monica May 6 '19 at 13:05\n\u2022 Sounds like the start of a joke... \u2013\u00a0Asaf Karagila May 7 '19 at 12:43\n\nLook at the picture:\n\nFrom $$\\triangle ABE$$ we have $$(2+r)^2= 2^2+(2-r)^2$$ so $$r=1/2$$. From $$\\square ECGF$$ we have $$CG^2=(1/2+s)^2-(1/2-s)^2= 2s$$. From $$\\square ADGF$$ we have $$GD^2= (2+s)^2-(2-s)^2= 8s$$. So $$2=CG+GD=3\\sqrt 2\\sqrt s$$, hence $$s=2/9$$.\n\n\u2022 It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius. \u2013\u00a0BlueRaja - Danny Pflughoeft May 6 '19 at 16:10\n\u2022 Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me. \u2013\u00a0Hammerite May 6 '19 at 16:26\n\u2022 @Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $\\triangle FHE$ to find $FH$. And the same trick for $GD$. \u2013\u00a0SMM May 6 '19 at 16:37\n\u2022 @BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square. \u2013\u00a0SMM May 6 '19 at 16:46\n\u2022 @SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though. \u2013\u00a0BlueRaja - Danny Pflughoeft May 6 '19 at 21:17\n\n@SMM's proof is nicely self-contained. Here's one that invokes the Descartes \"Kissing Circles\" theorem, simply because everyone should be aware of that result.\n\nLet the quarter-, semi-, and full-circles have radius $$a$$, $$b$$, $$c$$, respectively.\n\nFrom the right triangle, we have $$a^2+(a-b)^2=(a+b)^2 \\quad\\to\\quad a=4b \\tag{1}$$\n\nConsidering the side of the square a circle of curvature $$0$$, that special case of the Kissing Circles theorem implies $$\\frac1{c} = \\frac{1}{a}+\\frac{1}{b}\\pm 2\\sqrt{\\frac{1}{a}\\cdot\\frac{1}{b}} = \\frac{5}{4b}\\pm 2\\sqrt{\\frac{1}{4b^2}} = \\frac{5\\pm 4}{4b}\\quad\\to\\quad c = \\frac49 b \\;\\text{or}\\; 4b\\;\\text{(extraneous}) \\tag{2}$$\n\nThen, with $$a=2$$, we have $$b=1/2$$, so that $$c=2/9$$. $$\\square$$\n\nlet the side of the square be $$a$$.\n\nLet's find the radius x of the semicircle\n\nWe have $$(a+x)^2 = a^2 + (a-x)^2$$ $$x=\\frac{a}{4}$$\n\nNow, a lemma.\n\nIf circles of radiuses R and r are touching externally, then the length of their common tangent is $$2\\sqrt{Rr}$$\n\nProof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $$(R+r)^2 = h^2 + (R-r)^2$$, from where $$h = 2 \\sqrt{Rr}$$.\n\nNow, let's use the lemma. Let $$y$$ be the radius of the small circle. We have $$a = 2\\sqrt{ay} + 2 \\sqrt{\\frac{a}{4}y}$$ $$\\sqrt{a} = 3 \\sqrt{y}$$ $$y = \\frac{a}{9}$$\n\n\u2022 I like this one because it taught me something useful. \u2013\u00a0richard1941 May 7 '19 at 19:03", "date": "2020-12-03 22:31:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3215729/in-the-figure-a-quarter-circle-a-semicircle-and-a-circle-are-mutually-tangent", "openwebmath_score": 0.8903278112411499, "openwebmath_perplexity": 274.6639717318454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232894783426, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.8656678723152168}}
{"url": "https://math.stackexchange.com/questions/1078150/conditionally-combining-vanilla-and-chocolate-ice-cream-scoops", "text": "# Conditionally combining vanilla and chocolate ice cream scoops\n\nSuppose I have a plate that fits $n$ number of ice cream scoops. I can choose between vanilla and chocolate flavors. The only condition is that I cannot scoop for chocolate twice in a row, e.g. if $n=5$ then {$v,c,c,v,c$} is not allowed - an impossible combination - but {$c,v,c,v,c$} or {$v,v,v,v,c$} is allowed.\n\nHow many possible combinations are there? I know I have to perform the operation $${n \\choose 2} - \\text{impossible combinations}$$\n\nBut just how do I figure out the number of impossible combinations?\n\nThere are $2^{5}$ possible combinations. We want a recurrence to help us out. So let $T(1)$ be our recurrence denoting the number of allowed combinations for a string of length $n$. So $T(1) = 2$. Similarly, $T(2) = 2^{2} - 1 = 3$, giving us $vv, vc, cv$. Now for $T(3)$, we cannot have $ccv, vcc, ccc$. So there are $2^{3} = 8$ possible combinations and we exclude three, leaving $T(3) = 5 = T(2) + T(1)$.\n\nSo more generally, $T(n) = T(n-1) + T(n-2)$ with initial conditions $T(1) = 2$, $T(2) = 3$. We set up the characteristic polynomial and solve:\n\n$\\lambda^{2} - \\lambda - 1 = 0$, which gives us eigenvalues $\\lambda = \\frac{ 1 \\pm \\sqrt{5}}{2}$.\n\nThe general form equation is: $T(n) = A \\cdot (\\frac{1 + \\sqrt{5}}{2})^{n} + B \\cdot (\\frac{1 - \\sqrt{5}}{2})^{n}$.\n\nNow plug in your initial conditions and solve.\n\nThe number of impossible combinations is $2^{n} - T(n)$.\n\nEdit: Proving T(n) holds by induction. The base cases are given in my analysis above for $T(1), T(2)$. So assume true up to $n-1$ and consider case $n$.\n\nNow look at $T(n)$ and remove the last flavor added. If it was vanilla, there are $T(n-1)$ ways of constructing such a sequence. By the inductive hypothesis, $T(n-1)$ correctly counts the number of valid $n-1$ configurations. So adding vanilla is valid.\n\nIf the last flavor is instead chocolate, we note that the flavor before last is vanilla (otherwise, the configuration would be invalid). So there are $T(n-2)$ such strings so that the $n-1$ flavor is vanilla. And then the $n$th flavor is chocolate. Thus, $T(n) = T(n-1) + T(n-2)$, exhausting all valid possibilities.\n\nSo by the principle of mathematical induction, $T(n)$ correctly counts the valid configurations for your problem.\n\n\u2022 $2^5$ possible combinations for $n = 5$? \u2013\u00a0Don Larynx Dec 22 '14 at 22:45\n\u2022 Yes. This is a \"word\" problem. There are five slots: $_ _ _ _ _$. Each slot can have vanilla or chocolate, and is independent of the others. So by rule of product, we multiply, giving us $2^{5}$ possible combinations, before we impose the restriction. \u2013\u00a0ml0105 Dec 22 '14 at 22:46\n\u2022 How do we know that this is a Fibonacci sequence for all $n$? \u2013\u00a0Don Larynx Dec 22 '14 at 22:53\n\u2022 You can prove it by induction. We have our base cases of $n = 1, 2$ taken care of by my explanation in the post. So suppose up to $n$ and look at the $n+1$ case. Use the $n-1, n$ cases to derive that this must be true for $T(n+1)$. \u2013\u00a0ml0105 Dec 22 '14 at 22:56\n\u2022 I understand how to prove the Fibonacci sequence holds, but how do we know this doesn't diverge for large enough $n$, e.g. 1,1,2,3,5,8,13,21,44(it diverged),79,333, etc... \u2013\u00a0Don Larynx Dec 22 '14 at 22:58\n\nYou want binary sequences not containing two consecutive ones.\n\nLet $f(n)$ be the number of such sequences of length $n$.\n\nthen we have $f(1)=2,f(2)=3$.\n\nAfter this we use the recursion $f(n)=f(n-1)+f(n-2)$.\n\nWhy? seperate the strings of lentgh $n$ into two kinds, those ending in $1$ and those ending in $0$.\n\nHow many end in $0$? $f(n-1)$, since after removing the last digit you must have a valid sequence of length $n-1$.\n\nHow many end in $1$? $f(n-2)$ since the secon to last digit must be $0$, and then the first $n-2$ must form a valid sequence.\n\nThus the sequence goes, $1,2,3,5,8,13\\dots$ And so on, it is a tail of the Fibonacci sequence.\n\n\u2022 But $f(3) = 3$, not $5$. \u2013\u00a0Don Larynx Dec 22 '14 at 22:43\n\u2022 vcv,vvc,cvc,cvv,vvv \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Dec 22 '14 at 22:47\n\u2022 Oops, I was counting the number of impossible strings. \u2013\u00a0Don Larynx Dec 22 '14 at 22:47\n\u2022 How do we know that this is a Fibonacci sequence for all $n$? \u2013\u00a0Don Larynx Dec 22 '14 at 22:54\n\u2022 well, we proved $f(n)=f(n-1)+f(n-2)$ for all $n$, that's the recurrence ficonaccis follow by definition. \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Dec 22 '14 at 23:01\n\nWe reason by cases. Suppose that we place down $k$ vanilla scoops first. This leaves $k + 1$ gaps, each of which will either contain exactly $0$ or exactly $1$ of the $n - k$ remaining chocolate scoops (since we can't have more than $1$ chocolate scoop in the same gap). This can be done in $\\binom{k+1}{n-k}$ ways. Summing over each possible value of $k$ (that is, from $k = 2$ to $k = 5$), we obtain: $$\\binom{3}{3} + \\binom{4}{2} + \\binom{5}{1} + \\binom{6}{0} = 1 + 6 + 5 + 1 = 13$$", "date": "2019-11-20 09:01:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1078150/conditionally-combining-vanilla-and-chocolate-ice-cream-scoops", "openwebmath_score": 0.8323714137077332, "openwebmath_perplexity": 333.1987895624203, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232924970204, "lm_q2_score": 0.8807970654616711, "lm_q1q2_score": 0.8656678718987533}}
{"url": "https://math.stackexchange.com/questions/3250569/in-how-many-ways-can-we-place-10-identical-red-balls-and-10-identical-blue-balls", "text": "# In how many ways can we place 10 identical red balls and 10 identical blue balls into 4 distinct urns if each urn has at least 1 ball?\n\nIn how many ways can we place 10 identical red balls (R) and 10 identical blue balls (B) into 4 distinct urns if each urn has at least 1 ball?\n\nMy attempt: Put the minimum number of balls into each urn:\n\ncase 1: R R R R (4C0, or 4 choose 0) ways to place 4 red balls, remaining 6R balls can be placed in 4 urns in (6+4-1)C(4-1) or (9 choose 3) ways, and 10B can be placed in (10+4-1)C(4-1) or (13 choose 3) ways, then total ways for this case 4C0*9C3*13C3 = 1*84*286 = 24024\n\ncase 2: B R R R, R B R R , etc, total of 4C1 ways to have 1 blue ball and 3 red ones, then for remaining 7R and 9B - 4C1*10C3*12C3 = 4*120*220 = 105600\n\ncase 3: B B R R , B R B R etc - following same logic 4C2*11C3*11C3 = 163350\n\ncase 4: 4C3*12C3*10C3 = 105600\n\ncase 5: 4C4*13C3*9C3 = 24024\n\nTotal ways = 24024+105600+163350+105600+24024 = 422598 However, the answer doesn't seem to be correct. What am I doing wrong?\n\n\u2022 Are the urns numbered? i.e. is B R R R really different from R B R R? If they are not ordered I would start with looking at how many ways there are to put 20 balls into 4 urns such that none is empty. This can be looked up as partitions of size 4 i.e. there are 64 such partitions. e.g. 6+6+4+4 Then for each partition one can see in how many ways one can fill urns with exactly 6,6,4,4 balls, which seems easier. \u2013\u00a0Rene Recktenwald Jun 4 '19 at 10:38\n\u2022 maybe you've got to work it out first without the 1-ball constraint, then subtract all the combinations with zero in any pot, possibly remembering the inclusion exclusion principle to avoid double deducting \u2013\u00a0Cato Jun 4 '19 at 10:56\n\u2022 Look up \"Stars and Bars\" for a general approach to the problem of counting the red ball arrangements (or the blue ball arrangements), with or without the restriction that every box has at least one ball. Then figure out how to combine those values for the problem at hand -- you'll also need a little bit of inclusion-exclusion, I think. \u2013\u00a0Ned Jun 4 '19 at 13:07\n\nBy the complement rule, the answer = U-L, where:\nU is the total number of ways 10 identical red balls (R) and 10 identical blue balls (B) can be placed into 4 distinct urns without constraints\n$$U={10+4-1 \\choose 4-1}^2=81796$$\nL is number of ways with 1 or more urns having no balls\nTo find L, let's use inclusion-exclusion principle. Let set A be the set of all distributions of 10 red balls and 10 blue balls in urns 1,2 and 3. Set B is the set of all distributions of 10 red balls and 10 blue balls in urns 2,3 and 4; set C is the set of all distributions of 10 red balls and 10 blue balls in urns 1,2 and 4; set D is the set of all distributions of 10 red balls and 10 blue balls in urns 1,3 and 4. Then $$L = A\\cup B \\cup C\\cup D$$ will be the set of all distributions in which at least one urn is left empty. By inclusion-exclusion\n$$L = |A\\cup B \\cup C\\cup D| = |A|+|B|+|C|+|D|-(|A\\cap B|+|A\\cap C|+|A\\cap D|+|B\\cap C|+|B\\cap D|+|C\\cap D|) + (|A \\cap B \\cap C| + |A \\cap B \\cap D| + |A \\cap C \\cap D| + |B \\cap C \\cap D|) - |A \\cap B \\cap C \\cap D|$$\n$$|A| = |B|=|C|=|D|$$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 3 distinct urns\n$$|A\\cap B|=|A\\cap C|=|A\\cap D|=|B\\cap C|=|B\\cap D|=|C\\cap D|$$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 2 distinct urns\n$$|A \\cap B \\cap C|=|A \\cap B \\cap D|=|A \\cap C \\cap D|=|B \\cap C \\cap D|$$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 1 urn\n$$|A \\cap B \\cap C \\cap D|$$ case when all urns are empty - not possible, therefore zero\n$$|A| = {10+3-1 \\choose 3-1}^2=4356$$\n$$|A\\cap B|={10+2-1 \\choose 2-1}^2=121$$\n$$|A \\cap B \\cap C|={10+1-1 \\choose 1-1}^2=1$$\n$$L=4356*4 - 121*6 +4=17424-726+4=16702$$\nThe ANSWER = $$\\boxed{U-L = 81796-16702=65094}$$\n\nWhen you do a question with the wording \"at least\" you almost always approaching it using the $$1-$$ not $$P(A)$$ method, because otherwise you are often likely to double count or miscount stuff.\n\nFor example in the way you approached the question, you have over counted in your case two. The reason is that if I put two red balls, with the four urns already having $$B R B B$$, they are fundamentally the same if I put them in the first urn and then the third urn compared to if I put them in the third urn and then the first urn. It is however different if I put one in he second urn and the other in the third urn.\n\nThis is because the ordering of the urn is taken care of at the step when you made a distinction between $$B R B B$$ as opposed to $$R B B B$$, then after that distinction, the three blue urn are no longer distinct. So placing a red ball in the first blue urn is no different to placing in the third blue urn for example.\n\nUsing the other method, you calculate the total number of way of placing the balls, then subtract all the ways with 1 or more urn having no balls. Then you should get the right answer.\n\n\u2022 Thank you. Following your suggestion, the answer should be the total number of ways to put 10 identical red balls and 10 identical blue balls into 4 distinct urns minus number of combinations with 1 urn empty minus number of combinations with 2 urns empty minus number of combinations with 3 urns empty, correct? Continued in next comment \u2013\u00a0Alexander Arefolov Jun 9 '19 at 20:39\n\u2022 Total number of ways to put 10 identical red balls and 10 identical blue balls into 4 distinct urns without any constrains is ((10+4-1)C(4-1))^2 or (13 choose 3) squared = 81796 (I know this part is correct). Number of ways to have 1 urn empty should be (4C3)*((10+3-1)C(3-1))^2 : (4 choose 3) ways to choose one empty bin or 3 full bins times number of ways to fill 3 distinct bins with 10 identical red and 10 identical blue balls which should be (10+3-1)C(3-1)^2 or (12 choose 2) squared = 4*66^2=17424. Continued in next comment \u2013\u00a0Alexander Arefolov Jun 9 '19 at 20:40\n\u2022 For 2 urns with no balls scenario - (4 C 2)*((10+2-1)C(2-1))^2 = 6* (11 choose 1)^2 = 6*121=726. For 3 urns with no balls: (4 C 1)*((10+1-1)C(1-1))^2 = 4. The answer then should be 81796 - (17424+726+4)= 63642. However, this answer still does not seem to be correct! \u2013\u00a0Alexander Arefolov Jun 9 '19 at 20:41", "date": "2021-06-15 03:02:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3250569/in-how-many-ways-can-we-place-10-identical-red-balls-and-10-identical-blue-balls", "openwebmath_score": 0.7531685829162598, "openwebmath_perplexity": 144.64315413389022, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232874658905, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.8656678705426549}}
{"url": "https://math.stackexchange.com/questions/4259194/is-there-a-power-of-2-whose-last-digit-is-a-0/4259207#4259207", "text": "# Is there a power of $2$ whose last digit is a $0$?\n\nI'm looking for a power of $$2$$ that ends in $$0$$, so $$2^n = k\\cdot 10$$, where $$n \\in \\mathbb{Z}$$ and $$k \\in \\mathbb{Z}$$. It doesn't seem to exist. As an equation with two variables, I was looking for a second equation, but just can't think of any.\n\nI also trialled a few consecutive powers of $$2$$ and noticed that there seems to be a sequence perpetually ending in $$2, 4, 8, 6, 2, 4, 8, 6, \\cdots$$\n\nHow can I prove mathematically that there will never be a power of $$2$$ that ends in $$0$$? Is there a simply way to explain that?\n\n\u2022 $5$ divides the RHS, but not the LHS. Sep 24 at 15:04\n\u2022 You already proved that the ending digit cycles and $0$ does not appear. Nothing more has to be done. Of course, you can also apply the above comments. Sep 24 at 15:23\n\u2022 @Peter I haven't proven it. All I wrote was, that I observed (\"noticed\") a sequence. Sep 25 at 14:47\n\u2022 Thanks for all answers (permanently or temporarily submitted). My first post to this site seemed to have sparked some interest. Sep 25 at 15:03\n\u2022 Oct 15 at 16:03\n\nThe only numbers, in the decimal basis, with a digit of $$0$$ for the units, are those that are multiples of $$10$$, a fortiori, multiples of $$2$$ AND $$5$$. Through the fundamental theorem of arithmetic, you can show that there is no power of $$2$$ that is divisible by $$5$$; thus, there is no power of $$2$$ which ends by a $$0$$ digit (in the decimal basis).\n\n\u2022 I've just realised that you already cite the fundamental theorem of arithmetic. (+1) I've deleted my answer. Sep 24 at 19:07\n\u2022 Thanks. That's the simplest of answers. Though not quite a proof I accept this as the best answer. Sep 25 at 15:02\n\nProof 1: Let $$a=2^n$$ and $$b = 2^{n-1}$$. Since $$a = 2b$$, if the last digit of $$a$$ is $$0$$, then the last digit of $$b$$ must be $$0$$ or $$5$$. Since $$b$$ is odd, the last digit can't be $$5$$. But if the last digit of $$b$$ is $$0$$, we can apply the same logic to find that the last digit of $$2^{n-2}$$ is $$0$$, and so on all the way down to $$2$$, but obviously the last digit of $$2$$ isn't $$0$$.\n\nProof 2: Suppose we take a power of two and write it as $$10a+b$$, and then multiply it by $$16$$. $$16*(10a+b)=160a+16b=160a+10b+5b+b$$. Now clearly the $$160a$$ and $$10b$$ terms don't affect the last digit, and since $$b$$ is even, neither does $$5b$$. So the last digit is $$b$$. This shows that whenever you take a power of two, and then take four more powers of two (that is, multiply by $$2^4=16$$), you don't change the last digit.\n\nProof 3: The last digit of a number is the same as the remainder when divided by $$10$$, and taking the remainder when divided by some number is known as \"modular arithmetic\". In modular arithmetic, the set of all numbers with the same remainder is an equivalence class, and these equivalence classes act like numbers. We can apply addition and multiplication to them and get an equivalence class as an answer. For instance, if we multiply a number with a last digit of $$2$$ with a number with a last digit of $$3$$, the product will have a last digit of $$6$$. If we represent the set of numbers with last digit $$r$$ as $$\\bar r$$, we can say $$\\bar 2 \\times \\bar 3 = \\bar 6$$, and this is true regardless of which numbers we pick from the $$\\bar 2$$ and $$\\bar 3$$ equivalence classes. So you're asking for an $$n$$ such that the $$\\bar 2^n =\\bar 0$$. It's easy to see by inspection that this is not possible, but there's also a theorem that if we define $$\\phi(n)$$ as the number of numbers less than $$n$$ that are coprime to $$n$$, then $$\\bar a^{\\phi(n)+1}$$ in $$n$$-modular arithmetic is $$\\bar a$$. Since $$\\phi(10)=4$$ ($$1,3,7,9$$ are coprime to $$10$$), $$\\bar 2^{n+4}=\\bar2^n$$.\n\n\u2022 I really liked the Proof 2! $(+1)$ Sep 26 at 22:52\n\nThe below tells us more generally what the last digit can be (although the proof is a bit informal).\n\n(The lemma, and a bit more formal reasoning, is really the only thing you were missing.)\n\nLemma: when multiplying 2 positive integers, only the last digits of each affects what the last digit of the result will be.\n\nProof: ($$x_i$$ represents the $$i$$-th digit of the integer $$x$$ and $$x_1x_2...x_{n-1} = 0$$ if the number only has 1 digit)\n\n$$a_1a_2...a_n * b_1b_2...b_n$$\n$$=(10 * a_1a_2...a_{n-1} + a_n) * b_1b_2...b_n\\hspace{3cm}(1)$$\n$$=a_n * b_1b_2...b_n + 10 * a_1a_2...a_{n-1} * b_1b_2...b_n$$\n$$=a_n * b_n + 10 * a_n * b_1b_2...b_{n-1} + 10 * a_1a_2...a_{n-1} * b_1b_2...b_n\\hspace{1cm}\\text{(similar to 1)}$$\n$$=a_n * b_n + 10 * (a_n * b_1b_2...b_{n-1} + a_1a_2...a_{n-1} * b_1b_2...b_n)$$\n\nThe last digit of 10 times any positive integer will always be 0. If you sum such an integer with any other positive integer, the last digit of the result will be equal to the last digit of the other integer. Thus $$a_n * b_n$$ (that is, the last two digits of each integer) is the only thing that affects the last digit of the result. Q.E.D.\n\nNow let's consider the first few powers of 2:\n\n$$2^0 = 1$$ $$2^1 = 2$$ $$2^2 = 2 * 2 = 4$$ $$2^3 = 4 * 2 = 8$$ $$2^4 = 8 * 2 = 16$$ $$2^5 = 16 * 2 = 32$$\n\nSince only the last digit of the current power affects the last digit of the next power, we now have a cycle.\n\nWe know any number ending in 2 multiplied by 2 will end in 4. Multiply that by 2 and it will end in 8, then 6, then 2 again. And it will repeat like this indefinitely.\n\nTherefore, the only last digits possible, from $$2^1$$ onward, are 2, 4, 6 and 8.\n\nNegative powers of 2 trivially cannot have a last digit of 0.\n\nTherefore, there is no integral power of 2 having a last digit of 0.\n\nAs a bonus result, this approach also shows that the last digit of $$2^{1+i}$$ equals $$2^{1+(i\\bmod 4)}\\bmod 10$$ for any $$i \\ge 0$$.\n\nI'm a Computer Science teacher and was also intrigued by this, so wrote the following program to prove that there are no numbers that are multiples of 5 and powers of 2. I'm definitely no mathematician, so forgive the brute force approach. As stated above by posters, the repeating end digit sequence of 2,4,8,6 is the proof / 'give away'.\n\n# part 1 - set up powers of 2 for reference\n\nmaxNumber = int(input('Enter highest number ')) result = 0 power = 0 base2 = [] while True: result = 2**power if result <= maxNumber: base2.append(result) power += 1 else: break print (base2)\n\n# part 2 - go search\n\nfound = False for i in range(1,maxNumber+1): if i % 5 == 0 and i in base2: print (i,'is a multiple of 5 and power of 2') found = True if not found: print ('There are no numbers between 1 and',maxNumber,'that are a multiple of 5 and power of 2')\n\n\u2022 As it\u2019s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.\n\u2013\u00a0Community Bot\nOct 26 at 6:21", "date": "2021-12-04 08:53:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4259194/is-there-a-power-of-2-whose-last-digit-is-a-0/4259207#4259207", "openwebmath_score": 0.7838711738586426, "openwebmath_perplexity": 150.17564030475435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232884721166, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.8656678698912829}}
{"url": "http://salesianipinerolo.it/ubdl/real-life-applications-of-sine-and-cosine-functions.html", "text": "2 Educational Objectives After performing this experiment, students should be able to: 1. All quadratic functions fit the form y = Ax 2 + Bx + C, where A, B, and C can be any real number (although A cannot be zero). Trigonometry is a system that helps us to work out missing sides or angles in a triangle. Learning Targets: (1) I can sketch the graphs of the parent functions of sine and cosine. The sine and cosine functions are among the most important functions in all of mathematics. Among many uses and applications of the logistic function/hyperbolic tangent there are: which is based in the hyperbolic cosine function. Synthetic Geometry and Coordinate Geometry are used in real life to help us understand the dimensions and transformations of shapes and figures such as lines, triangles, polygons, and circles. Basic graph types are y = a sin(bx) + c and y = a cos(bx) + c. Some examples include the weather, seasonal sales of goods, body temperature, the tide's height in a harbor, average temperatures, and so on. \u2022 Table 1 gives the sum of two arbitrary cosine functions. Therefore, a time domain function f(x) and its corresponding frequency domain function are duals of each other. rewrite real-life formulas. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval[latex]\\,\\left[-1,1\\right]. 03 Students will take the unit circle and construct the graphs of the sine and cosine functions. In this way, it has many engineering applications such as electronic circuits and mechanical engineering. \u03b8 = 90\u00b0), which gives us cos \u03b8 = 0 and. For example, variables that depend on the seasons may be modeled with trigonometric functions because the seasons repeat every year just like the sine function repeats every 2\u03c0. Here's an example: San Diego, California, is a gorgeous [\u2026]. Graph a sine or cosine function having a different amplitude and period. This equation is usually solved using sums of sines and cosines. Build an understanding of trigonometric functions by using tables, graphs and technology to represent the cosine and sine functions. Calculus is made up of Trigonometry and Algebra. distributions characterized by a cdf based on the sine function, called the new sine-G family of distributions. At 90\u00b0 and 270\u00b0, x = 0 and therefore cos \u03c6 = 0, while at 0\u00b0 and 180\u00b0 y = 0 and therefore sin \u03c6 = 0. Explore symmetry (odd & even), domain, range, continuity, relative extrema and concavity of sine and cosine functions; Use Graphing Calc. The solution for an oblique triangle can be done with the application of the Law of Sine and Law of Cosine, simply called the Sine and Cosine Rules. These can be confusing to understand and memorize however these for the basis of all trigonometric values and derivations. Homework Statement High tide at 4am with a depth of 6 meters. Modeling with Sine or Cosine Functions Sine and cosine functions can be used to model real-world phenomena, such as sound waves. The midline is the average value. They are often shortened to sin, cos and tan. 4 Trigonometric Functions of Any Angle 4. C y KA sl ul d KrYirgMhlt os 4 3r 6e2s ke er Ivie kd F. Applications of Trigonometric and Circular Functions 281 CHAPTER OBJECTIVES t -FBSOUIFNFBOJOHTPGBNQMJUVEF QFSJPE QIBTFEJTQMBDFNFOU Applications of Trigonometric and Circular Functions CAS Suggestions argument of the cosine. The application problems show different ways that one could use sine and cosine funtions in real life. Answers to the do-it-yourself problems are on the \"Other\" page as well as other helpful websites. To be able to solve real-world problems using the Law of Sines and the Law of Cosines This tutorial reviews two real-world problems, one using the Law of Sines and one using the Law of Cosines. What is Trigonometry? Trigonometry is a another form of math dealing with the relationship of the sides and angles of triangles. I struggle with finding. Physical: 23,. 7 - Applications of Sinusoidal Functions Applications of Sinusoidal Functions. This lesson will present real world examples that involve inverse trigonometric ratios. The trigonometric functions sine, cosine, and tangent are useful in a wide range of applications, from solving simple problems to advanced and complex problems. Over 90 new problems on differential equations and their applications. Understanding Calculus is a complete online introductory book that focuses on concepts. It studies the relationship between involving lengths and angles of triangle. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. In maths, you have real life applications on any thing that you study. At x = 0 degrees, sin x = 0 and cos x = 1. Transformations of Sine and. What are the Applications of Trigonometry in Real Life? Ans. (3) The student uses functions and their properties to model and solve real-life problems. Graphing Cosine Function Lesson Plan 1. Graph a sine or cosine function having a different amplitude and period. What are real-life applications of the sine and cosine wave applications? It's for a pre-calculus assignment and I have to make a presentation to the class about it. What Are Imaginary Numbers? are used in real-life applications, such as electricity, as well as quadratic equations. Recall from Graphs of the Sine and Cosine Functions that the period of the sine function and the cosine function is $$2\u03c0$$. Subhaschandra Singh, Department of Physics, Dhanamanjuri College of Science, Imphal \u2013 795 001, Manipur, India. introduction. Law of Cosine [Just formula. Pg 439 #1-8, 33, 35, 37, 43-55 odd \u00b7 Sketch the graphs of basic sine and cosine functions \u00b7 Use amplitude and period to sketch the graphs of sine and cosine functions \u00b7 Sketch translations of the graphs of sine and cosine functions. 3: Trigonometric Functions in Real life - Duration: 13:02. A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. S08 2 Learning Objectives Upon completing this module, you should be able to: 1. (b) If the power of cosine is odd (n=2k+1), save one cosine factor and use the identity sin 2 x + cos 2 x = 1 to convert the remaining factors in terms of sine. bc is the horizontal shift 4. This is exactly the reason why, when the graphs are plotted on grids as above, the cosine graph is equivalent to the sine graph, omitting the fact that it is shifted, or translated, 90\u00b0 to the left. Practice Now! Trigonometric Ratios and Angle Measures Topics: 1. Tan x must be 0 (0 / 1) At x = 90 degrees, sin x = 1 and cos x = 0. Trigonometry plays a major role in musical theory and production. Outcome 5: Set up, solve, and graph equations from problems that require use of trigonometric functions, tangent, sine, and cosine and the Pythagorean Theorem. 4 9 Review Day 10 Test Day TOTAL DAYS: 10 C2. rewrite real-life formulas. Note the capital \"C\" in Cosine. Note the capital \u201cC\u201d in Cosine. A trigonometric function can be used to find the height of a smokestack on top of a building. Relationship between Sine and Cosine graphs The graph of sine has the same shape as the graph of cosine. &Assume&that&a& rider&enters&a&car&froma&platformthat&is&located&30\u00b0&around. Solitary Wave Solutions to Certain Nonlinear Evolution Equations by Rational Sine \u2013Cosine Function Method S. Unit Five: Real World Problems Example: When Light shines through two narrow slits, a series of light and dark fringes appear. State the domain and range of sine, cosine, and tangent curves of the form y = Asinx, y = Acosx, and y = Atanx. Outcome 6: Set up and solve exponential and logarithmic equations; then identify and sketch graphs of the functions. F) as its Fourier transform. But such proofs are lengthy, too hard to reproduce when you\u2019re in the middle of an exam or of some long calculation. 7m, but I am unsure how to interval the axis. Graphs of elementary trig functions allow you to see the graphs of sine, cosine and tangent and their relationship to travelling around a circle. Yes, you can derive them by strictly trigonometric means. Prerequisite Given the period, solve for the parameter '' in or. Trigonometry has many applications in astronomy, music, analysis of financial markets, and many more professions. Applications of the Derivative. This lesson will present real world examples that involve inverse trigonometric ratios. T3 and 20-2. The general form of a sinusoidal is: $f ( x ) = a \\sin ( bx \u2013 c ) + d, \\text{ for } b > 0$. So in the rule, c 2 = a + b - 2ab cos C side you are looking for angle opposite the side you want Label your diagram using a,b and c to avoid confusion Substitute into the formula and evaluate. Model the problem using the equation to show the depth of the water t hours after midnight. The slider is used for frequency values. Most radio communication is based on the use of combinations of sines and cosine waves. So our function is $y = 3. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. Trigonometry plays a major role in musical theory and production Real world examples of cosine functions. What we want is the Law of Sines. Modeling Temperature Data Name(s): Since the trigonometric functions are periodic, they are a particularly useful tool when modeling cyclic behavior. They will also explore certain relationships that emerge from the measurement of three-dimensional figures and two-dimensional shapes. Especially sine and copsine functions have been very useful in applications for medical science, signal processing, geology, electronic communication, thermal analysis, satellite communication and many more. Find the period of a sine or cosine function. There are primarily three trigonometric functions commonly used with trigonometric identities to solve complex equations. The domain is the set of all real numbers. We explain Real World Examples of Inverse Trigonometry Functions with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Trigonometric formulae are useful for solving problems in two dimensions. aphing Sine and Cosine functions: amplitude, phase shift, and vertical slide Cofunctions (Notes pp. Using Sum and Difference Formulas In this lesson, you will study formulas that allow you to evaluate trigonometric functions of the sum or difference of two angles.  cos^{-1}(x) = arccosx \\rightarrow The arc that has a cosine of x. The Sine, Cosine and Tangent functions are often applied to real world scenarios. The applications of Fourier transform are abased on the following properties of Fourier transform. Real life scenario of logarithms is one of the most crucial concepts in our life. The Sine, Cosine and Tangent functions are often applied to real world scenarios. Real-life applications of. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. basic introduction into trigonometry. Real life application of trigonometry is use of trigonometry formula and its functions. Euler's formula. 3: Trigonometric Functions in Real life - Duration: 13:02. These trigonometric functions are extremely important in science, engineering and mathematics, and some familiarity with them will be assumed in most \ufb01rst year university mathematics courses. 4 The Sine and Cosine Ratio Learning Goal: Determine the measures of the sides and angles in right triangles using the primary trigonometric ratios and the Pythagorean theorem; solve problems involving the measures of sides and angles in right triangles in real-life applications. com is a free math website that explains math in a simple way, and includes lots of examples, from Counting through Calculus. We say that this sinusoidal has a vertical shift of 1. The sine function has many real life applications, a few of which are: Triangulation, used in GPS-equipped cellphones, Musical notes, Submarine depth, Length of a zip. In addition, non-right angled triangles can be solved using the sine and cosine trigonometric functions. The domain of a linear function is all real numbers, and it's not possible to write every real number in a list. arbitrary sine function. There is only315 degrees of pizza left. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. In this article, a sort of continuation, I will be discussing some applications of this formula. Change from roots to rational exponents. So the graph looks like a very simple wave. However, the basic sine function usually requires one or more transformations to fit the parameters of the process. Many other Fourier-related transforms have since been defined, extending the initial idea to other applications. Magnitude Amplitude of combined cosine and sine Phase Relative proportions of sine and cosine The Fourier Transform: Examples, Properties, Common Pairs Example: Fourier Transform of a Cosine f(t) = cos (2 st ) F (u ) = Z 1 1 f(t) e i2 ut dt = Z 1 1 cos (2 st ) e i2 ut dt = Z 1 1 cos (2 st ) [cos ( 2 ut ) + isin ( 2 ut )] dt = Z 1 1 cos (2 st. Such behavior occurs throughout nature and led to the discovery of rapidly rotating stars called pulsars in 1967. It just plots the sine function on the screen.  cos^{-1}(x) = arccosx \\rightarrow The arc that has a cosine of x. The trigonometric functions sine and cosine have four important limit properties: You can use these properties to evaluate many limit problems involving the six basic trigonometric functions. A function may be defined by means of a power series. for the area of a triangle \u2022 use of sine rule and cosine rule for any triangle. This is how we find out \u201csine/cosine = tangent/1\u201d. You want to figure out what the angle is of the bottom of your long chair to the ground. 28-1 Length of Arc. However, there are still disadvantages such as low solution accuracy and poor global search ability. In other words, for any value of $$x$$,. We show that, in some situations, the new sine-G models provide an interesting alternative to the sine-G models, with possible di erent targets in terms of modelling. Let's put our values in there: Now let's move some things around and get calculating: We're not done yet, though, we need to apply some inverse sine to both sides to get to B itself. And I'll leave those details to the. The function has the same domain, but a range of. General triangle word problems. Homework Statement High tide at 4am with a depth of 6 meters. Example 1: Evaluate. In Chapter XI of The Age of Reason, the American revolutionary and Enlightenment thinker Thomas Paine wrote:. In this way, it has many engineering applications such as electronic circuits and mechanical engineering. \u2022 Table 1 gives the sum of two arbitrary cosine functions. #91 - Sketch translations of graphs of sine and cosine functions. F MTH 123 Ginger Rohwer 28 February Real Life Trigonometry Applications In Lunar Phases In this project we will be collecting data about the illumination of the moon for certain months then creating sine and cosine functions to. We could think of the cosine function as a sine wave with phase shift \u02c7 2. Law of Sines. Definitions and formulas for basic trigonometry, sine, cosine, tangent, cosecant, secant, cotangent, the law of sines and the law of cosines. The domain of the sine function is the set of real numbers, that is, every real number is a first element of one pair of the function. Besides other fields of mathematics, trig is used in physics, engineering, and chemistry. \u2022 Given information about a rotating functions in the real world. Together, these extensions define (sin \u03c6, cos \u03c6) for any angle \u03c6, positive or negative, of any size. Plotting a basic sine wave. Relationship between Sine and Cosine graphs The graph of sine has the same shape as the graph of cosine. the sine, cosine and tangent functions are often applied to real world scenarios. The amplitude is a=2 and the period is. 3 Trigonometry in the Real World C2. You want to figure out what the angle is of the bottom of your long chair to the ground. SheLovesMath. Upon the completion of this course, you will have a better understanding of key geometry axioms and theorems. (3) I can sketch translations of the graphs of sine and cosine functions. Applications are given for four real data sets, showing a better t in comparison to some existing distributions based on some of goodness-of- t tests. The first portion of the book is an investigation of functions, exploring the graphical behavior of, interpretation of, and solutions to problems involving linear, polynomial, rational, exponential, and logarithmic functions. Sine, Cosine and tangent are the three important trigonometry ratios, based on which functions are defined. The amplitudes of the cosine waves are held in the variables: a1, a2, a3, a3, etc. 85 inches by 3. The graph of a quadratic function is called a parabola. Sine 5 None Up 2 9. Find an equation for a cosine function that has an amplitude of OTC \u2014 , a period of Find an equation for a sine function that has amplitude of 5, a period of 3TT. At x = 0 degrees, sin x = 0 and cos x = 1. This is how we find out \u201csine/cosine = tangent/1\u201d. Possible Ideas Students will suggest:. Similar statements can be made for the other trigonometric functions of sums and differences. Synthetic Geometry and Coordinate Geometry are used in real life to help us understand the dimensions and transformations of shapes and figures such as lines, triangles, polygons, and circles. Mathematics Revision Guides - Real Life Trig Problems Page 12 of 14 Author: Mark Kudlowski Method 2 - Using the sine and cosine rules. Domain and Range of Sine and Cosine The domain of sine and cosine is all real. (Sine, Cosine, Secant, etc. The sine function takes an angle and tells the length of the y-component (rise) of that triangle. FIND the coordinates of A & C. The idea is to decompose any such function f(t) into an in nite sum, or series, of simpler functions. Hyperbolic functions show up in many real-life situations. We must now decide whether to use a sine function or a cosine function to get the phase shift. 11 (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e. Sine waves are very easy to produce electronically, and can be viewed on an oscilloscope, for. You decide to plug it into a sum and difference formula for sine. Use trigonometric functions to model and solve real-life problems. Graphs are shown in figure 4. SOH-CAH-TOA is a nice shortcut, but get a real understanding first! Gotcha: Remember Other Angles. It is analogous to a Taylor series, which represents functions as possibly infinite sums of monomial terms. 5b Translations of Sine and Cosine Curves Given the functions: y = a sin (bx \u2013 c) + d and y = a cos (bx \u2013 c) + d creates horizontal and vertical translations of the basic sine and cosine curves. The sine function is a set of ordered pairs of real numbers. In line 4 we use the properties of cosine (cos -x = cos x) and sine (sin -x = -sin x) to simplify the. Because angles are an intricate part of nature, sines, cosines and tangents are a few of the trigonometry functions ancient and modern architects use in their work. The period of tangent and cotangent are  \\pi . Like all functions, trigonometric functions can be transformed by changing properties like the period, midline, and amplitude of the function. Find the equation of the normal to the curve of y=tan^-1(x/2) at x=3. We must now decide whether to use a sine function or a cosine function to get the phase shift. 4 8 More Real World Problems C2. Signal Processing. 3 Trigonometry in the Real World C2. Real life scenario of trigonometry is studied in cbse class 10. Trigonometric formulae are useful for solving problems in two dimensions. 7 Inverse Trigonometric Functions 4. Here are the topics that She Loves Math covers, as expanded below: Basic Math, Pre-Algebra, Beginning Algebra, Intermediate Algebra, Advanced Algebra, Pre-Calculus, Trigonometry, and Calculus. The Fourier series synthesis equation creates a continuous periodic signal with a fundamental frequency, f, by adding scaled cosine and sine waves with frequencies: f, 2 f, 3 f, 4 f, etc. In this article, a sort of continuation, I will be discussing some applications of this formula. Unit 4 - Graphing & Writing Sine & Cosine Functions; Application Problems October 21 to November 5th, 2013 Date Topic Assignment Monday Gr 10/21 changes. functions to find a efficient navigational route for long trips if GPS or other radar equipment is unavailable. The book has as much to do with calculus as with philosophy. Plotting more points gives the full shape of the sine and cosine functions. 4 9 Review Day 10 Test Day TOTAL DAYS: 10 C2. The mathematical topics of Fourier series and Fourier transforms rely heavily on knowledge of. Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Pre-Cal 12 - Applications of Sinusoidal Functions New Project 5. FIND the coordinates of A & C. Part of the market-leading Graphing Approach Series by Larson, Hostetler, and Edwards, Precalculus Functions and Graphs: A Graphing Approach, 5/e, is an ideal student and instructor resource for courses that require the use of a graphing calculator. Substituting 0 for x, you find that cos x approaches 1 and sin x \u2212 3 approaches \u22123; hence, Example 2: Evaluate. Definition 1 is the simplest and most intuitive definition of the sine and cosine function. A real life example of the sine function could be a ferris wheel. ) Circles are an example of two sine waves. The Fourier Series, the founding principle behind the eld of Fourier Analysis, is an in nite expansion of a function in terms of sines and cosines. 3 The symmetries of the six trig functions Since the sine function is odd and the cosine function is even then tan( ) = sin( ) cos( ) = sin( ) cos( ) = tan( ) and so the tangent function is odd. At 90\u00b0 and 270\u00b0, x = 0 and therefore cos \u03c6 = 0, while at 0\u00b0 and 180\u00b0 y = 0 and therefore sin \u03c6 = 0. Return and go over Unit 2 Test. We have been studying how the equations of sinusoidal functions change when the graphs are shifted up and down and left and right and amplitude changes and period changes. In calculus, we are often. (See Example 3. The trigonometric functions sine and cosine have four important limit properties: You can use these properties to evaluate many limit problems involving the six basic trigonometric functions. GIVEN y x and B: 3cos2 0,3. The linear combination of a cosine and a sine of the same angle is an expression of the form: # a cos x + b sin x # That looks very much like the sum angle formula for sine or the difference angle formula for cosine:. Sine waves are a single note, whereas multiple sine waves make a chord. FREE (46) gnasher30 Uses of Trigonometry in Real Life. The book has as much to do with calculus as with philosophy. 3, P10Trig6. The PowerPoint is complete with custom animation so that pieces of the solving process are revealed, as well as the final answer. The Unit Circle is a tool used to find trigonometric functions. Basic Sine and Cosine Curves For 0\u2264 \u22642\ud835\udf0b, the sine function has its maximum point at. (Check your answer with your. The tangent function differs in its shape and range from its sine and cosine counterparts. 4 Part 2 : Applications of Trigonometric Functions Self Evaluation Solutions Modelling Real Life Situations Worksheet with key Review Worksheet and Review Answers. (See Example 5. Decide whether you will need Pythagoras theorem, sine, cosine or tangent. The dynamic used in calculation fit with maximum precision and overflow is. This equation is usually solved using sums of sines and cosines. Sine, cosine, and tangent functions are used for many real life applications. Sine and cosine are keys to the success of Fourier Transform because sound may be. 42 on the original plot we get: Looks as though we are almost there. ) Sine or cosine functions can used to the motion. Many real world situations can be modeled using the sine and cosine functions. It may not have direct applications in solving practical issues but used in the various field. The heat equation is used to model how things get hot (electronics, spacecraft, ovens, etc). In maths, you have real life applications on any thing that you study. 28-3 Centroids. Web browsers do not support MATLAB commands. We used the law of sines again, sin(B)/(b)=sin(C)/(c) and fill in everything we know, reduced it down, solve for (b) and give our answer in terms of miles. 5 Objective: To draw (using technology), sketch, and analyse the graphs of sine and cosine function whose periods are rational numbers. In this way you can find the size of any unknown angle of a right triangle if you just know 2 sides of. Sine and cosine functions are used primarily in physics and engineering to model oscillatory behavior, such as the motion of a pendulum or the current In an AC circuit, but these functions also arise m other sciences in this project, you will consider an application to biology - we use sine functions to model the populations of a predator and its prey. B = sin-1 (0. We closed the day with an application of what the applet taught us about sine to help us graph cosine functions with translations. What is the value. Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Pre-Cal 12 - Applications of Sinusoidal Functions New Project 5. Put the results in a table. The period of the function sin(x) is 2\u03c0. When the program returns, examine the stack for how many times the hyp sine was called and how many times hyp sine/cosine was called vs. For example, this transformed graph above would show which frequency sine and cosine functions to use to model our original function. Trigonometry Applications in Real Life. Applications of Trigonometric and Circular Functions 281 CHAPTER OBJECTIVES t -FBSOUIFNFBOJOHTPGBNQMJUVEF QFSJPE QIBTFEJTQMBDFNFOU Applications of Trigonometric and Circular Functions CAS Suggestions argument of the cosine. Stay safe and healthy. org are unblocked. The Sine Cosine Algorithms (SCA) has been recently proposed; it is a global optimization approach based on two trigonometric functions. The output of the transformation represents the image in the Fourier or frequency domain , while the input image is the spatial domain equivalent. In problems 12 & 13, the graphs of the sine and cosine functions are waveforms like the figure below. Area of a triangle 1. Hyperbolic functions show up in many real-life situations. & 3)&Theheight,\u210e,&in&meters,&of&the&tide&in&a&given&location&on&a&given&day&at&!&hours&after&midnight&can&be& modeled&using. I hope the following reasons and picture will help you understand why we study trigonometry. Describe that because they are functions, we need to be able to UNDO them (just like addition, subtraction, etc. In a right triangle, one angle is 90\u00ba and the side across from this angle is called the hypotenuse. m?X Law of Sines Substitute the given values. Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. Find and use reference angles to evaluate trigonometric functions. The hypotenuse is the longest side in a right triangle. The graphs of all sine and cosine functions are related to the graphs ofGOAL 2 Graph tangent y = sin x and y = cos xfunctions. Real-world Applications Using the Sine Function The table below shows the number of. Groupwork. The heat equation is used to model how things get hot (electronics, spacecraft, ovens, etc). Students will discover the area of triangles through the laws of sine and cosine. A1/2 ab sin c. Law of Cosine [Just formula. These can be confusing to understand and memorize however these for the basis of all trigonometric values and derivations. 1 Graphing Sine and Cosine Functions Focus on. Interpret the sine function as the relationship between the radian measure of an angle formed by the horizontal axis and a terminal ray on the unit circle and its y coordinate. PDF | In this paper, we propose a new hybrid algorithm called sine\u2013cosine crow search algorithm that inherits advantages of two recently developed | Find, read and cite all the research you. Basic functions in TI-83 Graphing Calculator. Take my hyperbolic sin/cos recursive function place the angle on a sine or cosine stack that represents a call to the sine or cosine. \" For example, \"an oscilloscope is an electronic instrument used to display changing electrical signals. 0333 170 -170 t e 0. On a graph together, they look like this: Tangent The third basic trigonometric function is called the tangent (tan for short), and it is defined as the ratio of the opposite and adjacent sides - that is: tan \u03b8 = y. By correctly labeling the coordinates of points A, B, and C, you will get the graph of the function given. Solution of triangles. In Chapter XI of The Age of Reason, the American revolutionary and Enlightenment thinker Thomas Paine wrote:. Time-saving lesson video on Sine and Cosine Functions with clear explanations and tons of step-by-step examples. 3 The symmetries of the six trig functions Since the sine function is odd and the cosine function is even then tan( ) = sin( ) cos( ) = sin( ) cos( ) = tan( ) and so the tangent function is odd. Find the period of a sine or cosine function. 5); \u2022 From sine and cosine functions, we obtain tangent easily: sin tan cos \u03b8 \u03b8 \u03b8 =; (3). Plotting a basic sine wave. Nov 2: Graphs of sine and cosine; functions of the form f(x) = Asin(Bx+C); their amplitude and period. Find an equation for a sine function that has amplitude of 4, a period of fl. We can relate sides and angles in an arbitrary triangle using two basic formulas known as the sine rule and the cosine rule. Essentially, if what is being measured relies on a sine or cosine wave. All we must do now is stretch the period of the sine function. By Victor Powell. Practice: General triangle word problems. This standard works in conjunction with the content standards. Real Life Examples. Use sigma notation. This is a powerpoint explaining some of the applications of trigonometry to answer that age old question 'But Miss, why would we need to know this?&' Conte. The sinusoidal functions provide a good approximation for describing a circuit's input and output behavior not only in electrical engineering but in many branches of science and engineering. Explore the graph of general sine functions interactively using an HTML 5 applet. (Sine, Cosine, Secant, etc. This is in comparison to a continuous function like a line. It is the application of sinusoidal modeling that makes its use in the secondary school environment so valuable. If working outside, choose a spot with two widely spaced (2-5 meters) and roughly parallel lines to define the \"river\" banks. Trigonometric functions have wide variety of applications in real life. Who is the Father of Trigonometry? Ans. By finding a few key points or aspects of the graph, any of the real-life problems we have today can be explained mathematically and much of the vibrations surrounding us can be better understood. In those tables, variables A and B are scalar constants, frequency \u03c9 is in radians/second, and variables \u03b1 and \u03b2 are phase angles measured. Thermal analysis. People board the ride at the ground (sinusoidal axis) and the highest and lowest heights you reach on the ride would be the amplitudes of the graph. 120 / 240 Vac sine Wave ac power distribution for residential application: The waveform of the electrical voltage distributed by the grid / the utility companies is like a sine wave. However, there are still disadvantages such as low solution accuracy and poor global search ability. introduction. Looks like MatPlotLib to me. Thefront panel of this instrument is 225 mm wide by 100 mm tall (8. However, in the real world all objects are three dimensional, so it is important that we extend the application of the area, sine and cosine formulae to three dimensional situations. The graphs of the sine and cosine functions are used to model wave motion and form the basis for applications ranging from tidal movement to signal processing which is fundamental in modern telecommunications and radio-astronomy. 2-5) Complete Notes pp. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval[latex]\\,\\left[-1,1\\right]. Hyperbolic Functions in Real Life. The exponential function, exp(X) or e^X, is a special function that comes from calculus. More Graphing Trigonometric Functions Worksheet Answers Sec 5. Possible Ideas Students will suggest:. The cosine rule: c2 + b2 \u20142abcosC. 5 -Graphs of Sine and Cosine Functions What You'll Learn: #89 - Sketch the graphs of basic sine and cosine functions. Essentially, if what is being measured relies on a sine or cosine wave. 28-3 Centroids. And I'll leave those details to the. Specific Objectives (measurable) Use and manipulate visual representations of unit circle and trigonometric functions. and \"use sine and cosine functions to model real-life data,\" iii. Some real life examples of periodic functions are the length of a day, voltage coming out of a wall socket and finding the depth of water at high or low tide. The general form of a sinusoidal is: \\[ f ( x ) = a \\sin ( bx \u2013 c ) + d, \\text{ for } b > 0$. study investigates the performances of these estimates. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. On a graph together, they look like this: Tangent The third basic trigonometric function is called the tangent (tan for short), and it is defined as the ratio of the opposite and adjacent sides - that is: tan \u03b8 = y. This lesson will present real world examples that involve inverse trigonometric ratios. 4 Trigonometric Functions of Any Angle 4. If either the imaginary or the real part of the input function is zero, this will result in a symmetric Fourier transform just as the even/odd symmetry does. Sine, Cosine, Tangent Applications. , surveying problems, resultant forces). docx from MATH 313B at K12. They are both expressed according to the triangle on the right, where each letter. At the end of. c) How many minutes, from t = 0, does it take the rider to reach the. Recall that the cosine function takes an angle x as input and returns the cosine of that angle as output: For example if 60\u00b0 is the input then 0. In these studies appear functions of the breast and cosine. Hyperbolic functions also satisfy identities analogous to those of the ordinary trigonometric functions and have important physical applications. Since the period of the sine function is 2, and the period of the temperature data is 52 weeks, we set b = 2 /52. ] Basic use of Geometer's Sketchpad. Find geometric and arithmetic formulas for sequences of numbers. \" As we point out and use functions in real-life settings, we can ask our students to keep alert for other input-output situations in the real world. More specifically, trigonometry is about right-angled triangles, where one of the internal angles is 90\u00b0. PPT TRIGONOMETRY in REAL LIFE - authorSTREAM Presentation. Trigonometry is not the work of any one person or nation. The function is assumed to be a function of time and the function values must be between -1 and 1. Crankshaft design (optimization of a function on a closed interval) Math 141 Projects, Spring 1998 1. When sound waves are produced by a musical instrument, they move in a reprising pattern and trigonometry functions such as cosine and sine can be used to represent them. In these trigonometry graphs, X-axis values of the angles are in radians, and on the y-axis its f (a), the value of the function at each given angle. The amplitudes of the cosine waves are held in the variables: a1, a2, a3, a3, etc. These include sine, cosine and tangent functions. &Assume&that&a& rider&enters&a&car&froma&platformthat&is&located&30\u00b0&around. The function has the same domain, range, and form as the sine function, but is offset by \u03c0/2 radians (90\u00b0). Fourier originally defined the Fourier series for real-valued functions of real arguments, and using the sine and cosine functions as the basis set for the decomposition. So our cosine curve will be shifted to the right by approximately `0. REAL LIFE APPLICATIONS AND REFLECTION; SUMMARY OF CONCEPT AND PROCESS - UNIT CIRCLE. There are many uses of sin,cos,tan in real life. REAL LIFE APPLICATIONS AND REFLECTION; SUMMARY OF CONCEPT AND PROCESS - UNIT CIRCLE. GIVEN y x and B: 3cos2 0,3. Identify sine, cosine, and tangent as trigonometric functions as well as relationships between side lengths in right triangles. #92 - Use sine and cosine functions to model real-life data. Learning resources you may be interested in. Our teacher asked up to come up with a \"real\" Example of Either a Sine Wave, or Cosine Wave. The student will evaluate trigonometric equations using trigonometric identities and special formulas. With the sine law. Pg 439 #1-8, 33, 35, 37, 43-55 odd \u00b7 Sketch the graphs of basic sine and cosine functions \u00b7 Use amplitude and period to sketch the graphs of sine and cosine functions \u00b7 Sketch translations of the graphs of sine and cosine functions. Many other Fourier-related transforms have since been defined, extending the initial idea to other applications. Real-life applications of. The next day, we continued by using the second applet (with some of the parameters changed) to walk us through the process of graphing sine functions with translations *and* changes to period/amplitude. Derived, but not applied. Applications of this branch of mathematics in real life are many and varied. Graphs are shown in figure 4. In other words, B measures the time it takes (how fast or slow) for our function to complete a full wave/cycle, as nicely stated by Khan Academy. Of the six functions in basic trigonometry, the sine, cosine and tangent are the most important to architecture because they allow the architect to easily find the opposite and adjacent values related to an angle or hypotenuse, translating a diagonal vector into horizontal and vertical vectors. The period of the function sin(x) is 2\u03c0. The dynamic used in calculation fit with maximum precision and overflow is. I need this last bit. In most encoders, this waveform is \"squared off\" inter-. For a low number of counts per revo-lution the waveform approaches a triangle while at higher counts the waveform becomes sinusoidal. This is how I like to introduce sine and cosine graphs this unit (after spending time with the unit circle and rotations it is a great way to see how we get the sinusoidal graph from a circle, see my blog post here for details ). In line 4 we use the properties of cosine (cos -x = cos x) and sine (sin -x = -sin x) to simplify the. The slope of a function Powers of sine and cosine; 3. Then ask learners to. Thermal analysis. basic sine and cosine functions Section 4. More specifically, trigonometry is about right-angled triangles, where one of the internal angles is 90\u00b0. 2a: Interpret the sine function as the relationship between the radian measure of an angle formed by the horizontal axis and a terminal ray on the unit circle and its y coordinate. Unit Five: Real World Problems Example: When Light shines through two narrow slits, a series of light and dark fringes appear. They are both expressed according to the triangle on the right, where each letter. Modeling Temperature Data Name(s): Since the trigonometric functions are periodic, they are a particularly useful tool when modeling cyclic behavior. Trigonometric Functions in Real Life There are many actions that complete some sort of regular cycle periodically and can be modeled by trigonometric functions. FREE (9) Popular paid. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. five bio-medical science dataset and one sine dataset problems. When I consider how to address the Precalculus objectives \u201cto solve real-life problems involving harmonic motion\u201d. 11 (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e. Domain, Range, and Period of the Sine Function. In this section, we explore transformations of the sine and cosine functions and use them to model real life situations. We will describe the numbered controls and their functions. You will see updates in your activity feed. As we know, in our maths book of 9th-10th class, there is a chapter named LOGARITHM is a very interesting chapter and its questions are some types that are required techniques to solve. Reply Pingback: Law of Cosines (Cont. circular functions. You and your friend had planned to walk to the movies together, but now it's five minutes to show time and no sign of him. Hyperbolic Functions in Real Life. Precalculus: An Investigation of Functions is a free, open textbook covering a two-quarter pre-calculus sequence including trigonometry. (See Example 3. Collapse menu 1 Analytic Geometry. This sheet describes the range, domain and period for each of the trig functions. My teacher gave me an example where in the sine curve was used in \"biorhythms\". At x = 0 degrees, sin x = 0 and cos x = 1. Introducing circular functions with. The graphs of all sine and cosine functions are related to the graphs of y =sinx and y=cos x which are shown below. 1 4 Sine Law C2. b: Apply the six trigonometric functions in relation to a right triangle to solve real-world applications and problems in mathematical settings. Real world uses of hyperbolic trigonometric functions. 1 Laboratory (Homework) Objective The objective of this laboratory is to learn basic trigonometric functions, conversion from rectangular to polar form, and vice-versa. Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. #92 - Use sine and cosine functions to model real-life data. We have already derived the derivatives of sine and. The relationship between trigonometric ratios; the circular functions sinx, cosx, and tanx; amplitude, their periodic nature, and their graphs. Line 1 just restates Euler\u2019s formula. Find area of a sector of a circle. 9 - 5 Applications to Navigation and Surveying Try the quiz at the bottom of the page! go to quiz We can use trigonometry to work with navigation problems as well as surveying problems. \u03b8 = 90\u00b0), which gives us cos \u03b8 = 0 and. The applications of Fourier transform are abased on the following properties of Fourier transform. 13 Check It Out! Example 2c Find the measure. Trigonometry plays a major role in musical theory and production. 2-5) Complete Notes pp. Angles: Real Life Applications of. Change from roots to rational exponents. 1 \u2013 solve problems, including those that arise from real-world applications (e. Understanding Calculus is a complete online introductory book that focuses on concepts. Title: Applications of Sine and Cosine Graphs Standard(s): MA3A3. 1 Laboratory (Homework) Objective The objective of this laboratory is to learn basic trigonometric functions, conversion from rectangular to polar form, and vice-versa. Similar statements can be made for the other trigonometric functions of sums and differences. An equation that can be used to model these data is of the form: y = A cos B(x - C) + D,. 5: Graphs of Sine & Cosine Functions KH\u00f6JLO (L Gra hs of the \"Parent\" Functions. A discrete function is a function where both the domain and range can be listed as distinct elements in a set. This provides a breathtaking example of how a simple idea involving geometry and ratio was abstracted and developed. Besides other fields of mathematics, trig is used in physics, engineering, and chemistry. Plotting a basic sine wave. Law of sines and cosines In most of the practical applications, related to trigonometry, we need to calculate the angles and sides of a scalene triangle and not a right triangle. This would be a Fourier series with only one term, and would return the desired function with the magnitude changed. Be able to split the limits in order to correctly find the area between a function and the x axis. If you can remember the graphs of the sine and cosine functions, you can use the identity above (that you need to learn anyway!) to make sure you get your asymptotes and x-intercepts in the right places when graphing the tangent function. The linear combination of a cosine and a sine of the same angle is an expression of the form: # a cos x + b sin x # That looks very much like the sum angle formula for sine or the difference angle formula for cosine:. Sin-1 is called \"inverse sine\" because it does the opposite of the sine function. sine and cosine Although sin(x) and cos(x) will create an n-petaled roses inscribed in the unit circle, what is the difference between them? The graph with the sine appears tangent to the positive x axis, while the cosine version has a petal centered at the positive x axis. Sine 5 None Up 2 9. In line 3 we plug in -x into Euler\u2019s formula. bc is the horizontal shift 4. real world examples of trigonometry Applications of Trigonometry in Real life. 15: Graphs of Sine and Cosine Functions. Many new applications from business, medicine, life and social sciences\u2014based on current real-world data. \u2013 Hipparchus was a Greek astronomer who lived between 190-120 B. As you know, our basic trig functions of cosine, sine, and tangent can be. Recall from Graphs of the Sine and Cosine Functions that the period of the sine function and the cosine function is $$2\u03c0$$. OVERVIEW The students will learn how to interpret and graph an inverse trig. This is exactly the reason why, when the graphs are plotted on grids as above, the cosine graph is equivalent to the sine graph, omitting the fact that it is shifted, or translated, 90\u00b0 to the left. Use special triangles to determine geometrically the values of sine, cosine, tangent for \u03c0/3, \u03c0/4 and \u03c0/6, and use the unit circle to express the values of sine, cosines, and tangent for x, \u03c0 + x, and 2\u03c0 \u2013 x in terms of their values for x, where x is any real number. We should now understand that any variable that is cyclical, harmonic, oscillating, or periodic in nature can be modeled graphically by a sine or cosine wave. You and your friends order a pizza. At the end of. Learn how to graph trigonometric functions and how to interpret those graphs. Angle in standard position. If you can remember the graphs of the sine and cosine functions, you can use the identity above (that you need to learn anyway!) to make sure you get your asymptotes and x-intercepts in the right places when graphing the tangent function. c) How many minutes, from t = 0, does it take the rider to reach the. I know for sure that pilots use the trig. These functions are most conveniently defined in terms of the exponential function, with sinh z = 1 / 2 (e z \u2212 e \u2212z) and cosh z = 1 / 2 (e z + e. Scroll down the page for examples and solutions. Sinusoidal functions graph wave forms. Date: 12/21/98 at 13:04:53 From: Doctor Santu Subject: Re: Trigonometry and music Dear Elizabeth, Certainly the functions sine and cosine have a connection to music. 2 to 5 Tuesday 10/22 Graphing Sine and Cosine functions cont'd. Applications of sinusoidal functions Description. Plotting a basic sine wave. Find the period of a sine or cosine function. View Notes - NOTESTrigonometry 3. Let's take a look at navigation. In addition, how do I know if this the graph of sine or cosine?. The midline is the average value. The applica-tions listed below represent a small sample of the applications. 1 4 Sine Law C2. 4 The Sine and Cosine Ratio Learning Goal: Determine the measures of the sides and angles in right triangles using the primary trigonometric ratios and the Pythagorean theorem; solve problems involving the measures of sides and angles in right triangles in real-life applications. (2) I can use amplitude and period to help sketch the graphs of sine and cosine functions. 3 sin 50 Use the inverse sine function to find m?X. Trigonometric Functions Arbitrary angles and the unit circle We\u2019ve used the unit circle to define the trigonometric functions for acute angles so far. Due to its exploration ability it has been applied to solve many real-life applications. Answers to the do-it-yourself problems are on the \"Other\" page as well as other helpful websites. In these trigonometry graphs, X-axis values of the angles are in radians, and on the y-axis its f (a), the value of the function at each given angle. The sine function just bounces back and forth (since you're basically tracing the outline of a circle). There is of course no point in simply copying this info here. The sine and cosine functions are among the most important functions in all of mathematics. \"The sine and cosine functions are defined for all real numbers, and these functions have many real-world applications. y= cos 2x SOLUTION a. So I differentiate the sine function twice and I get, I claim minus the sine function. By thinking of sine and cosine as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [\u22121,1]. We say that this sinusoidal has a vertical shift of 1. We then use the sine rule to find the side labelled a: sin12 40 sin16 a sin12. PPT TRIGONOMETRY in REAL LIFE - authorSTREAM Presentation. 0028 Open image in a new page Graph of e = 170 cos(120\u03c0t - \u03c0/3). The hypotenuse is the longest side in a right triangle. 2 to 5 Tuesday 10/22 Graphing Sine and Cosine functions cont'd. So that part, I hope, is clear. Just as with the sine function: \u2022 the domain for the cosine function is all real numbers; \u2022 the range of the cosine function is -1 \u2264 y \u2264 1; \u2022 the cosine function is periodic and will repeat this pattern over intervals of 2\u03c0. The Sine Ratio Passy's World of Mathematics. \u2022 Apply addition or subtraction identities for sine, cosine, and tangent. I know for sure that pilots use the trig. The main goal is to illustrate how this theorem can be used to evaluate various types of integrals of real valued functions of real variable. In this article, a sort of continuation, I will be discussing some applications of this formula. Analysis of beams in mechanics (polynomial integration and optimization of a function on a closed interval) 2. Law of Sine. As you see, $$y = 1 + \\sin x$$ merely raises the graph of sine one unit. The law of sines is a formula that helps you to find the measurement of a side or angle of any triangle. Double angle identities for sine and cosine. If it helps, you can draw a rough sketch to view this triangle, but this is optional. When we first learn the sine and cosine functions, we learn how to use them to find missing side-lengths & angles in right-angled triangles. The student is expected to: (A) use functions such as logarithmic, exponential, trigonometric, polynomial, etc. This would be a Fourier series with only one term, and would return the desired function with the magnitude changed. 2 to 5 Tuesday 10/22 Graphing Sine and Cosine functions cont'd. The sine and cosine graphs. In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. We check out triangle ABD and work out that angle ADB = 49\u00b0. and \"use sine and cosine functions to model real-life data,\" iii. Applications of this branch of mathematics in real life are many and varied. A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. 4)AFerris&wheel&has&a&diameter&of&20&mand&is&4&mabove&ground&level&at&its&lowest&point. graphs of sine and cosine functions. 42 on the original plot we get: Looks as though we are almost there. 03 Real-world Applications of Trigonometric Functions. He is considered the father of trigonometry. Let's take a look at navigation. Peggy and Raymond then present groups with several real-world situations to work with. 5 \u2013Graphs of Sine and Cosine Functions What You\u2019ll Learn: #89 - Sketch the graphs of basic sine and cosine functions. (sine or cosine) inputs. 6ep7yrwj7rao, o7e5upezd1u3u, gw9joobvhirz2sm, w5yw1hcyy3, hgc3hy119dy, x0bluc6ksm4p, nztz99cmdv1t, ihnhyyxs5u9, f6mhu9u964bou, z52phaeu284nuz, ue99gp4bhf, 3x0oekbpvdy, jm2sczb2mnj1, lr41lj4kntz, bknyyhwsu4v, xlgusjsmamqhbf, olijtkg53n, n0oq7e9mjj, lordvnua01, xat1ajnfbhtu1j, ct63zp7jwogh, odkeljziwabr, 2otnx4k3rxnfck8, b6d986skqf, sbxpdkzn2rrj, 1hpc8sts6g2up, k4lj0jtibfrei, 19ltt17gps", "date": "2020-05-25 20:54:06", "meta": {"domain": "salesianipinerolo.it", "url": "http://salesianipinerolo.it/ubdl/real-life-applications-of-sine-and-cosine-functions.html", "openwebmath_score": 0.754558265209198, "openwebmath_perplexity": 489.46033083010536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.990291521799369, "lm_q2_score": 0.8740772450055544, "lm_q1q2_score": 0.8655912851267503}}
{"url": "https://math.stackexchange.com/questions/2137223/solving-the-congruence-x2-equiv-4-mod-105-is-there-an-alternative-to-using/2137308", "text": "# Solving the congruence $x^2 \\equiv 4 \\mod 105$. Is there an alternative to using Chinese Remainder Theorem multiple times?\n\nI'm trying to solve $$x^2 \\equiv 4 \\mod 105.$$\n\nThis is of course equivalent to $$(x+2)(x-2) \\equiv 0 \\mod 105$$\n\nwhich is also equivalent to the system of congruences\n\n$$(x+2)(x-2) \\equiv 0 \\mod 3$$ $$(x+2)(x-2) \\equiv 0 \\mod 5$$ $$(x+2)(x-2) \\equiv 0 \\mod 7$$\n\nwhich have solutions of $$x \\equiv 1\\ \\text{or}\\ 2$$ $$x \\equiv 2\\ \\text{or}\\ 3,\\ \\text{and}$$ $$x \\equiv 2\\ \\text{or}\\ 5$$ respectively.\n\nNow, I could in principle take each combination of {$1,2$}, {$2,3$}, and {$2,5$} and use the Chinese Remainder Theorem to solve each system, but that seems incredibly tedious.\n\nIs there a simpler way?\n\nI'll note that this is a homework question, so it seems likely that there's a trick. Unfortunately I can't spot one.\n\n\u2022 no trick....................................At the same time, some of your writing suggests you don't know what happens. What are the values $\\pmod {15},$ just combine results for $3,5$ before going on \u2013\u00a0Will Jagy Feb 9 '17 at 21:14\n\u2022 I do understand that I can just use the values that I obtain $\\mod 15$ and then solve the system involving those values and the values $\\mod 7$. Is that what you're getting at? \u2013\u00a0Alex Feb 9 '17 at 21:19\n\u2022 sure. How many values $\\pmod {15},$ and what are they? \u2013\u00a0Will Jagy Feb 9 '17 at 21:21\n\nThere are a couple ways to optimize. First you need only compute half of the $8$ combinations since if $\\,x\\equiv (a,b,c)\\bmod (7,5,3)\\,$ then $\\,-x\\equiv (-a,-b,-c)\\bmod (7,5,3).\\$ Then use CRT to solve it for general $\\,a,b,c\\,$ to get $\\,x\\equiv 15a+21b-35c\\,$ Use that to compute those $4$ values. It's very easy\n\ne.g. note $\\ x\\equiv (2,\\color{#0a0}{-2},\\color{#c00}2)\\equiv 15(2)+21(\\color{#0a0}{-2})-35(\\color{#c00}2)\\equiv 23\\pmod{105}$\n\nNegating it $\\ (-2,\\color{#0a0}2,\\color{#c00}{-2})\\equiv -x\\equiv -23\\equiv 82\\pmod{105}.\\$ Of course $\\pm(2,2,2)\\equiv \\pm2.\\$\n\nDo as above for $\\,(-2,2,2),\\ (2,2,-2)\\$ to get the other$\\,4\\,$ solutions (a couple minutes work).\n\nRemark  There are also other ways we can exploit the negation symmetry on the solution space, i.e. if $x$ is a root so too is $-x$ since $\\,x^2\\equiv 2\\,\\Rightarrow\\, (-x)^2\\equiv x^2 \\equiv 2\\pmod{\\!105}.\\,$ Below is one such method, selected primarily because it reveals how to view Rob's answer in CRT language.\n\nBy CCRT = Constant case CRT: $\\,x\\equiv (2,2,2)\\pmod{\\!7,5,3}\\iff x\\equiv 2\\pmod{\\!105}.$ Its negation is $\\,(-2,-2,-2)\\,$ corresponding to $\\,-2\\pmod{\\!105}.\\,$ For other \"nontrivial\" solutions,  either $x$ or $-x$ has one entry $\\equiv -2$ and both others $\\equiv 2,\\,$ say $\\, x\\equiv (-2,2,2)\\pmod{p,q,r}.\\,$ Again by CCRT $\\,x\\equiv (2,2)\\pmod{q,r}\\iff x\\equiv 2\\pmod{qr}$ by $\\,q,r\\,$ coprime. So we reduce from $3$ to the $2$ congruences below. Solving them by Easy CRT, using $p$ coprime to $qr,\\,$ we get\n\n\\quad\\ \\ \\begin{align} x&\\equiv -2\\!\\pmod p\\\\ x&\\equiv\\ \\ \\,2\\!\\pmod{qr}\\end{align}\\!\\iff x\\equiv 2\\ +\\ qr\\left[\\,\\dfrac{-4}{qr}\\ \\bmod\\ p\\,\\right]\\pmod{pqr}\\,\n\n\\qquad \\qquad\\qquad\\qquad\\! \\begin{align} p=7\\,\\ \\Rightarrow\\,\\ x &\\equiv 2 + 3\\cdot 5(-4/(\\color{#c00}{3\\cdot 5)})\\bmod 7)\\equiv 2+3\\cdot 5(3)\\equiv 47\\\\[.2em] p=5\\,\\ \\Rightarrow\\,\\ x&\\equiv 2 + 3\\cdot 7(-4/(\\color{#c00}{3\\cdot 7}))\\bmod 5)\\equiv 2+3\\cdot 7(1)\\equiv 23\\\\[.2em] p=3\\,\\ \\Rightarrow\\,\\ x&\\equiv 2 + 5\\cdot 7{(-4/\\!\\!\\!\\underbrace{(\\color{#0a0}{5\\cdot 7})}_{\\large \\equiv\\ \\color{#c00}{1}\\ {\\rm or}\\ \\color{#0a0}{-1}}}\\!\\!\\!)\\bmod 3)\\equiv 2\\color{}{ +}5\\cdot 7(1)\\equiv 37\\\\ \\end{align}\n\nWe arranged the above to exploit easy inverses  (of $\\,\\color{#c00}{1}$ or $\\color{#0a0}{-1})\\,$ just as in the first solution (cf. my comment below). So $\\,47,23,37\\,$ and their negatives $\\,58,82,68\\,$ are all the nontrivial solutions.\n\nThe method in Rob's answer is essentially equivalent to the above (without the CRT language), except it doesn't take advantage of the easy inverses, instead solving the congruences by brute force (sometimes this may be quicker than general methods when the numbers are small enough).\n\nThere is also another CRT optimization used implicitly in Rob's answer. Namely a change of variables $\\ y = x\\!-\\!2\\,$ is performed to shift one of the congruences into the form $\\,y\\equiv 0,\\,$ which makes it easy to eliminate explicit use of CRT. We show how this works for the prior congruences, using the mod Distributive Law $\\,ca\\bmod cn =\\, c(a\\bmod n)\\quad\\qquad$\n\n$\\qquad qr\\mid x\\!-\\!2\\,\\ \\Rightarrow\\,\\ x\\!-\\!2\\bmod{pqr}\\, =\\, qr\\!\\!\\!\\!\\!\\!\\overbrace{\\left[\\dfrac{x\\!-\\!2}{qr}\\bmod p\\right]}^{\\large\\quad\\ \\ x\\ \\equiv\\ -2\\pmod{p}\\ \\ \\Rightarrow}\\!\\!\\!\\!\\!\\!\\! =\\, qr\\left[\\dfrac{-4}{qr}\\bmod p\\right]$\n\nThat's the same solution for $\\,x\\!-\\!2\\,$ that Easy CRT gave above. So the mod Distributive Law provides a \"shifty\" way to apply CRT in operational form - one that often proves handy.\n\n\u2022 Thanks. This seems like a bit more reasonable of an approach than just explicitly using CRT $8$ times. It's not entirely clear to me why we have $x \\equiv 15a+21b-35c$. I recognize that those are the three different ways we can multiply {$3,5,7$}, but why? \u2013\u00a0Alex Feb 10 '17 at 14:51\n\u2022 @Alex It follows by applying the general CRT formula, namely \\begin{align} x\\ &\\equiv\\ (a,b,c)\\!\\!\\pmod{7,5,3}\\\\[.3em] \\overset{\\rm CRT}\\iff x\\ &\\equiv\\ a(3\\cdot 5)\\left[\\dfrac{1}{3\\cdot 5}\\right]_7 + b(3\\cdot 7)\\left[\\dfrac{1}{3\\cdot 7}\\right]_5 + c(5\\cdot 7) \\left[\\dfrac{1}{5\\cdot 7}\\right]_3 \\pmod{105}\\\\[.3em] &\\equiv \\quad a\\,(15)\\,(1/1)_7\\ \\ +\\ \\ b\\,(21)\\,(1/1)_5\\ \\ +\\ \\ c\\,(35)\\,(1/(-1))_3\\\\[.3em] &\\equiv\\ \\ 15\\,a\\ +\\ 21\\,b\\ -\\ 35\\,c \\end{align} Note that the problem was designed to make this very easy, i.e. all the inverses are inverses of $\\pm1.$ $\\quad$ \u2013\u00a0Bill Dubuque Feb 10 '17 at 15:42\n\u2022 What you describe as a computation by \"brute force\" in my answer amounts to a quick pass over a list of $12$ natural numbers $n$ less than $100$ to decide whether $n\\pm4$ is a multiple of one of $3$, $5$, $7$. You seem to be doing much more calculation than I did. \u2013\u00a0Rob Arthan Feb 10 '17 at 23:05\n\u2022 @Rob First, \"brute force\" is a very common name for such exhaustive searches. It's not meant to denigrate such methods (which are often useful). Second, if you show all your work (as I do) then it will be of comparable length (if not more). Third, the above is efficient for any size numbers, but brute force searching is not. Finally, above I do not aim to optimize brevity, Rather, I aim to highlight the essence of the matter, from a more general viewpoint. \u2013\u00a0Bill Dubuque Feb 10 '17 at 23:14\n\u2022 @BillDubuque: first: the essence of this problem to me is \"how can a quadratic equation have more than two roots in a commutative ring?\". Second: tabulating the quick pass over a list of 12 small natural numbers is much quicker than any application of the CRT formula. \u2013\u00a0Rob Arthan Feb 10 '17 at 23:26\n\nHere's a method for avoiding CRT altogether (which comes from thinking about why there can be more than two roots to a quadratic in a commutative ring). If $x$ is any solution, we have $$(x - 2)(x + 2) \\equiv 0 \\bmod 105.$$ So in addition to the easy solutions $x = 2$ and $x = 103$, we get a solution $x = y + 2$ for any $y$ such that $$y(y + 4) \\equiv 0 \\bmod 105.$$ I.e., for any $y$ such that $y$ and $y + 4$ are a \"complementary\" pair of zero divisors in the ring $\\Bbb{Z}_{105}$. Any such pair has one of the following forms (possibly with the order of the factors reversed). \\begin{align*} 3m &\\cdot 35n & \\quad &\\mbox{with 1 \\le n \\le 2}\\\\ 15m &\\cdot 7n & \\quad &\\mbox{with 1 \\le m \\le 6}\\\\ 5m &\\cdot 21n & \\quad &\\mbox{with 1 \\le n \\le 4} \\end{align*} It doesn't take long to work through the $2+6+4$ possibilities for $y$ or $y + 4$ to get that $y(y+4)$ must be one of:\n\n$$21 \\cdot 25\\\\ 35 \\cdot 39 \\\\ 45 \\cdot 49\\\\ 56 \\cdot 60\\\\ 66 \\cdot 70\\\\ 80 \\cdot 84$$\n\nGiving 23, 37, 47, 58, 68 and 82 as the not-so-easy solutions.\n\n\u2022 As someone currently taking a ring theory class, I think this solution is really neat. Could you clarify why we have bounds on $n$ and $m$ in the third chunk of centred math? I don't exactly follow why a \"complementary pair of zero divisors\" must have one of those forms. It makes sense to me that you can split up the prime factors $3,5,7$ but I can't figure out why the $n$ and $m$ are necessary, and how they're chosen. \u2013\u00a0Alex Feb 9 '17 at 22:56\n\u2022 @Alex: I think you understand that to get a non-trivial solution to $ab=0$ in $\\Bbb{Z}_{105}$, you need to split the prime factors of $105$ between $a$ and $b$. Then you can multiply either $a$ or $b$ by any number you like, but as the equivalence class mod $105$ is all that matters, you only need to look at multiples between $0$ and $104$. My case analysis just picks the bigger of $a$ and $b$ to give the smallest number of cases. Then you just test whether the resulting multiple of $a$ (or $b$) $\\pm4$ is a multiple of the \"other\" factor. \u2013\u00a0Rob Arthan Feb 9 '17 at 23:09\n\u2022 @Alex Though CRT is not explicitly used above, in fact the method is a special case of general CRT optimizations, as I describe in the Remark in my answer. If one does many manual CRT calculations it proves handy to be familiar with these and related ideas. \u2013\u00a0Bill Dubuque Feb 10 '17 at 22:32", "date": "2019-12-13 10:50:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2137223/solving-the-congruence-x2-equiv-4-mod-105-is-there-an-alternative-to-using/2137308", "openwebmath_score": 0.8990423083305359, "openwebmath_perplexity": 532.8739708644362, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708045809529, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8655903564540146}}
{"url": "https://math.stackexchange.com/questions/3116473/selecting-2-cards-from-a-full-deck-of-cards", "text": "# Selecting $2$ cards from a full deck of cards.\n\nI thought of a problem earlier and I am quite clueless on how to solve it, or begin solving it, because I cannot find a way to easily compute the amount of combinations of $$2$$ cards that sum to a certain value $$x$$. Anyway, the first part of the problem is as follows\n\nWe have a full deck of $$52$$ cards, and randomly select $$2$$ cards from this deck. We look at the cards and compute the sum of the values of the cards, ace being $$1$$ and K, Q and J being $$13$$, $$12$$ and $$11$$ respectively. We then shuffle the cards back into the deck and randomly select $$2$$ cards once more. What is the probability that the sum of the value of these $$2$$ cards, is the same as the sum of the values of the first $$2$$ selected cards?\n\nThis brought me to think of another problem, which is comparable. It is as follows\n\nLet's say we have a full deck of $$52$$ cards, we randomly select $$2$$ cards, and we do this twice, yielding $$2$$ sets of $$2$$ cards. What is the probability that the sum of the values of the cards in the first set, equals the sum of the values of the cards in the second set?\n\nAgain, I'm quite confused about this problem, because I cannot think of an easy way to compute the amount of possible configurations of two cards, that sum to some value $$x$$.\n\nAny help on solving these problems is appreciated. Furthermore, what would be a good guesstimate for these probabilities that could be given without any computations?\n\n\u2022 Have you tried to work out the number of combinations for any specific sums? How many combinations add up to $2?$ What about $3?$ Try a few examples and look for pattern. \u2013\u00a0saulspatz Feb 17 at 17:34\n\u2022 I'm thinking a decent guesstimate is in the area of $1/338 = 1/(2 \\cdot 13^2)$. \u2013\u00a0Robert Shore Feb 17 at 17:35\n\u2022 @robjohn As a guesstimate or as a result of a computation? \u2013\u00a0S. Crim Feb 17 at 18:29\n\u2022 @S.Crim: It was a computation, but I had a replacement in there that shouldn't have been, so I need to compute again. \u2013\u00a0robjohn Feb 17 at 18:35\n\u2022 @saulspatz I've managed to solve the first problem, but am now stuck on the second problem because it is confusing to me what happens when the sum is unknown. \u2013\u00a0S. Crim Feb 17 at 18:36\n\nFor an interview, I'd try to compare the second scenario to the first.\n\nSay that you draw $$6$$ and $$3$$ the first time. If you drew from a separate deck, you'd have $$16$$ ways to make $$9,$$ but if you draw from the same deck, you only have $$9$$ ways. In the worst case you draw two aces, or two Kings, and the number of ways to succeed goes down from $$6$$ to $$1.$$ In the best case, you draw two $$7$$s and the number of ways goes down from 96 to 91. $$102$$ to $$97$$.\n\nOf course, the better cases, near the middle will occur more frequently than the worse cases, near the ends, so I would guess that the probability of success in the second case might be about $$\\frac34$$ of the probability of success in the first case.\n\nDrat! Now I'm going to have to work them out and see how close I came.\n\nEDIT\n\nIn response to the OP's that we don't have an explicit answer for the first problem either, in an interview, I would do the same kind of estimation. There are $${52\\choose2}={52\\cdot51\\over2}\\approx 1300$$ ways to chose two cards. For a sum of $$2$$ we must choose two Aces, and there are only $$6$$ ways to do that, so about half a percent chance of success. If the sum is $$14,$$ we have $$102$$ chances of success, so around $$8\\%.$$ Of course, the sums near $$14$$ will arise more frequently than the sums near $$2$$ or $$26$$ so they should be weighted more heavily. I'd guess around $$5$$ percent.\n\nWith one deck, the probability of success is $${328888\\over6497400}\\approx .05062$$ and with two decks, the probability of success is $${365392\\over7033104}\\approx.05195$$ I computed the exact probabilities with a script, and confirmed them through simulation.\n\n\u2022 Hah! That sounds good. However, for the first scenario, we cannot give an explicit probability, e.g. 60%, right? Because our probability depends on the outcome of the first sum, as you stated. So then how exactly would we compare the two situations in terms of probability of getting equal sums? \u2013\u00a0S. Crim Feb 17 at 20:30\n\u2022 @S.Crim I was assuming that assuming that you'd already somehow come up with answer for the first scenario. In one of your comments, you said \"I've managed to solve the first problem.\" I'll add some more to my answer. \u2013\u00a0saulspatz Feb 17 at 21:38\n\u2022 Yeah I solved the first problem to the point where, knowing the outcome of the sum, I can compute the probability of drawing this sum again after reshuffling the drawn pair back in. So I cannot compute some explicit formula or something for the probability yet. \u2013\u00a0S. Crim Feb 17 at 21:56\n\nHint:\n\n$$\\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \\hline + & \\color{red}{1} & \\color{red}{2} & \\color{red}{3} & \\color{red}{4} & \\color{red}{5} & \\color{red}{6} & \\color{red}{7} & \\color{red}{8} & \\color{red}{9} & \\color{red}{10} & \\color{red}{11} & \\color{red}{12} & \\color{red}{13} \\\\ \\hline \\color{red}{1} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\\\ \\hline \\color{red}{2} & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\\\ \\hline \\color{red}{3} & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\\\ \\hline \\color{red}{4} & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\\\ \\hline \\color{red}{5} & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\\\ \\hline \\color{red}{6} & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\\\ \\hline \\color{red}{7} & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\\\ \\hline \\color{red}{8} & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 \\\\ \\hline \\color{red}{9} & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\\\ \\hline \\color{red}{10} & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \\\\ \\hline \\color{red}{11} & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\\\ \\hline \\color{red}{12} & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 \\\\ \\hline \\color{red}{13} & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 \\\\ \\hline \\end{array}$$\n\nThe row of red numbers is the possible values for the 1st card, the column of red numbers is the possible values for the 2nd card. How many ways are there to get, say, a total sum of $$9$$?\n\n\u2022 I see, very helpful. This is probably a good tool for me to try and find the solution. Thanks alot! \u2013\u00a0S. Crim Feb 17 at 18:14\n\u2022 This allowed me to find the solution to first scenario, but I'm still stuck on the second one. Because if we do not know the value of the sum yet, then how can we compute the probability that the two sums are equal? I imagine we again have to look at possibilities or amounts of configurations but I'm not sure. \u2013\u00a0S. Crim Feb 17 at 18:20\n\u2022 Use the Law of Total Probability. \u2013\u00a0Graham Kemp Feb 18 at 22:46\n\nOn the problem you thought about\n\nWe select cards $$c_1, c_2, c_3, c_4$$ from the same deck; there are $$U:=52\\cdot 51\\cdot 50\\cdot 49= 6497400$$ possible cases, and we wish to count the cases for which $$N(c_1)+N(c_2)=N(c_3)+N(c_4) \\label{eq1}\\tag{1}.$$\n\nLet $$V$$ denote the count of cases that satisfy $$(\\ref{eq1})$$, and let us start by figuring out how many cases there are that satisfy $$N(c_1)+N(c_2) =n= N(c_3)+N(c_4),\\label{eq2}\\tag{2}$$ for some fixed $$n$$:\n\nAs robjohn says, there are $$W(n)=208-16\\,|n-14|-4\\,[\\,2\\mid n\\,]$$ ways we can select $$c_1, c_2$$ so that they sum to $$n$$. (The formula for $$W(n)$$ should be clear from the square provided by cansomeonehelpmeout.) Fix such a choice of $$c_1, c_2$$; if we want to continue selecting $$c_3, c_4$$ such that they sum to $$n$$ too, then they must be chosen from the remaining ways $$W(n)-Q,$$ where $$Q$$ is the number of ways to select two cards that sum to $$n$$ such that at least one of the two cards is either $$c_1$$ or $$c_2$$. When $$N(c_1)\\neq N(c_2)$$ (which is always true if $$n$$ is odd), contemplating on some specific cases, it does not take much effort to realize that $$Q$$ is always $$14$$. On the other hand, if $$n$$ is even, of all the choices we could have made for $$c_1, c_2$$, exactly $$12$$ of them have $$N(c_1)=N(c_2)$$, and for each one of these we have $$Q=10$$ and not $$14$$. Therefore, we let $$Q=14$$ and we add $$(14-10)\\cdot12$$ to the total when $$n$$ is even to compensate for the extras we have subtracted, and we get the number of cases satisfying $$(\\ref{eq2})$$ as $$W(n)(W(n)-14)+4\\cdot 12\\,[\\,2\\mid n\\,]. \\label{eq3}\\tag{3}$$\n\nAll we need to do now is to sum $$(\\ref{eq3})$$ over all possible $$n$$: $$V=\\sum_{n=2}^{26}\\left(W(n)(W(n)-14)+4\\cdot 12\\,[\\,2\\mid n\\,]\\right).$$ The probability that we get $$(\\ref{eq1})$$ will then be: $$\\frac{V}{U}.$$\n\nNote: The computation by hand is time consuming. I would use a program to calculate the sum and give a final answer, but the software I use is a very old version of Maple, and it is very cumbersome.\n\nWays to draw $$n$$ with $$2$$ cards: $$W(n)=\\overbrace{208-16\\,|n-14|}^{16\\,(13-|n-14|)}-4\\,[\\,2\\mid n\\,]$$ That is, there are $$13-|n-14|$$ ways to choose $$1\\le a,b\\le13$$ so that $$a+b=n$$. There are $$4\\cdot4=16$$ card choices for $$a$$ and $$b$$, except if $$2\\mid n$$ and $$a=b$$, when there are only $$4\\cdot3=12$$, so we subtract $$4$$ if $$2\\mid n$$. $$\\sum_{n=2}^{26}W(n)=2652$$ The probability of drawing $$n$$ is then $$\\frac{W(n)}{2652}$$ The probability of drawing the same thing twice in a row, replacing between draws, is then $$\\frac1{2652^2}\\sum_{n=2}^{26}W(n)^2=\\frac{22837}{439569}\\doteq0.051953$$\n\n\u2022 Thanks alot for the answer. Could you elaborate a little more on the formule you gave for $W(n)$? What does the $208$ mean, and what exactly does it mean that something only occurs when $2|n$? Perhaps I am just unfamiliar with notations. \u2013\u00a0S. Crim Feb 18 at 15:08\n\u2022 @S.Crim: I have hopefully made the explanation better. \u2013\u00a0robjohn Feb 19 at 8:05\n\u2022 Thanks alot, that indeed clarifies some things. \u2013\u00a0S. Crim Feb 19 at 14:45\n\u2022 @S.Crim: Is something still unclear? $[\\dots]$ are Iverson Brackets and $a\\mid b$ means that $a$ divides $b$; e.g. $[\\,2\\mid n\\,]$ is $1$ when $n$ is even and $0$ when $n$ is odd. \u2013\u00a0robjohn Feb 19 at 15:05\n\u2022 I think I get it now, yeah. Thanks for the clarifications on the notation. I was unfamiliar with the Iverson Brackets before this, and its been a while since I've worked with the $a|b$ notation. Thanks alot! \u2013\u00a0S. Crim Feb 19 at 17:45", "date": "2019-04-24 06:32:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3116473/selecting-2-cards-from-a-full-deck-of-cards", "openwebmath_score": 0.8519119620323181, "openwebmath_perplexity": 170.35556732043972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707999669627, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8655903523677196}}
{"url": "http://shey.madzuli.it/regular-polygon-questions.html", "text": "# Regular Polygon Questions\n\n24, 25 Homework Help Interior Angles of Regular Polygons Find the measure of an interior angle of the regular polygon. Polygon Questions and Answers - Math Discussion. The Area of a Trapezoid [1/27/1996] My daughter has been assigned to derive the formula for the area. Use your knowledge of the sums of the interior and exterior angles of a polygon to answer the following questions. 180(x-3) + 180 C. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Practice questions. 21-23 Example 4: Exs. Ask the class if they agree with where everyone placed their shapes or if they disagree to explain why they think a shape should be moved. Read each question carefully before you begin answering it. Use whatever is shorter in your language. Each interior angle of a regular polygon is the sum of the interior angles of the polygon divided by the number of sides. Is this a inductive reasoning or deductive re. Polygons are multi-sided shapes with different properties. Also, l == [1] * n (1 repeated n times) case is just the regular polygon untouched, which is a valid input. Poly- means \"many\" and -gon means \"angle\". Example: How many sides does a regular polygon have if one interior angle. Use Microsoft Paint to create a polygon picture. Name the quadrilaterals shown: rhombus, parallelogram, square, rectangle, or trapezoid. Polygons Questions and Answers - Math Discussion. You can only use the formula to find a single interior angle if the polygon is regular!. Buuuut then you had some questions about the quant section\u2014specifically question 13 of the second Quantitative section of Practice Test 1. Age 11 to 14 Short. Example of how to draw regular polygons using a \\foreach loop inside a path. Polygons are two-dimensional, closed, plane shapes composed of a finite number of straight sides that meet at points called vertices. Each student will receive an Area of Irregular Polygons Mini Lesson Examples. 360/x -10x Kudos for the right answer and explanation. Have students come up to the T-chart and place their shapes on the correct side. Enter the number of sides on the polygon. A polygon with the fewest number of sides is a triangle. The moral of this story- While you can use our formula to find the sum of. If you attempt to use geoshape or geotrace questions in forms authored prior to 3. Regular polygon. The problem concerns a polygon with twelve sides, so we will let n = 12. Return from the Polygon Game page to the Math Play homepage. 9 degrees are in the measure of an interior angle of a regular seven sided polygon. Regular Polygons Printable regular polygon sheet which includes picture and name of each of the 10 polygons from triangle, through to dodecagon. Read each question carefully before you begin answering it. Videos, worksheets, 5-a-day and much more. In this video, we'll talk about a few things you need to know about polygons for the GMAT, some of their basic names of regular polygons, and the sum of the interior angles of a polygon. Polygons can be regular or. Polygon diagonals. Find out which polygons are regular and which are not. Polygons problems with solutions for CAT exam. (The other option is to use shape=regular polygon from \\usetikzlibrary{decorations. 652 Responses. Download the set (5 Worksheets). The definition of a regular polygons is a polygon that is both equiangular and equilateral. Central angle = 20\u00b0 Supposing that the polygon is regular, that's all sides are of equal length and all central angles are equal. Formula to find the sum of interior angles of a n-sided polygon is = (n - 2) \u22c5 180 \u00b0 By using the formula, sum of the interior angles of the above polygon is = (9 - 2) \u22c5 180 \u00b0 = 7 \u22c5 180 \u00b0 = 126 0 \u00b0 Formula to find the measure of each interior angle of a n-sided regular polygon is. It is the point which is equidistant from each vertex. Unanswered Questions. Angles of Polygons Interior and Exterior Angles of a Polygon After all, there are only so many concepts that are \"fair game\" on the GMAT, so some questions will be designed to be more difficult by putting these concepts together in unexpected ways. Find the area of a regular hexagon with side length of 10 cm and apothem length of 8. You will have to read all the given answers and click over the correct answer. The number of these congruent triangles is the same as that of its sides. A regular polygon is a polygon that is both equiangular and equilateral. Quiz questions focus on the types of polygons and the. Vocabulary. A regular polygon is a closed, 2-dimensional figure consisting exclusively of line segments of equal length. 6-1 Polygons & Review 29 TEST Friday, 1/25/13 6-1: Properties and Attributes of Polygons I can name polygons with up to ten sides. Consider a random irregular convex polygon, for example, the 6-side polygon I want to define a function that, given a certain parameter r (roundness), rounds each. A regular polygon is a bounded, not self-intersecting polygon in which all edges have the same length and all interior angles are identically, say, $\\theta$. 3 of the tikz manual. Regular Polygons A regular polygon has lines that are all the same length and it also has all the same angles. This is part of our collection of Short Problems. Each interior angle of a regular polygon is the sum of the interior angles of the polygon divided by the number of sides. in a kite not all the sides are the same then its angles are not all equal then it is not a regular polygon. A regular polygon has all sides and angles equal (e. Area= (2)(apothem)(perimeter) is the formula for finding the area of a regular polygon. Try to get 100%. My goal is to build a map as this one, where the color represents the number of overlapping polygons, here from 0 to 6:. Level 1 Level 2 Level 3 Exam-Style Description Reminders More Angle Activities Fill in the table with the names and angle sizes of the first eight regular polygons. Practice questions. We can make \"pencilogons\" by aligning multiple, identical pencils end-of-tip to start-of-tip together without leaving any gaps, as shown above, so that the enclosed area forms a regular polygon (the example above left is an 8-pencilogon). THE NUMBER OF INTERSECTION POINTS MADE BY THE DIAGONALS OF A REGULAR POLYGON BJORN POONEN AND MICHAEL RUBINSTEIN Abstract. I can classify a polygon as concave or convex and regular or irregular. Matching Polygons Matching game for kindergarten, preschool and 1st grade. Includes a number of exercises for which solutions are in the slides. How to use polygon in a sentence. In the limit, a sequence of regular polygons with an increasing number of sides approximates a circle, if the perimeter or area is fixed, or a regular apeirogon (effectively a straight line. Attempt this quiz as many times as you like. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. Start studying True or False polygon questions. Have students come up to the T-chart and place their shapes on the correct side. How many vertices does the following polygon have? Gimme a Hint. You can start playing for free! Polygon Angles - Sample Math Practice Problems The math problems below can be generated by MathScore. all the space on a plane. For example if a polygon has 41 sides, it would be called a 41-gon. \u2026And how about the pentagon\u2026which kinda looks like a home plate. I also make them available for a student who wants to do focused independent study on a topic. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Last week we looked at questions on polygons inscribed in a circle. Convex or Concave Polygons A polygon is either convex or concave. the polygon is the same length and all of its angles have the same degree both equilateral and e Design a class named RegularPolygon that contains: A private int data field named n that defines the number of sides in the polygon. Both equilateral and equiangular conditions are required. Add ScratchPad. \u2020 Can any quadrilateral tile the plane? \u2020 Can any irregular polygon tile the plane?. Preview and details. Videos, worksheets, 5-a-day and much more. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Prepare a Presentation. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. The Overflow Blog How to develop a defensive plan for your open-source software project. Find the interior angle sum for each polygon. Now that we\u2019ve been through all of our polygon rules and formulas, let\u2019s look at a few different types of polygon questions you\u2019ll see on the SAT. Coloring Polygons Worksheets. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser 10. If you're seeing this message, it means we're having trouble loading external resources on our website. Find out which polygons are regular and which are not. Definition of Polygon explained with real life illustrated examples. org are unblocked. Read each question carefully before you begin answering it. A hexagon (six-sided polygon) can be divided into four triangles. Concave polygon. What is the value So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. The dodecagon would have sides of alternating lengths, with ratio $4:\\sqrt{12}$; as a specific size for the 30-60-90 triangle is not given. Being equilateral is not enough. Buuuut then you had some questions about the quant section\u2014specifically question 13 of the second Quantitative section of Practice Test 1. 30x+180/x D. Each Interior Angle of a Regular Polygon - MathHelp. Grade 5 Other Polygons CCSS: 5. A regular polygon is one that has equal sides. \u2020 Can any quadrilateral tile the plane? \u2020 Can any irregular polygon tile the plane?. A \"Polygon\" object contains the polygon outline. Examples of regular polygons include the equilateral triangle and square. Name the quadrilaterals shown: rhombus, parallelogram, square, rectangle, or trapezoid. User: Approximately how many degrees are in the measure of an interior angle of a regular seven sided polygon? 102. \ud83d\udc4dIf you like this resource, then please rate it and/or leave a comment\ud83d\udcac. Videos, worksheets, 5-a-day and much more. of sides = 360\u00b0/20\u00b0 = 18. We show a quadrilateral and a pentagon below. Regular polygon D. The vertices of an N-vertex polygon are located at the angles (2*Math. a square is a regular quadrilateral). Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. How to use polygon in a sentence. What is the probability that the center of the circle lies inside the. Instructions. Chapter 12. The Corbettmaths Practice Questions on Frequency polygons. The measure of an exterior angle of a convex regular polygon is 45. It is easy to show [39] that there are just five possibilities (vertex figure = triangle, square, tetrahedron, octahedron and cube) and each leads to a unique net. 30x+180/x D. A polygon is a shape that has any number of straight sides, such as a triangle, square or hexagon. Regular polygons have congruent edges and congruent. Central angle = 20\u00b0 Supposing that the polygon is regular, that's all sides are of equal length and all central angles are equal. I can classify a polygon as concave or convex and regular or irregular. provided a robust regular polygon detection method [4] [5], which derived the \u03b1 posteriori probability for a mixture of regular polygons and thus the probability density function. A Polygon is a closed figure made up of lines segments (not curves) in two-dimensions. So, if we know all the interior angles other than x, then we can find x. A hexagon has 6 sides. Return from the Polygon Game page to the Math Play homepage. There are 9 polygon maths worksheets. Length = Students who took this test also took : Parallelograms (8. (Must be 18 years old to sign up. Print full size. Area of Polygons Choose the best answer. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. Decagons may be irregular or regular. An octagon is any eight-sided polygon, and the sum of its angles is 1080\u00b0, as we saw above. If you are not sure about the answer then you can check the answer using Show Answer button. PRACTICE: Polygons \u2013 Assignment Worksheet Monday, 1/28/13 6-1: Properties and Attributes of Polygons. A minimum of three line segments are required for making a closed figure, thus a polygon with a minimum of three sides is known as Triangle. The problem concerns a polygon with twelve sides, so we will let n = 12. Selecting Objects Inside Polygon Is there a way within a source drawing to select all objects within a predetermined boundary/polygon? The regular select routine is limited to defining a polygon but not selecting an existing one. The Area of a Trapezoid [1/27/1996] My daughter has been assigned to derive the formula for the area. Similar polygons worksheet answers. Award Information. The simplest polygons are triangles (three sides), quadrilaterals (four sides), and pentagons (five sides). For this lesson, students will find the area of irregular polygons by separating it into smaller identifiable regular polygons. A regular polygon has equal length sides with equal angles between each side. You can now earn points by answering the unanswered questions listed. Instructions. Calculators may be used. Typical Polygon Questions. Irregular polygons are polygons that have unequal angles and unequal sides, as opposed to regular polygons which are polygons that have equal sides and equal angles. 18 sides 8. [1] 6) One interior angle of a regular polygon is 135\u00b0. An irregular polygons have sides of differing length and angles of differing measure. So, the above regular polygon has 9 sides. A \"Polygon\" object contains the polygon outline. Polygons are multi-sided shapes with different properties. In addition to the examples given by Karen: octagons (eight sides), nonagons (nine sides), decagons (ten sides), hendecagons (eleven sides), dodecagons (twelve sides), tridecagons (thirteen sides), tetradecagons (fourteen sides), pendecagons (fift. Do you really know everything about these 2-D polygons and. In these applets, a pentagon and a triangle will be used to illustrate this definition. Learn regular+polygon geometry with free interactive flashcards. How many diagonals does an octagon have?. Question: regular polygon Tags are words are used to describe and categorize your content. Identifying shape attributes, Understanding polygons. The word 'apothem' can also refer to the length of that line segment. Expressions are given for two side lengths. A regular polygon is always convex. Categories of polygons based on the number of sides, and whether they are concave or convex. Show Answer Example 3. Each page is designed with the latest design principles. So a rotation of 72, 144, 216, 288, 360,. Higher Tier. next solving the apothem of octagon. Sum of the interior angles of a polygon = (N - 2) x 180\u00b0 The number of diagonals in a polygon = 1/2 N(N-3) The number of triangles (when you draw all the diagonals from one vertex) in a polygon = (N - 2) Polygon Parts. Khan Academy is a 501(c)(3) nonprofit organization. Draw the lines of symmetry for each regular polygon, fill in the table including an expression for the number of lines of symmetry in a n-sided polygon. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A Lesson for Kindergarten, First, and Second Grade. In this worksheet, decide whether a polygon is regular or irregular. com, a math practice program for schools and individual families. Polygons are 2-dimensional shapes. All regular star polygons are non-convex polygons. If a regular polygon has x angles each measuring q degrees, then what is the value of q ? A. The Corbettmaths Practice Questions on Frequency polygons. For example, a triangle has 3 sides and 3 angles. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Like 0 Dislike 0 Reply Quote Follow. General Questions: 1. Common Core Standards: Grade 3 Geometry, Grade 4 Geometry. Polygon features One Click Demo Install, so demo content can be installed in minutes. Do you really know everything about these 2-D polygons and. MathScore EduFighter is one of the best math games on the Internet today. Videos, worksheets, 5-a-day and much more. First, take a look at the question itself: The hexagon above has interior angles whose measures are all equal. PRACTICE: Polygons \u2013 Assignment Worksheet Monday, 1/28/13 6-1: Properties and Attributes of Polygons. A triangle with all sides and angles equal is known as an equilateral triangle. The following diagram gives the formula to find the area of a regular polygon using the perimeter and the apothem. We also compute the number of regions formed by. I passed my intro programming class with a B because my prof gave us 20 min. 3 A polygon has 2 or more sides and angles. Examples: Regular: Not regular: More Geometry Subjects Circle Polygons Quadrilaterals Triangles Pythagorean Theorem Perimeter Slope Surface Area Volume of a Box or Cube Volume and Surface Area of a Sphere Volume and Surface Area of a. Use MathJax to format equations. Use your knowledge of the sums of the interior and exterior angles of a polygon to answer the following questions. 360/x -10x Kudos for the right answer and explanation. So if we know the sum of the interior angles, we just need to divide that sum by the number of sides to find out the measure of any angle of the polygon: 180(n-2)/n Exterior Angles. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. Convex polygon B. Do you really know everything about these 2-D polygons and. Properties of Decagon. 7) The area of a parallelogram with base of 10 ft and height of 2 ft is 20 square ft. Theorem: The area of a regular polygon. The vertices of an N-vertex polygon are located at the angles (2*Math. This quiz is incomplete! To play this quiz, please finish editing it. All sides are equal length placed around a common center so that all angles between sides are also equal. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Access the answers to hundreds of Polygons questions that are explained in a way that's easy for you to understand. In addition to the examples given by Karen: octagons (eight sides), nonagons (nine sides), decagons (ten sides), hendecagons (eleven sides), dodecagons (twelve sides), tridecagons (thirteen sides), tetradecagons (fourteen sides), pendecagons (fift. Many of these shapes, or polygons, can be described as flat closed figures with 3 or more sides. Hand touch global map form lines and polygonal. Here are your FREE materials for this lesson. See our pages on circles and. It is possible to do better than a hexagon, if an irregular polygon is acceptable. A regular polygon is one that has equal sides. A rectangle is a polygon. To know more about different data collection methods, and statistics download BYJU'S -The Learning App. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Polygons come in many shapes and sizes and you will have to know your way around them with confidence in order to ace those SAT questions on test day. Polygons come in many shapes and sizes and you will have to know them inside and out in order to take on the many different types of polygon questions the ACT has to offer. If the measures of a polygon's angles or side lengths differ, then the polygon is called an irregular polygon. Identifying and Describing Polygons All Classroom Lessons. Start by filling in the missing lengths of the sides. The sum of its angles will be 180\u00b0 \u00d7 4 = 720\u00b0 The sum of interior angles in a hexagon is 720\u00b0. Identifying shape attributes, Understanding polygons. Videos, worksheets, 5-a-day and much more. A polygon is a regular polygon if and only if all of its interior angles are equal and all of its sides are of equal length. In this polygon game you will classify different polygons based on their properties. Polygons are 2-dimensional shapes. Combine multiple words with dashes(-), and seperate tags with spaces. In this video I will take you through everything you need to know in order to answer basic questions about the angles of polygons. Irregular polygon C. The number of sides and the radius can be entered via a file, STDIN, or just a plain old variable. 5 (Geometry: n-sided regular polygon) An n-sided regular polygon's sides all have (i. This calculator is designed to give the angles of any regular polygon. Please leave workings so I understand how you get the answer. Let's say you have pixel bitmaps that look something like this: From this I can easily extract a contour, which will be a concave polygon defined by a set of 2D points. Make flat shapes using toothpicks and clay. Before we get too caught up on the excitement of quadrilaterals, let's take time to learn the names and basic properties of different polygons. Updated: Aug 5, 2019. - [Voiceover] So ,a few things\u2026that you should know about polygons,\u2026let's start with the most basic polygon\u2026which is our friend the triangle. Let\u2019s first find the measure of each one of those angles. all of these. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180\u00b0 \u00d7 (n - 2) degrees. As the concept of irregular polygons is extremely general, knowledge about this concept can be very valuable, both for problem solving and for simple problems in the real world. The simplest regular polygon is the equilateral triangle, which consists of three edges of equal length and three angles between each pair of edges to be 60 degrees. Consider, for instance, the ir regular pentagon below. How many sides does an octagon have? eight 2. Creating Polygons - investigating the Concept of Triangle and the Properties of Polygons ( from NCTM ) ; Figures and Polygons - definitions and examples of a large number of figures fiting the definition of polygon ; Identify Polygons - names of polygons by number of. For the regular polygons inscribed in the unit circle, what is the range of values that the apothem can have?. Higher Tier. Please leave workings so I understand how you get the answer. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. shapes}; see section 51. A polygon is regular, if all its sides have the same length and all its angles have the same measure. , AB, BC and CA are equal and there are three angles i. Abstract polygon. If a polygon is regular, there is a point which we can define as the center. Then ask guiding questions to see if they recognize the pattern (every time you add a side, the interior angle sum increases by 180\u00b0). Our online polygon trivia quizzes can be adapted to suit your requirements for taking some of the top polygon quizzes. A regular three-sided polygon is called an equilateral triangle. An Apothem is a line segment from center point of side of a regular polygon to the midpoint of the regular polygon. Categories & Grades. 1) heptagon 2) decagon 3) nonagon 4) hexagon 5) pentagon 6) nonagon 7) hexagon 8) nonagon State if each polygon is concave or convex. \u2020 Can any quadrilateral tile the plane? \u2020 Can any irregular polygon tile the plane?. A square is a rhombus. \" Knowing the names of the polygons is only the start. Feel free to always add Asymptote solution to any questions related to plotting, drawing, diagramming, etc because it is allowed by law in this site. Indeed, it is clear, that under the given conditions, a regular triangle is the only possible regular polygon with the smaller interior angle. Each student will receive an Area of Irregular Polygons Mini Lesson Examples. Choose from the following regular polygons: Triangle,. Test comprehension with the questions that follow. A regular polygon has sides which are all the same length as well as angles which are all the same. Regular Polygon. Browse other questions tagged tikz-pgf or ask your own question. A regular polygon is one that has equal sides. Polygons can be regular or. Khan Academy is a 501(c)(3) nonprofit organization. Abstract background with polygonal shapes. 25 Mocks + 30 Sectionals for Rs. 7) The area of a parallelogram with base of 10 ft and height of 2 ft is 20 square ft. You can tell, just by looking at the picture, that $$\\angle A and \\angle B$$ are not congruent. The shape on the left has equal sides and angles, so it is a regular polygon. \ud83d\udc4dIf you like this resource, then please rate it and/or leave a comment\ud83d\udcac. Questions for Students. A regular polygon is a closed, 2-dimensional figure consisting exclusively of line segments of equal length. Polygon abstract background. Exterior angle = 180-178 = 2\u00b0 because the interior and exterior angle together make a straight line. If a regular polygon has x sides, then the degree measure of each exterior angle is 360 divided by x. The sum of the exterior angles of any polygon is 360\u00b0. If none of the sides, when extended, intersects the polygon, it is a convex polygon; otherwise it is concave. What is the latest version of SketchUp that Keyframe Animation will work with?. In this polygon game you will classify different polygons based on their properties. The applets below illustrate what it means for any polygon to be classified as a regular polygon. 9 degrees are in the measure of an interior angle of a regular seven sided polygon. A Polygon is a 2D shape which is made up of straight line segments. In our example, it's equal to 5 in. Start studying Quadrilateral & Polygon Unit Review Questions. Fortunately, as you'll see in the following practice questions, there's a handy formula that you can use to find a missing interior angle in a [\u2026]. We can make \"pencilogons\" by aligning multiple, identical pencils end-of-tip to start-of-tip together without leaving any gaps, as shown above, so that the enclosed area forms a regular polygon (the example above left is an 8-pencilogon). Convex polygon B. The resulting polygon is guaranteed to be simple (it does not intersect or touch itself). It is the point which is equidistant from each vertex. Upgrade your subscription to get access to this quiz, more lessons, and more practice questions. polygon: a 2D or flat, closed shape with straight sides. A regular polygon is a polygon whose sides are equal length and whose angles are all the same value. Polygons are 2-dimensional shapes with straight sides. Area= (2)(apothem)(perimeter) is the formula for finding the area of a regular polygon. Work out the identity of the missing regular. Tutors Answer Your Questions about Polygons (FREE) Get help from our free tutors ===> Algebra. Name the quadrilaterals shown: rhombus, parallelogram, square, rectangle, or trapezoid. Upgrade your subscription to get access to this quiz, more lessons, and more practice questions. Here we look at Regular Polygons only. I can classify a polygon as concave or convex and regular or irregular. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180\u00b0 \u00d7 (n - 2) degrees. To do so, I would like to construct an irregular polygon, and from each of it's sides, construct a *regular* n-gon, on the outside of the irregular-gon. Sample Problem 4: Draw a figure that fits the description. Selecting Objects Inside Polygon Is there a way within a source drawing to select all objects within a predetermined boundary/polygon? The regular select routine is limited to defining a polygon but not selecting an existing one. This quiz will focus on different shapes classified as polygons. t = 360 o / 12 = 30 o; We now use the formula for the area when the side of the regular polygon is known Area = (1 / 4) n x 2 cot (180 o / n) Set n = 12 and x = 6 mm area = (1 / 4) (12) (6 mm) 2 cot (180 o / 12). One interior angle of a regular polygon - (n - 2). Diagonals of a Regular Octagon. Can be \u201cregular\u201d \u2013 all sides and all angles are equal to each other Two sides of equal length Three acute angles Sum of angles = 180\u00b0 All sides equal length Three acute angles Sum of angles = 180\u00b0 Is a regular polygon No sides are equal No angles are equal May have obtuse angle Sum of angles = 180 \u00b0 Opposite sides are parallel Opposite. A polygon is a shape that has any number of straight sides, such as a triangle, square or hexagon. Polygon or Not? This Is the Question : In this game, students will quickly decide whether a geometrical figure is a polygon or not a polygon. Round your answer to the nearest tenth if necessary. For the Independent Practice students will use the area formulas to help them complete the Finding the Finding the Area of Regular Polygons Worksheet. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Mathematics (Linear) - 1MA0 ANGLES: POLYGONS Answer the questions in the spaces provided - there may be more space than you need. A clock is constructed using a regular polygon with 60 sides the polygon rotates every minute how has the polygon rotated after 7 minutes. Using the area of regular polygon calculator: an example. Videos, worksheets, 5-a-day and much more. The angles have been calculated manually. Polygons are multi-sided shapes with different properties. Videos, worksheets, 5-a-day and much more. Solve these reasoning questions. 2019 23:53, kumarivishakha118. Always show your workings. A three-sided polygon is called a triangle. A regular polygon is both equilateral and equiangular. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. The formula is (n - 2) times 180 divided by n , where n is the number of sides. Find the area of each of the following regular polygons. The outline itself is matched with the InPolygon rule (which is what you want). Area of Polygons. parallel lines: two lines that are always the same distance apart and never touch. They should not be confused with regular polygons. Is this a inductive reasoning or deductive re. A \"Regular Polygon\" has: all sides equal and. The only constructible regular polygons with an odd number of sides are those for which this number is a product of distinct Fermat primes (so for instance 15 = 3 times 5, 51 = 3 times 17), and the only ones with an even number of sides are those obtained by repeatedly doubling these numbers (including 1), thus:- (1,2), 4, 8, 16, 32, 64,. true false. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180\u00b0 \u00d7 (n - 2) degrees. (Must be 18 years old to sign up. 1) 2) 3) regular 19-gon 4) regular 14-gon Find the measure of one exterior angle in each regular polygon. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. The area of the regular polygon is 297 m 2. 2D SHAPES > REVISION > GCSE QUESTIONS. If someone can help me out that would be awesome. Videos, worksheets, 5-a-day and much more. Unit 7 Test Polygons And Quadrilaterals Answer Key. Unanswered Questions. A brief description of the worksheets is on each of the worksheet widgets. A \"Polygon\" object contains the polygon outline. A diagonal of a polygon is any line segment that connects non-consecutive vertices of the polygon. Expressions are given for two side lengths. Award Information. Menu Skip to content. [1] 6) One interior angle of a regular polygon is 135\u00b0. The measure of an exterior angle of a convex regular polygon is 45. A polygon is regular, if all its sides have the same length and all its angles have the same measure. This figure shows the most common polygons. Improve your math knowledge with free questions in \"Interior angles of polygons\" and thousands of other math skills. Geometry Polygons questions and answers for CAT. 652 Responses. A polygon is irregular, if it is not regular. Hint: 360/No. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. I will be focusing on convex regular. Download the set (5 Worksheets). TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. You will have to read all the given answers and click over the correct answer. An n-sided regular polygon is rotationally symmetric around its centre. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. Tags: Question 40. A regular polygon is a polygon all of whose sides are of the same length (equilateral) and all of whose internal angles are the same (equiangular). A polygon that is not a regular polygon is called an irregular polygon. Classify the polygon by the number of sides. Chapter 12. all angles equal. As a child, there's a point in life where shapes and what we now have come to refer as \"geometry\" are the most fascinating things in the world. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. ) Sign Up More Info Close. disproportional perimeters. As an example: A pentagon has five sides, 360/5 =72. This figure shows the most common polygons. If you are not sure about the answer then you can check the answer using Show Answer button. Type all answers in the boxes provided. Polygons Exercises ; Topics Which of the following look like regular polygons? Gimme a Hint. The size of each of the interior angle of a regular polygon = (n-2) x 180\u00ba \u00f7 n Where n is the number of sides. You Try It! How many sides does a regular polygon have if each interior angle measures 160 ? Example 5: Solving Algebraic Problems Find the value of x. I can classify a polygon as concave or convex and regular or irregular. (Construct it from scratch, or use a script. You will have to read all the given answers and click over the correct answer. We can see from the above examples that the number of triangles in a polygon is always two less than the number of sides of the polygon. A brief description of the worksheets is on each of the worksheet widgets. Polygons Author: act Last modified by: act Created Date: 7/10/2008 4:49:14 PM Document presentation format: On-screen Show (4:3) Other titles: Corbel Arial Wingdings 2 Wingdings Wingdings 3 Calibri Module 1_Module 2_Module 3_Module 4_Module 5_Module 6_Module Polygons What is a polygon?. An octagon is any eight-sided polygon, and the sum of its angles is 1080\u00b0, as we saw above. You can tell, just by looking at the picture, that $$\\angle A and \\angle B$$ are not congruent. PI*K)/N where K goes from 0 to N-1, inclusive. The sum of the interior angles, in degrees, of a regular polygon is given by the formula 180(n \u2013 2), where n is the number of sides. Main Ideas/Questions POLYGON Notes/Examples A polygon isa figure formed by three or more line Date: Class: hexagon? EXAMPLES 9. 180(x-3)/x B. Formula for the sum of interior angles. A convex polygon has no angles pointing inwards. s = (160cm)/8 = 20 cm. Two congruent figures have equal areas. Print full size. Examples of regular polygons include the equilateral triangle and square. 652 Responses. What is the latest version of SketchUp that Keyframe Animation will work with?. To find the perimeter of a polygon, add the lengths of its sides. If none of the sides, when extended, intersects the polygon, it is a convex polygon; otherwise it is concave. This is part of our collection of Short Problems. Formula to find the sum of interior angles of a n-sided polygon is = (n - 2) \u22c5 180 \u00b0 By using the formula, sum of the interior angles of the above polygon is = (9 - 2) \u22c5 180 \u00b0 = 7 \u22c5 180 \u00b0 = 126 0 \u00b0 Formula to find the measure of each interior angle of a n-sided regular polygon is. There are multiple ways to found the Apothem of a Regular Polygon. I can find the measure of an interior angle of any regular polygon. Background polygon. You can now earn points by answering the unanswered questions listed. Background polygon. The applets below illustrate what it means for any polygon to be classified as a regular polygon. A polygon is irregular, if it is not regular. Along with it, the property of equal-length sides, this implies that every regular polygon also has an inscribed circle or incircle that is tangent to every side at the midpoint. (Construct it from scratch, or use a script. org are unblocked. Diagonals of a Regular Octagon. The vertices of an N-vertex polygon are located at the angles (2*Math. Identifying and Describing Polygons All Classroom Lessons. The sum of the exterior angles of any polygon is 360 degrees. What is the minimum interior angle possible for a regular polygon or why. A worksheet on angles in polygons. A Ogee curve is a _____ a) semi ellipse b) continuous double curve with convex and concave c) freehand curve which connects two parallel lines d) semi hyperbola View Answer. A polygon is a plane shape (two-dimensional) with straight sides. Tell whether the polygon is equilateral, equiangular, or regular. 2nd and 3rd Grades. Interact with these applets for a minute or two, then answer the questions that follow. of sides = Interior angle Attempt drawing a pentagon on your own first. A convex polygon has no angles pointing inwards. Concave polygon. A regular polygon is a polygon that is both equilateral and equiangular (having both. Polygon abstract background. All photos are included but can be easily changed to match your branding\u2019s style and personality. Assume that the regular polygon has n sides (or angles). A regular polygon is a polygon all of whose sides are of the same length (equilateral) and all of whose internal angles are the same (equiangular). Solution to Problem 2: Let t be the size of angle AOB, hence t = 360 o / 5 = 72 o; The polygon is regular and OA = OB. Add ScratchPad. For example if a polygon has 41 sides, it would be called a 41-gon. The Corbettmaths Practice Questions on Frequency polygons. \u2026And how about the pentagon\u2026which kinda looks like a home plate. Each question will change subtly every time you take. The number of sides is also the number of angles and, in a regular polygon, all the angles are the same. Expressions are given for two side lengths. (Remade and reuploaded 9th February 2020) Calculating interior and exterior angles of polygons. Making statements based on opinion; back them up with references or personal experience. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. Find the area of the regular polygon. Sample Problem 5: Each figure is a regular polygon. of sides = 360\u00b0/20\u00b0 = 18. So, if we know all the interior angles other than x, then we can find x. polygons Questions #1 Relates the number of sides, angles, and vertices of a polygon Question #2 Describes attributes of polygons Question #3 Wow! Thoroughly describes. Shape And Formula Of A Polygon Quiz 10 Questions | By Swords31535 | Last updated: Oct 11, 2017 | Total Attempts: 9314 All questions 5 questions 6 questions 7 questions 8 questions 9 questions 10 questions. You may also be interested in our longer problems on Angles, Polygons and Geometrical Proof Age 11-14 and Age 14-16. Regular Polygons Printable regular polygon sheet which includes picture and name of each of the 10 polygons from triangle, through to dodecagon. A polygon is a flat shape with straight sides that are joined to form a closed shape. You will have to read all the given answers and click over the correct answer. 1) 2) 3) regular 19-gon 4) regular 14-gon Find the measure of one exterior angle in each regular polygon. Although students should work independently, they may want to discuss or ask questions of their group. An n-sided regular polygon is rotationally symmetric around its centre. A regular polygon is always convex. I can solve reasoning questions about comparing, classifying and finding unknown angles in polygons. no space at all. Design a class named RegularPolygon that contains: Aprivate int data field named n that defines the number of sides in the polygon with default value 3. where n - number of sides of octagon. Irregular polygons are polygons that have unequal angles and unequal sides, as opposed to regular polygons which are polygons that have equal sides and equal angles. How many angles does a triangle have? three 3. While we are given two sides - the base and the height - we do not know the hypotenuse. com Regular Polygons Grade 3 Geometry Worksheet Answers: Color the regular quadrilateral. Questions and Facts \u2020 What are the only edge-to-edge regular tilings? That is, what regular polygons allow for edge-to-edge tiling of the plane? \u2020 Can we use more than one regular polygon, with different number of size, to tile the plane edge-to-edge? \u2020 Any triangle can tile the plane. Here we look at Regular Polygons only. Improve your math knowledge with free questions in \"Regular and irregular polygons\" and thousands of other math skills. Note that the polygon we are given is not a regular polygon, since the side lengths are not all equal. In a regular octagon, each angle = 1080\u00b0/8 = 135\u00b0 That angle is the supplement of a 45\u00b0 angle. To find the perimeter, multiply the length of one side by the total number of sides. In this video, we'll talk about a few things you need to know about polygons for the GMAT, some of their basic names of regular polygons, and the sum of the interior angles of a polygon. Draw all of the diagonals in each regular polygon. Two congruent figures have equal areas. angles of polygons section 8-1 jim smith jchs names of polygons sides triangle 3 quadrilateral 4 pentagon 5 hexagon 6 heptagon 7 octagon 8 nonagon 9 decagon 10 dodecagon 12 n - gon n interior angle sum each triangle has 180\u00b0 if n is the number of sides then: int angle sum = (n - 2 ) 180\u00b0 angles of polygons section 8-1 names of polygons sides triangle 3 quadrilateral 4 pentagon 5 hexagon. Restate for the class a clear definition of polygon: a 2D or flat, closed shape with straight sides. Formula for the sum of interior angles. The moral of this story- While you can use our formula to find the sum of. Polygon features One Click Demo Install, so demo content can be installed in minutes. I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. There can't be a better way to introduce polygons, than what kids love the most - coloring! Direct kids to follow the color key provided to identify polygons like triangles, quadrilaterals, pentagons and more. A regular polygon has an exterior angle of 20 degrees. com For a complete lesson on regular polygons, go to https://www. The simplest regular polygon is the equilateral triangle, which consists of three edges of equal length and three angles between each pair of edges to be 60 degrees. Find the area of each regular polygon. The shape on the left has equal sides and angles, so it is a regular polygon. ) The area formula for a regular polygon is 1/2ap. $$\\therefore$$ the area of one such triangle $$= \\frac{1}{2} \\times$$ sides of polygon $$\\times$$ length of perpendicular from the center to any side of. 2D SHAPES > REVISION > GCSE QUESTIONS. The vertical coordinate can be calculated as a sine of the angle times the radius of the circumcircle; the horizontal coordinate is calculated the same way, except you need to multiply the radius by the cosine of the angle. In this regular polygon lesson plan, 10th graders find the area of a regular polygon in terms of the apothem and the perimeter. Regular Polygon. The regular polygons were known to the ancient Greeks, and the pentagram, a non-convex regular polygon (star polygon), appears on the vase of Aristophonus, Caere, dated to the 7th century B. Using the area of regular polygon calculator: an example. t = 360 o / 12 = 30 o; We now use the formula for the area when the side of the regular polygon is known Area = (1 / 4) n x 2 cot (180 o / n) Set n = 12 and x = 6 mm area = (1 / 4) (12) (6 mm) 2 cot (180 o / 12). The sum of the exterior angles of any polygon is 360 degrees. A Polygon is a 2D shape which is made up of straight line segments. Browse other questions tagged qgis polygon intersection or ask your own question. Polygons maths worksheet 3 accurately completes the polygons. 6 9 52) 14 14. A regular polygon has sides of equal length, and all its interior angles are of equal size. A regular polygon have the number of lines of symmetry equal to the number of sides of the regular polygon. The dodecagon would have sides of alternating lengths, with ratio $4:\\sqrt{12}$; as a specific size for the 30-60-90 triangle is not given. Label the sides and angles on each polygon. Sum of interior angles = 180 (n \u2013 2) where n = the number of sides in the polygon. A polygon is a 2D shape with at least three sides. A regular polygon has sides which are all the same length as well as angles which are all the same. a positive number representing the interior space of a polygon. all of these. Categories & Grades. Kahoot! is a free game-based learning platform that makes it fun to learn - any subject, in any language, on any device, for all ages! Unit 7 test polygons and quadrilaterals answer key. 25 Mocks + 30 Sectionals for Rs. If a regular polygon has x angles each measuring q degrees, then what is the value of q ? A. In geometry, polygons cover a lot of ground, so you can bet that some questions on the ACT Math exam will involve polygons\u2014specifically, finding the interior angles of a polygon. This calculator is designed to give the angles of any regular polygon. Always show your workings. White neural texture abstract vector. Khan Academy is a 501(c)(3) nonprofit organization. Let's say you have pixel bitmaps that look something like this: From this I can easily extract a contour, which will be a concave polygon defined by a set of 2D points. of sides = 360\u00b0/20\u00b0 = 18. The sum of the exterior angles of any polygon is 360\u00b0. An irregular polygons have sides of differing length and angles of differing measure. Find the value of. The following figures are polygons. Find the length of one side. Support your method with examples. This quiz is incomplete! To play this quiz, please finish editing it. Updated: Aug 5, 2019. (n \u2013 2)*180 = sum of all interior angles (6 \u2013 2)*180 = 720. There are various ways to solve this question, but each takes a bit of effort. For a polygon to be 'regular' it must have. Each angle in a regular decagon is equal to 144\u00b0. How to find the interior angle of a regular polygon. Polygons are two-dimensional objects, not solids. tells you the sum of the interior angles of a polygon, where n represents the number of sides. Polygons are 2-dimensional shapes. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. x1, x2 ,y1 ,y2. com for more resources, like predicted GCSE Maths Papers, Topic Buster and challenge yourself against our Demon questions. The formula. This quiz will focus on different shapes classified as polygons. A triangle is a 3 sided polygon with interior angles that add to 180 degrees. In these applets, a pentagon and a triangle will be used to illustrate this definition. Question 1. Identifying and Describing Polygons All Classroom Lessons. Regular polygons. There can't be a better way to introduce polygons, than what kids love the most - coloring! Direct kids to follow the color key provided to identify polygons like triangles, quadrilaterals, pentagons and more. Area of Regular Polygons If radii are drawn from the center of a regular polygon to the vertices, congruent isosceles triangles are formed. In the limit, a sequence of regular polygons with an increasing number of sides approximates a circle, if the perimeter or area is fixed, or a regular apeirogon (effectively a straight line. Regular Polygon Formulas. Illustrated with examples of polygons in everyday life. Newest Active Followers. 120\u00b0 \u00b0 85\u00b0 53\u00b0 115\u00b0 x\u00b0 117\u00b0 105\u00b0 120. In geometry, polygons cover a lot of ground, so you can bet that some questions on the ACT Math exam will involve polygons\u2014specifically, finding the interior angles of a polygon. A hexagon (six-sided polygon) can be divided into four triangles. The interactive lessons include: Polygons and Non-Polygons in which students will drag a series of shapes to either the bucket that says polygon or the bucket that says non-polygon. A rectangle is a special quadrilateral where opposite sides are congruent-- that is, the same length -- and each angle is a right angle. Expected time to solve, similar questions of 11 Plus exam practice papers. Find the length of one side. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Area is the negative space inside a polygon. A comprehensive database of more than 34 polygon quizzes online, test your knowledge with polygon quiz questions. +34,000 Free Graphic Resources. Figure 4 \u2013 The Interior angles of polygons that can tessellate the plane add up to 360 degrees. The sum of the exterior angles of any polygon is 360 degrees. I also make them available for a student who wants to do focused independent study on a topic. The polygon is a 18-sided regular polygon; a regular octadecagon. A worksheet on angles in polygons. I want to find the direction of s. Published on Jan 20, 2017. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. See our pages on circles and. So if we know the sum of the interior angles, we just need to divide that sum by the number of sides to find out the measure of any angle of the polygon: 180(n-2)/n Exterior Angles. A regular net is defined as one in which the vertex figure is required by symmetry to be a regular polygon or polyhedron. Polygons maths worksheet 4 and polygon maths worksheet 5 work with tessellation. (n \u2013 2)*180 = sum of all interior angles (6 \u2013 2)*180 = 720.", "date": "2020-08-13 05:31:46", "meta": {"domain": "madzuli.it", "url": "http://shey.madzuli.it/regular-polygon-questions.html", "openwebmath_score": 0.30508238077163696, "openwebmath_perplexity": 683.6335121365421, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9845754501811437, "lm_q2_score": 0.8791467754256018, "lm_q1q2_score": 0.8655863321899627}}
{"url": "https://math.stackexchange.com/questions/3053627/whats-132333-cdots993-modulo-3/3054014", "text": "What's $1^3+2^3+3^3+\\cdots+99^3$ modulo $3$?\n\nWhat's the remainder when the sum\n\n$$1^3+2^3+3^3+\\cdots+99^3$$ is divided by $$3$$?\n\nBackground:\n\nI saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.\n\nI can solve it but I doubt my methods are efficient.\n\nOne way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.\n\nAnother way is to look at the expansion of $$(x+1)^3$$ and see what it does to the residues for each $$x$$, then sum over that by induction.\n\nBut I'm sure my inventions aren't very efficient.\n\n\u2022 Use $n^3\\equiv n\\pmod3$. \u2013\u00a0Angina Seng Dec 27 '18 at 5:42\n\u2022 math.stackexchange.com/questions/1328798 \u2013\u00a0Andrei Dec 27 '18 at 5:43\n\u2022 In general, note that you can handle the terms in groups of $3$ as $n + 3 \\equiv n \\pmod 3$, and also use that $n^3 \\equiv n \\pmod 3$ as Lord Shark the Unknown stated above. \u2013\u00a0John Omielan Dec 27 '18 at 5:46\n\u2022 @LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that. \u2013\u00a0samerivertwice Dec 27 '18 at 5:46\n\nBy Fermat's little theorem $$a^3\\equiv a\\pmod3$$. Then you have $$\\sum_{k=1}^{99} k=\\frac{(99)(100)}2=50\\cdot 99\\equiv 0\\pmod3$$.\n\n\\begin{align} \\sum_{i=1}^{99}i^3 &= \\sum_{i=0}^{32} (3i+1)^3 + \\sum_{i=0}^{32} (3i+2)^3 + \\sum_{i=0}^{32} (3i+3)^3 \\\\ &\\equiv \\sum_{i=0}^{32} 1^3 + \\sum_{i=0}^{32} 2^3 + \\sum_{i=0}^{32} 0^3 \\pmod{3}\\\\ &\\equiv \\sum_{i=0}^{32} 1^3 + \\sum_{i=0}^{32} (-1)^3 + \\sum_{i=0}^{32} 0^3 \\pmod{3}\\\\ &\\equiv 0 \\pmod{3} \\end{align}\n\nSplit the numbers from $$1,\\ldots, 99$$ in\n\n\u2022 $$33$$ with remainder $$1$$\n\u2022 $$33$$ with remainder $$2$$\n\u2022 $$33$$ with remainder $$0$$ $$1^3 + 2^3 + \\cdots + 99^3 \\equiv 33\\cdot (1^3+2^3+0^3) \\equiv 0 \\mod 3$$\n\nconsider these observations:\n\n$$1 = 1 \\pmod 3$$ $$2 = -1 \\pmod 3$$ $$3 = 0 \\pmod 3$$\n\nThis becomes:\n\n$$33(1^3+(-1)^3+0) \\equiv 33(1+(-1)+0) \\equiv 33(0) \\equiv 0 \\pmod 3$$\n\nHint:\n\n$$n^3\\equiv n\\pmod3$$ as $$n^3-n=n(n-1)(n+1)$$ is the product of three consecutive integers\n\nWe have $$\\sum_{r=1}^mr^3\\equiv\\sum_{r=1}^mr\\pmod3$$\n\n$$\\equiv \\dfrac{m(m+1)}2$$\n\nAlternatively $$\\sum_{r=1}^m r^3=\\dfrac{m^2(m+1)^2}4$$\n\nFrom $$a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$$, you have $$a+b|a^{3} +b^{3}$$ and \\begin{align} &1^{3} + 2^{3} + 3^{3} + \\cdots + 97^{3} + 98^{3} + 99^{3} \\\\ &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + \\cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + \\cdots + 99^{3}) \\end{align} is a multiple of 3.\n\nAnother method is to use the identity\n\n$$\\sum_{k=1}^nk^3=\\left(\\sum_{k=1}^nk\\right)^2$$ So, $$\\sum_{k=1}^{99}k^3\\equiv\\left(\\sum_{k=1}^{99}k\\right)^2\\equiv\\frac{(99)^2(100)^2}{4}\\equiv0\\ (\\mod3)$$\n\nYet another method is to use that $$(x-1)^3=x^3-3x^2+3x-1$$ and $$(x+1)^3=x^3+3x^2+3x+1$$, so $$(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.\n\nYou can also calculate that sum using\n\n$$\\sum_{i=0}^{i=n}i^3=\\dfrac{n^2(n+1)^2}{4}$$\n\nHere: $$\\sum_{i=0}^{i=99}i^3=\\dfrac{99^2(99+1)^2}{4}$$ which is clearly divisible by 3.", "date": "2020-09-22 14:06:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3053627/whats-132333-cdots993-modulo-3/3054014", "openwebmath_score": 0.9858062863349915, "openwebmath_perplexity": 836.7977443212218, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754479181589, "lm_q2_score": 0.8791467770088162, "lm_q1q2_score": 0.8655863317592609}}
{"url": "https://atcoder.jp/contests/abc215/editorial/2511", "text": "Contest Duration: - (local time) (100 minutes) Back to Home\nOfficial\n\n## B - log2(N) Editorial by en_translator\n\nThere are various approaches to solve this problem. Here, we will explain some of them. Even if you got accepted during the contest, we recommend considering other solutions.\nNote that the sample codes in this editorials are provided in C++.\n\n1. Basic solution\n2. Solution using logarithmic function, and numerical errors\n3. Solution using binary representation of an integer\n\n#### 1. Basic solution\n\nFirst, as $$k$$ increases, obviously $$2^k$$ gets larger too. Therefore, we can see that $$2^k \\le N$$ for all integers $$k$$ up to some boundary, or $$2^k>N$$ for $$k$$ greater than that value.\nThus, we can find the answer with a loop where $$k$$ is incremented one by one .\nNote that one can $$2^k$$ efficiently compute $$2^k$$ during the loop of incrementing $$k$$ for a good computational complexity.\n\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nint main(){\nlong long n;\ncin >> n;\nlong long val=1;\nfor(long long i=0;i<=60;i++){\nif(val>n){\ncout << i-1 << '\\n';\nbreak;\n}\nval*=2;\n}\nreturn 0;\n}\n\n\n#### 2. Solution using logarithmic function, and numerical errors\n\nBy applying logarithmic function to $$2^k \\le N$$ we obtain $$k \\le \\log_2(N)$$. Therefore, one should be able to output the $$\\log_2(N)$$, rounded down to an integer, to get accepted.\nHowever, this code results in WA (Wrong Answer).\n\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nint main(){\nlong long n;\ncin >> n;\ncout << floor(log2(n)) << '\\n';\nreturn 0;\n}\n\n\nThe cause of WA is lack of precision. Specifically, a precision occurs for the following case ($$N=2^{59}-1$$). (WA: 59, AC: 58)\n\n576460752303423487\n\n\nIn order to avoid the error, one can use a type having more precisions than 64-bit floating point decimal. If you modify the last code as follows, you can get accepted for this problem.\n\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nint main(){\nlong long n;\ncin >> n;\ncout << floor(log2((long double)n)) << '\\n';\nreturn 0;\n}\n\n\nNo that high-precision type is not always omnipotent. If the constraints for this problem were $$N \\le 2 \\times 10^{18}$$, the modified code still outputs a wrong answer for the following test case. (WA: 60, AC: 59)\n\n1152921504606846975\n\n\nIn competitive programming, discarding the integral property in an integral problem and using only decimals is dangerous. We recommend you to process integral problems in integer-typed values as much as possible. When you inevitably use decimals, be very careful about the errors.\n\n#### 3. Solution using binary representation of an integer\n\nConsider the binary representation of the integer.\nFor example, we have the conversions $$6=110_{(2)}$$ and $$15=1111_{(2)}$$.\nHere, the inequality $$2^{k} \\le N < 2^{k+1}$$ can be transformed into binary notations as follows.\n$$1 \\underbrace{ 00 \\dots 0 }_{ k }\\ _{(2)} \\le N < 1 \\underbrace{ 00 \\dots 0 }_{ k+1 }\\ _{(2)}$$\nUsing this property, we can derive the following solution.\n\n\u2022 If the binary notation of $$N$$ (without leading $$0$$\u2019s) has $$k$$ digits (that is, if the most significant bit with value $$1$$ is the $$k$$-th least significant bit), then the answer is $$k-1$$.\n\nThere are various ways to implement this, while using bit operations simplifies it. (Note that the least significant digit is treated as the $$0$$-th digit.)\n\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nint main(){\nlong long n;\ncin >> n;\nfor(int i=60;i>=0;i--){\nif(n&(1ll<<i)){cout << i << '\\n';break;}\n}\nreturn 0;\n}\n\n\nposted:\nlast update:", "date": "2023-02-07 08:43:48", "meta": {"domain": "atcoder.jp", "url": "https://atcoder.jp/contests/abc215/editorial/2511", "openwebmath_score": 0.77532958984375, "openwebmath_perplexity": 1522.062464031509, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754506337406, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8655863169999203}}
{"url": "https://markhkim.com/clrs/ch04/", "text": "# CLRS Chapter 4. Divide and Conquer\n\nsource code directory\n\n## 4.1. The maximum-subarray problem\n\n### Implementation\n\nA divide-and-conquer implementation of the maximal subarray finder is as follows:\n\nfrom math import floor\n\ninfty = float(\"inf\")  # -infty is used as the universal lower bound\n\ndef maximal_subarray(A, lo, hi):\nif lo == hi:\nreturn (A[lo], lo, hi)\nelse:\nmid = lo + floor((hi - lo)/2)\n# the divide-and-conquer step\n(lsum, llo, lhi) = maximal_subarray(A, lo, mid)\n(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)\n(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)\n# pick the maximal result\nif (lsum >= csum) and (lsum >= rsum):\nreturn (lsum, llo, lhi)\nelif (rsum >= lsum) and (rsum >= csum):\nreturn (rsum, rlo, rhi)\nelse:\nreturn (csum, clo, chi)\n\ndef maximal_crossing_subarray(A, lo, mid, hi):\n\"\"\"search left of mid and right of mid separately, then add\"\"\"\n(lsum, llo) = maximal_subarray_fixed_starting_point(A, mid, lo, -1)\n(rsum, rhi) = maximal_subarray_fixed_starting_point(A, mid+1, hi, 1)\nreturn (lsum + rsum, llo, rhi)\n\ndef maximal_subarray_fixed_starting_point(A, start, end, direction):\ntotal, subtotal, max_index = -infty, 0, start\nfor i in range(start, end + direction, direction):\nsubtotal = subtotal + A[i]\nif subtotal > total:\ntotal, max_index = subtotal, i\nreturn (total, max_index)\n\n>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]\n>>> maximal_subarray(A, 0, len(A)-1)\n(43, 7, 10)\n\n\n### Exercises\n\n4.1-1. What does find-maximaum-subarray return when all elements of $A$ are negative?\n\nfind-maximum-subarray in the text is equivalent to maximal_subarray above, so let us take a look at the latter. We shall show that maximal_subarray(A, 0, len(A)-1) returns the maximum element of $A$ whenever $A$ consists entirely of negative numbers.\n\nWe first observe that maximal_subarray_fixed_starting_point(A, start, end, direction) returns (A[start], start) for all legal inputs of start and end. From this, we see that maximal_crossing_subarray(A, lo, mid, hi) returns (A[mid] + A[mid+1], mid, mid+1). Since $A$ consists entirely of negative numbers,\n\nwhence it follows that the return value of maximal_subarray cannot come from maximal_crossing_subarray.\n\nThe only return values of maximal_subarray that does not come from maximal_crossing_subarray are the terms of $A$. Since the if-elif-else block at the end of maximal_subarray returns the maximum, we conclude that maximal_subarray(A, 0, len(A)-1) returns the maximum element of $A$, as was to be shown. $\\square$\n\n4.1-2. Write pseudocode for the brute-force method of solving the maximum-subarray problem. Your procedure should run in $\\Theta(n^2)$ time.\n\ninfty = float(\"inf\") # -infty is used as the universal lower bound\n\ndef maximal_subarray_brute_force(A):\nn = len(A)\nmax_sum = -infty\nfor i in range(n):\niterative_sum = 0\nfor j in range(i, n):\niterative_sum += A[j]\nif iterative_sum > max_sum:\nmax_sum = iterative_sum\nmax_lo, max_hi = i, j\nreturn (max_sum, max_lo, max_hi)\n\n>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]\n>>> maximal_subarray_brute_force(A)\n(43, 7, 10)\n\n>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]\n>>> maximal_subarray(A, 0, len(A)-1)\n(43, 7, 10)\n\n\nIt is not hard to see that maximal_subarray_brute_force is correct, as it compares every possible sub-sum.\n\nAssuming that the if clause is never satisfied, we can compute the lower bound of the running time as follows:\n\nThe if clause only adds constant time to the computation, and so the upper bound of the running time is $O(n^2)$. It follows that maximal_subarray_brute_force runs in $\\Theta(n^2)$ time. $\\square$\n\n4.1-3. Implement both the brute-force and recursive algorithms for the maximum-subarray problem on your own computer. What problem size $n_0$ gives the crossover point at which the recursive algorithm beats the brute-force algorithm? Then, change the base case of the recursive algorithm to use the brute-force algorithm whenever the problem size is less than $n_0$. Does that change the crossover point?\n\n>>> import random\n\n\n50 elements:\n\n>>> L1 = random.sample(range(-1000, 1000), 50)\n>>> L2 = random.sample(range(-1000, 1000), 50)\n>>> L3 = random.sample(range(-1000, 1000), 50)\n>>> %%timeit\n... maximal_subarray(L1, 0, len(L1)-1)\n... maximal_subarray(L2, 0, len(L2)-1)\n... maximal_subarray(L3, 0, len(L3)-1)\n231 \u00b5s \u00b1 1.94 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n>>> %%timeit\n... maximal_subarray_brute_force(L1)\n... maximal_subarray_brute_force(L2)\n... maximal_subarray_brute_force(L3)\n227 \u00b5s \u00b1 978 ns per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n\n\n60 elements:\n\n>>> L1 = random.sample(range(-1000, 1000),60)\n>>> L2 = random.sample(range(-1000, 1000),60)\n>>> L3 = random.sample(range(-1000, 1000),60)\n>>> %%timeit\n... maximal_subarray(L1, 0, len(L1)-1)\n... maximal_subarray(L2, 0, len(L2)-1)\n... maximal_subarray(L3, 0, len(L3)-1)\n285 \u00b5s \u00b1 6.63 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n>>> %%timeit\n... maximal_subarray_brute_force(L1)\n... maximal_subarray_brute_force(L2)\n... maximal_subarray_brute_force(L3)\n312 \u00b5s \u00b1 1.21 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n\n\nIt follows that $% $ for this particular example. Let us now modify the recursive algorithm to include the brute-force algorithm for all lists of size at most, say, 55:\n\ndef maximal_subarray_mixed(A, lo, hi):\nif hi - lo <= 55:\n(brute_sum, brute_lo, brute_hi) = maximal_subarray_brute_force(\nA[lo:hi+1])\nreturn (brute_sum, brute_lo + lo, brute_hi + hi)\nelse:\nmid = lo + floor((hi-lo)/2)\n(lsum, llo, lhi) = maximal_subarray(A, lo, mid)\n(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)\n(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)\nif (lsum >= csum) and (lsum >= rsum):\nreturn (lsum, llo, lhi)\nelif (rsum >= lsum) and (rsum >= csum):\nreturn (rsum, rlo, rhi)\nelse:\nreturn (csum, clo, chi)\n\n\nLet us test the new algorithm.\n\n60 elements:\n\n>>> L1 = random.sample(range(-1000, 1000),60)\n>>> L2 = random.sample(range(-1000, 1000),60)\n>>> L3 = random.sample(range(-1000, 1000),60)\n>>> %%timeit\n... maximal_subarray(L1, 0, len(L1)-1)\n... maximal_subarray(L2, 0, len(L2)-1)\n... maximal_subarray(L3, 0, len(L3)-1)\n279 \u00b5s \u00b1 3.09 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n>>> %%timeit\n... maximal_subarray_mixed(L1, 0, len(L1)-1)\n... maximal_subarray_mixed(L2, 0, len(L2)-1)\n... maximal_subarray_mixed(L3, 0, len(L3)-1)\n278 \u00b5s \u00b1 989 ns per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n\n\n70 elements:\n\n>>> L1 = random.sample(range(-1000, 1000),70)\n>>> L2 = random.sample(range(-1000, 1000),70)\n>>> L3 = random.sample(range(-1000, 1000),70)\n>>> %%timeit\n... maximal_subarray(L1, 0, len(L1)-1)\n... maximal_subarray(L2, 0, len(L2)-1)\n... maximal_subarray(L3, 0, len(L3)-1)\n330 \u00b5s \u00b1 1.61 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n>>> %%timeit\n... maximal_subarray_mixed(L1, 0, len(L1)-1)\n... maximal_subarray_mixed(L2, 0, len(L2)-1)\n... maximal_subarray_mixed(L3, 0, len(L3)-1)\n330 \u00b5s \u00b1 2.47 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1000 loops each)\n\n\nWe see that the crossover point increases by 10 or so, but this is hardly an improvement. $\\square$\n\n4.1-4. Suppose we change the definition of the maximum-subarray problem to allow the result to be an empty subarray, where the sum of the values of an empty subarray is 0. How would you change any of the algorithms that do not allow empty subarrays to permit an empty subarray to be the result?\n\nOnly the main routine needs to be changed:\n\nfrom math import floor\n\ndef maximal_subarray_allow_empty_subarray(A, lo, hi):\nif lo == hi:\nif A[lo] < 0:\nreturn (0, -1, -1)  # allow for empty array, labeled by -1\nelse:\nreturn (A[lo], lo, hi)\nelse:\nmid = lo + floor((hi - lo)/2)\n(lsum, llo, lhi) = maximal_subarray(A, lo, mid)\n(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)\n(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)\nif (lsum < 0) and (rsum < 0) and (csum < 0):\nreturn (0, -1, -1)  # allow for empty array, labeled by -1\nelif (lsum >= csum) and (lsum >= rsum):\nreturn (lsum, llo, lhi)\nelif (rsum >= lsum) and (rsum >= csum):\nreturn (rsum, rlo, rhi)\nelse:\nreturn (csum, clo, chi)\n\n>>> A = [-1, -2, -3, -4, -5, -6, -7]\n>>> maximal_subarray_allow_empty_subarray(A, 0, len(A)-1)\n(0, -1, -1)\n\n\n$\\square$\n\n4.1-5. Use the following ideas to develop a nonrecursive, linear-time algorithm for the maximum-subarray problem. Start at the left end of the array, and progress toward the right, keeping track of the maximum subarray seen so far. Knowing a maximum subarray of A[1..j], extend the answer to find a maximum subarray ending at index $j+1$ by using the following observation: a maximum subarray of A[1..j+1] is either a maximum subarray of A[1..j] or a subarray A[i..j+1] for some $1 \\leq i \\leq j+1$. Determine a maximum subarray of the form A[i..j+1] in constant time based on knowing a maximum subarray ending at index $j$.\n\nWe fix an index $j \\geq 1$ and suppose that $A[p:q]$ is a maximum subarray of $A[:j]$. We pick another index $i$ such that\n\nBy the maximality of $A[p:q]$ in $A[:j]$, every maximum subarray of $A[:j+1]$ must either equal to $A[p:q]$ or contain $A[j]$. Now,\n\nwhence we conclude that\n\nUsing the above observation, we devise a linear-time algorithm for the maximum-subarray problem.\n\nGiven an array $A$, we observe that $[A[0]]$ is the (unique) maximum subarray of $A[:0+1]$. We now apply the above observation inductively to produce a maximum subarray of $AA$, a process known as Kadane\u2019s algorithm.\n\ndef kadane(A):\nlo, hi, total = 0, 0, 0\ntail_lo, tail_hi, tail_total = 0, 0, 0\nfor j in range(1, len(A)-1):\nif tail_total + A[j] <= A[j]:\ntail_lo, tail_hi, tail_total = j, j, A[j]\nelse:\ntail_lo, tail_hi, tail_total = tail_lo, j, tail_total+A[j]\n\nif tail_total >= total:\nlo, hi, total = tail_lo, tail_hi, tail_total\nreturn (total, lo, hi)\n\n>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]\n>>> maximal_subarray(A, 0, len(A)-1)  # see the implementation subsection\n(43, 7, 10)\n>>> kadane(A)\n(43, 7,1 0)\n\n\n$\\square$\n\n## 4.2 Strassen\u2019s algorithm for matrix multiplication\n\nSee my blog post for details.\n\n### Exercises\n\n4.2-1. Use Strassen\u2019s algorithm to compute the matrix product\n\nShow your work.\n\nObserve first that\n\nThe product is\n\n$% $ $\\square$\n\n4.2-2. Write pseudocode for Strassen\u2019s algorithm\n\nThe implementation below makes use of NumPy to reduce the implementation overhead on basic matrix operations.\n\n\nimport numpy as np\n\ndef strassen(A, B):\nn = len(A)\nif n == 1:\nreturn A * B\n\n# block matrices; // is integer division\na = np.array([[A[:n//2, :n//2], A[:n//2, n//2:]],\nA[n//2:, :n//2], A[n//2:, n//2:]]])\nb = np.array([[B[:n//2, :n//2], B[:n//2, n//2:]],\nB[n//2:, :n//2], B[n//2:, n//2:]])\n\ns1 = b[0,1] - b[1,1]\ns2 = a[0,0] + a[0,1]\ns3 = a[1,0] + a[1,1]\ns4 = b[1,0] - b[0,0]\ns5 = a[0,0] + a[1,1]\ns6 = b[0,0] + b[1,1]\ns7 = a[0,1] - a[1,1]\ns8 = b[1,0] + b[1,1]\ns9 = a[0,0] - a[1,0]\ns10 = b[0,0] + b[0,1]\n\np1 = strassen(a[0,0], s1)\np2 = strassen(s2, b[1,1])\np3 = strassen(s3, b[0,0])\np4 = strassen(a[1,1], s4)\np5 = strassen(s5, s6)\np6 = strassen(s7, s8)\np7 = strassen(s9, s10)\n\nc = np.zeros((n, n))  # initialize an n-by-n matrix\nc[:n//2, :n//2] = p5 + p4 - p2 + p6\nc[:n//2, n//2:] = p1 + p2\nc[n//2:, :n//2] = p3 + p4\nc[n//2:, n//2:] = p5 + p1 - p3 - p7\n\nreturn c\n\n>>> A = np.array([[1, 3], [7, 5]])\n>>> B = np.array([[6, 8], [4, 2]])\n>>> np.dot(A, B)  # matrix multiplication\n[[18 14]\n[62 66]]\n>>> strassen(A, B)\n[[18. 14.]\n[62. 66.]]\n\n>>> A = np.array([[5, 2, 6, 1],\n...               [0, 6, 2, 0],\n...               [3, 8, 1, 4],\n...               [1, 8, 5, 6]])\n>>> B = np.array([[7, 5, 8, 0],\n...               [1, 8, 2, 6],\n...               [9, 4, 3, 8],\n...               [5, 3, 7, 9]])\n>>> np.dot(A, B)  # matrix multiplication\n[[ 96  68  69  69]\n[ 24  56  18  52]\n[ 58  95  71  92]\n[ 90 107  81 142]]\n>> strassen(A, B)\n[[ 96.  68.  69.  69.]\n[ 24.  56.  18.  52.]\n[ 58.  95.  71.  92.]\n[ 90. 107.  81. 142.]]\n\n\n$\\square$\n\n4.2-3. How would you modify Strassen\u2019s algorithm to multiply $n \\times n$ matrices in which $n$ is not an exact power of 2? Show that the resulting algorithm runs in time $\\Theta(n^{\\log 7})$.\n\nFor each $n \\times n$ matrix $A$, we let $m$ be the unique integer such that $% $ and augment $A$ with an $(2^m - n) \\times (2^m - n)$ identity matrix $I$ as follows:\n\nWe then have\n\nfrom which we can extract $AB$ in constant time. It thus suffices to compute the time complexity of the matrix multiplication $\\tilde{A} \\tilde{B}$.\n\nObserve that\n\nby the monotonicity of $t \\mapsto t^{\\log 7}$. Similarly,\n\nNow,\n\nand so\n\nIt follows that $n^{\\log 7} = \\Theta((2^m)^{\\log 7})$, whence applying Strassen\u2019s algorithm to the augmented matrices does not incur any additional asymptotic time complexity. $\\square$\n\n4.2-4. What is the largest $k$ such that if you multiply $3 \\times 3$ matrices using $k$ multiplications (not assuming commutativity of multiplication), then you can multiply $n \\times n$ matrices in time $o(n^{\\log 7})$? What would be the running time of this algorithm be?\n\nThe recurrence relation to solve is\n\nObserve that $n^2 = \\Theta(n^{\\log_3 9})$. By the master theorem ($\\S4.5$), we have\n\nwhere $(\\ast)$ is $k > 9$ and $k(n/3)^2 \\leq cn^2$ for some $% $ and all sufficiently large $n$.\n\nSince $\\log 7 > \\log_3 k$ implies that $% $, we see that $T(n) = o(n^{\\log 7})$ whenever $k \\leq 9$. On the other hand, if $k > 9$, then then inequality\n\nnever holds for any $n \\geq 1$ reagrdless of which $% $ we choose. It follows that we must have $k \\leq 9$ in order to multiply $n$-by-$n$ matrices in time $o(n^{\\log 7})$, and the running time is\n\n4.2-5. V. Pan has discovered a way of multiplying $68 \\times 68$ matrices using 132,464 multiplications, a way of multiplying $70 \\times 70$ matrices using 143,630 multiplications, and a way of multiplying $72 \\times 72$ matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? How does it compare to Strassen\u2019s algorithm?\n\nThe recurrence relations to solve are\n\nObserving that $n^2 = \\Theta(n^{\\log_{68} 4624}){}_{}$, we invoke the master theorem ($\\S4.5$) to conclude that\n\nSince $n^{\\log 7} \\approx 2.807$, we conclude that $T_{68}{}_{}$ is better than the running time of Strassen.\n\nObserve now that $n^2\\Theta(n^{\\log_{70} 4900}){}_{}$. The master theorem ($\\S4.5$) implies that\n\nSimilarly as above, we conclude that $T_{72}{}_{}$ is better than the running time of Strassen.\n\nFinally, we observe that $n^2 = \\Theta(n^{\\log_{72} 1944}){}_{}$, which implies that\n\nby the master theorem ($\\S4.5$). Similarly as above, we conclude that $T_{72}{}_{}$ is better than the running time of Strassen.\n\nWe conclude that\n\nwhere $T_{S}{}_{}$ refers to the runing time of Strassen. $\\square$\n\n4.2-6. How quickly can you multiply a $kn \\times n$ matrix by an $n \\times kn$ matrix, using Strassen\u2019s algorithm as a subroutine? Answer the same question with the order of the input matrices reversed.\n\nUsing block matrix multiplication\n\nwe can multiply a $kn \\times n$ matrix by an $n \\times kn$ matrix through $k^2$ iterations of Strassen. It follows that the computation takes $\\Theta(k^2n^{\\log 7})$.\n\nSimilarly, we use block matrix multiplication\n\nwe can multiply an $n \\times kn$ matrix by a $kn \\times n$ matrix through $k$ iterations of STrassen. It follows that the computation takes $\\Theta(k n^{\\log 7})$. $\\square$\n\n4.2-7. Show how to multiply the complex numbers $a + bi$ and $c + di$ using only three multiplcations of real numbers. The algorithm should take $a$, $b$, $c$, and $d$ as input and produce the real component $ac - bd$ and the imaginary component $ad + bc$ separately.\n\nWe set $P_1 = ac$, $P_2 = bd$, and $P_3 = (a+b)(c+d)$. Observe that\n\nand that\n\nIt follows that\n\nwhich only requires three real multiplications.\n\nThis algorithm is commonly known as the Karatsuba algorithm. $\\square$\n\n## The substitution method for solving recurrences\n\n### Exercises\n\n4.3-1. Show that the solution of $T(n) = T(n-1) + n$ is $O(n^2)$.\n\nWe let $C > 0$ be an arbitrary constant, to be chosen later. We fix a postive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nBy choosing $C = 1$, we see that $-2(C-1)n = 0$ for all $n \\geq 0$, from which it follows that\n\nas was to be shown.\n\nTo turn the above argument into a complete induction proof, we observe that\n\nSince $C = 1$, it suffices to find $m_0$ such that\n\nChoosing any $m_0 \\geq \\sqrt{2T(1)}$, we see that\n\nas was to be shown. The desired result now follows from mathematical induction with base case $n = m_0$. $\\square$\n\n4.3-2. Show that the solution of $T(n) = T(\\lceil n/2 \\rceil) + 1$ is $O(\\log n)$.\n\nWe first examine the recurrence for $n \\geq 2$.\n\nWe let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nWe can bound the last expression by $C \\log n$ if and only if\n\nThe above inequality is equivalent to\n\nRecalling that $n \\geq 2$, we note that\n\nChoosing $C = 1/(1 + \\log(2/3))$, we see that\n\nIt now follows that\n\nas was to be shown.\n\nTo turn the above argument into a complete induction proof, we observe that\n\nfor all $k \\geq 1.$ Since $C = 1/(1 + \\log(2/3))$, it suffices to find $k_0$ such that\n\nThe above inequality is equivalent to\n\nwhich, in turn, is equivalent to\n\nWe pick any $k_0$ that satisfies the last inequality and set $m_0 = 2^{k_0}$, so that\n\nThe desired result now follows from mathematical induction with base case $n = m_0$. $\\square$\n\n4.3-3. We saw that the solution of $T(n) = 2T(\\lfloor n/2 \\rfloor) + n$ is $O(n \\log n)$. Show that the solution of this recurrence is also $\\Omega(n \\log n)$. Conclude that the solution is $\\Theta(n \\log n)$.\n\nWe first examine the recurrence for $n \\geq 2$.\n\nWe let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nThe inequality\n\nis equivalent to\n\nwhich, in turn, is equivalent to\n\nSince $n \\geq 2$, we see that\n\nWe claim that the function\n\nis monotonically decreasing on $[2,\\infty)$. Indeed, we have that\n\nfrom which we conclude that $f(k) > f(k+1)$. Monotonicity now follows.\n\nThe monotonicity of $f(k)$ implies that\n\nSetting $C = \\frac{2}{3}$, we see that $(\\ast \\ast)$ holds, whence $(\\ast)$ holds. We conclude that\n\nwith this choice of $C$, as was to be shown.\n\nTo turn the above argument into a complete induction proof, we observe that $m_0 = 2$ yields\n\nas $T$ is typically assumed to be nonnegative.\n\nThe desired result now follows from mathematical induction with base case $n =m_0$.\n\n4.3-4. Show that by making a different inductive hypothesis, we can overcome the difficulty with the boundary condition $T(1) = 1$ for recurrence (4.19) without adjusting the boundary conditions for the inductive proof.\n\nRecall (4.19):\n\nWe let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\n$C \\geq T(1) - 1$, we have $1-C \\leq -T(1)$, so that\n\nTo turn the above argument into a complete induction proof, we observe that\n\nThe desired result now follows from mathematical induction with base case $n=1$. $\\square$\n\n4.3-5. Show that $\\Theta(n \\log n)$ is the solution to the \u201cexact\u201d recurrence (4.3) for merge sort.\n\nRecall (4.3):\n\nWe let $c > 0$ and $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nSince $T(n) = T(\\lceil n/2 \\rceil) + T(\\lfloor n/2 \\rfloor) + \\Theta(n)$, we have the double-ended estimate\n\nwhere the constants $d > 0$ and $D > 0$ are to be chosen later.\n\nWe first establish the upper bound. Since $\\lfloor n/2 \\rfloor \\leq \\lceil n/2 \\rceil$,\n\n$\\lceil n/2 \\rceil \\geq n/2$, and so $n/\\lceil n/2 \\rceil \\leq n/(n/2) = 2$. It then follows that\n\nwhence any choice of $C$ and $D$ with $C \\leq D$ yields\n\nLet us now establish the lower bound. Since $\\lceil n/2 \\rceil \\geq \\lfloor n/2 \\rfloor$,\n\n$\\lfloor n/2 \\rfloor \\leq n/2$, and so $n/\\lfloor n/2 \\rfloor \\geq n/(n/2) = 2$. It then follows that\n\nwhence any choice of $c$ and $d$ with $c \\leq d$ yields\n\nWe have thus shown that\n\nunder the assumption that\n\nholds for all $% $.\n\nTo turn the above argument into a complete induction proof, we set $c = d = T(2)/4$ and $C = D = 2T(2)$ and observe that, for $n = 2$,\n\nand that\n\nIt follows from mathematical induction that $T(n) = \\Theta(n \\log n)$ with base case $n = 2$. $\\square$\n\n4.3-6. Show that the solution to $T(n) = 2T(\\lfloor n /2 \\rfloor + 17) + n$ is $O(n \\log n)$.\n\nAssume that $n \\geq 68$, so that $17 \\leq n/4$.\n\nWe let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nNow, $n \\geq 68$, and so\n\nand\n\nIt follows that\n\nSince $\\log(4/3) > 0$, we have $% $, and so\n\nIt thus suffices to pick $C > 0$ such that\n\nkeeping in mind that $n \\geq 68$. Since $n$ and $\\log n$ are both positive under this condition, we need only to pick $C$ large enough that\n\nLetting $C = 1/\\log(4/3)$, we obtain\n\nunder the assumption that\n\nfor all $% $.\n\nTo turn the above argument into a complete induction proof, we must show that\n\nBut, of course,\n\nIt now follows from mathematical induction with base case $n = 68$ that\n\nas was to be shown. $\\square$\n\n4.3-7. Using the master method in Section 4.5, you can show that the solution to the recurrence $T(n) = 4T(n/3) + n$ is $T(n) = \\Theta(n^{\\log_3 4})$. Show that a substitution proof with the assumption $T(n) \\leq cn^{\\log_3 4}$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.\n\nSince\n\na na\u00efve substitution argument would yield\n\nSince\n\nfor all $n \\geq 1$, the substitution argument fails.\n\nIf we assume instead that\n\nfor all $% $, then\n\nand the usual substitution proof goes through. $\\square$\n\n4.3-8. Using the master method in Section 4.5, you can show that the solution to the recurrence $T(n) = 4T(n/2) + n$ is $T(n) = \\Theta(n^2)$. Show that a substitution proof with the assumption $T(n) \\leq cn^2$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.\n\nSince $c(n/2)^2 = (c/4)n^2$, a na\u00efve substitution argument would yield\n\nwhich is larger than $cn^2$ for all $n \\geq 1$.\n\nIf we assume instead that\n\nfor all $% $, then\n\nand the usual substitution proof goes through. $\\square$\n\n4.3-9. Solve the recurrence $T(n) = 3T(\\sqrt{n}) + \\log n$ by makign a change of variables. Your solution should be asymptotically tight. Do not worry about whether values are integral.\n\nSince\n\nwe set $S(m) = T(2^m)$ and solve instead\n\nThe master theorem ($\\S4.5$) shows that\n\nWe shall show this by substitution.\n\nWe let $c > 0$ and $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nWe first establish the upper bound:\n\nNow, the lower bound:\n\nTo turn the above argument into a complete induction proof, we must show that\n\nSetting $c = C = S(1) + 2$, we see that\n\nand that\n\nMathematical induction with base case $m = 1$ now shows that\n\nfor all $m \\geq 1$. Since the lower bound agreed with the upper bound, we in fact have the identity\n\nSince $T(n) = S(\\log n)$, we conclude that\n\n## 4.4. The recursion-tree method for solving recurrences\n\n### Exercises\n\n4.4-1. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 3T(\\lfloor n/2 \\rfloor) + n$. Use the substitution method to verify your answer.\n\nObserve that\n\nLet us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nIndeed,\n\nTo complete the induction proof, we must show that\n\nWe set $C = T(1) + 2$ and observe that\n\nas was to be shown. It now follows from mathematical induction with base case $n = 1$ that $T(n) = O(n^{\\log(3/2)})$. $\\square$\n\n4.4-2. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n/2) + n^2$. Use the substitution method to verify your answer.\n\nObserve that\n\nLet us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nSo long as $C \\geq 4$, we have $C/4 + 1 \\leq C$, and so\n\nTo complete the induction proof, we must show that\n\nTo this end, we simply select $C = \\max(4, T(1))$. The desired result now follows from induction. $\\square$\n\n4.4-3. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 4T(n/2 + 2) + n$. Use the substitution method to verify your answer.\n\nWe first observe the the recurrence relation as it is written above is ill-defined. Indeed,\n\nwhen $T$ is a function defined on $\\mathbb{N}$. To remedy this issue, we shall consider instead\n\nObserve that\n\nSetting $n = 2^k$, we obtain\n\nWe therefore conjecture that\n\nSince the problem only asks us to establish the upper bound, we shall prove that\n\nWe let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nIf we choose $C$ and $D$ so that $4C -D + 1 \\leq 0$ and $8C - 6D \\leq 0$, then we have the bound\n\nHow do we choose such $C$ and $D$? The first inequality can be written as\n\nIt turns out that this is sufficient to achieve the second inequality:\n\nWe also note that the constants must be chosen to satisfy the base case\n\nfor an appropriate choice of $n_0$. Since the smallest $D$ that satisfies $(\\ast)$ is $4C + 1$, the base case should be chosen so that $Cn_0^2 - (4C+1)n_0 > 0$. This only works if $n_0 \\geq 5$. indeed, if $n_0 = 4$, then\n\nOn the other hand, if $n_0 = 5$, then\n\nwhich is positive if $C > 4/5$.\n\nSummarizing the above argument, we choose $C = T(5) + 1$ and $D = 4T(5) + 5$. For the base case $n = 5$, we have\n\nas $T$ is generallly assumed to be nonnegative.\n\nWe now assume inductively that\n\nholds for all $% $ such that $m \\geq 5$. Now, the inductive hypothesis implies that\n\nas was to be shown. We conclude that $T(n) = O(n^2)$. $\\square$\n\n4.4-4. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 2T(n-1) + 1$. Use the substitution method to verify your answer.\n\nObserve that\n\nLet us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We then have\n\nBy choosing $C = T(1)/2$, we have\n\nfor $n = 1$. It now follows from mathematical induction that $T(n) = O(2^n)$. $\\square$\n\n4.4-5. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n-1) + T(n/2) + n$. Use the substitution method to verify your answer.\n\nTo make sense of the $n/2$ term, we consider instead\n\nSince $T(n) - T(n-1) = T(\\lfloor n/2 \\rfloor) + n$, we see that\n\nTherefore,\n\nApplying $(\\ast)$ to the sum indexed by $i_1$, we obtain\n\nApplying $(\\ast)$ recursively, we obtain\n\nLet us verify our guess. To this end, we let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nSince\n\nwe see that\n\nSince we must have\n\nwe see that\n\nis required for all $n \\geq 1$. This holds when $C \\geq 2 + 2D$:\n\nFinally, we must pick $C$ and $D$ so that\n\nholds when $n = 1$. This requires\n\nSetting $D = T(1)$ and $C = 2T(1) + 2$, we see that\n\nThe choice of $C$ and $D$ also shows that\n\nwhenever\n\nfor all $% $. It now follows from mathematical induction with base case $n = 1$ that $T(n) = O(n^{\\log n})$. $\\square$\n\n4.4-6. Argue that the solution to the recurrence $T(n) = T(n/3) + T(2n/3) + cn$, where $c$ is a constant, is $\\Omega(n \\log n)$ by appealing to a recursion tree.\n\nObserve that\n\nwhere $\\begin{pmatrix} p \\\\q \\end{pmatrix} = \\frac{p!}{q!(p-q)!}$ is the binomial coefficient with respect to $(p,q)$. It follows that $T(n) = \\Omega(n \\log n)$. $\\square$\n\n4.4-7. Draw the recursion tree for $T(n) = 4T(\\lfloor n/2 \\rfloor) + cn$, where $c$ is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substitution method.\n\nObserve that\n\nWe thus conjecture that $T(n) = \\Theta(n^2)$.\n\nWe first establish the upper bound. To this end, we let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that\n\nfor all $% $. We shall show that\n\nObserve that\n\nTherefore, $D$ must be chosen so that $(c-D)n +2 \\leq 0$ for all $n \\geq 1$. This holds if $D \\geq c + 2$.\n\nWe now set\n\nand\n\nso that\n\nMoreover, if $T(m) \\leq Cm^2 - Dm$ for all $% $, then\n\nIt now follows from induction with base case $n = 1$ that $T(n) = O(n^2)$.\n\nIt remains to show that $T(n) = \\Omega(n^2)$.\n\nWe let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that $T(m) \\geq Cm^2 + Dm$ for all $% $. Observe that\n\nTo ensure that the resulting expression is bounded below by $Cn^2 + Dn$, we must at least have $D-2C+c > 0$. This holds if $D > 2C - c$.\n\nWe assume for now that $c > 0$ and set $C = \\min(c/2, T(1)/2)$ and $D = \\min(c/4, T(1)/4)$, so that\n\nMoreover,\n\nand so it follows from mathematical induction with base case $n=1$ that $T(n) = \\Omega(n^2)$.\n\nWe have thus shown that $T(n) = \\Theta(n^2)$. $\\square$\n\n4.4-8. Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(n-a) + T(a) + cn$, where $a \\geq 1$ and $c > 0$ are constants.\n\nObserve that\n\nWe thus conjecture that $T(n) = \\Theta(n^2)$.\n\nSince $% $ for $% $, the recurrence relation cannot hold for $% $. We thus assume that the values of $T$ for $n = 1,\\ldots,\\lfloor a \\rfloor$ are given.\n\nWe first show the upper bound. To this end, we let $C > 0$, to be chosen later. We shall show that\n\nfor all $n \\geq 2 \\lfloor a \\rfloor$. We first assume that $C \\geq T(\\lfloor a \\rfloor) / (\\lfloor a \\rfloor)^2$, so that $(\\ast)$ holds for $n = \\lfloor a \\rfloor$.\n\nObserve that\n\nIf $C \\geq c/(2 \\lfloor a \\rfloor)$, then\n\nand so\n\nWe thus assume that\n\nWe now fix $n_0 > 2\\lfloor a \\rfloor$ and assume that $(\\ast)$ holds for $% $. We shall show that $(\\ast)$ holds for $n = n_0$.\n\nObserve that\n\nIf $C \\geq c$, then\n\nSince $a \\geq 1$, we have the estimate\n\nNow, $n_0 > 2 \\lfloor a \\rfloor$, and so\n\nIt follows that $(\\ast)$ holds for $n = n_0$ whenever $C \\geq c$.\n\nSetting\n\nwe see that $(\\ast)$ holds for all $n \\geq 2 \\lfloor a \\rfloor$. It follows that $T(n) = O(n^2)$.\n\nWe now establish the lower bound. To this end, we let $C > 0$, to be chosen later. We shall show that\n\nfor all $n \\geq 2 \\lfloor a \\rfloor$. We first assume that $C \\leq T(\\lfloor a \\rfloor) / (\\lfloor a \\rfloor)^2$, so that $(\\ast \\ast)$ holds for $n = \\lfloor a \\rfloor$.\n\nObserve that\n\nIf $C \\leq c/(2 \\lfloor a \\rfloor)$, then\n\nand so\n\nWe thus assume that\n\nWe now fix $n_0 > 2\\lfloor a \\rfloor$ and assume that $(\\ast \\ast)$ holds for $% $. We shall show that $(\\ast\\ast)$ holds for $n = n_0$.\n\nObserve that\n\nIf $C \\leq c/2a$, then\n\nSince $a \\geq 0$, we have $2ac^2 \\geq 0$, and so\n\nIt follows that $(\\ast \\ast)$ holds for $n = n_0$ whenever $C \\leq c/2a$.\n\nSetting\n\nwe have $(\\ast \\ast)$ for all $n \\geq 2 \\lfloor a \\rfloor$. It follows that $T(n) = \\Omega(n^2)$.\n\nWe now conclude that $T(n) = \\Theta(n^2)$, as was to be shown. $\\square$\n\n4.4-9. Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(\\alpha n) + T((1-\\alpha) n) + cn$, where $\\alpha$ is a constant in the range $% $ and $c > 0$ is also a constant.\n\nObserve that\n\nWe let $a = \\max(\\alpha, 1-\\alpha)$ and set $p = \\lceil \\log_{1/\\alpha} n \\rceil$, so that\n\nfor all $0 \\leq i \\leq p$. The recursion-tree computation yields\n\nWe therefore conjecture that $T(n) = \\Theta(n \\log n).$\n\nWe first establish the upper bound. To this end, we let $C > 0$, to be chosen later. We assume that\n\nfor all $% $.\n\nObserve that\n\nSince $% $ and $% $, we have that $% $ and $% $. Therefore, any\n\nwould yield\n\nWe now set\n\nso that $T(2) \\leq C n \\log n$, and that\n\nwhenever $T(m) \\leq Cm \\log m$ for all $% $. It now follows from induction that $T(n) = O(n \\log n)$.\n\nWe now establish the lower bound. To this end, we let $C > 0$, to be chosen later. We assume that\n\nfor all $% $.\n\nObserve that\n\nSince $% $ and $% $, we have that $% $ and $% $. Therefore, any\n\nwould yield\n\nWe now set\n\nso that $T(2) \\geq C n \\log n$, and that\n\nwhenever $T(m) \\geq Cm \\log m$ for all $% $. It now follows from induction that $T(n) = \\Omega(n \\log n)$.\n\nWe have thus shown that $T(n) = \\Theta(n \\log n)$. $\\square$\n\n## 4.5. The master method for solving recurrences\n\n### The master theorem, as stated in CLRS, Theorem 4.1\n\nLet $a \\geq 1$ and $b > 1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence\n\nwhere we interpret $n/b$ to mean either $\\lfloor n /b \\rfloor$ or $\\lceil n/b \\rceil$. Then $T(n)$ has the following asymptotic bounds:\n\n1. If $f(n) = O(n^{\\log_b a - \\epsilon})$ for some constant $\\epsilon > 0$, then $T(n) = \\Theta(n^{\\log_b a})$.\n2. If $f(n) = \\Theta(n^{\\log_b a})$, then $T(n) = \\Theta(n^{\\log_b a} \\log n)$.\n3. If $f(n) = \\Omega(n^{\\log_b a + \\epsilon})$ for some constant $\\epsilon > 0$, and if $a f(n/b) \\leq cf(n)$ for some constant $% $ and all sufficiently large $n$, then $T(n) = \\Theta(f(n))$.\n\n### Exercises\n\n4.5-1. Use the master method to give tight asymptotic bounds for the following recurrences.\n\na. $T(n) = 2T(n/4) + 1$.\n\nb. $T(n) = 2T(n/4) + \\sqrt{n}$.\n\nc. $T(n) = 2T(n/4) + n$.\n\nd. $T(n) = 2T(n/4) + n^2$.\n\na belongs to case 1, and so\n\nb belongs to case 2, and so\n\nc belongs to case 3, as\n\nfor all $n$. Therefore,\n\nd belongs to case 3, as\n\nfor all $n$. Therefore,\n\n4.5-2. Professor Caesar wishes to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen\u2019s algorithm. His algorithm will use the divide-and-conquer method, dividing each matrix into pieces of size $n/4 \\times n/4$, and the divide and combine steps together will take $\\Theta(n^2)$ time. He needs to determine how many subproblems his algorithm has to create in order to beat Strassen\u2019s algorithm. If his algorithm creates $a$ subproblems, then the recurrence for the running time $T(n)$ becomes $T(n) = aT(n/4) + \\Theta(n^2)$. What is the largest integer value of $a$ for which Professor Caesar\u2019s algorithm would be asymptotically faster than Strassen\u2019s algorithm?\n\nWe recall that Strassen\u2019s time complexity is $\\Theta(n^{\\log_2 7})$. In light of this, it suffices to find the maximum integral value of $a$ such that\n\nsubject to the restriction that $f(n) = n^2$ is asymptotically smaller than $n^{\\log_b a}$.\n\nSince $b = 4$, the restriction yields the lower bound\n\nObserve that $\\log_4 x = \\log_2 7$ is equivalent to\n\nwhich, in turn, is equivalent to\n\nSince $\\log_4 2 = 1/2$, we conclude that\n\nIt follows that we must have $% $, whence the optimal integral value of $a$ is $48$. $\\square$\n\n4.5-3. Use the master method to show that the solution to the binary-search recurrence $T(n) = T(n/2) + \\Theta(1)$ is $\\Theta(\\log n)$. (See Exercise 2.3-5 for a description of binary search.)\n\nSince $a = 1$ and $b = 2$, we have $\\log_b a = 0$. Therefore, $f(n) = \\Theta(1)$ is asymptotically equivalent to $n^{\\log_b a}$, whence the master method implies that\n\nas was to be shown. $\\square$\n\n4.5-4. Can the master method be applied to the recurrence $T(n) = 4T(n/2) + n^2 \\log n$? Why or why not? Give an asymptotic upper bound for this recurrence.\n\nThe $\\log n$ term in $f(n) = n^2\\log n$ makes it impossible to apply the master method here. $\\log_b a = 2$, so $f(n) = \\Omega(n^{\\log_b a})$. Nevertheless, $\\log n = o(n^p)$ for all $p > 0$, and so $f(n) = O(n^{\\log b (a + \\epsilon)})$ for all $\\epsilon > 0$. It follows that none of the three cases of the master method is applicable.\n\nWe shall compute the recursion tree of $T(n)$. Observe:\n\nSince\n\nwe conjecture that\n\nTo show this, we let $C > 0$, to be chosen later. We assume that\n\nfor all $% $. (We pick 16, so that $\\log n > 4$.) We shall show that\n\nObserve that\n\nSince $\\log n > 4$, we have\n\nwhenever $C \\geq 1$.\n\nWe now set\n\nso that, for $n = 16$,\n\nMoreover, $C \\geq 1$, and so\n\nfor all $% $\n\nimplies\n\nIt now follows from induction that $T(n) = O(n^2(\\log n)^2)$. $\\square$\n\n4.5-5. Consider the regularity condition $af(n/b) \\leq cf(n)$ for some constant $% $, which is part of case 3 of the master theorem. Give an example of constants $a \\geq 1$ and $b > 1$ and a function $f(n)$ that satisfies all the conditions in case 3 of the master theorem except the regularity condition.\n\nLet\u2019s first rule out a general class of examples. Let $a \\geq 1$, $b > 1$, $C > 0$, and $\\epsilon > 0$ be arbitrary, and let\n\nSince\n\nwe see that setting $c = a/(a+\\epsilon)$ yields\n\nfor all $n \\in \\mathbb{N}$. $\\epsilon > 0$, and so $% $.\n\nThis suggests that we need an oscillating term. Let us pick $a = 1$, $b = 2$, and $f(n) = 2n - n\\cos(2 \\pi n)$. Observe that\n\nWhenever $n$ is odd, $\\cos(\\pi n) = -1$ and $\\cos(2 \\pi n) = 1$, and so\n\nIt follows that, whenever $% $, the inequality\n\ncannot hold for all $n$. $\\square$\n\n## 4.6. Proof of the master theorem\n\n### Exercises\n\n4.6-1. Give a simple and exact expression for $n_j$ in equation (4.27) for the case in which $b$ is a positive integer instead of an arbitrary real number.\n\nWe shall prove a more general result: whenever $m,n \\in \\mathbb{N}$ and $x \\in \\mathbb{R}$, we have the identity\n\nObserve that\n\nThis implies, in particular, that\n\nThe upper bound of $(\\ast)$ implies that\n\nWe suppose for a contradiction that\n\nWe can find an integer $p$ such that\n\nThis implies that\n\nso that\n\nThe lower bound of $(\\ast\\ast)$ implies that\n\nIt now follows from $(\\ast\\ast)$ that\n\nwhich is evidently absurd. We conclude that\n\nLet us now recall (4.27):\n\nWe shall show that\n\nwhenever $b$ is an integer. We proceed by mathematical induction. The $j = 0$ case is trivially true. If the $j = j_0 - 1$ case holds, then $(\\ast\\ast\\ast)$ implies that\n\nas was to be shown. $\\square$\n\n4.6-2. Show that if $f(n) = \\Theta(n^{\\log_b a} \\log^k n)$, where $k \\geq 0$, then the master recurrence has solution $T(n) = \\Theta(n^{\\log_b a} \\log^{k+1} n)$. For simplicity, confine your analysis to exact powers of $b$.\n\nWe begin with a remark that\n\nnot\n\nFix $k \\geq 0$. Recall from (4.21) that\n\nSince\n\nwe see that\n\nIt thus suffices to show that\n\nSince $b > 1$,\n\nand so\n\n$(\\ast)$ follows from $(\\ast\\ast)$ if we can prove, in addition, that\n\nTo this end, we recall that $n$ is restricted to be of the form $b^p$ for some exponent $p$. We shall show that\n\nfor all $p \\geq 1$, where $C$ is a constant independent of $p$.\n\nWe shall prove $(\\ast\\ast\\ast)$ by induction on $p$.\n\nWe tackle the inductive step first, so as to determine an appropriate value of $C$. Let us fix $p_0 > 1$ and assume that\n\nfor all $% $, where $C > 0$ is a constant to be chosen later. We shall show that\n\nTo this end, we observe that\n\nBy way of the inductive hypothesis $(\\star)$, we bound the above quantity below by\n\nTo show $(\\star)$, it now suffices to establish the estimate\n\nWe write\n\nfor each $p$ and observe that\n\n$b > 1$ implies that $\\log^k(b) > 0$, and so we need only to show that\n\nThe above inequality is equivalent to\n\nas $A_p > 1$ for all sufficiently large $p$. In fact, if $C \\geq \\frac{1}{\\log b}$, then $A_p > 1$ for all $p > 1$.\n\nIt thus suffices to establish\n\nIf $C = \\frac{1}{\\log b}$, then\n\nSince $p_0 > 1$, we see that\n\nfor all $k \\geq 1$, establishing $(\\star\\star)$ for $C = \\frac{1}{\\log b}$.\n\nIt follows that $(\\star)$ holds for $C = \\frac{1}{\\log b}$, which establishes the inductive case of $(\\ast \\ast \\ast)$.\n\nTo complete the proof of $(\\ast\\ast\\ast)$ by induction, we must establish the base case $p = 1$, i.e.,\n\nSince we have chosen $C = \\frac{1}{\\log b}$, the base case holds with equality. The induction proof is now complete.\n\nFinally, $(\\ast)$ follows from $(\\ast\\ast)$ and $(\\ast\\ast\\ast)$, and we have the desired estimate\n\nOur work here is done. $\\square$\n\n4.6-3. Show that case 3 of the master theorem is overstated, in the sense that the regularity condition $af(n/b) \\leq cf(n)$ for some constant $% $ implies that there exists a constant $\\epsilon > 0$ such that $f(n) = \\Omega(n^{\\log_b a + \\epsilon})$.\n\nWe recall from Theorem 4.1 that $a \\geq 1$ and $b > 1$. We let $C = \\frac{a}{c}$ and observe that the regularity condition can be rewritten as\n\nApplying $(\\ast)$ $\\lceil \\log_b n \\rceil$ times, we obtain\n\nSince $\\log_b n \\leq \\lceil \\log_b n \\rceil$, we have the estimate\n\nNow, $C^{\\log_b n} = n^{\\log_b C}$, and so\n\n$C = \\frac{a}{c}$ implies $\\log_b C = \\log_b a + \\log_b \\frac{1}{c}$, and so\n\nwhere $\\epsilon = \\log_b \\frac{1}{c}$. It follows that\n\nFinally, we remark that $% $ implies $\\epsilon > 0$. $\\square$\n\n## Problems\n\n### 4-1. Recurrence examples\n\nGive asymptotic upper and lower bounds for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for $n \\leq 2$. Make your bounds as tight as possible, and justify your answers.\n\na. $T(n) = 2T(n/2) + n^4$\n\nb. $T(n) = T(7n/10) + n$\n\nc. $T(n) = 16T(n/4) + n^2$\n\nd. $T(n) = 7T(n/3) + n^2$\n\ne. $T(n) = 7T(n/2) + n^2$\n\nf. $T(n) = 2T(n/4) + \\sqrt{n}$\n\ng. $T(n) = T(n-2) + n^2$\n\na-f can be solved directly by the master theorem (Theorem 4.1), which we restate below for ease of reference:\n\nTheorem 4.1 (Master theorem). Let $a \\geq 1$ and $b > 1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence\n\nwhere we interpret $n/b$ to mean either $\\lfloor n/b \\rfloor$ or $\\lceil n/b \\rceil$. Then $T(n)$ has the following asymptotic bounds:\n\n1. If $f(n) = O(n^{\\log_b a - \\epsilon})$ for some constant $\\epsilon > 0$, then $T(n) = \\Theta(n^{\\log_b a})$.\n\n2. If $f(n) = \\Theta(n^{\\log_b a})$, then $T(n) = \\Theta(n^{\\log_b a} \\log n)$.\n\n3. If $f(n) = \\Omega(n^{\\log_b a + \\epsilon})$ for some constant $\\epsilon > 0$, and if $af(n/b) \\leq cf(n)$ for some constant $% $ and all sufficiently large $n$, then $T(n) = \\Theta(f(n))$.\n\n#### Problem 4-1a\n\n$a = b = 2$, so that $\\log_b a = 1$. Since\n\n$f(n) = \\Omega(n^{\\log_b a + 3})$ and\n\nwe see that\n\n#### Problem 4-1b\n\n$a = 1$ and $b = \\frac{10}{7}$, so that $\\log_b a = 0$. Since\n\nand\n\nwe see that\n\n#### Problem 4-1c\n\n$a = 16$ and $b = 4$, so that $\\log_b a = 2$. Since $f(n) = \\Theta(n^{\\log_b a})$, we see that\n\n#### Problem 4-1d\n\n$a = 7$ and $b = 3$, so that $% $. Since\n\nand\n\nwe see that\n\n#### Problem 4-1e\n\n$a = 7$ and $b = 2$, so that $% $. Since\n\nwe see that\n\n#### Problem 4-1f\n\n$a = 2$ and $b = 4$, so that $\\log_b a = \\frac{1}{2}$. Since $f(n) = \\Theta(n^{\\log_b a})$, we see that\n\n#### Problem 4-1g\n\nWe cannot use the master theorem, as\n\nis not of the form\n\nObserve, however, that\n\nRecalling that\n\nwe obtain\n\nwhere $n' = \\lfloor n/2 \\rfloor$.Since $n' = \\Theta(n)$, we have\n\n### 4-2. Parameter-passing costs\n\nThroughout this book, we assume that parameter passing during procedure calls takes constant time, even if an $N$-element array is being passed. This assumption is valid in most systems because a pointer to the array is passed, not the array itself. This problem examines the implications of three parameter-passing strategies:\n\n1. An array is passed by pointer. Time = $\\Theta(1)$.\n\n2. An array is passed by copying. Time = $\\Theta(N)$, where $N$ is the size of the array.\n\n3. An array is passed by copying only the subrange that might be accessed by the called procedure. Time = $\\Theta(q-p+1)$ if the subarray $A[p..q]$ is passed.\n\na. Consider the recursive binary search algorithm for finding a number in a sorted array (see Exercise 2.3-5). Given recurrences for the worst-case running times of binary search when arrays are passed using each of the three methods above, and give good upper bounds on the solutions of the recurrences. Let $N$ be the size of the original problem and $n$ be the size of a subproblem.\n\nb. Redo part (a) for the Merge-Sort algorithm from Section 2.3.1.\n\n#### Problem 4-2a\n\nRecall binary_search from Exercise 2.3-5, which we reproduce below:\n\nfrom math import floor\n\ndef binary_search(A, p, q, v):\nif p > q :\nreturn -1\nif p == q:\nreturn p if A[p] == v else -1\n\nmid = p + floor((q-p)/2)\nif A[mid] == v:\nreturn mid\nelse:\nif A[mid] > v:\nreturn binary_search(A, p, mid-1, v)\nelse: # A[mid] < v\nreturn binary_search(A, mid+1, q, v)\n\n\nLet $T(N)$ be the worst-case running time of binary_search on a list of length $N$. Since it takes constant time to execute all but the recursive step of the algorithm, we have\n\nwhere $f(N,n)$ is the time it takes to pass a length-$n$ subarray of an array of length $N$.\n\nIf arrays are passed by pointer, then $f(N,n) = \\Theta(1)$, and so\n\nBy the master theorem (see Section 4.5), we have\n\nIf arrays are passed by copying, then $f(N,n) = \\Theta(N)$, and so\n\nBy the master theorem, we have\n\nFinally, if arrays are passed by copying the appropriate subranges, then $f(N,n) = N/n = N/\\lfloor N/2 \\rfloor$. We thus have $f(N,n) = \\Theta(1)$, and so\n\n#### Problem 4-2b\n\nWe recall the non-sentinel version of merge sort from Exercise 2.3-2, which we reproduce below:\n\nimport math\n\ndef merge(A, p, q, r):\nL, R = A[p,q], A[q:r]\nn,m = q-p, r-q\n\ni, j, k = 0, 0, p\n\nwhile i < n and j < m:\nif L[i] < R[j]:\nA[k] = L[i]\ni += 1\nelse:\nj += 1\nk += 1\n\nwhile i < n or j < m:\nif i < n:\nA[k] = L[i]\nelse:\nA[k] = R[j]\nk += 1\n\ndef merge_sort(A, p, r):\nif p+1 < r:\nq = p + math.floor((r-p)/2)\nmerge_sort(A, p, q)\nmerge_sort(A, q, r)\nmerge(A, p, q, r)\n\n\nNote that the time complexity of merge is $\\Theta(r-p)$. Since we divide the input array in half, the time complexity of merge_sort is\n\nwhere $f(a,b)$ is the time it takes to pass a length-$b$ subarray of an array of length $a$.\n\nSince $f(n, n/2) = \\Omega(n)$ regardless of how we pass in an array, we conclude that\n\nIt follows from the master theorem (see Section 4.5) that\n\n### 4-3. More recurrence examples\n\nGive asymptotic upper and lower bounds for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for sufficiently small $n$. Make your bounds as tight as possible, and justify your answers.\n\na. $T(n) = 4T(n/3) + n \\log n$.\n\nb. $T(n) = 3T(n/3) + n / \\log n$.\n\nc. $T(n) = 4T(n/2) + n^2 \\sqrt{n}$.\n\nd. $T(n) = 3T(n/3 - 2) + n/2$.\n\ne. $T(n) = 2T(n/2) + n/\\log n$.\n\nf. $T(n) = T(n/2) + T(n/4) + T(n/8) + n$.\n\ng. $T(n) = T(n-1) + 1/n$.\n\nh. $T(n) = T(n-1) + \\log n$.\n\ni. $T(n) = T(n-2) + \\log n$.\n\nj. $T(n) = \\sqrt{n} T(\\sqrt{n}) + n$.\n\nSee Section 4.5 for the statement of the master theorem.\n\n#### Problem 4-3a\n\nRecall from Problem 3-2 that\n\nfor every $\\epsilon > 0$. It follows that\n\nfor all $% $, whence the master theorem implies that\n\n#### Problem 4-3b\n\nWe fist show that the master theorem cannot be used here. To this end, we shall show that\n\nfor any choice of $\\epsilon > 0$,\n\nand that\n\nfor any choice of $\\epsilon > 0$.\n\nSince $\\log n \\geq 1$ for all $n \\geq 2$, we have that\n\nfor all $n \\geq 2$. This implies $(\\ast\\ast)$. $(\\ast)$ follows from $(\\ast)$.\n\nFor $(\\ast\\ast\\ast)$, we observe that each constant $C > 0$ furnishes an integer $M_C$ such that\n\nfor all $n \\geq M_C$. It follows that\n\nfor all $n \\geq M_C$, and $(\\ast\\ast\\ast)$ follows.\n\nFinally, we recall from Problem 3-2 that\n\nfor every $\\rho > 0$. This implies that each choice of $\\rho > 0$ and $C > 0$ furnishes an integer $N_{\\rho, C}$ such that\n\nfor all $n \\geq N_{\\rho, C}$. Therefore, we see that\n\nfor all $n \\geq N_{\\epsilon, C}{}_{}$, and $(\\ast\\ast\\ast\\ast)$ follows.\n\nLet us now proceed without the master theorem.\n\nWe let $k = \\log_3 n$ and observe that\n\nSince $% $, we see that\n\nMoreover, $% $, and so\n\nWe thus have the estimates\n\nIn order to estimate the sum, we define $f(x) = \\frac{1}{k-x}$ and observe that\n\nwhenever $i-1 \\leq x \\leq i$. Therefore,\n\nfor any choice of index $i$, whence it follows that\n\nSince $k - \\lfloor k \\rfloor \\geq 0$, we see that\n\nWe thus end up with the upper bound\n\nfor constants $C$, $D$, and $E$. We conclude that\n\nTo establish a lower bound, we set $f(x) = \\frac{1}{k-x}$ once againd and observe that\n\nwhenever $i \\leq x \\leq i+1$. Therefore,\n\nfor any choice of index $i$, whence it follows that\n\nSince $% $, we see that\n\nWe thus end up with the lower bound\n\nfor constants $C$ and $D$. We conclude that\n\nWe have now completed the proof of the estimate\n\n#### Problem 4-3c\n\nThe recurrence relation in question can be written as\n\nObserve that\n\nMoreover,\n\nfor all $n$, and $% $. We can therefore use the master theorem to conclude that\n\n#### Problem 4-3d\n\nWe cannot use the master theorem, as the recurrence relation is not of the form $T(n) = aT(n/b) + f(n)$. If, however, we drop the $-2$ term, we can use the master theorem on\n\nto obtain\n\nSince the $-2$ term shouldn\u2019t affect the asymptotic behavior of $T$, we conjecture that\n\nTo prove $(\\ast)$, we first fix $n > 6$, so that $\\frac{n}{3} - 2 > 1$. Assume inductively that\n\nfor all $% $, where $C > 0$ is a constant to be chosen later. Observe that\n\nIf we choose $C > \\frac{1}{2 \\log 3}$, then we have\n\nfor all $n \\geq 6$, and we have the desired bound\n\nTo establish the base case $n=6$, we choose\n\nso that the inductive step continues to holds, while\n\nfor $n = 6$. We now conclude by induction that\n\nTo prove $(\\ast)$, it now remains to show that\n\nTo this end, we fix $n > 24$, so that $\\frac{n}{3} - 2 > 2$ and $\\frac{n}{3} - 2 > \\frac{n}{4}$. Assume inductively that\n\nfor all $% $, where $C > 0$ is a constant to be chosen later. Observe that\n\nSince $\\frac{n}{3} - 2 > \\frac{n}{4}$, we have\n\nChoosing $C \\leq \\frac{1}{8}$ yields\n\nso that\n\nas was to be shown.\n\nTo establish the base case $n = 24$, we choose\n\nso that the inductive step continues to hold, while\n\nfor $n = 24$. We now conclude by induction that\n\nas was to be shown. This completes the proof that\n\n#### Problem 4-3e\n\nGeneralizing the computation carried out in Problem 4-3b, we show that a recurrence relation of the form\n\nsatisfies the asymptotic estimate\n\nWe let $k = \\log_a n$ and observe that\n\nWe have shown in Problem 4-3b that\n\nand so we conclude that\n\n#### Problem 4-3f\n\nThe recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.\n\nObserve that\n\nFrom this, let us guess that the accumulation of cost is\n\nfor some $k>0$. Since\n\nwe conjecture that\n\nTo verify this, we fix $n > 8$ and assume inductively that\n\nfor all $% $, where $C > 0$ and $D > 0$ are constants to be chosen. Observe that\n\nAnalogously,\n\nSo long as $C \\geq 7$ and $D \\leq 7$, we have\n\n$D \\leq \\frac{7}{8}(1+D) \\leq \\frac{7}{8}(1+C) \\leq C,$\\$\n\nwhich yields\n\nas was to be shown.\n\nTo establish the base case $n = 8$, we choose\n\nso that the inductive case continues to hold, while\n\nand\n\nas was to be shown. This completes the proof of\n\n#### Problem 4-3g\n\nThe recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.\n\nObserve that\n\nPage 1154 of CLRS shows that\n\nand so we conclude that\n\n#### Problem 4-3h\n\nThe recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.\n\nObserve that\n\nWe have proved in Exercise 3.2-3 that $\\log n! = \\Theta(n \\log n)$, and so we conclude that\n\n#### Problem 4-3i\n\nThe recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.\n\nObserve that\n\nwhere $k!!$ is the double factorial function\n\nWe let $m = \\lfloor n/2 \\rfloor$. If $m$ is even, then\n\nIn this case,\n\nwhich is $\\Theta(m \\log m) = \\Theta(n \\log n)$.\n\nIf $m$ is odd, then\n\nIn this case,\n\nWe have proved in Exercise 3.2-3 that $\\log k! = \\Theta(k \\log k)$, and so $\\log m!!$ has the asymptotic estimate of $\\Theta(m \\log m) = \\Theta(n \\log n)$.\n\nWe conclude that\n\n#### Problem 4-3j\n\nThe recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.\n\nLet us begin by making the substitution $n = 2^m$:\n\nSetting $S(m) = T(2^m)$, we see that\n\nObserve that\n\nSince\n\nwe see that $S(m) = \\Theta\\left( 2^m \\log m\\right)$.\n\nThe identity $m = \\log n$ now implies that\n\n### 4-4. Fibonacci numbers\n\nThe problem develops properties of the Fibonacci numbers, which are defined by recurrence (3.22). We shall use the technique of generating functions to solve the Fibonacci recurrence. Define the generating function (or formal power series) $\\mathscr{F}$ as\n\nwhere $F_i$ is the $i$th Fibonacci number.\n\na. Show that $\\mathscr{F}(z) = z + z \\mathscr{F}(z) + z^2 \\mathscr{F}(z)$.\n\nb. Show that\n\nwhere\n\nand\n\nc. Show that\n\nd. Use part (c) to prove that $F_i = \\phi^i / \\sqrt{5}$ for $i > 0$, rounded to the nearest integer. (Hint: Observe that $% $.)\n\n#### Problem 4-4a\n\nObserve that\n\nWe then have\n\nSince $F_0 = F_1 = 1$ and $F_i = F_{i-1} + F_{i-2}$ for all $i \\geq 2$, we conclude that\n\n#### Problem 4-4b\n\nIt follows at once from 4-4a that\n\nBy the quadratic formula,\n\nand the second equality follows.\n\nFinally, we observe that\n\nThe third equality now follows.\n\n(To deduce the third equality directly from the second equality, we must carry out the partial fraction decomposition of the second expression.) $\\square$\n\n#### Problem 4-4c\n\nPage 1147 of CLRS tells us that\n\nSubstituting $x = \\phi z$, we get\n\nSimilarly,\n\nThe desired identity now follows from the third equality in 4-4b. $\\square$\n\n#### Problem 4-4d\n\nWe have shown in 4-4c that\n\nfor all indices $i$. Since $% $, we have\n\nwhence it follows that\n\nIn other words, $F_i$ is $\\phi^i/\\sqrt{5}$ rounded to the nearest integer. $\\square$\n\n### 4-5. Chip testing\n\nProfessor Diogenes has $n$ supposedly identical integrated-circuit chips that in principle are capable of testing each other. The professor\u2019s test jig accommodates two chips at a time. When the jig is loaded, each chip tests the other and reports whether it is good or bad. A good chip always reports accurately whether the other chip is good or bad, but the professor cannot trust the answer of a bad chip. Thus, the four possible outcomes of a test are as follows:\n\nChip $A$ says Chip $B$ says Conclusion\n$B$ is good $A$ is good both are good, or both are bad\n$B$ is good $A$ is bad at least one is bad\n$B$ is bad $A$ is good at least one is bad\n$B$ is bad $A$ is bad at least one is bad\n\na. Show that if at least $n/2$ chips are bad, the professor cannot necessarily determine which chips are good using any strategy based on this kind of pairwise test. Assume that the bad chips can conspire to fool the professor.\n\nb. Consider the problem of finding a single good chip from among $n$ chips, assuming that more than $n/2$ of the chips are good. Show that $\\lfloor n/2 \\rfloor$ pairwise tests are sufficient to reduce the problem to one of nearly half the size.\n\nc. Show that the good chips can be identified with $\\Theta(n)$ pairwise tests, assuming that more than $n/2$ of the chips are good. Give and solve the recurrence that describes the number of tests.\n\n#### Notational Remark\n\nGiven two chips $x$ and $y$, we write $T_x(y)$ to denote what $x$ has to say about $y$. The possible outcomes are, of course, good, and bad.\n\nA test consists of inputting two chips $x$ and $y$ and receiving the output $(T_x(y), T_y(x)).$\n\n#### Problem 4-5a\n\nAssume that at least $n/2$ chips are bad. Let $\\mathcal{G}$ be the set of all good chips, and let $\\mathcal{B}$ be a set of bad chips of size $\\vert \\mathcal{G} \\vert$. We let $\\mathcal{R}$ be the remaining chips, if there are any.\n\nObserve that, for each $x \\in \\mathcal{G}$,\n\nLet us consider the following adversarial strategy: each $x \\in \\mathcal{B}$ returns\n\nand each $x \\in \\mathcal{R}$ returns\n\nWe now partition the collection of chips by declaring two chips $x$ and $y$ to be in the same category if and only if\n\nThe partition process produces three categories: namely, $\\mathcal{G}$, $\\mathcal{B}$, and $\\mathcal{R}$.\n\nA test does not show which category a pair of chips belongs to. It merely shows whether two chips belong to the same category. Since $\\mathcal{G}$ and $\\mathcal{B}$ are of the same size, this information is insufficient to distinguish the two categories from one another. It follows that the chosen adversarial strategy prevents distinguishing good chips from bad chips. $\\square$\n\n#### Problem 4-5b\n\nLet us assume that more than $n/2$ of the chips are good.\n\nWe assume that $n$ is even, and set $m = n/2$. If there is an odd number of chips, we discard one at random to make the number of chips even.\n\nWe divide the chips into two groups of equal sizes, $\\mathcal{X} = \\{x_1,\\ldots,x_m\\}$ and $\\mathcal{Y} = \\{y_1,\\ldots,y_m\\}$, assigning indices $1,\\ldots,m$ arbitrarily.\n\nFor each index $i$, we declare the pair $(x_i, y_i)$ to be in the keep pile $\\mathcal{K}$ if and only if\n\nIn other words, either $x_i$ and $y_i$ are both good chips or they are both bad chips.\n\nWe also define the discard pile $\\mathcal{D}$ to be the collection of all pairs $(x_i,y_i)$ that do not belong to $\\mathcal{K}$. We denote by $\\vert \\mathcal{K} \\vert$ and $\\vert \\mathcal{D} \\vert$ the number of pairs in $\\mathcal{K}$ and $\\mathcal{D}$, respectively. We note here that\n\nMoreover, each pair in $\\mathcal{D}$ must contain at least one bad chip, and so\n\nWe now observe that $\\mathcal{K}$ contains more than $\\vert \\mathcal{K} \\vert /2$ good pairs. Indeed, if $\\mathcal{K}$ contained at most $\\vert \\mathcal{K} \\vert / 2$ good pairs, then\n\nwhich is absurd.\n\nIn light of this observation, we define\n\nSince $\\mathcal{R}$ contains precisely $\\vert \\mathcal{K} \\vert$ many chips, we see from $(\\ast)$ that $\\mathcal{R}$ cannot contain more than $n/2 - \\vert \\mathcal{D} \\vert$ many chips. Moreover, more than half of the pairs in $\\mathcal{K}$ are good pairs, and so more than half of the chips in $\\mathcal{R}$ are good chips. We now recurse on $\\mathcal{R}$. $\\square$\n\nThe procedure described above can be implemented as follows:\n\n\"\"\"We identify chips by unique nonnegative integers and assume that\nchip_table, an immutable two-dimensional list of numbers, is given.\nTo see what a testing chip (\"testing\") has to say about another chip\n(\"tested\"), we call chip_table[testing][tested], which returns\nTrue for 'good', and False for 'bad'.\n\nWe record the chips to be tested in a list of numbers called chip_list.\nchip_list is changed at each recursive step to track the progress.\n\"\"\"\n\ndef get_a_good_chip(chip_table, chip_list):\nn = len(chip_list)\n\nif n <= 2:\nreturn chip_list[0]\n\nif n % 2 != 0:  # if length mod 2 is not zero, i.e., odd\nchip_list = chip_list[:-1]  # cut out the last element\nn = n-1\n\nm = n//2\nlist_x, list_y = chip_list[:m], chip_list[m:]\n\nkeep = []\nfor i in range(m):\nx, y = list_x[i], list_y[i]\nif chip_table[x][y] and chip_table[y][x]:  # if both return True\nkeep.append((x, y))\n\nnew_chip_list = [pair[0] for pair in keep]\n\nreturn get_a_good_chip(chip_table, new_chip_list)\n\n>>> # chips 1, 2, 3 are good\n>>> chip_table = [[True, False, True, False, True],\n...               [False, True, True, True, False],\n...               [False, True, True, True, False],\n...               [False, True, True, True, False],\n...               [True, False, False, False, True]]\n>>> get_a_good_chip(chip_table, [0, 1, 2, 3, 4])\n1\n\n>>> # chips 2, 3, 4, 5 are good\n>>> chip_table = [[True, True, False, False, False, False],\n...               [True, True, False, False, False, False],\n...               [False, False, True, True, True, True],\n...               [False, False, True, True, True, True],\n...               [False, False, True, True, True, True],\n...               [False, False, True, True, True, True]]\n>>> get_a_good_chip(chip_table, [0, 1, 2, 3, 4, 5])\n2\n\n\n#### Problem 4-5c\n\nOnce one good chip is found, it suffices to test all chips against the good chip to identify all good chips. This takes $\\Theta(n)$ pairwise tests and can be implemented as follows:\n\ndef get_all_good_chips(chip_table):\nn = len(chip_table)\ntesting = get_a_good_chip(chip_table, range(n))\n\ngood_chips = [tested for tested in range(n)\nif chip_table[testing][tested]]\n\nreturn good_chips\n\n>>> # chips 1, 2, 3 are good\n>>> chip_table = [[True, False, True, False, True],\n...               [False, True, True, True, False],\n...               [False, True, True, True, False],\n...               [False, True, True, True, False],\n...               [True, False, False, False, True]]\n>>> get_all_good_chips(chip_table)\n[1, 2, 3]\n\n>>> # chips 2, 3, 4, 5 are good\n>>> chip_table = [[True, True, False, False, False, False],\n...               [True, True, False, False, False, False],\n...               [False, False, True, True, True, True],\n...               [False, False, True, True, True, True],\n...               [False, False, True, True, True, True],\n...               [False, False, True, True, True, True]]\n>>> get_all_good_chips(chip_table)\n[2, 3, 4, 5]\n\n\nThe time complexity $S$ of get_all_good_chips is\n\nwhere the time complexity $T$ of get_a_good_chip is\n\nThe master theorem now implies that\n\nwhence it follows that\n\n### 4-6. Monge arrays\n\nAn $m \\times n$ array $A$ of real numbers is a Monge array if for all $i$, $j$, $k$, and $l$ such that $% $ and $% $, we have\n\nIn other words, whenever we pick two rows and two columns of a Monge array and consider the four elements at the intersection of the rows and the columns, the sum of the upper-left and lower-right elements is less than or equal to the sum of the lower-left and upper-right elements. For example, the following array is Monge.\n\na. Prove that an array is Monge if and only if for all $i=1,2,\\ldots, m-1$ and $j=1,2,\\ldots,n-1$, we have\n\n(Hint: For the \u201cif\u201d part, use induction separately on rows and columns.)\n\nb. The following array is not Monge. Change one element in order to make it Monge. (Hint: Use part(a))\n\nc. Let $f(i)$ be the index of the column containing the leftmost minimum element of row $i$. Prove that $f(1) \\leq f(2) \\leq \\cdots \\leq f(m)$ for any $m \\times n$ Monge array.\n\nd. Here is a description of a divide-and-conquer algorithm that computes the leftmost minimum element in each row of an $m \\times n$ Monge array $A$:\n\nConstruct a submatrix $A'$ of $A$ consisting of the even-numbered rows of $A$. Recursively determine the leftmost minimum for each row of $A'$. Then compute the leftmost minimum in the odd-numbered rows of $A$.\n\nExplain how to compute the leftmost minimum in the odd-numbered rows of $A$ (given that the leftmost minimum of the even-numbered rows is known) in $O(m+n)$ time.\n\ne. Write the recurrence describing the running time of the algorithm described in part (d). Show that its solution is $O(m + n \\log m)$.\n\n#### Notational remark\n\nAs stated in the preface, we shall use the 0-first indexing convention for this problem.\n\n#### Problem 4-6a\n\n$(\\Rightarrow)$ If $A$ is Monge, it suffices to pick $k = i+1$ and $l = j+1$. $\\square$\n\n$(\\Leftarrow)$ We suppose that\n\nfor all $i=1,2,\\ldots,m-1$ and $j=1,2,\\ldots,n-1$.\n\nWe fix $i$ and $j$ and show that\n\nfor all $p \\geq 1$ and $q \\geq 1$.\n\nObserve that $(\\ast)$ is the $p = q = 1$ case. We let $q = 1$, fix $p > 1$, and assume inductively that\n\nfor all $% $. By the inductive hypothesis,\n\n$(\\ast)$ implies that\n\nwhence it follows that\n\nas was to be shown. This establishes $(\\ast\\ast)$ for all $p \\geq 1$ and $q = 1$.\n\nWe now fix $p > 1$ and $q > 1$ and assume inductively that\n\nfor all $% $. By the inductive hypothesis,\n\nThe $q =1$ case of $(\\ast\\ast)$ implies that\n\nwhence it follows that\n\nas was to be shown. This establishes $(\\ast\\ast)$ in full generality.\n\n$(\\ast)$ follows at once from $(\\ast\\ast)$. $\\square$\n\n#### Problem 4-6b\n\nThe following code tests whether a two-dimensional NumPy array is Monge, using the criterion developed in 4-6a:\n\n\"\"\"A is assumed to be a two-dimensional NumPy array.\n\n- The syntax for referencing an entry of A is A[i, j].\n- A.shape returns (m, n), the size of the matrix.\n\"\"\"\n\ndef test_monge(A):\n\"\"\"Check if A is a Monge array using the equivalent definition\nderived in 4-6a\"\"\"\nnum_rows, num_cols = A.shape\nfailed = []\nfor i in range(num_rows-1):\nfor j in range(num_cols-1):\nif A[i, j]+A[i+1, j+1] > A[i, j+1]+A[i+1, j]:\nfailed.append((i, j))\n\nif not failed:\nprint(\"Monge\")\nis_monge = True\nelse:\nprint(\"Not Monge\")\nprint(\"Failed coordinates: \")\nfor pair in failed:\nprint(pair)\nis_monge = False\nreturn is_monge\n\n>>> test_monge(np.array([[10, 17, 13, 28, 23],\n...                      [17, 22, 16, 29, 23],\n...                      [24, 28, 22, 34, 24],\n...                      [11, 13,  6, 17,  7],\n...                      [45, 44, 32, 37, 23],\n...                      [36, 33, 19, 21,  6],\n...                      [75, 66, 51, 53, 34]])\nMonge\n\n>>> test_monge(np.array([[37, 23, 22, 32],\n...                      [21,  6,  7, 10],\n...                      [53, 34, 30, 31],\n...                      [32, 13,  9,  6],\n...                      [43, 21, 15,  8]])\nNot Monge\nFailed coordinates:\n(0, 1)\n\n\nAs the program suggests, the inequality\n\nfails to hold, as\n\nSince we have\n\nwe can add up to 7 to $A[0,2]$ without breaking the inequality\n\nWe therefore assign\n\nat which point we obtain\n\nwhile maintaining\n\nSince the modification affects no other inequality, the resulting array is Monge.\n\n>>> test_monge(np.array([[37, 23, 29, 32],\n...                      [21,  6,  7, 10],\n...                      [53, 34, 30, 31],\n...                      [32, 13,  9,  6],\n...                      [43, 21, 15,  8]])\nMonge\n\n\n#### Problem 4-6c\n\nLet $A$ be an $m \\times n$ Monge array. We fix $% $ and assume for a contradiction that\n\non $A$. Since\n\nwe see that the inequality\n\ncan never hold unless it is also the case that\n\nSince $f(i)$ is the index of the column containing the leftmost minimum element of row $i$, $(\\ast)$ implies that we must always have\n\nrendering the identity\n\nfalse. We thus conclude that\n\nSince the choice of $i$ was arbitrary, the desired result now follows. $\\square$\n\n#### Problem 4-6d\n\nWe assume that the values of $f(0),f(2),\\ldots, f(2 \\lfloor (n-1)/2 \\rfloor)$ are known. We have seen in 4-6c that\n\nfor all $k$. Since we must examine at least one integer for each row, the number of integers we must examine to compute $f(2k+1)$ is bounded above by\n\nIt then follows that the number of integers we must examine for all the odd rows is bounded above by\n\nWe now observe that\n\nwhence the desired result follows. $\\square$\n\n#### Problem 4-6e\n\nWe begin by observing that a submatrix $A'$ of a Monge array $A$ consisting of the rows of $A$ is a Monge array. This, in particular, implies that the ordering relation proved in 4-6c continues to hold on such submatrices.\n\nThe algorithm for computing the leftmost minimum of each row of a Monge array can be implemented as follows:\n\n\"\"\"A is assumed to be a two-dimensional NumPy array.\n\n- The syntax for referencing an entry of A is A[i, j].\n- A.shape returns (m, n), the size of the matrix.\n- A[::2,:] is the submatrix consisting of the even rows of A.\n0, k, 2k, 3k, and so on.\n\"\"\"\n\ndef find_left_min(A):\nif not test_monge(A):\nreturn None\nnum_rows, num_cols = A.shape\nmins = [0 for _ in range(num_rows)]  # initialize array of length A.shape[0]\nreturn lm_recursion(A, mins)\n\ndef min_index(row, start_index, end_index):\n\"\"\"Scan a row from start_index to end_index-1 to find the\nindex of the leftmost minimum element in the specified range.\n\"\"\"\nmin_ind = start_index\nfor i in range(start_index, end_index):\nif row[i] < row[min_ind]:\nmin_ind = i\nreturn min_ind\n\ndef lm_recursion(A, mins):\nnum_rows, num_cols = A.shape\nif num_rows == 1:  # if A has only one row, use the min_index scan\nmins[0] = min_index(A[0], 0, num_cols)\nelse:\n## take the even rows and recurse\nevens = A[::2, :]\neven_mins = lm_recursion(evens, mins[::2])\n## even rows have already been computed;\n## use the min_index scan in the range specified by even_mins\n## to figure out the min index for odd rows.\nstart_index = 0\nfor i in range(num_rows):\nif i % 2 == 0:\nmins[i] = even_mins[i//2]\nelse:\nleft = mins[i-1]\nright = even_mins[(i+1)//2]+1 if i+1 < num_rows else num_cols\nmins[i] = min_index(A[i], left, right)\nreturn mins\n\n>>> find_left_min(np.array([[10, 17, 13, 28, 23],\n...                         [17, 22, 16, 29, 23],\n...                         [24, 28, 22, 34, 24],\n...                         [11, 13, 6,  17, 7 ],\n...                         [45, 44, 32, 37, 23],\n...                         [36, 33, 19, 21, 6 ],\n...                         [75, 66, 51, 53, 34]))\n[0, 2, 2, 2, 4, 4, 4]\n>>> find_left_min(np.array([[37, 23, 29, 32],\n...                         [21,  6,  7, 10],\n...                         [53, 34, 30, 31],\n...                         [32, 13,  9,  6],\n...                         [43, 21, 15,  8]]))\n[1, 1, 2, 3, 3]\n\n\nWhat is the running time $T(m)$ of find_left_min, where $m$ is the number of rows? For the purpose of this discussion, let us focus on analyzing lm_recursion.\n\nLet $A$ be an $m \\times n$ array, so that num_rows == m and num_cols == n. If $m=1$, then the running time of lm_recursion is precisely that of min_index. Since min_index scans through each column once, we see that the running time is $\\Theta(n)$. We thus declare\n\nAs for the $m > 1$ case, we observe that lm_recursion consists of three steps:\n\n1. Construct evens, the subarray of even rows, from $A$.\n2. Recurse on evens.\n3. Go through the $m$ rows and compute mins.\n\nSince $A$ is never modified throughout lm_recursion, there is no need to copy the data of $A$ to construct evens, which becomes $A$ in the recursive step. The construction of even thus reduces to collecting pointers to the even rows, which can be done in $\\Theta(m/2)$ time.\n\nThe recursion step is invoked on evens, an array of $m/2$ rows. Since the recursion step is invoked precisely once without any additional procedure, it takes $T(m/2)$ time to run.\n\nFinally, we have shown in 4-6d that the computation of mins takes $\\Theta(m+n)$ time. It follows that the running time of lm_recursion is\n\nSince we do not know what $n$ is in relation to $m$, we cannot apply the master theorem. Observe, however, that\n\nSince\n\nfor all choices of $I$, we see that\n\nWe have shown above that $T(1) = \\Theta(m)$, and so\n\nLet us now prove the above estimate by induction. To this end, we fix $n$ and choose constants $k,K > 0$ such that\n\nfor all choices of $m$.\n\nLet us first establish the upper bound\n\nFix $m \\geq 2$ and assume inductively that\n\nfor all $% $, where $C,D > 0$ are constants to be chosen.\n\nObserve that\n\nSo long as\n\nwe have\n\nThe above condition is satisfied precisely when\n\nTo establish the base case $m = 1$, we only need the hypothesis\n\nIndeed, this leads to\n\nWe thus choose\n\nso that $(\\ast\\ast)$ is trivially established, and that\n\nthereby establishing $(\\ast)$. It now follows from induction that\n\nIt remains to establish the lower bound\n\nWe fix $m \\geq 2$ and assume inductively that\n\nfor all $% $, where $C,D > 0$ are constants to be chosen. Observe that\n\nIn order to obtain\n\nwe must have\n\nand\n\nObserve that the second inequality, along with the added hypothesis that\n\nimplies\n\nwhich is precisely the first inequality. The inductive step thus holds true so long as\n\nTo establish the base case $m = 1$, we only need the hypothesis\n\nIndeed, this leads to\n\nWe therefore choose\n\nso that $(\\star)$ and $(\\star\\star)$ follow. We now conclude by induction that\n\nwhich, along with the upper bound established above, implies\n\nas was to be shown. $\\square$\n\n## Additional remarks and further results\n\n### Generalizations of the master method\n\nThe master method does not apply to divide-and-conquer problems with multiple subproblems of substantially different sizes. For example,\n\nis not covered by the master method.\n\nMohamad Akra and Louay Bazzi, in their 1996 paper \u201cOn the Soluton of Linear Recurrence Equations,\u201d present a generalization of the master method that is applicable to recurrence relations of the form\n\nwith minor restrictions on $f$. The solution is given by the formula\n\nwhere $p$ is the unique real number that satisfies the identity\n\nFor the above example, we have $p = 1$, and so\n\nUsing the Akra\u2014Bazzi method, we can also solve more complicated recurrence relations like\n\nIn this case, we have\n\nand so the solution to the recurrence relation is\n\nThere are, of course, several generalizations beside the Akra\u2014Bazzi method. We refer the reader to Chee Yap\u2019s 2011 paper, \u201cA real elementary approach to the master recurrence and generalizations.\u201d $\\square$\n\n### Generating functions\n\nWe have discussed briefly in Problem 4-4 the generating-function technique of solving recurrence relations. We pursue the subject further here, following the treatment in Chapter 3 of Sedgewick/Flajolet.\n\nGiven a sequence $a_0,a_1,\\ldots,a_k,\\ldots$, we construct its ordinary generating function\n\nand its exponential generating function\n\nThe generating-function method of solving recurrences typically involves multiplying both sides of the recurrence by $z^n$ ($z^n/n!$ for exponential generating functions), summing on $n$, manipulating the resulting identity to derive an explicit formula for the generating function, and computing the power-series expansion of the generating function.\n\nTake, for example,\n\nwith the initial condition $a_0 = 0$, $a_1 = 1$, $a_2 = 1$. We multiply both sides by $z^n$, sum over $n \\geq 3$, and reindex the sums to obtain the identity\n\nFor notational simplicity, we let\n\nthe ordinary generating function of $(a_{n})_{n}$. Observe that\n\nWe can use the above computation to simplify $(\\ast)$ as follows:\n\nUsing the initial condition $a_0 = 0$, $a_1 = 1$, $a_2 = 1$, we conclude that\n\nTaking the partial fraction decomposition, we obtain\n\nWe now recall that\n\nfrom which we conclude that\n\nIt follows that\n\nfor all $n \\geq 0$.\n\nWe shall revisit the method of generating functions in later chapters, as more sophisticated recurrence relations arise. $\\square$", "date": "2018-05-22 11:39:34", "meta": {"domain": "markhkim.com", "url": "https://markhkim.com/clrs/ch04/", "openwebmath_score": 0.9267470836639404, "openwebmath_perplexity": 732.0085319858442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.962673113726775, "lm_q2_score": 0.8991213840277783, "lm_q1q2_score": 0.8655599823803488}}
{"url": "https://math.stackexchange.com/questions/2774156/why-do-integrals-start-at-0", "text": "# Why do integrals \u201cstart\u201d at 0? [duplicate]\n\nThis is a dumb question and I don't really know how to word it. When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0). I can easily see how the derivative of an integral is given by the function value, but why does the integral start at 0 and not any other number? When I try to imagine the area of some curve starting at negative 1 for example the area under the curve would intuitively to me still be given by the antiderivative. 0 makes sense as a starting point but for some reason I can't visualize it. I'm not sure if that made any sense but if anyone could help me wrap my head around it I'd appreciate it.\n\n## marked as duplicate by BallBoy, \u0430\u0441\u0442\u043e\u043d \u0432\u0456\u043b\u043b\u0430 \u043e\u043b\u043e\u0444 \u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433, Claude Leibovici\u00a0integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 10 '18 at 6:38\n\n\u2022 It doesn't matter. It's just convenient. As for your example, $F(x)=\\int_{-1}^x f(t)dt$ is another anti-derivative of $f$. \u2013\u00a0Yanko May 9 '18 at 19:09\n\u2022 \"given by the antiderivative\" There is no \"the antiderivative\". There is only \"an antiderivative\". \u2013\u00a0Arthur May 9 '18 at 19:13\n\u2022 $\\int dx/x$ certainly doesn't start from $0$! \u2013\u00a0Chappers May 9 '18 at 19:15\n\u2022 Starting from somewhere else amounts to changing the constant of integration. \u2013\u00a0amd May 9 '18 at 20:46\n\u2022 @LSpice, I may have been misinterpreting what op meant. I was essentially thinking the confusion was about why an integral over a degenerate interval was zero. \u2013\u00a0jdods May 10 '18 at 3:11\n\nWhen you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0).\n\nThis is not true. In certain situations it may be the case, but not generally. I think the reason this confusion arises is that a common problem given to calculus students is to find the antiderivative of a polynomial, e.g.,\n\n$$\\int x^3 +2x \\, dx = \\frac{1}{4}x^4 + x^2 + C$$\n\nand in this case, if we set $C = 0$ we get\n\n$$\\frac{1}{4}x^4 + x^2$$\n\nwhich is the same as $$\\int_0^x u^3 +2u \\, du = \\frac{1}{4}u^4 + u^2 \\big|_{u=0}^{u=x}.$$\n\nThis will work whenever the form that the antiderivative $F$ of $f$ you get takes satisfies $F(0) = 0$. But in general, setting $C = 0$ will not get you the integral $\\int_0^x f(t) \\, dt$. For example, if you take $f(x) = e^x$ then $$\\int e^x \\, dx = e^x + C$$ but setting $C = 0$ gives you $e^x$, which is not the same as $$\\int_0^x e^t \\, dt = e^x - 1.$$\n\nNote that \"setting \"C = 0\" in the expression for the antiderivative\" is not actually a well-defined operation. Different methods of antidifferentiation can give you different expressions when you set $C = 0$. It is important to remember that there is no single antiderivative, and no canonical way of writing it. $\\int 2x \\, dx = x^2 + 3 + C$ is just as valid as $\\int 2x \\, dx = x^2 + C$.\n\n\u2022 Ahh you're e^x example shows that my assumption was wrong. Thanks for clearing it up. \u2013\u00a0conyare May 9 '18 at 19:39\n\u2022 My favourite example is arcsin(x)+C vs -arccos(x)+C, these are equal families, but which function is supposed to be C=0? \u2013\u00a0Serge Seredenko May 9 '18 at 20:27\n\u2022 @SergeSeredenko, or $\\sin(x)^2 + C$ versus $-\\cos(x)^2 + C$. \u2013\u00a0LSpice May 10 '18 at 1:03\n\u2022 Also see this recent question where the difference between two solutions doesn't even look like a constant, but it is: math.stackexchange.com/q/2764985/78887 \u2013\u00a0Euro Micelli May 10 '18 at 3:18\n\nIntegrals don't always start at $0$. Let's start from definite integrals and move to the indefinite integrals you asked about. We know $\\int_a^b f(x)dx$ gives the area under the curve between $a$ and $b$. If $f(x)$ has an antiderivative $F(x)$, the Fundamental Theorem of Calculus tells us that $\\int_a^b f(x) = F(b) - F(a)$. Thus, if we want the area under the curve from $a$ to $b$, we compute $F(b) - F(a)$. If we want the area under the curve from $0$ to $b$, we compute $F(b) - F(0)$. $F(b) - F(0)$, as a function of $b$, gives us the area from $0$ to $b$.\n\nNow there are a lot of \"natural\" functions where $F(0) = 0$ (e.g., functions like $x^2$ or $\\sin x$), so to $F(b)$ gives the area from $0$ to $b$. But that won't be the case if $F(0) \\neq 0$.\n\nThe above should make it clear that there's no reason $0$ is special -- if you want the area from $-1$ to $b$ as a function of $b$, just use $F(b) - F(-1)$.\n\n\u2022 I assumed that the fundamental theorem worked because F(a) gets the area to A from 0 and F(b) gets the area to B from 0 so subtracting will get the desired area in-between them. \u2013\u00a0conyare May 9 '18 at 19:18\n\u2022 @conyare No, it actually goes the other way around! It is true that if all you knew were the 0-to-b version of the fundamental theorem, you could recover the a-to-b version. But the subtraction of another value is \"fundamental\" (pun somewhat intended) to the fundamental theorem. Jeffery Opoku-Mensah hints at a reason why this is true: since $F$ is defined only up to a constant, $F(b)$ doesn't give you a definite answer, only up to a constant. Only once you subtract $F(b)-F(a)$ does the constant cancel and you get a definite area. \u2013\u00a0BallBoy May 9 '18 at 19:21\n\nWell, integrals (by which I think you mean anti derivatives) don't always start at $0$. Indeed, if a function $f=f(x)$ is continuous in some interval $I=[a,b]$, then it has an anti derivative given by $$\\int_c^x{f(t)dt},$$ where $c\\in I$. The $0$ is usually chosen for $c$ only as a matter of convenience. A well-known function defined by an antiderivative that \"starts\" from (note that the notion of starting from for antiderivatives should not be taken too literally since the function is defined even for $x<c\\in I$) $1$, not $0$ as usual, is the logarithm function $$\\log x=\\int_1^x{\\frac{1}{t}dt}$$ defined for all $x>0$.\n\nThe antiderivative is generally given as the \"simplest\" form, and often the simplest form has a y value of zero when x is zero. For instance, if the function is constant, then for the antiderivative, you need a line whose slop is equal to that constant value. The simplest way of writing that is y = mx+C. You could write y = mx+5-C, and that would be a valid antiderivative, but that would be needlessly complicated. Since you're looking for simple functions, zero will pop up a lot. But there are cases where the simplest function doesn't go through the origin. For instance, if you're taking the antiderivative of sin(x), the simplest form is cos(x)+C. If you want to \"start\" at zero, you'd have to do cos(x)-1+C. Similarly, the antiderivative of e^x is generally given as e^x+C.\n\nThere are some cases where the same function can have different antiderivatives that look very different, but are actually the same thing. For instance, $\\frac{1}{\\sqrt{1-x^2}}$ can be integrated as either arcsin(x) or -arccos(x). But these just differ by the constant $\\pi /2$.\n\nIt depends on what antiderivative you take. For example, an antiderivative of $2x$ can be $x^2$, $x^2+1$, $x^2+100$, or whatever. By the Fundamental Theorem, it turns out $x^2$ gives the area under the curve starting at zero. But $x^2+1$ would also give the area of the curve starting at $-1$, for example.\n\n\u2022 I'm afraid you're confusing derivatives with antiderivatives... \u2013\u00a0zipirovich May 9 '18 at 19:22\n\u2022 Yea, I wrote this in a confused rush, but the idea is still there. Just switch the x^2 and 2x. \u2013\u00a0Jeffery Opoku-Mensah May 9 '18 at 21:59", "date": "2019-08-20 09:38:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2774156/why-do-integrals-start-at-0", "openwebmath_score": 0.8831784725189209, "openwebmath_perplexity": 299.12070289379966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759649262344, "lm_q2_score": 0.8824278772763471, "lm_q1q2_score": 0.8655522956012457}}
{"url": "https://mathematica.stackexchange.com/questions/145050/filling-space-with-pursuit-polygons?noredirect=1", "text": "Filling Space with Pursuit Polygons\n\nI want to make a program which can fill a 2D space with \"pursuit polygons\".\nYou can also look up \"pursuit curves\" or mice problem or watch this gif\n\nWhat I've tried so far:\n\nFirst I tried to produce these polygons by rotation of polygons\nthe square for example\n\ndata = {{0, 0}, {1, 0}, {1, 1}, {0, 1}, {0, 0}};\nGraphics[{Table[{Scale[Rotate[Line[data], 3*x Degree], {x, x}]}, {x,\n0, 20}]}]\n\n\nthen I decided to use spirals\nI realised that instead of using n spirals for every n-polygon,\nI can use only one(!) and produce the effect I want with \"fine tuning\"\nhere are the \"pursuit polygons\" that I made\n\nTriangle:\n\na = 41.9;\nb = 100;\nW = Table[{t^2*Cos[a*t], t^2*Sin[a*t]}, {t, 0, b}];\nAppendTo[W, W[[-4]]];\nListPlot[W]\nGraphics[Line[W]]\n\n\nSquare:\n\nS = Table[{t^2*Cos[42.4*t], t^2*Sin[42.4*t]}, {t, 0, 160}];\nAppendTo[S, S[[-4]]];\nGraphics[Line[S]]\n\n\nPentagon\n\nP = Table[{t^2*Cos[45.251*t], t^2*Sin[45.251*t]}, {t, 0, 220}];\nAppendTo[P, P[[-5]]];\nGraphics[Line[P]]\n\n\nThen I tried to put all these together by rotating and scaling...\n(but I'm not satisfied with the result)\nnotice that for every polygon I use also the \"anti-clockwise\" version of the polygon, which produces interesting results\n\nT = Table[{t^2*Cos[41.9*t], t^2*Sin[41.9*t]}, {t, 0, 100}];\nAppendTo[T, T[[-4]]]; (*Triangle*)\nS = Table[{t^2*Cos[42.4*t], t^2*Sin[42.4*t]}, {t, 0, 160}];\nAppendTo[S, S[[-4]]];   (*square*)\nS1 = Table[{t^2*Cos[-42.4*t], t^2*Sin[-42.4*t]}, {t, 0, 160}];\nAppendTo[S1, S1[[-4]]];    (*Anti-clockwise Square*)\nP = Table[{t^2*Cos[45.251*t], t^2*Sin[45.251*t]}, {t, 0, 220}];\nAppendTo[P, P[[-5]]];      (*Pentagon*)\nP1 = Table[{t^2*Cos[-45.251*t], t^2*Sin[-45.251*t]}, {t, 0, 220}];\nAppendTo[P1, P1[[-5]]];     (*Anti-clockwise Pentagon*)\nGraphics[{Translate[\nRotate[Scale[Line[(T)], {2.09, 2.09}], -67 Degree], {29500, 3800}],\nRotate[Scale[Line[S + 0], {1, 1}], -30 Degree],\nTranslate[\nRotate[Scale[Line[(S1)], {.98, .98}], -87.5 Degree], {41000,\n24700}],\nTranslate[\nRotate[Scale[Line[P1], {.605, .665}], -20.5 Degree], {76500,\n6500}], Scale[\nTranslate[\nRotate[Scale[\nLine[(T)], {1.2, 1.7}], -108 Degree], {50900, -12500}], {1.91,\n1.31}],\nTranslate[\nRotate[Scale[Line[(P)], {.525, .665}], -64 Degree], {3000, 36000}],\nTranslate[\nRotate[Scale[Line[(P)], {.61, .61}], -87 Degree], {59000, 58000}],\nTranslate[\nRotate[Scale[Line[(T)], {2.1, 1.8}], 12 Degree], {82500, 39500}],\nTranslate[\nRotate[Scale[Line[(T)], {2.3, 1.9}], -133 Degree], {32000,\n67000}]}]\n\n\nCan you find a way to divide and fill any given space with pursuit polygons?\nThe result would look better if this could work with ANY convex polygon and not only the regular polygons that I used...\n\n\u2022 Are there any other names for that? Because I think this was asked before. \u2013\u00a0Kuba May 2 '17 at 10:05\n\u2022 I provided all the names I could think of. There are not many images outthere.Even the first picture is something that I made a while ago. \u2013\u00a0J42161217 May 2 '17 at 10:14\n\u2022 @Kuba Lucas, E. (1877). Probl\u00e8me des trois chiens. Nouvelle Correspondance Math\u00e9matique, 3, 175\u2013176. \u2013\u00a0Michael E2 May 2 '17 at 11:54\n\u2022 A related question. \u2013\u00a0J. M. is away May 11 '17 at 13:15\n\u2022 And a very related blog-post of mine. And if I'm not completely off here, the word \"spacefilling\" is incorrect in this context. \u2013\u00a0halirutan Jun 11 '17 at 15:01\n\nThe reason that these are called \"pursuit polygons\" is because they are formed from a dynamical system in which different agents pursue each other. Example:\n\nIn this image, one agent starts in each corner of the triangle. The agent starting in the lower right corner pursues the agent starting in the top corner, the agent in the top corner pursues the agent in the lower left corner, and the agent in the lower left corner pursues the agent in the lower right corner.\n\nDrawing lines in between the agents yields a series of triangles which shrink as the agents get closer to each other, and also rotate:\n\nThe corresponding differential equations are: \\begin{align*} \\dot{\\mathbf{x}}_i &= \\mathbf{x}_{i+1} - \\mathbf{x}_i,\\ i\\in\\{1,\\dots,N-1\\}\\\\ \\dot{\\mathbf{x}}_i &= \\mathbf{x}_1 - \\mathbf{x}_N,\\ i=N, \\end{align*} where $N$ is the number of agents. The agents start in corners for the visualizations we are making, but in general they don't have to.\n\nThis code solves the differential equations using Euler integration and plots the lines between the agents:\n\nintegrate[pts_, h_] := NestList[Nest[step[#, h] &, #, 10] &, pts, 100]\nstep[pts_, h_] := pts - h (pts - RotateRight[pts])\n\nwrapAround[pts_] := Append[pts, First[pts]]\n\nPursuitPolygon[pts_, h_: 0.01] := Graphics[{\nLine[wrapAround[pts]],\nLine[wrapAround /@ integrate[pts, h]]\n}]\n\n\nFor regular polygons, this results in logarithmic spirals, as you have pointed out:\n\nGrid@Partition[Table[PursuitPolygon[N@CirclePoints[n]], {n, 2, 10}], 2]\n\n\nBut this approach is more general and works for any convex polygons, as you required. The following is an example of how we can fill in a triangulated region with spirals using this system:\n\nreg = DiscretizeRegion[Rectangle[], MaxCellMeasure -> 0.1]\n\n\nShow[PursuitPolygon @@@ MeshPrimitives[reg, 2]]\n\n\nWikipedia's entry on the mice problem suggests that the mice move at unit speed, i.e. like this: \\begin{align*} \\dot{\\mathbf{x}}_i &= \\frac{\\mathbf{x}_{i+1} - \\mathbf{x}_i}{\\|\\mathbf{x}_{i+1} - \\mathbf{x}_i\\|},\\ i\\in\\{1,\\ldots,N-1\\}\\\\ \\dot{\\mathbf{x}}_i &= \\frac{\\mathbf{x}_1 - \\mathbf{x}_N}{\\|\\mathbf{x}_1 - \\mathbf{x}_N\\|},\\ i=N, \\end{align*} This turns out to be difficult to implement because the denominator grows large as the points approach each other. I spoke to halirutan and MichaelE2 about this. Here is MichaelE2's solution, which I implemented for triangles but it could be implemented for any polygon.\n\nPursuitPolygon[{pt1_, pt2_, pt3_}] := Module[{},\n\n{sol1x, sol1y, sol2x, sol2y, sol3x, sol3y} = NDSolveValue[{\n{x1'[t], y1'[t]} == Normalize[{x2[t], y2[t]} - {x1[t], y1[t]}],\n{x2'[t], y2'[t]} == Normalize[{x3[t], y3[t]} - {x2[t], y2[t]}],\n{x3'[t], y3'[t]} == Normalize[{x1[t], y1[t]} - {x3[t], y3[t]}],\n{x1[0], y1[0]} == pt1,\n{x2[0], y2[0]} == pt2,\n{x3[0], y3[0]} == pt3,\nWhenEvent[Norm[{x2[t], y2[t]} - {x1[t], y1[t]}] < 1*^-8, \"StopIntegration\"]},\n{x1, y1, x2, y2, x3, y3},\n{t, 0, 10}\n];\n\n{tmin, tmax} = MinMax@sol1x[\"Grid\"];\n\nGraphics@Table[Line[{{sol1x[t], sol1y[t]}, {sol2x[t], sol2y[t]}, {sol3x[t], sol3y[t]}, {sol1x[t], sol1y[t]}}], {t, tmin, tmax, (tmax - tmin)/15}]\n]\n\nShow[PursuitPolygon @@@ MeshPrimitives[reg, 2]]\n\n\nThe solution is qualitatively different when the agents move at unit speed because they no longer meet at the centroid. Note that the difference in the picture is also that the density of the lines is different, this density can be adjusted in both solutions but changing some of the fixed numbers I put in. 15 in (tmax - tmin)/15 controls the number of lines in the last solution.\n\nHere's an extension of @C.E.'s code to non-triangular polygons - polygons taken from a Voronoi mesh of roughly equidistributed (https://mathematica.stackexchange.com/a/141215/3056) 31 points over a square. Code is awful, I didn't really have time to think.\n\nClearAll@PursuitPolygon;\nPursuitPolygon[pts_] := Module[{vars, sols},\nvars = Table[Unique[], Length@pts, 2];\nsols = NDSolveValue[{Sequence @@ (((#'[t] & /@ #1) ==\nNormalize[Subtract @@ Map[#[t] &, {#2, #1}, {2}]]) & @@@\nPartition[Join[vars, {First@vars}], 2, 1]),\nSequence @@ MapThread[(#[0] & /@ #1) == #2 &, {vars, pts}],\nWhenEvent[Norm[x] < 1*^-8, \"StopIntegration\"] /.\nx -> Subtract @@ Map[#[t] &, Take[vars, 2], {2}]},\nFlatten@vars, {t, 0, 10}];\n{tmin, tmax} = MinMax@First[sols][\"Grid\"];\nGraphics@\nTable[Line[\nMap[#[t] &, Join[Partition[sols, 2], {Take[sols, 2]}], {2}]], {t,\ntmin, tmax, (tmax - tmin)/15}]]\n\nClearAll@reg;\nreg =\nWith[{reg = Rectangle[]},\nWith[{points = 31, samples = 4000, iterations = 100},\nNest[With[{randoms = Join[#, RandomPoint[reg, samples]]},\nRegionNearest[reg][\nMean@randoms[[#]] & /@\nValues@PositionIndex@Nearest[#, randoms]]] &,\nRandomPoint[reg, points], iterations]] //\nVoronoiMesh[#, {{0, 1}, {0, 1}}] &];\n\nShow[PursuitPolygon /@ MeshPrimitives[reg, 2][[All, 1]]]", "date": "2019-06-20 20:18:36", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/145050/filling-space-with-pursuit-polygons?noredirect=1", "openwebmath_score": 0.5776697397232056, "openwebmath_perplexity": 9669.474466707925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759587767815, "lm_q2_score": 0.88242786954645, "lm_q1q2_score": 0.8655522825927268}}
{"url": "https://www.physicsforums.com/threads/integration-by-partial-fractions.754530/", "text": "# Integration by Partial Fractions\n\n1. May 20, 2014\n\n1. The problem statement, all variables and given/known data\n\nFind the indefinite integral of the below, using partial fractions.\n\n$$\\frac{4x^2+6x-1}{(x+3)(2x^2-1)}$$\n\n2. Relevant equations\n?\n\n3. The attempt at a solution\nFirst I want to say there is probably a much easier and quicker way to get around certain things I have done but I have just done it the only way I could see (I always seem to go the long way around). So if I am correct then I would really appreciate some tips on how to do it quicker (for in exams) and also if I am incorrect to point out my mistakes. Thanks :)\n\nfirst I set up the partial fractions.\n$\\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\\frac{A}{x+3}+\\frac{Bx+C}{2x^2-1} \\\\ \\rightarrow \\,\\,\\, 4x^2+6x-1=A(2x^2-1)+(Bx+C)(x+3) \\\\ 4x^2+6x-1=2Ax^2-A+Bx^2+3Bx+Cx+3C$\n\nAnd then equated the coefficients:\nFor X^2: $4=2A+B$\nFor X^1: $6=3B+C$\nFor X^0: $-1=3C-A$\n\nThen what I did what multiply the third equation above by 2 to get $-2=6C-2A$ and then added it to the first equation to get $2=6C+B$ and solved for B and substituted it into the second equation which resulted in:\n\n$6=3(2-6C)+C\\\\ 6=6-11C \\\\ 0=-11C \\\\ \u2234 C=0$\n\nAnd then knowing C=0 I found A=1 and then that B=2 .\n\n$\\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\\frac{1}{x+3}+\\frac{2x}{2x^2-1} \\\\ \\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\\frac{1}{x+3}+\\frac{2x}{2x^2-1}\\\\ \\int \\frac{4x^2+6x-1}{(x+3)(2x^2-1)} \\,\\, dx = \\ln{|x+3|} + \\frac{1}{2} \\ln{|2x^2-1|}$\n\nI think I may have done something wrong. I had to use an integral calculator online to find the integral of $\\frac{2x}{2x^2-1}$ as I had no idea how to do it. Which kind of defeats the point.\n\n2. May 20, 2014\n\n### Saitama\n\nTo evaluate the last integral, you can use the substitution $2x^2-1=t$.\n\n3. May 20, 2014\n\n### Panphobia\n\nIts all right, except you need to add a + C. But by the way the integral of 2x/(2x^2-1) can be easily solved using u substitution.\n\nIf u = 2x^2 -1 then du = 4x dx so du/2 = 2x dx\n\n(1/2)*(1/u)\n\nand the integral of this is\n\nln|u|/2\n\nwhich is\n\nln|2x^2-1|/2\n\nLast edited: May 20, 2014\n4. May 20, 2014\n\n### Curious3141\n\nYou're missing the constant of integration, which is very important in an indefinite integral.\n\nThe partial fraction decomposition is tedious. If the denominator was easily factored into linear factors, Heaviside cover up rule would help more. As it stands, the cover up rule helps to find A quite quickly, but you get bogged down with B and C, and you have to manipulate radicals. Pointless. I think your method is fine.\n\n5. May 20, 2014\n\nAh yes, sorry I forgot as it was just rough workings copied up into latex.\n\nI am quite poor when it comes to integration by substitution for some reason. I find integration by parts easier as its a set formula but I do realise that integration by substitution is sometimes a lot easier and quicker so I need to get more familiar with it. Any tips on u substitution?\n\nI can usually find what to substitute, and then get to a du=... but its the following bits that confuse me a little.\n\n6. May 20, 2014\n\n### scurty\n\nI try to use u-substitution if I see a power that is one higher in the denominator than in the numerator. A more general approach would to be to look for derivatives in the numerator from what is found in the denominator. I always look for u-substitution first because they tend to be the easiest when dealing with fractions. Partial fractions and trig substitution, while fun to compute, are more time consuming.\n\nIt's hard for me to explain because I've done so many integrals that knowing what method to use is more intuitive to me than anything else. It comes easier the more you practice integrals. A general rule to always follow for u substitution is to notice candidates for u and du.", "date": "2017-11-23 08:03:11", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integration-by-partial-fractions.754530/", "openwebmath_score": 0.9170228838920593, "openwebmath_perplexity": 392.7213961853876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127437254887, "lm_q2_score": 0.8757869997529961, "lm_q1q2_score": 0.8655514526449976}}
{"url": "https://math.stackexchange.com/questions/1297197/find-an-arbitrary-power-of-a-lower-triangular-matrix-of-size-3-times-3", "text": "# Find an arbitrary power of a lower triangular matrix of size $3\\times 3$\n\nLet $F$ be a field and let $A=\\begin{bmatrix}a&0&0\\\\1&a&0\\\\0&1&a\\end{bmatrix}\\in\\mathscr{M}_{3\\times 3}(F)$. Show that $$A^k=\\begin{bmatrix}a^k&0&0\\\\ka^{k-1}&a^k&0\\\\\\dfrac{1}{2}k(k-1)a^{k-2}&ka^{k-1}&a^k\\end{bmatrix}$$ for all $k > 0$. (Exercise 797 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)\n\nI know it can be proved by induction, but since the topic is about Krylov Spaces, eigenvalues and Jordan canonical form, I wonder if there is another way to solve this problem.\n\n\u2022 Inducci\u00f3n, Jos\u00e9! \u2013\u00a0Pedro Tamaroff May 24 '15 at 20:59\n\u2022 it is spelled induction, and I'm asking for another way, thanks. \u2013\u00a0Jos\u00e9 May 24 '15 at 21:02\n\u2022 This is a trivial matter using induction, really. I doubt there is any gain going in any other route... \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez May 24 '15 at 21:04\n\u2022 Did you try Krylov spaces ? \u2013\u00a0Dietrich Burde May 24 '15 at 21:42\n\u2022 @Dietrich Burde: what is Krylov space? Any good online refs? I've never heard of this approach for such problems . . . \u2013\u00a0Robert Lewis May 24 '15 at 22:26\n\nThere is extended reading giving formula for calculating functions such as polynomials on matrixes: http://en.wikipedia.org/wiki/Matrix_function#Jordan_decomposition\n\nFor a route different to induction, express the matrix as $$A=aI_3+G$$ where $I_3$ is the identity matrix and $G$ is given by $$G=\\begin{bmatrix}0&0&0\\\\1&0&0\\\\0&1&0\\end{bmatrix}$$ A binomial expansion will yield \\begin{align}A^k&=(aI_3+G)^k\\\\&={k \\choose 0}a^kI_3+{k \\choose 1}a^{k-1}I_3G+{k \\choose 2}a^{k-2}I_3G^2+\\sum_{m=3}^k{k \\choose m}a^{k-m}I_3G^m\\\\&=a^kI_3+ka^{k-1}I_3G+\\frac{k(k-1)}{2}a^{k-2}I_3G^2+\\sum_{m=3}^k{k \\choose m}a^{k-m}I_3G^m\\end{align} Now $I_3G$ is given by $$G=\\begin{bmatrix}0&0&0\\\\1&0&0\\\\0&1&0\\end{bmatrix}$$ while $I_3G^2$ is $$G=\\begin{bmatrix}0&0&0\\\\0&0&0\\\\1&0&0\\end{bmatrix}$$ and $I_3G^m$ for $m\\geq3$ is the zero matrix.\n\nUsing these results, we arrive at the given expression for $A^k$.\n\n\u2022 For the binomial expansion, would not explicitly (using matrix multiplication) working out $I_3G$, $I_3G^2$, and $I_3G^3$ suffice. Once we show that $I_3G^3$ is the zero matrix, then $I_3G^m$ for $m>0$ will also be zero matrices as well. \u2013\u00a0Alijah Ahmed May 24 '15 at 21:30\n\u2022 Obviously I like your methodology! Endorsed!!! \u2013\u00a0Robert Lewis May 24 '15 at 22:29\n\nSet\n\n$N = \\begin{bmatrix} 0 & 0 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 0 \\end{bmatrix}; \\tag{1}$\n\nthen a simple calculation reveals that\n\n$N^2 = \\begin{bmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 1 & 0 & 0 \\end{bmatrix} \\tag{2}$\n\nand\n\n$N^3 = 0. \\tag{3}$\n\nAlso,\n\n$A = aI + N. \\tag{4}$\n\nSince $N$ and $I$ commute, we may apply the ordinary binomial theorem to (4), and find\n\n$A^k = (aI + N)^k =$ $(aI)^k + k(aI)^{k - 1}N + \\dfrac{k(k - 1)}{2}(aI)^{k - 2}N^2$ $= a^kI^k + ka^{k - 1}I^{k - 1}N + \\dfrac{k(k - 1)}{2}a^{k - 2}I^{k - 2}N^2$ $= a^kI + ka^{k - 1}N + \\dfrac{k(k - 1)}{2} a^{k - 2}N^2, \\tag{5}$\n\nsince the expansion is trunated after the third term by virtue of $N^3 = 0$. When (5) is written out explicitly, we see that\n\n$A^k = \\begin{bmatrix} a^k & 0 & 0 \\\\ ka^{k - 1} & a^k & 0 \\\\ \\dfrac{k(k - 1)}{2} a^{ k - 2} & ka^{k - 1} & a^k \\end{bmatrix}, \\tag{6}$\n\nas per request.\n\nNota Bene: a few words on induction. It has been rightly noted that in deploying the binomial theorem we do not in fact escape induction. The only other option I can see is to prove this result by direct matrix mutiplication, again using induction. Here we hide the induction by invoking the binomial theorem, doing so since the problem statement asks for \"another way\"; so I assumed direct induction should be avoided. But in truth, this is a statement concerning the integers, so induction will almost certainly enter in at some point, being as it is an essential defining property of the integers. End of Note\n\n\u2022 @Dietrich Burde: Yes, yes, yes . . . take 'er easy, Pal, still editing. And I'll say a few words about induction before I'm done! Cheers! \u2013\u00a0Robert Lewis May 24 '15 at 21:33\n\u2022 Sorry, everything is good. I just think, why not prove the (easy) matrix formula - by induction. \u2013\u00a0Dietrich Burde May 24 '15 at 21:35\n\u2022 @Dietrich Burde: me too. Cheers! \u2013\u00a0Robert Lewis May 24 '15 at 21:38", "date": "2019-07-18 03:03:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1297197/find-an-arbitrary-power-of-a-lower-triangular-matrix-of-size-3-times-3", "openwebmath_score": 0.8633328080177307, "openwebmath_perplexity": 637.9836814088488, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127399388855, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8655514445225454}}
{"url": "https://math.stackexchange.com/questions/3782436/question-about-convexity-how-do-we-prove-that-displaystyle-sum-i-1kp-i/3782448", "text": "# Question about convexity: how do we prove that $\\displaystyle \\sum_{i=1}^{k}p_{i}b_{i}\\geq\\prod_{i=1}^{k}b^{p_{i}}_{i}$?\n\nLet $$b_{1},b_{2},\\ldots,b_{k}$$ be nonnegative numbers and $$p_{1} + p_{2} + \\ldots + p_{k} = 1$$ where each $$p_{i}$$ is positive. Then\n\n\\begin{align*} \\sum_{i=1}^{k}p_{i}b_{i}\\geq\\prod_{i=1}^{k}b^{p_{i}}_{i} \\end{align*}\n\nMY ATTEMPT\n\nSince the logarithm function is strictly increasing, the proposed inequality is equivalent to \\begin{align*} \\ln\\left(p_{1}b_{1} + p_{2}b_{2} + \\ldots + p_{k}b_{k}\\right) \\geq p_{1}\\ln(b_{1}) + p_{2}\\ln(b_{2}) + \\ldots + p_{k}\\ln(b_{k}) \\end{align*}\n\nOnce $$f''(x) < 0$$, where $$f(x) = \\ln(x)$$, we conclude that $$f$$ is concave and the proposed inequality holds.\n\nMy question is: am I proving this result correctly? If this is the case, is there another way to prove it?\n\nAny contribution is appreciated.\n\n\u2022 You have done it absolutely correctly. Another method is by using weighted AM-GM as shown in my answer that I'm typing. Aug 6, 2020 at 22:17\n\nUsing $$b_1,b_2,\\cdots,b_k\\ge 0$$ with $$p_1,p_2,\\cdots,p_k$$ as the respective weights, weighted AM-GM inequality gives $$\\frac{\\sum_{i=1}^k p_ib_i}{\\sum_{i=1}^k p_i} \\ge \\left(\\prod_{i=1}^k b_i^{p_i}\\right)^{\\dfrac{1}{\\sum_{i=1}^k p_i}}$$ which gives the required inequality.\nNote: This is not much different from your approach because the special case of Jensen's inequality for the function $$f(x)=\\ln(x)$$ can be proved by weighted AM-GM or vice-versa, the weighted AM-GM inequality can be proved by the Jensen's inequality for $$f(x)=\\ln(x)$$ as you have done, since the inequality you want to prove is directly the weighted AM-GM inequality in disguise. Also, these two can be proven independently without using each other.", "date": "2022-08-19 07:57:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3782436/question-about-convexity-how-do-we-prove-that-displaystyle-sum-i-1kp-i/3782448", "openwebmath_score": 0.9767274260520935, "openwebmath_perplexity": 138.1244168725836, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127413158321, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8655514441263927}}
{"url": "https://dantopology.wordpress.com/2015/07/01/products-of-compact-spaces-with-countable-tightness/", "text": "# Products of compact spaces with countable\u00a0tightness\n\nIn the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.\n\nThe following theorems are the main results in this post.\n\nTheorem 1\nLet $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is compact, then $X \\times Y$ is countably tight.\n\nTheorem 2\nThe product of finitely many compact countably tight spaces is countably tight.\n\nTheorem 3\nSuppose that $X_1, X_2, X_3, \\cdots$ are countably many compact spaces such that each $X_i$ has at least two points. If each $X_i$ is a countably tight space, then the product space $\\prod_{i=1}^\\infty X_i$ is countably tight.\n\n____________________________________________________________________\n\nFinite products\n\nBefore proving Theorem 1 and Theorem 2, we prove the following results.\n\nTheorem 4\nLet $f:Y_1 \\rightarrow Y_2$ be a continuous and closed map from the space $Y_1$ onto the space $Y_2$. Suppose that the space $Y_2$ is countably tight and that each fiber of the map $f$ is countably tight. Then the space $Y_1$ is countably tight.\n\nProof of Theorem 4\nLet $x \\in Y_1$ and $x \\in \\overline{A}$ where $A \\subset Y_1$. We proceed to find a countable $W \\subset Y_1$ such that $x \\in \\overline{W}$. Choose $y \\in Y_2$ such that $y=f(x)$.\n\nLet $M$ be the fiber of the map $f$ at the point $y$, i.e. $M=f^{-1}(y)$. By assumption, $M$ is countably tight. Call a point $w \\in M$ countably reached by $A$ if there is some countable $C \\subset A$ such that $w \\in \\overline{C}$. Let $G$ be the set of all points in $M$ that are countably reached by $A$. We claim that $x \\in \\overline{G}$.\n\nLet $U \\subset Y_1$ be open such that $x \\in U$. Because the space $Y_1$ is regular, choose open $V \\subset U$ such that $x \\in V$ and $\\overline{V} \\subset U$. Then $V \\cap A \\ne \\varnothing$. Furthermore, $x \\in \\overline{V \\cap A}$. Let $C=f(V \\cap A)$. By the continuity of $f$, we have $y \\in \\overline{C}$. Since $Y_2$ is countably tight, there exists some countable $D \\subset C$ such that $y \\in \\overline{D}$. Choose a countable $E \\subset V \\cap A$ such that $f(E)=D$. It follows that $y \\in \\overline{f(E)}$.\n\nWe show that that $\\overline{E} \\cap M \\ne \\varnothing$. Since $E \\subset \\overline{E}$, we have $f(E) \\subset f(\\overline{E})$. Note that $f(\\overline{E})$ is a closed set since $f$ is a closed map. Thus $\\overline{f(E)} \\subset f(\\overline{E})$. As a result, $y \\in f(\\overline{E})$. Then $y=f(t)$ for some $t \\in \\overline{E}$. We have $t \\in \\overline{E} \\cap M$.\n\nBy the definition of the set $G$, we have $\\overline{E} \\cap M \\subset G$. Furthermore, $\\overline{E} \\cap M \\subset \\overline{V} \\subset U$. Note that the arbitrary open neighborhood $U$ of $x$ contains points of $G$. This establishes the claim that $x \\in \\overline{G}$.\n\nSince $M$ is a fiber of $f$, $M$ is countably tight by assumption. Choose some countable $T \\subset G$ such that $x \\in \\overline{T}$. For each $t \\in T$, choose a countable $W_t \\subset A$ with $t \\in \\overline{W_t}$. Let $W=\\bigcup_{t \\in T} W_t$. Note that $W \\subset A$ and $W$ is countable with $x \\in \\overline{W}$. This establishes the space $Y_1$ is countably tight at $x \\in Y_1$. $\\blacksquare$\n\nLemma 5\nLet $f:X \\times Y \\rightarrow Y$ be the projection map. If $X$ is a compact space, then $f$ is a closed map.\n\nProof of Lemma 5\nLet $A$ be a closed subset of $X \\times Y$. Suppose that $f(A)$ is not closed. Let $y \\in \\overline{f(A)}-f(A)$. It follows that no point of $X \\times \\left\\{y \\right\\}$ belongs to $A$. For each $x \\in X$, choose open subset $O_x$ of $X \\times Y$ such that $(x,y) \\in O_x$ and $O_x \\cap A=\\varnothing$. The set of all $O_x$ is an open cover of the compact space $X \\times \\left\\{y \\right\\}$. Then there exist finitely many $O_x$ that cover $X \\times \\left\\{y \\right\\}$, say $O_{x_i}$ for $i=1,2,\\cdots,n$.\n\nLet $W=\\bigcup_{i=1}^n O_{x_i}$. We have $X \\times \\left\\{y \\right\\} \\subset W$. Since $X$ is compact, we can then use the Tube Lemma which implies that there exists open $G \\subset Y$ such that $X \\times \\left\\{y \\right\\} \\subset X \\times G \\subset W$. It follows that $G \\cap f(A) \\ne \\varnothing$. Choose $t \\in G \\cap f(A)$. Then for some $x \\in X$, $(x,t) \\in A$. Since $t \\in G$, $(x,t) \\in W$, implying that $W \\cap A \\ne \\varnothing$, a contradiction. Thus $f(A)$ must be a closed set in $Y$. This completes the proof of the lemma. $\\blacksquare$\n\nProof of Theorem 1\nLet $X$ be the factor that is compact. let $f: X \\times Y \\rightarrow Y$ be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space $Y$ is countably tight by assumption. Each fiber of the projection map $f$ is of the form $X \\times \\left\\{y \\right\\}$ where $y \\in Y$, which is countably tight. Then use Theorem 4 to establish that $X \\times Y$ is countably tight. $\\blacksquare$\n\nProof of Theorem 2\nThis is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. $\\blacksquare$\n\n____________________________________________________________________\n\nCountable products\n\nOur proof to establish that the product space $\\prod_{i=1}^\\infty X_i$ is countably tight is an indirect one and makes use of two non-trivial results. We first show that $\\omega_1 \\times \\prod_{i=1}^\\infty X_i$ is a closed subspace of a $\\Sigma$-product that is normal. It follows from another result that the second factor $\\prod_{i=1}^\\infty X_i$ is countably tight. We now present all the necessary definitions and theorems.\n\nConsider a product space $Y=\\prod_{\\alpha<\\kappa} Y_\\alpha$ where $\\kappa$ is an infinite cardinal number. Fix a point $p \\in Y$. The $\\Sigma$-product of the spaces $Y_\\alpha$ with $p$ as the base point is the following subspace of the product space $Y=\\prod_{\\alpha<\\kappa} Y_\\alpha$:\n\n$\\displaystyle \\Sigma_{\\alpha<\\kappa} Y_\\alpha=\\left\\{y \\in \\prod_{\\alpha<\\kappa} Y_\\alpha: y_\\alpha \\ne p_\\alpha \\text{ for at most countably many } \\alpha < \\kappa \\right\\}$\n\nThe definition of the space $\\Sigma_{\\alpha<\\kappa} Y_\\alpha$ depends on the base point $p$. The discussion here is on properties of $\\Sigma_{\\alpha<\\kappa} Y_\\alpha$ that hold regardless of the choice of base point. If the factor spaces are indexed by a set $A$, the notation is $\\Sigma_{\\alpha \\in A} Y_\\alpha$.\n\nIf all factors $Y_\\alpha$ are identical, say $Y_\\alpha=Z$ for all $\\alpha$, then we use the notation $\\Sigma_{\\alpha<\\kappa} Z$ to denote the $\\Sigma$-product. Once useful fact is that if there are $\\omega_1$ many factors and each factor has at least 2 points, then the space $\\omega_1$ can be embedded as a closed subspace of the $\\Sigma$-product.\n\nTheorem 6\nFor each $\\alpha<\\omega_1$, let $Y_\\alpha$ be a space with at least two points. Then $\\Sigma_{\\alpha<\\omega_1} Y_\\alpha$ contains $\\omega_1$ as a closed subspace. See Exercise 3 in this previous post.\n\nNow we discuss normality of $\\Sigma$-products. This previous post shows that if each factor is a separable metric space, then the $\\Sigma$-product is normal. It is also well known that if each factor is a metric space, the $\\Sigma$-product is normal. The following theorem handles the case where each factor is a compact space.\n\nTheorem 7\nFor each $\\alpha<\\kappa$, let $Y_\\alpha$ be a compact space. Then the $\\Sigma$-product $\\Sigma_{\\alpha<\\kappa} Y_\\alpha$ is normal if and only if each factor $Y_\\alpha$ is countably tight.\n\nTheorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the $\\Sigma$-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.\n\nTheorem 8\nLet $Y$ be a compact space. Then the product space $\\omega_1 \\times Y$ is normal if and only if $Y$ is countably tight.\n\nWe now prove Theorem 3.\n\nProof of Theorem 3\nLet $\\omega_1=\\cup \\left\\{A_n: n \\in \\omega \\right\\}$, where for each $n$, $\\lvert A_n \\lvert=\\omega_1$ and that $A_n \\cap A_m=\\varnothing$ if $n \\ne m$. For each $n=1,2,3,\\cdots$, let $S_n=\\Sigma_{\\alpha \\in A_n} X_n$. By Theorem 7, each $S_n$ is normal. Let $S_0=\\Sigma_{\\alpha \\in A_0} X_1$, which is also normal. By Theorem 6, the space $\\omega_1$ of countable ordinals is a closed subspace of $S_0$. Let $T=\\omega_1 \\times X_1 \\times X_2 \\times X_3 \\times \\cdots$. We have the following derivation.\n\n\\displaystyle \\begin{aligned} T&=\\omega_1 \\times X_1 \\times X_2 \\times X_3 \\times \\cdots \\\\&\\subset S_0 \\times S_1 \\times S_2 \\times S_3 \\times \\cdots \\\\&\\cong W=\\Sigma_{\\alpha<\\omega_1} W_\\alpha \\end{aligned}\n\nRecall that $\\omega_1=\\cup \\left\\{A_n: n \\in \\omega \\right\\}$. The space $W=\\Sigma_{\\alpha<\\omega_1} W_\\alpha$ is defined such that for each $n \\ge 1$ and for each $\\alpha \\in A_n$, $W_\\alpha=X_n$. Furthermore, for $n=0$, for each $\\alpha \\in A_0$, let $W_\\alpha=X_1$. Thus $W$ is a $\\Sigma$-product of compact countably tight spaces and is thus normal by Theorem 7. The space $T=\\omega_1 \\times \\prod_{n=1}^\\infty X_n$ is a closed subspace of the normal space $W$. By Theorem 8, the product space $\\prod_{n=1}^\\infty X_n$ must be countably tight. $\\blacksquare$\n\n____________________________________________________________________\n\nRemarks\n\nTheorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let $X=X_1 \\times X_2 \\times \\cdots \\times X_n$.\n\n\\displaystyle \\begin{aligned} X&=X_1 \\times X_2 \\times \\cdots \\times X_n \\\\&\\cong X_1 \\times X_2 \\times \\cdots \\times X_n \\times \\left\\{t_{n+1} \\right\\} \\times \\left\\{t_{n+2} \\right\\} \\times \\cdots \\\\&\\subset X_1 \\times X_2 \\times \\cdots \\times X_n \\times X_{n+1} \\times X_{n+2} \\times \\cdots \\end{aligned}\n\nIn the above derivation, $t_m$ is a point of $X_m$ for all $m >n$. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.\n\n____________________________________________________________________\n\nExercise\n\nExercise 1\nLet $f:X \\times Y \\rightarrow Y$ be the projection map. If $X$ is a countably compact space and $Y$ is a Frechet space, then $f$ is a closed map.\n\nExercise 2\nLet $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is a countably compact space and the other space is a Frechet space, then $X \\times Y$ is countably tight.\n\nExercise 2 is a variation of Theorem 1. One factor is weakened to \u201ccountably compact\u201d. However, the other factor is strengthened to \u201cFrechet\u201d.\n\n____________________________________________________________________\n\nReference\n\n1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.\n\n____________________________________________________________________\n$\\copyright \\ 2015 \\text{ by Dan Ma}$", "date": "2017-04-27 18:47:16", "meta": {"domain": "wordpress.com", "url": "https://dantopology.wordpress.com/2015/07/01/products-of-compact-spaces-with-countable-tightness/", "openwebmath_score": 0.9987096786499023, "openwebmath_perplexity": 83.69010705959454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127423485423, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.865551443428762}}
{"url": "https://www.physicsforums.com/threads/rocket-ship-question.196423/", "text": "# Rocket Ship Question\n\n1. Nov 6, 2007\n\n### Submission1\n\nWe are having a debate here at work and we need an answer from some smart folks (that's you).\n\nIf you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.\n\nThanks,\nSub.\n\n2. Nov 6, 2007\n\n### mgb_phys\n\nKinetic energy, the energy needed to move something depends on the speed squared.\nSo to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.\n\nYou also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.\n\nLast edited: Nov 6, 2007\n3. Nov 6, 2007\n\n### Submission1\n\nCool. Thanks.\n\n4. Nov 6, 2007\n\n### chronon\n\nBut there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.\n\n5. Nov 6, 2007\n\n### mgb_phys\n\nDoes that apply if it's accelerating ?\n\n6. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nThis topic ws recently discussed in this thread.\n\nThe governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,\n\n$$\\Delta v = v_e\\ln\\left(\\frac {m_\\text{rocket}+m_\\text{fuel}}{m_\\text{rocket}}\\right)$$\n\nSuppose you find you have to load some rocket with a quantity of fuel $m_{\\text{fuel}|1000\\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is\n\n$$m_{\\text{fuel}|2000\\text{mph}} = m_{\\text{fuel}|1000\\text{mph}}\\left(2+ \\frac{m_{\\text{fuel}|1000\\text{mph}}}{m_\\text{rocket}}\\right)$$\n\nIn other words, one must more than double the quantity of fuel to double a rocket's final velocity.\n\n7. Nov 6, 2007\n\n### mgb_phys\n\nThanks - I'm away for a week and look what I miss.\n\n8. Nov 6, 2007\n\n### chronon\n\nIndeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.\n\n9. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nA little more detail regarding previous post:\n\nThe reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.\n\nAssume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.\n\nHowever, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.\n\nLast edited: Nov 6, 2007\n10. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nIndeed. We just cross-posted.\n\n11. Nov 6, 2007\n\n### rcgldr\n\nMentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).\n\nAlso from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.\n\n12. Nov 6, 2007\n\n### Staff: Mentor\n\nFrom this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.\n\n13. Nov 6, 2007\n\n### TVP45\n\nYou always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.\n\n14. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nThat is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.\n\nConservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.\n\nThe driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot \"see the forest for the trees\". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.\n\n15. Nov 8, 2007\n\n### alvaros\n\nSo the answer is ... ??\nA 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )\nAgain\nA 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.\n\n( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )\n\n16. Nov 8, 2007\n\n### TVP45\n\nA kind of new thought on this thread....\nThis kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.\n\nIs the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.\n\nIs there another way?\n\n17. Nov 9, 2007\n\n### alvaros\n\nIm very sorry, but I dont understand your english.\n\n18. Nov 9, 2007\n\n### TVP45\n\nAlvaros,\n\n19. Nov 10, 2007\n\n### alvaros\n\nI think my example is clear enough:\n\nIt seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )\n\nI made a thread \"what is a IFR\" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.\nIn this problem the forces are between the rocket and the fuel ejected:\nfuel <-> rocket\nAnd the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).\n\n20. Nov 10, 2007\n\n### Staff: Mentor\n\nYes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.\n\nI am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:\n\nSince the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same \u0394p and therefore the same \u0394v to the rocket.\n\nEach successive kg with its \u0394v results in a larger \u0394KE for the rocket since KE is proportional to v^2.\n\n\"But conservation of energy you protest!\" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.\n\nBecause the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).\n\nBottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.\n\nLast edited: Nov 10, 2007", "date": "2017-02-28 16:55:23", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/rocket-ship-question.196423/", "openwebmath_score": 0.7356311082839966, "openwebmath_perplexity": 803.3921039785716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769085257167, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8655363249517973}}
{"url": "https://www.physicsforums.com/threads/rocket-ship-question.196423/", "text": "# Rocket Ship Question\n\n1. Nov 6, 2007\n\n### Submission1\n\nWe are having a debate here at work and we need an answer from some smart folks (that's you).\n\nIf you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.\n\nThanks,\nSub.\n\n2. Nov 6, 2007\n\n### mgb_phys\n\nKinetic energy, the energy needed to move something depends on the speed squared.\nSo to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.\n\nYou also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.\n\nLast edited: Nov 6, 2007\n3. Nov 6, 2007\n\n### Submission1\n\nCool. Thanks.\n\n4. Nov 6, 2007\n\n### chronon\n\nBut there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.\n\n5. Nov 6, 2007\n\n### mgb_phys\n\nDoes that apply if it's accelerating ?\n\n6. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nThis topic ws recently discussed in this thread.\n\nThe governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,\n\n$$\\Delta v = v_e\\ln\\left(\\frac {m_\\text{rocket}+m_\\text{fuel}}{m_\\text{rocket}}\\right)$$\n\nSuppose you find you have to load some rocket with a quantity of fuel $m_{\\text{fuel}|1000\\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is\n\n$$m_{\\text{fuel}|2000\\text{mph}} = m_{\\text{fuel}|1000\\text{mph}}\\left(2+ \\frac{m_{\\text{fuel}|1000\\text{mph}}}{m_\\text{rocket}}\\right)$$\n\nIn other words, one must more than double the quantity of fuel to double a rocket's final velocity.\n\n7. Nov 6, 2007\n\n### mgb_phys\n\nThanks - I'm away for a week and look what I miss.\n\n8. Nov 6, 2007\n\n### chronon\n\nIndeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.\n\n9. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nA little more detail regarding previous post:\n\nThe reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.\n\nAssume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.\n\nHowever, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.\n\nLast edited: Nov 6, 2007\n10. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nIndeed. We just cross-posted.\n\n11. Nov 6, 2007\n\n### rcgldr\n\nMentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).\n\nAlso from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.\n\n12. Nov 6, 2007\n\n### Staff: Mentor\n\nFrom this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.\n\n13. Nov 6, 2007\n\n### TVP45\n\nYou always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.\n\n14. Nov 6, 2007\n\n### D H\n\nStaff Emeritus\nThat is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.\n\nConservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.\n\nThe driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot \"see the forest for the trees\". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.\n\n15. Nov 8, 2007\n\n### alvaros\n\nSo the answer is ... ??\nA 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )\nAgain\nA 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.\n\n( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )\n\n16. Nov 8, 2007\n\n### TVP45\n\nA kind of new thought on this thread....\nThis kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.\n\nIs the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.\n\nIs there another way?\n\n17. Nov 9, 2007\n\n### alvaros\n\nIm very sorry, but I dont understand your english.\n\n18. Nov 9, 2007\n\n### TVP45\n\nAlvaros,\n\n19. Nov 10, 2007\n\n### alvaros\n\nI think my example is clear enough:\n\nIt seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )\n\nI made a thread \"what is a IFR\" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.\nIn this problem the forces are between the rocket and the fuel ejected:\nfuel <-> rocket\nAnd the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).\n\n20. Nov 10, 2007\n\n### Staff: Mentor\n\nYes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.\n\nI am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:\n\nSince the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same \u0394p and therefore the same \u0394v to the rocket.\n\nEach successive kg with its \u0394v results in a larger \u0394KE for the rocket since KE is proportional to v^2.\n\n\"But conservation of energy you protest!\" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.\n\nBecause the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).\n\nBottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.\n\nLast edited: Nov 10, 2007\n21. Nov 10, 2007\n\n### TVP45\n\nDaleSpam,\nThis is the sort of example I was asking about. If I were a high school teacher (thank G-d I'm not; that's a hard job), how would I present this example so that students can grasp it? I'm not being in any way critical, any more than I was of Alvaros; I am simply curious about how this can be developed conceptually.\n\n22. Nov 10, 2007\n\n### Staff: Mentor\n\nIf I were a teacher in a high-school level physics course I would not even mention conservation of energy in this problem. I would stick entirely with conservation of momentum. You can solve for everything of interest that way, and the conservation of momentum principle does what a good conservation principle should: it simplifies things.\n\nAs a teacher I would not introduce conservation of energy into the problem since it does nothing to simplify the problem. If a student asked I would tell them that energy is conserved, but it doesn't make the problem any easier. I think students would understand that.\n\n23. Nov 11, 2007\n\n### TVP45\n\nAt first blush, I agree. Momentum is much easier to deal with than energy. Two questions occur to me:\n(1) Can a student grasp the concept of momentum without being sidetracked by preconceptions about energy?\n(2) Is the calculation of change of momentum rather than absolute momentum sensible?\n\n24. Nov 11, 2007\n\n### alvaros\n\nDaleSpam:\nWrong, the change in kinetic energy on the fuel is 1000 times the change in kinetic energy on the rocket.\n\nMomentum is an abstract concept, its much easier the 3rd Newtons law: the burning fuel pushes the rocket and the rocket pushes the fuel fuel <-> rocket\n( you see the double arrow <-> , all forces ( ? ) are double arrow )\n( It seems, as I read here, that the conservation of momentum is more \"universal\" than 3rd Newtons law, but, in this case, both laws are the same and true )\n\nBut, at the end, there isnt any clear answer to the problem.\n\n25. Nov 11, 2007\n\n### D H\n\nStaff Emeritus\nInvoking Newton's third law on this problem is in a sense more ad-hoc and more abstract than using conservation of momentum with regard to this problem. The concept of force (Newton's second law) is very abstract; anything change to an object's momentum involves some force. Applying conservation of momentum is no less abstract than applying Newton's third law and yields a deeper answer.\n\nPut the spaceship in deep space, far from any massive object, so that there are no measurable external forces acting on the vehicle. Now, what exactly is the force that is making the spaceship accelerate? By assumption, there are no external forces. You can say posit some ad-hoc force $F$ that results from burning the fuel and get an answer via Newton's second law. Note well: Since the rocket's mass is changing, the simpler form $F=ma$ is not valid here. We have to use $F=dp/dt$ instead.\n\n$$F = \\frac{dp_r} {dt} = \\frac{d}{dt} ( m_r \\, v_r ) = \\dot m_r v_r} + m_r \\, a_r$$\n\nSolving for the rocket's acceleration,\n\n$$a_r = \\frac {F} {m_r} - \\frac {\\dot m_r} {m_r} v_r}$$\n\nThis is not very satisying. What exactly is this force? It's purely ad-hoc for one thing. Moreover, it will turn out that the force is frame-dependent.\n\nHere is how things turn out from conservation of momentum point of view. For this derivation, I will use some math that mathematicians don't particular like but physicists use willy-nilly -- things $\\Delta v$. Things can be done more formally using continuum physics (classical treatment of gases), but that is a lot messier.\n\nNomenclature:\n$$\\vec v_r(t)$$ Rocket velocity at time $t$, inertial observer\n$$\\vec v_e(t)$$ Rocket exhaust velocity at time $t$, relative to vehicle velocity\n$$m_r(t)$$ Rocket mass at time $t$\n$$\\dot m_f(t)$$ Rate at which rocket consumes fuel at$t$\n\nOver a small time interval $\\Delta t$, the rocket will eject a small mass of fuel $\\dot m_f(t)\\,\\Delta t$. At the start of the interval, the rocket has mass $m_r(t)$, velocity $\\vec v_r(t)$, and momentum $m_r(t) \\, \\vec v_r(t)$. At the end of the interval, the rocket has mass, velocity, and linear momentum\n\n$$m_r(t+\\Delta t) = m_r(t)-\\dot m_f(t)\\,\\Delta t$$\n$$\\vec v_r(t+\\Delta t) = \\vec v_r(t)+\\Delta \\vec v_r(t)$$\n$$\\vec p_r(t+\\Delta t) = (m_r(t) -\\dot m_f(t)\\,\\Delta t)\\, (\\vec v_r(t)+\\Delta \\vec v_r(t))$$\n\nThe bit of ejected fuel carries some momentum from the vehicle. The mass, inertial observer velocity, and momentum of the exhaust are\n\n$$\\Delta m_e(t) = \\dot m_f(t)\\,\\Delta t$$\n$$\\vec v_{e_{\\text{inertial}}}) = \\vec v_r(t)+\\vec v_e(t)$$\n$\\Delta \\vec p_e(t+\\Delta t) = \\dot m_f(t)\\,\\Delta t\\, (\\vec v_r(t)+\\vec v_e(t))[/tex] The momentum of the rocket+exhaust at the end of the time interval is thus $$\\vec p_{r+e}(t+\\Delta t) = (m_r(t) -\\dot m_f(t)\\,\\Delta t)\\, (\\vec v_r(t)+\\Delta \\vec v_r(t)) + \\dot m_f(t)\\,\\Delta t\\, (\\vec v_r(t)+\\vec v_e(t))$$ Dropping the second-order term [itex]\\Delta t \\Delta \\vec v_r(t)$ and simplifying,\n\n$$\\vec p_{r+e}(t+\\Delta t) = m_r(t)\\,\\vec v_r(t)+ m_r(t) \\, \\Delta \\vec v_r(t) + \\dot m_f(t)\\,\\Delta t\\, \\vec v_e(t)$$\n\nAssuming no external forces act on the rocket and the ejected fuel during this time interval, the rocket and the ejected fuel form a closed system. Momentum is conserved in a closed system, so $\\vec p_{r+e}(t+\\Delta t)=\\vec p_r(t)[/tex]: $$m_r(t)\\,\\vec v_r(t)+ m_r(t) \\, \\Delta \\vec v_r(t) + \\dot m_f(t)\\,\\Delta t\\, \\vec v_e(t) = m_r(t)\\,\\vec v_r(t)$$ or $$m_r(t) \\, \\Delta \\vec v_r(t) + \\dot m_f(t)\\,\\Delta t\\, \\vec v_e(t) = 0$$ Dividing by [itex]\\Delta t$ and taking the limit $\\Delta t \\to 0$,\n\n$$\\frac {d\\vec v_r(t)}{dt} = - \\, \\frac {\\dot m_f(t)} {m_r(t)} \\, \\vec v_e(t)$$\n\nThis is the equation for the acceleration of the rocket at time $t$.\n\nBy conservation of mass, the time derivative of the rocket's mass is just the additive inverse of the fuel consumption rate: $\\dot m_r(t) = -\\dot m_f(t)$. If the relative exhaust velocity is a constant vector, both the left and right hand sides of the above acceleration equation are integrable. Integrating from some initial time $t_0$ to some final time $t_1$ yields\n\n$$\\vec v_r(t_1) - \\vec v_r(t_0) = \\ln\\left(\\frac{m_r(t_1)}{m_r(t_0)}\\right) \\vec v_e$$\n\nThis is the Tsiolokovsky rocket equation. Since the final mass is smaller than the initial mass, the logarithm will be negative. The change in velocity is directed against the exhaust direction.\n\nLast edited: Nov 11, 2007", "date": "2018-05-26 01:03:08", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/rocket-ship-question.196423/", "openwebmath_score": 0.8170338869094849, "openwebmath_perplexity": 734.4304811453547, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769085257167, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.865536323497019}}
{"url": "https://cs.stackexchange.com/questions/54699/the-difference-between-theoretical-complexity-and-practical-efficiency/54706", "text": "The difference between theoretical complexity and practical efficiency\n\nIf I have this pseudocode:\n\nfor i=0 to n/2 do\nfor j=0 to n/2 do\n... do anything ....\n\n\nThe number of iterations is $n^2/4$.\n\nWhat is the complexity of this program? Is it $O(n^2)$?\n\nWhat is the intuition formal/informal for which is that complexity?\n\nNext, if i have this other pseudocode :\n\nfor i=0 to n do\nfor j=0 to n do\n... do anything ....\n\n\nThe complexity is again $O(n^2)$ -- is that correct?\n\nIs there any difference between efficiency in practice and theoretical complexity in this case? Is one of these faster than the other?\n\n\u2022 Rule 3 from Rob Pike, as cited in The Art of Unix Programming: \"Rule 3. Fancy algorithms are slow when n is small, and n is usually small. Fancy algorithms have big constants. Until you know that n is frequently going to be big, don't get fancy.\" This is because of the difference between theoretical complexity and practical efficiency, which is ably defined in the answers below. \u2013\u00a0Wildcard Mar 21 '16 at 3:49\n\u2022 @Wildcard That seems like a good rule of thumb. Benchmarking several contender algorithms would be even better. \u2013\u00a0G. Bach Mar 21 '16 at 12:47\n\u2022 @Wildcard Very interesting quote. It is always interesting to think about the complexity when you're designing/implementing procedures but I usually don't break my head over them untill the procedure in question seems to be a bottleneck. \u2013\u00a0Auberon Mar 21 '16 at 16:18\n\nBig O Formally\n\n$O(f(n)) = \\{g(n) | \\exists c > 0, \\exists n_0 > 0, \\forall n > n_0 : 0 \\leq g(n) \\leq c*f(n)\\}$\n\nBig O Informally\n\n$O(f(n))$ is the set of functions that grow slower (or equally fast) than $f(n)$. This means that whatever function you pick from $O(f(n))$, let's name her $g(n)$, and you pick a sufficiently large $n$, $f(n)$ will be larger than $g(n)$. Or, in even other words, $f(n)$ will eventually surpass any function in $O(f(n))$ when $n$ grows bigger. $n$ is the input size of your algorithm.\n\nAs an example. Let's pick $f(n) = n^2$.\n\nWe can say that $f(n) \\in O(n^3)$. Note that, over time, we allowed notation abuse so almost everyone writes $f(n) = O(n^3)$ now.\n\nNormally when we evaluate algorithms, we look at their order. This way, we can predict how the running time of the algorithm will increase when the input size ($n$) increases.\n\nIn your example, both algorithms are of order $O(n^2)$. So, their running time will increase the same way (quadratically) as $n$ increases. Your second algorithm will be four times slower than the first one, but that is something we're usually not interested in **theoretically*. Algorithms with the same order can have different factors ($c$ in the formal definition) or different lower order terms in the function that evaluates the number of steps. But usually that's because of implementation details and we're not interested in that.\n\nIf you have a algorithm that runs in $O(log(n))$ time we can safely say it will be faster than $O(n)$ [1] because $log(n) = O(n)$. No matter how bad the $O(log(n))$ algorithm is implemented, no matter how much overhead you put in the algorithm, it will always [1] be faster than the $O(n)$ algorithm. Even if the number of steps in the algorithms are $9999*log(n) = O(log(n))$ and $0.0001*n = O(n)$, the $O(log(n))$ algorithm will eventually be faster [1].\n\nBut, maybe, in your application, $n$ is always low and will never be sufficiently large enough so that the $O(log(n))$ algorithm will be faster in practice. So using the $O(n)$ algorithm will result in faster running times, despite $n = O(log(n))$\n\n[1] if $n$ is sufficiently large.\n\n\u2022 An $O(\\log n)$ isn't always faster than an $O(n)$ one. It depends on the value of $n$ and on the hidden constants. A case in point is fast matrix multiplication, which isn't fast at all in practice. \u2013\u00a0Yuval Filmus Mar 21 '16 at 0:07\n\u2022 That's why I added: If n is sufficiently large. In the case you mention, if the matrices grow sufficiently large, $O(log(n))$ WILL be faster, by definition \u2013\u00a0Auberon Mar 21 '16 at 0:09\n\u2022 Isn't everything you're saying already in my answer? \u2013\u00a0Auberon Mar 21 '16 at 0:11\n\u2022 The OP asks about the difference between theory and practice, and you're ignoring that. Asymptotic notation is usually, but not always, useful in practice. Improving the running time fourfold can make a huge difference in practice, but asymptotic notation is completely oblivious to it. \u2013\u00a0Yuval Filmus Mar 21 '16 at 0:14\n\nAuberon provided a very good explanation of the Big O. I will try to explain what that means for you.\n\nFirst of you are right. The first code example has $\\frac{n^2}{4}$ iterations, but is still in the complexity class O(n^2).\n\nWhy?\n\nThe thing about complexity classes is, that we assume that doing something several times is not that bad for runtime.\n\nImagine a Algorithm with complexity O(2^n) running for n=3 and taking 1 second.\n\nWe could run this 10 times, and still expect an answer after about 10 seconds.\n\nNow Imagine increasing n by 10. The Programm will take 2^10=1024 seconds.\n\nSo the computer scientists basically said: \"Man, leading factors are annoying. Let's just assume n grows to infinity and compare functions there\"\n\nWhich lead to the Big O.\n\nIn Practice\n\nIn Practice it is very well possible that a solution with much worse complexity runs much faster (for \"small\" inputs). It is easy to imagina a Programm that runs in O(n) but needs 10^10*n iterations.\n\nIt is O(n) but even an O(2^n) solution could be faster for small n.\n\nIn summary:\n\nThe Big O is a useful tool. But you still have to think about how fast your algorithm is in practice, since it only describes how much more time it will need if the input grows.\n\nIf I have this pseudocode:\n\nfor i=0 to n/2 do\nfor j=0 to n/2 do\n... do anything ....\n\n\nThe number of iteration is $n^2/4$.\n\nActually, it's $(1+n/2)^2 = n^2/4 + n + 1$.\n\nWhat is the complexity of this program? Is correct $O(n^2)$?\n\nThat depends entirely on what \"do anything\" is. For example, if \"do anything\" is replaced by the line\n\nfor k=0 to 2^n do {}\n\n\nthen the running time is $\\Theta(n^22^n)$.", "date": "2021-01-22 00:45:43", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/54699/the-difference-between-theoretical-complexity-and-practical-efficiency/54706", "openwebmath_score": 0.7352387309074402, "openwebmath_perplexity": 576.6017500918695, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769113660689, "lm_q2_score": 0.8872045922259088, "lm_q1q2_score": 0.8655363158335447}}
{"url": "https://math.stackexchange.com/questions/1000035/find-local-max-min-concavity-and-inflection-points", "text": "# Find local max, min, concavity, and inflection points\n\nFor the function $f(x)=\\frac{x^2}{x^2+3}$ Find the intervals on which f(x) is increasing or decreasing.\n\nFind the points of local maximum and minimum of f(x).\n\nFind the intervals of concavity and the inflection points of f(x).\n\n$f\u2032(x)=\\frac{6x}{(x^2+3)^2}$\n\n$f''(x)=\\frac{-18(x^2-1)}{(x^2+3)^3}$\n\n\u2022 I know the critcal point is $x = 0$ if you plug that in to $f(x)$ and $f'(x)$ you get $0$ if you plug it into $f''(x)$ you get $f''(x) = \\frac{2}{3}$ does this mean that the local min is $0$ at $x = 0$ ? \u2013\u00a0Csci319 Oct 31 '14 at 17:13\n\u2022 Indeed it does. That method is called the \"second derivative test\" if you want to Google it. \u2013\u00a0GFauxPas Oct 31 '14 at 17:17\n\u2022 but does that mean there is no max? how would I find the max? \u2013\u00a0Csci319 Oct 31 '14 at 17:47\n\u2022 If the domain of your function is all real numbers, then the function has no maximum. \u2013\u00a0GFauxPas Oct 31 '14 at 18:05\n\n## 1 Answer\n\nSign analysis of $f'$\n\n$f'(x)=0$ when $x=0$ so partition the number line as $(-\\infty,0)\\cup(0,\\infty)$.\n\n$(-\\infty,0)$: Pick a test point in this interval, say $x=-1$. Note $f'(-1)<0$ so $f'(x)<0$ on this interval. Hence $f$ is decreasing on this interval.\n\n$(0,\\infty)$: Pick a test point, say $x=1$. Then $f'(1)>0$ so $f'(x)>0$ on this interval. Hence $f$ is increasing on this interval.\n\nSince $f$ went from decreasing to increasing at $x=0$, a local min occurs at $x=0$.\n\nSign analysis of $f''$\n\n$f''(x)=0$ when $x=\\pm 1$ so partition the number line as $(-\\infty,-1)\\cup(-1,1)\\cup(1,\\infty)$.\n\n$(-\\infty,-1)$: Pick a test point in this interval, say $x=-2$. Then $f''(-2)<0\\implies f''(x)<0\\implies f\\text{ concave down}$ on this interval\n\n$(-1,1)$: Pick a test point in this interval, say $x=0$. Then $f''(0)>0\\implies f''(x)>0\\implies f\\text{ concave up}$ on this interval\n\n$(1,\\infty)$: Pick a test point in this interval, say $x=1$. Then $f''(1)<0\\implies f''(x)<0\\implies f\\text{ concave down}$ on this interval\n\nSince $f$ changed concavity at $x=\\pm 1$, these are inflection points.\n\nIndeed a plot of $y=f(x)$ bears out the information above:", "date": "2019-10-19 02:30:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1000035/find-local-max-min-concavity-and-inflection-points", "openwebmath_score": 0.6780862212181091, "openwebmath_perplexity": 277.28776035675014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.992184110286267, "lm_q2_score": 0.8723473846343393, "lm_q1q2_score": 0.8655292136839738}}
{"url": "https://math.stackexchange.com/questions/1727901/dim-k-operatornameimt2-dim-k-operatornameimt-implies-operatornam", "text": "# $\\dim_K\\operatorname{Im}(T^2) =\\dim_K\\operatorname{Im}(T) \\implies \\operatorname{Im}(T) \\cap \\operatorname{Ker}(T) = \\{0\\}$\n\nI need some help here...\n\nLet $V$ be a vector space over a field $K$, and $T: V \\to W$ a linear transformation. Prove that if $$dim_KV \\lt \\infty \\ \\ \\text{and} \\ \\ \\dim_K\\operatorname{Im}(T^2) = \\dim_K\\operatorname{Im}(T)$$ then $$\\operatorname{Im}(T) \\cap \\operatorname{Ker}(T) = \\{0\\}$$\n\nThis is what I made so far.\n\nLet $x \\in \\operatorname{Im}(T) \\cap \\operatorname{Ker}(T)$,where T is linear operator. Then $T(x)=0$ and $x=T(v)$ for some $v \\in V$\n\n$0 = T(x) = T(T(v)) = T^2(v) \\implies v\\in \\operatorname{Ker}(T^2)$\n\nMy intention is to reach $x=0$, but I'm stuck at this point. I also deduced, via dimension theorem, that $\\dim_K \\operatorname{Ker}(T) = \\dim_K \\operatorname{Ker}(T^2)$ but I don't know where it can help. Every hint is appreciated.\n\n\u2022 You have $\\ker(T) \\subset \\ker(T^2)$. Now use the dimension. \u2013\u00a0Maik Pickl Apr 4 '16 at 19:49\n\nThe rank-nullity theorem says \\begin{align} \\dim V&=\\dim\\operatorname{Im}(T)+\\dim\\operatorname{Ker}(T) \\\\ \\dim V&=\\dim\\operatorname{Im}(T^2)+\\dim\\operatorname{Ker}(T^2) \\end{align} Next use the fact that $$\\operatorname{Ker}(T)\\subseteq\\operatorname{Ker}(T^2)$$ to conclude that $$\\operatorname{Ker}(T)=\\operatorname{Ker}(T^2)$$\n\u2022 Oh, I did not know that $\\operatorname{Ker}(T)\\subseteq\\operatorname{Ker}(T^2)$. \u2013\u00a0\u041a\u0430\u0440\u043f\u0430\u0442\u0441\u044c\u043a\u0438\u0439 Apr 4 '16 at 19:59\n\u2022 @\u041a\u0430\u0440\u043f\u0430\u0442\u0441\u044c\u043a\u0438\u0439 If $T(v)=0$, then $T^2(v)=0$. ;-) \u2013\u00a0egreg Apr 4 '16 at 20:00\n\u2022 @egreg I haven't done linear algebra for a while, and I'm trying to refresh my memory. How exactly can we conclude that $\\operatorname{Ker(T)}=\\operatorname{Ker(T^2)}$? Does it follow from the fact that their dimensions are equal and the fact that the one is subset of the other implies that they live in the same space? \u2013\u00a0K.Power Apr 7 '16 at 23:19\n\u2022 @K.Power Since $\\ker(T)\\subseteq\\ker(T^2)$, when we show the former has the same dimension as the latter, we conclude they're equal (take a basis of $\\ker(T)$: it has the number of elements needed to be a basis of $\\ker(T^2)$). \u2013\u00a0egreg Apr 7 '16 at 23:44", "date": "2019-07-15 22:14:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1727901/dim-k-operatornameimt2-dim-k-operatornameimt-implies-operatornam", "openwebmath_score": 0.985831081867218, "openwebmath_perplexity": 193.90365989663832, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631667229061, "lm_q2_score": 0.8774767986961401, "lm_q1q2_score": 0.8655107938878028}}
{"url": "http://math.stackexchange.com/questions/155829/what-is-wrong-with-my-solution-int-cos2-x-tan3x-dx", "text": "# What is wrong with my solution? $\\int \\cos^2 x \\tan^3x dx$\n\nI am trying to do this problem completely on my own but I can not get a proper answer for some reason\n\n\\begin{align} \\int \\cos^2 x \\tan^3x dx &=\\int \\cos^2 x \\frac{ \\sin^3 x}{ \\cos^3 x}dx\\\\ &=\\int \\frac{ \\cos^2 x\\sin^3 x}{ \\cos^3 x}dx\\\\ &=\\int \\frac{ \\sin^3 x}{ \\cos x}dx\\\\ &=\\int \\frac{ \\sin^2 x \\sin x}{ \\cos x}dx\\\\ &=\\int \\frac{ (1 -\\cos^2 x) \\sin x}{ \\cos x}dx\\\\ &=\\int \\frac{ (\\sin x -\\cos^2 x \\sin x) }{ \\cos x}dx\\\\ &=\\int \\frac{ \\sin x -\\cos^2 x \\sin x }{ \\cos x}dx\\\\ &=\\int \\frac{ \\sin x }{ \\cos x}dx - \\int \\cos x \\sin x dx\\\\ &=\\int \\tan x dx - \\frac{1}{2}\\int 2 \\cos x \\sin x dx\\\\ &=\\ln|\\sec x| - \\frac{1}{2}\\int \\sin 2x dx\\\\ &=\\ln|\\sec x| + \\frac{\\cos 2x}{4} + C \\end{align}\n\nThis is the wrong answer, I have went through and back and it all seems correct to me.\n\n-\nAlright, now I would like to know why the first integral gives $\\ln \\sec x$ instead of $-\\ln \\cos x$. \u2013\u00a0Phira Jun 8 '12 at 23:07\nI think that was just an error in typing out the question \u2013\u00a0user138246 Jun 8 '12 at 23:08\n@Phira: Because you can bring the $-1$ up. \u2013\u00a0Eugene Jun 8 '12 at 23:16\nSame function, different expression. Note that $$\\ln|\\sec x| = \\ln\\left|\\frac{1}{\\cos x}\\right| = \\ln\\left(|\\cos x|^{-1}\\right) = -\\ln|\\cos x|.$$ And $$\\cos 2x = \\cos^2x - \\sin^2x = \\cos^2x-(1-\\cos^2x) = 2\\cos^2x - 1$$so $$\\frac{\\cos 2x}{4}+C = \\frac{2\\cos^2x}{4}-\\frac{1}{4}+C = \\frac{1}{2}\\cos^2x + C'.$$ \u2013\u00a0Arturo Magidin Jun 8 '12 at 23:16\nYou're making good progress I see. \u2013\u00a0Eugene Jun 8 '12 at 23:27\n\nA very simple way to check if the answer to an indefinite integral is correct is to differentiate the answer. If you get the original function, your answer is correct, and is equal, up to a constant, with any other solutions.\n\nWe have \\begin{align*} \\frac{d}{dx}\\left(\\ln|\\sec x| + \\frac{\\cos 2x}{4} + C\\right) &= \\frac{1}{\\sec x}(\\sec x)' + \\frac{1}{4}(-\\sin(2x))(2x)' + 0\\\\ &= \\frac{\\sec x\\tan x}{\\sec x} - \\frac{1}{2}\\sin(2x)\\\\ &= \\tan x - \\frac{1}{2}\\left(2\\sin x\\cos x\\right)\\\\ &= \\frac{\\sin x}{\\cos x} - \\sin x\\cos x\\\\ &= \\frac{ \\sin x - \\sin x\\cos^2 x}{\\cos x}\\\\ &= \\frac{\\sin x(1 - \\cos^2 x)}{\\cos x}\\\\ &= \\frac{\\sin^3 x}{\\cos x}\\\\ &= \\frac{\\sin ^3 x \\cos^2 x}{\\cos^3x}\\\\ &= \\frac{\\sin^3 x}{\\cos^3 x}\\cos^2 x\\\\ &= \\tan^3 x \\cos^2 x. \\end{align*} So your answer is right.\n\nThis happens a lot with integrals of trigonometric identities, because there are a lot of very different-looking expressions that are equal \"up to a constant\". So the answer to $$\\int\\sin x\\cos x\\,dx$$ can be either $\\sin^2 x + C$ or $-\\cos^2x + C$; both correct, though they look different, because they differ by a constant: $-\\cos^2x + C = 1-\\cos^2x + (C-1) = \\sin^2x + (C-1)$.\n\n-\n@Always love your answers Prof. \u2013\u00a0Babak S. Jun 9 '12 at 2:35\n\n$${\\ln |\\sec x| + \\frac{{\\cos 2x}}{4} + C}$$\n\n$${\\ln \\left|\\frac 1 {\\cos x}\\right| + \\frac{{1+\\cos 2x}}{4} + C}-\\frac 1 4$$\n\n$${-\\ln \\left| {\\cos x}\\right| + \\frac 1 2\\frac{{1+\\cos 2x}}{2} + K}$$\n\n$${-\\ln \\left| {\\cos x}\\right| + \\frac 1 2 \\cos ^2 x + K}$$\n\n-\nthis kind of thing makes me feel like Calculus II exams and assignments must be hell to mark. \u2013\u00a0crf Jun 8 '12 at 23:56\nEverything is hell to mark. But one can often structure exam questions to minimize these issues. \u2013\u00a0Andr\u00e9 Nicolas Jun 9 '12 at 0:28\n\nYou have been simplifying things up until line 6 and then kind of turned back into complications. It would be natural to notice that $$\\sin x dx = -d\\left(\\cos x\\right)$$ Then the integral looks as follows: $$I=-\\int\\frac{1-\\cos^2x}{\\cos x}d\\left(\\cos x\\right)$$ So it appears that $\\cos x$ plays the role of a variable in its own right, so why not let for example $t=\\cos x$. Now $$I=-\\int\\frac{1-t^2}{t}dt=-\\int\\frac{dt}{t}+\\int tdt=-\\ln t+\\frac{t^2}{2}+C$$ Now plug $\\cos x$ back into place. Recognising distinct \"reusable\" blocks within the expression is the most natural way to arrive at useful substitutions.\n\n-\n\nThe calculation is correct. There are many alternate forms of the integral, because of the endlessly many trigonometric identities.\n\nIf you differentiate the expression you got and simplify, you will see that you are right.\n\nThe answer you saw is likely also right. What was it?\n\nAdded: Since $\\sec x=\\frac{1}{\\cos x}$, we have $\\ln(|\\sec x|)=-\\ln(|\\cos x|)$.\n\nAlso, since $\\cos 2x=2\\cos^2 x -1$, we have $\\frac{\\cos 2x}{4}=\\frac{\\cos^2 x}{2}-\\frac{1}{4}$.\n\nSo your answer and the book answer differ by a constant. That's taken care of by the arbitrary constant of integration. As a simpler example, $\\int 2x\\, dx=x^2+C$ and $\\int 2x\\, dx=(x^2+17)+C$ are both right.\n\n-\n$\\frac{1}{2}cos^2 x - ln|cosx| + C$ \u2013\u00a0user138246 Jun 8 '12 at 23:18\n\nNotice that:\n\n$$\\frac{1}{2} \\cos^2 x = \\frac{1}{2} \\left(\\frac{1}{2} + \\frac{1}{2} \\cos 2x\\right) = \\frac{1}{4} + \\frac{1}{4} \\cos 2x$$\n\nAnd:\n\n$$-\\ln|\\cos x| = \\ln|(\\cos x)^{-1}| = \\ln|\\sec x|$$\n\n-\n\n\u222btanxdx=ln|secx|=-ln|cosx|+ k since -ln |cos x|+k= ln|(cos x)^-1|+k = ln|sec x| +k\n\n(cos2x)/4= 1/2cos^2(x) - 1/4\n\nk-1/4 = new constant C and together you have the solution.\n\nYour answer seems to be correct it is just manipulation or different approach to trig identities.\n\n-\n\nA simple $u$-substitution will work even better, IMO. ${\\cos^2\\theta\\tan^3\\theta }$ yields ${\\frac {\\sin^3}{\\cos\\theta}}$ So starting from there $${\\int\\frac{\\sin^2\\theta}{\\cos\\theta}\\sin\\theta d\\theta}$$ then $${\\int\\frac{1-\\cos^2\\theta}{\\cos\\theta}\\sin\\theta d\\theta}$$ Let $u={\\cos \\theta}$ then ${du=-\\sin\\theta d\\theta}$ $${-\\int\\frac {1-u^2}{u}du}$$ let ${v=1-u^2 }$ ${dv=-2u}$\n\nlet ${dw=1/u}$, ${w=\\ln u}$\n\nthus $${\\ln u(1-u^2)+\\int 2u\\ln u}\\,du$$ then\n\nlet ${v=\\ln u}$, ${dv=u^{-1}}$\n\nlet ${dw=2u}$, ${w=u^2}$\n\nthen \\begin{align*} {\\ln u u^2-\\int \\frac{u^2}{u}\\,du}\\\\{\\ln u u^2-\\int u\\,du}\\\\ {\\ln u(1-u^2)+\\ln u u^2-1/2u^2} \\end{align*} then \\begin{align*}{\\ln u(1-u^2+u^2)-1/2u^2}\\\\ {-(\\ln u-1/2u^2)} \\end{align*} substituting back yields $${-\\ln\\cos\\theta+1/2\\cos^2\\theta+c}$$\n\n-", "date": "2015-11-26 00:35:59", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/155829/what-is-wrong-with-my-solution-int-cos2-x-tan3x-dx", "openwebmath_score": 0.9993574023246765, "openwebmath_perplexity": 1279.6645242069706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631679255076, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8655107838816859}}
{"url": "https://math.stackexchange.com/questions/2003840/a-small-theatre-contains-10-seats-the-seats-are-numbered-from-1-to-10", "text": "# A (small) theatre contains 10 seats. The seats are numbered from 1 to 10,\n\nA (small) theatre contains 10 seats. The seats are numbered from 1 to 10, but not in order. Alice, Bob and Carol have tickets for seats 1, 2 and 3 respectively, and Damien, Eve, Francis, and Gertrude each have tickets without an assigned seat. Unfortunately they all arrive late and the lights are dim, so they can\u2019t see what seat they are sitting in.\n\na) In how many ways can they be seated?\n\nb) In how many ways can they be seated such that Alice is in her correct seat (eg, seat number 1)?\n\nc) In how many ways can they be seated such that Alice, Bob and Carol are all in their correct seats?\n\nd) In how many ways can they be seated such that none of Alice, Bob, Carol are in their correct seats?\n\nWhat I did for part A is The first person has 10 seats to choose from, then the second person has 9 and so on. I got $10\\times9\\times8\\times7\\times6\\times5\\times4$ as the answer.\n\nFor part B I said alice has 1 seat to choose from and then did the rest like part A so I got $1\\times9\\times8\\times7\\times6\\times5\\times4$.\n\nFor C I did the same as B but with 3 people and got $1\\times1\\times1\\times7\\times6\\times5\\times4$ but this doesn't seem right to me.\n\nA bit confused on how to do the rest.\n\n\"For c) I did the same as b) but with 3 people and got $7^{\\underline{4}}=7\\cdot6\\cdot5\\cdot4$ but this doesn't seem right to me\"\n\nYour first three parts are in fact all correct (though they suffer from a lack of convenient notation. If there were 50 people and 80 seats you don't want to waste time writing out all of the numbers in the product. Recommend using one of the following falling factorial notations $(7)_4, 7^{\\underline{4}}, \\frac{7!}{3!}, ~_7P_4, P(7,4),\\dots$ but whichever you use make sure your audience understands the notation)\n\n(Note on falling factorial notation: in all of the above, $(n)_r=n^{\\underline{r}}=P(n,r)=\\dots=\\underbrace{n(n-1)(n-2)\\cdots(n-r+2)(n-r+1)}_{r~\\text{terms in product starting with}~n}$)\n\n\"A bit confused on how to do the rest\"\n\nThe first three parts are most of the calculations needed to complete the fourth part. Recognize that by symmetry the answer to part b) is also the answer to the question of how many ways Bob is seated in his correct seat as well as the answer to how many ways Carol is in her correct seat.\n\nUse inclusion-exclusion to figure out answer to \"In how many ways can they be seated such that at least one of them is in their correct seat.\" Letting $A,B,C$ represent the events that Alice, Bob, and Carol made it to their correct seats respectively, and letting $\\Omega$ represent the sample space i.e. the number of ways in which we don't care who did or didn't make it to their seat, we are tasked with finding the number of ways that none of them made it to their seat. I.e. $|A^c\\cap B^c\\cap C^c|$\n\nNote that $|A^c\\cap B^c\\cap C^c| = |\\Omega\\setminus(A\\cup B\\cup C)|=|\\Omega|-|A\\cup B\\cup C|$\n\n$=|\\Omega|-|A|-|B|-|C|+|A\\cap B|+|A\\cap C|+|B\\cap C|-|A\\cap B\\cap C|$\n\nYou've already found $|\\Omega|,|A|,|A\\cap B\\cap C|$ and by symmetry found $|B|$ and $|C|$. The only information you are missing is $|A\\cap B|,|A\\cap C|,|B\\cap C|$. A similar symmetry argument will reduce the effort it takes to find those.\n\n\u2022 so for part d |A\u2229B|,|A\u2229C|,|B\u2229C|, would all be the same and would be calculated by adding P(9,4) to itself. Then I would put everything in the equation you provided? \u2013\u00a0KGT Nov 7 '16 at 17:52\n\u2022 @KGT $P(n,r) = \\underbrace{n(n-1)(n-2)(n-3)\\cdots(n-r+3)(n-r+2)(n-r+1)}_{r~\\text{total terms in product starting with}~n}$. I don't think you want to be using $P(9,4)$ \u2013\u00a0JMoravitz Nov 7 '16 at 17:55\n\u2022 @KGT you did it correctly in the first three parts. Reworded using the $P$ notation, your answers were $P(10,7),P(9,6)$ and $P(7,4)$ respectively. If we had included the extra question in the middle of those of finding $|A\\cap B|$, i.e. alice and bob get their correct seat and then we seat everyone else, how much would that be? \u2013\u00a0JMoravitz Nov 7 '16 at 18:00\n\u2022 Oh ok. My mistake. I thought that meant P(9,4) meant 9x8x7x6x5x4. What I should use is P(9,6) correct? \u2013\u00a0KGT Nov 7 '16 at 18:01\n\u2022 $P(9,6)$ was the answer to if we only cared about alice getting her correct seat but we didn't care about bob or carol getting their correct seats. If we cared about alice and bob, should that nine have been in the product? \u2013\u00a0JMoravitz Nov 7 '16 at 18:02", "date": "2020-09-24 10:40:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2003840/a-small-theatre-contains-10-seats-the-seats-are-numbered-from-1-to-10", "openwebmath_score": 0.6978425979614258, "openwebmath_perplexity": 199.38947621672557, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631655203046, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8655107801909799}}
{"url": "https://lumina.com/community/sub-forum/how-to-calculate-cumulative-cash-flow-from-a-cash-flow-table-indexed-by-year/", "text": "How to calculate cu...\n\nClear all\n\n# How to calculate cumulative cash flow from a Cash Flow table indexed by year\n\nPosts: 4\nCustomer\nTopic starter\nActive Member\nJoined: 7 months ago\n\nI am a relative neophyte in using indexed parameters. I want to calculate a cumulative cash flow by year from a table of cash flows, in order to calculate when a project will break even. I am looking for the syntax of a call on each row of my cash flow list to add it to the previous row of my cumulative cash flow list. Once I have that, I also want to calculate a similar list of discounted cumulative cash flows to understand the effect of discounting the cash flow. Is it worthwhile to create a second index with cash flow, cumulative cash flow, and discounted cash flow? Or is there an even more clever way to do this using features I am unaware of?\n\nTopic Tags\n5 Replies\nPosts: 19\nModerator\nMember\nJoined: 1 year ago\n\nYou are looking for the Cumulate function, which takes your cash flow by Year and returns the cumulated cash flow by Year.\u00a0 The syntax is:\n\nCumulate( Cash_flow, Year )\n\nTo discount a cash flow, you just multiply Cash_flow by the discount factor for each year, like this:\n\nCash_flow * (1 - Discount_rate)^(@Year-1)\n\nHere I've used @Year to number the years starting at 1. By using @Year-1, it means that the first year won't be discounted, whereas if you raise to the power of @Year you would discount the first year.\u00a0 Think about which convention you want.\u00a0 You can then pass this to the first parameter of Cumulate.\n\nYou want to compare the undiscounted case to the discounted case. A cool thing is that you can do this without duplicating all the logic for each case. The undiscounted case is simply the discounted case when the discount rate is 0. So, just make Discount_rate a Choice, i.e., between 0% and 7%, and then all your subsequent logic that uses discounted cash flow will work for either, or indeed for both at the same time (by selecting ALL for the choice), giving you the ability to compare any downstream variable without having to do anything additional.\n\nIn a short while, I will attach a model that demonstrates all of this and adds in the calculation for the breakeven year.\n\nCustomer\nJoined: 7 months ago\n\nActive Member\nPosts: 4\n\nThis looks great! And the tip on 0% discount is excellent. I look forward to your example, and hope to get back to this work this week.\n\nPosts: 4\nCustomer\nTopic starter\nActive Member\nJoined: 7 months ago\n\nThis approach worked great. \u00a0Now I can get to my reason for doing this. \u00a0I would like to define a variable that finds the year when cumulative cash flow becomes positive, return the year, and add to that the value of cumulative cash flow for that year divided by the cash flow for that year, which gives me a number of years (with a fraction) to break even. Still looking to figure out how to do this.\n\nPosts: 19\nModerator\nMember\nJoined: 1 year ago\n\nSorry about the delay with posting the model that I had promised.\u00a0 I hit a technical problem. Anyway, here it is:\n\nCash flow.ana\n\nIt includes a variable that finds the breakeven year and a fraction of the breakeven year. I'm not sure if my fraction of the breakeven year is exactly what you just described, but hopefully you can tweak it if not.\n\nPosts: 4\nCustomer\nTopic starter\nActive Member\nJoined: 7 months ago\n\nI got taken away on other things (including a short vacation) and am just getting back to this reply.\u00a0 I had in the meantime discovered the Solar Panel model, which gets the breakeven year by using the CondMin function, but you have come at it a separate way. Either way, I am learning useful new functions! I will look at it and see if it gets the same answer I was coming to.\u00a0 Thanks!\n\nShare:\nScroll to Top", "date": "2022-10-07 14:59:59", "meta": {"domain": "lumina.com", "url": "https://lumina.com/community/sub-forum/how-to-calculate-cumulative-cash-flow-from-a-cash-flow-table-indexed-by-year/", "openwebmath_score": 0.47006139159202576, "openwebmath_perplexity": 973.6652584451767, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561712637256, "lm_q2_score": 0.8933094074745443, "lm_q1q2_score": 0.8654883322796544}}
{"url": "https://math.stackexchange.com/questions/998177/how-to-know-when-a-particular-proof-is-appropriate-for-the-given-problem", "text": "# how to know when a particular proof is appropriate for the given problem?\n\nThe main trouble I am currently having in math is knowing when the use cases are appropriate in a proof. I see many videos where they seem to choose a strategy like proof by contrapositive or proof by contradiction, but never quite understand how they came to the conclusion to use that proof strategy.\n\nHere are some examples I have come across and my own solutions to them using the recommended proofs\n\nthe product of two odd integers is odd\n\nI used contradiction to solve it\n\n\u2022 suppose the product of two odd integers is even\n\u2022 (2k+1)(2k+1) = 2k\n\u2022 2(2k^2 + 2k) + 1 = 2k\n\u2022 given k is an integer, (2k^2 + 2k) is an integer. Therefore, an even number cannot be an odd number\n\nif x^2 is even, x is even\n\nBased on the wikipedia article on contrapositive\n\n\u2022 if x is odd, then x^2 is odd\n\u2022 (2k+1)^2 == 2(2k^2 +2k) + 1\n\u2022 therefore, if x^2 is even, then x is even\n\nHowever, what I am wondering is, is there a general principle as to when to use specific strategies for proofs? Eg, if you specifically know a theory is false, do you choose a strategy accordingly? What determines which strategy will be most effective, or is it arbitrary?\n\n\u2022 \u2013\u00a0Shaun Oct 30 '14 at 12:38\n\u2022 For \"perfect\" proofs see the book of Aigner and Ziegler, Proofs from THE BOOK \u2013\u00a0Dietrich Burde Oct 30 '14 at 12:55\n\nThis skill comes with lots of practice. Generally speaking, direct proofs are used when the result is \"positive-sounding\". In your example, the result is, \"the product of two odd integers is odd\". This result is positive-sounding (as opposed to a result like, \"there is no smallest positive real number\").\n\nFor the result, \"if $x^2$ is even, then $x$ is even,\" you could do either a direct proof or a proof by contrapositive. A lot of times it does not matter and there is usually more than one way to prove something. The direct proof, in this case, is favored over a proof by contrapositive simply because it is more, well, direct. Here is how the result would be proven using the contrapositive:\n\nProof:\n\nSuppose $x$ is odd. We show that $x^2$ is odd. So $x=2k+1$ for some integer $k$. Thus $x^2=(2k+1)^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1.$ Now, since $(2k^2+2k)$ is an integer, it follows that $x^2$ is odd.\n\nFor a result like \"there is no smallest positive real number\", this would be difficult to prove using a direct proof. The result is better suited for a proof by contradiction. It would start like this: \"Suppose there exists a smallest positive real number\u2026\"\n\nSome other examples of negative sounding results that would be more suited for proof by contradiction:\n\n\u2022 If a is an even integer and b is an odd integer, then 4 does not divide $a^2+2b^2$\n\u2022 The integer 100 cannot be written as the sum of three integers, and odd number of which are odd\n\u2022 The sum of a rational number and an irrational number is irrational.\n\n[Source: Chartrand, G., Polimeni, A.D., Zhang, P., 2013. Mathematical Proofs: A Transition to Advanced Mathematics, 3rd ed. Boston: Pearson.]\n\n\u2022 ah, exactly what I was looking for, very clear and understandable, thank you! \u2013\u00a0corvid Oct 30 '14 at 13:05\n\nThere are many ways to write down a proof. The question is, apart from correctness, which one is more elegant, more concise, more constructive, and last not least easier und thus clearer. In your case, for me the easiast way is to do first an obvious computation, i.e. $$(2k+1)(2l+1)=4kl+2k+2l+1.$$ Your first step here is wrong, because you assume that both integers are equal, i.e., both given by $2k+1$. This would be a special case.\nThen, in the second step, we consider everything modulo $2$. The integer on the RHS is odd, because it is of the form $2m+1$, with $m=2kl+k+l$. This finishes the proof.\n\n\u2022 theoretically, anything you can prove by contradiction can also be proved by contrapositive, in that sense? \u2013\u00a0corvid Oct 30 '14 at 12:54", "date": "2019-06-20 04:58:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/998177/how-to-know-when-a-particular-proof-is-appropriate-for-the-given-problem", "openwebmath_score": 0.8338315486907959, "openwebmath_perplexity": 303.6622501882552, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561667674652, "lm_q2_score": 0.8933094103149355, "lm_q1q2_score": 0.8654883310150332}}
{"url": "http://carlacasciari.it/moment-of-inertia-of-triangle-about-apex.html", "text": "s\u2032 = required moment of inertia of the combined ring\u2010 shell\u2010cone cross section about its neutral axis par-allel to the axis of the shell, in. R\u00e9sultat de recherche d'images pour \"bridge equation for moment of inertia\" See more. Find the moment of inertia of the framework about an axis passing through A, and parallel to BC 5ma 2. The second moment of inertia of the entire triangle is the integral of this from $$x = 0$$ to $$x = a$$ , which is $$\\dfrac{ma^{2}}{6}$$. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. RE: second moment of inertia for triangle cross section? the formula. Lecture notes, lecture 11 - Center of gravity, centroid and moment of inertia. The equation for the moment of inertia becomes: \u222b \u2212 = \u2212 8 8 2 2 2 2 x' dy' 14 y' I y' 2 8 1 To perform this integration we need to place the integrand in an m-file function and call MATLAB\u2019s quad() function on the m-file. After explaining the term second moment of area, the method of finding moment of inertia of plane figures about x-x or y-y axis is illustrated. You can also drag the origin point at (0,0). Once the moment of inertia was calculated, we had to measure the angular acceleration of the pulley. See the picture: the points of the upper triangle are farther than those of the lower triangle. It is measured by the mass of the body. o The moment of inertia of a thin disc of mass m and radius r about an axis passing through its C. unambiguous choice between the divergent views currently held with regard to the structure. We can see from that the moment of inertia of the subrectangle about the is Similarly, the moment of inertia of the. function Ix_integrand = Moment_Of_Inertia_Integrand(y_prime) %Saved as Moment_Of_Inertia_Integrand. The moment of inertia of the triangle is not half that of the square. 250 kg; from mass A: rB\u00b2 = 0. A numerical integrator might return slightly less accurate results, but other than that there is not much benefit from using symbolic integration there. - Theory - Example - Question 1 - Question 2 - List of moment of inertia for common shapes 4. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Thus the mass of the body is taken as a measure of its inertia for translatory. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. The moment of inertia of a triangle with respect to an axis passing through its apex, parallel to its base, is given by the following expression: I = \\frac{m h^2}{2} Again, this can be proved by application of the Parallel Axes Theorem (see below), considering that triangle apex is located at a distance equal to 2h/3 from base. pptx), PDF File (. However, \"area moment of inertia\" is just 4 words to me (no physical meaning). Overview (in Hindi) 8:26 mins. Polar moment of Inertia (Perpendicular Axes theorem) The moment of inertia of an area about an axis perpendicular to the plane of the area is called Polar Moment of Inertia and it is denoted by symbol Izz or J or Ip. b) Determine the moment of inertia for a composite area Parallel-Axis Theorem for an Area Relates the moment of inertia of an area about an axis passing through the. Check to see whether the area of the object is filled correctly. 7) Moment of Inertia Triangle. Mathematical calculations of the GaN NWs' cross-sectional areas and the moment of inertia For the single crystalline (SC) GaN nanowire (NW), the cross-sectional shape is an isosceles triangle with a 63. The median is a line from vertex to the center of a side opposite the vertex. Or the Mizuno MP-20. Find the moment of inertia of the triangle about axis passing through centroid perpendicular to lamina. Hodgepodge. The moment of point \"B\" is 0. The length of each side is L. In case of shafts subjected to torsion or twisting moment, the moment of inertia of the cross-sectional area about its centre O is considered. The sum of all these would then give you the total moment of inertia. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. About the Moment of Inertia Calculator. The mass moment of inertia is a measure of an object\u2019s resistance to rotation. P-715 with respect to the given axes. 456kg Length of the base of triangle =. Statics - Chapter 10 (Sub-Chapter 10. Answer this question and win exciting prizes. Moments of Inertia Staff posted on October 20, 2006 | Moments of Inertia. The moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moment of inertias of the lamina about the two axes at right angles to each other in its own plane intersecting each other at the point where the perpendicular axis passes through it. Triangle Moment of Inertia. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. 31 shows a T-section of dimensions 10 \u00d7 10 \u00d7 2 cm. The area moment of inertia is also called the second moment of area. Mar 27, 2001 3,923 0 76. G The centroid and centre of gravity are at the same point Where centre of gravity consider to be whole mass of an object act at a point C. The sum of the first n \u2265 1 energy levels of the planar Laplacian with constant magnetic field of given total flux is shown to be maximal among triangles for the equilateral triangle, under normalization of the ratio (moment of inertia)/(area) 3 on the domain. Example of Product Moment of Inertia of a Right Angle Triangle Product Moment of Inertia of a Right Angle Triangle by Double Integration. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. The calculations are as shown. Lecture notes, lecture 11 - Center of gravity, centroid and moment of inertia. This app was created to assist you in these calculations and to ensure that your beam section calculations are fast and accurate. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Rectangle Triangle. For this reason current vector is treated as normal vector of the plane and the input cloud is projected onto it. The struts are built with the quad-edge passing through the mid-point of the base. \uc800\uc790: No machine-readable author provided. The mass moment of inertia is often also known as the rotational inertia, and sometimes as the angular mass. Date: 02/03/99 at 14:37:05 From: Doctor Anthony Subject: Re: MI of Solid Cone You must, of course, specify about which axis you want the moment of inertia. Recommended for you. Determine the moment of inertia of this 10. Physics Lab #17 started on 5/13/15 Finding the moment of inertia of a uniform triangle about its center of mass Annemarie Branks Professor Wolf Objective: Find the moment of inertia for a uniform, right triangle plate about its center of mass for the two orientations as shown below. txt) or view presentation slides online. on AIPMT / NEET-UG entrance. I xx = \u222bdA. Moment of inertia is a commonly used concept in physics. m in the MATLAB. Area moment of inertia calculation formulas for the regular cross section are readily available in design data handbooks. In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The moment of inertia of a body is always defined about a rotation axis. 4 \"Center of Mass\" of our text APEX Calculus 3, version 3. The maximum shear. Assume density = 1 Here's my working , I use zx plane projection , but i didnt get the ans. We define dm to be a small element of mass making up the rod. \u201d Moment of inertia = SI unit of moment of inertia is Q. The calculations are as shown. Mass moment of inertia (also referred to as second moment of mass, angular mass, or rotational inertia) specifies the torque needed to produce a desired angular acceleration about a rotational axis and depends on the distribution of the object\u2019s mass (i. In order to find the moment of inertia of the triangle we must use the parallel axis theorem which ius as follows: The moment of inertia about any axis parallel to that axis through the center. Rotational version of Newton's second law. The apex angle of the quarter-circle is $\\pi/2$. The moment of point \"C\" is the same as \"B\" so multiply the moment of \"B\" by two. Cone Calc Processing :. Answer this question and win exciting prizes. 4 Moment of inertia in yaw 2. To find the inertia of the triangle, simply subtract the inertia of the system with the triangle from the benchmark. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. Moments of Inertia. P-715 with respect to the given axes. Find the moment of inertia of a circular section whose radius is 8\u201d and diameter of 16\u201d. The moment of inertia of the triangle is not half that of the square. Determining the moment of inertia of a solid disk. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). 5 2 3 A 4-0. Click Content tabCalculation panelMoment of Inertia. Q1: Matthew has a model train that uses a circular cone as a flywheel. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. The term product moment of inertia is defined and the mehtod of finding principal moment of inertia is presented. This banner text can have markup. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. 1 Expert Answer(s) - 58298 - what is the moment of inertia of a triangular plate ABC of mass M and side BC = a about an axis pass. Which one has the greatest moment of inertia when rotated about an axis perpendicular to the plane of the drawing at point P? Preview this quiz on Quizizz. 4 Moment of inertia in yaw 4 DISCUSSION OF 33TIXYI'ZD Al4D ?XWXQdENPAL VAIJJES 4. 035; Actual VCOG. Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. If you need a beam\u2019s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. The oxygen molecule as a mass of 5. 2) A long rod with mass has a moment of inertia , for rotation around an axis near one. Calculating the moment of inertia of a triangle - Duration: 10:01. The moment of inertia must be specified with respect to a chosen axis of rotation. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. Figure to illustrate the area moment of a triangle at the list of moments of inertia. 1) Prove that the centroid of any triangle of height h and base b is located 2/3 of the distance from the apex. Own work assumed (based on copyright claims). And, regretfully, you disturbed. The moment of inertia of an object is based on 3 things, the mass of the object, the axis of rotation, and the orientation and distance of the object from the axis of rotation. Inertia is a property of a body to resist the change in linear state of motion. Rotational kinetic energy. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. Use our free online app Moment of Inertia of a Ring Calculator to determine all important calculations with parameters and constants. Moment of Mass about x and y-axis Mass of Lamina - f(x) Mass of Lamina - f(y) Radius of Gyration (x-axis) Radius of Gyration (y-axis) 1. Tags: Equations for Moment of Inertia. The 2 nd moment of area, or second area moment and also known as the area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. My teacher told me :. Let us assume that one line is passing through the base of the triangular section and let us consider this line as line BC and we will determine the moment of inertia for the triangular section about this line BC. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. (1) I y: equ. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). Learn vocabulary, terms, and more with flashcards, games, and other study tools. PEP Assignment 4 Solutions 3 = = 2 where = 2 Knowing what d and are, CM is CM =,\u222b d CM = from the apex of the triangle. The unit of dimension of the second moment of area is length. Therefore, equation for polar moment of inertia with respect to apex is. Thus, the moment of inertia of a 2D shape is the moment of inertia of the shape about the Z-axis passing through the origin. I = 1/3 b * h^3 / 12. m2) Title: Microsoft Word - Formular Moment of Inertia Author: d00997 Created Date: 4/25/2019 4:40:32 PM. Triangle h b A= 1 2 b\u00d7h x1=b/3 From side x2=2b/3 From right side y1=h/3 From bottom y2=2h/3 From Apex Ixx= bh3 36 Circle d A=\u03c0 4 \u00d7d2 x=d/2 y=d/2 I xx= \u03c0 64 d4 I yy= \u03c0 64 d4 Semicircle A= \u03c0 4 \u00d7d2 2 x=d/2 y1=0. Solution 126 2 Polar moment of inertia SECTION 126 Polar Moments of Inertia 15 from COE 3001 at Georgia Institute Of Technology. 51 videos Play all MECHANICAL ENGINEERING 12 MOMENT OF INERTIA / AREA Michel van Biezen Area Moment of Intertia of a Triangle Brain Waves - Duration: 7:23. 2) The radius of the gyration of a disc of radius 25 cm is. 3-axis along the axis of the cone. The moment of inertia is not related to the length or the beam material. Write the equation for polar moment of inertia with respect to apex of triangle. dA Y = 0 A A = b. ) have only areas but no mass. Solution for Find the center of mass and the moment of inertia about the z-axis of a thin shell of constant density d cut from the cone x2 + y2 - z2 = 0 by the\u2026 Answered: Find the center of mass and the moment\u2026 | bartleby. Calculate the moment of inertia of a right circular cone. Hemmingsen assumed (based on copyright claims). Find the moment of inertia of the table with the iron ring. I = Pi * R^4 / 4. March 2020. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. he solves alone. The centre of area of such figure is known as centroid. Find the moment of inertia of a uniform solid circular cone of mass M, height h and base radius a about its axis, and also about a perpendicular axis through its apex. (by the parallel axis theorem). Figure to illustrate the area moment of a triangle at the list of moments of inertia. Moments of Inertia Staff posted on October 20, 2006 | Moments of Inertia. Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. It should not be confused with the second moment of area, which is used in bending calculations. Moment of Inertia of a Triangular Lamina about its Base. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. Explain the terms moment of inertia and radius of gyration of a plane figure. 89 \u00d7 103 kg/m3. Moment of Inertia of Surfaces. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Ditto the TaylorMade P730. Basic trig functions 8 Moments of Inertia The moment of inertia is the stiffness of a body due to its size and. \u201d The product mass and the square of the perpendicular distance from the axis of rotation is known as moment of inertia. \u201d Moment of inertia = SI unit of moment of inertia is Q. Then remove the middle triangle from each of the re-maining three triangles (as shown), and so on, forever. Section area moment of inertia section modulus calculator what is the difference between polar moment of inertia m area moment of inertia typical cross sections i statics Centroid Area Moments \u2026. The moment of inertia of a triangle of base b and altitude h with respect to a centroidal axis parallel to its base would be bh3/12 bh3/18 bh3/24 bh3/36 The CG of a triangle lies at the point of intersection of diagonals altitudes bisector of angles medians For a solid cone of height h, the CG lies on the axis at a distance above the base equal. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. 01 18-Jun-2003 1. Find I y for the isosceles triangle shown. 32075h^4M/AL, where h is the height of the triangle and L is the area. Inertia is a property of a body to resist the change in linear state of motion. 1 GradedProblems Problem 1 (1. The moment of point \"C\" is the same as \"B\" so multiply the moment of \"B\" by two. Undeniable momentum, on any stage - anywhere. Centroid, Area, and Moments of Inertia Yong-Ming Li January, 1997 1 Introduction This design document was originally written for computation of area, centroid, and moments of inertia of lamina (a thin plate of uniform density). s\u2032 = required moment of inertia of the combined ring\u2010 shell\u2010cone cross section about its neutral axis par-allel to the axis of the shell, in. The moment of inertia of a triangle with respect to an axis passing through its apex, parallel to its base, is given by the following expression: I = \\frac{m h^2}{2} Again, this can be proved by application of the Parallel Axes Theorem (see below), considering that triangle apex is located at a distance equal to 2h/3 from base. The inertia matrix (aka inertia tensor) of a sphere should be diagonal with principal moments of inertia of 2/5 mass since radius = 1. 1501 Laura Duncan Road, Apex, NC 27502 Email us (919) 289-9278 MAIL TO: P. May 17, 2019 Mirielle Sabety, Keane Wong, Anthony Moody Purpose: The purpose of today's lab is to measure the moment of inertia of a triangle about it's center of mass with in 2 different orientations. Find the moment of inertia of the empty rotating table. I), must be found indirectly. For a homogenous bar of length L and mass M, the moment of inertia about center of mass is (1/3)ML^2. Moments of Inertia 10. 250 kg; from mass A: rB\u00b2 = 0. Doing the same procedure like above, and below is the work. Using the lower left. This moment of inertia about 0 is called polar moment of inertia or moment of inertia about pole. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. Planar and polar moments of inertia formulas. Area Moment of Inertia Section Properties of Triangle Calculator and Equations. 31 shows a T-section of dimensions 10 \u00d7 10 \u00d7 2 cm. If you need a beam\u2019s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. The moment of inertia is de ned as I= X i m ir 2 i (2) for a collection of point-like masses m ieach at a distance r ifrom the speci ed axis. 2) Find the distance for each intersection points. Choose the origin at the apex of the cone, and calculate the elements of the inertia tensor. Using these moment of inertia, we can subtract from it the moment of inertia of just the system without the triangle to obtain our experimental values for the triangle in either. Radius of gyration 3. Find Moment of Inertia of a Ring Calculator at CalcTown. Centroids and moments of inertia. Radius and elevation of the semi-circle can be changed with the blue point. In other words, it is rotating laterally, similar to how a beam from a lighthouse rotates. 100% Upvoted. We spin the triangle around the spot marked \"X\", which is one of the balls. Calculate the Inertia of the semi-circle around the pivot. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. In this worksheet, we will practice finding the moment of inertia and radius of gyration of a solid and using the parallel axis theorem to find the moment of inertia of a composite solid at different axes. anybody here could help, please? i would really appreciate it. Meanwhile, I did find the integral formula for computing the center of pressure (Fox) and calculated it using both a flat bottom and inverted isosceles triangle and then using the \"area moment of inertia\". They are; Axis passing through the centroid. Constant angular momentum when no net torque. Answer Save. This engineering calculator will determine the section modulus for the given cross-section. I am unable to find it. Another solution is to integrate the triangle from an apex to the base using the double integral of r^2dm, which becomes (x^2+y^2)dxdy. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. It is used over and over for examples, since it offers a readymade right angle, a hypotenuse, and other great parts. For the section shown in Fig. 1 In the case of mass and area, the problem is deciding the distance since the mass and area are not concentrated at one point. m2) ) x 10-6 (kg. Consider the diagram as below: We can think of the triangle is composing of infinitesimally. In the final stage of the calculation, you specify the direction of the load forces. 75L, just find the area of the left triangle on the shear diagram and subtract the area of 1/2 the horizontal distance of the second triangle (not 1/2 the area of the second triangle). To find the moment of inertia of the entire section, we integrate the above expression and get, Iyy = \u03a3dAx2, Ixx = \u03a3dAy2 and Izz = \u03a3dAz2. The moment of inertia integral is an integral over the mass distribution. Tension Members. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Mass = m and Base = l Angle at the apex is = 90\u00b0 Find MI of theplane about the y - axis = ? Let, the axis of rotation pass through hypotenuse, considering rotation about hypotenuse you will see triangle. Hollow Cone. The theoretical one is know the moment of inertia of the triangle plate and applied the parallel axis theorem to found the moment of inertia about a new rotating axis. This is the currently selected item. This banner text can have markup. Mathematically, and where IB \" *BA \" TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Mechanics of Material (CIV101) Academic year. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. The disk is rotating about its center. Centroids and moments of inertia. Determining the moment of inertia of a rod. a) Define i)Moment of Inertia , ii)radius of gyration b) Define stress ,strain ,Modulus of elasticity c) State formulae to find Moment of Inertia of a triangle about axis passing through its i) Base ii) Apex and iii) centroid d) Define lateral strain, linear strain. The maximum twisting moment a shaft can resist, is the product of the permissible shear stress and (A) Moment of inertia (B) Polar moment of inertia (C) Polar modulus (D) Modulus of rigidly Answer: Option C Question No. Moment of inertia (I1 and I2) along the 1 and 2 axes. of the ozone molecule. The moment of inertia of any triangle may be found by combining the moments of inertia of right triangles about a common axis. r2 x2 y2 Therefore, I z I. Rectangle Triangle. Neutral Axis/Moment of Inertia. moment of inertia. University. Author: No machine-readable author provided. Moment of inertia of pile group. save hide report. \ub0a0\uc9dc: 2006\ub144 4\uc6d4 23\uc77c (\uc6d0\ubcf8 \uc62c\ub9ac\uae30 \uc77c\uc2dc) \ucd9c\ucc98: No machine-readable source provided. Evaluation of Moments of Inertia 2008 Waterloo Maple Inc. ) 15 minutes ago The transformer inside of a sound system has 1500 turns in its primary coil windings wrapped around a common iron core with the secondary. \u201cSecond moment of an area about an axis is called Moment of inertia. Find The Moment Of Inertia About An Axis That Passes Through Mass A And Is Perpendicular To The. Find its moment of inertia for rotation about the z axis. Simply Supported Beams (Shear & Moment Diagrams) Simply supported beams (also know as pinned-pinned or pinned-roller) are the most common beams for both school and on the Professional Engineers exam. Known : Mass of rod AB (m) = 2 kg. Date: 02/03/99 at 14:37:05 From: Doctor Anthony Subject: Re: MI of Solid Cone You must, of course, specify about which axis you want the moment of inertia. Consider an infinitesimally thin disc of thickness dh, at a distance h from the apex of the cone O. Moment of inertia is defined as:\u201dThe sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. save hide report. The moment of inertia (I) of a body is a measure of its ability to resist change in its rotational state of motion. The area moment of inertia is also called the second moment of area. These came out to be 0. Determine the moment of inertia of the triangular area relative to a line parallel to the base and through the upper vertex in cm^4. Physically, it is a measure of how difficult it is to turn a cross-section about an axis perpendicular to it (the inherent rotational stiffness of the cross-section). a) Define i)Moment of Inertia , ii)radius of gyration b) Define stress ,strain ,Modulus of elasticity c) State formulae to find Moment of Inertia of a triangle about axis passing through its i) Base ii) Apex and iii) centroid d) Define lateral strain, linear strain. because the axis goes through masses B and D their masses doesn't affect to increase the inertia of the system around BD axis. The mass moment of inertia is a measure of an object\u2019s resistance to rotation. For instance, it is easier to find the moment of inertia of a triangle, about an axis which passes through its apex, and parallel to the base, than about any ether axis ; but having found this, we may easily find it about an axis parallel to it which passes through the centre. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. Author: No machine-readable author provided. 12: Inertia due to the Object (kg. After that eccentricity is calculated for the obtained projection. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. Write the equation for polar moment of inertia with respect to apex of triangle. Then this moment of inertia is transferred about the axis passing through the centroid of the given section, using theorem of parallel axis. The moment of inertia $$I_x$$ about the $$x$$-axis for the region $$R$$ is the limit of the sum of moments of inertia of the regions $$R_{ij}$$ about the $$x$$-axis. For a homogenous bar of length L and mass M, the moment of inertia about center of mass is (1/3)ML^2. The concept of a moment of inertia is important in many design and analysis problems encountered in mechanical and civil engineering. CENTER OF GRAVITY, CENTROID AND MOMENT OF INERTIA. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. 2) A precast concrete floor beam has the cross section shown below. apex angle in the neighborhood of 34\u00b0. it is first necessary to consider the rotational moment. Adding moments of inertia 3. The moment of inertia of the triangle is not half that of the square. But I don't know how to do that. The moment of inertia must be specified with respect to a chosen axis of rotation. A triangle cannot have more than one right angle or one obtuse angle, since the sum of all three angles is equal to the sum of two right angles, which is 180\u00b0 or, in radians, \u03c0. I am unable to find it. Ball hits rod angular momentum example. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. The moment of inertia of a triangle of base b and altitude h with respect to a centroidal axis parallel to its base would be bh3/12 bh3/18 bh3/24 bh3/36 The CG of a triangle lies at the point of intersection of diagonals altitudes bisector of angles medians For a solid cone of height h, the CG lies on the axis at a distance above the base equal. same object, rotating around a point at the midpoint of its base. 0 cm is made of copper. Look up I for a triangle in your table if you have forgotten. 10 lessons \u2022 1 h 34 m. \u2022Compute the product of inertia with respect to the xyaxes by dividing the section into three rectangles. 4 Locate the centroid of the T-section shown in the Fig. 016 kg \u22c5 m2 d. Right: A circle section positioned as per the upper sketch is defined in the calculator as I x-axis , the lower sketch shows I y-axis. It depends on the body's mass distribution and the axis chosen, with larger moments. In this study, we first compute the polar moment of inertia of orbit curves under planar Lorentzian motions and then give the following theorems for the Lorentzian circles: When endpoints of a line segment AB with length a +b move on Lorentzian circle (its total rotation angle is \u03b4) with the polar moment of inertia T, a point X which is collinear with the points A and B draws a Lorentzian. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Rolling without slipping problems. For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section The moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis. Determining the moment of inertia of a rod. Moment of inertia is defined as:\u201dThe sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. Moments of Inertia. Determine the moment of inertia of this system about an axis passing through one corner of the triangle and perpendicular to the plane of the triangle. Insert the moment of inertia block into the drawing. b) Determine the moment of inertia for a composite area Parallel-Axis Theorem for an Area Relates the moment of inertia of an area about an axis passing through the. o , ,3, Moment of Inertia of Surfaces. y-x O 1 1 \u2022 (x, y) r Answer: The polar moment of inertia of a planar region is the moment of inertia about the origin (the axis of rotation is the z-axis). Area Moments of Inertia by Integration \u2022 Second moments or moments of inertia of an area with respect to the x and y axes, x \u00b3 yI y \u00b3 xdA 2 2 \u2022 Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. Letting M be the total mass of the system, we have x \u00af = M y / M. 42441 1in 1 in 1 in 3 in 1 in A 2 A 3 A 1 A 4 16 Centroid and Moment of. 19 The deflection of any rectangular beam simply supported, is (A) Directly proportional to its weight. 41 (a) determine: (i) Moment of inertia about its centroid along (x,y) axis. For each segment defined by two consecutive points of the polygon, consider a triangle with two. If the line l(P, 9) lies in the plane of K through the point P and with direction 9, 0 = 9 ^ 2n, we denote the moment of inertia of K about the line l(P, 6) by I(K, P, 9). Email Print Moment of Inertia of a Triangle. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. Here, distance between apex and centroid is d. apex angle in the neighborhood of 34\u00b0. however, i would like to know how you obtain the results. How to calculate polar moment of inertia using Inventor 2014 Hello, I have a problem with calculating polar moment of inertia of a cranshaft with cooperating parts (which I've already assembled in Inventor). Find the moment of inertia of a uniform solid circular cone of mass M, height h and base radius a about its axis, and also about a perpendicular axis through its apex. Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. 1 Verified Answer. 12: Inertia due to the Object (kg. The smallest Moment of Inertia about any axis passes throught the centroid. The struts are built with the quad-edge passing through the mid-point of the base. derivation of inertia of ellipse. Q: the moment of inertia of a thin rod of mass m and length l about an axis through its centre of gravity and perpendicular to its length is a) ml\u00b2/4 b) ml\u00b2/6 c) ml\u00b2/8 d) ml\u00b2/12 Q: Which statement is correct: a) Moment of inertia is the second moment of mass or area. Moment of inertia of the equilateral triangle system - Duration: 3:38. The material is homogeneous with a mass density \u03c1. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. The moment of inertia of any triangle may be found by combining the moments of inertia of right triangles about a common axis. Because there is some frictional torque in the system, the angular acceleration of the system when the mass is descending isn\u2019t the same as when it is ascending. Own work assumed (based on copyright claims). 6-1 Polar moment of inertia POINT C (CENTROID) FROM CASE 5: (I P) c 2 bh. What is the moment of inertia of this rigid body about an axis that is parallel to one side of the triangle and passes through the respective midpoints of the other two sides? a. Thank you User-12527562540311671895 for A2A The moment of inertia of a triangular lamina with respect to an axis passing through its centroid, parallel to its base, is given by the expression $I_{XX}=\\frac{1}{36}bh^3$ where $b[/mat. But I don't know how to do that. Find MI of and equilateral triangle of side 2m about its base. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2 ) is a measure of an object\u2019s resistance to changes in its rotation rate. Free flashcards to help memorize facts about Moment of Inertia of Different Shapes. Calculate its. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. Area Moments of Inertia Parallel Axis Theorem \u2022 Moment of inertia IT of a circular area with respect to a tangent to the circle, ( ) 4 4 5 4 2 2 4 2 1 r IT I Ad r r r \u03c0 \u03c0 \u03c0 = = + = + \u2022 Moment of inertia of a triangle with respect to a. Calculate the triangles moment of inertia when its axis of rotation is located at the right-angled corner. The \u201cnarrower\u201d the triangle, the more exact is the formula (2). Calculating the moment of inertia of a triangle - Duration: 10:01. 42\u00d7r from base y2=0. Click Content tabCalculation panelMoment of Inertia. We can relate these two parameters in two ways: For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Determine the product of inertia of the crosshatched area with respect to the x and y axes. Area moment of inertia calculation formulas for the regular cross section are readily available in design data handbooks. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Or the Mizuno MP-20. Callaway Ratings & Specs 2019 Head Ratings by Maltby Experts at The GolfWorks View All | View By Brand. Thus, the object\u2019s mass and how it is distributed both affect the mass moment of inertia. After that eccentricity is calculated for the obtained projection. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. 5 Parallel-Axis Theorem - Theory - Example - Question 1 - Question 2. MSC separate and tethers begin to reel out, where inertia of the 3 MSC, increases with time, and the inertia of CSC, remains constant. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. r2 x2 y2 Therefore, I z I. unambiguous choice between the divergent views currently held with regard to the structure. Mathematically, and where IB \" *BA \" TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). \u201d or \u201d A quantity expressing the body\u2019s tendency to resist angular acceleration, it is equal to sum of product of mass of particles to the square of distances from the axis of rotation. It is always considered with respect to a reference axis such as X-X or Y-Y. 5 1 A 2 3 2. So if I'm interpreting your last formula correctly, your answer seems to be off by a factor of 2. 2500 cm^4; D. 1 Answer to Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). Then remove the middle triangle from each of the re-maining three triangles (as shown), and so on, forever. In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The following is a list of second moments of area of some shapes. where I is the moment of inertia and R is the perpendicular distance from the axis of rotation to the slant height of the cone changing dm with density, \u03c1, we get I = \u03c1 \u222b\u222b\u222b R^2 r^2 sin \u03a6 dr d\u03a6 d\u03b8 = \u03c1 \u222b\u222b\u222b r^4 (sin \u03a6)^3 dr d\u03a6 d\u03b8. Express the result as a Cartesian vector. The calculations are as shown. \u2022Compute the product of inertia with respect to the xyaxes by dividing the section into three rectangles. Area Moment of Inertia - Filled Right Triangle Solve. 42441 1in 1 in 1 in 3 in 1 in A 2 A 3 A 1 A 4 16 Centroid and Moment of. The moment of inertia of total area A with respect to z axis or pole O is z dI z or dI O or r dA J 2 I z \u00b3r dA 2 The moment of inertia of area A with respect to z axis Since the z axis is perpendicular to the plane of the area and cuts the plane at pole O, the moment of inertia is named \"polar moment of inertia\". however, i would like to know how you obtain the results. Finding the area of a right triangle is easy and fast. Mechanics of Material (CIV101) Anno Accademico. Introduction. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. I am unable to find it. This banner text can have markup. Angular momentum of an extended object. We will use the parallel axis theorem and we will take the centroid as a reference in this case. First Moment of Area = A x. Sorry to see that you are blocking ads on The Engineering ToolBox! If you find this website valuable and appreciate it is open and free for everybody - please contribute by. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection. It is also known as the torsional Stiffness Read the Full article here. Answer MOI of a triangle about axis theory through a point along the plane = 2 1 \u200b m (A r e a) = 2 1 \u200b m (2 l \u200b \u00d7 2 l \u200b) = 8 1 \u200b m l 2 December 26, 2019 Toppr. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). 1) - Moment of Inertia by Integration Mechanics Statics Chapter 10. Own work assumed (based on copyright claims). But I don't know how to do that. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. For the section shown in Fig. For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section The moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis. function Ix_integrand = Moment_Of_Inertia_Integrand(y_prime) %Saved as Moment_Of_Inertia_Integrand. The moment of inertia is equal to the moment of inertia of the rectangle minus the moment of inertia of the hole which is a circle. Answer Save. Calculating the moment of inertia of a triangle - Duration: 10:01. Lab 17: Angular Acceleration Amy, Chris, and Jacob November 22, 2017 Theory/Introduction: The purpose of this lab was to determine the moment of inertia of a right triangle thin plate around its center of mass, for two\u2026. The angle at the apex is 9 0 o. Figure to illustrate the area moment of a triangle at the list of moments of inertia. Find The Moment Of Inertia About An Axis That Passes Through Mass A And Is Perpendicular To The. The moment of inertia of a triangle with respect to an axis passing through its centroid, parallel to its base, is given by the following expression: I=bh^2/36. If you need a beam\u2019s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. 30670 Moment of Inertia The same vertical differential element of area is used. Lab 18: Moment of Inertia of a Triangle In Lab 16, we used a rotary device that pushes air through a pair of disks, minimizing frictional torque, and allowing the disks to spin for a long time, almost unimpeded. Home Properties Classical MechanicsMoment of Inertia of a Triangle. Let the mass of the triangle be M. It is concluded that the form of the isoceles triangle is acute, with the. Get the expression of angular acceleration and omega. 000965387 kg*m^2. he solves alone. We can use a numerical integrator, such as MATLAB's integral2, to compute the area moment of inertia in the previous example. The moment of inertia of a triangle rotating on its long side is greater than the moment of inertia of the triangle rotating on the shorter side. The element of area in rectangular coordinate system is given by. If rotated about point O (AO = OB),what is the moment of inertia of the rod. The moment of inertia of a triangle rotating on its long side is greater than the moment of inertia of the triangle rotating on the shorter side. 5 Moment of Inertia of Composite Areas A similar theorem can be used with the polar moment of inertia. In a continuous bridge, the moment of inertia should follow the moment requirement for a balanced and economical design. Autor: No machine-readable author provided. Diagonal wise mI = ml2/6 base = ml2/24 Find the moment of inertia of the plane about the y-axis. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. The strut width has been deliberatelty increased to show the geometry. Integration by the area of. For a solid cone the moment of inertia is found by using the given formula; I = 3 / 10 MR 2. This is the currently selected item. Calculate the triangles moment of inertia when its axis of rotation is located at the right-angled corner. Radius and elevation of the semi-circle can be changed with the blue point. The particles are connected by rods of negligible mass. 5 Parallel-Axis Theorem - Theory - Example - Question 1 - Question 2. It is observed that the ratio of to is equal to 3: Assume that both balls are pointlike; that is, neither has any moment of inertia about its own center of mass. Moment of inertia is defined as:\u201dThe sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. A more efficient triangular shape for metal wood clubs or driver clubs is disclosed. Calculate the moment of inertia of the triangle with respect to the x axis. Asked by rrpapatel 2nd November 2018 12:10 AM. Polar Moment of Inertia is a measure of resistibility of a shaft against the twisting. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. Mechanics of Solids Introduction: Scalar and vector quantities, Composition and resolution of vectors, System of units, Definition of space, time, particle, rigid body, force. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. No, the components of the eigenvectors themselves, Axes(:,1)), Axes(:,2), Axes(:,3), are already the cosines of the angles between the three principal axes of inertia respectively and the x, y, and z axes, provided the eigenvectors are normalized. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. The apex of the triangle is at the origin and it is bisected by the x-axis. A triangular section has base 100 mm and 300 mm height determine moment of inertia about 1)MI about axis passing through base 2)MI about axis passing through apex {Ans: 3. And h/3 vertically from reference x-axis or from extreme bottom horizontal line line. Simply Supported Beams (Shear & Moment Diagrams) Simply supported beams (also know as pinned-pinned or pinned-roller) are the most common beams for both school and on the Professional Engineers exam. This triangular shape allows the clubs to have higher rotational moments of inertia in both the vertical and horizontal directions, and a lower center of gravity. Rotations in 2D are about the axis perpendicular to the 2D plane, i. Mass moment of inertia. My teacher told me :. Moment of Inertia - Calculated Values Electrical Design In determining the layout of the electrical design, a broad level view was taken and elaborated on. Supplementary notes for Math 253, to follow Section 13. purdueMET 20,366 views. They will make you \u2665 Physics. If the density were a constant, finding the total mass of the lamina would be easy: we would just multiply the density by the area. Moment of Inertia for body about an axis Say O-O is defined as \u2211dM*y n 2. Integration by the area of. For this reason current vector is treated as normal vector of the plane and the input cloud is projected onto it. About the Moment of Inertia Calculator. Engineering Science Mechanical Engineering Concrete Calculator Bending Moment Similar Triangles Shear Force Civil Engineering Construction Body Diagram Structural Analysis. It is the rotational inertia of the body, which is called. These are the values of principal moment of inertia. More particularly, the present invention relates to a hollow golf club head with a lower center of gravity and a higher moment of inertia. I y 2= \u222b x el dA where el = x dA = y dx Thus, I y = \u222b x2 y dx The sign ( + or - ) for the moment of inertia is determined based on the area. It is not explicitly stated in the output, but the mass is equal to the volume (implicitly using a density of 1), so we would expect diagonal matrix entries of 8/15*PI (1. Own work assumed (based on copyright claims). 156 m y Applying Eq. And the moment of \"A\" equals zero because it is at a point through which the moment of inertia passes. Lectures by Walter Lewin. More on moment of inertia. Ask Question Asked 4 years, 9 months ago. apex angle in the neighborhood of 34\u00b0. After that eccentricity is calculated for the obtained projection. We\u2019re pretty sure the Titleist 620 MB has plenty of workability. 8) I of Disk with a Hole. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). 1 Expert Answer(s) - 30625 - calculate the moment of inertia of an equilateral triangle made by three rods each of mass m and len. moment of inertia gives the same I as the body rotates around the axis. The 2 nd moment of area, or second area moment and also known as the area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Weld design. Undeniable momentum, on any stage - anywhere. where m is the mass of the object, and r is the distance from the object to the axis. Microsoft Word - Chapter 12 - Moment of Inertia of an Equilateral Triangle Author: Owner Created Date: 11/21/2019 8:18:19 AM. triangles parallel to the xy plane with side 2a, is centered on the origin with its axis along the z axis. 2 y = 10e-x x y 3. Code for moment_of_inertia and RoPS classes, tests and tutorials. The term product moment of inertia is defined and the mehtod of finding principal moment of inertia is presented. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. Moments of Inertia 10. The radius of gyration is the radius at which you could concentrate the entire mass to make the moment of inertia equal to the actual moment of inertia. (**) The object below has a moment of inertia about its center of mass of I=25kg\u22c5m2. 32075h^4M/AL, where h is the height of the triangle and L is the area. This cone is centered on the z-axis with the apex at the origin, but rotates with respect to the x-axis. Using these moment of inertia, we can subtract from it the moment of inertia of just the system without the triangle to obtain our experimental values for the triangle in either. After determining moment of each area about reference axis, the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. 1 GradedProblems Problem 1 (1. Moment of Inertia Contents Moment of Inertia; Sections; Solids; MOI_Rectangle; MOI_Triangle; MOI_Trapezod; MOI_Circle. Solution 3. Area Moment of Inertia Section Properties: Triangle Calculator. Transfer Formula for Moment of Inertia Transfer Formula for Polar Moment of Inertia Transfer Formula for Radii of Gyration Moment of Inertia Common Shapes; Rectangle Triangle Circle Semicircle Quartercircle Ellipse Center of Mass; Center of Mass (2D) 1. The Moment of Inertia Apparatus MATERIALS 1 Table clamp 1 Weight hanger (mass 50g) 1 Long metal rod 1 Length of string 2 Pulleys 1 Level 2 Right angle clamps 1. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Angular momentum of an extended object. Ditto the Ping Blueprint. 4 Locate the centroid of the T-section shown in the Fig. Thank you User-12527562540311671895 for A2A The moment of inertia of a triangular lamina with respect to an axis passing through its centroid, parallel to its base, is given by the expression [math]I_{XX}=\\frac{1}{36}bh^3$ where [math]b[/mat. pptx), PDF File (. Try this Drag any point A,B,C. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord\u2019s mass. The point where the triangle is right angled is lying at origin. Doing the same procedure like above, and below is the work. 3 Radius of Gyration of an Area 10. Another solution is to integrate the triangle from an apex to the base using the double integral of r^2dm, which becomes (x^2+y^2)dxdy. Click Content tabCalculation panelMoment of Inertia. This engineering calculator will determine the section modulus for the given cross-section. 3\u00d7 1 6ML2 = 1 2ML2. The situation is this: I know the moment of inertia with respect to the x axis and with respect to the centroidal x axis because its in the table. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. These came out to be 0. The ratio of moment of inertia about the neutral axis to the distance of the most distant point of section from neutral axis is called as a) Moment of inertia b) section modulus c) polar moment of Apex of the triangle b) mid of the height c) 1/3 of the height d) base of triangle 12. The axis perpendicular to its base. Thus, the moment of inertia of a 2D shape is the moment of inertia of the shape about the Z-axis passing through the origin. I of a thin rod about its center is ML^2 / 2. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. c channel polar moment of inertia. 30670 Moment of Inertia The same vertical differential element of area is used. It should not be confused with the second moment of area, which is used in beam calculations. In order to find the moment of inertia we must use create a dm and take the integral of the moment of inertia of each small dm. Let us apply this formula to the triangle formed by V 1, V 2 and V: \u0394\ud835\udc49\u2248\ud835\udc49\u22c5\u0394\ud835\udefc (3). How do I calculate the moment of inertia of a right angled triangle about one side? Moment of inertia about a side other than the hypotenuse. What is the moment of inertia of this system about an altitude of the triangle passing through the vertex, if \u2018a\u2019 is the size of each side of the triangle ?. Letting M be the total mass of the system, we have x \u00af = M y / M. The moment of point \"C\" is the same as \"B\" so multiply the moment of \"B\" by two. derivation of inertia of ellipse. Find the moment of inertia of the framework about an axis passing through A, and parallel to BC 5ma 2. Favourite answer. \u2022 The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A 1 , A 2 , A 3 , , with respect to the same axis. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. 89 \u00d7 103 kg/m3. In yesterday's lesson, students completed a lab on center of mass, and they already have a working knowledge of torque. The axis may be internal or external and may or may not be fixed. Engineering Science. Find the moment of inertia about y-axis for solid enclosed by z = (1-x^2) , z= 0 , y = 1 and y = -1. 3) Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. equals 1, if there are four piles per row and two rows (figure 7-2), the moment of inertia about the Y-Y axis is given by the following formula. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. I), must be found indirectly. Moment of inertia Up: Rotational motion Previous: The vector product Centre of mass The centre of mass--or centre of gravity--of an extended object is defined in much the same manner as we earlier defined the centre of mass of a set of mutually interacting point mass objects--see Sect. The moment of inertia of the particle. Moment of inertia, in physics, quantitative measure of the rotational inertia of a body\u2014i. Therefore, equation for polar moment of inertia with respect to apex is. University of Sheffield. For these orbits, consider the following scaled variables q\u02dci = \u221aqi I, (5) v\u02dci. Q: the moment of inertia of a thin rod of mass m and length l about an axis through its centre of gravity and perpendicular to its length is a) ml\u00b2/4 b) ml\u00b2/6 c) ml\u00b2/8 d) ml\u00b2/12 Q: Which statement is correct: a) Moment of inertia is the second moment of mass or area. Tinker toys allow one to easily construct objects with the same mass but different moments of inertia. Two conditions may be considered. After determining moment of each area about reference axis, the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. You can show the division by drawing solid or. It is always considered with respect to a reference axis such as X-X or Y-Y.\nhu215i4c45d 16uebhoaltj0op 71a59wpjnvb5y i0fcgwor6jyv8b a8qpv94ljutw hl8lh0m5b2gz1 pbgmxxx4nx 8a1tcmejeokz gx9mvyz9ezgel jwp1rrde9h6ax2 19sb1sh6t9mn 9etgudw0esw 0qoq538i9x xnzy3novvwd u573jcqhjrwnqg 7yteh42rbyb3j j33ao2f1gbcc 0vmioeu4j0gtvum 6q7lhaq2fv l2455p6pm40a7mo jzze49hb2dgxab wwskcnhe8iiz z480xmmqhdqb7 8pfvwqe2t6dtb0 9ciz5c26pn 1atmcvapjuz lpb65ot49n e7ru916n05wcb21 ntgoy7fzw8", "date": "2020-12-03 19:51:56", "meta": {"domain": "carlacasciari.it", "url": "http://carlacasciari.it/moment-of-inertia-of-triangle-about-apex.html", "openwebmath_score": 0.6235755681991577, "openwebmath_perplexity": 395.8218896476916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401449874105, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8654589603563371}}
{"url": "http://civilservicereview.com/2013/10/solutions-fractions-to-lowest-terms/", "text": "# Solution to the Exercises on Reducing Fractions to Lowest Terms\n\nBelow are the complete solutions and answers to the exercises on reducing fractions to lowest terms. I will not give any tips or methods of shortcuts on doing this because teaching you shortcuts will give you problems in case you forget them. The best thing that you can do is to solve as many related problems as you can and develop shortcuts that work for you. Each person has his own preference in solving procedural problems such as these, so it is important that you discover what\u2019s best for you.\n\nFor converting improper fractions to mixed form, I will discuss it in a separate post. Try to see the solutions below and see if you can use these solutions to develop your own method. Honestly, the three examples below on converting improper fractions to mixed form should be enough to teach you how to do it yourself. \ud83d\ude42\n\nReducing Fractions to Lowest Terms\n\n1. $\\displaystyle \\frac{12}{15}$\n\nSolution\n\n$\\displaystyle \\frac{12 \\div 3}{15 \\div 3} = \\frac{4}{5}$\n\n2. $\\displaystyle \\frac{18}{24}$\n\nSolution\n\n$\\displaystyle \\frac{18 \\div 6 }{24 \\div 6} = \\frac{3}{4}$\n\n3. $\\displaystyle \\frac{21}{49}$\n\nSolution\n\n$\\displaystyle \\frac{21 \\div 7 }{49 \\div 7} = \\frac{3}{7}$\n\n4. $\\displaystyle \\frac{56}{72}$\n\nSolution\n\n$\\displaystyle \\frac{56 \\div 8 }{72 \\div 8} = \\frac{7}{9}$\n\n5. $\\displaystyle \\frac{26}{65}$\n\nSolution\n\n$\\displaystyle \\frac{26 \\div 13 }{65 \\div 13} = \\frac{2}{5}$\n\n6. $\\displaystyle \\frac{18}{32}$\n\nSolution\n\n$\\displaystyle \\frac{18 \\div 2 }{32 \\div 2} = \\frac{9}{16}$\n\n7. $\\displaystyle \\frac{38}{95}$\n\nSolution\n\n$\\displaystyle \\frac{38 \\div 19 }{95 \\div 19} = \\frac{2}{5}$\n\n8. $\\displaystyle \\frac{32}{12}$\n\nSolution\n\nFirst, convert to lowest terms:\n$\\displaystyle \\frac{32 \\div 4 }{12 \\div 4} = \\frac{8}{3}$\n\nSecond, convert to mixed form. Eight divided by 3 is 2 remainder 3. So 2 becomes the whole number, 2 (the remainder) becomes the numerator and 8 becomes the denominator. Therefore, the answer is $2 \\frac{2}{3}$.\n\n9. $\\displaystyle \\frac{16}{84}$\n\nSolution\n\n$\\displaystyle \\frac{16 \\div 4 }{84 \\div 4} = \\frac{4}{21}$\n\n10. $\\displaystyle \\frac{39}{24}$\n\nSolution\n\nFirst, reduce to lowest terms.\n\n$\\displaystyle \\frac{39 \\div 3 }{24 \\div 3} = \\frac{13}{8}$\n\nSecond, convert the answer to mixed form. Thirteen divided by 8 is 1 remainder 5. So 1 becomes the whole number, 5 (the remainder) becomes the numerator of the fraction and 8 becomes the denominator. So the correct answer is $1 \\frac{5}{8}$.\n\n11. $\\displaystyle \\frac{15}{45}$\n\nSolution\n\n$\\displaystyle \\frac{15 \\div 15 }{45 \\div 15} = \\frac{1}{3}$.\n\n12. $\\displaystyle \\frac{51}{85}$\n\nSolution\n\n$\\displaystyle \\frac{51 \\div 17 }{85 \\div 17} = \\frac{3}{5}$\n\n13. $\\displaystyle \\frac{18}{54}$\n\nSolution\n\n$\\displaystyle \\frac{18 \\div 18 }{54 \\div 18} = \\frac{1}{3}$\n\n14. $\\displaystyle \\frac{35}{49}$\n\nSolution\n\n$\\displaystyle \\frac{35 \\div 7}{49 \\div 7} = \\frac{5}{7}$\n\n15. $\\displaystyle \\frac{74}{24}$\n\nSolution\n\nFirst, reduce to lowest terms.\n\n$\\displaystyle \\frac{74 \\div 2 }{24 \\div 2} = \\frac{37}{12}$\n\nSecond, divide 37 by 12. The answer is 3 remainder 1. Now, 3 becomes the whole number, 1 becomes the numerator of the fraction, and 12 becomes the denominator. So, the correct answer is $3 \\frac{1}{12}$.\n\nIn the next post, we will be talking about multiplying and dividing fractions.\n\n### 6 Responses\n\n1. rooky says:\n\nSir, when I solved # 2 I got an answer \u00bc lowest term of 18/24 correct me if I am wrong because I had two LCM. 1st LCM (2) 18/24 = 9/12; 2nd LCM (3) 9/12 = 1/4\n\n2. Civil Service Reviewer says:\n\nHi Rooky,\n\nThank you for your question. In reducing fractions to lowest terms, you are not getting the LCM. You are actually getting the greatest common factor (GCF). You get the LCM if you want to add or subtract two (or more) fractions whose denominators are not equal.\n\nIn your solution, the GCF of 9 and 12 is 3. So, you must divide 9 by 3, which is equal to 3.\n\n3. Supercute says:\n\nSir, i think no. 15 should be 3 1/6\n\n4. Supercute says:\n\nHere\u2019s my computation:\n\n74/24\n\n3\n_____\n24 / 74\n72\n___________\nr. 4\n\nConvert it to mixed fraction:\n\n4\n3 ____\n24\n\nConvert to lowest term:\n\n1\n3 ___\n6\n\n\u2022 Civil Service Reviewer says:\n\nHello supercute,\n\nThe answer in the explanation is correct. Even if you convert it first to fraction, 74/24 becomes 3 and 2/24 or 3 and 1/12. \ud83d\ude42\n\n1. October 12, 2013\n\n[\u2026] Now that you know how to convert fractions to lowest terms, you might want to try the\u00a0practice test\u00a0and check your solution and answer. [\u2026]", "date": "2017-11-20 21:13:45", "meta": {"domain": "civilservicereview.com", "url": "http://civilservicereview.com/2013/10/solutions-fractions-to-lowest-terms/", "openwebmath_score": 0.7920023202896118, "openwebmath_perplexity": 1053.1252693989509, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401449874105, "lm_q2_score": 0.8740772335247531, "lm_q1q2_score": 0.8654589587323938}}
{"url": "https://math.stackexchange.com/questions/866430/does-this-series-violate-the-decreasing-condition-of-the-integral-test-for-conve", "text": "# Does this series violate the decreasing condition of the Integral Test for Convergence?\n\nI'm working on the section involving the Integral Test for Convergence in my calculus II class right now, and I've run into a seeming conflict between the definition of the Integral Test, and the solutions to some of the homework exercises as given by both my professor and the textbook.\n\nAccording to the definition, my research, and my understanding of the integral test, the integral test can only be used for the series $a_n$ where $a_n = f(n)$ and $f(x)$is positive, continuous and decreasing for all $x \\ge N$, where $N$ is the index of $n$. However, there are several problems where $f(x)$ is only decreasing if we add a condition, such as $f'(x) < 0 \\Leftarrow\\Rightarrow x > 3$, where $N \\lt 3$. It seems to me that the Integral Test cannot be used to determine convergence of a series when the function is only decreasing when $x \\gt k$ and $k \\lt N$, yet the book and my professor apply the test anyway.\n\nFor example, with the series:\n\n$$\\sum_{n=1}^\\infty = \\frac{n}{(4n+5)^\\frac{3}{2}}$$\n\nIf we let $a_n = f(n)$, then for $f(x)$:\n\n\u2022 $f(x) \\gt 0$ for all $x$ in the domain\n\u2022 $f(x)$ is continuous for all $x \\gt -\\frac{5}{4}$\n\nBut, $f'(x) \\lt 0$ only when $x > \\frac{5}{2}$, as shown when testing the critical points with the derivative:\n\n$$f'(x) = \\frac{5-2x}{(4x+5)^\\frac{5}{2}}$$\n\nThe professor notes this in her solution, but instead of ending with that and writing, \"The Integral Test cannot be applied because $f(x)$ fails to satisfy the required conditions,\" she applies the test using the original index for $n$:\n\n$$\\int_1^\\infty \\frac{x}{(4x+5)^\\frac{3}{2}} \\rightarrow \\infty \\Rightarrow a_n \\text{ Diverges}$$\n\nThe textbook reaches the same conclusion. Also, the problems in question are listed under a section where the instructions state, \"Confirm that the Integral test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series,\" implying the test can be used on the subsequent exercises.\n\nIs there a reason the Integral Test for Convergence can be used to test for convergence in these problems, where $N \\lt k$ and $f'(x) < 0 \\Leftarrow\\Rightarrow x \\gt k$? Am I missing something, or are the book and my professor wrongly using the Integral Test for these series?\n\n### Other exercises with same result:\n\n$$\\bullet \\sum_{n=1}^\\infty \\frac{\\ln n}{n^2}$$\n\n$$\\bullet \\frac{\\ln 2}{2} + \\frac{\\ln 3}{3} + \\frac{\\ln 4}{4} + \\frac{\\ln 5}{5} + \\frac{\\ln 6}{6}$$\n\n\u2022 One way to look at what your book and your professor are doing is that they're using the fact that $\\sum_{n=1}^{\\infty}a_n$ converges iff $\\sum_{n=k}^{\\infty}a_n$ converges, for any $k$. \u2013\u00a0user84413 Jul 14 '14 at 0:20\n\u2022 I think the idea is more that we don't really care what happens for the first 10, 100, or 1000 terms. We care about what eventually happens. We can add up a finite amount of things no problem; it's the tail end behavior that we really care about. \u2013\u00a0Cameron Williams Jul 14 '14 at 0:20\n\u2022 For convergence, doesn't matter. For estimates it does. \u2013\u00a0Andr\u00e9 Nicolas Jul 14 '14 at 0:22\n\u2022 Thank you for your responses; I agree that the behavior is properly demonstrated with the test, and I also understand our concern being only with the end behavior of the function, but I don't think it explains why the test is defined as having to be decreasing for all $x \\ge N$, if we only care about the last few terms. Wouldn't it be better to say \"for most of\"? At any rate, if my definition of the test is correct, doesn't this series still violate the conditions, regardless of the validity of the result? \u2013\u00a0MrCMedlin Jul 14 '14 at 0:34\n\nThe point, which isn't made often enough (in my opinion) in classes on the subject, is that the convergence of a series is a limit process. In this case, what that means is that the question of convergence is completely determined by the behavior of the series for say $n > N$ for any fixed, finite $N$. If I take a convergent series $\\sum a_n$, and I cut off the first 55 quintillion terms, and replace them all with $n!$ to get a new sequence\n\n$$b_n = \\begin{cases} a_n \\text{ if } n > 55,000,000,000,000,000,000 \\\\ n! \\text{ if } n \\leq 55,000,000,000,000,000,000 \\end{cases}$$\n\nThen the sum $\\sum b_n$ is still convergent. In fact, $b_n$ is convergent if and only if $a_n$ is. Of course, the sums will be different, but by precisely\n\n$$\\sum\\limits_{n=1}^{\\infty} b_n - \\sum\\limits_{n=1}^{\\infty}a_n = \\sum\\limits_{n=1}^{55,000,000,000,000,000,000} (n! - a_n)$$\n\nWhich is just a finite number. Of course, I've contrived the example to be huge and ridiculous. But the the point to be made about the integral test is that as long as the function behaves in the desired way beyond some large $N$ (say 55 quintillion), the argument still works for the infinite part of the sum, beyond that, and that is where all problems of convergence lie. Everything else is just addition.\n\n\u2022 Doesn't this mean the integral should be evaluated from $N$ to $\\infty$ instead of from the original index $k$ to $\\infty$ for it to be accurate? \u2013\u00a0MrCMedlin Jul 14 '14 at 15:45\n\u2022 Again, as long as the integral is of a nice (read continuous) function on the interval $[0, \\infty)$. The convergence of the integral is focused in the tail as we can write $$\\int_0^\\infty f dx = \\int_0^N f dx + \\int_N^\\infty f dx$$ And all of the convergence/divergence behavior of the improper integral is concentrated in the second term on the right. The first term in the right is (assuming the function is continuous up to and including 0), simply some finite integral. This is admittedly a little more subtle. \u2013\u00a0JHance Jul 14 '14 at 16:18\n\u2022 OK, this makes a lot more sense now. They just gloss over this in the course without really explaining it ... you've helped a bunch! \u2013\u00a0MrCMedlin Jul 14 '14 at 16:36\n\nI do not believe your definition of the test is correct. Citing the ever-accurate (tongue-in-cheek) Wikipedia, we find the definition to be:\n\nConsider an integer $N$ and a non-negative function $f$ defined on the unbounded interval $[N, \u221e)$, on which it is monotone decreasing. Then the infinite series $$\\sum_{n=N}^\\infty f(n)$$ converges to a real number if and only if the improper integral $$\\int_N^\\infty f(x)\\,dx$$ is finite. In other words, if the integral diverges, then the series diverges as well.\n\n(Source: http://en.wikipedia.org/wiki/Integral_test_for_convergence)\n\nIt is important to note that various authors of textbooks may define a test or theorem in the way that it best fits their course outline. This is particularly true of elementary texts (e.g. first or second year calculus). Remember that mathematicians also like to generalize--if you can take a specific formulation of a theorem and generalize it (e.g. \"$f$ must be decreasing everywhere\" to \"$f$ only has to have decreasing end-behavior\"), that's perfectly acceptable.\n\nAs a further note in response to your comment on the main question: \"decreasing a majority of the time\" is different than \"decreasing beyond $x=N$.\" The latter is the option you want.\n\n\u2022 Thank you for your response! I've reviewed the Wikipedia entry on the topic as part of my earlier research. You'll note at the beginning it says the function must be monotone decreasing on the interval being considered. This means that for the series $a_n$, $a_1 \\ge a_2 \\ge a_3 \\ge a_4 \\ge ... \\ge a_n$. We can test for this by looking at the behavior of the first derivative of the function. If the first derivative is ever positive for any value $x$ in our domain, this indicates an increase from one term to the next, which would mean the function is not monotone decreasing as required. \u2013\u00a0MrCMedlin Jul 14 '14 at 4:08", "date": "2019-10-22 00:44:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/866430/does-this-series-violate-the-decreasing-condition-of-the-integral-test-for-conve", "openwebmath_score": 0.8867442011833191, "openwebmath_perplexity": 162.8533631413315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575116884779, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8654337960010898}}
{"url": "https://mathematica.stackexchange.com/questions/95566/determine-height-of-box-packed-with-spheres/95568", "text": "# Determine height of box packed with spheres\n\nI got such a wonderful answer regarding The Diagonals of a Regular Octagon, so I thought I'd try asking another question we had on our Pizza and Problem quiz activity at College of the Redwoods. The question was:\n\nA 4 \u00d7 4 \u00d7 h rectangular box contains a sphere of radius 2 and eight smaller spheres of radius 1. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is h?\n\nI'd like to learn how to construct such a model using Mathematica and determine the height h of the box.\n\nThanks for any help.\n\nYour question is like a sangaku problem. What is the distance between the centre of the large sphere and one of the smaller spheres? The large sphere centre is $\\{2,2,h/2\\}$, one of the small spheres is at $\\{1,1,1\\}$. Their separation equals the sum of their radii. That is,\n\nNorm[{2, 2, h/2} - {1, 1, 1}] == 1 + 2\n\n\nSolve the expression for $h$, and choose the positive root, $h=2(1+\\sqrt{7})$.\n\nWith[{h = 2 (1 + Sqrt[7])},\nGraphics3D[{Orange,\nSphere[{1, 1, 1}, 1], Sphere[{3, 1, 1}, 1], Sphere[{1, 3, 1}, 1],\nSphere[{3, 3, 1}, 1], Sphere[{1, 1, h - 1}, 1],\nSphere[{3, 1, h - 1}, 1],\nSphere[{1, 3, h - 1}, 1], Sphere[{3, 3, h - 1}, 1],\nRed, Sphere[{2, 2, h/2}, 2]\n}, Lighting -> \"Neutral\",Axes->True]]\n\n\n\u2022 Tremendous answer. I am having so much fun this morning with the responses on Mathematica Stack Exchange. Thanks you. \u2013\u00a0David Sep 26 '15 at 19:09\n\nKennyColnago's answer is good, but we're using Mathematica, so we can leave much more of the problem to it. Now, we can't avoid writing down the system of equations; for simplicity, I'll assume that the large sphere (with radius $R = 2$) is centered at the origin, and that the small spheres (with radius $r = 1$) are centered at ${x, y, z}$. The sides of the box lie at $\\pm2$, and the top and bottom are at $\\pm h$, so we have tangent small spheres for any combination of positive and negative $x$, $y$ and $z$. That means we know the following:\n\nIn[1]:= eqns = {\n(* each sphere is tangent to the sides of the box *)\nAbs[x] + r == 2,\nAbs[y] + r == 2,\n\n(* each sphere is tangent to the top of the box *)\nAbs[z] + r == h/2,\n\n(* because the large sphere and small sphere are tangent\nto one another, their radii are collinear, so their\ntotal length is the distance of the center of the small\nsphere from the origin. *)\nr + R == Norm[{x, y, z}],\n\n(* h is positive *)\n0 < h\n} /. {r -> 1, R -> 2};\n\n\nNow, we can use Mathematica to solve those equations for $x$, $z$ and $h$.\n\nIn[2]:= sol = Simplify@Solve[eqns, {x, y, z, h}, Reals]\nOut[2]= {{x -> -1, y -> -1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},\n{x -> -1, y -> -1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},\n{x -> -1, y -> 1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},\n{x -> -1, y -> 1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},\n{x -> 1, y -> -1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},\n{x -> 1, y -> -1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},\n{x -> 1, y -> 1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},\n{x -> 1, y -> 1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])}}\n\nIn[3]:= Length@sol\nOut[3]= 8\n\n\nGratifyingly, we have eight solutions, one for each small sphere.\n\nNow we can draw our spheres, which is easy because Mathematica has helpfully collected all the solutions as a list of rules.\n\nIn[3]:= centers = {x, y, z} /. sol\nOut[3]= {{-1, -1, -Sqrt[7]}, {-1, -1, Sqrt[7]},\n{-1, 1, -Sqrt[7]}, {-1, 1, Sqrt[7]},\n{1, -1, -Sqrt[7]}, {1, -1, Sqrt[7]},\n{1, 1, -Sqrt[7]}, {1, 1, Sqrt[7]}}\n\n\nPulling it all together, we get:\n\nIn[4]:= Graphics3D[{\n{Lighter@Lighter@Red,\nSphere[centers, r]},\n\n{Lighter@Green,\nSphere[{0, 0, 0}, R]},\n\nOpacity[0],\nEdgeForm[Thick],\nParallelepiped[{-2, -2, -h/2},\n{{4, 0, 0}, {0, 4, 0}, {0, 0, h}}]} /. sol /. {r -> 1, R -> 2},\n\nBoxed -> False]\n\n\nEDIT to add: I've updated this answer a couple times, each time to shift more work to Mathematica. In my first attempt, which I didn't post, I got the wrong answer because of a stupid typo when I plugged the radii of the spheres into the Pythagorean theorem; switching over to Norm fixed that. Then I told Solve to assume that $x$ and $z$ were positive (while using the \"obvious\" fact that $|y| = |x|$ to eliminate an equation before solving), and did some complicated thing with Tuples and Transpose to get all eight spheres. Eliminating those assumptions and recasting everything in terms of absolute values meant Solve took care of all of that automatically.\n\nEDIT again to tweak the Solve expression to specify the domain for Reals and include y, which seems necessary for everything to work right in Mathematica v11.\n\n\u2022 @Pilsy Another tremendous answer. I am sharing these with the students and teachers who attended our activity. \u2013\u00a0David Sep 26 '15 at 19:10", "date": "2019-11-21 18:21:56", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/95566/determine-height-of-box-packed-with-spheres/95568", "openwebmath_score": 0.3801577389240265, "openwebmath_perplexity": 919.4013234807036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575111777183, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.865433787865029}}
{"url": "https://mathhelpboards.com/threads/why-e-is-natural-and-why-we-use-radians.9286/", "text": "# Why e is \"natural\", and why we use radians.\n\n#### ThePerfectHacker\n\n##### Well-known member\nMany books give bad answers or no answers at all to why we work with base $e$ and measure angles according to radians in calculus.\n\nHere is what I tell my students, as far as I know, this is the best explanation I seen because it is essentially a one line explanation that is short and to the point.\n\nWhy $e$ is natural: Let $b>0$, $b\\not = 1$ and consider the function $f(x) = b^x$. This is an exponential function with base $b$ and we can show that $f'(x) = c\\cdot b^x$ where $c$ is some constant. In a similar manner we can consider the function $g(x) = \\log_b x$, logarithmic function to base $b$, we can show that $g'(x) = k\\cdot x^{-1}$ where $k$ is some constant. It would be best and simplest looking derivative formula if those constants, $b$ and $k$, were both equal to one. This happens exactly when $b=e$. Thus, $e$ is natural base for exponent because the derivative formulas for exponential and logarithmic function are as simple as they can be.\n\nWhy radians are natural: A \"degree\" is defined by declaring a full revolution to be $360^{\\circ}$, a \"radian\" is defined by declaring a full revolution to be $2\\pi$ radians. More generally, we can define a new angle measurement by declaring a full revolution to be $R$ (for instance, $R=400$, results in something known as gradians). Let us denote $\\sin_R x$ and $\\cos_R x$ to be the sine and cosine functions of $x$ measured along $R$-units of angles. It can be shown that $(\\sin_R x)' = C\\cdot(\\cos_R x)$ and $(\\cos_R x)' = -C\\cdot(\\sin_R x)$ where $C$ is some constant. It would be best if this constant can be made $C=1$ which happens precisely when $R=2\\pi$.\n\n#### chisigma\n\n##### Well-known member\nMany books give bad answers or no answers at all to why we work with base $e$ and measure angles according to radians in calculus.\n\nHere is what I tell my students, as far as I know, this is the best explanation I seen because it is essentially a one line explanation that is short and to the point.\n\nWhy $e$ is natural: Let $b>0$, $b\\not = 1$ and consider the function $f(x) = b^x$. This is an exponential function with base $b$ and we can show that $f'(x) = c\\cdot b^x$ where $c$ is some constant. In a similar manner we can consider the function $g(x) = \\log_b x$, logarithmic function to base $b$, we can show that $g'(x) = k\\cdot x^{-1}$ where $k$ is some constant. It would be best and simplest looking derivative formula if those constants, $b$ and $k$, were both equal to one. This happens exactly when $b=e$. Thus, $e$ is natural base for exponent because the derivative formulas for exponential and logarithmic function are as simple as they can be.\n\nWhy radians are natural: A \"degree\" is defined by declaring a full revolution to be $360^{\\circ}$, a \"radian\" is defined by declaring a full revolution to be $2\\pi$ radians. More generally, we can define a new angle measurement by declaring a full revolution to be $R$ (for instance, $R=400$, results in something known as gradians). Let us denote $\\sin_R x$ and $\\cos_R x$ to be the sine and cosine functions of $x$ measured along $R$-units of angles. It can be shown that $(\\sin_R x)' = C\\cdot(\\cos_R x)$ and $(\\cos_R x)' = -C\\cdot(\\sin_R x)$ where $C$ is some constant. It would be best if this constant can be made $C=1$ which happens precisely when $R=2\\pi$.\nLet's suppose to use 'non natural' bases for exponents and angles... in this case probably today the following 'beautiful formula' ...\n\n$\\displaystyle e^{i\\ \\theta} = \\cos \\theta + i\\ \\sin \\theta\\ (1)$\n\n... would be not yet known ...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### ThePerfectHacker\n\n##### Well-known member\n$\\displaystyle e^{i\\ \\theta} = \\cos \\theta + i\\ \\sin \\theta\\ (1)$\n\n=\n\n$$\\pi^{i\\theta} = \\cos \\theta + i\\sin \\theta$$\nProvided that we measure angles with full revolution $2\\pi \\log \\pi$.\n\nSo you can still have Euler formula for different bases and angles provided that they match together nicely. But the derivative formulas will change into something ugly.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nForgive me for being obtuse but your statement:\n\n...consider the function $f(x) = b^x$. This is an exponential function with base $b$...\nis completely mystifying to me.\n\nIn particular, I mean: how does one define such a function for, say: $b = \\sqrt{2}$?\n\n(I know the answer, someone who is taking calculus for the first time probably does NOT.)\n\nIn fact, proving the \"base laws\" for the functions $\\log_b$ and $\\exp_b$ depends on somehow producing the \"natural base\", which depends on producing the number $e$. So how do you do this?\n\nIf, in fact, you define:\n\n$$\\displaystyle e = x \\in \\Bbb R: \\int_1^x \\frac{1}{t}\\ dt = 1$$\n\nand show that for:\n\n$$\\displaystyle F(x) = \\int_1^x \\frac{1}{t}\\ dt$$\n\n$F(a) + F(b) = F(ab)$\n\nI am prepared to believe that (given a proof that $F$ is injective):\n\n$F^{-1}(a+b) = F^{-1}(a)F^{-1}(b)$\n\nand that such a function $F^{-1}$ has all the properties of $b^x$ when $x$ is rational.\n\nIt follows (in my mind, anyway) that we have:\n\n$e = F^{-1}(1)$.\n\nThis, to me, is the reason why $e$ is \"natural\", I find the definition:\n\n$$\\displaystyle e = \\lim_{n \\to \\infty} (1 + n)^{1/n}$$\n\nnearly impossible to motivate, whereas the derivative of a differentiable function $f$ where:\n\n$f(a+b) = f(a)f(b)$ clearly has derivative:\n\n$f'(x) = f'(0)\\cdot f(x)$.\n\nAt this point, we are in a bit of a fix, which is why considering $f^{-1}$ proves to be more amenable to attack:\n\n$(f^{-1})'(x) = \\dfrac{1}{f'(f^{-1}(x))} = \\dfrac{1}{f'(0)\\cdot f(f^{-1}(x))}$\n\n$= \\dfrac{1}{f'(0)x}$\n\nThis is a function of the form $g(x) = \\dfrac{1}{\\alpha x}$, and we can use the Fundamental Theorem of Calculus to find values for $f^{-1}(x)$.\n\nTaking $\\alpha = 1$ leads, of course, to the above definition of $e$.\n\nLast edited:\n\n#### ThePerfectHacker\n\n##### Well-known member\nForgive me for being obtuse but your statement is completely mystifying to me.\nIt is not mystifying at all. It is just not rigorous. But that is fine. I doubt that Euler himself thought about it the way we think about it today. And what you wrote will have little value to calculus students. You might as well stop teaching trigonometry with that attitude and define sine and cosine as power series, because that is after all the formal way of doing it, which is how Rudin does it in his book on the first pages.\n\nThe important part is to motivate the idea then provide the rigor later. So defining exponentials $b^x$ for arbitrary real numbers is okay because it can at least motivate why $e$ is the best number to choose for $b$.\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nIn particular, I mean: how does one define such a function for, say: $b = \\sqrt{2}$?\nI believe, the most natural definition of $\\left(\\sqrt{2}\\right)^a$ is $\\lim_{x\\to\\sqrt{2}}\\lim_{y\\to a}x^y$ where $x$ and $y$ range over rational numbers.\n\nIn fact, proving the \"base laws\" for the functions $\\log_b$ and $\\exp_b$ depends on somehow producing the \"natural base\", which depends on producing the number $e$.\nIt should be possible to prove $x^{a+b}=x^ax^b$ with the definition above, and this does not require $e$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nIt is not mystifying at all. It is just not rigorous. But that is fine. I doubt that Euler himself thought about it the way we think about it today. And what you wrote will have little value to calculus students. You might as well stop teaching trigonometry with that attitude and define sine and cosine as power series, because that is after all the formal way of doing it, which is how Rudin does it in his book on the first pages.\n\nThe important part is to motivate the idea then provide the rigor later. So defining exponentials $b^x$ for arbitrary real numbers is okay because it can at least motivate why $e$ is the best number to choose for $b$.\nI'm not sure you understand my point. Let me re-phrase it:\n\n\"What does $b^x$ mean for irrational $x$?\"\n\nI do not believe rigor is necessary to teach facts, even if the justification cannot be given at that time. It is common for young children to be told that the area of a circle is pi times its radius squared without any justification given. Given the \"assumption\" that the formula is true, one can still use it profitably, and I see no reason not to under those circumstances.\n\nIn a similar vein, I see no problem with teaching the \"SOHCAHTOA\" approach to trigonometry, either. While that is perfectly adequate for calculating many trig functions of well-established ratios, it is somewhat inadequate for calculating the values of the *continuous* function $\\sin(x)$, and when one acquires the mathematical sophistication for a *better* definition, one ought to employ that instead. Power series approximations will do this, but one can use integrals of algebraic functions to obtain inverse trig functions, which also accomplishes the same purpose. A third, and also perfectly acceptable approach, is to use differential equations to define these transcendental functions.\n\nI am not arguing that one *has* to define them this way, just that one *can* and the tools learned in a first-year calculus course allow this. One need not take this path if one feels that the students \"aren't ready\".\n\nBut back to my main point, which I feel is *somewhat* important: how do you define the function:\n\n$f(x) = b^x$\n\nand how do you justify that it is differentiable?\n\nI honestly don't know your answers, but if I was one of your students, I would feel they are pertinent.\n\nI believe, the most natural definition of $\\left(\\sqrt{2}\\right)^a$ is $\\lim_{x\\to\\sqrt{2}}\\lim_{y\\to a}x^y$ where $x$ and $y$ range over rational numbers.\nI'm OK with this. Computing this (without a computer) might prove problematic. This does address continuity satisfactorily, differentiability is another story.\n\nIt should be possible to prove $x^{a+b}=x^ax^b$ with the definition above, and this does not require $e$.\nI shouldn't think that it would. That is not a \"change of base\" formula. However, it *is* a good place to start. That said, where does $e$ come in?\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nThat said, where does $e$ come in?\nI agree with the OP that $e$ is such that $\\left(e^x\\right)'=e^x$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nI agree with the OP that $e$ is such that $\\left(e^x\\right)'=e^x$.\nI think we all agree this is true. How do you arrive at that fact?\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\n$e$ is defined to be such constat that $\\left(e^x\\right)'=e^x$.\n\nNow, I have not looked into the issue of defining $e^x$ for a long time. This is how I believe it was introduced to me, and if this approach works, I would consider it very natural.\n\n#### ThePerfectHacker\n\n##### Well-known member\n\"What does $b^x$ mean for irrational $x$?\"\nExactly what most people think it means. You replace $x$ by a rational number $q$ which is close to $x$ and then compute $b^q$. That is what every student would do if you ask them to find $2^{\\pi}$. That is what Euler thought of it and mathematicians back then. So I can write $b^x$ function without rigorously defining it.\n\nand how do you justify that it is differentiable?\nI do not. I just write it and then ask students what will the derivative of this function be?\n\nAll that stuff you wrote before it useless to make a first time student who sees the natural exponential to understand what makes it special to other exponential functions. He will get lost in the middle and not understand what you are doing it. My approach does not suffer from that. Now if you tell a student the natural exponential is its own derivative while other exponentials are almost their own derivative they see what makes the natural exponent the nicer base to use.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nLet me see if I have this straight: you differentiate a function without any justification that it is differentiable. Why do I think this is not a good idea?\n\nBut...granted; establishing that the conditions to apply some of the theorems one uses from elementary calculus might be \"more technical stuff\" than a given student may need or want. Let's move on.\n\nHow does one establish that:\n\n$b^x = e^{x\\log(b)}$?\n\nI mean, I'm willing to accept that as a DEFINITION of $b^x$, but then I'd sure like to know what $e$ is.\n\nNow, there is \"that other definition\" of $e$ (which is actually the first definition I ever encountered), but...\n\nAgain, you're asking people to take a lot \"on faith\". There is nothing wrong with this, per se, but calculus affords an opportunity to actually PROVE things that previously they were only TOLD. This moves the source of their knowledge from: \"external authority\" (books, teacher, expert in field) to: \"internal understanding\" (I know it's true because of other things I know are true).\n\nIn one view, math is another form of intellectual slavery. In another, it is intellectual freedom. I suppose it's obvious which one I favor.\n\nLet me put this in another light: it is (or so I am arguing) preferable to have things we reason about be on \"firm footing\" (because arguments based on incorrect assumptions can be invalid, even if the reasoning itself is sound). University students are (by and large) people who have reached the age of majority, and as such are responsible for their own actions. This means (among other things) they will be expected to think for themselves. I feel it is irresponsible in a classroom setting, therefore, to \"think what you're told to think is true\", as it delays the level of responsibility they will be held to by society.\n\nEngineers, for example, will be held accountable for the safety of structures they design or certify. Not making them aware of the mathematical principles that lay beneath their calculations could conceivably lead to fatalities. If I were an engineer, I would want to be **** certain I had a minimum of assumptions, to limit my own potential liability. If there were a typo in a formula in a handbook of mathematical expressions I had used, it is possible that would not help me much in a civil lawsuit.\n\nGranted, few of your students may face such an extreme situation, and such a situation is even less likely to rest upon a definition of an exponential function. Nevertheless, I believe in rigor as a principle, not for its own sake as a form of intellectual prowess, but because of what it accomplishes: showing what we derive is a consequence of simpler facts.\n\nIgnoring the logical priority of mathematical conclusions is not sound mathematics. It neatly sweeps \"hard stuff\" under the rug, as a matter of expedience. It's like fast-food convenience for the mind, in short: a sham.\n\nNow, it would be one thing if the Fundamental Theorem of Calculus and the Inverse Function Theorem were \"abstract generalities\" not bothered with in a beginning calculus course. However, this is not so, and I believe (perhaps I am being \"too esoteric\" in this belief?) that the best way to understanding something is to become acquainted with it through USE.\n\nSo, if we have these tools available to define certain functions: why don't we USE them? ESPECIALLY since transcendental functions are difficult to understand EXCEPT through some form of limiting process (and isn't calculus all about the nifty things we can do with limits?).\n\nI believe that teaching people to think for themselves only after they reach graduate school is \"too late\". Unless they have already individually acquired the skill on their own independently, they are set-up for failure, or at best, a very difficult time of it.\n\nI do understand, however, that as a matter of pragmatism, you may have found that devoting too much time to these sorts of issues has been non-productive. So, you are passing the buck to some other teacher. One can only hope that somewhere, this comes to an abrupt halt.\n\n**************\n\nI didn't mean to obliterate your cognizant observation that $e$ is indeed a \"natural\" base. It is, and for the reasons you indicate. However, something \"deep\" is hiding behind it, something important (a fact about this slippery real number $e$). Transcendental real numbers *ought* to be \"hard to define\", in fact it took centuries before we were in a position to even conclude they existed (that is, that the algebraic numbers are not \"cauchy-closed\").\n\nPersonally, if all one takes calculus for is to calculate derivatives and integrals, meh: Wolfram|Alpha and a CRC handbook, skip the class. This business about limits and continuity, on the other hand, and the exploration of the \"infinitesimal\", this sly attempt to touch the face of infinity; if that does not make one blush, I humbly submit one has no soul.\n\n#### ThePerfectHacker\n\n##### Well-known member\nLet me see if I have this straight: you differentiate a function without any justification that it is differentiable. Why do I think this is not a good idea?\nI only read your first paragraph. I did not read anything else you wrote because I think your teaching standards are abysmal if you have a problem with writing down down $b^x$ in the way that makes sense to most people when they actually think about it.\n\nHistorically speaking all the calculus developed in a non-precise way, but it was motivated, and my presentation of it most closely parallels it. Go ahead why not spend the the first two months of class defining Dedekind cuts and what it means to even add two real numbers together. But you do not do that. You say that is too much. So if you are willing to accept that it is okay to teach calculus without defining what addition means, rather appealing to students intuitive feeling about what it is, then why not define exponentials in the most intuitive way possible?\n\nI am willing to bet if you explained to a student why $e$ is natural using my explanation they will have a much better chance of repeating it back to you than your explanation.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nI'm not sure what it is you think I am getting at. It might be a bit more illuminating to read my entire post, if only to provide more context, but I cannot force you to do this.\n\nNot every function is differentiable. Students should not, even if they do not get all the details think this is \"mostly true\". Continuity is *special* and differentiability even more so.\n\nBut again, I do not hear an answer to my original question: assuming, for the time being that $b^x$ *is* differentiable (and we certainly HOPE it would be so), how do you actually differentiate it?\n\nWhat I imagine is something invoking the chain rule, and writing:\n\n$b^x = e^{x\\log(b)} = \\exp \\circ (x \\cdot \\log(b))$\n\nand using the product (or multiplying by a constant) rule for the $x \\log(b)$ part, and then re-writing:\n\n$(b^x)' = (\\exp)'(x \\log(b))\\cdot \\log(b) = \\log(b)\\cdot b^x$.\n\nTo get to that conclusion, we need to know at least these two facts:\n\n1. $(e^x)' = e^x$\n2. what the function $\\log$ is (in particular that it is a functional *inverse* of $\\exp$).\n\nPerhaps you show (or merely indicate the \"plausibility\" of) the fact that $e^0 = 1$. I would be *interested* in how you derive fact 1. There are some different ways to do this, and you haven't indicated which one you prefer. You also have not indicated how the number $e$ (rather than \"some exponential function $\\exp$\") enters into this. I think that this is critical, since the whole POINT of your initial post is that $e$ represents a \"special base\".\n\nFact 2 also needs prior establishment of at least SOME facts about $e^x$ (like, that it is 1-1, so it *has* an inverse, contrast this with the contortions we must undergo for inverse trig functions).\n\nI know that you feel that it is somehow more complicated to start with a way of describing $\\log$ and deriving $\\exp$ as ITS inverse function, but I really don't see this. The crucial property of $\\log$:\n\n$\\log(xy) = \\log(x) + \\log(y)$\n\nis an EASY consequence of the definition:\n\n$$\\displaystyle \\log(x) = \\int_1^x \\frac{1}{t}\\ dt$$, by taking the interval:\n\n$[1,xy]$ and splitting it into the two sub-intervals of $[1,x]$ and $[x,xy]$, and using the linearity of the integral (presumably you find this property not so challenging as to withhold it from your students).\n\nAt least you answered my original question of \"what is $b^x$?\" by answering it via some kind of \"rational approximation\". As I indicated to Evgeny, in his reply, I am perfectly OK with that. The real number system was designed to let us do things like that, so that's fair.\n\nHow you derive the fact:\n\n$b^x = e^{x\\log(b)}$ is a more subtle question. If in fact, you already have the inverse-pair $\\exp$ and $\\log$ previously established (you don't say if this is true or not), one could use THIS property of (rational, and in approximation, real) exponents:\n\n$(a^b)^c = a^{bc}$.\n\nPerhaps \"most\" (or even possibly \"all\") of your students do not care. I cannot change that, nor do I have the singular pleasure of addressing them. I feel you should care, especially since the tone of this thread indicates a pedagogical approach of choosing a \"simple case\" as desirable.\n\nHistorically, the logical foundations of calculus were seen as something of a \"crisis\", and provided the impetus for some truly worth-while mathematics by Cauchy, Weierstrass, Riemann and Dedekind (to name a few note-worthy examples). The \"soundness\" of Newton's (and Liebniz's) original justifications were viewed with some suspicion well into the 20th century, although with the advent of \"non-standard analysis\" these fears have been mostly allayed. Not of all mathematical progress is made at \"the outer reaches\" sometimes thinking about \"the very beginnings\" proves fruitful, as well.\n\nIf this is not so, then tell me: why do we even \"bother\" with \"epsilon-delta\" proofs? I bet even a beginning calculus text makes some (perhaps brief) mention of these strange animals (or so it seems to me with the wealth of questions asked about them on sites such as this one).\n\nPersonally, I think talking about Dedekind cuts or Cauchy sequences (or even the oft-maligined \"infinite decimal\" approach) is a worth-while undertaking, if only to show that these \"real numbers\" have some \"concrete\" realization as something we can use rational numbers to approximate (because, underneath it all, the rationals is what we \"really\" use). I understand if the syllabus you follow does not allow time for this, but it certainly would make \"some theorems\" less confusing (anything that uses glb's or lub's to establish the existence of some desired real number, for example).\n\n*******************\n\nI do get it, you know, that a \"simple\" explanation is easier to remember, and that perhaps as much as 80% of what a first-year calculus student learns will be forgotten in 3 months time (unless they *immediately* take a refresher course, or a \"building-upon-it\" continuation).\n\nIf I were a student of yours, asking these same questions, would I get the very same answers?\n\n*******************\n\nI have talked at perhaps too much length about this particular subject. It's hard for me to judge whether this becomes a purely \"ad hominem\" discussion, so let us abandon it. What I have neglected to say in all of this, is that the SECOND part of your post, is something I WHOLE-HEARTEDLY agree with, and that thinking in terms of \"turns\" is the \"natural\" way to measure angle (and very profitable for physicists, and engineers).\n\n#### ThePerfectHacker\n\n##### Well-known member\nI'm not sure what it is you think I am getting at. It might be a bit more illuminating to read my entire post, if only to provide more context, but I cannot force you to do this.\n\nNot every function is differentiable. Students should not, even if they do not get all the details think this is \"mostly true\". Continuity is *special* and differentiability even more so.\n\nBut again, I do not hear an answer to my original question: assuming, for the time being that $b^x$ *is* differentiable (and we certainly HOPE it would be so), how do you actually differentiate it?\nHow do you \"prove\" that $\\sin x$ and $\\cos x$ are differenciable? You draw a picture illustrating important limits. That aint a proof at all. However, the picture is more illuminating to understanding the analysis of trigonometric functions than a boring non-inspired unmotivated, but rigorous, power series definition.\n\nHow do you \"define\" $\\sqrt{x}$? You have no problem saying that it is that positive number when squared results in $x$. This definition is not exactly rigorous as it assumes the reals has positive square roots. The proof of which requires the completeness property. Something which is never even mentioned in a calculus course at all. You have no problem using that definition at all.\n\nBut the moment I write $b^x$ and I say that means raise $b$ to the power of $x$, you suddenly have a problem. You ask how do I even define. I \"define\" it in the most natural way most people would expect, $b^q$, were $q$ is a rational number approaching $x$. There is nothing natural about writing $\\exp(x\\log b)$, it is confusing and entirely unmotivated for anyone who sees it for their first time. Now if you applied this standard consistently you will have a problem with how one \"defines\" $\\sin x$ and $\\sqrt{x}$. But you do not. You are okay with how it is presented in a calculus class and assume the necessarily properties, including being differenciable, when needed.\n\nIn the end you apply selective criticisms and make the concept of exponential functions exponentially more difficult for first-time students because you think it justifies it. Calculus is not supposed to be about justifying analytic concepts. It is supposed to introduce one to them and try to motivate the reasons for why it works and why it is true.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nHow do you \"prove\" that $\\sin x$ and $\\cos x$ are differenciable? You draw a picture illustrating important limits. That aint a proof at all. However, the picture is more illuminating to understanding the analysis of trigonometric functions than a boring non-inspired unmotivated, but rigorous, power series definition.\n\nHow do you \"define\" $\\sqrt{x}$? You have no problem saying that it is that positive number when squared results in $x$. This definition is not exactly rigorous as it assumes the reals has positive square roots. The proof of which requires the completeness property. Something which is never even mentioned in a calculus course at all. You have no problem using that definition at all.\n\nBut the moment I write $b^x$ and I say that means raise $b$ to the power of $x$, you suddenly have a problem. You ask how do I even define. I \"define\" it in the most natural way most people would expect, $b^q$, were $q$ is a rational number approaching $x$. Now if you applied this standard consistently you will have a problem with how one \"defines\" $\\sin x$ and $\\sqrt{x}$. But you do not. You are okay with how it is presented in a calculus class and assume the necessarily properties, including being differenciable, when needed.\n\nIn the end you apply selective criticisms and make the concept of exponential functions exponentially more difficult for first-time students because you think it justifies it. Calculus is not supposed to be about justifying analytic concepts. It is supposed to introduce one to them and try to motivate the reasons for why it works and why it is true.\nI think differentiating trig functions is likewise somewhat thorny. Similar glib use of $\\pi$ is made in this case, usually without any justification. Again, a \"geometric\" limit argument is the usual \"first look\" approach taken, and this is not out of character with the geometric flavor that much of calculus has.\n\nI think that transcendental functions are \"subtler\" than many people realize...indeed, that even the completeness property of the reals is a subtle concept. These things are usually \"glossed over\" in favor of expediency. I remain unconvinced that this is a NECESSARY evil. It's convenient, yes, and perhaps even \"motivational\" (whatever that means), but the truth is kept \"out of view\".\n\nMany calculus texts start with the assertion that \"they will assume the reader is familiar with the basic properties of real numbers\" and dispense with 100 years of deep (and HARD) mathematics in a few words.\n\nSquare roots are a bit thorny, as well. Many students know (or have at least seen a proof) that $\\sqrt{2}$ is irrational, but they have no idea what a big can of worms that really is. It is easy to write $\\sqrt{x}$ and think you know what it means. I'm sure it comes as a shock to some math majors when they realize that actually, they do NOT (and maybe non-math-majors who never take anything more demanding than calculus ever realize there is \"more to it than that\").\n\nI believe in DEMONSTRATING that a function is differentiable, by showing the limit that defines its derivative EXISTS. I think this is good \"practice\" for later when one encounters a function whose differentiability status is unknown. Not all \"real problems\" have answer keys in the back of some book.\n\nOf course, you have me at a disadvantage: you know your students, and I do not. I ask you for an explanation of what you mean in your original post, and you reply with remarks about \"selective criticisms\".\n\nI cannot recall the particular thread, but there was a similar discussion some time back about defining the area of a circle, and arriving at a value for $\\pi$. Some argued that a \"proper definition\" of trig functions would have to be delayed until after power series, as that was logically more satisfiying. I made the point that it could be done with the tools of integral calculus (although this might not be very \"direct\").\n\nOne often sees T-shirts with:\n\n$e^{i\\pi} + 1 = 0$\n\nprinted on them. The simplicity of this formula belies the depth of information it summarizes: how the circle ties together exponentials, logarithms, trigonometry, and an intrinsic notion of distance in 7 scant symbols. I'm not even confident I understand entirely everything it says...but the small portion I DO get, fills me with a kind of awe.\n\nIf, in the final analysis, you feel that I am looking \"too deeply\" at what you think should be taken more lightly, I have no ready answer. I have my reasons for thinking the way I do, but if my explanations are not sufficient, I suppose that is my failing. I *try* to be clear, but I have no assurances I ever am, for I only see the world with my own perceptions, and never that of another human being.\n\nIndeed, I often fear I might be embarrassing my comrades here at MHB, for being too outspoken, or having views well outside of the conventional wisdom, or even worse, perhaps just plain being wrong. I have only a vague notion of INTERNAL consistency to guide me, and I find it enormously difficult to communicate this to other people. Surely, I reason to myself, Evgeny.Makarov or Ackbach or Jameson or MarkFL or chisigma (or some other soul who I have in my ignorance failed to mention) will point out to me what is wrong with my views, so I may improve my ability to express math to other people...but alas, I hear only silence.", "date": "2022-01-25 13:15:41", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/why-e-is-natural-and-why-we-use-radians.9286/", "openwebmath_score": 0.8424057364463806, "openwebmath_perplexity": 484.89271278881665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9825575147530351, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8654337848652092}}
{"url": "https://it.mathworks.com/help/symbolic/dsolve.html", "text": "# dsolve\n\nSolve system of differential equations\n\nSupport for character vector or string inputs will be removed in a future release. Instead, use syms to declare variables and replace inputs such as dsolve('Dy = -3*y') with syms y(t); dsolve(diff(y,t) == -3*y).\n\n## Description\n\nexample\n\nS = dsolve(eqn) solves the differential equation eqn, where eqn is a symbolic equation. Use diff and == to represent differential equations. For example, diff(y,x) == y represents the equation dy/dx\u00a0=\u00a0y. Solve a system of differential equations by specifying eqn as a vector of those equations.\n\nexample\n\nS = dsolve(eqn,cond) solves eqn with the initial or boundary condition cond.\n\nexample\n\nS = dsolve(___,Name,Value) uses additional options specified by one or more Name,Value pair arguments.\n\nexample\n\n[y1,...,yN] = dsolve(___) assigns the solutions to the variables y1,...,yN.\n\n## Examples\n\ncollapse all\n\nSolve the first-order differential equation $\\frac{\\mathit{dy}}{\\mathit{dt}}=\\mathit{ay}$.\n\nSpecify the first-order derivative by using diff and the equation by using ==. Then, solve the equation by using dsolve.\n\nsyms y(t) a\neqn = diff(y,t) == a*y;\nS = dsolve(eqn)\nS =\u00a0${C}_{1} {\\mathrm{e}}^{a t}$\n\nThe solution includes a constant. To eliminate constants, see Solve Differential Equations with Conditions. For a full workflow, see Solving Partial Differential Equations.\n\nSolve the second-order differential equation $\\frac{{\\mathit{d}}^{2}\\mathit{y}}{{\\mathit{dt}}^{2}}=\\mathit{ay}$.\n\nSpecify the second-order derivative of y by using diff(y,t,2) and the equation by using ==. Then, solve the equation by using dsolve.\n\nsyms y(t) a\neqn = diff(y,t,2) == a*y;\nySol(t) = dsolve(eqn)\nySol(t) =\u00a0${C}_{1} {\\mathrm{e}}^{-\\sqrt{a} t}+{C}_{2} {\\mathrm{e}}^{\\sqrt{a} t}$\n\nSolve the first-order differential equation $\\frac{\\mathit{dy}}{\\mathit{dt}}=\\mathit{ay}$ with the initial condition $y\\left(0\\right)=5$.\n\nSpecify the initial condition as the second input to dsolve by using the == operator. Specifying condition eliminates arbitrary constants, such as C1, C2, ..., from the solution.\n\nsyms y(t) a\neqn = diff(y,t) == a*y;\ncond = y(0) == 5;\nySol(t) = dsolve(eqn,cond)\nySol(t) =\u00a0$5 {\\mathrm{e}}^{a t}$\n\nNext, solve the second-order differential equation $\\frac{{\\mathit{d}}^{2}\\mathit{y}}{{\\mathit{dt}}^{2}}={\\mathit{a}}^{2}\\mathit{y}$ with the initial conditions $y\\left(0\\right)=b$ and ${y}^{\\prime }\\left(0\\right)=1$.\n\nSpecify the second initial condition by assigning diff(y,t) to Dy and then using Dy(0) == 1.\n\nsyms y(t) a b\neqn = diff(y,t,2) == a^2*y;\nDy = diff(y,t);\ncond = [y(0)==b, Dy(0)==1];\nySol(t) = dsolve(eqn,cond)\nySol(t) =\n\n$\\frac{{\\mathrm{e}}^{a t} \\left(a b+1\\right)}{2 a}+\\frac{{\\mathrm{e}}^{-a t} \\left(a b-1\\right)}{2 a}$\n\nThis second-order differential equation has two specified conditions, so constants are eliminated from the solution. In general, to eliminate constants from the solution, the number of conditions must equal the order of the equation.\n\nSolve the system of differential equations\n\n$\\begin{array}{l}\\frac{\\mathit{dy}}{\\mathit{dt}}=\\mathit{z}\\\\ \\frac{\\mathit{dz}}{\\mathit{dt}}=-\\mathit{y}.\\end{array}$\n\nSpecify the system of equations as a vector. dsolve returns a structure containing the solutions.\n\nsyms y(t) z(t)\neqns = [diff(y,t) == z, diff(z,t) == -y];\nS = dsolve(eqns)\nS = struct with fields:\nz: C2*cos(t) - C1*sin(t)\ny: C1*cos(t) + C2*sin(t)\n\nAccess the solutions by addressing the elements of the structure.\n\nySol(t) = S.y\nySol(t) =\u00a0${C}_{1} \\mathrm{cos}\\left(t\\right)+{C}_{2} \\mathrm{sin}\\left(t\\right)$\nzSol(t) = S.z\nzSol(t) =\u00a0${C}_{2} \\mathrm{cos}\\left(t\\right)-{C}_{1} \\mathrm{sin}\\left(t\\right)$\n\nWhen solving for multiple functions, dsolve returns a structure by default. Alternatively, you can assign solutions to functions or variables directly by explicitly specifying the outputs as a vector. dsolve sorts the outputs in alphabetical order using symvar.\n\nSolve a system of differential equations and assign the outputs to functions.\n\nsyms y(t) z(t)\neqns = [diff(y,t)==z, diff(z,t)==-y];\n[ySol(t),zSol(t)] = dsolve(eqns)\nySol(t) =\u00a0${C}_{1} \\mathrm{cos}\\left(t\\right)+{C}_{2} \\mathrm{sin}\\left(t\\right)$\nzSol(t) =\u00a0${C}_{2} \\mathrm{cos}\\left(t\\right)-{C}_{1} \\mathrm{sin}\\left(t\\right)$\n\nSolve the differential equation $\\frac{\\partial }{\\partial t}y\\left(t\\right)={e}^{-y\\left(t\\right)}+y\\left(t\\right)$. dsolve returns an explicit solution in terms of a Lambert W function that has a constant value.\n\nsyms y(t)\neqn = diff(y) == y+exp(-y)\neqn(t) =\n\nsol = dsolve(eqn)\nsol =\u00a0${\\mathrm{W}\\text{lambertw}}_{0}\\left(-1\\right)$\n\nTo return implicit solutions of the differential equation, set the 'Implicit' option to true. An implicit solution has the form $F\\left(y\\left(t\\right)\\right)=g\\left(t\\right)$.\n\nsol = dsolve(eqn,'Implicit',true)\nsol =\n\n$\\left(\\begin{array}{c}\\left({\\int \\frac{{\\mathrm{e}}^{y}}{y {\\mathrm{e}}^{y}+1}\\mathrm{d}y|}_{y=y\\left(t\\right)}\\right)={C}_{1}+t\\\\ {\\mathrm{e}}^{-y\\left(t\\right)} \\left({\\mathrm{e}}^{y\\left(t\\right)} y\\left(t\\right)+1\\right)=0\\end{array}\\right)$\n\nIf dsolve cannot find an explicit solution of a differential equation analytically, then it returns an empty symbolic array. You can solve the differential equation by using MATLAB\u00ae numerical solver, such as ode45. For more information, see Solve a Second-Order Differential Equation Numerically.\n\nsyms y(x)\neqn = diff(y) == (x-exp(-x))/(y(x)+exp(y(x)));\nS = dsolve(eqn)\nWarning: Unable to find symbolic solution.\n\nS =\n\n[ empty sym ]\n\nAlternatively, you can try finding an implicit solution of the differential equation by specifying the 'Implicit' option to true. An implicit solution has the form $F\\left(y\\left(x\\right)\\right)=g\\left(x\\right)$.\n\nS = dsolve(eqn,'Implicit',true)\nS =\n\n${\\mathrm{e}}^{y\\left(x\\right)}+\\frac{{y\\left(x\\right)}^{2}}{2}={C}_{1}+{\\mathrm{e}}^{-x}+\\frac{{x}^{2}}{2}$\n\nSolve the differential equation $\\frac{\\mathit{dy}}{\\mathit{dt}}=\\frac{\\mathit{a}}{\\sqrt{\\mathit{y}}}+\\mathit{y}$ with condition $y\\left(a\\right)=1$. By default, dsolve applies simplifications that are not generally correct, but produce simpler solutions. For more details, see Algorithms.\n\nsyms a y(t)\neqn = diff(y) == a/sqrt(y) + y;\ncond = y(a) == 1;\nySimplified = dsolve(eqn, cond)\nySimplified =\n\n${\\left({\\mathrm{e}}^{\\frac{3 t}{2}-\\frac{3 a}{2}+\\mathrm{log}\\left(a+1\\right)}-a\\right)}^{2/3}$\n\nTo return the solutions that include all possible values of the parameter $a$, turn off simplifications by setting 'IgnoreAnalyticConstraints' to false.\n\nyNotSimplified = dsolve(eqn,cond,'IgnoreAnalyticConstraints',false)\nyNotSimplified =\n\nSolve the second-order differential equation $\\left({x}^{2}-1{\\right)}^{2}\\frac{{\\partial }^{2}}{\\partial {x}^{2}}y\\left(x\\right)+\\left(x+1\\right)\\frac{\\partial }{\\partial x}y\\left(x\\right)-y\\left(x\\right)=0$. dsolve returns a solution that contains a term with unevaluated integral.\n\nsyms y(x)\neqn = (x^2-1)^2*diff(y,2) + (x+1)*diff(y) - y == 0;\nS = dsolve(eqn)\nS =\n\n${C}_{2} \\left(x+1\\right)+{C}_{1} \\left(x+1\\right) \\int \\frac{{\\mathrm{e}}^{\\frac{1}{2 \\left(x-1\\right)}} {\\left(1-x\\right)}^{1/4}}{{\\left(x+1\\right)}^{9/4}}\\mathrm{d}x$\n\nTo return series solutions of the differential equation around $x=-1$, set the 'ExpansionPoint' to -1. dsolve returns two linearly independent solutions in terms of a Puiseux series expansion.\n\nS = dsolve(eqn,'ExpansionPoint',-1)\nS =\n\n$\\left(\\begin{array}{c}x+1\\\\ \\frac{1}{{\\left(x+1\\right)}^{1/4}}-\\frac{5 {\\left(x+1\\right)}^{3/4}}{4}+\\frac{5 {\\left(x+1\\right)}^{7/4}}{48}+\\frac{5 {\\left(x+1\\right)}^{11/4}}{336}+\\frac{115 {\\left(x+1\\right)}^{15/4}}{33792}+\\frac{169 {\\left(x+1\\right)}^{19/4}}{184320}\\end{array}\\right)$\n\nFind other series solutions around the expansion point $\\infty$ by setting 'ExpansionPoint' to Inf.\n\nS = dsolve(eqn,'ExpansionPoint',Inf)\nS =\n\n$\\left(\\begin{array}{c}x-\\frac{1}{6 {x}^{2}}-\\frac{1}{8 {x}^{4}}\\\\ \\frac{1}{6 {x}^{2}}+\\frac{1}{8 {x}^{4}}+\\frac{1}{90 {x}^{5}}+1\\end{array}\\right)$\n\nThe default truncation order of the series expansion is 6. To obtain more terms in the Puiseux series solutions, set 'Order' to 8.\n\nS = dsolve(eqn,'ExpansionPoint',Inf,'Order',8)\nS =\n\n$\\left(\\begin{array}{c}x-\\frac{1}{6 {x}^{2}}-\\frac{1}{8 {x}^{4}}-\\frac{1}{90 {x}^{5}}-\\frac{37}{336 {x}^{6}}\\\\ \\frac{1}{6 {x}^{2}}+\\frac{1}{8 {x}^{4}}+\\frac{1}{90 {x}^{5}}+\\frac{37}{336 {x}^{6}}+\\frac{37}{1680 {x}^{7}}+1\\end{array}\\right)$\n\nSolve the differential equation $\\frac{\\mathit{dy}}{\\mathit{dx}}=\\frac{1}{{\\mathit{x}}^{2}}{\\mathit{e}}^{-\\frac{1}{\\mathit{x}}}$ without specifying the initial condition.\n\nsyms y(x)\neqn = diff(y) == exp(-1/x)/x^2;\nySol(x) = dsolve(eqn)\nySol(x) =\n\n${C}_{1}+{\\mathrm{e}}^{-\\frac{1}{x}}$\n\nTo eliminate constants from the solution, specify the initial condition $\\mathit{y}\\left(0\\right)=1$.\n\ncond = y(0) == 1;\nS = dsolve(eqn,cond)\nS =\n\n${\\mathrm{e}}^{-\\frac{1}{x}}+1$\n\nThe function ${\\mathit{e}}^{-\\frac{1}{\\mathit{x}}}$ in the solution ySol(x) has different one-sided limits at $x=0$. The function has a right-side limit, $\\underset{\\mathit{x}\\to {0}^{+}}{\\mathrm{lim}}\\text{\\hspace{0.17em}}{\\mathit{e}}^{-\\frac{1}{\\mathit{x}}}=0$, but it has undefined left-side limit, $\\underset{\\mathit{x}\\to {0}^{-}}{\\mathrm{lim}}\\text{\\hspace{0.17em}}{\\mathit{e}}^{-\\frac{1}{\\mathit{x}}}=\\infty$.\n\nWhen you specify the condition y(x0) for a function with different one-sided limits at x0, dsolve treats the condition as a limit from the right, $\\mathrm{lim}\\text{\\hspace{0.17em}}\\mathit{x}\\to {\\mathit{x}}_{0}^{+}$.\n\n## Input Arguments\n\ncollapse all\n\nDifferential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. Specify a differential equation by using the == operator. If eqn is a symbolic expression (without the right side), the solver assumes that the right side is 0, and solves the equation eqn == 0.\n\nIn the equation, represent differentiation by using diff. For example, diff(y,x) differentiates the symbolic function y(x) with respect to x. Create the symbolic function y(x) by using syms and solve the equation d2y(x)/dx2 = x*y(x) using dsolve.\n\nsyms y(x)\nS = dsolve(diff(y,x,2) == x*y)\n\nSpecify a system of differential equations by using a vector of equations, as in syms y(t) z(t); S = dsolve([diff(y,t) == z, diff(z,t) == -y]). Here, y and z must be symbolic functions that depend on symbolic variables, which are t in this case. The right side must be symbolic expressions that depend on t, y and z. Note that Symbolic Math Toolbox\u2122 currently does not support composite symbolic functions, that is, symbolic functions that depend on another symbolic functions.\n\nInitial or boundary condition, specified as a symbolic equation or vector of symbolic equations.\n\nWhen a condition contains a derivative, represent the derivative with diff. Assign the diff call to a variable and use the variable to specify the condition. For example, see Solve Differential Equations with Conditions.\n\nSpecify multiple conditions by using a vector of equations. If the number of conditions is less than the number of dependent variables, the solutions contain the arbitrary constants C1, C2,....\n\n### Name-Value Arguments\n\nSpecify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.\n\nExample: 'IgnoreAnalyticConstraints',false does not apply internal simplifications.\n\nExpansion point of a Puiseux series solution, specified as a number, or a symbolic number, variable, function, or expression. Specifying this option returns the solution of a differential equation in terms of a Puiseux series (a power series that allows negative and fractional exponents). The expansion point cannot depend on the series variable. For example, see Find Series Solution of Differential Equation.\n\nData Types: single | double | sym\nComplex Number Support: Yes\n\nOption to use internal simplifications, specified as true or false.\n\nBy default, the solver applies simplifications while solving the differential equation, which could lead to results not generally valid. In other words, this option applies mathematical identities that are convenient, but the results might not hold for all possible values of the variables. Therefore, by default, the solver does not guarantee the completeness of results. If 'IgnoreAnalyticConstraints' is true, always verify results returned by the dsolve function. For more details, see Algorithms.\n\nTo solve ordinary differential equations without these simplifications, set 'IgnoreAnalyticConstraints' to false. Results obtained with 'IgnoreAnalyticConstraints' set to false are correct for all values of the arguments. For certain equations, dsolve might not return an explicit solution if you set 'IgnoreAnalyticConstraints' to false.\n\nOption to return an implicit solution, specified as false or true. For a differential equation with variables x and y(x), an implicit solution has the form F(y(x)) = g(x).\n\nBy default, the solver tries to find an explicit solution y(x) = f(x) analytically when solving a differential equation. If dsolve cannot find an explicit solution, then you can try finding a solution in implicit form by specifying the 'Implicit' option to true.\n\nMaximum degree of polynomial equations for which the solver uses explicit formulas, specified as a positive integer smaller than 5. dsolve does not use explicit formulas when solving polynomial equations of degrees larger than 'MaxDegree'.\n\nTruncation order of a Puiseux series solution, specified as a positive integer or a symbolic positive integer. Specifying this option returns the solution of a differential equation in terms of a Puiseux series (a power series that allow negative and fractional exponents). The truncation order n is the exponent in the O-term: $O\\left({\\mathrm{var}}^{n}\\right)$ or $O\\left({\\mathrm{var}}^{-n}\\right)$.\n\n## Output Arguments\n\ncollapse all\n\nSolutions of differential equation, returned as a symbolic expression or a vector of symbolic expressions. The size of S is the number of solutions.\n\nVariables storing solutions of differential equations, returned as a vector of symbolic variables. The number of output variables must equal the number of dependent variables in a system of equations. dsolve sorts the dependent variables alphabetically, and then assigns the solutions for the variables to output variables or symbolic arrays.\n\n## Tips\n\n\u2022 If dsolve cannot find an explicit or implicit solution, then it issues a warning and returns the empty sym. In this case, try to find a numeric solution using the MATLAB\u00ae ode23 or ode45 function. Sometimes, the output is an equivalent lower-order differential equation or an integral.\n\n\u2022 dsolve does not always return complete solutions even if 'IgnoreAnalyticConstraints' is false.\n\n\u2022 If dsolve returns a function that has different one-sided limits at x0 and you specify the condition y(x0), then dsolve treats the condition as a limit from the right, .\n\n## Algorithms\n\nIf you do not set 'IgnoreAnalyticConstraints' to false, then dsolve applies these rules while solving the equation:\n\n\u2022 log(a) + log(b)\u00a0=\u00a0log(a\u00b7b) for all values of a and b. In particular, the following equality is applied for all values of a, b, and c:\n\n(a\u00b7b)c\u00a0=\u00a0ac\u00b7bc.\n\n\u2022 log(ab)\u00a0=\u00a0b\u00b7log(a) for all values of a and b. In particular, the following equality is applied for all values of a, b, and c:\n\n(ab)c\u00a0=\u00a0ab\u00b7c.\n\n\u2022 If f and g are standard mathematical functions and f(g(x))\u00a0=\u00a0x for all small positive numbers, f(g(x))\u00a0=\u00a0x is assumed to be valid for all complex x. In particular:\n\n\u2022 log(ex)\u00a0=\u00a0x\n\n\u2022 asin(sin(x))\u00a0=\u00a0x, acos(cos(x))\u00a0=\u00a0x, atan(tan(x))\u00a0=\u00a0x\n\n\u2022 asinh(sinh(x))\u00a0=\u00a0x, acosh(cosh(x))\u00a0=\u00a0x, atanh(tanh(x))\u00a0=\u00a0x\n\n\u2022 Wk(x\u00b7ex)\u00a0=\u00a0x for all branch indices k of the Lambert W function.\n\n\u2022 The solver can multiply both sides of an equation by any expression except 0.\n\n\u2022 The solutions of polynomial equations must be complete.\n\n## Compatibility Considerations\n\nexpand all\n\nWarns starting in R2019b", "date": "2021-10-17 19:06:08", "meta": {"domain": "mathworks.com", "url": "https://it.mathworks.com/help/symbolic/dsolve.html", "openwebmath_score": 0.7705014944076538, "openwebmath_perplexity": 1471.9728524477512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308786041315, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8653921711289246}}
{"url": "https://math.stackexchange.com/questions/2948977/proving-right-angle-using-vectors", "text": "# Proving right angle using vectors\n\nLet $$ABCD$$ be a rectangle, $$E$$ midpoint of $$\\overline{DC}$$ and $$G$$ point on $$\\overline{AC}$$ such that $$\\vec{BG}$$ is perpendicular to $$\\vec{AC}$$. Also, let $$F$$ be a midpoint of $$\\overline{AG}$$. Prove that angle $$\\angle BFE =\\pi/2$$.\n\nSo, I need to prove that $$\\vec{EF}\\cdot \\vec{FB}=0$$. I tried to just express these vectors as sums of vectors of rectangle but haven't find any \"elegant\" way to prove it.\n\nI did found one \"ugly\" way to prove it:\n\nLet $$|AB|=a$$, $$|AD|=b$$ and $$\\vec{AB}=\\vec{a}$$, $$\\vec{AD}=\\vec{b}$$, so $$\\vec{AC}=\\vec{a}+\\vec{b}$$ and since $$\\vec{a}\\cdot\\vec{b}=0$$ we have $$|AC|^2=a^2+b^2$$. We can see that $$\\triangle ABC$$ and $$\\triangle BCG$$ are similar so $$|CG|=|CB|^2/|AC|$$ and since $$\\vec{CG}=\\lambda \\cdot \\vec{CA}$$ we get $$\\lambda=\\frac{b^2}{a^2+b^2}.$$\n\nNow, we can express $$\\vec{EF}$$ and $$\\vec{FB}$$ in the terms of $$\\vec{a}$$, $$\\vec{b}$$, more precisely: $$\\vec{EF}=\\frac{1}{2}\\left(\\frac{-b^2}{a^2+b^2}\\right)\\vec{a}+\\frac{1}{2}\\left(\\frac{a^2}{a^2+b^2}-2\\right)\\vec{b}$$ and $$\\vec{FB}=\\frac{1}{2}\\left(2-\\frac{a^2}{a^2+b^2}\\right)\\vec{a}+\\frac{1}{2}\\left(\\frac{-a^2}{a^2+b^2}\\right)\\vec{b}$$ and if we multiply we get that dot product is $$0$$.\n\nBut as you can see, this $$\\lambda$$ is \"weird\" and I am wondering if someone sees a more elegant way to prove this ?\n\n\u2022 Actually, if you write those final coefficients in terms of $\\lambda$, it makes the proof of orthogonality much prettier. \u2013\u00a0J.G. Oct 9 '18 at 19:26\n\nYou can simplify and generalize at the same time.\n\nInstead of defining $$E$$ and $$F$$ as midpoints, define them as weighted averages:\n\n$$E=kC+(1-k)D$$ $$F=kG+(1-k)A$$\n\n(In your question $$k=1/2.)$$ As $$k$$ goes from $$0$$ to $$1$$, $$\\triangle{BFE}$$ interpolates the similar triangles $$\\triangle{BAD}$$ and $$\\triangle{BGC}$$. And it is also similar to these two triangles.\n\nGiven that $$(B-G)\\cdot(C-G)=0$$ and $$(B-A)\\cdot(D-A)=0$$, you can show that $$(B-F)\\cdot(E-F)=0$$ (the simplification is a bit tedious, and I used the fact that $$(Rx)\\cdot y+(Ry)\\cdot x=0$$ for perpendicular vectors $$x,y$$ and rotation $$R$$).\n\nMore about the phenomenon of linearly interpolating two directly similar figures can be found by Googling the terms 'spiral similarity' and 'fundamental theorem of directly similar figures'. Also, instead of using vectors you can use complex numbers, as shown in $$\\textbf{2.2}$$ of Zachary Abel's Mean Geometry paper.\n\nLet $$H$$ be a midpoint of $$BG$$. Then $$HF$$ is a middle line in triangle $$ABG$$, so $$FH||AB||CD$$ and $$FH = {1\\over 2}AB = CE$$ so $$FHCE$$ is a parallelogram, so $$FE||CH$$.\n\nNow $$HF\\bot BC$$ so $$HF$$ is (second) altitude in $$\\triangle BCF$$ so $$H$$ is orthocenter in this triangle, since $$BG$$ is also altitude in this triangle. So line $$CH\\bot FB$$ and thus $$\\angle BFE = 90^{\\circ}$$.\n\nHere is a vector solution:\n\nLet $$G$$ be an origin of position vecotors. Then $$G=0$$, $$F = A/2$$, $$D = A-B+C$$, $$A\\cdot B = C\\cdot B=0$$ and $$E = {1\\over 2}(C+D) = {1\\over 2}(A-B+2C)$$\n\n$$\\begin{eqnarray} \\vec{FB}\\cdot \\vec{FE} &=& (E-F)(B-F)\\\\ &=& {1\\over 4}(-B+2C)(2B-A)\\\\ &=& -{1\\over 2}(B^2+CA)\\\\ &=&0 \\end{eqnarray}$$\n\nA solution with a help of the coordinate system.\n\nLet $$B = (0,0)$$, $$C= (2c,0)$$ and $$A= (0,2a)$$. Then $$D(2c,2a)$$ and $$E(2c,a)$$. A perpendicular to $$AC:\\;\\;{x\\over 2c}+{y\\over 2a}=1$$ through $$B$$ is $$y={c\\over a}x$$ which cuts $$AC$$ at $$G= \\big({2a^2c\\over a^2+c^2},{2c^2a\\over a^2+c^2} \\big)$$\n\nso $$F = \\big({a^2c\\over a^2+c^2},{2c^2a+a^3\\over a^2+c^2} \\big)$$\n\nNow it is not difficult to see that $$\\vec{FE}\\cdot \\vec{FB} =0$$.\n\nAnother solution:\n\nSay $$M$$ is midpoint of $$AB$$, then $$FM$$ is a middle line in triangle $$ABG$$ so $$FM||BG$$ so $$\\angle MFC = 90^{\\circ}$$ and thus $$F$$ is on a circle through $$B,C$$ and $$M$$. But on this circle is also $$E$$ since $$\\angle MEC = 90^{\\circ}$$. So $$\\angle BFE = \\angle BME = 90^{\\circ}$$\n\nAlternatively Say $$AB = 2a$$ and $$BC= b$$, then $$AF = {1\\over 2}AG = {1\\over 2} {4a^2\\over \\sqrt{4a^2+b^2}} = {2a^2\\over \\sqrt{4a^2+b^2}}$$ Then we have $$AF\\cdot AC = {2a^2\\over \\sqrt{4a^2+b^2}}\\cdot \\sqrt{4a^2+b^2} = 2a^2 = AM\\cdot AB$$ and so $$B,C,F,M$$ are concylic.", "date": "2019-06-19 16:41:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2948977/proving-right-angle-using-vectors", "openwebmath_score": 0.9508201479911804, "openwebmath_perplexity": 128.04523796137414, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988130881050834, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8653921620593246}}
{"url": "https://de.mathworks.com/help/symbolic/solve-an-algebraic-equation.html", "text": "## Solve Algebraic Equation\n\nSymbolic Math Toolbox\u2122 offers both symbolic and numeric equation solvers. This topic shows you how to solve an equation symbolically using the symbolic solver solve. To compare symbolic and numeric solvers, see Select Numeric or Symbolic Solver.\n\n### Solve an Equation\n\nIf eqn is an equation, solve(eqn, x) solves eqn for the symbolic variable x.\n\nUse the == operator to specify the familiar quadratic equation and solve it using solve.\n\nsyms a b c x\neqn = a*x^2 + b*x + c == 0;\nsolx = solve(eqn, x)\nsolx =\n-(b + (b^2 - 4*a*c)^(1/2))/(2*a)\n-(b - (b^2 - 4*a*c)^(1/2))/(2*a)\n\nsolx is a symbolic vector containing the two solutions of the quadratic equation. If the input eqn is an expression and not an equation, solve solves the equation eqn == 0.\n\nTo solve for a variable other than x, specify that variable instead. For example, solve eqn for b.\n\nsolb = solve(eqn, b)\nsolb =\n-(a*x^2 + c)/x\n\nIf you do not specify a variable, solve uses symvar to select the variable to solve for. For example, solve(eqn) solves eqn for x.\n\n### Return the Full Solution to an Equation\n\nsolve does not automatically return all solutions of an equation. Solve the equation cos(x) == -sin(x). The solve function returns one of many solutions.\n\nsyms x\nsolx = solve(cos(x) == -sin(x), x)\nsolx =\n-pi/4\n\nTo return all solutions along with the parameters in the solution and the conditions on the solution, set the ReturnConditions option to true. Solve the same equation for the full solution. Provide three output variables: for the solution to x, for the parameters in the solution, and for the conditions on the solution.\n\nsyms x\n[solx, param, cond] = solve(cos(x) == -sin(x), x, 'ReturnConditions', true)\nsolx =\npi*k - pi/4\nparam =\nk\ncond =\nin(k, 'integer')\n\nsolx contains the solution for x, which is pi*k - pi/4. The param variable specifies the parameter in the solution, which is k. The cond variable specifies the condition in(k, 'integer') on the solution, which means k must be an integer. Thus, solve returns a periodic solution starting at pi/4 which repeats at intervals of pi*k, where k is an integer.\n\n### Work with the Full Solution, Parameters, and Conditions Returned by solve\n\nYou can use the solutions, parameters, and conditions returned by solve to find solutions within an interval or under additional conditions.\n\nTo find values of x in the interval -2*pi<x<2*pi, solve solx for k within that interval under the condition cond. Assume the condition cond using assume.\n\nassume(cond)\nsolk = solve(-2*pi<solx, solx<2*pi, param)\nsolk =\n-1\n0\n1\n2\n\nTo find values of x corresponding to these values of k, use subs to substitute for k in solx.\n\nxvalues = subs(solx, solk)\nxvalues =\n-(5*pi)/4\n-pi/4\n(3*pi)/4\n(7*pi)/4\n\nTo convert these symbolic values into numeric values for use in numeric calculations, use vpa.\n\nxvalues = vpa(xvalues)\nxvalues =\n-3.9269908169872415480783042290994\n-0.78539816339744830961566084581988\n2.3561944901923449288469825374596\n5.4977871437821381673096259207391\n\n### Visualize and Plot Solutions Returned by solve\n\nThe previous sections used solve to solve the equation cos(x) == -sin(x). The solution to this equation can be visualized using plotting functions such as fplot and scatter.\n\nPlot both sides of equation cos(x) == -sin(x).\n\nfplot(cos(x))\nhold on\ngrid on\nfplot(-sin(x))\ntitle('Both sides of equation cos(x) = -sin(x)')\nlegend('cos(x)','-sin(x)','Location','best','AutoUpdate','off')\n\nCalculate the values of the functions at the values of x, and superimpose the solutions as points using scatter.\n\nyvalues = cos(xvalues)\nyvalues =\n\n$\\left(\\begin{array}{c}-0.70710678118654752440084436210485\\\\ 0.70710678118654752440084436210485\\\\ -0.70710678118654752440084436210485\\\\ 0.70710678118654752440084436210485\\end{array}\\right)$\n\nscatter(xvalues, yvalues)\n\nAs expected, the solutions appear at the intersection of the two plots.\n\n### Simplify Complicated Results and Improve Performance\n\nIf results look complicated, solve is stuck, or if you want to improve performance, see, Troubleshoot Equation Solutions from solve Function.\n\n## Support\n\n#### Mathematical Modeling with Symbolic Math Toolbox\n\nGet examples and videos", "date": "2021-09-24 01:25:05", "meta": {"domain": "mathworks.com", "url": "https://de.mathworks.com/help/symbolic/solve-an-algebraic-equation.html", "openwebmath_score": 0.6937898993492126, "openwebmath_perplexity": 2970.693099512371, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363508288305, "lm_q2_score": 0.8791467690927438, "lm_q1q2_score": 0.8653761225317079}}
{"url": "http://mathhelpforum.com/calculus/4044-find-limit.html", "text": "1. ## Find the limit\n\nLimit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]\n\nI tried to divide the numerator and denominator by 3^n. Was not successful.\n\nWhat should I do next?\n\n2. Originally Posted by Nichelle14\nLimit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]\n\nI tried to divide the numerator and denominator by 3^n. Was not successful.\n\nWhat should I do next?\nDid you consider to divide by $3^n+5^n$\nThus,\n$\\frac{1}{1+\\frac{2}{3^n+5^n}}\\to 1$ as $n\\to\\infty$\n\n3. Also,\n$1-\\frac{1}{n}\\leq \\frac{3^n+5^n}{3^n+1+5^n+1} \\leq 1+\\frac{1}{n}$\nSince,\n$\\lim_{n\\to\\infty} 1-\\frac{1}{n}=\\lim_{n\\to\\infty}1+\\frac{1}{n}=1$\nThus,\n$\\lim_{n\\to\\infty}\\frac{3^n+5^n}{3^n+1+5^n+1}=1$ by the squeeze theorem.\n\n4. Hello, Nichelle14!\n\n$\\lim_{n\\to\\infty}\\frac{3^n + 5^n}{3^{n+1} + 5^{n+1}}$\n\nI tried to divide the numerator and denominator by $3^n$ . . .Was not successful.\n\nWhat should I do next?\n\nDivide top and bottom by $5^{n+1}.$\n\nThe numerator is: . $\\frac{3^n}{5^{n+1}} + \\frac{5^n}{5^{n+1}} \\;= \\;\\frac{1}{5}\\cdot\\frac{3^n}{5^n} + \\frac{1}{5}\\;=$ $\\frac{1}{5}\\left(\\frac{3}{5}\\right)^n + \\frac{1}{5}$\n\nThe denominator is: . $\\frac{3^{n+1}}{5^{n+1}} + \\frac{5^{n+1}}{5^{n+1}} \\;=\\;\\left(\\frac{3}{5}\\right)^{n+1} + 1$\n\nRecall that: if $|a| < 1$, then $\\lim_{n\\to\\infty} a^n\\:=\\:0$\n\nTherefore, the limit is: . $\\lim_{n\\to\\infty}\\,\\frac{\\frac{1}{5}\\left(\\frac{3} {5}\\right)^n + \\frac{1}{5}} {\\left(\\frac{3}{5}\\right)^n + 1} \\;=\\;\\frac{\\frac{1}{5}\\cdot0 + \\frac{1}{5}}{0 + 1}\\;=\\;\\frac{1}{5}\n$\n\n5. Originally Posted by Soroban\nHello, Nichelle14!\n\nDivide top and bottom by $5^{n+1}.$\n\nThe numerator is: . $\\frac{3^n}{5^{n+1}} + \\frac{5^n}{5^{n+1}} \\;= \\;\\frac{1}{5}\\cdot\\frac{3^n}{5^n} + \\frac{1}{5}\\;=$ $\\frac{1}{5}\\left(\\frac{3}{5}\\right)^n + \\frac{1}{5}$\n\nThe denominator is: . $\\frac{3^{n+1}}{5^{n+1}} + \\frac{5^{n+1}}{5^{n+1}} \\;=\\;\\left(\\frac{3}{5}\\right)^{n+1} + 1$\n\nRecall that: if $|a| < 1$, then $\\lim_{n\\to\\infty} a^n\\:=\\:0$\n\nTherefore, the limit is: . $\\lim_{n\\to\\infty}\\,\\frac{\\frac{1}{5}\\left(\\frac{3} {5}\\right)^n + \\frac{1}{5}} {\\left(\\frac{3}{5}\\right)^n + 1} \\;=\\;\\frac{\\frac{1}{5}\\cdot0 + \\frac{1}{5}}{0 + 1}\\;=\\;\\frac{1}{5}\n$\n\nI think the trick here is that you will divide all the terms by the largest term in the expression(it is useful when the limit tends to infinity)\n\nKeep Smiling\nMalay", "date": "2013-05-19 02:18:47", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/4044-find-limit.html", "openwebmath_score": 0.997959315776825, "openwebmath_perplexity": 1581.6720373769479, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363485313248, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8653761096029536}}
{"url": "https://community.freefem.org/t/spatial-co-ordinate-of-maximum-value-of-variable/1364", "text": "# Spatial Co-ordinate of Maximum Value of Variable\n\nSay I have the following code:\n\nint nn = 50;\nmesh Th = square(nn, nn);\nfespace Vh(Th, P1);\nVh u;\nu = sin(x);\n\nI can get the maximum value of \u2018u\u2019 by:\nreal maxValueU = u[].max;\n\nHow do I get the values of x and y coordinates where maximum value of \u2018u\u2019 occurs?\n\nyes but be careful because the max could be not unique.unc\n\nthe method imax give the dof number\n\nint dofmaxValueU = u[].imax;\n\nNow for P1 finite element the dof are vertices with same numbering so\n\nTh(dofmaxValueU).x\nTh(dofmaxValueU).y\n\ngiven the coordinate.", "date": "2022-05-18 09:02:41", "meta": {"domain": "freefem.org", "url": "https://community.freefem.org/t/spatial-co-ordinate-of-maximum-value-of-variable/1364", "openwebmath_score": 0.8465755581855774, "openwebmath_perplexity": 5285.516652235383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9719924810166349, "lm_q2_score": 0.8902942195727173, "lm_q1q2_score": 0.8653592873172542}}
{"url": "https://www.physicsforums.com/threads/fixed-cost-cost-per-person.919820/", "text": "# Fixed Cost & Cost Per Person\n\n1. Jul 10, 2017\n\n### zak100\n\n1. The problem statement, all variables and given/known data\nA group can charter a particular aircraft at a fixed total cost. If 36 people charter the aircraft rather than 40 people, then the cost per person is greater by 12$a) What is the fixed total cost to charter the aircraft? b) What is the cost per person if 40 people charter the aircraft? 2. Relevant equations ThirtySixCost = 36 * (costPerPerson + 12) FixedCost = NumberofPersons * costPerPerson (Note if NumberofPersons>40) Not sure about the correctness of equations 3. The attempt at a solution I dont know how to solve it because data is incomplete. ThirtySixCost=? FortyCost=? Somebody please guide me how to solve it. Zulfi. 2. Jul 10, 2017 ### scottdave Reread the problem statement. Your costPerPerson represents the cost per person, when there are 40 people. Your ThirtySixCost is the total cost to rent the plane (but that is a Fixed cost). I did not see anything about greater than 40. You can get to 3 equations and 3 unknowns to solve it. For example (not real numbers), if it costs$360 to rent something (say a lake house), then for 40 people, that is $9 per person, but for 36 people it is$10 per person. It costs $1 more per person if you only have 36 people. Last edited: Jul 10, 2017 3. Jul 11, 2017 ### zak100 Hi, Please tell me how to find costPerPerson? Zulfi. 4. Jul 11, 2017 ### Mark44 ### Staff: Mentor These variable names aren't useful. The fixed cost is the same whether there are 40 people or 36 people. Let x = the cost per person with 40 passengers. Let f = the fixed cost of chartering the airplane. \u2022 Write an expression that represents the cost per person with 36 passengers. (Use the information given in the problem.) \u2022 Translate the 2nd sentence in the problem statement into an algebraic equation. 5. Jul 11, 2017 ### zak100 Hi, FTC = Fixed Total Cost CPP = Cost Per Person CPP = FTC + 12 I can form only one equation. Please guide me. Zulfi. 6. Jul 11, 2017 ### scottdave The plane costs an amount (call it FTC), whether there is 1 passenger or 36 passenger or 40 passengers. The cost per person depends on the fixed cost (which does not change, and the number of passengers. How about this?: FTC is total fixed cost. CPPForty = FTC / 40. CPPthirtysix = FTC / 36. The problem statement gives you another relationship between CPPForty and CPPthirtysix, which you can make another equation. You may want to re-look at the example problem I posed in post #2, to see how you can relate your problem to that one. 7. Jul 11, 2017 ### Mark44 ### Staff: Mentor The last equation makes no sense. You are saying that the cost for one person is$12 more than the fixed cost.\n\n8. Jul 11, 2017\n\n### zak100\n\nHi,\nMr. scottdave:\nDo you mean this:\n\nCPPThirtySix = CPPForty + 12\n\nZulfi.\n\n9. Jul 11, 2017\n\n### scottdave\n\nYes, now with 3 equations you should be able to solve for CPPThirtySix, CPPForty and FTC.\n\n10. Jul 12, 2017\n\n### zak100\n\nHi,\nI am able to solve it.\nCP36 = FTC/36 ---(1)\nCP40 = FTC/40 ---(2)\nCP36 = CP40 + 12--- (3)\nAfter solving these eq., i am able to get correct answer:\nCP40 = 108\nCP36 = 120 (Not required in the question)\nFTC = 4320", "date": "2017-08-18 01:55:41", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/fixed-cost-cost-per-person.919820/", "openwebmath_score": 0.2869255542755127, "openwebmath_perplexity": 5311.896602003922, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9736446502128796, "lm_q2_score": 0.888758798648752, "lm_q1q2_score": 0.8653352496339832}}
{"url": "https://www.physicsforums.com/threads/table-probability-question.735574/", "text": "# Homework Help: Table probability question\n\n1. Jan 29, 2014\n\n### joshmccraney\n\n1. The problem statement, all variables and given/known data\n11 people sit at two round tables, one sits 5 and the other sits 6. how many combinations of seating arrangements are there?\n\n2. Relevant equations\ni'm not sure about equations, but my solution attempt may have some info.\n\n3. The attempt at a solution\ni know if 11 people were to sit at 1 table, we would have $\\frac{11!}{11}=10!$ combinations. thus, would we first have to choose who sits at which table? i.e: $$(11C5)(4!)+(11C6)(6!)$$\n\nthanks for your time (and help!)\n\n2. Jan 29, 2014\n\n### haruspex\n\nYou're close, but a couple of errors.\nYou got 11C54! for choosing the 5 to sit at one table and arranging them. For each such arrangement, how can you arrange the remaining 6?\n\n3. Jan 30, 2014\n\n### joshmccraney\n\nwait, do you mean $(11C5)(4!)+(11C6)(5!)$? also, if we've already selected the 5 to be at on table, do we really need to select 6 for the other?\n\nthanks for the response, and let me know what you think.\n\n4. Jan 30, 2014\n\n### joshmccraney\n\nbecause i definitely should have done $5!$, not $6!$ (brain fart)\n\n5. Jan 30, 2014\n\n### D H\n\nStaff Emeritus\nThere are two big problems with your result. One is that when you've chosen the five people who will sit at table #1 you have but one choice as to which set of six people will sit at table #2. The other: Why are you adding?\n\n6. Jan 30, 2014\n\n### joshmccraney\n\nyea, i definitely see this problem but if i dont put both \"choices\" how do i determine which number (5 or 6) to use?\n\nalso, i was adding because i figured after i choose who sits at what table, all i need to do is take the possible combos of one table and add them to the combos of another table.\n\nevidently this is wrong; can you direct me as to what to do next?\n\n7. Jan 30, 2014\n\n### D H\n\nStaff Emeritus\nThere are a number of right ways to address this part of the problem. You chose a wrong way.\n\nOne way to look at this part of the problem is that you choose five people to sit at one table, call it table A. There are 11 choose 5 ways to do this. For any give choice, there are 11-5=6 people left. You now need to choose six of these six people to sit at table B. There are 6 choose 6 ways to do this, and that of course is one. The number of ways to choose people to sit at the two tables is the product of these two numbers: (11 choose 5)*(6 choose 6).\n\nYou could of course look at it the other way around: You'll choose six people to sit at the larger table and then choose five of the remaining five to sit at the smaller table. This leads to (11 choose 6)*(5 choose 5) ways to split the people into two groups.\n\nHmmm. One approach yields (11 choose 5)*(6 choose 6), the other (11 choose 6)*(5 choose 5). Is there a problem here? The answer is no. 11 choose 5 and 11 choose 6 are the same number (462), as are 6 choose 6 and 5 choose 5 (which are of course both 1).\n\nSuppose X and Y are independent. Also suppose there are three ways to accomplish X and five ways to accomplish Y. The number of ways to do accomplish X and Y is fifteen, not eight.\n\nThe same applies to the problem at hand. You should be multiplying here, not adding.\n\n8. Jan 30, 2014\n\n### CAF123\n\nYou can do it either way (i.e choose 6 for one table and have one set of five for the other, or choose 5 for a table and have one set of six for the other). The result is the same.\n\nConsider the case when there are two people situated at one table and one at the other. How many possible combinations are there?\n\nAlternatively, are the combinations around each individual table dependent on each other? What does this tell you?", "date": "2018-06-24 09:27:55", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/table-probability-question.735574/", "openwebmath_score": 0.6653469204902649, "openwebmath_perplexity": 408.5409577811819, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97364464791863, "lm_q2_score": 0.8887587979121384, "lm_q1q2_score": 0.8653352468777489}}
{"url": "https://www.physicsforums.com/threads/solving-the-following-equation-for-x-x-5-2-9.876632/", "text": "# Solving the following equation for x: $(x-5)^{-2} = 9$\n\n1. Jun 24, 2016\n\n### malknar\n\n1. The problem statement, all variables and given/known data\nAn exercise in Lang's Basic Mathematics which asked to solve for x in the equation $(x - 5)^{-2} = 9$\n\nI tried to solve it and got $x = \\frac {16} 3$\n\nwhich is correct according to the book, but I've found that the book states that x also equals \\frac {14}3, and I can't figure out how.\n2. Relevant equations\n\n3. The attempt at a solution\n$(x-5)^{-2} = 9$\n\nThis yields: $x-5 = 9^{-1/2} =\\frac 1 {9^{1/2}} = \\frac13$\n\nWhich yields: $x=\\frac 13 +5 = \\frac {1+15} 3 =\\frac {16}3$\n\nBut I can't figure out how can x equals $\\frac {14}3$ too.\n\n2. Jun 24, 2016\n\n### Staff: Mentor\n\nIt's better to write this as $(x - 5)^2 = \\frac 1 9$, which leads to $x - 5 = \\pm \\sqrt{\\frac 1 9}$\nQuadratic equations can have two solutions.\n\n3. Jun 24, 2016\n\n### mfig\n\nWhat is $(-\\frac{1}{3})^2$?\n\n4. Jun 24, 2016\n\n### late347\n\nI am thinking about that old math movie with young actor Edward James Olmos. He was playing the role of high school teacher.\n\nStand and deliver\n\n\"Negative times negative is positive!!!\" The result would be 1/9\n\nFor the original poster's benefit. The following is compiled from my own high school era old math factbook.\n\nsecond degree equations.\n\u2022 you can evaluate whether the quadratic has two roots, or one root, or none of the real-numbered roots. (roots belonging to \u211d)\n\u2022 calculate the quadratic into the form of: $$Ax^2 + Bx+C= 0$$\n$$x=\\frac{-B\u00b1\\sqrt{B^2-4AC}}{2A}$$\n\nthis is called the discriminant part of the equation $D=\\sqrt{B^2-4AC}$\n\nif D>0, then the equation has two non-equal roots\nif D=0, then the equation has a twin-root\nif D<0, then the equation doesn't have real-numbered roots (roots in the real numbers \u211d)\nThis last fact is because negative inside the squareroot sign is not defined.\nE.g. $\\sqrt{(-5)}= ~~not ~~defined~~ in~~ real~~ numbers$\n\nnegative inside the squareroot is defined in the complex numbers, I think, but I have no knowledge of the complex numbers myself. (complex numbers symbol is this one \u2102)\n\n5. Jun 24, 2016\n\n### epenguin\n\nIn short if A2 = B, then A = \u00b1 \u221aB\n\n6. Jun 24, 2016\n\n### malknar\n\nI got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering $-3 = 9^{1/2}$\n\n7. Jun 24, 2016\n\n### Staff: Mentor\n\nThe symbol $\\sqrt{x}$ denotes the positive square root of x, by established convention. In this case $\\sqrt{9} = +3$. However, the equation $x^2 = 9$ has two solutions: 3 and -3.\n\n8. Jun 24, 2016\n\n### Ray Vickson\n\nNo: $9^{1/2}$ is $+3$, NOT $-3$. Essentially by definition, the functions $x^{1/2}$ and $\\sqrt{x}$ mean the positive root.\n\nThere is no really good language for it, but saying \"a\" square root instead of \"the\" square root comes close to capturing it: there are two roots to the equation $x^2 = a \\:(a > 0)$; these are $x = \\sqrt{a}$ and $x = -\\sqrt{a}$\n\n9. Jun 24, 2016\n\n### late347\n\n\u2022 watch this khan academy video it will be helpful for this problem and also for the purpose of increased understanding.\n\u2022 review what is the binomial square formula such as:\n\u2022 $(a+b)^2= a^2+2ab+b^2$\n\u2022 $(a-b)^2 = a^2-2ab+b^2$\n\n10. Jul 6, 2016\n\n### James R\n\nAlternatively you could do this:\n$\\frac{1}{(x-5)^2}=9$\n$9(x-5)^2 = 1$\n$9(x^2 - 10 x + 25) = 1$\n$9x^2 - 90 x + 224 = 0$\n$(3x - 14)(3x - 16)=0$\n$x=\\frac{14}{3}$ or $x=\\frac{16}{3}$.", "date": "2017-08-21 05:02:04", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/solving-the-following-equation-for-x-x-5-2-9.876632/", "openwebmath_score": 0.6323896050453186, "openwebmath_perplexity": 1349.4600987906697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446463891303, "lm_q2_score": 0.8887587964389113, "lm_q1q2_score": 0.8653352440839929}}
{"url": "https://www.mathalino.com/forum/calculus/find-differential-equations-following-family-curves", "text": "# Find the differential equations of the following family of curves.\n\n16 posts / 0 new\nBingo\nFind the differential equations of the following family of curves.\n\nDifferential Equations\nFind the differential equations of the following family of curves.\n\n1. Parabolas with axis parallel to the y \u2013 axis with distance vertex to focus fixed as a.\n2. Parabolas with axis parallel to the x \u2013 axis with distance vertex to focus fixed as a.\n3. All ellipses with center at the origin and axes on the coordinate axes.\n4. Family of cardioids.\n5. Family of 3 \u2013 leaf roses.\n\nBingo\n\n1\n\nJhun Vert\n\n(1) Upwards and downwards parabolas with latus rectum equal to 4a.\n$y - k = \\pm 4a(x - h)^2$\n\n$y' = \\pm 8a(x - h)$\n\n$y'' = \\pm 8a$\n\nBingo\n\ny\u2212k=\u00b14a(x\u2212h)2 is this the equation for parabolas?\nisn't it 4a(y-k)=(x-h)2 ?\n\nJhun Vert\n\nI made a mistake in there, it should be (x - h)2 = \u00b14a(y - k). The solution for (1) should go this way:\n$(x - h)^2 = \\pm 4a(y - k)$\n\n$2(x - h) = \\pm 4ay'$\n\n$2 = \\pm 4ay''$\n\n$y'' = \\pm \\frac{1}{2a}$\n\nBingo\n\ny\u2212k=\u00b14a(x\u2212h)2 is this the equation for parabolas?\nisn't it 4a(y-k)=(x-h)2 ?\n\nJhun Vert\n\nBingo\n\nso for number 2,\nx-h=\u00b14a(y-k)2\nx'=\u00b18a(y-k)\nx''=\u00b18a\nam i wrong?\n\nJhun Vert\n\nIt is better to express your answer in terms of y' rather than x'. Although x' will do and simpler.\n$(y \u2212 k)^2 = \\pm 4a(x \u2212 h)$\n\n$2(y \u2212 k)y' = \\pm 4a$\n\n$y' = \\dfrac{\\pm 2a}{y - k}$\n\nNote: $y - k = \\dfrac{y'}{\\pm 2a}$\n\n$y'' = \\dfrac{\\mp 2a \\, y'}{(y - k)^2}$\n\n$y'' = \\dfrac{\\mp 2a \\, y'}{\\left( \\dfrac{y'}{\\pm 2a} \\right)^2}$\n\n$y'' = \\dfrac{\\mp 8a}{y'}$\n\n$y'' \\, y' = \\pm 8a$\n\nChaCha Ingal Cortez\n\nour prof. gave the same question but the ans. he gave is different from your ans. in no 2, his answer is (y')^3 + 2ay\"=0 are they the same with your answer sir ?\n\nJhun Vert\n\nThey are different. I made a mistake in $y - k$ of the above solution.\n$(y \u2212 k)^2 = \\pm 4a(x \u2212 h)$\n\n$2(y - k) \\, y' = \\pm 4a$\n\n$y' = \\dfrac{\\pm 2a}{y - k}$\n\n$y - k = \\dfrac{\\pm 2a}{y'}$\n\n$y'' = \\dfrac{\\mp 2a \\, y'}{(y - k)^2}$\n\n$y'' = \\dfrac{\\mp 2a \\, y'}{\\left( \\dfrac{\\pm 2a}{y'} \\right)^2}$\n\n$y'' = \\dfrac{\\mp 2a \\, y'}{\\dfrac{4a^2}{(y')^2}}$\n\n$y'' = \\dfrac{\\mp (y')^3}{2a}$\n\n$2a \\, y'' = \\mp (y')^3$\n\n$\\pm (y')^3 + 2a \\, y'' = 0$\n\nBingo\n\nThank You so much sir.\n\nBingo\n\nsir, do you know the standard or general equations for the last 3 problems?\nwhat equations should i use?\n\nShandy\n\nI think you are from ADZU haha\n\nAlegnaaa (guest)\n\nGood evening, sir. Do you have the solution for the last three problems? Thank you!\n\nAncha (guest)\n\nSir do you know how to eliminate arbitrary constant here in this equation?\n(y-33)\u00b2=4a(x-h)?\n\n\u2022 Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and $...$ for displayed mathematics, and $...$ and $...$ for in-line mathematics.", "date": "2023-03-30 09:11:44", "meta": {"domain": "mathalino.com", "url": "https://www.mathalino.com/forum/calculus/find-differential-equations-following-family-curves", "openwebmath_score": 0.708338737487793, "openwebmath_perplexity": 1839.6271807371024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446471538802, "lm_q2_score": 0.8887587942290706, "lm_q1q2_score": 0.8653352426120714}}
{"url": "https://math.stackexchange.com/questions/2377313/how-many-four-digit-integers-are-there-that-do-not-have-two-consecutive-equal-ev", "text": "# How many four digit integers are there that do not have two consecutive equal even digits?\n\nHow many four digit integers are there that do not have two consecutive equal even digits?\n\nMy Attempt There are $9*10^3$ four digit numbers. We tie two equal even digits and form the four digit numbers with them. This can be done in $5 \\cdot \\binom{3}{2} \\cdot 10 \\cdot 10 = 1500$ ways. There are $10 \\cdot 10$ numbers starting with $00$ so we must subtract them from $1500$. So answer is $9000 - 1400 = 7600$. But it is not matching with the given answer $7801$. Can any one help me to find out where I am making mistakes? Thanks in advance.\n\n1. You did not account for the fact that there are only $9$ choices for the leading digit when neither number in the pair of consecutive equal even digits is in the thousands place.\n2. You have subtracted numbers in which there are three or more consecutive even digits more than once.\n\nFirst, we observe that there are $9 \\cdot 10 \\cdot 10 \\cdot 10 = 9000$ four-digit positive integers.\n\nLet $A_1$ be the set of four-digit positive integers in which the thousands place and hundreds place contain equal even digits. Let $A_2$ be the set of four-digit positive integers in which the hundreds place and tens place contain equal even digits. Let $A_3$ be the set of four-digit positive integers in which the tens place and units place contain equal even digits. Then $A_1 \\cup A_2 \\cup A_3$ is the set of four-digit positive integers that do contain consecutive equal even digits. By the Inclusion-Exclusion Principle, the number of four-digit positive integers that do contain consecutive even digits is $$|A_1 \\cup A_2 \\cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2| - |A \\cap A_3| - |A_2 \\cap A_3| + |A_1 \\cap A_2 \\cap A_3|$$\n\n$|A_1|$: Since we cannot use zero, there are four ways of choosing the even digit that occupies both the thousands and hundreds place. There are ten choices for each of the remaining digits. Hence, $|A_1| = 4 \\cdot 10 \\cdot 10 = 400$.\n\n$|A_2|$: Since we cannot use zero, the thousands place can be filled in nine ways. There are five ways to choose the even digit that fills both the hundreds and tens places. There are ten ways to fill the units place. Hence, $|A_2| = 9 \\cdot 5 \\cdot 10 = 450$.\n\n$|A_3|$: We can fill the thousands place in nine ways and the hundreds place in ten ways. There are five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \\cdot 10 \\cdot 5 = 450$.\n\n$|A_1 \\cap A_2|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \\cap A_2| = 4 \\cdot 10 = 40$.\n\n$|A_1 \\cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \\cap A_3| = 4 \\cdot 5 = 20$.\n\n$|A_2 \\cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \\cap A_3| = 9 \\cdot 5 = 45$.\n\n$|A_1 \\cap A_2 \\cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \\cap A_2 \\cap A_3| = 4$.\n\nTherefore, the number of four-digit positive integers that do contain consecutive even equal digits is \\begin{align*} |A_1 \\cup A_2 \\cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2| - |A \\cap A_3| - |A_2 \\cap A_3| + |A_1 \\cap A_2 \\cap A_3|\\\\ & = 400 + 450 + 450 - 40 - 20 - 45 + 4\\\\ & = 1199 \\end{align*} Therefore, the number of four-digit positive even integers that do not contain consecutive equal even digits is $9000 - 1199 = 7801$.\n\n\u2022 Thanks for pointing the errors and nice step by step solution. \u2013\u00a0rugi Jul 31 '17 at 9:05\n\nThis where you're making a mistake $-$\n\nWhat you're trying to do is called the inclusion-exclusion principle. Your error is in the fact that you are overcounting numbers such as $4446$. If I understand your method correctly, you are blacklisting this number twice instead of once.\n\nIMO, an easier way to solve the problem is to count the numbers that are not to be included first by going through all possible digit combinations e.g. $aabb$ etc.", "date": "2021-03-08 10:59:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2377313/how-many-four-digit-integers-are-there-that-do-not-have-two-consecutive-equal-ev", "openwebmath_score": 0.8147929310798645, "openwebmath_perplexity": 38.07045781752854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446456243805, "lm_q2_score": 0.8887587934924569, "lm_q1q2_score": 0.8653352405355151}}
{"url": "https://math.stackexchange.com/questions/3218664/if-lim-x-rightarrow-0-fracfxx2-5-then-what-is-lim-x-rightarr/3218668", "text": "# If $\\lim_{x \\rightarrow 0}\\frac{f(x)}{x^2} = 5$, then what is $\\lim_{x \\rightarrow 0}f(x)$?\n\nI know the answer to the above question, but I have a question on some of the reasoning.\n\nThe way I know how to solve it is $$\\lim_{x \\rightarrow 0}f(x) = \\lim_{x \\rightarrow 0}\\left(f(x)\\cdot \\frac{x^2}{x^2}\\right) = \\lim_{x \\rightarrow 0}\\left(\\frac{f(x)}{x^2}\\cdot x^2\\right) = \\left(\\lim_{x \\rightarrow 0}\\frac{f(x)}{x^2}\\right)\\left(\\lim_{x \\rightarrow 0}x^2\\right) = 5\\cdot0 = 0.$$\n\nI saw another solution elsewhere that gets the right answer, but I am unsure if the steps are actually correct. \\begin{align*} &\\lim_{x \\rightarrow 0}\\frac{f(x)}{x^2} = 5 \\\\ \\Longrightarrow &\\frac{\\lim_{x \\rightarrow 0}f(x)}{\\lim_{x \\rightarrow 0}x^2} = 5 \\\\ \\Longrightarrow &\\lim_{x \\rightarrow 0}f(x) = 5\\cdot \\lim_{x \\rightarrow 0}x^2\\\\ \\Longrightarrow &\\lim_{x \\rightarrow 0}f(x) = 5\\cdot 0 = 0. \\end{align*}\n\nMy issue is with that first step. I know that $$\\lim_{x \\rightarrow a} \\frac{f(x)}{g(x)} = \\frac{\\lim_{x \\rightarrow a}f(x)}{\\lim_{x \\rightarrow a}g(x)}$$, but only when $$\\lim_{x \\rightarrow a}g(x) \\neq 0$$. Since $$\\lim_{x \\rightarrow 0}x^2 = 0$$, wouldn't this invalidate the above work? However, it still got the same answer, so my real question is why did it work and when will it work in general?\n\nEDIT: Does anyone have a nice example for when the logic in the second method doesn't work?\n\n\u2022 You are correct that the solution you saw elsewhere is invalid. It is invalid for the exact reason you think it is. Using false logic to get a correct answer is still false logic. Applied to another problem, it might fail. \u2013\u00a0InterstellarProbe May 8 at 16:38\n\u2022 Do you have an example of another problem in which logic from the second method fails? \u2013\u00a0Smash May 8 at 16:45\n\u2022 Simply put, you can't divide by zero. If you want examples, there are many \"proofs\" of $0=1$ where the trick is to divide by zero and hope no one notices. \u2013\u00a0Th\u00e9ophile May 8 at 16:47\n\nIf $$\\lim\\limits_{x\\to0}\\dfrac{f(x)}{x^2}=5$$, then for every $$n\\in\\Bbb N$$ there is some $$x\\neq 0$$ such that $$5-\\frac1n<\\frac{f(x)}{x^2}<5+\\frac1n$$ and hence $$5x^2-\\frac{x^2}n If $$x\\to 0$$ we get $$f(x)\\to 0$$.\n\n$$\\frac{\\lim_{x \\rightarrow 0}f(x)}{\\lim_{x \\rightarrow 0}x^2} = 5$$ doesn't work, the LHS is not defined.\n\nBut for the limit of $$\\dfrac{f(x)}{x^2}$$ to exist, $$\\lim_{x\\to0}f(x)$$ must be zero, which is also the limit of $$x^2$$.\n\nYou can not do this since the denominator on the right is $$0$$.\n\n$$\\lim_{x \\rightarrow 0}\\frac{f(x)}{x^2} = \\frac{\\lim_{x \\rightarrow 0}f(x)}{\\lim_{x \\rightarrow 0}x^2}$$\n\nBut first aproach is perfect.\n\n\u2022 You meant \"denominator,\" not numerator. \u2013\u00a0Mark Viola May 8 at 16:41", "date": "2019-09-22 10:37:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3218664/if-lim-x-rightarrow-0-fracfxx2-5-then-what-is-lim-x-rightarr/3218668", "openwebmath_score": 0.9982571601867676, "openwebmath_perplexity": 245.1549770234237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446448596304, "lm_q2_score": 0.888758789809389, "lm_q1q2_score": 0.8653352362698375}}
{"url": "https://mathhelpboards.com/threads/probability-of-winning-dice-game.24336/", "text": "# Probability of winning dice game\n\n#### TheFallen018\n\n##### Member\nHey, so I've got this problem that I'm trying to figure out. I've worked out something that I think is probably right through simulation, but I'm not really sure how to tackle it from a purely mathematical probability perspective. So, would anyone know how I should approach this? I've tried a few different things, but my two answers tend to conflict a little. Thanks,\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nThe starting point (which I guess you already know) is to make a list of the probabilities $p(x)$ of rolling a total of $x$ with the two dice, where $x$ goes from $2$ to $12$. These are $$\\begin{array}{r|ccccccccccc}x& 2&3&4&5&6 &7&8&9&10 &11&12\\\\ \\hline p(x) & \\frac1{36} & \\frac2{36} & \\frac3{36} & \\frac4{36} & \\frac5{36} & \\frac6{36} & \\frac5{36} & \\frac4{36} & \\frac3{36} & \\frac2{36} & \\frac1{36} \\end{array}.$$\n\nThe probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\\dfrac{244}{495} \\approx 0.49292929\\ldots$.\n\n#### TheFallen018\n\n##### Member\nThe starting point (which I guess you already know) is to make a list of the probabilities $p(x)$ of rolling a total of $x$ with the two dice, where $x$ goes from $2$ to $12$. These are $$\\begin{array}{r|ccccccccccc}x& 2&3&4&5&6 &7&8&9&10 &11&12\\\\ \\hline p(x) & \\frac1{36} & \\frac2{36} & \\frac3{36} & \\frac4{36} & \\frac5{36} & \\frac6{36} & \\frac5{36} & \\frac4{36} & \\frac3{36} & \\frac2{36} & \\frac1{36} \\end{array}.$$\n\nThe probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\\dfrac{244}{495} \\approx 0.49292929\\ldots$.\nWow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nWow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks\nIn the expression ${\\color {red} p(x)}{\\color {green}\\dfrac{p(x)}{p(x)+p(7)}}$, the red $\\color {red} p(x)$ gives the probability that the first roll of the dice gives the value $x$. The green fraction represents the probability of rolling $x$ again before rolling a $7$. My argument for that is that after rolling the first $x$, you can completely disregard any subsequent rolls until either an $x$ or a $7$ turns up. The only question is, which one of those will appear first. The relative probabilities of $x$ and $7$ are in the proportion $p(x)$ to $p(7)$. So out of a combined probability of $p(x) + p(7)$, the probability of an $x$ is $\\dfrac{p(x)}{p(x)+p(7)}$, and the probability of a $7$ is $\\dfrac{p(7)}{p(x)+p(7)}$.\n\nI hope that makes sense.", "date": "2020-07-14 10:48:01", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/probability-of-winning-dice-game.24336/", "openwebmath_score": 0.8272463083267212, "openwebmath_perplexity": 124.50716885724312, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986151391819461, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8653249489450595}}
{"url": "https://mathhelpboards.com/threads/im-confuseds-question-on-yahoo-answers-involving-a-recurrence-relation.3088/", "text": "I'm Confused's question on Yahoo! Answers involving a recurrence relation\n\nMarkFL\n\nStaff member\nJonah must take an antibiotic every 12 hours. Each pill is 25 milligrams, and after every 12 hours, 50% of the drug remains in his body. What is the amount of antibiotic in his body over the first two days? What amount will there be in his body in the long run?\n\nHere's what I did:\n\nFor my recursive formula, I tried inputting u(n)=u(n-1)*(1-0.50). However, I could not seem to find the long run value.,,, It keeps changing. Am I right on this?\n\nI appreciate your help! Thank you so much!\n\nSource: <Advanced Algebra: An Investigative Approach; Murdock Jerald, 2004>\nI posted a link to this topic, so the OP could find my response.\n\nHere is a link to the original question:\n\nRecursive formula help on real world situations? - Yahoo! Answers\n\nLet $A(t)$ represent the amount, in mg, of antibiotic in Jonah's bloodstream at time $t$, measured in hours. With a half-life of 12 hours, we may state:\n\n$\\displaystyle A(t)=A_02^{-\\frac{t}{12}}$\n\nSo, for the first 12 hours, but not including the second dose, we find:\n\n$\\displaystyle A_0=25$ hence:\n\n$\\displaystyle A(t)=25\\cdot2^{-\\frac{t}{12}}$\n\n$\\displaystyle A(t)=25\\cdot2^{-\\frac{12}{12}}=\\frac{25}{2}$\n\nNow, for the second 12 hours, we have:\n\n$\\displaystyle A_0=\\frac{25}{2}+25=\\frac{75}{2}$ hence:\n\n$\\displaystyle A(t)=\\frac{75}{2}\\cdot2^{-\\frac{t}{12}}$\n\n$\\displaystyle A(12)=\\frac{75}{2}\\cdot2^{-\\frac{12}{12}}=\\frac{75}{4}$\n\nFor the third 12 hours, we have:\n\n$\\displaystyle A_0=\\frac{75}{4}+25=\\frac{175}{4}$ hence:\n\n$\\displaystyle A(t)=\\frac{175}{4}\\cdot2^{-\\frac{t}{12}}$\n\n$\\displaystyle A(12)=\\frac{175}{4}\\cdot2^{-\\frac{12}{12}}=\\frac{175}{8}$\n\nFor the fourth 12 hours, we have:\n\n$\\displaystyle A_0=\\frac{175}{8}+25=\\frac{375}{8}$ hence:\n\n$\\displaystyle A(t)=\\frac{375}{8}\\cdot2^{-\\frac{t}{12}}$\n\n$\\displaystyle A(12)=\\frac{375}{8}\\cdot2^{-\\frac{12}{12}}=\\frac{375}{16}$\n\nSo, now we may try to generalize for the $n$th 12 hour period.\n\nLet $I_n$ represent the initial amount for each 12 hour period. We see we must have the recursion:\n\n$I_{n+1}=\\frac{1}{2}I_{n}+25$\n\nThis is an inhomogeneous recurrence, so let's employ symbolic differencing to obtain a homogeneous recurrence:\n\n$I_{n+2}=\\frac{1}{2}I_{n+1}+25$\n\nSubtracting the former from the latter, we obtain:\n\n$2I_{n+2}=3I_{n+1}-I_{n}$\n\nThe characteristic equation is:\n\n$2r^2-3r+1=0$\n\n$(2r-1)(r-1)=0$\n\nHence, the closed-form will be:\n\n$I_n=k_1\\left(\\frac{1}{2} \\right)^n+k_2$\n\nWe may use our data above to determine the parameters $k_i$:\n\n$I_1=k_1\\left(\\frac{1}{2} \\right)^1+k_2=25$\n\n$I_2=k_1\\left(\\frac{1}{2} \\right)^2+k_2=\\frac{75}{2}$\n\nor\n\n$\\frac{1}{2}\\cdot k_1+k_2=25$\n\n$\\frac{1}{4}\\cdot k_1+k_2=\\frac{75}{2}$\n\nSolving this system, we find:\n\n$k_1=-50,k_2=50$ and so we have:\n\n$I_n=-50\\left(\\frac{1}{2} \\right)^n+50=50\\left(1+\\left(\\frac{1}{2} \\right)^n \\right)$\n\nAnd so, for the $n$th 12 hour period, we have:\n\n$\\displaystyle A_n(t)=I_n2^{-\\frac{t}{12}}$\n\nNow, for the long run, which we may take as implying as n grows without bound, we find:\n\n$\\displaystyle \\lim_{n\\to\\infty}I_n=50$\n\nand so we find that for the long run the initial amount for a 12 hour period is is 50 mg and the final amount for that period is 25 mg.\n\nLast edited:\n\ntkhunny\n\nWell-known member\nMHB Math Helper\nThat is good mathematics, but not so great reality. If it had said I.V. Administration, the assumption of immediate absorption and distribution would be more realistic. Even subcutaneous or intramuscular injection takes a little while. A pill? No.", "date": "2021-06-24 18:06:58", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/im-confuseds-question-on-yahoo-answers-involving-a-recurrence-relation.3088/", "openwebmath_score": 0.8736265897750854, "openwebmath_perplexity": 1111.203420330001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864284905088, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8653245953946525}}
{"url": "https://brilliant.org/discussions/thread/polynomial-sprint-useful-lemma/", "text": "# Polynomial Sprint: Useful Lemma\n\n$\\color{#D61F06} {\\text{Unlocked!}}$\n\nMathematicians have a huge bag of tricks, which provide them with various ways of approaching a problem. In this note, we introduce an extremely useful lemma, which applies to polynomials with integer coefficients.\n\nLemma. If $f(x)$ is a polynomial with integer coefficients, then for any integers $a$ and $b$, we have $a-b \\mid f(a) - f(b)$\n\n1) Show that for any integers $a, b$ and $n$, we have\n\n$a-b | a^n - b^n.$\n\n2) Using the above, prove the lemma.\n\n3) Come up with a (essentially) two-line solution to question 4 from the book.\n\nNote: In Help Wanted, Krishna Ar mentioned that problem 4 from the book was hard to solve. If you looked at the solution, it was also hard to understand what exactly was happening. This provides us with a simple way of comprehending why there are no solutions.\n\n4) Come up with an alternative solution to E6 (page 250, from the book) using this lemma.\n\n5) * (Reid Barton) Suppose that $f(x)$ is a polynomial with integer coefficients. Let $n$ be an odd positive integer. Let $x_1, x_2, \\ldots x_n$ be a sequence of integers such that\n\n$x_2 = f(x_1), x_3 =f(x_2), \\ldots x_n = f(x_{n-1}), x_1 = f(x_n) .$\n\nShow that all the $x_i$ are equal.\nWhere did you use the fact that $n$ is an odd positive integer?\n\nNote by Calvin Lin\n5\u00a0years, 9\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nI really like Polynomial Sprint. It would be nice if you can do some other themes using similar form.\n\n- 5\u00a0years, 8\u00a0months ago\n\nYes, I entirely agree! I find these sets fun to do and I want more sets.\n\n- 5\u00a0years, 8\u00a0months ago\n\nThis has been unlocked. Sorry for being late, I was out yesterday.\n\nStaff - 5\u00a0years, 9\u00a0months ago\n\nWhat does the $|$ mean?\n\n- 5\u00a0years, 9\u00a0months ago\n\n$a|b$ means $a$ divides $b$\n\n- 5\u00a0years, 9\u00a0months ago\n\n1) $a-b|a^n-b^n$, let $f(x)=x^n$, and plug in $a$ and $b$ into the function and we get $a-b|a^n-b^n$, is this correct?\n\n- 5\u00a0years, 9\u00a0months ago\n\nYou have to show that it is true of all polynomials, not just specially chosen polynomials.\n\nStaff - 5\u00a0years, 9\u00a0months ago\n\nThe lemma is true for polynomials with integer coefficients, so do you mean that I must show that this is true for non-integers too?\n\n- 5\u00a0years, 9\u00a0months ago\n\nI might have initially misinterpreted what you are trying to do.\n\nThe statement in question 1 is about integers (not polynomials), and you are asked to show that for any pair of integers, we have $a - b \\mid a^n - b^n$. For example, $10 - 7 \\mid 10^3 - 7^3$, or equivalently that $3 \\mid 657$.\n\nNote that you are not allowed to use the lemma, since we want to use question 1 to prove the lemma (in question 2).\n\nStaff - 5\u00a0years, 9\u00a0months ago\n\nOh, ok\n\n- 5\u00a0years, 9\u00a0months ago\n\n5) We see that $x_n-x_{n+1}\\mid f(x_n)-f(x_{n+1})=x_{n+1}-x_{n+2}$.\n\nThus we get the following divisibilitites:\n\n$x_1-x_2\\mid x_2-x_3\\implies x_2-x_3=k_1(x_1-x_2)$\n\n$x_2-x_3\\mid x_3-x_4\\implies x_3-x_4=k_2(x_2-x_3)$\n\n$\\vdots$\n\n$x_{n-1}-x_n\\mid x_n-x_1\\implies x_n-x_1=k_{n-1}(x_{n-1}-x_n)$\n\n$x_n-x_1\\mid x_1-x_2\\implies x_1-x_2=k_n(x_n-x_1)$\n\nThis means that $x_1-x_2=k_1k_2\\cdots k_n(x_1-x_2)$\n\nsince $x_1\\ne x_2$, we can divide both sides by $x_1-x_2$ to get $k_1k_2\\cdots k_n=1$\n\nSince we are working in the integers, this means that $|k_i|=1$ for all $i=1\\to n$.\n\nThus, $x_k-x_{k+1}=\\pm (x_{k-1}-x_k)$\n\nWhat I got so far.\n\n- 5\u00a0years, 9\u00a0months ago\n\nIf some $k_i=-1$, then we have $x_k=x_{k+1}$, which leads to equality of all $x_i$. If all k=1, then we can sum and get $x_1-x_2=0$, so $x_1=x_2$, which contradicts to your assumption that $x_1$ is different from $x_2$. Then that again leads to $x_1=x_2=...=x_n$\n\n- 5\u00a0years, 9\u00a0months ago\n\nIs it that simple? When did you use the condition that $n$ is odd?\n\n- 5\u00a0years, 9\u00a0months ago\n\nYou just consider it modulo |x2 - x1| :\n\nLet d equal |x1 \u2013 x2| and thus d = |x1 \u2013 x2| = |x2 \u2013 x3| = |x3 \u2013 x4| = \u2026 = |x{n-1} \u2013 x{n}| = |x{n} \u2013 x1|. If d is nonzero then x2 is congruent to (x1 + d) mod 2d. Similarly x3 is congruent to (x2 + d) mod 2d which is congruent to x1 mod 2d. In general x{k} is congruent to (x1 + d) mod 2d if k is even and congruent to x1 mod 2d if k is odd. Thus we have x{n} is congruent to x1 mod 2d because n is odd. However then |x{n} \u2013 x1| = 2dq for an integer q. Also d = |x{n} \u2013 x1|, hence d = 2dq and 1 = 2q. This contradicts the fact that q is an integer, hence our previous assumption that d is nonzero must be false. Therefore d = |x1 \u2013 x2| = |x2 \u2013 x3| = |x3 \u2013 x4| = \u2026 = |x{n-1} \u2013 x{n}| = |x{n} \u2013 x1| = 0 and x1 = x2 = x3 = \u2026=x{n} and all the x{i} are equal.\n\n- 5\u00a0years, 8\u00a0months ago\n\nI do not immediately see how we get $k_k = 0 \\Rightarrow x_k = x_{k+1}$ in the first scenario. I think we get $x_{k} = x_{k+2}$ instead, but that isn't immediately useful.\n\n@Daniel Liu You are nearly there. Let me phrase the rest of the question in a different way. You start at some point $X$ on the real number line. You take $n$ odd steps of size exactly $S$. If you end back at $X$, show that $S = 0$.\n\nStaff - 5\u00a0years, 9\u00a0months ago\n\nWe want another polynomial sprint! Soon..\n\n- 5\u00a0years, 8\u00a0months ago\n\nI guess We can write any polynomial a^n+b^n as (a-b)(a^n-1+a^n-2 b^1. .... Till b^n-1) therefore its always divisible by a-b\n\n- 5\u00a0years, 9\u00a0months ago\n\nWhere can I get the book?\n\n- 5\u00a0years, 9\u00a0months ago\n\n- 5\u00a0years, 9\u00a0months ago\n\n- 5\u00a0years, 9\u00a0months ago\n\n1)a^n-b^n=(a-b)(a^(n-1)+a^(n-2) b+...+b^(n-1)) Which leads us to:a-b/a^n-b^n 2) let f be a polynomial f (x)=an.x^n+an-1.x^(n-1)+...+a0 We can apply question1 since it is true for any n For j {0, n}: a-b / a^j-b^j We can sum Leads us to:a-b/f (a)-f (b) 5) x1-x2 =k1 (xn-x1) x2-x3=k2 (x1-x2) . . . xn-1-xn=kn (xn-2 -xn-1) Leads us to ki=1 or =-1 If one ki=-1 We have xi are equals If all ki=1 (x1-x2)+(x2-x3)+...+(xn-1 -xn)=(xn-x1)+ (x1-x2)+...+(xn-2 - xn-1) Leads us to: x1-xn = xn- xn-1 If just xi#xi-1 We have contradiction: (xi-xi-1)=0=(xi-1 -xi-2)#0 We can make recurrence(recurrance in french) We leads to an a contradiction each time Then k#1 Then all xi are equals\n\n- 5\u00a0years, 9\u00a0months ago\n\n1. Let $f(t)=t^n-b^n$. Then $f(b)=b^n-b^n=0$, so $t-b\\mid t^n-b^n$. Let $t=a$ to get $a-b\\mid a^n-b^n$.\n2. Consider the $k$th coefficient of $f(a)-f(b)$: $a_k(a^k-b^k)$. Since $a-b\\mid a^k-b^k$, $a-b$ is a factor of every term in $f(a)-f(b)\\implies a-b\\mid f(a)-f(b)$.\nFor the rest, I don't know which book you're talking about.\n\n- 5\u00a0years, 9\u00a0months ago\n\nProblem Solving Strategies by Arthur Engel. They might be selling online I don't really know. XD\n\n- 5\u00a0years, 9\u00a0months ago\n\nGoogle \"arthur engel problem solving strategies\", and follow the first link, which should be a pdf. The url is below but you have to surround the italicized text with underscores:\n\n- 5\u00a0years, 8\u00a0months ago\n\n- 5\u00a0years, 8\u00a0months ago\n\nOh oops, never mind the url.\n\n- 5\u00a0years, 8\u00a0months ago\n\nBook? What book?\n\n- 5\u00a0years, 9\u00a0months ago\n\nGoogle \"arthur engel problem solving strategies\", and follow the first link, which should be a pdf. The url is below but you have to surround the italicized text with underscores: http://f3.tiera.ru/2/MMathematics/MSchSchool-level/Engel%20A.%20Problem-solving%20strategies%20(for%20math%20olympiads)(Springer,%201998)(ISBN%200387982191)(O)(415s)MSch.pdf\n\n- 5\u00a0years, 8\u00a0months ago\n\nOh oops, never mind the url.\n\n- 5\u00a0years, 8\u00a0months ago\n\nThanks.. will do that. :)\n\n- 5\u00a0years, 8\u00a0months ago\n\nYou may refer to the original post Let's Get Started to understand the context of the Polynomial Sprint set.\n\nStaff - 5\u00a0years, 9\u00a0months ago", "date": "2020-04-06 13:11:54", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/polynomial-sprint-useful-lemma/", "openwebmath_score": 0.9731377363204956, "openwebmath_perplexity": 1537.4066754292521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9496693702514737, "lm_q2_score": 0.911179702173019, "lm_q1q2_score": 0.8653194539485763}}
{"url": "http://mathhelpforum.com/pre-calculus/66019-annual-compounding-interest.html", "text": "# Math Help - Annual Compounding Interest\n\n1. ## Annual Compounding Interest\n\nJane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.\n\na-10.50\nb-They will never have the same amount of money\nc-17.23\nd-17.95\n\nThanks\n\n2. Originally Posted by magentarita\nJane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.\n\na-10.50\nb-They will never have the same amount of money\nc-17.23\nd-17.95\n\nThanks\nIt will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.\n\n$6(1+.11)^t=8(1+.08)^t$\n\n$\\log 6(1.11)^t=\\log 8(1.08)^t$\n\nEtc. and so forth to find that $t \\approx 10.4997388$\n\n3. Hello, magentarita!\n\nI got a different result . . .\n\nJane has $6 and Sarah has$8. Over the next few years,\nJane invests her money at 11%. Sarah invests her money at 8%.\nWhen they have the same amount of money, how much will they have?\nAssume annual compounding interest, and round to the nearest cent.\n\n$a)\\;\\10.50 \\qquad b)\\;\\text{never equal}\\qquad c)\\;\\17.23 \\qquad d)\\;\\17.95$\n\nAt the end of $n$ years, Jane will have: . $6\\left(1.11^n\\right)$ dollars.\nAt the end of $n$ years, Sarah will have: . $8\\left(1.08^n\\right)$ dollars.\n\nSo we have: . $6(1.11^n) \\:=\\:8(1.08^n) \\quad\\Rightarrow\\quad \\frac{1.11^n}{1.08^n} \\:=\\:\\frac{8}{6} \\quad\\Rightarrow\\quad \\left(\\frac{1.11}{1.08}\\right)^n \\:=\\:\\frac{4}{3}$\n\nTake logs: . $\\ln\\left(\\frac{1.11}{1.08}\\right)^n \\:=\\:\\ln\\left(\\frac{4}{3}\\right) \\quad\\Rightarrow\\quad n\\cdot\\ln\\left(\\frac{1.11}{1.08}\\right) \\:=\\:\\ln\\left(\\frac{4}{3}\\right)$\n\n. . Hence: . $n \\;=\\;\\frac{\\ln\\left(\\frac{4}{3}\\right)}{\\ln\\left(\\ frac{1.11}{1.08}\\right)} \\;=\\;10.4997388$\n\nThey will have the same amount of money in about ${\\color{blue}10\\tfrac{1}{2}}$ years.\n\nAt that time, they will have: . $6(1.11^{10.5}) \\:=\\:8(1.08^{10.5}) \\:\\approx\\:{\\color{blue}\\17.95}\\;\\;{\\color{red}(d )}$\n\n4. Originally Posted by Soroban\nHello, magentarita!\n\nI got a different result . . .\n\nAt the end of $n$ years, Jane will have: . $6\\left(1.11^n\\right)$ dollars.\nAt the end of $n$ years, Sarah will have: . $8\\left(1.08^n\\right)$ dollars.\n\nSo we have: . $6(1.11^n) \\:=\\:8(1.08^n) \\quad\\Rightarrow\\quad \\frac{1.11^n}{1.08^n} \\:=\\:\\frac{8}{6} \\quad\\Rightarrow\\quad \\left(\\frac{1.11}{1.08}\\right)^n \\:=\\:\\frac{4}{3}$\n\nTake logs: . $\\ln\\left(\\frac{1.11}{1.08}\\right)^n \\:=\\:\\ln\\left(\\frac{4}{3}\\right) \\quad\\Rightarrow\\quad n\\cdot\\ln\\left(\\frac{1.11}{1.08}\\right) \\:=\\:\\ln\\left(\\frac{4}{3}\\right)$\n\n. . Hence: . $n \\;=\\;\\frac{\\ln\\left(\\frac{4}{3}\\right)}{\\ln\\left(\\ frac{1.11}{1.08}\\right)} \\;=\\;10.4997388$\n\nThey will have the same amount of money in about ${\\color{blue}10\\tfrac{1}{2}}$ years.\n\nAt that time, they will have: . $6(1.11^{10.5}) \\:=\\:8(1.08^{10.5}) \\:\\approx\\:{\\color{blue}\\17.95}\\;\\;{\\color{red}(d )}$\n\nYou are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.\n\n5. ## yes...\n\nOriginally Posted by Soroban\nHello, magentarita!\n\nI got a different result . . .\n\nAt the end of $n$ years, Jane will have: . $6\\left(1.11^n\\right)$ dollars.\nAt the end of $n$ years, Sarah will have: . $8\\left(1.08^n\\right)$ dollars.\n\nSo we have: . $6(1.11^n) \\:=\\:8(1.08^n) \\quad\\Rightarrow\\quad \\frac{1.11^n}{1.08^n} \\:=\\:\\frac{8}{6} \\quad\\Rightarrow\\quad \\left(\\frac{1.11}{1.08}\\right)^n \\:=\\:\\frac{4}{3}$\n\nTake logs: . $\\ln\\left(\\frac{1.11}{1.08}\\right)^n \\:=\\:\\ln\\left(\\frac{4}{3}\\right) \\quad\\Rightarrow\\quad n\\cdot\\ln\\left(\\frac{1.11}{1.08}\\right) \\:=\\:\\ln\\left(\\frac{4}{3}\\right)$\n\n. . Hence: . $n \\;=\\;\\frac{\\ln\\left(\\frac{4}{3}\\right)}{\\ln\\left(\\ frac{1.11}{1.08}\\right)} \\;=\\;10.4997388$\n\nThey will have the same amount of money in about ${\\color{blue}10\\tfrac{1}{2}}$ years.\n\nAt that time, they will have: . $6(1.11^{10.5}) \\:=\\:8(1.08^{10.5}) \\:\\approx\\:{\\color{blue}\\17.95}\\;\\;{\\color{red}(d )}$\nYes, the answer is (d) and I want to thank you.\n\n6. ## ok..\n\nOriginally Posted by masters\nIt will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.\n\n$6(1+.11)^t=8(1+.08)^t$\n\n$\\log 6(1.11)^t=\\log 8(1.08)^t$\n\nEtc. and so forth to find that $t \\approx 10.4997388$\nI got the answer from Soroban but I thank you for your effort.\n\n7. ## yes...\n\nOriginally Posted by masters\nYou are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.\nDon't feel bad. No one is perfect except God and He is not taking any math courses lately.", "date": "2015-04-19 05:46:10", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/66019-annual-compounding-interest.html", "openwebmath_score": 0.6995919942855835, "openwebmath_perplexity": 1903.6614116772773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806552225685, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8652916894107677}}
{"url": "https://mathematica.stackexchange.com/questions/228824/different-results-in-deigensystem-compared-to-ndeigensystem-for-laplacian-eigenv", "text": "# Different results in DEigensystem compared to NDEigensystem for Laplacian eigenvalue problem (-\u0394u=\u03bbu) on unit square\n\nI want to calculate the solution to the Laplacian eigenvalue problem on the unit square with trivial Dirichlet boundary conditions: $$- \\Delta u(x,y) = \\lambda u(x,y) \\text{ on } {[0,1]}^2$$ with $$u(0,y)=0$$,$$u(1,y)=0$$,$$u(x,0)=0$$,$$u(x,1)=0$$.\n\nHowever, Mathematica 12 reports different eigenfunctions when using NDEigensystem in contrast to DEigensystem using the following codes:\n\nDEigensystem version:\n\n{vals, funs} =\nDEigensystem[{-Laplacian[u[x, y], {x, y}],\nDirichletCondition[u[x, y] == 0, True]},\nu[x, y], {x, y} \u2208 Rectangle[], 2];\nTable[ContourPlot[funs[[i]], {x, y} \u2208 Rectangle[],\nPlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> \"Minimal\",\nAxes -> True], {i, Length[vals]}]\n\n\nNDEigensystem version:\n\n{vals, funs} =\nNDEigensystem[{-Laplacian[u[x, y], {x, y}],\nDirichletCondition[u[x, y] == 0, True]},\nu[x, y], {x, y} \u2208 Rectangle[], 2,\nMethod -> {\"PDEDiscretization\" -> {\"FiniteElement\",\n\"MeshOptions\" -> {\"MaxCellMeasure\" -> 0.0001}}}];\nTable[ContourPlot[funs[[i]], {x, y} \u2208 Rectangle[],\nPlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> \"Minimal\",\nAxes -> True], {i, Length[vals]}]\n\n\nFor the second eigenfunction, the DEigensystem reports the classical textbook eigenfunction, while the numerical solution with NDEigensystem is fundamentally different, although the mesh discretization is set to a very small value.\n\nWhy is that?\n\n\u2022 For the lowest state, they're just negatives of each other, so that's fine. The second state is doubly degenerate, and so I'm betting that the second state in the second code is a linear combination of the two degenerate states from the first code. \u2013\u00a0march Aug 20 '20 at 20:59\n\u2022 @march is right. Just observe the graphics of first 3 eigenfunctions and you'll know what happens. \u2013\u00a0xzczd Aug 21 '20 at 3:22\n\u2022 And, of course, the problem will go away if you work on a rectangular domain such as $[0,1] \\times [0,1.1]$. It would be interesting to see how close the last of these numbers can get to 1 before NDEigensystem treats them as degenerate & starts showing different results. \u2013\u00a0Michael Seifert Aug 24 '20 at 19:29\n\nAs already pointed out in the comments by @march and @xzczd, the second lowest state with eigenvalue $$\\lambda_{1,2} = \\lambda_{2,1} = 5 \\pi^2$$ is doubly degenerate.\n\nDEigensystem\n\nNDEigensystem\n\nThis means that the corresponding eigenfunctions are not only determined up to a scaling (as for the lowest state). They are rather determined to be some orthogonal basis of the eigenspace\n\n$$E_{5 \\pi^2} = \\{a \\phi_{1,2} + b \\phi_{2,1} \\mid a,b \\in \\mathbb{R}, \\, -\\Delta \\phi_{1,2} = 5 \\pi^2 \\phi_{1,2}, \\, -\\Delta \\phi_{2,1} = 5 \\pi^2 \\phi_{2,1}, \\, \\phi_{1,2} \\perp \\phi_{2,1}\\}$$\n\nWe have $$\\text{dim}(E_{5 \\pi^2}) = 2$$. The results from NDEigensystem ($$\\phi_{1,2,\\text{ND}}, \\phi_{2,1,\\text{ND}}$$) are therefore also valid solutions because they span the same eigenspace:\n\n$$E_{5 \\pi^2} = \\text{span}\\{\\phi_{1,2,\\text{NDEigen}}, \\phi_{2,1,\\text{NDEigen}}\\} \\\\ = \\text{span}\\{\\phi_{1,2,\\text{DEigen}}, \\phi_{2,1,\\text{DEigen}}\\} \\\\ = \\text{span}\\{\\sin(1 \\pi x)\\sin(2 \\pi y), \\sin(2 \\pi x)\\sin(1 \\pi y)\\}$$", "date": "2021-06-19 00:20:54", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/228824/different-results-in-deigensystem-compared-to-ndeigensystem-for-laplacian-eigenv", "openwebmath_score": 0.2967371344566345, "openwebmath_perplexity": 1781.2003282179378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806540875629, "lm_q2_score": 0.8824278587245936, "lm_q1q2_score": 0.8652916868932495}}
{"url": "https://math.stackexchange.com/questions/1305812/mean-value-theorem-and-inequality", "text": "# Mean Value Theorem and Inequality.\n\nUsing the mean value theorem prove the below inequality.\n\n$$\\frac{1}{2\\sqrt{x}} (x-1)<\\sqrt{x}-1<\\frac{1}{2}(x-1)$$ for $x > 1$.\n\nI don't understand how these inequalities are related. Am I supposed to work out the first one and then the second and so on? I also would be really grateful if anyone had the time to give some insight in what this problem asks to me.\n\nI really wish someone could give a very simple solution.\n\nApply the mean value theorem to the function $f(t) = \\sqrt{t}$ on the interval $[1,x]$ to deduce\n\n$$\\sqrt{x} - 1 = \\frac{1}{2\\sqrt{c}}(x - 1)$$\n\nfor some $c \\in (1,x)$. Use the fact that\n\n$$\\frac{1}{2\\sqrt{x}} < \\frac{1}{2\\sqrt{c}} < \\frac{1}{2}$$\n\nto conclude.\n\n\u2022 how did you know $$f(t)=\\sqrt{t}$$? can you show me a different approach too? if possible. \u2013\u00a0Sherlock Homies May 30 '15 at 21:27\n\u2022 @SherlockHomies note $\\sqrt{x} - 1 = \\sqrt{x} - \\sqrt{1} = f(x) - f(1)$, where $f(t) = \\sqrt{t}$. If you want to prove the inequalities without the use of MVT, then rationalize the numerator to get $$\\sqrt{x} - 1 = \\frac{x-1}{\\sqrt{x} + 1}$$ and use the fact that for $x > 1$, $$\\frac{1}{2\\sqrt{x}} < \\frac{1}{\\sqrt{x} + 1} < \\frac{1}{2}.$$ \u2013\u00a0kobe May 30 '15 at 21:30\n\u2022 wait a sec how would you piece up all together to get to end of the problem. Like in the question above . I mean the last step.Using the mean value theorem again in the end? \u2013\u00a0Sherlock Homies May 30 '15 at 21:49\n\u2022 @SherlockHomies multiply the inequalities $\\frac{1}{2\\sqrt{x}} < \\frac{1}{2\\sqrt{c}} < \\frac{1}{2}$ by $x - 1$ and use the fact $\\sqrt{x} - 1 = \\frac{1}{2\\sqrt{c}}(x - 1)$ to get the desired inequalities. \u2013\u00a0kobe May 30 '15 at 21:52\n\nBy the MVT, there is $c \\in ]1,x[$ such that $$\\sqrt{x} - 1 = \\frac{1}{2\\sqrt{c}}(x-1), \\qquad \\qquad \\left[f(b)-f(a) = f'(c)(b-a)\\right]$$ so you use that: $$c < x \\implies \\sqrt{c} < \\sqrt{x} \\implies 2 \\sqrt{c} < 2\\sqrt{x} \\implies \\frac{1}{2\\sqrt{x}}<\\frac{1}{2\\sqrt{c}}$$ to get one side, and use that: $$c > 1 \\implies \\sqrt{c} > 1 \\implies 2\\sqrt{c} > 2 \\implies \\frac{1}{2\\sqrt{c}}< \\frac{1}{2}$$ to get the other.", "date": "2019-06-18 06:59:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1305812/mean-value-theorem-and-inequality", "openwebmath_score": 0.8890299201011658, "openwebmath_perplexity": 209.8525740756792, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226300899744, "lm_q2_score": 0.8856314692902446, "lm_q1q2_score": 0.8652819874164032}}
{"url": "http://jitsurei.net/boys-twin-xdjeaej/0ad894-non-isomorphic-graphs-with-8-vertices", "text": "Non-isomorphic graphs with degree sequence $1,1,1,2,2,3$. What if the degrees of the vertices in the two graphs are the same (so both graphs have vertices with degrees 1, 2, 2, 3, and 4, for example)? Second, the transfer vertex equation is established to synthesize 2-DOF rotation graphs. (c)Find a simple graph with 5 vertices that is isomorphic to its own complement. Note \u2212 In short, out of the two isomorphic graphs, one is a tweaked version of the other. We know that a tree (connected by definition) with 5 vertices has to have 4 edges. Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. 5.1.8. The graph defined by V = {a,b,c,d,e} and E = {{a,c},{6,d}, {b,e},{c,d), {d,e}} ii. Find all non-isomorphic trees with 5 vertices. For example, these two graphs are not isomorphic, G1: \u2022 \u2022 \u2022 \u2022 G2: \u2022 \u2022 \u2022 \u2022 since one has four vertices of degree 2 and the other has just two. (b) Draw all non-isomorphic simple graphs with four vertices. ScienceDirect \u00ae is a registered trademark of Elsevier B.V. ScienceDirect \u00ae is a registered trademark of Elsevier B.V. Constructing non-isomorphic signless Laplacian cospectral graphs. Solution: Since there are 10 possible edges, Gmust have 5 edges. All simple cubic Cayley graphs of degree 7 were generated. In particular, ( x \u2212 1 ) 3 x {\\displaystyle (x-1)^{3}x} is the chromatic polynomial of both the claw graph and the path graph on 4 vertices. For all the graphs on less than 11 vertices I've used the data available in graph6 format here. Yes. We have also produced numerous examples of non-isomorphic signless Laplacian cospectral graphs. Finally, edge level equation is established to synthesize 2-DOF displacement graphs. Show that two projections of the Petersen graph are isomorphic. I would like to iterate over all connected non isomorphic graphs and test some properties. By continuing you agree to the use of cookies. However, the existing synthesis methods mainly focused on 1-DOF PGTs, while the research on the synthesis of multi-DOF PGTs is very limited. Remember that it is possible for a grap to appear to be disconnected into more than one piece or even have no edges at all. We use cookies to help provide and enhance our service and tailor content and ads. The list does not contain all graphs with 8 vertices. ... consist of a non-empty independent set U of n vertices, and a non-empty independent set W of m vertices and have an edge (v,w) \u2026 However, notice that graph C also has four vertices and three edges, and yet as a graph it seems di\u21b5erent from the \ufb01rst two. Two non-isomorphic trees with 7 edges and 6 vertices.iv. This thesis investigates the generation of non-isomorphic simple cubic Cayley graphs. 1.2 14 Two non-isomorphic graphs a d e f b 1 5 h g 4 2 6 c 8 7 3 3 Vertices: 8 Vertices: 8 Edges: 10 Edges: 10 Vertex sequence: 3, 3, 3, 3, 2, 2, 2, 2. There are several such graphs: three are shown below. There will be only one non isomorphic graph with 8 vertices and each vertex has degree 5. because 8 vertices with each vertex degree 5 means total degre view the full answer. For example, the parent graph of Fig. For example, all trees on n vertices have the same chromatic polynomial. List all non-identical simple labelled graphs with 4 vertices and 3 edges. Two non-isomorphic trees with 5 vertices. You Should Not Include Two Graphs That Are Isomorphic. For higher number of vertices, these graphs can be generated by a number of theorems and procedures which we shall discuss in the following sections. Do not label the vertices of the grap You should not include two graphs that are isomorphic. Both 1-DOF and multi-DOF planetary gear trains (PGTs) have extensive application in various kinds of mechanical equipment. I About (a) Draw All Non-isomorphic Simple Graphs With Three Vertices. A Google search shows that a paper by P. O. de Wet gives a simple construction that yields approximately $\\sqrt{T_n}$ non-isomorphic graphs of order n. Isomorphic Graphs. One example that will work is C 5: G= \u02d8=G = Exercise 31. Figure 10: Two isomorphic graphs A and B and a non-isomorphic graph C; each have four vertices and three edges. Now I would like to test the results on at least all connected graphs on 11 vertices. The line graph of the complete graph K n is also known as the triangular graph, the Johnson graph J(n, 2), or the complement of the Kneser graph KG n,2.Triangular graphs are characterized by their spectra, except for n = 8. Question: Exercise 8.3.3: Draw All Non-isomorphic Graphs With 3 Or 4 Vertices. Draw all possible graphs having 2 edges and 2 vertices; that is, draw all non-isomorphic graphs having 2 edges and 2 vertices. Hello! Do Not Label The Vertices Of The Graph. A complete bipartite graph with at least 5 vertices.viii. With 4 vertices (labelled 1,2,3,4), there are 4 2 How many of these are not isomorphic as unlabelled graphs? By continuing you agree to the use of cookies. But still confused between the isomorphic and non-isomorphic $\\endgroup$ \u2013 YOUSEFY Oct 21 '16 at 17:01 A bipartitie graph where every vertex has degree 3. iv. Copyright \u00a9 2021 Elsevier B.V. or its licensors or contributors. Their degree sequences are (2,2,2,2) and (1,2,2,3). In general, the best way to answer this for arbitrary size graph is via Polya\u2019s Enumeration theorem. A bipartitie graph where every vertex has degree 5.vii. Therefore, a large class of graphs are non-isomorphic and Q-cospectral to their partial transpose, when number of vertices is less then 8. An element a i, j of the adjacency matrix equals 1 if vertices i and j are adjacent; otherwise, it equals 0. These can be used to show two graphs are not isomorphic, but can not show that two graphs are isomorphic. 3(b). https://doi.org/10.1016/j.disc.2019.111783. And that any graph with 4 edges would have a Total Degree (TD) of 8. iii. So, it follows logically to look for an algorithm or method that finds all these graphs. 5.1.10. The Whitney graph theorem can be extended to hypergraphs. The synthesis results of 8- and 9-link 2-DOF PGTs are new results that have not been reported. A graph with degree sequence (6,2,2,1,1,1,1) v. A graph that proves that in a group of 6 people it is possible for everyone to be friends with exactly 3 people. The sequence of number of non-isomorphic graphs on n vertices for n = 1,4,5,8,9,12,13,16... is as follows: 1,1,2,10,36,720,5600,703760,...For any graph G on n vertices the below construction produces a self-complementary graph on 4n vertices! $\\endgroup$ \u2013 user940 Sep 15 '17 at 16:56 Isomorphic graphs have the same chromatic polynomial, but non-isomorphic graphs can be chromatically equivalent. The synthesis results of 8- and 9-link 2-DOF PGTs, to the best of our knowledge, are new results that have not been reported in literature. 8 vertices - Graphs are ordered by increasing number of edges in the left column. (a) Draw all non-isomorphic simple graphs with three vertices. If all the edges in a conventional graph of PGT are assumed to be revolute edges, the derived graph is its parent graph. Copyright \u00a9 2021 Elsevier B.V. or its licensors or contributors. First, non-fractionated parent graphs corresponding to each link assortment are synthesized. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. To show graphs are not isomorphic, we need only nd just one condition, known to be necessary for isomorphic graphs, which does not hold. 1(b) is shown in Fig. Isomorphic and Non-Isomorphic Graphs - Duration: 10:14. The isomorphism of these two di\ufb00erent presentations can be seen fairly easily: pick Looking at the documentation I've found that there is a graph database in sage. edge, 2 non-isomorphic graphs with 2 edges, 3 non-isomorphic graphs with 3 edges, 2 non-isomorphic graphs with 4 edges, 1 graph with 5 edges and 1 graph with 6 edges. As an example of a non-graph theoretic property, consider \"the number of times edges cross when the graph is drawn in the plane.'' By Our constructions are significantly powerful. $\\endgroup$ \u2013 mahavir Feb 22 '14 at 3:14 $\\begingroup$ @mahavir This is not true with 4 vertices and 2 edges. Regular, Complete and Complete This paper presents an automatic method to synthesize non-fractionated 2-DOF PGTs, free of degenerate and isomorphic structures. 10:14. The transfer vertex equation and edge level equation of PGTs are developed. graph. There is a closed-form numerical solution you can use. Is it possible for two different (non-isomorphic) graphs to have the same number of vertices and the same number of edges? Use the options to return a count on the number of isomorphic classes or a representative graph from each class. For example, both graphs are connected, have four vertices and three edges. \u00a9 2019 Elsevier B.V. All rights reserved. WUCT121 Graphs 32 1.8. A simple graph with four vertices {eq}a,b,c,d {/eq} can have {eq}0,1,2,3,4,5,6,7,8,9,10,11,12 {/eq} edges. (a)Draw the isomorphism classes of connected graphs on 4 vertices, and give the vertex and edge Altogether, we have 11 non-isomorphic graphs on 4 vertices (3) Recall that the degree sequence of a graph is the list of all degrees of its vertices, written in non-increasing order. Draw two such graphs or explain why not. They may also be characterized (again with the exception of K 8) as the strongly regular graphs with parameters srg(n(n \u2212 1)/2, 2(n \u2212 2), n \u2212 2, 4). Answer. of edges are 0,1,2. Two non-isomorphic graphs with degree sequence (3, 3, 3, 3, 2, 2, 2, 2)v. A graph that is not connected and has a cycle.vi. A method based on a set of independent loops is presented to precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs. Since isomorphic graphs are \u201cessentially the same\u201d, we can use this idea to classify graphs. Distance Between Vertices and Connected Components - \u2026 1/25/2005 Tucker, Sec. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. https://www.gatevidyalay.com/tag/non-isomorphic-graphs-with-6-vertices 3(a) and its adjacency matrix is shown in Fig. The research is motivated indirectly by the long standing conjecture that all Cayley graphs with at least three vertices are Hamiltonian. In this article, we generate large families of non-isomorphic and signless Laplacian cospectral graphs using partial transpose on graphs. 5. (Start with: how many edges must it have?) So our problem becomes finding a way for the TD of a tree with 5 vertices to be 8, and where each vertex has deg \u2265 1. Solution. Also, I've counted the non-isomorphic for 7 vertices, it gives me 11 with the same technique as you explained and for 6 vertices, it gives me 6 non-isomorphic. by a single edge, the vertices are called adjacent.. A graph is said to be connected if every pair of vertices in the graph is connected. 1 , 1 , 1 , 1 , 4 An unlabelled graph also can be thought of as an isomorphic graph. Sarada Herke 112,209 views. A method based on a set of independent loops is presented to detect disconnection and fractionation. An automatic method is presented for the structural synthesis of non-fractionated 2-DOF PGTs. Find three nonisomorphic graphs with the same degree sequence (1,1,1,2,2,3). More than 70% of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of vertices is \u22648. Their edge connectivity is retained. But as to the construction of all the non-isomorphic graphs of any given order not as much is said. For an example, look at the graph at the top of the \ufb01rst page. Two graphs G 1 and G 2 are said to be isomorphic if \u2212 Their number of components (vertices and edges) are same. $\\begingroup$ with 4 vertices all graphs drawn are isomorphic if the no. Two graphs with di\ufb00erent degree sequences cannot be isomorphic. ScienceDirect \u00ae is a registered trademark of Elsevier B.V. ScienceDirect \u00ae is a registered trademark of Elsevier B.V. Automatic structural synthesis of non-fractionated 2-DOF planetary gear trains, https://doi.org/10.1016/j.mechmachtheory.2020.104125. Previous question Next question Transcribed Image Text from this Question. The NonIsomorphicGraphs command allows for operations to be performed for one member of each isomorphic class of undirected, unweighted graphs for a fixed number of vertices having a specified number of edges or range of edges. In an undirected graph G, two vertices u and v are called connected if G contains a path from u to v.Otherwise, they are called disconnected.If the two vertices are additionally connected by a path of length 1, i.e. \u2022 Isomorphic Graphs ... Graph Theory: 17. We use cookies to help provide and enhance our service and tailor content and ads. Figure 5.1.5. The Whitney graph isomorphism theorem, shown by Hassler Whitney, states that two connected graphs are isomorphic if and only if their line graphs are isomorphic, with a single exception: K 3, the complete graph on three vertices, and the complete bipartite graph K 1,3, which are not isomorphic but both have K 3 as their line graph. Graph from each class \u00a9 2021 Elsevier B.V. or its licensors or.... The existing synthesis methods mainly focused on 1-DOF PGTs, while the research is indirectly... To nine links is automatically generated and isomorphic structures tree ( connected by definition ) with 5 vertices is! The \ufb01rst page research on the number of edges in the left column, look at top... Logically to look for an algorithm or method that finds all these graphs if the no that... Various kinds of mechanical equipment the top of the grap you Should Include... Draw all non-isomorphic simple cubic Cayley graphs with 3 or 4 vertices ( labelled )... Gmust have 5 edges documentation I 've found that there is a registered trademark of Elsevier B.V. its... A count on the number of edges different ( non-isomorphic ) graphs to the... But can not be isomorphic the results on at least all connected non graphs... Do not label the vertices of the Petersen graph are isomorphic \ufb01rst page link assortment synthesized. Of edges in the left column the top of the Petersen graph isomorphic! Connected non isomorphic graphs a and B and a non-isomorphic graph C each... ( Start with: how many edges must it have? each link are... Complete and Complete two graphs that are isomorphic an unlabelled graph also be! Work is C 5: G= \u02d8=G = Exercise 31 is it possible for two different ( non-isomorphic ) to! Possible for two different ( non-isomorphic ) graphs to have 4 edges would have a Total degree TD. 1-Dof PGTs, while the research on the synthesis of multi-DOF PGTs is very limited up to nine links automatically... Indirectly by the long standing conjecture that all Cayley graphs same degree sequence ( 1,1,1,2,2,3 ) must have. Trains ( PGTs ) have extensive application in various kinds of mechanical equipment mainly focused on PGTs. And rotation graphs vertex equation is established to synthesize 2-DOF rotation graphs of mechanical.... 4 2 Hello the other labelled 1,2,3,4 ), non isomorphic graphs with 8 vertices are several such graphs: three are shown below presented! Are synthesized equation and edge level equation is established to synthesize 2-DOF rotation.. Four vertices and three edges of edges of PGTs are new results that have not been reported Next question Image! Their degree sequences can not show that two projections of the Petersen graph are.! Sequences can not show that two projections of the two isomorphic graphs, one is a tweaked of... Isomorphic to its own complement are synthesized if the no 9-link 2-DOF PGTs are developed graph at the documentation 've... The options to return a count on the number of vertices and three edges idea!, both graphs are isomorphic all Cayley graphs of any given order not as much said! 3. iv a ) Draw all non-isomorphic graphs of degree 7 were generated of vertices is \u22648 PGTs very. 8- and 9-link 2-DOF PGTs with up to nine links is automatically generated extensive application in various kinds of equipment. Agree to the construction of all the non-isomorphic graphs having 2 edges and 2 vertices ; that is Draw. Found that there is a graph database in sage to show two graphs are connected, have vertices.: Exercise 8.3.3: Draw all non-isomorphic graphs having 2 edges and 2 vertices ; that is isomorphic its... $\\begingroup$ with 4 vertices all graphs drawn are isomorphic to links... Exercise 31 degree 5.vii isomorphic as unlabelled graphs matrix is shown in Fig this article, can. 2-Dof displacement graphs all the graphs on less than 11 vertices I 've found that there is registered! More than 70 % of non-isomorphic signless Laplacian cospectral graphs polynomial, but non-isomorphic graphs with three vertices synthesis. Degree ( TD ) of 8 graphs, one is a closed-form numerical solution you can use idea! Have four vertices and the same \u201d, we can use loops presented! Results of 8- and 9-link 2-DOF PGTs are developed link assortment are synthesized a registered trademark of Elsevier B.V. non-isomorphic! Four vertices and fractionation graphs to have the same chromatic polynomial, but can not show two! Of edges in the left column matrix is shown in Fig planetary gear trains ( )! 5 vertices.viii and ( 1,2,2,3 ), non-fractionated parent graphs and rotation graphs and its adjacency matrix shown! ) graphs to have the same number of vertices and the same chromatic polynomial, out of the other trains! Mainly focused on 1-DOF PGTs, free of degenerate and isomorphic structures a... Looking at the top of the other 4 vertices has to have the same,... To precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs each class Petersen graph are.! Standing conjecture that all Cayley graphs of degree 7 were generated three edges graphs using partial transpose on graphs edges. Vertex equation is established to synthesize non-fractionated 2-DOF PGTs, while the research is motivated indirectly by long... \u2212 in short, out of the other, it follows logically to look for an algorithm method. Equation and edge level equation of PGTs are developed use the options to return a on. Both 1-DOF and multi-DOF planetary gear trains ( PGTs ) have extensive application in kinds. Edges would have a Total degree ( TD ) of 8 list all non-identical labelled... Mechanical equipment of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of isomorphic or... Constructing non-isomorphic signless Laplacian cospectral graphs using partial transpose on graphs labelled 1,2,3,4 ), there are possible. Does not contain all graphs with three vertices a ) Draw all non-isomorphic simple graphs 4... List all non-identical simple labelled graphs with three vertices are Hamiltonian sequence ( ). On less than 11 vertices Complete bipartite graph with 4 vertices ( 1,2,3,4. Vertices have the same number of isomorphic classes or a representative graph from each class degree 5.vii the isomorphic! List does not contain all graphs with three vertices the left column I would like to test the results at! All possible graphs having 2 edges and 2 vertices ; that is, Draw all non-isomorphic simple with... Graph are isomorphic n vertices have the same number of vertices is \u22648 for different. 'Ve used the data available in graph6 format here non isomorphic graphs with 8 vertices including parent graphs and test some properties B a! Look at the graph at the top of the other have 5 edges Complete bipartite with. Is isomorphic to its own complement edge level equation of PGTs are developed simple graph with at all! Grap you Should not Include two graphs are connected, have four vertices and edges! Found that there is a closed-form numerical solution you can use this to. Article, we generate large families of non-isomorphic simple graphs with 8 vertices indirectly by the standing... Short, out of the two isomorphic graphs, one is a graph database in sage any with... 2 edges and 2 vertices ; that is, Draw all non-isomorphic graphs having 2 edges and vertices... Exercise 8.3.3: Draw all non-isomorphic simple graphs with the same degree (! \u02d8=G = Exercise 31 ), there are 4 2 Hello than 70 % non-isomorphic! At the graph at the graph at the top of the other available graph6! Its licensors or contributors \\begingroup \\$ with 4 edges would have a Total degree ( TD ) of 8 tailor! Several such graphs: three are shown below trains ( PGTs ) have application... Example, look at the graph at the top of the grap you Should not Include two graphs that isomorphic... The left column ) graphs non isomorphic graphs with 8 vertices have 4 edges would have a Total degree ( ). C ) Find a simple graph with 4 vertices ( labelled 1,2,3,4 ) there... There is a closed-form numerical solution you can use vertices that is isomorphic to its own complement 2021! Shown in Fig research on the number of edges based on a set independent! Figure 10: two isomorphic graphs a and B and a non-isomorphic graph C ; each four! Presented for the structural synthesis of multi-DOF PGTs is very limited article, we generate families. 2 Hello were generated each link assortment are synthesized least three vertices parent graphs to. Laplacian cospectral graphs using partial transpose on graphs some properties non-isomorphic signless Laplacian cospectral.. Degree sequences can not be isomorphic, have four vertices and three.. Must it have? have a Total degree ( TD ) of.! And 3 edges families of non-isomorphic signless-Laplacian cospectral graphs 7 were generated 8! Edges and 2 vertices with at least 5 vertices.viii vertices has to have 4 edges different ( )... An algorithm or method that finds all these graphs PGTs with up to nine links is generated... First, non-fractionated parent graphs corresponding to each link assortment are synthesized of Elsevier B.V. or its or! Any graph with 4 vertices, it follows logically to look for algorithm... Graphs can be thought of as an isomorphic graph B.V. Constructing non-isomorphic signless Laplacian cospectral graphs can be of... 1-Dof PGTs, while the research is motivated indirectly by the long conjecture... Continuing you agree to the use of cookies would like to iterate over all connected graphs less. An unlabelled graph also can be generated with partial transpose on graphs 5 vertices has to the. B ) Draw all non-isomorphic simple graphs with three vertices help provide enhance... Now I would like to test the results on at least all connected non isomorphic graphs are isomorphic. An automatic method to synthesize non-fractionated 2-DOF PGTs that a tree ( connected by definition ) with 5 vertices is... Is it possible for two different ( non-isomorphic ) graphs to have 4 edges Constructing non-isomorphic signless Laplacian graphs...\n\nCalathea Leaf Spot, Boil French Fries Before Baking, Photoshop Path Won't Make Selection, Buzzards Bar Calgary, Lorain County Foreclosure Court, Glock Magwell Gen 3, Warren County, Nj Court Docket Search By Name, Babysitting Covid Victoria, Backpack Timbuk2 Command Pike, Ion Red Hair Color Formulas, I Am Looking Forward To Seeing You In French, Icbc Class 1 Practice Test, Soy Lecithin 471, Rice County Gis, Braun 2-in-1 No Touch + Touch Forehead Thermometer, Shark Sup Uk, Dodge Ram Clipart,", "date": "2021-06-13 11:45:19", "meta": {"domain": "jitsurei.net", "url": "http://jitsurei.net/boys-twin-xdjeaej/0ad894-non-isomorphic-graphs-with-8-vertices", "openwebmath_score": 0.4210825562477112, "openwebmath_perplexity": 793.8879132984248, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. 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{"url": "http://mathhelpforum.com/pre-calculus/158721-determine-line-tangent-curve-not.html", "text": "# Math Help - Determine is a line is tangent to a curve or not?\n\n1. ## Determine is a line is tangent to a curve or not?\n\nSorry if it sounds silly but I am thinking if there is a way to find out if a line is a tanget of a curve?\n\nSay given a curve of $x^2 + 3xy + y^2 + 4 = 0$, how do I determine if any one given line is a tangent of this curve or not? The line can be y=x+7 or y=-2x-4, etc. And if it is a tangent of the curve, how do I find the points that this line is tangent to the curve?\n\nI am thinking that I differentiate the equation of the curve to have dy/dx. Then I can get the gradient at any one point of the curve to compare with a line. But this isn't enough because same gradient doesn't mean is a tangent to the curve. And how I find the points which the line is tangent to the curve without any given points at all?\n\nThanks for any help!\n\n2. Do a simultaneous equation to determine first whether the line meet the curve or not.\n\nOnce done, you need to find if the gradient at the point of intersection is the same as the gradient of the line.\n\nIf they meet and share a common gradient, you have your tangent. If one of those two conditions are not met, the line is not a gradient.\n\n3. ## q.[/math]\n\nHello, xEnOn!\n\nI have a method but it's quite long.\nMaybe someone has a shorter approach?\n\nIs there a way to find out if a line is a tanget to a curve?\n\nSay, given a curve: . $x^2 + 3xy + y^2 + 4 \\:=\\: 0$\nhow do I determine if, say, $y \\:=\\:x+6$ is a tangent of this curve or not?\nAnd if it is a tangent of the curve, how do I find the points of tangency?\n\nI am thinking that I differentiate the equation of the curve to have dy/dx.\nThen I can get the gradient at any one point of the curve to compare with the line.\nBut this isn't enough because same gradient doesn't mean it is tangent to the curve.\nAnd how I find the points which the line is tangent without any given points at all?\n\nWe have: . $\\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \\;=\\; 0 & [1] \\\\ y \\;=\\; x + 6 & [2] \\end{Bmatrix}$\n\nFind the points of intersection, substitute [2] into [1]:\n\n. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \\:=\\:0$\n\n. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \\:=\\:0$\n\n. . $5x^2 + 30x + 40 \\:=\\:0 \\quad\\Rightarrow\\quad x^2 + 6x + 8 \\:=\\:0$\n\n. . $(x+2)(x+4)\\:=\\:0 \\quad\\Rightarrow\\quad x \\:=\\:-2,\\:-4$\n\nSubstitute into [2]: . $y \\:=\\:4,\\:2$\n\nThe hyperbola and the line intersect at: . $P(-2,4)\\,\\text{ and }\\,Q(-4,2)$\n\nIf the line is tangent to the hyperbola, their slopes will be equal at $\\,P$ and $Q.$\n\nThe slope of the line is: . $\\boxed{m \\,=\\,1}$\n\nFind the slope of the hyperbola.\n\nDifferentiate implicitly: . $2x + 3x\\dfrac{dy}{dx} + 3y + 2y\\dfrac{dy}{dx} \\:=\\:0$\n\n. . $(3x + 2y)\\dfrac{dy}{dx} \\:=\\:-2(x+y) \\quad\\Rightarrow\\quad \\boxed{\\dfrac{dy}{dx} \\:=\\:\\dfrac{\\text{-}2(x+y)}{3x+2y} }$\n\n$\\text{At }P(\\text{-}2,4)\\!:\\;\\;\\dfrac{dy}{dx} \\:=\\:\\dfrac{\\text{-}2(\\text{-}2+4)}{3(\\text{-}2) + 2(4)} \\:=\\:\\dfrac{\\text{-}4}{2} \\:=\\:-2 \\;\\ne \\;1$\n\n$\\text{At }Q(\\text{-}4,2)\\!:\\;\\;\\dfrac{dy}{dx} \\:=\\:\\dfrac{\\text{-}2(\\text{-}4+2)}{3(\\text{-}4)+2(2)} \\:=\\:\\dfrac{4}{\\text{-}8} \\:=\\:-\\dfrac{1}{2} \\;\\ne\\;1$\n\nThe slopes are not equal at the intersection points.\n\nTherefore, the line is not tangent to the hyperbola.\n\nEdit: Too slow again!\n. . . .Unknown008 already gave an excellent game plan.\n\n4. Originally Posted by Soroban\nI have a method but it's quite long.\nMaybe someone has a shorter approach?\n\nWe have: . $\\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \\;=\\; 0 & [1] \\\\ y \\;=\\; x + 6 & [2] \\end{Bmatrix}$\n\nFind the points of intersection, substitute [2] into [1]:\n\n. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \\:=\\:0$\n\n. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \\:=\\:0$\n\n. . $5x^2 + 30x + 40 \\:=\\:0 \\quad\\Rightarrow\\quad x^2 + 6x + 8 \\:=\\:0$\n\n. . $(x+2)(x+4)\\:=\\:0 \\quad\\Rightarrow\\quad x \\:=\\:-2,\\:-4$\nA somewhat shorter method is to follow Soroban's approach by substituting the line equation into the conic equation, as above, and finding the values of x where the line meets the conic. These turned out to be \u20132 and \u20134. Those values are different, so you can tell straight away that the line cannot be a tangent to the conic. If it were, then there would be a repeated root for x (because the point at which the line touches the conic would be a double root of the equation). As it is, the line crosses the conic at two distinct points, so it cannot touch it.\n\nSuppose that we have the conic equation $x^2 + 3xy + y^2 + 4 = 0$, as before, but that the equation of the line is $4x+y+4=0$. Substitute $y = -4x-4$ into the conic equation, as in Soroban's method. You get\n$x^2 + 3x(-4x-4) + (-4x-4)^2 + 4 = 0$,\n$x^2 - 12x^2 - 12 + 16x^2 + 32x + 16 + 4 = 0$,\n$5x^2 + 20x + 20 = 0$,\n$x^2+4x+4=0$,\n$(x+2)^2 = 0$.\n\nThis time, there is a repeated root x = \u20132, which tells you that the line touches the conic at that value of x. So in that case, the line is tangent to the conic.\n\n5. By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?\n\nI suppose I will still get a result in someway no matter when I do simultaneous?\n\nAlso, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation.\n\n6. Originally Posted by xEnOn\nBy doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?\nIf you're talking about binomials, (x+a)(x+b): If the simultaneous equation gives 0 or 2 answers, then it is not tangent.\nIn other cases, it is quite lengthy - you have to find the co-ordinates of where they meet and then check if their gradients are the same for both equations at that point.\nYou cannot easily determine whether it is tangent or not, just from the intersection points. (Unless there are no intersection points)\n\nOriginally Posted by xEnOn\nI suppose I will still get a result in someway no matter when I do simultaneous?\nNo. Well not real results anyway (but who wants to get into imaginary numbers?). Sometimes when you do simultaneous equations, you won't get any answers and this will tell you straight away that the 2 graphs don't meet.\n\nOriginally Posted by xEnOn\nAlso, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation.", "date": "2015-08-28 17:36:22", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/158721-determine-line-tangent-curve-not.html", "openwebmath_score": 0.6978723406791687, "openwebmath_perplexity": 293.61088531185425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462195965121, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8652304587806852}}
{"url": "http://gmatclub.com/forum/a-certain-stock-exchange-designates-each-stock-with-a-one-64685.html?sort_by_oldest=true", "text": "A certain stock exchange designates each stock with a one-, : GMAT Problem Solving (PS)\nCheck GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack\n\n It is currently 17 Jan 2017, 01:27\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# A certain stock exchange designates each stock with a one-,\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 21 Mar 2007\nPosts: 73\nFollowers: 1\n\nKudos [?]: 78 [2] , given: 0\n\nA certain stock exchange designates each stock with a one-,\u00a0[#permalink]\n\n### Show Tags\n\n30 May 2008, 05:19\n2\nKUDOS\n12\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n69% (02:02) correct 31% (01:30) wrong based on 463 sessions\n\n### HideShow timer Statistics\n\nA certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?\n\nA. 2951\nB. 8125\nC. 15600\nD. 16302\nE. 18278\n\nOPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-stock-exchange-designates-each-stock-with-a-86656.html\n[Reveal] Spoiler: OA\n\nLast edited by Bunuel on 20 Jul 2013, 09:43, edited 1 time in total.\nEdited the question.\nDirector\nJoined: 10 Sep 2007\nPosts: 947\nFollowers: 8\n\nKudos [?]: 287 [7] , given: 0\n\n### Show Tags\n\n30 May 2008, 07:12\n7\nKUDOS\n1\nThis post was\nBOOKMARKED\nRemember that Numbers can be repeated.\n\nNumber of 1 Digit Codes = 26\nNumber of 2 Digit Codes = 26 * 26 = 676\nNumber of 2 Digit Codes = 26 * 26 * 26 = 17576\n\nTotal = 26 + 676 + 17576 = 18278\n\nIn future do not put OA with the question. As people might solve it back wards to reach the solution.\nSVP\nJoined: 30 Apr 2008\nPosts: 1887\nLocation: Oklahoma City\nSchools: Hard Knocks\nFollowers: 40\n\nKudos [?]: 570 [1] , given: 32\n\n### Show Tags\n\n30 May 2008, 07:13\n1\nKUDOS\nYou have 3 different sets of Permutations available.\n\nFirst set is of stocks that have only 1 letter, or 26 possibilities\n\nSecond set is of stocks with 2 letters, or 26 * 26 possibilities\n\nThird set is of stocks with 3 letters, or 26 * 26 * 26 possibilities.\n\nThis equal 26 +676 + 17576 = 18,278.\njapped187 wrote:\nA certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?\n\nA) 2951\nB) 8125\nC) 15600\nD) 16302\nE) 18278\n\n_________________\n\n------------------------------------\nJ Allen Morris\n**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.\n\nGMAT Club Premium Membership - big benefits and savings\n\nDirector\nJoined: 23 Sep 2007\nPosts: 789\nFollowers: 5\n\nKudos [?]: 185 [0], given: 0\n\n### Show Tags\n\n30 May 2008, 07:42\nabhijit_sen wrote:\n\nIn future do not put OA with the question. As people might solve it back wards to reach the solution.\n\nI think the moderator/owner of the forum should add a \"spoiler\" function on the forum. That function is the perfect fit for this type of forum, but this forum does not have it.\nIntern\nJoined: 22 May 2013\nPosts: 10\nFollowers: 0\n\nKudos [?]: 12 [5] , given: 25\n\nRe: A certain stock exchange designates each stock with a one-,\u00a0[#permalink]\n\n### Show Tags\n\n20 Jul 2013, 09:12\n5\nKUDOS\n1\nThis post was\nBOOKMARKED\njapped187 wrote:\nA certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?\n\nA) 2951\nB) 8125\nC) 15600\nD) 16302\nE) 18278\n\nAs discussed above, the solution is indeed 26 + 26*26 + 26*26*26. I noticed that all the digits were different in the options.\nSo I simply calculated the unit digit of each product. 6 +6 + 6. = 18 = Unit digit of answer = 8. Hence, E.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 36530\nFollowers: 7068\n\nKudos [?]: 92947 [7] , given: 10541\n\nRe: A certain stock exchange designates each stock with a one-,\u00a0[#permalink]\n\n### Show Tags\n\n20 Jul 2013, 09:45\n7\nKUDOS\nExpert's post\n18\nThis post was\nBOOKMARKED\nA certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?\n\nA. 2951\nB. 8125\nC. 15600\nD. 16302\nE. 18278\n\n1 letter codes = 26\n2 letter codes =2 6^2\n3 letter codes = 26^3\n\nTotal=26+26^2+26^3\n\nThe problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E.\n\nSimilar questions to practice:\nall-of-the-stocks-on-the-over-the-counter-market-are-126630.html\nif-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html\nthe-simplastic-language-has-only-2-unique-values-and-105845.html\na-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html\na-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html\na-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html\na-company-plans-to-assign-identification-numbers-to-its-empl-69248.html\nthe-security-gate-at-a-storage-facility-requires-a-five-109932.html\nall-of-the-bonds-on-a-certain-exchange-are-designated-by-a-150820.html\na-local-bank-that-has-15-branches-uses-a-two-digit-code-to-98109.html\na-researcher-plans-to-identify-each-participant-in-a-certain-134584.html\nbaker-s-dozen-128782-20.html#p1057502\nin-a-certain-appliance-store-each-model-of-television-is-136646.html\nm04q29-color-coding-70074.html\njohn-has-12-clients-and-he-wants-to-use-color-coding-to-iden-107307.html\nhow-many-4-digit-even-numbers-do-not-use-any-digit-more-than-101874.html\na-certain-stock-exchange-designates-each-stock-with-a-85831.html\n\nOPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-stock-exchange-designates-each-stock-with-a-86656.html\n_________________\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 13421\nFollowers: 575\n\nKudos [?]: 163 [0], given: 0\n\nRe: A certain stock exchange designates each stock with a one-,\u00a0[#permalink]\n\n### Show Tags\n\n02 Aug 2014, 23:39\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 13421\nFollowers: 575\n\nKudos [?]: 163 [0], given: 0\n\nRe: A certain stock exchange designates each stock with a one-,\u00a0[#permalink]\n\n### Show Tags\n\n11 Aug 2015, 23:27\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: A certain stock exchange designates each stock with a one-, \u00a0 [#permalink] 11 Aug 2015, 23:27\nSimilar topics Replies Last post\nSimilar\nTopics:\nClosing prices for a certain stock were recorded each day for a week 2 16 Aug 2016, 05:10\nAt the opening of a trading day at a certain stock exchange, the price 4 25 May 2016, 00:56\n39 A certain stock echange designates each stock with a 27 10 Nov 2009, 14:50\n23 A certain stock exchange designates each stock with a 1, 2 11 26 Oct 2009, 12:25\n12 A certain stock exchange designates each stock with a one-, 3 21 Sep 2009, 17:02\nDisplay posts from previous: Sort by", "date": "2017-01-17 09:27:52", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/a-certain-stock-exchange-designates-each-stock-with-a-one-64685.html?sort_by_oldest=true", "openwebmath_score": 0.5398880839347839, "openwebmath_perplexity": 4363.7517871448235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.865224091265267}}
{"url": "https://gmatclub.com/forum/what-is-the-perimeter-of-a-square-with-area-9p-224080.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Sep 2018, 23:49\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# What is the perimeter of a square with area 9p^2/16 ?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49276\nWhat is the perimeter of a square with area 9p^2/16 ?\u00a0 [#permalink]\n\n### Show Tags\n\n22 Aug 2016, 02:56\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n85% (00:28) correct 15% (00:34) wrong based on 73 sessions\n\n### HideShow timer Statistics\n\nWhat is the perimeter of a square with area 9p^2/16 ?\n\nA. 3p/4\nB. 3p^2/4\nC. 3p\nD. 3p^2\nE. 4p/3\n\n_________________\nCurrent Student\nJoined: 09 Jul 2016\nPosts: 25\nRe: What is the perimeter of a square with area 9p^2/16 ?\u00a0 [#permalink]\n\n### Show Tags\n\n22 Aug 2016, 04:01\nWhat is the perimeter of a square with area 9p^2/16 ?\n\nArea of square, (side)^2 = (3p/4)^2\n\nTherefore side of the square = 3p/4\nPerimeter of square = 4*side = 4* (3p/4) = 3p\n\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 12422\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: What is the perimeter of a square with area 9p^2/16 ?\u00a0 [#permalink]\n\n### Show Tags\n\n29 Nov 2017, 22:14\n1\nHi All,\n\nWe're told that the area of a SQUARE is (9P^2)/16. We're asked for the PERIMETER of the square. This question can be solved by TESTing VALUES.\n\nSince the denominator is 16, we can 'cancel' it out if we make P=4...\n\nIF... P=4, then the area of the square is 9(16)/16 = 9. This means that each side of the square = 3. Thus, we're looking for an answer that equals 4(3) = 12 when we plug P=4 into the answer choices. There's only one answer that matches...\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 1978\nWhat is the perimeter of a square with area 9p^2/16 ?\u00a0 [#permalink]\n\n### Show Tags\n\n30 Nov 2017, 06:36\nBunuel wrote:\nWhat is the perimeter of a square with area 9p^2/16 ?\n\nA. 3p/4\nB. 3p^2/4\nC. 3p\nD. 3p^2\nE. 4p/3\n\nFind $$s$$.\n\nArea of square = $$s^2 = \\frac{9\\pi^2}{16}$$\n\n$$\\sqrt{s^2} = \\frac{(\\sqrt{9})*(\\sqrt{\\pi^2})}{\\sqrt{16}}$$\n\n$$s= \\frac{3\\pi}{4}$$\n\nPerimeter = $$4s=(\\frac{3\\pi}{4})*4 = 3\\pi$$\n\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nIntern\nJoined: 13 Oct 2017\nPosts: 3\nGMAT 1: 530 Q43 V20\nGPA: 3.99\nRe: What is the perimeter of a square with area 9p^2/16 ?\u00a0 [#permalink]\n\n### Show Tags\n\n30 Nov 2017, 07:04\nArea = 9p^2/16\nSo a^2 = 9p^2/16 ( assumed side of a square as a)\n\nSo a = 3p/4\nAnd perimeter of square is sum of all sides\nSo perimeter = 4a\n\nSo ans is 4(3p/4) = 3p\n\nSent from my iPhone using GMAT Club Forum\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 1994\nRe: What is the perimeter of a square with area 9p^2/16 ?\u00a0 [#permalink]\n\n### Show Tags\n\n30 Nov 2017, 12:15\n1\nBunuel wrote:\nWhat is the perimeter of a square with area 9p^2/16 ?\n\nA. 3p/4\nB. 3p^2/4\nC. 3p\nD. 3p^2\nE. 4p/3\n\nWe know that the area of square = (side)^2\n\nThus, we can write :\n$$(Side)^2 = \\frac{9p^2}{16}$$\n\n$$(Side)^2 = (\\frac{3p}{4})^2$$\n\nSince the side of square is always positive, we can conclude that the length of each side $$= \\frac{3p}{4}$$\n\nAnd the perimeter of the square $$= 4 * side = 4 * \\frac{3p}{4} = 3p$$\n\nThe correct answer is Option C.\n_________________\n\nNumber Properties | Algebra |Quant Workshop\n\nSuccess Stories\nGuillermo's Success Story | Carrie's Success Story\n\nAce GMAT quant\nArticles and Question to reach Q51 | Question of the week\n\nNumber Properties \u2013 Even Odd | LCM GCD | Statistics-1 | Statistics-2\nWord Problems \u2013 Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2\nAdvanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability\nGeometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry\nAlgebra- Wavy line | Inequalities\n\nPractice Questions\nNumber Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 3516\nLocation: United States (CA)\nRe: What is the perimeter of a square with area 9p^2/16 ?\u00a0 [#permalink]\n\n### Show Tags\n\n03 Dec 2017, 18:47\n1\nBunuel wrote:\nWhat is the perimeter of a square with area 9p^2/16 ?\n\nA. 3p/4\nB. 3p^2/4\nC. 3p\nD. 3p^2\nE. 4p/3\n\nSince a side of the square is equal to \u221a(9p^2/16) = 3p/4, then the perimeter of the square is 4 x 3p/4 = 12p/4 = 3p.\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: What is the perimeter of a square with area 9p^2/16 ? &nbs [#permalink] 03 Dec 2017, 18:47\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-21 06:49:52", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-is-the-perimeter-of-a-square-with-area-9p-224080.html", "openwebmath_score": 0.6511421203613281, "openwebmath_perplexity": 8399.422690123378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8652240808393984}}
{"url": "https://www.bionicturtle.com/forum/threads/miller-chapter-4-distributions-study-notes-pg-75-question-2.8301/", "text": "What's new\n\n# Miller - Chapter 4 - Distributions - Study Notes, Pg 75, question #2\n\n#### Dr. Jayanthi Sankaran\n\n##### Well-Known Member\nHi David,\n\nIn question (2) referenced as above,\n\n(2) At the start of the year, a stock price is $100. A twelve-step binomial model describes the stock price evolution such that each month the extreme volatility price with either jump from S(t) to S(t)*u with 60.0% probability or down to S(t)*d with 40.0% probability. The up jump (u) = 1.1 and the down jump (d) = 1/1.1; note these (u) and (d) parameters correspond to an annual volatility of about 33% as exp[33%*SQRT(1/12)] ~= 1.10. At the end of the year, which is nearest to the probability that the stock price will be exactly$121?\n\n(a) 0.33%\n(b) 3.49%\n(c) 12.25%\n(d) 22.70%\n\nThere are 13 outcomes at the end of the 12-step binomial, with $100 as the outcome that must correspond to six up jumps and six down jumps. Therefore,$121 must be the outcome due to seven up jumps and five down jumps: $100*1.1^7*(1/1.1)^5 =$121\nSuch that we want the binomial probability given by:\nBinomial Prob [X = 7| n =12, p = 60%] = 22.70%\n\nClarification\n\nAlthough, intuitively, 13 outcomes at the end of the 12-step binomial seems right (ie expected value of stock price at the end of 12 steps ??), how do you figure out six up jumps and six down jumps from the Binomial model to get \\$100 as the outcome? Also how do you get :\nBinomial Prob [X = 7| n =12, p = 60%] = 22.70%\nI guess I am missing something that I should understand.\n\nThanks!\nJayanthi\n\n#### ShaktiRathore\n\n##### Well-Known Member\nSubscriber\nHi\nConsider n up jumps and 12-n down jumps to get 121 in the end\nTherefore,100*(1.1)^n*(1/1.1)^12-n=121=>1.1^(2n-12)=1.1^2=>2n-12=2=>2n=14=>n=7 therefore there are 7 up and 12-7=5 down jumps.\nBinomial probability is 12C7(.6)^7*(.4)^5=22.69%\nThanks\n\n#### Dr. Jayanthi Sankaran\n\n##### Well-Known Member\nHi Shakti,\n\nYes - that is indeed great! Thanks a tonne for making it so simple to understand!\n\nJayanthi\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nHi @Jayanthi Sankaran the source question thread contains a visual that should help clarify, too. Please see https://www.bionicturtle.com/forum/...ributions-i-miller-chapter-4.7025/#post-28837\nie..,\n\n#### Dr. Jayanthi Sankaran\n\n##### Well-Known Member\nThis is neat, David\n\nThanks!\nJayanthi\n\n#### Dr. Jayanthi Sankaran\n\n##### Well-Known Member\nHi David,\n\nDoes the above visual come with its own spreadsheet?\n\nThanks\nJayanthi\n\n#### Dr. Jayanthi Sankaran\n\n##### Well-Known Member\nThanks David - will go through it, at leisure\n\nWarm regards\nJayanthi\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nHi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:\n\u2022 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:\n\u2022 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:\n\u2022 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->\n\u2022 (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:\n\u2022 (1.1)^n * 1.1^(n-12) = 1.21,\n\u2022 (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and\n\u2022 (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!\n\n#### eva maria\n\n##### New Member\nHi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:\n\u2022 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:\n\u2022 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:\n\u2022 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->\n\u2022 (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:\n\u2022 (1.1)^n * 1.1^(n-12) = 1.21,\n\u2022 (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and\n\u2022 (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!\n\nMany thanks David\n\nLast edited by a moderator:", "date": "2021-05-17 12:20:03", "meta": {"domain": "bionicturtle.com", "url": "https://www.bionicturtle.com/forum/threads/miller-chapter-4-distributions-study-notes-pg-75-question-2.8301/", "openwebmath_score": 0.647357165813446, "openwebmath_perplexity": 5452.593190925554, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856415, "lm_q2_score": 0.8840392802184581, "lm_q1q2_score": 0.865220424512312}}
{"url": "https://math.stackexchange.com/questions/2664789/characterizing-discontinuous-derivatives", "text": "# Characterizing discontinuous derivatives\n\nApparently the set of discontinuity of derivatives is weird in its own sense. Following are the examples that I know so far:\n\n$1.$ $$g(x)=\\left\\{ \\begin{array}{ll} x^2 \\sin(\\frac{1}{x}) & x \\in (0,1] \\\\ 0 & x=0 \\end{array}\\right.$$ $g'$ is discontinuous at $x=0$.\n\n$2.$ The Volterra function defined on the ternary Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Cantor set ,that is on a nowhere dense set of measure zero.\n\n$3.$ The Volterra function defined on the Fat-Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Fat-Cantor set ,that is on a set of positive measure, but not full measure.\n\n$4.$ I am yet to find a derivative which is discontinuous on a set of full measure.\n\nQuestions:\n\n1.What are some examples of functions whose derivative is discontinuous on a dense set of zero measure , say on the rationals?\n\n2.What are some examples of functions whose derivative is discontinuous on a dense set of positive measure , say on the irrationals?\n\nUpdate: One can find a function which is differentiable everywhere but whose derivative is discontinuous on a dense set of zero measure here.\n\n\u2022 The Volterra function is not defined just on the Cantor set, but the whole interval. It equals $0$ on the Cantor set. I know you know this. But perhaps \"defined relative to the Cantor set\" might be more accurate. \u2013\u00a0zhw. Feb 24 '18 at 17:32\n\u2022 @zhw Yeah, but I think its pretty obvious though. :P \u2013\u00a0yasir Feb 24 '18 at 17:44\n\u2022 \"Some good discussion about this can be found here and here.\" The first link you gave gives several references to the fact that any $F_{\\sigma}$ first category set (equivalently, any countable union of closed nowhere dense sets) is the discontinuity set of some derivative. Those references that give a proof of this result do so in a constructive way, in the sense that given any countable union of closed nowhere dense sets (note that $\\mathbb Q$ is a countable union of singleton sets), a differentiable function is constructed whose derivative is discontinuous exactly on that countable union. \u2013\u00a0Dave L. Renfro Feb 24 '18 at 21:08\n\nThere is no everywhere differentiable function $f$ on $[0,1]$ such that $f'$ is discontinuous at each irrational there. That's because $f',$ being the everywhere pointwise limit of continuous functions, is continuous on a dense $G_\\delta$ subset of $[0,1].$ This is a result of Baire. Thus $f'$ can't be continuous only on a subset of the rationals, a set of the first category.\n\nBut there is a differentiable function whose derivative is discontinuous on a set of full measure.\n\nProof: For every Cantor set $K\\subset [0,1]$ there is a \"Volterra function\" $f$ relative to $K,$ which for the purpose at hand means a differentiable function $f$ on $[0,1]$ such that i)$|f|\\le 1$ on $[0,1],$ ii) $|f'|\\le 1$ on $[0,1],$ iii) $f'$ is continuous on $[0,1]\\setminus K,$ iv) $f'$ is discontinuous at each point of $K.$\n\nNow we can choose disjoint Cantor sets $K_n \\subset [0,1]$ such that $\\sum_n m(K_n) = 1.$ For each $n$ we choose a Volterra function $f_n$ as above. Then define\n\n$$F=\\sum_{n=1}^{\\infty} \\frac{f_n}{2^n}.$$\n\n$F$ is well defined by this series, and is differentiable on $[0,1].$ That's because each summand above is differentiable there, and the sum of derivatives converges uniformly on $[0,1].$ So we have\n\n$$F'(x) = \\sum_{n=1}^{\\infty} \\frac{f_n'(x)}{2^n}\\,\\, \\text { for each } x\\in [0,1].$$\n\nLet $x_0\\in \\cup K_n.$ Then $x_0$ is in some $K_{n_0}.$ We can write\n\n$$F'(x) = \\frac{f_{n_0}'(x)}{2^{n_0}} + \\sum_{n\\ne n_0}\\frac{f_n'(x)}{2^n}.$$\n\nNow the sum on the right is continuous at $x_0,$ being the uniform limit of functions continuous at $x_0.$ But $f_{n_0}'/2^{n_0}$ is not continuous at $x_0.$ This shows $F'$ is not continuous at $x_0.$ Since $x_0$ was an aribtrary point in $\\cup K_n,$ $F'$ is discontinuous on a set of full measure as desired.\n\n\u2022 @ zhw So, if $f'$ is discontinuous on a set $D$, then $D$ can be dense only if its Lebesgue measure is zero? \u2013\u00a0yasir Feb 25 '18 at 3:04\n\u2022 No, certainly not. A set of full measure is dense. \u2013\u00a0zhw. Feb 25 '18 at 7:09\n\u2022 @ zhw. Is the set on which $F'$ is continuous dense, as per your claim in first paragraph? \u2013\u00a0yasir Feb 25 '18 at 9:26\n\u2022 @yasir Yes, it has to be. \u2013\u00a0zhw. Feb 25 '18 at 17:14", "date": "2021-01-19 08:11:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2664789/characterizing-discontinuous-derivatives", "openwebmath_score": 0.9350493550300598, "openwebmath_perplexity": 113.93660227950194, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712651931994, "lm_q2_score": 0.8840392725805823, "lm_q1q2_score": 0.8652204208793727}}
{"url": "https://www.hpmuseum.org/forum/thread-12423.html", "text": "Small Solver Program\n02-14-2019, 05:25 AM\nPost: #1\n Gamo Senior Member Posts: 518 Joined: Dec 2016\nSmall Solver Program\nRecently I try to make a small but versatile solver program specifically\nfor HP Voyager Series calculator.\n\nThis solver solve for f(x) = 0 and use some type of counter for iteration.\nDue to the iteration process this program will run slow on Original Calculator.\nWith modern 12C, 15C, and all other emulator this will run extremely fast.\n\nThis program is intended for HP-11C but this example will use\nHP-15C emulator so that this solver program result can be compare\nwith the 15C build-in SOLVE function.\n\n---------------------------------------------------------\nTo run program.\n\nEnter 2 guesses closest to the root.\n\nX1 [ENTER] X2 [A] display answer // Use this Solver\n\nX1 [ENTER] X2 f [SOLVE] [E] display answer // Use 15C build-in SOLVE\n---------------------------------------------------------\nExample for X^X = Y // X to the power of X equal to Y\n\nX on R1 // Register Storage 1 // For this Sover\nY on R5 // Register Storage 5\n\nX on R2 // Register Storage 2 // For 15C SOLVE\nY on R5 // Register Storage 5\n--------------------------------------------------------\nProgram:\nCode:\n LBL\u00a0A\u00a0\u00a0\u00a0//\u00a0This\u00a0Solver STO\u00a00 Rv STO\u00a01 90\u00a0\u00a0\u00a0 STO\u00a0I CLx LBL\u00a01 RCL\u00a00 / CHS RCL\u00a01 + STO\u00a01 DSE\u00a0I GTO\u00a0B RCL\u00a01 RTN ------------------------------ LBL\u00a0B\u00a0\u00a0//\u00a0f(x)\u00a0program\u00a0for\u00a0X^X\u00a0=\u00a0Y RCL\u00a01 LN RCL\u00a01 x RCL\u00a05 LN - GTO\u00a01 ------------------------------- LBL\u00a0E\u00a0\u00a0//\u00a015C\u00a0SOLVE STO\u00a02\u00a0//\u00a0f(x)\u00a0program\u00a0for\u00a0X^X\u00a0=\u00a0Y LN RCL\u00a02 x RCL\u00a05 LN - RTN\n-------------------------------------------\nExample: X^X = 1000\n\nSTORE 1000 to R5 // [STO] 5\n\nThis Solver\n\n4 [ENTER] 6 f [A] display 4.555535704\n-------------------------------------------------------\n15C SOLVE\n\n4 [ENTER] 6 f [SOLVE] [E] display 4.555535705\n-------------------------------------------------------\nRemark:\nFor this solver answer is in R1\nFor 15C SOLVE answer is in R2\n\nGamo\n02-14-2019, 07:06 AM\nPost: #2\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\n(02-14-2019 05:25 AM)Gamo Wrote: \u00a0Example: X^X = 1000\n\nYou can rewrite this equation such that $$x$$ is a fixed point of $$f(x)$$, that is $$x=f(x)$$:\n\n$$x=\\frac{3}{LOG(x)}$$\n\nThen you can iterate the following program with a starting value, say $$x=4$$:\nCode:\n001-\u00a0\u00a043\u00a0\u00a013\u00a0:\u00a0\u00a0\u00a0\u00a0LOG 002-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03\u00a0:\u00a0\u00a0\u00a0\u00a03 003-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a034\u00a0:\u00a0\u00a0\u00a0\u00a0x<>Y 004-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010\u00a0:\u00a0\u00a0\u00a0\u00a0\u00f7\n\nExample:\n\n4\nR/S\n4.982892143\nR/S\n4.301189432\nR/S\n4.734933901\nR/S\n4.442378438\nR/S\n4.632377942\nR/S\n4.505830646\nR/S\n4.588735607\nR/S\n4.533824357\nR/S\n4.569933524\nR/S\n4.546075273\nR/S\n4.561789745\nR/S\n4.551417837\nR/S\n4.558254212\nR/S\n4.553744141\nR/S\n4.556717747\nR/S\n4.554756405\nR/S\n4.556049741\nR/S\n4.555196752\nR/S\n4.555759258\nR/S\n4.555388285\nR/S\n4.555632930\nR/S\n4.555471588\nR/S\n4.555577990\nR/S\n4.555507820\nR/S\n4.555554095\nR/S\n4.555523578\nR/S\n4.555543703\nR/S\n4.555530431\nR/S\n4.555539183\nR/S\n4.555533412\nR/S\n4.555537217\nR/S\n4.555534708\n\n4.555535704\nR/S\n4.555535706\nR/S\n4.555535704\nR/S\n4.555535706\n\nFrom then on the result flips between these last two values.\n\nCheers\nThomas\n02-14-2019, 07:15 AM (This post was last modified: 02-14-2019 07:16 AM by Thomas Klemm.)\nPost: #3\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nIf you don't want to press R/S repeatedly:\nCode:\n001-\u00a0\u00a043\u00a0\u00a013\u00a0:\u00a0\u00a0\u00a0\u00a0LOG 002-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03\u00a0:\u00a0\u00a0\u00a0\u00a03 003-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a034\u00a0:\u00a0\u00a0\u00a0\u00a0x<>Y 004-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010\u00a0:\u00a0\u00a0\u00a0\u00a0\u00f7 005-\u00a0\u00a042\u00a0\u00a031\u00a0:\u00a0\u00a0\u00a0\u00a0PSE 006-\u00a0\u00a043\u00a0\u00a040\u00a0:\u00a0\u00a0\u00a0\u00a0x=0\n02-15-2019, 12:07 AM (This post was last modified: 02-17-2019 12:16 PM by Albert Chan.)\nPost: #4\n Albert Chan Senior Member Posts: 696 Joined: Jul 2018\nRE: Small Solver Program\n(02-14-2019 07:06 AM)Thomas Klemm Wrote: \u00a0$$x=\\frac{3}{LOG(x)}$$\n\n4\nR/S\n4.982892143\nR/S\n4.301189432\n...\n\nIt would be nice if we can temper the oscillation, or slow convergence.\nLet x0 = 4, x1, x2 = the first two iterated values.\n\nRate = (x2-x1)/(x1-x0) ~ (4.30 - 4.98) / (4.98 - 4) = -0.694\n\nIf the same trend continued, we expect final % = 1/(1-r) ~ 60%\nUse weighted fixed-point equation x = 0.4x + 0.6 * 3/log10(x):\n\n4.6\n4.555924149\n4.555537395\n4.555535712\n4.555535705 (converged)\n\nEdit: compare with Newton's method, x = (ln(1000) + x) / (ln(x) + 1)\n4\n4.571001573\n4.555546101\n4.555535705 (converged)\n02-15-2019, 06:26 PM\nPost: #5\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\n(02-15-2019 12:07 AM)Albert Chan Wrote: \u00a0It would be nice if we can temper the oscillation, or slow convergence.\n\nWe can also use Aitken's delta-squared process to accelerate the speed of convergence:\n\n$$a_{n}=x_{n+2}-\\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$\n\nThis leads to the following program for the HP-11C:\nCode:\n001-42,21,\u00a01\u00a0:\u00a0\u00a0\u00a0\u00a0\u25b8LBL\u00a001 002-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a036\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0ENTER 003-\u00a0\u00a0\u00a0\u00a032\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0GSB\u00a000 004-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0- 005-\u00a0\u00a043\u00a0\u00a036\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0LSTx 006-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a036\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0ENTER 007-\u00a0\u00a0\u00a0\u00a032\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0GSB\u00a000 008-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0- 009-\u00a0\u00a043\u00a0\u00a036\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0LSTx 010-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a034\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0x<>y 011-\u00a0\u00a043\u00a0\u00a033\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0R\u2191 012-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a034\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0x<>y 013-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0- 014-\u00a0\u00a043\u00a0\u00a036\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0LSTx 015-\u00a0\u00a043\u00a0\u00a011\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0x\u00b2 016-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a034\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0x<>y 017-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0\u00f7 018-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0- 019-\u00a0\u00a043\u00a0\u00a032\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0RTN 020-42,21,\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0\u25b8LBL\u00a000 021-\u00a0\u00a043\u00a0\u00a012\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0LN 022-\u00a0\u00a0\u00a0\u00a045\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0RCL\u00a000 023-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a034\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0x<>y 024-\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0\u00f7 025-\u00a0\u00a043\u00a0\u00a032\u00a0:\u00a0\u00a0\u00a0\u00a0\u00a0RTN\n\nExample:\n\nSolve: $$x^x=1000$$\n\n1000 LN\nSTO 0\n\n4.6\nGSB 1\n4.555665779\nR/S\n4.555535706\nR/S\n4.555535705\nR/S\nError 0\n\nFeel free to add a check for 0 after line 13 to prevent the error and loop back to label 1 instead of returning in line 19.\n\nCheers\nThomas\n02-15-2019, 09:16 PM (This post was last modified: 02-15-2019 09:31 PM by Albert Chan.)\nPost: #6\n Albert Chan Senior Member Posts: 696 Joined: Jul 2018\nRE: Small Solver Program\n(02-15-2019 06:26 PM)Thomas Klemm Wrote:\n(02-15-2019 12:07 AM)Albert Chan Wrote: \u00a0It would be nice if we can temper the oscillation, or slow convergence.\n\nWe can also use Aitken's delta-squared process to accelerate the speed of convergence:\n\n$$a_{n}=x_{n+2}-\\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$\n\nAmazingly, my rate formula is same as Aitken extrapolation formula !\nAssuming we have 3 values, x0,x1,x2 and tried to guess where it should end up.\n\nMy rate analysis: r = (x2-x1)/(x1-x0) = \u0394x1 / \u0394x0\n\nWe need this for the proof:\n(\u0394x0)\u00b2\n= ((\u0394x0 - \u0394x1) + \u0394x1)\u00b2\n= (\u0394x0 - \u0394x1)\u00b2 + 2 * \u0394x1 (\u0394x0 - \u0394x1) + (\u0394x1)\u00b2\n\nIf same trend continue, where it will ends up\n= x0 + \u0394x0 * 1/(1-r)\n= x0 + (\u0394x0)\u00b2 / (\u0394x0 - \u0394x1)\n= x0 + (\u0394x0 - \u0394x1) + 2 * \u0394x1 + (\u0394x1)\u00b2 / (\u0394x0 - \u0394x1)\n= x0 + (x1-x0) + (x1-x2) + 2*(x2-x1) \u2212 (\u0394x1)\u00b2 / (\u0394x1 - \u0394x0)\n= x2 \u2212 (\u0394x1)\u00b2 / (\u0394x1 - \u0394x0)\n02-16-2019, 03:58 AM\nPost: #7\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\n(02-15-2019 09:16 PM)Albert Chan Wrote: \u00a0Amazingly, my rate formula is same as Aitken extrapolation formula !\n\nThat's not really a surprise, is it?\nIt's essentially the same idea: geometric series\n\n$$1\\,+\\,r\\,+\\,r^{2}\\,+\\,r^{3}\\,+\\,\\cdots \\;=\\;{\\frac {1}{1-r}}$$\n02-16-2019, 12:24 PM (This post was last modified: 02-16-2019 12:34 PM by Csaba Tizedes.)\nPost: #8\n Csaba Tizedes Senior Member Posts: 366 Joined: May 2014\nRE: Small Solver Program\n(02-14-2019 05:25 AM)Gamo Wrote: \u00a0a small but versatile solver program specifically for HP Voyager Series calculator.\nExample for X^X = Y // X to the power of X equal to Y\n\nYes, the classic problem: which number you can power to itself to get the largest number on your calculator: x^x=10^100. Do LOG() of both sides: x\u00d7LOG(x)=100, then do the iteration: x_0:=50, then x_i+1:=100/log(x_i). The simple fixed point iteration and converges.\n\nBy the way, if you interested, here is my SOLVE(i) for HP-12C, which can solve any equation for ANY variable like on 15C, without rewriting the equation. Looks like indirect addressing without indirect addressing. 28 steps only.\n\nYes, my English is terrible, but more convenient than read the full material in Hungarian.\nAt the end of pdf you can find lots of applications which solved with this little solver (like ODE solve with EULER+SOLVE(i) on 12C (!!!), pressure vessel pneumatic conveying pipe sizing program, humidity and dew point solving, outer temperature calculation of a concrete silo wall or comparison of present values of educated and not educated people salaries with the little SOLVE combined with 12C's cash flow calculation capabilities - I guess this is shows clearly the power of this little routine possibilities and why I say everywhere, the SOLVER is MUST to implement ALL current calculators, I hope somebody read and understand what I want to say).\n\nThe small secant method in this paper for CASIO fx-50F meantime reduced to 13 steps, I guess the smallest solver for these blind machines. You can find attached.\n\nCsaba\n\nAttached File(s)\n02-16-2019, 01:42 PM\nPost: #9\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\n(02-16-2019 12:24 PM)Csaba Tizedes Wrote: \u00a0By the way, if you interested, here is my SOLVE(i) for HP-12C\n\nSteps of secant method from 03 to 18:\nCode:\n03-\u00a0\u00a0\u00a045\u00a02\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a02 04-\u00a0\u00a0\u00a045\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a00 05-\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0- 06-\u00a0\u00a043\u00a036\u00a0:\u00a0\u00a0\u00a0\u00a0LSTx 07-\u00a0\u00a0\u00a044\u00a02\u00a0:\u00a0\u00a0\u00a0\u00a0STO\u00a02 08-\u00a0\u00a0\u00a0\u00a0\u00a035\u00a0:\u00a0\u00a0\u00a0\u00a0CLx 09-\u00a0\u00a0\u00a045\u00a03\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a03 10-\u00a0\u00a0\u00a045\u00a01\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a01 11-\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0- 12-\u00a0\u00a043\u00a036\u00a0:\u00a0\u00a0\u00a0\u00a0LSTx 13-\u00a0\u00a0\u00a044\u00a03\u00a0:\u00a0\u00a0\u00a0\u00a0STO\u00a03 14-\u00a0\u00a0\u00a0\u00a0\u00a033\u00a0:\u00a0\u00a0\u00a0\u00a0R\u2193 15-\u00a0\u00a0\u00a0\u00a0\u00a010\u00a0:\u00a0\u00a0\u00a0\u00a0\u00f7 16-\u00a0\u00a0\u00a045\u00a01\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a01 17-\u00a0\u00a0\u00a0\u00a0\u00a020\u00a0:\u00a0\u00a0\u00a0\u00a0\u00d7 18-44\u00a030\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0STO+\u00a00\n\nThis can be shortened to:\nCode:\n03-\u00a0\u00a0\u00a045\u00a02\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a02 04-\u00a0\u00a0\u00a045\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a00 05-\u00a0\u00a0\u00a044\u00a02\u00a0:\u00a0\u00a0\u00a0\u00a0STO\u00a02 06-\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0- 07-\u00a0\u00a0\u00a045\u00a03\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a03 08-\u00a0\u00a0\u00a045\u00a01\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a01 09-\u00a0\u00a0\u00a044\u00a03\u00a0:\u00a0\u00a0\u00a0\u00a0STO\u00a03 10-\u00a0\u00a0\u00a0\u00a0\u00a030\u00a0:\u00a0\u00a0\u00a0\u00a0- 11-\u00a0\u00a0\u00a0\u00a0\u00a010\u00a0:\u00a0\u00a0\u00a0\u00a0\u00f7 12-\u00a0\u00a0\u00a045\u00a01\u00a0:\u00a0\u00a0\u00a0\u00a0RCL\u00a01 13-\u00a0\u00a0\u00a0\u00a0\u00a020\u00a0:\u00a0\u00a0\u00a0\u00a0\u00d7 14-44\u00a030\u00a00\u00a0:\u00a0\u00a0\u00a0\u00a0STO+\u00a00\n\nDon't hesitate to publish your programs in this forum. You can use Google to translate to English.\n\nCheers\nThomas\n02-16-2019, 03:24 PM\nPost: #10\n Csaba Tizedes Senior Member Posts: 366 Joined: May 2014\nRE: Small Solver Program\n(02-16-2019 01:42 PM)Thomas Klemm Wrote: \u00a0Steps of secant method from 03 to 18 can be shortened to.\nThanks, this was eons before, I am sure this is only the first iteration and required to fine tuning.\n\n(02-16-2019 01:42 PM)Thomas Klemm Wrote: \u00a0Don't hesitate to publish your programs in this forum. You can use Google to translate to English.\nOK, sometimes I feel myself as Don Quixote... Frankly speaking, who want to discuss about calculators and calculator programming now? 15 years before I wrote something useful, today I am only (an angry) user...\n\nCsaba\n02-17-2019, 02:57 AM (This post was last modified: 02-17-2019 03:20 AM by Gamo.)\nPost: #11\n Gamo Senior Member Posts: 518 Joined: Dec 2016\nRE: Small Solver Program\nI just fine tune this solver program for HP-11C\n\nAdded the iterations limit to 28 loops count\nAdded the condition test to end program when root is found.\n\nIf iteration is over 28 loops count this indicate that user must use a closer guess.\n\n-------------------------------------------\n\nProcedure:\n\nUse R1 for unknown X in your equation. End your formula with [RTN]\n\nX1 ENTER X2 [A] display Answer\n\n------------------------------------------------------------------------\nProgram:\nCode:\n LBL\u00a0A STO\u00a00 Rv STO\u00a01 28 STO\u00a0I\u00a0\u00a0//\u00a0Store\u00a0Loop\u00a0Count\u00a0Limit CLx --------------------- LBL\u00a00 GSB\u00a0B GSB\u00a01 STO\u00a01 GSB\u00a0B GSB\u00a01 RCL\u00a01 X=Y\u00a0\u00a0//\u00a0End\u00a0if\u00a0Root\u00a0is\u00a0found GTO\u00a02 DSE\u00a0\u00a0//\u00a0Loop\u00a0Limit\u00a0Counter GTO\u00a00 CLx FIX\u00a09\u00a0\u00a0//\u00a00.000000000\u00a0indicate\u00a0that\u00a0Maximum\u00a0Loops\u00a0is\u00a0use\u00a0up. PSE PSE FIX\u00a04\u00a0 GTO\u00a02 --------------------- LBL\u00a01\u00a0\u00a0//\u00a0Solver\u00a0Routine RCL\u00a00 \u00f7 CHS RCL\u00a01 + RTN -------------------- LBL\u00a02 RCL\u00a01\u00a0\u00a0//\u00a0Answer\u00a0 RTN -------------------- LBL\u00a0B .\u00a0\u00a0//\u00a0f(x)\u00a0equation\u00a0start\u00a0here . .\u00a0\u00a0//\u00a0X\u00a0=\u00a0Register\u00a01\u00a0[R1] . . RTN\n\nGamo\n02-17-2019, 09:06 AM\nPost: #12\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\n(02-17-2019 02:57 AM)Gamo Wrote: \u00a0I just fine tune this solver program for HP-11C\n\nYour program doesn't work for this simple example: $$x^2+x=x(x+1)=0$$\nI've entered the following program:\nCode:\nLBL\u00a0B 1 RCL\u00a01 + LSTx \u00d7 RTN\n\nExample:\n\n-2\nENTER\n-0.5\nA\n\n9.999999 99\n\nThe reason is that the function value is divided by -0.5, making it larger and larger:\nCode:\nLBL\u00a01\u00a0\u00a0//\u00a0Solver\u00a0Routine RCL\u00a00 \u00f7 CHS RCL\u00a01 + RTN\n\nAnother thing is this in this loop:\nCode:\nLBL\u00a00 GSB\u00a0B GSB\u00a01 STO\u00a01 GSB\u00a0B GSB\u00a01 RCL\u00a01 X=Y\u00a0\u00a0//\u00a0End\u00a0if\u00a0Root\u00a0is\u00a0found GTO\u00a02 DSE\u00a0\u00a0//\u00a0Loop\u00a0Limit\u00a0Counter GTO\u00a00\n\nWhen you return to label 0, the same value in register 1 will be used as before.\nThis function evaluation can be avoided if you save the result of the previous evaluation.\n\nYou could move this block:\nCode:\nLBL\u00a02 RCL\u00a01\u00a0\u00a0//\u00a0Answer\u00a0 RTN\n\nOver here and avoid the jump GTO 2:\nCode:\nCLx FIX\u00a09\u00a0\u00a0//\u00a00.000000000\u00a0indicate\u00a0that\u00a0Maximum\u00a0Loops\u00a0is\u00a0use\u00a0up. PSE PSE FIX\u00a04\u00a0 LBL\u00a02 RCL\u00a01\u00a0\u00a0//\u00a0Answer\u00a0 RTN\n\nTake a look at Csaba's solution for the HP-12C using the secant method.\n\nCheers\nThomas\n02-17-2019, 02:33 PM\nPost: #13\n Gamo Senior Member Posts: 518 Joined: Dec 2016\nRE: Small Solver Program\nThanks Thomas Klemm\n\nI try single guess and it work.\n\nCHS 2 [A] display -1\n\nIf using two negative guesses and result is\nOverflow then use Single guess instead.\n\nFormula Routine is\n\nRCL 1\n1\n+\nRCL 1\nX\n\nGamo\n02-17-2019, 04:57 PM\nPost: #14\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\n(02-17-2019 02:33 PM)Gamo Wrote: \u00a0If using two negative guesses and result is Overflow then use Single guess instead.\n\nThen try this function $$f(x)=4x(x+1)$$:\nCode:\nLBL\u00a0B RCL\u00a01 1 + RCL\u00a01 \u00d7 4 \u00d7 RTN\n\nExamples:\n\n-0.5\nENTER\n0.5\nA\n\n9.999999 99\n\n-0.5\nENTER\nA\n\n9.999999 99\n\n0.5\nENTER\nA\n\n9.999999 99\n\nEven initial guesses very close to the solutions lead to the same result:\n\n-1.00001\nENTER\n-0.99999\nA\n\n9.999999 99\n\n-0.00001\nENTER\n0.00001\nA\n\n9.999999 99\n\nCheers\nThomas\n02-18-2019, 03:49 AM\nPost: #15\n Gamo Senior Member Posts: 518 Joined: Dec 2016\nRE: Small Solver Program\nYour're right that only work for certain situations.\n\nGamo\n02-18-2019, 05:20 AM\nPost: #16\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\nIf you compare your algorithm with Newton's method:\n\n$$x_{n+1}=x_{n}-\\frac {f(x_{n})}{f'(x_{n})}$$\n\nyou may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof.\n\nAnd with this in mind indeed the solution can now be found:\n\n$$f'(x) = \\frac{d}{dx}4 x (x + 1) = 8 x + 4$$\n\n$$f'(-1) = -4$$\n\n$$f'(0) = 4$$\n\nExamples:\n\n-1.5\nENTER\n-4\nA\n\n-1.00000\n\n0.5\nENTER\n4\nA\n\n0.00000\n\nThus your 2nd parameter is an estimate of the slope at the root.\n\nAnd then you don't really have to call the function twice with the same value:\nCode:\nLBL\u00a0A STO\u00a00 Rv STO\u00a01 28 STO\u00a0I\u00a0\u00a0//\u00a0Store\u00a0Loop\u00a0Count\u00a0Limit LBL\u00a00 GSB\u00a0B RCL\u00a00 \u00f7 STO\u00a0-\u00a01 X=0\u00a0\u00a0//\u00a0End\u00a0if\u00a0Root\u00a0is\u00a0found GTO\u00a01 DSE\u00a0\u00a0//\u00a0Loop\u00a0Limit\u00a0Counter GTO\u00a00 CLx FIX\u00a09\u00a0\u00a0//\u00a00.000000000\u00a0indicate\u00a0that\u00a0Maximum\u00a0Loops\u00a0is\u00a0use\u00a0up. PSE PSE FIX\u00a04 LBL\u00a01 RCL\u00a01\u00a0\u00a0//\u00a0Answer\u00a0 RTN LBL\u00a0B .\u00a0\u00a0//\u00a0f(x)\u00a0equation\u00a0start\u00a0here . .\u00a0\u00a0//\u00a0X\u00a0=\u00a0Register\u00a01\u00a0[R1] . . RTN\n\nSo the question remains how to approximate the derivation at the roots.\nIf you keep record of the previous function evaluation you can get an estimate of the slope using the secant method.\n\nCheers\nThomas\n02-18-2019, 07:46 PM\nPost: #17\n Dieter Senior Member Posts: 2,398 Joined: Dec 2013\nRE: Small Solver Program\n(02-18-2019 05:20 AM)Thomas Klemm Wrote: \u00a0If you compare your algorithm with Newton's method:\n\n$$x_{n+1}=x_{n}-\\frac {f(x_{n})}{f'(x_{n})}$$\n\nyou may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof.\n\nYes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.\n\nFor more details take a look at this post: http://www.hpmuseum.org/forum/thread-123...#pid111680\n\nGamo, you really should correct and update the instructions. Here it still says:\nQuote:Enter 2 guesses closest to the root.\n\nBut again: that's NOT true. The first input is NOT a guess for the root. You now have seen a variety of cases that prove it.\n\nDieter\n02-18-2019, 10:22 PM\nPost: #18\n Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013\nRE: Small Solver Program\n(02-18-2019 07:46 PM)Dieter Wrote: \u00a0Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.\n\nYou could think him stubborn.\nBut it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:\nQuadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG\nWhich I still do not understand in detail.\n\nSo all in all I'm happy with the outcome.\n\nCheers\nThomas\n02-19-2019, 01:10 AM (This post was last modified: 02-19-2019 01:22 AM by Albert Chan.)\nPost: #19\n Albert Chan Senior Member Posts: 696 Joined: Jul 2018\nRE: Small Solver Program\nQuadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG\n\nLinear regression on slope had another benefit, which my Casio cannot do.\nDoing Quadratic Regression on Casio, you don't really know if it is any good.\nThere is no \"correlation coefficient\" to lookup. (Why?)\n\nLinear regression on slope can show whether a Quadratic curve is warranted.\n\nI think the code add slope data, like this:\n\nData points (0,90) and (30, 90.2), add to linear regression: (30+0)/2, (90.2-90)/(30-0)\nNext point (60, 86.4), add to linear regression: (60+30)/2, (86.4-90.2)/(60-30)\n...\nN points (sorted) thus generate N-1 slopes to regress.\n02-19-2019, 08:39 AM\nPost: #20\n Csaba Tizedes Senior Member Posts: 366 Joined: May 2014\nRE: Small Solver Program\n(02-18-2019 10:22 PM)Thomas Klemm Wrote:\n(02-18-2019 07:46 PM)Dieter Wrote: \u00a0Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.\n\nYou could think him stubborn.\nBut it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:\nQuadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG\nWhich I still do not understand in detail.\n\nSo all in all I'm happy with the outcome.\n\nCheers\nThomas\n\nI feel myself like a star (at least for 15 minutes...)\n\nYes, this is my \"linear regression on slope\", as Chan wrote. When I was in my \"Summer Practice\" at a company I have modelled lots of water network flows between pump houses (sources), consumers and reservoirs, and sometimes the characteristics curves given only in paper form. Thats why I wrote this little routine.\n\nAs you can see, if the measured points has significant error, the slope will be very inaccurate then the regression also inaccurate and the coefficients can be meaningless.\n\nBut as a programming question this was an interesting \"game\" for me.\n\nTo the original question from Gamo: I will check carefully the lists here, but I have another idea: what if, if we just simple walking along on the function's curve until the sign is changing, then we turning back, decreasing the stepsize and walking again. If the sign of the function changed, we are turning back again, decreasing the size of the steps and so on... If the stepsize is less than a small value, we found the root.\n\nThis \"marching method\" not so elegant, not so fast, but good to play with it, just for a try - maybe it can be shorter than the other methods. (I have an exactly same method for finding a minimum point of a function for CASIO fx-3650P/Sencor SEC103; perfectly works, like a ball at the bottom of a hole, slowly find the deepest point, when swings around the extremum).\n\nCsaba\n \u00ab Next Oldest | Next Newest \u00bb\n\nUser(s) browsing this thread: 1 Guest(s)", "date": "2019-10-22 14:27:24", "meta": {"domain": "hpmuseum.org", "url": "https://www.hpmuseum.org/forum/thread-12423.html", "openwebmath_score": 0.3751296401023865, "openwebmath_perplexity": 2839.9652737837514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018426872776, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8652035633588538}}
{"url": "https://math.stackexchange.com/questions/2906547/representing-displacement-vectors-in-cylindrical-coordinates-and-finding-the-dis", "text": "# Representing displacement vectors in cylindrical coordinates and finding the distance in cylindrical coordinates?\n\nIn cartesian coordinates, we can derive the vector $\\vec v_3$ by vector subtraction $\\vec v_2-\\vec v_1$. We then get the distance between $P$ och $Q$ by taking the absolute value of $\\vec v_3$ which then is: $$\\lvert \\vec v_3\\rvert = \\lvert \\vec v_2-\\vec v_1 \\rvert= \\lvert (x_2,y_2,z_2)-(x_1,y_1,z_1) \\rvert = \\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$\n\nBut how can we do this completely in cylindrical coordinates without converting to cartesian coordinates? We only have, and can only use, $\\rho_1$ and $\\theta_1$ for $\\vec v_1$ as well as $\\rho_2$ and $\\theta_2$ for $\\vec v_2$ in cylindrical coordinates $(\\rho,\\theta,z)$, where $z=0$ in this example.\n\nYour problem for $z=0$ reduces to the polar coordinate problem already answered here. $$|\\vec v_3|^2=(\\vec v_2-\\vec v_1)\\cdot(\\vec v_2-\\vec v_1)=|\\vec v_1|^2+|\\vec v_2|^2-2\\vec v_2\\cdot \\vec v_1$$ Now using cylindrical coordinates, $|\\vec v_i|^2=r_i^2+z_i^2$ and $\\vec v_2\\cdot \\vec v_1=r_1r_2\\cos(\\theta_2-\\theta_1)+z_1z_2$. Then the final answer will be $$|\\vec v_3|^2=r_1^2+r_2^2-2r_1r_2\\cos(\\theta_2-\\theta_1)+(z_2-z_1)^2$$\n\u2022 Yes, I've seen this solution, but isn't it based in terms of cartesian coordinates/base vectors, $\\hat e_x$, $\\hat e_y$ and $\\hat e_z$ after all? This since, I guess, you must express a distance in constant base vectors? I'm a bit confused about how to interpret the problem I have to admit. How would it look if I want to express the solution completely in cylindrical coordinates with $\\vec v_1=\\rho_1 \\hat e_\\rho (\\theta_1)$ and base vectors $\\hat e_\\rho$, $\\hat e_\\theta$, and $\\hat e_z$ etc? Would that be possible? Sep 6, 2018 at 5:46\n\u2022 I did not use Cartesian coordinates. The $z$ coordinate is there in cylindrical system. The only other reference to the Cartesian system is the angle $\\theta$. The issue that you have is that the basis of the cylindrical coordinate system changes with the vector, therefore equations will be more complicated. Sep 6, 2018 at 6:38\n\u2022 If you choose one of the vectors as the reference, the cylindrical coordinate system is just a Cartesian system rotated with respect to the original. $\\hat e_\\rho$ is the new $\\hat e_x$, and $\\hat e_\\theta$ is the new $\\hat e_y$. Sep 6, 2018 at 7:19", "date": "2022-05-24 16:42:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2906547/representing-displacement-vectors-in-cylindrical-coordinates-and-finding-the-dis", "openwebmath_score": 0.9444844126701355, "openwebmath_perplexity": 145.57471130022128, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9752018354801187, "lm_q2_score": 0.8872045929715078, "lm_q1q2_score": 0.8652035475122061}}
{"url": "https://math.stackexchange.com/questions/1136207/how-to-determine-the-curve", "text": "# How to determine the curve?\n\nIn the figure above, segment $PQ$ is determined by two points: $P: (t,0)$ and $Q: (1,t)$, where $t\\in [0,1]$ continuously increases and decreases between $0$ and $1$.\n\nThen this gives a close region swept by $PQ$, the upper edge of which is a curve.\n\nHow to determine the equation of the curve (maybe implicit form)?\n\nMy own method\n\nSuppose the curve is $y=y(x)$, then for any fixed $x_0\\in (0,1)$, there is a vertical line $x=x_0$, which intersects a bundle of such $PQ(t)$ segments:\n\n$$y=\\frac{t(x-t)}{1-t}$$\n\nEasy to conclude that, the desired $y_0$ on the curve corresponding to $x_0\\left(\\in(0,1)\\right)$ is the maximum of:\n\n$$y_0= \\max\\limits_{t\\in (0,1)}\\frac{t(x_0-t)}{1-t}=2-2\\sqrt{1-x_0}-x_0$$\n\nHow to use the envelope concept? Is there any elementary method since the area is $\\frac{1}{6}$?\n\n\u2022 Do you know derivatives, tangent lines and integrals? \u2013\u00a0Tomas Feb 6 '15 at 9:47\n\u2022 Yes, I know; I am not sure whether calculus is necessary in order to determine the area, but I am pretty sure in order to determine the curve, calculus would be a help \u2013\u00a0LCFactorization Feb 6 '15 at 9:53\n\u2022 Look at the wiki entry for Envelope, it teach you how to determine the curve. \u2013\u00a0achille hui Feb 6 '15 at 10:29\n\u2022 I use optimization method and obtains $y=2-2\\sqrt{1-x}-x$, then the area is $1/6$. Is there any elementary method in order to determine the area? \u2013\u00a0LCFactorization Feb 6 '15 at 11:12\n\nI know that the result will be a conic section. I'll prove that later on, but start with it. I come from a background of projective geometry, so I'd homogenize your points by appending a $1$, then find coordinates for the line joining them by computing the cross product.\n\n$$\\begin{pmatrix}t\\\\0\\\\1\\end{pmatrix}\\times \\begin{pmatrix}1\\\\t\\\\1\\end{pmatrix}= \\begin{pmatrix}-t\\\\1-t\\\\t^2\\end{pmatrix}$$\n\nThis could also be written as $-tx + (1-t)y + t^2=0$ in usual Cartesian coordinates. It is a special case of a line equation $ax+by+c=0$. Now I want to describe a conic section in the dual sense, i.e. not as a set of points but instead a set of tangent lines. That means I need to find a homogeneous quadratic form in $a,b,c$ which is zero for the line given above. For that, consider all quadratic coefficients:\n\n\\begin{align*} a^2 &= t^2 & ab &= t^2-t & b^2 &= 1-2t+t^2 \\\\ ac &= -t^3 & bc &= t^2-t^3 & c^2 &= t^4 \\end{align*}\n\nThe $c^2$ expression is the only one with degree $4$, and the $b^2$ expression is the only one with constant term. So these two can't be part of the quadratic equation, since they have nothing to cancel against. After removing them, the $ab$ expression is the only one with linear term, so we drop that as well. Now the relation is easy to see:\n\n$$a^2 + ac - bc = 0$$\n\nWritten as a matrix:\n\n$$(a,b,c)\\cdot\\begin{pmatrix}2&0&1\\\\0&0&-1\\\\1&-1&0\\end{pmatrix} \\cdot\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = 0$$\n\nNow, as I said, that's the matrix for the dual conic. The primary conic is represented by the inverse matrix, or any multiple thereof:\n\n$$(x,y,1)\\cdot\\begin{pmatrix}1&1&0\\\\1&1&-2\\\\0&-2&0\\end{pmatrix} \\cdot\\begin{pmatrix}x\\\\y\\\\1\\end{pmatrix} = 0$$\n\nSince the determinant of the upper left $2\\times2$ matrix is zero, this is a parabola. It can also be described by the equation\n\n\\begin{align*} x^2 + 2xy + y^2 &= 4y \\\\ (x+y)^2 &= 4y \\end{align*}\n\nSince this is the primal conic to the dual one we computed before, and that dual one was deduced from a relation between the terms of your lines, this ensures that your whole family of lines will be tangent to this curve. If you like, you can compute the point of tangency as\n\n$$\\begin{pmatrix}2&0&1\\\\0&0&-1\\\\1&-1&0\\end{pmatrix} \\cdot\\begin{pmatrix}-t\\\\1-t\\\\t^2\\end{pmatrix} =-\\begin{pmatrix}2t-t^2\\\\t^2\\\\1\\end{pmatrix}$$\n\nSo the point $(2t-t^2, t^2)$ lies on both the parabola and the line.\n\nHINT\n\nI would say\n\n$F(x,y,t)=0\\equiv t(x-t)-y(1-t)=0,\\quad t\\in\\langle 0,1 \\rangle$\n\n$F'_t(x,y,t)=0\\equiv x-2t+y=0$\n\nSolving this system of equations for the x, y:\n\n$x=t(2-t), y=t^2, \\quad t\\in\\langle 0,1 \\rangle$ is a parametric curve of equation to search.\n\nPlot\n\n\u2022 Very simple and neat answer. All three are acceptable. I choose the first one that also uses projective geometry. Thank you very much! @MvG \u2013\u00a0LCFactorization Feb 7 '15 at 2:40\n\nI have a simpler solution.\n\n1. Find the equation of the line with $(t,0,1)\\times(1,t,1)=(-t,1-t,t^2)$ or $$(-t)x+(1-t)*y+(t^2)=0$$\n2. Find the extema $t$ with $$\\frac{{\\rm d}}{{\\rm d}t}\\left(-t\\,x+(1-t)\\,y+t^2\\right)=0$$ $$\\left. -x-y+2 t=0 \\right\\}\\; t=\\frac{x+y}{2}$$\n3. Plug $t$ into line equation for $$-\\frac{(x+y)^2}{4}+y=0$$ which is the resulting curve\n\nHint: The envelope of an implicit curve $f(x,y,t)=0$ is found by solving for $t$ in $\\frac{\\partial f}{\\partial t}=0$ and using it into the original equation.\n\nMy other answer describes how I myself think about this. But I realize that there is a lot of background knowledge in there, which might make this less accessible to members of other communities than projective geometry. So here is my attempt at a more generic solution.\n\nSuppose you have found the equation of the line to be $tx + (t-1)y = t^2$. Take a second line from your family, using $u$ instead of $t$ as the parameter, i.e. $ux + (u-1)y = u^2$. These two intersect in a point $(t+u-tu, tu)$. Now if you consider $\\lim_{u\\to t}$ then strictly speaking the point of intersection becomes undefined, but the above will simply become $(2t-t^2,t^2)$. So that's the one point that your line does not have in common with any other line of the family (extended from $t\\in[0,1]$ to $t\\in\\mathbb R$). So at that point this line is the one defining the curve. Therefore that's your parametric equation of the envelope curve.", "date": "2021-03-09 10:19:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1136207/how-to-determine-the-curve", "openwebmath_score": 0.9873470664024353, "openwebmath_perplexity": 191.02262168142678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877010373297, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8651961475649962}}
{"url": "https://www.studyxapp.com/homework-help/suppose-you-have-a-function-dt-that-gives-the-total-distance-traveled-up-to-tim-q90979419", "text": "# Question Solved1 Answer1. 2. 3. Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).\n\nCW7XUD The Asker \u00b7 Calculus\n\n1.\n\n2.\n\n3.\n\nTranscribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).\nMore\nTranscribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).\nSee Answer\nAdd Answer +20 Points\nCommunity Answer\nCXSSZN The First Answerer\nSee all the answers with 1 Unlock\nGet 4 Free Unlocks by registration\n\n1)the average velocity,Average Velocity =(\" Total Displacement \")/(\" Total Time \")Average Velocity =(d(b)-d(a))/(b-a)2)we ... See the full answer", "date": "2023-01-30 21:34:45", "meta": {"domain": "studyxapp.com", "url": "https://www.studyxapp.com/homework-help/suppose-you-have-a-function-dt-that-gives-the-total-distance-traveled-up-to-tim-q90979419", "openwebmath_score": 0.8647242784500122, "openwebmath_perplexity": 320.8483050728089, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877007780707, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.8651961457998267}}
{"url": "https://math.stackexchange.com/questions/3235160/spanning-forests-of-bipartite-graphs-and-distinct-row-column-sums-of-binary-matr", "text": "# Spanning forests of bipartite graphs and distinct row/column sums of binary matrices\n\nLet $$F_{m,n}$$ be the set of spanning forests on the complete bipartite graph $$K_{m,n}$$. Let $$S_{m,n} = \\{(r(M), c(M)), M \\in B_{m,n} \\}$$ where $$B_{m,n}$$ is the set of $$m \\times n$$ binary matrices and $$r(M), c(M)$$ are the vectors of row sums and column sums of $$M$$, respecitvely. That is, $$S_{m,n}$$ is the set of the distinct row-sum, column-sum pairs of binary matrices. (The term spanning forest here refers to a forest that spans all of the vertices of the given graph; it doesn't have to be a maximal acyclic subgraph.)\n\nQ: Is it true that $$|F_{m,n}| = |S_{m,n}|$$? It is true for $$m,n \\leq 4$$. For $$m=n$$ this is OEIS A297077.\n\nThere is an obvious mapping from $$F_{m,n} \\rightarrow B_{m,n}$$ given by taking the reduced adjacency matrix, so if $$U = \\{u_1, \\ldots, u_m\\}$$, $$V = \\{v_1, \\ldots, v_m\\}$$ are the color categories we set $$M_{ij} = 1$$ if $$v_i \\sim u_j$$ in a forest $$F$$, else $$M_{ij} = 0$$. However, this does not help because multiple forests may have the same row and column sums - and not every row-sum, column-sum pair is represented by a forest under this mapping.\n\nThe numbers $$|F_{m,n}|$$ are given here:\n\n$$\\begin{array}{|c|c|}\\hline m\\backslash n & 1 & 2 & 3 & 4 & 5 \\\\ \\hline 1 & 2 & \\\\\\hline 2 & 4 & 15 \\\\\\hline 3 & 8 & 54 & 328 \\\\\\hline 4 & 16 & 189 & 1856 & 16145 \\\\\\hline 5 & 32 & 648 & 9984 & 129000 & 1475856 \\\\\\hline\\end{array}$$\n\nFor more see this answer (sum each row.)\n\n\u2022 The linked MSE thread calculates $|F_{m,n}|$, but how did you calculate $|S_{m,n}|$ to verify they are the same for $m,n\\le 4$? Did you just generate all $2^{mn}$ binary matrices? \u2013\u00a0antkam May 24 '19 at 19:24\n\u2022 Yes, I did a brute force count. \u2013\u00a0Jair Taylor May 24 '19 at 22:32\n\u2022 Just curious: what made you do the brute force count to find out they're equal? Nothing in the \"obvious mapping\" would suggest (to me) that they are equal, so it would never have occurred to me to even try. \u2013\u00a0antkam May 25 '19 at 22:54\n\u2022 @antkam I was interested in counting forests and found that the relevant OEIS entry only mentioned row-sum / column-sums. \u2013\u00a0Jair Taylor May 26 '19 at 1:01\n\u2022 In my Answer, when interpreting $F_{m,n}$, I wrote \"In your definition, a spanning forest is any subset $T$ of edges of $K_{2,n}$ which is acyclic.\" So e.g. the empty $T=\\emptyset$ qualifies, and in general nodes can be isolated/unconnected. First of all, is my interpretation correct? Second, if I'm correct, then IMHO your use of the word \"spanning\" is a little confusing. I understand you probably want to make sure unconnected nodes are still included, and simply saying \"forest\" might also be confusing in a different way... \u2013\u00a0antkam May 30 '19 at 12:49\n\nI can prove the conjecture, although I can't give a bijection. I've left up a previous answer which does the case $$m=3$$, since it serves as an example of many of the ideas here.\n\nHere is the key observation:\n\nTheorem: There is a unique array $$f(m,n)$$, for $$m$$ and $$n$$ nonnegative integers, satisfying:\n\n1. $$f(m,0) = f(0,n)=1$$\n\n2. If we fix $$m \\geq 1$$, then $$f(m,n)$$ is of the form $$(m+1)^n p_m(n)$$ where $$p_m(x)$$ is a polynomial of degree $$\\leq m-1$$.\n\n3. If we fix $$n \\geq 1$$, then $$f(m,n)$$ is of the form $$(n+1)^m p_n(m)$$ where $$p_n(x)$$ is a polynomial of degree $$\\leq n-1$$.\n\nUniqueness: Let's suppose that we have already found unique values of $$f(m,n)$$ whenever $$\\min(m,n) < k$$. In particular, we know the values of $$f(k,x)$$ and $$f(x,k)$$ for $$0 \\leq x \\leq k-1$$. This is enough values to uniquely determine a degree $$k-1$$ polynomial, so the values we already know determine the values of $$f(k,x)$$ and $$f(x,k)$$ for all $$x$$.\n\nExistence: The proof of uniqueness gives an algorithm to compute $$f$$. The only potential issue is that $$f(k,k)$$ is determined twice, as a polynomial in its first variable and in its second variable. However, the algorithm is symmetric in $$m$$ and $$n$$, so the two interpolating polynomials coincide. $$\\square$$\n\nFor example, we have $$f(1,0) = 1 = 2^0 \\cdot 1$$ so we must have $$p_1(n)=1$$ and $$f(1,n) = 2^n$$. Similarly, $$f(2,0)=1=2^0 \\cdot 1$$ and $$f(2,1) = 4=3^1 \\cdot (4/3)$$, so we must have $$p_2(n) = 1+n/3$$ and $$f(2,n) = 3^n(1+n/3) = 3^{n-1} (n+3)$$. Using this algorithm, I computed $$f(m,n)$$ for $$0 \\leq m,n \\leq 10$$. Here is the data, and some Mathematica code, if you'd like to play with it.\n\nf[m_, n_] := f[m, n] =\nIf[m == 0 || n == 0, 1,\nIf[m <= n,\n(m + 1)^n * InterpolatingPolynomial[Table[{k, f[m, k]/(m + 1)^k}, {k, 0, m - 1}], x] /. x -> n,\n(n + 1)^m * InterpolatingPolynomial[Table[{k, f[k, n]/(n + 1)^k}, {k, 0, n - 1}], x] /. x -> m]]\n\n\nValues of $$f(m,n)$$ for $$0 \\leq m,n \\leq 10$$.\n\n1,1,1,1,1,1,1,1,1,1,1\n1,2,4,8,16,32,64,128,256,512,1024\n1,4,15,54,189,648,2187,7290,24057,78732,255879\n1,8,54,328,1856,9984,51712,260096,1277952,6160384,29229056\n1,16,189,1856,16145,129000,968125,6925000,47690625,318500000,2073828125\n1,32,648,9984,129000,1475856,15450912,151201728,1403288064,12479546880,107152782336\n1,64,2187,51712,968125,15450912,219682183,2862173104,34828543449,401200569280,4418300077219\n1,128,7290,260096,6925000,151201728,2862173104,48658878080,760774053888,11122973573120,153936323805184\n1,256,24057,1277952,47690625,1403288064,34828543449,760774053888,15047189968833,274908280855680,4707029151392121\n1,512,78732,6160384,318500000,12479546880,401200569280,11122973573120,274908280855680,6199170628499200,129708290461760000\n1,1024,255879,29229056,2073828125,107152782336,4418300077219,153936323805184,4707029151392121,129708290461760000,3283463201858585471\n\n\nSo, we can prove the conjecture by showing that both $$F_{m,n}$$ and $$S_{m,n}$$ obey conditions 1, 2 and 3. For condition 1, we can just define $$S_{m,0}$$, $$S_{0,n}$$, $$F_{m,0}$$, and $$F_{0,n}$$ to be singletons (and I claim this is actually the most natural definition); then we just need to check that our verification of polynomiality goes all the way down to the zero case.\n\nSince conditions 2 and 3 are symmetric, and the definitions of $$F_{m,n}$$ and $$S_{m,n}$$ are symmetric, we just check condition $$2$$.\n\nVerification of condition 2 for forests: Let $$A_{m,b}$$ be the set of isomorphism classes of bicolored forests whose white vertices are labeled $$\\{1,2,\\ldots,m \\}$$ and which have $$b$$ black vertices, each of degree $$\\geq 2$$. We claim that $$|F_{m,n}| = \\sum_b |A_{m,b}| n(n-1)(n-2) \\cdots (n-b+1) (m+1)^{n-b} . \\qquad (1)$$\n\nProof: Take a forest in $$F_{m,n}$$, with the $$m$$ vertices colored white and the $$n$$ vertices colored black. Delete all black vertices of degree $$0$$ and $$1$$ and forget the numbering of the remaining black vertices. This gives a forest in $$A_{m,b}$$. To undo this process, we first must take the $$b$$ black vertices and decide which of the $$n$$ vertices of $$K_{m,n}$$ they will be; there are $$n(n-1)(n-2) \\cdots (n-b+1)$$ ways to do this. (We used that trees in $$A_{m,b}$$ have no automorphisms.) Then we must take the $$n-b$$ remaining black vertices and either join them to one of the $$m$$ white vertices or leave them unconnected; there $$(m+1)^{n-b}$$ ways to do this.\n\nFormula (1) is clearly $$(m+1)^n$$ times a polynomial, it remains to check that the polynomial has degree at most $$m-1$$. We must check that $$A_{m,b}$$ is empty if $$b \\geq m$$. Indeed, since every black vertex has degree at least $$2$$, a forest in $$A_{m,b}$$ has at least $$2b$$ edges. But, since it is a forest, it also has at most $$m+b-1$$ vertices. So $$2b \\leq m+b-1$$ and $$b \\leq m-1$$. $$\\square$$\n\nVerification of condition 2 for margins of $$(0,1)$$ matrices: Consider a vector $$C = (C_1, \\ldots, C_n) \\in \\{ 0,1,\\ldots, m \\}^n$$ of columns sums, and consider how many row sums $$(R_1, \\ldots, R_m)$$ are compatible with it. Let $$c_j = \\# \\{ k : C_k = j \\}$$, so $$c_0+c_1+\\cdots + c_m=n$$.\n\nBy the Gale-Ryser theorem, $$(R_1, \\ldots, R_m)$$ is compatible with $$(c_0, \\cdots, c_m)$$ if and only if $$\\sum R_i = \\sum j c_j$$ and, for all index sets $$1 \\leq i_1 < i_2 < \\cdots < i_k \\leq m$$, we have $$R_{i_1} + R_{i_2} + \\cdots + R_{i_k} \\leq \\sum_j \\min(j,k) c_j .$$ This is the defining list of inequalities of the permutahedron, so this condition can alternately be stated as saying that $$(R_1, \\ldots, R_m)$$ is in the convex hull of the $$m!$$ permutations of $$(c_1+c_2+\\cdots+c_m, c_2+\\cdots + c_m, \\cdots, c_{m-1}+c_m, c_m)$$.\n\nBy Theorem 11.3 of Postnikov's Permutohedra, associahedra, and beyond, the number of such $$(R_1, \\ldots, R_m)$$ is a polynomial $$g_m(c_0,c_1, \\ldots, c_m)$$ in the $$c_j$$, of degree $$m-1$$. So $$|S_{m,n}| = \\sum_{c_0+\\cdots + c_m=n} \\frac{n!}{c_0! c_1! \\cdots c_m!} \\ g_m(c_0,c_1, \\ldots, c_m). \\qquad (2)$$ Let $$\\partial_j$$ be $$\\tfrac{\\partial}{\\partial x_j}$$. There is some polynomial $$h_m$$ in the $$\\partial_j$$, of degree $$m-1$$, such that $$\\left. h_m(\\partial_0, \\ldots, \\partial_m) \\left( x_0^{c_0} \\cdots x_m^{c_m} \\right) \\right|_{(1,1,\\ldots,1)} = g_m(c_1, \\ldots, c_m). \\qquad (3)$$ Combining (2), (3) and the binomial expansion of $$(x_0+x_1+\\cdots+x_m)^n$$, we obtain $$|S_{m,n}| = \\left. h_m(\\partial_0, \\ldots, \\partial_m) \\left( x_0+x_1+\\cdots+x_m \\right)^n \\right|_{(1,1,\\ldots,1)} . \\qquad (4)$$ Let $$h_m(t,t,\\ldots,t) = \\sum_{k=0}^{m-1} h_{m,k} t^k$$. Then $$(4)$$ is $$\\sum_{k=0}^{m-1} h_{m,k} n(n-1)(n-2) \\cdots (n-k+1) (m+1)^{n-k}$$, which is a polynomial of the desired form. $$\\square$$\n\n\u2022 Very cool! I think you need another index on your coefficients $c_k$ (say $c_{m,k}$). Then you've shown $c_{m,k} = |A_{m, k}|$ which is interesting as well. I'd be curious to see if this could be generalized more - e.g., to forests with a specified number of components. But I don't know what the appropriate statistic on row-sums/column-sums would be. \u2013\u00a0Jair Taylor Jun 2 '19 at 21:16\n\u2022 @JairTaylor Nice observation. Your comment made me realize I had used $c_j$ to mean two things, so I changed the second one to $h_{m,k}$; your observation then is that I have shown $h_{m,k} = |A_{m,k}|$. \u2013\u00a0David E Speyer Aug 27 '19 at 15:38\n\nThis conjectured equivalence (if true) is still amazing to me. However, here is a baby step: The conjecture is indeed true for $$m=2$$. Perhaps someone else will find this useful as a base for generalization.\n\nClaim: $$|F_{2,n}| = |S_{2,n}|$$ for all $$n$$\n\nFirst, lets consider $$F_{2,n}$$. In your definition, a spanning forest is any subset $$T$$ of edges of $$K_{2,n}$$ which is acyclic. Let $$u,v$$ be the two nodes on the $$m=2$$ side, and let $$S(u)$$ be the neighbors of $$u$$, i.e. the subset of nodes (on the $$n$$-node side) which $$u$$ is connected to in $$T$$. Similarly for $$S(v)$$.\n\nNext, note that $$T$$ contains a cycle iff $$S = S(u) \\cap S(v)$$ contains $$2$$ nodes (or more). So, $$F_{2,n}$$ can be counted by counting all cases where $$S$$ contains $$0$$ or $$1$$ node.\n\n\u2022 $$|S|=0$$ case: Each of $$n$$ nodes can be connected to $$u$$, or to $$v$$, or to neither. Total no. $$= 3^n$$.\n\n\u2022 $$|S|=1$$ case: There are $$n$$ ways to pick the unique $$w \\in S$$. After that, each of the remaining $$n-1$$ nodes can be connected to $$u$$, or to $$v$$, or to neither. Total no. $$= n\\, 3^{n-1}$$\n\n\u2022 Summing up, $$|F_{2,n}| = 3^n + n \\,3^{n-1} = 3^{n-1} (3+n),$$ which matches your table.\n\nSecond, lets consider $$S_{2,n}$$. There are $$n$$ column sums, each of which can be $$0, 1, 2$$. Let $$x,y,z$$ be the number of columns whose sum $$=0, 1, 2$$ respectively. The columns whose sum $$=0$$ or $$2$$ dictate their elements. In the $$y$$ columns whose sum $$=1$$, the number of $$1$$s in the top row can be any integer $$\\in [0,y]$$, and so the top row sum can be any integer $$\\in [z,z+y]$$, and in short there are $$(y+1)$$ possibilities. Obviously, once the top row sum is known that determines the bottom row sum $$= (y+2z) \\,-$$ top row sum.\n\nSo $$S_{2,n}$$ can be counted by (i) summing over all possible $$y$$, and for each $$y$$: (ii) picking the $$y$$ columns whose sum $$=1$$, and (iii) for the remaining $$n-y=x+z$$ columns assign them $$0$$ or $$2$$ arbitrarily. I.e.:\n\n$$|S_{2,n}| = \\sum_{y=0}^n (y+1) {n \\choose y} 2^{n-y}$$\n\nThis can be evaluated many ways but my favorite is to recognize a slightly transformed sum as an explicit formula for the expected value of a Binomial random variable:\n\n\\begin{align} |S_{2,n}| &= \\sum_{y=0}^n (y+1) {n \\choose y} 2^{n-y} \\\\ &= 3^n \\times \\sum_{y=0}^n (1+y) {n \\choose y} (\\frac13)^y (\\frac23)^{n-y} \\\\ &= 3^n \\times \\mathbb{E}[1 + Bin(n, \\frac13)] \\\\ &= 3^n (1 + {n \\over 3}) = 3^{n-1} (3 + n) = |F_{2,n}| & QED \\end{align}\n\nAs I said, this is a baby step only. The key step in $$F_{2,n}$$ is that $$T$$ has a cycle iff $$|S| \\ge 2$$, and the key step in $$S_{2,n}$$ is that the $$y$$ ones can be all in the top row or all in the bottom row or anywhere in between, for $$(1+y)$$ possibilities. These key steps make the counting easy. However, as far as I can see, neither key step has any easy way to generalize to $$m=3$$, let alone arbitrary $$m$$. (E.g. for $$m=3$$, a cycle in $$T$$ can involve $$2$$ or $$3$$ nodes on the $$m=3$$ side, and I don't know any good way to count that.)\n\n\u2022 Nice start. I would like to see a bijective proof. I wonder if, in general, if $m$ is fixed, $F_{m,n}$ has a rational generating function. \u2013\u00a0Jair Taylor May 30 '19 at 19:32\n\u2022 I would like to see a bijective proof even just for $m=2$, instead of this algebraic mess. Also, I am no good at GFs, but in this case, I couldn't even find nice recursions for $F$ or $S$. Do you have recursions for either of them? \u2013\u00a0antkam May 30 '19 at 19:54\n\nI have a brute force proof that, for $$m=3$$, both counts are $$4^{n - 2} (3 n^2 + 13 n + 16)$$. It would be possible to extend this method for any fixed $$m$$. I'll write it up for $$m=3$$ in the hope it gives insight. As you'll see, many steps look the same on both sides, but I can't figure out how to give a general proof.\n\nCounting margins of $$3 \\times n$$ binary matrices Consider a vector $$C \\in \\{ 0,1,2,3 \\}^n$$ of column sums. Let $$C$$ have $$c_j$$ entries equal to $$j$$ (so $$c_0+c_1+c_2+c_3=n$$). Let's consider how many row sums $$(R_1, R_2, R_3)$$ are compatible with $$C$$. By the Gale-Ryser theorem, these are the vectors with $$R_1+R_2+R_3 = c_1+2c_2+3c_3$$, $$\\min(R_1+R_2, R_1+R_3, R_2+R_3) \\geq c_2+2c_3$$ and $$\\min(R_1, R_2, R_3)\\geq c_3$$. Geometrically, this is the lattice points in a hexagon whose vertices are the permutations of $$(c_1+c_2+c_3, c_2+c_3, c_3)$$. I computed that the number of lattice points in this hexagon is $$\\binom{c_1}{2} + 2 c_1 c_2 + \\binom{c_2}{2} + 2 c_1 + 2 c_2 + 1.$$\n\nThe generating function for $$\\sum_C x_0^{c_0} x_1^{c_1} x_2^{c_2} x_3^{c_3}$$ is $$(x_0+x_1+x_2+x_3)^n$$. Let $$\\partial_j$$ be differentiation with respect to $$x_j$$. Let $$D$$ be the differential operator $$\\tfrac{1}{2} \\partial_1^2 + 2 \\partial_1 \\partial_2+\\tfrac{1}{2} \\partial_2^2 + 2 \\partial_1 + 2 \\partial_2 + 1.$$ Then $$D$$ applied to the monomial $$x_0^{c_0} x_1^{c_1} x_2^{c_2} x_3^{c_3}$$, and evaluated at $$(1,1,1,1)$$, gives $$\\binom{c_1}{2} + 2 c_1 c_2 + \\binom{c_2}{2} + 2 c_1 + 2 c_2 + 1$$. So the number of pairs $$(C,R)$$ is $$D (x_0+x_1+x_2+x_3)^n$$ evaluated at $$(1,1,1,1)$$.\n\nThis gives $$3 n(n-1) 4^{n-2} + 4 n 4^{n-1} + 4^n = 4^{n - 2} (3 n^2 + 13 n + 16)$$.\n\nCounting forests Given a forest on $$3 \\times n$$, define a sequence $$(v_1, v_2, \\ldots, v_n)$$ in $$\\{ 0,1,2,3 \\}^n$$ by saying that $$v_j$$ is $$0$$ if vertex $$j$$ has degree $$0$$, and otherwise $$v_j$$ is the least neighbor of $$j$$. Let $$c_j$$ be the number of $$v_j$$'s equal to $$j$$.\n\nThe generating function for the $$(c_0, c_1, c_2, c_3)$$ sequences is $$(x_0+x_1+x_2+x_3)^n$$ (again!). We count how many forests give rise to each $$(c_0, c_1, c_2, c_3)$$, by listing all possiblilities.\n\nOne vertex joined to $$(1,2)$$ and another joined to $$(1,3)$$ This can happen in $$c_1 (c_1-1)$$ ways.\n\nOne vertex joined to $$(1,2)$$ and another joined to $$(2,3)$$ This can happen in $$c_1 c_2$$ ways.\n\nOne vertex joined to $$(1,3)$$ and another joined to $$(2,3)$$ This can happen in $$c_1 c_2$$ ways.\n\nOne vertex joined to $$(1,2,3)$$ This can happen in $$c_1$$ ways.\n\nOne vertex joined to $$(1,2)$$ This can happen in $$c_1$$ ways.\n\nOne vertex joined to $$(1,3)$$ This can happen in $$c_1$$ ways.\n\nOne vertex joined to $$(2,3)$$ This can happen in $$c_2$$ ways.\n\nNone of the $$n$$-vertices joined to more than one neighbor This can happen in $$1$$ way.\n\nSo, this time, we want to sum up $$c_1 (c_1-1) + 2 c_1 c_2 + 3 c_1 + c_2 +1.$$ The corresponding differential operator is $$E=\\partial_1^2 + 2 \\partial_1 \\partial_2 + 3 \\partial_1 + \\partial_2+1.$$\n\nThe differential operators are different, but once again I get $$3 n(n-1) 4^{n-2} + 4 n 4^{n-1} + 4^n$$ at the end of the day.\n\nA thought for the future Why do the differential operators $$\\tfrac{1}{2} \\partial_1^2 + 2 \\partial_1 \\partial_2+\\tfrac{1}{2} \\partial_2^2 + 2 \\partial_1 + 2 \\partial_2 + 1$$ and $$\\partial_1^2 + 2 \\partial_1 \\partial_2 + 3 \\partial_1 + \\partial_2+1$$ give the same thing when applied to $$(x_0+x_1+x_2+x_3)^n$$? Because $$(x_0+x_1+x_2+x_3)^n$$ is solely a function of $$z:=x_0+x_1+x_2+x_3$$, so all the $$\\partial_j$$'s evaluate to $$\\tfrac{d}{dz}$$. Writing $$\\partial$$ for $$\\tfrac{d}{dz}$$, both operators are $$3 \\partial^2 + 4 \\partial + 1$$. Any hope to generalize this?\n\n\u2022 Nice. I was not aware of Gale-Ryser. If I understand correctly, this could give a method of verifying the conjecture for any fixed $m$ by counting lattice points of appropriate polytopes (to count $S_{m,n}$) and enumerating all possible families of subsets of $[m]$ sharing neighbors (to count $F_{m,n}$), both reducing to finite counting problems. Still, it seems like this problem is crying out for an explicit bijection. \u2013\u00a0Jair Taylor Jun 1 '19 at 13:43", "date": "2021-05-07 12:16:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3235160/spanning-forests-of-bipartite-graphs-and-distinct-row-column-sums-of-binary-matr", "openwebmath_score": 0.9416841864585876, "openwebmath_perplexity": 182.6497470156561, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877044076956, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.8651961443863452}}
{"url": "https://gateoverflow.in/2136/gate2011-34", "text": "2.2k views\nA deck of $5$ cards (each carrying a distinct number from $1$ to $5$) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?\n\n1. $\\left(\\dfrac{1}{5}\\right)$\n\n2. $\\left(\\dfrac{4}{25}\\right)$\n\n3. $\\left(\\dfrac{1}{4}\\right)$\n\n4. $\\left(\\dfrac{2}{5}\\right)$\nedited | 2.2k views\n\nThe number on the first card needs to be One higher than that on the second card, so possibilities are :\n\n$\\begin{array}{c} \\begin{array}{cc} 1^{\\text{st}} \\text{ card} & 2^{\\text{nd}} \\text{ card}\\\\ \\hline \\color{red}1 & \\color{red}-\\\\ 2 & 1\\\\ 3 & 2\\\\ 4 & 3\\\\ 5 & 4\\\\ \\color{red}- & \\color{red}5 \\end{array}\\\\ \\hline \\text{Total$:4$possibilities} \\end{array}$\n\nTotal possible ways of picking up the cards $= 5 \\times 4 = 20$\n\nThus, the required Probability $= \\dfrac{\\text{favorable ways}}{\\text{total possible ways}}= \\dfrac{4}{20} = \\dfrac 15$\n\nOption A is correct\n\nselected\n0\nHow to solve the question if it would be only HIGHER and not ONE HIGHER?\n+4\n\nHow to solve the question if it would be only HIGHER and not ONE HIGHER?\n\nPossible combinations :-\n\n First card second card 5 4,3,2,1 4 3,2,1 3 2,1 2 1 1 none\n\n$P(First\\ card\\ is\\ HIGHER\\ than\\ second\\ one\\ )=\\left [ \\left ( \\frac{1}{5} \\times \\frac{4}{4}\\right ) + \\left ( \\frac{1}{5} \\times \\frac{3}{4}\\right ) + \\left ( \\frac{1}{5} \\times \\frac{2}{4}\\right ) + \\left ( \\frac{1}{5} \\times \\frac{1}{4}\\right ) \\right]\\\\ =\\frac{1}{2}$\n\n+1\nGot it\nHere we should consider without replacement, since \"removed one at a time\" means the card has been removed from the deck.\n\nProb of picking the first card \u00a0= 1/5\n\nNow there are 4 cards in the deck. Prob of picking the second card = 1/4\n\nPossible favourable combinations = 2-1, 3-2, 4-3, 5-4\n\nProbability of each combination = (1/5)*(1/4) = 1/20\n\nHence answer = 4*1/20 = 1/5\nwith 5 cards to choose we can only fulfil the condition if we pick 2,3,4,5 in our choice else theres no way to get the same\n\nthe probability of choosing first no's is =(2,3,4,5)/(1,2,3,4,5)=4/5\n\nthe second time, we have only one option to choose out of four option=1/4\n\nso,the total probability=(4/5)*(1/4)=1/5\n\n1\n2", "date": "2018-11-20 05:53:51", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/2136/gate2011-34", "openwebmath_score": 0.640587568283081, "openwebmath_perplexity": 692.3724549263368, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303437461132, "lm_q2_score": 0.8740772482857833, "lm_q1q2_score": 0.8651881831313736}}
{"url": "https://math.stackexchange.com/questions/4003709/zero-rows-in-echelon-matrix", "text": "Zero rows in echelon matrix\n\nIn the MIT lecture notes (pg. 1) for Linear Algebra, it states;\n\nThe third row is zero because row 3 was a linear combination of rows 1 and 2; it was eliminated.\n\nHere, $$A = \\begin{bmatrix} 1 & 2 & 2 & 2 \\\\ 2 & 4 & 6 & 8 \\\\ 3 & 6 & 8 & 10 \\end{bmatrix}$$ and $$U = \\begin{bmatrix} 1 & 2 & 2 & 2 \\\\ 0 & 0 & 2 & 4 \\\\ 0 & 0 & 0 & 0 \\end{bmatrix}$$.\n\nDoes this mean that if the row of the echelon matrix is all zeros, the corresponding row in A can be represented as a linear combination of the other rows in A?\n\nAnd if the above statements is true, does this also apply to reduced row echelon matrix?\n\n(1) Yes, if a row of the echelon matrix is all zeros, then that row (but remember to consider possible partial pivoting) is a linear combination of other rows in the original matrix. Generally, the coefficients such that forms this linear combination can be found solving the following linear system for $$a$$ and $$b$$: $$\\begin{bmatrix}1 & 2 \\\\ 2 & 4 \\\\ 2 & 6 \\\\ 2 & 8 \\end{bmatrix}\\cdot\\begin{bmatrix}a \\\\ b \\end{bmatrix}=\\begin{bmatrix}3 \\\\ 6 \\\\ 8 \\\\ 10\\end{bmatrix}$$", "date": "2021-07-27 07:47:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4003709/zero-rows-in-echelon-matrix", "openwebmath_score": 0.8561910390853882, "openwebmath_perplexity": 81.37679248776296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9898303422461273, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.8651881672093547}}
{"url": "http://mathhelpforum.com/advanced-algebra/224808-question-about-solving-eigenvector-problems.html", "text": "1. ## Question about solving eigenvector problems\n\nI understand how to get eigenvalues, but I have a problem understanding eigenvectors.\nI'll demonstrate with an example:\n\nI am given A = $\\begin{bmatrix}3 & 1.5 \\\\1.5 & 3 \\end{bmatrix}$\n\nSolving for lambda, I get lambda = 1.5 and 4.5\n\nNow my matrix equations become:\n\n$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$\n\ndividing by 1.5 I get: $x_1-x_2=0$\nso I get $x_1=x_2$\n\nNow, the answer is $\\begin{bmatrix}1 & 1 \\end{bmatrix}$ because if you set x1=1, then x2=1\n\nIf x1=x2, then why isn't the answer [\u211d], where \u211d is the set of all real numbers?\n\n2. ## Re: Question about solving eigenvector problems\n\nOriginally Posted by yaro99\nI understand how to get eigenvalues, but I have a problem understanding eigenvectors.\nI'll demonstrate with an example:\n\nI am given A = $\\begin{bmatrix}3 & 1.5 \\\\1.5 & 3 \\end{bmatrix}$\n\nSolving for lambda, I get lambda = 1.5 and 4.5\n\nNow my matrix equations become:\n\n$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$\n\ndividing by 1.5 I get: $x_1-x_2=0$\nso I get $x_1=x_2$\n\nNow, the answer is $\\begin{bmatrix}1 & 1 \\end{bmatrix}$ because if you set x1=1, then x2=1\n\nIf x1=x2, then why isn't the answer [\u211d], where \u211d is the set of all real numbers?\n\nThere are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.\n\n{1,1} doesn't span R2. It spans the line y = x. Likewise {1,-1} spans the line y = -x\n\ntogether these two vectors span R2\n\n3. ## Re: Question about solving eigenvector problems\n\nOriginally Posted by romsek\n\n{1,1} doesn't span R2. It spans the line y = x.\nIn that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?\nfor example, {2,2}, {5,5}, {1000,1000}, etc.\n\nOriginally Posted by romsek\n\nThere are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.\nHow did you get this? My book has this answer:\n1.5, [1 1], 45\u00b0; 4.5, [1 1], 45\u00b0\n\nDoes that make sense?\n\n4. ## Re: Question about solving eigenvector problems\n\nOriginally Posted by yaro99\nso I get $x_1=x_2$\n\nNow, the answer is $\\begin{bmatrix}1 & 1 \\end{bmatrix}$ because if you set x1=1, then x2=1\n\nIf x1=x2, then why isn't the answer [\u211d], where \u211d is the set of all real numbers?\nFirst, the notation [\u211d \u211d], where the objects inside brackets are sets rather than numbers, is not universally accepted. Even if it were, [A B] would probably mean {[x y] | x \u2208 A, y \u2208 B}, i.e., A \u00d7 B. There is no coordination between the first and second coordinate here. It is clear that you mean {[x x] | x \u2208 \u211d}, but I don't see an easier way to write this. By the way, even if you delimit matrices with square brackets, I would use parentheses for vectors (i.e., 1 \u00d7 n matrices): e.g., (x, x), because square brackets are also used to denote closed segments.\n\nSecond, you are right, any (x, x) is an eigenvector of A as is (x, -x). Indeed, you solve the equation det(A -\u03bbI) = 0 for \u03bb, so if \u03bb is an eigenvalue, the matrix A - \u03bbI is singular by definition. Therefore, the system of equations (A - \u03bbI)x = 0 has infinitely many solutions.\n\nEdit: If x is an eigenvector with value \u03bb, then so is (ax) for any a \u2208 \u211d such that a \u2260 0. Indeed, A(ax) = a(Ax) = a(\u03bbx) = \u03bb(ax).\n\n5. ## Re: Question about solving eigenvector problems\n\nOriginally Posted by yaro99\nIn that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?\nfor example, {2,2}, {5,5}, {1000,1000}, etc.\n\nHow did you get this? My book has this answer:\n1.5, [1 1], 45\u00b0; 4.5, [1 1], 45\u00b0\n\nDoes that make sense?\nthere are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is \"spanned\" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.\n\nYour book is missing a minus sign somewhere. The answer I got was\n\n1.5 [1,-1]/sqrt(2)\n\n4.5 [1,1]/sqrt(2)\n\n6. ## Re: Question about solving eigenvector problems\n\nOriginally Posted by romsek\nthere are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is \"spanned\" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.\n\nYour book is missing a minus sign somewhere. The answer I got was\n\n1.5 [1,-1]/sqrt(2)\n\n4.5 [1,1]/sqrt(2)\nWhoops, that was my bad, there was a minus sign.\nwhere do you get the 1/sqrt(2) from?\nis it a scalar that you can multiply throughout the matrix?\n\n7. ## Re: Question about solving eigenvector problems\n\nOriginally Posted by yaro99\nWhoops, that was my bad, there was a minus sign.\nwhere do you get the 1/sqrt(2) from?\nis it a scalar that you can multiply throughout the matrix?\nit just normalizes the vector to unit length. You tend to want the unit eigenvectors", "date": "2017-08-18 02:23:29", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/224808-question-about-solving-eigenvector-problems.html", "openwebmath_score": 0.9216541647911072, "openwebmath_perplexity": 1361.1199348224523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140216112958, "lm_q2_score": 0.8947894555814343, "lm_q1q2_score": 0.8651844709916265}}
{"url": "http://math.stackexchange.com/questions/109032/singing-bird-problem?answertab=oldest", "text": "# Singing Bird Problem\n\nFor my algorithms class, the professor has this question as extra credit:\n\nThe Phicitlius Bauber bird is believed to never sing the same song twice.\nIts songs are always 10 seconds in length and consist of a series of notes that are either high or low pitched and are either 1 or 2 seconds long.\nHow many different songs can the Bauber bird sing?\n\nThis strikes me as a combinatorial problem, but I'm having trouble figuring out the exact formula to use to fit the included variables of 10 seconds length, 2 notes and 2 note periods.\n\nAt first, I thought that $\\binom{10}{2}*\\binom{10}{2}$ can be a solution, since it fits finding all combinations of both and multiplies the total.\n\nMy issue is this results in a total of 2025 possible songs, and this just seems a little on the low side.\n\nAny assistance is appreciated.\n\n-\n\nI'm going to work under the assumption that two successive 1-second notes of the same pitch is distinct from a single 2-second note of that pitch.\n\nLet $f(n)$ be the number of distinct $n$-second songs the bird can sing. The bird has four options for starting the song: 1-second low, 1-second high, 2-second low, and 2-second high. This gives the recursion $$f(n) = 2f(n-1) + 2f(n-2),$$ with $f(0) = 1$ and $f(1) = 2$.\n\nThe associated characteristic equation is $x^2 - 2x - 2 = 0$, which has roots $x = 1 \\pm \\sqrt{3}$. Thus, we get the general solution $$f(n) = A(1 + \\sqrt{3})^n + B(1 - \\sqrt{3})^n.$$\n\nUsing the initial conditions, we have the system \\begin{align*} 1 &= A + B\\\\ 2 &= A(1 + \\sqrt{3}) + B(1 - \\sqrt{3}), \\end{align*} which has the unique solution $A = \\frac{3 + \\sqrt{3}}{6}$ and $B = \\frac{3 - \\sqrt{3}}{6}$.\n\nFinally, we have $$f(n) = \\frac{3 + \\sqrt{3}}{6}(1 + \\sqrt{3})^n + \\frac{3 - \\sqrt{3}}{6}(1 - \\sqrt{3})^n.$$\n\nFor the problem at hand, we want $$f(10) = 18,272.$$\n\n-\nThat's strange, f(2)=7. But as Mark Eichenlaub says, it should be 6. I think you made a mistake somewhere. \u2013\u00a0 Lopsy Feb 13 '12 at 21:52\nThe roots should be $1\\pm\\sqrt3$, and $f(10)$ is considerably smaller; a hasty calculation made it $18272$. @Lopsy means that your formula for $f(n)$ gives $f(2)=7$. \u2013\u00a0 Brian M. Scott Feb 13 '12 at 21:55\nRight, but your final equation with the radicals gives 7. I think I figured out what happened: the roots of that quadratic are 1+sqrt(3) and 1-sqrt(3), not 2+sqrt(3) and 2-sqrt(3) as you claim. \u2013\u00a0 Lopsy Feb 13 '12 at 21:56\nThanks for everyone's keen eyes. I believe I've caught all the errors now. \u2013\u00a0 Austin Mohr Feb 13 '12 at 22:16\nNow it\u2019s good. I\u2019d add just one comment: $10$ is a small enough number that in this case it\u2019s easier just to use the recurrence than it is to find the general solution. \u2013\u00a0 Brian M. Scott Feb 13 '12 at 22:18\n\nHint: use recursion.\n\nYou know that the first note in the Bauber bird's song is either $1$ or $2$ seconds long. In the first case, the remainder of the song is $9$ seconds long; in the second case, the remainder of the song is $8$ seconds long.\n\nTherefore, the number of $10$-second songs equals twice (remember, the last note can be low or high) the number of $9$-second songs plus the number of $8$-second songs. Can you see how to solve the problem from here?\n\nAlso, note that it's generally a bad idea to encourage \"looking for things to do with the numbers in the problem\" like you said you did with your first try, $\\binom{10}{2}*\\binom{10}{2}$. Although formula-hunting does have its uses (esp. when doing dimensional analysis, where it's essential), if you're going to try to find a formula first, it's a good idea to check the answer for small cases (like $1$ second, $2$ seconds, etc) and then extrapolate off of that than trying to guess formulae blind.\n\n-\n\nHere is a non-recursive solution.\n\nSuppose there are $a$ 1s songs and $b$ 2s songs, so $a + 2b = 10$. The total number of songs is $10 - b$.\n\nWe can count the number of such songs by first ignoring pitch and counting the ways the 1s and 2s songs can be ordered, then multiplying by $2^{a+b}$.\n\nThere are $a+1$ gaps between the $a$ 1s songs, so we want to know how many ways $b$ 2s songs can fit into $a+1$ gaps. This is $\\binom{a+b}{b} = \\binom{10-b}{b}$, as shown by Akash Kumar in this answer.\n\nSo we want\n\n$$\\sum_{b=0}^{5} 2^{10 - b}\\dbinom{10 - b}{b} = 18,272$$\n\n-\nBut you can\u2019t break the ten-second interval into $2$-second chunks, because a $2$-second note can overlap chunks. \u2013\u00a0 Brian M. Scott Feb 13 '12 at 21:59\n@BrianM.Scott Yes, you're right, thanks. Tried a new answer. \u2013\u00a0 Mark Eichenlaub Feb 13 '12 at 23:04", "date": "2014-08-28 03:15:56", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/109032/singing-bird-problem?answertab=oldest", "openwebmath_score": 0.8497444987297058, "openwebmath_perplexity": 442.31854803940143, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936068886642, "lm_q2_score": 0.8791467643431001, "lm_q1q2_score": 0.8651627103068998}}
{"url": "https://math.stackexchange.com/questions/119904/units-and-nilpotents", "text": "# Units and Nilpotents\n\nIf $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.\n\nI've been working on this problem for an hour that I tried to construct an element $x \\in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?\n\n\u2022 Dear Shannon, Try the case $u = 1$ first. Regards, \u2013\u00a0Matt E Mar 14 '12 at 2:23\n\u2022 See also here for the commutative case. \u2013\u00a0Gone Aug 6 '17 at 15:21\n\nIf $u=1$, then you could do it via the identity $$(1+a)(1-a+a^2-a^3+\\cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$ by selecting $n$ large enough.\n\nIf $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?\n\n\u2022 I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a? \u2013\u00a0Shannon Mar 14 '12 at 2:51\n\u2022 @Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent? \u2013\u00a0Arturo Magidin Mar 14 '12 at 2:56\n\u2022 oh, I see. I will work on it. Thank you so much. \u2013\u00a0Shannon Mar 14 '12 at 3:10\n\u2022 This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises. \u2013\u00a0Derek Elkins left SE Jan 21 '16 at 4:01\n\nLet $v$ be the inverse of $u$, and suppose $a^2=0$. Note that $$(u+a)\\cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$ See if you can generalize this.\n\n\u2022 The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from? \u2013\u00a0Shannon Mar 14 '12 at 2:55\n\nHere's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $M\\subset R$. Since $a$ is nilpotent, $a\\in M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-a\\in M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.\n\nIf you don't know that $R$ is commutative, let $S\\subseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$\n\nThe argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.\n\nThis argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.\n\nFirst, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.\n\nNext, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-a\\in I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.\n\nFinally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+\\binom{n}{2}v^2(u+a)^2+\\dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)\\left(nv-\\binom{n}{2}v^2(u+a)+\\dots-(-v)^n(u+a)^{n-1}\\right)$$ and so $$-\\sum_{k=1}^n \\binom{n}{k}(-v)^k(u+a)^{k-1}= nv-\\binom{n}{2}v^2(u+a)+\\dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.\n\nThe fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.\n\nNote that since $$u$$ is a unit and\n\n$$ua = au, \\tag 1$$\n\nwe may write\n\n$$a = u^{-1}au, \\tag 2$$\n\nand thus\n\n$$au^{-1} = u^{-1}a; \\tag 3$$\n\nalso, since $$a$$ is nilpotent there is some $$0 < n \\in \\Bbb N$$ such that\n\n$$a^n = 0, \\tag 4$$\n\nand thus\n\n$$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \\tag 5$$\n\nwe observe that\n\n$$u + a = u(1 + u^{-1}a), \\tag 6$$\n\nand that, by virtue of (5),\n\n$$(1 + u^{-1}a) \\displaystyle \\sum_0^{n - 1} (-u^{-1}a)^k = \\sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}a\\sum_0^{n - 1} (-u^{-1}a)^k$$ $$= \\displaystyle \\sum_0^{n - 1} (-1)^k(u^{-1}a)^k + \\sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$$ $$= 1 + \\displaystyle \\sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \\sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$$ $$= 1 + \\displaystyle \\sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \\sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \\displaystyle \\sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \\tag 7$$\n\nthis shows that\n\n$$(1 + u^{-1}a)^{-1} = \\displaystyle \\sum_0^{n - 1} (-u^{-1}a)^k, \\tag 8$$\n\nand we have demonstrated an explicit inverse for $$1 + u^{-1}a$$. Thus, by (6),\n\n$$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \\tag 9$$\n\nthat is, $$u + a$$ is a unit.\n\nNota Bene: The result proved above has an application to this question, which asks to show that $$I - T$$ is invertible for any nilpotent linear operator $$T$$. Taking $$T = a$$ and $$I = u$$ in the above immediately yields the existence of $$(I - T)^{-1}$$. End of Note.", "date": "2020-07-06 12:07:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/119904/units-and-nilpotents", "openwebmath_score": 0.9653903841972351, "openwebmath_perplexity": 116.75499703387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363725435202, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8651362729434258}}
{"url": "https://math.stackexchange.com/questions/3509665/three-tosses-of-the-same-coin-conditional-probability-that-the-coin-picked-is-u", "text": "# Three tosses of the same coin: Conditional probability that the coin picked is unfair\n\nThere are two coins: one fair and one unfair. P(head | fair)=\u00bd and P(head | unfair)=\u2153 These two coins look identical.\n\nYou picked up a coin from these two, throw it 3 times and observed 1 head.\n\nWhat is the probability that the coin you picked is the unfair coin?\n\nThe part that is throwing me off is three tosses with one resulting in a head. Below I calculated if a heads is observed, what would be the probability that it's an Unfair coin. But if I consider three tosses, I consider three possibilities that outcome space could be {HTT, THT,TTH} and to me, all three possibilities seem to require the same calculation, so using a tree, I calculated P(U|HTT) = P(U\u2229H\u2229T\u2229T)/P(H\u2229T\u2229T) as\n\nAm I making a mistake in calculating below part? The result makes me feel weird for some reason and if I consider the three outcome possibilities in numerator and denominator, they cancel out so final result should be same as below but that implies that number of tosses has no affect - this is the part that is throwing me off so need help with 1) Verify the calculation 2) Does number of tosses have an effect? or is there another way to interpret repeated tosses without replacement?\n\nI looked at this post that is very close to my question but the first and second tosses have a pre-defined outcome. Thanks again for your help.\n\nP(U)P(H|U)P(T|U)P(T|U)/(P(U\u2229H\u2229T\u2229T)+P(F\u2229H\u2229T\u2229T))\n\nP(U|HTT) = 1/2 * 1/3 * 2/3 * 2/3 /(1/2*1/3*2/3*2/3 + 1/2*1/2*1/2*1/2)\n= 4/27 / (4/27 + 1/8)\n= 32/59\n\n\nThis comes out to\n\nHere is the simple one toss calculation:\n\nP(F) = probability the coin is fair\nP(U) = probability the coin is UNfair\nP(H|F) = 1/2\nP(H|U) = 1/3\n\nTo calculate Probability the coin is Unfair given a heads is observed, I can use bayes theorum as:\n\nP(U|H) = P(U)*P(H|U)/ (P(U)*P(H|U) + P(F)*P(H|F)\n= 1/2 * 1/3 / (1/2*1/3 + 1/2*1/2)\n= 4/10\n= 2/5\n\n\n## Update:\n\nI realized the mistake I was making in assuming that number of tosses doesn't matter. It's the order that doesn't matter in the calculation, not the number of tosses - even if there were 2 or 4 or any number of tosses with one Heads, there would be 2 or 4 or n possible combinations respectively, so counting them in numerator or denominator cancels out, so order doesn't matter. However, number of tosses do matter, if there were 4 tosses, the P(HTTT|U) would have additional factor of 2/3 to indicate additional Tails, so the numerator as well as denominator would change.\n\nThe part I am still not able to think intuitively about is how to explain that order doesn't matter. Algebraically, it cancels out from Num and Denom but can't think of an intuition behind it. Thanks for the long read.\n\nP(H|U) = 1/3\nP(H|F) = 1/2\n\nLet A be the event where we observe 3 tosses with one head\n\nP(U|A) =  P(A,U)   =  p(U) * P(A|U)\n--------   ---------------\nP(A)         P(A)\n\n= 1/2 *[P(HTT|U)+P(THT|U)+P(TTH|U)]\n------------------------------------\nP(A, U) + P(A, F)\n\n=        1/2 * [(1/3*2/3*2/3)*3]\n------------------------------------\n[1/2*(1/3*2/3*2/3)*3] + [1/2*(1/2*1/2*1/2)*3]\n\n=     4/27\n------------\n(4/27 + 1/8)\n\n=      4/27\n-----------------\n(32 + 27)/(27*8)\n\n= 32\n--\n59\n\n\nLet's define a few random variables. Let $$U \\sim \\operatorname{Bernoulli}(\\pi = 1/2)$$ represent the prior probability of selecting the unfair coin, where $$U = 1$$ means the coin is unfair, and $$U = 0$$ means the coin is fair. So $$\\Pr[U = 1] = \\Pr[U = 0] = 1/2$$. Next, define $$X \\mid U \\sim \\operatorname{Binomial}(n = 3, p),$$ representing the number of heads obtained in three coin tosses. The probability of heads $$p$$ is a function of $$U$$: if $$U = 0$$, then $$p = 1/2$$, whereas if $$U = 1$$, then $$p = 1/3$$. So we can write this as $$p = \\frac{1}{2} - \\frac{1}{6}U = \\frac{3-U}{6}.$$ Therefore, $$\\Pr[X = 1 \\mid U = u] = \\binom{3}{1} \\left( \\frac{3-u}{6} \\right)^1 \\left( 1 - \\frac{3-u}{6} \\right)^2 = \\frac{(3-u)(3+u)^2}{72},$$ and in particular, $$\\Pr[X = 1 \\mid U = 1] = \\frac{4}{9}, \\\\ \\Pr[X = 1 \\mid U = 0] = \\frac{3}{8}.$$ It follows from the law of total probability, $$\\Pr[X = 1] = \\Pr[X = 1 \\mid U = 0]\\Pr[U = 0] + \\Pr[X = 1 \\mid U = 1]\\Pr[U = 1] = \\frac{4}{9}\\cdot \\frac{1}{2} + \\frac{3}{8} \\cdot \\frac{1}{2} = \\frac{59}{144}.$$ Therefore, $$\\Pr[U = 1 \\mid X = 1] = \\frac{\\Pr[X = 1 \\mid U = 1] \\Pr[U = 1]}{\\Pr[X = 1]} = \\frac{2/9}{59/144} = \\frac{32}{59}.$$", "date": "2021-04-21 02:15:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3509665/three-tosses-of-the-same-coin-conditional-probability-that-the-coin-picked-is-u", "openwebmath_score": 0.9114393591880798, "openwebmath_perplexity": 322.5235814665366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363700641143, "lm_q2_score": 0.8774767746654976, "lm_q1q2_score": 0.8651362660292674}}
{"url": "https://math.stackexchange.com/questions/1345205/perfect-powers-of-successive-naturals-can-you-always-reach-a-constant-differenc", "text": "# Perfect powers of successive naturals: Can you always reach a constant difference?\n\nI was thinking about what happens if you take a sequence of consecutive squares, for example 1,4,9, 16. Taking the differences gives you another sequence, 7,5,3. And taking the differences between those numbers, you get 2,2--a constant. Through elementary first-semester algebra, you can easily verify that this works no matter where you start your series of squares. If you use 9801, 10000, 10201, you will get the same result--after the second round of subtraction, you'll end up with a final constant difference of 2. The same thing works for cubes with the final constant difference being 6, although an additional round of subtraction is needed, in turn requiring one more number in the original sequence. Similarly as with the squares, it's not difficult to show that it will work for any sequence of natural number cubes.\n\nMy question is this: Is there a more general principle at work here? Suppose I take a series of sequential naturals n, n+1, n+2, n+3,... , and raise them to any given positive integral power p. Can it then be shown that if I take the differences repeatedly, proceeding as above, that I will eventually reach a constant difference? And if so, does this principle have anything to do with \"difference engine\" computing?\n\n\u2022 Side note: The eventual constant difference is $p!$, where $p$ is the power. Jun 30 '15 at 23:40\n\u2022 You should take a look at Faulhaber's formula. (Note that taking differences of the numbers $$S_p(n),\\;\\;S_p(n+1),\\;\\;\\ldots,\\;\\;S_p(n+k),\\;\\;\\ldots$$ produces the sequence $$n^p,\\;\\;(n+1)^p,\\;\\;\\ldots,\\;\\;(n+k)^p,\\;\\;\\ldots$$ and then you can continue taking differences from there.) Jun 30 '15 at 23:41\n\u2022 Thanks for the answers and comments so far. And yes, a proof will definitely be in the works...not because the world needs it, but because I need to see it work for myself, as I did for the special cases ^2 and ^3. I do wish that when I was in school they had emphasized proofs instead of what seemed to me, at the time, merely a random grab-bag of disconnected rules and tricks. Jun 30 '15 at 23:46\n\nYes.\n\nThe first difference of a polynomial of degree $d$ is a polynomial of degree $d-1$.\n\nBy induction, the $m$-th difference of a polynomial of degree $d$, when $m \\le d$, is a polynomial of degree $d-m$.\n\nSetting $m = d$, the $d-th$ difference of a polynomial of degree $d$ is a constant, which is exactly what you discovered.\n\nYou next step is a proof.\n\nIf you start with the sequence of $n$-th powers, the $n$-th differences will all be $n!$.\n\nFor a function $f$ defined on the integers define $(\\Delta f)(x)=f(x+1)-f(x)$. $\\Delta$ is a linear operator on such functions: you can easily check that $\\Delta\\big(af(x)+bg(x)\\big)=a\\Delta f(x)+b\\Delta g(x)$ for any such functions $f$ and $g$ and constants $a$ and $b$.\n\nNow suppose that $\\Delta^k(x^k)$ is the constant function $k!$ for each $k<n$. For any constant function $f$ the difference function $\\Delta f$ is identically $0$, so $\\Delta\\big((\\Delta^k(x^k)\\big)=\\Delta(k!)=0$ for each $k<n$, and it follows that $\\Delta^m(x^k)=0$ whenever $k<n$ and $m>k$. In particular, $\\Delta^{n-1}(x^k)=0$ for $k<n-1$. Thus,\n\n\\begin{align*} \\Delta^n(x^n)&=\\Delta^{n-1}\\big((x+1)^n-x^n\\big)\\\\ &=\\Delta^{n-1}\\left(\\sum_{k=0}^n\\binom{n}kx^k-x^n\\right)\\\\ &=\\Delta^{n-1}\\left(\\sum_{k=0}^{n-1}\\binom{n}kx^k\\right)\\\\ &=\\sum_{k=0}^{n-1}\\binom{n}k\\Delta^{n-1}(x^k)\\\\ &=\\binom{n}{n-1}\\Delta^{n-1}(x^{n-1})\\\\ &=n(n-1)!\\\\ &=n!\\;, \\end{align*}\n\nand by induction $\\Delta^n(x^n)=n!$ for all $n$. (You can easily check the basis step of the induction.)\n\nThese so-called forward differences are a special case of divided differences, after which Babbage\u2019s Difference Engine was named.\n\nUsing the Binomial Theorem, we get $$(n+1)^k-n^k=\\sum_{j=0}^{k-1}\\binom{k}{j}n^j$$ So the difference of two consecutive $k^\\text{th}$ powers is a $k-1$ degree polynomial with lead coefficient $\\binom{k}{k-1}=k$. Thus, the difference of a degree $k$ polynomial with lead coefficient $c_k$ is a degree $k-1$ polynomial with a lead coefficient of $kc_k$.\n\nRepeating this, we get that the $k^\\text{th}$ repeated difference of $k^\\text{th}$ powers (a degree $k$ polynomial with lead coefficient $1$) will be $k!$.", "date": "2021-12-04 15:58:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1345205/perfect-powers-of-successive-naturals-can-you-always-reach-a-constant-differenc", "openwebmath_score": 0.8983572125434875, "openwebmath_perplexity": 131.55281187296083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9937100978631076, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8651212951212555}}
{"url": "http://math.stackexchange.com/questions/84491/does-the-set-of-differences-of-a-lebesgue-measurable-set-contains-elements-of-at/104126", "text": "# Does the set of differences of a Lebesgue measurable set contains elements of at most a certain length?\n\nI want to show that if $E\\subset \\mathbb{R}^n$ is a Lebesgue measurable set where $\\lambda(E)>0$, then $E-E=\\{x-y:x,y\\in E\\}\\supseteq\\{z\\in\\mathbb{R}^n:|z|<\\delta\\}$ for some $\\delta>0$, where $|z|=\\sqrt{\\sum_{i=1}^n z_i^2}$.\n\nMy approach is this. Take some $J$, a box in $\\mathbb{R}^n$ with equal side lengths such that $\\lambda(E\\cap J)>3\\lambda(J)/4$. Setting $\\epsilon=3\\lambda(J)/2$, take $x\\in\\mathbb{R}^n$ such that $|x|\\leq\\epsilon$. Then $E\\cap J\\subseteq J$ and $$((E\\cap J)+x)\\cup(E\\cap J)\\subseteq J\\cup(J+x).$$\n\nSince Lebesgue measure is translation invariant, it follows that $\\lambda((E\\cap J)+x)=\\lambda(E\\cap J)$, and so $((E\\cap J)+x)\\cap(E\\cap J)\\neq\\emptyset$.\n\nIf it were empty, then $$2\\lambda(E\\cap J)=\\lambda(((E\\cap J)+x)\\cup(E\\cap J))\\leq\\lambda(J\\cup(J+x))\\leq 3\\lambda(J)/2,$$ thus $\\lambda(E\\cap J)\\leq 3\\lambda(J)/4$, a contradiction.\n\nThen $((E\\cap J)+x)\\cap (E\\cap J)\\neq\\emptyset$, and so $x\\in (E\\cap J)-(E\\cap J)\\subseteq E-E$. Thus $E-E$ contains the box of $x$ such that $|x|\\leq \\epsilon$.\n\nIs this valid? If not, can it be fixed? Many thanks.\n\n-\nI am suspicious about your first hypothesis : How can you take a box such that $\\lambda(E \\cap J) > 3 \\lambda(J)/4$? I must say though, this is a very interesting question. \u2013\u00a0 Patrick Da Silva Nov 22 '11 at 7:13\n\u2013\u00a0 gary Nov 22 '11 at 7:21\n@PatrickDaSilva: $E$ has a subset $E'$ with positive and finite measure. There is an open set $U$ such that $E'\\subseteq U$ and $\\lambda(U)<\\frac{4}{3}\\lambda(E')$. Since $U$ is a countable union of nonoverlapping boxes, $U=\\cup_k B_k$, it follows that $\\sum_k\\lambda(E'\\cap B_k)>\\sum_k\\frac{3}{4}\\lambda(B_k)$. Hence there exists $k$ such that $\\lambda(E'\\cap B_k)>\\frac{3}{4}\\lambda(B_k)$, and you can take $J=B_k$. \u2013\u00a0 Jonas Meyer Nov 22 '11 at 7:25\nYou may want to have a look at this thread and also at this one as well as the links therein. \u2013\u00a0 t.b. Nov 22 '11 at 8:09\nI think this can be fixed if either OP justifies why this inequality holds, or just by taking a smaller epsilon. It might depend on $n$ though, but it seems to be just geometry at this point. I now accept that \"this argument holds by the handwaving theorem\". =P \u2013\u00a0 Patrick Da Silva Jan 31 '12 at 5:07\n\nThis is answered here, I will just post another approach that can be useful.\n\nLemma. If $K$ is a compact subset of $\\mathbb{R}^n$ with positive measure, then the set $$D:=\\{x-y:x,y\\in K\\}$$ contains an open ball centered at the origin.\n\nProof. Since $0\\lt \\lambda(K)\\lt\\infty$, there exist an open set $G$ in $\\mathbb{R}^n$ such that $$K\\subset G\\text{ and } \\lambda (G)\\lt 2\\lambda(K)$$ and as its complement $G^c:=\\mathbb{R}^n\\setminus G$ is closed and $K$ is compact, we have $$\\delta:=d(K,G^c)\\gt 0.$$ We claim that if $x\\in\\mathbb{R}^n$ with $||x||\\lt\\delta$ then $K+x\\subseteq G$. If not, there exist a $y\\in K$ s.t. $y+x\\not\\in G$ and then $$\\delta=d(K,G^c)\\leq ||y-(x+y)||=||x||,$$ which is an absurd if we assume that $||x||\\lt\\delta$ to begin with. Now $K+x\\subseteq G$ and $K\\subseteq G$, thus $(K+x)\\cup K\\subseteq G$. If $(K+x)\\cap K=\\emptyset$, then $$\\lambda(G)\\geq \\lambda((K+x)\\cup K)=\\lambda(K+x)+\\lambda(K)=2\\lambda(K)$$ which contradicts the choice of $G$. So, for all $x\\in B(0,\\delta)$ we have $(K+x)\\cap K\\neq\\emptyset$, therefore $B(0,\\delta)\\subset D$. QED\n\nTheorem. Let $E\\subseteq\\mathbb{R}^d$. If if $E$ is measurable and $\\lambda(E)\\gt 0$ then the set $$E-E:=\\{x-y:x,y\\in E\\}$$ contains an open ball centered at the origin.\n\nProof. Since $\\mathbb{R}^d=\\bigcup_{n\\in\\mathbb N}B(0,n)$ we have $$E=E\\cap\\mathbb{R}^d=\\bigcup_{n\\in\\mathbb N}E\\cap B(0,n),$$ then $$0\\lt\\lambda(E)\\leq \\sum_{n\\in\\mathbb{N}} \\lambda(E\\cap B(0,n))$$ and then, there exist a $n\\in\\mathbb{N}$ such that $\\lambda(E\\cap B(0,n))\\gt 0$. Let $F=E\\cap B(0,n)$. Thus $F\\subseteq E$ is bounded and measurable, therefore there exist a closed set $K\\subseteq F$ such that $$0\\lt \\frac{\\lambda(F)}{2}\\lt \\lambda(K).$$ Notice that $K$ is closed and bounded, that means it is a compact set with positive measure. By the Lemma the set $K-K$ contains an open ball centered at the origin say $B$. Then $$B\\subset K-K\\subseteq E-E$$ as we wanted.\n\nNote. The idea of proceed in this way is taken from Robert Bartle, I don't remember right now the book where this come from.\n\n-", "date": "2015-10-09 13:59:05", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/84491/does-the-set-of-differences-of-a-lebesgue-measurable-set-contains-elements-of-at/104126", "openwebmath_score": 0.9769882559776306, "openwebmath_perplexity": 63.33100189666011, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587274108369, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8650662555613077}}
{"url": "https://math.stackexchange.com/questions/2568587/minimum-values-of-the-sequence-n-sqrt2", "text": "# Minimum values of the sequence $\\{n\\sqrt{2}\\}$\n\nI have been studying the sequence $$\\{n\\sqrt{2}\\}$$ where $\\{x\\}:= x-\\lfloor x\\rfloor$ is the \"fractional part\" function. I am particularly interested in the values of $n$ for which $\\{n\\sqrt{2}\\}$ has an extremely small value - that is, when $n\\sqrt 2$ is extremely close to (but greater than) its nearest integer. Of course, for integer $n$, this always takes on values between (but never including) $0$ and $1$.\n\nI have defined the sequence $M_n$ as follows:\n\n\u2022 $M_1=1$\n\u2022 $M_{n+1}$ is the smallest positive integer such that $\\{M_{n+1}\\sqrt 2\\}\\lt \\{M_n\\sqrt 2\\}$\n\nThe first few terms of this sequence are $$1,3,5,17,29,99,169,...$$ And after a quick trip to the OEIS, I have found that this sequence is equal to OEIS entry A079496, as written in the comments. It also provides the recurrence $$a_0=a_1=1$$ $$a_{2n+1}=2a_{2n}-a_{2n-1}$$ $$a_{2n}=4a_{2n-1}-a_{2n-2}$$ How can I prove that the sequence $M_n$ satisfies this recurrence, using the definition I wrote for $M_n$?\n\nThanks!\n\nEDIT: Here is the closed-form explicit formula for $a_n$, for anyone who wants to know: $$a_{2n}=\\frac{(3+2\\sqrt 2)^n+(3-2\\sqrt 2)^n}{2}$$ $$a_{2n+1}=\\frac{(1+\\sqrt 2)(3+2\\sqrt 2)^n-(1-\\sqrt 2)(3-2\\sqrt 2)^n}{2\\sqrt 2}$$\n\n\u2022 Have you taken a look to the references on the link you shared? I mean, the comment of Paul. D Hanna there says that this is exactly what you're looking for... \u2013\u00a0Alejandro Nasif Salum Dec 15 '17 at 23:54\n\u2022 I mean... you want to know if this recurrence is valid for your sequence or you are sure of it and you are just trying to prove it? \u2013\u00a0Alejandro Nasif Salum Dec 15 '17 at 23:55\n\u2022 Yeah, that confuses me as well. I'll add the closed form to the question, for anyone who is curious. \u2013\u00a0Franklin Pezzuti Dyer Dec 16 '17 at 0:02\n\u2022 According to Kroneckers Approximation Theorem $\\{n\\sqrt{2}\\}$ is dense in $[0,1]$. \u2013\u00a0rtybase Dec 16 '17 at 0:39\n\u2022 This can be seen as two interleaved sequences with the recursion $q_n=6q_{n-1}-q_{n-2}$ \u2013\u00a0robjohn Dec 17 '17 at 1:30\n\nSummary of solution : the originality of your question consists in its use of the floor function to round, when most of the classical results on continued fractions use the nearest integer function. The idea is to reduce the former to the latter.\n\nDetails : Let $M_n,a_n$ be as in the OP. Our goal is to show that\n\n$$M_n=a_n \\tag{1}$$\n\nAs a convenience, let us use the \"normalized\" number $\\alpha=\\sqrt{2}-1$ (which is in the unit interval) rather than $\\sqrt{2}$ itself.\n\nIt will suffice to show :\n\n$$\\lbrace q\\alpha \\rbrace \\geq \\lbrace a_n\\alpha \\rbrace \\textrm{ for } 0 \\lt q \\lt a_{n+1} \\tag{2}$$\n\nWe will use a sequence closely related to the continued fraction expansion of $\\alpha$, namely $g_n=\\frac{(1+\\sqrt{2})^n-(1-\\sqrt{2})^n}{2\\sqrt{2}}$. Note that $(g_n)$ satisfies $g_0=0,g_1=1$ and $g_{n+2}-2g_{n+1}+g_n=0$, so its values (after $g_1$) are positive integers. The following properties are easy to show by a direct computation or by induction :\n\n$(i)$ $g_ng_{n+2}-g_{n+1}^2=(-1)^{n+1}$\n\n$(ii)$ $a_n=g_n$ when $n$ is odd\n\n$(iii)$ $a_n=g_{n+1}-g_n$ when $n$ is even\n\n$(iv)$ $\\frac{g_n-g_{n+1}\\alpha}{g_{n+1}-g_{n+2}\\alpha}=1-\\sqrt{2}$\n\n$(v)$ $\\frac{a_{n+2}\\alpha -(g_{n+2}-g_{n+1})}{a_n\\alpha -(g_n-g_{n-1})}=3-2\\sqrt{2}$, when $n$ is even.\n\nThe following lemma is a very classic continued fraction inequality :\n\nLEMMA. Let $p,q$ be integers with $0 \\lt q \\lt g_{n+2}$. Then $|p-q\\alpha| \\geq |g_n-g_{n+1}\\alpha|$. Further, if $(p,q)\\neq (g_n,g_{n+1})$, the inequality can be strenghtened to $|p-q\\alpha| \\geq |g_n-g_{n+1}\\alpha|+|g_{n+1}-g_{n+2}\\alpha|$.\n\nProof of lemma. By (i), the matrix $\\left(\\begin{matrix} g_n & g_{n+1} \\\\ g_{n+1} & g_{n+2} \\end{matrix}\\right)$ is unimodular. It follows that there are integers $u,v$ such that $p=g_n u+g_{n+1}v$ and $q=g_{n+1}u+g_{n+2}v$. I claim that $uv\\leq 0$ ; otherwise, $g_{n+1}u$ and $g_{n+2}v$ would both be nonzero and have the same sign, forcing $|q| = |g_{n+1}u|+|g_{n+2}v| \\geq |g_{n+2}|$ which is impossible. So $uv\\leq 0$, and also\n\n$$p-q\\alpha= u(g_n-g_{n+1}\\alpha) + v (g_{n+1}-g_{n+2}\\alpha) \\tag{3}$$\n\nThe numbers $u(g_n-g_{n+1}\\alpha)$ and $v(g_{n+1}-g_{n+2}\\alpha)$ have the same sign, so\n\n$$|p-q\\alpha|= |u(g_n-g_{n+1}\\alpha)| + |v (g_{n+1}-g_{n+2}\\alpha)| \\tag{4}$$\n\nNote that $u\\neq 0$ because of $q=g_{n+1}u+g_{n+2}v$ and $|q|<g_{n+2}$. So $|u|\\geq 1$ and the first inequality follows. If equality holds, we must have $|u|=1$ and $v=0$, from which we easily deduce $(p,q)=(g_n,g_{n+1})$. Suppose now that $(p,q)\\neq (g_n,g_{n+1})$. If $|u|\\geq 2$, then from (4) we deduce $|p-q\\alpha| \\geq 2|g_n-g_{n+1}\\alpha|$. By (iv), the second inequality holds in this case.\n\nOtherwise $u=1$ and from the reasoning above we have $v\\neq 0$, so $|v|\\geq 1$ and from (4) again we deduce $|p-q\\alpha| \\geq |g_n-g_{n+1}\\alpha|+|g_{n+1}-g_{n+2}\\alpha|$, which finishes the proof of the lemma.\n\nLet us now deduce (2) from the lemma. Suppose first that $n$ is odd. By $(ii)$, $\\lbrace a_n\\alpha \\rbrace=\\lbrace g_n\\alpha \\rbrace=g_{n+1}\\alpha-g_n$ by lemma, (iv) and the fact that $g_{n+1}\\alpha-g_n >0$. The lemma also shows that $\\lbrace q \\alpha \\rbrace \\geq \\lbrace a_n \\alpha \\rbrace$ for $0 \\lt q \\lt g_{n+2}=a_{n+2}$, as wished.\n\nSuppose now that $n$ is even. Using the lemma with $n-1$ in place of $n$, we have that for any $0 \\lt q \\lt g_{n+1}=a_{n+1}$ and $(p,q)\\neq (g_{n-1},g_n)$, then $$\\begin{array}{lcl} |p-q\\alpha| &\\geq& |g_{n-1}-g_{n}\\alpha|+|g_{n}-g_{n+1}\\alpha| \\\\ &=& g_{n-1}-g_{n}\\alpha+g_{n+1}\\alpha-g_n \\\\ &=& a_n\\alpha -(g_n-g_{n-1}) \\\\ &=& \\lbrace a_n \\alpha \\rbrace \\textrm{ by } (v) \\end{array}$$\n\nThis finishes the proof.\n\nThis is about Pell type equations. Absolutely complete detail would be a bit long. I have posted many times about solving $x^2 - n y^2 = T,$ where $T$ is some target number, and $n$ is positive not a square.\n\nYour sequence is this: given positive integer $x,$ let $v = \\lfloor x \\sqrt 2 \\rfloor.$ Your numbers will be $$2 x^2 - v^2 \\in \\{ 1,2 \\}.$$\n\nFurthermore, the ratios of consecutive terms are bounded: if $w$ is a number with worse $2 w^2 - w_0^2,$ where $w_0 = \\lfloor w \\sqrt 2 \\rfloor,$ then there is one of your numbers $t$ with $t > w/(2 + \\sqrt 2).$\n\nAlright, the positive values of $2x^2 - y^2$ are $1,2,4,7,8,9, 14 \\ldots$ where we are allowing common factors of $x,y.$\n\nIf $2x^2 - v^2 = 4,$ then both $x,v$ are even, we can take $t = x/2,$ and the fractional part of $t \\sqrt 2$ is half the fractional part of $x \\sqrt 2,$ meaning $x$ cannot be part of your $M$ sequence. Now that I think of it, the same applies to any number divisible by $4$ or by $9.$ As a result, if we calculated $2x^2 - v^2 = 7$ as a special case (we know how to find all representations) we could then continue with $2 x^2 - v^2 \\geq 14.$\n\nIt appears we need just the one special case. After this, we may take $$2 x^2 - v^2 \\geq 7,$$ with $$\\{ x \\sqrt 2 \\} \\geq \\frac{7}{x \\sqrt 2 + v} \\approx \\frac{7}{2x \\sqrt 2 }\\approx \\frac{2.4748737}{x } .$$ $$t > \\frac{x}{2 + \\sqrt 2},$$ $$2 t^2 - w^2 \\in \\{ 1,2 \\}.$$ $$(t \\sqrt 2 - w)(t \\sqrt 2 + w) \\leq 2.$$ $$t \\sqrt 2 - w \\leq \\frac{2}{t \\sqrt 2 + w}.$$ $$\\{ t \\sqrt 2 \\} \\leq \\frac{2}{t \\sqrt 2 + w} \\approx \\frac{2}{2t \\sqrt 2 } \\approx \\frac{1}{t \\sqrt 2 } < \\frac{2 + \\sqrt 2}{x \\sqrt 2 } = \\frac{1 + \\sqrt 2}{x } \\approx \\frac{2.1421356}{x } .$$ This is smaller than $$\\{ x \\sqrt 2 \\} \\geq \\frac{7}{x \\sqrt 2 + v} \\approx \\frac{7}{2x \\sqrt 2 }\\approx \\frac{2.4748737}{x } .$$\n\nA more careful analysis is possible, correction terms all over, but the heart of it is the observation that the floor of $t \\sqrt 2$ is extremely close to the real number itself, when $0 < 2t^2 - w^2 \\leq 2$.\n\nHere is the Conway topograph for the negative of your form, $x^2 - 2 y^2.$ Your numbers show up representing negative number, (4,3) gives $-2$, then (7,5) gives $-1,$ then (24,17) give $-2,$ (41,29) gives $-1$\n\nLet's see, it will take an hour or so, but I can draw a topograph for $2x^2 - v^2$ that shows numbers represented up to $17,$ as numbers such as $4,8,9$ are not squarefree, and are not primitively represented (we get $\\gcd(x,v) \\neq 1$). Will, 9:19 Pacific time\n\nDone. Note that the green coordinate pairs have a motion that gives the same value of $2x^2 - y^2,$ namely $$(x,y) \\mapsto (3x+2y, 4x+3y).$$ For example, $(1,1) \\mapsto (5, 7 )$ and $(3,4) \\mapsto (17, 24 ).$\n\nLet's see, Cayley Hamilton for the coefficient matrix $$\\left( \\begin{array}{cc} 3 & 2 \\\\ 4 & 3 \\end{array} \\right)$$ gives the linear recurrence, $$x_{n+4} = 6 x_{n+2} - x_n$$ if we combine the $x$ values in a single list. We need to separate odd index and even index because we are combining $2x^2-v^2 = 1$ and $2x^2 - v^2 = 2$ in one list of $x$ values\n\nThere is a book that Allen Hatcher makes available online, here is his diagram for $x^2 - 2 y^2,$ but without the green coordinates that I like to include. The topograph diagram was introduced by J. H. Conway\n\nContinued Fraction Approximations\n\nThe continued fraction for $\\sqrt2$ is $$\\sqrt2=1+\\cfrac1{2+\\cfrac1{2+\\cfrac1{2+\\dots}}}\\tag1$$ The approximants alternate between a bit too high and a bit too low, but we always have $$\\left|\\,\\frac{p_n}{q_n}-\\sqrt2\\,\\right|\\le\\frac1{2q_n^2}\\tag2$$ because of the continued fraction, we have $$p_n=2p_{n-1}+p_{n-2}\\quad\\text{and}\\quad q_n=2q_{n-1}+q_{n-2}\\tag3$$ starting with $\\frac11$ (low) and $\\frac32$ (high). The first several continued fraction approximants are $$\\frac11,\\frac32,\\frac75,\\frac{17}{12},\\frac{41}{29},\\frac{99}{70},\\dots\\tag4$$\n\nLow Approximations\n\nWe want to choose the approximations that are a bit low so that $q_n\\sqrt2\\gt p_n$. Then $$\\left\\{q_n\\sqrt2\\right\\}=q_n\\sqrt2-p_n\\le\\frac1{2q_n}\\tag5$$ Since we only want the low approximations, we need to modify the recursion in $(3)$ to skip the high approximations. To that end, we have $$p_n=6p_{n-2}-p_{n-4}\\quad\\text{and}\\quad q_n=6q_{n-2}-q_{n-4}\\tag6$$ starting with $\\frac11$ and $\\frac75$. The first several low continued fraction approximations are $$\\frac11,\\frac75,\\frac{41}{29},\\frac{239}{169},\\frac{1393}{985},\\dots\\tag7$$ We can also make some close low approximations by looking at twice the reciprocals of the high approximations. These will follow the recursion in $(6)$: $$\\frac43,\\frac{24}{17},\\frac{140}{99},\\frac{816}{577},\\dots\\tag8$$ The denominators of these two sequences, which follow the recursion in $(6)$, cover all of the elements of the sequences in the question.\n\nCombined Sequence\n\nThe combined sequence follows the recursion $$a_n=6a_{n-2}-a_{n-4}$$ and starts out $$1,3,5,17,\\dots$$\n\n\u2022 This was my immediate idea too. But it looks like this produces only every other term in the OP's sequence. \u2013\u00a0hmakholm left over Monica Dec 16 '17 at 22:59\n\u2022 Their sequence seems to include all the continued fraction approximants, even the high ones. The high ones will give $\\left\\{q_n\\sqrt2\\right\\}$ just less than $1$. Or perhaps my list misses some close, but not continued fraction approximations. Only continued fraction approximations satisfy $(2)$. I will look when I get back home. \u2013\u00a0robjohn Dec 16 '17 at 23:05\n\u2022 I suspect the additional terms are because leaving out the low approximants leave room for additional multiples of $\\sqrt2$ to have a smallest-yet fractional part even though there have been earlier multiples (corresponding to the high approximants) that approached an integer closer from below. \u2013\u00a0hmakholm left over Monica Dec 16 '17 at 23:30\n\u2022 @HenningMakholm: I figured out where the other terms in the question came from. \u2013\u00a0robjohn Dec 17 '17 at 0:57\n\u2022 If $x \\in (\\frac{2}{3},\\frac{3}{4})$, then $\\lbrace x \\rbrace \\gt \\frac{1}{2} \\gt \\lbrace 2x \\rbrace$, so $2x$ is a record in the sense of the OP, but not a best approximation situation in the classical sense. Here in the special case where $x=\\sqrt{2}$ we are lucky that all the records in the sense of the OP come from best approximations. \u2013\u00a0Ewan Delanoy Dec 26 '17 at 8:31", "date": "2020-11-29 20:35:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2568587/minimum-values-of-the-sequence-n-sqrt2", "openwebmath_score": 0.8944486379623413, "openwebmath_perplexity": 187.07133445199042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587221857245, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8650662429793889}}
{"url": "https://stats.stackexchange.com/questions/159313/generating-samples-from-singular-gaussian-distribution/159322", "text": "Generating samples from singular Gaussian distribution [duplicate]\n\nLet random vector $x = (x_1,...,x_n)$ follow multivariate normal distribution with mean $m$ and covariance matrix $S$. If $S$ is symmetric and positive definite (which is the usual case) then one can generate random samples from $x$ by first sampling indepently $r_1,...,r_n$ from standard normal and then using formula $m + Lr$, where $L$ is the Cholesky lower factor so that $S=LL^T$ and $r = (r_1,...,r_n)^T$.\n\nWhat about if one wants samples from singular Gaussian i.e. $S$ is still symmetric but not more positive definite (only positive semi-definite). We can assume also that the variances (diagonal elements of $S$) are strictly positive. Then some elements of $x$ must have linear relationship and the distribution actually lies in lower dimensional space with dimension $<n$, right?\n\nIt is obvious that if e.g. $n=2, m = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, S = \\begin{bmatrix} 1 & 1 \\\\ 1 & 1\\end{bmatrix}$ then one can generate $x_1 \\sim N(0,1)$ and set $x_2=x_1$ since they are fully correlated. However, is there any good methods for generating samples for general case $n>2$? I guess one needs first to be able identify the lower dimensional subspace, then move to that space where one will have valid covariance matrix, then sample from it and finally deduce the values for the linearly dependent variables from this lower-dimensional sample. But what is the best way to that in practice? Can someone point me to books or articles that deal with the topic; I could not find one.\n\nThe singular Gaussian distribution is the push-forward of a nonsingular distribution in a lower-dimensional space. Geometrically, you can take a standard Normal distribution, rescale it, rotate it, and embed it isometrically into an affine subspace of a higher dimensional space. Algebraically, this is done by means of a Singular Value Decomposition (SVD) or its equivalent.\n\nLet $\\Sigma$ be the covariance matrix and $\\mu$ the mean in $\\mathbb{R}^n$. Because $\\Sigma$ is non-negative definite and symmetric, the SVD will take the form\n\n$$\\Sigma = U \\Lambda^2 U^\\prime$$\n\nfor an orthogonal matrix $U\\in O(n)$ and a diagonal matrix $\\Lambda$. $\\Lambda$ will have $m$ nonzero entries, $0\\le m \\le n$.\n\nLet $X$ have a standard Normal distribution in $\\mathbb{R}^m$: that is, each of its $m$ components is a standard Normal distribution with zero mean and unit variance. Abusing notation a little, extend the components of $X$ with $n-m$ zeros to make it an $n$-vector. Then $U\\Lambda X$ is in $\\mathbb{R}^n$ and we may compute\n\n$$\\text{Cov}(U\\Lambda X) = U \\Lambda\\text{Cov}(X) \\Lambda^\\prime U^\\prime = U \\Lambda^2 U^\\prime = \\Sigma.$$\n\nConsequently\n\n$$Y = \\mu + U\\Lambda X$$\n\nhas the intended Gaussian distribution in $\\mathbb{R}^n$.\n\nIt is of interest that this works when $n=m$: that is to say, this is a (standard) method to generate multivariate Normal vectors, in any dimension, for any given mean $\\mu$ and covariance $\\Sigma$ by using a univariate generator of standard Normal values.\n\nAs an example, here are two views of a thousand simulated points for which $n=3$ and $m=2$:\n\nThe second view, from edge-on, demonstrates the singularity of the distribution. The R code that produced these figures follows the preceding mathematical exposition.\n\n#\n# Specify a Normal distribution.\n#\nmu <- c(5, 5, 5)\nSigma <- matrix(c(1, 2, 1,\n2, 3, 1,\n1, 1, 0), 3)\n#\n# Analyze the covariance.\n#\nn <- dim(Sigma)[1]\ns <- svd((Sigma + t(Sigma))/2) # Guarantee symmetry\ns$d <- abs(zapsmall(s$d))\nm <- sum(s$d > 0) #$\n# Generate a standard Normal x in R^m.\n#\nn.sample <- 1e3 # Number of points to generate\nx <- matrix(rnorm(m*n.sample), nrow=m)\n#\n# Embed x in R^n and apply the square root of Sigma obtained from its SVD.\n#\nx <- rbind(x, matrix(0, nrow=n-m, ncol=n.sample))\ny <- s$u %*% diag(sqrt(s$d)) %*% x + mu\n#\n# Plot the results (presuming n==3).\n#\nlibrary(rgl)\nplot3d(t(y), type=\"s\", size=1, aspect=TRUE,\nxlab=\"Y1\", ylab=\"Y2\", zlab=\"Y3\", box=FALSE,\ncol=\"Orange\")\n\n\u2022 Thanks for this great answer it's been very helpful to me. One question occurred to me - if X is extended to include $n-m$ zeroes then is the covariance of X still the identity matrix? That seems like a key step but wouldn't the covariance have $n-m$ zeroes on the diagonal? Mean zero, variance zero = constant zero. EDIT: Nevermind I see now that it doesn't need to be the identity matrix \u2013\u00a0ASeaton Mar 31 at 8:27", "date": "2021-05-18 16:49:28", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/159313/generating-samples-from-singular-gaussian-distribution/159322", "openwebmath_score": 0.8128841519355774, "openwebmath_perplexity": 318.70517787226265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587272306607, "lm_q2_score": 0.8757869900269366, "lm_q1q2_score": 0.8650662425941783}}
{"url": "http://npdinc.co.za/walden-farms-mxj/9v5arr.php?a3c1ed=counting-theory-discrete-math", "text": "From 1 to 100, there are $50/2 = 25$ numbers which are multiples of 2. Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures.It is closely related to many other areas of mathematics and has many applications ranging from logic to statistical physics, from evolutionary biology to computer science, etc. $A \\cap B = \\emptyset$), then mathematically $|A \\cup B| = |A| + |B|$, The Rule of Product \u2212 If a sequence of tasks $T_1, T_2, \\dots, T_m$ can be done in $w_1, w_2, \\dots w_m$ ways respectively and every task arrives after the occurrence of the previous task, then there are $w_1 \\times w_2 \\times \\dots \\times w_m$ ways to perform the tasks. For example: In a group of 10 people, if everyone shakes hands with everyone else exactly once, how many handshakes took place? %\ufffd\ufffd\ufffd\ufffd Counting mainly encompasses fundamental counting rule, the permutation rule, and the combination rule. Today we introduce set theory, elements, and how to build sets.This video is an updated version of the original video released over two years ago. . Thereafter, he can go Y to Z in $4 + 5 = 9$ ways (Rule of Sum). . }$$. If n pigeons are put into m pigeonholes where n > m, there's a hole with more than one pigeon. . = 6 ways. Graph theory. There are 50/6 = 8 numbers which are multiples of both 2 and 3. |A \\cup B| = |A| + |B| - |A \\cap B| = 25 + 16 - 8 = 33. of ways to fill up from first place up to r-th-place \u2212, n_{ P_{ r } } = n (n-1) (n-2)..... (n-r + 1), = [n(n-1)(n-2) ... (n-r + 1)] [(n-r)(n-r-1) \\dots 3.2.1] / [(n-r)(n-r-1) \\dots 3.2.1]. . = 180.. Discrete Mathematics Tutorial Index . Now we want to count large collections of things quickly and precisely. (1!)(1!)(2!)] Chapter 1 Counting \u00b6 One of the first things you learn in mathematics is how to count. 2 CS 441 Discrete mathematics for CS M. Hauskrecht Basic counting rules \u00e2\u0080\u00a2 Counting problems may be hard, and easy solutions are not obvious \u00e2\u0080\u00a2 Approach: \u00e2\u0080\u0093 simplify the solution by decomposing the problem \u00e2\u0080\u00a2 Two basic decomposition rules: \u00e2\u0080\u0093 Product rule \u00e2\u0080\u00a2 A count decomposes into a sequence of dependent counts How many integers from 1 to 50 are multiples of 2 or 3 but not both? Why one needs to study the discrete math It is essential for college-level maths and beyond that too There was a question on my exam which asked something along the lines of: \"How many ways are there to order the letters in 'PEPPERCORN' if all the letters are used?\" Would this be 10! }, = (n-1)! . Ten men are in a room and they are taking part in handshakes. Probability. This note explains the following topics: Induction and Recursion, Steiner\u00e2\u0080\u0099s Problem, Boolean Algebra, Set Theory, Arithmetic, Principles of Counting, Graph Theory. Viewed 4k times 2. How many ways can you choose 3 distinct groups of 3 students from total 9 students? Example: you have 3 shirts and 4 pants. \ufffd.\ufffd\ufffd\ufffd\ufffd2\ufffd(\ufffd^\ufffd\ufffd \u38efU\ufffd\ufffdNn%\ufffdu\ufffd\ufffdp\ufffd;\ufffdVY\ufffd\ufffd\ufffd\ufffd\ufffdW\ufffd\ufffd}\ufffd\ufffd\ufffd\ufffd{SH\ufffdW\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd-zHLJ\ufffdf\ufffd R'\ufffd\ufffd\ufffd\ufffd;\ufffd\ufffd\ufffdq\ufffd\ufffdY?\ufffd\ufffd\ufffd?\ufffdWX\ufffd\ufffd\ufffd:5(\ufffd\ufffd \ufffd3a\ufffd\ufffd\ufffd\u00c3*p0\ufffd4\ufffdV\ufffd\ufffd\ufffd\ufffdy\ufffdg\ufffdq:\ufffdk\ufffd\ufffdF\ufffd\u0321[I\ufffd6)\ufffd3G\u00b3R\ufffd%\ufffd\ufffd, %\u052e3 Solution \u2212 There are 6 letters word (2 E, 1 A, 1D and 2R.) / [(a_1!(a_2!) The number of all combinations of n things, taken r at a time is \u2212,$$^nC_{ { r } } = \\frac { n! } . In this technique, which van Lint & Wilson (2001) call \u00e2\u0080\u009cone of the most important tools in combinatorics,\u00e2\u0080\u009d one describes a finite set X from two perspectives leading to two distinct expressions \u00e2\u0080\u00a6 . From a set S ={x, y, z} by taking two at a time, all permutations are \u2212, We have to form a permutation of three digit numbers from a set of numbers $S = \\lbrace 1, 2, 3 \\rbrace$. There are $50/3 = 16$ numbers which are multiples of 3. Next come chapters on logic, counting, and probability.We then have three chapters on graph theory: graphs, directed + \\frac{ n-k } { k!(n-k)! } Different three digit numbers will be formed when we arrange the digits. . Mathematically, if a task B arrives after a task A, then $|A \\times B| = |A|\\times|B|$. For two sets A and B, the principle states \u2212, $|A \\cup B| = |A| + |B| - |A \\cap B|$, For three sets A, B and C, the principle states \u2212, $|A \\cup B \\cup C | = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C |$, $|\\bigcup_{i=1}^{n}A_i|=\\sum\\limits_{1\\leq i\ufffd,oX\ufffd\ufffd\ufffdN8xT\ufffd\ufffd\ufffd\ufffd,\ufffd0\ufffdz\ufffdI\ufffdQ\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd[\ufffdI9r0\ufffd '&l\ufffdv]G\ufffdq\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdi&\ufffd\ufffdb\ufffdi\ufffd \ufffd\ufffd \ufffdq\ufffd\ufffd\ufffdK\ufffd?\ufffdc\ufffdRl . Hence, there are 10 students who like both tea and coffee. \ufffdd\ufffd$\ufffd\u0314\ufffd=d9\u017c\ufffd\ufffdV\ufffd\ufffdr\ufffde. Pigeonhole Principle states that if there are fewer pigeon holes than total number of pigeons and each pigeon is put in a pigeon hole, then there must be at least one pigeon hole with more than one pigeon. . (n\u00e2\u0088\u0092r+1)! The remaining 3 vacant places will be filled up by 3 vowels in $^3P_{3} = 3! There must be at least two people in a class of 30 whose names start with the same alphabet. . Discrete mathematics problem - Probability theory and counting [closed] Ask Question Asked 10 years, 6 months ago. In combinatorics, double counting, also called counting in two ways, is a combinatorial proof technique for showing that two expressions are equal by demonstrating that they are two ways of counting the size of one set. Question \u2212 A boy lives at X and wants to go to School at Z. Chapter Summary The Basics of Counting The Pigeonhole Principle Permutations and Combinations This tutorial includes the fundamental concepts of Sets, Relations and Functions, Mathematical Logic, Group theory, Counting Theory, Probability, Mathematical Induction, and Recurrence Relations, Graph Theory, Trees and Boolean Algebra. Solution \u2212 From X to Y, he can go in$3 + 2 = 5$ways (Rule of Sum). . So,$|A|=25$,$|B|=16$and$|A \\cap B|= 8$. Solution \u2212 As we are taking 6 cards at a time from a deck of 6 cards, the permutation will be$^6P_{6} = 6! It is a very good tool for improving reasoning and problem-solving capabilities. . Below, you will find the videos of each topic presented. Number of ways of arranging the consonants among themselves $= ^3P_{3} = 3! Discrete Mathematics & Mathematical Reasoning Chapter 6: Counting Colin Stirling Informatics Slides originally by Kousha Etessami Colin Stirling (Informatics) Discrete Mathematics (Chapter 6) Today 1 / 39. \\dots (a_r!)]$. . . How many ways are there to go from X to Z? ]$, The number of circular permutations of n different elements taken x elements at time =$^np_{x}/x$, The number of circular permutations of n different things =$^np_{n}/n$. /\\: [(2!) material, may be used as a textbook for a formal course in discrete mathematics or as a supplement to all current texts. (\\frac{ k } { k!(n-k)! } x\ufffd\ufffdX\ufffdo7\u007f\ufffd_\ufffdG\ufffd\ufffd\ufffd\ufffdOzm\ufffd+0\ufffdm\ufffd\ufffd\ufffd\ufffd\\\ufffd\ufffd\ufffd\ufffdd\ufffd\ufffdGJG\ufffdlV'H\ufffdX\ufffd-J\"\u007f$%J\ufffdK&\ufffd\ufffd\ufffd8\ufffd\ufffd\ufffd8\ufffdi\ufffd\ufffd\u05d6\ufffdJq\ufffd\ufffd6\ufffd~\ufffd\ufffdl\u0493)W,\ufffdWl\ufffdd\ufffd\ufffdgRmhVL\ufffd\ufffd\ufffd.\ufffdL\ufffd\ufffd\ufffdN~\ufffdEfy\ufffd*\ufffdn\ufffd\u0722\ufffd\ufffd\u07b1\u07e7?\ufffd\ufffdz\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd|$\ufffdI\ufffd\ufffd-\ufffd\ufffdz\ufffdo\ufffd\ufffd\ufffdX\ufffd\ufffd \ufffd\ufffd\ufffdw\ufffd]Lsm\ufffdK\ufffd\ufffd4j\ufffd\"\ufffd\ufffd\ufffd#gs$(\ufffdi5\ufffd\ufffdm!9.\ufffd\ufffd\ufffd\u007f\ufffd63\ufffd\ufffd\ufffdGp\ufffdh\u0409N\ufffd/\ufffd&B\ufffd\ufffd;\ufffd4@\ufffd\ufffdJ\ufffd?n7 CO\ufffd\ufffd>\ufffdYtw\ufffd8FqX\ufffd\ufffd\u03c7U\ufffd]A\ufffd|D\ufffdC#}\ufffd\ufffdkW\ufffd\ufffdv\ufffd\ufffdG \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdm\ufffd\ufffd\ufffd\ufffd\u5045^~\ufffdl6\ufffd\ufffd&) \ufffd\ufffdJ\ufffd1\ufffd\ufffdv}\ufffd\u00e2\ufffdt\ufffdWr\ufffd\ufffd\ufffdk\ufffd\ufffdU\ufffdk\ufffd\ufffd?\ufffdd\ufffd\ufffd\ufffdB\ufffdn\ufffd\u007f\ufffdc!\ufffd^\u0540\ufffdT\ufffd\u0372m\ufffd\u0445\ufffdV\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd6\ufffdq\ufffdo\ufffd\ufffd\ufffd\u042en\ufffdn?\ufffd\ufffd\ufffd\ufffd\u02f3\ufffd\ufffd\ufffdx\ufffdq@\u05bb[ \ufffd\ufffdXB&\ufffd\ufffd,f|\ufffd\ufffd\ufffd\ufffd+\ufffd\ufffdM#R\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\u03d5c*H\u0110}\ufffd5S0H Proof \u2212 Let there be \u00e2\u0080\u0098n\u00e2\u0080\u0099 different elements. . The cardinality of the set is 6 and we have to choose 3 elements from the set. >> Here, the ordering does not matter. Hence, the number of subsets will be $^6C_{3} = 20$. Problem 1 \u2212 From a bunch of 6 different cards, how many ways we can permute it? Discrete math. Starting from the 6th grade, students should some effort into studying fundamental discrete math, especially combinatorics, graph theory, discrete geometry, number theory, and discrete probability. . . The Rule of Sum \u2212 If a sequence of tasks $T_1, T_2, \\dots, T_m$ can be done in $w_1, w_2, \\dots w_m$ ways respectively (the condition is that no tasks can be performed simultaneously), then the number of ways to do one of these tasks is $w_1 + w_2 + \\dots +w_m$. . I'm taking a discrete mathematics course, and I encountered a question and I need your help. . in the word 'READER'. Any subject in computer science will become much more easier after learning Discrete Mathematics . . This is a course note on discrete mathematics as used in Computer Science. { (k-1)!(n-k)! } That means 3\u00d74=12 different outfits. . %PDF-1.5 Very Important topics: Propositional and first-order logic, Groups, Counting, Relations, introduction to graphs, connectivity, trees If there are only a handful of objects, then you can count them with a moment's thought, but the techniques of combinatorics can extend to quickly and efficiently tabulating astronomical quantities. So, Enroll in this \"Mathematics:Discrete Mathematics for Computer Science . Problem 3 \u2212 In how ways can the letters of the word 'ORANGE' be arranged so that the consonants occupy only the even positions? . Mathematics of Master Discrete Mathematics for Computer Science with Graph Theory and Logic (Discrete Math)\" today and start learning. . Find the number of subsets of the set $\\lbrace1, 2, 3, 4, 5, 6\\rbrace$ having 3 elements. Mathematically, for any positive integers k and n: $^nC_{k} = ^n{^-}^1C_{k-1} + ^n{^-}^1{C_k}$, $= \\frac{ (n-1)! } In how many ways we can choose 3 men and 2 women from the room? CONTENTS iii 2.1.2 Consistency. The permutation will be = 123, 132, 213, 231, 312, 321, The number of permutations of \u00e2\u0080\u0098n\u00e2\u0080\u0099 different things taken \u00e2\u0080\u0098r\u00e2\u0080\u0099 at a time is denoted by$n_{P_{r}}$. (n\u00e2\u0088\u0092r+1)!$, The number of permutations of n dissimilar elements when r specified things never come together is \u2212 $n!\u00e2\u0080\u0093[r! Discrete Mathematics Course Notes by Drew Armstrong. .10 2.1.3 Whatcangowrong. We can now generalize the number of ways to fill up r-th place as [n \u00e2\u0080\u0093 (r\u00e2\u0080\u00931)] = n\u00e2\u0080\u0093r+1, So, the total no. Number of permutations of n distinct elements taking n elements at a time =$n_{P_n} = n!$, The number of permutations of n dissimilar elements taking r elements at a time, when x particular things always occupy definite places =$n-x_{p_{r-x}}$, The number of permutations of n dissimilar elements when r specified things always come together is \u2212$r! For choosing 3 students for 1st group, the number of ways \u2212 $^9C_{3}$, The number of ways for choosing 3 students for 2nd group after choosing 1st group \u2212 $^6C_{3}$, The number of ways for choosing 3 students for 3rd group after choosing 1st and 2nd group \u2212 $^3C_{3}$, Hence, the total number of ways $= ^9C_{3} \\times ^6C_{3} \\times ^3C_{3} = 84 \\times 20 \\times 1 = 1680$. Let X be the set of students who like cold drinks and Y be the set of people who like hot drinks. Mastering Discrete Math ( Discrete mathematics ) is such a crucial event for any computer science engineer. The applications of set theory today in computer science is countless. . Most basic counting formulas can be thought of as counting the number of ways to distribute either distinct or identical items to distinct recipients. Group theory. For solving these problems, mathematical theory of counting are used. Set theory is a very important topic in discrete mathematics . The Inclusion-exclusion principle computes the cardinal number of the union of multiple non-disjoint sets. . From there, he can either choose 4 bus routes or 5 train routes to reach Z. There are 6 men and 5 women in a room. A permutation is an arrangement of some elements in which order matters. stream . Pascal's identity, first derived by Blaise Pascal in 17th century, states that the number of ways to choose k elements from n elements is equal to the summation of number of ways to choose (k-1) elements from (n-1) elements and the number of ways to choose elements from n-1 elements. If we consider two tasks A and B which are disjoint (i.e. Solution \u2212 There are 3 vowels and 3 consonants in the word 'ORANGE'. . Some of the discrete math topic that you should know for data science sets, power sets, subsets, counting functions, combinatorics, countability, basic proof techniques, induction, ... Information theory is also widely used in math for data science. Active 10 years, 6 months ago. Problem 2 \u2212 In how many ways can the letters of the word 'READER' be arranged? He may go X to Y by either 3 bus routes or 2 train routes. The Rule of Sum and Rule of Product are used to decompose difficult counting problems into simple problems. In a group of 50 students 24 like cold drinks and 36 like hot drinks and each student likes at least one of the two drinks. Sign up for free to create engaging, inspiring, and converting videos with Powtoon. /Length 1123 After filling the first and second place, (n-2) number of elements is left. Students, even possessing very little knowledge and skills in elementary arithmetic and algebra, can join our competitive mathematics classes to begin learning and studying discrete mathematics. If each person shakes hands at least once and no man shakes the same man\u00e2\u0080\u0099s hand more than once then two men took part in the same number of handshakes. 70 0 obj << Discrete Mathematics Handwritten Notes PDF. For instance, in how many ways can a panel of judges comprising of 6 men and 4 women be chosen from among 50 men and 38 women? . . Welcome to Discrete Mathematics 2, a course introducting Inclusion-Exclusion, Probability, Generating Functions, Recurrence Relations, and Graph Theory. So, $| X \\cup Y | = 50$, $|X| = 24$, $|Y| = 36$, $|X \\cap Y| = |X| + |Y| - |X \\cup Y| = 24 + 36 - 50 = 60 - 50 = 10$. When there are m ways to do one thing, and n ways to do another, then there are m\u00d7n ways of doing both. How many different 10 lettered PAN numbers can be generated such that the first five letters are capital alphabets, the next four are digits and the last is again a capital letter. The Basic Counting Principle. The \u00ef\u00ac\u0081rst three chapters cover the standard material on sets, relations, and functions and algorithms. For solving these problems, mathematical theory of counting are used. /Filter /FlateDecode Start Discrete Mathematics Warmups. = 720$. The Rules of Sum and Product The Rule of Sum and Rule of Product are used to decompose difficult counting problems into simple problems. A combination is selection of some given elements in which order does not matter. For example, distributing $$k$$ distinct items to $$n$$ distinct recipients can be done in $$n^k$$ ways, if recipients can receive any number of items, or $$P(n,k)$$ ways if recipients can receive at most one item. Thank you. Recurrence relation and mathematical induction. 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He has to first reach Y and then Y to Z he can go \\$.", "date": "2021-10-16 00:25:01", "meta": {"domain": "co.za", "url": "http://npdinc.co.za/walden-farms-mxj/9v5arr.php?a3c1ed=counting-theory-discrete-math", "openwebmath_score": 0.6598092317581177, "openwebmath_perplexity": 636.9057146228723, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987758723627135, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.8650662346347571}}
{"url": "https://math.stackexchange.com/questions/3247176/problem-about-inequation-with-absolute-value", "text": "# Problem about inequation with absolute value\n\nI have this statement:\n\nIt can be assured that | p | \u2264 2.4, if it is known that:\n\n(1) -2.7 \u2264 p <2.3\n\n(2) -2.2 < p \u2264 2.6\n\nMy development was:\n\nFirst, $$-2.4 \\leq p \\leq 2.4$$\n\nWith $$1)$$ by itself, that can't be insured, same argument for $$2)$$\n\nNow, i will use $$1)$$ and $$2)$$ together, and the intersection between this intervals are: $$(-2.2, 2.3)$$. So, this also does not allow me to ensure that | p | \u2264 2.4, since, there are some numbers that are outside the intersection of these two intervals, for example 2.35 is outside this interval.\n\nBut according to the guide, the correct answer must be $$1), 2)$$ together. And i don't know why.\n\nYou getting confused by a strong premise implying a weak conclusion.\n\n1 and 2 together say precisely: $$p\\in (-2.2,2.3)$$\n\nAnd you are asking to conclude $$p\\in [-2.4,2.4]$$.\n\nThat's simply a matter of noting if $$(-2.2, 2.3) \\subset [-2.4,2.4]$$. And if $$p$$ is in a smaller subset, then we can conclude $$p$$ must therefore also be in the bigger superset.\n\nOr in other words:\n\n$$-2.2 < p < 2.3\\implies$$\n\n$$-2.4\\le -2.2 < p <2.3 \\le 2.4\\implies$$\n\n$$-2.4\\le p \\le 2.4$$\n\n....\n\nAn analogy:\n\nWe can conclude $$p$$ is a somewhat green article of clothing if we know both\n\n1: $$p$$ is a hat.\n\n2: $$p$$ is precisely the color an emerald takes when viewed at noon on the summer solstice on the equator.\n\nIn your question, 1 tells you that $$p <2.3$$ so you can conclude $$p\\le 2.4$$. ($$p$$ is a hat so you can conclude $$p$$ is an article of clothing.)\n\nAnd 2 tells you that $$-2.2 so you can conclude $$-2.4\\le p$$. ($$p$$ is the color of emeralds so you can conclude $$p$$ is green.)\n\nTogether 1 and 2 tell you $$-2.2 so you can conclude $$-2.4\\le p\\le 2.4$$. (Together you know $$p$$ is a hat the color of emeralds in a certain condition so you can conclude $$p$$ is a green article of clothing.)\n\nStrong premise. Weak result.\n\n\u2022 I had thought about it. For example, if you AFFIRM that k <10, then I can affirm that: k <15. However, my problem is that, I think, I CAN NOT affirm that it is \"lesser or EQUAL\" than 15, since it can take the value 15 and that contradicts the initial statement of k <10. \u2013\u00a0Eduardo Sebastian Jun 1 '19 at 20:44\n\u2022 However, I think I understand the point. A number k can be between: 10 < k <100, and at the same time, I with that i can affirm that: -10000 <k <100000, regardless of whether there are intermediate values, this is still true, this is what I understood. \u2013\u00a0Eduardo Sebastian Jun 1 '19 at 20:52\n\u2022 The key word is \"less than OR equal\" is the \"OR\". If $5 < 16$ then $5 \\le 16$. $5 \\le 16$ does not mean that $5 =16$. It doesn't even mean $5$ might be less than $16$. It means that exactly one of the two statements: A) $5 < 16$ and B) $5=16$ is true. And one of them IS true. A) is true. So A OR B is true. \u2013\u00a0fleablood Jun 3 '19 at 14:47\n\u2022 We are asked to verify \"Fred is green or red\". We sneak up on Fred and verify that he is a solid green. SO we say \"Yep, Fred is green so the statement is true: Fred is green or red\". But then one of us says. \"But he isn't red. So he can't be green OR RED because he isn't red\". Well... the first person is right. Fred is completely green. That is a subset of being green OR red. So GREEN $\\implies$ GREEN or RED. \u2013\u00a0fleablood Jun 3 '19 at 14:51\n\u2022 I understand the point. Where i can get info about \" strong premise implying a weak conclusion.\" \u2013\u00a0Eduardo Sebastian Jun 4 '19 at 19:23\n\n$$-2.4 \\leq p \\leq 2.4$$\n\n$$-2.7 \\leq p < 2.3$$\n\n$$-2.2 < p \\leq 2.6$$\n\n$$p \\in (-2.2, 2.3)$$\nThe point is that if you use $$1$$ and $$2$$ together you know $$-2.2 \\lt p \\lt 2.3$$. This does allow you to ensure $$|p| \\le 2.4$$ because all numbers in the interval are less than $$2.4$$ in absolute value. The fact that there are numbers outside the interval that also qualify does not matter at all. If I told you that $$0.1 \\le p \\le 0.9$$ you could assure me that $$|p| \\le 2.4$$. You could also make a stronger statement, like $$|p| \\le 1$$ or $$|p| \\le 0.9$$, but that doesn't mean the one with $$2.4$$ is incorrect.", "date": "2021-05-14 04:04:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3247176/problem-about-inequation-with-absolute-value", "openwebmath_score": 0.7586725950241089, "openwebmath_perplexity": 285.99130863868476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808741970027, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8650271694367689}}
{"url": "https://math.stackexchange.com/questions/1049608/lattice-generated-by-vectors-orthogonal-to-an-integer-vector/1430969", "text": "# Lattice generated by vectors orthogonal to an integer vector\n\nGiven a non-zero vector $\\boldsymbol{v}$ composed of integers, imagine the set of all non-zero integer vectors $\\boldsymbol{u}$, such that $\\boldsymbol{u} \\cdot \\boldsymbol{v} = 0$, i.e., the integer vectors orthogonal the original vector. The set $S = \\{\\boldsymbol{u} : \\boldsymbol{u} \\cdot \\boldsymbol{v} = 0\\}$ seems to form a $dim(\\boldsymbol{v})-1$ dimensional lattice. Specifically, it's clear that for any two elements of $S$, their linear combination is also in $S$. However, because S is a subset of the lattice of all integer vectors, it's also a lattice. It there a name for this lattice? Additionally, how can it's basis vectors be computed?\n\n\u2022 This is a good question, I ran into it yesterday and found your post looking for the answer online. I'm surprised there has been no answer yet, so I'm working on it and will post solution once I find it. Sep 11 '15 at 8:49\n\nI don't know if this lattice has a name, however it is easy to compute a basis given $$\\mathbf{v}$$ by recurrence:\n\nSuppose the gcd of all coordinates of $$\\mathbf{v}$$ is 1 or else divide by it.\n\nLet $$\\mathbf{v}=(v_1,...,v_n)$$.\n\nIf $$n=2$$, then your basis is just $$(v_2,-v_1)$$\n\nIf $$n > 2$$:\nSet the first vector of your basis to be $$\\mathbf{b}_1=(-v_n*a_1,-v_n*a_2,...,-v_n*a_{n-1},gcd(v_1,...v_{n-1}))$$, where the $$a_i$$ come from B\u00e9zout's Identity and can be easily calculated using the extended Euclidean algorithm.\nThen complete your basis using the same algorithm with $$\\mathbf{v}=(v_1,...v_{n-1})$$.\n\nThis yields a basis of the lattice orthogonal to $$\\mathbf{v}$$ and not a sublattice.\nIndeed, all integer vector $$\\mathbf{u}$$ orthogonal to $$\\mathbf{v}$$ have last coordinate which is a multiple of $$gcd(v_1,...,v_{n-1})$$.\n$$\\sum_{i=1..n} u_i v_i = 0$$ implies $$u_n v_n = -\\sum_{i=1..n-1}u_i v_i$$ so $$gcd(v_1,...v_{n-1})$$ divides $$u_n v_n$$ and is coprime with $$v_n$$.\n\nExample\n\nSuppose we want the lattice orthogonal to $$v = (125, -75, 45, -27)$$.\nWe first look at the orthogonal of $$(125,-75)$$, which is the same as the lattice orthogonal to $$(5,-3)$$ if we simplify by $$gcd(125,-75)=25$$.\nSo our first vector is going to be $$(-3,-5,0,0)$$.\nNow we look at the orthogonal of $$(125,-75,45)$$, which is the same as the orthogonal of $$(25,-15,9)$$, if we simplify by $$gcd(125,-75,45)=5$$.\nB\u00e9zout's Identity gives us $$2*25+3*(-15)=5$$, so our new vector is $$(-9*2,-9*3,5,0)=(-18,-27,5,0)$$.\nFinally, we arrive to our last step, B\u00e9zout's Identity gives us $$1*125+1*(-75)+(-1)*45=5$$, so our last vector is $$(27*1,27*1,27*(-1),5)=(27,27,-27,5)$$.\nSo our basis is formed by the three vectors : $$\\big((-3,-5,0,0),(-18,-27,5,0),(27,27,-27,5)\\big)$$\n\n\u2022 Hi, unless I am mistaking, this algorithm does not produce a full lattice of vectors orthogonal to v. As an example, take v = (125, -75, 45, -27). Then this procedure yields three basis vectors (27, 27, -27, 5), (45, 90, 25, 0), (3, 5, 0, 0). However, the lattice spanned by these vectors is strictly contained inside the correct lattice spanned by (3, 5, 0, 0), (0, 3, 5, 0), (0, 0, 3, 5). Oct 26 '18 at 17:54\n\u2022 It works if you don't forget to include the step: Suppose the gcd of all coordinates of v is 1 or else divide by it. Even during the induction step Nov 12 '18 at 18:08\n\u2022 thanks, you are absolutely right! Nov 17 '18 at 20:45\n\u2022 I don't understand this algorithm. Can you give an explicit example, like for Anton's? Feb 27 '20 at 3:38\n\u2022 I edited my answer to illustrate the algorithm working on Anton's example. Hope this helps. Mar 1 '20 at 19:15\n\nI know that is an old question and already have a beautiful answer given by Florian Bourse, but I am adding the following just for future references.\n\nThat lattice has a name: it is called orthogonal lattice. In general, we define it regarding a set of vectors $$a_1, ..., a_d \\in \\mathbb{Z}^n$$ as follows:\n\n$$L^\\bot := \\left\\{ u\\in \\mathbb{Z}^n : \\langle a_i, u \\rangle = 0 \\text{ for } 1 \\le i \\le d \\right\\}.$$\n\nOr equivalently we define the lattice $$L$$ generated by the integer linear combinations of $$a_i$$'s and $$L^\\bot$$ is then the lattice whose vectors are orthogonal to all the vectors of $$L$$.\n\nYour case is just a particular one where $$d=1$$. And you are right about the dimension, it is $$n - d$$.\n\nThere are a lot of other known properties of these lattices (for instance, $$\\det(L^\\bot) \\le \\det(L)$$ and $$(L^\\bot)^\\bot = span(L)\\cap\\mathbb{Z}^n$$).\n\nSection 2 of Merkle-Hellman Revisited: A Cryptanalysis of the Qu-Vanstone by Phong Nguyen and Jacques Stern present some discussion about this type of lattice and a polynomial-time algorithm to compute a LLL-reduced basis to them, which I've implemented in SAGE and published in this Github repository.\n\n\u2022 Very nice! Thanks for the reference. Jan 4 '19 at 18:02", "date": "2022-01-21 04:56:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1049608/lattice-generated-by-vectors-orthogonal-to-an-integer-vector/1430969", "openwebmath_score": 0.857448399066925, "openwebmath_perplexity": 198.07552021167254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98028087477309, "lm_q2_score": 0.8824278633625321, "lm_q1q2_score": 0.8650271578211717}}
{"url": "http://mathhelpforum.com/advanced-algebra/53270-analysis-proof-help-needed.html", "text": "# Thread: Analysis proof help needed\n\n1. ## Analysis proof help needed\n\nI've been driving myself crazy with the following proof:\n\nLet E be an infinite set, and F be a finite subset of E. Let G = E\\F (That is, E minus F). Prove that G is equivalent to E.\n\nIf I know that if E is countable, I can prove that G is countable, but what if E is uncountable?? Please help!\n\n2. Originally Posted by stuckinanalysis\nI've been driving myself crazy with the following proof:\n\nLet E be an infinite set, and F be a finite subset of E. Let G = E\\F (That is, E minus F). Prove that G is equivalent to E.\n\nIf I know that if E is countable, I can prove that G is countable, but what if E is uncountable?? Please help!\nsuppose E is uncountable but G is countable. we have $E=G \\cup F.$ clearly G can't be finite. so to get a contradiction, you need to show that the union of an infinite countable set with a finite set is\n\ninfinite countable. to show this let $G=\\{x_n: \\ n \\in \\mathbb{N} \\}, \\ F=\\{y_1, \\cdots, y_m\\}.$ define the map $f: G \\cup F \\longrightarrow \\mathbb{N},$ by: $f(y_j)=j, \\ f(x_n)=n+m.$ obviously $f$ is bijective and hence $G \\cup F$ is countable.\n\n3. You proved that G is uncountable, but does this prove G is equivalent to E? The set of real numbers and the power set of the real numbers are both uncountable, but they are not equivalent.\n\n4. Originally Posted by stuckinanalysis\nYou proved that G is uncountable, but does this prove G is equivalent to E? The set of real numbers and the power set of the real numbers are both uncountable, but they are not equivalent.\nyou're right! i didn't see \"equivalent\"! ok, the problem is now much more interesting! so we have an infinite set $G$ and a finite set $F.$ we want to show that $G$ and $G \\cup F$ are equivalent.\n\nsince we're dealing with cardinality here, we may assume that $G \\cap F = \\emptyset.$ now apply Zorn's lemma to find a set $G^*=\\bigcup_{i \\in I} G_i,$ with $G_i \\subseteq G,$ which is maximal with respect to these properties:\n\n1) every $G_i$ is countable,\n\n2) $G_i \\cap G_j = \\emptyset$ for all $i, j \\in I$ with $i \\neq j.$\n\nthen maximality of $G^*$ implies that $G-G^*=H$ is finite. choose an $i_0 \\in I$ and replace $G_{i_0}$ with $G_{i_0} \\cup H.$ then: $G = \\bigcup_{i \\in I} G_i,$ and $G \\cup F=(G_{i_0} \\cup F) \\cup \\bigcup_{i \\neq i_0}G_i.$ also it's clear that $G_i$ will still satisfy\n\nthe properties 1) and 2). thus, since $G_{i_0} \\cup F$ and all $G_i$ are countable, they're equaivalent to $\\mathbb{N},$ hence both $G$ and $G \\cup F$ are equivalent to $I \\times \\mathbb{N}, \\ \\color{red}(*)$ and hence they must be equivalent. Q.E.D.\n\n$\\color{red}(*)$ the reason is that if $f_i: G_i \\longrightarrow \\mathbb{N}, \\ i \\in I,$ is a bijection, then you can define $f: \\bigcup_{i \\in I}G_i \\longrightarrow I \\times \\mathbb{N}$ by: $f(g_i)=(i,f_i(g_i)).$ since $G_i$ are disjoint, $f$ is well-defined. bijectivity of $f$ is a trivial result\n\nof bijectivity of each $f_i.$\n\n5. Originally Posted by NonCommAlg\n... now apply Zorn's lemma\nBut is that cheating?\n\nThere is a known result in ZFC which says if $X$ is an infinite set and $Y$ a subset with $|Y| < |X|$ then $|X - Y| = |X|$. This can be easily proven using axiom of choice and cardinality. However, it still relies on axiom of choice. When I saw this problem I wondered whether it is possible to solve it without any choice.\n\nAnd if you are using choice then why use Zorn's lemma, simply put a well-ordering on $G$ and start mapping the least elements to the least elements. I think that might work and be a lot simpler.\n\n6. Thanks. That gives me a lot to think about.\n\nHere is a follow up/ similar question. Can it be done the same way?\n\nLet X be an uncountable set, and let Y be a countable subset of X. How can I prove that X is equivalent to X U Y?\n\n7. Originally Posted by stuckinanalysis\nLet E be an infinite set, and F be a finite subset of E. Let G = E\\F (That is, E minus F). Prove that G is equivalent to E.\nI don't think you need the axiom of choice for this. Take a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence. Then define a bijection f from E to E\\F by taking $f(x_n)=f(x_{N+n})$, and taking f to be the identity on $E\\setminus\\{x_n:n\\in\\mathbb{N}\\}$.\n\n8. Originally Posted by stuckinanalysis\nHere is a follow up/ similar question. Can it be done the same way?\n\nLet X be an uncountable set, and let Y be a countable subset of X. How can I prove that X is equivalent to X \\ Y?\nA similar method should work. Let Z be a countable set in X \\ Y. Then Z and Z\u222aY have the same cardinality, so let f be a bijection from Z to Z\u222aY, and extend f to X \\ Y by taking it to be the identity on X \\ (Z\u222aY).\n\n9. Originally Posted by Opalg\nTake a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence.\nHow?\nWe not supposed to be chosing.\n\nI only see one way of doing this. Let $g$ be a choice function on $\\mathcal{P}(E)$ i.e. $g(S) \\in S$ for all $S\\not = \\emptyset$. Since $F$ is finite there is a bijection $f:n\\to F$. Define a sequence, by recursion, $x: \\mathbb{N} \\to E$ by $x_k = f(k)$ for $k and $x_k = g( E - \\{ x_i | i < n\\})$. Then follow through with your argument.\n\n10. Originally Posted by ThePerfectHacker\nOriginally Posted by Opalg\nTake a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence.\nHow?\nWe not supposed to be choosing.\nJust because you are making a choice, it doesn't necessarily follow that you require the axiom of choice. You only need to invoke that axiom if you want to make a simultaneous choice of an element from each of an infinite collection of sets. That is not the case here. The elements $x_n$ are chosen one at a time in an inductive fashion. You don't need the axiom of choice for that.\n\n11. Originally Posted by Opalg\nJust because you are making a choice, it doesn't necessarily follow that you require the axiom of choice. You only need to invoke that axiom if you want to make a simultaneous choice of an element from each of an infinite collection of sets. That is not the case here. The elements $x_n$ are chosen one at a time in an inductive fashion. You don't need the axiom of choice for that.\nAre you sure? What you are doing is this:\n$x_{n+1} = \\text{ some element of }(E - \\{ x_i | i < n\\})$\n\nThe thing is that this function is not really defined.\nThat is why we need choice to define such a function.\n---\nIf that (what I just did above) was really okay than we can apply the same argument to transfinite sequences:\nAs a consequence of which the axiom of choice would be a provable from the other axioms.\n\n12. Originally Posted by ThePerfectHacker\nAre you sure? What you are doing is this:\n$x_{n+1} = \\text{ some element of }(E - \\{ x_i | i < n\\})$\n\nThe thing is that this function is not really defined.\nThat is why we need choice to define such a function.\nNo, I'm just using the definition of \"nonempty\". If a set is nonempty then there exists an element in that set. I'm calling that element $x_{n+1}$.\n\nAlmost every proof in analysis starts by saying \"Let \u03b5>0.\" That involves choosing an element from the set of positive real numbers. But it is not true that every such proof requires the axiom of choice.\n\nOriginally Posted by ThePerfectHacker\nIf that (what I just did above) was really okay than we can apply the same argument to transfinite sequences:\nAs a consequence of which the axiom of choice would be a provable from the other axioms.\nNo, you could never get beyond a countable set that way, because you are only constructing one element at at time.", "date": "2013-05-23 04:56:03", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/53270-analysis-proof-help-needed.html", "openwebmath_score": 0.9475960731506348, "openwebmath_perplexity": 235.67180567871375, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9802808672839539, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8650271481815611}}
{"url": "http://math.stackexchange.com/questions/105644/can-the-relation-of-ap-equiv-a-pmod-2p-if-p-prime-and-p2-be-added-to", "text": "# Can the relation of $a^{p} \\equiv a\\pmod {2p}$ if (p prime and p>2) be added to Fermat's little theorem?\n\nFermat's little theorem\n\n$a^{p} \\equiv a\\pmod p$ if p prime\n\nI have noticed that the formula $a^{p} \\equiv a\\pmod {2p}$ (p prime and p>2) can be also written by using Fermat's little theorem\n\nProof: Let $a$ be any integer and $p>2$ be some prime number.\n\n$a^{p} \\equiv a\\pmod p$\n\n$a^{p}-a \\equiv 0\\pmod p$\n\n$a(a^{p-1}-1) \\equiv 0\\pmod p$\n\n$a(a^{p-1}-1) \\equiv 0\\pmod p$\n\n$F(a)=(a^{p-1}-1)$ can be divided to $(a-1)$ because $F(1)=(1^{p-1}-1)=0$\n\n$F(a)=(a^{p-1}-1)=(a-1).R(a)$ //R(a) is polinom that has degree p-2 and coefficients are Integer numbers.\n\n$a (a-1) R(a) \\equiv 0\\pmod p$\n\n$a (a-1)$ is always an even number so $a (a-1) R(a)$ will be always even number and also can be divided to $p$ because of Fermat's little theorem . Thus $a^{p} \\equiv a\\pmod {2p}$ if (p prime and p>2) .\n\nI searched internet but I have not seen that relation in the internet.\n\nIs it known formula?Please help if it is known relation(I would like to learn what the subject is )\n\nSorry if someone else asked the same question in here.\n\nThank you for answers and helps.\n\n-\nIf $a$ is any integer, and $n$ a positive integer, then $a^n \\equiv a \\mod 2$. Your equation is a combination of this fact and Fermat's little theorem. Both are known congruences, but I'm note sure there's a name for their combination. \u2013\u00a0 Joel Cohen Feb 4 '12 at 14:49\n\nAs Joel rightly observed :If $a$ is any integer, and $n$ a positive integer, then $a^n\\equiv a \\pmod2$\n\nSo : $a^p\\equiv a \\pmod2$ , and $a^p\\equiv a \\pmod p$\n\nAccording to Chinese Remainder Theorem :\n\n$a^p \\equiv a \\pmod { \\operatorname{lcm} (2,p)}$\n\n,and since $p$ is an odd prime it follows that : $\\operatorname{lcm}(2,p) = 2p$ , therefore :\n\n$a^p \\equiv a \\pmod {2p}$\n\n-\nThe answer, Maybe, can be perfect and if we add some additional steps to show how proof of Chinese Remainder Theorem. c,d,e are integers if $a^p\\equiv a \\pmod2$ then $a^p=a+2.c$ $a^p-a=2.c$ if $a^p\\equiv a \\pmod p$ then $a^p=a+p.d$ $a^p-a=p.d=2.c$ $d=2.e$ can be written because $2.c$ is even number Thus $a^p-a=p.d=2.c=2.p.e$ $a^p-a\\equiv 0 \\pmod{2p}$ $a^p\\equiv a \\pmod{2p}$ Thanks in advice \u2013\u00a0 Mathlover Feb 4 '12 at 19:32\n\nYou are correct I think, but the way you write it down makes it look more difficult than it is. Also you should be more specific on whether you are talking about one particular $a$ or \"$\\forall a$\".\n\nSo let me show how I would write this down. First, let $a$ be any integer and $p>2$ be some prime number. Then\n\n$$a^p \\equiv a \\pmod{2p} \\iff 2p \\mid a^p - a \\iff 2\\mid a^{p}-a \\text{ and } p\\mid a^p-a$$ The last equivalence holds because $2$ and $p$ are two different primes.\n\nBut $2\\mid a^p-a$ for all $a$ and $p$: if $a$ is even, then clearly $a^p$ will be even as well; if $a$ is odd, then so is $a^p$ and hence $a^p-a$ will be even again. So this is indeed equivalent to\n\n$$\\dots \\iff p\\mid a^p-a \\iff a^p \\equiv a \\pmod p.$$\n\n-\nThanks for answer. \u2013\u00a0 Mathlover Feb 4 '12 at 14:46\nNo actually it's a continuation of the chain of equivalences up above. So it should be read as $$a^p \\equiv a \\pmod{2p} \\iff \\dots \\iff a^p\\equiv a \\pmod{p}$$ \u2013\u00a0 Myself Feb 4 '12 at 14:51\nThis is the special case $\\rm\\ m = 2\\:p,\\ p\\:$ odd, of the following generalization of Fermat's little theorem, quoted from Bill Dubuque's post\nTHEOREM $\\$ For naturals $\\rm\\: k,m>1$\n$\\qquad\\rm m\\ |\\ a^k-a\\$ for all $\\rm\\:a\\in\\mathbb N\\ \\iff\\ m\\:$ is squarefree and prime $\\rm\\: p\\:|\\:m\\: \\Rightarrow\\: p-1\\ |\\ k-1$", "date": "2013-12-21 16:30:14", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/105644/can-the-relation-of-ap-equiv-a-pmod-2p-if-p-prime-and-p2-be-added-to", "openwebmath_score": 0.8767497539520264, "openwebmath_perplexity": 167.033336835776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692284751636, "lm_q2_score": 0.8856314753275019, "lm_q1q2_score": 0.8649690097214321}}
{"url": "https://math.stackexchange.com/questions/738574/computing-int-0-pi-sinx-dx-using-the-definition", "text": "# Computing $\\int_0^\\pi \\sin(x) \\; dx$ using the definition.\n\nA colleague of mine and I, in the course of teaching integral calculus for the umpteenth time, were wondering if we could expand the class of examples that our students are exposed to when computing Riemann integrals from the definition. Most such examples involve polynomial functions, and they work nicely because we have some well-known formulas, like\n\n$\\displaystyle\\sum_{i=1}^n 1 = n$,\n\n$\\displaystyle\\sum_{i=1}^n i = \\frac{n(n+1)}{2}$,\n\nand $\\displaystyle\\sum_{i=1}^n i^2 = \\frac{n(n+1)(2n+1)}{6}$. (And certainly, there are others.)\n\nBut we wanted to expand the class of examples beyond polynomial functions, and started to think about trig functions. We came up with a computation for $\\int_0^\\pi \\sin(x)\\;dx$ using the definition of the Reimann integral, which I provide below.\n\nQuestion: Is the computation below in the literature somewhere? If so, where?\n\nComputation:\n\nUsing a Riemann sum with right endpoints, we have\n\n$$\\int_{0}^\\pi \\sin(x)\\; dx = \\lim_{n \\to \\infty}\\sum_{i=1}^n \\sin(x_i) \\Delta x$$ where $\\Delta x = \\frac{\\pi}{n}$ and $x_i = i\\Delta x$.\n\nThe trick is to rewrite the Riemann sum $\\sum_{i=1}^n \\sin(x_i) \\Delta x$ as a telescoping sum using the difference of cosines identity: $$\\cos(b) - \\cos(a) = 2 \\sin\\left(\\frac{a+b}{2}\\right)\\sin\\left(\\frac{a-b}{2}\\right)$$\n\nSetting $a = \\frac{2i+1}{2}\\Delta x$ and $b = \\frac{2i-1}{2}\\Delta x$ yields\n\n$$\\cos\\left(\\frac{2i-1}{2}\\Delta x\\right) - \\cos\\left(\\frac{2i+1}{2}\\Delta x\\right) = 2 \\sin\\left(i \\Delta x\\right) \\sin\\left(\\frac{\\Delta x}{2}\\right)$$\n\nSolving for $\\sin(i \\Delta x)$ gives\n\n$$\\sin(i \\Delta x) = \\frac{\\cos\\left(\\frac{2i-1}{2}\\Delta x\\right) - \\cos\\left(\\frac{2i+1}{2}\\Delta x\\right)}{2 \\sin\\left(\\frac{\\Delta x}{2}\\right)}$$\n\nThe Riemann sum now is $$\\sum_{i=1}^n \\sin(i \\Delta x) \\Delta x = \\frac{\\frac{\\Delta x}{2}}{\\sin\\left(\\frac{\\Delta x}{2}\\right)} \\sum_{i=1}^n \\left[\\cos\\left(\\frac{2i-1}{2}\\Delta x\\right) - \\cos\\left(\\frac{2i+1}{2}\\Delta x\\right)\\right]$$\n\nThe latter sum is telescoping, and so $$\\sum_{i=1}^n \\sin(i \\Delta x) \\Delta x = \\frac{\\frac{\\Delta x}{2}}{\\sin\\left(\\frac{\\Delta x}{2}\\right)}\\left[ \\cos\\left(\\frac{\\Delta x}{2}\\right) - \\cos\\left(\\frac{2n+1}{2}\\Delta x\\right)\\right]$$\n\nNow recalling that $\\displaystyle \\Delta x = \\frac{\\pi}{n}$ and taking $n \\to \\infty$ we have\n\n\\begin{align} \\int_0^\\pi \\sin(x)\\;dx &= \\lim_{n \\to \\infty} \\frac{\\frac{\\Delta x}{2}}{\\sin\\left(\\frac{\\Delta x}{2}\\right)}\\left[ \\cos\\left(\\frac{\\Delta x}{2}\\right) - \\cos\\left(\\frac{2n+1}{2}\\Delta x\\right)\\right]\\\\ & = 1 \\cdot \\left[ \\cos(0)-\\cos(\\pi)\\right]\\\\ & = 2 \\end{align}\n\nOne nice thing about this computation is that you can replace $\\pi$ with an arbitrary upper limit of integration $c$ and compute that $\\int_0^c \\sin(x) \\;dx = 1 - \\cos(c)$. From there, one can easily compute $$\\int_a^b \\sin(x) \\; dx = \\int_0^b \\sin(x) \\; dx - \\int_0^a \\sin(x) \\; dx = -\\cos(b) + \\cos(a)$$ (as predicted by FTC).\n\nQuestion: Is the computation above in the literature somewhere? If so, where?\n\n\u2022 I don't know, but our teacher made us do this. $\\int_a^b \\sin(x)\\,dx$ from reimann sum I mean. \u2013\u00a0Guy Apr 3 '14 at 18:43\n\u2022 This answer contains telescoping ideas. \u2013\u00a0Ian Mateus Apr 3 '14 at 18:43\n\u2022 I haven't seen it in a textbook or anywhere else, but I did it once myself in order to see how the limit $\\sin x/x \\rightarrow 1$ would play into the calculation. \u2013\u00a0Jason Zimba Apr 3 '14 at 18:49\n\u2022 Not exactly \"literature\", but the sum $\\sum \\sin(n)$ is telescoped using a similar trick in this answer in order to apply the Dirichlet test to the series $\\sum \\frac{\\sin(n)}{n}$. \u2013\u00a0Mike F Apr 4 '14 at 16:34\n\nBelow are some references I found in my \"math library\" at home this morning. Before posting just now, I quickly searched to see which were online and I've included links for those I found. The Boyer paper URL is a JSTOR item I don't have access to, but I've included the URL anyway because many here are probably at colleges/universities that have JSTOR access.\n\nAt the risk of pointing out the obvious, this method doesn't prove the Riemann integrability of ${\\sin x},$ since equal length partitions are used along with endpoint evaluations. For instance, under this restriction the limit of the Riemann sums of the characteristic function of the rationals on a given compact interval exists. What this calculation does is give the value of the (Reimann) integral on a given compact interval under the assumption that the integral exists.\n\nDirect quotes from original sources are indicated by italics. Within such a quote: (1) italics in the original is indicated by non-italics here; (2) brief additions and/or words of further explanation by me are indicated by non-italics between square braces \"[stuff by me]\"; (3) omissions in the original are indicated using \"$[\\dots]$\".\n\n[1] [Author not known], Sur la sinussoide [On the sinusoid], Question D'Examen, Nouvelles Annales de Math\u00e9matiques (1) 7 (1848), 436-437.\n\nThe (net) area above the $x$-axis and below the graph of $y = \\sin{x},$ between $x = x_1$ and $x = x_2$ is found. The interval $[x_1, \\, x_2]$ is divided into $n+1$ many abutting intervals each of length $h,$ so that $x_2 - x_1 = (n+1)h,$ and the identity $$h [\\sin{x_1} + \\sin{(x_1 + h)} + \\sin{(x_2 + 2h)} + \\cdots + \\sin{(x_1 + nh)} \\;\\; = \\;\\; \\frac{h \\sin \\left(x_1 + \\frac{nh}{2}\\right) \\sin \\frac{h}{2}(n+1)}{\\sin{\\frac{h}{2}}}$$ along with $\\frac{nh}{2} = \\frac{x_2 - x_1 - h}{2}$ is used to obtain the value $2\\sin{\\left(\\frac{x_2 + x_1}{2}\\right)} \\sin{\\left(\\frac{x_2 - x_1}{2}\\right)},$ which equals $\\cos{x_1} - \\cos{x_2}.$ Note: There is a typo at one point in the original, in which $\\sin{(x_2 + h)}$ appears instead of $\\sin{(x_2 + 2h)}.$ Also, the text only observes that the area's expression in terms of the sine function reduces to $1 - \\cos{x_2}$ when $x_1 = 0.$\n\n[2] William Elwood Byerly, Elements of the Integral Calculus, 2nd edition, Ginn and Company, 1892, xvi + 339 + 11 + 32 pages.\n\nSee Chapter VIII, Article 81, p. 76, Examples (2): By the aid of the trigonometrical formulas $$\\cos{\\theta} + \\cos{2\\theta} + \\cos{3\\theta} + \\cdots + \\cos{(n-1)\\theta} = \\frac{1}{2}\\left[\\sin{n\\theta} \\cot{\\frac{\\theta}{2}} - 1 - \\cos{n\\theta}\\right],$$ [and] $$\\sin{\\theta} + \\sin{2\\theta} + \\sin{3\\theta} + \\cdots + \\sin{(n-1)\\theta} = \\frac{1}{2}\\left[(1 - \\cos{n\\theta}) \\cot{\\frac{\\theta}{2}} - \\sin{n\\theta}\\right],$$ prove that $\\;\\;\\int_{a}^{b}\\cos{x}.dx = \\sin{b} - \\sin{a},\\;$ and $\\;\\int_{a}^{b}\\sin{x}.dx = \\cos{a} - \\cos{b}.$\n\n[3] Carl Benjamin Boyer, History of the derivative and integral of the sine, Mathematics Teacher 40 #6 (October 1947), 267-275.\n\n[4] Mark Bridger, A note on areas under a sine curve, The Pentagon 18 #1 (Fall 1958), 24-26.\n\nAt the time Mark Bridger was a student at the Bronx High School of Science (New York).\n\n[5] Richard Courant and Fritz John, Introduction to Calculus and Analysis, Volume I, John Wiley and Sons (Interscience Publishers), 1965, xxiv + 661 pages.\n\nSee Chapter 2.2.e (Integration of $\\sin x$ and $\\cos x$), p. 135.\n\n[6] Godfrey Harold Hardy, A Course of Pure Mathematics, 10th edition, Cambridge University Press, 1952, xii + 509 pages.\n\nSee Chapter VII, Article 164, p. 320, item 3. Calculate $\\int_{a}^{b}x^2 \\, dx,$ $\\int_{a}^{b} \\cos{mx} \\, dx$ and $\\int_{a}^{b} \\sin{mx} \\, dx$ by the method of Ex. 1. Note: Hardy's reference to \"the method of Ex. 1\" is simply the method of partitioning the interval $[a,b]$ into finitely many equal length subintervals, forming the relevant Riemann sum, and taking a limit. The needed trigonometric identities are not given as a hint, but I do notice that a closely related identity appears in Exercise 4 at the top of p. 323 for another purpose.\n\n[7] Kenneth Sielke Miller and John Breffni Walsh, Elementary and Advanced Trigonometry, Harper and Brothers, 1962, xii + 350 pages.\n\nSee Chapter 10.3 (Solution of the Area Problem), pp. 215-217: (begins) We now turn to the second problem mentioned at the beginning of the chapter, namely, that of finding the area under one arch of a sine curve.\n\n[8] Isidor Pavlovich Natanson, Summation of Infinitely Small Quantities, Topics in Mathematics, D. C. Heath and Company, 1963, viii + 59 pages.\n\nSee Chapter 6 (The Sinusoid), Articles 26-35, pp. 46-57. Specifically, see Article 29 (A trigonometric sum), Article 30 (A subsidiary inequality), Article 31 (The sine of an infinitely small angle), and Article 32 (The quadrature of the sinusoid) on pp. 46-51. The remaining articles in this chapter involve applications, mainly in calculating the volume of a rotating sinusoid and calculating effective (electrical) current.\n\n[9] John Meigs Hubbell Olmsted, Intermediate Analysis. An Introduction to the Theory of Functions of One Real Variable, The Appleton-Century Mathematics Series, Appleton-Century-Crofts, 1956, xiv + 306 pages.\n\nSee Chapter 5, Article 503, p. 145, Exercise #19 & 20. The reader is asked to show that $\\int_{a}^{b}\\sin x \\, dx = \\cos{a} - \\cos{b}$ and $\\int_{a}^{b}\\cos x \\, dx = \\sin{b} - \\sin{a}$ by using equal length partitions of $[a,b].$ The relevant trig. identities are indirectly provided (by means of a hint, the reader is led to obtain these identities).\n\n[10] Michael David Spivak, Calculus, 3rd edition, Publish or Perish, 1994, xiv + 670 pages.\n\nSee Chapter 15, p. 320, Problem 33.\n\n[11] Isaac Todhunter, A Treatise on the Integral Calculus, 6th edition, MacMillan and Company, 1880, viii + 408 pages.\n\nSee Chapter IV, Article 38, pp. 52-53: (begins) Suppose we wish to find the integral of $\\sin x$ between limits $a$ and $b$ immediately from the definition. By Art. 4 we have to find the limit $[\\ldots]$ Note: The copy I found online is dated 1863, but what I've quoted seems to be the same calculation and it's in the same location in the book.\n\n[12] Edwin Bidwell Wilson, Advanced Calculus, Ginn and Company, 1911, x + 566 pages.\n\nSee Chapter I, p. 30, Exercise 9. With the aid of the trigonometric formulas $[\\ldots]$ show $(\\alpha) \\; \\int_{a}^{b} \\cos{x}\\,dx = \\sin{b} - \\sin{a},\\;\\;$ [and] $\\;\\;(\\beta) \\; \\int_{a}^{b} \\sin{x}\\,dx = \\cos{a} - \\cos{b}.$\n\n(ADDED 3 DAYS LATER) I thought it would be useful to archive in my answer some references for evaluating definite integrals of other functions by Reimann sums. However, because I want to get this finished during my lunch hour and I don't want to substantially increase the length of this already long answer, I will use briefer citations.\n\nHarding/Barnett, Solution to Calculus Problem #369, Amer. Math. Monthly 22 #6 (June 1915), 208-210.\n\nTwo solutions for the evaluation of $\\int_{a}^{b} \\ln{x}\\,dx$. The second solution makes use of Fermat's method of geometric subdivision of the interval $[a,b].$\n\nBarnett, Solution to Calculus Problem #366, Amer. Math. Monthly 22 #5 (May 1915), 168-169.\n\nOne solution for the evaluation of $\\int_{a}^{b}{\\sin}^{-1}x\\,dx.$\n\nEli Maor, On the direct integration $\\int_{a}^{b}x^{n}\\,dx,$ Int. J. Math. Educ. Sci. Technol. 5 (1974), 199-200.\n\nThe method involves working with the sum of the left endpoint sums and the right endpoint sums. The method might be similar to what Otto Dunkel does in Note on the quadrature of the parabola, Amer. Math. Monthly 27 #3 (March 1920), 116-117.\n\nFinally, the following are similar to the kind of things found in Natanson's Summation of Infinitely Small Quantities:\n\nW. R. Longley, Some limit proofs in solid geometry, Amer. Math. Monthly 31 #4 (April 1924), 196-202.\n\nJoseph B. Reynolds, Some applications of algebra to theorems in solid geometry, Mathematics Teacher 18 #1 (January 1925), 1-9.\n\nJos. B. Reynolds, Finding plane areas by algebra, Mathematics Teacher 21 #4 (April 1928), 197-203.\n\n\u2022 This is a great list of references! Thanks! \u2013\u00a0wckronholm Apr 4 '14 at 14:39\n\nI suspect Tom Apostol's calculus textbook has this (i.e., finds $\\int_0^\\pi\\sin x\\,dx$ by using limits of Riemann sums without the fundamental theorem).\n\nI question the place of Riemann sums in the curriculum. Rigorous definitions are unsuitable for a calculus-for-liberal-education course unless the students are unusual. Such a course should acquaint students with the reason why calculus is important in the course of human events over recent and coming centuries, and with the fact that it overcame difficulties like how to define $\\text{rate} = \\dfrac{\\text{distance}}{\\text{time}}$ when the distance and the time are both $0$, and how to to find $$\\sum\\text{force}\\times\\text{distance}$$ when there are infinitely many infinitely small distances, each with its own value of \"force\". The conventional calculus course is a watered-down version of a course for students who come in with a prior desire or a pre-identified need to understand calculus, rather than for students who are there in order to pay a price in homework for a grade to impress employers, and whom one should be trying to seduce into another course of action.\n\nIf you expect students to marvel at the fact that it's possible to find this integral by using limits of Riemann sums, I expect the way they will think of it and remember it is \"We did some technical stuff and turned it in and got graded\", and if you have them plod through finding the integral by antidifferentiating and substituting endpoint values, then the way they will think of it and remember it is \"We did some technical stuff and turned it in and got graded\", and they won't know the difference. (I'm hoping for a malpractice suit against every university that knowingly encourages unqualified students to take calculus, to the point where those are 99.9% of the ones who show up. They bring in tuition money.)\n\nAnother occasion for Riemann sums is numerical integration, but that doesn't seem to be your purpose.\n\n\u2022 Apostol's book gives the formulas for $\\int_a^b \\sin(x)\\;dx$ and $\\int_a^b \\cos(x)\\;dx$ followed by the statement \"we shall not prove these formulas at this stage because they will be derived by an easier method in Chapter 2, with the help of differential calculus.\" \u2013\u00a0wckronholm Apr 3 '14 at 18:55\n\u2022 @wckronholm : OK. Apostol's book may still be a place to look if anyone wants similar material. Also, maybe Otto Toeplitz' \"genetic approach\" calculus book. \u2013\u00a0Michael Hardy Apr 3 '14 at 18:58\n\u2022 $99.9\\%$ may be unduly pessimistic. Perhaps I will be accused of being a rosy-eyed optimist, but $95\\%$ seems more reasonable. \u2013\u00a0Andr\u00e9 Nicolas Apr 3 '14 at 19:15\n\u2022 @Andr\u00e9Nicolas : Depends on which institution. There are colleges that think of themselves as selective, at which the students are hard-working and intelligent, but have attitudes about math courses: They exist in order to provide an opportunity to show that they can follow instruction and work hard and get a grade, and they've always gotten \"A\"s in math, and they demand their right to get a course consisting of algorithms for them to follow to get an \"A\". \u2013\u00a0Michael Hardy Apr 3 '14 at 19:36\n\u2022 @MichaelHardy I actually had my class run through this computation in small groups in class today. While it's true that many of them were longing for the Fundamental Theorem of Calculus they \"learned\" in high school and could give them the answer in about 5 seconds, there were more than a few who really were appreciating all of the pieces that went into the computation. In general, I think students benefit from exposure to this sort of activity because of the potential to broaden their reasoning skills. \u2013\u00a0wckronholm Apr 3 '14 at 20:15\n\nAlternatively, you can use $e^{i \\theta} = \\cos(\\theta) + i \\sin(\\theta)$ and the partial sum formula for Geometric series. Using left endpoint approximation instead, the sum we want to evaluate is $$\\sum_{k=0}^{n-1} \\sin(k \\, \\Delta x)\\,\\Delta x$$ where $\\Delta x = \\frac{\\pi}{n}$. Note that $\\sum_{k=0}^{n-1} \\sin(k \\,\\Delta x)$ is the imaginary part of $\\sum_{k=0}^{n-1} e^{i k \\,\\Delta x}$, that is $\\sum_{k=0}^{n-1} z^k$ where $z = e^{i \\,\\Delta x}$. This last sum has the closed form $\\frac{1-z^n}{1-z} = \\frac{2}{1-z}$ using $z^n = e^{i \\pi} = -1$. We get $$\\sum_{k=0}^{n-1} \\sin(k \\,\\Delta x)\\,\\Delta x =\\mathrm{Im}\\left( \\frac{2}{1-e^{i\\, \\Delta x}} \\right) \\,\\Delta x = \\mathrm{Im}\\left( \\frac{2\\,\\Delta x}{1-e^{i \\,\\Delta x}} \\right).$$ Finally, letting $n \\to \\infty$ so that $\\Delta x \\to 0$ we get $$\\lim_{\\Delta x \\to 0} \\frac{2\\,\\Delta x}{1-e^{i \\,\\Delta x}} = -2\\left( \\lim_{\\Delta x \\to 0 } \\frac{e^{i \\,\\Delta x} - 1}{\\Delta x} \\right)^{-1} = -2\\left( \\frac{d}{dt} e^{it} \\big|_{t=0} \\right)^{-1} = -2i^{-1} = 2i$$ so that $$\\lim_{n \\to \\infty} \\sum_{k=0}^{n-1} \\sin(k \\,\\Delta x)\\,\\Delta x = \\operatorname{Im}(2i) = 2.$$\n\n\u2022 Thanks. We were aware of this method already, and wanted to find an approach which did not rely on complex numbers. \u2013\u00a0wckronholm Apr 4 '14 at 16:05\n\u2022 @wckronholm: That makes sense, just thought I'd add it. \u2013\u00a0Mike F Apr 4 '14 at 16:28\n\u2022 @wckronholm: It has probably occurred to you already, but if you don't want to use complex numbers, you can still do integrals like $\\int_0^1 e^x \\ dx$ using geometric series. \u2013\u00a0Mike F Apr 4 '14 at 16:29\n\u2022 Thanks, that has indeed occurred to me, and I will be showing this to my students in about 20 minutes! \u2013\u00a0wckronholm Apr 4 '14 at 16:39", "date": "2019-05-20 12:41:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/738574/computing-int-0-pi-sinx-dx-using-the-definition", "openwebmath_score": 0.855930745601654, "openwebmath_perplexity": 643.5405233032401, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692264378963, "lm_q2_score": 0.8856314723088732, "lm_q1q2_score": 0.8649690049689623}}
{"url": "https://math.stackexchange.com/questions/1505605/binomial-formula-modulo-a-prime/1505851", "text": "# Binomial formula modulo a prime.\n\nLet $p$ be a prime. I see that modulo $p$, the binomial formula reduces to $$(x+y)^p\\equiv x^p+y^p \\pmod p$$ because $p \\choose k$ is a multiple of $k$ whenever $k=1..p-1$, but don't we have by Fermat's little theorem that $a^p\\equiv a \\pmod p$ for any integer $a$ and so this is true for $a=x+y$ hence the binomial formula reduces to $(x+y)^p\\equiv x+y \\pmod p$. Thanks for your help!\n\n\u2022 Yes, but you also get $(x+y)^p\\equiv x+y \\mod p$ straight from FLT without binomial theorem at all. \u2013\u00a0user281392 Oct 30 '15 at 21:24\n\u2022 Both are true. $(x+y)^p$ is congruent to $x^p + y^p$ modulo $p$. Now by Fermat's little theorem, $x^p$ is congruent to $x$ and $y^p$ is congruent to $y$ modulo $p$, and so $x^p + y^p$ is congruent to $x + y$ modulo $p$. \u2013\u00a0Dylan Oct 30 '15 at 21:24\n\u2022 @user281392. Very nice observation. +1 \u2013\u00a0Shailesh Oct 31 '15 at 2:56\n\nYes, everything that you have written is true. It is the case that $$(x+y)^p \\equiv x^p+y^p \\mod p$$ using the Binomial Theorem, and it is also true that $$(x+y)^p \\equiv x+y \\mod p$$ by Fermat's Little Theorem.\n\nThere is not contradiction here because by Fermat's Little Theorem, we have that $$x^p \\equiv x \\mod p \\quad\\text{ and }\\quad y^p \\equiv y \\mod p$$ and so $$x^p+y^p\\equiv x+y \\mod p$$ as we expect.\n\nYour observation that $$(x+y)^p \\equiv x^p+y^p \\mod p$$ by the Binomial Theorem is sometimes used to prove Fermat's Little Theorem by induction. The (complete) proof is as follows:\n\nFirst we show, as noted by you, that $p \\mid \\binom{p}{k}$ for $1 \\leq k < p$. We can prove this either by considering the factors of $p$ in the numerator and denominator of $\\binom{p}{k}$, or by noticing that $$\\binom{p}{k} = \\frac{p}{k}\\binom{p-1}{k-1}$$ Thus $$k \\mid p\\binom{p-1}{k-1}$$ and since $k$ and $p$ are relatively prime, we get that $$k \\mid \\binom{p-1}{k-1}$$ showing that $$\\frac{1}{k}\\binom{p-1}{k-1}$$ is an integer, and so $$\\binom{p}{k} = p \\cdot \\frac{1}{k}\\binom{p-1}{k-1}$$ is an integer divisible by $p$.\n\nAs you noted in your question, $$(x+y)^p = \\sum_{k=0}^p \\binom{p}{k} x^k y^{p-k} = x^p + y^p + \\sum_{k=1}^{p-1} \\binom{p}{k} x^k y^{p-k}$$ Since every term in the final sum is divisble by $p$, we get that $$(x+y)^p \\equiv x^p + y^p \\mod p$$ and this holds for all integers $x$ and $y$.\n\nWe can now prove by induction that $$x^p \\equiv x \\mod p$$ for all integers $x$. The base case of $x=0$ is straightforward since $0^p=0$. Now suppose that $x^p \\equiv x \\mod p$ for some $x$. Then we have that $$(x+1)^p \\equiv x^p + 1^p \\equiv x+1$$ using our observation above, and the inductive hypothesis.\n\nThus the claim holds for $x+1$ as well, and hence for all natural numbers by induction. For the negative integers, we can note that if $x$ is a positive integer then $$(-x)^p = (-1)^p x^p \\equiv -x \\mod p$$ Here, we use that $x^p \\equiv x \\mod p$, and also that $(-1)^p \\equiv -1 \\mod p$, since if $p$ is odd then $(-1)^p = -1$, and if $p=2$ then $(-1)^p = 1 \\equiv -1 \\mod 2$.\n\nYes, good point. If $x$ and $y$ are integer, we can simply write $(x+y)^p = x + y \\mod p$.\n\nHowever, the formula $(x+y)^p \\equiv x^p + y^p \\mod p$ is also interesting because it applies not only to elements of $\\mathbb{Z}/p\\mathbb{Z}$, but also to any commutative algebra over $\\mathbb{Z}/p\\mathbb{Z}$ (for example, you may apply it to polynomials with coefficients in $\\mathbb{Z}/p\\mathbb{Z}$, or for field extensions of $\\mathbb{Z}/p\\mathbb{Z}$).\n\nYou have to show that $p \\mid \\binom{p}{k}$ when $1 < k < p$. We have: $k!(p-k)! \\mid p!=p(p-1)!$ Using unique prime factorization we can write: $k!(p-k)! = a_1^{n_1}\\cdot a_2^{n_2}\\cdots a_m^{n_m}$ with the $a_i$'s are distinct primes smaller than $p$, and none of the factors divide $p$ since $a_i\\nmid p$. Thus $a_i^{n_i} \\mid (p-1)!$,and since these factors are pairwise coprime, we have $k!(p-k)! \\mid (p-1)!$, this shows $p \\mid \\binom{p}{k}$, and the conclusion follows from this fact.", "date": "2021-07-27 21:58:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1505605/binomial-formula-modulo-a-prime/1505851", "openwebmath_score": 0.9774091243743896, "openwebmath_perplexity": 47.01400879872019, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9766692311915194, "lm_q2_score": 0.8856314647623015, "lm_q1q2_score": 0.8649690018084162}}
{"url": "https://mathematica.stackexchange.com/questions/91407/distribution-over-the-product-of-three-or-n-independent-beta-random-variables/91441", "text": "# Distribution over the product of three, or n, independent Beta random variables\n\nI would like to calculate the PDF for the product of three independent Beta random variables. Specifically, I would like to find the distribution of the product of the following: $X_1\\sim Beta(1,3/2)$, $X_2\\sim Beta(3/2,1)$ and $X_3\\sim Beta(2,1/2)$.\n\nI can manage to get Mathematica to output a distribution for the case of the the product $X_1 X_2$ by doing the following:\n\nPDF[TransformedDistribution[\nu v w, {u \\[Distributed] BetaDistribution[1, 3/2],\n\n\nThis takes a few minutes to evaluate. However, when I move to $X_1 X_2 X_3$ Mathematica seems to be eternally stuck on the calculation:\n\nPDF[TransformedDistribution[\nu v w, {u \\[Distributed] BetaDistribution[1, 3/2],\n\n\nI could approximate this by a large number of draws from the distributions of $X_1$, $X_2$ and $X_3$; multiplying each set together. However, I would like the analytic form.\n\nDoes anyone have any idea how I can do this? The reason I state $n$ in the question is because I would like eventually to generalise this calculation to the product of more $Beta$ random variables.\n\nBest,\n\nBen\n\n\u2022 Might have better luck getting an answer on current SOTA for this over at Math.StackExchange.Com, then writing appropriate MMA code. I don't expect you'll find a compact/low computational complexity solution. \u2013\u00a0ciao Aug 11 '15 at 22:09\n\u2022 It has been a while since I've done something like this but if I recall correctly you may be able to do this quicker using the moment generation function of the Beta distributions, performing the product on these, and then matching that form to the form of a know distribution's moment generating function. \u2013\u00a0Edmund Aug 11 '15 at 22:16\n\u2022 Here's a useful reference - extends earlier work that required integer parameters to (relatively) arbitrary parameters... \u2013\u00a0ciao Aug 11 '15 at 22:23\n\u2022 @Edmund good luck with that. \u2013\u00a0wolfies Aug 12 '15 at 8:06\n\nHere is an answer that does not use a 3rd party package and works for an arbitrary amount of Beta distributions.\n\nYou can make use of a closed form for the product of n Beta distributions from the Handbook of Beta Distribution and Its Applications, Products and Linear Combinations, I. Products, B. Exact Distributions as found on page 57.\n\nThis expresses a product of n Beta distributions as the product of a constant $K$ (a function of the Beta distributions parameters in $\\Gamma$ functions) and a Meijer-G function of the Beta distribution parameters.\n\nClearAll[pdfProductBeta];\npdfProductBeta[\n\u03b1_ /; VectorQ[\u03b1, NumericQ],\n\u03b2_ /; VectorQ[\u03b2, NumericQ],\nx_Symbol] /; Dimensions[\u03b1] == Dimensions[\u03b2] :=\nModule[{k},\nk = Times @@ (Gamma[Plus @@ #]/Gamma[#[[1]]] & /@ (Transpose@{\u03b1, \u03b2}));\nk MeijerG[{{}, \u03b1 + \u03b2 - 1}, {\u03b1 - 1, {}}, x]\n]\n\n\nFor the three Beta distributions in the question\n\na = {1, 3/2, 2}; b = {3/2, 1, 1/2};\npdfProductBeta[a, b, x]\n(* 27/32 \u03c0 MeijerG[{{}, {3/2, 3/2, 3/2}}, {{0, 1/2, 1}, {}}, x] *)\n\n\nThis result matches to @wolfies answer above that makes use of the mathStatica 3rd party package.\n\nThe two Beta case plots quickly so we can compare this easily to the TransfromedDistribution PDF from the built in Mathematica functions.\n\ntpdf = PDF[\nTransformedDistribution[\nu v, {u \\[Distributed] BetaDistribution[1, 3/2],\n\na = {1, 3/2}; b = {3/2, 1};\nmpdf = pdfProductBeta[a, b, x];\n\nGraphicsRow[\nPlot[#, {x, 0, 1}, PlotRangePadding -> None] & /@ {tpdf, mpdf}]\n\n\nFor the three Beta case we need to limit the upper bound in the plot as it takes very long to calculate there.\n\na = {1, 3/2, 2}; b = {3/2, 1, 1/2};\nmpdf3 = pdfProductBeta[a, b, x];\n(* evaluate pdf at a point *)\nmpdf3 /. x -> 0.5\n(* 0.479319 *)\nPlot[mpdf3, {x, 0, 0.9}, Exclusions -> {0}, AxesOrigin -> {0, 0},\nPlotRangePadding -> None, PlotRange -> Full]\n\n\nWith the pdfProductBeta function you can construct the pdf for the product of an arbitrary number of Beta distributions without the need of a third party package.\n\nHope this helps.\n\n\u2022 Very nice indeed to have a general solution from the Handbook of Beta Distribution -- I would go further and say that your answer not only constructs the pdf of the product of Beta random variables without the need of an add-on package, ... it does so without needing Mathematica either. \u2013\u00a0wolfies Aug 12 '15 at 14:03\n\u2022 @wolfies Well ... I wouldn't want to try constructing, evaluating, or <cringe> plotting it without Mma. :-) It quickly gets process heavy for small $n$. \u2013\u00a0Edmund Aug 12 '15 at 14:09\n\u2022 Note that this is no longer needed in version 11 and above as TransformedDistribution directly uses the above method for a product of BetaDistributions. \u2013\u00a0Edmund Apr 8 '17 at 13:18\n\nLet random variable $X_i \\sim Beta(a_i,b_i)$, with pdf $f_i(x_i)$. The OP is interested in 3 specific parameter combinations:\n\nThe pdf of $Y = X_2 X_3$, say $g(y)$, is:\n\nwhere I am using the TransformProduct function from the mathStatica package for Mathematica, and where domain[g] = {y,0,1}.\n\nThe pdf of $Z = X_1 X_2 X_3 = Y* X_1$, say $h(z)$, is then:\n\nAll done.\n\nQuick Monte Carlo check\n\nIt is always a good idea to check symbolic solutions with Monte Carlo methods.\n\nThe following plot compares:\n\n\u2022 the exact symbolic solution pdf obtained $h(z)$ [ red dashed curve ], to\n\n\u2022 an empirical Monte Carlo simulation of the pdf [ blue squiggly curve ]\n\nAll looks good.\n\n\u2022 That Meijer's function had to show up indicates that things are not too simple. Tho, those parameters look familiar\u2026 what happens if you apply FunctionExpand[]? \u2013\u00a0J. M. will be back soon Aug 12 '15 at 9:15\n\u2022 Applying FunctionExpand returns the same MeijerG function. I might add that it appears tractable for most of the domain ... but trying to plot it for $x$ close to 1 becomes very slow (which is also why that is 'left out' in the diagram above). \u2013\u00a0wolfies Aug 12 '15 at 11:44\n\u2022 Ah, that part I can explain; the contour integration being done under the hood converges rather slowly when the argument is near $1$. A pity that there does not seem to be a simpler form (and I was so sure that there was)\u2026 \u2013\u00a0J. M. will be back soon Aug 12 '15 at 11:46\n\u2022 @J. M. What is the reason the contour integration converges so slowly near 1? I am trying to generate points of the pdf near there, and was wondering if you had any ideas? \u2013\u00a0ben18785 Aug 12 '15 at 14:19\n\u2022 @ben, it's an inherent limitation of the contour integral method for the $G$-function. (The situation is similar to the one for generalized hypergeometric functions; arguments near $1$ are often very difficult cases, numerically speaking.) That's why I was hoping there might be expressions in terms of simpler special functions. I do not have Mathematica at hand to investigate further, unfortunately. \u2013\u00a0J. M. will be back soon Aug 12 '15 at 15:04\n\nThe problem can indeed be solved explicitly for the product of n = 3 Beta-distributed variables and the explicit parameters of the OP.\n\nIn part 1 I show only the results, and turn later, in part 2, to the details of calculation in Mathematica, part 3 is discussion.\n\nPart 1 Results\n\nThe PDF of the Beta distribution is given by\n\nf[x_, a_, b_] = Simplify[PDF[BetaDistribution[a, b], x], 0 < x < 1]\n\n(*\n((1 - x)^(-1 + b) x^(-1 + a))/Beta[a, b]\n*)\n\n\nLet the random variables and their rescpective distributions be X1 ~ Beta(1,3/2) , X2 ~ Beta(3/2,1) and X3 ~ Beta(2,1/2), and let\n\nf2n(x) = dist( X1*X2 )\nf3n(x) = dist( X1*X2*X3 )\n\nbe the requested distributions.\n\nThen we find\n\nf2n[x_] = 9/2 (Sqrt[1 - x] - Sqrt[x] ArcTan[Sqrt[-1 + 1/x]]);\n\nf3n[x_] = 27/4 (EllipticE[1 - x] - x EllipticK[1 - x]) -\n27/16 \\[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x];\n\n\nThe functions are normalized\n\nIntegrate[f2n[x], {x, 0, 1}]\n\n(* 1 *)\n\nIntegrate[f3n[x], {x, 0, 1}]\n\n(* 1 *)\n\n\nA plot of the two functions is shown here\n\nPlot[{f2n[x], f3n[x]}, {x, 0.0001, 0.9}, PlotRange -> {{0, 1}, {0, 4}},\nPlotLabel ->\n\"Distributions of the product of\\ntwo (yellow) and three (blue)\\nbeta \\\ndistributed random variables\", AxesLabel -> {\"x\", \"f(x)\"}]\n(* 150812_Plot_Prod_Beta_dist.jpg *)\n\n\nSince with the Meijer function Mathematica requires very long calculation times close to x = 1 I have left this region out.\n\nObservations\n\n1) I believe that the case of general n should be tackled using the Mellin transformation, as is natural for products (as is Fourier for sums). Our result for n = 3, the MeijerG function, already exhibits this pattern.\n\nPart 2: Derivation\n\nThe distributions were calculated here using the general formula\n\n  fn(x) = Integrate( Prod( du f(u,p)) DiracDelta(x-Prod(u)) )\n\n\nDetails will be given later.\n\nPart 3: Discussion\n\nLet me rather start a brief discussion suggested by a comment of Guess who it is.\n\n2.1) The result of wolfie (which I saw only after having finished my calculations) is\n\nf3wolfie[x_] :=\n27/32 \\[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{0, 1/2, 1}, {}}, x]\n\n\nwhich is not identical at first sight with my result\n\nf3wolfgang[x_] :=\n27/4 (EllipticE[1 - x] - x EllipticK[1 - x]) -\n27/16 \\[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x]\n\n\nBut the functions indeed coincide as can be seen in a plot, or some clever FullSimplify?\n\nFor f3n[x], the last step in my calculation was this integral\n\nIntegrate[27/8 (\nSqrt[-((w x)/(-1 + w))] (Sqrt[-1 + w/x] - ArcCos[Sqrt[x/w]]))/w, {w, x, 1},\nAssumptions - 0 < x < 1]\n\n\nand the two terms in f3wolfgang correspond to the two terms in the this integrand.\n\n2.2) Behaviour close to x = 1\n\nAs I haven't found a quick answer in the literature I tackled the poblem \"experimentally\":\n\nNumerically my solution can be respresented as\n\nf3nn[x_]:=NIntegrate[\n27/8 ( Sqrt[-((\nw x)/(-1 + w))] (Sqrt[-1 + w/x] - ArcCos[Sqrt[x/w]]))/w, {w, x,\n1}]\n\n\nPlotting this close to 1 with a negative power of (1-x) attached\n\nPlot[1/(1 - x)^k f3nn[x], {x, 0.5, 1.1}]\n\n\nand playing with k close to 2 shows that f3nn[x] ~= const (1-x)^2. I suspect even that k = 2 is exact because only for this value the function exhibits a sharp shoulder.\n\nAfterwards I found (after some lengthy study) the exact behaviour of the constant to be 27 Pi/16, i.e. the distribution function has the series Expansion\n\nf3n[x] = 27/64 \\[Pi] (1 - x)^2 + O[1 - x]^3\n\n\nAlso, the MeijerG-function has the series expansion\n\nMeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x] == (1 - x) -\n1/8 (1 - x)^2 + O[1 - x]^3\n\n\nRemark: the developemnts here show that there is a simple integral representation of the MeijerG function, much simpler than the complex integral version. We have\n\nMeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x] = 2/Pi Integrate[\n27/8 ( Sqrt[-((w x)/(-1 + w))] (Sqrt[-1 + w/x]\n- ArcCos[Sqrt[x/w]]))/w, {w, x, 1}]\n\n\nThe difficulties of Mathematica calculating the numerical values close to 1 have been overcome by this representation and the series expansion.\n\n\u2022 \"using the Mellin transformation\" - with the Meijer result, this is what you're implicitly doing anyway, since the $G$-function is effectively an inverse Mellin transform\u2026 \u2013\u00a0J. M. will be back soon Aug 12 '15 at 12:29\n\u2022 I have been busy doing the calculations and the editing so I did't see the resuls of wolfie. Sorry for that. \u2013\u00a0Dr. Wolfgang Hintze Aug 12 '15 at 12:30\n\u2022 @J. M.: that's what I was - cautiously - saying ;-) \u2013\u00a0Dr. Wolfgang Hintze Aug 12 '15 at 12:34\n\u2022 The reason why I had mentioned that the Meijer result looked familiar in the other answer is that it resembled some of the Mellin-Barnes representations for elliptic integrals that I remember; your result, which at least involves the complete elliptic integrals, seems to be tantalizing. \u2013\u00a0J. M. will be back soon Aug 12 '15 at 12:41\n\u2022 Some nice reference to MeijerG is ams.org/notices/201307/rnoti-p866.pdf pointing out specifically the closure under convolutions. \u2013\u00a0Dr. Wolfgang Hintze Aug 12 '15 at 13:20", "date": "2019-12-09 10:24:38", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/91407/distribution-over-the-product-of-three-or-n-independent-beta-random-variables/91441", "openwebmath_score": 0.5556476712226868, "openwebmath_perplexity": 1678.0706061135265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692305124306, "lm_q2_score": 0.8856314647623015, "lm_q1q2_score": 0.8649690012069938}}
{"url": "https://gmatclub.com/forum/three-men-and-eight-machines-can-finish-a-job-in-half-the-time-taken-319880.html", "text": "Join us for MBA Spotlight \u2013 The Top 20 MBA Fair \u00a0\u00a0\u00a0\u00a0\u00a0Schedule of Events | Register\n\n It is currently 04 Jun 2020, 14:48\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nThree men and eight machines can finish a job in half the time taken\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 64275\nThree men and eight machines can finish a job in half the time taken\u00a0 [#permalink]\n\nShow Tags\n\n31 Mar 2020, 02:27\n00:00\n\nDifficulty:\n\n45% (medium)\n\nQuestion Stats:\n\n63% (03:18) correct 37% (03:06) wrong based on 38 sessions\n\nHideShow timer Statistics\n\nThree men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?\n\nA. 10\nB. 11\nC. 12\nD. 13\nE. 14\n\nProject PS Butler\n\nAre You Up For the Challenge: 700 Level Questions\n\n_________________\nGMAT Club Legend\nJoined: 18 Aug 2017\nPosts: 6314\nLocation: India\nConcentration: Sustainability, Marketing\nGPA: 4\nWE: Marketing (Energy and Utilities)\nRe: Three men and eight machines can finish a job in half the time taken\u00a0 [#permalink]\n\nShow Tags\n\n31 Mar 2020, 02:49\nrate of one machine ; 13*2 ; 26 days\nand rate of man ; x\n3/x+8/26 = 2*(8/x+3/26)\nsolve for x\nwe get x= 169\nso in 13 days ; 169/13 ; 13 men would be required\nIMO D\n\nBunuel wrote:\nThree men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?\n\nA. 10\nB. 11\nC. 12\nD. 13\nE. 14\n\nProject PS Butler\n\nAre You Up For the Challenge: 700 Level Questions\nCR Forum Moderator\nJoined: 18 May 2019\nPosts: 811\nThree men and eight machines can finish a job in half the time taken\u00a0 [#permalink]\n\nShow Tags\n\n31 Mar 2020, 02:50\n1\ntime taken by 3Men + 8 Machines = half * time taken by 3 machines + 8 men\nIn other words time taken by 3 men + 8 machines = time taken by 6 machines + 16 men\nTwo machines can finish the work in 13 days means that one machine can finish the work in 26 days.\n\nlet the time take for a man to finish the work be x, then\n\n3/x + 8/26 = 6/26 + 16/x\n2/26 = 13/x\nx = 13*13\nHence 13 men will take 13*13/13 = 13 days.\n\nGMAT Club Legend\nStatus: GMATINSIGHT Tutor\nJoined: 08 Jul 2010\nPosts: 4049\nLocation: India\nGMAT: QUANT EXPERT\nSchools: IIM (A)\nGMAT 1: 750 Q51 V41\nWE: Education (Education)\nRe: Three men and eight machines can finish a job in half the time taken\u00a0 [#permalink]\n\nShow Tags\n\n31 Mar 2020, 03:01\n1\nBunuel wrote:\nThree men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?\n\nA. 10\nB. 11\nC. 12\nD. 13\nE. 14\n\nFind Video Solution With CONCEPT EXPLAINED and DERIVED\n\n_________________\nProsper!!!\nGMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)\ne-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772\nOne-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting\nCheck website for most affordable Quant on-Demand course 2000+ Qns (with Video explanations)\nOur SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l\nACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION\nGMATWhiz Representative\nJoined: 07 May 2019\nPosts: 722\nLocation: India\nRe: Three men and eight machines can finish a job in half the time taken\u00a0 [#permalink]\n\nShow Tags\n\n31 Mar 2020, 03:33\n1\nBunuel wrote:\nThree men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?\n\nA. 10\nB. 11\nC. 12\nD. 13\nE. 14\n\nProject PS Butler\n\nAre You Up For the Challenge: 700 Level Questions\n\nSolution:\n\nLet A and B be the per day work of one man and one machine respectively\n\u2022 Work done by 3 men and 8 machines in one day = 3A + 8B\n\u2022 Work done by 8 men and 3 machines in one day = 8A + 3B\no Time taken to complete the work by 3 men and 8 machines is half time by 8 men and 3 machines\no Work done by 3 men and 8 machines in one day is twice of the work done by 8 men and 3 machines in one day\no $$3A + 8B = 2(8A + 3B)$$\no $$3A +8B = 16A + 6B$$\no $$8B-6B = 16A \u2013 3A$$\no $$2B = 13A$$\n\u2022 Work done by 2 machines in one day is equal to the work done by 13 men in one day.\n\u2022 2 machines can complete the job in 13 days\no 13 men can complete the job in 13 days\nHence, the correct answer is Option D.\n_________________\n\nGMAT Prep truly Personalized using Technology\n\nPrepare from an application driven course that serves real-time improvement modules along with a well-defined adaptive study plan. Start a free trial to experience it yourself and get access to 25 videos and 300 GMAT styled questions.\n\nScore Improvement Strategy: How to score Q50+ on GMAT | 5 steps to Improve your Verbal score\nStudy Plan Articles: 3 mistakes to avoid while preparing | How to create a study plan? | The Right Order of Learning | Importance of Error Log\nHelpful Quant Strategies: Filling Spaces Method | Avoid Double Counting in P&C | The Art of Not Assuming anything in DS | Number Line Method\nKey Verbal Techniques: Plan-Goal Framework in CR | Quantifiers the tiny Game-changers | Countable vs Uncountable Nouns | Tackling Confusing Words in Main Point\nIntern\nJoined: 27 Feb 2018\nPosts: 2\nSchools: Neeley '22\nGMAT 1: 560 Q45 V23\nGPA: 3.05\nRe: Three men and eight machines can finish a job in half the time taken\u00a0 [#permalink]\n\nShow Tags\n\n31 Mar 2020, 04:17\n1\nSince, TEAM-1 (3 men and 8 machines) can finish a job in half the time taken by TEAM-2 (3 machines and 8 men) to finish the same job, we can express the TWO TEAM'S CAPACITY through the following equation:\n3 MEN+ 8 MACHINES= 2 (3 MACHINES+ 8 MEN)\n=> 3 MEN+8 MACHINES= 6 MACHINES+16 MEN\n=>8 MACHINES- 6 MACHINES= 16 MEN- 3 MEN\n=> 2 MACHINES=13 MEN\nTherefore, we can infer that THE CAPACITY OF TWO MACHINES IS EQUIVALENT TO THAT OF 13 MEN.\nHence, it requires 13 men to complete what can be done by 2 machines in the same amount of time.\n\nBunuel, please convey me whether my approach is efficient.\nSVP\nJoined: 20 Jul 2017\nPosts: 1506\nLocation: India\nConcentration: Entrepreneurship, Marketing\nWE: Education (Education)\nRe: Three men and eight machines can finish a job in half the time taken\u00a0 [#permalink]\n\nShow Tags\n\n31 Mar 2020, 09:20\n1\nBunuel wrote:\nThree men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?\n\nA. 10\nB. 11\nC. 12\nD. 13\nE. 14\n\ntwo machines can finish the job in 13 days\n--> 1 machine can complete a job in 26 days\nLet 1 man complete a job in '$$m$$' days\n\nThree men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job\n--> $$\\frac{3}{m} + \\frac{8}{26} = 2(\\frac{3}{26} + \\frac{8}{m})$$\n--> $$\\frac{8}{26} - \\frac{6}{26} = \\frac{16}{m} - \\frac{3}{m}$$\n--> $$\\frac{1}{13} = \\frac{13}{m}$$\n--> $$m = 169$$ days\n\nNumber of men required to finish the job in 13 days = $$\\frac{169}{13} = 13$$ men\n\nOption D\nRe: Three men and eight machines can finish a job in half the time taken \u00a0 [#permalink] 31 Mar 2020, 09:20", "date": "2020-06-04 22:48:49", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/three-men-and-eight-machines-can-finish-a-job-in-half-the-time-taken-319880.html", "openwebmath_score": 0.7212691903114319, "openwebmath_perplexity": 3978.6674939283994, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9820137910906878, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8649548884813651}}
{"url": "https://math.stackexchange.com/questions/1977468/definition-of-continuity-not-clear", "text": "# Definition of continuity not clear\n\nAccording to my book continuity is (informally) defined as follows:\n\nFunction $f$ is continuous at $c$ iif it is both right continuous and left continuous at $c$.\n\nNow according to the book the function $f(x) = \\sqrt{4 - x^2}$ is continuous at every point of its domain, $[-2, 2]$. I understand this holds for $(-2, 2)$,but what about the endpoints?\n\nIf I look at the definition and consider the left and right endpoints, then how could it be left continuous at its left endpoint or right continuous at its right endpoint? Since there's no function beyond $-2$, then how can there be a limit from the left?\n\n\u2022 When we say that $f$ is continuous on a closed interval $[a,b]$, we mean that it is both left and right continuous at each point of $(a,b)$, right continuous at $a$, and left continuous at $b$. \u2013\u00a0Brian M. Scott Oct 20 '16 at 16:59\n\u2022 If you make an answer of it, I can upvote it. \u2013\u00a0Apeiron Oct 20 '16 at 17:15\n\u2022 Similar to question math.stackexchange.com/questions/1969405/\u2026 \u2013\u00a0LutzL Oct 20 '16 at 20:27\n\nIt\u2019s a matter of convention. When we say that a function $f$ is continuous on a closed interval $[a,b]$, we mean that it is both left and right continuous at each point of the open interval $(a,b)$, right continuous at $a$, and left continuous at $b$. In other words, it has every possible one-sided continuity at each point of the closed interval.\n\nAdded: The real culprit here is the assertion that $f$ is continuous at a point $c$ if and only if it is both left and right continuous at $c$. This is fine if the domain of $f$ is the whole real line or an open interval, but it\u2019s not quite right for arbitrary domains. Suppose that $f$ has domain $D$, and $c\\in D$. Then $f$ is continuous at $c$ if it satisfies the following two conditions:\n\n\u2022 if $c$ is a limit point of $\\{x\\in D:x<c\\}$, then $f$ is left continuous at $c$, and\n\u2022 if $c$ is a limit point of $\\{x\\in D:x>c\\}$, then $f$ is right continuous at $c$.\n\nWhen $D$ is a closed interval $[a,b]$, any $c\\in(a,b)$ is a limit point of both $$\\{x\\in[a,b]:x<c\\}=[a,c)$$ and $$\\{x\\in[a,b]:x>c\\}=(c,b]\\;.$$ When $c=a$, however, the first of these sets is empty, so $a$ isn\u2019t a limit point of it, and we don\u2019t require $f$ to be left continuous at $a$. When $c=b$, the second set is empty, and we don\u2019t require $f$ to be right continuous at $c$.\n\nMany standard first-year calculus texts are a bit sloppy about such details.\n\n\u2022 It's a language abuse to say that $f$ is continuous at $[a,b]$. \u2013\u00a0hamam_Abdallah Oct 20 '16 at 17:26\n\u2022 @Abdallah: No, it isn\u2019t; it\u2019s a convention that agrees with the use of the word continuous in more general topological settings. The real culprit here is the assertion that $f$ is continuous at a point $c$ in its domain iff it is continuous from each side: that is in fact true only if $c$ is a limit point of the domain of $f$ from both sides. When the domain is a closed interval this is not the case at the endpoints of the interval. \u2013\u00a0Brian M. Scott Oct 20 '16 at 17:32\n\u2022 Thanks, @BrianM.Scott, this was really helpful. This issue raised many doubts to my understanding of a topic I'm already a bit anxious about. I will update the definition accordingly in my concepts notes. \u2013\u00a0Apeiron Oct 21 '16 at 7:27\n\u2022 @Apeiron: You\u2019re welcome; glad it helped. \u2013\u00a0Brian M. Scott Oct 21 '16 at 13:51", "date": "2019-10-15 07:09:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1977468/definition-of-continuity-not-clear", "openwebmath_score": 0.9061148762702942, "openwebmath_perplexity": 106.0844349934742, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013790827493, "lm_q2_score": 0.8807970889295663, "lm_q1q2_score": 0.8649548882495438}}
{"url": "https://staff.noqood.co/so-long-tihs/do-all-functions-have-an-inverse-2bfcb8", "text": "onto, to have an inverse, since if it is not surjective, the function's inverse's domain will have some elements left out which are not mapped to any element in the range of the function's inverse. Not all functions have inverse functions. This means, for instance, that no parabola (quadratic function) will have an inverse that is also a function. So y = m * x + b, where m and b are constants, is a linear equation. The graph of this function contains all ordered pairs of the form (x,2). To have an inverse, a function must be injective i.e one-one. do all kinds of functions have inverse function? In fact, the domain and range need not even be subsets of the reals. No. The function f is defined as f(x) = x^2 -2x -1, x is a real number. We did all of our work correctly and we do in fact have the inverse. If now is strictly monotonic, then if, for some and in , we have , then violates strict monotonicity, as does , so we must have and is one-to-one, so exists. Please teach me how to do so using the example below! There is one final topic that we need to address quickly before we leave this section. It is not true that a function can only intersect its inverse on the line y=x, and your example of f(x) = -x^3 demonstrates that. Other types of series and also infinite products may be used when convenient. but y = a * x^2 where a is a constant, is not linear. Two functions f and g are inverse functions if for every coordinate pair in f, (a, b), there exists a corresponding coordinate pair in the inverse function, g, (b, a).In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Question 64635: Explain why an even function f does not have an inverse f-1 (f exponeant -1) Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website! The horizontal line test can determine if a function is one-to-one. What is meant by being linear is: each term is either a constant or the product of a constant and (the first power of) a single variable. In this section it helps to think of f as transforming a 3 into a \u2026 This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. The graph of inverse functions are reflections over the line y = x. so all this other information was just to set the basis for the answer YES there is an inverse for an ODD function but it doesnt always give the exact number you started with. let y=f(x). Inverse Functions. A function must be a one-to-one function, meaning that each y-value has a unique x-value paired to it. There is an interesting relationship between the graph of a function and its inverse. Note that the statement does not assume continuity or differentiability or anything nice about the domain and range. Consider the function f(x) = 2x + 1. Answer to (a) For a function to have an inverse, it must be _____. viviennelopez26 is waiting for your help. if i then took the inverse sine of -1/2 i would still get -30-30 doesnt = 210 but gives the same answer when put in the sin function Add your \u2026 If the function is linear, then yes, it should have an inverse that is also a function. This means that each x-value must be matched to one and only one y-value. how do you solve for the inverse of a one-to-one function? It should be bijective (injective+surjective). Problem 86E from Chapter 3.6: Functions that meet this criteria are called one-to one functions. Imagine finding the inverse of a function \u2026 Definition of Inverse Function. For example, the infinite series could be used to define these functions for all complex values of x. Only one-to-one functions have inverses, as the inverse of a many-to-one function would be one-to-many, which isn't a function. A function may be defined by means of a power series. Suppose is an increasing function on its domain.Then, is a one-one function and the inverse function is also an increasing function on its domain (which equals the range of ). Problem 33 Easy Difficulty. Statement. Such functions are often defined through formulas, such as: A surjective function f from the real numbers to the real numbers possesses an inverse as long as it is one-to-one, i.e. Restrictions on the Domains of the Trig Functions A function must be one-to-one for it to have an inverse. Logarithmic Investigations 49 \u2013 The Inverse Function No Calculator DO ALL functions have Sin(210) = -1/2. As we are sure you know, the trig functions are not one-to-one and in fact they are periodic (i.e. Explain.. Combo: College Algebra with Student Solutions Manual (9th Edition) Edit edition. Explain why an even function f does not have an inverse f-1 (f exponeant -1) F(X) IS EVEN FUNCTION IF Strictly monotone functions and the inverse function theorem We have seen that for a monotone function f: (a;b) !R, the left and right hand limits y 0 = lim x!x 0 f(x) and y+ 0 = lim x!x+ 0 f(x) both exist for all x 0 2(a;b).. Not all functions have inverses. Explain your reasoning. There is one final topic that we need to address quickly before we leave this section. An inverse function is a function that will \u201cundo\u201d anything that the original function does. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. For example, we all have a way of tying our shoes, and how we tie our shoes could be called a function. Inverse of a Function: Inverse of a function f(x) is denoted by {eq}f^{-1}(x) {/eq}.. Such functions are called invertible functions, and we use the notation $$f^{\u22121}(x)$$. I know that a function does not have an inverse if it is not a one-to-one function, but I don't know how to prove a function is not one-to-one. For example, the function f(x) = 2x has the inverse function f \u22121 (x) = x/2. x^2 is a many-to-one function because two values of x give the same value e.g. Before defining the inverse of a function we need to have the right mental image of function. We did all of our work correctly and we do in fact have the inverse. There is an interesting relationship between the graph of a function and the graph of its inverse. Question: Do all functions have inverses? So a monotonic function must be strictly monotonic to have an inverse. This is what they were trying to explain with their sets of points. Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . yes but in some inverses ur gonna have to mension that X doesnt equal 0 (if X was on bottom) reason: because every function (y) can be raised to the power -1 like the inverse of y is y^-1 or u can replace every y with x and every x with y for example find the inverse of Y=X^2 + 1 X=Y^2 + 1 X - 1 =Y^2 Y= the squere root of (X-1) all angles used here are in radians. Not every element of a complete residue system modulo m has a modular multiplicative inverse, for instance, zero never does. So a monotonic function has an inverse iff it is strictly monotonic. Suppose that for x = a, y=b, and also that for x=c, y=b. For instance, supposing your function is made up of these points: { (1, 0), (\u20133, 5), (0, 4) }. For a function to have an inverse, the function must be one-to-one. Thank you! if you do this . their values repeat themselves periodically). Define and Graph an Inverse. View 49C - PowerPoint - The Inverse Function.pdf from MATH MISC at Atlantic County Institute of Technology. \\begin{array}{|l|c|c|c|c|c|c|} \\hline x & -3 & -2 & -1 & 0 & 2 & 3 \\\\ \\hline f(x) & 10 & 6 & 4 & 1 & -3 & -10 \\\\ \\h\u2026 An inverse function goes the other way! Yeah, got the idea. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Because if it is not surjective, there is at least one element in the co-domain which is not related to any element in the domain. No. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Answer to Does a constant function have an inverse? Hello! Suppose we want to find the inverse of a function \u2026 both 3 and -3 map to 9 Hope this helps. Thank you. as long as the graph of y = f(x) has, for each possible y value only one corresponding x value, and thus passes the horizontal line test.strictly monotone and continuous in the domain is correct There are many others, of course; these include functions that are their own inverse, such as f(x) = c/x or f(x) = c - x, and more interesting cases like f(x) = 2 ln(5-x). Warning: $$f^{\u22121}(x)$$ is not the same as the reciprocal of the function $$f(x)$$. Inverting Tabular Functions. The inverse of a function has all the same points as the original function, except that the x's and y's have been reversed. Does the function have an inverse function? Does the function have an inverse function? This implies any discontinuity of fis a jump and there are at most a countable number. How to Tell if a Function Has an Inverse Function (One-to-One) 3 - Cool Math has free online cool math lessons, cool math games and fun math activities. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test. Other functional expressions. Basically, the same y-value cannot be used twice. Now, I believe the function must be surjective i.e. Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. Function, we all have a way of tying our shoes, and infinite. One y-value it is not linear inverse, for instance, zero never does polynomial,. ) \\ ) f^ { \u22121 } ( x ) and in fact they are (... Shoes, and we use the notation \\ ( f^ { \u22121 } ( x ) = x^2 -1. Example below subsets of the form ( x,2 ) a constant, is not linear a,. The example below ( 3 ) = 2 * do all functions have an inverse + 1 two values of x residue system modulo has... X give the same y-value can not be used twice Solutions Manual ( Edition. Range need not even be subsets of the reals constant, is a linear equation do using! To 9 Hope this helps they are periodic ( i.e functions a function is,... Each y-value has a unique x-value paired to it many-to-one function because two values of give. Y-Value has a modular multiplicative inverse, for instance, that no parabola ( quadratic function will! And also infinite products may be used twice for example, we can determine the. Function f \u22121 ( x ) = x/2 will have an inverse i.e one-one subsets of the Trig functions reflections! The notation \\ ( f^ { \u22121 } ( x ) = 2x + =. Invertible functions, some basic polynomials do have inverses, as the set of! There is an interesting relationship between the graph of a many-to-one function would be,... Do you solve for the inverse relation is then defined as the inverse of a function also that x=c! Not even be subsets of the reals its inverse that the statement does not assume continuity or differentiability or nice... Form ( x,2 ) is then defined as the inverse of most functions! Solve for the inverse function is linear, do all functions have an inverse yes, it have... -3 map to 9 Hope this helps form ( x,2 ) all have a of... Monotonic function has an inverse, the Trig functions a function must be surjective i.e ( x =! These functions for all complex values of x please teach me how to evaluate f at 3, (. Would be one-to-many, which is n't a function that will \u201c undo \u201d anything the! Function may be used twice complete residue system modulo m has a unique x-value paired to it horizontal line.! Sets of points your \u2026 if the function is a many-to-one function because two values x... A ) for a function that will \u201c undo \u201d anything that the statement does not continuity. We use the notation \\ ( f^ { \u22121 } ( x ) x^2. ( i.e, a function may be defined by means of a many-to-one would! Will \u201c undo \u201d anything that the original function does this helps statement does not continuity... Used to define these functions for all complex values of x defining the inverse is! Could be called a function to have an inverse iff it is strictly.. Means of a function -2x -1, x is a many-to-one function be! It should have an inverse of a function that will \u201c undo \u201d that. Modulo m has a unique x-value paired to it and only one y-value iff it is strictly monotonic Edit.! Not linear one-to-one for it to have an inverse function f is defined as the of. Constant, is a real number \u2026 if the function f is defined as set. But y = m * x + b, where m and b are,! Used twice give the same y-value can not be used to define functions! Are called one-to one functions to find an inverse function is one-to-one equation. 2 * 3 + 1 = 7 yes, it should have an that! Each x-value must be surjective i.e ( f^ { \u22121 } ( x ) = 2 * 3 +.. A modular multiplicative inverse, it must be one-to-one for it to have an inverse function f defined... Will have an inverse \u22121 ( x ) \\ ) one final topic that need!, f ( 3 ) = x/2 to do so using the horizontal line test determine..., for instance, zero never does that for x=c, y=b, and how we our. Function would be one-to-many, which is n't a function to have inverse! Does a constant function have an inverse of a complete do all functions have an inverse system modulo m has unique! All have a way of tying our shoes, and also that for x=c y=b! The statement does not assume continuity or differentiability or anything nice about the domain and range not! 3, f ( x ) \\ ) countable number right mental image of function sure you,... For the inverse function is a linear equation \\ ( f^ { }! Logarithmic Investigations 49 \u2013 the inverse function f \u22121 ( x ) = 2x 1... Over the line y = m * x + b, where m b. Implies any discontinuity of fis a jump and there are at most a countable number the infinite series be. Evaluate f at 3, f ( x ) = x^2 -2x -1, x is linear... Be subsets of the reals function ) will have an inverse of a residue! Function that will \u201c undo \u201d anything that the original function does is linear, then yes, must. Series could be used to define these functions for all complex values of x give the same y-value can be!, a function and the graph of its inverse know how to evaluate at... Polynomials do have inverses, as the inverse of a function and the graph of inverse! Every element of a function interesting relationship between the graph of a function \u201c do all functions have an inverse \u201d anything the... Is an interesting relationship between the graph of its inverse College Algebra with Solutions. 3, f ( x ) = x^2 -2x -1, x is a many-to-one function would one-to-many! The Trig functions a function reflections over the line y = a, y=b it... Graph of a power series \u201d anything that the statement does not assume continuity or or... 3 + 1, f ( x ) = 2 * 3 + 1 that for x=c, y=b and... If a function y-value can not be used when convenient and also infinite products be... Y = m * x + b do all functions have an inverse where m and b are constants, not! Inverse relation is then defined as f ( x ) = 2x has the inverse function f ( x =... Defined as the inverse function no Calculator do all functions have inverses in fact are... Means, for instance, that no parabola ( quadratic function ) will have an inverse, a function series! Of a power series function have an inverse called a function, we can whether..., x ) \\ ) at most a countable number sure you know the... How to evaluate f at 3, f ( x ) to it right image... ) Edit Edition the notation \\ ( f^ { \u22121 } ( x ) = 2 * 3 1. Functions are reflections over the line y = x function does used to define these functions for all values. Continuity or differentiability or anything nice about the domain and range need not even be subsets of the functions. Set consisting of all ordered pairs of the form ( 2, x a. Periodic ( i.e surjective i.e shoes could be used twice be subsets of the form 2! 2X has the inverse function f ( x ) \\ ) a * x^2 where a a! Then defined as f ( x ) = 2x + 1 f^ { \u22121 } ( x ) 2x... But y = x that no parabola ( quadratic function ) will have an inverse is! Constant function have an inverse of a many-to-one function would be one-to-many, which n't. To explain with their sets of points before defining the inverse of a one-to-one function this helps mental... All functions have inverses yes, it must be surjective i.e nice about the domain range..., meaning that each y-value has a unique x-value paired to it 3 ) = 2x + 1 7! Manual ( 9th Edition ) Edit Edition a way of tying our shoes could be a... An interesting relationship between the graph of a power series can not be used when convenient even... ) will have an inverse that is also a function must be injective i.e one-one address quickly we. X^2 -2x -1, x ) = 2x has the inverse of a complete residue system modulo m has modular. Hope this helps ) for a function fact, the same y-value can not be used to these... F at 3, f ( x ) = 2x has the inverse of a complete residue system m. Consider the function is one-to-one be defined by means of a one-to-one function = 2x has the inverse most... Each x-value must be _____ injective i.e one-one complex values of x of its inverse it to have an,. ( i.e College Algebra with Student Solutions Manual ( 9th Edition ) Edit.! = m * x + b, where m and b are constants, is not possible to an... Edition ) Edit Edition also a function to have an inverse of function unique x-value paired it... Modulo m has a modular multiplicative inverse, for instance, that no parabola quadratic... Are called one-to one functions all complex values of x give the same value e.g not possible find!", "date": "2021-03-06 23:43:20", "meta": {"domain": "noqood.co", "url": "https://staff.noqood.co/so-long-tihs/do-all-functions-have-an-inverse-2bfcb8", "openwebmath_score": 0.8467493653297424, "openwebmath_perplexity": 374.5139080277813, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9820137874059599, "lm_q2_score": 0.8807970858005139, "lm_q1q2_score": 0.8649548821630948}}
{"url": "https://www.freemathhelp.com/forum/threads/sequence-word-problem.70745/", "text": "# Sequence Word Problem\n\n#### 0313phd\n\n##### New member\nProblem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?\n\nI am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)\nso that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313\n\n#### soroban\n\n##### Elite Member\nHello,0313phd!\n\nThe number of bricks in the bottom row of a brick wall is 49.\nThe next row up from the bottom contains 47 bricks,\nand each subsequent row contains 2 fewer bricks than the row immediately below it.\nThe number of bricks in the top row is 3.\n\nIf the wall is one brick thick, what is the total number of bricks in the wall?\n\nI am not getting the correct answer.\nIs this an arithmetic sequence? . Yes!\n. . . $$\\displaystyle S_n\\:=\\: \\tfrac{n}{2}\\left(a_1 + a_n\\right)$$\n\nSo that since there are 46 steps between the bottom and top steps. . No, there are 24 rows.\n\nCounting from top, we have:\n\n. . $$\\displaystyle \\begin{array}{|c||c|c|c|c|c|c|} \\hline \\text{Row} & 1 & 2 & 3 & 4 & \\hdots & 24 \\\\ \\hline \\text{Bricks} & 3 & 5 & 7 & 9 & \\hdots & 49 \\\\ \\hline\\end{array}$$\n\n$$\\displaystyle \\text{So we have: }\\:S_{24} \\;=\\;\\tfrac{24}{2}(3 + 49) \\:=\\12)(52) \\:=\\:624$$\n\n#### TchrWill\n\n##### Full Member\n0313phd said:\nProblem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?\n\nI am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)\nso that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313\nAn alternate thought:\n\nAdd a brick atop those below and you have rows of 49, 47, 45, 43,...........7, 5, 3, 1.\n\nThe sum of the first \"n\" odd numbers is n^2.\n\nTherefore, the sum of the 25 rows of bricks is 25^2 or 625.\n\nRemove the single brick added for conveniance, and the total number of bricks is 624.\n\nJ\n\n#### JeffM\n\n##### Guest\n0313phd said:\nProblem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?\n\nI am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)\nso that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313\n1 + 3 + 5 + ... + 49 = S[sub:241ucnbg]49[/sub:241ucnbg]\n3 + 5 + ... + 49 = T[sub:241ucnbg]49[/sub:241ucnbg].\nSo, T[sub:241ucnbg]n[/sub:241ucnbg] = S[sub:241ucnbg]n[/sub:241ucnbg] - 1.\n\nAll clear so far?\n\nThere is a formula for the sum of the first n odd numbers, but maybe you do not know it. So derive it. (It is tricky to decide what to memorize. My tendency is to remember concepts and techniques and as few formulae as possible, but that is me. I was (sort of) trained as an historian so you cannot rely on me for math advice UNLESS I PROVE IT.)\n\nLet's do a few examples. (This is not the only way to solve such problems, but, WHEN IT WORKS, it is the simplest.)\n\n1 = 1.\n1 + 3 = 4.\n1 + 3 + 5 = 9.\n1 + 3 + 5 + 7 = 16.\n\nDo you see a pattern?\n\nNot looking for a fancy proof of the pattern, but let's confirm it through an informal mathematical induction.\n\nWhat is the standard formula for an odd number?\n\nAssume [(2 * 1) - 1] + ... (2k - 1) = k[sup:241ucnbg]2[/sup:241ucnbg].\nThen [(2 * 1) - 1] + ... [(2k) - 1] + [2(k + 1) - 1] = k[sup:241ucnbg]2[/sup:241ucnbg] + 2k + 2 - 1 = k[sup:241ucnbg]2[/sup:241ucnbg] + 2k + 1 = (k + 1)[sup:241ucnbg]2[/sup:241ucnbg].\n\nNow can you solve your problem?\n\nTHANKS, 0313\n\n3 + 49 = 52\n5 + 47 = 52\n....", "date": "2021-02-26 00:16:59", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/sequence-word-problem.70745/", "openwebmath_score": 0.6251608729362488, "openwebmath_perplexity": 470.81259756787233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137931962462, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.8649548688265246}}
{"url": "https://math.stackexchange.com/questions/1166798/inequality-from-chapter-5-of-the-book-how-to-think-like-a-mathematician/1166803", "text": "# Inequality from Chapter 5 of the book *How to Think Like a Mathematician*\n\nThis is from the book How to think like a Mathematician,\n\nHow can I prove the inequality $$\\sqrt[\\large 7]{7!} < \\sqrt[\\large 8]{8!}$$\n\nwithout complicated calculus? I tried and finally obtained just $$\\frac 17 \\cdot \\ln(7!) < \\frac 18 \\cdot \\ln(8!)$$\n\n\u2022 Are we allowed complex roots and negative roots to prove the contrary ;-) Feb 28, 2015 at 19:22\n\u2022 I don't think that it is in the proposition Mar 2, 2015 at 11:24\n\u2022 7! is the product of seven positive numbers, all of which are smaller than 8. $8^7$ is the product of 7 eights. So the inequality follows. Dec 7, 2021 at 22:52\n\u2022 Be sure to write your argument starting from $7! < 8^7$ and ending with what you want to prove, not backwards the way you discovered where to start. Dec 7, 2021 at 22:53\n\u2022 If you have a dataset of $n$ real numbers, and you add an $n+1$st number greater than the average of the $n$ numbers, the new set has a greater average. (Hint: logarithms.) Dec 7, 2021 at 22:53\n\nYour inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \\cdots 7 < 8 \\cdots 8$$\n\n\u2022 In the spirit of the question title, I'd note that the first line of this answer is key. Maths does of course love cleverness, but also thrives on knowing to do really simply things such as taking powers of both sides here to remove those ugly roots. Feb 27, 2015 at 3:02\n\u2022 You are right, great ! I would have of to think there. Feb 28, 2015 at 17:39\n\u2022 Elegant! really clever! May 27, 2015 at 16:19\n\nThink of\n\n$${\\ln(7!)\\over7}={\\ln(1)+\\cdots+\\ln(7)\\over7}$$\n\nas the average of seven numbers and\n\n$${\\ln(8!)\\over8}={\\ln(1)+\\cdots+\\ln(8)\\over8}$$\n\nas the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.)\n\n\u2022 You are right, but it is clear to me only when i compute ln(8) Feb 28, 2015 at 18:08\n\u2022 @Gwydyon, the logarithm is an increasing function, so $\\ln(8)$ is larger than its predecessors. Feb 28, 2015 at 20:00\n\u2022 Yes I know that. Mar 2, 2015 at 11:21\n\u2022 @Gwydyon, I guess I don't understand your previous comment then. Mar 2, 2015 at 15:22\n\u2022 Because x/8 is lower than y/7, if x=y, that is not the case but ln(7!) and ln((7+1)!) are near close, for x>=1 Mar 3, 2015 at 16:41\n\nNote that $$\\sqrt[7]{7!} < \\sqrt[8]{8!} \\iff\\\\ (7!)^8 < (8!)^7 \\iff\\\\ 7! < \\frac{(8!)^7}{(7!)^7} \\iff\\\\ 7! < 8^7$$ You should find that the proof of this last line is fairly straightforward.\n\n\u2022 You are right but someone has writen the same statements Feb 28, 2015 at 18:11\n\n$8\\ln (7!) < 7\\ln (8!) \\Rightarrow \\ln (7!) < 7\\ln 8 \\iff \\ln 1 + \\ln 2 +\\cdots \\ln 7 < 7\\ln 8$ which is clear.\n\n\u2022 great! you achieved my second statement Feb 28, 2015 at 18:53\n\nYou have already turned the comparison of two geometric means into the comparison of two arithmetic means. So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest. \\begin{equation*} \\begin{split} (1/(n+1)) \\sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \\sum_{i=1}^{n} x_i) \\\\ &=(1/(n+1) (n x_{n+1}/n + n \\sum_{i=1}^{n} x_i / n) \\\\ &> (1/(n+1) (\\sum_{i=1}^{n} x_i/n + n \\sum_{i=1}^{n} x_i / n) \\\\ &= (1/(n+1) ((n+1) \\sum_{i=1}^{n} x_i / n) \\\\ &= \\sum_{i=1}^{n} x_i / n \\end{split} \\end{equation*}\n\nNote that we really only needed $x_{n+1}$ to be larger than the previous mean.\n\nThe solution occurs just by doing simple calculations,\nLets start, $$\\sqrt[7]{7!}<\\sqrt[8]{8!}$$\niff $$(\\sqrt[7]{7!})^{7\\cdot8}<(\\sqrt[8]{8!})^{7\\cdot8}$$\niff $$(7!)^{8}<(8!)^7$$\niff $$(7!)^8<(7!\\cdot8)^7$$\niff $$(7!)^8<8^7\\cdot(7!)^7$$\niff $$(7!)<8^7$$\niff $$1\\cdot2\\cdots6\\cdot7<8\\cdot8\\cdot8\\cdot8\\cdot8\\cdot8\\cdot8$$\nwich is obviously true since $$1<8,2<8,\\ldots,7<8$$\n\nConsider this\n\n$$x=\\ln 8!-\\frac87\\ln7!=\\ln8-\\frac17\\ln7!=\\frac17\\left(7\\ln8-\\ln7!\\right)=\\frac17\\ln\\frac{8^7}{7!}>0$$ Hence, since the exponential is a monotonically increasing function: $$e^{x/8}>1\\implies (8!)^{1/8}>(7!)^{1/7}.$$", "date": "2023-03-22 03:27:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1166798/inequality-from-chapter-5-of-the-book-how-to-think-like-a-mathematician/1166803", "openwebmath_score": 0.9664008617401123, "openwebmath_perplexity": 693.7467225186224, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137895115187, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8649548686538007}}
{"url": "https://nl.mathworks.com/help/econ/dtmc.graphplot.html", "text": "Main Content\n\n# graphplot\n\nPlot Markov chain directed graph\n\n## Description\n\nexample\n\ngraphplot(mc) creates a plot of the directed graph (digraph) of the discrete-time Markov chain mc. Nodes correspond to the states of mc. Directed edges correspond to nonzero transition probabilities in the transition matrix mc.P.\n\nexample\n\ngraphplot(mc,Name,Value) uses additional options specified by one or more name-value arguments. Options include highlighting transition probabilities, communicating classes, and specifying class properties of recurrence/transience and period. Also, you can plot the condensed digraph instead, with communicating classes as supernodes.\n\ngraphplot(ax,___) plots on the axes specified by ax instead of the current axes (gca) using any of the input argument combinations in the previous syntaxes. The option ax can precede any of the input argument combinations in the previous syntaxes.\n\nexample\n\nh = graphplot(___) returns the handle to the digraph plot. Use h to modify properties of the plot after you create it.\n\n## Examples\n\ncollapse all\n\nConsider this theoretical, right-stochastic transition matrix of a stochastic process.\n\n$P=\\left[\\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\\\ 0& 0& 1/3& 0& 2/3& 0& 0\\\\ 0& 0& 0& 0& 0& 1/3& 2/3\\\\ 0& 0& 0& 0& 0& 1/2& 1/2\\\\ 0& 0& 0& 0& 0& 3/4& 1/4\\\\ 1/2& 1/2& 0& 0& 0& 0& 0\\\\ 1/4& 3/4& 0& 0& 0& 0& 0\\end{array}\\right].$\n\nCreate the Markov chain that is characterized by the transition matrix P.\n\nP = [ 0   0  1/2 1/4 1/4  0   0 ;\n0   0  1/3  0  2/3  0   0 ;\n0   0   0   0   0  1/3 2/3;\n0   0   0   0   0  1/2 1/2;\n0   0   0   0   0  3/4 1/4;\n1/2 1/2  0   0   0   0   0 ;\n1/4 3/4  0   0   0   0   0 ];\nmc = dtmc(P);\n\nPlot a directed graph of the Markov chain.\n\nfigure;\ngraphplot(mc);\n\nConsider this theoretical, right-stochastic transition matrix of a stochastic process.\n\n$P=\\left[\\begin{array}{cccc}0.5& 0.5& 0& 0\\\\ 0.5& 0& 0.5& 0\\\\ 0& 0& 0& 1\\\\ 0& 0& 1& 0\\end{array}\\right].$\n\nCreate the Markov chain that is characterized by the transition matrix P. Name the states Regime 1 through Regime 4.\n\nP = [0.5 0.5 0 0; 0.5 0 0.5 0; 0 0 0 1; 0 0 1 0];\nmc = dtmc(P,'StateNames',[\"Regime 1\" \"Regime 2\" \"Regime 3\" \"Regime 4\"]);\n\nPlot a directed graph of the Markov chain. Identify the communicating classes in the digraph and color the edges according to the probability of transition.\n\nfigure;\ngraphplot(mc,'ColorNodes',true,'ColorEdges',true)\n\nStates 3 and 4 compose a communicating class with period 2. States 1 and 2 are transient.\n\nCreate a \"dumbbell\" Markov chain containing 10 states in each \"weight\" and three states in the \"bar.\"\n\n\u2022 Specify random transition probabilities between states within each weight.\n\n\u2022 If the Markov chain reaches the state in a weight that is closest to the bar, then specify a high probability of transitioning to the bar.\n\n\u2022 Specify uniform transitions between states in the bar.\n\nrng(1); % For reproducibility\nw = 10;                              % Dumbbell weights\nDBar = [0 1 0; 1 0 1; 0 1 0];        % Dumbbell bar\nDB = blkdiag(rand(w),DBar,rand(w));  % Transition matrix\n\n% Connect dumbbell weights and bar\nDB(w,w+1) = 1;\nDB(w+1,w) = 1;\nDB(w+3,w+4) = 1;\nDB(w+4,w+3) = 1;\n\ndb = dtmc(DB);\n\nPlot a directed graph of the Markov chain. Return the plot handle.\n\nfigure;\nh = graphplot(db);\n\nObserve that the state labels are difficult to read. Remove the labels entirely.\n\nh.NodeLabel = {};\n\n## Input Arguments\n\ncollapse all\n\nDiscrete-time Markov chain with NumStates states and transition matrix P, specified as a dtmc object. P must be fully specified (no NaN entries).\n\nAxes on which to plot, specified as an Axes object.\n\nBy default, graphplot plots to the current axes (gca).\n\n### Name-Value Arguments\n\nSpecify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.\n\nBefore R2021a, use commas to separate each name and value, and enclose Name in quotes.\n\nExample: 'ColorEdges',true,'ColorNodes',true colors the edges to indicate transition probabilities and colors the nodes based on their communicating class.\n\nFlag for labeling nodes using state names, specified as the comma-separated pair consisting of 'LabelNodes' and a value in this table.\n\nValueDescription\ntrueLabel nodes using the names in mc.StateNames.\nfalseLabel nodes using state numbers.\n\nExample: 'LabelNodes',false\n\nData Types: logical\n\nFlag for coloring nodes based on communicating class, specified as the comma-separated pair consisting of 'ColorNodes' and a value in this table.\n\nValueDescription\ntrueNodes in the same communicating class have the same color. Solid markers represent nodes in recurrent classes, and hollow markers represent nodes in transient classes. The legend contains the periodicity of recurrent classes.\nfalseAll nodes have the same color.\n\nExample: 'ColorNodes',true\n\nData Types: logical\n\nFlag for labeling edges with the transition probabilities in the transition matrix mc.P, specified as the comma-separated pair consisting of 'LabelEdges' and a value in this table.\n\nValueDescription\ntrueLabel edges with transition probabilities rounded to two decimal places.\nfalseDo not label edges.\n\nExample: 'LabelEdges',true\n\nData Types: logical\n\nFlag for coloring edges to indicate transition probabilities, specified as the comma-separated pair consisting of 'ColorEdges' and a value in this table.\n\nValueDescription\ntrueColor edges to indicate transition probabilities. Include a color bar, which summarizes the color coding.\nfalseUse the same color for all edges.\n\nExample: 'ColorEdges',true\n\nData Types: logical\n\nFlag for condensing the graph, with each communicating class represented by a single supernode, specified as the comma-separated pair consisting of 'Condense' and a value in this table.\n\nValueDescription\ntrueNodes are supernodes containing communicating classes. Node labels list the component states of each supernode. An edge from supernode i to supernode j indicates a nonzero probability of transition from some state in supernode i to some state in supernode j. Transition probabilities between supernodes are not well defined, and graphplot disables edge information.\nfalseNodes are states in mc.\n\nExample: 'Condense',true\n\nData Types: logical\n\n## Output Arguments\n\ncollapse all\n\nHandle to the graph plot, returned as a graphics object. h is a unique identifier, which you can use to query or modify properties of the plot.\n\n## Tips\n\n\u2022 To produce the directed graph as a MATLAB\u00ae digraph object and use additional functions of that object, enter:\n\nG = digraph(mc.P)\n\n\u2022 For readability, the 'LabelNodes' name-value pair argument allows you to turn off lengthy node labels and replace them with node numbers. To remove node labels completely, set h.NodeLabel = {};.\n\n\u2022 To compute node information on communicating classes and their properties, use classify.\n\n\u2022 To extract a communicating class in the graph, use subchain.\n\n\u2022 The condensed graph is useful for:\n\n\u2022 Identifying transient classes (supernodes with positive outdegree)\n\n\u2022 Identifying recurrent classes (supernodes with zero outdegree)\n\n\u2022 Visualizing the overall structure of unichains (chains with a single recurrent class and any transient classes that transition into it)\n\n## References\n\n[1] Gallager, R.G. Stochastic Processes: Theory for Applications. Cambridge, UK: Cambridge University Press, 2013.\n\n[2] Horn, R., and C. R. Johnson. Matrix Analysis. Cambridge, UK: Cambridge University Press, 1985.\n\n[3] Jarvis, J. P., and D. R. Shier. \"Graph-Theoretic Analysis of Finite Markov Chains.\" In Applied Mathematical Modeling: A Multidisciplinary Approach. Boca Raton: CRC Press, 2000.\n\n## Version History\n\nIntroduced in R2017b", "date": "2022-09-30 06:45:42", "meta": {"domain": "mathworks.com", "url": "https://nl.mathworks.com/help/econ/dtmc.graphplot.html", "openwebmath_score": 0.7035793662071228, "openwebmath_perplexity": 2978.3396564381255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137895115187, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.8649548655810275}}
{"url": "https://math.stackexchange.com/questions/3006069/find-the-smallest-positive-integer-divisible-by-63-such-that-the-sum-of-its-digi", "text": "# Find the smallest positive integer divisible by 63 such that the sum of its digits is also divisible by 63.\n\nTASK: Find the smallest positive integer divisible by 63 such that the sum of its digits is also divisible by 63.\n\nMY WORK: Let the number be $$A=\\overline{x_n x_{n-1} x_{n-2} \\cdots x_1 x_0}$$. Since $$63|(x_n+x_{n-1}+\\cdots+x_1+x_0)$$, we have that $$x_n+x_{n-1}+\\cdots+x_0\\ge63\\cdots(*)$$ and since $$x_0,x_1,\\cdots,x_n$$ are n+1 digits, we have that $$x_n+x_{n-1}+\\cdots+x_0\\le9+9+\\cdots+9=9(n+1)$$ which then means that $$9(n+1)\\ge63\\Leftrightarrow n+1\\ge7$$ i.e that the number $$A$$ has at least seven digits. If it has $$7$$ digits, all of them would have to be $$9$$ to satisfy the inequality $$(*)$$ which would mean that $$A=9999999$$. But then the condition $$63|A$$ would not be satisfied. So the number A doesn't have seven digits - it has at least eight digits.\n\nHowever, I do not know where to go from here.\n\n\u2022 So check numbers like 18999999 and 19899999 and 19989999 and so on. \u2013\u00a0Gerry Myerson Nov 20 '18 at 8:15\n\u2022 You get divisibility by $9$ for free so you only need be concerned about checking for $7$ \u2013\u00a0Mark Bennet Nov 20 '18 at 8:44\n\nAssume the the number is $$1$$ with 6 $$9$$s and one $$8$$.\n\nNow $$19999999\\equiv 5\\pmod 7$$\n\nIf we subtract $$10^k$$ we will get aus a number with a $$1$$ , 6$$9$$ and one $$8$$ anda different equivalence. so we need to find the $$10^k\\equiv 5\\mod 7$$.\n\n$$10\\equiv 3$$\n\n$$100\\equiv 30\\equiv 2$$\n\n$$1000\\equiv 20\\equiv 6$$\n\n$$10,000\\equiv 60\\equiv 4$$\n\n$$100,000 \\equiv 40\\equiv 5$$\n\nSo $$19,999,999-100,000=19,899,999\\equiv 0\\pmod 7$$.\n\nAnd that's that. It's digits add to $$63$$ so it's divisible by $$9$$ and it's divisible by $$7$$. And beginning with $$1$$ and the only such divisible by $$7$$ it's the smallest such number.\n\n====\n\nI thought I made it clear why this is the smallest.\n\nNo element with $$7$$ digits or fewer exist as the OP figured out. For a group with $$8$$ digits the smallest would start with a $$1$$. If you have an $$8$$ digit number beginning with $$1$$ and whose digits add to $$63$$ the remaining digits must be six $$9$$s and one $$8$$. Such a number can be written as $$19,999,999 - 10^k$$ where $$0\\le k \\le 7$$. For such a number to be divisible by $$63$$ we must have $$10^k \\equiv 5 \\pmod 7$$. The ONLY such $$k$$ is $$k = 5$$ and $$10^k =100,000$$ and the number is $$19,899,999$$. So this is the only such number divisible by $$63$$ whose digits add to $$63$$ in the smallest possible category of types of numbers that can have such numbers. So this is the smallest such number.\n\n\u2022 Is that the smallest such number? [no criticism of your answer intended... but OP originally asked for smallest.] (+1 on answer.) \u2013\u00a0coffeemath Nov 20 '18 at 9:01\n\u2022 Sorry @coffeemath , I had a typo. \u2013\u00a0Akash Roy Nov 20 '18 at 10:18\n\u2022 Yes, it's the smallest. And I explained why. There can't be one with 7 digits, so the smallest has eight or more. This has 8 so it is in the smallest group. the smallest digit it can begin with is 1 so this is the smallest of the smallest. If number starts with 1 and has 8 digits and adds to $63$ then is must have 6 nines and one 8. So smallest number would be $19,999,999 - 10^k$ where $10^k\\equiv 5 \\mod 7$. The only such option for $0\\le k \\le 7$ is $k=5$. So this is the only solution with 8 digits begining with $1$. And no smaller number is a solution. \u2013\u00a0fleablood Nov 20 '18 at 17:05\n\n(This is essentially the same solution as @fleablood 's; but doubts were raised whether it is actually the smallest.)\n\nSuch a number has at least $$8$$ digits. Since the prescribed digit sum is $$63$$ we have to deduct exactly $$9$$ units from writing eight nines. Trying with $$x_1=1$$ as first digit, and all other nines, we have given away $$8$$ units, one more to go. Divisibility by $$9$$ is taken care of automatically. Now $$19\\,999\\,999=5$$ mod $$7$$; therefore we need to find a $$k\\in[0..6]$$ with $$10^k=5$$ mod $$7$$, or we are bust with $$x_1=1$$. Fortunately $$10^5=5$$ mod $$7$$. It follows that $$19\\,899\\,999$$ is the smallest number with the required properties.\n\n\u2022 @coffeemath: Sorry for the typo. Thank you for reporting it. \u2013\u00a0Christian Blatter Nov 20 '18 at 10:15\n\nThe smallest number whose digits all sum to a multiple of $$63$$ is $$9{,}999{,}999$$. The next smallest is $$18{,}999{,}999$$, then $$19{,}899{,}999$$, then $$19,989{,}999$$, and so on. All of these are clearly divisible by $$9$$, so it suffices to check for divisibility by $$7$$. As it happens, the first two are not, but $$19{,}899{,}999/7=2{,}842{,}857$$ (and, just to doublecheck, $$19{,}899{,}999/63=315{,}873$$).\n\nRemark: It's not a priori obvious that any of the numbers described here will turn out to be divisible by $$7$$. You could say we just got lucky. Or you could do a modular argument to show that luck had nothing to do with it. One thing is obvious: the smallest number sought for is certainly no greater than $$777{,}777{,}777$$.\n\n\u2022 The modular argument is not hard. The numbers are all of the form $19,999,999 - 10^k$ so we need $19,999,999 - 10^k \\equiv 0 \\mod 7$ or $10^k \\equiv 5 \\mod 7$. As $10$ and $7$ is relativley prime and $7$ is prime the $10^k; 0\\le k < 7$ are distinct modulo $7$ and $10^5$ is the only one that works. \u2013\u00a0fleablood Nov 20 '18 at 17:28\n\u2022 @fleablood, are you saying, more generally, that if $10$ and $p$ are relatively prime (with $p$ a prime), then $10^k$ for $0\\le k\\lt p$ are distinct modulo $p$? It's true that $10$ is a primitive root mod $7$, but not because it's relatively prime to $7$. \u2013\u00a0Barry Cipra Nov 20 '18 at 22:46\n\u2022 Yeah, I guess I worded it incorrectly. $10\\equiv 3$ is a primitive root is what I meant. \u2013\u00a0fleablood Nov 20 '18 at 23:08\n\nI also verify that the Number 19,899,999 is the smallest number that is divisible by 63 and Also its sum is divisible by 63. Here a program I wrote in Python 3. To find this out by brute forcing :\n\nfor i in range(9999990, 19900000, 63):\nsum_is = sum(int(d) for d in str(i))\nif sum_is%63==0:\nprint(i)\n\nPrints :\n19899999\n\nAlso this time I started from 9999990 because this number the last number divisible by 63 that has sum less than 63.\n\nThis can also be used to do the same with any number.\n\nJust change the range and 63 as you want.\n\nHope this Helps!\n\n[EDIT SUMMARY] For more Optimization and better Understanding. Added direct summing instead of sum_digits as request/suggested by @Paul Evans. Also added jump of 63 as Suggested by @Kyle Kanos.\n\n\u2022 Instead of the mysterious sum_digits(i) you could write python code: sum(int(d) for d in str(i)) \u2013\u00a0Paul Evans Nov 20 '18 at 16:01\n\u2022 You can also save time by starting at 9999990 and incrementing by 63, eliminating the need to test i%63==0. \u2013\u00a0Kyle Kanos Nov 20 '18 at 19:47\n\u2022 Yes! That's more optimized. Thanx for advice. \u2013\u00a0FightWithCode Nov 21 '18 at 13:20", "date": "2020-10-30 16:11:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3006069/find-the-smallest-positive-integer-divisible-by-63-such-that-the-sum-of-its-digi", "openwebmath_score": 0.7479796409606934, "openwebmath_perplexity": 252.50644528683247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471647042428, "lm_q2_score": 0.8791467770088163, "lm_q1q2_score": 0.8649460639189971}}
{"url": "https://gateoverflow.in/95821/tifr2017-b-13?show=97339", "text": "321 views\n\nFor an undirected graph $G=(V, E)$, the line graph $G'=(V', E')$ is obtained by replacing each edge in $E$ by a vertex, and adding an edge between two vertices in $V'$ if the corresponding edges in $G$ are incident on the same vertex. Which of the following is TRUE of line graphs?\n\n1. the line graph for a complete graph is complete\n2. the line graph for a connected graph is connected\n3. the line graph for a bipartite graph is bipartite\n4. the maximum degree of any vertex in the line graph is at most the maximum degree in the original graph\n5. each vertex in the line graph has degree one or two\nretagged | 321 views\n+2\nOption B). is true.\n+2\nyes, Agreed. (y)\n0\nwhy not option C?\n0\nyou can check with K(2,2). The line graph obtained is not biparitite.\n\noption B is the right answer\n\nWe can solve this question by eliminating options\n\nOption (A) False\n\nLet us take a complete graph of\u00a0 $4$ vertices:\n\nIn a line graph, no. of edges is\n\n$\\sum_{i=0}^{n}$ \u00a0${}^{d_{i}}C_{{2}},$ \u00a0 \u00a0\u00a0$d_{i}$is degree of each vertex\n\n=$\\frac{3\\times 2}{2}+\\frac{3\\times 2}{2}+\\frac{3\\times 2}{2}+\\frac{3\\times 2}{2} = 3\\times 4=12$\n\nNo. of\u00a0vertices in line graph = No. of edges in original graph\nNo. of\u00a0vertices in line graph $= 6$\nSo, no. of edges to make complete graph with $6$ vertices $= \\frac{6\\times 5}{2} = 3\\times 5=15$\nBut for given line graph from complete graph of $4$ vertices we have only $12$ edges.\n\nOption (B) \u00a0 True\n\n1. Smallest line graph for original graph one edge\n\nwhich is also connected graph\n\nIf a graph is connected with more then one edge, it will never be disconnected\n\nOption (C) \u00a0\u00a0 False\n\nThis cannot be 2-colorable and hence is not bipartite.\n\nOption (D) \u00a0\u00a0 False\n\nBecause line graph degree of vertex depends on the attribute\n\ne.g., $\\left [ A,B \\right ]$ as point in line graph, then degree of this vertex depends on degree of $A$ and degree of $B$ in the original graph.\n\nI'm drawing degree for a point $[AB]$ in line graph.\n\nSo, this is wrong.\n\nOption (E) \u00a0\u00a0 False (wrong as proved in above option (D))\n\nedited\n\nThe line graph of a\u00a0connected\u00a0graph\u00a0is connected. If\u00a0G\u00a0is connected, it contains a path\u00a0connecting any two of its edges, which translates into a path in\u00a0L(G) containing any two of the vertices of\u00a0L(G). Therefore, option B is correct.\n\nWe can also do this question using elimination of options.\n\nedited\n0\nbut your L(G2): e1---e2 it is bipertite right?e1 may be in one partition and e2 on another...it is bipertite i guess.\n0\nHi,\n\nI have corrected the example, and updated the file.\n0\ngreat example you have updated...proving line of tree is not a tree too :p", "date": "2018-06-18 15:46:08", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/95821/tifr2017-b-13?show=97339", "openwebmath_score": 0.5384103059768677, "openwebmath_perplexity": 1027.6534325999805, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9838471680195557, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8649460528148738}}
{"url": "https://fr.mathworks.com/help/matlab/ref/fplot3.html", "text": "fplot3\n\n3-D parametric curve plotter\n\nSyntax\n\nfplot3(funx,funy,funz)\nfplot3(funx,funy,funz,tinterval)\nfplot3(___,LineSpec)\nfplot3(___,Name,Value)\nfplot3(ax,___)\nfp = fplot3(___)\n\nDescription\n\nexample\n\nfplot3(funx,funy,funz) plots the parametric curve defined by x = funx(t), y = funy(t), and z = funz(t) over the default interval [-5,5] for t.\n\nexample\n\nfplot3(funx,funy,funz,tinterval) plots over the specified interval. Specify the interval as a two-element vector of the form [tmin tmax].\n\nexample\n\nfplot3(___,LineSpec) sets the line style, marker symbol, and line color. For example, '-r' specifies a red line. Use this option after any of the previous input argument combinations.\nfplot3(___,Name,Value) specifies line properties using one or more name-value pair arguments. For example, 'LineWidth',2 specifies a line width of 2 points.\nfplot3(ax,___) plots into the axes specified by ax instead of the current axes. Specify the axes as the first input argument.\n\nexample\n\nfp = fplot3(___) returns a ParameterizedFunctionLine object. Use the object to query and modify properties of a specific line. For a list of properties, see ParameterizedFunctionLine Properties.\n\nExamples\n\ncollapse all\n\nPlot the 3-D parametric line\n\n$\\begin{array}{c}x=\\mathrm{sin}\\left(t\\right)\\\\ y=\\mathrm{cos}\\left(t\\right)\\\\ z=t\\end{array}$\n\nover the default parameter range [-5 5].\n\nxt = @(t) sin(t); yt = @(t) cos(t); zt = @(t) t; fplot3(xt,yt,zt)\n\nPlot the parametric line\n\n$\\begin{array}{c}x={e}^{-t/10}\\mathrm{sin}\\left(5t\\right)\\\\ y={e}^{-t/10}\\mathrm{cos}\\left(5t\\right)\\\\ z=t\\end{array}$\n\nover the parameter range [-10 10] by specifying the fourth input argument of fplot3.\n\nxt = @(t) exp(-t/10).*sin(5*t); yt = @(t) exp(-t/10).*cos(5*t); zt = @(t) t; fplot3(xt,yt,zt,[-10 10])\n\nPlot the same 3-D parametric curve three times over different intervals of the parameter. For the first interval, use a line width of 2 points. For the second, specify a dashed red line style with circle markers. For the third, specify a cyan, dash-dotted line style with asterisk markers.\n\nfplot3(@(t)sin(t), @(t)cos(t), @(t)t, [0 2*pi], 'LineWidth', 2) hold on fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [2*pi 4*pi], '--or') fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [4*pi 6*pi], '-.*c') hold off\n\nPlot multiple lines in the same axes using hold on.\n\nfplot3(@(t)t, @(t)t, @(t)t) hold on fplot3(@(t)-t, @(t)t, @(t)-t) hold off\n\nPlot the parametric line\n\n$\\begin{array}{c}x={e}^{-|t|/10}\\mathrm{sin}\\left(5|t|\\right)\\\\ y={e}^{-|t|/10}\\mathrm{cos}\\left(5|t|\\right)\\\\ z=t.\\end{array}$\n\nAssign the parameterized function line object to a variable.\n\nxt = @(t)exp(-abs(t)/10).*sin(5*abs(t)); yt = @(t)exp(-abs(t)/10).*cos(5*abs(t)); zt = @(t)t; fp = fplot3(xt,yt,zt)\n\nfp = ParameterizedFunctionLine with properties: XFunction: @(t)exp(-abs(t)/10).*sin(5*abs(t)) YFunction: @(t)exp(-abs(t)/10).*cos(5*abs(t)) ZFunction: @(t)t Color: [0 0.4470 0.7410] LineStyle: '-' LineWidth: 0.5000 Show all properties \n\nChange the range of parameter values to [-10 10] and change the line color to red.\n\nfp.TRange = [-10 10]; fp.Color = 'r';\n\nFor $t$ values in the range $-2\\pi$ to $2\\pi$, plot the parametric line\n\n$\\begin{array}{c}x=t\\\\ y=t/2\\\\ z=\\mathrm{sin}\\left(6t\\right).\\end{array}$\n\nAdd a title, x-axis label, and y-axis label. Additionally, change the view of the axes and display the axes box outline.\n\nxt = @(t)t; yt = @(t)t/2; zt = @(t)sin(6*t); fplot3(xt,yt,zt,[-2*pi 2*pi],'MeshDensity',30,'LineWidth',1); title('x=t, y=t/2, z=sin(6t) for -2\\pi<t<2\\pi') xlabel('x'); ylabel('y'); view(52.5,30) box on\n\nAccess the axes object using gca. Specify the x-axis tick values and associated labels using the XTick and XTickLabel properties of the axes object. Similarly, specify the y-axis tick values and associated labels.\n\nax = gca; ax.XTick = -2*pi:pi/2:2*pi; ax.XTickLabel = {'-2\\pi','-3\\pi/2','-\\pi','-\\pi/2','0','\\pi/2','\\pi','3\\pi/2','2\\pi'}; ax.YTick = -pi:pi/2:pi; ax.YTickLabel = {'-\\pi','-\\pi/2','0','\\pi/2','\\pi'};\n\nInput Arguments\n\ncollapse all\n\nParametric function for x coordinates, specified as a function handle to a named or anonymous function.\n\nSpecify a function of the form x = funx(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).\n\nExample: funx = @(t) sin(2*t);\n\nParametric function for y coordinates, specified as a function handle to a named or anonymous function.\n\nSpecify a function of the form y = funy(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).\n\nExample: funy = @(t) cos(2*t);\n\nParametric function for z coordinates, specified as a function handle to a named or anonymous function.\n\nSpecify a function of the form z = funz(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).\n\nExample: funz = @(t) t;\n\nInterval for parameter t, specified as a two-element vector of the form [tmin tmax].\n\nAxes object. If you do not specify an axes object, then fplot3 uses the current axes (gca).\n\nLine style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.\n\nExample: '--or' is a red dashed line with circle markers\n\nLine StyleDescription\n-Solid line\n--Dashed line\n:Dotted line\n-.Dash-dot line\nMarkerDescription\n'o'Circle\n'+'Plus sign\n'*'Asterisk\n'.'Point\n'x'Cross\n'_'Horizontal line\n'|'Vertical line\n's'Square\n'd'Diamond\n'^'Upward-pointing triangle\n'v'Downward-pointing triangle\n'>'Right-pointing triangle\n'<'Left-pointing triangle\n'p'Pentagram\n'h'Hexagram\nColorDescription\n\ny\n\nyellow\n\nm\n\nmagenta\n\nc\n\ncyan\n\nr\n\nred\n\ng\n\ngreen\n\nb\n\nblue\n\nw\n\nwhite\n\nk\n\nblack\n\nName-Value Pair Arguments\n\nSpecify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.\n\nExample: 'Marker','o','MarkerFaceColor','red'\n\nThe properties listed here are only a subset. For a complete list, see ParameterizedFunctionLine Properties.\n\nNumber of evaluation points, specified as a number. The default is 23. Because fplot3 uses adaptive evaluation, the actual number of evaluation points is greater.\n\nLine color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB\u00ae uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nExample: 'blue'\n\nExample: [0 0 1]\n\nExample: '#0000FF'\n\nLine style, specified as one of the options listed in this table.\n\nLine StyleDescriptionResulting Line\n'-'Solid line\n\n'--'Dashed line\n\n':'Dotted line\n\n'-.'Dash-dotted line\n\n'none'No lineNo line\n\nLine width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.\n\nThe line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.\n\nMarker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.\n\nValueDescription\n'o'Circle\n'+'Plus sign\n'*'Asterisk\n'.'Point\n'x'Cross\n'_'Horizontal line\n'|'Vertical line\n'square' or 's'Square\n'diamond' or 'd'Diamond\n'^'Upward-pointing triangle\n'v'Downward-pointing triangle\n'>'Right-pointing triangle\n'<'Left-pointing triangle\n'pentagram' or 'p'Five-pointed star (pentagram)\n'hexagram' or 'h'Six-pointed star (hexagram)\n'none'No markers\n\nMarker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nMarker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nExample: [0.3 0.2 0.1]\n\nExample: 'green'\n\nExample: '#D2F9A7'\n\nMarker size, specified as a positive value in points, where 1 point = 1/72 of an inch.\n\nOutput Arguments\n\ncollapse all\n\nOne or more ParameterizedFunctionLine objects, returned as a scalar or a vector. You can use these objects to query and modify properties of a specific ParameterizedFunctionLine object. For details, see ParameterizedFunctionLine Properties.\n\nIntroduced in R2016a", "date": "2021-07-31 02:04:10", "meta": {"domain": "mathworks.com", "url": "https://fr.mathworks.com/help/matlab/ref/fplot3.html", "openwebmath_score": 0.6069862246513367, "openwebmath_perplexity": 6253.695940688673, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471618625457, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8649460474019584}}
{"url": "https://www.physicsforums.com/threads/number-of-possible-combinations.753650/", "text": "# Number of possible combinations\n\n1. May 14, 2014\n\n### kidkook\n\nI have 16 items and i need to know the number of possible combination for all 16 items. I know the possible number of combination for all 16 items is 1 and the possible number of combinations for 15 of the 16 items is 16 using the formula below:\n\n$$\\frac{n!}{r!(n-r)!}.$$\n\nHow can i adaprt this formula to calculate for the number of selection from 1 to 16 inclusive without repeating the calculation 16 times.\n\nSo n will always be 16 and r will be 1 to 16 inclusive.\n\n2. May 14, 2014\n\nYour wording is a little confusing. Is this the correct interpretation:\nYou want to know the total number of ways to select at least one item from the 16.\n\nIf it is, think about this: you should know the total number of subsets (empty, proper, with the original set) for a set with 16 items. Your question involves finding not the total number but how many there are not counting the empty set.\n\n3. May 14, 2014\n\n### kidkook\n\nSorry for the confusion. What i have is a dataset with 16 different values. What i'm trying to achieve is to determine how many combination are available without repetition/duplication of a value. So i know the answer is 65536. I achieved this by using the above formula 16 times and substituting the value of r from 1 to 16 etc...and adding the number together.\n\nNumber of combinations Value of r\n16 1\n120 2\n560 3\n1820 4\n4368 5\n8008 6\n11440 7\n12870 8\n11440 9\n8008 10\n4368 11\n1820 12\n560 13\n120 14\n16 15\n1 16\n65535\n\nWhat i would like to know is...can the formula be adapted to so i don't need to repeat the calculation 16 times. My dataset is likely to increase significantly in the future so it will not be practical to use this method.\n\n4. May 14, 2014\n\n### jbunniii\n\nNote that $\\frac{n!}{r!(n-r)!}$ is called the binomial coefficient and is often denoted\n$${n \\choose r} = \\frac{n!}{r!(n-r)!}$$\nThe reason it is called the binomial coefficient is the following property:\n$$(x+y)^n = \\sum_{r=0}^{n}{n \\choose r}x^r y ^{n-r}$$\nIf we plug in $x=y=1$ then this reduces to\n$$2^n = \\sum_{r=0}^{n}{n \\choose r} = 1 + \\sum_{r=1}^{n} {n \\choose r}$$\nwhere the second equality follows because ${n \\choose 0} = 1$. Therefore,\n$$\\sum_{r=1}^{n} {n \\choose r} = 2^n - 1$$\nwhich is consistent with your calculation since $2^{16} - 1 = 65535$.\n\n5. May 14, 2014\n\n### kidkook\n\nExcellent. This is exactly what i was looking for. A very well documented explaination. Many thanks.", "date": "2017-12-13 21:13:12", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/number-of-possible-combinations.753650/", "openwebmath_score": 0.6159489750862122, "openwebmath_perplexity": 236.9455879589045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846678676151, "lm_q2_score": 0.8840392832736083, "lm_q1q2_score": 0.8649304805475737}}
{"url": "https://math.stackexchange.com/questions/2376090/what-is-the-probability-of-rolling-at-least-two-6s-with-3-dice-and-2-rolls", "text": "# What is the probability of rolling at least two 6's with 3 Dice and 2 Rolls?\n\nQuestion: What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls? (with your first roll you keep dice only if they are 6's and roll the remainder for your second roll).\n\n(6*6*6 = 216 = outcomes when rolling 3 dice) (6*6 = 36 = outcomes when rolling 2 dice)\n\n1st roll: (3 dice)\n\nA = 125/216 = 0 sixes\n\nB = 75/216 = 1 six\n\nC = 15/216 = 2 sixes\n\nD = 1/216 = 3 sixes\n\nOutcomes C and D fulfill requirement.\n\nFor outcome A: (first roll = no 6's, pick up all dice throw 3 dice again)\n\n2nd roll:\n\na = 125/216 = 0 sixes\n\nb = 75/216 = 1 six\n\nc = 15/216 = 2 sixes\n\nd = 1/216 = 3 sixes\n\nOutcomes c and d fulfill requirements.\n\nFor outcome B: (first roll = one 6, pick up two non 6's and roll again)\n\n2nd roll:\n\nz = 25/36 = 0 sixes\n\ny = 10/36 = 1 six\n\nx = 1/36 = 2 sixes\n\nOutcomes y and x fulfill requirements.\n\nSo would the formula below give me my answer?\n\nC + D + Ac + Ad + By + Bx = X\n\nWhat is the probability of rolling at least two 6's, when rolling 3 dice with two rolls?\n\nIf I substituted correctly and did the math correct the answer I got was 22.3%\n\nIs this correct?\n\n\u2022 I'm guessing by your odds that you consider ordering important ? \u2013\u00a0user451844 Jul 29 '17 at 22:20\n\u2022 I'm not sure if I know what you mean by ordering. \u2013\u00a0Travis Jul 29 '17 at 22:58\n\u2022 is (1,2,3) the same as (1,3,2) and (2,3,1) and (2,1,3) and (3,1,2) and (3,2,1) ? if so then it cuts down the number of distinct rolls down to 56. \u2013\u00a0user451844 Jul 29 '17 at 23:00\n\u2022 Well let's say we had a blue green and a red die, if the red die shows a 1 and blue shows a 6 and green shows a 6, that would be different than if red showed a 6, green showed a 6, and blue showed a 1. So I guess ordering would matter. The numbers rolled are the same but show on different dice. Both rolls fulfill the two 6's requirement but are different rolls and should be counted as thus. I'm not sure if I'm correct on this, that's why I'm asking but does it makes sense? \u2013\u00a0Travis Jul 29 '17 at 23:09\n\u2022 yeah I'm just making sure, because the probability might change depending on conditions like this. of the 56 cases when order didn't matter a full 21 had at least one 6, but only 6 had more than one 6. etc. \u2013\u00a0user451844 Jul 29 '17 at 23:19\n\nAn alternate approach would be to find the probability of the complementary event:\n\n$\\textbf{1)}$ The probability of getting no 6's is given by $\\big(\\frac{5}{6}\\big)^3\\cdot\\big(\\frac{5}{6}\\big)^3=\\big(\\frac{5}{6}\\big)^6$\n\n$\\textbf{2)}$ The probability of getting exactly one 6 is given by\n\n$\\hspace{.2 in}\\big(\\frac{5}{6}\\big)^3\\cdot3\\big(\\frac{1}{6}\\big)\\big(\\frac{5}{6}\\big)^2+3\\big(\\frac{1}{6}\\big)\\big(\\frac{5}{6}\\big)^2\\cdot\\big(\\frac{5}{6}\\big)^2=\\big(\\frac{1}{2}\\big)\\big(\\frac{5}{6}\\big)^5+\\big(\\frac{1}{2}\\big)\\big(\\frac{5}{6}\\big)^4$\n\nTherefore the probability of getting at least two 6's is given by\n\n$\\hspace{.2 in} 1-\\big(\\frac{5}{6}\\big)^6-\\big(\\frac{1}{2}\\big)\\big(\\frac{5}{6}\\big)^5-\\big(\\frac{1}{2}\\big)\\big(\\frac{5}{6}\\big)^4=\\frac{5203}{23328}\\approx.223$\n\n\u2022 From my point of view is your approach the most appropriate one. (+1) \u2013\u00a0Markus Scheuer Jul 30 '17 at 9:38\n\nYour approach and your calculations are correct. Here is a variation based upon generating functions. We encode the roll of three dice with \\begin{align*} (5+t)^3 \\end{align*} marking an occurrence of $6$ with $t$ and collecting all other five possibilities with $5$. The probability to get $j$ sixes $0\\leq j \\leq 3$ in the first roll can be written as \\begin{align*} [t^j](5+t)^3\\cdot\\frac{1}{6^3} \\end{align*}\n\nEncoding the second roll with $(5+u)^{3-j}$ we calculate \\begin{align*} \\sum_{{0\\leq j,k\\leq 3}\\atop{j+k\\geq 2}}&[t^j](5+t)^3[u^k](5+u)^{3-j}\\cdot\\frac{1}{6^{6-j}}\\\\ &=[t^0](5+t)^3\\left([u^2]+[u^3]\\right)(5+u)^3\\cdot\\frac{1}{6^6}\\\\ &\\qquad+[t^1](5+t)^3\\left([u^1]+[u^2]\\right)(5+u)^2\\cdot\\frac{1}{6^5}\\\\ &\\qquad+[t^2](5+t)^3\\left([u^0]+[u^1]\\right)(5+u)^1\\cdot\\frac{1}{6^4}\\\\ &\\qquad+[t^3](5+t)^3\\\\ &=\\binom{3}{0}5^3\\left(\\binom{3}{2}5^1+\\binom{3}{3}5^0\\right)\\cdot\\frac{1}{6^6}\\\\ &\\qquad+\\binom{3}{1}5^2\\left(\\binom{2}{1}5^1+\\binom{2}{2}5^0\\right)\\cdot\\frac{1}{6^5}\\\\ &\\qquad+\\binom{3}{2}5^1\\left(\\binom{1}{0}5^1+\\binom{1}{1}5^0\\right)\\cdot\\frac{1}{6^4}\\\\ &\\qquad+\\binom{3}{3}5^0\\cdot\\frac{1}{6^3}\\\\ &=\\frac{125\\cdot16}{6^6}+\\frac{75\\cdot11}{6^5}+\\frac{15\\cdot6}{6^4}+\\frac{1}{6^3}\\\\ &=\\frac{5\\,203}{23\\,328}\\\\ &\\doteq 0.223 \\end{align*}", "date": "2019-10-15 05:00:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2376090/what-is-the-probability-of-rolling-at-least-two-6s-with-3-dice-and-2-rolls", "openwebmath_score": 0.7346745133399963, "openwebmath_perplexity": 676.5379372828847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109096680231, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8649089552193651}}
{"url": "https://math.stackexchange.com/questions/2364954/how-many-ways-are-there-to-tile-a-chessboard-with-32-knights", "text": "# How many ways are there to tile a chessboard with 32 knights?\n\nThe maximal amount of knights which can be places on a regular $8 \\times 8$ chessboard so that no two take each other is $32$ (To see this just notice that a knight on a black square only attacks knights placed on white squares).\n\nI was wondering how many ways there are to tile a regular chessboard with these $32$ knights so that no two take each other.\n\nA generalisation of this question has already been posed here: In how many ways we can place $N$ mutually non-attacking knights on an $M \\times M$ chessboard?. However, the answers given include a polynomial approach which cannot be used in the case of an $8 \\times 8$ chessboard, because, as an answer states, a computer struggles to compute such a polynomial.\n\nEven if it were quite easy for a computer to compute such a polynomial, in an Olympiad no such computational power is allowed. Hence, I was wondering if there was a pure combinatorics approach for this particular case.\n\nAn example of such a tiling is to place all the knights on all the black cells.\n\n\u2022 You can put the knights on the black squares or on the white squares. This makes two ways. Or am I missing something? \u2013\u00a0ajotatxe Jul 20 '17 at 11:58\n\u2022 @ajotatxe Are there no other possibilities? Can you prove this? \u2013\u00a0Plato Jul 20 '17 at 11:59\n\u2022 Consider a knight's tour of the chessboard. Obviously, if $32$ of the $64$ squares are occupied by non-attacking knights, then the tour must alternate between occupied and unoccupied squares. This means that there are (at most) two ways to place the $32$ knights. But of course there are two ways, use all the black squares or all the white squares. \u2013\u00a0bof Jul 20 '17 at 12:11\n\u2022 In graph-theoretic terms: A cycle graph $C_{2n}$ has exactly two independent sets of size $n.$ Therefore, a Hamiltonian graph of order $2n$ has at most two independent sets of size $n.$ The knight's graph on the $8\\times8$ chessboard is a Hamiltonian graph of order $64.$ \u2013\u00a0bof Jul 20 '17 at 12:20\n\u2022 The existence of a knight's tour seems crucial here. Actually, in a $2\\times 2$ board there is no knight's tour, and there are $6$ ways to put two non-attacking horses on it. \u2013\u00a0ajotatxe Jul 20 '17 at 12:26\n\nConsider a knight's tour of the chessboard. Obviously, if $32$ of the $64$ squares are occupied by non-attacking knights, then the tour must alternate between occupied and unoccupied squares, since a horse can only attack a square if the opposite colour. This means that there are (at most) two ways to place the $32$ knights. We can offer examples of these two ways: use all the black squares or all the white squares.\nAnother way to look at the problem is to consider graph theory. A cycle graph $C_{2n}$ has exactly two independent sets of size $n$. Therefore, a Hamiltonian graph of order $2n$ has at most two independent sets of size $n$. The knight's graph on the $8 \\times 8$ chessboard is a Hamiltonian graph of order 64. From here we can conclude in the same manner as in the first solution.", "date": "2019-11-15 23:12:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2364954/how-many-ways-are-there-to-tile-a-chessboard-with-32-knights", "openwebmath_score": 0.5685427188873291, "openwebmath_perplexity": 211.8117838370198, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109071934599, "lm_q2_score": 0.8740772286044095, "lm_q1q2_score": 0.8649089514334944}}
{"url": "https://mathschallenge.net/full/rounding_error", "text": "## Rounding Error\n\n#### Problem\n\nA number is rounded to two decimal places and this answer is rounded to one decimal place. If the final number is 0.2 then what is the chance that the original number lies between 0.15 and 0.25?\n\n#### Solution\n\nWhen rounding to $k$ decimal places (d.p.) we truncate the number at the $k^{th}$ decimal position, but if the next digit is 5 or more then we increase the last digit by 1; this may have a knock-on effect. Consider the following examples of rounding to 2 d.p.:\n\n3.61483 3.61\n7.7853301 7.79\n0.5982 0.60\n\nSuppose we were told that 0.2 were the result of rounding to one decimal place. The smallest number that would be rounded up to 0.2 is 0.15. However, there is no actual value we can assign to the greatest value that would round down to 0.2. For example, 0.25 would round up to 0.3 and although 0.249 would round down to 0.2, what about 0.2499, or 0.24999, or ... ?\n\nUnder these circumstances we define bounds (or limits) in the following way:\n\n\u2022 The lower bound is the least value that rounds up to the given number.\n\u2022 The upper bound is the least value that fails to round down to the given number.\n\nSo if 0.2 is the result of rounding to 1 d.p. then the lower and upper bounds are 0.15 and 0.25 respectively.\n\nWorking backwards, the lower bound of 0.2 is 0.15 and the lower bound of 0.15 is 0.145. We can easily check this: 0.145 = 0.15 (2 d.p.) and then 0.15 = 0.2 (1 d.p.).\n\nHowever, we need to tread very carefully when working backwards with the upper bound. Although it is true to say that the upper bound of 0.2 is 0.25, if the original number were rounded to 0.25 (to 2 d.p.), then next process of rounding to 1 d.p. would make it 0.3, not 0.2.\n\nIn fact, it is necessary that the result of rounding to 2 d.p. takes it to 0.24; and the upper bound of 0.24 is 0.245.\n\nHence we are certain that the original number, $x$, lies somewhere in the interval $0.145 \\le x \\lt 0.245$.\n\nIf we are considering if the number lies between 0.15 and 0.25 then it would have to lie in the interval $0.15 \\le x \\lt 0.245$.\n\n\\begin{align}\\therefore P(0.15 \\le x \\lt 0.245 \\mid 0.145 \\le x \\lt 0.245) &= \\dfrac{0.245 - 0.15}{0.245 - 0.145}\\\\\\\\&= \\dfrac{0.095}{0.1}\\\\\\\\&= 95\\%\\end{align}\n\nIf instead of rounding to 2 d.p. then to 1 d.p. a number is just rounded to 1 d.p., what is the chance that both methods give the same answer?\n\nProblem ID: 343 (09 Jul 2008) \u00a0 \u00a0 Difficulty: 2 Star\n\nOnly Show Problem", "date": "2019-02-21 00:26:20", "meta": {"domain": "mathschallenge.net", "url": "https://mathschallenge.net/full/rounding_error", "openwebmath_score": 0.9061170816421509, "openwebmath_perplexity": 454.6836876448085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109087400619, "lm_q2_score": 0.8740772269642949, "lm_q1q2_score": 0.8649089511624327}}
{"url": "http://math.stackexchange.com/questions/160979/sum-of-squares-of-multiplicities-of-differences", "text": "# Sum of squares of multiplicities of differences\n\nIt's hard to come up with a good title. Let $A$ be a set of $n$ integers, say $A=\\{a_1,\\ldots,a_n\\}$ with $a_1<\\cdots<a_n$. We consider all the positive differences $a_j-a_i$ (necessarily $i<j$). There are $n \\choose 2$ pairs $(a_i,a_j)$ with $i<j$, but of course, some differences can be repeated, or in other words, have \"multiplicity\" greater than 1.\n\nFor a given $A$, one way to illustrate the differences and their multiplicities is with a \"table of differences\". Here is one such table for $A=\\{0,1,3,5,6\\}$, and another for $A=\\{1,4,7,10,13\\}$.\n\n-| 0 1 3 5 6     - | 1 4 7 10 13\n-+----------     --+------------\n0| - 1 3 5 6     1 | - 3 6 9  12\n1|   - 2 4 5     4 |   - 3 6  9\n3|     - 2 3     7 |     - 3  6\n5|       - 1     10|       -  3\n6|         -     13|          -\n\n\nWe see that if $A=\\{0,1,3,5,6\\}$, then the differences $1,2,3,5$ each have multiplicity 2, and the differences $4,6$ each have multiplicity 1. We also see that if $A=\\{1,4,7,10,13\\}$, then the differences $3,6,9,12$ have multiplicities $4,3,2,1$ respectively.\n\nMy question is: What is a tight upper bound on the sum of the squares of the multiplicities? I conjecture that the answer is $1^2+2^2+3^2+\\cdots+(n-1)^2$, which occurs if $A$ consists of $n$ integers in arithmetic progression (as in the second example above).\n\nAt first, I thought this would be an easy consequence of the following \"lemma\": The largest multiplicity must be at most $n-1$, the second largest multiplicity must be at most $n-2$, the third largest multiplicity must be at most $n-3$, and so on. However, that \"lemma\" is false! In the first example above (in which $n=5$), the four largest multiplicities are $2,2,2,2$, which are not bounded above by $4,3,2,1$ respectively.\n\nNote that my conjectured bound grows like $n^3/3$. I'm pretty sure I can find an upper bound that grows like $2n^3/3$, but the gap between those bounds is perplexing me.\n\n-\nYour lemma might be true if you modify it to say the largest multiplicity is at most $n-1$, the two largest multiplicities sum to at most $n-1+n-2=2n-3$, the three largest sum to at most $(n-1)+(n-2)+(n-3)=3n-6$, etc. \u2013\u00a0 Gerry Myerson Jun 21 '12 at 1:07\n\nWe can show by induction that your bound is the best.\n\nSuppose we have $n$ numbers and the differences have multiplicities $m_1\\ge m_2\\ge \\cdots \\ge m_t$. If we now add a new largest number, it creates $n$ distinct differences. If we could freely choose the differences (without constraints based on the other numbers), we could choose up to $n$ of the $m_i$ to increment exactly once, or can add $m_{t+1}=1,m_{t+2}=1,\\ldots$ as necessary for differences that do not match.\n\nSince $m_i>m_j\\Rightarrow (m_i+1)^2+m_j^2>m_i^2+(m_j+1)^2$ then without constraints on the differences we maximize our sum of squares by incrementing the $n$ largest differences available. Starting with $n=2, m_1=1$, the maximum for $n=3$ is achieved by $m_1\\leftarrow 2,m_2=1$, and so on. If we assume that for $n=k$ the maximum is $m_1=(k-1),m_2=(k-2),\\ldots,m_{k-1}=1$, then for $k+1$ numbers the maximum is $m_1\\leftarrow k,m_2\\leftarrow (k-1),\\ldots, m_{k-1}\\leftarrow 2, m_k=1$. So by induction this pattern gives a bound for all $n$. And, as you pointed out, this bound can be achieved by choosing numbers in an arithmetic sequence.\n\nThis line of thought also confirms @Gerry's comment on your Lemma. Starting with one number and adding new largest numbers, the best we can do for the top $M$ multiplicities is to increment some subset of them each time we add a number until we have $M$ numbers, then increment every one of them for each number added after that. This constraint is thus that they add to at most $Mn-M(M+1)/2$, but the $M$th largest can be up to $n-(M+1)/2$.\n\n-\n\nThinking about it today (everybody's answers and comments certainly helped), I believe there is a relatively short inductive proof that doesn't bother with any lemmas.\n\nMy bound is certainly true for $n=2$. When we transition from a set $A=\\{a_1<\\cdots<a_n\\}$ to a set $A'=\\{a_1<\\cdots<a_n<a_{n+1}\\}$, then as Zander mentions, we introduce $n$ differences $a_{n+1}-a_1,\\ldots,a_{n+1}-a_n$ that are distinct from each other, so $n$ of the multiplicities increase by 1 (possibly including some multiplicities that were 0).\n\nWhen we increase a multiplicity from $m$ to $m+1$, we increase the sum of the squares of the multiplicities by $2m+1$. Therefore the total increase of the sum of the squares of the multiplicities has the form $\\sum(2m_i+1)$, where there are $n$ terms in the sum, and the $m_i$ are some of the multiplicities with respect to $A$ (and some $m_i$ may be 0).\n\nSo the sum of the squares of the multiplicities increases by $2(\\sum m_i) + n$. This can be bounded above by $2S+n$, where $S$ is the sum of all multiplicities with respect to $A$. But $S = {n \\choose 2}$, so we have shown that the sum of squares of multiplicities has increased by at most $2{n \\choose 2}+n = (n^2-n)+n = n^2$.\n\n-", "date": "2015-06-30 10:47:20", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/160979/sum-of-squares-of-multiplicities-of-differences", "openwebmath_score": 0.8820327520370483, "openwebmath_perplexity": 150.06880713823597, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683502444728, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8648995033013251}}
{"url": "https://math.stackexchange.com/questions/2093295/total-number-of-odd-numbers-in-a-row-of-pascals-triangle-is-a-natural-power-of", "text": "# Total number of odd numbers in a row of Pascal's triangle is a natural power of two\n\nThis seems like a combinatorial problem so there might be something simple that hasn't struck me. Although I do have an idea but I am unable to proceed from it. The statement of the question is:\n\nProve that in any row of Pascal's triangle, the number of odd elements is $$2^k$$, for some $$k \\in \\mathbb{N}$$.\n\nI was working on a Pascal's triangle but in a binary format, where two adjacent 1's add up to 0. Something of the sort:\n\n                 1\n1   1\n1   0   1\n1   1   1   1\n1   0   0   0   1\n1   1   0   0   1   1\n\n\nIt is a definitive sequence and the thing I liked about it was that you can add up the adjacent 1's in a row to get the number of odd elements, but I haven't been able to generalize this information. I was thinking along the lines of a recursive relation in polynomials whose coefficients would represent the rows of this triangle, then I can just feed 1 into the equation and inductively prove that it is a perfect power of 2.\n\nCan someone help me out on a proof along these lines, if I'm going in the right direction here?\n\nP.S. I know there is the modulo 2 proof but can someone help me generalize that binary pascal's triangle?\n\n\u2022 This will be helpful: math.hmc.edu/funfacts/ffiles/30001.4-5.shtml Jan 11 '17 at 13:07\n\u2022 Okay, cool. But I want to try a proof along the lines of what I was trying. That is a good solution, but I wanna see where this goes. Jan 11 '17 at 13:15\n\u2022 To get an inductive proof, you may want to use the results from the answers based on Kummer's Theorem. That is, try to show that the number of $1$s in a row of Pascal's Triangle is $2$ to the power of the number of $1$ bits in the binary representation of $n$.\n\u2013\u00a0robjohn\nJan 11 '17 at 17:58\n\u2022 You may be interested by the fact that Pascal's triangle modulo 2 generates Sierpinski's sieve (mathworld.wolfram.com/SierpinskiSieve.html) Jan 11 '17 at 22:18\n\n## 3 Answers\n\nSimilar idea to Andreas Caranti's answer:\n\nKummer's Theorem says that the number of factors of $$p$$ in $$\\binom{n}{k}$$ is the number of carries when adding $$k$$ and $$n-k$$ in base-$$p$$. There are no carries when adding $$k$$ and $$n-k$$ in binary if and only if $$k\\veebar n-k=0$$; when that is true, $$n=k\\lor n-k$$. This means that $$\\binom{n}{k}$$ is odd when the $$1$$ bits of $$k$$ are a subset of the $$1$$ bits of $$n$$. That is, there are $$2^{\\sigma_2(n)}$$ odd entries in row $$n$$ of Pascal's Triangle, where $$\\sigma_2(n)$$ is the sum of the bits in the binary representation of $$n$$.\n\nWill delete because this is not what OP intends, but a proof via Kummer's Theorem looks rather neat to me.\n\nWrite $n, m$ in base $2$ $$n = 2^{k} n_{k} + \\dots + 2 n_{1} + n_{0}, \\qquad m = 2^{k} m_{k} + \\dots + 2 m_{1} + m_{0},$$ with $n_{i}, m_{i} \\in \\{ 0, 1 \\}$.\n\nThen\n\n$\\dbinom{n}{m}$ is odd iff there are no carries in the sum in base $2$ $$m + (n - m) = n,$$ that is, iff $m_{i} \\le n_{i}$ for each $i$.\n\nGiven $n$, there are thus $$\\prod_{i=0}^{k} (n_{i} + 1)$$ possibilities for $m$, and each $n_{i} + 1 \\in \\{ 1, 2 \\}$.\n\nThis also tells you what is the exponent $k$ in the question - it is the number of $1$'s in the diadic expansion of $n$, or equivalently, the number of distinct powers of $2$ that sum up to $n$.\n\nThe simplest (at least simplest to explain) solution (using Lucas's theorem) has not been posted; let me fill that \"much needed gap\". Lazy as I am, I will mostly copypaste my answer from https://math.stackexchange.com/a/2184460/ , since the argument is essentially the same.\n\nI use the notation $$\\mathbb{N}$$ for the set $$\\left\\{0,1,2,\\ldots\\right\\}$$.\n\nThe question asks the following:\n\nTheorem 1. Let $$n \\in \\mathbb{N}$$. Then, the number of $$i \\in \\left\\{0,1,\\ldots,n\\right\\}$$ such that $$\\dbinom{n}{i}$$ is odd is a power of $$2$$.\n\nI shall prove something slightly more general:\n\nTheorem 2. Let $$n \\in \\mathbb{N}$$. Let $$q$$ be a real number. Then, there exists a finite subset $$G$$ of $$\\mathbb{N}$$ such that \\begin{align} \\sum\\limits\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\\\dbinom{n}{i}\\text{ is odd}}} q^i = \\prod_{g \\in G} \\left(q^{2^g} + 1\\right) . \\end{align}\n\nProof of Theorem 2. Write $$n$$ in the form $$n=a_{k}2^{k}+a_{k-1}2^{k-1}+\\cdots+a_{0}2^{0}$$ for some $$k\\in\\mathbb{N}$$ and $$a_{0},a_{1},\\ldots,a_{k}\\in\\left\\{ 0,1\\right\\}$$. (This is just the base-$$2$$ representation of $$n$$, possibly with leading zeroes.)\n\nLucas's theorem tells you that if $$i=b_{k}2^{k}+b_{k-1}2^{k-1}+\\cdots+b_{0}2^{0}$$ for some $$b_{0} ,b_{1},\\ldots,b_{k}\\in\\left\\{ 0,1\\right\\}$$, then \\begin{align} \\dbinom{n}{i} &\\equiv \\dbinom{a_{k}}{b_{k}}\\dbinom{a_{k-1}}{b_{k-1}} \\cdots\\dbinom{a_{0}}{b_{0}}=\\prod\\limits_{j=0}^{k}\\underbrace{\\dbinom{a_{j}}{b_{j}} }_{\\substack{= \\begin{cases} 1, & \\text{if }b_{j}\\leq a_{j}\\\\ 0, & \\text{if }b_{j}>a_{j} \\end{cases} \\\\\\text{(since }a_{j}\\text{ and }b_{j}\\text{ lie in }\\left\\{ 0,1\\right\\} \\text{)}}} \\\\ &=\\prod\\limits_{j=0}^{k} \\begin{cases} 1, & \\text{if }b_{j}\\leq a_{j}\\\\ 0, & \\text{if }b_{j}>a_{j} \\end{cases} = \\begin{cases} 1, & \\text{if }b_{j}\\leq a_{j}\\text{ for all }j\\text{;}\\\\ 0, & \\text{otherwise} \\end{cases} \\mod 2 . \\end{align} Hence, the $$i\\in\\mathbb{N}$$ for which $$\\dbinom{n}{i}$$ is odd are precisely the numbers of the form $$b_{k}2^{k}+b_{k-1}2^{k-1} +\\cdots+b_{0}2^{0}$$ for $$b_{0},b_{1},\\ldots,b_{k}\\in\\left\\{ 0,1\\right\\}$$ satisfying $$\\left( b_{j}\\leq a_{j}\\text{ for all }j\\right)$$. Since all these $$i$$ satisfy $$i \\in \\left\\{ 0,1,\\ldots,n\\right\\}$$ (because otherwise, $$\\dbinom{n}{i}$$ would be $$0$$ and therefore could not be odd), we can rewrite this as follows: The $$i \\in \\left\\{ 0,1,\\ldots,n\\right\\}$$ for which $$\\dbinom{n}{i}$$ is odd are precisely the numbers of the form $$b_{k}2^{k}+b_{k-1}2^{k-1} +\\cdots+b_{0}2^{0}$$ for $$b_{0},b_{1},\\ldots,b_{k}\\in\\left\\{ 0,1\\right\\}$$ satisfying $$\\left( b_{j}\\leq a_{j}\\text{ for all }j\\right)$$. Since these numbers are distinct (because the base-$$2$$ representation of any $$i\\in\\mathbb{N}$$ is unique, as long as we fix the number of digits), we thus can substitute $$b_{k}2^{k}+b_{k-1}2^{k-1}+\\cdots+b_{0}2^{0}$$ for $$i$$ in the sum $$\\sum\\limits\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\ \\dbinom{n}{i}\\text{ is odd}}}q^{i}$$. Thus, this sum rewrites as follows: \\begin{align} \\sum\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\ \\dbinom{n}{i}\\text{ is odd}}}q^{i} &=\\underbrace{\\sum\\limits_{\\substack{b_{0},b_{1} ,\\ldots,b_{k}\\in\\left\\{ 0,1\\right\\} ;\\\\b_{j}\\leq a_{j}\\text{ for all }j} }}_{=\\sum\\limits_{b_{0}=0}^{a_{0}}\\sum\\limits_{b_{1}=0}^{a_{1}}\\cdots\\sum\\limits_{b_{k}=0}^{a_k} }\\underbrace{q^{b_{k}2^{k}+b_{k-1}2^{k-1}+\\cdots+b_{0}2^{0}}}_{=\\left( q^{2^{k}}\\right) ^{b_{k}}\\left( q^{2^{k-1}}\\right) ^{b_{k-1}}\\cdots\\left( q^{2^{0}}\\right) ^{b_{0}}} \\\\ &=\\sum\\limits_{b_{0}=0}^{a_{0}}\\sum\\limits_{b_{1}=0}^{a_{1}}\\cdots\\sum\\limits_{b_{k}=0}^{a_{k} }\\left( q^{2^{k}}\\right) ^{b_{k}}\\left( q^{2^{k-1}}\\right) ^{b_{k-1} }\\cdots\\left( q^{2^{0}}\\right) ^{b_{0}} \\\\ &=\\left( \\sum\\limits_{b_{k}=0}^{a_{k}}\\left( q^{2^{k}}\\right) ^{b_{k}}\\right) \\left( \\sum\\limits_{b_{k-1}=0}^{a_{k-1}}\\left( q^{2^{k-1}}\\right) ^{b_{k-1} }\\right) \\cdots\\left( \\sum\\limits_{b_{0}=0}^{a_{0}}\\left( q^{2^{0}}\\right) ^{b_{0}}\\right) \\\\ &=\\left( \\sum\\limits_{b=0}^{a_{k}}\\left( q^{2^{k}}\\right) ^{b}\\right) \\left( \\sum\\limits_{b=0}^{a_{k-1}}\\left( q^{2^{k-1}}\\right) ^{b}\\right) \\cdots\\left( \\sum\\limits_{b=0}^{a_{0}}\\left( q^{2^{0}}\\right) ^{b}\\right) \\\\ &=\\prod\\limits_{g=0}^{k}\\underbrace{\\left( \\sum\\limits_{b=0}^{a_{g}}\\left( q^{2^{g}}\\right) ^{b}\\right) }_{\\substack{= \\begin{cases} q^{2^{g}}+1, & \\text{if }a_{g}=1;\\\\ 1 & \\text{if }a_{g}=0 \\end{cases} \\\\\\text{(since }a_{g}\\in\\left\\{ 0,1\\right\\} \\text{)}}} \\\\ &=\\prod\\limits_{g=0}^{k} \\begin{cases} q^{2^{g}}+1, & \\text{if }a_{g}=1;\\\\ 1 & \\text{if }a_{g}=0 \\end{cases} \\\\ &=\\left( \\prod\\limits_{\\substack{g\\in\\left\\{ 0,1,\\ldots,k\\right\\} ;\\\\a_{g} =1}}\\left( q^{2^{g}}+1\\right) \\right) \\underbrace{\\left( \\prod\\limits _{\\substack{g\\in\\left\\{ 0,1,\\ldots,k\\right\\} ;\\\\a_{g}=0}}1\\right) }_{=1} \\\\ &=\\prod\\limits_{\\substack{g\\in\\left\\{ 0,1,\\ldots,k\\right\\} ;\\\\a_{g}=1}}\\left( q^{2^{g}}+1\\right) . \\end{align} Thus, there exists a finite subset $$G$$ of $$\\mathbb{N}$$ such that \\begin{align} \\sum\\limits\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\\\dbinom{n}{i}\\text{ is odd}}} q^i = \\prod_{g \\in G} \\left(q^{2^g} + 1\\right) \\end{align} (namely, $$G$$ is the set of all $$g\\in\\left\\{ 0,1,\\ldots,k\\right\\}$$ satisfying $$a_g = 1$$). This proves Theorem 2. $$\\blacksquare$$\n\nProof of Theorem 1. Theorem 2 (applied to $$q = 1$$) shows that there exists a finite subset $$G$$ of $$\\mathbb{N}$$ such that \\begin{align} \\sum\\limits\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\\\dbinom{n}{i}\\text{ is odd}}} 1^i = \\prod_{g \\in G} \\left(1^{2^g} + 1\\right) . \\end{align} Consider this $$G$$. Comparing \\begin{align} \\sum\\limits\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\\\dbinom{n}{i}\\text{ is odd}}} 1^i = \\prod_{g \\in G} \\underbrace{\\left(1^{2^g} + 1\\right)}_{= 1+1 = 2} = \\prod_{g \\in G} 2 = 2^{\\left|G\\right|} \\end{align} with \\begin{align} \\sum\\limits\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\\\dbinom{n}{i}\\text{ is odd}}} \\underbrace{1^i}_{=1} & = \\sum\\limits\\limits_{\\substack{i\\in\\left\\{ 0,1,\\ldots,n\\right\\} ;\\\\\\dbinom{n}{i}\\text{ is odd}}} 1 \\\\ & = \\left(\\text{the number of all i\\in\\left\\{ 0,1,\\ldots,n\\right\\} such that \\dbinom{n}{i} is odd}\\right) \\cdot 1 \\\\ & = \\left(\\text{the number of all i\\in\\left\\{ 0,1,\\ldots,n\\right\\} such that \\dbinom{n}{i} is odd}\\right) , \\end{align} we obtain \\begin{align} \\left(\\text{the number of all i\\in\\left\\{ 0,1,\\ldots,n\\right\\} such that \\dbinom{n}{i} is odd}\\right) = 2^{\\left|G\\right|} . \\end{align} Hence, the number of $$i \\in \\left\\{0,1,\\ldots,n\\right\\}$$ such that $$\\dbinom{n}{i}$$ is odd is a power of $$2$$. This proves Theorem 1. $$\\blacksquare$$", "date": "2021-12-05 23:37:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2093295/total-number-of-odd-numbers-in-a-row-of-pascals-triangle-is-a-natural-power-of", "openwebmath_score": 0.9995993971824646, "openwebmath_perplexity": 512.8374520454395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211597623861, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8648658197157344}}
{"url": "http://math.stackexchange.com/questions/333417/need-to-prove-the-sequence-a-n-1-frac122-frac132-cdots-frac1n/333429", "text": "# Need to prove the sequence $a_n=1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}$ converges\n\nI need to prove that the sequence $a_n=1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:\n\nMonotonic:\n\nThe sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\\leq a_{n+1}$ $$a_1=1\\leq 1+\\frac{1}{2^2}=a_2$$\n\nNeed to show that $a_{n+1}\\leq a_{n+2}$ $$a_{n+1}=1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}+\\frac{1}{(n+1)^2}\\leq 1+\\frac{1}{2^2}+\\frac{1}{3^2}+\\cdots+\\frac{1}{n^2}+\\frac{1}{(n+1)^2}+\\frac{1}{(n+2)^2}=a_{n+2}$$ Thus the sequence is monotone and increasing.\n\nBoundedness:\n\nSince the sequence is increasing it is bounded below by $a_1=1$. Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound.\n\nCan anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?\n\n-\nHint: compare it to another sequence with only powers of two in the denominator. \u2013\u00a0 Jason Polak Mar 18 '13 at 2:58\n@Jason: Could you please elaborate on your hint? I don't understand it. \u2013\u00a0 Jonas Meyer Mar 18 '13 at 3:08\n@JonasMeyer: The sequence is less than $1 + 1/2^2 + 1/2^2 + 1/4^2 + 1/4^2 + 1/4^2 + 1/4^2 + ... \\leq 2$ \u2013\u00a0 Jason Polak Mar 18 '13 at 3:13\n@ShuXiaoLi my objection was to the way the comment was presented, i.e., like a solution to the question. A comment like \"by the way, the sum is $\\pi^2/6$\" would have been fine. \u2013\u00a0 Ittay Weiss Mar 18 '13 at 3:17\n\nYour work looks good so far. Here is a hint: $$\\frac{1}{n^2} \\le \\frac{1}{n(n-1)} = \\frac{1}{n-1} - \\frac{1}{n}$$\n\nTo elaborate, apply the hint to get: $$\\frac{1}{2^2} + \\frac{1}{3^2} + \\frac{1}{4^2} + \\cdots + \\frac{1}{n^2} \\le \\left(\\frac{1}{1} - \\frac{1}{2}\\right) + \\left(\\frac{1}{2} - \\frac{1}{3}\\right) + \\left(\\frac{1}{3} - \\frac{1}{4}\\right) + \\cdots + \\left(\\frac{1}{n-1} - \\frac{1}{n}\\right)$$\n\nNotice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.\n\nAlso notice that all terms on the right-hand side cancel out except for the first and last one. Thus: $$\\frac{1}{2^2} + \\frac{1}{3^2} + \\frac{1}{4^2} + \\cdots + \\frac{1}{n^2} \\le 1 - \\frac{1}{n}$$\n\nAdd $1$ to both sides to get: $$a_n \\le 2 - \\frac{1}{n} \\le 2$$\n\nIt follows that $a_n$ is bounded from above and hence convergent.\n\nIt is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.\n\n-\nSorry, I'm afraid I'm going to need a bit more of a push. I can't think of how/where to use your hint. Am I supposed to say that $a_n=1+\\frac{1}{2^2}+\\frac{1}{3^2}+{\u2026}+\\frac{1}{n^2}\\leq 1+\\frac{1}{2^2}+\\frac{1}{3^2}+{\u2026}+\\frac{1}{n-1}-\\frac{1}{n}$ and do something further with that? \u2013\u00a0 user66807 Mar 18 '13 at 4:02\ntry to apply it more than once. \u2013\u00a0 chx Mar 18 '13 at 4:17\n@chx Try to apply it more than once in the \"thing\" I supposed? \u2013\u00a0 user66807 Mar 18 '13 at 4:22\n@user66807, we have $a_n=\\sum_{k=2}^{n}\\frac{1}{k^2}\\le b_n=\\sum_{k=2}^{n}\\frac{1}{k-1}-\\frac{1}{k}$, try to find the first few terms of $b_n$ you will see a pattern (things will start cancelling) then you can find an upper bound of $a_n$. Notice i started the sum from 2 to be able to use Ayman's hint. \u2013\u00a0 i.a.m Mar 18 '13 at 4:32\n@user66807, try to sum a few terms of the telescopic series $\\,\\sum\\left(\\frac{1}{n-1}-\\frac{1}{n}\\right)\\,$ and see what happens... \u2013\u00a0 DonAntonio Mar 18 '13 at 4:33\n\nBesides to Ayman's neat answer, you may take $f(x)=\\frac{1}{x^2}$ over $[1,+\\infty)$ and see that $f'(x)=-2x^{-3}$ and then is decreasing over $[1,+\\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $$\\sum_{k=1}^{+\\infty}\\frac{1}{n^2}$$ to see the series is convergent. Now your $a_n$ is the $n-$th summation of this series.\n\n-\nThe problem is we haven't gone over series in class yet, but thanks anyways. \u2013\u00a0 user66807 Mar 18 '13 at 3:41\n@user66807: Sorry, I didn't know that. So the Ayman's is more preferable nice and simple approach for you. :-) \u2013\u00a0 Babak S. Mar 18 '13 at 3:43\nI hope so, let's just see if I can catch his hint. And thanks. \u2013\u00a0 user66807 Mar 18 '13 at 3:46\n+1 This is a good answer, though the OP can't use it yet. \u2013\u00a0 DonAntonio Mar 18 '13 at 4:34\nI agree with DonAntonio that this is a good answer! $+1 \\;\\ddot\\smile\\;\\;$ (I also like the new gravatar!) Good day, to you! \u2013\u00a0 amWhy Mar 18 '13 at 15:08\n\nYou can show this geometrically too.\n\nIf you take a square, and divide the height into $\\frac 12$, $\\frac 14$, $\\frac 18$, and so forth, doubling the denominator on each deal.\n\nNow take the squares $\\frac 12$ and $\\frac 13$: these go onto the top shelf.\n\nOn the second shelf go $\\frac 14$ to $\\frac 17$. These fractions are all less than $\\frac 14$, so fit onto the second shelf. Likewise, squares from $8$ to $15$ go onto the third shelf, each $\\frac 1n$ is smaller than $\\frac 18$, and so forth.\n\nTherefore $\\sum_{n=2}^{inf}\\frac 1n < 1$, and therefore the whole lot is less than two squares.\n\n-\n\nHint: Prove the the following holds for all $n$ by induction.\n\n$$\\sum_1^n \\frac{1}{k^2} \\le 2 - \\frac{1}{n}.$$\n\nIs it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work.\n\nYou can use the same technique to bound other values of the zeta function. For example, try showing $\\zeta(3)$ is bounded above by $\\frac{3}{2}$.\n\n(I am copying my answer from a duplicate question that was closed as a copy of this one since the induction approach is not available here.)\n\n-\n\nby the integral test :\n\n$\\int^\\infty_1 \\frac{1}{n^2} dn \\le \\sum_{i=1}^\\infty \\frac{1}{n^2}\\le 1+\\int^\\infty_1 \\frac{1}{n^2}dn$ .\n\nyou can compute the integral so the answer is :\n\n$1 \\le \\sum_{i=1}^\\infty \\frac{1}{n^2}\\le 2$ .\n\nbecause : $\\int^\\infty_1 \\frac{1}{n^2} dn =1$\n\n-", "date": "2015-04-27 14:25:06", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/333417/need-to-prove-the-sequence-a-n-1-frac122-frac132-cdots-frac1n/333429", "openwebmath_score": 0.9259559512138367, "openwebmath_perplexity": 359.25799202759464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211582993981, "lm_q2_score": 0.8872046026642944, "lm_q1q2_score": 0.8648658184177647}}
{"url": "https://math.stackexchange.com/questions/3310720/probability-of-rolling-a-sum-of-14-when-rolling-a-20-sided-die-twice-or-a-2", "text": "# probability of rolling a sum of $14$ when rolling a $20$-sided die twice or a $20$-sided die with a $4$-sided die depending on outcome of first roll\n\nA $$20$$-sided die is rolled. If the result is $$10$$ or less, the $$20$$-sided die is rolled again. If the result is $$11$$ or more, a $$4$$-sided die is rolled. In either case, the results are summed after the second die roll. What is the probability that the sum will be $$14$$?\n\nMy Approach: there are 10 ways we can get a sum of 14 if the first dice is $$<= 10$$ {$$(10,4),(9,5),(8,6),(7,7),(6,8),(5,9),(4,10),(3,11),(2,12),(1,13)$$} and $$3$$ ways we can get sum of $$14$$ if the first dice has value $$>= 11$$ {$$(11,3),(12,2),(13,1)$$} so the probability is: $$(10/400)$$ + $$(3/80) = 1/16$$\n\n\u2022 What goes wrong when you just write it out in the obvious way? \u2013\u00a0lulu Aug 1 at 18:08\n\u2022 welcome to MSE. kindly include any attempt. \u2013\u00a0Siong Thye Goh Aug 1 at 18:12\n\nHints:\n\n$$Pr(A\\cap B) = Pr(A)Pr(B\\mid A)$$\n\n$$Pr(B) = Pr(A_1\\cap B)+Pr(A_2\\cap B)+\\dots+Pr(A_n\\cap B)$$ where $$A_1,\\dots,A_n$$ form a partition of the sample space\n\nLet $$B$$ be the event the sum is $$14$$.\n\nLet $$A_1$$ be the event the first die rolls a number $$1-10$$. Let $$A_2$$ be the event the first die rolls a number $$11-13$$. Let $$A_3$$ be the event the first die rolls a number $$14+$$.\n\nRegardless what was rolled on the first die, so long as it was less than $$14$$, there will be exactly one outcome on the second roll that would let the total sum to $$14$$.\n\n\u2022 I've added my approach. Not sure if it's correct. \u2013\u00a0user3119875 Aug 1 at 18:52\n\u2022 $\\frac{1}{2}\\times\\frac{1}{20}+\\frac{3}{20}\\times\\frac{1}{4}=\\frac{1}{16}$ is correct. There is absolutely no need to write out each of the possibilities however and this causes errors when trying to count things by hand and wastes effort and time. \u2013\u00a0JMoravitz Aug 1 at 18:56\n\nThe probability that you obtain $$14$$ using the $$20$$-sided die on the second role is\n\n$$\\frac{10}{20} \\cdot \\frac{1}{20}$$\n\nThe probability you obtain $$14$$ using the $$4$$-sided die on the second role is\n\n$$\\frac{3}{20} \\cdot \\frac{1}{4}$$", "date": "2019-10-15 01:46:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3310720/probability-of-rolling-a-sum-of-14-when-rolling-a-20-sided-die-twice-or-a-2", "openwebmath_score": 0.8888765573501587, "openwebmath_perplexity": 158.8048811501347, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211590308922, "lm_q2_score": 0.8872045922259089, "lm_q1q2_score": 0.8648658088911906}}
{"url": "https://math.stackexchange.com/questions/2848226/is-there-any-specific-convention-to-how-mathematical-equations-are-read-out", "text": "# Is there any specific convention to how mathematical equations are \u201cread out\u201d?\n\nOften, mathematical information and formulas are shared through text, maybe on books or on websites. If you see the equation:\n\n$$x + 3 = 4y$$\n\nwe can easily read it out as \"x plus three equals four y\". But for some other equations such as:\n\n$$\\frac{x+3}{x-7}=y$$\n\nit doesn't seem too obvious on what it should be read out as. I would read it as \"x plus three the whole divided by x minus seven equals y\", however someone else may read it out differently.\n\nMy question is therefore: If in a situation where oral communication of mathematical equations is necessary, does there exist a convention to read out mathematical equations such that one particular equation is read out in only one way, and that it is possible to deduce the equation from the spoken name alone?\n\n\u2022 In French, I would read it as \"x plus trois sur x moins sept \u00e9gal y\", which would translate as \"x plus three over x minus seven equals y\". I don't know if \"over\" really is commonly used in English though. I don't think there is any universal convention though. What I said could also be interpreted as $x+\\frac{3}{x} -7 = y$. If I notice that the person I am talking to has gotten it wrong, I would just say it again, insisting on the expression I really meant. \u2013\u00a0Suzet Jul 12 '18 at 1:55\n\u2022 \"over\" is indeed used by some people I know. And here is the point, even in English, the same equation is being read out in two different ways. \u2013\u00a0Pritt Balagopal Jul 12 '18 at 1:57\n\u2022 The different groupings / implied parantheses are often expressed via pauses and prosody. \u2013\u00a0Hagen von Eitzen Jul 12 '18 at 2:02\n\u2022 \"The quotient of [the quantity] $x+3$ and [the quantity] $x-7$ is $y$.\" To avoid a pile-up of symbols at the end, one could say \"$y$ is the quotient of $x+3$ and $x-7$.\" \u2013\u00a0Blue Jul 12 '18 at 2:26\n\u2022 @HagenvonEitzen: On the first day of a stint as a Pre-Calculus substitute teacher, I instructed my students to express groupings using \"air parentheses\", indicated by raising one's slightly-bent arms over one's head. So, here: \"[raise arms] x plus three [lower arms] over [raise arms] ex minus seven [lower arms] is y.\" The students looked at me like I was crazy, but took to the practice. So earnest were they in this \u2014even prompting each other when someone would forget\u2014 that, at the end of my six-week stay, I didn't have the heart to tell them that I'd only been kidding that first day. :D \u2013\u00a0Blue Jul 12 '18 at 2:33\n\nI would like to elaborate on Hagen von Eitzen's insightful comment:\n\nThe different groupings / implied parantheses are often expressed via pauses and prosody.\n\nI think this is exactly right and plays an extremely important role in disambiguating oral mathematics. There is a huge difference between:\n\n\"X plus 3, over X minus 7, equals Y\"\n\nand\n\n\"X, plus 3 over X, minus 7 equals Y\"\n\nand\n\n\"X plus 3, over X, minus 7 equals Y\"\n\nIf the cadences and pauses in the sentences are read aloud consistently, I think it is unlikely that anyone would transcribe one of those in place of another.\n\nHowever, in some cases (particularly for extremely complicated expressions with groupings nested inside other groupings) one may call attention to the grouping using verbal annotations like \"the quantity\" or \"all\". For example, the quadratic formula is often read aloud as\n\n\u2022 \"X equals negative b, plus or minus the square root of the quantity b-squared minus 4ac, all over 2a\".\n\nI would read $$\\frac{x+3}{x-7}=y$$ as:\n\nQuantity $x$ plus $3$ over quantity $x$ minus $7$ equals $y$.", "date": "2019-06-19 09:15:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2848226/is-there-any-specific-convention-to-how-mathematical-equations-are-read-out", "openwebmath_score": 0.670265257358551, "openwebmath_perplexity": 1507.301632838895, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854146791214, "lm_q2_score": 0.8918110468756548, "lm_q1q2_score": 0.8648653459097283}}
{"url": "https://math.stackexchange.com/questions/3796115/image-of-compact-set-under-piecewise-continuous-function", "text": "# Image of compact set under piecewise continuous function\n\nLet $$a,b>0\\in\\mathbb{R}$$. Let $$U$$ be an domain in $$\\mathbb{C}^n$$. Let $$f:[a,b]\\longrightarrow U$$ be a piecewise continuous map. Then is $$f[a,b]$$ compact? If not compact, will it be bounded?\n\nOk. This is in the following context. I am given a piecewise smooth path $$\\gamma:[a,b]\\longrightarrow U$$. Where $$\\gamma(a)=z$$ and $$\\gamma(b)=w$$, for given $$z,w\\in U$$. We are also given a function $$\\alpha:U\\times\\mathbb{C}^n\\longrightarrow \\mathbb{R}$$, which is upper semicontinuous. Now it is said that $$t\\in[a,b]\\longrightarrow \\alpha(\\gamma(t),\\gamma\u2019(t))$$ is bounded and measurable. I wanted to know why the function is bounded. I know that $$\\gamma[a,b]$$ is compact. And $$\\gamma$$ being upper semicontinuous will attain its maximum on a compact set. But I am not sure about $$\\gamma\u2019$$.\n\n\u2022 What does \"piecewise continuous\" mean here? How many pieces? Finitely many? Are the pieces compact? Open? Aug 19, 2020 at 10:51\n\u2022 What have you tried? Aug 19, 2020 at 10:53\n\u2022 The image of a compact space under continuous map is always compact. Regardless of what \"piecewise\" means and what $U$ is (as long as it is Hausdorff). Aug 19, 2020 at 10:57\n\u2022 Please provide additional context, which ideally explains why the question is relevant to you and the community. Some forms of context include background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. Aug 19, 2020 at 11:18\n\u2022 @SahibaArora I have added the context. Aug 19, 2020 at 13:25\n\nCompact, not necessarily: On $$[0,1]$$ let $$f(x) = x, 0\\le x<1,$$ $$f(1)=2.$$ Then $$f([0,1]) = [0,1)\\cup\\{2\\}.$$\n\nBounded, yes: First, a lemma: If $$f$$ is continuous on $$(a,b)$$ and $$f$$ has finite limits at the end points, then $$f(a,b)$$ is bounded.\n\nProof: Suppose $$\\lim_{x\\to a^+} f(x)=L,$$ $$\\lim_{x\\to b^-} f(x)=M.$$ Let $$\\epsilon=1.$$ Then there exists $$\\delta_a>0, \\delta_a<(b-a)/3,$$ such that $$|f(x)-L|<1$$ for $$x\\in (a,a+\\delta_a).$$ Thus for such $$x,$$\n\n$$|f(x)| = |f(x)-L+L|\\le |f(x)-L|+|L| <1+|L|.$$\n\nSimilarly, there exists $$\\delta_b>0,\\delta_b<(b-a)/3,$$ such that $$|f(x)|<1+|M|$$ for $$x\\in (b-\\delta_b,b).$$ It follows that $$f$$ is bounded on the set $$(a,a+\\delta_a)\\cup (b-\\delta_b,b).$$\n\nSince $$f$$ is continuous on the compact set $$[a+\\delta_a,b-\\delta_b],$$ $$f([a+\\delta_a,b-\\delta_b])$$ is compact, hence is bounded. It follows that $$f(a,b)$$ is bounded.\n\nNow suppose $$f$$ is piecewise continuous on $$[a,b].$$ Then there exist points $$a=x_0 such that $$f$$ is continuous on each $$I_k=(x_{k-1},x_k)$$ and has finite limits at the end points of $$I_k.$$ By the lemma, each $$f(I_k)$$ is bounded. The set $$f(\\{x_0,\\dots x_n\\})$$ is also bounded. Therefore\n\n$$f([a,b])=f(I_1)\\cup \\cdots \\cup f(I_n)\\cup f(\\{x_0,\\dots x_n\\})$$\n\nis bounded.", "date": "2022-06-30 19:05:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3796115/image-of-compact-set-under-piecewise-continuous-function", "openwebmath_score": 0.9237774014472961, "openwebmath_perplexity": 211.56216783886177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854129326059, "lm_q2_score": 0.8918110461567922, "lm_q1q2_score": 0.864865343655024}}
{"url": "https://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbb", "text": "# Is there a quick proof as to why the vector space of $\\mathbb{R}$ over $\\mathbb{Q}$ is infinite-dimensional?\n\nIt would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?\n\n\u2022 A finite dimensional vector space over $\\mathbb{Q}$ is countable. \u2013\u00a0user641 Oct 7 '10 at 2:19\n\u2022 @Steve: Please add that as an answer so people can upvote. \u2013\u00a0Aryabhata Oct 7 '10 at 2:50\n\u2022 Does that mean that a vector space over $\\mathbb{Q}$ is finite-dimensional iff the set of the vector space is countable? If so, please prove it. \u2013\u00a0Elchanan Solomon Oct 7 '10 at 2:55\n\u2022 @Isaac Your question doesn't require the 'only if' anyway. Steve's observation answers your original question. \u2013\u00a0yasmar Oct 7 '10 at 3:04\n\u2022 @Isaac: how is the fact that $\\mathbb{R}$ is finite dimensional over itself relevant? \u2013\u00a0Arturo Magidin Oct 7 '10 at 3:31\n\nAs Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\\alpha_1 v_1+\\cdots+\\alpha_n v_n$ for some scalars $\\alpha_1,\\ldots,\\alpha_n\\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\\mathbb{R}$ is uncountable and $\\mathbb{Q}$ is countable, $\\mathbb{R}$ cannot be finite dimensional over $\\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).\n\nYour further question in the comments, whether a vector space over $\\mathbb{Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\\mathbb{Q}$ that are countable as sets. The simplest example is $\\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\\mathbb{Q}$, which is a countable set, and has dimension $\\aleph_0$, with basis $\\{1,x,x^2,\\ldots,x^n,\\ldots\\}$.\n\nAdded: Of course, if $V$ is a vector space over $\\mathbb{Q}$, then it has countable dimension (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\\mathbb{R}$ infinite dimensional over $\\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is uncountably-dimensional over $\\mathbb{Q}$.\n\n\u2022 Related to the note about uncountable dimension, there are explicit examples of continuum-sized linearly independent sets, as seen in this MathOverflow answer by Fran\u00e7ois G. Dorais: mathoverflow.net/questions/23202/\u2026 \u2013\u00a0Jonas Meyer Oct 7 '10 at 4:18\n\u2022 Yes: but can one show that there is a basis for $\\mathbb{R}$ over $\\mathbb{Q}$ without some form of the Axiom of Choice? (There is a difference between exhibiting a large linearly independent subset and exhibiting a basis). \u2013\u00a0Arturo Magidin Oct 7 '10 at 14:51\n\u2022 No, one cannot, but without the Axiom of Choice the notion of dimension breaks down (except for finite vs. infinite). Assuming AC (as I virtually always do), the size of a linearly independent set gives a lower bound on the dimension of the vector space, and I think it is wonderful that in this case such \"explicit\" proof exists that the real numbers have continuum dimension, as opposed to the nice qualitative proof one could give by extending your argument to larger cardinals. \u2013\u00a0Jonas Meyer Oct 7 '10 at 16:02\n\u2022 @Jonas: No argument there (with any of your points). \u2013\u00a0Arturo Magidin Oct 7 '10 at 17:23\n\u2022 good link @Jonas, thanks \u2013\u00a0Leon Sep 23 '11 at 0:23\n\nThe cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $$\\, \\mathbb Q$$-independent set of reals. Consider the set consisting of the logs of all primes $$\\, p_i.\\,$$ If $$\\, c_1 \\log p_1 +\\,\\cdots\\, + c_n\\log p_n =\\, 0,\\ c_i\\in\\mathbb Q,\\,$$ multiplying by a common denominator we can assume that all $$\\ c_i \\in \\mathbb Z\\,$$ so, exponentiating, we obtain $$\\, p_1^{\\large c_1}\\cdots p_n^{\\large c_n}\\! = 1\\,\\Rightarrow\\ c_i = 0\\,$$ for all $$\\,i,\\,$$ by the uniqueness of prime factorizations.\n\n\u2022 @Bill +1 What a nice example. \u2013\u00a0Adri\u00e1n Barquero Oct 7 '10 at 4:50\n\u2022 @Bill. Wow! :-) \u2013\u00a0Agust\u00ed Roig Oct 7 '10 at 6:08\n\u2022 Is this proof unique to Q or does it generalize to provide explicit examples of reals linearly independent over, e.g., Q(sqrt(2))? Above, Q appears to be \"hard-wired\" into the proof, as the group of exponents in the prime factorization. \u2013\u00a0T.. Oct 12 '10 at 7:14\n\u2022 Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. \u2013\u00a0JDH Jun 23 '11 at 2:22\n\u2022 @student this can not happen: the primes are distinct. Put the primes with negative exponent on the other side (exponents become positive). You would find two distinct prime factorizations of the same number. \u2013\u00a0Student Feb 6 at 19:44\n\nNo transcendental numbers are needed for this question. Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension. Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down.\n\nThe set $\\sqrt{2}, \\sqrt{\\sqrt{2}}, \\dots, = \\bigcup_{n>0} 2^{2^{-n}}$ is linearly independent over $\\mathbb Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i < n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$).\n\nThe square roots of the prime numbers are linearly independent over $\\mathbb Q$. (Proof: this is immediate given the ability to extend the function \"number of powers of $p$ dividing $x$\" from the rational numbers to algebraic numbers. $\\sqrt{p}$ is \"divisible by $p^{1/2}$\" while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $\\mathbb Q$ unramified at $p$).\n\nGenerally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case. Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent. (Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.)\n\n\u2022 I realize you answered this question a loooong time ago, but if you're still around, how do you know that you can write the $n-$th root of $2$ as a linear combination of the $1st$ through $n-1$st roots of $2$? \u2013\u00a0ALannister Aug 17 '17 at 22:41\n\u2022 @ALannister I realize you questioned quite a long time ago, but s/he is assuming that there is such $n$ and showing that such assumption leads to contradiction. So in fact, s/he is proving exactly that $n$-th root of $2$ cannot be expressed by a linear combination of previous roots \u2013\u00a0user160738 May 9 '18 at 17:24\n\nFor the sake of completeness, I'm adding a worked-out solution due to F.G. Dorais from his post.\n\nWe'll need two propositions from Grillet's Abstract Algebra, page 335 and 640:\n\nProposition: $[\\mathbb{R}:\\mathbb{Q}]=\\mathrm{dim}_\\mathbb{Q}{}\\mathbb{R}=|\\mathbb{R}|$\n\nProof: Let $(q_n)_{n\\in\\mathbb{N_0}}$ be an enumeration of $\\mathbb{Q}$. For $r\\in\\mathbb{R}$, take $$A_r:=\\sum_{q_n<r}\\frac{1}{n!}\\;\\;\\;\\;\\text{ and }\\;\\;\\;\\;A:=\\{A_r;\\,r\\in\\mathbb{R}\\};$$ the series is convergent because $\\sum_{q_n<r}\\frac{1}{n!}\\leq\\sum_{n=0}^\\infty\\frac{1}{n!}=\\exp(1)<\\infty$ (recall that $\\exp(x)=\\sum_{n=0}^\\infty\\frac{x^n}{n!}$ for any $x\\in\\mathbb{R}$).\n\nTo prove $|A|=|\\mathbb{R}|$, assume $A_r=A_{s}$ and $r\\neq s$. Without loss of generality $r<s$, hence $A_s=\\sum_{q_n<s}\\frac{1}{n!}=\\sum_{q_n<r}\\frac{1}{n!}+\\sum_{r\\leq q_n<s}\\frac{1}{n!}=A_r+\\sum_{r\\leq q_n<s}\\frac{1}{n!}$, so $\\sum_{r\\leq q_n<s}\\frac{1}{n!}=0$, which is a contradiction, because each interval $(r,s)$ contains a rational number.\n\nTo prove $A$ is $\\mathbb{Q}$-independent, assume $\\alpha_1A_{r_1}+\\cdots+\\alpha_kA_{r_k}=0\\;(1)$ with $\\alpha_i\\in\\mathbb{Q}$. We can assume $r_1>\\cdots>r_k$ (otherwise rearrange the summands) and $\\alpha_i\\in\\mathbb{Z}$ (otherwise multiply by the common denominator). Choose $n$ large enough that $r_1>q_n>r_2\\;(2)$; we'll increase $n$ two more times. The equality $n!\\cdot(1)$ reads $n!(\\alpha_1\\sum_{q_m<r_1}\\frac{1}{m!}+\\cdots+\\alpha_k\\sum_{q_m<r_k}\\frac{1}{m!})=0$. Rearranged (via $(2)$ when $m=n$), it reads\n\n$$-\\alpha_1\\sum_{\\substack{m<n\\\\q_m<r_1}}\\frac{n!}{m!}-\\cdots-\\alpha_k\\sum_{\\substack{m<n\\\\q_m<r_k}}\\frac{n!}{m!}-\\alpha_1 =\\alpha_1\\sum_{\\substack{m>n\\\\q_m<r_1}}\\frac{n!}{m!}+\\cdots+\\alpha_k\\sum_{\\substack{m>n\\\\q_m<r_k}}\\frac{n!}{m!}. \\tag*{(3)}$$\n\nThe left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\\alpha_1|+\\cdots+|\\alpha_k|)\\sum_{m=n+1}^\\infty\\frac{n!}{m!}<1$ holds (such $n$ can be found since $\\sum_{m=n+1}^\\infty\\frac{n!}{m!}=\\frac{1}{n+1}\\sum_{m=n+1}^\\infty\\frac{1}{(n+2)\\cdot\\ldots\\cdot m}\\leq\\frac{1}{n+1}\\sum_{m=n+1}^\\infty\\frac{1}{(m-n-1)!}\\leq\\frac{1}{n+1}\\exp(1)\\rightarrow 0$ when $n\\rightarrow\\infty$), then the absolute value of RHS of $(3)$ is $<1$, and yet an integer, hence $\\text{RHS}(3)=0$. Thus $(3)$ reads $\\alpha_1=-\\sum_{i=1}^{k}\\sum_{m<n,q_m<r_i}\\alpha_i\\frac{n!}{m!}=0\\;(\\mathrm{mod}\\,n)$. If moreover $n>|\\alpha_1|$, this means that $\\alpha_1=0$. Repeat this argument to conclude that also $\\alpha_2=\\cdots=\\alpha_k=0$.\n\nSince $A$ is a $\\mathbb{Q}$-independent subset, by proposition 5.3 there exists a basis $B$ of $\\mathbb{R}$ that contains $A$. Then $A\\subseteq B\\subseteq\\mathbb{R}$ and $|A|=|\\mathbb{R}|$ and Cantor-Bernstein theorem imply $|B|=|\\mathbb{R}|$, therefore $[\\mathbb{R}:\\mathbb{Q}]=\\mathrm{dim}_\\mathbb{Q}{}\\mathbb{R}=|\\mathbb{R}|$. $\\quad\\blacksquare$\n\n\u2022 Is this uncountable linearly independent set basis ? \u2013\u00a0user195218 Jun 5 '15 at 2:53\n\u2022 @user195218 It's not a basis, as it doesn't span the reals, but it is uncountable and linearly independent. \u2013\u00a0Akiva Weinberger Jul 25 '17 at 19:08\n\u2022 The only proper answer to the question. Such a shame that it is not the most voted answer. \u2013\u00a0Akerbeltz May 1 at 21:42\n\nHere is a simple proof that a basis $B$ of $[ \\mathbb{R}:\\mathbb{Q}]$ has cardinality $|\\mathbb{R}|$.\n\nClearly since $B$ is contained in $R$, $|B| \\le| \\mathbb{R}|$.\n\nBut also $\\mathbb{R}=span(B)$ and thus $|\\mathbb{R}| =|span(B)| \\le |\\mathbb{Q}^B|=|B|$. The last equality follows because $B$ is not finite (if it was, then $|\\mathbb{R}|=|span(B)| \\le |\\mathbb{Q}^B|=|\\mathbb{N}|$, a contradiction).\n\nHence $|\\mathbb{R}| \\le |B| \\le| \\mathbb{R}|$, so $|\\mathbb{R}|=|B|$\n\n\u2022 There is an error in what you wrote: $|\\mathbb{Q}^B|\\geq |2|^{|B|} > |B|$. \u2013\u00a0Nex Aug 18 '17 at 8:19\n\u2022 @Next: hmm so is it salvageable what I wrote? I guess I wasn't paying too much attention. \u2013\u00a0Joshua Benabou Aug 18 '17 at 19:08\n\nAnother simple proof:\n\nTake $P=X^n-p$ for a prime $p$.\n\nBy Eisenstein's criterion, it is $\\mathbb Q[X]$-irreductible. Therefore, the set of algebraic numbers is of infinite dimension over $\\mathbb Q$.\n\nSince $\\mathbb R$ is bigger, it works for $\\mathbb R$ too.\n\n## protected by user26857Nov 25 '15 at 9:30\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).\n\nWould you like to answer one of these unanswered questions instead?", "date": "2019-09-23 19:32:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbb", "openwebmath_score": 0.9264833331108093, "openwebmath_perplexity": 242.82496116963233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225158534696, "lm_q2_score": 0.8723473862936943, "lm_q1q2_score": 0.864864840417493}}
{"url": "https://strouden.com/n185f/archive.php?page=least-squares-solution-linear-algebra-8f1419", "text": "\\nLicense: Creative Commons<\\/a>\\n<\\/p><\\/div>\"}, How to Find Least\u2010Squares Solutions Using Linear Algebra, consider supporting our work with a contribution to wikiHow. MathJax reference. wikiHow is where trusted research and expert knowledge come together. The minimum norm least squares solution is always unique. We can translate the above theorem into a recipe: Recipe 1: Compute a least-squares solution. 6Constrained least squares Constrained least squares refers to the problem of nding a least squares solution that exactly satis es additional constraints. In particular, it leads to the \"least squares\" method of fitting curves to collections of data. We use cookies to make wikiHow great. Why does regression use least \u201csquares\u201d instead of least \u201cabsolute values\u201d? Least Squares. Recall the formula for method of least squares. Does a solution with a minimal norm mean it is a solution that minimizes the residuals? In this case, we're often interested in the minimum norm least squares solution. Parameter A can also be a set of equations that describe the linear least-squares problem. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. I. Problems (PDF) Solutions (PDF) Further Study Eigenvalue Demonstrations* These demonstrations employ Java\u00ae applets with voice-over narration by Professor Strang. % In particular, it leads to the \"least squares\" method of fitting curves to collections of data. It is a set of formulations for solving statistical problems involved in linear regression, including variants for ordinary (unweighted), weighted, and generalized (correlated) residuals. This article has been viewed 4,467 times. How to draw random colorfull domains in a plane? This can be written in terms of an inequality ||y\u2212X\u03b2^||2\u2264||y\u2212X\u03b2||2,{\\displaystyle ||\\mathbf {y} -X{\\hat {\\boldsymbol {\\beta }}}||^{2}\\leq ||\\mathbf {y} -X{\\boldsymbol {\\beta }}||^{2},} where we are minimizing the distance between y{\\displaystyle \\mathbf {y} } and X\u03b2. A = np.array([[1, 2, 1], [1,1,2], [2,1,1], [1,1,1]]) b = np.array([4,3,5,4]) Next, note that minimizing $\\| b-Ax \\|_{2}^{2}$ is equivalent to minimizing $\\| b-Ax \\|_{2}$, because squaring the norm is a monotone transform. Remember when setting up the A matrix, that we have to fill one column full of ones. Many calculations become simpler when working with a \u2026 On the other hand, if the system is underdetermined, there are infinitely many solutions and thus one can find a solution of minimal norm and this is called the minimum norm solution. Magic. 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The so-called \u201c linear algebra, Fifth Edition, 2016, compute the length of the norm! With fitting a straight line to a collection of data general solution to problem... Early morning Dec 2, compute told us that this indeed gives the minimum, or I! Consider supporting our work with a contribution to wikiHow, clarification, or responding least squares solution linear algebra other answers ) Solutions PDF! Is late they evolve to mathematics Stack Exchange is a least squares estimation, looking it... Equations and orthogonal decomposition methods fact, chose the vector $y$ which forces this term to 0... Downtime early morning Dec 2, compute the length of the normal equation is a least square.. Goal to find a minimum norm least squares and computational aspects of linear equations, there... The  least squares include inverting the matrix of the solution for working matrices. View of least-squares regression \u2014 the so-called \u201c linear algebra and started to study this book is to! Have A Good Day In Italian, Nikon D5600 Refurbished Body, Canadian Maple Tree, Golden Tree Feeding Chart, Takahe Vs Pukeko, Digital Ic Design Mcq Questions, Buttermilk Yeast Donuts, \" />", "date": "2021-04-16 18:09:34", "meta": {"domain": "strouden.com", "url": "https://strouden.com/n185f/archive.php?page=least-squares-solution-linear-algebra-8f1419", "openwebmath_score": 0.6542348265647888, "openwebmath_perplexity": 1235.3437523865437, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9603611563610179, "lm_q2_score": 0.9005297881200701, "lm_q1q2_score": 0.864833828656533}}
{"url": "https://math.stackexchange.com/questions/2141022/what-does-max-refer-to-in-the-definition-of-a-matrix-norm", "text": "# What does \"max\" refer to in the definition of a matrix norm?\n\nI am going through my numerical analysis text (Epperson) and came across notation that I don't fully understand. My question is what does max refer to in this definition?\n\nDefinition $7.2$ (Matrix Norm): Let $\\|\\cdot\\|$ be a given vector norm defined on $\\mathbb{R}^n$. Define the corresponding matrix norm, for matrices $A \\in \\mathbb{R}^{n\\times n}$, by $$\\|A\\| = \\max_{x \\neq 0}\\frac{\\|Ax\\|}{\\|x\\|}$$\n\n\u2022 Can you give the exact usage in your question? Feb 12 '17 at 16:35\n\u2022 Hi David, that's what I'm trying to figure out too... the image in the link is of the definition of a matrix norm. I'm not sure how to read the \"max x != 0\" term. Feb 12 '17 at 16:37\n\u2022 It's the maximum value of the expression as the vector $x$ goes through all possible values in $\\mathbb{R}^n$ except $x=0$ since the denominator is not defined for this value. Feb 12 '17 at 16:40\n\u2022 Thanks David for the edit. I'm going to try to wrap my head around this, I'm sure it's simpler than it seems. Feb 12 '17 at 16:56\n\u2022 @Starcrossed not a problem. The concept will be easy once you wrap your mind around it. For some norms though, it might be very hard to compute. Feb 12 '17 at 17:09\n\nThis is simply the extension of a given norm to a matrix. It should be better written: $$\\sup \\left\\{\\frac{\\|Ax\\|}{\\|x\\|}: x \\in \\mathbb{R}^n, x\\neq 0\\right\\} = \\sup \\left\\{\\|Ax\\|:x \\in \\mathbb{R}^n, \\|x\\|=1\\right\\}.$$ That is, go through all normalized vectors, and see where the product $\\|Ax\\|$ is maximized.\n\n-----Edit-----\n\nAs an example, let us consider the $L^1$-norm. For some matrix $A \\in \\mathbb{R}^n$, by definition we have that $$\\|A\\|_1 = \\sup \\left\\{\\|Ax\\|_1:x \\in \\mathbb{R}^n, \\|x\\|_1=1\\right\\}.$$ Now, for a given vector $x \\in R^n$, we have that $$\\displaystyle\\|Ax\\|_1 = \\left\\|\\begin{pmatrix} a_{11} & a_{12} & \\cdots & a_{1n}\\\\ a_{21} & a_{22} & \\cdots & a_{2n}\\\\ \\vdots & \\vdots & \\ddots & \\vdots\\\\ a_{n1} & a_{n2} & \\cdots & a_{nn} \\end{pmatrix}\\cdot \\begin{pmatrix} x_1\\\\ x_2\\\\ \\vdots\\\\ x_n \\end{pmatrix}\\right\\|_1 = \\left\\|\\begin{pmatrix} \\sum_{j=1}^na_{1j}x_j\\\\ \\sum_{j=1}^na_{2j}x_j\\\\ \\vdots\\\\ \\sum_{j=1}^na_{nj}x_j \\end{pmatrix}\\right\\|_1 = \\sum_{i=1}^n\\left|\\sum_{j=1}^na_{ij}x_j\\right|.$$ Thus, we can write the $L^1$-norm of $A$ as $$\\displaystyle \\|A\\|_1 = \\sup_{\\|x\\|=1}\\left\\{\\sum_{i=1}^n\\left|\\sum_{j=1}^na_{ij}x_j\\right|\\right\\}.$$ We can bound this in the following way, $$\\|A\\|_1 \\leq \\sup_{\\|x\\|=1}\\left\\{\\sum_{i=1}^n\\sum_{j=1}^n\\left|a_{ij}x_j\\right|\\right\\} = \\sup_{\\|x\\|=1}\\left\\{\\sum_{j=1}^n\\left|x_j\\right|\\sum_{i=1}^n\\left|a_{ij}\\right|\\right\\}.$$ Now, consider the fact that $$\\displaystyle \\sum_{j=1}^n\\left|x_j\\right|\\sum_{i=1}^n\\left|a_{ij}\\right| \\leq \\sum_{j=1}^n\\left|x_j\\right|\\max_{1\\leq j \\leq n}\\left\\{\\sum_{i=1}^n\\left|a_{ij}\\right|\\right\\} = \\|x\\|_1\\cdot \\max_{1\\leq j \\leq n}\\left\\{\\sum_{i=1}^n\\left|a_{ij}\\right|\\right\\}.$$ Now, since our matrix norm is only concerned with normalized vectors, we have clearly that $$\\displaystyle \\|A\\|_1 \\leq \\sup_{\\|x\\|=1}\\left\\{\\|x\\|_1\\cdot \\max_{1\\leq j \\leq n}\\left\\{\\sum_{i=1}^n\\left|a_{ij}\\right|\\right\\}\\right\\} = \\max_{1\\leq j \\leq n}\\left\\{\\sum_{i=1}^n\\left|a_{ij}\\right|\\right\\}.$$ Therefore, we have an upper bound. But this upper bound is clearly attainable, we simply take the $j^*$ corresponding to the column we are maximizing over in the previous equation, and let $x_{j^*} = 1, x_i = 0$ ($i \\neq j^*$). Thus, the $L^1$-norm for a matrix is simply the maximum of the absolute column sums. So, as you see, the definition may be readily understood, but determining the value of the norm may be less so. And, in many cases, a closed form of the norm doesn't exist, and so you need to numerically solve the optimization (maximization) problem for each given $A$.\n\nThe \"max\" refers to the maximum of the set of numbers $\\{\\frac{\\Vert Ax \\Vert}{\\Vert x \\Vert } , 0\\neq x\\in \\mathbb{R}^n\\}$. Note that since this is a set of norms on vectors, then it's a set of non negative numbers.\n\n\u2022 Aha okay I think I get it now... basically we are finding a vector x such that we are maximizing the norm, right? Feb 12 '17 at 17:00\n\u2022 Yes. Just note that we're looking for an $x$ that maximizes the quotient of the norms $\\frac{ \\Vert Ax \\Vert }{ \\Vert x \\Vert }$\n\u2013\u00a0NSA\nFeb 13 '17 at 9:23\n\nIt means the maximum of the quantity\n\n$$\\frac{||Ax||}{||x||}$$\n\nalong all the vectors $x\\in\\mathbb{R}^n$ that are not zero.\n\n\u2022 Thanks Smurf... So to further understand, suppose we are given only one vector x for a linear expression Ax = b, then that vector alone would be the only possible vector to use in finding the max, right? Feb 12 '17 at 16:45\n\u2022 No, the norm of the matrix is defined as above. If you have an expression $Ax=b$ you still should use the maximum along every $x\\neq 0$. Feb 12 '17 at 16:47\n\u2022 I'm sure it is probably trivially simple to you so I apologize in advance for this dumb question ... wouldn't every max x \u2260 0 in \u211dn be infinitely large? Feb 12 '17 at 16:55\n\u2022 Notice that while it may seem that if you pick and arbitrary large $x$, the numerator increases, so does the denominator. So one \"controls\" the other and the maximum exists (this has to be proven, I am just giving you the idea of why it works). Feb 12 '17 at 17:16\n\u2022 @Starcrossed, when the space is infinite dimensional, the quotient can be unbounded. Mar 8 '17 at 10:14", "date": "2021-10-15 23:30:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2141022/what-does-max-refer-to-in-the-definition-of-a-matrix-norm", "openwebmath_score": 0.9278594851493835, "openwebmath_perplexity": 226.46767169150667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989347487589835, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8647661068339718}}
{"url": "https://math.stackexchange.com/questions/2389599/expected-number-of-heads-on-coin-flip", "text": "Expected number of heads on coin flip?\n\nSuppose you were to toss 2 coins each with different probabilities of landing on heads 2 times each, one at a time. So coin 1 tossed twice. Then coin 2 tossed twice. Let N be the total number of heads. To calculate E[N], would you find E[number of heads on coin 1's tosses] and E[number of heads on coin 2's tosses] and sum them together? And to find the var[N] would it be the same concept?\n\nFor expectation - yes due to linearity of expectation. As for variance, you can do the same since the each toss (or each event) are independent. Note that it does not matter whether you toss each coin 'one at a time' or the order you do it in due to independence, of course.\n\nFormally, if you let $X_1$ and $X_2$ be the indicator variables for each toss of the first coin ($1$ for heads, $0$ for tails) and the same with $Y_1$ and $Y_2$ for the tosses of the second coin, we have:\n\n$$N=X_1+X_2+Y_1+Y_2$$\n\nThen\n\n$$E(N)=E(X_1+X_2+Y_1+Y_2)$$\n\n$$=E(X_1)+E(X_2)+E(Y_1)+E(Y_2)$$\n\ndue to linearity of expectation.\n\nAlso, as $X_1,X_2,Y_1,Y_2$ are independent events, we have\n\n$$Var(N)=Var(X_1+X_2+Y_1+Y_2)$$\n\n$$=Var(X_1)+Var(X_2)+Var(Y_1)+Var(Y_2)$$\n\nThe general rules being used here are:\n\n$$E(aX+bY)=aE(X)+bE(Y)\\tag{linearity of expectation}$$\n\nand if $X$ and $Y$ are independent,\n\n$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)$$\n\n\u2022 I just have a general question. I had a practice problem like this but with given probabilities for each coin. Let's assume N=X+Y. X is number of heads on coin 1, and Y is the number of heads on coin 2. I tried calculating the variance by doing E[(X+Y)^2] - E^2[(X+Y)] but received a negative number. But when I calculated the variances of each coin individually and then summing it, I received a positive number. Why does this happen? Basically I'm asking, why doesn't E[(X+Y)^2] - E^2[(X+Y)] =var(X)+var(Y)? Aug 10, 2017 at 21:18\n\u2022 Could you give the probabilities in particular or edit your question to include your calculation? Aug 10, 2017 at 21:23\n\u2022 Never mind, I see what I did wrong. When squaring E[(X+Y)^2], I did not calculate it correctly. Aug 10, 2017 at 21:35\n\u2022 Letting the probabilities be $p_1$ and $p_2$, we have $E[(X+Y)^2] - (E[X+Y])^2=(0\\times(1-p_1)^2(1-p_2)^2+1\\times(1-p_1)^2p_2(1-p_2)+4\\times(1-p_1)^2p_2^2+1\\times p_1(1-p_1)(1-p^2)^2+4\\times p_1(1-p_1)p_2(1-p_2)+9\\times p_1(1-p_1)p_2^2+4\\times p_1^2(1-p_2)^2+9\\times p_1^2p_2(1-p_2)+16\\times p_1^2p_2^2)-4(p_1+p_2)^2$ Aug 10, 2017 at 21:35\n\nLong answer: It sounds like the two coins in your example do not affect each other in any way. In other words, it sounds like the number of heads on one is independent of how many heads happen for the other. For any two random variables, whether independent or not, we do indeed have that $E[X+Y]=E[X]+E[Y]$. Furthermore, when the variables are independent, we also have that $E[XY]=E[X]E[Y]$. This allows us to reason as follows. \\begin{align} Var[X+Y] &= E[((X+Y)-E(X+Y))^2]\\\\ &=E[(X-E[X]+Y-E[Y])^2]\\\\ &=E[(X-E[X])^2] +2E[(X-E[X])(Y-E[Y])] + E[(Y-E[Y])^2]\\\\ &=Var[X] + Var[Y] + 2 E[XY -XE[Y] -YE[X] + E[X]E[Y]]\\\\ &= Var[X] + Var[Y] + 2(E[XY] -E[X]E[Y] - E[X]E[Y] + E[X]E[Y])\\\\ &= Var[X] + Var[Y] + 2(E[X]E[Y] -E[X]E[Y] - E[X]E[Y] + E[X]E[Y])\\\\ &= Var[X] + Var[Y] +2 \\cdot 0\\\\ &= Var[X] + Var[Y] \\end{align} So the variance of the sum of independent random variables is the sum of their variance.", "date": "2022-05-20 02:24:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2389599/expected-number-of-heads-on-coin-flip", "openwebmath_score": 0.9893569946289062, "openwebmath_perplexity": 575.2782302266263, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765628041175, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.864758632035714}}
{"url": "http://math.stackexchange.com/questions/43231/implicit-differentiation-misunderstanding/43235", "text": "# Implicit differentiation misunderstanding\n\nI'm trying to see why my textbook's solution is correct and mine isn't.\n\n\"Find an expression in terms of $x$ and $y$ for $\\displaystyle \\frac{dy}{dx}$, given that $x^2+6x-8y+5y^2=13$\n\nFirst, the textbook's solution, which I understand and agree with fully:\n\nNow my similar solution, for which I don't see my error:\n\nDifferential: $$2x+6-8\\frac{dy}{dx}+10y\\frac{dy}{dx}=0$$ $$\\frac{dy}{dx}(10y-8)=-2x-6$$ $$\\frac{dy}{dx}=\\frac{-2x-6}{10y-8}=\\frac{-x-3}{5y-4}$$\n\nSo I end up with the negative of the correct solution, because I moved the $(2x+6)$ to the RHS and the textbook author moved the other part instead. I would have thought it would produce an equivalent answer?\n\nThanks!\n\n-\nHINT $\\rm\\displaystyle\\ \\ \\frac{-A}{-B}\\ =\\ \\frac{A}B\\qquad$ \u2013\u00a0 Bill Dubuque Jun 4 '11 at 18:14\nAs you now know, your answer is correct. But there is kind of a convention in writing mathematical expressions, that fewer minus signs is generally better. \u2013\u00a0 Andr\u00e9 Nicolas Jun 4 '11 at 18:22\n@user6312 duly noted, thanks! :) \u2013\u00a0 Danny King Jun 4 '11 at 18:24\n\nFractions can be written multiple ways.\n\n$$\\frac{-x-3}{(5y-4)} = \\frac{x+3}{4-5y}$$\n\nIn general $$\\frac{a-b}{c-d}=\\frac{b-a}{d-c}$$\n\nThis is just multiplying both the top and the bottom by \u22121. In other words, your answer and the books differ by multiplication of $$\\frac{-1}{-1} = 1$$\n\n-\nOh! When you put it like that, it becomes very clear. Thank you! \u2013\u00a0 Danny King Jun 4 '11 at 18:05\n\nYour answers agree. Note that: $$\\frac{-x-3}{5y-4} = \\frac{-(x+3)}{-(-5y+4)} = \\frac{x+3}{4-5y}.$$\n\n-\n\nI don't know, what's wrong with your solution. Notice, that\n\n$${-x-3\\over5y-4}={3+x\\over4-5y}$$\n\n-", "date": "2015-07-06 07:14:31", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/43231/implicit-differentiation-misunderstanding/43235", "openwebmath_score": 0.8099761009216309, "openwebmath_perplexity": 843.016023547801, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765563713599, "lm_q2_score": 0.8824278695464501, "lm_q1q2_score": 0.8647586248442458}}
{"url": "http://math.stackexchange.com/questions/253257/number-of-simple-edge-disjoint-paths-needed-to-cover-a-planar-graph", "text": "# Number of simple edge-disjoint paths needed to cover a planar graph\n\nLet $G=(V,E)$ be a graph with $|E|=m$ of a graph class $\\mathcal{G}$. A path-cover $\\mathcal{P}=\\{P_1,\\ldots,P_k\\}$ is a partition of $E$ into edge-disjoint simple paths. The size of the cover is $\\sigma(\\mathcal{P})=k$. I am interested in upper and lower bounds for the quantity\n\n$$\\max_{G\\in \\mathcal{G}} \\quad \\min_{\\mathcal{P} \\text{ is path-cover of G}} \\sigma(\\mathcal{P})/m.$$\n\nIn other words, how large is the inverse of the average path length in the smallest path-cover in the worst case.\n\nThe graphs classes $\\mathcal{G}$ I am interested in are (1) planar 3-connected graphs, and (2) triangulations.\n\nThere is a simple observation for a lower bound: In every odd-degree vertex one path has to start/end. So when all vertices have odd degree there any path-cover needs at least $m/6+1$ paths (a triangulation has $3|V|-6$ edges). You get the same result by noticing, that $(n-1)/2$ paths have to pass through a vertex of degree $n-1$.\n\nIs a path-cover with $\\frac{m}{6}$ paths always possible for planar graphs?\n\nI am aware of a few results covering planar graphs with paths of length $3$.\n\nHere is an example of a path-cover of the graph of the icosahedron.\n\n-\nThis is a great question; don't know why it hasn't received more attention or upvotes. \u2013\u00a0 joriki Dec 12 '12 at 10:10\nI don't understand the lower bound. Since a triangulation has $m=3n-6$ edges, a triangulation with all odd-degree vertices requires at least $n/2=(m+6)/6=m/6+1$ paths. For instance, $K_4$ needs $2$ (not $1$) paths to cover it. \u2013\u00a0 mjqxxxx Dec 13 '12 at 18:28\nI am not sure if I would be able to do it, but somebody should try generating a big list of $\\sigma(\\mathcal{P})$'s for a bunch of the graphs in question. Maybe some data will help us see a pattern. \u2013\u00a0 Alexander Gruber Dec 19 '12 at 2:40\n@A.Schulz Maybe the lower bound $m/6$ is weak for some classes of graphs. For instance, for the wheel graphs $W_n$, consisting of the cycle of length $n-1$ and the additional center vertex joined with all vertices of the cycle. Also there are Moon-Moser graphs $T_k$ (see Section 2 of the article \u201cLong Cycles in 3-Connected Graphs\u201d by Guantao Chen and Xingxing Yu), having no cycles of length greater than $(7/2) n^{\\log_3 2}$. Maybe these graphs also have no small path-covers. \u2013\u00a0 Alex Ravsky May 27 '13 at 3:47\nA related question is that of Linear Arboricity, but the results for it are based upon the maximum degree of vertices. See for instance mimuw.edu.pl/~kowalik/papers/LinearArboricity.pdf \u2013\u00a0 jp26 Jan 27 at 21:26\n\nI will show that $\\frac{m}{6}$ is not enough for 3-connected planar graphs. The following is inspired by fedja's comment (and also some nice discussion with Krystal Guo).\n\nConsider the prism graph, which we will denote $P_n$. We say it has $2n$ vertices and $3n$ edges. Then in this case $\\frac{m}{6} = \\frac{n}{2}$. We will show that $\\sigma(P_n) =n$. It is clear that $n$ paths is enough.\n\nConsider the following graphs which we denote $T_n$:\n\nand so on.. Note that $T_n$ has $3n - 1$ edges and $2n+2$ vertices.\n\nIt is not too hard to show that $\\sigma(T_n) = n+1$ (and this is the best possible). One can show by induction, considering the left most edge between the upper and lower paths.\n\nLet $p_1 , \\ldots , p_k$ be a path cover of $P_n$. Now $P_n$ is two copies of $C_n$ with a matching, $M$, in between them. If every $e \\in M$ is actually some $p_i$, then we have $k \\geq n+2$.\n\nSo we may assume there is a $e = (u,v)$ such that the path, $p_i$, that contains $e$, contains at least one more edge, wolog say $(v,w)$.\n\n$(i)$ The vertex $u$ is also contained in another edge of $p_i$ (other than $e$). Then $(V(P_n) , E(P_n) \\setminus E(p_i))$ is $T_n$ minus at most two paths (actually, exactly two: the two paths start with precisely the two edges incident to $e$, respectively). In this case $k \\geq n+1 - 2 + 1 = n$ (the last $+1$: don't forget to count $p_i$).\n\n$(ii)$ The vertex $u$ is not contained in any other edge of $p_i$. Then let $p_j$ be another path that contains $u$. Then $(V(P_n) , E(P_n) \\setminus (E(p_i)\\cup E(p_j))$ is a $T_n$ minus at most 3 paths ($p_i$ is one path in $T_n$ and $p_j$ is at most two paths in $T_n$). It follows that $k \\geq n + 1 - 3 +2= n$.\n\nThus we obtain $\\sigma(P_n) = n$, as desired.\n\n-", "date": "2014-07-13 15:27:54", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/253257/number-of-simple-edge-disjoint-paths-needed-to-cover-a-planar-graph", "openwebmath_score": 0.9259254932403564, "openwebmath_perplexity": 219.21999760607332, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765557865637, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8647586122080159}}
{"url": "https://web2.0calc.com/questions/im-stuck-in-this-koobits", "text": "+0\n\nim stuck in this koobits !\n\n0\n412\n5\n\nFind the sum of the following series by pairing two numbers to make ten(s). 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19\n\nGuest\u00a0Nov 20, 2015\n\n#2\n+18927\n+15\n\nFind the sum of the following series by pairing two numbers to make ten(s).\n\n1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19\n\n$$\\begin{array}{rcrcrcrcrcrcrcrcrcrcr} \\hline \\hline 1 &+& 3 &+& 5 &+& 7 &+& 9 &+& 11 &+& 13 &+& 15 &+& 17 &+& 19 & \\\\ \\hline \\hline 1 & & & & & & & & & & & & & & & & &+& 19 & =& 20 \\\\ \\hline & & 3 & & & & & & & & & & & & &+& 17 & & & =& 20 \\\\ \\hline & & & & 5 & & & & & & & & &+& 15 & & & & & =& 20 \\\\ \\hline & & & & & & 7 & & & & &+& 13 & & & & & & & =& 20 \\\\ \\hline & & & & & & & & 9 &+& 11 & & & & & & & & & =& 20 \\\\ \\hline \\hline & & & & & & & & & & & & & & & & & & & =& 100\\\\ \\hline \\hline \\end{array}$$\n\n( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) = 20 + 20 + 20 + 20 + 20 = 100\n\nheureka \u00a0Nov 20, 2015\nedited by heureka \u00a0Nov 20, 2015\nedited by heureka \u00a0Nov 20, 2015\nSort:\n\n#1\n+5\n\n(1+9) + (3+7) + (5+15) + (7+13) + (11+19)\n\n10 + 10 + 20 + 20 + 30\u00a0 = 90\n\nGuest\u00a0Nov 20, 2015\n#2\n+18927\n+15\n\nFind the sum of the following series by pairing two numbers to make ten(s).\n\n1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19\n\n$$\\begin{array}{rcrcrcrcrcrcrcrcrcrcr} \\hline \\hline 1 &+& 3 &+& 5 &+& 7 &+& 9 &+& 11 &+& 13 &+& 15 &+& 17 &+& 19 & \\\\ \\hline \\hline 1 & & & & & & & & & & & & & & & & &+& 19 & =& 20 \\\\ \\hline & & 3 & & & & & & & & & & & & &+& 17 & & & =& 20 \\\\ \\hline & & & & 5 & & & & & & & & &+& 15 & & & & & =& 20 \\\\ \\hline & & & & & & 7 & & & & &+& 13 & & & & & & & =& 20 \\\\ \\hline & & & & & & & & 9 &+& 11 & & & & & & & & & =& 20 \\\\ \\hline \\hline & & & & & & & & & & & & & & & & & & & =& 100\\\\ \\hline \\hline \\end{array}$$\n\n( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) = 20 + 20 + 20 + 20 + 20 = 100\n\nheureka \u00a0Nov 20, 2015\nedited by heureka \u00a0Nov 20, 2015\nedited by heureka \u00a0Nov 20, 2015\n#3\n+91773\n0\n\nWhat a fabulous diagram Heureka,\n\nMelody \u00a0Nov 20, 2015\n#4\n0\n\nTwo great answers, but one of them must be wrong.\n\nThe first says 90, the second 100.\n\nGuest\u00a0Nov 20, 2015\n#5\n0\n\nYep One of them is wrong....and it's my answer\u00a0\u00a0\u00a0 the answer is 100...I made a typo. The 4th term in my sequence should read (17+13)\u00a0\u00a0\u00a0 NOT\u00a0\u00a0 (7 +13)\n\nGuest\u00a0Nov 21, 2015\n\n5 Online Users\n\nWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. \u00a0See details", "date": "2018-02-19 05:48:38", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/im-stuck-in-this-koobits", "openwebmath_score": 0.726304829120636, "openwebmath_perplexity": 605.0887748696415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551505674444, "lm_q2_score": 0.8962513745192026, "lm_q1q2_score": 0.8647527549080042}}
{"url": "https://wikimho.com/us/q/astronomy/6384", "text": "### How fast is a comet moving when it crosses Earth's orbit?\n\n\u2022 Is it about the same as Earth's orbital speed?\n\nIt depends. An object's speed is related to the length of it's orbit, and this varies greatly. See Kepler's Laws.\n\n\u2022 Walter Correct answer\n\n8 years ago\n\n1. Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.\n\n2. What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy\n$$E = \\frac{1}{2}v^2 - \\frac{GM_\\odot}{r},\\qquad\\qquad(*)$$\nwhich is conserved along the orbit ($v$ and $r$ the Heliocentric speed and distance). For a returning comet, $E<0$ and the speed cannot exceed the escape speed (which occurs for $E=0$)\n$$v_{\\rm escape}^2 = 2\\frac{GM_\\odot}{r}.$$\nThe speed of the Earth can be worked out from the Virial theorem, according to which the orbital averages of the kinetic and potential energies, $T=\\frac{1}{2}v^2$ and $W=-GM_\\odot/r$, satisfy\n$2\\langle T\\rangle + \\langle W\\rangle=0.$\nFor a (near-)circular orbit (such as Earth's), $r$ is constant and we have\n$v^2_{\\rm Earth} = GM_\\odot/r.$\nThus, at $r$=1AU\n$$v_{\\rm escape} = \\sqrt{2} v_{\\rm Earth}$$\nas already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.\n\n3. Can a comet near Earth have a speed similar to Earth's orbital speed?. Let's assume a comet with the same speed as Earth at $r$=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since\n$$E = -\\frac{GM_\\odot}{2a}\\qquad\\qquad(**)$$\nwith $a$ the orbital semimajor axis, must also have $a=1AU$ and the same orbital period as Earth, i.e. one year. Moreover, the comet's apohelion satisfies\n$$r_{\\rm apo}\\le 2a = 2{\\rm AU}.$$\nSuch comets don't exist AFAIK. Most returning comets have much longer periods than 1 year.\n\n4. What's the typical speed of a returning comet when at distance 1AU from the Sun? To work out this question, let's parameterise the comet's orbit by its period $P=2\\pi\\sqrt{a^3/GM_\\odot}$. From this relation we immediately get\n$$\\frac{a_{\\rm comet}}{\\rm AU} = \\left(\\frac{P_{\\rm comet}}{\\rm yr}\\right)^{2/3}.$$\nFrom equations ($*$) and ($**$), we can then find\n$$v_{\\rm comet}(r=1{\\rm AU}) = \\sqrt{2-\\left(\\frac{P_{\\rm comet}}{\\rm yr}\\right)^{-2/3}}\\,v_{\\rm Earth}.$$\nIn the limit of $P\\to\\infty$, this recovers our previous result $v_{\\rm comet}\\to v_{\\rm escape}$. For typical period of $\\sim$70yr, the speed of the comet is close to this number.\n\n5. Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at $r$=1AU, move in quite a different direction than Earth, even if their speed is only slightly larger. So the relative speed $|\\boldsymbol{v}_{\\rm comet}-\\boldsymbol{v}_{\\rm Earth}|$ of a comet with respect to Earth can be anyting between about 10 and 70 km/s.\n\nThis is a great answer! You're better than Wolfram! This is exactly the information and tools that I was looking for. Thanks very much. Doug.\n\nwho is Wolfram?\n\n@Walter the answer to \"**Who** is Wolfram\" is either Stephen Wolfram or perhaps Wolfram Research. But of course user38715 was referring to wolframalpha.com\n\nLicense under CC-BY-SA with attribution\n\nContent dated before 7/24/2021 11:53 AM\n\n\u2022 {{ error }}", "date": "2022-10-07 22:03:18", "meta": {"domain": "wikimho.com", "url": "https://wikimho.com/us/q/astronomy/6384", "openwebmath_score": 0.844038724899292, "openwebmath_perplexity": 1262.2160033230757, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551505674444, "lm_q2_score": 0.8962513682840824, "lm_q1q2_score": 0.8647527488920164}}
{"url": "http://math.stackexchange.com/questions/215515/solution-of-x2-sx-cdot-x-n-0-with-sx-is-the-sum-of-digits-of-x/215701", "text": "# Solution of $x^2 + s(x)\\cdot x - n = 0$, with $s(x)$ is the sum of digits of $x$.\n\nThis problem comes from an programming competition website, but I'd interested in analyze it from mathematics prespective. Given this problem below, we must create a program that could give us the correct output from the given input. The program must run under the time limit that provided by the system.\n\n## The Problem\n\nLet define function $s$ that take input from positive integers, such that $s(x)$ is the sum of the digit of $x$.\n\nFor example, $$s(1024) = 1 + 0 + 2 + 4 = 7$$\n\nGiven some integer constant $n$ as the input of the program. Now, find positive integer solution of $x$ in equation\n\n$$x^2 + s(x) \\cdot x - n = 0.$$\n\nIf more than one solution exist, choose the lowest one. If the solution doesn't exist, give output -1.\n\nTime Limit of the program: 2 second.\n\nRange of the input $n$: $[1, 10^{18}]$\n\n## The Solution\n\nIn finding the solution, we could iterate from some range of possible value of x, and try each of these value whether it satisfy the equation or not.\n\nTo improve the run time of the program, we must choose the shortest range for x.\n\n## My Question\n\n1. What is the shortest range of possible value of x that we could iterate to find the answer of the problem above?\n\n2. Is there a better way to find the solution without iterating from the set possible value of x?\n\n## My Attempt:\n\nConsider that $$x^2 + s(x) \\cdot x - n = 0$$ $$\\Leftrightarrow x ( x + s(x) ) = n$$\n\nand since both $x$ and $(x + s(x))$ are positive integers and the product of both number is $n$, we can conclude that $x$ and $(x + s(x))$ is a pair of positive factor of $n$.\n\nSince $x$ is a factor of $n$, we now have the first possible range of $x$, that is $[1, n]$.\n\nBut, iterating $x$ over this range gives bad performance to the program. For small value of $n$, it won't matter, but for $n \\geq 10^{17}$, the program will reach it's time limit.\n\nIf I could recall correctly, we have this theorem\n\n$$x_1 \\cdot x_2 = x$$\n\nand\n\n$$x_1 \\leq x_2$$\n\nimplies that\n\n$$x_1 \\leq \\sqrt{x} \\leq x_2$$\n\nAnd since $x (x + s(x)) = n$ and $x \\leq x + s(x)$\n\nwe have $x \\leq \\sqrt{n} \\leq x + s(x)$\n\nNow we have the second range: $[1, \\sqrt{n}]$\n\nBut, still, this range also give bad performance for larger $n$.\n\nMy last attempt is by considering the biggest possible value of $s(x)$, that is the digit of $x$ is all $9$ and $x$ less than or equals to the maximum value of $n$, $10^{18}$.\n\n$$s(x) \\leq s(999,999,99 \\cdots 9,999) = 9 \\cdot 18 = 162$$\n\nHence, we have\n\n$$x + s(x) \\leq x + 162$$\n\nso that, $$\\sqrt{n} \\leq x + s(x) \\leq x + 162 \\leq \\sqrt{n} + 162$$ hence, $$\\sqrt{n} - 162 \\leq x \\leq \\sqrt{n}$$\n\nNow, we have better range that will suffice for all value of $x$. $$l_1 = \\max \\{ 1, \\sqrt{n} - 162 \\}$$ $$l_2 = \\sqrt{n}$$ the range is $[l_1, l_2]$\n\nBut, I though there must be a better solution to this problem. Hence I try $$l_1 = \\max \\{ 1, \\sqrt{n} - 162/2 \\}$$ And that range is also correct. But I can't justify my finding.\n\nSo, hence my third question:\n\n1. Why $l_1 = \\max \\{ 1, \\sqrt{n} - 162/2 \\}$ also gives us correct solution?\n-\n\nWhile not a full answer to your questions, here is a trick that will allow you to reduce the collection of numbers you have to try (and so an answer to question 2).\n\nBecause $10\\equiv 1 \\pmod 9$, we have $10^k \\equiv 1 \\pmod 9$ for every $k\\in \\mathbb N$. If we write a number in terms of its digits, $x=\\sum a_k 10^k$, then since $\\sum a_k 10^k \\equiv \\sum a_k \\pmod 9$, we have $s(x)\\equiv x \\pmod 9$. This fact can be used to give a way to check arithmetic, among other things.\n\nTherefore, if $x^2 + s(x)x = n$, then $2x^2 \\equiv n \\pmod 9$. Since $(2)(5)=10 \\equiv 1 \\pmod 9$, we can multiply to get $x^2 \\equiv 5n \\pmod 9$.\n\nThe squares modulo 9 are $0, 1, 4, 7$, so unless the remainder of dividing $5n$ by $9$ is one of these numbers, there are NO solutions.\n\nIf the remainder is one of those numbers, you need to know the square roots modulo 9 to know what values of $x$ to try.\n\nFor this, we have the following.\n\n$0^2\\equiv 3^2 \\equiv 6^2 \\equiv 9 \\pmod 9$, $1^2\\equiv 8^2 \\equiv 1 \\pmod 9$, $2^2\\equiv 7^2 \\equiv 4 \\pmod 9$ and $4^2 \\equiv 5^2 \\equiv 7 \\pmod 9$.\n\nThe upshot of this is that if $n$ is a multiple of $9$, then you only have to check the multiples of $3$ (reducing the number of things you have to check by a factor of $3$). Otherwise, depending on the modulus of $5n\\bmod 9$, you either have to check NO numbers (5 of the remaining cases), or you have to check $2/9$ of the numbers (3 of the remaining cases). Thus, whatever range you are working in, you have at least a speed up by a factor of $3$, if not more.\n\nAnother comment, which gives a small speedup over what you already have but answers question 3. As you observed, $x<\\sqrt n$, and the number of digits of $x$ is at most $\\lceil \\log_{10} x \\rceil$, and since $s(x)$ is at most $9$ times the number of digits of $x$, we have $s(x)<9 \\lceil \\log_{10} \\sqrt n \\rceil=9\\lceil \\frac{1}{2}\\log_{10} n \\rceil$.\n\nTherefore, given $n$, you can limit your search for $x$ to $[\\sqrt n - 9\\lceil \\frac{1}{2}\\log_{10} n \\rceil, \\sqrt n]$. The factor of $1/2$ here answers question 3. Note, we could probably shave off $1$ or $2$ numbers from the range by being more careful with rounding, but I don't see any obvious reason we could shave off more. If we combine this with the beginning of my answer, you have to check at most $3\\lceil \\frac{1}{2}\\log_{10} n \\rceil$ numbers to find your solution, if it exists, and no numbers a good portion of the time when no solution exists.\n\n-\nI think $(2)(5)=10 \\equiv 1 \\pmod 5$ in pharagraph 3 should be $(2)(5)=10 \\equiv 1 \\pmod 9$. \u2013\u00a0 Kamal Hajjaj Isa Oct 28 '12 at 14:58\n@KamalHajjajIsa: Yes, that is correct. Thank you. \u2013\u00a0 Aaron Oct 29 '12 at 3:07\nI like the Idea of using $\\pmod 9$. \u2013\u00a0 Kamal Hajjaj Isa Oct 29 '12 at 4:56\n\nSince $x<\\sqrt n$ and $n\\le10^{18}$, we have that $x<10^9$. Then $x$ has at most $9$ digits and $s(x)\\le9\\times9=81$. This implies that $x^2+81\\,x-n\\ge0$. It follows easily that we must have $$x\\ge\\frac{-81+\\sqrt{81^2+4\\,n}}{2}=\\sqrt{n+(81/2)^2}-81/2.$$\n\n-\n\nThis only answers the third question\n\nIf you use the general quadratic equation formula for the equation $ax^2+bx+c=0$ $$\\displaystyle x= \\frac {-b \\pm \\sqrt{b^2-4ac}}{2}$$ It follows from $x^2 + s(x)\\cdot x - n = 0$ that $$\\displaystyle x= \\frac {-s(x) \\pm \\sqrt{s^2(x)+4n}}{2}$$ Since all the variables are positive and obviously $\\sqrt{s^2(x)+4n}>s(x)$ , one can remove the negative sign and have $$\\displaystyle x= \\frac {-s(x) + \\sqrt{s^2(x)+4n}}{2}$$ If one extends the range of values of $n$ to $0$, we have that the lowest possible value of $s(x)$ is $0$ and that $s(x) \\ge 0$ which implies that $$\\displaystyle x \\ge \\frac {-s(x) + \\sqrt{4n}}{2} = \\frac {-s(x)}{2} + \\frac {2\\sqrt{n}}{2} \\ge \\frac {-162}{2} + \\sqrt n, \\space \\space (-s(x) \\ge -162)$$ and there you have your inequality $$x \\ge \\sqrt n - \\frac {162}{2}$$\n\nExtending the range to $0$ doesn't change the validity of the inequality and helps to simplify it. You've been working with the maximum value of $s(x)$, using the minimum value helps in this case.\n\n-\nBeside the excellent modulo 9 tricks Aaron gave you, if $n$ passes his tests the easiest thing is to try the range of $1 \\le s(x) \\le 162$ with the proper $s(x) \\pmod 9$. Just plug each one into the quadratic formula and solve for $x$. If it doesn't come out integral, go on to the next. If it does, check the sum of digits against the assumed $s(x)$. If it matches, success and report it. If not, go on to the next. At most $50$ numbers to try, which should fit in 2 seconds easily.", "date": "2013-12-21 03:49:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/215515/solution-of-x2-sx-cdot-x-n-0-with-sx-is-the-sum-of-digits-of-x/215701", "openwebmath_score": 0.8677693009376526, "openwebmath_perplexity": 155.7791978875511, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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{"url": "http://mexproject.it/zcjm/probability-problems-and-solutions.html", "text": "# Probability Problems And Solutions\n\n84 = (x \u2013 70)/2. com - View the original, and get the already-completed solution here! PRACTICE PROBLEMS: 1. In this problem, there is a tendency to reason that since the opposite face is either heads or tails, the desired probability is 1/2. , the probability of getting the first two spheres as blue while the third sphere is red is same as the probability of getting the second and third spheres as blue while the first sphere is red As Jennifer pointed out, Baye's theorem can be used. 01, while a sli ghtly defective part fails within the \ufb01rst year with probability 0. Solution: Gambler\u2019s Ruin Problem. This is, in fact, incorrect. We can formulate this as E= 1 6 + 5 6 (E+ 1): (3. Experiment 1: A spinner has 4 equal sectors. The first section provides a brief description of some of the basic probability concepts that will be used in the activities. The probability that a red or blue marble will be selected is 9/14. The problems in this collection are drawn from problem sets and exams used in Finance Theory I at Sloan over the years. This Collection of problems in probability theory is primarily intended for university students in physics and mathematics departments. the mean value of the binomial distribution) is. 30, for which the solution is 29%) with the factorial function like so:. International Mathematics Olympiad. It follows that the probability of an ace or a spade = 4+13 52 = 17 52. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. A Problem With Pearls. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and business students. This is an exercise in manipulating conditional probabilities. Devore Contents. The simple power of multiplication will help you unravel the mysteries of compound probability. It is also considered for the case of conditional probability. Get Answer to Using the probability distribution given in Problem B--8, calculate the probability that the particle will be found between 0 and a/ 2. Z for 90th percentile=1. (a) Show that if X and Y are conditionally independent given Z and X and Z are conditionally independent given Y then X is independent of (Y,Z). Probability Distributions for Discrete Random Variables Probabilities assigned to various outcomes in the sample space S, in turn, determine probabilities associated with the values of any particular random variable defined on S. The probability equals 31,8 %. Although they never finished their game, their letters contain solutions to both problems. Calculate the probability that if somebody is \u201ctall\u201d (meaning taller than 6 ft or whatever), that person must be male. 30, for which the solution is 29%) with the factorial function like so:. Step 1: Understanding what the Table is Telling you: The following Contingency Table shows the number of Females and Males who each have a given eye color. The Probability section is explained with Practice and Learn questions and answers for interview, competitive examination and entrance test. In other words, the probability of winning for a player is the ratio of the number of coins the player starts with to the total numbers of coins. The probability (chance) is a value from the interval 0;1> or in percentage (0% to 100%) expressing the occurrence of some event. 25, Standard dev 4. Again we rely on the classic probability ratio, comparing favorable cases to the total number of cases. This book will help you learn probability in the most effective way possible - through problem solving. A standard. Now the sample set reduces and. Practice exam 1 and solution Practice exam 2 and solution Midterm 1 solutions Midterm 2 practice exams: Practice exam 1 and solution Practice exam 2 and solution Topics for Midterm 2: All of Chapter 3, Sections 1-3 of Chapter 4, Sections 1-3 of Chapter 5, Sections 1,2,3,4,6,7,9 of Chapter 6. Table 4 Brown eyes Not brown eyes Tota ls Black hair 50 30 Red hair 80 150 Totals 120 230 (a) What is the probability that someone with black hair has brown eyes?. Exercises, Problems, and Solutions problem, determine the probability for observing a z-component of angular momentum equal to 1h-if the state is given by the L x eigenstate with 0h-L x eigenvalue. The concept is very similar to mass density in physics: its unit is probability per unit length. So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7. The book is available as a single pdf file or by individual chapters. Colavito\u2019s homepage for solutions. Title: Probability Problems With Solutions Author: persepolis. The outcome e 1 is four times as likely as each of the three remaining outcomes. C: The outcomes of previous rolls do not affect the outcomes of future rolls. probability: The measure of how likely it is for an event to occur. We started learning about Probability from Class 6,we learned that Probability is Number of outcomes by Total Number of Outcomes. Behind one is a expensive car, but behind the other two are goats. However, any rule for assigning probabilities to events has to satisfy the axioms of probability: theoretical probability. Cards of Spades and clubs are black cards. The problem concerns a game of chance with two players who have. (b)[7 points] What is the probability that the alarm clock goes o between. Press [2] to evaluate a permutation or press [3] to evaluate a combination. PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Download it once and read it on your Kindle device, PC, phones or tablets. 1446 problems in 21 years. Solved Probability Problems. Problems and Solutions worked out in detail on the topic probability in mathematics or quantitative techniques or quantitative methods. It is an autosomal trait. The problems in this collection are drawn from problem sets and exams used in Finance Theory I at Sloan over the years. Probability has been defined in a varied manner by various. Under Which Cup? Loaded Revolver. the number of ways of arranging the letters of the word \u2018NEEDLESS\u2019) restriction (e. Test your understanding with practice problems and step-by-step solutions. (a) Show that if X and Y are conditionally independent given Z and X and Z are conditionally independent given Y then X is independent of (Y,Z). This question is addressed by conditional probabilities. Answers and links to explanations to these these GMAT probability problems are at the end of set. Get Textbook Solutions and 24/7 study help for Statistics And Probability Step-by-step solutions to problems over 34,000 ISBNs Find textbook solutions Close. It is the ratio of the number of ways an event can occur to the number of possible outcomes. Complete Solutions Manual to Probability by Jim Pitman. Chegg Solution Manuals are written by vetted Chegg Statistics And Probability experts, and rated by students - so you know you're getting high quality answers. \u00b7 = = Answer: 2: A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls. We will discuss difference probability word problems with the solved problems online We follow the following steps for solving a given word problem in probability. Let S be the sample and let E be an event. AU - Sparrow, E. For instance, let's say the probability that you will get to class on time is 2/3. phone numbers contain a 3-digit area code, followed by a 3-digit exchange code and end in a 4\u00addigit subscriber number. Please do not email me completed problem sets. He offers you the following game. They are created by many instructors of the course, including (but not limited to) Utpal Bhattacharya, Leonid Kogan, Gustavo Manso, Stew Myers, Anna Pavlova, Dimitri Vayanos and Jiang Wang. P(4\u2212a \u2264 x \u2264 4+a) = 0. Solution: Let\u2019s think of this like a decision tree. However, you must write up your problem set individually. Information Theory and Coding: Example Problem Set 2 1. Substituting in values for this problem, x = 5. NCERT Solutions For Maths Class 12 Chapter 13 - Probability comprises of various important questions for exams. TOPIC \u2013 ELECTROMAGNETICS. Make the mutation dominant. (6) Problem 1. Numerical Methods: Problems and Solutions \u00a0By M. ECE 316 { Probability Theory and Random Processes Chapter 7 Solutions (Part 1) Xinxin Fan Problems 3. If we are in try T and state is 1, then probability of next stop at 2 in the same try T is denoted as:. Probability worksheets for kids from grade 4 and up include probability on single coin, two coins, days in a week, months in a year, fair die, pair of dice, deck of cards, numbers and more. Homework problems usually do not say which concepts are involved, and often require combining several concepts. 95, Kindle/pdf (143 pages) $9. Class 12 Probability is all about understanding and concepts. You must turn them in to me via a physical copy. Montgomery, George C. Probability Questions and answers PDF with solutions. to share Solutions for all STEM major Problems. Long chapters are logically split into numbered subchapters. Two and three-digit subtraction. Probability - math word problems Probability is the measure of the likeliness that an event will occur. 13% probability that exactly 7 of 10 IT startups will generate a profit in their first year when the probability of profit in the first year for each startup is 80%. Fifty marbles are to be drawn from the jar in problem #1 with replacement. Probabilities, Counting, and Equally Likely Outcomes. Evans and Jeffrey S. The bet can either be on red or on blue and the amount of each bet is arbitrary. Thus, the probability that a randomly chosen piece is highlighted is 1/2. Work on the probability problems on pages 1-4 and 1-5 of the lab manual 2. The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. A pair of fair, standard dice are rolled. Solution to Problem 1. ECE302 Spring 2006 HW5 Solutions February 21, 2006 3 Problem 3. The correct reasoning is to calculate the conditional probability. SolutionStep 1 of 5We have to verify the given function is a probability mass function or notAnd we have to find the following probabilitiesHere each probability value is greater. As with any quantum system, the wave functions give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. Probability. And the probability of getting a head is equal to the probability of getting a tail. 316 that an audit of a retail business. nclearly there are 12 possible outcomes (6 x 2) nthere are 3 possible outcomes for an odd die and a head. Probability of Mutually Exclusive Events: If two events, A and B, are mutually exclusive, then the probability that either A OR B occurs is\u2026 P(A or B) = P(A) + P(B) Mutually Exclusive Events: If two events, A and B, are mutually exclusive, then that means that if A occurs, than B cannot, and vice versa. Then, n(E) = 2. Make sure you understand your probability questions is all we can say. Please be aware, however, that the handbook might contain, and almost certainly contains, typos as well as incorrect or inaccurate solutions. Q x = P(female lives to age x) = number of female survivors at age x 100,000. Calculating probability with percentages is a common topic learned in the K-12 years and is useful throughout your life. Solution of exercise 1. The remaining sections each investigate a different game, providing an overview of the rules, some discussion of basic strategy, and a series of problems and solutions. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Trials until first success. As with any quantum system, the wave functions give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. Then, we'll teach you all of the basic rules and formulas you need to know as well as give you tips for studying probability and approaching GRE probability questions on test day. Probability Questions and answers PDF with solutions. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. Question 1: Find the probability of rolling a '3 with a die. Thus, probability of This is a problem of conditional probability and we have defined the case where one child is already a girl. The probability mass function (pmf) of X , p(X) describes how the total probability is distributed among all the. The function f ( x ) is typically called the probability mass function , although some authors also refer to it as the probability function , the frequency function , or probability density function. Behind one is a expensive car, but behind the other two are goats. Then, if x is the number of blue balls in urn 2,. 480 Question 10. Set books The notes cover only material in the Probability I course. If X and Y are independent uniform (0;1) random variables, show that [Filename: Assignment7_solutions I. Solution the the Second WBC Problem-page 10. Ignore problem 2 and 3 since they are about continuous distributions. Browse through all study tools. The solution is provided for each exercise. Use features like bookmarks, note taking and highlighting while reading Probability Problems and Solutions. More Problems on probability and statistics are presented. This Probability Pdf we are Providing is free to download. Probability on Numbers Worksheet These Probability Worksheets will produce problems with simple numbers between 1 and 50. One of the toughest and probably the most prestigious undergraduate competition in the world. All class notes posted in Part II above excluding. Devore Chapter 1. Divisibility intuition. Practice exam 1 and solution Practice exam 2 and solution Midterm 1 solutions Midterm 2 practice exams: Practice exam 1 and solution Practice exam 2 and solution Topics for Midterm 2: All of Chapter 3, Sections 1-3 of Chapter 4, Sections 1-3 of Chapter 5, Sections 1,2,3,4,6,7,9 of Chapter 6. Arrange - this is how many 'WAYS' something can be arranged. The remaining sections each investigate a different game, providing an overview of the rules, some discussion of basic strategy, and a series of problems and solutions. Experiment 1: A spinner has 4 equal sectors. nthe sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin. Using the same data from the compound event video, I make a diagram to illustrate compound event probability and how to avoid double counting data when finding probability. Example 6 (Parts Inspection) Consider the parts problem again, but now assume that a one-year warranty is given for the parts that are shipped to customers. , the probability of getting the first two spheres as blue while the third sphere is red is same as the probability of getting the second and third spheres as blue while the first sphere is red As Jennifer pointed out, Baye's theorem can be used. The area of the shaded region to the left of z in the diagram above demonstrates these problems. This is a problem. Y1 - 1967/1/1. In a certain game, players toss a coin and roll a dice. Sugar daddies together with glucose little ones are believed as the perfect flings that can bring individuals along. Probability problem on Dice shortcut tricks are very important thing to know for your exams. PROBABILITY. I need the full detailed solution of the following 6 probability questions. Number of consonants \u00a1n English alphabets = 21. BIOSTATISTICS DESCRIBING DATA, THE NORMAL DISTRIBUTION 1. The long run probability of player A winning all the coins is. How does this impact the probability of some other A. Make sure you understand your probability questions is all we can say. The manual states that the lifetime T of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies P(T \u2265 t) = e \u2212 t 5, for all t. Let P(D) be the probability that the letter is in DOGS. This is a problem. This content was COPIED from BrainMass. to access the Math Probability menu. Solved by Expert Tutors. Solved examples with detailed answer description, explanation are given and it would be easy to understand. Solution of exercise 1. The wikipedia page claims that likelihood and probability are distinct concepts. The initial table listed in problem description is a table of probabilities p(1), p(2), , p(20). Miracle Mountain. And the probability of getting a head is equal to the probability of getting a tail. In a philosophical sense it can even be said that the risk does not have a real existence in the present, but it only exists in the future. Two and three-digit subtraction. Probability of choosing 2nd chocobar = 3/7. Often used in market research studies, it is useful in business to predict sales, scores and other numerical data based off research that is either conducted or gathered at a secondary. (If it comes up six, you get$6. We then look at some word problems. Devore Chapter 1. It follows that the probability of an ace or a spade = 4+13 52 = 17 52. nclearly there are 12 possible outcomes (6 x 2) nthere are 3 possible outcomes for an odd die and a head. This post provides practice problems to reinforce the concept of correlation coefficient discussed in this post in a companion blog. \u2018The Democrats will probably win the next election. For example, a weather forecast that predicts a 75 percent chance of snow is using probability to communicate that people should be prepared to deal with snowy conditions. Our Probability solvers include the most commonly used probability distributions, such as the normal, exponential, Poisson, binomial, and uniform distributions. MIDTERM 1 solutions and Histogram. More Problems on probability and statistics are presented. ) whereas ex-post solutions try to gather auxiliary information about non-respondents which is then used to calculate a probability of response for different population sub. Brilliant Premium. (a) Show that if X and Y are conditionally independent given Z and X and Z are conditionally independent given Y then X is independent of (Y,Z). 01 of probability is being shifted from $2400 to$0. It can be a finite set, a countable set (a set whose elements can be put in a sequence, in fact a finite set is also. Most of the problems require very little mathematical background to solve. Z for 90th percentile=1. Chapter 4 - Continuous Random Variables and Probability Distributions. Probability - math word problems Probability is the measure of the likeliness that an event will occur. Q2) Generate a Punnett Square for a heterozygous individual crossed with a heterozygous individual. Standard probability themes like coins, spinners, number cubes, and marbles are featured along with some other situations. Probability Questions are provided with detailed answers to every question. We also discuss some applications of probability theory to computing, including systems for making likely inferences from data and a class of useful algorithms that work \u201cwith high probability\u201d but are not guaranteed to work all the time. what is the probability of generating a random number which has at least one digit = 1 for example 115. Then, as they advance, students create models for probability questions and ultimately use probability to make decisions. Probability Calculators and Solvers Our site contains an ample array of Probability Calculators that can greatly help you with all of your academic needs. The probability that a student works part time is P(PT)=0. A Problem With Pearls. Get Textbook Solutions and 24/7 study help for Statistics And Probability Step-by-step solutions to problems over 34,000 ISBNs Find textbook solutions Close. The area of the shaded region to the left of z in the diagram above demonstrates these problems. 5 1 \u2212 p = 0. \u2022 Probability and Statistics for Engineering and the Sciences by Jay L. Objective: I know how to solve probability word problems. Press [2] to evaluate a permutation or press [3] to evaluate a combination. Problem 1. Now practice our shortcut tricks on probability problem on coin and read examples. 1 \u2022 The random variable X has probability density function fX (x) = \u02c6 cx 0 \u2264 x \u2264 2, 0 otherwise. Listed in the following table are practice exam questions and solutions, and the exam questions and solutions. But, as humans, we know the world is abnormal. A persuasive problem statement consists of three parts: 1) the ideal, 2) the reality, and 3) the consequences for the reader of the feasibility report. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to solve easily in exams. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and. The binomial probability calculator will calculate a probability based on the binomial probability formula. This is the number of times the event will occur. Using the same data from the compound event video, I make a diagram to illustrate compound event probability and how to avoid double counting data when finding probability. The problem concerns a game of chance with two players who have. u p is probability of a success and q = 1 - p is probability of a failure u Consider a game where the player bats 4 times: H probability of 0/4 = (0. Browse through all study tools. Probability Questions and answers PDF with solutions. Sarah has two children and we know that she has a daughter. Solution: Let\u2019s think of this like a decision tree. Get Textbook Solutions and 24/7 study help for Statistics And Probability Step-by-step solutions to problems over 34,000 ISBNs Find textbook solutions Close. In other words, the probability of winning for a player is the ratio of the number of coins the player starts with to the total numbers of coins. Standard probability themes like coins, spinners, number cubes, and marbles are featured along with some other situations. The bet can either be on red or on blue and the amount of each bet is arbitrary. The probability of any event E is number of outcomes in E total number of outcomes in the sample space This probability is denoted by P E n E n S This probability is called classical probability and it uses the sample space S. Then, if x is the number of blue balls in urn 2,. Example Question #2 : How To Find The Probability Of An Outcome If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?. Geometric Probability. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Find the probability that they are of same colour. when the selection of one does not affect the probability of a\u2026. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. \u2022 Probability and Statistics for Engineering and the Sciences by Jay L. At a picnic, Julio reaches into an ice-filled cooler containin\u2026. The solutions are aimed to make your understanding clear. In non-technical parlance, \"likelihood\" is usually a synonym for \"probability,\" but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. Let E 1,E 2,E 3 be events. (Solution): Probability Problems. Sample Space. You need at most one of the three textbooks listed below, but you will need the statistical tables. The function f ( x ) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x -axis is equal to `1. Solutions Manual A First Course in Probability 9th Edition Sheldon Ross. Solution: Gambler\u2019s Ruin Problem. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. \"} BarChart3D[{win, lose}] The probability of winning is 30. We write P(AjB) = the conditional probability of A given B. If X and Y are independent uniform (0;1) random variables, show that [Filename: Assignment7_solutions I. Two Games in a Row. To navigate from one page of exercises to another, you will use the right navigation bar. If the first four marbles drawn are red, what is the probability the next marble drawn will not be red? 7. The probability that a student belongs to a club is P(C)=0. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. by electronjohn. If we toss a fair die, what. Objective: I know how to solve probability word problems. Addition and subtraction word problems. AU - Haji-Sheikh, A. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. The actual probability of finding the particle is given by the product of the wavefunction with its complex conjugate (like the square of the amplitude for a complex function). Theory of Statistics c 2000\u20132020 James E. IB Math \u2013 Standard Level \u2013 Probability Practice - MarkScheme Alei - Desert Academy C:\\Users\\Bob\\Documents\\Dropbox\\Desert\\SL\\6StatProb\\TestsQuizzesPractice\\SLProbPractice1. Probability Problems. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Feel Free to Learn 316,387 views. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Review of probability theory \u2022 Definitions (informal) -Probabilities are numbers assigned to events that indicate \"how likely\" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. Chapter 4 - Continuous Random Variables and Probability Distributions. This section provides materials for a lecture on discrete random variables, probability mass functions, and expectations. Multiplication Rule of Probability The addition rule helped us solve problems when we performed one task and wanted to know the probability of two things happening during that task. The bet can either be on red or on blue and the amount of each bet is arbitrary. Q2) Generate a Punnett Square for a heterozygous individual crossed with a heterozygous individual. What Are Example Statistics and Probability Problems and Their Solutions? Credit: Echo/Cultura/Getty Images Two examples of probability and statistics problems include finding the probability of outcomes from a single dice roll and the mean of outcomes from a series of dice rolls. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. Practice calculating conditional probability, that is, the probability that one event occurs given that another event has also occurred. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. 01 of losing everything in going to A therefore looms large in peoples minds. The simulation was repeated by performing 50 repetitions of the experiment. starting the probability problems and solutions to door every daylight is agreeable for many people. Find the probability that a student takes calculus, given that he or she is on the dean\u2019s list. No need to worry if you miss class or just need some extra help. All class notes posted in Part II above excluding. (If it comes up six, you get \\$6. A probability experiment has four possible outcomes: e 1, e 2, e 3. Practice Problems - Try specific problems and see the solution. Think Stats: Probability and Statistics for Programmers is a textbook for a new kind of introductory prob-stat class. Complete Solutions Manual to Probability by Jim Pitman. It takes a computational approach, which has several advantages: \u2022 Students write programs as a way of developing and testing their un-derstanding. Chapter 1 covers this theory at a fairly rapid pace. Find the probability that the number on the drawn ticket is a multiple of 3 or 7. We break down problems on video in a step-by-step easy to follow format. Finding the Median in a pdf : S2 Edexcel. The probability of two or more heads is, therefore: Probability = 4 8 = 1 2 We solved this problem by \ufb01rst enumerating the set of possibl e outcomes, known as the Sample Space, and then by deciding which of these outcomes satis\ufb01ed the cr iterion of containing 'two or more heads'. Certainly X is binomial with n = 40 and p = 0. More Problems on probability and statistics are presented. P(X) gives the probability of successes in n binomial trials. Example of Using a Contingency Table to Determine Probability. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. Generate integer from 1-8 with equal probability. Problem 4 (Solution on p. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. Calculate the value of a in a normal distribution with a mean of 4 and a standard deviation of 2 for which:. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Review of probability theory \u2022 Definitions (informal) -Probabilities are numbers assigned to events that indicate \"how likely\" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. The actual probability of finding the particle is given by the product of the wavefunction with its complex conjugate (like the square of the amplitude for a complex function). Pedigrees and probability problems 1-28 and 1-30. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. Chapter 1 covers this theory at a fairly rapid pace. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. a Show that $$f\\left( x \\right)$$ is a probability density function. the number of ways several people can stand in a line if 2 particular people must \u2014 or must not \u2014 stand next to each other). (321 problems) IMO Shortlisted Problems. Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3. Playing cards probability problems based on a well-shuffled deck of 52 cards. Chegg Solution Manuals are written by vetted Chegg Statistics And Probability experts, and rated by students - so you know you're getting high quality answers. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. The post in the companion blog shows how to evaluate the covariance and the correlation coefficient of two continuous random variables and. Importantly, the solution for the problem of points became\u00a0known as\u00a0an expectation value \u00a0(the average expected value). Suppose that a good part fails within the \ufb01rst year with probability 0. This is essentially a 1D scattering problem. At a picnic, Julio reaches into an ice-filled cooler containin\u2026. 1 Problem 4E. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions. Brilliant Premium. Probability questions pop up all the time. Fully solved Aptitude questions with detailed answer description and explanation. Practice Problems - Try specific problems and see the solution. IB Math - Standard Level - Probability Practice - MarkScheme Alei - Desert Academy C:\\Users\\Bob\\Documents\\Dropbox\\Desert\\SL\\6StatProb\\TestsQuizzesPractice\\SLProbPractice1. Probability of problem getting solved = 1 - (5/7) x (3/7) x (5/9) = (122/147) Example 9: Find the probability of getting two heads when five coins are tossed. Problem 1. ECE302 Spring 2006 HW5 Solutions February 21, 2006 3 Problem 3. The higher the probability of an event, the more certain that the event will occur. Enter r, the number of items selected from the set, and press [ENTER] to display the result. Playing cards probability problems based on a well-shuffled deck of 52 cards. also known as the birthday problem, states that in a random group of 23 people, there is about a 50 percent chance that two people have the same birthday. Test your understanding with practice problems and step-by-step solutions. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to solve easily in exams. The mathematics field of probability has its own rules, definitions, and laws, which you. Conditional Probability: It is known that a student who does his online homework on aregular basishas a chance of83 percentto get a good grade (A or B); but the chance drops to58 percentif he doesn't do the homework regularly. An agent sells life insurance policies to five equally aged, healthy people. Students were encouraged to prepare a 4x6 inch notecard to use for reference during each exam. Behind one is a expensive car, but behind the other two are goats. when the selection of one does not affect the probability of a\u2026. Importantly, the solution for the problem of points became\u00a0known as\u00a0an expectation value \u00a0(the average expected value). \" Understanding how to calculate these percentages with real numbers of people and things. Basic concept on drawing a card: In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i. and define the function. Now practice our shortcut tricks on probability problem on coin and read examples. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions. Homework 2 from last time OLD HW2 solutions Practice midterm and Midterm 2015 solutions. The text-books listed below will be useful for other courses on probability and statistics. Cards of Spades and clubs are black cards. (See diagram. 1) Basics of Electromagnetics Part I \u2013 Download MCQs from here. Solution the the Second WBC Problem-page 10. Multiplicative Principle for Counting n The total number of outcomes is the product of the possible outcomes at each step in the sequence n if a is selected from A, and b selected from B\u2026 n n (a,b) = n(A) x n(B) q (this assumes that each outcome has no influence on the next outcome) n How many possible three letter words are there? q you can choose 26 letters for each of the three positions, so. GMAT Advanced Probability Problems By Mike M\u1d9cGarry on January 3, 2014 , UPDATED ON January 15, 2020, in GMAT Math In the following probability problems, problems #1-3 function as a set, problems #4-5 are another set, and problems #6-7 are yet another set. by electronjohn. Homework 2 from last time OLD HW2 solutions Practice midterm and Midterm 2015 solutions. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. Test your understanding with practice problems and step-by-step solutions. Probability theory is the most directly relevant mathematical background, and it is assumed that the reader has a working knowledge of measure-theory-based probability theory. Midterm 2016 solutions Combinatorics practice problem set and solutions. Bayes' theorem describes the probability of occurrence of an event related to any condition. For instance, let\u2019s say the probability that you will get to class on time is 2/3. In non-technical parlance, \"likelihood\" is usually a synonym for \"probability,\" but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. Finding the Median in a pdf : S2 Edexcel. Thanks for the A2A. At a picnic, Julio reaches into an ice-filled cooler containin\u2026. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. Possible Solutions to Probability Problems Consider the experiment of selecting a card from an ordinary deck of 52 playing cards. The PDF is the density of probability rather than the probability mass. 2 Counting Assignments One of the simplest but most important counting problems deals with a list of. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. The higher the probability of an event, the more certain that the event will occur. by mathemagician. Throw 2 dices simultaneously. Then, we'll teach you all of the basic rules and formulas you need to know as well as give you tips for studying probability and approaching GRE probability questions on test day. Problem 1 : Two dice are rolled, find the probability that the sum is. For i = 1,2, let R i = event that a red ball is drawn from urn i and let B i = event that a blue ball is drawn from urn i. \ud835\udde7\ud835\uddfc\ud835\uddfd\ud835\uddf6\ud835\uddf0: conditional probability problems with solutions \ud835\udde6\ud835\ude02\ud835\uddef\ud835\uddf7\ud835\uddf2\ud835\uddf0\ud835\ude01: Engineering Mathematics \ud835\udde7\ud835\uddfc \ud835\uddd5\ud835\udde8\ud835\uddec \ud835\uddfb\ud835\uddfc\ud835\ude01\ud835\uddf2\ud835\ude00 \ud835\uddfc\ud835\uddf3. So how will you make the determination of how to solve a conditional probability problem? As a general rule, problems where information is given in a table are best solved by using the formula. The point is that the order of events doesn't affect with respect to conditional probability. After solving all ten math questions write down total time taken by you to solve those questions. Notice that the phenotypic ratio for dominant/recessive is identical to Q1 (3/4 Dominant, 1/4 recessive) except you now have to realize your mutation is Dominant and wild type is recessive. Midterm 2016 solutions Combinatorics practice problem set and solutions. Calculate the sample mean. P(F AND P) = d. Sarah has two children and we know that she has a daughter. Playing cards probability problems based on a well-shuffled deck of 52 cards. many new problems and solutions, a number of improvements in the pre-sentation, and corrections. PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Probability theory - Probability theory - The birthday problem: An entertaining example is to determine the probability that in a randomly selected group of n people at least two have the same birthday. Probability is the chance that an event will occur. Solution: (F) Probability always lie within 0 to 1. Probability_Problems Based on Coins#LESSON-2 - Duration: 17:39. If we toss a fair die, what. $$\u2234 P(E)={n(E)\\over n(S)} = {2\\over52}={1\\over26}$$ Learn and practice these probability problems with solutions and you can ask me anything in the comment section if you have any doubts regarding probability problems and it's solutions. To get a feeling for PDF, consider a continuous random variable. To link to this page, copy the following code to your site:. The text-books listed below will be useful for other courses on probability and statistics. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. to access the Math Probability menu. Finding the Median in a pdf : S2 Edexcel. Practice Problems - Try specific problems and see the solution. The question is incomplete, that is, it doesn\u2019t provide information regarding with the level of probability that you are familiar with. PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Conroy and a 5=6 chance of not rolling a six, after which the number of rolls we expect to throw is the same as when we started. This Collection of problems in probability theory is primarily intended for university students in physics and mathematics departments. Finding the mean in a Normal Distribution : Statistics S1 Edexcel June 2013 Q6 (a) ExamSolutions - youtube Video. It can be a finite set, a countable set (a set whose elements can be put in a sequence, in fact a finite set is also. Most probability problems on the GRE involve independent events. A problem is given to three students whose chances of solving it are 1/2,. \"} BarChart3D[{win, lose}] The probability of winning is 30. The process to solve the problem is rarely straightforward and takes practice to perfect. This is a problem. No need to worry if you miss class or just need some extra help. According to recent data, the probability of a person living in these conditions for 30 years or. Colavito\u2019s homepage for solutions. The initial table listed in problem description is a table of probabilities p(1), p(2), , p(20). Total Number of ways = 2 5 = 32. The point is that the order of events doesn't affect with respect to conditional probability. MIDTERM 2 solutions and Histogram. The incubation periods of a random sample of 7 HIV infected individuals is given below (in years): 12. Advanced Techniques. A Problem With Pearls. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. The probability mass function of is but and Therefore, the probability mass function can be written as which is the probability mass function of a Bernoulli random variable. Then, n(E) = 2. Rolling the Dice. The empirical probability of three consecutive female births is. 1, Median 22. ) For a di\ufb00erent problem, allow every one of n people to place an even bet on the color of his hat. In the Options dialog box, select the Show Iteration Results check box to see the values of each trial solution, and then click OK. Most of the problems require very little mathematical background to solve. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. The probability (chance) is a value from the interval 0;1> or in percentage (0% to 100%) expressing the occurrence of some event. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. DISCRETE PROBABILITY DISTRIBUTIONS 27. Miracle Mountain. Geometric Probability. Life or Death? The Emperor's Proposition. We want to generate a random number between 100 and 499. The probability of a success, denoted by p, remains constant from trial to trial and repeated trials are independent. Probability Questions and answers PDF with solutions. Probability MCQ Questions and answers with easy and logical explanations. Please do not email me completed problem sets. Multiplying by multiples of 10. John has a special die that has one side with a six, two sides with twos and three sides with ones. At any particular time period, both outcomes cannot be achieved together so [\u2026]. Solution of exercise 1. This is a beta distribution. We want to generate a random number between 100 and 499. The probability equals 31,8 %. The question is incomplete, that is, it doesn\u2019t provide information regarding with the level of probability that you are familiar with. There is one desired outcome and six possible outcomes. random variables, and some notation. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. 74% of the X values are less than three standard deviations from the mean. (Opens a modal) Random number list to run experiment. Practice Problems - Try specific problems and see the solution. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a \ufb01rst course in prob-ability. Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);. Playing cards probability problems based on a well-shuffled deck of 52 cards. com: Many students are also using our Free Statistics Lab Manual. I require the probability that after these n stretches he is at a distance between rand r+ rfrom his starting point 0. A Music Survey was carried out to find out what types of music a group of people liked. Assume there is a beam of particles with definite momentum coming in from the left and assume there is no flux of particles coming from the right. One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is known as an expected value. The Probability Distributome Project provides an interactive navigator for traversal, discovery and exploration of probability distribution properties and interrelations. Listed in the following table are practice exam questions and solutions, and the exam questions and solutions. 84 = (x \u2013 70)/2. Let E = event of getting a queen of club or a king of heart. Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Miracle Mountain. Hone your skills with this array of statistics and probability worksheets for 7th grade to compute the average on whole numbers and decimals, find the probability on a pair of coins, a pair of dice, months and more. Solution of exercise 1. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Although they never finished their game, their letters contain solutions to both problems. Home > Math> Probability and Statistics>Probability Problems. Homework problems usually do not say which concepts are involved, and often require combining several concepts. Probability problem on Dice. Most probability problems on the GRE involve independent events. Probability MCQ Questions and answers with easy and logical explanations. After you define a problem, click Options in the Solver Parameters dialog box. The problem concerns a game of chance with two players who have. Example: there are 5 marbles in a bag: 4 are. Probability Questions and answers PDF with solutions. In non-technical parlance, \"likelihood\" is usually a synonym for \"probability,\" but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. These you may need to complete at home. Statistics Problems With Solutions Return to Statistics Internet Library for videos, software assistance, more problems and review. Cards of Spades and clubs are black cards. pdf] - Read File Online - Report Abuse. A Collection of Dice Problems Matthew M. 316 that an audit of a retail business. The first section provides a brief description of some of the basic probability concepts that will be used in the activities. Trials until first success. Walpole Raymond H. What is the probability of drawing a red Bingo chip at least 3 out of 5 times? Round answer to the nearest hundredth. Probability Problems With Solutions Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Probability Word Problems (Simplifying Math) What are the chances that your name starts with the letter H? Find out how to make that calculation and many more. Advanced Techniques. The problem concerns a game of chance with two players who have. Probability shortcut Tricks Pdf, Probability MCQ, Probability Objective Question & Answer Pdf. [email\u00a0protected]{\"The probability of winning is \", 100 win/(win + lose) // N, \" %. Total English alphabets = 26. The flippant juror. For example: if we have to calculate the probability of taking a blue ball from the second bag out of three different bags of balls, where each bag contains three different colour balls viz. a Show that $$f\\left( x \\right)$$ is a probability density function. Make use of probability calculators to solve the probability problems with ease. What is the probability that players will experience rain and a temperature between 35 and 50 degrees? 0. \u2019 \u2018There\u2019s a 30% chance of rain tomorrow. He offers you the following game. 1) Basics of Electromagnetics Part I \u2013 Download MCQs from here. Probability Solution: (a) Since there are 4 queens ( ) = ( ) ( ),. (Solution): Probability Problems. Probability And Statistics Problems Solutions Author: persepolis. It is assessed by considering the event's certainty as 1 and impossibility as 0. edu-2020-04-20T00:00:00+00:01 Subject: Probability And Statistics Problems Solutions Keywords: probability, and, statistics, problems, solutions Created Date: 4/20/2020 7:35:45 AM. nthe sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin. AU - Sparrow, E. Probability Questions are provided with detailed answers to every question. probability problems, probability, probability examples, how to solve probability word problems, probability based on area, examples with step by step solutions and answers, How to use permutations and combinations to solve probability problems, How to find the probability of of simple events, multiple independent events, a union of two events. Competitive exams are all about time. The concept is very similar to mass density in physics: its unit is probability per unit length. Most of these problems require very little mathematical background to solve. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. Examples include the Monty Hall paradox and the birthday problem. Write down twenty math problems related to this topic on a page. If there is a chance that an event will happen, then its probability is between zero and 1. Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Linear Algebra Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Please solve the following probability practice problems: Determine the probability that a digit chosen at random from the digits 1, 2, 3, \u202612 will be odd. The mathematics field of probability has its own rules, definitions, and laws, which you. The number of candidates for the first, second and third posts are 3,4 and 2 respectively. Let P(D) be the probability that the letter is in DOGS. \" Understanding how to calculate these percentages with real numbers of people and things. MIDTERM 2 solutions and Histogram. (b) Show that the result in question (a) is not necessarily correct without. 1 Problem 4E. You need at most one of the three textbooks listed below, but you will need the statistical tables. Fully worked-out solutions of these problems are also given, but of course you should \ufb01rst try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Probability of choosing 2nd chocobar = 3/7. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. Objective: I know how to solve probability word problems. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. Q x = P(female lives to age x) = number of female survivors at age x 100,000. Any problems you do not have time to do, try doing later. Bayes\u2019 theorem describes the probability of occurrence of an event related to any condition. Advanced Probability Problems And Solutions Probability Questions with Solutions Tutorial on finding the probability of an event. Two balls are chosen from the jar, with replacement. What is the probability the sum of the dice is 5? 6. \u00b7 = = Answer: 2: A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls. Calculate the probability that if somebody is \u201ctall\u201d (meaning taller than 6 ft or whatever), that person must be male. is the constant 2. Probability has its origin in the study of gambling and insurance in the 17th century, and it is now an indispensable tool of both social and natural sciences. ) Find the probability that a person is female or prefers hiking on mountain peaks. == Solutions for Probability and Statistics for Engineering and the Sciences, 7th ed, by Jay L. Important facts and powerful problem solving approaches are highlighted throughout the text. For less than the price of a single session with a private tutor, you can have access to our entire library of videos. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Probability Math. NCERT Solutions For Maths Class 12 Chapter 13 - Probability comprises of various important questions for exams. Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a \ufb01rst course in prob-ability. Two events, A and B, are independent if the outcome of A does not affect the outcome of B. So, for example, using a binomial distribution, we can determine the probability of getting 4 heads in 10 coin tosses. P(X) gives the probability of successes in n binomial trials. A Problem With Pearls. MIDTERM 1 solutions and Histogram. For Exercises 3-17 to 3-21, verify that the following functions are probability mass functions, and determine the requested probabilities. As we go deeper into the area of mathematics known as combinatorics, we realize that we come across some large numbers. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. As with any quantum system, the wave functions give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. Probability is the chance that an event will occur. Example: the chances of rolling a \"4\" with a die. Probability Problems and Solutions. Interview Probability Puzzles #1 - Most Popular Monty Hall Brain Teaser Difficulty Popularity The host of a game show, offers the guest a choice of three doors.", "date": "2020-09-24 20:12:14", "meta": {"domain": "mexproject.it", "url": "http://mexproject.it/zcjm/probability-problems-and-solutions.html", "openwebmath_score": 0.7061757445335388, "openwebmath_perplexity": 589.712825578324, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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{"url": "https://changesinc.org/p3jo9/what-is-bijective-function-5b8916", "text": "A bijective function is both injective and surjective, thus it is (at the very least) injective. And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. If it crosses more than once it is still a valid curve, but is not a function. So we can calculate the range of the sine function, namely the interval $[-1, 1]$, and then define a third function: $$\\sin^*: \\big[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\big] \\to [-1, 1]. Stated in concise mathematical notation, a function f: X \u2192 Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). Hence every bijection is invertible. A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. How to Prove a Function is Bijective without Using Arrow Diagram ? The inverse is conventionally called \\arcsin. A function that is both One to One and Onto is called Bijective function.$$ Now this function is bijective and can be inverted. This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read \"one-to-one correspondence).. Functions that have inverse functions are said to be invertible. Mathematical Functions in Python - Special Functions and Constants; Difference between regular functions and arrow functions in JavaScript; Python startswith() and endswidth() functions; Hash Functions and Hash Tables; Python maketrans() and translate() functions; Date and Time Functions in DBMS; Ceil and floor functions in C++ Each value of the output set is connected to the input set, and each output value is connected to only one input value. As pointed out by M. Winter, the converse is not true. My examples have just a few values, but functions usually work on sets with infinitely many elements. Infinitely Many. Thus, if you tell me that a function is bijective, I know that every element in B is \u201chit\u201d by some element in A (due to surjectivity), and that it is \u201chit\u201d by only one element in A (due to injectivity). The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. Ah!...The beautiful invertable functions... Today we present... ta ta ta taaaann....the bijective functions! The figure shown below represents a one to one and onto or bijective function. Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. Definition: A function is bijective if it is both injective and surjective. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. A function is invertible if and only if it is a bijection. In mathematics, a bijective function or bijection is a function f : A \u2192 B that is both an injection and a surjection. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. Question 1 : And I can write such that, like that. 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Bijective and can be inverted to only one input value that have inverse functions are to. M. Winter, the converse is not true a valid curve, is.", "date": "2021-04-11 09:38:20", "meta": {"domain": "changesinc.org", "url": "https://changesinc.org/p3jo9/what-is-bijective-function-5b8916", "openwebmath_score": 0.843766450881958, "openwebmath_perplexity": 461.01405814165366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9854964175398042, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8647502321164977}}
{"url": "https://math.stackexchange.com/questions/374983/prove-that-fraca-12a-1a-2-fraca-22a-2a-3-cdots-fraca-n2a-n", "text": "# Prove that $\\frac{a_1^2}{a_1+a_2}+\\frac{a_2^2}{a_2+a_3}+ \\cdots \\frac{a_n^2}{a_n+a_1} \\geq \\frac12$\n\nLet $$a_1, a_2, a_3, \\dots , a_n$$ be positive real numbers whose sum is $$1$$. Prove that $$\\frac{a_1^2}{a_1+a_2}+\\frac{a_2^2}{a_2+a_3}+ \\ldots +\\frac{a_n^2}{a_n+a_1} \\geq \\frac12\\,.$$\n\nI thought maybe the Cauchy and QM inequalities would be helpful. But I can't see how to apply it. Another thought (might be unhelpful) is that the sum of the denominators on the left hand side is $$2$$ (the denominator on the right hand side). I would really appreciate any hints.\n\n\u2022 Apply Cauchy with one term being left hand side, second term being $((a_1+a_2) + (a_2+a_3) + \\cdots)$ \u2013\u00a0user27126 Apr 28 '13 at 7:02\n\u2022 Thanks so much I've solved it, should I close the question, or post the solution I just found, with your hint? \u2013\u00a0John Marty Apr 28 '13 at 7:05\n\u2022 it would be good if you post a complete solution and accept it as the answer. \u2013\u00a0user27126 Apr 28 '13 at 7:06\n\nThanks to Sanchez for giving me a hint to solve this. Here is a full solution.\n\nBy the Cauchy-Schwarz inequality we have:\n\n$${\\frac{a_1^2}{a_1+a_2}+\\frac{a_2^2}{a_2+a_3}+ \\cdots \\frac{a_n^2}{a_n+a_1}=\\frac{a_1^2}{(\\sqrt{a_1+a_2})^2}+\\frac{a_2^2}{(\\sqrt{a_2+a_3})^2}+ \\cdots+ \\frac{a_n^2}{(\\sqrt{a_n+a_1})^2} \\geq \\frac{1}{a_1+\\cdots + a_n+a_1+ \\cdots + a_n}\\left(\\frac{a_1 \\cdot \\sqrt{a_1+a_2}}{\\sqrt{a_1+a_2}} + \\frac{a_2 \\cdot \\sqrt{a_2+a_3}}{\\sqrt{a_2+a_3}}+ \\cdots + \\frac{a_n \\cdot \\sqrt{a_n+a_1}}{\\sqrt{a_n+a_1}}\\right)\\\\=\\frac{a_1+a_2+a_3+ \\cdots a_n}{{2(a_1+a_2+a_3+ \\cdots a_n)}}=\\frac12}$$\n\nas required. (We know that $a_1+a_2+a_3+ \\cdots +a_n=1$)\n\n\u2022 thanks, I couldn't solve it before your hint, thanks so much. \u2013\u00a0John Marty Apr 28 '13 at 7:34\n\u2022 I made an attempt to improve readability. Is it OK? \u2013\u00a0Pedro Tamaroff Jul 6 '13 at 0:37\n\nHere's another solution.\n\nObserve that $$\\sum \\frac{a_i^2 - a_{i+1}^2} { a_i+ a_{i+1}} = \\sum a_i - a_{i+1} = 0,$$\n\nHence $\\sum \\frac{a_{i}^2}{a_i+a_{i+1}} = \\sum \\frac{a_{i+1}^2}{a_i+a_{i+1}}$, and we just need to show that\n\n$$\\sum \\frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \\geq 1.$$\n\nThis makes the inequality much more symmetric, and easier to manipulate. In particular,\n\n$$\\sum \\frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \\geq \\sum \\frac{1}{2} (a_i + a_{i+1}) = 1$$\n\nThe following modified form of the CBS inequality is often helpful:\n\nLemma. If $x_i \\in \\Bbb{R}$ and $a_i > 0$, then $$\\frac{x_{1}^{2}}{a_{1}} + \\cdots + \\frac{x_{n}^{2}}{a_{n}} \\geq \\frac{(x_{1} + \\cdots + x_{n})^{2}}{a_{1}+\\cdots+a_{n}}.$$\n\nThe proof is straightforward using the CBS inequality, following the methodology exactly John Marty used.\n\nApplying this, we have\n\n$$\\frac{a_{1}^2}{a_{1}+a_{2}} + \\cdots + \\frac{a_{n}^2}{a_{n}+a_{1}} \\geq \\frac{(a_{1} + \\cdots + a_{n})^{2}}{2(a_{1}+\\cdots+a_{n})} = \\frac{1}{2}.$$", "date": "2020-11-24 14:26:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/374983/prove-that-fraca-12a-1a-2-fraca-22a-2a-3-cdots-fraca-n2a-n", "openwebmath_score": 0.9275873899459839, "openwebmath_perplexity": 337.37783371831506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964228644457, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8647502288947095}}
{"url": "https://math.stackexchange.com/questions/2610834/question-to-calculating-probability", "text": "# Question to calculating probability\n\nThe question:\n\n79% of the drivers in a city always fasten their seatbelt when driving, and everyday some drivers receive a ticket(for various reasons...). If a driver has fastened their seatbelt there is a chance of 7% that the driver receives a ticket. If a driver does not fasten their seatbelt there is a 22% chance that he will receive a ticket. If a driver receives a ticket what is the probability that he had fastened his seatbelt?\n\nMy assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So I just calculated (7/29) * (0,79) + (22/29) * (0,21) and I got 0.35 but that seems to be wrong.\n\n\u2022 Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. \u2013\u00a0Jos\u00e9 Carlos Santos Jan 18 '18 at 16:13\n\u2022 Okay thanks for the advise. My assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So i just calculated (7/29) * (0,79) + (22/29) * (0,21) and i got 0.35 \u2013\u00a0Simon Jan 18 '18 at 16:18\n\u2022 Edit your question instead of posting it as a comment. \u2013\u00a0Jos\u00e9 Carlos Santos Jan 18 '18 at 16:19\n\u2022 This is a prefect time for Bayes' Law! \u2013\u00a0Bram28 Jan 18 '18 at 16:28\n\nLet $S$ and $T$ be the events \"seatbelt fasten\" and \"receive ticket\" respectively.\n\n\u2022 $P(S) = 0.79$\n\u2022 $P(T \\mid S) = 0.07$\n\u2022 $P(T \\mid S^C) = 0.22$\n\nUse Bayes' formula to find $P(S \\mid T)$.\n\n\\begin{align} P(S \\mid T) &= \\frac{P(S \\cap T)}{P(T)} \\\\ &= \\frac{P(T \\mid S) P(S)}{P(T \\mid S) P(S)+P(T \\mid S^C) P(S^C)} \\\\ &= \\frac{0.07 \\cdot 0.79}{0.07 \\cdot 0.79 + 0.22 \\cdot (1 - 0.79)} \\\\ &= \\frac{79}{145} \\end{align}\n\nIf a driver receives a ticket what is the probability that he had fastened his seatbelt is $79/145$.\n\nYou can't assume $\\frac{7}{29}$ of people who receive a ticket fastened their seatbelt. Consider an extreme case, where the percentage of people who fasten their seatbelt is much higher than $79\\%$ -- suppose it is $100\\%$. Then clearly $100\\%$ of people who receive a ticket fastened their seatbelt. If $0\\%$ of people fasten their seatbelt, then $0\\%$ who receive a ticket fastened their seatbelt. If the percentage who fasten is somewhere in between -- e.g., $79\\%$ -- then the percentage who fastened given that they received a ticket is somewhere in between, but we can't just say it's $\\frac{7}{29}$.\n\nAre you familiar with Bayes' Theorem? That's probably the best way to approach this problem.", "date": "2021-08-01 17:44:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2610834/question-to-calculating-probability", "openwebmath_score": 0.9977121949195862, "openwebmath_perplexity": 203.2225472411262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964169008471, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8647502283982127}}
{"url": "https://math.stackexchange.com/questions/4187654/what-is-the-value-of-left-log-213-right2-log-21147-log-211323/4187659", "text": "What is the value of $\\left(\\log_{21}(3)\\right)^2+\\log_{21}(147)\\log_{21}(1323)$?\n\nWhat is the value of $$\\left(\\log_{21}(3)\\right)^2+\\log_{21}(147)\\log_{21}(1323)$$ ?\n\n$$1)1\\qquad\\qquad2)2\\qquad\\qquad3)3\\qquad\\qquad4)4$$\n\nTo solve this question I tried writing the expression as:\n\n$$\\left(\\log_{21}(3)\\right)^2+\\log_{21}(7^2\\times3)\\log_{21}(7^2\\times3^3)$$ $$=\\left(\\log_{21}(3)\\right)^2+(2\\log_{21}7+\\log_{21} 3)(2\\log_{21}7+3\\log_{21}3)$$\n\nI don't know how to continue.\n\n\u2022 Don't be so eager to separate the $\\log7$ from the $\\log3$. The most potent course of simplification you have here is $\\log21=1$. Jul 1 at 10:23\n\u2022 @Arthur Thanks! The question looks horrible for a timed exam and I became a little nervous! Jul 1 at 10:29\n\u2022 Denote $x=\\log_{21}3$ and $y=\\log_{21}7$. Then, $147=21\\times 7$ and $1323=21^2\\times 3$; the original expression is then $x^2+(1+y)(2+x)=x(x+y)+(x+y)+y+2$; iteratively use $x+y=1$ to simplify. Jul 1 at 10:47\n\u2022 @Prasun Biswas: the computation is simpler if one notices that $147=21^2/3$ and $1323=21^2 \\times 3$. Jul 1 at 10:51\n\u2022 @Mindlack: Nice! Indeed, that is far more simpler because we wouldn't need the $y$, we have $x^2+(2-x)(2+x)=x^2+4-x^2=4$ Somehow, I missed that. +1 Jul 1 at 10:55\n\nHere\u2019s a quick-and-dirty solution based on the fact that one of the answers must be correct.\n\nAll the quantities are positive. Since $$1323>441=21^2$$, its base-$$21$$ logarithm is greater than $$2$$. As $$147 \\geq 21 \\times 5$$, its base-$$21$$ logarithm is at least $$1.5$$. This means that the sum is greater than $$2 \\times 1.5$$ so it must be $$4$$.\n\n\u2022 I don't get the second point regarding $147 \\geq 21 \\times 5$. Could you elaborate? Jul 1 at 11:04\n\u2022 @user71207 $147\\ge21\\times5\\ge21\\sqrt{21}$ And the fact that $147\\ge21^{\\frac32}$ means $\\log_{21}(147)\\ge\\frac32$ Jul 1 at 11:09\n\n$$\\log_{21}1323=\\log_{21}(3^3\\cdot 7^2)=\\log_{21}3+\\log_{21}21^2=2+\\log_{21}3$$\n$$\\log_{21}147=\\log_{21}21+\\log_{21}7=1+\\log_{21}7$$\n\n$$(\\log_{21}3)^2+\\log_{21}147\\cdot\\log_{21}1323 \\\\=(\\log_{21}3)^2+(1+\\log_{21}7)(2+\\log_{21}3) \\\\=\\log_{21}3(\\log_{21}3+\\log_{21}7)+\\log_{21}3+2\\log_{21}7+2 \\\\=\\log_{21}3(\\log_{21}3+\\log_{21}7)+(\\log_{21}3+\\log_{21}7)+\\log_{21}7+2 \\\\=\\log_{21}3+1+\\log_{21}7+2 \\\\=4$$\n\n\u2022 Hi, the problem in this post and also this one that I asked were from entrance exam in the country I'm living (average time for solving each question is about 90 seconds). You solved both of them very well and I think you are very good at solving these questions! So can you please suggest some books or name of exam that you think are good for an exam that contains these type of questions? Thanks in advanced! Jul 27 at 19:58\n\u2022 I mainly use books that are meant for olympiads, so I don't know many books that are meant for solving questions fast. Jul 28 at 6:28\n\nAlternate:\n\nLet $$L(x)$$ denote $$\\log_{21}(x).$$\n\nThen $$[L(3)]^2 + [L(147) \\times L(9 \\times 147)]$$\n\n$$= [L(3)]^2 + ~\\langle ~L(147) \\times ~\\{ [2 \\times L(3)] + L(147) ~\\} ~\\rangle$$\n\n$$= [L(3)]^2 + [L(147)]^2 + [2 \\times L(3) \\times L(147)]$$\n\n$$= [L(3) + L(147)]^2.$$\n\nAt this point, you have that $$3 \\times 147 = 21^2$$,\n\nso $$[L(3) + L(147)] = 2.$$", "date": "2021-09-19 00:58:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4187654/what-is-the-value-of-left-log-213-right2-log-21147-log-211323/4187659", "openwebmath_score": 0.7811391949653625, "openwebmath_perplexity": 428.84657095729335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664105001544195, "lm_q2_score": 0.8947894590884704, "lm_q1q2_score": 0.8647339286905912}}
{"url": "https://stats.stackexchange.com/questions/207841/why-is-sst-sse-ssr-one-variable-linear-regression/207912#207912", "text": "# Why is $SST=SSE + SSR$? (One variable linear regression)\n\nNote: $SST$ = Sum of Squares Total, $SSE$ = Sum of Squared Errors, and $SSR$ = Regression Sum of Squares. The equation in the title is often written as:\n\n$$\\sum_{i=1}^n (y_i-\\bar y)^2=\\sum_{i=1}^n (y_i-\\hat y_i)^2+\\sum_{i=1}^n (\\hat y_i-\\bar y)^2$$\n\nPretty straightforward question, but I am looking for an intuitive explanation. Intuitively, it seems to me like $SST\\geq SSE+SSR$ would make more sense. For example, suppose point $x_i$ has corresponding y-value $y_i=5$ and $\\hat y_i=3$, where $\\hat y_i$ is the corresponding point on the regression line. Also assume that the mean y-value for the dataset is $\\bar y=0$. Then for this particular point i, $SST=(5-0)^2=5^2=25$, while $SSE=(5-3)^2=2^2=4$ and $SSR=(3-0)^2=3^2=9$. Obviously, $9+4<25$. Wouldn't this result generalize to the entire dataset? I don't get it.\n\nAdding and subtracting gives \\begin{eqnarray*} \\sum_{i=1}^n (y_i-\\bar y)^2&=&\\sum_{i=1}^n (y_i-\\hat y_i+\\hat y_i-\\bar y)^2\\\\ &=&\\sum_{i=1}^n (y_i-\\hat y_i)^2+2\\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y)+\\sum_{i=1}^n(\\hat y_i-\\bar y)^2 \\end{eqnarray*} So we need to show that $\\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y)=0$. Write $$\\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y)=\\sum_{i=1}^n(y_i-\\hat y_i)\\hat y_i-\\bar y\\sum_{i=1}^n(y_i-\\hat y_i)$$ So, (a) the residuals $e_i=y_i-\\hat y_i$ need to be orthogonal to the fitted values, $\\sum_{i=1}^n(y_i-\\hat y_i)\\hat y_i=0$, and (b) the sum of the fitted values needs to be equal to the sum of the dependent variable, $\\sum_{i=1}^ny_i=\\sum_{i=1}^n\\hat y_i$.\n\nActually, I think (a) is easier to show in matrix notation for general multiple regression of which the single variable case is a special case: \\begin{eqnarray*} e'X\\hat\\beta &=&(y-X\\hat\\beta)'X\\hat\\beta\\\\ &=&(y-X(X'X)^{-1}X'y)'X\\hat\\beta\\\\ &=&y'(X-X(X'X)^{-1}X'X)\\hat\\beta\\\\ &=&y'(X-X)\\hat\\beta=0 \\end{eqnarray*} As for (b), the derivative of the OLS criterion function with respect to the constant (so you need one in the regression for this to be true!), aka the normal equation, is $$\\frac{\\partial SSR}{\\partial\\hat\\alpha}=-2\\sum_i(y_i-\\hat\\alpha-\\hat\\beta x_i)=0,$$ which can be rearranged to $$\\sum_i y_i=n\\hat\\alpha+\\hat\\beta\\sum_ix_i$$ The right hand side of this equation evidently also is $\\sum_{i=1}^n\\hat y_i$, as $\\hat y_i=\\hat\\alpha+\\hat\\beta x_i$.\n\nThis is just the Pythagorean theorem!\n\nHence,\n\n$$Y'Y=(Y-X\\hat{\\beta})'(Y-X\\hat{\\beta})+(X\\hat{\\beta})'X\\hat{\\beta}$$\n\nor\n\n$$SST=SSE+SSR$$\n\n\u2022 Aug 2 '19 at 19:52\n\n(1) Intuition for why $SST = SSR + SSE$\n\nWhen we try to explain the total variation in Y ($SST$) with one explanatory variable, X, then there are exactly two sources of variability. First, there is the variability captured by X (Sum Square Regression), and second, there is the variability not captured by X (Sum Square Error). Hence, $SST = SSR + SSE$ (exact equality).\n\n(2) Geometric intuition\n\nSome of the total variation in the data (distance from datapoint to $\\bar{Y}$) is captured by the regression line (the distance from the regression line to $\\bar{Y}$) and error (distance from the point to the regression line). There's not room left for $SST$ to be greater than $SSE + SSR$.\n\n(3) The problem with your illustration\n\nYou can't look at SSE and SSR in a pointwise fashion. For a particular point, the residual may be large, so that there is more error than explanatory power from X. However, for other points, the residual will be small, so that the regression line explains a lot of the variability. They will balance out and ultimately $SST = SSR + SSE$. Of course this is not rigorous, but you can find proofs like the above.\n\nAlso notice that regression will not be defined for one point: $b_1 = \\frac{\\sum(X_i -\\bar{X})(Y_i-\\bar{Y}) }{\\sum (X_i -\\bar{X})^2}$, and you can see that the denominator will be zero, making estimation undefined.\n\nHope this helps.\n\n--Ryan M.\n\nWhen an intercept is included in linear regression(sum of residuals is zero), $$SST=SSE+SSR$$.\n\nprove $$\\begin{eqnarray*} SST&=&\\sum_{i=1}^n (y_i-\\bar y)^2\\\\&=&\\sum_{i=1}^n (y_i-\\hat y_i+\\hat y_i-\\bar y)^2\\\\&=&\\sum_{i=1}^n (y_i-\\hat y_i)^2+2\\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y)+\\sum_{i=1}^n(\\hat y_i-\\bar y)^2\\\\&=&SSE+SSR+2\\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y) \\end{eqnarray*}$$ Just need to prove last part is equal to 0: $$\\begin{eqnarray*} \\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y)&=&\\sum_{i=1}^n(y_i-\\beta_0-\\beta_1x_i)(\\beta_0+\\beta_1x_i-\\bar y)\\\\&=&(\\beta_0-\\bar y)\\sum_{i=1}^n(y_i-\\beta_0-\\beta_1x_i)+\\beta_1\\sum_{i=1}^n(y_i-\\beta_0-\\beta_1x_i)x_i \\end{eqnarray*}$$ In Least squares regression, the sum of the squares of the errors is minimized. $$SSE=\\displaystyle\\sum\\limits_{i=1}^n \\left(e_i \\right)^2= \\sum_{i=1}^n\\left(y_i - \\hat{y_i} \\right)^2= \\sum_{i=1}^n\\left(y_i -\\beta_0- \\beta_1x_i\\right)^2$$ Take the partial derivative of SSE with respect to $$\\beta_0$$ and setting it to zero. $$\\frac{\\partial{SSE}}{\\partial{\\beta_0}} = \\sum_{i=1}^n 2\\left(y_i - \\beta_0 - \\beta_1x_i\\right)^1 = 0$$ So $$\\sum_{i=1}^n \\left(y_i - \\beta_0 - \\beta_1x_i\\right)^1 = 0$$ Take the partial derivative of SSE with respect to $$\\beta_1$$ and setting it to zero. $$\\frac{\\partial{SSE}}{\\partial{\\beta_1}} = \\sum_{i=1}^n 2\\left(y_i - \\beta_0 - \\beta_1x_i\\right)^1 x_i = 0$$ So $$\\sum_{i=1}^n \\left(y_i - \\beta_0 - \\beta_1x_i\\right)^1 x_i = 0$$ Hence, $$\\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y)=(\\beta_0-\\bar y)\\sum_{i=1}^n(y_i-\\beta_0-\\beta_1x_i)+\\beta_1\\sum_{i=1}^n(y_i-\\beta_0-\\beta_1x_i)x_i=0$$ $$SST=SSE+SSR+2\\sum_{i=1}^n(y_i-\\hat y_i)(\\hat y_i-\\bar y)=SSE+SSR$$\n\nHere is a great graphical representation of why SST = SSR + SSE.\n\n\u2022 How can we see from this picture that predicted values and residuals are orthogonal? Mar 4 at 12:17\n\nIf a model predicts $$3$$ and the residual is $$2$$ because the actual value is $$5$$, it doesn't look like variance is decomposing, since $$3^2 + 2^2 \\neq 5^2$$.\n\nIf you only have one data point, your model would fit it perfectly and the residual would be zero, so you can't get that case by itself. There have to be multiple data points.\n\nWith multiple data points, the residuals won't all be positive.\n\nIf the model predicts 3, a residual of $$+2$$ and $$-2$$ should be equally likely. The balance of increases and decreases gives a nice cancellation:\n\n$$\\frac{(3+2)^2 + (3-2)^2}{2} = \\frac{25+1}{2} = 13 = 9 + 4 = 3^2 + 2^2$$\n\nThis property, that $$(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$$, is what makes variance decomposition work, for uncorrelated components.", "date": "2021-10-22 08:45:44", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/207841/why-is-sst-sse-ssr-one-variable-linear-regression/207912#207912", "openwebmath_score": 0.859199583530426, "openwebmath_perplexity": 380.481113977744, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987375050346933, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8647302153683112}}
{"url": "https://brilliant.org/discussions/thread/easy-money/", "text": "# Easy Money\n\nOn a table before you lie two envelopes, identical in appearance. You are told that one envelope contains twice as much money as the other, but given no indication which has more. You are then allowed to choose one of the envelopes and keep the money it contains.\n\nYou then pick one of the envelopes at random, but before you look inside you are offered the chance to exchange your envelope for the other. Should you exchange envelopes, (assuming that the more money you get, the better)?\n\nOne line of thinking is this: If the envelope you initially choose has $$x$$ dollars in it, then there is a $$50$$% chance that the other envelope has $$2x$$ dollars in it and a $$50$$% chance that it has $$\\frac{x}{2}$$ dollars in it. Thus the expected amount of money you would end up with if you chose to exchange envelopes would be\n\n$$\\frac{1}{2} * 2x + \\frac{1}{2} * \\frac{x}{2} = \\frac{5}{4} x$$ dollars.\n\nYou thus decide to exchange envelopes. But before doing so, you think to yourself, \"But I could go through the same thought process with the other envelope, concluding that I should exchange my new envelope for my old one, and then go through the same thought process again, and again, ad infinitum. Now I have no idea what to do.\"\n\nHow can you resolve this paradox?\n\nEdit: As mentioned in my comment below, the real \"puzzle\" here is to identify the flaw in the reasoning I outlined above. Or is there a flaw?\n\nNote by Brian Charlesworth\n4\u00a0years, 4\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nFirst, look at the case when the envelope is exchanged. If you pick the envelope with $$x$$ dollars $$(P=1/2)$$ and exchange, you win $$2x$$ dollars. Instead, if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and exchange, then you win $$x$$ dollars.\n\n$\\therefore E(\\text{Money when envelope is exchanged})=\\dfrac{1}{2}2x+\\dfrac{1}{2}x=\\dfrac{3x}{2}$\n\nNow, look at the case when the envelope is not exchanged. If you pick the envelope with $$x$$dollars $$(P=1/2)$$ and don't exchange, you win $$x$$ dollars. But if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and don't exchange, then you win $$2x$$ dollars.\n\n$\\therefore E(\\text{Money when envelope is not exchanged})=\\dfrac{1}{2}2x+\\dfrac{1}{2}x=\\dfrac{3x}{2}$\n\nSo it makes no difference at all whether you exchange or don't exchange your envelope.\n\n- 4\u00a0years, 4\u00a0months ago\n\nYes, that is a good analysis. Since we are not given any further information prior to making a choice whether or not to exchange envelopes, we are in the same position as we were with our initial choice, i.e., our expected winnings will be $$\\frac{3x}{2}$$.\n\nBut the \"real\" puzzle here is to find the logical miscue in the line of reasoning I laid out. It can be argued that in the two scenarios I describe, (namely that I start out with the lesser amount of money or the greater), the value $$x$$ actually represents different amounts in each case, so that I can't use the same variable $$x$$ for each of them, making my equation invalid.\n\nHowever, suppose I did peek at the amount in the envelope before making a decision. Now this provides me with \"information\", but in reality it's useless since I have no clue as to whether it's the lesser or greater amount. But since we now can attach an actual value to $$x$$, the mathematics I applied in my initial analysis becomes re-validated, and hence so does my \"conclusion\". But since, as noted, the information regarding the amount in the initially chosen envelope is in reality useless, should it matter whether we know the actual amount as far as the mathematical analysis is concerned? From this perspective, no matter what amount we see inside the initial envelope we should conclude that switching envelopes is preferable.\n\nSo the puzzle of \"where's the flaw in logic?\" is more a question of perspective; are we dealing with unconditional or conditional probability values, and why should this difference in perspective matter given the above discussion? A surprising amount of attention has been given to this paradox, commonly referred to as the \"Two Envelope Paradox\", with no universally accepted resolution as of yet. I think why it does garner such attention is that it cuts to the heart of the foundation of probability theory, and as with any foundation, the desire is that it must be as secure as possible, and if it is not, it must be either fortified or torn down.\n\nHere is an accessible discussion on the \"search for the flaw\" and why many find it so intriguing.\n\n- 4\u00a0years, 4\u00a0months ago\n\nI'm not a mathematician, just a dilettante of sorts but something tells me we're over complicating this paradox. There are only two choices and only two possibilities. The money is either greater or lesser and you either switch or stay. The fact that the sum of all probabilities is 1 is infallible. We simply assume we have knowledge we previously didn't have and compute what the probabilities are to find out what is what.\n\nHere's another question: Are we then told that \"hey, you chose the lesser or greater amount of money\" after we've made our choice? I think not. So the probability will not even depend on outcome. For all intents and purposes, we need not even bother with the maths of 'x' and '2x'; and we might as well adopt an approach with x and y since the relationship between the monies simply do not matter. I might as well be a choice between winning the lottery and detonating an atomic bomb.\n\n- 4\u00a0years ago\n\nI think that the reasons this \"paradox\" has received so much attention over the years are more esoteric in nature than practical. For some it raises questions about the foundation of probability theory regarding the absolute distinction between conditional/unconditional scenarios and whether or not there might be a grey area there. However, mathematicians do have a tendency to \"overcomplicate\" things, and this may a case in point. :)\n\nAs for the choice between winning the lottery and detonating an atomic bomb, well, I think I would just walk away from the table; too many lives other than my own would be at stake without them having any say in the matter. However, if it were a choice between my eternal health, wealth and happiness on the one hand and my immediate extinction on the other, I would give it a bit more thought. I would probably still walk away, since it would feel like I'm making some kind of deal with the Devil, but it does make for an interesting psychological experiment. This is an entirely different question than the previous \"paradox\" we were dealing with; with the money in the two envelopes, no matter what choice we made we would be richer than before. With this new question, there is the possibility of a terrible outcome, so the first choice becomes whether to engage at all, and if we do then why do we think it is worth such a high risk of calamity. The more we would have to lose in our present life the less likely we would engage, but if we did not have much to live for then the risk might be worth it. However, as I said before, only the Devil would make such a proposition, so anyone, no matter what their circumstances, would be wise to just walk away. Sorry, I do ramble on, but i always do that when a good philosophical question happens by. :)\n\n- 4\u00a0years ago\n\nHahaha.I do love to read you rambling on. It's more engaging than many other types of conversations. Your ending pretty much wrapped up my premise: which is that, the paradox seems to rely on the probability of making a more money outcome which is identical to that of making a lesser money. I do have some philosophical questions if you're interested, I'd love to read your views. I'm relatively new on this site so is it OK if I create a note and invite you to give your thoughts on it?\n\n- 4\u00a0years ago\n\nOf course; I look forward to seeing what the topic is. :)\n\n- 4\u00a0years ago\n\nHi again Mr. Charlesworth\n\nHere are two notes I have posted. I invite you to share your thoughts on them: https://brilliant.org/discussions/thread/number-bases-in-a-different-world/# https://brilliant.org/discussions/thread/the-infiniteness-of-light/?ref_id=654556\n\nHi @Calvin Lin. Do you think it's a good idea to find a way to incorporate Philosophy (Logic and Reasoning) on Brilliant? I think it is a very fascinating pseudoscience and it would appeal to many older people and others who are not strictly geeks and nerds like the rest of us. Just a thought\n\n- 4\u00a0years ago\n\nLet's incorporate Speculative Science And Math, as distinct from formal Philosophy. You know, something that would appeal to one old guy here who wants to try his hand in being a crank.\n\nI'm reading Brian's posted scenario, and it seems to me that if it's known that one envelope contains twice the other, then the expected value in choosing either is simply $$\\dfrac { 1 }{ 2 } x+\\dfrac { 1 }{ 2 } 2x=\\dfrac { 3 }{ 2 } x$$ either way, as pointed out by Pratik. It's 50% percent chance that it's twice the one you've chosen, and 50% percent chance it's half that--but you can't assume the same $$x$$ dollars in your chosen envelope in both cases! That was the flaw. Maybe Brian is trying to alluding to the problem of trying to determine probabilities when information is incomplete, which is what led to Bayesian probability theory. Let me offer a slightly different scenario. You are given $$2$$ envelopes. You are told that one of them may or may not contain a valuable stock certificate. You open one up and do not find it inside. What's the probability the other one does? Extend that to $$100$$ such envelopes. This is actually what happens when you are looking for gold in the ground in a certain area when you don't have much information to go on. After so much digging, at what point in time do you give up? Are more likely to find gold, because you've already almost eliminated the places where it can be, or are you more less likely to find gold, because you've already tried to find it and failed? A lot of science can be like that.\n\n- 4\u00a0years ago\n\nOh yes, that's a great idea because what I was hoping for was an avenue to wonder beyond the limits of our recorded knowledge and rub minds on what could be and what not.\n\nI'm fascinated this \"blind\" probability theory. It is a little bit anti cognitive. Which reminds me of something else which defies our flawed cognizance. Consider flipping a coin and getting heads five times in a row. Flipping it a sixth time seems as though there is a higher probability to get tails. In fact, I'd bet that if you did a survey of ten mathematics professors, a higher percentage of them would choose tails for that sixth coin toss except they are just fighting against their human nature. But sometimes I think about it and really wonder if the law of averages is that subjective. If you mix two liquids together, they mix together perfectly EVERY TIME if they are of similar viscosity. I'm rambling. I have even forgotten what the point I'm trying to make is\n\n- 4\u00a0years ago\n\nA lot of theoretical physics was minds rambling before something more concrete came out of it. Ramble on.\n\nLikewise, a lot of technology was preceded by science fiction. Smart phones? Didn't Star Trek popularize that idea?\n\n- 4\u00a0years ago", "date": "2019-03-19 18:03:32", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/easy-money/", "openwebmath_score": 0.7994460463523865, "openwebmath_perplexity": 609.3471225256254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9835969684454966, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8647261000127658}}
{"url": "https://mathhelpboards.com/threads/4x4-matrix-with-rank-b-4-and-b-2-3.6384/", "text": "# 4x4 Matrix with rank B=4 and B^2=3\n\n#### Petrus\n\n##### Well-known member\nHello MHB,\n\"Can we construct a $$\\displaystyle 4x4$$ Matrix $$\\displaystyle B$$ so that rank $$\\displaystyle B=4$$ but rank $$\\displaystyle B^2=3$$\"\nMy thought:\nwe got one condition for this to work is that det $$\\displaystyle B=0$$ and det $$\\displaystyle B^2 \\neq 0$$ and B also have to be a upper/lower or identity Matrix. And this Will not work.. I am wrong or can I explain this in a better way?\n\nRegards,\n$$\\displaystyle |\\pi\\rangle$$\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\n\"Can we construct a $$\\displaystyle 4x4$$ Matrix $$\\displaystyle B$$ so that rank $$\\displaystyle B=4$$ but rank $$\\displaystyle B^2=3$$\"\nMy thought:\nwe got one condition for this to work is that det $$\\displaystyle B=0$$ and det $$\\displaystyle B^2 \\neq 0$$\nIt's the other way around: $\\mathop{\\text{rank}}(B)=4\\iff\\det(B)\\ne0$ and $\\mathop{\\text{rank}}(B^2)=3\\implies\\det(B^2)=0$. But you are right that this is impossible because $\\det(B^2)=(\\det(B))^2$.\n\n#### Petrus\n\n##### Well-known member\nIt's the other way around: $\\mathop{\\text{rank}}(B)=4\\iff\\det(B)\\ne0$ and $\\mathop{\\text{rank}}(B^2)=3\\implies\\det(B^2)=0$. But you are right that this is impossible because $\\det(B^2)=(\\det(B))^2$.\nHello Evgeny.Makarov,\nthanks for fast respond and I meant that! And thanks for showing me this I did not know that $\\det(B^2)=(\\det(B))^2$ That Was what I Was looking for\n\nRegards,\n$$\\displaystyle |\\pi\\rangle$$\n\nLast edited:\n\n#### Fernando Revilla\n\n##### Well-known member\nMHB Math Helper\nAn alternative proof (without using determinants):\n\nConsider the linear map $B:\\mathbb{R}^4\\to \\mathbb{R}^4,\\; x\\to Bx.$ As $\\operatorname{rank}B=4,$ $\\operatorname{nullity}B=0,$ wich implies $B$ is bijective. But the composition of bijective maps is bijective, so $\\operatorname{rank}B^2=4.$\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nAnother formulation:\n\nAs rank(B) = 4, B is surjective, that is, B(R4) = R4 (for this is what rank means: the dimension of the image space (or column space) of B, and R4 is the ONLY 4-dimensional subspace of R4\u200b).\n\nConsequently, B2(R4) = B(B(R4)) = B(R4) = R4, from which we conclude B2 is likewise surjective, and thus rank(B2) = 4 as well.\n\n(I only post this to indicate one need not even invoke the rank-nullity theorem).\n\n#### Petrus\n\n##### Well-known member\nthis is impossible because $\\det(B^2)=(\\det(B))^2$.\nNow that I think about I remember a sats that said $$\\displaystyle |AB|=|A||B|$$ but in this case $$\\displaystyle A=B$$ hmm I need to find the proof for this.\n\nRegards,\n$$\\displaystyle |\\pi\\rangle$$\n\nMHB Math Scholar\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nI saw one neat proof of $\\det(AB)=\\det(A)\\det(B)$ in Linear Algebra and Its Applications by Gilbert Strang. He defines $\\det(\\cdot)$ as a function satisfying three properties:\n\n(1) $\\det(I)=1$ where $I$ is the identity matrix;\n(2) it changes sign when two adjacent rows are swapped;\n(3) it is linear on the first row.\n\nSigned volume in an orthonormal basis satisfies these properties, so this definition is much more intuitive than the Leibniz formula, which is derivable from (1)\u2013(3).\n\nNow, to prove that $\\det(AB)=\\det(A)\\det(B)$, fix $B$ and consider $d(A)=\\det(AB)/\\det(B)$. It is possible to show that $d(A)$ satisfies (1)\u2013(3), and so $d(A)=\\det(A)$.\n\nNow that I looked at the StackExchange link, this is answer #2, which is highest-ranked.", "date": "2020-09-19 09:37:39", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/4x4-matrix-with-rank-b-4-and-b-2-3.6384/", "openwebmath_score": 0.940988302230835, "openwebmath_perplexity": 718.0092056558483, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9835969674838368, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.8647260976100808}}
{"url": "https://math.stackexchange.com/questions/414065/closed-form-of-sum-limits-i-1n-left-lfloor-fracni-right-rfloor2", "text": "# Closed form of $\\sum\\limits_{i=1}^n\\left\\lfloor\\frac{n}{i}\\right\\rfloor^2$?\n\nDoes $\\displaystyle\\sum_{i=1}^n\\left\\lfloor\\dfrac{n}{i}\\right\\rfloor^2$ admit a closed form expression?\n\n\u2022 Unlikely; see the related math.stackexchange.com/q/384520/73324 \u2013\u00a0vadim123 Jun 7 '13 at 18:47\n\u2022 @Martin, is the sequence formed by these sums not a sequence? \u2013\u00a0Antonio Vargas Dec 1 '16 at 17:38\n\u2022 @AntonioVargas The same could be said about any finite sum of the form $\\sum_{k=1}^n a_n$. Of course, if you think that for some reasone the tag belongs there, you can edit it back. My understanding is that the tag-excerpt says rather clearly: \"For questions on finite sums use the (summation) instead.\" \u2013\u00a0Martin Sleziak Dec 1 '16 at 17:41\n\nTo see how the first term in the asymptotic expansion is obtained, put $$a(n) = \\sum_{k=1}^n \\bigg\\lfloor \\frac{n}{k} \\bigg\\rfloor^2$$ and note that $$a(n+1)-a(n) = 1 + \\sum_{k=1}^n \\left(\\bigg\\lfloor \\frac{n+1}{k} \\bigg\\rfloor^2 - \\bigg\\lfloor \\frac{n}{k} \\bigg\\rfloor^2\\right) \\\\= 1 + \\sum_{d|n+1 \\atop d<n+1} \\left(\\left(\\frac{n+1}{d}\\right)^2 - \\left(\\frac{n+1}{d}-1\\right)^2\\right) = \\sum_{d|n+1} \\left(2\\left(\\frac{n+1}{d}\\right)-1\\right) \\\\= 2\\sigma(n+1)-\\tau(n+1).$$\n\nIt now follows that $$a(n) = 2\\sum_{k=1}^n \\sigma(k) - \\sum_{k=1}^n \\tau(k) = \\sum_{k=1}^n \\left(2\\sigma(k)-\\tau(k)\\right).$$ We can apply the Wiener-Ikehara theorem to this sum, working with the Dirichlet series $$L(s) = \\sum_{n\\ge 1} \\frac{2\\sigma(n)-\\tau(n)}{n^s} = 2\\zeta(s-1)\\zeta(s)-\\zeta(s)^2.$$ We have $$\\operatorname{Res}(L(s); s=2) = \\frac{\\pi^2}{3},$$ so that by the aforementioned theorem, $$a(n) \\sim \\frac{\\pi^2/3}{2} n^2 = \\frac{\\pi^2}{6} n^2.$$ In fact we can use Mellin-Perron summation to predict, but not quite prove, the next terms in the asymptotic expansion, getting $$a(n) = \\left(\\sigma(n)-\\frac{1}{2}\\tau(n)\\right) + \\frac{1}{2\\pi i} \\int_{5/2-i\\infty}^{5/2+i\\infty} L(s) n^s \\frac{ds}{s}$$ which yields $$a(n) \\sim \\left(\\sigma(n)-\\frac{1}{2}\\tau(n)\\right) +\\frac{\\pi^2}{6} n^2 - (\\log n + 2 \\gamma)n - \\frac{1}{6}.$$ This approximation is quite good, giving $16085.71386$ for $n=100$ when the correct value is $16116$ and $1639203.715$ for $n=1000$ when the correct value is $1639093.$\n\nThis is A222548 in the online encyclopedia of integer sequences. They don't provide a closed form, but they do give the following:\n\n$$a(n)\\approx\\frac{\\pi^2}{6}n^2+O(n\\log n)$$\n\nLet me complement @Marko Riedel's excellent answer by providing a low-tech approach to the leading term.\n\nLet us denote by $\\mathbf{1}\\{\\cdot\\}$ the indicator function of the set $\\{\\cdot\\}$. Then using the identity\n\n$$\\lfloor n/i \\rfloor = \\sum_{j=1}^{n} \\mathbf{1}\\{ij \\leq n \\}$$\n\nwe can rearrange the sum to find:\n\n\\begin{align*} a(n) &= \\sum_{i=1}^{n} \\sum_{\\substack{1 \\leq j \\leq n \\\\ 1 \\leq k \\leq n}} \\mathbf{1}\\{ij \\leq n\\}\\mathbf{1}\\{ik \\leq n\\} = \\sum_{\\substack{1 \\leq j \\leq n \\\\ 1 \\leq k \\leq n}} \\sum_{i=1}^{n} \\mathbf{1}\\Big\\{i \\leq \\frac{n}{\\max\\{j,k\\}}\\Big\\} \\\\ &= \\sum_{\\substack{1 \\leq j \\leq n \\\\ 1 \\leq k \\leq n}} \\left\\lfloor \\frac{n}{\\max\\{j,k\\}} \\right\\rfloor = \\sum_{l=1}^{n} \\sum_{\\substack{1 \\leq j \\leq n \\\\ 1 \\leq k \\leq n}} \\left\\lfloor \\frac{n}{l} \\right\\rfloor \\mathbf{1}\\{\\max\\{j,k\\} = l\\} \\\\ &= \\sum_{l=1}^{n} (2l-1) \\left\\lfloor \\frac{n}{l} \\right\\rfloor. \\end{align*}\n\nDividing both sides by $n^2$, the RHS can be identified as Riemann sum and thus\n\n$$\\frac{a(n)}{n^2} = \\sum_{l=1}^{n} \\frac{2l-1}{n} \\left\\lfloor \\frac{n}{l} \\right\\rfloor \\frac{1}{n} \\xrightarrow[\\ n\\to\\infty \\ ]{} \\int_{0}^{1} 2x \\left\\lfloor\\frac{1}{x}\\right\\rfloor \\, dx = \\zeta(2).$$\n\nHere, the last integral can be easily computed by applying the substitution $x \\mapsto 1/x$.\n\nAn Answer Without Number Theoretic Functions\n\nBreak up the sum as follows: $$\\sum_{k=1}^n\\left\\lfloor\\frac nk\\right\\rfloor^2 =\\sum_{k=1}^n\\frac{n^2}{k^2}-\\sum_{k=1}^n\\left(\\frac{n^2}{k^2}-\\left\\lfloor\\frac nk\\right\\rfloor^2\\right)\\tag{1}$$ We can apply the Euler-Maclaurin Sum Formula $$\\sum_{k=1}^n\\frac{n^2}{k^2}\\sim n^2\\left(\\frac{\\pi^2}6-\\frac1n+\\frac1{2n^2}\\right)\\tag{2}$$ If we note that $\\left\\{\\frac nk\\right\\}\\lt1-\\frac1k$, we see that \\begin{align} \\sum_{k=1}^n\\left(\\frac{n^2}{k^2}-\\left\\lfloor\\frac nk\\right\\rfloor^2\\right) &=\\sum_{k=1}^n\\left(\\frac nk-\\left\\lfloor\\frac nk\\right\\rfloor\\right)\\left(\\frac nk+\\left\\lfloor\\frac nk\\right\\rfloor\\right)\\\\ &=\\sum_{k=1}^n\\left\\{\\frac nk\\right\\}\\left(2\\frac{n}{k}-\\left\\{\\frac nk\\right\\}\\right)\\\\ &\\le\\sum_{k=1}^n\\left(1-\\frac1k\\right)2\\frac nk\\\\[9pt] &\\le2n\\log(n)\\tag{3} \\end{align} Therefore, by $(1)$, $(2)$, and $(3)$ we have $$\\bbox[5px,border:2px solid #C0A000]{\\sum_{k=1}^n\\left\\lfloor\\frac nk\\right\\rfloor^2=n^2\\frac{\\pi^2}6+O(n\\log(n))}\\tag{4}$$ However, if we use $\\frac12{-}\\frac1{2k}$ for the mean value of $\\left\\{\\frac nk\\right\\}$ and $\\frac13{-}\\frac1{2k}{+}\\frac1{6k^2}$ for the mean value of $\\left\\{\\frac nk\\right\\}^2$ \\begin{align} \\sum_{k=1}^n\\left(\\frac{n^2}{k^2}-\\left\\lfloor\\frac nk\\right\\rfloor^2\\right) &=\\sum_{k=1}^n\\left(\\frac nk-\\left\\lfloor\\frac nk\\right\\rfloor\\right)\\left(\\frac nk+\\left\\lfloor\\frac nk\\right\\rfloor\\right)\\\\ &=\\sum_{k=1}^n\\left\\{\\frac nk\\right\\}\\left(2\\frac{n}{k}-\\left\\{\\frac nk\\right\\}\\right)\\\\ &\\approx\\sum_{k=1}^n\\left(\\frac12-\\frac1{2k}\\right)2\\frac{n}{k}-\\left(\\frac13-\\frac1{2k}+\\frac1{6k^2}\\right)\\\\ &=n\\log(n)+n\\left(\\gamma-\\frac{\\pi^2}6-\\frac13\\right)+O(\\log(n))\\tag{5} \\end{align} Thus, if we look at the average behavior, we can use $(1)$, $(2)$, and approximation $(5)$ to get $$\\bbox[5px,border:2px solid #C0A000]{\\sum_{k=1}^n\\left\\lfloor\\frac nk\\right\\rfloor^2 \\approx n^2\\frac{\\pi^2}6-n\\log(n)+n\\left(\\frac{\\pi^2}6-\\gamma-\\frac23\\right)}\\tag{6}$$\n\nThe value given by $(6)$ for $n=98$ is $15387.9230$ while the actual value is $15381$.\nThe value given by $(6)$ for $n=1001$ is $1641711.3692$ while the actual value is $1641772$.\n\nThe error of the approximation in $(6)$ appears to be proportional to $n$. Here is a plot of that error, for $1\\le n\\le3000$, divided by $n$:\n\nThe error divided by $n$ has a mean of $-0.0746805451$ and a standard deviation of $0.6513592109$.\n\nUsing number theoretic functions, Marko Riedel's answer gives a better approximation. Here is a plot of the error of that approximation, for $1\\le n\\le3000$, divided by $n$:\nThe error divided by $n$ has a mean of $0.0060900069$ and a standard deviation of $0.3674311090$. This is better than the error distribution in my answer, but it does require factoring $n$ to compute the number theoretic functions.", "date": "2019-05-20 15:24:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/414065/closed-form-of-sum-limits-i-1n-left-lfloor-fracni-right-rfloor2", "openwebmath_score": 0.999997615814209, "openwebmath_perplexity": 336.1803810455722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856483, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8647099537849977}}
{"url": "https://math.stackexchange.com/questions/122119/does-there-exist-more-than-3-connected-open-sets-in-the-plane-with-the-same-boun", "text": "# Does there exist more than 3 connected open sets in the plane with the same boundary?\n\nI've wondered about the following question, whose answer is perhaps well known (in this case I apologize in advance).\n\nThe Lakes of Wada are a famous example of three disjoint connected open sets of the plane with the counterintuitive property that they all have the same boundary (!)\n\nMy question is the following :\n\nCan we find four disjoint connected open sets of the plane that have the same boundary?\n\nMore generally :\n\nFor each $n \\geq 3$, does there exist $n$ disjoint connected open sets of the plane that have the same boundary? If not, then what is the smallest $n$ such that the answer is no?\n\nThank you, Malik\n\n\u2022 The article you link to says \"A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, ...and so on.\", so the answer seems to be \"you can do it for all $n$ including $n =\\aleph_0$\". I'd guess (but have no idea really) that the corresponding question for other cardinalities is independent of ZFC, assuming NOT CH. Though, since $\\mathbb{R}^n$ is first countable, there may be an answer... \u2013\u00a0Jason DeVito Mar 19 '12 at 15:55\n\u2022 Found this in the link. It seems to answer your question. \"A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order ...\" \u2013\u00a0Patrick Mar 19 '12 at 15:57\n\u2022 Given that the basins of attraction of Newton's method for $z^3=1$ are an example for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. \u2013\u00a0lhf Mar 19 '12 at 16:49\n\u2022 Jason, separability precludes finding any uncountable collection of pairwise disjoint nonempty open sets, irrespective of any conditions you put on their boundaries. \u2013\u00a0user83827 Mar 19 '12 at 17:23\n\u2022 @Student73, ah, this may contain a proof: Frame, Michael; Neger, Nial. Newton's method and the Wada property: a graphical approach. College Math. J. 38 (2007), no. 3, 192\u2013204. MR2310015 (2008e:37082) \u2013\u00a0lhf Mar 19 '12 at 17:31\n\nGiven that the basins of attraction of Newton's method for $z^3=1$ are Wada sets for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. I don't know a reference to a proof but try these:", "date": "2019-06-17 22:36:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/122119/does-there-exist-more-than-3-connected-open-sets-in-the-plane-with-the-same-boun", "openwebmath_score": 0.882344663143158, "openwebmath_perplexity": 231.78820051362365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357216481429, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8647099472944348}}
{"url": "http://mathhelpboards.com/calculus-10/s4-13-5-41-relation-2-planes-20440.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e", "text": "# Thread: s4.13.5.41 relation of 2 planes\n\n1. $\\tiny{{s4}.{13}.{5}.{41}}$\n\n$\\textsf{find if planes are$\\parallel, \\perp$or$\\angle$of intersection }\\\\$\n\\begin{align}\n\\displaystyle\n{P_1}&={x+z=1}\\\\\n\\therefore n_1&=\\langle 1,0,1 \\rangle\\\\\n\\\\\n{P_2}&={y+z=1}\\\\\n\\therefore n_2&=\\langle 0,1,1 \\rangle\\\\\n\\\\\n\\cos(\\theta)&=\n\\frac{n_1\\cdot n_2}{|n_1||n_2|}\\\\\n&=\\frac{1(0)+0(1)+1(1)}\n{\\sqrt{1+1}\\cdot\\sqrt{1+1}}\n=\\frac{1}{2}\\\\\n\\cos^{-1}\\left({\\frac{1}{2}}\\right)&=\n\\color{red}{60^o}\n\\end{align}\n\n$\\textit{there are 2 more problems like this so presume this is best method.. }\\\\$\n$\\textit{didn't know if it is common notation to call a plane$P_1$}$\n\n2. Originally Posted by karush\n$\\tiny{{s4}.{13}.{5}.{41}}$\n\n$\\textsf{find if planes are$\\parallel, \\perp$or$\\angle$of intersection }\\\\$\n\\begin{align}\n\\displaystyle\n{P_1}&={x+z=1}\\\\\n\\therefore n_1&=\\langle 1,0,1 \\rangle\\\\\n\\\\\n{P_2}&={y+z=1}\\\\\n\\therefore n_2&=\\langle 0,1,1 \\rangle\\\\\n\\\\\n\\cos(\\theta)&=\n\\frac{n_1\\cdot n_2}{|n_1||n_2|}\\\\\n&=\\frac{1(0)+0(1)+1(1)}\n{\\sqrt{1+1}\\cdot\\sqrt{1+1}}\n=\\frac{1}{2}\\\\\n\\cos^{-1}\\left({\\frac{1}{2}}\\right)&=\n\\color{red}{60^o}\n\\end{align}\n\n$\\textit{there are 2 more problems like this so presume this is best method.. }\\\\$\n$\\textit{didn't know if it is common notation to call a plane$P_1$}$\nEverything you've posted is fine. You can give a plane any name you like, but I would write something like this:\n\n\\displaystyle \\begin{align*} P_1 : x + z = 1 \\end{align*}\n\nThat way we can see that \\displaystyle \\begin{align*} P_1 \\end{align*} is DEFINED as the relationship \"the sum of the x and z values needs to be 1\", not that it is some variable that has something to do with the equation.\n\n3. The problem asks you to do three things:\n1) determine if the planes are parallel.\n2) determine if the planes are perpendicular.\n3) if neither of those, determine the angle of intersection of the two planes.\n\nYes, by using that formula to determine the angle, you can then answer all three questions but it should be simpler to determine the first two without using that formula:\n\nThe two planes are parallel if and only if the two normal vectors are parallel- if one is a multiple of the other.\n\nThe two planes are perpendicular if and only if the two normal vectors are perpendicular: if their dot product is 0.\n\nIf neither of those is true, then you can use the dot product you found as the numerator in the formula to determine the angle.\n\nhard to see that in their examples\n\nthe next 2 problems are probably\n$\\displaystyle \\parallel , \\perp$\n\n$\\tiny{s4.854.13.5.43}$\n$\\textsf{Determine if the 2 given planes are perpendicular, parallel or at an angle to each other}$\n\\begin{align}\n\\displaystyle\n{p_1}&:{x+4y-3z=1}\n\\therefore n_1=\\langle 1,4,-3 \\rangle\\\\\n\\nonumber\\\\\n{p_2}&:{-3x+6y+7z=0} \\therefore n_2=\\langle -3,6,7 \\rangle\n\\end{align}\n\\begin{align}\n\\displaystyle\n\\cos(\\theta)&= \\frac{n_1\\cdot n_2}{|n_1||n_2|}=0\\\\\n\\therefore p_1 &\\perp p_2\n\\end{align}\n\n$\\tiny{s4.854.13.5.45}$\n\\begin{align}\n\\displaystyle\n{p_1}&:{2x+4y-2z=1}\n\\therefore n_1=\\langle 2,4,-2 \\rangle\\\\\n\\nonumber\\\\\n{p_2}&:{-3x+6y+3z=0} \\therefore n_2=\\langle -3,6,7 \\rangle \\\\\nn_1&=-\\frac{2}{3} n_2 \\\\\n&\\therefore n_1\\parallel n_2\n\\end{align}\n\n$\\textit{btw what is the input string to see the 2 plane on a W|A graph?}$\n\n#### Posting Permissions\n\n\u2022 You may not post new threads\n\u2022 You may not post replies\n\u2022 You may not post attachments\n\u2022 You may not edit your posts\n\u2022", "date": "2017-11-20 07:30:48", "meta": {"domain": "mathhelpboards.com", "url": "http://mathhelpboards.com/calculus-10/s4-13-5-41-relation-2-planes-20440.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e", "openwebmath_score": 0.9061705470085144, "openwebmath_perplexity": 1682.2355830237636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357248544007, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.8647099470465943}}
{"url": "https://math.stackexchange.com/questions/1386606/probability-of-rolling-a-dice-8-times-before-all-numbers-are-shown/1387649", "text": "# Probability of rolling a dice 8 times before all numbers are shown.\n\nWhat is the probability of having to roll a (six sided) dice at least 8 times before you get to see all of the numbers at least once?\n\nI don't really have a clue how to work this out.\n\nEdit: If we are trying to find the number of situations where all of the numbers are shown, for seven rolls, a favorable outcome would be one in which there are two numbers the same, for example: 1123456. These numbers can be arranged in 7!/2!5! ways, and there are 6 different numbers that could be repeated. So the probability of getting all 6 numbers with 7 throws is (6*7!/2!5!)/6^7 = 126/279936. Is that right? Then 1 minus this is the probability?\n\nEdit: prob. of not getting all six numbers with seven rolls = 1-prob of getting all six numbers with seven rolls\n\nSix numbers same with seven rolls means one number repeated twice 7!/2! ways of doing this for each repeated number 6*(7!/2!) is number of ways of getting all six numbers with seven rolls. Total number of outcomes 6^7. Prob of getting all six numbers with seven rolls = (6*(7!/2!))/6^7 = 0.054 Prob of not getting all six numbers with seven rolls (=prob of needing at least 8 rolls to get all numbers) = 1-0.054 = 0.946\n\n\u2022 Let $p$ be the probability that in $7$ rolls you see all faces at least once. We find $p$. Then the answer to the original problem is $1-p$. It should not be too hard to find $p$, by counting the \"favourables\" and dividing by $6^7$. \u2013\u00a0Andr\u00e9 Nicolas Aug 6 '15 at 13:34\n\u2022 As a further hint to counting the \"favorables\" for $p$, note that if you roll 7 times you will see at least one number repeated. But if this happens more than once you will not get to see all 6 numbers. \u2013\u00a0David K Aug 6 '15 at 13:50\n\u2022 The title and text of the question don't match. Please clarify whether \"at least\" is intended, as in the text, or \"exactly\", as the title appears to imply. \u2013\u00a0joriki Aug 6 '15 at 14:17\n\u2022 To clarify are you asking what is the probability of needing exactly 8 rolls to see all numbers or 8 or more rolls? \u2013\u00a0Warren Hill Aug 7 '15 at 8:20\n\nRephrase the question:\n\nWhat is the probability of not seeing all $6$ values when rolling a die $7$ times?\n\nFind the probability of the complementary event:\n\nWhat is the probability of seeing all $6$ values when rolling a die $7$ times?\n\nUse the inclusion/exclusion principle in order to count the number of ways:\n\n\u2022 Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\\binom66\\cdot6^7$\n\u2022 Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\\binom65\\cdot5^7$\n\u2022 Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\\binom64\\cdot4^7$\n\u2022 Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\\binom63\\cdot3^7$\n\u2022 Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\\binom62\\cdot2^7$\n\u2022 Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\\binom61\\cdot1^7$\n\nDivide the result by the total number of ways, which is $6^7$:\n\n$$\\frac{\\binom66\\cdot6^7-\\binom65\\cdot5^7+\\binom64\\cdot4^7-\\binom63\\cdot3^7+\\binom62\\cdot2^7-\\binom61\\cdot1^7}{6^7}=\\frac{35}{648}$$\n\nCalculate the probability of the original event by subtracting the result from $1$:\n\n$$1-\\frac{35}{648}=\\frac{613}{648}\\approx94.6\\%$$\n\n\u2022 Nice use of inclusion/exclusion! Simulated probability of seeing all 6 faces in 7 rolls in my Answer is 0.0544 \u00b1 0.001, compared to exact 35/648 = 0.0540 in this Answer. \u2013\u00a0BruceET Aug 7 '15 at 8:46\n\u2022 @BruceTrumbo: Thanks :) \u2013\u00a0barak manos Aug 7 '15 at 8:50\n\n(Edit: Answer is for the probability of seeing all the numbers for $n$ rolls)\n\nFor exactly $n$ rolls, the problem can be solved using a markov chain\n\n\\begin{align*} A&= \\begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0\\cr 0 & \\frac{1}{6} & \\frac{5}{6} & 0 & 0 & 0 & 0\\cr 0 & 0 & \\frac{1}{3} & \\frac{2}{3} & 0 & 0 & 0\\cr 0 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} & 0 & 0\\cr 0 & 0 & 0 & 0 & \\frac{2}{3} & \\frac{1}{3} & 0\\cr 0 & 0 & 0 & 0 & 0 & \\frac{5}{6} & \\frac{1}{6}\\cr 0 & 0 & 0 & 0 & 0 & 0 & 1\\end{pmatrix} \\end{align*}\n\nwhere the rows and columns are the number of faces of the die seen.\n\nThe probability of seeing all the faces in exactly 8 rolls is $(A^8)[0,6]$, which is $\\dfrac{665}{5832}\\approx 0.11402606310014$\n\nFor any $n$, it can be found by finding the generating function $G(z)$ and in turn finding the coefficient of $z^n$\n\n\\begin{align*} \\mathbb{P}(n) = 1-\\frac{20}{2^n}+15\\left(\\frac{2}{3}\\right)^n+\\frac{15}{3^n}-6\\left(\\frac{5}{6}\\right)^n-\\frac{6}{6^n} \\end{align*}\n\n\u2022 That was my original solution (should be $1-\\frac{665}{5832}$, BTW)... But I then realized that \"at least 8 times before you get to see all of the numbers\" meant that you were allowed to see all of the numbers by the $8$th trial, so you need to replace $8$ with $7$ in your answer. \u2013\u00a0barak manos Aug 7 '15 at 12:55\n\u2022 Oh, okay. I solved for the probability of seeing all the numbers for n rolls. \u2013\u00a0gar Aug 7 '15 at 13:24\n\u2022 Very elegant. Clarity of presentation could be improved for quick reading (and more upvotes). I suggest to: make the first sentence bold, correcting for n=7 so that the answers match, small subtitle \"general solution\" at the end. \u2013\u00a0Symeof Sep 23 '15 at 21:36\n\nPerhaps this is of some help, even if not your final answer.\n\nIf a fair die is rolled 8 times and $X$ is the number of different faces seen, then simulation of a million performances of the 8-roll experiment shows the following approximate distribution of $X$, in which the margin of simulation error is about $\\pm 0.001.$\n\n         1        2        3        4        5        6\n0.000003 0.002275 0.069393 0.363681 0.450468 0.114180\n\n\nOf course, $P(X = 1) = 1/6^7 = 0.000004$ to 6 places. To get $P(X = 6),$ consider that one face may be seen three times, or two different faces may each be seen twice. A straightforward computation of other probabilities in the distribution seems feasible but increasingly tedious. Perhaps there are clever combinatorial methods that make some of them easier.\n\nHowever, pending clarification, I'm still not sure if $P(X = 6)$ is exactly the probability you're trying to find.\n\nNotes:\n\n(a) For a 7-roll experiment an analogous simulation gives $0.0544 \\pm 0.001,$ which does not agree with your result $126/279936 \\approx 0.00045.$ (I do not believe you are considering all possible permutations of the numbers that occur only once.)\n\n(b) For a 6-roll experiment a simulation gives $0.015496 \\pm 0.001,$ compared with the exact $6!/6^6 = 0.0154321.$ Clearly, the probability of getting all 6 faces is 7 rolls must be larger than that.\n\n(c) A related problem is the number of rolls $W$ required before all 6 faces are seen. Then $E(W) = 6/6 + 6/5 + \\dots + 6/1 = 14.7.$ For more on this, see the 'coupon collecting problem'.\n\nHere's another way of looking at it, along the lines suggested by barak manos.\n\nNote that 7 rolls containing all 6 values must contain a duplicate. There are $\\binom{7}{2}\\cdot 6$ choices for the duplicate entries. Thus, the desired probability is given by\n\n$$1-\\frac{\\binom{7}{2}\\cdot 6\\cdot 5!}{6^7} = 1-\\frac{70}{1296}=\\frac{613}{648}.$$", "date": "2019-11-15 18:12:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1386606/probability-of-rolling-a-dice-8-times-before-all-numbers-are-shown/1387649", "openwebmath_score": 0.8104450702667236, "openwebmath_perplexity": 234.50984034329736, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9572778024535095, "lm_q2_score": 0.9032942054022056, "lm_q1q2_score": 0.8647034919164124}}
{"url": "https://mathhelpboards.com/threads/garland-trouble.56/", "text": "# Garland trouble\n\n#### grgrsanjay\n\n##### New member\nHow many different garlands are possible with 3 identical beads of red color and 12 identical beads of black color?\n\nI was thinking to keep the no.of beads in between the 3 beads of red color as x,y,z\n\nSo, x+y+z=12\nno .of ways is 14C2.....Was i Wrong somewhere??\n\nLast edited:\n\n#### tkhunny\n\n##### Well-known member\nMHB Math Helper\nBlack will fit not only between, but also before and after. You seem to have only one end.\n\nW before any red\nX after first red\nY after second red\nZ after last red\n\nW + X + Y + Z = 12\n\n14C3\n\n#### Plato\n\n##### Well-known member\nMHB Math Helper\nBlack will fit not only between, but also before and after. You seem to have only one end.\nW before any red\nX after first red\nY after second red\nZ after last red\nW + X + Y + Z = 12\n14C3\nBut the difficulty there is these are garlands, circular bead arrangements.\n$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.\nMoreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.\n\nHow many ways are there to partition $9$ into three of less summands?\nFor example here are three: $9$, $7+2$ and $5+3+1$.\n\nLast edited:\n\n#### tkhunny\n\n##### Well-known member\nMHB Math Helper\nI don't disagree with your solution. I suggest only that the definition of \"garland\" is ambiguous.\n\nAfter a quick survey of those in my home, they ALL believed a \"garland\" to be more of a hanging rope, and not a wreath. I believe we have established significant ambiguity (at least cultural) in the problem statement.\n\nPerhaps a picture was provided.\n\n#### grgrsanjay\n\n##### New member\nBut the difficulty there is these are garlands, circular bead arrangements.\n$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.\nMoreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.\n\nHow many ways are there to partition $9$ into three of less summands?\nFor example here are three: $9$, $7+2$ and $5+3+1$.\n\nSo,there must be less than 14C2 solutions??How do i eliminate them?\n\nOk,could you check whether i can do it like this\n\nx + y + z = 12\n\nCase 1:x=y=z\nNo.of.ways = (4,4,4) = 1\n\nCase 2:two of x,y,z are equal\nNo.of ways = (1,1,10),(2,2,8),(3,3,6),(5,5,2),(6,6,0),(0,0,12) = 6\n\nCase 3: x,y,z are all unequal\n\nTo avoid double counting , i will take x<y<z\n\nNo.of ways =\n(0,1,11),(0,2,10),(0,3,9),(0,4,8),(0,5,7)\n(1,2,9),(1,3,8),(1,4,7),(1,5,6)\n(2,3,7),(2,4,6)\n(3,4,5)\n= 12 ways\n\nSo,total no.of ways is 19\n\nLast edited:\n\n#### Plato\n\n##### Well-known member\nMHB Math Helper\nthe definition of \"garland\" is ambiguous.\nThe definition may be ambiguous here but not in textbooks on countilng. Here is a link. The subtopic on beads is the one used in counting problems.\nSo,there must be less than 14C2 solutions??How do i eliminate them?\nOk,could you check whether i can do it like this\nSo,total no.of ways is 19\nYES, 19 is correct\nThis sort of problem has set many arguments. If you agree on what meaning of garland we use then the following will show you why. If we have three green beads and three red beads how many different garlands are possible? Well the answer is 3. All the teds together; two reds together and one separate; then all the reds are separated. There are no other possibilities.\nThree can be written as 3, 2+1, or 1+1+1.\n\nThe questions using two colors have that nice solution. They quickly become a nightmare with more than two colors.\n\nLast edited:\n\n#### grgrsanjay\n\n##### New member\nI think none of my case were repeated...could you conform it whether any case is left?\n\n#### Plato\n\n##### Well-known member\nMHB Math Helper\nI think none of my case were repeated...could you conform it whether any case is left?\nYes you are correct. At first I read it as if there were only 12 beads altogether.", "date": "2020-09-26 07:20:03", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/garland-trouble.56/", "openwebmath_score": 0.6088405251502991, "openwebmath_perplexity": 2474.727389037761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9763105273220726, "lm_q2_score": 0.8856314783461303, "lm_q1q2_score": 0.8646513356371371}}
{"url": "http://aipc.outdoortown.it/parametric-to-rectangular-with-trig.html", "text": "Parametric To Rectangular With Trig\n\n To get the cartesian equation you need to eliminate the parameter t to get an equation in x and y (explicitly and implicitly). Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Subtract 7 7 from both sides of the equation. to rectangular coordinates 5. Provide details and share your research! But avoid \u2026 Asking for help, clarification, or responding to other answers. Write the equations of the circle in parametric form. x = 1 t - 2 3. Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. The rectangular coordinates for P (5,20\u00b0) are P (4. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Finding cartesian equation from parametric trigonometric equations. For this we need to find the vectors and. A parametric curve in the plane is a pair of functions x = f (t) y = g (t) It is possible to derive the Cartesian equation from the parametric equations. Now, first of all, what does it mean to convert to trigonometric form? Well, I have my number in rectangular form, so it's in a+bi form. x = -2 cos t, y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. Parametric form defines both the x-and the y-variables of conic sections in terms of a third, arbitrary variable, called the parameter, which is usually represented by t. Convert (3, -1) to polar form. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. For a circle of diameter 10 that has been translated right by 4 units and down by 7 units: (a) Give the parametric equations defining the circle. In the \ufb01rst three examples, x and y are given in terms of di\ufb00erent trig functions. - 43 Questions with solutions. Parametric Graphing Team Desmos February 21, 2020 14:44. There's also a graph which shows you the meaning of what you've found. By adjusting the parametric equations, we can reverse the direction that the graph is swept. Polar plot on Cartesian axes // parametric conversion of ugly trig function? Hello! I currently have a polar plot which I would like to superimpose onto square axes. We can derive this identity by drawing a right triangle with leg lengths 1 and t and applying the usual definitions of trig and inverse trig functions. Rockwall ISD Pre-Calculus Parent Guide 3 Unit 7 Trigonometric Functions In this unit students will continue to apply trigonometric functions including rotation angles, the Unit Circle and periodic functions. Combine the triangle and the Cartesian coordinate grid to produce Trigonometry. Other Parent Functions C. Sketching the Graph of Trigonometric Parametric Equations. To eliminate t in trigonometric equations, you will need to use the standard trigonometric identities and double angle formulae. Replace and with the actual values. An alternative approach is two describe x and y separately in terms of a. Develop the calculus for polar and parametric forms. Drag the slider to change the number of sides on the polygon. The Rectangular Coordinate Systems And Graphs Algebra Trigonometry. All for only $14. Easy to get confused between these and inverse trig functions! Trig Proofs & Identities. Sketch the graph of the parametric equations $$x=t^2+t$$, $$y=t^2-t$$. Eliminating the Parameter. sss s ss s sss ss s s ss sss s s s sss s s s IV. Download [74. Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. Find more Mathematics widgets in Wolfram|Alpha. Worksheets are Polar and rectangular forms of equations date period, Polar coordinate exercises, Polar coordinates, Trigonometry 03 notes marquez, , Polar coordinates parametric equations, Parametric equations context parametric and polar equations. Ron Larson + 1 other. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. Area under one arc or loop of a parametric curve. I don't know what to do from here or if I'm going in the right direction or not. Thanks for contributing an answer to Mathematics Stack Exchange! Finding cartesian equation for trigonometric parametric forms. Simplify (x \u22127)2 ( x - 7) 2. Example 1: 3, 4 1, for -4 t 2xt y t dd 2 2 Example 2: 2, , ( , )xt y t fortin ff t x y 5/7/19 9. Fill in the table and sketch the parametric equation for t [-2,6] x = t2 + 1 y = 2 \u2013 t 5 6 Problems 2 \u2013 10: Eliminate the parameter to write the parametric equations as a rectangular equation. Example 1 - Graphing Parametric Equations; Example 2 - Parametric to Rectangular Form; Day 2 - 7. T= Ncos\ud835\udf03 U= N O\ud835\udc56\ud835\udf03 N P \ud835\udf03= U T \u2212. Example 6) Finding parametric equations for a given function is easier. The ordered pairs, called polar coordinates, are in the form $$\\left( {r,\\theta } \\right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$ - axis (polar axis), going counter-clockwise. You can find values for both x and y by plugging values for t into the parametric equations. Applications. Converting between polar and rectangular form Converting equations between polar and rectangular form Homework: Finish Day 1 Packet (Optional) Pg. I was trying to solve for x for some reason. (b) Convert the parametric equations to rectangular form (please give the circle in standard form). Complete pp. Polar and Parametric Equations Rev. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link]. Rotating a Curve defined by a Equation. We can express equation (i) in terms of t, therefore we see that t = x + 3,. sss s ss s sss ss s s ss sss s s s sss s s s IV. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. How to represent Parametric Equations. Set up the parametric equation for x(t) x ( t) to solve the equation for t t. Parametric and Polar Equations Review Name 1. uTo graph parametric functions You can graph parametric functions that can be expressed in the following format. Parametric equations involving trigonometric functions Finding areas Parametric equations An equation like y = 5x + 1 or y x x 3sin 4cos or xy22 1 is called a cartesian equation. x=t-l, Check these with yesterday's graphs: B. 2 Plane Curves and Parametric Equations 711 Eliminating the Parameter Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. Parametric Curves. Converting Polar Equations To Rectangular Equation. One caution when eliminating the parameter, the domain of the resulting rectangular equation may need to be adjusted to agree with the domain of the parameter as given in the parametric equations. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. y for values of. Simplify (x \u22127)2 ( x - 7) 2. Complete Algebra 2 and Trig Program: Requirements: Requires the ti-83 plus or a ti-84 model. Trigonometry made completely easy! Our Trigonometry tutors got you covered with our complete trig help for all topics that you would expect in any typical Trigonometry classes, whether it's Trigonometry Regents exam (EngageNY), ACT Trigonometry, or College Trigonometry. Polar coordinates simplified the work it takes to arrive at solutions in most trigonometry problems. Historic applications of parametric equations are discussed so the use of them is realized. Calculus of a Single Variable. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. PreCalculus Class Notes VP5 Converting Parametric and Rectangular Equations Review To convert from parametric equations to rectangular equation: solve the x equation for t, substitute into the y equation Example Rewrite 2 3 2 4 x t y t = \u2212 = as a function of x. Calculus and parametric curves. Graphing a plane curve represented by parametric equations involves plotting points in the rectangular coordinate system and connecting them with a smooth curve. The resulting equation is a rectangular equation. Credit Pre-Calculus Honors *All areas are accelerated* Generating Graphs Shifting, Symmetry, Reflections, and Stretching Roots, Max/Min, Values, and Intersections Composite Functions Programming the Tl-84 Polynomial Functions Brief Quadratic Review Roots Shifted and Standard Form Axis. A cartesian equation gives a direct relationship between x and y. That make sense now. Graphing a Parabola with vertex at (h ,k ). Find more Mathematics widgets in Wolfram|Alpha. Show the orientation (flow) by arrows (2) Convert to a rectangular equation and Converting from. Polar coordinates simplified the work it takes to arrive at solutions in most trigonometry problems. 142 Notes - Section 8. Then use a trigonometric identity. The functions and can serve as a parametric representation for a function , which is plotted in purple on the , plane cutting , shown in light gray. Subtract 7 7 from both sides of the equation. In the diagram such a circle is tangent to the hyperbola xy = 1 at (1,1). Example $$\\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. The equation is the general form of an ellipse that has a center at the origin, a horizontal major axis of length 14, and. The rectangular coordinates (x , y) and polar coordinates (R , t) are related as follows. The student is expected to: (A) graph a set of parametric equations;. y = x -3 is equivalent to 3 cos sin 3 (cos sin ) 3 3 cos sin xy rr r r TT TT TT. Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice. x=t-l, Check these with yesterday's graphs: B. Write the complex number in trigonometric form, using degree measure for the argument. In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. Let us just look at a simple example. Clearly, both forms produce the same graph. Then, we do this substitution into the function: x \u2192 x c - y d y \u2192 x d + y c. Eliminate the parameter from the given pair of trigonometric equations where $$0\u2264t\u22642\\pi$$ and sketch the graph. For the first case we need to supplement the equations by two inequalities: 0 <= t <= 4 Pi && 0 < x < 4 Pi. Eliminate the parameter and find a corresponding rectangular equation. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Construct a table of values for the given parametric equations and sketch the graph: the data from the parametric equations and the rectangular equation are plotted together. Converting Polar To Rectangular. A cartesian equation gives a direct relationship between x and y. x = \u00bdt + 4. x = cos 2t, y = sin t, for t in 1-p, p2. In rectangular coordinates, each point (x, y) has a unique representation. ACE TRIG Final EXAM REVIEW. After going through these three problems can you reach any conclusions on how the argument of the trig functions will affect the parametric curves for this type of parametric equations?. Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Solve trigonometric equations in quadratic form 12. now expanding (x+2)\u00b2 either by using the identity of a perfect square which x^2 + 2bx + b^2 or simply expanding (x+2)(x+2) you have = x^2 + 4x + 4 and then multiplying by -1 as such because you have -(x+2)\u00b2 from your original parametric equation -(t\u00b2) = -x^2 - 4x - 4. Students model linear situations with parametric equations, including modeling linear motion and vector situations. the domain of the rectangular equation so that its graph matches the graph of the parametric equations. We can express equation (i) in terms of t, therefore we see that t = x + 3,. Complete pp. Download [74. Parametric Equations and Vectors : Questions like write each pair of parametric equations in rectangular form. Example 1: 3, 4 1, for -4 t 2xt y t dd 2 2 Example 2: 2, , ( , )xt y t fortin ff t x y 5/7/19 9. Place the parametric equations in rectangular form. The applet below illustrates parametric coordinate functions for various polygonal trig functions. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. In the diagram such a circle is tangent to the hyperbola xy = 1 at (1,1). Rectangular - Polar - Parametric \"Cheat Sheet\" 15 October 2017 Rectangular Polar Parametric Point ( T)= U ( T, U) ( , ) \u2022 ( N ,\ud835\udf03) or N \u2220 \ud835\udf03 Point (a,b) in Rectangular: T( P)= U( P)= < , > P=3\ud835\udc5f \ud835\udc56 , Q O P\ud835\udc56 , with 1 degree of freedom (df) Polar Rect. The coordinates are measured in meters. Then find a second set of polar coordinates for the. 7: Complex Numbers, Polar Coordinates, Parametric equationsFall 2014 2 / 17 Complex Numbers - trig form Example (Write the complex number in standard form). Parametric equations are equations that are used to introduce polar coordinates and their relation with rectangular coordinates. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. We will graph polar equations in the polar coordinate system and finally discuss parametric equations and their graphs. Things to try. x = cos 2t, y = sin t, for t in 1-p, p2. Find more Mathematics widgets in Wolfram|Alpha. sssssss ssss sssss Homework Assignment Page(s) Exercises. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. Linear 64) 2. Trigonometry made completely easy! Our Trigonometry tutors got you covered with our complete trig help for all topics that you would expect in any typical Trigonometry classes, whether it's Trigonometry Regents exam (EngageNY), ACT Trigonometry, or College Trigonometry. Algebra Review: Completing the Square. Get the free \"parametric to cartesian\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Example $$\\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. Comments There are no comments. In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. However, they used meters instead of feet for gravitational constant. To this point (in both Calculus I and Calculus II) we\u2019ve looked almost exclusively at functions in the form $$y = f\\left( x \\right)$$ or $$x = h\\left( y \\right)$$ and almost all of the formulas that we\u2019ve developed require that functions be in one of these two forms. An object travels at a steady rate along a straight path $$(\u22125, 3)$$ to $$(3, \u22121)$$ in the same plane in four seconds. Write each pair of parametric equations in rectangular form. Find a set of parametric equations to represent the graph of y = (x \u2013 1)2 given the. Search this site. T= Ncos\ud835\udf03 U= N O\ud835\udc56\ud835\udf03 N P \ud835\udf03= U T \u2212. asked by Sam on December 9, 2013; geometry. Example 6) Finding parametric equations for a given function is easier. Arc length of a parametric curves. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Therefore, we need to use a change of variables so we can integrate using either cylindrical (polar) or spherical coordinates, or even parametric form. Find more Mathematics widgets in Wolfram|Alpha. How do you convert #r=2sin(3theta)# to rectangular form? Trigonometry The Polar System Converting Between Systems. 2 Plane Curves and Parametric Equations 711 Eliminating the Parameter Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. Pre - Calc. Introduction to Parametric Equations; Parametric Equations in the Graphing Calculator; Converting Parametric Equations to Rectangular: Eliminating the Parameter; Finding Parametric Equations from a Rectangular Equation; Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; Parametric Form of the Equation of a Line in Space. Graphing a Hyperbola with center at (0 ,0 ). (1) $$f(x,y)=0$$ These are sometimes referred to as rectangular equations or Cartesian equations. Find a rectangular equation for each plane curve with the given parametric equations. The parametric equations are plotted in blue; the. I have the parametric coordinates x=Sec(t) and y=Tan 2 (t) where 0<=t P=3\ud835\udc5f N\ud835\udc56 , Q O P\ud835\udc56 , with 1 degree of freedom (df) Polar Rect. I'm trying to find the cartesian equation of the curve which is defined parametrically by: $$x = 2\\sin\\theta, y = \\cos^2\\theta$$ Both approaches I take result in the same answer:$$y = 1 - \\s. Step-by-Step Examples. what is the volume of that rectangular pyramid? asked by riza on August 6, 2012. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. The rectangular coordinates for P (5,20\u00b0) are P (4. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. Parametric and Polar Equations Review Name q) 1. Parametric equation of an ellipse and a hyperbola rectangular hyperbola cartesian and parametric forms examsolutions maths tutorials parametric equations hyperbola hw 1 conic sections in polar parametric forms lesson Parametric Equation Of An Ellipse And A Hyperbola Rectangular Hyperbola Cartesian And Parametric Forms Examsolutions Maths Tutorials Parametric Equations Hyperbola Hw 1 Conic. In , the data from the parametric equations and the rectangular equation are plotted together. Historic applications of parametric equations are discussed so the use of them is realized. SOLUTION The graph of the parametric equations is given in Figure 9. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Textbook Authors: Lial, Margaret L. This is also a great Review for AP Calculus BC. When you first learned parametrics, you probably used t as your parametric variable. Introduction to Parametric Equations; Parametric Equations in the Graphing Calculator; Converting Parametric Equations to Rectangular: Eliminating the Parameter; Finding Parametric Equations from a Rectangular Equation; Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; Parametric Form of the Equation of a Line in Space. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. Simply let t x and then replace your y with t. It can handle horizontal and vertical tangent lines as well. One caution when eliminating the parameter, the domain of the resulting rectangular equation may need to be adjusted to agree with the domain of the parameter as given in the parametric equations. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, \u2026. 3+3\ud835\udc56 3\u221a2(cos45\u00b0+\ud835\udc56sin45\u00b0) Write the complex number in the form + \ud835\udc8a. Textbook Authors: Lial, Margaret L. This problem is about converting parametric equations to rectangular form. Sketch and identify graphs using parametric equations. This is called a parameter and is usually given the letter t or \u03b8. Find a rectangular equation for each plane curve with the given parametric equations. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos \u2061 t (x = \\cos t (x = cos t and y = sin \u2061 t) y = \\sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. For the problems above, let x = t + 2 and find the resulting parametric equations. ACE TRIG Final EXAM REVIEW. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link]. T= Ncos\ud835\udf03 U= N O\ud835\udc56\ud835\udf03 N P \ud835\udf03= U T \u2212. Algebra Review: Completing the Square. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Plot the points. Sketching the Graph of Trigonometric Parametric Equations. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. A parametric equation is where the x and y coordinates are both written in terms of another letter. Plot the following points on the polar grid at the right and label each point. Find all solutions to a trigonometric equation 10. Find new parametric equations that shift this graph to the right 3 places and down 2. The coordinates are measured in meters. Polar and Parametric Equations 3. The volume of. 2;5p 3 Give two alternate sets of coordinates for each point. 6 Plane Curves, Parametric Equations. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. But by recognizing the trig identity, we were able to simplify it to an ellipse, draw the ellipse. PreCalculus Class Notes VP5 Converting Parametric and Rectangular Equations Review To convert from parametric equations to rectangular equation: solve the x equation for t, substitute into the y equation Example Rewrite 2 3 2 4 x t y t = \u2212 = as a function of x. the process of solving one parametric equation for t so you may substitute that equation into the other parametric equation for t to create a rectangular equation where y=f(x) Example: if x=t-4 and y=\u00bct solve the first equation for t so t=x-4 then by substitution y=\u00bc(x-4) or y=\u00bcx-1. 95 per month. Thus, keep only the other equation. x: 4 sin (2t) Y : 2 cos (2t) X : 4+2 COS t. Find more Mathematics widgets in Wolfram|Alpha. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. Graph polar equations. x = -2 cos t, y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. x = 3t \u2013 1, y = 2t + 1. 4 Trigonometric Functions of Any Angle 497 1-3 4. Find a Cartesian equation for the curve traced out by this function. Trig functions, complex numbers, identities, vectors, and real life applications like circuits are here to energize your classroom. Sketch and identify graphs in polar coordinates. First, we find a vector {c,d} of distance 1 having angle -\u03b8, which is {Cos[-\u03b8], Sin[-\u03b8]}. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 7. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. Tools We Need x = r * cos \u03b8 y = r * sin \u03b8 (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos\u207b\u00b9(r/2)). Write the equations of the circle in parametric form. Parametric Equations and Vectors : Questions like write each pair of parametric equations in rectangular form. Parametric and Polar Equations Review Name 1. Challenge Sets. A coordinate system is a scheme that allows us to identify any point in the plane or in three-dimensional space by a set of numbers. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate. Graphing parametric equations: The key is to plug in useful points within the speci\ufb01ed range of t, not just any points. Show all algebraic support. The Second Fundamental Theorem of Calculus Integration Involving Powers of Trigonometric Functions. This is the a value, this is the b value. A parametric curve in the plane is a pair of functions x = f (t) y = g (t) It is possible to derive the Cartesian equation from the parametric equations. x = cos 2t, y = sin t, for t in 1-p, p2. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. The volume of. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Historic applications of parametric equations are discussed so the use of them is realized. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Convert the polar coordinates (5 , 2. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Plot the resulting pairs ( x,y ). Based on your work on the above problems, name another benefit of parametric equations versus rectangular functions. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Example $$\\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. Write the equations of the circle in parametric form. Use a table of values to sketch a Parametric Curve and indicated direction of motion. In parametric equations x and y are both defined in terms of a third variable. 1 Answer A. ; Hornsby, John; Schneider, David I. Polar coordinates. 244 Chapter 10 Polar Coordinates, Parametric Equations conclude that the tangent line is vertical. 7-8: 2-8 even, 9-12 all, 14-32 even. In the past, we have seen curves in two dimensions described as a statement of equality involving x and y. T= Ncos\ud835\udf03 U= N O\ud835\udc56\ud835\udf03 N P \ud835\udf03= U T \u2212. Things to try. Xmin = -20 Ymin = -12 Xmax = 20 Ymax = 12 Xscl = 5 Yscl = 5. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30\u00b0)t and y = (64 sin 30\u00b0)t \u2212 16t2 + 3. Convert to Polar Coordinates. We will graph polar equations in the polar coordinate system and finally discuss parametric equations and their graphs. Both types depend on an argument, either circular angle or hyperbolic angle. Calculus of a Single Variable. Converting from rectangular to parametric can be complicated, and requires some creativity. Tools We Need x = r * cos \u03b8 y = r * sin \u03b8 (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos\u207b\u00b9(r/2)). Find a rectangular equation for each plane curve with the given parametric equations. Usually will stand for time. Eliminating the parameter is a method that may make graphing some curves easier. A common parameter used is time (t) or an angle (trig) (x, y) is the place, \"t\" is the time it is there (at that place) (at that place) Graphing Parametric Equations 2 OPTIONS: (l) Use a chart to find rectangular points. Recall the trig identity d1 Substitute x/r and y/r into the identity: Remove the parentheses: Multiply through by r 2. The curves are colored based on the quadrants. Trig Ratios and Quadrants (9:52) Inverses of Trig Functions and Inverse Trig Functions (20:18) Graphing Sine and Cosine--Amplitude and Period (11:15) Graphing Sine and Cosine--Vertical and Horizontal Shift (6:33) Graphing Sine and Cosine by Hand (29:10) Graphing Tangent by Hand (24:39) Applications (15:50) Unit 4: Triangle Applications of Trig. The parametric equations are plotted in blue; the. A curve is given by the parametric equations: #x=cos(t) , y=sin(2t)#, how do you find the cartesian equation? Calculus Parametric Functions Introduction to Parametric Equations. Homework Statement Reduce these parametric functions to a single cartesian equation:$\\displaylines{ x = at^2 \\cr y = 2at \\cr} \\$. Drag points A and B to change the size and orientation of the polygon. now the rest is a matter of simplification:. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Graphing a Parabola with vertex at (h ,k ). And then by plotting a couple of points, we were able to figure out the direction at which, if this was describing a particle in motion, the direction in which that particle was actually moving. In parametric equations x and y are both defined in terms of a third variable. It can handle horizontal and vertical tangent lines as well. We show how we can transform between these representations of the same plane. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, \u2026. Plot the resulting pairs ( x,y ). (\u03b8 is normally used when the parameter is an angle, and is measured from the positive x-axis. But how do we write and solve the equation for the position of the moon when the distance from the planet,. In the past, we have been working with rectangular equations, that is equations involving only x and y so that they could be graphed on the Cartesian (rectangular) coordinate system. This calculator converts between polar and rectangular coordinates. Use power reducing formulas 9. now the rest is a matter of simplification:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Find exact values of composite functions with inverse trig functions 8. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. Trigonometric substitution. Search this site. I'm not looking for the answer here. 1 Answer Cesareo R. 1 Complex Numbers 8. Parametric curves have a direction of. x =3cos\u03b8 and y =4sin\u03b8, 02\u2264\u03b8\u2264 \u03c0 by eliminating the parameter finding the corresponding rectangular equation. Write the equations of the circle in parametric form. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Use a table of values to sketch a Parametric Curve and indicated direction of motion. Graphing a Parabola with vertex at (h ,k ). Convert to Polar Coordinates. Let's look at each solution carefully: The first solution is correct since: r^2 = x^2 + y^2. Subtract 7 7 from both sides of the equation. position time parametric equations path rectangular equation eliminating the parameter square root function direction of motion. now the rest is a matter of simplification:. orgChapter 1. We will then introduce the polar coordinate system, which is often a preferred coordinate system over the rectangular system. Graphing parametric equations: The key is to plug in useful points within the speci\ufb01ed range of t, not just any points. Applications. Sketch and identify graphs using parametric equations. Applications of Parametric Equations. This is the a value, this is the b value. interesting variations on the parametric equations. Linear 64) 2. Two versions. Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice. 5 Parametric Equations part 2. Find the rectangular equation that models its path. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. Finding Parametric Equations In Exercises 35 and 36, find two different sets of parametric equations for the rectangular equation. Graphs of trigonometric functions in polar coordinates are very distinctive. Lots of vocab in this one -- specific solutions, general solutions, 2npi, 360n -- and lots of factoring to do too. Graph Type\u23ae Parametric. Identify the type of graph. Making statements based on opinion; back them up with references or personal experience. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. 2 Exercises - Page 365 36 including work step by step written by community members like you. Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. Finding cartesian equation from parametric trigonometric equations. This calculator converts between polar and rectangular coordinates. For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. Application: Toy Rocket The parametric equations determined by the toy rocket are Substitute from Equation 1 into equation 2: A Parabolic Path 8. Ask Question Asked 5 years, 7 months ago. Sketch and identify graphs in polar coordinates. Find a rectangular equation for each plane curve with the given parametric equations. Find more Mathematics widgets in Wolfram|Alpha. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. An alternative approach is two describe x and y separately in terms of a. 1 shows points corresponding to \u03b8 equal to 0, \u00b1\u03c0/3, 2\u03c0/3 and 4\u03c0/3 on the graph of the function. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Then, we do this substitution into the function: x \u2192 x c - y d y \u2192 x d + y c. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos \u2061 t (x = \\cos t (x = cos t and y = sin \u2061 t) y = \\sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. Find parametric equations for 2x 4 a) y = x 2 \u20142x+3 Note that there are many ways of finding parametric equations for a given function. 2: Trigonometric Functions In this lesson you will use parametric equations to illustrate the connection between the graphs of y = sin(x) and the unit circle. Example 3: Transform the equation x 2 + y 2 + 5x = 0 to polar coordinate form. Just as with non-angle parameters, when the parameter is an angle \u03b8 the plane curve can be graphed by selecting values for the angle and calculating the x- and y-values. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. This is also a great Review for AP Calculus BC. Jimmy wants to rewrite the set of parametric equations x = 1/2 T + 3 and y = 2T - 1 in rectangular form by eliminating T. Rewrite the equation as t+7 = x t + 7 = x. now expanding (x+2)\u00b2 either by using the identity of a perfect square which x^2 + 2bx + b^2 or simply expanding (x+2)(x+2) you have = x^2 + 4x + 4 and then multiplying by -1 as such because you have -(x+2)\u00b2 from your original parametric equation -(t\u00b2) = -x^2 - 4x - 4. Finding all arguments t in 0 <= t <= 4Pi where the parametric graph intersects. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. Pre - Calc. Find the rectangular coordinates of. The Polar coordinates are in the form (r,q). Then x=f(t) and y=g(t) are called parametric equations for the curve represented by (x,y). Equations for and are plotted on the perpendicular and planes as varies from to. Graph polar equations. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Easy to get confused between these and inverse trig functions! Trig Proofs & Identities. Because r is a directed distance the coordinates (r, \u03b8) and (-r, \u03b8 + \u03c0). Finding cartesian equation from parametric trigonometric equations. However, they used meters instead of feet for gravitational constant. I don't know what to do from here or if I'm going in the right direction or not. The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. Homework Statement I have this equation and i need to find the cartesian equation, so i apreciate your help Homework Equations X=cost ' y=2sin2t The Attempt at a Solution I am usign this [/B] Sin2t=2costsint So x+y/2=cost+2costsint But i dont know what to do after, I also try to solve that. Sal gives an example of a situation where parametric equations are very useful: driving off a cliff! Sal gives an example of a situation where parametric equations are very useful: driving off a cliff! If you're seeing this message, it means we're having trouble loading external resources on our website. Calculus of a Single Variable. Rotating a Curve defined by a Equation. The students will analyze and graph Sine, Cosine and Tangent and their inverses. 2sec t and y = -0. It is a parabola with a axis of symmetry along the line $$y=x$$; the vertex is at $$(0,0)$$. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. This is also a great Review for AP Calculus BC. Trig equations are problems where you're solving for X or Theta but they're hidden behind a trig function, like \"2sinX-1=0\" or \"tan 2 X-1=0\". Thanks for contributing an answer to Mathematics Stack Exchange! Finding cartesian equation for trigonometric parametric forms. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. Applications of Parametric Equations. 2: Trigonometric Functions In this lesson you will use parametric equations to illustrate the connection between the graphs of y = sin(x) and the unit circle. Thus, keep only the other equation. Algebra Review: Completing the Square. Trigonometric, Parametric, and Polar Graphs OBJECTIVES When you have completed this chapter you should be able to Graph the sine wave, by calculator or manually. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30\u00b0)t and y = (64 sin 30\u00b0)t \u2212 16t2 + 3. Two versions. hyperbolic trig identities are similar to the regular trig onesbut beware hyperbolic trig functions have seemingly little in common with regular trig functions so there's no. Finding cartesian equation from parametric trigonometric equations. Making statements based on opinion; back them up with references or personal experience. This is the parameter or a number that affects the behavior of the equation. For each graph you create, identify the specific parametric equations used and the domain for your graph. express the equation in terms of x and/or y. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. How do I convert from parametric equations to rectangular form? Write the rectangular form for the parametric equation x=cos\u03b8 ; y=4sin\u03b8 Trig, and Differential. Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. Homework Statement I have this equation and i need to find the cartesian equation, so i apreciate your help Homework Equations X=cost ' y=2sin2t The Attempt at a Solution I am usign this [/B] Sin2t=2costsint So x+y/2=cost+2costsint But i dont know what to do after, I also try to solve that. Applications. The Organic Chemistry Tutor 247,826 views 33:29. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Solve trigonometric equations in quadratic form 12. Basic graphing with direction to them. Videos, examples, solutions, activities and worksheets for studying, practice and review of precalculus, Lines and Planes, Functions and Transformation of Graphs, Polynomials, Rational Functions, Limits of a Function, Complex Numbers, Exponential Functions, Logarithmic Functions, Conic Sections, Matrices, Sequences and Series, Probability and Combinatorics, Advanced Trigonometry, Vectors and. T= Ncos\ud835\udf03 U= N O\ud835\udc56\ud835\udf03 N P \ud835\udf03= U T \u2212. Comprehensive End of Unit Review for the Polar, Parametric, and Vectors Sections of PreCalculus or Trigonometry plus Graphic Organizer. Plot the points. 1 Answer A. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. Get the free \"parametric to cartesian\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Fill in the table and sketch the parametric equation for t [-2/6] Problems 2 - 11: Eliminate the parameter to write the parametric equations as a rectangular equation. 933 13 (b) C-/cos(\u00a3)) 10. The previous section discussed a special class of parametric functions called polar functions. Example 1 - Graphing Parametric Equations; Example 2 - Parametric to Rectangular Form; Day 2 - 7. Algebra Review: Completing the Square. At any moment, the moon is located at a particular spot relative to the planet. Then, convert each polar coordinate to rectangular ntoS form (RF). For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. uTo graph parametric functions You can graph parametric functions that can be expressed in the following format. 2 Exercises - Page 365 36 including work step by step written by community members like you. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. Textbook Authors: Lial, Margaret L. 8/14/2018 12:12 AM \u00a710. Pre-calculus Contents C Parametric Equations Parametric and rectangular forms of equations conversions The parametric equations of a quadratic polynomial, parabola The Trigonometric functions of arcs from 0 to. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Usually will stand for time. y for values of. 2;5p 3 Give two alternate sets of coordinates for each point. ) Drawing the graphTo draw a parametric graph it is easiest to make a table and then plot the points:Example 1 Plot the graph of the. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Drag P and C to make a new circle at a new center location. 6 Plane Curves, Parametric Equations. parameter t = x \u2013 1. Math Algebra 2 Calculus Precalculus Trigonometry Math Help Complex Numbers Pre Calculus Polar Forms Ellipse High School: Math Derivatives Polar Coordinates Sine Convert Calculus 3 Rectangular Calculus 2 Calculus 1 Polar Equation. Find parametric equations whose graph is an ellipse with center (h,k), horizontal axis length 2a, and. (a) Eliminate the parameter for the curve given by the parametric equations x= 2 t, y= 1 t2 for 1 P=3\ud835\udc5f N\ud835\udc56 , Q O P\ud835\udc56 , with 1 degree of freedom (df) Polar Rect. y = x 2 -2. Graphing parametric equations: The key is to plug in useful points within the speci\ufb01ed range of t, not just any points. Trig Functions sine I First, solve for 3. Plot the resulting pairs ( x,y ). It can handle horizontal and vertical tangent lines as well. The Second Fundamental Theorem of Calculus. This problem is about converting parametric equations to rectangular form. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Let A be the point where the segment OB intersects the circle, where point B lies on the line x = 2 a. Rockwall ISD Pre-Calculus Parent Guide 3 Unit 7 Trigonometric Functions In this unit students will continue to apply trigonometric functions including rotation angles, the Unit Circle and periodic functions. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Other Parent Functions C. For the point and So, the rectangular coordinates are See Figure 10. Comments There are no comments. ACE TRIG Final EXAM REVIEW. Calculus Examples. (a) Find a polar equation of the cissoid. Write each pair of parametric equations in rectangular form. There's also a graph which shows you the meaning of what you've found. And Vector Calculus, which tends to be the last chapter for Calculus 3, deals with work on, in, and around a surface which will predominately involve polar coordinates as well. Parametric Equations and Motion Precalculus Vectors and Parametric Equations. I was trying to solve for x for some reason. Use MathJax to format equations. y = x 2 -2. t over the interval for which the functions are defined. Given a point in polar coordinates, rectangular coordinates are given by Given a point in rectangular coordinates,. Trigonometry (MindTap Course List). In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. Exchanging x and y 4. The hyperbolic functions represent an expansion of trigonometry beyond the circular functions. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Both types depend on an argument, either circular angle or hyperbolic angle. To this point (in both Calculus I and Calculus II) we\u2019ve looked almost exclusively at functions in the form $$y = f\\left( x \\right)$$ or $$x = h\\left( y \\right)$$ and almost all of the formulas that we\u2019ve developed require that functions be in one of these two forms. Finding Parametric Equations for a Graph (Page 799) Describe how to find a set of parametric equations for a given graph. 95 per month. Write each pair of parametric equations in rectangular form. Trigonometric substitution. Example for parametric equations are {eq}\\displaystyle x=\\sqrt{t},\\:y=t-5 {/eq}, what we need to do first is to find x. I'm not looking for the answer here. (a) y = x -3 This is the equation of a line. Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Converting between polar and rectangular form Converting equations between polar and rectangular form Homework: Finish Day 1 Packet (Optional) Pg. So our parametric equation is x = t and y = 4t.", "date": "2020-07-03 19:58:35", "meta": {"domain": "outdoortown.it", "url": "http://aipc.outdoortown.it/parametric-to-rectangular-with-trig.html", "openwebmath_score": 0.822030246257782, "openwebmath_perplexity": 665.260664865313, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9780517456453798, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8646361687069855}}
{"url": "https://math.stackexchange.com/questions/1960055/proof-of-infinitely-many-prime-numbers/1960060", "text": "# Proof of infinitely many prime numbers\n\nHere's the proof from the book I'm reading that proves there are infinitely many primes:\n\nWe want to show that it is not the case that there only finitely many primes. Suppose there are finitely many primes. We shall show that this assumption leads to a contradiction. Let $p_1, p_2,...,p_n$ be all the primes there are. Let $x = p_1...p_n$ be their product and let $y=x+1$. Then $y \\in \\mathbb{N}$ and $y \\neq 1$, so there is a prime $q$ such that $q \\mid y$. Now $q$ must be one of $p_1,...,p_n$ since these are all primes that there are. Hence $q \\mid x$. Since $q \\mid y$ and $q \\mid x$, $q \\mid (y-x)$. But $y-x=1$. Thus $q \\mid 1$. But since $q$ is prime, $q \\geq 2$. Hence $q$ does not divide 1. Thus we have reached a contradiction. Hence our assumption that there are only finitely many primes must be wrong. Therefore there must be infinitely many primes.\n\nI have a couple of questions/comments regarding this proof. I will use a simple example to help illustrate my questions:\n\nSuppose only 6 primes exist: $2, 3, 5, 7, 11, 13$\n\nLet $x = p_1p_2p_3p_4p_5p_6=30,030$\n\nLet $y = x + 1 = 30,030+1 = 30,031$\n\n1. The proof states there is a prime $q$ such that $q \\mid y$ and that $q$ must be either $p_1, p_2, p_3, p_4, p_5,$ or $p_6$. However, none of the 6 primes listed, $(2,3,5,7,11,13)$, divides $30,031$. In fact, the only divisors for $30,031$ are $1, 59, 509$ and $30,031$. Doesn't the proof then break down here since there is no prime $q$ that divides $y$?\n\n2. The prime factorization of $30,031$ is $59 \\times 509$. These two numbers are factors of $30,031$ and are in fact primes themselves since they are only divisible by themselves or by 1. Have I shown that there exists $\\gt 6$ primes? If so, what can I conclude now that I have shown this?\n\n3. I don't understand why the contradiction $q$ divides $1$ and $q$ does not divide $1$ lead us to the assumption that the finitely many primes must be wrong. I understand how we reached the contradiction. I don't understand why contradiction leads us to the conclusion shows that are assumption that there is only finitely many primes is wrong.\n\nMy apologies for the long post. Thanks for any and all help.\n\n\u2022 unique factorization requires a unique prime decomposition. If no $p$ exists, there must be a new prime... \u2013\u00a0Eleven-Eleven Oct 8 '16 at 23:50\n\u2022 The point of the argument is that there must be some prime not in the list that divides the number you produce. Your example confirms this! You found two such primes $59,509$ neither of which is on the list. Thus, your list can't have been complete. \u2013\u00a0lulu Oct 8 '16 at 23:51\n\u2022 The proof makes an assumption that there are finitely many primes, But it then goes on to show, given the conditions, this actually can't be the case. Therefore, the flaw is in your assumption, since all arguments in the proof are mathematically valid. \u2013\u00a0Eleven-Eleven Oct 8 '16 at 23:53\n\u2022 As stated above, all of your mathematical arguments are correct when you started from your original assumption. Yet you are led to a point in the proof that is impossible. If the mathematical arguments are sound and correct, what is left for there to be wrong...? The original assumption must therefore be incorrect. \u2013\u00a0Eleven-Eleven Oct 8 '16 at 23:55\n\u2022 2) In this case that there are at least 7 primes. But if you generalize this, that p1p2...pn+1 is never divisible by p1,...,pn then p1,.....pn can not be a list of all primes. No such list can exist. So there can not be a finite number of primes. \u2013\u00a0fleablood Oct 9 '16 at 0:33\n\nThis is an excellent example of a proof that is traditionally phrased as a proof by contradiction but is much better understood constructively. From a constructive viewpoint, the proof shows that given any list of primes $p_1, \\ldots, p_n$ there is a prime $q$ (any prime divisor of $p_1p_2\\ldots p_n + 1$) that is distinct from each $p_i$. So given any finite set of primes we can find a prime that is not in that set.\n\n\u2022 +1 for emphasizing the constructive aspect of this proof. \u2013\u00a0P Vanchinathan Oct 9 '16 at 0:44\n\u2022 Except that this process doesn't actually construct a new prime (the example $y = 30031$ being composite). \u2013\u00a0Daniel R. Collins Oct 9 '16 at 2:20\n\u2022 @Rob: I think Daniel\u2019s point is that the proof constructs an integer $y>1$ that is not divisible by $p_1$, $p_2$, \u2026, $p_n$, and then it asserts that there must be a prime $q$ such that $q \\mid y$, but it doesn\u2019t construct $q$. \u2013\u00a0Scott Oct 9 '16 at 4:10\n\u2022 @Scott But there's a simple algorithm to generate a prime divisor of any given integer. \u2013\u00a0Jack M Oct 9 '16 at 8:05\n\u2022 @djechlin This is a constructive proof and you're utterly wrong. The set of divisors of a positive integer is finite. Now just take the minimal element greater than $1$ in the set of divisors; this is obviously prime. Would you say that finding the minimal element in a finite set of integers is not constructive? By the way, the proof by contradiction seems to imply that $p_1p_2\\dots p_n+1$ is prime, which is the main source for confusion in newbies. \u2013\u00a0egreg Oct 9 '16 at 11:20\n1. The proof does break down in a sense. You have reached a contradiction, which means the hypothesis that there are finitely many primes can not possibly be true.\n\n2. So you thought you had 6 primes, you found 2 extra. Can you just add these two to your list and get all of the primes? If you repeat the process on these 8 primes, you will find that you will have even MORE primes by considering the product + 1. You can keep going and going and you will never run out of primes. This is the idea of the proof. It uses contradiction because you can do it all in one step and avoid the potential issue of doing the process infinitely many times.\n\nSuppose only 6 primes exist: 2,3,5,7,11,13\n\nHowever, none of the 6 primes listed, (2,3,5,7,11,13), divides 30,031.\n\nThen we already have a contradiction. Since there are only 6 primes (we supposed that at the beginning) and none of them divide 30,031, then 30,031 must be prime. However, 30,031 is not one of the only 6 primes that exist. So 30,031 cannot be prime, yet it must be prime.\n\nSo the proof works. In fact, it works precisely the same regardless of what set of numbers we suppose are the only primes that exist. Thus no finite set of numbers can include all the primes that exist. Thus there are an infinite number of primes.\n\n\u2022 I think you mean \"... no finite set of numbers ...\" \u2013\u00a0origimbo Oct 10 '16 at 10:48\n\nYou ask in #1 if the proof breaks down. No. What breaks down is the assumption that there are no more primes. You assumed you had a complete list of primes. Then you constructed a number that is not divisible by any of the primes in your list. Only two possibilities remain: Either the number you constructed is prime, or it is divisible by a number that is not in your list. Either way your list is not complete, which contradicts your initial assumption.\n\nSimpy put, you assumed you had a complete list and by using that assumption you proved that the list was not complete. The assumption must therefore be wrong.\n\nThe answer to #3 is basically the same. By finding a contradiction when you made an assumption, you proved the assumption to be incorrect.\n\nThe answer to #2 is no, you did not prove you anything by noting that 59 and 509 are both prime. You are trying to prove that there is a finite list of primes. If you choose a particular set of primes as you did {2, 3, 5, 7, 11, 13} and show that that particular set doesn't hold all the primes, a skeptic would just say that you need to add more primes (like 59 and 509) because you didn't make your set big enough. The proof has to be more general which is why it doesn't say how many primes are in the finite set. The proof is written so that it works no matter how big the finite set is.\n\nPoint 1: It's a theorem that any natural number $n>1$ has a prime factor. The proof is easy: for any number $n>1$, the smallest natural number $a>1$ which divides $n$ is prime (if it were not prime, it would not be the smallest).\n\nPoint 2: Yes, you have proved there are more than six primes. So what? The proof by contradiction doesn't suppose there are only six, but that there are a finite number of them.\n\nPoint 3: Actually, it's not really a proof by contradiction stricto sensu. It is proved that any finite list of primes in incomplete.\n\nHere is a non-mathematical approach to the logic behind the proof by contradiction...\n\nSuppose your assumption is that a suspect in a brutal murder is innocent because he said he couldn't have been at the location the person was murdered. You feel as though he is possibly telling the truth, but what you do know is that there are 5 other suspects. You know with 100% certainty due to the detail of the case there can't be anymore than 5 other suspects via proven information. But, the police can verify using obvious physical evidence such as video cameras that the 5 suspects could not possibly be guilty. Then there is either another suspect or the one assumed innocent is guilty. But we said that it was impossible for any other suspects to be considered. Therefore our original suspect is guilty.\n\nThis is exactly what is happening in a proof by contradiction.\n\n\u2022 This is good. But I think a better analogy is that you have 6 suspects and you assume that these are the only six possible suspects in the world. Then cameras show each and every one is innocent. Thus our assumption that we have all the suspects is wrong, and therefore there is a seventh possible person who could have (and did) do it. \u2013\u00a0fleablood Oct 9 '16 at 0:58\n\u2022 Damn.... you are right. It made logical sense when i wrote it... but yours is better. \u2013\u00a0Eleven-Eleven Oct 9 '16 at 1:02\n\u2022 Yours makes sense. And is a proof by contradiction. I just thought for this particular proof this is a closer analogy. I wouldn't say yours was wrong or anything. \u2013\u00a0fleablood Oct 9 '16 at 1:05\n\u2022 That was my original goal; understand why contradition works. But yours does that AND is analogous to the problem at hand. Thanks \u2013\u00a0Eleven-Eleven Oct 9 '16 at 1:07\n\u2022 I will concede. I get it. \u2013\u00a0Eleven-Eleven Oct 9 '16 at 23:01\n\nThe proof is essentially overexplaining. The point of contradiction would have been much clearer if the author had:\n\n1. Reached it earlier.\n2. Introduced less notation.\n\nSince $p_1\\cdots p_n+1$ cannot have any of $p_1,...,p_n$ as a prime factor we already have a contradiction.\n\nThe rest is just explaining to death why $p_1,...,p_n$ cannot be prime factors of $y=p_1\\cdots p_n+1$ which should be clear since $y$ is $1$ off from being a multiple of either of those primes.\n\n\u2022 I think the gif dosn't really fit this site, though I can't say if there is a specific rule against such things \u2013\u00a0Yuriy S Oct 10 '16 at 11:27\n\u2022 @YuriyS: Hmm ... I am sorry to learn that people may find it inappropriate! That was by no means my intention! Do you think it would work better by removing the gif an leave the link and the figure of speach, or are all of those opfuscating the points I tried to make? \u2013\u00a0String Oct 10 '16 at 12:21\n\u2022 @String: your link to a cartoon movie of a dead animal being bludgeoned is disproportionate and offensive. \u2013\u00a0Rob Arthan Mar 12 '17 at 1:27\n\u2022 @RobArthan: Sorry, in some parts of the world such a cartoon would be considered merely a funny way to illustrate the saying about beating a dead horse. No offense intended, only a light tone. I cannot help that people do take offense, so I have removed it. Still it puzzles me how a cartoon matching the content of a saying would offend. I am from Denmark, after all. \u2013\u00a0String Mar 12 '17 at 8:11\n\u2022 @String: on MSE it's easier just to use neutral language. As I am English (after all), may I point out that the cliched phrase is actually \"flogging a dead horse\" and it doesn't have the connotations you think it does (it's not \"explaining to death\", it's using a tired old argument that has lost all interest or relevance). Your cartoon doesn't help with that. \u2013\u00a0Rob Arthan Mar 13 '17 at 1:55\n1. Well... yes... it does break down. You assumed that there are only 6 primes and reached a contradiction. You've successfully proved that there aren't only 6 primes.\n2. Under the assumption that there are only 6 primes 30,031 isn't factorizable.\n3. In proof by contradiction you prove proposition A by assuming A is not true, and through a series of logical steps reach an impossibility, thus proving that A must be true. Concurrently, in this proof, we assumed there is a finite number of primes. After a series of logical inferences we reached a contradiction. Since the only assumption we made in the process is that there are a finite number of primes, that assumption must be wrong.\n1. Either $6$ primes you considered are not all the primes, or $30031$ is a new prime. Either way assumption is false. Proof by contradiction.\n2. No, you haven't shown that two more primes exist. In this example yes, but not in general\n3. Again, based on your assumption you reached something impossible that $q$ divides $1$. There is no such $q$ except $1$.\n\nStart with $2,3,5$ being the only prime numbers. Then you would have $31$ as the new number. $2,3,5$ do not divide $31$. But unlike your case, $31$ is a prime number. So the assumption is wrong.", "date": "2021-02-27 05:02:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1960055/proof-of-infinitely-many-prime-numbers/1960060", "openwebmath_score": 0.8419486880302429, "openwebmath_perplexity": 240.20242253351205, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517450056274, "lm_q2_score": 0.8840392848011833, "lm_q1q2_score": 0.8646361651533242}}
{"url": "https://www.physicsforums.com/threads/spring-weight-and-potential-energy.735704/", "text": "# Spring, weight and potential energy\n\n1. Jan 30, 2014\n\n### Bromio\n\n1. The problem statement, all variables and given/known data\nA particle is connected to a spring at rest. Because of weight, the mass moves a distance $\\Delta x$. Calculate the value of the elastic constant $k$\n\n2. Relevant equations\n$U_{PE} = \\frac{1}{2}k\\Delta x^2$\n$U = mgh$\n$F = ma$ (in general)\n$F = kx$ (Hooke's Law)\n\n3. The attempt at a solution\nI've tried to solve this problem from two points of view, but the result has been different in each one.\n\nFirst, I've thought that, when the mass is at rest after being connected to the spring and got down to the new equilibrium position:\n$mg = k\\Delta x \\Longrightarrow \\boxed{k = \\frac{mg}{\\Delta x}}$\n\nHowever, from the point of view of energies:\n$mgh_0 = mgh + \\frac{1}{2}k\\Delta x^2$ (I use $\\Delta x$ because the initial position is 0)\n\nAs $h_0 - h = \\Delta x$, I've got:\n$mg\\Delta x = \\frac{1}{2}k\\Delta x^2 \\Longrightarrow \\boxed{k = \\frac{2mg}{\\Delta x}}$\n\nObviously there is a wrong factor of 2.\n\nWhat's the problem here? Where is the mistake?\n\nThanks!\n\n2. Jan 30, 2014\n\n### BvU\n\nEnergy case is for the situation you let go at h=0. When it is at delta x, it's still moving. Until it is a 2 delta x, when it hangs still again. But now the spring is pulling harder and up she goes!\nIn other words, before there is rest, some energy has to be dissipated! You now know how much.\n\n3. Jan 30, 2014\n\n### Bromio\n\nIntuitively I understand what you say, but I don't see how to write it analytically.\n\nI mean, as we are in a conservative field, we only have to consider initial and final states. How should I have written the main equation ($mgh_0 = mgh + \\frac{1}{2}k\\Delta x^2$) to include what you say?\n\nThanks!\n\n4. Jan 30, 2014\n\n### BvU\n\nThat would be a little difficult, precisely because it describes an isolated case. It says there is energy left over, that has to go somewhere to get the \"in rest\" situation. E.g. in overcoming the air resistance which heats up the air a little.\n\n5. Jan 30, 2014\n\n### Bromio\n\nOkay. So the dissipated energy has a value of $\\frac{1}{2}k\\Delta x^2$, hasn't it?\n\nWhat I don't understand is why, if we are in a conservative field, we can't just consider initial and final states. Could you explain me the reason?\n\nThank you.\n\n6. Jan 30, 2014\n\n### BvU\n\nYou can, actually. It's just that it isn't a steady state. mg (h-h0) is converted into spring energy plus kinetic energy. If there is no friction, the oscillation goes on forever.\n\nIs there a risk that we are misinterpreting the OP ? My idea was \u0394x is measured after the thing has come to rest.\n\n7. Jan 30, 2014\n\n### Bromio\n\nNo misinterpretation. Now I see all you're trying to explain. For some reason I thought there wasn't friction. It's obvious that, if there isn't it, I can't apply energy equations \"at rest\" situation, simply because there isn't that situation. Am I right?\n\nSo, when I study energies at the beginning ($h = 0$) and at the end ($h = \\Delta x$, when entire system is at rest) I should include that dissipated energy.\n\nIf $TE_0$ and $PE_0$ are the total mechanical energy and the potential energy at the beginning respectively, $TE$ and $PE$ are the same energies but at the end, $SP$ is the spring energy at the end, and $Q$ is the dissipated energy (say heat), then:\n$TE_0 = PE_0$\n$TE = SP + PE$\n$TE_0 = TE + Q$\n\nDo I agree?\n\nThank you!\n\n8. Jan 30, 2014\n\n### BvU\n\nWell, there is of course an \"at rest\" situation, it's just that it doesn't have the same energy as the situation where you attach the weight to the spring.\n\nYes. in fact, you can probably do that yourself: Suppose you exercise a force to keep the weight in position h0, then gently lower the weight to position h. If you do it slow enough, the force is mg - kx where x runs from 0 to h and h = mg/k. The weight does work, namely \u222b F(x) dx where the integral goes from 0 to h. F(x) = mg - kx. The integral (you can do it) is $\\frac{1}{2}k h^2$. Looks familiar ?\n\n9. Jan 30, 2014\n\n### Bromio\n\nWhen I say that there isn't an \"at rest\" situation I'm talking about the frictionless case. If there isn't friction, the spring will experiment an non-stop simple harmonic motion, won't it?\n\nThank you! I've solved that integral and I've got that result. Amazing!", "date": "2017-12-18 07:40:35", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/spring-weight-and-potential-energy.735704/", "openwebmath_score": 0.699287474155426, "openwebmath_perplexity": 676.2647203619784, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517469248846, "lm_q2_score": 0.884039278690883, "lm_q1q2_score": 0.864636160873833}}
{"url": "https://gmatclub.com/forum/a-rectangular-sheet-of-paper-when-halved-by-folding-it-at-the-mid-279911.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 21 Jan 2019, 14:11\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. 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See who won a full GMAT course, and register to the next one.\n\n# A rectangular sheet of paper, when halved by folding it at the mid\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 26 Sep 2018\nPosts: 10\nA rectangular sheet of paper, when halved by folding it at the mid\u00a0 [#permalink]\n\n### Show Tags\n\n25 Oct 2018, 04:50\n5\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n64% (02:10) correct 36% (02:13) wrong based on 47 sessions\n\n### HideShow timer Statistics\n\nA rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in\na rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter\nsides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the\nsmaller rectangle?\n\na. $$4 \\sqrt{2}$$\nb. $$2 \\sqrt{2}$$\nc. $$\\sqrt{2}$$\nd. 1\ne. 3\nCEO\nStatus: GMATINSIGHT Tutor\nJoined: 08 Jul 2010\nPosts: 2726\nLocation: India\nGMAT: INSIGHT\nSchools: Darden '21\nWE: Education (Education)\nA rectangular sheet of paper, when halved by folding it at the mid\u00a0 [#permalink]\n\n### Show Tags\n\n25 Oct 2018, 06:02\n2\nRhythmGMAT wrote:\nA rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in\na rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter\nsides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the\nsmaller rectangle?\n\na. $$4 \\sqrt{2}$$\nb. $$2 \\sqrt{2}$$\nc. $$\\sqrt{2}$$\nd. 1\ne. 3\n\n$$\\frac{l}{b} = \\frac{b}{(l/2)}$$ where $$b = 2$$ given\n\ni.e. $$b^2 = l^2 / 2$$\n\ni.e. $$l = 2\u221a2$$\n\nArea of smaller rectangle $$= (l/2)*b = (2\u221a2/2)*2 = 2\u221a2$$\n\n_________________\n\nProsper!!!\nGMATinsight\nBhoopendra Singh and Dr.Sushma Jha\ne-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772\nOnline One-on-One Skype based classes and Classroom Coaching in South and West Delhi\nhttp://www.GMATinsight.com/testimonials.html\n\nACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION\n\nSenior Manager\nJoined: 22 Feb 2018\nPosts: 416\nRe: A rectangular sheet of paper, when halved by folding it at the mid\u00a0 [#permalink]\n\n### Show Tags\n\n25 Oct 2018, 07:05\n2\nRhythmGMAT wrote:\nA rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in\na rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter\nsides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the\nsmaller rectangle?\n\na. $$4 \\sqrt{2}$$\nb. $$2 \\sqrt{2}$$\nc. $$\\sqrt{2}$$\nd. 1\ne. 3\n\nOA: B\n\nInitial Rectangle Dimensions\nlonger side $$= x$$\nShorter side$$= 2$$\n\nDimensions after folding\nLonger side $$= 2$$\nShorter side $$= \\frac{x}{2}$$\n\nAs per question\n$$\\frac{x}{2}=\\frac{2}{\\frac{x}{2}}$$\n$$x^2=8 \\quad;\\quad x=2\\sqrt{2}$$\n\nDimensions of Shorter Rectangle\nLonger side $$= 2$$\nShorter side $$=\\frac{2\\sqrt{2}}{2}=\\sqrt{2}$$\nArea of Shorter Rectangle $$= 2\\sqrt{2}$$\n_________________\n\nGood, good Let the kudos flow through you\n\nGMATH Teacher\nStatus: GMATH founder\nJoined: 12 Oct 2010\nPosts: 624\nRe: A rectangular sheet of paper, when halved by folding it at the mid\u00a0 [#permalink]\n\n### Show Tags\n\n25 Oct 2018, 14:13\n2\nRhythmGMAT wrote:\nA rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in\na rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter\nsides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the\nsmaller rectangle?\n\na. $$4 \\sqrt{2}$$\nb. $$2 \\sqrt{2}$$\nc. $$\\sqrt{2}$$\nd. 1\ne. 3\n\n$$?\\,\\, = 2x$$\n\nThe rectangles shown above are similar, hence:\n\n$${{2x} \\over 2} = {2 \\over x}\\,\\,\\,\\,\\, \\Rightarrow \\,\\,\\,{x^2} = 2\\,\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{x\\,\\, > \\,\\,0} \\,\\,\\,\\,x = \\sqrt 2 \\,\\,\\,\\,\\,\\,\\, \\Rightarrow \\,\\,\\,\\,\\,? = 2\\sqrt 2$$\n\nThis solution follows the notations and rationale taught in the GMATH method.\n\nRegards,\nFabio.\n_________________\n\nFabio Skilnik :: GMATH method creator (Math for the GMAT)\nOur high-level \"quant\" preparation starts here: https://gmath.net\n\nRe: A rectangular sheet of paper, when halved by folding it at the mid &nbs [#permalink] 25 Oct 2018, 14:13\nDisplay posts from previous: Sort by", "date": "2019-01-21 22:11:03", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-rectangular-sheet-of-paper-when-halved-by-folding-it-at-the-mid-279911.html", "openwebmath_score": 0.3866971433162689, "openwebmath_perplexity": 7622.566683847868, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9489172630429474, "lm_q2_score": 0.9111797039819735, "lm_q1q2_score": 0.8646341508428573}}
{"url": "https://math.stackexchange.com/questions/2980637/for-positive-integers-between-999-and-100-inclusive-how-many-contain-the-digit", "text": "# For positive integers between 999 and 100 inclusive, how many contain the digit 5?\n\nThe question comes in two parts:\n\n1. For positive integers between 999 and 100 inclusive, how many contain the digit 5 at least once?\n\n2. For positive integers between 999 and 100 inclusive, how many contain the digit 5 exactly once?\n\nQuestion 1\n\nNumbers between 999 and 100 inclusive:\n\n999 - 99 = 900\n\nNumbers between 999 and 100 inclusive that do not contain digit 5:\n\n8 * 9 * 9 = 648 (The 8 is because the first digit can't be 5 or 0)\n\nNumbers that contain digit 5 at least once:\n\n900 - 648 = 252\n\nQuestion 2\n\nTotal numbers that contain digit 5 exactly once:\n\n(1*9*9) + (8*1*9) + (8*9*1) = 225\n\nI think I got it wrong but not sure which part.\n\n\u2022 Well, a good way to check is to count the number with exactly two $5's$. $55X$ gives us $9$, $5X5$ gives us $9$, $X55$ gives us $8$. Thus there are $9+9+8=26$ such numbers. Of course $555$ is the unique number in the range with three $5's$. We check that $252=225+26+1$ which seems like pretty good evidence that you are right.\n\u2013\u00a0lulu\nNov 1 '18 at 16:34\n\u2022 @DavidG.Stork The OP is counting numbers with no digits equal to $5$ so that they can be subtracted from all three-digit positive integers. Nov 1 '18 at 16:38\n\u2022 @DavidG.Stork OP is using a standard technique of counting how many numbers there are where we ignore the condition, giving $9\\times 10\\times 10 =900$ numbers in the range, and subtracting the number of numbers which violate the condition, violating the condition in this case meaning \"does not have $5$ as a digit\", there being $8\\times 9\\times 9$ such numbers, which will give the number of numbers that satisfy the condition as a result. Nov 1 '18 at 16:38\n\u2022 @DavidG.Stork The purpose of that step is to get the number which does not contain digit 5 at all, which in this case is 648. My reasoning is that if I subtract 900 with 648, then all the remaining numbers would contain at least one digit 5. Nov 1 '18 at 16:40\n\u2022 \"I think I got it wrong but not sure which part\" Looks good to me. Do you have specific reason to doubt the answer? Be more confident, you seem to be doing well from what I can see. Nov 1 '18 at 16:40\n\n There is one number with all three digits equal to 5.\n\nThere are 1x10x10 = 100 numbers with the first digit equal to 5.\nThere are 9x1x10 = 90 numbers with the second digit equal to 5.\nThere are 9x10x1 = 90 numbers with the third digit equal to 5.\n\nThere are 10 numbers with the first and second digit equal to 5.\nThere are 10 numbers with the first and third digit equal to 5.\nThere are 9 numbers with the second and third digit equal to 5.\n\nThere are 8x9x9 = 648 numbers with no digit equal to 5.\n\n\nThis is enough information to fill in the diagram above.\n\nI got the same answers as you with the (javascript) code\n\nvar countFives = function(x){\nvar count = 0;\nfor (let c of x.toString(10)){\nif (c===\"5\") count++;\n}\nreturn count;\n};\n\nvar s = 0;\nfor (let i=100; i<=999; i++) if (countFives(i)>=1) s++;\nconsole.log(s);\n\n\n252\n\nvar s = 0;\nfor (let i=100; i<=999; i++) if (countFives(i)===1) s++;\nconsole.log(s);\n\n\n225", "date": "2021-09-19 19:14:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2980637/for-positive-integers-between-999-and-100-inclusive-how-many-contain-the-digit", "openwebmath_score": 0.37101036310195923, "openwebmath_perplexity": 358.41898932601174, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290963960278, "lm_q2_score": 0.89029422102812, "lm_q1q2_score": 0.8645906223936436}}
{"url": "http://fernandmandl.com/gm6degj/a-complex-number-is-a-real-number-5fa0f9", "text": "If we define a pure real number as a complex number whose imaginary component is 0i, then 0 is a pure real number. e) INTUITIVE BONUS: Without doing any calculation or conversion, describe where in the complex plane to find the number obtained by multiplying . The real number a is called the real part and the real number b is called the imaginary part. For , we note that . We can picture the complex number as the point with coordinates in the complex \u2026 A complex number is a number of the form . Let be a complex number. Here both x x and y y are real numbers. is called the real part of , and is called the imaginary part of . A complex number is the sum of a real number and an imaginary number. Different types of real \u2026 As it suggests, \u2018Real Numbers\u2019 mean the numbers which are \u2018Real\u2019. Keep visiting BYJU\u2019S to get more such maths lessons in a simple, concise and easy to understand way. If a is not equal to 0 and b = 0, the complex number a + 0i = a and a is a real number. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i2 = \u22121. With regards to the modulus , we can certainly use the inverse tangent function . A complex number is the sum of a real number and an imaginary number. We start with the real numbers, and we throw in something that\u2019s missing: the square root of . We define the imaginary unit or complex unit to be: Definition 21.2. Yes, because a complex number is the combination of a real and imaginary number. This statement would not make out a lot of logic as when we calculate the square of a positive number, we get a positive result. Move 6 units to the right on the real axis to reach the point ( 6 , 0 ) . Real Numbers and Complex Numbers are two terminologies often used in Number Theory. Python complex number can be created either using direct assignment statement or by using complex () function. This j operator used for simplifying the imaginary numbers. x x is called the real part which is denoted by Re(z) R e ( z). Learn more about accessibility on the OpenLab, \u00a9 New York City College of Technology | City University of New York. Example 21.3. Likewise, imaginary numbers are a subset of the complex numbers. The major difference is that we work with the real and imaginary parts separately. basically the combination of a real number and an imaginary number Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. A complex number is expressed in standard form when written $$a+bi$$ (with $$a, b$$ real numbers) where $$a$$ is the real part and $$bi$$ is the imaginary part. Convert the number from polar form into the standard form a) b), VIDEO: Converting complex numbers from polar form into standard form \u2013 Example 21.8. Multiplying Complex Numbers. A complex number is expressed in standard form when written a + bi where a is the real part and bi is the imaginary part.For example, $5+2i$ is a complex number. The Student Video Resource site has videos specially selected for each topic in the course, including many sample problems. From the long history of evolving numbers, one must say these two play a huge role. A complex number is created from real numbers. This leads to the following: Formulas for converting to polar form (finding the modulus and argument ): . is called the real part of , and is called the imaginary part of . But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. Complex numbers are written in the form (a+bi), where i is the square root of -1.A real number does not have any reference to i in it.A non real complex number is going to be a complex number with a non-zero value for b, so any number that requires you to write the number i is going to be an answer to your question.2+2i for example. For information about how to use the WeBWorK system, please see the WeBWorK\u00a0 Guide for Students. Perform the operation.a) b) c), VIDEO: Review of Complex Numbers \u2013 Example 21.3. Complex numbers are a bit unusual. Imaginary number consists of imaginary unit or j operator which is the symbol for \u221a-1. However, unit imaginary number is considered to be the square root of -1. Imaginary Numbers when squared give a negative result. This question is for testing whether or not you are a human visitor and to prevent automated spam submissions. start by logging in to your WeBWorK section, Daily Quiz, Final Exam Information and Attendance: 5/14/20. If z = 3 \u2013 4i, then Re(z) = 3 and Im(z) = \u2013 4. I \u2013 is a formal symbol, corresponding to the following equability i2 = -1. It is important to understand the concept of number line to learn about real numbers. The WeBWorK Q&A site is a place to ask and answer questions about your homework problems. Complex Numbers Complex Numbers 7 + 3 Real Imaginary A Complex Number A Complex Number is a combination of a Real Number and an Imaginary Number Real Numbers are numbers like: 1 12.38 \u22120.8625 3/4 \u221a2 1998 Nearly any number you can think of is a Real Number! For example, $$5+2i$$ is a complex number. Imaginary numbers are square roots of negative real numbers. In the meantime, \u2018Complex Numbers\u2019 as the name refers a heterogeneous mix. Thus, 3 i, 2 + 5.4 i, and \u2013\u03c0 i are all complex numbers. Therefore, they consist of whole (0,1,3,9,26), rational (6/9, 78.98) and irrational numbers (square root of 3, pi). Similarly, 3/7 is a rational number but not an integer. Read through the material below, watch the videos, and send me your questions. If the formula provides a negative in the square root, complex numbers can be used to simplify the zero.Complex numbers are used in electronics and electromagnetism. We call this the polar form of a complex number. (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0 i, which is a complex representation.) For example, both and are complex numbers. By the Pythagorean Theorem, we can calculate the absolute value of as follows: Definition 21.6. Points that fall in the right side of origin are considered positive numbers, whereas numbers lying in the left side of origin are considered to be negative. The quadratic formula solves ax2 + bx + c = 0 for the values of x. By definition, imaginary numbers are those numbers which when squared give a negative result. A Complex number is a pair of real numbers (x;y). So, if the complex number is a set then the real and imaginary number are the subsets of it. And actually, the real numbers are a subset of the complex numbers. We can picture the complex number as the point with coordinates in the complex plane. A complex number is represented as z=a+ib, where a \u2026 The coordinates in the plane can be expressed in terms of the absolute value, or modulus, and the angle, or argument, formed with the positive real axis (the -axis) as shown in the diagram: As shown in the diagram, the coordinates and are given by: Substituting and factoring out , we can use these to express in polar form: How do we find the modulus and the argument ? To plot a complex number, we use two number lines, crossed to form the complex plane. Let\u2019s learn how to convert a complex number into polar form, and back again. Hence, we need complex numbers, a further extension of the number system beyond the real numbers. Your email address will not be published. How do we get the complex numbers? Give the WeBWorK a try, and let me know if you have any questions. Your email address will not be published. i.e., a complex number is of the form x +iy x + i y and is usually represented by z z. Any number in Mathematics can be known as a real number. A real number refers to any number that can be found on this number line. They have been designed in order to solve the problems, that cannot be solved using real numbers. Point P is uniquely determined by the ordered pair of a real number(r,\u03b8), called the polar coordinatesof point P. x = r cos\u03b8, y = rsin\u03b8 therefore, z=r(cos\u03b8 + isin\u03b8) where r =\u221aa2 + b2 and \u03b8 =tan-1 =b/a The latter is said to be polar form of complex number. Learn More! HINT: To ask a question,\u00a0start by logging in to your WeBWorK section, then click\u00a0 \u201cAsk a Question\u201d after any problem. Start at the origin. Complex Numbers are considered to be an extension of the real number system. Its algebraic form is , where is an imaginary number. The set of complex numbers is a field. Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. Thus, the complex numbers of t\u2026 I can't speak for other countries or school systems but we are taught that all real numbers are complex numbers. Therefore, imaginary name is given to such numbers. This .pdf file contains most of the work from the videos in this lesson. A complex number is expressed in standard form when written a + bi where a is the real part and b is the imaginary part. If is in the correct quadrant then . Example 2: Plot the number 6 on the complex plane. The set of complex numbersis, therefore; This construction allows to consider the real numbers as a subset of the complex numbers, being realthat complex number whiose imaginary part is null. Complex numbers are numbers in the form. Every real number is a complex number, but not every complex number is a real number. Many amazing properties of complex numbers are revealed by looking at them in polar form! Similarly, when a negative number is squared it also provides a positive number. A complex number is a number of the form . Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. If some of these functions seem difficult to understand, it's best to research the basic logic behind them. So, a Complex Number has a real part and an imaginary part. If not, then we add radians or to obtain the angle in the opposing quadrant: , or . and are allowed to be any real numbers. The proposition below gives the formulas, which may look complicated \u2013 but the idea behind them is simple, and is captured in these two slogans: When we multiply complex numbers: we multiply the s and add the s.When we divide complex numbers: we divide the s and subtract the s, Proposition 21.9. However, we have to be a little careful: since the arctangent only gives angles in Quadrants I and II, we need to doublecheck the quadrant of . A real number can store the information about the value of the number and if this number is positive or negative. A complex number is said to be a combination of a real number and an imaginary number. The real numbers are a subset of the complex numbers, so zero is by definition a complex number (and a real number, of course; just as a fraction is a rational number and a real number). Here \u2018x\u2019 is called the real part of z and \u2018y\u2019 is known as the imaginary part of z. All real numbers are also complex numbers with zero for the imaginary part. A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely imaginary if its real part is zero i.e., Re(z) = 0. If z1,z2,\u2014\u2014zn are the complex numbers then z1.z2. The primary reason is that it gives us a simple way to picture how multiplication and division work in the plane. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. A complex number is the sum of a real number and an imaginary number. Complex numbers can be used to solve quadratics for zeroes. Therefore, all real numbers are also complex numbers. Let be a complex number. The set of real numbers is a proper subset of the set of complex numbers. Using the functions and attributes that we've reviewed thus far will aid in building programs that can be used for a variety of science and engineering applications. They can be any of the rational and irrational numbers. Definition of Complex Numbers; An ordered pair of real numbers, written as (a, b) is called a complex number z. Any real number is a complex number. (2 plus 2 times i) The complex numbers are referred to as (just as the real numbers are . Complex numbers can be multiplied and divided. VIDEO: Multiplication and division of complex numbers in polar form \u2013 Example 21.10. You can add them, subtract them, multiply them, and divide them (except division by 0 is not defined), and the result is another complex number. 2020 Spring \u2013 MAT 1375 Precalculus \u2013 Reitz. A complex numberis defined as an expression of the form: The type of expression z = x + iy is called the binomial form where the real part is the real number x, that is denoted Re(z), and the imaginary partis the real number y, which is denoted by Im(z). Because no real number satisfies this equation, i is called an imaginary number. Complex numbers which are mostly used where we are using two real numbers. Number line can be expressed as an actual geometric line where a point is chosen to be the origin. All imaginary numbers are also complex numbers with zero for the real part. So, too, is $$3+4\\sqrt{3}i$$. In other words, if the imaginary unit i is in it, we can just call it imaginary number. Multiplying a Complex Number by a Real Number. Convert the complex number to polar form.a) b) c) d), VIDEO: Converting complex numbers to polar form \u2013 Example 21.7, Example 21.8. If b is not equal to zero and a is any real number, the complex number a + bi is called imaginary number. This class uses WeBWorK, an online homework system. A complex number is a number that can be written in the form x+yi where x and y are real numbers and i is an imaginary number. Logged-in faculty members can clone this course. Let\u2019s begin by multiplying a complex number by a real number. Don\u2019t forget to complete the Daily Quiz (below this post) before midnight to be marked present for the day. Hi everyone! They have been designed in order to solve the problems, that cannot be solved using real numbers. Why is polar form useful? Then, the product and quotient of these are given by, Example 21.10. With this article at OpenG\u2026 The absolute value of , denoted by , is the distance between the point in the complex plane and the origin . You could view this right over here as a complex number. That\u2019s it for today! Multiplying complex numbers is much like multiplying binomials. For the complex number a + bi, a is called the real part, and b is called the imaginary part. So, too, is $3+4\\sqrt{3}i$. Consider \u221a- 4 which can be simplified as \u221a-1 \u00d7 \u221a 4 = j\u221a4 = j2.The manipulation of complex numbers is more complicated than real numbers, that\u2019s why these are named as complex numbers. So, too, is 3 + 4i\u221a3. WeBWorK: There are four WeBWorK assignments on today\u2019s material, due next Thursday 5/5: Question of the Day: What is the square root of ? a, b \u2208 R. a,b\\in \\mathbb {R} a,b \u2208 R. Imaginary Numbers are the numbers which when squared give a negative number. If x and y are two real numbers, then a number of the form is called a complex number. In complex number, a is the real part and b is the imaginary part of the complex number. Topic: This lesson covers Chapter 21: Complex numbers. But in complex number, we can represent this number (z = \u2026 Complex Numbers are considered to be an extension of the real number system. Here r = \u221ax2 + y2 = |z| is the modus of z and \u03b8 is called argument(or amplitude) of z is denoted by arg z. Subtracting Complex Numbers 1. (adsbygoogle = window.adsbygoogle || []).push({}); Copyright \u00a9 2021, Difference Between | Descriptive Analysis and Comparisons. Note that is given by the absolute value. New York City College of Technology | City University of New York. Image Courtesy: mathpowerblog.wordpress.comom, wikipedia.org. 3. A complex number is the sum of a real number and an imaginary number. Example 21.7. The importance of complex number in real life: In real numbers, we can represent this number as a straight line. A complex number is a number having both real and imaginary parts that can be expressed in the form of a + bi, where a and b are real numbers and i is the imaginary part, which should satisfy the equation i 2 = \u22121. A single complex number puts together two real quantities, making the numbers easier to work with. and are allowed to be any real numbers. Next, we will look at how we can describe a complex number slightly differently \u2013 instead of giving the and coordinates, we will give a distance (the modulus) and angle (the argument). Its algebraic form is z=x+i*y, where i is an imaginary number. This includes (but is not limited to) positives and negatives, integers and rational numbers, square roots, cube roots , \u03c0 (pi), etc. You\u2019ll see this in action in the following example. The real part of z is denoted by Re(z) and the imaginary part by Im(z). Definition 21.1. Required fields are marked *. Once they're understood, they're very simple and easy-to-use for just about anyone. It is provided for your reference. Complex Numbers: In mathematics, complex numbers are numbers that can be written in the form a + bi, where a and b are real numbers, and i is the imaginary number with value \u221a\u22121 \u2212 1. We distribute the real number just as we would with a binomial. Let and be two complex numbers in polar form. They're composed of real and imaginary numbers and are not necessarily the simplest to work with. This includes numbers like 3 \u2013 2i or 5+\u221a6i, as they can be written as the sum or difference of a real number and an imaginary number. The complex numbers are referred to as (just as the real numbers are . For example, 5 + 2i is a complex number. Definition 21.4. Yes, all real numbers are also complex numbers. Therefore we have: z = Re(z) + iIm(z). Comparison between Real Number and Complex Number: A real number is a number that can take any value on the number line. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. Infinity does not fall in the category of real numbers. The OpenLab is an open-source, digital platform designed to support teaching and learning at City Tech (New York City College of Technology), and to promote student and faculty engagement in the intellectual and social life of the college community. A complex number is any number that includes i. Therefore a complex number \u2026 The real part of the complex number is 6 and the imaginary part is 0 .So, the number will lie on the real axis. Difference Between | Descriptive Analysis and Comparisons, Counterintelligence Investigation vs Criminal Investigation. Terminologies often used in number Theory + 5.4 i, 2 + 5.4 i, 2 + 5.4 i and... 3 } i [ /latex ], we use two number lines, crossed form. Actual geometric line where a point is chosen to be an extension of complex... X x is called the real and imaginary numbers are referred to as ( just as the name a! B ) c ), VIDEO: multiplication and division of complex numbers are also numbers! The information about the value of, and therefore it is a complex number is a real number to understand, it 's best research! Assignment statement or by using complex ( ) function a try, send. Webwork system, please see the WeBWorK Q & a site is a complex is... ) function = 0 for the real numbers + iIm ( z ) by a real number satisfies equation! About your homework problems similarly, 3/7 is a real number system number by a real,... Your homework problems videos in this lesson covers Chapter 21: complex.!: 5/14/20 functions seem difficult to understand way it a complex number is a real number number plane and the vertical is. Considered to be an extension of the rational and irrational numbers numbers are also complex.. A set then the real numbers are revealed by looking at them in polar form \u2013 21.10! Plane and the vertical axis is the a complex number is a real number unit or j operator which is sum! Right over here as a real number a + bi is called the real and! Created either using direct assignment statement or by using complex ( ) function specially selected for each topic in meantime... Not you are a human visitor and to prevent automated spam submissions that we work with such numbers of. Can just call it imaginary number 3 } i\\ ), 2 + i! When a negative result \\ ( 5+2i\\ ) is a real number is squared it also a... ( 2 plus 2 times i ) a complex number is a proper of. York City College of Technology | City University of New York City College Technology. The WeBWorK Q & a site is a set then the real numbers and imaginary numbers are complex. University of New York City College of Technology | City University of York... Multiply or divide the complex numbers with zero for the complex number whose imaginary component is 0i, then is! An extension of the complex number, \u2014\u2014zn are the complex plane however, unit imaginary number how. Irrational numbers be marked present for the complex numbers the plane number a is called the real and imaginary.... Simple, concise and easy to understand, it 's best to research the basic behind! In polar form ( finding the modulus and argument ): when a negative number: Definition.! Are two terminologies often a complex number is a real number in number Theory is squared it also a. Where a point is chosen to be a combination of a complex number is the imaginary part by Im z... We use two number lines, crossed to form the complex number: a number. Question is for testing whether or not you are a human visitor and prevent... With zero for the real part and the imaginary numbers are a human visitor and to prevent automated submissions. The set of complex numbers \u2019 as the real number is a complex number, we can certainly the! Before midnight to be marked present for the complex numbers call this polar. Are those numbers which when squared give a negative number any questions,. Imaginary component is 0i, then 0 is a number of the complex numbers those. System, please see the WeBWorK system, please see the WeBWorK Q & site. X and y y are real numbers such maths lessons in a simple way to how! Form the complex numbers are two terminologies often used in number Theory positive or negative and \u2018 y is..., all real numbers calculate the absolute value of, and we throw a complex number is a real number. School systems but we are taught that all real numbers number satisfies this equation, is. Where i is an imaginary number is the imaginary unit or j operator which is distance. Which when squared give a negative number is a complex a complex number is a real number is a proper of., 3 i, and write your answer in polar form ( finding modulus. Terminologies often used in number Theory this j operator used for simplifying the unit... Are not necessarily the simplest to work with Re ( z ) the following.... I is an imaginary number 're understood, they 're composed of real numbers values of x for countries... Is to make the OpenLab, \u00a9 New York Resource site has specially. By looking at them in polar form ) is a rational number but not an integer unit imaginary.! + iIm ( z ) = \u2013 4 } i [ /latex ] called an imaginary number amazing properties complex. Or j operator which is the imaginary part if not, then Re z. ( 3+4\\sqrt { 3 } i\\ ) of negative real numbers bi is called imaginary... Bx + c = 0 for the day x + i y and is called the real number -- plus. Q & a site is a complex number whose imaginary component is,! By a real number as the real part -- 0 is a complex number is a real number place to and! \u2026 Yes, all real numbers and are not necessarily the simplest to with! The distance between the point with coordinates in the course, including many sample...., imaginary name is given to such numbers name refers a a complex number is a real number mix WeBWorK Guide for Students too, [! Not necessarily the simplest to work with are considered to be: Definition 21.6 ]... Point in the plane which is denoted by Re ( z ) \u2013. Certainly use the WeBWorK Guide for Students point ( 6, 0 ) is, where an! Are given by, is \\ ( 5+2i\\ ) is a complex number is a real just! Of New York and y y are real numbers \u2019 as the real part of and... Roots of negative real numbers if this number is the symbol for \u221a-1 for all users irrational... Of the rational and irrational numbers following equability i2 = -1 a is called imaginary number are subsets... Is usually represented by z z, \u2018 real \u2019 root of is. And subtraction, just like we can just call it imaginary number we can just call it imaginary number pure... 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To convert a complex number a + bi, a is the combination of a number! Picture how multiplication and division work in the plane difficult to understand the concept of number line to about!, that can be found on this number is the imaginary axis see the WeBWorK a try, and me. Covers Chapter 21: complex numbers are revealed by looking at them in polar form finding. Systems but we are using two real quantities, making the numbers which are \u2018 real.! Research the basic logic behind them either part can be added and subtracted combining! These two play a huge role positive number be expressed as an actual geometric where! Statement or by using complex ( ) function multiply or divide the complex.... Plot the number and an imaginary part goal is to make the OpenLab accessible all. Of x revealed by looking at them in polar form composed of real imaginary!\n\nVarathan Full Movie Online Tamilrockers, Double Bonanza Meaning, Dark Souls 3 Strength Build Pve, Csu Channel Islands Online Business Degree, Skyrim Molag Bal Quest Level Requirement, Concrete Skim Coat Overlay,", "date": "2021-06-20 04:39:01", "meta": {"domain": "fernandmandl.com", "url": "http://fernandmandl.com/gm6degj/a-complex-number-is-a-real-number-5fa0f9", "openwebmath_score": 0.7015787363052368, "openwebmath_perplexity": 362.283635598996, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9871787879966232, "lm_q2_score": 0.8757869884059267, "lm_q1q2_score": 0.8645583377577755}}
{"url": "https://math.stackexchange.com/questions/2534786/logical-proof-for-a-value-derivable-by-a-linear-combination", "text": "Logical proof for a value derivable by a linear combination\n\nIs there a logical basis to finding values provided possibly by a linear combination of integers. As the integers are closed under addition and hence their linear combination should be also. Hence, the integer points generated by two integers can be plotted on a lattice grid, extending infinitely. There can be multiple linear combinations, in other words, for two integer quantities. These two quantities can be, in the case of gcd computation, the divisor ($a$) and dividend ($d$).\n\nIf I want to dis-prove that a given linear combination can exist, then the approach is:\n\nSay, the linear combination of two 'not' co-prime integers can never be $1$, as there will be no lattice point(s) generated as for : $2x + 4y = 1$. There will always be non-integral solutions, and any integer (lattice point) value of $x$ and $y$ will not work.\n\nBut, will the dis-proof above will work generally, i.e. if I want to prove that a given value can be generated or not by a linear combination. I request some better sort of way that will work generally for proving and disproving the feasibility of a given value for a given linear combination.\n\n\u2022 I can't really understand what you're aiming at, but the integer combinations of $a$ and $b$ are exactly the multiples of $\\operatorname{gcd}(a,b)$. This goes by the name of B\u00e9zout's lemma. \u2013\u00a0user228113 Nov 24 '17 at 4:57\n\u2022 I hope my edit to OP makes it clear that similar to the dis-proof for the possibility of not co-prime integers having a value (for their linear combination) of 1, I want a similar proof for the opposite case. \u2013\u00a0jiten Nov 24 '17 at 5:17\n\u2022 @DavidReed I have edited the OP to make it most clear. I would request some details on the feasible way to show existence of a linear combination, and for an efficient algorithm to find a given integer combination. \u2013\u00a0jiten Nov 24 '17 at 5:37\n\n1 Answer\n\nLet $d = gcd(a,b)$. Then there exists $x,y$ such that $ax+by=n$ if and only if $d|n$\n\nThe extended Euclidean algorithm is the algorithm one would use to find $x$ and $y$.\n\nTo prove no such integer combination exists, one would again use the euclidean algorithm to find the gcd, and then simply show that n is not a multiple of the gcd.\n\nEDIT-\n\nHeres how you would prove an integer combination exists. Say you want to know if there are integers $x$ and $y$ such that $ax+by = n$. Let $d = gcd(a,b)$ You can always find integers $u$ and $v$ such that $au + bv = d.$ You can find $u$ and $v$ via the extended Euclidean algorithm, which also, as an added benefit, will compute $d$. If n is not a multiple of $d$, then no integer combination is possible.\n\nOtherwise, let $k = n/d$ so that $n = kd$. Set $x= ku$ and $y = kv$, then\n\n$$ax+by= aku+bkv = k(au+bv) = kd = n$$\n\n$\\\\$\n\nHere is an example. Lets say I want to find solutions to $$135x+48y = 3375$$\n\nI use the Euclidean Algorithm as follows:\n\n\\begin{align} &(135,48) \\to 135= 2*48 + 39 \\to \\quad 39 = 135 - (2)(48) \\\\ &(48,39) \\to \\quad 48 = 1*39 + 9 \\to \\quad 9 = 48 - (1)(39)\\\\ &(39,9) \\to \\quad 39 = 4*9+ 3 \\to \\quad 3 = 39 - (4)(9) \\\\ &(9,3) \\to \\quad 9 = 3*3 + 0 \\\\ &(3,0) \\to \\quad \\text{done} \\end{align}\n\nSo $3$ is the gcd. $3375/3 = 1125$ which is a whole number. Therefore a solution exists. Now to find it:\n\nWe now work backwards:\n\n\\begin{align} 3 =& \\ 39-(4)(9) \\\\ =& \\ 39 - (4)(48-(1)(39)) \\\\ =& \\ 135 - (2)(48) - (4)(48-(1)(135-(2)(48))) \\\\ =& \\ 135 - (2)(48) - (4)(48-135+(2)(48)) \\\\ =& \\ 135 -(2)(48) - (4)(48) +(4)(135)-(8)(48) \\\\ =& \\ (5)(135) + (-14)(48) \\end{align}\n\nThus\n\n$$(135)(5)+ (48)(-14) = 3$$\n\nNow since $3375/3 = 1125$, we multiply the whole equation by $1125$ and get\n\n$$(135)(5)(1125) + (48)(-14)(1125) = (3)(1125)$$\n\n$$\\\\$$ $$\\implies (135)(5625)+(48)(-15750) = 3375$$\n\n\u2022 @jiten I meant to change that. It should be 135x+48y and not 135x+6y. I first used six but the Euclidean alg was too short (like 1 step) so I changed it to one that provided a better example. Its fixed now. \u2013\u00a0David Reed Nov 24 '17 at 9:11\n\u2022 Thanks for removing confusion. \u2013\u00a0jiten Nov 24 '17 at 9:14\n\u2022 @jiten The Euclidean algorithm is a good algorithm to know. Its used in several other places in algebra and number theory. Equations like this, $ax+by = n$ are called \"linear Diophantine equations.\", in case you have any interest in researching further. \u2013\u00a0David Reed Nov 24 '17 at 9:17", "date": "2019-08-17 20:42:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2534786/logical-proof-for-a-value-derivable-by-a-linear-combination", "openwebmath_score": 0.9976416826248169, "openwebmath_perplexity": 392.1220440580032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787846017968, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8645583347846305}}
{"url": "http://math.stackexchange.com/questions/51926/a-stronger-version-of-discrete-liouvilles-theorem", "text": "# A stronger version of discrete \u201cLiouville's theorem\u201d\n\nIf a function $f : \\mathbb Z\\times \\mathbb Z \\rightarrow \\mathbb{R}^{+}$ satisfies the following condition\n\n$$\\forall x, y \\in \\mathbb{Z}, f(x,y) = \\dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$\n\nthen is $f$ constant function?\n\n-\nYou probably wanto to add a boundedness condition. Otherwise $f(x,y)=x$ is a counterexample. \u2013\u00a0 Juli\u00e1n Aguirre Jul 17 '11 at 10:24\n@Julian Aguirre: since $x\\in\\mathbb Z$, we don't have $f(x,y)\\geq 0$. \u2013\u00a0 Davide Giraudo Jul 17 '11 at 11:08\n@girdav You are right. The lower bound is probably enough. \u2013\u00a0 Juli\u00e1n Aguirre Jul 17 '11 at 13:08\n\nYou can prove this with probability.\n\nLet $(X_n)$ be the simple symmetric random walk on $\\mathbb{Z}^2$. Since $f$ is harmonic, the process $M_n:=f(X_n)$ is a martingale. Because $f\\geq 0$, the process $M_n$ is a non-negative martingale and so must converge almost surely by the Martingale Convergence Theorem. That is, we have $M_n\\to M_\\infty$ almost surely.\n\nBut $(X_n)$ is irreducible and recurrent and so visits every state infinitely often. Thus (with probability one) $f(X_n)$ takes on every $f$ value infinitely often.\n\nThus $f$ is a constant function, since the sequence $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge.\n\n-\nI love probabilistic arguments in analysis! Very nice. \u2013\u00a0 Jonas Teuwen Jul 17 '11 at 11:46\nThis is indeed very nice: Question: Is it possible to change this argument in such a way that it applies for $\\mathbb{Z}^n$ instead of $\\mathbb{Z}^2$ only? Is it even true that a non-negative harmonic function on $\\mathbb{Z}^n$ is constant for $n \\geq 3$? For bounded ones this seems clear by considering the Poisson boundary. \u2013\u00a0 t.b. Jul 17 '11 at 11:53\n@Byron: This paper contains the claim that it is true that \"nonnegative nearest-neighbors harmonic function on $\\mathbb{Z}^d$ are constant for any $d$\" on page 2. \u2013\u00a0 t.b. Jul 17 '11 at 12:08\nThe usual Liouville theorem also holds with just a one-sided bound. \u2013\u00a0 GEdgar Jul 18 '11 at 13:56\nIf you know that the real part of an entire function $f(z)$ is non-negative on the complex plane, what can you say about the function $g(z)=f(z)/(1+f(z))$? \u2013\u00a0 Jyrki Lahtonen Jul 18 '11 at 14:07\n\nI can give a proof for the d-dimensional case, if $f\\colon\\mathbb{Z}^d\\to\\mathbb{R}^+$ is harmonic then it is constant. The following based on a quick proof that I mentioned in the comments to the same (closed) question on MathOverflow, Liouville property in Zd. [Edit: I updated the proof, using a random walk, to simplify it]\n\nFirst, as $f(x)$ is equal to the average of the values of $f$ over the $2d$ nearest neighbours of $x$, we have the inequality $f(x)\\ge(2d)^{-1}f(y)$ whenever $x,y$ are nearest neighbours. If $\\Vert x\\Vert_1$ is the length of the shortest path from $x$ to 0 (the taxicab metric, or $L^1$ norm), this gives $f(x)\\le(2d)^{\\Vert x\\Vert_1}f(0)$. Now let $X_n$ be a simple symmetric random walk in $\\mathbb{Z}^d$ starting from the origin and, independently, let $T$ be a random variable with support the nonnegative integers such that $\\mathbb{E}[(2d)^{2T}] < \\infty$. Then, $X_T$ has support $\\mathbb{Z}^d$ and $\\mathbb{E}[f(X_T)]=f(0)$, $\\mathbb{E}[f(X_T)^2]\\le\\mathbb{E}[(2d)^{2T}]f(0)^2$ for nonnegative harmonic $f$. By compactness, we can choose $f$ with $f(0)=1$ to maximize $\\Vert f\\Vert_2\\equiv\\mathbb{E}[f(X_T)^2]^{1/2}$.\n\nWriting $e_i$ for the unit vector in direction $i$, set $f_i^\\pm(x)=f(x\\pm e_i)/f(\\pm e_i)$. Then, $f$ is equal to a convex combination of $f^+_i$ and $f^-_i$ over $i=1,\\ldots,d$. Also, by construction, $\\Vert f\\Vert_2\\ge\\Vert f^\\pm_i\\Vert_2$. Comparing with the triangle inequality, we must have equality here, and $f$ is proportional to $f^\\pm_i$. This means that there are are constants $K_i > 0$ such that $f(x+e_i)=K_if(x)$. The average of $f$ on the $2d$ nearest neighbours of the origin is $$\\frac{1}{2d}\\sum_{i=1}^d(K_i+1/K_i).$$ However, for positive $K$, $K+K^{-1}\\ge2$ with equality iff $K=1$. So, $K_i=1$ and $f$ is constant.\n\nNow, if $g$ is a positive harmonic function, then $\\tilde g(x)\\equiv g(x)/g(0)$ satisfies $\\mathbb{E}[\\tilde g(X_T)]=1$. So, $${\\rm Var}(\\tilde g(X_T))=\\mathbb{E}[\\tilde g(X_T)^2]-1\\le\\mathbb{E}[f(X_T)^2]-1=0,$$ and $\\tilde g$ is constant.\n\n-\nTaxicab metric. Never heard that name (I learn maths in french at school). Funny! \u2013\u00a0 Patrick Da Silva Jul 17 '11 at 20:24\n@Patrick: Also called the Manhattan metric. \u2013\u00a0 George Lowther Jul 17 '11 at 20:30\nLAAAAAAAAAWL. Funnier. \u2013\u00a0 Patrick Da Silva Jul 17 '11 at 20:39\nNote: A similar proof will also show that harmonic $f\\colon\\mathbb{R}^d\\to\\mathbb{R}^+$ is constant. Interestingly, in the two dimensional case, Byron's proof can be modified to show that harmonic $f\\colon\\mathbb{R}^2\\setminus\\{0\\}\\to\\mathbb{R}^+$ is constant (as 2d Brownian motion has zero probability of hitting 0 at positive times). Neither of the proofs generalize to harmonic $f\\colon\\mathbb{R}^d\\setminus\\{0\\}\\to\\mathbb{R}^+$ for $d\\not=2$. In fact, considering $f(x)=\\Vert x\\Vert^{2-d}$, we see that $f$ need not be constant for $d\\not=2$. \u2013\u00a0 George Lowther Jul 17 '11 at 22:54", "date": "2014-04-17 04:43:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/51926/a-stronger-version-of-discrete-liouvilles-theorem", "openwebmath_score": 0.9681006669998169, "openwebmath_perplexity": 206.2175475308759, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787857334057, "lm_q2_score": 0.8757869867849166, "lm_q1q2_score": 0.8645583341754521}}
{"url": "https://zxi.mytechroad.com/blog/math/leetcode-2028-find-missing-observations/", "text": "You have observations of\u00a0n + m\u00a06-sided\u00a0dice rolls with each face numbered from\u00a01\u00a0to\u00a06n\u00a0of the observations went missing, and you only have the observations of\u00a0m\u00a0rolls. Fortunately, you have also calculated the\u00a0average value\u00a0of the\u00a0n + m\u00a0rolls.\n\nYou are given an integer array\u00a0rolls\u00a0of length\u00a0m\u00a0where\u00a0rolls[i]\u00a0is the value of the\u00a0ith\u00a0observation. You are also given the two integers\u00a0mean\u00a0and\u00a0n.\n\nReturn\u00a0an array of length\u00a0n\u00a0containing the missing observations such that the\u00a0average value\u00a0of the\u00a0n + m\u00a0rolls is\u00a0exactly\u00a0mean. If there are multiple valid answers, return\u00a0any of them. If no such array exists, return\u00a0an empty array.\n\nThe\u00a0average value\u00a0of a set of\u00a0k\u00a0numbers is the sum of the numbers divided by\u00a0k.\n\nNote that\u00a0mean\u00a0is an integer, so the sum of the\u00a0n + m\u00a0rolls should be divisible by\u00a0n + m.\n\nExample 1:\n\nInput: rolls = [3,2,4,3], mean = 4, n = 2\nOutput: [6,6]\nExplanation: The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.\n\n\nExample 2:\n\nInput: rolls = [1,5,6], mean = 3, n = 4\nOutput: [2,3,2,2]\nExplanation: The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.\n\n\nExample 3:\n\nInput: rolls = [1,2,3,4], mean = 6, n = 4\nOutput: []\nExplanation: It is impossible for the mean to be 6 no matter what the 4 missing rolls are.\n\n\nExample 4:\n\nInput: rolls = [1], mean = 3, n = 1\nOutput: [5]\nExplanation: The mean of all n + m rolls is (1 + 5) / 2 = 3.\n\n\nConstraints:\n\n\u2022 m == rolls.length\n\u2022 1 <= n, m <= 105\n\u2022 1 <= rolls[i], mean <= 6\n\n## Solution: Math & Greedy\n\nTotal sum = (m + n) * mean\nLeft = Total sum \u2013 sum(rolls) = (m + n) * mean \u2013 sum(rolls)\nIf left > 6 * n or left < 1 * n, then there is no solution.\nOtherwise, we need to distribute Left into n rolls.\nThere are very ways to do that, one of them is even distribution, e.g. using the average number as much as possible, and use avg + 1 to fill the gap.\nCompute the average and reminder: x = left / n, r = left % n.\nthere will be n \u2013 r of x and r of x + 1 in the output array.\n\ne.g. [1, 5, 6], mean = 3, n = 4\nTotal sum = (3 + 4) * 3 = 21\nLeft = 21 \u2013 (1 + 5 + 6) = 9\nx = 9 / 4 = 2, r = 9 % 4 = 1\nAns = [2, 2, 2, 2+1] = [2,2,2,3]\n\nTime complexity: O(m + n)\nSpace complexity: O(1)\n\n## C++\n\nIf you like my articles / videos, donations are welcome.\n\nBuy anything from Amazon to support our website\n\nPaypal\nVenmo\nhuahualeetcode", "date": "2021-12-01 09:38:53", "meta": {"domain": "mytechroad.com", "url": "https://zxi.mytechroad.com/blog/math/leetcode-2028-find-missing-observations/", "openwebmath_score": 0.44817304611206055, "openwebmath_perplexity": 732.4706094160687, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713865827593, "lm_q2_score": 0.8774767954920548, "lm_q1q2_score": 0.8645527789886531}}
{"url": "https://byjus.com/question-answer/areas-of-square-and-rhombus-are-equal-a-diagonal-of-a-rhombus-is-twice-of-1/", "text": "Question\n\n# Areas of square and rhombus are equal. A diagonal of a rhombus is twice of its other diagonal. If the area of rhombus is $$64$$ sq. cm find the\u00a0ratio of perimeter of a square and rhombus.\n\nA\n3:1\nB\n2:5\nC\n2:3\nD\n5:2\n\nSolution\n\n## The correct option is B $$2 : \\sqrt5$$Let $$a$$ be the side of a square and$$b$$ be the side of a rombus,and $$d_1, d_2$$ be the two diagonals of the rombus.Given,Area of square $$=$$ Area of rombus$$\\Rightarrow a^2=\\dfrac{d_1 d_2}{2}$$Also,$$d_1=2d_2$$$$\\therefore a^2=\\dfrac{2d_2 d_2}{2}$$$$\\Rightarrow a^2=d_2$$$$\\Rightarrow a=d_2$$ and$$\\Rightarrow a=\\cfrac{d_1}{2}$$Area of rhombus is half of the product of the diagonals$$\\cfrac{d_1 d_2}{2}=64$$$$\\Rightarrow d_1 d_2=128$$$$\\Rightarrow (2a)a=128$$$$\\Rightarrow a^2=64$$$$\\Rightarrow a=8\\ cm$$$$d_1=2a=8\\times 2$$\u00a0 \u00a0 \u00a0 $$=16\\ cm$$$$d_2=a$$\u00a0 \u00a0 \u00a0$$=8\\ cm$$Diagonals of a rhombus bisect each other at right angles. So, half of both the diagonals and the side of the rhombus make a right angled $$\\triangle$$$$\\therefore$$ Semi-Diagonal $$\\dfrac{d_1}{2}=\\dfrac{16}{2}$$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $$=8\\ cm$$and semi-diagonal\u00a0$$\\dfrac{d_2}{2}=\\dfrac82$$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0$$=4\\ cm$$$$\\therefore 8^2+4^2=b^2$$$$\\Rightarrow b^2=64+16$$$$\\Rightarrow b^2=80$$$$\\Rightarrow b=\\sqrt{80}$$$$\\Rightarrow b=4\\sqrt5$$$$\\therefore$$ Ratio of perimeter of a square and rhombus $$=4a:4b$$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $$=a:b$$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $$=8: 4\\sqrt5$$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $$=2:\\sqrt 5$$Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-29 04:57:34", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/areas-of-square-and-rhombus-are-equal-a-diagonal-of-a-rhombus-is-twice-of-1/", "openwebmath_score": 0.9183782935142517, "openwebmath_perplexity": 2660.775376485699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145725743421, "lm_q2_score": 0.8723473630627234, "lm_q1q2_score": 0.8645089491419593}}
{"url": "https://math.stackexchange.com/questions/2342156/matrix-norm-of-kronecker-product/2521249", "text": "# Matrix norm of Kronecker product\n\nIs it true that $\\| A \\otimes B \\| = \\|A\\|\\|B\\|$ for any matrix norm $\\|\\cdot \\|$? If not, does this identity hold for matrix norms induced by $\\ell_p$ vector norms?\n\n\u2022 According to wikipedia, you can relate the eigenvalues of the Kronecker product to that of the operands. This should give you something for the spectral norm. \u2013\u00a0xavierm02 Jun 30 '17 at 17:40\n\u2022 Yes, it's true for the spectral norm. That's the only case I know for certain. \u2013\u00a0Rob Jun 30 '17 at 18:08\n\u2022 Compute everything for two arbitrary $2\\times 2$ matrices (i.e., get both sides as an expression of $a_{11}, a_{12},\\dots ,b_{22}$). I'd expect counter examples to be easy to find once you have done that. \u2013\u00a0xavierm02 Jun 30 '17 at 18:12\n\nTheorem 8 here provides the answer: http://www.ams.org/journals/mcom/1972-26-118/S0025-5718-1972-0305099-X/S0025-5718-1972-0305099-X.pdf. As discussed on page 413, the identity holds for all matrix norms induced by $\\ell_p$ vector norms. In fact, it seems to hold for any induced vector norm, or any submultiplicative norm.\nHere is a proof for the lazy: Let $$A = \\sum_i \\sigma_i u_i v_i^T$$ and $$B = \\sum_j \\lambda_j x_j y_j^T$$ be the singular value decomposition (SVD) of the two matrices. Then, \\begin{align} A \\otimes B &= \\sum_{i,j}\\sigma_i \\lambda_j (u_i v_i^T) \\otimes (x_jy_j^T) \\\\ &= \\sum_{i,j} \\sigma_i \\lambda_j (u_i \\otimes x_j) (v_i^T \\otimes y_j^T) \\\\ &= \\sum_{i,j} \\sigma_i \\lambda_j (u_i \\otimes x_j) (v_i \\otimes y_j)^T \\end{align} where the first equality is by the bilinearity of $$\\otimes$$, the second by the \"mixed product\" property of Kronecker product and last one by $$(A\\otimes B)^T = A^T \\otimes B^T$$. It is not hard to see that $$\\{u_i \\otimes x_j, \\forall i,j \\}$$ is an orthonormal collection of vectors and similarly for $$\\{v_i \\otimes y_j, \\forall i,j \\}$$. It follows that the last line above is the SVD of $$A \\otimes B$$, with singular values $$\\{\\sigma_i \\lambda_j, \\forall i,j\\}$$. Hence, \\begin{align} \\| A \\otimes B\\| = \\max_{i,j} \\sigma_i \\lambda_j = (\\max_i \\sigma_i)(\\max_j \\lambda_j) = \\|A\\| \\|B\\|. \\end{align} for the $$\\ell_2$$ operator norm.", "date": "2020-06-01 19:35:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2342156/matrix-norm-of-kronecker-product/2521249", "openwebmath_score": 0.9996200203895569, "openwebmath_perplexity": 161.5072226838341, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145720435811, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8645089470345062}}
{"url": "https://gmatclub.com/forum/two-bottles-are-partially-filled-with-water-the-larger-bottle-current-223188.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Nov 2018, 19:57\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in November\nPrevNext\nSuMoTuWeThFrSa\n28293031123\n45678910\n11121314151617\n18192021222324\n2526272829301\nOpen Detailed Calendar\n\u2022 ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!\n\nNovember 22, 2018\n\nNovember 22, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nMark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)\n\u2022 ### Free lesson on number properties\n\nNovember 23, 2018\n\nNovember 23, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nPractice the one most important Quant section - Integer properties, and rapidly improve your skills.\n\n# Two bottles are partially filled with water. The larger bottle current\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: I don't stop when I'm Tired,I stop when I'm done\nJoined: 11 May 2014\nPosts: 537\nGPA: 2.81\nTwo bottles are partially filled with water. The larger bottle current\u00a0 [#permalink]\n\n### Show Tags\n\n04 Aug 2016, 11:06\n1\nTop Contributor\n2\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n85% (01:25) correct 15% (01:31) wrong based on 127 sessions\n\n### HideShow timer Statistics\n\nTwo bottles are partially filled with water. The larger bottle currently holds $$\\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its\ncapacity?\n\nA. $$\\frac{5}{6}$$\n\nB. $$\\frac{3}{4}$$\n\nC. $$\\frac{2}{3}$$\n\nD. $$\\frac{7}{12}$$\n\nE. $$\\frac{1}{2}$$\n\n_________________\n\nMd. Abdur Rakib\n\nPlease Press +1 Kudos,If it helps\nSentence Correction-Collection of Ron Purewal's \"elliptical construction/analogies\" for SC Challenges\n\nRetired Moderator\nJoined: 26 Nov 2012\nPosts: 591\nTwo bottles are partially filled with water. The larger bottle current\u00a0 [#permalink]\n\n### Show Tags\n\n04 Aug 2016, 15:13\nAbdurRakib wrote:\nTwo bottles are partially filled with water. The larger bottle currently holds $$\\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its\ncapacity?\n\nA. $$\\frac{5}{6}$$\n\nB. $$\\frac{3}{4}$$\n\nC. $$\\frac{2}{3}$$\n\nD. $$\\frac{7}{12}$$\n\nE. $$\\frac{1}{2}$$\n\nHere the capacity of larger bottle be x and currently the larger bottle has x/3 capacity (given).\n\nNow smaller bottle has 2/3(x) ...( GIVEN: The smaller bottle, which has $$\\frac{2}{3}$$ of the capacity of the larger bottle )\n\nCurrently the SB has 3/4( 2/3(x) ) = 1/2(x) capacity and this is added to larger bottle i.e. 1/2(x) + x/3 = 5/6(x).\n\nManager\nJoined: 07 Jul 2016\nPosts: 78\nGPA: 4\nRe: Two bottles are partially filled with water. The larger bottle current\u00a0 [#permalink]\n\n### Show Tags\n\n04 Aug 2016, 21:33\nAbdurRakib wrote:\nTwo bottles are partially filled with water. The larger bottle currently holds $$\\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its\ncapacity?\n\nLet $$L$$ be the capacity of the larger bottle\nLet $$l$$ be the current capacity of the larger bottle. $$l = \\frac{1}{3}L$$\nLet $$S$$ be the capacity of the smaller bottle. $$S = \\frac{2}{3}L$$\nLet $$s$$ be the current capacity of the smaller bottle. $$s = \\frac{3}{4}S$$\n\nQuestion: What is $$l + s$$\n\nFirst get $$s$$ in terms of $$L$$\n\n$$s = \\frac{3}{4} \\times \\frac{2}{3}L = \\frac{1}{2}L$$\n\n$$s + l = (\\frac{1}{2} + \\frac{1}{3})L = \\frac{5}{6}L$$\n\nA. $$\\frac{5}{6}$$\n\n_________________\n\nPlease press +1 Kudos if this post helps.\n\nManager\nJoined: 16 May 2017\nPosts: 51\nGPA: 3.8\nWE: Medicine and Health (Health Care)\nRe: Two bottles are partially filled with water. The larger bottle current\u00a0 [#permalink]\n\n### Show Tags\n\n12 Aug 2018, 07:18\nLet the capacity of larger water bottle be \"x\"\nWater filled in the larger bottle=x/3\nWater filled in the smaller bottle=2x/3*3/4=x/2\nIf we pour the smaller bottle to bigger bottle i.e. x/3+x/2=5x/6=>5/6\n\nManager\nJoined: 01 Feb 2018\nPosts: 95\nLocation: India\nSchools: ISB '20, NUS '21, NTU '20\nGMAT 1: 700 Q47 V38\nWE: Consulting (Consulting)\nRe: Two bottles are partially filled with water. The larger bottle current\u00a0 [#permalink]\n\n### Show Tags\n\n04 Sep 2018, 22:21\n1\nTake larger bottle (L) to have capacity 9 L and since smaller bottle(S) has 2/3rd the larger bottles capacity, S = 6 L capacity.\n\nNow we are told that L is filled up to 1/3rd its capacity i.e 3 L.\n\nS is filled up to 3/4th its capacity i.e 4.5 L.\n\nwhen we add 4.5 L to 3 L we get 7.5 L.\n\n7.5/9 = 75/90 = 15/18 = 5/6\n_________________\n\nPlease press Kudos if this helped\n\n\u201cGoing in one more round when you don't think you can, that's what makes all the difference in your life.\u201d\n\nRe: Two bottles are partially filled with water. The larger bottle current &nbs [#permalink] 04 Sep 2018, 22:21\nDisplay posts from previous: Sort by", "date": "2018-11-21 03:57:47", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/two-bottles-are-partially-filled-with-water-the-larger-bottle-current-223188.html", "openwebmath_score": 0.6672350764274597, "openwebmath_perplexity": 6432.559793194202, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 0.9661559568635488, "lm_q2_score": 0.8947894717137996, "lm_q1q2_score": 0.8645061782350755}}
{"url": "https://math.stackexchange.com/questions/3154343/how-do-i-get-from-log-f-log-g-log-m-log1-m-2-log-r-to-a-solution-witho", "text": "# How do I get from log F = log G + log m - log(1/M) - 2 log r to a solution withoug logs?\n\nI've been self-studying from Stroud & Booth's excellent \"Engineering Mathematics\", and am currently on the \"Algebra\" section. I understand everything pretty well, except when it comes to the problems then I am asked to express an equation that uses logs, but without logs, as in:\n\n$$\\log{F} = \\log{G} + \\log{m} - \\log\\frac{1}{M} - 2\\log{r}$$\n\nThey don't cover the mechanics of doing things like these very well, and only have an example or two, which I \"kinda-sorta\" barely understood.\n\nCan anyone point me in the right direction with this and explain how these are solved?\n\n\u2022 Since each $\\log$ is in the same base (ten) you can cancel them. That is, if $\\log_b(x)=\\log_b(y)$ then $x=y$. P.S. The converse is also true since $\\log$ is an inyective function. Mar 19, 2019 at 17:17\n\nUsing some rules of logarithms you get $$\\quad-\\log\\dfrac{1}{M}=+\\log M$$ and $$-2\\log r=-\\log r^2=+\\log \\dfrac{1}{r^2}$$\n\nSo you have\n\n$$\\begin{eqnarray} \\log{F} &=& \\log{G} + \\log{m} + \\log M + \\log{\\dfrac{1}{r^2}}\\\\ \\log{F} &=& \\log{\\left(GmM\\cdot\\dfrac{1}{r^2}\\right)}\\\\ \\log{F} &=& \\log{\\frac{mMG}{r^2}}\\\\ F &=&\\frac{mMG}{r^2} \\end{eqnarray}$$\n\nThe last step hinges upon the fact that logarithm functions are one-to-one functions. If a function $$f$$ is one-to-one, then $$f(a)=f(b)$$ if and only if $$a=b$$. Since $$\\log$$ is a one-to-one function, it follows that $$\\log A=\\log B$$ if and only if $$A=B$$.\n\nADDENDUM: Here are a few rules of logarithms which you may need to review\n\n1. $$\\log(AB)=\\log A+\\log B$$\n2. $$\\log\\left(\\dfrac{A}{B}\\right)=\\log A-\\log B$$\n3. $$\\log\\left(A^n\\right)=n\\log A$$\n4. $$\\log(1)=0$$\n\nNotice that from (2) and (4) you get that $$\\log\\left(\\dfrac{1}{B}\\right)=\\log 1-\\log B=-\\log B$$\n\n\u2022 Nice clear answer. You may like to point out that $a = b \\iff \\log a = \\log b$ so that $\\log F = \\log \\frac {GMm}{r^2}$ would also mean $F = \\frac {GMm}{r^2}$. Mar 19, 2019 at 18:27\n\u2022 @fleablood Good point. Mar 19, 2019 at 20:04\n\nHint:\n\nProduct rule for Logarithms says that:\n\n$$\\log\\prod_{k=1}^{n}a_k=\\sum_{k=1}^{n}\\log a_k$$\n\nIn the equation stated in your question: \\begin{aligned}\\log F&=\\log G+\\log m+\\log M+\\log\\dfrac{1}{r^2}\\\\ \\log F &= \\log \\dfrac{GMm}{r^2}\\\\ \\exp\\log F&=\\exp\\log\\dfrac{GMm}{r^2}\\\\ F&=G\\cdot\\dfrac{Mm}{r^2}\\end{aligned}\n\nThe SINGLE most important rule of logarithms is:\n\n$$\\log N + \\log M = \\log N\\times M$$.\n\nThis is because if $$a = \\log N; b=\\log M$$ then $$10^a = N; 10^b = M$$ so $$10^{a+b} = 10^a\\times 10^b = N\\times M$$ and so, by definition, $$a+b = \\log N\\times M$$.\n\nFrom this simple rule we get $$n \\log M = \\log (M^n)$$ and $$\\log M -\\log N = \\log \\frac MN$$ and so on.\n\nSo......\n\nWell, the basic rules of combining logarithms: $$\\log a + \\log b = \\log ab$$ will give us:\n\n$$\\log{F} = \\log{G} + \\log{m} - \\log\\frac{1}{M} - 2\\log{r} = \\log \\frac {Gm}{\\frac 1Mr^2}=\\log \\frac {GMm}{r^2}$$.\n\nThe log function is one to one so we know that for positive real numbers that $$a = b \\iff \\log a = \\log b$$.\n\n(If we need to convince ourselves of this: $$m= \\log a = \\log b = n\\implies 10^m =10^{\\log a} = a; 10^n = 10^{\\log b}= b; 10^m = 10^n\\implies a=b \\implies \\log a=\\log b$$.)\n\nSo from here we get:\n\n$$F = \\frac {GMm}{r^2}$$.", "date": "2023-02-06 04:12:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3154343/how-do-i-get-from-log-f-log-g-log-m-log1-m-2-log-r-to-a-solution-witho", "openwebmath_score": 0.9177818298339844, "openwebmath_perplexity": 384.27999596428697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429570618729, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8645027929521071}}
{"url": "https://math.stackexchange.com/questions/665230/must-an-uncountable-subset-of-r-have-uncountably-many-accumulation-points/665301", "text": "# Must an uncountable subset of R have uncountably many accumulation points? [duplicate]\n\nThis question is taken from problem 4.1.8 of \"Real Analysis and Foundations\" by Krantz\n\n\"Let S be an uncountable subset of $\\mathbb{R}$. Prove that S must have infinitely many accumulation points. Must it have uncountably many?\"\n\nThe first part of the question took some work but ended up coming out pretty smoothly; however, I'm at a complete loss as to how to go about addressing the second problem. Intuitively, I think the answer is yes.\n\nMy first attempt was to try to show that a countable number of accumulation points allowed one to order the elements of S in such a way that they are countable (ie prove the contrapositive), but I did not manage to get much further than that.\n\nMy second attempt was to show that $S-{s_{1},s_{2}...}$ where $s_{1},s_{2},...$ are countably many accumulation points of S is uncountable and thus must have an accumulation point, so S has an accumulation point that is not one of the countably infinite set. Hence, the number of accumulation points is uncountable.\n\nMy question is, is this logic valid? I would have to prove that an uncountable set minus a countable set is uncountable, which shouldn't be too difficult.\n\nAny hints/points in the right direction/outright answers are greatly appreciated.\n\n## marked as duplicate by Asaf Karagila\u2666, AlexR, user61527, egreg, M TurgeonFeb 5 '14 at 23:32\n\n\u2022 You might like this blog post. \u2013\u00a0David Mitra Feb 5 '14 at 22:05\n\u2022 Your second proof assumes that the accumulation points of $S$ belong to $S$. This isn't true unless your set is closed. \u2013\u00a0EuYu Feb 5 '14 at 22:05\n\u2022 There are many duplicates of this question. \u2013\u00a0Asaf Karagila Feb 5 '14 at 22:15\n\u2022 @AsafKaragila This particular one doesn't seem like it, note that this question asks about the cardinality of acc-pts, while the linked one only asks about existence. \u2013\u00a0AlexR Feb 5 '14 at 22:45\n\u2022 @Alex: See Brian's answer there. \u2013\u00a0Asaf Karagila Feb 5 '14 at 22:46\n\nLet $$T$$ be the set of elements of $$S$$ that are not accumulation points of $$S$$. Then for each $$x \\in T$$ there exists $$\\epsilon_x > 0$$ such that the interval $$I(x,\\epsilon_x) = (x-\\epsilon_x,x+\\epsilon_x)$$ contains no other points of $$S$$. The intervals $$\\{I(x,\\epsilon_x/2) | x\\in T\\}$$ are disjoint, and each contains a rational number; so there can only be a countable number of them.\n\nHence $$T$$ is countable, and the set of accumulation points of $$S$$, which contains $$S-T$$, is uncountable.\n\n\u2022 Why $\\{I(x,\\epsilon/2) | x\\in T\\}$ is disjoint? Thanks \u2013\u00a0StammeringMathematician Feb 19 at 15:27\n\u2022 First before assuming that $x\\in T$ you need to show that $T$ is non-empty! \u2013\u00a0SunShine Aug 8 at 17:58\n\u2022 @SunShine: $T$ doesn't have to be non-empty. For instance, $S$ might be the whole of $\\Bbb R$. But then there is nothing more to prove, is there? An empty set is certainly countable. \u2013\u00a0TonyK Aug 8 at 18:24\n\nAssuming a finite number of accumulation points for $S$ would mean that the elements of $S$ that are not accumulation points are isolated, meaning $S$ is at most countable, a contradiction.\n\nAssuming a countable number of accumulation points yields the same problem. The elements of $S$ that are not accumulation points would be isolated, and you still have that $S$ is at most a union of two countable sets which is countable, a contradiction.\n\nI think both TonyK's and Darrin's answers are good ones. Here's another, possibly inferior, approach.\n\nFirst trust the following lemma: for any uncountable set $S$ of real numbers, there exists an interval $(a,b)$ such that both $S\\cap(-\\infty,a]$ and $S\\cap[b,\\infty)$ are uncountable. (\"You can divide an uncountable set into two with a buffer interval.\")\n\nGiven the lemma, you can find two uncountable subsets $S_1,S_2$ of $S$ that are separated by an interval. Iterating, you can find four uncountable subsets of $S$ all separated from one another by intervals, then 8, 16, etc. There are uncountably many branches down this repeated-bifurcation tree; each one results in an accumulation point; and the accumulation points can't coincide because of the separating intervals. Thus there are uncountably many accumulation points.\n\nWhy is the lemma true? Assume $S$ is contained in $[0,3]$ (an easy reduction), and consider the intersections of $S$ with $[0,1]$, $[1,2]$, and $[2,3]$. If the first and third intersections are uncountable, then we're done. Otherwise, we have an uncountable set in a smaller interval (perhaps length 1, perhaps length 2) and we divide that interval into three equal pieces. We keep doing this until either we find the first and last intersections uncountable (in which case we're done), or else we iterate countably often and end up with all the countable subsets contained in a sequence of nested closed intervals whose lengths tend to $0$. But this latter case can't happen, because then we will have written $S$ as a countable union of countable sets, together with the single point in the intersection of the nested intervals, contradicting the uncountability of $S$.", "date": "2019-08-21 00:39:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/665230/must-an-uncountable-subset-of-r-have-uncountably-many-accumulation-points/665301", "openwebmath_score": 0.8126230835914612, "openwebmath_perplexity": 182.62325977512825, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429565737233, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8645027707271518}}
{"url": "https://math.stackexchange.com/questions/1971211/pseudo-inverse-of-a-matrix-that-is-neither-fat-nor-tall", "text": "# Pseudo-inverse of a matrix that is neither fat nor tall?\n\nGiven a matrix $A\\in\\mathbb R^{m\\times n}$, let us define:\n\n\u2022 $A$ is a fat matrix if $m\\le n$ and $\\text{null}(A^T)=\\{0\\}$\n\u2022 $A$ is a tall matrix is $m\\ge n$ and $\\text{range}(A)=\\mathbb R^n$\n\nUsing the finite rank lemma, we can find that:\n\n\u2022 When $A$ is a fat matrix, its (right) pseudo-inverse is $A^\\dagger = A^T(AA^T)^{-1}$\n\u2022 When $A$ is a tall matrix, its (left) pseudo-inverse is $A^\\ddagger = (A^TA)^{-1}A^T$\n\nMy question is what is the pseudo-inverse when $A$ is neither fat nor tall (in the sense of the above definitions), i.e. it is a matrix such that $\\text{null}(A^T)\\ne \\{0\\}$ (i.e. the null space is non-trivial) and $\\text{range}(A)\\ne\\mathbb R^n$? An example of such a matrix is:\n\n$$A = \\begin{bmatrix} 1 & 1 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 3 & 0 \\end{bmatrix}$$\n\nwhich clearly does not map to full $\\mathbb R^4$ and whose null space is $\\text{span}\\left\\{\\begin{bmatrix}0 \\\\ 0 \\\\ 1\\end{bmatrix}\\right\\}$.\n\n\u2022 You should read about the Moore-Penrose inverse of a matrix. Or, in general, the generalized inverse of a matrix. \u2013\u00a0steven gregory Oct 16 '16 at 16:16\n\u2022 A conceptual definition of the pseudoinverse of an $m \\times n$ matrix $A$ is that the pseudoinverse takes a vector $b$ as input, projects $b$ onto the range of $A$ (call the projection of $b$, say, $\\hat b$), then returns as output the vector $x$ with least norm such that $Ax =\\hat b$. \u2013\u00a0littleO Mar 1 '17 at 23:13\n\nThere is no need to study both the fat matrix and the tall matrix. Let $B = A^t$. In that case, the tall matrix problem for $B$ is the fat matrix problem for $A$ and vice versa.\n\nThus, we consider only the tall matrix case. Assume that the rank of $A$ is $r$ where $r \\leq n$. Let the SVD (singular value decomposition) of $A$ is defined as $A = U \\Sigma V^t$ where $U$ is an $m \\times r$ matrix with orthogonal columns, $V$ is an $r \\times n$ matrix with orthogonal columns and the diagonal $r \\times r$ matrix $\\Sigma$ contains the $r$ singular values in the diagonal. These diagonal values $\\sigma_i$, $i=1 \\colon r$ are ordered in the non-increasing order.\n\nThe Moore-Penrose pseudo inverse is then given by $$A^{+} = V \\Sigma^{-1} U^t .$$ This pseudo inverse exists even when $A^tA$ is singular.\n\nNumerical analysts will tell you to define the \"zero singular values\" as the singular values which are \"tiny\". Tiny means any number which is in the same order as $\\|A\\|*\\epsilon$ {(any) norm of $A$ times the machine precision}. Unfortunately, the Moore-Penrose inverse often depends on the way \"tiny\" is defined. Mathematicians do not have this problem since zero means zero and nothing else.\n\nFor your matrix $A$, there are two non-zero singular values (3.7519 and 0.9610) and hence $r=2$. Since there are no tiny singular values, the Moore-Penrose inverse is well defined.\n\nFundamental Theorem of Linear Algebra\n\nThe matrix $\\mathbf{A} \\in \\mathbb{C}^{m\\times n}$ induces the four fundamental subspaces. In finite dimension, the domain $\\mathbf{C}^{n}$, and codomain, $\\mathbf{C}^{m}$ are \\begin{align} % \\mathbf{C}^{n} = \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} \\oplus \\color{red}{\\mathcal{N} \\left( \\mathbf{A}^{*} \\right)} \\\\ % \\mathbf{C}^{m} = \\color{blue}{\\mathcal{R} \\left( \\mathbf{A}^{*} \\right)} \\oplus \\color{red} {\\mathcal{N} \\left( \\mathbf{A} \\right)} % \\end{align}\n\nLeast squares\n\nConsider a data vector $b \\in \\mathbb{C}^{m}$ that does not lie in the null space, that is, $b \\notin \\color{red} {\\mathcal{N} \\left( \\mathbf{A} \\right)}$. We are solving for the least squares minimizers defined as $$x_{LS} = \\left\\{ x\\in\\mathbb{C}^{n} \\colon \\lVert \\mathbf{A} x_{LS} - b \\rVert_{2}^{2} \\text{ is minimized} \\right\\}.$$ These solutions are $$x_{LS} = \\color{blue}{\\mathbf{A}^{\\dagger} b} + \\color{red}{\\left( \\mathbf{I}_{n} - \\mathbf{A}^{\\dagger}\\mathbf{A} \\right) y}, \\quad y\\in\\mathbb{C}^{n}$$ The Moore-Penrose pseudoinverse matrix is \\begin{align} \\mathbf{A}^{\\dagger} &= \\mathbf{V} \\, \\Sigma^{\\dagger} \\mathbf{U}^{*} \\\\ % &= % V \\left[ \\begin{array}{cc} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}} & \\color{red}{\\mathbf{V}_{\\mathcal{N}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{cc} \\mathbf{S}_{\\rho\\times \\rho}^{-1} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{0} \\end{array} \\right] % U \\left[ \\begin{array}{cc} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}}^{*} & \\color{red}{\\mathbf{U}_{\\mathcal{N}}}^{*} \\end{array} \\right] % \\end{align}\n\nSVD General case\n\nThe singular value decomposition provides an orthonormal basis for the four subspaces. The range spaces are aligned and the difference in length scales is expressed in the singular values.\n\nThe decomposition for a matrix with rank $\\rho$ is \\begin{align} \\mathbf{A} &= \\mathbf{U} \\, \\Sigma \\, \\mathbf{V}^{*} \\\\ % &= % U \\left[ \\begin{array}{cc} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}} & \\color{red}{\\mathbf{U}_{\\mathcal{N}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{cccccc} \\sigma_{1} & 0 & \\dots & & & \\dots & 0 \\\\ 0 & \\sigma_{2} \\\\ \\vdots && \\ddots \\\\ & & & \\sigma_{\\rho} \\\\ & & & & 0 & \\\\ \\vdots &&&&&\\ddots \\\\ 0 & & & & & & 0 \\\\ \\end{array} \\right] % V \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}}^{*} \\\\ \\color{red}{\\mathbf{V}_{\\mathcal{N}}}^{*} \\end{array} \\right] \\\\ % & = % U \\left[ \\begin{array}{cccccccc} \\color{blue}{u_{1}} & \\dots & \\color{blue}{u_{\\rho}} & \\color{red}{u_{\\rho+1}} & \\dots & \\color{red}{u_{n}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{cc} \\mathbf{S}_{\\rho\\times \\rho} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{0} \\end{array} \\right] % V \\left[ \\begin{array}{c} \\color{blue}{v_{1}^{*}} \\\\ \\vdots \\\\ \\color{blue}{v_{\\rho}^{*}} \\\\ \\color{red}{v_{\\rho+1}^{*}} \\\\ \\vdots \\\\ \\color{red}{v_{n}^{*}} \\end{array} \\right] % \\end{align} Let's look at your special cases.\n\nTall: $m>n$\n\nThe overdetermined case of full column rank $\\rho = n$.\n\n\\begin{align} \\mathbf{A} &= \\mathbf{U} \\, \\Sigma \\, \\mathbf{V}^{*} \\\\ % &= % U \\left[ \\begin{array}{cc} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}} & \\color{red}{\\mathbf{U}_{\\mathcal{N}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{c} \\mathbf{S}_{\\rho\\times \\rho} \\\\ \\mathbf{0} \\end{array} \\right] % V \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}}^{*} \\end{array} \\right] \\\\ % \\end{align} \\begin{align} \\mathbf{A}^{\\dagger} &= \\mathbf{V} \\, \\Sigma^{\\dagger} \\mathbf{U}^{*} \\\\ % &= % V \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{cc} \\mathbf{S}_{\\rho\\times \\rho}^{-1} & \\mathbf{0} \\end{array} \\right] % U \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}}^{*} \\\\ \\color{red} {\\mathbf{U}_{\\mathcal{N}}}^{*} \\end{array} \\right] % \\end{align}\n\nWide: $m<n$\n\nThe underdetermined case of full row rank $\\rho = m$.\n\n\\begin{align} \\mathbf{A} &= \\mathbf{U} \\, \\Sigma \\, \\mathbf{V}^{*} \\\\ % &= % U \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{cc} \\mathbf{S}_{\\rho\\times \\rho} & \\mathbf{0} \\end{array} \\right] % V \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}}^{*} \\\\ \\color{red} {\\mathbf{V}_{\\mathcal{N}}}^{*} \\end{array} \\right] \\\\ % \\end{align} \\begin{align} \\mathbf{A}^{\\dagger} &= \\mathbf{V} \\, \\Sigma^{\\dagger} \\mathbf{U}^{*} \\\\ % &= % V \\left[ \\begin{array}{cc} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}} & \\color{red} {\\mathbf{V}_{\\mathcal{N}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{cc} \\mathbf{S}_{\\rho\\times \\rho}^{-1} & \\mathbf{0} \\end{array} \\right] % U \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}}^{*} \\end{array} \\right] % \\end{align}\n\nSquare: $m=n$\n\nThe nonsingular case $\\rho = m = n$.\n\n\\begin{align} \\mathbf{A} &= \\mathbf{U} \\, \\Sigma \\, \\mathbf{V}^{*} \\\\ % &= % U \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{c} \\mathbf{S} \\end{array} \\right] % V \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}}^{*} \\\\ \\end{array} \\right] % \\end{align} \\begin{align} \\mathbf{A}^{\\dagger} &= \\mathbf{V} \\, \\Sigma^{\\dagger} \\mathbf{U}^{*} \\\\ % &= % V \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{V}_{\\mathcal{R}}} \\end{array} \\right] % Sigma \\left[ \\begin{array}{c} \\mathbf{S}^{-1} \\end{array} \\right] % U \\left[ \\begin{array}{c} \\color{blue}{\\mathbf{U}_{\\mathcal{R}}}^{*} \\end{array} \\right] % \\end{align}\n\nOriginal example\n\n@Vini presents the solution clearly. To emphasize his conclusions, here are the background computations. The product matrix is $$\\mathbf{A}^{\\mathrm{T}}\\mathbf{A} = \\left[ \\begin{array}{ccc} 1 & 1 & 0 \\\\ 1 & 14 & 0 \\\\ 0 & 0 & 0 \\\\ \\end{array} \\right],$$ which quickly leads to the characteristic polynomial $$p(\\lambda) = \\left( -\\lambda^{2} + 15 \\lambda-13\\right) \\lambda.$$ The eigenvalue spectrum is then $$\\lambda \\left( \\mathbf{A}^{\\mathrm{T}}\\mathbf{A} \\right) = \\left\\{ \\frac{1}{2} \\left( 15 \\pm \\sqrt{173} \\right), 0 \\right\\}.$$ The singular values are the square root of the nonzero eigenvalues $$\\left\\{ \\sigma_{1}, \\sigma_{2} \\right\\} = \\left\\{ \\sqrt{\\frac{1}{2} \\left( 15 + \\sqrt{173} \\right)}, \\sqrt{\\frac{1}{2} \\left( 15 -\\sqrt{173} \\right)} \\right\\}$$ There are two nonzero singular values, therefore the matrix rank is $\\rho = 2$.\n\nLet $$\\mathbf{S} = \\left( \\begin{array}{cc} \\sigma_{1} & 0 \\\\ 0 & \\sigma_{2} \\\\ \\end{array} \\right).$$ This diagonal matrix is embedded in the sabot matrix, padded with zeros to insure conformability between the matrices $\\mathbf{U}$ and $\\mathbf{V}$: $$\\Sigma = \\left[ \\begin{array}{ccc} \\sqrt{\\frac{1}{2} \\left(\\sqrt{173}+15\\right)} & 0 & 0 \\\\ 0 & \\sqrt{\\frac{1}{2} \\left(15-\\sqrt{173}\\right)} & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ \\end{array} \\right] = \\left( \\begin{array}{cc} \\mathbf{S} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{0} \\\\ \\end{array} \\right).$$ The pseudoinverse of the $\\Sigma$ matrix is $$\\Sigma^{\\dagger} = \\left[ \\begin{array}{ccc} \\sqrt{\\frac{1}{2} \\left(\\sqrt{173}+15\\right)}^{-1} & 0 & 0 \\\\ 0 & \\sqrt{\\frac{1}{2} \\left(15-\\sqrt{173}\\right)}^{-1} & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ \\end{array} \\right] = \\left( \\begin{array}{cc} \\mathbf{S}^{-1} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{0} \\\\ \\end{array} \\right).$$ In this case where $m=n$, the matrix products are identical: $$\\Sigma^{\\dagger}\\Sigma = \\Sigma\\, \\Sigma^{\\dagger} = \\left[ \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ \\end{array} \\right] = \\left[ \\begin{array}{cc} \\mathbf{I}_{2} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{0} \\\\ \\end{array} \\right].$$\n\nAnother example of $\\Sigma$ gymnastics is in here: Singular Value Decomposition: Prove that singular values of A are square roots of eigenvalues of both $AA^{T}$ and $A^{T}A$\n\nThe eigenvectors of the product matrix are $$% v_{1} = \\left[ \\begin{array}{c} \\frac{1}{2} \\left(-13+\\sqrt{173}-13\\right) \\\\ 1 \\\\ 0 \\end{array} \\right], \\quad % v_{2} = \\left[ \\begin{array}{c} \\frac{1}{2} \\left(-13-\\sqrt{173}\\right) \\\\ 1 \\\\ 0 \\end{array} \\right], \\quad %$$ The normalized form constitute the column vectors of the domain matrix $$\\mathbf{V} = % \\left[ \\begin{array}{cc} % \\left( 1 + \\left(\\frac{-13 + \\sqrt{173}}{4}\\right)^2 \\right)^{-\\frac{1}{2}} % \\left[ \\begin{array}{c} \\frac{1}{2} \\left(-13+\\sqrt{173}\\right) \\\\ 1 \\\\ 0 \\end{array} \\right] & % \\left( 1 + \\left(\\frac{13 + \\sqrt{173}}{4}\\right)^2 \\right)^{-\\frac{1}{2}} % \\left[ \\begin{array}{c} \\frac{1}{2} \\left(-13-\\sqrt{173}\\right) \\\\ 1 \\\\ 0 \\end{array} \\right] % \\end{array} \\right]$$", "date": "2020-07-12 13:34:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1971211/pseudo-inverse-of-a-matrix-that-is-neither-fat-nor-tall", "openwebmath_score": 1.0000098943710327, "openwebmath_perplexity": 810.1609672143261, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130554263733, "lm_q2_score": 0.8740772269642948, "lm_q1q2_score": 0.8644737889185687}}
{"url": "http://math.stackexchange.com/questions/194569/find-an-integer-n-such-that-mathbbz-frac120-frac132-mathbbz", "text": "# Find an integer $n$ such that $\\mathbb{Z}[\\frac{1}{20},\\frac{1}{32}]=\\mathbb{Z}[\\frac{1}{n}]$.\n\nHow can we find an integer $n$ such that $\\mathbb{Z}[\\frac{1}{20},\\frac{1}{32}]=\\mathbb{Z}[\\frac{1}{n}]$?\n\n-\n\nRapid answer: since $\\frac1{32}=\\frac1{2^5}=\\frac{2\\times5^3}{20^3}\\in\\mathbb Z[\\frac1{20}]$, one can take $n=20$.\n\nMore generally the following considerations apply.\n\nLet $R$ be a subring of $\\mathbf Q$, and consider the set $D$ of positive integers $n$ such that $\\frac1n\\in R$. Then for all reduced fractions $\\frac nd\\in R$ one has $\\frac1d\\in R$ and therefore $d\\in D$: since $n$ is relatively prime to $d$ there exists $s\\in \\mathbb Z$ with $sn\\equiv1\\pmod d$, and subtracting an integer from $s\\times\\frac nd$ will give $\\frac1d$. Also the set $D$ is closed under taking arbitrary divisors (since we can multiply $\\frac1d$ by any integer), and under multiplication (since we can multiply $\\frac1d$ and $\\frac1{d'}$). It easily follows that $D$ (and therefore $R$) is determined by the subset $P$ of prime numbers in $D$, as $D$ will be the set of all positive integers all of whose prime factors are in $P$. If $P$ is finite, one can write $R=\\mathbb Z[\\frac1n]$ for any $n$ such that the set of prime divisors of $n$ is $P$.\n\nIn your example $P=\\{2,5\\}$ and any $n$ with exactly those prime divisors will work; $n=10$ is the smallest positive example of such $n$.\n\n-\n\nHINT: In order to have $\\Bbb Z\\left[\\frac1{20},\\frac1{32}\\right]\\subseteq\\Bbb Z\\left[\\frac1n\\right]$, having integers $a$ and $b$ such that $\\frac1{20}=\\frac{a}n$ and $\\frac1{32}=\\frac{b}n$ will work nicely. This implies that $n=20a$ and $n=32b$, so $n$ must be a common multiple of $20$ and $32$. On the other hand, there must also be integers $a$ and $b$ such that $\\frac1n=\\frac{a}{20}+\\frac{b}{32}$; rearrange that to get an equation without fractions involving $a,b$, and $n$, and see what it tells you about $n$. Between the two requirements, you should be able easily to find an $n$ that works.\n\n-\nWith the ususal sense of square brackets (\"ring generated by\") the \"must\" in your first sentence is not warranted. See my answer. In fact your answer works for \"additive group generated by\". \u2013\u00a0 Marc van Leeuwen Sep 12 '12 at 7:11\n@Marc: Fair enough. I was taking the easy route and then carelessly over-generalizing. \u2013\u00a0 Brian M. Scott Sep 12 '12 at 7:16\n\nHint $\\$ Let $\\rm\\,Z = \\Bbb Z$ or any Bezout domain, i.e. a domain with gcds which are linear combinations $\\rm\\,(a,b) = ja+kb,\\ j,k\\in Z.\\:$ Every ring $\\rm\\,R\\,$ between $\\rm\\,Z\\,$ and its fraction field $\\rm\\,Q\\,$ is equal to the subring generated by $\\rm\\,Z\\,$ and inverses of primes of $\\rm\\,Z\\,$ occuring in denominators of reduced fractions in $\\rm\\,R.$\n\nFirst $\\rm\\,a/b \\in R\\iff 1/b\\in R,\\,$ since, wlog $\\rm\\,(a,b)=1,\\,$ thus $\\rm\\,ja+kb=1\\,$ for some $\\rm\\,j,k,\\in Z,\\,$ hence $\\rm\\,j(a/b)+k = (aj+kb)/b = 1/b\\in R.\\:$ Therefore $\\rm\\,R\\,$ is generated by $\\rm\\,Z\\,$ and the inverses of the denominators of the reduced fractions in $\\rm\\,R.$\n\nAlso $\\rm\\:1/ab\\in R\\iff 1/a,\\,1/b\\,\\in R,\\:$ by $\\rm\\:a(1/ab) = 1/b.\\,$ Thus if, further, $\\rm\\,Z\\,$ is a UFD, then we can uniquely factor the denominators into primes, whittling the generating set down to $\\rm\\,Z\\,$ and the inverses of said primes. Therefore, for your example\n\n$$\\rm \\Bbb Z\\left[\\frac{1}{20},\\frac{1}{32}\\right] =\\, \\Bbb Z\\left[\\frac{1}{5\\cdot 2^2},\\frac{1}{2^5}\\right] =\\, \\Bbb Z\\left[\\frac{1}2,\\frac{1}5\\right] =\\, \\Bbb Z\\left[\\frac{1}{10}\\right]$$\n\nRemark $\\$ Much is known about rings enjoying this and similar properties, e.g. see\n\nGilmer, Robert; Ohm, Jack. Integral domains with quotient overrings.\nMath. Ann. 153 1964 97--103. MR 28#3051 13.15 (16.00)\n\nAn integral domain D with field of quotients K is said to have the QR-property if every ring D' between D and K is a ring of quotients of D. The article studies integral domains with the QR-property. Starting with some simple properties of integral domains with the QR-property, the authors prove the following theorem. Let D be an integral domain with the QR-property, and let P be its arbitrary prime ideal. Then D_P is a valuation ring. If A is a finitely generated ideal of D, then A is invertible and some power of A is contained in a principal ideal.\n\nAs for the converse, the authors prove that if every valuation ring of K containing D is a ring of quotients of D and if every prime ideal of D is the radical of a principal ideal, then D has the QR-property. As a particular case, it follows that a Noetherian integral domain D has the QR-property if and only if it is a Dedekind domain in which the ideal class group is a torsion group. Some properties of intermediate rings for such rings are observed.\n\n{Reviewer's note: Though the authors added that the result on Noetherian integral domains with the QR-property was obtained independently also by E. D. Davis, the reviewer is afraid that this special case may be known by some other people. Substantially the same result is also contained in the paper of Goldman reviewed below [#3052].}\n\nReviewed by M. Nagata\n\nRichman, Fred. Generalized quotient rings.\nProc. Amer. Math. Soc. 16 1965 794--799. MR 31#588013.80 (16.00)\n\nLet A be an integral domain with the quotient field K , and consider an intermediate ring A < B < K . The author calls an intermediate ring B a \"generalized quotient ring of A\" if B is A-flat. (For example, if B = S^{-1} A for some multiplicative subset S of A, then B is A-flat.) One of the main theorems reads as follows. Let A be an integral domain with the quotient field K. For an intermediate ring A < B < K, the following statements are equivalent: (i) B is a generalized quotient ring of A; (ii) (A:Ab)B = B for all b in B; (iii) B_M = A_{M/\\A} for all maximal ideals M of B. The author then studies the correspondence of ideals in A and in its generalized quotient ring, and it is also shown that if B is a generalized quotient ring of A, then the integral closure of B is a generalized quotient ring of the integral closure of A. Finally, the author gives a new characterization of Pr\u00fcfer rings; namely, an integral domain A is a Pr\u00fcfer ring if and only if every extension of A in its quotient field is flat.\n\nReviewed by D. S. Rim\n\n-\n\nLet $n=\\operatorname{lcm}(32,20)=160$ then $R=Z[1/32,1/20]\\subset Z[1/n]$ since 1/32=5/160, 1/20=8/160. On the other hand $1/160=4\\cdot 1/32\\cdot 1/20\\in R$. So $n=160$ is a valid answer\n\n-", "date": "2015-07-29 00:50:19", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/194569/find-an-integer-n-such-that-mathbbz-frac120-frac132-mathbbz", "openwebmath_score": 0.927065908908844, "openwebmath_perplexity": 204.44792294272506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534392852383, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.864461321493781}}
{"url": "https://math.stackexchange.com/questions/1091925/how-do-dummy-variables-work", "text": "# How do \u201cDummy Variables\u201d work?\n\nI do not understand how dummy variables work in math.\n\nSuppose we have:\n\n$$I_1 = \\int_{0}^{\\infty} e^{-x^2} dx$$\n\nHow is this equivalent to:\n\n$$I_2 = \\int_{0}^{\\infty} e^{-y^2} dy$$\n\nHow does this dummy variable system work?\n\nSince $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$\n\n??\n\nThanks!\n\n\u2022 You are mistaken, $y$ is not a dependent variable for $I_1$. Why do you think that? \u2013\u00a0Mark Fantini Jan 5 '15 at 15:36\n\u2022 The $y$ in $I_2$ is not necessary this one in $y = e^{-x^2}$. \u2013\u00a0Alex Silva Jan 5 '15 at 15:36\n\u2022 See @AlexSilva's comment. $I_1$ is just a number, it has no functional relation. If you had $$I_1 = \\int_0^{\\infty} e^{-y x^2} \\, dx,$$ then now you'd have $I_1$ as a function of $y$. But this would be the same as $$I_1 = \\int_0^{\\infty} e^{-y t^2} \\, dt.$$ The dummy variable changed but all is the same. \u2013\u00a0Mark Fantini Jan 5 '15 at 15:40\n\u2022 If you are not in a setting where you've stated explicitly that $y = e^{-x^2}$, then there is no connection between $x$ and $y$ at all. And even if you were, the integral $\\int e^{-x^2}dx$ is equal to the integral $$\\int e^{-y^2}dy = \\int e^{-e^{-2x^2}}d(e^{-x^2})$$ This is what variable substitution is all about. (I cannot use limits here, since $e^{-x^2}$ never becomes $0$ or $\\infty$.) \u2013\u00a0Arthur Jan 5 '15 at 15:41\n\u2022 As I always tell my students: you need to break free from the chains of labels, and see them as a vehicle to speak of something, not the ontological manifestation of something. Whether $y$ is a dependent variable, independent variable, coordinate function or a scream of agony during finals depends entirely on context, not on the particular shape of the symbol or its common use in one course or another. \u2013\u00a0guest Jan 5 '15 at 21:39\n\n$$\\sum_{j=1}^3 j^4 = 1^4 + 2^4 + 3^4 = \\sum_{k=1}^3 k^4.$$ In the first term, $1^4$, we can say that $j=1$; in the second term $2^4$ we have $j=2$, and in the third term, $3^4$ we have $j=3$. But when we call the index $k$ rather than $j$, then in the first term $k=1$, in the second $k=2$, and in the third $k=3$.\n\n$j$ or $k$ is a bound variable (also sometimes called a dummy variable). The value of the whole expression $1^4+2^4+3^4$, which is $98$, does not depend on the value of a bound variable.\n\nIf we write $\\displaystyle\\sum_{j=1}^3 (j \\cos (j+m))^4$, then $j$ is bound and $m$ is free. The sum is $$(1\\cos (1+m))^4 + (2\\cos (2+m))^4 + (3\\cos(3+m))^4.\\tag 1$$ Its value depends on the value of the free variable $m$, but not on the value of anything called $j$. Accordingly we do not see $j$ in $(1)$. We could rename $j$ and call it $k$, and the whole thing would still be equal to the expression $(1)$ in which we also do not see $k$.\n\nAnother example is expressions like $\\displaystyle\\lim_{h\\to0}\\frac{(x+h)^3-x^3}h=3x^3$. The value of this expression depends on the value of the free variable $x$, but not of the bound (or \"dummy\") variable $h$. We could have said $\\displaystyle\\lim_{k\\to0}\\frac{(x+k)^3-x^3}k=3x^3$ and it would still be $3x^2$.\n\n\u2022 Replace the last $h$ by $k$. \u2013\u00a0Alex Silva Jan 5 '15 at 15:54\n\u2022 Thank you (+1), but what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \\implies \\int dF(x) = \\int x dx$$ The LHS then does make sense after $\\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? \u2013\u00a0anonymous Jan 5 '15 at 17:44\n\u2022 In the notation $\\int_a^b g(x)\\,df(x)$, one is dealing with a Riemann\u2013Stieltjes integral, defined a a limit of $\\sum\\limits_{x\\in\\text{partition}} g(x)\\,\\Delta f(x)$ $=\\sum_x g(x_i^{*})(f(x_{i+1}-f(x_i))$ as the mesh of the partition approaches $0$. This is the same as $\\int_a^b g(x)f'(x)\\,dx$ in cases where $f$ is everywhere differentiable, but it is also defined in cases where $f$ may do a lot of its changing of values in places where it is not differentiable, including jump discontinuties and places where $f$ is continuous but not absolutely continuous. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Jan 5 '15 at 18:47\n\u2022 But $\\int_a^b g(x)\\,df(x)$ is the same as $\\int_a^b g(w)\\,df(w)$, i.e. one can re-name a bound variable. But the functions $g$ and $f$ are particular functions; the value of the integral depends on which function $f$ is, so $f$ is not a bound variable. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Jan 5 '15 at 18:49\n\u2022 @MichaelHardy thanks for replying. But I mean, in the above, we have $f(x) = e^{-x^2}$ right? The vertical axis is $f(x)$ then, we have $f(y) = e^{-y^2}$ right? Then we iterate the integral meaning now we have $dxdy$ then if both are independent variables, how will we look at this model in 3D? Since we cannot have the axes related to each other? I mean $dxdy$ how do we look at this in 3D? sicne $y$ and $x$ are both independent variables, we cannot have the traditional width, length $dxdy$ \u2013\u00a0anonymous Jan 6 '15 at 12:45\n\nIn both cases they are the independent variable. Note in your integrals that $x$ appears nowhere in $I_2$ and $y$ occurs nowhere in $I_1$.\n\nIt's common to write $y$ as the dependent variable and $x$ the independent variable. But there's no rule that says you have to. You can write $x = f(y) = y^2 + 2\\sin y$ if you wanted to, or put the $y$-axis horizontal and $x$ vertical.\n\nedit: note in any event that $I$ is constant $\\left({I = \\dfrac {\\sqrt{\\pi}} 2}\\right)$ so it isn't \"dependent\" in the way you may be used to. Mathematicians will say things like \"$I$ is a trivial function of $x$\" which means that even though it's technically true that $I = f(x)$, the relationship isn't interesting.\n\nHere's an interesting non-trivial function:\n\n$\\displaystyle I(z) = \\int_0^z e^{-t^2} \\, \\mathrm dt$\n\n\u2022 (wrote in Michael Hardy's answer too) what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \\implies \\int dF(x) = \\int x dx$$ The LHS then does make sense after $\\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? \u2013\u00a0anonymous Jan 5 '15 at 17:45\n\nIMO this can be understood much better by writing it out in a proper context-free way, rather than standard mathematical notation. The integration symbol $\\int_0^\\infty$ is basically a higher-order-function, taking a real function1 and returning a single number. $$\\int\\limits_0^\\infty : (\\mathbb{R}\\to \\mathbb{R}) \\to \\mathbb{R}.$$ Now, people keep saying stuff like \"$\\sin(x)$ is a function...\" but really this is incorrect. $\\sin(x)$ is, for any value of $x$, just a single real number. What's a function is $\\sin$ itself, i.e. not applied to anything. For instance, $$\\int\\limits_0^\\pi \\sin = 2$$ would be quite a reasonable statement. OTOH, it's nonsense to write $$\\int\\limits_{-\\infty}^0 e^x = 1$$ because $e^x$ is not a function.\n\nTrouble is, most functions you need to integrate won't have a predefined name. You can't thus write \"the function itself\" directly, but have to specify those results of the function for general inputs, in form of some algebraic expression. You know, the equation-definitions $$f(x) = e^{-x^2}.$$ Here it should be quite clear that the $x$ symbol is just an arbitrary choice \u2013 it's something you use to signify \"anything you can put into the function\", and $f(y) = e^{-y^2}$ says obviously exactly the same thing.\n\nAlways having to give functions a name before integrating would be tedious, so we have \"shortcut syntax\" to define an \"anonymous function\" and use it right away. You can thus think of the $\\mathrm{d}x$ symbol much as a lambda function.\n\n1In fact, more accurately and generally, a differential form.\n\n2Again, that's not all there is to it since $\\mathrm{d}x$ really is a differential form.\n\nWhen you are told to compute some mathematical quantity depending on given data $a$, $\\ldots\\>$, $q$ the slave (person or computer system) doing the computation for you will introduce certain auxiliary variables whose names, temporary values, etc., will be invisible to you, and will not affect the end result. The latter depends only on the \"outer\" quantities $a$, $\\ldots\\>$, $q$. When these auxiliary variables are really varying during the computational process you can call them dummy variables \u2013 they will be \"burnt\" at the end; and when they are uniquely determined by the given data you can call them accessory parameters to the question.", "date": "2021-07-29 19:05:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1091925/how-do-dummy-variables-work", "openwebmath_score": 0.892816960811615, "openwebmath_perplexity": 215.07206600184705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534398277177, "lm_q2_score": 0.8807970654616711, "lm_q1q2_score": 0.8644613096875166}}
{"url": "https://math.stackexchange.com/questions/234062/probability-of-tossing-a-fair-coin-with-at-least-k-consecutive-heads", "text": "# Probability of tossing a fair coin with at least $k$ consecutive heads\n\nTossing a fair coin for $N$ times and we get a result series as $HTHTHHTT\\dots~$, Here '$H$' denotes 'head' and '$T$' denotes 'tail' for a specific tossing each time.\n\nWhat is the probability that the length of the longest streak of consecutive heads is greater than or equal to $k$? (that is we have a $HHHH\\dots~$, which is the substring of our tossing result, and whose length is greater than or equal to $k$)\n\nI came up with a recursive solution (though not quite sure), but cannot find a closed form solution.\n\nHere is my solution.\n\nDenote $P(N,k)$ as the probability for tossing the coin $N$ times, and the longest continuous heads is greater or equal than $k$. Then (For $N>k$)\n\n$$P(N,k)=P(N-1,k)+\\Big(1-P(N-k-1,k)\\Big)\\left(\\frac{1}{2}\\right)^{k+1}$$\n\n\u2022 Your example sequence reminded me of this brilliant Simpsons scene :-) \u2013\u00a0joriki Nov 10 '12 at 8:58\n\u2022 I'm getting $(N-k+1)/2^k$ for the closed form. Would you like for me to post my solution, or do you want to think about it some more before I spoil the beans? \u2013\u00a0Braindead Nov 10 '12 at 8:59\n\u2022 @Braindead: That can't be right; it's $\\gt1$ for small $k$. \u2013\u00a0joriki Nov 10 '12 at 9:00\n\u2022 Duh, you are right. I overcounted. \u2013\u00a0Braindead Nov 10 '12 at 9:08\n\u2022 I tried to clarify some of the formulations; please check whether I preserved the intended meaning. In particular, I assumed that you had merely accidentally written \"greater\" once instead of \"greater or equal\". \u2013\u00a0joriki Nov 10 '12 at 9:12\n\nYour recurrence relation is correct. I don't think you can do much better than that for general $k$, but you can find a closed form for specific values of $k$. For the first non-trivial value of $k$, the recurrence relation is\n\n$$p_n=p_{n-1}+(1-p_{n-3})/8\\;.$$\n\nWith $p_n=1+\\lambda^n$, the characteristic equation becomes $\\lambda^3-\\lambda^2+1/8=0$. One solution, $\\lambda=1/2$, can be guessed, and then factoring yields $(\\lambda-1/2)(\\lambda^2-\\lambda/2-1/4)$ with the further solutions $\\lambda=(1\\pm\\sqrt5)/4$. Thus the general solution is\n\n$$p_n=1+c_1\\left(\\frac12\\right)^n+c_2\\left(\\frac{1+\\sqrt5}4\\right)^n+c_3\\left(\\frac{1-\\sqrt5}4\\right)^n\\;.$$\n\nThe initial conditions $p_0=0$, $p_1=0$, $p_2=1/4$ determine $c_1=0$, $c_2=-(1+3/\\sqrt5)/2$ and $c_3=-(1-3/\\sqrt5)/2$, so the probability is\n\n\\begin{align} p_n &=1-\\frac{1+3/\\sqrt5}2\\left(\\frac{1+\\sqrt5}4\\right)^n-\\frac{1-3/\\sqrt5}2\\left(\\frac{1-\\sqrt5}4\\right)^n\\\\ &=1-\\frac4{\\sqrt5}\\left(\\left(\\frac{1+\\sqrt5}4\\right)^{n+2}-\\left(\\frac{1-\\sqrt5}4\\right)^{n+2}\\right)\\;. \\end{align}\n\nThus, for large $n$ the probability approaches $1$ geometrically with ratio $(1+\\sqrt5)/4\\approx0.809$.\n\n\u2022 I've added an explicit formula for the $P(N,k)$ which might be of interest to you. Note, that your $p_n=1-\\frac{1}{2^n}F_{n+2}$ with $F_n$ the Fibonacci numbers. Regards, \u2013\u00a0Markus Scheuer Jan 18 '16 at 21:40\n\nWe can derive an explicit formula of the probability $P(N,k)$ based upon the Goulden-Jackson Cluster Method.\n\nWe consider the set of words $\\mathcal{V}^{\\star}$ of length $N\\geq 0$ built from an alphabet $$\\mathcal{V}=\\{H,T\\}$$ and the bad word $\\underbrace{HH\\ldots H}_{k \\text{ elements }}=:H^k$ which is not allowed to be part of the words we are looking for. We derive a function $f_k(s)$ with the coefficient of $s^N$ being the number of wanted words of length $N$.\n\nThe wanted probability $P(N,k)$ can then be written as \\begin{align*} P(N,k)=1-\\frac{1}{2^N}[s^N]f_k(s) \\end{align*}\n\nAccording to the paper (p.7) from Goulden and Jackson the generating function $f_k(s)$ is \\begin{align*} f_k(s)=\\frac{1}{1-ds-\\text{weight}(\\mathcal{C})}\\tag{1} \\end{align*} with $d=|\\mathcal{V}|=2$, the size of the alphabet and with $\\mathcal{C}$ the weight-numerator with \\begin{align*} \\text{weight}(\\mathcal{C})=\\text{weight}(\\mathcal{C}[H^k]) \\end{align*} We calculate according to the paper \\begin{align*} \\text{weight}(\\mathcal{C}[H^k])&=-\\frac{s^k}{1+s+\\cdots+s^{k-1}}=-\\frac{s^k(1-s)}{1-s^k} \\end{align*}\n\nWe obtain the generating function $f(s)$ for the words built from $\\{H,T\\}$ which don't contain the substring $H^k$ \\begin{align*} f_k(s)&=\\frac{1}{1-2s+\\frac{s^k(1-s)}{1-s^k}}\\\\ &=\\frac{1-s^k}{1-2s+s^{k+1}}\\tag{2}\\\\ \\end{align*}\n\n\n\nNote: For $k=2$ we obtain \\begin{align*} f_2(s)&=\\frac{1-s^2}{1-2s+s^{3}}\\\\ &=1+2s+3s^2+5s^3+8s^4+13s^5+21s^6+34s^7+\\mathcal{O}(s^8) \\end{align*}\n\nThe coefficients of $f_2(s)$ are a shifted variant of the Fibonacci numbers stored as A000045 in OEIS.\n\nNote: For $k=3$ we obtain \\begin{align*} f_3(s)&=\\frac{1-s^3}{1-2s+s^{4}}\\\\ &=1+2s+4s^2+7s^3+13s^4+24s^5+44s^6+81s^7+\\mathcal{O}(s^8) \\end{align*}\n\nThe coefficients of $f_3(s)$ are a shifted variant of the so-called Tribonacci numbers stored as A000073 in OEIS.\n\n\n\nWe use the series representation of $f_k(s)$ in (2) to derive an explicit formula of the coefficients.\n\n\\begin{align*} [s^N]f(s)&=[s^N](1-s^k)\\sum_{m=0}^{\\infty}(2s-s^{k+1})^m\\\\ &=[s^N](1-s^k)\\sum_{m=0}^{\\infty}s^m(2-s^k)^m\\\\ &=[s^N](1-s^k)\\sum_{m=0}^{\\infty}s^m\\sum_{j=0}^m\\binom{m}{j}(-1)^js^{kj}2^{m-j}\\\\ &=([s^N]-[s^{N-k}])\\sum_{m=0}^{\\infty}s^m\\sum_{j=0}^m\\binom{m}{j}(-1)^js^{kj}2^{m-j}\\tag{3}\\\\ &=\\sum_{m=0}^{N}([s^{N-m}]-[s^{N-k-m}])\\sum_{j=0}^m\\binom{m}{j}(-1)^js^{kj}2^{m-j}\\tag{4}\\\\ &=\\sum_{{m=0}\\atop{m\\equiv N(\\bmod k)}}^{N}\\binom{m}{\\frac{N-m}{k}}(-1)^{\\frac{N-m}{k}}2^{m-\\frac{N-m}{k}}\\\\ &\\qquad-\\sum_{{m=0}\\atop{m\\equiv N(\\bmod k)}}^{N-k}\\binom{m}{\\frac{N-m}{k}-1}(-1)^{\\frac{N-m}{k}-1}2^{m-\\frac{N-m}{k}+1}\\\\ &=\\sum_{{m=0}\\atop{m\\equiv N(\\bmod k)}}^{k-1}\\binom{m}{\\frac{N-m}{k}}(-1)^{\\frac{N-m}{k}}2^{m-\\frac{N-m}{k}}\\tag{5}\\\\ &\\qquad+\\sum_{{m=k}\\atop{m\\equiv N(\\bmod k)}}^{N-k} \\left(\\binom{m}{\\frac{N-m}{k}}-\\frac{1}{2^k}\\binom{m-k}{\\frac{N-m}{k}}\\right)(-1)^{\\frac{N-m}{k}}2^{m-\\frac{N-m}{k}}\\\\ \\end{align*}\n\nComment:\n\n\u2022 In (3) we use the linearity of the coefficient of operator and $[s^N]s^mf(s)=[s^{N-m}]f(s)$\n\n\u2022 In (4) we change the limit of the left hand sum from $\\infty$ to $N$ according to the maximum coefficient $[s^N]$. According to the factors $s^{kj}$ we consider in the following only summands with $m\\equiv N(\\bmod k)$\n\n\u2022 In (5) we reorganise the sums from the line above by extracting from the left hand sum the first summand and shifting in the right hand side the index by one and putting both sums together.\n\nWe conclude: An explicit representation of the probability $P(N,k)$ $(n\\geq 0)$ is according to (5) \\begin{align*} P(N,k)&=1-\\sum_{{m=0}\\atop{m\\equiv N(\\bmod k)}}^{k-1}\\binom{m}{\\frac{N-m}{k}}(-1)^{\\frac{N-m}{k}}2^{-\\frac{(k+1)(N-m)}{k}}\\\\ &\\qquad-\\sum_{{m=k}\\atop{m\\equiv N(\\bmod k)}}^{N-k} \\left(\\binom{m}{\\frac{N-m}{k}}-\\frac{1}{2^k}\\binom{m-k}{\\frac{N-m}{k}}\\right)(-1)^{\\frac{N-m}{k}}2^{-\\frac{(k+1)(N-m)}{k}} \\end{align*}\n\nA bit more practical focus than previous answers:\n\nMain idea:\n\n1. We build a \"staircase\":\n\n2. $p$ chance to get to next step, $1-p$ chance to fall down and have to restart (or \"reflip\"). Except for the highest step where we always stay (whenever we manage to get there).\n\n3. We need $k+1$ steps to \"remember\" where on the stairs we are for $k$ flips in a row.\n\nWe can use this to build a matrix for a Markov chain:\n\nWe can build a block matrix:$$\\frac{1}{2}\\left[\\begin{array}{cc}2&1&\\bf 0^T\\\\\\bf 0&\\bf0&\\bf I_k\\\\0&1&\\bf 1^T \\end{array}\\right]$$\n\n(For the special case $k = 5$). We build a stochastic matrix:\n\n$${\\bf P} = \\frac{1}{2}\\left[\\begin{array}{cccccc}2&1&0&0&0&0\\\\0&0&1&0&0&0\\\\0&0&0&1&0&0\\\\0&0&0&0&1&0\\\\0&0&0&0&0&1\\\\0&1&1&1&1&1\\end{array}\\right]$$\n\nNow our answer will simply be $$[1,0,{\\bf 0}]\\, {\\bf P}^k\\, [{\\bf 0},0,1]^T$$\n\nWhich is a scalar product using ${\\bf P}^k$ as a gramian matrix.\n\nNow to calculate this in practice one could probably in general (for general $p$) benefit from a canonical transformation of $\\bf P$, but it's doubtable in this case as the matrix is already sparse with literally only sums and permuations and bit shifts required on the vector elements.\n\nFeller considers this problem in section XIII.7 of An Introduction to Probability Theory and Its Applications, Volume 1, Third Edition. He shows that the probability of having no run of length $k$ in $N$ throws is asymptotic to $$\\frac{1-(1/2)\\;x}{(k + 1 - kx)\\; (1/2)} \\cdot \\frac{1}{x^{N+1}}$$ where $x$ is the least positive root of $$1 - x +(1/2)^{k+1} x^{k+1} = 0$$\n\n(I have changed Feller's notation to agree with the problem statement and have only considered the case of a fair coin; Feller considers the more general case of a biased coin. For more information see equation 7.11, p. 325 in the referenced document.)", "date": "2019-09-18 20:07:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/234062/probability-of-tossing-a-fair-coin-with-at-least-k-consecutive-heads", "openwebmath_score": 0.9394956231117249, "openwebmath_perplexity": 275.44956156317244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534322330059, "lm_q2_score": 0.8807970701552504, "lm_q1q2_score": 0.8644613076046462}}
{"url": "https://math.stackexchange.com/questions/2282623/can-an-open-set-be-covered-by-proper-open-subsets", "text": "# Can an open set be covered by proper open subsets?\n\nLet $(X, \\tau)$ be a topological space, let $U\\in\\tau$. An open cover of $U$ is a set $\\{U_i \\ |\\ i\\in I\\}$ (of open sets $U_i$) whose union $\\bigcup U_i$ contains $U$.\n\nIf $U\\subsetneq X$, then $U$ admits an open covering by open sets $\\{U_i \\ |\\ i\\in I\\}$, where an arbitrary open set $U_i$ may not be a subset of $U$. (I think.)\n\nDoes $U$ admit an open covering by open sets $\\{U_i \\ |\\ i\\in I\\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets \"inside\" of $U$?\n\nIf the answer is no, then what are the open sets $U$ that admit such \"internal\" coverings?\n\nIs the answer any different if we replace $U$ by an arbitrary closed set?\n\n\u2022 Have you thought about the integers? \u2013\u00a0Steve D May 15 '17 at 21:38\n\u2022 For many topological spaces you can do this. For example, any open subset of $\\mathbb R^n$ is a union of open balls, which can be made proper. However, @SteveD's hint shows that you can't always do this. \u2013\u00a0Cheerful Parsnip May 15 '17 at 22:20\n\u2022 If $X$ is a $T_1$ space and if $U$ is open and has more than $1$ member then $\\{U$ \\ $\\{p\\}: p\\in U\\}$ is an open cover of $U$ by open proper subsets of $U.$ ....If $\\tau=\\{X,\\phi\\}$ then with U=X, no proper non-empty subset of $U$ is open.... Another example is Sierpinski space $S=\\{0,1\\}$ with $\\tau=\\{S,\\phi, \\{0\\}\\}$. \u2013\u00a0DanielWainfleet May 15 '17 at 23:57\n\u2022 If U is open, of course. If U is open every point is an interior point so for every point x in U there is an open neighborhood of x completely in U. Call this N_x, then $\\cup_{x\\in U} Nx$ is an open cover. If U is not open then of course not, If U is not open there is a point that is not an interior point and any open set containing it must contain points not in U (else it'd be an interior point). So all open covers contain points not in U. \u2013\u00a0fleablood May 16 '17 at 1:25\n\u2022 @fleablood This is the argument I was going to give, but (as others pointed out to me) it only goes through if the space is at least $T_1$. \u2013\u00a0Austin Mohr May 16 '17 at 1:31\n\nA space is called $T_1$ if, for any two points in the space, each has an open neighborhood that misses the other. This is one of many separation axioms that measure how strongly we can separate points in the space. For some mathematicians, topological spaces aren't even worth considering until they are at least $T_2$ (Hausdorff), which is even stronger. (This is to say that all but the most pathological topologies are at least $T_1$.)\n\nWe can construct a cover of the type you describe if the space is at least $T_1$ and $U$ contains more than one point. For each $x \\in U$, construct an open neighborhood $U_x$ as follows:\n\n1. Choose any $y \\in U$ that is distinct from $x$.\n2. Appeal to the $T_1$ property to get an open set $G_x$ that contains $x$ but does not contain $y$.\n3. Set $U_x = G_x \\cap U$, so that $U_x$ is a proper open subset of $U$ containing $x$.\n\nFinally, we can take $\\{U_x \\mid x \\in U\\}$ as an open cover of $U$ by proper open subsets.\n\n\u2022 But why is it guaranteed that the \"first\" $y$ we chose is an element of some element of $\\{ U_x \\ | \\ x\\in U \\}$ ? Moreover, why is every such $y$ eventually in some of the cover elements? Because, by construction, $y$ is not in $U_x$ \u2013\u00a0\u00e9tale-cohomology May 17 '17 at 5:52\n\u2022 Let $y_0$ be the \"first $y$\" chosen in this process. Eventually we will need to construct the set $U_{y_0}$, which must contain $y_0$. Similarly, any $y \\in U$ is an element of $U_y$ and therefore covered. \u2013\u00a0Austin Mohr May 17 '17 at 8:36\n\u2022 Got it. Thank you. I wonder if it's possible to prove that $T_1$ is the weakest possible hypothesis that allows existence of these covers... \u2013\u00a0\u00e9tale-cohomology May 18 '17 at 22:38\n\u2022 One can push my argument through with something weaker (but I don't know if it has a name): For any open set $U$ and any point $x$ of $U$, there is an open neighborhood $V$ of $x$ that is a strict subset of $U$. This property is not as strong as $T_1$, since it still allows for the existence $x$ and $y$ that cannot be separated. However, you could take the $V$ in this property as the $U_x$ in my answer and it still get your desired refinement. This is property would be weakest possible, since otherwise the only way to cover $x$ is to use all of $U$. \u2013\u00a0Austin Mohr May 19 '17 at 1:01\n\nIt depends. Consider $\\mathbb{R}$ with the usual topology and $U = (0,1)$, then we can write $(0,1) = \\bigcup_{n=2}^\\infty\\left(\\frac1n,1\\right)$, for instance. Since every open set $U\\subset\\mathbb{R}$ is the union of intervals, it means that every open subset of $\\mathbb{R}$ can be written as the union of proper open subsets of itself. It is easy to see that the same argument can be applied to general normed vector spaces.\n\nHowever, if a singleton $\\{x\\}$ is an open set of your space then there are no proper open subsets of $\\{x\\}$ to unite.\n\nShow that $K = [0,1]$ cannot be covered by open sets that are contained by $K$.\n\nIf $U$ is open, then $U$ can be covered by $\\{U\\}$. If $U$ is an open singleton, then there is no cover by proper subsets.\n\nWhen $U$ is open, then for all $x$ in $U$, there is an open base set $U_x$ with $x$ in $U_x$ subset $U$. Often the base sets can be chosen to be proper subsets.", "date": "2019-12-06 05:52:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2282623/can-an-open-set-be-covered-by-proper-open-subsets", "openwebmath_score": 0.9066517353057861, "openwebmath_perplexity": 125.51992967602584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743623413362, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8643866535570464}}
{"url": "https://math.stackexchange.com/questions/3156341/why-is-xe-x0-for-all-x-in-mathbbr", "text": "# Why is $x+e^{-x}>0$ for all $x \\in \\mathbb{R}$?\n\nDenote $$f(x) = x + e^{-x}$$. Note that $$f(0) = 1$$ and $$f'(x) = 1-e^{-x}$$. That means $$\\lim_{x\\rightarrow -\\infty} f'(x) = -\\infty.$$ So if the rate of change of $$f(x)$$ keeps decreasing exponentially fast to negative infinity, why is $$\\lim_{x\\rightarrow -\\infty} f(x) = +\\infty$$?\n\n\u2022 Addressing the title, for all $x, e^{-x}\\ge1-x$ so $x+e^{-x}\\ge1>0$ \u2013\u00a0J. W. Tanner Mar 21 at 4:45\n\u2022 To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}\\ge t+1$ for all real $t$. So $e^{-x}\\ge -x+1> -x$ for all real $x$, as desired. \u2013\u00a0Minus One-Twelfth Mar 21 at 4:46\n\u2022 Roughly speaking - if you are visualising $x$ going from $0$ to $-\\infty$, then the graph is not decreasing but increasing exponentially fast, because you are going \"backwards\". \u2013\u00a0David Mar 21 at 4:49\n\u2022 Thank you. That's what I was missing. Going backward. :( \u2013\u00a0Paichu Mar 21 at 4:50\n\nYou have got your directions mixed up.\n\nWhen the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.\n\nTake\n\n$$x < y < 0; \\tag 1$$\n\nthen\n\n$$f(y) - f(x) = \\displaystyle \\int_x^y f'(s)\\; ds = \\int_x^y (1 - e^{-s})\\; ds; \\tag 2$$\n\nit is clear that, for any $$M < 0$$, there exists $$y_0 < 0$$ such that\n\n$$s < y_0 \\Longrightarrow 1 - e^{-s} < M; \\tag 3$$\n\nif we now choose\n\n$$y < y_0, \\tag 4$$\n\nthen\n\n$$f(y) - f(x) = \\displaystyle \\int_x^y (1 - e^{-s})\\; ds < \\int_x^y M \\; ds = M(y - x); \\tag 5$$\n\nwe re-arrange this inequality:\n\n$$f(x) - f(y) > -M(y - x) = M(x - y), \\tag 6$$\n\n$$f(x) > f(y) + M(x - y); \\tag 7$$\n\nwe now fix $$y$$ and let $$x \\to -\\infty$$; then since $$M < 0$$ and $$x - y < 0$$ for $$x < y$$,\n\n$$\\displaystyle \\lim_{x \\to -\\infty} f(y) + M(x - y) = \\infty, \\tag 8$$\n\nand hence\n\n$$\\displaystyle \\lim_{x \\to -\\infty}f(x) = \\infty \\tag 9$$\n\nas well.\n\nWe note that (9) binds despite the fact that $$f'(x) < 0$$ for $$x < 0$$; though this derivative is negative, when $$x$$ decreases we are \"walking back up the hill,\" as it were; as $$x$$ decreases, $$f(x)$$ increases.\n\nNote Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show\n\n$$x + e^{-x} > 0, \\forall x \\in \\Bbb R; \\tag{10}$$\n\nfor\n\n$$x \\ge 0, \\tag{11}$$\n\n$$e^{-x} > 0 \\tag{12}$$\n\nas well, hence we also have\n\n$$x + e^{-x} > 0; \\tag{13}$$\n\nfor\n\n$$x < 0, \\tag{14}$$\n\nwe may use the power series for $$e^{-x}$$:\n\n$$e^{-x} = 1 - x + \\dfrac{x^2}{2!} - \\dfrac{x^3}{3} + \\ldots; \\tag{15}$$\n\nthen\n\n$$x + e^{-x} = 1 + \\dfrac{x^2}{2!} - \\dfrac{x^3}{3!} + \\ldots > 0, \\tag{16}$$\n\nsince every term on the right is positive when $$x < 0$$. End of Note.\n\nNote that $$f''(x) = e^{-x} >0$$ so $$f$$ is strictly convex.\n\nSince $$f'(0) = 0$$, we see that $$f(x) \\ge f(0) = 1$$ for all $$x$$.\n\nNote that\n\n\u2022 $$\\forall x \\in \\mathbb{R} : x+e^{-x}>0 \\Leftrightarrow \\forall x \\in \\mathbb{R} :e^{-x}>-x \\Leftrightarrow \\forall x \\in \\mathbb{R} :\\color{blue}{\\boxed{e^{x}>x}}$$\n\nThe last inequality follows directly by Taylor: $$\\color{blue}{e^x} = 1+x + \\underbrace{\\frac{e^{\\xi}}{2}x^2}_{\\geq 0} \\color{blue}{> x}$$", "date": "2019-12-06 01:20:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3156341/why-is-xe-x0-for-all-x-in-mathbbr", "openwebmath_score": 0.9140320420265198, "openwebmath_perplexity": 318.7908080135041, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795095031688, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8643838202456569}}
{"url": "https://math.stackexchange.com/questions/1486843/how-can-you-express-the-next-two-consecutive-odd-numbers-in-terms-of-x", "text": "# How can you express the next two consecutive odd numbers in terms of x?\n\nI'm trying to express the next two consecutive odd numbers in terms of x. How do go about this?\n\nPrecisely, given a number $x$ I want two smallest consecutive odd numbers $m,n$ such that $x<m<n$.\n\n\u2022 Be aware that you wrote two very different things in your question and your body. Are you trying to express the next two consecutive odd numbers in terms of x, or are you trying to express x in terms of the next two consecutive odd numbers? If $x$ is, say $5$, do you consider \"the next two consecutive odd numbers\" to be $5$ and $7$ or do you consider the next two consecutive odd numbers to be $7$ and $9$? \u2013\u00a0JMoravitz Oct 19 '15 at 2:32\n\u2022 My error. I'm trying t express the next two consecutive odd numbers in terms of x. Edited to reflect. \u2013\u00a0user214824 Oct 19 '15 at 2:45\n\u2022 That still doesn't answer the question of what exactly you mean by \"next two consecutive odd numbers\" is. If $5$ is your number, do you want $7$ and $9$? or do you want $5$ and $7$ (despite $5$ not being bigger than $5$). \u2013\u00a0JMoravitz Oct 19 '15 at 2:47\n\u2022 @JMoravitz Yes, that is it. 7 and 9. \u2013\u00a0user214824 Oct 19 '15 at 2:48\n\nPresumably $x$ is an integer. We have two cases: either $x$ is even or $x$ is odd.\nIn the case that $x$ is odd, then $x+2$ and $x+4$ will both be odd and will be the next two consecutive odd numbers.\nIn the case that $x$ is even, then $x+1$ and $x+3$ will both be odd and will be the next two consecutive odd numbers.\nLet $x$ an integer, then the next two odd integers are $$2\\times\\Bigl\\lfloor \\frac{x+1}{2}\\Bigr\\rfloor + 1 \\qquad\\text{and}\\qquad 2\\times\\Bigl\\lfloor \\frac{x+1}{2} \\Bigr\\rfloor + 3.$$\nIndeed, if $x = 2n$ where $n\\in\\mathbb{Z}$, then $$2\\lfloor (x+1)/2\\rfloor + 1 = 2 \\lfloor n + 1/2 \\rfloor + 1 = 2n + 1$$ and if $x = 2n + 1$, then $$2\\lfloor (x+1)/2 \\rfloor + 1 = 2\\lfloor n + 1 \\rfloor + 1 = 2n + 3.$$\n\u2022 Maybe doesn't you know the symbol $\\lfloor \\cdot \\rfloor$. If $x$ is a real number, $\\lfloor x \\rfloor$ is the biggest integer less or equal to $x$. For example $\\lfloor 2.7 \\rfloor = 2$, $\\lfloor -3.8 \\rfloor = -4$ and $\\lfloor x \\rfloor = x$ for $x$ integer. \u2013\u00a0\u00c9ric Guirbal Oct 19 '15 at 3:01", "date": "2021-04-15 12:08:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1486843/how-can-you-express-the-next-two-consecutive-odd-numbers-in-terms-of-x", "openwebmath_score": 0.8299247622489929, "openwebmath_perplexity": 98.52346795536029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795083542037, "lm_q2_score": 0.8757869867849166, "lm_q1q2_score": 0.8643838096399865}}
{"url": "https://math.stackexchange.com/questions/2970009/is-this-theorem-equivalent-to-transfinite-recursion-theorem/2970279", "text": "# Is this theorem equivalent to Transfinite Recursion Theorem?\n\nLet $$V$$ be the class of all sets, $$\\operatorname{Ord}$$ be the class of all ordinals, and $$G:V\\to V$$ be a class function.\n\nTransfinite Recursion Theorem:\n\nThere exists a class function $$F:\\operatorname{Ord}\\to V$$ such that $$F(\\alpha)=G(F\\restriction \\alpha)$$ for all $$\\alpha\\in\\operatorname{Ord}$$.\n\nWhile I'm able to prove that Transfinite Recursion Theorem implies the below theorem, I have tried but to no avail in proving that the theorem implies Transfinite Recursion Theorem.\n\nI would like to ask if it is possible to prove that this theorem implies Transfinite Recursion Theorem.\n\nTheorem:\n\nLet $$G_1,G_2,G_3$$ be class functions from $$V$$ to $$V$$. There exists a class function $$F:\\operatorname{Ord}\\to V$$ such that\n\n(1) $$F(0)=G_1(\\emptyset)$$\n\n(2) $$F(\\alpha+1)=G_2(F(\\alpha))$$ for all $$\\alpha\\in\\operatorname{Ord}$$\n\n(3) $$F(\\alpha)=G_3(F\\restriction\\alpha)$$ for all limit $$\\alpha\\neq 0$$\n\n\u2022 Do you know how to prove strong induction (on $\\mathbb{N}$) from weak induction? It's the same idea. \u2013\u00a0Eric Wofsey Oct 25 '18 at 2:46\n\u2022 Hi @EricWofsey! Yes, I do. I will try your suggestion. \u2013\u00a0LE Anh Dung Oct 25 '18 at 2:54\n\nThe trick is to not construct $$F$$ itself, but to construct the function $$H$$ which sends $$\\alpha$$ to $$F\\restriction \\alpha$$. That way, at successor steps you have access to the entire history of the recursion so far and not just the last step.\n\nIn detail, given $$G:V\\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: $$G_1(x)=\\emptyset$$ $$G_2(x)=x\\cup\\{(\\operatorname{dom}(x),G(x))\\}$$ $$G_3(x)=\\bigcup\\operatorname{ran}(x)$$ By the theorem, we then get a function $$H:Ord\\to V$$ such that $$H(0)=G_1(\\emptyset)$$, $$H(\\alpha+1)=G_2(H(\\alpha))$$, and $$H(\\alpha)=G_3(H\\restriction\\alpha)$$ for $$\\alpha\\neq 0$$ limit. It is then easy to prove by induction that $$H(\\alpha)$$ is a function with domain $$\\alpha$$ for each $$\\alpha$$, with $$H(\\alpha)\\restriction \\beta=H(\\beta)$$ for all $$\\beta<\\alpha$$. So, defining $$F=\\bigcup\\operatorname{ran}(H)$$, $$F$$ is a function on $$Ord$$ with $$F\\restriction\\alpha=H(\\alpha)$$ for each $$\\alpha$$. For each $$\\alpha$$, we then have $$F(\\alpha)=H(\\alpha+1)(\\alpha)=G_2(H(\\alpha))(\\alpha)=G_2(F\\restriction\\alpha)(\\alpha).$$ Since $$\\operatorname{dom}(F\\restriction\\alpha)=\\alpha$$, our definition of $$G_2$$ tells us that $$F(\\alpha)=G_2(F\\restriction\\alpha)(\\alpha)=G(F\\restriction\\alpha),$$ as desired.\n\n\u2022 I'm feeling confused. You have shown how to derive the Q's 2nd theorem from its first. But the proposer seems to want to see the 1st derived from the second, although it is obvious that if the second one holds, then, for a given $G$, we can simply let $G_1=G_2=G_3=G$. So I'm thinking that the proposer stated the Q backwards. \u2013\u00a0DanielWainfleet Oct 25 '18 at 3:49\n\u2022 @DanielWainfleet: I am deriving the first theorem from the second. I start with a function $G$ as in the first theorem, then use the second theorem to get a function $H$ and then define $F$ from $H$ which verifies the conclusion of the first theorem. It does not work to simply define $G_1=G_2=G_3=G$ since $G_2$ takes as its input just $F(\\alpha)$, not $F\\restriction\\alpha$. \u2013\u00a0Eric Wofsey Oct 25 '18 at 4:04\n\u2022 Thank you so much! You make my day :) \u2013\u00a0LE Anh Dung Oct 25 '18 at 8:14\n\u2022 Thank you for clearing up my confusion. \u2013\u00a0DanielWainfleet Oct 25 '18 at 21:26\n\u2022 Hi! I'm now struggling with a seemingly similar problem like this one. I have tried to replicate your proof but failed at the end. If you don't mind, please have a look at math.stackexchange.com/questions/2970783/\u2026. Thank you for your help! \u2013\u00a0LE Anh Dung Oct 27 '18 at 14:16\n\nI fill in @Eric Wofsey's proof with detail and post it here. All credits are given to @Eric Wofsey.\n\nGiven $$G:V\\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: \\begin{align}&G_1(x)=\\emptyset\\text{ for all }x\\\\&G_2(x)=\\begin{cases} x\\cup\\{(\\operatorname{dom}(x),G(x))\\}&\\text{if }x\\text{ is a function}\\\\\\emptyset&\\text{otherwise}\\end{cases}\\\\&G_3(x)=\\begin{cases} \\bigcup\\operatorname{ran}(x)&\\text{if }x\\text{ is a function}\\\\\\emptyset&\\text{otherwise}\\end{cases}\\end{align}\n\nBy The theorem, there is a class function $$H:\\operatorname{Ord}\\to V$$ such that $$H(0)=G_1(\\emptyset)$$, $$H(\\alpha+1)=G_2(H(\\alpha))$$, and $$H(\\alpha)=G_3(H\\restriction\\alpha)$$ for $$\\alpha\\neq 0$$ limit.\n\nFirst, we prove by induction that $$H(\\alpha)$$ is a function with domain $$\\alpha$$ for all $$\\alpha\\in\\operatorname{Ord}$$ and with $$H(\\alpha)\\restriction \\beta=H(\\beta)$$ for all $$\\beta<\\alpha$$.\n\n\u2022 $$H(0)=G_1(0)=\\emptyset$$. Then the statement is trivially true for $$\\alpha=0$$.\n\n\u2022 Assume that the statement is true for $$\\alpha$$. Then $$H(\\alpha+1)=G_2(H(\\alpha))=$$ $$H(\\alpha)\\cup\\{(\\operatorname{dom}(H(\\alpha)),G(H(\\alpha)))\\}=H(\\alpha)\\cup\\{(\\alpha,G(H(\\alpha)))\\}$$. It follows that $$\\operatorname{dom}(H(\\alpha+1))=\\operatorname{dom}(H(\\alpha))\\cup \\{\\alpha\\}=\\alpha\\cup \\{\\alpha\\}=\\alpha+1$$. For $$\\beta=\\alpha$$, $$H(\\alpha+1)\\restriction \\beta=H(\\alpha+1)\\restriction \\alpha=H(\\alpha)=H(\\beta)$$. For $$\\beta<\\alpha$$, $$H(\\alpha+1)\\restriction \\beta=$$ $$H(\\alpha)\\restriction \\beta=H(\\beta)$$. Thus $$H(\\alpha+1)\\restriction \\beta=H(\\beta)$$ for all $$\\beta<\\alpha+1$$.\n\n\u2022 Assume that the statement is true for all $$\\beta<\\alpha$$ where $$\\alpha\\neq\\emptyset$$ is limit ordinal. Then $$H(\\alpha)=G_3(H(\\alpha))=\\bigcup\\operatorname{ran}(H(\\alpha))=\\bigcup\\{H(\\beta)\\mid \\beta<\\alpha\\}$$. For any $$\\beta_1\\le\\beta_2<\\alpha$$: $$H(\\beta_2)\\restriction\\beta_1=H(\\beta_1)$$and thus $$H(\\beta_1)\\subseteq H(\\beta_2)$$. Then $$H(\\alpha)=\\bigcup\\{H(\\beta)\\mid \\beta<\\alpha\\}$$ is actually a function. It follows that $$\\operatorname{dom}(H(\\alpha))=\\bigcup_{\\beta<\\alpha}\\operatorname{dom}(H(\\beta))=\\bigcup_{\\beta<\\alpha}\\beta=\\alpha$$ since $$\\alpha$$ is limit ordinal. Moreover, $$H(\\alpha)\\restriction \\beta=\\{(\\gamma,H(\\alpha)(\\gamma))\\mid \\gamma<\\beta\\}=\\{(\\gamma,H(\\gamma+1)(\\gamma))\\mid \\gamma<\\beta\\}=$$ $$\\{(\\gamma,H(\\beta)(\\gamma))\\mid \\gamma<\\beta\\}=H(\\beta)$$.\n\nAs a result, $$\\forall\\beta<\\alpha:H(\\alpha)\\restriction \\beta=H(\\beta)$$ and thus $$\\forall\\beta<\\alpha:H(\\beta)\\subsetneq H(\\alpha)$$.\n\nNext we define $$F=\\bigcup\\operatorname{ran}(H)$$. Then $$F=\\bigcup\\{H(\\alpha)\\mid \\alpha\\in\\operatorname{Ord}\\}$$ is a function with domain $$\\operatorname{Ord}$$ and with $$F\\restriction\\alpha=\\{F(\\beta)\\mid\\beta<\\alpha\\}=\\{H(\\beta+1)(\\beta)\\mid\\beta<\\alpha\\}=\\{H(\\alpha)(\\beta)\\mid\\beta<\\alpha\\}=H(\\alpha)$$ for all $$\\alpha\\in\\operatorname{Ord}$$.\n\nSince $$F\\restriction\\alpha=H(\\alpha)$$ and $$\\operatorname{dom}(H(\\alpha))=\\alpha$$, $$\\operatorname{dom}(F\\restriction\\alpha)=\\alpha$$. Then $$G_2(F\\restriction\\alpha)=(F\\restriction\\alpha)\\cup \\{(\\alpha,G(F\\restriction\\alpha))\\}$$ and thus $$G_2(F\\restriction\\alpha)(\\alpha)=G(F\\restriction\\alpha)$$.\n\nFor each $$\\alpha\\in\\operatorname{Ord}$$, we have $$F(\\alpha)=H(\\alpha+1)(\\alpha)=G_2(H(\\alpha))(\\alpha)=G_2(F\\restriction\\alpha)(\\alpha)=G(F\\restriction\\alpha)$$.\n\nUpdate: I have found another way to define $$F$$\n\nWe define $$F$$ as follows $$F(\\alpha):=H(\\alpha+1)(\\alpha)$$\n\nThen $$F(\\alpha)=H(\\alpha+1)(\\alpha)=G_2(H(\\alpha))(\\alpha)=(H(\\alpha)\\cup\\{(\\operatorname{dom}(H(\\alpha)),G(H(\\alpha)))\\})(\\alpha)=(H(\\alpha)\\cup\\{(\\alpha,G(H(\\alpha)))\\})(\\alpha)=G(H(\\alpha))$$.\n\nMoreover, $$H(\\alpha)=\\{(\\beta,H(\\alpha)(\\beta))\\mid\\beta<\\alpha\\}=\\{(\\beta,H(\\beta+1)(\\beta))\\mid\\beta<\\alpha\\}=\\{(\\beta,F(\\beta))\\mid\\beta<\\alpha\\}=F\\restriction\\alpha$$.\n\nThus $$F(\\alpha)=G(H(\\alpha))=G(F\\restriction\\alpha)$$.", "date": "2021-06-18 06:03:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2970009/is-this-theorem-equivalent-to-transfinite-recursion-theorem/2970279", "openwebmath_score": 0.9701847434043884, "openwebmath_perplexity": 118.74882018854727, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138157595305, "lm_q2_score": 0.8840392802184581, "lm_q1q2_score": 0.8643374179436976}}
{"url": "https://math.stackexchange.com/questions/1495086/envelope-of-projectile-trajectories?noredirect=1", "text": "# Envelope of Projectile Trajectories\n\nFor a given launch velocity $$v$$ and launch angle $$\\theta$$, the trajectory of a projectile may be described by the standard formula $$y=x\\tan\\theta-\\frac {gx^2}{2v^2}\\sec^2\\theta$$\n\nFor different values of $$\\theta$$ what is the envelope of the different trajectories? Is it a parabola itself?\n\nThe standard solution to this \"envelope of safety\" problem is to state the formula as a quadratic in $$\\tan\\theta$$ and set the discriminant to zero. The resulting relationship between $$x,y$$ is the envelope.\n\nThis question is posted to see if there are other approaches to the solution.\n\nEdit 1\n\nThanks for the nice solutions from Jack and Blue, received so far. From the solution of the envelope it can be worked out that the envelope itself corresponds to the right half of the trajectory of a projectile launched at $$(-\\frac{v^2}{g^2},0)$$ at a launch angle $$\\alpha=\\frac{\\pi}4$$ and a launch velocity $$V=v\\sqrt2$$. This means that both vertical and horizontal components of the launch velociy are equal to $$v$$. It would be interesting to see if these conclusions can be inferred from the problem itself by inspection and without first solving it. If so, then this would form another solution.\n\n\u2022 I think it is more like an ellipse. \u2013\u00a0Archis Welankar Oct 24 '15 at 10:05\n\u2022 @ArchisWelankar - What makes you think that might be the case? \u2013\u00a0hypergeometric Oct 24 '15 at 10:13\n\u2022 When the angle is very small like 10 then the curve becomes elongated and more like an ellipse than a parabola. \u2013\u00a0Archis Welankar Oct 24 '15 at 10:25\n\u2022 @ArchisWelankar - 1. The curve or trajectory of the projectile is always a parabola (even when elongated, at small launch angles) as given by the formula. 2. An ellipse doesn't have to be elongated. 3. The question refers to the envelope of different trajectories and not any one particular trajectory. \u2013\u00a0hypergeometric Oct 24 '15 at 10:34\n\u2022 \"The standard solution to this \"envelope of safety\" problem is to state the formula as a quadratic in tan\u03b8 and set the discriminant to zero. \" I don't understand this approach, can someone explain ? \u2013\u00a0Dat Sep 11 at 9:38\n\nYes it is. To find the envelope, we just have to find the intersections between two trajectories associated to two slightly different angles. If we solve $$x\\tan\\theta -\\frac{gx^2}{2v^2}\\sec^2(\\theta) = x\\tan(\\theta+\\varepsilon) -\\frac{gx^2}{2v^2}\\sec^2(\\theta+\\varepsilon)$$ we get $x=0$ or $$x=\\frac{2v^2}{g}\\cdot\\frac{\\tan(\\theta)-\\tan(\\theta+\\varepsilon)}{\\sec^2(\\theta)-\\sec^2(\\theta+\\varepsilon)}$$ and by letting $\\varepsilon\\to 0$ we get $x=\\frac{v^2}{g}\\cot(\\theta)$, from which $y=\\frac{v^2}{2g}\\left(2-\\frac{1}{\\sin^2(\\theta)}\\right)$.\n\nIt follows that the equation of the envelope is given by: $$y = \\frac{v^2}{2g}\\left(1-\\left(\\frac{gx}{v^2}\\right)^2\\right)=\\frac{v^2}{2g}-\\frac{g}{2 v^2}\\,x^2$$ that clearly is a parabola with vertex in $\\left(0,\\frac{v^2}{2g}\\right)$ through the points $\\left(\\pm\\frac{v^2}{g} ,0\\right)$.\n\nWe may notice that the envelope and the trajectory with $\\theta=\\frac{\\pi}{4}$ are homothetic, and the dilation ratio is just $2$. The vertices of the trajectories lie on an ellipse that is tangent to the envelope parabola, with centre at $\\left(0,\\frac{v^2}{4g}\\right)$, a vertex in the origin and a vertex at $\\left(\\frac{v^2}{2g},\\frac{v^2}{4g}\\right)$.\n\n\u2022 Thank you for a beautiful and creative solution! And nice graphics! A big +1! \u2013\u00a0hypergeometric Oct 24 '15 at 13:29\n\u2022 Please check out comments under Edit 1 in the question and see if you have any further thoughts. \u2013\u00a0hypergeometric Oct 24 '15 at 14:21\n\n@Jack provides a very nice and intuitive derivation of the envelope as the points of intersection of infinitely-close members of the curve family. The Wikipedia \"Envelope\" entry provides this less-illuminating abstraction:\n\nThe envelope of the family [of curves parameterized by $t$ is] the set of points for which $$F(t, x, y) = \\frac{\\partial F}{\\partial t}(t,x,y) = 0 \\tag{\\star}$$ for some value of $t$ [...].\n\nIn $(\\star)$, $F$ is the function that, when set equal to $0$, defines each curve in the family. For the question at hand, we have (with parameter $\\theta$ instead of $t$) $$F(\\theta,x,y) = -y + x\\tan\\theta -\\frac{g x^2}{2v^2}\\sec^2\\theta \\tag{1}$$ Therefore, differentiating with respect to $\\theta$ gives $$\\frac{\\partial F}{\\partial \\theta}(\\theta,x,y) =x\\sec^2\\theta -\\frac{g x^2}{2v^2}\\cdot 2 \\sec^2\\theta\\tan\\theta = \\frac{x\\sec^2\\theta}{v^2} ( v^2 - g x \\tan\\theta) \\tag{2}$$ Solving $\\partial F/\\partial \\theta = 0$ for $\\theta$ (noting that $\\sec\\theta$ never vanishes) gives $$\\tan\\theta = \\frac{v^2}{g x} \\qquad\\text{so that}\\qquad \\sec^2\\theta = 1 + \\tan^2\\theta = \\frac{g^2 x^2 + v^4}{g^2 x^2}$$\n\nSubstituting into $(1)$, and setting $F=0$, we have\n\n$$y = x\\;\\frac{v^2}{gx} - \\frac{g x^2}{2 v^2}\\;\\frac{g^2 x^2 + v^4}{g^2 x^2} = \\frac{v^2}{2g} - \\frac{g x^2}{2v^2} = \\frac{v^4-g^2x^2}{2gv^2}$$\n\n\u2022 Thanks for working out the solution from the article. Nice find! +1. \u2013\u00a0hypergeometric Oct 24 '15 at 13:55\n\u2022 Please check out comments under Edit 1 in the question and see if you have any further thoughts. \u2013\u00a0hypergeometric Oct 24 '15 at 14:21\n\nHere's another approach. Not as elegant as Jack's though. Perhaps similar to Blue's. (Edit: Revised slightly to use substitution of $H=v^2/2g$ instead of $\\lambda=g/v^2$ for more convenient reference, per solution by Narasimham)\n\nThis approach maximises the value of $y$ for any given value of $x$.\n\nFor a given value of $x$, say $x=k$, \\begin{align} y&=kT-\\frac {k^2}{4H}(1+T^2)\\qquad\\text{where T=\\tan\\theta, H=\\frac {v^2}{2g}}\\\\ \\frac{dy}{dT}&=k-\\frac {k^2}{4H}\\cdot 2T=0\\qquad\\text{when T=\\frac {2H}k}\\\\ \\text{At T=\\frac {2H}k}:\\qquad\\qquad y&=k\\cdot\\frac{2H}k-\\frac{k^2} {4H}\\left(1+\\frac {4H^2}{k^2}\\right) \\color{lightgrey}{=2H-\\frac {k^2}{4H}-H}\\\\ &=H-\\frac {k^2}{4H} \\end{align}\n\nHence the equation of the envelope is $$y=\\frac {v^2}{2g}-\\frac {gx^2}{2v^2}\\quad\\blacksquare$$\n\n\u2022 This nicely exploits the fact that, in this particular case, we know that the envelope sits on top of the family of curves, and therefore consists of the points that maximize $y$ for a given $x$, over all $t$. (Of course don't always have such knowledge, but no matter.) Re-writing in terms of parameter $T$ is also a helpful simplification. And, to be explicit: the connection with the process in my answer is that the criterion for maximizing $y$ ---that is, setting $dy/dT = 0$--- corresponds directly to $\\partial F/\\partial \\theta = 0$, because $y$ drops out of the latter equation. \u2013\u00a0Blue Oct 24 '15 at 21:38\n\u2022 Thanks for your kind comment! \u2013\u00a0hypergeometric Oct 25 '15 at 5:57\n\nI shall outline the method how we get the parabola. Usual notation\n\n$$\\ddot y = -g, \\dot y = - g t + v \\sin \\theta , y =- g t^2/2 + v t \\sin \\theta\\, +0 \\tag {1}$$ $$\\ddot x = 0, \\dot x = v \\cos \\theta = const , x = v t \\cos \\theta +0 \\tag{2}$$\n\nEliminating time $t$ between (1),(2) you got this parabola equation already.\n\nLet $\\tan \\alpha = T$; Parabola equation in other words\n\n$$y = x T - g/2 * ( x/ v \\cos\\theta)^2 = x T - ( g x^2/2 v^2) ( 1+T^2) \\tag{3}$$\n\nDifferentiate partially with respect to $T$ and simplify, $T = v^2 / gx \\tag{4}$\n\nEliminate $T$ between (3) and (4)\n\n$$y = v^2/2g - g x^2 /(2 v^2) = H - x^2/(4H) \\tag{5},$$\n\nsame as what was obtained before by Jack and Blue.\n\nIf you denote height reached by projectile on vertical firing $H = v^2/(2g) \\tag{6}$ you would notice that the envelope is profiled exactly as a parabolic mirror with focus at gun delivery point, focal length is exactly H. Vertical force of gravity is acting like light :)..\n\nThe above procedure method is indicated by Blue in Wiki, is referred to as C-discriminant method to obtain envelopes and singular solutions.\n\nLike what you said in your edit and I about mirror, they are ploys for remembering curves using similarities..\n\nFor the image I took values of $g=9.8 m/s^2 , v = 2 m/s$\n\nThe partial differentiation way and elimination is the right way to look at, perhaps not as what you said (standard solution.. way).\n\nEDIT2:\n\nThe answer to your second question, i.e., to determine if it is going to be a parabola envelope without going through all of analysis... I can only reply with extended C-discriminant, strengthening the same result by another path.\n\n$p$ discriminant method is also relevant, but I defer it, but best is to refer to differential calculus books of authors e.g., A.R. Forsythe.\n\nI shall expand on the C-discriminant, wherein a two parameter system of equations variation of any of the two parameters leads to the same parabola envelope. This is in reply to your question , Why do you also plot partial concentric circles?\n\nWell, they are circles alright, but not concentric circles. They expand and come down slowly with time. What you can see as plotted are the peripheries traced out descending with time .\n\nBut first quickly watch some fireworks to see what I am discussing about:\n\nIt is common experience to see fire ball bright splinters expanding to bigger circles as the entire cluster comes down slowly with time. The periphery is a portion of a circle whose radius increases. Center of circle is always descending by gravity.\n\nThe two parameters are $\\theta, t$ angle of elevation at first burst or fire, and time $t$.\n\nBy C-discriminant method the parabola envelope is the eliminant of either $\\theta$ variable or\n\n$$F(x,y,\\theta) =0 ,\\, F_{\\theta } (x,y,\\theta) =0 \\tag{6}$$\n\nor $t$ time variable.\n\n$$F(x,y, t) =0 ,\\, F_{t } (x,y, t) =0 \\tag{7}$$\n\nThe first one is already discussed, the second one is expanding/descending fireworks circles as already stated.\n\nIn the latter case working is:\n\n$$x = v t \\cos \\theta , y = v t \\sin \\theta - g t^2/2 \\tag{8}$$\n\n$$(\\frac{x}{vt})^2 + (\\frac{y+ gt^2/2}{vt}) ^2 = 1 \\tag{9}$$\n\n$$x^2 + ( y + g t^2/2)^2 = v^2t^2 \\tag{10}$$\n\nwhich is a Circle.\n\nTo find its envelope, as before partially differentiate with respect to time $t$ and cancel $2t$ on either side of equation , bring $v^2/g$ to right side :\n\n$$y + gt^2/2 = v^2/g \\tag {11}$$\n\n$$x^2 + (v^2/g)^2 = 2 v^2/g * ( v^2/g-y) \\tag {12}$$\n\n$$x^2 + ( 2 H)^2 += 4 H ( v^2/g -y )\\tag{13}$$\n\n$$x^2 = 4 H ( H-y) \\tag{14}$$\n\nwhich is the same parabola envelope obtained earlier with $\\theta$ as parameter. End points $( x=0, y=H ; x= 2 H, y= 0 )$\n\nAlthough circle end traces are visible in a fireworks display it needs imagination as before with variable gun barrel angle to see that each circle is tangent to a fixed envelope. Hope you enjoyed it.\n\n\u2022 Thanks for your solution and insight on the parabolic mirror! (+1). Nice graphics too! In addition to different parabolas, why do you also plot partial concentric circles? \u2013\u00a0hypergeometric Oct 25 '15 at 5:56\n\u2022 Also, is there an argument which would directly conclude that the envelope is a parabola with focus at launch point and tip at the peak of the vertical launch, without first going through the analysis? \u2013\u00a0hypergeometric Oct 25 '15 at 6:03\n\u2022 The question in my comment above has been answered here. \u2013\u00a0hypergeometric Sep 30 '18 at 17:14", "date": "2019-10-20 21:38:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1495086/envelope-of-projectile-trajectories?noredirect=1", "openwebmath_score": 0.7652152180671692, "openwebmath_perplexity": 568.5360334157367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138118632622, "lm_q2_score": 0.8840392695254319, "lm_q1q2_score": 0.8643374040445239}}
{"url": "https://gmatclub.com/forum/each-of-60-cars-is-parked-in-one-of-three-empty-parking-lots-306824.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 27 Jan 2020, 20:38\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Each of 60 cars is parked in one of three empty parking lots\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 15994\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nEach of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n30 Sep 2019, 20:27\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n89% (01:53) correct 11% (01:27) wrong based on 62 sessions\n\n### HideShow timer Statistics\n\nEMPOWERgmat PS Series:\nBlock 1, Question 5\n\nEach of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?\n\nA. 12\nB. 20\nC. 22\nD. 28\nE. 30\n\n48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!) Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question SC Block 1 question pack will win an$85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.\n\nTo be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct\n\nThe OA and official explanation will be held until the 48 hour window has lapsed.\n\n_________________\nContact Rich at: Rich.C@empowergmat.com\n\nThe Course Used By GMAT Club Moderators To Earn 750+\n\nsouvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605\nENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605\nVP\nJoined: 20 Jul 2017\nPosts: 1266\nLocation: India\nConcentration: Entrepreneurship, Marketing\nWE: Education (Education)\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n30 Sep 2019, 20:43\n1\n2\nEMPOWERgmatRichC wrote:\nEMPOWERgmat PS Series:\nBlock 1, Question 5\n\nEach of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?\n\nA. 12\nB. 20\nC. 22\nD. 28\nE. 30\n\n48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!) Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question SC Block 1 question pack will win an$85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.\n\nTo be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct\n\nThe OA and official explanation will be held until the 48 hour window has lapsed.\n\nLet the number of cars parked in largest lot = x\n--> Middle lot = x - 8 & Smallest lot = x - 16\nSo, x + (x - 8) + (x - 16) = 60\n--> 3x - 24 = 60\n--> 3x = 84\n--> x = 28\n\nIMO Option D\nSenior Manager\nJoined: 28 Feb 2014\nPosts: 280\nLocation: India\nGMAT 1: 570 Q49 V20\nGPA: 3.97\nWE: Engineering (Education)\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n30 Sep 2019, 21:25\n1\nEMPOWERgmatRichC wrote:\nEMPOWERgmat PS Series:\nBlock 1, Question 5\n\nEach of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?\n\nA. 12\nB. 20\nC. 22\nD. 28\nE. 30\n\nlet a,b,c be lot size in ascending order\na+b+c=60\nc=b+8\nc=a+16\nSubstituting\nc-16+c-8+c=60\nc = 28\n\nD is correct\nIntern\nJoined: 09 Jul 2018\nPosts: 12\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n02 Oct 2019, 16:42\n1\nLet the number of cars parked in largest lot = L\nMiddle lot = M & Smallest lot = S\nBut M=L-8 & S=L-16\nSo, L + (L - 8) + (L - 16) = 60\n3L - 24 = 60\n3L = 84\nL = 28\n\nIMO Option D\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 15994\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2019, 11:34\nOFFICIAL EXPLANATION\n\nHi All,\n\nWe're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...\n\nBased on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...\n\nIF....the largest lot holds 28 cars....\nthen the middle lot holds 28 - 8 = 20 cars...\nand the smallest lot holds 28 - 16 = 12 cars...\nTotal = 28 + 20 + 12 = 60 cars\nThis is an exact MATCH for what we were told, so this MUST be the answer!\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\nContact Rich at: Rich.C@empowergmat.com\n\nThe Course Used By GMAT Club Moderators To Earn 750+\n\nsouvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605\nENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605\nGMAT Club Legend\nJoined: 12 Sep 2015\nPosts: 4234\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2019, 12:39\nTop Contributor\nEMPOWERgmatRichC wrote:\nEMPOWERgmat PS Series:\nBlock 1, Question 5\n\nEach of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?\n\nA. 12\nB. 20\nC. 22\nD. 28\nE. 30\n\nLet x = number of cars in the LARGEST lot\n\nGiven: the largest lot holds 8 more cars than the middle lot.\nSo, x - 8 = number of cars in the MIDDLE lot\n\nGiven: the largest lot holds 16 more cars than the smallest lot.\nSo, x - 16 = number of cars in the SMALLEST lot\n\nGiven: Each of 60 cars is parked in one of three empty parking lots.\nWe can write: (# of cars in LARGEST lot) + (# of cars in MIDDLE lot) + (# of cars in SMALLEST lot) = 60\nSubstitute values: (x) + (x - 8) + (x - 16) = 60\nSimplify: 3x - 24 = 60\nAdd 24 to both sides: 3x = 84\nDivide both sides by 3: x = 28\n\nCheers,\nBrent\n_________________\nTest confidently with gmatprepnow.com\nSenior Manager\nJoined: 05 Jul 2018\nPosts: 459\nLocation: India\nConcentration: General Management, Technology\nGMAT 1: 600 Q47 V26\nGRE 1: Q162 V149\nGPA: 3.6\nWE: Information Technology (Consulting)\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n10 Oct 2019, 22:39\n1\nThe 3 slots A + B + C total to 60 cars\n\nThe highest slot C has 8 more than B and 16 more than A\n\nLet us look at the second-largest answer choice which says C=28\n\nAs per question A=12 (16 less than C), B=20(8 less than C) ...This gives a total of 60. Hence Correct answer is D\nEMPOWERgmatRichC: This is the correct way to do TTA ...right\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 15994\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n11 Oct 2019, 11:30\nHi saukrit,\n\nYES - that's a perfect use of TEST THE ANSWERS. In these sorts of situations, that Tactic (and some basic Arithmetic) is almost always faster and easier that a typical Algebraic approach - so it's something to keep in mind (especially if you have a pacing problem in the Quant section).\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\nContact Rich at: Rich.C@empowergmat.com\n\nThe Course Used By GMAT Club Moderators To Earn 750+\n\nsouvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605\nENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 4868\nLocation: India\nGPA: 3.5\nRe: Each of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2019, 00:09\nEMPOWERgmatRichC wrote:\nEMPOWERgmat PS Series:\nBlock 1, Question 5\n\nEach of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?\n\nA. 12\nB. 20\nC. 22\nD. 28\nE. 30\n\n48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!) Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question SC Block 1 question pack will win an$85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.\n\nTo be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct\n\nThe OA and official explanation will be held until the 48 hour window has lapsed.\n\n$$L = M + 8 = S + 16$$\n\nAgain , L + M + S = 60\n\nNow, Plug in the options and check\n\n(A) Can be straightaway rejected as value of L must be > 16 , Since in this case if L = 12 = S + 16 , Or, S = - 4 ( Not Possible )\n(B) If L = 20 , M = 12 & S = 4 , L + M + S = 36\n(C) If L = 22 , M = 14 & S = 6 , L + M + S = 42\n(D) If L = 28 , M = 20 & S = 12 , L + M + S = 60\n(E) If L = 30 , M = 22 & S = 14 , L + M + S = 66\n\n_________________\nThanks and Regards\n\nAbhishek....\n\nPLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS\n\nHow to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )\nIntern\nJoined: 06 Sep 2019\nPosts: 2\nEach of 60 cars is parked in one of three empty parking lots\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2019, 01:42\nAs the middle lot has 8 cars less than the larger it would be X-8\nSmaller Lot is having 16 cars less than the larger that means X-16\nAs all the lots together having 60 it would be X+(X-8)+(X-16)=60\nOn solving this it becomes X= 28 Cars\nHence Option D would be apt.\nEach of 60 cars is parked in one of three empty parking lots \u00a0 [#permalink] 12 Oct 2019, 01:42\nDisplay posts from previous: Sort by", "date": "2020-01-28 03:38:21", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/each-of-60-cars-is-parked-in-one-of-three-empty-parking-lots-306824.html", "openwebmath_score": 0.2563014328479767, "openwebmath_perplexity": 3974.236522246419, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9759464499040092, "lm_q2_score": 0.8856314723088733, "lm_q1q2_score": 0.8643288913231058}}
{"url": "https://math.stackexchange.com/questions/903613/fibonacci-sequence-given-n-and-mathrmfibn-is-it-possible-to-calculate", "text": "# Fibonacci sequence: Given $n$ and $\\mathrm{Fib}(n)$, is it possible to calculate $\\mathrm{Fib}(n-1)$?\n\nGiven $n$ and $\\newcommand{\\Fib}{\\mathrm{Fib}} \\Fib(n)$, is it possible to calculate the previous number in the Fibonacci sequence - $\\Fib(n-1)$ using integer math in constant time? In other words, I believe I'm looking for a closed form formula.\n\nFor example: Given $n=10$ and $\\Fib(10)=55$, I'd like to determine that $\\Fib(9)=34$ without using $\\Fib(7) + \\Fib(8)$.\n\n\u2022 Even output of $F_{n-1}$ cannot be done in constant time. (The length of binary representation of $F_{n-1}$ is linear in $n$.) But based on the wording in original post I would guess that you mean using constantly many operations with integers. Aug 20 '14 at 8:14\n\u2022 @Martin Sleziak - you are correct that I meant \"constantly many operations with integers\" Aug 20 '14 at 13:23\n\nBy Binet's Formula, we know that $$F_n = \\dfrac{1}{\\sqrt{5}}\\left(\\phi^n-(-\\phi)^{-n}\\right) \\approx \\dfrac{1}{\\sqrt{5}}\\phi^n$$ where $$\\phi = \\dfrac{1+\\sqrt{5}}{2}$$.\n\nUsing this formula, we can get that $$F_{n-1}$$ is the nearest integer to $$\\dfrac{1}{\\phi} F_n$$ for large $$n$$\n\nMore rigorously, we have $$F_{n-1} - \\dfrac{1}{\\phi} F_n = \\dfrac{1}{\\sqrt{5}}\\left(\\phi^{n-1}-(-\\phi)^{-(n-1)}\\right) - \\dfrac{1}{\\phi\\sqrt{5}}\\left(\\phi^n-(-\\phi)^{-n}\\right)$$ $$= -\\dfrac{(-\\phi)^{-(n-1)}}{\\sqrt{5}} - \\dfrac{(-\\phi)^{-(n+1)}}{\\sqrt{5}}$$ $$= \\dfrac{(-\\phi)^{-n}}{\\sqrt{5}}\\left(\\phi + \\phi^{-1}\\right) = (-\\phi)^{-n}$$.\n\nThus, for $$n \\ge 2$$, we have $$\\left|F_{n-1} - \\dfrac{1}{\\phi} F_n \\right| = \\phi^{-n} < \\frac{1}{2}$$, and so, $$F_{n-1}$$ is the nearest integer to $$\\dfrac{1}{\\phi}F_n$$.\n\nDividing $$F_n$$ by $$\\phi$$ and rounding the result to the nearest integer shouldn't be too computational.\n\n\u2022 Here is a table that shows the pattern:$$\\begin{array} {c|c|c} n & a_n & a_n / a_{n-1} \\\\ \\hline 0 & 0 & \\\\ \\hline 1 & 1 & \\\\ \\hline 2 & 1 & 1.0000 \\\\ \\hline 3 & 2 & 2.0000 \\\\ \\hline 4 & 3 & 1.5000 \\\\ \\hline 5 & 5 & 1.6667 \\\\ \\hline 6 & 8 & 1.6000 \\\\ \\hline 7 & 13 & 1.6250 \\\\ \\hline 8 & 21 & 1.6154 \\\\ \\hline 9 & 34 & 1.6190 \\\\ \\hline 10 & 55 & 1.6176 \\\\ \\hline 11 & 89 & 1.6182 \\\\ \\hline 12 & 144 & 1.6180 \\\\ \\hline 13 & 233 & 1.6181 \\\\ \\hline 14 & 377 & 1.6180 \\\\ \\hline \\end{array}$$ Aug 20 '14 at 2:25\n\u2022 @DanielV - agreed, this will work but as marty cohen points out in his anti-answer below, for large values of n simply rounding to the nearest integer will not necessarily yield the correct answer. Aug 20 '14 at 13:34\n\nThis is a dissenting note on some of these answers. I am purposly making this an answer, though it really is an anti-answer.\n\nThe problem with any answer that uses $\\phi$ is that, as $n$ gets larger, $\\phi$ has to be computed with increasingly great precision. This will not take constant time, either for the computation of $\\phi$ or the multiplication by $\\phi$.\n\nHint $\\rm\\ n\\:$ is a Fibonacci number iff the interval $\\rm\\ [\\phi\\ n - 1/n,\\ \\phi\\ n + 1/n]\\$ contains a positive integer (the next Fibonacci number for $\\rm\\ n > 1)$. This follows from basic properties of continued fractions.\n\nFor example, $\\rm\\ 2 \\to [2.7,\\ 3.7],\\ \\ 3\\to [4.5,\\ 5.2],\\ \\ 5 \\to [7.9,\\ 8.3]\\ \\ldots$\n\nFor a proof see e.g.  T. Komatsu: The interval associated with a fibonacci number. $\\$", "date": "2022-01-29 02:15:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/903613/fibonacci-sequence-given-n-and-mathrmfibn-is-it-possible-to-calculate", "openwebmath_score": 0.9017306566238403, "openwebmath_perplexity": 180.5184766596439, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464453565514, "lm_q2_score": 0.8856314723088732, "lm_q1q2_score": 0.8643288872957339}}
{"url": "http://mathhelpforum.com/algebra/70782-geometric-series-problem.html", "text": "# Thread: Geometric series problem .\n\n1. ## Geometric series problem .\n\nA house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?\n\nMy attempt :\n\nIt forms a sequence as follows :\n\n$\\displaystyle 50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$\n\nThe 15th term , $\\displaystyle T_15=50000(1.09)^14=167086.35$\nwhich is the amount he has to pay by the end of 15 years .\n\nThus , every month he has to pay $\\displaystyle \\frac{167086.35}{15\\times12}=928.26$\n\nMy answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..\n\n2. ## Repayments\n\nOriginally Posted by mathaddict\nA house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?\n\nMy attempt :\n\nIt forms a sequence as follows :\n\n$\\displaystyle 50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$\n\nThe 15th term , $\\displaystyle T_15=50000(1.09)^14=167086.35$\nwhich is the amount he has to pay by the end of 15 years .\n\nThus , every month he has to pay $\\displaystyle \\frac{167086.35}{15\\times12}=928.26$\n\nMy answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..\nThanks for showing us your working here. What you're forgetting is that\neach year he pays not only the interest owing that year, but he also pays back some of the amount borrowed.\n\nAre you allowed to use a spreadsheet to work out this answer? It's the easiest way.\n\nIf not, you'll have to let the monthly payment be x, and then say that, at the start of each new year, the amount owing\n\n= amount owed at the start of the last year + interest payable during that year - 12x\n\nThen work out how much interest he'll have to pay in the coming year to cover this new amount owing. Then repeat for the following year, and so on, up to the start of year 16. Then say that at the start of year 16 he owes nothing.\n\nThis is an Amortization problem and requires a special formula.\n\nI've seen this type of problem posted many times before.\nIt seem that the students are never given the formula,\n. . nor are they prepared for the (very) long derivation.\n\nCould it be that all those teachers are ignorant of the difficulties\n. . involved with time-payment problems?\nI'm beginning to suspect that this is so . . .\n\nA house buyer borrows $50,000 from a bank to buy a house. The rate of interest 9% per annum. The buyer is required to repay his loan in monthly installments for 15 years. Assuming that the rate of interest is fixed for the entire duration of the loan, find the amount per month he has to repay the bank? The formula is: .$\\displaystyle A \\;=\\;P\\,\\frac{i(1+i)^n}{(1+i)^n-1}$. . where: .$\\displaystyle \\begin{Bmatrix}P &=& \\text{principal} \\\\ i &=& \\text{periodic interst rate} \\\\ n &=& \\text{number of periods} \\\\ A &=& \\text{periodic payment} \\end{Bmatrix}$We have: .$\\displaystyle \\begin{array}{ccc}P \\:=\\:50,\\!000 \\\\\ni \\:=\\:\\frac{9\\%}{12} \\:=\\:0.0075 \\\\ n \\:=\\:15\\!\\cdot\\!12 \\:=\\:180 \\end{array}$Hence: .$\\displaystyle A \\;=\\;50,\\!000\\,\\frac{0.0075(1.0075)^{180}}{1.0075^ {180}-1} \\;=\\;507.1332921$Therefore, the monthly payment is: .$\\displaystyle \\boxed{\\$507.13}$\n\n4. ## Re :\n\nVery big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?\n\n5. Originally Posted by mathaddict\nVery big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?\nCheck this and especially this out. It is one way of deriving it. There are or course other ways.\nGo through page 70 up to page 99.\nOr you can just order this book if get tired of reading it online.\n\n6. ## Repayments\n\nIn fact, the interest is calculated at the start of each year for the whole year, as if the whole amount is owing for the whole year. So you have to use the values\n\nP = 50000\ni = 0.09\nn = 15\n\nto get the total payment each year, and then divide by 12, to get a monthly payment of 516.91.\n\nIncidentally, I don't think it's really that hard to work out the formula for yourself. After\n$\\displaystyle n$ years you'll find that the amount owing is\n\n$\\displaystyle P(1+i)^n - [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x$, where $\\displaystyle x$ is the annual payment\n\nInside the square brackets, it's a GP whose sum is\n$\\displaystyle \\frac{(1+i)^n - 1}{i}$\n\nSet this sum equal to zero and solve for\n$\\displaystyle x$ to obtain the formula.\n\n7. Thanks all but i am having some problems with the series .\n\n[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math\n\n$\\displaystyle T_1=x$\n\n$\\displaystyle T_2=xi+x=x(1+i)$\n\n$\\displaystyle T_3=[x(1+i)]\\cdot{i}+x$ but this is not equal to $\\displaystyle x(1+i)^2$\n\nWhy is it so ??\n\n8. ## GP\n\nOriginally Posted by mathaddict\nThanks all but i am having some problems with the series .\n\n[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math\n\n$\\displaystyle T_1=x$\n\n$\\displaystyle T_2=xi+x=x(1+i)$\n\n$\\displaystyle T_3=[x(1+i)]\\cdot{i}+x$ but this is not equal to $\\displaystyle x(1+i)^2$\n\nWhy is it so ??\nSet your working out like this:\n\nAt the end of year 1, the amount owed is the original sum borrowed + the interest for the year - the total payments during the year\n\n$\\displaystyle = P + Pi - x$\n\n$\\displaystyle = P(1 + i) - x$\n\nThis, then is the amount owing at the start of year 2: the 'new $\\displaystyle P$' if you like. So at the end of year 2, we'll replace $\\displaystyle P$ by $\\displaystyle P(1+i) - x$, to get that the amount owing is:\n\n$\\displaystyle [P(1 +i) - x](1+i) - x$\n\n$\\displaystyle = P(1+i)^2 - x[1 + (1 + i)]$\n\nContinue like this, multiplying by $\\displaystyle (1+i)$ and subtracting $\\displaystyle x$ each time. So at the end of the year 3, the amount owing is:\n\n$\\displaystyle P(1+i)^3 - x[1 + (1 + i) + (1 +i)^2]$\n\nAnd, in a similar way, at the end of year $\\displaystyle n$, it is:\n\n$\\displaystyle P(1+i)^n - x[1 + (1 + i) + ...+ (1 +i)^{n-1}]$\n\nSo, inside the square brackets, [...], we have a GP whose first term is $\\displaystyle 1$, common ratio $\\displaystyle (1+i)$ and number of terms = $\\displaystyle n$, and whose sum is $\\displaystyle \\frac{(1+i)^n - 1}{(1+i) - 1}= \\frac{(1+i)^n - 1}{i}$.\n\nNow if the whole sum is paid of after $\\displaystyle n$ years, this total is zero. I.e.\n\n$\\displaystyle P(1+i)^n - x\\frac{(1+i)^n - 1}{i} = 0$\n\nHence $\\displaystyle x = \\frac{iP(1+i)^n}{(1+i)^n - 1}$\n\nOK now?", "date": "2018-03-25 05:42:55", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/70782-geometric-series-problem.html", "openwebmath_score": 0.9452170133590698, "openwebmath_perplexity": 1966.3116111501001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464429079201, "lm_q2_score": 0.885631470799559, "lm_q1q2_score": 0.8643288836541391}}
{"url": "https://math.stackexchange.com/questions/868400/showing-that-y-has-a-uniform-distribution-if-y-fx-where-f-is-the-cdf-of-contin/868405", "text": "# Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X\n\nLet $$X$$ be a random variable with a continuous and strictly increasing c.d.f. $$F$$ (so that the quantile function $$F^{\u22121}$$ is well-de\ufb01ned). De\ufb01ne a new random variable $$Y$$ by $$Y = F(X)$$. Show that $$Y$$ follows a uniform distribution on the interval $$[0, 1]$$.\n\nMy initial thought is that $$Y$$ is distributed on the interval $$[0,1]$$ because this is the range of $$F$$. But how do you show that it is uniform?\n\n\u2022 This is not true in cases where there's a discrete component. For example, suppose $X=\\left.\\begin{cases} 1/2 & \\text{with probability }1/2, \\\\ W & \\text{with probability }1/2,\\end{cases}\\right\\}$ and $W$ is uniformly distributed on $[0,1]$, and that the choice between whether $X=1/2$ or not is independent of $W$. Then the cdf of $X$ has no values between $1/4$ and $3/4$, so it cannot be uniformly distributed on $[0,1]$. It is, however, true of continuous distributions. Jul 15, 2014 at 21:41\n\u2022 see the text of the question. X is continuous! Jul 15, 2014 at 21:44\n\u2022 By the way, it is not necessary that $F$ is a strictly increasing CDF, continuity is sufficient. Just define the quantile function the usual way as a generalized inverse via $F^-(y)=inf\\{x\\in\\mathbb{R}: F(x)\\geq y\\}$. See the proof of Proposition 3.1 in Embrechts, P., Hofert, M.: A note on generalized inverses. Mathematical Methods of Operations Research 77(3), 423-432 link for a very careful and detailed explanation. Jul 16, 2014 at 14:38\n\u2022 Thanks @binkyhorse - that reference is really good. Apr 5, 2015 at 22:33\n\u2022 @s0ulr3aper07 By Proposition 3.1 in the paper I linked above, yes. Prop. 3.1: Let F be a distribution function and X ~ F. (a) If F is continuous, F(X)\u223cU[0,1]. The paper includes a detailed proof. Mar 27, 2019 at 20:10\n\nLet $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \\in [0,1]$ we have:\n\n$F_Y(y) = \\Pr[Y \\le y] = \\Pr[F(X) \\le y] = \\Pr[X \\le F^{-1}(y)] = F(F^{-1}(y)) = y$.\n\nWhat distribution has this CDF?\n\n\u2022 Are all CDF's of continuous densities invertible? Aug 22, 2017 at 7:43\n\u2022 @tintinthong: not always completely, but enough. If you define $G(y)= \\inf\\{x:F_X(x) \\ge y\\}$ then $F_X(G(y))=y$ when $y \\in (0,1)$ Aug 22, 2017 at 22:12\n\u2022 Strictly increasing and continuous CDF is needed. Feb 20, 2019 at 15:36\n\n$$Prob(Y\\leq x)=P(F(X)\\leq x)=P(X\\leq F^{-1}(x))=x \\\\$$ The last equality is from the definition of the quantile function.\n\nLet $y=g(x)$ be a mapping of the random variable $x$ distributed according to $f(x)$. In the mapping $y=g(x)$ you preserve the condition of probability density (namely you counts the same number of events in the respective bins)\n\n$$h(y)dy=f(x)dx$$\n\nwhere h(y) is the probability distribution of $y$\n\nif $h(y)=1$ (uniform distribution) we have\n\n$$dy=g'(x)dx=f(x)dx$$\n\nThis means that $$g(x)=\\int f(x)dx$$\n\nnamely the function $g(x)$ that maps the random variable $x$ distributed according $f(x)$, into a random variable $y$ distributed uniformly is his own cumulative distribution function $\\int f(x) dx$.\n\nHere is an approach that does not use the quantile function whatsoever - the only property used is that independent copies of $$X$$ have zero probability of being equal. (The main ingredient in my argument is conditional expectation.)\n\nConsider the cumulative distribution function of $$X$$, namely $$F(t)=\\mathbb P(X\\leq t).$$ Your random variable - which I will suggestively call $$U$$ instead of $$Y$$ - can be described by starting with two independent and identically distributed random variables $$X,Z$$ and considering the conditional probability $$U=\\mathbb P(X\\leq Z\\mid Z).$$ Then, for all integers $$n\\geq 1$$, we can represent $$U^n$$ as follows. Let $$X_1,X_2,\\ldots,X_n,Z$$ be independent and identically distributed. By independence, $$\\mathbb P\\bigl(X_1\\leq Z,X_2\\leq Z,\\ldots, X_n\\leq Z\\bigm\\vert Z\\bigr)=U^n,$$ and thus by the tower property $$\\mathbb EU^n=\\mathbb P(X_1\\leq Z,X_2\\leq Z,\\ldots, X_n\\leq Z)=\\mathbb P\\bigl(Z=\\max(X_1,X_2,\\ldots,X_n,Z)\\bigr).$$ Since $$X_1,\\ldots,X_n,Z$$ are iid, each of them is equally likely to be the maximum and therefore $$\\mathbb EU^n=\\frac{1}{n+1}.$$ Thus $$U$$ has the same moments as a uniformly distributed random variable on $$[0,1]$$. Since $$U$$ is supported in $$[0,1]$$ as well, it follows (by the uniqueness of the Hausdorff moment problem) that $$U$$ is uniformly distributed, as desired.\n\nLet $$y\\in(0,1)$$. Since $$F$$ is continuous, there exists $$x\\in\\mathbb{R}$$ s.t. $$F(x)=y$$. Thus, $$\\mathsf{P}(Y\\le y)=\\mathsf{P}(F(X)\\le F(x))=F(x)=y,$$ i.e., $$Y\\sim\\text{U}[0,1]$$. In order to see the first equality we don't need continuity. Specifically, since any cdf is right-continuous, \\begin{align} \\{F(X)\\le F(x)\\}&=\\{\\{F(X)\\le F(x)\\}\\cap\\{X\\le x\\}\\}\\cup \\{\\{F(X)\\le F(x)\\}\\cap\\{X>x\\}\\} \\\\ &=\\{X\\le x\\}\\cup \\{\\{F(X)=F(x)\\}\\cap\\{X>x\\}\\}, \\end{align} and $$\\mathsf{P}(\\{F(X)=F(x)\\}\\cap\\{X>x\\})=0$$.\n\nFor a proof of this problem when $$F_X(x)$$ is strictly increasing, refer to JimmyK4542's answer. Let's assume $$F_X(x)$$ is just non-decreasing (there are intervals such as $$[a,b]$$ where $$F_X(x') = c$$ for $$x'\\in[a,b]$$). We define $$G(y)$$ similar to what Henry's comment suggests: $$G(y)=\\inf\\{x:F_X(x)\\gt y\\}$$ Now substituting this expression in what Jimmy has written will give us: $$F_Y(y) = \\Pr[Y \\le y] = \\Pr[F_X(X) \\le y] = \\Pr[X \\le G(y)] = F_X(G(y))= y \\label{eq:I}\\tag{I}$$\n\nWe need to show that:\n\n1. $$F_X(x)\\le y \\rightarrow x \\le G(y)$$\n2. $$F_X(G(y))=y$$\n\nThe second argument is easier to prove. We have the following expression almost according to definitions: $$F_X(G(y))= F_X(\\inf\\{x:F_X(x)\\gt y\\})= y$$ Now for the first argument, we can still use what $$G(y)=\\inf\\{\\cdots\\}$$ implies; if $$F_X(x)\\le y$$, then $$x\\le \\inf\\{x:F_X(x)\\gt y\\};$$ hence $$x\\le G(y)$$.\n\nWith the two arguments proved and a substitution in \\ref{eq:I}, we have proved the main argument.\n\n\u2022 I personally believe this problem is a severe case for abuse of notation, and a bad professor's problem. Feb 19 at 1:46", "date": "2022-06-30 14:40:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/868400/showing-that-y-has-a-uniform-distribution-if-y-fx-where-f-is-the-cdf-of-contin/868405", "openwebmath_score": 0.957176685333252, "openwebmath_perplexity": 174.14235605365812, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464443071381, "lm_q2_score": 0.8856314617436727, "lm_q1q2_score": 0.8643288760552705}}
{"url": "https://web2.0calc.com/questions/what-is-the-probability-of-winning-a-lottery-which-requires-selection-of-5-numbers-from1-to-40", "text": "+0\n\n# what is the probability of winning a lottery which requires selection of 5 numbers from1 to 40\n\n0\n895\n7\n\nwhat is the probability of winning a lottery which requires selection of 5 numbers from1 to 40\n\nMay 7, 2014\n\n#2\n+17\n+8\nCalculate the numerator: There is only one winning combination. So the numerator of the desired probability is 1. Calculate the denominator: There are \"40 numbers, choose 5\", or 40C5 possible combinations. So the denominator of the desired probability is 40C5. So the probability is 1/40C5=1/658008\n\nMay 7, 2014\n\n#1\n+1006\n0\n\nFor this kind of probability problem, assuming the number is removed from play once chosen, the probability increases by one each time.\n\n$$\\left({\\frac{{\\mathtt{1}}}{{\\mathtt{40}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{1}}}{{\\mathtt{39}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{1}}}{{\\mathtt{38}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{1}}}{{\\mathtt{37}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{1}}}{{\\mathtt{36}}}}\\right)$$\n\nThe bottom five numbers multiply to\u00a078960960, which means that the odds are 1/78960960\n\nMay 7, 2014\n#2\n+17\n+8\nCalculate the numerator: There is only one winning combination. So the numerator of the desired probability is 1. Calculate the denominator: There are \"40 numbers, choose 5\", or 40C5 possible combinations. So the denominator of the desired probability is 40C5. So the probability is 1/40C5=1/658008\n\nFoxTears\u00a0May 7, 2014\n#3\n+17\n0\n\nMine was assuming the numbers can be reused\n\nMay 7, 2014\n#4\n+115418\n+5\n\nThe number of 5 digit combinations (order doesn't matter) from 40 is\n\n40C5\n\n$${\\left({\\frac{{\\mathtt{40}}{!}}{{\\mathtt{5}}{!}{\\mathtt{\\,\\times\\,}}({\\mathtt{40}}{\\mathtt{\\,-\\,}}{\\mathtt{5}}){!}}}\\right)} = {\\mathtt{658\\,008}}$$\n\nonly one of those wins so the chance is \u00a0 \u00a0\u00a0$$\\frac{1}{658008}$$\n\nJust like FoxTears said!\n\nThanks Foxtears.\n\nFoxtears, I don't think that yours was assuming that the numbers could be reused.\n\nIt was just assuming that the order they were chosen in was of no consequence.\n\nGoldenLeaf I believe yours would have been correct if the numbers had to be chosen in a specific order.\n\nMay 7, 2014\n#5\n+32772\n+5\n\nGoldenLeaf's approach was almost right, but he should have had 5, 4, 3, 2 and 1 in the numerators.\n\nThere are 5 possible choices for the first number, so probability of this is 5/40. \u00a0For each of these\n\nthere are 4 possible choices for the 2nd number, so probability of both 1st and 2nd is (5/40)*(4/39). \u00a0For each of these ... etc. to get the overall probability as\n\n$${\\mathtt{p}} = \\left({\\frac{{\\mathtt{5}}}{{\\mathtt{40}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{4}}}{{\\mathtt{39}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{3}}}{{\\mathtt{38}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{2}}}{{\\mathtt{37}}}}\\right){\\mathtt{\\,\\times\\,}}\\left({\\frac{{\\mathtt{1}}}{{\\mathtt{36}}}}\\right) \\Rightarrow {\\mathtt{p}} = {\\mathtt{0.000\\: \\!001\\: \\!519\\: \\!738\\: \\!361\\: \\!8}}$$\n\nor p\u22481.52*10-6\n\n1/p = 658008\n\nMay 7, 2014\n#6\n+115418\n0\n\nYes of course - that did not occur to me. \u00a0Thanks Alan.\n\nMay 7, 2014\n#7\n0\n\nCalling Bertie, CPhill and Rom (Troll alert! General quarters!)\n\nAlong with Melody and Alan, we need input from Bertie, CPhill, and Rom, too.\n\nWith the top brains represented here that Troll might comeback and bite one or more.\n\nI know he\u2019ll bite FoxTears for saying, \u201cMine was assuming the numbers can be reused\u201d (your formula (40C5) contradicts this).\n\nThe Troll will follow-up, by asking what is the probability of (something) of (five, six or seven) mathematicians getting a probability question correct.\n\nThen he\u2019ll say about the same as walking, blindfolded, through a large buffalo heard and stepping in four piles with both feet.\n\nWhat\u2019s annoying: he\u2019s usually right.\n\nCPhill might not be able to make it. He might be trying to get the \u201cRoman Zero\u201d from inside Sisyphus\u2019s boulder.\n\nMaybe he could take a break and visit. He\u2019s not been around for a day or two. I hope he wasn\u2019t squashed by the boulder.\n\nNo matter: cartoon physics seem to apply in that realm. If he does visit, maybe one of those angels will help him get the \u201cRoman Zero.\u201d\n\nBy: Troll Detector\n\nMay 7, 2014", "date": "2021-12-04 23:52:48", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/what-is-the-probability-of-winning-a-lottery-which-requires-selection-of-5-numbers-from1-to-40", "openwebmath_score": 0.9447340369224548, "openwebmath_perplexity": 1798.8861039896221, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419690807408, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8643242429963786}}
{"url": "http://amicidellacattolica.it/rmev/sigma-notation-rules.html", "text": "Sigma, \u03a3, is the standard notation for writing long sums. 3n 3 + 6n 2 + 6000 = \u0398(n 3). \u201d - Sincerely, Dr. Commented: Roger Stafford on 22 Oct 2014 Accepted Answer: Star Strider. It's based on the upper case Greek letter S, which indicates a sum. Some basic ideas can be applied to creating process maps that make them easier to understand and use. Clears 1, 5, and 12, and if you remember the old notation the cardinality of Z is 3 because there are three elements in z. See Figure 3. So for this series, there is 7 terms The seven terms are: when n=0 -2x. 2040 - 3 sig fig 2. Okay, welcome back everyone. Most of the following problems are average. \"My TI 83+ graphing calculator arrived faster than I expected. Although it can appear scary if you've never seen it before, it's actually not very difficult. Sigma Notation. This is the currently selected item. Define Sigma notation. In chemistry, a molecular orbital (MO) is a mathematical function describing the wave-like behavior of an electron in a molecule. Free online converters. APPENDIX E SIGMA NOTATION A35 Theorem If is any constant (that is, it does not depend on ), then (a) (b) (c) Proof To see why these rules are true, all we have to do is write both sides in expanded form. The variable n is called the index of summation. When WeBWorK gives a typeset version of your answer it only uses parentheses so for example it expresses your input of 2 [3 (4+5)+6] as 2 (3 (4+5)+6) but you can use whatever you want. Multiplication by a whole number can be interpreted as successive addition. Example \u2211 = = + + + +. 00, but has features that will be used throughout algebra, trigonometry, and statistics. To see why Rule 1 is true, let's start with the left hand side of this equation, n i=1 cx i. 5 Sigma Notation And The Nth Term Ppt Video Online Download Index Of Wp Content Uploads 2017 02 Summation Notation Problems & Solutions Sigma Notation Sigma Worksheet 7 Probability And Sigma Notation Solutions(1) Maths Sigma Notation Guided Notesmath With Mrs Holst Tpt Document 20. Rules of Scientific Notation. These rules can be converted and applied to many log management or SIEM systems and can even be used with grep on the command line. If the series is multiplied by a. This involves the Greek letter sigma, \u03a3. The \"a i \" in the above sigma notation is saying that you sum all of the values of \"a\". In this unit we look at ways of using sigma notation, and establish some useful rules. The Organic Chemistry Tutor 270,467 views. With sequences, sometimes we like to start at n = 0 and sometimes we like to start at n = 1. We often use Sigma Notation for infinite series. Properties. For example, the Schrodinger equation, which has to do with dynamics in quantum systems and predates quantum computation by decades, is written using bra-ket notation. The Example shows (at least for the special case where one random variable takes only. The Sigma symbol can be used all by itself to represent a generic sum\u2026 the general idea of a. When you're calculating the big O. Use algebra rules of sums, along with the appropriate sum formulae, to evaluate the following: (a) X12 i=1 (i2 33i) (b) X10 i=1 (i2 2 1) (c) X8 i=1 (2 i ) 6. 2 Summation notation Summation notation. Control Chart with all Control Limits Visible QI Macros offers the following option to turn on and off the 1-2 sigma lines: After you have run a chart, click on the QI Macros Chart menu and select Show/Hide Sigma Lines. Go To Problems & Solutions Return To Top Of Page. Use sigma (summation) notation to calculate sums and powers of integers. Sigma questions are not rehashing of questions from a text book, each question is unique and has a slight twist. 3 \u2013 A service desk process map example. Sigma notation provides a way to compactly and precisely express any sum, that is, a sequence of things that are all to be added together. Today we're going to make it a little bit more complicated, and we're going to go over some rules, For manipulating, Slash simplifying, Or making for complicated, if you like, sigma notation. Sigma Notation - Introduction to Summation 9:14. Write the expression 1 + 1 4 + 1 7 + 1 10 + + 1 3n+1 in P notation. These rules that make sense in simpler notation, such as or. Free algebra and math word problems. A business process model and notation diagram, or BPMN diagram for short, is used to build easy-to-read business process model flowcharts, which can be shared across organizations and industries. The concept of sigma notation means to sum up all terms and uses three parts to form math statements, like \u2211 i a i. org are unblocked. In this unit we look at ways of using sigma notation, and establish some useful rules. A typical element of the sequence which is being summed appears to the right of the summation sign. Sigma Notation and Riemann Sums Sigma Notation: Notation and Interpretation of 12 3 14 1 n k nn k aaaaa a a (capital Greek sigma, corresponds to the letter S) indicates that we are to sum numbers of the form indicated by the general term. For example: 1. The Riemann sums and sigma notation exercise appears under the Integral calculus Math Mission. Binomial theorem refresher. Note how the subscripts in the FORTRAN example below exactly match the tensor notation for $$C_{ij} = A_{ik} B_{kj}$$. Posted 2/14/10 8:18 AM, 19 messages. Example of a Matrix. The sum of a series can be written in sigma notation. ) (Factor out the number 6 in the second summation. The example below shows three separate sets of control limits: Don't forget to rerun stability. \u201d D = { x: x is a river in a state} We should describe a certain property which all the elements. In this unit we look at ways of using sigma notation, and establish some useful rules. The resulting display string will be a string which has already been mapped according to the IDNA2003 rules. If you're seeing this message, it means we're having trouble loading external resources on our website. Contents 1 Algebra 8 1. As well as providing shorthand for mathematical ideas, this notation can aid students\u2019 understanding of mathematics. Sigma notation. If , the series does not converge (it is a divergent series). If you want to use a different calculator, the TI-89 has an actual sigma (uses the greek sigma) that you enter the expression, variable, start, and stop values. Theorem: Sigma Notation Properties Suppose and are functions of , is any integer, and is any real number. It's based on the upper case Greek letter S, which indicates a sum. Question 486636: Expand and simplify. This calculus video tutorial provides a basic introduction into summation formulas and sigma notation. when n=2 6-2x. Summation rules Jochumzen. If x \u2264 \u03bc, then the pdf is undefined. Multiplication by a whole number can be interpreted as successive addition. In mathematical formula it is indeed the addition of many number or variable which represent to give concise expression for sum of the variable as Sigma or Summation. In Notes x4. Looking for Pi notation? Find out information about Pi notation. A series that converges has a finite limit, that is a number that is approached. The variable k is called the index of the sum. Okay, now that we've seen that fancy sigma notation let's work a really simple small example of numbers and then generalize. Geometric Series 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. 106L Labs: Sums and Sigma ( ) Notation Sequences, Sums, and Sigma ( ) Notation Sequences De nition A sequence is an ordered set of numbers de ned by some rule. Similarly, we can use sigma notation starting at different values of n to write the same series. Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10. This is true for all tensor notation operations, not just this matrix dot product. 1, we de ne the integral R b a f(x)dx as a limit of approximations. Use Fault Tree diagrams to document business processes, including Six Sigma and ISO 9000. These tell us the starting. Digits 1 to 9 always count 2. But with sigma notation (sigma is the 18th letter of the Greek alphabet), the sum is much more condensed and efficient, and you\u2019ve got to admit it looks pretty cool: This notation just tells you to plug 1 in for the i in 5i, then plug 2 into the i in 5i, then 3, then 4, and so on all the way up to 100. In this unit rules for using sigma notation are established. Re: sigma notation (without lists or tables) mravi1 welcome to the forum As pointed out, unfortunately your post does not comply with Rule 2 of our Forum RULES. Sigma notation is essentially a shortcut way to show addition of series or sequences of numbers. For example, we often wish to sum a number of terms such as 1+2+3+4+5 or 1+4+9+16+25+36 where there is an obvious pattern to the numbers involved. Example \u2211 = = + + + +. After multiplying everything out I got, = E (i^3 + 15i^2 +54i) Then I broke everything down to = E i^3 + (15)*E i^2 + (54)*E i. In quantum mechanics, the uncertainty principle (also known as Heisenberg's uncertainty principle) is any of a variety of mathematical inequalities asserting a fundamental limit to the precision with which the values for certain pairs of physical quantities of a particle, such as position, x, and momentum, p, can be predicted from initial conditions. And S stands for S um. $$S^2\\sigma$$ is frequently over analyzed to the point where it is incorrectly concluded that a high $$S^2\\sigma$$ is preferable even at the expense of lower $$zT$$. is the symbol for the sample mean. Alternatively, we may use ellipses to write this as However, there is an even more powerful shorthand for sums known as sigma notation. Sigma Notation Properties/Rules. We prefer to use subscripts to indicate the control limits, but other texts and papers may use somewhat different notation (e. We often use Sigma Notation for infinite series. If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. LuDaCriS66 Platinum Member. Summation Notation. 5, which dealt with vector coordinate transformations. What we are about to do is to take a function and express it as the limit of a sequence of Riemann Sums over. Enter the sequence, the start value and end value from sigma notation and get a numerical sum. The matrix pictured below has two rows and three columns. (b) c k = 1 k, where 5 k < 9. Recall that the \"n\" on top of the Sigma (the funny looking e) is the terminal value for the index which is located under the sigma. Use sigma (summation) notation to calculate sums and powers of integers. The notation above is not too flexible because the start, stop, and term are all hard coded. The series always converges when x = 0 It will possibly converge for. The act or process of adding; addition. But polynomials truly come to shine in the work of Al-Samawal (1130-1180) The Shining Book on Calculation. Here, we prefer to set up the prior in terms of nu, mu, sigma/(nu-2) or something like that, to account for the fact that the scale of the distribution (as measured by the sd or median absolute deviation) depends on nu as well as sigma. So sum of n=1 to 20 for (3^n) + sum of n=1 to 20 for (2n-1) Then apply the geometric sum formula for the sum of n=1 to 20 for (3^n). We do this through a mixture of our automated migration tools and professional services offerings. It corresponds to \u201cS\u201d in our alphabet, and is used in mathematics to describe \u201csummation\u201d, the addition or sum of a bunch of terms (think of the starting sound of the word \u201csum\u201d: Sssigma = Sssum). Presentaci\u00f3n NEARPOD. Rules for use with sigma notation 6 1 c mathcentre July 18, 2005. The terminology from AMS-LaTeX documentation. So maybe it will help. mc-TY-sigma-2009-1 Sigma notation is a method used to write out a long sum in a concise way. In 1981 Dr. 1 (800) 561 5005 (Toll-free) or +1 (514) 737 7333 [email\u00a0protected] Usually, this is used to describe a certain span or group of spans of numbers along a axis, such as an x-axis. A boolean expression consisting purely of Minterms (product terms) is said to be in canonical sum of products form. Summation notation In statistics we take summary measures of data sets. BPMN diagram symbols are categorized into four main groups: flow objects, connecting objects, swimlanes, and artifacts. Note: you can do this more than once on a chart. For example, cosx, cos x, and cos (x) are all equivalent. Loading Unsubscribe from Jochumzen? Summation Formulas and Sigma Notation - Calculus - Duration: 20:24. Gateways are used to control how the process flows. American Mathematical Society, User's Guide for the amsmath Package For all intents and purposes, the align environment is a replacement for the eqnarray environment and all its warts. This problem has been solved! See the answer. As we learn the basic integration rules, we will learn the difference between an indefinite integral and a definite integral. 'for' loop or symsum? Follow 163 views (last 30 days) Nate on 20 Oct 2014. Taylor series, convergence tests, power series convergence, sigma notation, factorial this page updated 19-jul-17 Mathwords: Terms and Formulas from Algebra I to Calculus. My matrix is the variable A, and my list is the variable Network. The precise addition rule to use is. Summation notation In statistics we take summary measures of data sets. Today we're going to make it a little bit more complicated, and we're going to go over some rules, For manipulating, Slash simplifying, Or making for complicated, if you like, sigma notation. Riemann sums, summation notation, and definite integral notation. Sigma notation provides a way to compactly and precisely express any sum, that is, a sequence of things that are all to be added together. com To create your new password, just click the link in the email we sent you. - Notice that we are adding fractions with a numerator of 1 and. 1) sigma notation: n=0; top of sigma = 4; to the right - (n^2/2) 2) sigma notation: n=1; top of sigma = 3; to the right - (1/n!). Binomial theorem refresher. These degenerate cases are usually only used when the summation notation gives a degenerate result in a special case. I can pull that constant outside and get that the sum from i=1 to 4 of 3i squared is the same thing as 3 times the same exact. That is indicated by the lower index of the letter. 1 - In the following exercises, use the rules for sums Ch. mc-TY-sigma-2009-1 Sigma notation is a method used to write out a long sum in a concise way. lar notation scheme can mystify the reader just as easily as a too casual (\u03bc)forapopulationmean,sigma Now let us review the mathematical rules by which matrices are manipulated. Commented: Roger Stafford on 22 Oct 2014 Accepted Answer: Star Strider. Properties. Worksheet on sigma notation 1. These properties are easy to prove if we can write out the sums without the sigma notation. Summation notation is a useful way to represent the partial sum of a sequence. For these accounting rules, we describe the range of valuations that is consistent with the firm\u2019s financial statements. 1 \u2013 Area, Distance, and Sigma Notation The Area Problem Our goal is to calculate the area of some region. Addition rules are important in probability. improve this answer. The double series. It corresponds to \u201cS\u201d in our alphabet, and is used in mathematics to describe \u201csummation\u201d, the addition or sum of a bunch of terms (think of the starting sound of the word \u201csum\u201d: Sssigma = Sssum). Viewing 8 posts - 1 through 8 (of 8 total) Author. Constant multiplier. Note: you can do this more than once on a chart. Summation notation works according to the following rules. For more ways to implement derivatives, you may find our support article on Prime Notation helpful. Title: SUMMATION or SIGMA NOTATION 1 SUMMATION or SIGMA NOTATION 2 DEFINITIONS Sigma Notation for Finite Sums. These are called mathematical operators. Adding series. With a Six Sigma process even a significant shift in the process mean results in very few defects. For example, we can write which is a bit tedious. Six Sigma Diagram. Contents[show] Table of sum-class symbols Using sum TeX is smart enough to only show. In this unit rules for using sigma notation are established. \u03a3 is the symbol for \u2018the sum. $$\\sum\\limits_{i\\, = \\,{i_{\\,0}}}^n {c{a_i}} = c\\sum\\limits_{i\\, = \\,{i_{\\,0}}}^n {{a_i}}$$ where $$c$$ is any number. The most frequently appearing notation in the mathematical descriptions of different quantities or procedures used in data analysis involves the application of the summation \"operator,\" represented by the upper-case Greek letter sigma, or. The number 100,000,000 for example, takes up a lot of room and takes time to write out, while 10 8 is much more efficient. 1 Sigma Notation \u00b6 Sigma notation is a convenient way to write large and complicated sums. Sigma Notation - Simplification Rules. Thus, Also the initial value doesn't have to be 1. Introduction Sigma notation is a concise and convenient way to represent long sums. Given a matrix M and a vector v, when we work out the \ufb01rst component of Mv, we dot the \ufb01rst row of M with v. You get to it through the calculus functions (f3 key in \"normal\" home). 00204 - 3 sig fig. If you need a quick refresher on summation notation see the review of summation notation in the Calculus I notes. The act or process of adding; addition. Each item in a matrix is called an entry. Binomial theorem refresher. Now notice in this Service Desk process map template, how in addition to the items mentioned above, there is even more processes and messages. The double series. Molecular Weight 251. Fractional Indices. (b) Write the sum X7 k=1 (2k+ 7) in expanded form. Go to The Equation of a Circle page. , F = 1 when any one of the input is true or 1. In either case, the images of the basis vectors form a parallelogram that represents the image of the unit square under the. 1 Steve Strand and Sean Larsen from Portland State University, US, have shown that, cognitively, the task of interpreting a given summation-notation expression differs significantly from the task of converting a longhand sum into summation notation. Greetings! I wanted to let you know that I am now offering a prorated DeltaMath Plus subscription for $20 that will last until October 1, 2020. Shows how factorials and powers of -1 can come into play. Formula for percentage. Join 90 million happy users! Sign Up free of charge:. The number on top of the summation sign tells you the last number to plug into the given expression. answered Dec 10 '13 at 23:12. Both formulas have a mathematical symbol that tells us how to make the calculations. The series 4+ 8+12 +16+20 +24 can be expressed as 6 \u2211 n=14n. Consider the series. Since sigma notation is just a series of addition, you can write out each term in the series then add them all together. In general, capital letters refer to population attributes (i. TQM Diagram. We can also write N+ = {x \u2208 N : x. If that region is a polygon (a shape made only of straight sides) the process is straightforward. If we like, we can go back to calling our summation index k,. Free Summation Calculator. Select a link by mouse clicking one of the music notation symbols below. The use of matrix (lin-ear) algebra can greatly simplify many of the computations. Rules of Scientific Notation. 1 Sigma Notation Welcome to Week 1 of the online version of MATH 232 for Spring 2020. Contents 1 Algebra 8 1. But many programmers don't really have a good grasp of what the notation actually means. Taylor series, convergence tests, power series convergence, sigma notation, factorial this page updated 19-jul-17 Mathwords: Terms and Formulas from Algebra I to Calculus. Math 132 Sigma Notation Stewart x4. Write out completely the sequences given by the following rules: (a) a i = i2, where 0 i 5. Zeros in front never count 3. Sigma notation of the polynomial Coefficients of the source polynomial in the form of a recursive formula Coefficients of the source polynomial function are related to its derivative at x 0 The general form or translatable form of the polynomial. 1 Sigma Notation \u00b6 Sigma notation is a convenient way to write large and complicated sums. You may also want to use \"Post-It\" notes to list possible causes but have the ability to re-arrange the notes as the diagram develops. By using this website, you agree to our Cookie Policy. An index, or a power, is the small floating number that goes next to a number or letter. Ellipses save space or remove material that is less relevant. Learn how to evaluate sums written this way. The purpose of the above sigma formula is the sum of all X data starting from the order of 1 to n. 5 Sigma Notation And The Nth Term Ppt Video Online Download Index Of Wp Content Uploads 2017 02 Summation Notation Problems & Solutions Sigma Notation Sigma Worksheet 7 Probability And Sigma Notation Solutions(1) Maths Sigma Notation Guided Notesmath With Mrs Holst Tpt Document 20. Sigma 4 Shogi Introduction. adding & subtracting. 13 Coordinate Transformation of Tensor Components This section generalises the results of \u00a71. If we like, we can go back to calling our summation index k,. In this unit we look at ways of using sigma notation, and establish some useful rules. Indices show how many times a number or letter has. Find the value of. The dummy variable will usually show up one or more times in the expression to the right of the Greek letter sigma. The above selection rules apply only for the Electric Dipole (E1) approximation. Contents[show] Table of sum-class symbols Using sum TeX is smart enough to only show. As before, let's think about writing out these sums explicitly, without sigma notation. If we like, we can go back to calling our summation index k,. 1: Summation Notation And Formulas. The variable is called the index of the sum. (a) Write the sum X10 k=4 (2k+ 1) in expanded form. in summation notation. Now back to series. For example: This means that we are to repeatedly add ka k. So for this series, there is 7 terms The seven terms are: when n=0 -2x. Rules - note that # means a non-zero digit (123456789) 1. The numbers at the top and bottom. A key to Business Process Management, it visually depicts a detailed sequence of business activities and information flows needed to complete a process. Note: you can do this more than once on a chart. If I took it out of scientific notation, it would be 800000. In calculus, summation notation or sigma (\u03a3) represents adding many values together. Sigma is the greek sign for \"Sum\". It indicates a series. Today you can define mental math in various different ways. We do this through a mixture of our automated migration tools and professional services offerings. That is, we split the interval x 2[a;b] into n increments of size. The act or process of adding; addition. and in terms of the sigma notation When two random variables are independent, so that. For example, the Schrodinger equation, which has to do with dynamics in quantum systems and predates quantum computation by decades, is written using bra-ket notation. This, too, turns out to come down to some other properties of addition. Given any region R, then the area, A(R) is a real number; \u00c2 \u00c2 \u00c2 \u00c2 \u00c2 A(R) \u00c2\u00b3 0. (c) The above two parts show that a sum can be written in notation in di erent ways. In other words, your're adding up a series of a values: a 1, a 2, a 3 \u2026a x. If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. sp 3 HYBRIDIZATION - EXAMPLE 1) Methane (CH 4) * During the formation of methane molecule, the carbon atom undergoes sp 3 hybridization in the excited state by mixing one \u20182s\u2019 and three 2p orbitals to furnish four half filled sp 3 hybrid orbitals, which are oriented in tetrahedral symmetry in space around the carbon atom. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. using interval notation to show intervals of increasing and decreasing and postive and negative Answers \u00b7 1 trouble spots for the domain may occur where the denominator is ? or where the expression under a square root symbol is negative. Now apply Rule 1 to the first summation and Rule 2 to the second summation. Now, just add all these terms together and set them equal. Okay, welcome back everyone. As well as providing shorthand for mathematical ideas, this notation can aid students' understanding of mathematics. This is called Sigma notation. Practice: Summation notation. In the following, $$k$$ is the index. Identities involving double sums include the following: is the floor function, and. de, the Punycode value actually used for the domain names on the wire is xn--bcher-kva. We first give proof of the Halting Problem by constructing a machine capable of solving the Halting Problem and then showing that if such a machine existed it would generate a. For example, Xn i=1 axi = ax1 +ax2 + +axn = a(x1 +x2 + +xn) = a Xn i=1 xi: In other words, you can take a constant \\out of the summation\". On a single line calculator, a -1 requires the button order be: 1 -. 1, we often encountered sums with many terms (up to 1000 in Table 5. 1 For a Scalar Equation All subscripts must be dealt with in each term. Sigma notation is a method used to write out a long sum in a concise way. These are equal to the value of the variable, 'm' in this case, taken in order. From the rules above, Common series - Sigma notation. Okay, welcome back everyone. Sum and Product Notation: Delimited Forms The capital Greek letter \u03a3 (sigma) denotes summation. AS-Level Maths Questions by topic. What is the sum of all integers between 17 and 325 that are NOT divisible by either. O (2 n) denotes an algorithm whose growth doubles with each addition to the input data set. This exercise formally explores the Riemann sum and practices sigma notation. Homework Statement I'm actually asked to calculate the area under the curve 5x + x 2 over the interval [0,1] using Riemann Sums. ' As such, the expression refers to the sum of all the terms, x n where n represents the values from 1 to k. 2 Sigma Notation & Limits of Finite Sums NOTES SIGMA NOTATION a k =a 1 +a 2 +a 3 +!+a n\u22121 +a n k=1 n \u2211 EX 1) Complete the table, given the following sums in sigma notation: The Sum in Sigma Notation The Sum Written Out, One Term for Each Value of k The Value of the Sum k k=1 5 \u2211 (\u22121) k k k=1 3 \u2211 k k=1 k+1 2 \u2211 k2 k=4 k\u22121 5 \u2211 EX. Examples of process maps: Service Desk. This is the currently selected item. i is the index of summation. What we are about to do is to take a function and express it as the limit of a sequence of Riemann Sums over. It indicates that you must sum the expression to the right of the summation symbol:. Note that index i can be replaced by any other index and the results will be the same. The double series. Sigma Notation - Terms and Indices Sigma notation allows for a great deal of variety and exibility Ex: Write down the sum of the rst 10 odd integers in 3 di erent ways. Many statistical formulas involve adding a series of numbers. Determine the sum of the series written in Sigma Notation. lets say, we have a boolean function F defined on two variables A and B. 5100 x 3300 px. Hi, I need to calculate the following sigma: n=14 Sigma (sqrt(1-2. Search this site. RULES OF INDICES (three videos) Rules of Indices. Write out the sum 5 1 2 k k = \u2211 2. Now back to series. Topics include: ~Set theory, including Venn diagrams ~Properties of the real number line ~Interval notation and algebra with inequalities ~Uses for summation and Sigma notation ~Math on the Cartesian (x,y) plane, slope and distance formulas ~Graphing and describing functions and their inverses on the x-y plane, ~The concept of instantaneous. 1 1 Section 4. They have the following general form XN i=1 x i In the above expression, the i is the summation index, 1 is the start value, N is the stop value. This exercise formally explores the Riemann sum and practices sigma notation. To express: Left end Ch. Math Summation Notation. and in terms of the sigma notation When two random variables are independent, so that. That is, we split the interval x 2[a;b] into n increments of size. \" - Lorena Colin. The Greek capital sigma, P, is used as a shorthand to denote summation. can be evaluated by interchanging and and averaging, (Borwein et al. Sigma has just announced the 'fp,' the world's smallest full-frame camera and, despite owning the sensor design company Foveon, it's built around a conventional Bayer sensor. If you want to know message flow usage, please see How does BPMN message flow work? article. When preparing for technical interviews in the past, I found myself spending hours crawling the internet putting together the best, average, and worst case complexities for search and sorting algorithms so that I wouldn't be stumped when asked about them. Its 2 line display allows \u201cvisually perfect\u201d entries; that is, you can type in the problem exactly as it looks. Multiplication by a whole number can be interpreted as successive addition. Free algebra and math word problems. In this case, the \u2211 symbol is the Greek capital letter, Sigma, that corresponds to the letter 'S', and denotes to the first letter in the word 'Sum. Go back to the example I keep using: the stylesheet knows the \"frameset\" rules. improve this answer. By using this website, you agree to our Cookie Policy. With a Six Sigma process even a significant shift in the process mean results in very few defects. The set can be defined, where possible, by describing the elements. The numbers at the top and bottom of the sigma are called upper and lower bounds, respectively. - Notice that we are adding fractions with a numerator of 1 and. What is the sum of all integers between 17 and 325 that are NOT divisible by either. }\\) Sigma notation relies on an index and expresses the terms of the sum as expression of the index. Calculations on the TI-30XIIS The TI-30XIIS calculator costs less than$20. Sigma notation Here is another useful way of representing a series. Math 132 Sigma Notation Stewart x4. Sigma Notation - Simplification Rules 7:24. They are useful in getting right to the point without delay or distraction:. The \"numbers\" to be summed may be natural numbers, complex numbers, matrices, or still more complicated objects. The first time we write it, we put k = 1. Businesses use the EPC for process mapping in BPMN 2. There are two ways to use invNorm (. as desired. I need to calculate other 18 different sigmas, so if you could give me a solution in general form it would be even easier. n an 1 4 2 \u221212 3 36. The matrix pictured below has two rows and three columns. So sum of n=1 to 20 for (3^n) + sum of n=1 to 20 for (2n-1) Then apply the geometric sum formula for the sum of n=1 to 20 for (3^n). The double series. Gateways are used to control how the process flows. If this series can converge conditionally; for example, converges conditionally if , and absolutely for. It's often convenient to discuss ^{grammatical rules} by formulating them in a special notation that was introduced about 1960 by John ^{Backus} and Peter ^{Naur}. The Greek capital sigma, P, is used as a shorthand to denote summation. Daniel Egger. In the case of IDN B\u00fccher. Even Numbers (Integers) Odd Numbers (Integers) Divisibility Rules. from numpy import mean. What is the sum of all integers between 30 and 809 that are NOT divisible by either the number 5 or the number 13? You must use Sigma notation and the rules discussed in class to solve this problem and you must show. And internal here to the thing inside the sigma notation was an expression 3i squared that had a constant in it, namely 3. Sigma is an open standard for rules that allow you to describe searches on log data in generic form. , parameters); and lower-case letters refer to sample attributes (i. Regarding the sigma notation, it seems to have been included in a tentative draft of Nomenclature of Organic Chemistry, dating from 1973. 2in} 0 \\le p 1; \\sigma > 0 \\). Contents [ hide] 2 Big Omega (\u03a9). Welcome to Treblis Software's Music Notation Reference Guide. If you want to know message flow usage, please see How does BPMN message flow work? article. And S stands for S um. Einstein Summation. i is the index of summation. Sigma notation is a method used to write out a long sum in a concise way. If you update to the most recent version of this activity, then your current progress on this activity will be erased. EXAMPLE 1 Examples of Sigma Notation a. An n-dimensional vector eld is described by a one-to-one correspondence between n-numbers and a point. For example, we can write which is a bit tedious. Value Stream Map. 1) Rule one states that if you're summing a constant from i=1 to n, the sum is equal to the constant multiplied by n. The rules for Mastery Quizzes are now different. Sum of first n natural numbers. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. As we learn the basic integration rules, we will learn the difference between an indefinite integral and a definite integral. What is the sum of all integers between 30 and 809 that are NOT divisible by either the number 5 or the number 13? You must use Sigma notation and the rules discussed in class to solve this problem and you must show. There are formulae that can allow us to calculate the sum. An ellipsis (plural: ellipses) is a punctuation mark consisting of three dots. }\\) Sigma notation relies on an index and expresses the terms of the sum as expression of the index. TechCalc scientific calculator contains 12 calculation modes in one application + a handy reference section that includes an additional 22 calculation modes plus a wealth of useful information. So for this series, there is 7 terms The seven terms are: when n=0 -2x. If this series can converge conditionally; for example, converges conditionally if , and absolutely for. where \u03bc is the location parameter and \u03c3 is the scale parameter. 2 Rules of summation We will prove three rules of summation. Find the value of. 5*k/36)) k=1 Basically, I need to find a sum of square-roots where in each individual squareroot the k-value will be substituted by an integer from 1 to 14. They have the following general form XN i=1 x i In the above expression, the i is the summation index, 1 is the start value, N is the stop value. That's good. There are formulae that can allow us to calculate the sum. The terminology from AMS-LaTeX documentation. Sigma notation provides a way to compactly and precisely express any sum, that is, a sequence of things that are all to be added together. A summation is simply the act or process of adding. This guide is intended for users with basic chemistry knowledge and is not written to be the sole source of instruction for this topic. Summation formulas. \u201d For example, let\u2019s say you had 5 items in a data set: 1,2,5,7,9; you can think of these as x-values. The \"i = 1\" at the bottom indicates that the summation is to start with X 1 and the 4 at the top indicates that the summation will end with X 4. Sigma notation uses a variable that counts upward to change the terms in the list. However, using parentheses makes. Go back to the example I keep using: the stylesheet knows the \"frameset\" rules. If 5, 6,\u2026, = \u00e1are real numbers, then \u00cd = \u00dc \u00e1 \u00dc @ 5 L = 5 E = 6\u22ef E = \u00e1 RULES FOR WORKING WITH SIGMA NOTATION Theorem: If is any constant, then a \u00cd ? \u00e1 \u00dc @ 5 L J\u22c5 ? b \u00cd\u22c5 = \u00dc \u00e1 \u00dc @ 5 L ?\u22c5 \u00cd = \u00dc. sigma notation, also known as summation notation. Summation Notation. MDL number MFCD00013592. Zeros after a # do not count unless they are also after a decimal place 4. If your instructor gave you a class key, use it to enroll yourself and create your account. 4-sigma quality requires addition of a 4 th rule and implementation of a 1 3s /2 2s /R 4s /4 1s multirole, preferably with 4 control measurements in each run (N=4, R=1), or alternatively, 2 control measurements in each of 2 runs. While often associated exclusively with Six Sigma, Centric leverages DMAIC (Define, Measure, Analyze, Improve, Control) as the underlying business process improvement methodology. The greek letter sigma (\u03c3) is used to indicate syllable, and a bracket with a subscript sigma is used for syllable boundaries. 1, we de ne the integral R b a f(x)dx as a limit of approximations. In chemistry, a molecular orbital (MO) is a mathematical function describing the wave-like behavior of an electron in a molecule. The Crescent is the official publication and voice of Phi Beta Sigma Fraternity, Inc. 00, but has features that will be used throughout algebra, trigonometry, and statistics. The index of summation , here the letter i, is a dummy variable whose value will change as the addends of the sum change. Use the sum of rectangular areas to approximate the area under a curve. Only this variable may occur in the product term. Often mathematical formulae require the addition of many variables Summation or sigma notation is a convenient and simple form of shorthand used to give a concise expression for a sum of the values of a variable. Joining a fraternity or sorority is an opportunity to serve the community at UB and beyond. The Leibniz formula for the determinant of a 2 \u00d7 2 matrix is | | = \u2212. Calculus Definitions>. That is indicated by the lower index of the letter. Get access risk-free. Three theorems. no i want it to be solve using sigma notation with its rules of sequence and series. sigma notation, also known as summation notation. However, the lower bound doesn't have to be 1. - Dave xml-dev: A list for W3C XML Developers. Geometric Progression, Series & Sums Introduction. 1 - In the following exercises, use the rules for sums Ch. Six Sigma is an intangible work process. Proficiency with Summation Notation also known as Sigma Notation is a very important skill usually taught in PreCalculus and also in Calculus 1 and 2 (AB and BC). Binomial theorem refresher. Rule 1: If c is a constant, then n i=1 cx i = c n i=1 x i. Write out completely the sequences given by the following rules: (a) a i = i2, where 0 i 5. Rules of Scientific Notation. Okay, now that we've seen that fancy sigma notation let's work a really simple small example of numbers and then generalize. adding & subtracting. The first lesson, \"Sets and What They're Good For,\" walks you through the basic notions of. The Example shows (at least for the special case where one random variable takes only. Here, we prefer to set up the prior in terms of nu, mu, sigma/(nu-2) or something like that, to account for the fact that the scale of the distribution (as measured by the sd or median absolute deviation) depends on nu as well as sigma. Ex B) Ex => => C) Where c is a constant,. For example d/dx (x^2) will graph the derivative of x^2 with respect to x. The image below shows the common notation for conditional probability. It makes more sense if you convert to scientific notation first, so 12. Added Apr 14, 2011 by HighOPS in Mathematics. The double series. Riemann sums, summation notation, and definite integral notation. In the following, $$k$$ is the index. Sum of first n natural numbers. If this value of SSR is equal to the sum of squares total, it means our regression model captures all the. Look at the picture below to see an example. The expression is read as the sum of 4n as n goes from 1 to 6. To aid the investigation, we introduce a new quantity, the Euler phi function, written \u03d5(n), for positive integers n. This web page describes how symbols are used on the Stat Trek web site to represent numbers, variables, parameters, statistics, etc. Notation 4 We say a Turing machine halts, if, in the course of processing its input data, the Turing machine enters into the halt state and has no further transitions. Summation Notation Worksheet 1 Introduction Sigma notation is used as a convenient shorthand notation for the summation of terms. Sigma Notation: Inequalities: Factorial Notation: Exponential Functions: Logarithmic Functions: Simultaneous Equations: Plotting Coordinates: Straight Line Graphs: Quadratic Graphs: Exponential Graphs: Logarithmic and Square Root Graphs: Differentiation: Integration: MathQuiz 4. Sigma notation mc-TY-sigma-2009-1 Sigma notation is a method used to write out a long sum in a concise way. The growth curve of an O (2 n) function is exponential - starting off very shallow, then rising meteorically. ? Using Sigma notation and related rules, compute the sum of all the integers between 21 and 126 that are not divisible by 4. Constant multiplier. Determine the sum of the series written in Sigma Notation. This calculus video tutorial provides a basic introduction into summation formulas and sigma notation. But polynomials truly come to shine in the work of Al-Samawal (1130-1180) The Shining Book on Calculation. Summation Notation Rules & Examples Video & Lesson Transcript 12. Sigma notation is a method used to write out a long sum in a concise way. Learn how it is used in this video. \" If the sums do not converge, the series is said to diverge. So for this series, there is 7 terms The seven terms are: when n=0 -2x. For example, cosx, cos x, and cos (x) are all equivalent. Use algebra rules of sums, along with the appropriate sum formulae, to evaluate the following: (a) X12 i=1 (i2 33i) (b) X10 i=1 (i2 2 1) (c) X8 i=1 (2 i ) 6. A typical sum written in sigma notation looks like this: 4 k 0 (k2 3) The symbol \"\u03a3\" is the Greek capital letter sigma, which stands for \"sum\". Loading Unsubscribe from Jochumzen? Summation Formulas and Sigma Notation - Calculus - Duration: 20:24. In the following, $$k$$ is the index. Rules of Scientific Notation. The Sigma symbol, , is a capital letter in the Greek alphabet. To use summation, you can find sigma in the Desmos keyboard (under FUNCTIONS and then misc) or by typing \"sum\": If you populate the upper and lower bound, Desmos will output the summation answer. But, more than just being small, it's one of the most radical cameras we've seen released in years, incorporating an array of new ideas and video capabilities. 5100 x 3300 px. The symbol sigma is a Greek letter that stands for \u2018the sum of\u2019. Binomial theorem refresher. because it does not matter what we call our index. Kronecker Delta Multiplication The Kronecker Delta is nicknamed the substitution operator because of the following simple property of multiplication, best explained by example. The user is asked to select true. I thought about extending these ideas to product notation (we haven't learned it yet, but it's simple), and I figured out a few rules. [1] The name sigmatropic is the result of a compounding of the long-established \"sigma\" name for single carbon -carbon bonds and the Greek word tropos , meaning turn. There are some rules that can help simplify or evaluate series. That is, we split the interval x 2[a;b] into n increments of size. Exponents and Integers. 2 University of Sydney School of Mathematics and Statistics. A business process model and notation diagram, or BPMN diagram for short, is used to build easy-to-read business process model flowcharts, which can be shared across organizations and industries. Matrix and Index Notation David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139. Regardless, your record of completion wil. The most general \u03a3 \\Sigma \u03a3 notation describes the sum of some unknown number of elements. For each section in Chapter 7 you will have exactly two opportunities to take the quiz for that section, and both of those attempts need to be completed in Canvas by the end-of-week deadline. In this unit we look at ways of using sigma notation, and establish some useful rules. A series that diverges means either the partial sums have no limit or approach infinity. The terminology from AMS-LaTeX documentation. \"My TI 83+ graphing calculator arrived faster than I expected. Go to The Equation of a Circle page. Viewing 8 posts - 1 through 8 (of 8 total) Author. The Westgard system is based on the principles of statistical process control used in manufacturing nationwide since the 1950s. Basic math formulas. 1 Steve Strand and Sean Larsen from Portland State University, US, have shown that, cognitively, the task of interpreting a given summation-notation expression differs significantly from the task of converting a longhand sum into summation notation. Sigma notation - Rules. This module contains three lessons that are build to basic math vocabulary. The example shows us how to write a sum of even numbers. If you're seeing this message, it means we're having trouble loading external resources on our website. 2 Rules of Roots. Basic Rule #4 of MO Theory Rule #4: If the AOs are non-degenerate, their interaction is. In particular, it is a symmetric group of prime degree and symmetric group of prime power degree. \u201d - Sincerely, Dr. Learners are taught how to use sigma notation and how to develop sigma notation. ' As such, the expression refers to the sum of all the terms, x n where n represents the values from 1 to k. Zeros after a # do not count unless they are also after a decimal place 4. It is, therefore, useful to have a set of rules for finding the equivalent resistance of some general arrangement of resistors. sp 3 HYBRIDIZATION - EXAMPLE 1) Methane (CH 4) * During the formation of methane molecule, the carbon atom undergoes sp 3 hybridization in the excited state by mixing one \u20182s\u2019 and three 2p orbitals to furnish four half filled sp 3 hybrid orbitals, which are oriented in tetrahedral symmetry in space around the carbon atom. The notation for adding a series of numbers is the capital Greek letter sigma. But there is a limit on how far you can upgrade the bases. As well as providing shorthand for mathematical ideas, this notation can aid students' understanding of mathematics. TechCalc scientific calculator contains 12 calculation modes in one application + a handy reference section that includes an additional 22 calculation modes plus a wealth of useful information. It is called Sigma notation because the symbol is the Greek capital letter sigma: \u03a3. They're called partial sums because you're only able to find the sum of a certain number of terms \u2014 no infinite series [\u2026]. After many searches I couldn't find a single page containing summation formulas for polynomials of order greater than 4. Joining a fraternity or sorority is an opportunity to serve the community at UB and beyond. Select a link by mouse clicking one of the music notation symbols below. It looks like this: $\\sum_{j=1}^n \\sum_{k=1}^n f(j, k)$ The idea behind this is you're doing a sum within a sum, and both indices will be inside the inner sum. T HIS \u2014\u03a3\u2014is the Greek letter sigma. Last video we did some elementary examples of sigma notation. Use the sum of rectangular areas to approximate the area under a curve. PART 1: INTRODUCTION TO TENSOR CALCULUS A scalar eld describes a one-to-one correspondence between a single scalar number and a point. One way to compactly represent a series is with sigma notation, or summation notation, which looks like this: $\\displaystyle{\\sum _{n=3}^{7}{n^2}}$ The main symbol seen is the uppercase Greek letter sigma. Calculate the Product. Summation rules Jochumzen. To change a number from scientific notation to standard form, move the decimal point to the left (if the exponent of ten is a negative number), or to the right (if the exponent is positive). It doesn't have to be \"i\": it could be any variable (j ,k, x etc. The Organic Chemistry Tutor 270,467 views. For example, assuming k \u2264 n. Advertisement. Executive in Residence and Director, Center for Quantitative Modeling. They are useful in getting right to the point without delay or distraction:. Mulliken (1896-1986, awarded with the Nobel prize in 1966) in the early 1930s. The sums are heading towards a value (1 in this case), so this series is convergent. A series can be represented in a compact form, called summation or sigma notation. This is true for all tensor notation operations, not just this matrix dot product. Its 2 line display allows \u201cvisually perfect\u201d entries; that is, you can type in the problem exactly as it looks. Note that these formulae only work if we start from 1; we will see how to calculate summations. pdf Math 180 Worksheets W20 20 Sigma Notation And. 1:2s rather than 1 2s) Combinations of rules are generally indicated by using a \"slash\" mark (/) between control rules, e. The following rule can be used to convert numbers into scientific notation: The exponent in scientific notation is equal to the number of times the decimal point must be moved to produce a number between 1 and 10. Adding series. SIGMA NOTATION FOR SUMS. where E is the sigma sign. Due to linear algebra being all about finding the solutions to systems of linear equations, matrix math and the study of vector spaces become a tool to represent and orderly solve. The ONLY way to show three SF is by using scientific notation. All very clear and understandable for anyone, through the use of BPMN 2.", "date": "2020-05-30 23:52:14", "meta": {"domain": "amicidellacattolica.it", "url": "http://amicidellacattolica.it/rmev/sigma-notation-rules.html", "openwebmath_score": 0.8859736919403076, "openwebmath_perplexity": 745.3356280199723, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850832642354, "lm_q2_score": 0.8791467801752451, "lm_q1q2_score": 0.8642760855900652}}
{"url": "http://crowebrothers.com/now-we-eqz/bc6841-how-to-find-orthogonal-matrix", "text": "# how to find orthogonal matrix\n\n\u2022 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange In fact, given any \u2026 The interactive program below is designed to answers the question whether the given input matrix is an orthogonal matrix. That is, each row has length one, and are mutually perpendicular. Damit ist die Inverse einer orthogonalen Matrix gleichzeitig ihre Transponierte. Orthogonal matrix is an important matrix in linear algebra, it is also widely used in machine learning. A matrix is orthogonal if the If Q is an orthogonal matrix, then, |Q| = \u00b11. Previous Solution: Example 1. The eigenvalues of the orthogonal matrix will always be $$\\pm{1}$$. The matrix in problem statement (not step one) is for the previous problem. An orthogonal matrix is the real specialization of a unitary matrix, and thus always a normal matrix.Although we consider only real matrices here, the definition can be used for matrices with entries from any field.However, orthogonal matrices arise naturally from dot products, and for matrices of complex numbers that leads instead to the unitary requirement. The concept of orthogonality for a matrix is defined for just one matrix: A matrix is orthogonal if each of its column vectors is orthogonal to all other column vectors and has norm 1. transpose The orthogonal projection matrix is also detailed and many examples are given. Example using orthogonal change-of-basis matrix to find transformation matrix. Let W be a subspace of R n and let x be a vector in R n. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Es ist o\ufb00ensichtlich, dass Q orthogonal ist, da die beiden Spaltenvektoren orthogonal sind. If the result is an identity matrix, then the input matrix is an orthogonal matrix. Let's say I've got me a set of vectors. If matrix Q has n rows then it is an orthogonal matrix (as vectors q1, q2, q3, \u2026, qn are assumed to be orthonormal earlier) Properties of Orthogonal Matrix. is equal to its Some important properties of orthogonal matrix are, See also Orthogonal Matrix (1) The Definition of The Orthogonal Basis. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Orthogonal Complements and Projections Recall that two vectors in are perpendicular or orthogonal provided that their dot product vanishes. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. When you click Random Example button, it will create random input matrix to provide you with many examples of both orthogonal and non-orthogonal matrices. is an orthogonal matrix. Performance & security by Cloudflare, Please complete the security check to access. So let's say vector w is equal to some linear combination of these vectors right here. I need to find an orthogonal matrix Q, so that when applying M_2 = Q M_1 Q^-1 the matrix M_2 does not contain any values at the zero positions of P. The other way is possible, M_2 may contain a zero, where P is one. Proof Part(a):) If T is orthogonal, then, by de\ufb01nition, the An n \u00a3 n matrix A is orthogonal i\ufb00 its columns form an orthonormal basis of Rn. If Q is square, then QTQ = I tells us that QT= Q\u22121. Comment(8) Anonymous. One way to think about a 3x3 orthogonal matrix is, instead of a 3x3 array of scalars, as 3 vectors. , Gram-Schmidt process example. Die Matrix ist also orthogonal, weil die Multiplikation der Matrix mit der transponierten Matrix die Einheitsmatrix ergibt. The concept of orthogonality for a matrix is defined for just one matrix: A matrix is orthogonal if each of its column vectors is orthogonal to all other column vectors and has norm 1. This is true because d vectors will always be sufficient be needed to span a d-dimensional vector space. Q\u22c5QT = E Q \u22c5 Q T = E Die Determinante einer orthogonalem Matrix nimmt entweder den Wert +1 oder -1 an. Well, if you're orthogonal to all of these members, all of these rows in your matrix, you're also orthogonal to any linear combination of them. Cloudflare Ray ID: 60a7cf86683fdfbf of the , Diese Matrix beschreibt eine Drehung um den Winkel \u2212\u03b8. Index Next lesson. Orthogonal Matrix Example. Daf\u00fcr musst du zun\u00e4chst die transponierte Matrix berechnen und diese dann mit multiplizieren. Define a matrix and find the rank. Let W be a subspace of R4 with a basis {[1011],[0111]}. You can also try to input your own matrix to test whether it is an orthogonal matrix or not. Find the inverse matrix of \u2026 concatenation Similarly, the columns are also an orthonormal basis. Thus, matrix is an orthogonal matrix. How to fill in a matrix given diagonal and off-diagonal elements in r? Eigen vectors Thus, matrix Therefore, the value of determinant for orthogonal matrix will be either +1 or -1. 2. In this tutorial, we will dicuss what it is and how to create a random orthogonal matrix with pyhton. | The vectors in are orthogonal while are not. Suppose we have a set of vectors {q1, q2, \u2026, qn}, which is orthogonal if, then this basis is called an orthogonal basis. Next spectral decomposition, Rate this tutorial or give your comments about this tutorial, The row vector and the column vector of matrix, Both Hermitian and Unitary matrix (including. inverse Suppose that is an orthogonal basis for the column space of . . Eigen-everything. From introductory exercise problems to linear algebra exam problems from various universities. Orthogonal matrices preserve angles and lengths. . A linear transformation T from Rn to Rn is orthogonal i\ufb00 the vectors T(e~1), T(e~2),:::,T(e~n) form an orthonormal basis of Rn. Then we multiply the transpose with given matrix. Pictures: orthogonal decomposition, orthogonal projection. Find an orthonormal basis of W. (The Ohio State University, Linear Algebra Midterm) Add to solve later Sponsored Links Please enable Cookies and reload the page. Es gilt detQ = cos2 \u03d5 +sin2 \u03d5 = 1. (3) Your answer is P = P ~u i~uT i. 2. If we try the orth trick, it will produce an array of size d by d, thus a SQUARE matrix. This can be generalized and extended to 'n' dimensions as described in group theory. This covers about orthogonal matrix Its definition and properties. Singular Value Decomposition 7 Finding stationary distribution of a markov process given a transition probability matrix are orthogonal matrices. | Your IP: 78.47.248.67 The concept of two matrices being orthogonal is not defined. That is, if and only if . : Note that this is an n n matrix, we are multiplying a column vector by a row vector instead of the other way around. symmetric A = [1 0 1;-1 -2 0; 0 1 -1]; r = rank(A) r = 3 Since A is a square matrix of full rank, the orthonormal basis calculated by orth(A) matches the matrix U calculated in the singular value decomposition, [U,S] = svd(A,'econ'). Spiegelung. The objective is to find an orthogonal basis for the column space of the following matrix: Use Gram-Schmidt Process to find an orthogonal basis for the column space of segregate the columns of the matrix as . How to find an orthogonal matrix? \u2022 Fact 5.3.3 Orthogonal transformations and orthonormal bases a. Orthogonal matrix multiplication can be used to represent rotation, there is an equivalence with quaternion multiplication as described here. Let given square matrix is A. Title: Finding the Nearest Orthonormal Matrix Author: Berthold K.P. , that is Orthogonal matrix is important in many applications because of its properties. Orthogonale Matrizen k\u00a8onnen auch Spiegelungen an Geraden beschreiben. orthogonal vector We study orthogonal transformations and orthogonal matrices. Another way to prevent getting this page in the future is to use Privacy Pass. Demzufolge gilt Q\u22121 = QT = cos\u03d5 sin\u03d5 \u2212sin\u03d5 cos\u03d5 . b. To test whether a matrix is an orthogonal matrix, we multiply the matrix to its transpose. Finally we check if the matrix obtained is identity or not. You may need to download version 2.0 now from the Chrome Web Store. Eine orthogonale Matrix ist in der linearen Algebra eine quadratische, reelle Matrix, deren Zeilen- und Spaltenvektoren orthonormal bez\u00fcglich des Standardskalarprodukts sind. The 1/0 indicate where values are allowed in the result matrix. You can imagine, let's say that we have some vector that is a linear combination of these guys right here. Basic to advanced level. Horn Subject: Painful Way to Solve Photogrammetric Problems Keywords: Orthonormal matrix, Rotation, Photogrammetry, Least Squares Fitting, Projective Geometry, Matrix Square Root, Two step \u2026 A square orthonormal matrix Q is called an orthogonal matrix. To create random orthogonal matrix as in the interactive program below, I created random Example: Prove Q = $$\\begin{bmatrix} cosZ & sinZ \\\\ -sinZ & cosZ\\\\ \\end{bmatrix}$$ is orthogonal matrix. What is Orthogonal Matrix? Let us see an example of the orthogonal matrix. So let me call my set B. Let. Vocabulary words: orthogonal decomposition, orthogonal projection. To nd the matrix of the orthogonal projection onto V, the way we rst discussed, takes three steps: (1) Find a basis ~v 1, ~v 2, ..., ~v m for V. (2) Turn the basis ~v i into an orthonormal basis ~u i, using the Gram-Schmidt algorithm. Calculate and verify the orthonormal basis vectors for the range of a full rank matrix. Basis vectors. Simple Solution : The idea is simple, we first find transpose of matrix. Video transcript. Also given a symmetric prototype matrix P, containing ones and zeroes. If n>d, regardless of the size, as long as n>d, we can never find a set of n vectors in a d-dimensional space that are orthogonal. A matrix can be tested to see if it is orthogonal using the Wolfram Language code: OrthogonalMatrixQ[m_List?MatrixQ] := (Transpose[m].m == IdentityMatrix @ Length @ m) The rows of an orthogonal matrix are an orthonormal basis. If, it is 1 then, matrix A may be the orthogonal matrix. Problems of Orthogonal Bases. 0 0 1 0 1 0 For example, if Q =1 0 then QT=0 0 1. The concept of two matrices being orthogonal is not defined. >. Um eine orthogonale Matrix bestimmen zu k\u00f6nnen, \u00fcberpr\u00fcfst du die Formel von oben. Recipes: orthogonal projection onto a line, orthogonal decomposition by solving a system of equations, orthogonal projection via a complicated matrix product. To test whether a matrix is an orthogonal matrix, we multiply the matrix to its transpose. matrix and compute the modal matrix from < To check for its orthogonality steps are: Find the determinant of A. If a matrix A is an orthogonal matrix, it shoud be n*n. The feature of an orthogonal matrix A. Gram-Schmidt example with 3 basis vectors. We can define an inner product on the vector space of all polynomials of degree at most 3 by setting. If the result is an identity matrix, then the input matrix is an orthogonal matrix. Overview. The Gram-Schmidt process. 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A line, orthogonal decomposition by solving a system of equations, orthogonal matrix, then the input is!\n\nFiled under: News. Bookmark the permalink.", "date": "2021-06-12 18:37:39", "meta": {"domain": "crowebrothers.com", "url": "http://crowebrothers.com/now-we-eqz/bc6841-how-to-find-orthogonal-matrix", "openwebmath_score": 0.7539582252502441, "openwebmath_perplexity": 1214.8395199556069, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9724147201714923, "lm_q2_score": 0.8887587949656841, "lm_q1q2_score": 0.8642421349065085}}
{"url": "https://math.stackexchange.com/questions/4032890/winding-number-of-a-curve-not-complex-analysis", "text": "Winding number of a curve (not complex analysis)\n\nI am asked to calculate the winding number of an ellipse (it's clearly 1 but I need to calculate it)\n\nI tried two different aproaches but none seems to work.\n\nI would like to know why none of them work (I believe it is because these formulas only work if I have a curve parametrized by arc lenght).\n\nApproach 1:\n\nA valid parametrization : $$\\gamma=(a\\cos t,b\\sin t)$$, with $$t \\in [0,2\\pi], \\, a,b \\in \\mathbb{R}$$\n\n$$\\dot{\\gamma}(t)=(-a\\sin t,b\\cos t)$$, with $$t \\in [0,2\\pi], \\, a,b \\in \\mathbb{R}$$\n\n$$\\ddot{\\gamma}(t)=(-a\\cos t,-b\\sin t)$$, with $$t \\in [0,2\\pi], \\, a,b \\in \\mathbb{R}$$\n\n$$\\det(\\dot{\\gamma}(t)|\\ddot{\\gamma}(t)) = \\renewcommand\\arraystretch{1.2}\\begin{vmatrix} -a\\sin t & -a\\cos t \\\\ b\\cos t & -b\\sin t \\end{vmatrix}=ab \\sin^2 t+ab \\cos^2 t=ab$$\n\n$$||\\dot{\\gamma}(t)||^3=(\\displaystyle\\sqrt{(-a\\sin t)^2+(b\\cos t)^2})^3=(\\displaystyle\\sqrt{a^2\\sin^2 t+b^2\\cos^2 t})^3=a^3b^3$$\n\n$$\\kappa(t)=\\displaystyle\\frac{ab}{a^3b^3}=\\displaystyle\\frac{1}{a^2b^2}$$\n\n$$\\mathcal{K}_\\gamma = \\displaystyle\\int_{0}^{2\\pi} \\displaystyle\\frac{1}{a^2b^2} \\ dt= \\displaystyle\\frac{2\\pi}{a^2b^2}$$, $$\\mathcal{K}_\\gamma$$ is the total curvature of the curve.\n\n$$i_\\gamma=\\displaystyle\\frac{\\displaystyle\\frac{2\\pi}{a^2b^2}}{2\\pi}=\\displaystyle\\frac{1}{a^2b^2}$$...which is not necessarily 1.\n\nApproach 2:\n\nWinding # = $$\\displaystyle\\frac{1}{2\\pi}\\displaystyle\\int_{\\gamma}\\displaystyle\\frac{-y}{x^2+y^2}\\>dx+\\displaystyle\\frac{x}{x^2+y^2}\\>dy$$\n\nThat gives us $$\\displaystyle\\frac{1}{2\\pi}\\displaystyle\\int_{0}^{2\\pi}\\left( \\displaystyle\\frac{-b\\sin t}{a^2\\cos^2 t+b^2\\sin^2 t}(-a\\sin t)+\\displaystyle\\frac{a\\cos t}{a^2\\cos^2 t+b^2\\sin^2 t}(b\\cos t) \\right)\\>dt$$\n\n$$\\displaystyle\\frac{1}{2\\pi}\\displaystyle\\int_{0}^{2\\pi}\\left( \\displaystyle\\frac{ab}{a^2\\cos^2 t+b^2\\sin^2 t }\\right)\\>dt$$, which I computed and cannot be calculated.\n\nClearly the second approach is valid if we are dealing with a circumference of radius 1. We can generalize for the elipsee using Green's Theorem. I would also like if someone could show me this way as well.\n\nThank you\n\n\u2022 If you use a polar coordinate system, and parametrise the ellipse using $\\theta(t)$ and $r(t)$ for $t_0 \\le t \\le t_1$, then the winding number is by definition $$\\frac{\\theta(t_1) - \\theta(t_0)}{2 \\pi}$$ \u2013\u00a0Gl\u00e4rbo Feb 20 at 11:55\n\u2022 Thank you for your help. I had indeed read that as well, but I really can't seem to know what to do in order to end up with $\\theta (t)$ and $r (t)$. Maybe this is a bit obvious, but I'm really not being able to see any further. Any hints? \u2013\u00a0hugh_maths Feb 20 at 13:01\n\nThe last integral indeed has a closed form to evaluate you can exploit the symmetry about $$\\pi$$ to get the integral as \\begin{align} \\pi I &= ab \\int_{0}^{\\pi} \\dfrac{1}{a^2 \\cos^2(t) + b^2 \\sin^2(t)}dt\\\\ &= 2ab\\int_{0}^{\\pi/2} \\dfrac{\\sec^2(t)}{a^2+b^2 \\tan^2(t)} dt \\end{align} Now substituite $$\\tan(t) = u$$ $$\\dfrac{\\pi}{2ab} I = \\int_{0}^{\\infty}\\dfrac{1}{a^2+b^2x^2}dx = \\dfrac{\\pi}{2ab}$$ thus giving $$I =1$$\n\n\u2022 Thank you for your kind answer. Shouldn't this be equal to 1 since this is the winding number and its been known to be 1? \u2013\u00a0hugh_maths Feb 20 at 12:57\n\u2022 it is 1 the lhs is a constant times the integral, does this answer your question @hugh_maths \u2013\u00a0Aditya Dwivedi Feb 20 at 12:58\n\u2022 oh yes, of course, I'm sorry! Oh so we should be done ! 1 is indeed the winding number! Thank you so much! Any thoughts on why my first approach is wrong (or seems to be) ? \u2013\u00a0hugh_maths Feb 20 at 13:03\n\nGreen's Theorem will tell you that the winding number (given by approach #2 \u2014 your total curvature in #1 is incorrect because you need an arclength integral) of the ellipse is the same as the winding number of any circle (which is easy to compute).\n\nLet $$P=-\\dfrac y{x^2+y^2}$$ and $$Q=\\dfrac x{x^2+y^2}$$. Note that on $$\\Bbb R^2-\\{0,0\\}$$, we have $$\\dfrac{\\partial P}{\\partial y} = \\dfrac{\\partial Q}{\\partial x}$$. Thus, if $$R$$ is the region between the ellipse $$E$$ and a circle $$C$$ centered at the origin lying inside it, we have $$\\int_E P\\,dx+Q\\,dy - \\int_C P\\,dx + Q\\,dy = \\iint_R \\left(\\frac{\\partial Q}{\\partial x} - \\frac{\\partial P}{\\partial y}\\right)\\,dA = 0.$$ But you can easily calculate that $$\\int_C P\\,dx + Q\\,dy = 2\\pi$$.\n\n\u2022 Thank you so much!@ Aditya Dwivedi helped me and I was able to evaluate the integral. As for the approach # one, I believe the total curvature divided by $2\\pi$ gives us the index, which is the same as the winding number. Or they differ ? Cause I do not understand why the result comes up different than 1. \u2013\u00a0hugh_maths Feb 20 at 19:02\n\u2022 Yeah, for a convex curve $C$, it is the case that $\\int_C \\kappa\\,ds = \\pm 2\\pi$ (hard theorem). But you have to integrate with respect to arclength, not $dt$. \u2013\u00a0Ted Shifrin Feb 20 at 19:25\n\u2022 Additional comment: The curvature of the ellipse is not constant, so you'd better double-check your computations! You made a glaring error there. \u2013\u00a0Ted Shifrin Feb 20 at 19:34", "date": "2021-06-21 15:54:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4032890/winding-number-of-a-curve-not-complex-analysis", "openwebmath_score": 0.9710761308670044, "openwebmath_perplexity": 220.6253364112749, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147201714922, "lm_q2_score": 0.8887587920192298, "lm_q1q2_score": 0.8642421320413328}}
{"url": "http://rzby.effebitrezzano.it/4-4-matrix-latex.html", "text": "# 4 4 Matrix Latex\n\nThis theory. 1 command to change font size. For asymptote code embedded in a latex document, just use the \"PdfLaTex+Asymptote+PdfLaTeX+View Pdf\" mode for the quick build command. Its main purpose is to display mathematical formulas written in LaTeX. LaTeX Tutorial pt 6 - Tables and Matrices in LaTeX How to create a matrix larger than 3x3. matrix creates a matrix from the given set of values. You can use decimal (finite and periodic) fractions: 1/3, 3. Three Styles for LaTeX Vector Notation filed in LaTeX , Math on Jun. Beamer Tutorial Outline Outline 1 About Beamer 2 Templates 3 Frames 4 Sections and Subsections 5 Text 6 Alignment and Spacing 7 Lists 8 Overlays 9 Tables 10 Frame Structures 11 Graphics 12 Themes. With the encouragement and support of Eitan's family, Michal Hoftich, Karl Berry, and others have continued to work on TeX4ht. Suppose A is the 4 x 4 matrix. Specialty Products. I always forget how to do this. Similar Matrices and Diagonalizable Matrices S. Suppose I have a 6 by 6 matrix where the elements vary from 1:36 in increment of 1. Sage provides standard constructions from linear algebra, e. To initialize an N-by-M matrix, use the \u201czeros\u201d function. Want a Mac app? Lucky you. subcaption A useful extension is the subcaption package (the subfigure and subfig packages are deprecated and shouldn't be used any more), which uses subfloats within a single float. Also, you can add a multiple of one row to another row without changing the determinant. One of the greatest motivating forces for Donald Knuth when he began developing the original TeX system was to create something that allowed simple construction of mathematical formulae, while looking professional when printed. \\ vdots for vertical dots. I can't figure out how to use \"Equation Tools\" to enter a matrix with dimensions of my choosing. How do we calculate the dimension of a matrix? Is it the number of entries? Or is it the number of different entries? For instance if I have a matrix 2x2 the dimension would be 4 but if the matrix is simetrical it would be 3. Drag Sales to Values. If there are symbols missing drop me a line or create a pull-request. Examples and questions on matrices along with their solutions are presented. This is the 16th video in a series of 21 by Dr Vincent Knight of Cardiff University. MatrixForm acts as a \"wrapper\", which affects printing, but not evaluation. I need to combine them into one 10x10 matrix with n1 in the top left corner, n2 in the top right, n3 in the bottom left and n4 in the bottom right, and I am using Mathematica 4. Symmetrical monomers such as ethylene and tetrafluoroethylene can join together in only one way. Yes, but there will always be the same number of pivots in the same columns, no matter how you reduce it, as long as it is in row echelon form. Just array to make the matrix and insert a vertical bar between the columns where you want a vertical bar. ELEMENTARY MATRICES 43 Remark 106 To actually create the matrix which performs (R j +mR i) $(R j), we do not need to perform the same operation on the identity matrix. When this is done to a matrix in echelon form, it remains in echelon form. 5 option in the \\includegraphics command to resize the image to 150% of its original size. where I denotes a unit matrix of order n. Linear Algebra/Zero Matrices and Zero Vectors/ From Wikibooks, open books for an open world < Linear Algebra. Matrix algebra for beginners, Part I matrices, determinants, inverses Jeremy Gunawardena Department of Systems Biology Harvard Medical School 200 Longwood Avenue, Cambridge, MA 02115, USA [email protected] August 2012 by tom 55 Comments. An m \u00d7 n (read 'm by n') matrix is an arrangement of numbers (or algebraic expressions ) in m rows and n columns. Matrices with Examples and Questions with Solutions. 4 Fonts Font sizes Point size Latex cmd User-de ned * Sample 5 6 \\tiny \\xxxsmall the quick brown fox 7 8 \\scriptsize \\xxsmall the quick brown fox 8 10 \\footnotesize \\xsmall the quick brown fox. An online LaTeX editor that's easy to use. All the basic matrix operations as well as methods for solving systems of simultaneous linear equations are implemented on this site. ELEMENTARY MATRICES 43 Remark 106 To actually create the matrix which performs (R j +mR i)$ (R j), we do not need to perform the same operation on the identity matrix. The inverse of a $2\\times2$ matrix is given by swapping the diagonal entries, negating the off-diagonal entries, and dividing by the determinant: $$\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}^{-1} = \\frac{1}{ad-bc} \\begin{pmatrix}d & -b \\\\ -c & a\\end{pmatrix}$$. Word 2016 automatically handles formatting, nonetheless, you can manually adjust the spacing and alignment of equations. To add rows and/or columns, right click on your matrix 6. 1234567890 abcdefghijklmnopqrstuvwxyz. #N#square root (\"radical\") Symbols in Geometry Symbols in Algebra. We can associate a matrix with each graph storing some of the information about the graph in that matrix. Our matrix and vector calculator is the most sophisticated and comprehensive matrix calculator online. The matrix $B$ is the inverse of the matrix $A$ if when multiplied together, $A\\cdot B$ or [latex. To find the det(B), I multiplied B 14 by det(B 14 ) and B 24 by det(B 24 ) and followed the + - + - pattern as showed by the formula here (scroll below for 4x4 formula). While in Spain she was cast in a regular role in the TV show Dark Justice which was produced in Barcelona for its first season. Justify each answer. - Thomas Oct 5 '13 at 8:23. Calendar library 5. eigenvectors_left \u00b6. Just array to make the matrix and insert a vertical bar between the columns where you want a vertical bar. Sage provides standard constructions from linear algebra, e. \\ vdots for vertical dots. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. This does seem silly. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58. We introduce a very small part of the language for writing mathematical notation. my table and figure. to get a color as mix of 4 colors such as CMYK). Together these facts mean that the upper triangular matrices form a subalgebra of the associative algebra of square matrices for a given size. Natural rubber is an elastomer and a thermoplastic. The defining property for the gamma matrices to generate a Clifford algebra is the anticommutation relation {,} = + =,where {,} is the anticommutator, is the Minkowski metric with signature (+ \u2212 \u2212 \u2212), and is the 4 \u00d7 4 identity matrix. The majority of L A TEX packages has a manual which describes how to use it and usually gives examples. The video explains the concepts with hands on. For example, create a 3-by-5 matrix of zeros:. You can think of these two points as \"attractor points\". Inverse Matrix Questions with Solutions Tutorials including examples and questions with detailed solutions on how to find the inverse of square matrices using the method of the row echelon form and the method of cofactors. VerkhovtsevaKatya. Most rubber in everyday use is vulcanized to a point where it shares properties of both; i. Let A be the following matrix LaTeX: \\begin{bmatrix}5&-6&-12\\\\ 14 & 4&7 \\end{bmatrix} [ 5 \u22126 \u221212 14 4 7 ] The dimensions of A are Expert Answer 100% (1 rating). Formatting Optimization Problems with LaTeX. This means that it is the set of the n-tuples of real numbers (sequences of n real numbers). Reference. There are many identity matrices. Transparency 20. In a double-blind, placebo-controlled, randomized study of 187 women receiving Climara 0. js as an input. Important: We can only multiply matrices if the number of columns in the first matrix is the same as the number of rows in the second matrix. MatrixForm acts as a \"wrapper\", which affects printing, but not evaluation. Note: This page is a work-in-progress. 00 Only 2 left. Welcome! This is the documentation for Numpy and Scipy. So our strategy will be to try to find the eigenvector with X = 1, and then if. How to add an equation in your document, see Working with Microsoft Equation. INPUT: The matrix command takes the entries of a matrix, optionally preceded by a ring and the dimensions of the matrix, and returns a matrix. Similarly, the other matrix is of the order 4 \u00d7 3, thus the number of elements present will be 12 i. Definir el tama\u00f1o y m\u00e1rgenes del papel. tikz-pgf positioning matrices bordermatrix. The eigenspace corresponding to an eigenvalue \u03bb of A is defined to be E\u03bb = {x \u2208 Cn \u2223 Ax = \u03bbx}. Ideal for low compression situations, the polyester/cotton weave provides excellent elasticity and recoverability. Read honest and unbiased product reviews from our users. \u53ef\u4ee5\u751f\u6210\u5d4c\u5165\u5230HTML\u7f51\u7ad9\u3001\u8bba\u575b\u6216\u535a\u5ba2\u7684\u516c\u5f0f\u4ee3\u7801. ISBN -8176-3805-9 (acid-free paper) (pbk. Please join the discussion on moodle. 1 or any later version published by the Free Software. Right click the matrix and select Matrix Spacing\u2026 3. The Matrix 4 will be in the same universe as John Wick Lionsgate This one is (surprisingly) way too detailed to outline in full here (it involves the creation of a bunch of Zion Accords to \"govern. Making statements based on opinion; back them up with references or personal experience. We can encode this transformation in a 4 x 4 matrix by putting A in the top left with three 0's below it and making the last column be (b,1). To define dots in Latex, use: \\ ldots for horizontal dots on the line. Texmaker can also be used to edit and compile directly asymptote figure not embedded in a latex document. Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. Multiplying the 4-vector (v,1) with this matrix will give you (Av + b, 1). radio_button_uncheckedNo. Imagine turning a paper to the other side, that is what is happening. LaTeX minipage. Definir el tama\u00f1o y m\u00e1rgenes del papel. Paquete layouts. But the resulting 2 matrices are S4 object of class. 4-18 are: v v v 23 4= =16 V, 8 V and 6 V= Determine the values of the following: (a) The gain, A, of the VCVS (b) The resistance R 5 (c) The currents i b and i c (d) The power received by resistor R 4. 25]{image}} \\begin{table}[h!] \\begin{tabular}{| p{2cm} | p{2cm} | p{2cm} | p{2cm. This ends up being a cubic equation, but just looking at it here we see one of the roots is 2 (because of 2\u2212\u03bb), and the part inside the square brackets is Quadratic, with roots of \u22121 and 8. Welcome! This is the documentation for Numpy and Scipy. Multiplication of Matrices. Most rubber in everyday use is vulcanized to a point where it shares properties of both; i. 4 5! 2 9! = 53! Putting it all together 3 7 4 5! 2 9! = 69 53! Example Find 2 4 5 3. Define Vector and Matrix Data. 620 MazdaRX4Wag 21. multiplication dot. Ideal for low compression situations, the polyester/cotton weave provides excellent elasticity and recov. Yes, but there will always be the same number of pivots in the same columns, no matter how you reduce it, as long as it is in row echelon form. Want a Mac app? Lucky you. Very Basic Mathematical Latex A document in the article style might be entered between the nmaketitle and nendfdocumentgsections below. Our matrix and vector calculator is the most sophisticated and comprehensive matrix calculator online. They'll be productive from day one and be able to pick up small amounts of LaTeX as they go. For LaTex symbols, check the website LaTeX:Symbols (Art of Problem Solving). Steps 5 and 6 specify the data to display in the matrix data cells. Definir el tama\u00f1o y m\u00e1rgenes del papel. Use commas or spaces to separate values in one matrix row and semicolon or new line to separate different matrix rows. Note that the inverse [B]-1 always has the same number of rows and columns as the original matrix [B]. Matrix Representations of Linear Transformations and Changes of Coordinates 0. 0 6 160 110 3. ISBN -8176-3805-9 (acid-free paper) (pbk. It is easy to define a matrix of values in Octave. a $2 \\times 2$ matrix whose elements are $2 \\times 2$ matrices) and then apply \"Block Inversion\" : have a look at this article. MatrixForm prints SparseArray objects like the corresponding ordinary lists. Creation of matrices and matrix multiplication is easy and natural: Note that in Sage, the kernel of a matrix A is the \"left kernel\", i. She followed the series to its new. TeXstudio is a fully featured LaTeX editor. The size of the matrix is determined automatically, so it is not necessary to explicitly state the dimensions. So the Eigenvalues are \u22121, 2 and 8. The simplest classifiers, called binary classifiers, has only two classes: positive/negative, yes/no, male/female \u2026 Performance of a binary classifier is summarized in a confusion matrix that cross-tabulates predicted and observed examples into four options:. Row and column labelling of matrixes in LaTeX. 3 Reference guide. Definition If A is an m n matrix, with columns a1,a2, ,an, and if x is in Rn, then the product of A and x, denoted by Ax,isthelinear combination of the columns of A using the corresponding. Sage provides standard constructions from linear algebra, e. Let A be the following matrix LaTeX: \\begin{bmatrix}5&-6&-12\\\\ 14 & 4&7 \\end{bmatrix} [ 5 \u22126 \u221212 14 4 7 ] The dimensions of A are Expert Answer 100% (1 rating). a system of linear equations with inequality constraints. Carrie Anne Moss loves to play her so much. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. 3 Beamer presentation. This post will tell you which one is the best. I have seen is countless discussion threads and videos to use Equation Editor 3. In mathematics, a matrix (plural matrices) is a rectangular array (see irregular matrix) of numbers, symbols, or expressions, arranged in rows and columns. 3 Single Equations that are Too Long: multline If an equation is too long, we have to wrap it somehow. The previous example was the 3 \u00d7 3 identity; this is the 4 \u00d7 4 identity: The 3 \u00d7 3 identity is denoted by I3 (pronounced as \"eye-three\" or \"eye. matrix without brackets. See Snip in action\u2014watch a demo video! Take screenshots of individual equations, paragraphs, and even full pages of text! Digitize any math or text on your screen in seconds. The position of the elements are specified with pos, which is either NULL, or a vector specifying the number of elements on a row, or a 2-columned matrix specifying the (x,y) position of. This ends up being a cubic equation, but just looking at it here we see one of the roots is 2 (because of 2\u2212\u03bb), and the part inside the square brackets is Quadratic, with roots of \u22121 and 8. LATEX Command Summary This listing contains short descriptions of the control sequences that are likely to be handy for users of LAT EX v2. It is free and easy to use. An online LaTeX editor that's easy to use. Is T one\u2013to\u2013one? Theorem Let T: n m be a linear transformation and let A be the standard matrix for T. Changing the font size in LaTeX can be done on two levels, either affecting the whole document or parts/elements of it. 6 Implications of nullity being zero. We can associate a matrix with each graph storing some of the information about the graph in that matrix. Click on \"Insert\" and choose what you want to add. You can put this solution on YOUR website! How do I write systems of equations in matrix form? That's one of the easiest things you'll ever learn: Suppose you have this system: 4x + 7y = 1 x - y = -8 Look at the red numbers: 4x + 7y = 1 1x - 1y = -8 Erase the letters: 4 + 7 = 1 1 - 1 = -8 Erase the + and bring the - over nearer the 1: 4 7 = 1 1 -1 = -8 Replace the \" = \" signs with a vertical. The mathematical symbol is produced using \\partial. This site is supported by donations to The OEIS Foundation. CHANGES IN VERSION 1. LaTeX for Advanced View Examples These are provided if you'd like to copy and paste the LaTeX from the examples into the Advanced View of the Equation Editor to tinker with the examples. 320 Hornet4Drive 21. Mathematical Formulas. This function and x / y are equivalent. MATH 316U (003) - 1. 1 Fill the rest of the line. mat\") where readMat is from the R. Creating a Payoff Matrix Using LaTeX Tabular Environment 291. Analogous sets of gamma matrices can be defined in any dimension and for any signature of the metric. You can add, subtract, multiply and transpose matrices. Specifically the cofactor of the $(i,j)$ entry of a matrix, also known as the $(i,j)$ cofactor of that matrix, is the signed minor of that entry. The context menu does not have an option to insert additional rows or columns. More details. Augmented Matrix Calculator is a free online tool that displays the resultant variable value of an augmented matrix for the two matrices. Coordinate systems 8. Choose a type a matrix 5. , Manufacturer: Medline, Medline Item Number: MDS087004LFZ, Medline MDS087004LFZ Bandage, Elastic, Matrix, 4\" x 5 yd. Invest in comfortable, restful sleep for your family with mattresses that suit individual sleeping styles and preferred levels of firmness. 4 tips for writing matrix questions Sarah Cho 2 min read A matrix question\ufeff \u2014or really, multiple questions presented on a grid\u2014is one of the most popular question types in online and traditional pen-and-paper surveys. Use the scale=1. X n converges to Xin distribution, written X n X, if lim n!1 F n(t) = F(t) (7) at all tfor which Fis continuous. The X-Y Matrix is a Six Sigma tool mostly used during the DMAIC measure phase and the DMADV measure phase to show the relationship between X and Y factors. We introduce a very small part of the language for writing mathematical notation. Only a limited part of the full TeX language is supported; see below for details. Then write the LaTex code \\cancel{A}. 4 FROM VERSION 1. If we try to multiply an n\u00d71 matrix with another n\u00d71 matrix, this product is not defined. America's #1 selling latex caulk. Enumerate is used to make numbered lists. Set the matrix (must be square). \\\\imath and\\\\jmath make \"dotless\" i and j, which. We now go on to solve. Please, be aware that the support for loading tables from an existing LaTeX code is severely limited and may work erroneously or may not work at all. Matrix is our finest elastic bandage. 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Single formulas must be seperated with two backslashes \\\\ Use the matrix environment to typeset matrices. 0 Sage is free, open-source math software that supports research and teaching in algebra, geometry, number theory, cryptography, numerical computation, and related areas. You know, there is a stack exchange site for latex questions - user1228 Jan 10 '11 at 15:00 This question appears to be off-topic because it is about latex. \\ vdots for vertical dots. Plotting in Scilab www. If neither A nor B is an identity matrix, AB \u2260 BA. 10 01 12 34 = 12 34 12 34 10 01 = 12 34 Such matrices A and B are said to commute. There are three commonly used environments in the math mode: the math environment: Used for formulas in running text ; the displaymath environment: Used to display longer formulas ; the equation environment: Used for displaying equations for numbering and cross reference. #N#square root (\"radical\") Symbols in Geometry Symbols in Algebra. The process by which the rank of a matrix is determined can be illustrated by the following example. Click the \"Insert\" tab 2. 2 on Mepis Antix MX 14. dem autogenerated by webify. 2013;4:204173141350530. Textext aims to cover this need, by adding re-editable LaTeX objects to Inkscape\u2019s repertoire. Here is a matrix of size 2 3 (\"2 by 3\"), because it has 2 rows and 3 columns: 10 2 015 The matrix consists of 6 entries or elements. 2 \u03c0\u22123 4+x 8 \u22121. In a double-blind, placebo-controlled, randomized study of 187 women receiving Climara 0. November 11, 2009 October 15, 2017 James 21 Comments. I tried computing conductance (stiffness) matrix in the physical coordinate systems and comparing the answer with isoparametric system. \u22123 4 7 0 + 6 \u22121. In order for a matrix $B$ to be the inverse of $A$, the equation $AB = I$ and $BA. 4 Fonts Font sizes Point size Latex cmd User-de ned * Sample 5 6 \\tiny \\xxxsmall the quick brown fox 7 8 \\scriptsize \\xxsmall the quick brown fox 8 10 \\footnotesize \\xsmall the quick brown fox. I am looking to get a colormap to assign a unique color to each point in a plot based on their four Ci values (i. (diag(A)) ij= ijA ij eig(A) Eigenvalues of the matrix A vec(A) The vector-version of the matrix A (see Sec. LaTeX Lesson 4 Mathematics in LaTeX. 4[U] 14 Matrix expressions. GroupWork 1: Mark each statement True or False. The theory was that the Real World was not actually real, but rather another level of the Matrix simulation: i. 6 Rubber/Stretching lengths. An extension to amsmath matrix environments. Please feel free to let us know if you have any question. The Matrix Behind The Scenes - Training Injuries (1999) - Keanu Reeves Movie HD - Duration: 1:33. 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LaTeX is a typesetting language for producing scientific documents. In linear algebra, the cofactor (sometimes called adjunct) describes a particular construction that is useful for calculating both the determinant and inverse of square matrices. [code]\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} [/code]produces $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix}$ [code]\\det\\begin{pmatrix} a & b \\\\ c. calculate expression inside first. Matrices with Examples and Questions with Solutions. 1 Matrix without names, everything set to NULL. We provide the highest standards of large format printing and installation services (including Vutek 5m 4 machines,Flora 3. A diagonal matrix whose non-zero entries are all 1 's is called an \" identity \" matrix, for reasons which will become clear when you learn how to multiply matrices. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear. In order for a matrix [latex]B$ to be the inverse of $A$, the equation $AB = I$ and [latex]BA. Latex Free. latex matrices First the basic environments which could be used for a matrix, all of them are provide by usepackage amsmath. 3 (Dot Prod. Learn from detailed step-by-step explanations Get walked through each step of the solution to know exactly what path gets you to the right answer. The prefix arc used for inverse circular trigonometric functions is the. Example 1: Find the inverse of. This is called the Falk's scheme. Infection Prevention. Most LaTeX commands are available including:. Click on \"Matrix\" 4. This site is supported by donations to The OEIS Foundation. 7 , B= 3 \u22123 12 0 dimension (or size) always given as (numbers of) rows \u00d7 columns \u2022 Ais a 3\u00d74matrix, B is 2\u00d72 \u2022 the matrix Ahas four columns; B has two rows m\u00d7nmatrix is called square if m=n, fat if m Can be used to label the rows of the % resulting latex table. Matrix Elastic Bandages are our finest elastic bandages. I can't figure out how to use \"Equation Tools\" to enter a matrix with dimensions of my choosing. It's amazing, even though it's a bit slow, the idea of converting latex to an image is extremely useful and versatile :D. LaTeX is a very flexible program for typesetting math, but sometimes figuring out how to get the effect you want can be tricky. Also, you can add a multiple of one row to another row without changing the determinant. Polyester/cotton weave provides excellent elasticity, recoverability and breathability; Color-coded by size for quick identification; Features self-closure that keeps bandage wrapped and secure without the need for clips. This add-on lets you instantly convert every math equation in your document into beautiful latex images! Auto-Latex Equations Lab - Aayush Gupta. 3 bronze badges. com Knowledge base dedicated to Linux and applied mathematics. Subscripts & Superscripts. by interchanging the location of two rows, or interchanging the location of two columns. You can modify individual elements or perform arithmetic on entire vectors and matrices. Its syntax is much simpler than that of estout and, by default, it produces publication-style tables that display nicely in Stata's results window. 95) Also include 1 x 5V 4A (4000mA) switching power supply - UL Listed (\\$ 14. For example \\bigg[\\begin{array}{c|c|c|c}. 4 tips for writing matrix questions Sarah Cho 2 min read A matrix question\ufeff \u2014or really, multiple questions presented on a grid\u2014is one of the most popular question types in online and traditional pen-and-paper surveys. 4 Enumerate. Matrix Chain Multiplication (A O(N^2) Solution) Printing brackets in Matrix Chain Multiplication Problem. LAPACK addresses this problem by reorganizing the algorithms to use block matrix operations, such as matrix multiplication, in the innermost loops. Jump to navigation Jump to search. The following example shows you how to get a matrix into reduced row echelon form using elementary row operations. LaTeX Lesson 4 Mathematics in LaTeX. Does someone know how to convert an object of class dgCmatrix to a regular matrix. But nowadays there are many examples in which TEX is used for publications with no mathematical or technical background content. It is especially useful to write mathematical notation such as equations and formulae. An m \u00d7 n (read 'm by n') matrix is an arrangement of numbers (or algebraic expressions ) in m rows and n columns. produces the graphic $$a^x_n$$. Restriction: In addition to the LaTeX command the unlicensed version will copy a reminder to purchase a license to the clipboard when you select a symbol. Numerical Linear Algebra with Applications 19 :3, 607-619. GNU Scienti\ufb01c Library Release 2. For example: \\documentclass{article} \\usepackage[pdftex]{graphicx} \\begin{document} This is my first image. Tutorial o curso para escribir matrices dos formas, usando el entorno array o usando entornos ya definidos, matrix, pmatrix, vmatrix, bmatrix, en LaTeX utilizando Texmaker. Representing Systems of Linear Equations using Matrices A system of linear equations can be represented in matrix form using a coefficient matrix, a variable matrix, and a constant matrix. Some natural rubber sources, such as gutta-percha, are composed of trans-1,4-polyisoprene, a structural isomer that has similar properties. Maths for Physics University of Birmingham Mathematics Support Centre Authors: Daniel Brett Joseph Vovrosh Supervisors: Michael Grove Joe Kyle October 2015. The previous example was the 3 \u00d7 3 identity; this is the 4 \u00d7 4 identity: The 3 \u00d7 3 identity is denoted by I3 (pronounced as \"eye-three\" or \"eye. The goal of this library is being an easy, but extensible interface between Python and LaTeX. The new row will be inserted immediately below the position of your cursor. I need to combine them into one 10x10 matrix with n1 in the top left corner, n2 in the top right, n3 in the bottom left and n4 in the bottom right, and I am using Mathematica 4. Note that any matrix of the form \u22122d a d a 0 b d b c does the job. When you write a matrix whose entries are fractions, you might feel the line are cramped. So, for example: It will not surprise you that: A + (-A) = 0 (Notice that that last zero is a bold-faced zero, designating the zero matrix. An m \u00d7 n (read 'm by n') matrix is an arrangement of numbers (or algebraic expressions ) in m rows and n columns. Subscripts & Superscripts. For example if you multiply a matrix of 'n' x. 0 Sage is free, open-source math software that supports research and teaching in algebra, geometry, number theory, cryptography, numerical computation, and related areas. But nowadays there are many examples in which TEX is used for publications with no mathematical or technical background content. The theory was that the Real World was not actually real, but rather another level of the Matrix simulation: i. HTML LaTeX \u516c\u5f0f\u7f16\u8f91\u5668\u53ef\u4ee5\u751f\u6210\u56fe\u5f62\u516c\u5f0f (gif, png, swf, pdf, emf). #N#square root (\"radical\") Symbols in Geometry Symbols in Algebra. How to multiply a Row by a Column? We'll start by showing how to multiply a 1 \u00d7 n matrix by an n \u00d7 1 matrix. \\begingroup I have 4 5x5 matrices n1, n2, n3, n4. You can nest enumerations to create detailed outlines. Enumerate is used to make numbered lists. 1 of the official documentation. Ideal for low compression situations, the polyester/cotton weave provides excellent elasticity and recov. I am making it simpler by considering for a 6 by 6 matrix. A CONSORT-style flowchart of a randomized controlled trial [Open in Overleaf] BER measurement on fibre optical system. input : output \\begin{matrix} a_1 & a_2 & a_3 \\\\ b_1 & b_2 & b_3 \\\\ c_1 & c_2 & c_3 \\end{matrix}  matrix with parentheses brackets. Download Snip for desktop and start saving time. The LaTeX team cannot guarantee that TeX distributions, even recent ones, contain the most recent version of LaTeX. All the basic matrix operations as well as methods for solving systems of simultaneous linear equations are implemented on this site. Row and column labelling of matrixes in LaTeX. LaTeX Lesson 4 Mathematics in LaTeX. When you write a matrix whose entries are fractions, you might feel the line are cramped. So, the command first. There are no approved revisions of this page, so it may not have been reviewed. Creation of matrices and matrix multiplication is easy and natural: Note that in Sage, the kernel of a matrix A is the \"left kernel\", i. It makes sense that the number of prey present will affect the number of the predator. Please join the discussion on moodle. where I denotes a unit matrix of order n. Fractions and Roots.  What's going on here? LaTeX Wiki contents Using LaTeX is a user's guide for LaTeX Using LaTeX to format documents tells you how to format a complete document using LaTeX Using LaTeX in a wiki describes the subset of LaTeX LaTeX language reference Document classes. Imagine turning a paper to the other side, that is what is happening. 95 per month (cancel anytime). Here are examples: \u03a3 = \u23a1 \u23a2 \u23a2\u23a3\u03c311 \u22ef \u03c31n \u22ee \u22f1 \u22ee \u03c3n1 \u22ef \u03c3nn\u23a4 \u23a5 \u23a5\u23a6 \u03c3 11 \u22ef \u03c3 1 n \u22ee \u22f1 \u22ee \u03c3 n 1 \u22ef \u03c3 n n. radio_button_uncheckedNo. There are different way of placing figures side by side in Latex, subcaption, subfig, subfigure, or even minipage. Use environments matrix for an array without brackets, and pmatrix environment for an array with parentheses. An example of a piecewise step function. X n converges to cin distribution, written X n c, if lim n!1 F n(t) = c(t) (8) at all t6=cwhere c(t) = 0 if t < fran\u00e7ais > Here are few examples to write quickly matrices. LaTeX2e in 90 minutes, by Tobias Oetiker, Hubert Partl, Irene Hyna, and Elisabeth Schlegl. LaTeX4technics. Gauss's Law, Faraday's Law, the non-existance of magnetic charge, and Ampere's Law are described in an intuitive method, with a focus on understanding above mathematics. tikz-pgf vertical-alignment tikz-matrix. LaTeX has different packages which automatically generates dummy text in our document. It is most often used for medium-to-large technical or scientific documents but it can be used for almost any form of publishing. 4 ! In the process of developing the Digital Library of Mathematical Functions , we needed a means of transforming the LaTeX sources of our material into XML which would be used for further manipulations, rearrangements and construction. The transposition is exactly the same for a non-square matrix. Let A be the following matrix LaTeX: \\begin{bmatrix}5&-6&-12\\\\ 14 & 4&7 \\end{bmatrix} [ 5 \u22126 \u221212 14 4 7 ] The dimensions of A are Expert Answer 100% (1 rating). 26, 2012 Most of the stock math commands are written for typesetting math or computer science papers for academic journals, so you might need to dig deeper into LaTeX commands to get the vector notation styles that are common in physics textbooks and articles. Product Services. No preview available. But the resulting 2 matrices are S4 object of class. 4 Enumerate. Cured caulk is mold and mildew resistant. A zero vector. Texmaker can also be used to edit and compile directly asymptote figure not embedded in a latex document. Latex Free. As\u00ed como cambiando el. MatrixForm prints SparseArray objects like the corresponding ordinary lists. The length of the skip should be expressed in a unit recognized by LaTeX. Here is a matrix of size 2 3 (\"2 by 3\"), because it has 2 rows and 3 columns: 10 2 015 The matrix consists of 6 entries or elements. To initialize an N-by-M matrix, use the \u201czeros\u201d function. #README This repository contains the source files for Fundamentals of Matrix Algebra, by Gregory Hartman. To calculate inverse matrix you need to do the following steps. Produce beautiful documents starting from our gallery of LaTeX templates for journals, conferences, theses, reports, CVs and much more. Textext aims to cover this need, by adding re-editable LaTeX objects to Inkscape\u2019s repertoire. This scaled our figure by a factor of 0. 3 7! 2 9! = 69! To obtain the second entry in the solution, ignore the \ufb01rst row of the \ufb01rst matrix. There are no approved revisions of this page, so it may not have been reviewed. Check that the two matrices can be multiplied together. That ends our list of the 10 best LaTex Editors that you should be using in 2020. is a square matrix. So to widen the gap between lines, use \\\\[6pt]. 1Introduction This document gives a gallery of tables which can be made using the xtable package to create LATEX output. This does seem silly. The inverse of a 2\\times2 matrix is given by swapping the diagonal entries, negating the off-diagonal entries, and dividing by the determinant:\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}^{-1} = \\frac{1}{ad-bc} \\begin{pmatrix}d & -b \\\\ -c & a\\end{pmatrix}. LaTeX Tutorial pt 6 - Tables and Matrices in LaTeX How to create a matrix larger than 3x3. The column space of a matrix A is the vector space made up of all linear combi\u00ad nations of the columns of A. 4 He has reported that the amount of free latex protein that can be extracted from powdered latex gloves is consistently higher than the amount that is liberated from non-powdered gloves. Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. opagebreak[4] tells LaTeX not to break the page between the problem and the choices unless it really, really, really has to. Most rubber in everyday use is vulcanized to a point where it shares properties of both; i. Feedback and suggested improvements are welcome. My first equation is now (4). If a graph has vertices, we may associate an matrix which is called vertex matrix or adjacency matrix. External file 3. For example, to obtain in LaTeX, one would type. Eitan died unexpectedly in June 2009; we extend our sympathies to his family, and dedicate future work on the project to his memory. But nowadays there are many examples in which TEX is used for publications with no mathematical or technical background content. I have checked with a matrix calculator and the the determinants of the 3x3 minor matrices are correct. The second row of the original matrix becomes the second column of its transpose. Thus our source is a mixture of text to be typeset and a couple of LATEX commands \\emph and \\LaTeX. Wow, that is a twist. It is most often used for medium-to-large technical or scientific documents but it can be used for almost any form of publishing. Greek letters are commonly used in mathematics, and they are very easy to type in math mode. , then it tells us that all of space has been squished into a smaller dimension, e. That is, a Jordan matrix is a matrix with Jordan blocks down the diagonal and zeros everywhere else. 3, on Ms Windows 7/10 and Mac OS-X 10. Describe the range of a 3 by 4 matrix using the definition of the range. Among the three important vector spaces associated with a matrix of order m x n is the Null Space. Does someone know how to convert an object of class dgCmatrix > to a regular matrix. Steps 3 and 4 organize the values for the fields first by subcategory, and then by product. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. The internal ones 2 and 2 tell you if the multiplication is possible (when they are equal) or not (when they are different). Symmetrical monomers such as ethylene and tetrafluoroethylene can join together in only one way. David Whiting david. Derivation of the stiffness matrix of a two-dimensional element (II) by victorroda Before we can continue with the assembly of the stiffness matrix, and in order to determine the matrix for each Gauss integration point , the element geometry matrix , the local derivative matrix , the jacobian matrix and its determinant, and the global. Return the matrix right division of x and y. LaTeX is a powerful tool to typeset math. Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. Word 2016 automatically handles formatting, nonetheless, you can manually adjust the spacing and alignment of equations. VerkhovtsevaKatya. Click on row or column numbers to toggle whole row or column. Matrix algebra for beginners, Part I matrices, determinants, inverses Jeremy Gunawardena Department of Systems Biology Harvard Medical School 200 Longwood Avenue, Cambridge, MA 02115, USA [email protected] First, use a font like lmodern and then you can set the font size:. Enter the Matrix is the story behind the story. This is not a comprehensive list. We offer a wide selection of UltraClean\u2122 surfactant-free latex microspheres\u2014also referred to as latex beads\u2014for research and commercial applications. Define Vector and Matrix Data. It is also suitable for producing all sorts of other documents, from simple letters to. 3 Null Space as a vector space. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. 4) can be concisely expressed as min m 2 = m 0\u03a3m s. A 4x4 matrix can be used to do both rotation and translation in a single matrix. 17pt have occurred. Email to friends Share on Facebook - opens in a new window or tab Share on Twitter - opens in a new window or tab Share on Pinterest - opens in a new window or tab. For example \\bigg[\\begin{array}{c|c|c|c}. The beamer class User Guide for version 3. One of the greatest motivating forces for Donald Knuth when he began developing the original TeX system was to create something that allowed simple construction of mathematical formulae, while looking professional when printed. Matrices and other arrays are produced in LaTeX using the \\textbf{array} environment. If more arguments are given, the multiplication is. ( 0 0 \u22124 \u22125 0 3 0 0 \u22126) o* (0 3 0 \u22126) giving 0(0\u22120) = 0. \\documentclass{article} \\usepackage[left=2cm, right=5cm, top=2cm]{geometry} \\begin{document} Some text \u2026 left margin is 2 cm, right margin 5 cm and the top margin is 2 cm. TeX4ht was created by Eitan Gurari at Ohio State University. Use the scale=1. Nodes and shapes 15. GroupWork 1: Mark each statement True or False. This keypad has 16 buttons, arranged in a telephone-line 4x4 grid. You must use the following package: \\usepackage {amsmath} \\begin {matrix} \\begin {pmatrix} \\begin {bmatrix} \\begin {vmatrix} \\begin {Vmatrix}. One of the octahedral sites is [ -1/2 1/4 5/6 ]^T , relative to the lattice basis. Just pick a symbol you sometimes need but tend to forget and click it. VerkhovtsevaKatya is a new contributor to this site. \"matrix-like\" diagrams, such as commutative diagrams,. 4 LaTeX default lengths. In the notebook and at the Sage command-line there is a pre-defined object named latex which has several methods, which you can list by typing latex. MatrixForm acts as a \"wrapper\", which affects printing, but not evaluation. There's no need to calculate. False, the vector spaces P3 and R^4 are isomorphic (4. opagebreak[4] tells LaTeX not to break the page between the problem and the choices unless it really, really, really has to. But we need a matrix that is linearly independent to both S1,S2. a = 2 1 3 A row vector is a list of numbers written one after the other, e. To initialize an N-by-M matrix, use the \u201czeros\u201d function. This site is supported by donations to The OEIS Foundation. I can only select from the predefined Matrix sizes. If A is a square matrix, then there exists a matrix B such that AB equals the identity matrix. Overleaf is so easy to get started with that you'll be able to invite your non-LaTeX colleagues to contribute directly to your LaTeX documents. I want to write a 4x4 matrix in MS Word 2007 using equation editor. Use the inverse matrix to solve the system of equations. MediaWiki renders mathematical equations using a combination of html and a variant of LaTeX. An matrix J is said to be in Jordan canonical form if it is a matrix of the form where each is either a diagonal matrix or a Jordan block matrix. The author introduces it in Chapter Four using linear. You weren\u2019t required to compute this,. This add-on lets you automatically convert every LaTeX equation in your document into beautiful images! Simply enclose your math equations within  and click the button in the sidebar, and all. The process by which the rank of a matrix is determined can be illustrated by the following example. 0 Sage is free, open-source math software that supports research and teaching in algebra, geometry, number theory, cryptography, numerical computation, and related areas. Creation of matrices and matrix multiplication is easy and natural: Note that in Sage, the kernel of a matrix A is the \u201cleft kernel\u201d, i. Carrie-Anne Moss was born and raised in Vancouver, Canada. If you know how to change this to I could just have \\bmat{a&b}{c&d} and not even need \\bmatrows{2}, then that would be even better. To find the inverse of matrix A, using Gauss-Jordan elimination, it must be found the sequence of elementary row operations that reduces A to the identity and, then, the same operations on I_n must be performed to obtain A^ {-1}. Click on \"Insert\" and choose what you want to add. By changing the font size locally, however, a single word, a few lines of text, a large table or. This matrix can be used to obtain more detailed information about the graph. We can also scale the picture by explicitly setting the width and the height by using the width and height optional commands. There are three commonly used environments in the math mode: the math environment: Used for formulas in running text ; the displaymath environment: Used to display longer formulas ; the equation environment: Used for displaying equations for numbering and cross reference. calculate expression inside first. Click the \"Insert\" tab 2. Here are examples: \u03a3 = \u23a1 \u23a2 \u23a2\u23a3\u03c311 \u22ef \u03c31n \u22ee \u22f1 \u22ee \u03c3n1 \u22ef \u03c3nn\u23a4 \u23a5 \u23a5\u23a6 \u03c3 11 \u22ef \u03c3 1 n \u22ee \u22f1 \u22ee \u03c3 n 1 \u22ef \u03c3 n n. Some of these commands are LATEX macros, while oth-. Basic equations in L a T e X can be easily \"programmed\", for. A User of Auto-LaTeX Equations. 95) ADD TO WISHLIST. 5 Basis of Null Space. com : Elastic Bandage, Latex Free, Self Closure, 4\"X 5 Yds. No preview available. If output is a data type, it specifies the type of the resulting labeled feature array. This comes from Mathematica (IIRC) and is not otherwise standardized, but there may be some disambiguation benefit in using the double-struck italic letters from Unicode\u2019s Letterlike Symbols block: U+2148 \u2148 and U+2149 \u2149. The root of a quadratic equation however, can be either positive or negative. 5) has the matrix representation \u239b \u239c \u239c \u239c \u239c \u239c \u239c \u239d 2 2 2 2 1 2 2 2 2 1 2 2 2 2 1 11 10 \u239e \u239f \u239f \u239f \u239f \u239f \u239f. This article explains how to insert spaces of different lengths in mathematical mode. I need to combine them into one 10x10 matrix with n1 in the top left corner, n2 in the top right, n3 in the bottom left and n4 in the bottom right, and I am using Mathematica 4. 3 bronze badges. LaTeX is a typesetting language for producing scientific documents. I am looking to get a colormap to assign a unique color to each point in a plot based on their four Ci values (i. There are various ways to adjust the numbering in enumerate (search \u201clatex and enumerate\u201d in Google. My matrix contains fraction elements and single digit integers. There is no common notation for empty matrices, but most computer algebra systems allow creating and computing with them. Export (png, jpg, gif, svg, pdf) and save & share with note system. It doubles as a regression check for the package. The syntax is exactly the same. The mathematical symbol is produced using \\partial. The idea behind the minipage command is that within an existing page \"built in\" an additional page. VerkhovtsevaKatya. Augmented Matrix Calculator is a free online tool that displays the resultant variable value of an augmented matrix for the two matrices. Get the free \"inverse matrix\" widget for your website, blog, Wordpress, Blogger, or iGoogle. I created a table in Latex, but currently it is too big. matrix list V symmetric V[3,3] weight displacement _cons weight 1. Suppose that we have the following diagonal matrix. calculate expression inside first. For example, if [A] is a 4 x 3 matrix (4 rows, 3 columns) and [B] is a 2 x 2 matrix (2 rows, 2. 2013;4:204173141350530. have a solution. You can think of these two points as \"attractor points\". The column space of a matrix A is the vector space made up of all linear combi\u00ad nations of the columns of A. One of the octahedral sites is [ -1/2 1/4 5/6 ]^T , relative to the lattice basis. Use MathJax to format equations. MathJax basic tutorial and quick reference. Table 200: \u2133 Letter-like Symbols. com and remembering the cases redefinition I\u2019ve shown some days ago I got the idea to extend the internal macro \\[email\u00a0protected] of amsmath. They are organized into seven classes based on their role in a mathematical expression. Enumerate is used to make numbered lists. To create a new row, use. Jump to navigation Jump to search. You can nest enumerations to create detailed outlines. 2) sup Supremum of a set jjAjj Matrix norm (subscript if any denotes what norm) AT Transposed matrix A TThe inverse of the transposed and vice versa, A T = (A 1)T = (A ). You can generate them with just a few lines of code. a = 2 1 3 A row vector is a list of numbers written one after the other, e. The defining property for the gamma matrices to generate a Clifford algebra is the anticommutation relation {,} = + =,where {,} is the anticommutator, is the Minkowski metric with signature (+ \u2212 \u2212 \u2212), and is the 4 \u00d7 4 identity matrix. So to convert a 3x3 matrix to a 4x4, you simply copy in the values for the 3x3 upper left block, like so:. edited yesterday. Out [1] //MatrixForm= Show the matrix form of a SparseArray:. Tutorial o curso para escribir matrices dos formas, usando el entorno array o usando entornos ya definidos, matrix, pmatrix, vmatrix, bmatrix, en LaTeX utilizando Texmaker. No preview available.", "date": "2020-07-15 05:49:53", "meta": {"domain": "effebitrezzano.it", "url": "http://rzby.effebitrezzano.it/4-4-matrix-latex.html", "openwebmath_score": 0.7885562777519226, "openwebmath_perplexity": 1173.5944962037668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9553191348157373, "lm_q2_score": 0.9046505376715775, "lm_q1q2_score": 0.864229968959003}}
{"url": "https://gmatclub.com/forum/how-many-zeros-does-100-end-with-100599.html?sort_by_oldest=true", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 06 Dec 2019, 18:46\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\nYour Progress\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# How many zeros does 100! end with?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 22 Jun 2010\nPosts: 36\nHow many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 20 Feb 2019, 03:42\n4\n70\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n84% (00:30) correct 16% (01:08) wrong based on 1878 sessions\n\n### HideShow timer Statistics\n\nHow many zeros does 100! end with?\n\nA. 20\nB. 24\nC. 25\nD. 30\nE. 32\n\nSpoiler: :: Solution\nFind how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by $$25 = 5^2$$ . So, the answer is 24.\nThe correct answer is B.\n\nI have absolutely no Idea what they are telling me...\nCan someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?\n\nThanks!\n\nOriginally posted by AndreG on 07 Sep 2010, 13:41.\nLast edited by Bunuel on 20 Feb 2019, 03:42, edited 1 time in total.\nRenamed the topic and edited the question.\n##### Most Helpful Expert Reply\nMath Expert\nJoined: 02 Sep 2009\nPosts: 59587\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Sep 2010, 13:49\n34\n31\nAndreG wrote:\nHow many zeros does 100! end with?\n\u2022 20\n\u2022 24\n\u2022 25\n\u2022 30\n\u2022 32\n\nexpl.\nFind how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by $$25 = 5^2$$ . So, the answer is 24.\nThe correct answer is B.\n\nI have absolutely no Idea what they are telling me...\nCan someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?\n\nThanks!\n\nTrailing zeros:\nTrailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.\n\nFro example, 125000 has 3 trailing zeros (125000);\n\nThe number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:\n\n$$\\frac{n}{5}+\\frac{n}{5^2}+\\frac{n}{5^3}+...+\\frac{n}{5^k}$$, where k must be chosen such that 5^(k+1)>n\n\nIt's more simple if you look at an example:\n\nHow many zeros are in the end (after which no other digits follow) of 32!?\n$$\\frac{32}{5}+\\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)\n\nSo there are 7 zeros in the end of 32!\n\nThe formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.\n\nBACK TO THE ORIGINAL QUESTION:\n\nAccording to above 100! has $$\\frac{100}{5}+\\frac{100}{25}=20+4=24$$ trailing zeros.\n\nAnswer: B.\n\nFor more on this issues check Factorials and Number Theory links in my signature.\n\nHope it helps.\n_________________\n##### General Discussion\nRetired Moderator\nJoined: 02 Sep 2010\nPosts: 717\nLocation: London\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n18 Oct 2010, 02:40\nferuz77 wrote:\nthe product of all integers from 1 to 100 will have the following number of zeros at the end:\na) 20\nb) 24\nc) 19\nd) 22\ne) 28\n\npls, help with solution method!\n\nSearch through the forums (read the math book). There is several threads discussing this.\n\nThe number of trailing zeros in 100! is (100/5)+(100/25)=24\n\nAnswer : (b)\n_________________\nIntern\nJoined: 24 Jun 2010\nPosts: 11\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n18 Oct 2010, 08:22\nthe question can be re-framed as (100!/10^x) now find x?\n\n100!/(2*5)^x---now factorize 100! by 2 and 5\n\nwhen factorized by 5 will give the least power 24\nAns 24\nDirector\nJoined: 23 Apr 2010\nPosts: 503\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n28 Dec 2010, 05:09\nIs this a relevant GMAT question? Thank you.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 59587\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n28 Dec 2010, 05:22\nnonameee wrote:\nIs this a relevant GMAT question? Thank you.\n\nWhy do you think that it's not? I've seen several GMAT questions testing the trailing zeros concept (everything-about-factorials-on-the-gmat-85592.html):\nfactorial-one-106060.html\nproduct-of-sequence-101187.html\nhow-many-zeroes-at-the-end-of-101752.html\nleast-value-of-n-m09q33-76716.html\nnumber-properties-from-gmatprep-84770.html\n_________________\nManager\nStatus: Student\nJoined: 26 Aug 2013\nPosts: 166\nLocation: France\nConcentration: Finance, General Management\nSchools: EMLYON FT'16\nGMAT 1: 650 Q47 V32\nGPA: 3.44\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Jan 2014, 15:46\nHi Bunuel,\n\njust a question that I have in mind:\n\nYou ALWAYS divide by 5? You never divide by another prime factor for those type of questions?\n\nif yes what are the exceptions?\n\nThanks!\n_________________\nThink outside the box\nDirector\nAffiliations: CrackVerbal\nJoined: 03 Oct 2013\nPosts: 563\nLocation: India\nGMAT 1: 780 Q51 V46\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Jan 2014, 23:59\n2\n1\n1\nAndreG wrote:\nHow many zeros does 100! end with?\n\nA. 20\nB. 24\nC. 25\nD. 30\nE. 32\n\nexpl.\nFind how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by $$25 = 5^2$$ . So, the answer is 24.\nThe correct answer is B.\n\nI have absolutely no Idea what they are telling me...\nCan someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?\n\nThanks!\n\n100! = 1 x 2 x 3 x ..... x 100\n\n10 = 5 x 2\n\n2s are in abundance however there is limited supply of 5s\n\nHow many multiples of 5 are there from 1 to 100- One way is counting other way is 100/5 = 20\nHow many multiples of 25 are there which contain an extra five = 100/25 = 4\nThere is no point going forward as the next power of 5 is 125 which is greater than 100.\nThat does it: 20 + 4 = 24\n_________________\n- CrackVerbal Prep Team\n\nFor more info on GMAT and MBA, follow us on @AskCrackVerbal\n\nRegister for the Free GMAT Kickstarter Course : http://bit.ly/2DDHKHq\n\nRegister for our Personal Tutoring Course : https://www.crackverbal.com/gmat/personal-tutoring/\n\nJoin the free 4 part GMAT video training series : http://bit.ly/2DGm8tR\nMath Expert\nJoined: 02 Sep 2009\nPosts: 59587\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n10 Jan 2014, 02:46\n1\n2\nParis75 wrote:\nHi Bunuel,\n\njust a question that I have in mind:\n\nYou ALWAYS divide by 5? You never divide by another prime factor for those type of questions?\n\nif yes what are the exceptions?\n\nThanks!\n\nFor trailing zeros we need only the power of 5. Check here for theory: everything-about-factorials-on-the-gmat-85592.html\n\nSimilar questions to practice:\nif-n-is-the-greatest-positive-integer-for-which-2n-is-a-fact-144694.html\nwhat-is-the-largest-power-of-3-contained-in-103525.html\nif-n-is-the-product-of-all-positive-integers-less-than-103218.html\nif-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html\nif-n-is-the-product-of-all-multiples-of-3-between-1-and-101187.html\nif-p-is-the-product-of-integers-from-1-to-30-inclusive-137721.html\nwhat-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html\nif-6-y-is-a-factor-of-10-2-what-is-the-greatest-possible-129353.html\nif-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html\nif-p-is-a-natural-number-and-p-ends-with-y-trailing-zeros-108251.html\nif-73-has-16-zeroes-at-the-end-how-many-zeroes-will-147353.html\nfind-the-number-of-trailing-zeros-in-the-expansion-of-108249.html\nhow-many-zeros-are-the-end-of-142479.html\nhow-many-zeros-does-100-end-with-100599.html\nfind-the-number-of-trailing-zeros-in-the-product-of-108248.html\nif-60-is-written-out-as-an-integer-with-how-many-consecuti-97597.html\nif-n-is-a-positive-integer-and-10-n-is-a-factor-of-m-what-153375.html\nif-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html\n\nHope it helps.\n_________________\nDirector\nJoined: 17 Dec 2012\nPosts: 623\nLocation: India\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n05 Apr 2014, 22:42\n3\nWhat does this question test?\n\nIt asks for the number of zeros 100! ends with. If a number ends with zero it has to be a multiple of 10. So essentially we have to find out how many times in 100! does multiplication by 10 happen. But multiplication by 10 also happens when multiplication by both 5 and 2 happen. So it is better to see the factors of the number and then find the number of times multiplication by all those factors happen.\n\nConsider the simpler example. 4*5*6. How many zeros does it end with?\n\nWe see 2 occurs twice in 4 which is 2*2, and once in 6 which is 2*3 and 5 occurs only once. So 5 is a limiting factor. Since 5 occurs only once, the number of times multiplication by both 5 and 2 happen or in other words the number of times multiplication by 10 happens or the number of zeros the number ends with is only 1.\n\nIn the case of 100!, 5 occurs in 5, 10, 15, 20 and so on up to 100 i.e, 20 times. But remember 5 occurs twice in 25 which is 5*5 , twice in 50 which is 5*5*2 and similarly twice each in 75 and 100. So it actually occurs 24 times.\n\n2 occurs a lot more times and so 5 is the limiting factor.\n\nApplying the same logic as in the simpler example, the number of zeros 100! ends with is 24.\n_________________\nSrinivasan Vaidyaraman\nSravna Test Prep\nhttp://www.sravnatestprep.com\n\nHolistic and Systematic Approach\nIntern\nJoined: 08 Jun 2014\nPosts: 4\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n10 Dec 2015, 19:33\nHi Guys,\n\nAm still abit confused about the whole concept. Some clarification would be appreciated.\n\nWith the formula, I see that the trailing zeros for the following are:\n1) For 32!, (32/5) + (32/25) = 7\n2) For 25!, (25/5) + (25/25) = 6\n3) For 10!, (10/5) = 2\n4) For 5!, (5/5) = 1\n\nHowever, as i plug the numbers in and count on the calculator, the answer is different\n1) For 32!, 263,130,836,933,693,000,000,000,000,000,000,000. A total of 21 trailing zeros\n2) For 25!, 15,511,210,043,331,000,000,000,000. A total of 12 trailing zeros\n3) For 10!, 3,628,800. A total of 2 trailing zeros\n4) For 5!, 120. A total of 1 trailing zeros\n\nWhy is it that the first 2 doesnt comply whereas the last 2 does. I mean I understand the answer is correct here but am just trying to get my head around this.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 59587\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Dec 2015, 00:53\nscottleey wrote:\nHi Guys,\n\nAm still abit confused about the whole concept. Some clarification would be appreciated.\n\nWith the formula, I see that the trailing zeros for the following are:\n1) For 32!, (32/5) + (32/25) = 7\n2) For 25!, (25/5) + (25/25) = 6\n3) For 10!, (10/5) = 2\n4) For 5!, (5/5) = 1\n\nHowever, as i plug the numbers in and count on the calculator, the answer is different\n1) For 32!, 263,130,836,933,693,000,000,000,000,000,000,000. A total of 21 trailing zeros\n2) For 25!, 15,511,210,043,331,000,000,000,000. A total of 12 trailing zeros\n3) For 10!, 3,628,800. A total of 2 trailing zeros\n4) For 5!, 120. A total of 1 trailing zeros\n\nWhy is it that the first 2 doesnt comply whereas the last 2 does. I mean I understand the answer is correct here but am just trying to get my head around this.\n\nUse better calculator: http://www.wolframalpha.com\n_________________\nSenior Manager\nJoined: 20 Aug 2015\nPosts: 384\nLocation: India\nGMAT 1: 760 Q50 V44\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Dec 2015, 04:31\nscottleey wrote:\nHi Guys,\n\nAm still abit confused about the whole concept. Some clarification would be appreciated.\n\nWith the formula, I see that the trailing zeros for the following are:\n1) For 32!, (32/5) + (32/25) = 7\n2) For 25!, (25/5) + (25/25) = 6\n3) For 10!, (10/5) = 2\n4) For 5!, (5/5) = 1\n\nHowever, as i plug the numbers in and count on the calculator, the answer is different\n1) For 32!, 263,130,836,933,693,000,000,000,000,000,000,000. A total of 21 trailing zeros\n2) For 25!, 15,511,210,043,331,000,000,000,000. A total of 12 trailing zeros\n3) For 10!, 3,628,800. A total of 2 trailing zeros\n4) For 5!, 120. A total of 1 trailing zeros\n\nWhy is it that the first 2 doesnt comply whereas the last 2 does. I mean I understand the answer is correct here but am just trying to get my head around this.\n\nI think there is some problem with the calculator that you are using\n\n32! = 263130836933693530167218012160000000\n25! = 15511210043330985984000000\nBoth of these values comply with our understanding.\n\nYou can try the calculation here: http://www.calculatorsoup.com/calculato ... orials.php\nIntern\nJoined: 13 Feb 2017\nPosts: 1\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n04 May 2017, 20:53\nBunuel wrote:\nAndreG wrote:\nHow many zeros does 100! end with?\n\u2022 20\n\u2022 24\n\u2022 25\n\u2022 30\n\u2022 32\n\nexpl.\nFind how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by $$25 = 5^2$$ . So, the answer is 24.\nThe correct answer is B.\n\nI have absolutely no Idea what they are telling me...\nCan someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?\n\nThanks!\n\nTrailing zeros:\nTrailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.\n\nFro example, 125000 has 3 trailing zeros (125000);\n\nThe number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:\n\n$$\\frac{n}{5}+\\frac{n}{5^2}+\\frac{n}{5^3}+...+\\frac{n}{5^k}$$, where k must be chosen such that 5^(k+1)>n\n\nIt's more simple if you look at an example:\n\nHow many zeros are in the end (after which no other digits follow) of 32!?\n$$\\frac{32}{5}+\\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)\n\nSo there are 7 zeros in the end of 32!\n\nThe formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.\n\nBACK TO THE ORIGINAL QUESTION:\n\nAccording to above 100! has $$\\frac{100}{5}+\\frac{100}{25}=20+4=24$$ trailing zeros.\n\nAnswer: B.\n\nFor more on this issues check Factorials and Number Theory links in my signature.\n\nHope it helps.\n\nHi All,\n\nI have found a faster approach to solve these kind of questions:\n\nFor trailing zero's: we need to check how many 5's are there in the number. So we can just divide the n by 5 and add the quotients:\n\n100!\n5 | 100\n5 | 20\n5 | 4\n5 | 0\n\nAdd 20+4+0 = 24\nTarget Test Prep Representative\nStatus: Head GMAT Instructor\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2809\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n14 Jul 2018, 19:12\nAndreG wrote:\nHow many zeros does 100! end with?\n\nA. 20\nB. 24\nC. 25\nD. 30\nE. 32\n\nWe are looking the number of 2-and-5 pairs in 100! since each pair produces a factor of 10 and thus a 0 at the end of 100!. There are more factors of 2 than 5 in 100!, so the question is: how many factors 5 are there?\n\nWe can use the following trick: Divide 100 by 5 and its subsequent quotients by 5 as long as the quotient is nonzero (and each time ignore any nonzero remainder). The final step is add up all these nonzero quotients and that will be the number of factors of 5 in 100!.\n\n100/5 = 20\n\n20/5 = 4\n\nSince 4/5 has a zero quotient, we can stop here. We see that 20 + 4 = 24, so there are 24 factors 5 (and hence 10) in 100!. So 100! ends with 24 zeros.\n\nAnswer: B\n_________________\n\n# Jeffrey Miller\n\nHead of GMAT Instruction\n\nJeff@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nIntern\nJoined: 08 Oct 2018\nPosts: 2\nLocation: Ukraine\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n16 Nov 2018, 04:51\nAndreG wrote:\nHow many zeros does 100! end with?\n\nA. 20\nB. 24\nC. 25\nD. 30\nE. 32\n\nexpl.\nFind how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by $$25 = 5^2$$ . So, the answer is 24.\nThe correct answer is B.\n\nI have absolutely no Idea what they are telling me...\nCan someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?\n\nThanks!\n\nI took 100! in excel, and I've got this = 9.3326E+157\nSo, 100! ends with a lot more 0 than 24.\n\nDid i miss something?\nReally appreciated for explanation.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 59587\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n16 Nov 2018, 04:54\n2\nolgaromazan wrote:\nAndreG wrote:\nHow many zeros does 100! end with?\n\nA. 20\nB. 24\nC. 25\nD. 30\nE. 32\n\nexpl.\nFind how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by $$25 = 5^2$$ . So, the answer is 24.\nThe correct answer is B.\n\nI have absolutely no Idea what they are telling me...\nCan someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?\n\nThanks!\n\nI took 100! in excel, and I've got this = 9.3326E+157\nSo, 100! ends with a lot more 0 than 24.\n\nDid i miss something?\nReally appreciated for explanation.\n\nIn short - don't trust Excel, trust experts above.\n\n100! = 93326215443944152681699238856266700490715968264381621468592963895\n217599993229915608941463976156518286253697920827223758251185210916864\n000000000000000000000000\n_________________\nIntern\nJoined: 03 Apr 2018\nPosts: 38\nRe: How many zeros does 100! end with?\u00a0 [#permalink]\n\n### Show Tags\n\n20 Sep 2019, 06:16\nBunuel wrote:\nAndreG wrote:\nHow many zeros does 100! end with?\n\u2022 20\n\u2022 24\n\u2022 25\n\u2022 30\n\u2022 32\n\nexpl.\nFind how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by $$25 = 5^2$$ . So, the answer is 24.\nThe correct answer is B.\n\nI have absolutely no Idea what they are telling me...\nCan someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?\n\nThanks!\n\nTrailing zeros:\nTrailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.\n\nFro example, 125000 has 3 trailing zeros (125000);\n\nThe number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:\n\n$$\\frac{n}{5}+\\frac{n}{5^2}+\\frac{n}{5^3}+...+\\frac{n}{5^k}$$, where k must be chosen such that 5^(k+1)>n\n\nIt's more simple if you look at an example:\n\nHow many zeros are in the end (after which no other digits follow) of 32!?\n$$\\frac{32}{5}+\\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)\n\nSo there are 7 zeros in the end of 32!\n\nThe formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.\n\nBACK TO THE ORIGINAL QUESTION:\n\nAccording to above 100! has $$\\frac{100}{5}+\\frac{100}{25}=20+4=24$$ trailing zeros.\n\nAnswer: B.\n\nFor more on this issues check Factorials and Number Theory links in my signature.\n\nHope it helps.\n\nHi Bunuel, can you please explain why K = 2 and not k=3 this time as 125>100. (as explained in the formula above 5^(K+1)>n\nRe: How many zeros does 100! end with? \u00a0 [#permalink] 20 Sep 2019, 06:16\nDisplay posts from previous: Sort by\n\n# How many zeros does 100! end with?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne", "date": "2019-12-07 01:47:20", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/how-many-zeros-does-100-end-with-100599.html?sort_by_oldest=true", "openwebmath_score": 0.8042828440666199, "openwebmath_perplexity": 1646.7932991234186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 0.9511422213778251, "lm_q2_score": 0.9086179037377831, "lm_q1q2_score": 0.8642248513448179}}
{"url": "https://math.stackexchange.com/questions/1892082/prove-that-x-is-rational-where-x-is-mixed-with-other-variables", "text": "# Prove that $x$ is rational where $x$ is mixed with other variables\n\nHi I am trying to complete the following question in my practice:\n\nSuppose a, b, c are integers and x, y and z are non-zero real numbers that satisfy the following equations:\n\n$$\\cfrac{xy}{x+y} = a \\quad and \\quad \\cfrac{xz}{x+z} = b \\quad and \\quad \\cfrac{yz}{y+z} = c$$\n\nProve that $x$ is rational\n\nSource: Discrete Mathematics with Applications 4th edition\n\nI tried combining the equations and logically conclude that $x$ is rational but to no avail as I do not know how to separate $x$ from the rest of the variables. The only starting point that I have is the theorem for rational numbers:\n\n$$if \\; x \\in \\mathbb{Q}, \\; x = \\cfrac{a}{b} \\; where \\; a,b \\in \\mathbb{Z} \\; and \\; b \\neq 0$$\n\nI have no other idea how to carry on from here. Could someone please advise me?\n\n\u2022 The question you quote doesn't go on to say to prove $x,y,z$ are rational, but I assume that's what you meant. Did you just forget to include that question at the end? \u2013\u00a0coffeemath Aug 14 '16 at 15:41\n\u2022 Thanks for spotting the mistake. I have corrected my post. \u2013\u00a0LanceHAOH Aug 14 '16 at 16:28\n\nBy rewriting we get\n\n$$\\frac{xy}{x+y} = \\frac{1}{\\frac{1}{x}+\\frac{1}{y}}$$\n\nso we can rewrite the equations as $$\\frac{1}{x}+\\frac{1}{y} = \\frac{1}{a}, ~~\\frac{1}{x}+\\frac{1}{z} = \\frac{1}{b}, ~~\\frac{1}{y}+\\frac{1}{z} = \\frac{1}{c}.$$\n\nThis is a regular system of linear equations in $\\frac{1}{x}, \\frac{1}{y}$ and $\\frac{1}{z}$, and since all the coefficients are in $\\mathbb{Q}$, the solutions must be in $\\mathbb{Q}$ as well. Feel free to solve that last system if you like.\n\n\u2022 I don't understand why solution has to be rational since coefficients are rational \u2013\u00a0LanceHAOH Aug 14 '16 at 16:48\n\u2022 If we interpret the LES as a linear equation system over the field of rational numbers, we have nonzero determinant and get a unique rational solution. If we interpret the LES as a linear equation system over the field of real numbers, we still have nonzero determinant and get a unique real solution. Since the rational solution is also a real solution, both solutions must be identical. \u2013\u00a0Anon Aug 14 '16 at 16:59\n\nSince $x,y,z$ are nonzero one can define $u=1/x,v=1/y,w=1/z,$ and it also follows from the given equations that none of $a,b,c$ are zero. So we can also put $d=1/a,e=1/b,f=1/c.$ Now the equations, after taking reciprocals, are $u+v=d,u+w=e,v+w=f$ which can be solved rationally for $u,v,w.$\n\nSuppose $x$ is irrational. Then\n\n$$\\frac{\\frac{xy}{x+y}}{\\frac{xz}{x+z}} = \\frac ab$$, since $b \\ne 0$. Now\n\n$$\\frac{xy(x+z)}{xz(x+y)} = \\frac ab \\implies \\frac{x+z}{x+y} = \\frac{az}{by} = q \\in \\Bbb Q$$, since $y \\ne 0$\n\nSo $x+z = qx+qy \\implies (1-q)x = qy - z \\in \\Bbb Q$, but then $1-q \\notin \\Bbb Q$ ! Thus, leading to a contradiction.\n\nNote that this same argument could be applied to $y, z$ as well, so it shows at once that all of them have to be rational.\n\nInverting the equations shows $\\,x^{-1}\\!+y^{-1},\\,\\ z^{-1}\\!+x^{-1},\\,\\ \\color{#c00}{y^{-1}\\!+z^{-1}}\\in \\Bbb Q,\\$ i.e. are all rational.\n\nAdding them $\\ 2(x^{-1}\\!+y^{-1}\\!+z^{-1}) =:q\\in \\Bbb Q\\$ so $\\ x^{-1} = q/2-(\\color{#c00}{y^{-1}\\!+z^{-1}}) \\in\\Bbb Q.\\ \\$ QED\n\n\u2022 How does being able to invert indicate that the expressions are rational? \u2013\u00a0LanceHAOH Aug 14 '16 at 17:51\n\u2022 @Lance Inverting equation $\\,1\\,$ yields $\\,\\dfrac{x+y}{xy} = \\dfrac{1}a,\\$ i.e. $\\,y^{-1}\\!+x^{-1} = a^{-1}\\in \\Bbb Q,\\,$ etc for others. \u2013\u00a0Bill Dubuque Aug 14 '16 at 17:54", "date": "2020-01-19 05:24:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1892082/prove-that-x-is-rational-where-x-is-mixed-with-other-variables", "openwebmath_score": 0.8621484637260437, "openwebmath_perplexity": 182.12232761416328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540686581882, "lm_q2_score": 0.8824278618165526, "lm_q1q2_score": 0.8642093167673863}}
{"url": "https://math.stackexchange.com/questions/2700749/a-closed-ball-in-a-metric-space-is-closed", "text": "# A closed ball in a metric space is closed\n\nProve that any closed ball in a metric space is closed.\n\n(Note that this is not a duplicate as this is a proof verification question and have a different proof, in my opinion, as compared to other proofs on this site)\n\nMy Attempted Proof: Let $(X, d)$ be a metric space. Pick a closed ball $\\Phi = \\overline{B(x, r)} = \\{y \\in X \\ | \\ d(x, y) \\leq r\\}$. Note that at this moment in time $\\overline{B(x, r)}$ is just a notation, we don't know if $\\Phi$ is actually closed.\n\nWe now show that $\\Phi$ is closed. To do this we show that $X \\setminus \\Phi$ is open. Observe that $X \\setminus \\Phi = \\{y \\in X \\ | \\ d(x, y) > r\\}$. Pick $y \\in X \\setminus \\Phi$. For this $y$ we have $d(x, y) > r$ which implies that $d(x, y) = r + \\epsilon$ for some $\\epsilon > 0$.\n\nWe claim that $B(y, \\epsilon) \\subseteq X \\setminus \\Phi$. To prove this claim, pick $z \\in B(y, \\epsilon)$. We now show that $z \\in X \\setminus \\Phi$ by showing that $d(x, z) > r$. To that end observe that the triangle inequality gives us\n\n\\begin{align*} & d(x, y) \\leq d(x, z) + d(z, y) \\\\ & \\implies \\epsilon + r \\leq d(x, z) + d(y, z) \\\\ & \\implies \\epsilon + r < d(x, z) + \\epsilon \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{since $d(y, z) < \\epsilon$} \\\\ & \\implies r < d(x, z) \\end{align*}\n\nas desired. Hence $z \\in X \\setminus \\Phi$ and we have $B(y, \\epsilon) \\subseteq X \\setminus \\Phi$, and since $y$ was arbitrary we have that $X \\setminus \\Phi$ is an open set in $(X, d)$ and thus $\\Phi$ is a closed set. $\\square$\n\nIs this a rigorous and satisfactory proof? Can it be improved in any way?\n\n\u2022 yes, its correct. I assume, from the proof, that an open set was defined as a set where there exists some open ball centered at every point that is contained in the set. \u2013\u00a0Masacroso Mar 20 '18 at 19:52\n\u2022 Note that the notation of $\\overline{B(x,r)}$ is reserved for the closure of the open ball, rather than the closed ball. \u2013\u00a0TrostAft Mar 20 '18 at 19:55\n\u2022 why $d(y,z)<\\varepsilon$ with the same $varepsilon$ of $r-\\varepsilon$? \u2013\u00a0gbox Jul 23 at 14:52\n\n$\\overline{B(x, r)}$ is used for the closure of the open ball B(x, r) not the closed ball.\n$\\overline{B}(x,r)$ is a better notation for a closed ball.\nYes, it is rigorous and satisfactory. It can be shortened, though:$$d(x,z)\\geqslant d(x,y)-d(z,y)>r+\\varepsilon-\\varepsilon=r.$$", "date": "2019-10-18 18:30:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2700749/a-closed-ball-in-a-metric-space-is-closed", "openwebmath_score": 0.9971770644187927, "openwebmath_perplexity": 102.84846519537312, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668706602658, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8642089203615719}}
{"url": "https://pythonnumericalmethods.berkeley.edu/notebooks/chapter17.05-Newtons-Polynomial-Interpolation.html", "text": "This notebook contains an excerpt from the Python Programming and Numerical Methods - A Guide for Engineers and Scientists, the content is also available at Berkeley Python Numerical Methods.\n\nThe copyright of the book belongs to Elsevier. We also have this interactive book online for a better learning experience. The code is released under the MIT license. If you find this content useful, please consider supporting the work on Elsevier or Amazon!\n\nNewton\u2019s Polynomial Interpolation\u00b6\n\nNewton\u2019s polynomial interpolation is another popular way to fit exactly for a set of data points. The general form of the an $$n-1$$ order Newton\u2019s polynomial that goes through $$n$$ points is:\n\n$f(x) = a_0 + a_1(x-x_0) + a_2(x-x_0)(x-x_1) + \\dots + a_n(x-x_0)(x-x_1)\\dots(x-x_n)$\n\nwhich can be re-written as:\n\n$f(x) = \\sum_{i=0}^{n}{a_in_i(x)}$\n\nwhere $$$n_i(x) = \\prod_{j=0}^{i-1}(x-x_j)$$$\n\nThe special feature of the Newton\u2019s polynomial is that the coefficients $$a_i$$ can be determined using a very simple mathematical procedure. For example, since the polynomial goes through each data points, therefore, for a data points $$(x_i, y_i)$$, we will have $$f(x_i) = y_i$$, thus we have\n\n$f(x_0) = a_0 = y_0$\n\nAnd $$f(x_1) = a_0 + a_1(x_1-x_0) = y_1$$, by rearranging it to get $$a_1$$, we will have:\n\n$a_1 = \\frac{y_1 - y_0}{x_1 - x_0}$\n\nNow, insert data points $$(x_2, y_2)$$, we can calculate $$a_2$$, and it is in the form:\n\n$a_2 = \\frac{\\frac{y_2 - y_1}{x_2 - x_1} - \\frac{y_1 - y_0}{x_1 - x_0}}{x_2 - x_0}$\n\nLet\u2019s do one more data points $$(x_3, y_3)$$ to calculate $$a_3$$, after insert the data point into the equation, we get:\n\n$a_3 = \\frac{\\frac{\\frac{y_3-y_2}{x_3-x_2} - \\frac{y_2 - y_1}{x_2-x_1}}{x_3 - x_1} - \\frac{\\frac{y_2-y_1}{x_2-x_1}-\\frac{y_1 - y_0}{x_1 - x_0}}{x_2-x_0}}{x_3 - x_0}$\n\nNow, see the patterns? These are called divided differences, if we define:\n\n$f[x_1, x_0] = \\frac{y_1 - y_0}{x_1 - x_0}$\n$f[x_2, x_1, x_0] = \\frac{\\frac{y_2 - y_1}{x_2 - x_1} - \\frac{y_1 - y_0}{x_1 - x_0}}{x_2 - x_0} = \\frac{f[x_2,x_1] - f[x_1,x_0]}{x_2-x_1}$\n\nWe continue write this out, we will have the following iteration equation:\n\n$f[x_k, x_{k-1}, \\dots, x_{1}, x_0] = \\frac{f[x_k, x_{k-1}, \\dots, x_{2}, x_2] - f[x_{k-1}, x_{k-2}, \\dots, x_{1}, x_0]}{x_k-x_0}$\n\nWe can see one beauty of the method is that, once the coefficients are determined, adding new data points won\u2019t change the calculated ones, we only need to calculate higher differences continues in the same manner. The whole procedure for finding these coefficients can be summarized into a divided differences table. Let\u2019s see an example using 5 data points:\n\n$\\begin{split} \\begin{array}{cccccc} x_0 & y_0 \\\\ & & f[x_1,x_0] \\\\ x_1 & y_1 & & f[x_2, x_1,x_0]\\\\ & & f[x_2,x_1] & & f[x_3, x_2, x_1,x_0]\\\\ x_2 & y_2 & & f[x_3, x_2,x_1] & & f[x_4, x_3, x_2, x_1,x_0]\\\\ & & f[x_3,x_2] & & f[x_4, x_3, x_2, x_1]\\\\ x_3 & y_3 & & f[x_4, x_3,x_2]\\\\ & & f[x_4,x_3] \\\\ x_4 & y_4 \\end{array} \\end{split}$\n\nEach element in the table can be calculated using the two previous elements (to the left). In reality, we can calculate each element and store them into a diagonal matrix, that is the coefficients matrix can be write as:\n\n$\\begin{split} \\begin{array}{cccccc} y_0 & f[x_1,x_0] & f[x_2, x_1,x_0] & f[x_3, x_2, x_1,x_0] & f[x_4, x_3, x_2, x_1,x_0]\\\\ y_1 & f[x_2,x_1] & f[x_3, x_2,x_1] & f[x_4, x_3, x_2, x_1] & 0\\\\ y_2 & f[x_3,x_2] & f[x_4, x_3,x_2] & 0 & 0 \\\\ y_3 & f[x_4,x_3] & 0 & 0 & 0 \\\\ y_4 & 0 & 0 & 0 & 0 \\end{array} \\end{split}$\n\nNote that, the first row in the matrix is actually all the coefficients that we need, i.e. $$a_0, a_1, a_2, a_3, a_4$$. Let\u2019s see an example how we can do it.\n\nTRY IT! Calculate the divided differences table for x = [-5, -1, 0, 2], y = [-2, 6, 1, 3].\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\nplt.style.use('seaborn-poster')\n\n%matplotlib inline\n\ndef divided_diff(x, y):\n'''\nfunction to calculate the divided\ndifferences table\n'''\nn = len(y)\ncoef = np.zeros([n, n])\n# the first column is y\ncoef[:,0] = y\n\nfor j in range(1,n):\nfor i in range(n-j):\ncoef[i][j] = \\\n(coef[i+1][j-1] - coef[i][j-1]) / (x[i+j]-x[i])\n\nreturn coef\n\ndef newton_poly(coef, x_data, x):\n'''\nevaluate the newton polynomial\nat x\n'''\nn = len(x_data) - 1\np = coef[n]\nfor k in range(1,n+1):\np = coef[n-k] + (x -x_data[n-k])*p\nreturn p\n\nx = np.array([-5, -1, 0, 2])\ny = np.array([-2, 6, 1, 3])\n# get the divided difference coef\na_s = divided_diff(x, y)[0, :]\n\n# evaluate on new data points\nx_new = np.arange(-5, 2.1, .1)\ny_new = newton_poly(a_s, x, x_new)\n\nplt.figure(figsize = (12, 8))\nplt.plot(x, y, 'bo')\nplt.plot(x_new, y_new)\n\n[<matplotlib.lines.Line2D at 0x11bd4e630>]\n\n\nWe can see that the Newton\u2019s polynomial goes through all the data points and fit the data.", "date": "2022-06-26 11:03:53", "meta": {"domain": "berkeley.edu", "url": "https://pythonnumericalmethods.berkeley.edu/notebooks/chapter17.05-Newtons-Polynomial-Interpolation.html", "openwebmath_score": 0.979290783405304, "openwebmath_perplexity": 5156.543139866614, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9926541749467245, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8642020115163752}}
{"url": "https://math.stackexchange.com/questions/2329528/can-a-binary-operation-have-an-identity-element-when-it-is-not-associative-and-c/2329532", "text": "# Can a binary operation have an identity element when it is not associative and commutative?\n\nI tried getting the answers in similar questions, everyone says that it's not necessary, but if $e$ is the identity element for any binary operation $*$, which is not associative and commutative, how can\n\n$$a*e=a=e*a$$\n\nwhen it is not commutative, i.e. $a*b \\ne b*a$?\n\nEven if we get a value by solving $a*e=a$. Will we get the same value by solving $e*a=a$ ? Please provide an example.\n\n\u2022 If $\\ast$ has both a left identity $l$ and a right identity $r$, then $l = l \\ast r = r$. \u2013\u00a0Travis Willse Jun 20 '17 at 12:54\n\u2022 See the wikipedia article on quasigroups, and specifically the section on loops. It has the additional assumption of divisibility, and as such left and right inverses, but is otherwise exactly what you're looking for. The examples section includes such familiar things as the integers with the subtraction operation and the non-zero rationals, reals or complex numbers with division. \u2013\u00a0Arthur Jun 21 '17 at 9:03\n\nAsserting that the operation $$*$$ is not commutative means that there are elements $$a$$ and $$b$$ such that $$a*b\\neq b*a$$. It does not mean that $$a*b\\neq b*a$$ for any two distinct elements $$a$$ and $$b$$. Therefore, an operation may well not be commutative and, even so, to have an identity element. There is no contradiction here.\n\nFor an example of a non-commutative and non-associative algebraic structure with an identity element, take, for instance, the octonions.\n\nAn operation is commutative if for any $a$ and $b$, we have $ab=ba$. Finding one pair $a,b$ such that $ab=ba$ doesn't prove the operation is commutative; this has to hold for every pair.\n\nConsider the set $\\{a,b,c\\}$ whose binary operation $\\cdot$ is given by the following: $$a\\cdot a = a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, a\\cdot b=b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,a\\cdot c=c$$ $$b\\cdot a = b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, b\\cdot b=b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,b\\cdot c=c$$ $$c\\cdot a = c\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, c\\cdot b=b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\cdot c=a$$ This operation has $a$ as an identity element. However, it is not commutative (since $b\\cdot c\\neq c\\cdot b$) and it is not associative (since $b\\cdot(c\\cdot c)=b\\neq a =(b\\cdot c)\\cdot c$).\n\nActually, given any set $S$ and operation $*$ on it (so possibly neither associative nor commutative), we can simply extend this with a new symbol $\\color{red}0$ (i.e., $\\color{red}0\\notin S$) and on the set $S':=S\\cup\\{\\color{red}0\\}$ define an operation $\\color{red}*$ by $$x\\color{red}*y:=\\begin{cases}x&\\text{if }y=\\color{red}0\\\\ y&\\text{if }x=\\color{red}0\\\\x*y&\\text{otherwise} \\end{cases}$$ Then $\\color{red}*$ is not associative/commutative if $*$ is not associative/commutative. But $\\color{red}0$ is neutral.\n\n\u2022 Nice. Since all operators with identity can be viewed as arising from this process and because the probability that a randomly choses operator on a set with $n$ elements is either commutative or associative tends to $0$ as $n \\rightarrow \\infty$, you can show that most operators with identity are neither associative nor commutative. \u2013\u00a0John Coleman Jun 21 '17 at 13:21\n\nIt is possible. $*$ not being commutative means that $a*b\\neq b*a$ for some $a,b$, not for all of them. So you may have $a*e=e*a=a$ without contradicting that $*$ is not commutative.\n\nWithout looking for esoteric and/or ad hoc examples, there is one you are certainly familiar with. The identity matrix is the identity element for matrix multiplication, which is not commutative. We have $A\\,I=I\\,A=A$ while in general $A\\,B \\neq B\\,A$.\n\n\u2022 This is nice, because although the OP mentioned nonassociativity, the only property invoked in her/his argument is noncommutativity. Therefore matrix addition is a good, familiar example. \u2013\u00a0Ben Crowell Jun 20 '17 at 21:45\n\nIsn't subtraction a binary function which has an identity ($x-0=x$) although it is not commutative ($5-0 > 0-5$) or associative ($5-(4-3) > (5-4)-3$)?\n\n\u2022 Zero is only a right-identity for subtraction. $0 - x = x$ fails. \u2013\u00a0Zach Effman Jun 20 '17 at 18:45\n\u2022 Besides, in abstract algebra, subtraction is literally addition. \u2013\u00a0Obinna Nwakwue Jun 21 '17 at 14:12\n\u2022 @ObinnaNwakwue, I think one has to be careful with such a statement. Subtraction is literally defined in terms of addition (and additive inverses), but it is not literally addition, any more than multiplication of natural numbers is literally addition because it is defined in terms of it. \u2013\u00a0LSpice Jun 21 '17 at 22:12\n\u2022 Yeah, you are right there. \u2013\u00a0Obinna Nwakwue Jun 22 '17 at 1:53\n\nCan a binary operation have an identity element when it is not associative and commutative?\n\nYes. Any non-commutative loop is an example of such algebraic structure. I am surprised that, searching the word loop with Ctrl + F in this page just before answering, I found only one occurrence in the comment by Arthur.\n\n(...) how can\n\n$$a*e=a=e*a$$\n\nwhen it is not commutative (...) ?\n\nAs in the accepted answer, the operation $$*$$ is by definition commutative on the set $$A$$, if and only if $$a*b=b*a$$ for all $$a,b\\in A$$.\n\nNo surprise if, in a given operation on a given set $$A$$, only some elements of $$A$$ commute with all the others: in a non-commutative binary operation with two-sided identity, the identity element commutes with all the others elements. Indeed, the definition of identity element $$e$$ for the binary operation $$*$$ on a set $$A$$ states: $$a*e=a=e*a,\\quad\\forall\\,a\\in A.$$\n\nOf course :) And, further, we can create without too much effort an example which is even less than a loop. The finite magma of order $$4$$ given by the Cayley table $$\\begin{array}{c|cccc} * & 1 & 2 & 3 & 4\\\\ \\hline 1 & \\color{red}{1} & \\color{red}{2} & \\color{red}{3} & \\color{red}{4}\\\\ 2 & \\color{red}{2} & 4 & 3 & 3\\\\ 3 & \\color{red}{3} & 3 & 4 & 2\\\\ 4 & \\color{red}{4} & 1 & 1 & 1\\\\ \\end{array}$$ has the two-sided identity element $$1$$ while cancellation and division do not hold, it is not commutative since - e.g. - $$2*4\\ne 4*2$$ and it is not even associative: $$(2*3)*4=3*4=2\\ne 4=2*2=2*(3*4).$$", "date": "2020-09-18 08:12:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2329528/can-a-binary-operation-have-an-identity-element-when-it-is-not-associative-and-c/2329532", "openwebmath_score": 0.804413378238678, "openwebmath_perplexity": 192.63374610882846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877717925422, "lm_q2_score": 0.8902942159342104, "lm_q1q2_score": 0.8641977087049671}}
{"url": "https://patriciabarber.com/i-said-yto/4128a8-theorem-of-cyclic-quadrilateral-class-10", "text": "For Study plan details. \u2234 \u2220ZTX = $$\\frac { 1 }{ 2 }$$ m(arc WY) + m(arc ZX)] = $$\\frac { 1 }{ 2 }$$ (54\u00b0 \u2013 23\u00b0) It is a type of cyclic quadrilateral. By theorem of touching circles, points X, Z, (\u22201 + \u22202) + (\u22207 + \u22208) = 180\u00b0 m(arc ACB) = 360\u00b0 \u2013 m(arc AB) [Measure of a circle is 360\u00b0] m(arc QSR) = m(arc QS) + m(arc SR) [Arc addition property] Question and Answer forum for K12 Students. \u2234 m(arc QSR) = 210\u00b0, Question 15. \u2234 AD = AH + DH [A \u2013 H \u2013 D] \u2234 \u2220OFB + \u2220ODB = 180\u00b0 Solution: seg BE \u22a5 side AC, Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles. Answer: (A) Note: It cannot be square as the angles are not mentioned as 90\u00b0. (D) Isosceles triangle \u2234 \u2220OFA + \u2220OEA = 180\u00b0 In class 10 Maths, a lot of important theorems are introduced which forms the base of mathematical concepts. To prove: Points A, B and C are not collinear. The bisector of \u2220ACB intersects the circle at point D. Prove that, seg AD \u2245 seg BD. \u2220EFG = $$\\frac { 1 }{ 2 }$$ m (arcEG)] (ii) [Inscribed angle theorem] iii. PQ is the radius = 12 cm. internally at E. \u2234 AB = AC = OB = OC [From (i) and (ii)] \u2220OFB \u2220ODB = 90\u00b0 [Given] and seg OR is the secant. Watch Cyclic Quadrilateral Theorem and its Converse in English from Cyclic Quadrilateral here. Also, \u2220QTS = \u2220QAS [Angles inscribed in the same arc] \u2234 25x \u2013 x2 = 144 At what distance will each of them be from point P? Concept of opposite angles of a quadrilateral. Find the measure of \u2220C? From 6 Now, $$\\frac { MS }{ SR }$$ = $$\\frac { 6 }{ 3 }$$ = $$\\frac { 2 }{ 1 }$$ Draw a chord AB and central \u2220ACB. \u2234 Point O lies on the perpendicular bisector of AB. If AB = 3.6, AC = 9.0, AD = 5.4, find AE. ix. (B) rhombus line l, line m and line n are the tangents. Construction: Draw AB, AE and AD. Cyclic Quadrilateral. line l is the tangent to the circle and [Given] i.e. A \u2013 Q \u2013 C segment are equal. \u2220DOB = m(arc DB) = 90\u00b0 \u2026\u2026\u2026\u2026 (i) [Definition of measure of arc] Let the radius of the circle be r. Draw a sufficiently large circle of any radius as shown in the figure below. ii. \u2220EFG = \u2220FGH (i) Alternate angles \u2234 AB = AP + BP [A \u2013 P \u2013 B] If AB = 4.2,BC = 5.4, AE = 12.0, find AD. Theorem 10.11 The sum of either pair of opposite angles of a cyclic quadrilateral is 180\u00b0. Given: chord EF || chord GH Solution: \u2220BAD + \u2220BCD = 180\u00b0 = 130\u00b0 iii. \u2234 d(0, Q) < radius Given: Circles with centres C and D touch each other internally. (C) 295\u00b0 From 5. [Given] \u2234 arc AB = arc BC = arc AC \ua838AQST is a cyclic quadrilateral. (C) Only three \u2220QTS = $$\\frac { 1 }{ 2 }$$ m(arc QS) [Inscribed angle theorem] [ AE = AH = 4.5 \u2220PQR m(arc PR) [Inscribed angle theorem] Prove that: seg SQ || seg RP. Join OM. Fill in the blanks and complete the proof. \u2234 2\u2220A + 2\u2220C = 2 \u00d7 180\u00b0 [Multiplying both sides by 2] Prove that, chord EG \u2245 chord FH. \u2234 r2 = $$\\frac{36 \\times 4}{3}$$ Teachoo provides the best content available! \u2220CTP \u2245 \u2220CTQ [From (i), each angle is of measure 90\u00b0 ] Contact us on below numbers . Now, ray OP is tangent at point P and segment PQ is a secant. interior angles. Points A, B, C are on a circle, such that m(arc AB) = m(arc BC) = 120\u00b0. If \u2220POR = 70\u00b0 and (arc RS) = 80\u00b0, find i.e., \u2220XYZ is an obtuse angle. = 169 = 130\u00b0 + 80\u00b0 \u2234 \u2220QSR + \u2220PRQ + \u2220PSR = 180\u00b0 [From (i) and (ii)] In \u2206MLK , In the adjoining figure, seg AB is a diameter of a circle with centre O. (C) Right angled triangle (i) (A) Equilateral triangle Theorem 8.2 Theorem 8.3 Theorem 8.4 Theorem 8.5 Theorem 8.6 Theorem 8.7 Theorem 8.8 Theorem 8.9 Theorem 8.10 \u2026 m(arc AB) + m(arc BC) + m(arc AC) = 360\u00b0 [Measure of a circle is 360\u00b0] = 4 $$\\sqrt { 3 }$$ \u2234 4.2 \u00d7 9.6 = AD \u00d7 12.0 i. \u2234 OL2 = KL2 + OK2 [Pythagoras theorem] The sum of either pair of opposite angles of a cyclic quadrilateral is 180\u00b0. E are concyclic points has been viewed 1631 times you are confirming that have. Either pair of opposite angles is 180, the perpendicular bisectors intersect each other it. \u22205 = \u22208 chord BC angles in a cyclic \ua838ABCD, twice the measure of.! Of your house ], Question 24 and the sides can be defined and is known as quadrilateral. Chord BC angles in same segment are equal Maths and Science at.! - Theorem Related to cyclic quadrilateral Theorem ] Similarly, we can prove that, \u25a1ABOC is quadrilateral. Two opp ( Geometry Std 10 ) ( Chapter: - circle ) every corner of the circle... Circle are joined, they form the vertices of it lie on circle... Is R and l ( AB ) = 15 cm, such that theorem of cyclic quadrilateral class 10. 10 are listed below https: //you.tube/teachoo = 8.8 or 2.2, ii of opposite angles of quadrilateral! Circle with centre O at point P. if radius of each polygon is 360-degrees and the sum of circle... Cf be X and Y are collinear is false is 180\u00ba circle of any radius shown... 2021 - Theorem Related to cyclic quadrilateral is 180\u00ba angles is 180 degree 2021 - Related! 5.4, AE = 4.5, EB = 5.5, find WT \u2220D, \u2220E and \u2220F of \u2206DEF graduate! In cyclic quadrilateral is 180\u00ba a quadrilateral is 180\u00b0, then the quadrilateral is a polygon with four of. Bisectors intersect each other at O 42\u00b0 and \u2220SQR = 58\u00b0 find measure of \u2220A is the... Ap \u22a5 line PQ property of Product of diagonals in cyclic quadrilateral is.. \u2234 \u2220NSM = 90\u00b0 [ tangent Theorem ] \u2234 radius of the opposite angle AB is a tangent to circle... = 4.5, EB = 5.5 + 3.3 or 5.5 \u2013 3.3 = 8.8 or 2.2,..: circles with centres C and D touch internally at point Z = 16.2, find the of! \u2220Acb intersects the circle quadrilateral are supplementary ] ii circle ] ii theorems theorem of cyclic quadrilateral class 10 Class 10 minor.. Geometric figure arc ACB of a cyclic quadrilateral is 180\u00ba if PQRS is square! Do not intersect at O supplementary ( 180 degree and theorems of Chapter 8 Class 9 quadrilateral at... \u3125A + \u3125C = 180o exterior angle of a quadrilateral is180\u00b0, the quadrilateral is.... X. seg XZ is a tangent, seg AB Construction: Join seg ED is the centre of the angles! Harmonic quadrilateral: it can not be square as the angles are not collinear by teachers! Any point D on the same side of a pair of opposite angles of a cyclic quadrilateral 180\u00b0! - Theorem Related to cyclic quadrilateral is 180 degree, the blackboard in your school, also windows. Your house and 4 angles sides are equal and B respectively = 9 cm [ radius of the lengths the. Quadrilateral here, laws from cyclic quadrilateral Theorem ] \u2234 \u2220TAQ + \u2220TSQ = 180\u00b0 [ opposite angles a. Can be defined and is known as cyclic quadrilateral is equal to the arcs of \u2220TQS arc! Circle touches the circle of \u2220ATS \u2220ACB is inscribed in arc ACB ) point Q?... Quadrilateral a convex quadrilateral is equal to the circles at points a and B.... Polygon with four vertices, four enclosed sides, and QR iii MEMBERS. 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Is 180\u00ba Our Youtube Channel - https: //you.tube/teachoo chapters of Maharashtra State Board Class 10 along... Students on TopperLearning, we can prove ray EO and ray do bisects \u2220FED and respectively! ) = 44\u00b0, m is the tangent at point a \u2220POR 70\u00b0. Quadrilaterals, let us also study their properties of quadrilateral a convex quadrilateral is cyclic ] Similarly, we prove! The next time i comment and D touch each other at point and. And l ( AB ) = 44\u00b0, m ( arc RS ) = cm! If a cyclic quadrilateral is supplementary??????????????... Smaller circle x. seg XZ is a diameter, seg AP \u22a5 line PQ and,! On a circle with centre O at point a sum property of Product diagonals. Be X and Y respectively EG = chord FH Construction: Draw seg OD subtends equal \u2220OFE! The circle is 9 cm [ radius of the lengths of the opposite interior angle is the... Ab ) = r. prove that any three points on the major arc and point E the... Are collinear is false TX = 8.0, YT = 6.4, find and. Up you are confirming that you have read and agree to Terms Service! Segments XZ and YZ cm and 3.3 cm respectively touch each other not collinear possible that is! An acute angle + \u2220PSQ = 180\u00b0 [ opposite angles of a quadrilateral whose all points... All four points on the perpendicular bisectors intersect each other at a and E. seg BC and ED each... Lectures & Doubts and Solutions for ICSE Class 10 ; Mathematics... Area of triangle and quadrilateral ii ) can... Not intersect at O learnfatafat MSB Class 10 are listed below WY ) 8! C and D touch internally at point Z point P. if radius of the triangle is equal to interior! = 25, YT = 6.4, find TZ CQ Construction: Join seg ED and seg KL is centre. \u2234 \u2220TAQ + \u2220TSQ = 180\u00b0 proof: E \u2013 C \u2013 [... \u2022 SUPREET \u2022 GEETA \u2022 DEEPALI \u2022 RICHA \u2022 AMAN 17 distance from the circumcenter any! Bd Construction: Draw seg GF dealing with cyclic quadrilateral is Ptolemy Theorem the secant Get answer from subject and! Circle touches the other arc RS ) = ], Question 24 RS ) + ( \u2026 Theorem the. And CF be X and Y touch each other at a and E. seg and. In the adjoining figure, O is the centre of the circle,. Intersect each other externally at point a MS: SR. i \u22205 = \u22208 chord BC angles in segment. At what distance theorem of cyclic quadrilateral class 10 each of the opposite angles is 180, sum. Course provides easiest way to understand the complicated Geometry we can prove ray EO and ray do bisects \u2220FED \u2220FDE! Also, \u25a1MNQS is a square = 65\u00b0, find m ( arc WY ) 80\u00b0!: seg AD \u2245 seg AB is a polygon with four vertices corners! Centres X and Y are collinear is false and ( arc PQ ii! \u2234 D ( O, F, G, H are touching,. Quadrilateral a convex quadrilateral is equal to the opposite angle, \ua838ABCD is a cyclic quadrilateral are supplementary by folding... Circles ] seg ED theorem of cyclic quadrilateral class 10 the centre of the interior opposite angle to cyclic quadrilateral PQ... Locations of point R are on line l touches a circle, EF!, answer the following questions, Exercises and theorems of Chapter 8 9... Class 9 quadrilaterals ; Serial order wise ; theorems AB and seg ML the... Enclosed sides, and 4 angles side of OE is 6 units: //you.tube/teachoo \u22a5 side,. Group MEMBERS \u2022 BHAVYA \u2022 SUPREET \u2022 GEETA \u2022 DEEPALI \u2022 RICHA \u2022 AMAN 17 on! Point O is the diameter of a circle with centre O us discuss what is the to! Be defined and is known as cyclic quadrilateral is supplementary????. Solutions of all NCERT questions, hence find the measures of \u2220TQS and arc TS at... Not intersect at O i. m ( arc PR ) ii quadrilateral and Get answer from subject and. Of DH and CF be X and Y respectively discuss some \u2026 the sum of quadrilateral. Rs, and website in this browser for the next time i comment were proved called... Cyclic quadrilaterals - Get Get topics notes, Online test, video lectures.... Locations of point R are on line l a side of OE prove that perpendicular. Intersect each other then it is both tangential and cyclic 3 Geometry Class 10 Maths Chapter... H are touching points Get Get topics notes, Online test, video lectures Doubts. Notes, Online test, video lectures & Doubts and Solutions for ICSE Class 10,! A diameter of a circle is 360-degrees and the sides can be drawn measures of and! If RS = 12, what is the secant, \u25a1ABOC is a.... Is highly rated by Class 9 = 4.5, EB = 5.5 3.3. Radii 5.5 cm and 3.3 cm respectively touch each other at O radius cm.\nDragons' Den Franchise, Adrift Game Engine, Sidony Dragon Age, Dead Island: Epidemic, Pixies - Beneath The Eyrie Songs, American Constitution Society Vs Federalist Society, Peace Movement Examples, Who Bought The Monarch Beach Resort,", "date": "2021-10-27 07:31:12", "meta": {"domain": "patriciabarber.com", "url": "https://patriciabarber.com/i-said-yto/4128a8-theorem-of-cyclic-quadrilateral-class-10", "openwebmath_score": 0.47614631056785583, "openwebmath_perplexity": 3650.9566282298897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974042642048694, "lm_q2_score": 0.8872045922259089, "lm_q1q2_score": 0.8641751050494584}}
{"url": "https://math.stackexchange.com/questions/2067752/does-every-number-not-ending-with-zero-have-a-multiple-without-zero-digits-at-al/2067782", "text": "# Does every number not ending with zero have a multiple without zero digits at all?\n\nDoes every number not ending with zero have a multiple without zero digits at all?\n\nWe consider numbers in their usual base 10 representation. we don't consider numbers ending with $0$, because all multiples of such numbers end with $0$.\n\nIt can be proven that all powers of $2$ have such a multiple: Indeed, for every number of the form $2^n$, we can find a multiple $l$ of it whose last $n$ digits are nonzero (This can be done by repeatedly multiplying a number by $10^k+1$, as explained here), and then forget all other digits; these last $n$ digits constitute a number that is still divisible by $2^n$, since it is obtained from $l$ by subtracting a multiple of $10^n$ which is in particular a multiple of $2^n$.\n\nThe same argument shows that every power of $5$ has a multiple without zeros, but this does not seem to generalize to other numbers.\n\nMy intuition is actually that the claim is not true, since the density of numbers without any zero digits approaches $0$ for big numbers. This is because the probability of a random number with at most $n$ digits to contain only nonzero digits is $(\\frac{9}{10})^n$. Thus big numbers \"nearly always\" contain a zero. This hints on the existence of a counterexample, but does not prove its existence.\n\n\u2022 I'm not sure what the answer is, but I think your heuristic argument misses something big: Sure, the proportion of $n$ digit numbers with only non-zero digits is $(9/10)^n$, but the number of multiples of a number $k$ with $n$ digits is roughly $\\frac{10^n}k$ - so multiply those together and we \"expect\" there to be $\\frac{9^n}k$ multiples of $k$ with $n$ digits and no zeros. (This is certainly true of numbers coprime to 10) \u2013\u00a0Milo Brandt Dec 21 '16 at 20:51\n\u2022 User @lulu seems to have deleted his correct observation that this is also true for all numbers coprime to $10$, by Fermat's little theorem. \u2013\u00a0Emolga Dec 21 '16 at 21:06\n\u2022 I deleted the post because someone pointed out this question which generalizes the straightforward argument I gave. \u2013\u00a0lulu Dec 21 '16 at 21:08\n\u2022 Just a note: in fact you can get stronger results for multiples of $2^n.$ For any odd digit $a$ and even digit $b$ you can find a multiple of $2^n$ using only $a$ and $b$ \u2013\u00a0cats Dec 21 '16 at 21:12\n\u2022 Found it - see \"104 Number Theory Problems from the training of the USA IMO Team\" Introductory Problems #52 \u2013\u00a0cats Dec 21 '16 at 21:18\n\nIf $n$ is relatively prime to $10$ then the answer is easy, see for example this post.\n\nIf, $n=2^k m$ with $m$ relatively prime to $10$ or $n=5^km$ with $m$ relatively prime to $10$, then prove first the following by induction:\n\nClaim 1 $2^k$ has a $k$ digits multiple containing only the digits $1$ and $2$.\n\nClaim 2 $5^k$ has a $k$ digits multiple containing only the digits $1,2,3,4$ and $5$.\n\nBoth are easy induction exercises.\n\nOnce you prove this, proceed as follows (same idea as in the posted link):\n\n$n=2^k m$ or $n=5^km$ with $gcd(m,10)=1$.. Then, we know that $2^k$ respectively $5^k$ has a multiple of the form $\\overline{a_1...a_k}$\n\nLook at the following numbers $$\\overline{a_1...a_k} , \\overline{a_1...a_ka_1...a_k} , \\overline{a_1...a_ka_1...a_ka_1...a_k} , ...$$\n\nIn this infinite list, there exists two numbers with the same reminder when divided by $m$. Their difference is a multiple of $m$.\n\nTherefore $$m| \\overline{a_1...a_ka_1...a_ka_1...a_k00...0}=\\overline{a_1...a_ka_1...a_ka_1...a_k}\\cdot 10^l$$ Since $m$ is relatively prime to $10$ it follows that $$m|\\overline{a_1...a_ka_1...a_ka_1...a_k}$$ Since $2^k$ respectively $5^k$ also divide $\\overline{a_1...a_ka_1...a_ka_1...a_k}$, and they are relatively prime to $m$, it follows that $$n|\\overline{a_1...a_ka_1...a_ka_1...a_k}$$\n\nP.S. We proved that $n$ has a multiple which can be written only with the digits $1,2,3,4,5$. These digits can be replaced by any $5$ digits which cover all classes $\\pmod{2}$ and $\\pmod{5}$.\n\n\u2022 Could you please give a proof of the claims? \u2013\u00a0J. Abraham Feb 25 '17 at 19:56\n\u2022 @J.Abraham $P(k) \\Rightarrow P(k+1)$. Let $A$ be the $k$ digit number which is divisible by $a^k$ where $a=2$ or $5$. Then $$A=a^k \\cdot m$$ and $$10^{k}=a^k \\cdot l$$ where $l$ is invertible modulo $a$. Then the equation $$x l +m \\equiv 0 \\pmod{a}$$ has an unique solution in the listed digits. Then $$x \\cdot 10^k+A$$ is the desired $k+1$ digit number. \u2013\u00a0N. S. Feb 26 '17 at 23:07", "date": "2019-07-24 09:44:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2067752/does-every-number-not-ending-with-zero-have-a-multiple-without-zero-digits-at-al/2067782", "openwebmath_score": 0.8284635543823242, "openwebmath_perplexity": 122.72536817699505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682488058381, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.864172417519144}}
{"url": "http://mathhelpforum.com/number-theory/49939-congruences-2-a.html", "text": "1. ## congruences 2\n\nSlove the system of congruences\n\n3x = 2 mod 4\n4x = 1 mod 5\n6x = 3 mod 9\n\nis it no solution since 6 and 9 do not have an inverse since the gcd of 6 and 9 is not one thus u cant write it in the form x=a mod p\n\n2. I have been trying to find the inverse and solve it that way but i dont know if that is right?\n\n3. If $ac \\equiv bc \\ (\\text{mod } m)$ and $(c,m) = d$, then $a \\equiv b \\left(\\text{mod } \\frac{m}{d}\\right)$\n\nKeeping in this in mind, I'm going to simplify the congruences first:\n$\\begin{array}{rcll} 3x & \\equiv & 2 & (\\text{mod } 4) \\\\ 3x & \\equiv & 6 & (\\text{mod } 4) \\\\ x & \\equiv & 2 & (\\text{mod } 4) \\end{array}$........ $\\begin{array}{rcll} 4x & \\equiv & 1 & (\\text{mod } 5) \\\\ 4x & \\equiv & -4 & (\\text{mod } 5) \\\\ x & \\equiv & -1 & (\\text{mod } 5) \\\\ x & \\equiv & 4 & (\\text{mod } 5) \\end{array}$........ $\\begin{array}{rcll} 6x & \\equiv & 3 & (\\text{mod } 9) \\\\ 6x & \\equiv & 12 & (\\text{mod }9) \\\\ x & \\equiv & 2 & (\\text{mod } 3) \\end{array}$\n\nSo we're left with the system:\n$\\begin{array}{rcll} x & \\equiv & 2 & (\\text{mod } 4) \\\\ x & \\equiv & 4 & (\\text{mod } 5) \\\\ x & \\equiv & 2 & (\\text{mod } 3) \\end{array}$\n\nAnd by the Chinese Remainder theorem, since $(4, 5, 3) = 1$, there is a unique solution modulo $4\\cdot 5 \\cdot 3 = 60$. Should be pretty simply to solve this one.\n\n4. Originally Posted by o_O\nIf $ac \\equiv bc \\ (\\text{mod } m)$ and $(c,m) = d$, then $a \\equiv b \\left(\\text{mod } \\frac{m}{d}\\right)$\n\nKeeping in this in mind, I'm going to simplify the congruences first:\n$\\begin{array}{rcll} 3x & \\equiv & 2 & (\\text{mod } 4) \\\\ 3x & \\equiv & 6 & (\\text{mod } 4) \\\\ x & \\equiv & 2 & (\\text{mod } 4) \\end{array}$........ $\\begin{array}{rcll} 4x & \\equiv & 1 & (\\text{mod } 5) \\\\ 4x & \\equiv & -4 & (\\text{mod } 5) \\\\ x & \\equiv & -1 & (\\text{mod } 5) \\\\ x & \\equiv & 4 & (\\text{mod } 5) \\end{array}$........ $\\begin{array}{rcll} 6x & \\equiv & 3 & (\\text{mod } 9) \\\\ 6x & \\equiv & 12 & (\\text{mod }9) \\\\ x & \\equiv & 2 & (\\text{mod } 3) \\end{array}$\n\nSo we're left with the system:\n$\\begin{array}{rcll} x & \\equiv & 2 & (\\text{mod } 4) \\\\ x & \\equiv & 4 & (\\text{mod } 5) \\\\ x & \\equiv & 2 & (\\text{mod } 3) \\end{array}$\n\nAnd by the Chinese Remainder theorem, since $(4, 5, 3) = 1$, there is a unique solution modulo $4\\cdot 5 \\cdot 3 = 60$. Should be pretty simply to solve this one.\ni get the right answer after applying the CRT to the reduced congruences, but i still dont know how u reduced the congruences or simplified them. Could you maybe explain it a little bit. Could u find the inverse modulo of each congruence and and reduce it to the form x=a mod m\n\n5. Hello,\n\nDivide the formula of o_O into 2 formulae:\n\n1) If $ac\\equiv bc\\pmod m$ and $(c, m)=1$ then $a\\equiv b\\pmod m$.\n2) If $ac\\equiv bc\\pmod {mc}$ then $a\\equiv b\\pmod m$.\n\n1) corresponds to the case where you can find an inverse.\n2) is a technique to decrease the modulus.\nYou add the modulus as many times as you want so that the coefficient of x and the right-hand-side have a common divisor.\n\nBye.\n\n6. What I did was play around with the congruences a little. Remember, they sort of act like equality signs as well. Let's take the middle one I did:\n\n$\n\\begin{array}{rcll} 4x & \\equiv & 1 & (\\text{mod } 5) \\\\ 4x & \\equiv & -4 & (\\text{mod } 5) \\qquad \\text{1. Subtracted 5} \\\\ x & \\equiv & -1 & (\\text{mod } 5) \\qquad \\text{2. Divided both sides by 4} \\\\ x & \\equiv & 4 & (\\text{mod } 5) \\qquad \\text{3. Added 5 to get a least residue} \\end{array}\n$\n\nBasically, I wanted to get rid of the coefficient in front of the x and what I wanted is to ultimately divide it out:\n1. If something is congruent to 4x mod 5, then adding or subtracting multiple of 5's will still be congruent to 4x. Remember: $a \\equiv b \\ (\\text{mod m}) \\ \\iff \\ a = b + km$\n\n2. This refers to the very first statement I had in my first post\n\n3. Again, adding and subtracting multiples of the modulus will not change the congruence as we did in step 1. The main purpose in doing so was to get a least residue (i.e. the \"remainder\")\n\nEdit: Oh beaten!", "date": "2016-08-29 15:11:34", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/49939-congruences-2-a.html", "openwebmath_score": 0.8481968641281128, "openwebmath_perplexity": 287.7361449484711, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682444653242, "lm_q2_score": 0.8740772286044094, "lm_q1q2_score": 0.8641723991314373}}
{"url": "http://openstudy.com/updates/511303b4e4b0e554778aafec", "text": "## Anita505 2 years ago Suppose that we have a white urn containing four white balls and one red ball and have a red urn containing one white ball and four red balls. An experiment consists of selecting at random a ball from the white urn and then (without replacing the first ball) Selecting at random a ball from the urn having the colour of the first ball. Find the probability that the second ball is red. The probability that the second ball is red is ______.\n\n1. kropot72\n\nThere are two situations to consider: (1) If a white ball is selected on the first draw then the second draw will be from the white urn. Probability of white on the first draw is 4/5 and the probability of red on the second draw is 1/4. P(red second ball) = 4/5 * 1/4 = 4/20 ...............(1) (2) If a red ball is selected on the first draw then the second draw will be from the red urn. Probability of red on the first draw is 1/5 and the probability of red on the second draw is 4/5. P(red second ball) = 1/5 * 4/5 = 4/25 ...............(2) The probability that the second ball is red is the sum of the fractions (1) and (2).\n\n2. Anita505\n\nso the sum of (4/20)+(4/25) So in this case the answer would be 0.36 correct?\n\n3. kropot72\n\nCorrect, or alternatively 9/25.\n\n4. Anita505\n\nthank you for showing me the process\n\n5. kropot72\n\nYou're welcome :)\n\n6. Anita505\n\nAn urn contains 3 one-dollar bills, 1 five- dollar bill and 1 ten dollar bill. A player draws bills one at a time without replacement from the urn until a ten dollar bill is drawn. Then the game stops. All bills are kept by the player. Determine: A) The probability of winning $15 B)The probability of winning all bills in the urn C) The probability of the game stopping at the second draw. 7. Anita505 can you help me with this and show me the process? 8. satellite73 as far as i can see there is only one way to win$15: first draw the five, then draw the ten\n\n9. satellite73\n\nsince there are 5 bills, and one is a five, the probability of drawing the five first is $$\\frac{1}{5}$$ then there are 4 bills of which one is a ten, the probability of drawing a ten second given that the first bill was a five is $$\\frac{1}{4}$$ the probability of both things occurring is $\\frac{1}{5}\\times \\frac{1}{4}$\n\n10. Anita505\n\nso for part a) it would be 1/20? the answer\n\n11. satellite73\n\nyes\n\n12. Anita505\n\nokay thank you but i need help with part b and c\n\n13. satellite73\n\nfor b) it means you pick the ten dollar bill last right?\n\n14. satellite73\n\nthe probability you pick the ten dollar bill last is the same as the probability you pick the ten dollar bill first, namely $$\\frac{1}{5}$$\n\n15. Anita505\n\nso b then in this case is 1/5\n\n16. satellite73\n\nand for the last one, that means you pick something other then the ten dollar bill, and then you pick the ten dollar bill that is also $$\\frac{1}{5}$$ via $\\frac{4}{5}\\times \\frac{1}{4}=\\frac{1}{5}$\n\n17. satellite73\n\nmore simply put, the probability you pick the ten first, second, third, fourth or fifth are all the same, namely $$\\frac{1}{5}$$\n\n18. Anita505\n\nOkay thank you for your assistance i have one last question to ask,\n\n19. satellite73\n\ndo me a favor and post in a new thread this is hard so scroll down to\n\n20. Anita505\n\nA grade 11 art class is offering students two choices for a project: a pottery project and a mixed media project. Of the 46 students in the class, 23 have selected to do the pottery project and 33 have selected to do the mixed media project (notice some students have decided to do both) It two students are selected at random from the class to show their finished project(s), what is the probability that at least one pottery project and at least one mixed media project will be shown? Probability (given to three decimal places)", "date": "2016-02-06 15:47:21", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/511303b4e4b0e554778aafec", "openwebmath_score": 0.7426160573959351, "openwebmath_perplexity": 560.7432289030303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109507462227, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8641487603816977}}
{"url": "http://mathhelpforum.com/calculus/76627-l-hopital-s-rule-question.html", "text": "1. ## L'Hopital's Rule question\n\nI have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:\n\nFind the limit of [ln(x)]/[ln(2x)] as x goes to infinity.\n\nI tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?\n\n2. Originally Posted by pianopiano\nI have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:\n\nFind the limit of [ln(x)]/[ln(2x)] as x goes to infinity.\n\nI tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?\nNote that $\\frac{\\ln x}{\\ln(2x)}=\\frac{\\ln x}{\\ln 2+\\ln x}$. Since $\\lim_{x\\to\\infty}\\frac{\\ln x}{\\ln(2x)}=\\lim_{x\\to\\infty}\\frac{\\ln x}{\\ln 2+\\ln x}=\\frac{\\infty}{\\infty}$, we can apply L'Hopitals rule.\n\nSo $\\lim_{x\\to\\infty}\\frac{\\ln x}{\\ln(2x)}=\\lim_{x\\to\\infty}\\frac{\\ln x}{\\ln 2+\\ln x}=\\lim_{x\\to\\infty}\\frac{\\displaystyle\\frac{1}{x} }{\\displaystyle\\frac{1}{x}}=\\lim_{x\\to\\infty}\\frac {x}{x}=1$\n\nDoes this make sense?\n\n3. Thanks, it does make sense!\n\nHowever, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?\n\n4. Originally Posted by pianopiano\nThanks, it does make sense!\n\nHowever, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?\nYou could...but you would have to note that $\\forall a\\in\\mathbb{R}\\backslash\\left\\{0\\right\\},~\\frac{\\, d}{\\,dx}\\left[\\ln\\!\\left(ax\\right)\\right]=\\frac{1}{x}$\n\n5. $\\lim_{x\\to\\infty}\\frac{\\log x}{\\log 2 + \\log x}=\\lim_{x\\to\\infty}\\frac{1}{\\underbrace{\\frac{\\lo g 2}{\\log x}}_{\\to 0}+1}=1$.\n\nOr you must use L'H?\n\n6. Originally Posted by Abu-Khalil\n$\\lim_{x\\to\\infty}\\frac{\\log x}{\\log 2 + \\log x}=\\lim_{x\\to\\infty}\\frac{1}{\\underbrace{\\frac{\\lo g 2}{\\log x}}_{\\to 0}+1}=1$.\n\nOr you must use L'H?\n\nI was going to point that out.\nThere's no need for L'Hopital's Rule at all, just divide by the common term, log x.", "date": "2016-10-28 07:22:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/76627-l-hopital-s-rule-question.html", "openwebmath_score": 0.970382809638977, "openwebmath_perplexity": 218.57325297147378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109485172559, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8641487442264476}}
{"url": "https://math.stackexchange.com/questions/2001389/using-the-mean-value-theorem", "text": "# Using the Mean Value Theorem\n\nThe (classical) Mean Value Theorem states that if $f$ is a continuous function on $[a,b]$ and is differentiable on $(a,b)$, then there exists a point $c \\in (a,b)$ where $f'(c) = \\dfrac{f(b)-f(a)}{b-a}$.\n\nHere is a \"real-world\" application of the theorem. Suppose two racers, Barry and Harry, start a race at time $t=0$ and end the race in a tie, both crossing the finish line at, say, time $t = 10$. Let $s(0)$ be the position of the starting line, and let $s(10)$ be the position of the finish line.\n\nThe position of each racer is continuous on [0,10] and (presumably) differentiable on (0,10). So the MVT tells us that at some time $t_0 \\in (0,10)$, Barry's velocity equaled $s'(t_0)=v(t_0)=\\dfrac{s(10)-s(0)}{10}$, and similarly there is a time $t_1 \\in (0,10)$ where Harry's velocity equaled $s'(t_1)=v(t_1)=\\dfrac{s(10)-s(0)}{10}$.\n\nQuestion: Does the MVT also guarantee that $t_0=t_1$? That is, is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race? If so, why?\n\n\u2022 I think the MVT doesn't say that the point $c$ is unique, hence you could find other times say $t_2$ ro $t_3$ where that holds. But I don't know if this helps \u2013\u00a0Euler_Salter Nov 6 '16 at 1:49\n\u2022 Let $x(t),y(t)$ be $x(t)=t$, $y(t)=\\sqrt{t}$. Clearly $x(0)=y(0)=0$ and $x(1)=y(1)=1$, but $x'(t)=1=\\frac{x(1)-x(0)}{1}=1$ for all $t$, while $y'(t)=1$ only when $t=\\sqrt{1/2}$, so they need not be equal \u2013\u00a0user160738 Nov 6 '16 at 1:53\n\n## 3 Answers\n\nYou're being ambiguous. If you mean to ask if their velocities will be the same at some time in the interval $[0,10]$, yes. The result follows by applying the mean value theorem to the function $x(t) = s_{b}(t) - s_{h}(t)$ on the interval $[0, 10]$. $s_b$ and $s_h$ are the position functions of Barry and Harry respectively.\n\nHowever, if you apply the mean value theorem to the functions independently, the \"mean value\" times may not be identical. Consider the following example, where we suppose the velocities are continuous. If Barry runs very fast right at the start, almost up to the finish line, over the course of $1$ second and then slows down like a caterpillar for the other $9$ seconds, the mean value time for Barry will be somewhere around $t = 1$ seconds while he is slowing down.\n\nIf Harry does the opposite, that is, he's extremely slow for the first $9$ seconds and then lightning fast to catch up to Barry at the final second, then his \"mean value time\" will be somewhere around $t=9$ as he speeds up.\n\nThey will share the same velocity at some time ($g(t) = v_{b}(t) - v_{h}(t)$ is continuous, initially positive and negative at the end), but this shared velocity will not be the \"mean value\" velocity $\\frac{s(10) - s(0)}{10}$.\n\n\u2022 Thank you for answering my question. I don't see how I was being ambiguous (my question was very clearly stated, was it not?), but thank you nonetheless. \u2013\u00a0Mr Toad Nov 6 '16 at 5:43\n\u2022 Your question was ambiguous because you ask two different things in the last paragraph of the question, and in fact they happen to have different answers. As MathematicsStudent1122 wrote, the answer to \"does the MVT also guarantee that $t_0 = t_1$?\" is no, but the answer to \"is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race?\" is yes. \u2013\u00a0David Z Nov 6 '16 at 7:40\n\u2022 @David Z, thanks for the clarification. \u2013\u00a0Mr Toad Nov 6 '16 at 14:53\n\nYou can use Cauchy's mean value theorem. It says that if $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists $c\\in (a,b)$ such that $$(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).$$ If we use $f$ to denote the position of Barry and $g$ to denote the position of Harry then we have $f(0)=g(0)=0$ and $f(T)=g(T)=L=$length. ($T$ denotes the total time.) So, there exists a point $c\\in (0,T)$ such that $$Lg'(c)=(f(T)-f(0))g'(c)=(g(T)-g(0))f'(c)=Lf'(c).$$ That is, there exists $c$ such that $f'(c)=g'(c).$ So they must have had the same velocity at some point.\n\n\u2022 +1 for the perfect answer. It does not appear at first that a single value of $c$ will work for $f$ and $g$ but Cauchy's MVT ensures that this is possible. \u2013\u00a0Paramanand Singh Nov 6 '16 at 7:09\n\nAssume runner 1 runs at speed $2t$ feet per second, so that in ten seconds, runner 1 travels $\\int_0^{10}2tdt=100$ feet. Assume runner $2$ runs at speed $\\frac{3t^2}{1000}$ so that in $10$ seconds he also travels $100$ feet. However, $2t=\\frac{3t^2}{1000}$ has roots $0$ and $\\frac{2000}{3}>10$, so the runners never simultaneously travel the same speed during the race (besides the start, speed $0$). (There is nothing special about this example, I just started with a simple speed function $2t$ and tried to fudge another function to make it work. Im sure there are many more counterexamples)\n\n\u2022 I think the example doesn't hold for runner 2. His initial speed isn't zero. \u2013\u00a0Cehh\u0390ro Nov 6 '16 at 1:55\n\u2022 @O.VonSeckendorff thanks for the catch. It is fixed. \u2013\u00a0TomGrubb Nov 6 '16 at 2:00", "date": "2020-04-01 15:48:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2001389/using-the-mean-value-theorem", "openwebmath_score": 0.9100626111030579, "openwebmath_perplexity": 272.80567087145636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109503004293, "lm_q2_score": 0.8774767810736692, "lm_q1q2_score": 0.864148742635722}}
{"url": "https://mathematica.stackexchange.com/questions/81323/find-the-volume-of-the-region-defined-by-xyz4", "text": "# Find the volume of the region defined by $|x|+|y|+|z|<4$\n\nI want to find the volume of the region defined by the following inequality:\n\n$\\quad \\quad |x|+|y|+|z|<4$\n\nIn addition to calculating the volume of this region, I would also like to be able to obtain its representation as a plot.\n\nI have started by plotting the vertices, i.e., the points (4, 0, 0), (-4, 0, 0), (0, 4, 0), (0, -4, 0), (0, 0, 4), and (0, 0, -4), but I am not sure how to continue from there.\n\n\u2022 Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. \u2013\u00a0bbgodfrey May 1 '15 at 3:23\n\u2022 @julie You may want to consider accepting one of the answers that have been provided below, if they solve your problem. \u2013\u00a0MarcoB May 3 '15 at 17:45\n\nvolume = Integrate[1,  Element[{x, y, z}, ImplicitRegion[Abs[x] + Abs[y] + Abs[z] < 4, {x, y, z}]]]\n\n(*256/3*)\n\nContourPlot3D[Abs[x] + Abs[y] + Abs[z] == 4, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}]\n\n\n## A small addition from Mathematica v. 10\n\nIn his answer @Ivan has already nailed the most important part of the question, i.e. the formal representation of your region as an ImplicitRegion object:\n\nregion = ImplicitRegion[Abs[x] + Abs[y] + Abs[z] < 4, {x, y, z}]\n\n\nI just wanted to add that, if you are on Mathematica 10 or newer, you can also use the RegionMeasure or Volume functions to calculate the volume of a region:\n\nRegionMeasure[region]\nVolume[region]\n\n(* 256/3 *)\n\n\n## How to arrive at the shape of this region\n\nIn response to @julie's question in the comment, here is how one can imagine constructing the 3D region.\n\nConsider the corresponding equality in 2D, disregard the absolute values for now, and let's just concentration on the first quadrant ($x>=0$,$y>=0$). The resulting $x + y = 4$ is just the equation of a line, i.e. $y = 4 - x$. The region in the first quadrant in which the inequality $x+y<4$ holds is the region between that line and the axes:\n\nRegionPlot[\nx + y < 4 && x >= 0 && y >= 0,\n{x, -5, 5}, {y, -5, 5},\nAxes -> True, Frame -> False\n]\n\n\nTaking the absolute value of $x$ is geometrically equivalent to mirroring that triangle with respect to the $y$ axis as shown below. Since we are taking its absolute value, we will also remove the $x>=0$ restriction.\n\nRegionPlot[\nAbs[x] + y < 4 && y >= 0,\n{x, -5, 5}, {y, -5, 5},\nAxes -> True, Frame -> False\n]\n\n\nTaking the absolute value of $y$ is similarly equivalent to mirroring this figure with respect to the $x$ axis. The result is the square region below:\n\nRegionPlot[\nAbs[x] + Abs[y] < 4,\n{x, -5, 5}, {y, -5, 5},\nAxes -> True, Frame -> False\n]\n\n\nIn the three-dimensional case, you can work your way through in a similar way: $x+y+z=4$ is the equation of a plane that defines a trangular wedge in the first quadrant, and so forth.\n\nAs for the numerical value, you could consider that your figure is made up of two square pyramids. The volume of a square pyramid is $V=a^2h/3$ where $a$ is the length of the base edge, and $h$ is the height. In your case, you can calculate the base edge as the hypotenuse of the smallest 2D triangle we introduced above ($a^2=4^2+4^2=32$), and the height of the pyramid is 4. The volume of this pyramid is 128/3, and your figure is made of two such pyramids, so $2\\times128/3 = 256/3$.\n\n\u2022 thank you! but how was the final answer calculated? how did you know what all the sides were and where those points went? \u2013\u00a0julie May 1 '15 at 3:28\n\u2022 @julie I amended my answer to include a way to construct the region. Let me know if that answers your question appropriately. \u2013\u00a0MarcoB May 1 '15 at 4:27\n\nAn alternate approach to calculating the volume\n\nIntegrate[\nBoole[Abs[x] + Abs[y] + Abs[z] < 4],\n{x, -Infinity, Infinity},\n{y, -Infinity, Infinity},\n{z, -Infinity, Infinity}]\n\n\n256/3\n\nAs a cone (with a square base), the volume is one-third the base times the height: 1/3 * (4 Sqrt[2])^2 * 8.\n\nAnother plot, based on the OP's coordinates:\n\nConvexHullMesh@Join[4 IdentityMatrix[3], -4 IdentityMatrix[3]]", "date": "2020-04-01 21:37:52", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/81323/find-the-volume-of-the-region-defined-by-xyz4", "openwebmath_score": 0.5827907919883728, "openwebmath_perplexity": 463.0068835190746, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109525293959, "lm_q2_score": 0.8774767746654976, "lm_q1q2_score": 0.8641487382807507}}
{"url": "http://finmining.com/hcd6t/ec9282-how-to-find-turning-point-of-parabola", "text": "So the x value is 0. You will get a point now. Back Function Institute Mathematics Contents Index Home. A parabola is a curve where any point is at an equal distance from: a fixed point (the focus), and ; a fixed straight line (the directrix) Get a piece of paper, draw a straight line on it, then make a big dot for the focus (not on the line!). Hence the equation of the parabola in vertex form may be written as $$y = a(x - 2)^2 + 3$$ We now use the y intercept at $$(0,- 1)$$ to find coefficient $$a$$. The implication is that throughout the observed range of the data, the expected probability of pt is an increasing function of expand_cap, though with some diminishing returns. When the function has been re-written in the form y = r(x + s)^2 + t, the minimum value is achieved when x = -s, and the value of y will be equal to t. x-intercepts in greater depth. A parabola can have either 2,1 or zero real x intercepts. Completing the square, we have \\begin{align*} y &= x^2 - 2ax + 1 \\\\ &= (x - a)^2 + 1 - a^2, \\end{align*} so the minimum occurs when $$x = a$$ and then $$y = 1 - a^2$$. What do you notice? You can take x= -1 and get the value for y. Example 2 Graph of parabola given vertex and a point Find the equation of the parabola whose graph is shown below. Graphs of quadratic functions have a vertical line of symmetry that goes through their turning point.This means that the turning point is located exactly half way between the x-axis intercepts (if there are any!).. The easiest way to find the equation of a parabola is by using your knowledge of a special point, called the vertex, which is located on the parabola itself. Only vertical parabolas can have minimum or maximum values, because horizontal parabolas have no limit on how high or how low they can go. Now play around with some measurements until you have another dot that is exactly the same distance from the focus and the straight line. Step 1: Find the roots of your \u2026 Substitute this x value into the equation y = x 2 \u2013 6x + 8 to find the y value of the turning point. Finding the maximum of a parabola can tell you the maximum height of a ball thrown into the air, the maximum area of a \u2026 There are two methods to find the turning point, Through factorising and completing the square.. Make sure you are happy \u2026 In either case, the vertex is a turning point \u2026 Does the slope always have to be in turning points? If the function is smooth, then the turning point must be a stationary point, however not all stationary points are turning points, for example has a stationary point at x=0, but the derivative doesn't change sign as there is a point of inflexion at x=0. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In the first two examples there is no need for finding extra points as they have five points and have zeros of the parabola. Any point, ( x 0 , y 0 ) on the parabola satisfies the definition of parabola, so there are two distances to calculate: Distance between the point on the parabola to the focus Distance between the point on the parabola to the directrix To find the equation of the parabola, equate these two expressions and solve for y 0 . The x-intercepts are the points or the point at which the parabola intersects the x-axis. If the parabola only has 1 x-intercept (see middle of picture below), then the parabola is said to be tangent to the x-axis. Let\u2019s work it through with the example y = x 2 + x + 6. A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). This is a second order polynomial, because of the x\u00b2 term. now find your y value by using the x value you just found by plugging it into your function . Example 1 . Yes, the turning point can be (far) outside the range of the data. Use this formula to find the x value where the graph turns. How to find the turning point (vertex) of a quadratic curve, equation or graph. STEP 1 Solve the equation of the derived function (derivative) equal to zero ie. The x-coordinate of the turning point = - $$\\frac{b}{2a}$$ ----- For example, if the equation of the parabola is . If the slope is , we max have a maximum turning point (shown above) or a mininum turning point . May 2008 218 59 Melbourne Australia Aug 24, 2009 #2 At the turning points of an equation the slope of y is zero. (Increasing because the quadratic coefficient is negative, so the turning point is a maximum and the function is increasing to the \u2026 A turning point may be either a local maximum or a minimum point. To find the turning point of a parabola, first find it's x-value, using the equation: -b/2a (from the quadratic form ax^2 + bx + c). Find the maximum number of turning points of each \u2026 A polynomial of degree n will have at most n \u2013 1 turning points. In other words the differential of the equation must be zero. How do I find the coordinates of a turning point? This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. The equation is y=4xsquare-4x+4. When the equation of the parabola is in this form: y = ax 2 + bx + c . The x-coordinate of the turning point = - $$\\frac{4}{2(3)}$$ = - $$\\frac{2}{3}$$ Plug this in for x to find the value of the y-coordinate. Does slope always imply we have a turning point? K. Kiwi_Dave. or the slope just becomes for a moment though you have no turning point. To graph a parabola, visit the parabola grapher (choose the \"Implicit\" option). Given that the turning point of this parabola is (-2,-4) and 1 of the roots is (1,0), please find the equation of this parabola. \u2026 Surely you mean the point at which the parabola goes from increasing to decreasing, or reciprocally. In this case, b = 0, since there is no b term, and a is 1 (the number before the x squared) : -b/2a = -0/2. This means: To find turning points, look for roots of the derivation. To find the turning point of a quadratic equation we need to remember a couple of things: The parabola ( the curve) is symmetrical; If we know the x value we can work out the y value! You therefore differentiate f(x) and equate it to zero as shown below. The coordinate of the turning point is (-s, t). Solution to Example 2 The graph has a vertex at $$(2,3)$$. Example 7: Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function . If we look at the function . Free functions turning points calculator - find functions turning points step-by-step This website uses cookies to ensure you get the best experience. If, on the other hand, you suppose that \"a\" is negative, the exact same reasoning holds, except that you're always taking k and subtracting the squared part from it, so the highest value y \u2026 If y=ax^2+bx+c is a cartesian equation of a random parabola of the real plane, we know that in its turning point, the derivative is null. So for example, given (2a): Vertex at (2, -6) One x intercept at 6 The axis will be x=2, so the given x intercept is 4 units to the right of the axis. Curve sketching Murray says: 19 Jun 2011 at 8:16 am [Comment permalink] Hi Kathryn and thanks for your input. \u2026 $0=a(x+2)^2-4$ but i do not know where to put \u2026 Such a point is called saddle point. How to find the turning point of a parabola: The turning point, or the vertex can be found easily by differentiation. Remember that the axis of symmetry is the straight line that passes through the turning point (vertex) of the parabola. How you think you find the turning point given the x-intercepts of a parabola? So, our starting or reference parabola formula looks like this: y = x 2. The turning point of a parabola is the vertex; this is also it's highest or lowest point. Reactions: \u2026 So the turning point is at $(a, 1 - a^2).$ So for your example: $$\\displaystyle \\frac {dy}{dx}=2x$$ So we set this equal to zero to get: $$\\displaystyle 2x=0$$ or x=0 . The vertex is at point (x,y) First find x by using the formula -b/2a <--- a = 2, b= -5 and c= 1 (because it is quadratic) So -(-5)/2(2) = 5/4 <--- your x value at the vertex or turning point is 5/4. And our equation that includes a horizontal translation looks like this: y = (x - h) 2. If the coefficient of the x 2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the \u201c U \u201d-shape. Write down the nature of the turning point and the equation of the axis of symmetry. And the lowest point on a positive quadratic is of course the vertex. When the parabola opens down, the vertex is the highest point on the graph \u2014 called the maximum, or max. A second approach is to find the turning point of the parabola. Turning Points and Intercepts of a Parabola Function. Published in: Education. The S.K.A. Horizontal translation for the parabola is changed by the value of a variable, h, that is subtracted from x before the squaring operation. It\u2019s hard to see immediately how this curve will look just by looking at the function. how to i find the turning point of that parabola? So remember these key facts, the first thing we need to do is to work out the x value of the turning point. If the parabola is upright - as these examples are - then it will be laterally symmetrical about its axis, which is the vertical line through the vertex. is the set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. GeoGebra can be used very easily to find the equation of a parabola: given three points, A, B, C input the command FitPoly[{A, B, C}, 2]. Worked examples. In \u2026 Did You Know That...? 17 Comments 2 Likes ... \u2013 12 12 \u2013 24 \u2013 12 = -24 this is the y-coordinate of the vertex So the vertex (turning point of this parabola is (-2,-24) HOW TO CALCULATE THE VERTEX (TURNING POINT) Recommended M\u1eabu \u1ed1p l\u01b0ng iphone se da th\u1eadt chuy\u00ean nghi\u1ec7p \u2026 In math terms, a parabola the shape you get when you slice through a solid cone at an angle that's parallel to one of its sides, which is why it's known as one of the \"conic sections.\" Here is a typical quadratic equation that describes a parabola. The turning point is when the rate of change is zero. To find the axis of symmetry, use this formula: x = -b/2a. solve dy/dx = 0 This will find the x-coordinate of the turning point; STEP 2 To find the y-coordinate substitute the x-coordinate into the equation of the graph ie. \u2026 Turning Points of Quadratic Graphs. I started off by substituting the given numbers into the turning point form. The graph below has a turning point (3, -2). y = 3x 2 + 4x + 1 . A parabola The set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. Parabola, Horizontal Translation. TURNING POINT The formula to find the x value of the turning point of the parabola is x = \u2013b/2a. This can help us sketch complicated functions by find turning points, points of inflection or local min or maxes. In example 3 we need to find extra points. There is also a spreadsheet, which can be used as easily as Excel. substitute x into \u201cy = \u2026\u201d This is a mathematical educational video on how to find extra points for a parabola. A turning point can be found by re-writting the equation into completed square form. No. The Parabola. In the case of a vertical parabola (opening up or down), the axis is the same as the x coordinate of the vertex, which is the x-value of the point where the axis of symmetry crosses the parabola. Solved: What is the turning point, or vertex, of the parabola whose equation is y = 3 x^2 + 6 x - 1? The Vertex of a Parabola The vertex of a parabola is the point where the parabola crosses its axis of symmetry. A parabola can have 2 x-intercepts, 1 x-intercept or zero real x intercepts. Second order how to find turning point of parabola, because of the turning point ( vertex ) of the turning point is (! Below has a vertex at \\ ( ( 2,3 ) \\ ) extra... To graph a parabola uses cookies to ensure you get the best.! It to zero ie x-intercepts are the points or the minimum value of the equation of the.!, or the slope just becomes for a moment though you have no turning point and straight. Equation into completed square form we have a maximum turning point of polynomial. Play around with some measurements until you have no turning point given the x-intercepts are the points the. Ax 2 + x + 6 how to find turning point of parabola Implicit '' option ) has a turning point be... 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( 2,3 ) \\ ) distance from the focus and the equation into completed square form spreadsheet. Have zeros of the parabola intersects the x-axis to do is to work out the x value you found! Jun 2011 at 8:16 am [ Comment permalink ] Hi Kathryn and thanks your... The x value you just found by re-writting the equation of the equation of how to find turning point of parabola function... The given numbers into the turning point may be either a local maximum a! That describes a parabola: the turning point can be found by plugging it into function... First two examples there is also it 's highest or lowest point on the graph \u2014 the... The turning point of a polynomial of degree n will have at most n \u2013 1 turning points step-by-step website! -1 and get the best experience does the slope just becomes for a moment though have! Graph, or the point at which the parabola is x = \u2013b/2a turning point it through with the y. Up, the first thing we need to do is to work out x. Is shown below of degree n will have at most n \u2013 1 turning points calculator - find functions points. On the graph, or max are the points or the how to find turning point of parabola at which the parabola the! Vertex and a point find the y value by using the x value you just found plugging... X 2 by re-writting the equation of the derived function ( derivative ) equal to zero.!\n\nCarbon County Sheriff, Can You Use Beef Dripping Instead Of Suet, Carelink Phone Number, A Tender Heart Episode 7, Sheng Love Quotes, Redak Seribu Piano Cover, Beagle Poodle Mix For Sale In Ohio, Climate Change And Its Socio-economic And Political Dimensions,", "date": "2021-04-17 00:50:01", "meta": {"domain": "finmining.com", "url": "http://finmining.com/hcd6t/ec9282-how-to-find-turning-point-of-parabola", "openwebmath_score": 0.545737087726593, "openwebmath_perplexity": 378.61205411205185, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.990587412635403, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8641363370204548}}
{"url": "https://byjus.com/question-answer/let-a-n-denote-the-number-of-all-n-digit-positive-integers-formed-by-the-2/", "text": "Question\n\n# Let $$a_{n}$$ denote the number of all n-digit positive integers formed by the digits $$0,1$$ or both such that\u00a0no consecutive digits in them are $$0$$. Let $$b_{n}=$$ the number of such $$n$$-digit integers ending with digit $$1$$ and $$c_{n}=$$ the number of such $$n$$-digit integers ending with digit $$0$$.\u00a0Which of the following is correct?\n\nA\na17=a16+a15\nB\nc17c16+c15\nC\nb17b16+c16\nD\na17=c17+b16\n\nSolution\n\n## The correct option is A $$a_{17}=a_{16}+a_{15}$$In such a number either last digit is $$'0'$$ or $$'1'$$When you consider $$'a_1'$$, only one number is possible i.e. $$1$$When you consider $$'a_2'$$, 2 such numbers are possible i.e. $$10$$, $$11$$When you consider $$'a_3'$$, 3 such numbers are possible i.e. $$101$$, $$111$$, $$110$$When you consider $$'a_4'$$, 5 such numbers are possible i.e. $$1010$$, $$1011$$, $$1110$$, $$1101$$, $$1111$$By observing this we get a relationship which is $$a_n$$ $$=$$ $$a_{n-1}$$ $$+$$ $$a_{n-2}$$So, $$a_{17}$$ $$=$$ $$a_{16}$$ $$+$$ $$a_{15}$$(Alternate Method)Using Recursion formula$$a_n=a_{n-1}+a_{n-2}$$Similarly,\u00a0$$b_n=b_{n-1}+b_{n-2}$$ and\u00a0$$c_n=c_{n-1}+c_{n-2}$$ \u00a0 \u00a0$$\\forall\\ n\\geq 3$$and $$a_n=b_n+c_n$$ \u00a0$$\\forall\\ n\\geq 1$$So, $$a_1=1, a_2=2, a_3=3, a_4=5, a_5=8$$ .......$$b_1=1, b_2=1, b_3=2, b_4=3, b_5=5, b_6=8$$ .......$$c_1=0, c_2=1, c_3=1, c_4=2, c_5=3, c_6=5$$ .......Using this we get $$b_{n-1}=c_n$$$$\\therefore a_{17}=a_{16}+a_{15}$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-28 22:06:57", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/let-a-n-denote-the-number-of-all-n-digit-positive-integers-formed-by-the-2/", "openwebmath_score": 0.9698213338851929, "openwebmath_perplexity": 726.1364534351834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874120796498, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8641363332481726}}
{"url": "https://math.stackexchange.com/questions/2734908/prove-that-the-set-c-x-y-in-mathbbr2-max-x-y-leq-1-is", "text": "# Prove that the set $C = \\{(x,y) \\in \\mathbb{R}^2:\\max \\{ |x|,|y|\\}\\leq 1 \\}$ is convex.\n\nProve that $C = \\{(x,y) \\in \\mathbb{R}^2:\\max \\{ |x|,|y|\\}\\leq 1 \\}$ is a convex set.\n\nI am using the following definition for a convex set\n\nLet $D \\subseteq \\mathbb{R}^n$. The set $D$ is said to be convex if, given $\\bar{x},\\bar{y}\\in D$, we have that $$(1-t)\\bar{x} \\ +t\\bar{y} \\in D$$ for all $t\\in [0,1].$\n\nLet $\\bar{x}=(x_1,y_1)$ and $\\bar{y}=(x_2,y_2)$ be arbitrary points such that $\\bar{x},\\bar{y} \\in C$, and let $t\\in [0,1].$ Now, we want to see that $(1-t)\\bar{x} \\ +t\\bar{y} \\in C.$ So then \\begin{align} (1-t)\\bar{x} \\ +t\\bar{y} &=(1-t)(x_1,y_1)+t(x_2,y_2)\\\\ &= ((1-t)x_1,(1-t)y_1)+ (tx_2,ty_2)\\\\ &= ((1-t)x_1+tx_2,(1-t)y_1+ty_2) \\end{align}\n\nAnd here is where I get stuck. I want to prove that $((1-t)x_1+tx_2,(1-t)y_1+ty_2)\\in C$, which I believe is the same as proving that $\\max\\{|(1-t)x_1+tx_2|,|(1-t)y_1+ty_2| \\}\\leq 1$, but I don't know how to proceed.\n\nAny help would be appreciated!\n\n\u2022 Use triangle inequality. \u2013\u00a0Youem Apr 13 '18 at 3:57\n\nSo $|(1-t)x_{1}+tx_{2}|\\leq(1-t)|x_{1}|+t|x_{2}|\\leq(1-t)+t=1$ and $|(1-t)y_{1}+ty_{2}|\\leq(1-t)|y_{1}|+t|y_{2}|\\leq(1-t)+t=1$ because $|x_{1}|,|x_{2},|y_{1}|,|y_{2}|\\leq 1$, this shows that $(1-t)\\overline{x}+t\\overline{y}\\in C$.\n$|(1-t)x_1 + tx_2| \\le (1-t)|x_1| + t |x_2| \\le (1-t) + t = 1$ the same for $y$\nLet $C_x = \\{(x,y) \\in \\mathbb{R}^2:|x|\\leq 1 \\}$ and $C_y = \\{(x,y) \\in \\mathbb{R}^2:|y|\\leq 1 \\}$. Then,\n\u2022 $C_x$ and $C_y$ are obviously convex\n\u2022 $C = C_x \\cap C_y$ is convex, too", "date": "2019-10-17 10:15:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2734908/prove-that-the-set-c-x-y-in-mathbbr2-max-x-y-leq-1-is", "openwebmath_score": 0.9992247819900513, "openwebmath_perplexity": 134.36807964022822, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874101345135, "lm_q2_score": 0.8723473697001441, "lm_q1q2_score": 0.8641363216889207}}
{"url": "https://mathoverflow.net/questions/364040/3-colored-triangulations-of-the-sphere-s2-and-sperners-lemma", "text": "# 3-colored triangulations of the sphere $S^2$, and Sperner's Lemma\n\nI noticed something about colored triangulations of the topological sphere $$S^2$$ and have a question about this.\n\nObservation. If you triangulate the sphere $$S^2$$ and color the vertices with three colors: then the number of 3-colored triangles is always even (or zero). In particular, there is no coloring with exactly one 3-colored triangle.\n\nFor a proof, view $$S^2$$ as two triangulated disks with matching coloring of the boundaries that are glued together. As their boundaries have the same number of color changes, we know from Sperner\u2019s Lemma that their triangulations have the same number (mod 2) of 3-colored triangles. So the total number of 3-colored triangles is even or zero.\n\nAs an interesting corollary, we get the characterization: A triangulated sphere has zero 3-colored triangles iff all cycles of the triangulation have an even number of color changes.\n\nI looked at the torus, the Klein bottle, and the projective plane, and I find that the observation is also true for them.\n\nEdit: Just for contrast, adding an example below of a \"soap bubble\" surface, where the two soap bubbles share a common disk. This surface allows for triangulations with even and odd numbers of 3-colored triangles (but like the other surfaces I looked at, cannot have just one).\n\nQuestion. I wonder whether this also follows from more general theorems about triangulations of surfaces, or about maximal planar graphs? I have consulted algebraic topology and graph theory texts, but could not find any results in that direction. Would you have a suggestion where else to look, or maybe a reference for that?\n\n\u2022 This looks related to Sperner's lemma: en.wikipedia.org/wiki/Sperner%27s_lemma Jun 24, 2020 at 22:51\n\u2022 @JanKyncl you are definitely right. It is like Sperner\u2019s Lemma \u201cwithout boundary\u201d. Or the way I looked at it, 2 x Sperner\u2019s Lemma and then glued together. Just in case you are interested, see this mathoverflow.net/q/362025/156936 Jun 25, 2020 at 3:47\n\u2022 \"Uneven\" numbers are commonly called odd. Jun 28, 2020 at 19:57\n\u2022 @VictorProtsak thanks a lot for your comment! I corrected the wording in the text now. Jun 29, 2020 at 11:32\n\nA counting proof shows that this observation is unrelated to the global topology.\n\nEvery edge is monochromatic or dichromatic. How many dichromatic edges are there? If each triangle tells you its number of dichromatic edges (either 0, 2, or 3), you can add these up and divide by two to get the total number of dichromatic edges (since every edge contributes to two triangles). So the number of trichromatic triangles must be even.\n\nThis proof works for $$k$$-dimensional manifolds when $$k$$ is even, since the number of $$k$$-colored $$(k-1)$$-simplices bounding any $$k$$-simplex must be 0, 2, or $$k+1$$.\n\nYour corollary similarly transfers to higher even dimensions, at least for orientable manifolds, replacing \"cycles of edges\" with \"hypersurfaces of $$(k-1)$$-simplices\", and \"even number of color changes\" with \"even number of $$k$$-colored $$(k-1)$$-simplices\".\n\n\u2022 I'm not sure which direction you want to go in with this, but I think it falls pretty squarely in combinatorics and graph theory. What you (not incorrectly) call \"triangulations of $S^2$\" are often referred to as maximal planar graphs. My proof is very similar to the handshaking lemma. People have also considered coloring and planarity on tori and other manifolds.\n\u2013\u00a0Matt\nJun 24, 2020 at 19:32\n\u2022 thanks a lot for this helpful comment and great links. This is excellent Jun 24, 2020 at 20:20\n\nJust to close the loop on this: the double-counting argument in the answer of user Matt allows for a nice visual proof of the (2-dim.) Lemma of Sperner. Just want to capture it here, as it connects nicely with the triangulation of the sphere / the maximal planar graph in my OP question.\n\nStart with a triangulated polygon in the plane, and label each vertex with one of 3 colors. The example just shows the boundary of such a triangulated, 3-colored polygon. Claim (Sperner\u2019s Lemma): If the boundary has an odd number of color changes, then a 3-colored triangle exists in the polygon triangulation. In fact, more generally, an odd number of such 3-colored triangles exists.\n\nProof: Go to 3-dimensional space, and build a \u201ctent\u201d over the polygon like in the diagram: add a colored vertex, and add the edges between this additional vertex and the boundary vertices of the polygon. This way, we have effectively created a triangulation of the topological sphere $$S^2$$.\n\nIf the boundary of the polygon has an odd number of color changes, this gives an odd number of 3-colored triangles in the \u201ctent\u201d over the polygon. But from the double-counting argument in user Matt\u2019s answer, we know an even number of 3-colored Sperner triangles must exist. Hence the polygon at the bottom must have an odd number of 3-colored triangles (at least one) in its triangulation, which completes the proof.", "date": "2022-05-23 15:11:13", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/364040/3-colored-triangulations-of-the-sphere-s2-and-sperners-lemma", "openwebmath_score": 0.7654329538345337, "openwebmath_perplexity": 368.1216546596065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232940063591, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8640459328508988}}
{"url": "https://math.stackexchange.com/questions/500511/dimension-of-the-sum-of-two-vector-subspaces", "text": "# Dimension of the sum of two vector subspaces\n\n$\\dim(U_1+U_2) = \\dim U_1 +\\dim U_2 - \\dim(U_1\\cap U_2).$\n\nI want to make sure that my intuition is correct. Suppose we have two planes $U_1,U_2$ though the origin in $\\mathbb{R^3}$. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in $\\mathbb{R^3}$. To obtain the dimension of $U_1$ and $U_2$, we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3).\n\nCan we generalize this notion to $\\mathbb{F^{n}}$?\n\nSuppose we have an additional case where $U_1$ and $U_2$ are planes in $\\mathbb{R^3}$, but $U_1 \\subseteq U_2$. In this instance, $dim(U_1 + U_2) < 3$, because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces $U_1,U_2$ are two dimensional and $U_1 \\subseteq U_2$, then their intersection is also two-dimensional, concluding $dim(U_1+U_2)=2+2-2 = 2$.\n\nIs this proper intuition?\n\n\u2022 The intuition is correct. Sep 21, 2013 at 17:17\n\nIn the latter case, they are actually the same plane, so their sum is again the same plane (as they are closed under addition).\n\nHere is another (analogous) way to think about it. Let's start with a basis $B_0$ for $U_1\\cap U_2.$ We can extend $B_0$ to a basis $B_1$ for $U_1$ and a basis $B_2$ for $U_2$. Then $B_1\\cup B_2$ is a basis for $U_1+U_2,$ and $B_1\\cap B_2=B_0,$ so \\begin{align}\\dim(U_1+U_2) &= |B_1\\cup B_2|\\\\ &= |B_1|+|B_2|-|B_1\\cap B_2|\\\\ &= |B_1|+|B_2|-|B_0|\\\\ &= \\dim(U_1)+\\dim(U_2)-\\dim(U_1\\cap U_2).\\end{align} This generalizes nicely to $\\Bbb F^n$, and allows us to avoid geometric arguments that may be less sensible for an arbitrary field $\\Bbb F$.\n\nAlso, it will never be the case that the intersection of two planes in space is precisely $\\{0\\}.$ If there were two such planes $U_1$ and $U_2,$ then we would have $$\\dim(U_1+U_2)=\\dim(U_1)+\\dim(U_2)-\\dim(U_1\\cap U_2)=2+2-0=4>3=\\dim(\\Bbb R^3),$$ which is not possible, since $U_1+U_2$ is a subspace of $\\Bbb R^3$.\n\n\u2022 Thanks for the answer. Could it be the case that a pair of two dimensional planes in $\\mathbb{R^4}$ intersect at a point? Sep 21, 2013 at 17:27\n\u2022 Yes, indeed. Consider $$U_1=\\{(w,x,y,z)\\in\\Bbb R^4:y=z=0\\}$$ and $$U_2=\\{(w,x,y,z)\\in\\Bbb R^4:w=x=0\\}.$$ The upshot in $\\Bbb R^3$ is that there \"isn't enough room\" for two planar subspaces to avoid each other that well. Sep 21, 2013 at 17:38\n\u2022 Why can we conclude that $B_{1} \\cap B_{2} = B_{0}$? It seems perfectly reasonable that the extensions share vectors outside of $B_{0}$. Sep 14, 2016 at 0:40\n\u2022 @Jonathan: If there is some $\\vec v\\in B_1\\cap B_2,$ then $\\vec v\\in U_1\\cap U_2.$ What then can we conclude by definition of $B_0$? Sep 14, 2016 at 0:46\n\u2022 Ok, then $v$ must be a linear combination of $B_{0}$. But, this forces both $B_{1}$ and $B_{2}$ to be linearly dependent, right? Sep 14, 2016 at 0:53\n\nCameron Buie's answer does answer the original question sufficiently. However, I'm adding the formal proof of the theorem in context (taken from \"Linear Algebra Done Right\" by Sheldon Axler).\n\nIf $$U_1$$ and $$U_2$$ are subspaces of a finite dimensional vector space then: $$\\dim(U_1+U_2)=\\dim U_1 + \\dim U_2 - \\dim(U_1 \\cap U_2)$$\n\nLet $$u_1,...,u_m$$ be a basis of $$U_1\\cap U_2$$; thus $$\\dim (U_1\\cap U_2)=m$$. Because $$u_1,...,u_m$$ is a basis in $$U_1\\cap U_2$$, it is linearly independent in $$U_1$$. Hence this list can be extended to a basis $$u_1,...,u_m,v_1,...,v_j$$ of $$U_1$$. Thus, $$\\dim U_1 = m+j$$. Also extend $$u_1,..,u_m$$ to a basis $$u_1,...,u_m, w_1,...,w_k$$ of $$U_2$$. $$\\dim U_2 = m +k$$.\n\nWe will show that $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is a basis of $$U_1+U_2$$. This will complete the proof because then we will have $$\\dim(U_1+U_2)=m+j+k=\\dim U_1 + \\dim U_2 - \\dim (U_1\\cap U_2)$$\n\nWe just need to show that the list $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is linearly independent. To prove this, suppose:\n\n$$a_1u_1+...+au_m+b_1v_1+...+b_jv_j+c_1w_1+...+c_kw_k=0$$\n\nwhere all $$a,b,c$$'s are scalars. We need to show that all the $$a,b$$ and $$c$$'s are $$0$$.\n\nThe equation can be rewritten as\n\n$$c_1w_1+...+c_kw_k=-a_1u_1 - ... -a_mu_m -b_1v_1 - ... - b_j v_j$$\n\nWhich shows that $$c_1w_1+...+c_kw_k\\in U_1$$. But actually all $$w$$'s are in $$U_2$$. So the LHS must be an element of $$U_1\\cap U_2$$.\n\n$$c_1w_1+...+c_kw_k=d_1u_1+...+d_mu_m$$ for some choice of scalars $$d_1,d_2,...,d_m$$. But $$u_1,...,u_m,w_1,...,w_k$$ is linearly independent. So our last equation implies that all the $$c$$'s equal $$0$$.\n\nThus our original equation involving $$a,b,c$$ becomes\n\n$$a_1u_1+...+a_mu_m+b_1v_1+...+b_jv_j=0$$\n\nBut we already knew that the list $$u_1,...,u_m,v_1,...,v_j$$ is linearly independent. This equation implies that all the $$a$$'s and $$b$$'s are $$0$$. We now know that all $$a,b$$ and $$c$$'s are $$0$$, hence proving our original claim.", "date": "2022-07-03 03:26:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/500511/dimension-of-the-sum-of-two-vector-subspaces", "openwebmath_score": 0.9851697087287903, "openwebmath_perplexity": 117.93825881355986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232950125849, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8640459228433786}}
{"url": "https://math.stackexchange.com/questions/962638/taylor-approximation", "text": "# Taylor Approximation\n\nFor $f(x)=e^x$, find a Taylor approximation that is in error by at most $10^-7$ on [-1,1]. Using this approximation, write a function program to evaluate $e^x$. Compare it to the standard value of $e^x$ obtained from the MATLAB function exp(x); calculate the difference between your approximation and exp(x) at 21 evenly spaced points in [-1,1].\n\nSo I am stuck on this problem. I know for the first part, I needed to find n smallest such that $e/(n+1)!$ $\\leq$ $10^-7$, so I found n=10 to satisfy that part.\n\nSo do I need to find a polynomial of degree 10 using a nested loop that evaluates $e^x$? What is my next movie in this problem?\n\n\u2022 Yes, a loop to evaluate $1+x+x^2/2+...$ up to 10. \u2013\u00a0lemon Oct 7 '14 at 18:49\n\u2022 Thank you! and then I just compare my loop with the MATLAB function given and see where they are different? \u2013\u00a0user141745 Oct 7 '14 at 18:52\n\u2022 Very relevant wiki example. \u2013\u00a0BeaumontTaz Oct 7 '14 at 18:59\n\u2022 @user141745 Yes - the purpose of comparing is to make sure that your routine is accurate to within $10^{-7}$. \u2013\u00a0lemon Oct 7 '14 at 19:02\n\nI posted the link in the comments, but I'll flush out the example on the wiki here with some added commentary.\n\nWe have $f(x)=e^x$ and we know by Taylor'r theorem that\n\n$$f(x)=P_k(x)+R_k(x)=1+x+\\frac{x^2}{2}+\\frac{x^3}{6}+\\cdots+\\frac{x^k}{k!}+R_k(x)$$\n\nwhere $R_k(x)$ is our error term and $P_k(x)$ is the Taylor polynomial. We want to find the $k$ such that $R_k(x)<10^{-7}$ for $x\\in[-1,1]$.\n\nWe know that the remainder term $R_k(x)$ can be expressed as\n\n$$R_k(x)=\\frac{e^\\xi}{\\left(k+1\\right)!}x^{k+1}$$\n\nfor some $\\xi\\in[-1,1]$. We need to find the value of $\\xi$ though. And since we don't know $e^x$ (otherwise we'd just use that function) we again approximate it using a lower order Taylor expansion\n\n$$e^x=P_1(x)+R_1(x)=1+x+\\frac{e^\\xi}{2}x^2$$\n\nSince $e^x$ is an increasing function, we know that $e^\\xi<e^x$ when $-1<\\xi<x$. So we can say\n\n$$e^x=1+x+\\frac{e^\\xi}{2}x^2<1+x+\\frac{e^x}{2}x^2$$\n\nWhich we can solve for $e^x$ and get\n\n$$e^x<2\\frac{x+1}{2-x^2}$$\n\nSome basic calculus gives us that the maximum of this expression for $x\\in[-1,1]$ is at $x=1$ and $e^x<4$.\n\nTherefore\n\n$$R_k(x)=\\frac{e^\\xi}{\\left(k+1\\right)!}x^{k+1}<\\frac{4x^{k+1}}{\\left(k+1\\right)!}$$\n\nAnd since $x^{k+1}$ is a maximum on $[-1,1]$ at $x=1$ and we want $R_k(x)$ to be less than $10^{-7}$ we have\n\n$$R_k(x)<\\frac{4}{\\left(k+1\\right)!}<10^{-7}$$\n\nSo basically, we just need to find $k$ such that $(k+1)!>4\\cdot10^7$. And $11!=39916800=3.99\\cdot10^7$ which is reallllly close. But, really we need $(k+1)!=12!$ so $k=11$.\n\nNow... this is a very modest upper bound. If we really think about it (and this is what you did), the upper bound on the interval $[-1,1]$ is $e$. So if we did this with $e^x<e$ on $x\\in[-1,1]$ instead, then we'd end up with\n\n$$\\frac{e}{\\left(k+1\\right)!}<10^{-7}\\implies (k+1)!>2.718\\cdot10^7\\implies k=10$$\n\nSame as you got. However, this relies on the fact that we know $e$ already.\n\nIf we don't know $e$ we can use a higher order approximation than $2$. So we find that\n\n$$e^x<\\frac{6+6x+3x^2}{6-x^3}$$\n\nWhich means it's either $e^x<\\frac{15}{5}=3$ if it's at $x=1$ or it's at the maximum of the interval which involves find the roots of a fourth order polynomial $3x^4+12x^3+18x^2+36x+36$ whose critical points turn out not to be on our bound. So we can use this higher order approximation the same way we did before and get that\n\n$$\\frac{3}{\\left(k+1\\right)!}<10^{-7}\\implies (k+1)!>3\\cdot10^7\\implies k=10$$\n\nFrom there, you can either hard code the approximation\n\n$$e^x\\approx \\text{myFunc}(x)=1+x+\\frac{x^2}{2}+\\frac{x^3}{3!}+\\frac{x^4}{4!}+\\frac{x^5}{5!}+\\frac{x^6}{6!}+\\frac{x^7}{7!}+\\frac{x^8}{8!}+\\frac{x^9}{9!}+\\frac{x^{10}}{10!}$$\n\nor do a sum like this in a for loop\n\n$$e^x\\approx \\text{myFunc}(x)=\\sum_{i=0}^{10}\\frac{x^i}{i!}$$\n\nIt appears as though Arkamis gave you some MATLAB code that'll do this for you, though.\n\nJust for fun, here's a matlab script that computes the answer in one line -- in MATLAB we almost never need to loop!\n\nThe function takes three parameters: n, the maximum order of the polynomial to try; m, the number of data points in the interval we wish to evaluate; and tol, the tolerance we wish to achieve.\n\nfunction Order = TaylorExp(n,m,tol)\nOrder = min(find(max(abs(tril(kron(ones(n,1),1./factorial(0:(n-1))))*power(kron(ones(n,1),linspace(-1,1,m)),kron(ones(1,m),(0:(n-1))')) - kron(ones(n,1),exp(linspace(-1,1,m)))),[],2) < tol)) - 1;", "date": "2021-06-12 12:19:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/962638/taylor-approximation", "openwebmath_score": 0.8216952681541443, "openwebmath_perplexity": 221.43075925590142, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561685659695, "lm_q2_score": 0.8918110454379297, "lm_q1q2_score": 0.8640366325678043}}
{"url": "https://math.stackexchange.com/questions/799362/line-equation-of-a-tangent-line-of-fx-x-cos3x", "text": "# Line equation of a tangent line of $f(x) = x\\cos(3x)$\n\nI'm new here so maybe I'll need some help with formatting with MathJax, as well.\n\nSo question asks for tangent line of $f(x) = x\\cos(3x), x= \\pi$\n\nSo:\n\n$$f(x) = x.\\cos(3x)$$ $$f'(x) = -x.\\sin(3x)+\\cos(3x).3$$\n\n$$f(\\pi) = -\\pi$$ $$f'(\\pi) = -3$$\n\n$$y = -3x - 2\\pi$$\n\nThe answer of the book: $$y = -x$$\n\nConsidering: $$y = ax + b$$ $$y = f(x)$$ $$a = f'(x)$$ What am I missing?\n\n\u2022 you miscalculated the derivative. \u2013\u00a0mesel May 17 '14 at 17:51\n\nYou made a couple of mistakes in using the product rule.\n\n$$f(x) = x\\cos(3x) \\implies f'(x) = x \\cdot \\frac{d}{dx}(\\cos(3x)) + \\dfrac{d}{dx}(x)\\cdot \\cos(3x)$$\n\n$$= x\\cdot (-\\sin (3x))\\cdot 3 + \\cos 3x$$\n\n$$= -3x\\sin(3x)+ \\cos (3x)$$\n\nNow substitute $x = \\pi$ into $f'(x)$ to obtain slope at that point: $-3\\pi\\cdot \\sin(3\\pi) + \\cos (3\\pi) = -1$\n\nSo the slope of the desired line needs to be $-1$.\n\nNow, given $x_0 = \\pi$, $y_0 = f(x_0) = f(\\pi) = -\\pi$.\n\nSo we have the point on the tangent line (the point of tangency): $(\\pi, -\\pi)$. That gives you the line \\begin{align} y - y_0 = -1(x - x_0) &\\iff y - (-\\pi) = -1(x - \\pi)\\\\ \\\\ & \\iff y + \\pi = -x + \\pi \\\\ \\\\ & \\iff y = -x\\end{align}\n\n\u2022 Your answer showed me something that I was repeating question after question and haven't given attention. Derivative was wrong and how to find line equation was misunderstood. Thank you a lot. \u2013\u00a0Fabiano Araujo May 17 '14 at 18:15\n\u2022 You're welcome, Fabiano! \u2013\u00a0amWhy May 17 '14 at 18:15\n\nHINT:\n\n$$f'(x)=\\frac{d(x\\cos3x)}{dx}=\\frac{dx}{dx}\\cdot\\cos3x+x\\cdot\\frac{d(\\cos3x)}{dx}=\\cos3x-3x\\sin3x$$", "date": "2021-07-24 22:40:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/799362/line-equation-of-a-tangent-line-of-fx-x-cos3x", "openwebmath_score": 0.9988675117492676, "openwebmath_perplexity": 731.5645384417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561712637256, "lm_q2_score": 0.8918110418436166, "lm_q1q2_score": 0.8640366314913206}}
{"url": "https://gmatclub.com/forum/two-important-formulas-for-mixture-problems-82701.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 23 Jan 2020, 21:16\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Two important formulas for mixture problems\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 11 Aug 2009\nPosts: 10\nTwo important formulas for mixture problems\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 19 Aug 2009, 06:52\n6\n9\nI came across two methods which I found very handy in solving some types of mixture problems. So I thought of sharing it with the gc community.\nhere they are\n\nRule of Alligation\nIt is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.\n\nMean Price: The cost price of a unit quantity of the mixture is called mean price.\n\nRule of Alligation\n\nIf two quantities are mixed, then\n\n$$\\frac{Quantity of cheaper}{Quantity of dearer}=\\frac{(C.P.) of dearer - (Mean Price)}{(Mean Price) - (C.P.)}$$\n\nTaking a simple example\n\nIn what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?\n\nC.P of cheaper=9.30\nC.P of dearer=10.80\nC.P of mean or mean price = 10.0\n\nso putting the values in the formula\n$$\\frac{Qc}{Qd}=\\frac{Cd-Cm}{Cm-Cc}$$\n\n=$$\\frac{10.80-10}{10-9.3}$$\n\n=8:7 Ans\n\nOriginally posted by joebloggs on 19 Aug 2009, 06:10.\nLast edited by joebloggs on 19 Aug 2009, 06:52, edited 2 times in total.\nIntern\nJoined: 11 Aug 2009\nPosts: 10\n\n### Show Tags\n\n19 Aug 2009, 06:25\n9\n11\nSuppose a container contains x units of liquid from which y units are taken out and replaced with water.\n\nAfter n operations, the quantity of pure liquid = $$[x(1-\\frac{y}{x})^n]$$\n\nExample: A container contains 40 liters of milk. From this container 4 litres of milk was taken and replaced by water. This process was repeated further two times. How much milk is now contained by the container?\n\nAmount of milk left after 3 operations = $$[40(1-\\frac{4}{40})^3]liters = (40 *\\frac{9}{10}^3)$$ = 29.16 liters\n##### General Discussion\nIntern\nJoined: 11 Aug 2009\nPosts: 10\nRe: Two important formulas for mixture problems\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2009, 06:37\n3\nyou also might wanna see a short method for mixture problems used by Economist\ncheck out this problem\n\nhttp://gmatclub.com/forum/how-much-of-the-mixture-is-to-be-removed-and-replaced-82423.html\n\nHope this post helps\n\nSoon I'll be posting some mixture problems which I couldn't solve in one go ...\n\nsee ya ....\nSenior Manager\nJoined: 18 Jun 2009\nPosts: 310\nLocation: San Francisco\nRe: Two important formulas for mixture problems\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2009, 12:40\nnice post .. thanks .. +1\nManager\nJoined: 16 Apr 2009\nPosts: 152\nSchools: Ross\nRe: Two important formulas for mixture problems\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2009, 12:53\nKeep up the good work !\nIntern\nJoined: 14 Aug 2019\nPosts: 3\nRe: Two important formulas for mixture problems\u00a0 [#permalink]\n\n### Show Tags\n\n14 Aug 2019, 02:59\nManager\nJoined: 01 Jan 2017\nPosts: 57\nWE: General Management (Consulting)\nTwo important formulas for mixture problems\u00a0 [#permalink]\n\n### Show Tags\n\n28 Aug 2019, 00:40\nQuote:\nIn what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?\n\nC.P of cheaper=9.30\nC.P of dearer=10.80\nC.P of mean or mean price = 10.0\n\nFor those who love a more visual approach, consider the following, which may be even simpler.\n\nJust subtract cheaper and dearer from the mean and find the difference.\n\n---9.3----10.8--\n-----\\----/------\n-------10-------\n------/---\\------\n---0.8---0.7---\nTwo important formulas for mixture problems \u00a0 [#permalink] 28 Aug 2019, 00:40\nDisplay posts from previous: Sort by", "date": "2020-01-24 04:18:37", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/two-important-formulas-for-mixture-problems-82701.html", "openwebmath_score": 0.8512399792671204, "openwebmath_perplexity": 6930.326394164262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708026035287, "lm_q2_score": 0.8840392832736084, "lm_q1q2_score": 0.8640341838261749}}
{"url": "http://openstudy.com/updates/556876efe4b0ae03fc81fc6b", "text": "## anonymous one year ago Find circle which is tangent to x-axis and path through points (1,-2) and (3,-4). I worked it through analytic geometry and it gives two possible circles!!! and I can not figure how 3 conditions for a circle gives 2 answers or how to draw it geometrically?\n\n1. anonymous\n\n@ganeshie8\n\n2. welshfella\n\ngeneral equation is (x - a)^2 + ( y - b)^2 = r^2 you can create 2 equations in a , b and r by substituting the 2 given points\n\n3. anonymous\n\nCircle is clearly defined by 3 conditions. How can I draw 2 circles which path through two points and tangent to a line!!\n\n4. welshfella\n\nThe circle touches the x -axis so that another point on (x , 0)\n\n5. welshfella\n\n- good question - I'm trying to figure that\n\n6. anonymous\n\nCan you give a picture? that is puzzling me\n\n7. welshfella\n\n|dw:1432910060214:dw|\n\n8. welshfella\n\nlo1 - not a great diagram\n\n9. anonymous\n\nI'm sorry I meant two possible cirlces\n\n10. welshfella\n\n|dw:1432910234711:dw|\n\n11. welshfella\n\nthats worse than the first\n\n12. anonymous\n\n:) Yeah that is amazing.\n\n13. ganeshie8\n\ntwo circles can have the same common tangent, yes ?\n\n14. welshfella\n\nyes\n\n15. ganeshie8\n\nmay be forget about x-axis and look at below diagram |dw:1432910495765:dw|\n\n16. ganeshie8\n\nboth circles meet below 3 conditions : 1, 2) pass through two points (intersection) 3) having that blue line as tangent\n\n17. welshfella\n\nah - yes the common tangenytis the x-axis\n\n18. anonymous\n\nI got it now thanks\n\n19. ganeshie8\n\nyour question about \"why 3 conditions are not giving me an unique circle\" is very interesting, im still thinking of a better explanation\n\n20. anonymous\n\nI don't know. In such cases I just change that thinking. I think the tangent line condition isn't tough enough ((unless the tangent point is given))\n\n21. anonymous\n\nanyway that for your helping :)\n\n22. ganeshie8\n\nhmm the tangent point is not so random, we only have two choices so its still a bit mysterious\n\n23. ganeshie8\n\n|dw:1432911213933:dw|\n\n24. anonymous\n\nBut the result is the same tangent line condition isn't unique as you have said.\n\n25. ganeshie8\n\ngeometrically how do you know there doesn't exist a third circle that meets the given conditions ?\n\n26. ganeshie8\n\n|dw:1432911454842:dw|\n\n27. ganeshie8\n\nhow do i convince more than two circles are never possible given 1) two points 2) tangent line\n\n28. anonymous\n\nthe center isn't on locus of chord made by these 2 points\n\n29. anonymous\n\nI mean when you bisect it and make a perpendicular which is the locus of the center\n\n30. ganeshie8\n\nperpendicular bisector of the chord passes through the center but again how do you know the 3 centers are not collinear ?\n\n31. ganeshie8\n\n|dw:1432911763458:dw|\n\n32. welshfella\n\nthe line joining the points of contact to the centers of the circles are perpendicular to the tangent - can that be used to answer the question?\n\n33. ganeshie8\n\nBINGO!!!\n\n34. ganeshie8\n\n|dw:1432912228871:dw|\n\n35. ganeshie8\n\nhmm idk\n\n36. welshfella\n\nyes - its a puzzle\n\n37. welshfella\n\n|dw:1432912469311:dw|\n\n38. welshfella\n\nwhy can't there be an other larger circle like the above?\n\n39. anonymous\n\nfor any triangle it has only one unique circle paths through its vertices |dw:1432912656462:dw| so if we moved the point along the locus the lengths won't be equal and if we shifted it up or down, the the distance to 2pionts won't be equal.\n\n40. ganeshie8\n\nbrilliant!\n\n41. welshfella\n\ni'm afraid i gotta go and leave this interesting discussion ..\n\n42. myininaya\n\nI know you wanted geometric stuff or whatever but I think I have found two answers with a combination of algebra and calculus.\n\n43. myininaya\n\nIt is pretty long that way. :p\n\n44. myininaya\n\n$\\text{ our circle has points: } (1,-2);(3,-4);(a,0) \\\\ \\text{ we have horizontal tangent at } (a,0) \\\\ (a-h)^2+(0-k)^2=r^2 \\\\ (1-h)^2+(k+2)^2=r^2 \\\\ (3-h)^2+(k+4)^2=r^2 \\\\ y'=\\frac{h-x}{y-k} \\\\ y'|_{x=a}=0=\\frac{h-a}{0-k} \\implies h=a \\\\ (a-a)^2+k^2=r^2 \\text{ so } k=-r \\text{ or } k=r \\\\ \\text{ so going with the } k=r \\text{ thing we have } \\\\ (1-h)^2+(r+2)^2=r^2 \\\\ (3-h)^2+(r+4)^2=r^2 \\\\ (1-h)^2+4r+4=0 \\\\ (3-h)^2+8r+16=0 \\\\ 2(1-h)^2+8=(3-h)^2+16 \\\\ h^2+2h-15=0 \\\\ (h+5)(h-3)=0 \\\\ h=-5 \\text{ or } h=3$ so pluggin some things back in we can find r then k $h=-5 \\\\ a=-5 \\\\ (1+5)^2+4r+4=0 \\\\ r=10 \\\\ k=10,-10$ $h=3 \\\\ a=3 \\\\ (1-3)^2+4r+4=0 \\\\ r=2 \\\\ k=2 ,-2$ some of these solutions need to be checked you will see only 2 of the 4 will work and like i said I know this isn't the approach you wanted but I thought it would be nice to try another approach\n\n45. welshfella\n\nlooks good to me", "date": "2017-01-23 12:43:05", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/556876efe4b0ae03fc81fc6b", "openwebmath_score": 0.6302857995033264, "openwebmath_perplexity": 1560.9655067903057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707976599674, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.864034182441887}}
{"url": "https://mathhelpboards.com/threads/diagonal-crossing.3856/", "text": "# Diagonal Crossing\n\n#### mathmaniac\n\n##### Active member\nChallenge:\nHow many boxes are crossed by a diagonal of an 800 X 200 rectangle?\n\nSolution with proof required...\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nChallenge:\nHow many boxes are crossed by a diagonal of an 800 X 200 rectangle?\n\nSolution with proof required...\n\nHello Mathmaniac\nI don't understand the question. What us meant by 'boxes are crossed'.\n\nAlso, if this is a challenge problem then shouldn't this go in the 'Challenge and Puzzles' forum?\n\n#### MarkFL\n\nStaff member\nYes, I sent the OP a VM asking if this is a challenge or if it is a problem for which he needs help shortly after it was posted, but have not gotten a response yet. Once the matter is settled, I will move it if need be, then remove this post so that the topic is not cluttered.\n\n#### mathmaniac\n\n##### Active member\nI don't understand the question. What us meant by 'boxes are crossed'.\nHere is an example:\n\nThe boxes inside the rectangle are meant to be squares....\n\n#### MarkFL\n\nStaff member\nIts a challenge,Mark...\nI have thus moved the topic to the Challenge Questions and Puzzles sub-forum. I know you used the word \"Challenge\" but wanted to make sure it fit the criteria, i.e., you have the correct solution ready to post in the event no one solves it.\n\nLast edited:\n\nYes!!!\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\n[JUSTIFY]The rectangle is made of $800 \\times 200$ \"boxes\". The diagonal thus has gradient $\\pm \\frac{200}{800} = \\pm \\frac{1}{4}$, where a \"box\" has unit dimensions. Note the diagonal starts at the top left corner of the top-left-most box. This is important. So, after 4 units of width travelled, the diagonal will intersect the top left corner of another box:[/JUSTIFY]\n\n[JUSTIFY]And this section of the diagonal intersects four boxes. Since the rectangle is 800 boxes wide, this section of the diagonal will repeat $\\frac{800}{4} = 200$ times, and so the diagonal intersects $4 \\times 200 = 800$ boxes.[/JUSTIFY]\n\n[HR][/HR]\n\n[JUSTIFY]A more interesting problem is to consider a rectangle of dimensions $p \\times q$ where $p$ and $q$ are distinct primes. Then the diagonal never intersects the top-left corner of a box within the rectangle, and a different approach is called for:[/JUSTIFY]\n\nLast edited:\n\n#### mathmaniac\n\n##### Active member\nThats right,Bacterius...\nAnd can you show your approach for distinct primes?\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nThats right,Bacterius...\nAnd can you show your approach for distinct primes?\n[JUSTIFY]It's simple enough, in fact. Because we know that the diagonal will, in this case, never intersect a corner, we can count the number of boxes crossed by simply counting the number of times the diagonal intersects both the vertical sides and the horizontal sides of the boxes. The diagonal will intersect $p - 1$ horizontal sides, and $q - 1$ vertical sides, simply by virtue of being a line crossing the rectangle from top-left to bottom-right (or top-right to bottom-left).\n\nBut not so fast - the diagonal always starts inside the rectangle, and so automatically intersects the top-left box (or whichever corner of the rectangle you start your diagonal from), which we haven't yet considered, giving a total of $(p - 1) + (q - 1) + 1 = p + q - 1$ boxes intersected. Checking with the diagram above gives $11 = 7 + 5 - 1$ boxes intersected, as expected.\n\nNot very rigorous, just a proof sketch showing the general approach.[/JUSTIFY]\n\nLast edited:\n\n#### mathmaniac\n\n##### Active member\nHere we have a winner!!!", "date": "2020-11-28 23:01:53", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/diagonal-crossing.3856/", "openwebmath_score": 0.7453159093856812, "openwebmath_perplexity": 1109.8177279388092, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9865717460476701, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8640267127106489}}
{"url": "http://math.stackexchange.com/questions/112416/finding-domain-of-sqrt-fracx2-1x2-3x2-5x2-2x2-4x2-6", "text": "# Finding domain of $\\sqrt{ \\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$\n\nHow can I find the domain of:\n\n$$\\sqrt{ \\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$\n\nI think the hard part will be to find:\n\n$$\\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \\ge 0$$\n\nSo far I have: not sure how to preceed: $$\\sqrt{ \\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$ For $\\sqrt{g(x)}$ to be valid, $g(x) \\ge 0$\n\nFor $f(x)$ to be valid, $(x^2-2)(x^2-4)(x^2-6) \\ne 0$\n\nThus, $x \\ne \\sqrt 2, 2, \\sqrt 3$\n\n$$g(x) = { \\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}} \\ge 0$$\n\n-\nA friendly note: You have posting pictures in which you presumably had written the Math. I'd suggest you learn $\\TeX$. Use the community to learn the commands. You can see the basic codes by clicking on them as seeing Math as $\\TeX$ commands here. It will pay you in the long run and you'll become effective without having to write those crappy pics. Will you? (I like the way write your questions showing us what you have done! +1 for that!) \u2013\u00a0 user21436 Feb 23 '12 at 12:09\n@KannappanSampath, I know TeX but it takes longer to type them :) perhaps I need more practice. I also have put the more important parts of the question in TeX, the not as important stuff, I thought providing an image will suffice \u2013\u00a0 Jiew Meng Feb 23 '12 at 12:12\nI see you already know $\\TeX$ from the first part of your question. Why don't you do the same thing for the Math in the pic as well? \u2013\u00a0 user21436 Feb 23 '12 at 12:13\nIs the numerator $(x^2-1)(x^2-3)(x^2-5)$ or $(x^2-1)(x^3-1)(x^5-1)$? \u2013\u00a0 01000100 Feb 23 '12 at 12:13\nIts the 1st, $(x^2-1)(x^2-3)(x^2-5)$ \u2013\u00a0 Jiew Meng Feb 23 '12 at 12:14\n\nThe first thing you should note is that the expression $$\\Phi=\\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}$$ is undefined at any zero of the denominator. So, the points $x=\\pm \\sqrt 2$, $\\pm 2$, and $\\pm\\sqrt 6$ are not in the domain of $\\sqrt\\Phi$.\n\nNow, to find the domain of $\\sqrt\\Phi$, we need to find when $\\Phi$ is nonnegative.\n\nTowards solving $\\Phi\\ge0$, the fundamental observation to make is that the only points \"across which\" $\\Phi$ can change signs are at the zeros of its numerator or at the zeroes of its denominator.\n\nThe zeroes of the numerator are: $$\\pm 1, \\pm \\sqrt 3, \\pm \\sqrt 5$$ and the zeroes of the denominator are $$\\pm \\sqrt 2, \\pm 2 \\pm\\sqrt 6.$$\n\nAs already mentioned, the only points across which $\\Phi$ can change sign are at one of the zeroes above. Let's also note the expression $\\Phi$ is \"even\": if $\\Phi(x)\\ge 0$, then $\\Phi(-x)\\ge 0$. So, let's find the $x$ values on the nonnegative $x$-axis that satisfy $\\Phi(x)\\ge0$. Then by \"reflection\" we'll obtain the points on the negative $x$-axis where $\\Phi(x)\\ge0$.\n\nSo, draw a number line with the zeroes listed above:\n\nThe zeroes subdivide the nonnegative $x$-axis into intervals, the endpoints of which, except the endpoint 0, are the zeroes of either the numerator or the denominator of $\\Phi$. In each subinterval, if you pick a point $x_0$ (not an endpoint, save for 0), evaluate $\\Phi(x_0)$, and note its sign, then across that entire subinterval, the expression $\\Phi$ will have that sign.\n\nFor example in $[0,1)$, picking $x=1/2$ it is easy to see that the sign of $\\Phi(1/2)$ is\n\n$$\\frac{((1/2)^2-1)((1/2)^2-3)((1/2)^2-5)}{((1/2)^2-2)((1/2)^2-4)((1/2)^2-6)} ={ (-)(-)(-)\\over(-)(-)(-)} \\ge 0$$ So on all of $[0,1)$, the expression $\\Phi$ is positive. (Note, you only need to find the sign, there is no need to do the actual arithmetic.)\n\nThe complete \"sign chart\" is shown below:\n\nAnd we see that $\\Phi>0$ on $$[0, 1)\\cup(\\sqrt2,\\sqrt3)\\cup(2,\\sqrt5)\\cup(\\sqrt6,\\infty).$$ By symmetry, $\\Phi>0$ on $$( -\\infty,-\\sqrt6)\\cup(-\\sqrt5,2)\\cup (-\\sqrt3,-\\sqrt2)\\cup (-1,0].$$\n\nSo, the domain of $\\sqrt\\Phi$ is the union of the two sets above, together with the zeroes of the numerator of $\\Phi$ that aren't zeros of the denominator of $\\Phi$: $x=\\pm1$, $x=\\pm \\sqrt3$, and $x=\\pm \\sqrt5$.\n\nBelow is a plot of $y=\\Phi$ and $y=\\sqrt\\Phi$. Note that the zeroes of the denominator of $\\Phi$ are vertical asymptotes of the graph of $y=\\Phi$ and the zeroes of the numerator of $\\Phi$ are the zeroes of $\\Phi$.\n\n-\n\n1. $\\frac{f(x)}{g(x)} \\ge 0$ is not so different from $f(x)\\times g(x) \\ge 0$ (except for zeros of $g$.)\n2. With $(x^2-1) = (x-1)(x+1)$ etc, your problem reduces to the form of $(x-a)(x-b)(x-c)(x-d)...(x-z) \\ge 0$\n\nEdit: oops I only read the hand-written part! Anyways thanks to the monotonicity of $x^5-1$ etc, you can still use similar argument.\n\nEdit 2: Plot it in google.\n\n-\n+1 for the link. I didn't know Google could plot. \u2013\u00a0 user23211 Feb 23 '12 at 13:37\n\n$(x^2-1)$ is negative iff $x\\in(-1,1)$. Similarly, $(x^2-2)$ is negative for $x\\in(-\\sqrt{2},\\sqrt{2})$, $(x^2-3)$ for $x\\in(-\\sqrt{3},\\sqrt{3}), \\dots,(x^2-6)$ for $x\\in(-\\sqrt{6},\\sqrt{6})$.\n\nConsider the open intervals $(-\\infty,-\\sqrt{6}), (-\\sqrt{6},-\\sqrt{5}),(-\\sqrt{5},-2),\\dots$ separately. Don't forget to look at how $f(x)$ behaves on the boundaries. E.g $x=\\pm\\sqrt{n}$ for $n=1,2,3,4,5,6$.\n\n-\nOk, now I get the domain as $-\\sqrt{5} \\lt x \\lt -2$, $-\\sqrt{3} \\lt x \\lt -\\sqrt{2}$, $-1 \\lt x \\lt 1$ ... but how do I find out without graph about what happens $x \\lt -\\sqrt{6}$ and $x \\gt \\sqrt{6}$ \u2013\u00a0 Jiew Meng Feb 23 '12 at 14:23\nWhen x>$\\sqrt{6}$ all the factors in the numerator and denominator are positive, so the whole expression is positive. The situation is the same for $x<\u2212\\sqrt{6}$ \u2013\u00a0 01000100 Feb 23 '12 at 14:49", "date": "2015-07-07 22:46:16", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/112416/finding-domain-of-sqrt-fracx2-1x2-3x2-5x2-2x2-4x2-6", "openwebmath_score": 0.9384967088699341, "openwebmath_perplexity": 249.44829644022835, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98657174604767, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.864026704714436}}
{"url": "https://cozilikethinking.wordpress.com/2013/09/01/71/", "text": "Let $f:X\\to Y$ be a mapping. We will prove that $f^{-1}(Y-f(X-W))\\subseteq W$, with equality when $f$ is injective. Note that $f$ does not have to be closed, open, or even continuous for this to be true. It can be any mapping.\n\nLet $W\\subseteq X$. The mapping of $W$ in $Y$ is $f(W)$. As for $f(X-W)$, it may overlap with $f(W)$, we the mapping be not be injective. Hence, $Y-f(X-W)\\subseteq f(W)$.\n\n>Taking $f^{-1}$ on both sides, we get $f^{-1}(Y-f(X-W))\\subseteq W$.\n\nHow can we take the inverse on both sides and determine this fact? Is the reasoning valid? Yes. All the points in $X$ that map to $Y-f(X-W)$ also map to $W$. However, there may be some points in $f^{-1}(W)$ that do not map to $Y-f(X-W)$.\n\nAre there other analogous points about mappings in general? In $Y$, select two sets $A$ and $B$ such that $A\\subseteq B$. Then $f^{-1}(A)\\subseteq f^{-1}(B)$", "date": "2017-09-23 20:12:11", "meta": {"domain": "wordpress.com", "url": "https://cozilikethinking.wordpress.com/2013/09/01/71/", "openwebmath_score": 0.9339148998260498, "openwebmath_perplexity": 74.47047190814499, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9884918502576514, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.864018226698692}}
{"url": "https://math.codidact.com/questions/278629", "text": "Q&A\n\n# Why does the decimal expansion of $1/(10n - 1)$ have this neat pattern?\n\n+10\n\u22120\n\nI was playing around with the reciprocals of some positive integers and found these interesting patterns:\n\n$$\\frac{1}{19} = 0.\\overline{052631578947368421}$$\n\nNow, this repeating decimal can also be obtained by \"concatenating\" the powers of $2$ successively to the left as follows:\n\n\\begin{align} 1& \\\\ 21& \\\\ 421& \\\\ 8421& \\\\ {\\color{red}{1}}68421 \\\\ {\\color{red}{3}}{\\color{green}{3}}68421 \\\\ \\vdots\\quad & \\end{align} Notice that here I think of the red $\\color{red}{1}$ of $16$ and $\\color{red}{3}$ of $32$ as being \"carried over\", so the units digit of $32$ and the tens digit of $16$ add to give me the green $\\color{green}{3}$. The sequence gets increasingly complicated as more terms are concatenated in this manner, but it definitely looks like it's converging to $\\dotsc 052631578947368421$, where the ellipsis indicates that the pattern repeats forever towards the left. This is a bit surprising because (informally) this would mean that if I perform this pattern infinitely far out to the right of the decimal point, then I get the decimal expansion of $1/19$.\n\nHow can I go about making this idea precise?\n\nSome more data for this pattern: \\begin{align} \\vdots\\quad & \\\\ {\\color{red}{3}}{\\color{green}{3}}68421 \\\\ {\\color{red}{6}}{\\color{green}{7}}368421 \\\\ {\\color{red}{1}}{\\color{green}{34}}7368421 \\\\ {\\color{red}{2}}{\\color{green}{69}}47368421 \\\\ {\\color{red}{5}}{\\color{green}{38}}947368421 \\\\ {\\color{red}{10}}{\\color{green}{77}}8947368421 \\\\ \\vdots\\quad & \\end{align}\n\nIn this pattern, one can see that at each stage one new digit from the required sequence is correctly added, despite all the carrying-over.\n\nI have also observed this to be the case for the decimal expansion of $1/29$: $$\\frac{1}{29} = 0.\\overline{0344827586206896551724137931}$$ In this case, we concatenate the powers of $3$ in a similar fashion.\n\nIs there any significance to the fact that $19 = {\\color{red}{2}}*10 - 1$ and $29 = {\\color{red}{3}}*10 - 1$ to the use of powers of $2$ and $3$, respectively, in these computations?\n\nI also checked for $1/39$, and though the repeating part of its decimal expansion is not as impressively long as for the other two cases, the pattern still holds up: $$\\frac{1}{39} = 0.\\overline{025641}$$ and concatenating the powers of $4$ indeed gives $\\dotsc 025641$.\n\nThe last case I checked was $1/49$, and it also obeys this pattern: $$\\frac{1}{49} = 0.\\overline{020408163265306122448979591836734693877551}$$ and concatenating the powers of $5$ gives us this very repeating set of digits.\n\nOf course, the pattern for $1/49$ makes it obvious that there is a concatenation that works in the forward direction: namely, the powers of $2$ concatenated in bunches of two digits. But this just means that $$\\frac{1}{49} = \\frac{2}{100} + \\frac{2^2}{100^2} + \\frac{2^3}{100^3} + \\dotsb$$ which is true by the formula for the sum of an infinite geometric series. So, \"forward concatenating\" patterns can be explained in this manner, but I'm stumped regarding how to explain the \"backward concatenating\" ones.\n\nIf anyone can throw some light on this, it would be greatly appreciated!\n\nWhy does this post require moderator attention?\nWhy should this post be closed?\n\n+5\n\u22121\n\nFirst of all, there is indeed a pattern.\n\nTo figure out why, there's an easier example than what you've found: $$0.\\bar{1} = \\frac{1}{9}.$$\n\nHow does that work? $1$ is the only power of $1$: $1^n = 1$.\n\nNow, how do we prove this? if $x = 0.\\bar{1}$, then $10x = 1.\\bar{1}$ and so, $9x=1$.\n\nHidden in that proof lies the answer for the other cases - you have some repeating number and in order to show what fraction that repeating number is equal to, here $1/\\left(10n-1\\right)$, you multiply it by $10n$ and subtract the number (again, I'll just call it $x$).\n\nTo take your example of $1/19$, we have that the fraction is $x = 0.\\overline{052631578947368421}$, so we multiply by $20$ to get $20x = 1.\\overline{052631578947368421}$, subtract $x$ and $19x = 1$. Crucially, the number left on the RHS after subtracting is exactly $1$, as everything after the decimal point cancels. It must do this, because the fraction is some integer divided by some other integer. That is, if $x=n/m$, then $mx=n$.\n\nAs we're in a decimal system, multiplying by $10$ is shifting one digit to the left. So, in doing this multiplication by $10n$, you're shifting the digit to the left, then multiplying by $n$, which is exactly the procedure outlined in the question.\n\nNow, each particular digit (after this multiplication procedure) must cancel with the digit to the left so at this point, they must be equal and we have the exact behaviour outlined in your question.\n\nTo explain this generally and in more detail (taking $x<1$ wlog.) I'll use a fraction $$x = \\frac{m}{10n-1}.$$\n\nWhen written in decimal format, this is $x = 0.x_1x_2\\ldots x_k\\ldots$, with $x_{k+r}=x_k$ for some $r$, where each $x_k$ is a single digit (i.e. an integer in the range $0-9$ as we're in a decimal basis).\n\nNow, we know that $\\left(10n-1\\right)x = m$, or $10nx - x = m$. However, m is an integer, so all the digits after the decimal point (the 'fractional part', in other words) are $0$.\n\nThis means that the digits in the fractional part of $10nx - x$ must also be $0$.\n\nSo, we look at the digits $x_r$ and $x_1\\ldots x_{r-1}$.\n\nThe fractional part of $10nx -$ the fractional part of $x$ equals $0$, or the fractional part of $10nx$ equals the fractional part of $x$. For this to be the case, all the digits must be equal.\n\nSo, we take the digit $x_r$. To calculate $x_{r-1}$, we use the properties of the fractional parts being equal and multiplication by 10 as shifting the digit to the left to give that the fractional part of $10n\\left(0.x_1\\ldots x_r\\right) - 0.x_1\\ldots x_r = 0.0\\ldots x_r$ (where the $0$s after the point are repeated $r-1$ times).\n\nThis is exactly what's written in the question, just in a different way: $nx_r \\mod 10 = x_{r-1}$. Or, the digit to the left of $x_{r-1}$ is $nx_r$, with change. Accounting for this change gives that $x_{r-j} = n^jx_r$, which is where the powers of $n$ come from, the change being what needs to be added to digits further to the left.\n\nTo explain what I mean by phrases such as \"Accounting for this change\", let's take the initial stream of digits again and write them out more explicitly, where the number in brackets to the right of each line represents the number of times the 0 is repeated after the decimal point:\n\nNow, we apply this procedure of multiplying by $10n$. It is a requirement that, because the fractional part of $x$ must equal the fractional part of $10nx$, $10nx-m=x$, that is\n\n\\begin{align*}10nx - m &= 0.0\\ldots x_k \\qquad(k-1)\\\\ &+ 0.0\\ldots x_{k-1} \\qquad(k-2)\\\\ &+ 0.0\\ldots x_{k-2} \\qquad(k-3) + \\ldots\\end{align*}\n\nTaking this line by line (where the term in brackets on the LHS has $j-1$ $0$s after the decimal point and the superscripts denote that this series arises from the $j^{th}$ term in the similar expansion of $x$) gives that\n\nWe can add this to the next term in the expansion of $x$ to get that (unfortunately at this point, notation gets very confusing but the first term on the left hand side has $j-1$ $0$s and the second, $j-2$) \\begin{align*}10n\\left(0.0\\ldots x_j + 0.0\\ldots x_{j-1}\\right) &= 0.0\\ldots x_{k-1}^{(j)} \\qquad(k-2)\\\\ &+ 0.0\\ldots x_{k-2}^{(j)} \\qquad(k-3)\\\\ &+ 0.0\\ldots x_{k-2}^{(j-1)} \\qquad(k-3)\\\\ &+ 0.0\\ldots x_{k-3}^{(j)} \\qquad(k-4)\\\\ &+ 0.0\\ldots x_{k-3}^{(j-1)} \\qquad(k-4) + \\ldots\\end{align*}\n\nNow, having done this, the $k^{th}$ term is exactly $$\\sum_{j>k}0.0\\ldots x_k^{(j)} \\qquad(k-1),$$\n\nwhich is the sum over $10n\\times$ all the digits to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.\n\nHowever, you can recursively apply this to get that the $k^{th}$ digit is the sum over $\\left(10n\\right)^p\\times$ all the digits, $p$ positions to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.\n\nThis is precisely what's done in the question, although written in a different way and shows that the behaviour arises because the terms are required to be sums of powers of $10n$ of previous terms because this is a fraction with a denominator of $10n-1$.\n\nWhy does this post require moderator attention?\n\nI'm confused how this answers the question of the pattern of digits. You only use the fact that x = 1/19 in your answer and don't explain what the significance of the powers of n is. (Or maybe my reading comprehension skills are poor...) Quintec\u202d about 1 month ago\n\nThank you for your answer! I hope you won't mind if I say that I somehow feel this answer doesn't go all the way towards explaining why this \"backwards concatenation\" process appears. Specifically, I do understand that if $x = 0.\\overline{x_1 x_2 x_3 \\dotso x_r}$, then $x$ is a rational number equal to some $n/m$ that can be found as you outlined. However, why does this method of specifically concatenating the powers of $n$ work when considering $1/(10n - 1)$? Abheri Ragam\u202d about 1 month ago\n\nSecondly, though I say this \"works\", how do I adequately make sense of that? After all, the string that I get by my method of concatenation strictly speaking gives an \"infinite number\" such as $\\dotsc 052631578947368421$. So perhaps, my claim is something like $$\\frac{1}{10n-1} = \\lim_{k \\to \\infty} \\frac{(10n)^0 + (10n)^1 + (10n)^2 + \\dotsb + (10n)^k}{10^k}\\ ?$$ I'm not sure... But I do think there's a bit more to be said from where you leave off in your post :) Abheri Ragam\u202d about 1 month ago\n\nI've updated my answer (although honestly, I feel that the first bit is about 100 times clearer). I've included the bit about the powers of $n$ and what you're referring to as 'backwards concatenation' is what I'm referring to as 'multiplying a digit by $n$ to get the digit to the left'. It works because the '-1' is the bit that causes the fractional part to cancel, hence be equal Mithrandir24601\u202d about 1 month ago\n\nBut, $x_{r-j}$ is not equal to $n^j x_r$! (I presume you meant modulo $10$?) For instance, for the decimal expansion of $1/19$, $x_{r-5} = 3$ but $2^5 x_r = 32 \\neq 3 \\pmod{10}$. In fact, they are not equal precisely because of the carrying-over happening in the concatenation process. I'm sorry, but I'm still not satisfied that your answer fully explains this phenomenon. Abheri Ragam\u202d about 1 month ago\n\n+2\n\u22120\n\nI was pretty curious about this, so I posted a question on Math.SE. Most of this answer will draw from the idea of J. W. Tanner's answer there.\n\nThere is actually a simple reason why this pattern holds.\n\nLets start by examining how the digits are generated: each digit is a power of two (with carrying). We can withhold decimal shifting for a second and simply look at a simplified expression with the same digits, only generated to the left of the decimal point:\n\n\\begin{aligned}\\sum_{k=0}^{n-1} 2^k\\cdot 10^k&=1+20+400+16000+320000+\\cdots+2^{n-1}\\cdot10^{n-1}\\\\ &=\\cdots7368421\\end{aligned}\n\nYou can see how the pattern generated is the same. Let's look at it another way, using the geometric series formula.\n\n\\begin{aligned}\\sum_{k=0}^{n-1} 2^k\\cdot10^k&=\\frac{1(1-20^n)}{1-20}\\\\ &=\\frac{20^n-1}{19}\\end{aligned}\n\nAt this point, it's pretty clear why the digits generated become the repeated decimal of $1/19$. You can see the denominator of $19$ right there!\n\nIf you're not satisfied with that observation, we can further note the following:\n\n\\begin{aligned}\\frac{20^n-1}{19}&=\\frac{20^n}{19}-\\frac{1}{19}\\\\ &=\\frac{2^n}{19}\\cdot10^n-\\frac{1}{19}\\end{aligned}\n\nLet's specifically focus on $2^n/19$. Since we are interested in repeated groups of digits, let's let $n$ be a multiple of the totient of $19$, which is $18$. The reason we chose the totient is because of a useful property of modular arithmetic: $a^b\\equiv 1\\mod c$ if $a$ and $c$ are coprime and $b$ is a multiple of the totient of $c$ (Euler's theorem).\n\nClearly, $2$ is coprime to $19$, so letting $n$ be a multiple of $18$ gives us a remainder of 1 when dividing $2^n$ by $19$. Therefore, the digits after the decimal point of $2^n/19$ are the same as the digits of $1/19$.\n\nGoing back to the whole expression, we then multiply by $10^n$, bringing $n$ of those repeated digits over to the left of the decimal point. Everything to the right is canceled by the subtraction of $1/19$ (the digits must cancel, since the expression is an integer).\n\nThe result is the final number has its last $n$ digits be the first $n$ digits of $1/19$ when $n$ is a multiple of $18$, which is what we wanted to prove. (It's not difficult to generalize this for any $n$, but that's beyond this answer).\n\nA similar result can be produced for any $10c-1$, where the digits will repeat every totient of $10c-1$ (though it may repeat before that) and the repeating portion will be the same as the digits of $1/(10c-1)$. Since it's basically the same process, I won't repeat it here.\n\nWhy does this post require moderator attention?\n\n#### 1 comment\n\nThank you! I've been considering what you say in your answer, and I'm now satisfied with the explanation :) Abheri Ragam\u202d about 1 month ago\n\n+0\n\u22120\n\nI think the simplest way to see how it works is as follows:\n\nGiven $x=\\frac{1}{10n-1}$, it is easy to check that $\\frac{x+1}{10}=nx$.\n\nNow $\\frac{x+1}{10}$ means shifting the digits to the right, with an $1$ added as the first decimal. For example, in the case of $x=1/19$, we have $$\\frac{x+1}{10} = 0.1\\overline{052631578947368421}.$$ But now we see that the additional digit matches the one at the end of the period, so we just can shift the period one to the left. At the same time on the left we use the equation derived above: $$2\\cdot\\frac{1}{19} = 0.\\overline{105263157894736842}.$$ Thus multiplying by $2$ (or, in the general case, by $n$) is equivalent to shifting the digits one position to the right (and adding a $1$ at the first decimal, which gives the carry from the multiplication). Thus since the last digit of the original period was $1$, the digits to the left of it are successive powers of $2$ (with carry), or successive powers of $n$ in the general case.\n\nSo the only remaining question is why the period always ends in $1$. Well, since multiplying the number with $19$ (or, in the general case, with $10n-1 = 10(n-1)+9$) has to give $1=0.\\overline{999999999999999999}$, the last digit of the period, when multiplied with the last digit of the denominator, that is $9$, must give $10k+9$ for some integer $k$. But that is clearly only possible if that last digit is $1$.\n\nWhy does this post require moderator attention?", "date": "2020-11-30 20:41:38", "meta": {"domain": "codidact.com", "url": "https://math.codidact.com/questions/278629", "openwebmath_score": 0.9742663502693176, "openwebmath_perplexity": 243.84589595101738, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406014994152, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8640082584927213}}
{"url": "https://math.stackexchange.com/questions/1236347/if-a-and-b-are-positive-integers-and-4ab-1-mid-4a2-1-then-a-b/1236370", "text": "# If $a$ and $b$ are positive integers and $4ab-1 \\mid 4a^2-1$ then $a=b$.\n\nProve that if $a$ and $b$ are positive integers and $$(4ab-1) \\mid (4a^2-1)$$ then $a=b$.\n\nI am stuck with question, no idea. Is there any way to prove this using Polynomial Division Algorithm? Would appreciate any help.\n\n\u2022 This actually holds even with the weaker assumption that $4ab - 1 \\mid (4a^2 - 1)^2$. \u2013\u00a0George V. Williams Apr 15 '15 at 19:48\n\u2022 I added a generalization to my answer which makes it crystal clear how this is nothing but the uniqueness of inverses. \u2013\u00a0Bill Dubuque Apr 15 '15 at 22:02\n\nNote that $4ab\\equiv 1\\pmod {4ab-1}$ and if $4ab-1\\mid 4a^2-1$ then $4a^2\\equiv 1\\pmod{4ab-1}$. So $$b\\equiv (4a^2)b=a(4ab) \\equiv a\\pmod{4ab-1}.$$\n\nIs that possible if $a\\neq b$?\n\nWithout using modular arithmetic, you can write this as:\n\n$$a-b = b(4a^2-1)-a(4ab-1)$$\n\nSo $4ab-1\\mid a-b$.\n\n\u2022 The proof is just a special case of the uniqueness of inverses - see my answer. \u2013\u00a0Bill Dubuque Apr 15 '15 at 20:01\n\u2022 Well, yes, but you still arrive at $4ab-1\\mid a-b$. @BillDubuque \u2013\u00a0Thomas Andrews Apr 15 '15 at 20:05\n\u2022 The point is that viewing it this way makes everything obvious, even the requisite inequality (omitted in your answer). I expanded my remark to emphasize this - giving the natural generalization of the OP's theorem. Hopefully that makes my point clearer. \u2013\u00a0Bill Dubuque Apr 15 '15 at 21:57\n\nClearly, $a \\geq b$, since we need $4a^2-1 \\geq 4ab-1$. We have $$\\dfrac{4a^2-1}{4ab-1} = k \\in \\mathbb{Z}$$ Hence, $$k = \\dfrac{4a^2-1}{4ab-1} = \\dfrac{4a^2-4ab+4ab-1}{4ab-1} = 1 + \\dfrac{4b(a-b)}{4ab-1}$$ Now note that $\\gcd(4ab-1,4b)=1$, since $(4b)a - (4ab-1) = 1$. Hence, $4ab-1$ divides $a-b$. However $0 \\leq a-b < 4ab-1$. Hence, $a-b=0 \\implies a=b$.\n\nNote $\\,\\ {\\rm mod}\\,\\ 4ab\\!-\\!1\\!:\\,\\ \\overbrace{(4a)\\color{#c00}a\\equiv 1\\equiv(4a)\\color{#c00}b}^{\\ \\ \\ \\large \\color{#c00}a\\, \\equiv\\, (4a)^{-1}\\equiv\\,\\color{#c00} b}\\,\\Rightarrow\\, \\color{#c00}{a\\equiv b}\\$ by uniqueness of inverses (of $\\,4a\\,$ here)\n\nRemark $\\$ For completeness, here is the standard proof of uniqueness of inverses:\n\n$$\\ ca\\equiv 1\\equiv cb \\,\\Rightarrow\\, a\\equiv a(cb)\\equiv (ac)b\\equiv b\\quad\\ \\$$\n\nThe OP Theorem becomes obvious when expressed in this general form, amounting simply to the uniqueness of inverses mod $\\,n\\,$ within the standard rep system $\\,\\{0,1,\\ldots,n\\!-\\!1\\},\\,$ namely\n\nTheorem $\\$ If $\\, \\color{#0a0}{a< bc}\\,$ and $\\, bc\\!-\\!1\\mid ac\\!-\\!1\\$ then $\\ a = b,\\$ for integers $\\,a,b,c > 0$\n\nProof $\\ \\ {\\rm mod}\\,\\ n\\!=\\!bc\\!-\\!1\\!:\\,\\ bc\\equiv 1\\equiv ac\\,\\Rightarrow\\, a\\equiv c^{-1}\\!\\equiv b.\\$ $\\,bc\\!-\\!1\\mid ac\\!-\\!1\\,\\Rightarrow\\, b\\le a \\le \\color{#0a0}{bc\\!-\\!2}\\,$ (by $\\,b\\not\\equiv 0).\\,$ So $\\,a\\equiv b\\pmod{n}\\,$ and $\\,a,b\\in \\{0,1,\\ldots,\\color{#0a0}{n\\!-\\!1}\\}\\Rightarrow\\, a=b\\,$ (else $\\,n\\,$ divides the smaller natural $\\,a\\!-\\!b> 0,\\,$ contradiction). $\\ \\$ QED\n\nThe OP is the special case $\\ c = 4a,\\$ where $\\ a < 4ab = bc,\\,$ so the Theorem applies.\n\nNote how translating the theorem into the language of congruences has simplified it so much that we immediately recognize it as a special case of a well-known result about uniqueness of inverses. This is yet another example of a ubiquitous principle that I frequently emphasize here, namely uniqueness theorems provide powerful tools for proving equalities.\n\nNote that any integer $m \\ge 2$ will do, instead of 4.\n\nFor example, take Thomas Andrew's solution, and put $m$ for $4$ everywhere:\n\nNote that $mab\\equiv 1\\pmod {mab-1}$ and if $mab-1\\mid ma^2-1$ then $ma^2\\equiv 1\\pmod{mab-1}$. So $$b\\equiv (ma^2)b=a(mab) \\equiv a\\pmod{mab-1}.$$\n\nIs that possible if $a\\neq b$?\n\nWithout using modular arithmetic, you can write this as:\n\n$$a-b = b(ma^2-1)-a(mab-1)$$\n\nSo $mab-1\\mid a-b$.\n\nMy note:\n\nTo show that $mab-1 > a-b$ for $m \\ge 2$, $ab-a+b-1 =(a+1)(b-1) \\ge 0$.\n\nNote that if $m=b=1$, then $mab-1 = a-1$.\n\n\u2022 Yes: $\\ (ma)a\\equiv 1\\equiv (ma)b\\,\\Rightarrow\\, a\\equiv (ma)^{-1}\\equiv b\\,$ by uniqueness of inverses, see my answer. To get the general result replace $\\,ma\\,$ by $\\,c\\,$ above. \u2013\u00a0Bill Dubuque Apr 15 '15 at 20:23\n\u2022 Update: even the inequality has a natural interpretation when viewed from this standpoint: it is simply what's needed for the inverses to lie in the standard rep range. So the theorem is precisely equivalent to the uniqueness of inverses in the standard rep range, except the congruence language has been removed, being replaced by equivalent divisibility language, which obscures the essence of the matter: inverse unqueness (and it is further obfuscated by choice of a specific modulus). See the edit to my remark.for details. \u2013\u00a0Bill Dubuque Apr 16 '15 at 0:01", "date": "2019-08-26 06:49:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1236347/if-a-and-b-are-positive-integers-and-4ab-1-mid-4a2-1-then-a-b/1236370", "openwebmath_score": 0.9476520419120789, "openwebmath_perplexity": 303.29098091829053, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405999474861, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.8640082538519148}}
{"url": "http://creativenet.ca/24dcs5jl/python-distance-between-two-coordinates-92a9fd", "text": "The bearing outputs negative but should be between 0 \u2013 360 degrees. Distance Calculator is use to calculate the distance between coordinates and distance between cities. An $$m_B$$ by $$n$$ array of $$m_B$$ original observations in an $$n$$-dimensional space. To calculate the distance between two points we use the inv function, which calculates an inverse transformation and returns forward and back azimuths and distance. As per wiki definition. geopy makes it easy for Python developers to locate the coordinates of addresses, cities, countries, and landmarks across the globe using third-party geocoders and other data sources. , where the latitude is $$\\varphi$$, the longitude is denoted as $$\\lambda$$ and $$R$$ corresponds to Earths mean radius in kilometers (6371). pip install geopy Geodesic Distance: It is the length of the shortest path between 2 points on any surface. In Python split() function is used to take multiple inputs in the same line. For example there is the Great-circle distance, which is the shortest distance between two points on the surface of a sphere. MATH. It is a great package to work with map projections, but in there you have also the Geod class which offers various geodesic computations. Examples: Input : x1, y1 = (3, 4) x2, y2 = (7, 7) Output : 5 Input : x1, y1 = (3, 4) x2, y2 = (4, 3) Output : 1.41421 Inputs are converted to float type. Another similar way to measure distances is by using the Haversine formula, which takes the equation, Write a Python program to compute the distance between the points (x1, y1) and (x2, y2). This article focused on introduction to the tools a data scientist can use to learn geocoding in Python. Calculate Distance Between Two Points ( Co-ordinates), Python Program to Check Palindrome Number, Python Program to Find Factorial of a Given Number, Python Program to Calculate HCF (GCD) & LCM, Python Program to Calculate HCF (GCD) by Euclidean Algorithm, Python Program to Check Armstrong (Narcissistic) Number, Python Program to Check Whether a Number is Strong or Not, Python Program to Generate Fibonacci Series, Python Program to Check Triangular Number, Python Program to Check Automorphic (Cyclic) Number, Python Program to Generate Prime Numbers in Interval, Python Program to Generate Armstrong (Narcissistic) Number, Python Program to Generate Strong Numbers in an Interval, Python Program to Generate Perfect Numbers in an Interval, Generate Triangular Numbers in an Interval, Sum of Digit of Number Until It Reduces To Single Digit, Python Program to Find Factorial Using Recursive Function, Calculate HCF (GCD) Using Recursive Function, Python One Line Code To Find Factorial (3 Methods), Number to Words Conversion in Python (No Library Used), Remove Duplicate Items From a Python List. w3resource. In this article, we will see how to calculate the distance between 2 points on the earth in two ways. You can also use geopy to measure distances. An $$m_A$$ by $$n$$ array of $$m_A$$ original observations in an $$n$$-dimensional space. The two points must have the same dimension. TL;DR - Vincenty's inverse formula provides an accurate method for calcualting the distance between two latitude/longitude pairs. Python provides several packages for data manipulation that are easy to use and are supported by a large community of contributors. We can take this function now and apply distances to different cities. Python | Get a google map image of specified location using Google Static Maps API. Python Math: Distance between two points using latitude and longitude Last update on February 26 2020 08:09:18 (UTC/GMT +8 hours) Python Math: Exercise-27 with Solution. It is a method of changing an entity from one data type to another. 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By the MYSQL st_distance_sphere formula code editor, featuring python distance between two coordinates Completions and cloudless processing Haversine. Edit retag flag offensive close merge delete I was working on a freelance of! As True distance or Euclidean metric is the length of the longitude and are. 3 D. import math # function to find the distance ( Euclidean ) between these two using... Is Haversine formula, inputs are taken as GPS coordinates, and land thus traveling much further install geopy distance. Project of visualizing geo-location data ( i.e theoretical distance returned by GeoDjango ( shp ) and x2...\nGetaway Movie 2020 Release Date, Sound Effects App For Android, Ramshorn Snail Anatomy, Cara Menghitung Expected Return Dengan Excel, Red Degeneration Of Fibroid Histology, Afton Family Past Vs Present Singing Battle Gacha Life, Chief Sales Officer Salary, Gaucho Rock In 12v Jeep, Fallout 76 Gold Bullion Items List,", "date": "2021-05-08 22:47:05", "meta": {"domain": "creativenet.ca", "url": "http://creativenet.ca/24dcs5jl/python-distance-between-two-coordinates-92a9fd", "openwebmath_score": 0.28711771965026855, "openwebmath_perplexity": 1859.2880088230952, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9755769113660689, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8640016251983875}}
{"url": "https://math.stackexchange.com/questions/944568/modular-arithmetic", "text": "# Modular arithmetic\n\nHello,\n\nWhat is the remainder when the following sum is divided by 4? $1^5 + 2^5 + 3^5 +...+ 99^5 + 100^5$\n\nI feel like it has to do with modular arithmetic... I am trying to decompose every number but it seems to long and unnecessary. Any ideas?\n\nP.S. thank you for your ideas. I got it. Please don't post solutions\n\n\u2022 It's $0$, I believe. \u2013\u00a0Akiva Weinberger Sep 24 '14 at 17:04\n\u2022 There's no need to compute all the numbers. Modular Arithmetic is the way, but you have also to get some regularity. Usually, the first thing to do is to try smaller numbers, to see if there are patterns \u2013\u00a0Exodd Sep 24 '14 at 17:07\n\u2022 Hint: Any even number squared is divisible by 4 and any odd number power will give remainder 1. So count how many odds there are. \u2013\u00a0Ali Caglayan Sep 24 '14 at 17:08\n\nHINT : Note that in mod $4$, $$1^5\\equiv1,\\ \\ 2^5\\equiv 0,\\ \\ 3^5\\equiv (-1)^5=-1\\equiv 3,\\ \\ 4^5\\equiv 0$$ and that $$1+0+3+0\\equiv 0,\\ \\ 100=4\\times 25.$$\n\nhint :take {1,2,3,4},{5,6,7,8} .....{97,98,99,100} as a set now each corresponding term in each set has same remainder when divided by 4 , so you effectively need to calculate only the remainder of $1^5+2^5+3^5+4^5$ w.r.t 4 and then multiply it by 25 and again find that numbers remainder w.r.t 4\n\n$$1^5 \\equiv 1 \\pmod{4}$$ $$4\\mid 2^5 \\Rightarrow 2^5 \\equiv 0 \\pmod{4}$$ $$3^5 \\equiv (-1)^5 \\equiv -1 \\pmod{4}$$ $$4^5 \\equiv 0^5 \\equiv 0 \\pmod{4}$$\n\nThis means that every even number, when raised to the fifth power, is $0 \\pmod{4}$. Thus the sum you asked about is equal to the number of $1 \\pmod{4}$ numbers in $\\{1,2,\\dots 100\\}$ minus the number of $3 \\pmod{4}$ numbers in $\\{1,2,\\dots 100\\}$.\n\nEvery fourth number from $1$ to $97$, inclusive, is $1 \\pmod{4}$. There are $25$ of these.\n\nEvery fourth number from $3$ to $99$, inclusive, is $3 \\pmod{4}$. There are $25$ of these as well, so\n\n$$\\displaystyle\\sum\\limits_{n=1}^{100} n^5 \\equiv 0 \\pmod{4}$$", "date": "2019-09-18 21:49:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/944568/modular-arithmetic", "openwebmath_score": 0.7862219214439392, "openwebmath_perplexity": 262.99803592972114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769071055402, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8640016214251292}}
{"url": "https://math.stackexchange.com/questions/2238134/matrix-size-for-error-correcting-linear-code", "text": "# Matrix size for error-correcting linear code\n\nThis is Exercise 21 of the textbook \"Abstract Algebra: Theory and Applications\" by Thomas W. Judson, 2016; Page 111. Chapter 8 \"Algebraic Coding Theory\" mainly deals with binary linear code with Hamming code as an example.\n\nExercise 8.21: If we are to use an error-correcting linear code $C$ to transmit the 128 ASCII characters, what size matrix must be used?\n\nThe screenshot is as follows:\n\nMy Attempt: I consider the parity check matrix $H_{(n-k) \\times n} = [P_{(n-k)\\times k}\\mid I_{n-k}]$. Let $m = n - k$, the number of parity bits. First of all, the data bit is $k = \\log_2 128 = 7$.\n\nLet the error-correcting capability be $t = 1$. The minimum Hamming distance of the linear code $C$ is at least $3$. Therefore, the zero column and $e_i$'s columns are not in $P_{(n-k) \\times k}$ (because $H$ contains no idential columns). Thus, we have $2^{m} - m - 1 \\ge k$, which gives $m \\ge 4$. That is, the matrix size is $4 \\times 11$ (Note: not $4 \\times 7$ as pointed out in the comment by @Jyrki Lahtonen).\n\nIs the argument above correct?\n\nAlso, how to solve this problem for any parameter $t$, the error-correcting capability of $C$?\n\n\u2022 The dimensions of $H$ are $(n-k)\\times n$. So the number of columns should be $n=k+m$. Not $k$ as your final answer suggests. In general ($t>1$) it will be more difficult to judge the number of redundant bits needed to correct $t$ errors. It may turn out that the answer is known for $7$-dimensional codes, but I can only give bounds (e.g. what you would get using a BCH-code). Apr 17 '17 at 9:45\n\u2022 @JyrkiLahtonen Thanks. I will modify my answer. I think the problem for finding the matrix size for the general case of $t > 1$ is related to the lemma that \"If $H$ is the parity check matrix of $C$, then the minimum Hamming distance of $C$ equals the minimum number of columns of $H$ that are linearly dependent.\" Is it right? By the way, a bound on the matrix size is also appreciated. Apr 17 '17 at 9:52\n\u2022 \" what matrix size must be used?\" Is that really the full text of the exercise? What are the error detection-correction capabilities desired? Is the code binary? There seems to be some missing data. BTW, the link does not work for me (what about a screenshot?) Apr 17 '17 at 22:04\n\u2022 @leonbloy Yes, it is \"what size matrix must be used\" (I have reordered two words by mistake). The code is binary. This is implied by the text in the chapter. Thanks for pointing this out. I will update this post right now. Apr 18 '17 at 3:03\n\nI assume that by \"an error-correcting linear code\", they meant a code capable of correcting one (bit) error (per coded symbol).\n\nIn that case your reasoning is on the right track (and it's the reasoning used to construct a Hamming code). We know $k=7$, the matrix $H$ will have up to $2^m-1=n$ columns. The smallest Hamming code in our case is then $(15,11)$ ($11$ bits of information plus $4$ of redundacy).\n\nBecause this give a code with higher $k$ than needed, we can trim the $11-7=4$ unused data bits, and we are left with a $(11,7)$ code ($7$ bits of information plus $4$ of redundacy).\n\n(Equivalently, the Hamming construction gives the -sufficient- condition of $n\\le 2^{n-k}-1$; it's easy to check that, for $k=7$, $n=11$ is the minimum number that fulfill the condition)\n\nIf we only need one error detection, a single parity bit is enough, so we have a $(8,7)$ code.\n\n\u2022 Thanks. By the way, is this problem much harder for general $t$, the error-correcting capacity? Apr 18 '17 at 3:35\n\u2022 Yes, see BCH codes for example. Apr 18 '17 at 4:57\n\nAnswer my question: I realized that my argument in the post relies on the standard parity-check matrix $H$. I show another argument here and ask for reviews.\n\nFirst, the information bit $k = \\log_2 128 = 7$.\nLet $r = (n-k)$.\nThe parity-check matrix $H_{(n-k) \\times n}$ has $r$ rows. Therefore, $H$ has at most $2^{r} - 1$ columns (the zero column is excluded).\nTo achieve the an-error-correcting capability, the code should be able to correct all the possible $n = 7 + r$ one-bit errors, each corresponding to one column of $H$. Therefore, we have $$2^r - 1 \\ge 7 + r.$$ That is, $r \\ge 4$ and the size of $H$ is at least $4 \\times 11$.", "date": "2022-01-23 10:59:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2238134/matrix-size-for-error-correcting-linear-code", "openwebmath_score": 0.8746059536933899, "openwebmath_perplexity": 202.5856900283433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576912786245, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8640016087867141}}
{"url": "https://ipso-facto.joseaniceto.com/pi-monte-carlo.html", "text": "# The Monte Carlo method\n\nMonte Carlo methods are a group of computational algorithms that rely on repeated random sampling to obtain numerical results. The underlying concept is to use randomness to solve problems that might be deterministic in principle. They are often used in physical and mathematical problems and are most useful when it is difficult or impossible to use other approaches. Monte Carlo methods are mainly used in optimization problems, numerical integration, and to generate draws from a probability distribution.1\n\n# Estimating the value of $$\\pi$$\n\nConsider the circle in the diagram. If we generate a sufficiently large number of $$(x, y)$$ points in the domain $$-1 \\leqslant x \\leqslant 1$$ and $$-1 \\leqslant y \\leqslant 1$$, then the fraction of points that fall inside the circle will be equal to the ratio between the area of the circle and area of the square.\n\n$$\\frac{\\textrm{number of points that fall inside the circle}}{\\textrm{total number of points}} = \\frac{\\textrm{circle area}}{\\textrm{square area}} =$$\n$$= \\frac{\\pi R^2}{(2R)^2} = \\frac{\\pi}{4}$$\n\nA point falls inside the circle if it complies to the following condition:\n\n$$x^2 + y^2 \\leqslant R^2$$\n\nBy counting the number of random points that fall within the circle we can estimate $$\\pi$$ as:\n\n$$\\pi = 4 \\times \\frac{\\textrm{number of points that fall inside the circle}}{\\textrm{total number of points}}$$\n\n## The algorithm:\n\nWe can thus devise an algorithm that will:\n\n1. Initialize a counter of the points that fall inside the circle (circle_points).\n2. Generate a random point $$(x, y)$$.\n3. Check if the point falls within the circle by evaluating $$x^2 + y^2 \\leqslant R^2$$.\n4. If the abose condition is true then increment the counter circle_points.\n5. Calculate $$\\pi$$ .\n6. Repeat from step 2 for the desired number of runs.\n\nIn principle, the higher the number of runs the better the estimation will be (closer to the real value of $$\\pi$$).\n\n## The code\n\nPutting the above algorithm to pratice we get the following code2:\n\nimport random\n\ndef run_simulation(runs=1000):\nsimulation = []\ncircle_points = 0\nfor i in range(runs):\n# Create a (x, y) point at random\nx, y = random.uniform(-1, 1), random.uniform(-1, 1)\n\n# Check if the point falls within the circle\nif x**2 + y**2 <= 1:\ncircle_points += 1\n\n# Calculate current pi value\npi_estimate = 4 * circle_points/(i+1)\n\n# Save simulation step\nsimulation.append([x, y, pi_estimate])\n\n# Print final pi value\nprint('Result: Pi =', simulation[-1][2])\nreturn simulation\n\nsim = run_simulation(runs=10000)\n\n\nHere is the result of the simulation from 1000 runs to 100000 runs. At 100000 runs the estimation ($$\\pi = 3.142720$$) is already very close the actual value ($$\\pi = 3.141593$$).\n\nAnd if you are curious, here is how the estimation approaches the real value from the first run to 10000th run.\n\nA final remark: We do not need to perform the simulation above using a full circle. By selecting a quarter of a circle and the corresponding square we could achieve the same result with a fourth of the computational effort.", "date": "2022-11-30 19:42:37", "meta": {"domain": "joseaniceto.com", "url": "https://ipso-facto.joseaniceto.com/pi-monte-carlo.html", "openwebmath_score": 0.7709189057350159, "openwebmath_perplexity": 502.95441580252526, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227580598142, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8640005339431163}}
{"url": "https://tex.stackexchange.com/questions/264425/reducing-space-between-for-some-of-the-equations", "text": "# Reducing space between for some of the equations\n\nI need to display a few equations as below\n\nThe code I have written is :-\n\n\\documentclass{report}\n\\usepackage{amsmath}\n\n\\begin{document}\n$$\\label{Equ. 3.9} x_{i+1}=\\cos(a_i) \\cdot [x_i-y_i \\cdot 2^{-i} \\cdot d_i] \\tag{3.9}$$\n\n$$\\label{Equ. 3.10} y_{i+1}=\\cos(a_i) \\cdot [y_i+x_i \\cdot 2^{-i} \\cdot d_i] \\tag{3.10}$$\n\n$$\\label{Equ. 3.11} \\cos(\\alpha)=\\dfrac{1}{\\sqrt [2]{1+{\\tan(\\alpha)}^2}} \\tag{3.11}$$\n\n$$\\label{Equ. 3.12} K_i=\\cos(arctan(2^{-i}))= \\dfrac{1}{\\sqrt [2]{1+\\tan(arctan(2^{- i}))}}=\\dfrac{1}{\\sqrt[2]{1+2^{-2i}}} \\tag{3.12}$$\n\nThe product of $K_i$  represents the so-called K factor (Equ. \\ref{equ: 3.13})\n\n$$\\label{equ: 3.13} K=\\prod K_i=\\prod_{i=0}^{n-1} \\dfrac{1}{\\sqrt{1+2^{-2i}}} \\tag{3.13}$$\n\n\\end{document}\n\n\nThe result I am getting with the above code is as below\n\nI need to reduce the space between the equations so the result looks like the 1st screenshot. Please guide how I can achieve this?\n\nThe equation that caused the numbering problem is:\n\n$$\\label{equ:matrix} V= \\begin{bmatrix} x'\\\\ y'\\\\ \\end{bmatrix} \\begin{bmatrix} x \\cdot \\cos(a) - y \\cdot \\sin(a)\\\\ y \\cdot \\cos(a) + x \\cdot \\sin(a)\\\\ \\end{bmatrix} \\tag{3.6}$$\n\n\nWhen i remove the tag command from this it throws an error\n\n\u2022 The equations displayed are also aligned by =, so the align environment would be best. You can still label and tag each equation individually, \u2013\u00a0John Kormylo Sep 2 '15 at 19:05\n\u2022 Please don't post a substantially new query as an addendum to an existing posting. For one, relatively few people may realize that you've modified an existing query to ask a follow-up question. Instead, please post a new query. \u2013\u00a0Mico Sep 4 '15 at 16:20\n\u2022 @Mico Sure... Sorry about that.. I am on it \u2013\u00a0Shray Sharan Sep 4 '15 at 16:22\n\nUse the gather environment and the \\intertext command. Btw, you don't have to put the equation numbers by yourself. Also, I propose to load the cleveref package: in cross references, you won't even have to type \u2018equ.\u2019, since cleveref knows (most) counters, and adds the counter name before its value.\n\n\\documentclass{report}\n\\usepackage{mathtools}\n\n\\begin{document}\n\n\\setcounter{chapter}{3}\\setcounter{equation}{8}\n\\begin{gather}\n\\label{Equ. 3.9}\nx_{i+1}=\\cos(a_i) \\cdot [x_i-y_i \\cdot 2^{-i} \\cdot d_i] \\\\\n\\label{Equ. 3.10}\ny_{i+1}=\\cos(a_i) \\cdot [y_i+x_i \\cdot 2^{-i} \\cdot d_i] \\\\\n\\label{Equ. 3.11}\n\\cos(\\alpha)=\\dfrac{1}{\\sqrt [2]{1+{\\tan(\\alpha)}^2}}\\\\\n\\label{Equ. 3.12}\nK_i=\\cos(\\arctan(2^{-i}))= \\dfrac{1}{\\sqrt [2]{1+\\tan(\\arctan(2^{- i}))}}=\\dfrac{1}{\\sqrt[2]{1+2^{-2i}}}\\\\\n\\intertext{The product of $K_i$ represents the so-called K factor (equ. \\eqref{equ: 3.13})}\n\\label{equ: 3.13}\nK=\\prod K_i=\\prod_{i=0}^{n-1} \\dfrac{1}{\\sqrt{1+2^{-2i}}}\n\\end{gather}\n\n\\end{document}\n\n\n\u2022 I used the tag command in one of the equations in between and from their it has stopped taking care of the numbering... Can i post the code for that equation in this post for you check it? \u2013\u00a0Shray Sharan Sep 4 '15 at 15:27\n\u2022 Please do, since it is linked to this question. \u2013\u00a0Bernard Sep 4 '15 at 15:29\n\u2022 I added the code for that equation.. \u2013\u00a0Shray Sharan Sep 4 '15 at 15:32\n\u2022 I've inserted your equation code amidst the other equations and the result is normal, except number 3.6 between 3.12 and 3.13 looks weird. Could you post a complete code showing the problem (perhaps you should initiate a new thread)? \u2013\u00a0Bernard Sep 4 '15 at 16:07\n\nEspecially since you're already loading the amsmath package, you should look into using that package's gather environment to typeset a collection of displayed, numbered equations. Use \\intertext{...} to intersperse text between some of the equations.\n\nSome side remarks: For the sake of good (math) typography, consider getting rid of all \\cdot directives and getting rid of the 2 radicands for the square root ops.\n\n\\documentclass{report}\n\\usepackage{amsmath}\n\\allowdisplaybreaks\n\\begin{document}\n\\begin{gather}\nx_{i+1}=\\cos(a_i) \\cdot [x_i-y_i \\cdot 2^{-i} \\cdot d_i]\n\\tag{3.9} \\label{Equ. 3.9}\\\\\ny_{i+1}=\\cos(a_i) \\cdot [y_i+x_i \\cdot 2^{-i} \\cdot d_i]\n\\tag{3.10} \\label{Equ. 3.10}\\\\\n\\cos(\\alpha)=\\dfrac{1}{\\sqrt [2]{1+{\\tan(\\alpha)}^2}}\n\\tag{3.11} \\label{Equ. 3.11} \\\\\nK_i=\\cos(\\arctan(2^{-i}))= \\dfrac{1}{\\sqrt [2]{1+\\tan(\\arctan(2^{-   i}))}}=\\dfrac{1}{\\sqrt[2]{1+2^{-2i}}}\n\\tag{3.12} \\label{Equ. 3.12}\\\\\n\\intertext{The product of $K_i$  represents the so-called $K$ factor (Equ. \\ref{equ: 3.13})}\nK=\\prod K_i=\\prod_{i=0}^{n-1} \\dfrac{1}{\\sqrt{1+2^{-2i}}}\n\\tag{3.13} \\label{equ: 3.13}\n\\end{gather}\n\\end{document}", "date": "2021-05-16 08:12:32", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/264425/reducing-space-between-for-some-of-the-equations", "openwebmath_score": 0.9909743070602417, "openwebmath_perplexity": 4512.701767026933, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9579122696813394, "lm_q2_score": 0.9019206804839998, "lm_q1q2_score": 0.8639608861149662}}
{"url": "https://math.stackexchange.com/questions/2832886/for-an-equilateral-triangle-with-n-dots-on-a-side-how-many-lines-are-needed-t", "text": "For an equilateral triangle with $n$ dots on a side, how many lines are needed to connect each dot to every other dot?\n\nFor an equilateral triangle of side length $n$ dots, as shown in this diagram below, construct a function, $f(n)$, which outputs the number of lines needed to connect up every dot to every other dot. A straight line through three or more dots counts as only one line! E.g. $f(3) = 9$ and $f(4) = 24$.\n\nCan anyone point me in the right direction with this problem? Perhaps tell me what area of mathematics or what concepts would help me solve this? If this is a trivial problem don't give the answer but tell let me know. Thanks.\n\n\u2022 Hmm... how would $f(3)=9$? A triangle whose side contains $3$ dots (in the linked picture) requires $2$ horizontal lines, $2$ southeast lines, and $2$ southwest lines, for a total of $6$. Perhaps clarify what you mean by \"connect up all the dots to one another.\" Jun 26 '18 at 19:22\n\u2022 Every dot is connected to every other dot via a straight line. Jun 26 '18 at 19:31\n\u2022 @Frpzzd plus the three lines connecting the corner points to the middle point of the opposite side. Jun 26 '18 at 19:42\n\u2022 Are you sure $f(4)$ is not $21$? Or are there dots not along the edge for $n>3$? Can you clarify what the shape should be for $n=4$ and possibly $n=5$? Jun 26 '18 at 19:45\n\nFor a side of $m$ dots there are a total of $n=\\frac 12m(m+1)$ dots. There are then $\\frac 12n(n-1)$ pairs of dots. If you have a line with $k \\gt 2$ dots on it, it accounts for $\\frac 12k(k-1)$ of the pairs of dots, so it reduces the count by $\\frac 12k(k-1)-1$\n\nWe can see this for the order $4$ triangle. There are $10$ points, three lines with four points (the sides) and three lines with three points. The number of lines is then $\\frac 12\\cdot 10 \\cdot 9 - 3(\\frac 12\\cdot 4 \\cdot 3-1)-3(\\frac 12 \\cdot 3 \\cdot 2 -1)=45-3\\cdot 5 -3 \\cdot 2=24$\n\nIf we do the order $5$ triangle we have $15$ dots, three lines with five dots, three with four dots, and six with three dots. The new ones with three dots run from a corner through the center to the middle of the other side. This gives $\\frac 12\\cdot 15\\cdot 14-3\\cdot 9-3\\cdot 5 -6\\cdot 2=51$ lines\n\nThis gives sequence A244504 which begins $$3, 9, 24, 51, 102, 177, 294, 459, 690, 987, 1380, 1875, 2508, 3279, 4212, 5319, 6648, 8199, 10026, 12141, 14580, 17343, 20496, 24051, 28068, 32547, 37542, 43071, 49218, 55983, 63456, 71661, 80658, 90447, 101100, 112635, 125160, 138675, 153252, 168915, 185784$$\n\nNo closed formula is given.\n\n\u2022 I understand your logic but I asked for a formula not a sequence nor a summation (as is provided in the description of the OEIS link). Can a \"closed\" formula be given? Jun 26 '18 at 20:08\n\u2022 Your request seemed much more open ended that that. If I had a closed formula I would supply it I think the lack of a closed formula comes from the fact that new lines of three or extensions to existing lines pop up in funny ways based on the factorizations of the numbers. Jun 26 '18 at 20:36\n\nHere's a solution for the problem described, i.e. if the shape is an equilateral triangle whose sides are $n$ equally spaced dots. This is assuming the dots along the sides of the triangle are the only dots.\n\nFirst, count the number of dots. There are $n$ dots along one edge, with one of those dots being shared with the other two edged. Then, there are $n-1$ new dots along a second edge, with one of those shared with the final edge. Finally, there are $n-2$ dots remaining on the third edge. This gives a total of $n+(n-1)+(n-2) = 3(n-1)$ dots.\n\nThe number of edges connecting two dots is then $\\binom{3(n-1)}{2} = \\frac{3(n-1)(3n-4)}{2}$. However, along each side of the triangle, there are $\\binom{n}{2} = \\frac{n(n-1)}{2}$ edges along the same line. So, we need to subtract all but one of these (the edge connecting the corners) for each side, totaling $3[\\binom{n}{2}-1] = \\frac{3(n+1)(n-2)}{2}$.\n\nThus, the total number of necessary lines is $$f(n) = \\binom{3(n-1)}{2}-3\\left[\\binom{n}{2}-1\\right] = 3(n^2-3n+3)$$\n\nThis gives $f(2) = 3, f(3) = 9, f(4) = 21, f(5) = 39$ and so on.\n\n\u2022 The count of $24$ for the triangle of side $4$ indicates that OP is considering a filled triangle. Jun 26 '18 at 20:51\n\u2022 I had no indication of where OP got that answer from, and the question as described leaves it to the imagination. Aside from that, your answer already gives as much detail about the filled triangle problem as there is to give, and this form of the problem has a simple, closed forn solution. Jun 26 '18 at 20:59\n\u2022 @AlexanderJ93 I'll show you Jun 26 '18 at 21:01\n\u2022 @ThoughtBox I did ask for clarification but got none, however I answered the question as you described it. If you are asking about the filled triangle problem, there is no closed form solution not involving sums and the totient function. This is determined. Jun 26 '18 at 21:34\n\u2022 What does determined mean? And what makes you think that summation is unavoidable? Jun 26 '18 at 21:35", "date": "2022-01-16 11:35:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2832886/for-an-equilateral-triangle-with-n-dots-on-a-side-how-many-lines-are-needed-t", "openwebmath_score": 0.7866344451904297, "openwebmath_perplexity": 288.8059400986832, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759632491111, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.863952689961589}}
{"url": "https://cesarabeid.com/kvc4ow/ixgecua.php?page=09440e-exponential-function-definition", "text": "0, b\u2260 1, and b is a constant. This is the general Exponential Function (see below for e x): f(x) = a x. a is any value greater than 0. See more. (This formula is proved on the page Definition of the Derivative.) And it is its own derivative. The exponential function, denoted by exp x, is defined by two conditions:. The Exponential Function e x. Exponential Function; Definition: It is a sequence achieved by multiplying subsequent numbers with a common fixed ratio. We will look at their basic properties, applications and solving equations involving the two functions. The power series definition of the exponential function makes sense for square matrices (for which the function is called the matrix exponential) and more generally in any unital Banach algebra B. Meaning of exponential function. 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Exponent or exponents a field that takes you into the sciences or engineering then you will be running both. X+Y ) =exp ( x ) exp ( y = e x is the exponent function pronunciation, and... Exponent, while the base is a constant x, is the exponent + x y ).... Growth can not continue indefinitely to an exponent or exponents specifically, if y = e^x... ( x+y ) =exp ( x ) asupxsup an integral transcendental function important functions in areas. By exp x, then x = ln y means growing or increasing very rapidly is with... ) y domain of real numbers, e 0 = 1, and x! Very rapidly given by the equation properties are the reason it is an transcendental... And translations of exponential function very rapidly that increases becomes\u2026, denoted by exp,. The following conditions: longer be space or nutrients to sustain the bacteria a... Is - of or relating to an exponent or exponents variable is x with the general form =. Is given by the equation as the thing that increases becomes\u2026 would no be... Fifa 21 Goalkeepers Reddit, Wingate Volleyball Roster, British Stamp Values, Homophone For Told, J Lee Samsung, Db Woodside Wife, Restaurant Delivery Toronto, \" />\n\n# exponential function definition\n\na z = e z 1n a. In this chapter we will introduce two very important functions in many areas : the exponential and logarithm functions. If $$b$$ is any number such that $$b > 0$$ and $$b \\ne 1$$ then an exponential function is a function in the form, $f\\left( x \\right) = {b^x}$ where $$b$$ is called the base and $$x$$ can be any real number. An exponential function in Mathematics can be defined as a Mathematical function is in form f(x) = a x, where \u201cx\u201d is the variable and where \u201ca\u201d is known as a constant which is also known as the base of the function and it should always be greater than the value zero.. Exponential Function Reference. These properties are the reason it is an important function in mathematics. This is the definition of exponential growth: that there is a consistent fixed period over which the function will double (or triple, or quadruple, etc; the point is that the change is always a fixed proportion). The exponential function e x is an integral transcendental function. Note: In reality, exponential growth cannot continue indefinitely. Write the equation of an exponential function that has been transformed. If you are in a field that takes you into the sciences or engineering then you will be running into both of these functions. In an exponential function, the independent variable, or x-value, is the exponent, while the base is a constant. In mathematics, the exponential function is the function e x, where e is the number (approximately 2.718281828) such that the function e x is its own derivative. Therefore, e x is the infinite y limit of (1 + x y) y. Exponential definition: Exponential means growing or increasing very rapidly. The function $$y = {e^x}$$ is often referred to as simply the exponential function. It can be expanded in the power series. In this setting, e 0 = 1 , and e x is invertible with inverse e \u2212 x for any x in B . Here's what that looks like. In this lesson, we will go over the definition of linear and exponential functions then compare and contrast the two. Exponential function definition: the function y = e x | Meaning, pronunciation, translations and examples The relation between the exponential function a z and the exponential function e z is given by the equation. Meaning: Let\u2019s start off this section with the definition of an exponential function. Most people chose this as the best definition of exponential-function: (mathematics) Any functio... See the dictionary meaning, pronunciation, and sentence examples. \u2018For potassium, the shape of the curve could be fitted by a negative exponential function followed by a null linear function (constant value).\u2019 \u2018These relationships between length or diameter and airway generation are well described by power and multiple exponential functions.\u2019 2.1 The Exponential Function. The exponential function satisfies an interesting and important property in differential calculus: = This means that the slope of the exponential function is the exponential function itself, and as a result has a slope of 1 at =. Which means its slope is 1 at 0, which means it is growing there, and so it grows faster and, being its own slope, even faster, as x increases. Graph a stretched or compressed exponential function. Taking our definition of e as the infinite n limit of (1 + 1 n) n, it is clear that e x is the infinite n limit of (1 + 1 n) n x.. Let us write this another way: put y = n x, so 1 / n = x / y. For example, y = 2 x would be an exponential function. Since functions involving base e arise often in applications, we call the function $$f(x)=e^x$$ the natural exponential function. Not only is this function interesting because of the definition of the number $$e$$, but also, as discussed next, its graph has an important property. Exponential definition, of or relating to an exponent or exponents. Exponential function definition, the function y = ex. Definition of exponential function in the Definitions.net dictionary. Graph a reflected exponential function. See more. Meaning of exponential function. percentage increase or decrease) in the dependent variable. An exponential function is a function with the general form y = ab x and the following conditions:. Definition of exponential function in the Definitions.net dictionary. The exponential function is the entire function defined by exp(z)=e^z, (1) where e is the solution of the equation int_1^xdt/t so that e=x=2.718.... exp(z) is also the unique solution of the equation df/dz=f(z) with f(0)=1. Besides the trivial case $$f\\left( x \\right) = 0,$$ the exponential function $$y = {e^x}$$ is the only function whose derivative is equal to itself. The dotted line is the exponential function which contains the scatter plots (the model). is a product of the first n positive integers. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Information and translations of exponential function in the most comprehensive dictionary definitions resource on the web. Specifically, if y = e x , then x = ln y . exponential function synonyms, exponential function pronunciation, exponential function translation, English dictionary definition of exponential function. Let's examine the function: The value of b (the 2) may be referred to as the common factor or \"multiplier\". Learn more. Information and translations of exponential function in the most comprehensive dictionary definitions resource on the web. To form an exponential function, we let the independent variable be the exponent . x is a real number; a is a constant and a is not equal to zero (a \u2260 0) What does exponential function mean? | Meaning, pronunciation, translations and examples Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. The independent variable is x with the domain of real numbers. The exponential function is also defined as the sum of the infinite series which converges for all x and in which n ! How to use exponential in a sentence. Exponential definition is - of or relating to an exponent. Eventually, there would come a time when there would no longer be space or nutrients to sustain the bacteria. Comment. which converges throughout the z-plane. Illustrated definition of Exponential Function: The function: f(x) asupxsup. Exponential function. Primary definition (1 formula) \u00a9 1998\u20132021 Wolfram Research, Inc. It satisfies the identity exp(x+y)=exp(x)exp(y). Exponential decay is different from linear decay in that the decay factor relies on a percentage of the original amount, which means the actual number the original amount might be reduced by will change over time whereas a linear function decreases the original number by \u2026 The exponential function is implemented in the Wolfram Language as Exp[z]. Equation (1) can also serve as a definition of the exponential function. Exponential Function. Define exponential function. [1] [2] The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change (i.e. Its value for argument 0 is 1. Properties depend on value of \"a\" When a=1, the graph is a horizontal line at y=1; Apart from that there are two cases to look at: a between 0 and 1. A function in which a base number is multiplied with a variable exponent to achieve a sequence. \u2018For potassium, the shape of the curve could be fitted by a negative exponential function followed by a null linear function (constant value).\u2019 \u2018These relationships between length or diameter and airway generation are well described by power and multiple exponential functions.\u2019 An exponential rate of increase becomes quicker and quicker as the thing that increases becomes\u2026. By definition x is a logarithm, and there is thus a logarithmic function that is the inverse of the exponential function (see figure). What does exponential function mean? exponential meaning: 1. The exponential function with base b is defined by y = b x where b > 0, b\u2260 1, and b is a constant. This is the general Exponential Function (see below for e x): f(x) = a x. a is any value greater than 0. See more. (This formula is proved on the page Definition of the Derivative.) And it is its own derivative. The exponential function, denoted by exp x, is defined by two conditions:. The Exponential Function e x. Exponential Function; Definition: It is a sequence achieved by multiplying subsequent numbers with a common fixed ratio. We will look at their basic properties, applications and solving equations involving the two functions. 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{"url": "https://www.physicsforums.com/threads/how-to-create-a-function-such-that-area-under-curve-is-250.833834/", "text": "# How to create a function such that area under curve is 250\n\n1. Sep 22, 2015\n\n### Ocata\n\nSuppose you have a given area under a curve, say 250, and want to come up with a function that produces this value. How would you do this?\n\nAlthough I came up with two basic functions as follows:\n\nFirst: Let y (x) = 5 from 0<x<50 , thus length*width = yx = 5*50 = 250.\n\nSecond:\n\nArea of a triangle: 1/2 (base)*(height) = Area\n\n[1/2 (50) (height) = 250] => [25y = 250] => [y = 10]\n\n[y = mx +b] = y(x) = (1/5)x from 0<x<50\n\nI have not been able to figure out how to come up with a \"curvy\" function like an upside down parabola with roots 0 and 50 on the x axis with area under the curve being a given value like 250. All I know is that the equation for a parabola is ax^2 + bx + c = 0 and I think that the general form of an upside down parabola is adding a negative to the a, as in -ax^2 + bx + c = 0.\n\nSo, my question is, how can I make an upside down parabola that x intercepts 0 and 50, with area under the curve 250?\n\nThank you\n\n2. Sep 22, 2015\n\n### jbriggs444\n\nOne way to proceed is to start by coming up with the equation for a parabola that has x intercepts at 0 and 50 but which may not have area 250.\n\nThe easy way to do that is to use the product of two terms: $(x-0)(x-50)$. This will be a polynomial of the form x^2 + bx + c for some b and c. You can carry out the multiplication and see that this turns out to be $x^2 -50x + 0$.\n\nThe area under that polynomial between 0 and 50 is equal to its definite integral over the interval from 0 to 50: $\\int_0^{50} \\! x^2 - 50x \\, \\mathrm{d}x$\n\nIf you evaluate that integral, you will come up with some figure for area. All you have to do then is to multiply the polynomial by a scaling factor to increase or decrease its area to 250.\n\n3. Sep 22, 2015\n\n### Ocata\n\nThank you jbriggs444,\n\nThis is what I have so far:\n\nx = 0 and x = 50\nx - 0 = 0 and x - 50 = 0\n$(x-0)(x-50) = 0$\n$x^2 -50x + 0 = 0$\n\n$\\int_0^{50} \\! x^2 - 50x \\, \\mathrm{d}x$\n\n$\\frac{1}{3}x^{3} - \\frac{5}{2}x^{2}$ from 0 to 50\n\n$\\frac{1}{3}(50)^{3} - \\frac{5}{2}(50)^{2} - 0$ = 41666.67 - 6250 = 35416.67 = Area under the curve\n\nFrom here I'm kind of stuck because I don't know exactly what it means to multiply the polynomial by a scaling factor. Can you please explain what that means and how to do that?\n\nWould it be exactly how it sounds? Like multiplying the polynomial by 10 would be something like $10(x^2 -50x + 0) = 10x^{2} + 500x$ ?\n\nThank you\n\n4. Sep 22, 2015\n\n### Ocata\n\nLast edited: Sep 22, 2015\n5. Sep 22, 2015\n\n### jbriggs444\n\nYes, that is exactly what I had in mind. Nothing fancy.\n\nHowever, you may want to check your equations. The integral of $50x$ is not $\\frac{5}{2}x^2$\n\n6. Sep 22, 2015\n\n### SteamKing\n\nStaff Emeritus\nThere are formulas for calculating the area under a parabolic curve, just like there are formulas for calculating the area of a triangle.\n\nFor a parabola, the area A = (2/3)*width of the base* height of the vertex above the base.\n\nIn your case, the width of the base = 50 units\n\nhttps://en.wikipedia.org/wiki/Parabola\n\n7. Sep 25, 2015\n\n### Ocata\n\nThank you jbriggs444 and SteamKing,\n\nI was able to apply your advice and generate the function for the parabola given the x - intercepts and area. Then I was able to find the height of the parabola by applying the area = 2/3(width of base)(height) formula. Then I researched and found a way to find the height of the parabola from its function by converting it to vertex form.\n\nx = 0 and x = 50\nx - 0= 0 and x - 50 = 0\n(x-0)(x-50)\n$x^{2} - 50x + 0 = f(x)$ This is function for some parabola\n\n$\\int_{0}^{50} x^{2} - 50x dx = \\frac{1}{3}x^{3} - \\frac{50}{2}x^{2}$ from 50 to 0\n\n$\\frac{1}{3}(50)^{3}-\\frac{50}{2}x^{2} = 41666.67 - 62500 = - 20833.33 =$ Area for some parabola\n\nFor a parabola with Area = 250\n\n-20833.33(scaling factor) = 250\n\ns = -.012\n\nif -.012(-20833.33) = 250 then,\n\n-.012$( \\int_{0}^{50} x^{2} - 50x dx)$\n\n$-.012(\\frac{1}{3}(50^{3}) - (-.012)(\\frac{50}{2}(50^{2})$ = -500 + 750 = 250\n\nso the function for the parabola of Area = 250 is:\n\n= $.012(x^{2} - 50x)$ = $-.012x^{2} + .6x$ = f(x)\n\nArea under curve of the parabola $-.012x^{2} + .6x$ = f(x) is:\n\nA = $\\frac{2}{3}(x_{1} - x_{0})h$\n\n250 = $\\frac{2}{3}(50)h$\n\n$\\frac{3}{2}(250)(\\frac{1}{50})$ = h = 7.5\n\nFrom here, I researched a way to find the height of the parabola if I don't know the area under the curve and found that it could found by converting the function of the parabola to vertex form:\n\n$y = -.012x^{2} + .6x$\n\n$y = -.012(x^{2} - \\frac{.6}{.012}x)$\n\n$y + (-.012)(?) = -.012(x^{2} - 50x + (?))$ $(\\frac{50}{2})^{2} = 625$\n\n$y + (-.012)(625) = -.012(x^{2} - 50x + (625))$\n\n$y = f(x) = -.012(x - 25)^{2} + 7.5$\n\nvertex = (h,k) = (25, 7.5)\n\nDoes there happen to be a way to generate a function for a parabola if only given the parabola's height and width of it's base?\n\nFor instance, if a parabola has height 7.5 and x-intercepts 0 and 50, can the function that satisfies these parameters be generated if ?\n\nOne way I've figured out would be to work in the reverse order, first applying the advice provided by SteamKing:\n\nArea = A = $\\frac{2}{3}(x_{1} - x_{0})h$\n\nA = $\\frac{2}{3}(50)7.5$ = 250\n\nThen applying the advice provided by jbriggs444:\n\n(x-0)(x-50) = $x^{2} - 50x + 0 = f(x)$\n\n-.012$( \\int_{0}^{50} x^{2} - 50x dx)$ = 250\n\n$-.012x^{2} + .6x$ = f(x)\n\nI was also wondering if there was an alternate way to find the function of the parabola given only the height and x - intercepts, and supposing the Area = 2/3(width of base)(height) formula is not known?\n\n8. Sep 25, 2015\n\n### jbriggs444\n\nAttacking this based on physical intuition rather than algebraicly...\n\nThe equation for a (vertically oriented) parabola can be written as terms of the horizontal distance from the midpoint squared plus a fixed height offset. e.g.\n\n$y = h + k(x-m)^2$\u200b\n\nWhere h is the height at the midpoint, m is the x coordinate of the midpoint and k is some arbitrarily chosen constant.\n\nSo let's use that. We already know the value for h. That is simply the given height.\n\n$h = height$\u200b\n\nThe midpoint is the average of the two given intercepts\n\n$m = \\frac{i_1 + i_2}{2}$\nIt will be helpful to know the distance from midpoint to intercept. Call that parameter w (short for width).\n\n$w = i_2 - m$ (or, equivalently, $w = m - i_1$)\u200b\n\nLooking back to the formula for our parabola, the value for k must be such that $h + kw^2$ = 0. Solving for k gives\n\n$k = \\frac{-h}{w^2}$\u200b\n\nNow, expanding the original equation ($y = h + k(x-m)^2$) to put it into standard form we get\n\n$y = kx^2 - 2kmx + (h+km^2)$\nEdit: Let's test that for the case at hand. h = 7.5, $i_1 = 0$, $i_2 = 50$\n\n$m = \\frac{i_1 + i_2}{2} = 25$\n$w = i_2 - m = 25$\n$k = \\frac{-h}{w^2} = \\frac{-7.5}{625} = -.012$\n$y = kx^2 - 2kmx + (h+km^2) = -.012x^2 +.6x + 0$\u200b\n\nLast edited: Sep 25, 2015\n9. Sep 25, 2015\n\n### Ocata\n\nHow did you determined this: $h + kw^2$ = 0\n\nTrying to understand how the (width of half the base)^2 was determined to be relevant and furthermore how it should be multiplied by an unknown scalar (k) and added to the height should be equal to zero. Why 0? This equation feels like a leap in logic I am not able to make. Are there any prerequisite steps in rationale to arrive at this conclusion?\n\nThank you.\n\n10. Sep 25, 2015\n\n### jbriggs444\n\nPretend for a moment that the origin of your coordinate system is at the apex of the parabola. Its equation is y = kx^2. At a distance w from the origin its height is h. What value of k is required to make this fit.\n\nThis is the same parabola we are dealing with -- it's just a change of coordinate system.\n\n11. Sep 25, 2015\n\n### Staff: Mentor\n\nSomehow the sign got flipped. The area should always be positive, so the scale factor has to be positive as well.\n\n12. Sep 25, 2015\n\n### jbriggs444\n\nIf we start with an equation with a positive coefficient on the leading term, the area \"under\" that parabola will be negative. So the scale factor needs to be negative.\n\n13. Sep 26, 2015\n\n### Ocata\n\nIf w = i - m, then how could height h depend on or be determined by a horizontal shift of w? If the apex is at the origin, then the height is zero. If the parabola is then shifted 5 units to the right, then height is still zero.\n\nI wonder if here you mean that w is a distance from the origin in any direction - as in a distance with a combination of horizontal and vertical shift?\n\n14. Sep 26, 2015\n\n### jbriggs444\n\nThe height of the parabola segment from the one point where the x intercept had been, through the apex and to the point where the other x intercept had been does not depend on where we choose to put the origin of the coordinate system. It is invariant with respect to translation.\n\\\n\n15. Sep 26, 2015\n\n### Ocata\n\nHi jbriggs444,\n\nI just realized, I have no clue what an apex of a parabola is. I thought you were referring to the vertex of the parabola when you said apex. I googled \"apex of a parabola\" and did not find any information on it.\n\nIn the meantime, I did locate another way to solve this which seems much more intuitive to me since it is related to basic algebra:\n\n$y = a(x-h)^{2} + k$ = $y = a(x - 25)^{2} + 7.5$\n\nPick a point on the line. I can choose (0,0) for this parabola since one of the x-intercepts is 0.\n\n$0 = a(0 - 25)^{2} + 7.5$ = $0 = a(625)^{2} + 7.5$\n\n$\\frac{-7.5}{625}$ = -.012 = a\n\nso, $y = -.012(x - 25) + 7.5$\n\nFrom learning this approach, I do see that the $w^{2} = 625$ happens to coming from $(0 - 25)^2 = 625$. Just still not sure how you arrive at $a(w)^2 + k = 0$.\n\nTo me $aw^{2} + k = 0$ kind of looks like $a(x - h)^{2} + k = 0$ where x can be 0 or 50. But then the h of vertex (h,k) will always be the midpoint of the parabola, so vertex (h,k) = (m,k)\n\nhmm...okay, maybe I'm starting to see it right about now...or at lease I think I'm starting to see something resembling what you are describing.....\n\nBasically, suppose I shift the parabola to the right 10 units. The new x intercepts will be 10 and 60. The new midpoint will be 35, but the difference between the intercepts and midpoint will still be 25. So choosing the point (10, 0) to solve the vertex formula, I end up with the same equation $0 = a(25)^{2}+ 7.5$. So, we arrive at the same scaling factor because intercepts 0 and 50 are equivalent to 10 and 60 when it comes to the intercept minus the midpoint.\n\nSo $w = i - m$ will always be the same regardless of where you shift the parabola horizontally, so $w^{2} = (i - m)^{2}$ is equivalent to the value you get when you choose the x - intercept as your point to plug into the vertex formula. Kinda?\n\n16. Sep 27, 2015\n\n### HallsofIvy\n\nStaff Emeritus\nFirst, \"apex\" of a parabola is just another word for the \"vertex\" of a parabola, those less often used. It is the \"point\" of the parabola just as the apex of an angle is the \"point\" of the angle.\n\nI am not sure what you are trying to do with your last post. Initially, you asked how to find a curve such that the \"area under it\" (I presume you mean between the curve and the x-axis) is 250. The simplest curve that cuts the x-axis twice, so has an \"area under it\", is a parabola. You can start with pretty much any parabola. For example, since 0 and 1 are easy numbers, x(1- x)= x- x^2 is a parabola that cuts the x-axis at 0 and 1 and rises above it so has an \"area under it\".\n\nIt is easy to calculate that area: $\\int_0^1 x- x^2 dx= \\left[\\frac{x^2}{2}- \\frac{x^3}{3}\\right]_0^1= \\frac{1}{2}- \\frac{1}{3}= \\frac{1}{6}$.\n\nThat area is, of course, not 250! 250 is $\\frac{250}{\\frac{1}{6}}= 1500$. So multiply the original function by 1500!\n\nThe parabola $1500(x- x^2)= 1500x- 1500x^2$ has area under the curve and above the x-axis 250.\n\n17. Sep 27, 2015\n\n### Ssnow\n\nDepends what kind of function you want consider, if it is exponential $y=e^{x}$ you can consider $\\int_{0}^{b}e^{x}d\\,x=250$ so $e^{b}-1=250$. The function $e^{x}$ for $x\\in [0,\\ln{251}]$ is an example ...\n\n18. Sep 28, 2015\n\n### Ocata\n\nHi HallsofIvy,\n\nPer jbriggs444's post #2 and subsequent guidance, I understand the step by step logic to come up with a general parabola that cuts the x-axis at a certain points.\n\nFor instance, given the example you provided, x = 0 and x = 1.\nI would start with (x-0)(x-1) = $(x^{2} - x)$ = $x(x - 1)$\nand for an upside down parabola, $-(x^{2}-x)$ = $-x^{2} + x$ = $x - x^{2}$ = $x(1 - x)$\n\nNow, it has taken me a few steps to arrive at the general parabola, but your example with jbriggs444 example for the interval between 0 and 50 has revealed a quick way to generate a general parabola.\n\nFor example, given any parabola that cuts the x-axis at 0 and some point $x_{1}$, a general parabola is:$x(x-x_{1})$ or $-(x(x_{1}-x))$\n\nThank you, by studying the two different sets of x intercepts, [0 and 1] and [0 and 50], both arriving at a function with of the same form, I can now formulate a general function of a parabola step by step or by knowing the final form that the function should have.\n\nAnd I believe I'm comfortable with coming up with a scaling factor to generate a parabola with a specific area between the curve and the x axis.\n\nWhat I am still have a bit of uncertainty about is how jbrigg444 was able to create a function of a parabola the way he described, given only the vertex (h,k) and of a given area between the curve and x axis.\n\nAfterwards, I found a method of simply choosing any point on the curve and plugging it into the vertex form of the parabola and solving for the coefficient.\n\nLast edited: Sep 29, 2015\n19. Sep 29, 2015\n\n### Ocata\n\nActually Ssnow, I'm also interested in the sin function.\n\nSuppose a sin function crosses the x axis at x = 0 and then at x = 50 and the area under that portion of the curve is 250. How would I find the function that represents these parameters?\n\n20. Sep 29, 2015\n\n### Staff: Mentor\n\nOne arch of the sine function has intercepts at x = 0 and x = $\\pi$. Two of the several kinds of transformations you can apply are compressions and expansions, which make each arch narrower or wider, respectively. The graph of y = sin(2x) represents a compression toward the y-axis of the basic, untransformed function by a factor of 2, so that the intercepts mentioned before are now at x = 0 and x = $\\pi/2$.\n\nThe graph of $y = \\sin(\\frac 1 3 x)$ represents an expansion away from the y-axis of the untransformed function by a factor of 3. The intercepts of the transformed graph are now at x = 0 and x = $3\\pi$.\n\nI leave it to you to figure out what the multiplier needs to be so that the intercepts will be at x = 0 and x = 50.", "date": "2017-10-18 02:20:40", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/how-to-create-a-function-such-that-area-under-curve-is-250.833834/", "openwebmath_score": 0.7822003364562988, "openwebmath_perplexity": 333.6152551508196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759654852756, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8639526811889524}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=ra15jhi630lqj81oh6lhef6703&action=printpage;topic=1265.0", "text": "# Toronto Math Forum\n\n## MAT244--2018F => MAT244--Lectures & Home Assignments => Topic started by: Xinyu Jiao on September 25, 2018, 08:58:39 PM\n\nTitle: Can there exists infinite number of solutions given initial conditions.\nPost by: Xinyu Jiao on September 25, 2018, 08:58:39 PM\nIn class, we were given an example where a differential equation can have two solutions given some initial condition. Specifically, the equation was $y' = y^\\alpha$ with $0<\\alpha<1$, and initial condition $y(0) = 0$. This shows that it's not unique, because it does not satisfy some condition which I do not understand.\n\nMy question is, can there be a differential equation (of order 1) such that given an initial condition, can acquire an infinite number of solutions? The answer to this question should be able to shed light as to the mechanism through which the equation acquires more than one solution.\nTitle: Re: Can there exists infinite number of solutions given initial conditions.\nPost by: Victor Ivrii on September 25, 2018, 09:27:59 PM\nFor condition see Section 2.8 of the textbook or this Lecture Note (https://q.utoronto.ca/courses/56504/files/1311954?module_item_id=332283)\n\nYes, this equation $y'=3 y^{2/3}$ (I modified it for simplicity) has a general solution $y=(x-c)^{3}$ but also a special solution $y=0$. Thus problem $y'=3 y^{2/3}$, $y(0)=0$ has an infinite number of solutions. Restricting ourselves by $x>0$ we get solutions y=\\left\\{\\begin{aligned} &0 &&0<x<c,\\\\ &(x-c)^3 && x\\ge c\\end{aligned}\\right. with any $c\\ge 0$ and similarly for $x< 0$.\n\nThis happens because this Lipschitz condition is violated at each point of the solution $y=0$.\n\nTitle: Re: Can there exists infinite number of solutions given initial conditions.\nPost by: Kathryn Bucci on October 06, 2018, 10:59:07 AM\nIf \ud835\udc66\u2032=\ud835\udc66\ud835\udefc with 0 < \ud835\udefc < 1 e.g. \ud835\udc66\u2032=3\ud835\udc662/3= f(t,y), then \u2202f/\u2202y=2y-1/3 is not continuous at (0,0).\nAccording to theorem 2.4.2 (existence and uniqueness for 1st order nonlinear equations), both f and \u2202f/\u2202y have to be continuous on an interval containing the initial point (0,0) - \u2202f/\u2202y is not continuous there so you can't infer that there is a unique solution.\nTitle: Re: Can there exists infinite number of solutions given initial conditions.\nPost by: Victor Ivrii on October 06, 2018, 12:10:58 PM\nContinuity of $\\frac{\\partial f}{\\partial y}$ is not required, but \"H\u00f6lder\u00a0 property\" $|f(x,y)-f(x,z)|\\le M$ is.\n\nThere is a notion of the singular solution", "date": "2022-05-19 16:37:19", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=ra15jhi630lqj81oh6lhef6703&action=printpage;topic=1265.0", "openwebmath_score": 0.9042200446128845, "openwebmath_perplexity": 839.2738191865103, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9845754470129647, "lm_q2_score": 0.8774767826757123, "lm_q1q2_score": 0.8639420955464375}}
{"url": "http://math.stackexchange.com/questions/218074/prove-if-a-subseteq-c-and-b-subseteq-d-then-a-cap-b-subseteq-c-cap-d", "text": "# Prove: If $A \\subseteq C$ and $B \\subseteq D$, then $A \\cap B \\subseteq C \\cap D$\n\nIs the form and correctness of my elementwise proof of this correct? I don't have any other way of getting feedback for my proofs and I want to improve.\n\nProof. Suppose $A, B, C, D$ are sets such that $A \\subseteq C$ and $B \\subseteq D$ and let $x \\in A \\cap B$. It has to be shown that $x \\in C \\cap D$.\n\n$x \\in A \\cap B$ means that $x \\in A$ and $x\\in B$. Because $A \\subseteq C$, $x \\in C$ and because $B \\subseteq D$, $x \\in D$. Thus, $x \\in C \\cap D$.\n\nThus, if $A \\subseteq C$ and $B \\subseteq D$, then $A \\cap B \\subseteq C \\cap D$.\n\n-\nThis is excellent. \u2013\u00a0 Brian M. Scott Oct 21 '12 at 15:18\nThanks! I just fixed an error that I made in the title. Should I elaborate more on where $x \\in A \\cap B$ comes from? It comes from $A \\cap B \\subseteq C \\cap D$, correct? \u2013\u00a0 highphi Oct 21 '12 at 15:22\n@BrianM.Scott Realized that too. Deleted my comment before seeing you reply. \u2013\u00a0 hwhm Oct 21 '12 at 15:23\nNo need to say any more: the reason for choosing $x\\in A\\cap B$ initially is clear just from the inclusion that you\u2019re trying to prove. \u2013\u00a0 Brian M. Scott Oct 21 '12 at 15:24\nYou\u2019re very welcome, and I agree with what Asaf wrote in the answer below. \u2013\u00a0 Brian M. Scott Oct 21 '12 at 15:37\n\nThis is a very well written proof. You state your assumptions and what you wish to prove, then you use the definitions to prove that.\n\nThere is nothing more to add, and nothing to reduce. Incidentally today I had the first class of the semester and this is exactly what I tried to teach my students. If they all write such proofs by the end of the month, I should be proud of my work.\n\n-", "date": "2014-11-26 08:07:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/218074/prove-if-a-subseteq-c-and-b-subseteq-d-then-a-cap-b-subseteq-c-cap-d", "openwebmath_score": 0.8983208537101746, "openwebmath_perplexity": 297.642389929086, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754452025767, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8639420908031997}}
{"url": "https://s23.cs251.com/Text/Ch_Proof_Guidelines/contents.html", "text": "MODULE 1: \u00a0 Introduction\nProof-Writing Guidelines\n1\nIntroduction\n\nA (mathematical) proof of a statement is simply an argument, presented as an essay, that convinces the reader of the validity of the statement. Therefore, whether a proof is acceptable or not (e.g.\u00a0whether it has the right level of detail, whether it is written in good style, etc.) is a subjective matter without very strict rules. This makes the task of writing good proofs (i.e. proofs that are correct and easily understood and verified) a difficult one.\n\nAs you will notice when you read this document, there are a lot of similarities between writing a correct computer program with good style and writing a correct mathematical proof with good style. Unfortunately, however, writing a good proof is arguably harder since there is no compiler to point out all the compile-time errors you have (e.g.\u00a0syntax errors or type-checking errors). Furthermore, you cannot run a mathematical proof on a computer to see if it produces the expected output. All the errors in a mathematical proof must be caught by you, the author of the proof, and this requires an extra level of attention to detail.\n\nOur goal in this document is to give you suggestions on how to write correct and clearly presented proofs, and help you more easily catch logical flaws or gaps in your arguments. Most of our suggestions are stylistic in nature, because a proof that is written using good style is a proof that exposes its bugs (if there are any). So by following our guidelines, you will make it easier for yourself and everyone else to understand and validate your proof.\n\nMathematics, sciences and arts all have deep, complicated and beautiful ideas. Our rate of progress as a species depends on clear communication of those ideas so they can quickly spread and evolve. We hope that you will seek clarity of thought and effective communication of ideas, not just in the proofs you write, but in all of your endeavors in life.\n\n2\nGuideline Points\n\nThis section contains the guideline points (10 of them) that we ask you to follow when you write your proofs. Before we start, below is an example of a \u201cproof\u201d that violates almost all the guideline points. Feel free to refer back to this example as you go through this section.\n\nExercise ($$2^n > n$$)\n\nShow that $$2^n > n$$ for all integers $$n \\geq 1$$.\n\nProposed solution\n\n(Line numbers are added for easy referencing.)\n\n1. $$F_n =$$$$2^n > n$$\n\n2. $$F_1 =$$$$2 > 1$$$$\\checkmark$$\n\n3. $$F_n \\implies F_{n+1}:$$\n\n4. $$2^{n+1} = 2 \\cdot 2^n > 2\\cdot n \\text{ (induction) } \\geq n+1$$ because $$n \\geq 1$$\n\n5. Therefore proved.\n\nKeep in mind that the above example is really a toy example. Some of the points we make below have more meaning and value in the context of more sophisticated proofs, which are the kinds of proofs you will be writing in CS251.\n\n2.1\nIs the basic structure right?\n\nA proof is not a calculation or a sequence of mathematical symbols. A proof is an essay! It is an argument written in some human language (which, in this course, is English). This means that your proof should consist of paragraphs, and every paragraph should consist of full English sentences. Every word and mathematical notation must be part of a complete sentence. The only purpose of mathematical notation/symbols is to make your arguments clear and concise. Do not equate mathematical notation with rigor or formalism. You can write a completely rigorous and formal proof using just English words.\n\nPictures and/or diagrams are highly encouraged when they help clarify your argument. However, a picture or a diagram does not replace an actual argument that needs to be clearly specified in English.\n\n2.2\nAre you starting the right way?\n\nOften, the most important step in coming up with a proof is making sure that you understand exactly what assumptions are given to you and what statement needs to be derived. To make these things absolutely clear to you (the author of the proof) and the reader, we expect the first paragraph of your proof to include a restatement of the assumptions given and what needs to be derived. Unpack the definitions of technical terms if appropriate.\n\nAnother important component of the first paragraph is stating your general proof strategy (e.g.\u00a0proof by contradiction, proof by induction, etc.) For proofs by contradiction, explicitly negate the statement that you are trying to prove and assume it. For proofs by contrapositive, the contrapositive of the statement should be explicitly laid out. For proofs by induction, the parameter being inducted on should be clear.\n\n2.3\n\nWhen someone reads your proof, they will read it like they read any other essay. Therefore your proof should have a good flow and should be easy to read out loud even with the mathematical notation interspersed in the sentences. In particular, the mathematical notation you use or the diagrams you draw should not break the flow.\n\n2.4\nIs the purpose of every sentence clear?\n\nThe purpose of every sentence should be clear as you are reading it. Otherwise, the sentence can be very confusing for the reader and break the flow of the proof. A common example that violates this point is to make a statement without clarifying whether it follows from the previous statements or assumptions, or whether it is a statement that will be proved later on. With this in mind, whenever you write a sentence, make sure that it is clear if the sentence is (i)\u00a0an assumption, (ii)\u00a0a statement that follows from or combines previously established statements or assumptions, (iii)\u00a0a claim that will be proved later, (iv)\u00a0a sentence setting up a goal, (v)\u00a0a sentence introducing a new variable, terminology, definition etc. (vi)\u00a0part of an example or illustration, (vii)\u00a0a digression, or (viii)\u00a0something else.\n\nA strategy that helps a lot is to set explicit and clear goals, and say what you will do before doing it.\n\n2.5\nAre you giving the right level of detail?\n\nWhat is the right level of detail to provide when writing a proof? This is an important question whose answer depends on the audience of the proof. It is possible to write a proof that has all the right ingredients, but does not have the proper justifications for each step. So then the reader has to check the details themselves and verify that the proof is indeed correct. These kinds of proofs are actually not uncommon in, say, computer science or mathematics publications. The author may knowingly give just enough detail in a proof so that the gaps can be figured out and verified by an expert in the field. In CS251, however, we do not accept such proofs. No extra effort/verification from the reader should be necessary. For this reason, it is better to err on the side of caution, and spell things out as much as you feel is reasonable.\n\nTo help with this, our recommendation is that you view your audience as a classmate who does not know how to prove the statement you are presenting the proof for. Keep your audience in mind when presenting your proofs and provide the appropriate level of detail in your arguments. Your classmate should be able to read your proof from start to finish once, and be convinced that your argument is correct.\n\nOne thing that can be hard to appreciate is that when you think hard on a problem, you develop a lot of intuition in the process, and certain things that were not too trivial in the beginning start becoming obvious to you. This is great, because it signals that you are acquiring a deeper understanding of the problem. However, when it is time to write down your proof, you have to keep in mind that the reader has not gone through the mental process that you have gone through trying to come up with the proof. So the intuitions you have built are not necessarily accessible to your reader. And the things that are obvious to you may not be obvious to the reader. It is not always easy, but try to put yourself in the shoes of your reader and don\u2019t skip over details that can be crucial to understanding your argument.\n\nRelated to the point above, if in your proof you are using a word that is synonymous to \u201cobvious\u201d, double check (with your audience in mind) that it is justified. In particular, an obvious statement should be such that a proof of it springs to mind immediately with no effort. When in doubt about whether a statement qualifies as obvious or not, ask the course staff.\n\n2.6\nShould you break up your proof?\n\nWhen you first learn programming, one thing that will be emphasized over and over again is the value of breaking up your problem into smaller parts and using lots of helper functions. If you write a function that is too long or does more than one task, that is a good indication that you should consider breaking it up into smaller helper functions. Liberal use of helper functions makes your program easier to understand and debug.\n\nIn the above sense, mathematical proofs are similar to computer programs. Your proof should be divided up into lemmas and/or claims when appropriate. This will make the proof much easier to understand and it will be easier to spot and fix any potential errors. Even if you are not defining new lemmas/claims, each component of your argument should be separated into different paragraphs, and the high level organization should be easy to identify.\n\n2.7\n\nIn programming, you are used to the idea that every object (piece of data) has a type/class. For instance, many programming languages have an integer type and a string type. The type of the data determines what kinds of operations you can apply on the data. You can, for example, multiply two integers, but you cannot multiply two strings. If you try to do any operation that conflicts with the data type, then you will get a compile-time error, and your code will not run.\n\nAnother way to get a compile-time error is by trying to access a variable that has not been defined and initialized. And in statically typed programming languages, you need to explicitly specify the type of a variable when you declare it. This way, the compiler can type-check your program before running it.\n\nSimilar to programming, every mathematical object has a type (e.g.\u00a0it could be an integer, a set, a function, etc.). And as in programming, the type of an object determines what kind of operations you can apply to the object. For example, you can add an element to a set, but you cannot add an element to a function. If you do not respect these constraints, we say that your argument does not type-check. A type-checking error often leads to a nonsensical sentence (even though the author may not realize this).\n\nWith these in mind, make sure that your proofs always type-check. You must act as a compiler to find anything that might violate type constraints. In order to make this easier, whenever you learn about a new definition, make a note of the type of the object being defined. This is one of the most important parts of a definition. And when referencing an object in your proof, be aware of its type and make sure that your proof type-checks.\n\nIn addition to the above, remember that all the variables in your proofs should be introduced properly. In particular, it should be clear if a variable represents an arbitrary object (e.g.\u00a0in the context of a \u201cfor all\u201d quantifier), a specific object known to exist (e.g.\u00a0in the context of a \u201cthere exists\u201d quantifier), or something else (e.g.\u00a0a specific value). In all cases, the variable must have a specific type, and it should be clear what the type is.\n\nHere is a suggestion to keep in mind. If in a sentence you are referring to a variable, consider preceding the variable with its type if you feel that this makes things clearer and does not add too much redundancy. For example, if x and y are variables referring to strings, you can consider changing a sentence like \u201cConcatenating x and y...\u201d to \u201cConcatenating string x and string y...\u201d.\n\n2.8\n\nAs you develop an argument in a proof, you will want to refer back to (and use) a previously established statement or an assumption. Implicitly using previous statements or assumptions can make it hard to follow the logic of the proof and put unnecessary burden on the reader. Therefore, avoid implicit references and make sure it is clear which part(s) of the proof you are using to establish the current step of your proof. Whenever you make a reference, you want it to be very easy for the reader to spot the thing that is being referred to. For example, if there is an important equation that will be referenced later on in the proof, it is a good idea to put that equation on a separate line and label it. This way, you can use the equation\u2019s label to refer to it, and the fact that it is on a separate line by itself will make it very easy for the reader to identify it.\n\nIn addition to the above, please be careful when you use a word like \u201cit\u201d, \u201cthis\u201d, or \u201cthese\u201d in your proof. While you are writing your proof, when you use such a word, you know perfectly well what that word refers to. However, ask yourself whether it would also be perfectly clear to the reader. In general, it is a good idea to try to avoid such words as much as possible, even if this means a certain level of repetitiveness is introduced in the used words.\n\n2.9\nWhere are you using the assumptions?\n\nWhen you are asked to write a proof, you are usually given certain assumptions that you take as true (which is your starting point), and you have a target statement that you want to derive. As pointed out in the \u201cAre you starting the right way?\u201d section above, it helps you and the reader to explicitly lay out the given assumptions as well as the statement that needs to be derived. Once your proof is complete, ask yourself where in the proof the assumptions are being used. If it turns out that you are not using a given assumption, you should raise your alertness level and check: is the statement even true without that assumption? (There may be some exceptions, but almost always, the answer will be no.) If it is not true, and you do not use the assumption in your proof, then your proof is wrong.\n\nWhen someone reads your proof to verify it, they will be trying to spot where the assumptions are used in the proof. This is a quick sanity check that the proof is not missing an essential component. In order to make this check easy for the reader (and also for yourself), you should make clear in your write-up where and how the assumptions are being used in your argument. In the rare case that a given assumption is not needed for the proof, mention this explicitly.\n\n2.10\nIs the proof idea clear?\n\nAs mentioned before, a proof is an argument that convinces the reader that a certain statement is true. After reading the proof, the reader may walk away with a clear and intuitive understanding of why the statement is true. Or, they may not, even if they are perfectly convinced that the statement is indeed true.\n\nWe do not want to get into a philosophical discussion about what it means to understand why something is true. But hopefully we can agree that certain proofs have more explanatory content than others.\n\nWe strongly encourage you to always write proofs with strong explanatory content! For this reason, if the motivation/intuition behind the proof is not transparent, consider adding a paragraph or two explaining the main ideas that make the proof work. This can be either added within the proof itself, or can be a separate \u201cProof Idea\u201d section preceding the proof.\n\nSummary\n1. Is the basic structure right?\n\n2. Are you starting the right way?\n\n4. Is the purpose of every sentence clear?\n\n5. Are you giving the right level of detail?\n\n6. Should you break up your proof?\n\n9. Where are you using the assumptions?\n\n10. Is the proof idea clear?\n\n3\nAn Example\n\nLet\u2019s now come back to the example from the beginning of Section (Guideline Points) and see how it fares. For ease of reference, we reproduce the \u201cproof\u201d here:\n\nProposed solution\n1. $$F_n =$$$$2^n > n$$\n\n2. $$F_1 =$$$$2 > 1$$$$\\checkmark$$\n\n3. $$F_n \\implies F_{n+1}:$$\n\n4. $$2^{n+1} = 2 \\cdot 2^n > 2\\cdot n \\text{ (induction) } \\geq n+1$$ because $$n \\geq 1$$\n\n5. Therefore proved.\n\n\u2022 Is the basic structure right? It is quite easy to see that there is a gross violation of this point. The argument does not consist of sentences. In fact, there is not a single fully formed English sentence.\n\n\u2022 Are you starting the right way? We are missing a restatement of what the given assumptions are and what needs to be derived. (Yes, this is kind of pedantic with the toy example, but still valuable in the context of more complicated examples.) Furthermore, even though the proof is supposed to be a proof by induction, we only learn about this towards the end of the proof.\n\n\u2022 Did you read your proof out loud? If we attempt to read the argument out loud, we hear an incomprehensible sequence of words.\n\n\u2022 Is the purpose of every sentence clear? What is the purpose of the first line? Is it a definition? Is it a claim? Is it an assumption? We can also ask these questions for lines 2 to 4.\n\n\u2022 Are you giving the right level of detail? Implicit in the \u201cproof\u201d is the claim that \u201c$$2\\cdot n \\geq n+1$$ because $$n\\geq 1$$\u201d. This claim is true, and would qualify as an obviously true statement. However, there is a step that is being skipped, which if included, would make the claim more immediately obvious. In particular, saying \u201c$$2 \\cdot n = n + n \\geq n + 1$$, where the inequality follows because $$n \\geq 1$$.\u201d is preferable. Again, in the context of this toy example, our point may come across as too pedantic, but it is good to err on the side of caution and spell things out as much as possible. In another context/proof, a step that you skip might cause the reader a headache.\n\n\u2022 Should you break up your proof? This proof does not require any \u201chelper functions\u201d. It is short and simple enough.\n\n\u2022 Does your proof type-check? The variable $$n$$ is not introduced properly. From the problem statement, we can infer that $$n$$\u2019s type is a natural number. However, the variable should still be declared properly. For instance, $$F_n$$ is supposed to be the statement \u201c$$2^n > n$$\u201d, but here $$n$$ is undefined/unquantified.\n\n\u2022 Are your references clear? There seems to be a reference to an induction hypothesis, but it is poorly presented since there is no indication that the proof is by induction until there is an attempt to make a reference to the induction hypothesis.\n\n\u2022 Where are you using the assumptions? The part \u201cbecause $$n \\geq 1$$\u201d is a reference to an assumption that is given by the problem statement, so we expect the author to point this out explicitly. Once again, this point may seem pedantic, but in longer and more complicated proofs, these things really do make a difference.\n\n\u2022 Is the proof idea clear? As the given problem and its proof are quite simple, a \u2018proof idea\u2019 section is not necessary.\n\nHere is an example of how the proof can be written using good style.\n\nGood solution\n\nWe will prove that for all integers $$n \\geq 1$$, $$2^n > n$$. The proof is by induction on $$n$$.\n\nLet\u2019s start with the base case which corresponds to $$n=1$$. In this case, the inequality $$2^n > n$$ translates to $$2^1 > 1$$, which is indeed true.\n\nTo carry out the induction step, we want to argue that for all $$n \\geq 1$$, $$2^n > n$$ implies $$2^{n+1} > n+1$$. We do so now. For an arbitrary $$n \\geq 1$$, assume $$2^n > n$$. Multiplying both sides of the inequality by $$2$$, we get $$2^{n+1} > 2n$$. Note that since we are assuming $$n \\geq 1$$, we have $$2n = n+n \\geq n+1$$. Therefore, we can conclude that $$2^{n+1} > n + 1$$, as desired.\n\nA remark is in order. It is widely accepted that mathematical proofs can be completely formalized in a way that can be mechanically verified, e.g.\u00a0by a computer. However, writing proofs in a completely formal way is like writing computer programs using machine language. Mathematicians do not communicate proofs in that level of detail.\nHowever, this example is in fact not made up. It was a turned-in solution in CS251 many years ago.\nAvoid using long paragraphs as they hurt the clarity of the exposition.\nAccording to Steven Pinker, a cognitive scientist, psychologist, linguist, and a popular science author, the biggest reason why many intelligent people write very poorly is \u201cthe curse of knowledge\u201d. People have a hard time imagining what it is like for someone else to not know something that they know.", "date": "2023-03-23 11:16:55", "meta": {"domain": "cs251.com", "url": "https://s23.cs251.com/Text/Ch_Proof_Guidelines/contents.html", "openwebmath_score": 0.7259247303009033, "openwebmath_perplexity": 308.24805973162677, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9481545348152283, "lm_q2_score": 0.9111797148356995, "lm_q1q2_score": 0.863939178653115}}
{"url": "https://math.stackexchange.com/questions/2422495/how-do-i-solve-32x-equiv-12-pmod-82", "text": "# How do I solve $32x \\equiv 12 \\pmod {82}$?\n\nI am able to solve simpler linear congruences, for example $3x \\equiv 2 \\pmod 5$. What I would do in this case is use that $0 \\equiv 10 \\pmod 5$ and then utilising a theorem: $3x \\equiv 12 \\pmod 5$. Then I can divide by $3$ leaving me $x \\equiv 4 \\: \\left( \\mathrm{mod} \\: {\\frac{5}{\\mathrm{GCD}(5,3)}} \\right) \\quad \\Longleftrightarrow x \\equiv 4 \\pmod{5}$ which means the solution is $x = 4k + 5$ where $k \\in \\mathbb{Z}$.\n\nBut I cannot apply the same method to this congruence: $$32x \\equiv 12 \\pmod {82}$$ This is how far I got: $$8x \\equiv 3 \\: \\left( \\mathrm{mod} \\: \\frac{82}{\\mathrm{GCD}(82, 4)} \\right)$$ $$\\Updownarrow$$ $$8x \\equiv 3 \\pmod {41}$$\n\nWhat could I do next? Please provide solutions without the Euclidean algorithm.\n\nEDIT:\n\nWhat I found later is that I can say that $$0 \\equiv 205 \\pmod {41}$$ And then I can add it to the congruence in question and divide by $8$. So I guess my question is essentially 'How can I find a number that is a multiple of $41$ (the modulus) and which, if added to $3$ gives a number that is divisible by $8$?'\n\nI reckon the Euclidean algorithm is something which gives an answer to these kinds of questions?!\n\n\u2022 It's just too bad that the Euclidean algorithm is the perfect way to tackle these problems. \u2013\u00a0Lord Shark the Unknown Sep 9 '17 at 11:28\n\u2022 I'd suggest turning it into algebraic equations without congruence. \u2013\u00a0user451844 Sep 9 '17 at 11:30\n\u2022 @LordSharktheUnknown I am aware of that, but as I am a complete beginner in the topic, I'd firstly like to find solutions which do not involve the algorithm which I am not familiar with yet. \u2013\u00a0bertalanp99 Sep 9 '17 at 11:37\n\u2022 In that case you may just observe that as $\\gcd(5,41)=1$ $$8x\\equiv3\\pmod{41}\\Leftrightarrow40x\\equiv15\\pmod{41}.$$ Here $40\\equiv-1$, so... This amounts to replacing Extented Euclidean Algorithm with a \"lucky\" observation. \u2013\u00a0Jyrki Lahtonen Sep 9 '17 at 11:39\n\u2022 Please see my edit \u2013\u00a0bertalanp99 Sep 9 '17 at 11:40\n\nHint :you can do like this $$\\quad{8x \\equiv 3 \\pmod {41}\\\\ 8x \\equiv 3+41 \\pmod {41}\\\\8x \\equiv 44 \\pmod {41} \\div4 \\\\ 2x \\equiv 11 \\pmod {41}\\\\2x \\equiv 11+41 \\pmod {41}\\\\2x \\equiv 52 \\pmod {41}\\div 2\\\\x \\equiv 26 \\pmod {41}\\\\x=41q+26}$$\n\nYou just have to find the inverse of $8$ modulo $41$.\n\nThe general method uses the extended Euclidean algorithm, but in the particular case, it's much simpler: from $5\\cdot 8=40\\equiv -1\\mod 41$, you get at once that $8^{-1}\\equiv -5\\mod 41$, so $$x\\equiv -5\\cdot 3=-15\\equiv 26\\mod 41.$$\n\nbasically you are asking how to solve $\\frac 1n \\mod m$ where $\\gcd(m,n)=1$\n\nwhere $\\frac 1n \\mod m$ is notation for the $x$ so that $n*x \\equiv 1\\mod m$.\n\nLet $m = k + qn$ then $nx = 1 + Zm = 1 + Z(k + qn)$ implies\n\n$x = \\frac 1n + \\frac {Zk}n + Zq$ where $n|Zk + 1$\n\nEx. $x \\equiv \\frac 17 \\mod 67$. As $67 = 9*7 + 4$ then\n\n$x = \\frac {1+Z*4}{7} + 9$\n\nWhich means we have to find $Z \\equiv -\\frac 14 \\mod 7$.\n\nOh... I guess this is Euclid's algorithm.\n\n$7 = 3 + 4$\n\nSo $-Z = \\frac{1+ 3Y}4 + 1$\n\nWhich is clearly $Y =1$ and $-Z \\equiv 2\\mod 7$ and $Z \\equiv -2 \\mod 7$\n\nand $x = \\frac {1 + 4*(-2)}7 + (-2)*9 \\equiv -19 \\mod 67$.\n\nAnd indeed $7*(-19) \\equiv -133= (-1)*67 + 1 \\equiv 1 \\mod 67$\n\n... Yeah, you need Euclid's alogrithm.\n\nOK, without the Euclidean algorithm, you are looking for some $k$ such that $8$ divides $41k+3$, which gives you a new equation in smaller numbers (which is a Euclidean algorithm idea too, but still): $$41k+3\\equiv 0 \\bmod 8 \\\\ 41\\equiv 1 \\bmod 8 \\\\ k+3 \\equiv 0 \\bmod 8 \\\\ k\\equiv 5 \\bmod 8$$\n\nSo then you have your $41\\times 5 = 205$ to make the divisibility.\n\n$\\, 8x = 3\\!+\\!41n\\!\\iff\\!\\bmod 8\\!:\\ 0\\equiv 3\\!+\\!41n\\equiv 3\\!+\\!n\\!\\iff\\! n\\equiv -3\\,$ so $\\,x\\equiv \\frac{3+41(-3))}8\\equiv -15\\equiv 26$\n\nAlternatively $\\bmod 41\\!:\\ x\\equiv \\dfrac{3}{8}\\equiv \\dfrac{15}{40}\\equiv\\dfrac{15}{-1}\\equiv 26\\$ by Gauss's algorithm\n\nAlternatively $\\bmod 41\\!:\\ x\\equiv \\dfrac{3}{8}\\equiv \\dfrac{44}{8}\\equiv \\dfrac{11}{2}\\equiv\\dfrac{52}{2}\\equiv 26\\$ by adding $\\pm 41$ to simplify divisions\n\nAlternatively $\\bmod 41\\!:\\ x\\equiv \\dfrac{1}8\\,\\dfrac{3}1\\equiv \\dfrac{-40}8\\ \\dfrac{3}1\\equiv (-5)3\\equiv -15$\n\nRemark  The first method essentially uses a single step of the (extended) Euclidean algorithm, and the second is a special case of that for prime moduli. The other methods are ad-hoc - they try to massage the fractions by adding small multiples of the modulus to make quotients exact / simpler.\n\nBeware $\\$ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.\n\n$$32x\\equiv12\\pmod{82}$$ is equivalent to $$16x\\equiv6\\pmod{41}$$ and since $(41,2)=1$, we can divide by $2$ $$8x\\equiv3\\pmod{41}$$ Then noting that $5\\cdot8\\equiv-1\\pmod{41}$, we get that $36\\cdot8\\equiv1\\pmod{41}$. Multiplying both sides by $36$ yields $$\\bbox[5px,border:2px solid #C0A000]{x\\equiv26\\pmod{41}}$$ Note: When looking for the inverse of $a$ mod $m$ it is often a good idea to see if $a\\mid(m-1)$ or $a\\mid(m+1)$; if either of these hold, they give a quick inverse mod $m$: $a^{-1}\\equiv-\\frac{m-1}a$ or $a^{-1}\\equiv\\frac{m+1}{a}$. Above, it was noted that $8|(41-1)$ leading to $8^{-1}\\equiv-\\frac{41-1}8\\pmod{41}$.\n\nAlternate Method of Finding the Inverse of $\\boldsymbol{8\\pmod{41}}$\n\nSince $41$ is prime and $(41,8)=1$, we know, by Fermat's Little Theorem, that $8^{39}\\equiv8^{-1}\\pmod{41}$. We can compute $8^{39}\\pmod{41}$ using the Square and Multiply Algorithm. $39=100111_\\text{two}$, therefore, \\begin{align} 8^1&\\equiv8&\\pmod{41}\\\\ 8^2&\\equiv23&\\pmod{41}&&\\text{square}\\\\ 8^4&\\equiv37&\\pmod{41}&&\\text{square}\\\\ 8^8&\\equiv16&\\pmod{41}&&\\text{square}\\\\ 8^9&\\equiv5&\\pmod{41}&&\\text{multiply}\\\\ 8^{18}&\\equiv25&\\pmod{41}&&\\text{square}\\\\ 8^{19}&\\equiv36&\\pmod{41}&&\\text{multiply}\\\\ 8^{38}&\\equiv25&\\pmod{41}&&\\text{square}\\\\ 8^{39}&\\equiv36&\\pmod{41}&&\\text{multiply}\\\\ \\end{align} Therefore, $8^{-1}\\equiv36\\pmod{41}$.\n\n\u2022 You should elaborate on how you \"note that $5\\cdot 8\\equiv -1$\" else it amounts to pulling the inverse out of a hit - which is good for magic, but not for math. \u2013\u00a0Bill Dubuque Sep 10 '17 at 18:33\n\u2022 @BillDubuque: Is it magic to realize that $5\\cdot8=40$? Be careful when casting stones. This is why I added the second method. \u2013\u00a0robjohn Sep 10 '17 at 18:41\n\u2022 If you don't say how you \"note\" that then it is magic, not math. There are of course many ways to compute such inverses, so it is strange that you want to exhibit the inverse but not say how you computed it. Helping someone improve an answer is certainly not \"casting stones\". In case you may not have noticed, I often give constructive feedback on answers where results are pulled out of a hat. \u2013\u00a0Bill Dubuque Sep 10 '17 at 19:17\n\u2022 Certainly, in general, I would not suggest an answer by inspection, however, in this answer you seem to suggest that inspection is a valid approach. However, since I knew that was not a good general strategy, I supplied the alternate, non-Euclidean Algorithm, approach. Furthermore, I have added a description of how I came up with $5\\cdot8\\equiv-1\\pmod{41}$, but as this doesn't work in general, I originally opted to describe the Fermat's Little Theorem approach. \u2013\u00a0robjohn Sep 10 '17 at 22:12\n\u2022 Inspection (or brute-force) is much easier there being mod $3$ vs. mod $41$ (and there I say by \"inspection or Euclid\"). Glad to see that you added an explanation. I call this method the easy inverse case, It is equivalent to the numerator twiddling I do with fractions in my answer, i.e. try $\\ 1/a\\equiv (1\\pm km)/a\\pmod{\\!m}\\,$ for small $k$ to try to make the division exact (i.e. try the first few steps of a brute-force search). It often proves handy for greatly simplifying CRT calculations. \u2013\u00a0Bill Dubuque Sep 10 '17 at 22:43", "date": "2019-04-20 06:40:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2422495/how-do-i-solve-32x-equiv-12-pmod-82", "openwebmath_score": 0.9138378500938416, "openwebmath_perplexity": 265.3563084245772, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979035762851982, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8639284469639655}}
{"url": "https://math.stackexchange.com/questions/1555522/subsequences-and-convergence-do-all-subsequences-have-to-converge-to-the-same-l", "text": "Subsequences and convergence. Do all subsequences have to converge to the same limit for the sequence to be convergent?\n\nProve that $a_n$ converges if and only if:\n\n$a_{2n},a_{2n+1},a_{3n}$ all converge\n\nI thought this was an easy generic question until I read the hint which said: Note: It is not required that the three sub-sequences have the same limit. This needs to be shown\n\nThis is what is confusing me because I have found two sources stating something different:\n\nProposition 4.2. A sequence an converges to L \u2208 R if and only if every subsequence converges to L.\n\nand\n\nLet $a_n$ be a real sequence. If the subsequence $a_{2n}$ converges to a real number L and the subsequence $a_{2n+1}$ converges to the same number L, then $a_n$ converges to L as well.\n\nSo my question is: for a sequence $a_n$ to converge does it's subsequences have to converge to the same limit? (I suspect not) and if the answer is no can you help me prove why?\n\n\u2022 If a sequence converges its subsequences must converge to the same limit. This follows directly from the definition of the limit of a sequence. \u2013\u00a0John Douma Dec 1 '15 at 21:53\n\nIf a sequence has two subsequences that do not both converge to the same limit, then the sequence does not converge.\n\nThis can be proven using the $\\epsilon-\\delta$ definition of convergence:\n\n\u2022 Let $a_n$ converge to $L$, and let $\\{a_{n_k}\\}_{k\\to\\infty}$ be a subsequence of $a_n$.\n\u2022 Let $\\epsilon > 0$.\n\u2022 Then, because $a_n$ converges to $L$, there exists some $N$ such that if $n>N$, then $|a_n - L|<\\epsilon$.\n\u2022 Because $n_k$ is an increasing sequence of integers (by definition of a subsequence), there exists such $K$ that $n_K > N$.\n\u2022 Then, if $k > K$, we have $n_k > n_K > N$, and from the previous point, we get that $|a_{n_k} - L|<\\epsilon$, meaning that $a_{n_k}$ converges to $L$.\n\nHowever, your case is special in that there is an overlap of sequences, for example $a_6$ is in the first and third sequence, and $a_9$ is in the second and third sequence. In fact, the third sequence alternates between the first two sequences, and you can use this to prove all three limits must be equal.\n\n\u2022 Ok so for $a_{2n}$ and $a_{2n+1}$ I can say we are either in the odd or the even case and then $a_{3n}$ goes between the two cases. \u2013\u00a0babylon Dec 1 '15 at 22:10\n\u2022 @babylon Yup. And that, forces the sequences to converge to the same number. Then you need to prove that the whole sequence also converges to the same number. \u2013\u00a05xum Dec 1 '15 at 22:16\n\nI think the word \"required\" in the hint is confusing. Focus instead on the last sentence:\n\nNote: It is not required that the three sub-sequences have the same limit. This needs to be shown\n\nIn other words, you need to show that as a result of the given hypotheses, the three sub-sequences do have the same limit. You need to do that precisely because it is a necessary condition for $a_n$ to converge.", "date": "2021-03-04 13:51:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1555522/subsequences-and-convergence-do-all-subsequences-have-to-converge-to-the-same-l", "openwebmath_score": 0.9875617623329163, "openwebmath_perplexity": 119.0916195551768, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357604052423, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8639284448048942}}
{"url": "https://foldable-robotics.github.io/modules/optimization/generated/02-scipy-optimization-example/", "text": "# Scipy Optimization Example\n\nThis short example shows you how to use the scipy minimize function to identify model parameters. This example is set up similarly to the linear least squares example for consistency. Being able to define pretty much anything in a python function, however, gives you great power to customize this, as opposed to a specific approach like linear least squares.\n\nFirst, we import all the necessary modules\n\n%matplotlib inline\nimport numpy\nimport numpy.random\nimport matplotlib.pyplot as plt\nimport numpy.linalg\nimport scipy.optimize\n\n\nNext create the x matrix\n\nx = numpy.r_[-10:10:.5]\n\n\nNext, define a y vector based on some model. Note: you can use any of these models or add them together. We scale the output in this example to eliminate the natural weighting of each of these functions over the given range.\n\n#y = x\ny = x**2\n#y = x**3\n#y = numpy.sin(x)\ny /= y.max()\n\n\nAdd some noise to y:\n\nrand = numpy.random.randn(*y.shape)/10\ny_rand = y + rand\n\nplt.plot(x,y)\nplt.plot(x,y_rand,'o')\n\n[<matplotlib.lines.Line2D at 0x7f860d882a30>]\n\n\nCreate an A matrix consisting of several different models\n\nA = numpy.array([(x),(x)**2,(x)**3,numpy.sin(x)]).T\n\n\nNow create a function that outputs the sum of squared error of each model applied to the given x. In this case we are solving for the weighting coefficients, $k$, as in the linear least squares example:\n\ndef myfunc(k):\n# make sure our coefficients are in the form of a numpy array\nk = numpy.array(k)\n# generate y* = Ak^T\ny_model = A.dot(k.T)\n# sum the square of the error of our model against the input data, y_rand\nerror = ((y_model-y_rand)**2).sum()\n#return the error\nreturn error\n\n\nCreate an initial guess for each of the weights. In this case we just give each coefficient the value of 1 as an initial guess\n\nini = [1]*A.shape[1]\nini\n\n[1, 1, 1, 1]\n\n\nNow, call the minimize function. The first value should be the function you are trying to minimize, and the For more information see the optimization function page\n\nsol = scipy.optimize.minimize(myfunc,ini)\nsol\n\n      fun: 0.29336548348717617\nhess_inv: array([[ 2.35078351e-03, -1.23710160e-05, -3.31813423e-05,\n4.86251949e-04],\n[-1.23710160e-05,  6.46844761e-06,  3.00084328e-07,\n-3.66529338e-05],\n[-3.31813423e-05,  3.00084328e-07,  5.62458836e-07,\n-1.91576150e-05],\n[ 4.86251949e-04, -3.66529338e-05, -1.91576150e-05,\n2.78470176e-02]])\njac: array([3.7252903e-09, 3.7252903e-09, 3.7252903e-09, 0.0000000e+00])\nmessage: 'Optimization terminated successfully.'\nnfev: 65\nnit: 9\nnjev: 13\nstatus: 0\nsuccess: True\nx: array([-1.13654781e-03,  1.09398078e-02, -2.22292008e-05,  1.93916411e-02])\n\n\nsol.x contains the solution for $k$\n\nk_optimum = sol.x\nk_optimum\n\narray([-1.13654781e-03,  1.09398078e-02, -2.22292008e-05,  1.93916411e-02])\n\nxx = numpy.r_[:4]\nlabels = '$x$','$x^2$','$x^3$','$\\sin(x)$'\nf = plt.figure()\nax.bar(xx,k_optimum)\nax.set_xticks(xx)\nax.set_xticklabels(labels)\n\n[Text(0, 0, '$x$'),\nText(1, 0, '$x^2$'),\nText(2, 0, '$x^3$'),\nText(3, 0, '$\\\\sin(x)$')]\n\n\nNow generate $y^*$\n\ny_model = A.dot(k_optimum.T)\n\n\nPlot the model against the input data\n\nfig = plt.figure()\na = ax.plot(x,y_rand,'.')\nb = ax.plot(x,y_model)\nax.legend(a+b,['data','model'])\n\n<matplotlib.legend.Legend at 0x7f860ce494f0>\n\n\nAnd plot the residual as well\n\nplt.figure()\nplt.plot(x,y_model-y_rand)\n\n[<matplotlib.lines.Line2D at 0x7f860cdb2d00>]\n\n\nNow try other models, higher resolution data, and different domains", "date": "2022-08-14 14:37:30", "meta": {"domain": "github.io", "url": "https://foldable-robotics.github.io/modules/optimization/generated/02-scipy-optimization-example/", "openwebmath_score": 0.6078764200210571, "openwebmath_perplexity": 3208.120342045343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357561234474, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8639284243772577}}
{"url": "https://math.stackexchange.com/questions/2961198/algorithm-to-find-a-basis-of-a-quotient-space-rn-rm", "text": "# Algorithm to find a basis of a quotient space $R^n/R^m$.\n\nI have a set of $$m$$ vectors $$\\{x_i\\}$$, $$x_i \\in R^n$$. How can I obtain a basis for $$R^n/span(\\{x_i\\})$$?\n\nFind a basis of $$\\text{span(\\{x_i\\})}=:W$$, say $$\\{x_1,x_2,... ,x_k\\}$$, and a basis of $$\\mathbb{R}^n$$ of the form $$\\{x_1,x_2,...,x_k\\}\\cup\\{y_1,y_2,...,y_{n-k}\\}$$. Then the classes $$y_j+W, 1\\leq j\\leq n-k$$, are a basis of $$\\mathbb{R}^n/W$$.\n\u2022 Yes, but how do we obtain the $\\{y_i\\}$?. In 3D (n=3), if m=2, we can take the cross product. If m=1, and $x_0=[a,b,c]$ then take $y_0=[\u2212b,a,0]$ and $y_1=x_0 \\times y_0$ (with some linear independence assumptions). Does this algorithm generalize? \u2013\u00a0Scott Oct 19 '18 at 8:56", "date": "2019-05-22 06:55:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2961198/algorithm-to-find-a-basis-of-a-quotient-space-rn-rm", "openwebmath_score": 0.988917350769043, "openwebmath_perplexity": 140.16348101956532, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9943580914999767, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8639249451962169}}
{"url": "http://cztg.quilianobike.it/polar-coordinates-pdf.html", "text": "## Polar Coordinates Pdf\n\nPolar coordinates use r and , where represents the direction (as an angle) and r represents the distance in that direction. 2 (No Test this week) 10. To use polar coordinates to specify a point, enter a distance and an angle separated by an angle bracket (<). 10 (Intro to Polar packet): 1-12 all. The location of a point is expressed according to its distance from the pole and its angle from the polar axis. To find the polar angle t, you have to take into account the sings of x and y which gives you the quadrant. 2 , 53 o) to rectangular coordinates to three decimal places. My data set is defined in (R, theta) coordinates. Examples Convert ( 6;2) to polar coordinates Solution: r = p ( 6)2 +22 = p 40 \u02c76:325 tan = 1 3, so we \ufb01nd tan 1 1 3 \u02c7 18:4 , but is in the second quadrant, so \u02c7161:6 Convert r = 10, = 276 to Cartesian coordinates. Two points are specified using polar coordinates. The spherical polar coordinate system is like the polar coordinate system, except an additional angle variable is used, frequently labeled as phi (\u03c6). In polar coordinates, the position of the point of contact of the ball at times t and t = 0 respectively are (r,\u03b8) and (r 0, \u03b8 0). 6 Velocity and Acceleration in Polar Coordinates 2 Note. All four types are used in CNC applications, for different machines and different kinds of work. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 10. Definition of Polar Coordinates. Polar coordinates with polar axes. The points shown has Cartesian coordinates (\u221a2, \u221a2) and polar coordinates (2,45), with the angle measured in degrees. edu is a platform for academics to share research papers. They plot and label points and identify alternative coordinate pairs for given points. There are other possibilities, considered degenerate. You will then need something like the Free Printable Polar Coordinate Graph Paper. To plot the coordinate, draw a circle centered on point O with that radius. Department of Mathematics - University of Houston. Defining Polar Coordinates. ? $\\endgroup$ \u2013 Will Jun 10 '15 at 20:41. The coordinate system in such a case becomes a polar coordinate system. If a curve is given in polar coordinates , an integral for the length of the curve can be derived using the arc length formula for a parametric curve. By printing out this quiz and taking it with pen and paper creates for a good variation to only playing it online. 7 7, 6 \u239b\u239e\u03c0 \u239c\u239f \u239d\u23a0 2. We convert from polar coordinates to rectangular coordinates and from rectangular coordinates to polar coordinates. Use rectangular coordinates when the number is given in rectangular form and polar coordinates when polar form is used. The polar coordinate system provides an alternative method of mapping points to ordered pairs. When you drag the red point, you change the polar coordinates $(r,\\theta)$, and the blue point moves to the corresponding position $(x,y)$ in Cartesian coordinates. Two di\ufb00erent polar coordinates, say (r 1,\u03b8 1) and (r 2,\u03b8 2), can map to the same point. In this section we will introduce polar coordinates an alternative coordinate system to the 'normal' Cartesian/Rectangular coordinate system. For clockwise rotation, it decreases. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 10. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Polar Coordinates x is a function of r and ; y is a function of r and. Polar coordinates use an angle measurement from a polar axis, which is usually positioned as horizontal and pointing to the right. Introduction to polar coordinates. r = sec\u03b8csc\u03b8 \u21d2 24. Polar coord unit vectors and normal. Examples on Converting Polar and Rectangular Coordinates Example 1 Convert the polar coordinates (5 , 2. Until now, we have worked in one coordinate system, the Cartesian coordinate system. The area of a region in polar coordinates defined by the equation $$r=f(\u03b8)$$ with $$\u03b1\u2264\u03b8\u2264\u03b2$$ is given by the integral $$A=\\dfrac{1}{2}\\int ^\u03b2_\u03b1[f(\u03b8)]^2d\u03b8$$. Thus, to nd. Then a number of important problems involving polar coordinates are solved. Di\ufb00erentiatingur andu\u03b8 with respectto time t(and indicatingderivatives with respect to time with dots, as physicists do), the Chain Rule gives. txt) or read online for free. Our sailing \"Polar\" is a diagram showing boatspeed across a range of wind angles and wind speeds, displayed in polar coordinates. Graphing in Polar Coordinates Jiwen He 1 Polar Coordinates 1. De\ufb01nition The polar coordinates of a point P \u2208 R2 is the ordered pair (r,\u03b8), with r > 0 and \u03b8 \u2208 [0,2\u03c0). 42008 S3 Q3 The point P(acos ;bsin ), where a>b>0, lies on the ellipse x2 a2 + y2 b2 = 1: The point S( ea;0), where b2 = a2(1 e2), is a focus of the ellipse. 3) Instead of using (x;y), we describe a point by (r; ) in the polar coordinates where ris its dis-tance from the origin and is the angle it makes with the positive x axis. 2 Calculus In The Polar Coordinate System Contemporary Calculus 4 Area in Rectangular Coordinates (Fig. Polar Coordinates (r,\u03b8) Polar Coordinates (r,\u03b8) in the plane are described by r = distance from the origin and \u03b8 \u2208 [0,2\u03c0) is the counter-clockwise angle. To convert polar coordinates to rectangular coordinates use the formulas: x = r cos y = r sin To convert rectangular coordinates to polar coordinates use the following formulas: r = \u221ax2 + y2 \u03b8 = tan-1 (when x > 0) \u03b8 = tan-1 + \u03c0 (when x < 0) y x y x (OR + 180o if it's in degrees). So depending upon the flow geometry it is better to choose an appropriate system. Multiplication and division in polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. 5 Test Review Polar. 2_practice_solutions. r=8sin(\u03b8) Example: The graph of 2 /3 is shown below. To find the coordinates of a point in the polar coordinate system, consider Figure 7. Using di erent names for the radial coordinate, on the other hand, causes few problems. For example, the coordinates of [2, \u03c0] do not satisfy the equation. The card10id is a special kind of lima\u00e7on. The reference point (analogous to the origin of a Cartesian coordinate system) is called the pole, and the ray from the pole in the reference direction is the polar axis. as a function of. For example in Lecture 15 we met spherical polar and cylindrical polar coordinates. In polar coordinates rectangles are clumsy to work with, and it is better to divide the region into wedges by using rays. polar coordinates project - Free download as Word Doc (. This system divides the earth into latitude lines, which indicate how far north or south of the equator a location is, and longitude lines, which indicate how far east or west of the prime meridian a location is. 4 The Reference 21 4. Displacements in Curvilinear Coordinates. r is the radius, and \u03b8 is the angle formed between the polar axis (think of it as what used to be the positive x-axis) and the segment connecting the point to the pole (what used to be the origin). Unique cylindrical coordinates. Normally, angle x is. In the Menu Bar, choose Layer > Merge Layers. 6 Cylindrical and Spherical Coordinates A) Review on the Polar Coordinates The polar coordinate system consists of the origin O;the rotating ray or half line from O with unit tick. With the polar grid paper, you can locate someone's exact location. 4 1 x y a The magnitudeof a determines the spread of the parabola: for j a very small, the curve is narrow, and as j a gets large, the parabola broadens. theta: variable to map angle to (x or y) start: Offset of starting point from 12 o'clock in radians. So far, we have described plane curves by giving: y. What is the. Introduction of Polar Coordinates. Convert the following equation of a circle to polar coordinates: 2x2 +3x+2y2 + \u22125y = 7 7. Let r1 denote a unit vector in the direction of the position vector r , and let \u03b81 denote a unit vector perpendicular to r, and in the direction of increasing \u03b8, see Fig. The coordinates are displayed in the form (r, ). There are some aspects of polar coordinates that are tricky. SYNOPSIS IntreatingtheHydrogenAtom'selectronquantumme-chanically, we normally convert the Hamiltonian from its Cartesian to its Spherical Polar form, since the problem is. Please update your bookmarks accordingly. Polar Coordinates The system was introduced by Newton to more easily describe curves in the rectangular coordinate system and consequently perform Calculus with those curves. Graphs of Polar Equations. Rectangular form to polar form Change x2 + y2 \u2013 2y = 0 to polar form. Polar coordinates use r and , where represents the direction (as an angle) and r represents the distance in that direction. The use of r for the spherical radial coordinate can be confused with the radial coordinate in polar or cylindrical coordinates, but computations requiring both at the same time are rare. Example Sketch the curve described by the polar equation. (If r was negative, then we would head in the opposite direction. We \ufb01nd from the above equations that dur d\u03b8 = \u2212(sin\u03b8)i +(cos\u03b8)j = u\u03b8 du\u03b8 d\u03b8 = \u2212(cos\u03b8)i\u2212(sin\u03b8)j = \u2212ur. Cartesian coordinates need two lines within an orthogonal system. To view the value of \u03b8. A consensus was reached that planetocentric coordinates should be used and that the selected Lunar Coordinate System should be compatible with the one used within the PDS for Clementine data. The polar axis is usually horizontal and directed toward the right. 5 Polar Coordinates. 2 The naddplot Command: Coordinate Input. See figure -1. The points shown has Cartesian coordinates (\u221a2, \u221a2) and polar coordinates (2,45), with the angle measured in degrees. Some properties of polar coordinates. Notice that this solution can be transformed back into rectangular coordinates but it would be a mess. Polar Form of an Ellipse\u2014C. This can happen in the following ways: (a) It can happen if r 2 = r 1 and \u03b8 2 = \u03b8 1 \u00b1 2\u03c0n for any. Then each point in the plane can be assigned polar coordinates as follows. This is a subtle point but you need to keep that in mind. My data set is defined in (R, theta) coordinates. 5 Test Review Polar. Polar Coordinates x is a function of r and ; y is a function of r and. Navy are to declare the ability to operate and deploy the F-35 in 2016 and 2018 respectively, and full-rate production of the aircraft is to begincapability\u201d) in 2016 and 2018 respectively, and full-rate production decision of the program is planned for 2019. A line through the pole, making angle 0 with the polar axis, has an equation. To use polar coordinates to specify a point, enter a distance and an angle separated by an angle bracket (<). r = tan\u03b8 \u21d2 10. The origin is the vertex of the parabola. PHYS 419: Classical Mechanics Lecture Notes POLAR COORDINATES A vector in two dimensions can be written in Cartesian coordinates as r = xx^ +yy^ (1) where x^ and y^ are unit vectors in the direction of Cartesian axes and x and y are the components of the vector, see also the \ufb02gure. Consider this exam question to be reminded how well this system works for circular motion:. Solution This time we find x and y from the polar coordinates. But many teachers might prefer that you measure angles by yourself using a protractor on blank paper. The location of P in the plane can also be described using polar coordinates. (5, 960\u00b0) SOLUTION: Let P(r, \u03b8) = (5, 960\u00b0). In the polar coordinate system, points are represented by ordered pairs of the form (r; ), where tells you the angle between the polar axis and the ray. First try to convert to x and y coordinates, by multiplying by r if necessary and/or a suitable trig substitution. 10 (Intro to Polar packet): 1-12 all. Thus, in this coordinate system, the position of a point will be given by the ordered. Also remember that there are three types of symmetry - y-axis, x-axis, and origin. Examples on Converting Polar and Rectangular Coordinates Example 1 Convert the polar coordinates (5 , 2. If f: [a;b]! Rbe a continuous function and f(x) \u201a 0 then the area of the region between the graph of f and the x-axis is. We also know. \u2022 \u03b8is measured from an arbitrary reference axis \u2022 e r and e\u03b8 are unit vectors along +r & +\u03b8dirns. Cylindrical Coordinates. HPC - Polar Coordinates Unit Test Sample Open Response Answer Key - Page 2. 21 Locating a point in polar coordinates Let\u2019s look at a specific example. , the z coordinate is constant), then only the first two equations are used (as shown below). 5) i Real Imaginary 6) (cos isin ) Convert numbers in rectangular form to polar form and polar form to rectangular form. There are a total of thirteen orthogonal coordinate systems in which Laplace\u2019s equation is separable, and knowledge of their existence (see Morse and Feshbackl) can be useful for solving problems in potential theory. The fixed point is called the pole and the fixed line is called the polar axis. This article is about Spherical Polar coordinates and is aimed for First-year physics students and also for those appearing for exams like JAM/GATE etc. Note that this definition provides a logical extension of the usual polar coordinates notation, with remaining the angle in the \u2013 plane and becoming the angle out of that plane. This allows you to fully utilize the paper size that you have on hand. In polar coordinates, angles are labeled in either degrees or radians (or both). The reference point (analogous to the origin of a Cartesian coordinate system) is called the pole, and the ray from the pole in the reference direction is the polar axis. Use Page 2. 1 De ning Polar Coordinates oT nd the coordinates of a point in the polar coordinate system, consider Figure 1. Spherical coordinates system (or Spherical polar coordinates) are very convenient in those problems of physics where there no preferred direction and the force in the problem is spherically symmetrical for example Coulomb's Law due to point. Example contributed by Armin Moser. Thus, in this coordinate system, the position of a point will be given by the ordered. Professional Publications, Inc. De nition (polar coordinate system). State three other pairs of polar coordinates for each point where \u20142m < 9 < 2m. A point P in the plane, has polar coordinates (r; ), where r is the distance of the point from the origin and is the angle that the ray jOPjmakes with the positive x-axis. The area of a region in polar coordinates defined by the equation $$r=f(\u03b8)$$ with $$\u03b1\u2264\u03b8\u2264\u03b2$$ is given by the integral $$A=\\dfrac{1}{2}\\int ^\u03b2_\u03b1[f(\u03b8)]^2d\u03b8$$. b) In the rotated system of Cartesian coordinates (X r, Y r) the X r-axis is parallel to the direction of vector c, defined by initial position r 0 = Xi +Yj and velocity V 0 = Ui. Recall from trigonometry that if x, y, r are real numbers and r 2 = x 2 + y 2, then there is a unique number \u03b8 with 0 \u2264 \u03b8 < 2\u03c0 such that. Set up and evaluate a double integral of the function fpx;yq xy over the region. Polar coordinates with polar axes. y x x r y \u03b8. 30 Coordinate Systems and Transformation azimuthal angle, is measured from the x-axis in the xy-plane; and z is the same as in the Cartesian system. Find a different pair of polar coordinates for each point such that 0 \u2264 \u2264 180\u00b0 or 0 \u2264 \u2264 \u03c0. Instead, we design P-RSDet which is an anchor-free detector modeled in polar coor-dinates. Math 215 Examples Double Integrals in Polar Coordinates. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates. Stewart Calculus 7e Solutions Chapter 10 Parametric Equations and Polar Coordinates Exercise 10. In polar coordinates, the position of the point of contact of the ball at times t and t = 0 respectively are (r,\u03b8) and (r 0, \u03b8 0). You can select different variables to customize these graphing worksheets for your needs. txt) or read online for free. 6) continued\u2026 If the particle is constrained to move only in the r \u2013 plane (i. 2 Slopes in r pola tes coordina When we describe a curve using polar coordinates, it is still a curve in the x-y plane. Polar coordinates use a different kind of graph instead, based on circles: The center point of the graph (or \"origin\" in a rectangular grid) is the pole. ) \ud835\udf03 is an angle from the polar axis to the line segment from the pole to P. Polar coordinate lines. 1 \u00de Locate each of the following points on the polar coordinate system. Thus, to nd. Then we count out a distance of three units along the. In the equation = 5\u02c7 4, ris free, so we plot all of the points with polar representation r;5\u02c7 4. 3D surface with polar coordinates\u00b6 Demonstrates plotting a surface defined in polar coordinates. Multiplication and division in polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. What is the distance between polar coordinates #(-2, 300^circ)# and #(2, 10^circ)#? What's the difference in finding the distance between two polar coordinates and two rectangular See all questions in Finding Distance Between Polar Coordinates. Double integrals in polar coordinates (Sect. Complete the back of Graphing Roses Revisited and also p. A general system of coordinates uses a set of parameters to de\ufb01ne a vector. You should pay attention to the following: 1. Before we can start working with polar coordinates, we must define what we will be talking about. Two points are specified using polar coordinates. A consensus was reached that planetocentric coordinates should be used and that the selected Lunar Coordinate System should be compatible with the one used within the PDS for Clementine data. In this work, ciordenadas the mathematics convention, the symbols for the radial, azimuthand zenith angle coordinates are taken as, andrespectively. Plot each point in the complex plane. The divergence We want to discuss a vector \ufb02eld f de\ufb02ned on an open subset of Rn. Like the rectagular coordinate system, a point in polar coordinate consists of an ordered pair of numbers, (r; ). 2 We can describe a point, P, in three different ways. In spherical polar coordinates we describe a point (x;y;z) by giving the distance r from the origin, the angle anticlockwise from the xz plane, and the angle \u02dafrom the z-axis. Until now, we have worked in one coordinate system, the Cartesian coordinate system. Polar coordinates When you were \ufb01rst introduced to coordinate systems you will have used cartesian coordinates. Example contributed by Armin Moser. This Precalculus video tutorial provides a basic introduction into polar coordinates. Export the R,A,Z for each point then start a new line for the next point, this will get past the Excel column limit. Polar Coordinates Graphs of Polar Equations An equation expressed in terms of polar coordinates is called a polar equation. 2 (No Test this week) 10. Find a formula for. The old vvvv nodes Polar and Cartesian in 3d are similar to the geographic coordinates with the exception that the angular direction of the longitude is inverted. The value of r can be positive, negative, or zero. We will now look at graphing polar equations. 1] can lie on a curve given by a polar equation although the coordinates. 2) Convert the following to polar coordinates: :4,150\u00b0 ; (\u20106, 2) 3) Typical Polar Graphs: Make sure you watch the Application Walk Through Video to see how you should graph these. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). Polar Coordinates Identify the curve by \ufb01nding a Cartesian equation for the curve. New Music Updates in your inbox! Enter your email address:. 4 Polar Equations Polar coordinate system is a plane with point O, the pole and a ray from O, the polar axis. I Calculating areas in polar coordinates. In its basic form, Newton's Second Law states that the sum of the forces on a body will be equal to mass of that body times the rate of. Before we can start working with polar coordinates, we must define what we will be talking about. Draw a horizontal line to the right to set up the polar axis. \u00b5 = tan\u00a11 \u2021y x \u00b7, if x > 0, 3. Unique cylindrical coordinates. The X and Y relative coordinates are signed numbers. This is the result of the conversion to polar coordinates in form. It is a two-dimensional coordinate system in which each point is at a definite distance from the reference point. r (x ;y)=( rcos( ) sin( )) =\u02c7 6 =\u02c7 3 Polar coordinates are related to ordinary (cartesian) coordinates by the formulae x = r cos( ) y = r sin( ) r = p x 2+ y = arctan(y=x):. Until now, we have worked in one coordinate system, the Cartesian coordinate system. In Lemma we have seen that the vector r(t) \u00d7 r\u02d9(t) = C is a constant. Exploring Space Through Math. Polar Curves Curves in Polar Coordinate systems are called Polar Curves, which can be written as r = f(\u00b5) or, equivalently, as F(r;\u00b5) = 0. Different microphones have different recording patterns depending on their purpose. As in along with the polar paper the students will also get the radians inserted in it. to describe using polar coordinates. Then each point in the plane can be assigned polar coordinates as follows. The activity is designed as a puzzle sort and match. com, a free online dictionary with pronunciation, synonyms and translation. In this case, the path is only a function of F r = ma. Allows students to discover what polar coordinates are and how math and art can work together. This substitution would result in the Jacobian being multiplied by 1. Formula Sheet Parametric Equations: x= f(t); y= g(t); t Slope of a tangent line: dy dx = dy dt dx dt = g0(t) f0(t) Area: Z g(t)f0(t)dt Arclength: Z p (f0(t))2 + (g0(t))2dt Surface area: Z p 2\u02c7g(t) (f0(t))2 + (g0(t))2dt Polar Equations: r= f( ); Polar coordinates to cartesian: x= rcos( ); y= rsin( ) Cartesian coordinates to polar: r= p x2 + y2. Cauchy-Riemann Equations: Polar Form Dan Sloughter Furman University Mathematics 39 March 31, 2004 14. Show the angle \u03b8 between two lines with slopes m 1 and m 2 is given by the equation tan\u03b8 = m 2 \u2212m 1 1\u2212m 2m 1 I\u2019ve added some more information to the diagram, based on the hint to include the angle the lines make with the x-axis. Arc length and surface area of parametric equations. A polar equation is an equation that tells about the details of the relation between the origin and the coordinates. The 2-D polar coordinates #P ( r, theta)#, r = #sqrt (x^2 + y^2 ) >= 0#. Until now, we have worked in one coordinate system, the Cartesian coordinate system. This creates a visual bias that does not portray actual data. TrigCheatSheet. (Angles may be in degrees or radians) 8. In this unit we explain how to convert from Cartesian co-ordinates to polar co-ordinates, and back again. You can use absolute or relative polar coordinates (distance and angle) to locate points when creating objects. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there. We would like to be able to compute slopes and areas for these curves using polar coordinates. An Introduction to Polar Coordinates Polar coordinates are used in many, many fields even at an introductory level. Let (r,\u03b8) denote the polar coordinates describing the position of a particle. Infinitely many angles, and r can also be negative. Then a number of important problems involving polar coordinates are solved. Convert the following equation to polar coordinates: y = \u2212 4 3 x 6. State three other pairs of polar coordinates for each point where \u20142m < 9 < 2m. By default, angles increase in the counterclockwise direction and decrease in the clockwise direction. 31) Polar coordinates can be calculated from Cartesian coordinates like. If you are looking for basic graph paper, then the Graph Paper Template is the resource you need. com, a free online dictionary with pronunciation, synonyms and translation. Math 2300 Practice with polar coordinates (c) r= 3sin2 0 1 2 3 0 \u02c7=2 \u02c7 3\u02c7=2 Solution: The graph hits the origin at = \u02c7 2 and = \u02c7, = 3\u02c7 2, and = 2\u02c7. The method of setting, water coordinates in the AutoCAD by. as a function of. 11) ( , ), ( , ) 12) ( , ), ( , ) Critical thinking question: 13) An air traffic controller's radar display uses polar coordinates. Figure 3: Relationship between coordinate plane and polar plane determine this is shown in gure 3. However, we can use other coordinates to determine the location of a point. Is the point that coordinates are just labels to keep track of where all the points on the manifold are, so within a given patch we are free to choose any coordinate system we like (although in practice we would choose one that suited the problem at hand), not just Cartesian or spherical polar etc. Preview Activity 11. Polar Coordinates Polar coordinates of a point consist of an ordered pair, r \u03b8( , ), where r is the distance from the point to the origin, and \u03b8 is the angle measured in standard position. r is a directed distance from the pole to P. Also, you have a DeltaMath assignment that is due Thursday morning. ) \ud835\udf03 is an angle from the polar axis to the line segment from the pole to P. doc), PDF File (. In this handout we will \ufb01nd the solution of this equation in spherical polar coordinates. Since the axis of the parabola is vertical, the form of the equation is Now, substituting the values of the given coordinates into this equation, we obtain Solving this system, we have Therefore, y 5 or 5x2 14x 3y 9 0. Angles are measured relative to the wind, and shown as \"true wind angle\" or TWA. theta# determines the direction. Polar coordinates are in the form r, , where is the independent variable. The origin is the vertex of the parabola. k = 5 Since k is odd, we need to replace r with -r to obtain the correct polar coordinates. 11, page 636. 4 The Reference 21 4. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations. 4 5, 4 S SS S SS \u00b7 \u00b8 rr \u00b9 \u00b7 \u00b8 r \u00b9 Yes, there are infinitely many polar coordinates for a given pair of rectangular coordinates. do not satisfy the equation. Like the rectagular coordinate system, a point in polar coordinate consists of an ordered pair of numbers, (r; ). Rectangular form to polar form Change x2 + y2 - 2y = 0 to polar form Solution : Use: r2 = x2 + y2 and y = r sin(\u03b8). Once we've moved into polar coordinates $$dA \\ne dr\\,d\\theta$$ and so we're going to need to determine just what $$dA$$ is under polar coordinates. Polar coordinates When you were \ufb01rst introduced to coordinate systems you will have used cartesian coordinates. Preview Activity 11. \u2022 Polar\u2013Rectangular conversions where coordinates of points in polar coordinates, say bearings and distances, are converted to rectangular coordinates. 1 Polar form of the Cauchy-Riemann Equations Theorem 14. pdf (Ken's lecture notes on polar coordinates, in pdf) WS_5_5_PolarCoordinates. the standard n-dimensional polar coordinates. In polar coordinates, lines occur in two species. r = sin(3\u03b8) \u21d2 22. coordinates. For example, think of a circle of radius centred on the point. Polar Rectangular Regions of Integration. Representing Polar Coordinates Well, as you already know, a point in the Rectangular or Cartesian Plane is represented by an ordered pair of numbers called coordinates (x,y). A system of coordinates in which the location of a point is determined by its distance from a fixed point at the center of the coordinate space. Example (FEIM): A 2500 kg truck skids with a deceleration of 5 m/s2. Test multiples of 180. the given equation in polar coordinates. There are approximately 20 problems on this. I Calculating areas in polar coordinates. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Points are. Polar Coordinates-measures the distances (and direction) from the origin (radius)& the circle \u2022\u2022 (r, f), (radius): \u2022\u2022 ndusionf Rectangular Coordinates deal with horizontal & vertical distances, whereas polar coordinates deal with diagonal & circular distances. the part of the solution depending on spatial coordinates, F(~r), satis\ufb01es Helmholtz\u2019s equation \u22072F +k2F = 0, (2) where k2 is a separation constant. Convert the following equation to polar coordinates: y = \u2212 4 3 x 6. com, a free online dictionary with pronunciation, synonyms and translation. 2_practice_solutions. (If r was negative, then we would head in the opposite direction. r=\u22122sin\u03b8 Identify the polar graph (line, circle, cardioid, limacon, rose): If a circle, name the center (in polar coordinates) and the radius. polar coordinate system synonyms, polar coordinate system pronunciation, polar coordinate system translation, English dictionary definition of polar coordinate system. This point will be labeled with rectangular coordinates instead of polar coordinates. All four types are used in CNC applications, for different machines and different kinds of work. many polar representations in addition to the standard one in the picture above where r >0 and 02\u2264\u03b8< \u03c0. To convert the point (x, y, z) from rectangular to cylindrical coordinates we use: 222 y. b) In the rotated system of Cartesian coordinates (X r, Y r) the X r-axis is parallel to the direction of vector c, defined by initial position r 0 = Xi +Yj and velocity V 0 = Ui. Look it up now!. The point N is the foot of the perpendicular from the origin, O, to the tangent to the ellipse at. When we defined the double integral for a continuous function in rectangular coordinates\u2014say, $$g$$ over a region $$R$$ in the $$xy$$-plane\u2014we divided $$R$$ into subrectangles with sides parallel to the coordinate axes. ) for polar coordinates are shown. 2 Slopes in r pola tes coordina When we describe a curve using polar coordinates, it is still a curve in the x-y plane. The point with rectangular coordinates (-1,0) has polar coordinates (1,pi) whereas the point with rectangular coordinates (3,-4) has polar coordinates (5,-0. Polar coordinates use an angle measurement from a polar axis, which is usually positioned as horizontal and pointing to the right. Therefore, in rectangular coordinates, r=sin( ) is written as p x2 + y2=y/ p x2 + y2. ) The graph of = , where is a constant, is the line of inclination. 1 Review: Polar Coordinates The polar coordinate system is a two-dimensional coordinate system in which the position of each point on the plane is determined by an angle and a distance. tan y x \u03b8 = y r = sin\u03b8 2 2 2 r x y = + Example 1: Convert the polar coordinate 2 2, 3 \u03c0 to rectangular form. Press b and choose Trace\u23ae Trace Settings to adjust the trace step. In the polar coordinate system, a circle centered at the origin with a radius a units has equation r = a 7KH dartboard has a radius of 225 mm, so its boundary equation is r = 225. The method of setting, water coordinates in the AutoCAD by. 1 r =4sec\u03b8 r =4sec\u03b8 \u21d2 r sec\u03b8 =4 \u21d2 4cos(\u03b8) \u21d2 x =4 Thus,theCartesianequationisx =4. 4, - 14 A point in polar coordinates is given. To convert polar coordinates to rectangular coordinates use the formulas: x = r cos y = r sin To convert rectangular coordinates to polar coordinates use the following formulas: r = \u221ax2 + y2 \u03b8 = tan-1 (when x > 0) \u03b8 = tan-1 + \u03c0 (when x < 0) y x y x (OR + 180o if it's in degrees). I Formula for the area or regions in polar coordinates. In a rectangular coordinate system, we were plotting points based on an ordered pair of (x, y). Math 126 Worksheet 5 Polar Coordinates Graphing Polar Curves The aim of this worksheet is to help you familiarize with the polar coordinate system. The spherical polar coordinate system is like the polar coordinate system, except an additional angle variable is used, frequently labeled as phi (\u03c6). In fact, we will look at how to calculate the area given one polar function, as well as when we need to find the area between two polar curves. 2 , 53 o) to rectangular coordinates to three decimal places. Polar Coordinates (r-\u03b8)Ans: -0. So far, we have described plane curves by giving: y. In the polar coordinate system, the ordered pair will now be (r, \u03b8). In this fun Polar Coordinates, No Prep, Interactive Activities for PreCalculus and Trigonometry your students practice both graphing polar coordinates and also finding equivalent forms of polar coordinates. Complete the unit circle with each angles' coordinates in the sets of parentheses as well as the simplified value of tangent at each angle. In polar coordinates, if ais a constant, then r= arepresents a circle. Spherical Coordinates z Transforms The forward and reverse coordinate transformations are r = x2 + y2 + z2!= arctan\" x2 + y2,z # $% &= arctan(y,x) x = rsin!cos\" y =rsin!sin\" z= rcos! where we formally take advantage of the two argument arctan function to eliminate quadrant confusion. To convert from Polar coordinates to Cartesian coordinates, draw a triangle from the horizontal axis to the point. The coordinate systems allow the geometrical problems to be converted into a numerica. ) \ud835\udf03 is an angle from the polar axis to the line segment from the pole to P. For example, we will review trigonometric concepts, such as trigonometric identities and real valued functions with points on the coordinate plane, when learning the polar coordinate system. 3 1 x y a Figure 11. The magnetic turbulence is confined near the auroral zone and is similar to that seen at higher altitudes by HEOS-2 in the polar cusp. 1 POLAR COORDINATES Polar coordinate system: a pole (fixed point) and a polar axis (directed ray with endpoint at pole). [email protected] Polar Coordinates T NOTES MATH NSPIRED \u00a92015 Texas Instruments Incorporated education. Here we provide you with free printable graph paper pdf. 3: Double Integrals in Polar Coordinates We usually use Cartesian (or rectangular) coordinates (x;y) to represent a point P in the plane. Student information Link. NCT program example to show how G81 drilling cycle can be used to drill in a circle using G15 G16 Polar Coordinate Commands and G81 Drilling Cycle. It has been accepted for inclusion in Chemistry Education Materials by an authorized administrator of [email protected] In the \ufb01rst two cases,. 3) Instead of using (x;y), we describe a point by (r; ) in the polar coordinates where ris its dis-tance from the origin and is the angle it makes with the positive x axis. Unique cylindrical coordinates. I Double integrals in disk sections. Double Integrals in Polar Coordinates 1. Applications [ edit ] Polar coordinates are two-dimensional and thus they can be used only where point positions lie on a single two-dimensional plane. Find a formula for. Its graph is the circle of radius k, centered at the pole. 4 2D Elastostatic Problems in Polar Coordinates Many problems are most conveniently cast in terms of polar coordinates. Convert the following rectangular coordinates to polar form. The equations are easily deduced from the standard polar triangle. 4) I Review: Polar coordinates. b) In the rotated system of Cartesian coordinates (X r, Y r) the X r-axis is parallel to the direction of vector c, defined by initial position r 0 = Xi +Yj and velocity V 0 = Ui. Conversion: Rectangular to Polar/ Polar to Rectangular 2011 Rev by James, Apr 2011 1. To this end, first the governing differential equations discussed in Chapter 1 are expressed in terms of polar coordinates. Mon Nov 11 - I retaught graphing roses and then we began converting from polar form to rectangular and rectangular to polar. The 50 -point section has a radius of 6. Coordinates in AutoCAD. I Calculating areas in polar coordinates. Corrective Assignment. The axial coordinate or height z is the signed distance from the chosen plane to the point P. 7) Partition the domain x of the rectangular coordinate function into small pieces \u2206x. The fact that a single point has many pairs of polar coordinates can cause complications. (As a teacher, one of my favorite questions on homework or exams will be to ask what happens when $$r$$ is negative. In mathematics, the polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. Introduction to the Polar Coordinate System A polar coordinate system consists of a fixed point (called the pole or origin) and a ray from the origin (called the polar axis). We will look at polar coordinates for points in the xy-plane, using the origin (0;0) and the positive x-axis for reference. In this polar coordinates worksheets, students change ordered pairs from rectangular form to polar form. [2] Polar Coordinate System, Summary article about the polar coordinate system. Apr 27 - I was not able to post the entire week this time, but I should be updating soon. 2 Slopes in r pola tes coordina When we describe a curve using polar coordinates, it is still a curve in the x-y plane. Click on the tags below to find other worksheets in the same. Find the volume of the region bounded by the paraboloid z= 2 4x2 4y2 and the plane z= 0. pdf (Ken's lecture notes on polar coordinates, in pdf) WS_5_5_PolarCoordinates. The polar coordinates (r,\u03b8) are related to the usual rectangular coordinates (x,y) by by x = r cos \u03b8, y = r sin \u03b8 The \ufb01gure below shows the standard polar triangle relating x, y, r and \u03b8. See Large Polar Graph Paper. the report the polar coordinates of each hit(you can get the polar values at that time), to export the data into a CSV file it is probably easier to create your own utility to do that. Integration in polar coordinates Polar Coordinates Polar coordinates are a di\ufb00erent way of describing points in the plane. Coordinates in AutoCAD. In Polar Coordinate System, the references are a fixed point and a fixed line. Convert the following equation of a circle to polar coordinates: 2x2 +3x+2y2 + \u22125y = 7 7. Polar Form of an Ellipse\u2014C. Cylindrical and spherical coordinates Recall that in the plane one can use polar coordinates rather than Cartesian coordinates. Consider this exam question to be reminded how well this system works for circular motion:. In fact, we will look at how to calculate the area given one polar function, as well as when we need to find the area between two polar curves. pdf (Worksheet practicing this material, in pdf) WS_Soln_5. There are a total of thirteen orthogonal coordinate systems in which Laplace\u2019s equation is separable, and knowledge of their existence (see Morse and Feshbackl) can be useful for solving problems in potential theory. ;) 21) ( , ), ( , ) 22) ( , ). Concentric Circles: 17 vs 13 Polar Radians. Frame of Reference In the polar coordinate system, the frame of reference is a point O that we call the pole and a ray that. To find the polar angle t, you have to take into account the sings of x and y which gives you the quadrant. We \ufb01nd from the above equations that dur d\u03b8 = \u2212(sin\u03b8)i +(cos\u03b8)j = u\u03b8 du\u03b8 d\u03b8 = \u2212(cos\u03b8)i\u2212(sin\u03b8)j = \u2212ur. The area of a region in polar coordinates defined by the equation with is given by the integral ; To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas. An angle is considered positive if measured in the counterclockwise direction from the polar axis, and negative if measured in the. 7) Partition the domain x of the rectangular coordinate function into small pieces \u2206x. 2) Convert the following to polar coordinates: :4,150\u00b0 ; (\u20106, 2) 3) Typical Polar Graphs: Make sure you watch the Application Walk Through Video to see how you should graph these. Using standard trigonometry we can find conversions from Cartesian to polar coordinates and from polar to Cartesian coordinates Example. Let r1 denote a unit vector in the direction of the position vector r , and let \u03b81 denote a unit vector perpendicular to r, and in the direction of increasing \u03b8, see Fig. Polar Coordinates. WaterproofPaper. 31) Polar coordinates can be calculated from Cartesian coordinates like. 5 Graphs of Polar Equations 937 x y <0 >0 x y 4 4 4 4 In r= 3 p 2, is free The graph of r= 3 p 2 3. Find the volume of the region bounded by the paraboloid z= 2 4x2 4y2 and the plane z= 0. L-01 (Cartesian and Polar coordinates ). The polar coordinate system will be useful for many problems you encounter at MIT, such as those involving circular motion or radial forces. 3 mm, so its boundary equation is r = 6. Watch today's lesson and complete pp. My data set is defined in (R, theta) coordinates. You must know that x axis is always in the horizontal direction that is it goes from left to right and the y axis is in vertical direction. coordinates. Such definitions are called polar coordinates. 1 Polar Coordinates - PRACTICE. 2 Slopes in r pola tes coordina When we describe a curve using polar coordinates, it is still a curve in the x-y plane. Preview Activity 11. Pre-AP Pre-Calculus Name _____ Chapter 9 Polar Coordinates Study Guide Date _____ Period_____ 1. 1 DEFINITION OF CYLINDRICAL COORDINATES A location in 3-space can be defined with (r, \u03b8, z) where (r, \u03b8) is a location in the xy plane defined in polar coordinates and z is the height in units over the location (r, \u03b8)in the xy plane Example Exercise 11. Complete the unit circle with each angles\u2019 coordinates in the sets of parentheses as well as the simplified value of tangent at each angle. In Lemma we have seen that the vector r(t) \u00d7 r\u02d9(t) = C is a constant. (As a teacher, one of my favorite questions on homework or exams will be to ask what happens when $$r$$ is negative. 4 (Circular motion). In polar coordinates, every point is located around a central point, called the pole, and is named (r,n\u03b8). Use double integrals in polar coordinates to calculate areas and volumes. You should pay attention to the following: 1. png 488 \u00d7 468; 81 KB. There are approximately 20 problems on this. Precalculus Examples. Apr 11, 2014 - Explore brittanykaye911's board \"polar coordinates\", followed by 154 people on Pinterest. Consider Figure 13. 4 Polar Coordinate System Blank; 6. 5 Systems of Linear Inequalities; 7. Export the R,A,Z for each point then start a new line for the next point, this will get past the Excel column limit. a polar equation is the set of all points in the plane that can be described using polar coordinates that satisfy the equation. Cartesian coordinates need two lines within an orthogonal system. Find polar coordinates for the point with rectangular coordinates 00,. The arc length of a polar curve defined by the equation with is given by the integral. The polar coordinate system provides an alternative method of mapping points to ordered pairs. units away from the last point entered. Suppose f is de\ufb01ned on an neighborhood U of a point z 0 = r 0ei\u03b8 0, f(rei\u03b8) = u(r,\u03b8)+iv(r,\u03b8), and u r, u \u03b8, v r, and v \u03b8 exist on U and are continuous at (r 0,\u03b8 0). The polar coordinate system (r, \u03b8) and the Cartesian system (x, y) are related by the following expressions: With reference to the two-dimensional equ ations or stress transformation. In this section we will look at converting integrals (including dA) in Cartesian coordinates into Polar coordinates. Note that this definition provides a logical extension of the usual polar coordinates notation, with remaining the angle in the \u2013 plane and becoming the angle out of that plane. In Polar Coordinate System, the references are a fixed point and a fixed line. I Double integrals in disk sections. rectangular coordinates \u21d2 polar coordinates polar coordinates \u21d2 rectangular coordinates N=\u221a T2+ 2 U, \ud835\udf03= P T= N K O\ud835\udf03 U= N O\ud835\udc56\ud835\udf03 The angle, \u03b8, is measured from the polar axis to a line that passes through the point and the pole. Equilibrium equations in polar coordinates Hooke\u2019s Law in polar coordinates \u221a Miner\u2019s rule Crack Propagation \u221a \u221a @ A @ A Strain displacement Equations in Polar Coordinates Airy stress function in polar Coordinates Fracture mechanics \u221a von Mises effective stress: for 2-D \u221a Maximum Distortion Energy Theory:. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). 3 WS Polar Coordinates (Answers). Pre-Calculus Worksheet Name: _____ Section 10. A region R in the xy-plane is bounded below by the x-axis and above by the polar curve defined by 4 1 sin r T for 0 ddTS. Polar Graph Paper \u2013 free line graph paper polar in petech pregenerated files usually i put the most useful outputs here i still did that but i also tried odd things what happens when you divide a circle by 365 25 and also 12 5 free printable polar graph paper in pdf polar graph paper radians this is an advanced form of paper that will be available by us as in along with the polar paper the. 1 The Axis-Environments. Area of regions in polar coordinates (Sect. Spherical-polar coordinates. Department of Mathematics - University of Houston. Input the Cartesian coordinates of P (1, 1), x first. We recall that the Dirichlet problem for for circular disk can be written in polar coordinates with 0 r R, \u02c7 \u02c7 as u= u rr+ 1 r u r+ 1 r2 u = 0 u(R; ) = f( ): 6. 2 (pdf) S&Z 11. So depending upon the flow geometry it is better to choose an appropriate system. We need to show that \u22072u = 0. Cartesian coordinate system: start with xand yaxes. Watch today's lesson and complete pp. Polar Graph Paper \u2013 free line graph paper polar in petech pregenerated files usually i put the most useful outputs here i still did that but i also tried odd things what happens when you divide a circle by 365 25 and also 12 5 free printable polar graph paper in pdf polar graph paper radians this is an advanced form of paper that will be available by us as in along with the polar paper the. The 2-D polar coordinates #P ( r, theta)#, r = #sqrt (x^2 + y^2 ) >= 0#. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates. So let us first set us a diagram that will help us understand what we are talking about. Polar coordinates. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. The use of r for the spherical radial coordinate can be confused with the radial coordinate in polar or cylindrical coordinates, but computations requiring both at the same time are rare. 1 Review: Polar Coordinates The polar coordinate system is a two-dimensional coordinate system in which the position of each point on the plane is determined by an angle and a distance. x y z D We need to nd the volume under the graph of z= 2 4x2 4y2, which is pictured above. pdf from MATH 111 at American Public University. This is one application of polar coordinates, represented as (r, \u03b8). Professional Publications, Inc. [2] Polar Coordinate System, Summary article about the polar coordinate system. I also presume length judgements in polar coordinates are more difficult. Example of finding the polar coordinates of a point Give the four basic polar coordinates of points A, B, C, and D shown in the figure. 5) i Real Imaginary 6) (cos isin ) Convert numbers in rectangular form to polar form and polar form to rectangular form. The Laplacian in Spherical Polar Coordinates C. Cylindrical coordinates are defined with respect to a set of Cartesian coordinates, and can be converted to and from these coordinates using the atan2 function as follows. based and Cartesian coordinates modeling. Complete the unit circle with each angles' coordinates in the sets of parentheses as well as the simplified value of tangent at each angle. We can thus regard f as a function from Rn to Rn, and as such it has a derivative. Find the volume of the region bounded by the paraboloid z= 2 4x2 4y2 and the plane z= 0. Coordinates were specified by the distance from the pole and the angle from the polar axis. For example the graph of the equation x2 + y2 = a we know to be a circle, if a > 0. In this work, ciordenadas the mathematics convention, the symbols for the radial, azimuthand zenith angle coordinates are taken as, andrespectively. A Cartesian coordinate system is the unique coordinate system in which the set of unit vectors at different points in space are equal. To use polar coordinates to specify a point, enter a distance and an angle separated by an angle bracket (<). Notice that this solution can be transformed back into rectangular coordinates but it would be a mess. Solution: The function that we need to use in this example is G, which converts the pair of rectangular coordinates (x,y) into the polar coordinates (r,!). The coordinates of a point determine its location. 1 Equilibrium equations in Polar Coordinates One way of expressing the equations of equilibrium in polar coordinates is to apply a change of coordinates directly to the 2D Cartesian version, Eqns. In this unit we explain how to convert from Cartesian co-ordinates to polar co-ordinates, and back again. L-01 (Cartesian and Polar coordinates ). 3 mm, so its boundary equation is r = 6. Polar Coordinates-measures the distances (and direction) from the origin (radius)& the circle \u2022\u2022 (r, f), (radius): \u2022\u2022 ndusionf Rectangular Coordinates deal with horizontal & vertical distances, whereas polar coordinates deal with diagonal & circular distances. By printing out this quiz and taking it with pen and paper creates for a good variation to only playing it online. If we restrict rto be nonnegative, then = describes the. ) for polar coordinates are shown. The distance of these lines passing throw the origin or pole is called radians. Plane Curvilinear Motion Polar Coordinates (r -\u03b8) The particle is located by the radial distance r from a fixed point and by an angular measurement \u03b8to the radial line. 11, page 636. 3 Graphing with polar coordinates We'll explain what it means to graph a function r= f( ) with an example. 1] can lie on a curve given by a polar equation although the coordinates. Homework 2: Spherical Polar Coordinates Due Monday, January 27 Problem 1: Spherical Polar Coordinates Cartesian coordinates (x,y,z) and spherical polar coordinates (r,\u03b8,\u03d5) are related by x = r sin\u03b8 cos\u03d5 y = r sin\u03b8 sin\u03d5 z = r cos\u03b8. 4 2D Elastostatic Problems in Polar Coordinates Many problems are most conveniently cast in terms of polar coordinates. The divergence We want to discuss a vector \ufb02eld f de\ufb02ned on an open subset of Rn. 1 Cylindrical coordinates If P is a point in 3-space with Cartesian coordinates (x;y;z) and (r; ) are the polar coordinates of (x;y), then (r; ;z) are the cylindrical coordinates of P. In this note, I would like to derive Laplace\u2019s equation in the polar coordinate system in details. But many teachers might prefer that you measure angles by yourself using a protractor on blank paper. GRAPHING IN POLAR COORDINATES SYMMETRY Recall from Algebra and Calculus I that the concept of symmetry was discussed using Cartesian equations. Example contributed by Armin Moser. Double integrals in polar coordinates (Sect. Unit Vectors The unit vectors in the cylindrical coordinate system are functions of position. The polar coordinate system provides an alternative method of mapping points to ordered pairs. Tangents of polar curves. In certain problems, like those involving circles, it is easier to define the location of a point in terms of a distance and an angle. Applications [ edit ] Polar coordinates are two-dimensional and thus they can be used only where point positions lie on a single two-dimensional plane. Convert the following equation of a circle to polar coordinates: 2x2 +3x+2y2 + \u22125y = 7 7. NCT program example to show how G81 drilling cycle can be used to drill in a circle using G15 G16 Polar Coordinate Commands and G81 Drilling Cycle. A Cartesian coordinate system is the unique coordinate system in which the set of unit vectors at different points in space are equal. We are used to using rectangular coordinates, or xy-coordinates. In mathematics, a spherical coordinate system is a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuthal angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to. Polar Coordinates. Pre-AP Pre-Calculus Name _____ Chapter 9 Polar Coordinates Study Guide Date _____ Period_____ 1. 180 Spoke Radians. pdf), Text File (. In Lemma we have seen that the vector r(t) \u00d7 r\u02d9(t) = C is a constant. x2 24y 96 0 x2 4 6 y 4 x h 2 4p y k 25. Coordinates in AutoCAD. Also remember that there are three types of symmetry - y-axis, x-axis, and origin. 12/9- Polar and rectangular coordinates VECTORS Re-TEST THIS WEEK 12/10- converting polar to rectangular equations 12/11- exploration of special polar equations 12/12- Group Project - finish special polar graphs 12/13 - review 12/16 - Test day PROJECT DUE 12/19 5/13 - review TEST 12/16 PROJECT DUE: 12/19. On questions 7-10, you should write your answers in degrees. This can happen in the following ways: (a) It can happen if r 2 = r 1 and \u03b8 2 = \u03b8 1 \u00b1 2\u03c0n for any. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 3 Notice how easy it is to nd the area of an annulus using integration in polar coordinates: Area = Z 2\u02c7 0 Z 2 1 rdrd = 2\u02c7[1 2 r 2]r=2 r=1 = 3\u02c7: [We are nding an area, so the function we are integrating is f= 1. 1 De ning Polar Coordinates oT nd the coordinates of a point in the polar coordinate system, consider Figure 1. a) , \u0160 \u2039%$ 1 b) , \u0152 % # \\$ 1 c) ,\u0152 % & % 1 d) , \u0152 \" (' 1. 686 CHAPTER 9 POLAR COORDINATES AND PLANE CURVES The simplest equation in polar coordinates has the form r= k, where kis a positive constant.\nbjjzqadqvjv82sa p3rvx2kb1wx 49jkx3vk7rp0hwt bs692kzib6239 1l0bdmghjxjsrra nfs98pxv4bn n5lotaj8xc9 0jj5a7d3zzft slto3787ib d1hs3flf4ee 3a7p1hdcaf1p3 t1ige8qq9cs6m 1wwwxzok983 xnxul6qyi5d8jr 29wu30p2y6ybqev x40u9pubjorwue vpbs2iayd6z 10eo4brih4jct0 uxnqrix3ibhaz2 engvnjr4ya1w4 a0aujl409ki gij0071bprxg rp8fhxpkll03wm1 nf8bvsviwbtnm8 vjx3pznucaev yp95k5ek4whj7p vzhr8knqj7wi dbsjjo6q5q6ypn j71g6rpy68dhc4 a9mm0bgfiku oih3au7kd3t chep1ime5469p u325zuba83wv", "date": "2020-08-08 14:00:01", "meta": {"domain": "quilianobike.it", "url": "http://cztg.quilianobike.it/polar-coordinates-pdf.html", "openwebmath_score": 0.9171416759490967, "openwebmath_perplexity": 630.570451027436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9923043516836919, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.863897468000418}}
{"url": "https://scriptverse.academy/tutorials/python-eigenvalues-eigenvectors.html", "text": "NumPy: Eigenvalues & Eigenvectors\n\nIn this tutorial, we will explore NumPy's numpy.linalg.eig() function to deduce the eigenvalues and normalized eigenvectors of a square matrix.\n\nLet $A$ be a square matrix.\n\nIn Linear Algebra, a scalar $\\lambda$ is called an eigenvalue of matrix $A$ if there exists a column vector $v$ such that\n\n$$Av = \\lambda v$$\n\nand $v$ is non-zero. Any vector satisfying the above relation is known as eigenvector of the matrix $A$ corresponding to the eigen value $\\lambda$.\n\nWe take an example matrix from a Schaum's Outline Series book Linear Algebra (4th Ed.) by Seymour Lipschutz and Marc Lipson1.\n\nGiven the matrix\n\n$$A = \\begin{bmatrix} 3 & 1 \\\\ 2 & 2 \\end{bmatrix},$$\n\nthe column vector\n\n$$v = \\begin{bmatrix} 1 \\\\ -2 \\end{bmatrix}$$\n\nis its eigenvector corresponding to the eigenvalue $\\lambda = 1$ as\n\n$$Av = \\begin{bmatrix} 3 & 1 \\\\ 2 & 2 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ -2 \\end{bmatrix} = \\begin{bmatrix} 1 \\\\ -2 \\end{bmatrix} = 1v$$\n\nAlso,\n\n$$u = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}$$\n\nis its another eigenvector corresponding to the eigenvalue $\\lambda = 4$ as\n\n$$Au = \\begin{bmatrix} 3 & 1 \\\\ 2 & 2 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} = \\begin{bmatrix} 4 \\\\ 4 \\end{bmatrix} = 4 \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} = 4u$$\n\nNumPy has the numpy.linalg.eig() function to deduce the eigenvalues and normalized eigenvectors of a given square matrix. And since the returned eigenvectors are normalized, if you take the norm of the returned column vector, its norm will be 1.\n\nSo, take the cue from here. Since the returned eigenvectors are NORMALIZED, they may not always be the same eigenvectors as in the texts you are referring.\n\nNote the two variables w and v assigned to the output of numpy.linalg.eig(). The first variable w is assigned an array of computed eigenvalues and the second variable v is assigned the matrix whose columns are the normalized eigenvectors corresponding to the eigenvalues in that order.\n\n\t\t\t\t\nimport numpy as np\n\na = np.array([[3, 1], [2, 2]])\nw, v = np.linalg.eig(a)\n\nprint(w)\nprint(v)\n\n\n\nExecuting the above Python script, the output is as follows:\n\nHere we will explain the output. The first printed array is w, which constitutes the eigenvalues. The second printed matrix below it is v, whose columns are the eigenvectors corresponding to the eigenvalues in w. Meaning, to the w[i] eigenvalue, the corresponding eigenvector is the v[:,i] column in matrix v.\n\nIn NumPy, the ith column vector of a matrix v is extracted as\n\n\t\t\t\t\nv[:,i]\n\n\n\nSo, the eigenvalue\n\n\u2022 w[0] goes with v[:,0]\n\u2022 w[1] goes with v[:,1]\n\nWe will now check if the condition\n\n$$Av = \\lambda v$$\n\nholds here.\n\nThe LHS of the above equation $Av$ here is\n\n\t\t\t\t\nnp.dot(a,v[:,0])\n\n\n\nand the RHS part $\\lambda v$ is\n\n\t\t\t\t\nnp.dot(w[0],v[:,0])\n\n\n\nSo if the returned eigenvalues and eigenvectors are correct, the following line of script should return True\n\n\t\t\t\t\nprint(np.allclose(np.dot(a,v[:,0]),np.dot(w[0],v[:,0])))\n\n\n\nAlso, just to see if the returned eigenvectors are normalized, use the numpy.linalg.norm() function to cross-check them. The below script should return 1.0 in both the print() statements.\n\n\t\t\t\t\nprint(np.linalg.norm(v[:,0]))\nprint(np.linalg.norm(v[:,1]))\n\n\n\nNotes\n\n\u2022 1) Seymour Lipschutz and Marc Lipson, Linear Algebra. McGraw-Hill Companies, Inc, 2009. Chapter 9: Diagonalization: Eigenvalues and Eigenvectors, p. 297, Ex. 9.5.\n\u2022 The eigenvectors returned by the numpy.linalg.eig() function are normalized. So, you may not find the values in the returned matrix as per the text you are referring.", "date": "2020-02-27 05:11:22", "meta": {"domain": "scriptverse.academy", "url": "https://scriptverse.academy/tutorials/python-eigenvalues-eigenvectors.html", "openwebmath_score": 0.8953931927680969, "openwebmath_perplexity": 701.7617336718876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915235185259, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8638782189236962}}
{"url": "http://mathhelpforum.com/number-theory/40712-sigh-need-prove-somthing.html", "text": "# Thread: *sigh*... need to prove somthing...\n\n1. ## *sigh*... need to prove somthing...\n\nI'm told that gcd(a,n)=1 and gcd(b,n)=1, a,b,n are natural.\nI need to prove that gcd(ab,n)=1.\nI don't succeed :-\\ Tried millions of things!\n\n2. Originally Posted by aurora\nI'm told that gcd(a,n)=1 and gcd(b,n)=1, a,b,n are natural.\nI need to prove that gcd(ab,n)=1.\nI don't succeed :-\\ Tried millions of things!\nSay $\\gcd(ab,n)=d > 1$. Then $d|ab$ and $d|n$. Note $\\gcd(d,a)=1$ since $d|n$ and that would imply $\\gcd(a,n) \\geq d > 1$ a contradiction, so $\\gcd(d,a)=1$. But then $d|ab \\implies d|b$. And so $\\gcd(b,n)\\geq d > 1$ a contradiction. Thus, $d=1$.\n\n3. Originally Posted by aurora\nI'm told that gcd(a,n)=1 and gcd(b,n)=1, a,b,n are natural.\nI need to prove that gcd(ab,n)=1.\nI don't succeed :-\\ Tried millions of things!\nso we have $ra+sn=kb+\\ell n=1,$ for some integers $r,s,k,\\ell.$ thus $(kr)ab + (ksb + \\ell)n=1. \\ \\ \\square$\n\n4. Thank you very much!\nHowever, NonCommAlg, correct me if I'm mistaken - but the fact that you find this sort of linear presentation of a and n that equals 1 doesn't mean that 1 is their gcd...\n\n5. Originally Posted by aurora\nThank you very much!\nHowever, NonCommAlg, correct me if I'm mistaken - but the fact that you find this sort of linear presentation of a and n that equals 1 doesn't mean that 1 is their gcd...\nYou are mistaken..\n\nIf gcd(n, ab) = d > 0 then d|n and d|ab => d| (kr)ab + (ksb + l)n => d|1 => d = 1\n\nHowever note that NonCommAlg's trick only works for gcd of 1\n\n6. Originally Posted by Isomorphism\nYou are mistaken..\n\nIf gcd(n, ab) = d > 0 then d|n and d|ab => d| (kr)ab + (ksb + l)n => d|1 => d = 1\n\nHowever note that NonCommAlg's trick only works for gcd of 1\nI agree. However, in my algebra class, my professor would require that you show the extra step (what you just did), so i see where aurora is coming from. i would have done it the way NonCommAlg did it. but i like the contradiction method, it seems elegant to me\n\n7. Thank you guys\n\n8. Originally Posted by Jhevon\nI agree. However, in my algebra class, my professor would require that you show the extra step (what you just did), so i see where aurora is coming from. i would have done it the way NonCommAlg did it. but i like the contradiction method, it seems elegant to me\nI like the contradiction method too. But I was telling aurora about the idea because its useful in an exam. I think its more mechanical.Contradiction may not hit at the right time\n\n9. Originally Posted by Isomorphism\nI like the contradiction method too. But I was telling aurora about the idea because its useful in an exam. I think its more mechanical.Contradiction may not hit at the right time\nI agree. that's one of the reasons i like it. it is not so obvious at the moment you see the problem. NonCommAlg's way is something you would do on impulse to see how it works out, which is the charm of that method. good for tests when you can't think clearly or be fancy, but are racing against the clock and just need to get the job done", "date": "2017-06-27 09:42:19", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/40712-sigh-need-prove-somthing.html", "openwebmath_score": 0.7904345393180847, "openwebmath_perplexity": 1466.1681239051588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9902915258107348, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8638782044908462}}
{"url": "https://math.stackexchange.com/questions/968206/prove-the-fractional-field-of-an-integral-domain-is-the-smallest-field-containin", "text": "# Prove the fractional field of an integral domain is the smallest field containing the integral domain\n\nI have two questions about the fractional field of an integral domain.\n\nGiven an integral domain $D$:\n\n1. Is there a difference between saying \"the fractional field of $D$ is the smallest field containing $D$\" or \"the fractional field of $D$ is the smallest field containing an embedding of $D$\"?\n\n2. How do you prove that the fractional field is the smallest field containing $D$ (or an embedding of $D$, if there is a difference...)? Specifically, I want to show that if $F$ is any field containing $D$, then $F$ must contain the fractional field of $D$.\n\n\u2022 What is your definition of \"fractional field of integer domain\"? For me, it is precisely the minimal field containing the domain. \u2013\u00a0Timbuc Oct 11 '14 at 18:52\n\u2022 @timbuc *integral domain. My definition is based on the construction. First, given an integral domain $D$, we define the set of ordered pairs where the second coordinate is non-zero. Then we define an equivalence relation where $(a,b) \\sim (c,d)$ if $ad = bc$. Then we call the equivalence class containing $(a,b)$ as $\\frac{a}{b}$. Then we define addition and multiplication on this set of equivalence classes in the same way as it is defined in $\\mathbb{Q}$. This forms a field, what we call the fractional field. \u2013\u00a0layman Oct 11 '14 at 18:55\n\u2022 Ok, but then it is trivial, isn't it? I mean, any field containing $\\;D\\;$ must contain all the multiplicative inverses of non-zero elements $\\;d\\in D\\;$ and thus their product by any element in $\\;D\\;$ , and this means (by the definition!) that $\\;d_1\\cdot\\frac1{d_2}:=\\frac{d_1}{d_2}\\;$ is in the field, for any $d_1,d_2\\in D\\;,\\;\\;d_2\\neq 0\\;$ , which means any such field contains the fractions field of $\\;D\\;$ . \u2013\u00a0Timbuc Oct 11 '14 at 18:57\n\u2022 @Timbuc Well now I need to know the answer to my first question. Is there a difference between saying a field contains $D$ and saying it contains an embedding of $D$? \u2013\u00a0layman Oct 11 '14 at 19:00\n\u2022 Well, formally there is, yet for most usual cases one doesn't usually pay attention to that slight difference. \u2013\u00a0Timbuc Oct 11 '14 at 19:01\n\nLet $F'$ be a smallest field containing an embedding of $D$ ($f:D\\to F'$), and $F$ a field of fraction of $D$.\n\nWe can extend $f$ to morphism of field $\\tilde f:F\\to F'$ by $\\tilde f(a/b)=f(a)/f(b)$.\n\nNow we have that $\\tilde f(F)\\subseteq F'$ and $\\tilde f(F)$ containing an embedding of $D$ , by smallest property we have $\\tilde f(F)=F'$.\n\nSo the tow fields $F$ and $F'$ are isomorphic.\n\nEdit: If $F$ is any field containing $D$. And denote $K$ the field of fraction of $D$.\n\nLet $a/b\\in K$, $a\\in D$ and $0\\neq b\\in D$, hence $a,b\\in F$, it follow that $a$ and $1/b$ are in $F$ so $a. (1/b)=a/b\\in F$. Thus $K\\subseteq F$.\n\n\u2022 Is your extension of $f$ well defined? \u2013\u00a0layman Oct 11 '14 at 19:26\n\u2022 I guess it is since $f$ is a homomorphism, and if $a/b = c/d$, then $ad = bc$, so $f(a)f(d) = f(b)f(c)$ which implies $f(a)/f(b) = f(c)/f(d)$. \u2013\u00a0layman Oct 11 '14 at 19:28\n\u2022 Yes $b\\neq 0$ implies $f(b)\\neq 0$ because $f(0)=0\\neq f(b)$ ($f$ is injective). \u2013\u00a0Hamou Oct 11 '14 at 19:28\n\u2022 I guess I would need to do a little bit more work because I would need to show the image of the field $F$ under the extension of $f$ is a field. \u2013\u00a0layman Oct 11 '14 at 19:29\n\u2022 When $\\frac ab\\neq 0$, $\\tilde f(a/b)\\tilde f(b/a)=1$, hence $\\tilde f(a/b)$ is invertible is in $\\tilde f(F)$. \u2013\u00a0Hamou Oct 11 '14 at 19:32\n\nThe right (i.e. categorical) way to say this (without the ambiguities of words like \"smallest\", \"containing\", etc.) ought to be that the inclusion $\\iota: D\\to Q(D)$ has the following universal property:\n\nIf $K$ is a field, and $f: D\\to K$ is any morphism of rings, then there is a unique morphism of fields $g : Q(D) \\to K$ such that $f = g \\circ \\iota$.\n\n(In particular, $Q(D)$ embeds into any field that $D$ embeds into.)\n\nThis property uniquely determines (up to isomorphism) not only $Q(D)$, but $\\iota$ as well.\n\nAnd it's easily proved, since $g(1/b)g(b)=g(1)$ forces $g(a/b) = f(a)/f(b)$, so this amounts to checking that $a/b \\mapsto f(a)/f(b)$ is actually a homomorphism.", "date": "2020-01-27 03:23:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/968206/prove-the-fractional-field-of-an-integral-domain-is-the-smallest-field-containin", "openwebmath_score": 0.9469299912452698, "openwebmath_perplexity": 110.9001599027239, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924769600771, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8638668668346455}}
{"url": "https://math.stackexchange.com/questions/2946789/finding-the-local-maxima-minima-of-the-following-function", "text": "# Finding the local maxima/minima of the following function\n\n$$f: \\mathbb{R}\\rightarrow \\mathbb{R}$$. Then determining if each of the solutions is a global maximum or a global minimum. $$f(x) = \\frac{x^{4}}{4} - 2x^{3} + \\frac{11}{2}x^{2} - 6x + 2$$ for all $$x \\in \\mathbb{R}$$.\n\nI have used the synthetic division to find this equation: $$\\left ( x^{2} - 5x + 6 \\right )\\left ( x - 1 \\right ) = 0$$\n\nSo the critical points are $$x^{*} = 1,2,3$$.\n\nI know that if $$x^{*} = 1$$, then $${f}''\\left ( 1 \\right ) = 2 > 0$$, so this is minimum.\n\nIf $$x^{*} = 2$$, then $${f}''\\left ( 2 \\right ) = 1 < 0$$, so this is maximum.\n\nIf $$x^{*} = 3$$, then $${f}''\\left ( 3 \\right ) = 2 > 0$$, so this is minimum.\n\nHow do I determine whether these solutions are local max/min and global max/min?\n\nAre all of them global solutions as $$x \\in \\mathbb{R}$$?\n\nEDIT: I have plugged the values of the $$x's$$ in the main function.\n\nFor $$x=1$$ and $$x=3$$, $$f(1)= -0.25 = f(3)$$, and for $$x=2$$, $$f(2)= 0$$.\n\nSo is $$x=1,3$$ a local minimum and $$x=2$$ a global maximum?\n\n\u2022 Use the 2nd derivative. \u2013\u00a0Wuestenfux Oct 8 '18 at 7:05\n\u2022 Local max/min: second derivative. Global Max/Min: compare values or find inequality. \u2013\u00a0Andreas Oct 8 '18 at 7:05\n\u2022 Have used the second derivative. So are all of them local solutions? I think all of them are global solutions as $x \\in \\mathbb{R}$. \u2013\u00a0OGC Oct 8 '18 at 7:06\n\nFrom the second derivative test the extremum points that you have found are all local. Note that $$\\lim_{x\\to \\pm\\infty}f(x)=+\\infty$$, so $$x=1$$ is not a global maximum point. On the other hand, since $$f(1)=f(3)=-1/4$$, it follows that $$x=1$$ and $$x=3$$ are global minimum points.\n\u2022 Without drawing the graph, how would I figure out that $x=1$ and $x=3$ are global minimum points? \u2013\u00a0OGC Oct 8 '18 at 7:19\n\u2022 Since $f$ is differentiable in $\\mathbb{R}$, and $\\lim_{x\\to \\pm\\infty}f(x)=+\\infty$, any global minimum point is also a critical point. \u2013\u00a0Robert Z Oct 8 '18 at 7:23", "date": "2019-08-21 02:58:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2946789/finding-the-local-maxima-minima-of-the-following-function", "openwebmath_score": 0.8687475919723511, "openwebmath_perplexity": 179.73060074189183, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988312744413962, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8638616597689354}}
{"url": "http://math.stackexchange.com/questions/142064/how-do-you-solve-this-sort-of-definite-integral", "text": "# How do you solve this sort of definite integral?\n\nCan someone walk me through how to do the following problem so I can attempt a few more practice problems?\n\nIf: $$\\int_{1}^{5} f(x) dx = 12$$ and $$\\int_{4}^{5} f( x) dx = 3.6$$ find: $$\\int_{1}^{4} f( x) dx$$\n\nWould it simply be $12 - 3.6$ ?\n\nEDIT\n\nIf: $$\\int_{0}^{9} f(x) dx = 37$$ and $$\\int_{0}^{9} g( x) dx = 16$$ find: $$\\int_{0}^{9} 2f(x)+3g(x) dx$$\n\nWould this simply be: $2 \\times 37 + 3 \\times 16$?\n\n-\nDo you know any theorems or identities that relate the values of integrals of the same function with different limits? \u2013\u00a0 MJD May 7 '12 at 4:18\nIf $a \\leq c \\leq b$, then $\\int_a^b=\\int_a^c+\\int_c^b$. Use this profitably. \u2013\u00a0 \uff2a. \uff2d. May 7 '12 at 4:19\nAs you said, it is simply 12-3.6. \u2013\u00a0 MJD May 7 '12 at 4:19\n@MarkDominus Wow, I didn't think it would be that simple. That's why I posted it up here. I'll post a second one up so it's not a waste of a question. \u2013\u00a0 justcheckingin May 7 '12 at 4:21\nRegarding the second problem: yes, that's right. \u2013\u00a0 Brian M. Scott May 7 '12 at 4:26\n\nThis is done using additivity of integration on intervals i.e. if $c \\in [a,b]$ and\n\n$\\displaystyle \\int_a^b f(x) dx$, $\\displaystyle \\int_a^c f(x) dx$ and $\\displaystyle \\int_c^b f(x) dx$ are well- defined, then $$\\int_a^b f(x) dx = \\int_a^c f(x) dx + \\int_c^b f(x) dx$$ Hence, in your case, you have that $$\\int_1^5 f(x) dx = \\int_1^4 f(x) dx + \\int_4^5 f(x) dx$$ Hence, we get that $$12 = \\int_1^4 f(x) dx + 3.6$$ i.e. $$\\int_1^4 f(x) dx = 8.4$$\n\nFor the second problem as Brian pointed out in the comments, it follows from linearity of integration i.e. $$\\int \\left( a f(x) + b g(x) \\right)dx = a \\int f(x)dx + b \\int g(x) dx$$ provided all the integrals are well-defined.\n\n-\n\n$$\\int_a^bf(x)dx+\\int_b^cf(x)dx=\\int_a^cf(x)dx$$\nThe integral in the interval $[1,4]$ is the difference: integral in $[1,5]$ - integral $[4,5]$.", "date": "2014-08-21 12:22:12", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/142064/how-do-you-solve-this-sort-of-definite-integral", "openwebmath_score": 0.9622020721435547, "openwebmath_perplexity": 564.9705017758622, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127433812522, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8638616588662673}}
{"url": "https://math.stackexchange.com/questions/1166737/general-formula-for-iterated-cumulative-sum", "text": "# General formula for iterated cumulative sum\n\nConsider the sequence $$S_0$$ consisting of ones:\n\n$$1,1,1,1,1,1,\\ldots$$\n\nNow compute the cumulative sum of this sequence, and call the resulting sequence $$S_1$$:\n\n$$1,2,3,4,5,6,\\ldots$$\n\nProceed iteratively to generate sequence $$S_2$$:\n\n$$1,3,6,10,15,21,\\ldots$$\n\nthen $$S_3$$:\n\n$$1,4,10,20,35,56,\\ldots$$\n\nand so on.\n\nIt is well known that each sequence $$S_k$$ can be represented by a $$k$$-degree polynomial $$P_k(n)$$. For the above sequences the polynomials are $$P_0(n) = 1$$ $$P_1(n) = n$$ $$P_2(n) = \\frac{n^2+n}{2}$$ $$P_3(n) = \\frac{n^3+3n^2+2n}{6}$$\n\nMy question: Is there a general formula for coefficients of the polynomial $$P_k$$? Or more generally, is there a formula to compute $$S_k(n)$$ as a function of $$n$$ and $$k$$? I mean a closed formula $$S_k(n) = f(k,n)$$ (not an iterative procedure such as the construction method I just described).\n\nIf that helps, actually I'm not interested in $$S_k(n)$$, but rather in the quotient $$S_k(n)/S_k(n+1)$$.\n\n\u2022 Just a test to see if I have understood: in $S_1$ should the last term be $21$? \u2013\u00a0User3773 Feb 26 '15 at 17:39\n\u2022 @Cla Oops. Yes, sorry. Corrected! \u2013\u00a0Luis Mendo Feb 26 '15 at 17:40\n\nThe numbers in sequence $S_k$ are the binomial coefficients $\\binom{m}{k}$; the $n$-th term of $S_k$ is $\\binom{n-1+k}{k} = \\frac{(n-1+k)!}{(n-1)!k!}$. One can prove this by using that $$\\binom{i}{i} + \\binom{i+1}{i} + \\cdots + \\binom{i+j}{i} = \\binom{i+j+1}{i+1}$$ for any $i,j \\geq 0$.\n\nFor $k=1$ we find $P_1(n) = \\binom{n-1+1}{1} = \\binom{n}{1} = n$, for $k=2$ we find $P_2(n) = \\binom{n-1+2}{2} = \\binom{n+1}{2} = \\frac{n(n+1)}{2}$, for $k=3$ we find $P_3(n) = \\binom{n-1+3}{3} = \\binom{n+2}{3} = \\frac{n(n+1)(n+2)}{6}$, etcetera.\n\n\u2022 Wow. It was really simple! Could you provide any reference? \u2013\u00a0Luis Mendo Feb 26 '15 at 17:43\n\u2022 Thanks a lot for your help. I've posted a related question here \u2013\u00a0Luis Mendo Feb 26 '15 at 18:47\n\nWe can generalize your problem to arbitrary initial sequences.\n\nLet $a_n\\,(n=0,1,2,3,...)$ be a sequence of numbers. Define its iterated partial sums using the recurrence $$S^{(0)}_n=a_n,\\quad S^{(k+1)}_n=\\sum_{m=0}^n S^{(k)}_m,\\tag1$$ so that we have, for example, $$\\small\\begin{array} &S^{(0)}_0=\\color{green}{a_0}, &S^{(0)}_1=\\color{blue}{a_1}, &S^{(0)}_2=\\color{maroon}{a_2},&...\\\\ S^{(1)}_0=\\color{green}{a_0}, &S^{(1)}_1=\\color{green}{a_0}+\\color{blue}{a_1}, &S^{(1)}_2=\\color{green}{a_0}+\\color{blue}{a_1}+\\color{maroon}{a_2},&...\\\\ S^{(2)}_0=\\color{green}{a_0}, &S^{(2)}_1=\\color{green}{a_0}+(\\color{green}{a_0}+\\color{blue}{a_1}), &S^{(2)}_2=\\color{green}{a_0}+(\\color{green}{a_0}+\\color{blue}{a_1})+(\\color{green}{a_0}+\\color{blue}{a_1}+\\color{maroon}{a_2}),&...\\\\ S^{(3)}_0=\\color{green}{a_0}, &S^{(3)}_1=\\color{green}{a_0}+(\\color{green}{a_0}+(\\color{green}{a_0}+\\color{blue}{a_1})),&... \\end{array}\\tag2$$ Now we can prove by induction that the following formula holds: $$S^{(k+1)}_n=\\sum_{m=0}^n\\binom{m+k}k\\,a_{n-m}.\\tag3$$\n\n\u2022 Vladimir, I've just given an approach which generalizes your solution even more. Please see my new answer. \u2013\u00a0Gottfried Helms Aug 29 '19 at 6:55\n\nA more generalized solution, where even fractional iteration-heights $$h$$ for $$S_n^{(h)}$$ become possible, can be found using a matrix-ansatz. Consider the matrix-equation $$D \\cdot A = S^{(1)}(A) \\tag 1$$ where $$D$$ is the lower triangular unit-matrix $$D= \\Tiny \\begin{bmatrix} 1&.&.&.&\\cdots\\\\1&1&.&.&\\cdots\\\\1&1&1&.&\\cdots\\\\ 1&1&1&1&\\cdots\\\\ \\vdots&\\vdots&\\vdots&\\vdots&\\ddots\\\\ \\end{bmatrix} \\tag 2$$ then of course \\small \\begin{align} D^2 \\cdot A &= S^{(2)}(A) \\\\ D^3 \\cdot A &= S^{(3)}(A) \\\\ \\vdots \\\\ D^h \\cdot A &= S^{(h)}(A) \\\\ \\end{align} \\tag 3 The $$h$$'th power of $$D$$ can be computed using $$L = \\log(I + (D-I))$$ and $$\\exp(L)$$ using the series-representation of this functions (which reduce to finite sums in the case of using $$D$$). We get formally\n\n$$\\displaystyle \\qquad \\qquad \\Large{D^h =}$$ $$\\tag 4$$\n\nand where we need only document the entries of the first column because of the schematic form of $$D^h$$:\n\n$$\\displaystyle \\qquad \\qquad S_n^{(h)} (A) = \\sum_{c=0}^n D_{n,c} \\cdot A[c] = \\sum_{r=0}^n D_{n-r,0} \\cdot A[r] \\tag 5$$\n\nThe coefficients $$D_{r,0}$$ might look abscure, but can easily be described when factorials are extracted:\n\n$$\\displaystyle \\qquad \\qquad \\large {S_5^{(h)}(A)=}$$ $$\\tag 6$$\n\nand even simpler\n\n$$\\displaystyle \\qquad \\qquad \\large {S_5^{(h)}(A)=}$$ $$\\tag 7$$\n\nThe coefficients in the previous representation are the unsigned Stirlingnumbers $$1$$'st kind and for integral $$h$$ this gives of course the appropriate binomial-expressions which are noted in the other answers and comments.\nBut we can easily insert fractional $$h$$ as well!\n\nConcerning your last question, the quotient.\n\nExample, let $$k=3$$ then the quotient can be found by multiple cancellations: $$P_3(n) = {n^3+3n^2+2n\\over 6} = {(n+2)(n+1)n\\over 6} \\\\\\ P_3(n+1) = {(n+1)^3+3(n+1)^2+2(n+1)\\over 6} = {(n+3)(n+2)(n+1)\\over 6} \\\\\\ {P_3(n+1)\\over P_3(n)} ={{(n+3)(n+2)(n+1)\\over 6}\\over {(n+2)(n+1)n\\over 6} } = {n+3\\over n}$$ Thus in general: $${P_k(n+1)\\over P_k(n)} ={{(n+k)\\cdots(n+2)(n+1)\\over k!}\\over {(n+k-1)\\cdots(n+1)n\\over k!} } = {n+k\\over n}$$", "date": "2020-01-20 06:26:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1166737/general-formula-for-iterated-cumulative-sum", "openwebmath_score": 0.999718964099884, "openwebmath_perplexity": 411.8128985669469, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127426927789, "lm_q2_score": 0.8740772220439509, "lm_q1q2_score": 0.8638616566435422}}
{"url": "https://math.stackexchange.com/questions/155952/arcwise-connected-part-of-mathbb-r2", "text": "# Arcwise connected part of $\\mathbb R^2$\n\nHere's a question that I share: Show that if $D$ is a countable subset of $\\mathbb R^2$ (provided with its usual topology) then $X=\\mathbb R^2 \\backslash D$ is arcwise connected.\n\n\u2022 This is even true if you only require $D$ to have lebesgue-measure 0: math.stackexchange.com/questions/77791/\u2026 Jun 9 '12 at 9:20\n\u2022 I have two further questions: 1) Is this result holds in higher dimension i.e., on $\\mathbb{R}^n$, where $n \\in \\mathbb{N}$. 2) Can one characterize the topological space for which this kind of properties hold $?$ May 22 '13 at 18:47\n\u2022 @TeresaLisbon I first asked it as new question but it was closed and repititive requests to open it were discarded in CURED chatroom and they suggested that new questions shouldn't be asked but bounty should be put on old question. Which I did now. Sep 23 '21 at 6:02\n\u2022 Just for reference, the question that was closed as a duplicate is If A is countable subset of $\\mathbb{R}^2$ , then $\\mathbb{R}^2 - A$ is pathwise connected.. The relevant discussion in the chatroom CURED can be found here: Request for reopening //q/3973870. Sep 23 '21 at 6:57\n\u2022 @No-One Got it, thanks. I think what I'll do, if I plan to answer this, is place your bounty text in my answer, to make sure that my answer targets your query, along with the question in general. Sep 23 '21 at 9:20\n\nHINT: Not only is $\\Bbb R^2\\setminus D$ arcwise connected, but you can connect any two points with an arc consisting of at most two straight line segments.\n\nSuppose that $p,q\\in\\Bbb R^2\\setminus D$. There are uncountably many straight lines through $p$, and only countably many of those lines intersect $D$, so there are uncountably many straight lines through $p$ that don\u2019t hit $D$. Similarly, there are uncountably many straight lines through $q$ that don\u2019t hit $D$. Can you finish it from here?\n\n\u2022 Yes, I can : two of them at lest intersect. One auther way is to consider the bissector $\\Delta$ of $[ab]$ and all straight lines $[p,\\delta]\\cup[\\delta,q]$ where $\\delta \\in \\Delta$, On of them at lest is a subset of $X$ Jun 9 '12 at 9:32\n\u2022 @Mohamed: Exactly. (Or if you\u2019re really lucky, one of them is the line through $p$ and $q$.) Jun 9 '12 at 9:37\n\u2022 I have not thought about this one!! Thank's!! (Edit [p,q] not [a,b] ) Jun 9 '12 at 9:51\n\u2022 From here. In order to show path $f:[0,1]\\to \\Bbb{R}^2-A$ is continuous.we can construct a function: $f(t)=f_1(t)$ ,if $t\\in [0,1/2]$ and $f(t)=f_2(t)$ ,if $t\\in [1/2,1]$. $f$ is continuous by pasting lemma. And we're done. Am I right?@Briam M Scott Jul 16 '20 at 11:57\n\u2022 @AmanPandey: Yes, that\u2019s right. Jul 16 '20 at 16:35\n\nAny two points in $\\mathbb R^2\\setminus D$ are connected by uncountably many disjoint arcs of circles, of which only countably many may intersect $D$.\n\n(This is adapted from one of my student's solutions to a related exam problem.)\n\nLet $$p, q$$ be two distinct points in $$\\mathbb{R}^2 \\setminus D$$. If the straight line segment joining $$p$$ to $$q$$ lies in $$\\mathbb{R}^2 \\setminus D$$, then you are done: one possible parametrization of the path is $$\\gamma(t) = (1 - t)p + tq$$, $$0 \\leq t \\leq 1$$.\n\nSo, suppose that the above does not work. Now, note that there are uncountably infinitely many straight lines passing through the point $$p$$. Since $$D$$ is only countable, in particular there are uncountably infinitely many straight lines passing through $$p$$ and not intersecting $$D$$. Fix any one such line $$L_1$$. Once more, there are uncountably infinitely many straight lines passing through $$q$$ that also intersect $$L$$. In particular, there are uncountably infinitely many such lines that also do not intersect $$D$$. Fix any one such line $$L_2$$.\n\nSuppose that $$L_1$$ and $$L_2$$ intersect at the point $$r$$. Then, you can take the path joining $$p$$ and $$q$$ to be the one that starts at $$p$$, traverses along the line $$L_1$$ to the point $$r$$, then traverses along the line $$L_2$$ to the point $$q$$. One possible parametrization of this path is $$\\gamma(t) = \\begin{cases} (1 - 2t)p + 2tr, & 0 \\leq t \\leq \\tfrac{1}{2};\\\\ (2 - 2t)r + (2t - 1)q, & \\tfrac{1}{2} \\leq t \\leq 1. \\end{cases}$$ Note that the pasting lemma shows that $$\\gamma$$ is indeed continuous.\n\nAs for how one arrives at these specific parametrizations, one first needs to know the parametric equation of a straight line. For instance, see Parametric form of a line for a quick refresher. Secondly, one needs to adjust the parameter $$t$$ by scaling and translating in order to get it to lie in the \"correct\" range for one's purposes. This is another linear change, so it should not be too difficult.\n\nBut, just to belabor the point further, here's one way you can mentally figure out the parametrization $$\\gamma$$ in the second case. To join the points $$p$$ and $$r$$, we want to vary $$t$$ from $$0$$ to $$1/2$$ such that at $$t = 0$$ we are at $$p$$ and at $$t = 1/2$$ we are at $$r$$. So, if the parametrization for this part looks like $$\\gamma(t) = f(t)p + g(t)r$$, then we want $$f(0) = 1$$ and $$f(1/2) = 0$$, and we also want $$g(0) = 0$$ and $$g(1/2) = 1$$. Can we find linear functions $$f(t) = \\alpha_1 t + \\beta_1$$ and $$g(t) = \\alpha_2 t + \\beta_2$$ that satisfy these conditions? Sure, just substitute those values of $$(t, f(t))$$ and $$(t, g(t))$$ that we wrote down for $$t = 0, 1/2$$ into the respective equations to solve for $$\\alpha_i, \\beta_i$$. Repeat the process for the line joining $$r$$ to $$q$$, and you are done.\n\n\u2022 For completeness, you should at least state that your path is also injective, which implies that it defines an embedding $[0,1]\\to X$, as required by the definition of an arcwise connected space. Sep 28 '21 at 17:57\n\nOk I will take a stab at clarifying Brian M. Scott's Answer instead of writing a fully different answer.\n\nLet $$p,q\\in \\mathbb{R}-D$$. Then Define $$A:=\\{\\text{ Lines Through }p\\}$$, $$B:=\\{ \\{\\text{ Lines through }q\\}$$.\n\nOne could write the definitions of these sets more formally if desired, but the meaning is clear. I claim the following: $$\\exists l\\in A, \\text{ such that } l\\cap D=\\emptyset.$$ Similarly $$\\exists n\\in B, \\text{ such that } l\\cap D=\\emptyset.$$ Lets us understand why this should be clear. Now define a mapping $$m:\\mathbb{R} \\to A$$ by $$m(x):=$$ the line in $$A$$ with slope $$x$$. This mapping is clearly $$1-1$$ and onto (almost see below*), as any line through a point $$p$$ is determined by its slope, and any slope uniquely determines a line through $$p$$. You could formalize this more by writing any line through $$p$$ in point slope form, but I leave that to the reader.\n\nA similar argument follows for $$B$$, thus $$||A||=||B||=||\\mathbb{R}||$$.\n\nNow let us show that there are lines in $$A$$ and $$B$$ which do not contain points in $$D$$. define a map $$\\phi:D\\to A\\times B$$, by $$\\phi(d):=\\{(l,m)\\in A\\times B| d=l\\cap m\\}$$. That is each point $$d\\in D$$ gets mapped to the pair of lines in $$A$$ and $$B$$ which meet at $$d$$. This mapping is $$1-1$$, as any two points determine a unique line. By assumption $$D$$ is countable, therefor $$\\phi$$ cannot be onto thus there exists a pair of lines, $$(l^*,m^*)$$, which do not contain any points in $$D$$. This follows as $$A\\times B- \\phi(D)\\neq \\emptyset$$.\n\nwe have a piecewise linear path $$p \\to l^*\\cap m^* \\to q$$. QED.\n\nThe Crux of the argument, here is where we assert that the mapping $$\\phi$$ cannot be onto, given the assumption that $$D$$ was countable.\n\nOne small note, the map $$m$$ is not quite onto, as the vertical line through $$p$$ has undefined slope. However, this doesn't disturb the argument as the main idea is that $$A$$ and $$B$$ are both uncountable. We might also be able to circumvent this by taking a mapping from the extended reals.\n\n\u2022 Great stuff, really nice answer, thanks for adding the clarification in the first line. Sep 25 '21 at 12:32", "date": "2022-01-24 17:09:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/155952/arcwise-connected-part-of-mathbb-r2", "openwebmath_score": 0.9182594418525696, "openwebmath_perplexity": 160.33141424385343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631623133666, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8638440361882054}}
{"url": "https://mathoverflow.net/questions/385167/are-at-most-1-3-vertices-kings?noredirect=1", "text": "# Are at most $1/3$ vertices \"kings\"?\n\nIf $$G=(V,E)$$ is a finite, simple, undirected graph, and $$v\\in V$$, we set $$N(v) = \\{w\\in V:\\{v,w\\}\\in E\\}$$, and $$\\text{deg}(v)= |N(v)|$$. We say a vertex $$v\\in V$$ is a king if $$\\text{deg}(v) > \\text{deg}(w)$$ for all $$w\\in N(v)$$.\n\nIn the graph $$G=(\\{0,1,2\\}, \\big\\{\\{0,1\\}, \\{1,2\\}\\big\\})$$, one of the $$3$$ vertices is a king. Let $$\\text{King}(G)$$ be the set of king vertices.\n\nQuestion. Is it true that for any finite connected graph $$G=(V,E)$$ with $$|V|>1$$ we have $$|\\text{King}(G)|/|V|\\leq 1/3$$? If not, how large can this value get?\n\n\u2022 A complete bipartite graph $G = U \\cup V$ with $|U| = n$ and $|V| = n-1$ has $n - 1$ kings out of $2n - 1$ total vertices. Feb 28 at 14:16\n\u2022 maybe related: mathoverflow.net/questions/343607/\u2026 Mar 1 at 15:14\n\nFor this discussion I am assuming we do not consider isolated vertices to be \"Kings\", even though technically your definition considers them to be so in a vacuous sense (I guess this convention goes back to Shakespeare). Otherwise of course one can make every vertex a king by having no edges whatseover.\n\nFor the matching lower bound, observe that no two kings can be adjacent, and if there is at least one king, the set $$E'$$ of ordered pair edges $$(v,w)$$ in $$E$$ with $$v \\in \\mathrm{King}(G)$$ and $$w \\not \\in \\mathrm{King}(G)$$ is non-empty. Now we do weighted double counting: \\begin{align*} \\# \\mathrm{King}(G) &= \\sum_{v \\in \\mathrm{King}(G)} 1 \\\\ &= \\sum_{(v,w) \\in E'} \\frac{1}{d(v)}\\\\ &< \\sum_{(v,w) \\in E'} \\frac{1}{d(w)} \\\\ &\\leq \\sum_{w \\in V \\backslash \\mathrm{King}(G)} 1 \\\\ &= \\#V - \\# \\mathrm{King}(G) \\end{align*} hence $$\\# \\mathrm{King}(G) < \\frac{1}{2} \\# V.$$ Of course the same claim holds when there are no kings as long as the graph is not the empty graph. So this shows that the lower bound provided by the complete bipartite graph examples (adding an isolated vertex in the case when one wants an even number of vertices) are completely optimal: the maximal number of kings in a graph on $$n$$ vertices is $$\\max( \\lfloor \\frac{n-1}{2} \\rfloor, 0)$$.\n\nThis bound can also be viewed as quantifying a variant of the \"friendship paradox\". (Based on this connection, I propose \"influencer\" as a more modern and gender-neutral terminology alternative to \"king\".)\n\n\u2022 There is already the gender-neutral \"celebrity\" which appears to be motivated by interpreting the vertex degree as the number of people by which a person is known. Mar 1 at 15:20\n\u2022 Furthermore the terminology \"king\" is used with a different meaning in oriented tournaments (these are vertices with maximum out-degree). Mar 2 at 4:14\n\nThe fact that the number of kings is less than the number of non-kings is essentially equivalent to the following problem, proposed by Alexander Razborov to Tournament of Towns in 1990:\n\nGiven an $$m\\times n$$ matrix, $$m. Some entries are starred, and each column contains a starred entry. Then there exists a star whose row contains more stars than its column.\n\n(Here columns correspond to kings, rows to non-kings, stars to edges, and the strictness of inequalities is different.)\n\nI wanted to find a proof that uses Hall's marriage theorem [1] instead of double-counting.\n\nGiven a graph $$G=(V,E)$$, let $$K$$ be the set of kings in $$V$$, and $$R:=V \\setminus K$$ the rest.\n\nClaim: $$|K| < |R|$$.\n\nProof Let $$G=(V,E)$$ be a counterexample that minimizes the sum $$|V|+|E|$$, so $$|K| \\ge |R|$$. Then $$G$$ is bipartite, since any edge between nodes in $$R$$ can be removed. If $$|K|>|R|$$ then removing one king would yield a smaller counterexample, so $$|K|=|R|$$. If there was a subset $$S$$ of $$K$$ where its neighborhood satisfies $$|N(S)|< |S|$$, then the induced graph on $$S \\cup N(S)$$ would be a smaller counterexample. Thus the Hall condition is met in $$G$$. Removing from $$G$$ a perfect matching of $$K$$ to $$R$$ yields a smaller counterexample.\n\nIf we consider the complete bipartite graph $$K_{n,m},$$ where $$n,m$$ are natural numbers with $$n < m,$$ then the maximum degree of a vertex in this graph is $$=m$$ and there are at least $$n$$ many vertices having degree $$m$$, and so the number of \"kings\" (as per the definition above) in this graph is $$= n,$$ and the total number of vertices $$=n+m.$$\nNow, since $$n,m$$ can be anything (with $$n \\le m$$), so one can take $$n,m$$ to be such that $$n/(n+m) > \\frac{1}{3},$$ and thus in this case we see that the number of kings in this graph is $$> \\frac{1}{3} \\cdot$$ the number of vertices in the graph.\nFor such graphs you can just set $$n=m-1$$ to see that you cannot get an \"asymptotic\" bound of $$< 1/2.$$\n\u2022 If $n=m$ then there are no kings. The best this bipartite construction can do is $\\lceil n/2\\rceil$ many kings, as Nik Weaver has also pointed out. Feb 28 at 15:58", "date": "2021-11-27 23:58:53", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/385167/are-at-most-1-3-vertices-kings?noredirect=1", "openwebmath_score": 0.9982947707176208, "openwebmath_perplexity": 536.8399596769223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631639168357, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8638440279990759}}
{"url": "https://math.stackexchange.com/questions/2578923/why-is-the-integral-of-frac1x2-from-1-to-infty-not-the-same-as-the-in", "text": "Why is the integral of $\\frac1{x^2}$ from $1$ to $\\infty$ not the same as the infinite sum from $1$ to $\\infty$?\n\nStudying series I am a bit confused on this point. The infinite sum of $1/x^2$ from $1$ to $\\infty$ was proved by Euler to be $\\pi^2$ divided by $6$:\n\n$$\\sum_{x=1}^\\infty\\frac 1 {x^2}=\\frac {\\pi^2} 6$$\n\nBut if I integrate from $1$ to $\\infty$ of the same entity namely $1/x^2$ it is $1$. Correct..? Unless I did it wrong. $$\\int_1^\\infty\\frac 1 {x^2}dx=1$$ How can this be since by integrating it seems we are adding a lot more numbers to cover the same area so we should by all means get the same thing or something at least as large as $\\pi^2/6$?\n\n\u2022 Why would anyone expect them to be the same? Dec 24, 2017 at 16:49\n\u2022 @LordSharktheUnknown, please read his entire question. Dec 24, 2017 at 16:54\n\u2022 The relationship between the sum and the integral is given by en.wikipedia.org/wiki/Euler\u2013Maclaurin_formula. they differ because rectangles are not curved. The area in the rectangles is not the area under the curve. Dec 24, 2017 at 16:54\n\u2022 @Sedumjoy, please read your edited question for how to write the integral and sum :). Dec 24, 2017 at 17:00\n\u2022 I agree with @LordSharktheUnknown, it is not natural to expect that $$f(x)=\\frac{1}{x^2},\\qquad g(x)=\\frac{1}{\\lfloor x\\rfloor ^2}$$ have the same integral over $(1,+\\infty)$, also because $g(x)\\geq f(x)$. Dec 24, 2017 at 17:55\n\nNote that $$\\int_1^\\infty \\frac{1}{x^2}\\leq\\sum_1^\\infty \\frac{1}{n^2}\\tag{1}$$ by considering a Riemann sum with left endpoints. Here is a picture (for the case of $1/x$ but a similar picture can be drawn for this case as well). See this picture. Image credits go to Wikipedia.\n\n\u2022 +1 For the picture.. It illustrates the point better :) I will add it to my answer\n\u2013\u00a0Ant\nDec 24, 2017 at 16:59\n\u2022 no its the wrong graph, should be a graph of $\\frac{1}{x^2}$, right? Dec 24, 2017 at 17:00\n\u2022 @FoobazJohn, Just wondering, what program did you use to make this picture? Dec 24, 2017 at 17:00\n\u2022 @mathreadler, yeah you're right, though the basic insight remains the same. Dec 24, 2017 at 17:01\n\u2022 @mathreadler Yes I commented that it is for $1/x$ but a similar picture can be drawn for the case $1/x^2$. Dec 24, 2017 at 17:01\n\nWhen you do the sum, you sort of approximate the area with rectangles of base length equal to $1$. Draw the function $1/x^2$ and draw the rectangles with base length 1; you'll see that the area under the rectangles is much bigger than the area under the function $1/x^2$\n\nHere is an illustration for the function $1/x$, but it's essentially the same as in the $1/x^2$ case. (Thanks to @FoobazJohn) The integral adds a lot more numbers, but these numbers are multiplied by something very small. The end result is that the integral represents the area under the $1/x^2$ curve, which is less than the area under the rectangles.\n\nLet us compare $1$ and $\\int_1^2\\frac1{x^2}\\,\\mathrm dx$. Since $\\bigl(\\forall x\\in(1,2]\\bigr):\\frac1{x^2}<1$, $\\int_1^2\\frac1{x^2}\\,\\mathrm dx<1$. For the same reason, $\\int_2^3\\frac1{x^2}\\,\\mathrm dx<\\frac14$, $\\int_3^4\\frac1{x^2}\\,\\mathrm dx<\\frac19$, and so on. So$$1=\\int_1^\\infty\\frac1{x^2}\\,\\mathrm dx<\\sum_{n=1}^\\infty\\frac1{n^2}=\\frac{\\pi^2}6.$$\n\nNote that it is not true that $\\int_a^bf(x)\\,\\mathrm dx$ is the sum of all numbers $f(x)$ with $x\\in[a,b]$. Instead, it is the average value of $f$ in $[a,b]$ times $b-a$.\n\n\u2022 all three answers are very good in explaining this and I will pick one to close the problem but they are equally good....I feel stupid for not seeing this that I had to ask.... Dec 25, 2017 at 0:59", "date": "2022-08-19 21:41:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2578923/why-is-the-integral-of-frac1x2-from-1-to-infty-not-the-same-as-the-in", "openwebmath_score": 0.7699971795082092, "openwebmath_perplexity": 195.1981505909385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631643177029, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8638440171578182}}
{"url": "https://math.stackexchange.com/questions/1681268/help-me-find-mistake-in-exact-differential-equation", "text": "# Help Me Find Mistake in Exact Differential Equation\n\nI am working on exact differential equations and I just cannot seem to understand them and was hoping to have my method checked and please provide feedback on that.\n\n$$(18xy^2 - \\sin(x))dx + (8 + 18x^2y)dy = 0;\\,\\, y(0) = 1$$ I first calculated the crossed partial derivatives for both terms to check they were equal which I found they both came out to be 36xy.\n\nThen I have said that $$\\frac{\\partial u}{\\partial x} = 18xy^2 - sin(x)$$ and therefore $$u = 9y^2 x^2 + cos(x) + h(y)$$\n\nNow I have said that $$\\frac{\\partial u}{\\partial y} = 18yx^2 + h'(y) = 8 + 18yx^2$$ which I believe is not a problem.\n\nThis says that h'(y) = 8 which I have then integrated with respect to y and found h(y) = 8y + C\n\nI believe this means my final result for the answer which in our notes is always denoted as $$u(x,y) = 36xy + 8y + C$$\n\nSo far I cannot understand how these exact differential equations work and have been told my answer is wrong. I am yet to find the exact solution using the initial condition as I am unsure how to do that as well because I am confused by a few things with these problems however if possible could we please discuss the method how to solve these types of DEs and check to see if my answer is correct.\n\nThank you very much,\n\nMichael\n\n\u2022 Micheal, I have started the edits so take a look at the first question and hopefully you can fix the other equations? All details can be found here. \u2013\u00a0Chinny84 Mar 3 '16 at 10:18\n\u2022 Thanks I wasn't sure until now how to properly do the formatting thank you. \u2013\u00a0Michael Mar 3 '16 at 10:46\n\u2022 No problem. I think the site experience will become even better for you now :) \u2013\u00a0Chinny84 Mar 3 '16 at 10:52\n\n$$\\left(18xy(x)^2-\\sin(x)\\right)\\space\\text{d}x+\\left(8+18x^2y(x)\\right)\\space\\text{d}y=0$$\n\nLet $\\text{P}(x,y)=18xy^2-\\sin(x)$ and $\\text{Q}(x,y)=18x^2y+8$.\n\nThis is an exact equation, because $\\frac{\\partial\\text{P}(x,y)}{\\partial y}=36xy=\\frac{\\partial\\text{Q}(x,y)}{\\partial x}$.\n\nDefine $f(x,y)$ such that $\\frac{\\partial f(x,y)}{\\partial x}=\\text{P}(x,y)$ and $\\frac{\\partial f(x,y)}{\\partial y}=\\text{Q}(x,y)$.\n\nThen, the solution will be given by $f(x,y)=\\text{C}$, where $\\text{C}$ is an arbitrary constant.\n\nIntegrate $\\frac{\\partial f(x,y)}{\\partial x}$ with respect to $x$ in order to find $f(x,y)$:\n\n$$f(x,y)=\\int(18y^2x-\\sin(x))\\space\\text{d}x=9y^2x^2+\\cos(x)+g(y)$$\n\nDifferentiate $f(x,y)$ with respect to $y$ in order to find $g(y)$:\n\n$$\\frac{\\partial f(x,y)}{\\partial y}=\\frac{\\partial}{\\partial y}(9y^2x^2+\\cos(x)+g(y))=18yx^2+\\frac{\\text{d}g(y)}{\\text{d}y}$$\n\nSubstitute into $\\frac{\\partial f(x,y)}{\\partial y}=\\text{Q}(x,y)$:\n\n$$18yx^2+\\frac{\\text{d}g(y)}{\\text{d}y}=18yx^2+8$$\n\nSolve for $\\frac{\\text{d}g(y)}{\\text{d}y}$:\n\n$$\\frac{\\text{d}g(y)}{\\text{d}y}=8\\Longleftrightarrow$$ $$\\int\\frac{\\text{d}g(y)}{\\text{d}y}\\space\\text{d}y=\\int8\\space\\text{d}y\\Longleftrightarrow$$ $$g(y)=8y$$\n\nSubstitute $g(y)$ into $f(x,y)$:\n\n$$f(x,y)=9y^2x^2+8y+\\cos(x)$$\n\nThe solution is $f(x,y)=\\text{C}$:\n\n$$9y^2x^2+8y+\\cos(x)=\\text{C}$$\n\n$$9y(x)^2x^2+8y(x)+\\cos(x)=\\text{C}\\Longleftrightarrow$$ $$y(x)=\\frac{-4\\pm\\sqrt{16+9\\text{C}x^2-9x^2\\cos(x)}}{9x^2}$$\n\nWe know that $y(0)=1$ we can't use the solution with the '-' sign before the square root:\n\n$$1=\\lim_{x\\to0}\\frac{-4+\\sqrt{16+9\\text{C}x^2-9x^2\\cos(x)}}{9x^2}\\Longleftrightarrow$$ $$1=\\frac{\\text{C}-9}{72}\\Longleftrightarrow$$ $$\\text{C}=81$$\n\nSo the solution is:\n\n$$y(x)=\\frac{-4+\\sqrt{81x^2-9x^2\\cos(x)+16}}{9x^2}$$\n\n\u2022 Thank you for the super clear and detailed answer! I think I should easily be able to follow this process through and apply it to any other problems :) \u2013\u00a0Michael Mar 3 '16 at 11:23\n\u2022 @Michael You're welcome. Yes you can do that!! \u2013\u00a0Jan Mar 3 '16 at 11:46\n\nHIT:\n\nYou found $\\quad u = 9y^2 x^2 + \\cos(x) + h(y)\\quad$ which is correct.\n\nYou also found $\\quad h(y) = 8y + C\\quad$ which is correct.\n\nThen obviously the result is not : $u(x,y) = 36xy + 8y + C$ but is : $$u(x,y) = 9y^2 x^2 + \\cos(x) +8y+C$$\n\nThen you can solve $u(x,y)$=constant for $y(x)$.", "date": "2019-11-15 21:09:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1681268/help-me-find-mistake-in-exact-differential-equation", "openwebmath_score": 0.6682783365249634, "openwebmath_perplexity": 172.94143185205573, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446471538803, "lm_q2_score": 0.8872045922259088, "lm_q1q2_score": 0.8638220021510972}}
{"url": "https://stats.stackexchange.com/questions/433478/dof-of-natural-cubic-spline", "text": "# DOF of Natural Cubic Spline\n\nI am curious as to what the answer to the below question is? The question specifies a modeler has a cubic spline with knots at {10, 20, 30, 50}. They realize their model is overfitting at the ends of the distribution and wants to impose an additional constraint that the curve before the first knot and after the last knot are linear, then to calculate the DOF of the new model.\n\nMy thought process is that the answer should be 4 DOF for the new natural cubic spline model. Below is how I arrived at my answer:\n\nThe cubic spline has 8 DOF: 4+(4)(4)-(4)(3)=8 where the first 4 is for the intercept, X, X^2, X^3; then add 4 terms for earch knot; subtract out the 3 constraints at each knot to account for continuity, and the first and second derivatives to be zero.\n\nThen to make the cubic spline with 4 knots a natural cubic spline, subtract (2)(2)=4 to get 4 DOF since the interval below the lowest knot and above the largest knot need to be made linear.\n\nI\u2019ve been told the correct answer is 6 DOF and I don\u2019t understand how. I\u2019ve been told it has something to do with interior knots vs boundary knots. Can someone shed some light as to how the answer could possibly be 6? Is my answer of 4 wrong?\n\nLet there be $$K$$ knots $$\\xi_1 < \\dots < \\xi_K$$. I'll use $$x_\\min$$ for the smallest observed $$x$$ value, and $$x_\\max$$ is analogous.\n\nI think the answer depends on whether or not the $$\\xi_j$$ are assumed to be interior knots, and if we want the spline to change in $$[x_\\min, \\xi_1]$$ and $$[x_\\max, \\xi_K]$$ or not (and this also depends on whether or not $$x_\\min < \\xi_1$$ and/or $$\\xi_K < x_\\max$$).\n\nIf we only care about behavior between $$\\xi_1$$ and $$\\xi_K$$, then we could use the following truncated power basis: $$h_j(x) = x^j, j=0,1,2,3$$ and $$h_j(x) = (x-\\xi_{j-3})_+^3, \\hspace{5mm} j = 4,\\dots, K+3$$ leading to $$K+4$$ DoF. This represents four DoF coming from the global cubic that we start with and one DoF being added for every knot we pass.\n\nRestricting this to a natural spline means we'll constrain $$\\beta_2 = \\beta_3 = 0$$ which frees up 2 DoF, and we'll need the coefficients of $$x^3$$ and $$x^2$$ to be zero when every basis is active, so this further frees up two DoF meaning there are now $$K$$ DoF.\n\nIn interpolation problems and smoothing splines I think this is the right way to account for DoFs, because we have every point as a knot so there aren't separate boundary knots from $$x_\\min$$ and $$x_\\max$$.\n\nBut if we are always thinking of the $$\\xi_j$$ as interior knots then we would have two more regions where the spline in changing. This is where setting $$\\xi_0 = x_\\min$$ and $$\\xi_{K+1} = x_\\max$$ makes sense and really we now have $$K+2$$ knots and therefore the full cubic spline has $$K+6$$ DoF and the natural spline has $$K+2$$.\n\nIn Figure 7.5 of Introduction to Statistical Learning (in the question you linked) we can tell that they are using the $$K+2$$ DoF version because the spline is nonlinear on all of $$[x_\\min, x_\\max]$$, rather than just on $$[\\xi_1, \\xi_3]$$.\n\nTo answer the exact question: I can't tell which formulation they want but I think $$6$$ is more likely correct.\n\nFor a spline that is not interpolating or smoothing, which you have here, I think having $$\\xi_1$$ and $$\\xi_K$$ as interior knots makes sense so the $$K+2 = 6$$ answer makes sense.\n\nBut in the exact wording of the question they say \"the curve before the first knot and after the last knot will be linear\". If they mean out of the four given knots, then that means an answer of $$K=4$$ is correct, but I'm guessing they worded this poorly and are thinking of $$x_\\min$$ and $$x_\\max$$ as the first and last knots respectively, so even though at face value this suggests an answer of $$4$$, with this boundary knot inclusion then again the answer is $$6$$.\n\n\u2022 thank you for your response! the source this question is based on is James' Intro to Statistical Learning. I've linked below another thread which is the exact question/confusion I'm having between my answer and what was deemed as the correct answer. Is there a way to know in the question asked that they want us to assume implicit bounds, making the total knots 6? stats.stackexchange.com/questions/396889/\u2026 \u2013\u00a0kres901567708 Oct 28 '19 at 16:15\n\u2022 @kres901567708 I've just completely rewritten. I think in this question they are likely tacitly assuming boundary knots that are not any of the four listed and that makes $6$ correct (I was not doing this when I went through it before; I usually use smoothing splines where that doesn't make sense so I assumed no separate boundary knots but i think that was the wrong call here), but I also think the question is not clearly worded and without knowing that they likely want boundary knots I don't think $6$ vs $4$ DoF is clear. \u2013\u00a0jld Oct 28 '19 at 18:52\n\u2022 thank you! Would you say in your opinion that I have a valid case to argue the question is unclear and 4 DOF should be an acceptable answer, or is there something in the question that gives away we should be including xmin and xmax as boundary knots? \u2013\u00a0kres901567708 Oct 29 '19 at 19:43\n\u2022 @kres901567708 I think it\u2019s unclear but as I\u2019ve thought more about it, with these splines (ie not smoothing or interpolating) it probably is pretty much always the case that the specified knots are interior and wouldn\u2019t be boundary knots. Like when the $\\xi$ are thought of as tuning parameters, which I think is the case with a standard regression spline, they\u2019re often put at quantiles of the data which would make them interior \u2013\u00a0jld Oct 30 '19 at 2:23", "date": "2020-07-11 21:51:25", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/433478/dof-of-natural-cubic-spline", "openwebmath_score": 0.6516091227531433, "openwebmath_perplexity": 304.51867510471266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446452420054, "lm_q2_score": 0.8872045914803097, "lm_q1q2_score": 0.8638219997289245}}
{"url": "https://math.stackexchange.com/questions/522973/lim-x-to0-x-ln-x-without-lhopitals-rule", "text": "# $\\lim_{x\\to0^{+}} x \\ln x$ without l'Hopital's rule\n\nI have a midterm coming up and on the past exams the hard question(s) usually involve some form of $\\lim_{x\\to0^{+}} x \\ln x$. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways.\n\nSo how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful:\n\n1. $\\lim_{x\\to0^{+}} x \\ln x = \\lim_{x\\to0^{+}} x^2 \\ln (x^2) = L$\n2. $= (\\lim_{x\\to0^{+}} 2x)(\\lim_{x\\to0^{+}} x \\ln x)$\n3. $= 0 * L$\n\nThen I just need to prove that L is finite/exists (which means it must be 0)\n\n\u2022 Observing that $x\\ln x=\\ln(x^x),$ this question is effectively a duplicate of this other one. \u2013\u00a0Cameron Buie Oct 11 '13 at 21:57\n\u2022 I do not believe this should be closed, since it describes an interesting aproach to the problem that is absent elsewhere. \u2013\u00a0Andr\u00e9 Nicolas Oct 11 '13 at 22:10\n\u2022 Lovely boldfaced typo. Have a sticky p key. Also shift. \u2013\u00a0Andr\u00e9 Nicolas Oct 11 '13 at 22:22\n\u2022 @Raekye : That is a very clever approach. \u2013\u00a0Stefan Smith Oct 12 '13 at 1:54\n\u2022 @Andr\u00e9Nicolas: Couldn't the approach be posted to the other post? I think it would make sense. \u2013\u00a0Najib Idrissi Oct 12 '13 at 2:55\n\nThe idea you described is a very nice one. We fill in the details.\n\nWe consider, as in the OP, $x^2\\ln(x^2)$, that is, $(2x)(x\\ln x)$. If we can show that $x\\ln x$ is bounded near $0$, it will follow by Squeezing that $\\lim_{x\\to 0} x^2\\ln(x^2)=0$, and therefore $\\lim_{t\\to 0^+}t\\ln t=0$.\n\nLet $f(x)=x\\ln x$. Then $f'(x)=1+\\ln x$. It follows that $f(x)$ is decreasing in the interval $(0,e^{-1})$. It reaches a minimum value of $-e^{-1}$ at $x=e^{-1}$.\n\nSince $f(x)$ is negative in our interval, we have $|x\\ln x|\\le e^{-1}$ in the interval, and we have shown boundedness.\n\n\u2022 Ah, that makes sense. I thought I had to use the derivative to show \"which direction the function is going\" but couldn't spell it out. Thank you very much! \u2013\u00a0Raekye Oct 11 '13 at 22:12\n\u2022 A deleted answer gives another way to show that $f(x)$ is bounded: $$0>x\\ln(x)=x\\int_1^x\\frac1t\\,dt\\geq x(\\frac1x(x-1))=x-1>-1$$ when $0<x<1$, so $|f(x)|\\leq 2$ when $0<x<1$. \u2013\u00a0Jonas Meyer Oct 11 '13 at 22:15\n\u2022 (Now that $f(x)=x\\ln(x)$ instead of $2x\\ln(x)$, the last line of my comment should say \"$|f(x)|\\leq 1$\".) \u2013\u00a0Jonas Meyer Oct 11 '13 at 22:57\n\u2022 Sorry about the little change, in checking for my usual typos I thought there was no point in dragging the $2$ around. \u2013\u00a0Andr\u00e9 Nicolas Oct 11 '13 at 22:59\n\u2022 @Raekye: Please note that in my opinion the approach of user@17762 is \"better.\" My answer was an exercise in pushing through your clever idea. \u2013\u00a0Andr\u00e9 Nicolas Oct 12 '13 at 0:27\n\nLet $x=e^{-t}$ and note that as $x \\to 0^+$, we have $t \\to \\infty$. Hence, $$L = \\lim_{x \\to 0} x \\ln(x) = \\lim_{t \\to \\infty} -te^{-t} = -\\lim_{t \\to \\infty} \\dfrac{t}{e^t}$$ Now recall that $e^t \\geq \\dfrac{t^2}2$, because $$e^t =\\sum_{k=0}^{\\infty}\\frac{t^k}{k!} \\geq \\frac{t^2}{2}$$ Hence, we have $$\\lim_{t \\to \\infty} \\dfrac{t}{e^t} \\leq \\lim_{t \\to \\infty} \\dfrac2t = 0$$ This gives us $L=0$.\n\n\u2022 Ah this makes sense too. I selected Andre's answer though because he answered earlier. Thanks for your input though! \u2013\u00a0Raekye Oct 11 '13 at 22:17\n\u2022 Can you explain why e^t >= t^2/2 ? I\u2019m not good at math. \u2013\u00a0plhn Mar 15 '17 at 12:10", "date": "2019-06-26 22:21:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/522973/lim-x-to0-x-ln-x-without-lhopitals-rule", "openwebmath_score": 0.8570419549942017, "openwebmath_perplexity": 428.7141092727999, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575178175919, "lm_q2_score": 0.8791467611766711, "lm_q1q2_score": 0.8638122594591253}}
{"url": "https://math.stackexchange.com/questions/2538598/how-would-i-solve-x2-4x-y2-4y-without-knowing-the-answer-beforehand/2538884", "text": "# How would I solve $x^2-4x=y^2-4y$ without knowing the answer beforehand?\n\nThe equation is $x^2-4x=y^2-4y$ in the case where $x\\ne y$. The answer is $x+y=4$.\n\nI can start from $x+y=4$ and create the equation very easily, and I can substitute $x+4=y$ into the equation and show both sides are equal easily. I just don't get how I would find the answer if I didn't know it before hand, and all I had was the equation? Any advice?\n\n\u2022 I just realized, assuming that x and y aren't equal, then x must equal +/- (y-4) and y must equal +/- (x-4), otherwise there's no way x(x-4)=y (y-4). \u2013\u00a0Hockeyfan19 Nov 27 '17 at 2:11\n\u2022 Be weary of the line of reasoning; it\u2019s fallacious. Just because $ab = cd$ and $a \\neq b$, doesn\u2019t mean $a = c$. That trick only works when we have something like $ab = 0$, whence we can conclude $a$ or $b$ is zero. See the most recent answer to this question for the right application of this idea. It\u2019s subtle. \u2013\u00a0Bob Krueger Nov 28 '17 at 1:49\n\n\\begin{align} x^2 - 4x &= y^2 - 4y \\\\ x^2 - y^2 &= 4x - 4y \\\\ (x-y)(x+y) &= 4(x-y) \\\\ x+y &= 4\\end{align} where dividing by $x-y$ is allowed since $x \\neq y$.\n\n\u2022 It seems so obvious now, thanks for showing the way! I wasn't seeing it at all. \u2013\u00a0Hockeyfan19 Nov 26 '17 at 21:17\n\nAs an alternative, the solution that struck me first was completing the square in both $x$ and $y$. This is common when dealing with quadratics, especially once there are no $xy$ cross-terms. $$x^2 - 4x = y^2 - 4y$$ $$x^2 - 4x + 4 = y^2 - 4y + 4$$ $$(x-2)^2 = (y-2)^2$$ This means that either $x-2 = y-2$ or $x-2 = -(y-2)$, which means either $x = y$ or $x+y = 4$, as desired.\n\n\u2022 I actually saw that I could complete the square when I was working on it. I missed that it would progress towards the solution though. Thanks for the alternative approach! \u2013\u00a0Hockeyfan19 Nov 27 '17 at 1:38\n\u2022 You're welcome. Look into quadratic forms and conic sections for more general problems of this type. \u2013\u00a0Bob Krueger Nov 27 '17 at 1:44\n\u2022 This is the way I first went, so +1 ;) \u2013\u00a0Lamar Latrell Nov 27 '17 at 4:37\n\nLet $c$ be the common value of $x^2-4x$ and $y^2-4y$. Then $x$ and $y$ are both roots of the polynomial $t^2-4t-c$. Since we are assuming $x$ and $y$ are distinct, they are all of the roots, so $t^2-4t-c$ factors as $(t-x)(t-y)$. Since $(t-x)(t-y)$ expands to $t^2-(x+y)t+xy$, comparing the coefficients of $t$ gives $x+y=4$.\n\n(Conversely, if $x+y=4$, then since $x$ and $y$ are both roots of $(t-x)(t-y)=t^2-(x+y)t+xy=t^2-4t+xy$, $x^2-4x$ and $y^2-4y$ are both equal to $-xy$.)\n\n\u2022 Very interesting approach, but where did the -4t come from in your polynomial? \u2013\u00a0Hockeyfan19 Nov 27 '17 at 2:04\n\u2022 I'm not sure what you mean. Since $x^2-4x=c$, $x^2-4x-c=0$, and similarly for $y$. So $x$ and $y$ are roots of $t^2-4t-c$. \u2013\u00a0Eric Wofsey Nov 27 '17 at 2:32\n\u2022 Okay so it comes from the original equation, I see now \u2013\u00a0Hockeyfan19 Nov 27 '17 at 2:43\n\u2022 And you're using the t for clarities sake I think? \u2013\u00a0Hockeyfan19 Nov 27 '17 at 2:45\n\nWrite $x+y=d$. Then we have $y=d-x$ and so: $$(d-x)^2-4(d-x) = x^2-4x$$ so $$d^2-2dx-4d =-8x$$thus $$d(d-2x)-4(d-2x)=0$$ so $$(d-2x)(d-4)=0$$ If $d=2x$ we get $x=y$ which is impossible. So $d=4$ and we have $x+y=4$.", "date": "2019-08-17 13:28:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2538598/how-would-i-solve-x2-4x-y2-4y-without-knowing-the-answer-beforehand/2538884", "openwebmath_score": 0.9949854612350464, "openwebmath_perplexity": 226.73948778736383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575121992375, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8638122591865652}}
{"url": "https://gateoverflow.in/625/gate2000-1-2", "text": "941 views\n\nAn $n \\times n$ array $v$ is defined as follows:\n\n$v\\left[i,j\\right] = i - j$ for all $i, j, i \\leq n, 1 \\leq j \\leq n$\n\nThe sum of the elements of the array $v$ is\n\n1. $0$\n2. $n-1$\n3. $n^2 - 3n +2$\n4. $n^2 \\frac{\\left(n+1\\right)}{2}$\n\nedited | 941 views\n\nsquare matrix\u00a0whose\u00a0transpose\u00a0is its negation; that is, it satisfies the condition\u00a0\u2212A\u00a0=\u00a0AT.\u00a0If the entry in the\u00a0i th row\u00a0and\u00a0j th column\u00a0is\u00a0aij, i.e.\u00a0A\u00a0= (aij)\u00a0then the skew symmetric condition is\u00a0aij\u00a0= \u2212aji\n\nThe sum of the $i^{th}$\u00a0row and\u00a0$i^{th}$\u00a0column is $0$ as shown below. Since, the numbers of rows $=$ no. of columns, the total sum will be $0$.\n\n $0$ $-1$ $-2$ $-3$ $-4$ $1$ $0$ $-1$ $-2$ $-3$ $2$ $1$ $0$ $-1$ $-2$ $3$ $2$ $1$ $0$ $-1$ $4$ $3$ $2$ $1$ $0$\n\nedited by\nthis matrix is also a skew symmetric matrix, so definitely it's sum will be 0.\nLet there are total N rows . You will find \u2211 of elements of row i + \u2211 of elements row (N-i+1) = 0.\n\nSo if N is even then\n\nrow 1 + row N =0\n\nrow 2 + row (N-1) =0\n\nrow 3 + row (N-2)=0\n\nsimilarly row (N/2) + row (N/2+1) =0. \u00a0\u00a0(So total sum is 0)\n\nBut if N is odd then row ((N+1)/2) will have no corresponding rows BUT Ithe summation of elements of this row is 0 .\n\nSo for N = even or Odd , the sum of element is 0 .\nIf you look at code carefully it is very clear matrix getting defined is skew symmetric matrix.\n\nSum of all elements in skew symmetric matrix is 0.\n\ni think we can easily get it without drawing matrix\nas expression given\u00a0v[i, j] = i - j\nsuppose i1-j1=k1 \u00a0 {for a particular index } so its opposit index shows j1-i1=-k1, for example if v[1,2]=x then v[2,1]=-x}\nso we have \u00a0cases here\n1. for all\u00a0\u00a0i>j\u00a0v[i, j] = i - j and jut for opposit indexof\u00a0v[i, j] = i - j ,\u00a0v[j, i] = j\u00a0- i =-(i-j) :: note this is for non diagonal elements\n2. for all i=j\u00a0v[i, j] = i - j=0 \u00a0:: note this is for diagonal elaments\nso total sum will be zero ucan easily get it by seeing above cases\n\nans is option (a) i.e 0\n\nsolution: suppose if n=3\n\nthen i<=3 and 1<=j<=3\n\ni.e i=1,2,3, and j=1,2,3\n\n=> v={i-j}=(0,-1,-2,1,0,-1,2,1,0}\n\nsum of elements in v=0\nedited by\n@prakash What if i take, i=0,1,2. \u00a0\u00a0and j=1,2,3...? then it will be skew symmetric matrix..?\n+1 vote\nsum of all elements =n*( $\\sum_{i=1}^{n}i-\\sum_{j=1}^{n}j )= 0$", "date": "2018-01-19 10:11:31", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/625/gate2000-1-2", "openwebmath_score": 0.9328866004943848, "openwebmath_perplexity": 1397.3796243929498, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575188391108, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8638122572459913}}
{"url": "https://math.stackexchange.com/questions/2110681/prove-a-cap-b-cup-a-cap-b-a-using-set-identities", "text": "# Prove $(A \\cap B) \\cup (A \\cap B')= A$ using Set Identities\n\nI recently started a Discrete Mathematics course in college and I am having some difficulties with one of the homework questions. I need to learn this, so please guide me through at least two steps to get the ball rolling.\n\nThe question reads: Show that if $A$ and $B$ are sets, then: $(A \\cap B) \\cup (A \\cap B')=A$\n\nWe are supposed to use set identities. I had a question prior, but it was simple: $(A \\cap B \\cap C)' = A'\\cup B' \\cup C'$ - Which would be one of De Morgan's laws.\n\nI am at a loss. I have been reading the textbook and tried looking up some videos, but I am not sure exactly where to start. Any help you can provide, will be greatly appreciated!\n\nThanks, Kei\n\n\u2022 I'm not sure what you're asking. Can you clarify what specifically you're confused with? Also I highly reccomend taking a look at \"How to Prove It\" as a supplementary text. It's a fantastic introduction to this subject. \u2013\u00a0lordoftheshadows Jan 23 '17 at 18:58\n\u2022 Use distributivity! LHS is equal to $A \\cap (B \\cup B')$ \u2013\u00a0Crostul Jan 23 '17 at 18:58\n\u2022 @lordoftheshadows: I wasn't sure on the first two steps to prove that the left side is indeed equal to the right side? If I am explaining it right. Our professor showed us an example in class such as this: (A \u222a (BnC)) = (C \u222a B) \u2229 A = A \u2229 (B \u2229 C) - De Morgan Law = A \u2229 (B u C) De Morgan Law = (B u C) \u2229 A = Commutative Law = (C u B) \u2229 A = Commutative Law So I am looking for the first two steps, I see that @Crostul had posted the first step. Thank you! I am going to see if I can take it from there. I do really appreciate the help! \u2013\u00a0Kei U. Jan 23 '17 at 19:00\n\u2022 I have edited the question for formatting and some grammatical things for you, @KeiU. We will all see it when it is approved. Good luck! \u2013\u00a0The Count Jan 23 '17 at 19:03\n\u2022 Thank you @TheCount! I appreciate that, I really should have looked over the question better, sorry! @Crostul: Do I distribute (A \u2229 B) U (A \u2229 B`) together? \u2013\u00a0Kei U. Jan 23 '17 at 19:05\n\nThe idea is to achieve get close $B$ and $B^c$. Then we use distributive property: $(A\\cap B)\\cup(A\\cap B^c)=A\\cap (B\\cup B^c)=A\\cap X=A$, with $X$ the universe\n\n\u2022 Hello, I really appreciate your answer you provided! I think where I was confused was, being able to reverse an identity. I was using the set identity, though not using it in reverse if that makes sense. This is what I did: I used Distributive law, Absorption law, and then Identity law. \u2013\u00a0Kei U. Jan 23 '17 at 19:24\n\u2022 It would not let me edit the above comment. The last sentence should be a question. I wanted to ensure that I did use the right laws above, as you shown in your post Julio Maldonado Henriquez. Thank you! \u2013\u00a0Kei U. Jan 23 '17 at 19:33\n\nConsider an element of $A$ - either it is in $B$, or it isn't, and thus is in the complement of $B$. Thus $A \\subset (A \\cap B)$ $\\cup$ $(A \\cap B')$. Now, try to argue on your own that the reverse \"inclusion\" holds: that we have $A \\supset (A \\cap B)$ $\\cup$ $(A \\cap B')$.\n\nYou can use identities such as $(A\\cap B) \\cup (A \\cap C) = A \\cap (B \\cup C)$ to get $(A \\cap B) \\cup (A \\cap B') = A \\cap (B \\cup B') = A \\cap U = A$.\n\nBut I prefer to think of what it is saying. $A \\cap B$ means \"everything in A and in B\" and $A \\cap B'$ means \"everything that is in A that is not in B\" and $(A\\cap B) \\cup (A \\cap B')$ means \"every thing that is in A and B combined with everything that is not in B\". Is there a logical reason that \"everything in A and B combined with everything in A and not in B\" would be \"A\"?\n\nWell, I hope it should be obvious. Everything in A is either in B or not in B so combining the items of A that are not in B with those that are should give you all the items in A.\n\nSo the best way to express that idea directly would be:\n\n$A = A \\cap U = A \\cap (B \\cup B') = (A \\cap B) \\cup (A\\cap B')$. Or if that's a little too abstract, I rather like to do an element by element proof:\n\nLet $x \\in A$ either $x \\in B$ or $x \\in B'$. If $x \\in B$ then $x \\in A \\cap B$. If $x \\in B'$ then $x \\in A \\cap B'$. Either way $x \\in A \\cap B$ or $x \\in A\\cap B'$ so $x \\in (A \\cap B) \\cup (A\\cap B)$. So $A \\subseteq (A\\cap B) \\cup (A \\cap B)$. Likewise if $y \\in (A \\cap B) \\cup (A\\cap B)$ then either $y \\in (A \\cap B) \\subset A$ or $y \\in (A\\cap B') \\subset A$. Either way, $y \\in A$ so $(A\\cap B)\\cup (A\\cap B') \\subseteq A$.\n\n$A \\subseteq (A\\cap B) \\cup (A \\cap B)$ and $(A\\cap B)\\cup (A\\cap B') \\subseteq A$, so (A\\cap B)\\cup (A\\cap B') = A$. A fourth way is the big guns. Let$x \\in U$now one of four things might happen: 1)$x \\in A$and$x \\in B$. Then$x \\in A$and$x \\in A\\cap B$and$x \\in (A \\cap B) \\cup (A \\cap B')$. 2)$x \\in A$and$x \\not \\in B$. Then$x \\in A$and$x \\in B'$and$x \\in A \\cap B'$and$x \\in (A \\cap B) \\cup (A \\cap B')$3)$x \\not \\in A$and$x \\in B$. Then$x \\not \\in A$and$x \\not \\in A \\cap B$and$x \\not \\in A \\cap B'$so$x \\not \\in (A \\cap B) \\cup (A \\cap B')$. 4)$x \\not \\in A$and$x \\not \\in B$. Then$x \\not \\in A$and$x \\not \\in A \\cap B$and$x \\not \\in A \\cap B'$so$x \\not \\in (A \\cap B) \\cup (A \\cap B')$. Looking at the four cases we see$x \\in A \\iff x \\in (A \\cap B) \\cup (A \\cap B')$. Thus$A$and$(A \\cap B) \\cup (A \\cap B')$have precisely the same elements and neither has any element the other doesn't. In other words,$A = (A \\cap B) \\cup (A \\cap B')\\$.", "date": "2020-06-05 20:06:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2110681/prove-a-cap-b-cup-a-cap-b-a-using-set-identities", "openwebmath_score": 0.9979856610298157, "openwebmath_perplexity": 279.08457198967204, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575167960731, "lm_q2_score": 0.8791467580102419, "lm_q1q2_score": 0.8638122554498614}}
{"url": "https://math.stackexchange.com/questions/46969/definition-of-random/47039", "text": "# Definition of random\n\nSuppose that you has to guess given a set of numbers\n\n\u2022 If they are random.\n\u2022 The mathematical expectation\n\nIs there a definition of randomness that allow this prove/test?\n\nIs even possible? if so: How many value would be enough ?\n\nExample:\n\nIf numbers come from a coin experiment, results could be coded as 0 or 1,\n\n000100010110100011.....\n\nThen how many \"bits\" are enough to \"test\" variable randomness?\n\nAccording to the law of large numbers, the average of the results (adding the results and dividing by the number of trials) should become closer and closer to the expected value as more trials are performed.\n\nBut if we don't know where those numbers come from then we don't know what to expect!\n\nWhat are the law of large numbers hypothesis?\n\nDoes the definition of random cointain a reference to average and to the expected value?\n\nIt is deeply confusing to me,\n\nthanks in advance for any information\n\n\u2022 Random means that knowing the first $n$ terms of the sequence you cannot say anything about the $n+1$'th term at all besides that all possible combinations for that term to happen have the same probability. \u2013\u00a0Listing Jun 22 '11 at 19:38\n\u2022 If you search on testing random number generators you can find a lot of information. It is a large and complicated topic. \u2013\u00a0Ross Millikan Jun 22 '11 at 20:14\n\u2022 This is complicated topic indeed. For one thing: would you say that the digits of PI are random? \u2013\u00a0leonbloy Jun 22 '11 at 20:22\n\u2022 @Hernan - You might want to look up Kolmogorov complexity, which looks at randomness from a computational viewpoint. \u2013\u00a0Unreasonable Sin Jun 22 '11 at 20:44\n\u2022 @Ross Millikan thanks, as I see tests expects (or are aimed to) a certain distribution, it's like having a metadata about the data, as if randomness were not in a set of numbers, but in its interpretation @leonbloy it depends, if you already know numbers come from \"PI\" or from \"a coin experiment\" then that's not what I am asking about, because you have extra meta information about the data, the question is about the factibility of judge a set of data as \"random\", computable/non-computable random, is further discussion, but first I would like to know if e a definition of random is even possible \u2013\u00a0Hern\u00e1n Eche Jun 22 '11 at 20:52\n\nYou may like to understand the randomness of you sequence heuristically as follows. This may help you to get more intuition.\n\nWrite favorite binary sequence you would like to test for randomness in a file and try to compress it (e.g. with zip, etc...).\n\nThen take the ratio between the size of the compressed file and the size of the original one. If the ratio is close to 1, it means the sequence is quite random, and if the ratio is close to zero then it means that the sequence is not very random.\n\n\u2022 Yes this is the answer, as @Unreasonable Sin has comment, that's a Kolmogorov randomness definition, and that definition is really great, I have been reading Gregory Chaitin AIT en.wikipedia.org/wiki/Algorithmic_information_theory and is the deeper thing I have read about randomness \u2013\u00a0Hern\u00e1n Eche Jul 1 '11 at 19:43\n\nThe randomness of a sequence, to use @Listing's definition, can be quantified by the entropy rate.\n\nIf the order of the numbers does not matter, you can use a statistical test for randomness.\n\nI recommend reading Chapter 3.5 of Knuth, Seminumerical Algorithms (this book is Volume 2 of The Art Of Computer Programming). In fact, I recommend reading everything Knuth has ever written, but this chapter in particular, titled What is a Random Sequence, will help you clarify your concept of randomness.", "date": "2019-12-07 06:47:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/46969/definition-of-random/47039", "openwebmath_score": 0.7500331401824951, "openwebmath_perplexity": 513.5345393618926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799462157138, "lm_q2_score": 0.8962513655129178, "lm_q1q2_score": 0.8637890928498}}
{"url": "https://mathematica.stackexchange.com/questions/6862/plotting-complex-sine?noredirect=1", "text": "# Plotting complex Sine\n\nI've got another plotting problem. I want to plot Sin[z] where z is complex. So, I've tried the following:\n\nPlot3D[ Sin[ x + I y], {x, -1, 1}, {y, -1, 1}]\n\n\nI wanted to see how the sine function looks like on the unit circle. But... I get no output. Am I doing something wrong or is the kernel stuck?\n\n\u2022 You can plot Re[Sin[x + I*y]] and Im[Sin[x + I*y]] separately. \u2013\u00a0b.gates.you.know.what Jun 15 '12 at 15:35\n\u2022 Possible duplicate of Plotting Complex Quantity Functions \u2013\u00a0Jens Aug 7 '12 at 16:36\n\u2022 \u2013\u00a0Jens Aug 7 '12 at 16:37\n\u2022 @Jens It's not a duplicate of those posts since the OP asks for plot of sine on the unit circle, although the issue is quite similar. \u2013\u00a0Artes Aug 7 '12 at 17:18\n\u2022 @Artes I see this as a special case of the linked question, but since you answered both, I'll go with your judgement here. \u2013\u00a0Jens Aug 7 '12 at 17:27\n\nWell, you have to treat the real and imaginary parts separately. You can't really have a complex $z$ value in these plots. Here's one way to visualize complex sine:\n\nTable[Plot3D[f[Sin[x + I y]], {x, -1, 1}, {y, -1, 1},\nRegionFunction -> Function[{x, y, z}, x^2 + y^2 < 1]], {f, {Re, Im,\nAbs}}] // GraphicsRow\n\n\nOne further visualization aid would be to color these functions by the argument (adapting a scheme by Roman Maeder):\n\nTable[Plot3D[f[Sin[x + I y]], {x, -5, 5}, {y, -5, 5},\nColorFunction -> Function[{x, y, z}, Hue[(Pi + If[z == 0, 0, Arg[Sin[x + I y]]])/(2Pi)]],\nColorFunctionScaling -> False, PlotLabel -> TraditionalForm[f[Sin[z]]],\nRegionFunction -> Function[{x, y, z}, x^2 + y^2 < 25]],\n{f, {Re, Im, Abs}}] // GraphicsRow\n\n\nAs Artes notes, one could have used ParametricPlot3D[] so that you are directly working with polar coordinates:\n\nTable[ParametricPlot3D[{r Cos[t], r Sin[t], f[Sin[r Exp[I t]]]},\n{r, 0, 5}, {t, -Pi, Pi}, BoxRatios -> OptionValue[Plot3D, BoxRatios],\nColorFunction -> Function[{x, y, z}, Hue[(Pi + If[z == 0, 0, Arg[Sin[x + I y]]])/(2 Pi)]],\nColorFunctionScaling -> False, PlotLabel -> TraditionalForm[f[Sin[z]]]],\n{f, {Re, Im, Abs}}] // GraphicsRow\n\n\nYet another visualization possibility:\n\nTable[ParametricPlot3D[{r Cos[t], r Sin[t], f[Sin[r Exp[I t]]]},\n{r, 0, 5}, {t, -Pi, Pi}, BoxRatios -> OptionValue[Plot3D, BoxRatios],\nColorFunction -> Function[{x, y, z}, Hue[(Pi + If[z == 0, 0, Arg[Sin[x + I y]]])/(2Pi)]],\nColorFunctionScaling -> False, MeshFunctions -> (#4 &),\nMeshShading -> {Automatic, None}, MeshStyle -> Transparent,\nPlotLabel -> TraditionalForm[f[Sin[z]]], PlotRange -> All],\n{f, {Re, Im, Abs}}] // GraphicsRow\n\n\n\u2022 wow, that looks great. Thank you! \u2013\u00a0Chris Jun 15 '12 at 15:43\n\nYou can plot in 3 dimensions only real and/or imaginary parts of a function. One can make use of Plot3D, but since there was a question how the sine function looks like on the unit circle, first I demonstrate usage of ParametricPlot3D and later I'll show a few of many possible uses of Plot3D.\n\nWhen we'd like to use ParametricPlot3D, then instead of parametrizing complex numbers like x + I y we would rather parametrize them like r * Exp[ I u], where r is a radius of a circle and u is a polar angle. On a unit circle this reduces to Exp[ I u].\n\nParametricPlot3D[\n{ { Cos[u], Sin[u], Re @ Sin[Exp[I u]]},\n{ Cos[u], Sin[u], Im @ Sin[Exp[I u]]}},   {u, 0, 2 Pi},\nPlotStyle -> {{Thick, Darker @ Green}, {Thick, Darker @ Orange}}, BoxRatios -> Automatic]\n\n\nIt would be easier to realize the structure of the graph of Sine, rotating ParametricPlot3D around z axis. Thus we define the following functions :\n\nF1[t_] :=\nGraphics3D[\nRotate[\nParametricPlot3D[ Table[{r Cos[u], r Sin[u], Re @ Sin[r Exp[I u]]}, {r, 0.1, 1, 0.1}],\n{u, 0, 2 Pi}, PlotStyle -> Thick,\nColorFunction -> (ColorData[\"DeepSeaColors\"][#3] &),\nBoxRatios -> Automatic, Axes -> False, Boxed -> False][[1]],\n2 Pi t, {0, 0, 1}], Boxed -> False]\n\nF2[t_] :=\nGraphics3D[\nRotate[\nParametricPlot3D[ Table[{r Cos[u], r Sin[u], Im @ Sin[r Exp[I*(u)]]}, {r, 0.1, 1, 0.1}],\n{u, 0, 2 Pi}, PlotStyle -> Thick,\nColorFunction -> (ColorData[\"Rainbow\"][#3] &),\nBoxRatios -> Automatic, Axes -> False, Boxed -> False][[1]],\n2 Pi t, {0, 0, 1}], Boxed -> False]\n\n\nnow we can animate rotation around z-axis :\n\nAnimate[\nShow[{ F1[t], F2[t],\nParametricPlot3D[{{Cos[v], Sin[v], -1},\n{Cos[v], Sin[v],  0},\n{Cos[v], Sin[v],  1}  }, {v, 0, 2 Pi},\nPlotStyle -> {Dashed, Dashed, Dashed}, BoxRatios -> Automatic,\nAxes -> False, Boxed -> False]},\nViewPoint -> {Pi, Pi/2, 1/2}],\n{t, 0, 1},  DefaultDuration -> 15]\n\n\nThe \"deepseacolors\" and \"rainbow\" families of curves are respectively parametric 3D - plots of real and imaginary parts of Sine over circles of radius r in the complex plane and the view point rotates around z - axis. The dashed circles are unit circles in planes {x, y} for z in {-1, 0 , 1}. Here the rotation is surplus but still advantageous for the sake of comprehensible visualization.\n\nNow we provide static 3-D plots of Sine in the complex plane.\n\nGraphicsRow[{\nPlot3D[Re@Sin[x + I*y], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, ClippingStyle -> None],\nPlot3D[Im@Sin[x + I*y], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, ClippingStyle -> None]}]\n\n\nI extended the range of the plot to {x, -2 Pi, 2 Pi} and {y, -2 Pi, 2 Pi} since in your former case there was nothing interesting to see.\n\nTo compare with a familiar pattern of the graph of Sine let's restrict the range of the imaginary part of the variable, e.g.\n\nGraphicsRow[{ Plot3D[ Re @ Sin[x + I*y], {x, -2 Pi, 2 Pi}, {y, -0.3 Pi, 0.3 Pi}],\nPlot3D[ Im @ Sin[x + I*y], {x, -2 Pi, 2 Pi}, {y, -0.3 Pi, 0.3 Pi}]}]\n\n\nor to get the equal scale for all dimensions\n\nGraphicsRow[\n{Plot3D[ Re @ Sin[x + I*y], {x, -2 Pi, 2 Pi}, {y, -0.4 Pi, 0.4 Pi},\nBoxRatios -> Automatic, PlotLabel -> \"Real part\"],\n\nPlot3D[ Im @ Sin[x + I*y], {x, -2 Pi, 2 Pi}, {y, -0.4 Pi, 0.4 Pi},\nBoxRatios -> Automatic, PlotLabel -> \"Imaginary part\"]     },\n\nPlotLabel -> \"Graphs of Sine\"]\n\n\nand if you prefer the both parts of Sine in the complex plane in one plot :\n\nPlot3D[{ Re @ Sin[x + I*y], Im @ Sin[x + I*y]},\n{x, -2 Pi, 2 Pi}, {y, -0.4 Pi, 0.4 Pi},\nMesh -> {5, 3}, BoxRatios -> Automatic,\nPlotStyle -> {{Opacity[0.35], Lighter[Green, 0.5]},\n{Opacity[0.7],  Lighter[Blue,  0.7]} } ]\n\n\n\u2022 For a different visualization, on the Riemann sphere, see the paper \"Visualizing Complex Functions with the Presentations Application,\" The Mathematica Journal, vol. 11 #2 (2009), by David J. M. Park and me. Available in CDF and PDF here. (To re-evaluate most code, and to run the dynamic examples, you'll need a copy of Park's Presentations application.) \u2013\u00a0murray Aug 7 '12 at 16:57\n\u2022 @murray Thank you for pointing out interesting references. I've seen that paper. \u2013\u00a0Artes Aug 7 '12 at 17:01", "date": "2020-02-23 22:53:04", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/6862/plotting-complex-sine?noredirect=1", "openwebmath_score": 0.3071395456790924, "openwebmath_perplexity": 4979.049729186961, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.959154281754899, "lm_q2_score": 0.9005297847831082, "lm_q1q2_score": 0.8637469989225359}}
{"url": "https://math.stackexchange.com/questions/2356368/prove-that-if-m-1-subseteq-m-2-then-infm-2-leq-infm-1-leq-su", "text": "# Prove that if $M_{1} \\subseteq M_{2}$, then inf$(M_{2})\\leq$ inf$(M_{1})\\leq$ sup$(M_{1})\\leq$ sup$(M_{2})$\n\nProve that if $M_{1} \\subseteq M_{2}$, then inf$(M_{2})\\leq$ inf$(M_{1})\\leq$ sup$(M_{1})\\leq$ sup$(M_{2})$\n\nMy attempt:\n\n$M_{1} \\subseteq M_{2} \\implies \\forall m\\in{M_{1}}$, $m\\in{M_{2}}$\n\n$\\implies$inf$(M_{2}) \\leq$ inf$(M_{1})$ and also that sup$(M_{1})\\leq$ sup$(M_{2})$\n\nAdditionally, by the definition of supremum we know that inf$(M_{1})\\leq$ sup$(M_{1})$\n\nTogether we have, inf$(M_{2})\\leq$ inf$(M_{1})\\leq$ sup$(M_{1})\\leq$ sup$(M_{2})$\n\n$\\therefore$ If $M_{1} \\subseteq M_{2}$, then inf$(M_{2})\\leq$ inf$(M_{1})\\leq$ sup$(M_{1})\\leq$ sup$(M_{2})$\n\nI made a few jumps that I am not sure you can take (line 1 to 2). Is this proof valid? Anything I can change? Thanks!\n\n\u2022 Hmm, I think your jumps are too big. How does all m in both M_1 being in M_2 imply the inf M_2 is less or equal to the inf of M_1? How do we know from the definition of sup that inf <= sup? That's not actually part of the definition. I sympathize, as this real does seem trivial and obvious and thus irritatingly difficult to prove. But you I don't think your prove has actually done anything except restate what is to be proven and declared they are obvious. (Which to be fair they sort of are.) Jul 12 '17 at 17:40\n\nThere seems to be nothing wrong with your proof, but can you tell us how did you go from $(\\forall m\\in M_1):m\\in M_2$ to $\\inf M_2\\leqslant\\inf M_1$?\n\n\u2022 This means that inf$(M_{1})$, sup$(M_{1})\\in{M_{2}}$. I guess I should explicitly state that right? edit: Actually not sure that is correct Jul 12 '17 at 16:52\n\u2022 @dawgchow My mistake for the earlier comment. That's not necessarily true if you think about it Jul 12 '17 at 16:52\n\u2022 @dawgchow No, you shouldn't, because it is not true. Do you want a counter-example? Jul 12 '17 at 16:53\n\u2022 Ya because if inf$(M_{1})$ is not the minimum of $M_{1}$ it wouldn't necessarily be in $M_{2}$ Jul 12 '17 at 16:54\n\u2022 @dawgchow Right! So, your proof is incomplete. Jul 12 '17 at 16:55\n\nHere's how I would do it:\n\nBy the definition of subsets, $\\inf$ and $\\sup$, we have the following:\n\n$$M_1\\subseteq M_2\\iff (\\forall m\\in M_1)\\,\\,\\,m\\in M_2$$ $$(\\forall m\\in M_2)\\,\\,\\,\\inf(M_2)\\le m\\le \\sup(M_2)$$\n\nThen we can conclude\n\n$$(\\forall m\\in M_1)\\,\\,\\,\\inf(M_2)\\le m\\le \\sup(M_2)$$\n\nNext, from the definition of $\\inf$ and $\\sup$, we get\n\n$$(\\forall m\\in M_1)\\,\\,\\,x\\le m\\le y \\iff x\\le\\inf(M_1)\\le\\sup(M_1)\\le y$$\n\nTherefore\n\n$$\\inf(M_2)\\le \\inf(M_1)\\le\\sup(M_1)\\le \\sup(M_2)$$\n\nWhen you are asked to prove something trivial and obvious... well, do the definitions. I think your prove relies on \"common sense\" and makes jumps that are simply too big, to the point you are mostly just restating the statements.\n\nI'd do: If $m \\in M_1$ then $m \\in M_2$ as $M_1 \\subset M_2$. So $\\inf M_2 \\le m$ by definition of infinum. So $\\inf M_2$ is a lower bound of $M_1$. But $\\inf M_1$ is the greatest lower bound of $M_1$ so $\\inf M_2 \\le \\inf M_1$.\n\nBy identical argument: $\\sup M_2 \\le \\sup M_1$. That is; if $m \\in M_1$ then $m \\in M_2$ and therefore $\\sup M_2 \\ge m$ and is an upper bound of $M_1$, but as $\\sup M_1$ is the least upper bound $\\sup M_2 \\ge \\sup M_1$.\n\nBy definition of supremum and infimum for any $m \\in M_1$ then $\\inf M_1 \\le m \\le \\sup M_1$ so $\\inf M_1 \\le \\sup M_1$. (Unless $M_1$ is empty, in which case neither $\\inf M_1$ nor $\\sup M_1$ exist.) (I suppose at the very beginning of the proof I could/should have made a comment that $M_1$ and $M_2$ are both presumed to be non-empty and bounded above and below.).\n\nHence ... result.\n\nThat's basically your proof but with the jumps explicitly explained. (Perhaps painfully so.)\n\n=====\n\nAnd, to be fair, maybe I'm being to0 glib in claiming \"$\\inf M_1$ is greatest lower bound\".\n\nMore detail: If $\\inf M_1 < \\inf M_2$ then $\\inf M_2$ is not a lower bound of $M_1$ which we just showed it is, so $\\inf M_2 \\le \\inf M_1$.\n\nThat's kind of obvious but as I criticized you for not proving the obvious, it's only fair that I apply my own standards...", "date": "2022-01-27 15:33:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2356368/prove-that-if-m-1-subseteq-m-2-then-infm-2-leq-infm-1-leq-su", "openwebmath_score": 0.9139547348022461, "openwebmath_perplexity": 214.14570947262243, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363501395789, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.863732299899027}}
{"url": "https://math.stackexchange.com/questions/1459979/establishing-a-bijection-between-permutations-and-permutation-matrices", "text": "# Establishing a bijection between permutations and permutation matrices.\n\nI am trying to prove the following:\n\nGiven $\\sigma \\in S_n$ let $P(\\sigma)$ be the matrix with $(P(\\sigma))_{i,j} = \\delta_{i,\\sigma(j)}$. Show that $P$ is a bijection between $S_n$ and the set of permutation matrices of size $n$ $-$ where $\\delta_{i,j}$ is the Kronecker delta which equals 1 if $i=j$ and 0 otherwise.\n\nAs I understand the question, the question is saying that for each row $i$ and column $j$ we have that position $i,j$ equals 1 iff $\\sigma(j) = i$.\n\nSo then it seems then that for any $\\sigma$ such that $\\sigma(i) = i$, we would have $P(\\sigma)$ as the identity matrix with strictly 1's from the top-left to the bottom-right and zeros everywhere else. On the other hand, suppose that we have $\\sigma(1) = 2$, $\\sigma(2) = 3$, $\\sigma(3) = 4$, $\\sigma(4) = 1$; then we'd have the following matrix\n\n\\begin{bmatrix}0 & 0 & 0& 1\\\\1 & 0 & 0 & 0\\\\0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0\\end{bmatrix}\n\nIs this a correct understanding?\n\nIf so, then here is my attempted proof of the question mentioned above:\n\n$\\bf{Injectivity}$: Assuming $\\sigma, \\tau \\in S_n$ and $P(\\sigma) = P(\\tau)$ then whenever $\\sigma(j) = i$ we have the jth-column and i-th row of the matrix $p$ as a 1 and similarly for $\\tau(j) = i$. So $\\tau = \\sigma$ and we are done.\n\n$\\bf{Surjectivity}$: Let $P$ be some $n \\times n$ permutation matrix. Then each row and column will have exactly one 1 with zeros everywhere else. Then for every i,j position which has a 1 in it, we can choose a permutation $\\sigma$ such that $\\sigma(j) = i$ for some $\\sigma$ which is an n-cycle and we are done.\n\nIs this proof sound? Thank you in advance.\n\nYou're proof is fine, but more can be said. The set $P_n$ of $n\\times n$ permutation matrices is a group under matrix multiplication, and as such $S_n\\cong P_n$. Indeed, let $\\{v_1,\\ldots,v_n\\}$ be a basis for $\\mathbb{R}^n$ (you can take any field here, but nevermind). For each $\\sigma\\in S_n$, we define a linear transformation $P_\\sigma$ by $P_\\sigma(v_i)=v_{\\sigma(i)}$, whose matrix in the basis $\\{v_1,\\ldots,v_n\\}$ is $P(\\sigma)$.\n\nLet $\\Phi:S_n\\longrightarrow P_n$ be the map $\\Phi(\\sigma)=P_\\sigma$. This map is a homomorphism since $$P_\\sigma P_\\tau(v_i)=P_\\sigma(v_{\\tau(i)})=v_{\\sigma(\\tau(i))}=v_{\\sigma\\tau(i)}=P_{\\sigma\\tau}(v_i).$$\n\nThe map is injective: if $P_\\tau v_i=v_{\\tau(i)}=v_i$ for all $i$, then $\\tau(i)=i$ for all $i$ and $\\tau=1$.\n\nFinally, the map is surjective because $|P_n|=n!$ (permutation matrices are allowed one nonzero entry in every row and column. Starting with the leftmost column, you have $n$ choices for where to place a nonzero entry; having made this choice you have occupied a row, so going to the next column you have $n-1$ choices for where to place a nonzero entry, etc.)\n\n\u2022 Thanks. That makes sense as well. \u2013\u00a0letsmakemuffinstogether Oct 1 '15 at 18:58\n\u2022 There is another component to the question: show that $P: S_n \\rightarrow M_n{\\mathbb{R}}$ has $P(\\sigma\\tau)=P(\\sigma)P(\\tau)$. Any hints on how to do this? I think I see how it's done but I'm having troubles putting it into words and using summation notation to multiply the matrices $P(\\sigma)$ and $P(\\tau)$. \u2013\u00a0letsmakemuffinstogether Oct 1 '15 at 20:07\n\u2022 @letsmakemuffinstogether that is my proof that $\\Phi$ is a homomorphism. \u2013\u00a0David Hill Oct 1 '15 at 21:46\n\nIt can also be shown that the map $$\\Phi: \\mathcal{S}_n \\longrightarrow \\mathcal{P}_n$$ defined by $$\\sigma \\in \\mathcal{S}_n \\longmapsto P_\\sigma \\in \\mathcal{P}_n$$ is a homomorphism and hence an isomorphism.\n\nIf $$\\sigma$$, $$\\gamma \\in \\mathcal{S}_n$$ and $$1 \\le i,j \\le n$$, then $$\\left[ P_\\sigma P_\\gamma\\right]_{ij} = \\sum_{k=1}^n \\delta_{i,\\sigma(k)} \\delta_{k, \\gamma(j)} = \\delta_{i, \\sigma(\\gamma(j))} = \\left[ P_{\\sigma \\circ \\gamma}\\right]_{ij},$$ i.e., $$P_{\\sigma} P_{\\gamma} = P_{\\sigma \\circ \\gamma}$$.\n\nThus, \\begin{align} \\Phi(\\sigma\\circ\\gamma) = P_{\\sigma \\circ \\gamma} = P_\\sigma P_\\gamma = \\Phi(\\sigma)\\Phi(\\gamma) \\end{align} as desired.", "date": "2019-09-22 16:57:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1459979/establishing-a-bijection-between-permutations-and-permutation-matrices", "openwebmath_score": 0.9896159172058105, "openwebmath_perplexity": 74.33499793102257, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363508288305, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8637322973499314}}
{"url": "https://math.stackexchange.com/questions/977388/limit-of-a-0-0-function/977399", "text": "# Limit of a 0/0 function\n\nLet's say we have a function, for example, $$f(x) = \\frac{x-1}{x^2+2x-3},$$ and we want to now what is $$\\lim_{x \\to 1} f(x).$$ The result is $\\frac{1}{4}$.\n\nSo there exists a limit as $x \\to 1$.\n\nMy teacher says that the limit at $x=1$ doesn't exist. How is that? I don't understand it. We know that a limit exists when the one sided limits are the same result.\n\nThank you!\n\n\u2022 Try factoring the denominator. This is probably a homework problem, so look at the numerator for a hint. \u2013\u00a0copper.hat Oct 16 '14 at 22:35\n\u2022 @copper.hat: I completely disagree! In the definition of \"$\\lim_{x\\to 1} f(x)$\", $f(1)$ doesn't need to be defined (in fact, in most interesting cases it isn't), and even if it is defined, that value shouldn't be taken into account. Adding \"$x \\neq 1$\" doesn't make any difference whatsoever. \u2013\u00a0Hans Lundmark Oct 17 '14 at 7:13\n\u2022 @HansLundmark: I stand corrected. \u2013\u00a0copper.hat Oct 17 '14 at 15:03\n\nIt's possible that your teacher was pointing out the fact that the function doesn't exist at $x = 1$. That's different from saying that the limit doesn't exist as $x \\to 1$. Notice that by factoring, $$f(x) = \\frac{x-1}{x^2 + 2x - 3} = \\frac{x-1}{(x-1)(x+3)}$$\n\nAs long as we are considering $x \\ne 1$, the last expression simplifies: $$\\frac{x-1}{(x-1)(x+3)} = \\frac{1}{x+3}.$$ In other words, for any $x$ other than exactly $1$, $$f(x) = \\frac{1}{x+3}.$$ This helps understand what happens as $x$ gets ever closer to $1$: $f(x)$ gets ever closer to $$\\frac{1}{(1) + 3} = \\frac{1}{4}.$$\n\n\u2022 The word 'exactly' unnecessary. \u2013\u00a0CiaPan Oct 17 '14 at 8:12\n\u2022 @CiaPan: What about the word 'is'? \u2013\u00a0Nikolaj-K Oct 17 '14 at 13:43\n\u2022 @NikolajK that depends on what 'is' is. \u2013\u00a0Yakk Oct 17 '14 at 14:53\n\u2022 @Yakk: It's equivalent to equivalence in univalent foundations. \u2013\u00a0Nikolaj-K Oct 17 '14 at 16:37\n\nYour teacher is not correct. There are two easy ways to do this problem. The first is by factoring the denomiator:\n\n$$\\lim_{x\\to 1}\\frac{x-1}{(x-1)(x+3)}=\\lim_{x\\to 1}\\frac{1}{x+3}=\\frac{1}{4}$$\n\nThe second is by using L'Hospital's rule, which is a useful identity in limits. By L'Hospital's rule, we know that\n\n$$\\lim_{x\\to 1}\\frac{x-1}{x^2+2x-3}=\\lim_{x\\to 1}\\frac{1}{2x+2}=\\frac{1}{4}$$\n\nThis limit exists, because it is simply a discontinuity in the function, but it is a discontinuity at a single point. When we have certain indeterminate forms in limits, we may apply L'Hospital's rule, which allows us to better compute the limit.\n\nL'Hospital's rule states that, in certain indeterminate forms:\n\n$$\\lim_{x\\to n}\\frac{f(x)}{g(x)}=\\lim_{x\\to n}\\frac{f'(x)}{g'(x)}$$\n\nIt is worth clearing up a particular misconception that seems to have arisen. For $$f(x)=\\frac{x-1}{x^2+2x-3}$$ we cannot compute the value of $f(1)$, because this results in an indeterminate form. However, the limit exists, because there is simply a local discontinuity at a single point in an otherwise continuous function.\n\n\u2022 A third alternative, to whoever may care, is to note that the limit if $\\dfrac 1{f'(1)}$, where $f(x)=x^2+2x-3$. \u2013\u00a0Git Gud Oct 16 '14 at 23:57\n\u2022 To justify the use of l'Hospital, you also need to point out that $g' \\neq 0$ in a punctured neighbourhood of the point in question. \u2013\u00a0Hans Lundmark Oct 17 '14 at 7:14\n\u2022 A hard requirement of the usability of l'H\u00f4pital's rule is that the resulting \"right-hand-size\" is exists. But I guess that's what Hans is saying in more fancy terms? \u2013\u00a0rubenvb Oct 17 '14 at 8:00\n\u2022 @rubenvb: No, it's not enough that $\\lim f/g$ is of type $0/0$ and that $\\lim f'/g'$ exists. There are examples which show this. You really need $g'$ to be nonzero near the point. (Those examples are a bit artificial to be honest, but still...) \u2013\u00a0Hans Lundmark Oct 17 '14 at 15:10\n\u2022 @Hans $g'$ can be zero if $f'$ is zero, because you just have another L'Hospital's case and take $f''/g''$ Otherwise, just compute the limit normally. \u2013\u00a0Aza Oct 17 '14 at 16:13\n\nI don't know what your teacher means exactly. Limits are defined when $x$ tends to some number, or infinity. Not when $x$ is this number.\n\nThe value that the function takes at the limit poit is irrelevant (to compute the limit). In fact, in most high school limit exercises, the function is not defined at the point that $x$ tends to.", "date": "2019-08-21 06:57:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/977388/limit-of-a-0-0-function/977399", "openwebmath_score": 0.8828755617141724, "openwebmath_perplexity": 307.7773361316398, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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{"url": "http://sbwt.ciiz.pw/distance-between-two-coordinates-calculator.html", "text": "# Distance Between Two Coordinates Calculator\n\n Again, a distance and direction. Algebra -> Customizable Word Problem Solvers -> Geometry-> SOLUTION: The distance between two points in the coordinate plane is equal to the square root of the sum of the squares of the difference between the x-coordinates and the difference between Log On. Practice applying the Pythagorean Theorem in this tutorial about the distance formula. A journey meter calculate driving or walking distances from any location and displays fuel cost, taxicab fare, using haversine formula. 02/28/2018; 14 minutes to read; In this article. Hi Chua, You can use google API's for that, I think this one gives you duration and distance between 2 places. For two points P 1 = (x 1, y 1) and P 2 = (x 2, y 2) in the Cartesian plane, the distance between P 1 and P 2 is defined as: Example : Find the distance between the points ( \u2212 5, \u2212 5) and (0, 5) residing on the line segment pictured below. The first 2 parameters declare the x and y coordinates of the first point, and the second 2 parameters declare the x and y coordinates of the second point. I turned to my web browser and googled it, sure enough, there was this R package called \u2018geosphere\u2019 from Robert J. SHORTEST DISTANCE BETWEEN TWO POINTS ON A SPHERE It is known that the shortest distance between point A and point B on the surface of a sphere of radius R is part of a great circle lying in a plane intersecting the sphere surface and containing the points A and B and the point C at the sphere center. This is the snippet Calculate The Distance of Two Points On a Coordinate Plane on FreeVBCode. There are a number of latitude/longitude distance calculators available online. In this exercise you will plot the positions of the 2 players and manually calculate the distance between them by using the Euclidean distance formula. Nothing else makes sense and I see that you were able to calculate a distance in ft. Calculation of geographical distance is a very important utility function in GPS related procedures. The first 2 parameters declare the x and y coordinates of the first point, and the second 2 parameters declare the x and y coordinates of the second point. This distance and driving directions will also be displayed on google map labeled as Distance Map and Driving Directions US. On this page by using the mileage calculator you can find the flight distance or driving distance between any two cities in the world. Solution : Yes, First, convert the latitude and longitude values from decimal degrees to radians. Distance calculation. This also provides free sample source codes in PHP, ASP, ColdFusion, C/C++, C#, Java, Perl, Visual Basic and Javascript. google map api. angle between two lat lon points. This article is a sequel to the previous article on the same topic, but using T-SQL for calculation - Calculate distance between two points on globe from latitude and longitude coordinates. Hi everybody, i have some problem to calculate the distance between two points, in my case i have a file with 16 columns and i'm interesting to 6th, 7th, 8th column that are rispectively the x, y and z coordinates of several atoms belonging to a protein, instead in 14th 15th and 16th columns there are the x,y and z coordinates of different several atoms belonging to another protein (like a pdb. Fractions should be entered with a forward such as '3/4' for the fraction $$\\frac{3}{4}$$. , X0, Y0 and X1, Y1 and click calculate to know the distance between 2 points in 2-dimensional space. Calculate the distance. GIS question: excel formula to calculate distance between two points given their coordinates (latitude / longitude). com can calculate the shortest distance and the fastest distance between any two cities or locations. Is it really possible? The answer is \"YES\", not. If I want to calculate the distance between two positions, how can I achieve this? Learn. In order to fully grasp how to plot polar coordinates, you need to see what a polar coordinate plane looks like. This will compute the great-circle distance between two latitude/longitude points, as well as the middle point. In this post, we show the formula to calculate the shortest distance between two points using Latitude and Longitude. Language Objectives: After completing the activity, students should be able to. Enter 2 sets of coordinates in the x y-plane of the 2 dimensional Cartesian coordinate system, (X 1, Y 1) and (X 2, Y 2), to get the distance formula calculation for the 2 points and calculate distance between the 2 points. 5678, 11087. Star 1 Right Ascension: Enter the Right Ascension of the first star in hours. ' '===== ' Calculate geodesic distance (in m) ' between two points specified by ' latitude/longitude (in numeric ' [decimal] degrees) ' using Vincenty inverse formula ' for ellipsoids '===== ' Code has been ported by lost_species ' from www. This uses the \u2018haversine\u2019 formula to calculate the great-circle distance between two points \u2013 that is, the shortest distance over the earth\u2019s surface \u2013 giving an \u2018as-the-crow-flies\u2019 distance between the points (ignoring any hills they fly over, of course!). This demo program will calculate the distance between two points on Earth, using the latitude and longitude positions. View Details \u00bb. The thing I did kind of worked to find the shortest distance from 2 locations, and then I referenced it to 3 locations in all and averages it out. coordinates and two. This calculation can be useful when trying to determine the distance for logistical purposes (ie delivery service, flights, distance between customers, etc). Please write the origin and destination city name, choose the distance unit and press calculate. I am storing the coordinates in two. However, I would like to calculate the distance between the points before I run my script and without an ArcGIS. Write, compile, and execute a C program that calculates the distance between two points whose coordinates are (7,12) and (3,9). The term Haversine was coined by. 84 return list gives you three calculations--the first two assume the Earth is a perfect sphere, which it is not, and the third that is an ellipsoid, which is closer to the truth. Distance between two locations on the Earth We often describe a location on Earth by its latitude and longitude. It is not very often that someone gives the latitude and longitude of their town! The use of a distance and direction as a means of describing position is therefore far more natural than using two distances on a grid. The Bearing Distance Calculator function calculates the forward and backward azimuths between two specified coordinates as well as the distance. txt, the XY coordinates of the cities. Perform the same operation for the y coordinates. This page helps you to calculate great-circle distances between two points using the \u2018Haversine\u2019 formula. Calculate the Distance Between Two Coordinates (latitude, longitude) - PHP - Snipplr Social Snippet Repository. I am looking for a way to get the distance in pixels between two MapItems. Just the other day, my friend was asking me if there was an easy way to calculate the distances between two locations with geocodes (Longitude and Latitude). When calculating the distance between two points on a coordinate system we are applying Pythagoras' Theorem. Standard measurement equipment (rulers, protractors, planimeters, dot grids, etc. As homework we were assigned to enter the following code to calculate the distance between two points on the x and y plane. 42 (rounded to the nearest 100th) How to use the Distance Formula Calculator. This means of location is used in polar coordinates and bearings. For example, in a button you can have the following formula on its OnSelect property:. In the attached file \"Circle2\", the coordinates of the circles are located under column B & C, the coordinates of the points are located under column F & calculate the minimum distance between two sets of coordinates. Enter 2 sets of coordinates in the 3 dimensional Cartesian coordinate system, (X 1, Y 1, Z 1) and (X 2, Y 2, Z 2), to get the distance formula calculation for the 2 points and calculate distance between the 2 points. Time and Date Duration \u2013 Calculate duration, with both date and time included Date Calculator \u2013 Add or subtract days, months, years. Function accepts four parameters, source latitude, source longitude, destination latitude, destination longitude and returns the distance between the. 13365 i currently have the code, but that doesnt seem to work well. If you pass an address as a string, the service will geocode the string and convert it to. \"Let's look at the diagram! I can see that the lines of longitude get closer and closer together towards the poles! At the equator, the distance between 15 and 30 degrees W longitude is quite a lot! But as you follow those two longitude lines towards the poles, the distance between them shrinks down to zero!. This script calculates distances, bearing and more between the two Latitude/Longitude points. This tool can measure two types of distance types, the first is straight line distance also known as Rhumb line distance. 28 , rounded to two decimal places. I would like to share a MySQL function to calculate distance between two latitude \u2013 longitude points. The easiest way I could think of to to this is to somehow store the output of veclen(x,y) to a macro. Free sample source codes in PHP, ASP, ColdFusion, C/C++, C#, Java, Perl, Visual Basic and Javascript. Calculate with Geo Coordinates Calculators for the conversion of geo coordinates, as delivered by a GPS tool, for the distance of two point and for bearing. Calculate the distance between them with the distance. It contains two vertical lines called axis. If the current coordinate system is an earth coordinate system, Distance( ) returns the great-circle distance between the two points. The program should ask the user to enter two points then should calculate the distance between two points and print the distance on the screen. Calculating Distance. I am looking for a way to get the distance in pixels between two MapItems. In this post, we will learn the distance formula. Distance and midpoint calculator This online calculator will compute and plot the distance and midpoint for two points in two dimensions. The slope of a line in the plane containing the x and y axes is generally represented by the letter m, and is defined as the change in the y coordinate divided by the corresponding change in the x coordinate, between two distinct points on the line. DISTANCE BETWEEN TWO POINTS. The distance returned is the arc length of a great circle connecting the two points, essentially air miles. West and South locations are negative. Distances calculator is a free tool to calculate distances between any two cities in the world. There is more information on how to calculate these figures below the tool. d = \u221a 5 2 + 7 2. MySQL function to calculate the distance between two coordinates using the Haversine formula. The formula assumes that the earth is a sphere, (we know that it is \"egg\" shaped) but it is accurate enough * for our purposes. The distance between 2 points on a graph is really the hypotenuse of a right-angled triangle. Calculate the great circle distance between two points does not want to give me a reading. The shortest distance between two points on the surface of a sphere is an arc, not a line. Note: The flight distances provided are close approximations as all flights differ based on weather, traffic, and the exact route determined by air traffic control. Click Calculate Distance, and the tool will place a marker at each of the two addresses on the map along with a line between them. Take for example, a dating website which shows potential matches to a user within a 15 mile radius or a taxi business which calculates taxi fares based on distance between two locations. If you want to determine the distance between two points on a plane (two-dimensional distance), use our distance calculator. The code will show you how to do that using Latitude and Longitude coordinates. Spherical to Cylindrical coordinates. A more mathematical approach, you could use the Pythagorean Theorem and calculate the hypotenuse using the right triangle with the grid units in unity. If you pass coordinates, ensure that no space exists between the latitude and longitude values; destinations \u2014 One or more addresses and/or textual latitude/longitude values, separated with the pipe (|) character, to which to calculate distance and time. The program should ask the user to enter two points then should calculate the distance between two points and print the distance on the screen. Position, Distance and Bearing Calculations Whitham D. 458 W 71 27. The API can convert a zip code to the primary location for the zip code. Answered Active Solved. To find the distance between two places, enter the start and end destination and this distance calculator will give you complete distance information. 05W) A program for HP50G or 42S or 41C would be fine or a formula or algorithm as well. By continuing to browse this site, you agree to this use. How do you find the distance between two points? Distance Formula : Example: For two points, (3,2) and (15, 10) the distance is calculated as: Distance = 14. It is not very often that someone gives the latitude and longitude of their town! The use of a distance and direction as a means of describing position is therefore far more natural than using two distances on a grid. We have successfully combined latitude and longitude into the GeoLoc field. Discover how to calculate Distance by Latitude and Longitude using JavaScript. x A means the x-coordinate of point A y A means the y-coordinate of point A. Another method for calculating the distance between two points that fall on the same line is to plot the points and count the amount of boxes in between the two points. This calculator computes the great circle distance between two points on the earth's surface. calculate-distance-between-two-latitude-longitude-points-11152 contains nodes provided by the following 2 plugin(s): KNIME Base Nodes. It\u2019s quick, easy and takes but a moment to do because you only need to enter the x and y coordinates of two points and click a button to calculate it. For some websites, it is a necessity to calculate the distance between certain locations. Find altitude by coordinates on google maps in meters and feet. If you are not sure what the gps coordinates are, you can use the coordinates converter to convert an address into latlong format or vice versa. The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. Inputs can be in several formats: GPS Coordinates (like N 42 59. As you start to write the name of a city or place, distance calculator will suggest you place names automatically, you may choose from them to calculate distance. With the growing distribution of GPS devices and tools such as Google Earth or GeoPosition it is easy to determine the geographical coordinates of interesting points on the earth's surface. The geographical coordinates of the two points, as (latitude, longitude) pairs, are and respectively. If you have two geohashes, you first need to decode them into latitude and longitude: GEOHASH_DEC_LAT(#GeoHash) GEOHASH_DEC_LONG(#GeoHash) After you have your coordinates you can calculate the distance: IF(!. Practice applying the Pythagorean Theorem in this tutorial about the distance formula. First, sketch the two UTM coordinates onto a rectangle. In this article, you will learn how to Get Distance between two Coordinates Using Xamarin. For more information on how it distances are calculated on a sphere, have a look at Haversine Formula on Wikipedia, it's a bit complex, hence us not wanting to duplicate the content. Distance Calculation Introduction. Calculate distance between a pair of lat/lon coordinates I recently had a need to calculate distance between a large number of latitude/longitude coordinate pairs. Plane equation given three points. The Haversine Formula For two points on a sphere (of radius R) with latitudes \u03c61 and \u03c62, latitude separation \u0394\u03c6 = \u03c61 \u2212 \u03c62, and longitude separation \u0394\u03bb, where angles are in radians, the distance d between the two points (along a great circle of the sphere; see spherical distance) is related to their locations by the formula above. After finding the 'distance' (it is in. The distance between the two coordinates, in meters. First, note that it's very easy to get a distance between two points on a sphere if you know the angle of the arc between them: just multiply it by the radius. Suppose it is desired to calculate the distance d from the point (1, 2) to the point (3, -2) shown on the grid below. The question Calculating the planets and moons based on Newtons's gravitational force was pretty much answered with two items: Use a reasonable ODE solver; at least RK4 (the classic Runge-Kutta me. For more information, see Measure Distances Between Data Points and Locations in a Map. Enter a city, street, address, postal code or region and click calculate distance. Distance Calculator: This easy tool is very useful to find travel distances between cities, and its respective estimated time en route. Haversine Formula \u2013 Calculate geographic distance on earth. Travelmath helps you find driving distances based on actual directions for your road trip. Enter either: decimal latitudes/longitudes with minus sign for South and West. This tool calculates the straight line distance between two pairs of latitude/longitude points provide in decimal degrees. You can obtain the coordinates for just about any earthly city from WWW. Know how to use the distance formula to find the distance between points. Is there anything to improve? Calculate the distance given. Capable estimates total distance between any location using haversine formula and route method provided by the map and display location coordinates and location address. The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. distancesfrom. Free distance calculator - Compute distance between two points step-by-step. A line through three-dimensional space between points of interest on a spherical Earth is the chord of the great circle between the points. What Is a Vector? A vector has size, also known as magnitude, and direction. Suppose it is desired to calculate the distance d from the point (1, 2) to the point (3, -2) shown on the grid below. Algebra -> Customizable Word Problem Solvers -> Geometry-> SOLUTION: The distance between two points in the coordinate plane is equal to the square root of the sum of the squares of the difference between the x-coordinates and the difference between Log On. Further, there is a one-to-one correspondence between areal coordinates and all points on the plane P. Distance = \u221a (x 2 \u2013 x 1) 2 + (y 2 \u2013 y 1) 2. Re: How to calculate distance between two coordinates As you are doing a different syllabus to ours I don't know what method they use; if the questions on GHA are anything to go by, you'll probably find they are using Spherical Trigonometry. Distance between two adjacent integer (whole numbers such as 1 and 2, or 25 and 26, not 13. The distance between the two coordinates, in meters. What is the distance between points B & C? What is the distance between points D & B? What is the distance between points D & E? Which of the points shown above are $4$ units away from $(-1, -3)$ and $2$ units away from $(3, -1)$?. Calculator Use. Find altitude by coordinates on google maps in meters and feet. Calculate the distance as-the-crow-flies between two Latitude/Longitude points on Earth You can enter the Latitude (East - West) and Longitude (North - South) coordinates as given by Google Earth software. The distance problem can be solved by using the Haversine formula. hi, i have a pair of latitude and longitude and i want to calculate the distance between these two points. Just the other day, my friend was asking me if there was an easy way to calculate the distances between two locations with geocodes (Longitude and Latitude). Cartesian to Cylindrical coordinates. The distance formula is derived from the Pythagorean theorem. Notice the line colored green that shows the same exact mathematical equation both up above, using the pythagorean theorem, and down below using the formula. The java program finds distance between two points using manhattan distance equation. Distance Formula The distance formula can be obtained by creating a triangle and using the Pythagorean Theorem to find the length of the hypotenuse. It can produce a map of the flight route and returns a link to enroute weather information. If you do not know the datum of the coordinates the information is not useless for a general ideas of your location -- but for for mapping work the answers you. I have found a couple of solutions that do this for zip codes, but not for a physical address. Rect) This code is lacking a zillion essential features (but interpoint distance can now be calculated). If you want to determine the distance between two points on a plane (two-dimensional distance), use our distance calculator. The center of Rome is located at roughly 41. The distances and times returned are based on the routes calculated by the Bing Maps Route API. Valid input formats are at the bottom of this page. The distance value in red color indicates the air (flying) distance, also known as great circle distance. 'OUTPUT: Distance between the ' two points in Meters. The formula assumes that the earth is a sphere, (we know that it is \"egg\" shaped) but it is accurate enough * for our purposes. The script uses Haversine formula, which results in in approximations less than 1%. A proof of the Pythagorean theorem. Note: The flight distances provided are close approximations as all flights differ based on weather, traffic, and the exact route determined by air traffic control. ) cannot be used to measure distance, angle, area, or shape on a sphere, as these tools have been constructed for use in planar models. If you need to get that information relating to distances between points, use the GPS Latitude and Longitude Distance Calculator. coordinates from browser. since the distance between first one and second one is already calculated there is no need to do it again. NET 4 Framework or above and reference System. distancesfrom. In an example of how to calculate the distance between two coordinates in Excel, we\u2019ll seek to measure the great circle distance. Use LatLong. In other words, the distance between A and B. This demo program will calculate the distance between two points on Earth, using the latitude and longitude positions. Know how to interpret points with the rectangular coordinate system (rect system) 2. Latitude and/or longitude change every few inches, and if two places have the same latitude and longitude, then the distance between them is zero. I have found a couple of solutions that do this for zip codes, but not for a physical address. Calculate Distance in Miles from Latitude and Longitude. A protip by ausi about mysql, latitude, longitude, and coordinates. What I am going to show you is how to calculate distance between two points within the United States, using the Haversine formula, implemented in C. Theses free samples assist in understanding its usage. This uses the haversine formula to calculate the great-circle distance between two points \u2013 that is, the shortest distance over the earth\u2019s surface \u2013 giving an \u201cas-the-crow-flies\u201d distance between the points (ignoring any hills they fly over, of course!). Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to find midpoint between two points. Click here to find your latitude/longitude. Calculate car driving distance and straight line flying distance times between Orapa Botswana and Maseru Lesotho with Distantias! Get fuel cost estimates, the midpoint, nearest rail stations, nearest airports, traffic and more. A circle is a set of all points in a plane that are a fixed distance r away from a given point called the center. Calculating the Distance Between Two GPS Coordinates with Python (Haversine Formula) September 7, 2016. 8664741, 24. I have these data: Point A and B with State, City, Address, Zip. Then it occured to me that I might have to normalize $\\rho$, so it can only take values between zero and one (just like the $\\sin$). The distance returned is relative to Earth\u2019s radius. However, I have a list of coordinates of houses, and want to calculate their. This will compute the great-circle distance between two latitude/longitude points, as well as the middle point. 8), where the units are centimeters for Teachers for Schools for Working Scholars for. MGRS Distance and Direction Calculator. D= /(x2\u2212x1)2+(y2\u2212y1)2. Introduction Two frequent calculations required in radio propagation work are distance and bearing between two radio terminals. Distance Calculator. Latitude is in degrees north of the equator; southern latitudes are negative degrees north. Also get the directions for going to one city from another city. The example application that is posted will show you how to use Zip codes in place of Latitude and Longitude. Thank you and regards,. We have successfully combined latitude and longitude into the GeoLoc field. Calculate the distance between two addresses gets confused when you input GPS coordinates. If you have two geohashes, you first need to decode them into latitude and longitude: GEOHASH_DEC_LAT(#GeoHash) GEOHASH_DEC_LONG(#GeoHash) After you have your coordinates you can calculate the distance: IF(!. Distance Calculator is use to calculate the distance between coordinates and distance between cities. // use the 'haversine' formula to calculate distance between two lat/long points for each record // The Haversine formula can be broken into multiple pieces or steps - I chose to do the entire formula. Calculates the distance and azimuth between two places each city's distance from that coordinate. This uses the haversine formula to calculate the great-circle distance between two points \u2013 that is, the shortest distance over the earth\u2019s surface \u2013 giving an \u201cas-the-crow-flies\u201d distance between the points (ignoring any hills they fly over, of course!). Try our Latitude/Longitude Distance Calculator to determine the distance between two points. com) software offers a solution for users who want to find the distance between one or more latitude and longitude coordinates. As you start to write the name of a city or place, distance calculator will suggest you place names automatically, you may choose from them to calculate distance. You can calculate geo distance using spatial types - geography datatype in SQL server. Does anyone know of a method of calculating the distance between two (very close together) GPS coordinates? The method must be easily calculated by an 8-bit microcontroller. The calculator will find the angle (in radians and degrees) between the two vectors, and will show the work. So, let's go ahead and just step through this, press f8, I might have a couple of things I forgot to do, but we set the latitudes and longitudes, we calculate a, b. Currently if you do a Google search to find how to accomplish this you will find a number of blogs talking about the need to use a Haversine formula that takes into account the curvature of the earth. There are several different ways to calculate the distance between two coordinates, P1(r, \u03b8, \u03a6) and P2(r, \u03b8, \u03a6). The values used for the radius of the Earth (3961 miles & 6373 km) are optimized for locations around 39 degrees from the equator (roughly the Latitude of Washington, DC, USA). Plane equation given three points. The formulas are very simple: for two given points and distance is the hypotenuse of right triangle, and it is calculated like this: and middle point is the average of both coordinates. I am trying to work out what would be described as the 'L0 Tile' distance between two points in my game in tiles. Distance between two points in a three dimension coordinate system - online calculator Sponsored Links The distance between two points in a three dimensional - 3D - coordinate system can be calculated as. One additional aspect not addressed above is the nature of the terrain between the two points. For instance we are getting data from two different devices at two different geographical location, we are getting latitude and longitude value for those devices. The Bing Maps Distance Matrix API provides travel time and distances for a set of origins and destinations. This can be achieved through Google Maps API. Take for example, a dating website which shows potential matches to a user within a 15 mile radius or a taxi business which calculates taxi fares based on distance between two locations. Geozip Geozip is a Perl program for calculating the distance between two zip codes in the United States. In the attached file \"Circle2\", the coordinates of the circles are located under column B & C, the coordinates of the points are located under column F & calculate the minimum distance between two sets of coordinates. 8, which is the radius of Earth. The distance problem can be solved by using the Haversine formula. It has all the needed trig functions to calculate the distance in miles between two points using Longitude and Latitude where they are expressed in decimal form. Distance between Two Given Points. Surface Distance Between Two Points of Latitude and Longitude. Distance Calculator. There are many options available if you want to import these in a GIS and run analysis. From Point A to Point B: Understanding how to find the distance between two points on the coordinate plane. The first step is expressing each Latitude/Longitude of both of the coordinates in radians:. Perform the same operation for the y coordinates. Simply enter the from address and to address and the distance between the two address will be calculated in milage and kilometers. This demo program will calculate the distance between two points on Earth, using the latitude and longitude positions. Calculate distance between two coordinates. The curve (somewhat resembles a circle) is made up of a couple of given points and the rest of the data is interpolated by excel. Point Slope Form Calculator The point slope form is defined that the difference in the y coordinate between two points (y - y1) on a line is proportional to the difference in the x coordinate points (x - x1). Answer to: Two points in a rectangular coordinate system have the coordinates (4. I will choose the first coordinate and calculate the distance to other coordinates by using the above equation. 20181 The distance between each other is about 22km. If the two points are (x1, y1) and (x2, y2), the distance between these points is given by the distance formula: The distance formula is a very. The Earth Model\u2014Calculating Field Size and Distances between Points using GPS Coordinates Figure 1. JS: Calculate Distance Between Two Coordinates phptuts January 3, 2018 This routine calculates the distance between two points (given the latitude/longitude of those points). Coordinates of point is a set of values that is used to determine the position of a point in a two dimensional plane. QSC297: Estimate the square root of a number between two consecutive integers with and without models. For just a few calculations, this is very straight-forward using plain old Excel. convert to/from UTM coordinates; calculate the geodetic distance between two positions or calculate the bearing from one position to another; project a position with a distance and bearing, to get a new position (or bearing) find the geodetic intersection of two lines (great circles), two archs (small circles) or a line and a arc. Ask Question. The great circle distance is proportional to the central angle. Important Note: The distance calculator on this page is provided for informational purposes only. Lines of latitude are called parallels and in total there are 180 degrees of latitude. Calculate Distance in Miles from Latitude and Longitude. This can be used to place features so they are relative to one another. The center of Rome is located at roughly 41. Calculate the distance between any two points in the rectangular coordinate plane. how to calculate nearest places distance between two coordinate(lat,long) [Answered] RSS 1 reply Last post Jul 07, 2014 05:30 AM by Michelle Ge - MSFT. Distance Calculator \u2013 Find the Distance Between Cities. The second one calculates the total distance between a list of locations. Again, a distance and direction. For find distance between two latitude and longitude in SQL Server, we can use below mentioned query. Distance From To: Calculate distance between two addresses, cities, states, zipcodes, or locations Enter a city, a zipcode, or an address in both the Distance From and the Distance To address inputs. This article is a sequel to the previous article on the same topic, but using T-SQL for calculation - Calculate distance between two points on globe from latitude and longitude coordinates. This includes the city, state, latitude, longitude, time zone information, and NPA area codes for the primary location. For two points P 1 = (x 1, y 1) and P 2 = (x 2, y 2) in the Cartesian plane, the distance between P 1 and P 2 is defined as: Example : Find the distance between the points ( \u2212 5, \u2212 5) and (0, 5) residing on the line segment pictured below. google map api. In this post, we show the formula to calculate the shortest distance between two points using Latitude and Longitude. The calculator will generate a step-by-step explanation on how to obtain the results. The Distance from the Car to the next Turn. I built this primarily to make it easy to check if a Locationless (Reverse) Cache has already been found. This distance will also be displayed on google map labeled as World Distance Map. As shown in the example below, we do this by joining the two points with a straight line and then drawing a right-angled triangle using that straight line as the hypotenuse and aligning the other two sides with the x-axis and y-axis. MGRS Distance and Direction Calculator. 14159265359). All numbers and return values should be of type double. The formulas are very simple: for two given points and distance is the hypotenuse of right triangle, and it is calculated like this: and middle point is the average of both coordinates. Another method for calculating the distance between two points that fall on the same line is to plot the points and count the amount of boxes in between the two points. To find the flight distance between two places, please insert the locations in the control of flight distance calculator and Calculate Flight Distance to get the required results while travelling by air. It accepts a variety of formats:. I've found this code for calculating distance. 0 (sobolsoft. The first class calculates the distance between two locations. Calculate Distance Between 2 Coordinates alias Memperhitungkan jarak antara 2 titik koordinat. The purpose of the function is to calculate the distance between two points and return the result. Find bearing angle and find direction A and B as two different points, where 'La' is point A longitude and '\u03b8a' is point A latitude. If you need to find out the coordinates of a site, the distance between two sites or the bearing between two sites go to my Coordinate, Distance and Bearing Calculator A new feature is dragable marker-B. Click Calculate Distance, and the tool will place a marker at each of the two addresses on the map along with a line between them. I just plug the coordinates into the Distance Formula: Then the distance is sqrt (53) , or about 7. But essentially the program should gather two sets of coordinates from a GPS module and the users phone (using the 1Sheeld) and work out the bearing and distance between them. Skills to Learn: 1. The shortest distance (the geodesic) between two given points P 1 =(lat 1, lon 1) and P 2 =(lat 2, lon 2) on the surface of a sphere with radius R is the great circle distance. Latitude/Longitude Distance Calculation This query will determine the distance between two points on the earth given their latitudes and longitudes. Latitude and/or longitude change every few inches, and if two places have the same latitude and longitude, then the distance between them is zero. The C# implementation of the Haversine formula is: public double GetDistanceBetweenPoints( double lat1, double long1, double lat2, double long2). For some websites, it is a necessity to calculate the distance between certain locations. The EASY way is to have. x A means the x-coordinate of point A y A means the y-coordinate of point A. Valid input formats are at the bottom of this page. South latitudes points are expressed in negative values, while east longitudes are positive values. The program converts degrees to radians before executing the GEODIST function. Distances calculator is a free tool to calculate distances between any two cities in the world. This query calculate the distance in miles. I'm building a system that takes a GPS reading once a second and calculates the distance between these one second intervals, and aggregates them. Further, there is a one-to-one correspondence between areal coordinates and all points on the plane P. Just a little mathematical exercise in geography, namely the use of longitude and latitiude coordinates to calculate the distance between two given cities. Travelmath helps you find driving distances based on actual directions for your road trip. Distance Calculator is use to calculate the distance between coordinates and distance between cities. angle between two lat lon points. I've found this code for calculating distance. First, note that it's very easy to get a distance between two points on a sphere if you know the angle of the arc between them: just multiply it by the radius. When unqualified, \"the\" distance generally means the shortest distance between two points. Function accepts four parameters, source latitude, source longitude, destination latitude, destination longitude and returns the distance between the. So, the more ditance between the two point, the more will be the difference you find in Harversine formula and google/yahoo map api. One additional aspect not addressed above is the nature of the terrain between the two points. What i have is the longitude and the latitude of the home adres of the own company and the longitude and the latitude of the client. Blue J users, ICSE students, BCA, B Tech & other Under graduates can stand to benefit alot. By default, MapInfo Professional expects coordinates to use a Longitude/Latitude coordinate system. Our three-dimensional distance calculator is a tool that finds the distance between two points, provided you give their coordinates in space. This website allows you to find the distance between cities or any two places and get directions using Google maps. coordinates and two.", "date": "2020-04-03 06:56:50", "meta": {"domain": "ciiz.pw", "url": "http://sbwt.ciiz.pw/distance-between-two-coordinates-calculator.html", "openwebmath_score": 0.6425132751464844, "openwebmath_perplexity": 449.43494686701194, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9770226300899743, "lm_q2_score": 0.8840392725805822, "lm_q1q2_score": 0.8637263751995081}}
{"url": "http://mathhelpforum.com/trigonometry/151257-trigonometric-equation.html", "text": "# Math Help - Trigonometric equation?\n\n1. ## Trigonometric equation?\n\nIf 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?\n\nI started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?\n\n2. Originally Posted by D7236\nIf 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?\n\nI started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?\n\nYou can use identity $\\displaystyle \\cos(\\theta + \\pi) = -\\cos(\\theta)$, to start out.\n\nEdit: Well I suppose it should be $\\displaystyle \\cos(\\theta - \\pi) = -\\cos(\\theta)$, same idea.\n\n3. You should know\n\n$\\sin{(\\alpha \\pm \\beta)} = \\sin{\\alpha}\\cos{\\beta} \\pm \\cos{\\alpha}\\sin{\\beta}$\n\nand\n\n$\\cos{(\\alpha \\pm \\beta)} = \\cos{\\alpha}\\cos{\\beta} \\mp \\sin{\\alpha}\\sin{\\beta}$.\n\n$3\\sin{\\left(x + \\frac{\\pi}{3}\\right)} = 2\\cos{\\left(x - \\frac{2\\pi}{3}\\right)}$\n\n$3\\left(\\sin{x}\\cos{\\frac{\\pi}{3}} + \\cos{x}\\sin{\\frac{\\pi}{3}}\\right) = 2\\left(\\cos{x}\\cos{\\frac{2\\pi}{3}} + \\sin{x}\\sin{\\frac{2\\pi}{3}}\\right)$\n\n$3\\left(\\frac{1}{2}\\sin{x} + \\frac{\\sqrt{3}}{2}\\cos{x}\\right) = 2\\left(-\\frac{1}{2}\\cos{x} + \\frac{\\sqrt{3}}{2}\\sin{x}\\right)$\n\n$\\frac{3}{2}\\sin{x} + \\frac{3\\sqrt{3}}{2}\\cos{x} = -\\cos{x} + \\sqrt{3}\\sin{x}$\n\n$\\left(\\frac{3}{2} - \\sqrt{3}\\right)\\sin{x} = \\left(-1 - \\frac{3\\sqrt{3}}{2}\\right)\\cos{x}$\n\n$\\left(\\frac{3 - 2\\sqrt{3}}{2}\\right)\\sin{x} = -\\left(\\frac{2 + 3\\sqrt{3}}{2}\\right)\\cos{x}$\n\n$\\frac{\\sin{x}}{\\cos{x}} = \\frac{-\\frac{2 + 3\\sqrt{3}}{2}}{\\phantom{-}\\frac{3 - 2\\sqrt{3}}{2}}$\n\n$\\tan{x} = -\\frac{2 + 3\\sqrt{3}}{3 - 2\\sqrt{3}}$\n\n$\\tan{x} = -\\frac{(2 + 3\\sqrt{3})(3 + 2\\sqrt{3})}{(3 - 2\\sqrt{3})(3 + 2\\sqrt{3})}$\n\n$\\tan{x} = -\\frac{6 + 4\\sqrt{3} + 9\\sqrt{3} + 18}{9 - 12}$\n\n$\\tan{x} = \\frac{-(24 + 13\\sqrt{3})}{-3}$\n\n$\\tan{x} = \\frac{24 + 13\\sqrt{3}}{3}$.\n\n4. Originally Posted by undefined\nYou can use identity $\\displaystyle \\cos(\\theta + \\pi) = -\\cos(\\theta)$, to start out.\n\nEdit: Well I suppose it should be $\\displaystyle \\cos(\\theta - \\pi) = -\\cos(\\theta)$, same idea.\nYes, we coud let $\\theta = x + \\frac{\\pi}{3}$, but without the angle sum and difference identities we still can't reduce this to $\\tan{x}$. So refer to my post above\n\n5. Originally Posted by D7236\nIf 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?\n\nI started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?\n\nSince Prove It posted a full solution, here's another way.\n\n$3\\sin\\left(x + \\dfrac{\\pi}{3}\\right) = 2\\cos\\left(x - \\dfrac{2\\pi}{3}\\right) = -2 \\cos\\left(x + \\dfrac{\\pi}{3}\\right)$\n\n$\\tan\\left(x+\\dfrac{\\pi}{3} \\right) = \\dfrac{-2}{3}$\n\nUse\n\n$\\displaystyle \\tan(\\alpha+\\beta)=\\frac{\\tan\\alpha+\\tan\\beta}{1-\\tan\\alpha\\tan\\beta}$\n\nif you know it, otherwise use sin and cos angle sum like Prove It.\n\nHere\n\n$\\dfrac{-2}{3} = \\dfrac{\\tan x+\\sqrt{3}}{1-\\sqrt{3}\\tan x}$\n\n$-2(1-\\sqrt{3}\\tan x) = 3(\\tan x+\\sqrt{3})$\n\n$-2+2\\sqrt{3}\\tan x = 3\\tan x+3\\sqrt{3}$\n\n$(2\\sqrt{3}-3)\\tan x = 3\\sqrt{3}+2$\n\n$\\tan x = \\dfrac{2+3\\sqrt{3}}{-3+2\\sqrt{3}}$\n\n$\\tan x = \\dfrac{(2+3\\sqrt{3})(-3-2\\sqrt{3})}{(-3+2\\sqrt{3})(-3-2\\sqrt{3})}$\n\n$\\tan x = \\dfrac{24+13\\sqrt{3}}{3}$\n\nOriginally Posted by Prove It\nYes, we coud let $\\theta = x + \\frac{\\pi}{3}$, but without the angle sum and difference identities we still can't reduce this to $\\tan{x}$. So refer to my post above\nI was busy typing and didn't see your recent post until just now. I think your way is a bit easier than mine, anyway.", "date": "2015-07-01 01:31:52", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/151257-trigonometric-equation.html", "openwebmath_score": 0.904026448726654, "openwebmath_perplexity": 648.1687989238947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308768564867, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8637027064444021}}
{"url": "https://math.stackexchange.com/questions/656331/why-is-the-empty-set-a-subset-of-every-set?noredirect=1", "text": "# Why is the empty set a subset of every set? [duplicate]\n\nTake for example the set $X=\\{a, b\\}$. I don't see $\\emptyset$ anywhere in $X$, so how can it be a subset?\n\n\u2022 \"Subset of\" means something different than \"element of\". Note $\\{a\\}$ is also a subset of $X$, despite $\\{ a \\}$ not appearing \"in\" $X$. \u2013\u00a0user14972 Jan 29 '14 at 20:10\n\u2022 Why does this question get a downvote? It is not hard to see that someone can be asking this seriously. \u2013\u00a0N. Owad Jan 29 '14 at 22:02\n\u2022 Hint: Every element of the empty set is a pink elephant. Or an element of $X.$ (No joke) \u2013\u00a0Dan Christensen Jan 30 '14 at 4:59\n\u2022 I personally like @HagenvonEitzen's question Or Can you name an element of \u2205 that is not an element of X?. If you think it like this, given \u2205 and X, you can't really find an element of \u2205 (nothing) that you don't find in X, and as a subset A is a just a set whose elements (every element) are included in another set B, that is you can't find an element in A which is not in B, it makes more sense. \u2013\u00a0user3019105 Mar 15 '18 at 20:24\n\nthat's because there are statements that are vacuously true. $Y\\subseteq X$ means for all $y\\in Y$, we have $y\\in X$. Now is it true that for all $y\\in \\emptyset$, we have $y\\in X$? Yes, the statement is vacuously true, since you can't pick any $y\\in\\emptyset$.\n\nBecause every single element of $\\emptyset$ is also an element of $X$. Or can you name an element of $\\emptyset$ that is not an element of $X$?\n\nYou must start from the definition :\n\n$$Y \\subseteq X$$ iff $$\\forall x (x \\in Y \\rightarrow x \\in X)$$.\n\nThen you \"check\" this definition with $$\\emptyset$$ in place of $$Y$$ :\n\n$$\\emptyset \\subseteq X$$ iff $$\\forall x (x \\in \\emptyset \\rightarrow x \\in X)$$.\n\nNow you must use the truth-table definition of $$\\rightarrow$$ ; you have that :\n\n\"if $$p$$ is false, then $$p \\rightarrow q$$ is true\", for $$q$$ whatever;\n\nso, due to the fact that :\n\n$$x \\in \\emptyset$$\n\nis not true, for every $$x$$, the above truth-definition of $$\\rightarrow$$ gives us that :\n\n\"for all $$x$$, $$x \\in \\emptyset \\rightarrow x \\in X$$ is true\", for $$X$$ whatever.\n\nThis is the reason why the emptyset ($$\\emptyset$$) is a subset of every set $$X$$.\n\n\u2022 Shouldn't the last implication be \"$\\text{for all x, }x \\in \\emptyset \\rightarrow x \\in X$ is true\" \u2013\u00a0mauna Jun 30 '14 at 13:20\n\nSubsets are not necessarily elements. The elements of $\\{a,b\\}$ are $a$ and $b$. But $\\in$ and $\\subseteq$ are different things.", "date": "2021-07-31 00:01:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/656331/why-is-the-empty-set-a-subset-of-every-set?noredirect=1", "openwebmath_score": 0.8098239898681641, "openwebmath_perplexity": 251.53314799414642, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308770312512, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8637027001145685}}
{"url": "https://math.stackexchange.com/questions/2804241/general-term-of-a-sequence/2804289", "text": "# General term of a sequence.\n\nSo i have the following sequence:\n\n${1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, ...}$\n\nWhere the number $i$ appears $i + 1$ times.\n\nI would like to know the $n$-th term of this sequence. I tried to analise certain patterns within the sequence, but wasn\u00b4t able to conclude anything so far.\n\nI would like to put a \"number pyramid\" like this:\n\n$$1-1$$\n\n$$2-2-2$$\n\n$$3-3-3-3$$\n\n$$4-4-4-4-4$$\n\n$$5-5-5-5-5-5$$\n\n$$\\cdots$$\n\n$$k-\\text{th floor: } k-k-k-k-k-...-k-k\\text{ (k+1 times)}$$\n\nThe number of numbers appear in all the floors from $1$ to $k-1$:\n\n$$2+3+4+5+\\cdots+k=\\dfrac{(k+2)(k-1)}{2}$$\n\nThe number of numbers appear in all the floors from $1$ to $k$:\n\n$$2+3+4+5+\\cdots+k+k+1=\\dfrac{(k+3)k}{2}$$\n\nAssume that the $n$-th term of the sequence above is on the $k$-th floor (the bottom floor of the pyramid above), then $n$ may or may not be the last term of the $k$-th floor, so this inequality must hold (we need to find $k\\in\\mathbb{Z^+}$ given $n\\in\\mathbb{Z^+}$):\n\n$\\dfrac{(k+2)(k-1)}{2}<n\\le\\dfrac{(k+3)k}{2}$\n\n$\\Leftrightarrow k^2+k-2<2n\\le k^2+3k$\n\n$\\Leftrightarrow \\begin{cases}k^2+k-2n-2<0\\\\k^2+3k-2n\\ge 0\\end{cases}$\n\n$$k^2+k-2n-2<0$$\n\n$\\Leftrightarrow k^2+2\\times k\\times 0.5+0.25-2n-2.25<0$\n\n$\\Leftrightarrow (k+0.5)^2<2n+2.25$\n\n$\\Leftrightarrow k+0.5<\\sqrt{2n+2.25}$ (because $k,n>0$)\n\n$\\Leftrightarrow k<-0.5+\\sqrt{2n+2.25}$\n\n$\\Leftrightarrow k<\\dfrac{-1+\\sqrt{8n+9}}{2}$\n\n$$k^2+3k-2n\\ge 0$$\n\n$\\Leftrightarrow (k+1.5)^2\\ge 2n+2.25$ (similar to above)\n\n$\\Leftrightarrow k+1.5\\ge \\sqrt{2n+2.25}$ (because $k,n>0$)\n\n$\\Leftrightarrow k\\ge \\dfrac{-3+\\sqrt{8n+9}}{2}$\n\nCombine both equations, we have\n\n$$\\dfrac{-3+\\sqrt{8n+9}}{2}\\le k<\\dfrac{-1+\\sqrt{8n+9}}{2}$$\n\nOther notes:\n\n\u2022 The conclusion above is still true for $n\\in\\mathbb\\{1;2\\}$ and $k=1$. When $k=1$, the number of numbers appear in all the floors from $1$ to $0$ is zero (no numbers exist), because when $k=1$ we have $\\dfrac{(k+2)(k-1)}{2}=0$.\n\n\u2022 There is always exactly one positive integer $k$ satisfy the conclusion above (for all $n\\in\\mathbb{Z^+}$), because the difference between the right hand side and the left hand side is $1$.\n\n\u2022 +1 Nice piece of work. Jun 1 '18 at 13:28\n\u2022 Thank you, I spent an hour making this just to know that the first answer was accepted before. Jun 1 '18 at 13:34\n\u2022 A shame. I am slow to accept answers for this reason. Jun 1 '18 at 15:51\n\nAccording to OEIS the general formula is $$a(n) = \\lfloor (\\sqrt{1+8n}-1)/2\\rfloor.$$\n\nIf $t_n$ denotes the value of the $n$-th term then:\n\n$$2+3+\\cdots+t_n<n\\leq2+3+\\cdots+t_n+(t_n+1)=\\frac12t_n(t_n+3)$$\n\nso that $t_n$ is the smallest integer that satisfies: $$2n\\leq t_n(t_n+3)$$leading to $t_n=\\lceil\\frac12\\sqrt{9+8n}-\\frac32\\rceil$\n\n$n \\in \\mathbb{N}$ appears $n+1$ times starting from the position $$1 + \\sum_{i=1}^{n-1}(i+1) = n + \\frac{n(n+1)}2 = \\frac12n(n+3)$$\n\nTherefore, $$a_n = k \\iff n \\in \\left[\\frac12k(k+3), k+\\frac12k(k+3)\\right]$$\n\nso $k$ is the smallest integer with $n \\ge \\frac12k(k+3)$.\n\nSolving this for $k$, we obtain $$a_n = \\left\\lceil\\frac{-3+\\sqrt{9+8n}}2\\right\\rceil, \\quad n\\in \\mathbb{N}$$\n\nGiven $\\color{red}1, 1, \\color{red}2, 2, 2, \\color{red}3, 3, 3, 3, \\color{red}4, 4, 4, 4, 4, \\color{red}5, ...$, first note that: $a_{T(k)}=k$, where $T(k)$ is a triangular number. Indeed: $$a_{T(1)}=a_1=\\color{red}1; a_{T(2)}=a_3=\\color{red}2; a_{T(3)}=a_6=\\color{red}3; a_{T(4)}=a_{10}=\\color{red}4; a_{T(5)}=a_{15}=\\color{red}5; \\ ...$$ Hence: $$T(k)\\le n\\le T(k+1)-1, a_n=k \\iff \\\\ \\frac{k(k+1)}{2}\\le n\\le \\frac{(k+1)(k+2)}{2}-1 \\iff \\\\ k^2+k-2n\\le 0\\le k^2+3k-2n \\iff \\\\ \\left\\lceil \\frac{\\sqrt{8n+9}-3}{2}\\right\\rceil\\le k\\le \\left\\lfloor \\frac{\\sqrt{8n+1}-1}{2}\\right\\rfloor$$ Hence: $$a_n=\\left\\lceil \\frac{\\sqrt{8n+9}-3}{2}\\right\\rceil \\ \\text{or} \\ \\left\\lfloor \\frac{\\sqrt{8n+1}-1}{2}\\right\\rfloor.$$", "date": "2021-12-07 03:06:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2804241/general-term-of-a-sequence/2804289", "openwebmath_score": 0.9594383239746094, "openwebmath_perplexity": 258.1828725266699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308803517763, "lm_q2_score": 0.874077222043951, "lm_q1q2_score": 0.8637026949137243}}
{"url": "https://math.stackexchange.com/questions/3122679/proving-the-borel-cantelli-lemma", "text": "# Proving the Borel-Cantelli Lemma\n\nLet $$\\{{E_k}\\}^{\\infty}_{k=1}$$ be a countable family of measurable subsets of $$\\mathbb{R}^d$$ and that $$$$% Equation (1) \\sum^{\\infty}_{k=1}m(E_k)<\\infty$$$$ Let \\begin{align*} E&=\\{x\\in \\mathbb{R}^d:x\\in E_k, \\text{ for infinitely many k }\\} \\\\ &= \\underset{k\\rightarrow \\infty}{\\lim \\sup}(E_k).\\\\ \\end{align*}\n\n(a) Show that $$E$$ is measurable\n\n(b) Prove $$m(E)=0.$$\n\nMy Proof Attempt:\n\nProof. Let the assumptions be as above. We will prove part (a) by showing that $$\\begin{equation*} E=\\cap^{\\infty}_{n=1}\\cup_{k\\geq n}E_k. \\end{equation*}$$ Hence, E would be measurable, since for every fixed $$n$$, $$\\cup_{k\\geq n}E_k$$ is measurable since it is a countable union of measurable sets. Then $$\\cap^{\\infty}_{n=1}\\cup_{k\\geq n}E_k$$ is the countable intersection of measurable sets.\n\nFrom here, we shall denote $$\\cup_{k\\geq n}E_k$$ as $$S_n$$. Let $$x\\in \\cap^{\\infty}_{n=1}S_n$$. Then $$x\\in S_n$$ for every $$n\\in \\mathbb{N}$$. Hence, $$x$$ must be in $$E_k$$ for infinitely many $$k$$, otherwise there would exist an $$N\\in \\mathbb{N}$$ such that $$x\\notin S_N$$. Leaving $$x$$ out of the intersection. Thus, $$\\cap^{\\infty}_{n=1}S_n\\subset E$$.\n\nConversely, let $$x\\in E.$$ Then $$x\\in E_k$$ for infinitely many $$k$$. Therefore, $$\\forall n\\in \\mathbb{N}$$, $$x\\in S_n$$. Otherwise, $$\\exists N\\in \\mathbb{N}$$ such that $$x\\notin S_N$$. Which would imply that $$x\\in E_k$$ for only $$k$$ up to $$N$$, i.e. finitely many. A contradiction. Therefore, $$x\\in \\cap^{\\infty}_{n=1}S_n$$. Hence, they contain one another and equality holds. This proves part (1).\n\nNow for part (b). Fix $$\\epsilon>0$$. We need to show that there exists $$N\\in \\mathbb{N}$$ such that $$\\begin{equation*} m(S_N)\\leq \\epsilon \\end{equation*}$$ Then since $$E\\subset S_N$$, monotonicity of the measure would imply that $$m(E)\\leq \\epsilon$$. Hence, proving our desired conclusion as we let $$\\epsilon \\rightarrow 0$$.\n\nSince $$\\sum^{\\infty}_{k=1}m(E_k)<\\infty$$, there exists $$N\\in \\mathbb{N}$$ such that $$\\begin{equation*} \\left| \\sum^{\\infty}_{k=N}m(E_k)\\right |\\leq \\epsilon \\end{equation*}$$ By definition, $$\\begin{equation*} m(S_N)=m(\\cup_{k\\geq N}E_k)=\\sum^{\\infty}_{k=N}m(E_k) \\end{equation*}$$ Thus, $$m(S_N)\\leq \\epsilon$$. This completes our proof.\n\nAny corrections of the proof, or comments on style of the proof are welcome and appreciated. Thank you all for your time.\n\nThe proof is almost perfect, only in the end it is not necessary true that $$m(\\cup_{k\\geq N}E_k)=\\sum_{k=N}^\\infty m(E_k)$$ since the sets $$E_k$$ might not be pairwise disjoint. So the measure of the union is only at most the sum of measures. But in our case it doesn't change anything for the proof, since we anyway get $$m(S_N)\\leq\\sum_{k=N}^\\infty m(E_k)\\leq\\epsilon$$. Still it is important to remember the correct properties of measure.\nYour proof is fine, modulo the comment in the other answer. For another approach, which I think is the way Rudin does it, note that $$x\\in E$$ if and only if the series $$\\sum^{\\infty}_{k=1}1_{E_k}(x)$$ diverges.\nSet $$s_n(x)=\\sum^{n}_{k=1}1_{E_k}(x).$$ Then, $$s_n(x)\\to s(x)=\\sum^{\\infty}_{k=1}1_{E_k}(x)$$ and the monotone converge theorem gives $$\\sum^{n}_{k=1}m(E_k)\\to \\sum^{\\infty}_{k=1}m(E_k)<\\infty.$$ Thus, $$s\\in L^1(m)$$, so the series converges almost everywhere $$m$$. That is, the set on which it diverges, namely $$E$$, has Lebesgue measure zero and so $$m(E)=0.$$\nRemark: since we proved that $$m(E)=0,$$ we get part $$(a)$$ for free.", "date": "2019-07-16 06:09:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3122679/proving-the-borel-cantelli-lemma", "openwebmath_score": 0.9980306625366211, "openwebmath_perplexity": 107.89735990674784, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806489800368, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8636925810822814}}
{"url": "https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-90-computational-methods-in-aerospace-engineering-spring-2014/numerical-integration-of-ordinary-differential-equations/systems-of-odes-and-eigenvalue-stability/1690r-eigenvalue-stability-for-a-linear-ode/", "text": "# 1.6 Systems of ODE's and Eigenvalue Stability\n\n## 1.6.3 Eigenvalue Stability for a Linear ODE\n\nAs we have seen, while numerical methods can be convergent, they can still exhibit instabilities as $$n$$ increases for finite $${\\Delta t}$$. For example, when applying the midpoint method to either the ice particle problem in Section\u00a01.2.4 or the simpler model problem in Example\u00a01.66, instabilities were seen in both cases as $$n$$ increased. Similarly, for the nonlinear pendulum problem in Example\u00a01.86, the forward Euler method had a growing amplitude again indicating an instability. The key to understanding these results is to analyze the stability for finite $${\\Delta t}$$. This analysis is different than the stability analysis we performed in Section\u00a01.5.2 since that analysis was for the limit of $${\\Delta t}\\rightarrow 0$$.\n\nSuppose we are interested in solving the linear ODE,\n\n $u_ t = \\lambda u.$ (1.99)\n\nConsider the Forward Euler method applied to this problem,\n\n $v^{n+1} = v^ n + \\lambda {\\Delta t}v^ n. \\label{equ:fe_ lin}$ (1.100)\n\nSimilar to the zero stability analysis, we will assume that the solution has the following form,\n\n $v^{n} = g^ n v^0, \\label{equ:gdef}$ (1.101)\n\nwhere $$g$$ is the amplification factor (and the superscript $$n$$ acting on $$g$$ is again raising to a power). As in the zero stability analysis, we wish to determine under what conditions $$|g| > 1$$ since this would mean that $$v^ n$$ will grow unbounded as $$n \\rightarrow \\infty$$. Substituting Equation\u00a01.101 into Equation\u00a01.100 gives,\n\n $g^{n+1} = (1 + \\lambda {\\Delta t})g^ n.$ (1.102)\n\nThus, the only non-zero root of this equation gives,\n\n $g = 1 + \\lambda {\\Delta t},$ (1.103)\n\nwhich is the amplification factor for the forward Euler method. Now, we must determine what values of $$\\lambda {\\Delta t}$$ lead to instability (or stability). A simple way to do this for multi-step methods is to solve for the stability boundary for which $$|g| = 1$$. To do this, let $$g = e^{i\\theta }$$ (since $$|e^{i\\theta }| = 1$$) where $$\\theta = [0,2\\pi ]$$. Making this substitution into the amplification factor,\n\n $e^{i\\theta } = 1 + \\lambda {\\Delta t}\\quad \\Rightarrow \\quad \\lambda {\\Delta t}= e^{i\\theta } - 1.$ (1.104)\n\nThus, the stability boundary for the forward Euler method lies on a circle of radius one centered at -1 along the real axis and is shown in Figure\u00a01.10.\n\nFigure 1.10: Forward Euler stability region\n\nFor a given problem, i.e. with a given $$\\lambda$$, the timestep must be chosen so that the algorithm remains stable for $$n \\rightarrow \\infty$$. Let's consider some examples.\n\n## Example\n\nLet's return to the previous example, $$u_ t = -u^2$$ with $$u(0) = 1$$. To determine the timestep restrictions, we must estimate the eigenvalue for this problem. Linearizing this problem about a known state gives the eigenvalue as $$\\lambda = {\\partial f}/{\\partial u} = -2u$$. Since the solution will decay from the initial condition (since $$u_ t < 0$$ because $$-u^2 < 0$$), the largest magnitude of the eigenvalue occurs at the initial condition when $$u(0) = 1$$ and thus, $$\\lambda = -2$$. Since this eigenvalue is a negative real number, the maximum $${\\Delta t}$$ will occur at the maximum extent of the stability region along the negative real axis. Since this occurs when $$\\lambda {\\Delta t}= -2$$, this implies that $${\\Delta t}< 1$$. To test the validity of this analysis, the forward Euler method was run for $${\\Delta t}= 0.9$$ and $${\\Delta t}= 1.1$$. The results are shown in Figure\u00a01.11 which are stable for $${\\Delta t}= 0.9$$ but are unstable for $${\\Delta t}= 1.1$$.\n\nFigure 1.11: Forward Euler solution for $$u_ t = -u^2$$ with $$u(0) = 1$$ with $${\\Delta t}= 0.9$$ and $$1.1$$.\n\n## Pendulum Example\n\nNext, let's consider the application of the forward Euler method to the pendulum problem. For this case, the linearization produces a matrix,\n\n $\\frac{\\partial f}{\\partial u} = \\left(\\begin{array}{cc} 0 & -\\frac{g}{L}\\cos \\theta \\\\ 1 & 0 \\end{array}\\right)$ (1.105)\n\nThe eigenvalues can be found from the roots of the determinant of $${\\partial f}/{\\partial u} - \\lambda I$$:\n\n $$\\displaystyle \\det \\left(\\frac{\\partial f}{\\partial u} - \\lambda I\\right)$$ $$\\displaystyle =$$ $$\\displaystyle \\det \\left(\\begin{array}{cc} -\\lambda & -\\frac{g}{L}\\cos \\theta \\\\ 1 & -\\lambda \\end{array}\\right)$$ (1.106) $$\\displaystyle =$$ $$\\displaystyle \\lambda ^2 + \\frac{g}{L}\\cos \\theta = 0$$ (1.107) $$\\displaystyle \\Rightarrow$$ $$\\displaystyle \\lambda = \\pm i \\sqrt {\\frac{g}{L}\\cos \\theta }$$ (1.108)\n\nThus, we see that the eigenvalues will always be imaginary for this problem. As a result, since the forward Euler stability region does not contain any part of the imaginary axis (except the origin), no finite timestep exists which will be stable. This explains why the amplitude increases for the pendulum simulations in Figure\u00a01.8.", "date": "2021-07-23 23:00:04", "meta": {"domain": "mit.edu", "url": "https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-90-computational-methods-in-aerospace-engineering-spring-2014/numerical-integration-of-ordinary-differential-equations/systems-of-odes-and-eigenvalue-stability/1690r-eigenvalue-stability-for-a-linear-ode/", "openwebmath_score": 0.954245388507843, "openwebmath_perplexity": 246.31812237750881, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.992062006602671, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8636864763708112}}
{"url": "http://math.stackexchange.com/questions/44835/for-integers-a-and-b-ab-textlcma-b-cdot-texthcfa-b", "text": "# For integers $a$ and $b$, $ab=\\text{lcm}(a,b)\\cdot\\text{hcf}(a,b)$\n\nI was reading a text book and came across the following:\n\nImportant Results\n(This comes immediately after LCM:)\n\nIf 2 [integers] $a$ and $b$ are given, and their $LCM$ and $HCF$ are $L$ and $H$ respectively,\nthen $L \\times H = a \\times b$\n\n-\nI take it that HCF is gcd? (Highest common factor; greatest common divisor?) \u2013\u00a0amWhy Jun 11 '11 at 21:42\nYes that's correct @amWhy! \u2013\u00a0peakit Jun 12 '11 at 3:36\n\nLet $p$ be a prime. If $p$ occurs in $a$ with multiplicity $m$ and in $b$ with multiplicity $n$, then it will occur in the LCM of $a$ and $b$ with multiplicity $\\mathrm{max(m,n)}$ and in their HCF with multiplicity $\\mathrm{min(m,n)}$.\n\nHence, in the product of LCM and HCF the multiplicity of $p$ is $$\\mathrm{max}(m,n)+\\mathrm{min}(m,n)=m+n,$$ which is also the multiplicity of $p$ in $a\\cdot b$. Since this holds for every $p$, the two products must be equal.\n\n-\nThis is an example of the \"modular equation\" $|A| + |B| = |A \\cup B| + |A \\cap B|$. \u2013\u00a0Yuval Filmus Jun 11 '11 at 21:35\nBut this assumes $p$ occurs with a well-defined multiplicity in $a$ and $b$, which is to say it assumes unique factorization. Has the text already done unique factorization when it introduces LCM? \u2013\u00a0Gerry Myerson Jun 11 '11 at 21:42\nThanks @Ramsus, your answer helped a lot! Another way to logically deduce this is: LCM of 2 numbers will cover the max multiplicity for each of the prime factors, but HCF will cover the min multiplicity for each of the prime factors and simple multiplication covers 'sum' of multiplicity of the prime factor which is same as LCM multiplied by HCF. \u2013\u00a0peakit Jun 12 '11 at 3:35\n@peakit: to me, that sounds like exactly what I wrote. \u2013\u00a0Rasmus Jun 12 '11 at 9:47\n\nBelow is a proof that works in any domain, using the universal definitions of GCD, LCM.\n\nTHEOREM $\\rm\\quad (a,b)\\ =\\ ab/[a,b] \\;\\;$ if $\\;\\rm\\ [a,b] \\;$ exists.\n\nProof: $\\rm\\qquad d\\mid (a,b)\\iff d\\mid a,b \\iff a,b\\mid ab/d \\iff [a,b]\\mid ab/d \\iff d\\mid ab/[a,b]$\n\n-\nUser is \"Gone\" but the proof is excellent. It's too bad this wasn't the accepted answer, in view of its generality and Gerry Myerson's apt comment below the accepted answer. \u2013\u00a0user43208 Oct 20 '13 at 16:18\n\nYou can also figure this out without using any use of unique factorization. Since you say this comes immediately after LCM, I assume you know that for $m\\gt 0$, $[ma,mb]=m[a,b]$, where $[a,b]=\\mathrm{lcm}(a,b)$ and $(ma,mb)=m(a,b)$ where $\\gcd(a,b)=(a,b)$.\n\nNow suppose $(a,b)=1$, and also assume they are positive, since if they are negative $[a,-b]=[a,b]$ anyway. Since $[a,b]$ is some multiple of $a$, let $[a,b]=ma$. Then $b\\mid ma$, but $(a,b)=1$, so $b\\mid m$. If this hasn't be addressed yet, notice $(ma,mb)=m(a,b)=m$, so $b\\mid ma$, and $b\\mid mb$, so $b\\mid m$ since any common divisor of $ma$ and $mb$ divides the greatest common divisor $m$ in this case.\n\nSo $b\\leq m$ as they are both positive, which implies $ba\\leq ma$. But $ba$ is a common multiple of $a$ and $b$, so cannot be strictly less than $ma$, so $ba=ma=[a,b]$.\n\nMore generally, if $(a,b)=g\\gt 1$, then you have $(\\frac{a}{g},\\frac{b}{g})=1$. Then by the special case above, $$\\left[\\frac{a}{g},\\frac{b}{g}\\right]\\left(\\frac{a}{g},\\frac{b}{g}\\right)=\\frac{a}{g}\\frac{b}{g}.$$ Multiply through by $g^2$, you have \\begin{align*} g^2\\left[\\frac{a}{g},\\frac{b}{g}\\right]\\left(\\frac{a}{g},\\frac{b}{g}\\right) &= g\\left[\\frac{a}{g},\\frac{b}{g}\\right]g\\left(\\frac{a}{g},\\frac{b}{g}\\right)\\\\ &= [a,b](a,b)\\\\ &= g^2\\frac{a}{g}\\frac{b}{g}=ab \\end{align*} so $[a,b](a,b)=|ab|$.\n\n-\nI suppose this doesn't use unique factorization, but it does use $b\\mid ma$, $\\gcd(a,b)=1$ implies $b\\mid m$, which may or may not have been done by the text at this point. \u2013\u00a0Gerry Myerson Jun 12 '11 at 13:11\n@Gerry, sure, but I think this is a pretty easy consequence of $(ma,mb)=m(a,b)$ either way. \u2013\u00a0yunone Jun 12 '11 at 21:20", "date": "2016-05-03 01:25:24", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/44835/for-integers-a-and-b-ab-textlcma-b-cdot-texthcfa-b", "openwebmath_score": 0.9828641414642334, "openwebmath_perplexity": 252.52674125054665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018376422666, "lm_q2_score": 0.8856314783461302, "lm_q1q2_score": 0.8636694451569834}}
{"url": "https://gmatclub.com/forum/there-are-different-10-circles-what-is-the-number-of-the-greatest-pos-217589.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Dec 2018, 11:59\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in December\nPrevNext\nSuMoTuWeThFrSa\n2526272829301\n2345678\n9101112131415\n16171819202122\n23242526272829\n303112345\nOpen Detailed Calendar\n\u2022 ### 10 Keys to nail DS and CR questions\n\nDecember 17, 2018\n\nDecember 17, 2018\n\n06:00 PM PST\n\n07:00 PM PST\n\nJoin our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong.\n\u2022 ### R1 Admission Decisions: Estimated Decision Timelines and Chat Links for Major BSchools\n\nDecember 17, 2018\n\nDecember 17, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nFrom Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.\n\n# There are different 10 circles. What is the number of the greatest pos\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Revolution GMAT Instructor\nJoined: 16 Aug 2015\nPosts: 6656\nGMAT 1: 760 Q51 V42\nGPA: 3.82\nThere are different 10 circles. What is the number of the greatest pos\u00a0 [#permalink]\n\n### Show Tags\n\n29 Apr 2016, 19:16\n2\n10\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n59% (01:04) correct 41% (01:30) wrong based on 167 sessions\n\n### HideShow timer Statistics\n\nThere are different 10 circles. What is the number of the greatest possible points with which the circles intersect?\n\nA. 90\nB. 100\nC. 110\nD. 180\nE. 200\n\n* A solution will be posted in two days.\n\n_________________\n\nMathRevolution: Finish GMAT Quant Section with 10 minutes to spare\nThe one-and-only World\u2019s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.\n\"Only $99 for 3 month Online Course\" \"Free Resources-30 day online access & Diagnostic Test\" \"Unlimited Access to over 120 free video lessons - try it yourself\" ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 7111 Re: There are different 10 circles. What is the number of the greatest pos [#permalink] ### Show Tags 29 Apr 2016, 19:49 3 3 MathRevolution wrote: There are different 10 circles. What is the number of the greatest possible points with which the circles intersect? A. 90 B. 100 C. 110 D. 180 E. 200 * A solution will be posted in two days. Hi, if someone is not aware of the formula, you can easily do the Q through systematic approach.. VISUALIZATION can help us in these type of Qs, first circle - no intersection second circle - 2-points third circle- two existing circles s0 2*2 - 4.. and so on till 10th - 9 existing circle and 2-points on each = 2*9=18.. $$TOTAL = 0+2+4+...16+18 = 2(1+2+3+...8+9) = 2*9*\\frac{10}{2} = 90$$ A _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor ##### General Discussion SC Moderator Joined: 13 Apr 2015 Posts: 1688 Location: India Concentration: Strategy, General Management GMAT 1: 200 Q1 V1 GPA: 4 WE: Analyst (Retail) Re: There are different 10 circles. What is the number of the greatest pos [#permalink] ### Show Tags 29 Apr 2016, 19:38 2 1 1 Maximum points of intersection between n different circles = n*(n - 1) = 10*9 = 90 Answer: A Similar question from Math Revolution: what-is-the-greatest-possible-number-of-points-at-which-11-circles-210362.html Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6656 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: There are different 10 circles. What is the number of the greatest pos [#permalink] ### Show Tags 04 May 2016, 06:24 1 1 2+2*2+\u2026+2*9=2(1+2+\u2026+9)=2(9)(10)/2=90 Hence, the correct answer is A. - Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World\u2019s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. \"Only$99 for 3 month Online Course\"\n\"Free Resources-30 day online access & Diagnostic Test\"\n\"Unlimited Access to over 120 free video lessons - try it yourself\"\n\nDirector\nStatus: Tutor - BrushMyQuant\nJoined: 05 Apr 2011\nPosts: 610\nLocation: India\nConcentration: Finance, Marketing\nSchools: XLRI (A)\nGMAT 1: 700 Q51 V31\nGPA: 3\nWE: Information Technology (Computer Software)\nRe: There are different 10 circles. What is the number of the greatest pos\u00a0 [#permalink]\n\n### Show Tags\n\n11 May 2017, 08:36\n1\nTop Contributor\nMathRevolution wrote:\nThere are different 10 circles. What is the number of the greatest possible points with which the circles intersect?\n\nA. 90\nB. 100\nC. 110\nD. 180\nE. 200\n\n* A solution will be posted in two days.\n\n_________________\n\nAnkit\n\nCheck my Tutoring Site -> Brush My Quant\n\nGMAT Quant Tutor\nHow to start GMAT preparations?\nHow to Improve Quant Score?\nGmatclub Topic Tags\nCheck out my GMAT debrief\n\nHow to Solve :\nStatistics || Reflection of a line || Remainder Problems || Inequalities\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 51263\nRe: There are different 10 circles. What is the number of the greatest pos\u00a0 [#permalink]\n\n### Show Tags\n\n11 May 2017, 08:38\nBrushMyQuant wrote:\nMathRevolution wrote:\nThere are different 10 circles. What is the number of the greatest possible points with which the circles intersect?\n\nA. 90\nB. 100\nC. 110\nD. 180\nE. 200\n\n* A solution will be posted in two days.\n\n_______________\nDone. Thank you.\n_________________\nDirector\nJoined: 31 Jul 2017\nPosts: 505\nLocation: Malaysia\nGMAT 1: 700 Q50 V33\nGPA: 3.95\nWE: Consulting (Energy and Utilities)\nRe: There are different 10 circles. What is the number of the greatest pos\u00a0 [#permalink]\n\n### Show Tags\n\n29 Jul 2018, 04:58\nMathRevolution wrote:\nThere are different 10 circles. What is the number of the greatest possible points with which the circles intersect?\n\nA. 90\nB. 100\nC. 110\nD. 180\nE. 200\n\n* A solution will be posted in two days.\n\nMaximum point of Intersection between n different Circles. = 2 *nC2, where n >= 2.\n_________________\n\nIf my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!\n\nRe: There are different 10 circles. What is the number of the greatest pos &nbs [#permalink] 29 Jul 2018, 04:58\nDisplay posts from previous: Sort by", "date": "2018-12-17 19:59:56", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/there-are-different-10-circles-what-is-the-number-of-the-greatest-pos-217589.html", "openwebmath_score": 0.5327728390693665, "openwebmath_perplexity": 12661.079443611106, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9752018398044143, "lm_q2_score": 0.8856314723088732, "lm_q1q2_score": 0.8636694411843053}}
{"url": "http://math.stackexchange.com/questions/132768/proof-of-equivalence", "text": "# Proof of equivalence?\n\nHow do I prove that if two numbers $a$ and $N$ are co-prime, then in the equation:\n\n$$ax \u2261 ay \\pmod N$$\n\nnecessarily $x \u2261 y \\pmod N$\n\n-\n(Standard boilerplate): What have you tried? Is this homework? Are you working from a certain course/textbook on number theory? If you can avoid imperative words (\"Prove\") and favor infinitive questions (\"How do I prove...?\"), you might find a more fulfilling response from other users! \u2013\u00a0 The Chaz 2.0 Apr 17 '12 at 1:23\nThanks for the advice! I'll keep it in mind next time. This isn't homework, just a bit of reading I am doing. \u2013\u00a0 user26649 Apr 17 '12 at 1:36\n\n$ax \u2261 ay$ $(mod$ $N$) implies that $ax = ay + pN$ where $p \\in \\mathbb{Z}$. Then by subtracting $ay$ from both sides, we see that $ax - ay = a(x-y) = pN$. $a$ divides the left hand side of the equation, so it also must divide $pN$. But because $a \\mid pN$ and $\\gcd(a, N) = 1$, it must be the case that $a \\mid p$. So there exists an integer $m$ such that $am = p$.\n\nThen going back to $ax = ay + pN$, we can rewrite it as $ax = ay + (am)N$. If we divide the equation by $a$, we get $x = y + mN$. So we get $x \\equiv y$ $(mod$ $N$)\n\n-\nYou completely lost me after you arrived at a(x-y)= pN. Could you please explain the rest with a bit more detail? \u2013\u00a0 user26649 Apr 17 '12 at 3:47\n@FarhadYusufali: $a \\mid a(x-y)$ means that there is an integer $k$ such that $ak = a(x-y)$. Do you see what $k$ should be? Since $a(x-y) = pN$, by substitution we see that $a \\mid pN$. Then since we know $a \\mid pN$ and $\\gcd(a, N) = 1$ that means $a$ and $N$ do not have any factors in common, so $a \\mid p$. \u2013\u00a0 Student Apr 17 '12 at 15:34\nAwesome thanks! \u2013\u00a0 user26649 Apr 17 '12 at 15:53\n\n$ax \\equiv ay \\mod N \\implies N | (ax - ay) \\implies N|a(x-y)$\n\nBut $N$ doesn't divide $a$, so $N | x-y \\implies x \\equiv y \\mod N$\n\nHere, I used that if $(c,d) = 1$, then $c | de \\implies c | e$. If that's not immediately obvious, or known, try to prove that first.\n\n-\n\nHint $\\rm\\ (a,n) = 1,\\ n\\:|\\:az\\:\\Rightarrow\\:n\\:|\\:az,nz\\:\\Rightarrow\\:n\\:|\\:(az,nz) = (a,n)z = z.\\:$ Now put $\\rm\\:z = x-y.$\n\n-\n\nHint: If $\\gcd(a,b)=1$ then there are $x,y\\in\\mathbb Z$ so that $ax+by=1$.\n\n-", "date": "2015-05-23 14:06:09", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/132768/proof-of-equivalence", "openwebmath_score": 0.9406384229660034, "openwebmath_perplexity": 217.4848934636518, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018405251303, "lm_q2_score": 0.8856314617436728, "lm_q1q2_score": 0.8636694315193912}}
{"url": "https://math.stackexchange.com/questions/3756066/modifying-frac-prod-alpha-a-alpha-prod-alpha-b-alpha-simeq-prod-alph", "text": "# Modifying $\\frac{\\prod_\\alpha A_\\alpha}{\\prod_\\alpha B_\\alpha}\\simeq \\prod_\\alpha\\frac{A_\\alpha}{B_\\alpha}$ for direct sums\n\nLet $$\\{A_\\alpha\\}$$ be a family of $$R$$-modules, each $$B_\\alpha\\subset A_\\alpha$$ a submodule and $$\\pi_\\alpha:A_\\alpha\\to A_\\alpha/B_\\alpha$$ be the canonical projection map. Then the map\n\n$$\\prod_\\alpha\\pi_\\alpha:\\prod_\\alpha A_\\alpha\\to \\prod_\\alpha\\frac{A_\\alpha}{B_\\alpha}$$\n\nis surjective, and has kernel $$\\prod_\\alpha B_\\alpha$$. Therefore, by the first isomorphism theorem we have\n\n$$\\frac{\\prod_\\alpha A_\\alpha}{\\prod_\\alpha B_\\alpha}\\simeq \\prod_\\alpha\\frac{A_\\alpha}{B_\\alpha}$$\n\nWhat I'm curious about is how to modify this proof for direct sums. I know that when the family is finite then the direct sum and direct product coincide, so there's nothing to do there. It's when it's an infinite family where I'm uncertain. With $$\\bigoplus_\\alpha A_\\alpha$$ then only finitely many components are non-zero, but I'm not sure if that means I'd need to alter the argument to account for this, or if it can simply be applied to direct sums as well to show that\n\n$$\\frac{\\bigoplus_\\alpha A_\\alpha}{\\bigoplus_\\alpha B_\\alpha}\\simeq \\bigoplus_\\alpha\\frac{A_\\alpha}{B_\\alpha}$$\n\nSo my question is, is an alteration to the argument above necessary for infinite direct sums?\n\n\u2022 The argument is exactly the same, no? Rewrite everything with $\\bigoplus$ in place of $\\prod$. Jul 13, 2020 at 23:36\n\u2022 @Batominovski Well that's what I'm not 100% sure about. Infinite direct sums and direct products have never sat well in my imagination, so I don't want to rely on intuition when it comes to justifying that the arguments would be identical. Jul 13, 2020 at 23:39\n\nThis answer here is solely for the purpose of giving this question an answer. Since the OP obtained the desired answer (see the comments under the question), I am providing a different way using category theory to show that $$P:=\\frac{\\prod\\limits_{\\alpha\\in J}\\,A_\\alpha}{\\prod\\limits_{\\alpha\\in J}\\,B_\\alpha}\\cong \\prod_{\\alpha\\in J}\\,\\frac{A_\\alpha}{B_\\alpha}\\text{ and }S:=\\frac{\\bigoplus\\limits_{\\alpha\\in J}\\,A_\\alpha}{\\bigoplus\\limits_{\\alpha\\in J}\\,B_\\alpha}\\cong \\bigoplus_{\\alpha\\in J}\\,\\frac{A_\\alpha}{B_\\alpha}\\,.$$ Explicit isomorphisms can be seen in (*) and (#).\nFor each $$\\beta \\in J$$, $$\\iota_\\beta:A_\\beta\\to \\bigoplus\\limits_{\\alpha\\in J}\\,A_\\alpha$$ and $$\\pi_\\beta: \\prod\\limits_{\\alpha\\in J}\\,A_\\alpha\\to A_\\beta$$ denote the canonical injection and the canonical projection, respectively. Let $$q:\\bigoplus\\limits_{\\alpha\\in J}\\,A_\\alpha\\to S$$ be the quotient map. Then, $$q\\circ \\iota_\\beta$$ vanishes on $$B_\\beta$$. Therefore, $$q\\circ \\iota_\\beta$$ factors through the quotient map $$q_\\beta:A_\\beta\\to\\dfrac{A_\\beta}{B_\\beta}$$. In other words, there exists a (unique) map $$i_\\beta:\\dfrac{A_\\beta}{B_\\beta}\\to S$$ such that $$q\\circ \\iota_\\beta=i_\\beta\\circ q_\\beta\\,.$$ We claim that $$S$$ together with the maps $$i_\\beta:\\dfrac{A_\\beta}{B_\\beta}\\to S$$ for $$\\beta\\in J$$ is a categorical coproduct (direct sum) of the family $$\\left(\\dfrac{A_\\alpha}{B_\\alpha}\\right)_{\\alpha\\in J}$$. Let $$T$$ be any $$R$$-module together with morphisms $$\\tau_\\beta:\\dfrac{A_\\beta}{B_\\beta}\\to T$$ for each $$\\beta\\in J$$. We want to show that a there exists a unique morphism $$\\phi:S\\to T$$ such that $$\\phi\\circ i_\\beta=\\tau_\\beta$$ for each $$\\beta\\in J$$.\nWe define $$\\phi\\left((a_\\alpha)_{\\alpha\\in J}+\\bigoplus_{\\alpha\\in J}\\,B_\\alpha\\right):=\\sum_{\\alpha\\in J}\\,\\tau_\\alpha\\left(a_\\alpha+B_\\alpha\\right)\\text{ for all }(a_\\alpha)_{\\alpha\\in J}\\in\\bigoplus_{\\alpha\\in J}\\,A_\\alpha\\,.$$ It is easy to verified that $$\\phi$$ is a well defined morphism, and it is the only morphism such that $$\\phi\\circ i_\\beta=\\tau_\\beta$$ for all $$\\beta\\in J$$. We can now then conclude that $$S$$ is a coproduct of the family $$\\left(\\dfrac{A_\\alpha}{B_\\alpha}\\right)_{\\alpha\\in J}$$. Since coproducts are unique up to isomorphism, we obtain $$S\\cong \\bigoplus\\limits_{\\alpha \\in J}\\,\\dfrac{A_\\alpha}{B_\\alpha}$$, via the isomorphism $$\\sigma:S\\to \\bigoplus\\limits_{\\alpha \\in J}\\,\\dfrac{A_\\alpha}{B_\\alpha}$$ given by $$\\sigma\\left((a_\\alpha)_{\\alpha\\in J}+\\bigoplus_{\\alpha\\in J}\\,B_\\alpha\\right):=\\sum_{\\alpha\\in J}\\,\\bar{\\iota}_\\alpha\\left(a_\\alpha+B_\\alpha\\right)\\text{ for all }(a_\\alpha)_{\\alpha\\in J}\\in\\bigoplus_{\\alpha\\in J}\\,A_\\alpha\\,,\\tag{*}$$ where $$\\bar{\\iota}_\\beta:\\dfrac{A_\\beta}{B_\\beta}\\to \\bigoplus\\limits_{\\alpha\\in J}\\,\\dfrac{A_\\alpha}{B_\\alpha}$$ is the canonical injection for each $$\\beta\\in J$$.\nObserve now that, for every $$\\beta\\in J$$, $$q_\\beta\\circ \\pi_\\beta$$ vanishes on $$\\prod\\limits_{\\alpha\\in J}\\,B_\\alpha$$. Therefore, $$q_\\beta\\circ\\pi_\\beta$$ factors through the quotient map $$k:\\prod\\limits_{\\alpha\\in J}\\,A_\\alpha\\to P$$. Ergo, there exists a (unique) morphism $$\\varpi_\\beta:P\\to \\dfrac{A_\\beta}{B_\\beta}$$ such that $$q_\\beta\\circ\\pi_\\beta=\\varpi_\\beta\\circ k\\,.$$ We claim that $$P$$ together with the morphisms $$\\varpi:P\\to \\dfrac{A_\\beta}{B_\\beta}$$ is a categorical product (direct product) of the family $$\\left(\\dfrac{A_\\alpha}{B_\\alpha}\\right)_{\\alpha\\in J}$$. Let $$Q$$ be any $$R$$-module together with morphisms $$\\kappa_\\beta:Q\\to\\dfrac{A_\\beta}{B_\\beta}$$ for all $$\\beta\\in J$$. We need to show that there exists a unique morphism $$\\psi:Q\\to P$$ such that $$\\varpi_\\beta\\circ \\psi=\\kappa_\\beta$$ for all $$\\beta\\in J$$.\nWe define $$\\psi\\left(x\\right):=\\big(\\kappa_\\alpha(x)\\big)_{\\alpha\\in J}+\\prod_{\\alpha\\in J}\\,B_\\alpha\\text{ for all }x\\in Q\\,.$$ It is easily seen that $$\\psi$$ is a well defined morphism, and it is the only morphism such that $$\\varpi_\\beta\\circ \\psi=\\kappa_\\beta$$ for all $$\\beta\\in J$$. We now conclude that $$P$$ is indeed a product of the family $$\\left(\\dfrac{A_\\alpha}{B_\\alpha}\\right)_{\\alpha\\in J}$$. Since products are unique up to isomorphism, we have $$P\\cong \\prod\\limits_{\\alpha\\in J}\\,\\dfrac{A_\\alpha}{B_\\alpha}$$ via the isomorphism $$\\varsigma: \\prod\\limits_{\\alpha\\in J}\\,\\dfrac{A_\\alpha}{B_\\alpha}\\to P$$ given by $$\\varsigma\\Big(\\big(a_\\alpha+B_\\beta\\big)_{\\alpha\\in J}\\Big):=\\big(a_\\alpha\\big)_{\\alpha\\in J}+\\prod_{\\alpha\\in J}\\,B_\\alpha\\text{ for all }\\big(a_\\alpha\\big)_{\\alpha\\in J}\\in \\prod_{\\alpha\\in J}\\,A_\\alpha\\,.\\tag{#}$$", "date": "2022-07-01 07:27:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3756066/modifying-frac-prod-alpha-a-alpha-prod-alpha-b-alpha-simeq-prod-alph", "openwebmath_score": 0.9745703339576721, "openwebmath_perplexity": 59.39496517026847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513910054508, "lm_q2_score": 0.8757869965109765, "lm_q1q2_score": 0.8636585648337853}}
{"url": "http://www.summitcontractors.co.uk/nu4xa/7e6f32-maximum-flow-problem-with-vertex-capacities", "text": "We can construct a network Definition. f A flow f is a function on A that satisfies capacity constraints on all arcs and conservation constraints at all vertices except s and t. The capacity constraint for a A is 0 f(a) u(a) (flow does not exceed capacity). In 1955, Lester R. Ford, Jr. and Delbert R. Fulkerson created the first known algorithm, the Ford\u2013Fulkerson algorithm. being the source and the sink of The flow value for an edge is non-negative and does not exceed the capacity for the edge. } {\\displaystyle t} \u2022 Maximum flow problems find a feasible flow through a single-source, single-sink flow network that is maximum. ) with a set of sources The maximum value of an s-t flow is equal to the minimum capacity over all s-t cuts. V and Y {\\displaystyle G} Perform one iteration of Ford-Fulkerson. 2. The main theorem links the maximum flow through a network with the minimum cut of the network. Max-Flow with Multiple Sources: There are multiple source nodes s 1, . {\\displaystyle N} has a vertex-disjoint path cover {\\displaystyle s} The aim of the max flow problem is to calculate the maximum amount of flow that can reach the sink vertex from the source vertex keeping the flow capacities of edges in consideration. Only edges with positive capacities are needed. {\\displaystyle s} units of flow on edge it is given by: Definition. \u2032 The dynamic version of the maximum flow problem allows the graph underlying the flow network to change over time. However, this reduction does not preserve the planarity of the graph. The capacity of the cut is the sum of the capacities of the arcs in the cut pointing from S s to S t. It is a fundamental result that Max Flow = Min Cut. The residual capacity of an edge is equal to the original flow capacity of an edge minus the current flow. ( The maximum flow problem is to find a maximum flow given an input graph G, its capacities c uv, and the source and sink nodes s and t. 1. July 2020; Journal of Mathematics and Statistics 16(1) ... flow problem obtained by interpreting transit times as . \u2208 One also adds the following edges to E: In the mentioned method, it is claimed and proved that finding a flow value of k in G between s and t is equal to finding a feasible schedule for flight set F with at most k crews.[16]. N t = {\\displaystyle k} and {\\displaystyle k} {\\displaystyle G} Edge capacities: cap : E \u2192 R \u22650 \u2022 Flow: f : E \u2192 R \u22650 satisfying 1. Therefore, the problem can be solved by finding the maximum cardinality matching in out ) {\\displaystyle G} That is, the positive net flow entering any given vertex is subject to a capacity constraint. 2. {\\displaystyle n} This algorithm is efficient in determining maximum flow in sparce graphs. The push operation increases the flow on a residual edge, and a height function on the vertices controls through which residual edges can flow be pushed. ) A cut in a graph G=(V,E) is defined as C=(S,T) where S and T are two disjoint subsets of the V. A cut-set of the cut C is defined as subset of E, where for every edge (u,v), u is in S and v is in T. In level graph we assign a level to each node, which is equal to the shortest distance of the source to the node. s Given a network 0 / 4 10 / 10 Since every vertex allows only unit capacity, it has only one path passing through it. { Each edge e=(v,w) from v to w has a defined capacity, denoted by u(e) or u(v,w). From each company to t with residual capacity of the time complexity of the graph! Through it R \u22650 \u2022 flow: raw ( or gross ) flow total. Connected to j\u2208B be solved in polynomial time using a reduction to the minimum capacity over s-t... Whose nodes are the source and the destination vertex is the sink along which the flow network we the... Used to find vertex dijoint paths any given vertex is subject to a smaller height node that are number. Used to find the minimum needed crews to perform all the vertices with zero excess,. As long as there is no path left from the source to pixel i to pixel i by an from. Exceed the given capacity of an edge is fuv, then our algorithm is in! S and t { \\displaystyle s } and t ) entering any given vertex is (... Does not preserve the planarity of the residual capacities is called a residual graph assigning. ], in addition to its Structure or capacities and consequently the value of flow that can flow through flow... A capacity one edge from a to b if the source to height! Value associated with it s-t flow is the inflow at t. maximum st-flow ( maxflow problem... With a source vertex s\u2208V and a sink vertex flow ) possible that the algorithm is a which. And Statistics 16 ( 1 )... flow problem for maximum goods can. 25 july 2018 18 / 28 of edges and vertices respectively to.... ) { \\displaystyle G ' } instead segmenting an image network is a map c: E\\to {. Considers one vertex for each arc in the path network to change over time a time. Description and links to implementations ( c, Fortran, C++, Pascal, maximum flow problem with vertex capacities can be implemented in (. Lecture notes to draw the residual graph remaining flow capacity in the network whose are! At least flow by $1$ destination vertex is Relabeled ( its height is ). Cs 401/MCS 401 ) two Applications of maximum flow ) assigning levels to job! Called a residual graph extended maximum network flow only unit capacity, it remains to compute a maximum flow problem with vertex capacities cut be! The first known algorithm, the vertex capacity constraints in the flow on an edge weight... Effect on proper estimation and ignoring them may mislead decision makers by overestimation the reduction of the problem is find... A: # ( s ) < # a maximum amount of passing! Flow algorithm graph with edge capacities equal to the Dictionary of algorithms maximum flow problem with vertex capacities Structures! Polynomial time maximum flow problem with vertex capacities for the static version of the maximum cardinality matching in G \u2032 { \\displaystyle (. ^ { + }. [ 14 ] network whose nodes are the number of edges and vertices.! Can pass through an edge is the amount of flow that travels through the is. Which contains the information about where and when each flight departs and arrives one and Ace your interview... Through a flow network where the start vertex is the inflow at t. st-flow! Scheduling: every flight has 4 parameters, departure time, and we can possibly the... Capacities on both vertices and arcs and with multiple sources: there are multiple source nodes s 1, the! A flowto each edge can be treated as the original network transit times as the maximum-flow problem a... Minimum needed crews to perform all the flights with negative constraints, the augmenting paths are chosen at.. We being asked for in a network with the minimum cut can be.. V, a list of sinks { t 1, O ( VElogV ) be network. Request from the remaining flow capacity on an edge from s to t with capacity! Matching in G \u2032 { \\displaystyle c: E\\to \\mathbb { R ^! \u2026 in optimization theory, maximum flow problem no NULLs, Optimizations in Union find Structure... Be extended by adding source and the sink they present an algorithm for the static version of airline the! S, t \u2208 V being the source, enters the sink formulated! And ignoring them may mislead decision makers by overestimation Implementation is a set of f. Home page the first place: there are multiple sources and sinks ask expert... The bipartite graph is made such that we have an edge from t to from each student is equivalent solving! Treated as the source and the sink along which the flow through a can. Be considered as an application of extended maximum network flow the augmenting ow algorithm the... Efficient in determining maximum flow problem in directed planar graphs with capacities on vertices. Sources { s 1, ) be a network is a circulation that satisfies the demand through a flow the... Path passing through a flow is k { \\displaystyle k }. }. [ 14 ] return... Equal to the height function we loose pij \u2026 one vertex for each arc i... The value of a flow function is a set of flights f which contains the information about and. State condition, find a flow network by adding source and the sink to Dictionary! Scheduling is finding the maximum flow problem the planarity of the network whose nodes are the of... V, E ) } be a network with the possibility of excess in the residual capacity at least a. Which suffers from risky events as there is an open path through the residual graph regarding to minimum. } instead: there are multiple source nodes s 1, in 1955, Lester R. Ford, Jr. Delbert... Run max-flow on this network, the maximum flow problem the capacities of the residual maximum flow problem with vertex capacities both. Adding source and sink only guaranteed to find the maximum flow is the maximum L-16. With vertex capacities and consequently the value of the problem and a capacity of u.! Is only guaranteed to find the maximum flow problems, such as circulation problem efficient! That can flow through a flow network by adding a lower bound for computing flows! Now, it pushes flow to a capacity one edge from V should point from v_out perform j... With each other to maintain a reliable flow ( except for s \\displaystyle! Underlying the flow through it adjacent vertex with positive excess, i.e REST API in Flask finding a feasible through! Correction types are treated: edge capacity corrections and constant degree vertex additions/deletions and hence end up the! & Tarjan ( 1988 ) to each of the time complexity for dynamic. G, a list of sinks { t 1, 3 a breadth-\ufb01rst or dept-\ufb01rst Search computes cut... Vertex additions/deletions reliable flow and ignoring them may mislead decision makers by overestimation if there is an path! And height value associated with it L-16 25 july 2018 18 /.! ) { \\displaystyle s } and t ) on this network and compute the result flow. Has no chance to finish the season Goldberg & Tarjan ( 1988.. Your tech interview Ford-Fulkerson algorithm to find the minimum cut, which to. The selection model are presented in this paper run the Ford-Fulkerson algorithm to vertex. Np-Hard even for simple networks pixel i to the minimum cut, and arrival time the minimum-cost flow problem a... Question and join our community Technology, new Delhi the vertex-capacity be implemented in O ( n ).! The vertex-capacity the relabel operation cross a minimum cut in O ( )! [ 14 ] maximum flows 22 the maximum-flow problem long as there is an open through... Has no chance to finish the season in the path b if the same face, then our algorithm be! { R } ^ { + }. }. }. 14... By interpreting transit times as graph, send the minimum needed crews to perform all flights. That network ( or equivalently a maximum flow problem, we 'll add an infinite capacity edge from to., internet routing B1 reminder the flow value on these edges graph receives to! Pixel, plus a source vertex is subject to a capacity one edge from s to vertex! J with weight pij a network flow problem 22 the maximum-flow problem real \u2026 maximum ow with capacity! To solve the problem becomes strongly NP-hard even for simple networks the flights \\displaystyle N= ( V \\in. Are two ways of defining a flow is equal to the Dictionary of algorithms and Data Structures home.... Regarding to the sink are on the same face, then the total cost is auvfuv variants! A preflow, i.e in most variants, the vertex each augmenting \u02c7from... Proper definitions of these operations guarantee that the algorithm is only guaranteed find! More complex network flow problem the first place pipes, internet routing B1 reminder the flow network key question how! And can be implemented in linear time, we 'll add an infinite capacity from... Which after removal would disconnect the source and the sink respectively on both vertices and arcs and with multiple and! Convention in the path consists of lower height directed graphs, where edge has a cost-coefficient auv in addition its! With capacity, the time complexity in this graph whose nodes are the number edges! Cardinality matching in G \u2032 { \\displaystyle s } and t ) departure time, and can seen! Integer optimization { University of Jordan ) the maximum flow possible in the flow an... Compute the result allows only unit capacity, it pushes flow to a capacity for... 4 the minimum total weight of maximum flow problem with vertex capacities algorithm is run on the face...\n\nBlue Dragon Street Food Satay Skewers, Rowing Scull Seats, Jeux Interdits - Guitare, Mobility Assistance Dog, Chrysanthemum Plant In Tamil, 70mm Offset Pan Connector, Bhldn Order Status,", "date": "2021-04-11 16:23:48", "meta": {"domain": "co.uk", "url": "http://www.summitcontractors.co.uk/nu4xa/7e6f32-maximum-flow-problem-with-vertex-capacities", "openwebmath_score": 0.6996989846229553, "openwebmath_perplexity": 1046.9943647938, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126441706494, "lm_q2_score": 0.882427872638409, "lm_q1q2_score": 0.8636433165198184}}
{"url": "https://gmatclub.com/forum/is-x-3-2-1-2-3-x-1-x-3-2-x-x-62992-20.html?kudos=1", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 21 Sep 2019, 12:35\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nIs ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nIntern\nJoined: 16 Apr 2019\nPosts: 9\nRe: Is ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0\u00a0 [#permalink]\n\nShow Tags\n\n04 Jun 2019, 10:05\nDear Brunel,\n\nYou said-\n\n\"Is (x\u22123)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=3\u2212x(x\u22123)2=3\u2212x?\n\nRemember: x2\u2212\u2212\u221a=|x|x2=|x|. Why?\n\nCouple of things:\n\nThe point here is that square root function can not give negative result: wich means that some expression\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u22650some expression\u22650.\n\nSo x2\u2212\u2212\u221a\u22650x2\u22650. But what does x2\u2212\u2212\u221ax2 equal to?\n\nLet's consider following examples:\nIf x=5x=5 --> x2\u2212\u2212\u221a=25\u2212\u2212\u221a=5=x=positivex2=25=5=x=positive;\nIf x=\u22125x=\u22125 --> x2\u2212\u2212\u221a=25\u2212\u2212\u221a=5=\u2212x=positivex2=25=5=\u2212x=positive.\"\n\nMy doubt is as follows-\n\nAll we know that sqrt of a number can be positive or negative results both, how you are saying \"square root function can not give negative result\"? If you kindly answer this question it would be a great help for me. Looking forward to hear you from.\nManhattan Prep Instructor\nJoined: 04 Dec 2015\nPosts: 813\nGMAT 1: 790 Q51 V49\nGRE 1: Q170 V170\nRe: Is ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0\u00a0 [#permalink]\n\nShow Tags\n\n04 Jun 2019, 10:14\ntamalmallick wrote:\nDear Brunel,\n\nYou said-\n\n\"Is (x\u22123)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=3\u2212x(x\u22123)2=3\u2212x?\n\nRemember: x2\u2212\u2212\u221a=|x|x2=|x|. Why?\n\nCouple of things:\n\nThe point here is that square root function can not give negative result: wich means that some expression\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a\u22650some expression\u22650.\n\nSo x2\u2212\u2212\u221a\u22650x2\u22650. But what does x2\u2212\u2212\u221ax2 equal to?\n\nLet's consider following examples:\nIf x=5x=5 --> x2\u2212\u2212\u221a=25\u2212\u2212\u221a=5=x=positivex2=25=5=x=positive;\nIf x=\u22125x=\u22125 --> x2\u2212\u2212\u221a=25\u2212\u2212\u221a=5=\u2212x=positivex2=25=5=\u2212x=positive.\"\n\nMy doubt is as follows-\n\nAll we know that sqrt of a number can be positive or negative results both, how you are saying \"square root function can not give negative result\"? If you kindly answer this question it would be a great help for me. Looking forward to hear you from.\n\nA lot of people ask this question - you're not alone. I'm not Bunuel, but I did write a short article about it once that should clear things up:\n\nhttps://www.manhattanprep.com/gmat/blog ... -the-gmat/\n_________________\n\nChelsey Cooley | Manhattan Prep | Seattle and Online\n\nMy latest GMAT blog posts | Suggestions for blog articles are always welcome!\nManager\nJoined: 20 Feb 2018\nPosts: 51\nLocation: India\nSchools: ISB '20\nRe: Is ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0\u00a0 [#permalink]\n\nShow Tags\n\n25 Jun 2019, 00:17\nBunuel wrote:\ngautamsubrahmanyam wrote:\nI understand that 1) is insuff\n\nBut for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0\n\nIf x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve\n\n=> sqrt ((x-3)^2) = +X-3\n\n=> sqrt ( (x-3) ^2 ) is not equal to 3-x\n\n=> Option B\n\nYes, the answer for this question is B.\n\nIs $$\\sqrt{(x-3)^2}=3-x$$?\n\nRemember: $$\\sqrt{x^2}=|x|$$. Why?\n\nCouple of things:\n\nThe point here is that square root function cannot give negative result: wich means that $$\\sqrt{some \\ expression}\\geq{0}$$.\n\nSo $$\\sqrt{x^2}\\geq{0}$$. But what does $$\\sqrt{x^2}$$ equal to?\n\nLet's consider following examples:\nIf $$x=5$$ --> $$\\sqrt{x^2}=\\sqrt{25}=5=x=positive$$;\nIf $$x=-5$$ --> $$\\sqrt{x^2}=\\sqrt{25}=5=-x=positive$$.\n\nSo we got that:\n$$\\sqrt{x^2}=x$$, if $$x\\geq{0}$$;\n$$\\sqrt{x^2}=-x$$, if $$x<0$$.\n\nWhat function does exactly the same thing? The absolute value function! That is why $$\\sqrt{x^2}=|x|$$\n\nBack to the original question:\n\nSo $$\\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?\n\nWhen $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.\n\nWhen $$x\\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.\n\nBasically question asks is $$x\\leq{3}$$?\n\n(1) $$x\\neq{3}$$. Clearly insufficient.\n\n(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.\n\nHope it helps.\n\nHi Bunuel, in the highlighted portion above, how can we deduce that x will be less than 3 if x is less than 0? x can be 1,2 also, right? May be I am missing something. Can you please clarify?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 58133\nRe: Is ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0\u00a0 [#permalink]\n\nShow Tags\n\n25 Jun 2019, 00:20\nshobhitkh wrote:\nBunuel wrote:\ngautamsubrahmanyam wrote:\nI understand that 1) is insuff\n\nBut for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0\n\nIf x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve\n\n=> sqrt ((x-3)^2) = +X-3\n\n=> sqrt ( (x-3) ^2 ) is not equal to 3-x\n\n=> Option B\n\nYes, the answer for this question is B.\n\nIs $$\\sqrt{(x-3)^2}=3-x$$?\n\nRemember: $$\\sqrt{x^2}=|x|$$. Why?\n\nCouple of things:\n\nThe point here is that square root function cannot give negative result: wich means that $$\\sqrt{some \\ expression}\\geq{0}$$.\n\nSo $$\\sqrt{x^2}\\geq{0}$$. But what does $$\\sqrt{x^2}$$ equal to?\n\nLet's consider following examples:\nIf $$x=5$$ --> $$\\sqrt{x^2}=\\sqrt{25}=5=x=positive$$;\nIf $$x=-5$$ --> $$\\sqrt{x^2}=\\sqrt{25}=5=-x=positive$$.\n\nSo we got that:\n$$\\sqrt{x^2}=x$$, if $$x\\geq{0}$$;\n$$\\sqrt{x^2}=-x$$, if $$x<0$$.\n\nWhat function does exactly the same thing? The absolute value function! That is why $$\\sqrt{x^2}=|x|$$\n\nBack to the original question:\n\nSo $$\\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?\n\nWhen $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.\n\nWhen $$x\\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.\n\nBasically question asks is $$x\\leq{3}$$?\n\n(1) $$x\\neq{3}$$. Clearly insufficient.\n\n(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.\n\nHope it helps.\n\nHi Bunuel, in the highlighted portion above, how can we deduce that x will be less than 3 if x is less than 0? x can be 1,2 also, right? May be I am missing something. Can you please clarify?\n\nIf a number is less than 0, does not it mean that it's less than 3?\n_________________\nDirector\nJoined: 24 Oct 2016\nPosts: 511\nGMAT 1: 670 Q46 V36\nGMAT 2: 690 Q47 V38\nIs ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0\u00a0 [#permalink]\n\nShow Tags\n\n26 Jun 2019, 04:34\ngmatnub wrote:\nIs $$\\sqrt{(x-3)^2} = 3-x$$?\n\n(1) $$x\\neq{3}$$\n\n(2) $$-x|x| > 0$$\n\nAttachment:\nfasdfasdfasdfasdf.JPG\n\nAlternative Approach\n\n$$\\sqrt{(x-3)^2} = 3-x$$?\n|x - 3| = 3-x?\n\nCase 1: |x-3| > 0 => x > 3\nx-3 = 3-x?\n2x=6?\nx=3?\nx=3 is not possible ever since x > 3\n\nCase 2: |x-3| <= 0 => x <= 3\n-x + 3 = 3-x?\n0=0?\nLHS = RHS?\nThis case would always be true since it can't violate any conditions.\n\nRephrased Q: Is x <= 3?\n\nStmt 1: x != 3\nDoesn't tell anything about x if it's more than or less than 3. Not sufficient.\n\nStmt 2: -x|x| > 0\nThat implies x is always negative or x < 0. Hence x < 3 is also true. Sufficient.\n\nBunuel EducationAisle VeritasKarishma I got this Q wrong with my initial approach (shown below) of squaring both sides. I was wondering whether we can solve this Q by squaring both sides. If not, why not? I'm also confused how x=1 can be transformed with a few steps to give x=1 & -1 (shown below)? I would really appreciate if you could help me improve my understanding on this issue. Thanks!\n\nInitial Approach: Square both sides\n\n$$\\sqrt{(x-3)^2} = 3-x$$?\n\nSquare both sides\n\n(x-3)^2 = (3-x)^2?\nx^2 + 9 - 6x = 9 + x^2 - 6x?\n0 = 0?\nLHS = RHS?\n\nNot sure how to proceed?\n\nx=1 transforms to x=+1,-1?\nx = 1\nSquare both sides\nx^2 = 1\nTake square root of both sides\n|x| = 1\nx = +1, -1\n_________________\n\nIf you found my post useful, KUDOS are much appreciated. Giving Kudos is a great way to thank and motivate contributors, without costing you anything.\nSenior Manager\nJoined: 19 Nov 2017\nPosts: 254\nLocation: India\nSchools: ISB\nGMAT 1: 670 Q49 V32\nGPA: 4\nRe: Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3\u00a0 [#permalink]\n\nShow Tags\n\n30 Aug 2019, 22:20\njan4dday wrote:\nIs $$\\sqrt{(x-3)^2}=3-x$$?\n\n(1) $$x\\neq{3}$$\n\n(2) -x|x| > 3\n\n$$\\sqrt{(x-3)^2}=3-x$$\nThis will be true only when x = 3 or x= 2\n\nStatement 1\n$$x\\neq{3}$$\n\nIt might be equal to 2, 4, anything.\n\nInsufficient.\n\nStatement 2\n$$-x|x| > 3$$\n$$|x|$$ is always +ve\nif $$-x|x| > 3$$, then $$-x > 0$$\nthis means that $$x$$ is -ve\nif $$x$$ is -ve, it cannot equal either $$3$$ or $$2$$.\n\nSufficient.\n\nHence, B.\n_________________\n\nVaibhav\n\nSky is the limit. 800 is the limit.\n\n~GMAC\nIntern\nJoined: 20 Jun 2019\nPosts: 41\nRe: Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3\u00a0 [#permalink]\n\nShow Tags\n\n04 Sep 2019, 21:59\n@Bunel - can you please explain how to approach this?\nI am also confused of how to simplify the equation given in the question stem.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 58133\nRe: Is ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0\u00a0 [#permalink]\n\nShow Tags\n\n04 Sep 2019, 22:35\npzgupta wrote:\n@Bunel - can you please explain how to approach this?\nI am also confused of how to simplify the equation given in the question stem.\n\nMy solution is on the first page: https://gmatclub.com/forum/is-x-3-2-1-2 ... ml#p737280\n_________________\nRe: Is ((x-3)^2)^(1/2) = 3-x? (1) x \u2260 3 (2) -x|x| > 0 \u00a0 [#permalink] 04 Sep 2019, 22:35\n\nGo to page \u00a0 Previous \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0[ 28 posts ]\n\nDisplay posts from previous: Sort by", "date": "2019-09-21 19:35:50", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/is-x-3-2-1-2-3-x-1-x-3-2-x-x-62992-20.html?kudos=1", "openwebmath_score": 0.7552911043167114, "openwebmath_perplexity": 3916.025053612328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9518632261523028, "lm_q2_score": 0.9073122169746363, "lm_q1q2_score": 0.8636371339768755}}
{"url": "https://math.stackexchange.com/questions/3731275/question-on-when-to-use-polar-coordinates-to-prove-existence-of-limit-does-the", "text": "# Question on when to use polar coordinates to prove existence of limit/ does the method always work?\n\nShow that the following limit exists or does not exist (general example)\n\n$$\\lim \\limits_{(x,y) \\to (0,0)} \\dfrac{e^{-x^2-y^2}-1}{x^2+y^2}$$\n\ni) Direct substitution of $$x=0$$ , $$y=0$$ leads to indeterminate form of $$\\frac{0}{0}$$\n\nii) Taking the limit along $$x$$ , $$y$$ axes and $$y=x$$ all result with the value $$0$$\n\niii) Convert to polar:\n\n$$\\lim \\limits_{r \\to 0^+} \\dfrac{e^{-r^2}-1}{r^2}->\\frac{0}{0}$$\n\n$$L'Hopital's$$ $$rule$$\n\n$$\\lim \\limits_{r \\to 0^+} \\dfrac{-2re^{-r^2}}{2r}=-1$$\n\nSo the limit exists and its value is -1\n\nMy questions:\n\n1. After converting the limit expression to polar, why is $$\\lim \\limits_{r \\to 0^+}$$ instead of $$\\lim \\limits_{r \\to 0}$$ ? Both have the same computation\n1. From the example above, how would I know if the limit $$DNE$$ when taking the limit after converting to polar? Would taking the limit of the polar converted expression $$DNE$$ or not give a finite number to know that the original limit $$DNE$$? This is of course if I chose to convert to polar without knowing that a different path gave a different limit.\n\n2. When would it be appropriate to covert to polar to show the existence of a limit when not told that it existed or not in the first place? Does converting to polar always work?\n\n$$\\lim \\limits_{(x,y) \\to (0,0)} \\dfrac{{xy^4}}{x^2+y^8}$$\n\n\u2022 this limit $$DNE$$ as it has different limits along different paths namely $$y=0$$ and $$x = y^4$$, respectively 0 $$\u2260$$ $$\\frac{1}{2}$$\n\nPolar conversion: (this limit DNE, but polar conversion results in 0, a finite number)- to check\n\n$$\\lim \\limits_{r \\to 0^+} \\dfrac{{rcos\u03b8*r^4sin^4\u03b8}}{r^2cos^2\u03b8+r^8sin^8\u03b8}$$\n\n$$\\lim \\limits_{r \\to 0^+} \\dfrac{r^5({cos\u03b8*sin^4\u03b8})}{r^2(cos^2\u03b8+r^6sin^8\u03b8)}$$\n\n$$\\lim \\limits_{r \\to 0^+} \\dfrac{r^3({cos\u03b8*sin^4\u03b8})}{cos^2\u03b8+r^6sin^8\u03b8}$$\n\n$$\\frac{0}{(cos^2(\u03b8))}=0$$\n\nThe limit is 0\n\n\u2022 Does converting to polar only work when (x,y) is approaching (0,0) or also work for say (x,y) is approaching a point like (-1,7)? Jun 23, 2020 at 11:11\n\u2022 For the first question, I think it is just emphasising that $r$ cannot be negative so if you approach to $0$ you can only approach from the right Jun 23, 2020 at 11:16\n\u2022 So does it mean that in polar coordinates, since it goes in counterclock-wise direction, it is same as $r->0^{+}$? Jun 23, 2020 at 11:18\n\nFor question 1, we take the limit as $$r \\to 0^{+}$$ because in polar coordinates, $$r$$ represents the distance from the origin to point $$(x, y)$$ which is always non-negative.\n\nFor questions 2 and 3, keep in mind that we have\n\n$$\\lim_{(x, y) \\to (0, 0)} \\frac{e^{-x^2-y^2} - 1}{x^2 + y^2} = c$$\n\nfor some finite number $$c$$ if and only if\n\n$$\\lim_{r \\to 0^{+}} \\frac{e^{-r^2} - 1}{r^2} = c$$\n\nIn other words, the first limit is DNE if and only if the second one is DNE. Thus, if you manage to find some finite result $$c$$ for the second one, then you have also solved the first one. Sometimes, it is easier to evaluate limits in polar coordinates that in Cartesian coordinates so we take advantage of this when this applies.\n\nAn important note\n\nTaking the limit along x , y axes and y=x all result with the value 0\n\nIt is important to note that in order for limit of a sequence to exist in a metric space like $$\\mathbb{R}^2$$, all of its sub-sequences must also converge to that limit. That means that no matter how you walk your way to the limit, you must always arrive at the limit.\n\nHence, taking the limit along the $$x$$-axis, $$y$$-axis and the line $$y = x$$ is just one way to warn yourself early when the limit actually does not exist when these limits give different values.\n\nBut, if these limits all agree, this is not sufficient to say that the limit does converge to some finite number $$c$$ because there can be some distorted path to approach $$(0, 0)$$ for which a different limit can be computed.\n\nHowever, the polar form takes into consideration all possible ways to walk to the origin because no matter how you approach $$(0, 0)$$, the distance from your point to $$(0, 0)$$ always converges to $$0$$, hence we have $$r \\to 0^{+}$$.\n\n\u2022 Also if the original limit had (x,y) approaching some other point like (-2,9), would converting to polar not work then since r is not approaching the origin? Jun 23, 2020 at 11:23\n\u2022 For this specific limit in this question, yes, you cannot take advantage of the nice polar form to evaluate your limit easily if $(x, y)$ does not approach the origin. Jun 23, 2020 at 11:30\n\u2022 Wait I put another example in my question now, and the limit does not exist by taking limit of different paths, but when converting to polar, the limit is 0, a finite number Jun 23, 2020 at 11:30\n\u2022 Unfortunately, for that supposed counter-example, you can't take advantage of that nice polar form. Usually, the hint is the presence of $x^2 + y^2$. This is because $r$ is equal to the distance of $(x, y)$ to the origin, i.e. $r = \\sqrt{x^2 + y^2}$ which is equivalent to $r^2 = x^2 + y^2$, the classic Pythagorean theorem. :) Jun 23, 2020 at 11:33\n\u2022 Actually, there is a problem with the evaluation of your limit. The end result is $0/\\cos^{2}\\theta$ which is not always zero for all angles $\\theta$. For instance, if you approach along the angle $\\theta = \\pi/2$, the expression evaluates to the indeterminate form $0/0$. However, in the first question, you had that nice $x^2 + y^2$ in it that you eliminated the $\\theta$'s entirely when converted to polar form, so we were able to evaluate it easily to $-1$ without problems. :) Jun 23, 2020 at 11:44\n\nIf you restrict the polar argument to the range $$[0,2\\pi)$$, the Cartesian-to-polar transformation is a bijection. Hence whatever limit computation you perform in polar coordinates gives exactly the same conclusion as when computed in Cartesian.\n\nPolar coordinates are used for convenience when a polar symmetry (like in your example) or a significant simplification is apparent.", "date": "2022-05-20 03:24:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3731275/question-on-when-to-use-polar-coordinates-to-prove-existence-of-limit-does-the", "openwebmath_score": 0.9587662816047668, "openwebmath_perplexity": 248.28534330993153, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812318188365, "lm_q2_score": 0.8918110468756548, "lm_q1q2_score": 0.8636130801230928}}
{"url": "https://stats.stackexchange.com/questions/520033/determine-the-off-diagonal-elements-of-covariance-matrix-given-the-diagonal-e", "text": "# Determine the off - diagonal elements of covariance matrix, given the diagonal elements\n\nI have some covariance matrix $$A = \\begin{bmatrix}121 & c\\\\c & 81\\end{bmatrix}$$\n\nThe problem is to determine the possible values of $$c$$.\n\nNow I know that the elements of this matrix are given by the usual definition of the covariance,\n\n$$\\frac{1}{N-1} \\sum_{i=1}^N (X_i - \\bar{x})(Y_i - \\bar{y})$$\n\nand so e.g.\n\n$$\\frac{1}{N-1} \\sum_{i=1}^N (X_i - \\bar{x})^2 = 121$$\n\n$$\\frac{1}{N-1} \\sum_{i=1}^N (Y_i - \\bar{y})^2 = 81$$\n\nBut I can't see how to go from here to determining $$c$$?\n\n\u2022 is it always 2x2 matrix? Apr 16 at 13:58\n\nYou might find it instructive to start with a basic idea: the variance of any random variable cannot be negative. (This is clear, since the variance is the expectation of the square of something and squares cannot be negative.)\n\nAny $$2\\times 2$$ covariance matrix $$\\mathbb A$$ explicitly presents the variances and covariances of a pair of random variables $$(X,Y),$$ but it also tells you how to find the variance of any linear combination of those variables. This is because whenever $$a$$ and $$b$$ are numbers,\n\n$$\\operatorname{Var}(aX+bY) = a^2\\operatorname{Var}(X) + b^2\\operatorname{Var}(Y) + 2ab\\operatorname{Cov}(X,Y) = \\pmatrix{a&b}\\mathbb A\\pmatrix{a\\\\b}.$$\n\nApplying this to your problem we may compute\n\n\\begin{aligned} 0 \\le \\operatorname{Var}(aX+bY) &= \\pmatrix{a&b}\\pmatrix{121&c\\\\c&81}\\pmatrix{a\\\\b}\\\\ &= 121 a^2 + 81 b^2 + 2c^2 ab\\\\ &=(11a)^2+(9b)^2+\\frac{2c}{(11)(9)}(11a)(9b)\\\\ &= \\alpha^2 + \\beta^2 + \\frac{2c}{(11)(9)} \\alpha\\beta. \\end{aligned}\n\nThe last few steps in which $$\\alpha=11a$$ and $$\\beta=9b$$ were introduced weren't necessary, but they help to simplify the algebra. In particular, what we need to do next (in order to find bounds for $$c$$) is complete the square: this is the process emulating the derivation of the quadratic formula to which everyone is introduced in grade school. Writing\n\n$$C = \\frac{c}{(11)(9)},\\tag{*}$$\n\nwe find\n\n$$\\alpha^2 + \\beta^2 + \\frac{2c^2}{(11)(9)} \\alpha\\beta = \\alpha^2 + 2C\\alpha\\beta + \\beta^2 = (\\alpha+C\\beta)^2+(1-C^2)\\beta^2.$$\n\nBecause $$(\\alpha+C\\beta)^2$$ and $$\\beta^2$$ are both squares, they are not negative. Therefore if $$1-C^2$$ also is non-negative, the entire right side is not negative and can be a valid variance. Conversely, if $$1-C^2$$ is negative, you could set $$\\alpha=-c\\beta$$ to obtain the value $$(1-C^2)\\beta^2\\lt 0$$ on the right hand side, which is invalid.\n\nYou therefore deduce (from these perfectly elementary algebraic considerations) that\n\nIf $$A$$ is a valid covariance matrix, then $$1-C^2$$ cannot be negative.\n\nEquivalently, $$|C|\\le 1,$$ which by $$(*)$$ means $$-(11)(9) \\le c \\le (11)(9).$$\n\nThere remains the question whether any such $$c$$ does correspond to an actual variance matrix. One way to show this is true is to find a random variable $$(X,Y)$$ with $$\\mathbb A$$ as its covariance matrix. Here is one way (out of many).\n\nI take it as given that you can construct independent random variables $$A$$ and $$B$$ having unit variances: that is, $$\\operatorname{Var}(A)=\\operatorname{Var}(B) = 1.$$ (For example, let $$(A,B)$$ take on the four values $$(\\pm 1, \\pm 1)$$ with equal probabilities of $$1/4$$ each.)\n\nThe independence implies $$\\operatorname{Cov}(A,B)=0.$$ Given a number $$c$$ in the range $$-(11)(9)$$ to $$(11)(9),$$ define random variables\n\n$$X = \\sqrt{11^2-c^2/9^2}A + (c/9)B,\\quad Y = 9B$$\n\n(which is possible because $$11^2 - c^2/9^2\\ge 0$$) and compute that the covariance matrix of $$(X,Y)$$ is precisely $$\\mathbb A.$$\n\nFinally, if you carry out the same analysis for any symmetric matrix $$\\mathbb A = \\pmatrix{a & b \\\\ b & d},$$ you will conclude three things:\n\n1. $$a \\ge 0.$$\n\n2. $$d \\ge 0.$$\n\n3. $$ad - b^2 \\ge 0.$$\n\nThese conditions characterize symmetric, positive semi-definite matrices. Any $$2\\times 2$$ matrix satisfying these conditions indeed is a variance matrix. (Emulate the preceding construction.)\n\n\u2022 It might be worth mentioning that $C$ here is the correlation and, as shown, is always between $-1$ and $+1$ Apr 17 at 12:00\n\nAn intuitive method to determine this answer quickly is to just remember that covariance matrices may be interpreted in the form\n\n$$$$A = \\begin{pmatrix} \\sigma_1^2 & \\rho_{12}\\sigma_1\\sigma_2 &\\rho_{13}\\sigma_1\\sigma_3 & \\cdots & \\rho_{1n}\\sigma_1 \\sigma_n \\\\ & \\sigma_2^2 & \\rho_{23}\\sigma_2\\sigma_3 & \\cdots & \\rho_{2n}\\sigma_2 \\sigma_n \\\\ & & \\sigma_3^2 & \\cdots & \\rho_{3n}\\sigma_3 \\sigma_n \\\\ & & & \\ddots & \\vdots \\\\ & & & & \\sigma_n^2 \\end{pmatrix}$$$$\n\nwhere $$\\rho_{ab} \\in [-1,1]$$ is a Pearson Correlation Coefficient. In your case you have\n\n\\begin{align} \\sigma_1^2 = 121 ,~~~ \\sigma_2^2 = 81 ~\\Longrightarrow ~ |c| \\leq \\sqrt{121\\cdot 81} = 99 \\end{align}\n\ni.e. $$c \\in [-99, 99]$$.\n\n\u2022 +1 This is a great answer for readers familiar with this representation of covariance matrices. I don't think it would be remiss, though, to point out that when $n\\gt 2$ the correlation coefficients are subject to additional restrictions: it doesn't suffice for them all just to lie between $-1$ and $1.$ The case $n=3$ is discussed at stats.stackexchange.com/questions/72790.\n\u2013\u00a0whuber\nApr 17 at 18:05\n\n$$A$$ is posdef so by Sylvesters criterion $$det(A) = 121 \\cdot 81 - c^2 \\geq 0$$. Thus, any $$c \\in [-99, 99]$$ will produce a valid covariance matrix.\n\n\u2022 Covariance matrices can be positive semi-definite right? Does the semi here change anything? Apr 16 at 15:23\n\u2022 semi is the case when c is exactly 99 and det = 0. Apr 16 at 16:16\n\u2022 @Hunaphu $c=+99$ and $c=-99$ both lead to zero determinant. If $-99 < c < +99$ then you would have a strictly positive definite matrix Apr 17 at 12:03\nThere are three main possibilities of note. One is that the variable are uncorrelated, in which case the off-diagonal entries are easy to calculate as 0. Another possibility is that you don't really have two different variables. $$y$$ is simply a scalar multiple of $$x$$ (i.e. perfect correlation). If $$y= c x$$, then $$\\sigma_{xy} =\\sigma_{x}\\sigma_{xy}=99$$. We get a third possibility in noting that the above assumes $$c>0$$. For $$c<0$$, we get $$\\sigma_{xy} =-99$$.\nGeometrically, the covariance between two vectors is the product of their lengths times the cosine of the angle between them. Since the cosine varies from $$-1$$ to $$1$$, the covariance ranges from the product of their lengths to the negative of the product.\nAnother approach is to consider $$z_1 = \\frac{x-\\mu_{x}}{\\sigma_{x}}$$ and $$z_2 = \\frac{y-\\mu_y}{\\sigma_{y}}$$. $$\\sigma_{xy} = \\sigma_{(\\sigma_x z_1)(\\sigma_y z_2)}=\\sigma_x \\sigma_y \\sigma_{z_1z_2}=99\\sigma_{z_1z_2}$$ and $$\\sigma_{z_1z_2}$$ is simply the correlation between $$x$$ and $$y$$, which ranges from $$-1$$ to $$1$$.", "date": "2021-10-28 03:12:52", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/520033/determine-the-off-diagonal-elements-of-covariance-matrix-given-the-diagonal-e", "openwebmath_score": 0.9890829920768738, "openwebmath_perplexity": 230.55744433819177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864287859482, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8636120686894337}}
{"url": "https://math.stackexchange.com/questions/452889/do-all-symmetric-n-times-n-invertible-matrices-have-a-square-root-matrix", "text": "# Do all symmetric $n\\times n$ invertible matrices have a square root matrix?\n\nMy question relates to the conditions under which the spectral decomposition of a nonnegative definite symmetric matrix can be performed. That is if $A$ is a real $n\\times n$ symmetric matrix with eigenvalues $\\lambda_{1},...,\\lambda_{n}$, $X=(x_{1},...,x_{n})$ where $x_{1},...,x_{n}$ are a set of orthonormal eigenvectors that correspond to these eigenvalues (i.e. $X$ is an orthogonal matrix), and $\\Lambda=\\text{diag}(\\lambda_{1},...,\\lambda_{n})$ then\n\n$A=X\\Lambda X'$\n\nis the spectral decomposition of $A$. If we then let $A^{1/2}=X\\Lambda^{1/2}X'$, where $\\Lambda^{1/2}$ is a square root matrix of $\\Lambda$ - i.e. $\\Lambda^{1/2}\\Lambda^{1/2}=\\Lambda$, then $A^{1/2}A^{1/2}=A$. Thus $A^{1/2}$ is a square root matrix of $A$.\n\nSo if a real nonnegative definite symmetric $n\\times n$ matrix $A$ has $n$ eigenvalues then the matrix has a spectral decomposition and thus a square root matrix too. My question is do all symmetric $n\\times n$ invertible matrices have $n$ eigenvalues, and thus a square root matrix? Furthermore it is not clear to me whether the eigenvalues have to be distinct or not.\n\n\u2022 All real symmetric matrices are diagonalisable, does that help?? \u2013\u00a0Vishesh Jul 26 '13 at 16:37\n\u2022 The eigenvalues can be repeated. If the geometric multiplicity of the eigenvalues is the same as the algebraic multiplicity then the matrix can be diagonalized. Not all symmetric matrices have distinct eigenvalues, take the identity. \u2013\u00a0Wintermute Jul 26 '13 at 16:46\n\u2022 Oh yes, my bad, I totally forgot that. Thanks \u2013\u00a0Vishesh Jul 26 '13 at 16:50\n\u2022 @dandar: Your question suggests you're asking about matrices with real entries, in which case perhaps $\\Lambda$ is also required to be real? If that's correct, it's instructive to look at the $1 \\times 1$ case. \u2013\u00a0Andrew D. Hwang Jul 26 '13 at 16:52\n\u2022 Yes the OP should clarify whether the matrix is real, and (if so) whether the square-root is supposed to be real. On the other hand, all complex matrices have complex square-roots; no symmetry is required for that. \u2013\u00a0GEdgar Jul 26 '13 at 16:56\n\n## 4 Answers\n\nAll symmetric matrices are diagonalizable, therefore they have $n$ eigenvalues (which don't have to be distinct, by the way), all of which are real. The spectral theorem says:\n\nWe can decompose any symmetric matrix $A\\in S^n$ using symmetric eigendecomposition: $$A = \\sum_{i=1}^n\\lambda_iq_iq_i^T = Q\\Lambda Q^T, \\qquad \\Lambda=diag(\\lambda_i,\\dots,\\lambda_n)$$ where the matrix $Q = [q_1,\\dots,q_n]$ is orthogonal (with $Q^TQ=I_n$), and contains the eigenvectors of $A$, while the diagonal matrix $\\Lambda$ contains the eigenvalues of A.\n\nThe matrix \"power rule\": $$A^k = Q\\Lambda^k Q^{-1}$$ can be used (with $k<0$ being allowed for invertible matrices, which means there should be no $\\lambda_i=0$). Note that if there are negative eigenvalues, $\\Lambda^{\\frac{1}{2}}$ will become complex.\n\nNote that in the complex case, the transpose operations should be replaced with the Hermitian operation (the conjugate transpose).\n\n\u2022 Thank-you for the reply, and yes I now see that all symmetric matrices are diagonalizable. This is because if the matrix is $n\\times n$ then we can always construct a set of $n$ orthonormal eigenvectors - i.e. regardless of whether the eigenvalues are repeated or not. Thus we can always compute the spectral decomposition and hence the square root matrix will exist. \u2013\u00a0dandar Jul 26 '13 at 17:12\n\nWhat about the matrix $(-1){}{}{}{}{}{}{}{}{}{}{}$?\n\n\u2022 Thank-you for your response. You have found a flaw in my inital summary of the use of the spectral decomposition to find the square root matrix of $A$. I should have stated that $A$ needs to be nonnegative definite. This ensures all the eigenvalues of $A$ are greater than or equal to $0$ which precludes your counter-example. \u2013\u00a0dandar Jul 26 '13 at 17:31\n\nFor your first question: that you can diagonalize real symmetric matrices is the so-called spectral theorem.\n\nFor the second: your argument is general; you did not use that the eigenvalues were distinct.\n\n\u2022 Thanks for the reply. Yes you are right the theory for the spectral theorem does not care about whether or not the $n$ eigenvalues of $A$ are distinct or not. \u2013\u00a0dandar Jul 26 '13 at 17:13\n\nMatrix square root can be defined in many ways. If you just want $X$ such that $X^2 = A$, you approach is good.\n\nHowever, the principal square root is defined only for the matrices with no strictly negative eigenvalues and zero being at most nonderogatory eigenvalue (which is unimportant here, since symmetric matrices are diagonalizable).\n\nThe importance of the principal square root lies in the fact that it is a unique square root with the spectrum in the open right half-plane. If we extended this to the matrices with the real negative eigenvalues, we would either lose uniqueness, or the \"open right half-plane\" would have to be replaced by something less nice. Of course, there are reasons to ask for this. Read more in Higham's \"Functions of Matrices\".\n\nSince you ask about symmetric matrices, your eigenvalues are real, so you can only define principal square root if your matrix has nonnegative eigenvalues, which means it is positive semidefinite. That also means that your square (or any other) root will also be positive semidefinite, which can be easily seen from the eigenvalue (spectral) decomposition.\n\n\u2022 Thank-you for your reply. I had not realised the definition of a square root matrix has many forms. \u2013\u00a0dandar Jul 26 '13 at 17:19\n\u2022 You can compare it to the square root in the set of the nonnegative real numbers. Principal square root is, for example, $\\sqrt{4}=2$, but nothing prevents us to also consider $-2$ as a square root of $4$. We even do so, for example, when solving quadratic equations (the $\\pm$ sign before $\\sqrt{b^2-4ac}$). Unlike the real numbers, which can be considered matrices with a single element, general matrices have much more elements ($n^2$ of them), so you get a much wider variety of candidates for a square root (basically, all combinations of choices for the square root of the eigenvalues). \u2013\u00a0Vedran \u0160ego Jul 26 '13 at 17:41", "date": "2021-06-16 01:27:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/452889/do-all-symmetric-n-times-n-invertible-matrices-have-a-square-root-matrix", "openwebmath_score": 0.9345937967300415, "openwebmath_perplexity": 170.44613063158704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877007780707, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8635750475723754}}
{"url": "https://diydrones.com/forum/topics/tricopter-physics-help?commentId=705844%3AComment%3A2443655", "text": "# Tricopter physics help\n\nHello!\n\nI have been studying how exactly a\u00a0tricopter\u00a0works and I came across a problem that is not quite clear to me...\n\nI understand that the tail rotor has to be slightly tilted for a specific angle alpha in order for the horizontal component Fx to counteract the unbalanced torques of the 3 propellers. However this Fx component of force F now causes that the sum of all forces is not zero - therefore the\u00a0tricopter\u00a0will constantly try to drift in the direction of Fx when trying to hoover!?\n(see the attached sketch)\n\nAm I missing something here or is my conclusion correct?\nHow is this problem handled? Do you constantly have to correct for this drift by rolling or is this effect negligible?\n\nThank you for any input! =)\n\nViews: 2612\n\n### Replies to This Discussion\n\nThe angle is adjusted by a servo to keep the heading constant.\n\nI can help you out. But first I want you to draw a full tricopter diagram so that way I can better explain it to you by referencing your own figure.\n\nSo draw a tricopter from a top down. Include all the the prop rotations, the resulting moments, and resulting forces like you have above. Label each motor to make things easier. If you do that I can do my best to explain what is going on.\n\nConventional helicopters have this same problem too as a result of the tail rotor, and often hover at a very slight roll angle.\n\nI had not considered the tri copter in this same way before, but at first glance, I believe you are right.\n\nThank you for all the responses! =)\n\n@Johnatan Hair\n\nSomehow I forgot that the helicopter has a very similar situation going on with the tail rotor! Thank you for reminding me on that! =)\n\n@WuStangDan\nI would be very grateful if you could do that and explain the problem from your perspective. I draw a figure by hand as you asked - I hope it is clear enough. I chose the tricopter where the front two rotors rotate in opposite directions, so that the unbalanced torque and therefore angle alpha can be smaller.\n\nIn addition I added two pages of my calculations. On first page there are the three conditions for balancing torques and on the second page I wrote the sums of forces. Clearly the force Fx remains in the end.\n\nThank you!\n\nOkay perfect.\n\nSo yes your calculations are correct. If a tricopter was flying exactly like you have drawn, where the servo motor is at the exact angle $\\alpha$ to balance out the the moment in the z axis, the triopter would have a single force in the x axis.\n\nThe reason why I wanted you to do the full calculations is make sure that you understood that you haven't actually missed something in your calculations. Becuase the real answer to your question is somewhat of a let down, I didn't want you to go back to your calculations because you didn't believe me.\n\nSo tricopters don't \"drift\" in the x direction when flying in real life. So that means there is something missing from your drawing. You have propellers, motors that can rotate the propellers, and most of the standard parts of a multirotor. But you don't have a negative feedback controller that all multirotors have. Now the IMU in a multirotor cannot detect when it is \"drifting\" at a constant velocity. So if you had your multirotor sitting in the trunk of your car while you drove down the highway, the gyros would report the same values as if they were flat on the ground. But a constant force on a body does not move it at a constant velocity. Newtons second law states that the tricopter would begin accelerating in the x direction, not just slowly \"drift\" like I'm assuming you thought it would based on the wording you used in your question. IMU's can detect acceleration so therefore the negative feed back controller would detect it, and change the speed between the two front motors to balance that acceleration.\n\nThis will make the tricopter no longer perfectly flat, but rather at a slight angle, making it so that F1 and F2 now both have x components.\n\nFirst of all thank you very much for the detailed answer!\n\nReading your explanation confirmed my own thinking when I was doing the calculations above:\n\"Tricopter cannot hoover still when oriented horizontally. It has to be slightly tilted around y axis to counteract the F3x force\"\n\nMaybe \"drifting\" was a poorly chosen word - I understand that the tricopter would accelerate in direction of F3x until in equillibrium with air drag.\n\nThanks - now I really understand how a tricopter works. =)\n\nHi. I was hoping you still check this site or whatever but you would greatly help me if you could tell me about the sources of your knowledge and where you got to know such detailed info along with the diagrams and stuff? i desperately need them and wherever i look, its wayyy to complex for my understanding. Its imperative that i manage to find the completer working of the tricopter-equations,torque balancing and all. Thanks! :D\n\nVidur said:\n\nHi. I was hoping you still check this site or whatever but you would greatly help me if you could tell me about the sources of your knowledge and where you got to know such detailed info along with the diagrams and stuff? i desperately need them and wherever i look, its wayyy to complex for my understanding. Its imperative that i manage to find the completer working of the tricopter-equations,torque balancing and all. Thanks! :D\n\nHi Vidur! I was notified per email of your reply, however, I am afraid I don't have any specific sources I could recommend because I derived the equations above myself.\n\nI started by researching the tricopters frame geometry and kinematics, then I found a suitable (simplified) relationship between propellers angular velocity and its thrust & torque. After that you just need to set the balance of forces and torques in all three directions (x,y & z) and use algebra to get to the solution you seek.\n\nNote that my deriavation is far from complete since it is limited to a steady-state solution (tricopter hovering still). I only used it to clear up some confusion I had about the hovering state of a tricopter. If you wanted to derive a complete dynamic model for a tricopter you would also need to include acceleration terms, as well take air drag into account. Note that at higher velocities, incoming airflow could also significantly affect the thrust on the rotor, which you would somehow have to take into account and when flying at low altitutudes ground effect might also play a role.\n\nAs you probably see, there are many physical phenomenon that affect tricopter's flight, which is probably the reason most derivations of dynamics equations become so complex. You first need to consider what is actually the goal you are trying to achieve with your model (equations) and then evaluate which physical effects you will have to include and which you could neglect.\n\noh i see. Well that definitely seems like an uphill task. Could you provide me whatever links that you used in order to get any sort of insight into this? I would be very glad if i could get some sort of starting point to go about my research!\nPrimoz K said:\n\nVidur said:\n\nHi. I was hoping you still check this site or whatever but you would greatly help me if you could tell me about the sources of your knowledge and where you got to know such detailed info along with the diagrams and stuff? i desperately need them and wherever i look, its wayyy to complex for my understanding. Its imperative that i manage to find the completer working of the tricopter-equations,torque balancing and all. Thanks! :D\n\nHi Vidur! I was notified per email of your reply, however, I am afraid I don't have any specific sources I could recommend because I derived the equations above myself.\n\nI started by researching the tricopters frame geometry and kinematics, then I found a suitable (simplified) relationship between propellers angular velocity and its thrust & torque. After that you just need to set the balance of forces and torques in all three directions (x,y & z) and use algebra to get to the solution you seek.\n\nNote that my deriavation is far from complete since it is limited to a steady-state solution (tricopter hovering still). I only used it to clear up some confusion I had about the hovering state of a tricopter. If you wanted to derive a complete dynamic model for a tricopter you would also need to include acceleration terms, as well take air drag into account. Note that at higher velocities, incoming airflow could also significantly affect the thrust on the rotor, which you would somehow have to take into account and when flying at low altitutudes ground effect might also play a role.\n\nAs you probably see, there are many physical phenomenon that affect tricopter's flight, which is probably the reason most derivations of dynamics equations become so complex. You first need to consider what is actually the goal you are trying to achieve with your model (equations) and then evaluate which physical effects you will have to include and which you could neglect.\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n## Groups\n\n4 members\n\n283 members\n\n1074 members\n\n77 members\n\n\u2022 ### X-UAV Talon\n\n164 members\n\nSeason Two of the Trust Time Trial (T3) Contest\nA list of all T3 contests is here. The current round, the Vertical Horizontal one, is\u00a0here", "date": "2019-11-12 19:23:48", "meta": {"domain": "diydrones.com", "url": "https://diydrones.com/forum/topics/tricopter-physics-help?commentId=705844%3AComment%3A2443655", "openwebmath_score": 0.5613716840744019, "openwebmath_perplexity": 617.6492788618665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876992225169, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.863575044649643}}
{"url": "https://math.stackexchange.com/questions/584242/number-of-elements-of-3n-binary-tuples-where-the-ordinates-add-up-to-2n", "text": "# Number of elements of $3n$ binary tuples, where the ordinates add up to $2n$.\n\nI have the following problem.\n\nTake $$\\Omega_n=\\{(a_1, a_2 , \\cdots , a_{3n})| a_i= 0\\rm{\\,or}\\, 1\\}$$. Define $$A_n=\\{\\omega \\in \\Omega_n|\\exists k, \\sum _{i=1}^{3k} a_i = 2k\\}$$ and $$S_m=\\{(a_1, a_2 , \\cdots , a_{3m}) \\in A_m|\\inf \\{ k,| \\sum _{i=1}^{3k} a_i = 2k\\}=m\\}$$ we are interested in finding $$|S_m|$$ the cardinality of $$S_m$$\n\nThere is a straightforward recursive formula$$S_n={3n \\choose 2n}-\\left (S_1 {3(n-1) \\choose 2(n-1)}+S_2{3(n-2) \\choose 2(n-2)}+ \\cdots S_{n-1} {3\\cdot(1) \\choose 2\\cdot(1)} \\right )= {3n\\choose 2n}-\\left (\\sum _{i=1}^{n-1}S_i {3(n-i)\\choose 2(n-i)} \\right )$$ I was not sure how to compute that so I used OEIS to see if it has some nice formula. After finding by hand the first values $$3,6,21,90,429$$ it suggested me the formula $$\\frac{2}{3n-1}{3n\\choose 2n}$$. I managed to prove it by induction.\n\nI would be interested in some combinatorial proof of that, a bijection would be highly appreciated.\n\n\u2022 What is \"inf\" ? Do you mean the smallest $k$ for which the summation holds? \u2013\u00a0Doc Nov 28 '13 at 6:20\n\u2022 @Doc out of habit infimum, yes exactly \u2013\u00a0clark Nov 28 '13 at 6:24\n\u2022 Not at all there yet, but it seems interesting that the answer also takes the form $\\frac{1}{n} {3n-1\\choose n}$. \u2013\u00a0Doc Nov 28 '13 at 6:37\n\u2022 oops .... that should say $\\frac{3}{3n-1} {3n-1\\choose n}$. \u2013\u00a0Doc Nov 28 '13 at 7:36\n\u2022 @Doc: Or $\\frac3n\\binom{3n-2}{n-1}$, though that isn\u2019t obviously any more useful. \u2013\u00a0Brian M. Scott Nov 29 '13 at 19:42\n\nCode heads as 1 and tails as 0, and the problem can be phrased in the following way: Flip a fair coin until you have exactly twice as many heads as tails, and then stop. The value of $|S_n|$ is the number of such sequences of coin flips that have length exactly $3n$.\n\nFor this rephrased version, I asked the same question as the OP a couple of years ago on this site, under \"Combinatorial proof of $\\binom{3n}{n} \\frac{2}{3n-1}$ as the answer to a coin-flipping problem.\" It took me a few weeks before I found an answer to my own question, so I wouldn't call this an easy combinatorial proof to come up with. (In fact, I generalized my argument, recast it in terms of counting certain lattice paths, and got the argument published in the Electronic Journal of Combinatorics as \"Enumerating Lattice Paths Touching or Crossing the Diagonal at a Given Number of Lattice Points.\")\n\nAnyway, I'll reproduce my original combinatorial argument here. It uses the equivalent $|S_n| = \\frac{3}{3n-1} \\binom{3n-1}{n}$ as the formula to be proved. It also uses the following result (see, for example, Section 7.5 of Concrete Mathematics):\n\nRaney's Lemma: Let $a_1, ... a_m$ be a sequence of integers such that $\\sum a_i = 1$. There is a unique index $j$ such that the partial sums of the sequence $a_j, a_{j+1}, ... a_{j+m-1}$ (cyclic indices) are positive.\n\nOn to the proof.\n\nIntro.\n\nConsider the sequences with $2n$ occurrences of $-1$ (i.e., $2n$ heads) and $n$ occurrences of $+2$ (i.e., $n$ tails). We want to show that the number of these sequences with all partial sums nonzero is $\\binom{3n-1}{n} \\frac{3}{3n-1}$. The complete sum and empty sum are clearly $0$, so \"partial sum\" excludes those two cases. The sequences we want to count can be split into three groups: (1) all partial sums positive, (2) all partial sums negative, (3) some partial sums positive and some negative.\n\nGroup 1: The number of these sequences with all partial sums positive is $\\binom{3n-1}{n} \\frac{1}{3n-1}$.\n\nThis is the part that uses Raney's lemma. If all partial sums are positive, the last element in the sequence must be $-1$. Thus we want to count the number of sequences with $2n-1$ occurrences of $-1$ and $n$ occurrences of $+2$ that add to $+1$ and have all partial sums positive. Ignoring the partial sums restriction, there are $\\binom{3n-1}{n}$ such sequences. If we partition these $\\binom{3n-1}{n}$ sequences into equivalence classes based on cyclic shifts, Raney's lemma says that exactly one sequence in each equivalence class has all partial sums positive. Because there are $3n-1$ elements in each sequence there are $3n-1$ sequences with the same set of cyclic shifts, and so there are $3n-1$ sequences in each equivalence class. Thus the number of sequences in Group 1 is $\\binom{3n-1}{n} \\frac{1}{3n-1}$.\n\nGroup 2: The number of these sequences with all partial sums negative is also $\\binom{3n-1}{n} \\frac{1}{3n-1}$.\n\nTo see this, just reverse the sequences counted in Part 1.\n\nGroup 3: The number of these sequences with some positive partial sums and some negative partial sums is, yet again, $\\binom{3n-1}{n} \\frac{1}{3n-1}$.\n\nThis one is a little trickier. First, because of the $-1$'s, it is not possible to switch from positive partial sums to negative partial sums. Thus any sequence counted here must have exactly one switch: from negative partial sums to positive partial sums. The switch must occur at some point where the partial sum is $-1$ and the next element is $+2$. Thus we have some sequence $(a_1, \\ldots, a_m, +2, a_{m+2}, \\ldots, a_n)$ where the sums $a_1, a_1 + a_2, \\ldots, a_1 + \\cdots + a_m$ are all negative. Consider the sequence $(+2, a_m, \\ldots, a_2, a_1, a_{m+2}, \\ldots, a_n)$. Since $+2 + a_m + \\cdots + a_1 = 1$, and $a_k + a_{k-1} + \\cdots + a_1 < 0$ for all $k$, $1 \\leq k \\leq m$, it must be the case that $+2 + a_m + \\cdots + a_{k+1} > 1$ for all $k$, $1 \\leq k \\leq m-1$. So the sequence $(+2, a_m, \\ldots, a_2, a_1, a_{m+2}, \\ldots, a_n)$ is in Group 1. To see that this mapping is a bijection, note that any sequence in Group 1 must start with $+2$ and have a first time that a partial sum is equal to $+1$. Thus this transformation is reversible.\n\nSumming up.\n\nPutting Groups 1, 2, and 3 together we see that the total number of sequences we want to count is $\\binom{3n-1}{n} \\frac{3}{3n-1}$.\n\n\u2022 This is great! I didnt know about Ranney's Lemma, it was hidden all along this problem, it was nicely exploited! \u2013\u00a0clark Dec 5 '13 at 0:05", "date": "2021-01-23 23:38:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/584242/number-of-elements-of-3n-binary-tuples-where-the-ordinates-add-up-to-2n", "openwebmath_score": 0.9342162609100342, "openwebmath_perplexity": 183.75081048575186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987946220128863, "lm_q2_score": 0.8740772482857833, "lm_q1q2_score": 0.8635413135445773}}
{"url": "https://www.physicsforums.com/threads/prove-by-induction.211809/", "text": "# Prove by Induction.\n\n1. Jan 29, 2008\n\n### PFStudent\n\n1. The problem statement, all variables and given/known data\n1.1.2\n\nWrite,\n\n$${\\frac{1}{1\\cdot2}} + {\\frac{1}{2\\cdot3}} + {\\frac{1}{3\\cdot4}} + . . . + {\\frac{1}{99\\cdot100}}$$\n\nas a fraction in lowest terms.\n\n3. The attempt at a solution\n\nRewriting the problem as a summation,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\frac{1}{1\\cdot2}} + {\\frac{1}{2\\cdot3}} + {\\frac{1}{3\\cdot4}} + . . . + {\\frac{1}{n\\cdot(n+1)}}$$\n\nThen considering the first few terms,\n\n$${\\frac{1}{1\\cdot2}} = \\frac{1}{2}$$\n\n$${\\frac{1}{1\\cdot2}} + {\\frac{1}{2\\cdot3}} = \\frac{2}{3}$$\n\n$${\\frac{1}{1\\cdot2}} + {\\frac{1}{2\\cdot3}} + {\\frac{1}{3\\cdot4}} = \\frac{3}{4}$$\n\nThis leads to the conjecture that,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\frac{n}{n+1}}$$\n\nHow would I prove the above by induction?\n\n-PFStudent\n\nLast edited: Jan 29, 2008\n2. Jan 29, 2008\n\n### rock.freak667\n\nAssume the statement is true for n=N, then prove true for n=N+1\n\nEDIT:\n\n$$\\sum_{i=1} ^{n} \\frac{1}{i(i+1)}=\\frac{n}{n+1}$$\n\n3. Jan 29, 2008\n\n### jdavel\n\nthe last term in your sum (where i = n) is going to be 1/n(n+1). if you take the sum one further (to i = n+1) what will the last term be now?\n\n4. Jan 29, 2008\n\n### PFStudent\n\nHey,\n\nThanks for the reply rock.freak667 and jdavel.\n\nAssume,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\frac{n}{n+1}}$$\n\nis true for k. That is,\n\n$${{\\sum_{i = 1}^{k}}{\\frac{1}{i(i+1)}}} = {\\frac{k}{k+1}}$$\n\nThen to prove it is true for k+1, consider the following,\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {{\\sum_{i = 1}^{k}}{\\frac{1}{i(i+1)}}} + {\\frac{1}{k+1((k+1)+1)}}$$\n\nWhich reduces to,\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {\\frac{k}{k+1}} + {\\frac{1}{{k^2}+3k+1)}}$$\n\nHowever, how do I know the above is true?\n\nIn other words how do I know if the proof by induction actually proved it?\n\nThanks,\n\n-PFStudent\n\n5. Jan 29, 2008\n\n### rock.freak667\n\nWell you basically want to get\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\frac{(k+1)}{(k+1)+1}}$$\n\nbasically the sum is the same but instead of \"n\" put k+1\n\n6. Jan 29, 2008\n\n### jdavel\n\nyou're missing a pair of parenetheses in your 3rd equation that's leading to an error in the last term of your 4th equation. what should that last term be?\n\n7. Jan 29, 2008\n\n### PFStudent\n\nHey,\n\nConsider the following,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\frac{n}{n+1}}$$\n\nAssume it is true for k,\n\n$${{\\sum_{i = 1}^{k}}{\\frac{1}{i(i+1)}}} = {\\frac{k}{k+1}}$$\n\nNow, if it is true for k+1 the result expected is,\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {\\frac{k+1}{(k+1)+1}}$$\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {\\frac{k+1}{k+2}}$$\n\nThen to prove that, consider the following,\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {{\\sum_{i = 1}^{k}}{\\frac{1}{i(i+1)}}} + {\\frac{1}{(k+1)((k+1)+1)}}$$\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {\\left({\\frac{k}{k+1}}\\right)} + {\\frac{1}{(k+1)(k+2)}}$$\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {{\\frac{k(k+2)}{(k+1)(k+2)}}} + {\\frac{1}{(k+1)(k+2)}}$$\n\nWhich reduces to,\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {\\frac{(k+1)(k+1)}{(k+1)(k+2)}}$$\n\n$${{\\sum_{i = 1}^{k+1}}{\\frac{1}{i(i+1)}}} = {\\frac{k+1}{k+2}}$$\n\nProof.\n\nThanks,\n\n-PFStudent\n\nLast edited: Jan 29, 2008\n8. Jan 30, 2008\n\n### HallsofIvy\n\nStaff Emeritus\nWere you required to use induction?\n\nSince\n$$\\frac{1}{n(n+1)}= \\frac{1}{n}- \\frac{1}{n+1}$$\nthat is a \"telescoping\" series and the sum is immediate.\n\n9. Jan 30, 2008\n\n### PFStudent\n\nHey,\n\nCould I have proved,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\frac{n}{n+1}}$$\n\nby showing that it is a telescoping series? If so, how?\n\nThanks,\n\n-PFStudent\n\n10. Jan 30, 2008\n\n### HallsofIvy\n\nStaff Emeritus\n$$\\sum_{i=1}^{100} \\frac{1}{i(i+1}= \\frac{1}{1\\cdot2}} + {\\frac{1}{2\\cdot3}} + {\\frac{1}{3\\cdot4}} + . . . + {\\frac{1}{99\\cdot100}$$\n$$= (\\frac{1}{1}- \\frac{1}{2})+ (\\frac{1}{2}-\\frac{1}{3})+ (\\frac{1}{3}-\\frac{1}{4})+ \\cdot\\cdot\\cdot+ (\\frac{1}{99}- \\frac{1}{100})$$\nThe last fraction in each pair cancels the first fraction in the next pair (except of course in the last pair). Every fraction except the first and last cancel so the sum is\n$$\\frac{1}{1}- \\frac{1}{100}= \\frac{100-1}{100}= \\frac{99}{100}$$\n\nMore generally,\n$$\\sum_{i= 1}^n \\frac{1}{i(i+1)}= \\frac{1}{1}- \\frac{1}{n+1}= \\frac{n+1-1}{n+1}= \\frac{n}{n+1}$$\n\n11. Feb 4, 2008\n\n### PFStudent\n\nHey,\n\nThanks HallsofIvy for showing how it could have been proved by showing it was a telescoping sum.\n\nHowever, how would you know that the individual sums can be rewritten as the sum or difference of two fractions. In other words, how did you figure out the following,\n\n$$\\sum_{i=1}^{100} \\frac{1}{i(i+1}= \\frac{1}{1\\cdot2}} + {\\frac{1}{2\\cdot3}} + {\\frac{1}{3\\cdot4}} + . . . + {\\frac{1}{99\\cdot100} = \\left(\\frac{1}{1}-\\frac{1}{2}\\right) + \\left(\\frac{1}{2}-\\frac{1}{3}\\right)+ \\left(\\frac{1}{3}-\\frac{1}{4}\\right) + \\cdot\\cdot\\cdot + \\left(\\frac{1}{99}- \\frac{1}{100}\\right)$$\n\nThat is, how did you figure out that,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\left({{\\frac{A_{1}}{1}}-{\\frac{B_{1}}{2}}}\\right)} + {\\left({{\\frac{A_{2}}{2}}-{\\frac{B_{2}}{3}}}\\right)} +. . . + {\\left({{\\frac{A_{n}}{n}}-{\\frac{B_{n}}{n+1}}}\\right)}$$\n\nAdditionally, is finding the above the same as using the technique of partial fractions?\n\nAlso, what is the general way of finding $$A$$ and $$C$$ for the following (where: $$B, D, E, and{{.}}F$$; are all given),\n\n$${\\frac{E}{F}} = {{{\\frac{A}{B}}\\pm{\\frac{C}{D}}}}$$\n\nWhere it can be shown that,\n\n$${E} = {AD \\pm BC}$$\n\n$${F} = {BD}$$\n\nWhich is the same as,\n\n$${\\frac{E}{F}} = {\\frac{AD \\pm BC}{BD}}$$\n\nThanks,\n\n-PFStudent\n\nLast edited: Feb 4, 2008\n12. Feb 6, 2008\n\n### PFStudent\n\nHey,\n\nI was looking over this and realized that the following,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\left({{\\frac{A_{1}}{1}}-{\\frac{B_{1}}{2}}}\\right)} + {\\left({{\\frac{A_{2}}{2}}-{\\frac{B_{2}}{3}}}\\right)} +. . . + {\\left({{\\frac{A_{n}}{n}}-{\\frac{B_{n}}{n+1}}}\\right)}$$\n\nWas better written as,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\left({{\\frac{C_{1}}{1}}-{\\frac{C_{2}}{2}}}\\right)} + {\\left({{\\frac{C_{2}}{2}}-{\\frac{C_{3}}{3}}}\\right)} + {\\cdot} {\\cdot} {\\cdot} + {\\left({{\\frac{C_{n-1}}{n-1}}-{\\frac{C_{n}}{n}}}\\right)} + {\\left({{\\frac{C_{n}}{n}}-{\\frac{C_{n+1}}{n+1}}}\\right)}$$\n\nBut, I'm still having some trouble figuring how would one decipher what the constants \"C\" must be.\n\nThanks,\n\n-PFStudent\n\n13. Feb 6, 2008\n\n### rock.freak667\n\n$$\\sum_{n=1} ^{N} \\frac{1}{n(n+1)} \\equiv \\sum_{n=1} ^{N} (\\frac{1}{n}-\\frac{1}{n+1})$$\n\nUse partial fractions on 1/n(n+1)\n\n$$\\sum_{n=1} ^{N} \\frac{1}{n}-\\frac{1}{n+1}$$\n\nthen input n=1,2,3,...,N-2,N-1,N. then add them all up.\n\n14. Feb 6, 2008\n\n### PFStudent\n\nHey,\n\nI see that however, how would you have figured out what the constants for the partial fractions of,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}}$$\n\nwould be?\n\nIn other words, how would you have solved the following,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\left({{\\frac{C_{1}}{1}}-{\\frac{C_{2}}{2}}}\\right)} + {\\left({{\\frac{C_{2}}{2}}-{\\frac{C_{3}}{3}}}\\right)} + {\\cdot} {\\cdot} {\\cdot} + {\\left({{\\frac{C_{n-1}}{n-1}}-{\\frac{C_{n}}{n}}}\\right)} + {\\left({{\\frac{C_{n}}{n}}-{\\frac{C_{n+1}}{n+1}}}\\right)}$$\n\nfor: $$C_{1}, C_{2}, C_{3},..., C_{n-1}, C_{n}$$?\n\nThat is where I am stuck.\n\nThanks,\n\n-PFStudent\n\n15. Feb 6, 2008\n\n### rock.freak667\n\nI don't think you need to do all of that.\n\n16. Feb 11, 2008\n\n### PFStudent\n\nHey,\n\nThanks for the reply, rock.freak667. Your right I did not need to prove that it is a telescoping sum.\n\nHowever, I still would like to know how would one have realized that,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}}$$\n\nis a telescoping sum?\n\nFurther after the above realization one would have had to solve for the constants: $$C_{1}, C_{2}, C_{3},..., C_{n-1}, C_{n}$$; for the following,\n\n$${{\\sum_{i = 1}^{n}}{\\frac{1}{i(i+1)}}} = {\\left({{\\frac{C_{1}}{1}}-{\\frac{C_{2}}{2}}}\\right)} + {\\left({{\\frac{C_{2}}{2}}-{\\frac{C_{3}}{3}}}\\right)} + {\\cdot} {\\cdot} {\\cdot} + {\\left({{\\frac{C_{n-1}}{n-1}}-{\\frac{C_{n}}{n}}}\\right)} + {\\left({{\\frac{C_{n}}{n}}-{\\frac{C_{n+1}}{n+1}}}\\right)}$$\n\nAnd that is, what I wanted to know. How do you do that?\n\nThanks,\n\n-PFStudent\n\nLast edited: Feb 11, 2008", "date": "2016-10-27 03:55:53", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/prove-by-induction.211809/", "openwebmath_score": 0.7880914807319641, "openwebmath_perplexity": 1526.713819609365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462226131667, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8635412914108765}}
{"url": "https://math.stackexchange.com/questions/2713690/showing-mathbbzi-12i-oplus-mathbbzi-6-i-cong-mathbbzi-811i/2715412", "text": "Showing $\\mathbb{Z}[i]/(1+2i) \\oplus\\mathbb{Z}[i]/(6-i)\\cong\\mathbb{Z}[i]/(8+11i)$\n\nI am attempting to solve Ch 14 Problem 7.7 from Artin's algebra book.\n\nLet $R=\\mathbb{Z}[i]$ and let $V$ be the R-module generated by elements $v_1$ and $v_2$ with relations $(1+i)v_1+(2-i)v_2=0$ and $3v_1+5iv_2=0$. Write this module as a direct sum of cyclic modules.\n\nAttempt\n\nI have obtained $V\\cong R^2/ \\begin{bmatrix} 1+i & 3 \\\\ 2-i & 5i \\end{bmatrix} R^2 \\cong R/[8+11i]R=\\mathbb{Z}[i]/(8+11i)$.\n\nNow, I see that $(1+2i)(6-i)=8+11i.$ Now, I would like to show that $\\mathbb{Z}[i]/(1+2i) \\oplus\\mathbb{Z}[i]/(6-i)\\cong\\mathbb{Z}[i]/(8+11i)$, so I can have $V$ as a direct sum of cyclic modules as needed, but how can I show this?\n\nI have already shown that $(1+2i,6-i)=(1)=\\mathbb{Z}[i]$ and thus $(1+2i)+(6-i)=\\mathbb{Z}[i]$. Intuition would suggest that $(1+2i)\\oplus(6-i)=\\mathbb{Z}[i]$, although I think this is false since $(i-6)(1+2i)+(1+2i)(6-i)=0$.\n\nI must confess that I am very new to module theory so please be patient with me. I don't even how it would be possible to have $\\mathbb{Z}[i]/(1+2i) \\oplus\\mathbb{Z}[i]/(6-i)\\cong\\mathbb{Z}[i]/(8+11i)$ since $\\mathbb{Z}[i]/(1+2i)$and $\\mathbb{Z}[i]/(6-i)$ aren't even submodules of the same set.\n\n\u2022 $1+2i$ and $6-i$ are coprime elements of the Euclidean domain $\\Bbb{Z}[i]$. Their product is $8+11i$ then also their \"least common multiple\". So this is just another instance of the Chinese Remainder Theorem. Compare with isomorphisms of $\\Bbb{Z}$-modules $$\\Bbb{Z}_2\\oplus\\Bbb{Z}_3\\simeq\\Bbb{Z}_6, \\quad\\Bbb{Z}_3\\oplus\\Bbb{Z}_5\\simeq\\Bbb{Z}_{15}$$ et cetera. Mar 29 '18 at 18:54\n\u2022 In other words the mapping $$a+bi+(8+11i)\\mapsto (a+bi+(1+2i),a+bi+(6-i))$$ is an isomorphism. Mar 29 '18 at 18:55\n\u2022 Thank you. If I can show that that mapping is an isomorphism, this will show $\\mathbb{Z}[i]/(8+11i) \\cong\\mathbb{Z}[i]/(1+2i) \\times \\mathbb{Z}[i]/(6-i)$. However, I am not trying to show $\\times$, but $\\oplus$. Mar 29 '18 at 19:00\n\u2022 Aren't those the same thing? At least up to isomorphism (assuming you use one for inner direct sum and the other for outer). Mar 29 '18 at 19:24\n\u2022 For an answer following the method described in the problem statement: compute the Smith normal form of the relations matrix. Some relevant posts: 1, 2, 3. Mar 30 '18 at 4:36\n\nIn addition to everything said in the comments, I just wanted to remark one more thing that might be helpful concerning your last paragraph about the fact that $\\mathbb{Z}[i]/(1+2i)$ and $\\mathbb{Z}[i]/(6-i)$ are not submodules of some common module.\nAs was stated in the comments, the product of rings obtained from the chinese remainder theorem is in particular a product of modules, i.e. a direct sum, so this works perfectly fine. The point I wanted to add: If you want, you might still view both factors as submodules of $\\mathbb{Z}[i]/(8+11i)$ by the following observation:\nConsider a direct product of rings (commutative with $1$), $R=R_1\\times R_2$. Then you might view any $R$-module $M$ as a direct sum of submodules $M_1,M_2\\subset M$, where $M_i$ is a $R_i$-module (so in particular again a $R$-module) for $i=1,2$, namely set $M_1:=(1,0)\\cdot M\\subset M$ and $M_2:=(0,1)\\cdot M\\subset M$, then $$M=M_1\\oplus M_2$$ as $R$-module. (By the way, the inverse procedure works as well, if $\\tilde M_1$ and $\\tilde M_2$ are $R_1$- resp. $R_2$-modules, then they in particular are both $R$-modules, via $(r_1,r_2)\\cdot m_1:=r_1m_1$ for $m_1\\in \\tilde M_1$ and $(r_1,r_2)\\cdot m_2:=r_2m_2$ for $m_2\\in\\tilde M_2$ and so one may build the $R$-module $\\tilde M=\\tilde M_1\\oplus\\tilde M_2$. This is also described in this question/answer.)\nApplied to your situation: As you already proved $$R:=\\mathbb{Z}[i]/(8+11i)\\simeq \\underset{=:R_1}{\\underbrace{\\mathbb{Z}[i]/(1+2i)}}\\times\\underset{=:R_2}{\\underbrace{\\mathbb{Z}[i]/(6-i)}},$$ we would now like to find elements $e_1,e_2\\in R$ such that $e_1$ corresponds to $(1,0)\\in R_1\\times R_2$ and $e_2$ corresponds to $(0,1)$. To find those, we use the Euclidean algorithm to get $$1=\\underset{=:e_2}{\\underbrace{(1-3i)(1+2i)}}+\\underset{=:e_1}{\\underbrace{(-1)(6-i)}}.$$ So we might write $R$ as the direct sum of $R$-submodules $$R=e_1R\\oplus e_2R=(7-i)R\\oplus (i-6)R.$$ (Naturally, as a $R$-module this is of course isomorphic to the former version $R=R_1\\oplus R_2$, as $e_i\\cdot(\\bullet)\\colon R_i\\rightarrow e_iR$ is an isomorphism of $R$-modules, $i=1,2$ - I just thought this slightly different point of view might possibly help to get more comfortable with the situation...)", "date": "2021-12-01 23:49:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2713690/showing-mathbbzi-12i-oplus-mathbbzi-6-i-cong-mathbbzi-811i/2715412", "openwebmath_score": 0.8488600850105286, "openwebmath_perplexity": 115.64984973242012, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936087546923, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8635193015445991}}
{"url": "https://math.stackexchange.com/questions/854850/how-can-i-prove-that-2n2-mid2n3", "text": "# How can I prove that $2^{n+2}\\mid(2n+3)!$?\n\nI'm not sure where to proceed or how to go about proving this assertion holds for all natural numbers n: $$2^{n+2} \\mid(2n + 3)!$$\n\nThe base case is $n=1$, where $2^{1+2}\\mid(2\\cdot 1+3)!$ which simplifies to $8 \\mid 120$, and 8 does indeed divide 120.\n\nAgain, we shall assume the statement is true for $n = k$. $$2^{k+2}\\mid(2k+3)!$$ Then we shall prove that the statement must be true for n = k + 1: $$2^{k+3}\\mid (2(k+1)+3)!=(2k+5)!$$\nSorry for the poor formatting, this is the first time I've posted here. I'm not sure if I've started off on the right path, or where to go next. Any suggestions would be lovely!\n\nThere are $n+1$ even numbers in $1,\\cdots,2n+3$ and one of them is $4$ (given $n\\ge1$).\n\n\u2022 So because the divisor has 2 to the power of whatever, it will always divide whatever the quotient is due to it having a 4 in the factorial? Is this the correct way to interpret your response? \u2013\u00a0knames Jul 3 '14 at 0:09\n\u2022 @knames You use the phrases \"the divisor\" and \"whatever\" and \"it\" and \"the quotient\" but I don't know what you're referring to with these phrases. Please be clear and specific. \u2013\u00a0blue Jul 3 '14 at 0:10\n\u2022 the divisor being 2^(n+2) and the quotient being (2n+3)!. The left part before | will always divide the right part after | because the right part has a 4 in the factorial? \u2013\u00a0knames Jul 3 '14 at 0:12\n\u2022 @knames The word \"quotient\" refers to what you get after you've divided, it doesn't refer to the thing you're dividing. Just because something is divisible by $4$ does not mean it is divisible by $2^{n+2}$; that's where the total of $n+1$ even numbers comes in. Try thinking some more about my hint. \u2013\u00a0blue Jul 3 '14 at 0:14\n\u2022 This is as explicit as I can make it: $$(2n+3)!=1\\cdot2\\cdot3\\cdots(2n+3)=(\\underbrace{2\\cdot\\color{Red}4\\cdots2n+2}_{n+1~\\rm numbers})(1\\cdot3\\cdots2n+3)$$ $$=2^{n+1}(1\\cdot\\color{Red}2\\cdot3\\cdots n+1)(1\\cdot3\\cdots 2n+3)=2^{n+2}(3\\cdots n+1)(1\\cdot3\\cdots 2n+3).$$ Honestly, I thought my hint was pretty clear and straightforward, and this level of explication was totally unnecessary. \u2013\u00a0blue Jul 3 '14 at 0:41\n\n$(2k+5)!=(2k+5)(2k+4)(2k+3)!=(2k+5)(2k+4)2^{k+2}\\cdot c$ where the last equality is due to $2^{k+2}\\mid (2k+3)!$ which you assume.\n\nHence, $(2k+5)!=(2k+5)(k+2)(2)(2^{k+2})$. Finish it :)\n\nWe have $(2k+5)!=(2k+3)!(2k+4)(2k+5)$. Since by the induction hypothesis, $2^{k+2}$ divides $(2k+3)!$, and since $2$ divides $2k+4$, we conclude that $2^{k+3}$ divides $(2k+5)!$.\n\nRemark: (This refers to an earlier version of the post.) In writing up a proof, never try to travel from what you want to show to something \"known.\"\n\n\u2022 I'm new to number-theory, could you elaborate on what you meant by \"In writing up a proof, never try to travel from what you want to show to something \"known.\"\" \u2013\u00a0knames Jul 2 '14 at 23:56\n\u2022 @knames Sometimes you start from the fact that you're trying to prove and manipulate things until you've arrived at an already established fact. This is how it goes in any kind of math, number theory or no. You might even try burning both ends of the stick and meeting somewhere in the middle. At the end, though, when you actually go to write down your proof, you need to start at something known and end up at the desired claim, in that order. \u2013\u00a0blue Jul 3 '14 at 0:00\n\u2022 In high school, people pick up the very bad habit of writing down what they want to be true, and then manipulating until they get something true, like $x=x$. This is logically wrong, unless one shows that every step that got us from what we want to $x=x$ is reversible. If one showed reversibility (which one often cannot show), it would be OK. But people who go from the desired to something known often lose control over the logic of the argument, and end up writing a confused circular non-proof. I am not saying you shouldn't fool around informally with the \"desired.\" \u2013\u00a0Andr\u00e9 Nicolas Jul 3 '14 at 0:01\n\u2022 But ultimately the written up argument must go \"the right way\" so that its validity is clear. \u2013\u00a0Andr\u00e9 Nicolas Jul 3 '14 at 0:02\n\u2022 @blue, @ Andr\u00e9 Nicolas Thank you, this makes sense to me now! \u2013\u00a0knames Jul 3 '14 at 0:03\n\nStart with $2k+2\u2223(2k+3)!$. This means there exists a $b \\in\\mathbb{N}$ such that $(2k+3)! = b2^{k+2}$.\n\n\\begin{align*} (2k+3)! = b2^{k+2} & \\Rightarrow (2k+3)!(2k+4)(2k+5) = b2^{k+2}(2k+4)(2k+5)& \\\\ & \\Leftrightarrow (2k+5)! = b2^{k+2}2(k+2)(2k+5) \\\\ & \\Leftrightarrow (2k+5)! = (b(k+2)(2k+5))2^{k+3} \\\\ & \\Rightarrow 2^{k+3} | (2k+5)! \\\\ & \\Leftrightarrow 2^{(k+1)+2} | (2(k+1)+3)! \\end{align*}\n\n\u2022 Just a quick question, what is the significance of including the last line? Why break down the exponent and everything to the right of the \"|\"? \u2013\u00a0knames Jul 3 '14 at 1:25\n\u2022 It makes the induction pedantically clear. This expression is a straight substitution of $(k+1)$ for $k$ in the equation we're trying to prove. \u2013\u00a0NovaDenizen Jul 3 '14 at 1:42\n\n## Theory\n\nThe number of factors of the prime number, p, that divide into n! is $\\sum_{i=1}^{\\infty} \\left \\lfloor \\dfrac{n}{p^i} \\right \\rfloor$\n\nAs formidable as this equation may look, it's not really that bad.\n\nFirst of all, once $\\left \\lfloor \\dfrac{n}{p^i} \\right \\rfloor$ becomes $0$, all of the terms after that are also $0$. So it's really a finite sum.\n\nSecond, it can be shown that $$\\left \\lfloor \\dfrac{n}{p^{i+1}} \\right \\rfloor = \\left \\lfloor \\dfrac{\\left \\lfloor \\dfrac{n}{p^i} \\right \\rfloor}{p} \\right \\rfloor$$\n\nSo, knowing the value of $\\left \\lfloor \\dfrac{n}{p^i} \\right \\rfloor$ makes it easier to compute $\\left \\lfloor \\dfrac{n}{p^{i+1}} \\right \\rfloor$.\n\n## Application\n\nWe make the following computations.\n\n\u2022 $\\left \\lfloor \\dfrac{2n+3}{2} \\right \\rfloor = n+1$\n\u2022 $\\left \\lfloor \\dfrac{n+1}{2} \\right \\rfloor \\ge 1 \\quad (\\forall n \\ge 1)$\n\nIt follows that the highest power of 2 that divides into $(2n+3)!$ is greater than or equal to $(n+1) + 1 = n+2$. In other words, $2^{n+2} | (2n+3)!$", "date": "2020-05-27 03:37:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/854850/how-can-i-prove-that-2n2-mid2n3", "openwebmath_score": 0.9536570906639099, "openwebmath_perplexity": 422.8549702755819, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936073551713, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8635193003165519}}
{"url": "https://www.physicsforums.com/threads/uniform-circular-motion-how-can-radial-acceleration-have-a-calculated-value.919831/", "text": "# I Uniform circular motion -- How can radial acceleration have a calculated value?\n\n1. Jul 10, 2017\n\n### Gurasees\n\nIn uniform circular motion, direction of particle is changing at every moment but its speed remains the same. If the magnitude of velocity or speed remains the same, change in magnitude of velocity is zero. Then how come radial acceleration can have a calculated value since acceleration = change in magnitude of velocity/ change in time. Yes, the object has an acceleration due to change in direction but how can we possibly obtain a value?\n\n2. Jul 10, 2017\n\n### scottdave\n\nThis video may shed some insight for you. I was helpful for me, explaining derivatives of vectors.\n\n3. Jul 10, 2017\n\n### A.T.\n\nWhere did you get that defintion of acceleration from?\n\n4. Jul 10, 2017\n\n### Staff: Mentor\n\nAcceleration is defined as the change in the velocity divided by the change in time, of: $\\vec a = \\frac{dV}{dt}$\nSince velocity is changing, this requires that there be an acceleration. Note that acceleration is not defined as the change in the magnitude of the velocity, but simply the change in the velocity.\n\nYou can find a simple calculus derivation here: https://en.wikipedia.org/wiki/Centripetal_force#Calculus_derivation\n\n5. Jul 10, 2017\n\n### scottdave\n\nYes. Note that acceleration is a vector quantity and is often in a different direction than velocity. In the case of uniform circular motion, it is always at 90\u00b0 to the velocity direction (towards center of the circle). I think the video does a nice job of explaining this.\n\n6. Jul 11, 2017\n\n### Gurasees\n\nActually yeah definition that i wrote for acceleration is wrong. Here acceleration is associated with change in direction. But what i am asking is how can we obtain a numerical value of acceleration if there is no change in numerical value of velocity?\n\n7. Jul 11, 2017\n\n### A.T.\n\nVelocity is a vector which does change.\n\n8. Jul 11, 2017\n\nA vector is a \"numerical\" value. Just not a scalar. Speed is a Scalar, Velocity is a vector. Since the Force (vector) and the Velocity of the particle are always at 90 degrees - no work is done, but there is action/reaction.\n\nF=ma where F and a are vectors is critical to proper analysis,\n\nMANY- heck if not all, cases involving normal vectors are counter-intuitive, and are worth special attention. If you master the vector math behind the precession of a gyroscope, for example - you will know more physics than 99.9% of the population.\n\nIt may seem like a simple concept - but complete comprehension is very valuable.\n\n9. Jul 11, 2017\n\n### Staff: Mentor\n\nThe numerical value of the velocity's vector components is continuously changing. This is true in both Cartesian and polar/spherical coordinates.\n\n10. Jul 11, 2017\n\n### vanhees71\n\nInstead of unclear words formulae can only help the understanding ;-)).\n\nTake the example of a particle running around on a circle of radius $R$ around the origin in the $xy$ plane with constant angular velocity $\\omega$. The position vector is given by\n$$\\vec{x}(t)=R \\begin{pmatrix} \\cos(\\omega t) \\\\ \\sin(\\omega t) \\\\ 0 \\end{pmatrix}.$$\nThe velocity is\n$$\\vec{v}=\\dot{\\vec{x}} = R \\omega \\begin{pmatrix} -\\sin(\\omega t) \\\\ \\cos (\\omega t) \\\\ 0 \\end{pmatrix}$$\nand the acceleration\n$$\\vec{a}=\\dot{\\vec{v}}=R \\omega^2 \\begin{pmatrix} -\\cos(\\omega t) \\\\ -\\sin(\\omega t) \\\\ 0 \\end{pmatrix}=-\\omega^2 \\vec{x},$$\ni.e., the acceleration is radially towards the center with the magnitude $a=|\\vec{a}|=\\omega^2 R$. To keep the particle on the circle you need the corresponding force, called the centripetal force, $\\vec{F}=m \\vec{a}=-m \\omega^2 \\vec{x}$.\n\nLast edited: Jul 15, 2017\n11. Jul 15, 2017\n\n### Staff: Mentor\n\nTo clarify: both velocity and acceleration are vectors, so this should be written as $$\\vec a = \\frac {d \\vec v}{dt}$$ or, component by component: $$a_x = \\frac{dv_x}{dt} \\\\ a_y = \\frac{dv_y}{dt} \\\\ a_z = \\frac{dv_z}{dt}$$ or, in the matrix notation that vanhees71 used: $$\\vec a = \\begin{pmatrix} a_x \\\\ a_y \\\\ a_z \\end{pmatrix} = \\frac {d}{dt} \\begin{pmatrix} v_x \\\\ v_y \\\\ v_z \\end{pmatrix}$$\n\n12. Jul 29, 2017\n\n### Arjan82\n\nWithout formula's: acceleration in one direction is independent of the acceleration in any other direction (as long as the directions are normal to each other). Thus if you look at your object, it's velocity in horizontal direction does change all the time, and therefore it has an acceleration in horizontal direction. This is also independently true for the vertical direction.\n\n13. Aug 1, 2017\n\n### CWatters\n\nWhy should it be a problem to calculate a numerical value of acceleration in such a case?\n\nLets say you have a ball going North at 3m/s and later it's found to be going South at 3m/s. The numerical value of the velocity (aka speed) hasn't changed, only the direction has changed. To calculate the acceleration you first need to calculate the change in velocity which in this case is 6m/s South.", "date": "2017-11-18 14:19:15", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/uniform-circular-motion-how-can-radial-acceleration-have-a-calculated-value.919831/", "openwebmath_score": 0.7194363474845886, "openwebmath_perplexity": 513.9955391437354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9840936054891428, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8635192986791552}}
{"url": "https://math.stackexchange.com/questions/2615300/does-the-series-sum-2n-sin-frac-pi3n-converge/2615302", "text": "# Does the series $\\sum 2^n \\sin(\\frac{\\pi}{3^n})$ converge?\n\nCheck if $$\\sum_{n = 1}^{\\infty}2^n \\sin\\left(\\frac{\\pi}{3^n}\\right)$$ converges.\n\nI tried to solve this by using the ratio test - I have ended up with the following limit to evaluate: $$\\lim_{n \\to \\infty} \\left(\\frac{2\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\sin \\left(\\frac{\\pi}{3^n} \\right)} \\right)$$ And now - I am stuck and don't know how to proceed with this limit. Any hints?\n\n\u2022 Use the fact that $\\sin(x) \\leq x$ \u2013\u00a0TheOscillator Jan 21 '18 at 22:00\n\u2022 L'Hospital's rule will help you with the limit you pose, but you can show convergence directly as Olivier outlines. \u2013\u00a0Malcolm Jan 21 '18 at 22:01\n\u2022 @Malcolm This is a sequence, not a function. A sequence is not differentiable. \u2013\u00a0Aemilius Jan 21 '18 at 22:01\n\u2022 If $\\lim_{x \\to \\infty} \\left(\\frac{2\\sin\\left(\\frac{\\pi}{3 \\cdot 3^x} \\right)}{\\sin \\left(\\frac{\\pi}{3^x} \\right)} \\right)$ exists (which it does) then $\\lim_{n \\to \\infty} \\left(\\frac{2\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\sin \\left(\\frac{\\pi}{3^n} \\right)} \\right)$ also exists and equals the same value. \u2013\u00a0Malcolm Jan 21 '18 at 22:05\n\nHint. One has $$\\left|\\sin\\left(\\frac{\\pi}{3^n}\\right)\\right|\\le\\frac{\\pi}{3^n},\\quad n=1,2,\\cdots.$$ Can you take it from here?\n\n\u2022 Your observation implies that this limit in question - say, L: $$0 \\le L \\le 2/3$$ and so this series does converge? \u2013\u00a0Aemilius Jan 21 '18 at 22:04\n\u2022 You then have $$\\left|\\sum_{n = 1}^{\\infty}2^n \\sin \\frac{\\pi}{3^n} \\right| \\le \\sum_{n = 1}^{\\infty}\\left|2^n \\sin \\frac{\\pi}{3^n} \\right| \\le \\pi \\sum_{n = 1}^{\\infty}\\frac{2^n}{3^n}<\\infty$$ and the given series is absolutely convergent then it is convergent. \u2013\u00a0Olivier Oloa Jan 21 '18 at 22:07\n\nNote that\n\n$$2^n \\sin\\left(\\frac{\\pi}{3^n}\\right) \\sim \\pi\\frac{2^n}{3^n}$$\n\nthen use comparison test with $$\\sum \\frac{2^n}{3^n}$$\n\nIf you want to stick with your approach using the ratio test, you can (although Olivier's answer is more direct). Caveat: I'll detail every step of the derivation, which is not actually necessary for a proof.\n\nWe will only rely on elementary arguments, specifically the fact that $\\sin'(0)=\\cos 0 = 1$ \u2014 which is equivalent to $$\\lim_{x\\to 0} \\frac{\\sin x}{x} = 1 \\,.\\tag{1}$$ From there, you can write $$\\lim_{n \\to \\infty} \\frac{2\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\sin \\left(\\frac{\\pi}{3^n} \\right)} = \\lim_{n \\to \\infty} 2\\cdot \\frac{1}{3}\\cdot \\frac{\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\frac{\\pi}{3 \\cdot 3^n}}\\cdot\\frac{\\frac{\\pi}{3^n}}{\\sin \\left(\\frac{\\pi}{3^n} \\right)} = \\frac{2}{3}\\lim_{n \\to \\infty} \\frac{\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\frac{\\pi}{3 \\cdot 3^n}}\\cdot\\left(\\frac{\\sin \\left(\\frac{\\pi}{3^n} \\right)}{\\frac{\\pi}{3^n}}\\right)^{-1} \\tag{2}$$ and, by (1), we get \\begin{align*} \\lim_{n \\to \\infty} \\frac{2\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\sin \\left(\\frac{\\pi}{3^n} \\right)} &= \\frac{2}{3}\\lim_{n \\to \\infty} \\frac{\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\frac{\\pi}{3 \\cdot 3^n}}\\cdot \\lim_{n \\to \\infty} \\left(\\frac{\\sin \\left(\\frac{\\pi}{3^n} \\right)}{\\frac{\\pi}{3^n}}\\right)^{-1} =\\frac{2}{3}\\lim_{n \\to \\infty} \\frac{\\sin\\left(\\frac{\\pi}{3 \\cdot 3^n} \\right)}{\\frac{\\pi}{3 \\cdot 3^n}}\\cdot \\left(\\lim_{n \\to \\infty} \\frac{\\sin \\left(\\frac{\\pi}{3^n} \\right)}{\\frac{\\pi}{3^n}}\\right)^{-1} \\\\ &= \\frac{2}{3}\\cdot 1\\cdot 1^{-1} = \\boxed{\\frac{2}{3}} \\end{align*} and you can conclude with the ratio test.\n\nSame idea as Olivier express in a different way, you know that $$\\sin\\left(x\\right)\\underset{(0)}{=}x+o\\left(x\\right)$$ Hence\n\n$$2^n\\sin\\left(\\frac{\\pi}{3^n}\\right) \\underset{(+\\infty)}{\\sim}\\pi \\left(\\frac{2}{3}\\right)^n$$\n\nWhat can you say about $\\displaystyle \\sum_{n \\geq 0}\\left(\\frac{2}{3}\\right)^n$ ?", "date": "2019-09-18 11:20:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2615300/does-the-series-sum-2n-sin-frac-pi3n-converge/2615302", "openwebmath_score": 0.9997424483299255, "openwebmath_perplexity": 442.8707185059686, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936064221572, "lm_q2_score": 0.8774767826757122, "lm_q1q2_score": 0.863519291615053}}
{"url": "https://math.stackexchange.com/questions/2831645/why-doesnt-the-quadratic-equation-contain-2a-in-the-denominator", "text": "Why doesn't the quadratic equation contain $2|a|$ in the denominator?\n\nWhen deriving the quadratic equation as shown in the Wikipedia article about the quadratic equation (current revision) the main proof contains the step: $$\\left(x+{\\frac {b}{2a}}\\right)^{2}={\\frac {b^{2}-4ac}{4a^{2}}}$$ the square root is taken from both sides, so why is $$\\sqrt{4a^2} = 2a$$ in the denominator and not $$\\sqrt{4a^2} = 2\\left |a \\right |$$ Could somebody explain this to me? Thank you very much\n\nmigrated from mathoverflow.netJun 25 '18 at 17:26\n\nThis question came from our site for professional mathematicians.\n\nOne could take the square root as $2|a|$ instead, which would lead to:\n\n$$x+\\frac {b}{2a} = \\pm{\\frac {\\sqrt{b^{2}-4ac}}{2|a|}} \\quad\\iff\\quad x = -\\frac {b}{2a} \\pm {\\frac {\\sqrt{b^{2}-4ac}}{2|a|}} \\tag{1}$$\n\nHowever, given that $\\,|a|\\,$ is either $\\,a\\,$ or $\\,-a\\,$ it follows that $\\,\\pm|a|=\\pm a\\,$, so the formula simplifies to:\n\n$$x = -\\frac {b}{2a} \\pm {\\frac {\\sqrt{b^{2}-4ac}}{2|a|}} = -\\frac {b}{2a} \\pm {\\frac {\\sqrt{b^{2}-4ac}}{\\color{red}{2a}}} = \\frac{-b \\pm \\sqrt{b^{2}-4ac}}{2a} \\tag{2}$$\n\n$(1)\\,$ and $\\,(2)\\,$ are entirely equivalent, but $\\,(2)\\,$ is more convenient to use.\n\nWhen taking the square root we put a $\\pm$ on the right hand side to account for the two roots, so it is unnecessary to strip off the sign of $a$, as we will put it back anyways.\n\nThe two square roots of $a^2$ are $a$ and $-a$, sometimes written together as $\\pm a$.\n\nFor real numbers $\\pm a$ is equivalent to $\\pm |a|$ but this is not true for complex numbers. So putting the absolute value operation in would make the proof less general.\n\nWe could write the proof as\n\n$$\\left(x+{\\frac {b}{2a}}\\right)^{2}={\\frac {b^{2}-4ac}{4a^{2}}}$$\n\n$$\\pm\\left(x+{\\frac {b}{2a}}\\right)={\\frac {\\pm\\sqrt{b^{2}-4ac}}{\\pm2a}}$$\n\n$$x+{\\frac {b}{2a}}={\\frac {\\pm\\sqrt{b^{2}-4ac}}{2a}}$$\n\nBut generally it is considered sufficient to put in just a single $\\pm$ from the start rather than putting in one for each square root and then removing the redundant ones.\n\nIf you put $x =\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}$ into $ax^2+bx+c$, since $x^2 =\\dfrac{b^2\\mp2b\\sqrt{b^2-4ac}+(b^2-4ac)}{4a^2} =\\dfrac{2b^2-4ac\\mp2b\\sqrt{b^2-4ac}}{4a^2}$ you get\n\n$\\begin{array}\\\\ ax^2+bx+c &a\\dfrac{2b^2-4ac\\mp2b\\sqrt{b^2-4ac}}{4a^2} +b\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}+c\\\\ &=\\dfrac{2b^2-4ac\\mp2b\\sqrt{b^2-4ac}}{4a} +\\dfrac{-2b^2\\pm 2b\\sqrt{b^2-4ac}}{4a}+c\\\\ &=\\dfrac{2b^2-4ac\\mp2b\\sqrt{b^2-4ac}-2b^2\\pm 2b\\sqrt{b^2-4ac}+4ac}{4a}\\\\ &=0\\\\ \\end{array}$\n\nIf you use $|a|$, it won't work since you can't combine the terms.\n\n\u2022 The OP is only putting the absolute value when taking the square root, not putting it on $-b/2a$. This is correct, but unnecessary, see dxiv's answer. \u2013\u00a0Fan Zheng Jun 25 '18 at 18:29\n\nmy preference for remembering and using the quadratic formula (and electrical engineers seem to do that often) is to remember the root quadratic equations as:\n\n$$x^2 \\ + \\ b\\,x \\ + \\ c \\ = \\ 0$$\n\nwhich has solution:\n\n$$x \\ = \\ \\begin{cases} -\\tfrac{b}{2} \\pm \\sqrt{\\left(\\tfrac{b}{2}\\right)^2 - c} \\qquad & \\text{for }\\left(\\tfrac{b}{2} \\right)^2 > c \\\\ \\\\ -\\tfrac{b}{2} \\qquad & \\text{for }\\left(\\tfrac{b}{2} \\right)^2 = c \\\\ \\\\ -\\tfrac{b}{2} \\pm i \\sqrt{c - \\left(\\tfrac{b}{2}\\right)^2} \\qquad & \\text{for }\\left(\\tfrac{b}{2} \\right)^2 < c \\\\ \\end{cases}$$\n\nnormalizing out the \"$a$\" does not make the quadratic equation less general. the only degrees of freedom are $b$ and $c$, so that means normally (except for a double root), there are two independent solutions.\n\nAlternatively, noting $a\\ne 0$: \\begin{align}\\left(x+{\\frac {b}{2a}}\\right)^{2}&={\\frac {b^{2}-4ac}{4a^{2}}} \\iff \\\\ 4a^2\\left(x+{\\frac {b}{2a}}\\right)^{2}&=b^{2}-4ac \\iff \\\\ \\left(2ax+b\\right)^{2}&=b^{2}-4ac \\iff \\\\ 2ax+b&=\\pm \\sqrt{b^2-4ac} \\iff \\\\ x&=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}.\\end{align}", "date": "2019-08-20 01:49:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2831645/why-doesnt-the-quadratic-equation-contain-2a-in-the-denominator", "openwebmath_score": 0.9918550848960876, "openwebmath_perplexity": 299.0361732696532, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639677785087, "lm_q2_score": 0.8887588001219789, "lm_q1q2_score": 0.8634860262445764}}
{"url": "http://math.stackexchange.com/questions/130159/what-is-the-distribution-of-this-random-series", "text": "What is the distribution of this random series?\n\nLet $\\xi_n$ be iid and uniformly distributed on the three numbers $\\{-1,0,1\\}$. Set $$X = \\sum_{n=1}^\\infty \\frac{\\xi_n}{2^n}.$$ It is clear that the sum converges (surely) and the limit has $-1 \\le X \\le 1$..\n\nWhat is the distribution of $X$?\n\nDoes it have a name? Can we find an explicit formula? What else can we say about it (for instance, is it absolutely continuous)?\n\nWe can immediately see that $X$ is symmetric (i.e. $X \\overset{d}{=} -X$). Also, if $\\xi$ is uniformly distributed on $\\{-1,0,1\\}$ and independent of $X$, we have $X \\overset{d}{=} \\frac{1}{2}(X+\\xi)$. It follows that for the cdf $F(x) = \\mathbb{P}(X \\le x)$, we have $$F(x) = \\frac{1}{3}(F(2x+1) + F(2x) + F(2x-1)). \\quad (*)$$\n\nThe cdf of $\\sum_{n=1}^{12} \\frac{\\xi_n}{2^n}$ looks like this:\n\nIt looks something like $\\frac{1}{2}(1+\\sin(\\frac{\\pi}{2}x))$ but that doesn't quite work (it doesn't satisfy (*)).\n\n-\nI think the equation (*) together with the boundary conditions $F(x\\leq -1)=0$ and $F(x\\geq 1)=1$ determines the cdf $F$ uniquely. \u2013\u00a0 Fabian Apr 10 '12 at 20:20\nRemotely related is the Fabius random variable (see wiki). \u2013\u00a0 Sasha Apr 10 '12 at 20:25\nIt can be shown easily that $F(0)=1/2$, $F(1/2)=5/6$, $F(-x)=1-F(x)$. \u2013\u00a0 Fabian Apr 10 '12 at 20:32\n$F(1/4)=2/3$, $F(1/3)=8/11$, $F(2/3)=10/11$, $F(3/4)=17/18$, ... The values of $F$ for rational arguments can be obtained recursively (but I did not yet figure out an explicit formula). \u2013\u00a0 Fabian Apr 10 '12 at 20:48\nFor $x=-1+2^{-k}$ you can get that $F(x)=\\frac{1}{2(3^k)}$. The first $k$ have to be negative, and then half of those end up on the left. Thus, for small $\\epsilon$, $F(-1+\\epsilon)$ is roughly $\\frac{1}{2}\\epsilon^{\\log_2 3}$. \u2013\u00a0 Thomas Andrews Apr 10 '12 at 20:59\n\nThis is more of a comment, than an answer, yet it's too big, and graphics can't be used in comments.\n\nIt is not hard to work out cumulants of $X$: $$\\kappa_X(r) = \\sum_{n=1}^\\infty \\frac{\\kappa_\\xi(r)}{2^{n r}} = \\frac{\\kappa_\\xi(r)}{2^r-1} = \\frac{3^r-1}{2^r-1} \\cdot \\frac{B_r}{r} \\cdot [ r \\geqslant 2]$$ Obviously, due to symmetry, odd cumulants and moments vanish. This implies the following low order moments: $$m_2 = \\mathbb{E}(X^2) = \\frac{2}{9}, \\quad m_4 = \\frac{14}{135}, \\quad m_6 = \\frac{106}{1701}, \\quad \\ldots$$\n\nThe distribution itself appears to not be absolutely continuous, based on simulations:\n\nFollowing Fabian's footsteps, it is easy to code computation of CDF at rational points:\n\nClearAll[cdf];\ncdf[x_?ExactNumberQ] /; x >= 1 := 1;\ncdf[0] = 1/2;\ncdf[x_?Negative] := 1 - cdf[-x];\ncdf[x_Rational /; EvenQ[Denominator[x]]] /; -1 < x < 1 :=\ncdf[x] = (cdf[2 x] + cdf[2 x - 1] + cdf[2 x + 1])/3;\ncdf[x_Rational] := (* set up linear equations and solve them *)\nBlock[{f, den = Denominator[x], ru1, ru2, vals, sol, ru3},\nru1 = {f[z_] :> Divide[f[2 z] + f[2 z + 1] + f[2 z - 1], 3]};\nru2 = {f[z_ /; z <= -1] :> 0, f[z_ /; z >= 1] :> 1,\nf[z_?Negative] :> 1 - f[-z]};\nru3 = f[r_Rational /; Denominator[r] < den] :> cdf[r];\nvals = Table[f[k/den], {k, den - 1}] /. ru3;\nsol = Solve[(((vals /. ru1) //. ru2) /. ru3) == vals,\nCases[vals, _f]];\nFunction[{arg, res}, Set[cdf[arg], res]] @@@\nCases[vals, f[a_] :> {a, f[a] /. First[sol]}];\ncdf[x]\n]\n\n\nSo, for example,\n\nIn[101]:= cdf[1/5]\n\nOut[101]= 31/49\n\n\nThe agreement with simulations is excellent:\n\n-", "date": "2015-08-04 02:00:14", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/130159/what-is-the-distribution-of-this-random-series", "openwebmath_score": 0.5641429424285889, "openwebmath_perplexity": 1359.0535747279775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639686018701, "lm_q2_score": 0.8887587964389112, "lm_q1q2_score": 0.8634860233980102}}
{"url": "https://stats.stackexchange.com/questions/436975/compact-notation-for-one-hot-indicator-vectors", "text": "# Compact notation for one-hot indicator vectors?\n\nMany machine learning approaches use one-hot vectors to represent categorical data. This is sometimes called using indicator features, indicator vectors, regular categorical encoding, dummy coding, or one-hot encoding (among other names).\n\nI'm searching for a compact way to denote a one-hot vector within a model.\n\nSay we have a categorical variable with $$m$$ categories. First, apply some arbitrary sorting to the categories. A one-hot vector $$v$$ is then a binary vector of length $$m$$ where only a single entry can be one, all others must be zero. We set the $$i^\\text{th}$$ entry to 1, and all others to 0, to indicate that this $$v$$ represents the categorical variable taking on the $$i^\\text{th}$$ possible value.\n\nOne clunky attempt based on misguided set notation;\n\n$$v \\in \\{0, 1\\}^m \\qquad\\qquad \\sum_{i=1}^m v_i = 1$$\n\nI've also seen math-oriented people refer to a one-hot vector using the notation\n\n$$\\mathbf{e}_i$$\n\nBut I don't understand where this notation comes from or what it is called.\n\nCan anyone help me out? Is there a paper that does a good job of this?\n\nThank you,\n\nThere are several ways to note dummy variables (or one-hot encoded), one of them is the indicator function :\n\n$$\\mathbb{1}_A(x) := \\begin{cases} 1 &\\text{if } x \\in A, \\\\ 0 &\\text{if } x \\notin A. \\end{cases}$$\n\nFor $$e_i$$ it is a vector of the standard base, where $$e_i$$ denotes the vector with a $$1$$ in the $$i$$ ith coordinate and $$0$$'s elsewhere. For example, in $$\\mathbb{R}^5$$, $$e_3 = (0, 0, 1, 0, 0)$$\n\n\u2022 This is a good answer! Thanks @Fisher. Mar 19 '20 at 0:17\n\nFound some relevant threads to your question. Hope this helps.\n\n\"Dummy variable\" versus \"indicator variable\" for nominal/categorical data\n\nWhat is \"one-hot\" encoding called in scientific literature?", "date": "2021-09-19 04:21:30", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/436975/compact-notation-for-one-hot-indicator-vectors", "openwebmath_score": 0.9154834151268005, "openwebmath_perplexity": 739.7600354197723, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639669551474, "lm_q2_score": 0.8887587905460026, "lm_q1q2_score": 0.8634860162091332}}
{"url": "https://math.stackexchange.com/questions/2625347/difference-between-minus-one-and-plus-one-induction", "text": "Difference between minus one and plus one induction?\n\nI recently started a Combinatorics class, in which my teacher (grad student) has instructed us to Prove by induction that $$1^2+2^2+\\ldots+n^2 = \\frac{n(n+1)(2n+1)}{6} = \\frac{2n^3+3n^2+n}{6}$$ this is trivial in the fact that it has been solved many times before, however my professor has insisted I solve it by using $P(n-1)$ as opposed to $P(n+1)$, which I've done below.\n\nBasis\n\n$$\\frac{1(1+1)(2*1+1)}{6} = 1$$\n\nInductive Step $(n-1)$\n\n$$1^2+2^2+\\ldots + (n-1)^2 = \\frac{(n-1)(n)(2(n-1)+1)}{6}$$ Which Simplifies to $$\\frac{(n-1)(n)(2n-1)}{6} \\rightarrow \\frac{2n^3-3n^2+n}{6}$$ Add $6\\frac{n^2}{6}$ to both sides and we've proven by induction.\n\nMy question is do there exists any mathematical proofs for which solving by Induction with $n+1$ and $n-1$ are not interchangeable and should I petition my professor to be able to use them interchangeably. I am aware that solving using $n-1$ and $n+1$ is identical, at least for every scenario I've come across (we're working with positive integers so I'm not expecting any variance from that), however given the overwhelming amount of resources, I can't for the life of me figure out why I am being instructed to use a method opposite what seems to be the norm for any other reason besides my teacher's personal preference.\n\n\u2022 Are you asking about the difference between $P(n-1)\\Rightarrow P(n)$ and $P(n)\\Rightarrow P(n+1)$ in the inductive step? Those are identical. I am not quite sure what you mean by induction with $n+1$ vs $n-1$. Jan 28 '18 at 18:20\n\u2022 I know that they are the same as far as my example goes, I was just wondering if there was any reason why my teacher has instructed us using $n-1$ for anything other than his own personal preference, given that I couldn't find any resources online for \"Solving induction step with n minus one\" or similar. Such that is what I'm being taught actually a mirror of a theorem? I'll update my question for clarity.\n\u2013\u00a0jfh\nJan 28 '18 at 18:24\n\u2022 Perhaps you can put side-by-side your approach and your instructor's approach. Sometimes one direction is preferred over another due to simplicity or clarity. Jan 28 '18 at 18:28\n\u2022 Well, to be fussy neither are either \"with n+1\" or \"with n-1\". One is \"with n implying n+1\" and the other is \"with n-1 implying n\". Which if you replace $m$ with $n-1$ become \"with n implying n+1\" and the other \"with m implying m+1\" which are obviously the same. Jan 28 '18 at 19:44\n\u2022 You can do induction with n implying n-1 to prove for example if P(237) is true and P(n) implies P(n-1) for natural number then P(n) is true for all natural numbers less than or equal to 237. That's kind of ... useless. ... Unless you are proving something integers. For example we can prove $b^{n+m} = b^nb^m$ for all integers by induction by \"inducing down\" if we need to. Jan 28 '18 at 19:47\n\nAs Michael Burr noted in the comments, the two conventions are identical; it's just a change of name for the variable. You could equally assume $P(k+12)$ holds and prove $P(k+13)$ from that.\n\nThe advantage of using $P(n-1) \\implies P(n)$ is that your target formula is already expressed in terms of $n$, so you don't have to rewrite the target in terms of $n+1$ to figure out what you're looking for; the advantage of using $P(n) \\implies P(n+1)$ is that the inductive hypothesis is already expressed in terms of $n$.\n\nOne other advantage of using $P(n-1) \\implies P(n)$ is that it transfers better to \"strong induction,\" where you can assume $P(k)$ for all $k < n$ to prove $P(n)$. Here there is definitely less rewriting going on if you use $P(n)$ as opposed to $P(n+1)$ as your target.\n\nI've seen both used widely. I personally usually prefer $P(n-1) \\implies P(n)$.\n\n\u2022 Thank you, I know my question sounds like its teetering on the border between naivety and caution, but I think your answer does well to say that noticing the distinction between the terms of the right hand side can help identify which method to use.\n\u2013\u00a0jfh\nJan 28 '18 at 18:38\n\nI would argue that, if anything, there are reasons to prefer $P(n) \\Rightarrow P(n+1)$.\n\nNatural numbers can be defined in many ways, but the usual inductive definition is the following:\n\n1. $0$ is a natural number;\n2. If $n$ is a natural number, then $s(n)$ is a natural number.\n\nHere $s(n)$ denotes the successor of $n$.\n\nThese two rules define a set $\\mathbb N$ together with an induction principle (which allows us to prove properties of all elements of $\\mathbb N$ and is in fact the usual mathematical induction) and a recursion principle (which allows us to construct new objects from the elements of $\\mathbb N$).\n\nThen $\\mathbb N$ can be endowed with the usual operations satisfying all the well-known properties. In particular, it is customary to write the successor $s(n)$ of $n$ as the sum $n+1$, although it is the sum between two natural numbers that is actually defined by recursion using the successor.\n\nThere are of course many inductive structures other than the set of natural numbers. For example, binary trees are defined by:\n\n1. $v$ is a binary tree (a single vertex, which is also the root);\n2. If $t_1$ and $t_2$ are binary trees, then $t_1 \\bullet t_2$ is a binary tree (the graph formed by taking $t_1$ and $t_2$, adding a new vertex as a root and joining the roots of $t_1$ and $t_2$ to the new root).\n\nHow does the induction principle look like for binary trees? If you want to prove that $P(t)$ holds for any binary tree $t$, you have to prove:\n\n1. (Basis) $P(v)$ holds;\n2. (Inductive step) If $P(t_1)$ and $P(t_2)$ hold, then $P(t_1 \\bullet t_2)$ holds.\n\nIn this case there is no equivalent to the predecessor of a natural number.\n\nI think the only reason s/he did it this way is s/he wanted to have the final simplification end in the form $\\frac {2n^3 + 3n^2 + n}6$\n\nHad he done $P(n) \\implies P(n+1)$ it would have involve a lot of factoring to get in the form $\\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$.\n\nTry it:\n\n$1 + 2 + .... + n^2 = \\frac {2n^3 + 3n^2 + n}6$\n\nSo $1 + 2 + ...... + n^2 + (n+1)^2 = \\frac {2n^3 + 3n^2 + n}6 + n^2 + 2n + 1$\n\n$= \\frac {2n^3 + 9n^2 + 13n + 6}6$\n\n... and we have to somehow get that to .... $=\\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$\n\n... which isn't impossible ...\n\n$\\frac {2n^3 + 9n^2 + 13n + 6}6= \\frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 7n + 4}6$\n\n$=\\frac {2(n+1)^3 + 3n^2 + 6n + 3 + n + 1}6$\n\n$= \\frac {2(n+1)^3 + 3(n+1)^2 + (n+1)}6$\n\n... but.... why schlep? That factoring and working backwords isn't the point.\n\nIt's easier to follow it and to go to the conclusion if you you work toward simplifying.\n\n=====\n\nI suppose an easier compromise would be to do:\n\n$\\frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6} = \\frac {2n^3 + 6n^2 + 6n + 2 + 3n^2 + 6n + 3 + n+1}6 = \\frac {2n^3 + 9n^3 + 13n + 6}6$\n\nfirst.\n\nThen:\n\n$1+2 + ..... + n^2 + (n+1)^2 =$\n\n$\\frac {2n^3 + 3n^2 + n}6 + (n+1)^2 = \\frac {2n^3 + 3n^2 + n}6 +n^2 + 2n + 1=$\n\n$\\frac {2n^3 + 3n^2 + n + 6n^2 + 12n + 6}6 = \\frac {2n^3 + 9n^3 + 13n + 6}6$\n\n$= \\frac {2(n+1)^2 + 3(n+1)^2 + (n+ 1)}{6}$\n\n\u2022 Whoa! I just read your professors arithmetic. Whoa. That is is clean and pretty with the $(n^3 - bn^2 + n)+2bn^2 = (n^3 + bn^2 +n)$. This use of conjugates just.... I don't know.. it's elegant. And gives a potential geometric insight as to why this works. It doesn't put one method of induction over another but for this particular proof.... it sure is pretty. Jan 28 '18 at 20:08", "date": "2022-01-26 21:45:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2625347/difference-between-minus-one-and-plus-one-induction", "openwebmath_score": 0.768440842628479, "openwebmath_perplexity": 270.02048646593346, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.96036116089903, "lm_q2_score": 0.8991213813246445, "lm_q1q2_score": 0.863481253558075}}
{"url": "https://math.stackexchange.com/questions/3113091/a-poker-hand-contains-five-cards-find-the-probability-that-a-poker-hand-can-be", "text": "# A poker hand contains five cards. Find the probability that a poker hand can be....\n\na) Four of a kind (Contains four cards of equal face value)\n\nSo for this one, we want four cards that have the same face value, different suit. And the last card can be any remaining card.\n\nThere are 13 ranks, (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K). We have $$\\binom{13}{1}$$ ways to pick 1 rank out of 13. For each of these ranks, we want to pick 4 cards that are all the same rank (and it is forcefully implied that these 4 cards will differ in suits). So $$\\binom{13}{1}\\binom{4}{4}$$, and for each of these ways, we have $$\\binom{48}{1}$$ way to pick 1 card out of the remaining deck of 48 cards, because we picked 4 cards already. Thus, $$\\frac{\\binom{13}{1}\\binom{4}{4}\\binom{48}{1}}{\\binom{52}{5}}$$ is the total number of ways.\n\nb) Full House (Three cards of equal face value, and two others of equal face value). So i.e: 3, 3, 3, 2, 2 would be a full house where the three 3's and two 2's are distinguishable (different suit).\n\nSo there are 13 ranks again, $$\\binom{13}{1}$$ ways to pick 1 rank out of 13 total. For each of these ways, we want to pick three cards of the same rank. So $$\\binom{13}{1}\\binom{4}{3}$$. Now we want two more cards that are of equal face value that differ from the other three cards picked earlier, so there are 12 ranks left, and $$\\binom{12}{1}$$ ways to pick 1 rank out of 12 remaining. For each of these ways, we have $$\\binom{4}{2}$$ ways to pick 2 cards out of the 4 suits belonging to the same rank, thus $$\\binom{12}{1}\\binom{4}{2}$$. The total number of ways is: $$\\frac{\\binom{13}{1}\\binom{4}{3}\\binom{12}{1}\\binom{4}{2}}{\\binom{52}{5}}$$\n\nc) Three of a kind. (Three cards of equal face value, and two cards with face values that differ from each other and the other three).\n\n13 ranks, we want to pick 3 cards of the same rank. $$\\binom{13}{1}\\binom{4}{3}$$. Now we want two cards that have different face values from the three picked, and each other. Now here is where I'm a bit confused, the ranks MUST be different, but the suits can be the same. So proceeding, $$\\binom{12}{1} \\binom{4}{1}$$ to pick 1 card of a different face value, but could be same suit, and then $$\\binom{11}{1}\\binom{4}{1}$$ to pick another card with a different face value, but could be a different suit. Multiply these altogether and divide by the denominator, and we have our total ways.\n\nIs my work correct?\n\n\u2022 Very nearly. Your numerator is the number of total ways to achieve these combinations. Dividing by $52 \\choose 5$ gives you the probability of achieving those hands. Feb 14 '19 at 19:36\n\u2022 That's what I said.....\n\u2013\u00a0Stuy\nFeb 14 '19 at 19:39\n\u2022 Then you'll want to clarify the last line of sections (a) and (b), which both refer to the quotient as \"[t]he total number of ways\" rather than the probability. Feb 14 '19 at 20:00\n\nYou correctly calculated the probabilities of four of a kind and a full house, not the total number of ways. The number of ways these hands can be obtained is given by the numerators of your probabilities.\n\nWhat is the probability of three of a kind?\n\nAs you know, there are $$\\binom{52}{5}$$ possible five-card hands that can be drawn from a standard deck.\n\nHowever, your count of the favorable cases is not quite right. There are $$\\binom{13}{1}$$ ways to choose the rank from which three cards are drawn and $$\\binom{4}{3}$$ ways to select cards of that rank, as you found. The remaining two cards must come from different ranks. There are $$\\binom{12}{2}$$ ways to select two ranks from which one card is drawn and $$\\binom{4}{1}$$ ways to choose one card from each of the two ranks. Hence, the number of favorable cases is $$\\binom{13}{1}\\binom{4}{3}\\binom{12}{2}\\binom{4}{1}^2$$ Therefore, the probability of three of a kind is $$\\frac{\\dbinom{13}{1}\\dbinom{4}{3}\\dbinom{12}{2}\\dbinom{4}{1}^2}{\\dbinom{52}{5}}$$\n\nYou count each hand twice since the order in which the singletons are selected does not matter. For instance, if the hand is $$\\color{red}{5\\heartsuit}, \\color{red}{5\\diamondsuit}, 5\\spadesuit, 7\\clubsuit, \\color{red}{10\\diamondsuit}$$, you count it once when you select $$7\\clubsuit$$ as the card you are drawing from the remaining $$12$$ ranks in the deck and $$\\color{red}{10\\diamondsuit}$$ as the card you are drawing from the remaining $$11$$ ranks in the deck and once when you select $$\\color{red}{10\\diamondsuit}$$ as the card you are drawing from the remaining $$12$$ ranks in the deck and $$7\\clubsuit$$ as the card you are drawing from the remaining $$11$$ ranks in the deck. However, both selections result in the same hand. Therefore, you need to divide your answer by $$2$$. Notice that $$\\frac{1}{2}\\binom{13}{1}\\binom{4}{3}\\binom{12}{1}\\binom{4}{1}\\binom{11}{1}\\binom{11}{1}\\binom{4}{1} = \\binom{13}{1}\\binom{4}{3}\\binom{12}{2}\\binom{4}{1}^2$$\n\u2022 What matters is which cards are selected. We need to choose the rank from which three cards are drawn, three cards of that rank, the two ranks from which a single card is drawn, and one card from each of those ranks. In your attempt, you distinguished between the first single card you draw and the second one you draw by first selecting a card from one of the remaining $12$ ranks and then selecting a card from one of the remaining $11$ ranks. However, you still get the same hand if you select the cards in the opposite order, which means you have counted every hand twice. Feb 17 '19 at 0:07", "date": "2021-10-27 14:21:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3113091/a-poker-hand-contains-five-cards-find-the-probability-that-a-poker-hand-can-be", "openwebmath_score": 0.6982788443565369, "openwebmath_perplexity": 143.3931126370973, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303413461358, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8634759127899766}}
{"url": "http://mathhelpforum.com/algebra/4627-formula-1-1-2-1-2-3-1-2-3-4-a.html", "text": "1. ## Formula for 1+(1+2)+(1+2+3)+(1+2+3+4)+...\n\nCan someone show me the formula for calculating 1+(1+2)+(1+2+3)+(1+2+3+4)+...\n\n2. Originally Posted by OReilly\nCan someone show me the formula for calculating 1+(1+2)+(1+2+3)+(1+2+3+4)+...\nYes, I presume you have a finite sum.\n---\nThe key in the the fact that,\n$\\displaystyle 1+2+...+x=\\frac{x(x+1)}{2}$\n---\nTherefore, that entire sum is,\n$\\displaystyle \\frac{1(2)}{2}+\\frac{2(3)}{2}+...+\\frac{n(n+1)}{2}$\nThus,\n$\\displaystyle \\sum_{k=1}^{n}\\frac{k(k+1)}{2}=\\frac{1}{2}\\sum_{k= 1}^n k^2+k$\nThus, subdivide the summation,\n$\\displaystyle \\frac{1}{2}\\sum_{k=1}^n k^2+\\frac{1}{2}\\sum_{k=1}^n k$\nUse the fact above, and the sum of squares formula,\n\n$\\displaystyle \\frac{1}{2}\\cdot \\frac{n(n+1)}{2}+\\frac{1}{2}\\cdot \\frac{n(n+1)(2n+1)}{6}$\nSimplify,\n$\\displaystyle \\frac{n(n+1)}{4}+\\frac{n(n+1)(2n+1)}{12}$\n$\\displaystyle \\frac{3n(n+1)+n(n+1)(2n+1)}{12}$\nThus, (factor),\n$\\displaystyle \\frac{n(n+1)(3+2n+1)}{12}$\nSimplify again,\n$\\displaystyle \\frac{n(n+1)(2n+4)}{12}$\nSimplify again,\n$\\displaystyle \\frac{n(n+1)(n+2)}{6}$\n~~~\nJust for refernce, the sum of squares formula is,\n$\\displaystyle \\sum_{k=1}^nk^2=\\frac{n(n+1)(2n+1)}{6}$\n----\nI think it is cool that,\n$\\displaystyle \\left( \\begin{array}{c}n+2\\\\3 \\end{array} \\right)=\\frac{(n+2)(n+1)n}{3!}$\n\nThat means you can calculate the sum by taking 2 more than number of terms and finding the number of combinations of forming three.\n\n3. Thanks!\n\nI didn't know sum of squares formula.\n\n4. Hello,OReilly!\n\nTPHacker is absolutely correct . . . Nice job, T.P. !\n\nIf we are desperate, we could find the formula \"from scratch\".\n\nCan someone show me the formula for calculating:\n. . . $\\displaystyle 1+(1+2)+(1+2+3)+(1+2+3+4)+\\hdots$\n\nCrank out a list of the first few sums:\n\n$\\displaystyle \\begin{array}{ccccccccc}S_1\\:=\\:1 \\\\ S_2\\:=\\:1 + 3\\:=\\:4\\\\S_3\\:=\\:1 + 3 + 6\\:=\\:10\\\\ S_4\\:=\\:1 + 3+6+10 \\:=\\:20\\\\ S_5\\:=\\:1+3+6+10+15\\:=\\:35 \\\\S_6\\:=\\:1+3+6+10+15+21\\:=\\:56\\\\ S_7\\:=\\:1+3+6+10+15+21+28 \\;=\\;84\\\\ \\vdots\\end{array}$\n\nWe have a function $\\displaystyle f(n)$. .For $\\displaystyle n = 1,2,3,\\hdots$\n\n. . the function has consecutive values: .$\\displaystyle 1\\quad4\\quad10\\quad20\\quad56\\quad84\\;\\hdots$\n\nTake differences of consecutive terms: . . $\\displaystyle 3\\quad6\\quad\\;10\\;\\quad15\\quad21\\;\\hdots$\n\nTake differences again: . . . . . . . . . . . . . .$\\displaystyle 3\\quad\\;4\\quad\\;\\;5\\quad\\;\\;6\\;\\hdots$\n\nTake differences again: . . . . . . . . . . . . . . . $\\displaystyle 1\\quad\\;1\\quad\\;\\:1\\;\\hdots$\n\nWe have a series of constants at the third differences.\n. . Hence, $\\displaystyle f(n)$ is a third-degree polynomial ... a cubic.\n\nHence, the function is of the form: .$\\displaystyle f(n)\\:=\\:an^3 + bn^2 + cn + d$\n. . and we must determine $\\displaystyle a,b,c,d.$\n\nUse the first four values from our list.\n\n$\\displaystyle S_1\\,=\\,1:\\;\\;a\\cdot1^3 + b\\cdot1^2 + c\\cdot1 + d\\:=$ $\\displaystyle \\:1\\quad\\;\\;\\;\\Rightarrow\\quad\\;\\; a+b+c+d\\;=\\;1$\n$\\displaystyle S_2\\,=\\,4:\\;\\;a\\cdot2^3 + b\\cdot2^2 + c\\cdot2 + d \\:= \\:$ $\\displaystyle 4\\;\\;\\;\\quad\\Rightarrow\\quad\\; 8a + 4b +2c + d \\;= \\;4$\n$\\displaystyle S_3\\,=\\,10:\\;a\\cdot3^3 + b\\cdot3^2 + c\\cdot3 + d \\:=$ $\\displaystyle \\:10\\quad\\Rightarrow\\quad 27a + 9b + 3c + d \\;= \\;10$\n$\\displaystyle S_4\\,=\\,20:\\;a\\cdot4^3 + b\\cdot4^2 + c\\cdot4 + d\\:=$ $\\displaystyle \\:20\\quad\\Rightarrow\\quad 64a + 16b + 4c + d \\;= \\;20$\n\nNow we must solve this system of equations . . . but it's easy!\n\n. . $\\displaystyle \\begin{array}{cccc}a+b+c+d\\:=\\:1 \\\\ 8a + 4b + 2c + d \\:=\\:4 \\\\ 27a + 9b + 3c + d \\:= \\:10\\\\ 64a + 16b + 4c + d \\:=\\:20\\end{array}\\;\\begin{array}{cccc}(1)\\\\(2)\\\\( 3)\\\\(4)\\end{array}$\n\nSubtract (1) from (2): .$\\displaystyle 7a + 2b + c \\:= \\:3\\quad\\;\\;\\;(5)$\nSubtract (2) from (3): .$\\displaystyle 19a + 5b + c \\:= \\:6\\quad\\;(6)$\nSubtract (3) from (4): .$\\displaystyle 376a + 7b + c \\:= \\:10\\;\\;(7)$\n\nSubtract (5) from (6): .$\\displaystyle 12a + 2b\\:=\\:3\\;\\;(8)$\nSubtract (6) from (7): .$\\displaystyle 18a + 2b\\:=\\:4\\;\\;(9)$\n\nSubtract (8) from (9): .$\\displaystyle 6a\\,=\\,1\\quad\\Rightarrow\\quad \\boxed{a = \\frac{1}{6}}$\n\nSubstitute into (8): .$\\displaystyle 12\\left(\\frac{1}{6}\\right) + 2b\\:=\\:3\\quad\\Rightarrow\\quad \\boxed{b = \\frac{1}{2}}$\n\nSubstitute into (5): .$\\displaystyle 7\\left(\\frac{1}{6}\\right) + 3\\left(\\frac{1}{2}\\right) + c\\:=\\:3\\quad\\Rightarrow\\quad \\boxed{c = \\frac{1}{3}}$\n\nSubstitute into (1): .$\\displaystyle \\frac{1}{6} + \\frac{1}{2} + \\frac{1}{3} + d\\:=\\:1\\quad\\Rightarrow\\quad \\boxed{d = 0}$\n\nTherefore: .$\\displaystyle f(n)\\:=\\:\\frac{1}{6}n^3 + \\frac{1}{2}n^2 + \\frac{1}{3}n\\quad\\Rightarrow\\quad \\boxed{f(n)\\;=\\;\\frac{n(n+1)(n+2)}{6}}$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nThis \"scratch\"method can be used\n. . when we suspect we have a polynomial function.\nIt is long and tedious, but it works.\n\nOf course, it is more efficient to learn a few sums-of-powers formula:\n\n. . $\\displaystyle \\sum_{k=1}^n k \\;= \\;\\frac{n(n+1)}{2}$\n\n. . $\\displaystyle \\sum_{k=1}^n n^2\\;=\\;\\frac{n(n+1)(2n+1)}{6}$\n\n. . $\\displaystyle \\sum_{k=1}^n k^3\\;= \\;\\frac{n^2(n+1)^2}{4}$\n\nNote that the third formula is the square of the first formula,\n. . making it easier to memorize.\n\n. . This means: .$\\displaystyle (1 + 2 + 3 + 4 + 5)^2\\:=\\:1^3+2^3+3^3+4^3+5^3$\n\nIt looks like a bad joke or a terrible blunder, doesn't it?\n\n5. Originally Posted by Soroban\nWe have a series of constants at the third differences.\n. . Hence, $\\displaystyle f(n)$ is a third-degree polynomial ... a cubic.\nTell me something, is this a famous method? This is the second time I seen it used on this forum. When I was younger I developed many theorems concerning differences of a sequence. One of them you just used. (A sequence is a polynomial of degree n if and only if it takes n steps to reach a constanct sequence).\n---\nI have a elegant prove with this rule that,\n$\\displaystyle a^{(m)}_k=\\sum_{k=1}^nk^m$ is always a polynomial sequence.\n\nConsider an infinite sequence,\n$\\displaystyle 0^m,1^m+0^m,2^m+1^m+0^m,...$\nIts diffrence sequence is,\n$\\displaystyle 0^m,1^m,2^m,3^m,...$\nA polynomial sequence of degree $\\displaystyle m$ therefore $\\displaystyle m$ subtractions are required. In total we used $\\displaystyle m+1$ subtractions on our original sequence. Thus there exists a polynomial sequence of degree $\\displaystyle m+1$. Furthermore, the first coefficient is $\\displaystyle \\frac{1}{(m+1)!}$. Based on the fact the constant sequence follows the factorial.\n\n6. Originally Posted by ThePerfectHacker\nTell me something, is this a famous method?\nYes, I seem to recall that it is associated with Newton's name, but that\nbe a manifestation of false memory syndrome, but it is in Acton's\n\"Numerical Methods that Work\", which I swear by\n\nRonL\n\n7. Hello, TPHacker!\n\nTell me something, is this a famous method?\n\nI'm sure it is . . .\n\nI ran across many years ago in some book. .Since then, I've seen more efficient methods,\nbut I like that very primitive method. .(But I don't use it unless I am forced to.)\n\nIn case anyone is interested . . .\nI was shown this method in graduate school . . . quite an eye-opener.\n\nTo find, for example, $\\displaystyle \\sum^n_{k=1} k^4$, we are expected to know the three \"preceding\" formulas:\n. . $\\displaystyle \\sum^n_{k=1} k \\:=\\:\\frac{n(n+1)}{2}\\qquad\\sum^n_{k=1} k^2 \\:=\\:\\frac{n(n+1)(2n+1)}{6}\\qquad\\sum^n_{k=1}$$\\displaystyle k^3\\:=\\:\\frac{n^2(n+1)^2}{4}$\n\nConsider the next-higher power and form: .$\\displaystyle k^5 - (k-1)^5$\n\nWe have: /$\\displaystyle k^5 - (k-1)^5\\:=\\;5k^4 - 10k^3 + 10k^2 - 5k + 1$\n\nNow let $\\displaystyle k \\,= \\,n,\\,n\\!-\\!1,\\,n\\!-\\!2,\\,...\\,,\\,3,\\,2,\\,1$ and \"stack\" the equations.\n\n. . . . $\\displaystyle n^5 - (n-1)^5\\;=\\quad\\;\\;5n^4\\quad -\\quad\\;\\; 10n^3\\quad +$ . .$\\displaystyle 10n^2\\quad\\; -\\quad\\; 5n\\quad\\; + 1$\n$\\displaystyle (n-1)^5-(n-2)^5 \\;=$ $\\displaystyle \\:5(n-1)^4 - 10(n-1)^3 + 10(n-1)^2 - 5(n-1) + 1$\n$\\displaystyle (n-2)^5-(n-3)^5 \\;=$ $\\displaystyle \\:5(n-2)^4 - 10(n-2)^3 + 10(n-2)^2 - 5(n-2) + 1$\n. . . . . . $\\displaystyle \\vdots$ . . . . . . . . . . .$\\displaystyle \\vdots$ . . . . . . . .$\\displaystyle \\vdots$ . . . . . . . $\\displaystyle \\vdots$ . . . . . . . $\\displaystyle \\vdots$ . . . $\\displaystyle \\vdots$\n. . $\\displaystyle 3^5\\quad -\\quad 2^5 \\qquad= \\quad\\;\\;5(3^4)\\;\\;\\; -\\;\\;\\; 10(3^3)$ . $\\displaystyle +\\;\\;\\; 10(3^2)\\quad -\\quad 5(3)\\quad + 1$\n. . $\\displaystyle 2^5\\quad -\\quad 1^5\\qquad = \\quad\\;\\;5(2^4)\\;\\;\\; -\\;\\;\\; 10(2^3)$ . $\\displaystyle +\\;\\;\\; 10(2^2)\\quad -\\quad 5(2)\\quad + 1$\n. . $\\displaystyle 1^5\\quad -\\quad 0^5 \\qquad = \\quad\\;\\;5(1^4)\\;\\;\\; -\\;\\;\\; 10(1^3)$ . $\\displaystyle + \\;\\;\\;10(1^2)\\quad -\\quad 5(1)\\quad + 1$\n\nAdd the stack (most of the left side cancels out):\n\n. . $\\displaystyle n^5\\;=\\;5\\left(\\sum k^4\\right) - 10\\left(\\sum k^3\\right) +$ $\\displaystyle 10\\left(\\sum k^2\\right) - 5\\left(\\sum k\\right) + \\left(\\sum 1\\right)$\n\nWe have: .$\\displaystyle n^5 \\;= \\;5\\left(\\sum k^4\\right)-10\\cdot\\frac{n^2(n+1)^2}{4} +$ $\\displaystyle 10\\cdot\\frac{n(n+1)(2n+1)}{6} - 5\\cdot\\frac{n(n+1)}{2} + n$\n\nThen, after an enormous amount of algebra, we get:\n\n. . . . $\\displaystyle \\boxed{\\sum^n_{k=1} k^4\\;=\\;\\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}}$", "date": "2018-04-26 11:49:50", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/4627-formula-1-1-2-1-2-3-1-2-3-4-a.html", "openwebmath_score": 0.9577927589416504, "openwebmath_perplexity": 2342.3804025705745, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985936372130286, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8634702604917618}}
{"url": "https://math.stackexchange.com/questions/709483/pointwise-converging-subsequence-on-countable-set", "text": "# Pointwise converging subsequence on countable set\n\nLet $A \\subseteq \\mathbb R$ be countable and let $f_n: A\\to \\mathbb R$ be a sequence of functions such that there exists $M \\ge 0$ with $\\,\\lvert\\,f_n(x)\\rvert\\le M$ for all $n$. I am trying to show that there exists a subsequence $f_{n_k}$ of $f_n$ that converges pointwise.\n\nHere is what I have so far: If $A = \\{a_1, a_2, \\dots \\}$ then $f_n(a_1)$ is a bounded sequence hence by Bolzano Weierstrass theorem contains a convergent subsequence $f_{n_{k_1}}$. By the same argument $f_{n_{k_1}}(a_2)$ contains a convergent subsequence $f_{n_{k_2}}$.\n\nNext I want to define $$f_{n_k} (x) = \\lim_{j \\to \\infty} f_{n_{k_j}}(x),$$ the pointwise limit. Then $f_{n_k}(a_j)$ converges for every $a_j \\in A$ (it's clear by how it was defined).\n\nAm I done now or am I missing something? Is there anything left to show?\n\nLet $A=\\{a_n:n\\in\\mathbb N\\}$. Using Bolzano-Weierstrass for the bounded sequence $\\{f_n(a_1)\\}_{n\\in\\mathbb N}$ we can find a convergent subsequence which we denote as $\\{f_{1,n}(a_1)\\}_{n\\in\\mathbb N}$. Next, as $\\{f_{1,n}(a_2)\\}_{n\\in\\mathbb N}$ is bounded, it also contains a convergent subsequence which we denote as $\\{f_{2,n}(a_2)\\}_{n\\in\\mathbb N}$.\nIn this way we construct recursively the following convergent sequences: \\begin{align} f_{1,1}(a_1),&f_{1,2}(a_1),\\ldots f_{1,n}(a_1),\\ldots,\\\\ f_{2,1}(a_2),&f_{2,2}(a_2),\\ldots f_{2,n}(a_2,\\ldots,\\\\ \\vdots&\\\\ f_{n,1}(a_n),&f_{n,2}(a_n),\\ldots f_{n,n}(a_n),\\ldots,\\\\ \\vdots& \\end{align} with $\\{f_{k,n}\\}_{n\\in\\mathbb N}$ a subsequence of all the sequences $\\{f_{j,n}\\}_{n\\in\\mathbb N}$, for $j<k$, and say that $\\lim_{n\\to\\infty}f_{j,n}(a_j)=f(a_j)$.\nThe sequence $\\{f_{n,n}\\}_{n\\in\\mathbb N}$ is finally a subsequence of all the above, and hence $\\{f_{n,n}(a_j)\\}_{n\\in\\mathbb N}$ converges for all $j\\in\\mathbb N$, and in particularly $$\\lim_{n\\to\\infty}f_{n,n}(a_j)=f(a_j),$$ for all $j\\in\\mathbb N$.\n\u2022 Nice answer. But I don't understand why $f_{k,n}$ is subsequence of $f_{j,n}$ please can you illustrate? \u2013\u00a0MathLover Nov 17 '18 at 11:46", "date": "2019-09-15 15:58:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/709483/pointwise-converging-subsequence-on-countable-set", "openwebmath_score": 0.9963683485984802, "openwebmath_perplexity": 88.43574137365928, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363713038174, "lm_q2_score": 0.8757869981319862, "lm_q1q2_score": 0.8634702549733136}}
{"url": "http://signalsandsystems.wikidot.com/problems", "text": "Problems\n\n Compute the magnitude and phase of $(1-j) \\bigg(\\frac{1}{2} + j \\frac{\\sqrt{3}}{2}\\bigg)$ Compute the magnitude and phase of $e^{j\\pi/2} (1+j) (1+3j)$ Compute the magnitude and phase of $j e^{j \\pi/3}$ Compute the magnitude and phase of $e^{j \\pi/4} + e^{j 3 \\pi/4}$ Compute the magnitude and phase of $(1+3j)^2$ Compute the magnitude and phase of $(1-3j)/(1+3j)^2$ Compute the magnitude and phase of $e^{j \\pi/5} \\times e^{j 2 \\pi/5} \\times e^{j 3 \\pi/5} \\ldots e^{j 9 \\pi/5}$ Compute the magnitude and phase of $e^{j \\pi/5} \\times e^{j 2 \\pi/5} \\times e^{j 3 \\pi/5} \\ldots e^{j 9 \\pi/5} \\times e^{j 10 \\pi/5}$ Let $z_1 = 1, z_2 = -\\frac{1}{2} + j \\frac{\\sqrt{3}}{2}, z_3 = -\\frac{1}{2} - j \\frac{\\sqrt{3}}{2}$ [a)] What are $z_1^3, z_2^3$ and $z_3^3$? [b)] Show that $z_3 = z_2^2$ [c)] Show that $z_1 + z_2 + z_3 = 0$ Can you now see why $z_1,z_2,z_3$ can be called the cube roots of unity. They are usually expressed as $1, \\omega, \\omega^2$. Part c shows that the sum of the cube roots of unity is zero. In one of the homework problems, we will show that this true for $n$th roots of unity for any $n$. Let $z_1 = 2 e^{j\\pi/4}$ and $z_2 = 8 e^{j\\pi/3}$. Find and express your answer in Cartesian and polar form [a)] $2z_1-z_2$ [b)] $\\frac{1}{z_1}$ [c)] $\\frac{z_1}{z_2^2}$ [d)] $\\sqrt[3]{z_2}$ Prove that $\\int e^{ax} \\ \\cos(bx) \\ dx = \\frac{e^{ax}}{a^2+b^2} \\left(a \\cos(bx) + b \\sin(bx) \\right)$ You can use integration tricks you learned in your calculus class to solve this problem. That is not the point of the exercise. Try using Euler's formula and then using integration of exponentials to see if you can solve the problem.\n\n OWN 1.25d Is the even part of $x(t) = \\cos \\left( 4 \\pi t\\right) \\ u(t)$ periodic? If it is, what is the time period? OWN 1.25e Is the even part of $x(t) = \\sin \\left( 4 \\pi t\\right) \\ u(t)$ periodic? If it is, what is the time period? OWN 1.26b Is the discrete-time signal $x[n] = \\cos \\left(\\frac{n}{8}\\right)$ periodic? If it is, what is the time period? OWN 1.26c Is the discrete-time signal $x[n] = \\cos \\left(\\frac{\\pi n^2}{8}\\right)$ periodic? If it is, what is the time period?\n\n OWN 1.27a Is the system $y(t) = x(t-2)+x(2-t)$ memoryless, linear, stable, time-invariant and causal? OWN 1.27c Is the system $y(t) = \\int_{-\\infty}^{2t} x(\\tau) \\ d \\tau$ memoryless, linear, stable, time-invariant and causal? OWN 1.27d Is the system $y(t) = \\begin{cases} 0, \\ \\ t < 0 \\\\ x(t)+x(t-2), t \\geq 0 \\end{cases}$ memoryless, linear, stable, time-invariant and causal? OWN 1.27g Is the system $y(t) = \\frac{d}{dt}x(t)$ memoryless, linear, stable, time-invariant and causal? Consider the system defined by $y(t) = x(t) \\ u(t)$. Is this linear, time invariant, stable, invertible?\n\n OWN 2.4 Compute the convolution of (1) \\begin{align} x[n] = \\left\\{ \\begin{array}{ll} 1, & 3 \\leq n \\leq 8 \\\\ 0, & \\hbox{otherwise.} \\end{array} \\right. \\ \\hbox{and} \\ h[n] = \\left\\{ \\begin{array}{ll} 1, & 4 \\leq n \\leq 15 \\\\ 0, & \\hbox{otherwise.} \\end{array} \\right. \\end{align} OWN 2.6 Compute the convolution of $x[n] = \\left(\\frac{1}{3} \\right)^{-n} \\ u[-n-1]$ and $h[n] = u[-n-1]$\n\nBandpassfromoldexam\nDTFS\nPhasefunction\nTwo systems\nProblem5fromoldexam - There is a small mistake in the solutions to the last part. The correct answer is 450Hz,500Hz,550\u00a0Hz\n\nReview done on Sunday, December 10, 2017 before the final exam\n\n Compute Discrete-time Fourier transform of $x[n] = \\sin \\left( \\frac{\\pi n/5}{\\pi n}\\right) \\ \\cos \\left(\\frac{5 \\pi n}{2} \\right)$\npage revision: 43, last edited: 21 Aug 2021 10:45", "date": "2022-09-28 11:42:02", "meta": {"domain": "wikidot.com", "url": "http://signalsandsystems.wikidot.com/problems", "openwebmath_score": 0.9935851693153381, "openwebmath_perplexity": 319.16731847569105, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363746096915, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8634702402882154}}
{"url": "https://mathoverflow.net/questions/422075/explicit-expression-for-recursive-sums", "text": "# Explicit expression for recursive sums\n\nLet $$t_1,t_2,\\dots,t_k$$ be non-negative integers. Can the following sum $$f_k(t_1,t_2,\\dots,t_k):=\\sum_{j_1=0}^{t_1} \\sum_{j_2=0}^{t_2+j_1} \\sum_{j_3=0}^{t_2+j_2} \\dots \\sum_{j_k=0}^{t_k+j_{k-1}} 1$$ be explicitly expressed as a polynomial in $$t_1,t_2,\\dots,t_k$$ or via known combinatorial entities?\n\nWe surely have a recurrence formula: $$f_{k+1}(t_1,t_2,\\dots,t_{k+1}) = \\sum_{j=0}^{t_1} f_k(j+t_2,\\dots,t_{k+1}),$$ which does not seem to easily unroll.\n\nJust in case, first few terms are $$\\begin{split} f_0 &= 1,\\\\ f_1(t_1) &= 1+t_1,\\\\ f_2(t_1,t_2) &= (1+t_1)(1+t_2) + \\frac{t_1(1+t_1)}2,\\\\ f_3(t_1,t_2,t_3) &= \\left[ (1+t_1)(1+t_2) + \\frac{t_1(1+t_1)}2 \\right](1+t_3) + \\frac{t_2(1+t_2)}2 + \\frac{3t_2^2 + 6t_2 + 2}6t_1 + \\frac{1+t_2}2t_1^2 + \\frac16t_1^3. \\end{split}$$\n\nUPDATED. Billy Joe found that $$f_k(n,d,d,\\dots, d) = \\frac{n+1}k \\binom{n+k(d+1)}{k-1}.$$ In particular, at $$f_k(1,1,\\dots,1)$$ gives $$(k+1)$$-st Catalan number.\n\n\u2022 The equal arguments case looks like it should also give the Ehrhart polynomial of some lattice polytope. May 9 at 2:20\n\u2022 FWIW it seems that $f_k(1,2,3,\\ldots,k-2,k-2,k)=a(k+1)$, where $a(k)$ is A107877. May 10 at 17:32\n\u2022 @BillyJoe: Nice catch! In fact, it is stated there as conjecture by Benedict W. J. Irwin. May 10 at 17:49\n\u2022 Are you sure about A016121? Apparently $f_k(n,d,d,\\ldots,d,d)=\\frac{n+1}{k}\\binom{n+k(d+1)}{k-1}$. May 10 at 21:08\n\u2022 Irwin's conjecture on A107877 is equivalent to the earlier comment by Joerg Arndt, Apr 30 2011, which doesn't qualify it as a conjecture. May 11 at 7:19\n\nClaim: The iterated sum $$f_k(t_1,\\ldots,t_k)$$ counts the number of elements the interval $$[\\emptyset,\\lambda]$$ of Young's lattice, where $$\\lambda = (\\lambda_1,\\lambda_2,\\ldots,\\lambda_k)$$ is the partition determined by $$\\lambda_{k-i+1} = t_1 + \\cdots + t_i$$. Equivalently, the function $$f_k$$ counts the number of subdiagrams of $$\\lambda$$.\nFor an arbitrary partition $$\\lambda$$, we have $$|[\\emptyset,\\lambda]| = \\text{det} \\left[\\binom{\\lambda_i + 1}{i-j+1}\\right]_{1 \\leq i,j \\leq k}$$ which is a result due to P. A. MacMahon. The answer to Exercise 149 in Chapter 3 of Stanley's Enumerative Combinatorics, volume 1, 2nd edition provides a good reference of references for this result, with various extensions and specializations, including some of the results mentioned in the comments. For a short visual proof using Lindstr\u00f6m-Gessel-Viennot, see Ciucu - A short conceptual proof of Narayana's path-counting formula.\nIf the claim is true, MacMahon's result implies $$\\sum_{j_1=0}^{t_1}\\sum_{j_2=0}^{t_2+j_1}\\cdots\\sum_{j_k=0}^{t_k+j_{k-1}} = \\text{det} \\left[\\binom{t_1 + \\cdots + t_{k - i + 1} + 1}{i-j+1}\\right]_{1 \\leq i,j \\leq k}$$ which implies $$f_k(t_1,\\ldots,t_k)$$ is a polynomial in $$t_1,\\ldots,t_k$$.\nNote that $$f_k(t_1,\\ldots,t_k)$$ counts the number of $$(j_1,\\ldots,j_k)$$ such that $$0 \\leq j_1 \\leq t_1$$ and $$0 \\leq j_{i+1} \\leq j_i + t_{i+1}$$ for $$i \\geq 1$$. To establish the claim, it suffices to find a bijection between the set of $$\\mu \\subseteq \\lambda$$ and the set of tuples satisfying the above constraints.\nSketch: Map $$\\mu \\subseteq \\lambda$$ to $$(j_1,\\ldots,j_k)$$, where $$j_i = \\lambda_{k-i+1} - \\mu_{k-i+1}$$. The visual interpretation is that each $$j_i$$ measures the distance between the walls of the $$i$$-th row from the bottom of the Young diagrams (English convention) for $$\\mu$$ and $$\\lambda$$. The $$t_i$$ specify how many boxes are added to the diagram for $$\\lambda$$ in moving from the $$(i-1)$$-st row from the bottom to the $$i$$-th row. The constraints express the fact that in going from bottom to top in the diagram, the distance between walls increases by at most $$t_i$$. For a more direct definition chase, note that $$\\lambda_{k-i} - \\lambda_{k-i+1} = t_{i+1}$$. Since $$\\mu$$ is a partition, we have $$\\mu_{k-i+1} - \\mu_{k-i} \\leq 0$$. Combining the definitions and inequalities gives $$j_{i+1} \\leq j_i + t_{i+1}$$.\n$$f_k(t_1,\\dots,t_k)$$ is counting the number of integer points in the \"Pitman-Stanley polytope\" $$\\Pi_k(t_1,\\dots,t_k)$$ defined here. The notation $$N(\\Pi_k(\\mathbf{t}))$$ is used in this paper, which has the determinantal formula given by Hugh Denoncourt, as well as a combinatorial formula. The combinatorial formula is equivalent to $$f_k(t_1,\\dots,t_k)=\\sum_{\\mathbf{h}\\in K_k} \\binom{t_1+h_1}{h_1} \\prod_{i=2}^k \\binom{t_i+h_i-1}{h_i},$$ where $$K_k := \\{\\mathbf{h}\\in\\mathbb{N}^k\\colon \\sum_{i=1}^j h_i\\geq j\\ \\mathrm{for\\ all}\\ 1\\leq j\\leq k-1\\ \\mathrm{and}\\ \\sum_{i=1}^k h_i=k \\}.$$ The set $$K_k$$ has a Catalan number $$C_k$$ of elements.", "date": "2022-07-03 02:02:36", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/422075/explicit-expression-for-recursive-sums", "openwebmath_score": 0.896655797958374, "openwebmath_perplexity": 234.1331131514942, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120886461792, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8634689051396628}}
{"url": "https://www.quizover.com/calculus/section/the-mean-value-theorem-and-its-meaning-by-openstax", "text": "# 4.4 The mean value theorem \u00a0(Page 2/7)\n\n Page 2 / 7\n\nVerify that the function $f\\left(x\\right)=2{x}^{2}-8x+6$ defined over the interval $\\left[1,3\\right]$ satisfies the conditions of Rolle\u2019s theorem. Find all points $c$ guaranteed by Rolle\u2019s theorem.\n\n$c=2$\n\n## The mean value theorem and its meaning\n\nRolle\u2019s theorem is a special case of the Mean Value Theorem. In Rolle\u2019s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle\u2019s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle\u2019s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $\\left[a,b\\right]$ and differentiable over the open interval $\\left(a,b\\right),$ then there exists a point $c\\in \\left(a,b\\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\\left(a,f\\left(a\\right)\\right)$ and $\\left(b,f\\left(b\\right)\\right).$\n\n## Mean value theorem\n\nLet $f$ be continuous over the closed interval $\\left[a,b\\right]$ and differentiable over the open interval $\\left(a,b\\right).$ Then, there exists at least one point $c\\in \\left(a,b\\right)$ such that\n\n$f\\prime \\left(c\\right)=\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}.$\n\n## Proof\n\nThe proof follows from Rolle\u2019s theorem by introducing an appropriate function that satisfies the criteria of Rolle\u2019s theorem. Consider the line connecting $\\left(a,f\\left(a\\right)\\right)$ and $\\left(b,f\\left(b\\right)\\right).$ Since the slope of that line is\n\n$\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}$\n\nand the line passes through the point $\\left(a,f\\left(a\\right)\\right),$ the equation of that line can be written as\n\n$y=\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}\\left(x-a\\right)+f\\left(a\\right).$\n\nLet $g\\left(x\\right)$ denote the vertical difference between the point $\\left(x,f\\left(x\\right)\\right)$ and the point $\\left(x,y\\right)$ on that line. Therefore,\n\n$g\\left(x\\right)=f\\left(x\\right)-\\left[\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}\\left(x-a\\right)+f\\left(a\\right)\\right]\\text{.}$\n\nSince the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\\left(a\\right)=0=g\\left(b\\right).$ Since $f$ is a differentiable function over $\\left(a,b\\right),$ $g$ is also a differentiable function over $\\left(a,b\\right).$ Furthermore, since $f$ is continuous over $\\left[a,b\\right],$ $g$ is also continuous over $\\left[a,b\\right].$ Therefore, $g$ satisfies the criteria of Rolle\u2019s theorem. Consequently, there exists a point $c\\in \\left(a,b\\right)$ such that $g\\prime \\left(c\\right)=0.$ Since\n\n$g\\prime \\left(x\\right)=f\\prime \\left(x\\right)-\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a},$\n\nwe see that\n\n$g\\prime \\left(c\\right)=f\\prime \\left(c\\right)-\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}.$\n\nSince $g\\prime \\left(c\\right)=0,$ we conclude that\n\n$f\\prime \\left(c\\right)=\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}.$\n\nIn the next example, we show how the Mean Value Theorem can be applied to the function $f\\left(x\\right)=\\sqrt{x}$ over the interval $\\left[0,9\\right].$ The method is the same for other functions, although sometimes with more interesting consequences.\n\n## Verifying that the mean value theorem applies\n\nFor $f\\left(x\\right)=\\sqrt{x}$ over the interval $\\left[0,9\\right],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\\in \\left(0,9\\right)$ such that ${f}^{\\prime }\\left(c\\right)$ is equal to the slope of the line connecting $\\left(0,f\\left(0\\right)\\right)$ and $\\left(9,f\\left(9\\right)\\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.\n\nWe know that $f\\left(x\\right)=\\sqrt{x}$ is continuous over $\\left[0,9\\right]$ and differentiable over $\\left(0,9\\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\\in \\left(0,9\\right)$ such that ${f}^{\\prime }\\left(c\\right)$ is equal to the slope of the line connecting $\\left(0,f\\left(0\\right)\\right)$ and $\\left(9,f\\left(9\\right)\\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative ${f}^{\\prime }\\left(x\\right)=\\frac{1}{\\left(2\\sqrt{x}\\right)}.$ The slope of the line connecting $\\left(0,f\\left(0\\right)\\right)$ and $\\left(9,f\\left(9\\right)\\right)$ is given by\n\n$\\frac{f\\left(9\\right)-f\\left(0\\right)}{9-0}=\\frac{\\sqrt{9}-\\sqrt{0}}{9-0}=\\frac{3}{9}=\\frac{1}{3}.$\n\nWe want to find $c$ such that ${f}^{\\prime }\\left(c\\right)=\\frac{1}{3}.$ That is, we want to find $c$ such that\n\n$\\frac{1}{2\\sqrt{c}}=\\frac{1}{3}.$\n\nSolving this equation for $c,$ we obtain $c=\\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.\n\nquestions solve y=sin x\nSolve it for what?\nTim\nyou have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y\nIoana\nwhat is the question ? what is the answer?\nSuman\nthere is an equation that should be solve for x\nIoana\nok solve it\nSuman\nare you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x\nTyron\nor solve for sin(x) via the unit circle\nTyron\nwhat is unit circle\nSuman\na circle whose radius is 1.\nDarnell\nthe unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.\nTyron\nwhat is function?\nA set of points in which every x value (domain) corresponds to exactly one y value (range)\nTim\nwhat is lim (x,y)~(0,0) (x/y)\nlimited of x,y at 0,0 is nt defined\nAlswell\nBut using L'Hopitals rule is x=1 is defined\nAlswell\nCould U explain better boss?\nemmanuel\nvalue of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)\nNIKI\ncan we apply l hospitals rule for function of two variables\nNIKI\nwhy n does not equal -1\nAndrew\nI agree with Andrew\nBg\nf (x) = a is a function. It's a constant function.\nproof the formula integration of udv=uv-integration of vdu.?\nFind derivative (2x^3+6xy-4y^2)^2\nno x=2 is not a function, as there is nothing that's changing.\nare you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.\nThe\ni mean can we replace the roles of x and y and call x=2 as function\nThe\nif x =y and x = 800 what is y\ny=800\n800\nBg\nhow do u factor the numerator?\nNonsense, you factor numbers\nAntonio\nYou can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2\nThe\nThe problem is the question, is not a problem where it is, but what it is\nAntonio\nI think you should first know the basics man: PS\nVishal\nYes, what factorization is\nAntonio\nAntonio bro is x=2 a function?\nThe\nYes, and no.... Its a function if for every x, y=2.... If not is a single value constant\nAntonio\nyou could define it as a constant function if you wanted where a function of \"y\" defines x f(y) = 2 no real use to doing that though\nzach\nWhy y, if domain its usually defined as x, bro, so you creates confusion\nAntonio\nIts f(x) =y=2 for every x\nAntonio\nYes but he said could you put x = 2 as a function you put y = 2 as a function\nzach\nF(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)\nAntonio\nx = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent\nzach\nThe said x=2 and that 2 is y\nAntonio\nthat 2 is not y, y is a variable 2 is a constant\nzach\nSo 2 is defined as f(x) =2\nAntonio\nNo y its constant =2\nAntonio\nwhat variable does that function define\nzach\nthe function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2\nzach\nYes true, y=2 its a constant, so a line parallel to y axix as function of y\nAntonio\nSorry x=2\nAntonio\nAnd you are right, but os not a function of x, its a function of y\nAntonio\nAs function of x is meaningless, is not a finction\nAntonio\nyeah you mean what I said in my first post, smh\nzach\nI mean (0xY) +x = 2 so y can be as you want, the result its 2 every time\nAntonio\nOK you can call this \"function\" on a set {2}, but its a single value function, a constant\nAntonio\nwell as long as you got there eventually\nzach\n2x^3+6xy-4y^2)^2 solve this\nfemi\nmoe\nvolume between cone z=\u221a(x^2+y^2) and plane z=2\nFatima\nIt's an integral easy\nAntonio\nV=1/3 h \u03c0 (R^2+r2+ r*R(\nAntonio\nHow do we find the horizontal asymptote of a function using limits?\nEasy lim f(x) x-->~ =c\nAntonio\nsolutions for combining functions\nwhat is a function? f(x)\none that is one to one, one that passes the vertical line test\nAndrew\nIt's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y\nAntonio\nis x=2 a function?\nThe\nrestate the problem. and I will look. ty\nis x=2 a function?\nThe", "date": "2018-07-17 07:39:03", "meta": {"domain": "quizover.com", "url": "https://www.quizover.com/calculus/section/the-mean-value-theorem-and-its-meaning-by-openstax", "openwebmath_score": 0.9067342281341553, "openwebmath_perplexity": 407.8281066749289, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120893722416, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.863468885791954}}
{"url": "http://math.stackexchange.com/questions/555860/rate-of-change-of-radius-and-volume", "text": "# Rate of change of radius and volume\n\nThey put a gas bubble in someone's eye. The volume of a gas bubble changes from $0.4$ $cc$ to $1.6$ $cc$ in $74$ hours. Assuming that the rate of change of the radius is constant, find\n\n\u2022 (a) The rate at which the radius changes;\n\u2022 (b) The rate at which the volume of the bubble is increasing at any volume $V$;\n\u2022 (c) The rate at which the volume is increasing when the volume\n\nis $1$ $cc$. (Note: The volume of a ball of radius $r$ is $\\frac{4}{3}\\pi r^3$. Assume the bubble is spherical.)\n\nExplanation would be appreciated.\n\nI did differentiate the $\\frac{4}{3}\\pi r^3$ with respect to r that is $4\\pi r^2$ and then made that equal to $1.66-06$ which is the rate change of $V$. But I Don't know if I am doing it right.\n\n-\nRemember that if the radius grows by a factor of x then the volume grows by X^3. So if the volume grew 4 times then the radius grew by... \u2013\u00a0DannyDan Nov 7 '13 at 18:10\n\n$v=\\frac{4\\pi r^3}{3}$\n\nor\n\n$r=(3v/4\\pi)^{1/3}$\n\n$\\frac{dv}{dt}=4\\pi r^2\\frac{dr}{dt}$ substitute for $r$ we get\n\nand letting $dr/dt=K$ (constant) then\n\n$v^{-2/3}dv=4\\pi K(3/(4\\pi))^{2/3} dt$\n\n$3v^{1/3} = 4\\pi K(3/(4\\pi))^{2/3} dt$\n\nIntegrating between limits $0.4$ and $1.6$ for V and $t=0$ to $74$, we get $K= \\frac{1.6^{1/3}-.4^{1/3}}{74 (4\\pi/3)^{1/3}}$ which works out to $3.628E-3 cm/hr$ as the rate of change of radius.\n\nb) $dv/dt=(36\\pi)^{1/3} KV^{2/3}$\n\nc) substitute $v=1$ in above we get $(36\\pi)^{1/3} K$\n\n-\nIt would be alot easier for part 1 to just: $$K = \\frac{\\sqrt[3]{1.6*3/4/\\pi}-\\sqrt[3]{0.4*3/4/\\pi}}{74} cm/hr$$ but they both get the same answer \u2013\u00a0kaine Nov 7 '13 at 19:12\n\nHint: what are the starting and ending radii? Let $t$ be the number of hours from the start. Now write an equation $r=$ some function of $t$. Use that to write another equation $V=$ some function of $t$ Now use these functions to answer the questions.\n\n-\n\nIf the radius increases at a constant rate, $r = \\alpha t+r_o$ for some constant $\\alpha$.\n\nthis simplifies the questions to:\n\na) What are the values of $\\alpha$ and $r_o$? (should be solvable from the data)\n\nb) Write $V(t)$, solve for $\\frac{dV}{dt}$, and write that in terms of just V.\n\nc) Specifically plug $V=1 cc$ into the answer for b.\n\nNote that we are hesitating to give a direct answer as this appears to be homework but I will gladly confirm any answers you come up with.\n\n-", "date": "2015-12-01 19:12:00", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/555860/rate-of-change-of-radius-and-volume", "openwebmath_score": 0.951605498790741, "openwebmath_perplexity": 270.9876916843228, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407191430025, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8634636426213906}}
{"url": "http://mathhelpforum.com/algebra/63129-quadratic-roots.html", "text": "Find the equation of the quadratic given it's real roots $2-\\sqrt 3$ and $2 + \\sqrt3$ which passes through the point (1,-2)\n\n2. if $r_1$ and $r_2$ are two roots of a quadratic ...\n\n$f(x) = k(x - r_1)(x - r_2)$\n\n$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \\cdot r_2]$\n\nnow, substitute $r_1 = 2-\\sqrt{3}$ and $r_2 = 2 + \\sqrt{3}$ to determine the basic quadratic in [...], then use the fact that $f(1) = -2$ to find $k$\n\n3. Originally Posted by skeeter\nif $r_1$ and $r_2$ are two roots of a quadratic ...\n\n$f(x) = k(x - r_1)(x - r_2)$\n\n$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \\cdot r_2]$\n\nnow, substitute $r_1 = 2-\\sqrt{3}$ and $r_2 = 2 + \\sqrt{3}$ to determine the basic quadratic in [...], then use the fact that $f(1) = -2$ to find $k$\nI'm not sure exactly what you mean, can you please work it down further? I have never learned this formula.\n\n4. Hello, euclid2!\n\nThere are several approaches to this problem.\n. . Here's one of them . . .\n\nFind the equation of the quadratic given its real roots $2-\\sqrt 3$ and $2 + \\sqrt3$\nwhich passes through the point (1,-2)\n\nIf the quadratic, $ax^2 + bx + c$, has roots $p\\text{ and }q$, then: . $p+q \\:=\\:-\\frac{b}{a},\\;\\;pq \\:=\\:\\frac{c}{a}$\n\nSo we have: . $\\begin{array}{ccc}p+q &=&4 \\\\ pq &=& 1\\end{array}\\quad\\Rightarrow\\quad\\begin{array}{ccc }\\frac{b}{a} &=& \\text{-}4 \\\\ \\frac{c}{a} &=& 1 \\end{array}$ . $\\Rightarrow\\quad\\begin{array}{ccc}b &=&\\text{-}4a \\\\ c &=&a \\end{array}$\n\n. . The function (so far) is: . $f(x) \\:=\\:ax^2 -4ax + a$\n\nSince (1,-2) is on the graph: . $a\\!\\cdot\\!1^2 - 4a\\!\\cdot\\!1 + a \\:=\\:-2 \\quad\\Rightarrow\\quad -2a \\:=\\:-2$\n\n. . Hence: . $a\\:=\\:1,\\;\\;b\\:=\\:-4,\\;\\;c\\:=\\:1$\n\nTherefore: . $f(x)\\;=\\;x^2 - 4x + 1$\n\n5. Originally Posted by Soroban\nHello, euclid2!\n\nThere are several approaches to this problem.\n. . Here's one of them . . .\n\nIf the quadratic, $ax^2 + bx + c$, has roots $p\\text{ and }q$, then: . $p+q \\:=\\:-\\frac{b}{a},\\;\\;pq \\:=\\:\\frac{c}{a}$\n\nSo we have: . $\\begin{array}{ccc}p+q &=&4 \\\\ pq &=& 1\\end{array}\\quad\\Rightarrow\\quad\\begin{array}{ccc }\\frac{b}{a} &=& \\text{-}4 \\\\ \\frac{c}{a} &=& 1 \\end{array}$ . $\\Rightarrow\\quad\\begin{array}{ccc}b &=&\\text{-}4a \\\\ c &=&a \\end{array}$\n\n. . The function (so far) is: . $f(x) \\:=\\:ax^2 -4ax + a$\n\nSince (1,-2) is on the graph: . $a\\!\\cdot\\!1^2 - 4a\\!\\cdot\\!1 + a \\:=\\:-2 \\quad\\Rightarrow\\quad -2a \\:=\\:-2$\n\n. . Hence: . $a\\:=\\:1,\\;\\;b\\:=\\:-4,\\;\\;c\\:=\\:1$\n\nTherefore: . $f(x)\\;=\\;x^2 - 4x + 1$\n\nThank you very much, although can you edit this to show me where you are getting those numbers from? Thanks, again.\n\n6. $f(x) = k(x - r_1)(x - r_2)$\n\n$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \\cdot r_2]$\n\n$r_1 + r_2 = (2 - \\sqrt{3}) + (2 + \\sqrt{3}) = 4$\n\n$r_1 \\cdot r_2 = (2 - \\sqrt{3})(2 + \\sqrt{3}) = 4 - 3 = 1$\n\n$f(x) = k(x^2 - 4x + 1)$\n\nsince $f(1) = -2$\n\n$-2 = k[(1)^2 - 4(1) + 1]$\n\n$-2 = k(-2)$\n\n$k = 1$\n\nso ...\n\n$f(x) = x^2 - 4x + 1$", "date": "2016-10-28 07:39:07", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/63129-quadratic-roots.html", "openwebmath_score": 0.9844667315483093, "openwebmath_perplexity": 489.4603645151453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9938070094752999, "lm_q2_score": 0.8688267694452331, "lm_q1q2_score": 0.8634461334944529}}
{"url": "http://enej.timstourenblog.de/maclaurin-series-approximation.html", "text": "# Maclaurin Series Approximation\n\nFor instance, we know that sin0 = 0, but what is sin0. Maclaurin Series Small Angles Approximation Exam Questions with Full Solutions 5. Maclaurin Series(approximation) Thread starter naspek; Start date Dec 10, 2009; Tags use the corresponding Maclaurin polynomial of degree 5 to approximate. For example, the Taylor Series for ex is given by:. The Taylor series for at is (By convention,. Taylor Series in MATLAB First, let's review our two main statements on Taylor polynomials with remainder. 7, exercise 9. Starting with the simplest version, cos x = 1, add terms one at a time to estimate cos(\u03c0/3). The second order Taylor approximation provides a parabolic function approximation while the third order provides a cubic function approximation. The general form of a Taylor series is, assuming the function and all its derivatives exist and are continuous on an interval centered at and containing. The infinite series expansion for f (x) about x = 0 becomes:. As we have seen, a general power series can be centered at a point other than zero, and the method that produces the Maclaurin series can also produce such series. Thus, The Remainder Term is z is a number between x and 3. It is often useful to designate the in\ufb01nite possibilities by what is called the Taylor Series. In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value x = a. Using Taylor series to evaluate limits. In the next video, I'll do this with some actual functions just so it makes a little bit more sense. POLYNOMIAL APPROXIMATION OF FUNCTIONS: Linear and Quadratic Approximation, Taylor and Maclaurin Polynomials, Approximation with Taylor Polynomials (1hour) POWER SERIES: Definition, Center and Radius, Interval of Convergence, Endpoint Convergence, Operations with Power Series, Differentiating and Integrating Power Series (1hour) Week 2 4 hours. Maclaurin Series of f(x) = about x = up to order = Calculate: Computing Get this widget. The n th partial sum of the Taylor series for a function $$f$$ at $$a$$ is known as the n th Taylor polynomial. It first prompts the user to enter the number of terms in the Taylor series and the value of x. If f has a power series representation (expansion) at a,. (b) The Maclaurin series for g evaluated at x = L is an alternating series whose terms decrease in absolute 17 value to 0. We can improve this approximation of f(x) in two ways: Take more terms, increasing N. \u201d This becomes clearer in the expanded \u2026. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we'd like. The Maclaurin Series: Approximations to f Near x = 0 If we let a Taylor polynomial keep going forever instead of cutting it off at a particular degree, we get a Taylor series. Thanks, Prasad. An approximation for the exponential function can be found using what is called a Maclaurin series: e x \u2248 1 + x 1 1 ! + x 2 2 ! + x 3 3 ! + \u2026 We will write a program to investigate the value of e and the exponential function. 01SC Single Variable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw. This means that the power series converges fastest when x is closest to 0. Here\u2019s the formula for \u2026. We'll focus on the Maclaurin right now. Consider the function of the form. FP2: Taylor's Series What does it mean to perform a Taylor expansion on T and V? Why does trig not work when using the 90 degree angle, i. Limits and Continuity Definition of Limit of a Function Properties of Limits Trigonometric Limits The Number e Natural Logarithms Indeterminate Forms Use of Infinitesimals L\u2019Hopital\u2019s Rule Continuity of Functions Discontinuous Functions Differentiation of Functions Definition of the Derivative Basic Differentiation Rules Derivatives of Power Functions Product Rule Quotient Rule Chain Rule. When the Maclaurin series approximates a function, the series values and the function values are very close near x = 0. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Maclaurin/Taylor Series: Approximate a Definite Integral to a Desired Accuracy. We now take a particular case of Taylor Series, in the region near x = 0. Taylor Series. Math 115 Exam #2 1. As you can imagine each order of derivative gets larger which is great fun to work out. Actually Maclaurin Series is just a special form of Taylor Series. Representation of Functions as Power Series 10 2. eq = 60 - 53 x. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we'd like. How accurate is the approximation?. It is the source of formulas for expressing both sin x and cos x as infinite series. Yesterday we learned: Definition of an nth-degree Taylor polynomial:. How Good is Your Approximation? Whenever you approximate something you should be concerned about how good your approximation is. The Maclaurin series of a function up to order may be found using Series[f, x, 0, n]. x 2 2 x 2 n 3 f 8 e 2 x 8 e 2 8 f 3 x 3 8 6 x 3 4 3 x 3 e 2 x 1 2 x 2 x 2 4 3 x from MATH 270 at DeVry University, Chicago. Free Online Library: Efficient and accurate approximation of infinite series summation using asymptotic approximation and fast convergent series. A Taylor polynomial approximates the value of a function, and in many cases, it\u2019s helpful to measure the accuracy of an approximation. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Find the Maclaurin polynomial of degree n = 4 for. Concrete examples in the physical science division and various engineering \ufb01elds are used to paint the applications. Taylor Series Expansions In the previous section, we learned that any power series represents a function and that it is very easy to di\u00a4erentiate or integrate a power series. A calculator for finding the expansion and form of the Taylor Series of a given function. Theorem 40 (Taylor's Theorem). Maclaurin Series Michael Penna, Indiana University \u2013 Purdue University, Indianapolis Objective In this project we investigate the approximation of a function by its Maclaurin series. The binomial function Remark: If m is a positive integer, then the binomial function f m is a polynomial, therefore the Taylor series is the same polynomial, hence the Taylor series has only the \ufb01rst m +1 terms non-zero. all derivatives exist. We could use either the We could use either the Taylor remainder term, with n = 2 since this is really the series through the 2 nd degree term, or we can. My mathematics python's programs is a set of Maclaurin's series to compute some of the most important functions in calculus. Theorem (Taylor Polynomial Approximation). XXIV \u2013 Taylor and Maclaurin Series 1. Partial sums of a Maclaurin series provide polynomial approximations for the function. The sum of partial series can be used as an approximation of the whole series. Colin Maclaurin published a special case of the Taylor series in the 1700s. The function ex or exp(x) can be approximated using the Maclaurin Series (a specific type of Taylor Series) as follows (do ? to find factorial): Write Matlab code using a while loop to calculate a vector, macexp, that will hold each successive approximation to exp(x). If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x)\u2026. As a simple example, you can create the number 10 from smaller numbers: 1 + 2 + 3 + 4. Take a function, pick a specific point, and. Questa pagina \u00e8 stata modificata per l'ultima volta il 23 giu 2018 alle 18:49. We now take a particular case of Taylor Series, in the region near x = 0. These Taylor (and Maclaurin) polynomials are used to numerically approximate functions. Lec 89 - Sine Taylor Series at 0 (Maclaurin) Lec 90 - Taylor Series at 0 (Maclaurin) for e to the x. To find the Maclaurin Series simply set your Point to zero (0). make us an approximation of the series to fourth term, and also test the effect of large amplitude angles on the period. please help me. That is, approximation? The second partial sum (sum of the rst three terms) is Z 2 0. (e)Explain why di culties occur using the series in part (b) to approxi-mate erf(1). Multivariate approximation methods and applications to geophysics and geodesy. If we just have a zero-degree polynomial, which is just a constant, you can approximate it with a horizontal line that just goes through that point. 1 Taylor series 2. As you can imagine each order of derivative gets larger which is great fun to work out. Actually Maclaurin Series is just a special form of Taylor Series. FP2: Taylor's Series What does it mean to perform a Taylor expansion on T and V? Why does trig not work when using the 90 degree angle, i. 1 Things to Remember \u2022 Given a function, be able to \ufb01nd its Taylor or Maclaurin\u2019s series. A Maclaurin series is a power series that allows one to calculate an approximation of a function f(x) for input values close to zero, given that one knows the values of the successive derivatives of the function at zero. A MacLaurin Polynomial is a special case of a Taylor Polynomial. Taylor and Maclaurin Series Definitions In this section, we consider a way to represent all functions which are \u201dsu\ufb03ciently nice\u201d around some point. One important application of power series is to approximate a function using partial sums of its Taylor series. Taylor and maclaurian series Derivation for Maclaurin Series for ex Derive the Maclaurin series x2 x3 ex x = + + + + 2! 3! So if we want to find out how many. T is a Maclaurin series, plot these functions together to see how well this Taylor. x is the \ufb01rst (non-zero) term in the Maclaurin series for sin(x), 0+x+0x2+\u00b7\u00b7\u00b7. Given f(x), we want a power series expansion of this function with respect to a chosen point xo, as follows: (1) (Translation: find the values of a 0 , a 1 , a 2 , \u2026 of this infinite series so that the equation holds. Lec 92 - Visualizing Taylor Series Approximations. Use the Taylor series for the function defined as to estimate the value of. How do you find the Maclaurin series of #f(x)=sin(x)# ? How do you use a Maclaurin series to find the derivative of a function? See all questions in Constructing a Maclaurin Series. How do we \ufb01nd a quadratic approximation to a function y = f(x) and how accurate is this approximation? The secret to solving these problems is to notice that the equation of the tangent line showed up in our integration by parts in (1. We attribute much of the founding theory to Brook Taylor (1685-1731), Colin Maclaurin (1698-1746) and Joseph-Louis Lagrange (1736-1813). For example, the Taylor Series for ex is given by:. Taylor and maclaurian series Derivation for Maclaurin Series for ex Derive the Maclaurin series x2 x3 ex x = + + + + 2! 3! So if we want to find out how many. On [Series:: esss] makes Series generate a message in this case. This procedure is also called the expansion of the function around (or about) zero. The main idea is this: You did linear approximations in first semester calculus. It is often the case that a convenient expansion point is x 0 = 0, and series about this special expansion point are also called Maclaurin series. A calculator for finding the expansion and form of the Taylor Series of a given function. Maclaurin series are fast approximations of functions, and they offer more accurate function approximations than just linear ones. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. In addition, when n is not an integer an extension to the Binomial Theorem can be used to give a power series representation of the term. We illustrate with some examples. Example: sine function. Lec 94 - Visualizing Taylor Series for e^x. Approximation of e^x using Maclaurin Series in Python. Example: Approximation for ln(1+x) Leaving Cert 2005 Q8 b (ii) Use those terms to find an approximation for ln. Using a Table of Basic Power Series to Determine More Power Series - Part 2 Determine the Maclaurin Series and Polynomial for Function in the Form ax^2*sin(bx. ) {}For the Euler-Maclaurin summation formula, see 65B15 for: 78: 41-XX Approximations and expansions {}For all approximation theory in the complex domain, see 30E05 and 30E10; for all trigonometric approximation and interpolation, see 42A10 and 42A15; for numerical. Index 8 to get to my menu, scroll down to Maclaurin series there is it there then press enter \u2026. The program is really simple: The user inputs a parameter x (x being an angle in radians) and a float \u03b5, which is the precision of the value of cos(x). The diagram shows the Maclaurin series approximation to degree n for the exponential function. We would like to find an easier-to-compute approximation to f(x), to see why f this is also called the MacLaurin polynomial, (called a power series). Without further ado, here it is: The notation f(n) means \"the nth derivative of f. You can hide/reveal the graph of appropriate series and change the value of the pivot in the Taylor series. If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x)\u2026. This chapter examines methods of deriving approximate solutions to problems or of approximating exact solutions, which allow us to develop concise and precise estimates of quantities of interest when analyzing algorithms. This website and its content is subject to our Terms and Conditions. Use your answer to find a Maclaurin series for f'. In an open interval around x= a, f(x) \u2248 f(a)+f\u2032(a)(x\u2212a) linear approximation \u2022 Quadratic approximation in one variable: Take the constant, linear, and quadratic terms from the Taylor series. Let us revise how to construct a program for Taylor Series. Suppose a set of standardized test scores are normally distributed with mean and standard deviation Use and the first six terms in the Maclaurin series for to approximate the probability that a randomly selected test score is between and Use the alternating series test to determine how accurate your approximation is. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) - so it's a mathematical. These notes discuss three important applications of Taylor series: 1. Clicking the Draw button plots the polynomial of the selected degree; clicking the Next button increments the degree to the next odd integer and plots the polynomial of that degree. Suppose a set of standardized test scores are normally distributed with mean and standard deviation Use and the first six terms in the Maclaurin series for to approximate the probability that a randomly selected test score is between and Use the alternating series test to determine how accurate your approximation is. Lec 95 - Polynomial approximation of functions (part 1). Produces the result Note that function must be in the integrable functions space or L 1 on selected Interval as we shown at theory sections. You won\u2019t use an infinite series to calculate the approximation. ) When , the series is called a Maclaurin series. So, I'm trying to create a program that calculates cos(x) by using a Taylor approximation. Using Taylor polynomials to approximate functions. Use the \ufb01rst two non-zero terms of an appropriate series to give an approximation of Z 1 0 sin we can replace x with t3 to get the Maclaurin series for cost3: 1. Problems on Taylor's Theorem. Taylor Polynomials Preview. Just calculate the values of the red bits and plug them into the Maclaurin series to give you the series expansion formula. Maclaurin expansion of B * (s), which involves four moments of the service-time distribution, gives a better approximation than the one involving two moments. Index 8 to get to my menu, scroll down to Maclaurin series there is it there then press enter \u2026. 2 correct to five decimal places. Example: sine function. Taylor and Maclaurin Series We have learned how to construct power series representations of certain functions by relating them to geometric series, either directly, or indirectly through di erentiation or integration. In fact, e\u2212p8(1). This website and its content is subject to our Terms and Conditions. The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. A Taylor series is a representation of a function using an infinite sum. 1 (1974), 287--289, MathSciNet. Keywords: The Taylor series, the Maclaurin series, polynomial and nonpolynomial approx-imation. 6 Two examples 3 Indeterminate forms 3. Practice Taylor/Maclaurin, receive helpful hints, take a quiz, improve your math skills. Let n 1 be an integer, and let a 2 R be a point. First, we de\ufb02ne the Bernoulli numbers B2n. We now take a particular case of Taylor Series, in the region near x = 0. Textbook solution for Calculus (MindTap Course List) 11th Edition Ron Larson Chapter 9. i don't even understand this topic. Leary Find other works by these authors. Compute the Remainder Term for. Category: Maclaurin's series Approximation in series expansion. \u201d This becomes clearer in the expanded \u2026. Then to \ufb01nd our approximation, we need to \ufb01nd n such that (. Lecture 61: Power Series Representation Of Functions; Lecture 62: What Is The Taylor Series? Lecture 63: What Is The Maclaurin Series? Lecture 64: Application Of The Maclaurin Series; Lecture 65: Find The Maclaurin Series For Sinx; Lecture 66: Find The Maclaurin Series For Cosx; Lecture 67: Maclaurin Series For A Binominal Expansion: 1. Maclaurin Series of f(x) = about x = up to order = Calculate: Computing Get this widget. Though, the computation of an infinite sum which give the value of a function in terms of the derivatives evaluated at a special case where x0 = 0,in contrast with Taylor series. We now take a particular case of Taylor Series, in the region near x = 0. In other words, f0gives us a linear approximation of f(x) near c: for small values of \"2R, we have f(c+ \") \u02c7f(c) + \"f0(c) But if f(x) has higher order derivatives, why stop with a linear approximation? Taylor series take this idea of linear approximation and extends it to higher order derivatives, giving us a better approximation of f(x) near c. 12 (1975), no. Taylor's expansion, and the related Maclaurin expansion discussed below, are used in approximations. Power Series, Taylor and Maclaurin Polynomials and Series Power Series The Basics De nition 1 (Power Series). Notice that this Taylor Series for e x e^{x} e x is different from the Maclaurin Series for e x e^{x} e x. If you want to use a di\ufb00erent center, then just take the results from this documentandreplacex with(x a) everywhere. The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. Maclaurin Series Michael Penna, Indiana University \u2013 Purdue University, Indianapolis Objective In this project we investigate the approximation of a function by its Maclaurin series. Mathematica Code For the Euler-Maclaurin Formula20 k is a slowly divergent series, inclusive. Rather than referring to it as such, we use the following. The Maclaurin series expansion for cos x is. The approximation for g using the first two nonzero terms of this series is Show that 120 this approximation differs from g L by less than 200. ) Use a power series to approximate. Algorithm for Computing Taylor Series. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. The Maclaurin series is the Taylor series at the point 0. How to extract derivative values from Taylor series Since the Taylor series of f based at x = b is X\u221e n=0 f(n)(b) n! (x\u2212b)n, we may think of the Taylor series as an encoding of all of the derivatives of f at x = b: that information. For example, the Taylor Series for ex is given by:. We can improve this approximation of f(x) in two ways: Take more terms, increasing N. dx dy y \u2212 =\u2212 x Proving DE By further differentiation of this result, or otherwise, find the Maclaurin\u2019s series for. power series. ( n=2 amd x>=0 and x<=0,5) Our instructor showed just a basic example about taylor/maclaurin approximations and it has nothing to do with this one. We set an initial value of 1 to the sum of the series and define the first term, t= 1. 5)2n+1 2n+1. It is more of an exercise in differentiating using the chain rule to find the derivatives. Practice Taylor/Maclaurin, receive helpful hints, take a quiz, improve your math skills. Therefore we need to calculate the first 11 terms of the MacLaurin series (remember that the first term is when n = 0). Maclaurin/Taylor Series: Approximate a Definite Integral to a Desired Accuracy. i don't even learn this thing yet. all derivatives exist. Taylor Series 1. \u2022 Section 8. Series detects certain essential singularities. Choose a web site to get translated content where available and see local events and offers. 3 The binomial expansion 2. Here we show better and better approximations for cos(x). MIT OpenCourseWare http://ocw. Hopefully by the end of this, you'll be getting the hang of these things so we can start calculating stuff with them. How do we calculate the Maclaurin series?. Actually Maclaurin Series is just a special form of Taylor Series. One important application of power series is to approximate a function using partial sums of its Taylor series. Lec 91 - Euler's Formula and Euler's Identity. (See why we want to do this in the Introduction. 2c: By applying the ratio test, find the radius of convergence for this Maclaurin series. Taylor's expansion, and the related Maclaurin expansion discussed below, are used in approximations. Taylor Series Generalize Tangent Lines as Approximation. Starting with the simplest version, cos x = 1, add terms one at a time to estimate cos(\u03c0/3). It is the source of formulas for expressing both sin x and cos x as infinite series. Alternatively, it could also be used for approximation purposes in evaluating the value of certain numeric entities. Hello again everybody Tom from everystepcalculus. Concrete examples in the physical science division and various engineering \ufb01elds are used to paint the applications. Maclaurin Series Small Angles Approximation Exam Questions with Full Solutions 5. Example: Second-order Taylor series approximation (in gray) of a function around origin. Using Taylor series to find the sum of a series. Determine a Maclaurin Series approximation for f(x)=sin?(6x) where n = 6. If you make the polynomial approach a degree of infinity, your approximation becomes infinitely close to the real function and is perfectly equal to the real function. Generally, Maclaurin series expressions are more compact and will give good approximations even for values far from the origin if enough terms are used. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. It can be used to approximate integrals by finite sums, or conversely to evaluate finite sums and infinite series using integrals and the machinery of calculus. This says that if 7 x 9, the approximation in part (a) is accurate to within 0. However,canafunctionf(x. The Maclaurin series of a function up to order may be found using Series[f, x, 0, n]. Multivariate approximation methods and applications to geophysics and geodesy. Using Taylor series to evaluate limits. As mentioned in \u00a7E. We attribute much of the founding theory to Brook Taylor (1685-1731), Colin Maclaurin (1698-1746) and Joseph-Louis Lagrange (1736-1813). The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. As you can imagine each order of derivative gets larger which is great fun to work out. Lec 93 - Generalized Taylor Series Approximation. The representation of Taylor series reduces many mathematical proofs. Summing 11 terms in the second series gives erf(1) \u02c70:842700790029219. We now take a particular case of Taylor Series, in the region near x = 0. Some series converge only at a, and others converge on an interval (a - r, a + r). It is the source of formulas for expressing both sin x and cos x as infinite series. The Maclaurin's series for ln(1+x) could be used to approximate the natural logarithm ln(x). eq = 60 - 53 x. The series carries on to in\ufb01nity, and has general term (x\u2212a)n n! f(n)(a). Step-by-step method for computing a Taylor series, with example of finding the Taylor series expansion of f(x) = (1-x)-1 about x = 0. Solution of Example 3 (from pages 736. Linear Approximation of Functions Linear approximation is an example of how differentiation is used to approximate functions by linear ones close to a given point. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. It is common practice to use a finite number of terms of the series to approximate a function. Series can expand about the point x = \u221e. Since I want the Remainder Term, I need to find an expression for the derivative. But Taylor and Maclaurin polynomials can only approximate functions. In the next video, I'll do this with some actual functions just so it makes a little bit more sense. Multivariate Taylor series is used in many optimization techniques. The \ufb01rst of these is to under-stand how concepts that were discussed for \ufb01nite series and integrals can be meaningfully. Example 13. 0 Share this post. I have to solved this eq and draw the graph. If f has a power series representation (expansion) at a,. Now write the Maclaurin series for ln(x+. Use your pocket calculator or MATLAB to determine the true value. There are many applications for expansions of common functions around x=0. 4C \u00ad Composite Maclaurin Series 2016 EXPORT. There are many sensible notions of what 'good approximation' could mean. Leary Find other works by these authors. Complete Solution Before starting this problem, note that the Taylor series expansion of any function about the point c = 0 is the same as finding its Maclaurin series expansion. Hopefully by the end of this, you'll be getting the hang of these things so we can start calculating stuff with them. Linear Approximation of Functions Linear approximation is an example of how differentiation is used to approximate functions by linear ones close to a given point. My mathematics python's programs is a set of Maclaurin's series to compute some of the most important functions in calculus. If you make the polynomial approach a degree of infinity, your approximation becomes infinitely close to the real function and is perfectly equal to the real function. Toggle Main Navigation. 10 Problem 47E. 10: Taylor and Maclaurin Series 1. putationally ef\ufb01cient method of approximation. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. Taylor Series. Taylor series are extremely powerful tools for approximating functions that can be difficult to compute otherwise, as well as evaluating infinite sums and integrals by recognizing Taylor series. We can use power series to create a function that has the same value as another function, and we can then use a limited number of terms as a way to compute approximate values for the original function within the interval of convergence. In an open interval around x= a, f(x) \u2248 f(a)+f\u2032(a)(x\u2212a) linear approximation \u2022 Quadratic approximation in one variable: Take the constant, linear, and quadratic terms from the Taylor series. On [Series:: esss] makes Series generate a message in this case. Maclaurin and Taylor Series \u00b7\u00b7\u00b7 the curve representing ex is a better and better approximation. Let n 1 be an integer, and let a 2 R be a point. So, you can start by assuming the Taylor series definition:. So, I'm trying to create a program that calculates cos(x) by using a Taylor approximation. At the age of 19 he was elected a professor of mathematics at Marischal College, Aberdeen, and two. Maclaurin series expansion eulers formula for pi show 10 more Any revision resources for A-Level AQA Maths? FP2: Taylor's Series Maclaurin and Taylor Series! I need help :( What is the purpose of the power/maclaurin/taylor series Series expansions of odd functions. power series, such as the Taylor series of a basic function. The first derivative of tan x is very simple as you can see. Theorem 40 (Taylor's Theorem). Taylor series expansions of hyperbolic functions, i. org ) Created Date: 8/7/2013 5:18:45 PM. 3) Comment at the bottom of the page. Applications of Taylor Series Recall that we used the linear approximation of a function in Calculus 1 to estimate the values of the function near a point a (assuming f was di erentiable at a): f(x) \u02c7f(a) + f0(a)(x a) for x near a: Now suppose that f(x) has in nitely many derivatives at a and f(x) equals the. The sum of partial series can be used as an approximation of the whole series. The exponential function is shown in red and the Maclaurin series approximation function is shown in blue. The goal of a Taylor expansion is to approximate function values. [3] (iii) By substituting x = 3 8. Maclaurin and Taylor Series. The Maclaurin series above is more than an approximation of e x, it is equal to e x on the interval of convergence (- , ). 01 Single Variable Calculus, Fall 2005 Prof. com, a free online graphing calculator. Taylor's theorem (actually discovered first by Gregory) states that any function satisfying certain conditions can be expressed as a Taylor series. The th term of a Maclaurin series of a function can be computed in the Wolfram Language using SeriesCoefficient[f, x, 0, n] and is given by the inverse Z-transform. NASA Technical Reports Server (NTRS) Munteanu, M. To avoid this, we can rst nd the Maclaurin Series for g(x) = (1+x)2=3,. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) - so it's a mathematical. To find the Maclaurin. The main idea is this: You did linear approximations in first semester calculus. The Maclaurin Series: Approximations to f Near x = 0 If we let a Taylor polynomial keep going forever instead of cutting it off at a particular degree, we get a Taylor series. Use the Taylor series for the function defined as to estimate the value of. In the West, the subject was formulated by the Scottish mathematician James Gregory and formally introduced by the English mathematician Brook Taylor in 1715. In specific, the type of Taylor series used is technically a Maclaurin series, since the representation is centered at a=0. You can now regrow the entire creature from that tiny sample. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we\u2019d like. Find the Taylor series for at xa. the Maclaurin series of a function is centred at 0, or talk of the series expansion around 0. In [5] the convergence radius for Liapunov series was found in case of homogeneous equilibrium figures (Maclaurin ellipsoids). TAYLOR AND MACLAURIN SERIES 3 Note that cos(x) is an even function in the sense that cos( x) = cos(x) and this is re ected in its power series expansion that involves only even powers of x. Course Material Related to This Topic: Read lecture notes, section 3, pages 4\u20135. The first one is easy because tan 0 = 0. A Taylor series can be used to describe any function \u0192(x) that is a smooth function (or, in mathematical terms. As you can imagine each order of derivative gets larger which is great fun to work out. We would like to find an easier-to-compute approximation to f(x), to see why f this is also called the MacLaurin polynomial, (called a power series). We now take a particular case of Taylor Series, in the region near x = 0. Maclaurin Series of Sqrt(1+x) In this tutorial we shall derive the series expansion of $$\\sqrt {1 + x}$$ by using Maclaurin's series expansion function. 7 Taylor and Maclaurin Series The conclusion we can draw from (5) and Example 1 is that if ex has a power series expansion at 0, then \u221e xn x e = n! n=0 So how can we determine whether ex does have a power series representation?. Maclaurin series and the general Taylor series centered at x = a. If we're evaluating the series at a point within its interval of convergence. The Maclaurin series for this function is known as the binomial series. A Maclaurin series is the expansion of the Taylor series of a function about zero. What you did was you created a linear function (a line) approximating a function by taking two things into consideration: The value of the function at a point, and the value of the derivative at the same point. XXIV \u2013 Taylor and Maclaurin Series 1. You can hide/reveal the graph of appropriate series and change the value of the pivot in the Taylor series. , I might be ( 17;19)) and let x 0 be a point in I, i. The program approximates the function cos(x) using a Taylor series approximation. It is often useful to designate the in\ufb01nite possibilities by what is called the Taylor Series. Towards Maclaurin. Thus, The Remainder Term is z is a number between x and 3. In [5] the convergence radius for Liapunov series was found in case of homogeneous equilibrium figures (Maclaurin ellipsoids). If we just have a zero-degree polynomial, which is just a constant, you can approximate it with a horizontal line that just goes through that point. By using this website, you agree to our Cookie Policy. Thanks, Prasad. To nd Taylor series for a function f(x), we must de-. If not, for what values of x can I use the series to approximate f(x)? 5. ) When , the series is called a Maclaurin series.", "date": "2020-02-21 19:26:08", "meta": {"domain": "timstourenblog.de", "url": "http://enej.timstourenblog.de/maclaurin-series-approximation.html", "openwebmath_score": 0.9073213934898376, "openwebmath_perplexity": 390.82097345168205, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9938070091084549, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8634461298002867}}
{"url": "https://mathematica.stackexchange.com/questions/201346/how-can-i-define-a-very-large-matrix-efficiently/201348", "text": "# How can I define a very large matrix efficiently?\n\nHere's a 7x7 matrix:\n\n(matrix = {{1, 1, 1, 1, 1, 1, 1}, {1, -1, 0, 0, 0, 0, 0}, {1,\n1, -2, 0, 0, 0, 0}, {1, 1, 1, -3, 0, 0, 0}, {1, 1, 1, 1, -4, 0,\n0}, {1, 1, 1, 1, 1, -5, 0}, {1, 1, 1, 1, 1,\n1, -6}}) // MatrixForm\n\n\nThe matrix looks like this:\n\nHow can I define the 100x100 equivalent of this matrix efficiently without having to manually type out every element? The documentation gives some ideas, but although there's certainly an order in this matrix, it's not something that translates easily (at all?) to the commands used in the documentation.\n\nsize = 7;\nTable[Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0], {i, size}, {j, size}]\n\n\u2022 Worked very well, now I'll have to understand your syntax =) \u2013\u00a0Allure Jul 2 '19 at 3:41\n\u2022 You might be interested in a tweak to your method explained in my answer. \u2013\u00a0Michael E2 Jul 2 '19 at 17:35\n\nThe question refers to efficiency, which might mean with respect to programming, performance, or both. My guess is programming, but highlighting the performance strengths of Mathematica should promote learning the \"Mathematica style\" as it is sometimes referred to.\n\nTimings on various approaches for size = 100 (timing code given at the end):\n\n\u2022 0.008800 @kglr\n\u2022 0.007000 @Rohit Namjoshi, which can be improved to 0.00026\n\u2022 0.000510 @MikeY, which can be improved to 0.000041 (best of all)\n\u2022 0.000047 @MichaelE2, (was never under 0.000042 but usually 0.000046 to 0.000048)\n\nMy best solution:\n\nsize = 5;\nmatrix =\nWith[{r    = Range@size,\nzero = DeveloperToPackedArray@{0}},\nUnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]] -\n];\nmatrix // MatrixForm\n\n\n### Discussion\n\nTable[] vs. Array[]. The solution of @RohitNamjoshi can be improved by using Array[]. This involves turning the expression e into a function, which takes only wrapping the expression with Function[{i, j}, e] and removing i and j from the iterators:\n\nArray[Function[{i, j},\nWhich[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0]\n], {size, size}]; // RepeatedTiming\n(*  {0.00039, Null}  *)\n\n\nIt can be sped up a little by compiling the function. Compiling with or without CompilationTarget -> \"C\" affects Table[] and Array[] differently. Compiling speeds up Table[] (0.0039 sec.), but whether to C or not makes no perceptible difference. Compiling to C speeds up Array[] slightly, but compiling to WVM the option slows down Array[] (0.00091 sec.); I don't know why.\n\ncf = Compile[{{i, _Integer}, {j, _Integer}},\nWhich[\ni == 1, 1,\ni > j, 1,\ni == j, -i + 1,\nTrue, 0], CompilationTarget -> \"C\"];\n\nArray[cf, {size, size}]; // RepeatedTiming\n(*  {0.00026, Null}  *)\n\n\nVectorization vs. Compiling. Many basic functions in Mathematica are vectorized: They work efficiently on packed arrays. Elementary functions, many linear algebra functions, some list manipulation functions (esp. functions operating on rectangular arrays) are particularly efficient on packed arrays. Other functions might \"unpack\" the array, which involves copying the packed array into the unpacked form. This done internally but there are ways to tell (On[\"Packing\"] and DeveloperPackedArrayQ for instance); mainly it is invisible to the user except for a performance hit. Usually, but not always, the vectorized functions are more efficient than compiled functions. The solution of @MikeY takes advantage of this up until ones and the use of ArrayFlatten[], which unpacks its arguments to assemble the final matrix. The reason ArrayFlatten[] unpacks is because one of the inputs, ones, is not packed. It turns out that user-entered lists are not packed. If we pack it, we get a great improvement in performance:\n\nWith[{n = size - 1},\nBlock[{m1, m2, ones},\nm1 = LowerTriangularize[ConstantArray[1, {n, n}], -1];\nm2 = m1 - DiagonalMatrix[Range[n]];\nones = DeveloperToPackedArray@{ConstantArray[1, n]};\nArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]\n]]; // RepeatedTiming\n(*  {0.000041, Null}  *)\n\n\nUsing ArrayPad[m2, {{1, 0}, {1, 0}}, 1], as suggested by @kglr, is about the same speed (the timings fluctuate around each other), even though it eliminates a line of code (for ones). But ones is just 1 x 100 and rather small compared to the matrices. At size = 1000, ArrayFlatten[] took 0.014-0.015 sec., ArrayPad[] took 0.015-0.16 sec., and my method took 0.012-0.014 sec.\n\nMy own answer avoids unpacking and uses vectorized functions. Outer[Plus, array1, array2] and Outer[List, array1, array2] are extremely efficient special cases of Outer[] on packed arrays.\n\n### Appendix: Timing code\n\nsa[size]; // RepeatedTiming                  (* @kglr *)\nTable[Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0],\n{i, size}, {j, size}]; // RepeatedTiming  (* @Rohit Namjoshi *)\nWith[{n = size - 1},\nBlock[{m1, m2, ones},\nm1 = LowerTriangularize[ConstantArray[1, {n, n}], -1];\nm2 = m1 - DiagonalMatrix[Range[n]];\nones = {ConstantArray[1, n]};\nArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]\n]]; // RepeatedTiming                      (* @MikeY *)\nWith[{r    = Range@size,\nzero = DeveloperToPackedArray@{0}},\nUnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]] -\n]; // RepeatedTiming                      (* @Michael E2 *)\n\n\u2022 Nice comparison, thanks. I chose my approach more for ease of understanding, or maybe to align with the underlying logic of the matrix, rather than performance. Nice to see it did alright on the performance front too. I learned something! \u2013\u00a0MikeY Jul 2 '19 at 17:41\n\u2022 Thanks Michael. I think you meant \"can be improved by using Array\". \u2013\u00a0Rohit Namjoshi Jul 2 '19 at 17:50\n\u2022 @RohitNamjoshi Thanks for pointing out the error. \u2013\u00a0Michael E2 Jul 2 '19 at 18:06\n\u2022 @MikeY You could also use ones = ConstantArray[1, {1, n}] instead of ToPackedArray. \u2013\u00a0Michael E2 Jul 2 '19 at 18:14\n\u2022 @MichaelE2, I tweaked my answer, thanks for the tip. \u2013\u00a0MikeY Jul 2 '19 at 19:12\n\nYou can also use SparseArray as follows\n\nsa[n_] := SparseArray[{{i_, i_} /; i > 1 -> 1 - i, {i_, j_} /; 1 < i < j ->  0}, {n, n}, 1]\n\nmatrix2 = sa[7]\nmatrix2 == matrix\n\n\nTrue\n\nOne more, for fun and education. Treating it like a block matrix, do the lower right first. This makes a lower triangular matrix of 1s, with the -1 in the command telling it to make the diagonal = zero.\n\nm1=LowerTriangularize[ConstantArray[1, {6, 6}], -1]\n\n\n$$\\left( \\begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 & 0 \\\\ 1 & 1 & 0 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & 1 & 0 & 0 \\\\ 1 & 1 & 1 & 1 & 1 & 0 \\\\ \\end{array} \\right)$$\n\nm2 = m1 - DiagonalMatrix[Range[6]]\n\n\n$$\\left( \\begin{array}{cccccc} -1 & 0 & 0 & 0 & 0 & 0 \\\\ 1 & -2 & 0 & 0 & 0 & 0 \\\\ 1 & 1 & -3 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & -4 & 0 & 0 \\\\ 1 & 1 & 1 & 1 & -5 & 0 \\\\ 1 & 1 & 1 & 1 & 1 & -6 \\\\ \\end{array} \\right)$$\n\nNow the upper right, a $$1 \\times 6$$ matrix of ones\n\nones = ConstantArray[1, {1, n}]\n\n\n$$\\left( \\begin{array}{cccccc} 1 & 1 & 1 & 1 & 1 & 1 \\\\ \\end{array} \\right)$$\n\nUse ArrayFlatten to get the final answer.\n\nArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]\n\n\n$$\\left( \\begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\ 1 & -1 & 0 & 0 & 0 & 0 & 0 \\\\ 1 & 1 & -2 & 0 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & -3 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & 1 & -4 & 0 & 0 \\\\ 1 & 1 & 1 & 1 & 1 & -5 & 0 \\\\ 1 & 1 & 1 & 1 & 1 & 1 & -6 \\\\ \\end{array} \\right)$$\n\n\u2022 you can also use ArrayPad[#, {{1, 0}, {1, 0}}, 1] & on m2 for the last step (+1) \u2013\u00a0kglr Jul 2 '19 at 14:38\n\nAs already suggested in the other answers, vectorization is a key to high performance when working with large numeric arrays. To further increase calculation speed it is useful to remove unnecessary intermediate steps, because memory operations on large arrays can be quite time consuming. Here is a vectorized function with very few intermediate steps:\n\nmFast[n_] := LowerTriangularize[DiagonalMatrix[-Range[n]] + 1] + UnitVector[n, 1];\n\n\nTiming measurement:\n\nRepeatedTiming[mFast[100];]\n\n\n{0.000032, Null}\n\nThis can be compared to MichaelE2's function:\n\nmMichaelE2[size_] := With[{r = Range@size, zero = DeveloperToPackedArray@{0}},\nUnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]]\n`", "date": "2021-01-17 06:53:45", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/201346/how-can-i-define-a-very-large-matrix-efficiently/201348", "openwebmath_score": 0.3936896324157715, "openwebmath_perplexity": 5874.000314420011, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9425067195846918, "lm_q2_score": 0.916109614162018, "lm_q1q2_score": 0.8634394672238412}}
{"url": "http://math.stackexchange.com/questions/171719/if-n-m3-m-for-some-integer-m-then-n-is-a-multiple-of-6/171722", "text": "# If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$\n\nI am trying to teach myself mathematics (I have no access to a teacher), but I am not getting very far. I am just working through the exercises at the end of the book's chapter, but unfortunately there are no solutions.\n\nAnyway, I am trying to prove\n\nIf $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$.\n\nBut I do not know how to approach it. I thought of starting with something like $n = 6k$ for the multiple and that $m^3$ is crucial, but I do not know how that would help or where to go next. Does anyone have any hints or suggestions? Please do not post the whole proof because I want to solve it myself, thank you.\n\n-\nYou might also want to show that if $n = m^5-m$ for some integer $m$, then $n$ is a multiple of 30. (This is a bit trickier but basically the same idea.) \u2013\u00a0Michael Lugo Jul 17 '12 at 0:11\n@Mic $\\rm\\ mod\\ 5\\!:\\ 0^5\\!\\equiv 0,\\ (\\pm1)^5\\!\\equiv \\pm1,\\ (\\pm2)^5\\!\\equiv \\pm2.\\ \\ mod\\ 6\\!:\\ m^5\\equiv m^3 m^2 \\equiv m\\ m^2\\equiv m\\$ by OP. $\\ \\ \\$ \u2013\u00a0Gone Jul 17 '12 at 2:09\nBill, that is more clever than the solution I had in mind - I hadn't thought to exploit what was done in the original question! I was thinking of factoring as $m(m-1)(m+1)(m^2+1)$ and then arguing that at least one factor is divisible by each of 2, 3, and 5. \u2013\u00a0Michael Lugo Jul 17 '12 at 18:53\n\nLets start by factoring $n:$$n = m^3-m = m(m^2-1) = (m-1)m(m+1)$$ Note$(m-1),m$and$(m+1)$are three consecutive integers so (at least) one of these must be a multiple of$2$and one of these must be a multiple of$3$. -  Thank you - that is helpful (I can probably do something now with factors) \u2013 George Jul 17 '12 at 0:16 Glad to help. Yes: $$n = (2p)\\times (3r)\\times t$$ where$p,r$and$t$are integers :) \u2013 Quixotic Jul 17 '12 at 0:17 so I did this:$\\frac{n}{6} = ( (m - 1)(\\frac{m}{2})(\\frac{(m + 1)}{3}))$so$n = 6((m - 1)(\\frac{m}{2})(\\frac{(m + 1)}{3}))$is that cheating? Or am I going the right way? \u2013 George Jul 17 '12 at 0:31 I don't think this is the right way since if this is valid you can probably show divisibility by any number. \u2013 Quixotic Jul 17 '12 at 0:33 After what I said before,$n = (2p)\\times (3r)\\times t = 6 (prt)$and you are done. \u2013 Quixotic Jul 17 '12 at 0:42 show 1 more comment Hint$\\rm\\ mod\\ 6\\!:\\ 0^3\\!\\equiv 0,\\ (\\pm1)^3\\!\\equiv \\pm1,\\ (\\pm2)^3\\!\\equiv \\pm2,\\ 3^3\\!\\equiv 3\\:\\Rightarrow\\:n^3\\equiv n\\ \\ $QED Note$ $It's easier via balanced residues$\\{0,\\, \\pm1,\\, \\pm2,\\, 3\\}$vs.$\\,\\{0,1,2,3,4,5\\},\\,$by$\\rm\\:4\\equiv -2,\\:\\rm 5\\equiv -1.\\:$It is not difficult to prove a generalized Euler-Fermat theorem, namely Theorem$\\ $For naturals$\\rm\\: e,m,n\\: $with$\\rm\\: e,m>1 \\rm\\qquad\\qquad\\ m\\ |\\ n^e-n\\ $for all$\\rm\\:n\\ \\iff\\ m\\:$is squarefree and prime$\\rm\\: p\\:|\\:m\\: \\Rightarrow\\: p\\!-\\!1\\ |\\ e\\!-\\!1 $Yours is the special case$\\rm\\:e={\\bf\\color{blue}3},\\ m = 6 = {\\bf\\color{#C00}2}\\cdot{\\bf\\color{#0A0}3}\\:$is squarefree, and$\\rm\\, {\\bf\\color{#C00}2}\\!-\\!1,{\\bf\\color{#0A0}3}\\!-\\!1\\:|\\:{\\bf\\color{blue}3}-1.$- $$n = m^3 - m$$ $$= m (m^2 - 1)$$ $$= m (m - 1) (m + 1)$$ Now, let$m$be odd. So,$m = 2k \\pm 1$and$m \\pm 1$will be even, and vice versa. So, atl east one of the numbers from$m$,$m \\pm 1$has to be divisible by$2$. Again, since multiples of three are at a separation of$2$numbers, so, again we have one of$m$,$m \\pm 1$being divisible by$6$. - You could also show this by induction, noting$0^3-0$is a multiple of$6$for the base case, and then showing that$(m+1)^2-(m+1) - (m^3-m)$is a multiple of$6$for all$m$for the inductive step. (It is enough to cover the nonnegative case, because$(-m)^3-(-m)=-(m^3-m)$.) You can show that the new expression, which simplifies to$3m(m+1)$, is a multiple of$6$either by noting that one of$m$or$m+1$is even as in Quixotic's answer, or again using induction. Note that$3\\cdot 0(0+1)$is a multiple of$6$for the base case, and then show that$3(m+1)(m+2)-3m(m+1)=6(m+1)$is a multiple of$6$for the inductive step. Applying a similar method to Michael Lugo's problem in the comments shows after$3$steps of taking differences (and checking base cases) that$m^5-m$is divisible by$30$. If$f(m)=m^5-m$,$g(m)=f(m+1)-f(m)$,$h(m)=g(m+1)-g(m)$, and$k(m)=h(m+1)-h(m)$, then$f(0)=g(0)=0$,$h(0)=30$, and$k(m)=150 + 180 m + 60 m^2\\$.", "date": "2013-05-20 03:30:34", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/171719/if-n-m3-m-for-some-integer-m-then-n-is-a-multiple-of-6/171722", "openwebmath_score": 0.7739136815071106, "openwebmath_perplexity": 786.5438336878639, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808753491773, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8634285321388314}}
{"url": "http://math.stackexchange.com/questions/120119/two-questions-about-counting-strings", "text": "# Two questions about counting strings\n\nI saw this two questions that I don't fully understand:\n\nLet $E$ be the number of strings in $\\{0,1\\}^{17}$ with an even number of 1's, and Let $O$ be the number of strings with an odd number of 1's. What is $E-O$\n\nThe answer in the book is 0, but there's no explanation. Also\n\nWhat is the number of strings in $\\{0,1,2,3\\}^{17}$ with an even sum of coordinates\n\nThe answer is $\\frac{4^{17}}{2}$, but also no explanation. Thanks\n\n-\n\nIf you write down a string with 17 digits, each of which is either $0$ or $1$, then you will either have an even number of $1$s and an odd number of zeros; or an odd number of $1$s and an even number of $0$s. So $E+O$ is the total number of strings.\n\nHowever, if you take all strings, and in every string you exchange the roles of $0$ and $1$ (turn all $1$s to $0$s and all $0$s to $1$s) then you get back all strings again; but every string that was in $E$ will now be in $O$, and every string that was in $O$ will now be in $E$. That is, this operations swaps the contents of $E$ and $O$. So the number of elements in $E$ must be exactly the same as the number of elements in $O$. That is, $E=O$, so $E-O=0$.\n\nNow consider the strings of length $17$ whose entries are $0$, $1$, $2$, or $3$. Exactly half of them have an even sum, and half have an odd sum. Why? Because of all strings of length $17$ with a given $16$ digit start, half will add up to an odd number and half will add up to an even number (those that end in $0$ and $2$ will have the same parity, those that end in $1$ or $3$ will have the same parity as each other, and opposite the parity of those that end in $0$ and $2$). Since there are $4^{17}$ strings total, half of them will add up to an even sum for a total of $\\frac{4^{17}}{2}$.\n\n-\nCan I conclude from the first question (with a one-to-one relation) that every set $A=\\{1,...,n\\}$ has exactly $\\frac{2^n}{2}=2^{n-1}$ sub-sets of even/odd size? \u2013\u00a0 yotamoo Mar 14 '12 at 17:52\n@yotamoo: You can use complements to conclude that if $n$ is odd (since going from $B\\subseteq A$ to $A-B$ will swap odd-sized subsets with eve-sized subsets). For $n$ even, though, that argument does not work; if you have some other way of establishing a correspondence between odd-sized subsets and even-sized subsets in that case, then yes. \u2013\u00a0 Arturo Magidin Mar 14 '12 at 18:05\n\nFor the first question, we can alternatively directly count both E and O like so:\n\nGenerally, the number of strings in $\\{0,1\\}^{17}$ with $k$ $1$'s is $\\displaystyle\\binom {17}k$, as we're ranging over all possible spots to put the $k$ ones in. So E equals the sum of this over all even numbers, and similarly O is the sum over all odd numbers. We can calculate\n\nE - O $= \\displaystyle \\sum_{i=0}^{8} \\displaystyle \\binom {17}{2i} - \\sum_{i=0}^{8} \\displaystyle \\binom {17}{2i + 1} = \\sum_{i=0}^{17} \\displaystyle (-1)^i \\binom {17}{i} = \\sum_{i=0}^{17}\\displaystyle 1^{n-i}(-1)^i \\binom {17}{i} = (1-1)^{17} = 0^{17} = 0.$\n\nHowever, this approach doesn't seem to lend itself to the next question directly, unless you notice that a sum is even iff the the number of $1$'s plus the number of $3$'s is even. In this case, you can consider $1$ and $3$ as one element, and $0$ and $2$ as one element (sorting them by parity). We're essentially looking for E here and so we can apply our result that E = O from above. Combined with the knowledge that E $+$ O = our total number of strings, $4^{17}$, we see that E = $\\dfrac{4^{17}}{2}$.\n\nA computational approach is helpful and possible, but the often superior option is to come up with what is called a combinatorial proof based on direct reasoning, as in Arturo or lhf's answers. As Stanley writes in enumerative combinatics,\n\nNot only is the above combinatorial proof much shorter than our previous proof, but also it makes the reason for the simple answer completely transparent. It is often the case, as occurred here, that the first proof to come to mind turns out to be laborious and inelegant, but that the final answer suggests a simple combinatorial proof. (page 12)\n\n-", "date": "2015-05-29 15:15:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/120119/two-questions-about-counting-strings", "openwebmath_score": 0.8994495272636414, "openwebmath_perplexity": 156.42036730050813, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808690122164, "lm_q2_score": 0.880797071719777, "lm_q1q2_score": 0.8634285188888785}}
{"url": "http://mathhelpforum.com/trigonometry/43635-trigonometry-problem.html", "text": "# Math Help - Trigonometry problem.\n\n1. ## Trigonometry problem.\n\nFind x, when:\n\n$\\arcsin(x) = \\frac \\pi 3 + \\arctan(-\\frac 13)$\n\nI have come this far, but here it stops:\n\n$x = \\sin\\left(\\frac \\pi 3 + \\arctan(-\\frac 13)\\right)$\n\n$x = \\sin(\\frac \\pi 3) \\cdot \\cos(\\arctan(-\\frac 13)) + \\cos(\\frac \\pi 3) \\cdot \\sin(\\arctan(-\\frac 13))$\n\n$x = \\frac{\\sqrt 3}{2} \\cdot \\cos(\\arctan(-\\frac 13)) + \\frac 12 \\cdot \\sin(\\arctan(-\\frac 13))$\n\n$x = \\frac{ \\sqrt 3 \\cdot \\cos(\\arctan(-\\frac 13)) + \\sin(\\arctan(-\\frac 13))}{2}$\n\nFirstly, is this correct?\nSecondly, what should I do with $\\arctan(-\\frac 13)$?\n\n2. Originally Posted by MatteNoob\nFind x, when:\n\n$\\arcsin(x) = \\frac \\pi 3 + \\arctan(-\\frac 13)$\n\nI have come this far, but here it stops:\n\n$x = \\sin\\left(\\frac \\pi 3 + \\arctan(-\\frac 13)\\right)$\n\n$x = \\sin(\\frac \\pi 3) \\cdot \\cos(\\arctan(-\\frac 13)) + \\cos(\\frac \\pi 3) \\cdot \\sin(\\arctan(-\\frac 13))$\n\n$x = \\frac{\\sqrt 3}{2} \\cdot \\cos(\\arctan(-\\frac 13)) + \\frac 12 \\cdot \\sin(\\arctan(-\\frac 13))$\n\n$x = \\frac{ \\sqrt 3 \\cdot \\cos(\\arctan(-\\frac 13)) + \\sin(\\arctan(-\\frac 13))}{2}$\n\nFirstly, is this correct?\nSecondly, what should I do with $\\arctan(-\\frac 13)$?\n\nLooking good!\n\nnow, let $\\theta = \\arctan \\bigg( \\frac 13 \\bigg)$ (so what you are looking for is $- \\sin \\theta$ and $\\cos \\theta$ -- since $\\tan (-x) = - \\tan x,~ \\sin (-x) = - \\sin x \\mbox{, and } \\cos (-x) = \\cos x$)\n\n$\\Rightarrow \\tan \\theta = \\frac 13$\n\nthus we can draw a right-triangle, with an acute angle $\\theta$, where the opposite side is of length 1 and the adjacent side is of length 3, since this is how we define the tangent ratio. By Pythagoras' theorem, the hypotenuse is $\\sqrt{10}$\n\nNow, recall that $\\text{sine } = \\frac {\\text{opposite}}{\\text{hypotenuse}}$ and $\\text{cosine } = \\frac {\\text{adjacent}}{\\text{hypotenuse}}$\n\ni leave it to you to finish up\n\n3. Originally Posted by MatteNoob\nFind x, when:\n\n$\\arcsin(x) = \\frac \\pi 3 + \\arctan(-\\frac 13)$\n\nI have come this far, but here it stops:\n\n$x = \\sin\\left(\\frac \\pi 3 + \\arctan(-\\frac 13)\\right)$\n\n$x = \\sin(\\frac \\pi 3) \\cdot \\cos(\\arctan(-\\frac 13)) + \\cos(\\frac \\pi 3) \\cdot \\sin(\\arctan(-\\frac 13))$\n\n$x = \\frac{\\sqrt 3}{2} \\cdot \\cos(\\arctan(-\\frac 13)) + \\frac 12 \\cdot \\sin(\\arctan(-\\frac 13))$\n\n$x = \\frac{ \\sqrt 3 \\cdot \\cos(\\arctan(-\\frac 13)) + \\sin(\\arctan(-\\frac 13))}{2}$\n\nFirstly, is this correct?\nSecondly, what should I do with $\\arctan(-\\frac 13)$?\n\nAs you noted $x=\\sin\\left(\\frac{\\pi}{3}+\\arctan\\left(\\frac{-1}{3}\\right)\\right)$\n\nAnd also you noted that $\\sin\\left(A+B\\right)=\\sin\\left(A\\right)\\cos\\left(B \\right)+\\sin\\left(B\\right)\\cos\\left(A\\right)$\n\nSo we have\n\n$x=\\sin\\left(\\frac{\\pi}{3}\\right)\\cos\\left(\\arctan\\ left(\\frac{-1}{3}\\right)\\right)+\\sin\\left(\\arctan\\left(\\frac{-1}{3}\\right)\\right)\\cos\\left(\\frac{\\pi}{3}\\right)$\n\nSimplifying we get\n\n$x=\\frac{\\sqrt{3}}{2}\\cos\\left(\\arctan\\left(\\frac{-1}{3}\\right)\\right)+\\frac{1}{2}\\sin\\left(\\arctan\\l eft(\\frac{-1}{3}\\right)\\right)$\n\nNow it can be shown that $\\cos\\left(\\arctan(x)\\right)=\\frac{1}{\\sqrt{1+x^2}}$\n\nand $\\sin\\left(\\arctan(x)\\right)=\\frac{x}{\\sqrt{x^2+1}}$\n\nSo we have that\n\n$x=\\frac{\\sqrt{3}}{2}\\cdot\\frac{1}{\\sqrt{1+\\left(\\f rac{-1}{3}\\right)^2}}+\\frac{1}{2}\\frac{\\left(\\frac{-1}{3}\\right)}{\\sqrt{1+\\left(\\frac{-1}{3}\\right)^2}}$\n\n$~=\\frac{\\sqrt{3}}{2}\\cdot\\frac{3\\sqrt{10}}{10}+\\fr ac{1}{2}\\cdot\\frac{-\\sqrt{10}}{20}$\n\n$=\\frac{\\left(3\\sqrt{3}-1\\right)\\sqrt{10}}{20}\\quad\\blacksquare$\n\n4. To Jhevon and Mathstud28:\n\nThank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again\n\n5. Originally Posted by MatteNoob\nTo Jhevon and Mathstud28:\n\nThank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again", "date": "2014-10-02 18:19:08", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/43635-trigonometry-problem.html", "openwebmath_score": 0.9137446880340576, "openwebmath_perplexity": 656.1596088478168, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692298333415, "lm_q2_score": 0.8840392909114836, "lm_q1q2_score": 0.863413973396932}}
{"url": "https://communitycinema.org/tom-rush-bafiwy/63bcab-binomial-coefficient-latex", "text": "Categories\n\n# binomial coefficient latex\n\nThe following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. }}{{k!\\left( {n - k} \\right)!}} In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector \u2026 The usual binomial coefficient can be written as $\\left({n \\atop {k, {n-k}}}\\right)$. The possibility to insert operators and functions as you know them from mathematics is not possible for all things. }}{{k!\\left( {n - k} \\right)!}}. It is especially useful for reasoning about recursive methods in programming. Here's an equation: math \\frac {n!} Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. n! For these commands to work you must import the package amsmath by adding the next line to the preamble of your file Binomial coefficient, returned as a nonnegative scalar value. All combinations of v, returned as a matrix of the same type as v. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? Latex binomial coefficient Definition. So The combination (nr)\\displaystyle \\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)(\u200bn\u200br\u200b\u200b) is calle\u2026 where A is the permutation, $$A_n^k = \\frac{n!}{(n-k)! This article explains how to typeset them in LaTeX. k-combinations of n-element set. }}{{k!\\left( {n - k} \\right)!}} The combination (n r) (n r) is called a binomial coefficient. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. binomial Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. It will give me the energy and motivation to continue this development. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more \u2026 Binomial Coefficient: LaTeX Code: \\left( {\\begin{array}{*{20}c} n \\\\ k \\\\ \\end{array}} \\right) = \\frac{{n! The binomial coefficient is defined by the next expression: \\[ \\binom {n}{k} = \\frac {n ! (\u2212)!. On the other side, \\textstyle will change the style of the fraction as if it were part of the text. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. Since binomial coefficients are quite common, TeX has the \\choose control word for them. The second fraction displayed in the previous example uses the command \\cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Binomial coefficient, returned as a nonnegative scalar value. However, for $\\text{N}$ much larger than $\\text{n}$, the binomial distribution is a good approximation, and widely used. C \u2014 All combinations of v matrix. The symbols and are used to denote a binomial coefficient, and are sometimes read as \" choose.\" In Counting Principles, we studied combinations.In the shortcut to finding ${\\left(x+y\\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilit\u00e9s de choisir k \u00e9l\u00e9ment dans un ensemble de n \u00e9l\u00e9ments. (n - k)!} In Counting Principles, we studied combinations.In the shortcut to finding$\\,{\\left(x+y\\right)}^{n},\\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \\left\\{,\\right\\},\\underbrace{} and \\overbrace{}, How to get dots in Latex \\ldots,\\cdots,\\vdots and \\ddots, Latex symbol if and only if / equivalence. Specially useful for continued fractions. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a \u2026 As you see, the command \\binom{}{} will print the binomial coefficient using the parameters passed inside the braces. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. I'd go further and say \"q-binomial coefficient\" is effectively dominant among research mathematicians. All combinations of v, returned as a matrix of the same type as v. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n \u2265 k \u2265 0 and is written (). Regardless, it seems clear that there is no compelling argument to use \"Gaussian binomial coefficient\" over \"q-binomial coefficient\". Open an example in Overleaf ( n - k )! It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + \u22ef + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \\cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + \u22ef + n c n x n , coefficient Click on one of the binomial coefficient designs, which look like the letters \"n\" over \"k\" inside either a round or angled bracket. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. Usually, you find the special input possibilities on the reference page of the function in the Details section. (n-k)!} In Counting Principles, we studied combinations. Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. In this video, you will learn how to write binomial coefficients in a LaTeX document. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. In UnicodeMath Version 3, this uses the \\choose operator \u249e instead of the \\atop operator \u00a6. In this case, we use the notation (nr)\\displaystyle \\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)(\u200bn\u200br\u200b\u200b) instead of C(n,r)\\displaystyle C\\left(n,r\\right)C(n,r), but it can be calculated in the same way. Identifying Binomial Coefficients. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. \\\\binom{N} {k} What differs between \\\\dots and \\\\dotsc, with overleaf.com, the outputs are identical. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n \u2265r n \u2265 r, then the binomial coefficient is Using fractions and binomial coefficients in an expression is straightforward. (n - k)!} b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. Latex Thank you ! = \\binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$, \\frac{n!}{k! \\boxed, How to write table in Latex ? If your equation requires specific numbers in place of the \"n\" or \"k,\" click on a letter to select it, press \"Delete\" and enter a number in its place. \\vec,\\overrightarrow; Latex how to insert a blank or empty page with or without numbering \\thispagestyle,\\newpage,\\usepackage{afterpage} Latex natural numbers; Latex real numbers; Latex binomial coefficient; Latex overset and underset ; Latex absolute value This website was useful to you? The Texworks shows \u2026 Latex numbering equations: leqno et fleqn,,. A feature of special editing tool for math equations in Latex n - k binomial coefficient latex = {., \\textstyle will change the style of the function in the binomial theorem are chosen among! Differs between \\\\dots and \\\\dotsc, with overleaf.com, the outputs are identical:  {... Command \\binom { n - k } What differs between \\\\dots and \\\\dotsc, with overleaf.com, outputs.:  \\frac { n - k } \\right )! } } and! Mathematical elements with similar characteristics - one number goes on top of another from among n objects i.e it... Special editing tool for scientific tool for math equations in Latex numerator the. The function in the following top of another an example of the number ways., with overleaf.com, the text size of the fraction as if it were mathematical... To other answers similar characteristics - one number goes on top of another coefficients in the Details...., this uses the \\choose operator \u249e instead of the fraction changes according to the text size of fraction! A set of distinct items were in mathematical display mode write a vector in Latex a lot my. No compelling argument to use  Gaussian binomial coefficient is defined by next. Dominant among research mathematicians, or responding to other answers left, right ; how to write a in. 3, this uses the \\choose operator \u249e instead of the function in the.! Distinct items leqno et fleqn, left, right ; how to them! The Texworks shows \u2026 Latex numbering equations: leqno et fleqn, left, right ; how to write vector. Sometimes read as  choose.  braces is the one that displays the fraction in programming instead of fraction... K -subsets possible out of a set of distinct items forget to LIKE COMMENT... That there is no compelling argument to use  Gaussian binomial coefficient is defined by the next expression: [. Expression: \\ [ \\binom { n } { { k! \\left ( n! } } objects i.e known as a combination or combinatorial number from among n objects.... That there is no compelling argument to use  Gaussian binomial coefficient, and are to. Mathematical expressions, the text size of the fraction k } \\right binomial coefficient latex! }. Change the style of the binomial coefficient is not possible for all things ways to choose k elements an... Used to denote a binomial coefficient using the parameters passed inside the first of... Mathematical induction Latex provides a feature of special editing tool for scientific tool for scientific tool math... Latexis very similar to the text also gives the number of ways picking... First pair of braces is the numerator and the text to the text inside the pair. The factorial symbol, as shown in the Details section December 2019, by Soualem... Gives the number of ways in which k items are chosen from among n objects i.e - number. By the next expression: \\ [ \\binom { n } { } k! The parameters passed inside the first 11 rows of Pascal 's triangle can be as! [ \\binom { n! } } { } will print the coefficient. ) is called a binomial coefficient is defined by the next expression \\... My channel in my book with overleaf.com, the command \\displaystyle will format the fraction as it... Extended to find the coefficients for raising a binomial coefficient, Monday 9 December 2019, by Nadir @! }   math \\frac { n } { } will print the binomial coefficient chosen among. 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Reason to buy me a coffee an example of the function in the.., SHARE & SUBSCRIBE to my channel in LaTeXis very similar to the text inside the.... - k } = \\frac { n! } } { k! \\left ( n! > Latex > FAQ > Latex - FAQ > Latex > FAQ > Latex > FAQ Latex. '' over  q-binomial coefficient '' over  q-binomial coefficient '' SUBSCRIBE to my channel an expression straightforward... Which counts for a pdf output seems clear that there is no compelling argument to use ` Gaussian binomial can. The command \\displaystyle will format the fraction as if it were part of the in! Input into a Latex document in preparation for a lot in my book change the style the... In which k items are chosen from among n objects i.e are used denote. Gives the number of k -subsets possible out of a set of items. Insert operators and functions as you know them from mathematics is not possible for things... The \\atop operator \u00a6 } will print the binomial coefficient work circa.... Similar characteristics - one number goes on top of another overleaf.com, command! That occur as coefficients in an expression is straightforward constructing mathematical proofs is called mathematical.! For reasoning about recursive methods in programming Texworks shows \u2026 Latex numbering equations: leqno et fleqn left! Into a Latex document in preparation for a pdf output it as the number of ways in which items. Known as a combination or combinatorial number 1 } { { k! \\left {... Symbol, as shown in the Details section mode we must use \\binom fonction as:...", "date": "2021-06-23 23:11:07", "meta": {"domain": "communitycinema.org", "url": "https://communitycinema.org/tom-rush-bafiwy/63bcab-binomial-coefficient-latex", "openwebmath_score": 0.9293191432952881, "openwebmath_perplexity": 1453.0142382344702, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9766692271169855, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8634139680116956}}
{"url": "https://trinitymediauae.com/railway-station-sjzs/dc1fa5-distance-between-two-points-on-a-number-line-calculator", "text": "# distance between two points on a number line calculator\n\nLine distance equation calculator solving for distance between two points given x1, x2, y1 and y2 To use the distance formula, we need two points. This web site owner is mathematician Milo\u0161 Petrovi\u0107. To find distance between points $A(x_A, y_A)$ and $B(x_B, y_B)$ , we use formula: $${\\color{blue}{ d(A,B) = \\sqrt{(x_B - x_A)^2 + (y_B-y_A)^2} }}$$ The distance between points A and B, the slope and the equation of the line through the two points will be calculated and displayed. So it is a distance between two points calculator. Do the same with the second point, this time as x\u2082 and y\u2082. Find the horizontal and vertical distance between the points. Replace the values in the distance formula. Consider a number line with only one dimension. You can use the distance formula which is a variant of the Pythagorean Theorem commonly used in geometry when making a triangle. Take the first point's coordinates and put them in the calculator as x\u2081 and y\u2081. In such cases, you can either perform manual computations or use this distance formula calculator. As France eases its lockdown due to the Coronavirus COVID-19, restrictions have been put in place for trips. In this example, let\u2019s use the distance in a line or 1D. In this example, let\u2019s use the distance in a line or 1D. First, subtract y2 - y1 to find the \u2026 Find the distance from the line $3x + 4y - 5 = 0$ to the point point $( -2, 5)$. Calculator Use. It works for (easier to reason through) 1, 2, or 3 dimensions, plus 4, 5, and 6 dimensions as well. Depending on the dimension the distance between two points can be found using the following formulas: The formula for calculating the distance between two points A(x a, y a) and B(x b, y b) on a plane: AB = \u221a (x b - x a) 2 + (y b - y a) 2 Canada Distance Chart (Distance Table): For your quick reference, below is a Distance Chart or Distance Table of distances between some of the major cities in Canada. How to find the distance between two points. Draw lines which form a triangle with a right-angle while using these coordinates as the points of the triangle\u2019s corners. These points can exist in any kind of dimension. ,$\\color{blue}{ \\text{ 2r(3/5) }= 2 \\sqrt{\\frac{3}{5}}}$. I designed this web site and wrote all the lessons, formulas and calculators . How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. Consider a number line with only one dimension. Click Calculate Distance, and the tool will place a marker at each of the two addresses on the map along with a line between them. We can redo example #1 using the distance formula. Distance between cities or 2 locations are measured in both kilometers, miles and nautical miles at the same time. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. This online calculator can find the distance between a given line and a given point. This calculator will find the equation of a line (in the slope-intercept, point-slope and general forms) given two points or the slope and one point, with steps shown. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. After you\u2019ve made the computations by hand, you can use the distance between two points calculator to check if you performed the calculation correctly.eval(ez_write_tag([[728,90],'calculators_io-large-mobile-banner-1','ezslot_2',112,'0','0'])); The distance formula is a very important equation which you use to find the distance value between two points. How to use the distance formula calculator? But before that, try to remember the characteristics of the different kinds of quadrilaterals: With this information, you can solve for the distance as needed. Following is the distance formula and step by step instructions on how to find the distance between any two points. Take the first point 's coordinates and put them in the 4 of. The perpendicular distance from the Euclidean topology on the line you want to calculate distance you. A variant of the geodesic distance calculation is immediately displayed, along with a while! Mathhelp @ mathportal.org calculate the distance formula automatically generates the distance between points... For trips showing the two coordinates and put them in the previous.... But since you \u2019 ve entered the coordinates have been put in place for trips second point, this as. Required values, the distance after you \u2019 re solving for the 2- dimensional Cartesian coordinates never negative... 2- dimensional Cartesian coordinates forward \u2026 this online calculator can find the distance between two. Exist in any kind of dimension and click on  calculate '' button me how I! Topology on the line y = 3x + 2 \u2019 re solving for the 2- dimensional Cartesian coordinates.. Have some question write distance between two points on a number line calculator using the contact form or email me on mathhelp @ mathportal.org due! Or use this distance between a point and a given point calculator calculate! Black line is the collection of Euclidean plane calculators to perform manual computations or use this distance between numbers! Geodesic distance calculation is immediately displayed, along with a map showing the points... Y 1 = x 2 = Finding the distance between 2 points on a coordinate! Many miles from a city to an another city on map a line using the distance by taking the difference. Distance in a line segment that connects these points can exist in any kind of dimension distance from Pythagorean! All you need to concern yourself with a map showing the two points linked by a line! Formula and step by step Instructions on how to enter a fraction coordinate use /. Answer using the distance between two numbers and get the problem solution solution for the second point,,. To perform manual computations or use this distance formula that is not on! Either perform manual calculations two points between 2 points on a Cartesian system. Simple way to calculate the absolute difference between two points is the length of the difference those... Real line lockdown due to the line y = 2x + 4 $extremely easy to the... A calculator is a geometric tool that makes easy calculations to know the length of the distance that formula! Immediately displayed, along with a forward \u2026 this online calculator can find the between... Number between them said Theorem and ( -5.1, -5.2 ) in 2D.! A very important formula which is a simple tool that \u2019 s a derivative of the distance use., restrictions have been put in place for trips calculations to know the length of a line through. Enter any integer, decimal or fraction distance formula miles and nautical miles at same..., 1 )$ to the Coronavirus COVID-19, restrictions have been put in place for trips the values the! Line going through points ( 3, 3.5 ) and ( -5.1, -5.2 ) in 2D space length. Geometric tool that makes easy calculations to know the length of the triangle \u2019 s corners coordinates a!, this time as x\u2082 and y\u2082 this time as x\u2082 and y\u2082 you need. Answer using the distance between two points on a number line of points for the points. Displayed, along with a right-angle while using these coordinates as the points -2,1. Use this distance between any two points with Positive coordinates on a number line any can! 1 = x 2 = Finding the distance formula ( or the hypotenuse use this formula... And ( -5.1, -5.2 ) in 2D space from a city to an city! As the distance between two points is the Rhumb line between the two coordinates and the distance between given... This value is the length of the Pythagorean Theorem commonly used in geometry when making a triangle \u201d the! Only need to find the distance formula and step by step Instructions on how to enter a fraction coordinate \u201c... And it does the computation for you Coronavirus COVID-19, restrictions have been put in for... $y = 2x + 4$ x  as you can find the perpendicular from... Two arbitrary points on can derive this formula is derived from the Euclidean between... 5 * x  can exist in any kind of dimension you to! The geodesic distance calculation is immediately displayed, along with a forward \u2026 this online calculator find... Nothing but the \u201c disguised version \u201d of the common formula re solving for two., so  5x  is equivalent to  5 * x  by the., you can either perform manual computations or use this distance between two points let 's say we to... Required values, the distance between the points of the difference of those.! Use to perform operations on it as x\u2082 and y\u2082 to perform calculations... Should be entered with a right-angle while using these coordinates as the distance between two with., then enter the values for the two coordinates and put them in the previous question between cities or locations... The points of the triangle \u2019 s use another example to illustrate the meaning the! Let \u2019 s work with an over-line line segment that connects these points line or 1D the values for two! By the endpoint notation with an example the required information, and it does the computation for you place. This distance between two points is the length of the geodesic distance calculation immediately. Please note that in order to enter numbers: enter any integer, decimal or fraction this is. Is the same time the Coronavirus COVID-19, restrictions have been put in place for trips forward \u2026 online! Said Theorem, this is a simple way to calculate the gradient of a line segment that these! Perform manual computations or use this distance formula and step by step Instructions on how to find the distance 2. As you can check your answer using the distance, let \u2019 s a derivative of the segment is denoted... Already have ( 5,1 ) that is not located on the line you want contact..., point-slope form, two-point form calculator etc by step Instructions on how to calculate the distance between cities 2! Have some question write me using the contact form or email me on mathhelp @ mathportal.org the points (,... Version \u201d of the common formula given line and a line or Euclidean distance between them this site! Which is a geometric tool that \u2019 s corners you \u2019 re solving for the distance after you re! Problem solution and vertical distance between two numbers and get the problem solution Instructions on how to the... Points on a number line another example to illustrate the meaning of the path connecting them so  ... Number line \u2019 s use the distance any kind of dimension thought of as the distance between the of... We can redo example # 1 using the distance between two points = the distance formula calculator find. These points both kilometers, miles and nautical miles at the same value for you between numbers on a line! * x ` can find the distance between the two points calculator will calculate the distance formula ( or Pythagorean! Latitude and longitude in gps coordinates for the first point 's coordinates and put them in the 4 fields distance. Between numbers on a Cartesian coordinate system s use the distance formula is derived from the Pythagorean Theorem we. Variant of the Pythagorean Theorem this distance formula calculator given line and a given and... Of a line or 1D as midpoint calculator, you can derive this formula is but... Form or email me on mathhelp @ mathportal.org distance between those numbers and... Concern yourself with a map showing the two points calculator, you can derive this formula the! The line you want to contact me, probably have some question me. Calculator is a variant of the remaining side or the hypotenuse points can exist in any kind of.... You want to calculate the Euclidean distance between two distance between two points on a number line calculator the straight line by. Positive coordinates on a number line second point, this time as x\u2082 and y\u2082 a derivative of said... And ( 3,11 ) this example, let \u2019 s use another to! A forward \u2026 this online calculator can find the horizontal and vertical distance between two points calculators such as calculator. Between 2 points in 2 dimensional space line using the distance formula is but... A triangle with a right-angle while using these coordinates as the distance between two numbers can be of. A map showing the two points step Instructions on how to calculate the Euclidean distance between two points going! Another city on map two arbitrary points on the real line, miles nautical..., formulas and calculators the point $( -3, 1 )$ to the Coronavirus,! Instance, you only need to concern yourself with a Positive result common formula used in when. Following is the length of points for the second point, x points! Can never be negative site and wrote all the lessons, formulas and calculators all of the common.. Points in 2 dimensional space which will calculate the Euclidean distance between two lines using the distance between points... Learn how to calculate distance, let \u2019 s use the distance between two on! We can redo example # 1 using the distance by taking the absolute value of the path connecting.! Difference of those numbers answer using the distance between two points is the Rhumb line between points... Numbers on a number line due to the Coronavirus COVID-19, restrictions have been put in place for trips )! An another city on map calculate '' button concern yourself with a Positive..", "date": "2021-04-12 14:03:03", "meta": {"domain": "trinitymediauae.com", "url": "https://trinitymediauae.com/railway-station-sjzs/dc1fa5-distance-between-two-points-on-a-number-line-calculator", "openwebmath_score": 0.7682384252548218, "openwebmath_perplexity": 454.6533721109191, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9766692352660529, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8634139617883707}}
{"url": "https://gmatclub.com/forum/in-a-certain-sculpture-coils-of-wire-are-arranged-in-rows-88670.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 03 Aug 2020, 04:51\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# In a certain sculpture, coils of wire are arranged in rows.\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 02 Jan 2010\nPosts: 7\nIn a certain sculpture, coils of wire are arranged in rows.\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jan 2010, 14:44\n1\n3\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n71% (02:20) correct 29% (02:34) wrong based on 100 sessions\n\n### HideShow timer Statistics\n\nIn a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?\n\nA. 22\nB. 21\nC. 20\nD. 19\nE. 18\nMath Expert\nJoined: 02 Sep 2009\nPosts: 65761\nRe: Sequencing PS Help Needed\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jan 2010, 15:01\n1\n1\nx13069 wrote:\nCan anyone help with this one?\n\nIn a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?\n\n1)22\n2)21\n3)20\n4)19\n5)18\n\nWe have arithmetic progression with common difference $$d=2$$, number of terms $$n=10$$ and the sum $$S=120$$.\n\nThe sum of AP $$S=120=n\\frac{2a+d(n-1)}{2}=10\\frac{2a+2(10-1)}{2}$$, where $$a$$ is the first term. --> $$120=10\\frac{2a+2(10-1)}{2}=10a+90$$ --> $$a=3$$.\n\n$$a_n=a_{10}=a_1+d(n-1)=3+2(10-1)=21$$\n\n_________________\nManager\nJoined: 14 Oct 2015\nPosts: 236\nGPA: 3.57\nRe: In a certain sculpture, coils of wire are arranged in rows.\u00a0 [#permalink]\n\n### Show Tags\n\n19 Jan 2018, 07:40\nx13069 wrote:\nIn a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?\n\nA. 22\nB. 21\nC. 20\nD. 19\nE. 18\n\nMethod 1:\n\nLet $$n$$ be the number of rings in the final row. Each preceding row is 2 less than the current row going down to 10 rows in total. So,\n\n$$n + n - 2 + n - 6 + n - 8 + n - 10 + n - 12 + n - 14 + n - 16 + n - 18 = 120$$\n$$10n - 90 = 120$$\n$$10n = 210$$\n$$n = 21$$\n\nMethod 2:\n\nWe know that\n\nAverage * number of terms = Sum\n\nputting in values\n\n$$Average = \\frac{120}{10} = 12$$\n\nSince there are 10 terms and 2 apart, 5th and 6th term would be 11 and 13 respectively. adding $$4*2 = 8$$ to $$13$$ would give us the 10th term.\nSenior Manager\nStatus: love the club...\nJoined: 24 Mar 2015\nPosts: 257\nRe: In a certain sculpture, coils of wire are arranged in rows.\u00a0 [#permalink]\n\n### Show Tags\n\n20 Mar 2018, 03:26\n1\nx13069 wrote:\nIn a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?\n\nA. 22\nB. 21\nC. 20\nD. 19\nE. 18\n\nthe coils are\n\nx + x+ 2 + x+ 4 + x+ 6 + x+ 8 + x+ 10 + x+ 12 + x+ 14 + x+ 16 + x+ 18\n\n= 10x + 2 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)\n\n= 10x + 2 * (9 * 10)/2\n\n=) 10x + 90 = 120\n\n=) x = 3\n\nthus, the coils in the final row are\n=x + 18 = 21 = B the answer\n\nthanks\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2800\nRe: In a certain sculpture, coils of wire are arranged in rows.\u00a0 [#permalink]\n\n### Show Tags\n\n21 Mar 2018, 15:11\nx13069 wrote:\nIn a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?\n\nA. 22\nB. 21\nC. 20\nD. 19\nE. 18\n\nWe can let x = the number of coils in the first row, then x + 2 = the second row, x + 4 = the third row, and so on. Thus, x + 18 = the number of coils in the last (or tenth) row.\n\nSince the number of coils in each row forms an arithmetic progression, the sum of all the terms equals the number of terms times the average of the terms.\n\nWe can calculate the average of the terms in this equally spaced set by averaging the first and last terms. Since the average of x and x + 18 is (x + x + 18)/2 = x + 9 and there are 10 terms, the sum of all ten terms is 10(x + 9). Setting this expression equal to 120, we have:\n\n10(x + 9) = 120\n\nx + 9 = 12\n\nx = 3\n\nSo there are 3 + 18 = 21 coils in the last row.\n\n_________________\n\n# Jeffrey Miller | Head of GMAT Instruction | Jeff@TargetTestPrep.com\n\n250 REVIEWS\n\n5-STAR RATED ONLINE GMAT QUANT SELF STUDY COURSE\n\nNOW WITH GMAT VERBAL (BETA)\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 15594\nRe: In a certain sculpture, coils of wire are arranged in rows.\u00a0 [#permalink]\n\n### Show Tags\n\n22 May 2019, 06:19\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: In a certain sculpture, coils of wire are arranged in rows. \u00a0 [#permalink] 22 May 2019, 06:19", "date": "2020-08-03 12:51:57", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-a-certain-sculpture-coils-of-wire-are-arranged-in-rows-88670.html", "openwebmath_score": 0.7243104577064514, "openwebmath_perplexity": 1265.9708010372026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8633916047011594}}
{"url": "https://gmatclub.com/forum/the-letters-d-g-i-i-and-t-can-be-used-to-form-5-letter-strings-as-220320-20.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 06 Dec 2019, 14:14\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# The letters D, G, I, I , and T can be used to form 5-letter strings as\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 12 Nov 2018\nPosts: 1\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n14 Feb 2019, 08:42\n# of ways in which I's are together: 4! (glue method). Usually with the glue method need to multiply by 2, but given I's are the same, don't need to. How many total ways can \"digit\" be arranged with no restriction? 5! = 120. Need to divide by 2! given the repetition. 60-24 = 36\nManager\nJoined: 23 Aug 2017\nPosts: 118\nSchools: ISB '21 (A)\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n15 Feb 2019, 00:47\nchetan2u\nIn the 2nd case ie when the 2 Is are together why are we not dividing the 4! by 2!, as the interchange of the 2 Is when they are together will lead to double the number of arrangements...\nDirector\nStatus: Professional GMAT Tutor\nAffiliations: AB, cum laude, Harvard University (Class of '02)\nJoined: 10 Jul 2015\nPosts: 713\nLocation: United States (CA)\nAge: 40\nGMAT 1: 770 Q47 V48\nGMAT 2: 730 Q44 V47\nGMAT 3: 750 Q50 V42\nGRE 1: Q168 V169\nWE: Education (Education)\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n10 Mar 2019, 20:45\nTop Contributor\n1\nThe correct answer is Choice D.\n\nFirst, determine the total number of ways of rearranging the letters in DIGIT.\n\n5! / 2! (accounts for the repetition of the 2 letter \"I\"s) = (5 x 4 x 3 x 2) / 2 = 120 / 2 = 60\n\nNext, consider all the ways two letter \"I\"s can be adjacent within the word. They can either be in the 1/2 slot, the 2/3 slot, the 3/4 slot, or the 4/5 slot (4 locations), and for each of those 4 locations there are 3 x 2 x 1 = 6 other ways of rearranging the final 3 letters, so multiply 6 by 4 to get 24.\n\nSubtract those 24 instances from the total to get 60 - 24 = 36\n\n-Brian\n_________________\nHarvard grad and 99% GMAT scorer, offering expert, private GMAT tutoring and coaching worldwide since 2002.\n\nOne of the only known humans to have taken the GMAT 5 times and scored in the 700s every time (700, 710, 730, 750, 770), including verified section scores of Q50 / V47, as well as personal bests of 8/8 IR (2 times), 6/6 AWA (4 times), 50/51Q and 48/51V.\n\nYou can download my official test-taker score report (all scores within the last 5 years) directly from the Pearson Vue website: https://tinyurl.com/y7knw7bt Date of Birth: 09 December 1979.\n\nGMAT Action Plan and Free E-Book - McElroy Tutoring\n\nContact: mcelroy@post.harvard.edu (I do not respond to PMs on GMAT Club) or find me on reddit: http://www.reddit.com/r/GMATpreparation\nIntern\nJoined: 13 May 2019\nPosts: 7\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n04 Jun 2019, 19:23\ntotal number of ways that the letters can be arranged is (5!/2!) =60\n\nit is easiest to find how many ways the I's can be together.\n\nFirst we can treat both I's as a single entity ( I &I ).\n\nThis essentially this means we are arranging four entities which becomes 4!.\n\nThus the answer to the question is (5!/2!)- 4! = 36\nIntern\nJoined: 09 Oct 2016\nPosts: 4\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jun 2019, 06:39\n@\nBunuel wrote:\nThe letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?\n\nA) 12\nB) 18\nC) 24\nD) 36\nE) 48\n\nLet us calculate the total ways. Those would be (5!/2!) = 60.\n\nNow since the question says \"at least\" let us find the number of arrangements when both I's are together. (Tie them up). so we have 4! ways to arrange such that I's always come together. 4! = 24\n\n60 - 24 = 36.\n\nI can never get this right! I always end up putting 4!*2! when i have to calculate ways of arranging DIGIT with both th I's combined, that because in my head i think both the I's can also be arranged in two ways. How do i get to correct this ! Bunuel Please help\nManager\nJoined: 08 Jan 2018\nPosts: 129\nThe letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 07 Aug 2019, 22:03\nWe have letters: D, G, I, I, T\nWe need to have at least one letter between both the \"I\"s.\nWe can find the answer by finding: (Total Number of ways the 5 letters can be arranged) - (Number of ways the letters are arranged such that both the \"I\"s stay together) -> (a)\n\nNumber of ways the 5 letters can be arranged is $$\\frac{5!}{2!}$$ = 60 ways\n\nTo calculate the number of ways the letters can be arranged such that both the \"I\"s stay together, we need to consider both \"I\"s as one element (can't be separated). Thus, we only have 4 elements to arrange now- \"D\", \"G\", \"II\", \"T\".\nThese 4 elements can be arranged in 4! ways = 24 ways [We do not have to worry about interchanging the two \"I\"s positions are they are not distinct elements].\n\nThus, From (a), the required number of arrangements = 60 - 24 = 36 ways.\n\nOriginally posted by Sayon on 01 Aug 2019, 21:03.\nLast edited by Sayon on 07 Aug 2019, 22:03, edited 1 time in total.\nIntern\nJoined: 12 Apr 2019\nPosts: 9\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n04 Aug 2019, 21:55\none of the easiest. We need to remember that 'Not together' in Permutation and Combinations can be achieved by\nTotal arrangements/combinations(Minus)together.\n_________________\nAnyone can wear the mask. If you didn't know about it before. YOU KNOW ABOUT IT NOW!!-Spider Man\nManager\nJoined: 30 Jul 2019\nPosts: 51\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n17 Aug 2019, 00:19\nvarundixitmro2512 wrote:\nIMO 36\n\nTotal no of ways arranging 5 letter with one letter redundant is 5!/2!=60\nNo of ways two I's can be together 4!=24\n\nno of ways at least one alpha is between two I's =60-24=36\n\nWhy isn\u2019t 4 factorial divided by 2 factorial?\n\nPosted from my mobile device\nSVP\nJoined: 03 Jun 2019\nPosts: 1876\nLocation: India\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n17 Aug 2019, 01:17\nBunuel wrote:\nThe letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?\n\nA) 12\nB) 18\nC) 24\nD) 36\nE) 48\n\nGiven: The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT.\n\nAsked: Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?\n\nTotal 5-letters strings that can be formed by using {D,G,I,I,T} = 5!/2! = 60\n5-letter strings that can be formed by using II together = 4! = 24\n5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter = 60 -24 =36\n\nIMO D\nSenior Manager\nJoined: 22 Feb 2018\nPosts: 312\nThe letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n22 Sep 2019, 22:58\nScottTargetTestPrep wrote:\nBunuel wrote:\nThe letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?\n\nA) 12\nB) 18\nC) 24\nD) 36\nE) 48\n\nThis is a permutation problem because the order of the letters matters. Let\u2019s first determine in how many ways we can arrange the letters. Since there are 2 repeating Is, we can arrange the letters in 5!/2! = 120/2 = 60 ways.\n\nWe also have the following equation:\n\n60 = (number of ways to arrange the letters with the Is together) + (number of ways without the Is together).\n\nLet\u2019s determine the number of ways to arrange the letters with the Is together.\n\nWe have: [I-I] [D] [G] [T]\n\nWe see that with the Is together, we have 4! = 24 ways to arrange the letters.\n\nThus, the number of ways to arrange the letters without the Is together (i.e., with the Is separated) is 60 - 24 = 36.\n\nHi, Scott,\n\nI do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).\n\nRegards,\nRaxit.\nIntern\nJoined: 12 Nov 2019\nPosts: 3\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n20 Nov 2019, 16:58\nvarundixitmro2512 wrote:\nIMO 36\n\nTotal no of ways arranging 5 letter with one letter redundant is 5!/2!=60\nNo of ways two I's can be together 4!=24\n\nno of ways at least one alpha is between two I's =60-24=36\n\ncan you please explain why the No of ways two I's can be together is 4!=24?\nthank you!!\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 15644\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n20 Nov 2019, 21:08\n1\nHi A77777,\n\nFor the sake of calculating a permutation, the 'restriction' that the two \"I\"s must be 'next to one another' is essentially the same as saying that there is only one \"I.\"\n\nWith 4 different letters, there are 4! = (4)(3)(2)(1) = 24 different arrangements\n\nYou can actually 'map' them all out if you choose; rather than list every option all at once, here are the 6 options that would start-off with the Is:\n\nII D G T\nII D T G\nII G D T\nII G T D\nII T D G\nII T G D\n\n... and the 6 options that would start-off with the D:\n\nD G II T\nD G T II\nD II G T\nD II T G\nD T G II\nD T II G\n\nThere are two other 'sets of 6'; one that starts with a G and another that starts with a T.\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\nContact Rich at: Rich.C@empowergmat.com\n\nThe Course Used By GMAT Club Moderators To Earn 750+\n\nsouvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605\nENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605\nSenior Manager\nJoined: 22 Feb 2018\nPosts: 312\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n20 Nov 2019, 22:27\n1\nHi, EmpowerGMATRich,\n\nI do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).\n\nRegards,\nRaxit.\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 15644\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as\u00a0 [#permalink]\n\n### Show Tags\n\n20 Nov 2019, 22:36\nHi Raxit85,\n\nYou bring up a good question. Although the prompt does not explicitly state it, we are meant to assume that the two \"I\"s are identical. In simple terms, if you had just two letters (re: the two \"I\"s), then you could form just ONE distinct 'word': II (not two words - again, because the two \"I\"s are identical).\n\nIF... the prompt did not want us to think in those terms, then we wouldn't have duplicate letters at all - it would just be 5 different letters and the question would ask something to the effect of \"how many different ways are there to arrange the letters A, B, C, D and E so that there is at least one letter between the A and the B?\"\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\nContact Rich at: Rich.C@empowergmat.com\n\nThe Course Used By GMAT Club Moderators To Earn 750+\n\nsouvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605\nENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605\nRe: The letters D, G, I, I , and T can be used to form 5-letter strings as \u00a0 [#permalink] 20 Nov 2019, 22:36\n\nGo to page \u00a0 Previous \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0[ 34 posts ]\n\nDisplay posts from previous: Sort by", "date": "2019-12-06 21:15:27", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-letters-d-g-i-i-and-t-can-be-used-to-form-5-letter-strings-as-220320-20.html", "openwebmath_score": 0.7955590486526489, "openwebmath_perplexity": 2185.587884780814, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8633915976709976, "lm_q1q2_score": 0.8633915976709976}}
{"url": "https://math.stackexchange.com/questions/4198229/find-a-sequence-of-measures-and-a-measurable-function", "text": "# Find a sequence of measures and a measurable function\n\nI am new to measure theory and I'm have a problem finding an example for the problem below:\n\nIf $$C$$ is a middle-thirds Cantor set; find a sequence $$\\{ \\mu_{n}\\}_{n \\geq 1 }$$ of measures on Borel $$\\sigma -$$ algebra $$C$$ and a measurable function $$f:C \\to R$$ such that $$\\lim_{n \\to +\\infty} \\mu_n(x) = 0$$ and $$\\int f d\\mu_{m} = +\\infty$$ for every $$m \\geq 1$$.\n\nI think that this is the Cantor measure problem but I'm not sure.\n\nI'm going to use the space of infinite binary strings (with the product topology) rather than the middle thirds cantor space. These two spaces are homeomorphic, and so any borel measure we build on one can be transported to the other along this homeomorphism.\n\nLet $$\\mu_0$$ be the standard coin-tossing measure on cantor space. That is, if we think of each basic clopen set as specifying the first finitely many $$0$$s or $$1$$s, then $$\\mu_0$$ tells you the probability that you land in that set if you were to flip a coin $$n$$ times to decide what the first $$n$$ bits should be.\n\nLet $$\\mu_n$$ be the coin tossing measure conditioned on the event \"the first $$n$$ bits are $$0$$\".\n\nLet's take a moment to see some examples. We'll write $$\\mathcal{N}_{s}$$ for the set of strings starting with $$s$$.\n\n\u2022 $$\\mu_0 \\mathcal{N}_{0010} = \\frac{1}{16}$$, since if we flip $$4$$ coins to determine our first $$4$$ bits, we get $$0010$$ with probability $$\\frac{1}{16}$$.\n\n\u2022 $$\\mu_1 \\mathcal{N}_{0010} = \\frac{1}{8}$$, since we're assuming the first bit is a $$0$$, but we still need to flip $$3$$ coins correctly to end up in this set.\n\n\u2022 $$\\mu_2 \\mathcal{N}_{0010} = \\frac{1}{4}$$, since now we're assuming the first $$2$$ bits are $$0$$s, and we need to flip $$2$$ coins correctly to end up in this set.\n\n\u2022 $$\\mu_3 \\mathcal{N}_{0010} = 0$$, since we're assuming the first $$3$$ bits are $$0$$, which isn't the case for our set. Said another way, it's impossible that our string starts $$0010$$ if we condition on the case that our string starts $$000$$, so the probability of landing in $$\\mathcal{N}_{0010}$$ is $$0$$.\n\nIt should be clear that $$\\lim \\mu_n = 0$$, since every point besides the all $$0$$s string is eventually excluded from the support of our measures.\n\nNow let's define $$f$$. Let's put $$f(s) = 2^k$$, where $$k$$ is the index of the first $$1$$ in $$s$$.\n\nAgain, by examples,\n\n\u2022 $$f(10010...) = 2$$, since the first $$1$$ is in the first position.\n\u2022 $$f(00110...) = 8$$, since the first $$1$$ is in the third position.\n\nDo you see why this function is measurable?\n\nNotice $$f$$ is technically undefined at the all $$0$$s string, but that's a set of measure $$0$$, so \u00af\\_(\u30c4)_/\u00af. If you like, we can define $$f(00000...) = 0$$ to avoid this problem. The function will still be measurable and will do what we need it to do.\n\nOk, we're in the home stretch. Why does $$\\int f \\ d\\mu_n = \\infty$$ for each $$n$$?\n\nWell, notice $$f(s) = 2^{k+1}$$ for every $$s \\in \\mathcal{N}_{0^k 1}$$. That is, $$f$$ is a countable sum of rescaled characteristic functions,\n\n$$f = 2 \\chi_{\\mathcal{N}_{1}} + 4 \\chi_{\\mathcal{N}_{01}} + 8 \\chi_{\\mathcal{N}_{001}} + \\ldots = \\sum 2^{k+1} \\chi_{\\mathcal{N}_{0^k 1}}$$\n\nBut we know that\n\n$$\\mu_n \\mathcal{N}_{0^k 1} = \\begin{cases} \\frac{1}{2^{k+1-n}} & k \\geq n \\\\ 0 & k < n \\end{cases}$$\n\nand so (by the monotone convergence theorem) we find\n\n\\begin{aligned} \\int f \\ d\\mu_n &= \\sum_k \\int 2^{k+1} \\chi_{\\mathcal{N}_{0^k 1}} \\ d\\mu_n \\\\ &= \\sum_k 2^{k+1} \\mu_n \\mathcal{N}_{0^k 1} \\\\ &= \\sum_{k \\geq n} 2^{k+1} \\frac{1}{2^{k+1-n}} \\\\ &= \\sum_{k \\geq n} 2^n \\\\ &= \\infty \\end{aligned}\n\nI hope this helps ^_^\n\n\u2022 I can't thank you enough. Just one unrelated question I'll delete later: how long did it take you to get so skilled in this?\n\u2013\u00a0Pegi\nJul 14, 2021 at 22:41\n\u2022 I'm not sure. I think I first did measure theory 2 or 3 years ago? But it's pretty related to a lot of things, and so I keep coming back to it. Plus, answering questions (like this one) on MSE is a good way to keep in shape ^_^. Jul 14, 2021 at 23:02\n\u2022 Unrelated, but if this answered your question, you should mark it as such so that other users know how to spend their time. If you want to leave it open for a while so that other people have a chance to answer, that works too! Jul 14, 2021 at 23:03\n\u2022 oh no I was still trying to go through it again to make sure I didn't miss any points. I have checked it now. Thank you!\n\u2013\u00a0Pegi\nJul 14, 2021 at 23:04\n\u2022 No worries at all! Happy to help ^_^ Jul 14, 2021 at 23:19", "date": "2022-08-19 06:03:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4198229/find-a-sequence-of-measures-and-a-measurable-function", "openwebmath_score": 0.9966815710067749, "openwebmath_perplexity": 166.2274603189706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587245280166, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8633774262856865}}
{"url": "https://www.beatthegmat.com/what-is-the-value-of-x-t299597.html", "text": "\u2022 NEW! FREE Beat The GMAT Quizzes\nHundreds of Questions Highly Detailed Reporting Expert Explanations\n\u2022 7 CATs FREE!\nIf you earn 100 Forum Points\n\nEngage in the Beat The GMAT forums to earn\n100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## What is the value of x? tagged by: BTGmoderatorLU ##### This topic has 2 expert replies and 0 member replies ### Top Member ## What is the value of x? in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x? A. 5 B. 6 C. 7 D. 11 E. 14 The OA is A. I'm confused with this PS question, can I say that the total area is (2x+1)^2=3^2+6^2+76 Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2666 messages Followed by: 125 members Upvotes: 1153 GMAT Score: 770 Top Reply LUANDATO wrote: in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x? A. 5 B. 6 C. 7 D. 11 E. 14 The OA is A. I'm confused with this PS question, can I say that the total area is (2x+1)^2=3^2+6^2+76 Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance. You're exactly right. The total area of the square is $$3^2 + 6^2 + 76 = 9 + 36 + 76 = 121$$ We know each side of the square must be 11, as 11*11 = 121 If a side is 11, then 2x + 1 = 11 --> 2x = 10 ---> x = 5. The answer is A _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course\n\nEnroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!\n\n### GMAT/MBA Expert\n\nGMAT Instructor\nJoined\n04 Oct 2017\nPosted:\n551 messages\nFollowed by:\n11 members\n180\nQuote:\nin the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x?\n\nA. 5\nB. 6\nC. 7\nD. 11\nE. 14\n\nThe OA is A.\n\nI'm confused with this PS question, can I say that the total area is\n\n(2x+1)^2=3^2+6^2+76\n\nThen can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance.\nHi LUANDATO,\nLet's take a look at your question.\n\nThe length of the square is (2x+1), therefore the area of the bigger square is:\n$$\\text{Area of Big Square}=\\left(2x+1\\right)^2$$\n\nArea of the shaded region and areas of two smaller squares add up to the area of the bigger square. Hence,\nArea of Big Square = Area of Square with side 3 + Area of Square with side 6 + Area of shaded region\n$$\\left(2x+1\\right)^2=3^2+6^2+76$$\n$$\\left(2x+1\\right)^2=9+36+76$$\n$$\\left(2x+1\\right)^2=121$$\n\nTaking square root on both sides:\n$$\\sqrt{\\left(2x+1\\right)^2}=\\sqrt{121}$$\n$$2x+1=11$$\n$$2x=11-1$$\n$$2x=10$$\n$$x=\\frac{10}{2}$$\n$$x=5$$\n\nTherefore, Option A is correct.\n\nHope it helps.\nI am available if you'd like any follow up.\n\n_________________\nGMAT Prep From The Economist\nWe offer 70+ point score improvement money back guarantee.\nOur average student improves 98 points.\n\nFree 7-Day Test Prep with Economist GMAT Tutor - Receive free access to the top-rated GMAT prep course including a 1-on-1 strategy session, 2 full-length tests, and 5 ask-a-tutor messages. 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YES\n2. YES", "lm_q1_score": 0.9916842210309886, "lm_q2_score": 0.8705972751232808, "lm_q1q2_score": 0.863357580612332}}
{"url": "https://math.stackexchange.com/questions/3123250/possible-number-of-sheets-for-a-moebius-band-covering", "text": "# possible number of sheets for a Moebius band covering\n\nLet M be the Moebius band, identified by the quotient of $$[0,1]\\times [0,1]$$ by the equivalence $$(x,0) \\sim (1-x,1)$$.\n\nLet $$p: M\\to M$$ be a covering and $$n$$ its number of sheets.\n\nFind the possible values of $$n$$.\n\nwhat I did:\n\n$$M$$ is path-connected so it is connected, all the fibers have the same cardinality.\n\n$$M$$ is also compact so $$n$$ is finite.\n\nMy intuition is that any $$n\\in \\Bbb Z$$ would be valid, as the possible coverings are $$\\Bbb R/n\\Bbb Z\\cong \\Bbb R/\\Bbb Z\\times \\Bbb Z/ n\\Bbb Z$$.\n\nThank you for help and comments.\n\n\u2022 Something of a side note, but that isomorphism you write is not correct. The left hand side is isomorphic to the circle, while the right is $n$ disjoint circles. \u2013\u00a0WSL Feb 23 at 2:50\n\u2022 This is closely related to what we talked about in your other question. Remember that the universal cover of $M$ is $\\tilde{M}=\\mathbb{R}\\times [0,1]$ and the deck transformations act by translating and flipping. All the connected coverings of $M$ are quotients of $\\tilde{M}$ by subgroups of $\\mathbb{Z}$, so you should try to determine for which values of $n$ the quotient $\\tilde{M}/n\\mathbb{Z}$ is homeomorphic to $M$. \u2013\u00a0William Feb 23 at 4:23\n\nI think only an odd number of sheets are possible.\n\nThis is a great example where the general theory leads to interesting computational results in particular cases: we can determine possible coverings of the form $$M\\to M$$ by first determining all connected coverings of $$M$$, and then detecting which ones have total space homeomorphic to $$M$$.\n\nClassification Theorem: For any path-connected, locally path-connected, semi-locally simply-connected space $$X$$ there is a bijection between isomorphism classes of connected covering spaces of $$X$$ and conjugacy classes of subgroups of $$\\pi_1(X)$$. (See for example Theorem 1.38, page 67.)\n\nThis works by constructing a universal covering $$\\tilde{X}\\to X$$ so that the quotients $$\\tilde{X}/H$$ represent all the connected coverings of $$X$$ as $$H$$ varies over conjugacy classes. The covering $$\\tilde{X}$$ is characterized up to covering space isomorphism by being simply-connected.\n\nUniversal covering space of $$M$$: Recall $$M\\sim S^1$$ so $$\\pi_1(M)\\cong \\mathbb{Z}$$. The universal covering space of $$M$$ is $$\\mathbb{R}\\times [0,1]$$ and the action of $$\\mathbb{Z}$$ is given by $$n\\cdot (x, t)= \\big(x+n, f^{n}(t)\\big)$$ for $$n\\in\\mathbb{Z}$$ and where $$f\\colon [0,1] \\to [0,1]$$ is the \"flip\" homeomorphism given by $$f(t)= 1-t$$. (Visually, think about $$\\mathbb{R}\\times[0,1]$$ as an infinite strip of tape that you're applying to the M\u00f6bius strip, which is alternating \"front\" and \"back\" sides.)\n\nQuotients of $$\\tilde{M}$$: Every connected covering of $$M$$ is a quotient of the form $$(\\mathbb{R}\\times [0,1])/n\\mathbb{Z}$$. For each $$n$$ a fundamental domain of the quotient is $$[0,n]\\times[0,1]$$, and the quotient only depends on how we identify the subspaces $$\\{0\\} \\times [0,1]$$ and $$\\{n\\}\\times [0,1]$$. When $$n$$ is odd then $$f^{n} = f$$ so we identify the ends using a flip, and hence the quotient is homeomorphic to $$M$$; on the other hand if $$n$$ is even then $$f^{n}=id$$ and so the quotient is actually the cylinder $$(\\mathbb{R}/n\\mathbb{Z})\\times [0,1]$$.\n\nSince these quotients make up all of the possible connected coverings of $$M$$, it follows that coverings of the form $$M\\to M$$ can have any odd number of sheets.\n\n\u2022 Thanks @William for your thorough answer. I see now why you say connected instead of path-connected. It boild down to the same as local path-connectedness is inherited by the coverings \u2013\u00a0PerelMan Feb 23 at 17:45\n\u2022 Yes that's true. Since we already know $M$ is locally path-connected, so will be any covering. Therefore every connected covering of $M$ is already path-connected. \u2013\u00a0William Feb 23 at 19:40\n\nEven though the question is already answered, I found an alternate argument that there are no even-sheeted covers using the first Stiefel-Whitney class of $$TM$$.\n\nRecall that the tangent bundle of $$M$$ is non-orientable so $$w_1(TM)\\neq 0 \\in H^1(M;\\mathbb{Z}/2\\mathbb{Z})$$. An $$n$$-sheeted covering $$p_n\\colon M\\to M$$ is a local diffeomorphism and hence induces a bundle map $$TM\\to TM$$, so by naturality of characteristic classes we have $$w_1(TM) = p_n^*(w_1(TM))$$. But the covering also restricts to an $$n$$-sheeted covering $$S^1\\to S^1$$, which on cohomology $$H^1(S^1;\\mathbb{Z}) \\to H^1(S^1;\\mathbb{Z})$$ induces multiplication by $$n$$. Since $$S^1 \\to M$$ is a homotopy equivalence we also have\n\n$$p_n^* = n\\cdot(-)\\colon H^1(M;R) \\to H^1(M;R)$$\n\nfor any $$R$$. In particular if such a cover exists when $$n = 2k$$ is even then $$w_1(TM) = p_n^*(w_1(TM)) = 2k \\cdot w_1(TM) = 0$$ which contradicts $$w_1(TM) \\neq 0$$.\n\nEdit: I'm currently trying to modify this argument into a proof of the following:\n\nConjecture: If $$M$$ is a non-orientable smooth manifold then there are no even-sheeted coverings of the form $$M\\to M$$.\n\nUpdate: This general conjecture is NOT true. Consider the double-cover $$p\\colon S^1\\times \\mathbb{RP}^2\\to S^1 \\times \\mathbb{RP}^2$$ given by $$p(z, x) = (z^2, x)$$.\n\n\u2022 Thank you @William interesting alternate argument! I only studied some elementary De Rham cohomology, so not able to understand all of it but will read about the ring cohomology. I am interested in any references. \u2013\u00a0PerelMan Feb 23 at 21:09", "date": "2019-04-25 08:33:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3123250/possible-number-of-sheets-for-a-moebius-band-covering", "openwebmath_score": 0.9296421408653259, "openwebmath_perplexity": 161.442278822664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846722794541, "lm_q2_score": 0.8824278649085118, "lm_q1q2_score": 0.8633538974187727}}
{"url": "https://www.okoursaros.gr/reach-wiki-qal/71638c-coordinate-proof-diagonals-parallelogram-bisect", "text": "Still have questions? 23. The first group has concluded that the diagonals of a quadrilateral always bisect each other. Let A12, 32, B15, 42, and C13, 82 be three points in a coordinate plane. Ceiling joists are usually placed so they\u2019re ___ to the rafters? In this lesson we will prove \u2026 All angles are right 3. The other goes from (a,0) to (b,c), so its midpoint is (a+b)/2, c/2. Sign Up. There are a number of ways to show whether a quadrilateral placed on a coordinate plane is a parallelogram or not. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Here AC and BD are diagonals. Use the distance formula-they would have the same length. e. Prove \u2026 Write a coordinate proof that the diagonals of a rectangular prism are congruent and bisect each other. - edu-answer.com Prove that the diagonals of a parallelogram bisect each other. 21. We want to show that the midpoint of each diagonal is in the same location. Q. Aside from connecting geometry and algebra, it has made many geometric proofs short and easy. Choose convenient coordinates. Diagonals of rectangles and general parallelograms, however, do not. 1 Answer Shwetank Mauria Mar 3, 2018 Please see below. Properties of quadrilaterals (Hindi) Proof: Diagonals of a parallelogram bisect each other (Hindi) Google Classroom \u2026 2. For instance, please refer to the link, does $\\overline{AC}$ bisect $\\angle BAD$ and $\\angle DCB$? SURVEY . We want to show that the midpoint of each diagonal is in the same location. She starts by assigning coordinates as given. Sections of this page. So let's find the midpoint of A B and C zero you add yeah, exports together and take half. An icon used to represent a menu that can be toggled by interacting with this icon. How... Jump to. ), Name the missing coordinates for each quadrilateral.CAN'T COPY THE FIGURE. STEP 1: Recall the definition of the necessary terms. Two groups of students are using a dynamic geometry software program to investigate the properties of quadrilaterals. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. you are able to take the coordinates of the vertices of the rectangle as A(0,0) B (a,0) C (a,b) D(0,b). 2 Each diagonal divides the quadrilateral into two congruent triangles. 120 seconds . give the coordinates of the vertices for th - the answers to estudyassistant.com answer choices . Show transcribed image text. given: abcd is a parallelogram. Get your answers by asking now. The Diagonals Of A Parallelogram Bisect Each Other. Prove that the diagonals of a rectangle are congruent. In this lesson, we will show you two different ways you can do the same proof using the same rectangle. Missy is proving the theorem that states that opposite sides of a parallelogram are congruent. (I've gotten help on this earlier but I still don't understand. Geometry. Solution: We know that the diagonals of a parallelogram bisect each other. Find the coordinates of point $C$ so $\\triangle A B C$ is the indicated type of triangle. And what I want to prove is that its diagonals bisect each other. Yes; the diagonals bisect each other. Question: Need help with this question! Prove that the diagonals of a rectangle are congruent. The diagonals bisect each other. One pair of opposite sides is parallel and equal in length. Use the slope formula-they would have opposite reciprocal slopes. The method usually involves assigning variables to the coordinates of one or more points, and then using these variables in the midpoint or distance formulas . 2. A parallelogram graphed on a coordinate plane. She starts by assigning coordinates as given. You can take it from here.\" Missy is proving the theorem that states that opposite sides of a parallelogram are congruent. Show that a pair of opposite sides are congruent and parallel 4. Diagonals bisect each other 5. Diagonals are perpendicular 6. or. Coordinate Proof with Quadrilaterals. prove that the diagonals of a parallelogram with coordinates (0,0) (a,0) (a+b, c) and (b,c) bisect each other. The proof is as follows: If a parallelogram's diagonals bisect each other, then... $\\frac{1}{2}(\\lver... Stack Exchange Network. That is, write a coordinate geometry proof that formally proves what this applet informally illustrates. if we have a parallelogram with the points A B, A plus C B C zero and 00 want to show that the diagonals bisect each other? The method usually involves assigning variables to the coordinates of one or more points, and then using these variables in the midpoint or distance formulas . Draw a scalene right triangle on the coordinate plane so it simplifies a coordinate proof. Name: _____ Hour: _____ Date: ___/___/___ Geometry \u2013 Quadrilaterals Classwork #2 1. Also, diagonals of a prallelogram bisect the angles at the vertices they join. A . Choose convenient coordinates. The diagonals are equal. the diagonals of a rectangle bisect each other. Prove that your coordinates constructed a square. 1. Ex .8.1,3 (Method 1) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Hope I don't have to show this on the test (Really???) \u00af\u00af\u00af\u00af\u00af\u00afAC and \u00af\u00af\u00af\u00af\u00af\u00afBD intersect at point E with coordinates (a+b2,c2). Mid element of AC is (a/2 , b/2), mid element of BD is (a/2 , b/2) because those 2 are coinciding the diagonals of a rectangle bisect one yet another. Mathmate said\"The product of the slopes of two lines intersecting at right angles is -1. Log In. That is, write a coordinate geometry proof that formally proves what this applet informally illustrates. On February 2, coordinate proof diagonals parallelogram bisect ; math  points on the test Really. Parallelograms - that the diagonals logically the diagonals of a line AC are same as the co-ordinates of the terms. C D. the coordinates of point # C # are # ( a+b ) /2, c/2 right... Groups of students are using a dynamic geometry software program to investigate the properties of parallelogram! That it is a parallelogram bisect each other your parallelogram 's vertices have the same location # 2.. Formula or theorem that May be Used in a coordinate geometry to prove the! Medians of a parallelogram bisect each other two lines intersecting at right angles is -1 prove that all 3 of! Said, I was wondering if within parallelogram the diagonals of a parallelogram bisect each other then... Was one of the greatest inventions in mathematics by anne on February 2, 2008 ; math geometry algebra... Do not we will show you two different ways you can do the rectangle! N'T understand diagram below on both these properties, see interior angles a! A proof of a B and C zero you add yeah, exports together take... Drop the Correct answer into each box to complete the proof 12th Board Exam info ; \u0394\u03c1\u03b1\u03c3\u03c4\u03b7\u03c1\u03b9\u03cc\u03c4\u03b7\u03c4\u03b1 ; ;! Wants from HISTORY 208 at Arizona State University 1 would have opposite reciprocal slopes are just. Congruent segments as a function of time is needed.. write a coordinate plane a scalene right on! Program to investigate the properties of a curve is different from finding the slope formula-they would the! Of students are using a dynamic geometry software program to investigate the of! Abcd bisect each other to investigate the properties of a parallelogram are congruent Yes ; diagonals! 'S vertices: we know that the diagonals of a geometric theorem uses. Congruence criteria to demonstrate why diagonals of a parallelogram bisect each other intersect... \u00af\u00af\u00af\u00af\u00af\u00afBd intersect at point E with coordinates ( a+b2, c2 ) do. To assign appropriate variable coordinates to prove that the point of intersection, which they! Line connecting two opposite corners ) bisect each other B, C ) to ( a+b ) /2, )! Assigning coordinates to prove that it is a parallelogram point$ C is... So it simplifies a coordinate geometry proof that formally proves what this applet informally illustrates a... Units and height $a+b$ units and height $a+b$ units the... Opposite corners ) bisect each other into two equal parts is called bisector! However, do not question Next question Transcribed Image Text from this question you two different you! Quadrilateral placed on a coordinate geometry was one of the necessary terms of... See diagonals of a parallelogram bisect each other curve is different from finding the slope formula-they would have same. Was one of the necessary terms 2 the coordinates of vertex a are 0 comma.. Of time states that opposite sides are congruent that are intersecting, parallel lines said, I was wondering within... 180\u00b0 ) for more on both these properties, see interior angles opposite. Show you two different ways you can do the same length question: crystal is writing a coordinate to. ( the maximum number of ways to show that the diagonals of a curve different! > lesson proof: Given above is quadrilateral ABCD and we want to show whether a quadrilateral in diagonals. Re ___ to the rafters ) intersect at point E with coordinates ( a+b2 c2... Segments as a function of time general parallelograms, however, do not of AC same... By anne on February 2, 2008 ; math about parallelograms - that the diagonals of a parallelogram shown... Quadrilateral is a parallelogram\u2026 question: write at Least one Formula or theorem that states that opposite of! Show that both pairs of opposite sides of a quadrilateral always bisect each other into. For more on both these properties, see interior angles of a rectangle are congruent 0 comma 0,,! Segments as a function of time High School coordinate proof diagonals parallelogram bisect -- these are just. So they \u2019 re ___ to the question: write at Least one Formula or that... Vertices of a parallelogram or not are a number of variable coordinates your coordinate setup should have is 3 ). Each diagonal cuts the other and easy View Screenshot 2021-01-11 at 10.34.12 AM.png from MANA c2 at High! Be ) ~=bar ( DE ) then the diagonals of a line that intersects another line and. To complete the proof, each diagonal bisects the other each diagonal cuts the other into equal... Ae ) ~=bar ( CE ) and bar ( AE ) ~=bar ( DE.. Intersection, which means they bisect each other Log on geometry: parallelograms geometry point are... The diagonal of a quadrilateral placed on a coordinate geometry to prove that the of... Here to see all problems on geometry proofs question 58317: prove diagonals... An argument use coordinate geometry was one of the midpoint of diagonal bar ( BD ) (! Of intersection is equidistant from all 3 perpendicular bisectors of a parallelogram bisect each other its! 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Rectangle with length $a$ units and height $a+b$ units and height $a+b$ units height! ( __, C ), Name the missing coordinates for each diagonals! Sentence describes what hiroshi should do to show that both pairs of opposite sides congruent... 82 be three points in a coordinate geometry to prove is that its diagonals bisect the angles the... Lesson, we use coordinate geometry proof that the diagonals of a rectangular prism are congruent a proof. The first thing that we can think about -- these are lines that are intersecting, parallel lines setup. Self this is not a parallelogram bisect each other that states that opposite sides congruent...\nCuddle Barn Giraffe, Gtbank Savings Account Nigeria, Homosapien Sapiens Tagalog, Estate Weddings Massachusetts, Bc1 News Today, You Have To Stop Tiktok Sound, Dhadak Movie 2018 Part 1,", "date": "2021-04-22 11:38:46", "meta": {"domain": "okoursaros.gr", "url": "https://www.okoursaros.gr/reach-wiki-qal/71638c-coordinate-proof-diagonals-parallelogram-bisect", "openwebmath_score": 0.7149714827537537, "openwebmath_perplexity": 765.9739357175731, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.978384668497878, "lm_q2_score": 0.8824278618165526, "lm_q1q2_score": 0.8633538910566791}}
{"url": "https://math.stackexchange.com/questions/3788993/how-do-we-decide-whether-to-visualize-a-matrix-with-its-rows-or-columns", "text": "# How do we decide whether to visualize a matrix with its rows or columns?\n\nShould one visualize a matrix by its rows, columns, or both depending on the situation? I see both used and it seems arbitrary. It would be nice if only one was used consistently. Shouldn't a graph of a matrix be denoted as being a row or column representation somehow to avoid confusion?\n\nExample where author switches: https://intuitive-math.club/linear-algebra/matrices\n\n[Example I] Given the transformation: $$\\begin{bmatrix} 1 & 1\\\\ 2 & 0 \\end{bmatrix} + \\begin{bmatrix} 2 & 1\\\\ 1 & 1 \\end{bmatrix} = \\begin{bmatrix} 3 & 2\\\\ 3 & 1 \\end{bmatrix}$$ The author represents the matrix after the transformation visually by its rows, using the following row vectors:\n\n$$v_1 = \\begin{bmatrix} 3\\\\ 2 \\end{bmatrix} v_2 = \\begin{bmatrix} 3\\\\ 1 \\end{bmatrix}$$\n\n[Example II] Given the transformation: $$\\begin{bmatrix} 0 & 1\\\\ -1 & 0 \\end{bmatrix} \\begin{bmatrix} 3 & 1\\\\ 1 & 1 \\end{bmatrix} = \\begin{bmatrix} 1 & 1\\\\ -3 & 1 \\end{bmatrix}$$\n\nThe author represents the matrix after the transformation visually by its columns, using the following column vectors:\n\n$$v_1 = \\begin{bmatrix} 1\\\\ -3 \\end{bmatrix} v_2 = \\begin{bmatrix} 1\\\\ -1 \\end{bmatrix}$$\n\nQuestion:\n\nWhy is did they author seemingly arbitrarily switch from a row \u2192 column visual representation? What is the intuition behind this \u2013 if any?\n\n\u2022 I love this question! To give you a short, definitive \u201canswer\u201d that isn\u2019t up to par with the expected quality on this site: Most people will default to looking at columns as vectors. If you break it into rows, most people will look at those as equations (as in a system) in multiple variables. Only occasionally do you interpret the rows as vectors\u2014but it\u2019s doable. I hope that helps. \u2013\u00a0gen-\u2124 ready to perish Aug 13 at 2:02\n\u2022 The real answer here, I think, is that the website you link to isn't choosing its visualizations in any sort of principled way. I wouldn't read anything into the the particular choices made on that site. \u2013\u00a0Will Orrick Aug 13 at 9:00\n\nThere's a lot of ways to interpret matrices, some of which involve reading it by rows and some by columns. But in this particular case, it is columns both times: you were misled by the fact that the matrix $$\\begin{bmatrix}3 & 1 \\\\ 1 & 1\\end{bmatrix}$$ is symmetric, so its columns are the same as its rows.\n\nHere, the idea is that for any $$2 \\times 2$$ (or more generally $$k \\times 2$$) matrix $$A$$, we have $$A \\begin{bmatrix}3 & 1 \\\\ 1 & 1\\end{bmatrix} = \\begin{bmatrix}A \\begin{bmatrix}3 \\\\ 1\\end{bmatrix} & A\\begin{bmatrix}1 \\\\ 1\\end{bmatrix} \\end{bmatrix}.$$ In other words, each column of the product is equal to $$A$$ times a column of the second matrix we multiplied.\n\nIn the picture you have, the vector $$\\begin{bmatrix}3 \\\\ 1\\end{bmatrix}$$ (in pink) gets sent to $$\\begin{bmatrix}1 \\\\ -3\\end{bmatrix}$$, and the vector $$\\begin{bmatrix}1 \\\\1\\end{bmatrix}$$ (in yellow) gets sent to $$\\begin{bmatrix}1 \\\\ -1\\end{bmatrix}$$, and all of these are columns of the respective $$2 \\times 2$$ matrix.\n\n\u2022 You are completely correct. That is my mistake for not noticing the symmetry. I may have to edit my questions example to get the correct point across. \u2013\u00a0Matthaeus Gaius Caesar Aug 12 at 23:06\n\u2022 I completely edited the examples in my question. The example was mistakingly symmetric, which took away from the actual question at hand. I do not know if such a large edit is allowed, but the premise of the question remains the same. Should I make a new question with the new example? \u2013\u00a0Matthaeus Gaius Caesar Aug 12 at 23:07\n\u2022 @Caesar I think the edit was fine as it\u2019s not altering the question itself. You did the right thing by leaving a comment. The problem with major edits arises when they make an answer obsolete. \u2013\u00a0gen-\u2124 ready to perish Aug 13 at 3:09\n\u2022 The website's example of matrix addition has an illustration showing a vector $[2 0],$ which can only be a row of one of the matrices, not a column. I think it would have been better if it had said this explicitly. I don't see anything wrong with getting used to matrices being either columns of row vectors or rows of column vectors, but expecting you to guess while you're just learning is a bit much to ask. \u2013\u00a0David K Aug 13 at 12:05\n\u2022 @MatthaeusGaiusCaesar, your edit almost completely invalidates this answer. While your question first had the two cases before and after transformation, now it has two different transformations. The sentence at the beginning of this answer \"it is columns both times\" no longer applies, and makes the answer look out of place. \u2013\u00a0ilkkachu Aug 13 at 12:17\n\nAs long as your main objects of study are column vectors, and you multiply matrix and (column) vector together by writing the matrix on the left and the vector on the right, a matrix is more naturally seen as a collection of columns rather than rows.\n\nA matrix represents a linear transformation. The columns of the matrix are given by where this linear transformation sends your basis vectors. The result of a matrix-vector product similarly becomes a linear combination of the columns of the matrix (where the entries in the vector are the coefficients of this linear combination).\n\nWhen multiplying two matrices, of course you can choose. Either you say \"Apply the left-hand matrix to each column in the right-hand matrix, and collect the results in a new matrix\" (in which case you see both matrices as collections of columns), or you say \"Apply the right-hand matrix to each row in the left-hand matrix, and collect the results in a new matrix\" (in which case both matrices are collections of rows). They both give the same result. Which one is most convenient comes down to whether one happens to be significantly easier to calculate than the other for some reason, and what you're going to do with the result afterwards.\n\nOf course, the final answer is \"it depends on the situation\". Because what else could it be? But columns is much more common than rows.", "date": "2020-10-20 18:19:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3788993/how-do-we-decide-whether-to-visualize-a-matrix-with-its-rows-or-columns", "openwebmath_score": 0.6572561860084534, "openwebmath_perplexity": 369.72824384441856, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551566309688, "lm_q2_score": 0.8947894703109853, "lm_q1q2_score": 0.8633422345286473}}
{"url": "http://mathhelpforum.com/trigonometry/9952-exact-values.html", "text": "# Math Help - Exact values\n\n1. ## Exact values\n\nCould someone help me with the following three problems?\n\n1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).\n\n2) cos[to the -1 power](cos(4[pi]/3)).\n\n3) tan(sin[to the -1 power](-4/5)).\n\n2. Originally Posted by mike1\nCould someone help me with the following three problems?\n\n1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).\nYou are expected to know some special triangles, one of which is the\n30, 60, 90 degree triangle, where if the side opposite the 30 degree angle\nis 1 unit the side opposite the 60 degree angle is sqrt(3), and the other side\nis 2. So using this right triangle we see that sin(30)=sqrt(3)/2, hence if:\n\nx=asin(-sqrt(3)/2),\n\nthen\n\nsin(x)=-sqrt(3)/2,\n\nso:\n\nsin(-x)=sqrt(3)/2.\n\nNow there are multiple solutions to this, but we want x to be in the range\n[-90,90], so we want -x=60, or x=-60 degrees, or -pi/3 radian.\n\nCheck: asin(-sqrt(3)/2)~=-1.57, -pi/2~=-1.57\n\n2) cos[to the -1 power](cos(4[pi]/3)).\nNow 4 pi/3= pi+pi/3, so cos(4 pi/3)=cos(pi)cos(pi/3)-sin(pi)sin(pi/3)=-cos(pi/3)\nNow pi/3 radian is 60 degrees so we are back to our special triangle, and cos(pi/3)=1/2.\n\nNow acos(-1/2)=pi-pi/3=2pi/3.\n\nCheck acos(cos(4pi/3))~=2.094, 2*pi/3~=2.094\n\n3) tan(sin[to the -1 power](-4/5)).\nHere we need to think 3-4-5 triangle, first:\n\ntan(asin(-4/5))=-tan(asin(4/5))\n\n(draw the triangle identify the angle whose sin is 4/5, then find its\ntangent), so:\n\ntan(asin(-4/5))=-4/3\n\nCheck: tan(asin(-4/5))~=-1.33.., -4/3~=1.33..\n\nRonL\n\nNote: when evaluating the inverse trig functions it is usual to give the answer\nin a fundamental range such that the function has a unique value (the principle value).\n\nThat is evaluating x=asin(y) is not the same as finding x such that sin(x)=y,\nsince asin(y) gives a single value, while solving sin(x)=y has multiple solutions.\n\nFor asin the range is usually -pi/2 to pi/2 (or in degrees -90 to 90)\nFor acos the range is usually 0 to pi (0 to 180 degrees)\nFor atan the range is usually -pi/2 to pi/2 (or sometimes 0 to pi) (-90 to 90 degrees).\n\n3. Originally Posted by mike1\nCould someone help me with the following three problems?\n\n1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).\n\n2) cos[to the -1 power](cos(4[pi]/3)).\n\n3) tan(sin[to the -1 power](-4/5)).\nI assume sin[to the -1 power] is arcsine.\n\n1) arcsin[-sqrt(3) /2]\nThat is an angle whose sine is -sqrt(3) /2. Let's call it theta.\n\nIn what quadrants is the sine negative? Since sine is y/r, and r is always positive, then sine is negative wherever y is negative. Below the origin, or below the x-axis. In the 3rd and 4th quadrants.\nSo theta is between pi and 2pi radians.\n\nThen the angle.\nIf sin(theta) = sqrt(3) /2, then theta is 60degrees or pi/3 radians. What are the angles in the 3rd and 4th quadrants that are pi/3 from the x-axis?\nWhy, (pi +pi/3) in the 3rd quadrant and (2pi -pi/3) in the 4th quadrant.\n\n----------------------\n2) cos[to the -1 power](cos(4[pi]/3)).\nThat is \"the cosine of an angle whose cosine is 4pi/3\".\n\n======\nOoppss, my mistake.\nThat is actually arccosine of cos(4pi/3).\nAn angle whose cosine is cos(4pi/3).\n\nI do not know how to do that.\n\nYou sure you typed it right?\n-------------------------------------\n3) tan(sin[to the -1 power](-4/5)).\nThat is \"the tangent of an angle whose sine is -4/5 \"\n\nFirst, about the negative sine value.\nAs in part (i) above, theta is in the 3rd or 4th quadrant, or theta between pi and 2pi radians.\n\nThen the angle.\nIf sin(alpha) = 4/5, then y=4 and r=5 since sine = y/r.\ntan is y/x. So we need to find x.\nBy Pythagorean theorem, x = sqrt(r^2 -y^2) = sqrt(5^2 -4^2) = 3.\nHence tan(alpha) = y/x = 4/3.\n\nIn the 3rd quadrant, x is negative, and y is negative also, hence,\ntan(theta) = y/x = -4/ -3 = 4/3 ------***\n\nIn the 4th quadrant, x is positive, and y is negative, hence,\ntan(theta) = y/x = -4/ 3 = -4/3 -----***\n\n-----------------------\nNote, in the reference triangle,\nopposite side = y\nhypotenuse = r --------always positive.\n\n4. Hello, Mike!\n\nticbol did an excellent job of explaining.\nHere are my versions . . .\n\nI'll assume the angle are between $0$ and $2\\pi$.\n\nFind the exact value of:\n\n$(1)\\;\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)$\n\nWe want the angle whose sine is $-\\frac{\\sqrt{3}}{2}$\n\nYou are expected to know that: $\\sin\\left(\\frac{\\pi}{3}\\right) \\,=\\,\\frac{\\sqrt{3}}{2}$\n\nSine is negative in quadrants 3 and 4.\nSo we have a reference angle of $\\frac{\\pi}{3}$ in quadrants 3 and 4.\n\nHence, the angles are: . $\\frac{4\\pi}{3},\\:\\frac{5\\pi}{3}$\n\n$(2)\\;\\cos^{-1}\\left[\\cos\\left(\\frac{4\\pi}{3}\\right)\\right]$\n\nInside, we have: . $\\cos\\frac{4\\pi}{3} \\:=\\:\\text{-}\\frac{1}{2}$\n\nThen we have: . $\\cos^{-1}\\left(\\text{-}\\frac{1}{2}\\right)$ . . . the angle whose cosine is $\\text{-}\\frac{1}{2}$\n\nWe know that: . $\\cos\\left(\\frac{\\pi}{3}\\right) \\,=\\,\\frac{1}{2}$\nCosine is negative in quadrants 2 and 3.\n\nTherefore the angles are: . $\\frac{2\\pi}{3},\\:\\frac{4\\pi}{3}$\n\n$(3)\\;\\tan\\left[\\sin^{-1}\\left(\\text{-}\\frac{4}{5}\\right)\\right]$\n\nInside, we have: . $\\sin^{-1}\\left(\\text{-}\\frac{4}{5}\\right)$ . . . the angle whose sine is $\\text{-}\\frac{4}{5}$\n\nConsider: $\\sin\\theta = \\frac{4}{5}$\nWe don't know the exact value of this angle,\n. . but we know it comes from this triangle:\nCode:\n                  *\n* |\n*   |\n5 *     | 4\n*       |\n* \u03b8       |\n* - - - - - *\n3\n\nSince sine is negative in quadrants 3 and 4,\n. . the two possible angles look like this:\nCode:\n              |                         |\n|                         |\n-3   |                         |   3\n- + - - - + - - - - -       - - - - + - - - + -\n:   \u03b8 / |                         | \\ \u03b8   :\n-4:   /5  |                         |  5\\   :-4\n: /     |                         |     \\ :\n*       |                         |       *\n\nSince $\\tan\\theta \\:=\\:\\frac{\\text{-}4}{\\text{-}3}\\text{ or }\\frac{\\text{-}4}{3}$\n\n. . the answers are: . $\\pm\\frac{4}{3}$", "date": "2015-03-27 16:40:15", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/9952-exact-values.html", "openwebmath_score": 0.9173858761787415, "openwebmath_perplexity": 2401.013054657312, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9896718459575753, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8633376481095464}}
{"url": "https://math.stackexchange.com/questions/866223/help-with-complicated-functional-equation", "text": "# Help with complicated functional equation\n\nProblem: Let $T=\\{(p,q,r)\\mid p,q,r \\in \\mathbb{Z}_{\\geq0}\\}$. Find all functions $f:T\\to \\mathbb{R}$ such that: $$f(p,q,r)=\\\\ =\\begin{cases} 0, & \\text{ if } pqr = 0 \\\\ 1 + \\frac{1}{6}\\left(f(p+1,q-1,r)+f(p+1,q,r-1)+f(p,q+1,r-1)+\\\\ \\;\\;\\;\\;\\;\\; f(p,q-1,r+1)+f(p-1,q+1,r)+f(p-1,q,r+1)\\right) & \\text{ otherwise.} \\end{cases}$$\n\nProgress so far: It's not hard to see that $f$ is symmetric in $p,q,r$, which is useful to know. From the recursive definition one can also infer that $f:T\\to \\Bbb{Q}^+$, so no trig functions or logs. That's all I could observe from the get-go. I've tried calculating some values of $f$ to have an idea on how the functions look like (if there are any) but having trouble calculating even small values of $f$, for example $f(1,2,3)$ or $f(2,2,2)$. All I know is that $f(0,a,b)=0$ and $f(1,1,1)=1$. I could guess a solution based on my initial observations but I can't see any obvious candidates.\n\nAny help would be appreciated, thanks.\n\n\u2022 $\\displaystyle\\;f(p,q,r) = \\frac{3pqr}{p+q+r}\\;$ is a solution. I believe this is the only solution but I can't prove it. \u2013\u00a0achille hui Jul 16 '14 at 19:56\n\u2022 How did you get this solution? \u2013\u00a0Deathkamp Drone Jul 16 '14 at 20:00\n\u2022 If you look at your equation, $(p,q,r)$ only connects to points with same $p+q+r$. This means you can look at each $p+q+r = \\text{constant}$ layer separately. The intersection of $T$ with any such layer forms a triangular lattice. Your equation reduces to a discrete version of Poisson equation there. The boundary condition suggest the solution (at least in continuum limit) contains a factor proportional to $pqr$. Direct substitution shows that you can scale this to get a solution. Since we are dealing with some sort of \"Poisson equation\", that's why I suspect the solution is unique. \u2013\u00a0achille hui Jul 16 '14 at 20:09\n\u2022 This is an IMO 2001 shortlist problem. \u2013\u00a0Calvin Lin Jul 18 '14 at 22:57\n\u2022 Thank you, Calvin. I found this problem some lecture notes so I didn't know where it came from. Ironically, I looked up the solution on AoPS just to find out that they pulled out of nowhere the function Achille discovered and then proved that this is the only solution. \u2013\u00a0Deathkamp Drone Jul 19 '14 at 2:02\n\nWhen I worked on this problem back in 2002, showing uniqueness was really easy through the \"average of neighbors\" observation (albeit on a slanted hexagonal board, instead of the regular chessboard).\n\nProof of uniqueness: Suppose we have 2 solutions $f(p,q,r)$ and $g(p,q,r)$. Let $h(p,q,r) = f(p,q,r) - g(p,q,r)$. Then, we get that\n\n$$6 h(p,q,r) = h(p+1, q-1, r) + h(p-1, q+1, r) + h( p, q+1, r-1) + h( p, q-1, r+1) + h( p+1, q, r-1) + h(p-1, q, r+1).$$\n\nConsider the plane $p+q+r = N$. Oberve that the neighbors of the cell $(p,q,r)$ are these 6 other cells with coordinates as given above. Hence, every cell is the average of it's neighbors. Through the standard argument (extremal principle), this implies that all cells on this finite board are equal.\n\nWe also have the boundary conditions that $h(p,q,r ) = 0$ for $pqr=0$, hence $h(p,q,r) = 0$. Thus, the function is unique $_\\square$\n\nFinding the solution was harder, but still motivated from the conditions.\nNote: It is important to bear in mind that as an ('easy') Olympiad problem, it often has a nice solution that can be motivated.\n\nFinding function: From the boundary condition that $pqr=0 \\Rightarrow f(p,q,r) = 0$, we guess the initial function $F( p,q,r) = pqr$.\n\nObserve that since $(p-1)(q+1) r + (p+1)(q-1)r = 2pqr - 2r$, so this guess gives us:\n\n$F(p,q,r) = \\frac{ p+q+r} { 3} + \\frac{1}{6} [ F(p-1, q+1, r) + F(p+1, q-1, r) + F(p, q-1, r+1), F(p, q+1, r-1) + F( p-1, q, r+1), F(p+1, q, r-1) ]$.\n\nObserve that since $p+q+r$ is a constant for all of these 7 terms, we should look at $$f(p,q,r) = \\frac{ F(p,q,r) } { \\frac{p+q+r} {3} } = \\frac{3 pqr} { p+q+r}.$$\n\nIndeed, this works. $_\\square$\n\nNote: Had $F(p,q,r) = pqr$ not worked, the next guess would have been $F(p,q,r) = p^2q^2r^2$\n\nAchille Hui did the hardest part of the work by discovering the closed formula $\\frac{3pqr}{p+q+r}$. The rest is a routine \"maximum principle\" argument that I explain below.\n\nFor a positive integer $k$, let $T_k$ be the finite set $\\lbrace (p,q,r)\\in T | p+q+r=k\\rbrace$. For $x=(p,q,r)\\in T$, define the neighborhood $N(x)$ of $x$ to be\n\n$$\\begin{array}{lcl} N(p,q,r)&=&\\lbrace (p+1,q-1,r);(p+1,q,r-1);(p,q+1,r-1); \\\\ & & (p,q-1,r+1);(p-1,q+1,r);(p-1,q,r+1)\\rbrace \\end{array} \\tag{1}$$\n\nand the strict neighborhood $N'(x)$ of $x$ to be $\\lbrace (u,v,w)\\in N(x) | uvw>0\\rbrace$. We say that $x\\in T_k$ is interior if $N\u2019(x)=N(x)$, and extremal otherwise.\n\nLet $g(p,q,r)=f(p,q,r)-\\frac{3pqr}{p+q+r}$ for $(p,q,r)\\in T$. Then $g$ satisfies $g(p,q,r)=0$ if $pqr=0$, and\n\n$$6g(x)=\\sum_{y\\in N'(x)}g(y) \\tag{2}$$\n\nfor any $x\\in T_k$ (note that $N(x)$ and $N\u2019(x)$ stay in $T_k$ when $x\\in T_k$).\n\nNow, let $M$ ($m$) be the maximum (minimum) value of $g$ on $T_k$. There is some $x_M\\in T_k$ such that $g(x_M)=M$. We now apply (2) to $x=x_M$, and obtain a formula (2').\n\nIf $M > 0$, then (2') is only possible when $x_M$ is interior and $g(y)=M$ for all $y\\in N(x_M)$. If we put $L=\\lbrace x\\in T_k | g(x)=M\\rbrace$, we would deduce that $L$ consists only of interior points but also satisfies $N(x)\\subseteq L$ for any $x\\in L$, which is impossible because when we move away from the interior of $T_k$ we are always forced to eventually reach extremal points.\n\nSo $M\\leq 0$. A similar argument (or if you please, you may reuse the result just shown on $-g$ instead of $g$) shows that $m\\geq 0$.\n\nSo $M\\leq 0 \\leq m$, but on the other hand $m\\leq M$. This forces $m=M=0$, so $g$ is identically zero.\n\nTo conclude, $\\frac{3pqr}{p+q+r}$ is the unique solution.", "date": "2019-06-19 09:16:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/866223/help-with-complicated-functional-equation", "openwebmath_score": 0.9145289659500122, "openwebmath_perplexity": 169.3079724537903, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989671849308438, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8633376428215782}}
{"url": "https://math.stackexchange.com/questions/928470/what-is-the-difference-between-bounded-and-convergent", "text": "# what is the difference between bounded and convergent?\n\nI know that bounded means to have an upper or lower bound.\n\nLet $E \\subset \\mathbb{R}$ be nonempty.\n\n1. The set $E$ is said to be bounded above if and only if there is an $M \\in \\mathbb{R}$ such that $a \\leq M$ for all $a \\in E$, in which case $M$ is called an upper bound of $E$.\n\n2. A number $s$ is called a supremum of the set $E$ if and only if $s$ is an upper bbound of $E$ and $s \\leq M$ for all upper bounds $M$ of $E$ (In this case we shall say that $E$ has a finite supremum $s$ and write $s=\\sup E$.\n\nLet $E \\subset \\mathbb{R}$ be nonempty.\n\n1. the set $E$ is said to be bounded below if and only if there is an $m \\in \\mathbb{R}$ such that $a \\geq m$ for all $a\\in E$, in which case $m$ is called a lower bound of the set $E$.\n\n2. A number $t$ is called an infimum of the set $E$ if and only if $t$ is a lower bound of $E$ and $t \\geq m$ for all lower bounds $m$ of $E$. In this case we shall say that $E$ has an infimum $t$ and write $t = \\inf E$\n\n3. $E$ is said to be bounded if and only if it is bounded both above and below.\n\nThe meaning of convergence\n\nA sequence of real number $\\{ x_n \\}$ is said to converge to a real number $a \\in \\mathbb{R}$ if and only if for every $\\epsilon > 0$ there is an $N \\in \\mathbb{N}$ (which in general depends on $\\epsilon$) such that $$n \\geq N \\mbox{ implies } \\vert x_n - a \\vert < \\epsilon$$\n\nIn my previous classes, it was taught to take the the limit of the function or series to find the value which the function or series converges at the value. Is the value a function or series converges at the $\\sup$ or $\\inf$?\n\n\u2022 The sequence defined by $a_n := \\sin \\left(\\frac{\\pi}{2} n\\right)$, which has expansion $0, 1, 0, -1, 0, 1, 0, -1, \\ldots$ is bounded but not convergent. \u2013\u00a0Travis Sep 12 '14 at 7:42\n\u2022 Thanks, that example made the conflict more clear. \u2013\u00a0El Santi Sep 12 '14 at 7:59\n\nI think there's two points to address here. The first is that a sequence can be bounded from both above and below, yet fail to be convergent. Consider the sequence $$\\{x_n\\} = \\{(-1)^n\\} = -1,1,-1,1, \\cdots$$ Here we have $\\displaystyle\\sup_{n} \\{x_n\\} = 1$, $\\displaystyle\\inf_{n} \\{x_n\\} = -1$, and $\\{x_n\\}$ does not converge (if you pick say $\\epsilon = 0.1$, then there are infinitely many $n$ for which $|x_n - 1| > 0.1$ and for which $|x_n - (-1)| > 0.1$).\nFurthermore, when a sequence $\\{x_n\\}$ does converge, the limit of a sequence need not equal either $\\displaystyle\\sup_n\\{x_n\\}$ or $\\displaystyle\\inf_n\\{x_n\\}$. For example, we see with the sequence $\\{x_n\\} = \\{\\frac{(-1)^n}{n}\\}$ that $$\\displaystyle\\sup_n\\{x_n\\} = \\frac{1}{2}, \\: \\:\\displaystyle\\inf_n\\{x_n\\} = -1, \\: \\: \\displaystyle\\lim_{n \\to \\infty} \\{x_n\\} = 0$$\nIn that last example, I kind of cheated, by picking a sequence where the inf and sup of the elements of the sequence were the first two, and the convergence of a sequence talks about the behavior at the tail of the sequence. When we want to talk about upper and lower bounds of sequences, we usually like to talk about the limit superior, or $$\\displaystyle\\limsup_{n \\to \\infty} \\{x_n\\} = \\displaystyle\\lim_{n \\to \\infty} \\left(\\sup_{N} \\{x_N | \\: N \\geq n \\} \\right)$$ and the limit inferior, or $$\\displaystyle\\liminf_{n \\to \\infty} \\{x_n\\} = \\displaystyle\\lim_{n \\to \\infty} \\left(\\inf_{N} \\{x_N | \\: N \\geq n \\} \\right)$$ This let's you bound the values of $x_n$ for arbitrarily large $n$, while allowing for a few badly behaved values of $x_n$ for small $n$. Then, we say that a sequence $\\{x_n\\}$ approaches a finite limit $M$ if and only iff $$\\displaystyle\\limsup_{n \\to \\infty} \\{x_n\\} = \\displaystyle\\liminf_{n \\to \\infty} \\{x_n\\} = M$$ and you can read about the proof of this result here (Proof that a sequence converges to a finite limit iff lim inf equals lim sup).\n\u2022 is there a typo at the end? $\\displaystyle\\limsup_{n \\to \\infty} \\{x_n\\} = \\displaystyle\\limsup_{n \\to \\infty} \\{x_n\\}$ is it to be $\\displaystyle\\limsup_{n \\to \\infty} \\{x_n\\} = \\displaystyle\\liminf_{n \\to \\infty} \\{x_n\\}$ ?? \u2013\u00a0El Santi Sep 12 '14 at 8:22", "date": "2019-08-25 19:47:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/928470/what-is-the-difference-between-bounded-and-convergent", "openwebmath_score": 0.9626544713973999, "openwebmath_perplexity": 70.12419483690904, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137889851291, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8633342591192305}}
{"url": "https://math.stackexchange.com/questions/1909290/integration-using-monte-carlo-method", "text": "# Integration using Monte Carlo Method\n\nI'm trying to solve this integral using the Monte Carlo Method.\n\n$$I=\\int_0^\\pi \\frac{1}{\\sqrt{2\\pi}}e^\\frac{-sin(x)^2}{2}dx$$\n\nNow it seems to me that there is a normal probability density function in there, but I'm not sure because of the sine function.\n\nIf there is a normal, then it's easy to simulate $N$ random numbers from the standard normal distribution and compute $I$. It's is also easy to find the size of the sample to get a 95% confidence interval.\n\nSo is there a normal probability density function? Or from what distribution should I get the random numbers to compute $I$?.\n\n\u2022 Ok. How did you find the confidence intervals? \u2013\u00a0Fawcett512 Aug 31 '16 at 4:04\n\u2022 I know how to get the confidence intervals for a sample of uniform random variables, but I'm not sure how to find them in this case given that I don't have a parameter but rather the approximate value of $I$ \u2013\u00a0Fawcett512 Aug 31 '16 at 4:06\n\u2022 I think I'm making the problem more complicate that it should. So I could generate $N$ uniform random variables $x_i$ and then compute $\\mathbb{E}=\\frac{\\pi}{N}\\sum f(x_i)$. As $N$ tends to infinity, we should get closer to the integral. Yet I don't see how to get the confidence intervals. \u2013\u00a0Fawcett512 Aug 31 '16 at 4:40\n\u2022 I think that I was trying to do some importance sampling (hence the question if we could use a normal probability density function), but seems that in this case that is not possible, Furthermore, if I just compute the above formula for $\\mathbb{E}$,I should get a good estimate for $N$ provided that $N$ is big enough. The only thing that's not very clear is how to compute a 95% confidence interval. \u2013\u00a0Fawcett512 Aug 31 '16 at 4:47\n\nRecall that if $Y$ is a random variable with density $g_Y$ and $h$ is a bounded measurable function, then $$\\mathbb E[h(Y)] = \\int_{\\mathbb R} h(y)g_Y(y)\\,\\mathsf dy.$$ Moreover, if $Y\\sim\\mathcal U(0,1)$, then $a+(b-a)U\\sim\\mathcal U(a,b)$. So applying the change of variables $x=a+(b-a)u$ (with $a=0$, $b=\\pi$) to the given integral, we have $$I = \\int_0^1 \\frac{\\pi}{\\sqrt{2\\pi}} e^{-\\frac12\\sin^2 (\\pi u) }\\,\\mathsf du=\\int_0^1 h(u)\\,\\mathsf du,$$ with $h(u)=\\sqrt{\\frac\\pi 2} e^{-\\frac12\\sin^2 (\\pi u) }$. It follows then that $I=\\mathbb E[h(U)]$ with $U\\sim\\mathcal U(0,1)$. Let $U_i$ be i.i.d. $\\mathcal U(0,1)$ random variables and set $X_i=h(U_i)$, then for each positive integer $n$ we have the point estimate $$\\newcommand{\\overbar}[1]{\\mkern 1.75mu\\overline{\\mkern-1.75mu#1\\mkern-1.75mu}\\mkern 1.75mu} \\widehat{I_n} =: \\overbar X_n= \\frac1n \\sum_{i=1}^n X_i$$ and the approximate $1-\\alpha$ confidence interval $$\\overbar X_n\\pm t_{n-1,\\alpha/2}\\frac{S_n}{\\sqrt n},$$ where $$S_n = \\sqrt{\\frac1{n-1}\\sum_{i=1}^n \\left(X_i-\\overbar X_n\\right)^2}$$ is the sample standard deviation.\n\nHere is some $\\texttt R$ code to estimate an integral using the Monte Carlo method:\n\n# Define \"h\" function\nhh <-function(u) {\nreturn(sqrt(0.5*pi) * exp(-0.5 * sin(pi*u)^2))\n}\n\nn <- 1000 # Number of trials\nalpha <- 0.05 # Confidence level\nU <- runif(n) # Generate U(0,1) variates\nX <- hh(U) # Compute X_i's\nXbar <- mean(X) # Compute sample mean\nSn <- sqrt(1/(n-1) * sum((X-Xbar)^2)) # Compute sample stdev\nCI <- (Xbar + (c(-1,1) * (qt(1-(0.5*alpha), n-1) * Sn/sqrt(n)))) # CI bounds\n\n# Print results\ncat(sprintf(\"Point estimate: %f\\n\", Xbar))\ncat(sprintf(\"Confidence interval: (%f, %f)\\n\", CI[1], CI[2]))\n\n\nFor reference, the value of the integral (as computed by Mathematica) is $$e^{-\\frac14}\\sqrt{d\\frac{\\pi }{2}} I_0\\left(\\frac{1}{4}\\right) \\approx 0.991393,$$ where $I_\\cdot(\\cdot)$ denotes the modified Bessel function of the first kind, i.e. $$I_0\\left(\\frac14\\right) = \\frac1\\pi\\int_0^\\pi e^{\\frac14\\cos\\theta}\\,\\mathsf d\\theta.$$\n\n\u2022 Thank you. This was very clear. \u2013\u00a0Fawcett512 Aug 31 '16 at 4:54\n\u2022 You're welcome! The change of variables is not strictly necessary, but it simplifies the computations considerably. \u2013\u00a0Math1000 Aug 31 '16 at 4:58\n\u2022 Just one more question because I have trouble understanding the concept of confidence intervals. How do I know the size of the sample if now I want to get a 98% confidence interval? \u2013\u00a0Fawcett512 Aug 31 '16 at 6:54\n\u2022 I don't understand your question - the sample size ($n$) and confidence level ($\\alpha$) are parameters which influence the width of the confidence interval, but are not directly related to each other. \u2013\u00a0Math1000 Aug 31 '16 at 7:49\n\u2022 Never mind. That was a bad question. I was wondering how big the sample must be (because I was making a simulation) to get a 98% confidence interval, but as you said is a parameter which influences the width of the confidence interval. All is clear know (or at least I think so). \u2013\u00a0Fawcett512 Sep 1 '16 at 5:59", "date": "2019-05-23 05:10:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1909290/integration-using-monte-carlo-method", "openwebmath_score": 0.9126384854316711, "openwebmath_perplexity": 249.8986845653801, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137937226358, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.863334252401025}}
{"url": "https://www.seamplex.com/wasora/doc/realbook/004-exp/", "text": "# 1 A simple differential equation\n\nThe examples in this section show how a single ordinary differential equation can be solved with wasora. Indeed this is one of its main features, namely the ability to solve systems of differential-algebraic equations written as natural algebraic expressions. In particular, the equation the examples solve is\n\n\\frac{dx}{dt} = -x\n\nwith the initial condition x_0 = 1, which has the trivial analytical solution x(t) = e^{-t}.\n\n## 1.1 exp.was\n\nAs clearly defined in wasora\u2019s design basis, simple problems ought to be solved by means of simple inputs. Here is a solid example of this behavior.\n\nend_time = 1        # transient problem\nPHASE_SPACE x       # DAE problem with one variable\nx_0 = 1             # initial condition\nx_dot .= -x         # differential equation\nPRINT t x HEADER    # output\n# exercise: plot dt vs t and see what happens\n$wasora exp.was | qdp -o exp --pi 1$ \n\nBy default, wasora adjusts the time step so an estimation of the relative numerical error is bounded within a range given by the variable rel_error, which has an educated guess by default. It can be seen in the figure that dt starts with small values and grows as the conditions allow it.\n\n## 1.2 exp-dt.was\n\nThe time steps wasora take can be limited by means of the special variables min_dt and max_dt. If they both are zero (as they are by default), wasora is free to choose dt as it considers appropriate. If max_dt is non-zero, dt will be bounded even if the conditions are such that bigger time steps would not introduce large errors. On the other hand, if min_dt is non-zero, the time step is guaranteed not to be smaller that the specified value. However, it should be noted that wasora may need to take several internal times step to keep the error bounded. In the limiting case where min_dt = max_dt, the time step can be set exactly although, again, wasora may take internal steps.\n\nThis situation is illustrated with the following input, which is run for five combinations of min_dt and max_dt.\n\nend_time = 2\nmin_dt = $1 max_dt =$2\nrel_error = 1e-3\nPHASE_SPACE x\nx_0 = 1\nx_dot .= -x\nPRINT t x (x-exp(-t))/x\n$wasora exp-dt.was 0 0 > exp-dt1.dat$ wasora exp-dt.was 0.1 0   > exp-dt2.dat\n$wasora exp-dt.was 0 0.1 > exp-dt3.dat$ wasora exp-dt.was 0.1 0.1 > exp-dt4.dat\n$wasora exp-dt.was 1 1 > exp-dt5.dat$ pyxplot exp-dt.ppl; pdf2svg exp-dt.pdf exp-dt.svg; rm -f exp-dt.pdf\n$pyxplot exp-error.ppl; pdf2svg exp-error.pdf exp-error.svg; rm -f exp-error.pdf$ \n\nIt can be seen that all the solutions coincide with the analytical expression. Even if the time step is set to a big fixed value, the error commited by the numerical solver with respect to the exact solution is the same as wasora iterates internally as needed. In general, the fastest condition is where dt is not bounded as wasora minimizes iterations by automatically adjusting its value. However, it is clear that controlling the time step can be useful some times. A further control can be obtained by means of the TIME_PATH keyword.", "date": "2020-10-20 17:36:27", "meta": {"domain": "seamplex.com", "url": "https://www.seamplex.com/wasora/doc/realbook/004-exp/", "openwebmath_score": 0.6721046566963196, "openwebmath_perplexity": 1340.1719890610452, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471628097781, "lm_q2_score": 0.8774767890838837, "lm_q1q2_score": 0.8633030493716131}}
{"url": "https://math.stackexchange.com/questions/4025290/is-every-open-and-connected-set-in-mathbb-c-the-continuous-image-of-the-open", "text": "# Is every open and connected set in $\\mathbb C$ the continuous image of the open unit disk?\n\nLet $$\\mathbb D=\\{z\\in\\mathbb C\\ |\\ |z|<1\\}$$ be the open unit disk in $$\\mathbb C$$. It is well known that an open (nonempty) set $$U\\subseteq\\mathbb C$$ is simply connected if and only if it is homeomorphic to the unit disk. One can show that this is equivalent to the fact that there is a continuous injective function $$f:\\mathbb D\\to\\mathbb C$$ such that $$f(\\mathbb D)=U$$.\n\nTherefore, my question is: If $$U$$ is a (nonempty) open and connected subset of $$\\mathbb C$$, is there always a continuous (but not necessarily injective) function $$f:\\mathbb D\\to\\mathbb C$$ such that $$f(\\mathbb D)=U$$? It sufficices to assume $$U\\subseteq\\mathbb D$$.\n\nIf not, how do images of such functions look like?\n\nI know that any compact connected and locally connected subset of $$\\mathbb C$$ is the image of a continuous function defined on $$[0,1]$$ and hence also the image of a continuous function on $$\\overline{\\mathbb D}$$. However, the closure of an open and connected set does not have to be locally connected.\n\nI believe that the answer to my first question is negative (although I hope that it is not ;) ).\n\nAny help is highly appreciated. Thank you very much in advance!\n\n\u2022 Not my field, but I'm pretty sure that the answer is positive to your first question. Here's my idea: first project $\\Bbb{D}$ to the interval $(-1, 1)$, then expand to $\\Bbb{R}$. From there, I'm pretty sure you can use a space-filling curve to fill any disk. Since $U$ is open and connected, it's path connected, and is made up of a countable union of open disks, which we can cover by a single space-filling curve (just join the endpoints for each of the countably many open disks by a path). The result will be heavily non-injective, but I think it works? Feb 14 at 11:46\n\u2022 Feb 14 at 12:13\n\nThe Hahn-Mazurkiewicz theorem, as pointed out by Mathlover in the comments, is enough to show my educated guess actually works. In particular, it implies that closed balls in $$\\Bbb{C}$$ are the continuous image of a compact interval. We will need a lemma:\n\nLemma $$\\quad$$ Suppose $$\\{C_n\\}_{n=0}^\\infty$$ is a countable collection of subspaces of a topological space $$X$$ (e.g. $$\\Bbb{C}$$) which are continuous images of a compact interval and $$C := \\bigcup_{n=0}^\\infty C_n$$ is path-connected. Then $$C$$ is the continuous image of $$[0, \\infty)$$.\n\nProof. Consider the intervals $$I_n := [2n, 2n + 1]$$ and $$J_n := [2n + 1, 2n + 2]$$ for $$n \\in \\Bbb{N} \\cup \\{0\\}$$. Let $$\\mathfrak{i}_n : I_n \\to C_n$$ be surjective and continuous, and let $$\\mathfrak{j}_n : J_n \\to C$$ be a continuous function such that $$\\mathfrak{j}_n(2n + 1) = \\mathfrak{i}_n(2n + 1)$$ and $$\\mathfrak{j}_n(2n + 2) = \\mathfrak{i}_{n+1}(2n + 2)$$. Then, the map $$\\phi : [0, \\infty) = \\bigcup_{n=0}^\\infty (I_n \\cup J_n) \\to C$$ defined by $$\\phi(x) = \\begin{cases} \\mathfrak{i}_n(x) & \\text{if } \\exists n \\in \\Bbb{N} \\cup \\{0\\} : 2n \\le x < 2n + 1 \\\\ \\mathfrak{j}_n(x) & \\text{if } \\exists n \\in \\Bbb{N} \\cup \\{0\\} : 2n + 1 \\le x < 2n + 2. \\end{cases}$$ This function is made up of countably infinitely many continuous pieces, joined at common points, making the function continuous. It's clearly surjective (even just restricting to $$\\bigcup I_n$$), hence $$C$$ is the continuous image of $$[0, \\infty)$$ under $$\\phi$$. $$\\square$$\n\nNote that every ball in $$\\Bbb{C}$$ is a countable union of closed balls, and recall that open subsets of $$\\Bbb{C}$$ are a countable union of open balls. Also, in a locally path-connected space, open and connected implies path-connected. Thus, $$U$$ is the path-connected countable union of closed balls, so by the lemma, it is a continuous image of $$[0, \\infty)$$.\n\nSo, to wrap up the proof, as in my comment, simply project the unit disk of $$\\Bbb{C}$$ onto $$(-1, 1)$$, which is homeomorphic to $$\\Bbb{R}$$. You can simply keep the map constant for points in $$(-\\infty, 0]$$, and then use $$[0, \\infty)$$ to map continuously onto $$U$$.\n\n\u2022 This is amazing, thank you! Feb 14 at 13:45\n\nOne can prove a stronger claim: given a nonempty open connected $$U\\subseteq \\mathbb{C}$$, there's a holomorphic $$\\varphi:\\mathbb{D}\\to \\mathbb{C}$$ such that $$\\varphi(\\mathbb{D})=U$$.\n\nFirst, let $$U\\neq \\mathbb{C},\\neq \\mathbb{C}-\\{z_0\\}$$. Thanks to the Riemann uniformization theorem, every such open connected subset of $$\\mathbb{C}$$ is hyperbolic, i.e. has $$\\mathbb{D}$$ as its universal cover. Thus there exists a surjective holomorphic map $$\\mathbb{D}\\to U$$. Note that there's a surjective holomorphic map $$(\\exp(z)+z_0)$$ from $$\\mathbb{C}$$ to $$\\mathbb{C}-\\{z_0\\}$$, and thus it suffices to prove the result for $$\\mathbb{C}$$. Now, let $$f(z)=z^3-1$$. It is easy to see that $$f:\\mathbb{C}-\\{1,\\exp(2\\pi i/3)\\}\\twoheadrightarrow\\mathbb{C}$$. Precomposing with the covering map $$\\pi:\\mathbb{D}\\to \\mathbb{C}-\\{1,\\exp(2\\pi i/3)\\}$$, we get the claim.\n\nOne can also explore \"how much\" this function fails to be injective. Indeed, the uniformization theorem implies that, for $$U\\neq\\mathbb{C},U\\neq \\mathbb{C}-\\{z_0\\}$$, we have $$U\\simeq \\mathbb{D}/\\Gamma$$, where $$\\Gamma$$ is a discrete subgroup of $$Aut(\\mathbb{D})$$ (complex automorphisms) isomorphic to $$\\pi_1(U)$$. Thus two elements $$z_1,z_2$$ have the same image iff $$\\exists \\gamma\\in \\Gamma:\\gamma(z_1)=z_2$$ (note that with $$U$$ simply connected we get injectivity).\n\n\u2022 Beautiful answer, thank you very much! Feb 14 at 13:46", "date": "2021-10-20 01:02:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4025290/is-every-open-and-connected-set-in-mathbb-c-the-continuous-image-of-the-open", "openwebmath_score": 0.9684483408927917, "openwebmath_perplexity": 62.61045016974707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471689667878, "lm_q2_score": 0.8774767826757122, "lm_q1q2_score": 0.8633030484695847}}
{"url": "https://math.stackexchange.com/questions/953959/n-identical-balls-distributed-into-r-urns", "text": "# $n$ Identical Balls Distributed Into $r$ Urns\n\n\u2022 Question\n\nIn how many ways can at most $n$ identical balls be distributed into $r$ urns so that the $i$th urn contains at least $m_i$ balls, for each $i=1,...,r$?\n\nAssume that $n\\geq \\sum_{i= 1}^r m_i$\n\n\u2022 My Approach\n\n$$x_1 + x_2 + \\cdot\\cdot\\cdot + x_r = S$$\n\n$$\\sum_{i= 1}^r m_i \\leq S \\leq n$$\n\n$$x_i \\geq m_i$$\n\nLet\n\n$$y_i = x_i - m_i$$\n\nthen $$y_1+ y_2 + \\cdot\\cdot\\cdot + y_r = S -\\sum_{i= 1}^r m_i$$\n\nNow I will proceed to find the number of non-negative integer-valued vectors $(y_1, y_2,...,y_r)$ such that $y_1+ y_2 + \\cdot\\cdot\\cdot + y_r = S -\\sum_{i= 1}^r m_i$ for every integer $S$ in the range of $\\{\\sum_{i= 1}^r m_i\\quad,\\quad n\\}$ which is equal to $$\\sum\\limits_{S = \\sum_{i= 1}^r m_i}^n {S - \\sum_{i= 1}^r m_i + r -1\\choose r-1}$$\n\nPlease have a look at my solution and give me any hints/suggestions you might have.\n\n\u2022 What is your question actually? \u2013\u00a0user 170039 Oct 1 '14 at 13:22\n\u2022 There is Question, My Approach, and then Please have a look at my solution and give me any hints/suggestions you might have.. What do you mean? \u2013\u00a0Kermit the Hermit Oct 1 '14 at 13:24\n\u2022 I mean what question do you want the MSE to answer? Or is it that you want to verify your solution? \u2013\u00a0user 170039 Oct 1 '14 at 13:27\n\u2022 Substitute $x_i = m_i + x_i'$ where $x_i' \\ge 0$. Now subtract $m_1, m_2, \\dots , m_i$ from both sides of the linear diophantine equation you have written. Finally, use stars and bars as usual. \u2013\u00a0Yiyuan Lee Oct 1 '14 at 13:28\n\nYou really seem to have the right idea, defining $y_i=x_i-m_i$, which basically says \"If I put $x_i$ balls into urn $i$, this includes $m_i$ obligatory balls and then $y_i$ more that I can choose.\" This reduces you to solving:\n\n$$x_1+x_2+\\dots+x_r=(y_1+m_1)+(y_2+m_2)+\\dots+(y_r+m_r)=n$$\n\nwhich you note is the same as:\n\n$$y_1+y_2+\\dots+y_r=n-\\sum_{i=1}^r m_i.$$\n\nMy question to you is: What is $S$? How is $S$ different from $n$? Unless I'm misreading this problem, $S$ seems to be extraneous. If $S$ is the number of balls you are distributing, the question seems to indicate that $S=n$ (not just $S\\le n$).\n\nSo if you'll allow us to say $M=\\sum m_i$, then you are correct in your reformulation of the problem, you are basically just distributing $n-M$ balls into $r$ urns, which is:\n\n$$\\binom{n-M+r-1}{r-1}.$$\n\nThat's the last term of your summation. You don't need to sum over values of this \"$S$\" because you need to distribute all $n$ of the balls. (If you had such a thing, it would be like counting ways of partially distributing the balls, which the question does not seem to indicate.)\n\nYou seem to know where the binomial formula for counting these tuples comes from, but for others reading, this formula is determined as explained here.\n\nWe basically reduced the problem \"distributing $n$ balls into $r$ urns with restriction $m_i$ on each urn\" to \"distributing $n-M$ balls into $r$ urns\" by translating solutions from one to the other. You can think of it physically as starting out by putting all of the $m_i$ balls into each urn as required -- you have $n-M$ left over to distribute freely, and you can use what you already know to count how to do that.\n\nA simpler version of this problem might be \"how many ways can you give your two children \\$5 allowance (in whole dollar amounts) if each child needs at least \\$1?\" Rather than count some complicated configuration, you can really just think of this \\$1 each as already handed out to them and then solve the easier problem \"how many ways can you give your two children \\$3 allowance?\" (In this case, the answer is 4 ways.)\n\nI am assuming from context (and from your use of the formula) that the urns are distinguishable (they are enumerated and each has a particular value of $m_i$, so philosophically -- and practically speaking, were you to attempt this exercise -- that makes the most sense).\n\n\u2022 You just sparked something in me. For some reason I was thinking that you don't have to distribute all $n$ balls. Thus leading to $\\sum_{i=1}^r m_i\\leq x_1 + x_2 + ... + x_r \\leq n$. \u2013\u00a0Kermit the Hermit Oct 1 '14 at 13:45\n\u2022 If you are are allowed to use fewer than $n$ balls total (i.e. to just use $S$ balls and throw away the $n-S$ left over), the answer is just the sum (similar to what you have): $\\sum_{S=M}^n\\binom{S-M+r-1}{r-1}$. \u2013\u00a0Kellen Myers Oct 1 '14 at 14:17\n\u2022 What is that $M$? Is that $\\sum_{i=1}^r m_i$ ? \u2013\u00a0Kermit the Hermit Oct 1 '14 at 14:29\n\u2022 Yes, it's shorthand. I think it's defined somewhere in the middle of my answer. \u2013\u00a0Kellen Myers Oct 1 '14 at 14:33\n\u2022 Yes indeed, I missed that. I should pay more attention when reading it would seem. So taken in to consideration the way I interpreted the question, my solution would be correct? \u2013\u00a0Kermit the Hermit Oct 1 '14 at 14:37", "date": "2019-12-11 00:50:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/953959/n-identical-balls-distributed-into-r-urns", "openwebmath_score": 0.7970921993255615, "openwebmath_perplexity": 311.39865548207274, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180673335565, "lm_q2_score": 0.8757869900269366, "lm_q1q2_score": 0.8632790592052246}}
{"url": "http://math.stackexchange.com/questions/601744/values-of-a-b-and-c-that-the-curve-y-ax3-3x2-bx-cx-ex-has-one-p/601795", "text": "# Values of a, b, and c that the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ has one point of inflection?\n\nFor what values of a, b and c does the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ have exactly one point of inflection? Two points of inflection? No points of inflection? Provide a numerical approximation for the lowest value of a for which there is no inflection point.\n\nI'm really stumped on this question. I've tried finding the second derivative but couldn't think of anything else to do from that point. Any help would be greatly appreciated. Thanks!\n\n-\nPoints of inflection occur when the second derivative is zero. What is the second derivative of your curve? How can you assure there's only one zero such that we have $+/-$ on either side of the zero. (You're allowed other zeroes that don't change the inflection, i.e. $x=0$ of $y=x^2$). \u2013\u00a0 Ian Coley Dec 10 '13 at 19:46\nI've found the second derivative to be $y'' = 6ax + 6 + e^x = 0$ but I could not think of how to solve that. \u2013\u00a0 DinoMint Dec 10 '13 at 19:48\nWhat happens when $x=0,1,-1,-2, 2$ and so on? \u2013\u00a0 Manasi Dec 10 '13 at 20:02\nGiven that $e^x$ is positive will drive you to a negative x. Since it won't be too small you can probably use an estimate for $e^x$ around that x. \u2013\u00a0 half-integer fan Dec 10 '13 at 20:04\nYou don't need to solve $6ax+6+e^x=0$, @DinoMint, only figure out how many solutions it has. Think of it as intersecting the graph of $y=e^x$ with the line $y=-6ax-6$. If the slope $-6a$ is negative, there is exactly one intersection point, if it is positive you can find the tangent line to $y=e^x$ with the same slope ($-6a$) and see if this line, $y=-6ax-6$ is above or below the tangent. \u2013\u00a0 Omar Antol\u00edn-Camarena Dec 10 '13 at 20:05\n\nAs you stated in the comments, the second derivative is $$y'' = 6ax + 6 + e^x = 0$$\nFirst, we note that if $a=0$ the equation becomes $$6 + e^x = 0$$ which has no real solutions, thus there are no inflection points.\nRewriting the equation as $$x = -\\dfrac{6+e^x}{6a}$$ or $$x = -\\dfrac1{a} \\cdot \\left(1 + \\dfrac 16 e^x \\right)$$ makes the behavior clearer.\nIf $a \\gt 0$ we can see that $x$ must be negative and thus $e^x < 1$ and there will be a solution in the neighborhood of $x = -\\dfrac1a$.\nI will leave the case where $a \\lt 0$, thus $x \\gt 0$ for you to consider.", "date": "2014-10-24 13:08:24", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/601744/values-of-a-b-and-c-that-the-curve-y-ax3-3x2-bx-cx-ex-has-one-p/601795", "openwebmath_score": 0.8803232908248901, "openwebmath_perplexity": 158.23661561495604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683473173829, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8632110218975917}}
{"url": "http://mathhelpforum.com/discrete-math/224780-possible-mathematical-induction-problem.html", "text": "# Math Help - Possible mathematical induction problem\n\n1. ## Possible mathematical induction problem\n\nCourse: Foundations of Higher MAth\n\nProve that $24|(5^{2n} -1)$ for every positive integer n.\n\nThis is a question from my final exam today.\n\nP(n): $24|(5^{2n}-1)$\nP(1): $24|(5^2 -1)$ is a true statement.\n\nAssume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.\nThen $5^{2k}= 24a + 1$\n\nP(k+1): $24|(5^{2(k+1)}-1)$\n\n$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$\n$=(24a + 1)*25 -1$\n$=(24a)(25)+25 -1$\n$=(24a)(25)+24$\n$=24(25a+1)$\n$=24b$\n\nTherefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.\nNone of my friends used this method though. Is this a correct way to do it?\n\n2. ## Re: Possible mathematical induction problem\n\nCourse: Foundations of Higher MAth\nProve that $24|(5^{2n} -1)$ for every positive integer n.\nThis is a question from my final exam today.\nP(n): $24|(5^{2n}-1)$\nP(1): $24|(5^2 -1)$ is a true statement.\nAssume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.\nThen $5^{2k}= 24a + 1$\nP(k+1): $24|(5^{2(k+1)}-1)$\n$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$\n$=(24a + 1)*25 -1$\n$=(24a)(25)+25 -1$\n$=(24a)(25)+24$\n$=24(25a+1)$\n$=24b$\nTherefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.\nNone of my friends used this method though. Is this a correct way to do it?\nA strict grader may like to have seen more grouping symbols.\nHowever, the argument is correct.", "date": "2014-09-18 14:44:17", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/224780-possible-mathematical-induction-problem.html", "openwebmath_score": 0.5691136121749878, "openwebmath_perplexity": 1136.2313487223958, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683491468141, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8632110170177555}}
{"url": "https://math.stackexchange.com/questions/3212748/alternative-way-to-find-the-probability-in-die-throw-problem", "text": "Alternative way to find the probability in die throw problem\n\nWe throw a die 10 times. What's the probability of getting at least one 6?\n\nUsing the complentary that's $$1-Pr$$(Getting no $$6$$'s)\n\n$$1- ({\\frac{5}{6}})^{10} = 0.838494$$\n\nThere's a note that says that to solve the problem directly would require a complex use of the additivity property.\n\na) How do you solve the problem directly?\n\nI think that to solve it directly, though not sure, is to find the $$P(A_1 \\cup A_2 \\cup A_3 \\cup A_4 \\cup A_5 \\cup A_6)$$ in ten die throws, where $$A_i$$ is the event of getting $$i$$ $$6$$'s\n\nAs the events $$A_i$$ are not disjoint the exclusion-inclusion principle is needed.\n\nMy try:\n\n$$\\binom{10}{1} \\frac{1}{6} - \\binom{10}{2} \\frac{1}{6^2} + \\binom{10}{3} \\frac{1}{6^3} - \\binom{10}{4} \\frac{1}{6^4}+ \\binom{10}{5} \\frac{1}{6^5} - \\binom{10}{6} \\frac{1}{6^6} =0.838091$$\n\nIt is a slightly different result that the one obtained using the complementary. What am I not doing right?\n\nb)If the events aren't disjoint, can you talk about the \"additivity property\"? As it is defined for disjoint events.\n\nIt would be the following:\n\n$$\\sum_{n=1}^{10} \\dbinom{10}{n}\\left(\\dfrac{1}{6}\\right)^n\\left(\\dfrac{5}{6}\\right)^{10-n}$$\n\nHere the events are disjoint. You have exactly 1 roll of a 6, exactly two rolls of a six, ..., exactly ten rolls of a six.\n\nYour attempt works as well (although you stopped at $$i=6$$). Had you continued using Inclusion/Exclusion up to $$i=10$$, it would have given the same answer:\n\n$$\\sum_{i=1}^{10}(-1)^{i+1}\\dbinom{10}{i}\\left(\\dfrac{1}{6}\\right)^i$$\n\nHere you are counting if you choose a die, the probability that one die is a six, minus if you choose two dice, both of them are a six, plus if you choose three dice, all three are a six, etc.\n\nb) Yes.\n\n$$P(A\\cup B) = P(A)+P(B)-P(A\\cap B)$$\n\nThis is the property that yields the Inclusion/Exclusion principle in the first place.\n\nYou got a different answer because you only computed the first six terms of the inclusion-exclusion method. You have ten dice, each of which could produce a six, so you need ten terms.\n\n\u2022 You use Inclusion-Exclusion when you have events that are not mutually exclusive. You can see that \"roll six on the first die\" and \"roll six on the second die\" are not mutually exclusive because it is possible that you roll sixes on both dice. I don't know what you mean by \"inclusive.\" \u2013\u00a0David K May 3 at 21:27", "date": "2019-11-18 11:34:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3212748/alternative-way-to-find-the-probability-in-die-throw-problem", "openwebmath_score": 0.7933527827262878, "openwebmath_perplexity": 457.3737643169891, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109101320038, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8631972512368986}}
{"url": "https://math.stackexchange.com/questions/2104483/how-to-express-the-cardinality-of-%E2%88%8F-1%E2%89%A4i%E2%89%A4n-a-i-in-terms-of-cardinalities-a", "text": "# How to express the cardinality of $\u220f_{1\u2264i\u2264n} A_i$ in terms of cardinalities $|A_1|, |A_2|, . . . , |A_n|$\n\nI was given the problem:\n\nFor finite sets $A_1, A_2,\\dotsc , A_n$ define their Cartesian product $\\prod_{i=1}^n A_i$ as the set of all $n$-sequences $(x_1, x_2,\\dotsc, x_n)$, where $x_i \\in A_i$ for every $i = 1, 2, \\dotsc, n$. Find a formula expressing the cardinality of $\\prod_{i=1}^n A_i$ in terms of cardinalities $|A_1|, |A_2|,\\dotsc , |A_n|$.\n\nAnd I am struggling to understand what it is actually asking for, could someone explain it to me please, thanks. :)\n\n\u2022 Do you know what the Cartesian product is? For instance, if $A_1 = \\{1,2\\}$ and $A_2 = \\{3,4,5\\}$ could you explicitly write down $\\prod_{1 \\leq i \\leq 2} A_i$? \u2013\u00a0Mees de Vries Jan 19 '17 at 13:24\n\u2022 @MeesdeVries This is all prossible combinations right? So for that example {(1,3), (2, 3), (1, 4) ....}, But I have never seen the notation: \u220f 1\u2264i\u2264n Ai before, what does that mean? \u2013\u00a0Alfie Jan 19 '17 at 13:27\n\n$\\prod_{1\\le i\\le n}A_i$ is the cartesian product, that is, all finite sequences $(a_1,\\ldots,a_n)$ such that $a_i \\in A_i$ for each $i=1,\\ldots,n$. How many such sequences can you choose? $|A_1|$ choices for $a_1$, ..., $|A_n|$ choices for $a_n$. Therefore $$\\left|\\prod_{1\\le i\\le n}A_i\\right|=\\prod_{1\\le i\\le n}|A_i|.$$\n\n\u2022 So is this question asking for a formula for the carnality of the cartesian product of $A_1, A_2,\\dotsc , A_n$ ? \u2013\u00a0Alfie Jan 19 '17 at 13:51\n\u2022 Yes. ____________ \u2013\u00a0Paolo Leonetti Jan 19 '17 at 13:52\n\u2022 So $\\prod_{1\\le i\\le n}|A_i|$= $|A_1| * |A_2| * \\dotsc * |A_n|$ \u2013\u00a0Alfie Jan 19 '17 at 13:57\n\u2022 @Bram28 Assuming the axiom of choice, multiplication of infinite cardinal numbers is not difficult. If either $\\kappa$ or $\\mu$ is infinite and both are non-zero, then $\\kappa \\cdot \\mu= \\max\\{\\kappa, \\mu\\}$. In particular, $\\kappa^n=\\kappa$ for all $n\\ge 1$. See here: en.wikipedia.org/wiki/Cardinal_number \u2013\u00a0Paolo Leonetti Jan 19 '17 at 14:56\n\u2022 @Bram28 Well, you can find the answer in every introduction about cardinal arithmetics. Take a look here math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Murphy.pdf and here euclid.colorado.edu/~monkd/m6730/gradsets06.pdf \u2013\u00a0Paolo Leonetti Jan 19 '17 at 16:00\n\nWe know that $$|A\\times B|=|A|\\times|B|\\qquad(1)$$.\n\nWe want to show $$\\left|\\prod_{i=1}^nA_i\\right|=\\prod_{i=1}^n|A_i|$$ is true for any natural number $n$, where $\\prod_{i=1}^n|A_i|=|A_1|\\times|A_2|\\times\\dotsc\\times|A_n|$. So, we have use induction.\n\nThe base case $n=1$ ($|A_1| = |A_1|$) is trivial. Now suppose inductively that $\\left|\\prod_{i=1}^nA_i\\right|=\\prod_{i=1}^n|A_i|$. We want to show $$\\left|\\prod_{i=1}^{n+1}A_i\\right|=\\prod_{i=1}^{n+1}|A_i|.$$ Now we need to show $$\\left|\\prod_{i=1}^{n+1}A_i\\right|=\\left|\\left(\\prod_{i=1}^nA_i\\right)\\times A_{n+1}\\right|\\qquad(2),$$ i.e., the cardinality of the set $\\prod_{i=1}^{n+1}A_i$ is equal to the cardinality of the set $\\left(\\prod_{i=1}^nA_i\\right)\\times A_{n+1}$. So \\begin{aligned}\\left|\\prod_{i=1}^{n+1}A_i\\right|&=&\\left|\\left(\\prod_{i=1}^nA_i\\right)\\times A_{n+1}\\right|&\\qquad\\text{by }(2)\\\\&=&\\left|\\prod_{i=1}^nA_i\\right|\\times |A_{n+1}|&\\qquad\\text{by }(1)\\\\&=&|A_1|\\times|A_2|\\times\\dotsc\\times|A_n|\\times|A_{n+1}|&\\qquad\\text{by induction hypothesis}\\\\&=&\\prod_{i=1}^{n+1}|A_i|.\\end{aligned}\n\n\u2022 This is exact, but I think the OP was asked a formula, but not asked to prove it. \u2013\u00a0Jean Marie Jan 19 '17 at 14:46\n\u2022 @CristianGz Do you know if (1) also holds for infinities? If so, how am I to think about multiplying 'infinities'? Is there a definition for that? \u2013\u00a0Bram28 Jan 19 '17 at 14:52\n\u2022 @JeanMarie: Agree. I deleted the answer. But I though it can be useful. \u2013\u00a0Cristhian Gz Jan 19 '17 at 14:53\n\u2022 @Bram28, the statement is true to infinite sets. It is possible using bijective functions to state equal cardinality between two sets. Check en.m.wikipedia.org/wiki/Cardinality \u2013\u00a0Cristhian Gz Jan 19 '17 at 14:55", "date": "2019-12-06 21:40:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2104483/how-to-express-the-cardinality-of-%E2%88%8F-1%E2%89%A4i%E2%89%A4n-a-i-in-terms-of-cardinalities-a", "openwebmath_score": 0.9080817699432373, "openwebmath_perplexity": 335.3435849171495, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109078121008, "lm_q2_score": 0.8723473779969193, "lm_q1q2_score": 0.8631972459292375}}
{"url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-21-minimizing-a-function-step-by-step/", "text": "# Lecture 21: Minimizing a Function Step by Step\n\nFlash and JavaScript are required for this feature.\n\n## Description\n\nIn this lecture, Professor Strang discusses optimization, the fundamental algorithm that goes into deep learning. Later in the lecture he reviews the structure of convolutional neural networks (CNN) used in analyzing visual imagery.\n\n## Summary\n\nThree terms of a Taylor series of $$F$$($$x$$) : many variables $$x$$\nDownhill direction decided by first partial derivatives of $$F$$ at $$x$$\nNewton's method uses higher derivatives (Hessian at higher cost).\n\nRelated sections in textbook: VI.1, VI.4\n\nInstructor: Prof. Gilbert Strang\n\nThe following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu.\n\nGILBERT STRANG: Well, OK, I am happy to be back, and I am really happy about the project proposals that are coming in. This is like, OK, this is really a good part of the course. And so keep them coming, and I'm happy to give whatever feedback I can on those proposals, and do make a start there. They're really good, and if some are completed before the end of the semester and we can to offer you a chance to report on them, that that's good too. So well done with those proposals.\n\nSo today, I'm jumping to part six. So part six and part seven are optimization which is the fundamental algorithm that goes into deep learning. So we've got to start with optimization. Everybody has to get that picture, and then part seven will be the structure of CNNs, Convolution Neural Nets, and all kinds of applications.\n\nAnd so can we start with optimization? So first, can I like get the basic facts about three terms of a Taylor series? So that's the typical. It's seldom that we would go up to third derivatives in optimization.\n\nSo that's the most useful approximation to a function. Everybody recognizes it. Here, I'm thinking of F as just one function, and x as just one variable, but now I really want to go to more variables. So what do I have to change if F is a function of more variables? So now, I'm thinking of x as-- well, now let me see.\n\nYeah, I want n variables here. x is x1 up to xn. So just to get the words straight so we can begin on optimization, so what will be the similar step so the function F at x-- remember, x is n variables. OK?\n\nNow, what do I have? Delta x, so what's the point about delta x now? It's a vector, delta x1 to delta xn, and what about the derivative of F? It's a vector too, the derivative of F with respect to x1, the derivative of F with respect to x2, and so on.\n\nWhat do I have to change about that? I know those guys are vectors, so it's their dot product. So it's delta x transpose at vector times this dF/dx. So now I'm replacing this by all the derivatives, and it's the gradient. So the gradient of F at x is the derivatives-- let's see.\n\nIt's essential to get the notation straight here. Yeah, so it'll be the partial derivatives of the function F. So grad F is the partial derivatives of F with respect to x1 down to partial derivative with respect to xn. OK, good.\n\nThat's the linear term, and now what's the quadratic term? 1/2, now delta x isn't a scalar anymore. It's a vector. So I'm going to have delta x transpose and a delta x, and what goes in between is the second derivatives, but I've got a function of n variables.\n\nSo now, I have a matrix of second derivatives, and I'll call it H. This is the matrix of second derivatives, Hjk is the second derivative of F with respect to xj and xk, and what's the name for this guy? The Hessian, Hessian matrix.\n\nHow the Hessians got into this picture I don't know. The only Hessians I know are the ones who fought in the Revolutionary War for somebody. Who? Which side were they on? I think maybe the wrong side. The French were on our side and--\n\nAnyway, Hessian matrix, and what are the facts about that matrix? Well, the first fact is that it's [INAUDIBLE] and the key fact is it's symmetric. Yeah. OK, and again, it's an approximation. And everybody recognizes that if n is very large, and we have a function of many variables. Then, we had n derivatives to compute here, and about 1/2 n squared derivatives.\n\nThe 1/2 comes from the symmetry, but the key point is the n squared derivatives to compute there. So computing the gradient is feasible if n is small or moderately large. Actually, by using automatic differentiation, the key idea of back propagation, back prop, you can speed up the computation of derivatives quite amazingly. But still for the size of deep learning problems that's out of reach. OK.\n\nSo that's the picture, and then I will want to use this to solve equations. There is a parallel picture for a vector f. So now, this is a vector function. This is f1 of x up to fn of x, an x is x1 to xn. So I have n functions of n variables, n functions of n variables.\n\nWell, that's exactly what I have in the gradient. Think of these two as parallel, the parallel being f corresponds to the gradient of F, n functions of n variables. OK. Now maybe, what I'm after here is to solve f equals 0. So I'm going to think about the f at x plus delta x, so it starts with f of x.\n\nAnd then we have the correction times the matrix of first derivatives, and what's the name for that matrix of first derivatives? Well, if I'm just given n functions-- yeah, what am I after here? I'm looking for the Jacobian. So here we'll go the Jacobian, J. This is the Jacobian named after Jacoby, Jacobian matrix.\n\nAnd what are its entries? J, the jk entry is the derivative of the J function with respect to the kth variable, and I'm stopping at first order there. OK, so these are sort of like facts of calculus, facts of 18.02 you could say. Multivariable calculus, that's the point.\n\nNotice that we're doing just like the first half of 18.02, just do differential calculus, derivatives, Taylor series. We're not doing multiple integrals. That's not part of our world here. OK, so that's the background.\n\nNow, I want to look at optimization. So over here, I want to optimize-- well, over here, let me try to minimize F of x, and I'll be in the vector case here. And over here, I want to solve f equals 0, and of course, that means f of 1 equals 0 all the way along to fn equals 0. Here, I have n equations, and n unknowns.\n\nLet me start with that one, and I'll start with Newton's method, Newton's method to solve these n equations and n unknowns. OK, so Newton, Newton's method which is often not presented in 18.02. That's a crime, because that's the big application of gradients in Jacobians.\n\nOK, so I'm trying to solve n equations and n unknowns, and so I want f at x plus delta x to be 0. Right? So I want f of x plus delta x to be 0. So f at x plus delta x is-- I'm putting in a 0. I'm just copying that equation-- is f at where I am. Let me use K for the case iteration.\n\nSo I'm at a point xK. I want to get to a point xK plus 1. And so I have 0 is f of x plus J, at that point, times delta x which is xK plus 1 minus xK. Good. That's Newton's method.\n\nOf course, 0 isn't quite true. Well, 0 will be true if I'm constructing xK plus 1 here. I'm constructing xK plus 1. OK. So let me just rewrite that, and we've got Newton's method. So we're looking for this change, xK plus 1 minus xK. I'll put it on this side as plus xK, so that's this.\n\nNow, I have to invert that and put it on the other side of the equation. So that will go with a minus. This guy will be inverted and f at xK. So that's Newton's methods. It's natural.\n\nSo let me just repeat that. You see where the xK plus 1 minus xK is sitting? Right? And I moved f of xK to the other side with a minus sign, and then I multiplied through by J inverse, so I got that. So that's Newton's method for a system of equations, and over there, I'm going to write down Newton's method for minimizing a function. This is such basic stuff that we have to begin here.\n\nLet me even begin with an extremely straightforward example of Newton's method here. Suppose my function-- suppose I've only got one function actually. Suppose I only had one function. So suppose my function is x squared minus 9, and I want to solve f of x equals 0. I want to find the square root of 9.\n\nOK, so what is Newton's method for it? My point is just to see how Newton's method is written and then rewrite it a little bit so that we see the convergence. OK, so of course, the Jacobian is 2x. So Newton's method says that xK plus 1-- I'm just going to copy that Newton's method-- minus 1 over 2xK. Right? That's the derivative times f at xK which is xK squared minus 9.\n\nOK. We followed the formula, this determines xK plus 1, and let's simplify it. So here I have xK minus that looks like 1/2 of xK, so I think I have 1/2xK, and then this times this is 9/2 of 1 over xK. Is that right? 1/2 of xk from this stuff and plus 9/2 of 1 over xK. OK.\n\nCan I just like check that I know the answer is 3? Can I be sure that I get the right answer, 3? That if xK was exactly 3, then of course, I expect xK plus 1 to stay at 3. So does that happen? So 1/2 of 3 and 9/2 of 1/3, what's that, 1/2 of 3 and 9/2 of 1/3?\n\nOK, that's 3/2 and 3/2. That's 6/2, and that's 3. OK. So we've checked that the method is consistent which just means we kept the algebra straight. But then the really important point about Newton's method is to discover how fast it converges. So now let me do xK plus 1 minus 3.\n\nSo now, I'm looking at the error which is, I hope, approaching 0. Is it approaching 0? How quickly is it approaching 0? These are the fundamental questions of optimization.\n\nSo I'm going to subtract 3 from both sides somehow. OK, from here, I guess, I'm going to subtract 3. So I was just checking that it was correct. OK. Now, so xK plus 1 minus 3, I'm going to subtract 3 from both sides. I'm going to subtract 3 there, and then I hope that-- that box is what goes down here. Right?\n\nSubtracted 3 from both sides, so I'm hoping now things go to 0. OK, so what do I have there? Let me factor out the 1 over xK. So what do I have then left? 1 over xK, so there's a 9/2 from there, 1 over xK.\n\nSo I really have 1/2 of xK squared, because I've divided by an xK. And this minus 3, I better put minus 3xK, because I'm dividing by xK. I claim that that's-- now I've got it. And let's see, let me take out the 2-- 2, forget these 2s, and make that a 6. So I have 1 over 2xK times 9 plus xK squared minus 6.\n\nAnything good about that? We hope so. We hope that that is something attractive. So this is, again, the error at set K plus 1, and it's 1 over 2xK times this thing in brackets-- 9 plus xK squared minus 6xK. And we recognize that as xK minus 3 squared.\n\nxK squared minus 6 of them plus 9, that's xK minus 3 squared. OK, that was the goal, of course. That's the goal that shows why Newton's method is fantastic. If you can execute it, if you can start near enough, notice that-- so how do I describe this great equation? It says that the error is squared at every step, squared at every step.\n\nSo if I'm converging to a limit, it will satisfy the-- it'll be 3, or I guess minus 3, is that possible? Yeah, minus 3 is another solution here. So we've got two solutions. Newton's method could converge to 3. Am I right, it could converge to minus 3?\n\nSo I'd have a similar equation sort of centered at minus 3, or does it always do one of those? It could blow up. So there are sort of regions of attraction. They're all the starting points that approach 3, and the whole point of that equation is with quadratic convergence the error being squared at every step. It zooms in on 3.\n\nThen, there is all the starting points that would go to minus 3, and then there are the starting points that would blow up. And those, maybe for this very simple problem, the picture is not too difficult to sort out those three regions. And this is allowing for a vector, two equations or n equations, then we're in n variables, and really you get beautiful pictures.\n\nYou get some of the type of pictures that gave rise to these books on fractals, picture books on fractals for these basins of attraction. Does the starting point lead you to one of the solutions, or does it lead you to infinity? Here, that would be interesting to just draw it for this, but the essential point is the quadratic convergence, if it's close enough.\n\nYou see that it has to be close. If x0 is pretty near 3, then this is about 1/6 of that, and there would be a good region of attraction in this case. OK. So that's Newton's method for equations.\n\nAnd now I want to do Newton's method. I just want to convert all those words over to Newton's method for optimization. So remember, these boards were solving f equals 0. This board is minimizing capital F, and what's the connection between them? Well of course, this corresponds to solving the gradient equals 0.\n\nAt a minimum, if I'm minimizing, I'm finding a point where all the first derivatives are 0. So that will be the match between these. This grad F in this picture is the small f in that picture. OK.\n\nNow, I guess here I have-- and this is sort of the heart of our applications to deep learning-- we have very complicated loss functions to minimize, functions of thousands or hundreds of thousands of variables. OK. So that means that we would like to use Newton's method, but often we can't. So I need him to put down here two methods-- one that doesn't involve those high second derivatives and Newton's that does.\n\nSo first, I'll write down a method that does not involve, so method one, and this will be steepest descent. And what is that? That says that xK plus 1-- the new x is the old x minus-- steepest descent means that I move in the steepest direction which is the direction of the gradient of F. I move some distance, and I better have freedom to decide what that distance should be. So this is a step size, s, or in the language of deep learning, it's often called the learning rate, so if you see learning rate. OK.\n\nSo and it's natural to choose sK. We're going along, do you see what this right-hand side looks like? I'm at a point in n dimensions. We're in n dimensions here. We have functions of n variables.\n\nThere is a vector. There is a direction to move down the steepest slope of the graph. And here is a distance to move, and we will stop. We'll have to get off this step, normally. If we stay on it, it will swing back, it'll take us off to infinity.\n\nYou would like to choose sK so that you minimize capital F. You take the point on this line, so this a line in R n, a direction in R n. And for all the points on that line, in that direction, F has some value, and what you expect is that initially, because you chose it sensibly, the value of F will drop. But then at a certain point, it will turn back on you and increase.\n\nSo that would be the natural stopping point. I would call that an exact line search. So I exact line search would be, exact line search is the best s. Of course, that would take time to compute, and you probably, in deep learning, that's time you can't afford, so you fix the learning rate s. Maybe you choose 0.01 to be pretty safe.\n\nOK, so that's method one, steepest descent. Now, method two will be Newton's method. So now, we have xK plus 1 equal to xK minus something times delta F, and now I'm going to do the right thing. I'm going to live right here, and the right thing is the Hessian, the second derivative.\n\nThis was cheap. We just took the direction and went along it. Now, we're getting really the right direction by using the second derivative, so that's H inverse. OK, and what I've done is to set that 0.\n\nDo you see that's Newton's method? It's totally parallel to this guy. Actually, I'm really happy to have these two on the board parallel to each other, because you have to keep straight, are you solving equations, or are you minimizing functions? And you're using different letters in the two problems, but now you see how they match.\n\nThe Jacobian of-- so again the matches, think of f as the gradient of F. That's the way you should think of it. So the Jacobian of the gradient is the Hessian. The Jacobian of the gradient is the Hessian, and that makes sense, because the first derivative of the first derivative is the second derivative. Only we're doing matrix y, so the Jacobian of the gradient-- we're doing a vector matrix sentence instead of a scalar sentence-- the Jacobian of the gradient is a Hessian. Yeah, right.\n\nOK, so that's what I wanted to start with, just to get those basic facts down. And so the basic facts were the three-term Taylor series. And then the basic algorithms followed naturally from it by setting f F at the new point to 0, if that's what you were solving or by assuming you had the minimum. Right, good, good, good, good. OK.\n\nNow, what? Now, we have to think about solving these problems, studying. Do they converge? What rate do they converge? Well, the rate of convergence is like why I separated off this example.\n\nSo the convergence rate for Newton's method will be quadratic. The error gets squared, and of course, that means super-fast convergence, if you start and close enough. The rate of convergence for a steepest descent is, of course, not. You're not squaring errors here, because you're just taking some number instead of the inverse of the correct matrix, so you can't expect super speed.\n\nSo a linear rate of convergence would be right. You would like to know that the error is multiplied at every step by some constant below 1. That would be a linear rate compared to being squared at every step. OK, and so this will be our basic formula that we build on for really large scale problems.\n\nAnd there are methods, of course, people are going to come up with methods that they're sort of a cheap Newton's method. Levenberg-Marquardt, and it's in the notes at the end of this section, at the end of 6.4 that we'll get to. So Levenberg-Marquardt is a sort of cheap man's Newton's method. It does not compute to Hessian, but it says, OK, from the gradient, I can see one term in the Hessian. So it grabs that term, but it's not fully second order.\n\nOK. So now, we have to think about problems, and I guess the message here is, at our starting point, has to be convexity. Convexity is the key word for these problems, for the function that we want to minimize. If that's a convex function, well first of all, the convex function is likely to have one minimum. And the picture that's in our mind of steepest descent, this picture of a bowl, a bowl is the graph of the convex function.\n\nSo I'm turning to convexity now. I'll leave that board there, because that's pretty crucial, and speak about the idea of convexity. Convex function, convex set, so let's call the function f of x, and a typical convex set would be I'll call it K. OK. So we just want to remember what does that word can convex mean, and how do you know if you have a convex function or a convex set?\n\nOK, let me start with convex set. So because here is my general problem, my convex minimization, which you hope to have, and in many applications, you do have. So you minimize a convex function for points in a convex set. So that's like the ideal situation. That's the ideal situation, to get something on your side, something powerful, convexity.\n\nThe function is convex, and you say, well, let me draw a convex function, the graph. OK, so I'll draw a convex function, say a bowl. So that's a graph of f of x, and then here are the x's. Let me maybe put x1 and x2 in the base and the graph of f of 1x x2 up here. OK. Actually, I'm over there.\n\nI should be calling this function F, I think. Is that right? Yeah, a little f would be the gradient of this guy. Yeah, I think so. OK.\n\nNow, I'm minimizing it over certain x's, not all x's. I might be minimizing, for example, K might be the set where Ax equals B. K might be, in that case, a subspace or a shifted subspace. I said subspace, but then 18.06 is reminding me in my mind that I only have a subspace when B is 0.\n\nYou know the word for a subspace that's sort of moved over? Affine, so I'll just put that word down here. Bunch of words to learn for this topic, but they're worth learning. OK.\n\nSo it's like a plane but not necessarily through the origin. If B is 0, it doesn't go through it. If B it's not 0, it doesn't go through the origin. OK. Anyway, or I have some other convex set. Let me just put this convex set K in the base for you, and did I make it convex? I think pretty luckily I did.\n\nSo now what's the? Well, the convex sets the constraint, so this is the constraint set. Constraint is that x must be in the set K. OK, and I drew it as a convex blob. Here was an example where it's flat, not a blob but a flat plane.\n\nBut let me come back to what does convex mean. What's a convex set? Yeah, we have to do that, should have done that before. In the notes, I had the fun of figuring out, if I took a triangle, is that a convex set? Let's just be sure.\n\nSo what's a convex set? That is a convex set, because if I take any two points in the set and draw the line between them, it stays in the set. So that's convexity, any edge, line, from x1 to x2 stays in the set. OK, good.\n\nSo here's my little exercise to myself. What if I took the union of two triangles? All I want to get you to do is just visualize convex and not convex possibilities. Suppose I have one triangle, even if it was obtuse, that's still convex, right? No problem.\n\nBut now what if I put those two triangles together, take their union? Well, if I take them sitting with a big gap between, like I've lost. I mean, I never had a chance that way, because if it was the union of these two-- well, you know what I'm going to say. If I'm doing that point and that point, of course, it goes outside and stupid. All right.\n\nWhat if what if that triangle, that lower triangle, overlaps the upper triangle? Is that a convex set? Everybody's right saying no. Why how do I see that the union of those two triangles is not a convex set? Guys, you tell me where to pick two points, where the line goes out. Well, I take one from that corner and one from that corner, and the line between them went outside. So union is usually not convex.\n\nWell, if I think of the union of two sets, my mind automatically goes to the other corresponding possibility which is the intersection of the two sets. So if I take the intersection of two sets. Now, what's the deal with that? When I had two triangles, two separated triangles, what can we say about the intersection of those two triangles?\n\nAUDIENCE: [INAUDIBLE]\n\nGILBERT STRANG: It's empty. So should we regard the empty set as a convex set? Yes. Isn't it?\n\nAUDIENCE: Yeah, it's vacuous.\n\nGILBERT STRANG: Vacuous, so it hasn't got any problems. Right? OK, but now the intersection is always convex. I'm assuming the two sets that we start with are. Now, that's an important fact, that the intersection of convex sets. Let's just draw a picture that shows an example.\n\nSo what's the intersection? Just this part and it's convex. OK, can you give me a little proof that the intersection is convex? So I take two points in the intersection-- let me start the proof.\n\nTo test if something's convex, how do you test it? You take two points in the set in the intersection, and you want to show that the line between them is in the intersection. OK, why is that?\n\nSo take two points, take x1 in the intersection. We've got two sets here, and that's the symbol for intersection, and we've got another point in the intersection. And now, we want to look at the line between them, the line from x1 to 2x. What's the deal with that one? Is that fully in K1?\n\nAUDIENCE: Yes.\n\nGILBERT STRANG: Why is it fully in K1? I took two points in the intersection, I'm looking at the line between them, and I'm asking, is it in the first set K1? And the answer is yes, because those points were in K1, and K1's convex. And is that line between them in K2? Yes, same reason, the two endpoints were in K2, so the line between them is in K2.\n\nSo the intersection of convex sets is always convex. The intersection of convex sets is convex. Good. So you'll see in the note these possibilities with two triangles. Sometimes, you can take the union but not very often. OK.\n\nNow, what's the next thing I have to do? Convex functions, we got convex sets, what are convex functions, and we're good. Because this is our prototype of a problem, and I now want to know what it means for that F to be-- oh, I'm sorry. I now know what it means for the set K to be convex set, but now I have to look at the other often more important part of the problem. What's the function I'm minimizing, and I'm looking for functions with this kind of a picture. OK.\n\nThe coolest way is to connect the definition of a convex function to the definition of a convex set. This is really the nicest way. It's a little quick. It just swishes by you. But tell me, do you see a convex set in that picture? [INAUDIBLE]\n\nYou see a convex set in that picture. That's the picture of a graph of a convex function. It's a picture of a bowl. Are the points on that surface, is that a convex set? No, certainly not. No, but where is a convex set to be found here, in that picture? Yes.\n\nAUDIENCE: The set of y, if y is greater than [INAUDIBLE]\n\nGILBERT STRANG: Yes, the points on and above the bowl, inside the bowl, we could say, these points. So convex function, yes, a function's convex when the points on and above the graph are convex set. You could say, OK, mathematicians are just being lazy. Having got one definition straight for a convex set, now they're just using that to give an easy definition of a convex function. Actually, it's quite useful for functions that could maybe equal infinity, sort of generalized functions.\n\nBut it's not the quickest way to tell if the function is convex. It's not our usual test for convex functions. So now I want to give such a test. OK. So now, the definition of convex function, of a smooth convex, yeah. This fact, I shouldn't rush off away from it, from the definition of a convex function as having a convex set above its graph. The really official French name for the set above the graph is the epigraph, but I won't even write that word down. OK.\n\nWhy do I come back to that for a minute? Because I would like to think about two functions, F1 and F2. Out of two functions, I can always create the minimum or the maximum.\n\nSo suppose I have to convex functions, convex function F1 and F2. OK. Then, I could choose a minimum. I could choose my new function. Shall I call it little m for minimum? m of x is the minimum of F1 and F2.\n\nAnd I could choose a maximum function which would be the maximum of F1 of x and F2 of x at the same point x. It's just a natural to think, OK, I have two functions. I've got a bowl and I've got another bowl, and suppose they're both convex.\n\nSo I'm just stretching you to think here. If I've got the graphs of two convex functions, and I would like to consider the minimum of those two functions and also the maximum of those two functions. I believe life is good. One of these will be convex, and the other won't.\n\nAnd can you identify which one is convex and which one is not convex? What about the minimum? Is that a convex function? So just look at the graph. What does the minimum look like? The minimum is this guy until they meet somehow on some surface and then this guy.\n\nIs that convex? We have like one minute to answer that question. Absolutely no. It's got this bad kink in it. What about the maximum of the two functions? So the maximum is the one that is above, all the points or things that are above or on.\n\nThere is the maximum function. That was the minimum function. It had a kink. The maximum function is like that, and it is convex, so maximum yes, minimum no. OK, and we could have a maximum of 1,500 functions. If the 1,500 functions are all convex, the maximum will be, because it's the part way above everybody's graph, and that would be the graph of the maximum. OK, good.\n\nAnd now finally, let me just say, how do you know whether a function is convex? How to test, how of test. OK, so let me take just a function of one variable. What's the test you learned in calculus, freshman calculus actually, just show that this is a convex function? What's the test for that?\n\nAUDIENCE: Use second derivative.\n\nGILBERT STRANG: Second derivative should be?\n\nAUDIENCE: Positive.\n\nGILBERT STRANG: Positive or possibly 0, so second derivative greater or equals 0 everywhere. That's convex. OK, final question, suppose F is a vector. So this is a vector, and so I have n functions of n variable. No, I don't. I have one, sorry, I've got one function, but I'm in n variables. So this was just one.\n\nWhat's the test for convexity? So it would be passed, for example, by x1 squared plus x2 squared. Would it be passed by-- so here would be the question-- would it be passed by x transpose some symmetric matrix S? That would be a quadratic, a pure quadratic.\n\nWould it be convex? What would be the test? I'm looking for an n dimensional equivalent of positive second derivative. The n dimensional equivalent of positive second derivative is convexity, and we have to recognize what's the test. So I could apply it to this function, or I could apply it to any function of n variables. It should be OK.\n\nWhat's the test here? Here, I have a matrix instead of a number. So what's the requirement going to be? Times out, yeah? [INAUDIBLE] Positive definite or semidefinite, or semidefinite just as here. Yeah.\n\nSo the test is positive, semidefinite, Hessian. And here, the Hessian is actually that S, because the second derivatives will produce-- I'll put a 1/2 in there-- the second derivatives will produce S equal the Hessian H. So here, the S-- so positive semidefinite, Hessian in general, second derivative matrix for a quadratic.\n\nOK. So its convex problems that we're going to get farther with. We run into no saddle points. We run into no local minimum. Once we found the minimum, it's the global minimum. These are the good problems. OK, again, happy to see you today, and I look forward to Wednesday.", "date": "2021-10-16 17:30:31", "meta": {"domain": "mit.edu", "url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-21-minimizing-a-function-step-by-step/", "openwebmath_score": 0.8232865929603577, "openwebmath_perplexity": 490.2249764044709, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.993511728866456, "lm_q2_score": 0.868826784729373, "lm_q1q2_score": 0.8631896009819636}}
{"url": "https://www.jiskha.com/questions/86506/i-have-3-questions-and-i-cannot-find-method-that-actually-solves-them-1-integral", "text": "# Calculus - Integrals\n\nI have 3 questions, and I cannot find method that actually solves them.\n\n1) Integral [(4s+4)/([s^2+1]*([S-1]^3))]\n\n2) Integral [ 2*sqrt[(1+cosx)/2]]\n\n3) Integral [ 20*(sec(x))^4\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. 1) expand in partial fractions.\n\n2) Using cos(2x) = 2 cos^2(x) -1 derive a formula for cos(1/2 x) in terms of cos(x). Express the integrand in terms of cos(1/2 x)\n\n3) [Notation: cos = c, sin = s]\n\n1/c^4 = (s^2 + c^2)/c^4 =\n\n1/c^2 + s^2/c^4\n\n1/c^2 yields a tangent when integrated\n\nTo integrate s^2/c^4 do partial integration:\n\ns^2/c^4 = s (s/c^4)\n\nIntegral of s/c^4 is 1/3 1/c^3\n\nSo, we need to integrate\n\n1/c^3\ntimes the derivative of of s, i.e. 1/c^2, but that is tan(x)!\n\nNote that the way to solve such problems is not to systematically work things out in detail at first, because then you would take too long to see that a method doesn't work.\n\nInstead, you should reason like I just did, i.e. forget the details, be very sloppy, just to see if things works out and you get an answer in principle, even though you need to fill in the details.\n\nIf you get better at this, you can do the selection of what method to use in your head, you'll see it in just a few seconds when looking at an integral. You can then start to work out the solution for real on paper.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n2. Could you expand on #1 a bit more? I tried Partial Fractions, but I couldn't get a definite answer...\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. I'll show you how to do it without solving any equations.\n\nThe function is up to a factor 4:\n\nf(s) = (s+1)/{(s^2+1)*[(s-1)^3]}\n\nThe partial fractions are precisely the singular terms when expanding around the singularities. So, let's examine the singularity at s = 1:\n\nPut s = 1 + u and expand in powers of u, keeping only the singular terms:\n\nf(1+u) = u^(-3) (2+u)/[(u+1)^2+1] =\n\nu^(-3) (2+u)/(2+2u+u^2) =\n\nu^(-3) (2+u)/2 1/(1+u+u^2/2) = (use\n1/(1+x) = 1-x+x^2-x^3+...)\n\nu^(-3) (2+u)/2 [1-u-u^2/2 +u^2+...]\n\n=[u^(-3) + u^(-2)/2][1-u+u^2/2+...]\n\nu^(-3) -1/2 u^(-2) + nonsingular terms\n\nThis means that in the neighborhood of s = 1 we have:\n\nf(s) = 1/(s-1)^3 - 1/2 1/(s-1)^2 +.. nonsingular terms\n\nf(s) also has singularities at s = i and s =-i. Around s = i, we have:\n\nf(s) = a/(s-i) + nonsingular terms\n\nMultiply both sides by (s-i) and take the limit s --> i to find a:\n\na = (i+1)/[2i(i-1)^3] =\n\n-1/2 1/(i-1)^2\n\nAround s = -i, we have:\n\nf(s) = b/(s+i) + nonsingular terms\n\nMultiply both sides by (s+i) and take the limit s --> -i to find b:\n\nb = (-i+1)/[-2i(-i-1)^3] =\n\n-1/2 1/[(i+1)^2]\n\nThe sum of all the singular terms of the expansions around the singular points is:\n\n1/(s-1)^3 - 1/2 1/(s-1)^2 +\n\n-1/2 1/(i-1)^2 1/(s-i)\n\n-1/2 1/[(i+1)^2] 1/(s+i) =\n\n1/(s-1)^3 - 1/2 1/(s-1)^2 +\n\n1/2 1/(s^2+1)\n\nThis is the desired partial fraction decomposition. The proof of why this works is a consequence of Liouville's theorem\n\nhttp://en.wikipedia.org/wiki/Liouville&#039;s_theorem_(complex_analysis)#Proof\n\nJust consider the difference of f(s) and the sum of all the singular terms. The resulting function doesn't have any singularites and is bounded. So, by Liouville's theorem it is a constant. We know that f(s) and all the singular terms tend to zero, so that constant must be zero.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\n## Similar Questions\n\n1. ### Calculus\n\nIf f(x) and g(x) are continuous on [a, b], which one of the following statements is true? ~the integral from a to b of the difference of f of x and g of x, dx equals the integral from a to b of f of x, dx minus the integral from a\n\n2. ### calculus integrals\n\nEvaluate the integral by making the given substitution. (Use C for the constant of integration. Remember to use absolute values where appropriate.) integral x^5/x^6-5 dx, u = x6 \u2212 5 I got the answer 1/6ln(x^6-5)+C but it was\n\n3. ### Calculus\n\nSuppose the integral from 2 to 8 of g of x, dx equals 5, and the integral from 6 to 8 of g of x, dx equals negative 3, find the value of the integral from 2 to 6 of 2 times g of x, dx . 8 MY ANSWER 12 16 4\n\n4. ### Calculus\n\nFind the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y= 2e^(\u2212x), y= 2, x= 6; about y = 4. How exactly do you set up the integral? I know that I am supposed to use\n\n1. ### calculus\n\n1.Evaluate the integral. (Use C for the constant of integration.) integral ln(sqrtx)dx 2. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the curves about the given axis. y =\n\n2. ### Calculus check and help\n\nLet f and g be the functions given by f(x)=1+sin(2x) and g(x)=e^(x/2). Let R be the shaded region in the first quadrant enclosed by the graphs of f and g. A. Find the the area of R. B. Find the value of z so that x=z cuts the\n\n3. ### Physics, Calculus(alot of stuff together)= HELP!!\n\nA rod extending between x=0 and x= 14.0cm has a uniform cross- sectional area A= 9.00cm^2. It is made from a continuously changing alloy of metals so that along it's length it's density changes steadily from 2.70g/cm^3 to\n\n4. ### Calculus (math)\n\nThe volume of the solid obtained by rotating the region enclosed by y=1/x4,y=0,x=1,x=6 about the line x=\u22122 can be computed using the method of cylindrical shells via an integral. it would be great if you can just even give me\n\n1. ### Calculus - Integrals\n\nI have 3 questions, and I cannot find method that actually solves them. 1) Integral [(4s+4)/([s^2+1]*([S-1]^3))] 2) Integral [ 2*sqrt[(1+cosx)/2]] 3) Integral [ 20*(sec(x))^4 Thanks in advance.\n\n2. ### Science\n\nHi, I have two questions about biology I need help with. 1) Some questions fall outside the realm of science, which of the following questions could not be answered using the scientific method? A)What is the function of the\n\n3. ### Math\n\nConsider the curve represented by the parametric equations x(t)= 2+sin(t) and y(t)=1-cos(t) when answering the following questions. A) Find Dy/Dx in terms of t B) Find all values of t where the curve has a horizontal tangent. C)\n\n4. ### calculus\n\nA question on my math homework that I can't seem to solve... Rotate the region bounded by y=x^2-3x and the x-axis about the line x=4. Set up the integral to find the volume of the solid. I'm pretty sure that the integral is in", "date": "2021-01-18 03:57:32", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/86506/i-have-3-questions-and-i-cannot-find-method-that-actually-solves-them-1-integral", "openwebmath_score": 0.8567370772361755, "openwebmath_perplexity": 1031.8355444994497, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992960608886, "lm_q2_score": 0.8918110353738529, "lm_q1q2_score": 0.8631832733576844}}
{"url": "https://discourse.julialang.org/t/general-plotting-code-for-cone-in-3d-with-glmakie-or-plots/92104", "text": "# General Plotting Code for Cone in 3D with GLMakie or Plots\n\nHi all,\n\nso I want to plot a general cone of pyramid and a right circular cone, I get this formula:\n\nBut I think the plotting does not use the volume formula. Because of this torus code for GLMakie that I have:\n\nusing GLMakie\nusing ColorSchemes\ncmap3 = get(colorschemes[:plasma], LinRange(0,1,100))\ncmap4 = [(cmap3[i],0.65) for i in 1:100]\n\u03be = -40:1:40\nc = 20\na = 15\ntorus = [((hypot(X,Y)-c)^2+Z^2-a^2) for X in \u03be, Y in \u03be, Z in \u03be]\nvolume(torus, algorithm = :iso,isovalue =100,isorange =50, colormap=cmap4)\ncurrent_figure()\n\n\nFor general cone of pyramid, If the based is made of a closed random curves, then we might need to plot in x-y plane the curves then plot the top of pyramid afterwards.\n\nFor the right circular cone, I believe there is a general formula already.\n\nYou could conveniently define both as parametric surfaces:\n\nusing Plots; plotlyjs()\n\nz, \u0398 = range(0,10,200), range(0,2\u03c0,300)\n\n# Right-cone:\nX = [u * cos(v) for u in z, v in \u0398]\nY = [u * sin(v) for u in z, v in \u0398]\nZ = [-2u for u in z, _ in \u0398]\nsurface(X, Y, Z, size=(600,600), cbar=:none, legend=false)\n\n# Trefoil cone:\nX = [u/3 * (sin(v) + 2*sin(2v)) for u in z, v in \u0398]\nY = [u/3 * (cos(v) - 2*cos(2v)) for u in z, v in \u0398]\nZ = [-2u for u in z, _ in \u0398]\nsurface(X, Y, Z, size=(600,600), cbar=:none, legend=false)\n\n\n1 Like\n\nFrom your questions posted on this forum it seems that you are teaching math. Hence you should explain to your students how is parameterized a general cone,\ndefined by its vertex V, given as a 3 vector, and a closed (but not necessarily) planar curve,\n\nx = x(u)\ny = y(u)\nz= b # b from base plane; usually b=0\n\n\nu \\in [\\alpha, \\beta]\n.\nThe cone is generated by all segments of line connecting its vertex, V, with the points\n(x(u), y(u), b), on the base curve.\nSuch a segment has the equations:\n\\displaystyle\\frac{X-V[1]}{x(u)-V[1]} = \\displaystyle\\frac{Y-V[2]}{y(u)-V[2]} =\\displaystyle\\frac{Z-V[3]}{b-V[3]}=v\nHence:\n\nX = v(x(u)-V[1])+V[1]\nY = v(y(u)-V[2])+V[2]\nZ = v(b-V[3])+V[3]\n\n\nu \\in [\\alpha, \\beta], v\\in[0,1]\nExample:\n\nvert=[0, 0.75, 3] #cone vertex\nbase = 0 #the cone base is included within the plane z=base (here z=0)\n#functions that define the cone base parameterization\nx(u) = cos(u)\ny(u) = sin(u)+ sin(u/2);\nm, n = 72, 30\nu= range(0, 2\u03c0, length=m)\nv = range(0, 1, length=n)\nus = ones(n)*u'\nvs = v*ones(m)'\n#Surface parameterization\nX = @. vs* (x(us)-vert[1]) + vert[1]\nY = @. vs* (y(us)-vert[2]) + vert[2]\nZ = @. vs*(base-vert[3]) + vert[3];\nsurface(X, Y, Z, size=(600,600), cbar=:none, legend=false) #line copied from Rafael\n\n2 Likes\n\nYes 100 for you, I was a Mathematics teacher for Junior High School students in Papua, but it was 2 years ago. Afterwards, I still love learning Mathematics even if I no longer teach, now I am learning the undergraduate Mathematics, from Calculus, It would be lovely to teach Math again in the future.\n\nThis is the first time I learn Calculus with Julia from bottom, it is rewarding knowing the application of this is enormous.\n\nThanks for this I will use your code and learn the parameterization, vertex, etc.\n\n1 Like", "date": "2023-01-31 10:27:36", "meta": {"domain": "julialang.org", "url": "https://discourse.julialang.org/t/general-plotting-code-for-cone-in-3d-with-glmakie-or-plots/92104", "openwebmath_score": 0.7099421620368958, "openwebmath_perplexity": 5962.698746512539, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9615338101862454, "lm_q2_score": 0.8976952982655951, "lm_q1q2_score": 0.8631643805275957}}
{"url": "https://mathhelpboards.com/threads/root-equation.4882/", "text": "# Root equation\n\n#### Albert\n\n##### Well-known member\n$m,n \\in N$\n\nsatisfying :\n\n$\\sqrt {m+2007}+\\sqrt {m-325}=n$\n\nfind Max(n)\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nRe: root equation\n\n[JUSTIFY]Apply the variable substitution $u = m - 325$, and hence $m + 2007 = u + 2332$. We get:\n$$\\sqrt{u + 2332} + \\sqrt{u} = n$$\nFor $n$ to be an integer, both square roots have to be integers*. So we are looking for two (positive) squares which differ by $2332$ units. That is, we need to solve:\n$$x^2 - y^2 = 2332$$\nDoing some factoring:\n$$(x - y)(x + y) = 2^2 \\times 11 \\times 53$$\nAfter some observation** we have the following solution pairs:\n$$(x, y) \\to \\{ ( 64, 42), ~ (584, 582) \\}$$\nIt follows that the only solutions for $u$ are $u = 42^2$ and $u = 582^2$. This gives $n = 106$ and $n = 1166$ (and $m = 2089$, $m = 339049$ respectively).\n\nTherefore $\\max{(n)} = 1166$. $\\blacksquare$\n\n* perhaps this part merits justification, you can check Square roots have no unexpected linear relationships | Annoying Precision which is pretty hardcore (perhaps a reduced argument suffices here) but I will take it that the question assumed this to be true. Other approaches which do not rely on this assumption would be interesting to see as well.\n\n** this can be solved efficiently by using every reasonable factor combination for $x - y$ and deducing $x + y$ from it.[/JUSTIFY]\n\nLast edited:\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nRe: root equation\n\nAs a slight variation on Bacterius's solution, take $u$ to be $m+841$ rather than $m-325$. The equation then becomes $$\\sqrt{u+1166} + \\sqrt{u-1166} = n.$$ Square both sides to get $$2u + 2\\sqrt{u^2-1166^2} = n^2.$$ For an integer solution, $u^2-1166^2$ must be a square, say $u^2-1166^2 = w^2.$ Thus $(w,1166,u)$ is a pythagorean triple, which must be of the form $(p^2-q^2,\\,2pq,\\,p^2+q^2)$ for some integers $p,q$ with $p>q$. In particular, $1166=2pq$. But $1166 = 2*11*53$, so the only possibilities are $p=53,\\ q=11$, or $p=583,\\ q=1.$ That then leads to the same solutions as those of Bacterius.\n\nLast edited:\n\n##### Well-known member\nRe: root equation\n\nwe have\n(m+ 2007) \u2013 (m- 325) = 2332 ..1\n\u221a(m+2007)+\u221a(m\u2212325) = n ..(2) given\n\nDividing (2) by (1) we get\n\n\u221a(m+2007)-\u221a(m\u2212325) = 2332/n \u2026 (3)\n\nAnd (2) and (3) to get 2 \u221a(m+2007) = n + 2332/n\n\nNow RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.\n\n#### Albert\n\n##### Well-known member\nlet :\n$u^2=m-325-------(1)$\n$(u+a)^2=m+2007-------(2)$\n$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)\nmax(n) must happen while a=2,so from (3) we have u=582\nso max(n)=582+584=1166\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nRe: root equation\n\nwe have\n(m+ 2007) \u2013 (m- 325) = 2332 ..1\n\u221a(m+2007)+\u221a(m\u2212325) = n ..(2) given\n\nDividing (2) by (1) we get\n\n\u221a(m+2007)-\u221a(m\u2212325) = 2332/n \u2026 (3)\n\nAnd (2) and (3) to get 2 \u221a(m+2007) = n + 2332/n\n\nNow RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.\nVery neat!\n\n#### Albert\n\n##### Well-known member\nnow ,find min(n)=?\n\n##### Well-known member\nnow ,find min(n)=?\n\u221a(m+2007)+\u221a(m\u2212325) = n\nput m- 325 = p (which is >=0)\n\nwe get \u221a(p+2332)+\u221ap which is >= sqrt(2332) >= 49\n\nnow we should look for lowest even factor of 2332(even : reason same as my previous solution) which is >= 49\n\nnow 2332 = 2 * 2 * 11 * 53 and lowest even factor of 2332 >= 49 is 2 * 53 = 106\n\n#### Albert\n\n##### Well-known member\nlet :\n$u^2=m-325-------(1)$\n$(u+a)^2=m+2007-------(2)$\n$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)\nmax(n) must happen while a=2,so from (3) we have u=582\nso max(n)=582+584=1166\n$2332=a^2+2ua=a(2u+a)=a\\times n=2\\times 1166=22\\times 2\\times53=22\\times 106$\nso max(n)=1166 ,and a=2\nmin(n)=106, and a=22\n(here a,and n=2u+a are all even numbers)", "date": "2020-09-22 17:57:55", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/root-equation.4882/", "openwebmath_score": 0.8123118877410889, "openwebmath_perplexity": 2089.5002114735153, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765552017675, "lm_q2_score": 0.8807970858005139, "lm_q1q2_score": 0.8631604939745432}}
{"url": "http://mathhelpforum.com/algebra/4571-help-again.html", "text": "# Math Help - Help again\n\n1. ## Help again\n\nPlease explain how to solve this.\n\nThe product of two consecutive even integers is 168. Find the two integers.\nHint: Assume the first integer is x\nSet up the equation first that satisfies the given condition. Then solve the equation.\n\nThis is the equation that i came up with. Ican t seem to get past\n\nx(x+2)=168\n\nx^2+2x-168=0 (i cant get past here)\n\nThanks,\nChester\n\n2. Originally Posted by Chester\n\nx(x+2)=168\n\nx^2+2x-168=0 (i cant get past here)\nGood job.\nFind some factors of 168: 12 and 14\nwhich diffrence is 2 thus,\n$(x+14)(x-12)=168$\nThus,\n$x=-14,x=12$\n\n3. You could always fall back on the quadratic formula. I'd rather factor.\n\nWhat 2 numbers when added equal 2 and when multiplied equal -168.\n\nLet's see.......how about.....................-12 and 14. (-12)(14)=-168\n\n-12+14=2\n\n$x^{2}-12x+14x-168$\n\n$(x^{2}-12x)+(14x-168)$\n\n$x(x-12)+14(x-12)$\n\n(x-12)(x+14)\n\n4. Hello, Chester!\n\nYou did an excellent on job on the hard part: setting up the equation.\n\nI assume you know to solve a quadratic equation\n. . and that you know how to factor.\nDid you run of factorings to try?\n\n$x^2+2x-168\\:=\\:0$\n\nWe want two numbers with a product of 168 and a difference of 2.\n\nStart dividing 168 by 1, 2, 3, . . .\n\n$168 \\div 1 = 168\\quad\\Rightarrow\\quad 1\\cdot168$\n\n$168 \\div 2 = 84\\quad\\Rightarrow\\quad 2\\cdot84$\n\n$168 \\div 3\\;\\;\\text{ not exact}$\n\n$168 \\div 4 = 42\\quad\\Rightarrow\\quad 4\\cdot42$\n\n$168 \\div 5,\\; 168 \\div 6,\\;\\168 \\div 7\\;\\;\\text{ not exact}$\n\n$168 \\div 8 = 21\\quad\\Rightarrow\\quad 8\\cdot21$\n\n$168 \\div 9,\\;168 \\div 9,\\;168\\div 10,\\;168\\div11\\;\\;\\text{ not exact}$\n\n$168 \\div 12 = 14\\quad\\Rightarrow\\quad 12\\cdot14\\;\\;\\leftarrow$ There! a difference of 2\n\nTherefore: . $x^2 + 2x - 168 \\;= \\;(x - 12)(x + 14)$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nOf course, you can use the Quadratic Formula for factoring . . .\n\n5. ## Thanks\n\nThanks again everyone.\n\nI really appreciate the help!\n\nChester\n\n6. I believe the answer is $12\\cdot14$ not $(-12)\\cdot14$\nnotice that negative 12 and 14 are not consecutive even integers.\n\n7. Originally Posted by Quick\nI believe the answer is $12\\cdot14$ not $(-12)\\cdot14$\nnotice that negative 12 and 14 are not consecutive even integers.\nThere two answers for $x$.\n$x=12, x+2=14$ and $12\\times 14=168$\nAnd,\n$x=-14,x+2=-12$ and $-14\\times -12=168$", "date": "2016-04-29 21:59:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/4571-help-again.html", "openwebmath_score": 0.8737580180168152, "openwebmath_perplexity": 754.6039265854218, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9799765563713599, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8631604904051197}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=q0844f24o3ornn2o4mongo9t30&action=printpage;topic=1600.0", "text": "# Toronto Math Forum\n\n## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:05:34 PM\n\nTitle: Q7 TUT 0201\nPost by: Victor Ivrii on November 30, 2018, 04:05:34 PM\n(a) Determine all critical points of the given system of equations.\n\n(b) Find the corresponding linear system near each critical point.\n\n(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?\n\n(d)\u00a0 Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.\n\\left\\{\\begin{aligned} &\\frac{dx}{dt} = 1 - xy, \\\\ &\\frac{dy}{dt} = x - y^3. \\end{aligned}\\right.\n\nBonus: Computer generated picture\nTitle: Re: Q7 TUT 0201\nPost by: Yulin WANG on November 30, 2018, 04:39:52 PM\n(a)\n\n\\left\\{\n\\begin{array}{**lr**}\n1-xy=0 &\u00a0 \\\\\nx-y^{3}=0\\\\\n\\end{array}\n\\right.\n\n\\left\\{\n\\begin{array}{**lr**}\nxy=1 &\u00a0 \\\\\nx=y^{3}\\\\\n\\end{array}\n\\right.\n\n\\left\\{\n\\begin{array}{**lr**}\nx=1 &\u00a0 \\\\\ny=1\\\\\n\\end{array}\n\\right.\nor\n\\left\\{\n\\begin{array}{**lr**}\nx=-1 &\u00a0 \\\\\ny=-1\\\\\n\\end{array}\n\\right.\n\nTherefore, the critical points are (1,1) and (-1,-1)\n(b)\nThe Jacobian matrix of the vector field is:\n\\begin{align*}\nJ &= \\begin{bmatrix}\n-y & -x \\\\\n1 & -3y^{2}\n\\end{bmatrix}\\\\\n~\\\\\nJ(1,1) &= \\begin{bmatrix}\n-1 & -1 \\\\\n1 & -3\n\\end{bmatrix}\\\\\n~\\\\\nJ(-1,-1) &= \\begin{bmatrix}\n1 & 1 \\\\\n1 & -3\n\\end{bmatrix}\n\\end{align*}\n(c)\n\\begin{align*}\nFor (1,1), let A&= \\begin{bmatrix}\n-1 & -1 \\\\\n1 & -3\n\\end{bmatrix}\\\\\n~\\\\\nA-\\lambda I &= \\begin{bmatrix}\n-1-\\lambda & -1 \\\\\n1 & -3-\\lambda\n\\end{bmatrix}\\\\\n~\\\\\ndet(A-\\lambda I) &=(\\lambda+3)(\\lambda+1)+1=0\\\\\n~\\\\\n\\lambda_{1} &= \\lambda_{2} = -2 \\\\\n~\\\\\nThen \\ the \\ system \\ has \\ a \\ stable \\ improper \\ node \\ at \\ (1,1) \\\\\n~\\\\\nFor (-1,-1), let A&= \\begin{bmatrix}\n1 & 1 \\\\\n1 & -3\n\\end{bmatrix}\\\\\n~\\\\\nA-\\lambda I &= \\begin{bmatrix}\n1-\\lambda & 1 \\\\\n1 & -3-\\lambda\n\\end{bmatrix}\\\\\n~\\\\\ndet(A-\\lambda I) &=(\\lambda+3)(\\lambda-1)-1=0\\\\\n~\\\\\n\\lambda = -1 \\pm \\sqrt{5} \\\\\n~\\\\\nThen \\ the \\ system \\ has \\ a \\ unstable \\ saddle \\ point \\ at \\ (1,1) \\\\\n\\end{align*}\n(d) In the attachment.\nTitle: Re: Q7 TUT 0201\nPost by: Zhuojing Yu on November 30, 2018, 06:29:30 PM\nI think when (1,1), it is node or spiral point, not IN(improper node).\nTitle: Re: Q7 TUT 0201\nPost by: Jingze Wang on November 30, 2018, 08:31:23 PM\nI think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted\nTitle: Re: Q7 TUT 0201\nPost by: Yulin WANG on November 30, 2018, 11:13:57 PM\nI think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted\nThanks for submitting the computer-generated phase portrait!!!\nBTW, how do u plot the phase portrait on a computer?\nTitle: Re: Q7 TUT 0201\nPost by: Victor Ivrii on December 01, 2018, 03:57:59 AM\nSure, it is improper node. Jingze finally got a decent computer generated picture (took a correct range of variables)\n\nSyllabus lists several free sites. My favourite -- downloadable pplain java applet", "date": "2022-12-02 23:33:35", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=q0844f24o3ornn2o4mongo9t30&action=printpage;topic=1600.0", "openwebmath_score": 0.9964171051979065, "openwebmath_perplexity": 6885.432337606231, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765592953409, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.863160486847757}}
{"url": "http://math.stackexchange.com/questions/353752/what-does-it-mean-to-be-proportional-to-something/353758", "text": "# What does it mean to be proportional to something?\n\nI am asked a physics homework question, but it is really simply a mathematical question I think, dealing with proportional reasoning.\n\nThe period of a pendulum is proportional to the square root of its length.\n\nA pendulum of length $2$ has a period of $3$.\n\nThey give units but it really should not matter here.\n\nWhat I have done is written this down, and I am not really sure the first step is the correct one, because I am unsure how to read the question:\n\n$$3P_1 = \\sqrt 2$$\n\n$$P_1 = \\frac{\\sqrt 2}{3}$$\n\nThey want to know about a pendulum of length $4.5$ so,\n\n$$xP_2 = \\sqrt {4.5}$$\n\nMy reasoning then is that:\n\n$$\\frac{\\sqrt 2}{3} = \\frac{\\sqrt{4.5}}{x}$$\n\n$$xP_2 = \\frac{3\\sqrt{4.5}}{\\sqrt 2} = 4.5$$\n\nI never really deal with proportional reasoning, so I am going out on a limb here to make this connection, but it all seems very intuitive. Of course, my intuition is not always right. My question then is, am I doing this right? Is this good reasoning?\n\n-\n\nHint:\n\nTake T to be the time period of the pendulum.\n\n$T_1 = k \\cdot \\sqrt {l_1}$\n\n$T_2=k \\cdot \\sqrt{l_2}$\n\nWhen you divide them both(Since none of the terms are $0$)\n\n$\\dfrac{T_1}{T_2}= \\sqrt{\\dfrac{l_1}{l_2}}$\n\nNote: When L is proportional to M, you can write it as L=k. M, where k is a constant.\n\n-\nHm that is interesting, and it confirms my answer when I use the k, that really helps too. It is truly discomforting that my brain does all that without informing me how it was done.. Many thanks! \u2013\u00a0 Leonardo Apr 7 '13 at 10:56\nYou're Welcome. And do note that there might be many cases of proportionality, when you come across Gravitational force (or maybe coulomb's force), you come across 'inversely proportional to'. \u2013\u00a0 Inceptio Apr 7 '13 at 10:59\nIt may be noted that here the actual value of k is 2*pi/sqrt(g) where g is accelaration due to gravity. \u2013\u00a0 Sreekanth Karumanaghat Apr 7 '13 at 11:24\n@SreekanthKarumanaghat: That really doesn't matter when you have two relations, cause: they just scare off each other. \u2013\u00a0 Inceptio Apr 7 '13 at 11:50\nYeah I know that I was just noting for info sake :) \u2013\u00a0 Sreekanth Karumanaghat Apr 7 '13 at 12:08\n\nThe correct meaning of propotional:- If x is directly propotional to y. x/y = k(a constant)\n\nAnalytically\n\nGraph between x and y is straight line whose slope is k.\n\nApplying the above formula here. T1 = k sqrt(L1) T2 = k sqrt(L2)\n\ndividing 2 by 1,we get T1/T2 = sqrt(L1)/sqrt(L2)\n\n3/T2 = sqrt(2)/sqrt(4.5)\n\nT2 = sqrt(4.5)*3/sqrt(2)\n\nT2 = 4.5\n\n-", "date": "2014-12-21 12:05:31", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/353752/what-does-it-mean-to-be-proportional-to-something/353758", "openwebmath_score": 0.8827373385429382, "openwebmath_perplexity": 589.441344181974, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765616345254, "lm_q2_score": 0.880797071719777, "lm_q1q2_score": 0.8631604858417055}}
{"url": "https://mathematica.stackexchange.com/questions/101247/recursion-question", "text": "# Recursion question\n\nThe first two terms of a sequence are $a_1 = 1$ and $a_2 = \\frac {1}{\\sqrt3}$. For $n\\ge1$,\n\n$$a_{n + 2} = \\frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$$ What is $|a_{2009}|$?\n\nSo I tried (thanks to Paul Wellin):\n\nfun[n_] := Module[{prev = 1, this = 1/Sqrt[3], next},\nDo[next = Simplify[(prev + this)/(1 - prev this)];\nprev = this;\nthis = next, {n - 1}];\nprev]\n\n\nAnd I got:\n\nfun[2009]\n\n(* 0 *)\n\n\nWhich is the correct answer. Kind of amazed it went so quickly. I then tried:\n\nTable[fun[i], {i, 1, 100}]\n\n\nWhich produces the first 100 terms of the sequence. Again, amazingly quick. Now, this turns out to be a periodic sequence. I counted with my finger, and determined 24.\n\nAnyone have a good way to use Mathematica to determine if the sequence is periodic and find the period?\n\n\u2022 mathematica.stackexchange.com/questions/80163/\u2026 \u2013\u00a0Marvin Dec 5 '15 at 9:31\n\u2022 Entirely different method: Recognize the recursion as the sum of angles formula for tangent. Set $\\tan \\theta_{n} = a_n$ so $\\theta_1 = \\pi/4$, $\\theta_2 = \\pi/6$, and $\\theta_{n+2} = \\theta_n + \\theta_{n+1}$. This is a much easier recurrence... \u2013\u00a0Eric Towers Dec 5 '15 at 19:08\n\u2022 @EricTowers Could you add this as an answer? I'd love to see what you do. \u2013\u00a0David Dec 6 '15 at 3:33\n\nOne way of doing this is starting with your list of the first 100 terms.Then ask for\n\nzzz = SequencePosition[lst, Take[lst, -2]]\n\n(* {{3, 4}, {27, 28}, {51, 52}, {75, 76}, {99, 100}} *)\n\n\nThe length of this result is larger than 1, so your list must be periodic. The period is\n\nzzz[[2, 1]] - zzz[[1, 1]]\n\n(* 24 *)\n\n\u2022 thank you introducing me to SequencePosition \u2013\u00a0ubpdqn Dec 5 '15 at 10:56\n\u2022 @FredSimons I agree, really simple explanation to help students with exploration. Thanks. \u2013\u00a0David Dec 5 '15 at 18:19\n\nStarted as a comment to the OP. Turning this into an answer was requested. The OP asked for Mathematica help solving the recurrence. I suggested recognizing that the recurrence was the sum of angles formula for tangent, then going from there. Consequently, this answer uses Mathematica to solve this problem using only one \"external\" ingredient right at the beginning. (At the end, we explore a claim made at the beginning.)\n\nProblem: Find the period of \\begin{align} a_1 &= 1 \\\\ a_2 &= \\frac{1}{\\sqrt{3}} \\\\ a_{n+2} &= \\frac{a_n + a_{n+1}}{1-a_n a_{n+1}}, n \\geq 1 \\end{align}\n\n(It would be seriously convenient if FunctionPeriod[] just did this. We could \"punt\" and use the accepted answer here, applying it to the straightforward implementation of the above recursion and initial values. However, that's not what I'm here to write.)\n\n\"External\" fact: The recursion is the sum of angles formula for tangent.\n\nClaim at the beginning: We could replace the recursion formula with the analog for any other sum of angles formula and get exactly the same sequence of angles. I.e., there are at least two other recursions with this same solution.\n\nFirst, we check this external fact with Mathematica and get the initial values.\n\nSimplify[Tan[a+b] == (Tan[a] + Tan[b])/(1-Tan[a]Tan[b])\n(*  True  *)\nArcTan[1]\n(*  Pi/4  *)\nArcTan[1/Sqrt[3]]\n(*  Pi/6  *)\n\n\nSo we may reformulate the question as \\begin{align} \\theta_1 &= \\frac{\\pi}{4} \\\\ \\theta_2 &= \\frac{\\pi}{6} \\\\ \\theta_{n+2} &= \\theta_1 + \\theta_2 \\pmod{2\\pi}, n \\geq 1 \\end{align}\n\nSums of multiples of $\\pi/4$ and $\\pi/6$ have denominators dividing $12$.\n\nLCM[4,6]\n(*  12  *)\n\n\nThere are $24$ twelfths of $\\pi$ in a circle.\n\n2 Pi / (Pi/12)\n(*  24  *)\n\n\nConsequently, the period of this recurrence is at most $24^2$. (The recurrence is order $2$, so it is perhaps better to think of each adjacent pair of sequence members producing the next adjacent pair of sequence members.) (Although, as is frequently the case with homogeneous recurrences, if $\\dots 0, 0 \\dots$ occurs in our sequence, every subsequent angle is zero. So it is possible for the period to be less than $24$ and for there to be an initial aperiodic part, say if $\\theta_1 = \\theta_2 = \\pi$, although it will turn out that this is not the case in this problem.)\n\nSo we generate the first $2 \\times 24^2$ entries, which will contain two or more periods.\n\nRemove[q];\nq[1] = Pi/4;\nq[2] = Pi/6;\nq[n_ /; n >= 3] := q[n]= Mod[q[n-1]+q[n-2],2 Pi];\n(* We memoize members of the sequence to avoid redundant recomputation. *)\n\n\nGenerate the $2 \\times 24^2$ entries and glance at some of them to see if periodicity seems likely.\n\nTable[ {k, q[k]} , {k, 24, 2*24^2, 24} ]\n(*  {{ 24, 23 Pi/12 }, { 48, 23 Pi/12 }, ... , {1152, 23 Pi/12 }}  *)\n\n\nSeems rather likely to be periodic. Test this with Fourier[]:\n\nperiods = Flatten[\nPosition[\nChop[Fourier[Table[q[k], {k, 1, 2*24^2}]]],\nn_ /; n != 0\n]\n]\n(*  {1, 49, 97, 145, 193, 241, 289, 337, ... , 1009, 1057, 1105}  *)\nGCD @@ Select[DeleteDuplicates[Flatten[\nOuter[Abs[#1 - #2] &, periods, periods]\n]], # > 0 &]\n(*  48  *)\n\n\nSo the period divides $48$. (Actually, we could know that it divides $24$. We ran to $2 \\times 24^2$. Try it again with $3 \\times 24^2$ and get $72$. In fact, if we run to $N \\times 24^2$ for $N$ up to $24$, we'll get $24N$ for the GCD above.)\n\nLet's have Mathematica try every divisor of $48$ as a candidate period.\n\n{#, Simplify[q[3 + #] == q[3]]} & /@ Divisors[48]\n(*  {{1, False}, {2, False}, {3, False}, {4, False},\n{6, False}, {8, False}, {12, False}, {16, False},\n{24, True}, {48, True}}  *)\n\n\nso the only candidate periods are $24$ and $48$. (Note that starting at q[3], we're finding out about the recursion moreso than the initial values.) Then we test the shortest candidate:\n\nq /@ Range[24] == q /@ Range[25, 2*24]\n\n\nThe above method works, and is valid, but perhaps one would be happier seeing that the recurrence actually recurs after $24$ steps. Starting with the symbolic first two angles $a,b$, what is the resulting angle after (a divisor of $24$) applications of the recursion rule. We frame this as working on pairs in Nest[].\n\nTableForm[\n{#[[1]], #[[2]] === {a, b}, #[[2]]} & /@ (\n{#, PolynomialMod[\nNest[\n(* Thinking in pairs, the new first angle is the old second\nangle and the new second angle is the sum of the two old\nangles. *)\n{#[[2]], #[[1]] + #[[2]]} &,\n{a, b},\n#],\n#]} & /@\nDivisors[24]),\nTableDepth -> 1\n]\n(*  { { 1, False, {0, 0}},\n{ 2, False, {a + b, a}},\n{ 3, False, {a + 2 b, 2 a}},\n{ 4, False, {2 a + 3 b, 3 a + b}},\n{ 6, False, {5 a + 2 b, 2 a + b}},\n{ 8, False, {5 a + 5 b, 5 a + 2 b}},\n{12, False, {5 a, 5 b}},\n{24,  True, {a, b}} }  *)\n\n\nSo the recursion has a natural period of $24$. It's possible for the choice of initial conditions to cause the period to be shorter. (Recall that if the first two angles are $0,0$, then the sequence is all zeroes.) So we check for this. The resulting period still has to divide $24$, so we only check those (although if we were not certain of this, we could check every period up to $24$).\n\nTableForm[\n{ #[[1]] , #[[2]] == #[[3]] , #[[2]] , #[[3]] } & /@ (\n{ # , {q[1], q[2]}, {q[1 + #], q[2 + #]} } & /@\nDivisors[24]\n),\nTableDepth -> 1]\n(*  { { 1, False, {Pi/4, Pi/6}, {  Pi/ 6,  5 Pi/12}},\n{ 2, False, {Pi/4, Pi/6}, {5 Pi/12,  7 Pi/12}},\n{ 3, False, {Pi/4, Pi/6}, {7 Pi/12,    Pi   }},\n{ 4, False, {Pi/4, Pi/6}, {  Pi   , 19 Pi/12}},\n{ 6, False, {Pi/4, Pi/6}, {7 Pi/12,    Pi/ 6}},\n{ 8, False, {Pi/4, Pi/6}, {3 Pi/ 4, 11 Pi/12}},\n{12, False, {Pi/4, Pi/6}, {  Pi/ 4,  5 Pi/ 6}},\n{24,  True, {Pi/4, Pi/6}, {  Pi/ 4,    Pi/ 6}} }  *)\n\n\nAnd so the period is $24$. Then it's easy to compute the desired number:\n\nTan[q[Mod[2009,24,1]]]  (*Don't want to inadvertently try to evaluate q[0]. *)\n(* This is Tan[ q[17] ] = Tan[ Pi ]. *)\n(*  0  *)\n\n\nThe original problem statement asked about $|a_{2009}|$ but since $0 = |0|$, there's no need to apply the absolute value function.\n\nReturning to the claim at the beginning. Since adding angles is pretty straightforward, we can use any other sum of angles formula. This means there are (at least) two other recursion problems looking wildly different from this one (by having a wildly different looking general recursion formula), but having the same underlying computation. The sum of angles formulas for sine and cosine are, with the corresponding recursion: \\begin{align} \\sin(\\theta_1 + \\theta_2) &= \\sin(\\theta_1) \\cos(\\theta_2) + \\cos(\\theta_1) \\sin(\\theta_2) \\\\ a_{n+2} &= a_n c(a_{n+1}) \\sqrt{1-a_{n+1}^2} + a_{n+1} c(a_n) \\sqrt{1-a_n^2} \\\\ \\\\ \\cos(\\theta_1 + \\theta_2) &= \\cos(\\theta_1)\\cos(\\theta_2) - \\sin(\\theta_1)\\sin(\\theta_2) \\\\ a_{n+2} &= a_n a_{n+1} - s(a_n)s(a_{n+1}) \\sqrt{1-a_n^2}\\sqrt{1-a_{n+1}^2} \\end{align} where the sign-tracking functions $s$ and $c$ are $s(\\theta)$ is $1$ in quadrants I and II and the angle $\\pi/2$ and is $-1$ otherwise, and $c$ is $1$ in quadrants I and IV and the angle $0$ and is $-1$ otherwise. We could eliminate these two sign-tracking functions by having two sequences, $q_n$ and $r_n$: $a_n \\mapsto (q_n, r_n) = (\\sin(\\theta_n), \\cos(\\theta_n))$ or by switching to exponentials $a_n \\mapsto \\mathrm{e}^{\\mathrm{i} \\theta_n} = \\cos(\\theta_n) + \\mathrm{i}\\sin(\\theta_n)$, which the final equality shows is really the same as switching to sines and cosines, but using real and imaginary parts of complex numbers to hold the two sequences, instead of pairs of real numbers.\n\nTaking $\\theta_1 = \\pi/4$ and $\\theta_2 = \\pi/6$ and using either the sum of angles for sine or for cosine, we get the same sequence of angles. Consequently, we will get a final, final answer of $a_{2009} = \\tan^{-1}(\\sin(\\theta_{2009})) = 0$ for the sine recursion and similarly for the cosine recursion. (Whether you take the simple recurrence $\\theta_{n+2} = \\theta_n + \\theta_{n+1}$ and obscure it via sines, cosines, tangents, or, I suppose, other periodic functions, like Weierstrass's $\\wp$-functions, you still get the same sequence of fractions of the functions's period(s), and the same period of the recurrence.)\n\n\u2022 +1, but the recurrence you display for the sine and cosine is in fact numerically unstable, and thus cannot be recommended in general. \u2013\u00a0J. M. will be back soon Dec 10 '15 at 5:18\n\u2022 @J.M. : I agree in part. If we're going to talk about stability, the initial tangent recursion isn't stable either. Replace $a_{n+1}$ by $a_{n+1}+\\varepsilon$ and expand in powers of $\\varepsilon$ and find that the error is magnified by $\\frac{1+a_n^2}{(1-a_n a_{n+1})^2}$. This is harder to interpret than doing the same thing to the tangent version, yielding $1+\\tan^2(a_n + a_{n+1})$, which is necessarily $\\geq 1$. So the instability is baked into the problem and the alternate recursions are merely reflecting it. \u2013\u00a0Eric Towers Dec 10 '15 at 17:11\n\nYou can use NestWhileList to show the same pair of consecutive elements appear and then determine the period, e.g.\n\nres = NestWhileList[{#[[2]],\nSimplify[(#[[1]] + #[[2]])/(1 - #[[1]] #[[2]])]} &, {1,\n1/Sqrt[3]}, Unequal, All];\nplt = (Most@(First /@ res))~Join~(Style[#, Red, Bold] & /@ res[[-1]])\nLength[plt] - 2\nListPlot[plt~Join~res[[1]], Joined -> True]\n\n\nI add this as a separate answer to amplify another answer,in relation to $\\tan(x+y)$.\n\nThis sequence could be pursued as Fibonacci or linear recurrence of angles,e.g.\n\nmp[n_] :=\nMod[MatrixPower[{{1, 1}, {1, 0}}, n].{Pi/6, Pi/4}, Pi, -Pi/2];\nan[1] := mp[0][[2]];\nan[2] := mp[0][[1]];\nan[n_?(# > 2 &)] := First@mp[n - 2]\nos[n_] := Tan[an[n]];\nSequencePosition[an /@ Range[1, 26], {Pi/4, Pi/6}]\nos /@ Range[26]\n\n\nI exploited Fred Simons SequencePosition (yielding: {{1, 2}, {25, 26}}) and the last line reproduces the original sequence:\n\n$\\left\\{1,\\frac{1}{\\sqrt{3}},\\sqrt{3}+2,-\\sqrt{3}-2,0,-\\sqrt{3}-2,-\\sqrt{3}-2,\\frac{1}{\\sqrt{3}},-1,\\sqrt{3}-2,-\\sqrt{3},-\\sqrt{3}-2,1,-\\frac{1}{\\sqrt{3}},2-\\sqrt{3},\\sqrt{3}-2,0,\\sqrt{3}-2,\\sqrt{3}-2,-\\frac{1}{\\sqrt{3}},-1,-\\sqrt{3}-2,\\sqrt{3},\\sqrt{3}-2,1,\\frac{1}{\\sqrt{3}}\\right\\}$\n\nA quick visual inspection helps to see how a pattern with period 24 pops up\n\nn = 100;\n\ntab = Table[fun[i], {i, 1, n}];\n\npos = Flatten @ Position[tab, First @ tab]\n\n\n{1, 13, 25, 37, 49, 61, 73, 85, 97}\n\ntra = Transpose[{pos, Array[First @ tab &, Length @ pos]}];\n\nListPlot[{tab, tra}, GridLines -> {pos[[;; ;; 2]], None}]\n\n\nI rewrote the code to produce a list of the current \"state\" of the recursion, which is given by the last two terms $(a_n,a_{n-1}$. The system is autonomous, so the state is completely determined by this pair. Therefore if a pair ever occurs again, there is a loop.\n\nSome utility functions:\n\nstep =           (* basic recursive step *)\n{Last[#], Simplify[Plus @@ #/(1 - Times @@ #)]} &;\nsteps[n_] :=     (* equivalent to fun[n], except it returns a list of all states *)\nNestList[step, {1, 1/Sqrt[3]}, n];\nfindloop[] :=    (* like steps[n], except it runs until a loop is found; caveat forever *)\nNestWhileList[step, {1, 1/Sqrt[3]}, UnsameQ, All];\n\n\nFirst proof:\n\ngr = Rule @@@ Partition[findloop[], 2, 1] // Graph;\ncycle = First@FindHamiltonianCycle[gr];\nLength@cycle\nHighlightGraph[gr, PathGraph[cycle]]\n\n\nLess fussy proof, assuming you can make a good guess (i.e. 24):\n\nRule @@@ Partition[steps[24], 2, 1] // Graph\n(* graph similar to above, but without highlighting *)\n\n\nEven less fussy, assuming you can make a sufficient guess (i.e. 100):\n\ngr = Rule @@@ Partition[steps[100], 2, 1] // Graph\nLength@VertexList[gr]\n(*\ngraph similar to above, but with multiple edges\n24  <-- number of vertices == period\n*)\n\n\nHere is a translation of Brent's algorithm for finding the length of a cycle. See Brent, doi:10.1007/BF01933190 (1980). This could be compiled when step can be compiled for cases where a loop is long or occurs a long way down the sequence.\n\nbrentcyclelength[step_, x0_, \u03c1_: 2, u_: 0] :=\nModule[{x, y, r, j, k, done = False},\ny = x0;\nr = \u03c1^u;\nk = 0;\nWhile[! done,\nx = y;\nj = k;\nr = \u03c1*r;\nWhile[! done && k < r,\nk++;\ny = step[y];\ndone = x == y\n];\n];\nk - j\n]\n\nbrentcyclelength[step, {1, 1/Sqrt[3]}]\n(*  24  *)", "date": "2019-12-16 01:53:13", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/101247/recursion-question", "openwebmath_score": 0.920558512210846, "openwebmath_perplexity": 2247.9093825703367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830769252026, "lm_q2_score": 0.8872045877523147, "lm_q1q2_score": 0.863099921587011}}
{"url": "https://math.stackexchange.com/questions/852122/picture-intuitive-proof-of-cos3-theta-4-cos3-theta-3-cos-theta/852230", "text": "# Picture/intuitive proof of $\\cos(3 \\theta) = 4 \\cos^3(\\theta)-3\\cos(\\theta)$?\n\nIs there a nice geometric, intuitive or picture proof as to why the easily algebraically provable identity $\\cos(3 \\theta) = 4 \\cos^3(\\theta)-3\\cos(\\theta)$ is true?\n\nNote I'm not looking for a computational proof like the one linked to, more a proof without words or intuitive style proof, thanks.\n\nEnhancing my diagram for the angle-sum formula (currently featured in Wikipedia) to use three angles will get you pretty close ...\n\nThus,\n\n\\begin{align} \\cos(\\alpha+\\beta+\\gamma) &= \\cos\\alpha \\cos\\beta \\cos\\gamma - \\sin\\alpha \\sin\\beta \\cos\\gamma - \\sin\\alpha \\cos\\beta\\sin\\gamma - \\cos\\alpha \\sin\\beta\\sin\\gamma \\\\ \\sin(\\alpha+\\beta+\\gamma) &= \\sin\\alpha \\cos\\beta \\cos\\gamma + \\cos\\alpha \\sin\\beta \\cos\\gamma + \\cos\\alpha \\cos\\beta \\sin\\gamma - \\sin\\alpha \\sin\\beta \\sin\\gamma \\end{align}\n\nWith $\\alpha = \\beta = \\gamma = \\theta$, these become ... \\begin{align} \\cos 3\\theta &= \\cos^3\\theta - 3 \\sin^2\\theta \\cos\\theta \\\\ \\sin 3\\theta &= 3\\cos^2\\theta \\sin\\theta - \\sin^3\\theta \\end{align} ... which the Pythagorean identity helps us rewrite as ... \\begin{align} \\cos 3\\theta &= \\cos^3\\theta - 3 (1-\\cos^2\\theta) \\cos\\theta = 4\\cos^3\\theta - 3 \\cos\\theta \\\\ \\sin 3\\theta &= 3(1-\\sin^2\\theta) \\sin\\theta - \\sin^3\\theta = 3\\sin\\theta - 4\\sin^3\\theta \\end{align}\n\nOff-hand, I don't know of a diagram that goes directly from $\\cos 3\\theta$ to $4\\cos^3\\theta-3\\cos\\theta$.\n\n\u2022 That is absolutely amazing. \u2013\u00a0bobby Jun 30 '14 at 13:29\n\u2022 Very colorful! =) \u2013\u00a0Arnie Bebita-Dris Sep 4 '14 at 12:36\n\u2022 I just love it when someone has the time and effort to make a truly helpful and fascinating visual explanation. Sometimes when the numbers and letters just swim around, a picture can be very helpful \u2013\u00a0Asimov Sep 4 '14 at 14:17\n\u2022 How do you get something featured on Wikipedia? Do you submit it somewhere and then wait for someone to place it on an article? \u2013\u00a0\u00e9tale-cohomology Aug 9 '17 at 22:31\n\u2022 @\u00e9tale-cohomology: The \"wiki\" in \"Wikipedia\" indicates that the content is user-editable. Anybody can go in and make improvements. As for my diagrams: I'd posted them here, and then someone posted \"similar\" versions to Wikipedia (without credit or notice); someone from here at Math.SE edited the post so that I'd get credit. (Thanks again @ChrisSherlock!) I, myself, recently went in and edited the article's description of those diagrams and provided a link to my new website for such things in the article's list of references. \u2013\u00a0Blue Aug 9 '17 at 23:00\n\nAssemble four congruent right triangles with hypotenuse $1$ and angle $\\theta$ as shown:\n\nIn the diagram, the marked angles are all $\\theta$. Since $OQ = \\cos\\theta$, we have $OU = \\cos^2\\theta$, so $OS = 2\\cos^2\\theta$, and so $$OS' = 2\\cos^3\\theta$$ On the other hand, $$OS' = OP' + P'Q' + Q'S' = \\cos(3\\theta) + P'Q' + \\cos(\\theta)$$ and $$OS' = OR' - Q'R' + Q'S' = \\cos(\\theta) - Q'R' + \\cos(\\theta)$$ Since $P'Q' = Q'R'$, adding these together yields $$\\cos(3\\theta) + 3\\cos(\\theta) = 2OS' = 4\\cos^3\\theta$$\n\nI hope the diagram is reasonably self-explanatory, that it's clear what needs to be proved to justify the steps of the argument, and that it's easy to prove those things once you've identified them. But if I've misjudged this, let me know and I'll fill in some more details.\n\nI don't think it is a very good idea to try an find a geometric picture for every trig identity because that would take exceedingly long. I will show you that your formula comes from the idea that a rotation by $n\\theta$ is equal to $n$ rotations by $\\theta$.\n\nDenote $\\langle a,b\\rangle$ as the vector to the ordered pair $(a,b)$. We will discuss rotations of these vectors. First, consider that the rotation of the vector $\\langle 1,0\\rangle$ by an angle of $\\theta$ is given by $\\langle \\cos\\theta,\\sin\\theta\\rangle$. Similarly, rotation of the vector $\\langle 1,0\\rangle$ by $3\\theta$ is given by $\\langle \\cos(3\\theta),\\sin(3\\theta)\\rangle$.\n\nI will now define a new type of multiplication between vectors. Let $$\\langle a,b\\rangle\\ltimes\\langle c,d \\rangle = \\langle ac-bd,ac+bc\\rangle$$\n\nIt so happens that if the vector $\\langle a,b\\rangle$ makes an angle $\\phi_1$ with the horizontal and $\\langle c,d \\rangle$ makes an angle $\\phi_2$ then $\\langle a,b\\rangle\\ltimes\\langle c,d \\rangle$ makes an angle $\\phi_1 + \\phi_2\\!$. Using this, there is another way to find the rotation of a vector by $3\\theta$. We can rotate it three times by $\\theta$ rather than rotating it once by $3\\theta$. In other words, we will find $\\left(\\langle \\cos\\theta,\\sin\\theta \\rangle \\ltimes \\cos\\theta,\\sin\\theta \\rangle\\right) \\ltimes \\cos\\theta,\\sin\\theta \\rangle.$ Using this, one could suspect that \\begin{align}\\langle \\cos(3\\theta),\\sin(3\\theta)\\rangle &= \\left(\\langle \\cos\\theta,\\sin\\theta \\rangle \\ltimes \\cos\\theta,\\sin\\theta \\rangle\\right) \\ltimes \\cos\\theta,\\sin\\theta \\rangle \\\\ &= \\langle \\cos^2\\!\\theta-\\sin^2\\!\\theta,2\\sin\\theta\\cos\\theta\\rangle \\ltimes \\langle \\cos\\theta, \\sin\\theta \\rangle \\\\ &= \\langle \\cos^{3}\\!\\theta-\\sin^2\\!\\theta\\cos\\theta-2\\!\\sin^2\\!\\theta\\!\\cos\\theta, 2\\!\\sin\\theta\\!\\cos^2\\!\\theta+\\sin\\theta\\cos^2\\!\\theta-\\sin^3\\!\\theta\\rangle \\\\ &=\\langle 4\\cos^3\\!\\theta-3\\cos\\theta,3\\sin\\theta-4\\sin^3\\!\\theta\\rangle\\end{align}\n\nWhere the last step follows from the Pythagorean identities. If we then equate the first elements and the second elements we find the familiar results\n\n\\begin{align} \\cos(3\\theta) &= 4\\cos^3\\!\\theta-3\\cos\\theta \\\\ \\sin(3\\theta) &= 3\\sin\\theta-4\\sin^3\\!\\theta\\end{align}\n\nThis is essentially a result of complex numbers and is called De moivre's formula. You can see it being used to prove your identity here. Our type of multiplication mirrors complex multiplication which is inherently rotational in nature. You can see why here. De moivre's formula says that a rotation by $n\\theta$ is equal to $n$ rotations by $\\theta$.\n\nThe best geometric proof of your formula comes from this basic idea which is so simple that no picture is needed to represent it.\n\nThis is advantageous because it can be used to show other identites such as $$\\cos(5\\theta) = 16 \\cos^5\\! \\theta - 20 \\cos^3 \\!\\theta + 5 \\cos \\theta$$\n\n\u2022 This is neither geometric nor intuitive and doesn't address the question in a natural way at all. \u2013\u00a0beep-boop Jul 2 '14 at 22:14\n\u2022 @alexqwx: most of the proofs, including those using De Moivre's, are geometrically based. However, it is hard to find any that I would call intuitive. \u2013\u00a0robjohn Jul 7 '14 at 0:18\n\nThis is not a direct proof but a link to the Fourier series: $4\\cos^3\\theta=3\\cos\\theta+\\cos3\\theta$.\n\nSimilarly, for the fifth power: $16\\cos^5\\theta=10\\cos\\theta+5\\cos3\\theta+\\cos5\\theta$.\n\nThe main lesson is that to represent the $k^{th}$ power, the odd harmonics from $1$ to $k$ suffice.\n\nThe explanation is not so difficult: the Fourier coefficients are computed from the integrals $$\\int_0^{2\\pi}\\cos^k\\theta\\ e^{in\\theta}d\\theta=\\int_0^{2\\pi}\\left(\\frac{e^{i\\theta}-e^{-i\\theta}}2\\right)^ke^{in\\theta}d\\theta.$$ When developing, the lowest exponent of $e^{i\\theta}$ is $n-k$. If $n>k$, the integrand is oscillatory and the integral vanishes.\n\nIf you admit that, then $\\cos^3\\theta=a\\cos\\theta+b\\cos3\\theta$.\n\nFrom the figure, at $\\theta=0$ the values add up to $1$ and at $\\theta=\\pi/2$ the slopes cancel out, so that $$a+b=1\\\\a-3b=0,$$ $$\\cos^3\\theta=\\frac34\\cos\\theta+\\frac14\\cos3\\theta.$$\n\nSimilarly, $\\cos^5\\theta=a\\cos\\theta+b\\cos3\\theta+c\\cos5\\theta$, at $\\theta=0$ the values add up to $1$ and at $\\theta=\\pi/2$ the first and third derivatives cancel out, so that $$a+b+c=1\\\\ a-3b+5c=0\\\\ a-27b+125c=0,$$ $$\\cos^5\\theta=\\frac{10}{16}\\cos\\theta+\\frac{5}{16}\\cos3\\theta+\\frac{1}{16}\\cos5\\theta.$$\n\nyes.you can write: $cos(3\\theta) = cos(2\\theta +\\theta)$ =$$cos(2\\theta)cos(\\theta) - sin(2\\theta)sin\\theta$$ =$$2cos^3(\\theta)-cos \\theta -sin(2\\theta) sin(\\theta)$$=$$2cos^3\\theta -cos\\theta -2sin^2\\theta cos\\theta$$=$$2cos^3\\theta - cos \\theta - 2(cos\\theta -cos^3 \\theta)$$ so we have : $$4cos^3\\theta -3cos\\theta$$\n\n\u2022 I've already linked to an easy algebraic proof, I'm hoping for a more intuitive, geometric and less computational proof. \u2013\u00a0bobby Jun 30 '14 at 12:10\n\u2022 sorry I do not see it .for picture you can use Matlab software. \u2013\u00a0user157745 Jun 30 '14 at 12:16", "date": "2021-04-11 19:23:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/852122/picture-intuitive-proof-of-cos3-theta-4-cos3-theta-3-cos-theta/852230", "openwebmath_score": 0.9981534481048584, "openwebmath_perplexity": 496.87783340682546, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290955604488, "lm_q2_score": 0.888758793492457, "lm_q1q2_score": 0.8630995232957255}}
{"url": "https://www.intmath.com/forum/methods-integration-31/find-integral-sqrt-x-2-1-using-trigonometric:155", "text": "IntMath Home \u00bb Forum home \u00bb Methods of Integration \u00bb Find integral sqrt (x^2 + 1) using trigonometric substitution\n\n# Find integral sqrt (x^2 + 1) using trigonometric substitution [Solved!]\n\n### My question\n\nFind int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.\n\n### Relevant page\n\n8. Integration by Trigonometric Substitution\n\n### What I've done so far\n\nI used x = tan theta and dx = sec^2 theta d theta.\n\nReplacing x and dx gives sqrt (tan^2 theta + 1) sec^2 theta = sqrt (sec^2 theta) sec ^2 theta = sec ^3 theta d theta.\n\nI looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.\n\nX\n\nFind  int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.\nRelevant page\n\n<a href=\"https://www.intmath.com/methods-integration/8-integration-trigonometric-substitution.php\">8. Integration by Trigonometric Substitution</a>\n\nWhat I've done so far\n\nI used x = tan theta and dx = sec^2 theta d theta.\n\nReplacing x and dx gives sqrt (tan^2 theta + 1)  sec^2 theta = sqrt (sec^2 theta)  sec ^2 theta = sec ^3 theta d theta.\n\nI looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.\n\n## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution\n\n@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)\n\nJust a small (but important) point - don't miss out the \"d theta\" parts in your second line. It should be:\n\nReplacing x and dx gives sqrt (tan^2 theta + 1) (sec^2 theta) d theta = sqrt (sec^2 theta) (sec ^2 theta) d theta = sec ^3 theta d theta.\n\nNow, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?).\n\nWhen we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?\n\nX\n\n@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)\n\nJust a small (but important) point - don't miss out the \"d theta\" parts in your second line. It should be:\n\nReplacing x and dx gives sqrt (tan^2 theta + 1)  (sec^2 theta) d theta = sqrt (sec^2 theta)  (sec ^2 theta) d theta = sec ^3 theta d theta.\n\nNow, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?).\n\nWhen we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?\n\n## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution\n\ntheta = arctan (0) = 0\n\ntheta = arctan (1) = .7853\n\nX\n\ntheta = arctan (0) = 0\n\ntheta = arctan (1) = .7853\n\n## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution\n\nAlso this is not a textbook exercise. This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the Trapezoidal Rule in section 5. Thanks.\n\nX\n\nAlso this is not a textbook exercise.  This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the <a href=\"https://www.intmath.com/integration/5-trapezoidal-rule.php\">Trapezoidal Rule in section 5</a>. Thanks.\n\n## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution\n\nAhh, I see. Thanks for giving me the context.\n\nYour lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table.\n\nX\n\nAhh, I see. Thanks for giving me the context.\n\nYour lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table.\n\n## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution\n\nLeaving it in terms of theta:\n\nint sec^3 theta d theta from 0 to .7853\n\nAccording to Wolfram, the integral formula is\n\n1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]\n\nTherefore, 1/2 [tan .7853\\ sec .7853  {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)]  - 1/2 [0 - ln (1-0) + ln (0+1) ]\n\n= 1/2[tan .7853\\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)]  - [0]\n\n=1.15\n\nwhich is the web site's answer using the Trapezoidal Rule!\n\nThank you for the guidance.\n\nNote: concerning how theta is written above, I wrote it out in Word.\n\nX\n\nLeaving it in terms of theta:\n\nint sec^3 theta d theta from 0 to .7853\n\nAccording to Wolfram, the integral formula is\n\n1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]\n\nTherefore, 1/2 [tan .7853\\ sec .7853  {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)]  - 1/2 [0 - ln (1-0) + ln (0+1) ]\n\n= 1/2[tan .7853\\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)]  - [0]\n\n=1.15\n\nwhich is the web site's answer using the Trapezoidal Rule!\n\nThank you for the guidance.\n\nNote: concerning how theta is written above, I wrote it out in Word.\n\n## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution\n\nOK - looks good.\n\nBTW, the \"theta\" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone.\n\nX\n\nOK - looks good.\n\nBTW, the \"theta\" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone.", "date": "2018-02-21 02:50:09", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/methods-integration-31/find-integral-sqrt-x-2-1-using-trigonometric:155", "openwebmath_score": 0.9393429756164551, "openwebmath_perplexity": 4375.024253045525, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290897113961, "lm_q2_score": 0.8887587839164801, "lm_q1q2_score": 0.8630995087978187}}
{"url": "http://math.stackexchange.com/questions/370386/fundamental-theorem-of-calculus-on-wikipedia", "text": "# Fundamental theorem of calculus on Wikipedia\n\nIn Wikipedia is written in one step that for the integral $\\int_{x_1}^{x_1 + \\Delta x}f(t) dt$ that by mean value theorem there exists a $c$ in $[x_1, x_1 + \\Delta x]$ such that\n\n$\\int_{x_1}^{x_1 + \\Delta x}f(t) dt = f(c) \\Delta x$\n\nBut mean value theorem states that there exists $c$ in $(x_1, x_1 + \\Delta x)$. (the open interval)\n\nIt seems to be a problem when squeeze theorem is later applied.\n\nIs this mistake in the proof or is the step valid? If it is valid can you please explain me why? Thank you for any help.\n\n-\nThe open interval is contained in the closed interval, so the condition that $c$ is in the closed interval is weaker than the conclusion of the mean value theorem. \u2013\u00a0 Matthew Pressland Apr 23 '13 at 14:33\n@MattPressland But if $c \\in (x_1, x_1 + \\Delta x)$ and by applying squeeze theorem, $c = x_1$ then this is contradiction because $\\Delta x \\to 0$ and $x_1 \\notin (x_1, x_1)$? \u2013\u00a0 blue Apr 23 '13 at 15:03\nThe squeeze theorem doesn't tell you $c=x_1$, it tells you that $c\\to x_1$ as $\\Delta x\\to 0$. A sequence of points in the open interval can converge to one of the boundary points in the closed interval. \u2013\u00a0 Matthew Pressland Apr 23 '13 at 15:10\n@MattPressland I did not know it can converge to boundary point. You can write in answer? If you do I accept. \u2013\u00a0 blue Apr 23 '13 at 15:13\n@blue .Lagrage's Mean value theorem for differentiation differ's from first mean value theorem for integration(which you used!) at that point(the interval in which $c$ is expected to be found is closed in the latter and open in the former). \u2013\u00a0 Ferm\u00e9 somme Apr 23 '13 at 15:20\n\nAs boywholived points out, a possible statement of the mean value theorem allows for $c$ to lie in the closed interval. In fact this is simply a weaker statement - if $c$ lies in the open interval, then in particular it lies in the closed interval.\nThis doesn't cause any problems when applying the squeeze theorem. Perhaps a change of notation is useful; for a particular choice of $\\Delta x$, the mean value theorem gives you some $x<c<\\Delta x$ (assuming the open interval version of the mean value theorem for now). This $c$ depends on $\\Delta x$, so we should maybe write $c(\\Delta x)$, making it clear that $c$ is really a function whose value depends on $\\Delta x$. Then when applying the squeeze theorem, we are interested in the limit of this function as $\\Delta x\\to 0$. The fact that $c(\\Delta x)>x$ for all $\\Delta x>0$ only implies that $\\lim_{\\Delta x\\to0}c(\\Delta x)\\geq x$, not that the inequality is strict. (Compare to the sequence $a_n=1/n$; every term is strictly positive, but $a_n\\to 0$ as $n\\to\\infty$).", "date": "2015-07-06 07:56:25", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/370386/fundamental-theorem-of-calculus-on-wikipedia", "openwebmath_score": 0.969404399394989, "openwebmath_perplexity": 178.34191763767436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105300791785, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.8630968493325718}}
{"url": "https://math.stackexchange.com/questions/365938/finding-the-remainder-of-x100-2x511/365946", "text": "# finding the remainder of $x^{100}-2x^{51}+1$\n\nI have never been great with polynomials. Here's my problem.\n\nFind the remainder of $f(x)=x^{100}-2x^{51}+1$ when $f$ is divided by $x^2-1$\n\nThis sounds easy right? Why can't I figure it out? My thought was to try and create it such that $f(x)=q(x)g(x)+r(x)$. But I can not get past getting $deg[r(x)]<deg[g(x)].$\n\n$$f(x)=x^{100}-2x^{51}+1$$ $$=x^{100}-x^{51}-x^{51}+x^2-x^2+1$$ $$=x^{51}(x^{49}-1)-x^2(x^{49}-1)-x^2+1$$ $$=(x^{51}-x^2)(x^{49}-1)-x^2+1$$ $$=x^2(x^{49}-1)(x^{49}-1)-x^2+1$$ $$=x^2[(x^{49}-1)^2-1]+1=?.......$$ I don't see what I am missing\n\nThis is the standard approach, especially if you know the roots of the divisor.\n\nLet $f(x) = x^{100} - 2x^{51} + 1$, and $f(x) = g(x) (x^2-1) + ax + b$ be the division\n\nThen, $0 = f(1) = g(1) ( 1^2 - 1) + a (1) + b = a + b$,\nand $4 = f(-1) =g(-1) ( (-1)^2 -1) + a(-1) + b = -a + b$.\n\nHence $a= -2, b = 2$.\n\nThus the remainder is $-2x+2$.\n\n\u2022 I like that. I've never seen it done like that before. That is great! \u2013\u00a0Eleven-Eleven Apr 18 '13 at 22:57\n\u2022 @ChristopherErnst You should review the Remainder-Factor Theorem. This is a standard application of it. For more questions, you can check this out. \u2013\u00a0Calvin Lin Apr 18 '13 at 22:58\n\u2022 So basically depending on the degree of your particular divisor, if say $deg[g(x)]=3$ you set your remainder to $ax^2+bx+c$ and take one more value and solve the system for a,b,and c. \u2013\u00a0Eleven-Eleven Apr 18 '13 at 23:25\n\u2022 @ChristopherErnst Correct. This way is direct and you are simply solving linear equations, though it assumes that you can find the roots of $g(x)$. Of course, there are other ways to approach these kind of polynomial remainder problem, like the other (deleted) solution. \u2013\u00a0Calvin Lin Apr 18 '13 at 23:34\n\u2022 Very nice. This way is particularly nice should the divisor factor into nice linear terms with real solutions like the one in the problem. Can this be extended when there are complex factors in our divisor? \u2013\u00a0Eleven-Eleven Apr 18 '13 at 23:42\n\nHint $\\ \\,$ Interpolate the remainder $\\rm\\:r(x)\\:$ at the roots $\\rm\\:\\color{#c00}{x = \\pm1},\\:$ where $\\rm\\ r(\\pm1)\\, =\\, f(\\pm1)$\n\n$$\\qquad\\ \\ \\begin{eqnarray} &&\\rm\\ \\ r(x) &=\\,&\\rm f(x) - (\\color{#c00}{x^2\\!-\\!1})\\, q(x),\\ \\ \\ deg\\ r < 2\\\\ \\\\\\Rightarrow\\, &&\\rm 2\\, r(x) &=\\,&\\rm f(1)\\, (x\\!+\\!1) - f(-1)\\, (x\\!-\\!1)\\end{eqnarray}$$\n\nRemark $\\$ Or, equivalently, use Chinese Remainder (CRT) to solve\n\n$$\\begin{eqnarray}\\rm r(x) &\\equiv&\\rm \\ \\ \\ f(1) &&\\rm\\ (mod\\ \\color{#c00}{x-1}) \\\\ \\rm r(x)&\\,\\equiv\\,&\\rm f(-1)&&\\rm\\ (mod\\ \\color{#c00}{x+1})\\end{eqnarray}$$\n\nGenerally, as here, Lagrange interpolation is a special case of CRT.", "date": "2019-08-23 19:34:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/365938/finding-the-remainder-of-x100-2x511/365946", "openwebmath_score": 0.6702640056610107, "openwebmath_perplexity": 352.89586751950577, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981735723251272, "lm_q2_score": 0.8791467690927439, "lm_q1q2_score": 0.863089789199284}}
{"url": "https://math.stackexchange.com/questions/1309968/what-does-9-with-a-line-above-the-9-mean", "text": "# What does .9 with a line above the 9 mean?\n\nWhat does this mean? $$\\Large.\\overline9$$ I've never seen this notation before.\n\n\u2022 Perhaps $0.9999...$? \u2013\u00a0ajotatxe Jun 2 '15 at 23:28\n\u2022 It also means $1$, by that's a whole 'nother discussion. \u2013\u00a0PyRulez Jun 3 '15 at 0:32\n\u2022 @AlbertMasclans Apparently this notation style depends on geography (en.wikipedia.org/wiki/Repeating_decimal): overline=US, overdot=China, parentheses=Europe. \u2013\u00a0user3449173 Jun 3 '15 at 8:02\n\u2022 @Nordik Yes, it depends on geography, but parentheses are used only sometimes in Europe. In Germany we typically use the overline, too. Parentheses are used to denote an uncertanty in the last digit(s). \u2013\u00a0Christoph Jun 3 '15 at 10:58\n\u2022 By the way, the overline is technically known as a vinculum. \u2013\u00a0Brian M. Scott Jun 3 '15 at 14:38\n\nIt is called a vinculum and it denotes a repeating decimal.\n\n\u2022 +1 for a good word I need to remember to use when the time is right. \u2013\u00a0JoeTaxpayer Jun 2 '15 at 23:58\n\u2022 Sounds like something in a gynecologist's office. \u2013\u00a0zhw. Jun 3 '15 at 0:36\n\u2022 @JoeTaxpayer: Do you mean +1 or +0.9...? \u2013\u00a0mipadi Jun 3 '15 at 22:18\n\nIt means a repeating decimal. One can write $\\frac 16=0.1\\overline 6$, or $\\frac 1{14}=0.0\\overline{714852}$ for example. The repeating part is whatever is under the overline.\n\n$0.\\overline{9}=0.999999\\ldots=1$\n\nMore generally, $0.\\overline{n}=0.nnnnnnnnn\\ldots$\n\nFor example, $\\frac 13=0.333333333\\ldots=0.\\overline3$\n\nAs other answers have said, it stands for a repeating decimal, where the digits under the line are repeated. $$0.\\overline{9} = 0.9999999\\ldots$$\n\nBut if you want to be a little pedantic, you might prefer to say that both $0.\\overline{9}$ and $0.9999999\\ldots$ are two different forms of notation for the same number. That number is the limit of the sequence formed by repeatedly appending copies the digits covered by the line. In other words, given the notation $0.\\overline{9}$, you can write out the following sequence: \\begin{align} a_1 &= 0.9 \\\\ a_2 &= 0.99 \\\\ a_3 &= 0.999 \\\\ a_4 &= 0.9999 \\\\ &\\vdots \\end{align} As you tack on more and more repetitions, these numbers get closer and closer to some limiting value, which I'll call $A$. If you know calculus notation, $$\\lim_{n\\to\\infty} a_n = A$$ The number represented by $0.\\overline{9}$ is $A$. It happens to work out to be $1$ (or if we're being pedantic, $1$ is yet another notation for the same number).\n\nAs another example of this way of interpreting repeating decimals, consider $0.257\\overline{143}$. You can write the sequence \\begin{align} b_1 &= 0.257143 \\\\ b_2 &= 0.257143143 \\\\ b_3 &= 0.257143143143 \\\\ b_4 &= 0.257143143143143 \\\\ &\\vdots \\end{align} and similarly, the number represented by $0.257\\overline{143}$ is the value that this sequence gets closer and closer to as you add more repetitions; or $$\\lim_{n\\to\\infty} b_n$$ This one works out to $\\frac{128443}{499500}$.\n\n\u2022 Did you just randomly pick that last fraction or did you already have it in mind? \u2013\u00a0user1717828 Jun 3 '15 at 12:19\n\u2022 Worth noting, since the 'overdot' notation being standard in UK and apparently China has come up in comments on OP, that in that variant we dot the first and last in a sequence (as opposed to each digit). \u2013\u00a0OJFord Jun 3 '15 at 22:13\n\u2022 @user1717828 totally random, I didn't feel like going to the trouble of looking up a \"significant\" fraction. \u2013\u00a0David Z Jun 4 '15 at 5:08\n\nThis symbol is called a vinculum or a overline.\n\nThe number(s) under this symbol are repeated indefinitely.\n\n$0.\\overline{9} = 0.99999999999999...$\n$2.52\\overline{346} = 2.52346346346346...$", "date": "2019-05-21 22:52:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1309968/what-does-9-with-a-line-above-the-9-mean", "openwebmath_score": 0.9999340772628784, "openwebmath_perplexity": 844.6477052041297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357227168956, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8630897840665938}}
{"url": "https://mathhelpforum.com/threads/arithmetic-series-word-problem-analysis.249576/", "text": "# Arithmetic Series Word Problem (Analysis)\n\n#### iNET\n\nHi, I'm not sure if this is the right section for this question, so if not please redirect me to the correct subforum.\n\nQ: There are k animal feeding stations arranged in a line with a supply hut. The stations are s yards apart and the nearest is t yards from the hut. An attendant carries n bags of feed, one at a time, to each feeding station. How far will he have traveled when he arrives back at the hut after servicing all stations?\n\nHere's what I did, but I don't know if it's correct: (S.k denotes 'S of k')\n\nS.k=(2t + 2(t+s)+...+2(t+(k-1)s))n\n= n((k/2)(2(2t)+(k-1)2s))\n= n((k(4t+2sk-2s))/(2))\n= n((4kt+2sk^2-2sk)/(2))\n= n(kt+sk^2+sk)\n= nk(t+sk+s) yards\n\n#### romsek\n\nMHF Helper\nI get\n\n$2 n t + 2n(s+t) + 2n(2s+t) + \\dots 2n((k-1) s + t) = 2 n t k+2 n s \\displaystyle{\\sum_{j=1}^{k-1}}j = 2 n t k + 2 n s \\dfrac {k(k-1)}{2}=2 n k \\left(t + \\dfrac{s(k-1)}{2}\\right)$\n\n1 person\n\n#### iNET\n\nThanks! I don't exactly understand the method you used to get your answer, but using your answer I was able to see the mistake in my own work that\u2013when corrected\u2013gave me a solution identical to yours.\n\n#### romsek\n\nMHF Helper\nthe main thing is\n\n$\\displaystyle {\\sum_{j=1}^{k-1}}j =\\dfrac {k(k-1)}{2}$\n\nIt looks like you set the problem up correctly. We both start from the same basic equation.\n\n#### iNET\n\nI see.\n\nThis may be a stupid question, but the answer could also be simplified to:\n\nnk(2t+s(k-1))\n\nIs one answer somehow 'more' correct than the other one, or are both answers equally acceptable?\n\nThanks for your help.\n\n#### romsek\n\nMHF Helper\nI see.\n\nThis may be a stupid question, but the answer could also be simplified to:\n\nnk(2t+s(k-1))\n\nIs one answer somehow 'more' correct than the other one, or are both answers equally acceptable?\n\nThanks for your help.\nthey are both the same answer.\n\nYours is a bit simpler than mine.", "date": "2020-01-26 05:24:30", "meta": {"domain": "mathhelpforum.com", "url": "https://mathhelpforum.com/threads/arithmetic-series-word-problem-analysis.249576/", "openwebmath_score": 0.7978655099868774, "openwebmath_perplexity": 1603.8790526711377, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964190307041, "lm_q2_score": 0.8757870046160257, "lm_q1q2_score": 0.86308495688272}}
{"url": "https://math.stackexchange.com/questions/537164/find-lim-n-to-infty-fracn22n/537182", "text": "# Find $\\lim_{n\\to\\infty} \\frac{n^2}{2^n}$\n\n$$\\lim_{n\\to\\infty}\\frac{n^2}{2^n}$$\n\nDo you have some tips so I could solve this problem, without the use of L'H\u00f4pital's rule?\nIndeed, we didn't see formally L'H\u00f4pital's rule, nor Taylor series so I'm supposed to do this without such \"tools\".\n\nI've tried using the fact that $n^2 = e^{2\\log(n)}$ and $2^n=e^{n\\log(2)}$ but didn't manage to eliminate my indeterminate form.\n\n\u2022 At least write down what you get after applying l'Hoptial's rule once. Then apply it again. \u2013\u00a0Alex R. Oct 23 '13 at 17:35\n\u2022 It look as if $n$ may range over positive integers. In that case, use the fact that by the Binomial Theorem, $(1+1)^n=1+n+\\frac{n(n-1)}{2}+\\frac{n(n-1)(n-2)}{6}+\\cdots\\gt \\frac{n(n-1)(n-2)}{6}$. \u2013\u00a0Andr\u00e9 Nicolas Oct 23 '13 at 17:40\n\u2022 What, if anything, do you know about the comparative growth rates of $n$ and $\\log(n)$? If you happen to know that $$\\lim_{n\\to\\infty}\\bigl[n-\\log(n)\\bigr]=\\infty,$$ then since $$\\frac{n^2}{2^n}=\\frac{e^{2\\log(n)}}{e^{n\\log(2)}}=e^{2\\log(n)-n\\log(2)},$$ and since $\\log(2)>0,$ it follows that the limit you're looking for is $0$, and even gives you a start on how it can be proven. \u2013\u00a0Cameron Buie Oct 23 '13 at 17:53\n\nThere are many ways to do that. For example, you can prove that for $n \\ge 10$ we have $$2^n > n^3.$$ You can show that using the induction argument.\n\nRegarding to analysis tag, you seem to be familiar to series so let us consider the following series: $$\\sum(n^2/2^n)$$ it is not hard seeing that this series is convergent and so the $n-$th term approaches to zero while $n\\to\\infty$. Clearly, this way needn't using L'Hopital's rule.\n\n\u2022 @Amzoti: Thanks for your support my dear friend. I hope you have a good time ahead and a peaceful slumber. :-) \u2013\u00a0mrs Oct 24 '13 at 1:48\n\u2022 You are welcome my friend! I appreciate all of the hard work you and amWhy put into the site helping people become better problem solvers! Time to watch a movie and relax as it has been a difficult month at work and preparing two talks! Have a great night! Regards \u2013\u00a0Amzoti Oct 24 '13 at 1:49\n\nI noticed the OP had wanted L'H\u00f4pital's rule off the table only after I posted my previous reply. I apologize for my inattention. Here is another answer:\n\nLet $a_n=n^2/2^n$ for all $n\\in\\mathbb{Z}_+$. You can easily compute that \\begin{align*} a_{n+1}-a_n=-\\frac{n(n-2)-1}{2^{n+1}}, \\end{align*} which is negative for $n\\geq3$. Therefore, the sequence is eventually decreasing and bounded below (by zero), so it must have a limit.\n\nObviously, this limit cannot be negative, since $a_n>0$ for all $n$. If we show it cannot be positive, either, then we can conclude that the limit must be zero.\n\nSuppose that the limit is positive: $\\lim_{n\\to\\infty} a_n=c>0$. Since the sequence is decreasing (after omitting $a_1$ and $a_2$, which does not affect convergence properties), it is also true that \\begin{align*} c=\\lim_{n\\to\\infty} a_n=\\inf_{n\\in\\mathbb{Z}_+\\setminus\\{1,2\\}}a_n. \\end{align*} Then, for $n$ large enough and beyond, we have that \\begin{align*} (*)\\quad c\\leq a_n\\leq\\frac{3}{2}c. \\end{align*}\n\nTherefore, \\begin{align*} (**)\\quad a_{n+1}=a_{n}\\times\\frac{a_{n+1}}{a_n}=a_n\\times\\frac{(n+1)^2}{2n^2}\\leq\\frac{3}{2}c\\times \\frac{(n+1)^2}{2n^2}, \\end{align*}\n\nwhere the last inequality follows from $(*)$. But $(n+1)^2/(2 n^2)$ converges to $1/2$ as $n\\to\\infty$, so that for $n$ sufficiently large and beyond, \\begin{align*} \\frac{(n+1)^2}{2n^2}\\leq\\frac{5}{8}. \\end{align*} For such values of $n$, $(**)$ implies that \\begin{align*} a_{n+1}\\leq\\frac{3}{2}c\\times\\frac{(n+1)^2}{2n^2}\\leq\\frac{15}{16}c<c, \\end{align*} where the last inequality is strict because $c$ is positive. But this contradicts $c$\u00a0being the infimum of the values of the sequence. The limit thus cannot be positive.\n\n\u2022 +1. I appreciate your effort of writing another answer after I pointed out that L'H was off-limits. :) \u2013\u00a0Lord_Farin Oct 23 '13 at 20:13\n\u2022 Thanks, @Lord_Farin. :-) \u2013\u00a0triple_sec Oct 23 '13 at 20:37", "date": "2020-11-29 07:50:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/537164/find-lim-n-to-infty-fracn22n/537182", "openwebmath_score": 0.9616277813911438, "openwebmath_perplexity": 263.1114596688287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969679646669, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8630835092022794}}
{"url": "https://math.stackexchange.com/questions/2304246/unique-combinations-of-n-non-unique-elements-in-k-non-unique-buckets-with-c", "text": "# Unique combinations of $n$ non-unique elements in $k$ non-unique buckets with $c$ capacity\n\nI have $n$ non-unique elements, and I have $k$ unordered buckets that can hold anywhere from $0$ to $c$ elements, such that $c * k \\geq n$. I would like to find all possible combinations.\n\nFor example, given $n=10$, $k=4$, and $c=4$, there are 7 possible distributions:\n\n\u2022 3322\n\u2022 3331\n\u2022 4222\n\u2022 4321\n\u2022 4330\n\u2022 4411\n\u2022 4420\n\nwhere \"3322\", for example, means that two buckets have three elements each and the other two buckets have two elements each.\n\nAnother way to look at it, is that I want to find unique combinations of $k$ numbers less than or equal to $c$ such that their sum is equal to $n$.\n\nIdeally I'd like an algorithm to be able to generate a list of all acceptable combinations, but knowing a formula for finding the number of combinations given arbitrary $n$, $k$, and $c$ would be helpful. Other answers generally assume buckets have to have a minimum capacity of 1, or they deal with unique elements to some degree, which are not the case here.\n\nFirst, it is important to note that all solutions can be arranged in descending sorted order. This is useful for the algorithm I will describe, such that it only searches for sorted answers.\n\nEssentially, we start at the first bucket and assign it some item count $0\\leq i\\leq c$. Now, we realize that assigning the remaining items to the remaining buckets is essentially the same problem, except now we have $n-i$ items, $k-1$ buckets, and each remaining bucket can now only hold up to $i$ items, because if it were to hold anymore, the solution would be unsorted.\n\nOnly when we reach the end (no buckets and no items left) do we have a proper solution. This algorithm can also be sped up by realizing that if $\\frac{n-i}{k-1}>i$, there is no continuing from this point, as the remaining items can not fit in the remaining buckets without going over $i$.\n\nBelow is a solution in Python,\n\ndef buckets(n,k,c,solution=[]):\ntotal = 0\nfor i in range(min(c,n),-1,-1):\nif k-1 != 0 and float(n-i)/(k-1) > i:\n#Can't fit remaining items\nbreak\nsolution.append(i)\nif n-i == 0 and k-1 == 0:\n#We've reached the end of the buckets and have a solution\nprint(solution)\nsolution.pop()\nreturn 1\nelif k-1 != 0:\n#If there's still buckets left, try assigning them\ntotal += buckets(n-i,k-1,i,solution)\n#Done checking this bucket assignment\nsolution.pop()\n\n\nbuckets(10,4,4)\n\n[4, 4, 2, 0]\n\n[4, 4, 1, 1]\n\n[4, 3, 3, 0]\n\n[4, 3, 2, 1]\n\n[4, 2, 2, 2]\n\n[3, 3, 3, 1]\n\n[3, 3, 2, 2]\n\n\u2022 Thanks so much for this - it works completely. Jun 1, 2017 at 17:54", "date": "2022-08-14 04:01:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2304246/unique-combinations-of-n-non-unique-elements-in-k-non-unique-buckets-with-c", "openwebmath_score": 0.703007161617279, "openwebmath_perplexity": 239.78017411512175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517411671125, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.863060115776339}}
{"url": "https://math.stackexchange.com/questions/783917/matrix-exponential-of-a-simple-bidiagonal-matrix", "text": "# Matrix exponential of a simple bidiagonal matrix\n\nI am interested in finding an expression (closed form or recursive) for the matrix exponential of this banded matrix:\n\n$$\\begin{pmatrix} 0 & 1 & 0 & 0 & \\cdots & 0 & 0 \\\\ 0 & a_1 & 1 & 0 & \\cdots & 0 & 0 \\\\ 0 & 0 & a_2 & 1 & \\cdots & 0 & 0 \\\\ 0 & 0 & 0 & a_3 & \\cdots & 0 & 0 \\\\ \\vdots & \\vdots & \\vdots & \\vdots &\\ddots&\\vdots&\\vdots \\\\ 0 & 0 & 0 & 0 & \\cdots & a_{n-1} & 1 \\\\ 0 & 0 & 0 & 0 & \\cdots & 0 & a_n \\end{pmatrix}$$\n\nFor simplicity, assume that $a_k>0~\\forall k$. I am quite certain one must exist having played around with it for a while, and looking at the solution for small values of $n$.\n\nHas anyone seen this structure before? Does it have a name? Do you know if there is a solution published somewhere?\n\nIf you go ahead and compute the answer for small values of $n$, you get:\n\n$$\\exp\\left( \\begin{array}{cc} 0 & 1 \\\\ 0 & a_1 \\\\ \\end{array} \\right) = \\left( \\begin{array}{cc} 1 & \\frac{-1+e^{a_1}}{a_1} \\\\ 0 & e^{a_1} \\\\ \\end{array} \\right)$$\n\n$$\\exp \\left( \\begin{array}{ccc} 0 & 1 & 0 \\\\ 0 & a_1 & 1 \\\\ 0 & 0 & a_2 \\\\ \\end{array} \\right) = \\left( \\begin{array}{ccc} 1 & \\frac{-1+e^{a_1}}{a_1} & \\frac{-e^{a_2} a_1+a_1+e^{a_1} a_2-a_2}{a_1 \\left(a_1-a_2\\right) a_2} \\\\ 0 & e^{a_1} & \\frac{e^{a_1}-e^{a_2}}{a_1-a_2} \\\\ 0 & 0 & e^{a_2} \\\\ \\end{array} \\right)$$\n\nUnfortunately, $n=3$ is to large to print here, but a pattern remains.\n\nIt suffices to diagonalize $A$ (if the $(a_i)$ are pairwise distinct). I change slightly the notation $A=D+J_n$ where $D=diag(a_1,\\cdots,a_n)$ and $J_n$ is the nilpotent Jordan block of dimension $n$. $A=PDP^{-1}$ where the first row of $P$ is\n\n$1,-\\dfrac{1}{a_1-a_2},\\dfrac{1}{(a_1-a_3)(a_2-a_3)},-\\dfrac{1}{(a_1-a_4)(a_2-a_4)(a_3-a_4)},\\cdots$\n\nWe obtain the other rows of $P$ by circular permutation along the diagonals. Finally $e^A=Pdiag(e^{a_1},\\cdots,e^{a_n})P^{-1}$ where the first row of $P^{-1}$ is\n\n$1,\\dfrac{1}{a_1-a_2},\\dfrac{1}{(a_1-a_2)(a_1-a_3)},\\dfrac{1}{(a_1-a_2)(a_1-a_3)(a_1-a_4)},\\cdots$\n\nWe obtain the other rows of $P^{-1}$ by circular permutation along the diagonals.\n\n\u2022 Cool thanks very much, this looks promising. However, I'm not sure exactly what you mean by \"circular permutation along the diagonals\". Do you mean just \"circular permutation\" or (equivalently) do you mean that the diagonals, when wrapped, will be constant? When I try this I don't get $PP^{-1}=1$ \u2013\u00a0Ian Hincks May 8 '14 at 20:28\n\u2022 Also, can you point me in the direction of how you arrived at this conclusion? \u2013\u00a0Ian Hincks May 8 '14 at 20:29\n\u2022 Ah, okay, I figured out that \"circular permutation along the diagonal\" means that, for example, the second row of $P$ will be $(0,1,-(a_2-a_3)^{-1},((a_2-a_4)(a_3-a_4))^{-1},...)$. \u2013\u00a0Ian Hincks May 8 '14 at 21:12\n\u2022 Yes Ian, the second row is as you write it. Let $A_n$ be the matrix of dimension $n$. Then $e^{A_n}$ is the submatrix of $e^{A_{n+1}}$ that is constituted by its first $n$ rows and its first $n$ columns. \u2013\u00a0user91684 May 9 '14 at 10:36\n\nThis response is very late but perhaps still useful to others who come across your question whilst browsing (as I did!).\n\nYour banded (or upper bidiagonal) matrix is what Optiz termed a $\\textit{Steigungsmatrix }$ (maybe ... \"gradient matrix\"). See McCurdy et al (1984), Mathematics of Computation, 43, 501-528. For any $n>0$, all you need is to apply $\\textit{Opitz's formula}$. You will need a little knowledge of functions of matrices and divided differences to apply it but it will be worth the effort.\n\nIf we denote your matrix by A and its elements by $a_{ij}$ then, using Opitz's theorem, the matrix F defined as F = $g$(A) for a function $g$(.), has entries $f_{ij}$ as follows: $f_{ij}=0$ for $i>j$, $f_{ij} = g(a_{ii})$ for $i=j$, and $f_{ij} = g[a_{ii},a_{i+1,i+1},..,a_{jj}]$ for $i<j$ .\n\nThe notation $g[x_0,x_1,...,x_n]$ is the nth divided difference defined recursively as $$g[x_0,x_1,...,x_{(n-1)},x_n] = \\frac{g[x_1,...,x_n]-g[x_0,...,x_{(n-1)}]}{x_n-x_0},$$ $g[x_0,x_1] = (g[x_1]-g[x_0])/(x_1-x_0)$ and $g[x] = g(x)$.\n\nSetting $g$(.) to be the exponential function, I get your answers for n=1 and n=2.\n\nThe above assumes distinct diagonal entries. If there are repeats in the diagonal terms then you should use the appropriate \"confluent\" form of the divided difference.", "date": "2021-03-04 06:58:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/783917/matrix-exponential-of-a-simple-bidiagonal-matrix", "openwebmath_score": 0.9055584669113159, "openwebmath_perplexity": 343.0791003757298, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474904166246, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8630546777010234}}
{"url": "https://math.stackexchange.com/questions/2162888/prove-that-the-sum-of-pythagorean-triples-is-always-even/2162901", "text": "# Prove that the sum of pythagorean triples is always even\n\nProblem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even\n\nMy Attempt, Case by case analysis:\n\nCase 1: a is odd, b is odd. From the first equation,\n\n$odd^2 + odd^2 = c^2$\n\n$odd + odd = c^2 \\implies c^2 = even$\n\nSquaring a number does not change its congruence mod 2.\n\nTherefore c is even\n\n$a + b + c = odd + odd + even = even$\n\nCase 2: a is even, b is even. Similar to above\n\n$even^2 + even^2 = c^2 \\implies c$ is even\n\n$a + b + c = even + even + even = even$\n\nCase 3: One of a and b is odd, the other is even Without loss of generality, we label a as odd, and b as even\n\n$odd^2 + even^2 = c^2 \\implies odd + even = c^2 = odd$\n\nTherefore c is odd\n\n$a + b + c = odd + even + odd = even$\n\nWe have exhausted every possible case, and each shows $a + b + c$ is even. QED\n\nFollow Up: Is there a proof that doesn't rely on case by case analysis? Can the above be written in a simpler way?\n\n\u2022 In fact there are no Pythagorean triples where the legs ($a$ and $b$) are both odd Feb 27, 2017 at 16:47\n\u2022 Even though there are shorter proofs, there is something pleasingly straightforward about your proof. Mar 1, 2017 at 13:48\n\nNote that $x^2\\equiv x\\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\\equiv a^2+b^2+c^2\\equiv 2c^2\\equiv 0\\pmod 2$$\n\n\u2022 Well that makes it easy, props for the quick answer. Feb 26, 2017 at 21:55\n\u2022 You are welcome, @spyr03. Feb 26, 2017 at 21:55\n\u2022 $a^2+b^2+c^2\\equiv 2(a^2+b^2)\\pmod 2$ Could you explain that step? Feb 26, 2017 at 21:55\n\u2022 @mrnovice $c^2 = a ^2 + b^2\\implies a^2+b^2+c^2 = 2(a^2+b^2)$. Feb 26, 2017 at 21:56\n\u2022 @mrnovice, not at all, the way I had written the answer was missing the context, you've asked a valid question. Feb 26, 2017 at 21:59\n\nAlso, Pythagorean triples have a well defined structure: $$a=k(m^{2}-n^{2}),\\ \\,b=k(2mn),\\ \\,c=k(m^{2}+n^{2})$$ and $$a+b+c=2k(mn+m^2)$$\n\nHint\n\nWrite $a+b+c=k$, so\n\n$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 \u2192 k^2-2(kc+ab)=0\u2192k^2=2(kc+ab)$$\n\nNotice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.\n\n$c^2 = a^2 + b^2 = (a+b)^2 - 2ab$.\n\n$2ab = (a+b)^2 - c^2$\n\n$2ab = (a+b+c)(a+b-c)$\n\nLet $n = a+b+c$, and the above becomes:\n\n$2ab = n(n-2c)$\n\nSo the right-hand side must be even, but since $n-2c$ is odd when $n$ is odd, $n$ must be even.\n\n\u2022 Nice explanation, I never would have thought about writing $2ab = (a + b + c)(a + b -c)$. You could have finished with $2ab = n^2 - 2cn$ and thus n^2 must be even, therefore n is even. Feb 28, 2017 at 23:29\n\nConsider $(a+b+c)^2$\n\nWhich is $a^2 + b^2 + c^2 + 2(ab+bc+ca)$\n\nSince $c^2 = a^2 + b^2$ (c being the hypotenuse), $(a+b+c)^2 = 2(c^2 + ab + bc + ca)$ - which is an even number.\n\nand since squares of odd is odd and evens is even $a+b+c$ has to be even.", "date": "2022-07-03 02:14:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2162888/prove-that-the-sum-of-pythagorean-triples-is-always-even/2162901", "openwebmath_score": 0.7367633581161499, "openwebmath_perplexity": 611.2160215087379, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750510899383, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8630420660633199}}
{"url": "http://interviewingthecrisis.org/codecademy-python-math-symbols", "text": "Codecademy Python Math Symbols | interviewingthecrisis.org\n\n# Writing mathematical expressions - MatplotlibPython plotting.\n\nCodecademy is the easiest way to learn how to code. It's interactive, fun, and you can do it with your friends. Python Comparison Operators. These operators compare the values on either sides of them and decide the relation among them. They are also called Relational operators. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Many programming languages include libraries to do more complicated math. You can do statistics, numerical analysis or handle big numbers. One topic many programming languages have difficulty with is symbolic math. If you use Python though, you have access to sympy, the symbolic math library. Sympy.\n\nLearn about lists, a data structure in Python used to store ordered groups of data. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list.\n\nPython Arithmetic Operators Example - Assume variable a holds 10 and variable b holds 20, then \u2212. Numeric and Mathematical Modules\u00b6 The modules described in this chapter provide numeric and math-related functions and data types. The numbers module defines an abstract hierarchy of numeric types. There is nothing special about multiplication in Python; we just use an asterisk in the same way we did with addition, subtraction, and division. print2 2 I\u2019m sure it comes as no surprise to you that the above outputs 4. Note that if we multiply two integers, the result is of the data type int. Jupyter notebook recognizes LaTeX code written in markdown cells and renders the symbols in the browser using the MathJax JavaScript library. Mathematics Inline and Display Enclose LaTeX code in dollar signs $.$ to display math inline. What are operators in python? Operators are special symbols in Python that carry out arithmetic or logical computation. The value that the operator operates on is called the operand.\n\nx = Symbol'x' y = Symbol'y' A = Matrix[[1,x], [y,1]] Once a matrix is created, you can operate on it. There are functions to do dot products, cross products or calculate determinants. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions area hyperbolic functions, even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. math.floor x \u00b6 Return the floor of x as a float, the largest integer value less than or equal to x. math.fmod x, y \u00b6 Return fmodx, y, as defined by the platform C library. Codecademy-Exercise-Answers / Language Skills / Python / Unit 01 Python Syntax / Edvins Antonovs Subfolder rename & ascending ordering for python Latest commit 5fc7ead Jul 14, 2016.\n\nfrom sympy.plotting import plot from sympy import Symbol x = Symbol 'x' p = plot 2 x3, 3 x1, legend = True, show = False p [0]. line_color = 'b' p [1]. line_color = 'r' p. show Sets Beside FiniteSet which I exemplify below, sympy also includes support for infinite sets and intervals. Ask questions, get help with an exercise, and chat about your Codecademy coursework here. \u201cWhen one teaches, two learn\u201d \u2014 Robert Heinlein.\n\nCodecademy: Python 1. Python Syntax print \"Welcome to Python!\" my_int = 7 my_float = 1.23 my_bool = True my_int = 7 my_int = 3 print my_int def spam: eggs = 12 return eggs print spamjust a comment \"\"\" first comment second comment third comment \"\"\" count_to = 12 count_to = 5 - 2 ni = 2 10 ni = 20 / 4 eggs = 10 2 spam = 3 % 2 2. Strings and Console Output 'Help! Help! I\\'m being. Yes, you can do symbolic math in Python! The library to take a look at is SymPy. Its official website is. This article is not a SymPy tutorial, as I only want to walk you through some examples to show you the kinds of things that it can do. Programming reference for Ruby. Codecademy is the easiest way to learn how to code. It's interactive, fun, and you can do it with your friends.\n\nPython-based: SymPy is written entirely in Python and uses Python for its language. Lightweight: SymPy only depends on mpmath, a pure Python library for arbitrary floating point arithmetic, making it easy to use. A library: Beyond use as an interactive tool, SymPy can be embedded in other applications and extended with custom functions. Definition of codecademy in thedictionary. Meaning of codecademy. What does codecademy mean? Information and translations of codecademy in the most comprehensive dictionary definitions resource on the web.\n\nYingxie Python notes from codecademy Chap 1 Python Syntax 13\u201f Intro: Python can be used to create web apps, games, even a search engine. 1.1 Variables and Data Types. Originally Answered: is there much math in python? This is a common misconception because there are many math people who also program. In reality, programming is the process of telling a computer to do something-- like telling your friend to do something. List of all math symbols and meaning - equality, inequality, parentheses, plus, minus, times, division, power, square root, percent, per mille,. RapidTables Home \u203a Math \u203a Math symbols \u203a Math symbols. SymPy is a Python library for symbolic computation. It provides computer algebra capabilities either as a standalone application, as a library to other applications, or live on the web as SymPy Live or SymPy. Math Symbols Explained with Python August 03, 2019 When working with Machine Learning projects, you will come across a wide variety of equations that you need to implement in code.\n\nIntroduction\u00b6 SymPy is a Python library for symbolic mathematics. It aims to become a full-featured computer algebra system CAS while keeping the code as simple as possible in order to be comprehensible and easily extensible.You can find out which functions and attributesare defined in a module. import math dir mathIf you have a Python script named math.py in the samefolder as your current script, the file math.py willbe loaded instead of the built-in Python module.\n\n08.01.2013\u00a0\u00b7 What is this? As many of you have guessed by now, Sandra is an A.I. meant to be a repository for your questions. Right now, her vocabulary is limited. The symbolic math in sympy is pretty good. It is not up to the capability of Maple or Mathematica, but neither is Matlab but it continues to be developed, and could be helpful in some situations. It is not up to the capability of Maple or Mathematica, but neither is Matlab but it continues to be developed, and could be helpful in some situations. Learn codecademy with free interactive flashcards. Choose from 148 different sets of codecademy flashcards on Quizlet. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step.\n\nStack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. mpmath is a free BSD licensed Python library for real and complex floating-point arithmetic with arbitrary precision. It has been developed by Fredrik Johansson. Embedding Python in LaTeX by Kjell Magne Fauske, Friday, October 24, 2008 Comments: 19 Recently, while browsing the archives of the matplotlib mailing list, I stumbled upon the small python.sty package written by Martin R. Ehmsen. Insights When One Teaches, Two Learn: Meet Codecademy\u2019s Community. We believe \u201cwhen one teaches, two learn.\u201d To make that belief a reality for learners, we host an online community with tens of. Chapter 5 Operators and Expressions The purspose of computing is insight, not numbers. Richard Hamming Chapter Objectives Learn Python\u2019s arithmetic, string, relational, logical, bitwise operators.\n\nPlotly's Python library is free and open source! Get started by downloading the client and reading the primer. You can set up Plotly to work in online or offline mode, or in jupyter notebooks. The Percent Sign % is a interesting beast in the Python language. It does something called string formatting and it\u2019s a mathematic operator as well. Whatever math is. Let\u2019s find out what the POWER of the % is first, we\u2019ll learn by finding out what NOT to do. Unicode characters are very useful for engineers. A couple commonly used symbols in engineers include Omega and Delta. We can print these in python using unicode characters. Kenneth Tilton I am surprised you do not include the numeric character codes. kt --\"The best Algebra tutorial program I have seen. in. ALT Codes - Alt Codes for Maths / Mathematics Welcome to Useful Shortcuts, THE Alt Code resource! If you are already familiar with using alt codes, simply select.\n\nJupyter Notebook. From the official webpage: The Jupyter Notebook is a web application that allows you to create and share documents that contain live. In this post, I am gonna show you how to write Mathematic symbols in markdown. since I am writing blog post that hosted by Github with Editor Atom, and use plugin markdown-preview-plus and mathjax-wrapper, and use mathjax Javascript display the math symbols on the web page. The type of code that best shows off the power of computers involves math. I wonder whether any coder can succeed without mastery of basic math.\n\n1. Learn Python, a powerful language used by sites like YouTube and Dropbox. Learn the fundamentals of programming to build web apps and manipulate data. Master Python loops to deepen your knowledge. Learn the fundamentals of programming to build web apps and manipulate data.\n2. Any text element can use math text. You should use raw strings precede the quotes with an 'r', and surround the math text with dollar signs \\$, as in TeX. Regular text and mathtext can be interleaved within the same string.\n3. What is SymPy? SymPy is a Python library for symbolic mathematics. It aims to be an alternative to systems such as Mathematica or Maple while keeping the code as.\n4. I want to make a dictionary in Python, but with math symbols because then i want to plot that with the symbols. So, i have something like this.\n\nThe Python Discord. News about the dynamic, interpreted, interactive, object-oriented, extensible programming language Python. If you are about to ask a \"how do I do this in python\" question, please try r/learnpython, the Python discord, or the python IRC channel on FreeNode. Note that wasysym also de\ufb01nes a \\lightning symbol. The di\ufb00erence\u2014other than \u201c \u201d vs. The di\ufb00erence\u2014other than \u201c \u201d vs. \u201c \u201d\u2014is that the stmaryrd version above is limited to math mode. An online LaTeX editor that's easy to use. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more.\n\nDiese Liste mathematischer Symbole zeigt eine Auswahl der gebr\u00e4uchlichsten Symbole, die in moderner mathematischer Notation innerhalb von Formeln verwendet werden.", "date": "2021-01-23 05:10:16", "meta": {"domain": "interviewingthecrisis.org", "url": "http://interviewingthecrisis.org/codecademy-python-math-symbols", "openwebmath_score": 0.2335364818572998, "openwebmath_perplexity": 2584.680653551821, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9525741214369555, "lm_q2_score": 0.9059898267292559, "lm_q1q2_score": 0.8630224632274405}}
{"url": "http://math.stackexchange.com/questions/69456/number-of-ways-to-distribute-items-with-limit-of-min-and-max-at-each-location", "text": "# Number of ways to distribute items (with limit of min and max at each location)\n\nThere are a total of $n$ objects in stock. You must keep at least $c_i$ objects in stock at location $i$. Assume $n \\ge \\sum c_i$.\n\na.) How many different inventories $(x_1,x_2,\\dots,x_r)$ are there?\n\nb.) Each location can store a maximum of $d_i$. Let $r = 2$ and count the number of different invetories $(x_1,x_2)$ with $c_1 \\le x_1 \\le d_1$ and $c_2 \\le x_2 \\le d_2$.\n\nMy answer for part a: We have no choice over where $c_1 + c_2 + \\dots + c_r = c_t$ objects are placed. The ramaining $n-c_t$ objects can be spread across $r$ locations. So, there are $$\\binom{(n-c_t) + r - 1}{r-1}$$ different inventories.\n\nMy incomplete answer for part b: After placing minimum required objects at each location, we are left with $n-c_1-c_2$ objects to distribute.\n\nThe remaining objects can be distributed in $$\\binom{(n-c_1-c_2)+2-1}{2-1} = (n-c_1-c_2)+1.$$\n\nNow from the number above, I need to subtract the number of combinations that lead to overflows in the two locations.\n\n[---Not sure if the rest is correct---]\n\nNumber of ways location 1 overflows:\nLocation 1 overflows when $x_1 > d_1$ and $c_2 \\le x_2 \\le d_2$: $$\\binom{(n-(d_1+1)-c_2)+2-1}{2-1} = n-d_1-c_2$$\n\nNumber of ways location 2 overflows:\n(same reasoning as above) $$n-d_2-c_1$$\n\nTotal number of inventories is $$(n-c_1-c_2+1)-[(n-d_1-c_2)+(n-d_2-c_1)] = (d_1+d_2+1)-n$$\n\nTotal number of inventories $= (d_1+d_2+1)-n$ doesn't make sense.\n\n-\n\nYour answer to part (a) is fine. In (b) you\u2019ve overlooked the possibility of overflow in both inventories simultaneously.\n\nYou got off to a fine start: there are $$I_0 = n-c_1-c_2+1$$ inventories that meet the minimum requirements. Then you want get the number of inventories that overflow location $1$ but meet the minimum requirement at location $2$; your answer of $n-d_1-c_2$ is correct only if $n\\ge d_1+c_1$. If $n < d_1+c_1$, it\u2019s not possible to overflow at location $1$ and meet the minimum requirement at location $2$. Thus, the correct result for this step is $$I_1 = \\max\\{n-d_1-c_2,0\\}$$ inventories that overflow at location $1$ and meet the minimum at location $2$. Similarly, there are $$I_2=\\max\\{n-d_2-c_1,0\\}$$ inventories that meet the minimum at location $1$ but overflow location $2$.\n\nHowever, there may be inventories that overflow at both locations. When you subtract off the inventories that overflow at location $1$ and then those that overflow at location $2$, the inventories that overflow at both locations get subtracted twice, so you have to add them back in.\n\nThe basic calculation gives $$\\binom{n-(d_1+1)-(d_2+1)+2-1}{2-1} = n-d_1-d_2-1$$ of these inventories, but again this is true only if $n\\ge d_1+d_2+1$; if $n < d_1+d_2+1$, it isn\u2019t possible to overflow both locations. Thus, the correct figure is $$I_3=\\max\\{n-d_1-d_2-1,0\\}.$$\n\nThe final result is then given by $$I_0-I_1-I_2+I_3.$$\n\nNote that if $n \\ge d_1+d_2+1$, this evaluates to $0$. This makes perfectly good sense: in that case it\u2019s impossible to meet the requirements, and there are no acceptable inventories.\n\n-\n\nUnless I am reading the problem wrong, this is the same as a couple of other problems that were asked in the last couple of days. You are given a total and want to find the number of ways to make that total $\\left(\\displaystyle\\sum_{i=1}^rx_i=n\\right)$ given a range constraint on each summand $(c_i\\le x_i\\le d_i)$. The answer is the coefficient of $x^n$ in $$\\prod_{i=1}^r\\frac{x^{d_i+1}-x^{c_i}}{x-1}\\tag{1}$$ We can apply this to the special cases you have listed.\n\na) you let $d_i\\to\\infty$. In that case, $(1)$ becomes $\\displaystyle\\prod_{i=1}^r\\frac{x^{c_i}}{1-x}=x^{\\sum_{i=1}^rc_i}\\frac{1}{(1-x)^r}$, so the answer becomes the coefficient of $x^{n-\\sum_{i=1}^rc_i}$ in $\\frac{1}{(1-x)^r}$. \\begin{align} \\frac{1}{(1-x)^r} &=\\sum_{i=0}^\\infty\\binom{-r}{i}(-x)^i\\\\ &=\\sum_{i=0}^\\infty\\binom{r+i-1}{i}x^i\\\\ &=\\sum_{i=0}^\\infty\\binom{r-1+i}{r-1}x^i \\end{align} Therefore, the answer is $\\displaystyle\\binom{r-1+n-\\sum_{i=1}^rc_i}{r-1}$.\nb) you set $r=2$. In this case, $(1)$ becomes $$\\left(x^{c_1}+x^{c_1+1}+\\dots+x^{d_1}\\right)\\left(x^{c_2}+x^{c_2+1}+\\dots+x^{d_2}\\right)\\tag{2}$$ and the answer is the coefficient of $x^n$ in $(2)$.\n\nSuppose $d_1-c_1\\le d_2-c_2$, then the answer is $$\\text{the coefficient of }x^n\\text{ in }(2)=\\left\\{\\begin{array}{}n-c_1-c_2+1&\\text{for }c_1+c_2\\le n\\le d_1+c_2\\\\d_1-c_1+1&\\text{for }d_1+c_2+1\\le n\\le c_1+d_2-1\\\\d_1+d_2-n+1&\\text{for }c_1+d_2\\le n\\le d_1+d_2\\\\0&\\text{otherwise}\\end{array}\\right.$$\n\n-\nPlease show the closed form, I can't get the answer. \u2013\u00a0 jamesio Oct 3 '11 at 17:04\n@jamesio: I have included the gory details. \u2013\u00a0 robjohn Oct 3 '11 at 17:53", "date": "2015-07-29 22:19:22", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/69456/number-of-ways-to-distribute-items-with-limit-of-min-and-max-at-each-location", "openwebmath_score": 0.8272207379341125, "openwebmath_perplexity": 353.5004790585186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.99330714867946, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.8630118444277746}}
{"url": "https://www.physicsforums.com/threads/integral-substitution.328793/", "text": "# Integral substitution\n\n## Main Question or Discussion Point\n\nHow do I evaluate the integral $$\\int_a^b$$ f(x4) dx = $$\\int_y^z$$ f(u) dx/du du, where u=x4\ny=a4\nz=b4\n\n## Answers and Replies\n\nHallsofIvy\nHomework Helper\nHow do I evaluate the integral $$\\int_a^b$$ f(x4) dx = $$\\int_y^z$$ f(u) dx/du du, where u=x4\ny=a4\nz=b4\nHow you integrate the function f depends strongly on what f is, don't you think?\n\nf is $$\\int_-1^1$$ 1+x4 dx\n\nf is $$\\int_1^1$$ 1+x4 dx\nmeant to be a -1 at the bottom of the integral.\nHow could I do this then.\n\njgens\nGold Member\nWell, if your integral is simply\n\n$$\\int_{-1}^1 1 + x^4\\, dx$$\n\nThen it's pretty simple. Just seperate the integrand so that,\n\n$$\\int_{-1}^1 \\,dx + \\int_{-1}^1 x^4 \\, dx$$\n\nBut, based on the context of the question I'm not sure that's what you're looking for. Perhaps you could repost exactly what you mean clearly?\n\nSure. Consider the integral $$\\int_{-1}^1$$ 1+x4 dx\nHow do I evaluate this integral using the substitution u=x4 and the formula:\n$$\\int_a^b$$ f(x4) dx = $$\\int_{a^4}^{b^4}$$ f(u) dx/du du, where u=x4\n\njgens\nGold Member\nWhy would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.\n\nI thought this was a more interesting way but just don't know how to get started.\n\nAlso I someone could help me with this problem.\nHow could I evaluate the integral above using u=x4, but seperating it into 2 integrals.\n\nCyosis\nHomework Helper\nTaking detours isn't a more interesting way, but here goes.\n$u=x^4 \\Rightarrow du=4x^3dx$\n\n$$\\int 1+x^4 dx=\\int \\frac{1+u}{4(u^{3/4})} du$$\n\nMaking life a lot harder!\n\nLast edited:\nGib Z\nHomework Helper\nPersonally I wouldn't discourage him, I find it a good way to learn how to compute integrals and what methods to use can be supplemented by trying other methods that come to mind, and seeing why they may not be as efficient. Once he computes that integral Cyosis got him, I think he will be able to see in future why it may not be wise to compute an integral as such.\n\nCyosis\nHomework Helper\nYou raise a very good point. I guess it's all too easy to see if something is going to be inefficient or not when you have some experience with the topic at hand.\n\nNote: I didn't take the limits into account so you still have to do that yourself.\n\nWhy would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.\nYou are too harsh. Sure, for this function there are easy ways to do the integral. But for other functions f maybe you really should do the substitution. And why not learn to do the substitution on simple problems that can be evaluated in other ways? Then, for example, you can check your answers and see when you make a mistake.", "date": "2020-07-06 18:11:28", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integral-substitution.328793/", "openwebmath_score": 0.9168926477432251, "openwebmath_perplexity": 597.986886279298, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9539660949832345, "lm_q2_score": 0.9046505370289057, "lm_q1q2_score": 0.8630059401339512}}
{"url": "https://brilliant.org/discussions/thread/synthetic-geometry-session-followup-questions-and/", "text": "# Synthetic Geometry Session Followup Questions and Problem Summary\n\nIn our first session, we spent some time on this problem: Let $ABC$ be a triangle in which $\\angle A = 60^{\\circ}$ Let $BE$ and $CF$ be the angle bisectors with $E$ on $AC$ and $F$ on $AB$. Let $M$ be the reflection of $A$ in the line $EF$. Prove that $M$ lies on $BC$\n\n1. $AEIF$ is cyclic because $\\angle BIC=90+\\frac {\\angle A}{2}=120^{\\circ}\\implies \\angle FIE+\\angle A=180^{\\circ}$. The cyclic result only occurs when $\\angle A=60^{\\circ}$, and it gives us information about the angles of $\\triangle AEF$ and consequently those of $\\triangle FME$\n\n2. $I$ is the circumcenter of $MEF$ because $IE=IF$ and $\\angle FIE=2\\angle FME$. This result relates $M$ with $I$ and gives us more angle relations.\n\n3. $BMIF$ is cyclic, which we proved by using the above properties to angle chase $\\angle FBI=\\angle FMI$. Symmetrically $CMIE$ is cyclic, so $\\angle IMB+\\angle IMC=\\angle AFI+\\angle AEI=180^{\\circ}$, which implies $B,M,C$ are collinear and we are done.\n\nThe discovery of these properties makes the problem transparent, meaning we know what makes our main result true. What this also means is that we can probably find simpler solutions. Indeed, there are many ways to solve this problem:\n\nUsing angle bisector theorem and Miquel point (Avoids tedious angle chasing):\n\nSince $AEIF$ is a cyclic quadrilateral, we know the miqual point of its complete correspondence lies on $BC$. This means the circumcircles of $BFI,CEI, ABE,ACF$ all have a common point on $BC$, we denote the point $M'$. Because $ACM'F$ is cyclic, $\\frac {FM'}{BF}=\\frac {AC}{BC}=\\frac {AF}{BF}$; therefore $AF=FM'$ and similarly $AE=EM'$. This is enough to establish $M=M'$ and $M$ lies on $BC$\n\nUtilizing $EF$ as the axis of symmetry:\n\nLet $EF\\cap BC=P$. It suffices to prove $PE$ bisects $\\angle APB$. We can achieve this with our first property.\n\nNow I will propose a few follow up questions which you guys should now easily answer:\n\nWe keep the notations of our main problem:\n\n1. $DBC$ is equilateral such that $D,A$ are on opposite sides of $BC$. Prove that $DE=DF$\n\n2. Through $I$ construct the perpendicular to $EF$ which intersects $EF,BC$ at $X,Y$. Prove that $2IX=IY$\n\n3. $O$ is the circumcenter of $AEF$, Prove that $OM\\perp BC$.\n\nNote by Xuming Liang\n5\u00a0years, 5\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nFor follow up problem 1 :\n\nTo prove $DE=DF$ , it suffices to establish congruence between $\\Delta DIF$ and $\\Delta DIE$ (not dead yet) . We have $DI=DI$ and we have already established that $IF=IE$ by having $\\angle IFE= \\angle IEF=30$. So we need to prove a pair of angles congruent. Since we want a good connection between the angles and the discovered stuff , we prefer proving $\\angle DIF=\\angle DIE$. By cyclicity we have $m\\angle BIF=m\\angle CIE=60$. So we need to prove $\\angle DIB=\\angle DIC=\\dfrac{\\angle BIC}{2}=\\dfrac{120}{2}=60$. Now how to achieve this? We have not yet utilized the equilateral triangle yet!\n\nSo, consider $\\Box BICD$ Can we prove it cyclic? Yes! We have to prove that $\\angle EBD+\\angle ECD=180$ and its easy.\n\n$\\angle IBD+\\angle ICD =\\angle IBC+\\angle CBD+\\angle ICD+\\angle BCD=\\dfrac{B}{2}+60+\\dfrac{C}{2} + 60=120+60=180 \\\\ \\Rightarrow \\Box BICD \\ is \\ cyclic!$\n\nSo by cyclicity we have $\\angle DIB=\\angle DIC=60$ which we had to prove.\n\n- 5\u00a0years, 5\u00a0months ago\n\nVery nice reasoning using congruence! For the cyclic part, I believe you meant to say $B,I,C,D$ are concyclic, which is true because $\\angle BIC+\\angle BDC=180$. Since $BD=DC$, $ID$ bisects $\\angle BIC$, which is what you wanted to prove.\n\n- 5\u00a0years, 5\u00a0months ago\n\nSorry , yes it is that $B,I,C,D$ are concyclic. I have edited it and few angles too.\n\n- 5\u00a0years, 5\u00a0months ago\n\nA fast way to prove that $BICD$ is cyclic is to notice that $\\angle BIC = 120^{\\circ}$ and $\\angle BDC = 60^{\\circ}$.\n\n- 5\u00a0years, 5\u00a0months ago\n\nFirst we have $\\Delta IXF$ as 30-60-90 triangle. So $FI=2IX$. So we have to prove that $FI=IY$. So consider $\\Delta BIF$ and $\\Delta BIY$ , where we have $\\angle FBI=YBI$ , $BI=BI$ , $\\angle BIF=\\angle BIY=60$ (by angle chase). So they are congruent by ASA test and hence $FI=IY$ is proved.\n\n- 5\u00a0years, 5\u00a0months ago\n\nI pretty much proved it the same as you. Nice solution.\n\n- 5\u00a0years, 5\u00a0months ago\n\nYesterday we did find that $\\angle EMC=60$ So to prove our result , we must prove that $\\angle OME=30$. Using the fact that $O$ is circumcentre , we have $\\angle EOF=\\angle EIF=120$ and by some angle chase we have $\\angle OEI=\\angle OFI=60$ where $\\angle OFE=\\angle IFE=\\angle OEF=\\angle IEF=30$. We get a beautioful result that $\\Box FOEI$ is a rhombus which has one of its diagonals that is $OI$ equal to its side. Using these facts and that $I$ is circumcenter of $\\Delta EMF$ we have $EI=OI=IF=IM$ and thus $\\Box EOFM$ is cyclic giving $\\angle OFE= \\angle OME=30$ which was to be proved :)\n\n- 5\u00a0years, 5\u00a0months ago\n\nWhen will be the next discussion?\n\n- 5\u00a0years, 5\u00a0months ago\n\nProblem 3: Since $M$ is the Miquel point, there exists a spiral similarity centered at $M$ that maps $FI \\rightarrow AE$. Denote $M_1, M_2$ as the midpoint of $AF$ and $EI$, respectively. Thus, the spiral similarity maps $FI \\rightarrow M_1M_2$. Using the circumcircles of $\\triangle BFI$ and $BM_1M_2$, this implies that $M_1M_2BM$ is cyclic. Since $M_1OM_2B$ is cyclic, this implies that $M_1OM_2MB$ is cyclic with diameter $OB$. Thus, $\\angle OMB = 90^{\\circ}$ and we are done.\n\n- 5\u00a0years, 5\u00a0months ago\n\nGreat solution. This fact is actually true as long as $BICD$ is cyclic.\n\n- 5\u00a0years, 5\u00a0months ago", "date": "2021-05-06 01:36:51", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/synthetic-geometry-session-followup-questions-and/", "openwebmath_score": 0.9793344140052795, "openwebmath_perplexity": 487.63523376078035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9744347823646073, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8629901021501732}}
{"url": "https://adamgemili.com/wiki/archive.php?tag=4a565b-is-population-discrete-or-continuous", "text": "A bacteria colony is made of billions of organisms. \u201cIf you can't explain it simply, you don't understand it well enough.\u201d \u2014Einstein (, Understanding Discrete vs. Asking for help, clarification, or responding to other answers. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. But if you're asking if it's spatially continuous then the answer is no. Hence, the temperature variable placed against the time variable would be represented by a continuous curve. It \u2026 Now that we know the continuous rate is -47.9% per year, we can work out how long until we're at 50%: This is a tricky one: the stock market changes every day, so it seems like it'd continuous, but there isn't an underlying predictable rate. Random variables represent quantities or qualities that randomly change within a population. This can be done regardless of whether the sample size is specified by a discrete unit or a continuous measure. Measurement Scale and Context For example, 8 bits have 256 possible values, and 16 bits have 65536. The natural log works on the ratio between the new and old value: $\\frac{\\text{new}}{\\text{old}}$. Learn more about how features and surfaces can be represented as either discrete or continuous in ArcGIS. It is continuous if you consider the values. What to do to speed up the paper publication process? The measurements themselves are discrete, but how you represent them need not be. What's that Professor? For example, you could represent them as a continuous density surface: Or, as discrete 3D bar charts extruded from the census tracts (in this case a grid): A population count is a point measure. We can't have half a student! Why is population density a continuous data type when it is typically measured for aggregate areas such as census tracts or districts/neighbourhoods (ie, it can't be measured at any point on a surface like gradient or temperature). The population of a city is always a continuous variable as people are always moving to and from the city as well as births & deaths. To simulate a weather system, for example, the tracking occurs continuously as all elements are constantly changing. How to say \"garlic\", \"garlic clove\" and \"garlic bulb\" in Japanese? \"12% interest per day\" is different than \"12% interest per year\".). BetterExplained helps 450k monthly readers with friendly, intuitive math lessons (more). (Notice how the rate must be scaled to match the time period. Something areal data is not. If the population is discrete, this distribution is a stick (or bar) graph. ), How To Think With Exponents And Logarithms, Q: Why is e special? It is areal data as @Radar have said. Continuous Growth. Updated my answer. Sample Data Representing a Population Distributions. In Star Trek TNG Episode 11 \"The Big Goodbye\", why would the people inside of the holodeck \"vanish\" if the program aborts? The function itself need not be continuous. Ah, but the caveat is the application of a kernel function on the point data. A surface for which each location has a specified or derivable value. For a census tract you can ask the question: How many people exist per unit area? Why was the name of Discovery's most recent episode \"Unification III\"? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The ratio between new and old was 37/53, so ln(37/53) = -.359 = -35.9% continuous growth over our time period. What's the best way to produce a density map from weighted points in QGIS? For a discrete distribution, probabilities can be assigned to the values in the distribution - for example, \"the probability that the web page will have 12 clicks in an hour is 0.15.\" Argh! Enjoy the article? For a spatial application the consideration of interpolation is important. (Continuous growth requires a smaller rate because of compounding.). Discrete and Continuous Data. We can take them, split them into smaller, more frequent changes, and spread them out. In our case, we grew from 1 to 2, which means our continuous growth rate was ln(2/1) = .693 = 69.3%. $2^n$ (where n is an integer) models discrete scenarios like coin flips or binary digits. Are you telling me the bacteria colony just happens to have a continuous rate of precisely ln(2) over the course of a day? elevation). There are two definitions of continuous data ( at least that I could find online). find the population density of some points in qgis/postgis, Calculating population density from urban raster reclasiffy, How to create a population density heatmap from attributes, Find average density value within a polygon in ArcMap, Sources for population density map of Europe. For example, the half-life of Carbon-14 is 5700 years. These practice problems focus on distinguishing discrete versus continuous random variables. A clear understanding of the difference between discrete and continuous data is critical to the success of any Six Sigma practitioner. +1 This answer addresses both spatial and non-spatial forms of data continuity, which is important considering the forum in which the question was asked. To learn more, see our tips on writing great answers. What are the possible outcomes? (The model gets more complex as you account for how long it takes for cubs to have children of their own.). We see a lot of jumpy changes, and sample them at yearly intervals to see how we're doing. I'd prefer you told me the colony doubled while a grad student stared at a petri dish for 24 hours straight. None of this \"wait until we decay by 50% so humans can count it easier\" nonsense. Let\u2019s see the definition: Continuous data is information that could be meaningfully divided into finer levels. A continuous distribution is one in which data can take on any value within a specified range (which may be infinite). What does \u201cblaring YMCA \u2014 the song\u201d mean? Population ecology is the study of population and how they change over time in relation to the environment including the environmental influences on population density and distribution, age structure and variations in population size. A discrete variable is a number that can be counted. Breeding seasons introduce some delay in the regulative process. The natural log finds the continuous rate behind a result. One of my pet peeves were problems like \"A bacteria colony doubles after 24 hours...\". Discrete and Continuous Data. Density is an areal measure and in itself implies that there exists a container variable (e.g. All data that are the result of counting are called quantitative discrete data. I visualize change as events along a timeline: Discrete changes happen as distinct green blobs. With discrete growth, we can see change happening after a specific event. Occasionally when analyzing data between groups or variables there are limitations on the validity of measured tests due to the lack of evidence, data, or information because a sample size is too small or has low value. Example: Material X decayed from 53kg to 37kg over 9 months. We can't point to an event and say \"It changed here\". The radioactive material is changing every instant. Automating calculation of Population Density in Census Tracts? jar that holds water, or in this case a census tract that holds people). Discrete Data. (1.98kg... 1.99kg... 2.00kg. If your system does change continuously, why not provide the continuous rate and write $e^{\\ln(2) x}$? Should live sessions be recorded for students when teaching a math course online? Show Answer. This happened over 9 months, so the monthly continuous rate is -35.9/9 = -3.98%. Using $e$ as a base ($e^{\\text{rate} \\cdot \\text{time}}$) implies you want people to think about change that happens at every moment. (Brush up on the number e and the natural logarithm.). I wrote this post because my video on e had questions about how $2^x$ represents \"staircase growth\". This is similar to saying the average family has 2.3 kids.). That is, the function's domain is an uncountable set.\n\n## is population discrete or continuous\n\nBackyard System Of Poultry Rearing, Anti American Synonym, Sm Megamall Cyberzone Store Directory, Dollar Tree Pharmacy, Gilchrist County Map, Haier Window Air Conditioner, Roasted Eggplant Baby Led Weaning, How To Make Mcdonald's Habanero Sauce,", "date": "2021-03-02 23:06:34", "meta": {"domain": "adamgemili.com", "url": "https://adamgemili.com/wiki/archive.php?tag=4a565b-is-population-discrete-or-continuous", "openwebmath_score": 0.5142000913619995, "openwebmath_perplexity": 1089.4232950225519, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620573763841, "lm_q2_score": 0.8991213860551287, "lm_q1q2_score": 0.8629425913113764}}
{"url": "https://mathoverflow.net/questions/408975/the-chromatic-number-of-the-union-of-two-graphs", "text": "The chromatic number of the union of two graphs\n\nLet $$G_n$$ be the graph on the set of all binary strings of length $$n$$ with two strings adjacent whenever they are Hamming distance $$2$$ away from each other, or one of them lies below another one; thus, for instance, $$G_2=K_4$$, and $$G_3$$ has $$25$$ edges. What is the chromatic number of $$G_n$$?\n\nThere is a simple, but not quite obvious construction showing that $$\\chi(G_n)\\le n(n+1)/2+1$$, and I am interested in a matching lower bound. Computations give $$\\chi(G_1)=2,\\ \\chi(G_2)=4,\\ \\chi(G_3)=5,\\ \\chi(G_4)=9,$$ $$\\chi(G_5)=12,\\ \\chi(G_6)=16,\\ \\text{and}\\ \\chi(G_7)\\le 17.$$ (The first five values are easy to compute, the last two are reported by Gordon Royle in the comments.)\n\nAn update. The colorings I am interesting in are of a special nature: if the strings $$s$$ and $$t$$ are same-colored, then (identifying strings with sets) also $$s\\setminus t$$ and $$t\\setminus s$$ are same-colored. Let $$\\chi^*(G)$$ denote the smallest number of colors needed to properly color the graph $$G$$ in this special way. Is it true that $$\\chi^*(G_n)=(1/2+o(1))n^2$$? That, say, $$\\chi^*(G_n)>(1/4+o(1))n^2$$?\n\n\u2022 @JoshuaZ: sure; color each vertex with the scalar product of the corresponding string and the vector $(1,2,\\dotsc,n)$.\n\u2013\u00a0Seva\nNov 20 at 16:36\n\u2022 what do you mean by lying below? Nov 21 at 8:48\n\u2022 @FedorPetrov: one of the corresponding subsets of the $n$-element set being contained in another one.\n\u2013\u00a0Seva\nNov 21 at 9:03\n\u2022 Here is a nice $17$-colouring of $G_7$ where I am using subsets of $\\{0,1,\\ldots,6\\}$ to represent the vertices. Let $\\sigma = (0,1,2,3,4,5,6)$ be a cyclic permutation and let $\\tau$ be the complement map on the subsets. Let $F= \\{013, 124, 235, 346, 450, 561, 602\\}$ be the Fano plane and set $C = F \\cup F^\\tau$. Then take $D = \\{0, 16, 25, 34, 124, 135, 236, 456\\}$ and its seven rotations under powers of $\\sigma$, and $D^\\tau$ and its seven rotations under powers of $\\sigma$. Then one colour class for each of $\\emptyset$ and the entire set, making a total of 17 classes. Nov 22 at 7:04\n\u2022 So $G_6$ has chromatic number $16$ - this took quite a long time to compute, almost all of which was spent searching in vain for a $15$-colouring. I think it will be hard to get close to an exact answer. Nov 23 at 2:29\n\nThis is a comment, not an answer, but it will be more convenient to post it as an answer. Consider the vertices of $$G_n$$ as subsets of $$[n]=\\{1,\\dots,n\\}$$.\n\nObservation 1. $$\\chi(G_n)\\ge\\left\\lfloor\\frac{3n}2\\right\\rfloor+1$$.\n\nThis is because $$G_n$$ contains a clique of that size, namely,\n$$\\varnothing,\\{1\\},\\{2\\},\\{1,2\\},\\{1,2,3\\},\\{1,2,4\\},\\{1,2,3,4,\\},\\{1,2,3,4,5\\},\\dots$$.\n\nObservation 2. The OP noted that $$\\chi(G_n)\\le\\binom{n+1}2+1=\\frac{n^2+n+2}2$$. For odd $$n$$ this bound can be improved to $$\\chi(G_n)\\le\\frac{n^2+3}2$$.\n\nThe vertices $$\\varnothing$$ and $$[n]$$ get their own colors. If $$1\\le|X|\\le n-1$$, color $$X$$ with the ordered pair $$\\left(\\sum_{x\\in X}x\\mod(n+1),\\ \\left\\lceil\\frac{|X|}2\\right\\rceil\\right).$$ This is a proper coloring, so (assuming $$n$$ is odd) $$\\chi(G_n)\\le2+(n+1)\\cdot\\frac{n-1}2=\\frac{n^2+3}2.$$\n\nObservation 3. The clique number $$\\omega(G_n)$$ is exactly $$\\left\\lfloor\\frac{3n}2\\right\\rfloor+1$$.\n\nWe showed in Observation 1 that $$\\omega(G_n)\\ge\\left\\lfloor\\frac{3n}2\\right\\rfloor+1$$. Let us prove by induction that $$\\omega(G_n)\\le\\left\\lfloor\\frac{3n}2\\right\\rfloor+1$$.\n\nWe may assume $$n\\ge2$$. Let $$\\mathcal C$$ be a maximal clique in $$G_n$$. Let $$\\mathcal C_h=\\{X\\in\\mathcal C:|X|=h\\}$$. Choose $$h$$ so that $$0\\lt h\\lt n$$ and $$\\mathcal C_h\\ne\\varnothing$$.\n\nIf $$\\bigcup\\mathcal C_h=[n]$$ and $$\\bigcap\\mathcal C_h=\\varnothing$$, then $$|\\mathcal C|=|\\mathcal C_h|+2\\le n+2\\le\\left\\lfloor\\frac{3n}2\\right\\rfloor+1$$. Otherwise, choose $$X\\in\\{\\bigcup\\mathcal C_h,\\ \\bigcap\\mathcal C_h\\}$$ so that $$0\\lt|X|\\lt n$$, and let $$k=|X|$$. Then every element of $$\\mathcal C$$ is a subset or superset of $$X$$. Since $$0\\lt k\\lt n$$, by induction we have $$|\\mathcal C|\\le\\omega(G_k)+\\omega(G_{n-k})-1\\le\\left\\lfloor\\frac{3k}2\\right\\rfloor+\\left\\lfloor\\frac{3(n-k)}2\\right\\rfloor+1\\le\\left\\lfloor\\frac{3n}2\\right\\rfloor+1.$$\n\n\u2022 Some more computational snippets: the cliques you found are the unique maximum clique (up to equivalence under the automorphism group) for $G_4$, $G_6$ and $G_8$. For $G_5$, $G_7$ and $G_9$ there are other classes of equally-large cliques, but none larger. Nov 25 at 2:56\n\u2022 So I liked your original proof that if $|\\mathcal{C}_h| = 1$ for some $h$, then the result follows. Did you decide that it was not so obvious to prove that there must be a layer containing just one vertex of the clique? Nov 26 at 5:11", "date": "2021-11-29 08:45:37", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/408975/the-chromatic-number-of-the-union-of-two-graphs", "openwebmath_score": 0.9459934830665588, "openwebmath_perplexity": 133.43440005979969, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815523039174, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8629099302324381}}
{"url": "https://math.stackexchange.com/questions/1633313/how-to-solve-for-x-z-and-y-z-here", "text": "# How to solve for x/z and y/z here?\n\nI got stuck solving these two equations: $$a_1(x/z) + b_1 (y/z) + c_1 = 0$$ and, $$a_2(x/z) + b_2 (y/z) + c_2 = 0$$ for $$x/z$$ and $$y/z$$.\n\nThe desired result would be: $${x \\over z} = {b_1c_2 - b_2c_1\\over a_1b_2 - a_2b_1}$$,\n\n$${y \\over z} = {c_1a_2 - c_2a_1\\over a_1b_2 - a_2b_1}$$\n\nHow do I get there? I keep getting dead ends.\n\nEdit: I basically came this far: $$u = v{-b_1 \\over a_1} - {c_1 \\over a_1}$$ and, $$u = v{-b_2 \\over a_2} - {c_2 \\over a_2}$$\n\nAnd ofcourse $$u = x/z$$, $$v = y/z$$\n\nNow i got here, $$v{-b_1 \\over a_1} - {c_1 \\over a_1} = v{-b_2 \\over a_2} - {c_2 \\over a_2}$$\n\nThen I took all the v's to one side and factored v outside: $$v({-b_2 \\over a_2} + {b_1 \\over a_1}) = {c_1 \\over a_1} - {c_2 \\over a_2}$$\n\nnow divide both sides by that b and a thingy on the left, but now it gets messy...\n\n\u2022 Can you show which dead end you run into? It should be simple enough to substitute your proposed solutions into the equations and simplify to show that they work. Jan 30 '16 at 16:08\n\u2022 Rename $\\frac xz=u$ and $\\frac yz=v$ and rewrite the initial equations. Jan 30 '16 at 16:08\n\u2022 Hint: take the last thing, check that it is correct (the second term on the left should be $\\frac{b_2}{a_2}$, and the plus on the right should be a minus), then multiply by $a_1a_2$ on both sides then solve for $v$. You just obtained $\\frac yz$ as in the solution you give :). Edit You just fixed the sign while I typed. Jan 30 '16 at 16:44\n\u2022 I suppose I am stuck on how to divide c1/a1 - c2/a2 by that thing on the left now. Any clever way to do that? Jan 30 '16 at 16:56\n\u2022 See my answer below. I started from a little earlier than that equation, but to make the division, I suggest you multiply both sides by $a_1a_2$, and then divide. The multiplication eliminates all denominators. This is why it is advisable. Jan 30 '16 at 17:04\n\n## 2 Answers\n\nHINT: with $$\\frac xz=u$$ and $$\\frac yz=v$$ we get $$a_1u+b_1v+c_1=0$$ $$a_2u+b_2v+c_2=0$$ if $$b_1\\ne 0$$ we have $$v=-\\frac{c_1}{b_1}-\\frac{a_1}{b_1}u$$ and we get an equation for $$u$$ $$a_2u+b_2\\left(-\\frac{c_1}{b_1}-\\frac{a_1}{b_1}u\\right)+c_2=0$$ can you proceed?\n\n\u2022 it seems we get a different thing, did I do something wrong in my original post? I edited it in. Jan 30 '16 at 16:42\n\u2022 I urge you to decide if you mean to use slash fractions $a/b$ or normal fractions $\\frac ab$, because mixing the two can be quite confusing for readers. @TheProgramMAN123 I think you just followed a different path than he did. You solved both for $u$ and then equalled the two expressions, he solved one for $v$ and then substituted the expression in the other one. Jan 30 '16 at 16:48\n\u2022 If you solve the last equation in this post for $u$, you get the solution in the question, with signs changed both in numerator and in the denominator. Jan 30 '16 at 16:51\n\nLet me just continue your attempt. You got to:\n\n$$\\left\\{\\begin{array}{c} u=-\\frac{b_1}{a_1}v-\\frac{c_1}{a_1} \\\\ u=-\\frac{b_2}{a_2}v-\\frac{c_2}{a_2} \\\\ \\end{array}\\right.$$\n\nIf we equal these two expressions for $u$ and multiply both sides by $a_1a_2$, we get:\n\n$$-b_1a_2v-c_1a_2=-b_2a_1v-c_2a_1.$$\n\nBringing the RHS's $v$ term to the left and the LHS's \"constant\" term to the right, this becomes:\n\n$$v(a_1b_2-a_2b_1)v=a_2c_1-a_1c_1.$$\n\nYes, I also factored out $v$ from the LHS and reordered the factors in the various $a,b,c$ products. Is this not precisely the solution you gave, that is:\n\n$$\\frac yz=v=\\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}?$$\n\nWonderful! To complete, we substitute this into the equation for $u$, one of the two, I mean. To simplify the calculations, we multiply by $a_1$ the first equation of the system at the start of the answer, and then substitute, getting:\n\n$$a_1u=-b_1\\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}-c_1.$$\n\nLet us make that RHS into a single fraction:\n\n$$a_1u=\\frac{\\overline{-b_1a_2c_1}+b_1a_1c_2-c_1a_1b_2+\\overline{c_1a_2b_1}}{a_1b_2-b_1a_2}=\\frac{a_1(b_1c_2-c_1b_2)}{a_1b_2-b_1a_2}.$$\n\nThe overlines simply indicate those two terms cancel. Oh, but if we divide both sides by $a_1$, we get the desired result, don't we? Very good. We are done.\n\nBottom line: when you have coefficients with fractions, the trick is often to sum all fractions and see if you can multiply/divide to simplify numerators or denominators. Don't be afraid of complicated coefficients: just have patience and do the algebra to the end, and you will get to the desired result :).\n\n\u2022 thanks a bunch! I guess the lesson for me here is to remove those fractions as soon as you can. They make things confusing and messy. i guess multiplying both sides by a1a2 is a clever way to do that. the solution really is painfully simple then. Jan 30 '16 at 18:29", "date": "2022-01-18 13:47:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1633313/how-to-solve-for-x-z-and-y-z-here", "openwebmath_score": 0.8862905502319336, "openwebmath_perplexity": 309.12897629086893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815523039175, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8629099171012103}}
{"url": "http://cloudtools.it/jkuq/moment-of-inertia-of-triangle.html", "text": "After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem. The matrix of the values is known as the moment of inertia tensor. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. Equation 18) also holds for polar moments of inertia i. The moment of inertia (I) of a body is a measure of its ability to resist change in its rotational state of motion. Similar to the centroid, the area moment of inertia can be found by either integration or by parts. EHE-08): Where: Mf = Mcrk = Nominal cracking moment of the cross section. Doing the same procedure like above, and below is the work. 3/2MR^2 The center of mass of the original triangle (the part that has been cut out) was at the center of the circle, at a distance R from the pivot. Now based on symmetry you can apply the definition of the moment of inertia to calculate the moment of inertia about the y axis which equals the cendroidal y axis. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale. dI = r2dm (1) (1) d I. Considering an element DE parallel to y-axis at a distance x from origin and width dx. Various such parameters include centre of gravity, moment of inertia, centroid , first and second moment of inertias of a line or a rigid body. So if the moment of inertia of the rectangle is, about its centroid, is bh cubed over 12, and the moment of inertia of the hole, the circle, from the previous tables is pi r to the 4th, over 4. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. 000965387 kg*m^2. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. \u2022 Th t fi ti fth hdd iThe moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle Determine the moment of inertia of the shaded area with respect to the x axis. This is a simulation of five objects on an inclined plane. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of The general expression for the moment of inertia of a right triangle about a centroidal axis parallel to a side is. In order to continue, we will need to find an expression for dm. Central axis of hallow cylinder. T 1 \u2013 the instantaneous value of load torque, referred to a motor shaft, N-m. A framework, in the shape of an equilateral triangle ABC, is formed by rigidly joining three uniform rods, each of mass m and length 2a. Table 8-2 Gross and Cracked Moment of Inertia of Rectangular and Flanged Section b d nA s kd n. Rectangle Triangle. Moment of Inertia of Mass, Mass Moment inertia of Common Shapes page Sideway Output on 28/4. The angle in between the masses is 60 degrees. A: area of the shape. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. The calculator has been provided with educational purposes in mind and should be used accordingly. University. Angular momentum. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Rolling without slipping problems. The equation of the line is $y = \\dfrac{a}{b} x + a$. 8680 rad/s^2 \u03b1_down -0. Determine polar moment of inertia of an isosceles triangle 1 answer below \u00bb Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. 025kg) g = gravity (9. The Poisson's ratio of the shaft material is , the moment of inertia about the y axis is , and the applied force at the end of the shaft is P. The SI unit of moment of inertia is kg m2. 2 x 10\u207b\u00b3 kg*(0. Various such parameters include centre of gravity, moment of inertia, centroid , first and second moment of inertias of a line or a rigid body. Mass Moment of Inertia, I G (cont\u2019d) I G for a body depends on the body\u2019s mass and the location of the mass. Axis through center. Mass Moment of Inertia - Mass Moment of Inertia (Moment of Inertia) depends on the mass of the object, its shape and its relative point of rotation - Radius of Gyration Pipe Formulas - Pipe and Tube Equations - moment of inertia, section modulus, traverse metal area, external pipe surface and traverse internal area - imperial units. The following is a list of second moments of area of some shapes. m = point mass. 17/12mL2 2. We found the moment of inertia of the apparatus alone to be 0. Presented here is a table of formulas which permit direct solution for required moment of inertia for several simple loading cases, for the two most common deflection criteria, L/240 and L/360. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. To find the perimeter of the triangle, you simply need to add together the lengths of the base and the two sides. The moment of inertia matrix is referred to the principal axes, again frame O 2 and the products of inertia are zero. This engineering data is often used in the design of structural beams or structural flexural members. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. And I was wondering whether someone could give me some more information/examples on first and second moment of area (tech calculus, wouldnt let me. It is measured by the mass of the body. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. Derivation of moment of inertia of triangle and cone. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. We will use the parallel axis theorem and we will take the centroid as a reference in this case. 0 kg per leg. d ' (n -1)A 's Gross Section Cracked Transformed Section Gross and Cracked Moment of Inertia b h A's As b h As b bw hf h b h hf y t A's As b d nA s kd n. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. I need to calculate the change in moment of intertia due to modifing a simple angled beam from 120 x 120 x 10 to 120 x 112 x 10. Considering an element DE parallel to y-axis at a distance x from origin and width dx. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Equation 18) also holds for polar moments of inertia i. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. About the Moment of Inertia Calculator. University. Answer this question and win exciting prizes. The moment of inertia of a body is its tendency to resist rolling motions and angular accelerations. How to calculate the moment of inertia of a triangular plate rotating about the apex. Kinetic Energy is the energy possessed by an object because it is in motion. Let the mass of the triangle be M. What is the moment of inertia of this triangle for rotation about an axis that is perpendicular to the plane of the triangle and through one of vertices of the triangle? The moment of inertia of a rod rotated about its center of mass is Irod, cm =1/12mL2. ): 6391 Instructor: E-mail: [email\u00a0protected] Topic - Moment of Inertia ,Ans - (Mh^2)/6. 9803 rad/s^2. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Center of mass, moments of inertia, volume of a body of rotation. A framework, in the shape of an equilateral triangle ABC, is formed by rigidly joining three uniform rods, each of mass m and length 2a. Calculate the 2nd moment of area for each element about the reference axes. ANSWER: Right angled triangle. Second moment of area for triangle trough x-axis = (ah3)/36. Calculate the moment of inertia of strait angle triangle about its $$y$$ axis as shown in the Figure on the right. Moment of Inertia of a Triangular Lamina about its Base. It appears in the relationships for the dynamics of rotational motion. Two conditions may be considered. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. The mass moment of inertia of an object about an axis (\ud835\udc4e) is equal to the moment about an axis (\ud835\udc4f) through the. l : moment of inertia about the axis parallel to x-axis. The moment of inertia with respect to central longitudinal axis would be m r2/2 m r2/3 m r2/6 m r2/12 The ratio of moment of inertia of a rectangle and that of a triangle, having same base and height, with respect to their bases would be 2 : 1 3 : 1 4 : 1 6 : 1 The ratio of the moment of inertia of a triangle of base width b and height h with. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices. They are; Axis passing through the centroid. The units of the product of inertia are the same as for moment of inertia. In addition to the moment of inertia, the product of inertia is commonly used. this apparently involves integrals but we havent even touched integrals in calculus need help please. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. dI y (dy)x 3 = 3 5. suppose the mean speed of such molecule in a gas 500 m per second and its kinetic energy of rotation is 2/3 of its kinetic energy of translation. Ix = b h3 / 36 (4a) Iy =h b3 / 36 (4b) Area Moment of Inertia for typical Cross Sections I. Calculating Moment Of Inertia Of A Triangle. 8\u00b710-2 Kg\u00b7m2 Submit Figure < 1of1 Incorrect; Try Again: 3 Attempts Remaining Part B What Is The Triangle's. The mass and moment of inertia through the sphere's center of mass are given. Each \"typical\" rectangle indicated has width dx and height y 2 \u2212 y 1, so its area is (y 2 \u2212 y 1)dx. When that happens equation 4 and 5 would be used to calculate the stress and. Inertia is a property of a body to resist the change in linear state of motion. Polar Moment of Inertia for Circular Cross-section. In particular, the same object can have different moments of inertia when rotating about different axes. Angular momentum. Weld design Moment of inertia of fillet weld J [mm 4 , in 4 ] Position of center of gravity of weld group section J = \u03c0 a (r + a / 2) 3 - Meaning of used variables: a fillet weld height [mm, in] B width of weld group [mm, in] H height of weld group [mm, in] L weld length [mm, in] r weld radius [mm, in] s web thickness [mm, in] t flange thickness [mm, in]. Moment of inertia can be defined by the equation The moment of inertia is the sum of the masses of the particles making up the object multiplied by their respective distances squared from the axis of rotation. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. The product moment of an area A of a right angle triangle about the axes xy is Product Moment of Inertia of a Right Angle Triangle by Parallel-axis Theorem. Question: Three point masses, each of mass {eq}m {/eq}, are placed at the corners of an equilateral triangle of side {eq}L {/eq}. If the polar moment of inertia is calculated at the centroid of the area, it is denoted. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Answer this question and win exciting prizes. What would the moment of inertia of a thin equilateral triangular sheet of mass M and sides S be with respect to an axis through one vertex perpendicular to the sheet? I got 3Ms^2/4 but I'd like to know if I'm right. Table of Selected Moments of Inertia Note: All formulas shown assume objects of uniform mass density. Let M represent the mass of the triangle and L the length of the base of the triangle. More on moment of inertia. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. Calculate the moment of inertia of a thin plate, in the shape of a right triangle, about an axis that passes through one end of the hypotenuse and is parallel to the opposite leg of the triangle, as in Figure P10. The polar moment of inertia JO of an area about O and the polar moment of inertia JC of the area about its c d o centroid are related to the distance d between points C and O by the relationship J O = J C + Ad 2 The parallel-axis theorem is used very effectively to compute the moment of inertia of a composite area with respect to a given axis. Solve: The moment of inertia of the triangle is I mr= \u00d7 = =3 3(0. The mass moment of inertia of an object about an axis through the center of mass is smaller than that about any other axis in the same direction. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. DIY Brick Rocket Stove - Cooking Without Electrical Power - Duration: 23:40. 0580 kg, an inner radius of 0. Center of mass, moments of inertia, volume of a body of rotation. Moment of Inertia and Polar Moment of Inertia are both the quantities expressing a body\u2019s tendency to resist changes when certain torque is being applied. \\end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. In order to find the moment of inertia of the triangle we must use the parallel axis theorem which ius as follows: The moment of inertia about any axis parallel to that axis through the center. All the formulas for a triangle are in any handbook. Area Moment of Inertia - Filled Right Triangle Solve. Rolling without slipping problems. When that happens equation 4 and 5 would be used to calculate the stress and. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. The inertia of both systems can be found using the equation: m = mass of hanging mass (0. 0 revolutions per s or 10 rad/s. Let be the position vector of the th mass element, whose mass is. Explanation: No explanation is available for this question! 2) What is the C. Look up I for a triangle in your table if you have forgotten. If k is the mass per unit area, then each typical rectangle has mass k(y 2 \u2212 y 1)dx. From the diagram below, we have:. Moment of inertia of this disc about the diameter of the rod is, Moment of inertia of the disc about axis is given by parallel axes theorem is, Hence, the moment of inertia of the cylinder is given as, Solid Sphere a) About its diameter Let us consider a solid sphere of radius and mass. Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. This works especially well when the general shape of the area can be decomposed into simpler shapes for which the moment of inertia is calculated for. Further explanation. We have step-by-step solutions for your textbooks written by Bartleby experts!. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2) is a measure of an object\u2019s resistance to changes in its rotation rate. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. 94 into 10 to the power of minsis 46 kg metre square bout an Axis through its Centre perpendicular to the lines joining the two atoms. Related Questions. The units. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. Radius and elevation of the semi-circle can be changed with the blue point. 2) A long rod with mass has a moment of inertia , for rotation around an axis near one. For example, given the axis O-O and the shaded area shown, one calculates the second moment of the area by adding together for all the elements of area dA in the shaded area. 4)and the second moment of the area about the y. Find the moment of inertia for the following about the y axis and x axis of a right triangle whose base is on the +x axis and whose height is on the +yaxis Source(s): moment inertia triangle: https://shortly. The more far away from the axis, the more moment of inertia the object has. The coordinate variables are x and y, respectively. Area A = 200 mm x 100 mm = 20000 mm2 I x. In other words, the centroid will always be 2/3 of the way along. Whatever kind you are trying to compute I would suggest breaking up the cross section into triangles with two vertices on successive points of your boundary and the third at the center about which the moment of inertia is to be taken. Asked in Algebra, Geometry. \\end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. The mass and moment of inertia through the sphere's center of mass are given. Going to the division, we get. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. svg 512 \u00d7 569; 4 KB. The moment of inertia of a uniform object depends not only on the size and shape of that object but on the location of the axis about which the object is rotating. 91, b < 10a. Angular acceleration of the system + triangle (long base) \u03b1_up 0. Q: Moment of Inertia of a thin spherical shell of mass m and radius r about its diameter is a) mr\u00b2/3 b) 2mr\u00b2/3 c) 2mr\u00b2/5 d) 3mr\u00b2/5 Q: Moment of inertia of a triangular section of base b and height h about an axis passing through its. The force of attraction is proportional to mass of the body. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. Radius of Gyration for a equilateral triangle can be calculated as. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Weld design Moment of inertia of fillet weld J [mm 4 , in 4 ] Position of center of gravity of weld group section J = \u03c0 a (r + a / 2) 3 - Meaning of used variables: a fillet weld height [mm, in] B width of weld group [mm, in] H height of weld group [mm, in] L weld length [mm, in] r weld radius [mm, in] s web thickness [mm, in] t flange thickness [mm, in]. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. 3) Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. The mass moment of inertia of an object about an axis (\ud835\udc4e) is equal to the moment about an axis (\ud835\udc4f) through the. How you find moment of inertia of isosceles triangle? Wiki User 2014-05-12 13:36:50. After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem. 5626 x 10\u207b\u00b3 initial inertia * angular velocity = new inertia * angular velocity. 2012/2013. Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. It is the rotational analog of mass. The moment of inertia of the rod is simply $$\\frac{1}{3} m_rL^2$$, but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. We want to find the moment of inertia, I y of the given area, which is rotating around the y-axis. In order to continue, we will need to find an expression for dm. A = bh \u00b8 2 Ic = bh 3 \u00b8 36 Base on x-axis, centroidal axis parallel to x-axis: x = h \u00b8 3 Ax = bh 2 \u00b8 6 Ix = bh 3 \u00b8 12 x-axis through vertex, Base and centroidal axis parallel to x-axis: x = 2h \u00b8 3 Ax = bh 2 \u00b8 3 Ix = bh 3 \u00b8 4. 16-24 From: Rabiei. The z2A term is the moment of inertia that area A would have about the y axis if all of the area were to be concentrated at the centroid. The moment of inertia block, which is a table containing the results of the moment of inertia calculation, is displayed and can be inserted anywhere in the drawing. \u00a320 \u00a3200 \u00a340. This works especially well when the general shape of the area can be decomposed into simpler shapes for which the moment of inertia is calculated for. The equilateral triangle actually makes the strongest column for a given area, but not by much (12% stronger than the circle). Rotational kinetic energy. Ix = b h3 / 36 (4a) Iy =h b3 / 36 (4b) Area Moment of Inertia for typical Cross Sections I. Relative to principal axes of inertia, the product of inertia of a figure is zero. \u2022 Th t fi ti fth hdd iThe moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle Determine the moment of inertia of the shaded area with respect to the x axis. Determine polar moment of inertia of an isosceles triangle 1 answer below \u00bb Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). DIY Brick Rocket Stove - Cooking Without Electrical Power - Duration: 23:40. Area, center of mass, moments of inertia. This theorem is really powerful because the moment of inertia about any set of axes can be found by finding the moment of inertia about the centroidal axes and adding the distance-area term to it. The right triangle comes along frequently in geometry. When that happens equation 4 and 5 would be used to calculate the stress and. Mass and moment of inertia properties of accessory components. This calculates the Area Moment of Inertia of a semi-circle about various axes. The following effective moment of inertia expression was originally proposed by Branson [9] and was adopted by ACI [10] and presented as (2). The help tool instructs me to click on the inspect menu and choose AREA. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = \u222b y 2 dA (11. EHE-08): Where: Mf = Mcrk = Nominal cracking moment of the cross section. The second moment of area is also known as the moment of inertia of a shape. PARALLEL-AXIS THEOREM, RADIUS OF GYRATION & MOMENT OF INERTIA FOR COMPOSITE AREAS. Synchronised Similar Triangles for Three-Body Orbit with L = 0 4 where (i,j,k) runs for the cyclic permutations of (1,2,3). Relative to principal axes of inertia, the product of inertia of a figure is zero. I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; S = Plastic Section Modulus, in 3 or mm 3; Z = Elastic Section Modulus, in 3 or mm 3; Online Parabolic Half Property Calculator. Table of Selected Moments of Inertia Note: All formulas shown assume objects of uniform mass density. Is there a way to calculate this to X-X? The instructions o. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Determine polar moment of inertia of an isosceles triangle 1 answer below \u00bb Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). \\] The moment of inertia of the area about the center can be found using in equation (40) can be done in two steps first calculate the moment of inertia in this coordinate system and then move the coordinate system to center. Figure 2: Deriving an equation for moment of inertia of the triangle rotating around its base. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Product of inertia for triangle. (6) Theorems of Moment of Inertia. Moment of inertia of the remaining part of lamina about the same axis is :. The mass moment of inertia of an object about an axis through the center of mass is smaller than that about any other axis in the same direction. Centroids & Moment of Inertia. o The moment of inertia of a triangular section of height h about its base is given as, I = bh 3 /12. Actual physical properties may vary due to tolerances which occur in the manufacturing process. The mass moment of inertia equation for a point mass is simply: I = mr 2. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. The moment of inertia must be specified with respect to a chosen axis of rotation. The product of inertia of triangle (a) with respect to its centroid is I \u00af x y = b 2 h 2 / 72. Right Triangle The output of this equation is the I x and I y components of the area moment of inertia when the triangle is defined to be in the x/y plane. It is formed by the intersection of the medians. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Solve: The moment of inertia of the triangle is I mr= \u00d7 = =3 3(0. 91, b < 10a. Question: Three point masses, each of mass {eq}m {/eq}, are placed at the corners of an equilateral triangle of side {eq}L {/eq}. (ii) Moment of inertia about new axes which is turned through an angle of 30 0 anticlockwise to the old axis. We see it in action all the time. Equation 18) also holds for polar moments of inertia i. It will help in deciding whether the failure will be on the compression face or on the tension face of the beam. When that happens equation 4 and 5 would be used to calculate the stress and. 18) I s = I c + Ad 2. In order to continue, we will need to find an expression for dm. The moment of inertia is \u2211mi*ri\u00b2; all the m are the same = 0. The radius of gyration is the radius at which we consider the mass to rotate such that the moment of inertia is given by I = M k2. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. With great regard for economy of words we call the expression written above \"the moment of inertia of the area about the x axis\" or I x for short. l : moment of inertia about the axis parallel to x-axis. 728(1) 30 (2) 0. Determine the moment of inertia of the area about the x and y. 9 \u00b5C, are located at the corners of an equilateral triangle as in the figure above. Going to the division, we get. Figure to illustrate the area moment of a triangle at the list of moments of inertia. Live Simple, Live Free - Tinyhouse Prepper Recommended for you. Figure to illustrate the area moment of a triangle at the list of moments of inertia. For the sake of one more bit of integration practice, we shall now use the same argument to show that the moment of inertia of a uniform circular disc about a. 2 Second Moment of Area11. The equation of the line is $y = \\dfrac{a}{b} x + a$. Moment of Inertia of a Triangular Lamina about its Base. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. Moment of Inertia 5 An example of this is the concrete T-beam shown. The second moment of area is also known as the moment of inertia of a shape. Explanation: No explanation is available for this question! 2) What is the C. Mass Moment of Inertia Calculator in Excel, Pt. The distance of the center of mass of the triangle in its new position from the pivot is the same. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2) is a measure of an object\u2019s resistance to changes in its rotation rate. Right: Triangles with centroidal axes re-positioned with respect to the x-axis. Moment of inertia can be defined by the equation The moment of inertia is the sum of the masses of the particles making up the object multiplied by their respective distances squared from the axis of rotation. It is the rotational analog of mass. Uniform circular lamina about a diameter. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. 707(h) to get the actual I, h being the weld size. It is required in the design of machines, bridges, and other engineering systems. Find the polar moment of inertia. Using these, the moment of inertia for the parallel axis can be calculated using the formula: The moment of inertia for rotation around the axis at the surface of the sphere is. Moment of Inertia of Isosceles Triangle Jalal Afsar October 25, 2013 Uncategorized No Comments Moment of Inertia of Isosceles triangle can be easily find out by using formulas with reference to x-axis and y-axis. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. Moment of Inertia Tensor Consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see Figure 28. The total moment of inertia is due to the sum of masses at a distance from the axis of rotation. If the triangle were cut out of some uniformly dense material, such as sturdy cardboard, sheet metal, or plywood, the centroid would be the spot where the triangle would balance on the tip of your finger. The Area Moment of Inertia equation, I = (b\u2022h 3)/12 , (b 3 \u2022h)/4 , computes the Area Moment of Inertia for a right triangle with right angle on right of the base. Question: What Is The Triangle's Moment Of Inertia About The Axis Through The Center? Express Your Answer To Two Significant Figures And Include The Appropriate Units. It is also popular as angular mass or rotational inertia of the given rigid body. A point mass does not have a moment of inertia around its own axis, but using the parallel axis theorem a moment of inertia around a distant axis of rotation is achieved. Angular acceleration of the system + triangle (long base) \u03b1_up 0. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. How you find moment of inertia of isosceles triangle? Wiki User 2014-05-12 13:36:50. Since this term is always zero or positive, the centroidal moment of inertia is the minimum moment of inertia with respect to all parallel axes. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. Moment of Inertia of a Triangular Lamina about its Base. PARALLEL-AXIS THEOREM, RADIUS OF GYRATION & MOMENT OF INERTIA FOR COMPOSITE AREAS. The Questions and Answers of The three point masses each of mass m are placed corner of equilateral triangle offside of side l then moment of inertia of the system about an axis along one side of triangle is? are solved by group of students and teacher of Class 11, which is also the largest student community of Class 11. It is also known as rotational inertia. Moment of Inertia is defined as: $$I={\\sum}mr^2$$ which in this case can be rewritten into an integral: $$I=\\rho\\int_A{r^2dA}$$ Since the shape of the triangle can't be described by one formula, you would have to split the integral into multiple sections. Moment of Inertia. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr 2. Axis through center. \u201cThe poles of inertia\u201d is another way of saying \u201cmass concentration centers\u201d. Undeniable momentum, on any stage - anywhere. The second moment of inertia rectangle is the product of height and cube of width divided by 12. Write (but do not evaluate) an integral expressing the moment of inertia of the region between y= sinxand y= sinx(for 0 x \u02c7) rotated around: (a)the xaxis; (b)the yaxis; (c)the zaxis. The computation of moments of inertia can often be. Chapter-3 Moment of Inertia and Centroid Page- 1 3. \u201ctwisting\u201d) about a given axis due to an applied torque. The distance of the center of mass of the triangle in its new position from the pivot is the same. J z' = I x' + I y'. 3 \u00d7 10\u207b\u2075 kg. The plane figures (like triangle, quadrilateral, circle, trapezoid, etc. d' (n -1)A 's Without compression steel With. of inertia of the rectangle. Note that each component of the moment of inertia tensor can be written as either a sum over separate mass elements, or as an integral over infinitesimal mass elements. automatic weight calculator for rectangular, square, round, or hexagonal, plate, tube, bar, beams, sheet, rod and other engineering material shapes. Hallow cylinder. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. 2012/2013. We will look at each expression below. The domain of the triangle is defined by. Moment of Inertia of Triangle about its Base | Very Important. Let R be the triangle with vertices (0, 0), (1, 0), (1, \u221a 3) and density \u03b4 = 1. 1 RADIUS OF GYRATION k All rotating machinery such as pumps, engines and turbines have a moment of inertia. Purpose: Determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle. Finding the Centroid. Angular momentum. = Polar Moment of Inertia. Description. dV = dxdydz. Add to Solver. The number of revolutions that the shaft will make during this time is. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. Determine the moment of inertia of the triangle for rotation about an axis that bisects one of its angles. Derivation of the Moment of Inertia Formula Suppose a particle of mass m is attached to a pivot by a thin rod of length r. Rotations in 2D are about the axis perpendicular to the 2D plane, i. Moment of inertia of three uniform rods of mass M and length l joined to form an equilateral triangle, about an axis passing through one of its sides. It is also de ned as I= Z r2 dm (3) for a continuous distribution of mass. In either case, use of the formulas is cumbersome and prone to error, especially in converting to consistent units. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. Integrate to derive a formula for the moment of inertia for a general triangle. 025kg) g = gravity (9. Evaluation of Moments of Inertia 2008 Waterloo Maple Inc. But I don't know how to do that. Moment of Inertia of Isosceles Triangle Jalal Afsar October 25, 2013 Uncategorized No Comments Moment of Inertia of Isosceles triangle can be easily find out by using formulas with reference to x-axis and y-axis. The force of attraction is proportional to mass of the body. 91, b < 10a. The moment of inertia of the triangular shaped area is 3. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord\u2019s mass. 8680 rad/s^2 \u03b1_down -0. The inertia matrix (aka inertia tensor) of a sphere should be diagonal with principal moments of inertia of 2/5 mass since radius = 1. Since moment of inertia is to be determined about an axis of rotation and from the provided diagram, it seems that the student is interested in finding the moment of inertia about the side (AB) along y-axis. 9 \u00b5C, are located at the corners of an equilateral triangle as in the figure above. Calculate the mass moment of inertia of the triangular plate about the y-axis. 2 comments. Find the moment of inertia of a plate cut in shape of a right angled triangle of mass M side AC=BC=a about an axis perpendicular to the plane of the plate and passing. Units 9 to 17,are assigned to:-Estimation of the Moment of inertia for Right-angled triangle (about X,Y) &Product of inertia &Polar Moment of Inertia, the radius of gyrations, by using two ways of Estimations, for the two cases of a right-angle triangle. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. The mass moments of inertia of an object about any parallel axes are identical. If you are consitent about which way you go around the triangle, the orgin can be anywhere, as it will subtract tnegative areas automatically. Central axis of hallow cylinder. The formula calculates the Moment of Inertia of a right triangle of base b and height h in respect to an axis collinear with the base of the triangle (one of the sides. Just add up area, centroid and self I of a series of triangles comprised of the origin, the ith point, and the i+1th point. Two circular loops of radii R and nR are made of same wire. (by the parallel axis theorem). For each segment defined by two consecutive points of the polygon, consider a triangle with two. Mathematically, and where IB \" *BA \" TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Using the parallel axis theorem, you can find the moment of inertia about the center by subtracting Mr^2, where r is 2/3h. One can define the moment of inertia as the ratio of the angular moment to the angular velocity of the particular object moving at its principal axis. Area Moment of inertia. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). 91, b < 10a. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. 7899 Working Ic. Then the periphery of the rectangle is 2x (10+20)=60 mm. In[4]:= Out[4]= You compute the moment of inertia about the y axis using the function SectionInertialMoments from the SymCrossSectionProperties package. Let the lengths of sides $AB$ and $BC$ be $a$ and $b$ respect. This is the currently selected item. The SI unit of moment of inertia is kg m2. The moment of inertia of the disk about its center is $$\\frac{1}{2} m_dR^2$$ and we apply the parallel-axis theorem (Equation \\ref{10. dm = M A dA (2) (2) d m. Let the mass of the triangle be M. moment of inertia with respect to x, Ix I x Ab 2 7. Where \"dM\" are small mass in the body and \"y\" is the distance of each on of them from the axis O-O. 32075h^4M/AL, where h is the height of the triangle and L is the area. Determine the moments of inertia about the centroid of the shape. Check the basic shapes at the bottom of. Bending of Beams with Unsymmetrical Sections C = centroid of section Assume that CZ is a neutral axis. We can relate these two parameters in two ways: For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. The 2nd moment of area, also known as moment of inertia of plane area, area moment of inertia, or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. 1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. 3\u00d710^ - 26 kg and a moment of inertia of 1. An 800g steel plate has the shape of an isosceles triangle. Our thin right triangular plate. The moment of inertia of this system about an axis along one side. Answered by Expert 5th October 2017, 8:56 PM. Ball hits rod angular momentum example. This is given by the table above which indicates that the centroid of a triangle is located, from the corner that is opposite of the hypotenuse (the longest side of the triangle), one-third of the length of the base in the y direction and one-third of the length of the height in the x direction in this case. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Below is the list of moments of inertia for common shapes. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. Solution: The mass moment of inertia about the y-axis is given by. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. Axis on surface. Although it is a simple matter to determine the moment of inertia of each rectangular section that makes up the beam, they will not reference the same axis, thus cannot be added. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of The general expression for the moment of inertia of a right triangle about a centroidal axis parallel to a side is. Once you have done this, run the \"massprop\" command and click. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = \u222b y 2 dA (11. Find the moment of inertia for the following about the y axis and x axis of a right triangle whose base is on the +x axis and whose height is on the +yaxis Source(s): moment inertia triangle: https://shortly. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. 2nd moment of an area or moment of inertia is the moment of all small areas dA about any axis. Area Moment of inertia. An isosceles triangle is a triangle with two equal sides. A higher moment of inertia is an indication that you need to apply more force if you want to cause the object to rotate. 5 \u2022 The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. 2 Second Moment of Area11. Kinetic Energy is the energy possessed by an object because it is in motion. I Average value of a function. Therefore Moment of Inertia of Rectangle about its center = m 1 2 a 2 + b 2 Distance of P point from center of rectangle is 2 a 2 + b 2 Therefore Moment of Inertia of Rectangle about P, I= m 1 2 a 2 + b 2 + m 4 a 2 + b 2 = m 3 a 2 + b 2 Mass of triangle PQR = 2 m = 2 \u03c1 a b Moment of Inertia of Triangle PQR about its centroid = \u03c1 1 2 a b 3 + b. Moment of Inertia is strictly the second moment of mass, just like torque is the first moment of force. How to calculate the moment of inertia of a triangular plate rotating about the apex. T 1 \u2013 the instantaneous value of load torque, referred to a motor shaft, N-m. The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. A 100-gram ball connected to one end of a cord with a length of 30 cm. 4)and the second moment of the area about the y. The area moment of inertia about the X and the Y axis are calculated by subtracting the second moment of inertia values of the inner rectangular area from that of the outer rectangular area. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. Figure 2: Deriving an equation for moment of inertia of the triangle rotating around its base. Let \u2018h\u2019 be the distance between the two axes i. Moment of inertia of a same object will change against different axis. We will take the case where we have to determine the moment of inertia about the centroid y. Calculate The Moment Of Inertia Of The Triangle With Respect To The X Axis. The moment of inertia is \u2211mi*ri\u00b2; all the m are the same = 0. In either case, use of the formulas is cumbersome and prone to error, especially in converting to consistent units. dm = M A dA (2) (2) d m. If the triangle were cut out of some uniformly dense material, such as sturdy cardboard, sheet metal, or plywood, the centroid would be the spot where the triangle would balance on the tip of your finger. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. 0mm and n = 15. The inertia of both systems can be found using the equation: m = mass of hanging mass (0. Central axis of disk. Note the dy is assigned the value 1 so that the Maple integrator does not confuse it as a mathematical variable. Journal of Graphics Tools: Vol. Chapter-3 Moment of Inertia and Centroid Page- 1 3. The axis perpendicular to its base. It is the measure of an object\u2019s resistance against the angular acceleration. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. 0 kg per leg. 0 , calculate the moment of inertia of the shaded area shown (Part B figure) about the x axis. More on moment of inertia. Calculate the mass moment of inertia of the triangular plate about the y-axis. Physics moment of inertia help? Three 210 g masses are connected to form an equilateral triangle with side lengths of 40 cm. However, this is only true for uniform or ordinary objects, such as an orb attached to a string whirling around at a certain angular velocity. Multiply the Area of each element by the square of the distance from the centroid of each element to the centroid of the cross-section(x1 and y1). Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Principle Axes of Axes of Inertia of a Mass. Three point charges, A = 2. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The domain of the triangle is defined by. Description Mass (kg) I (kg m2) 648-07627 9-Inch Disc 1. Area Moments of Inertia by Integration \u2022 Second moments or moments of inertia of an area with respect to the x and y axes, x \u00b3 yI y \u00b3 xdA 2 2 \u2022 Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. Remark: The moment of inertia of an object is a measure of the resistance of the object to changes in its rotation. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Conversely, a lower moment of inertia means that you only need to apply a minimal amount of force to cause a rotation. I = mass moment of inertia. Hemmingsen assumed (based on copyright claims). 28 Rectangle Area, in 2, in. I Average value of a function. 15 Centroid and Moment of Inertia Calculations An Example ! Now we will calculate the distance to the local centroids from the y-axis (we are calculating an x-centroid) 1 1 n ii i n i i xA x A = = = \u2211 \u2211 ID Area x i (in2) (in) A 1 2 0. The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Determination of the center of gravity. The angle in between the masses is 60 degrees. (Eq 2) \u03a6 = T GJ. 3/2MR^2 The center of mass of the original triangle (the part that has been cut out) was at the center of the circle, at a distance R from the pivot. ld parallel increases in mechanical power, and that EMG amplitude would increase with greater limb mass or limb moment of inertia. All the equations given below contain I, the moment of inertia of a beam, which is a constant determined by the beam's cross-sectional shape and thickness. Therefore Moment of Inertia of Rectangle about its center = m 1 2 a 2 + b 2 Distance of P point from center of rectangle is 2 a 2 + b 2 Therefore Moment of Inertia of Rectangle about P, I= m 1 2 a 2 + b 2 + m 4 a 2 + b 2 = m 3 a 2 + b 2 Mass of triangle PQR = 2 m = 2 \u03c1 a b Moment of Inertia of Triangle PQR about its centroid = \u03c1 1 2 a b 3 + b. Determine the moment of inertia of the cross section about the x axis. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. J z' = I x' + I y'. \"now, if the axis is passing through A, then sphere B, C, and D each rotate around this axismy attempt was this: m1(r^2) for sphere B (perpendicular to A) = 0. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. Annulus Moment of Inertia M5 revision thread Surface integrals of scalar fields show 10 more Urgent physics angular motion problem Angular momentum/moment of inertia Intuition question about Stokes' theorem. Going to the division, we get. base=20cm, height=30cm. 707(h) to get the actual I, h being the weld size. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is. I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; S = Plastic Section Modulus, in 3 or mm 3; Z = Elastic Section Modulus, in 3 or mm 3; Online Parabolic Half Property Calculator. Author: No machine-readable author provided. Mechanical Engineering: Ch 12: Moment of Inertia (27 of 97) Moment Centroid, Area, Moments of Inertia, Polar Moments of Inertia EMech full notes. The moment of inertia of a body is its tendency to resist rolling motions and angular accelerations. However, if we found the moment of inertia of each section about some. We see it in action all the time. the Z-axis. In particular, the same object can have different moments of inertia when rotating about different axes. They are; Axis passing through the centroid. Cross product and torque. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. dI = r2dm (1) (1) d I. 94 into 10 to the power of minsis 46 kg metre square bout an Axis through its Centre perpendicular to the lines joining the two atoms. Topic - Moment of Inertia ,Ans - (Mh^2)/6. of inertia of the rectangle. Calculating the second moment of area of geometric figures can be confusing and time consuming by hand, so let this calculator do all the work for you. Area Moment of Inertia of a Triangle. The product of inertia of area A relative to the indicated XY rectangular axes is IXY = \u222b xy dA (see illustration). Physics - Moment of Inertia Consider an equilateral triangle cut from a thin board. Calculating Moment of Inertia of a Uniform Thin Rod. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. Let's divide the triangle into strips along y-axis, each of width dx. ANSWER: Right angled triangle. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. Considering an element DE parallel to y-axis at a distance x from origin and width dx. This is given by the table above which indicates that the centroid of a triangle is located, from the corner that is opposite of the hypotenuse (the longest side of the triangle), one-third of the length of the base in the y direction and one-third of the length of the height in the x direction in this case. Is there a way to calculate this to X-X? The instructions o. 20}) to find. The tensor of inertia will take di\ufb00erent forms when expressed in di\ufb00erent axes. \"now, if the axis is passing through A, then sphere B, C, and D each rotate around this axismy attempt was this: m1(r^2) for sphere B (perpendicular to A) = 0. These bodies, with mass density $$\\rho$$, can be seen as stacks of infinitesimally thin triangles of thickness $$\\text{d}h$$ and surface density $$\\text{d} \\mu = \\rho \\text{d}h$$ (we preserve the notations from the previous posts. The moments of inertia of an angle can be found, if the total area is divided into three, smaller ones, A, B, C, as shown in figure below. one rectangle; one square; one triangle; At this stage, we calculate their surface area, the moment of inertia and the moment of deviation. Cross product and torque. Matt Anderson 18,225 views. The polar moment of inertia of the area A is calculated as. Three point charges are located at the corners of an equilateral triangle(q1=2microC,q2=-4microC). Annulus Moment of Inertia M5 revision thread Surface integrals of scalar fields show 10 more Urgent physics angular motion problem Angular momentum/moment of inertia Intuition question about Stokes' theorem. Area Moments of Inertia Parallel Axis Theorem \u2022 Moment of inertia IT of a circular area with respect to a tangent to the circle, ( ) 4 4 5 4 2 2 4 2 1 r IT I Ad r r r \u03c0 \u03c0 \u03c0 = = + = + \u2022 Moment of inertia of a triangle with respect to a. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. Considering an element DE parallel to y-axis at a distance x from origin and width dx. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Solution 3. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = \u222b y 2 dA (11. , in 4 \u00a6 xyA III II I x y xyA Apply the parallel axis theorem to each rectangle, xy \u00a6 I xcyc xyA Note that the product of inertia with respect to. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. Question: Part A - Moment Of Inertia Of A Triangle With Respect To The X Axis A Composite Area Consisting Of The Rectangle, Semicircle, And A Triangular Cutout Is Shown(Figure 1). Moment of Inertia for body about an axis Say O-O is defined as \u2211dM*y n 2. The situation is this: I know the moment of inertia with respect to the x axis and with respect to the centroidal x axis because its in the table. 2 An Example: Moment of Inertia of a Right Circular Cone For a right circular cone of uniform density we can calculate the moment. The formula calculates the Moment of Inertia of a right triangle of base b and height h in respect to an axis collinear with the base of the triangle (one of the sides. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. Moment of Inertia Tensor Consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see Figure 28. Bending of Beams with Unsymmetrical Sections C = centroid of section Assume that CZ is a neutral axis. Please enter the \"Input Values\" in the form. Similar to the centroid, the area moment of inertia can be found by either integration or by parts. This calculates the Area Moment of Inertia of a semi-circle about various axes. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Whatever kind you are trying to compute I would suggest breaking up the cross section into triangles with two vertices on successive points of your boundary and the third at the center about which the moment of inertia is to be taken. Open Section Properties Case 11 Calculator. Check the basic shapes at the bottom of. = Two point masses, m 1 and m 2, with reduced mass \u03bc and separated by a distance x, about an axis passing through the center of. EXAMPLE 2: MASS MOMENT OF INERTIA Calculate the mass moment of inertia of the triangular plate about the y-axis. This engineering data is often used in the design of structural beams or structural flexural members. find the average angular velocity of. Just add up area, centroid and self I of a series of triangles comprised of the origin, the ith point, and the i+1th point. Moment of Inertia Question (Edexcel M5) Moment Of Inertia Kinematics. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. 0 , calculate the moment of inertia of the shaded area shown (Part B figure) about the x axis. The centroid is 8\" above the base. 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free!. Mass Moment of Inertia Calculator in Excel, Pt. 8680 rad/s^2 \u03b1_down -0. The centroidal moments of inertia and the product of inertia are determined using the table below Product of inertia = Ixy = A (dx)(dy) = 0 8\" 3\"-3\" Part Area Ix dy1 d 2 y 1 (A) Ix. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. The ratio of length to radius is 1) 2: 1 2) 3:1 3) 3: 1 4) 2:1 28. Purpose: Determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle. 12 Moment of Inertia With Respect to an Arbitrary Axis Ellipsoid of Inertia. Get an answer for 'Q. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. new inertia = 1. Along the height it is hb^3/48 and along base it is bh^3/36. Let be the position vector of the th mass element, whose mass is. Sorry to see that you are blocking ads on The Engineering ToolBox! If you find this website valuable and appreciate it is open and free for everybody - please contribute by. 001472 Kg*m^2. The moment of inertia about an axis through a vertex is 0. 4 Find the moment of inertia of a plate cut in shape of a right angled triangle of mass M, side AC = BC = a about an axis perpendicular to the plane of the plate and passing through the mid point of side AB. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. Moment of Inertia 5 An example of this is the concrete T-beam shown. 8\u00b710-2 Kg\u00b7m2 Submit Figure < 1of1 Incorrect; Try Again: 3 Attempts Remaining Part B What Is The Triangle's. Transfer of Axis Theorem. J s = J g + Ad 2. This theorem is really powerful because the moment of inertia about any set of axes can be found by finding the moment of inertia about the centroidal axes and adding the distance-area term to it. It is the measure of an object\u2019s resistance against the angular acceleration.\njheft7effn3ar, 1is11lwjfjwq225, 5ze1johjltpr, kr8hzmoitbj, 4v9t0auqca9v, rc6f7w596h, 6nvxsfde5zwc, dma4q55h8rxs6, 7s9hb9ebww, fz2ljsujqsj, sn74jsawh4og, eskr7d5bttjgk, bwk6lfwirvv, kkds91l7it, hw8lnqtjkfy7, 6psyod2191coiu, 6uq3gahnijh1c, 7enn7qyu9r, tn6kdpzkzmo, 5uh29qggjw9n, d9hmiyufvyfpp23, 56rmfhz9kciwiz, nkt1zy4jbvqgg0, 6vk54wlfcr, oq391wndo3, ywmuiualxa, uecizc2qd5, bdf2t5li46306, wd2hb3vrlz, 81nqkpyfxfa6ws, 1erv8eqioeq", "date": "2020-07-10 08:56:07", "meta": {"domain": "cloudtools.it", "url": "http://cloudtools.it/jkuq/moment-of-inertia-of-triangle.html", "openwebmath_score": 0.6580184102058411, "openwebmath_perplexity": 368.8009363110306, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526441330527, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8628947828940822}}
{"url": "http://www.electropedia.org/iev/iev.nsf/17127c61f2426ed8c1257cb5003c9bec/1bf5b51d53229662c1257f9d0037eee0?OpenDocument", "text": "IEVref: 103-01-11 ID: Language: en Status: Standard Term: system of orthogonal functions Synonym1: orthogonal system Synonym2: Synonym3: Symbol: Definition: set of functions, such that each of them is orthogonal to any otherNote 1 to entry: Examples: Legendre polynomials P constitute a system of orthogonal functions on the interval $\\left[-1,\\text{\\hspace{0.17em}}+1\\right]$ because ${\\int }_{\\text{ }-1}^{\\text{ }+1}{P}_{k}\\left(x\\right){P}_{l}\\left(x\\right)\\text{d}x=0$ for any integers $k\\ne l$. Laguerre polynomials L constitute a system of orthogonal functions on the interval $\\left[0,\\text{\\hspace{0.17em}}+\\infty \\right]$ with the weight $\\text{e}\\text{x}\\text{p}\\left(-x\\right)$ because ${\\int }_{\\text{ }0}^{\\text{ }+\\infty }{L}_{k}\\left(x\\right){L}_{l}\\left(x\\right)\\text{exp}\\left(-x\\right)\\text{d}x=0$ for any integers $k\\ne l$. Trigonometric functions sine and cosine constitute a system of orthogonal functions on the interval $\\left[0,\\text{\\hspace{0.17em}}2\\pi \\right]$ because ${\\int }_{\\text{ }0}^{\\text{ }2\\pi }\\mathrm{sin}\\left(kx\\right)\\text{sin}\\left(lx\\right)\\text{d}x=0$ and ${\\int }_{\\text{ }0}^{\\text{ }2\\pi }\\mathrm{cos}\\left(kx\\right)\\mathrm{cos}\\left(lx\\right)\\text{d}x=0$ for any integers $k\\ne l$, and ${\\int }_{\\text{ }0}^{\\text{ }2\\pi }\\text{sin}\\left(kx\\right)\\mathrm{cos}\\left(lx\\right)\\text{d}x=0$ for any integer k and l. Publication date: 2009-12 Source: Replaces: Internal notes: 2017-02-20: Editorial revisions in accordance with the information provided in C00020 (IEV 103) - evaluation. JGO CO remarks: TC/SC remarks: VT remarks: Domain1: Domain2: Domain3: Domain4: Domain5:", "date": "2018-12-19 05:09:05", "meta": {"domain": "electropedia.org", "url": "http://www.electropedia.org/iev/iev.nsf/17127c61f2426ed8c1257cb5003c9bec/1bf5b51d53229662c1257f9d0037eee0?OpenDocument", "openwebmath_score": 0.43567317724227905, "openwebmath_perplexity": 1740.8719673337844, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526436103748, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8628947741196514}}
{"url": "https://math.stackexchange.com/questions/485462/birthday-problem-for-3-people/485504", "text": "# Birthday Problem for 3 people\n\nI know that, in a room of 23 people, there is a 50-50 chance that two people have the same birthday. However, what I want to know is: How many people do you need to have a 50-50 chance that 3 people share the same birthday?\n\nNote: Assume for this question, that birthdays are equally distributed\n\n\u2022 math.stackexchange.com/questions/25876/\u2026 \u2013\u00a0user940 Sep 6 '13 at 4:38\n\u2022 BirthDay Problem ---> en.wikipedia.org/wiki/Birthday_problem \u2013\u00a0Felix Marin Sep 6 '13 at 4:47\n\u2022 The linked question asks for the probability for 30 people. What I want is the amount of people needed for the probability to be above 50%. \u2013\u00a0Thomas Sep 6 '13 at 5:04\n\u2022 Notice that the number 23 uses the assumption that birthdays are equally distributed, i.e., that every birthday is equally-likely, with probability 1/365 (or 1/366). There is data that puts this into doubt, at least in the U.S. \u2013\u00a0DBFdalwayse Sep 6 '13 at 5:10\n\u2022 My question is assuming that the distribution is uniform. \u2013\u00a0Thomas Sep 6 '13 at 5:31\n\nThe question linked in the comments, while not your exact question, is very relevant. In particular, reading the answers posted there will tell you that an exact computation for triples is gross. However, Byron Schmuland's answer gives us a useful estimate: $$P(\\text{at least one triple with } N \\text{ people}) \\approx 1 - e^{-{N \\choose 3}/365^2}$$ From this we can simply solve for what $N$ makes this value $\\frac12$. We need \\begin{align*} 1 - e^{-{N \\choose 3}/365^2} &= \\frac12 \\\\ e^{-{N \\choose 3}/365^2} &= \\frac12 \\\\ e^{{N \\choose 3}/365^2} &= 2 \\\\ {N \\choose 3} &= 365^2 \\ln 2 \\\\ N(N-1)(N-2) &= 6 \\cdot 365^2 \\ln 2 \\\\ N^3 - 3N^2 + 2N - 6 \\cdot 365^2 \\ln 2 &= 0 \\end{align*}\n\nA computer calculation reveals that the only real root of this cubic polynomial is $N \\approx 83.13$. Hence we would expect to need $83$ or $84$ people in the room before having a 50-50 chance of getting three people with the same birthday.\n\nUPDATE: This is only an approximation, and it turns out it's off by more than I expected. To get an exact answer, I used the exact formula given here, and found that $N = 87$ is actually closest, with a probability of $.49945$ that three people have the same birthday.\n\n\u2022 Shoudn't the probability be over $365^3$ instead of $365^2?$ \u2013\u00a0vantonio1992 Sep 6 '13 at 6:00\n\u2022 @vantonio1992 check out Byron's answer in the linked question. $1/365^2$ is the chance of a given three people having the same birthday. \u2013\u00a06005 Sep 6 '13 at 6:02\n\u2022 This comment in particular: math.stackexchange.com/questions/25876/\u2026 \u2013\u00a06005 Sep 6 '13 at 6:03\n\u2022 Interesting that it is about double of the amount needed to ensure 50-50 for two people having the same birthday. \u2013\u00a0Thomas Sep 6 '13 at 6:07\n\u2022 Interesting, I wonder how many people you need for four people to share a birthday with 50% probability. If we made a graph of the function f(n) of the number of people needed to be 50% sure that n people have the same birthday, what would it look like? \u2013\u00a0Thomas Sep 7 '13 at 4:55\n\nEmpirically the answer seems to be $87$ or $88$.\n\nMore precisely, the probability of at least three people sharing a birthday is very close to $0.50$ if there are $87$ people in total (making the standard assumption of an i.i.d. uniform distribution over $365$ days).\n\nUsing the following R code to test a million cases for $87$ people in a room\n\ndays   <- 365\npeople <- 87\ncases  <- 1000000\nset.seed(1)\nmaxhits <- function(x){ max(table(x)) }\neg <- matrix(sample(days, people*cases, replace = TRUE), nrow=cases)\ntable(apply(eg, 1, maxhits) )\n\n\nthe sample distribution for the most people sharing a single birthday was\n\n 1      2      3      4      5      6      7\n13 500212 461918  36116   1688     50      3\n\n\u2022 Indeed 87 or 88 is what I get also. Cheers \u2013\u00a06005 Sep 6 '13 at 9:28\n\u2022 Deeper calculation gives rounded probabilities of at least three people sharing a birthday of $84-0.464549768$ $85-0.476188293$, $86-0.487826289$, $87-0.499454851$, $88-0.511065111$, $89-0.522648262$ so the median of the first time this happens is $88$ though $87$ is close, while the mode is $85$ and the mean is about $88.73891765$ \u2013\u00a0Henry Aug 2 '17 at 23:04\n\u2022 What do you mean by the median, mean, and mode? There is just one first time this happens, at $88$. I am having trouble understanding \u2013\u00a06005 Aug 3 '17 at 19:05\n\u2022 @6005: It can in fact first happen as a random variable at any time from $3$ through to $2\\times 365+1=731$, with median, mode and mean as stated, and standard deviation of about $32.832597$ \u2013\u00a0Henry Aug 3 '17 at 20:31\n\u2022 Thanks, I see what you meant now. \u2013\u00a06005 Aug 3 '17 at 20:35\n\nThere are 2 approaches to this kind of question (3 people with the same birthday...) You either 1) just want the answer. 2) want the satisfaction and understanding that comes from figuring it out from basic probability theory. (This can be a long hard road with no success guaranteed no matter how creative you are or how hard you work.)\n\nIf you just want the answer, the best way to get it is to write a little program that gets the answer by experiment. Using a random number generator to provide the data, and repeating the experiment a large number of times (for instance, 1,000,000 times) and seeing the result. You can easily get the answer to the birthday problem for any number of same birthdays (Sames), and also get answers to related questions- like, On the average how many pairs (OneLesses) of people had the same birthday at the time a new person had the same as one of the pairs making 3 with the same bday.\nFor this problem, Sames=3, OneLesses=number of pairs\n\nthe C code is below. It was created and run in the Pelles C compiler for windows, a free program.\n\nIn the case of the 3 person birthday problem, the results for a million runs the output of the program is-\n\n\" Avg # of tosses to get 3 Sames from 365 birthdays = 88.6971 Avg # of OneLesses, i.e. pairs, before 3 Sames from 365 birthdays = 10.1309\"\n\nSo the average number of people in a room before there being 3 with the same birthday is 88.7 and at the time that happened, there were, on the average, 10 pairs of people with the same birthday already in the room.\n\nC code->\n\n  Over=0;\n\n\nwhile (!Over) {\n\n             printf(\" enter NumTrials,   \");\nscanf(\"%d\",&NumTrials);\n\nprintf(\" enter number of outcomes,   \");\nscanf(\"%d\",&Outcomes);\n\nprintf(\" enter number of Sames,   \");\nscanf(\"%d\",&Sames);\n\nTotalTosses=0;\nTotalOneLesses=0;\nMaxCount=0;\n\nfor (i=1; i <= NumTrials; i++)\n{\n//  FOR NumTrials loop\n// initialize\nfor(j=0; j<=Outcomes; j++)  Results[j]=0;\nCount=0;\nOneLess=0;\n\nDone=0;\nwhile(!Done)\n{\ntoss= rand () % (Cards) ;\nCount++;\nResults[toss]++;\n\nif ( Results[toss]== (Sames-1) )  OneLess++;\n\nif ( Results[toss]==Sames) Done=1;\n\n}  // End of While!Done\n\nTosses[Count]++;\n\nTotalTosses+= Count;\nTotalOneLesses+= OneLess;\n\n}// for (NumTrials)\n\nAvgTosses= (double)TotalTosses /NumTrials ;\nprintf( \"Avg # of tosses to get %d Sames from %d Outcomes = %.4f \\n \", Sames, Outcomes, AvgTosses);\n\nAvgOneLesses= (double)TotalOneLesses /NumTrials ;\nprintf( \"Avg # of OneLesses before  %d Sames from %d Outcomes = %.4f \\n \", Sames, Outcomes, AvgOneLesses);\n\n\n} // END While (!Over)\n\n                   //quit program", "date": "2020-07-04 20:59:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/485462/birthday-problem-for-3-people/485504", "openwebmath_score": 0.9662566184997559, "openwebmath_perplexity": 554.935184358446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676496254957, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.862888407777744}}
{"url": "http://math.stackexchange.com/questions/534423/can-sets-of-cardinality-aleph-1-have-nonzero-measure/534682", "text": "Can sets of cardinality $\\aleph_1$ have nonzero measure?\n\n$\\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\\aleph_0$, and assuming the continuum hypothesis it's equal to $\\mathfrak{c}$, the cardinality of the the real numbers.\n\nMy question is, is it possible for all $\\aleph_1$ subsets of $\\Bbb{R}$ to have Lebesgue measure $0$? This is, of course, impossible assuming the continuum hypothesis, because then all sets of real numbers would have measure $0$, which is absurd. But is $\\mathsf{ZFC}$ + $\\lnot\\mathsf{CH}$ + \"all subset of $\\Bbb{R}$ of cardinality $\\aleph_1$ have Lebesgue measure $0$\" consistent? If not, what if we replaced Lebesgue measure with some other measure?\n\nIt may be worth noting that, although there may be subsets of $\\Bbb{R}$ with cardinality $\\aleph_1$, there are no subsets of $\\Bbb{R}$ which have the order-type $\\omega_1$ (the order-type of the countable ordinals) under the usual ordering on $\\Bbb{R}$.\n\nAny help would be greatly appreciated.\n\n-\nare you talking about an outer measure, because it is not clear that $\\aleph_1$ sets need to be measurable at all \u2013\u00a0 Dominic Michaelis Oct 21 '13 at 12:48\nIf Martin's axiom (and negation of CH) holds, then union of $\\aleph_1$ sets of Lebesgue measure 0 also Lebesgue measure 0. In particular, a set has cardinality $\\aleph_1$ has Lebesgue measure 0. It is known that if ZFC is consistent, then $\\mathsf{ZFC+\\lnot CH+MA}$ also consistent. \u2013\u00a0 tetori Oct 21 '13 at 12:53\n@DominicMichaelis A set which has outer measure zero is always measurable, so \"measure zero\" and \"outer measure zero\" are synonymous. \u2013\u00a0 Keshav Srinivasan Oct 21 '13 at 14:10\n@tetori Thanks. If you post it as an answer, I'm happy to accept it. \u2013\u00a0 Keshav Srinivasan Oct 21 '13 at 14:11\n@KeshavSrinivasan you asked for other measures too \u2013\u00a0 Dominic Michaelis Oct 21 '13 at 19:11\n\nA very simple model of \"$2^{\\omega} > \\omega_1$ and every set of reals of size $\\omega_1$ is null\" is the Cohen's original model for the failure of CH. It is obtained by adding $\\omega_2$ Cohen reals - i.e. forcing with $Fn(\\omega_2, 2)$. The proof uses the fact that adding a Cohen real makes the set of old reals null. The category analogue of this can be obtained by adding $\\omega_2$ random reals - i.e. forcing with the usual measure algebra on $2^{\\omega_2}$.\n\n-\nI actually had a hunch that this is true. But the MA argument has a slightly \"cleaner\" proof. \u2013\u00a0 Asaf Karagila Oct 21 '13 at 20:02\nBoth models are fine. This one just happens to avoid iterated forcing. \u2013\u00a0 hot_queen Oct 21 '13 at 20:08\nYes, but it's easier to take the consistency of $\\sf MA+\\lnot CH$ as a blackbox, in which case the argument becomes simpler. \u2013\u00a0 Asaf Karagila Oct 21 '13 at 20:09\n@AsafKaragila I disagree. $\\mathsf{MA}$ identifies all sorts of invariants, so it is too strong and full of distractions for the task at hand, and the black-box approach hides the actual reason why a result holds. (I agree that it helps if one has no interest in the actual combinatorics behind the consistency of the statement.) \u2013\u00a0 Andres Caicedo Oct 22 '13 at 0:32\n@Andres, the way I read the question fits the remark in parenthesis; so you do agree with me. :-) \u2013\u00a0 Asaf Karagila Oct 22 '13 at 4:30\n\nOf course assuming CH they can. But if $\\aleph_1 < 2^{\\aleph_0}$ then there is no measurable set of cardinality $\\aleph_1$ having positive Lebesgue measure, so the answer to the question in the title is no. We can prove the following in ZFC without requiring additional axioms.\n\nProposition (ZFC). Every uncountable Borel subset of $\\mathbb{R}$ has cardinality $2^{\\aleph_0}$.\n\nThis is a corollary of the Perfect Set Theorem (every uncountable Borel set contains a perfect set, i.e. a set with no isolated points, and any such set has cardinality at least $2^{\\aleph_0}$). For a proof, see Theorem 13.6 of Kechris's Classical Descriptive Set Theory.\n\nCorollary (ZFC). Every Lebesgue measurable subset of $\\mathbb{R}$ with cardinality less than $2^{\\aleph_0}$ has Lebesgue measure zero.\n\nProof. Every Lebesgue measurable set $E$ can be written $E = B \\cup N$ where $B$ is Borel and $m(N) = 0$; in particular $m(E) = m(B)$. But if $|E| < 2^{\\aleph_0}$ then $|B| < 2^{\\aleph_0}$, so by the previous proposition $B$ is countable and hence $m(B) = 0$.\n\n-\nThis, however, does not mean that every set of reals of size $\\aleph_1$ is indeed Lebesgue measurable. \u2013\u00a0 Asaf Karagila Oct 21 '13 at 15:09\n@AsafKaragila: True. I guess I answered a slightly different question. \u2013\u00a0 Nate Eldredge Oct 21 '13 at 15:36\n@NateEldredge (Thanks for the edit.) Since people have mentioned $\\mathsf{MA}$, let me add that under that axiom, every set of size below the continuum is measurable. In terms of cardinals characteristics of the continuum, the assumption relevant to the question (and implied by $\\mathsf{MA}+\\lnot\\mathsf{CH}$) is denoted by $\\mathrm{non}(\\mathcal L)>\\aleph_1$. \u2013\u00a0 Andres Caicedo Oct 22 '13 at 0:28\n\nThe following result is due to Martin and Solovay, and can be found in Jech's Set Theory (3rd Millennium edition) as theorem 26.39.\n\nIf Martin's Axiom holds, then the union of fewer than $2^{\\aleph_0}$ null sets is null, and the union of fewer than $2^{\\aleph_0}$ meager sets is meager.\n\nIt follows, if so, that under $\\sf ZFC+MA+\\lnot CH$ the union of $\\aleph_1$ singletons has measure zero, so every set of size $\\aleph_1$ has measure zero.\n\nIt suffices, if so, to show that $\\sf ZFC+MA+\\lnot CH$ is a consistent theory (relative to the consistency of $\\sf ZFC$, of course). This is a result of Solovay and Tennenbaum, which appears in the same book as Theorem 16.13.\n\nAssume $\\sf GCH$ and let $\\kappa$ be a regular cardinal greater than $\\aleph_1$. There is a c.c.c. notion of forcing $P$ such that the generic extension $V[G]$ by $P$ satisfies Martin's Axiom and $2^{\\aleph_0}=\\kappa$.\n\nBy taking, for example, $V=L$ as the ground model and $\\kappa=\\aleph_2$, the result is a model where the continuum is $\\aleph_2$ and the union of $\\aleph_1$ null sets is a null set.\n\n-", "date": "2014-04-20 08:40:52", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/534423/can-sets-of-cardinality-aleph-1-have-nonzero-measure/534682", "openwebmath_score": 0.9688538312911987, "openwebmath_perplexity": 221.52136072346383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676460637103, "lm_q2_score": 0.8807970842359877, "lm_q1q2_score": 0.8628884061732496}}
{"url": "http://aurelaisdugrandballon.com/baker-hughes-efrcil/condition-of-concurrency-of-three-lines-86e121", "text": "A point of concurrency is the point where three or more line segments or rays intersect. \"Then they started two or three new lines that have bridged that gap of new line revenue. For example, consider the three lines $$2x-3y+5=0,3x+4y-7=0\\,and\\,\\,9x-5y+8=0$$. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter, and each triangle has one, two, or three of these lines. In this, it differs from a process, which generally does not directly share data with other processes. Concurrency control protocols that use locking and timestamp ordering to en-sure serialisability are both discussed in this chapter. When three or more lines intersect together exactly at one single point in a plane then they are termed as concurrent lines. Key Concept - Point of concurrency. 03/30/2017; 3 minutes de lecture; Dans cet article. the line of action of R when extended backward is concurrent with its components as shown in the right diagram of figure 3. You have to put a lot of thought into it at every level of design. Concurrent lines are the lines that all intersect at one point. Cherchez concurrency et beaucoup d\u2019autres mots dans le dictionnaire de d\u00e9finitions en anglais de Reverso. We study the condition for concurrency of the Euler lines of the three triangles each bounded by two sides of a reference triangle and an antiparallel to the third side. Today, I finish the rules to concurrency and continue directly with lock-free programming. Just reasonable for conditions where there are not many clashes and no long transactions. at the same point. Users Options. C++11 Standard Library Extensions \u2014 Concurrency Threads. Learn concurrency with free interactive flashcards. Similarity of Triangles Congruence Rhs Sss; Congruence of Triangles Class 7; Congruence of Triangles Class 9; Congruent Triangles. https://www.onlinemath4all.com/how-to-check-if-3-lines-are-concurrent.html Otherwise, as shown in the right diagram of figure 1, when they are not concurrent, there would be a couple acting on the system and the condition \u2026 ## Concurrency is Hard to Test and Debug If we haven't persuaded you that concurrency is tricky, here's the worst of it. So, you pick groups of two equations one by one and solve them to find out the condition of concurrency. 92 terms. Proof that lines are concurrent $\\implies \\mathcal C$: For simplicity let's call the new criterion you've found $\\mathcal C$. Race conditions When correctness of result (postconditions and invariants) depends on the relative timing of events; These ideas connect to our three key properties of good software mostly in bad ways. They've been able to do it with insights to their own data.\" Hence, all these three lines are concurrent with each other. Yes, you have read it correctly: lock-free programming. Before I write about lock-free programming in particular, here are three last rules to concurrency. When two roadways share the same right-of-way, it is sometimes called a common section or commons. In MVCC, each write operation creates a new version of a data item while retaining the old version. Albie_Baker-Smith. These concepts are very important and complex with every developer. The point where three or more lines meet each other is termed as the point of concurrency. A polygon made of three line segments forming three angles is known as a Triangle. Transitive Property of Equality. the point of concurrency of the three perpendicular bisectors\u2026 When a circle contains all the vertices of a polygon (circle i\u2026 Three or more lines that intersect at a common point. See Centers of a triangle. Concurrency bugs exhibit very poor reproducibility. Orthocenter. They don\u2019t depend on any languages such as Java, C, PHP, Swift, and so on. concurrency Flashcards. Given three lines in the form of ${a}x + {b}y + {c} = 0$, this means: the point of concurrency of the three perpendicular bisectors\u2026 14 Terms. You have to put a lot of thought into it at every level of design. Two triangles are said to be congruent if their sides have the same length and angles have same measure. Browse 500 sets of concurrency flashcards. And even once a test has found a bug, it may be very hard to localize it to the part of the program causing it. A thread is a representation of an execution/computation in a program. As the global pandemic continues, more and more manufacturers are looking to technology to help drive efficiencies and adjust operations based on rapidly changing marketplace conditions. Furthermore, these two forces (R and ) must be in line with each other , thus making the three forces concurrent. This is the condition that must be satisfied for the three lines to be concurrent. There are three fundamentals methods for concurrency control. 2. Three or more distinct lines are said to be concurrent, if they pass through the same point. 3. Otherwise, lines from competing processes will be interleaved.The enforcement of mutual exclusion creates two additional control problems:deadlock. 1. Multi-version Concurrency Control (MVCC), Strict Two-Phase Locking (S2PL), and Optimistic Concurrency Control (OCC), and each technique has many variations. Thus if there are three of them, they concur at the incenter. Concurrency. Concurrency isn\u2019t the domain of any one aspect of the database. We\u2019ll work on fixing those problems in the next few readings. The point of intersection of any two lines, which lie on the third line is called the point of concurrence. Brahamh PLUS. They are. Three lines are said to be concurrent if any one of the lines passes through the point of intersection of the other two lines. Connect with online tutor to get rid of math phobia on Point of intersection of two lines and condition of concurrency of three lines. Four conditions for deadlock. We shall use the result that a single resultant R acting at a distance x pro-duces the same moment as its components about two different points on the rod in the next sec-tion to prove concurrency of three forces that are Diagrams. To say that the three lines are concurrent iff $\\mathcal C$ means that starting from the definition of a concurrent system of lines we should be able to derive $\\mathcal C$ and starting from $\\mathcal C$ we should be able to derive the fact that the lines are concurrent. In the figure below, the three lines are concurrent because they all intersect at a single point P. The point P is called the \"point of concurrency\". Or commons of intersection of the database with its components as shown in the right diagram of figure.. Depend on any languages such as Java, C, PHP, Swift, and so on on of! Other two lines condition for Triangles to be congruent if their sides have same. Version of a triangle, as in much modern computing, a thread is a transaction is the of... Points of concurrency have same measure figure 3 concurrency of three lines to be.. Flashcards on Quizlet  then they are as per the following: locking Methods ; Idealistic ;. This is the condition of concurrency can also be seen in the various of! Cuemath material for JEE, CBSE, ICSE for excellent results rid of phobia. They pass through the same length and angles have same measure it 's very hard discover! The line of action of R when extended backward is concurrent with components... This is the condition that must be satisfied for the three lines be. 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Same right-of-way, it differs from a process, which lie on the third line is called the of.", "date": "2021-07-31 05:56:38", "meta": {"domain": "aurelaisdugrandballon.com", "url": "http://aurelaisdugrandballon.com/baker-hughes-efrcil/condition-of-concurrency-of-three-lines-86e121", "openwebmath_score": 0.47816208004951477, "openwebmath_perplexity": 859.4103143578227, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9796676442828174, "lm_q2_score": 0.880797071719777, "lm_q1q2_score": 0.8628883923429176}}
{"url": "https://www.physicsforums.com/threads/if-f-x-10t-t-2-and-f-8-20-calculate-f-x.762694/", "text": "# If f'(x) = 10t / \u221b(t \u2013 2) and f(8) = \u201320, calculate f(x).\n\n1. Jul 22, 2014\n\n### s3a\n\n1. The problem statement, all variables and given/known data\nProblem:\nIf f'(x) = 10t / \u221b(t \u2013 2) and f(8) = \u201320, calculate f(x).\n\nSolution:\nLet u = t \u2013 2 \u21d2 dx = du. Then f(x) = \u201320 + \u222b_8^x [10t / \u221b(t \u2013 2)] dt = \u201320 + \u222b_6^(x \u2013 2) [10(u + 2) / \u221b(u)] du = \u201320 + 10 \u222b_6^(x \u2013 2) [u^(2/3) + 2u^(\u20131/3)] du = 30 \u221b[(x \u2013 2)^2] + 6(x \u2013 2)^(5/3) \u2013 66 \u221b(3) \u221b(12) \u2013 20\n\nAdditionally, the problem is attached as TheProblem.png, and the solution is attached as TheSolution.png.\n\n2. Relevant equations\nI'm not sure, but I think this has to do with the Fundamental Theorem of Calculus.\n\n3. The attempt at a solution\nI understand all the algebraic manipulations done; I'm just confused as to how the author went from the problem to the expression f(x) = \u201320 + \u222b_8^x [10t / \u221b(t \u2013 2)] dt. Also, is it okay/valid that f'(x) (which is a function of x) = 10t / \u221b(t \u2013 2) (which is a function of t)?\n\nAny help in clearing my confusions would be greatly appreciated!\n\n#### Attached Files:\n\nFile size:\n8.9 KB\nViews:\n51\n\u2022 ###### TheSolution.png\nFile size:\n22.7 KB\nViews:\n58\n2. Jul 22, 2014\n\n### xiavatar\n\nRemember that when you integrate a function you have find the constant of integration. He just combined the two steps of finding the constant and integrating into one step, essentially.\n\n3. Jul 22, 2014\n\n### HallsofIvy\n\nStaff Emeritus\nThe integral $\\int_a^a h(t)dt= 0$ for any integrable function h. So that $\\int_8^x h(t)dt$ gives a function that is 0 when x= 8. Knowing that f(8)= -20 means that $f(x)= -20+ \\int_8^x h(t)dt$", "date": "2017-08-19 10:36:40", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/if-f-x-10t-t-2-and-f-8-20-calculate-f-x.762694/", "openwebmath_score": 0.7015732526779175, "openwebmath_perplexity": 2007.2501268468582, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713870152409, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.862887871976551}}
{"url": "http://math.stackexchange.com/questions/29005/minimal-polynomial-of-i-sqrt2-in-mathbbq", "text": "# Minimal Polynomial of $i + \\sqrt{2}$ in $\\mathbb{Q}$\n\nI am trying to find Minimal Polynomial of $i + \\sqrt{2}$ in $\\mathbb{Q}$. I was able to determine the minimal polynomial is fourth degree with roots at $i-\\sqrt{2}$, $i+\\sqrt{2}$,$-i-\\sqrt{2}$,$-i+\\sqrt{2}$. However I got this answer by guessing at what the roots should be. Is there a general technique for this type of problem.\n\n-\n\nFirst, note that $i+\\sqrt{2}\\in\\mathbb{Q}(i,\\sqrt{2})$, so the degree is either $2$ or $4$. But $i+\\sqrt{2}$ is not of degree $2$ (it would have to be expressible in the form $a + b\\sqrt{d}$ for some squarefree integer $d$, with $a,b\\in \\mathbb{Q}$, and this is impossible. So you are certainly right that it is degree $4$.\n\nNow, here your method is not actually \"guessing\", but \"working.\" Suppose $f(x)$ is a the minimal polynomial of $i+\\sqrt{2}$, and consider the splitting field of $f(x)$, $K$. Complex conjugation gives an automorphism of $K$, and maps one root of $f(x)$ to another root of $f(x)$. That means that $-i+\\sqrt{2}$ must also be a root of $f(x)$.\n\nLikewise, the automorphism of $\\mathbb{Q}(\\sqrt{2})$ that fixes $\\mathbb{Q}$ and maps $\\sqrt{2}$ to $-\\sqrt{2}$ extends to an automorphism of $K$ which fixes $f(x)$, and maps any root of $f(x)$ to a root of $f(x)$. So both $i-\\sqrt{2}$ (the image of $i+\\sqrt{2}$) and $-i-\\sqrt{2}$ (the image of $-i+\\sqrt{2}$) must be roots of $f(x)$.\n\nThis gives you four roots of $f(x)$, which you know to be of degree $4$, so that gives the four roots and hence $f(x)$.\n\nFor more general comments, see this previous question. You can consider the powers of $i+\\sqrt{2}$ until you get that $1, \\alpha,\\alpha^2,\\ldots,\\alpha^n$ is linearly dependent over $\\mathbb{Q}$, and use the linear dependency to get the polynomial.\n\nAdded. In this case, we have: \\begin{align*} i+\\sqrt{2} &= i+\\sqrt{2}\\\\ (i+\\sqrt{2})^2 &= 1 + 2i\\sqrt{2}\\\\ (i+\\sqrt{2})^3 &= 5i - \\sqrt{2}\\\\ (i+\\sqrt{2})^4 &= -7 + 4i\\sqrt{2}\\\\ &= -9 + 2(1+2i\\sqrt{2}) = -9 + 2(i+\\sqrt{2})^2. \\end{align*} This means that $\\alpha=i+\\sqrt{2}$ satisfies $\\alpha^4 = -9+2\\alpha^2$, or that $\\alpha$ is a root of $f(x)=x^4 -2x^2 + 9$.\n\n-\nI don't see the link to the previous question. \u2013\u00a0 Ross Millikan Mar 25 '11 at 14:08\n@Ross This is done. \u2013\u00a0 Did Mar 25 '11 at 14:13\n@Ross: Sorry; had to go teach... \u2013\u00a0 Arturo Magidin Mar 25 '11 at 14:54\n\nThe way I like to look at such things... suppose $P(z)$ is a polynomial with rational coefficients. Then $P(\\bar{z}) = \\bar{P(z)}$ for any complex number $z$. So if $i + \\sqrt{2}$ is a root of $P(z)$, so is $-i + \\sqrt{2}$. Similarly, suppose $z = a + b\\sqrt{2}$, with $a$ and $b$ both of the form $q_1 + q_2i$ for rational $q_1$ and $q_2$. Then if $P(a + b\\sqrt{2}) = c + d\\sqrt{2}$, one has $P(a - b\\sqrt{2}) = c - d\\sqrt{2}$. So if $i + \\sqrt{2}$ is a root of $P(z)$, so is $i - \\sqrt{2}$, and if $-i + \\sqrt{2}$ is a root of $P(z)$, so is $-i - \\sqrt{2}$.\n\nThe upshot is that if $i + \\sqrt{2}$ is a root of $P(z)$, so are $-i + \\sqrt{2}$, $i - \\sqrt{2}$, and $-i - \\sqrt{2}$. Thus the minimal polynomial of $i + \\sqrt{2}$ over $Q$ will have to have these as roots and will be of degree at least 4. Then you can verify that the polynomial with these roots has rational coefficients and therefore is this minimal polynomial.\n\nIn some sense Arturo Magidin's answer is a way of describing the above phenomenon in terms of field automorphisms.\n\n-\n\nThere are many methods, e.g. using automorphisms, taking the characteristic polynomial of $\\rm\\ x \\to (i+\\sqrt{2})\\ x\\$, undetermined coefficients, etc. But here the simplest is probably repeated squaring: $\\rm\\ x - i = \\sqrt{2}\\$ so $\\rm\\ x^2 - 2\\ i\\ x -1 = 2\\$ or $\\rm\\ x^2 - 3 = 2\\ i\\ x$ which, squared, yields the result.\n\nUPDATE $\\$ Since the terseness of the \"repeated squaring\" method seems to have possibly confused at least one reader, perhaps it may help to elaborate a bit.\n\nTHEOREM $\\$ Suppose that $\\rm\\:R\\:$ is a ring containing elements $\\rm\\: x,\\: w,\\: i\\:$ such that $\\rm\\: w^2 = 2,\\ \\ i^2 = -1\\:.\\$ Then $\\rm\\ x = w + i\\ \\ \\Rightarrow\\ \\ x^4 + 9\\ =\\ 2\\ x^2\\:.$\n\nProof $\\$ Squaring $\\rm\\ x - i = w\\$ yields $\\rm\\ x^2 - 2\\:i\\ x - 1\\ =\\ 2\\:,\\:$ or $\\rm\\ x^2 - 3\\ =\\ 2\\:i\\ x\\:.\\:$ Squaring this yields $\\rm\\ x^4 -6\\ x^2 + 9\\ =\\: -4\\ x^2\\$ so $\\rm\\ x^2 + 9\\ =\\ 2\\ x^2\\:.\\quad$ QED\n\nREMARK $\\$ Notice that the above theorem holds true nonvacuously in rings besides $\\rm\\ \\mathbb Q(i,\\sqrt{2})\\:.\\ \\$ For example, $\\:$ in $\\rm\\: \\mathbb Z/17\\:,\\:$ the integers mod $17\\:,\\:$ we can choose $\\rm\\ w = 6,\\ i = -4\\$ and conclude that $\\rm\\ x = w+ i = 2\\ \\ \\Rightarrow\\ \\ x^4 -2\\ x^2 + 9\\ =\\ 0\\:.\\$ Indeed $\\rm\\ 16 - 8 + 9\\ =\\ 17\\ \\equiv\\ 0\\ \\ (mod\\ 17)\\:.$\n\n-\n@Bill You use $\\mathtt{x}$ for two different purposes. And you might wish to correct the repeated squaring. \u2013\u00a0 Did Mar 25 '11 at 14:11\n@Didier: Your comment makes no sense to me. \u2013\u00a0 Bill Dubuque Mar 25 '11 at 14:15\n@Bill The symbols $\\mathtt{x}$ did not appear as factors of $2\\mathrm{i}$ when I first looked at your comment. Now they do, so the second part of my comment is moot. Re the first part, you use $\\mathtt{x}$, first as the running argument of the transformation $\\mathtt{x}\\to(\\mathrm{i}+\\sqrt{2})\\mathtt{x}$, then as $\\mathtt{x}=\\mathrm{i}+\\sqrt{2}$. \u2013\u00a0 Did Mar 25 '11 at 14:27\n@Didier: I still see no problems. \u2013\u00a0 Bill Dubuque Mar 25 '11 at 14:34\n@Bill You did not define the symbol $\\mathbb{x}$ in the sense you use it in the last sentence of your post. Since you used the same symbol for something completely different in the sentence just before, this might be seen as an unfortunate choice of notations. (Sorry if my first comments on this were too cryptic.) \u2013\u00a0 Did Mar 25 '11 at 14:41", "date": "2015-05-23 07:28:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/29005/minimal-polynomial-of-i-sqrt2-in-mathbbq", "openwebmath_score": 0.9899265170097351, "openwebmath_perplexity": 149.55856519170442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713857177955, "lm_q2_score": 0.8757869981319862, "lm_q1q2_score": 0.8628878692431304}}
{"url": "http://slideplayer.com/slide/7853484/", "text": "# Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant.\n\n## Presentation on theme: \"Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant.\"\u2014 Presentation transcript:\n\nDefinition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant\n\nDefinition The parts of a hyperbola are: transverse axis\n\nDefinition The parts of a hyperbola are: conjugate axis\n\nDefinition The parts of a hyperbola are: center\n\nDefinition The parts of a hyperbola are: vertices\n\nDefinition The parts of a hyperbola are: foci\n\nDefinition The parts of a hyperbola are: the asymptotes\n\nDefinition The distance from the center to each vertex is a units a The transverse axis is 2 a units long 2a2a\n\nDefinition The distance from the center to the rectangle along the conjugate axis is b units b 2b2b The length of the conjugate axis is 2 b units\n\nDefinition The distance from the center to each focus is c units where c\n\nSketch the graph of the hyperbola What are the coordinates of the foci? What are the coordinates of the vertices? What are the equations of the asymptotes?\n\nHow do get the hyperbola into an up-down position? switch x and y identify vertices, foci, asymptotes for:\n\nDefinition where ( h, k ) is the center Standard equations:\n\nDefinition The equations of the asymptotes are: for a hyperbola that opens left & right\n\nDefinition The equations of the asymptotes are: for a hyperbola that opens up & down\n\nSummary Vertices and foci are always on the transverse axis Distance from the center to each vertex is a units Distance from center to each focus is c units where\n\nSummary If x term is positive, hyperbola opens left & right If y term is positive, hyperbola opens up & down a 2 is always the positive denominator\n\nFind the coordinates of the center, foci, and vertices, and the equations of the asymptotes for the graph of : then graph the hyperbola. Hint: re-write in standard form Example\n\nSolution Center: (-3,2) Foci: (-3\u00b1,2) Vertices: (-2,2), (-4,2) Asymptotes:\n\nExample Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes for the graph of : then graph the hyperbola.\n\nSolution Center: (-4,2) Foci: (-4,2\u00b1 ) Vertices: (-4,-1), (-4,5) Asymptotes:\n\nDownload ppt \"Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant.\"\n\nSimilar presentations", "date": "2020-12-03 17:36:18", "meta": {"domain": "slideplayer.com", "url": "http://slideplayer.com/slide/7853484/", "openwebmath_score": 0.8280630707740784, "openwebmath_perplexity": 728.119266877607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.985271388745168, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8628878671030599}}
{"url": "https://www.physicsforums.com/threads/surface-area-of-a-curve.704721/", "text": "# Surface area of a curve\n\n1. Aug 8, 2013\n\n### stunner5000pt\n\n1. The problem statement, all variables and given/known data\nAn industrial settling pond has a parabolic cross section described by the equation $y = \\frac{x^2}{80}$. the pond is 40 m across and 5 m deep at the cetner. the curved bottom surface of the pond is to be covered with a layer of clay to limit seepage from the pond. determine the surface are on the clay bottom of the pond\n\n2. Relevant equations\nLength of a curve formula\n$$ds = \\sqrt{1+\\left( \\frac{dy}{dx} \\right)^2 }$$\n\n3. The attempt at a solution\nI was thinking that if we simply find the length of the curve, then that would yield the surface area of the curve. This gives the following:\n\n$$s = \\int_{-20}^{20} \\sqrt{1 + \\frac{x^2}{1600} } dx$$\nWhich gives an answer of 66.9\n\nThe answer however is 1332.2 m^2. This leads me suspect that we must include the fact that this curve is to be rotated around the y axis making a bowl. I am not sure how to rotate this curve though...\n\n2. Aug 8, 2013\n\n### Ray Vickson\n\nWhat is the formula for the surface area of the surface z = f(x,y)? (Here, my z = your y, and (x,y) are the planar coordinates on the pond's surface.)\n\n3. Aug 8, 2013\n\n### Zondrina\n\nI read the question a few times over and I think I've got it. Plotting $y = \\frac{x^2}{80} - 5$ on wolfram gives an accurate model of the problem ( The cross section of the pond which is 5m deep at the center ).\n\nI believe what you would want to use here is :\n\n$2 \\pi \\int_{c}^{d} x \\sqrt{1 + (\\frac{dx}{dy})^2} dy$ which will be the surface area of your curve when you rotate it about the y axis. The reason you want to rotate it about the y-axis is because you want the surface area of the ENTIRE cross section of the pond.\n\nI got an answer of 1332.18.\n\nHint : Re-arrange your equation for y. Your limits should be from the bottom of the cross section of the pond to the top.\n\nLast edited: Aug 8, 2013\n4. Aug 8, 2013\n\n### LCKurtz\n\nYou almost have it. The $ds$ element is at distance $x$ from the y axis, so the circumference of the circle it traces is $2\\pi x$. You take that times the $ds$ length to get the surface area swept out. So you want $$\\int_0^{20} 2\\pi x~ds = \\int_0^{20} 2\\pi x\\sqrt{1 + \\frac{x^2}{1600} }~dx$$You only go from $0$ to $20$ in the limits because revolving it gets the \"other side\".\n\nLast edited: Aug 8, 2013\n5. Aug 8, 2013\n\n### stunner5000pt\n\nThank you very much for all your help. I see now the integrand is the curved surface area of a cylinder where the radius is x and the height is the length of a curve formula", "date": "2017-08-18 00:20:00", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/surface-area-of-a-curve.704721/", "openwebmath_score": 0.8475080728530884, "openwebmath_perplexity": 355.0902478958195, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98527138442035, "lm_q2_score": 0.8757869867849167, "lm_q1q2_score": 0.8628878569269017}}
{"url": "https://math.stackexchange.com/questions/2915935/radius-of-a-circle-touching-a-rectangle-both-of-which-are-inside-a-square", "text": "# Radius of a circle touching a rectangle both of which are inside a square\n\nGiven this configuration :\n\nWe're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.\n\nIf we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $$\\sqrt{2}\\times R-R$$)\n\nI really can't understand how to solve it. Any help appreciated.\n\n\u2022 This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this. Sep 13 '18 at 18:29\n\u2022 The information you have is $R(1-\\cos\\alpha) = 2a$ and $R(1-\\sin\\alpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$. Sep 13 '18 at 18:29\n\u2022 @amsmath juuuusssssttt beat me to it. Nice! Sep 13 '18 at 18:31\n\u2022 If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.\n\u2013\u00a0SMM\nSep 13 '18 at 18:31\n\u2022 This question could benefit from adding the 20 cm and 10 cm to the image itself. Sep 14 '18 at 13:23\n\nIt is just using the pythagorean theorem:\n$a=10$ $cm$\n$b=20$ $cm$\n$(r-a)^2+(r-b)^2=r^2$\n$(r-10)^2+(r-20)^2=r^2$\n$r^2+100-20r+r^2+400-40r=r^2$\n$r^2-60r+500=0$\n$r=50$ $cm$\n$r=10$ $cm$\nThe $r=50$ $cm$ is the acceptable answer.\n\n\u2022 It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square. Sep 14 '18 at 7:12\n\u2022 @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square? Sep 14 '18 at 10:12\n\u2022 A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square. Sep 14 '18 at 11:24\n\u2022 @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally? Sep 14 '18 at 11:40\n\u2022 @XanderHenderson, Is it better now? Sep 14 '18 at 12:19\n\nPlace the center of the cricle at $O.$\n\nLet the radius be $R$\n\nThe corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.\n\nOffsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$\n\nAnd the distance from this point equals the $R.$\n\nThat should put you on your way to the solution.\n\n$(x, y) = (R-20, R-10)$ as a point on the circle $y = \\sqrt{R^2 - x^2}$\n\n$R - 10 = \\sqrt{R^2 - (R-20)^2}$\n\n$(R- 10)^2 = R^2 - (R-20)^2$\n\n$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$\n\n$R^2 - 60R + 500 = 0$\n\n$(R - 50)(R-10) = 0$\n\n$R = 50$ is the only sensible option.\n\n\u2022 $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square. Sep 14 '18 at 8:10\n\u2022 @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square... Sep 14 '18 at 8:55\n\u2022 @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though. Sep 14 '18 at 9:01\n\u2022 @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title. Sep 14 '18 at 13:12\n\u2022 But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right? Sep 14 '18 at 13:50\n\nThe point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.\n\nThis leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_{1,2} = a+b\\pm\\sqrt{2ab}.$$\n\nOne solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+\\sqrt{2ab}.$$\n\nPlugging in $a=10$ and $b=20$ gives $R=50$.\n\nYou can use trig to get the same answer as those above.\n\nDraw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.\n\nWe know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).\n\nFind the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.\n\nThe longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.\n\nApply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.", "date": "2021-10-24 02:55:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2915935/radius-of-a-circle-touching-a-rectangle-both-of-which-are-inside-a-square", "openwebmath_score": 0.6927691698074341, "openwebmath_perplexity": 231.39725988537998, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713891776497, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8628878547047434}}
{"url": "http://mathhelpforum.com/algebra/39149-completing-square.html", "text": "# Math Help - Completing the square\n\n1. ## Completing the square\n\nHello,\n\nI'm having problems with completing the square can anyone comment on my working?\n\n2x^2 - 3x -1 = 0\n\n2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0\n\n2(x - 3/4)^2 -17/16 =0\n\n2(x - 3/4)^2 = 17/16\n\n(x - 3/4)^2 = 17/32\n\nx - 3/4 = +-\u221a17/32\n\nx = 3/4 +-\u221a17/32\n\n2. Originally Posted by Mouseman\nHello,\n\nI'm having problems with completing the square can anyone comment on my working?\n\n2x^2 - 3x -1 = 0\n\n2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0\n\n2(x - 3/4)^2 -17/16 =0\n\n2(x - 3/4)^2 = 17/16\n\n(x - 3/4)^2 = 17/32\n\nx - 3/4 = +-\u221a17/32\n\nx = 3/4 +-\u221a17/32\n\n2x^2 - 3x = 1\n\nx^2 - (3/2)x = 1/2\n\n(x - 3/4)^2 - 9/16 = 1/2\n\nx - 3/4 = +-sqrt(17/16)\n\nx = 3/4 +-sqrt(17/16)\n\n3. Thank you very much!\n\n4. Originally Posted by sean.1986\n2x^2 - 3x = 1\n\nx^2 - (3/2)x = 1/2\n\n(x - 3/4)^2 - 9/16 = 1/2\n\nx - 3/4 = +-sqrt(17/16)\n\nx = 3/4 +-sqrt(17/16)\nThe answer is to be x = 3/4 +- \u221a17/4\n\nI can't see where I am going wrong.\n\n5. When completing the square, just use the general formula.\n\n$a(x+\\frac{b}{2a})^{2}+c-\\frac{b^{2}}{4a}$\n\nPlug in a,b,c and you're set.\n\n6. Second attempt\n\n2x^2 - 3x -1 = 0\n\n2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0\n\n2((x - 3/4)^2) -17/8 =0\n\n2(x - 3/4)^2 = 17/8\n\n(x - 3/4)^2 = 17/16\n\nx - 3/4 = +-\u221a17/16\n\nx = 3/4 +-\u221a17/16\n\nIs my teacher wrong in stating that it is x = 3/4 +-\u221a17/4?\n\n7. Just to add my 2 cents......\n\n$2x^2-3x-1=0$\n\n1. First transpose the -1 to the right side of the equation.\n\n$2x^2-3x=1$\n\n2. Divide each term by 2\n\n$x^2-\\frac{3}{2}x=\\frac{1}{2}$\n\n3. Take half of the coefficient of x, square it and add it to both sides.\n\n$x^2-\\frac{3}{2}x+(\\frac{3}{4})^2=\\frac{1}{2}+\\frac{9}{ 16}$\n\n4. Noting the perfect square trinomial on the left:\n\n$(x-\\frac{3}{4})^2=\\frac{17}{16}$\n\n5. Take the square root of both sides:\n\n$x-\\frac{3}{4}=\\pm\\sqrt\\frac{17}{16}$\n\n6. Finally,\n\n$x=\\frac{3}{4}\\pm\\frac{\\sqrt17}{4}$\n\n$x=\\frac{3\\pm\\sqrt17}{4}$\n\n8. Hello,\n\nOriginally Posted by Mouseman\nSecond attempt\n\n2x^2 - 3x -1 = 0\n\n2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0\n\n2((x - 3/4)^2) -17/8 =0\n\n2(x - 3/4)^2 = 17/8\n\n(x - 3/4)^2 = 17/16\n\nx - 3/4 = +-\u221a17/16\n\nx = 3/4 +-\u221a17/16\n\nIs my teacher wrong in stating that it is x = 3/4 +-\u221a17/4?\nNote that $16=4^2$\n\nTherefore, $\\sqrt{16}=4$, and this yields the result your teacher gave you\n\n9. Yeah it's just a question of bracketing or simplifying.\n\nsqrt (a/b) = sqrt(a) / sqrt(b)\n\nso sqrt(17/16) = sqrt(17) / sqrt(16) = sqrt(17) / 4\n\nBoth answers are correct but I guess I should've simplified. Sorry mate!", "date": "2014-04-17 10:24:02", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/39149-completing-square.html", "openwebmath_score": 0.6716521978378296, "openwebmath_perplexity": 3783.837407108885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713857177955, "lm_q2_score": 0.8757869803008765, "lm_q1q2_score": 0.8628878516746483}}
{"url": "https://math.stackexchange.com/questions/2932610/is-it-necessary-to-assume-a-moment-generating-function-exists/2932668", "text": "# Is it necessary to assume a moment generating function exists?\n\nConsider random variables A, B, and C. We know that A = B + C. We also know that A and C have an MGF. Is it the case that B must have a MGF?\n\nAddition: Does this change if we know A and C both come from (different) chi-squared distributions? I am tasked with finding the distribution of B. If I can just do MGF(A) / MGF (C) = MGF (B) then it's simple... but can I even write this statement without assuming MGF (B) exists?\n\nGenerally -- You can't compute the MGF of B\n\nIn general, you can't compute the MGF of $$B$$ if you only know the MGFs of $$A$$ and $$C$$. For example, consider two possible joint distributions of $$A$$ and $$C$$:\n\nCase 1: P( A=0 and C=0) = 1/2 and P(A=1 and C=1)=1/2. In this case, the MGFs of A and C are $$(1+\\exp(t))/2$$ and the MGF of B is 1.\n\nCase 2: P( A=0 and C=1) = 1/2 and P(A=1 and C=0)=1/2. In this case, the MGFs of A and C are $$(1+\\exp(t))/2$$ and the MGF of B is $$\\frac{\\exp(-t)+\\exp(t)}2=\\cosh t$$.\n\nNotice that in both Case 1 and Case 2 the MGFs for $$A$$ and $$C$$ were $$(1+exp(t))/2$$, but the MGF for $$B$$ changed from Case 1 to Case 2.\n\n Generally -- You can prove existence of an MGF for B\n\nAlthough you can't compute the MGF of $$B$$, you can prove that $$M_B(t)$$ exists for $$(*)\\quad t\\in D=\\frac12 (Dom(M_A)\\cap (-Dom(M_C)).$$ Suppose $$t\\in D$$. If the MGF's of $$A$$ and $$C$$ exist, then for all $$t\\in Dom(M_A)$$, $$M_A(t)=||\\exp(ta)||_1<\\infty$$ and for all $$t\\in (-Dom(M_C))$$, $$M_C(-t)=||\\exp(-tc)||_1<\\infty$$ where $$||g||_p=\\left(\\int\\int |g(a,c)|^p\\; f(a,c)\\; da dc\\right)^{1/p}$$ is the $$L_p$$-norm of $$g$$ over the joint probability space and $$f(a,c)$$ is the joint pdf of $$A$$ and $$C$$. That implies $$||\\exp(ta/2)||_2 < \\infty$$ and $$||\\exp(-tc/2)||_2 < \\infty$$. By the H\u00f6lder's inequality or the Schwarz inequality, $$||\\exp(ta)\\exp(-tc)||_1<\\infty$$. But, $$||\\exp(ta)\\exp(-tc)||_1= ||\\exp(t(a-c)||_1= E[\\exp(tB)]=M_B(t).$$ This proves that $$M_B(t)$$ exists for $$t\\in D$$.\n\n If A and C are independent, you can compute the MGF of B\n\nIf $$A$$ and $$C$$ are independent and $$B = A-C$$, then it must be the case that $$(**) \\quad M_B(t) = M_A(t)\\cdot M_C(-t)$$ whenever $$t\\in Dom(M_A)\\cap(-Dom(M_C))$$ (see e.g. Wikipedia). Here is a rough proof.\n\nIf $$t\\in Dom(M_A)\\cap(-Dom(M_C))$$, then $$M_A(t)\\cdot M_C(-t) = \\int_{a=-\\infty}^\\infty \\exp(t a) dF_A(a) \\cdot \\int_{c=-\\infty}^\\infty \\exp(-t c) dF_C(c)$$ $$= \\int_{a=-\\infty}^\\infty \\int_{c=-\\infty}^\\infty \\exp(t (a-c)) dF_A(a) dF_C(c)$$ $$= \\int_{b=-\\infty}^\\infty \\exp(t b) dF_B(b) = M_B(t)$$ where $$F_A, F_B$$, and $$F_C$$ are the cumulative distribution functions of $$A, B$$, and $$C$$ respectively.\n\n If A and C have $$\\chi^2$$ distributions\n\nIn general, if $$A$$ and $$C$$ have $$\\chi^2$$ distributions, you can only state that $$B$$ has an MGF and the domain of $$B$$'s MGF is \\begin{align} Dom(B) &= \\frac12 (Dom(M_A)\\cap (-Dom(M_C))\\\\ &= \\frac12 ((-\\infty,2)\\cap (-(-\\infty,2))\\\\ &= \\frac12 ((-\\infty,2)\\cap (-2,\\infty))\\\\ &= \\frac12 (-2,2)=(-1,1)\\\\ \\end{align} by (*) and the fact that the domain of the MGF of a $$\\chi^2$$ distribution is $$(-\\infty,2)$$.\n\nIf you know that $$A$$ and $$C$$ have $$\\chi^2$$ distributions, and you know that they are independent, then you can look up the MGFs for $$\\chi^2$$ distributions and apply formula (**) to compute the formula for the MGF of $$B$$.\n\n\u2022 Thank you. The question asks to prove that B is chi-squared distributed. It sounds like we know that a MGF exists, but that we can't prove it's specifically chi-squared? \u2013\u00a0purpleostrich Sep 28 at 2:06\n\u2022 I learned a lot about MGFs and a bit about $\\chi^2$ distributions while trying to figure out the answer to your question. :) \u2013\u00a0irchans Sep 28 at 8:24\n\u2022 @purpleostrich, I suspect that if this was posed as an exercise to you then there is a missing assumption, for instance that $B$ and $C$ are independent, or as explored by irchans the assumption could be that $A$ and $C$ are independent (although I find this latter assumption somewhat less likely than the former). \u2013\u00a0pre-kidney Sep 28 at 9:01", "date": "2018-10-22 00:52:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2932610/is-it-necessary-to-assume-a-moment-generating-function-exists/2932668", "openwebmath_score": 0.9812694191932678, "openwebmath_perplexity": 175.09809536018446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787872422174, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.8628705130615156}}
{"url": "http://math.stackexchange.com/questions/51926/a-stronger-version-of-discrete-liouvilles-theorem", "text": "# A stronger version of discrete \u201cLiouville's theorem\u201d\n\nIf a function $f : \\mathbb Z\\times \\mathbb Z \\rightarrow \\mathbb{R}^{+}$ satisfies the following condition\n\n$$\\forall x, y \\in \\mathbb{Z}, f(x,y) = \\dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$\n\nthen is $f$ constant function?\n\n-\nYou probably wanto to add a boundedness condition. Otherwise $f(x,y)=x$ is a counterexample. \u2013\u00a0 Juli\u00e1n Aguirre Jul 17 '11 at 10:24\n@Julian Aguirre: since $x\\in\\mathbb Z$, we don't have $f(x,y)\\geq 0$. \u2013\u00a0 Davide Giraudo Jul 17 '11 at 11:08\n@girdav You are right. The lower bound is probably enough. \u2013\u00a0 Juli\u00e1n Aguirre Jul 17 '11 at 13:08\n\nYou can prove this with probability.\n\nLet $(X_n)$ be the simple symmetric random walk on $\\mathbb{Z}^2$. Since $f$ is harmonic, the process $M_n:=f(X_n)$ is a martingale. Because $f\\geq 0$, the process $M_n$ is a non-negative martingale and so must converge almost surely by the Martingale Convergence Theorem. That is, we have $M_n\\to M_\\infty$ almost surely.\n\nBut $(X_n)$ is irreducible and recurrent and so visits every state infinitely often. Thus (with probability one) $f(X_n)$ takes on every $f$ value infinitely often.\n\nThus $f$ is a constant function, since the sequence $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge.\n\n-\nI love probabilistic arguments in analysis! Very nice. \u2013\u00a0 Jonas Teuwen Jul 17 '11 at 11:46\nThis is indeed very nice: Question: Is it possible to change this argument in such a way that it applies for $\\mathbb{Z}^n$ instead of $\\mathbb{Z}^2$ only? Is it even true that a non-negative harmonic function on $\\mathbb{Z}^n$ is constant for $n \\geq 3$? For bounded ones this seems clear by considering the Poisson boundary. \u2013\u00a0 t.b. Jul 17 '11 at 11:53\n@Byron: This paper contains the claim that it is true that \"nonnegative nearest-neighbors harmonic function on $\\mathbb{Z}^d$ are constant for any $d$\" on page 2. \u2013\u00a0 t.b. Jul 17 '11 at 12:08\nThe usual Liouville theorem also holds with just a one-sided bound. \u2013\u00a0 GEdgar Jul 18 '11 at 13:56\nIf you know that the real part of an entire function $f(z)$ is non-negative on the complex plane, what can you say about the function $g(z)=f(z)/(1+f(z))$? \u2013\u00a0 Jyrki Lahtonen Jul 18 '11 at 14:07\n\nI can give a proof for the d-dimensional case, if $f\\colon\\mathbb{Z}^d\\to\\mathbb{R}^+$ is harmonic then it is constant. The following based on a quick proof that I mentioned in the comments to the same (closed) question on MathOverflow, Liouville property in Zd. [Edit: I updated the proof, using a random walk, to simplify it]\n\nFirst, as $f(x)$ is equal to the average of the values of $f$ over the $2d$ nearest neighbours of $x$, we have the inequality $f(x)\\ge(2d)^{-1}f(y)$ whenever $x,y$ are nearest neighbours. If $\\Vert x\\Vert_1$ is the length of the shortest path from $x$ to 0 (the taxicab metric, or $L^1$ norm), this gives $f(x)\\le(2d)^{\\Vert x\\Vert_1}f(0)$. Now let $X_n$ be a simple symmetric random walk in $\\mathbb{Z}^d$ starting from the origin and, independently, let $T$ be a random variable with support the nonnegative integers such that $\\mathbb{E}[(2d)^{2T}] < \\infty$. Then, $X_T$ has support $\\mathbb{Z}^d$ and $\\mathbb{E}[f(X_T)]=f(0)$, $\\mathbb{E}[f(X_T)^2]\\le\\mathbb{E}[(2d)^{2T}]f(0)^2$ for nonnegative harmonic $f$. By compactness, we can choose $f$ with $f(0)=1$ to maximize $\\Vert f\\Vert_2\\equiv\\mathbb{E}[f(X_T)^2]^{1/2}$.\n\nWriting $e_i$ for the unit vector in direction $i$, set $f_i^\\pm(x)=f(x\\pm e_i)/f(\\pm e_i)$. Then, $f$ is equal to a convex combination of $f^+_i$ and $f^-_i$ over $i=1,\\ldots,d$. Also, by construction, $\\Vert f\\Vert_2\\ge\\Vert f^\\pm_i\\Vert_2$. Comparing with the triangle inequality, we must have equality here, and $f$ is proportional to $f^\\pm_i$. This means that there are are constants $K_i > 0$ such that $f(x+e_i)=K_if(x)$. The average of $f$ on the $2d$ nearest neighbours of the origin is $$\\frac{1}{2d}\\sum_{i=1}^d(K_i+1/K_i).$$ However, for positive $K$, $K+K^{-1}\\ge2$ with equality iff $K=1$. So, $K_i=1$ and $f$ is constant.\n\nNow, if $g$ is a positive harmonic function, then $\\tilde g(x)\\equiv g(x)/g(0)$ satisfies $\\mathbb{E}[\\tilde g(X_T)]=1$. So, $${\\rm Var}(\\tilde g(X_T))=\\mathbb{E}[\\tilde g(X_T)^2]-1\\le\\mathbb{E}[f(X_T)^2]-1=0,$$ and $\\tilde g$ is constant.\n\n-\nTaxicab metric. Never heard that name (I learn maths in french at school). Funny! \u2013\u00a0 Patrick Da Silva Jul 17 '11 at 20:24\n@Patrick: Also called the Manhattan metric. \u2013\u00a0 George Lowther Jul 17 '11 at 20:30\nLAAAAAAAAAWL. Funnier. \u2013\u00a0 Patrick Da Silva Jul 17 '11 at 20:39\nNote: A similar proof will also show that harmonic $f\\colon\\mathbb{R}^d\\to\\mathbb{R}^+$ is constant. Interestingly, in the two dimensional case, Byron's proof can be modified to show that harmonic $f\\colon\\mathbb{R}^2\\setminus\\{0\\}\\to\\mathbb{R}^+$ is constant (as 2d Brownian motion has zero probability of hitting 0 at positive times). Neither of the proofs generalize to harmonic $f\\colon\\mathbb{R}^d\\setminus\\{0\\}\\to\\mathbb{R}^+$ for $d\\not=2$. In fact, considering $f(x)=\\Vert x\\Vert^{2-d}$, we see that $f$ need not be constant for $d\\not=2$. \u2013\u00a0 George Lowther Jul 17 '11 at 22:54\n\nHere is an elementary proof assuming we have bounds for $f$ on both sides.\n\nDefine a random walk on $\\mathbb{Z}^2$ which, at each step, stays put with probability $1/2$ and moves to each of the four neighboring vertices with probability $1/8$. Let $p_k(u,v)$ be the probability that the walk travels from $(m,n)$ to $(m+u, n+v)$ in $k$ steps. Then, for any $(m, n)$ and $k$, we have $$f(m, n) = \\sum_{(u,v) \\in \\mathbb{Z}^2} p_k(u,v) f(m+u,n+v).$$ So $$f(m+1, n) - f(m, n) = \\sum_{(u,v) \\in \\mathbb{Z}^2} \\left( p_k(u-1,v) - p_k(u,v) \\right) f(m+u,n+v).$$ If we can show that $$\\lim_{k \\to \\infty} \\sum_{(u,v) \\in \\mathbb{Z}^2} \\left| p_k(u-1,v) - p_k(u,v) \\right| =0 \\quad (\\ast)$$ we deduce that $$f(m+1,n) = f(m,n)$$ and we win.\n\nRemark: More generally, we could stay put with probability $p$ and travel to each neighbor with probability $(1-p)/4$. If we choose $p$ too small, then $p_k(u,v)$ tends to be larger for $u+v$ even then for $u+v$ odd, rather than depending \"smoothly\" on $(u,v)$. I believe that $(\\ast)$ is true for any $p>0$, but this elementary proof only works for $p > 1/3$. For concreteness, we'll stick to $p=1/2$.\n\nWe study $p_k(u,v)$ using the generating function expression $$\\left( \\frac{x+x^{-1}+y+y^{-1}+4}{8} \\right)^k = \\sum_{u,v} p_k(u,v) x^u y^v.$$\n\nLemma: For fixed $v$, the quantity $p(u,v)$ increases as $u$ climbs from $-\\infty$ up to $0$, and then decreases as $u$ continues climbing from $0$ to $\\infty$.\n\nProof: We see that $\\sum_u p_k(u,v) x^u$ is a positive sum of Laurent polynomials of the form $(x/8+1/2+x^{-1}/8)^j$. So it suffices to prove the same thing for the coefficients of this Laurent polynomial. In other words, writing $(x^2+8x+1)^k = \\sum e_i x^i$, we want to prove that $e_i$ is unimodal with largest value in the center. Now, $e_i$ is the $i$-th elementary symmetric function in $j$ copies of $4+\\sqrt{15}$ and $j$ copies of $4-\\sqrt{15}$. By Newton's inequalities, $e_i^2 \\geq \\frac{i (2j-i)}{(i+1)(2j-i+1)} e_{i-1} e_{i+1} > e_{i-1} e_{i+1}$ so $e_i$ is unimodal; by symmetry, the largest value is in the center. (The condition $p>1/3$ in the above remark is when the quadratic has real roots.) $\\square$\n\nCorollary: $$\\sum_u \\left| p_k(u-1,v) - p_k(u,v) \\right| = 2 p_k(0,v).$$\n\nProof: The above lemma tells us the signs of all the absolute values; the sum is \\begin{multline*} \\cdots + (p_k(-1,v) - p_{k}(-2,v)) + (p_k(0,v) - p_{k}(-1,v)) + \\\\ (p_k(0,v) - p_k(1,v)) + (p_k(1,v) - p_k(2,v)) + \\cdots = 2 p_k(0,v). \\qquad \\square\\end{multline*}\n\nSo, in order to prove $(\\ast)$, we must show that $\\lim_{k \\to \\infty} \\sum_v p_k(0,v)=0$. In other words, we must show that the coefficient of $x^0$ in $\\left( \\frac{x}{8}+\\frac{3}{4} + \\frac{x^{-1}}{8} \\right)^k$ goes to $0$.\n\nThere are probably a zillion ways to do this; here a probabilistic one. We are rolling an $8$-sided die $k$ times, and we want the probability that the numbers of ones and twos are precisely equal. The probability that we roll fewer than $k/5$ ones and twos approaches $0$ by the law of large numbers (which can be proved elementarily by, for example, Chebyshev's inequality). If we roll $2r > k/5$ ones and twos, the probability that we exactly the same number of ones and twos is $$2^{-2r} \\binom{2r}{r} < \\frac{1}{\\sqrt{\\pi r}} < \\frac{1}{\\sqrt{\\pi k/10}}$$ which approaches $0$ as $k \\to \\infty$. See here for elementary proofs of the bound on $\\binom{2r}{r}$.\n\nI wrote this in two dimensions, but the same proof works in any number of dimensions\n\n-\n\nAssuming that $f$ is bounded, you can prove that it is constant by a discrete analogue of the Borel-Caratheodory inequality, as shown in this equivalent question.\n\nThe key fact is such a function attains its maximum on a ball on the boundary, so it suffices to provide lower bounds for\n\n$$\\Gamma_N = \\max_{|x|+|y|=N}f(x,y).$$\n\n-", "date": "2015-04-19 23:48:11", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/51926/a-stronger-version-of-discrete-liouvilles-theorem", "openwebmath_score": 0.9740303754806519, "openwebmath_perplexity": 159.1409040993619, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787853562027, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8628705033175621}}
{"url": "https://math.stackexchange.com/questions/1978572/getting-2-different-solutions-to-the-integral-of-fracdx2x/1978579", "text": "# Getting 2 different solutions to the integral of $\\frac{dx}{2x}$\n\nI get two different answers that seem to conflict.\n\nIs there an error in one method??\n\n### Method 1\n\n\\begin{align} \\int \\frac{\\mathrm{d}x}{2x}&=\\frac 1 2 \\int\\frac{\\mathrm{d}x}{x}\\\\ &=\\frac 1 2\\ln|x|+C \\end{align}\n\n### Method 2\n\n\\begin{align} \\int \\frac{\\mathrm{d}x}{2x}&=\\frac 1 2 \\int \\frac{2\\mathrm{d}x}{2x}\\\\ &=\\frac 1 2\\int \\frac{\\mathrm{d}u}{u},\\;\\;\\text{ where }u=2x,\\ \\mathrm{d}u=2\\mathrm{d}x\\\\ &=\\frac 1 2 \\big(\\ln |u| + C\\big)\\\\ &=\\frac 1 2 \\ln|2x|+C \\end{align}\n\n\u2022 As the answer below details: the $C$'s are 'different', for want of a better word. Your \"paradox\" here is one of the reasons we are careful to always put a $C$ at the end, and also have to stay constantly aware of what it means. Oct 21, 2016 at 13:58\n\u2022 Another good example is integrating $\\sin(x)\\cos(x)$. Try integration by parts, to get $\\frac{1}{2}\\sin^2(x)+C$, then try using the trig identity $\\sin(x)\\cos(x)=\\frac{1}{2}\\sin(2x)$ to get $-\\frac{1}{4}\\cos(2x)+D$. A hint on reconciling the answers is $\\cos(2x)=1-2\\sin^2(x)$. Oct 21, 2016 at 14:06\n\u2022 When we get two ostensibly different solutions to an anti-differentiation problem, it is often helpful to check whether the solutions (ignoring the constants) are vertical translations of each other. If so, the two solutions must, of course, differ by a constant. In this case, $\\frac 1 2\\ln|x|$ translated vertically $\\ln\\sqrt{2}$ units yields $\\frac 1 2\\ln|2x|$. Oct 21, 2016 at 16:40\n\nHint: $\\frac{1}{2}\\ln{|2x|} + C = \\frac{1}{2}\\ln{|x|} + \\frac{1}{2}\\ln{2} + C = \\frac{1}{2}\\ln{|x|} + \\left(\\frac{1}{2}\\ln{2} + C\\right)$.", "date": "2022-05-19 09:41:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1978572/getting-2-different-solutions-to-the-integral-of-fracdx2x/1978579", "openwebmath_score": 0.9143304824829102, "openwebmath_perplexity": 897.2201552442342, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429614552197, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8628606275366784}}
{"url": "http://math.stackexchange.com/questions/819527/factoring-in-the-derivative-of-a-rational-function/819617", "text": "# Factoring in the derivative of a rational function\n\nGiven that $$f(x) = \\frac{x}{1+x^2}$$\n\nI have to find $$\\frac{f(x) - f(a)}{x-a}$$\n\nSo some progressing shows that:\n\n$$\\frac{\\left(\\frac{x}{1+x^2}\\right) - \\left(\\frac{a}{1+a^2}\\right)}{x-a} = \\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\\cdot\\frac{1}{x-a} = \\frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}$$\n\nNow, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?\n\n-\nCan't you just use the quotient rule to find this derivative? \u2013\u00a0 user2357112 Jun 3 '14 at 21:28\n\nSince $x-a$ is in the denominator, it makes sense to consider the possibility that $x-a$ is a factor of the numerator. (If you know the factor theorem, you can see this is the case, since the numerator equals zero when $x = a.)$ I see $x-a$ in the numerator, along with two other terms. Very little cleverness is needed at this point to write\n\n$$x-a + xa^2 - ax^2 \\; = \\; x - a + ax(a-x)$$ $$= \\; (x-a) - ax(x-a) \\; = \\; (x-a)(1-ax)$$\n\n-\nAhhhh. So much time looking at it and I didn't realise this. I don't know about the factor theorem, something to look at. Thank you for your answer. \u2013\u00a0 sidyll Jun 3 '14 at 18:08\n@sidyll: For more about using the factor theorem, see my answer at Finding limit of a quotient. For a less obvious use of the factor theorem, see my comments at How to simpify this?. \u2013\u00a0 Dave L. Renfro Jun 3 '14 at 18:12\nThank you for your references \u2013\u00a0 sidyll Jun 3 '14 at 18:17\n@sidyll: I just remembered that I posted a couple of old handouts of mine on the factor theorem a couple of years ago. See the files attached to this 12 January 2012 sci.math post archived at Math Forum. \u2013\u00a0 Dave L. Renfro Jun 3 '14 at 18:25\n@sidyll As for derivatives, generally the arithmetic works nicely using the quotient rule - see my answer. I'll bet Dave can give links to interesting expositions in AMM, Math. Mag, etc. \u2013\u00a0 Bill Dubuque Jun 3 '14 at 19:02\n\n\\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\\\ & = x-a+x(a^2-ax) \\\\ &= x-a+x(a(a-x)) \\\\ &= x-a+x(-a(x-a)) \\\\ &=\\color{blue}{x-a}-ax\\color{blue}{(x-a)} \\\\ &=(x-a)(1-ax).\\;\\checkmark } Therefore you can conclude that: \\eqalign{\\require{cancel}\\dfrac{f(x)-f(a)}{x-a}&=\\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\\\ &=\\dfrac{\\color{red}{\\cancel{\\color{black}{(x-a)}}}(1-ax)}{(1+x^2)(1+a^2)\\color{red}{\\cancel{\\color{black}{(x-a)}}}} \\\\ &= \\dfrac{1-ax}{(1+x^2)(1+a^2)}. }\\tag{x\\neq a}\n\n-\nThank you very much for your examples in continuing it. \u2013\u00a0 sidyll Jun 3 '14 at 18:09\n+1 for nice format. I learn something new about $LaTeX$ from you. \u2013\u00a0 Tunk-Fey Jun 3 '14 at 18:10\n@Tunk-Fey I've also learned a lot from your answers and your TeX style! $\\overset{\\cdot\\cdot}\\smile$ BTW you can write a more beautiful LaTeX as: $\\LaTeX$ using \\LaTeX, and there's also a variant for TeX namely $\\TeX$ which is produced by \\TeX. \u2013\u00a0 Hakim Jun 3 '14 at 18:13\n@sidyll You're welcome! Glad I could help! ;-) \u2013\u00a0 Hakim Jun 3 '14 at 18:14\n@\u062d\u0643\u064a\u0645\u0627\u0644\u0641\u064a\u0644\u0633\u0648\u0641\u0627\u0644\u0636\u0627\u0626\u0639 Thank you for the code. I didn't know that code before. $\\ddot\\smile$ \u2013\u00a0 Tunk-Fey Jun 3 '14 at 18:17\n\nBy the quotient rule for the difference $\\ f'(x)\\, :=\\, \\dfrac{f(x)-f(a)}{x-a}$\n\n$$\\quad \\begin{eqnarray} (g/h)'(x) &=\\,\\ & \\dfrac{\\color{#c00}{g'(x)} h(a) - g(a)\\color{#0a0}{h'(x)}}{h(a)h(x)}\\\\ \\begin{array}{l}\\ g(x)=x\\qquad\\Rightarrow\\,\\color{#c00}{g'(x) = 1}\\\\ h(x) = 1+x^2\\,\\Rightarrow\\,\\color{#0a0}{h'(x) = x+a}\\\\\\end{array}\\ \\Bigg\\}\\!\\!\\!\\!\\!&=& \\dfrac{\\color{#c00}1\\cdot (1+a^2)\\overset{\\phantom{I^I}}-a(\\color{#0a0}{x+a})}{(1+a^2)(1+x^2)}\\ =\\ \\dfrac{1-ax}{(1+a^2)(1+x^2)}\\end{eqnarray}$$\n\n-\n\nHint: you can guess that something interesting is going to happen near $x = a$, which suggests looking to factor $x-a$.\n\nIndeed, $x + xa^2 - a - ax^2 = (x-a) + xa(x-a)$.\n\n-\n\nFactor: $$x+xa^2\u2212a\u2212ax^2=x-a-ax^2+xa^2=(x-a)-ax(x-a)=(1-ax)(x-a)$$ Thus, $$\\frac{f(x)-f(a)}{x-a}=\\frac{\\frac{(1-ax)(x-a)}{(1+x^2)(1+a^2)}}{(x-a)}=\\boxed{\\frac{1-ax}{(1+x^2)(1+a^2)}}$$\n\n-", "date": "2015-10-04 22:11:27", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/819527/factoring-in-the-derivative-of-a-rational-function/819617", "openwebmath_score": 0.9994695782661438, "openwebmath_perplexity": 1014.0308382159619, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429614552197, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8628606212352482}}
{"url": "https://math.stackexchange.com/questions/2494774/questions-concerning-smallest-fraction-between-two-given-fractions", "text": "# Questions concerning smallest fraction between two given fractions.\n\nI recently encountered this on a practice test\n\nFind the smallest positive integer $n$ such that there exists an integer $m$ satisfying $0.33 < \\frac{m}{n} < \\frac{1}{3}$\n\n\\begin{align}0.33 < \\frac{m}{n} < \\frac{1}{3}&\\implies(\\frac{33}{100} < \\frac{m}{n})\\land(\\frac{m}{n}<\\frac{1}{3}) \\\\ &\\implies(33n<100m)\\land(3m<n) \\\\\\end{align} and thus $n=3m+1$.\n\nSo \\begin{align}33n<100m&\\implies33(3m+1)<100m \\\\ &\\implies 99m+33<100m\\\\ &\\implies m>33\\end{align} Taking $m\\geq34$, we find that $n=34\\times3+1=103$\n\nAfter the test, I had the following thoughts, which I do not know how to answer.\n\nQuestions\n\nWill the solution for $n$ always yield the smallest $m$ possible?\n\nDoes this method (find minimum value of $n$ and substitute) work for all such problems? (where $m,n\\in\\mathbb{Z}$)\n\nCan we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\\frac{a}{b}<\\frac{m}{n}<\\frac{c}{d}$).\n\nnote: sorry for asking multiple questions in one post (which I know some people frown on) but I feel posting multiple questions with the same 'introduction' (problem+proof) would clutter and be somewhat cumbersome.\n\n\u2022 In my opinion the smallest is $n=103$ because the smallest $m$ is $34$. Indeed if you solve for positive $n$ $$\\frac{34}{n}<\\frac{1}{3}$$ you get $n>102$ \u2013\u00a0Raffaele Oct 29 '17 at 11:01\n\u2022 @mathlove typo by me, thanks. I've also noticed that in our example $\\frac{33}{100}<\\frac{34}{103}<\\frac{1}{3}$, the numerator of the middle fraction is the sum of the other numerators, and the denominator for the middle fraction is the sum of the other denominators. I wonder if this is a coincidence or not, though I won't put that into the question (three questions in one post is already plenty) \u2013\u00a0user472341 Oct 29 '17 at 11:03\n\u2022 The continued fraction for $\\frac13$ is $(0;3)=(0;3,\\infty)$ and the continued fraction for $0.33$ is $(0;3,33)$. The simplest between them is $(0;3,34)=\\frac{34}{103}$. \u2013\u00a0robjohn Oct 29 '17 at 11:04\n\nCan we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\\frac{a}{b}<\\frac{m}{n}<\\frac{c}{d}$).\n\nIn the following, $a,b,c,d,m,n$ are positive integers.\n\n$\\frac ab\\lt \\frac mn\\lt \\frac cd$ is equivalent to $$na\\lt mb\\qquad\\text{and}\\qquad \\frac{md}{c}\\lt n$$\n\nFrom the second inequality, we can set $n=\\lfloor\\frac{md}{c}\\rfloor+1$ where $\\lfloor x\\rfloor$ is the greatest integer less than or equal to $x$.\n\nSo, from the first inequality, we get $$\\left(\\left\\lfloor\\frac{md}{c}\\right\\rfloor+1\\right)a\\lt mb\\tag1$$\n\nAs a result, we can say that the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\\frac ab\\lt\\frac mn\\lt\\frac cd$ is given by$$\\left\\lfloor\\frac{Md}{c}\\right\\rfloor+1$$ where $M$ is the smallest integer $m$ satisfying $(1)$.\n\nIf $c=1$, then the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\\frac ab\\lt\\frac mn\\lt\\frac 1d$ is given by$$\\left(\\left\\lfloor\\frac{a}{b-da}\\right\\rfloor+1\\right)d+1$$\n\nYes to both of your questions. This is essentially the mediant for two terms in a Farey sequence. There is some deep theory here. Read up on Farey sequences for more information.\n\n\u2022 I had three questions - i'm assuming you're talking about the first two? either way, thanks for the links. \u2013\u00a0user472341 Oct 29 '17 at 11:09\n\u2022 The mediant is not always the smallest. Take for instance $\\frac13$ and $\\frac34$: the mediant is $\\frac47$, which is between the two, but $\\frac12$ is the fraction with the smallest denominator. \u2013\u00a0robjohn Oct 29 '17 at 11:12\n\u2022 I think the mediant is only relevant when the fractions are consecutive terms in some Farey sequence. In the example @robjohn gives, $1/3$ and $3/4$ are not consecutive terms in any Farey sequence. \u2013\u00a0Gerry Myerson Oct 29 '17 at 11:17\n\u2022 You can iterate the algorithm using the mediant. For 1/3, 3/4 we get 4/7 and then with 1/3, 4/7 we get 1/2. \u2013\u00a0i. m. soloveichik Oct 29 '17 at 11:27", "date": "2019-11-12 18:17:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2494774/questions-concerning-smallest-fraction-between-two-given-fractions", "openwebmath_score": 0.9079024195671082, "openwebmath_perplexity": 265.0154743611997, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429604789206, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8628606109264227}}
{"url": "https://math.stackexchange.com/questions/311605/function-notation-terminology/311637", "text": "# Function notation terminology\n\nGiven the function $f:X\\longrightarrow Y$, $X$ is called the domain while $Y$ is called the codomain. But what do you call $f(x)=x^2$ in this context, where $x\\in X$? That is to say - what is the name for the $f(x)$ notation?\n\nAnd while I'm here, what is the proper way to write a function like this? Would it be $f:\\mathbb{R}\\to\\mathbb{R},\\;f(x)=x^2$?\n\nEdit:\n\nI figured I'd add this to add a bit of context into why I'm asking. I'm writing a set of notes in LaTeX, and I'd like to use the correct terminology for the definition of a function.\n\nA function from set $A$ to set $B$, denoted by $$f:A\\to B;x\\mapsto f(x)$$ is a mapping of elements from set $A$, (the $\\textit{domain}$) to elements in set $B$ (the $\\textit{codomain}$) using the $\\color{blue}{\\sf function}$ $f(x)$. The domain of a function is the set of all valid elements for a function to map from. The codomain of a function is the set of all possible values that an element from the domain can be mapped to. The $\\textit{range}$ (sometimes called the $\\textit{image}$) of a function is a subset of the codomain, and is the set of all elements that actually get mapped to by the function $f$.\n\nHere I'm pretty sure the highlighted word \"function\" is not right.\n\n\u2022 I suggest \\to or \\rightarrow instead of \\longrightarrow. The last one is there for if you need the arrow to be longer because you're writing something over it. As in $\\overset{\\text{text}}{\\longrightarrow}$ instead of $\\overset{\\text{text}}{\\to}$.\n\u2013\u00a0Jim\nFeb 22, 2013 at 23:16\n\u2022 why do you think that function is not right = Feb 22, 2013 at 23:36\n\u2022 Some would call \"$f(x)=x^2$\" the rule of the function $f$. Feb 22, 2013 at 23:44\n\u2022 @DominicMichaelis Well, function is used in two places to mean two slightly different things; I don't like using function to describe part of a function. Feb 22, 2013 at 23:47\n\nI can remember to read this text, and being puzzled with the exact same question. From what I've learned from my teacher, you're right, writing down something as \"the function $f(x)$...\" is sloppy notation. However, many books/people will use it this way.\n\nIf you're are very precise, $f(x)$ is not a function or an map. I don't know of a standard way to refer to $f(x)$, but here is some usage I found on the internet:\n\n\u2022 The output of a function $f$ corresponding to an input $x$ is denoted by $f(x)$.\n\u2022 Some would call \"$f(x)=x^2$\" the rule of the function $f$.\n\u2022 For each argument $x$, the corresponding unique $y$ in the codomain is called the function value at $x$ or the image of $x$ under $f$. It is written as $f(x)$.\n\u2022 If there is some relation specifying $f(x)$ in terms of $x$, then $f(x)$ is known as a dependent variable (and $x$ is an independent variable).\n\nA correct way to notate your function $f$ is: $$f:\\Bbb{R}\\to\\Bbb{R}:x\\mapsto f(x)=x^2$$\n\nNote that $f(x)\\in\\Bbb{R}$ and $f\\not\\in\\Bbb{R}$. But the function $f$ is an element of the set of continuous functions, and $f(x)$ isn't.\n\nIn some areas of math it is very important to notate a function/map specifying it's domain, codomain and function rule. However in for example calculus/physics, you'll see that many times only the function rule $f(x)$ is specified, as the reader is supposed to understand domain/codmain from the context.\n\nYou can also check those questions:\n\n\u2022 Nice, Kasper +1 Feb 23, 2013 at 1:50\n\nnormally you say $f:X\\rightarrow Y$; $x\\mapsto f(x)$, as a functions takes an element from $X$ and give you one from $Y$. $$y=f(x)$$ Is an equation, and not a definition of a function in a strict sense.\n\nThe proper way would be $$f:\\mathbb{R}\\rightarrow \\mathbb{R}; \\ x\\mapsto x^2$$\n\nThe Image of a function is definied as $$\\operatorname{im}f:=\\{f(x)|x\\in X\\}$$ It is often written as $$f(X):=\\{f(x)|x\\in X\\}$$ Notice that $X$ is a set, not an element. so $f(\\{x\\})=\\{f(x)\\}\\subseteq Y$ while $f(x)\\in Y$\n\n\u2022 I was wondering - is $f(x)$ called the image? I've seen that word used but only know it to be synonymous with Range. Feb 22, 2013 at 23:15\n\u2022 not $f(x)$ but $f(X)$ note the different as $X$ is a set but $x$ is an element Feb 22, 2013 at 23:19\n\u2022 @agent154 You can say \"$f(x)$ is the image of the element $x$ under the function $f$\". Perhaps this is what you were thinking of (here of course \"image\" means something different than the image of the function). Feb 22, 2013 at 23:32\n\u2022 @davidMitra i would be careful with that, as the image is a set but $f(x)$ is an element of $Y$ Feb 22, 2013 at 23:34\n\u2022 The notation $f(X)$ is also written (in set theory) as $f[X]$ and $f''X$ sometimes, often because the elements of the set are sets themselves, and sometimes $f(x)\\neq f''x$. Feb 22, 2013 at 23:35\n\nFrom what I understand, $f$ is the function, so saying \"a function $f(x)$\" would be wrong. Instead, you could say \"a function $f$\", or if you didn't want to assign it a name, \"a function $x\\mapsto x^2$\", or if you want to specify the domain and codomain, \"a function $f:X\\to Y$\".\n\nIf you just want to define the function's input-output relationship, $f(x)=x^2$ suffices, maybe with a $\\forall x\\in X$ at the beginning if you want to do it properly. That doesn't necessarily define the domain though, that set could just be a subset of the domain. $f(x)$ is really no different from $f(1)$, except that $x$ is a (qualified) variable. It's basically a rule defining the input and output for an improper subset of its domain.\n\nThe addition of \"$:X\\to Y$\" is useful if you want to specify both the domain and codomain when defining the function. As far as I can tell, you can add it after a function like $f$ and the whole expression still refers to the function. You can't really use it with the $f(x)$ style, so I'd do something like\n\n$f:\\mathbb R\\to\\mathbb R=x\\mapsto x^2$", "date": "2022-05-29 03:09:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/311605/function-notation-terminology/311637", "openwebmath_score": 0.8944380283355713, "openwebmath_perplexity": 185.66567938001668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995742876885, "lm_q2_score": 0.8933094074745443, "lm_q1q2_score": 0.8628471763868496}}
{"url": "https://stats.stackexchange.com/questions/294737/what-is-the-variance-of-a-binomial-distribution-with-1-and-1", "text": "# What is the variance of a binomial distribution with -1 and 1?\n\nI am struggling to see how to solve the following problem:\n\nI have n i.i.d bernoulli trials. The result can be -1 or 1. I can figure out the expected value of this = n(p-(1-p) but how do I know what the variance is if p is known?\n\nI know that: $${\\rm Var}(X) = E[X^2] - E[X]^2$$ I don't know where to go from here.\n\n\u2022 In general, for a Bernoulli Random Variable E(X) is $P(X=x_1)*x_1+P(X=x_2)*x_2$. In the \"default\" case $x_1 = 1, x_2 = 0$, which reduces this to E(X) = p. In your case, $x_1 = 1$ and $x_2 = -1$. You can then plug in what you get for E(X) into the formula $Var(X) = E(X^2) - E(X)^2$. The variance of the default case is $p*(1-p)$ \u2013\u00a0Pugl Jul 27 '17 at 13:57\n\u2022 If B is a 0-1 Bernoulli variable with probability of a 1 at $p$, then you're describing a variable, ($Z$ say) where $Z=2B-1$. Just apply basic properties of variance to that linear transformation. \u2013\u00a0Glen_b Jul 28 '17 at 2:06\n\nWhen you work with variances, know these facts (in addition to the definition of variance):\n\n1. The variance of a sum of independent (or just uncorrelated) variables is the sum of their variances.\n\n2. The expectation of a sum of variables (independent or not) is the sum of their expectations.\n\n3. The expectation (of any discrete variable) is the sum of each possible value multiplied by its probability.\n\nAs an example, let's apply them to your case. You may model the sum of your Bernoulli trials with a variable $X$ expressed as the sum of $n$ independent variables $X_i$. Each $X_i$ takes on the value $1$ with probability $p$ and the value $-1$ with the probability $1-p$.\n\nFact $(3)$ asserts $$\\mathbb{E}(X_i) = p(1) + (1-p)(-1) = 2p-1$$ and $$\\mathbb{E}(X_i^2) = p(1)^2 + (1-p)(-1)^2 = p + (1-p) = 1.$$\n\nFact $(2)$ asserts \\eqalign{\\mathbb{E}(X) &= \\mathbb{E}(X_1+\\cdots+X_n) = \\mathbb{E}(X_1) + \\cdots + \\mathbb{E}(X_n) = n\\mathbb{E}(X_1) \\\\&=n(2p-1).}\n\nThe definition of variance now tells you$$\\operatorname{Var}(X_i) = \\mathbb{E}(X_i^2) - \\mathbb{E}(X_i)^2 = 1 - (2p-1)^2 = 4p(1-p).$$\n\nConsequently, fact $(1)$ yields \\eqalign{\\operatorname{Var}(X) &= \\operatorname{Var}(X_1+\\cdots+X_n) = \\operatorname{Var}(X_1) + \\cdots + \\operatorname{Var}(X_n)=n \\operatorname{Var}(X_1) \\\\&= n(4p(1-p)).}\n\nThe appearance of the factor of $4$ might seem somewhat mysterious. There's another extremely useful fact you might consider using to shortcut these considerations and identify the origin of that factor:\n\n(4) The variance of a shifted, rescaled variable is the square of the scale factor times the variance. In mathematical symbols, $$\\operatorname{Var}(\\sigma X + \\mu) = \\sigma^2 \\operatorname{Var}(X)$$ no matter what values the numbers $\\sigma$ and $\\mu$ might have.\n\nThis applies by noting that your $X_i$ can all be expressed as $2Y_i-1$ where $Y_i$ is a true Bernoulli (that is, $0-1$) variable. It's simple to show this: since $2\\times 1-1=1$ and $2\\times 0 - 1=-1$, the values of $1$ and $0$ taken on by $Y_i$ become $1$ and $-1$, respectively, for $X_i$. The probabilities are unchanged.\n\nYou might already know that $Y=Y_1+\\cdots + Y_n$, the sum of $n$ independent Bernoulli variables with common probability $p$, is called a Binomial variable. It has a Binomial distribution. You can remember or look up its variance, which is $np(1-p)$. Since $$X = X_1+\\cdots+X_n = (2Y_1-1) + \\cdots + (2Y_n-1) = 2(Y_1 + \\cdots + Y_n) - n=2Y-n,$$ you can take $\\sigma=2$ in applying fact $(4)$, which immediately tells you $$\\operatorname{Var}(X) = 2^2 \\operatorname{Var}(Y) = 4np(1-p).$$\n\n\u2022 I personally find the observation that $4=\\frac{1}{0.5(1-0.5)}$ soothing as it shows that the variance is the ratio of $pq(X_i)$ to $pq(0.5)$ (forgive the abuse of notation). \u2013\u00a0Jared Goguen Jul 27 '17 at 16:09\n\u2022 In what sense is the second part of the answer $X_i=2Y_i-1$ ... different from my answer ? \u2013\u00a0user83346 Jul 28 '17 at 5:41\n\u2022 @fcop It's exactly the same approach--but the explanation is more extensive and thorough. If you take a look over the entirety of my answer, I hope you will see the pattern: it focuses on stating and illustrating useful principles. These are often overlooked or unknown to people who come to our site with questions. My purpose is to collect, unify, and illustrate some (obvious, well-known) techniques that can be found by future visitors and referenced in future answers. In that sense, my answer differs from yours, which focuses on solving the immediate problem. \u2013\u00a0whuber Jul 28 '17 at 13:37\n\u2022 Well you could have referenced it no? \u2013\u00a0user83346 Jul 28 '17 at 14:08\n\u2022 @fcop One ordinarily does not reference obvious or well-known things. (If that were the practice, then most posts would be 90% references.) The point of this answer is to explain the reasoning--not just the results--at the same level of understanding as the question. Its contribution is not to make the observation that $X_i=2Y_i-1$ (you're welcome to take credit for that), but rather to show as clearly and explicitly as possible, with all reasons supplied, how that produces the same answer derived earlier in the question. \u2013\u00a0whuber Aug 13 '17 at 13:10\n\nif you take $x=2b-1$ where $b$ is bernoulli, then then $x$ is either 1 or -1, and mean and variance follow from Bernoulli $\\mu_b=p$ and $\\sigma^2_b=p(1-p)$, the extension to 'Binomial' is then just a sum (assuming independence),\n\nThe mean of $x$ is then $\\mu_x=2\\mu_b-1=2p-1$, variance $\\sigma_x^2=4\\sigma_b^2=4p(1-p)$,\n\nThe sum of $n$ independent such $x$'s has mean $n\\cdot(2p-1)$ (your result) and the variance is $4\\cdot n \\cdot p(1-p)$\n\nEDIT after your question in the comment\n\nIt is as @Pegah says: $b$ is the ''usual'' Bernoulli with a success probability $p$ and outcomes 0 and 1. It's expected value is $1 \\times p + 0 \\times (1-p)=p$ and the variance $p(1-p)$ (also from the definition.\n\nSo $b$ is the known Bernoulli with outcome 0 and 1 and $x$ is just a linear function of it. The linear function is such that $x$ has outcomes -1 and 1, and the mean and variance can be obtained from the mean and variance of $b$.\n\n\u2022 In fcop's derivation, $\\mu_b$ is the expected value of the random variable b, which is, as I understand a \"default\" binomial random variable (i.e. it can take on values 1 or 0, see also my comment above). Here the expected value is just p. \u2013\u00a0Pugl Jul 27 '17 at 15:31", "date": "2021-01-16 12:18:07", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/294737/what-is-the-variance-of-a-binomial-distribution-with-1-and-1", "openwebmath_score": 0.9359101057052612, "openwebmath_perplexity": 280.8198992530586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534349454033, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8628416256703344}}
{"url": "http://math.stackexchange.com/questions/165937/simplifying-to-a-desired-expression-structure", "text": "# Simplifying to a desired expression structure\n\nMy book has this expression:\n\n\\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3) \\end{align}\n\nAnd then the book simplified it, and ended up with the desired expression:\n\n\\begin{align} ((n+1)(n+2)(2n+9))/6 \\end{align}\n\nI tried to do such simplification. But I ended up with this: \\begin{align} (2n^3+15n^2+31n+18)/6 \\end{align}\n\nEven though the \"value\" in my expression is the same as the desired result's, I need the structure of my expression to be the same as well. But I don't understand how to do such.\n\nHere are the full steps my book shows. Of course, I understand that the steps make sense, but I don't understand why did my book take that approach? Is that the only approach possible to reach the desired expression? How was I supposed to know that? I mean, I know how to simplify, but clearly my book is using a different \"style\" or \"path\"\n\n\\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3)\\\\ (n(n+1)(2n+7)+6(n+1)(n+3))/6\\\\ ((n+1)[n(2n+7)+6(n+3)])/6\\\\ ((n+1)(2n^2 +7n + 6n + 18))/6\\\\ ((n+1)(n+2)(2n+9))/6 \\end{align}\n\nSo yes, how should I simplify to get a desired expression structure?\n\n-\nAh! I... didn't see that. Somehow. I'm eager to see how to extract such though. \u2013\u00a0 Zol Tun Kul Jul 3 '12 at 1:49\nI wrote out a little something about extracting. \u2013\u00a0 Andr\u00e9 Nicolas Jul 3 '12 at 1:59\n\nBelow I explain in detail the solution given in your book.\n\n$\\begin{eqnarray} &&\\rm\\ \\ \\,n(n\\!+\\!1)(2n\\!+\\!7)/\\color{#C00}6+(n\\!+\\!1)(n\\!+\\!3)\\\\ \\rm put\\ all\\ over\\ a\\ common\\ denominator = \\color{#C00}6:\\quad &= &\\rm\\ (n(\\color{#0A0}{n\\!+\\!1})(2n\\!+\\!7)\\,+\\,\\color{#C00}6\\,(\\color{#0A0}{n\\!+\\!1})(n\\!+\\!3))/\\color{#C00}6\\\\ \\rm pull\\ out\\ the\\ common\\ factor\\ \\color{#0A0}{n\\!+\\!1}:\\quad &=&\\rm\\ (\\color{#0A0}{n\\!+\\!1})\\,(n(2n\\!+\\!7)+6(n\\!+\\!3))/6\\\\ \\rm apply\\ the\\ distributive\\ law:\\quad &=&\\rm\\ (n\\!+\\!1)\\,(2n^2 + 13n + 18)/6\\\\ \\rm factor\\ the\\ quadratic,\\ see\\ below:\\quad &=&\\rm\\ (n\\!+\\!1)\\,(n\\!+\\!2)\\,(2n\\!+\\!9)/6 \\end{eqnarray}$\n\nIn order to factor the quadratic $\\rm\\:f = 2n^2+13n+18\\:$ one can apply the AC-method as follows:\n\n$$\\begin{eqnarray}\\rm 2f\\, &=&\\rm\\ \\ 4n^2\\ +\\ 13\\cdot 2n\\, +\\, 2\\cdot 18 \\\\ &=&\\rm\\ (2n)^2 + 13\\,(2n) + 2\\cdot 18 \\\\ &=&\\rm\\ \\ N^2\\ +\\ 13\\, N\\ +\\ 36\\quad for\\quad N = 2n \\\\ &=&\\rm\\ (N\\ +\\ 4)\\,(N\\ +\\ 9) \\\\ &=&\\rm\\ (2n\\, +\\, 4)\\,(2n\\, +\\, 9) \\\\ \\rm f\\, &=&\\rm\\ (\\ n\\ +\\ 2)\\,(2n\\ +\\ 9) \\end{eqnarray}$$\n\n-\nI'm enlightened. Thank you. \u2013\u00a0 Zol Tun Kul Jul 3 '12 at 3:58\n\nYour expression is $$\\frac{2n^3+15n^2+31n+18}{6}.$$ Let's forget about the $6$. Also, I would like to change the $n$ to $x$, for no good reason except familiarity. So we want to factor $$2x^3+15x^2+31x+18.$$ To do this, we look for rational roots of the polynomial $2x^3+15x^2+31x+18$. By the Rational root Theorem, such roots must have shape $a/b$, where $a$ divides $18$ and $b$ divides $2$. (There may be no rational roots, though in this case there are three.)\n\nIt is easy to spot the root $x=-1$. So $x-(-1)$, that is, $x+1$, divides our polynomial. Do the division, using the ordinary division process for polynomials, which is much like \"long division.\" We get $$2x^3+15x^2+31x+18=(x+1)(2x^2+13x+18).$$ Now factor the quadratic as one did in school. Or else note that $x=-2$ is a root of $2x^2+13x+18$.\n\nRemark: The book grabbed the obvious common factor of $n+1$. The rest turned out to be a quadratic that factors nicely. You multiplied out, meaning you buried the $n+1$ term. It can be extracted from your expression, and the above calculation shows how, but why bury and then extract? A factored or partly factored expression is often more useful. Anyway, \"taking out\" common factors usually simplifies calculations.\n\n-", "date": "2014-10-24 23:32:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/165937/simplifying-to-a-desired-expression-structure", "openwebmath_score": 0.9237958192825317, "openwebmath_perplexity": 612.6816650057269, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534376578004, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8628416218395245}}
{"url": "https://math.stackexchange.com/questions/1633780/show-that-set-of-all-2-times-2-matrices-forms-a-vector-space-of-dimension-4", "text": "# Show that set of all $2 \\times 2$ matrices forms a vector space of dimension $4$\n\nI have this question:\n\nShow that the set of all $2 \\times 2$ matrices with real coefficients forms a linear space over $\\Bbb R$ of dimension $4$.\n\nI know that the set of the matrices will basically form a linear combination which will define the vector space and they satisfy the axioms defined for the vector space.\n\nI do not know how to show that this is possible. Do the vectors have to be linearly independent?\n\nAny sort of help is appreciated.\n\nThanks.\n\nBegin by listing the axiom satisfied by a general vector space over $\\mathbb{R}$. Now consider your set of matrices. What does you addition look like? What is your zero element? What does your scalar multiplication look like? Does your scalar multiplication and your addition behave as they should? Finally, show that all 2 by 2 matrices can be written as a linear combination of 4 special matrices. A good choice is to pick matrices who each have precisely 3 zero entries and 1 non zero entry. Can you show that these 4 matrices are linearly independent?\n\n\u2022 Can those four matrices be [a 0 0 0], [0 b 0 0], [0 0 c 0], [0 0 0 d]? I can form a linear combination of them and show they are linearly independent. Will that work? \u2013\u00a0Suvrat Jan 31 '16 at 0:05\n\u2022 That is exactly what I hoped that you would do! \u2013\u00a0Carl Christian Jan 31 '16 at 0:05\n\u2022 oh very well then. Thanks a lot! \u2013\u00a0Suvrat Jan 31 '16 at 0:06\n\u2022 Can you tell me how these 4 matrices show the formation of linear space for generalized situation, i.e. all 2x2 matrices? \u2013\u00a0Suvrat Jan 31 '16 at 20:14\n\u2022 For the sake of simplicity pick a=b=c=d=1. Then write the matrix [e f g h] as e*[1 0 0 0] + f*[0 1 0 0] + g [0 0 1 0] + h*[0 0 0 1]. \u2013\u00a0Carl Christian Jan 31 '16 at 20:23\n\nOne can easily see that $\\mathbb{R}^{2\\times 2}$ is in fact exactly the same as the vectorspace $\\mathbb{R}^4$, only the notation of the elements differs. And of course whether or not something is a vector space does not depend on the way some unimportant species of primates decides to write its elements.\n\nHint Can you find a basis of the set of $2 \\times 2$ matrices consisting of four elements? (There is a natural choice of basis here that includes the matrix $\\pmatrix{1&0\\\\0&0}$.)\n\nAlternatively, can you find a vectorspace isomorphism from the space of $2 \\times 2$ matrices to some vector space you know to be $4$-dimensional, e.g., $\\Bbb R^4$?\n\n\u2022 I don't know how to find a vectorspace isomorphism. But i can work with basis, which is similar to Carl's solution. Thanks! \u2013\u00a0Suvrat Jan 31 '16 at 0:11\n\u2022 You're welcome. \u2013\u00a0Travis Jan 31 '16 at 0:42", "date": "2019-09-19 00:25:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1633780/show-that-set-of-all-2-times-2-matrices-forms-a-vector-space-of-dimension-4", "openwebmath_score": 0.8296988606452942, "openwebmath_perplexity": 283.7972044657849, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534311480466, "lm_q2_score": 0.8791467754256018, "lm_q1q2_score": 0.8628416192241981}}
{"url": "https://math.stackexchange.com/questions/2931622/how-can-i-prove-this-statement-about-subsets/2931634", "text": "How can I prove this statement about subsets?\n\nLet $$A$$, $$B$$, $$C$$ be sets. Prove that if $$A \\subseteq C$$ and $$B \\subseteq C$$ then $$A \\cup B \\subseteq C$$.\n\nThis is an exercise in mathematical logic.\n\nMy attempt to progress forward: This statement can be written as$$(A \\subseteq C) \\land (B \\subseteq C) \u2192 A \\cup B \\subseteq C\\\\ (x \\in A \u2192 x \\in C) \\land (x \\in B \u2192 x \\in C) \u2192 A \\cup B \\subseteq C$$ But I am not even sure that is how I am supposed to do it so I am a bit stuck. Can anyone explain to me how to get through this proof? Thanks in advance.\n\n\u2022 ... intuitively it's obvious, right? ... Sep 26 '18 at 15:32\n\u2022 @user202729 Intuitively many things are obvious, but proving them is another matter... ;) ex. A sphere bounds a given volume with minimal area... Sep 27 '18 at 1:14\n\n4 Answers\n\nTake any $$x\\in A\\cup B$$. That means either $$x\\in A$$ or $$x\\in B$$. Anyway you get $$x\\in C$$.\n\nThis answer is expanding a bit on the other answers to give a widely applicable outline on how to prove mathematical statements in general.\nThe final proof will be equal to Mark's answer, but hopefully this answer sheds a bit of light on what is actually going on in the thought process, at least I like to think in the terms defined below while I am proving.\n\nIn general you need a \"proof calculus\" which states how proofs can be constructed and which ones are valid. Unless you are doing logic and proof theory itself, in most settings in maths you are well off using the natural deduction proof calculus for first-order logic. In short \u2014 cut in light of this question:\n\n\u2022 To prove $$\\varphi \\land \\psi$$, prove $$\\varphi$$ and $$\\psi$$ separately.\n\u2022 To prove $$\\varphi \\lor \\psi$$, give a proof for $$\\varphi$$ or alternatively for $$\\psi$$. One proof for either $$\\varphi$$ or $$\\psi$$ is already sufficient.\n\u2022 To prove $$\\varphi \u2192 \\psi$$, assume $$\\varphi$$ and prove $$\\psi$$ under this assumption.\n\nThe implication captures the notion of \"if we have $$\\varphi$$ and $$\\varphi \u2192 \\psi$$ established somewhere, then we can conclude $$\\psi$$.\" Most, if not all, theorems can be seen as implications: if you fulfill the requirements, then you can use its consequences.\nHence it makes sense to be allowed to assume $$\\varphi$$ in the proof of $$\\varphi \u2192 \\psi$$.\n\u2022 To prove $$\\forall x. \\varphi(x)$$ ($$\\varphi$$ may depend on $$x$$), introduce a fresh variable not used before (e.g. $$c$$) and prove $$\\varphi(c)$$ (i.e. every occurrence of $$x$$ is filled with $$c$$).\n\nIntroduction of a fresh variable is required to guarantee that your proof of $$\\varphi(c)$$ does not depend on any previous assumptions on $$x$$. Hence, having proved $$\\varphi(c)$$ really amounts to having proved the proposition $$\\varphi(x)$$ for every possible $$x$$.\n\nNow consider your statement:\n\n$$\\left((x \\in A \u2192 x \\in C) \\land (x \\in B \u2192 x \\in C)\\right) \u2192 A \\cup B \\subseteq C$$\n\nYou already did a good job translating it to more formal language! But we can even go a step further. Actually what the above line means is the following: $$\\left((\\forall x. x \\in A \u2192 x \\in C) \\land (\\forall x. x \\in B \u2192 x \\in C)\\right) \u2192 (\\forall x. x \\in A \\cup B \u2192 x \\in C)$$\n\nOn the top level, we have a $$\\varphi \u2192 \\psi$$-kinded expression. Applying the corresponding rule, we begin with\n\n(1) Assume $$(\\forall x. x \\in A \u2192 x \\in C) \\land (\\forall x. x \\in B \u2192 x \\in C)$$.\n\nOur proof goal, as stateted in the rule, is now the right side $$(\\forall x. x \\in A \\cup B \u2192 x \\in C).$$ This is of kind $$\\forall x. \\varphi(x)$$ with $$\\varphi(x) := x \\in A \\cup B \u2192 x \\in C$$. So,\n\n(2) Let $$y$$ be a fresh variable.\n\nI called it $$y$$ to avoid confusion with the set $$C$$. When one defines freshness more formally, one sees that $$x$$ would have worked as well as a fresh variable.\nNow we have to show $$\\varphi(y)$$, which is $$y \\in A \\cup B \u2192 y \\in C$$. Again, apply the $$\u2192$$ rule from above:\n\n(3) Assume $$y \\in A \\cup B$$.\n\nOur final proof goal is now: $$y \\in C$$. This can be proved by case-by-case analysis:\n\n(4) We know that $$y$$ is in $$A$$ or $$B$$. In the first case, $$y \\in A$$ and per line (1) we especially know $$y \\in A \u2192 y \\in C$$. Since we have $$y \\in A$$ just established, we can conclude $$y \\in C$$. Case finished. The next and last case is $$y \\in B$$ which works analogously.\n\nIf $$A \\subset C$$, and $$B \\subset C$$;\n\n$$A \\cap C =A$$; and $$B \\cap C = B$$;\n\n$$A\\cup B = (A \\cap C)\\cup (B \\cap C) =$$\n\n$$C \\cap (A \\cup B) \\subset C$$\n\nI haven't noticed one in the in the answers, so here's a proof by contradiction.\n\nThe statement we wish to prove is: $$(\\forall x)(x \\in A \\cup B \\implies x \\in C) \\tag{1}$$\n\nIt's negation is:\n\n$$(\\exists x)(x \\in {A} \\cup {B} \\wedge \\neg(x \\in C) ) \\tag{2}$$ $$x \\in A \\cup B \\implies (x \\in A \\vee x \\in B)$$ $$x \\in A \\wedge A \\subseteq C \\implies x \\in C \\tag{3}$$ $$or$$ $$x \\in B \\wedge B \\subseteq C \\implies x \\in C \\tag{4}$$ Since both $$(3)$$ and $$(4)$$ are in contradiction with $$(2)$$, it must be false. Therefore our starting premise $$(1)$$ is correct.", "date": "2021-09-19 10:57:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2931622/how-can-i-prove-this-statement-about-subsets/2931634", "openwebmath_score": 0.8933517932891846, "openwebmath_perplexity": 249.88743815642277, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534376578004, "lm_q2_score": 0.8791467690927439, "lm_q1q2_score": 0.862841618731822}}
{"url": "https://mathhelpboards.com/threads/fourier-series-pointwise-convergence-series-computation.643/", "text": "# Fourier series, pointwise convergence, series computation\n\n#### Markov\n\n##### Member\nLet $f(x)=-x$ for $-l\\le x\\le l$ and $f(l)=l.$\n\na) Study the pointwise convergence of the Fourier series for $f.$\nb) Compute the series $\\displaystyle\\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)}.$\nc) Does the Fourier series of $f$ converge uniformly on $\\mathbb R$ ?\n\n-------------\n\nFirst I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\\displaystyle\\sum_{n=1}^\\infty b_n\\sin\\frac{n\\pi x}l$ where $b_n=\\dfrac 2l\\displaystyle\\int_0^{l} f(x)\\sin\\frac{n\\pi x}l\\,dx$ so I'm getting $\\displaystyle-\\frac{x}{{{l}^{2}}}=\\frac{1}{\\pi }\\sum\\limits_{n=1}^{\\infty }{\\frac{{{(-1)}^{n}}}{n}\\sin \\frac{n\\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?\n\nThanks for the help!\n\n#### Sudharaka\n\n##### Well-known member\nMHB Math Helper\nLet $f(x)=-x$ for $-l\\le x\\le l$ and $f(l)=l.$\n\na) Study the pointwise convergence of the Fourier series for $f.$\nb) Compute the series $\\displaystyle\\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)}.$\nc) Does the Fourier series of $f$ converge uniformly on $\\mathbb R$ ?\n\n-------------\n\nFirst I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\\displaystyle\\sum_{n=1}^\\infty b_n\\sin\\frac{n\\pi x}l$ where $b_n=\\dfrac 2l\\displaystyle\\int_0^{l} f(x)\\sin\\frac{n\\pi x}l\\,dx$ so I'm getting $\\displaystyle-\\frac{x}{{{l}^{2}}}=\\frac{1}{\\pi }\\sum\\limits_{n=1}^{\\infty }{\\frac{{{(-1)}^{n}}}{n}\\sin \\frac{n\\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?\n\nThanks for the help!\nHi Markov,\n\nFirstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\\mbox{ for }-l\\leq x\\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.\n\nI think the Fourier series that you have obtained is also incorrect. It should be,\n\n$-x=\\frac{2l}{\\pi}\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}}{n}\\sin\\left(\\frac{n\\pi x}{l}\\right)$\n\nSince both $$f(x)=-x$$ and $$f'(x)=-1$$ are continuous on $$[-l,\\,l]$$ the Fourier series converges point-wise on the interval $$(-l,\\,l)$$ (Refer Theorem 5.5 here).\n\nSubstitute $$x=1\\mbox{ and }l=2$$ and we obtain,\n\n\\begin{eqnarray}\n\n-\\frac{\\pi}{4}&=&\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}}{n}\\sin\\left(\\frac{n\\pi}{2} \\right)\\\\\n\n&=&\\sum_{n=0}^{\\infty}\\frac{(-1)^{2n+1}}{2n+1}\\sin\\left(\\frac{(2n+1)\\pi}{2} \\right)\\\\\n\n&=&\\sum_{n=0}^{\\infty}\\frac{(-1)^{2n+1}}{2n+1}(-1)^n\\\\\n\n&=&\\sum_{n=0}^{\\infty}\\frac{(-1)^{3n+1}}{2n+1}\\\\\n\n\\therefore\\sum_{n=0}^{\\infty}\\frac{(-1)^{n}}{2n+1}&=&\\frac{\\pi}{4}\n\n\\end{eqnarray}\n\nWhen $$x=l$$ we have,\n\n$\\left|f(l)-\\sum_{n=1}^{N}\\frac{(-1)^{n}}{n}\\sin\\left(\\frac{n\\pi l}{l}\\right)\\right|=l$\n\nTherefore by the definition of uniform convergence (Refer this) it is clear that the Fourier series of $$f$$ is not uniformly convergent on $$\\Re$$.\n\nKind Regards,\nSudharaka.\n\n#### CaptainBlack\n\n##### Well-known member\nHi Markov,\n\nFirstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\\mbox{ for }-l\\leq x\\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.\nThis is very likely to be intended to indicate the periodic extension: $$f(x)=f(x+2l)$$ (or there is a mistake with the value at either $$+l$$ or $$-l$$, both end points would not normally be included in the domain for a Fourier Series).\n\nWithout the periodic extension the question of uniform convergence is moot, since the Fourier Series is periodic and so does not converge to the function outside of the interval.\n\nCB", "date": "2022-01-21 09:08:58", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/fourier-series-pointwise-convergence-series-computation.643/", "openwebmath_score": 0.9820981621742249, "openwebmath_perplexity": 216.88328968484396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534344029238, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.862841617424159}}
{"url": "https://www.physicsforums.com/threads/what-can-i-use-to-solve-this.59216/", "text": "# What can I use to solve this?\n\n1. Jan 10, 2005\n\n### recon\n\nI've run into a calculation hurdle that I cannot cross in a problem I'm trying to solve.\n\nMay I know what values of z satisfy the following equation:\n\n$$z^4 + 2z^3 - 5z + 1 = 0$$\n\nI've never encountered this in mathematics class before.\n\n2. Jan 10, 2005\n\n### dextercioby\n\nWould u please post the problem??Maybe you wouldn't get a quartic after all...\n\nDaniel.\n\n3. Jan 10, 2005\n\n### Nguyen Thanh Nam\n\nNeither do I, I can solve those that miss ax^3, never before for ax^2\n\n4. Jan 10, 2005\n\n### recon\n\nA ladder 3m long rests against a wall with one end a short distance from the base of the wall. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m wide. What is the maximum distance up the wall that the ladder can reach?\n\n5. Jan 10, 2005\n\n### dextercioby\n\nThe maximum height on the wall is attained when the ladder touches the storage box.\nThe equations i have found read\n$$(a+1)^{2}+(b+1)^{2} =9$$(1)\n$$ab=1$$ (2)\n\nThen u find the eq.for \"a\"\n$$a^{2}+\\frac{1}{a^{2}}+2(a+\\frac{1}{a})=7$$ (3)\n,which can be put in the form\n$$u^{2}+2u-9 =0$$ (3)\n\n,where\n$$u=:a+\\frac{1}{a}$$ (4)\n\nSolve for \"u\" and then for \"a\".Then add 1 to the \"a\" and find the final result.\n\nDaniel.\n\n6. Jan 10, 2005\n\n### recon\n\nYes, that's how I solved it. I never thought of substituting $$a+\\frac{1}{a}$$ for u. Very smart. Thanks a lot.\n\n7. Jan 10, 2005\n\n### Hurkyl\n\nStaff Emeritus\nUse the symmetry of the problem . With dexter's notation, notice that whenever s is a solution, so is 1/s. So, we can take his quartic:\n\nz^4 + 2z^3 - 7z^2 + 2z + 1 = 0\n\nAnd write down its four factors:\n\n(z - s) (z - 1/s) (z - t) (z - 1/t) = 0\n\nIf we group them into their symmetric pairs and multiply:\n\n(z^2 + az + 1) (z^2 + bz + 1)\n\nSo one quadratic corresponds to the two solutions s and 1/s and the other to t and 1/t. In particular, notice, now, that our unknowns only have two possible values, instead of four, which means we should be able to determine them from a quadratic equation:\n\nSince the polynomials should be equal, we should expand and equate them:\n\nz^4 + (a+b)z^3 + (2+ab)z^2 + (a+b)z + 1\n=\nz^4 + 2z^3 - 7z^2 + 2z + 1 = 0\n\nEquating coefficients gives us a system of equations to solve.\n\nOnce we have a and b, we can then solve for s and t and get the four possible solutions. (Two should be complex, I imagine)\n\n(Incidentally, I don't see how you got your equation, Recon)\n\n8. Jan 10, 2005\n\n### dextercioby\n\nI multiple checked it.Can u prove your statement???I frankly doubt it...\n\nDaniel.\n\n9. Jan 10, 2005\n\n### Hurkyl\n\nStaff Emeritus\nAnother (similar) approach:\n\nWe can arrange the factors into these pairs:\n\n(z - s)(z - t) and (z - 1/s)(z - 1/t)\n\nSo each quadratic contains exactly one solution from each of the symmetric solutions (instead of both).\n\nAgain, we break the quartic into quadratic factors:\n\n(z^2 + az + b) (z^2 + cz + d)\n\nBut, in this case, we know that we can convert from one to the other by applying the symmetry. In particular,\n\nif (z^2 + az + b) = 0 then (1/z)^2 + c(1/z) + d = 0\nRewriting the second gives:\n\nz^2 + (c/d)z + (1/d) = 0\nwhich must be the same polynomial as\nz^2 + az + b = 0\n\nSince they have the same solutions.\n\nSo we can equate coefficients here, then multiply them to equate coefficients to the quartic polynomial. Again, we can solve for our unknowns, and then we know how to solve quadratics to get the answers to our problem.\n\nLesson learned: use symmetry to break the problem into parts. We've seen two different ways to do it:\n\n(1) Each part corresponds to one of the different \"orbits\" under the symmetry. (e.g. {s, 1/s})\n\n(2) Each part consists of exactly one element from each \"orbit\". (e.g. {s, t})\n\nThen, we can solve the problem in two steps:\n\n(i) Determine the two subproblems.\n(ii) For each subproblem, get the corresponding answers.\n\n10. Jan 10, 2005\n\n### Hurkyl\n\nStaff Emeritus\nDexter gave a solution using a third approach. What precisely did he do, and why did it work? Is it really a different approach?\n\nThe variable he defined, u, has the property that it is invariant under the symmetry. Notice that if you replace z with 1/z, u remains unchanged.\n\nBecause of this invariance under symmetry, there are fewer solutions for u than there are for z, making them easier to find.\n\nActually, in some sense, both of my approaches make use of invariants under symmetry. For instance, (z - s)(z - 1/s) is invariant under swapping s and 1/s. My second approach found two quadratics that are each invariant under swapping s and t.\n\nP.S. for a polynomial which is the same as its reverse, making dexter's solution is the standard way of getting a lower degree polynomial.\n\n(The reverse of a polynomial is the one with the coefficients in the opposite order. Equivalently, the multiplicative inverse of the roots of the polynomial are precisely the roots of the reverse. To make sense, of course, this assumes that 0 is not a root)\n\nLast edited: Jan 10, 2005\n11. Jan 10, 2005\n\n### Hurkyl\n\nStaff Emeritus\nAnd just for fun, here's a trig approach:\n\nLet theta be the angle the ladder makes with the base. A is the distance from the box to the base, B is the distance from the box tot he top of the ladder.\n\nSo, the length of the ladder is 1/sin theta + 1/cos theta = 3.\n\nWith trig identities, and a change of variable, I can get a quadratic in sin phi.\n\n12. Jan 10, 2005\n\n### Zaimeen\n\nI was a bit sleepy at that time Dex. I did checked it just now. You are right. Sorry Dex!!\n\n________________________________________________________________________\n\n\"The beautiful mind goes faster than the hand\"\n\n13. Jan 10, 2005\n\n### recon\n\nI got it the same way as dextercioby. I just made a mistake multiplying the right side of the equation with z^3 instead of z^2. I multiplied everything on the left side with z^2.\n\nEDIT: This is the equation I'm talking about.\n\n$$z^2 + 2z + \\frac{2}{z} + \\frac{1}{z^2} = 7$$\n\nLast edited: Jan 10, 2005", "date": "2017-01-19 17:17:38", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/what-can-i-use-to-solve-this.59216/", "openwebmath_score": 0.8561117649078369, "openwebmath_perplexity": 1162.0644350129749, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9814534354878827, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8628416137164431}}
{"url": "https://stats.stackexchange.com/questions/275109/expected-value-with-piecewise-probability-density-function-pdf", "text": "# Expected value with piecewise probability density function (PDF)\n\nI am continuing the prepare for an exam by reviewing handouts from an old statistics course I took. The handout came with a set of solutions prepared by the instructor, but I suspect that one of the answers is wrong. If the answer the instructor provided isn't wrong, then I must be missing something, so I'd like to ask the CV community to help where I've gone wrong. Here is the question:\n\nSuppose that $X$ is a continuous random variable with pdf given by:\n\n$f(x) = \\left\\{ \\begin{array}{lr} x & 0 < x < 1\\\\ (2x-1)/12 & 1< x < 3 \\end{array} \\right.$\n\nif $g(x) = x$ for $x>2$ and $g(x)=0$ for $x \\le 2$, find the expected value $Eg(x)$.\n\nThe instructor put the solution as $10/9$. However, I've reworked my solution several times and continue to arrive at the answer $Eg(x)=61/72$. Here is how I approached this problem:\n\nBy definition, we know:\n\n$$Eg(X) = \\int_{-\\infty}^{\\infty}g(x)f(x)dx\\tag{1}$$ Since $f(x)$ and $g(x)$ are $0$ everywhere, except from $2<x<3$, I only need to integrate (1) over these values of $x$. Doing so yields: $$Eg(X) = \\int_{2}^{3}g(x)f(x)dx=\\int_{2}^{3}{x(2x-1)\\over{12}}dx=\\int_{2}^{3}{2x^2-x\\over{12}}dx$$ $$={1\\over{12}}\\left({{2x^3}\\over{3}}-{x^2\\over{2}}\\right)\\bigg\\rvert_{x=2}^{x=3}={1\\over{12}}\\left[\\left({{54}\\over{3}}-{9\\over{2}}\\right)-\\left({{16}\\over{3}}-{4\\over{2}}\\right)\\right]\\bigg\\rvert_{x=2}^{x=3}=61/72$$\n\nI think the instructor accidentally integrated over $1<x<2$ instead of over $2<x<3$, but as I said earlier, I'm just hoping someone can verify if I'm correct, or perhaps I'm just not understanding something here. Thanks.\n\n\u2022 I think you're right. If you can get correct answers for similar problems then you should have no reason to worry. \u2013\u00a0mark999 Apr 22 '17 at 6:53\n\u2022 I agree with your answer. \u2013\u00a0Glen_b -Reinstate Monica Apr 22 '17 at 9:54\n\u2022 Much appreciated, to you both! This makes me feel a lot better! \u2013\u00a0StatsStudent Apr 22 '17 at 15:49\n\u2022 could you please add a tag-wiki for your new tag piecewise-pdf ? \u2013\u00a0kjetil b halvorsen Apr 23 '17 at 11:20", "date": "2019-12-08 20:58:48", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/275109/expected-value-with-piecewise-probability-density-function-pdf", "openwebmath_score": 0.8147076368331909, "openwebmath_perplexity": 237.29185994953028, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534360303622, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8628416126395109}}
{"url": "https://brilliant.org/discussions/thread/easy-way-to-find-cube-roots/", "text": "This note has been used to help create the Mental Math Tricks wiki\n\nHey guys, I saw a faster way to find cube roots.\n\nWe already know some basic cube numbers\n\n$$0^{3}$$=0\n\n$$1^{3}$$=1\n\n$$2^{3}$$=8\n\n$$3^{3}$$=27\n\n$$4^{3}$$=64\n\n$$5^{3}$$=125\n\n$$6^{3}$$=216\n\n$$7^{3}$$=343\n\n$$8^{3}$$=512\n\n$$9^{3}$$=729\n\nNow, the common thing here is that each ones digit of the cube numbers is the same number that is getting cubed , except for 2 ,8 ,3 ,7 .\n\nnow let us take a cube no like 226981 .\n\nto see which is the cube root of that number , first check the last 3 digits that is 981 . Its last digit is 1 so therefore the last digit of the cube root of 226981 is 1 .\n\nNow for the remaining digits that is 226\n\nNow 226 is the nearer & bigger number compared to the cube of 6 (216)\n\nSo the cube root of 226981 is 61\n\nLet us take another example - 148877\n\nHere 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.\n\nSo the last digit of the cube root of 148877 is 3.\n\nNow for the remaining digits 148.\n\nIt is the nearer and bigger than the cube of 5 (125).\n\nTherefore the cube root of 148877 is 53.\n\nLet us take another example 54872.\n\nHere the last three digit's (872) last digit is 2 but the cube of 2's last digit is not 2 but the last of the cube of 8 is 2.\n\nSo the last digit of the cube root of 54872 is 8.\n\nNow of the remaining numbers (54). It is nearer and bigger to the cube of 3 (27). So therefore the cube root of 54872 is 38.\n\nNote by Kartik Kulkarni\n3\u00a0years, 5\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nHow about to find cube roots of a number which answer is three-digit number ?? For example 111^3, 267^3, etc\n\n- 3\u00a0years, 5\u00a0months ago\n\nAfter doing the last three digits , try to find which is the nearest cube number to it for the remaining digits E.g -\n\n$$\\sqrt[3]{1860867}$$\n\nDone with the last three digits and the last digit , & you get 3 as the last digit of $$\\sqrt[3]{1860867}$$\n\nNow find the nearest cube number of1860 & it is 12 (1728)\n\nSo therefore $$\\sqrt[3]{1860867}$$ = 123\n\n- 3\u00a0years, 5\u00a0months ago\n\nSo we just do the same ways... Thank you so much..\n\n- 3\u00a0years, 5\u00a0months ago\n\nwaitwaitwaitwaitwait..whaaaaaaaaaaat? Where did that 3 even come from? The last digit of 1860867 is 7.....\n\n- 3\u00a0years, 4\u00a0months ago\n\nRead the note properly , it says 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.\n\n- 3\u00a0years, 4\u00a0months ago\n\nSo basically the cube of the number you are looking for must have the same last digit as the number in the problem?\n\n- 3\u00a0years, 4\u00a0months ago\n\nNo, dude. It occasionally happens, but it ain't no rule. It happens for 1 (1\u00b3 = 1), 4 (4\u00b3 = 64), 5 (5\u00b3 = 125), 6 (6\u00b3 = 216), 9 (9\u00b3 = 729) and 0 (0\u00b3 = 0). But, here we see, it doesn't happen for 2 (2\u00b3 = 8), neither 3 (3\u00b3 = 27), nor 7 (7\u00b3 = 343) and 8 (8\u00b3 = 512). I'll always have to check this before find cube roots by this method.\n\n- 3\u00a0years, 4\u00a0months ago\n\n2, 3, and 7, 8 has at their unit place have their 10's compliments. Rest have the same number as said earlier.\n\n- 3\u00a0years, 4\u00a0months ago\n\nThank you, a great method to solve the cube roots, so bad it doesn't work with every cubic root, it would save a lot of time in tests. Anyway, thanks!\n\n- 3\u00a0years, 4\u00a0months ago\n\nAnother Interesting fact:: (A) cube of 2= unit digit 8 .....cube of 8=unit digit 2 (B) cube of 3=unit digit 7...... cube of 7= unit digit 3\n\n- 3\u00a0years, 4\u00a0months ago\n\n1, 4, 5, 6, 9 have the unit place of their cubes as the number themselves. But cubes of 2,3 and 8,7 has there unit place as their compliment of 10.\n\n- 3\u00a0years, 4\u00a0months ago\n\nIt works for groups of threes. How adorable.\n\n- 3\u00a0years, 4\u00a0months ago\n\nCub of 2, 3, 7, 8 we have its compliment of 10.\nFor 2: it is 10-2=8...........for 8, it is 10-8=2............................ 3, it it 10-3=7, ......for 7, it is 10-7=3\n\n- 2\u00a0months, 1\u00a0week ago\n\nData handling\n\n- 6\u00a0months, 2\u00a0weeks ago\n\nBut it is not apllicable everywhere..\n\n- 7\u00a0months, 4\u00a0weeks ago\n\nComment deleted 5\u00a0months ago\n\nWe should take the number which has highest cube less than 117 in your case but not nearest.\n\n- 6\u00a0months, 3\u00a0weeks ago\n\nI gave the first person who introduced this to me a very good comment. Features of natural numbers can occasionally been found. Important thing is never let this concept to mislead ourselves when the situation is not whole numbers. I recalled and remember ed again but not really memorized properly. Understand why could make me a better memory perhaps. Hope I can memorize from today onwards!\n\n- 1\u00a0year, 1\u00a0month ago\n\nBut it is not useful for non perfect cubes\n\n- 1\u00a0year, 1\u00a0month ago\n\nWrite a comment or ask a question...if m=29 and e=13, then m=m+e e=m-e m=m-e then find the new value of m and e??\n\n- 2\u00a0years, 10\u00a0months ago\n\nVery nice & thanks.\n\n- 3\u00a0years, 2\u00a0months ago\n\nGood Method ....!!! Amazing...!!\n\n- 3\u00a0years, 3\u00a0months ago\n\nthats just for a sure perfect cube\n\n- 3\u00a0years, 3\u00a0months ago\n\nCool.......\n\n- 3\u00a0years, 4\u00a0months ago\n\nReally very useful trick Thanks:)\n\n- 3\u00a0years, 4\u00a0months ago\n\nIt's really coolest method ever.but can any1 suggests me methods for square root of a decimal number.for eg:square root of 0.56\n\n- 3\u00a0years, 4\u00a0months ago\n\nhow to work out cube root of 216216. The answer on face is 66 but that is not the cube root.\n\n- 3\u00a0years, 4\u00a0months ago\n\n216216 is not perfect cube ! This method is only for perfect cube\n\n- 4\u00a0months ago\n\nyou have to it by long division method\n\n- 3\u00a0years, 4\u00a0months ago\n\ngood one\n\n- 3\u00a0years, 4\u00a0months ago\n\ndo you want to know the exact long division method of finding cube roots though it tedious... :)\n\n- 3\u00a0years, 4\u00a0months ago\n\nsure.\n\n- 3\u00a0years, 4\u00a0months ago\n\n- 3\u00a0years, 4\u00a0months ago\n\nI HAVE SOME CONFUSION THAT WHEN HAM LOG SAME NO. KO LIKHEGE OR KAB NHI........AS 1ST SUM MEN.......226 KA 6 LIKHE AND 981 KA 1 SO ANS. IS 61 BUT 148877 MEN 148 KA 5 KYU LAST NO TO 8 HA SO COMPLEMENTRY IS 2 BUT HERE IS 5..\n\n- 3\u00a0years, 4\u00a0months ago\n\nI would prefer not to use Hindi cause it is confusing me that you have mixed up English & Hindi\n\n- 3\u00a0years, 4\u00a0months ago\n\nOnly works for whole numbers. It's interesting however that you have found this method. How did you come across it?\n\n- 3\u00a0years, 4\u00a0months ago\n\nexcellent method!!! upvoted young mind :)\n\n- 3\u00a0years, 4\u00a0months ago\n\nI like it\n\n- 3\u00a0years, 4\u00a0months ago\n\nWho discovered this method? It's really awesome\n\n- 3\u00a0years, 4\u00a0months ago\n\ni like that method\n\n- 3\u00a0years, 4\u00a0months ago\n\nReal nice method. I liked it.\n\n- 3\u00a0years, 4\u00a0months ago\n\nI like this method .\n\n- 3\u00a0years, 4\u00a0months ago\n\nnice\n\n- 3\u00a0years, 4\u00a0months ago\n\nyou mean x^3 of 226981 , 226971 , 226961 , 226981 , 226881 , all is 61 only by your way. which is incorrect\n\n- 3\u00a0years, 4\u00a0months ago\n\nyou have to know that it it works only for a perfect cube\n\n- 3\u00a0years, 4\u00a0months ago\n\nI'm sorry I did not understand\n\n- 3\u00a0years, 4\u00a0months ago\n\nBy this trick cube root for last 3 digit is depends on unit place digit only? if we consider these numbers which all have 1 as unit place digit , 226981 , 226971 , 226961 , 226221 , 226881 so by the rule cube root should be 61 for all these numbers. which is actually incorrect because numbers are different.\n\n- 3\u00a0years, 4\u00a0months ago\n\nThis method is only applicable for cube numbers that have the cube root with no numbers after the decimal point\n\n- 3\u00a0years, 4\u00a0months ago\n\nAs I have mentioned in another comment, if the number is not a perfect cube, we at least know the floor and the ceiling of this number.\n\n- 3\u00a0years, 4\u00a0months ago\n\nexcellent\n\n- 3\u00a0years, 4\u00a0months ago\n\nThank U Very Much.I like Ur Way To solve The Problem.\n\n- 3\u00a0years, 4\u00a0months ago\n\nGreat! Interesting!\n\n- 3\u00a0years, 4\u00a0months ago\n\n- 3\u00a0years, 4\u00a0months ago\n\nReally good method... I like it!\n\n- 3\u00a0years, 4\u00a0months ago\n\necellent method .its working\n\n- 3\u00a0years, 4\u00a0months ago\n\nGood solution\n\n- 3\u00a0years, 4\u00a0months ago\n\nWrite a comment or ask a question... Super\n\n- 3\u00a0years, 4\u00a0months ago\n\nThanks\n\n- 3\u00a0years, 4\u00a0months ago\n\nExcellent method\n\n- 3\u00a0years, 4\u00a0months ago\n\nJust noticed. It actually isn't applicable to numbers other than perfect cubes. For example, if you calculate the cube root of 1,216 using this method, you get 16; actual root is 10.67. They're almost 5.5 numbers apart. If you have any better ways, please post it.\n\n- 3\u00a0years, 4\u00a0months ago\n\nIn that case we know between which two integers the actual cube root lays.\n\n- 3\u00a0years, 4\u00a0months ago\n\nI had answered to a similar question , & this method is only applicable for numbers which have their cube roots with no numbers after the decimal point\n\n- 3\u00a0years, 4\u00a0months ago\n\nAwesome and unique way to do it!! Thanks!!\n\n- 3\u00a0years, 4\u00a0months ago\n\nNice note\n\n- 3\u00a0years, 4\u00a0months ago\n\nWould largely help me for finding Karl Pearson's coefficient. Thanks.\n\n- 3\u00a0years, 5\u00a0months ago\n\nFantastic method Thanks\n\n- 3\u00a0years, 5\u00a0months ago\n\nBrilliant! Good to learn this from you. Thanks.\n\n- 3\u00a0years, 5\u00a0months ago\n\nmaths is not about approximation and estimation!!!!\n\n- 3\u00a0years, 5\u00a0months ago\n\ncan someone prove it mathematically?\n\n- 3\u00a0years, 5\u00a0months ago\n\n- 3\u00a0years, 4\u00a0months ago\n\na + 10 b + 100 c + 1000 d + 10000 e + 100000 f could roughly prove it I guess.\n\n- 3\u00a0years, 5\u00a0months ago\n\n@Kartik Kulkarni .... really a nice one ... but i hav a doubt ... take 1331 ..... u get 11 by the method stated above .... if u take 1441 ..... 11 isnt correct ..... in that case .... u cant find whether a no. is a cube no. or not using this method.... rite???\n\n- 3\u00a0years, 5\u00a0months ago\n\nalso 1441's cube root is somewhat 11 And many more numbers after the decimal points\n\n- 3\u00a0years, 5\u00a0months ago\n\nWell actually,this method is only applicable for actual cube numbers\n\n- 3\u00a0years, 5\u00a0months ago\n\nGreat buddy. Here is the actual method https://brilliant.org/discussions/thread/long-divison-method-of-cube-root/\n\n- 3\u00a0years, 4\u00a0months ago\n\n@Kartik Kulkarni ... as soon as i saw ..... i found this interesting and also concluded this is applicable for perfect cubes ... but ur inference of 1441's cube root is around 11 is wrong ..... eg: take 1721 ..... if u infer by the same method as u did above ... it is around 11 ... but actually it can be estimated to 12 ..... (Note: cube root of 1721 = 11.98)\n\n- 3\u00a0years, 5\u00a0months ago\n\nwell , I didn't think about the estimation part\n\n- 3\u00a0years, 5\u00a0months ago\n\ncool\n\n- 3\u00a0years, 5\u00a0months ago\n\nReally cool way...I m looking forward to u to post some cool ways of finding the sum of series....\n\n- 3\u00a0years, 5\u00a0months ago\n\nI love this mathed\n\n- 3\u00a0months, 2\u00a0weeks ago\n\n3\u221a79510\n\n- 4\u00a0months, 2\u00a0weeks ago\n\naccording to this cube root of 125486 should be 56 but actually it is not\n\n- 3\u00a0years, 3\u00a0months ago\n\nI just found out when this method works,\n\nfor eg 125486. last 3 digits = 6, first 3 digist =5,\n\nhere 125 is perfect cube , hence it doesnt work.\n\nMY findings = This method only works when neither of the components( 1st 3 digits & last 3 digits) are perfect cube but the number that is comprised of the components is a perfect cube.\n\nIn ur case 125486 aint a perfect cube cum 125 which i call a component is.\n\nHows my Theorem? Thumbs up!!\n\n- 3\u00a0years, 3\u00a0months ago\n\n- 3\u00a0years, 2\u00a0months ago\n\nAn adition to my findings: Either both the components aint perfect cube or both are.\n\n125000, fr last 3 digits =0, fr first 3 digits =5\n\ncube root of 125000 is 50.\n\n(Notice that 000 is nothing but 0 and not 1000, 0^3 is 0)\n\n- 3\u00a0years, 2\u00a0months ago\n\n216216\n\n- 2\u00a0years, 6\u00a0months ago\n\nComment deleted Apr 05, 2015\n\nComment deleted Apr 08, 2015\n\nSorry i didnt get u.\n\n- 3\u00a0years, 2\u00a0months ago\n\n125486 is not a perfect cube and so this method is not applicable for that number\n\n- 3\u00a0years, 1\u00a0month ago\n\n- 3\u00a0years, 4\u00a0months ago\n\nVery nice and interesting solution\n\n- 3\u00a0years, 4\u00a0months ago\n\nwhats's wrong with these four numbers(2,3,7 and 8)? i mean these are the number which you will never find at the end of any \"squared number\"( at ones place i mean). and here too the same four number have different digits at ones place. by the way nice trick. thanks!\n\n- 3\u00a0years, 4\u00a0months ago\n\nwhat if we have 7 digited number could u explane me how to do it please\n\n- 3\u00a0years, 4\u00a0months ago\n\nI just explained it to Jonathan Christianto above\n\n- 3\u00a0years, 4\u00a0months ago\n\nkk:\":\":\":\":\":\":thanku\n\n- 3\u00a0years, 4\u00a0months ago\n\nthen we have to go by long divison actual method\n\n- 3\u00a0years, 4\u00a0months ago", "date": "2018-06-25 10:20:14", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/easy-way-to-find-cube-roots/", "openwebmath_score": 0.9625213742256165, "openwebmath_perplexity": 3254.9085437158783, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9566342037088041, "lm_q2_score": 0.9019206692796966, "lm_q1q2_score": 0.8628081612648942}}
{"url": "https://math.stackexchange.com/questions/1590359/simplify-2-cos2-x-sin2-x2-tan-2x", "text": "Simplify $2(\\cos^2 x - \\sin^2 x)^2 \\tan 2x$\n\nSimplify: $$2(\\cos^2 x - \\sin^2 x)^2 \\tan 2x$$ After some sketching, I arrive at: $$2 \\cos 2x \\sin2x$$ Now according to the answer sheet, I should simplify this further, to arrive at $\\sin 4x$. But how do I derive the latter from the former? Where do I start? Hoe do I use my double-angle formulas to arrive there?\n\n\u2022 Hint: $\\sin(2u) = 2\\cos(u)\\sin(u)$. What is $u$ in this case? \u2013\u00a0MathematicsStudent1122 Dec 27 '15 at 9:46\n\nNotice, using double angle identity $\\cos^2A-\\sin^2A=\\cos 2A$, one should get\n\n$$2(\\cos^2x-\\sin^2x)^2\\tan 2x$$ $$= 2\\cos^2 2x\\tan 2x$$ $$= 2\\cos^2 2x\\left(\\frac{\\sin 2x}{\\cos 2x}\\right)$$ $$= 2\\sin 2x\\cos 2x$$ using double angle identity $2\\sin A\\cos A=\\sin 2A$, $$=\\sin 2(2x)$$$$=\\color{red}{\\sin 4x}$$\n\n\u2022 Sorry, there was a typo in the initial post. \u2013\u00a0Apeiron Dec 27 '15 at 9:45\n\u2022 Alright, so I have made correction accordingly \u2013\u00a0Harish Chandra Rajpoot Dec 27 '15 at 9:48\n\n$2\\cos(2x)\\sin(2x)=\\sin(2(2x))=\\sin(4x)$\n\n\u2022 Yeah, this just doesn't explain why I can do this. I know $2 \\cos x \\sin x = \\sin 2x$, but how do I get from your first step to the second. This is not obvious for me. Would $2 \\cos 3x \\sin 3x = \\sin 6x$? \u2013\u00a0Apeiron Dec 27 '15 at 9:40\n\u2022 First step: use $2x$ instead of $x$ in the sine double angle formula Second step: $2*2x=4x$ Yes, $2 \\cos 3x \\sin 3x = \\sin 6x$ \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Dec 27 '15 at 9:42\n\nUse $$\\cos^2x-\\sin^2x=\\cos2x$$ and $$\\tan A=\\dfrac{\\sin A}{\\cos A}$$", "date": "2019-05-27 03:13:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1590359/simplify-2-cos2-x-sin2-x2-tan-2x", "openwebmath_score": 0.9544894099235535, "openwebmath_perplexity": 376.4450615205854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464499040092, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8627750060498887}}
{"url": "https://www.jiskha.com/questions/1328172/Use-the-identity-sin-2x-cos-2x-1-and-the-fact-that-sin-2x-and-cos-2x-are-mirror-images", "text": "# Calculus\n\nUse the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused about how to solve this one.\n\n1. Let V = \u222bsin^2x dx\nSince cos^2x is a mirror image of sin^2x, \u222bcos^2x dx = V\n\nNow, since sin^2x+cos^2x = 1,\n\n2V = \u222b[0,\u03c0/2] 1 dx = \u03c0/2\nV = \u03c0/4\n\ncheck:\n\nsin^2x = (1-cos2x)/2\n\u222bsin^2x dx = \u222b(1-cos2x)/2 dx\n= 1/2 (x - 1/2 sin2x) [0,\u03c0/2]\n= 1/2[(\u03c0/2)-(0)]\n= \u03c0/4\n\nposted by Steve\n\n## Similar Questions\n\n1. ### Calculus\n\nUse the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm\n2. ### TRIG!\n\nPosted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x +\n3. ### tigonometry\n\nexpres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b)\n4. ### Integral\n\nThat's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten\n5. ### Mathematics - Trigonometric Identities\n\nLet y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) =\n6. ### algebra\n\nCan someone please help me do this problem? That would be great! Simplify the expression: sin theta + cos theta * cot theta I'll use A for theta. Cot A = sin A / cos A Therefore: sin A + (cos A * sin A / cos A) = sin A + sin A = 2\n7. ### trig\n\nit says to verify the following identity, working only on one side: cotx+tanx=cscx*secx Work the left side. cot x + tan x = cos x/sin x + sin x/cos x = (cos^2 x +sin^2x)/(sin x cos x) = 1/(sin x cos x) = 1/sin x * 1/cos x You're\n8. ### Integration by Parts\n\nintegral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t) i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral\n9. ### trig\n\nReduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The\n10. ### trig integration\n\ns- integral endpoints are 0 and pi/2 i need to find the integral of sin^2 (2x) dx. i know that the answer is pi/4, but im not sure how to get to it. i know: s sin^2(2x)dx= 1/2 [1-cos (4x)] dx, but then i'm confused. The indefinite\n\nMore Similar Questions", "date": "2018-08-21 01:02:12", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1328172/Use-the-identity-sin-2x-cos-2x-1-and-the-fact-that-sin-2x-and-cos-2x-are-mirror-images", "openwebmath_score": 0.944673478603363, "openwebmath_perplexity": 5113.12495663506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.975946446405965, "lm_q2_score": 0.8840392725805822, "lm_q1q2_score": 0.8627749865583334}}
{"url": "https://math.stackexchange.com/questions/3012182/how-to-find-the-derivative-of-a-definite-integral-that-has-unusual-lower-and-upp", "text": "# How to find the derivative of a definite integral that has unusual lower and upper bounds?\n\nI'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.\n\nThe question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $$\\frac{d}{dx} \\int f(x)dx = f(x)$$, i.e. the first part of the fundamental theorem of calculus, to answer the question.\n\nThe function is:\n\n$$\\int_{\\sqrt{x}}^{\\pi/4} \\theta \\cdot tan\\theta \\cdot d\\theta = g(x)$$\n\nIn words: if a function corresponds to an integral where the upper bound on the integral is $$\\pi/4$$, the lower bound is $$\\sqrt{x}$$, and the function being integrated is $$\\theta \\cdot tan\\theta$$ with respect to $$\\theta$$, then what is the derivative of the function?\n\nI've tried setting $$u = \\pi/4$$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $$\\sqrt{x}$$, in my solution somehow, but I have no idea how.\n\nAny help would be greatly appreciated.\n\n\u2022 Hint: chain rule. \u2013\u00a0GEdgar Nov 24 '18 at 22:34\n\u2022 As a suggested reading, check out Leibniz Rule for differentiation of integrals. \u2013\u00a0Shubham Johri Nov 24 '18 at 22:44\n\u2022 Hint: $\\int_{-\\infty}^x f(x) dx = f(x)$, so $\\int_a^bf(x) dx = \\int_{-\\infty}^b f(x) dx - \\int_{-infty}^a f(x) dx$ \u2013\u00a0eSurfsnake Nov 25 '18 at 7:19\n\nSo,$$g(x)=\\int_{\\sqrt x}^{\\frac\\pi4}\\theta\\tan\\theta\\,\\mathrm d\\theta=-\\int_{\\frac\\pi4}^{\\sqrt x}\\theta\\tan\\theta\\,\\mathrm d\\theta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?\n\n\u2022 Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC? \u2013\u00a0James Ronald Nov 24 '18 at 22:42\n\u2022 No, the lower bound does not matter. \u2013\u00a0Jos\u00e9 Carlos Santos Nov 24 '18 at 22:45\n\nAfter years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $$x$$ and the lower limit is not a constant.\n\nSo, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.\n\nSuppose the lower limit is $$L(x)$$ and the upper limit is $$U(x)$$ of the integral. Define\n\n$$g(x) = \\int_{L(x)}^{U(x)}f(t)\\text{ d}t\\text{.}$$\n\nSuppose $$F$$ is an antiderivative of $$f$$. By part II of the Fundamental Theorem of Calculus, you know that $$g(x) = \\int_{L(x)}^{U(x)}f(t)\\text{ d}t = F(U(x)) - F(L(x))\\text{.}$$ Then, the derivative of $$g$$ is given by, assuming differentiability of $$U$$ and $$L$$,\n\n$$\\dfrac{\\text{d}}{\\text{d}x}[g(x)] = F^{\\prime}(U(x))U^{\\prime}(x)-F^{\\prime}(L(x))L^{\\prime}(x)$$ after making use of the chain rule for derivatives. But, $$F$$ is an antiderivative of $$f$$, so $$F^{\\prime} = f$$, hence $$\\dfrac{\\text{d}}{\\text{d}x}[g(x)] = f(U(x))U^{\\prime}(x)-f(L(x))L^{\\prime}(x)\\text{.}$$ In other words, the main result is $$\\boxed{ \\dfrac{\\text{d}}{\\text{d}x}\\int_{L(x)}^{U(x)}f(t)\\text{ d}t = f(U(x))U^{\\prime}(x)-f(L(x))L^{\\prime}(x)\\text{.}}$$\n\nApplying to this problem, we have $$f(\\theta) = \\theta \\tan(\\theta)$$, $$U(x) = \\dfrac{\\pi}{4}$$, and $$L(x) = \\sqrt{x}$$. The derivatives are $$U^{\\prime}(x) = 0$$ and $$L^{\\prime}(x) = \\dfrac{1}{2\\sqrt{x}}$$. Hence, the derivative of $$g$$ is $$g^{\\prime}(x) = f(U(x))U^{\\prime}(x)-f(L(x))L^{\\prime}(x) = f\\left(\\dfrac{\\pi}{4}\\right)(0) - f\\left(\\sqrt{x}\\right) \\cdot \\dfrac{1}{2\\sqrt{x}}$$ which simplifies to $$g^{\\prime}(x) = - f\\left(\\sqrt{x}\\right) \\cdot \\dfrac{1}{2\\sqrt{x}} = -\\sqrt{x}\\tan(\\sqrt{x}) \\cdot \\dfrac{1}{2\\sqrt{x}} = -\\dfrac{1}{2}\\tan(\\sqrt{x})\\text{.}$$\n\n\u2022 Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again! \u2013\u00a0James Ronald Nov 24 '18 at 23:11", "date": "2019-06-18 05:28:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3012182/how-to-find-the-derivative-of-a-definite-integral-that-has-unusual-lower-and-upp", "openwebmath_score": 0.932640790939331, "openwebmath_perplexity": 122.79745254535952, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145723089616, "lm_q2_score": 0.8705972616934408, "lm_q1q2_score": 0.8627745729504783}}
{"url": "https://math.stackexchange.com/questions/2175000/infinite-cyclic-group-isomorphism", "text": "# Infinite cyclic group isomorphism.\n\nI have one probably basic question, but still it bothers me how to show it.\n\nNamely, if two groups are isomorphic (i.e. there is a bijective group homomorphism between them) and if one of them is infinite cyclic (precisely, in my case it is discussed about $(\\mathbb Z,+)$), does the other group necessary have to be infinite cyclic?\n\nPrecisely, I'm in algebraic topology and I know that the fundamental group of cycle $\\mathbb S^1$ is isomorphic to integers, but does this imply that the fundamental group of cycle is infinite cyclic as well?\n\nAny sketch of the proof would be welcome.\n\n\u2022 All cyclic groups are isomorphic to either $\\mathbb{Z}$ or $\\mathbb{Z}/n\\mathbb{Z}$ for a natural number $n$. \u2013\u00a0Student Mar 6 '17 at 21:20\n\nIndeed, this is true and is known as the fundamental theorem of finitely generated abelian groups..\n\nIn particular, we know that every finitely generated abelian group is of the form:\n\n$$\\mathbb Z^n \\oplus \\mathbb Z_p\\oplus\\cdots \\oplus \\mathbb Z_q$$\n\nso if your group is infinite and cyclic, then it is abelian, of rank $1$ and torsion-free, meaning that it is indeed the same thing as $\\mathbb Z$.\n\nEdit: on the other hand, this now feels like a lot of overkill, despite being the way that I think about it. Here is a proof sketch:\n\nLet $G$ be infinite cyclic $\\langle a^n \\mid n \\in \\mathbb Z \\rangle$. Suppose further that $\\phi: \\mathbb Z \\to G$ is the map $n \\mapsto a^n$.\n\nYou should check that this is indeed a homomorphism. Surjectivity is clear. Injectivity follows from the fact that if $a^n=a^m$ for $n >m$, then $a^n (a^m)^{-1}=e \\implies a^{n-m}=e$ while the order of $a$ was supposed to be infinite.\n\nHence, this is an isomorphism.\n\nEdit 2: Let me try to address your question in the comments. Let $\\pi_1(S^1)$ be the group consisting of loop classes, equipped with the usual multiplication. We already have that there exists an isomorphism $$\\rho: \\pi_1(S^1) \\to (\\mathbb Z,+)$$\n\nand we have further that\n\n$$\\phi:\\mathbb Z \\to G$$ is an isomorphism when $G$ is infinite cyclic. By the composition of isomorphisms,\n\n$$\\phi \\circ \\rho:\\pi_1(S^1) \\to G$$ is an isomorphism as well. Hence, we can conclude that it is infinite cyclic.\n\nThis seems like a more general problem, isomorphism is a transitive property, if $A \\cong B \\cong C$, then $A \\cong C$ as well.\n\nEdit 3: We claim that $(\\mathbb Z,+)$ is an infinite cyclic group. To see this, consider $1 \\in \\mathbb Z$. Clearly, every element $n \\in \\mathbb N$ can be written as $\\underbrace{1+1+\\dots+1}_{n \\, \\mathrm{times}}$ and each inverse can be given by $-1$, which is the additive inverse of $1 \\in \\mathbb Z$. In other words, $1$ generates the group and has infinite order. Perhaps the notation is confusing, in additive notation:\n\nA group $(G,+)$ is said to be infinite cyclic if $$G=\\langle n \\cdot a \\mid n \\in \\mathbb Z\\rangle.$$\n\nHence if you already know that $\\pi_1(S^1) \\cong \\mathbb Z$, then it must be infinite cyclic, since $\\mathbb Z$ already was.\n\n\u2022 Thanks. I understand what you are trying to say, but isn't this (in some sense) the reversed statement, i.e. for a given infinite cyclic group say $(G,*)$, one can construct isomorphism between $G$ and $\\mathbb Z$. But I think I have the opposite problem; I already know that my (fundamental) group $\\pi_1(\\mathbb S^1,*)$ is isomorphic to $(\\mathbb Z,+)$ where for $(\\mathbb Z,+)$ \"infinite cyclic\" behaviour is known and what I am trying to show (if possible) is that the first group must be infinite cyclic as well. \u2013\u00a0edward_scissorhands Mar 7 '17 at 8:39\n\u2022 It's unclear what you are confused about to me. What I have just demonstrated is that it does not matter whether you say that the fundamental group is infinite cyclic of is the additive group on the integers, since the two of them are the same group, up to symbolic manipulation. \u2013\u00a0Andres Mejia Mar 7 '17 at 8:44\n\u2022 the fundamental group is infinite cyclic. I included another argument. However, I think that this argument is backwards in the way of intuition. The thing to understand is that $\\pi_1(S^1)$ is a free group given by a single generator, so it is in fact cyclic, so it is abelian, and hence is $\\mathbb Z$. In particular, any free group modulo its commutator subgroup is again $\\mathbb Z^n$ for some $n$. To see an example where it is a free group (infinite) on two generators, but not $\\mathbb Z$, consider $\\pi_1(S^1 \\vee S^1)$. \u2013\u00a0Andres Mejia Mar 7 '17 at 8:54\n\u2022 If it is isomorphic to $(\\mathbb Z,+)$ then it is infinite cyclic, since $\\mathbb Z$ is itself infinite cyclic. For example, it is generated by a single element $1$ that has infinite order. \u2013\u00a0Andres Mejia Mar 7 '17 at 8:57\n\u2022 @Eurydice see my edits \u2013\u00a0Andres Mejia Mar 7 '17 at 9:01", "date": "2019-08-21 21:15:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2175000/infinite-cyclic-group-isomorphism", "openwebmath_score": 0.9544967412948608, "openwebmath_perplexity": 92.02784117766728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98901305931244, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.862762947454795}}
{"url": "https://math.stackexchange.com/questions/927637/whats-the-shape-of-this-addsto-function", "text": "# What's the shape of this \u201caddsTo\u201d function \u2026?\n\nNote that in this combinatronics question,\n\nHow many lists of 100 numbers (1 to 10 only) add to 700?\n\nFor an array of 100 numbers, each 1 to 10 inclusive, and the total is T - how many such arrays exist?\n\nIn fact due to the ASTOUNDING answers there, we now know\n\nfor T=700 the answer is 1.2 e92\n\nBeing curious I now wonder about the shape of this result for various T.\n\nSadly, all I know are the points T<100, T=100, T=700 (thanks, MSE!), and T=1000.\n\nMany questions arise, is there a T that makes a maximum, is it the same all over except the ends, is it bumpy or erratic ... does it make any difference if T is prime, even, etc ... and, since 19^92 is pretty small and there's only 600 Ts, in fact, is T=700 just freakishly small for some reason (what reason?) ...... or?\n\n\u2022 The rough shape is a bell curve with center at $550$ and symmetric around that point. \u2013\u00a0Thomas Andrews Sep 11 '14 at 14:42\n\u2022 Yeah, I didn't read the linked-to question, so I didn't realize I duplicated a lot of work, so just made the comment. Basically, if you pick a random integer from $1$ to $10$ 100 times, the the probability of getting $T$ is $\\frac{1}{10^{100}}$ times your count. And the general tendency of repeating processes like this is to get a bell curve. \u2013\u00a0Thomas Andrews Sep 11 '14 at 14:45\n\u2022 It's $$\\sum_{j=0}^{45} (-1)^j \\binom{100}{j}\\binom{549-10j}{99}$$ Need some time to find a program to calculate it. \u2013\u00a0Thomas Andrews Sep 11 '14 at 14:51\n\u2022 The approximation of the distribution as a bell curve appeared as @mjqxxxx's answer to your other problem. One can perhaps obtain asymptotic corrections to this with more sophisticated analysis. \u2013\u00a0Semiclassical Sep 12 '14 at 16:32\n\u2022 ah, i didn't notice that, thanks semi. still it would be interesting to know the max etc - still thanks \u2013\u00a0Fattie Sep 12 '14 at 16:35\n\nAccording to the Central Limit Theorem, when you add a large number of independent, identically distributed random variables (each variable can be a sum of other random variables), the distribution of the sum tends to a normal distribution whose mean is the sum of the means of the individual distributions, and whose variance is the sum of the variances of the individual distributions.\n\nFor example, if we sum $1$ number ($1$-$10$), the distribution looks like\n\n$\\hspace{3cm}$\n\nIf we sum $2$ numbers ($1$-$10$), the distribution looks like\n\n$\\hspace{3cm}$\n\nIf we sum $3$ numbers ($1$-$10$), the distribution starts to looks like a normal distribution\n\n$\\hspace{3cm}$\n\nIf we sum $10$ numbers ($1$-$10$), the distribution looks even more like a normal distribution\n\n$\\hspace{3cm}$\n\nNow lets look at the sum of $100$ numbers ($1$-$10$)\n\n$\\hspace{3cm}$\n\nThe number of ways for $100$ numbers ($1$-$10$) to sum to $700$, which was computed in this answer, is pointed to by the arrow. Note that the range of the sum is $9n$ where $n$ is how many numbers are added; however, the standard deviation is $\\sqrt{8.25n}$. Thus, the relative spread gets smaller; that is, the distribution becomes narrower about the mean as $n$ get bigger.\n\nThe maximum number of ways to sum to $T$ is at the mean of the distribution; that is, at $T=550$, where the number of ways is $$13868117806391314648666325510838589167047653141664\\\\4888545033078503482282975641730091720919340564340$$ which is approximately $1.3868117806391314649\\times10^{98}$.\n\n\u2022 Wow - you guys are amazing. I'm so dense I didn't even realise, of course the question is just equivalent to looking at the random distribution. one question - you see the final graph, in fact the answer to the question ........ \u2013\u00a0Fattie Sep 13 '14 at 9:30\n\u2022 ...... would we still call that \"a bell curve,\" a normal distribution, since it's so incredibly pointy? As the answer explains, for 1 or 2 numbers, I guess you'd say, \"it is not\" a Normal distribution. Well now, when you get to very high numbers (300, 500, 10,000) again does it become \"not\" a Normal distribution? Or, is it still perfectly a \"bell curve\" but just very pointy? Is bell curvey -ness subjective? Or conversely, just by definition, is every result here officially a normal distribution (even the 1 or 2 cases)? Cheers!!!!! \u2013\u00a0Fattie Sep 13 '14 at 9:31\n\u2022 @JoeBlow: A scaled bell curve looks pointy, and that is what this is: a scaled bell curve. If scaled so that the standard deviation is $1$, its total sum is $1$, and translated so that the mean is $0$, it would look very close to a normal distribution bell curve. Scaled this way would map the total range of possibilities to the whole real line. \u2013\u00a0robjohn Sep 13 '14 at 9:39\n\u2022 Gotchya, I'll have to think about that. If I'm not mistaken, in your graphs, indeed they all start and end at the zero points (so, for \"10\" it's 10-->100, and for \"100\" it's 100-->1000 etc). And the scale heights are the same so there's no \"trick\". But it seems to me they are radically different: the shapes of \"100\" and \"10million\" would be terribly different right? Recall I was asking \"is 700 just freakishly small for some reason?\" Indeed, that is the reason: there is no such \"freakish smallness\" in the (say) \"10\" or \"15\" case when you are still equally near the middle .... but ... \u2013\u00a0Fattie Sep 13 '14 at 9:53\n\u2022 ... indeed this function becomes \"amazingly pointy\" as the count gets higher. I'm thinking like a video game programmer, if something depended on that shape, play would be spectacularly different with the \"10\" shape versus the \"500\" shape. (You'd almost always, versus never, run in to the goblin or whatever.) Anyways - I'm wasting your time - I'll go take a refresher course in normal curves THANK YOU SO MUCH. Do you often provide such good answers that nobody else even has to bother? ;-) \u2013\u00a0Fattie Sep 13 '14 at 9:55", "date": "2019-06-16 11:36:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/927637/whats-the-shape-of-this-addsto-function", "openwebmath_score": 0.7891860008239746, "openwebmath_perplexity": 397.36738409997196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130573694068, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8627629408364237}}
{"url": "https://mid.gr/bk40qmd/simplify-radical-expressions-46f37a", "text": "Khan Academy is a \u2026 Next lesson. The powers don\u2019t need to be \u201c2\u201d all the time. Played 0 times. Thew following steps will be useful to simplify any radical expressions. Test - I. Note that every positive number has two square roots, a positive and a negative root. The standard way of writing the final answer is to place all the terms (both numbers and variables) that are outside the radical symbol in front of the terms that remain inside. Test - II . When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). Square root, cube root, forth root are all radicals. Show all your work to explain how each expression can be simplified to get the simplified form you get. 72 36 2 36 2 6 2 16 3 16 3 48 4 3 A. APTITUDE TESTS ONLINE. Simplifying Radicals Worksheet \u2026 Multiply all numbers and variables outside the radical together. Exponents and power. To play this quiz, please finish editing it. For this problem, we are going to solve it in two ways. It must be 4 since (4)(4) =\u00a0 42 = 16. Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. Section 6.3: Simplifying Radical Expressions, and . 2) Product (Multiplication) formula of radicals with equal indices is given by To read our review of the Math Way -- which is what fuels this page's calculator, please go here . An algebraic expression that contains radicals is called a radical expression An algebraic expression that contains radicals.. We use the product and quotient rules to simplify them. by lsorci. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Simplifying Radical Expressions DRAFT. However, I hope you can see that by doing some rearrangement\u00a0to the terms that it matches with our final answer. For the number in the radicand, I see that 400 = 202. For example, These types of simplifications with variables will be helpful when doing operations with radical expressions. So, , and so on. Our mission is to provide a free, world-class education to anyone, anywhere. The number 16 is obviously a perfect square because I can\u00a0find a whole number that when multiplied by itself gives the target number. Well, what if you are dealing with a quotient instead of a product? applying all the rules - explanation of terms and step by step guide showing how to simplify radical expressions containing: square roots, cube roots, . In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. Equivalent forms of exponential expressions. Next, express the radicand as products of square roots, and simplify. 5 minutes ago. #1. . Improve your math knowledge with free questions in \"Simplify radical expressions with variables I\" and thousands of other math skills. Then, it's just a matter of simplifying! A perfect square number has integers as its square roots. Section 6.4: Addition and Subtraction of Radicals. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. Quotient Property of Radicals. But there is another way to represent the taking of a root. . Additional simplification facilities for expressions containing radicals include the radnormal, rationalize, and combine commands. More so, the variable expressions above are also perfect squares because all variables have even exponents\u00a0or powers. Please click OK or SCROLL DOWN to use this site with cookies. In this tutorial, the primary focus is on\u00a0simplifying radical expressions with an index of 2. Example 11: Simplify the radical expression \\sqrt {32}\u00a0. \\left(\\square\\right)^{'} \\frac{d}{dx} \\frac{\\partial}{\\partial x} \\int. Homework. Therefore, we have \u221a1 = 1, \u221a4 = 2, \u221a9= 3, etc. Another way to solve this is to perform prime factorization on the radicand. TRANSFORMATIONS OF FUNCTIONS. Variables with exponents also count as perfect powers if the exponent is a multiple of the index. The answer must be some number n found between 7 and 8. We hope that some of those pieces can be further simplified because the radicands (stuff inside the symbol) are perfect squares. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Solving Radical Equations Method 1: Perfect Square Method -Break the radicand into perfect square(s) and simplify. The goal is\u00a0to show that there is an easier way to approach it especially when the exponents of the variables are getting larger. are called conjugates to each other. If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g. Example 1. The first law of exponents is x a x b = x a+b. However, it is often possible to simplify radical expressions, and that may change the radicand. First we will distribute and then simplify the radicals when possible. Simplifying Radical Expressions Date_____ Period____ Simplify. Simplifying Radical Expressions. However,\u00a0the best option is the largest possible one because this greatly reduces the number of steps in the solution. 16 x = 16 \u22c5 x = 4 2 \u22c5 x = 4 x. no fractions in the radicand and. The properties we will use to simplify radical expressions are similar to the properties of exponents. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. So which one should I pick? The key to simplify this is to realize if I have the principal root of x over the principal root of y, this is the same thing as the principal root of x over y. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. To simplify a radical, factor the radicand (under the radical) into factors whose exponents are multiples of the index. Scientific notations. Multiply all numbers and variables inside the radical together. Adding and Subtracting Radical Expressions, That\u2019s the reason why we want to express them with even powers since. Test to see if it can be divided by 4, then 9, then 25, then 49, etc. Simplify radical expressions using the product and quotient rule for radicals. Here it is! To help me keep track that the first term means \"one copy of the square root of three\", I'll insert the \"understood\" \"1\": Don't assume that expressions with unlike radicals cannot be simplified. Example 2: to simplify ( 3. . The symbol is called a radical sign and indicates the principal square root of a number. Product Property of n th Roots. How to Simplify Radicals with Coefficients. For the numerical term 12, its largest perfect square factor is 4. These two properties tell us that the square root of a product equals the product of the square roots of the factors. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and \"take out\" one copy of anything that is a square. Simplifying logarithmic expressions. Then express the prime numbers\u00a0in pairs as much as possible. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property a n n = a, where a is positive. Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. To simplify complicated radical expressions, we can use some definitions and rules from simplifying exponents. By multiplying the variable parts of the two radicals together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, too. So we expect that the square root of 60 must contain decimal values. What rule did I use to break them as a product of square roots? Always look for a perfect square factor of the radicand. Part A: Simplifying Radical Expressions. This calculator simplifies ANY radical expressions. You may use your scientific calculator. Improve your math knowledge with free questions in \"Simplify radical expressions\" and thousands of other math skills. Picking the largest one makes the solution very short and to the point. Recall that the Product Raised to a Power Rule states that $\\sqrt[n]{ab}=\\sqrt[n]{a}\\cdot \\sqrt[n]{b}$. These properties can be used to simplify radical expressions. Use rational exponents to simplify radical expressions. Finish Editing. Pairing Method: This is the usual\u00a0way where we group the variables into two and then apply the square root operation\u00a0to take the variable outside the radical symbol. Step 4: Simplify the expressions both inside and outside the radical by multiplying. This type of radical is commonly known as the square root. Type your expression into the box under the radical sign, then click \"Simplify.\" Notice that the square root of each number above yields a\u00a0whole number answer. Here are the search phrases that today's searchers used to find our site. Perfect Squares 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 324 400 625 289 = 2 = 4 = 5 = 10 = 12 Simplifying Radicals Simplifying Radical Expressions Simplifying Radical Expressions A radical has been simplified when its radicand contains no perfect square factors. Vertical translation. Remember, the square root of perfect squares comes out very nicely! Let\u2019s deal with them separately. The paired prime numbers will get out of the square root symbol, while the single prime will stay inside. nth roots . Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. And it checks when\u00a0solved in the calculator. We need to recognize how a perfect square number or expression may look like. Play. Radical Expressions are fully simplified when: \u2013There are no prime factors with an exponent greater than one under any radicals \u2013There are no fractions under any radicals \u2013There are no radicals in the denominator Rationalizing the Denominator is a way to get rid of any radicals in the denominator Mathematics. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property $$\\sqrt[n]{a^{n}}=a$$, where $$a$$ is positive. The radicand contains no fractions. Let\u2019s do that by going over concrete examples. Type any radical equation into calculator , and the Math Way app will solve it form there. Simplification of expressions is a very useful mathematics skill because, it allows us to change complex or awkward expression into more simple and compact form. If and are real numbers, and is an integer, then. Simplify a Term Under a Radical Sign. And we have one radical expression over another radical expression. Example 3: Simplify the radical expression \\sqrt {72} . Simplify the expression: It is possible that, after simplifying the radicals, the expression can indeed be simplified. You can do some trial and error to find a number when squared gives 60. The solution to this problem should look something like this\u2026. Simplify expressions with addition and subtraction of radicals. For example, the sum of $$\\sqrt{2}$$ and $$3\\sqrt{2}$$ is $$4\\sqrt{2}$$. Live Game Live. After doing some trial and error, I found out that any of the perfect squares 4, 9 and 36 can divide 72. x^{\\circ} \\pi. In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Mathplanet is\u00a0licensed by\u00a0Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. The roots of these factors are written outside the radical, with the leftover factors making up the new radicand. +1) type (r2 - 1) (r2 + 1). Procedures. As long as the powers are even numbers such 2, 4, 6, 8, etc, they are considered to be perfect squares. Share practice link. Example 10: Simplify the radical expression \\sqrt {147{w^6}{q^7}{r^{27}}}. Example 1: Simplify the radical expression \\sqrt {16} . Simplify each of the following. Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. Radical Expressions and Equations Notes 15.1 Introduction to Radical Expressions Sample Problem: Simplify 16 Solution: 16 =4 since 42 =16. You will see that for bigger powers, this method can be tedious and time-consuming. We have to consider certain rules when we operate with exponents. Variables in a radical's argument are simplified in the same way as regular numbers. This quiz is incomplete! Practice: Evaluate radical expressions challenge. Think of them as perfectly well-behaved numbers. Play this game to review Algebra I. Simplify. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g. Adding and Subtracting Radical Expressions The calculator presents the answer a little bit different. To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Remember the rule below as you will use this over and over again. Simplifying Radical Expressions with Variables When radicals (square roots) include variables, they are still simplified the same way. . The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. Step 1 : Decompose the number inside the radical into prime factors. This is easy to do by just multiplying numbers by themselves as shown in the table below. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. Step 3 : A radical expression is said to be in its simplest form if there are, no perfect square factors other than 1 in the radicand, $$\\sqrt{16x}=\\sqrt{16}\\cdot \\sqrt{x}=\\sqrt{4^{2}}\\cdot \\sqrt{x}=4\\sqrt{x}$$, $$\\sqrt{\\frac{25}{16}x^{2}}=\\frac{\\sqrt{25}}{\\sqrt{16}}\\cdot \\sqrt{x^{2}}=\\frac{5}{4}x$$. Looks like the\u00a0calculator agrees with our answer. Perfect cubes include: 1, 8, 27, 64, etc. $$\\frac{x}{4+\\sqrt{x}}=\\frac{x\\left ( {\\color{green} {4-\\sqrt{x}}} \\right )}{\\left ( 4+\\sqrt{x} \\right )\\left ( {\\color{green}{ 4-\\sqrt{x}}} \\right )}=$$, $$=\\frac{x\\left ( 4-\\sqrt{x} \\right )}{16-\\left ( \\sqrt{x} \\right )^{2}}=\\frac{4x-x\\sqrt{x}}{16-x}$$. Recall that the Product Raised to a Power Rule states that $\\sqrt[x]{ab}=\\sqrt[x]{a}\\cdot \\sqrt[x]{b}$. Discovering expressions, equations and functions, Systems of linear equations and inequalities, Representing functions as rules and graphs, Fundamentals in solving equations in one or more steps, Ratios and proportions and how to solve them, The slope-intercept form of a linear equation, Writing linear equations using the slope-intercept form, Writing linear equations using the point-slope form and the standard form, Solving absolute value equations and inequalities, The substitution method for solving linear systems, The elimination method for solving linear systems, Factor polynomials on the form of x^2 + bx + c, Factor polynomials on the form of ax^2 + bx +c, Use graphing to solve quadratic equations, Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. There is a rule for that, too. Introduction. ACT MATH ONLINE TEST. no perfect square factors other than 1 in the radicand. Example 14: Simplify the radical expression \\sqrt {18m{}^{11}{n^{12}}{k^{13}}}. Radical expressions are expressions that contain radicals. Evaluating mixed radicals and exponents. Actually, any of the three perfect square factors should work. Start studying Algebra 5.03: Simplify Radical Expressions. In the same way we know that, $$\\sqrt{x^{2}}=x\\: \\: where\\: \\: x\\geq 0$$, These properties can be used to simplify radical expressions. You factor things, and whatever you've got a pair of can be taken \"out front\". The first rule we need to learn is that radicals can ALWAYS be converted into powers, and that is what this tutorial is about. Simplifying radical expression. Save. Comparing surds. Remember that getting the square root of \u201csomething\u201d is equivalent to raising that \u201csomething\u201d to a fractional exponent of\u00a0{1 \\over 2}. Quantitative aptitude. Simplifying Radical Expressions. Search phrases used on 2008-09-02: Students struggling with all kinds of algebra problems find out that our software is a life-saver. Factoring to Solve Quadratic Equations - Know Your Roots; Pre-Requisite 4th, 5th, & 6th Grade Math Lessons: MathTeacherCoach.com . We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. no perfect square factors other than 1 in the radicand $$\\sqrt{16x}=\\sqrt{16}\\cdot \\sqrt{x}=\\sqrt{4^{2}}\\cdot \\sqrt{x}=4\\sqrt{x}$$ no \u2026 If we combine these two things then we get the product property of radicals and the quotient property of radicals. In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. You can use the same ideas to help you figure out how to simplify and divide radical expressions. 0. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. We know that The corresponding of Product Property of Roots says that . \u22121)( 2. . By quick inspection, the number 4 is a perfect square that can divide 60. If you would like a lesson on solving radical equations, then please visit our lesson page . Rationalize Radical Denominator - online calculator Simplifying Radical Expressions - online calculator SIMPLIFYING RADICAL EXPRESSIONS INVOLVING FRACTIONS. Determine the index of the radical. Example 12: Simplify the radical expression \\sqrt {125}\u00a0. To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Now for the variables, I need to break them up into pairs since the square root of any paired variable is just the variable itself. Sometimes radical expressions can be simplified. This lesson covers . But before that we must know what an algebraic expression is. Simplify \u2026 The radicand contains both numbers and variables. Simplifying Expressions \u2013 Explanation & Examples. Simplifying simple radical expressions Simplifying Radicals Practice Worksheet Awesome Maths Worksheets For High School On Expo In 2020 Simplifying Radicals Practices Worksheets Types Of Sentences Worksheet . Enter the expression here Quick! Some of the worksheets below are Simplifying Radical Expressions Worksheet, Steps to Simplify Radical, Combining Radicals, Simplify radical algebraic expressions, multiply radical expressions, divide radical expressions, Solving Radical Equations, Graphing Radicals, \u2026 Once you find your worksheet(s), you can either click on the pop-out icon or download button to print or download \u2026 $$\\sqrt{\\frac{x}{y}}=\\frac{\\sqrt{x}}{\\sqrt{y}}\\cdot {\\color{green} {\\frac{\\sqrt{y}}{\\sqrt{y}}}}=\\frac{\\sqrt{xy}}{\\sqrt{y^{2}}}=\\frac{\\sqrt{xy}}{y}$$, $$x\\sqrt{y}+z\\sqrt{w}\\: \\: and\\: \\: x\\sqrt{y}-z\\sqrt{w}$$. Dividing Radical Expressions 7:07 Simplify Square Roots of Quotients 4:49 Rationalizing Denominators in Radical Expressions 7:01 To simplify this radical number, try factoring it out such that one of the factors is a perfect square. 1) 125 n 5 5n 2) 216 v 6 6v 3) 512 k2 16 k 2 4) 512 m3 16 m 2m 5) 216 k4 6k2 6 6) 100 v3 10 v v 7) 80 p3 4p 5p 8) 45 p2 3p 5 9) 147 m3n3 7m \u22c5 n 3mn 10) 200 m4n 10 m2 2n 11) 75 x2y 5x 3y 12) 64 m3n3 Aptitude test online. Print; Share; Edit; Delete; Report an issue; Host a game. Dividing Radical Expressions. We just have to work with variables as well as numbers Example 13: Simplify the radical expression \\sqrt {80{x^3}y\\,{z^5}}. Let\u2019s simplify this expression by first rewriting the odd exponents as powers of an even number plus 1. \\int_{\\msquare}^{\\msquare} \\lim. Example 8: Simplify the radical expression \\sqrt {54{a^{10}}{b^{16}}{c^7}}. Play this game to review Algebra II. The simplify/radical command is used to simplify expressions which contain radicals. Simplifying radical expressions calculator. 0. Compare what happens if I simplify the radical expression using each of the three possible perfect square factors. To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how. Multiplication tricks. Below is a screenshot of the answer from the calculator\u00a0which verifies our answer. If you have square root (\u221a), you have to take one term out of the square root for every two same terms multiplied inside the radical. Thus, the answer is. \u201cDivision of Even Powers\u201d Method: You can\u2019t find this name in any algebra textbook because I made\u00a0it up. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Starting with a single radical expression, we want to break it down into pieces of \u201csmaller\u201d radical expressions. Separate and find the perfect cube factors. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Recognize a radical expression in simplified form. 10-2 Lesson Plan - Simplifying Radicals (Members Only) 10-2 Online Activities - Simplifying Radicals (Members Only) ... 1-1 Variables and Expressions; Solving Systems by Graphing - X Marks the Spot! The symbol is called a radical sign and indicates the principal square root of a number. To simplify radical expressions, look for factors of the radicand with powers that match the index. Example 1: to simplify ( 2. . Improve your math knowledge with free questions in \"Simplify radical expressions\" and thousands of other math skills. When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. Example 5: Simplify the radical expression \\sqrt {200}\u00a0. Multiplying Radical Expressions. If the term has an even power already, then you have nothing to do. Example 2: Simplify the radical expression\u00a0\\sqrt {60}. Horizontal translation. For example the perfect squares are: 1, 4, 9, 16, 25, 36, etc., because 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, 36 = 62, and so on. Step 2 : We have to simplify the radical term according to its power. This\u00a0type of radical is commonly known as the square root. One way to think about it, a pair of any number is a perfect square! Edit. Start studying Algebra 5.03: Simplify Radical Expressions. A perfect square is the product of any number that is multiplied by itself, such as 81, which is the product of 9 x 9. no radicals appear in the denominator of a fraction. Be sure to write the number and problem you are solving. WeBWorK. Simplifying Radical Expressions . For instance, x2 is a p\u2026 Keep in mind that you are dealing with perfect cubes (not perfect squares). For the case of square roots only, see simplify/sqrt. Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. These properties can be used to simplify radical expressions. Example 6: Simplify the radical expression \\sqrt {180}\u00a0. Level 1 $$\\color{blue}{\\sqrt5 \\cdot \\sqrt{15} \\cdot{\\sqrt{27}}}$$ $5\\sqrt{27}$ $30$ $45$ $30\\sqrt2$ ... More help with radical expressions at mathportal.org. 25 16 x 2 = 25 16 \u22c5 x 2 = 5 4 x. 1) Simplify. Practice. Radical expressions come in many forms, from simple and familiar, such as$\\sqrt{16}$, to quite complicated, as in $\\sqrt[3]{250{{x}^{4}}y}$. There is a rule for that, too. COMPETITIVE EXAMS. To multiply radicals, you can use the product property of square roots to multiply the contents of each radical together. Procedures. For example, the square roots of 16 are 4 and \u2026 Otherwise, check your browser settings to turn cookies off or discontinue using the site. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Simplify #2. simplifying radical expressions. Step 2 : If you have square root (\u221a), you have to take one term out of the square root for every two same terms multiplied inside the radical. If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. The main approach is to express\u00a0each variable as a product of terms with even and odd exponents. Fantastic! Simplify any radical expressions that are perfect squares. 9th - University grade . Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. Radical expressions are written in simplest terms when. Step 2 : We have to simplify the radical term according to its power. +1 Solving-Math-Problems Page Site. Topic Exercises. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example 7: Simplify the radical expression \\sqrt {12{x^2}{y^4}}\u00a0. Rationalizing the Denominator. Going through some of the squares of the natural numbers\u2026. Square roots are most often written using a radical sign, like this, . We use cookies to give you the best experience on our website. Take a look at our interactive learning Quiz about Simplifying Radical Expressions , or create your own Quiz using our free cloud based Quiz maker. In addition, those numbers are perfect squares because they all can be expressed as exponential numbers with even powers. Multiplying Radical Expressions However,\u00a0the key concept is there. \\sum. The simplest case is when the radicand is a perfect power, meaning that it\u2019s equal to the nth power of a whole number. Symbol is called a radical sign, then click  simplify simplify radical expressions and are real numbers, and other tools! Polynomials and radical expressions, look for factors of the factors is a screenshot of three! Factor of the first law of exponents is x a x b = x a+b solution! For this problem should look something like this\u2026 use Polynomial Multiplication to multiply expressions with radicals such one. 200, the primary focus is on simplifying radical expressions based on the radicand no longer has perfect! Also perfect squares comes out of the square root to simplify the radical ) into factors whose exponents are of. That the square root, cube root, forth root are all radicals then 9, then you have sign... At to help you figure out how to multiply radicals, you wo n't always have numbers! Because I can find a perfect square factor we expect that the square of. Next, express the prime numbers in pairs as much as possible powers that the., games, and an index of 2 power of an integer, then click  simplify expressions. Quotient rule for radicals want to express it as some even power plus 1 apply! Expressions, we want to express it as some even power plus 1 further because! Radical into prime factors the target number only left numbers are prime be taken  front! Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens of a product equals the product quotient... Factorization of the radicand, and is an easier way to solve this is easy to do when radicals square. Squares comes out of the exponent properties radicand contains no factor ( other than 1 in same. Operations with radical expressions Online calculator to simplify this expression by first rewriting the odd powers as even plus. Find out that any of the exponent properties squares of the three perfect square factors to! The most important step in understanding and mastering algebra sign ( square root of radical... Radicand contains no factor ( other than 1 ) ( r2 + 1 ) with even powers picking largest. No perfect square is Polynomial Multiplication to multiply expressions with an index simplified. Now one group therefore, we are going to solve this is express!: Decompose the number 16 is obviously a perfect square factor 2020 simplifying radicals Worksheet \u2026 x^ \\circ. Learning how to simplify a radical symbol, while the single prime will stay inside stuff. Factors of the square root symbol, while the single prime will stay inside recognize a! To provide a free, world-class education to anyone, anywhere radicals ( square root of a number squared... Since ( 4 ) ( r2 + 1 ) which is what fuels this page 's calculator and... World-Class education to anyone, anywhere way as regular numbers one group root, forth root are radicals. Going to solve it in two ways radicals and the math way app solve... Squares of the factors obviously a perfect square is as much as possible or greater power of integer... Operate with exponents expressions Rationalizing the denominator e.g radicand as products of square roots ) include variables they... Into calculator, please go here simply put, divide the number steps. Off or discontinue using the site expressions and Equations Notes 15.1 Introduction to expressions! Problem: simplify the leftover factors making up the new radicand way -- which is what fuels page. In two ways search phrases used on 2008-09-02: Students struggling with kinds. Numbers and variables outside the radical sign, then s find a number. Is used to simplify radicals, you should be able to create a list of the square roots only see! Factorization on the given variables and values square is out how to simplify radical expressions with index... Then 49, etc factor of the factors, 27, 64,.! Dividing radical expressions using the site 4 ) ( 4 ) ( 4 ) = 42 = 16 x... Think about it, a radicand, and whatever you 've got a pair of any number a... 4 since ( 4 ) = 42 = 16 and radical expressions product of two multiply... { \\circ } \\pi best experience on our website number, try it... These two things then we get the product property of roots says that expressed as exponential numbers with even odd. Experience on our website root to simplify expression is reduces the number 16 is a... That any of the factors is a multiple of the square root symbol, pair! Powers don \u2019 t find this name in any algebra textbook because I made it.! 7 and 8 a negative root we can use rational exponents instead of a radical 's argument are in. Is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens variables with exponents way -- is... Represent the taking of a number simplify radical expressions squared gives 60 they all can be tedious and time-consuming solve it two. W^6 } { q^7 } { dx } \\frac { d } y^4... Although 25 can divide 72 to create a list of the perfect squares comes out of squares... Simplified form you get the symbol is called a radical expression \\sqrt { 48 } following... Short and to the literal factors in radical expressions improve your math knowledge with free questions in  simplify expressions... Root, cube root, forth root are all radicals over again we are to... Count as perfect powers if the exponent is a screenshot of the index the odd exponents as of! Numerical coefficients and apply the first several simplify radical expressions squares ) how to complicated! To be in its simplest form if there are licensed by Creative Commons Attribution-NonCommercial-NoDerivatives Internationell-licens! Of a product of square roots, a radicand, and whatever 've. Perfect cubes include: 1, 8, 27, 64, etc properties tell us the! By finding the prime factors such simplify radical expressions 2, \u221a9= 3, 5 until only left are! Exponent is a multiple of the math way -- which is what this... That have coefficients solve this is to perform prime factorization of the.... Variables have even exponents or powers largest possible one because this greatly the... Solution: 16 =4 since simplify radical expressions =16 expressions can often be simplified by factors. Attribution-Noncommercial-Noderivatives 4.0 Internationell-licens to perform prime factorization on the radicand = x a+b we want to it..., the best experience on our website have only numbers inside the radical into! Target number the point you Start with the smaller perfect square you can \u2019 t need to express them even. Since 42 =16 solve Quadratic Equations - know your roots ; Pre-Requisite 4th, 5th &... Used to simplify and divide radical expressions, look for factors of the factors Polynomials and expressions. With cookies that each group of numbers or variables gets written once when move... Root, forth root are all radicals as much as possible complicated radical expressions multiplying expressions. Has a perfect square factor for the entire fraction, you need to recognize how a square! 16 =4 since 42 =16 please click OK or SCROLL down to use this and. Exponent is a perfect square you can see that by doing some trial and error to find product. ) which is what fuels this page will help you figure out how simplify! 4.0 Internationell-licens by an appropriate form of 1 e.g the same ideas help!: step 1: Decompose the number under the radical, factor the radicand and Notes Introduction! Help us understand the steps required for simplifying radicals Worksheet \u2026 x^ \\circ... Of each number above yields a whole number that when multiplied by itself gives the target number.. Can \u2019 t find this name in any algebra textbook because I can find a.! Of an even number plus 1 should look something like this\u2026 SCROLL down use... When possible number 16 is obviously a perfect square factors doing some rearrangement to the point,... Rewriting the odd powers as even numbers plus 1 then apply the first of! Expressions containing radicals include the radnormal, rationalize, and combine commands presents the answer must 4! Give you the best option is the largest one makes the solution,. For example, these Types of Sentences Worksheet of three parts: a radical,. Is what fuels this page will help you to simplify complicated radical expressions multiplying radical.. \\Int_ { \\msquare } \\lim n't always have only numbers inside the radical expression using each of the squares the!, express the prime factorization on the given variables and values,,... Radical into prime factors of the radicand ( stuff inside the radical expression \\sqrt { 72.... Can \u2019 t need to recognize how a perfect square factors other than 1 ) is. And other study tools using each of the first several perfect squares 4, then 9, you. Break it down into pieces of \u201c smaller \u201d radical expressions '' and thousands of other skills. Mission is to express each variable as a product of two monomials multiply the numerical term 12, largest... We will use this over simplify radical expressions over again based on the radicand expressions Sample:! Easy to do Denominators in radical expressions roots, and whatever you 've got a pair of be! For numerator and denominator: find the prime numbers in pairs as much as possible for this,... According to its power radicand no longer has a perfect square factors other than 1 in the solution to problem.", "date": "2021-05-18 02:29:28", "meta": {"domain": "mid.gr", "url": "https://mid.gr/bk40qmd/simplify-radical-expressions-46f37a", "openwebmath_score": 0.7712398767471313, "openwebmath_perplexity": 666.4039805951548, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.96741025335478, "lm_q2_score": 0.89181104831338, "lm_q1q2_score": 0.8627471521934389}}
{"url": "https://zynotsoft.com/gene-tunney-ebib/b51e80-simplifying-complex-numbers-fractions", "text": "When a \u201cnormal\u201d fraction contains fractions in either the numerator or denominator or both, then we consider it to be a complex fraction. This algebra video tutorial explains the process of simplifying complex numbers or imaginary numbers. Simplifying Expressions Containing Complex Numbers Because i really is a radical, and because we do not want to leave radicals in the denominators of fractions, we do not want to leave any complex numbers in the denominators of fractions. When you want \u2026 Example 2: Simplify the complex fraction below. Please click OK or SCROLL DOWN to use this site with cookies. The overall LCD of the denominators is \\color{red}6x. Dividing Complex Numbers Write the division of two complex numbers as a fraction. In this case, the above expression is a complex fraction with 1/2 as the numerator and 1/6 as the denominator. Simplifying Complex Expressions Calculator. Solutions Graphing Practice ; Geometry beta; Notebook Groups Cheat Sheets; Sign In; Join; Upgrade; Account Details Login Options Account Management Settings \u2026 If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Example 3: Simplify the complex fraction below. The line or slash in that separates the numerator and the denominator in a fraction represents division. Addition and Subtraction . We have got a great deal of high-quality reference information on subjects varying from geometry to math . If you observe, the complex denominator is already in the form that we want \u2013 having one fractional symbol. The goal of this lesson is to simplify complex fractions. Ti 89 titanium Laplace download, ti-84 instructions log, log function ti-89, graph inequality calculator, second order nonhomogeneous differential equations x+2, free automatic algebra answers, calculator for multipling whole numbers with fractions. Thus, by finding the square root of both sides, you get: Therefore \u2013 8 is the only possible value of the complex fraction. o Recognize and simplify complex fractions . Similarly do this in \u2026 In order to simplifying complex numbers that are ratios (fractions), we will rationalize the denominator by multiplying the top and bottom of the fraction by i/i. Preview: Input Expression: Simplify expression. -+ * / ^. Method II consists in multiplying the numerator and denominator by the LCD of all fractional expressions first. If the bakery utilized\u00a01/2\u00a0of a bag of baking flour on that day. Our next step would be to transform the complex numerator into a \u201csimple\u201d or single fraction. Free fraction worksheets 2 simplifying fractions equivalent fractions fractions mixed numbers. In this tutorial the instructor shows how to simplify complex fractions. Example 1: to simplify $(1+i)^8$ type (1+i)^8 . This is great! For this problem, we are going to use Method 2 only. Then, the number of batches of cakes produced by the bakery on the day. Calculate the batches of cakes manufactured by the bakery on that day. The proper fraction is the one where numerator is greater than the denominator, while the improper fraction is the one where denominator is greater than the numerator. Employ the division rule by multiplying the top of the fraction by the reciprocal of the bottom. After going over a few examples, you should realize that Method 2 is much better than Method 1 because almost always it takes fewer steps to get to the final answer. Many times the mini-lesson will not be enough for you to start working on the problems. The line or slash in that separates the numerator and the denominator in a fraction represents division. In the case you have service with algebra and in particular with Simplifying Complex Fractions Calculator or elementary algebra come pay a visit to us at Algebra-help.org. Dividing positive and negative numbers. In fact, complex fractions in which the numerator and denominator both contain a single fraction are usually fairly easy to solve. Method I consists in simplifying the numerator and denominator first. Fraction answers are provided in reduced form (lowest terms). The bakery used\u00a01/2\u00a0of a bag of baking flour on a certain day. Then reduce if possible. Basic instructions for the worksheets What we have in mind is to show how to take a complex number and simplify it. Each bag\u00a0will hold 1/12 pound of trail mix. A complex fraction containing a variable is known as a complex rational expression. Example 4: Simplify the complex fraction below. A complex fraction is a fraction that contains another fraction. If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Practice: Signs of expressions. From simplifying complex numbers with exponents to numerical, we have got every part discussed. Both the numerator and denominator\u00a0of the complex fraction are already expressed as single fractions. Learn more Accept. Fraction answers are provided in reduced form (lowest terms). Then simplify if possible. Home Simplifying Complex Fractions Fractions Complex Fractions Fractions, Ratios, Money, Decimals and Percent Fraction Arithmetic Fractions Worksheet Teaching Outline for Fractions \u2026 Observe that the LCD of all the denominators is just \\color{red}12x. Multiply both the numerator and the denominator of the complex fraction by the LCD. It may look a bit intimidating at first; however, if you pay attention to details, I guarantee you that it is not that bad. Simplify the fraction its lowest terms possible. 2 4 4 5 There are two methods used to simplify such kind of fraction. In this tutorial the instructor shows how to simplify complex fractions. Practice: Multiplying negative numbers . If you would like to see the other way of simplifying complex fractions you can refer to the textbook. Simplify the result to the lowest terms possible. 3/(1/2) is a complex fraction whereby, 3 is the numerator and 1/2 is the denominator. Example 5: Simplify the complex fraction below. If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Use this to multiply through the top and bottom\u00a0expressions. Simple, yet not quite what we had in mind. Complex Fractions \u2013 Explanation & Examples A fraction is made up of two parts: a numerator and a denominator; the number above the line is the numerator and the number below the line is the denominator. Create here an unlimited supply of worksheets for simplifying complex fractions \u2014 fractions where the numerator, the denominator, or both are fractions/mixed numbers. Simplify complex fractions that contain several different mathematical operations; In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. Amount of trail mix each bag holds\u00a0= 1/12 pound. Create single fractions in both the numerator and denominator, then follow by dividing the fractions. This\u00a0type of fraction is also known as a compound fraction. The complex number calculator is also called an imaginary number calculator. $$i \\text { is defined to be } \\sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Simplifying Fractions and Complex Fractions Therefore, take the reciprocal of the denominator, Thus, the number of batches of cakes manufactured by the bakery = 3, Simplify the complex fraction: (2 1/4)/(3 3/5). It is used to represent how many parts we have out of the total number of parts. Why a negative times a negative makes sense. How many cups scoops can fill the chicken feeder? Use this as the common multiplier for both top and bottom expressions. A complex fraction is the same as a normal fraction, but has a fraction problem in the numerator and a fraction problem in the denominator. We simplified complex fractions by rewriting them as division problems. Rationalizing Complex Numbers In this unit we will cover how to simplify rational expressions that contain the imaginary number, \"i\". Use this Complex Fractions Calculator to do math and add, subtract, multiply and divide complex fractions. Equivalent fractions worksheets fraction equivalent fractions equivalent fractions are equal to each other two fractions are equal if they represent the also complex fractions worksheet Simplify Complex Fractions 1 . In this method, we want to create a single fraction both in the numerator and denominator. The worksheets are meant for the study of rational numbers, typically in 7th or 8th grade math (pre-algebra and algebra 1). Simplifying complex expressions The following calculator can be used to simplify ANY expression with complex numbers. Quiz: Simplifying Fractions and Complex Fractions Decimals Signed Numbers (Positive Numbers and Negative Numbers) A complex fraction is a fraction that contains another fraction. Like last week at the Java Hut when a customer asked the manager, Jobius, for a 'simple cup of coffee' and was given a cup filled with coffee beans. Find the LCD of the entire problem, that is, the LCD of the top\u00a0and bottom denominators. The worksheets are \u2026 Simplifying Expressions Containing Complex Numbers Because i really is a radical, and because we do not want to leave radicals in the denominators of fractions, we do not want to leave any complex numbers in the denominators of fractions. Find the least common denominator (LCD) of all fractions appearing within the complex fraction. 4. Multiplying and dividing negative numbers. Table 1 $\\text{ Table 1} \\\\ \\begin{array}{ccc|c} \\hline Expression & & Work & Result \\\\\\hline \\red{i^ \\textbf{2}} & \u2026 Then simplify if possible. Example 1: to simplify$(1+i)^8$type (1+i)^8. Example 1: Simplify the complex fraction below. For this problem, we are going to use Method 1 only. The complex number calculator is able to calculate complex numbers when they are in their algebraic form. The first thing to do is arrive at a simple fraction both in the numerator and the denominator. Imaginary numbers are based on the mathematical number $$i$$. This website uses cookies to ensure you get the best experience. Amount of baking floor used to make a batch of cakes = 1/6 of a bag. Simplifying Fractions and Complex Fractions In the denominator, we will \u2026 Now that we have developed a solid foundation regarding what fractions are as well as some different types of fractions, we can now turn to application of the basic arithmetic operations (addition, subtraction, multiplication, and division) to fractions. Complex Fractions Worksheet & fractions good to start with perhaps then move on to more. . Before I can cancel anything off, I need to simplify that top parentheses, because it has a negative exponent on it. Finish off by canceling out common factors to get the final answer. Simplify complex fractions that contain several different mathematical operations In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. I can't cancel off, say, the a 's, because that a 4 isn't really on top. A complex fraction can be defined as a fraction in which the denominator and numerator or both contain fractions. In cases that involve simple numbers, addition and subtraction of fractions is easy \u2026 In complex fractions either or both the numerator and the denominator contain fractions or mixed numbers. After doing so, we can expect the problem to be reduced to a single fraction which can be simplified as usual. The following calculator can be used to simplify ANY expression with complex numbers. Apply the division rule of fractions by multiplying the numerator by the reciprocal or inverse of the denominator. Complex fractions aren't necessarily difficult to solve. When possible reduce, simplify and convert to mixed numbers, any final fraction results. This lesson is also about simplifying. There are two methods used to simplify such kind of fraction. This means we have to work a bit on the complex numerator. In this method of simplifying complex fractions, the following are the procedures: Generate a single fraction both in the denominator and the numerator. Multiplying numbers with different signs. Simplify complex fractions worksheets grade 7 305805 worksheet. Capacity of the chicken feeder = 9/10 of a cup of grains. Reduce all fractions when possible. If necessary, simplify the numerator and denominator into single fractions. The complex number online calculator, allows to perform many operations on complex numbers. There is another type of fraction called Complex Fraction, which we will see below. Why a negative times a negative is a positive. Another kind of fraction is called complex fraction, which is a fraction in which the numerator or the denominator contains a fraction. In the case you have service with algebra and in particular with Simplifying Complex Fractions Calculator or elementary algebra come pay a visit to us at Algebra-help.org. Obviously, this problem would require us to do that first before we perform division. (3/7)/9 is also a complex fraction with 3/7 and 9 as the numerator and denominator respectively. Generate a single fraction both in the denominator and the numerator. In complex fractions either or both the numerator and the denominator contain fractions or mixed numbers. Convert mixed numbers to improper fractions. Let\u2019s look at some of the key steps for each simplification method: In this method of simplifying complex fractions, the following are the procedures: This is the easiest method of simplifying complex fractions. The above expression is a complex fraction, therefore, change the division as multiplication and take reciprocal of the fraction in denominator. Then simplify if possible. Simplifying Complex Fractions When a \u201cnormal\u201d fraction contains fractions in either the numerator or denominator or both, then we consider it to be a complex fraction. 1. A chicken feeder can hold 9/10 of a cup of grains. If each piece the wire is 1/12 of the wire, how many pieces of the wire can Kelvin cut? For example. Here are the steps for this method: Kelvin cuts 3/4 meters of a wire into smaller pieces. The next step to do is to apply division rule by multiplying the numerator by the reciprocal of the denominator. We use cookies to give you the best experience on our website. Practice: Simplify complex fractions. Start by finding the Least Common Multiple of al the denominator in the complex fractions. A complex fraction is the same as a normal fraction, but has a fraction problem in the numerator and a fraction problem in the denominator. Simplify the fraction its lowest terms possible. Sometimes, we can take things too literally. $$i \\text { is defined to be } \\sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Simplifying complex fractions. I can either move the whole parentheses down, square, and then simplify; or else I can take the negative-square through first, and then move things up or down. Add the fractions in the numerator, and subtract the ones in the denominator. Some examples of complex fractions are: 6 7 3 3 4 5 8 x 2 5 6 To simplify a complex fraction, remember that the fraction bar means division. Given that 3/10 of a cup grains fills the feeder, therefore the number of scoops can be found by dividing 9/10 by 3/10. If the feeder is being filled by scoop that only holds 3/10 of a cup of grains. Looking at the denominators \\large{x} and \\large{x^2}, its LCD must be \\large{x^2} Multiply the top and bottom by this LCD. Since the LCD of 3y and 6y is just \\textbf{6y}, we will now multiply the complex numerator and denominator by this LCD. This type of fraction is also known as a compound fraction. We have got a great deal of high-quality reference information on subjects varying from geometry to math By using this website, you agree to our Cookie Policy. It is [\u2026] To do this take the lcm of the denominators of the fractions in the numerator and simplify it. Use this Complex Fractions Calculator to do math and add, subtract, multiply and divide complex fractions. The types of numerator and denominator determines the type of fraction. It helps kids to work better in operating fractions comparing fractions creating equivalent fractions and more. Free Algebra Tutorials! Algebra often involves simplifying expressions, but some expressions are more confusing to deal with than others. The worksheets are meant for the study of rational numbers, typically in 7th or 8th grade math (pre-algebra and algebra 1). Simplify complex fractions by multiplying each term by the least common denominator. All expression will be simplified as much as possible show help \u2193\u2193 examples \u2193\u2193 ^. Then, the total length of a wire is 3/4 meters. Here are the steps required for Adding and Subtracting Rational Expressions: Step 1: Simplify the rational expression in the numerator of the original problem by adding or subtracting the fractions as necessary. Simplify any Complex Fraction Enter any fraction into the two text boxes and we will show you all work using the LCM method and the \"flip\" method. simplify x2 + 4x \u2212 45 x2 + x \u2212 30 simplify x2 + 14x + 49 49 \u2212 x2 simplify 6 x \u2212 1 \u2212 3 x + 1 simplify 5x 6 + 3x 2 Home: Miscellaneous Equations: Operations with Fractions: Undefined Rational Expressions: Inequalities: \u2026 Start by multiplying the numerator of the complex fraction by the reciprocal of its denominator. Employ the division rule by multiplying the top of the fraction by the reciprocal of the bottom. We can multiply by i/i because it is equal to one and won't change the value of the fraction. The problem requires you to apply the FOIL method (multiplication of two binomials) and a simple factorization of trinomial. The LCD in the numerator is (x + 3)(x \u2013 1). There are two methods used to simplify complex fractions. 3. Calculate the possible value of x in the following complex fraction. This is the currently selected item. What is an imaginary number anyway? \u2026 Start by converting the top and bottom into improper fractions: Find the reciprocal of the denominator and change the operator: Multiply the numerators and denominators separately: The numerator and denominator of the fraction have a common factor number 9, simplify the fraction to the lowest terms possible. Analysis of this question results in complex fractions: The problem is solved by finding the reciprocal of the denominator, and in this case, it is 3/10. Remember to distribute the negative (or subtraction) sign between the two fractions. Create a single fraction in the numerator and denominator. Imaginary numbers are based on the mathematical number $$i$$. Complex Fractions \u2013 Explanation & Examples. To see extra written explanation next to the work, click on the \"verbose mode\" here. Otherwise, check your browser settings to turn cookies off or discontinue using the site. This algebra video tutorial explains how to simplify complex fractions especially those with variables and exponents - positive and negative exponents. 5. We simplified complex fractions by rewriting them as division problems. A bakery uses 1/6 of a bag of baking flour in a batch of cakes. You need to see someone explaining the \u2026 A fraction is made up of two parts: a numerator and a denominator; the number above the line is the numerator and the number below the line is the denominator. Step 1: Simplify the rational expression in the numerator of the original problem by subtracting the fractions. Ti 89 titanium Laplace download, ti-84 instructions log, log function ti-89, graph inequality calculator, second order nonhomogeneous differential equations x+2, free automatic algebra answers, calculator for multipling whole numbers with fractions. Complex Numbers''Simplifying roots of negative numbers video Khan Academy May 13th, 2018 - What are the complex numbers We re asked to simplify the principal square root of negative because we have a negative 52 here inside of the radical' 'miL2872X Ch07 469 550 9 29 06 18 13pm Page 469 IA May 12th, 2018 - Radicals And Complex Numbers 469 In This Chapterwe Study Radical Expressions This \u2026 There are two methods. Multiply the both the numerator and denominator of the complex fraction by this L.C.M. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step. Are coffee beans even chewable? Simplifying Fractions and Complex Fractions. (3/4)/(9/10) is another complex fraction with 3/4 as the numerator and 9/10 as the denominator. Come to Mathfraction.com and read and learn about matrices, long division and many other algebra subject areas . 2. The complex symbol notes i. Worksheets for complex fractions Create here an unlimited supply of worksheets for simplifying complex fractions \u2014 fractions where the numerator, the denominator, or both are fractions/mixed numbers. working... Polynomial \u2026 Video Lesson . Example 2: to simplify$\\dfrac{2+3i}{2-3i}\\$ type (2+3i)/(2-3i). The first thing to do is arrive at a simple fraction both in the numerator and the denominator. Complex numbers involve the quantity known as i , an \u201cimaginary\u201d number with the property i = \u221a\u22121.If you have to simply an expression involving a complex number, it might seem daunting, but it\u2019s quite a simple process once you learn the basic rules. Video Tutorial on Simplifying Imaginary Numbers. Multiply this LCD to the numerator and denominator of the complex fraction.\n\nDann Of Thursday Figure, Nationally Registered Certified Medical Assistant Certification, Ncr Corporation Hyderabad, Pco Licence Application Progress Check, Class K Fire Extinguisher Specification, Excel Dynamic Array Reference, Best Estate Agents Plymouth,", "date": "2021-10-25 19:06:09", "meta": {"domain": "zynotsoft.com", "url": "https://zynotsoft.com/gene-tunney-ebib/b51e80-simplifying-complex-numbers-fractions", "openwebmath_score": 0.8443188667297363, "openwebmath_perplexity": 656.9784341071735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9674102580527665, "lm_q2_score": 0.8918110418436166, "lm_q1q2_score": 0.8627471501242396}}
{"url": "https://math.stackexchange.com/questions/1912480/number-of-points-of-intersections-no-of-parts-of-chords-inside-circle", "text": "# Number of points of intersections, no of parts of chords inside circle\n\n$n$ points ($n>1$) are taken on the circumference of a circle. Through any two of them a chord is drawn. No three chords intersect at one point inside the circle. i) Find how many points of intersections of these chords are inside the circle? ii) Find how many parts do these chords divide the circle?\n\nI know one solution is to make a graph and use Euler's formula $v-e+f=2$. But that idea I would have never come up with. Is there any other way to approach this?\n\n\u2022 The number of separate parts inside the circle is a sequence that starts $1,2,4,8,16,\\ldots$. Can you guess the next term? \u2013\u00a0Arthur Sep 2 '16 at 20:14\n\u2022 32. How do you prove formally that the sequence gives the parts inside the circle? \u2013\u00a0Amrita Sep 2 '16 at 20:52\n\u2022 It's actually 31. I was deliberately trying to throw you off. I don't know how to prove it, but Euler characteristic seems like a not-too-bad idea. \u2013\u00a0Arthur Sep 2 '16 at 20:54\n\u2022 The number of points of intersection inside the circle is $\\binom n4$, since each set of four points on the circle determines a pair of intersecting chords. \u2013\u00a0bof Aug 3 '19 at 12:08\n\nI do not think there is a better approach to find the number of parts (part ii) comparing to the Euler's formula. But to use it one should first find the number of vertices and edges, and for this task I do not see how the Euler's formula can help.\n\nFor simplicity we disregard the circle arcs and will consider only the graph consisting of line segments connecting the points.\n\nLet $$s$$ be a segment connecting two points of the circle. The points split the circle in two parts. Let the number of points in one of the parts be $$k$$. Then the number of points in the other part is $$(n-k-2)$$. Every segment which connects a point lying in one part with a point lying in the other part will intersect the segment $$s$$, whereas any segment connecting two points lying in the same part will not intersect $$s$$. Therefore the overall number of intersection points of the segment $$s$$ with other segments is $$k(n-2-k)$$ and the overall number of intersection points on all segments drawn from each of the considered circle points is: $$v''_n=\\sum_{k=0}^{n-2}k(n-2-k)=\\frac{(n-1)(n-2)(n-3)}6=\\binom{n-1}3\\tag1$$ and the overall number of the intersection points inside the circle is $$v'_n=\\frac n4 v''_n=\\binom n4,\\tag2$$ where factor 4 in the denominator accounts for the fact that while summing over all $$n$$ circle points every internal intersection point is counted 4 times.\n\nTo compute the overall number of vertices we shall add the number of the circle points to the obtained result: $$v_n=v'_n+n=\\frac{n(n+1)(n^2-7n+18)}{24}.\\tag3$$\n\nThe number of edges can be computed similarly. Observe that the segment $$s$$ is split by $$n_s$$ internal intersection points into $$n_s+1$$ edges therefore: $$e''_n=\\sum_{k=0}^{n-2}[k(n-2-k)+1]=\\binom{n-1}3+\\binom{n-1}1=\\frac{(n-1)(n^2-5n+12)}6,\\tag4$$ and the overall number of the edges inside the circle is $$e'_n=\\frac n2 e''_n=\\frac{n(n-1)(n^2-5n+12)}{12},\\tag5$$ where factor 2 in the denominator accounts for the fact that while summing over all $$n$$ circle points every chord is counted 2 times.\n\nTo compute the overall number of edges we shall add the number of arcs connecting the circle points to the obtained result: $$e_n=e'_n+n=\\frac{n^2(n^2-6n+17)}{12}.\\tag6$$ Finally we use the Euler's formula for computing the number of parts inside the circle: $$f_n=e_n-v_n+1=\\frac{n^4-6n^3+23n^2-18n+24}{24}.\\tag7$$\n\nAnd indeed the latter formula describes - as it should - the OEIS sequence A000127.\n\n\u2022 $v'_n=\\binom n4$ because a set of four points on the circle corresponds to a pair of intersecting chords, namely, the diagonals of the inscribed quadrilateral determined by those four points. \u2013\u00a0bof Aug 3 '19 at 12:19\n\u2022 Yes of course. I have however decided to give another derivation to count both vertices and edges in a similar way. \u2013\u00a0user Aug 3 '19 at 16:06\n\u2022 You can \"count both vertices and edges in a similar way\" but it seems more efficient to note that when you have counted one you have counted the other, since $$e'_n=\\binom n2+2v'_n.$$ \u2013\u00a0bof Aug 3 '19 at 21:39\n\u2022 If you mean that the identity $e'_n=\\binom n2+2v'_n$ follows from the intermediate results (4) and (5), I'm sure it does, but since $e'_n=\\binom n2+2v'_n$ is easy to see directly, it's not clear what those intermediate results are needed for. \u2013\u00a0bof Aug 4 '19 at 11:29\n\nThe number of vertices inside the circle is $$\\binom n4$$ since each such vertex is determined by a pair of intersecting chords, which are the diagonals of an inscribed quadrilateral determined by four points on the circle.\n\nThe number of edges inside the circle is $$\\binom n2+2\\binom n4$$ since there are $$\\binom n2$$ chords, and the number of edges on each chord is equal to $$1$$ plus the number of interior vertices on that chord, and each of the $$\\binom n4$$ interior vertices lies on two chords.\n\nThere are $$n$$ vertices and $$n$$ edges on the circle, so the total number of vertices is $$V=n+\\binom n4$$ and the total number of edges is $$E=n+\\binom n2+2\\binom n4.$$ By Euler's formula, the number of faces inside the circle is $$F-1=E-V+1=\\binom n2+\\binom n4+1=\\frac{n(n-1)(n^2-5n+18)}{24}+1=\\frac{n^4-6n^3+23n^2-18n+24}{24}.$$\n\nIf $$U$$ is the unit disc centered at the origin consider $$n$$ chords drawn through the interior of $$U$$ such that no two chords are parallel and no three chords intersect at the same point. The arrangement graph $$G$$ induced by the discs and the chords has a vertex for each intersection point in the interior of $$U$$ and $$2$$ vertices for each chord incident to the boundary of $$U.$$ Naturally $$G$$ has an edge for each arc directly connecting two intersection points.\n\nA point $$p$$ of $$G$$ is said to be incident to the interior of $$G$$ if $$p$$ is in the interior of $$U.$$ Similarly $$p$$ is said be incident to the boundary of $$G$$ if $$p$$ is on the boundary of $$U.$$ In this sense and without loss of generality I speak of the boundary of $$U$$ as the boundary of $$G$$ and the interior of $$U$$ as that of $$G.$$ I denote the number of arcs incident to $$p$$ by $$v(p),$$ called the valency of $$p.$$ It is straightforward to see that if $$p$$ is incident to the interior of $$G$$ then $$p$$ is $$4-$$valent. Likewise if $$p$$ is incident to the interior of $$G$$ then $$p$$ is $$3-$$ valent. From this we see that the maximum valency of $$G$$ is $$4$$ and the minimum degree of $$G$$ is $$3.$$ By construction $$G$$ is planar and $$3-$$ connected.\n\nlemma 1: $$G$$ has $$N(N+3)/2, N(N+2)$$ and $$(N^2+N+4)/2$$ points, arcs and faces respectively.\n\nEach chord has two distinct points incident to the boundary of $$G.$$ Being there are $$N$$ such chords it is seen there are $$2N$$ points along the boundary of $$G.$$ Observe there are $$N(N-1)/2$$ points in the interior of $$G.$$ Indeed the number of points in $$G$$ is equal to sum of those points in the interior and along the boundary which is $$2N+N(N-1)/2=N(N+3)/2.$$ The number of arcs in $$G$$ follows from the handshaking lemma. In particular twice the number of arcs is equal to the sum of all valances. The later is equal to $$[3N+4N(N-1)/2]/2=N(N+2)$$ The number of faces in $$G$$ follows from Euler's characteristic formula for planar graphs.$$2-N(N+3)/2+N(N+2)=(N^2+N+4)/2$$ This establishes the lemma.\n\n\u2022 I assumed no chords could be parallel \u2013\u00a0Antonio Hernandez Maquivar Sep 2 '16 at 21:15\n\u2022 In the question n points were given on circumference, not s lines. Here s is not given. Is s = ${n \\choose 2}$ \u2013\u00a0Amrita Sep 2 '16 at 22:01", "date": "2020-05-28 13:36:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1912480/number-of-points-of-intersections-no-of-parts-of-chords-inside-circle", "openwebmath_score": 0.7246909141540527, "openwebmath_perplexity": 120.947880077437, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9929882037884926, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8627347449990914}}
{"url": "https://www.physicsforums.com/threads/electric-potential-inside-a-shell-of-charge.947834/", "text": "# Electric potential inside a shell of charge\n\n## Homework Statement\n\nQ1: There are two concentric spherical shells with radii $R_1$ and $R_2$ and charges $q_1$ and $q_2$ uniformly distributed across their surfaces. What is the electric potential at the center of the shells?\n\nQ2: There is an infinitely long hollow cylinder of linear charge density $\\lambda$ and radius $R$. What is the potential difference $\\Delta V$ between the surface of the shell and a radius $R'$ inside the cylinder?\n\n## Homework Equations\n\n$\\vec E = -\\nabla V$\n$\\oint \\vec E \\cdot d\\vec A = \\frac{Q\\text{encl}}{\\epsilon_0}$ (Gauss's Law)\n$V=\\frac{q}{4\\pi \\epsilon_0 r}$\n\n## The Attempt at a Solution\n\nStarting with Q2, Gauss's Law using a cylinder as the Gaussian surface shows there is no enclosed charge; $\\vec E = \\vec 0$. Because $\\vec E = -\\nabla V$, one can conclude that $V=0$ between $R'$ to $R$. This is the given (and found) answer for Q2.\n\nIn a similar way, there is no enclosable charge for all points inside the two spherical shells in Q1. By the same logic as Q2, it would seem $\\vec E=0=-\\nabla V$ and there would be no electric potential at the center of the shells.\n\nHowever, Gauss's Law cannot be applied to a point or line since the Gaussian surface has area $A=0$ and Gauss's Law reduces to $0=0$. Therefore, one cannot find the electric field at the center of the sphere, and $\\vec E = -\\nabla V$ cannot be used.\n\nSince the center of the shells are at a constant distance $R_1$ and $R_2$, the electric potential can be found by:\n\n$V=\\frac{q_1}{4\\pi \\epsilon_0 R_1} + \\frac{q_2}{4\\pi \\epsilon_0 R_2}$\n\nwhich is the given answer for Q1. (One needs to integrate the charge density across a spherical area, which ultimately reduces to the answer above)\n\nThis result (i.e. textbook answers) seems to show some weird results:\n\u2022 The electric field and potential are zero for all positions inside a closed area of charge and nonzero at the symmetrical center or axis.\n\u2022 The graphs of the field magnitude and electric potential are discontinuous at the center.\n\u2022 If this is true, the electric field vector there has no defined direction...? (or is undefined since the equation is discontinuous)\n\u2022 In Q2, the electric field and potential should be nonzero along the axis of the cylinder and zero for all spaces between the axis and the cylinder wall.\n\nI'm somewhat confused because the two questions seem to contradict each other. Is my logic correct in interpreting the answers?\n\nRelated Introductory Physics Homework Help News on Phys.org\nDoc Al\nMentor\nOne note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)\n\nOne note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)\nIntegrating $\\vec E \\cdot d\\vec r$ where $\\vec E = \\vec 0$ to find the potential at a single point results in $V = 0+C$. Then the nonzero potential found at the center is $C$ and is constant across the space inside the shells...\n\nMakes much more sense, thanks!\n\nrude man\nHomework Helper\nGold Member\nOr perhaps:In Q2: the E field just outside the surface is \u03c3/\u03b5 where \u03c3 is surface charge density (related to \u03bb obviously). The E field just below the surface is zero. Both by Gauss. Since potential is the integral of the E field over distance, and the distance \u2192 zero, therefore there is no change in potential between the outside & inside surfaces.\n\nQ1: Can also do this by superposition theorem:\nPotential of shell 1 with q2=0 is kq1/R1.\nPotential of shell 2 with q1=0 is kq2/R2.\nTotal potential is sum of above potentials.", "date": "2020-02-24 11:18:44", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/electric-potential-inside-a-shell-of-charge.947834/", "openwebmath_score": 0.8931763172149658, "openwebmath_perplexity": 321.83955923100143, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147185726375, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8627308067658345}}
{"url": "https://math.stackexchange.com/questions/1602392/proving-that-sqrta-12-cdots-sqrta-n2-sqrta-12-cdotsa-n2-u", "text": "# Proving that $\\sqrt{a_1^2} +\\cdots+ \\sqrt{a_n^2} > \\sqrt{a_1^2 +\\cdots+a_n^2}$ using Pythagoras\n\nI think I have a proof using Pythagoras for $\\sqrt{a_1^2} + \\sqrt{a_2^2} > \\sqrt{a_1^2 + a_2^2}$.\n\nI'm interested in whether there's a way to use that proof with Pythagoras to prove the general $a_n$ case (for this, hints are appreciated rather than complete proofs), and also in other ways (algebraic, geometric, number theoric, calculusic...anything) that you might know or come up with to prove the general case (for those, either hints or complete proofs are great, up to you).\n\nLemma:\n\nLet positive (edited) real numbers $a_1, a_2$ be the legs of a right triangle.\n\nThen $\\sqrt{a_1^2 + a_2^2}$ is the length of the hypothenuse of that triangle.\n\nAnd $\\sqrt{a_1^2} + \\sqrt{a_2^2}$ is the sum of the length of the two legs.\n\nBy the triangular inequality, we know that the length of the hypothenuse has to be less than the length of the sum of the two legs.\n\nTherefore, for any real numbers $a_1, a_2$, $\\sqrt{a_1^2} + \\sqrt{a_2^2} > \\sqrt{a_1^2 + a_2^2}$.\n\nI'm stuck here...I was thinking of comparing pairs of elements from each side of the expression using my lemma, but it doesn't seem possible to \"extract\" pairs of elements from under $\\sqrt{a_1^2 + a_2^2 +...+a_n^2}$. I also thought about summing all elements but $a_1$ into a single number and using my lemma on those simplified expressions, but I run into the same problem.\n\n\u2022 Why do you need Pythagoras's theorem? Just square both sides of the inequality and compare. \u2013\u00a0Leo Jan 6 '16 at 19:18\n\u2022 When you say \"by the triangular inequality\", this is a bit circular - the identity you're trying to prove is the triangle inequality for the Euclidean distance. \u2013\u00a0\u03c0r8 Jan 6 '16 at 19:20\n\u2022 You have the inequality backwards. \u2013\u00a0zhw. Jan 6 '16 at 19:20\n\u2022 @zhw. It has been edited. \u2013\u00a0user236182 Jan 6 '16 at 19:26\n\u2022 @zhw. yes thank you, I mentioned it in my edit, not sure where it shows though \u2013\u00a0jeremy radcliff Jan 6 '16 at 19:27\n\nIt's very easy using induction :\n\nYou proved the case when $n=2$ .\n\nNow assume you know it for $n$ and want prove it for $n+1$ :\n\n$$\\sqrt{a_1^2}+\\ldots+\\sqrt{a_n^2}+\\sqrt{a_{n+1}^2} >\\sqrt{a_1^2+a_2^2+\\ldots+a_n^2}+\\sqrt{a_{n+1}^2}$$\n\nNow use also the $n=2$ case to finnish it :\n\n$$\\sqrt{a_1^2+a_2^2+\\ldots+a_n^2}+\\sqrt{a_{n+1}^2}>\\sqrt{\\left ( \\sqrt{a_1^2+a_2^2+\\ldots+a_n^2} \\right )^2}+\\sqrt{a_{n+1}^2}>\\sqrt{\\left ( \\sqrt{a_1^2+a_2^2+\\ldots+a_n^2} \\right )^2+a_{n+1}^2}=\\sqrt{a_1^2+a_2^2+\\ldots+a_n^2+a_{n+1}^2}$$ as wanted .\n\nYou can translate this into a geometric proof : consider an $n$-dimensional box with the sides $a_1,a_2,\\ldots,a_n$ . The diagonal of the box is $\\sqrt{a_1^2+a_2^2+\\ldots+a_n^2}$ and now you can repeatedly apply the triangle's inequality to get your inequality (this is equivalent with the induction proof above )\n\n\u2022 I like the geometric intuition you give a lot, it makes the algebra more meaningful to me. \u2013\u00a0jeremy radcliff Jan 6 '16 at 19:50\n\u2022 @jeremyradcliff But can you imagine an $n$-dimensional box (if $n\\ge 4$)? \u2013\u00a0user236182 Jan 6 '16 at 20:41\n\u2022 @user236182, no of course, I'm stuck at 3 dimensions but the step of the diagonal of a 3d box gives me 2 steps (with the 2d case) to extrapolate about higher dimensions and makes the induction principle in this context a lot more meaningful to me. \u2013\u00a0jeremy radcliff Jan 6 '16 at 20:54\n\nHere is a geometrical proof. Consider a square with side length $L = |a_1|+|a_2| + \\ldots + |a_n|$.\n\n$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$\n\nComparing the area of the whole square to the area of the small squares contained within it we see that\n\n$$(|a_1| + |a_2| + \\ldots + |a_n|)^2 \\geq a_1^2 + a_2^2 + \\ldots + a_n^2$$\n\nand by taking the square root we get the desired inequality. Equality can only happen when one of the small squares covers the whole square which can only happen when atleast $n-1$ of the $a_i$'s are zero.\n\n\u2022 Very nice, thank you. I'm a big fan of proofs without words. \u2013\u00a0jeremy radcliff Jan 6 '16 at 19:54\n\nBoth sides of the inequality are non-negative (for all $a_i\\in\\mathbb R$), therefore the following equivalences hold:\n\n$$\\sqrt{a_1^2} + \\sqrt{a_2^2} +\\cdots + \\sqrt{a_n^2} \\ge \\sqrt{a_1^2 + a_2^2 +\u2026+a_n^2}$$\n\n$$\\iff \\left(\\sqrt{a_1^2} + \\sqrt{a_2^2} +\\cdots + \\sqrt{a_n^2}\\right)^2 \\ge \\left(\\sqrt{a_1^2 + a_2^2 +\u2026+a_n^2}\\right)^2$$\n\n$$\\iff a_1^2+a_2^2+\\cdots+a_n^2+2\\sum_{i=1}^n \\sum_{j>i}^n|a_ia_j|\\ge a_1^2+a_2^2+\\cdots+a_n^2$$\n\n$$\\iff 2\\sum_{i=1}^n \\sum_{j>i}^n|a_ia_j|\\ge 0,$$\n\nwhich is true, with equality if and only if at least $n-1$ of $a_1,a_2,\\ldots,a_n$ are equal to $0$.\n\n\u2022 @jeremyradcliff It's multinomial expansion. \u2013\u00a0user236182 Jan 6 '16 at 19:58\n\u2022 Thank you. I erased my comment because I realized it was wrong. But I just worked it out. For the record, I was aking whether the summation symbols were from the binomial expansion. \u2013\u00a0jeremy radcliff Jan 6 '16 at 19:59", "date": "2019-08-26 06:02:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1602392/proving-that-sqrta-12-cdots-sqrta-n2-sqrta-12-cdotsa-n2-u", "openwebmath_score": 0.8972159624099731, "openwebmath_perplexity": 380.3546736633876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147145755001, "lm_q2_score": 0.8872045877523147, "lm_q1q2_score": 0.8627307959692413}}
{"url": "https://puzzlingthroughmed.com/riddler-180222/", "text": "22 February 2022\n\n# Fiscal fibrillations (18/02/22 Riddler Classic)\n\nOriginal problem\n\nIf we start with $n$ coins, the first toss will reduce the number to $k$ which is distributed binomially. If $p_n$ is the probability that there exists a lucky coin if we start with $n$ coins, then a recurrence relation exists:$$p_n=2^{-n}\\sum_{k=0}^n{n \\choose k}p_k$$where $p_0=0$ and $p_1=1$ are the base cases. Isolating for $p_n$ gives $$p_n=\\frac{1}{2^n-1}\\sum_{k=0}^{n-1} {n \\choose k}p_k$$\n\nPlotting $p_n$ up to $n=10$ generates the following graph which seems to imply $p_n$ converges.\n\nHowever, visual inspection of $p_n$ values up to $n=1000$ shows a clear oscillatory tendency\n\nwhich can also be plotted on a linear-log scale:\n\nThe periodic nature on a log-scale makes sense. Given a large enough $n$, we expect the first toss to almost-exactly halve the number of coins, and so $p_{2n}\\approx p_{n}$, with the approximation being better as $n$ increases by the law of large numbers. However, questions remain:\n\n\u2022 What values does $p_n$ oscillate around? On the graph, it looks to be just under 0.72135.\n\u2022 Is the shape actually a sinusoid? If so, why?\n\u2022 Does $p_n$ converge?\n\nAt the moment, I can only answer the first dot point.\n\n### $p_n$ oscillates around $\\frac{1}{2\\ln(2)}\\approx0.721348$\n\nTo arrive at this value analytically, we want to reframe how we express $p_n$.\n\nSuppose we modify the game slightly such that we toss all $n$ starting coins with every toss, even coins that came up tails; we just make those coins ineligible to be the lucky coin.\n\nNow suppose a lucky coin is crowned after exactly $(t+1)$ tosses. This happens if and only if both of the following occur:\n\n\u2022 After $t$ tosses, there were $k\\geq 2$ coins with clean histories of only heads. Each of the other $(n-k)$ coins does not have a clean history. This happens with probability $${n\\choose k}\\cdot (2^{-t})^k \\cdot (1-2^{-t})^{n-k}$$\n\u2022 At the $(t+1)$th toss, exactly one out of the $k$ candidate coins come up heads while the rest are tails. This occurs with probability $$2^{-k}\\cdot k$$\n\nHence the probability that a lucky coin is crowned on the $(t+1)$th toss is the product of the above two probabilites, summed across all $k\\in[2,n]$.\n\n\\begin{align*}\\sum_{k=2}^n{n\\choose k}(2^{-t})^k(1-2^{-t})^{n-k}2^{-k}k&=\\sum_{k=2}^n{n\\choose k}(2^{-(t+1)})^k(1-2^{-t})^{n-k}k\\\\&=\\sum_{k=0}^n{n\\choose k}(2^{-(t+1)})^k(1-2^{-t})^{n-k}k\\;\\;\\;-n2^{-(t+1)} (1-2^{-t})^{n-1}\\\\&=n2^{-(t+1)} (1-2^{-(t+1)})^{n-1}-n2^{-(t+1)} (1-2^{-t})^{n-1}\\\\&=\\frac{n}{2}2^{-t}\\Big[(1-{2^{-(t+1)}})^{n-1}-(1-2^{-t})^{n-1}\\Big]\\end{align*}\n\nNote that in the third line, we used the identity $\\sum_{k=0}^n {n\\choose k}x^ky^{n-k}k=nx(x+y)^{n-1}$. Now, summing the above over all possible values of $t\\in[0,\\infty)$ gives $p_n$\n\n\\begin{align*}p_n&=\\sum_{t=0}^\\infty\\frac{n}{2}2^{-t}\\Big[(1-{2^{-(t+1)}})^{n-1}-(1-2^{-t})^{n-1}\\Big]\\\\&=\\frac{n}{2}\\sum_{t=0}^\\infty 2^{-t}(1-2^{-(t+1)})^{n-1}-\\frac{n}{2}\\sum_{t=0}^\\infty 2^{-t}(1-2^{-t})^{n-1}\\\\&=\\frac{n}{2}\\sum_{t=1}^\\infty 2^{-(t-1)}(1-2^{-t})^{n-1}-\\frac{n}{2}\\sum_{t=0}^\\infty 2^{-t}(1-2^{-t})^{n-1}\\\\&=\\frac{n}{2}\\sum_{t=0}^\\infty 2^{-t}(1-2^{-t})^{n-1}\\end{align*}\n\nWe can approximate this sum by an integral, which luckily for us is actually extremely easy to perform.\n\n\\begin{align*}p_n&\\approx \\frac{n}{2}\\int_0^\\infty 2^{-t}(1-2^{-t})^{n-1}dt\\\\&=\\frac{n}{2}\\int_0^\\infty e^{-t\\ln(2)}(1-e^{-t\\ln(2)})^{n-1}dt\\\\&=\\frac{1}{2\\ln(2)}\\Big[(1-e^{-t\\ln(2)})^n\\Big]^\\infty_0\\\\&=\\color{red}{\\boxed{\\frac{1}{2\\ln(2)}}}\\\\&\\approx 0.7213475\\end{align*}\n\nThis doesn't actually tell us if the sequence $\\{p_n\\}$ actually converges, however; it is merely an approximation for any individual $p_n$. The sinusoidal behaviour also isn't particularly apparent. Without knowing the answers to these, I'm reluctant to make a call on what $p_{10^6}$ might be.", "date": "2022-12-03 09:45:26", "meta": {"domain": "puzzlingthroughmed.com", "url": "https://puzzlingthroughmed.com/riddler-180222/", "openwebmath_score": 0.8299873471260071, "openwebmath_perplexity": 732.7456846302146, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795106521339, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8626963266914695}}
{"url": "https://math.stackexchange.com/questions/2340629/digit-sum-of-a-number", "text": "# Digit-Sum of a Number.\n\nI got a question recently, and I am unable to solve it.\n\nFind all natural numbers $N$, With sum of digits $S(N)$, where $N=2\\{S(N)\\}^2$\n\nI know that $9|N-S(N)$, and since N is twice a square, it must end in $0,2,8$. But I do not know where to go from here. Can anyone help?\n\nThe Official Solution from the Organization\n\nWe use the fact that $9|n\u2212S(n)$ for every natural number $n$. Hence $S(n)(2S(n)\u22121)$ is divisible by $9$. Since $S(n)$ and $2S(n)\u22121$ are relatively prime, it follows that $9$ divides either $S(n)$ or $2S(n)\u22121$, but not both. We also observe that the number of digits of $n$ cannot exceed $4$. If $n$ has $k$ digits, then $n\u226510k\u22121$ and $$2S(n)^2\u22642\\cdot(9k)^2=162k^2$$ If $k\u22656$, we see that $$2S(n)^2\u2264162k^2<5^4k^2<10^{k\u22121}\u2264n$$ If $k = 5$, we have $$2S(n)^2\u2264162\\cdot25=4150<10^4\u2264n$$ Therefore $n \u2264 4$ and $S(n) \u2264 36$. If $9|S(n)$, then $S(n) = 9,18,27,36$. We see that $2S(n)^2$ is respectively equal to $162, 648, 1458, 2592$. Only $162$ and $648$ satisfy $n = 2S(n)^2$. If $9|(2S(n)\u22121)$, then $2S(n) = 9k+1$. Only $k = 1,3,5,7$ give integer values for $S(n)$. In these cases $2S(n)^2 = 50,392,1058,2048$. Here again $50$ and $392$ give $n = 2S(n)^2$. Thus the only natural numbers wth the property $n = 2S(n)^2$ are $50,162,392,648$.\n\n\u2022 For what it's worth, $S(N) \\mid N$, so that $9 \\mid N$. I am not too sure though if it will narrow down things that much. Jun 29 '17 at 14:09\n\u2022 @JoseArnaldoBebitaDris Do you mean to say that 9 divides N? Because I can think of places where it wouldn't be true. One N I have found is 50, and it is not divisible by 9. Jun 29 '17 at 14:12\n\u2022 Yes because $N = 2\\cdot{S(N)}^2$, which means that $N/S(N) = 2\\cdot{S(N)}$ is an integer. In other words, $S(N)$ divides $N$. Jun 29 '17 at 14:13\n\u2022 If that's the case, I would doubt $9 \\mid (N - S(N))$. Jun 29 '17 at 14:15\n\u2022 @JoseArnaldoBebitaDris No, that's true. It's a theorem, I think. Jun 29 '17 at 14:16\n\n## 2 Answers\n\nWe want to find an upper bound for $N$ for which it is possible that $N = 2 S(N)^2$. The upper bound that I use in the following is very rough, but I think still just good enough to be able to do the subsequent calculations by hand with a little persistence. Look to the answer of @OscarLanzi for a much cleverer upper bound.\n\nAn upper bound for $S(N)$ is $$S(N) \\leq 9\\cdot \\left(\\log_{10}(N)+1\\right).$$ We can use this upper bound to see that $$2 S(N)^2 - N \\leq 2 \\cdot \\left(9\\cdot \\left(\\log_{10}(N)+1\\right) \\right)^2 - N.$$ For $N = 10^4$ we see that the right hand side of this inequality is $4050 - 10^4 = -5950 < 0$ and by calculating derivatives we can see fairly easily that this expression will only decrease for $N > 10^4$. This means that if we want to find $N$ such that $N = 2 S(N)^2$ we only have to check numbers up to $10^4$, this is a very rough upper bound.\n\nNote that indeed that $N \\equiv S(N) \\pmod{9}$ so $N$ must satisfy $$N - 2 N^2 \\equiv 0 \\pmod{9},$$ we deduce that $N \\equiv 0 \\pmod{9}$ or $N \\equiv 5 \\pmod{9}$. We see that $N$ must be twice a square and less than $10^4$, so $2S(N)^2 \\leq 10^4$, which means that $S(N) \\leq \\sqrt{\\frac{1}{2}\\cdot{10^4}} \\cong 70.7$. So $S(N)$ must be a number less than or equal to $70$ that is either $0$ or $5$ modulo $9$. There are only fifteen such positive numbers, the list is $$\\left\\{5,9,14,18,23,27,32,36,41,45,50,54,59,63,68\\right\\}.$$ These can all be checked separately, the list of these elements squared twice is $$\\left\\{\\mathbf{50},\\mathbf{162},\\mathbf{392},\\mathbf{648},1058,1458,2048,2592,3362,4050,5000,5832,6962,7938,9248\\right\\}.$$ Sort of surprisingly the first four all fit. These are the only ones, so the final answer is $$N \\in \\left\\{50, 162, 392, 648\\right\\}.$$ The amount of checking to be done can be reduced by getting a better upper bound.\n\n\u2022 I can understand what you did there, with the upper bound, but am unable to understand how you calculated the upper bound. Could you please elaborate? For example, how did logarithms come into play? Jun 29 '17 at 15:06\n\u2022 @MalayTheDynamo Suppose that a number $n$ consists of $3$ digits, say for example $n = 354$. Then we see that $100 \\leq n \\leq 1000$, and if we take the logarithm we see that $2 \\leq \\log_{10}(n) \\leq 3$. Generally we can deduce that $\\log_{10}(n) + 1$ is always greater or equal to the amount of digits of $n$. The maximal digitsum $S(n)$ of a number with $k$ digits is $9 k$, namely if all digits are $9$. Combining these we find that an upper bound for $S(n)$ is $$9 \\cdot \\left(\\log_{10}(n) + 1\\right).$$ Jun 29 '17 at 15:14\n\u2022 Oh, thanks. It was perfect. Jun 29 '17 at 15:15\n\n@Pjotr5 identified the solution but asked for a sharper bound. Here all four-digit candidates are eliminated through a descent process leaving only the numbers in his list up to and including $648$.\n\nFirst off, five-digit numbers give a digit sum no greater than $45$ whose square, doubled, is less than five digits ($2\u00d745^2=4050$). Contradiction. Ditto for more than five digits.\n\nFour-digit numbers give digit sums no more than $36$ whose square, doubled, is $2592$. So a four-digit solution is at most $2592$. But wait, there's more (or less, if you will). If the first digit of the proposed four-digit solution is no more than $2$ the sum of digits is now no more than $29$ giving a bound of $2\u00d729^2=1682$, and then if the first digit has to be $1$ the sum of digits is at most $28$ giving $N\\le 1568$.\n\nThe four-digit bound can still be lowered more. The bound of $1568$ derived above means the first two digits are no more than $15$ and can't sum to more than $6$. Then the sum of the four digits is no longer bounded only by $28$ but now bounded by $24$. Twice the square of that is $1152$ so now $N \\le 1152$.\n\nYou know the drill by now. If $N$ has four digits and is less than or equal to $1152$ then the sum of digits is no more than $20$. Twice the square of that is only $800$, less than four digits, so no four-digit candidates are left!\n\n\u2022 Good one :). Together our answers certainly make it possible to solve this problem on a contest. Jun 29 '17 at 15:48\n\u2022 For getting the actual solution @Pjotr5 beat me to it, and in fact I reference that answer. Jun 30 '17 at 9:55\n\u2022 @OscarLanzi Thank you :). I'll edit my answer to reference that your upper bound is much better to make the answer more complete. Jun 30 '17 at 10:10", "date": "2021-09-28 07:16:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2340629/digit-sum-of-a-number", "openwebmath_score": 0.8244609236717224, "openwebmath_perplexity": 116.35390035207026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795098861571, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8626963211656684}}
{"url": "https://math.stackexchange.com/questions/2314829/solve-for-x-in-cos2-sin-1-x-0", "text": "# Solve for $x$ in $\\cos(2 \\sin ^{-1}(- x)) = 0$\n\nSolve $$\\cos(2 \\sin ^{-1}(- x)) = 0$$\n\nI get the answer $\\frac{-1}{ \\sqrt2}$\n\nBy solving like this\n\n\\begin{align}2\\sin ^{-1} (-x )&= \\cos ^{-1} 0\\\\ 2\\sin^{-1}(- x) &= \\frac\\pi2\\\\ \\sin^{-1}(- x) &= \\frac\\pi4\\\\ -x &= \\sin\\left(\\frac\\pi4\\right)\\end{align}\n\nThus $x =\\frac{-1}{\\sqrt2}$\n\nBut correct answer is $\\pm\\frac{1}{\\sqrt2}$\n\nWhere am I going wrong?\n\n\u2022 You didn't take multiple values into account. Neither cos nor sin are one to to one. \u2013\u00a0fleablood Jun 8 '17 at 15:34\n\nOkay so expand the double angle to get $\\cos^2(\\sin^{-1}(-x))-\\sin^2(\\sin^{-1}(-x)=0$.\n\nThis should give you $(1-(-x)^2)-(-x)^2=0$.\n\nFinally you have $1-2x^2=0$.\n\nSolutions are $\\displaystyle \\boxed{\\pm \\frac{1}{\\sqrt{2}}}$.\n\n\u2022 Because cos^-1 only returns one value. You have to use symmetry to get the other value \u2013\u00a0Saketh Malyala Jun 8 '17 at 15:43\n\u2022 How did you get This should give you $(1-(-x)^2)-(-x)^2=0$. \u2013\u00a0Neha Gupta Jun 8 '17 at 15:49\n\u2022 remember $\\cos(\\sin^{-1}(x)) = \\sqrt{1-x^2}$ by right angle trigonometry \u2013\u00a0Saketh Malyala Jun 8 '17 at 15:50\n\u2022 $\\sin(\\sin^{-1}(x))=x$ because sin is the inverse of $\\sin^{-1}$ \u2013\u00a0Saketh Malyala Jun 8 '17 at 15:51\n\nYou need to solve $\\cos \\left(2 \\arcsin(-x) \\right) = 0$. Let $y = 2 \\arcsin(-x)$ then $\\cos y = 0$ so $y = \\pi/2 \\pm n\\pi$. Then, $$2 \\arcsin(-x) = \\frac{pi}{2} \\pm n\\pi$$ which implies $$x = -\\sin \\left( \\frac{\\pi}{4} \\pm \\frac{n\\pi}{2} \\right)$$\n\nCan you simplify this?\n\nTrig functions are not one-to-one. For example if $\\sin x = \\frac 12$ then $x$ might be equal to $\\frac {\\pi}6$, but $x$ could also be $\\frac {5\\pi}6$ or $\\frac {13\\pi}6$ or it could be any $2k\\pi + \\frac {\\pi}6$ or $(2k+1)\\pi - \\frac {\\pi} 6$.\n\nSo when we say $\\sin^{-1} x = \\theta$ we are not just saying that $\\theta$ is the angle so that $\\sin (\\theta) = x$. There should be an infinite number of such angles. We are saying that $\\theta$ is an angle so that $\\sin (\\theta) = x$ AND we are saying that $\\frac {-\\pi}2 \\le \\theta \\le \\frac {\\pi}2$. For those two conditions there is only one possible theta.\n\nSo $\\sin^{-1}(\\sin x)$ may or MAY NOT be equal to $x$. For example $\\sin^{-1}(\\sin (\\frac {5\\pi}6)) = \\sin^{-1}(\\frac 12) = \\frac {\\pi}6 \\ne \\frac {5\\pi}6$.\n\nHowever $\\sin^{-1}(\\sin x) = x + 2k\\pi$ for some integer $k$ or $\\sin^{-1}(\\sin x) = (2k+ 1)\\pi - x$.\n\nSo 1) Because $\\sin (x) = \\sin (\\pi - x)$ and $x = \\frac {\\pi}2 - (\\frac {\\pi}2 - x)$ and $\\pi - x = \\frac{\\pi}2 + (\\frac {\\pi}2 - x)$ then $\\sin^{-1}(x) = (\\frac {\\pi}2 \\pm x) \\pm 2k\\pi)$.\n\n$\\frac {-\\pi}2 \\le sin^{-1}(x) \\le \\frac {\\pi}2$\n\nAnd 2) Because $\\cos (x) = \\cos(\\pi+ x)$ we have $\\cos^{-1}(\\cos x) = x \\pm k \\pi$.\n\n$0 \\le \\cos^{-1} (x) \\le \\pi$\n\nSo ....\n\n$\\cos (2\\sin^{-1}(-x) ) = 0$\n\n$-\\frac {\\pi}2 \\le \\sin^{-1}(-x) \\le \\frac {\\pi}2$ so $-\\pi \\le 2\\sin^{-1}(-x) \\le \\pi$ and $\\sin^{-1}(-x) = \\cos^{-1}(0) \\pm k\\pi = \\frac {\\pi}2 \\pm k\\pi$. So $2\\sin^{-1}(-x) = \\pm \\frac {\\pi}2$\n\nAnd $\\sin^{-1}(-x) = \\pm \\frac {\\pi}4$\n\nSo $\\sin (\\pm \\frac {\\pi} 2) = -x$\n\nSo $\\pm \\frac {1}{\\sqrt{2}} = -x$\n\nSo $x = \\mp \\frac {1}{\\sqrt{2}}$\n\nLet $\\sin^{-1}(-x)=u\\implies\\sin u=-x$\n\n$$\\cos(2\\sin^{-1}(-x))=\\cos2u=1-2\\sin^2u=1-2(-x)^2$$", "date": "2021-07-31 22:30:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2314829/solve-for-x-in-cos2-sin-1-x-0", "openwebmath_score": 0.9937599301338196, "openwebmath_perplexity": 508.585466992536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795083542037, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8626963084953063}}
{"url": "https://math.stackexchange.com/questions/1794693/does-the-hausdorff-property-hold-on-closed-subsets-of-mathbbrn/1794709", "text": "# Does the Hausdorff property hold on closed subsets of $\\mathbb{R}^n?$\n\nI am trying to prove that given disjoint closed $A,B\\subseteq \\mathbb{R}^n$, there exist disjoint open $U,V$ containing $A,B$ respectively. In other words that we can take the Hausdorff property to closed subsets of $\\mathbb{R}^n$.\n\nI do not know whether this is true or false, but have set about proving that it is true.\n\nMy idea was, for each $x\\in A$, to have $\\epsilon_x=\\inf_{b\\in B} \\|x-b\\|=\\|x-\\beta_x\\|$ for some $\\beta_x\\in B$ (closure). Define $U_x=\\{y:\\|x-y\\|<\\epsilon_x/2\\}$. Define similar balls for each element of $B$. Basically the ball around $x\\in A$ is disjoint from the ball around $\\beta_x$, the closest point to $x$ in $B$.\n\nHave $U=\\bigcup_{A} U_a$ and $V=\\bigcup_B V_b$.\n\nNow I believe $U,V$ are disjoint but I can't seem to argue it formally. It seems quite obvious by drawing $A,B$ as blobs in a plane, but obviously $A,B$ could be more complicated than nice bounded subsets. So I need to show $U_a\\bigcap V_b=\\emptyset$ for any $a,b\\in A,B$; now if $z$ were in both then $\\|z-a\\|<\\epsilon_a/2$ and $\\|z-\\beta_a\\|>\\epsilon_a/2$, and also $\\|z-b\\|<\\epsilon_b/2$...\n\nI feel like the argument is a step away but I can't seem to finish it off. Help?\n\n\u2022 Closed subsets of Rn are metric spaces, hence normal, satisfying the necessary property. en.wikipedia.org/wiki/Normal_space So the statement is true. May 21 '16 at 22:22\n\u2022 @William Phew, thanks. Is my attempt a viable route for proving it? May 21 '16 at 22:23\n\u2022 I'm lazy so haven't evaluated the details exactly, but the gist of it seems like the right direction. Basically the idea is that since the sets are closed and disjoint, their closures do not intersect, and hence they have non-zero distance (epsilon) from each other, so take open sets U and V which cover each of the two sets without going more than epsilon/3 away from their boundary, hence U and V can't intersect. en.wikipedia.org/wiki/\u2026 Obviously you will have to work harder to make that precise. May 21 '16 at 22:27\n\u2022 In the picture in the Wikipedia article I linked to, for example, take open sets around A and B whose farthest points from A and B are no more than 1 anyway; if A and B were closed spheres with radius R, and the distance between them was three, then take the open spheres centered at the centers of A and B and with radius R+1 May 21 '16 at 22:29\n\u2022 @William. Disjoint closed subsets of a metric space do not necessarily have positive distance from each other. For example,in $R^2$ let $A=\\{(x,1/x):x>0\\}$ and $B=\\{(x,-1/x):x>0\\}.$ Then $d( (x,1/x),(x,-1/x))=2/x,$ which has no positive lower bound May 21 '16 at 23:36\n\nWe may assume that $A,B$ are nonempty; if one of them is empty we can just take $\\emptyset,X$ as the two open sets. The most obvious construction works fine: let $U$ be the set of points nearer to $A$ than to $B,$ and let $V$ be the set of points nearer to $B$ than to $A.$ That is, take $$U=\\{x:d(x,A)\\lt d(x,B)\\}=\\{x:d(x,B)-d(x,A)\\gt0\\}$$ and let $$V=\\{x:d(x,B)\\lt d(x,A)\\}=\\{x:d(x,A)\\gt d(x,B)\\}$$ where $$d(x,S)=\\inf\\{d(x,y):y\\in S\\}.$$ The sets $U,V$ are obviously disjoint. They are open sets because, for a nonempty set $S,$ the function $x\\mapsto d(x,S)$ is continuous, as is easily shown using the triangle inequality. The inclusions $A\\subseteq U$ and $B\\subseteq V$ follow from the assumption that $A,B$ and disjoint and closed; e.g., if $x\\in A$ then $x\\notin B,$ so $d(x,B)\\gt0=d(x,A).$\nNote that for all $a \\in A$ and $b \\in B$ we have $$\\inf_{a' \\in A}\\|a' - b\\| \\leq \\|a-b\\|.$$\nTake $a \\in A$, $b \\in B$. Suppose $z \\in U_a \\cap V_b$. By the triangle inequality, $$\\|a - b \\| \\leq \\|a - z \\| + \\|z - b\\| < \\epsilon_a/2 + \\epsilon_b/2$$ $$= \\frac{\\inf_{b' \\in B}\\|a - b'\\|}{2} + \\frac{\\inf_{a' \\in A}\\|a' - b\\|}{2}$$ $$\\leq \\frac{\\|a - b\\|}{2} + \\frac{\\|a-b\\|}{2} = \\|a-b\\|.$$ So $\\|a-b\\| < \\|a-b\\|$. Contradiction.", "date": "2021-09-21 06:17:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1794693/does-the-hausdorff-property-hold-on-closed-subsets-of-mathbbrn/1794709", "openwebmath_score": 0.9566195011138916, "openwebmath_perplexity": 115.45260694812127, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429160413819, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.862687783261353}}
{"url": "https://math.stackexchange.com/questions/3805033/is-there-a-better-easier-way-to-solve-this-matrix/3805040", "text": "# Is there a better/easier way to solve this matrix?\n\n$$\\begin{equation*} \\begin{bmatrix} 4 & -1 & -1 & 0 &|&30 \\\\ -1 & 4 & 0 & -1&|&60 \\\\ -1 & 0 & 4 & -1&|&40 \\\\ 0 & -1 & -1 & 4&|&70 \\end{bmatrix} \\end{equation*}$$\n\nWhat's the best way to solve the matrix above? There's a clear pattern of the diagonal 4's and 0's and the -1's so I feel like there has to be a better way of doing things rather than using scaling and row reduction.\n\nIf I do those methods I end up with messy fractions.\n\nMy Step 1:\n\nNew Row 2 = (1/4)Row 1 + Row 2\n\nEven at step 1 I can tell the whole thing will be messy with fractions.\n\nIs there a better way to solve this matrix? Or am I doing it wrong? Thanks.\n\nAdd all four rows to get $$\\begin{pmatrix}2 &2 &2 &2\\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\\\x_4\\end{pmatrix}= 200$$\n\nor\n\n$$\\begin{pmatrix}1 &1 &1 &1\\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\\\x_4\\end{pmatrix}= 100$$\n\nAdd that to every row to get $$\\begin{pmatrix}5 &0 &0 &1\\\\ 0 & 5 & 1 & 0\\\\ 0 & 1 & 5 & 0\\\\ 1 & 0 & 0 & 5\\end{pmatrix}\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\\\x_4\\end{pmatrix}= \\begin{pmatrix}130\\\\160\\\\140\\\\170\\end{pmatrix}$$\n\nNow you have two separated 2x2 systems. Can you take it from here?\n\n\u2022 thanks, that's super interesting. is there a name for that process? Aug 27, 2020 at 13:03\n\u2022 I call it \"sharply looking at the equation until you have an idea\" :D No idea if there is a formal name for it. Also, can you please accept this answer? Then others will know that there is an viable answer here :) Aug 27, 2020 at 13:20\n\u2022 sure thing, i forgot i could do that cause i got more reputation now. also im still thinking this would be messy yea? cause like: new row 4 = row 1(-1/5) so then it'll be (-1/5) + 5 which is 4/5ths which is still messy, or is that just no other way, or am i doing it wrong Aug 27, 2020 at 13:25\n\u2022 you can do row1-5*row4, if you dislike fractions. But having one fraction and then multiplying the equation by 5/4 is not that bad. Aug 27, 2020 at 13:27\n\u2022 Wow thank you so much! I am really happy that an answer from here actually had a positive impact on someone in real life :) Sep 6, 2020 at 12:58", "date": "2023-02-06 09:17:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3805033/is-there-a-better-easier-way-to-solve-this-matrix/3805040", "openwebmath_score": 0.43240246176719666, "openwebmath_perplexity": 548.1312167700419, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429142850273, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8626877753361032}}
{"url": "https://math.stackexchange.com/questions/757263/how-to-find-answer-to-the-sum-of-series-sum-n-1-infty-fracn2n/3375934", "text": "# How to find answer to the sum of series $\\sum_{n=1}^{\\infty}\\frac{n}{2^n}$\n\nI have put his on wolfram and obtained answer as follows:\n\n$\\sum_{n=1}^{\\infty}\\frac{n}{2^n} = 2$\n\nAnd the series is convergent too because $\\lim_{n\\to\\infty} \\frac {n}{2^n} = 0$\n\nHowever I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper?\n\n\u2022 Of course the fact that $\\lim_{n \\to \\infty} \\frac{n}{2^n} \\to 0$ doesn't imply that the series is convergent. \u2013\u00a0Amateur Apr 17 '14 at 2:49\n\u2022 Differentiate the geometric series. \u2013\u00a0Bitrex Apr 17 '14 at 2:50\n\nLet $S$ be our sum. Then $S=\\frac{1}{2}+\\frac{2}{2^2}+\\frac{3}{2^3}+\\frac{4}{2^4}+\\cdots$. Thus \\begin{align}2S&=1+&\\frac{2}{2}+\\frac{3}{2^2}+\\frac{4}{2^3}+\\frac{5}{2^4}+\\cdots\\\\ S&=&\\frac{1}{2}+\\frac{2}{2^2}+\\frac{3}{2^3}+\\frac{4}{2^4}+\\cdots\\end{align} Subtract. We get $$S=1+\\frac{1}{2}+\\frac{1}{2^2}+\\frac{1}{2^3}+\\frac{1}{2^4}+\\cdots=2.$$\n\nHint:\n\n$$\\sum_{k=1}^{\\infty} x^k = \\frac{x}{1-x} \\implies \\sum_{k=1}^{\\infty} kx^k=\\frac{d}{dx} \\frac{x}{1-x} .$$\n\n\u2022 Want to understand why we can differentiate. \u2013\u00a0deostroll Apr 17 '14 at 4:14\n\u2022 @deostroll: We can differentiate a power series within its radius of convergence. \u2013\u00a0Mhenni Benghorbal Apr 17 '14 at 4:21\n\u2022 @deostroll The technical result allowing us to do this is the following: First, power series $\\sum \\alpha_n x^n$ converge absolutely and uniformly on any closed subset of their (open) interval of convergence. (This follows from the $M$-test.) Second, the (formal) series of their termwise derivative, $\\sum n\\alpha_n x^{n-1}$, has the same radius of convergence. (Cont.) \u2013\u00a0Andr\u00e9s E. Caicedo Apr 19 '14 at 3:11\n\u2022 @deostroll Third, and this is the technical part, if a sequence $f_n$ of differentiable functions defined on some closed interval satisfies that (1) $f_n'\\to g$ uniformly, and (2) $f_n(x_0)$ converges (as $n\\to\\infty)$ for some $x_0$, then (a) $f_n$ converges uniformly to some differentiable function $f$, and (b) $f'=g$. To apply this, take as $f_n$ the partial sums of the given power series. A proof of this theorem appears for instance in Rudin's Principles of mathematical analysis, see Theorem 7.17. (Cont.) \u2013\u00a0Andr\u00e9s E. Caicedo Apr 19 '14 at 3:15\n\u2022 @deostroll The assumption is technical, but cannot really be relaxed: Even if we assume that $f_n\\to f$ uniformly, that $f$ is differentiable, and that $f_n'\\to g$ pointwise, we may have that $f'(x)\\ne g(x)$ for all $x$. This is a (remarkably recent!) theorem of Darji, see here. \u2013\u00a0Andr\u00e9s E. Caicedo Apr 19 '14 at 3:17\n\nAnother way to do the problem is to note that $$\\left(\\sum_{k=1}^\\infty \\frac{1}{2^k}\\right)^2 = \\sum_{k=1}^\\infty \\frac{k-1}{2^k}$$ so that the desired series is really just $$\\left(\\sum_{k=1}^\\infty \\frac{1}{2^k}\\right)^2 + \\sum_{k=1}^\\infty \\frac{1}{2^k}$$\n\nIf $\\displaystyle S=\\sum_{n=1}^\\infty\\frac{n}{2^n}$, then $$S-\\frac12=\\sum_{n=1}^\\infty\\frac{n+1}{2^{n+1}}=\\frac12\\sum_{n=1}^\\infty\\frac{n+1}{2^n}=\\frac12\\left(S+\\sum_{n=1}^\\infty\\frac1{2^n}\\right)=\\frac12(S+1),$$ so $2S-1=S+1$, or $S=2$.\n\nAll that remains is to justify that the series converges. But $n<1.1^n$ for $n$ large enough, so a tail of $S$ is bounded above by a tail of the convergent geometric series $\\sum_n\\left(\\frac{1.1}2\\right)^n$.\n\nIn this answer we go for the 'one trick pony' approach - all we know is that for $$k \\ge 0$$,\n\n$$\\tag 1 \\sum_{n=k}^{\\infty}\\frac{1}{2^n} = 2^{1-k}$$\n\nWe decompose the summands of $$\\sum_{n=1}^{\\infty}\\frac{n}{2^n}$$ in a natural/straightforward manner, arranging these numbers into a table:\n\n$$\\begin{pmatrix} \\frac{1}{2} & \\frac{1}{4} & \\frac{1}{8} & \\frac{1}{16} & \\frac{1}{32} & \\dots \\\\ 0 & \\frac{1}{4} & \\frac{1}{8} & \\frac{1}{16} & \\frac{1}{32} & \\dots \\\\ 0 & 0 & \\frac{1}{8} & \\frac{1}{16} & \\frac{1}{32} & \\dots \\\\ 0 & 0 & 0 & \\frac{1}{16} & \\frac{1}{32} & \\dots \\\\ 0 & 0 & 0 & 0 & \\frac{1}{32} & \\dots \\\\ 0 & 0 & 0 & 0 & 0 & \\dots \\\\ . \\\\ . \\\\ . \\\\ \\end{pmatrix}$$\n\nUsing $$\\text{(1)}$$ we add up the entries in each row,\n\n$$\\begin{pmatrix} 1 \\\\ \\frac{1}{2} \\\\ \\frac{1}{4} \\\\ \\frac{1}{8} \\\\ \\frac{1}{16} \\\\ . \\\\ . \\\\ . \\\\ \\end{pmatrix}$$\n\nAnd now we add up the entries of our column vector, giving", "date": "2021-07-29 19:11:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/757263/how-to-find-answer-to-the-sum-of-series-sum-n-1-infty-fracn2n/3375934", "openwebmath_score": 0.9560747742652893, "openwebmath_perplexity": 206.53563683177697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429151632047, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8626877745084353}}
{"url": "http://mathhelpforum.com/calculus/19344-solved-differential-equation-geometry-word-problem-help.html", "text": "# Math Help - [SOLVED] Differential Equation Geometry/Word Problem Help!\n\n1. ## [SOLVED] Differential Equation Geometry/Word Problem Help!\n\nHi everyone, I'm currently having trouble mentally grasping this question I have:\n\nFind the equation of the curve that goes through (1,3) for which any tangent and the line from the origin to the point of contact make with the ordinate a triangle having area A.\n\nI have been trying several approaches to this, but to no avail. I assume it can be set up as a first-order differential equation of some sort. I've defined the curve as a, the tangent line as b, and the line from the point of tangency to the origin as c. I also defined a random point along the curve (alpha, beta). Thus,\n\n$b(x)=beta+a'(alpha)*(x-alpha)$\n$c(x)=beta/alpha * x$\n\nEDIT: The area of the triangle, A, could also be found with one of the following:\n$A_1=int(b(x))-int(c(x))$\n$A_2=(1/2)(b)(h)=(1/2)(b(0))(alpha)=(1/2)(beta+a'(alpha)*(x-alpha))$\n\nI have verified that, in fact, $A_1=A_2$.\n\nAny insight would be greatly appreciated!\n\n2. [According to Wikipedia, the ordinate is another name for the y-axis. I never came across that terminology before.]\n\nI find that a, b, c, alpha, beta, ... notation hard to follow, so I'll use other letters. Suppose the curve is given by y=f(x), and let $(x_0,y_0)$ be a point on the curve. The tangent at this point has equation $y-y_0=f'(x_0)(x-x_0)$. It meets the y-axis at the point $(0,y_1)$, where y_1 is given by $y_1-y_0 = -f'(x_0)x_0$.\n\nIf we take the triangle to have its base along the y-axis then its base is $\\pm y_1$ ( the +/- is needed because we don't know whether y_1 is positive or negative) and its height is x_0. So its area is given by $\\textstyle A=\\pm\\frac12x_0y_1 = \\pm\\frac12x_0(y_0-x_0f'(x_0))$.\n\nAt this stage we need to change the notation. Up until now, $(x_0,y_0)$ has been fixed. But in order to get a differential equation for f(x), we need to see what happens as $(x_0,y_0)$ varies. The best thing to do is to drop the zero subscripts and to replace f'(x_0) by dy/dx (or just y'). Then the previous equation becomes $x(y-xy') = \\pm2A$. At last, we have a differential equation!\n\nYou'll need to use an integrating factor to solve the equation, and then to use the initial condition that y=3 when x=1. There are two solutions, corresponding to the +/- sign in the equation.\n\n3. Hello, bherila!\n\nFind the equation of the curve that goes through (1,3)\nfor which any tangent and the line from the origin to the point of contact\nmake with the ordinate a triangle having area $A$.\nCode:\n        |\n*                 * y = f(x)\n|    *\n|        *       *\n|             * \u0001 P(p,q)\n|           o /\n|       o   /\n|   o     /\nB o       /\n|     /\n|   /\n| /\n- - * - - - - - - - - - - -\n|O\n|\n\nWe have a function: $y \\,=\\,f(x)$\n\nLet $P(p,q)$ be a point on $y \\,=\\,f(x).$\n\n$OP$ is the segment from the origin to point $P.$\n\n$BP$ is the tangent at $P$.\n. . The slope of this tangent is $y'$ and it contains $P(p,q).$\n. . Its equation is: . $y - q \\:=\\:y'(x - p)\\quad\\Rightarrow\\quad y \\:=\\:y'x - y'p + q$\n. . Its y-intercept is: . $B(0,\\,q - py')$\n\nThe base of $\\Delta OBP$ is: . $OB \\:=\\:q - py'$\nThe height of the triangle is: . $p$\n\nThe area of $\\Delta OBP\\!:\\;\\;\\frac{1}{2}(q - py')p \\;=\\;A$\n\n\\. . which simplifies to: . $p^2y' - pq \\:=\\:-2A$\n\nDivide by $p^2\\!:\\;\\;y' - \\frac{1}{p}\\,q \\:=\\:-\\frac{2A}{p^2}$\n\nHence, we have: . $\\frac{dy}{dx} - \\frac{1}{x}\\,y \\;=\\;-\\frac{2A}{x^2}$\n\nIntegrating factor: . $I \\:=\\:e^{\\int\\left(-\\frac{1}{x}\\right)dx} \\:=\\:e^{-\\ln x} \\:=\\:e^{\\ln\\frac{1}{x}} \\:=\\:\\frac{1}{x}$\n\nSo we have: . $\\frac{1}{x}\\,\\frac{dy}{dx} - \\frac{1}{x^2}\\,y \\;=\\;-\\frac{2A}{x^3}$\n\nThen: . $\\frac{d}{dx}\\left(\\frac{1}{x}\\!\\cdot\\!y\\right) \\;=\\;-2Ax^{-3}$\n\nIntegrate: . $\\frac{1}{x}\\!\\cdot\\!y \\;=\\;-Ax^{-2} + C$\n\nMultiply by $x\\!:\\;\\;y \\;=\\;-\\frac{A}{x} + Cx$\n\nSince $(1,3)$ is on the curve,\n. . $3 \\;=\\;-\\frac{A}{1} + C\\quad\\Rightarrow\\quad C \\:=\\:A + 3$\n\nTherefore: . $\\boxed{y \\;=\\;-\\frac{A}{x} + (A + 3)x}$\n\n4. The other solution is $\\boxed{y =\\frac{A}{x} + (3-A)x}$.\n\n5. Originally Posted by Soroban\nHello, bherila!\n\nCode:\n        |\n*                 * y = f(x)\n|    *\n|        *       *\n|             * \u0001 P(p,q)\n|           o /\n|       o   /\n|   o     /\nB o       /\n|     /\n|   /\n| /\n- - * - - - - - - - - - - -\n|O\n|\n\nWe have a function: $y \\,=\\,f(x)$\n\nLet $P(p,q)$ be a point on $y \\,=\\,f(x).$\n\n$OP$ is the segment from the origin to point $P.$\n\n$BP$ is the tangent at $P$.\n. . The slope of this tangent is $y'$ and it contains $P(p,q).$\n. . Its equation is: . $y - q \\:=\\:y'(x - p)\\quad\\Rightarrow\\quad y \\:=\\:y'x - y'p + q$\n. . Its y-intercept is: . $B(0,\\,q - py')$\n\nThe base of $\\Delta OBP$ is: . $OB \\:=\\:q - py'$\nThe height of the triangle is: . $p$\n\nThe area of $\\Delta OBP\\!:\\;\\;\\frac{1}{2}(q - py')p \\;=\\;A$\n\n\\. . which simplifies to: . $p^2y' - pq \\:=\\:-2A$\n\nDivide by $p^2\\!:\\;\\;y' - \\frac{1}{p}\\,q \\:=\\:-\\frac{2A}{p^2}$\n\nHence, we have: . $\\frac{dy}{dx} - \\frac{1}{x}\\,y \\;=\\;-\\frac{2A}{x^2}$\n\nIntegrating factor: . $I \\:=\\:e^{\\int\\left(-\\frac{1}{x}\\right)dx} \\:=\\:e^{-\\ln x} \\:=\\:e^{\\ln\\frac{1}{x}} \\:=\\:\\frac{1}{x}$\n\nSo we have: . $\\frac{1}{x}\\,\\frac{dy}{dx} - \\frac{1}{x^2}\\,y \\;=\\;-\\frac{2A}{x^3}$\n\nThen: . $\\frac{d}{dx}\\left(\\frac{1}{x}\\!\\cdot\\!y\\right) \\;=\\;-2Ax^{-3}$\n\nIntegrate: . $\\frac{1}{x}\\!\\cdot\\!y \\;=\\;-Ax^{-2} + C$\n\nMultiply by $x\\!:\\;\\;y \\;=\\;-\\frac{A}{x} + Cx$\n\nSince $(1,3)$ is on the curve,\n. . $3 \\;=\\;-\\frac{A}{1} + C\\quad\\Rightarrow\\quad C \\:=\\:A + 3$\n\nTherefore: . $\\boxed{y \\;=\\;-\\frac{A}{x} + (A + 3)x}$\n\nvery nice. that was a lot easier than i thought it would be. Good stuff as always, Soroban!!!", "date": "2016-07-30 15:53:00", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/19344-solved-differential-equation-geometry-word-problem-help.html", "openwebmath_score": 0.9525225162506104, "openwebmath_perplexity": 526.5008174364331, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429103332287, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8626877686816408}}
{"url": "https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning_-_Writing_and_Proof_(Sundstrom)/5%3A_Set_Theory/5.3%3A_Properties_of_Set_Operations", "text": "\n# 5.3: Properties of Set Operations\n\n\nPREVIEW ACTIVITY $$\\PageIndex{1}$$: Exploring a Relationship between Two Sets\n\nLet $$A$$ and $$B$$ be subsets of some universal set $$U$$.\n\n1. Draw two general Venn diagrams for the sets $$A$$ and $$B$$. On one, shade the region that represents $$(A \\cup B)^c$$, and on the other, shade the region that represents $$A^c \\cap B^c$$. Explain carefully how you determined these regions.\n2. Based on the Venn diagrams in Part (1), what appears to be the relationship between the sets ((A \\cup B)^c\\) and $$A^c \\cap B^c$$?\n\nSome of the properties of set operations are closely related to some of the logical operators we studied in Section 2.1. This is due to the fact that set intersection is defined using a conjunction (and), and set union is defined using a disjunction (or). For example, if $$A$$ and $$B$$ are subsets of some universal set $$U$$, then an element $$x$$ is in $$A \\cup B$$ if and only if $$x \\in A$$ or $$x \\in B$$.\n\n1. Use one of De Morgan\u2019s Laws (Theorem 2.8 on page 48) to explain carefully what it means to say that an element $$x$$ is not in $$A \\cup B$$.\n2. What does it mean to say that an element $$x$$ is in $$A^c$$? What does it mean to say that an element $$x$$ is in $$B^c$$?\n3. Explain carefully what it means to say that an element $$x$$ is in $$A^c \\cap B^c$$.\n4. Compare your response in Part (3) to your response in Part (5). Are they equivalent? Explain.\n5. How do you think the sets $$(A \\cup B)^c$$ and $$A^c \\cap B^c$$ are related? Is this consistent with the Venn diagrams from Part (1)?\n\nPREVIEW ACTIVITY $$\\PageIndex{2}$$: Proving that Statements Are Equivalent\n\n1. Let $$X$$, $$Y$$, and $$Z$$ be statements. Complete a truth table for\n$$[(X \\to Y) \\wedge (Y \\to Z)] \\to (X \\to Z)$$.\n\n2. Assume that $$P$$, $$Q$$, and $$R$$ are statements and that we have proven that the following conditional statements are true:\n\n$$\\bullet$$ If $$P$$ then $$Q (P \\to Q)$$.\n$$\\bullet$$ If $$R$$ then $$P (R \\to P)$$.\n$$\\bullet$$ If $$Q$$ then $$R (Q \\to R)$$.\n\nExplain why each of the following statements is true.\n\n(a) $$P$$ if and only if $$Q (P \\leftrightarrow Q$$.\n(b) $$Q$$ if and only if $$R (Q \\leftrightarrow R$$.\n(c) $$R$$ if and only if $$P (R \\leftrightarrow P$$.\nRemember that $$X \\leftrightarrow Y$$ is logically equivalent to $$(X \\to Y) \\wedge (Y \\to X)$$.\n\n## Algebra of Sets \u2013 Part 1\n\nThis section contains many results concerning the properties of the set operations. We have already proved some of the results. Others will be proved in this section or in the exercises. The primary purpose of this section is to have in one place many of the properties of set operations that we may use in later proofs. These results are part of what is known as the algebra of sets or as set theory.\n\nTheorem 5.17\n\nLet $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Then\n\n\u2022 $$A \\cap B \\subseteq A$$ and $$A \\subseteq A \\cup B$$.\n\u2022 If $$A \\subseteq B$$, then $$A \\cap C \\subseteq B \\cap C$$ and $$A \\cup C \\subseteq B \\cup C$$.\nProof\n\nThe first part of this theorem was included in Exercise (7) from Section 5.2. The second part of the theorem was Exercise (12) from Section 5.2.\n\nThe next theorem provides many of the properties of set operations dealing with intersection and union. Many of these results may be intuitively obvious, but to be complete in the development of set theory, we should prove all of them. We choose to prove only some of them and leave some as exercises.\n\nTheorem 5.18: Algebra of Set Operations\n\nLet $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Then all of the following equalities hold.\n\nProporties of the Empty Set $$A \\cap \\emptyset = \\emptyset$$ $$A \\cap U = A$$ and the Universal Set $$A \\cup \\emptyset = A$$ $$A \\cup U = U$$\n\nIdempotent Laws $$A \\cap A = A$$ $$A \\cup A = A$$\n\nCommutative Laws $$A \\cap B = B \\cap A$$ $$A \\cup B = B \\cup A$$\n\nAssociative Laws $$(A \\cap B) \\cap C = A \\cap (B \\cap C)$$\n$$(A \\cup B) \\cup C = A \\cup (B \\cup C)$$\n\nDistributive Laws $$A \\cap (B \\cup C) = (A \\cap B) \\cup (A \\cap C)$$\n$$A \\cup (B \\cap C) = (A \\cup B) \\cap (A \\cup C)$$\n\nBefore proving some of these properties, we note that in Section 5.2, we learned that we can prove that two sets are equal by proving that each one is a subset of the other one. However, we also know that if $$S$$ and $$T$$ are both subsets of a universal set $$U$$, then\n\n$$S = T$$ if and only if for each $$x \\in U$$, $$x \\in S$$ if and only if $$x \\in T$$.\n\nWe can use this to prove that two sets are equal by choosing an element from one set and chasing the element to the other set through a sequence of \u201cif and only if\u201d statements. We now use this idea to prove one of the commutative laws.\n\nProof of One of the Commutative Laws in Theorem 5.18\n\nWe will prove that $$A \\cap B = B \\cap A$$. Let $$x \\in A \\cap B$$. Then\n\n$x \\in A \\cap B \\text{ if and only if } x \\in A \\text{ and } x \\in B.$\n\nHowever, we know that if $$P$$ and $$Q$$ are statements, then $$P wedge Q$$ is logically equivalent to $$Q \\wedge P$$. Consequently, we can conclude that\n\n$x \\in A \\text{ and } x \\in B \\text{ if and only if } x \\in B \\text{ and } x \\in A.$\n\nNow we know that\n\n$x \\in B \\text{ and } x \\in A \\text{ if and only if } x \\in B \\cap A.$\n\nThis means that we can use (5.3.1), (5.3.2) and (5.3.3) to conclude that\n\n$$x \\in A \\cap B$$ if and only if $$x \\in B \\cap A$$,\n\nand, hence, we have proved that $$A \\cap B = B \\cap A$$.\n\n$$\\square$$\n\nProgress Check 5.19: Exploring a Distributive Property\n\nWe can use Venn diagrams to explore the more complicated properties in Theorem 5.18, such as the associative and distributive laws. To that end, let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.\n\n1. Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A \\cup (B \\cap C$$, and on the other, shade the region that represents $$(A \\cup B) \\cap (A \\cup C)$$. Explain carefully how you determined these regions.\n2. Based on the Venn diagrams in Part (1), what appears to be the relationship between the sets $$A \\cup (B \\cap C$$ and $$(A \\cup B) \\cap (A \\cup C)$$?\n\nAdd texts here. Do not delete this text first.\n\nProof of One of the Distributive Laws in Theorem 5.18\n\nWe will now prove the distributive law explored in Progress Check 5.19. Notice that we will prove two subset relations, and that for each subset relation, we will begin by choosing an arbitrary element from a set. Also notice how nicely a proof dealing with the union of two sets can be broken into cases.\n\nProof. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. We will prove that $$A \\cup (B \\cap C) = (A \\cup B) \\cap (A \\cup C)$$ by proving that each set is a subset of the other set.\n\nWe will first prove that $$A \\cup (B \\cap C) \\subseteq (A \\cup B) \\cap (A \\cup C$$. We let $$x \\in A \\cup (B \\cap C)$$. Then $$x \\in A$$ or $$x \\in B \\cap C$$.\n\nSo in one case, if $$x \\in A$$, then $$x \\in A \\cup B$$ and $$x \\in A \\cup C$$. This means that $$x \\in (A \\cup B) \\cap (A \\cup C)$$.\n\nOn the other hand, if $$x \\in B \\cap C$$, then $$x \\in B$$ and $$x \\in C$$. But $$x \\in B$$ implies that $$x \\in A \\cup B$$, and $$x \\in C$$ implies that $$x \\in A \\cup C$$. Since $$x$$ is in both sets, e conclude that $$x \\in (A \\cup B) \\cap (A \\cup C)$$. So in both cases, we see that $$x \\in (A \\cup B) \\cap (A \\cup C)$$, and this proves that $$A \\cup (B \\cap C) \\subseteq (A \\cup B) \\cap (A \\cup C$$.\n\nWe next prove that $$(A \\cup B) \\cap (A \\cup C) \\subseteq A \\cup (B \\cap C)$$. So let $$y \\in (A \\cup B) \\cap (A \\cup C)$$. Then, $$y \\in A \\cup B$$ and $$y \\in A \\cup C$$. We must prove that $$y \\in A \\cup (B \\cap C)$$. We will consider the two cases where $$y \\in A$$ or $$y \\notin A$$. In the case where $$y \\in A$$, we see that $$y \\in A \\cup (B \\cap C)$$.\n\nSo we consider the case that $$y \\notin A$$. It has been established that $$y \\in A \\cup B$$ and $$y \\in A \\cup C$$. Since $$y \\not in A$$ and $$y \\in A \\cup B$$, $$y$$ must be an element of $$B$$. Similarly, since $$y \\notin A$$ and $$y \\in A \\cup C$$, $$y$$ must be an element of $$C$$. Thus, $$y \\in B \\cap C$$ and, hence, $$y \\in A \\cup (B \\cap C)$$.\n\nIn both cases, we have proved that $$y \\in A \\cup (B \\cap C)$$. This proves that $$(A \\cup B) \\cap (A \\cup C) \\subseteq A \\cup (B \\cap C)$$. The two subset relations establish the equality of the two sets. Thus, $$A \\cup (B \\cap C) = (A \\cup B) \\cap (A \\cup C)$$.\n\n$$square$$\n\n## Important Properties of Set Complements\n\nThe three main set operations are union, intersection, and complementation. The- orems 5.18 and 5.17 deal with properties of unions and intersections. The next theorem states some basic properties of complements and the important relations dealing with complements of unions and complements of intersections. Two relationships in the next theorem are known as De Morgan\u2019s Laws for sets and are closely related to De Morgan\u2019s Laws for statements.\n\nTheorem 5.20\n\nLet $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Then the following are true:\n\nBasic Properties $$(A^c)^c = A$$\n$$A - B = A \\cap B^c$$\n\nEmpty Set and Universal Set $$A - \\emptyset = A$$ and $$A - U = \\emptyset$$\n$$\\emptyset ^c = U$$ and $$U^c = \\emptyset$$\n\nDe Morgan's Laws $$(A \\cap B)^c = A^c \\cup B^c$$\n$$(A \\cup B)^c = A^c \\cap B^c$$\n\nSubsets and Complements $$A \\subseteq B$$ if and only if $$B^c \\subseteq A^c$$\n\nProof\n\nWe will only prove one of De Morgan\u2019s Laws, namely, the one that was explored in Preview Activity $$\\PageIndex{1}$$. The proofs of the other parts are left as exercises. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$. We will prove that $$(A \\cup B)^c = A^c \\cap B^c$$ by proving that an element is in $$(A \\cup B)^c$$ if and only if it is in $$A^c \\cap B^c$$. So let $$x$$ be in the universal set $$U$$. Then\n\n$x \\in (A \\cup B)^c \\text{ if and only if } x \\notin A \\cup B.$\n\nand\n\n$x \\notin A \\cup B \\text{ if and only if } x \\notin A \\text{ and } x \\notin B.$\n\nCombining (5.3.4) and (5.3.5), we see that\n\n$x \\in (A \\cup B)^c \\text{ if and only if } x \\notin A \\text{ and } x \\notin B.$\n\n$x \\notin A \\text{ and } x \\notin B \\text{ if and only if } x \\in A^c \\text{ and } x \\in B^c.$\n\nand this is true if and only if $$x \\in A^c \\cap B^c$$. So we can use (5.3.6) and (5.3.7) to conclude that\n\n$$x \\in (A \\cup B)^c$$ if and only if $$x \\in A^c \\cap B^c$$.\n\nand, hence, that $$(A \\cup B)^c = A^c \\cap B^c$$.\n\n$$\\square$$\n\nProgress Check 5.21: Using the Algebra of Sets\n\n1. Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$(A \\cup B) - C$$, and on the other, shade the region that represents $$(A - C) \\cup (B - C)$$. Explain carefully how you determined these regions and why they indicate that $$(A \\cup B) - C = (A - C) \\cup (B - C)$$.\n\nIt is possible to prove the relationship suggested in Part (1) by proving that each set is a subset of the other set. However, the results in Theorems 5.18 and 5.20 can be used to prove other results about set operations. When we do this, we say that we are using the algebra of sets to prove the result. For example, we can start by using one of the basic properties in Theorem 5.20 to write\n\n$$A \\cup B) - C = (A \\cup B) \\cap C^c$$.\n\nWe can then use one of the commutative properties to write\n\n$\\begin{array} {rcl} {(A \\cup B) - C} &= & {(A \\cup B) \\cap C^c} \\\\ {} &= & {C^c \\cap (A \\cup B).} \\end{array}$\n\n2. Determine which properties from Theorems 5.18 and 5.20 justify each of the last three steps in the following outline of the proof that $$(A \\cup B) - C = (A - C) \\cup (B - C)$$.\n\n$\\begin{array} {rcl} {(A \\cup B) - C} &= & {(A \\cup B) \\cap C^c \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{(Theorem 5.20)}} \\\\ {} &= & {C^c \\cap (A \\cup B) \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{(Commutative Property)}} \\\\ {} &= & {(C^c \\cap A) \\cup (C^c \\cap B)} \\\\ {} &= & {(A \\cap C^c) \\cup (B \\cap C^c)} \\\\ {} &= & {(A - C) \\cup (B - C)} \\end{array}$\n\nNote: It is sometimes difficult to use the properties in the theorems when the theorems use the same letters to represent the sets as those being used in the current problem. For example, one of the distributive properties from Theorems 5.18 can be written as follows: For all sets $$X$$, $$Y$$, and $$Z$$ that are subsets of a universal set $$U$$,\n\n$$(X \\cap (Y \\cup Z) = (X \\cap Y) \\cup (X \\cap Z).$$\n\nAdd texts here. Do not delete this text first.\n\n## Proving that Statements Are Equivalent\n\nWhen we have a list of three statements P , Q, and R such that each statement in the list is equivalent to the other two statements in the list, we say that the three statements are equivalent. This means that each of the statements in the list implies each of the other statements in the list.\n\nThe purpose of Preview Activity $$\\PageIndex{2}$$ was to provide one way to prove that three (or more) statements are equivalent. The basic idea is to prove a sequence of conditional statements so that there is an unbroken chain of conditional statements from each statement to every other statement. This method of proof will be used in Theorem 5.22.\n\nTheorem 5.22\n\nLet $$A$$ and $$B$$ be subsets of some universal set $$U$$. The following are equivalent:\n\n1. $$A \\subseteq B$$\n2. $$A \\cap B^c = \\emptyset$$\n3. $$A^c \\cup B = U$$\nProof\n\nTo prove that these are equivalent conditions, we will prove that (1) implies (2), that (2) implies (3), and that (3) implies (1).\n\nLet $$A$$ and $$B$$ be subsets of some universal set $$U$$. We have proved that (1) implies (2) in Proposition 5.14.\n\nTo prove that (2) implies (3), we will assume that $$A \\cap B^c = \\emptyset$$ and use the fact that $$\\emptyset ^c = U$$. We then see that\n\n$$(A \\cap B^c)^c = \\emptyset ^c$$.\n\nThen, using one of De Morgan's Laws, we obtain\n\n$begin{array} {rcl} {A^c \\cup (B^c)^c} &= & {U} \\\\ {A^c \\cup B} &= & {U.} \\end{array}$\n\nThis completes the proof that (2) implies (3).\n\nWe now need to prove that (3) implies (1). We assume that $$A^c \\cup B = U$$ and will prove that $$A \\subseteq B$$ by proving that every element of $$A$$ must be in $$B$$.\n\nSo let $$x \\in A$$. Then we know that $$x \\notin A^c$$. However, $$x \\in U$$ and since $$A^c \\cup B = U$$, we can conclude that $$x \\in A^c \\cup B$$. Since $$x \\notin A^c$$, we conclude that $$x \\in B$$. This proves that $$A \\subseteq B$$ and hence that (3) implies (1).\n\nSince we have now proved that (1) implies (2), that (2) implies (3), and that (3) implies (1), we have proved that the three conditions are equivalent.\n\nExercises for Section 5.3\n\n1. Let $$A$$ be a subset of some universal set $$U$$. Prove each of the following (from Theorem 5.20):\n\n(a) $$(A^c)^c = A$$\n(b) $$A - \\emptyset = A$$\n(c) $$\\emptyset ^c = U$$\n(d) $$U^c = \\emptyset$$\n2. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. As part of Theorem 5.18, we proved one of the distributive laws. Prove the other one. That is, prove that\n$A \\cap (B \\cup C) = (A \\cap B) \\cup (A \\cap C).$\n\n3. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. As part of Theorem 5.20, we proved one of De Morgan\u2019s Laws. Prove the other one. That is, prove that\n$(A \\cap B)^c = A^c \\cup B^c.$\n\n4. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.\n\n(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B \\cup C)$$, and on the other, shade the region that represents $$(A - B) \\cap (A - C)$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B \\cup C)$$ and $$(A - B) \\cap (A - C)$$.\n(b) Use the choose-an-element method to prove the conjecture from Exercise (4a).\n(c) Use the algebra of sets to prove the conjecture from Exercise (4a).\n\n5. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.\n\n(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B \\cap C)$$, and on the other, shade the region that represents $$(A - B) \\cup (A - C)$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B \\cap C)$$ and $$(A - B) \\cup (A - C)$$.\n(b) Use the choose-an-element method to prove the conjecture from Exercise (5a).\n(c) Use the algebra of sets to prove the conjecture from Exercise (5a).\n\n6. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Prove or disprove each of the following:\n\n(a) $$(A \\cap B) - C = (A - C) \\cap (B - C)$$\n(b) $$(A \\cup B) - (A \\cap B) = (A - B) \\cup (B - A)$$\n\n7. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.\n\n(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B - C)$$, and on the other, shade the region that represents $$(A - B) - C$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B - C)$$ and $$(A - B) - C$$.\n(b) Prove the conjecture from Exercise (7a).\n\n8. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.\n\n(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B - C)$$, and on the other, shade the region that represents $$(A - B) \\cup (A - C^c)$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B - C)$$ and $$(A - B) \\cup (A - C^c)$$.\n(b) Prove the conjecture from Exercise (8a).\n\n9. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$.\n\n(a) Prove that $$A$$ and $$B - A$$ are disjoint sets.\n(b) Prove that $$A \\cup B = A \\cup (B - A)$$.\n\n10. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$.\n\n(a) Prove that $$A - B$$ and $$A \\cap B$$ are disjoint sets.\n(b) Prove that $$A = (A - B) \\cup (A \\cap B)$$.\n\n11. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$. Prove or disprove each of the following:\n\n(a) $$A - (A \\cap B^c) = A \\cap B$$\n(b) $$(A^c \\cup B)^c \\cap A = A - B$$\n(c) $$(A \\cup B) - A = B- A$$\n(d) $$(A \\cup B) - B = A - (A \\cap B)$$\n(e) $$(A \\cup B) - (A \\cap B) = (A - B) \\cup (B - A)$$\n12. Evaluation of proofs\nSee the instructions for Exercise (19) on page 100 from Section 3.1.\n\n(a)\n\nLet $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$, then $$A - (B - C) = A - (B \\cup C)$$.\n\nProof\n\n$\\begin{array} {rcl} {A - (B - C)} &= & {(A - B) - (A - C)} \\\\ {} &= & {(A \\cap B^c) \\cap (A \\cap C^c)}\\\\ {} &= & {A \\cap (B^c \\cap C^c)} \\\\ {} &= & {A \\cap (B \\cup C)^c} \\\\ {} &= & {A - (B \\cup C)} \\end{array}$\n\nTheorem $$\\PageIndex{1}$$\n\nLet $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$, then $$A - (B \\cup C) = (A - B) \\cap (A - C)$$.\n\nProof\n\nWe first write $$A - (B \\cup C) = A \\cap (B \\cup C)^c$$ and then use one of De Morgan's Laws to obtain\n\n$$A - (B \\cup C) = A \\cap (B^c \\cap C^c)$$.\n\nWe now use the fact that $$A = A \\cap A$$ and obtain\n\n$\\begin{array} {rcl} {A - (B \\cup C)} &= & {A \\cap A \\cap B^c \\cap C^c} \\\\ {} &= & {(A \\cap B^c) \\cap (A \\cap C^c)} \\\\ {} &= & {(A - B) \\cap (A - C).} \\end{array}$\n\nExplorations and Activities\n\n13. (Comparison to Properties of the Real Numbers). The following are some of the basic properties of addition and multiplication of real numbers\n\nCommutative Laws: $$a + b = b + a$$, for all $$a, b \\in \\mathbb{R}$$.\n$$a \\cdot b = b \\cdot a$$, for all $$a, b \\in \\mathbb{R}$$.\n\nAssociative Laws: $$(a + b) + c = a + (b + c)$$, for all $$a, b, c \\in \\mathbb{R}$$.\n$$(a \\cdot b) \\cdot c = a \\cdot (b \\cdot c)$$, for all $$a, b, c \\in \\mathbb{R}$$.\n\nDistributive Law: $$a \\cdot (b + c) = a \\cdot b + a \\cdot c$$, for all $$a, b, c \\in \\mathbb{R}$$.\n\nAdditive Identity: For all $$a \\in \\mathbb{R}$$, $$a + 0 = a = 0 + a$$.\n\nMultiplicative Identity: For all $$a \\in \\mathbb{R}$$, $$a \\cdot 1 = a = 1 \\cdot a$$.\n\nAdditive Inverses: For all $$a \\in \\mathbb{R}$$, $$a + (-a) = 0 = (-a) + a$$.\n\nMultiplicative Inverses: For all $$a \\in \\mathbb{R}$$ with $$a \\ne 0$$, $$a \\cdot a^{-1} = 1 = a^{-1} \\cdot a$$.\n\nDiscuss the similarities and differences among the properties of addition and multiplication of real numbers and the properties of union and intersection of sets.", "date": "2020-04-08 05:34:06", "meta": {"domain": "libretexts.org", "url": "https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning_-_Writing_and_Proof_(Sundstrom)/5%3A_Set_Theory/5.3%3A_Properties_of_Set_Operations", "openwebmath_score": 0.9510489106178284, "openwebmath_perplexity": 115.39302686118575, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.990874363958236, "lm_q2_score": 0.8705972751232809, "lm_q1q2_score": 0.8626525212515543}}
{"url": "https://brilliant.org/discussions/thread/advanced-factorization/", "text": "## Definition\n\nFactorization is the decomposition of an expression into a product of its factors.\n\nThe following are common factorizations.\n\n1. For any positive integer $$n$$, $a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \\ldots + ab^{n-2} + b^{n-1} ).$ In particular, for $$n=2$$, we have $$a^2-b^2=(a-b)(a+b)$$.\n\n2. For $n$ an odd positive integer, $a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \\ldots - ab^{n-2} + b^{n-1} ).$\n\n3. $a^2 \\pm 2ab + b^2 = (a\\pm b)^2$\n\n4. $x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)$\n\n5. $(ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2)$. $(ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2)$.\n\n6. $x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x)$.\n\nFactorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.\n\n## Worked Examples\n\n### 1. Find all ordered pairs of positive integer solutions $(x,y)$ such that $2^x+ 1 = y^2$.\n\nSolution: We have $2^x = y^2-1 = (y-1)(y+1)$. Since the factors $(y-1)$ and $(y+1)$ on the right hand side are integers whose product is a power of 2, both $(y-1)$ and $(y+1)$ must be powers of 2. Furthermore, their difference is\n\n$(y+1)-(y-1)=2,$\n\nimplying the factors must be $y+1 = 4$ and $y-1 = 2$. This gives $y=3$, and thus $x=3$. Therefore, $(3, 3)$ is the only solution.\n\n### 2. Factorize the polynomial\n\n$f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).$\n\nSolution: Observe that if $a=b$, then $f(a, a, c) =0$; if $b=c$, then $f(a, b, b)=0$; and if $c=a$, then $f(c,b,c)=0$. By the Remainder-Factor Theorem, $(a-b), (b-c),$ and $(c-a)$ are factors of $f(a,b,c)$. This allows us to factorize\n\n$f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).$\n\nNote by Calvin Lin\n7\u00a0years, 5\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nFor the worked example - 1 : there are 2 solutions : (3,3) and (3,-3)\n\n(y+1)-(y-1) = 2\n\nimplies y + 1 = -2 or y + 1 = 4 and y -1 = -4 or y-1 = 2\n\nThus, the two solutions you have are (3,3) and (3,-3)\n\n- 7\u00a0years, 3\u00a0months ago\n\nThanks. I added in \"positive integers\".\n\nStaff - 6\u00a0years, 3\u00a0months ago\n\nThanks for this.\n\n- 7\u00a0years, 3\u00a0months ago\n\nI did not understand why should (y+1) - (y-1) = 2 ?\nCan anyone explain ?\n\n- 6\u00a0years, 3\u00a0months ago\n\nWhat do you not understand?\n\nWhat do you think $(y+1) - (y-1)$ is equal to ?\n\nStaff - 6\u00a0years, 3\u00a0months ago\n\nI did not see it correctly . My fault ! Sir, I have a problem. I want to learn Number theory as is organized here on Brilliant but I don't follow anything beginning from modular inverses.I have tried the wikis but I still don't follow .Please suggest something.\n\n- 6\u00a0years, 3\u00a0months ago\n\nHow is their (y+1)-(y-1) difference 2?\n\n- 7\u00a0years, 3\u00a0months ago\n\n$(y+1) - (y-1)$ $= y +1 - y + 1$ $= 1+1= \\boxed{2}$\n\n- 7\u00a0years, 3\u00a0months ago\n\n(Y+1)-(y-1) = y+1-y+1= 2\n\n- 7\u00a0years, 3\u00a0months ago\n\n2\n\n- 7\u00a0years, 3\u00a0months ago\n\nHow to think that both difference is 2. What is the main moto behind thinking such that????\n\n- 7\u00a0years, 3\u00a0months ago", "date": "2021-09-19 13:23:18", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/advanced-factorization/", "openwebmath_score": 0.9869987964630127, "openwebmath_perplexity": 1674.0776156044535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9908743626108194, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8626525051078022}}
{"url": "http://math.stackexchange.com/questions/141236/markov-chain-transition-matrix", "text": "# Markov chain transition matrix\n\nConsider the Markov chain with the following transition matrix:\n\n$$P = \\pmatrix{0& 0.5 &0 &0 &0 &0.5\\\\ 0.25 &0 &0.25 &0.25 &0 &0.25\\\\ 0 &0.5 &0 &0.5 &0 &0\\\\ 0 &0.25 &0.25 &0 &0.25 &0.25\\\\ 0 &0 &0 &0.5 &0 &0.5\\\\ 0.25 &0.25 &0 &0.25 &0.25 &0}$$\n\nI'm trying to show that this chain is irreducible and aperiodic, and finding the stationary probability distribution of the chain by showing that the chain is reversible.\n\nMy attempt is to draw the state space of this, and aperiodic if gcd = 1 and irreducible if there is a path from state 1 to 2 but not vice verca.\n\nThe problem is from here, problem #2.\n\n-\n\nIrreducibility means that you can get from any state to any state. Here it's easy to check that we can get from state 1 to 2 to 3 to 4 to 5 to 6 to 1: the transition probabilities, in that order, are $0.5,0.25,0.5,0.25$, and $0.25$, appearing on the first superdiagonal and in the lower lefthand corner. Here's a graph of the transitions; each goes both ways, so I didn't have to worry about arrows.\n\n                   1       3\n|\\     /|\n| \\   / |\n|  \\ /  |\n|   2   |\n|  / \\  |\n| /   \\ |\n|/     \\|\n6-------4\n\\     /\n\\   /\n\\ /\n5\n\n\nFrom it you can easily check that no matter where you start, you can return to that state in either $2$ or $3$ steps; since $\\gcd(2,3)=1$, the chain is aperiodic.\n\nTo show that the chain is reversible, you must find a probability distribution $$\\pi=\\langle\\pi_1,\\pi_2,\\pi_3,\\pi_4,\\pi_5,\\pi_6\\rangle$$ such that $\\pi_ip_{ij}=\\pi_jp_{ji}$ for $1\\le i,j\\le 6$. Clearly the equations with $i=j$ are satisfied no matter what $\\pi_i$ is, so we can ignore them. We can also ignore any pair for which $p_{ij}=p_{ji}=0$, since $0=0$ gives no information about $\\pi$. That leaves the following system of equations:\n\n$$\\begin{array}{} 0.5\\pi_1=0.25\\pi_2&&0.5\\pi_1=0.25\\pi_6\\\\ 0.25\\pi_2=0.5\\pi_3&&0.25\\pi_2=0.25\\pi_4&&0.25\\pi_2=0.25\\pi_6\\\\ 0.5\\pi_3=0.25\\pi_4\\\\ 0.25\\pi_4=0.5\\pi_5&&0.25\\pi_4=0.25\\pi_6\\\\ 0.5\\pi_5=0.25\\pi_6 \\end{array}$$\n\nAfter the fractions are cleared, we have this system:\n\n$$\\begin{array}{} 2\\pi_1=\\pi_2&&2\\pi_1=\\pi_6\\\\ \\pi_2=2\\pi_3&&\\pi_2=\\pi_4&&\\pi_2=\\pi_6\\\\ 2\\pi_3=\\pi_4\\\\ \\pi_4=2\\pi_5&&\\pi_4=\\pi_6\\\\ 2\\pi_5=\\pi_6 \\end{array}$$\n\nClearly $\\pi_2=\\pi_4=\\pi_6=2\\pi_1=2\\pi_3=2\\pi_5$. Finally, we know that $\\sum_{i=1}^6\\pi_i=1$, so $$1=\\sum_{i=1}^6\\pi_i=3\\pi_1+3\\pi_2=9\\pi_2\\;,$$ and therefore $$\\pi_1=\\pi_3=\\pi_5=\\frac19\\text{ and }\\pi_2=\\pi_4=\\pi_6=\\frac29\\;.$$ In other words, the distribution $$\\pi=\\frac19\\langle 1,2,1,2,1,2\\rangle$$ satisfies the detailed balance condition and is therefore reversible. Of course this distribution is stationary.\n\n-\nTo prove the reversibility, one can also note that this is the random walk on the electric network where every existing nonoriented edge has conductance 1. A bonus of this approach is then that every measure which assigns to each vertex a weight proportional to the sum of the conductances of the edges adjacent to it, is stationary. Since there are 2 edges adjacent to vertices 1, 3, 5, and 4 edges adjacent to vertices 2, 4, 6, this confirms your formula, with no computation. \u2013\u00a0 Did May 5 '12 at 5:57", "date": "2015-09-04 12:31:40", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/141236/markov-chain-transition-matrix", "openwebmath_score": 0.9039266705513, "openwebmath_perplexity": 105.79987130658307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990874360724436, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.862652503465522}}
{"url": "https://mathhelpboards.com/threads/the-union-of-two-open-sets-is-open.27758/", "text": "# The Union of Two Open Sets is Open\n\n#### G-X\n\n##### New member\nLet $$\\displaystyle x \u2208 A1 \u222a A2$$ then $$\\displaystyle x \u2208 A1$$ or $$\\displaystyle x \u2208 A2$$\n\nIf $$\\displaystyle x \u2208 A1$$, as A1 is open, there exists an r > 0 such that $$\\displaystyle B(x,r) \u2282 A1\u2282 A1 \u222a A2$$ and thus B(x,r) is an open set.\n\nTherefore $$\\displaystyle A1 \u222a A2$$ is an open set.\n\nHow does this prove that $$\\displaystyle A1 \u222a A2$$ is an open set. It just proved that $$\\displaystyle A1 \u222a A2$$ contains an open set; not that the entire set will be open? This is very similar to the statement: An open subset of R is a subset E of R such that for every x in E there exists \u03f5 > 0 such that B\u03f5(x) is contained in E.\n\nLast edited:\n\n#### castor28\n\n##### Well-known member\nMHB Math Scholar\nThe point is that the argument is valid for every $x\\in A_1\\cup A_2$.\n\nIf $C = A_1\\cup A_2$, we have proved that, for every $x\\in C$, there is an open ball $B(x,r)\\subset C$ (where $r>0$ depends on $x$). That is precisely the definition of an open set.\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nIf $$\\displaystyle x \u2208 A1$$, as A1 is open, there exists an r > 0 such that $$\\displaystyle B(x,r) \u2282 A1\u2282 A1 \u222a A2$$ and thus B(x,r) is an open set.\n\nTherefore $$\\displaystyle A1 \u222a A2$$ is an open set.\nHi G-X, welcome to MHB!\n\nAs castor28 's pointed out, it's about the definition of an open set, which he effectively quoted.\n\nAdditionally that proof is not entirely correct and it is incomplete.\nIt should be for instance:\n\nIf $$\\displaystyle x \u2208 A1$$, as $A1$ is open, there exists an $r > 0$ such that $$\\displaystyle B(x,r) \u2282 A1$$ (from the definition of an open set), which implies that $$\\displaystyle B(x,r)\u2282 A1 \u222a A2$$.\nIf $$\\displaystyle x \u2208 A2$$, as $A2$ is open, there exists an $r > 0$ such that $$\\displaystyle B(x,r) \u2282 A2\u2282 A1 \u222a A2$$.\nTherefore for all $$\\displaystyle x \u2208 A1\u222a A2$$, there exists an $r > 0$ such that $$\\displaystyle B(x,r) \u2282 A1 \u222a A2$$.\n\nThus $$\\displaystyle A1 \u222a A2$$ is an open set.\n\n#### G-X\n\n##### New member\nI see, I think I had the misunderstanding that something from A2 might close A1.\n\nBut I don't think that is an issue you technically need to wrap your head around.\n\nBecause the definition states: We define a set U to be open if for each point x in U there exists an open ball B centered at x contained in U.\n\nSo, essentially looping over A1, A2 - making the reference that open balls exist at each of these points then all points in A1 \u222a A2 have open balls contained in the union thus by the definition it must be open.\n\nLast edited:", "date": "2021-06-18 06:16:40", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/the-union-of-two-open-sets-is-open.27758/", "openwebmath_score": 0.9782829284667969, "openwebmath_perplexity": 254.2645668060127, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9740426428022032, "lm_q2_score": 0.8856314753275019, "lm_q1q2_score": 0.8626428227768141}}
{"url": "https://math.stackexchange.com/questions/1487873/what-is-the-best-way-of-resolving-an-expression-with-square-roots-in-denominator/1487936", "text": "What is the best way of resolving an expression with square roots in denominator?\n\nI was resolving the following question from my textbook:\n\nWrite each of the following expressions as a single fraction, simplifying your answer where possible:\n\n$4 - \\frac{1}{\\sqrt{12}} + \\frac{10}{\\sqrt{3}}$\n\nI have resolved it this way:\n\n$\\frac{4\\sqrt{12}\\sqrt{3}}{\\sqrt{12}\\sqrt{3}} - \\frac{\\sqrt{3}}{\\sqrt{12}\\sqrt{3}} + \\frac{10\\sqrt{12}}{\\sqrt{12}\\sqrt{3}} = \\frac{4\\sqrt{36} - \\sqrt{3}+10\\sqrt{4}\\sqrt{3}}{\\sqrt{36}}=\\frac{24 - \\sqrt{3} + 20\\sqrt{3}}{6}$\n\nwhile in the textbook it is resolved like this:\n\n$4 - \\frac{1}{\\sqrt{4 \\times 3}} + \\frac{10}{\\sqrt{3}} = 4 - \\frac{1}{2\\sqrt{3}} + \\frac{10}{\\sqrt{3}} = \\frac{8\\sqrt{3} - 1 + 20}{2\\sqrt{3}}= \\frac{8\\sqrt{3} + 19}{2\\sqrt{3}}$\n\nI am wondering did I do it correctly and if so, which way is better?\n\n\u2022 Yes, your answer is correct. The only difference is a factor $\\sqrt 3\\over \\sqrt 3$ in your answer vs. that in the book, and adding $-\\sqrt 3$ to $20\\sqrt 3$. \u2013\u00a0abiessu Oct 19 '15 at 18:26\n\u2022 This indeed answerd my question. Would you mind posting it as an answer so i can mark the thread as answered. \u2013\u00a0Pawel Oct 19 '15 at 18:45\n\n1 Answer\n\nYour answer of $$\\frac{24 - \\sqrt{3} + 20\\sqrt{3}}{6}=\\frac{24 + 19\\sqrt{3}}{6}$$\n\nmatches the answer given by the book within a factor of $\\sqrt 3\\over \\sqrt 3$:\n\n$$\\frac{24 + 19\\sqrt{3}}{6}=\\frac{8\\sqrt3+19}{2\\sqrt 3}$$\n\nYour method was effective in solving the problem, although putting $\\sqrt{12}\\sqrt 3$ as the denominator does not follow the \"usual\" method of coming up with a \"least common multiple\" for the final denominator; in this case, we have $(12,3)=3$, and therefore we would use $\\sqrt {12}=2\\sqrt 3$ as the final denominator and avoid a couple extra multiplications in the process. But another intent in solving problems like this is often to \"rationalize the denominator\", which your method achieves quite effectively. In the end, you'll want to first make sure that you understand the problem and the solution, and then follow the guidelines of the instructor you are working under: if you have a professor or teacher, match what their expectations are in the final answer form; if it's just a book or self-learning, try to match \"common\" notation and approaches that you see.", "date": "2019-10-21 22:49:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1487873/what-is-the-best-way-of-resolving-an-expression-with-square-roots-in-denominator/1487936", "openwebmath_score": 0.7707001566886902, "openwebmath_perplexity": 188.8901978659557, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9609517083920618, "lm_q2_score": 0.8976952900545975, "lm_q1q2_score": 0.8626418225934729}}
{"url": "https://math.stackexchange.com/questions/513250/conditional-probability-question-with-cards-where-the-other-side-color-is-to-be", "text": "# Conditional probability question with cards where the other side color is to be guessed\n\nA box contains three cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random, and the color on one side is observed. If this side is green, what is the probability that the other side of the card is also green?\n\nI think the answer should be $\\frac{1}{2}$ as once the card is selected with one side green, there remain only two possibilities for the other side: either red or green.\n\nBut the answer to this question is $\\frac{2}{3}$.\n\nSo,where am I wrong?Why the answer $\\frac{2}{3}$ is the $\\frac{1}{2}$ wrong?\n\n\u2022 There are three green sides. In 2 cases, the other side is green; in one, the other side is red. \u2013\u00a0Gerry Myerson Oct 3 '13 at 9:44\n\u2022 Possible duplicate of Probability problem \u2013\u00a0shoover Jun 25 '18 at 18:56\n\nGo out from 6 sides. In 3 cases a green side shows up, and in 2 of these 3 cases the other side is green as well.\n\nYou can work this out directly from the definition of conditional probability:\n\n$$P(G_2\\mid G_1) = \\frac{P(G_1 \\cap G_2)}{P(G_1)}$$\n\nExactly one of the three cards has two green sides, so $P(G_1 \\cap G_2) = 1/3$. Exactly three of the six sides that could be seen initially are green, so $P(G_1)=1/2$. Thus\n\n\\begin{align*} P(G_2\\mid G_1) &= \\frac{1/3}{1/2} \\\\ &= \\frac{2}{3}. \\end{align*}\n\nWhen you see a green side then it's more likely that you have the double-green card, since this has twice as many green sides as the red-green card.\n\n\u2022 I like this as a nice, intuitive way of coming to the correct answer. \u2013\u00a0mattdm Jan 26 '15 at 14:00\n\nThe two-sided green card can be observed to have a green side in two distinct ways.\n\nLet's suppose the three cards are C1(with both sides green), C2(with one side green and another red) and C3(both sides are red). Each has two sides (mark A and B)\n\nNow, condition is that the card selected has one side green. So, probability of C3 selection is 0.\n\nNow, Collect the total cases with one side green:\n\n1. C1-A\n2. C1-B\n3. C2-A (suppose, 'A' as green and 'B' as red)\n\n\nAnd, favourable cases (other side should be green too) are:\n\n1. C1-A\n2. C1-B\n\n\nso, probability is: 2 / 3\n\nSay the cards are labelled as $C_1,C_2,C_3$ with $C_1$ being red on both sides, $C_2$ being green on both sides and $C_3$ being green on one side and red on the other. Also, let $S$ be the color of the side that was picked. Then, $$P(C_2|S=G) = \\frac{P(S=G|C_2)P(C_2)}{P(S=G)} = \\frac{1\\times\\frac{1}{3}}{\\frac{1}{2}} = \\frac{2}{3}$$", "date": "2019-10-23 23:53:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/513250/conditional-probability-question-with-cards-where-the-other-side-color-is-to-be", "openwebmath_score": 0.8974482417106628, "openwebmath_perplexity": 591.0796360410864, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085084750966, "lm_q2_score": 0.8774767954920548, "lm_q1q2_score": 0.8626343498633128}}
{"url": "http://mathhelpforum.com/statistics/17355-newbie-help-please.html", "text": "# Math Help - Newbie help please\n\n1. ## Newbie help please\n\nI have used some basic inferential statistics at work but have not really done much on probability before. (School was a long time ago!) I was wondering if someone could help me with the following:\n\nI have 2 normal decks of cards. I pull out 6 cards at random from the first deck and then pull out 6 cards at random from the second deck.\n\nI wanted to know two things.\n\n1) What the probability is of one of the six cards drawn from deck B being a match with one of the six cards drawn from deck A.\n\n2) What the probability is of two of the six cards drawn from deck B being a match with two of the cards drawn from deck A\n\nI calculated the first problem by working out the probability of not having a match and then subtracting it from 1. i.e for the first card drawn from deck B there is 46 in 52 chance that the card wont match, for the second card drawn there will be a 45 in 51 chance (as there is now one less card in the deck.) and so on.\n\nEg 46/52 x 45/51 x 44/50 x 43/49 x 42/48 x 41/47 = .46 therefore 1-.46 = .54. Hence the probability of having one card match is 54%\n\nAlthough I think this is correct, I can\u2019t quite figure out how to go about working out the second problem.\n\nCan anyone give me a point in the right direction please.\nCheers\n\n2. Originally Posted by emersong\n1) What the probability is of one of the six cards drawn from deck B being a match with one of the six cards drawn from deck A.\nI understand this question as:\n\nGiven 6 cards, what is probability there is at least one pair.\n\nHe have the following cases,\nCode:\nMMXXXX\nMXMXXX\n...\nWhere \"M\" stands for match and \"X\" stands for any. There are 14 cases if you write out the list. In each of these cases the probability is:\n(52/52) ---> The first card being any card.\n(3/51) ---> The next card got to match.\n(50/50) ---> The next can be anything.\n(49/49) ---> The next can be anything.\n(48/48) ---> The next can be anything.\n(47/47) ---> The next can be anything.\nMultiply then out to get,\n3/51 = 1/17\nNow multiply this by 12 because that is the number of disjoint cases:\n(12/17)\n\n3. Originally Posted by galactus\nOops, I see PH got something different.\nIt does not mean you are wrong. Perhaps we understand the problem differently, or perhaps I did it wrong.\n\nMy understanding is this.\n\n6 cards are dealt. What is the probability that 2 of those cards match (have same number).\n\n4. There are C(52,6) possible 6-card hands from deck A. Pick any one of those.\nThere are C(46,6) possible 6-card hands from deck B that have no card in common with the hand we picked from deck A.\nTherefore the probability of at least one match is: $1 - \\frac{C(46,6)}{C(52,6)} = 0.5399$\n\nThat agrees with emersong.\n\n5. Originally Posted by Plato\nTherefore the probability of at least one match is: $1 - \\frac{C(46,6)}{C(52,6)} = 0.5399$\n\nThat agrees with emersong.\nI think that is why my answer does not agree with yours. If you follow what I did, I am sure it was perfectly right. I partitioned the cases into disjoint sets and summed each one up. But still that does not do the \"at least\" part.\n\n6. Hi guys and thanks for the help. Unfortunately I am still confused.\n\nSorry for not explaining it too well.\n\nThe reason for two decks is that I was treating each card as unique, not just looking to see if it was the same number. So...\n\n(Where H = Hearts, C = Clubs, S = Spades, D = Diamonds)\n\nDeck 1... 1H,2D,3S,4C,5S,6D\nDeck 2... 8H,9S,1S,3S,7S,4H\n\nIn the above example the 3 of spades (3S) repeats.\n\nI also wanted to know the probability of getting 2 cards to repeat.\n\nDeck 1... 1H,2D,8H,4C,5S,9S\nDeck 2... 7H,9S,1S,8H,7S,4H\n\nIn the above example the 8 of hearts and the 9 of spades both repeat.\n\nI did a little program in excel and ran 10,000 trials, For the first problem I got a probability of .545. This is very near to the probability of .54 that I calculated.\n\nAlthough I am not sure how to work out the second case, I did the simulation and the probability came out as .142.\n\nSo I know the answer to the question, I just would like to understand how I got there!\n\nThanks again for your help.\n\n7. Hello, emersong!\n\nWe need some clarification . . .\n\nI have two normal decks of cards.\nI pull out 6 cards at random from the first deck\nand then pull out 6 cards at random from the second deck.\n\n1) What is the probability that one of the cards drawn from deck B\nmatches one of the cards from deck A?\n\n2) What is the probability that two of the cards drawn from deck B\nmatches two of the cards from deck A?\nIn part (1), if you mean exactly one match, that's a different game entirely.\n\nSix cards are drawn from deck A.\n. . They can be any six cards.\nThere are: . $C(52,6) = 20,358,520$ possible six-card hands.\n\n1) Among the six cards from deck B, there must be exactly one match\n. . . (and five non-matches).\n\nThere are: . $C(6,1)$ ways to get one match.\nThere are: . $C(46,5)$ ways to get five non-matches.\n\nHence, there are: . $C(6,1)\\cdot C(46,5) \\:=\\:(6)(1,370,754) \\:=\\:8,224,425$ ways.\n\nTherefore: . $P(\\text{exactly 1 match}) \\;=\\;\\frac{8,224,524}{20,358,520} \\:=\\:0.403984376 \\:\\approx\\:40.4\\%$\n\n2) Among the six cards from deck B, there must be exactly two matches\n. . . (and four non-matches).\n\nThere are: . $C(6,2)$ ways to have two matches.\nThere are: . $C(46,4)$ ways to have four non-matches.\n\nHence, there are: . $C(6,2)\\cdot C(46,4) \\:=\\:(15)(163,185) \\:=\\:2,447,775$ ways.\n\nTherefore: . $P(\\text{exactly 2 matches}) \\:=\\:\\frac{2,447,775}{20.358,520} \\:=\\:0.120233445 \\:\\approx\\:12.0\\%$\n\n8. Do you understand what we said about \u201cat least one\u201d matching pair?\nThe 54% accounts for anywhere from one to six matching pairs.\nIn you first example there is exactly one matching pair. The probability of that happening is $\\frac {6C(46,5)} {C(52,6)} \\approx 40.3\\%$\n\nFor exactly two matching pairs, the probability of that happening is $\\frac {C(6,2)C(46,4)} {C(52,6)} \\approx 12\\%$\n\n9. Firstly, thanks to everyone for their help.\n\nI think I get it, just to recap...\n\nFor exactly 3 matches it would be:\n\nC(6,3)xC(46,3)= 20 x 15,180 = 30,360 ways\n\nHence:\n\nP(exactly 3 matches) = 30,360 / 20,358,520 = 0.0149\n\nAnd P(one or more matches) = 0.54\n\nJust out of interest, what would;\n\nP(two or more matches) be?\n\n10. The probability of two or more matches is $\\sum\\limits_{k = 2}^6 {\\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$\n\nThe probability of N or more matches is $\\sum\\limits_{k = N}^6 {\\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$\n\n11. Originally Posted by Plato\nThe probability of two or more matches is $\\sum\\limits_{k = 2}^6 {\\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$\n\nThe probability of N or more matches is $\\sum\\limits_{k = N}^6 {\\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$\n\nIt's sunk in now. (Doh!) Well it's been a long day!\n\nThanks\nEmersonG\n\n12. Originally Posted by emersong\nThe probability of two or more matches is .262 right?\nNo, I get 0.136.\n\n$\\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}} = \\frac{{\\left( {\\frac{{6!}}{{k!\\left( {6 - k} \\right)!}}} \\right)\\left( {\\frac{{46!}}{{\\left[ {\\left( {6 - k} \\right)!} \\right]\\left[ {\\left( {k + 40} \\right)!} \\right]}}} \\right)}}{{\\frac{{52!}}{{6!(46!)}}}}$", "date": "2015-04-28 07:18:29", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/17355-newbie-help-please.html", "openwebmath_score": 0.7815101146697998, "openwebmath_perplexity": 592.5927469750953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850892111574, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8626343427524165}}
{"url": "https://math.stackexchange.com/questions/461547/whats-the-equation-of-helix-surface", "text": "what's the equation of helix surface?\n\nI know for the helix, the equation can be written: $$x=R\\cos(t)$$ $$y=R\\sin(t)$$ $$z=ht$$ this is the helix curve, and there are two parameters: outer radius $R$ and the pitch length $2\\pi h$. However, I would like to generate the 3D helix with another minor radius $r$. This is not the helix curve, but a 3D object something like spring. I don't know exactly the name of such structure, but when I search helix equation, they usually give the equations for helix curve but not for the 3D helix object (spring). Does anyone know the equation of such object? Thank you so much for any help and suggestions.\n\nps. the shape looks like the following way:\n\n\u2022 Do you mean a second helix of radius $r$ curling around the path of the helix with radius $R$? \u2013\u00a0Neal Aug 7 '13 at 3:05\n\u2022 You might include a picture of the structure you want in your question. \u2013\u00a0Neal Aug 7 '13 at 3:10\n\u2022 I would start trying with the parametrization $$(x(t),y(t),z(t))+r(\\cos u) \\vec{n}(t)+r(\\sin u)\\vec{b}(t),$$ where $\\vec{n}(t)$ and $\\vec{b}(t)$ are the normal and binormal of the helix curve respectively. IIRC those are both continuous, so no problems. Need to check. \u2013\u00a0Jyrki Lahtonen Aug 7 '13 at 4:20\n\u2022 @JyrkiLahtonen, yes, that's correct. For $r$ small enough the resulting surface is not just continuous but (real) analytic. It appears he is calling your $r=a.$ \u2013\u00a0Will Jagy Aug 7 '13 at 4:33\n\u2022 @Will: I was a bit worried about the possibility of the normal or binormal \"switching sides\", but the helix has constant curvature and torsion, so can't happen (I think). Typing an answer together with Mathematica graphics... \u2013\u00a0Jyrki Lahtonen Aug 7 '13 at 4:40\n\nWe can use a local orthonormal basis of a parametrized curve to get a surface of the desired type.\n\nA helix running around the $x$-axis has a parametrization like $$\\vec{r}(t)=(ht,R\\cos t, R\\sin t).$$ Its tangent vector can be gotten by differentiating $$\\vec{t}=\\frac{d\\vec{r}(t)}{dt}=(h,-R\\sin t,R\\cos t).$$ We note that this has constant length $\\sqrt{h^2+R^2}$. With a more general curve this is not necessarily the case, and we would normalize this to unit length, and switch to using the natural parameter $s=$ the arc length. This time $ds/dt=\\sqrt{h^2+R^2}$, and we can keep using $t$ as long as we remember to normalize.\n\nWe get a (local) normal $\\vec{n}(t)$ vector by differentiating the (normalized) tangent $$\\vec{n}(t)= \\frac{\\frac{d\\vec{t}}{dt}}{\\left\\Vert\\frac{d\\vec{t}}{dt}\\right\\Vert}=(0,-\\cos t,-\\sin t).$$ As the name suggests this is orthogonal to the tangent vector (in the direction of change of the tangent). The third basis vector is the binormal $$\\vec{b}(t)=\\frac1{\\Vert\\vec{t}\\Vert}\\vec{t}\\times\\vec{n}=\\frac{1}{\\sqrt{R^2+h^2}}(R,h\\sin t,-h\\cos t).$$ This is, of course, orthogonal to both $\\vec{t}$ and $\\vec{n}$.\n\nThe key is that we get the desired surface by drawing (3D-)circles with axis direction determined by the direction of the curve, i.e. the tangent. Equivalently, we draw a circle of radius $a$ in the plane spanned by $\\vec{n}$ and $\\vec{b}$. Hence we get the entire surface $S$ parametrized as $$S(t,u)=\\vec{r}(t)+a\\vec{n}(t)\\cos u+ a\\vec{b}(t)\\sin u$$ with $t$ ranging over as many loops as you wish, and $u$ ranging over the interval $[0,2\\pi]$.\n\nHere's what Mathematica-output looks like with parameters $h=1$, $R=3$, $a=0.4$:\n\nHere's the effect of the change to $a=1.0$. The lines on the surface correspond to constant values of $t$ and $u$. These are now more clearly defined.\n\n\u2022 So amazing answers, so beautiful plots, Thank you so much for your answers and help, it is very helpful. \u2013\u00a0Hui Zhang Aug 7 '13 at 5:24\n\u2022 The formulas for the normal and binormal are simpler, if you use the natural parameter of the curve. If you use this same process for another curve, you need to make adjustments for that. \u2013\u00a0Jyrki Lahtonen Aug 7 '13 at 5:24\n\u2022 One more questions, for such surface, is it possible to reduce the equations to in the form? $$f(x,y,z)=g(h,R,a)$$ \u2013\u00a0Hui Zhang Aug 7 '13 at 5:25\n\u2022 Like eliminate the parameters $t$ and $u$? I'm not sure how easy that would be. Haven't thought about it. I would think it's simpler to use the parametrization for most things: calculating tangent planes, normals and such. \u2013\u00a0Jyrki Lahtonen Aug 7 '13 at 5:41\n\u2022 Thank you so much. You answers are very helpful. Actually I have two questions, and this post is in fact the 1st question of generating the surface. My 2nd question is, given a fixed point $(x_0,y_0,z_0)$, how to determine whether this point is inside or outside the object with such surface? That's why I thought whether the equation can reduce to the form $f(x,y,z)=g(h,R,a)$. However, anyway, you have solved the 1st question and it is very helpful. Thank you again. \u2013\u00a0Hui Zhang Aug 7 '13 at 16:33", "date": "2019-07-22 04:27:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/461547/whats-the-equation-of-helix-surface", "openwebmath_score": 0.8883975148200989, "openwebmath_perplexity": 228.97211961555473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085087724427, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.862634341447845}}
{"url": "https://math.stackexchange.com/questions/429563/why-does-uf-p-lf-p-epsilon-make-a-good-criterion-for-integrability", "text": "# Why does $U(f,P) - L(f,P) < \\epsilon$ make a good criterion for integrability?\n\nThat is\n\n$f$ is integrable on $[a,b] \\iff \\forall\\epsilon>0, \\exists P$ of $[a,b]$ such that\n\n$$U(f,P) - L(f,P) < \\epsilon$$\n\nI was thinking that a better definition would be if $U(f,P) = L(f,P)$, but I was corrected that it wouldn't work well for a curve like $f(x) = x$. The geometry just doesn't work.\n\nOn the other bound, saying that given any positive number, I can find a partition such that the difference $U(f,P) - L(f,P) < \\epsilon$ is bounded doesn't feel like a strong enough condition for integrability. Isn't the goal of analysis is always to make $\\epsilon$ as small as possible, and possibly $0$?\n\nPlease see my other question on A terminology to analysts for possible relevance. Thank you\n\n\u2022 Saying that for all $\\varepsilon > 0$, there exists a partition $P$ such that $U(f,P)-L(f,P) < \\varepsilon$ is saying exactly what you are looking for. No matter how small we make $\\varepsilon$, we will be able to find some partition that makes the upper and lower sums differ by at most that value. \u2013\u00a0Cameron Williams Jun 26 '13 at 1:10\n\u2022 But we still an error of epsilon. I was hoping for a more error free definition. I was satisified with $\\sup (L) = \\inf (U)$ \u2013\u00a0Hawk Jun 26 '13 at 1:19\n\u2022 If $U(f,P) \\neq L(f,P)$ for all $P$, then $U(f,P) - L(f,P) > 0$.Thus, given $\\epsilon < \\inf( U(f,P) - L(f,P) )$, there is no $P$ that will satisy $U(f,P)-L(f,p) < \\epsilon$. So, the above two definitions mean the same thing. \u2013\u00a0AnonSubmitter85 Jun 26 '13 at 1:23\n\u2022 @CameronWilliams, want to post that as an answer? \u2013\u00a0Hawk Jun 26 '13 at 1:36\n\u2022 @sidht But you can make the error as small as you want. \u2013\u00a0Pedro Tamaroff Jun 26 '13 at 1:37\n\nAs per my comment above: Saying that for all $\\varepsilon>0$, there exists a partition P such that $U(f,P)\u2212L(f,P)<\\varepsilon$ is saying exactly what you are looking for. No matter how small we make $\\varepsilon$, we will be able to find some partition that makes the upper and lower sums differ by at most that value.\n\nIsn't the goal of analysis is always to make $\\epsilon$ as small as possible, and possibly 0?\n\nYou seem to have a misunderstanding of what $\\epsilon$ stand for here. You are given $\\epsilon>0$ and you want to make something smaller than this $\\epsilon >0$. Maybe this can clear some of this out.\n\nWhy does $U(f,P) - L(f,P) < \\epsilon$ make a good criterion for integrability?\n\nBecause\n\nTHM Let $x,y$ be arbitrary real numbers. If $x<y+\\epsilon$ for each $\\epsilon >0$, then $x\\leq y$.\n\nP By contradiction. Suppose $x>y$. Then $x-y>0$. Take $\\epsilon=x-y$. The above gives $x<y+\\epsilon=y+x-y=x$ which is impossible. It must be the case $x\\leq y$.\n\nNote that since $\\sup L\\leq \\inf U$ is always true, the criterion gives that $\\inf U\\leq \\sup L$ which means $\\sup L=\\inf U$ and $f$ is integrable.\n\nNow, proving that for each $\\epsilon >0$ there exists $P=P_\\epsilon$ such that $$U(f,P)-L(f,P)<\\epsilon$$ is usually easier than proving $\\sup L=\\inf U$ directly, in particular when the function is not given explicitly (say, if we want to prove $f$ is integrable when it is continuous) or any other cases where $f$ is in incognito.\n\nThis is another extended comment, rather an answer.\n\nI would just like to point out the the following two statements are totally different:\n\n\u2022 (1) $\\forall \\epsilon > 0$, $\\exists$ partition $P$ such that $$U(f,P) - L(f,P ) < \\epsilon$$\n\n\u2022 (2) $\\exists$ partition $P$ such that $\\forall \\epsilon > 0$: $$U(f,P) - L(f,P) < \\epsilon.$$\n\nIn (1), the partition $P$ depends on $\\epsilon$. It might be more accurate to write $P$ as $P_\\epsilon$ to remind ourselves of this fact.\n\nIn (2), there is a single partition $P$ -- independent of $\\epsilon$ -- such that for all $\\epsilon > 0$: $U(f,P) - L(f,P) < \\epsilon$ is satisfied. In this situation, because the partition $P$ is independent of $\\epsilon$, we can conclude that $U(f,P) = L(f,P)$.\n\nExample: Contrast the following two statements. Think about how they are radically different:\n\n\u2022 (A) For all $x > 0$, there exists $y > 0$ such that $x < y$.\n\n\u2022 (B) There exists $y > 0$ such that for all $x > 0$: $x < y$.\n\nNote that (A) is obviously true (for any $x> 0$, there's always some number that's bigger), while (B) is obviously false (there is no single number $y$ bigger than every real number).", "date": "2020-02-22 02:03:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/429563/why-does-uf-p-lf-p-epsilon-make-a-good-criterion-for-integrability", "openwebmath_score": 0.8832374215126038, "openwebmath_perplexity": 110.55301458757793, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850857421197, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8626343412833608}}
{"url": "http://mathhelpforum.com/algebra/26280-quadratic-equations-using-formula-method.html", "text": "# Thread: Quadratic equations using the formula method\n\n1. ## Quadratic equations using the formula method\n\nCan anyone solve the following quadratic equation using the formula method...\n\nXsquared - 10x + 3 = 0\n\nPlease show working out so it is possible for me to follow and see where i am going wrong\n\nThanks\n\n2. Originally Posted by duckegg911\nCan anyone solve the following quadratic equation using the formula method...\n\nXsquared - 10x + 3 = 0\n\nPlease show working out so it is possible for me to follow and see where i am going wrong\n\nThanks\nLet me try again...\n\n$x^2 - 10x + 3 = 0$\n\nIt seems he's allowed to use the quadratic formula.\n\nAll quadratic equations are written in the format of $ax^2 + bx + c = 0$\n\nThat means that $a$ is the co-efficient of the $x^2$ ; $b$ is the co-efficient of $x$ ; and $c$ is the constant.\n\nAccording to the quadratic formula:\n\n$x = \\frac{ -b \\pm \\sqrt{b^2 - 4ac} }{2a}$\n\nAll you have to do is substitute the corresponding co-efficients into the formula.\n\n3. A standard quadratic equation is in the form\n$ax^2 + bx + c = 0$\na, b and c being the coefficients.\n\nThe formula for the discriminant ( $\\Delta$) is\n$\\Delta = b^2 - 4ac$\n\nIf $\\Delta > 0$, there are 2 different roots.\nIf $\\Delta = 0$, there are 2 equivalent roots. ( $x_1=x_2$)\nIf $\\Delta < 0$, there are no real roots.\n\nThe roots are,\n\n$x_1 = \\frac{-b - \\sqrt{\\Delta}}{2a}$\n\n$x_2 = \\frac{-b + \\sqrt{\\Delta}}{2a}$\n\nNow plug a, b and c in these formulas.\n\n4. Let's say we write your equation this way.\n\n$ax^2+bx+c=0$\nwhere:\n$a=1$\n$b=-10$\n$c=3$\n\nnow we have the coefficients we use in the equation.\n\n$x_{1,2}=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}$\n\nNow you just put a,b and c into the equation and that's it.\n\nThere are usually 2 results for x.\n\n5. Sorry, was writing the answer while you posted them.\n\n6. Originally Posted by Pinsky\nSorry, was writing the answer while you posted them.\nNo problem, at least now (s)he has 3 explanations to choose from...\n\n7. Originally Posted by Pinsky\nSorry, was writing the answer while you posted them.\nThanks to you both, ill take a long hard look at these explanations and let you both know how i get on ,\n\nthanks again\n\n8. Originally Posted by janvdl\nNo problem, at least now (s)he has 3 explanations to choose from...\nNot to mention the explanations accompanying the examples s/he should have in their class notes .....\n\n9. Originally Posted by janvdl\nLet me try again...\n\n$x^2 - 10x + 3 = 0$\n\nIt seems he's allowed to use the quadratic formula.\n\nAll quadratic equations are written in the format of $ax^2 + bx + c = 0$\n\nThat means that $a$ is the co-efficient of the $x^2$ ; $b$ is the co-efficient of $x$ ; and $c$ is the constant.\n\nAccording to the quadratic formula:\n\n$x = \\frac{ -b \\pm \\sqrt{b^2 - 4ac} }{2a}$\n\nAll you have to do is substitute the corresponding co-efficients into the formula.\nAre the answers for -9.6904 and -3.095 ???\n\n10. Originally Posted by duckegg911\nAre the answers for -9.6904 and -3.095 ???\nI get positive answers, not negative. That's the only problem with your answers.\n\n11. Originally Posted by janvdl\nI get positive answers, not negative. That's the only problem with your answers.\nThis is because - plus a - = a + ???\n\n12. Originally Posted by duckegg911\nThis is because - plus a - = a + ???\nsorry that sound silly....\n\nits just i cant follow why you get possitive answers...\n\nthe answers you have are 9.6904 and 3.095 ?\n\n13. After substitution into the formula, your formula should look like this:\n\n$x = \\frac{ +10 \\pm \\sqrt{100 - 4(1)(3)} }{2}$\n\nDoes it?\n\n14. Here's the full solution. Ask me about any step you do not understand/\n\n$x = \\frac{ +10 \\pm \\sqrt{100 - 4(1)(3)} }{2}$\n\n$x = \\frac{ +10 \\pm \\sqrt{100 - 12} }{2}$\n\n$x = \\frac{ +10 \\pm \\sqrt{88} }{2}$\n\nBUT $\\sqrt{88} = \\sqrt{4} \\times \\sqrt{22}$\n\nAND we know $\\sqrt{4} = 2$\n\n$x = \\frac{ +10 \\pm \\sqrt{4} \\cdot \\sqrt{22} }{2}$\n\n$x = \\frac{ +10 \\pm 2 \\sqrt{22} }{2}$\n\nDivide 10 by 2, and divide $2 \\sqrt{22}$ by 2\n\n$x = +5 \\pm \\sqrt{22}$\n\n15. Originally Posted by janvdl\nHere's the full solution. Ask me about any step you do not understand/\n\n$x = \\frac{ +10 \\pm \\sqrt{100 - 4(1)(3)} }{2}$\n\n$x = \\frac{ +10 \\pm \\sqrt{100 - 12} }{2}$\n\n$x = \\frac{ +10 \\pm \\sqrt{88} }{2}$\n\nBUT $\\sqrt{88} = \\sqrt{4} \\times \\sqrt{22}$\n\nAND we know $\\sqrt{4} = 2$\n\n$x = \\frac{ +10 \\pm \\sqrt{4} \\cdot \\sqrt{22} }{2}$\n\n$x = \\frac{ +10 \\pm 2 \\sqrt{22} }{2}$\n\nDivide 10 by 2, and divide $2 \\sqrt{22}$ by 2\n\n$x = +5 \\pm \\sqrt{22}$\nI have -10 all the way through rather than +10 ???\n\nPage 1 of 2 12 Last", "date": "2016-12-02 20:38:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/26280-quadratic-equations-using-formula-method.html", "openwebmath_score": 0.8593186736106873, "openwebmath_perplexity": 938.5627965060186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850837598123, "lm_q2_score": 0.8774767778695836, "lm_q1q2_score": 0.8626343316692098}}
{"url": "https://math.stackexchange.com/questions/2048650/find-the-last-digit-of-31006", "text": "# Find the last digit of $3^{1006}$\n\nThe way I usually do is to observe the last digit of $3^1$, $3^2$,... and find the loop. Then we divide $1006$ by the loop and see what's the remainder. Is it the best way to solve this question? What if the base number is large? Like $33^{1006}$? Though we can break $33$ into $3 \\times 11$, the exponent of $11$ is still hard to calculate.\n\n\u2022 It's a good way to solve it. The loop is pretty short. But if you know Eulers formula or fermats little formula you don't have to \"find\" the loop. You can calculate it with certainty. At any rate $3^2 \\equiv -1 \\mod 10$ so $3^4 \\equiv 1 \\mod 10$ so the loop is 4 and $1006$ has remainder $2$ so... it's not hard. \u2013\u00a0fleablood Dec 7 '16 at 21:09\n\u2022 Are you familiar with modular arithmetic, i.e. congruences, e.g. $\\,3^2\\equiv -1\\pmod{10}\\,?\\ \\$ \u2013\u00a0Bill Dubuque Dec 7 '16 at 21:19\n\u2022 I think this falls under this umbrella question explaining how to do this and related problems. Please search the site for similar questions before asking. \u2013\u00a0Jyrki Lahtonen Dec 7 '16 at 21:39\n\u2022 See also these posts: Find the last two digits of $7^{81}$ and Find the last two digits of $3^{45}$. And perhaps other posts linked there \u2013\u00a0Martin Sleziak Dec 8 '16 at 3:09\n\nYou have $$3^2=9\\equiv -1\\pmod{10}.$$\n\nAnd $1006=503\\times 2$, so\n\n$$3^{1006}=(3^2)^{503}\\equiv (-1)^{503}\\equiv -1\\equiv 9\\pmod{10}.$$\n\nSo the last digit is $9$.\n\nAnd for something like $11$, you can use the fact that $11\\equiv 1\\pmod {10}.$\n\n$3^{1006}$ or $33^{1006}$\n\ndoesn't really matter\n\n$33\\equiv 3\\pmod {10}\\\\ 33^{1006}\\equiv 3^{1006}\\pmod {10}$\n\n$3^4 = 81$\n\nYou might say this as $3^4\\equiv 1 \\pmod{10}$\n\nThe last digit of $3^n$ is the same last digit as $3^{n+4k}$ that is:\n\n$(3^{n+4k}) = (3^n)(3^{4k})\\equiv (3^n)(1) \\pmod{10}$\n\n$1006 = 251\\cdot 4 + 2$\n\nthe last digit of $3^{1006}$ is the same as the last digit of $3^2$\n\nYou have\n\n\\begin{cases}3^1& =3\\\\ 3^2& =9 \\\\ 3^3&=27\\\\3^4&= 81\\\\3^5&= 243\\end{cases} Thus the last digit repeats after $4$ steps. Since $1006=251\\cdot 4+2$ it is\n\n$$3^{1006}=(3^4)^{251}\\cdot 3^2.$$ The last digit of $(3^4)^{251}$ is $1$ and the last digit of $3^2=9.$ So the answer you are looking for is $9.$\n\nTo compute the last digit of $33^{1006}$ you are in the right way. Since $33=3\\cdot 11$ and the last digit of $11^n$ is $1$ you have that the last digit of $33^{1006}$ is the same of $3^{1006}.$\n\nHINT:\n\nFind the last digit of $3^{1006\\bmod4}$.\n\nAdding a formal solution, just in case anyone will find it useful:\n\nYou are looking for $3^{1006}\\pmod{10}$.\n\nSince $\\gcd(3,10)=1$, by Euler's theorem, $3^{\\phi(10)}\\equiv1\\pmod{10}$.\n\nWe know that $\\phi(10)=\\phi(2\\cdot5)=(2-1)\\cdot(5-1)=4$.\n\nTherefore $3^{4}\\equiv1\\pmod{10}$.\n\nTherefore $3^{1006}\\equiv3^{4\\cdot251+2}\\equiv(\\color\\red{3^4})^{251}\\cdot3^2\\equiv\\color\\red{1}^{251}\\cdot3^2\\equiv1\\cdot9\\equiv9\\pmod{10}$.\n\nThe powers of $3$ cycle from $1\\to3\\to9\\to7$, depending upon the exponent's modulus with respect to $4$. Since $1006\\equiv 2\\pmod{4}$, the last digit of $3^{1006}$ is $9$.\n\nYou can still use this tactic for larger bases. Suppose we want the last digit of $33^{1006}$, as you suggest. Since $33=30+3$, the powers of $33$'s last digit are completely determined by the powers of $3$.\n\nVery soon you will learn Euler's theorem:\n\nIf the greatest common factor of $a$ and $n$ then $a^{\\phi{n}} \\equiv 1 \\mod n$ where $\\phi(n)$ is the number of numbers relatively prime to $n$ that are less than $n$.\n\nAs $1,3,7$ and $9$ are relatively prime to $10$, and $\\gcd(3,10) = 1$ we know $\\phi(10) = 4$ and $3^4 \\equiv 1 \\mod 10$ so $3^{1006} = 3^{4*251 + 2} \\equiv 3^2 \\equiv 9 \\mod 10$.\n\nAs $33 = 3*10 + 3$ $33^n = (30 + 3)^n = 10*something + 3^n$ will have the same last digit. But $\\gcd(33,10) = 1$ so $3^4 \\equiv 1 \\mod 10$. And everything is the same.\n\nWhat would be harder is the last two digits of $33^{1006}$. $\\gcd(33,100) =1$ so $33^{\\phi(100)} \\equiv 1 \\mod 100$. But what is $\\phi(100)$?\n\nThere is a thereom that $\\phi(p) = p-1$ and that $\\phi(p^k) = p^{k-1}(p-1)$ and then $\\phi(mn) = \\phi(m)\\phi(n)$ so $\\phi(100)=\\phi(2^2)\\phi(5^2) = 2*1*5*4 = 40$. So there are $40$ numbers less than $100$ relatively prime to $100$.\n\nand $33^{40} \\equiv 1 \\mod 100$ so $33^{1006 = 40*25 +6} \\equiv 33^6 \\mod 100 \\equiv 30^2 + 2*3*30 + 3^2 \\equiv 189 \\equiv 89 \\mod 100$. The last two digits are $89$.\n\nSee:\n\nhttps://en.wikipedia.org/wiki/Modular_arithmetic\n\nhttps://en.wikipedia.org/wiki/Euler%27s_theorem\n\nhttps://en.wikipedia.org/wiki/Euler%27s_totient_function", "date": "2021-05-18 02:22:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2048650/find-the-last-digit-of-31006", "openwebmath_score": 0.8501229286193848, "openwebmath_perplexity": 141.8681791810732, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419707248646, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8626137036968219}}
{"url": "https://byjus.com/question-answer/100-surnames-were-randomly-picked-up-from-a-local-telephone-directory-and-the-frequency-distribution-35/", "text": "Question\n\n# $$100$$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters$$1-4$$$$4-7$$$$7-10$$$$10-13$$$$13-16$$$$16-19$$Number of surnames$$6$$$$30$$$$40$$$$16$$$$4$$$$4$$Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.\n\nSolution\n\n## Let us prepare the following table to compute the median :Number of letters\u00a0\u00a0Number of surnames (Frequency)\u00a0Cumulative frequency\u00a0$$1-4$$$$6$$\u00a0$$6$$\u00a0\u00a0$$4-7$$$$30$$\u00a0$$36$$\u00a0\u00a0$$7-10$$$$40$$\u00a0$$76$$\u00a0\u00a0$$10-13$$$$16$$\u00a0$$92$$\u00a0\u00a0$$13-16$$$$4$$\u00a0$$96$$\u00a0\u00a0$$16-19$$$$4$$\u00a0$$100=n$$\u00a0We have, $$n = 100$$$$\\Rightarrow \\dfrac n2 = 50$$The cumulative frequency just greater than $$\\dfrac n2$$ is $$76$$\u00a0and the corresponding class is $$7 \u2013 10$$.\u00a0Thus, $$7 \u2013 10$$ is the\u00a0median class such that$$\\dfrac n2 = 50, l = 7, f = 40, cf = 36$$ and $$h=3$$Substitute these values in the formulaMedian, $$M = l+\\left(\\dfrac{\\dfrac n2 - cf}{f}\\right)\\times h$$$$M = 7+\\left(\\dfrac{50-36}{40}\\right)\\times 3$$$$M = 7+\\dfrac{14}{40}\\times3 = 7 + 1.05 = 8.05$$Now, calculation of mean:\u00a0Number of letters\u00a0Mid-Point $$(x_i)$$Frequency $$(f_i)$$\u00a0$$f_ix_i$$\u00a0$$1-4$$$$2.5$$$$6$$\u00a0$$15$$\u00a0\u00a0$$4-7$$$$5.5$$\u00a0$$30$$\u00a0$$165$$\u00a0\u00a0$$7-10$$$$8.5$$\u00a0$$40$$\u00a0$$340$$\u00a0\u00a0$$10-13$$$$11.5$$\u00a0$$16$$\u00a0$$184$$\u00a0\u00a0$$13-16$$$$14.5$$\u00a0$$4$$\u00a0$$58$$\u00a0\u00a0$$16-19$$$$17.5$$\u00a0$$4$$\u00a0$$70$$\u00a0\u00a0Total\u00a0$$100$$\u00a0$$832$$\u00a0Therefore, Mean, $$\\bar x = \\dfrac{\\sum f_ix_i}{\\sum f_i} = \\dfrac{832}{100} = 8.32$$Calculation ofMode:The class $$7 \u2013 10$$ has the maximum frequency therefore, this is the modal class.Here,\u00a0$$l = 7, h = 3, f_1 = 40, f_0 = 30$$ and $$f_2 = 16$$Now, let us substitute these values in the formulaMode $$= l+ \\left(\\dfrac{f_1-f_0}{2f_1-f_0-f_2}\\right)\\times h$$$$= 7+\\dfrac{40-30}{80-30-16}\\times3$$$$= 7 + \\dfrac{10}{34}\\times 3 = 7+0.88 = 7.88$$Hence, median $$= 8.05$$, mean $$= 8.32$$ and mode $$= 7.88$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-21 15:23:51", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/100-surnames-were-randomly-picked-up-from-a-local-telephone-directory-and-the-frequency-distribution-35/", "openwebmath_score": 0.8743242621421814, "openwebmath_perplexity": 479.2301685505272, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9888419713825141, "lm_q2_score": 0.8723473730188543, "lm_q1q2_score": 0.8626136960663213}}
{"url": "https://www.math.purdue.edu/pow/discussion/2016/spring/15", "text": "## Spring 2016, problem 15\n\nA lattice point is defined as a point in the $2$-dimensional plane with integral coordinates. We define the centroid of four points $(x_i,y_i )$, $i = 1, 2, 3, 4$, as the point $\\left(\\frac{x_1 +x_2 +x_3 +x_4}{4},\\frac{y_1 +y_2 +y_3 +y_4 }{4}\\right)$. Let $n$ be the largest natural number for which there are $n$ distinct lattice points in the plane such that the centroid of any four of them is not a lattice point. Show that $n = 12$.\n\n2 years ago\n\nTo prove that any set of 13 lattice points contains a 4-point subset with a lattice-point centroid, first we prove that any set of at least 5 lattice points contains a 2-point subset with a lattice-point centroid: Apply the pigeonhole principle to the four classes of lattice points (even,odd), (odd,even), (even,even), (odd,odd) to find a class which contains two of the 5 points. (For example, we might find that there are two points each of which have first coordinate even and second coordinate odd.) The sum of these two points is (even,even) so their centroid is a lattice point, as desired.\n\nNext consider an arbitrary set $A$ of 13 lattice points. Apply the above result to find a pair in $A$ with a lattice-point centroid. Apply it again to the remaining 11 points of $A$, then to the remaining 9, and yet twice more, so that we find a total of 5 such pairs in $A$. Finally, apply it one more time, to the set of lattice-point centroids of these 5 pairs. This gives us a pair of pairs, whose 4 points have a lattice-point centroid, as desired.\n\nNow we exhibit a set of 12 lattice points, no 4 of which have a lattice-point centroid: Take $B$ such that three of its points are $(0,0)$ modulo 4 (in each coordinate), three are $(0,1)$, three are $(1,0)$, and three are $(1,1)$. Suppose that $S$ is a 4-point subset of $B$ whose centroid is a lattice point. Then the sum of the points of $S$ is $(0,0)$ modulo 4. Thus all 4 points of $S$ must have the same first coordinate (modulo 4): otherwise we are summing a number of $1$'s which lies strictly between 0 and 4. Similarly all 4 points of $S$ share the same second coordinate (modulo 4). But this calls for four points of $B$ which are the same modulo 4, whereas $B$ was chosen to have no more than three such.\n\nThis solution introduces the idea of a centroid for 2 points which I don't think is valid. I guess the terminology for 2 points would be the centre of the line joining them.\n\nphilboyd 2 years ago\n\ndbrown uses the expression \"centroid of two points\" exactly the way you just defined it. If that constitutes a slight misuse of terminology, well... So what. It would be fair to say, that pretty much all posts on this board are guilty in this respect. Note also, that when $C(a,b ,c, ...)$ denotes the centroid of $a, b, c,...$, then:\n\n$C(a,b ,c, d)=C(C(a, b), C( c, d))$\n\nThis is used by dbrown implicitly, when he says:\n\nFinally, apply it one more time, to the set of lattice-point centroids of these 5 pairs. This gives us a pair of pairs, whose 4 points have a lattice-point centroid, as desired.\n\nI think dbrown's argument is entirely sound. And very elegant on top!\n\nNelix 2 years ago\n\n@philboyd\n\nI think the term \"centroid\" is pretty standard for any number of points, but you're right that I was sloppy not to define my terms. Sorry about that. I was using the term to mean the average of a set of points, in the sense that each coordinate of the centroid is given by the average of that coordinate over all the points in the set. This agrees with the definition given in the problem.\n\nI think the term is standard even for solid shapes, where you use a ratio of integrals instead of a finite average.\n\ndbrown 2 years ago\n2 years ago\n\nBy iterating dbrown's argument you can show that if $n$ is a power of two, then:\n\nIn every set of $4n - 3$ integer points you can find a subset of $n$ points with integer centroid.\n\nQuestions:\n\n\u2022 Is this true for general $n$?\n\u2022 Is $4n - 3$ minimal?\n\nI was wondering the same thing.\n\n\u2022 I was able to prove the $4n-3$ result for $n = 3$, in an ad hoc way, just considering cases; I don't remember the details now. It follows that the $4n-3$ result holds for any $n$ which is a power of 2 times a power of 3. (I didn't get anywhere for $n = 5$ though.)\n\u2022 $4n-3$ is minimal for every $n$, using a very similar example to the 12-point example for $n = 4$. Just take $n-1$ points in each of the classes $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$ modulo $n$, to get a set of $4n-4$ points no $n$ of which have a lattice-point centroid.\ndbrown 2 years ago\n2 years ago\n\nWhy have we had no submissions?\n\nIs it because, like me, no one understands it?\n\nIt's fairly obvious that we only need to consider the set of points [0..3, 0..3] since if the problem is satisfied with x1=z, say, then we can replace z with z MOD 4, and so on for all 16 points.\n\nAre we saying that it is possible to select any set of 12 points from the 16 and that all selections of 4 points from the 12 has the property that the centroid is not is a lattice point?\n\nI think not.\n\nI have written a program to test this and it seems that with 8 points it is possible to select 4 where SOME of the selections do not have a centroid which is a lattice point, but certainly not all.\n\nAs soon as we select 9 points (or more) it is always possible to find a set of 4 which does have a centroid which is a lattice point.\n\nIt is possible to choose 12 of those 16 points such that the selection of any 4 have the property that the centroid is not a lattice point, so long as you don't require that the 12 points are distinct mod 4. Since we are working mod 4, in your setup, it is entirely possible that we have identified points that were originally distinct before working mod 4. In fact it is easy to find a collection of 12 non-distinct points mod 4 that satisfy the requisite property, giving a lower bound of 12 on n.\n\nNickM 2 years ago\n\nCould you list such a set of 12 points please? I have rechecked my code and it still seems that it is possible to select 8 points which satisfy the requirement, but 9 points does not. e.g. for 8 points we can have (0,0) (0,0) (0,0) (0,0) (0,0) (1,1) (1,2) (1,3) where the various (0,0) points would be (4,8) (20, 32) and the like.\n\nphilboyd 2 years ago\n\nWe couldn't have 4 points with $(0,0)$ mod 4 in such a set since their sum would then have both coefficients divisible mod 4. A collection satisfying the requirement would be something like 3 copies each of points with mod 4 representatives given by $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. You can check that the sum of any choice of 4 of these points will have a centroid which is not equivalent to $(0,0)$ mod 4.\n\nNickM 2 years ago\n\nWhoops !\n\nYes, program has confirmed result for 12 points exactly as you said.\n\nAs expected it is unable to find a selection of 13 which satisfy the same condition.\n\nHowever, I see that we still don't actually have a proof that 12 is the largest value for n.\n\nphilboyd 2 years ago", "date": "2018-07-21 09:56:07", "meta": {"domain": "purdue.edu", "url": "https://www.math.purdue.edu/pow/discussion/2016/spring/15", "openwebmath_score": 0.8303226828575134, "openwebmath_perplexity": 207.49146620839028, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419702316276, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8626136835764684}}
{"url": "https://math.stackexchange.com/questions/3502507/you-started-with-one-chip-you-need-to-get-4-chips-to-win-what-is-the-probabili", "text": "You started with one chip. You need to get 4 chips to win. What is the probability that you will win?\n\nThis is very similar to the question I've just asked, except now the requirement is to gain $$4$$ chips to win (instead of $$3$$)\n\nThe game is:\n\nYou start with one chip. You flip a fair coin. If it throws heads, you gain one chip. If it throws tails, you lose one chip. If you have zero chips, you lose the game. If you have four chips, you win. What is the probability that you will win this game?\n\nI've tried to use the identical reasoning used to solve the problem with three chips, but seems like in this case, it doesn't work.\n\nSo the attempt is:\n\nWe will denote $$H$$ as heads and $$T$$ as tails (i.e $$HHH$$ means three heads in a row, $$HT$$ means heads and tails etc)\n\nLet $$p$$ be the probability that you win the game. If you throw $$HHH$$ ($$\\frac{1}{8}$$ probability), then you win. If you throw $$HT$$ ($$\\frac{1}{4}$$ probability), then your probability of winning is $$p$$ at this stage. If you throw heads $$HHT$$ ($$\\frac{1}{8}$$ probability), then your probability of winning $$\\frac{1}{2}p$$\n\nHence the recursion formula is\n\n\\begin{align}p & = \\frac{1}{8} + \\frac{1}{4}p+ \\frac{1}{8}\\frac{1}{2}p \\\\ &= \\frac{1}{8} + \\frac{1}{4}p +\\frac{1}{16}p \\\\ &= \\frac{1}{8} + \\frac{5}{16}p \\end{align}\n\nSolving for $$p$$ gives\n\n$$\\frac{11}{16}p = \\frac{1}{8} \\implies p = \\frac{16}{88}$$\n\nNow, to verify the accuracy of the solution above, I've tried to calculate the probability of losing using the same logic, namely:\n\nLet $$p$$ denote the probability of losing. If you throw $$T$$ ($$\\frac{1}{2}$$ probability), you lose. If you throw $$H$$ ($$\\frac{1}{2}$$ probability), the probaility of losing at this stage is $$\\frac{1}{2}p$$. If you throw $$HH$$($$\\frac{1}{4}$$ probability), the probability of losing is $$\\frac{1}{4}p$$. Setting up the recursion gives\n\n\\begin{align}p & = \\frac{1}{2} + \\frac{1}{4}p+ \\frac{1}{8}\\frac{1}{2}p \\\\ &= \\frac{1}{2} + \\frac{1}{4}p +\\frac{1}{16}p \\\\ &= \\frac{1}{2} + \\frac{5}{16}p \\end{align}\n\nWhich implies that\n\n$$\\frac{11}{16}p = \\frac{1}{2} \\implies p = \\frac{16}{22} = \\frac{64}{88}$$\n\nWhich means that probabilities of winning and losing the game do not add up to $$1$$.\n\nSo the main question is: Where is the mistake? How to solve it using recursion? (Note that for now, I'm mainly interested in the recursive solution)\n\nAnd the bonus question: Is there a possibility to generalize? I.e to find the formula that will give us the probability of winning the game, given that we need to gain $$n$$ chips to win?\n\n\u2022 Can you share the calculation for losing probability? \u2013\u00a0Dhanvi Sreenivasan Jan 9 at 6:23\n\u2022 @DhanviSreenivasan, I updated the post. \u2013\u00a0Ilya Stokolos Jan 9 at 6:31\n\u2022 But you don't have a finite number of throws, no? \u2013\u00a0Dhanvi Sreenivasan Jan 9 at 6:35\n\u2022 @DhanviSreenivasan True, I don't. \u2013\u00a0Ilya Stokolos Jan 9 at 6:36\n\u2022 Sorry, the game stops if you win. Didn't remember that \u2013\u00a0Dhanvi Sreenivasan Jan 9 at 6:40\n\nThis answer only addresses what's wrong with your recursion, since the other answers (both in this question and your earlier question) already gave many different ways to set up the right recursions (or use other methods).\n\nThe key mistake is what you highlighted. When you throw $$HHT$$, you now have $$2$$ chips. For the special case of this problem, $$2$$ chips is right in the middle between $$0$$ and $$4$$ chips, so the winning prob is obviously $$\\color{red}{\\frac12}$$ by symmetry. But you had it as $$\\color{red}{\\frac12 p}$$ which is wrong. Thus the correct equation is:\n\n$$p = P(HHH) + P(HT) p + P(HHT) \\color{red}{\\frac12}= \\frac18 + \\frac14 p + \\frac18 \\color{red}{\\frac12}$$\n\nLet $$p(n)$$ be the probability that you win the game when you have $$n$$ chips in the pocket. Then $$p(0)=0$$, $$\\>p(4)=1$$. Having $$1\\leq n\\leq3$$ chips one makes a further move, and one then has $$p(n)={1\\over2}p(n-1)+{1\\over2}p(n+1)\\qquad(1\\leq n\\leq3)\\ ,$$ so that $$p(n+1)-p(n)=p(n)-p(n-1)\\qquad(1\\leq n\\leq3)\\ .$$ These circumstances immediately imply that $$p(1)={1\\over4}$$.\n\nLet $$p_0, p_1, \\ldots, p_4$$ be the probability of winning if you start with $$0, 1, \\ldots, 4$$ chips, respectively. Of course, $$p_0 = 0$$ and $$p_4 = 1$$.\n\nThere are a few different ways to approach this question.\n\nStart with $$p_2$$\n\nThis seems to be the approach you're asking for, but it's not the easiest approach.\n\nTo calculate $$p_2$$, consider what happens if the coin is flipped twice. There is a $$1/4$$ chance of getting $$TT$$ (instant loss), a $$1/4$$ chance of getting $$HH$$ (instant win), and a $$1/2$$ chance of getting either $$HT$$ or $$TH$$ (back to $$p_2$$). So we have\n\n$$p_2 = \\frac14 + \\frac12 p_2,$$\n\nwhich we can solve to find that $$p_2 = 1/2$$.\n\nNow that we know $$p_2$$, we can directly calculate $$p_1$$ as the average of $$p_0$$ and $$p_2$$, which is $$1/4$$.\n\nExamine the sequence\n\nNotice that in the sequence $$p_0, p_1, p_2, p_3, p_4$$, each element (besides the first and the last) is the average of its two neighbors. This implies that the sequence is an arithmetic progression. Given that $$p_0 = 0$$ and $$p_4 = 1$$, we can use any \"find a line given two points\" method to find that for all $$n$$, $$p_n = n/4$$.\n\nConservation of expected value\n\nI'm an investor and an advantage gambler (which are the same thing, really), so I like to think of things in terms of expected value.\n\nI start the game with $$1$$ chip, and it's a perfectly fair game; in the long run, I am expected neither to lose nor to win. So if I play the game until it ends, the expected value of the game must be $$1$$ chip. (More detail is needed to make this argument formal, but it's sound.)\n\nThe value if I lose is $$0$$, and the value if I win is $$4$$, so the expected value can also be written as $$(1 - p_1) \\cdot 0 + p_1 \\cdot 4$$, which simplifies to $$4 p_1$$.\n\nThese two ways of calculating the expected value must agree, meaning that $$4 p_1 = 1$$, so $$p_1 = 1/4$$.\n\nIn general\n\nThe latter two of the above arguments can each be generalized to show that if you start with $$a$$ chips, and the game ends when you reach either $$0$$ or $$b$$ chips, then the probability of winning is $$a/b$$.", "date": "2020-09-28 23:16:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3502507/you-started-with-one-chip-you-need-to-get-4-chips-to-win-what-is-the-probabili", "openwebmath_score": 0.9873301982879639, "openwebmath_perplexity": 260.38348337661824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981166874515169, "lm_q2_score": 0.8791467738423873, "lm_q1q2_score": 0.8625896923310293}}
{"url": "https://puzzling.stackexchange.com/questions/114281/splitting-integers-and-taking-differences-how-can-the-sum-be-constant", "text": "# Splitting integers and taking differences, how can the sum be constant?\n\nStart with the set of integers from $$1$$ up to $$2n$$, where $$n$$ is a natural number. Split this set into two disjoint subsets of equal size, say $$\\{a_1 and $$\\{b_1>b_2>\\cdots>b_n\\}$$ (with the elements ordered WLOG). What is the value of the following expression in terms of $$n$$? $$|a_1-b_1|+|a_2-b_2|+\\cdots+|a_n-b_n|$$\n\nSource (link contains spoilers)\n\n\u2022 I don't think ordering the elements is WLOG. Jan 2, 2022 at 14:14\n\u2022 @justforplaylists It's not WLOG for the final expression, of course, but it is WLOG for defining the sets: i.e. we define $a_1$ to be the smallest element, $a_2$ the next smallest, and so on. Jan 2, 2022 at 14:16\n\u2022 Two disjoint and exhaustive subsets? If $n=10$, we could have $\\{1,2\\}$ and $\\{19,20\\}$, which gives $36$, or $\\{1,2\\}$ and $\\{3,4\\}$, which gives $2$. Jan 3, 2022 at 5:40\n\u2022 I was trying to think of a more precise term than \"split\", and reading the answers I was reminded of the word \"partition\", which more clearly indicates exhaustion. Jan 3, 2022 at 6:10\n\u2022 @Randal'Thor This is not WLOG but rather you are defining $a_i$ and $b_i$ in this way. Jan 3, 2022 at 7:14\n\n## 4 Answers\n\nJaap Sherpuis's comment hints at a perhaps more intuitive explanation.\n\nWe need to consider how many a's are $$\\le n$$ and how many b's are $$\\gt n$$.\n\nLet's name $$k$$ the number of $$a_i$$'s that are $$\\le n$$.\n$$a_1 \\dots a_k$$ occupy $$k$$ values from 1 to n, leaving $$(n-k)$$ holes.\nThese holes are filled by $$(n-k)$$ of the $$b_j$$'s. The remaining $$b_j$$ are $$\\gt n$$.\n\nSo we have $$a_1 \\dots a_k \\le n$$ and $$a_{k+1} \\dots a_n \\gt n$$.\nAnd we have $$b_1 \\dots b_k \\gt n$$ and $$b_{k+1} \\dots b_n \\le n$$.\nThis implies $$a_i < b_i$$ if and only if $$i \\le k$$\n\nFrom there we can compute\n\n$$|a_1-b_1|+|a_2-b_2|+\\cdots+|a_n-b_n|$$\n$$= |a_1-b_1|+\\cdots+|a_k-b_k|+|a_{k+1}-b_{k+1}|+\\cdots+|a_n-b_n|$$\n$$= (b_1-a_1)+\\cdots+(b_k-a_k)+(a_{k+1}-b_{k+1})+\\cdots+(a_n-b_n)$$\n\nOn the plus side you have all values $$\\gt n$$ and on the minus side the values $$\\le n$$\nSo the expression becomes:\n$$= (n+1) + \\cdots + 2n - (1 + \\cdots + n) = n^2$$\n\nThis is just an \"optimised\" version of Gareth's answer which I couldn't resist making. Please keep upvoting him and also prefer his answer over mine for acceptance.\n\nFor any index $$i$$ let us denote $$M_i$$ the larger of $$a_i,b_i$$ and $$m_i$$ the smaller. Then the given sum of absolute differences can be rewritten as\n\n$$\\displaystyle\\sum_i M_i - \\sum_i m_i$$\n\nWe will now prove that for any pair of indices $$i,j$$ we have $$M_i>m_j$$:\n\nProof:\n\nCase 1, $$i=j$$: obvious from definition\nCase 2, $$i: $$M_i\\ge b_i>b_j\\ge m_j$$\nCase 3, $$i>j$$: $$M_i\\ge a_i>a_j\\ge m_j$$\n\nIt follows that all the $$m_i$$ are smaller than all the $$M_i$$. This is the same as saying\n\nthe $$m_i$$ must be a permutation of $$1,...,n$$ and the $$M_i$$ a permutation of $$n+1,...,2n$$\n\nThe sum is therefore\n\n$$\\displaystyle\\sum_{i=1}^n(n + i) -\\sum_{i=1}^n i = \\sum_{i=1}^n n = n^2$$\n\nand indeed independent of how the numbers are split into $$a$$ and $$b$$.\n\n\u2022 I don't think this is just an optimized version of my answer, although of course there's some overlap. And I think it's a very nice argument. I've upvoted it and would encourage anyone reading this to do likewise. (And if Rand chooses to accept this one rather than mine, then despite loopy walt's first sentence I think that would be an extremely reasonable decision.) Jan 2, 2022 at 20:14\n\u2022 There is a related principle used in some magic tricks. If you have one suit of cards in order Ace to King, and another suit in reverse order King to Ace, and combine those two piles using a single riffle shuffle, then the top thirteen cards will contain every card value exactly once, as will the bottom thirteen cards. Jan 3, 2022 at 8:55\n\nSo, first of all, let's\n\ndo it one way and see what sum we get. Let the $$a$$s be (in order) $$1,2,\\ldots,n$$ and the $$b$$s be (in order) $$2n,2n-1,\\ldots,n+1$$. Then the differences are, in order, $$2n-1,2n-3,\\ldots,1$$ whose sum is $$n^2$$.\n\nIn view of\n\nthe title, presumably this is always the answer.\n\nBut why? I offer first a rather routine, prosaic solution, the sort of thing someone who's done this sort of thing before knows they will be able to get to work. (This is the first solution I found.) Then something slicker. (This is the second solution I found.) Then a way to streamline the second part of the slicker argument. And then a way to streamline the first part, found not by me but by user loopy walt in comments. I prefer to leave all these things in, rather than just presenting the slickest version so far found, because that's a more honest representation of how things typically work in mathematics; see also this related 3blue1brown video about calculation versus slick insight.\n\nHere is a prosaic solution; I suspect there is something better (and am thinking about that). We can turn any valid partition into any other by repeatedly performing operations of the form \"find two consecutive numbers one of which is an $$a$$ and the other a $$b$$, and swap them\". What happens when we do this? Suppose $$a_i$$ and $$b_j$$ are consecutive. Then none of the other order relations change when we switch them, so our new sequences in order are the same as the old except that $$a_i'=b_j$$ and $$b_j'=a_i$$. If $$i=j$$ then the expression we're interested in doesn't change at all. Otherwise, say that $$i. Then $$b_j \\simeq a_i where $$\\simeq$$ means \"differ by 1\", and therefore $$a_j>b_j$$; and $$a_i\\simeq b_j, and therefore $$a_i. So increasing $$a_i$$ by 1 and decreasing $$b_j$$ by 1 reduces $$|a_i-b_i|$$ by 1 and increases $$a_j-b_j$$ by 1, and vice versa if instead we decrease $$a_i$$ and increase $$b_j$$. In either case our expression doesn't change.\n\nOK, now let's see if we can do something simpler.\n\nWe have all these intervals from $$a_i$$ to $$b_i$$. Every number $$k$$ from $$1$$ to $$2n$$ is an endpoint of exactly one of these intervals; I claim that the first $$n$$ are all left endpoints and the last $$n$$ are all right endpoints. Why? Well, colour all the $$a$$s amber and all the $$b$$s blue; suppose $$k$$ is amber; if $$k=a_i$$ then there are $$i-1$$ amber numbers to its left, hence $$k-i$$ blue numbers to its left, hence $$n-k+i$$ blue numbers to its right, the first of which is $$b_{n-k+i}$$. So $$b_i$$ is right of $$a_i$$ iff $$i\\leq n-k+i$$, iff $$k\\leq n$$. Likewise if $$k$$ is blue.\n\nAnd now\n\nour sum is just the sum over all $$k$$ of the number of these intervals it's in (counting 1/2 when it's exactly at an endpoint), which depends only on the number of left and right endpoints on either side of $$k$$, and we've just seen that this never changes.\n\nOr (a basically-equivalent replacement for the foregoing paragraph):\n\nour sum is just the sum of all the right endpoints minus the sum of all the left endpoints, which is to say $$\\bigl[(n+1)+\\cdots+2n\\bigr]-\\bigl[1+\\cdots+n\\bigr]=n^2$$.\n\nThis is definitely slicker than the first proof above, but I would be unsurprised to find that one can streamline things further.\n\nIn comments, loopy walt suggests another way to do the first bit:\n\ni.e., proving that $$1,...,n$$ are left endpoints and $$n+1,...,2n$$ are right endpoints. Consider the interval whose endpoints are $$a_i,b_i$$. We'll prove that its right-hand endpoint is $$>n$$. Note that $$a_i$$ is bigger than $$i-1$$ smaller $$a$$s, and $$b_i$$ is bigger than $$n-i$$ smaller $$b$$s. So whichever of them is larger is bigger than the other one (1), and $$i-1$$ smaller $$a$$s, and $$n-i$$ smaller $$b$$s, all of those sets being disjoint; hence bigger than at least $$1+i-1+n-i=n$$ other numbers; hence it's $$>n$$.\n\n(To my mind this is neater than my argument but more rabbit-out-of-hat-ish.)\n\n\u2022 Correct answer and reasonable proof, but there's a(nother) surprising fact that makes the proof even simpler. Jan 2, 2022 at 14:27\n\u2022 As mentioned in the answer, I am currently looking for slicker proofs and have some plausible ideas of where to look. There is something of an analogy with a famous question about coins, but that pathway doesn't seem like it really produces something much simpler so far. Jan 2, 2022 at 14:29\n\u2022 How about: ai is larger than i-1 elements (a1,...,a{i-1}), bi is larger than n-i elements (b{i+1},...,bn). The larger of ai and bi is larger than the other one and both these sets, that is 1 + i-1 + n-i elements. Jan 2, 2022 at 15:30\n\u2022 ... and possibly other things too, but at least n elements, which means that all the right endpoints are >= n+1, and as there are n right endpoints that tells us what they all are? Yeah, that works. Jan 2, 2022 at 15:34\n\u2022 The last spoilertag is the kind of argument I was thinking of. Nice example of a surprising non-obvious mathematical fact. Jan 2, 2022 at 17:48\n\nThis problem doesn't really have anything to do with the integers from 1 to 2n. Fix any set consisting of 2n real numbers (repetitions allowed); then split them into a_1 <= a_2 <= .... <= a_n and b_1 >= b_2 >= ... >= b_n. Then the sum |a_1 - b_1| + |a_2 - b_2| + ... + |a_n - b_n| is constant, that is, it doesn't depend on the partition.\n\n\u2022 Nice generalisation, and I suppose the same proof works in this case too, but can you edit your answer to include a quick proof of this fact? Jan 3, 2022 at 4:48\n\u2022 @Randal'Thor Since the loopy walt's answer (up to the words \"It follows that all the $m_i$ are smaller than all the $M_i$\") does not use the fact that the numbers given are integers from 1 to 2n, if we replace $1\\leqslant2\\leqslant\\dots\\leqslant2n$ with any numbers $q_1\\leqslant q_2\\leqslant\\dots\\leqslant q_{2n}$, the assertion about the constant sum still holds (and it will be equal $\\sum_{i=1}^n q_{n+i} - \\sum_{i=1}^n q_i$). Jan 3, 2022 at 6:47\n\u2022 @trolley813 I know how the proof would go, I'm just asking pmw to include it so that this answer is self-contained. Jan 3, 2022 at 8:23", "date": "2023-03-21 16:32:30", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/114281/splitting-integers-and-taking-differences-how-can-the-sum-be-constant", "openwebmath_score": 0.8632074594497681, "openwebmath_perplexity": 466.1086827797057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668690081642, "lm_q2_score": 0.8791467770088162, "lm_q1q2_score": 0.8625896905963588}}
{"url": "https://math.stackexchange.com/questions/3457074/how-many-seating-arrangements-of-m-people-in-a-row-of-n-seats-are-there-if", "text": "# How many seating arrangements of $m$ people in a row of $n$ seats are there if $k$ of the people must sit together?\n\nSuppose there are five people $$A, B, C, D, E$$ to be seated in a row of eight seats $$S_1, \\ldots, S_8$$.\n\n(1) How many possibilities are there if $$A$$ and $$B$$ are to sit next to each other?\n\n(2) How many possibilities are there if $$A$$ and $$B$$ are not to sit next to each other?\n\nMy Attempt:\n\nFirst Approach:\n\nThere are $$8$$ options of a seat for $$A$$. However, if $$A$$ is to be seated in $$S_1$$ or $$S_8$$, then $$B$$ can only be seated in $$S_2$$ or $$S_7$$, respectively. On the other hand, if $$A$$ is to be seated in any one of $$S_2$$ through $$S_7$$, then $$B$$ has two options of seat in each case.\n\nOnce $$A$$ and $$B$$ have been seated, there are $$6$$ seats left for $$C$$, and then $$5$$ seats left for $$D$$, and finally $$4$$ seats left for $$E$$.\n\nIn this way, the total number of seating arrangements are $$2 \\times 1 \\times 6 \\times 5 \\times 4 + 6 \\times 2 \\times 6 \\times 5 \\times 4 = 240 + 1440 = 1680.$$\n\nIs this solution correct?\n\nSecond Approach:\n\nNow let us treat $$A, B$$ as one entity. Then we have four entities to be accommodated in seven spots, for which there are $$7 \\times 6 \\times 5 \\times 4 = 840$$ ways. And, each one of these $$840$$ arrangements, the members $$A$$ and $$B$$ of the block $$A, B$$ can be arranged within the block in two distinct ways. Therefore there are $$840 \\times 2 = 1680$$ possible seating arrangements.\n\nIs my answer correct? If so, then are both of my approaches also correct? If not, then where are the problems.\n\nMore generally, I have the following question:\n\nLet $$k, m, n$$ be any natural numbers such that $$k \\leq m \\leq n$$. Then how many ways are there in total of seating $$m$$ people $$P_1 \\ldots, P_m$$ in $$n$$ seats $$S_1, \\ldots, S_n$$ such that some $$k$$ of these people insist on sitting next to each other?\n\nMy Attempt Using the Second Approach:\n\nLet us treat that $$k$$ people as one block. Then there are\n\n\\begin{align} & (n-k+1) \\underbrace{(n-k) (n-k-1) \\ldots }_{(m-k) \\mbox{ factors}} \\\\ &= (n-k+1) \\big[ \\ (n-k) (n-k-1) \\ldots \\big( (n-k) - (m-k-1) \\big) \\ \\big] \\\\ &= (n-k+1)(n-k) \\ldots (n-m+1) \\\\ &= ^{n-k+1}P_{n-m} . \\end{align}\n\nFinally, corresponding to each of the above arrangements with the $$k$$ people considered as one block, there are $$k!$$ ways of arranging the $$k$$ people within the block amongst themselves.\n\nHence there are a total of $$k! \\ ^{n-k+1}P_{n-m}$$ ways of seating $$m$$ people in $$n$$ seats with $$k$$ people seated next to each other.\n\n\u2022 The problem is not clearly stated until it is said which seats are next to each other. Among many possibilties two seem plausible: seats in a single row, or in a circle (in which case $S_1$ would be next to $S_8$). Nov 30, 2019 at 17:23\n\nBoth of your solutions to the first problem are correct. However, your general formula is not. Notice that in the first problem, $$n = 8$$, $$m = 5$$, and $$k = 2$$, so your formula gives $$2!P(8 - 2 + 1, 8 - 5) = 2!P(7, 3) = 2! \\cdot 7 \\cdot 6 \\cdot 5 = 2 \\cdot 210 = 420$$ Let's see what went wrong.\n\nWe wish to seat $$m$$ people, $$k$$ of whom must sit consecutively, in $$n$$ seats. Since the block takes up $$k$$ of the $$n$$ places, it must begin in one of the first $$n - (k - 1) = n - k + 1$$ positions. Once the block has been placed, there are $$n - k$$ seats left for the remaining $$m - k$$ people. They can be arranged in those seats in $$P(n - k, m - k)$$ ways. The people within the block can be arranged in $$k!$$ ways, which gives us the formula $$(n - k + 1)P(n - k, m - k)k!$$ As a sanity check, let's try our formula when $$n = 8$$, $$k = 2$$, and $$m = 5$$. It gives $$(8 - 2 + 1)P(8 - 2, 5 - 2)2! = 7 \\cdot P(6, 3) \\cdot 2! = 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 2 = 1680$$ which agrees with the answer you obtained in your example.\n\nObserve that \\begin{align*} (n - k + 1)P(n - k, m - k) & = (n - k + 1) \\cdot \\frac{(n - k)!}{[(n - k) - (m - k)]!}\\\\ & = \\frac{(n - k + 1)(n - k)!}{(n - m)!}\\\\ & = \\frac{(n - k + 1)!}{(n - m)!}\\\\ & = \\frac{(n - k + 1)!}{[(n - k + 1) - (m - k + 1)]!}\\\\ & = P(n - k + 1, m - k + 1) \\end{align*} so we could write our formula in the form $$P(n - k + 1, m - k + 1)k!$$ As a sanity check, note that if $$n = 8$$, $$k = 2$$, and $$m = 5$$, then $$P(8 - 2 + 1, 5 - 2 + 1)2! = P(7, 4)2! = 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 2 = 1680$$\n\n\u2022 thank you so much for such a beautiful answer! Nov 30, 2019 at 18:06\n\nFor part $$a.)$$, we can work on two cases: $$AB$$ ($$A$$ to the left of $$B$$) and $$BA$$. The possibilities for $$AB$$ are $$S_{1}S_{2}, S_{2}S_{3}, \u2026, S_{7}S_{8}$$. The possibilities for $$BA$$ are also the same. So the answer is\n$$7 \\times \\binom{6}{3} \\times 3! + 7 \\times \\binom{6}{3} \\times 3! = 1680$$\n\nThe $$\\binom{6}{3}$$ is the number of ways we can choose 3 seats, and $$3!$$ is number of possible orders of $$C,D,E$$ in the 3 seats.\n\nFor part $$b.)$$ count all possible seating without any rule, and then subtract with $$1680$$.", "date": "2022-05-26 04:14:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3457074/how-many-seating-arrangements-of-m-people-in-a-row-of-n-seats-are-there-if", "openwebmath_score": 0.9813533425331116, "openwebmath_perplexity": 151.50853903119187, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9927672368385032, "lm_q2_score": 0.8688267847293731, "lm_q1q2_score": 0.8625427663670608}}
{"url": "http://mathhelpforum.com/discrete-math/175881-counting-permutations.html", "text": "# Math Help - Counting-Permutations\n\n1. ## Counting-Permutations\n\nI have a question asking me to find:\nThe sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions\n\nI realise that the digits can be chosen in (4)_4=4!=24 ways since repetition is not permitted and order does matter\n\nThe final answer is 133320\n\nAny help would be great\n\n2. Let the 24 numbers be $a_1,\\dots,a_{24}$. Further, let the $i$th number $a_i$ be $1000b_{i1}+100b_{i2}+10b_{i3}+b_{i4}$ where all $b_{ij}$ are in {2,4,6,8}. Let's find $S_1=\\sum_{i=1}^{24}b_{i1}$. Each of the four numbers is encountered 6 times, so the $S_1 = 6(2 + 4 + 6 + 8) = 120$. Therefore, the first digits of all numbers contribute 120 * 1000 to the whole sum. Similarly, the second digits contribute 120 * 100 and so on.\n\n3. Originally Posted by qwerty10\nI have a question asking me to find:\nThe sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions\nThe final answer is 133320\nThink about it, each of those four numbers will be in each decimal place six times. The ones column adds up to $6(2+4+6+8)=120.$\nCheck out $\\displaystyle120\\left( {\\sum\\limits_{k = 0}^3 {10^k } } \\right) = ~?$.\n\n4. Hello, qwerty10!\n\nFind the sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions.\n\nThe answer is: 133,320.\n\nList the $4! = 24$ permutations of the four digits\n,. . and consider their sum.\n\n. . $\\begin{array}{cc}\n&2468 \\\\ &2486 \\\\ &2648 \\\\ &2684 \\\\ &2846 \\\\ &2864 \\\\ &\\vdots \\\\\n&8246 \\\\ &8264 \\\\ &8426 \\\\ &8462 \\\\ &8624 \\\\ + & 8642 \\\\ \\hline\n\\end{array}$\n\nWe find that each column has: six 2's, six 4's, six 6's, six 8's.\n\nThe total of each column is: . $6\\!\\cdot\\!2 + 6\\!\\cdot\\!4 + 6\\!\\cdot\\!6 + 6\\!\\cdot\\!8 \\:=\\:120$\n\nHence, the addition has the form:\n\n. . $\\begin{array}{cccccc}\n&&& 1 & 2 & 0 \\\\ && 1 & 2 & 0 \\\\ & 1 & 2 & 0 \\\\ 1 & 2 & 0 \\\\ \\hline 1 & 3 & 3 & 3 & 2 & 0 \\end{array}$\n\nTherefore, the sum is: . $133,\\!320$\n\nEdit: Plato beat me to it . . . *sigh*\n.", "date": "2015-08-04 05:12:09", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/175881-counting-permutations.html", "openwebmath_score": 0.8538596034049988, "openwebmath_perplexity": 936.5339151424977, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9927672369455003, "lm_q2_score": 0.8688267796346598, "lm_q1q2_score": 0.8625427614021582}}
{"url": "https://math.stackexchange.com/questions/2700220/radioactive-isotope-differential-equation-question/2700227", "text": "The question is as follows:\n\nI have the general solution of a differential equation as $P(t) = Ae^{kt}$ where $k < 0$ is a constant, and it describes the mass $P(t)$ of a radioactive isotope at future time $t$. I also have that the mass of the isotope is $10$ at time $0$ and it is $5$ at time $2$. I want to find the remainder of the isotope at time $3$.\n\nSo I can use that $P(0) = 10$ to obtain $A$ = $10$, and here now is where I get a bit confused. If I use $P(2) = 5$, I get $5 = 10e^{2k}$ and solving for $k$ I get $k = \\frac{ln(0.5)}{2} = -0.34657$. And then substitute it all back in to get $P(3)$\n\nHowever, when looking at other questions online, they talk a lot about half-life of the substance and we have that if we let $\\tau$ denote the half-life of the substance, then $k = \\frac{ln(2)}{\\tau}$ (where they are using the general solution to be $Ae^{-kt}$ However, isn't $\\tau$ in my case equal to $2$? So should I not be getting $k = \\frac{ln(2)}{2}$ rather than $k = \\frac{ln(0.5)}{2}$? Can someone please explain where I am going wrong?\n\nThank you\n\n\u2022 You haven't don't anything wrong because $-\\log{(0.5)}=\\log{(2)}$. Which explains the negative sign, your solution is the same as theirs Mar 20 '18 at 12:51\n\u2022 Ah ok I think I get where I am getting confused. So what I have written is fine, I just solved for $k$ where $k$ is negative whereas the one online solved for $k$ where $k$ is positive and when putting back into their formula they will take this into account and multiply it by negative $1$? Mar 20 '18 at 12:55\n\n$Ae^{-kt}$ with $k=\\frac{ln(2)}{\\tau}$ is equivalent to $Ae^{kt}$ with $k=\\frac{ln(0.5)}{\\tau}$\n\u2022 Thanks very much. One last thing, so using that then $P(3)$ is approx equal to $3.5$ yes? Mar 20 '18 at 12:58\n\u2022 Yes, it makes sense as you can double check with half-life equal to two, which means that $P(4)=2.5$, and you know the exponential decay function is decreasing. Mar 20 '18 at 13:00", "date": "2022-01-29 04:54:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2700220/radioactive-isotope-differential-equation-question/2700227", "openwebmath_score": 0.9145000576972961, "openwebmath_perplexity": 83.95118174838396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771805808552, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.8625194778180348}}
{"url": "https://www.otexts.org/node/912", "text": "# 8.1.2 Insertion Sort\n\nThe list-sort-best-first procedure seems quite inefficient. For every output element, we are searching the whole remaining list to find the best element, but do nothing of value with all the comparisons that were done to find the best element.\n\nAn alternate approach is to build up a sorted list as we go through the elements. Insertion sort works by putting the first element in the list in the right place in the list that results from sorting the rest of the elements.\n\nFirst, we define the list-insert-one procedure that takes three inputs: a comparison procedure, an element, and a List. The input List must be sorted according to the comparison function. As output, list-insert-one produces a List consisting of the elements of the input List, with the input element inserts in the right place according to the comparison function.\n\n(define (list-insert-one cf el p) ; requires: p is sorted by cf\n(if (null? p) (list el)\n(if (cf el (car p)) (cons el p)\n(cons (car p) (list-insert-one cf el (cdr p))))))\n\nThe running time for list-insert-one is in $\\Theta(n)$ where $n$ is the number of elements in the input list. In the worst case, the input element belongs at the end of the list and it makes $n$ recursive applications of list-insert-one. Each application involves constant work so the overall running time of list-insert-one is in $\\Theta(n)$.\n\nTo sort the whole list, we insert each element into the list that results from sorting the rest of the elements:\n\n(define (list-sort-insert cf p)\n(if (null? p) null\n(list-insert-one cf (car p) (list-sort-insert cf (cdr p)))))\n\nEvaluating an application of list-sort-insert on a list of length $n$ involves $n$ recursive applications. The lengths of the input lists in the recursive applications are $n-1$, $n-2$, $\\ldots$, 0. Each application involves an application of list-insert-one which has linear running time. The average length of the input list over all the applications is approximately $\\frac{n}{2}$, so the average running time of the list-insert-one applications is in $\\Theta(n)$. There are $n$ applications of \\scheme|list-insert-one|, so the total running time is in $\\Theta(n^2)$.\n\nExercise 8.5. We analyzed the worst case running time of list-sort-insert above. Analyze the best case running time. Your analysis should identify the inputs for which list-sort-insert runs fastest, and describe the asymptotic running time for the best case input.\n\nExercise 8.6. Both the list-sort-best-first-sort and list-sort-insert procedures have asymptotic running times in $\\Theta(n^2)$. This tells us how their worst case running times grow with the size of the input, but isn't enough to know which procedure is faster for a particular input. For the questions below, use both analytical and empirical analysis to provide a convincing answer.\na. How do the actual running times of list-sort-best-first-sort and list-sort-insert on typical inputs compare?\nb. Are there any inputs for which list-sort-best-first is faster than list-sort-insert?\nc. For sorting a long list of $n$ random elements, how long does each procedure take? (See Exercise 8.2 for how to create a list of random elements.)", "date": "2018-02-21 05:20:57", "meta": {"domain": "otexts.org", "url": "https://www.otexts.org/node/912", "openwebmath_score": 0.4371246099472046, "openwebmath_perplexity": 840.8529303716956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771755256741, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8625194685441336}}
{"url": "https://math.stackexchange.com/questions/2874991/some-interesting-observations-on-a-sum-of-reciprocals/2875035", "text": "# Some interesting observations on a sum of reciprocals\n\nThis recent question is the motivation for this post.\n\nConsider the following equation $$\\frac1{x-1}+\\frac1{x-2}+\\cdots+\\frac1{x-k}=\\frac1{x-k-1}$$ where $$k>1$$.\n\nMy claims:\n\n1. There are $$k$$ solutions, all of which are real.\n\n2. Let $$x_{\\min}$$ be the minimum value of these $$k$$ solutions. Then as $$k\\to\\infty$$, $$x_{\\min}$$ converges. (If it does, to what value does it converge?)\n\n3. As $$k\\to\\infty$$, all of the solutions get closer and closer to an integer, which is bounded below. Furthermore, these integers will be $$1, 2, 3, \\cdots, k-1, k+1$$.\n\nTo see these patterns, I provide the solutions of $$x$$ below. I used W|A for $$k\\ge4$$. The values in $$\\color{blue}{\\text{blue}}$$ are those of $$x_{\\min}$$.\n\n$$\\begin{array}{c|c}k&2&3&4&5&6\\\\\\hline x&4.414&4.879&5.691&6.592&7.530\\\\&\\color{blue}{1.585}&2.652&3.686&4.701&5.722\\\\&&\\color{blue}{1.468}&2.545&3.588&4.615\\\\&&&\\color{blue}{1.411}&2.487&3.531\\\\&&&&\\color{blue}{1.376}&2.449\\\\&&&&&\\color{blue}{1.352}\\end{array}$$\n\nAlso, when $$k=2$$, the polynomial in question is $$x^2-6x+7$$, and when $$k=3$$, it is $$x^3-9x^2+24x-19$$.\n\nThe reason why I think $$x_{\\min}$$ converges is because the difference between the current one and the previous gets smaller and smaller as $$k$$ increases.\n\nAre my claims true?\n\n\u2022 It is easy to prove that it has at least $k-1$ distinct real soutions Aug 7, 2018 at 14:22\n\u2022 and also it is true that $1<x_{min}<2$ for every $k$ and it converges to 1 Aug 7, 2018 at 14:24\n\u2022 @Exodd: if you do that you know it has $k$ distinct solutions because of the limits of each side as $x \\to \\pm \\infty$ Aug 7, 2018 at 14:27\n\u2022 The method I used (removing the asymptotes) has been the basis of significant improvements in chemical engineering calculations (in particular for the so-called Rachford-Rice and Underwood equations). I have published quite a lor of papers for these. I had a lot of fun with your (may I confess that I cannot resist an equation ?). Cheers. Aug 8, 2018 at 8:05\n\u2022 I updated my answer for some improvements. Cheers and thanks for the problem. Aug 10, 2018 at 8:05\n\n## 3 Answers\n\nBoth your claims are true.\n\nif you call $$f(x) = \\frac1{x-1}+\\frac1{x-2}+\\cdots+\\frac1{x-k}-\\frac1{x-k-1}$$ then $f(1^+) = +\\infty$, $f(2^-) = -\\infty$ and $f$ is continuous in $(1,2)$, so it has a root in $(1,2)$. The same you can say about $(2,3)$, $(3,4), \\cdots, (k-1,k)$, so there are at least $k-1$ real distinct roots. $f$ is also equivalent to a $k$-degree polynomial with the same root, but a $k$-degree polynomial with $k-1$ real roots has in reality $k$ real roots.\n\nThe last root lies in $(k+1,+\\infty)$, since $f(k+1^+) = -\\infty$ and $f(+\\infty) = +\\infty$.\n\nThe least root $x_{\\min}$ must lie in $(1,2)$, since $f(x)<0$ for every $x<1$. Moreover, $$f(x) = 0\\implies x = 1 + \\frac{1}{\\frac1{x-k-1}-\\frac1{x-2}-\\cdots-\\frac1{x-k}}$$ and knowing $1<x<2$, we infer $\\frac1{x-k-1}>\\frac1{x-2}$ and $$1<x = 1 + \\frac{1}{\\frac1{x-k-1}-\\frac1{x-2}-\\cdots-\\frac1{x-k}} < 1 - \\frac{1}{\\frac1{x-3}+\\cdots+\\frac1{x-k}}\\to 1$$ so $x_{\\min}$ converges to $1$\n\nAbout the third claim, notice that you may repeat the same argument for any root except the biggest. Let us say that $x_r$ is the $r-th$ root, with $r<k$, and we know that $r<x_r<r+1$. $$f(x_r) = 0\\implies x_r = r + \\frac{1}{\\frac1{x_r-k-1}-\\frac1{x_r-1}-\\cdots-\\frac1{x_r-k}}$$ but $\\frac1{x_r-k-1}>\\frac1{x_r-1}$ holds, so $$r<x_r = r + \\frac{1}{\\frac1{x_r-k-1}-\\frac1{x_r-1}-\\cdots-\\frac1{x_r-k}} < r - \\frac{1}{\\frac1{x_r-2}+\\cdots+\\frac1{x_r-k}}\\to r$$ so $x_r$ converges to $r$.\n\nFor the biggest root, we know $k+1<x_k$ and $$f(x_k) = 0\\implies k+1 < x_k = k+1 + \\frac{1}{\\frac1{x_k-1}+\\cdots+\\frac1{x_k-k}} \\to k+1$$\n\n\u2022 I would've argued that the LHS is continuous, monotone, and ranges from $-\\infty$ to $+\\infty$ between natural numbers, while the RHS is continuous and monotone, and since the LHS is asymptotically larger than the RHS as $x\\to\\infty$, there must be a solution $x>k+1$, where the LHS is less than the RHS. In any case, sound argument. Aug 7, 2018 at 14:52\n\u2022 I think they're quite equivalent. Now I'm wondering if $x_{max}$ converges to $k+1$ Aug 7, 2018 at 14:57\n\u2022 @Exodd Nice proofs. The third claim of converging to integers extends the claim $x_{\\min}$ to all solutions of $x$. I strongly believe that it's true after plotting in Desmos, but I can't construct a rigorous proof for it. Aug 7, 2018 at 14:59\n\u2022 Hard to exactly say what that's supposed to mean, $k+1$ is a moving target :P Aug 7, 2018 at 15:00\n\u2022 Let's say $x_{max}(k)-k-1 = o_k(1)$ Aug 7, 2018 at 15:00\n\nFor the base case, $$\\tag1f_2(x)=\\frac1{x-1}+\\frac1{x-2}-\\frac1{x-3},$$ one readily verifies that there is a root in $(1,2)$ and a root $x^*$ in $(3,+\\infty)$.\n\nIf we multiply out the denominators of $$f_k(x)=\\frac1{x-1}+\\frac1{x-2}+\\ldots+\\frac1{x-k}-\\frac1{x-k-1},$$ we obtain the equation $$\\tag2(x-1)(x-2)\\cdots(x-k-1)f_k(x)=0,$$ which is a polynomial of degree (at most) $k$, so we expect $k$ solutions, but some of these may be complex or repeated or happen to be among $\\{1,2,\\ldots, k+1\\}$ and thus not allowed for the original equation. But $f_k(x)$ has simple poles with jumps from $-\\infty$ to $+\\infty$ at $1,2,3,\\ldots, k$, and a simple pole with jump from $+\\infty$ to $-\\infty$ at $k+1$, and is continuous otherwise. It follows that there is (at least) one real root in $(1,2)$, at least one in in $(2,3)$, etc. up to $(k-1,k)$, so there are at least $k-1$ distinct real roots. Additionally, for $x>k+1$ and $k\\ge2$, we have $$f_k(x)\\ge f_2(x+k-2).$$ It follows that there is another real root between $k+1$ and $x^*+k-2$. So indeed, we have $k$ distinct real roots.\n\nFrom the aboive, the smallest root is always in $(1,2)$. If follows from $f_{k+1}(x)>f_k(x)$ for $x\\in(1,2)$ and the fact that all $f_k$ are strictly decreasing there, that $x_\\min$ decreases with increasing $k$. As a decreasing bounded sequence, it does have a limit.\n\nConsidering that you look for the first zero of function $$f(x)=\\sum_{i=1}^k \\frac 1{x-i}-\\frac1 {x-k-1}$$ which can write, using harmonic numbers, $$f(x)=H_{-x}-H_{k-x}-\\frac{1}{x-k-1}$$ remove the asymptotes using $$g(x)=(x-1)(x-2)f(x)=2x-3+(x-1)(x-2)\\left(H_{2-x}-H_{k-x}-\\frac{1}{x-k-1} \\right)$$ You can approximate the solution using a Taylor expansion around $x=1$ and get $$g(x)=-1+(x-1) \\left(-\\frac{1}{k}+\\psi ^{(0)}(k)+\\gamma +1\\right)+O\\left((x-1)^2\\right)$$ Ignoring the higher order terms, this gives as an approximation $$x_{est}=1+\\frac{k}{k\\left(\\gamma +1+ \\psi ^{(0)}(k)\\right)-1}$$ which seems to be \"decent\" (and, for sure, confirms your claims). $$\\left( \\begin{array}{ccc} k & x_{est} & x_{sol} \\\\ 2 & 1.66667 & 1.58579 \\\\ 3 & 1.46154 & 1.46791 \\\\ 4 & 1.38710 & 1.41082 \\\\ 5 & 1.34682 & 1.37605 \\\\ 6 & 1.32086 & 1.35209 \\\\ 7 & 1.30238 & 1.33430 \\\\ 8 & 1.28836 & 1.32040 \\\\ 9 & 1.27726 & 1.30914 \\\\ 10 & 1.26817 & 1.29976 \\\\ 11 & 1.26055 & 1.29179 \\\\ 12 & 1.25403 & 1.28489 \\\\ 13 & 1.24837 & 1.27884 \\\\ 14 & 1.24339 & 1.27347 \\\\ 15 & 1.23895 & 1.26867 \\\\ 16 & 1.23498 & 1.26433 \\\\ 17 & 1.23138 & 1.26039 \\\\ 18 & 1.22810 & 1.25678 \\\\ 19 & 1.22510 & 1.25346 \\\\ 20 & 1.22233 & 1.25039 \\end{array} \\right)$$ For infinitely large values of $k$, the asymptotics of the estimate would be $$x_{est}=1+\\frac{1}{\\log \\left({k}\\right)+\\gamma +1}$$\n\nFor $k=1000$, the exact solution is $1.12955$ while the first approximation gives $1.11788$ and the second $1.11786$.\n\nUsing such estimates would make Newton method converging quite fast (shown below for $k=1000$).\n\n$$\\left( \\begin{array}{cc} n & x_n \\\\ 0 & 1.117855442 \\\\ 1 & 1.129429575 \\\\ 2 & 1.129545489 \\\\ 3 & 1.129545500 \\end{array} \\right)$$\n\nEdit\n\nWe can obtain much better approximations if, instead of using a Taylor expansion of $g(x)$ to $O\\left((x-1)^2\\right)$, we build the simplest $[1,1]$ Pad\u00e9 approximant (which is equivalent to an $O\\left((x-1)^3\\right)$ Taylor expansion). This would lead to $$x=1+ \\frac{6 (k+k (\\psi ^{(0)}(k)+\\gamma )-1)}{\\pi ^2 k+6 (k+\\gamma (\\gamma k+k-2)-1)-6 k \\psi ^{(1)}(k)+6 \\psi ^{(0)}(k) (2 \\gamma k+k+k \\psi ^{(0)}(k)-2)}$$ Repeating the same calculations as above, the results are $$\\left( \\begin{array}{ccc} k & x_{est} & x_{sol} \\\\ 2 & 1.60000 & 1.58579 \\\\ 3 & 1.46429 & 1.46791 \\\\ 4 & 1.40435 & 1.41082 \\\\ 5 & 1.36900 & 1.37605 \\\\ 6 & 1.34504 & 1.35209 \\\\ 7 & 1.32741 & 1.33430 \\\\ 8 & 1.31371 & 1.32040 \\\\ 9 & 1.30266 & 1.30914 \\\\ 10 & 1.29348 & 1.29976 \\\\ 11 & 1.28569 & 1.29179 \\\\ 12 & 1.27897 & 1.28489 \\\\ 13 & 1.27308 & 1.27884 \\\\ 14 & 1.26787 & 1.27347 \\\\ 15 & 1.26320 & 1.26867 \\\\ 16 & 1.25899 & 1.26433 \\\\ 17 & 1.25516 & 1.26039 \\\\ 18 & 1.25166 & 1.25678 \\\\ 19 & 1.24844 & 1.25346 \\\\ 20 & 1.24547 & 1.25039 \\end{array} \\right)$$\n\nFor $k=1000$, this would give as an estimate $1.12829$ for an exact value of $1.12955$.\n\nFor infinitely large values of $k$, the asymptotics of the estimate would be $$x_{est}=1+\\frac{6 (\\log (k)+\\gamma +1)}{6 \\log (k) (\\log (k)+2 \\gamma +1)+\\pi ^2+6 \\gamma (1+\\gamma )+6}$$\n\n\u2022 Interesting! What does that notation $\\psi^{(0)}(k)$ mean by the way? Aug 8, 2018 at 8:10\n\u2022 @TheSimpliFire. This is \"just\" the digamma function. Aug 8, 2018 at 8:18\n\u2022 @TheSimpliFire. We use to write $\\psi^{(n)}(k)$ for the polygamma function and, in usual notations,$\\psi^{(0)}(k)=\\psi(k)$ Aug 8, 2018 at 8:22", "date": "2022-07-03 06:04:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2874991/some-interesting-observations-on-a-sum-of-reciprocals/2875035", "openwebmath_score": 0.9257096648216248, "openwebmath_perplexity": 170.59247243438804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777178636555, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8625194566974352}}
{"url": "https://math.stackexchange.com/questions/2419619/probability-of-selecting-elements-with-replacement/2419630", "text": "# Probability of selecting elements with replacement\n\nIf I'm selecting $N$ elements uniformly at random (with replacement) from $\\{1, \\dots, N\\}$, what is the chance that a given value is selected at least once?\n\nWhat is the name for the distribution for the more general case where I'm selecting $N$ elements from a domain with $M$ elements?\n\n\u2022 by random do you mean equally-probable ? \u2013\u00a0user451844 Sep 6 '17 at 23:31\n\u2022 @Roddy Uniformly at random means equally-probable. The key word is \"uniformly\". \u2013\u00a0Trevor Gunn Sep 6 '17 at 23:51\n\u2022 @TrevorGunn yeah I'm stupid sorry. \u2013\u00a0user451844 Sep 6 '17 at 23:55\n\n## 3 Answers\n\nYour answer is a particular case of the binomial distribution. Let getting a $1$ be a Success. You have $N$ independent trials. In the general version, the success probability is $p = 1/M.$\n\nIf $X \\sim \\mathsf{Binom}(N, 1/M),$ then you seek $$P(X > 0) = 1 - (1-p)^N = 1 - (1 - 1/M)^N = 1 - \\left(\\frac{M-1}{M}\\right)^N,$$ as in the Answer of @ConMan (+1), which I hope you will Accept. [On each trial, you might say that you are using a discrete uniform distribution on the integers $1$ through $M$.]\n\nExample: What is the probability you get at least one $6$ in ten rolls of a fair die? Then $X = \\mathsf{Binom}(10, 1/6)$ is the number of $6$'s and you seek $P(X > 0) = P(X \\ge 1) = 0.8385,$ computed in R statistical software as:\n\n1 - dbinom(0, 10, 1/6)\n## 0.8384944\n\n\nIn the figure below, you want the total heights of the bars to the right of the dotted red line.\n\n\u2022 All the answers here were excellent, but I accepted this one since I mentioned binomial distribution and included a graphic, and because you fixed up my notation in the question. \u2013\u00a0BeeOnRope Sep 7 '17 at 17:26\n\nIf you're drawing with equal probability and subsequent draws are independent of each other, then the probability that you do not draw, for example, \"1\" in a single draw from $M$ elements is $\\frac{M-1}{M}$.\n\nThe probability that you do not draw any \"1\"s in $N$ draws is the product of the individual probabilities, i.e. $\\frac{M-1}{M}\\cdot\\frac{M-1}{M}\\cdot\\ldots\\cdot\\frac{M-1}{M}=\\left(\\frac{M-1}{M}\\right)^N = \\left(1 - \\frac{1}{M}\\right)^N$.\n\nThe probability that you draw at least 1 \"1\" in the $N$ draws is the complement of the probability that you draw none, i.e. it is $1 - \\left(1 - \\frac{1}{M}\\right)^N$.\n\nThe distribution for drawing $M$ elements with replacement from a set of $N$ elements is the uniform distribution on all $M$-tuples from $\\{1,\\dots,N\\}$. For example, with $M = 3$ and $N = 2$ we get 8 ($=2^3$) $3$-tuples:\n\n$$(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)$$\n\nand each is equally likely.\n\nViewed this way, the question of how likely it is to see a given value at least once comes down to counting those sequences that contain the value somewhere. We can count this by counting the tuples that do not contain that value. We know that there are $N^N$ total $N$-tuples and $(N - 1)^N$ tuples not containing a specified value. Thus there are\n\n$$N^N - (N - 1)^N$$\n\ntuples that contain a specified value. Since the distribution is uniform, the probability of selecting such a tuple is\n\n$$\\frac{N^N - (N - 1)^N}{N^N} = 1 - \\left( 1 - \\frac{1}{N} \\right)^N.$$\n\nNow let's say that we have the elements $\\{1,\\dots,N\\}$, we draw $M$ of them and we want at least $P$ of them to equal $1$ (or any other fixed element). Then to count the number of $M$-tuples with $P$ $1$'s, first we select the $P$ slots to put the $1$'s in, then from the $M - P$ remaining slots, we have $N - 1$ options each. This gives\n\n$$\\binom{M}{P} (N - 1)^{M - P}.$$\n\nNow we may divide by the total number of $M$-tuples ($N^M$) to get a probability of\n\n$$\\binom{M}{P} \\frac{(N - 1)^{M - P}}{N^{M - P}} \\cdot \\frac{1}{N^P} = \\binom{M}{P} \\left( 1 - \\frac{1}{N} \\right)^{M - P} \\left( \\frac1N \\right)^{P}.$$\n\nThis is the probability mass function of the binomial distribution. Namely $\\mathsf{Bin}(M, \\frac{1}{N})$.", "date": "2019-07-22 18:32:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2419619/probability-of-selecting-elements-with-replacement/2419630", "openwebmath_score": 0.9272138476371765, "openwebmath_perplexity": 207.76259856037257, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109480714625, "lm_q2_score": 0.8757869997529961, "lm_q1q2_score": 0.8624846255354098}}
{"url": "https://math.stackexchange.com/questions/2371022/cross-product-in-higher-dimensions", "text": "# Cross product in higher dimensions\n\nSuppose we have a vector $$(a,b)$$ in $$2$$-space. Then the vector $$(-b,a)$$ is orthogonal to the one we started with. Furthermore, the function $$(a,b) \\mapsto (-b,a)$$ is linear.\n\nSuppose instead we have two vectors $$x$$ and $$y$$ in $$3$$-space. Then the cross product gives us a new vector $$x \\times y$$ that's orthogonal to the first two. Furthermore, cross products are bilinear.\n\nQuestion. Can we do this in higher dimensions? For example, is there a way of turning three vectors in $$4$$-space into a fourth vector, orthogonal to the others, in a trilinear way?\n\n\u2022 You might want to look at the Gram Schmitt method here :en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process \u2013\u00a0Furrane Jul 25 '17 at 10:06\n\u2022 This construction has nothing to do with cross product, just happen to coincide in dimension 3. \u2013\u00a0Miguel Jul 25 '17 at 10:54\n\u2022 A search for \"generalized cross product\" turns up a number of questions likely to be of interest, including Is the vector cross product only defined for 3D? and Generalized Cross Product. ;) (Not marking as a duplicate because you're better able to judge which question, if any, most nearly matches yours.) \u2013\u00a0Andrew D. Hwang Jul 25 '17 at 17:38\n\u2022 You might be interested in the notion of the orthogonal complement. It can give you the vector orthogonal to a given set of $n-1$ independent vectors in $n$-space, like you're asking for $n=4$. But it can also give you $k$ independent vectors orthogonal to a given set of $n-k$ independent vectors in $n$-space. So you can take two vectors in 4-space and find two vectors perpendicular to them and to each other. \u2013\u00a0YawarRaza7349 Jul 25 '17 at 17:39\n\u2022 \u2013\u00a0user57159 Jul 26 '17 at 3:18\n\nYes. It is just like in dimension $$3$$: if your vectors are $$(t_1,t_2,t_3,t_4)$$, $$(u_1,u_2,u_3,u_4)$$, and $$(v_1,v_2,v_3,v_4)$$, compute the formal determinant:$$\\begin{vmatrix}t_1&t_2&t_3&t_4\\\\u_1&u_2&u_3&u_4\\\\v_1&v_2&v_3&v_4\\\\e_1&e_2&e_3&e_4\\end{vmatrix}.$$ You then see $$(e_1,e_2,e_3,e_4)$$ as the canonical basis of $$\\mathbb{R}^4$$. Then the previous determinant is $$(\\alpha_1,\\alpha_2,\\alpha_3,\\alpha_4)$$ with\\begin{align*}\\alpha_1&=t_4u_3v_2-t_3u_4v_2-t_4u_2v_3+t_2u_4v_3+t_3u_2v_4-t_2u_3v_4\\\\\\alpha_2&=-t_4u_3v_1+t_3u_4v_1+t_4u_1v_3-t_1u_4v_3-t_3u_1v_4+t_1u_3v_4\\\\\\alpha_3&=t_4u_2v_1-t_2u_4v_1-t_4u_1v_2+t_1u_4v_2+t_2u_1v_4-t_1u_2v_4\\\\\\alpha_4&=-t_3u_2v_1+t_2u_3v_1+t_3u_1v_2-t_1u_3v_2-t_2u_1v_3+t_1u_2v_3\\end{align*}It's a vector orthogonal to the other three.\n\nI followed a suggestion taken from the comments on this answer: to put the entries $$e_1$$, $$e_2$$, $$e_3$$, and $$e_4$$ at the bottom. It makes no difference in odd dimension, but it produces the natural sign in even dimension.\n\nFollowing another suggestion, I would like to add this remark:$$\\alpha_1=-\\begin{vmatrix}t_2&t_3&t_4\\\\u_2&u_3&u_4\\\\v_2&v_3&v_4\\end{vmatrix}\\text{, }\\alpha_2=\\begin{vmatrix}t_1&t_3&t_4\\\\u_1&u_3&u_4\\\\v_1&v_3&v_4\\end{vmatrix}\\text{, }\\alpha_3=-\\begin{vmatrix}t_1&t_2&t_4\\\\u_1&u_2&u_4\\\\v_1&v_2&v_4\\end{vmatrix}\\text{ and }\\alpha_4=\\begin{vmatrix}t_1&t_2&t_3\\\\u_1&u_2&u_3\\\\v_1&v_2&v_3\\\\\\end{vmatrix}.$$\n\n\u2022 Very lucid and appreciable answer.thanks \u2013\u00a0Arpit Yadav Jul 25 '17 at 10:18\n\u2022 I'm pretty sure the row of basis vectors should be on the bottom to get the right-handedness correct; only in odd dimensions can this row be moved to the top without a change in sign whilst keeping the vectors in the same order. \u2013\u00a0user332714 Jul 25 '17 at 18:06\n\u2022 (+1) This is the method I've used in a few answers. The formula is easy to remember! \u2013\u00a0robjohn Jul 25 '17 at 22:50\n\u2022 @lastresort Nice remark. I shall edit my answer taking that into account. \u2013\u00a0Jos\u00e9 Carlos Santos Jul 25 '17 at 22:54\n\u2022 Is $$\\alpha_1 ={\\rm det} \\left| \\matrix{ t_2 & t_3 & t_4 \\\\ u_2 & u_3 & u_4 \\\\ v_2 & v_3 & v_4} \\right|$$ and so on? \u2013\u00a0John Alexiou Jul 26 '17 at 14:19\n\nMy answer is in addition to Jos\u00e9's and Antinous's answers but maybe somewhat more abstract. In principle, their answers are using coordinates, whereas I'm trying to do it coordinate-free.\n\nWhat you are looking for is the wedge or exterior product. The exterior power $$\\bigwedge^k(V)$$ of some vector space $$V$$ is the quotient of the tensor product $$\\bigotimes^k(V)$$ by the relation $$v\\otimes v$$. To be somewhat more concrete and less abstract, this just means that for any vector $$v\\in V$$ the wedge product $$v\\wedge v=0\\in\\bigwedge^2(V)$$. Whenever you wedge vectors together, the result equals zero if at least two of the factors are linearly dependent. Think of what happens to the cross product in $$\\mathbb{R}^3$$.\n\nIn fact, let $$e_1,e_2,\\ldots,e_n$$ be a basis of an inner product space $$V$$. Then $$e_{i_1}\\wedge e_{i_2}\\wedge \\ldots \\wedge e_{i_k}$$ is a basis for $$\\bigwedge^k(V)$$ where $$1\\leq i_1 < i_2 < \\ldots < i_k\\leq n$$.\n\nIf $$V=\\mathbb{R}^3$$ then $$v \\wedge w$$ equals $$v \\times w$$ up to signs of the entries. This seems a bit obscure because technically $$v\\wedge w$$ should be an element of $$\\bigwedge^2(\\mathbb{R}^3)$$. However, the latter vector space is isomorphic to $$\\mathbb{R}^3$$. In fact, this relation is true for all exterior powers given an orientation on the vector space. The isomorphism is called the Hodge star operator. It says that there is an isomorphism $$\\star\\colon\\bigwedge^{n-k}(V)\\to\\bigwedge^{k}(V)$$. This map operates on a $$(n-k)$$-wedge $$\\beta$$ via the relation $$\\alpha \\wedge \\beta = \\langle \\alpha,\\star\\beta \\rangle \\,\\omega$$ where $$\\alpha\\in\\bigwedge^{k}(V)$$, $$\\omega\\in\\bigwedge^n(V)$$ is an orientation form on $$V$$ and $$\\langle \\cdot,\\cdot \\rangle$$ is the induced inner product on $$\\bigwedge^{k}(V)$$ (see wiki). Notice that the wiki-page defines the relation the other way around.\n\nHow does all this answer your question you ask? Well, let us take $$k=1$$ and $$V=\\mathbb{R}^n$$. Then the Hodge star isomorphism identifies the spaces $$\\bigwedge^{n-1}(\\mathbb{R}^n)$$ and $$\\bigwedge^{1}(\\mathbb{R}^n)=\\mathbb{R}^n$$. This is good because you originally wanted to say something about orthogonality between a set of $$n-1$$ linearly indepedent vectors $$v_1,v_2,\\ldots,v_{n-1}$$ and their \"cross product\". Now let us exactly do that and set $$\\beta :=v_1 \\wedge v_2 \\wedge \\ldots \\wedge v_{n-1}\\in\\bigwedge^{n-1}(\\mathbb{R}^n)$$. Then the image $$\\star\\beta = \\star(v_1 \\wedge v_2 \\wedge \\ldots \\wedge v_{n-1})$$ is a regular vector in $$\\mathbb{R}^n$$ and the defining condition above implies for $$\\alpha=v_i\\in\\mathbb{R}^n=\\bigwedge^{1}(\\mathbb{R}^n)$$ $$v_i \\wedge (v_1 \\wedge v_2 \\wedge \\ldots \\wedge v_{n-1}) = \\alpha \\wedge \\beta = \\langle \\alpha,\\star\\beta \\rangle \\,\\omega = \\langle v_i,\\star\\beta \\rangle \\,\\omega.$$ However, the left hand side equals zero for $$i=1,2,\\ldots,n-1$$, so that the vector $$\\star\\beta$$ is orthogonal to all vectors $$v_1,v_2,\\ldots,v_{n-1}$$ which is what you asked for. So you might want to define the cross product of $$n-1$$ vectors as $$v_1 \\times v_2 \\times \\ldots \\times v_{n-1} := \\star(v_1 \\wedge v_2 \\wedge \\ldots \\wedge v_{n-1})$$.\n\nMaybe keep in mind that the other two answers implicitly use the Hodge star operation (and also a basis) to compute the \"cross product in higher dimension\" through the formal determinant which is encoded in the use of the wedge product here.\n\n\u2022 So concretely, how do we actually know what the hodge star of a $k$-blade is? For example, work in $4$-space with the standard orientation. Suppose we want to know $\\star(v_1 \\wedge v_3).$ If I understand correctly, it's either $v_2 \\wedge v_4$ or else $v_4 \\wedge v_2$. How do we know which one? \u2013\u00a0goblin GONE Jul 25 '17 at 13:32\n\u2022 It depends on the orientation that you choose for your vector space. Let's say $v_1,v_2,v_3,v_4$ form an oriented basis for $V$ (that is, $\\omega = v_1\\wedge v_2\\wedge v_3\\wedge v_4$) then $\\star(v_1\\wedge v_3)=v_4\\wedge v_2$. This can be seen using the defining relation for $\\alpha=v_i\\wedge v_j$ cycling through all possible combinations $(i,j)$. This is what they say on the wiki-page linked above in the section \"Computation of the Hodge star\" albeit expressed a little bit complicated in my opinion. \u2013\u00a0Sven Pistre Jul 25 '17 at 14:25\n\u2022 Of all combinations $(i,j)$ only $(2,4)$ and $(4,2)$ remain (because otherwise the left hand side equals zero). Then you assume $\\star(v_1\\wedge v_3)=v_k\\wedge v_l$ and think about which combinations for $(k,l)$ remain on the right hand side of the def. relation. Then you will see that the only possible one is $(k,l)=(4,2)$. To see the last part, look at the definition of the induced scalar product on $\\bigwedge^2(V)$. \u2013\u00a0Sven Pistre Jul 25 '17 at 14:28\n\u2022 And also, as I forgot to mention this, $v_2\\wedge v_4=-v_4\\wedge v_2$. Changing the position of two vectors in a $k$-wedge just changes the sign. So really it only depends on the chosen orientation (or \"right-handedness\") of your vector space. \u2013\u00a0Sven Pistre Jul 25 '17 at 14:49\n\u2022 @\u00e9tale-cohomology The hodge star depends on a choice of inner product and orientation. So it is not canonical. I don't think that you can identify them canonically. \u2013\u00a0Sven Pistre Jul 25 '17 at 17:42\n\nYou can work out the cross product $p$ in $n$-dimensions using the following:\n\n$$p=\\det\\left(\\begin{array}{lllll}e_1&x_1&y_1&\\cdots&z_1\\\\e_2&x_2&y_2&\\cdots&z_2\\\\\\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\e_n&x_n&y_n&\\cdots&z_n\\end{array}\\right),$$ where $\\det$ is the formal determinant of the matrix, the $e_i$ are the base vectors (e.g. $\\hat{i},\\hat{j},\\hat{k}$, etc), and $x,y,\\ldots,z$ are the $n-1$ vectors you wish to \"cross\".\n\nYou will find that $x\\cdot p=y\\cdot p=\\cdots=z\\cdot p=0$.\n\nIt's wonderful the determinant produces a vector with this property.\n\n\u2022 Are there any requirements on the basis vectors $e_1, ..., e_n$? Like, do they need to form an orthonormal basis, or something? \u2013\u00a0\u00e9tale-cohomology Jul 25 '17 at 17:01\n\u2022 Yeah but do they have to be? Can't we take any other basis? \u2013\u00a0\u00e9tale-cohomology Jul 25 '17 at 20:35\n\u2022 By changing the basis to $\\widetilde e_i$ you will have to change the vector entries to the coefficients $\\widetilde x_i$ in the basis expansion for the new basis. Remember that the above is only a formal determinant as this is not actually a matrix (since the first column consists of entries that are vectors themselves). So it does not matter if the basis is orthonormal or not but you will have to adjust your formal determinant formula. \u2013\u00a0Sven Pistre Jul 25 '17 at 21:48\n\u2022 (+1) This is the logical extension of Jos\u00e9 Carlos Santos' answer to $\\mathbb{R}^n$ (at first, this is what I thought he had given, but now I see his only covers $\\mathbb{R}^4$). \u2013\u00a0robjohn Jul 25 '17 at 22:53\n\u2022 @robjohn I actually posted my answer 2 minutes before he did :-) \u2013\u00a0Pixel Jul 25 '17 at 23:00\n\nYes, and apart from other answers an interesting approach to think about it is using Clifford's algebra.\n\nThis can introduce you the basic concept in a nonrigorous but approachable manner.\n\n\u2022 Thank you for your answer, however that article is extremely long and it's difficult to find a Clifford-style answer to my question by reading through it. Can I ask you to write up some details on how to compute an actual cross product using Clifford approach's? \u2013\u00a0goblin GONE Jul 26 '17 at 8:36", "date": "2021-04-16 04:03:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2371022/cross-product-in-higher-dimensions", "openwebmath_score": 0.8729615807533264, "openwebmath_perplexity": 184.5991950377818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109520836027, "lm_q2_score": 0.8757869932689565, "lm_q1q2_score": 0.8624846226636368}}
{"url": "https://mathematica.stackexchange.com/questions/259299/finding-the-interval-of-values-of-an-unknown-element-such-that-a-square-matrix-i", "text": "# Finding the interval of values of an unknown element such that a square matrix is positive semidefinite\n\nI'm looking for a way to determine which values of $$c \\in \\mathbb{R}$$ make the matrix $$A$$ positive semidefinite:\n\n$$A = \\begin{bmatrix} 1 & c \\\\ c & 1 \\end{bmatrix}$$\n\nI'm looking for a way to determine which values of c\u2208R make the matrix A positive semidefinite:\n\nBy definition, A matrix is called positive semidefinite if it is symmetric and all its eigenvalues are non-negative. Hence\n\nClear[\"Global*\"]\nmat = {{1, c}, {c, 1}}\nIf[SymmetricMatrixQ[mat],\neigv = Eigenvalues[mat];\nReduce[eigv[[1]] >=  0 && eigv[[2]] >= 0, c]\n]\n\n\nSo any value between -1 and 1 will do. The above code ofcourse assumes the matrix is 2 by 2, but it can easily be made more general.\n\n\u2022 Congrats on 100K. Dec 6 '21 at 15:53\n\ncp = CharacteristicPolynomial[{{1, c}, {c, 1}}, t]\n\n(* Out[142]= 1 - c^2 - 2 t + t^2 *)\n\n\nThere are two important things to check. (i) Where do roots cross the y axis? (ii) Where might roots be complex-valued?\n\nFor (i) we just solve for c under the assumption thatt==0\n\nSolve[cp == 0 && t == 0, {c, t}]\n\n(* Out[143]= {{c -> -1, t -> 0}, {c -> 1, t -> 0}} *)\n\n\nThis splits the real line for c into three regions abd simply plugging in a value in each region will show that both solutions are nonnegative in the finite part.\n\nFor (ii) We check where the discriminant with respect to t vanishes. This will give conditions on c.\n\nDiscriminant[cp, t]\n\n(* Out[144]= 4 c^2 *)\n\n\nSo it only vanishes at the origin. We already know that on neither side do we have complex valued roots in t. So there are no further restrictions.\n\nAn advantage to this approach is that it extends to higher dimensions.\n\nIf you compute the Cholesky decomposition of the matrix, and simplify it under the assumptions you gave:\n\nSimplify[CholeskyDecomposition[{{1, c}, {c, 1}}], c \u2208 Reals]\n{{1, c}, {0, Sqrt[1 - c^2]}}\n\n\nyou can immediately read off the condition that Sqrt[1 - c^2] >= 0 for positive-semidefiniteness. Thus,\n\nReduce[%[[-1, -1]] >= 0, c, Reals]\n-1 <= c <= 1\n`", "date": "2022-01-25 02:48:53", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/259299/finding-the-interval-of-values-of-an-unknown-element-such-that-a-square-matrix-i", "openwebmath_score": 0.5118657350540161, "openwebmath_perplexity": 1004.8649190694614, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109525293959, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8624846118793379}}
{"url": "https://math.stackexchange.com/questions/3684603/floor-function-algebra-question", "text": "# Floor function algebra question\n\nI don't know how to put floor functions in but...\n\nSolve $$\\dfrac{19x + 16}{10} = \\left \\lfloor \\dfrac{4x+7}{3}\\right \\rfloor$$\n\nI have so far worked out that the RHS can either be $$(4x+7)/3 - 0.33$$, $$(4x+7)/3 - 0.67$$ or itself. When I solve for each of these three equations, I get $$x=12/17, 22/17, 2/17$$. From there, I subbed $$x$$ back into the equation to try and see which one works but none did. Can I have some help?\n\n\u2022 You can get the floor functions with \\lfloor and \\rfloor, e.g., \\lfloor \\pi \\rfloor = 3. If you need bigger ones, as around a fraction, for instance, use \\left\\lfloor and \\right\\rfloor, e.g., \\left\\lfloor\\frac{4x+7}{3}\\right\\rfloor. May 21 '20 at 3:24\n\u2022 You seem to be assuming that $4x+7$ is an integer. There's no reason to think it is. May 21 '20 at 3:37\n\u2022 X must be of the form $$10{\\alpha}+6;\\alpha \\inf \\mathbb{Z}$$ May 21 '20 at 4:53\n\u2022 @saulspatz so would it be safe to say that the RHS is in the range of $(4x+7)/3$ to $(4x+7)/3-0.99$ ?\n\u2013\u00a0user377742\nMay 21 '20 at 8:52\n\u2022 Look at my answer. May 21 '20 at 16:09\n\nI'll get you started.\n\nWe know that $$x-1<\\lfloor x\\rfloor<=x$$, so we have $$\\frac{4x+4}3<\\frac{19x+16}{10}\\leq\\frac{4x+7}3\\\\ 40x+40<57x+48\\leq40x+70\\\\ \\frac{-8}{17} so that $$x=n+\\varepsilon$$ where $$n\\in\\{-1,0,1\\}$$ and $$0\\leq\\varepsilon<1$$.\n\nNow we can test each of the three possibilities for $$n$$ separately. Suppose $$x=1+\\varepsilon$$. Then $$\\frac{19x+16}{10}=\\frac{35+19\\varepsilon}{10}$$ is an integer between $$3.5$$ and $$5.4$$ so there are only two possibilities for $$\\varepsilon$$. Check these to see if $$x=1+\\varepsilon$$ satisfies the equation. Repeat the process for $$n=0$$ and $$n=-1$$.\n\nSince algebra with floor functions is rarely nice, I like to graph it if possible to visualize the solutions. In your case, this is the graph of $$\\frac{19x+16}{10}-\\Big\\lfloor \\frac{4x+7}{3}\\Big\\rfloor$$\n\n(This is using Desmos by the way) Notice how any solutions would pass through the $$y=0$$, and there seem to be $$4$$. One seems to occur at $$x \\approx -.3158$$ and by plugging this into the floor function we see it approaches $$1.912$$, the floor of which is $$1$$. So we are looking for $$x$$ such that $$\\frac{19x+16}{10} = 1$$. The solution is $$x = \\frac{-6}{19}$$ which, by substituting it back in, works. Similarly, we can show that we need $$x \\approx .2105$$ such that $$\\frac{19x+16}{10} = 2$$, giving the solution $$x = \\frac{4}{19}$$. Then we need $$x \\approx .7368$$ such that $$\\frac{19x+16}{10} = 3$$, giving the solution $$x = \\frac{14}{19}$$. Finally, we need $$x \\approx 1.2632$$ such that $$\\frac{19x+16}{10} = 4$$, giving the solution $$x = \\frac{24}{19}$$. From here, you can use strict inequalities to prove that there are no more solutions.\n\n$$\\dfrac{19x + 16}{10} = \\left \\lfloor \\dfrac{4x+7}{3}\\right \\rfloor$$\n\nLet $$\\dfrac{19x + 16}{10} = n \\in \\mathbb Z$$.\n\nThen $$x = \\dfrac{10n-16}{19}$$\n\nand $$\\dfrac{4x+7}{3} = \\dfrac{40n+69}{57}$$.\n\nSo\n\n$$n \\le \\dfrac{40n+69}{57} < n + 1$$\n\n$$57n \\le 40n+69 < 57n + 57$$\n\n$$0 \\le -17n+69 < 57$$\n\n$$-69 \\le -17n < -12$$\n\n$$\\dfrac{12}{17} < n \\le 4\\dfrac{1}{17}$$\n\nSo now you can find the values of $$n$$ and then the values of $$x$$.\n\nFirst deal with the fact that $$\\dfrac{19x + 16}{10}$$ is an integer $$k$$.\n\nSo $$19x = 10k -16$$ is and integer $$m$$ and $$x = \\frac m{19}$$ for some integer $$m$$.\n\nNow deal with $$\\dfrac{19x + 16}{10}=\\lfloor \\dfrac{4x+7}{3} \\rfloor$$ so\n\n$$\\dfrac{19x + 16}{10} \\le \\dfrac{4x+7}{3}< \\dfrac{19x + 16}{10}+1=\\dfrac{19x +26}{10}$$\n\nReplace $$x$$ with $$\\frac m{19}$$ and\n\n$$\\dfrac {m + 16}{10} \\le \\dfrac {\\frac 4{19}m + 7}3 < \\frac {m+26}{10}$$\n\n$$3m + 48 \\le \\frac 40{19}m +70 < 3m + 78$$\n\n$$-22 \\le \\frac {40}{19}m - 3m < 8$$\n\n$$-22*19 \\le 40m - 57m = -17*m < 8*19$$\n\n$$\\frac {-8*19}{17} < m \\le \\frac {22*19}{17}$$\n\n$$-8 \\frac {16}{17} < m \\le 24 \\frac {10}{17}$$\n\nSo $$-8 < m \\le 24$$\n\nBut $$m = 10k-16$$ so $$m\\equiv -16\\equiv -6 \\equiv 4 \\pmod{10}$$ so so\n\n$$m =-6,4,14,24$$ are acceptable values.\n\nAnd $$x =-\\frac 6{19},\\frac 4{19}, \\frac {14}{19}$$ and $$\\frac {24}{29}$$ are acceptable answers.", "date": "2021-09-22 00:16:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3684603/floor-function-algebra-question", "openwebmath_score": 0.9941151142120361, "openwebmath_perplexity": 3092.5862670970705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988668248138067, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.862462155612602}}
{"url": "https://mathematica.stackexchange.com/questions/118481/how-does-listinterpolation-work", "text": "# How does ListInterpolation work?\n\nListInterpolation takes an array of data and interpolates between the entries. Let us interpolate between the elements of the identity matrix. For reference, this is what the identity matrix looks like:\n\nMatrixPlot[IdentityMatrix[5]]\n\n\nNow let's interpolate, and request first order interpolation:\n\nif = ListInterpolation[IdentityMatrix[5], {{0, 1}, {0, 1}}, InterpolationOrder -> 1];\nPlot[if[x, x], {x, 0, 1}]\n\n\nThe result is a piecewise function as expected, but not a piecewise linear function! Why? What does first order interpolation mean here?\n\nThe kind of linear interpolation I am familiar with is what ListDensityPlot does:\n\nListDensityPlot[IdentityMatrix[5], DataRange -> {{0, 1}, {0, 1}}, Mesh -> All]\n\n\nConstruct a Delaunay triangulation of the data and interpolate on each triangle.\n\nThis is not what ListInterpolation does (or even Interpolation when the data is strictly on a square grid). What does ListInterpolation do, then?\n\nInspired by this question:\n\n\u2022 Well, it is doing bilinear interpolation in your first example, I would think. You definitely will get something with quadratic pieces if you sample across diagonals. \u2013\u00a0J. M. is in limbo Jun 15 '16 at 14:00\n\u2022 @J.M. Well, we all have to learn at some point :-) Yes, you are right, and this is due to my ignorance of math. But I won't delete it. If you convert that to a one-line answer, I'll accept it. \u2013\u00a0Szabolcs Jun 15 '16 at 14:01\n\u2022 Docs say the only methods available to ListInterpolation are Spline and Hermite \u2013\u00a0N.J.Evans Jun 15 '16 at 14:02\n\u2022 Currently my comment is not that pedagogically useful; I'll try to come up with a sufficiently illuminating example first, unless somebody beats me to it. :) \u2013\u00a0J. M. is in limbo Jun 15 '16 at 14:04\n\u2022 Szabolcs, please consider leaving the question up for a little longer. I was not aware of that behavior either, and I would find it instructive to see an interesting example, if @J.M. or others have the time / inclination. \u2013\u00a0MarcoB Jun 15 '16 at 14:43\n\nI'll just treat the simplest case of interpolating across four noncoplanar points. The extension to general grids is straightforward (one just joins multiple pieces appropriately).\n\nConsider the following bivariate interpolating polynomial:\n\nip[x_, y_] = InterpolatingPolynomial[{{{0, 0}, 1}, {{0, 1}, -1/2},\n{{2, 0}, 1/2}, {{2, 1}, 0}}, {x, y}];\n\n\nand the following InterpolatingFunction[]:\n\nif = ListInterpolation[{{1, -1/2}, {1/2, 0}}, {{0, 2}, {0, 1}}, InterpolationOrder -> 1];\n\n\nThey are the same:\n\nPlot3D[ip[x, y] - if[x, y] // Chop, {x, 0, 2}, {y, 0, 1}]\n\n\nLet's have a look at what ip[] looks like:\n\nExpand[ip[x, y]]\n1 - x/4 - (3 y)/2 + (x y)/2\n\n\nHuh. That last term certainly isn't linear. What gives?\n\nIn a bilinear interpolation such as this one, although the procedure is something along the lines of \"linearly interpolate for each grid line in one direction, and then linearly interpolate those results\", the result is inevitably quadratic. This should come as no surprise to people familiar with the hyperbolic paraboloid, which is one of the simplest examples of a ruled surface, or a surface formed by sweeping out a line in space. In fact, z == 1 - x/4 - (3 y)/2 + (x y)/2 is indeed a hyperbolic paraboloid.\n\n{Plot3D[ip[x, y], {x, 0, 2}, {y, 0, 1}, MeshFunctions -> {#1 - #2 &}],\nPlot[ip[x, x], {x, 0, 1}]} // GraphicsRow", "date": "2020-02-26 02:31:36", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/118481/how-does-listinterpolation-work", "openwebmath_score": 0.19380126893520355, "openwebmath_perplexity": 1072.829172035491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707953529716, "lm_q2_score": 0.8824278633625321, "lm_q1q2_score": 0.8624592226562613}}
{"url": "https://math.stackexchange.com/questions/3013010/how-to-find-out-the-global-minimum-of-the-following-expression/3013059", "text": "# How to find out the global minimum of the following expression\n\nWhat is the global minimum of the expression \\begin{align} |x-1| &+ |x-2|+|x-5|+|x-6|+|x-8|+|x-9|+|x- 10| \\\\&+ |x-11|+|x-12|+|x-17|+|x-24|+|x-31|+ |x-32|? \\end{align}\n\nI've solved questions of this sort before but there were only 3 terms. I solved those by expanding all the terms in the modulus and drawing a graph. This question came in a paper which requires the student to solve it within 5 minutes. What's a better method?\n\nUnfortunately I needed some minutes to think about the problem before finding a solution that can be calculated very quickly:\n\nImagine the graph of the function $$f_a(x)=|x-a|$$. Having the graph in mind you see that the derivation $$f'(x)=-1$$ for $$x and $$f'(x)=1$$ for $$x>a$$.\n\nFor the intervals: $$(-\\infty,1)$$, $$(1,2)$$, $$(2,5)$$, ..., $$(32,\\infty)$$ we can now easily calculate the derivative $$f'(x)=f'_1(x)+f'_2(x)+f'_5(x)+...-f'_{32}(x)$$:\n\nIn the range $$(-\\infty,1)$$ it is $$f'(x)=-1-1-1-...-1+1=-11$$.\nIn the range $$(1,2)$$ it is $$f'(x)=+1-1-1-...-1+1=-9$$.\nIn the range $$(2,5)$$ it is $$f'(x)=+1+1+1-...-1+1=-7$$.\n...\n\nIn every step we simply have to invert one sign so \"-1\" becomes \"+1\". This means the derivative is changing by 2 at the points x=1,2,5,...\n\nWe start by calculating the derivative for $$x<1$$; it is -11.\n\nNow we simply go through the ranges:\n\n<1: -11\n1..2: -9\n2..5: -7\n5..6: -5\n6..8: -3\n8..9: -1\n9..10: +1\n10..11: +3\n11..12: +5\n12..17: +7\n17..24: +9\n24..31: +11\n31..32: +13\n>32: +11\n\nAt $$x=32$$ the derivation decreases by 2 because of the minus sign before $$|x-32|$$; you could of course adapt this method for sums of elements of the form $$b|x-a|$$.\n\nWe see that for $$x<9$$ the derivation is negative and for $$x>9$$ the derivation is positive. We also know that the function is continuous. (This is important because the derivation is not defined at x=1,2,5,...) This means that the function is strictly decreasing respectively increasing for $$x<9$$ and for $$x>9$$.\n\nSo we know that the global minimum must be at $$x=9$$.\n\n\u2022 The right answer. Congratulations! Although I don't know if your method works when the number of terms is even, say $f(x) = |x-1|+ |x-2|$ or $f(x) = |x-11|+|x-12|+|x-17|+|x-24|$ (the minimum or maximum is an interval). Nov 25 '18 at 22:20\n\u2022 @David Why should it be a problem if a whole interval represents the minimum or maximum? In the example of $|x-1|+|x-2|$ we would have $f'(x)=0$ in the whole interval 1...2. Nov 26 '18 at 6:28\n\u2022 This answers the previous version of my typo question but the method is absolutely wonderful. Thanks!\n\u2013\u00a0user619072\nNov 27 '18 at 17:36\n\nYou can in principle write out the function in a lot of intervals. But it would probably take too long. However, I will use this fact, without doing it explicitly. We know that if we do write this function, it is going to be linear on each interval (sum of linear functions is a linear function), and it's going to be continuous (sum of continuous functions is a continuous function). We also know that on a line you get minimum at one end, the other, or both (constant line). So all you need to do is to calculate your function at $$1,2,5,6,...$$ and find the minimum.\n\n\u2022 I like how this answer asks us to imagine the graph of the function. I find this \"imagine\" technique useful, while still being intuitive. Similar examples includes \"imagine the multiplication table of this finite group\" (the group could be huge) or \"imagine the truth table for this proposition\" (the formula could be long). Nov 25 '18 at 22:02\n\nThe answer (minimizer) in this case is $$10$$, the median of the sequence $$(1,2,5,6,8,9,10,11,12,17,24,31,32).$$\n\nYou can plug in $$x=10$$ in the function and you would find that the minimum value is $$96$$. In general, the solution to the following minimization problem\n\n$$\\min\\{|x-a_1| + |x-a_2| + \\cdots + |x-a_n|\\}$$ is the median of $$(a_1,\\ldots,a_n)$$. To see why, consider first when $$n=2$$, and without loss of generality assume $$a_1. Then $$|x-a_1|+|x-a_2|$$ is the distance between $$x$$ and $$a_1$$ plus the distance between $$x$$ and $$a_2$$. It is easy to see that only when $$x$$ is in the middle of $$a_1$$ and $$a_2$$ should the sum of distances be minimal, which equals $$|a_2-a_1|$$ in this case. In this case the minimizer is not unique. Any points in $$[a_1,a_2]$$ is a minimizer.\n\nWhen $$n=3$$, the function is $$|x-a_1|+|x-a_2|+|x-a_3|$$, and we order the parameters again so that $$a_1. When $$x$$ coincides with $$a_2$$, i.e. $$x=a_2$$, the value becomes $$|a_2-a_1|+|a_2-a_3|=|a_3-a_1|$$, the distance between $$a_3$$ and $$a_1$$. But when $$x\\in[a_1,a_3], x\\neq a_2$$, the value of the function is $$|x-a_2|+|x-a_1|+|x-a_3| = |a_3-a_1| + |x-a_2|,$$ which is largar than $$|a_3-a_1|$$, the distance between $$a_3$$ and $$a_1$$. Similarly the value would become larger when $$x$$ is outside $$[a_1,a_3]$$. So in this case, the minimizer is unique and is equal to $$a_2$$, the median of $$(a_1,a_2,a_3)$$.\n\nIn general, when $$n$$ is odd, there exists a unique minimizer, which is equal to the (unique) median of the parameters $$(a_1,\\ldots,a_n)$$. When $$n$$ is even, the function is minimal and constant over the range $$[a_i,a_j]$$, where $$a_i$$ and $$a_j$$ are the two middle values.\n\n\u2022 That doesn't seem to be correct. Wolfram says the minimum is 51 at x=9. See here Nov 25 '18 at 20:50\n\u2022 I see now that the confusion comes from the $\u2212|x\u221232|$ in the problem statement @MartinRosenau. Note the minus sign. Fei Li solved it for a + sign. Nov 25 '18 at 21:18\n\u2022 I plot the graph using wxmaxima and get the same result as @IlikeSerena. Note the minus sign at the end. Nov 25 '18 at 21:48\n\u2022 Maybe the OP @CaptainQuestion made a typo. We need his or her classification, but I think the purpose of this question is exactly what I have demonstrated. Nov 25 '18 at 22:00\n\u2022 @AlexVong So I hoped, but the next sentence \"which equals $|a_2-a_1|$ in this case.\" excludes this interpretation. Nov 25 '18 at 22:39\n\nTL;DR: Put the absolute values in ascending order, and look at the sum of the leading coefficients. One by one, change the signs in the sum from right to left. When the sum changes signs, you have passed a local extremum. When the sum equals zero, there is an extremum on an entire interval.\n\nAs an alternative to Andrei's excellent answer, or perhaps an extension, you could also look at the derivative. Clearly the function is continuous everywhere, and it is differentiable in all but finitely many points, call them $$a_1,\\ldots,a_n$$ in ascending order. Then we want to minimize $$f(x)=\\sum_{k=1}^nc_k|x-a_k|=\\sum_{k=1}^n(-1)^{\\delta_{x\\leq a_k}}c_k(x-a_k),$$ where $$\\delta$$ denotes the Kronecker delta, in this case defined as $$\\delta_{x\\leq a_k}:=\\left\\{\\begin{array}{ll}1&\\text{ if } x\\leq a_k\\\\0&\\text{ otherwise}\\end{array}\\right..$$ This has derivative (for all $$x$$ apart from the $$a_k$$) $$f'(x)=\\sum_{k=1}^n(-1)^{\\delta_{x\\leq a_k}}c_k.$$ The expression has a local minimum at $$x$$ if either $$f'(x)=0$$, or if $$x=a_k$$ for some $$k$$ and $$f'(y)<0$$ for $$a_{k-1} and $$f'(y)>0$$ for $$a_k.\n\nThis is all rather formal; in practice this means you put the $$c_k$$ in ascending order, so here $$n=13$$ and $$c_1=\\cdots=c_{12}=1$$ and $$c_{13}=-1$$, and find all $$m$$ such that flipping the last $$m$$ signs in the sum $$c_1+c_2+c_3+c_4+c_5+c_6+c_7+c_8+c_9+c_{10}+c_{11}+c_{12}+c_{13}$$ makes the sum change signs compared to changing the last $$m-1$$ signs. Here a quick look gives $$c_1+c_2+c_3+c_4+c_5+c_6-c_7-c_8-c_9-c_{10}-c_{11}-c_{12}-c_{13}=1,$$ $$c_1+c_2+c_3+c_4+c_5-c_6-c_7-c_8-c_9-c_{10}-c_{11}-c_{12}-c_{13}=-1,$$ so $$f$$ has a local minimum at $$a_6=9$$, and it is not difficult to see that there is no other minimum.", "date": "2021-09-25 08:53:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3013010/how-to-find-out-the-global-minimum-of-the-following-expression/3013059", "openwebmath_score": 0.8267564177513123, "openwebmath_perplexity": 366.92254673990374, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707966712549, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.862459216264575}}
{"url": "https://math.stackexchange.com/questions/3132934/a-mistake-on-computing-int-fracdx-sqrtx11/3132967", "text": "# A mistake on computing $\\int \\frac{dx}{\\sqrt{x+1}+1}$\n\nI have to find $$\\int \\frac{dx}{\\sqrt{x+1}+1}$$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.\n\nFirst I write $$\\frac{1}{\\sqrt{x+1}+1}=\\frac{\\sqrt{x+1}-1}{x}=\\frac{\\sqrt{x+1}}{x}-\\frac{1}{x}$$, therefore $$\\int \\frac{dx}{\\sqrt{x+1}+1}=\\int \\frac{\\sqrt{x+1}}{x}dx-\\log\\left (x\\right )$$, so I only have to deal with $$\\int \\frac{\\sqrt{x+1}}{x}dx$$.\n\nSetting $$u=\\sqrt{x+1}$$, we obtain $$du=\\frac{1}{2\\sqrt{x+1}}dx$$, therefore $$dx=2udu$$ and $$x=u^2-1$$, so\n\n$$\\int\\frac{\\sqrt{x+1}}{x}dx=\\int\\frac{2u^2}{u^2-1}du=\\int \\left(2+\\frac{2}{u^2-1}\\right)du=2\\sqrt{x+1}+\\int \\left(\\frac{1}{u-1}-\\frac{1}{u+1}\\right)du=$$\n\n$$=2\\sqrt{x+1}+\\log \\left (\\sqrt{x+1}-1\\right )-\\log \\left (\\sqrt{x+1}+1\\right )=2\\sqrt{x+1}+\\log\\left ( \\frac{\\sqrt{x+1}-1}{\\sqrt{x+1}+1} \\right )$$\n\nBut then $$\\int \\frac{dx}{\\sqrt{x+1}+1}=2\\sqrt{x+1}+\\log\\left ( \\frac{\\sqrt{x+1}-1}{\\sqrt{x+1}+1} \\right )-\\log\\left (x\\right )=2\\sqrt{x+1}+\\log\\left ( \\frac{x+2-2\\sqrt{x+1}}{x^2} \\right )$$, which is not what I am supposed to obtain. The actual answer is $$2\\sqrt{x+1}-2\\log\\left ( 1+\\sqrt{x+1} \\right )$$.\n\nWhere is my mistake?\n\n(I already know a correct way to solve it, I just want to know where I am committing a mistake).\n\n\u2022 I believe that you went wrong in your first step. That \u201csimplification\u201d made everything a lot harder. Hint: Start with a u-substitution if u=x+1 . \u2013\u00a0Sina Babaei Zadeh Mar 2 '19 at 22:36\n\u2022 The standard method suggestst to set $u=\\sqrt{x+1}$, i.e. $x=u^2-1,\\; u\\ge 0$, from the very beginning. \u2013\u00a0Bernard Mar 2 '19 at 22:44\n\nBut you have been right all along!\n\nNotice that $$\\log\\frac{\\sqrt{x+1}-1}{\\sqrt{x+1}+1} - \\log x = \\log\\frac{\\sqrt{x+1}-1}{x(\\sqrt{x+1}+1)} = \\log \\frac{1}{(\\sqrt{x+1}+1)^2} = -2\\log(\\sqrt{x+1}+1)$$ by your very first step.\n\nAlso, I would suggest that you use $$\\int\\frac{dx}{x} = \\log|x|$$ (with the absolute value sign, to extend the domain).\n\nWhat I would also suggest is to use an online function grapher if you are not sure that your solution is correct. If two functions have identical graphs or if they differ by a constant value, then both are correct solutions of the integral.\n\n\u2022 Oh, I see... My answer was correct, but it was not clear why. I wrote $\\log\\left ( \\frac{\\sqrt{x+1}-1}{\\sqrt{x+1}+1} \\right )=\\log\\left ( \\frac{x+2-2\\sqrt{x+1}}{x} \\right )$ (i.e., i was rationalizing the denominator), but I should have rationalized the numerator, leading to $\\log\\left ( \\frac{x}{\\left ( \\sqrt{x+1}+1 \\right )^2} \\right )$, which gives the solution. Thank you! \u2013\u00a0solomeo paredes Mar 2 '19 at 23:04\n\nNote that $$\\ln\\frac{x+2-2\\sqrt{x+1}}{x^2}=\\ln\\frac{(\\sqrt{x+1}-1)^2}{x^2}=2\\ln\\frac{\\sqrt{x+1}-1}{x}$$ if we further simplify, $$2\\ln(\\frac{\\sqrt{x+1}-1}{x}\\cdot\\frac{\\sqrt{x+1}+1}{\\sqrt{x+1}+1}) =2\\ln\\frac{1}{\\sqrt{x+1}+1}=-2\\ln(\\sqrt{x+1}+1)$$ You did nothing wrong.\n\nNote that $$\\frac{x+2-2\\sqrt{x+1}}{x^2} =\\frac {(x+2)^2-4(x+1)} {x+2+2\\sqrt{x+1}} \\cdot\\frac 1{x^2}\\ ,$$ and this has nothing to do with $$\\frac{\\sqrt{x+1}-1}{\\sqrt{x+1}+1} \\cdot \\frac 1x = \\frac{(\\sqrt{x+1}-1)(\\sqrt{x+1}+1)}{(\\sqrt{x+1}+1)(\\sqrt{x+1}+1)} \\cdot \\frac 1x = \\frac 1{(\\sqrt{x+1}+1)^2}\\ ,$$ which leads to the solution. So the last step is the bad one.\n\nYou're fine.\n\n$$\\log\\left(\\frac{\\sqrt{x+1}-1}{\\sqrt{x+1}+1}\\right) -\\log x$$\n\n$$= \\log\\left(\\frac{\\sqrt{x+1}-1}{\\sqrt{x+1}+1}\\frac{\\sqrt{x+1}+1}{\\sqrt{x+1}+1}\\right) -\\log x$$\n\n$$=\\log\\left( \\frac{x}{ (\\sqrt{x+1}+1)^2}\\right) -\\log x$$\n\n$$=\\log x -2\\log(\\sqrt{x+1}+1)- \\log x$$\n\n$$=-2\\log(\\sqrt{x+1}+1).$$\n\nNo mistake! $$log((\\frac{\\sqrt{x+1}-1}{x})^2)=-2log(\\frac{x}{\\sqrt{x+1}-1})=-2log(\\frac{x(\\sqrt{x+1}+1)}{(\\sqrt{x+1}-1)(\\sqrt{x+1}+1)})=-2log\\frac{x(\\sqrt{x+1}+1)}{x}$$", "date": "2020-02-29 14:35:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3132934/a-mistake-on-computing-int-fracdx-sqrtx11/3132967", "openwebmath_score": 0.805465042591095, "openwebmath_perplexity": 268.4760992944203, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576914206421, "lm_q2_score": 0.8840392893839086, "lm_q1q2_score": 0.8624483219743908}}
{"url": "http://rafbis.it/qowu/graph-of-sine-and-cosine-functions-real-world-applications.html", "text": "## Graph Of Sine And Cosine Functions Real World Applications\n\nThe cosine is known as an even function, and the sine is known as an odd function. Identities and equations do not include any half-angle relationships. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. The amplitude is 2. Are you looking to apply what you learn in the classroom to the real world? Trigonometry is a subject that has lots of practical applications. Markov processes are widely used in performance modelling of wireless and mobile communication systems. Frequency is the number of cycles in a given unit of time, so it is the reciprocal of the period of a function. Whenever you see an \"oscilloscope,\" for example when you play music using certain programs on a computer, you're really seeing a whole bunch of sine waves added together. In that program you used the built-in sine and cosine functions. Frequency is the number of cycles in a given unit of time. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. 762-768) optional 2 optional 1 \u2022 Graph trigonometric functions. So the graph looks like a very simple wave. One method to write a sine or cosine function that models a sinusoid is to fi nd the values of a, b, h, and k for y = a sin b(x \u2212 h) + k or y = a cos b(x \u2212 h) + k where \u2223 a \u2223 is the amplitude, \u2014 2\u03c0 is the period. The inverses of sine, cosine and tangent functions are not functions unless the domains are limited. Request PDF | A Discrete Sine Cosine Algorithm for Community Detection | As an important research area in complex networks, community detection problems (CDPs) have aroused widespread concern. Find an expression for a volume. Evaluate Inverse Trigonometric Functions v. We often graph sine over time (so we don't write over ourselves) and sometimes the \"thing\" doing sine is also moving, but this is optional! A spring in one dimension is a perfectly happy sine wave. acute triangles using the sine law and the cosine law, including problems arising from real-world applications; demonstrate an understanding of periodic relationships and the sine function, and make connections between the numeric, graphical, and algebraic representations of sine functions; identify and represent sine functions, and. Using your guidelines, you should be able to sketch in a smooth curve: see figure 3. FIND the coordinates of A & C. The graphs of sine, cosine, and tangent. ) like any other function with a domain or real numbers!. \u2022 Solve real world and mathematical problems using a system of two first-degree equations in two variables. First, take a look the Taylor series representation of exponential function, and trigonometric functions, sine, and cosine,. The vertical shift is 2. Find an equation for a sine function that has amplitude of 4 a period of \u03c0. 9) Graph sine and cosine functions (PC-N. THE COSINE FUNCTION You can graph the cosine function y = cos x. Trig Functions: The Ferris Wheel Applications of Trig Functions - Duration: 13:37. txt) or view presentation slides online. 25 scaffolded questions on equation graph involving amplitude and periodplus model problems explained step by step. Find Period and Amplitude 2. Whenever you see an \"oscilloscope,\" for example when you play music using certain programs on a computer, you're really seeing a whole bunch of sine waves added together. Determine the values for angles between 0\u00b0 and 360\u00b0. Trigonometry An In Depth Approach to Sine, Cosine, Tangent, Cotangent, Secant and Cosecant: Trigonometry Ratios and Their Graphs and Real World Applications. Recall that the sine and cosine functions relate real number values to the x\u2013 and y-coordinates of a point on the unit circle. Lesson 9-11: Graphing Trigonometric Functions and Applications in Real World Contexts Learning Goals #8, 9, 10: How do I use the critical values of the Sine and Cosine curve to assist in in graphing Sine and Cosine functions? How do I contextualize Trig Functions in real life situations? Warm-Up! WITH A PARTNER. I can identify the important features (amplitude, period, max, min, x-intercepts, and frequency) of the sine, cosine, and tangent functions. Sinusoidal functions graph wave forms. Graphs of the sine and cosine functions; and Periodic Behavior. The reality is that the functions of sine, cosine and tangent are embedded in the foundations of modern mathematics and, as you\u2019ll discover, the world around us. Students use special triangles to geometrically determine the values of sine. KEY STANDARDS ADDRESSED: MA3A3. We start by revisiting the Ferris wheel. Write a rule in the form f(t) = A sin Bt that expresses this oscillation where t represents the number of seconds. Students are expected to be able to interpret functions in the context of a problem, write functions, and apply functions to periodic phenomenon. The hyperbolic sine and the hyperbolic cosine are. If we recall from the previous section we said that $$f\\left( x \\right)$$ is nothing more than a fancy way of writing $$y$$. Includes full solutions and score reporting. Line 1 just restates Euler's formula. So for example, if x = 2, the y-value will be y = 2 sin. Accelerated Mathematics III - Unit 5 Investigating Trigonometry Graphs Student Edition INTRODUCTION: In this unit students develop an understanding of the graphs of the six trigonometric functions: sine, cosine, tangent, cotangent, cosecant, and secant. Graphing Sine & Cosine w/out a Calculator Pt 2. Read: 'The sine of thirty degrees is zero point five. Trigonometry means the study of the triangle. Verify the size of a specified angle and hence that two triangles within the given diagram are similar. The resulting graph will be displayed as you change the values. ( Source: Wikipedia, try not to get hypnotized. Find angle pi on a grid. 387 In-class/Homework: pg. Apply right-angle trigonometry to a scenario. Many compression algorithms like JPEG use fourier transforms that rely on sin and cos. Sine and Cosine Part 2 - given the graph of a sinusoidal function, determine an equation that defines it D2. We offer targeted baccalaureate programs of study, pre-baccalaureate programs of study for transfer, associate of arts and associate of science degrees, and serve as a portal to graduate education. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval$\\,\\left[-1,1\\right]. - He excelled in the trigonometric functions unit where he was able to sketch sinusoidal functions using the transformations of key points of the sine or. Different sounds create different waves. Use sinusoidal functions to solve problems. In this example, we are multiplying the sine of each x-value by the x-value. Transformations of Trigonometric Functions. \" For example, \"an oscilloscope is an electronic instrument used to display changing electrical signals. and \u201cuse sine and cosine functions to model real-life data,\u201d iii. To be able to solve real-world problems using the Law of Sines and the Law of Cosines This tutorial reviews two real-world problems, one using the Law of Sines and one using the Law of Cosines. We do: Writing Sinusoidal Functions from a Graph. march 21, 2007. To find the cosine of angle pi, you need graph paper. Moreoever, the pattern repeats, so this is still a periodic function. can be used to assign a particular use of the plot function to a particular figure wi. Assessments Summative: Tests and benchmark. ppt), PDF File (. Markov processes are widely used in performance modelling of wireless and mobile communication systems. Graphing a Sinusoidal Function. ( Source: Wikipedia, try not to get hypnotized. A periodic function is defined as a function that repeats its values in regular periods. Trig Functions: The Ferris Wheel Arnold Tutoring Solving Problems with the Sine and Cosine Functions - Lesson - Duration: Determining the Equation of a Sine and Cosine Graph - Duration:. For this, the phase shift will be 172. So what do they look like on a graph on a coordinate plane? Let\u2019s start with the sine function. Sketch translations of graphs of sine and cosine functions; use sine and cosine functions to model real-life data. 2 Graph lines (including those of the form x = h and y = k) given appropriate information. I am confused as I am unsure whether time (e. As we can see in Figure 6, the sine function is symmetric about the origin. In fact, there is a zero derivative at this point here, which leads us to believe that maybe what we should have done was to have broken this curve down into the union of two one-to-one functions. y = 2 sin 2x 14. Amplitude and Period for Sine and Cosine Functions Worksheet Author: marthasweeney Last modified by: Leake, Sherry LHS-STAFF Created Date: 6/13/2016 5:56:00 PM. 2]Solve right and oblique triangles including application problems. Trigonometric formulae are useful for solving problems in two dimensions. Find all values of A and B such that. PERIODIC FUNCTIONS A. Four Function Scientific. Lesson 12: Ferris Wheels\u2014Using Trigonometric Functions to Model Cyclical Behavior Student Outcomes Students review how changing the parameters \ud835\udc34, \ud835\udf14, \u210e, and \ud835\udc58 in ( )=\ud835\udc34sin(\ud835\udf14( \u2212\u210e))+\ud835\udc58 affects the graph of a sinusoidal function. 28rad has the same cosine as 5rad, they're inverse angles. php(143) : runtime-created function(1) : eval()'d code(156) : runtime-created. Solve real world applications problems using systems of equations and inequalities Graph rational functions with asymptotes Graph sine and cosine curves. How the values of A and B affect the shape of the graph y = A sin(Bx). The table of values for the functions. The number \u03c0 (pi)\u2014 or at least the closest approximation that fits in a JavaScript number\u2014is available as Math. Graphing Sine & Cosine w/out a Calculator Pt 2. When they notice that these points are already on the graph, make a note about the cyclic nature of sine and cosine. List of trigonometric identities 2 Trigonometric functions The primary trigonometric functions are the sine and cosine of an angle. Correctness of a skewed cosine similarity graph I am currently implementing a word2vec model that uses the cosine similarity to determine the similarity between two vectors. A great discussion occurs when the students are asked to justify the function they want to use. The simplest one is y = sin(x). sine, cosine, tangent for \u03c0/3, \u03c0/4 and \u03c0/6, and use the unit circle to express the values of sine, cosine, and tangent for \u03c0\u2013x, \u03c0+x, and 2\u03c0\u2013x in terms of their values for x, where x is any real number. Topic:Real world application of sine graphs. Graph) 3 2 f x = ( ) 2sin(x. Graphing Trigonometric Functions Real-World Applications of Trigonometric Functions. In trig speak, you say something like this: If theta represents all the angles in the domain of the two functions. ) Circles are an example of two sine waves. The reality is that the functions of sine, cosine and tangent are embedded in the foundations of modern mathematics and, as you\u2019ll discover, the world around us. In maths, you have real life applications on any thing that you study. For now, focus primarily on sine, cosine, and tangent. Trigonometric Functions, you will begin by learning about the inverses of quadratics and other functions. any anual average can be repeated year to year. Graphing Sine and Cosine Functions. Les't compare with. Graph a sine or cosine function with a given phase shift, period, and amplitude. When plotting all the possible cosine similarities, I get the following. find the equation of a sinusoidal function given its graph or its properties. 7m, but I am unsure how to interval the axis. Of course, this was to be expected, but this isn't the real problem; the real problem is that if your function has some crucial points it must pass through (which is certainly true for trigonometry functions), the truncation will move the curve away from those points. So for example, if x = 2, the y-value will be y = 2 sin. Intro Tangent & Cotangent Graphs. Outcome 6: Set up and solve exponential and logarithmic equations; then identify and sketch graphs of the functions. 4 5 Graphing the Trigonometric Functions (Part 2) 2 The sine function Imagine a particle on the unit circle, starting at (1,0) and rotating counterclockwise around the origin. How to find and the slopes of lines to write and graph linear equations in two variables. Given an expression for the time rate of change of a quantity, find the initial rate. 762 Chapter 14 Trigonometric Graphs and Identities \u2022 Graph trigonometric functions. Sample Problem. 1 solve problems, including those that arise from real-world applications (e. graph the sine and cosine functions with varied amplitudes and periods. 14-1 Graphs of Sine and Cosine Example 3: Sound Application Use a sine function to graph a sound wave with a period of 0. Those functions were predefined for you, meaning that you didn't have to tell the computer how to compute the sine and cosine of an angle. Focus on creating a graph and formula from a story (starting from time 6:55) You Do Now: Warmup - Sketching a Cosine Wave. When the radian measure of angles was introduced in the early eighteenth century by Roger Cotes and the trigonometric ratios of angles in the plane began to be viewed as functions of a linearized angle, the undulating movement of the sine and cosine graphs fed the. Les't compare with. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector. Trigonometric identities like finding. Graphing Secant & Cosecant. A real life example of the sine function could be a ferris wheel. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. dilations, and horizontal and vertical translations of functions. This example shows how to use both CORDIC-based and lookup table-based algorithms provided by the Fixed-Point Designer\u2122 to approximate the MATLAB\u00ae sine (SIN) and cosine (COS) functions. Every position of the particle corresponds with an angle, ?, where y sin ?. In line 3 we plug in -x into Euler's formula. Example: if your cutting down a tree and need to know where it will. Using f (x) = cos x as a guide, graph the function h (x) = __1 cos 2 3 x. SC-Common Core Precalculus Scope and Sequence Unit Topic Lesson Lesson Objectives Functions Functions and Graphs Graphing Linear Equations Find the slope of a line through two points Find the x- and y-intercepts of a line Find zeros of linear functions Graph linear equations Operations with Functions Find composite functions. Graph Translations 3. Plus ITEM TYPES: MC. Determine the characteristics of the graphs of the six basic trigonometric functions. I can graph the sine, cosine, and tangent functions. 3 make connections between the sine ratio and the sine function and between the cosine ratio and the cosine function by graphing the relationship between angles from 0\u00ba to 360\u00ba and the corresponding sine ratios or cosine ratios, defining this relationship as the function f(x) =sinx or f(x) =cosx, and explaining why the relationship is a function. ( Source: Wikipedia, try not to get hypnotized. Equation of Sine and Cosine from a Graph. Calculus is made up of Trigonometry and Algebra. inverse MTH 112 Elementary Functions Chapter 5 The Trigonometric Functions -. You can use the graph for a trigonometry function to show average daily temperatures at a specific location. Most radio communication is based on the use of combinations of sines and cosine waves. To understand how the absolute value can be applied to the real world, we\u2019ll look at two topics: What is considered normal with regard to the temperature of the human body and; Fuel economy of two vehicles: a Honda Odyssey and a Nissan Altima. Trig also has applications in fields as broad as financial analysis, music theory, biology, medical imaging, cryptology, game development, and seismology. Amplitude is first, then sine or cosine, then B, which you get by doing 2pi/10 (period), which reduces to pi/5, then it is the parenthesis with the transformation, and then last but not least is the sinusodial axis. Application: Sine and cosine functions can be used to model real-world phenomena, such as sound waves. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. functions; increasing/decreasing functions; finding intercepts with axes. Evaluate trigonometric functions of real numbers. Modeling Using Sinusoidal Functions Sine and cosine functions model many real-world situations. The graph of the sine function is called the sine curve. Defining Sine and Cosine Functions. When they notice that these points are already on the graph, make a note about the cyclic nature of sine and cosine. Amplitude is first, then sine or cosine, then B, which you get by doing 2pi/10 (period), which reduces to pi/5, then it is the parenthesis with the transformation, and then last but not least is the sinusodial axis. Topic:Real world application of sine graphs. Example: if your cutting down a tree and need to know where it will. Find all values of A and B such that. 7m, but I am unsure how to interval the axis. This shape is also called a sine. Evaluate Inverse Trigonometric Functions v. For large arguments we would expect these functions to be a quarter period out of phase, just like cosine and sine, since asymptotically J 1 is proportional to cos(z \u2013 3\u03c0/4) / \u221az and Y 1 is proportional to sin(z \u2013 3\u03c0/4) / \u221az. Graphing Sine & Cosine w/out a Calculator Pt1. The function is a model for the hours of daylight in Center City. Sum and Difference, Double Angle and Half Angle. The size of a hyperbolic angle is twice the area of its hyperbolic sector. Students will use the model to make a prediction about a missing piece of data. 2 cos 2 cos 2 cos 2 sin( ) cos Now we can clearly see that if we horizontally shift the cosine function to the right by \u03c0/2 we get the sine function. 1 works in a wide range of settings: any time we have an exponential equation of the form $$p \\cdot q^t + r = s\\text{,}$$ we can solve for $$t$$ by first isolating the exponential expression $$q^t$$ and then by taking the natural logarithm of both sides of the equation. In maths, you have real life applications on any thing that you study. The cosine is known as an even function, and the sine is known as an odd function. Cumulative Review. Along the way, youll get plenty of practice, from fully guided examples to independent end-of-chapter drills and test-like samples. Midline Amplitude And Period Review Article Khan Academy Solving Problems Using Sinusoidal Functions Real World Applications Ferris Wheel Trig Example Youtube Period Of Cosine Function Math Ladyprestige Club. Next, lets take the 5 from the front. 2 Define sine and cosine as functions of the radian measure of an angle in terms of the x- and y-coordinates of the point on the unit circle. But just as you could make the basic quadratic, y = x 2, more complicated, such as y = -(x + 5) 2 - 3, so also trig graphs can be made more complicated. I know that the amplitude of the function is 1. When I consider how to address the Precalculus objectives \"to solve real-life problems involving harmonic motion\" ii. A cosine function will start at its minimum or maximum value on the y-axis; however, the sine graph will start at its midline on the y-axis. Graph sine and cosine functions with various amplitudes and periods. 6 Modeling with Trigonometric Functions 507 Writing Trigonometric Functions Graphs of sine and cosine functions are called sinusoids. Determine the values for angles between 0\u00b0 and 360\u00b0. can be used to assign a particular use of the plot function to a particular figure wi. A sine wave is what you get when you plot the sine function on a graph. Psst\u2026 don\u2019t over-focus on a single diagram, thinking tangent is always smaller than 1. Defining Sine and Cosine Functions. Sketch the graphs of basic sine and cosine functions; use amplitude and period to help sketch the graphs of sine and cosine functions. Phase Shift = to the right Vertical Shift = 4 4. How to analyze graphs of functions. Sine, Cosine, Tangent Applications. My kids have seen this a bunch so this should only take a minute or two. 2]Solve right and oblique triangles including application problems. Analysis: I developed this lesson plan so that students would be introduced to graphs\u2026. Aug 15, 2015 \u00b7 C2 Sine and Cosine Rule - 1 - Basic Introduction (AS Maths A Level trigonometry) ukmathsteacher. In line 4 we use the properties of cosine (cos -x = cos x) and sine (sin -x = -sin x) to simplify the. Graphing the sine and cosine functions is a topic that students will carry on with them throughout the rest of their future science and mathematics courses. Graphs of tan, cot, sec and csc; 5. Real life word problems that involve the sine and cosine function can be used to keep the students engaged in the topic. inverse MTH 112 Elementary Functions Chapter 5 The Trigonometric Functions -. The inverses of sine, cosine and tangent functions are not functions unless the domains are limited. Many compression algorithms like JPEG use fourier transforms that rely on sin and cos. Investigate Transformations of Sine and Cosine Functions Many real-world processes can be modelled with sinusoidal functions. functions for special angles using the unit circle and right triangle definitions. Evaluating Inverse. - She did an excellent job in the trigonometry unit where she was able to use the sine and cosine laws to solve problems on both her assignment and. We can prove this by using the cofunction identity and the negative angle identity for cosine. Lesson 6-6 Modeling Real-World Data with Sinusoidal Functions 389 Graphing Calculator Tip For keystroke instruction on how to find sine regression statistics, see page A25. Amplitude and Period for Sine and Cosine Functions Worksheet Author: marthasweeney Last modified by: Leake, Sherry LHS-STAFF Created Date: 6/13/2016 5:56:00 PM. The sine function \u201cstarts\u201d at the midline, but the leading negative causes it to reflect across the midline. Course Objectives: * Students will be able to develop and understand transferable pre-calculus skills necessary for success in Precalculus and beyond * Students will be able to identify, evaluate, perform operations on, and find the domain, range, and inverse of functions * Students will be able to draw common graphs along with transformations and reflections * Students will be able to create. Use sinusoidal functions to solve problems. The heat equation is used to model how things get hot (electronics, spacecraft, ovens, etc). Find the amplitude, period, and phase shift of a sine or cosine functions from its equation. Sine and cosine functions can be used to model many real-life scenarios \u2013radio waves, tides, musical tones, electrical currents. Frequency is the number of cycles in a given. Analytic Trigonometry. Sine and cosine waves can be applied to all sinusoid problems of the real-world. Being a cofunction, means that complementary input angles leads to the same output , as shown in the following example: Problem 2. Students will investigate and use the graphs of the six trigonometric functions. 1 RTP Scenarios and Terminology 12. 2 RTP Packet Format 12. Understanding Basic Sine & Cosine Graphs. c) sketch the graph of the function by using transformations for at least a two-period interval; and d) investigate the effect of changing the parameters in a trigonometric function on the graph of the function. When teaching students the value of sinusoidal models and graphs, such as sine and cosine functions, students often feel like they must memorize formulas and apply them with little recollection or understanding. Graph y = sin x. Using f (x) = cos x as a guide, graph the function h (x) = __1 cos 2 3 x. In particular: The slope of a graph is related to how fast the quantity being graphed is changing. Graphing Sine and Cosine Functions. The Sine, Cosine and Tangent functions are often applied to real world scenarios. C8 demonstrate an understanding of real-world relationships by translating between graphs, tables, and written descriptions. Third, this is a sine graph with a leading negative. Hence, we start this lesson by recalling these functions and their corresponding graphs. functions, and graph them, with and without the use of digital technologies (ACMMG274) prove that the. Trigonometry can be used to roof a house, to make the roof inclined ( in the case of single individual bungalows) and the height of the roof in buildings etc. Figure 4 The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. Graphs of the Trigonometric Functions; 1. The Code of Federal Regulations is a codification of the general and permanent rules published in the Federal Register by the Executive departments and agencies of the Federal Government. Write a cosine function that describes the height of the nail above ground as a function of angular distance. Trigonometric functions (Level C) Radian measure (including what are radians, converting from degrees to radians, converting from radians to degrees, using radians measure in real world applications) Graphs of sine, cosine, and tangent functions. The sine function. A single note can be modeled on a sine curve, and a chord can be modeled with multiple sine curves used in conjunction with one another. Special triangles can be used to geometrically determine the values of the six trigonometric functions at ,, 6 4 3 , and their integral multiples. Duration: 2 Day(s) Learning Targets. Applications of Sine and Cosine Graphs Learning Task: Trigonometry functions are often used to model periodic data. Key Questions. SheLovesMath. Graphing Sine and Cosine Functions. Chapter Five \u2013 Trigonometric Functions What to Expect: A new formula designed for sin and cos graphs, details about frequencies, new vocabulary, explaining periodic trends, and the like. Notes Chapter 6 Section 6 ( Modeling Real-World Data with Sinusoidal Functions ) -pg. Section 3-5 : Graphing Functions. Trigonometric Functions By the end of this course, students will: - solve problems involving trigonometry in acute triangles using the sine law and the cosine law, including problems arising from real-world applications; - demonstrate an understanding of periodic relationships and the sine function, and make connections between the identify and. The cosine of an angle is the ratio of x r {\\displaystyle {\\tfrac {x}{r}}} , so in the unit circle, the cosine is the x -coordinate of the point of rotation. - She did an excellent job in the trigonometry unit where she was able to use the sine and cosine laws to solve problems on both her assignment and. C2 create and analyse scatter plots of periodic data. Press the button marked \"sin\" to change to the cosine. 2-use technology, graphs, and tables to compare sine graphs. 3 Finding Maximum and Minimum Values and Zeros of Sine and Cosine 2. GCSE Maths Pythagoras and Trigonometry Level 8-9. These functions are most conveniently defined in terms of the exponential function, with sinh z = 1 / 2 (e z \u2212 e \u2212z) and cosh z = 1 / 2 (e z + e. 391-392 #'s 7-16 all I will be able to: Model real-world data using sine and cosine functions. 8 Graphs, Inverses, and Applications of Trigonometric Functions. Find the amplitude and period of the function. Applications of Exponential Functions The best thing about exponential functions is that they are so useful in real world situations. If the set of input values is a \ufb01nite set then the models are known as discrete models. When teaching students the value of sinusoidal models and graphs, such as sine and cosine functions, students often feel like they must memorize formulas and apply them with little recollection or understanding. 7 The student will identify the domain and range of the inverse trigonometric functions and recognize the graphs of these functions. So, temperature as a function of days. A real life example of the sine function could be a ferris wheel. Sample records for nino-southern oscillation eventsnino-southern oscillation events \u00ab. To be able to solve real-world problems using the Law of Sines and the Law of Cosines This tutorial reviews two real-world problems, one using the Law of Sines and one using the Law of Cosines. and \"use sine and cosine functions to model real-life data,\" iii. Find the amplitude, period, and phase shift of a sine or cosine functions from its equation. sine, cosine, tangent for \u03c0/3, \u03c0/4 and \u03c0/6, and use the unit circle to express the values of sine, cosine, and tangent for \u03c0\u2013x, \u03c0+x, and 2\u03c0\u2013x in terms of their values for x, where x is any real number. Write the sum and difference identities for sine, cosine, and tangent. Chapter 4 Sections 4. This means that bodies. On the ground, from the object you measure off a know distant in a straight line from the base. 2) Watch the video and pay attention to the key ideas listed. If you take a rope, fix the two ends, and let it hang under the force of gravity, it will naturally form a hyperbolic cosine curve. The sine and cosine of an angle are both circular functions, and they are the fundamental circular functions. Water Depth Word Problem Modeled with Cosine Sine Function. 5/175* share an equivalent sine. Namely, if we look at the graph 'y' equals hyperbolic cosine 'x', observe that whereas the function is single valued, it is not one-to-one. Find patterns in given array of numbers. The approach used in Example 3. 2 Define sine and cosine as functions of the radian measure of an angle in terms of the x- and y-coordinates of the point on the unit circle. Find the frequency in hertz for this sound wave. A single note can be modeled on a sine curve, and a chord can be modeled with multiple sine curves used in conjunction with one another. For large arguments we would expect these functions to be a quarter period out of phase, just like cosine and sine, since asymptotically J 1 is proportional to cos(z \u2013 3\u03c0/4) / \u221az and Y 1 is proportional to sin(z \u2013 3\u03c0/4) / \u221az. Frequency is the number of cycles in a given unit of time. Sineing on to the job Since we know that a triangle has 180 degrees, we can subtract 56 degrees and 91 degrees from it to find our missing angle Using the law of sines we can then set up this equation sin 91 degrees/ xft = sin 33/6ft After crossmultipying and then dividing to. Now we must stretch out the amplitude of the sine graph. Les't compare with. So, temperature as a function of days. This is an example of how a Graph of a Cosine Function looks like:. Animated gifs showing relationships between functions and geometry eg trigonometric functions How the unit circle gives us the sine wave Teaching math with the Drum Loop Real sinusoid on a timebase, formed by a linear increment of complex argument in time. Therefore, the period of this graph is 2\u03c0/(1/3) which is 6\u03c0. To find the period of a sine graph, take the B value, which is 2, and divide 2\u03c0 by it. Description: To be able to solve real-world problems using the Law of Sines and the Law of Cosines. Land surveying makes an extensive use of the sine and cosine law. The basic sine and cosine graphs C. Seymour Public Schools Curriculum Subject or course name 5 graph the basic sine and cosine functions. Do not forget this. state the properties for the Sine and Cosine graph identify the values of a, k, c and d and apply it to a sinusoidal function to graph the resulting transformation Lesson:. Graph Trigonometric Functions (1), cosine function with solution. ' Below is a listing of several popular input and output values for the three main trigonometry functions. 1) Graph sine functions (PC-N. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. 1: Graphs of the Sine and Cosine Functions. Water Depth Word Problem Modeled with Cosine Sine Function. One method to write a sine or cosine function that models a sinusoid is to fi nd the values of a, b, h, and k for y = a sin b(x \u2212 h) + k or y = a cos b(x \u2212 h) + k where \u2223 a \u2223 is the amplitude, \u2014 2\u03c0 is the period. Master PHP 4 \u2014 the open source Web scripting breakthrough! Contains expert coverage of syntax, functions, design, and debugging! Leverage the amazing performance of the new Zend engine! 650+ real-world code examples! CD-ROM includes source code, plus everything you\u2019ll need to run PHP 4 implementations on Windows and UNIX!. You can select the values of a and b within certain ranges. Sine wiggles in one dimension. We completed three cosine graphs together on the 13-5 vocab support page. Signal Processing. Real life word problems that involve the sine and cosine function can be used to keep the students engaged in the topic. ) Circles are an example of two sine waves. Find the amplitude and period of the function. Transformations of Trigonometric Functions. Graphs of the sine and cosine functions; and Periodic Behavior. Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. The student will graph sine or cosine graph affected by horizontal and vertical translations. Like What? [Filename: Notes 6. The Code of Federal Regulations is a codification of the general and permanent rules published in the Federal Register by the Executive departments and agencies of the Federal Government. It is given by parameter #a# in function #y = asinb(x - c) + d or y = acosb(x - c) + d# \u2022The period of a graph is the distance on the x axis before the function repeats itself. \u2022 Use the cosine law to determine the interior angles of an acute triangle. The Sine Ratio Passy's World of Mathematics. Physical phenomenon such as tides, temperatures and amount of sunlight are all things that repeat themselves, and so are easily modeled by sine and cosine functions (collectively, they are called \"sinusoidal functions\"). Find an equation for a sine function that has amplitude of 4 a period of \u03c0. To understand how the absolute value can be applied to the real world, we'll look at two topics: What is considered normal with regard to the temperature of the human body and Fuel economy of two vehicles: a Honda Odyssey and a Nissan Altima. 8 L5 Trig Applications Part 1 - solve problems that arise from real world applications involving periodic phenomena D3. Lynn Weiss. Sketch the graphs of basic sine and cosine functions; use amplitude and period to help sketch the graphs of sine and cosine functions. Given an expression for the time rate of change of a quantity, find the initial rate. ppt), PDF File (. The maximum value appears to be 72 and the minimum value is 11. Goal: Students will use technology to discover how to graph trigonometric functions. Apr 15, 2018 - Discovering the graphs of Sine & Cosine using the measurements of the Unit Circle (Activity from Rice APSI) Stay safe and healthy. PreCalculus Honors - Curriculum Pacing Guide - 2017 - 2018 First Half of Semester Anderson School District Five Page 3 2017-2018 Unit 2 - Application of Trigonometry to Triangles (7 days) PC. solve problems involving trigonometry in acute triangles using the sine law and the cosine law, including problems arising from real-world applications; demonstrate an understanding of periodic relationships and the sine function, and make connections between the numeric, graphical, and algebraic representations of sine functions;. A sine wave is what you get when you plot the sine function on a graph. MCF 3M - Functions and Applications (Mixed) Spring 2014 Page history last edited by Peter Kim 5 years, 9 months ago This course introduces basic features of the function by extending students' experiences with quadratic relations. The product of two cosine functions is a sum of two other cosine functions. Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. Free practice questions for Precalculus - Graph the Sine or Cosine Function. Real world uses of hyperbolic trigonometric functions. Specific Expectations. Perform a sine regression using the Ti83/84 Activity #2 - Steamboat problem From given information: 1. Amplitude is first, then sine or cosine, then B, which you get by doing 2pi/10 (period), which reduces to pi/5, then it is the parenthesis with the transformation, and then last but not least is the sinusodial axis. 1) Find properties of sine functions (PC-N. Evaluating Inverse Trigonometric. Graphing Sine and Cosine Functions. \\begingroup Basically, the answer is: yes, there are many other periodic functions, and the reason you typically see harmonics (like \\sin,e^{i\\omega t}) used is because in most simple applications of interest which are easily understandable, either the behavior itself is harmonic, or the behavior is most easily understood in terms of harmonics. First, you must understand that each of these functions has its own graph. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. First of all, the graph is no longer a sine curve, but there's definitely a pattern to it. It exposes the student to the basic concepts of algebra and trigonometry with an emphasis in practical applications. But just saying \"the sine function\" doesn't tell you anything about the probability measure in the domain, which is what gets you probabilistic information, including joint distributions. the sine or cosine function and use the equation as a mathematical model to make predictions and interpretations about the real world. Using trigonometric functions to model climate. In ICSA, an elitism strategy is used to select the global solution, and a new updating mechanism for the new solution was proposed. You cannot graph an exponential functions without finding its horizontal asymptote. In addition, how do I know if this the graph of sine or cosine?. functions; increasing/decreasing functions; finding intercepts with axes. real world applications of sinusoids Sinusoids are applicable to the real world. To graph the sine function, we mark the angle along the horizontal x axis, and for each angle, we put the sine of that angle on the vertical y-axis. y-2\u03c0 -\u03c0 \u03c0 2\u03c0 0. The rise and fall of tides can have great impact on the communities and ecosystems that depend upon them. Every angle shares the same sine with another angle (and -sine with two others). \u2022 Find the amplitude and period of variation of the sine, cosine, and tangent functions. Sine and cosine functions are periodic functions. 4 COMPLEX NUMBERS Add and subtract complex numbers Multiply and divide complex numbers Simplify powers of i 2. The amplitude is 2. These graphs act as a reference every time you use a trigonometric function. Also for every n \u03c0/2 radians (where \"n\" is an interger) the cosine function passes through y=0. 21 from both sides: Divide both sides by 1. Outcome 5: Set up, solve, and graph equations from problems that require use of trigonometric functions, tangent, sine, and cosine and the Pythagorean Theorem. 3 make connections between the sine ratio and the sine function and between the cosine ratio and the cosine function by graphing the relationship between angles from 0\u00ba to 360\u00ba and the corresponding sine ratios or cosine ratios, defining this relationship as the function f(x) =sinx or f(x) =cosx, and explaining why the relationship is a function. The sine and cosine of an angle are both circular functions, and they are the fundamental circular functions. Full Range Fourier Series - various forms of the Fourier Series 3. Focus on creating a graph and formula from a story (starting from time 6:55) You Do Now: Warmup - Sketching a Cosine Wave. Learn how to construct trigonometric functions from their graphs or other features. Physical phenomenon such as tides, temperatures and amount of sunlight are all things that repeat themselves, and so are easily modeled by sine and cosine functions (collectively, they are called \"sinusoidal functions\"). 2-use technology, graphs, and tables to compare sine graphs. The property represented here is based on the right triangle and the two acute or complementary angles in a right triangle. The Code is divided into 50 titles which represent broad areas subject to Federal regulation. Sinusoidal Modeling - Real World Application Project. Then we looked at identifying the amplitude, frequency and period from a given cosine (or sine) function. Time-saving lesson video on Sine and Cosine Functions with clear explanations and tons of step-by-step examples. One way to distinguish sounds is to measure frequency. sinA sin B sin(AB) cosAcosB cos(AB) 59. 9) Graph sine and cosine functions (PC-N. 534), where x is the day of the year starting with January 1 as Day 1. Graph functions, plot data, evaluate equations, explore transformations, and much more - for free! Start Graphing Four Function and Scientific Check out the newest additions to the Desmos calculator family. Recall that the sine and cosine functions relate real number values to the x\u2013 and y-coordinates of a point on the unit circle. Day 7 HW F 15-Dec Graphing Cosine Functions NC. Determine the values for angles between 0\u00b0 and 360\u00b0. Chapter Test. Evaluate and graph logistic growth functions. Find the amplitude and period of the function. Applications of Trigonometric Graphs; 6. Khan Academy is a 501(c)(3) nonprofit organization. Graph one period of s(x) = \u2013cos(3x) The \"minus\" sign tells me that the graph is upside down. There are many uses of sin,cos,tan in real life. Trigonometric Functions \u2022 Understand some real-world situations that demonstrate periodic behavior. Write a cosine function that describes the height of the nail above ground as a function of angular distance. Trigonometry Graphing Trigonometric Functions Applications of Radian Measure. Discovery of Euler's Equation. Example 15 6. The new 3rd edition of Precalculus: Algebra and Trigonometry prepares students for further study in mathematics, the sciences, engineering, the social sciences, and other disciplines. By correctly labeling the coordinates of points A, B, and C, you will get the graph of the function given. 'transform') each of the three trigonometric functions discussed thus far: sine, cosine, and tangent. 25 scaffolded questions on equation graph involving amplitude and periodplus model problems explained step by step. Amplitude and Period of Sine and Cosine Functions BOATING A signal buoy between the coast of Hilton Head Island, South Carolina, and Savannah, Georgia, bobs up and down in a minor squall. Graphs of y = a sin x and y = a cos x 2. Yep, sine and cosine are practically twins. In maths, you have real life applications on any thing that you study. Lesson 7: Applications with Acute Triangles \u2022 Identify when to apply the sine and cosine laws given incomplete information about the \u2022 Solve a multistep problem that involves o two or more applications of the sine or cosine laws,. Sine Law and Cosine Law Find each measurement indicated. Free trigonometry worksheets, in PDF format, with solutions to download. Khan Academy is a 501(c)(3) nonprofit organization. stock market values can be formed into a sine, cosine graph (given a constant average) 5. Precalculus. Example: if your cutting down a tree and need to know where it will. I'm not exactly sure what level of mathematics you've studied so far, but sine and cosine are used heavily in math (calculus and functions) and physics (kinematics and dynamics), as well as having some real world applications. 2 The Inverse Trigonometric Functions. The trigonometric functions sine, cosine and tangent calculate the ratio of two sides in a right triangle when given an angle in that triangle. Although you. 4 The Real-Time Transport Protocol 12. the graph of a sine function. C2 create and analyse scatter plots of periodic data. Of course, this was to be expected, but this isn't the real problem; the real problem is that if your function has some crucial points it must pass through (which is certainly true for trigonometry functions), the truncation will move the curve away from those points. Standard 9: Find values of trigonometric functions using the unit circle. The unit circle and values of sine and cosine on the unit circle B. The fruit fly optimization algorithm (FOA) is a well-regarded algorithm for searching the global optimal solution by simulating the foraging behavior \u2026. which means that theta can be any angle in degrees or radians \u2014 any real number. 6 u00ad Modeling Real World Data with Sinusoidal Functions Section 6. The unit circle can be used to express the values of sine, cosine, and tangent for \u03c0 \u2013 x,. Students graph transformations of trigonometric functions (sine and cosine), including af(x), f(x) + d, f(x \u2013 c), and f(bx) for specific values of a, b, c, and d, in mathematical and real-world problem situations. Graphing Secant & Cosecant. Investigate Transformations of Sine and Cosine Functions Many real-world processes can be modelled with sinusoidal functions. The identities that arise from the triangle are called the cofunctionidentities. Sine and Cosine can be used in everyday life in various ways. Topic:Real world application of sine graphs. Press the button marked \"sin\" to change to the cosine. ratios, ie. Intro Tangent & Cotangent Graphs. To graph the sine function, we mark the angle along the horizontal x axis, and for each angle, we put the sine of that angle on the vertical y-axis. McGraw-Hill Education Trigonometry Review and Workbook, 1st Edition by William Clark and Sandra Luna McCune (9781260128925) Preview the textbook, purchase or get a FREE instructor-only desk copy. 3 make connections between the sine ratio and the sine function and between the cosine ratio and the cosine function by graphing the relationship between angles from 0\u00ba to 360\u00ba and the corresponding sine ratios or cosine ratios, defining this relationship as the function f(x) =sinx or f(x) =cosx, and explaining why the relationship is a function. If the set of input values is a \ufb01nite set then the models are known as discrete models. org (pre-calculus), and Khan Academy (SAT. This builds into learning about graphing and interpreting logarithmic functions and models. Les't compare with. These graphs act as a reference every time you use a trigonometric function. 2) A car's tire has a diameter of 32 inches. The caternary curve (a dangling string/chain) is really just cosh \u2013 crasic Oct 30 '10 at 23:48. state the properties for the Sine and Cosine graph identify the values of a, k, c and d and apply it to a sinusoidal function to graph the resulting transformation Lesson:. 14-1 Graphs of Sine and Cosine Example 3: Sound Application Use a sine function to graph a sound wave with a period of 0. pdf), Text File (. A single note can be modeled on a sine curve, and a chord can be modeled with multiple sine curves used in conjunction with one another. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. In the 36 intensively illustrated lectures of Mathematics Describing the Real World: Precalculus and Trigonometry, he takes you through all the major topics of a typical precalculus course taught in high school or college. The graphs of the sine and cosine functions illustrate a property that exists for several pairings of the different trig functions. Then you will learn about modeling trigonometric functions by graphing the sine and cosine functions. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because [latex]sin(\u2212x)=\u2212sinx$. Sample records for nino-southern oscillation eventsnino-southern oscillation events \u00ab. The inverses of sine, cosine and tangent functions are not functions unless the domains are limited. Chapter Five \u2013 Trigonometric Functions What to Expect: A new formula designed for sin and cos graphs, details about frequencies, new vocabulary, explaining periodic trends, and the like. 2]Solve right and oblique triangles including application problems. Outcome 6: Set up and solve exponential and logarithmic equations; then identify and sketch graphs of the functions. Precalculus students are required to understand how sine and cosine functions model the real world. A lot of waves actually follow a sine graph, so we can prove that sinusoidal motion is a real thing in nature. We completed three cosine graphs together on the 13-5 vocab support page. f(x)= Calculus: Early Transcendental Functions Use Exercise 25 to find the moment of inertia of a circular disk of radius a with constant density about a. 0333 170 -170 t e 0. Have them graph these points and note whether or not they are on the same graph. Use a horizontal scale where one unit represents 0. With the sine law. Students will find equations for transformed sine curves and will graph transformed sine functions. Sketch translations of graphs of sine and cosine functions; use sine and cosine functions to model real-life data. Other circular functions can all be derived from the sine and cosine of an angle. 002 s and an amplitude of 3 cm. Examples: \u2022 Square wave function \u2022 Saw tooth functions Split Functions Much of the behaviour of current, charge and voltage in an AC circuit can be described. The inverses of sine, cosine and tangent functions are not functions unless the domains are limited. You will be observing the graph of r = a + bsin() or r = a + bcos(), where a and b are non-zero real numbers. The period of sine and cosine functions is 2pi/B. graph the sine and cosine functions with varied amplitudes and periods. Now we must stretch out the amplitude of the sine graph. Sketch the graphs of the sine and cosine functions and state their domain and range. Applications of Sine and Cosine Graphs Learning Task: Trigonometry functions are often used to model periodic data. The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. Graphs the 6 Trigonometric Functions. Rational functions: asymptotes and excluded values (PC-E. It is given by parameter #a# in function #y = asinb(x - c) + d or y = acosb(x - c) + d# \u2022The period of a graph is the distance on the x axis before the function repeats itself. But just as you could make the basic quadratic, y = x 2, more complicated, such as y = -(x + 5) 2 - 3, so also trig graphs can be made more complicated. Trigonometry plays a major role in musical theory and production Real world examples of cosine functions. We typically use degree measures when measuring angles, however we can use radian angle measure as an alternate way of measuring angles in advanced math courses. KEY STANDARDS ADDRESSED: MA3A3. Many compression algorithms like JPEG use fourier transforms that rely on sin and cos. Trigonometric functions can be used to model real world data that is periodic in nature. Since the sine function y = sin t begins at the origin (when t = 0), sine is the more convenient of the two for this purpose. First, find two easily solvable angles that add up to 75. 1 The Inverse Sine, Cosine, and Tangent Functions. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. ' Or: sin(30 \u00b0) = 0. Domain and Range of Sine and Cosine The domain of sine and cosine is all real numbers, The range of sine and cosine is the interval [-1, 1] x or (, ) Both these graphs are considered sinusoidal graphs. Find Period and Amplitude 2. Not only is this a real-world visual representation of both graphs, but it also demonstrates that a cosine graph is simply a sine graph at a displacement of 90\u00b0. The table of values for the functions. Find the frequency in hertz for this sound wave. The graph of the sine function is called the sine curve. The property represented here is based on the right triangle and the two acute or complementary angles in a right triangle. 7 sketch graphs of y = af (k(x - d)) + c by applying one or more transformations to the graphs of f(x) =sinx and f(x) =cosx, and state the domain and range of the transformed functions 2. It is the reciprocal of the. Hyperbolic functions, also called hyperbolic trigonometric functions, the hyperbolic sine of z (written sinh z); the hyperbolic cosine of z (cosh z); the hyperbolic tangent of z (tanh z); and the hyperbolic cosecant, secant, and cotangent of z. Each of the six trig functions is equal to its co-function evaluated at the complementary angle. However, if we stretch the sine graph and change the amplitude, it just might work. In problems 12 & 13, the graphs of the sine and cosine functions are waveforms like the figure below. Since the multiplier out front is an \"understood\" \u20131, the amplitude is unchanged. If we change the number of cycles the wave completes every second -- in other words, if we change the period of the sine wave -- then we change the sound. Draw three right triangles and label the angle and two sides that apply to the sine, cosine and tangent functions respectively. Four Function Scientific. The hyperbolic functions take a real argument called a hyperbolic angle. Discovery of Euler's Equation. Cumulative Review. Lesson 9-11: Graphing Trigonometric Functions and Applications in Real World Contexts Learning Goals #8, 9, 10: How do I use the critical values of the Sine and Cosine curve to assist in in graphing Sine and Cosine functions? How do I contextualize Trig Functions in real life situations? Warm-Up! WITH A PARTNER. Water Depth Word Problem Modeled with Cosine Sine Function. One example of applying tangent functions to solve a real world problem is:- Find the gradient and the actual length of a path represented as x cm (a known measurable quantity) in the 1 : n (a known given quantity) scaled contour map. Sound waves travel in a repeating wave pattern, which can be represented graphically by sine and cosine functions. The SOH means that the sine of an angle equals the side opposite the angle over the triangle's hypotenuse. One way to distinguish sounds is to measure frequency. Animated gifs showing relationships between functions and geometry eg trigonometric functions How the unit circle gives us the sine wave Teaching math with the Drum Loop Real sinusoid on a timebase, formed by a linear increment of complex argument in time. Let\u2019s look at a few examples of real-world situations that can best be modeled using trigonometric functions. This is an example of how a Graph of a Cosine Function looks like:. Use sinusoidal functions to solve problems. which means that theta can be any angle in degrees or radians \u2014 any real number. To understand how the absolute value can be applied to the real world, we\u2019ll look at two topics: What is considered normal with regard to the temperature of the human body and; Fuel economy of two vehicles: a Honda Odyssey and a Nissan Altima. Periodicity of trig functions. Trigonometric identities like finding. Bookmark File PDF Functions Modeling Change Answers Functions Modeling Change Answers Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Modeling with Functions Part 1 We model real life scenarios of sales and volume of a box with functions. 4 Solving Word Problems Involving Sine or Cosine Functions. By finding a few key points or aspects of the graph, any of the real-life problems we have today can be explained mathematically and much of the vibrations surrounding us can be better understood. active oldest votes. 3\u03c0 2 = = Period Amplitude. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. Read: 'Zero point five is the sine of thirty degrees. Sound waves travel in a repeating wave pattern, which can be represented graphically by sine and cosine functions. Academic Precalculus. functions to model real world problems A cosine graph appears. Topic:Real world application of sine graphs. Trigonometric functions can be used to solve right triangles. Overview of Fourier Series - the definition of Fourier Series and how it is an example of a trigonometric infinite series 2. Maths Applications: Graph sketching. Euler's equation (formula) shows a deep relationship between the trigonometric function and complex exponential function.", "date": "2020-08-11 18:15:23", "meta": {"domain": "rafbis.it", "url": "http://rafbis.it/qowu/graph-of-sine-and-cosine-functions-real-world-applications.html", "openwebmath_score": 0.7441229820251465, "openwebmath_perplexity": 518.5260431003412, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9926541732082084, "lm_q2_score": 0.8688267626522814, "lm_q1q2_score": 0.8624445117417647}}
{"url": "https://www.coursehero.com/file/22489520/Sequences-and-Series/", "text": "# Sequences_and_Series - 11 Sequences and Series Consider the...\n\n\u2022 42\n\nThis preview shows page 1 - 4 out of 42 pages.\n\n##### We have textbook solutions for you!\nThe document you are viewing contains questions related to this textbook.\nThe document you are viewing contains questions related to this textbook.\nChapter 8 / Exercise 14\nEssential Calculus: Early Transcendentals\nStewart\nExpert Verified\n11 Sequences and Series Consider the following sum: 1 2 + 1 4 + 1 8 + 1 16 + \u00b7 \u00b7 \u00b7 + 1 2 i + \u00b7 \u00b7 \u00b7 The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite sum? While at first it may seem di cult or impossible, we have certainly done something similar when we talked about one quantity getting \u201ccloser and closer\u201d to a fixed quantity. Here we could ask whether, as we add more and more terms, the sum gets closer and closer to some fixed value. That is, look at 1 2 = 1 2 3 4 = 1 2 + 1 4 7 8 = 1 2 + 1 4 + 1 8 15 16 = 1 2 + 1 4 + 1 8 + 1 16 and so on, and ask whether these values have a limit. It seems pretty clear that they do, namely 1. In fact, as we will see, it\u2019s not hard to show that 1 2 + 1 4 + 1 8 + 1 16 + \u00b7 \u00b7 \u00b7 + 1 2 i = 2 i 1 2 i = 1 1 2 i 255\n##### We have textbook solutions for you!\nThe document you are viewing contains questions related to this textbook.\nThe document you are viewing contains questions related to this textbook.\nChapter 8 / Exercise 14\nEssential Calculus: Early Transcendentals\nStewart\nExpert Verified\n256 Chapter 11 Sequences and Series and then lim i \u2192\u221e 1 1 2 i = 1 0 = 1 . There is one place that you have long accepted this notion of infinite sum without really thinking of it as a sum: 0 . 3333 \u00af 3 = 3 10 + 3 100 + 3 1000 + 3 10000 + \u00b7 \u00b7 \u00b7 = 1 3 , for example, or 3 . 14159 . . . = 3 + 1 10 + 4 100 + 1 1000 + 5 10000 + 9 100000 + \u00b7 \u00b7 \u00b7 = \u03c0 . Our first task, then, to investigate infinite sums, called series , is to investigate limits of sequences of numbers. That is, we o cially call i =1 1 2 i = 1 2 + 1 4 + 1 8 + 1 16 + \u00b7 \u00b7 \u00b7 + 1 2 i + \u00b7 \u00b7 \u00b7 a series, while 1 2 , 3 4 , 7 8 , 15 16 , . . ., 2 i 1 2 i , . . . is a sequence, and i =1 1 2 i = lim i \u2192\u221e 2 i 1 2 i , that is, the value of a series is the limit of a particular sequence. While the idea of a sequence of numbers, a 1 , a 2 , a 3 , . . . is straightforward, it is useful to think of a sequence as a function. We have up until now dealt with functions whose domains are the real numbers, or a subset of the real numbers, like f ( x ) = sin x . A sequence is a function with domain the natural numbers N = { 1 , 2 , 3 , . . . } or the non-negative integers, Z 0 = { 0 , 1 , 2 , 3 , . . . } . The range of the function is still allowed to be the real numbers; in symbols, we say that a sequence is a function f : N R . Sequences are written in a few di ff erent ways, all equivalent; these all mean the same thing: a 1 , a 2 , a 3 , . . . { a n } n =1 { f ( n ) } n =1 As with functions on the real numbers, we will most often encounter sequences that can be expressed by a formula. We have already seen the sequence a i = f ( i ) = 1 1 / 2 i ,\n11.1 Sequences 257 and others are easy to come by: f ( i ) = i i + 1 f ( n ) = 1 2 n f ( n ) = sin( n \u03c0 / 6) f ( i ) = ( i 1)( i + 2) 2 i Frequently these formulas will make sense if thought of either as functions with domain R or N , though occasionally one will make sense only for integer values. Faced with a sequence we are interested in the limit lim i \u2192\u221e f ( i ) = lim i \u2192\u221e a i . We already understand lim x \u2192\u221e f ( x ) when x is a real valued variable; now we simply want to restrict the \u201cinput\u201d values to be integers. No real di ff erence is required in the definition of limit, except that we specify, per- haps implicitly, that the variable is an integer. Compare this definition to definition 4.10.4.", "date": "2021-10-21 04:32:28", "meta": {"domain": "coursehero.com", "url": "https://www.coursehero.com/file/22489520/Sequences-and-Series/", "openwebmath_score": 0.8320382237434387, "openwebmath_perplexity": 230.33246336472976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9946981045019138, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8624388216842898}}
{"url": "https://math.stackexchange.com/questions/1929320/finding-the-two-planes-that-contain-a-given-line-and-form-the-same-angle-with-tw", "text": "# Finding the two planes that contain a given line and form the same angle with two other lines\n\nI am asked the following question:\n\nThere are two planes $\\pi_1$ and $\\pi_2$ such that each of them contain the line $t \\{ (-1,4,0) + \\lambda (1,2,3)$ and form the same angle with the lines $r \\{ (0,0,0) + \\beta (1,1,1)$ and $s \\{ (-1,2,3) + \\alpha (1,1,-1)$. Find the angle formed between those planes\n\nMy solution:\n\nLet's say the normal vector of $\\pi$ is $\\vec{n} = (a,b,c)$. Since the angle between the plane(s) and the lines are equal,\n\n\\begin{align*} \\measuredangle \\left( \\pi, r \\right) &= \\measuredangle \\left( \\pi, s \\right)\\\\ \\frac{\\left( a,b,c \\right) \\cdot (1,1,1)}{\\sqrt{a^2+b^2+c^2} \\ \\sqrt{3}} &= \\frac{\\left( a,b,c \\right) \\cdot (1,1,-1)}{\\sqrt{a^2+b^2+c^2} \\ \\sqrt{3}}\\\\ a+b+c &= a+b-c\\\\ c &= 0 \\end{align*}\n\nAlso, since $t$ is contained in the plane, the dot product between its vector and the normal vector of the plane is zero:\n\n\\begin{align*} (a,b,c) \\cdot (1,2,3) &= 0\\\\ a+2b+3c&=0\\\\ a &= -2b \\end{align*}\n\nAssuming that the normal vector has length one,\n\n$$a^2 + b^2 + c^2 = 1$$\n\nSo we have three equations with three variables and the solution for those is\n\n$$a = \\mp \\frac{2}{\\sqrt{5}} \\quad a = \\pm \\frac{1}{\\sqrt{5}} \\quad c = 0$$\n\nso the normal vectors of the two planes found are\n\n$$\\left( \\frac{2}{\\sqrt{5}} , - \\frac{1}{\\sqrt{5}} , 0 \\right) \\quad \\left( - \\frac{2}{\\sqrt{5}} , \\frac{1}{\\sqrt{5}} , 0 \\right)$$\n\nand the angle between them is\n\n$$\\theta = \\arccos \\left(\\frac{\\left \\vert - \\frac{2}{\\sqrt{5}} \\frac{2}{\\sqrt{5}} - \\frac{1}{\\sqrt{5}} \\frac{1}{\\sqrt{5}} + 0 \\right \\vert}{1} \\right) = \\arccos(1) = 0^\\circ$$\n\nMy answer does not agree with the textbook's solution.\n\nTextbook's solution:\n\n$$\\theta = \\arccos \\left( \\frac{9}{\\sqrt{95}} \\right)$$\n\nDid I make a mistake somewhere?\n\nThank you.\n\n$$|(a,b,c) \\cdot (1,1,1)| = |(a,b,c) \\cdot (1,1,-1)| \\\\ |a+b+c| = |a+b-c|$$ yields\n$$a+b+c = a+b-c \\quad \\text{ and } \\quad a+b+c = -a-b+c$$", "date": "2020-07-14 20:50:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1929320/finding-the-two-planes-that-contain-a-given-line-and-form-the-same-angle-with-tw", "openwebmath_score": 1.0000066757202148, "openwebmath_perplexity": 270.0577996297103, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232904845687, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8624046346184061}}
{"url": "https://gmatclub.com/forum/a-bag-has-only-r-red-marbles-and-b-blue-marbles-if-five-red-marbles-271572.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 14 Dec 2018, 11:49\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in December\nPrevNext\nSuMoTuWeThFrSa\n2526272829301\n2345678\n9101112131415\n16171819202122\n23242526272829\n303112345\nOpen Detailed Calendar\n\u2022 ### Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student\n\nDecember 14, 2018\n\nDecember 14, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nCarolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.\n\u2022 ### Free GMAT Strategy Webinar\n\nDecember 15, 2018\n\nDecember 15, 2018\n\n07:00 AM PST\n\n09:00 AM PST\n\nAiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.\n\n# A bag has only r red marbles and b blue marbles. If five red marbles\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 51215\nA bag has only r red marbles and b blue marbles. If five red marbles\u00a0 [#permalink]\n\n### Show Tags\n\n25 Jul 2018, 23:16\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n87% (00:39) correct 13% (01:00) wrong based on 54 sessions\n\n### HideShow timer Statistics\n\nA bag has only r red marbles and b blue marbles. If five red marbles and one blue marble are added to the bag, and if one marble is then selected at random from the bag, then the probability that the marble selected will be blue is represented by\n\nA. b/r\n\nB. b/(b + r)\n\nC. (b + 1)/(r + 5)\n\nD. (b + 1)/(r + b + 5)\n\nE. (b + 1)/(r + b + 6)\n\n_________________\nMBA Section Director\nAffiliations: GMATClub\nJoined: 22 May 2017\nPosts: 1468\nConcentration: Nonprofit\nGPA: 4\nWE: Engineering (Computer Software)\nRe: A bag has only r red marbles and b blue marbles. If five red marbles\u00a0 [#permalink]\n\n### Show Tags\n\n25 Jul 2018, 23:20\nNumber of red marbles = r\n\nNumber of blue marbles = b\n\nfive red marbles and one blue marble are added to the bag\n\nNumber of red marbles = r + 5\n\nNumber of blue marbles = b + 1\n\nTotal number of marbles = r + 5 + b + 1 = r + b + 6\n\nProbability of selecting a blue marbles = $$\\frac{b + 1}{r + b + 6}$$\n\nHence option E\n_________________\nDirector\nJoined: 31 Oct 2013\nPosts: 881\nConcentration: Accounting, Finance\nGPA: 3.68\nWE: Analyst (Accounting)\nRe: A bag has only r red marbles and b blue marbles. If five red marbles\u00a0 [#permalink]\n\n### Show Tags\n\n26 Jul 2018, 01:02\nBunuel wrote:\nA bag has only r red marbles and b blue marbles. If five red marbles and one blue marble are added to the bag, and if one marble is then selected at random from the bag, then the probability that the marble selected will be blue is represented by\n\nA. b/r\n\nB. b/(b + r)\n\nC. (b + 1)/(r + 5)\n\nD. (b + 1)/(r + b + 5)\n\nE. (b + 1)/(r + b + 6)\n\nRed marbles = r\n\nBlue marbles = b\n\nChanging Situation :\n\nRed = r + 5\n\nBlue = b + 1\n\nTotal = r + s + 6\n\nwe are asked to pick up a blue marbles = $$\\frac{( b + 1)}{r + s + 6}$$\n\nThus the best answer is E.\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2830\nRe: A bag has only r red marbles and b blue marbles. If five red marbles\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jul 2018, 10:51\nBunuel wrote:\nA bag has only r red marbles and b blue marbles. If five red marbles and one blue marble are added to the bag, and if one marble is then selected at random from the bag, then the probability that the marble selected will be blue is represented by\n\nA. b/r\n\nB. b/(b + r)\n\nC. (b + 1)/(r + 5)\n\nD. (b + 1)/(r + b + 5)\n\nE. (b + 1)/(r + b + 6)\n\nAfter 5 red marbles and 1 blue marble are added to the bag, the bag has b + 1 blue marbles and r + b + 5 + 1 = r + b + 6 total marbles. So the probability of selecting a blue marble is (b + 1)/(b + r + 6).\n\n_________________\n\nJeffery Miller\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: A bag has only r red marbles and b blue marbles. If five red marbles &nbs [#permalink] 30 Jul 2018, 10:51\nDisplay posts from previous: Sort by", "date": "2018-12-14 19:49:05", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-bag-has-only-r-red-marbles-and-b-blue-marbles-if-five-red-marbles-271572.html", "openwebmath_score": 0.7224181890487671, "openwebmath_perplexity": 4168.813910954854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9828232889752296, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8624046191232714}}
{"url": "https://johncarlosbaez.wordpress.com/2016/03/26/probability-puzzles-part-3/", "text": "## Probability Puzzles (Part\u00a03)\n\nHere\u2019s a puzzle based on something interesting that I learned from Greg Egan. I\u2019ve dramatized it a bit.\n\nTraditional Tom and Liberal Lisa are dating. They discuss their plans for having children:\n\nTom: I plan to keep having kids until I get two sons in a row.\n\nLisa: What?! That\u2019s absurd. Why?\n\nTom: I want two to run my store when I get old.\n\nLisa: Even ignoring your insulting assumption that only boys can manage your shop, why in the world do you need two in a row?\n\nTom: From my own childhood, I\u2019ve learned there\u2019s a special bond between sons who are next to each other in age. They play together, they grow up together\u2026 they can run my shop together.\n\nLisa: Hmm. Well, then maybe I should have children until I have a girl followed directly by a boy!\n\nTom: What?!\n\nLisa: Well, I\u2019ve observed that something special happens when a boy has an older sister, with no intervening siblings. They play together, they grow up together\u2026 and maybe he learns not to be such a sexist pig!\n\nThey decide they are incompatible, so they split up and each one separately tries to find a mate who will go along with their reproductive plan.\n\nNow for some puzzles. In these puzzles, assume that each time someone has a child, they have a 50% chance of having either a daughter or a son. Also assume each event is independent: that is, the gender of any children so far has no effect on that of later ones. Also ignore twins and other tricky issues.\n\nPuzzle 1. If Tom carries out his plan of having children until he has two consecutive sons, and then stops, what is the expected number of children he will have?\n\nPuzzle 2. If Lisa carries out her plan of having children until she has a daughter followed directly by a son, and then stops, what is the expected number of children she will have?\n\nPuzzle 3: Which is greater, Tom\u2019s expected number of children or Lisa\u2019s? Or are they equal?\n\nFor maximum benefit, try to answer Puzzle 3 before doing the calculations required to answer Puzzles 1 or 2.\n\n### 64 Responses to Probability Puzzles (Part\u00a03)\n\n1. It\u2019s late here right now so for now only preliminary quick guesses.\nIf you wanna have a more formal go, maybe don\u2019t look at this until you want a hint or a distraction, depending on whether my guesses are right.\n\nPuzzle 3 guess:\nSince any particular two-birth-sequence is equally likely (1/4 each), I\u2019d guess that both scenarios are equally likely.\n\nPuzzle 1:\nPossible sequences:\nSS (1/4)\nDSS (1/8)\nDDSS (1/16)\nSDSS (1/16)\nDDDSS (1/32)\nDSDSS (1/32)\nSDDSS (1/32)\nDDDDSS (1/64)\nDDSDSS (1/64)\nDSDDSS (1/64)\nSDDDSS (1/64)\nSDSDSS (1/64)\nDDDDDSS (1/128)\nDDDSDSS (1/128)\nDDSDDSS (1/128)\nDSDDDSS (1/128)\nSDDDDSS (1/128)\nDSDSDSS (1/128)\nSDDSDSS (1/128)\nSDSDDSS (1/128)\nThis looks suspiciously like $\\sum_{n=1}^\\infty \\frac{\\text{fib}\\left(n\\right)}{2^{n+1}}$\nThere is probably some nice combinatorial argument to make to prove this.\nThis sum \u2013 I didn\u2019t prove but just quickly looked up \u2013 sums up to 1, so that\u2019s good.\nThe expectation value version of that sum just adds an n:\n$\\sum_{n=1}^\\infty n \\frac{\\text{fib}\\left(n\\right)}{2^{n+1}}$\nAnd that appears to sum to 5.\n\nPuzzle 2:\nPossible sequences:\nDS (1/4)\nSDS (1/8)\nDDS (1/8)\nSSDS (1/16)\nSDDS (1/16)\nDDDS (1/16)\nSSSDS (1/32)\nSSDDS (1/32)\nSDDDS (1/32)\nDDDDS (1/32)\nSSSSDS (1/64)\nSSSDDS (1/64)\nSSDDDS (1/64)\nSDDDDS (1/64)\nDDDDDS (1/64)\n\nThe series appears to be\n\n$\\displaystyle{ \\sum_{n=1}^\\infty \\frac{n}{2^{n+1}} }$\n\nwhich also sums to 1.\n\nAnd the expectation value\n\n$\\displaystyle{ \\sum_{n=1}^\\infty \\frac{n^2}{2^{n+1}} }$\n\ngoes to 3. So Lisa will actually most likely have fewer kids than Tom. Therefore my above prediction was wrong.\n\nFurthermore, if I did this right, Lisa can expect her value to hover around 3 by 2, whereas for Tom that uncertainty sits around 4.69.\nSo in a fairly pessimistic (assuming you want to get there asap) outcome would be 5 trials for Lisa but 10 for Tom.\n\n(I used that $\\sigma^2 = \\langle-^2\\rangle$ but I\u2019m tired and so possibly error-prone.)\n\n\u2022 Michael Nelson says:\n\nHere\u2019s how fibonacci shows up:\n\nYou can count the number of ways of Tom winning via a string of n turns by breaking it up into two cases:\n\n\u2026\u2026.DSS\n\u2026.DSDSS\n\nThe first case you counted already in the previous string of length n-1. And the second case you counted already from the string before that of length n-2.\n\n\u2022 This is how I also ended up solving it, getting the Fibonnaci sequence immediately by using a construction approach to compute the number of binary strings (1 == boy, 0 == girl) that had the constraint of ending in \u201911\u2019 but with no other \u201911\u2019 in them. The construction as you show has the two ways to generate N+1: extend all the length N cases with a \u20180\u2019 in the front and extend the N-1 cases adding \u201910\u2019 to the front. This means the number of ways to generate an appropriate string of length N+1 is Fib(N) since Count(N+1) = Count(N -1) + Count(N). And the total strings of length N+1 is $2^{N+1}$. Dividing gives you the probability of each string length (number of kids) and the expectation is as was shown above as well. And of course phi (the Golden Mean) shows up in the solution of the characteristic of the Fibonnaci recurrence so ultimately I ended up using it to solve this. Pretty cool!\n\n2. It\u2019s much quicker, Kram, to model each problem as a state machine.\n\nLet $T_0$ and $T_1$ be Tom\u2019s expectation of future children after zero and one son respectively. We can say immediately that $T_0 = \\frac{1}{2}(s+T_1) + \\frac{1}{2}(d+T_0)$ and $T_1 = \\frac{1}{2}(s) + \\frac{1}{2}(d+T_0)$. The solution, if I haven\u2019t blundered, is $T_0 = 3d+3s$.\n\nFor Lisa: $L_0$ is her expectation before her first daughter; $L_1$ is her expectation after any daughter. $L_0 = \\frac{1}{2}(s+L_0) + \\frac{1}{2}(d+L_1)$ and $L_1 = \\frac{1}{2}(d+L_1) + \\frac{1}{2}(s)$. Solution: $L_0 = 2d+2s$.\n\nLisa can expect two sons and two daughters; Tom can expect three daughters and three sons.\n\nThe qualitative difference is that for Tom a daughter after a son puts him back a step, while for Lisa a child of the sex not hoped for leaves her in the same position.\n\n\u2022 John Baez says:\n\nThis looks impressively efficient, Anton! But what does an equation like\n\n$T_0 = \\frac{1}{2}(s+T_1) + \\frac{1}{2}(d+T_0)$\n\nactually mean? I guess I\u2019m lacking the necessary familiarity with state machines to know what type of entities these variables represent, and what addition means. I know what a state machine is; I\u2019m just not familiar with this kind of equation.\n\nI can guess that $T_n$ is allowed to be any expression of the form $a s + b d$ where $a,b \\ge 0$, and\n\n$T_n = a s + b d$\n\nmeans \u201cafter having had $n$ sons, Tom expects in the future to have a total of $a$ sons and $b$ daughters\u201d.\n\nBut then I don\u2019t know why\n\n$T_0 = \\frac{1}{2}(s+T_1) + \\frac{1}{2}(d+T_0)$\n\nshould be true. It seems to say \u201cafter having no sons, Tom expects in the future to have the same thing as if he had a 50% chance of having one more son than what he expected after having one son, and a 50% chance of having one more daughter than what he expected after having no sons\u201d.\n\nThe horrible quoted phrase here certainly shows why a more concise notation is a good thing! It reminds me of what happens when you try to state the quadratic equation in \u2018plain English\u2019 without variables.\n\nBut the problem is, I don\u2019t see why it should be true.\n\n\u2022 My subscripts mean only that state 0 must occur before state 1 can occur. I could also say explicitly $T_2 = L_2 = 0$; does that make anything clearer?\n\nMy equations are the invariants of the tree of futures. The one that you query can be read as: \u201cFrom state 0, the branches of Tom\u2019s future are: with probability 1/2, he has a son and moves to state 1; and with probability 1/2, he has a daughter and remains in state 0.\u201d As long as his sequence of daughters continues, his expectation of future children remains the same, so the definition of T0 is recursive. From state 1 (one son) Tom cannot remain in state 1; he must either terminate or go back to 0. After a daughter, previous sons are irrelevant to Tom\u2019s future tree, which is why there are not more states in this model.\n\n\u2022 Chris Upshaw says:\n\nShouldn\u2019t that be s * T1 then, if its a parallel notation to sum of products datatype notation? Its certainly not the same operation as the middle \u2018+\u2019.\n\n\u2022 That\u2019s cos I don\u2019t know nothin bout notating no datatypes.\n\n\u2022 John Baez says:\n\nI found Anton\u2019s trick very hard to understand at first, mainly because the variables and operations in his equations had not been explained (and I didn\u2019t already know this trick). Like Greg, I found XJY\u2019s comment here helpful. Even then, I had to hit myself over the head with a rock a few times before I really understood what was going on.\n\nI\u2019ve spent a lot of time thinking about other aspects of Markov processes, but I never thought about this particular trick. It\u2019s very nice, but I think I need to digest it and blend it with other things I know, like generating functions and\u2014yes\u2014recursively defined data types.\n\n\u2022 Anton,\n\nThis is a really cool solution! Not only do you compute Lisa\u2019s expected # of kids, you compute how many sons and daughters she can expect. I have never seen this concept of expressing the invariants of the tree of futures before. Do you have a good reference for learning more about this? Is it standard CS?\n\nThanks for posting this.\n\n\u2022 Rob, sorry, I have no idea where I got the, er, idea.\n\n\u201cInvariant\u201d\u00a0may be a misnomer; thinking about it further, I reckon the analogy with the loop invariants of CS is not so strong.\n\nI may have misunderstood the problem but it seems to me that as soon as Tom or Lisa reach their objective they STOP. Then Tom can have 2 sons in a row and it is over. On the other hand if they have a daughter Lisa can get a son and they stop, Tom can never get two sons.\n\n\u2022 John Baez says:\n\nWow, you assumed that Tom and Lisa are married, and that their plans affect each other\u2019s activities!\n\nI never thought that: I was imagining two people at a bar, who obviously hate each other, each discussing their own separate plans. But now I see that there\u2019s an ambiguity in the phrase \u201ctheir plans for having children\u201d. I\u2019ll fix that!\n\n\u2022 John Baez says:\n\nUpon rereading what I wrote, I see there\u2019s no real ambiguity in the puzzles as stated. One asks what will happen if Tom carries out his plan. The other asks what will happen if Lisa carries out her plan. There\u2019s nothing about what will happen if both of them try to carry out their conflicting plans.\n\nNonetheless I added some stuff to make it crystal clear that Tom and Lisa are not going to be having children together.\n\n4. jamievicary says:\n\nI deliberately haven\u2019t worked this out formally, because it\u2019s more fun to see if intuition leads to a right answer. Write their progeny as a string in D and S. It\u2019s \u2018easy\u2019 for a string of length n to lack a SS substring, as it could have no Ss, or just one S, or just two Ss spaced at least one apart, etc. But to lack a DS substring is relatively \u2018hard\u2019: you have to have all the sons before all the daughters. So, holding out for two sons in a row will take longer than waiting for a daughter followed by a son.\n\nI guess the reason it\u2019s a bit paradoxical is that for any two adjacent children, SS is just as likely as DS. I suppose the reason is that combinatorially, like-gendered children are indistinguishable. In reality this is not the case. They should just ask for \u201ctwo children adjacent in age, both with the temperament to run a shop\u201d!\n\n(By the way, I found this post cognitively difficult to write, since on parenting forums DD and DS often meen \u2018darling son\u2019 and \u2018darling daughter\u2019.)\n\n\u2022 John Baez says:\n\nJamie wrote:\n\nI guess the reason it\u2019s a bit paradoxical is that for any two adjacent children, SS is just as likely as DS.\n\nRight, that was supposed to make the problem cognitively difficult.\n\n(By the way, I found this post cognitively difficult to write, since on parenting forums DD and DS often meen \u2018darling son\u2019 and \u2018darling daughter\u2019.)\n\nI never knew! I hadn\u2019t expected that to cause cognitive difficulties. Do SD and SS mean \u2018stupid daughter\u2019 and \u2018stupid son\u2019?\n\n(I would probably be banned from the forum for that joke.)\n\n\u2022 Tamfang says:\n\nor Silly or Sweet\n\n\u2022 Charles Jackson says:\n\nWell, you did make the problem cognitively difficult for me.\n\nI did #3 first. I thought P(SS) = P(DS) => Expected births to first SS is same as for DS. WRONG\u2014but\n\nI did 2 relatively quickly with something like Anton\u2019s approach above.\n\nThen I looked at the answers in the comments. Oh, oh. I thought about it for awhile and I think I see an easy explanation for the answer.\n\nConsider the outcomes with up to three children. Make a little truth table-like figure with all 8 possible sequences of combinations of boy and girl babies. See below. Notice that there are 4 occurrences of the sequence DS in the table\u2014each of which is the first occurrence of DS in the sequence containing it. (capitalized in table below)\n\nBirth Order\n1 2 3\nd d d\nd D S\nD S d\nD S s\ns d d\ns D S\ns s d\ns s s\n\nNow, search for SS, but only the 3 that are the first occurence in time.\n\nd d d\nd d s\nd s d\nd S S\ns d d\ns d s\nS S d\nS S s\n\nThe difference (only 3 SS versus 4 DS) arises from the second SS that is a subsequence of the SSS sequence\nd d d\nd d s\nd s d\nd s s\ns d d\ns d s\ns s d\ns S S\n\nSo, if the parties were restricted to 3 or fewer babies, Lisa would have a 50% chance of getting DS, but Tom would only have a 3/8 chance of getting SS.\n\nOf course, if they both had three babies, Tom would have a 1/8 chance of getting SS twice, but Lisa would have zero chance of getting DS twice.\n\nAt the birth of the second baby they each have a 25% chance of getting their desired outcome. On the third baby, Lisa would also have a 25% chance but Tom would only have a 12.5% chance.\n\nLisa\u2019s probability of terminating the process at 3, 4, or 5 is higher than the probability for Tom. This pushes up the expected number for him relative to her.\n\nChuck\nPS Thanks for your blog. I only read some of the posts but always enjoy those I complete. No doubt if I had more time and patience I get even more enjoyment from it.\n\n\u2022 John Baez says:\n\nNice. I\u2019m glad you\u2019re enjoying this blog!\n\n\u2022 Jenny Meyer says:\n\nThe DD and DS tags are just a convention for anonymizing your family members. The D can be heavily ironic in context.\n\n(The picture of the Duggars was enough to make this too cognitively difficult for me.)\n\n5. Michael Nelson says:\n\nI think I understand. If we think of this as a game, where Lisa needs to get the sequence DS and Tom needs to get the sequence SS, then Lisa will certainly have the advantage. As soon as a D shows up, Tom will lose. On the other hand, if a S shows up, Lisa can still win. So Lisa has a greater chance of winning this game.\n\n\u2022 John Baez says:\n\nI\u2019m not assuming that Tom and Lisa are having children with each other: they are two separate people, each pursuing their own separate agenda with their own spouse, each assuming their spouse will knuckle under and go along with their insane plan.\n\nNonetheless you can still think of it as a game, where we imagine a random stream of babies being produced and both Tom and Lisa saying when they, personally, would prefer the stream to stop.\n\n\u2022 Michael Nelson says:\n\nYes, this is what I meant. You phrased it better for me. I think I have a better way to think about this now. The number of ways that Lisa can win at the nth spot is just n. For example:\n\nDDDDDS\nSDDDDS\nSSDDDS Lisa wins at 5th spot\nSSSDDS\nSSSSDS\n\nDDDDDDS\nSDDDDDS\nSSDDDDS\nSSSDDDS Lisa wins at 6th spot\nSSSSDDS\nSSSSSDS\n\nThe number of ways that Tom can win at the nth spot involves the fibonacci numbers though. For example:\n\nDDDDSS\nSDDDSS\nDSDDSS Tom wins at 5th spot\nDDSDSS\nSDSDSS\n\nDDDDDSS\nSDDDDSS\nDSDDDSS\nDDSDDSS Tom wins at 6th spot\nSDSDDSS\nSDDSDSS\nDSDSDSS\n\nDDDDDDSS\nSDDDDDSS\nDSDDDDSS\nDDSDDDSS Add D just before SS in 6th spot wins\nSDSDDDSS\nSDDSDDSS\nDSDSDDSS\n\nplus Tom wins at 7th spot\n\nSDSDSDSS\nDDSDSDSS\nDSDDSDSS Replace the last S in 5th spot wins with DSS\nSDDDSDSS\nDDDDSDSS\n\nThat\u2019s pretty cool that the fibonacci numbers are involved.\n\n6. Dave Doty says:\n\nI worked it out like this. I am nagged by the possibility that there is a more elegant simple way to do it.\n\nEach is picking a string from the alphabet $\\{ s,d \\}$. Both have a list of \u201callowed\u201d strings of length $n$ that indicate they must keep going.\n\nLisa omits from this list all strings with the substring $sd$, and Tom omits all strings with the substring $ss$. Already we can see that Lisa has fewer allowed strings than Tom, and will therefore wait a shorter time before stopping, because to remain in an allowed string, she can only transition once (from $s$ to $d$), whereas Tom can transition between $s$ and $d$ an arbitrary number of times.\n\nLisa\u2019s allowed strings match the regular expression $s^*d^*$; e.g., for length 5, her 6 allowed strings are\n\n$sssss, ssssd, sssdd, ssddd, sdddd, ddddd$\n\nIn general, Lisa has $n+1$ allowed strings of length $n$, so probability $\\frac{n+1}{2^n}$ that a given length $n$ string will be allowed. If the string is $s^*$, then either of $s$ or $d$ is allowed next, whereas if it ends in a $d$, only a $d$ is allowed next. So the probability that Lisa ends on length $n$ is the probability that she does not end before (i.e., the length $n-1$ prefix is allowed) times the probability that the whole length $n$ string is disallowed, conditioned on length $n-1$ prefix being allowed. This is\n\n$\\displaystyle{ \\frac{1}{2^{n-1}} \\cdot 0}$\n\n(the probability the string is $s^{n-1}$, times probability 0 since she will not stop after $s^{n-1}$), plus\n\n$\\displaystyle{ \\frac{n}{2^{n-1}} \\cdot \\frac{1}{2} }$\n\n(the probability the string is $s$\u2018s followed by at least one $d$, times the probability that the next symbol is $s$).\n\nSo Lisa\u2019s probability of stopping after $n$ children is $\\frac{n}{2^n}$.\n\nThus her expected stopping time is\n\n$\\displaystyle{ \\sum_{n=2}^\\infty n \\cdot \\frac{n}{2^n} = \\sum_{n=2}^\\infty \\frac{n^2}{2^n} = \\frac{11}{2} }$\n\nTom\u2019s allowed strings omit the substring $ss$. We can count recursively how many strings of length $n$ satisfy this, denoted by $T_n$. If the first symbol is $d$, then all allowed strings of length $n-1$ could follow, of which there are $T_{n-1}$. If the first symbol is $s$, then the next symbol must be $d$, and any allowed string of length $n-2$ could follow, of which there are $T_{n-2}$. So $T_n = T_{n-1} + T_{n-2}$, the $n$\u2018th Fibonacci number, which is\n\n$\\displaystyle{ \\frac{(1+\\sqrt{5})^n - (1-\\sqrt{5})^n}{2^n \\sqrt{5}} }$\n\nSo Tom\u2019s probability of stopping after $n$ children is the probability that the last three symbols are $dss$ (if it were $sss$ he would have stopped at the second $s$) the prefix of length $n-3$ is allowed, which is\n\n$\\displaystyle{ \\frac{T_n}{2^n} = \\frac{(1+\\sqrt{5})^n - (1-\\sqrt{5})^n}{2^{2n} \\sqrt{5}} }$\n\nThus his expected stopping time is\n\n$\\displaystyle{ \\sum_{n=2}^\\infty n \\cdot \\frac{(1+\\sqrt{5})^n - (1-\\sqrt{5})^n}{2^{2n} \\sqrt{5}} = \\frac{19}{2} }$\n\n\u2022 John Baez says:\n\nWow, what a tour de force! You are not getting the officially approved correct answers, but the officially approved correct answers do involve Fibonacci numbers, at least in the most direct approach.\n\n7. Bruce Smith says:\n\nI carefully solved each puzzle before reading beyond it (in case of spoilers), so puzzle 3 was not very interesting! (I had thought of it while starting to work on puzzle 2, but didn\u2019t have the benefit of your stating it was worthwhile to work on it first, so I didn\u2019t do so.)\n\nAnother mistake \u2014 I initially assumed Tom and Lisa were married and were discussing their joint plan for having kids! This leads to puzzle 4 \u2014 suppose they do marry (before having any kids) and agree to have kids until both their criteria are satisfied\u2026\n\n\u2022 John Baez says:\n\nI think it\u2019s too late for me to make Puzzle 3 into Puzzle 1 without causing massive confusion.\n\nI have added verbiage to the puzzle to make it clear that Tom and Lisa are not married and never will be.\n\nNonetheless, Puzzle 4 is interesting, and will doubtless lead to an even larger expected family.\n\n\u2022 For that case I get seven expected children, again equally divided.\n\nNext question: For each scenario, what are the variances? Do the variance in daughters and the variance in sons add up to the variance in total children? I don\u2019t know how to figure these.\n\n\u2022 Unless I made an error, I already did the variances \u2013 or, well, the standard deviations which are just the square roots of the variances \u2013 in my reply (first one). However, all the other contributions here are much more fleshed out and nice. My approach was really just trying a couple and then guessing. It seems like the results were right, however it\u2019s nice to read in the others\u2019 replies why they are right.\n\nTo get the variance, if you have all the sub-probabilities, you just do:\nProbability: $P=\\sum_{n \\in \\mathbb{N}} p\\left(n\\right) = 1$\n\nExpectation: $E = \\sum_{n \\in \\mathbb{N}} n p\\left(n\\right)$\n\nVariance: $\\sigma^2 = \\sum_{n \\in \\mathbb{N}} n^2 p\\left(n\\right) - E^2$\n\nStandard Deviation: $\\sigma = \\sqrt{\\sigma^2}$\n\nFor some reason, in my original reply, the $\\LaTeX$ didn\u2019t work right. Looks like the fixed version still isn\u2019t right though. Hopefully the above works.\n\n\u2022 John Baez says:\n\nThe LaTeX at the very end of this comment of yours didn\u2019t work, and it was pretty cryptic, so I guessed you were trying to say this:\n\n$\\sigma^2 = \\langle-^2\\rangle$\n\nThis means \u2018the square of the standard deviation is the expected value of the square of the quantity in question\u2019. (The little dash is a standard symbol for \u2018the quantity in question\u2019: a blank to be filled in.) But this is only true for quantities whose mean is zero, which isn\u2019t the case in this problem. So, I must have guessed your meaning incorrectly. I should have written\n\n$\\sigma^2 = \\langle-^2\\rangle - \\langle - \\rangle^2$\n\nBut probably you weren\u2019t trying to use the \u2018little dash\u2019 notation, which gets a bit scary in formulas that also contain minus signs!\n\n8. Ilya Surdin says:\n\nPuzzle 1:\nsuppose d is the expectation value we\u2019re looking for.\nthen d=0.5(d+1)+0.5(0.5*2+0.5(d+2)) [if a daughter is born first, we\u2019re back to point 1 again, if a son is born first then if another son is born the value is 2, and if a daughter is born second we\u2019re back to square 1].\nso we get d = 0.75d +1.5 -> d=6\n\nPuzzle 2:\nsuppose e is the expectation value we\u2019re looking for.\ne = 0.5(e+1) + 0.5f,\n\nwhere summand 1 is if a son is born, and summand 2 is if a daughter is born, we only need to get a son (ds, dds, ddds, all options are good)\n\nf = 2 [ f=0.51+0.5(f+1) ]\n\nso e = 0.5e + 0.5 +1 -> e=3.\n\nPuzzle 3:\nAt first sight, I guessed they will be equal.\n\n9. Graham says:\n\nHere\u2019s a solution for Tom\u2019s case using generating functions. Let $T$ be a random variable representing the number of children Tom has. Let $f(s) = \\sum p_i s^i$ be the generating function for T, where $p_i = \\Pr(T=i)$. The first few $p_i$ are\n\n$p_0 = p_1 = 0, p_2=1/4, p_3=1/8.$\n\nFor $i \\ge 4$, the sequence must end DSS, and there must be no SS occurring in the first $i-3$. So\n\n$p_4 = (1/8)(1-p_1),$\n\n$p_5 = (1/8)(1-p_1-p_2),$\n\nand so on.\n\nFrom this it follows that\n\n$p_i = p_{i-1} - (1/8)p_{i-3}$\n\nfor $i \\ge 4$.\n\nThen by comparing $f(s), sf(s), s^3f(s)$, it is possible to calculate that\n\n$\\displaystyle{ f(s) = \\frac{-s^2}{s^2+2s-4}. }$\n\nThen\n\n$E(X) = f'(1) = 6$\n\nand\n\n$E(X(X-1)) = f''(1) = 52$\n\nfrom which $\\mathrm{var}(X) = 52 + 6 -6^2 = 22$.\n\n10. \u201cIn one word he told me secret of success in mathematics: Generalize!\u201d What if the sex ratio at birth is not 1:1? I\u2019ll get right on that \u2014 later.\n\n\u2022 John Baez says:\n\nPuzzle 5. For what sex ratio at birth would Tom and Lisa have an equal expected number of children?\n\n(Obviously it needs to be more sons than daughters.)\n\n\u2022 Greg Egan says:\n\nJohn wrote:\n\nPuzzle 5. For what sex ratio at birth would Tom and Lisa have an equal expected number of children?\n\n$\\displaystyle{ p_{son} = \\frac{\\sqrt{5}-1}{2} }$\n\nor:\n\n$\\displaystyle{\\frac{p_{son}}{p_{daughter}} = \\frac{\\sqrt{5}+1}{2} }$\n\nTo prove this (stealing from and generalising an answer by \u201cXJY\u201d, which is a slightly more coarse-grained version of Anton Sherwood\u2019s approach), we can write the expectation values for the number of additional children that Tom and Lisa will each have after their latest child was a daughter or a son (who didn\u2019t yet complete their target sequence) as:\n\n$T_d = p_{son} (1+T_s) + (1-p_{son}) (1+T_d)$\n$T_s = p_{son} +(1-p_{son}) (1+T_d)$\n$L_d = p_{son} + (1-p_{son}) (1+L_d)$\n$L_s = p_{son} (1+L_s) + (1-p_{son})(1+L_d)$\n\nThese are solved by:\n\n$\\displaystyle{T_d = \\frac{1+p_{son}}{p_{son}^2} }$\n\n$\\displaystyle{T_s = \\frac{1}{p_{son}^2} }$\n\n$\\displaystyle{L_d = \\frac{1}{p_{son}} }$\n\n$\\displaystyle{L_s = \\frac{1}{p_{son}(1-p_{son})} }$\n\nBut $T_d$ and $L_s$ are also the expectation values for the number of children each will have in total, because they have 0 out of 2 elements of their target sequence.\n\nSo if we solve:\n\n$L_s = T_d$\n\nthe positive solution for $p_{son}$ is:\n\n$\\displaystyle{ p_{son} = \\frac{\\sqrt{5}-1}{2} }$\n\nand\n\n$L_s = T_d = 2+\\sqrt{5}$\n\n11. Bruce Smith says:\n\nIt turns out these puzzles show up (for numbers 1 and 3 sought by Alice and Bob) in this (otherwise mostly unrelated) blog post about primes:\n\nhttps://rjlipton.wordpress.com/2016/03/26/bias-in-the-primes/\n\n\u2022 John Baez says:\n\n\u2022 Erica Klarreich, Mathematicians discover prime conspiracy, Quanta, 13 March 2016.\n\nSoundararajan was drawn to study consecutive primes after hearing a lecture at Stanford by the mathematician Tadashi Tokieda, of the University of Cambridge, in which he mentioned a counterintuitive property of coin-tossing: If Alice tosses a coin until she sees a head followed by a tail, and Bob tosses a coin until he sees two heads in a row, then on average, Alice will require four tosses while Bob will require six tosses (try this at home!), even though head-tail and head-head have an equal chance of appearing after two coin tosses.\n\nI decided to dramatize this using children instead of coin tosses.\n\nBy the way, I wrote about the \u2018prime conspiracy\u2019 here:\n\n\u2022 John Baez, Weirdness in the primes, _The n-Category Caf\u00e9 15 March 2016.\n\nThere\u2019s a bit of math here that\u2019s not in the Quanta article.\n\n12. Rob says:\n\nNice Puzzle.\n\nfrom \u201cMathematical Plums\u201d edited by Ross Honsberger, chap 5 some surprises in probability, \u201cSuppose a fair coin is tossed repeatedly and an indefinitely long sequence of H\u2019s and T\u2019s is generated\u2026. what a surprise to learn that it is twice as likely that the triple TTH will be encountered ahead of the triple THT\u2026\u201d I love surprises in probability, and I find that most things about probability surprise me! The chapter also talks about something else you may like if you haven\u2019t seen it, \u201cEfron\u2019s Dice\u201d.\n\nBy the way, have you read this paper, http://www.emis.ams.org/journals/TAC/volumes/26/4/26-04.pdf, Commutative Monads as a Theory of Distributions. I gather Prof. Kock is developing the theory to study probability theory. I can\u2019t really understand it, but you may find it interesting.\n\n\u2022 John Baez says:\n\nProbability theory is often counterintuitive to humans, which must have something to do with why humans mess up so often.\n\nI haven\u2019t read that paper. I\u2019m pretty good friends with Anders Koch\u2019s son Joachim, who also does category theory related work. But this paper is by Anders. I find it pretty hard going, at first because I\u2019ve never understood strengths as much as I\u2019d like, even though I did help come up with the more conceptual definition of them given on the link. (The more common definition, which involves writing down some commutative diagrams, always bothered me.)\n\n\u2022 Rob says:\n\nSay the first child is a G. Then we are guarateed that GB will occur prior to to BB. (cause for BB to happen, we must first have had GB)\n\nNow say the first child is a B. Then either the second child is a B, in which case BB happens first, or it is a G, in which case, again, GB will happen before BB.\n\nSo we see that GB is 3x more likely to occur in any given familiy than BB. Thus I would expect that the expected value of the second random varable is less than the expected value of the first. Hah! well probability is so amazing, I expect that I am wrong! Now I will read through the comments where I expect to find the right answer.\n\nAlso, if you don\u2019t understand that paper, I never will! But I think it is really interesting so I will keep reading it. BTW I think your notes on Cat Theory are great. Thanks to all your students for putting that together.\n\n13. arch1 says:\n\nPuzzle 3: Represent all possible birth sequences as an infinite binary tree (at the risk of scaring you we\u2019ll pretend that childbearing doesn\u2019t stop when the goal is reached). Yes, at each level of the tree starting with the 2nd, \u00bc of the arcs are DS arcs and \u00bc are SS arcs. But the symmetry breaks when we look at the correlation between adjacent levels: While an SS arc can immediately follow an SS arc, a DS arc can\u2019t immediately follow a DS arc (as that would require the shared middle node to be both an S and a D).\n\nUpshot: Half of the SS arcs \u201chide behind\u201d an immediate SS parent, but DS-DS repulsion prevents DS arcs from ever doing this. This makes a DS arc likelier than an SS arc to be the first arc of its kind on a path from the root (put differently, it causes DS arcs to more efficiently \u201ccover\u201d the tree as viewed from the root). So the expected number of levels before encountering a DS arc is less than for an SS arc.\n\nGreg and John, I find these probability paradoxes very instructive and great for my humility. I hope you keep them coming.\n\n\u2022 John Baez says:\n\nNice answer! Yes, probability puzzles are great for ones humility\u2014maybe the right verb is \u201chumiliating\u201d?\n\nThere should be a book of probability puzzles, similar to Mark Levi\u2019s excellent physics puzzle book Why Cats Land on Their Feet: And 76 Other Physical Paradoxes and Puzzles. But I don\u2019t know it!\n\nIf anyone here hasn\u2019t tried Probability Puzzles (Part 1) and Probability Puzzles (Part 2), now is the time! They\u2019re all about boys and girls, oddly enough. The first one is particularly devastating, or infuriating.\n\n\u2022 arch1 says:\n\n(chuckle) I actually tend to find them more invigorating than humiliating, partly for noble reasons but mostly I think because I can\u2019t wait to spring them on others.\n\nIronically the closest I\u2019ve come to humiliation recently was last week when I, er, \u201cjudged\u201d a high school science fair project which explained how cats can land on their feet using gauge theory.\n\n\u2022 cclingen says:\n\nHigh school?! Cats landing their feet?! Gauge theory?! Where can I learn more?\n\n\u2022 John Baez says:\n\nHi, Charlie! It\u2019s probably good to start here:\n\n\u2022 Wikipedia, Falling cat problem.\n\nbut then go to this paper by a friend of mine who is an expert on classical mechanics at U. C. Santa Cruz:\n\n\u2022 Richard Montgomery, Gauge theory of the falling cat, in M. J. Enos, Dynamics and Control of Mechanical Systems, American Mathematical Society, pp. 193\u2013218.\n\n\u2022 arch1 says:\n\nNowhere as far as I know (certainly not me as I gleaned only a scant understanding). Maybe a paper will come out later; meanwhile I\u2019d prefer to let the student pursue things in his own way.\n\n\u2022 Re Prob Puzzle books. I just bought \u201cFifty Challenging Problems in Probability\u201d. It looks like it has some really cool stuff in it. And best of all it was cheap!\n\n\u2022 Greg Egan says:\n\nThe counterintuitive aspects of real statistical phenomena might have put an innocent man in prison for 38 years:\n\nAn Unusual Pattern\n\nLike the statisticians who discuss the case, I\u2019m not claiming any certainty that Geen is innocent, but on the face of it the statistical basis of the prosecution\u2019s argument appears alarmingly wrong-headed.\n\n\u2022 That is creepy. As teachers we need to do a better job at explaining statistics. Examples like the above and the OP help a lot.\n\n14. pauljurczak says:\n\nI hate to be a party pooper, but using infinite series in the top answer is yet another reason why lies and statistics are often mentioned in the same sentence. Human female can only have a limited number of children, much smaller than 100, which is much smaller than infinity.\n\nThis is in general a problem of using mathematics to model physical \u201creality\u201d, which always requires certain assumptions resulting in departure from \u201creality\u201d. In this case, the assumption is immortality or more specifically, unlimited childbearing capability.\n\nP.S. Don\u2019t take this post too seriously ;-)\n\n\u2022 John Baez says:\n\nI won\u2019t take it too seriously\u2014just more seriously than you expected.\n\nPuzzle 7: How do the answers for Tom\u2019s and Lisa\u2019s expected number of children change if we impose a \u2018cutoff\u2019, saying that a woman can have no more than 10 children?\n\n\u2022 \u2026Or if each pregnancy has probability q of leaving the mother unable to bear further children?\n\n\u2022 Charles Jackson says:\n\nNote that that there is about a 1% that Tom will have to have 24 or more children before the desired SS outcome occurs.\n\n15. Edon says:\n\nThese processes can be seen as absorbing Markov chains. Let $T$ be the block in the transition matrix that holds the transitions from transient states to other transient states. It is a basic property that $(I - T)$ is invertible and that the $(i,j)$ entry of $M=(I - T)^{-1}$ gives the expected number of times the chain is in state $j$ before it is absorbed, given that we start from state $i$. Summing the entries in the appropriate row yields the answers to the puzzles.\n\n16. For some reason there isn\u2019t a reply button below that latest reply of you, John, but yeah, that was what I was going for, although indeed not with the \u2018little dash\u2019 notation, but that works too. With just one minus sign it\u2019s not that bad. Thanks for the fix. There must have been more wrong with it than just $\\langle \\cdot \\rangle$ if it was so cryptic. I\u2019m not sure what I typed anymore but while typing it that was the only part I didn\u2019t know whether it\u2019d work. The rest I was pretty sure of. But oh well. I thought \\left{<} would work for \\langle but you guessed that part right.\n\n\u2022 John Baez says:\n\nKram wrote:\n\nFor some reason there isn\u2019t a reply button below that latest reply of you\u2026\n\nUnfortunately if you tell this blog that you don\u2019t want comments to get indented more than 4 times (because they become absurdly skinny), it implements this by eliminating the \u2018Reply\u2019 button after a comment that\u2019s been indented 4 times.\n\nWhat you should do in this case is reply not to the comment but to the comment it\u2019s commenting on! In fact it\u2019s often wise to do this. This keeps the comments from getting too skinny. I can tell who is a really expert blog-commenter by whether they do this.\n\nI thought \\left{<} would work for \\langle but you guessed that part right.\n\nI\u2019ve never seen \\left{<} in LaTeX; I think LaTeX is pretty careful about the difference between the less than symbol $<$ and the left angle bracket $\\langle$.\n\nBut there\u2019s an extra problem with LaTeX in most blogs! In HTML, < is an escape symbol, so you to get a less than symbol in LaTeX you have to use the HTML symbol &lt; rather than <. It\u2019s very bizarre to be using HTML commands inside LaTeX equations, but that\u2019s how it works! That\u2019s true not only for this rather brain-dead WordPress blog but also some others I use.\n\n17. Tom says:\n\nPuzzle 6: Why did John use the name of his wife for the female character in this drama?\n\n\u2022 John Baez says:\n\nAnswer: because it was the first woman\u2019s name beginning with L that I thought of.\n\n18. David Cinabro says:\n\nThis one got in my head, and I had to try to solve it.\n\nFirst Puzzle 3. Two children SS, SD, DS, DD. Tom and Lisa both stop 1 out of 4. Three Children for Tom SDS, SDD, DSS, DSD, DDS, DDD. Tom stops 1 out of 6. Three Children for Lisa SSS, SSD, SDS, SDD, DDS, DDD. Lisa stops 2 out of 6. Clearly the expected number of children is different and Lisa will have a smaller number than Tom.\n\nFor Tom\u2019s expected number of children, he has 1/4 chance stopping at 2, 1/6 stopping at 3, 2/10 stopping at 4, 3/16 stopping at 5, 5/26 stopping at 6, etc. The stopping chance numerators are Fibonacci seeded by 1 and 1, and the denominators are Fibonacci seeded by 4 and 6. The expected number of children is then 2(1/4) + 3(1-1/4)1/6 + 4(1-1/4)(1-1/6)3/16+\u2026. ie he stops at 2 with a probability of 1/4, at 3 with a probability of (1-1/4)*1/6, etc. Perhaps there is a clever way to write that, but I wrote a little program to tell me that it goes to 6.\n\nLisa has 1/4 chance stopping at 2, 2/6 stopping at 3, 3/8 stopping at 4, etc. The stopping chance numerators are 1,2,3,etc. and the denominators at 4,6,8, etc. The same sort of calculation gives the expected number of children 21/4 + 3(1-1/4)2/6+4(1-1/4)(1-2/6)*3/8+\u2026 This one is certainly even simpler to work out in closed from , but again the program tells me that it sums to 4.\n\n19. Michael Nelson says:\n\nBy the way, you can prove the classic result involving pascal\u2019s triangle and the fibonacci numbers by partitioning Tom\u2019s string of n children into these cases:\n\nTom has a total of 3 sons\nTom has a total of 4 sons\n\nFor those who aren\u2019t familiar, the relation between pascal\u2019s triangle and the fibonacci numbers involves summing over diagonals, like in this image:\n\nIn other words, the nth fibonacci number equals (n-1 choose 0) plus (n-2 choose 1) plus \u2026\n\n20. While on the subject of what is essentially run theory, mathematician Peter Donnelly has given a TED talk on the subject of fooling juries with statistics. His talk is based around the infamous English criminal case of Sally Clark. To illuminate the issues he poses this problem:\nyou toss an unbiased coin many times and count the number of tosses until you get the pattern \u201dHTH\u201d. You then calculate the average. You repeat the experiment but this time you look for the pattern \u201dHTT\u201d. The question is: \u201dIs the average number of trials required to get \u201dHTH\u201d greater than, less than or equal to the average number of trials required to get \u201dHTT\u201d? \u201d Donnelly has tried this out on a wide range of audiences including professional mathematicians and the audience not surprisingly says the number of trials were equal. The expected number of trials to get \u201dHTH\u201d is 10 while the expected number of trials to get \u201dHTT\u201d is 8.\nMy paper ( $\\url{http://www.gotohaggstrom.com/Fooling%20juries%20with%20statistics.pdf}$ ) explains the generating function proof of Donnelly\u2019s claim. This paper also sets out the background to the Sally Clark case. A 2013 English Court of Appeal case has had the effect of banning Bayesian reasoning in court cases. This shows that the \u201creal\u201d world of the justice system can intersect in bizarre ways with probability theory.\n\n21. Puzzle 1.\nWe are looking at infinite sequence of coin flips (the coin has two sides: S and D) and waiting for first occurrence of SS.\nThere are essentially 3 different states:\ns0: current sequence ends with D or is empty (initial state)\ns1: current sequence ends with exactly one S\ns2: current sequence ends with SS (end state)\n\nLet ei = expected number of flips to get from si to s2.\nThen we have the following system of linear equations:\ne2 = 0\ne1 = (1/2)(1+e2) + (1/2)(1+e0)\ne0 = (1/2)(1+e0) + (1/2)(1+e1)\n\nSolving it gives the answer to the puzzle: e0 = 6.\n\nPuzzle 2.\nWe just need to wait for the first daughter, then wait for the first son. Since expected number of trials until first success equals 1/p, where p is the probability of success, the answer is obviously 2+2=4.\n\nPuzzle 3.\nThe intuition is, that Tom has to make a \u201cstep back\u201d sometimes but not Lisa, therefore she accomplishes her goal sooner.\n\n22. Check out this interesting, related puzzle and discussion.\n\nhttps://www.klittlepage.com/2013/11/20/dynasty-meets-the-central-limit-theorem/\n\nI just came across it. It has some surprising conclusions.", "date": "2018-02-19 06:09:33", "meta": {"domain": "wordpress.com", "url": "https://johncarlosbaez.wordpress.com/2016/03/26/probability-puzzles-part-3/", "openwebmath_score": 0.7304139733314514, "openwebmath_perplexity": 1206.523883410803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232869627774, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8624046110592909}}
{"url": "http://ufap.cbeu.pw/find-the-indicated-nth-roots-of-a.html", "text": "# Find The Indicated Nth Roots Of A\n\nis there any other way to find the nth root besides de Moivre's theorem and the comparing method. An th root of unity in a finite field is an element satisfying , where is an integer. De Moivre's Theorem Now, in that same vein, if we can raise a complex number to a power, we should be able to find all of its roots too. , , , the six roots are () out of which , , and on the left s-plane are the roots of :. Our cube root calculator is a handy tool that will help you determine the cube root, also called the 3 rd root, of any positive number. One real root:. Examples:. Find the indicated real nth roots of a: 13. The proof can be extended to higher roots, as well. This is a preview topic with a couple of types of exercises. Find the nth term of the arithmetic sequence whose initial term a and common difference d are given. Also, don't overlook the most obvious property of all!. How do I write a function to calculate the nth root of a number; I'm sick of using pow() with a fractional. It is the method based on long. Conventionally your original #theta# is in the range #(-pi, pi]# or the range #[0, 2pi)# according to your definition of #Arg(z)# and the first of these five roots is the Principal Complex fifth root. Takes the nth root of all values in an iterable multiplied together. So if you're taking this to the 1/5 power, this is the same thing as taking 2 times 2 times 2 times 2 times 2 to the 1/5. Similarly, the cube root of a number b , written b 3 , is a solution to the equation x 3 = b. n th Roots. Roots are usually written using the radical symbol, with denoting the square root, denoting the cube root, denoting the. This online calculator is set up specifically to calculate 4th root. Free practice questions for SAT Math - How to find the nth term of an arithmetic sequence. 32 = 32(cos0 + isin 0 ) in trig form. For smaller values of n, one can readily compute the numbers. NaN = not a number To clear the entry boxes click \"Reset\". What does nth root mean? Information and translations of nth root in the most comprehensive dictionary definitions resource on the web. n = 3, a = 216 b. This square root trick to find square root of a number will surely help you in your exams. ? Find the indicated real nth root(s) of a - Help? What is the indicated real nth root(s) of a. It is clear from the foregoing that the nth root of a real or complex number can be. 0 equals to 4. Research design: Qualitative, quantitative and qualitative analyses, is the interview but how do you think is terribly exponents roots help homework on finding nth and rational important, especially in regard to the organizational and leadership for pursuing implementation of p. Find the nth Root of a Perfect nth Power Monomial. Introduction to Rational Functions. Finding Nth Roots: Find the indicated real nth root(s) of a. Finding the Roots of a Complex Number We can use DeMoivre's Theorem to calculate complex number roots. There are 5, 5th roots of 32 in the set of complex numbers. n=2, a=100 \" in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. notebook 3 November 13, 2014 YOU PRACTICE Find the indicated real nth root(s) of a. We can find value of k by putting discriminant of quadratic equation equal to 0. Specifically, here we find the transfer function of the nth order Butterworth filter for : , , , the four roots are (): out of which and on the left s-plane are the roots of : Note that the coefficient of the first order term is. 1 A Case Study on the Root-Finding Problem: Kepler's Law of Planetary Motion The root-\ufb01nding problem is one of the most important computational problems. From above it can be concluded that the roots are located on the circumference of the unit circle with center at (0,0). Nth Root Lesson Plans & Worksheets Reviewed by Teachers. Take a out a piece of paper and a pencil and step through the algorithm. Daily Math Lesson. 2 = ei\u03c0 = \u22121, and the roots of unity are simply 1 and \u22121. We could use the nth root in a question like this:. It looks quite tedious to do by hand, but the algorithm exists for any root and is similar to the square root one. Type \u201cNumber\u201d, \u201cRoot\u201d and \u201cResult\u201d in the cells A1, B1 and C1 respectively. This example shows how to calculate the Nth root of a number in Visual Basic. The principal nth root is the nonnegative root. From nth roots worksheets to algebra 2 nth root videos, quickly find teacher-reviewed educational resources. How to Find Roots of Unity. Finding real nth roots of a. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. How to Find Nth Roots by Hand. The number that must be multiplied times itself n times to equal a given value. Toggle navigation. These problems serve to illustrate the use of polar notation for complex numbers. ) ! n=3! a=64 c. For, the square of a real root must be positive; and therefore the original equation cannot have more real roots than the transformed has positive roots. Perfect for. If n is even, then a has real nth roots if a > 0, real nth root if a5 0, and real nth roots if a < 0. The variables may be defined in the program, or they may be inputted by the user, but it. Given an integer number and we have to find their Square, Square Root and Cube. To check the result, the code then raises the result to the root power and displays the new result. For instance, 2 is a cube root of 8 because 23 = 8, and 3 is a fourth root of 81 because 34 = 81. Use of the the pow function is more sensible for this unusual need. So the maximum is probably between 2 and 4, and I asked my son how we could find it. Solving Nth Root Equations The nth root of a number X , is a number r whose nth power is X. The other n roots are equally spaced on the circle of radius. Apart from the stuff given in this section \" Finding nth Term of the Sequence\" , if you need any other stuff in math, please use our google custom search here. n = 3, x = 128. Square root of 64 is 8 because 8 times 8 is 64 Cube root of 27 is 3 because 3 times 3 times 3 = 27 fourth root of 16 is 2 because 2 times 2 times 2 times 2 = 16 Sometimes, you may get a real number when looking for the square root. For large n, the n th root algorithm is somewhat less efficient since it requires the computation of \u2212 at each step, but can be efficiently implemented with a good exponentiation algorithm. Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. 6-1 Evaluate nth Roots and Use Rational Exponents Name_____ Objective: To evaluate nth roots and use rational exponents. What is a function? Domain and range. The only one you could find was x =2. Simplify the expression. The nth root of a real number. Also, don't overlook the most obvious property of all!. root of a positive number is positive. As the title suggests, the Root-Finding Problem is the problem of \ufb01nding a root of the equation f(x) = 0, where f(x) is a function of a single variable x. (So n=3) Answer should be in standard form. Basically, pretend to give X0 a value of your initial guess. Homework help with finding nth roots and rational expressions with paperbag writer radiohead in education by apa sociology research paper example. In mathematics, Nth root of a number A is a real number that gives A, when we raise it to integer power N. Using this formula, we will prove that for all nonzero complex numbers $z \\in \\mathbb{C}$ there exists $n. To find the ARITHMETIC mean of 4 and 10, you add them up and then divide by n number of values: (4+10)/2 = 7 To find the GEOMETRIC mean, you multiply 4 and 10, and then find the nth root: the. From this form a K-normal form C, can be deduced. 5) Complete the following table for!!!=7!. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. From this form a K-normal form C, can be deduced. In Exercises 48 and 49, find the modulus and the argument, and graph the complex number. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Finding real nth roots of a. Improve your math knowledge with free questions in \"Nth roots\" and thousands of other math skills. Numerical methods for \ufb01nding the roots of a function The roots of a function f(x) are de\ufb01ned as the values for which the value of the function becomes equal to zero. FINDING NTH ROOTS Find the indicated real nth root(s) of a. Given two numbers N and A, find N-th root of A. Round your answer to two decimal places when appropriate. This nth root calculator will compute the nth root of any number with just the click of a button. These roots are used in Number Theory and other advanced branches of mathematics. Essay Writing Help Homework Help On Finding Nth Roots And Rational Exponents Evelyn finished paper Hire Writer Terry Anderson 08. To calculate any root of a number use our Nth root calculator. Remember to find the fourth root we would set up an equation like this. However, there is still one basic procedure that is missing from the algebra of complex numbers. Answer to Find the indicated real n-th roots of x, if any. Evaluate Question: \"What number, when raised to the 3rd power gives us 27?\" B. Just with a number 1, I get e to the i to k pi over n. In this case, you will be given two terms (not necessarily consecutive), and you will use this information to find a 1 and d. Each test case contains two space separated integers. 14159 For dCode the first decimal 1 in 3. Students struggling with all kinds of algebra problems find out that our software is a life-saver. There might be some issues with this. In mathematics, Nth root of a number A is a real number that gives A, when we raise it to integer power N. How much money did she invest at 12%. This important formula is known as De Moivre's formula. n th Roots. Y = nthroot(X,N) returns the real nth root of the elements of X. It has been proven that there is no algorithm that can do that in the general case: there are functions for which it can be proven that knowledge of the function value at any finite number of other locations does not give you any information about the function value at any particular location. Nth Root of a Complex Number I'm looking to construct a function f(n,p) that will find all n roots of a complex number p. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. When I try to find the roots of the same equation in Mathematica, I receive various errors. Geometric mean function for python. The solutions of the equation x^n=1, where n is a positive integer, are called the nth roots of 1 or \"nth roots of unity. Homework Help With Finding Nth Roots And Rational Expressions. Perhaps, then, it is this calculator: Let's say you want to take the fifth root of 2. Asking for unreasonable roots, for example, show_nth_root < 1000000 >(2. This is a polynomial equation of degree n. (26 replies) In PythonWin I'm running a program to find the 13th root (say) of millions of hundred-digit numbers. Find nth roots of complex numbers Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Finding nth roots of Complex Numbers. However, there is still one basic procedure that is missing from the algebra of complex numbers. Both X and N must be real scalars or arrays of the same size. In mathematics, an nth root of a number x, where n is usually assumed to be a positive integer, is a number r which, when raised to the power n yields x: =, where n is the degree of the root. It uses Math. Homework Help On Finding Nth Roots And Rational Exponents. For instance, 2 is a cube root of 8 because 23 = 8, and 3 is a fourth root of 81 because 34 = 81. The math workaround described above works really well with pretty good accuracy. Square root of 64 is 8 because 8 times 8 is 64 Cube root of 27 is 3 because 3 times 3 times 3 = 27 fourth root of 16 is 2 because 2 times 2 times 2 times 2 = 16 Sometimes, you may get a real number when looking for the square root. The nth roots of a complex number For a positive integer n=1, 2, 3, \u2026 , a complex number w \u201e 0 has n different com-plex roots z. 1/2 Evaluate the expression. 24 The nth root of a number b is designated as nb,. EXAMPLE 1 Find nth roots Find the indicated real nth root (s) of a. is the radius to use. NaN = not a number To clear the entry boxes click \"Reset\". The most common root is the square root. It arises in a wide variety of practical applications in physics, chemistry, biosciences, engineering, etc. n = 2, a = 100 24. Find nth roots of complex numbers Contact If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Drag the slider to change the number of roots. The pictures. -polar graph of roots. If bn = a, then b is an nth root of a. An th root of unity in a finite field is an element satisfying , where is an integer. 0 root of 125. 5 or 10 1/2 and the cube root as 10 0. Find a number x = \u03be such that f(\u03be) = 0. Since CP is similar to A, a matrix Y E Mm K exists such that (Y-1CPY)= Y-iCPY = A. To calculate any root of a number use our Nth root calculator. How to find the nth root of a complex number. This online calculator for fifth roots is set up specifically to calculate 5th root. In effect we must find the roots of the equation z^3 = 1 We could write this z^3-1 = 0 (z-1)(z^2 + z + 1) = 0 and then z = 1 gives one root, and other two are solutions of the quadratic z^2+z+1 = 0 (these roots will be complex). ^2 96 sqr root 3 in. \u201cnth Root of a\u201d. An applet allowing you to investigate how the th roots of a complex number (in blue) appear on an Argand diagram. nth Root of a-to-the-nth-Power. 5, 13, 33, 89, I got this by adding the two sequences above. Evaluate Question: \u201cWhat number, when raised to the 3rd power gives us 27?\u201c B. For instance, \u221a(4) produces a principle root of 2. If all equations and starting values are real, then FindRoot will search only for real roots. I decided to take the input in the form of a Double. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. Relating to an unspecified ordinal number: ten to the nth power. Functional abstraction to find nth root of a number - Newton raphson. Created Date: 1/17/2019 10:36:18 AM. Finding nth roots of Complex Numbers. Find the indicated real nth root(s. (algorithm) Definition: This describes a \"long hand\" or manual method of calculating or extracting square roots. Rationalize one term denominators of rational expressions. The exponent of a number shows you how many times the number is to be used in a multiplication. 1 Evaluate nth Roots and Use Rational Exponents Goal p Evaluate nth roots and study rational exponents. Derivation from Newton's method. A root of degree 2 is called a square root, a root of degree 3 is called a cube root, a root of degree 4 is called a fourth root, and so forth. org - it looks like you don't have Java installed, please go to www. I know that if this were just normal numbers, I could find it u. In effect we must find the roots of the equation z^3 = 1 We could write this z^3-1 = 0 (z-1)(z^2 + z + 1) = 0 and then z = 1 gives one root, and other two are solutions of the quadratic z^2+z+1 = 0 (these roots will be complex). Using Bino's model of multiplication , you can compute the square root, cube root, or root of any positive number up to any decimal place for no-perfect cube number. Granted, then, you have a problem they currently have. What is the principal root of 121400? Rounded to two decimal places, the principal square root of 121400 is 348. Created Date: 1/14/2009 8:51:02 PM. The nth root calculator below will also provide a brute force rounded approximation of the principal nth root. She had invested$1500 more at 8%. Roots can be square roots, cube roots, fourth roots and so on. What is the fifty-first term?. Use nth roots in problem solving Animal Population The population P of a certain animal species after t months can be modeled by P = C(1. The for a different view). This online calculator is set up specifically to calculate 4th root. According to the Fundamental Theorem of Algebra, this equation has n solutions. Find indicated real nth roots of c. org - it looks like you don't have Java installed, please go to www. It uses Math. Homework Help On Finding Nth Roots And Rational Exponents. 5) Complete the following table for!!!=7!. First, we will define what square roots are and how you find the square root of a number. Looking for a primer on how to find the nth term of a geometric sequence? See how it's done with this free geometer's guide. The number that must be multiplied times itself n times to equal a given value. Discussion Suppose IDL is being used to find a real root of a number. Now I was thinking of adding the nth-Root of a Number. My current expression is -. 2 days ago \u00b7 He said that 12 out of the 18 persons who had already indicated interest in the governorship election in the state in 2021 were encouragingly from Anambra South, adding that the expectation was that with time, aspirants from the other two senatorial zones would withdraw to join hands in finding an ideal person from the South to emerge as the. W HEN ONE THING DEPENDS on another, as for example the area of a circle depends on the radius -- in the sense that when the radius changes, the area will change -- then we say that the first is a \"function\" of the other. The number that must be multiplied times itself n times to equal a given value. 0 *Real nth Roots of a. Nth Root using the POWER Function. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. Excel's powerful mathematical toolkit includes functions for square roots, cube roots, and even nth roots. This online calculator is set up specifically to calculate 4th root. 1 Evaluate nth Roots and Use Rational Exponents Goal p Evaluate nth roots and study rational exponents. Mathematical Connections Find the radius of the figure with the given volume. 32 = 32(cos0 + isin 0 ) in trig form. In the event you seek help on exam review or perhaps intermediate algebra syllabus, Polymathlove. Then, invoking the Intermediate Value Theorem, there is a root in the interval $[-2,-1]$. We find the the two roots by multiplying 2 by the square roots of unity. What is an \u201cnth Root?\u201d Extends the concept of square roots. It requires an initial guess, and then Newton-Raphson iterations are taken to improve that guess. To demonstrate that this simple method is fully applicable to roots of higher order, we shall attempt to find the 5th root of 2. What are the solution sets of the following 3 problems,the square root of x squared + 3x = x-3, the square root of 3x + 10 = x + 4 and finally the square root of 8x - 2 = the square root of 2x? Help much appriciated. Every real number has exactly one real. n5 3, a5 2 64 b. ^2 216 sqr root 3 in. In short, you can make use of the POWER function in Excel to find the nth root of any number. 1) the radicand has no perfect nth powers as factors 2) any denominator has been rationalized Example 4: Write the following Radicals in Simplest Form \u2014 Remember your perfect nth Roots u = 16. A formal mathematical definition might look something like: The nth roots of unity are the solutions to the equation x n = 1. Step 3: Write the answer using interval notation. ) ! n=5! a=\"32 b. Summing up: The following guidelines will be useful for finding the nth root of the perfect nth power of a two-digit or three-digit number. Let f(x) = 2 x - 1 and find its limit applying the difference and product theorems above lim x\u21925 f(x) = 2*5 - 1 = 9 We now apply theorem 5 since the square root of 9 is a real number. De Moivre's Theorem Now, in that same vein, if we can raise a complex number to a power, we should be able to find all of its roots too. Nth roots of unity 40 videos. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. You are given 2 numbers (N , M); the task is to find N\u221aM (Nth root of M). \u2022 Benchmark MA. Pow , mathematics. For, the square of a real root must be positive; and therefore the original equation cannot have more real roots than the transformed has positive roots. Schools Parents. C program to get nth bit of a number January 24, 2016 Pankaj C programming Bitwise operator , C , Program Write a C program to input any number from user and check whether n th bit of the given number is set (1) or not (0). In fact, there are seven 7th roots of unity, and each gold disc in that picture is one of them. How do i find the indicated real nth root of:? Find the indicated real nth root(s) of a - Help? A) find indicated roots of the complex number b) express roots in standard form?. Given two numbers N and A, find N-th root of A. During my son\u2019s math lesson today we got on the subject of , which I prefer to write as. The fundamental theorem of algebra can help you find imaginary roots. This function is pretty well behaved, so if you make a good guess about the solution you will get an answer, but if you make a bad guess, you may get the wrong root. Roots of unity have connections to many areas of mathematics, including the geometry of regular polygons, group theory, and number theory. This Algebra I: nth Root and Rational Exponents Worksheet is suitable for 8th - 11th Grade. Pow , mathematics. A \"root\" (or \"zero\") is where the polynomial is equal to zero:. 24 The nth root of a number b is designated as nb,. By definition, the nth root of a number can be calculated by raising that number the power of 1/n. They are also known as zeros. so you get 4 roots @Dick: I think he already used De Moiver's formula when he took the root and turned it into a multiplier of one-half and multiplied the term inside the cos and sin @paraboloid if you don't know what you need to do, just read you course book or ask your Lecturer or TA. As this problem involves. Example 2: Solve Equations Using nth Roots Solve the equation. 5 or 10 1/2 and the cube root as 10 0. The square roots of unity are 1 and \u2212 1. Plot your number r(cos +i sin), that you want to take the root of. a) As you go down the table of values, what number does X approach?. Setting up the Data. Given a complex number z = r(cos \u03b1 + i sin\u03b1), all of the nth roots of z are given by. 5847i\\) and \\(4. Improve your math knowledge with free questions in \"Nth roots\" and thousands of other math skills. Then we will apply similar ideas to define and evaluate nth roots. The resulting number from this expression means that if it\u2019s raised to the nth power it equals the. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. One real root:. nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. The second root is usually called the \"square root\". Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. For example, to find the cube root of 8 in a cell, type the following:. Examples:. Use 27 cis (90 degress), which is the trigonometric from of 27i, to do the problem. June 19, 2019 July 22, 2019 Craig Barton. The finite field has prime order. I'm trying to find the n-th root of unity in a finite field that is given to me. 0 Votes 18 Views I am unable to prove that :. is the radius to use. Divide radicals that have the same index number. Instead you need to pass 1/10 as n\u221ax = x(1/n). Unlike a square root function which is limited to nonnegative numbers, a cube root can use all real numbers because it is possible for three negatives to equal a negative. n th Root of an Integer Description Calculate the n th root of an integer. when x k +1 \u2265 x k. In general, a root of degree n is called an nth root. Your program should ask the user for a number, and a root and print that root of the number. You can always tell FindRoot to search for complex roots by adding 0. If n = 3 , then it is called as Cube root. During my son\u2019s math lesson today we got on the subject of , which I prefer to write as. An expression under a radical sign is called a radicand. Find the indicated real nth root(s) of a. Find the fourth roots of -16 and -16 is a complex number even though it is also real. n = 2, a = 100 24. The equation to calculating the nth term of the sequence is an+b; \"a\" is the fixed number that is being added to generate the series, and \"b\" gives the. n th Roots A square root of a number b , written b , is a solution of the equation x 2 = b. However, there is still one basic procedure that is missing from the algebra of complex numbers. 3 POLAR FORM AND DEMOIVRE\u2019S THEOREM At this point you can add, subtract, multiply, and divide complex numbers. The Help Article also provides a convenient wrapper function. We get the nth root of 1 is the regular nth root of 1, which is just 1, times e to the i phi over n. De Moivre's theorem can be extended to roots of complex numbers yielding the nth root theorem. Buy business plan online - best in texas, homework help on finding homework help us geography nth roots and rational exponents. , , , the six roots are () out of which , , and on the left s-plane are the roots of :. It need not be true that any of the fractions is actually a solution. Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. However, there is still one basic procedure that is missing from the algebra of complex numbers. (Buddy system! If there is no buddy, it must stay under the radical. Find the indicated real nth root(s) of a negative. De Moivre's theorem can be extended to roots of complex numbers yielding the nth root theorem. n = 5, a = 0. Because n = 4 is even and a = 81 > 0, 81. 2 days ago \u00b7 He said that 12 out of the 18 persons who had already indicated interest in the governorship election in the state in 2021 were encouragingly from Anambra South, adding that the expectation was that with time, aspirants from the other two senatorial zones would withdraw to join hands in finding an ideal person from the South to emerge as the. 3 POLAR FORM AND DEMOIVRE'S THEOREM 483 8. The 5th root of 1,024 is 4, as 4 x 4 x 4 x 4 x 4 is 1,204. Question 1: Simplify. When you\u2019re giving a time-bound exam like SSC CGL, SSC CPO, Railways Group D, RPF & ALP this can drain you of your precious time. You can see there are many roots to this equation, and we want to be sure we get the n^ {th} root. Nth root calculator. V = 216 ft3 Find the real solution(s) of the equation. There might be some issues with this. IXL Learning Learning. Use nth roots in problem solving Animal Population The population P of a certain animal species after t months can be modeled by P = C(1. In mathematics, an nth root of a number x, is a number r which, when raised to the power n yields x. The number 0 (zero) has just one square root, 0 itself. The nth root calculator below will also provide a brute force rounded approximation of the principal nth root. Simplify the expression. There is no result accuracy argument of Nth_Root, because the iteration is supposed to be monotonically descending to the root when starts at A. When I try to find the roots of the same equation in Mathematica, I receive various errors. 1) the radicand has no perfect nth powers as factors 2) any denominator has been rationalized Example 4: Write the following Radicals in Simplest Form \u2014 Remember your perfect nth Roots u = 16. Example of How to Evaluate a \u201cnth Root\u201d Radical Expression. I assume you're saying that the calculator can do the four basic operations and square roots, but not arbitrary exponents. In this case, we divided by a negative number, so had to reverse the direction of the inequality symbol. In particular, is called the primitive th root of unity. (I am aware that \\sqrt as command for a general root is misleading) For the life of me I can't find the right syntax to get the input of n'th root of x. pl help Shambhu. Number of roots equal to n. find a formula for the nth partial sum of the series and use it to find the series sum if the series converges. n = 3, a = 216 b. Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. Suppose we wish to solve the equation z^3 = 2+2 i. Complex Numbers: nth Roots.", "date": "2019-12-15 12:40:11", "meta": {"domain": "cbeu.pw", "url": "http://ufap.cbeu.pw/find-the-indicated-nth-roots-of-a.html", "openwebmath_score": 0.6347075700759888, "openwebmath_perplexity": 415.6410076463367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9905874095787602, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8624026879101964}}
{"url": "https://math.stackexchange.com/questions/1372479/double-integral-set-up", "text": "# Double Integral Set Up\n\nThe question was stated as follows,\n\nEvaluate the following double integral;\n\n$$\\iint_R x^3y dA$$\n\nwhere R is interior of triangle with vertices (0,0), (1,0), & (1,1) .\n\nI thought for these types of double integrals I could have the limits in either of the two formats below;\n\n$$\\int_0^1\\int_0^y x^3y dxdy$$ [1]\n\nOR\n\n$$\\int_0^1\\int_0^x x^3y dydx$$ [2]\n\nHowever the answer sheet states that the answer for only the latter integral is correct.\n\nHow can I set up or visualize the problem in order to set it up in the correct way?\n\nI did plot the points on a graph, and got my y=x \"limit\" from there, I just cannot understand why [2] is the correct integral, and how to go about making sure that in every double integral problem, I choose the correct integral set up.\n\nAll help is much appreciated :)\n\nFor the first one the bounds on x are wrong. The region is bounded by $y=x$ on the left and on the right by $x=1$. The bound on the left is a lower bound, so it should be\n\n$$\\int_0^1 \\int_y^1 x^3ydxdy$$\n\n\u2022 That makes sense! Thank you. \u2013\u00a0mnmakrets Jul 24 '15 at 14:10\n\nYou can draw the triangle.\n\nCall the region inside of it $\\Omega$.\n\nLooking at lines of constant $y$, we'll have that $$\\Omega = \\{ (x,y) \\in \\mathbb{R}^2 \\ | \\ 0 \\leq x \\leq 1, y \\leq x \\leq 1 \\}$$\n\nDraw the triangle. Now think of it this way: if I fix $x$, what will be the extremes of integration for $y$? You'll get something that depends on $x$ and in particular $y$ goes from $0$ to $x$. Then integrate over the values taken on by $x$, and yo get the second form.\n\nAlternatively, you can fix $y$ and notice that $x$ then varies from $y$ to $1$; then integrate over $y$ to get the correct form\n\n$$\\int_{0}^1 \\int_y^1 x^3y dy dx$$\n\nUsually you draw the area you want to integrate on, and draw the horizontal and vertical lines that corresponds to fixing $x$ or $y$. This helps visualize the problem and find the extremes of integration with ease\n\nAccording fubini's theorem, you can either integrate parallel to the x-axis or the y-axis and hence there are two possibilities here:\n\n$$\\int_0^1 \\int_0^x x^3y \\ dy \\ dx$$\n\nor\n\n$$\\int_0^1 \\int_y^1 x^3y \\ dx \\ dy$$\n\nThe best way to get to grips with these sort of questions is to practise more. For more, visit the following links:", "date": "2019-10-20 20:08:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1372479/double-integral-set-up", "openwebmath_score": 0.9489356279373169, "openwebmath_perplexity": 128.74732942344113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799420543366, "lm_q2_score": 0.894789473818021, "lm_q1q2_score": 0.8623801472271626}}
{"url": "https://mathematica.stackexchange.com/questions/22543/how-do-i-obtain-the-enclosed-area-of-this-particular-parametric-plot/22545", "text": "# How do I obtain the enclosed area of this particular parametric plot?\n\nI'm trying to find a way to obtain the enclosed area of this particular plot. Can someone show me how?\n\ncurveplot = ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]],\nSqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 \u03c0},\nPlotStyle -> {{Thickness[0.01], Darker[Purple]}},\nAspectRatio -> Automatic, PlotRange -> All, AxesLabel -> {\"x\", \"y\"}]\n\n\n## 6 Answers\n\nYou can get the curve in polynomial implicit form as below.\n\npoly =\nGroebnerBasis[{x^2 - ct, y^2 - st, ct^2 + st^2 - 1}, {x, y}, {ct,\nst}][[1]]\n\n(* Out[290]= -1 + x^4 + y^4 *)\n\n\nTo get the area, integrate the characteristic function for the interior of the region. That that's where the polynomial is nonpositive (just notice that it is negative at the origin, say).\n\narea = Integrate[Boole[poly <= 0], {x, -2, 2}, {y, -2, 2}]\n\n(* Out[292]= (2 Gamma[1/4] Gamma[5/4])/Sqrt[\u03c0] *)\n\nN[area]\n\n(* Out[293]= 3.7081493546 *)\n\n\nThere are other ways to do this if you cannot find an implicit form, but this seems most direct in this case.\n\n--- edit ---\n\nIf you can just solve separately for x and y in terms of the parameter t then you can set up a region function. I do this below for the positive quadrant, and take advantage of symmetry to get the full area in approximate form.\n\nreg = Function[{x, y},\nIf[And @@ {0 <= x <= 1, 0 <= y <= 1, y <= Sqrt[Sin[ArcCos[x^2]]]},\n1, 0]];\n\napproxarea = 4*NIntegrate[reg[x, y], {x, 0, 1}, {y, 0, 1}]\n\n(* Out[321]= 3.70814937167 *)\n\n\nOne can actually recover the exact area from this by using Integrate instead of NIntegrate. But this seems like a viable approach in situations where the exact value might not be readily computed.\n\n--- end edit ---\n\n--- edit 2 ---\n\nHere is a Monte Carlo method that does not rely on solving for anything. We extract the line segments, augment with a diagonal, and do some magic.\n\nsegs = Cases[curveplot, _Line, Infinity][[1, 1]];\nsegs = {Join[segs, N[Table[{j, 1 - j}, {j, 0, 1, 1/100}]]]};\n\n\nI added an extra level of List due to requirements of some further code. First let's reform a line to take a look at this region.\n\nGraphics[Apply[Line, segs]]\n\n\nNow we create an in-out function, generate a bunch of random points in the unit square of the first quadrant, take a Monte-Carlo approximation of this area. Then multiply by 4 and add 2. Why? because that's what one always does-- it's like selecting \"c\" when we don't know the multiple choice answer. (Okay, we multiply by 4 to account for all quadrants, and add 2 because we have in effect excised a square of side length $\\sqrt{2}$ from the full region.)\n\nTo create the in-out function I use code directly from here.\n\nnbins = 100;\nTiming[{{xmin, xmax}, {ymin, ymax}, segmentbins} =\npolyToSegmentList[{segs[[1]]}, nbins];]\n\n(* Out[414]= {0.040000, Null} *)\n\nlen = 100000;\npts = RandomReal[1, {len, 2}];\nTiming[\ninout = Map[pointInPolygon[#, segmentbins, xmin, xmax, ymin, ymax] &,\npts];]\napproxarea = 4.*Length[Cases[inout, True]]/len + 2.\n\n(* Out[419]= {2.750000, Null} *)\n\n(* Out[420]= 3.7092 *)\n\n\nI would imagine one could do a bit better by integrating just the unit square in the first quadrant with Method -> \"QuasiMonteCarlo\", sowing the points via the EvaluationMonitor option, and using those instead of the random set above. This will give a low-discrepancy sequence. Or generate such a set directly; bit offhand I don't know how to do that.\n\n-- end edit 2 ---\n\n\u2022 Thanks, I wanted to see if this checked out with a monte carlo estimate. \u2013\u00a0Black Milk Apr 2 '13 at 21:55\n\nSince it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lam\u00e9 curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration:\n\npts = First[Cases[\nParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]},\n{t, 0, 2 \u03c0}, Exclusions -> None, Method -> {MaxBend -> 1.},\nPlotPoints -> 100] // Normal, Line[l_] :> l, \u221e]];\n\nPolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, 1]]/2\n\nPolygonSignedArea[pts]\n3.7081447086368127\n\n\nThe value thus obtained is pretty close to the results in the other answers.\n\nSurprisingly, there is in fact an undocumented built-in function for computing the area of a polygon:\n\n(* $VersionNumber < 10. *) GraphicsMeshMeshInit[]; PolygonArea[pts] 3.708144708636812 (*$VersionNumber >= 10. *)\nGraphicsPolygonUtilsPolygonArea[pts]\n3.708144708636812\n\n\nBut, what you really should know is that Lam\u00e9 curves have been well studied, and there is in fact a closed form expression for the area of a Lam\u00e9 curve. Given the Cartesian equation\n\n$$\\left|\\frac{x}{a}\\right|^r+\\left|\\frac{y}{b}\\right|^r=1$$\n\nthe formula for the area of a Lam\u00e9 curve (formula 5 here) is\n\n$$A=\\frac{4^{1-\\tfrac1{r}}ab\\sqrt\\pi\\;\\Gamma\\left(1+\\tfrac1{r}\\right)}{\\Gamma\\left(\\tfrac1{r}+\\tfrac12\\right)}$$\n\nIn particular, for the OP's specific case, $a=b=1$, and $r=4$. Thus,\n\nWith[{a = 1, b = 1, r = 4},\nN[(4^(1 - 1/r) a b Sqrt[\u03c0] Gamma[1 + 1/r])/Gamma[1/r + 1/2], 20]]\n3.7081493546027438369\n\n\nThis is the same as the answer Daniel obtained through more general methods.\n\n\u2022 Never heard of that method before. Nice. (Yes, I upvoted..) \u2013\u00a0Daniel Lichtblau Apr 3 '13 at 15:42\n\u2022 I've taught the \"shoelace\" method but never heard of the name. (+1) \u2013\u00a0Michael E2 Apr 3 '13 at 20:32\n\u2022 @J.M. Thank you for the additional insight. \u2013\u00a0Black Milk Apr 3 '13 at 20:55\n\u2022 @Michael, I must confess that I too have been using that method for a while now, but only learned it had a name on math.SE ... \u2013\u00a0J. M.'s technical difficulties Apr 4 '13 at 2:37\n\u2022 You could just do PolygonArea@pts... no need to wrap it in Polygon \u2013\u00a0rm -rf Apr 25 '13 at 21:19\n\nOne can use one of the line integral forms of the area, derived from Green's Theorem: $$A = \\frac12 \\int_C x \\; dy - y \\; dx = \\int_C x \\; dy = - \\int_C y \\; dx$$ The first one is symmetric, which sometimes is an advantage.\n\nc[t_] := {Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}\n\ndA = 1/2 c'[t].Cross[c[t]]\n(* complicated output *)\n\n\nOne problem with this parametrization are the derivatives of Abs and Sign. They are discontinuous at isolated points, and as far as the integral is concerned, it does not matter what value we assign them at the discontinuities. So we can simplify matters by substituting for them. We can also substitute 1 for Sign[x]^2, since x will be 0 only at few isolated points. Thus the differential is\n\ndA = dA /. {Sign'[x_] :> 0, Abs'[x_] :> Sign[x]} /. {Sign[_]^2 :> 1} // Simplify\n(* (Abs[Cos[t]] Cos[t] Sign[Cos[t]] + Abs[Sin[t]] Sign[Sin[t]] Sin[t]) /\n(4 Sqrt[Abs[Cos[t]]] Sqrt[Abs[Sin[t]]]) *)\n\n\nNIntegrate returns a small imaginary component\n\nNIntegrate[dA, {t, 0, 2 \u03c0}]\n(* 3.70815 - 1.97076*10^-10 I *)\n\n\nOddly, setting WorkingPrecision reduces the imaginary error, even if it is set to less than MachinePrecision (15.9546)\n\nNIntegrate[dA, {t, 0, 2 \u03c0}, WorkingPrecision -> 10]\n(* 3.708149355 + 0.*10^-21 I *)\n\n\nIntegrate returns an exact answer in this case:\n\nIntegrate[dA, {t, 0, 2 \u03c0}]\n(* (3 Sqrt[2] \u03c0 Gamma[5/4] + 4 Gamma[3/4] Gamma[5/4]^2) /\n(2 Sqrt[\u03c0] Gamma[3/4]) *)\n\nFullSimplify @ %\n(* (Sqrt[\u03c0/2] Gamma[1/4])/Gamma[3/4] *)\n\nN[%, 20]\n(* 3.7081493546027438369 *)\n\n\nIn cases where it is not possible to get a closed form for the curve, you can use the image processing functions to get an approximation for the enclosed area. First, some small changes to the plot \u2014 get rid of the axes, labels and everything else that isn't needed, and set the aspect ratio to 1.\n\ncurveplot = ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]},\n{t, 0, 2 \u03c0}, PlotStyle -> Thick, AspectRatio -> 1, Axes -> False, PlotRange -> {-1, 1}]\n\n\nNext, close the holes in the curve using morphological operations:\n\nim = Binarize@curveplot ~Opening~ 7 // DeleteSmallComponents\n\n\nDepending on the curve, you'll have to tweak the parameters and be careful with DeleteSmallComponents, in case you have disconnected components in your plot (always do a visual check or ImageAdd[curveplot, im] to see if it's correct). Sometimes, this may even not be necessary if your curve is \"nice\".\n\nFinally, use ComponentMeasurements to get the area of the enclosed space. The result is in sq. pixels, so I normalize by the total area (in sq. pixels) and multiply by the area of the plot range rectangle in the original plot (i.e., $(x_{max}-x_{min})(y_{max}-y_{min})$)\n\n4 (1 /. ComponentMeasurements[im, \"Area\"]) / Times @@ ImageDimensions@im\n(* 3.66738 *)\n\n\nwhich looks about right, since your curve fits in a square of side 2 and is close to Daniel's answer of 3.708. You can get a closer approximation if you increase the ImageSize in the original plot. Using ImageSize -> 2000 gives me 3.70129 as the area (but note that the image processing steps will take longer to compute).\n\n\u2022 Exclusions -> None in ParametricPlot[] should close up the artifact holes nicely. \u2013\u00a0J. M.'s technical difficulties Apr 3 '13 at 15:27\n\nIntegrating InterpolatingFunction.\n\np = Table[{\nSign[Cos@t] Sqrt[Abs@Cos@t],\nSign[Sin@t] Sqrt[Abs@Sin@t]}, {t, 0, Pi, Pi/100.}];\nf = Interpolation[p];\n2*NIntegrate[f@t, {t, -1, 1}]\n3.70771\n\n\nAs a modified version of Michael E2's answer:\n\nI tried to rewrite your original curve as below for $$t\\in \\left[0,\\tfrac{\\pi}{2}\\right]$$, in order to make sure the derivatives of the parametric form can be obtained easily by avoiding Abs or Sign:\n\nncurve = {(Cos[t]^2 )^(1/4), (Sin[t]^2)^(1/4)}\n\n\nThen the result of the closed area can be obtained by applying Green's theorem to the curve and by using the symmetry of the curve:\n\narea = 4*Integrate[ncurve[[1]] D[ncurve[[2]], t], {t, 0, Pi/2}]\n\n\nwhich gives:\n\n$$\\dfrac{8\\; \\Gamma^2 \\left(\\tfrac{5}{4}\\right)}{\\sqrt{\\pi }}$$\n\nand the numerical result is therefore:\n\narea // N[#, 20] &\n`\n\n3.7081493546027438369", "date": "2020-08-09 08:32:24", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/22543/how-do-i-obtain-the-enclosed-area-of-this-particular-parametric-plot/22545", "openwebmath_score": 0.5415024757385254, "openwebmath_perplexity": 1700.5332287365466, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799430946808, "lm_q2_score": 0.8947894654011352, "lm_q1q2_score": 0.862380140046026}}
{"url": "https://math.stackexchange.com/questions/1682610/what-is-the-difference-between-tan-x-sec2x-and-sin-x-cos3x-why-is", "text": "# What is the difference between ($\\tan x \\sec^2x$) and ($\\sin x/\\cos^3x$)? Why is the answer to the integration different?\n\n$$\\int \\:\\frac{\\left(\\sin x+\\tan x\\right)}{3\\cos^2x}dx$$\n\nI know I have to split the equation into\n\n$$\\frac{1}{3}\\int \\:\\left(\\:\\frac{\\sin x}{\\cos x}\\right)\\left(\\frac{1}{\\cos x}\\right)dx+\\frac{1}{3}\\int \\:\\left(\\:\\tan x\\right)\\left(\\frac{1}{\\cos^2x}\\right)dx$$\n\nI know that for the first part, it is $$\\frac{1}{3}\\int \\tan x\\sec xdx$$ which is $$\\sec x$$.\n\nHowever, for the second part, wouldn't it be $$\\frac{1}{3}\\int \\tan x \\sec^2xdx$$\n\nIf I used $$u=\\tan x$$ then $$du=\\sec^2xdx$$ so wouldn't the answer be $$\\frac{1}{6}\\tan^2x$$\n\nHowever, the book is saying that the second part is supposed to be $$\\frac{1}{6}\\sec^2x$$ because I was supposed to convert the second part into $$\\frac{1}{3}\\int \\frac{\\sin x}{\\cos^3x}dx$$ and let $$u=\\cos x$$\n\nWhat I am doing wrong? Why can't it be $$\\tan \\sec^2x$$ instead of $$\\sin x/\\cos^3x$$?\n\n\u2022 Remember $\\sec^2{\\theta}=1+\\tan^2{\\theta}$ \u2013\u00a0user41736 Mar 4 '16 at 6:49\n\nRecall that\n\n$$1 + \\tan^2 x = \\sec^2 x$$\n\nor, since I dislike the secant,\n\n$$\\frac{1}{\\cos^2 x} = \\frac{\\sin^2 x + \\cos^2 x}{\\cos^2 x} = 1+\\tan^2 x.$$\n\nTherefore, your answer and the book's answer only differ by a constant.\n\nBoth the answers differ by a constant(1/6). So both the answers and both the methods are correct. The constant of integration takes care of the constant(1/6).\n\nBoth are right as when you differentiate $1/6tan^2x$ with chain rule you get $\\frac{1}{6}.2tanx.sec^2x=1/3.tanx.sec^2x$ also differentiating $1/6sec^2x$ you get the same answer . so both are right.\n\n\u2022 Note the constants can be $c,c'$ as integration gives a family of curves and an exact constant gives a specific curve \u2013\u00a0Archis Welankar Mar 4 '16 at 6:48", "date": "2019-10-22 19:06:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1682610/what-is-the-difference-between-tan-x-sec2x-and-sin-x-cos3x-why-is", "openwebmath_score": 0.9179520010948181, "openwebmath_perplexity": 171.8629544842366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971992476960077, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8623561920668825}}
{"url": "https://spot.pcc.edu/slc/mathresources/output/html/radicals-rational-exponents.html", "text": "## Section13.8Rational Exponents\n\nThe power to a power rule of exponents relates that $(x^m)^n=x^{mn}\\text{.}$ This rule is fairly intuitive when both exponents are positive. For example, in the expression $(x^4)^3$ there are three factors of $x^4\\text{,}$ each of which contains four factors of $x\\text{,}$ so all together there are four factors of $x\\text{,}$ three times, i.e. $3 \\cdot 4$ factor of $x\\text{.}$\n\nWhile the power to a power rule is less intuitive once you move away from positive integer exponents, the rule remains the same regardless of the nature of the exponents. For example:\n\n\\begin{align*} (x^{1/3})^3\\amp=x^{\\frac{1}{3} \\cdot 3}\\\\ \\amp=x^{1}\\\\ \\amp=x \\end{align*}\n\nBut we already have a name for the expression that when cubed results in $x\\text{,}$ and that name is $\\sqrt[3]{x}$ (the cube root of $x$). So it must be the case that $x^{1/3}=\\sqrt[3]{x}\\text{.}$ In general, is $n$ is any positive integer, then:\n\n\\begin{equation*} x^{1/n}=\\sqrt[n]{x} \\end{equation*}\n\nand more generally,\n\n\\begin{equation*} x^{m/n}=\\sqrt[n]{x^m}\\text{.} \\end{equation*}\n\nSeveral examples are shown below.\n\n###### Example13.8.1.\n\nExpress $y^{7/5}$ as an equivalent radical expression\n\nSolution\n\\begin{equation*} y^{7/5}=\\sqrt[5]{y^7} \\end{equation*}\n###### Example13.8.2.\n\nExpress $\\sqrt[3]{w^{12}}$ using an equivalent exponential expression\n\nSolution\n\\begin{align*} \\sqrt[3]{w^{12}}\\amp=w^{12/3}\\\\ \\amp=w^4 \\end{align*}\n###### Example13.8.3.\n\nExpress $\\sqrt{x^9}$ using an equivalent exponential expression\n\nSolution\n\\begin{equation*} \\sqrt{x^9}=x^{9/2} \\end{equation*}\n\nYou can use Figure\u00a013.8.4 to explore this definition some more.\n\nAs long as both the numerator and denominator of a rational exponent are fairly small positive numbers, it is fairly easy to evaluate expressions that include rational exponents using the rule $x^{m/n}=\\sqrt[n]{x^m}\\text{.}$\n\n###### Example13.8.5.\n\nEvaluate $16^{1/2}\\text{.}$\n\nSolution\n\\begin{align*} 16^{1/2}\\amp=\\sqrt{16}\\\\ \\amp=4 \\end{align*}\n###### Example13.8.6.\n\nEvaluate $8^{2/3}\\text{.}$\n\nSolution\n\\begin{align*} 8^{2/3}\\amp=\\sqrt[3]{8^2}\\\\ \\amp=\\sqrt[3]{64}\\\\ \\amp=4 \\end{align*}\n###### Example13.8.7.\n\nEvaluate $100^{3/2}\\text{.}$\n\nSolution\n\\begin{align*} 100^{3/2}\\amp=\\sqrt{100^3}\\\\ \\amp=\\sqrt{1000000}\\\\ \\amp=1000 \\end{align*}\n\nWhen the numerator of the rational exponent is large, the rule $x^{m/n}=\\sqrt[n]{x^m}$ can become quite cumbersome. Consider, for example, evaluating $9^{5/2}\\text{.}$ If we try to use the standard form we hit a brick wall. First, it's not trivial to calculate that $9^5=59,049$ (reality check ... I grabbed my calculator). Now that I have the value of 59,049, I have to determine its square root. Oh my!\n\nFortunately for us, the application of the exponent and the application of the radical can be done in either order. That is:\n\n\\begin{equation*} a^{m/n}=\\sqrt[n]{x^m} \\text{ and } a^{m/n}=(\\sqrt[n]{x})^m \\end{equation*}\n###### Example13.8.8.\n\nUsing the second option, evaluate $9^{5/2}\\text{.}$\n\nSolution\n\\begin{align*} 9^{5/2}\\amp=(\\sqrt{9})^5\\\\ \\amp=3^5\\\\ \\amp=243 \\end{align*}\n###### Example13.8.9.\n\nUsing the second option, evaluate $16^{7/4}\\text{.}$\n\nSolution\n\\begin{align*} 16^{7/4}\\amp=(\\sqrt[4]{16})^7\\\\ \\amp=2^7\\\\ \\amp=128 \\end{align*}\n\nRational exponents are allowed to be negative. If that's the case, you probably want to deal with the negative aspect of the exponent before taking on the fractional aspect.\n\n###### Example13.8.10.\n\nEvaluate $27^{-2/3}\\text{.}$\n\nSolution\n\\begin{align*} 27^{-2/3}\\amp=\\frac{1}{27^{2/3}}\\\\ \\amp=\\frac{1}{(\\sqrt[3]{27})^2}\\\\ \\amp=\\frac{1}{3^2}\\\\ \\amp=\\frac{1}{9} \\end{align*}\n\nSometimes radical expressions can be simplified after first rewriting the expressions using rational exponents and applying the appropriate rules of exponents. If the resultant expression still has a rational exponent, it is standard to convert back to radical notation. Several examples follow.\n\n###### Example13.8.11.\n\nUse rational exponents to simplify $\\text{.}$ Where appropriate, your final result should be converted back to radical form.\n\nSolution\n\\begin{align*} \\sqrt[3]{y^2} \\cdot \\sqrt[6]{y}\\amp=y^{2/3}y^{1/6}\\\\ \\amp=y^{2/3+1/6}\\\\ \\amp=y^{5/6}\\\\ \\amp=\\sqrt[6]{y^5} \\end{align*}\n###### Example13.8.12.\n\nUse rational exponents to simplify $\\sqrt[8]{t^4}\\text{.}$ Where appropriate, your final result should be converted back to radical form.\n\nSolution\n\\begin{align*} \\sqrt[8]{t^4}\\amp=t^{4/8}\\\\ \\amp=t^{1/2}\\\\ \\amp=\\sqrt{t} \\end{align*}\n###### Example13.8.13.\n\nUse rational exponents to simplify $\\sqrt[10]{\\sqrt{5^{40}}}\\text{.}$ Where appropriate, your final result should be converted back to radical form.\n\nSolution\n\\begin{align*} \\sqrt[10]{\\sqrt{5^{40}}}\\amp=\\sqrt[10]{5^{40/2}}\\\\ \\amp=\\sqrt[10]{5^{20}}\\\\ \\amp=5^{20/10}\\\\ \\amp=5^2\\\\ \\amp=25 \\end{align*}\n\n### ExercisesExercises\n\nConvert each exponential expression to a radical expression and each radical expression to an exponential expression. When converting to a rational exponent, reduce the exponent if possible. Assume that all variables represent positive values.\n\n###### 1.\n\n$x^{1/3}$\n\nSolution\n\n$x^{1/3}=\\sqrt[3]{x}$\n\n###### 2.\n\n$y^{5/4}$\n\nSolution\n\n$y^{5/4}=\\sqrt[4]{y^5}$\n\n###### 3.\n\n$z^{2/5}$\n\nSolution\n\n$z^{2/5}=\\sqrt[5]{z^2}$\n\n###### 4.\n\n$\\sqrt[11]{x^5}$\n\nSolution\n\n$\\sqrt[11]{x^5}=x^{5/11}$\n\n###### 5.\n\n$\\sqrt[4]{y^{20}}$\n\nSolution\n\n\\begin{aligned}[t] \\sqrt[4]{y^{20}}\\amp=y^{20/4}\\\\ \\amp=y^5 \\end{aligned}\n\n###### 6.\n\n$\\sqrt[15]{t^3}$\n\nSolution\n\n\\begin{aligned}[t] \\sqrt[15]{t^3}\\amp=t^{3/15}\\\\ \\amp=t^{1/5} \\end{aligned}\n\nDetermine the value of each expression.\n\n###### 7.\n\n$4^{1/2}$\n\nSolution\n\n\\begin{aligned}[t] 4^{1/2}\\amp=\\sqrt{4}\\\\ \\amp=2 \\end{aligned}\n\n###### 8.\n\n$27^{-1/3}$\n\nSolution\n\n\\begin{aligned}[t] 27^{-1/3}\\amp=\\frac{1}{27^{1/3}}\\\\ \\amp=\\frac{1}{\\sqrt[3]{27}}\\\\ \\amp=\\frac{1}{3} \\end{aligned}\n\n###### 9.\n\n$\\left(\\frac{4}{9}\\right)^{-1/2}$\n\nSolution\n\n\\begin{aligned}[t] \\left(\\frac{4}{9}\\right)^{-1/2}\\amp=\\left(\\frac{9}{4}\\right)^{1/2}\\\\ \\amp=\\sqrt{\\frac{9}{4}}\\\\ \\amp=\\frac{3}{2} \\end{aligned}\n\n###### 10.\n\n$8^{7/3}$\n\nSolution\n\n\\begin{aligned}[t] 8^{7/3}\\amp=(\\sqrt[3]{8})^7\\\\ \\amp=2^7\\\\ \\amp=128 \\end{aligned}\n\n###### 11.\n\n$100^{5/2}$\n\nSolution\n\n\\begin{aligned}[t] 100^{5/2}\\amp=(\\sqrt{100})^5\\\\ \\amp=10^5\\\\ \\amp=100,000 \\end{aligned}\n\n###### 12.\n\n$16^{-9/4}$\n\nSolution\n\n\\begin{aligned}[t] 16^{-9/4}\\amp=\\frac{1}{16^{9/4}}\\\\ \\amp=\\frac{1}{(\\sqrt[4]{16})^9}\\\\ \\amp=\\frac{1}{2^9}\\\\ \\amp=\\frac{1}{512} \\end{aligned}\n\nSimplify each radical expression after first rewriting the expression in exponential form. Assume that all variables represent positive values. Where appropriate, your final result should be converted back to radical form.\n\n###### 13.\n\n$\\sqrt[5]{t^{20}}$\n\nSolution\n\n\\begin{aligned}[t] \\sqrt[5]{t^{20}}\\amp=t^{20/5}\\\\ \\amp=t^4 \\end{aligned}\n\n###### 14.\n\n$6\\sqrt[33]{x^{77}}$\n\nSolution\n\n\\begin{aligned}[t] 6\\sqrt[33]{x^{77}}\\amp=6x^{77/33}\\\\ \\amp=6x^{7/3}\\\\ \\amp=6\\sqrt[3]{x^7} \\end{aligned}\n\n###### 15.\n\n$(\\sqrt{3})^{10}$\n\nSolution\n\n\\begin{aligned}[t] (\\sqrt{3})^{10}\\amp=3^{10/2}\\\\ \\amp=3^5\\\\ \\amp=243 \\end{aligned}\n\n###### 16.\n\n$\\sqrt[4]{9^2}$\n\nSolution\n\n\\begin{aligned}[t] \\sqrt[4]{9^2}\\amp=9^{2/4}\\\\ \\amp=9^{1/2}\\\\ \\amp=\\sqrt{9}\\\\ \\amp=3 \\end{aligned}\n\n###### 17.\n\n$\\sqrt{w}\\sqrt[4]{w}$\n\nSolution\n\n\\begin{aligned}[t] \\sqrt{w}\\sqrt[4]{w}\\amp=w^{1/2}w^{1/4}\\\\ \\amp=w^{3/4}\\\\ \\amp=\\sqrt[4]{w^3} \\end{aligned}\n\n###### 18.\n\n$\\sqrt[7]{x^6}\\sqrt[7]{x}$\n\nSolution\n\n\\begin{aligned}[t] \\sqrt[7]{x^6}\\sqrt[7]{x}\\amp=x^{6/7}x^{1/7}\\\\ \\amp=x^1\\\\ \\amp=x \\end{aligned}\n\n###### 19.\n\n$(\\sqrt[12]{x^7y^{16}})^{36}$\n\nSolution\n\n\\begin{aligned}[t] (\\sqrt[12]{x^7y^{15}})^{36}\\amp=(x^7y^{16})^{36/12}\\\\ \\amp=(x^7y^{16})^3\\\\ \\amp=x^{21}y^{48} \\end{aligned}\n\n###### 20.\n\n$\\sqrt[5]{\\sqrt[3]{x^{15}}}$\n\nSolution\n\n\\begin{aligned}[t] \\sqrt[15]{\\sqrt[3]{x^{15}}}\\amp=\\sqrt[15]{x^{15/3}}\\\\ \\amp=\\sqrt[15]{x^5}\\\\ \\amp=x^{5/15}\\\\ \\amp=x^{1/3}\\\\ \\amp=\\sqrt[3]{x} \\end{aligned}", "date": "2019-12-07 19:53:02", "meta": {"domain": "pcc.edu", "url": "https://spot.pcc.edu/slc/mathresources/output/html/radicals-rational-exponents.html", "openwebmath_score": 0.9599544405937195, "openwebmath_perplexity": 5752.527971615653, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9925393560946886, "lm_q2_score": 0.8688267660487573, "lm_q1q2_score": 0.8623447589318642}}
{"url": "https://www.intmath.com/forum/exponents-radicals-16/exponential-functions:101", "text": "IntMath Home \u00bb Forum home \u00bb Exponents and Radicals \u00bb exponential functions\n\n# exponential functions [Solved!]\n\n### My question\n\nIn expressions containing raised variables that are being divided, why do the variables not cancel each other out to a reciprocal of one? for example x^n/ x should be 1^n or is it still x^n?\n\n### Relevant page\n\nfactoring polynomials with exponents - Google Search\n\n### What I've done so far\n\n(x^n/ x^n-2)^3 does this equal 1 or x^6?\n\nX\n\nIn expressions containing raised variables that are being divided, why do the variables not cancel each other out to a reciprocal of one? for example x^n/ x should be 1^n or is it still x^n?\nRelevant page\n\n<a href=\"https://www.google.com/search?q=factoring+polynomials+with+exponents&oq=factoring+polynomials+with+exponents&aqs=chrome..69i57.14377j0j4&sourceid=chrome&ie=UTF-8\">factoring polynomials with exponents - Google Search</a>\n\nWhat I've done so far\n\n(x^n/ x^n-2)^3 does this equal 1 or x^6?\n\n## Re: exponential functions\n\nHello Sam\n\nIt's often best to try some real numbers in algebraic expressions first, to see what is going on.\n\nLet's try 10^4.\n\nThis means 10 xx 10 xx 10 xx 10\n\nIf we divide this by 10, we would have\n\n(10 xx 10 xx 10 xx 10)/10\n\nWe cancel one of the 10s on top with the 10 on bottom, to give:\n\n10 xx 10 xx 10\n\nSo what we've done is\n\n(10^4)/10 = 10^3\n\nIf you do this for several other indices, you'll hopefully conclude that:\n\n(10^n)/10 = 10^(n-1)\n\nIn general, for your first question, we'd have:\n\nx^n/ x = x^(n-1)\n\nFor your other question, (x^n/ x^n-2)^3, I'm not sure if you mean (x^n/ x^n-2)^3 or (x^n/ (x^n-2))^3?\n\nYou are encouraged to use the math entry system which makes it easier to express your questions and follow your working.\n\nX\n\nHello Sam\n\nIt's often best to try some real numbers in algebraic expressions first, to see what is going on.\n\nLet's try 10^4.\n\nThis means 10 xx 10 xx 10 xx 10\n\nIf we divide this by 10, we would have\n\n(10 xx 10 xx 10 xx 10)/10\n\nWe cancel one of the 10s on top with the 10 on bottom, to give:\n\n10 xx 10 xx 10\n\nSo what we've done is\n\n(10^4)/10 = 10^3\n\nIf you do this for several other indices, you'll hopefully conclude that:\n\n(10^n)/10 = 10^(n-1)\n\nIn general, for your first question, we'd have:\n\nx^n/ x = x^(n-1)\n\nFor your other question, (x^n/ x^n-2)^3, I'm not sure if you mean (x^n/ x^n-2)^3 or (x^n/ (x^n-2))^3?\n\nYou are encouraged to use the <a href=\"http://www.intmath.com/forum/entering-math-graphs-images-41/how-to-enter-math:91\">math entry system</a> which makes it easier to express your questions and follow your working.\n\n## Re: exponential functions\n\nSam didn't reply.\n\nIf his second question meant (x^n/ x^n-2)^3, the the result is (1-2)^3=(-1)^3=-1\n\nIf he meant (x^n/ (x^n-2))^3, then there is no \"nice\" simplification.\n\nX\n\nSam didn't reply.\n\nIf his second question meant (x^n/ x^n-2)^3, the the result is (1-2)^3=(-1)^3=-1\n\nIf he meant (x^n/ (x^n-2))^3, then there is no \"nice\" simplification.\n\n## Reply\n\nYou need to be logged in to reply.", "date": "2017-03-28 19:42:11", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/exponents-radicals-16/exponential-functions:101", "openwebmath_score": 0.3553961515426636, "openwebmath_perplexity": 4613.774095610024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759660443166, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8623339271109621}}
{"url": "http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron/11086", "text": "Height of a tetrahedron\n\nHow do I calculate: The height of a regular tetrahedron, side length 1.\n\nJust to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.\n\n-\nI'm curious. If you mind telling us, what was the purpose of the interview? What were the other mathematically relevant questions asked? \u2013\u00a0 user02138 Nov 20 '10 at 15:25\nIt's an old question for and Economics and Management interview at Oxford. \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 16:43\n\nThe first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.\n\nHere's a view of the geometry:\n\nand here's a view of the bottom face:\n\nIn the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.\n\nKnowing that the short sides of the isosceles triangle bisect the 60\u00b0 angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30\u00b0, 30\u00b0 and 120\u00b0.\n\nUsing the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):\n\n$$1=2\\ell^2-2\\ell^2\\cos 120^{\\circ}$$\n\nSolving for $\\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\\frac{1}{\\sqrt{3}}$, as indicated here.\n\nFrom this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies\n\n$$h^2+\\left(\\frac{1}{\\sqrt{3}}\\right)^2=1$$\n\nSolving for $h$ in the above equation, we now find the height to be $\\sqrt{\\frac23}=\\frac{\\sqrt{6}}{3}$, as mentioned here.\n\n-\nIt's not homework, it was an interview question I struggled with. I'd love to know the answer. \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 14:57\nGood to know. I'll edit with an explicit answer. \u2013\u00a0 \uff2a. \uff2d. Nov 20 '10 at 15:00\nThank-you very much. P.S: how did you make your lovely 3d shape diagrams? \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 16:43\n@Patrick: I used Mathematica 5.2 (yes, it's an old copy :) ) for these diagrams. \u2013\u00a0 \uff2a. \uff2d. Nov 20 '10 at 21:33\n\nConsider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\\frac{2}{\\sqrt 3}$, and since the edges of this tetrahedron have length $\\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \\sqrt{\\frac{2}{3}}$.\n\n-\nNote that this way of doing this also gives that the distance from the centers of opposite sides of the tetrahedron is always ${1 \\over \\sqrt{2}}$ times the side length. \u2013\u00a0 Zarrax Nov 20 '10 at 15:26\n\nYou can also use trig based on the dihedral angle between two faces of the tetrahedron.\n\nWriting $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,\n\n$$\\text{height of tetrahedron} = |OK| = |OM|\\sin{M}$$\n\n$OM$ is the height of the (equilateral) face $OBC$, measuring $\\frac{\\sqrt{3}}{2}s$, where $s$ is the length of a side.\n\nAs for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:\n\n$$\\begin{eqnarray} |OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\\cos{M} \\\\ s^2 &=& \\left(\\frac{\\sqrt{3}}{2}s\\right)^2 + \\left(\\frac{\\sqrt{3}}{2}s\\right)^2 - 2 \\left(\\frac{\\sqrt{3}}{2}s\\right)\\left(\\frac{\\sqrt{3}}{2}s\\right) \\cos{M} \\\\ s^2 &=& \\frac{3}{4} s^2 + \\frac{3}{4}s^2 - 2 \\frac{3}{4} s^2 \\cos{M} \\\\ 1 &=& \\frac{3}{2} - \\frac{3}{2} \\cos{M} \\\\ \\frac{-1}{2} &=& - \\frac{3}{2} \\cos{M} \\\\ \\frac{1}{3} &=& \\cos{M} \\;\\;\\; (**)\\\\ \\Rightarrow \\sqrt{1-\\left(\\frac{1}{3}\\right)^2} = \\frac{\\sqrt{8}}{3} =\\frac{2\\sqrt{2}}{3}&=& \\sin{M} \\end{eqnarray}$$\n\nTherefore,\n\n$$\\text{height of tetrahedron} = |OK| = |OM|\\sin{M} = \\frac{\\sqrt{3}}{2} s \\cdot \\frac{2\\sqrt{2}}{3} = \\frac{\\sqrt{6}}{3}s$$\n\n(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):\n\n$$\\begin{eqnarray} \\cos\\left({\\text{angle between two sides of a regular triangle}}\\right) &=& \\frac{1}{2}\\\\ \\cos\\left({\\text{angle between two faces of a regular tetrahedron}}\\right) &=& \\frac{1}{3}\\\\ \\cos\\left({\\text{angle between two facets of a regular n-simplex}}\\right) &=& \\frac{1}{n} \\end{eqnarray}$$\n\n(Who would've suspected, upon first encountering it, that the \"$2$\" in \"$\\cos{60^{\\circ}}=\\frac{1}{2}$\" was actually a reference to the dimension of the triangle?)\n\n-\nThank-you v. much! What's a simplex? \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 17:14\nA simplex is the analog of \"triangle\" in any dimension: the simplest possible shape. It's what you get when you join $(n+1)$ points in $n$-dimensional space. A triangle is a \"$2$-simplex\" ($3$ points in $2$ dimensions); a tetrahedron is a \"$3$-simplex\" ($4$ points in $3$ dimensions); going upward in dimensions, one generally just says \"$4$-\", \"$5$-\", ..., \"$n$-simplex\"; going downward, one can say the line segment is a \"1-simplex\" (determined by $2$ points) and the point is a \"0-simplex\" ($1$ point!). See en.wikipedia.org/wiki/Simplex and mathworld.wolfram.com/Simplex.html \u2013\u00a0 Blue Nov 20 '10 at 17:30\nVery nice! I never knew about this pattern until now, thanks! :D \u2013\u00a0 \uff2a. \uff2d. Nov 20 '10 at 21:37", "date": "2014-08-30 10:29:59", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron/11086", "openwebmath_score": 0.7827844023704529, "openwebmath_perplexity": 417.45181368116533, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759654852754, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8623339266194829}}
{"url": "https://math.stackexchange.com/questions/2429631/put-7-balls-into-7-cells-probability-that-exactly-2-cells-containing-3", "text": "# Put $7$ balls into $7$ cells. Probability that exactly $2$ cells containing $3$ balls?\n\n1.46. Seven balls are distributed randomly into seven cells. What is the probability that the number of cells containing exactly $3$ balls is $2$?\n\nI am getting different answer from this solution manual. My argument is the following:\n\nIn order for us to have an arrangement where there are exactly $2$ cells with $3$ balls, we can follow the following procedure. We first decide which $3$ balls we want to put them together among all $7$ balls, and then decide to which cell we want to put them in. This gives us $\\binom{7}{3} \\binom{7}{1}$. Now, among the left $4$ balls, we choose $3$ balls to put them together, and choose one cell among the remaining $6$ empty cells. This gives us $\\binom{4}{3} \\binom{6}{1}$. Finally, we are left with one ball, and we have $5$ choices regarding where to put it. Putting them together, we have\n\n$$P(X_3 = 2) = \\frac{\\binom{7}{3} \\binom{7}{1} \\binom{4}{3} \\binom{6}{1} 5}{7^7}$$\n\nBut the solution manual says the answer should be\n\n$$\\frac{\\binom{7}{2}\\binom{7}{3}\\binom{4}{3}5}{7^7}$$\n\nWho is wrong and why?\n\n\u2022 I think, that the problem is, that you must divide by 2, because in your case you specify, that the first 3 goes to the one and the other 3 goes to the other cell. However, the order of cells doesn't matter. Therefore, divide by 2 and get it. Jan 21, 2018 at 14:36\n\nThe manual is right. The difference between your solution and the books solution is a factor of $2.$ That is ${7\\choose1}{6\\choose1} = 2{7\\choose 2}$", "date": "2022-08-09 00:44:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2429631/put-7-balls-into-7-cells-probability-that-exactly-2-cells-containing-3", "openwebmath_score": 0.8188285827636719, "openwebmath_perplexity": 137.51072174272278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357591818726, "lm_q2_score": 0.8807970904940926, "lm_q1q2_score": 0.8623318481770684}}
{"url": "http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron", "text": "# Height of a tetrahedron\n\nHow do I calculate: The height of a regular tetrahedron, side length 1.\n\nJust to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.\n\n-\nI'm curious. If you mind telling us, what was the purpose of the interview? What were the other mathematically relevant questions asked? \u2013\u00a0 user02138 Nov 20 '10 at 15:25\nIt's an old question for and Economics and Management interview at Oxford. \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 16:43\n\nThe first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.\n\nHere's a view of the geometry:\n\nand here's a view of the bottom face:\n\nIn the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.\n\nKnowing that the short sides of the isosceles triangle bisect the 60\u00b0 angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30\u00b0, 30\u00b0 and 120\u00b0.\n\nUsing the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):\n\n$$1=2\\ell^2-2\\ell^2\\cos 120^{\\circ}$$\n\nSolving for $\\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\\frac{1}{\\sqrt{3}}$, as indicated here.\n\nFrom this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies\n\n$$h^2+\\left(\\frac{1}{\\sqrt{3}}\\right)^2=1$$\n\nSolving for $h$ in the above equation, we now find the height to be $\\sqrt{\\frac23}=\\frac{\\sqrt{6}}{3}$, as mentioned here.\n\n-\nIt's not homework, it was an interview question I struggled with. I'd love to know the answer. \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 14:57\nGood to know. I'll edit with an explicit answer. \u2013\u00a0 Guess who it is. Nov 20 '10 at 15:00\nThank-you very much. P.S: how did you make your lovely 3d shape diagrams? \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 16:43\n@Patrick: I used Mathematica 5.2 (yes, it's an old copy :) ) for these diagrams. \u2013\u00a0 Guess who it is. Nov 20 '10 at 21:33\n\nConsider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\\frac{2}{\\sqrt 3}$, and since the edges of this tetrahedron have length $\\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \\sqrt{\\frac{2}{3}}$.\n\n-\nNote that this way of doing this also gives that the distance from the centers of opposite sides of the tetrahedron is always ${1 \\over \\sqrt{2}}$ times the side length. \u2013\u00a0 Zarrax Nov 20 '10 at 15:26\n\nYou can also use trig based on the dihedral angle between two faces of the tetrahedron.\n\nWriting $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,\n\n$$\\text{height of tetrahedron} = |OK| = |OM|\\sin{M}$$\n\n$OM$ is the height of the (equilateral) face $OBC$, measuring $\\frac{\\sqrt{3}}{2}s$, where $s$ is the length of a side.\n\nAs for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:\n\n$$\\begin{eqnarray} |OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\\cos{M} \\\\ s^2 &=& \\left(\\frac{\\sqrt{3}}{2}s\\right)^2 + \\left(\\frac{\\sqrt{3}}{2}s\\right)^2 - 2 \\left(\\frac{\\sqrt{3}}{2}s\\right)\\left(\\frac{\\sqrt{3}}{2}s\\right) \\cos{M} \\\\ s^2 &=& \\frac{3}{4} s^2 + \\frac{3}{4}s^2 - 2 \\frac{3}{4} s^2 \\cos{M} \\\\ 1 &=& \\frac{3}{2} - \\frac{3}{2} \\cos{M} \\\\ \\frac{-1}{2} &=& - \\frac{3}{2} \\cos{M} \\\\ \\frac{1}{3} &=& \\cos{M} \\;\\;\\; (**)\\\\ \\Rightarrow \\sqrt{1-\\left(\\frac{1}{3}\\right)^2} = \\frac{\\sqrt{8}}{3} =\\frac{2\\sqrt{2}}{3}&=& \\sin{M} \\end{eqnarray}$$\n\nTherefore,\n\n$$\\text{height of tetrahedron} = |OK| = |OM|\\sin{M} = \\frac{\\sqrt{3}}{2} s \\cdot \\frac{2\\sqrt{2}}{3} = \\frac{\\sqrt{6}}{3}s$$\n\n(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):\n\n$$\\begin{eqnarray} \\cos\\left({\\text{angle between two sides of a regular triangle}}\\right) &=& \\frac{1}{2}\\\\ \\cos\\left({\\text{angle between two faces of a regular tetrahedron}}\\right) &=& \\frac{1}{3}\\\\ \\cos\\left({\\text{angle between two facets of a regular n-simplex}}\\right) &=& \\frac{1}{n} \\end{eqnarray}$$\n\n(Who would've suspected, upon first encountering it, that the \"$2$\" in \"$\\cos{60^{\\circ}}=\\frac{1}{2}$\" was actually a reference to the dimension of the triangle?)\n\n-\nThank-you v. much! What's a simplex? \u2013\u00a0 Patrick Beardmore Nov 20 '10 at 17:14\nA simplex is the analog of \"triangle\" in any dimension: the simplest possible shape. It's what you get when you join $(n+1)$ points in $n$-dimensional space. A triangle is a \"$2$-simplex\" ($3$ points in $2$ dimensions); a tetrahedron is a \"$3$-simplex\" ($4$ points in $3$ dimensions); going upward in dimensions, one generally just says \"$4$-\", \"$5$-\", ..., \"$n$-simplex\"; going downward, one can say the line segment is a \"1-simplex\" (determined by $2$ points) and the point is a \"0-simplex\" ($1$ point!). See en.wikipedia.org/wiki/Simplex and mathworld.wolfram.com/Simplex.html \u2013\u00a0 Blue Nov 20 '10 at 17:30\nVery nice! I never knew about this pattern until now, thanks! :D \u2013\u00a0 Guess who it is. Nov 20 '10 at 21:37\nbtw, $\\cos\\left({\\text{angle between two faces of a regular tetrahedron}}\\right) = -\\frac{1}{3}$ \u2013\u00a0 Narasimham Mar 16 at 0:07\n@Narasimham: The angle between two faces of a regular tetrahedron is acute, so its cosine must be positive. \u2013\u00a0 Blue Mar 16 at 0:45\n\nI'd like to offer a slightly simpler approach to part of the 1st answer above. We know that the equilateral triangle of the base of the tetrahedron has sides of 1, 1, and 1, and we know we can split that in half, creating two right triangles having sides of hypotenuse=1, base=1/2, and perpendicular=\u221a3/2, along with angles of 30, 60, and 90 degrees.\n\nNow consider the 2nd diagram of the 1st answer, which shows a solid-line triangle having angles of 30, 30, and 120 degrees. That triangle could be divided in half, creating two right triangles having angles of 30, 60, and 90 degrees. If we consider the base of each triangle to be its shortest side, then the perpendicular of either one of those triangles has a length of 1/2. We can now use the power of ratios to compute the other two sides:\n\n(1/2):(\u221a3/2):(1) --triangle 1: half of tetrahedral face, angles of 30, 60 & 90 degrees.\n\n( ):( 1/2):( ) --triangle 2: has unknown base & hypotenuse, but is proportionate to triangle 1.\n\nWe really only need to compute the hypotenuse of triangle 2, because that is the desired distance from the corner to the center of the tetrahedron's base:\n\nMultiply triangle-1-hypotenuse by triangle-2-perpendicular; divide by triangle-1-perpendicular.\n\n[(1/2)(1)]/(\u221a3/2) = (1/2)(2/\u221a3) = 1/\u221a3, as the 1st answer also computes in a more complicated way.\n\nFor the sake of completeness, since in any 30-60-90-degree triangle the base is simply half the length of the hypotenuse, the second unknown is 1/(2\u221a3) or \u221a3/6 (although it could also have been figured by using ratios, as above). The reader is invited to verify that the square of (\u221a3/6) plus the square of 1/2 equals the square of (1/\u221a3).\n\n-\n\nThe normal height ($H_{n}$) of any regular tetrahedron having edge length $a$ is equal to the sum of radii of its inscribed & circumscribed spheres which is given as follows $$H_{n}=\\frac{a}{2\\sqrt{6}}+\\frac{a}{2}\\sqrt{\\frac{3}{2}}=\\frac{4a}{2\\sqrt{6}}=a\\sqrt{\\frac{2}{3}}$$ Hence, the normal height ($H_{n}$) of regular tetrahedron with edge length $a$ is generalized by the formula $$\\bbox[4pt, border: 1px solid blue;] {H_{n}=a\\sqrt{\\frac{2}{3}}}$$ As per given value of edge length $a=1$ in the question, the normal height of tetrahedron is $\\sqrt{\\frac{2}{3}}$\n\nNote: for derivation & detailed explanation, kindly go through HCR's Formula for Regular n-Polyhedrons\n\n-", "date": "2015-05-27 06:48:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron", "openwebmath_score": 0.8091424107551575, "openwebmath_perplexity": 425.8305622993112, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357616286122, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8623318411417867}}
{"url": "https://math.stackexchange.com/questions/2134265/can-endpoints-be-local-minimum", "text": "# Can endpoints be local minimum?\n\nMy textbook defines local maximum as follows:\n\nA function $$f$$ has local maximum value at point $$c$$ within its domain $$D$$ if $$f(x)\\leq f(c)$$ for all $$x$$ in its domain lying in some open interval containing $$c$$.\n\nThe question asks to find any local maximum or minimum values in the function\n\n$$g(x)=x^2-4x+4$$ in the domain $$1\\leq x<+\\infty$$.\n\nThe answer at the back has the point $$(1,1)$$, which is the endpoint.\n\nAccording to the definition given in the textbook, I would think endpoints cannot be local minimum or maximum given that they cannot be in an open interval containing themselves. (ex: the open interval $$(1,3)$$ does not contain $$1$$). Where am I wrong?\n\n\u2022 Your question is an excellent and important one. It is common to include endpoints in these types of calculations, and note that every point is in an open interval, just not necessarily an open interval in the domain of the function. Be sure to also check these definitions with and instructor or professor for total clarification. \u2013\u00a0The Count Feb 8 '17 at 2:15\n\u2022 This is mainly just a matter of convention - lacking a deep mathematical basis for preferring inclusion or exclusion of these points. I have seen professors and texts that disagree on this point. What really matters is that a convention is decided upon and used consistantly. I would agree that you ought to ask your professor for clarification on the convention they want to use. \u2013\u00a0David Feb 8 '17 at 3:19\n\u2022 Yes, it is a matter of convention. I personally think that the definition where \"open\" means open in $D$ makes more sense. Apparently, so does your textbook. However, the majority of Calculus textbooks that I've seen specifically exclude endpoints from the definition of local extrema, i.e. they treat \"open\" as being open in $\\mathbb{R}$. \u2013\u00a0zipirovich Feb 8 '17 at 4:46\n\nActually, the question is settled by reading the definition you provided carefully:\n\nA function $f$ has local maximum value at point $c$ within its domain $D$ if $f(x)\\leq f(c)$ for all $x$ in its domain lying in some open interval containing $c$.\n\nI.e., the points $x$ for which the condition must hold are required to both be in the open interval and in $D$.\n\nTo see that $(1,1)$ is a local maximum, consider the open interval $(0, 2)$. If $x \\in (0, 2)$ and $x$ is also in the domain $[1,\\infty)$, then $1 \\le x < 2$. Now $g(x) = x^2-4x + 4 = (2 - x)^2$. So $g(1) = (2 - 1)^2 = 1^2 = 1$, but if $x > 1$, then $0 < 2 - x < 1$, so $0 < (2-x)^2 = g(x) < 1$. So for $x$ in the open interval $(0,2)$ and also in the domain $[1,\\infty)$, we have that $g(x) \\le g(1)$.\n\n\u2022 For a different perspective, this is not how my calculus professor taught it recently. She said that if you are at an endpoint, you cannot compare the values outside of the interval, so you would not include it as a local extremum. \u2013\u00a0Max Li Nov 18 '17 at 20:58\n\u2022 @MaxLi - your calculus professor said WHAT!? Either you misunderstood your what your calculus professor was saying (which I unfortunately have to rate at only a 90% probability instead of the 100% probability I would prefer to), or else I advise you to carefully double-check everything she tells you against reliable sources and avoid taking any more classes from her in the future. Some things I can shrug off as differences in taste (is $0 \\in \\Bbb N$ or not?), and some things as pointless but not really harmful (is $0^0 = 1$ or undefined?). But this violates the basic meaning of the words. \u2013\u00a0Paul Sinclair Nov 19 '17 at 0:52\n\u2022 @PaulSinclair, it's actually common for textbooks to take this perspective. For example, the book with scans posted on this question (which is Stewart Calculus, if I'm not mistaken). \u2013\u00a0PersonX Feb 21 '18 at 2:28\n\u2022 @PersonX - Just because my comment has a wider application than one person does not at all make me recant. \u2013\u00a0Paul Sinclair Feb 21 '18 at 13:33\n\u2022 @MattBrenneman - $[a,c)$ is indeed an open neighborhood of $a$ in $D$. And it is exactly because \"local extremum\" is defined on topological spaces as \"extremum when restricted to some neighborhood\" that I take exception to Max Li's professor's remark. Under this definition, $a$ is a local extremum, just as it is under the explanation I gave in the post (which essentially amounts to the same thing, but without the subspace terminology). Max Li's professor would deny it that status, even though it is greater (or lesser) than everything near it in the domain. \u2013\u00a0Paul Sinclair Oct 24 '18 at 16:22\n\nI think fundamentally the comments are right, and you should speak with your teacher to confirm definitions and expectations. But there's also a point to make about topology here, which could justify the book's definition and answer as consistent.\n\nThe definition of local maximum you gave is:\n\nA function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \\leq f(c)$ for all $x$ in its domain lying in some ** open ** interval containing $c$.\n\nIf you interpret this as saying that the interval can come from $\\mathbb{R}$, and is not restricted to $D$, then you have no problem, as others have pointed out. But like you I am thinking about being restricted to $D$ and my instinct is to think only about intervals in $D$. This can still be ok, if we just alter our interpretation of \"open\" a little bit (in a natural way)...\n\nNow, whenever we say \"open\" we're really saying \"open with respect to ** insert topology here ** .\" A lot of the time it's obvious from context or the textbook has established a practice of contextual implication, but in this case (without knowing your book) I'd argue there are two reasonable interpretations:\n\n1. We might be talking open intervals with respect to the standard topology on $\\mathbb{R}$ (which is what you've probably been using in your class), but\n2. since we're restricting our attention to a domain $D \\subset \\mathbb{R}$, it's also pretty normal to talk about a different topology, called the subset topology on $D$ (induced by the standard topology on $R$).\n\nIn the subset topology on $D \\subset \\mathbb{R}$ (induced by the standard topology), a set $S$ is open if and only if $S$ is the intersection $D \\cap X$, with $X$ open in $\\mathbb{R}$ with respect to the standard topology on $\\mathbb{R}$.\n\nWe're often more interested in the subset topology than the usual topology on the whole space just because of situations like the one you're in, in which a definition doesn't work quite like you expect when $D \\not= \\mathbb{R}$.\n\nSo let's work with a slightly different definition of local maximum:\n\nA function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \\leq f(c)$ for all $x$ in its domain lying in some interval $I$ containing $c$ such that $I$ is open with respect to the subset topology on $D$.\n\nNow back to your case. Let $D = [1, \\infty)$. For any $a > 1$, we have that $$[1,a) = D \\cap (-a,a)$$ Since $(-a,a)$ is open in $\\mathbb{R}$ with respect to the standard topology, $[1,a)$ is open in $D$ with respect to the subset topology on $D$. This intuitively makes sense, because if you were an ant walking on $f(D)$, when you came to $f(1)$ you'd have nowhere to go but down.\n\n\u2022 I am going to have to spend some time trying to understand your answer, but thank you nonetheless. \u2013\u00a0Phil Feb 11 '17 at 23:13", "date": "2021-03-08 11:21:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2134265/can-endpoints-be-local-minimum", "openwebmath_score": 0.899631917476654, "openwebmath_perplexity": 217.52022104564307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357585701875, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8623318323210263}}
{"url": "http://thevelveetaroom.com/don-barclay-xgwmngs/definition-of-rational-numbers-for-class-7-d81286", "text": "If the numerator and the denominator both are either positive integers or negative integers, then the rational number is positive. The set of all rational numbers, often referred to as \"the rationals\" , the field of rationals or the field of rational numbers is usually denoted by a boldface Q (or blackboard bold $${\\displaystyle \\mathbb {Q} }$$, Unicode /\u211a); it was thus denoted in 1895 by Giuseppe Peano after quoziente, Italian for \"quotient\". Solution: Our notes of Chapter 1 Rational numbers are prepared by Maths experts in an easy to remember format, covering all syllabus of CBSE, KVPY, NTSE, Olympiads, NCERT & other Competitive Exams. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. Solution: Let the required rational number be x. The integers which are in the form of p/q where q is not equal to 0 are known as Rational Numbers. It is in the form of and q \u2260 0. Yes, it is a rational number. Copyright \u00a9 2021 Applect Learning Systems Pvt. Every integer is a rational number: for example, 5 = 5/1. we get, 14 x 15 = 210. and 12 x 21 = 252. A number that can be made by dividing two integers (an integer is a number with no fractional part). Advantages Of CBSE NCERT Class 8 Rational Numbers Worksheets . You can also register Online for Class 7 Science tuition on Vedantu.com to score more marks in CBSE board examination. It is something related to the ratios. \u201cAny number which can be expressed in the form , where p and q are integers and, is called a rational number.\u201d For example, is a rational number in which the numerator is 15 and the denominator is 19. Examples of Rational Numbers. (i) Absolute value = -12171=12171 What Is A Rational Number? Definition of Rational Numbers The integers which are in the form of p/q where q \u2260 0 are known as Rational Numbers. Understand the impact of properties of operations on rational numbers. Get Revision notes of Class 8th Mathematics Chapter 1 Rational numbers to score good marks in your Exams. NCERT CBSE Class 7 Maths Rational Numbers Solutions provide detailed explanations to the Class 7 Maths Chapter 9 exercises. (ii) 1219 Solution: Question 10. The examples of rational numbers are 6/5, 10/7, and so on. On Cross Multiplying the given Rational numbers. Thus, rational numbers can be defined as follows. In this article, we\u2019ll discuss the rational number definition, give rational numbers examples, and offer some tips and tricks for understanding if a number is rational or irrational. c) Daily Practice Sheets will help to develop a regular schedule of studies . Conventions used for writing a rational number: We know that in a rational number, the numerator and denominator both can be positive or negative. A rational number is defined as a number that can be expressed in the form p/q, where p and q are integers and. On solving equations like 3x + 5 = 0, we get the solution as x = -5/3. The absolute value of a rational number is its numerical value regardless of its sign. If the product of two rational numbers is $$\\frac { -9 }{ 16 }$$ and one of them is $$\\frac { -4 }{ 15 }$$, find the other number. Rational Numbers - Solution for Class 7th mathematics, NCERT solutions for Class 7th Maths. A rational number which has either the numerator negative or the denominator negative is called the negative rational number. Class 8 Maths Rational Numbers: Closure Properties: Properties of the types of numbers - Closure . So, 14/21 \u2260 12/15. A rational number is defined as a number that can be expressed in the form $$\\frac{p}{q}$$, where p and q are integers and q\u22600. The solution -5/3 is \u2026 Example 3. Get Textbook solutions for maths from evidyarthi.in You have studied fractional numbers in your earlier classes. These numbers are also known as rational numbers. Let us now write the given numbers according to the convention. Our Solutions contain all type Questions with Exe-1 A, Exe-1 B, Exe-1 \u2026 Thus, we can say that every integer is a rational number. Rational number 1 is the multiplicative identity for all rational numbers because on multiplying a rational number with 1, its value does not change. A set of numbers is said to be closed for a specific mathematical operation if the result obtained when an operation is performed on any two numbers in the set, is itself a member of the set. According to convention, the given number should be written as . Conventionally, rational numbers are written with positive denominators. On solving equations like 3x + 5 = 0, we get the solution as x = -5/3. Any rational number can be called as the positive rational number if both the numerator and denominator have like signs. NCERT CBSE Class 7 Maths Rational Numbers Solutions provide detailed explanations to the, Chapter 9 exercises. Some examples of fractional numbers are. Also, understand how to represent rational numbers on a number line with the support of our ICSE Class 8 Maths learning materials. The numerator and the denominator of a rational number will be integers. Write each of the following rational numbers according to the convention. The examples of rational numbers will be 1/4, 2/7, - 3/10, 34/7, etc. Sorry!, This page is not available for now to bookmark. They meet the requirements of students to make their basics strong in the subject. Examples: p q p / q = 1: 1: 1/1: 1: 1: 2: 1/2: 0.5: 55: 100: 55/100: 0.55: 1: 1000: 1/1000: 0.001: 253: 10: 253/10: 25.3 : 7: 0: 7/0: No! The absolute value of a rational number\u00a0pq is denoted as pq. In short, rational number represents a ratio of two integers. and experience Cuemath's LIVE Online Class with your child. For example, are positive rational numbers. In other words, the rational number is defined as the ratio of two numbers (i.e., fractions). This slideshow lesson is very animated with \u2026 Dive into the topic of rational numbers with TopperLearning\u2019s ICSE Class 8 Maths Chapter 1 study materials. \"q\" can't be zero! If are two rational numbers such that q > 0 and s > 0 then it can be said that if ps > qr. The rational number is represented using the letter \u201cQ\u201d. This document is highly rated by Class 9 students and has been viewed 25980 times. A rational number is a number that can be written in the form p q, where p and q are integers. Rational Numbers ICSE Class-8th Concise Selina Solutions Chapter-1. Many people are surprised to know that a repeating decimal is a rational number. Rational Numbers 1. These solutions give students an advantage over their peers because it helps them to prepare well for their examination. Properties of rational number: There is infinite number \u2026 \u221234 can be written as. Here, \u201cp\u201d is a numerator and \u201cq\u201d is a denominator. Question 8. More formally we say: A rational number is a number that can be in the form p/q where p and q are integers and q is not equal to zero. Definition of Rational Numbers: A rational number is any number that can be expressed as the quotient or fraction of two integers, with the denominator q not equal to zero. Definition of Rational Numbers. (c) q \u2260 1 (d) q \u2260 0. Students learn the definition of rational number, and they write rational numbers as ratios of integers and as repeating or terminating decimals. Class 7 Maths Chapter 9: Rational Numbers. Have a Query? Key Concepts . Jan 11, 2021 - Rational and Irrational Numbers - Number Systems, Class 9, Mathematics | EduRev Notes is made by best teachers of Class 9. Absolute Value of a Rational Number: Like real numbers, the Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. Since, 14 x 15 \u2260 12 x 21. In ratios, the numerator and denominator both are positive numbers while in rational numbers, they can be negative also. In NCERT solutions for class 7 Maths Chapter 9, you will study the concept of rational numbers along with their addition, subtraction, multiplication and division operations, their need, types of rational number, representation on the number line, standard form, comparison of rational numbers, rational between two rational numbers. In Mathematics, a rational number is defined as a number, which is written in the form p/q, where, q \u2260 0. 3. Mathematics NCERT Grade 7, Chapter 9: Rational Numbers- As the name suggests, the chapter deals with rational numbers.A detailed explanation about rational numbers is given in the chapter.Following key points and the topics are discussed in the first section of the chapter rational numbers: . Make Studies fun with over 9000+ Animated Videos, Make Homework stress free with Guaranteed Homework Help, Ace your exams with our Accurate Sample Papers. In our daily lives, we use some quantities which are not whole numbers but can be expressed in the form of $$\\frac{p}{q}$$. Need for rational numbers This means that 0 can also be a rational number as a rational number can be represented as \u2026 A rational number is a number that can be expressed as a ratio of p/q, where p and q are integers, and q does not equal to zero. Therefore, -32=32,\u00a012-7=127 etc. Yes, it is a rational number. Now, let us go through the given example. If in a rational number, either the numerator or the denominator is a negative integer, then the rational number is negative. a) NCERT CBSE Class 8 Rational Numbers Worksheets will help the students to clear concepts and get more score in examinations. As, 1 2, 4 5, 5 7 are in standard form. b) These printable worksheets for Rational Numbers Class 8 will help to improve problem solving and analytical skills. A rational number is said to be in the standard form if its denominator is a positive integer and the numerator and denominator have no common factor other than 1. We will give you a call shortly, Thank You, Office hours: 9:00 am to 9:00 pm IST (7 days a week). A rational number is a number that is of the form $$\\dfrac{p}{q}$$ where: $$p$$ and $$q$$ are integers $$q \\neq 0$$ The set of rational numbers is denoted by $$Q$$. Example: Examples : 5/8; -3/14; 7/-15; -6/-11 3. After an educational video defines the term, show how to compare and order them with guided examples. NCERT Solutions for Class 7 Maths Chapter 9: Two rational numbers with different denominators are added by first taking the LCM of the two denominators and then converting both the rational numbers to their equivalent forms having the LCM as the denominator. Properties of Rational Numbers 1) Closure Property 2) Associative Property 3) Distributive Law 4) Additive Inverse 5) Multiplicative Inverse The best techniques and approach to each question on NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers are discussed in great detail by the subject matter experts. These solutions give students an advantage over their peers because it helps them to prepare well for their examination. Solution : (d) By definition, a number that can be expressed in the form of p/q, where p and q are integers and q\u22600, is \u2026 MCQ Questions for Class 7 Maths: Ch 9 Rational Numbers. We will call you right away. Rational numbers: The numbers which can be expressed as ratio of integers are known as rational numbers. We provide step by step Solutions of Exercise / lesson-1 Rational Numbers for ICSE Class-8 Concise Selina Mathematics. According to convention, the given number should be written as . For example, is a rational number in which the numerator is 15 and the denominator is 19. Explained well with examples. CBSE Class 7 Maths Chapter 9 \u2013 Rational Numbers. Rational Numbers Class 7 Extra Questions Short Answer Type. These decimal numbers are also rational numbers as these can be written as. The best techniques and approach to each question on, for Class 7 Maths Chapter 9 Rational Numbers are discussed in great detail by the subject matter experts. Use this lesson plan to teach your students about rational numbers. Examples : 5/8; -3/14; 7/-15; -6/-11 Hence, 14/21 and 12/15 are not equivalent Rational Number. Now, is \u221234 a rational number? According to the convention used in rational numbers, the denominator must be a positive number. Words, the numerator or the denominator both are either positive integers or integers! Each of the following rational numbers analytical skills to 0 are known as rational numbers are 6/5, 10/7 and! Class 9 students and has been viewed 25980 times numerical value regardless of its sign numbers according the. 3/10, 34/7, etc number, and so on if the numerator negative or the denominator is 19 3! Following rational numbers, r and s, we get, 14 x 15 \u2260 12 x.. Q is not equal to 0 are known as rational numbers with absolute value of a rational is... Working with rational numbers q > 0 then it can be negative also be defined as a that. Denominator negative is called the negative rational number ( b ) q =.... ( c ) q \u2260 1 ( d ) q \u2260 1 ( d ) \u2260... Negative or the denominator of a rational number is positive ; -6/-11 3 c Daily! Denominator is a rational number notes of Class 8th Mathematics Chapter 1 rational numbers throughout unit... Numbers \\frac { 2 } \u2026 rational numbers as ratios of integers and ; 7/-15 -6/-11. 'S LIVE Online Class with your child to the convention, the numerator or the both. Each of the following rational numbers will be 1/4, 2/7, - 3/10,,. Integers which are in standard form: let the required rational number be x Daily Practice will! P, q, r and s > 0 then it can be: q... Write rational numbers 1 5 7 are in standard definition of rational numbers for class 7 line with the of... For example, is a rational number which has either the numerator or denominator... Problem solving and analytical skills, 10/7, and so on the Class Maths... Problem solving and analytical skills, etc 1 ( d ) q \u2260 0 are known as rational.. ) Daily Practice Sheets will help to develop a regular schedule of studies help to problem. Integer, then the rational number: for example, is a denominator students., they can be expressed in the form of and q \u2260 are... Class 8th Mathematics Chapter 1 rational numbers with absolute value of a rational number is negative will help improve. ) these printable Worksheets for rational numbers number determined by the ratio 4:5 can be represented in form... 8 Maths learning materials decimal is a rational number hence, 14/21 and 12/15 are not equivalent rational number always. 10/7, and so on marks in CBSE board examination ) Daily Practice Sheets will help develop.: for example, is a numerator and \u201c q \u201d in your earlier classes examples., \u201c p \u201d is a rational number, and so on according! Get the solution as x = -5/3, then the rational number as denominators! = 210. and 12 x 21 = 252 always non-negative solutions for 7! Of our ICSE Class 8 Maths learning materials comes first to your mind you! Form p/q, where p and q \u2260 0 are known as rational numbers to score good marks in earlier! Will be integers p/q where q is not equal to 0 are known as rational numbers regardless! Is positive: the absolute value of any rational number written in the form of and \u2260... As the ratio of two integers form p q, where p q! On a number that can be negative also to bookmark that can be written in the of... Your mind when you hear the word rational available for now to bookmark properties of operations on numbers... P/Q, where p and q are integers and as repeating or terminating decimals the of!, - 3/10, 34/7, etc x 21 is denoted as pq Short Type... Numbers, the Use this lesson plan to teach your students about rational numbers 1... 3X + 5 = 5/1 { 2 } \u2026 rational numbers throughout unit.!, this page is not available for now to bookmark to convention, the Use lesson., show how to represent rational numbers Class 8 rational numbers form p q: q! Also rational numbers \\frac { 2 } \u2026 rational numbers Worksheets will help to improve problem solving and skills... Say that every integer is a number determined by the ratio of two integers number should written. Of properties of operations on rational numbers as these can be represented in the form of where... Of p/q where q \u2260 0 9 \u2013 rational numbers, they can be also! Pq is denoted as pq, etc a ) NCERT CBSE Class 8 Maths learning materials, let go!", "date": "2021-04-12 01:20:22", "meta": {"domain": "thevelveetaroom.com", "url": "http://thevelveetaroom.com/don-barclay-xgwmngs/definition-of-rational-numbers-for-class-7-d81286", "openwebmath_score": 0.749920666217804, "openwebmath_perplexity": 493.7412368347948, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9790357616286122, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8623318304196959}}
{"url": "https://bassamshakir.com/3g7tho/szudzik-pairing-function-588218", "text": "Wolfram Science Conference NKS 2006. \\right.$$However, a simple transformation can be applied so that negative input can be used. In this ramble we will cover two different pairing functions: Cantor and Szudzik. Szudzik pairing function accepts optional boolean argument to map Z x Z to Z. Another JavaScript example: Szudzik can also be visualized as traversing a 2D field, but it covers it in a box-like pattern. As such, we can calculate the max input pair to Szudzik to be the square root of the maximum integer value. But for R the Axiom of Choice is not required. For the Cantor function, this graph is traversed in a diagonal function is illustrated in the graphic below. September 17, 2019 2:47 AM. The Rosenberg-Strong Pairing Function. 5 0 obj %\ufffd\uc3e2 For example, cantor(33000, 33000) = 2,178,066,000 which would result in an overflow. For the Szudzik pairing function, the situation is only slightly more complicated. Examples 148 VIEWS. Other than that, the same principles apply. This can be easily implemented in any language. In: Wolfram Research (ed.) -2x - 1 & : x < 0\\\\ b^2 + a & : a < b\\\\ /// /// So, if user didn't make something stupid like overriding the GetHashCode() method with a constant, /// we will get the same unique number for the same row and column every time. \\right.$$, $$c(a,b) = \\left\\{\\begin{array}{ll} Essentially any time you want to compose a unique identifier from a pair of values. \\end{array} a^2 + a + b & : a \\ge b Given two points 8u,v< and 8x,y<, the point 8u,v< occurs at or before 8x,y< if and only if PairOrderedQ@8u,v<,8x,y= 0 The performance between Cantor and Szudzik is virtually identical, with Szudzik having a slight advantage. You can then map the row to an X axis, the column to an Y axis. Yes, the Szudzik function has 100% packing efficiency. /// 2- We use a pairing function to generate a unique number out of two hash codes. \\end{array} It should be noted though that all returned pair values are still positive, as such the packing efficiency for both functions will degrade. a^2 + a + b & : a \\ge b The formula for calculating mod is a mod b = a - b[a/b]. od_id* functions take two vectors of equal length and return a vector of IDs, which are unique for each combination but the same for twoway flows. the Szudzik pairing function, on two vectors of equal length. -c - 1 & : (a < 0 \\cap b \\ge 0) \\cup (a \\ge 0 \\cap b < 0) For a 32-bit unsigned return value the maximum input value for Szudzik is 65,535. 62 no 1 p. 55-65 (2007) \u2013 In this paper, some results and generalizations about the Cantor pairing function are given. (Submitted on 1 Jun 2017 ( v1 ), last revised 28 Jan 2019 (this version, v5)) Abstract: This article surveys the known results (and not very well-known results) associated with Cantor's pairing function and the Rosenberg-Strong pairing function, including their inverses, their generalizations to higher dimensions, and a discussion of a few of the advantages of the Rosenberg \u2026 a * a + a + b : a + b * b; where a, b >= 0 Generate ordered ids of OD pairs so lowest is always first This function is slow on large datasets, see szudzik_pairing for faster alternative Usage od_id_order(x, id1 = names(x)[1], id2 = names(x)[2]) function pair(x,y){return y > x ? More than 50 million people use GitHub to discover, fork, and contribute to over 100 million projects. See Also. A pairing function is a mathematical function taking two numbers as an argument and returning a third number, which uniquely identifies the pair of input arguments. x\ufffd\ufffd\\[\ufffdEv\ufffd\ufffd\ufffd\u0787~\ufffd\u06eb.\ufffd~1\ufffd\u00c2\ufffd ^\"\ufffda\u0607\ufffd \u0695f@B\ufffd\ufffd\ufffd;y=Y\ufffd53\ufffd;\ufffdZUy9y\ufffdw\ufffd\ufffdY\ufffd\ufffd\ufffd\"w\ufffd\ufffd+\u007f\ufffd\ufffd\ufffd\ufffd:\ufffd\ufffdL\ufffd\u05fb\ufffd\ufffd\ufffd\ufffd\u0747\ufffdh\"\ufffdN\ufffd\ufffd\ufffd\ufffd3\ufffd\ufffd\ufffd\ufffdV;e\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\u007f?\u007f\ufffd/\ufffd\ufffd#U|kw\ufffd/\ufffd\ufffd^\ufffd\ufffd\ufffd_w;v\u007f\ufffd\ufffdFo\ufffd\u007f;\ufffd\ufffd\ufffd\ufffd3\ufffd=\ufffd\ufffd~Q\ufffd\ufffd.S)w\u0499\ufffd\uc734\ufffdv4\ufffd\ufffd\ufffdZ\ufffdq*\ufffd9\ufffd\ufffd\ufffd\ufffd\ufffd>\ufffd4hd\ufffd\ufffd\ufffdb\ufffdpq\u007f\ufffd\ufffd^['\ufffd\ufffd\ufffdLm<5D'\ufffd\ufffd\ufffd\ufffd\ufffd\"\ufffdU\ufffd'\ufffd\ufffd They may also differ in their performance. (yy+x) : (xx+x+y);} function unpair(z){var q = Math.floor(Math.sqrt(z)), l = z - \u2026 <> However, cantor(9, 9) = 200. \\end{array} Matthew P. Szudzik. One nice feature about using the Szudzik pairing function is that all values below the diagonale are actually subsequent numbers. Pairing library using George Cantor (1891) and Matthew Szudzik (2006) pairing algorithms that reversibly maps Z \u00d7 Z onto Z*. Simple C# class to calculate Cantor's pairing function - CantorPairUtility.cs. The full results of the performance comparison can be found on jsperf. Nothing really special about it. The cantor pairing function can prove that right? It should be noted that this article was adapted from an earlier jsfiddle of mine. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. c & : (a < 0 \\cap b < 0) \\cup (a \\ge 0 \\cap b \\ge 0)\\\\ The algorithms have been modified to allow negative integers for tuple inputs (x, y). In mathematics, a pairing function is a process to uniquely encode two natural numbers into a single natural number.. Any pairing function can be used in set theory to prove that integers and rational numbers have the same cardinality as natural numbers. F{\ufffd\ufffd\ufffd\ufffd+\ufffd\ufffdj#,\ufffd\ufffd{\"1Ji\ufffd\ufffd+p@{\ufffdax\ufffd/q+M\ufffd\ufffdB\ufffdH\ufffd\ufffd\u0440\ufffd\ufffd\ufffd DQ\ufffdP\ufffd\ufffd\ufffd\ufffd\ufffdK\ufffd\ufffd\ufffd\ufffd\ufffdo\ufffd\ufffd\ufffd \ufffdu\ufffd\ufffdZ\ufffd\ufffdx\ufffd\ufffd>\ufffd \ufffd-_\ufffd\ufffd2B\ufffd\ufffd\ufffd\u007f\ufffd\ufffd;\ufffd\ufffd \ufffdu\u0591. \\end{array}$$index = \\left\\{\\begin{array}{ll} Comparing against Cantor we see: Yes, the Szudzik function has 100% packing efficiency. cantor pairing function inverse. An example in JavaScript: How Cantor pairing works is that you can imagine traversing a 2D field, where each real number point is given a value based on the order it which it was visited. Cantor pairing function: (a + b) * (a + b + 1) / 2 + a; where a, b >= 0 The mapping for two maximum most 16 bit integers (65535, 65535) will be 8589803520 which as you see cannot be fit into 32 bits. Ask Question Asked 1 year, 2 months ago. A quadratic bijection does exist. - pelian/pairing Different pairing functions known from the literature differ in their scrambling behavior, which may impact the hashing functionality mentioned in the question. 1. ambuj_kumar 16. Let's not fail silently! It returns a vector of ID numbers. Abstract This article surveys the known results (and not very well-known re- sults) associated with Cantor\u2019s pairing function and the Rosenberg-Strong pairing function, including their inverses, their generalizations to higher dimensions, and a discussion of a few of the advantages of the Rosenberg- Strong pairing function over Cantor\u2019s pairing function \u2026 Like Cantor, the Szudzik function can be easily implemented anywhere. k cursive functions as numbers, and exploits this encoding in building programs illustrating key results of computability. $$b = \\left\\{\\begin{array}{ll} \ufffd\ufffd\ufffd ^a\ufffd\ufffd\ufffd0\ufffd\ufffd4\ufffd\ufffdq\u007f\ufffd\ufffdNXk\ufffd_d\ufffd\ufffdz\ufffd}k\ufffd; \ufffd\ufffd\ufffd\u05ec\ufffdHUf A\ufffd\ufffd|Pv \u0445\ufffdEk\ufffd\ufffd\ufffdRA\ufffd\ufffd\ufffd\ufffd\ufffd@\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdx\ufffd\ufffd kP[Z\ufffd\ufffde \ufffd\\\ufffdUW6JZi\ufffd\ufffd\ufffd_\ufffd\ufffdD\ufffdQ;)\ufffdhI\ufffd\ufffd\ufffdB\\\ufffd\ufffdaG\ufffd\ufffdK\ufffd\ufffd\u04c3^dd\ufffd\ufffd\ufffdZ\ufffd\ufffd\ufffd\ufffd\ufffdV\ufffd8\ufffd\ufffd\"( \ufffd|\ufffdN\ufffd(\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd/x\ufffd\u0162U \ufffd\ufffd\ufffd\ufffda\ufffd\ufffd\ufffd\ufffd[\ufffdE\ufffdg\ufffd\ufffd\ufffd\ufffdb\ufffd\"\ufffd\ufffd\ufffd&\ufffd>\ufffdB\ufffd*e\ufffd\ufffdX\ufffd\u00cfD\ufffd\ufffd{pY\ufffd\ufffd\ufffd\ufffd#\ufffdg\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdV\ufffdU}\ufffd\ufffd\ufffdI\ufffd\ufffd\ufffd\ufffd@\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\u007f\ufffdq\ufffdPX\u0493\ufffdd%=\ufffd{\ufffd\ufffd\ufffd\ufffdzp\ufffd.B{\ufffd\ufffd\ufffd\ufffd\"\ufffd\ufffdY\ufffd\ufffd!\ufffd\ufffd\ufffd\u05b0\ufffd\ufffd\ufffd\ufffdG)I\ufffdPi\ufffd\ufffd\u049b\ufffdXB\ufffdK(\ufffdW! , To find x and y such that \u03c0(x, y) = 1432: The graphical shape of Cantor's pairing function, a diagonal progression, is a standard trick in working with infinite sequences and countability. And as the section on the inversion ends by saying, \"Since the Cantor pairing function is invertible, it must be one-to-one and onto.\" \\right.$$, https://en.wikipedia.org/wiki/Pairing_function. function(x, y, z) { max = MAX(x, y, z) hash = max^3 + (2 * max * z) + z if (max == z) hash += MAX(x, y)^2 if (y >= x) hash += x + y else hash += y return hash} This pairing function only works with positive numbers, but if we want to be able to use negative coordinates, we can simply add this to the top of our function: x = if x >= 0 then 2 * x else -2 * x - 1 -2y - 1 & : y < 0\\\\ An Elegant Pairing Function Matthew Szudzik Wolfram Research Pairing functions allow two-dimensional data to be compressed into one dimension, and they play important roles in the arrangement of data for exhaustive searches and other applications. A pairing function is a function which maps two values to a single, unique value. Additional space can be saved, giving improved packing efficiency, by transferring half to the negative axis. Value. The pairing function then combines two integers in [0, 226-2] into a single integer in [0, 252). The primary downside to the Cantor function is that it is inefficient in terms of value packing. Enter Szudzik's function: a >= b ? Matthew P. Szudzik 2019-01-28. Active 1 year, 2 months ago. Pairing functions with square shells, such as the Rosenberg-Strong pairing function, are binary perfect. Two pairing functions are \u2026 I found Cantor's and Szudzik's pairing function to be very interesting and useful, however it is explicitly stated that these two functions are to be used for natural numbers. A pairing function for the non-negative integers is said to be binary perfect if the binary representation of the output is of length 2k or less whenever each input has length k or less. In[13]:= PairOrderedQ@8u_,v_<,8x_,y_= b ? It is always possible to re-compute the pair of arguments from the output value. We quickly start to brush up against the limits of 32-bit signed integers with input values that really aren\u2019t that large. Viewed 40 times 0. Szudzik M (2006) An elegant pairing function. Source. Cantor pairing function: (a + b) * (a + b + 1) / 2 + a; where a, b >= 0 The mapping for two maximum most 16 bit integers (65535, 65535) will be 8589803520 which as you see cannot be fit into 32 bits. b^2 + a & : a < b\\\\ /// 3- We use the unique number as the key for the entry. 2y & : y \\ge 0 The inverse function is described at the wiki page. Use a pairing function for prime factorization. \\right.$$,$$a = \\left\\{\\begin{array}{ll} %PDF-1.4 Szudzik, Matthew P. Abstract This article surveys the known results (and not very well-known results) associated with Cantor's pairing function and the Rosenberg-Strong pairing function, including their inverses, their generalizations to higher dimensions, and a discussion of a few of the advantages of the Rosenberg-Strong pairing function over Cantor's pairing function in practical applications. Java : 97% speed and 66.67% memory : using Szudzik's Pairing Function and HashSet. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. Proof. This relies on Cantor's pairing function being a bijection. Usage. \\end{array} Szudzik, M. (2006): An Elegant Pairing Function. I used Matthew Szudzik's pairing function and got this: $(p - \\lfloor\\sqrt{p}\\rfloor^2)\\cdot\\lfloor\\sqrt{p}\\rfloor = n$ There, we need to make a distinction between values below the diagonale and those above it. This is useful in a wide variety of applications, and have personally used pairing functions in shaders, map systems, and renderers. 2x & : x \\ge 0 x^2 + x + y & : x \\ge y \\right.$$Trying to bump up your data type to an unsigned 32-bit integer doesn\u2019t buy you too much more space: cantor(46500, 46500) = 4,324,593,000, another overflow. In elementary set theory, Cantor's theorem is a fundamental result which states that, for any set, the set of all subsets of (the power set of , denoted by ()) has a strictly greater cardinality than itself. stream This means that all one hundred possible variations of ([0-9], [0-9]) would be covered (keeping in mind our values are 0-indexed). \\end{array} PREREQUISITES. In theoretical computer science they are used to encode a function defined on a vector of natural numbers : \u2192 into a new function : \u2192$$index = {(x + y)(x + y + 1) \\over 2} + y. The limitation of Cantor pairing function (relatively) is that the range of encoded results doesn't always stay within the limits of a 2N bit integer if the inputs are two N bit integers. T\u00e5ngav\u00e4gen 5, 447 34 V\u00e5rg\u00e5rda info@futureliving.se 0770 - 17 18 91 If you want to have all paris x, y < 2 15, then you can go with the Szudzik's function: \u03c3 (x, y) = { x 2 + x + y if x \u2265 y x + y 2 otherwise Algorithms have been modified to allow negative integers for tuple inputs ( x, y.! The maximum input value for Szudzik is virtually identical, with Szudzik having a slight advantage \\over 2 } y. Covers it in a wide variety of applications, and renderers function - CantorPairUtility.cs return. Map Z x Z to Z number as the key for the first 100 combinations, an efficiency of %. Earlier jsfiddle of mine of computability a 32-bit signed return value, we need to make a between. Both functions will degrade and HashSet return value, we can calculate the max input pair Szudzik! Slight advantage of mine the full results of the maximum input value for Szudzik is 65,535 half to the function... With that unordered pair function being a bijection functions will degrade from an earlier jsfiddle of mine functions: and! Functions in shaders, map systems, and renderers positive, as such the efficiency... Natively with negative input can be understood as an ordering of the performance between Cantor Szudzik! In [ 0, 252 ) an y axis can calculate the max pair... \\Over 2 } + y  index = { ( x y. That really aren \u2019 t that large /// 3- we use 200 pair values for the Cantor,! Programs illustrating key results of computability unique identifier from a pair of arguments the. Cover two different pairing functions work natively with negative input values that really aren \u2019 t that large be.! 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A slight advantage is not required full results of the points in the plane Z x Z to Z than!, are binary perfect % packing efficiency for both functions will degrade an Elegant pairing to! Question Asked 1 year, 2 months ago 33000 ) = 200 from an jsfiddle! Row to an x axis, the column to an y axis with Szudzik having a slight advantage diagonal! To compose a unique identifier from a pair of values transformation can be applied so that negative input be... 2 } + y ) key for the Cantor function, this graph is traversed a... Are binary perfect an Elegant pairing function then combines two integers in [ 0, 252.. Be easily implemented anywhere on Cantor 's pairing function then combines two integers in [ 0, ). Scrambling behavior, which may impact the hashing functionality mentioned in the plane for visually. A box-like pattern using the Szudzik function can be saved, giving improved packing efficiency by... Differ in their scrambling behavior, which may impact the hashing functionality in! 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Will cover two different pairing functions in shaders, map systems, and contribute to over 100 million.. Uniquely associated with that unordered pair input can be applied so that negative input values is a mod b a! You can then map the row to an x axis, the Szudzik can! Subsequent numbers to calculate Cantor 's pairing function for prime factorization \u2019 t that.. Have the maximum input value for Szudzik is 65,535 pairing functions in shaders, map systems, and renderers mine...  Q\ufffdP\ufffd\ufffd\ufffd\ufffd\ufffdK\ufffd\ufffd\ufffd\ufffd\ufffdo\ufffd\ufffd\ufffd \ufffdu\ufffd\ufffdZ\ufffd\ufffdx\ufffd\ufffd > \ufffd \ufffd-_\ufffd\ufffd2B\ufffd\ufffd\ufffd\u007f\ufffd\ufffd ; \ufffd\ufffd \ufffdu\u0591 outputs a single non-negative integer that is uniquely associated with unordered. A perfectly efficient function we would expect the value of pair ( x + y ) { return y x. Negative integers for tuple inputs ( x, y ) { return >! Outputs a single non-negative integer that is uniquely associated with that unordered pair easily implemented anywhere of!", "date": "2021-09-23 05:29:33", "meta": {"domain": "bassamshakir.com", "url": "https://bassamshakir.com/3g7tho/szudzik-pairing-function-588218", "openwebmath_score": 0.7441880106925964, "openwebmath_perplexity": 1349.1221608188678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9702399043329856, "lm_q2_score": 0.8887587964389112, "lm_q1q2_score": 0.8623092496319886}}
{"url": "https://brilliant.org/discussions/thread/how-to-count-rectangles/", "text": "# How to Count Rectangles!\n\nHi there! Me again. *munches*\n\nSorry, I still haven't finished my hamburger since the last time we count squares, I hope you don't mind that.\n\nAnyway, we are here to see how to count rectangles of a certain size in a grid.\n\nConsider this question: How many $$3\\times2$$ rectangles are there in the $$4\\times5$$ grid below?\n\nI'm a little bit picky here, note that dimensions here are expressed as width$$\\times$$height, so in this case I only want $$3\\times2$$ rectangles (that is, a rectangle of width 3 and height 2) and not the $$2\\times3$$ ones.\n\nIf we mark an 'X' on every bottom-right corner of every $$3\\times2$$ rectangle, we would see this:\n\nHey, that means counting the number of $$3\\times2$$ rectangles in the original grid is the same as counting how many X's are there, or in other words, it is the same as counting the number of unit squares in an $$2\\times4$$ grid.\n\nThere are $$2\\times4=8$$ unit squares in a $$2\\times4$$ grid, therefore, the number of $$3\\times2$$ rectangles in the original grid is 8! Hooray!\n\n*munches*\n\nMmm... so in general, how many $$x\\times y$$ rectangles ($$x$$ is the width of the rectangle, $$y$$ is the height of the rectangle) are there in an $$a\\times b$$ grid ($$a$$ is the width of the grid, $$b$$ is the height of the grid), given that $$x\\leqslant a$$ and $$y\\leqslant b$$?\n\nAgain imagine marking an 'X' on every bottom-right corner of every $$x\\times y$$ rectangle, we will get a grid full of X's.\n\nThe width of the \"X grid\" is $$a-x+1$$, while the height of the \"X grid\" is $$b-y+1$$.\n\nSince the number of $$x\\times y$$ rectangles in an $$a\\times b$$ grid equals the number of unit squares in the \"X grid\", hence the number of $$x\\times y$$ rectangles in an $$a\\times b$$ grid is $(a-x+1)(b-y+1)$\n\n*munches*\n\nThe number of $$x\\times y$$ ($$x$$ is the width of the rectangle, $$y$$ is the height of the rectangle) rectangles (the exact $$x\\times y$$ rectangle, not including rotated variations such as $$y\\times x$$) in an $$a\\times b$$ grid ($$a$$ is the width of the grid, $$b$$ is the height of the grid) provided that $$x\\leqslant a$$, $$y\\leqslant b$$ is $(a-x+1)(b-y+1)$\n\n##### This is one part of Quadrilatorics.\n\nNote by Kenneth Tan\n2\u00a0years, 10\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nI think another way to do it is breaking them up. For example, if we want to count the number ot rectangles in $$10\u00d710$$ grid, is first counting the number of rectangles in $$10\u00d71$$ grid then xounting the number of rectangles in $$10\u00d72$$ grid, then in $$10\u00d73$$, and so in, in this way we would be able to establish a pattern in the number of rectangles, and by arithmetic/geometric progression, we would het the answer.\n\n- 2\u00a0years, 10\u00a0months ago\n\nYes, but in this case I only want to count rectangles of a specific size.\n\n- 2\u00a0years, 10\u00a0months ago\n\nWe can do in that case too. Your questions are awesome.\n\n- 2\u00a0years, 10\u00a0months ago\n\nYou may see the solution of my problem to see my approach.\n\n- 2\u00a0years, 10\u00a0months ago", "date": "2019-01-23 06:20:32", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/how-to-count-rectangles/", "openwebmath_score": 0.9728492498397827, "openwebmath_perplexity": 876.5467964245305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9702399008908847, "lm_q2_score": 0.8887587831798665, "lm_q1q2_score": 0.8623092337083369}}
{"url": "http://pacianca.it/improper-integrals-comparison-test.html", "text": "# Improper Integrals Comparison Test\n\nIntegrals over unbounded intervals. Improper integrals (Sect. Let f and g. Show graphically that the benchmark you chose satisfies the conditions of the Comparison Test. Comparison tests for convergence. The book gives students the prerequisites and tools to understand the convergence, principal value, and evaluation of the improper/generalized Riemann integral. , find the values of p for which this integral converges, and the values for which it diverges. Calculus of parametric curves. Comparison Tests. Math 185 - Calculus II. This articles needs some examples, better references, a discussion of the relation with other convergence tests, an explanation of the comparison test for improper integrals, and so forth. We want to prove that the integral diverges so if we find a smaller function that we know diverges the area analogy tells us that there would be an infinite amount of area under the smaller function. (right) Comparison of test prediction accuracy when using ourIDKkernel to a numerical estimation of the kernel integral using random features, as a function of the number of features used for estimation. Partial Fractions. Does Z 1 0 1 p 1 x2 dxconverge? Notice that the function p1 1 x2 has a vertical asymptote at x= 1, so this is an improper integral and we will need to consider the. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. 1 x2 3 + 3 4. Problem 4 (15 points) For p >0 consider the improper integral Derive the p -test for these integrals, i. Where k is going to be some positive number. The comparison test. However, for x>1 we have x2 >x, so that 1/ex2 > 1/ex. Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Select the second example from the drop down menu, showing Use the same guidelines as before, but include the exponential term also: The limit of the ratio seems to converge to 1 (the \"undefined\" in the table is due to the b terms getting so small that the algorithm thinks it is dividing by 0), which we can verify: The limit comparison test says that in this. aa g x dx f x dx \u00b3\u00b3 ff When we cannot evaluate an integral directly, we first try to determine whether it converges or diverges by comparing it to known integrals. Frequently we aren't concerned along with the actual value of these integrals. Theorem 1 (Comparison Test). We are covering these tests for definite integrals now because they serve as a model for similar tests that check for convergence of sequences and series -- topics that we will cover in the next chapter. For example, one possible answer is BF, and another one is L. Partial Fractions 32 1. Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. Page 632 Numbers 74. By de\ufb01nition, these integrals can only be used to compute areas of bounded regions. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. This is , \u2203 a psoitive number ,independent of ,such that \u222b + < M, 0< < \u2212. If Maple is unable to calculate the limit of the integral, use a comparison test (either by plotting for direct comparison or by limit comparison if applicable). 7 Review of limits at infinity:-2 -1 0 1 2 2 4 6 x y y ex 1 2 3-4-2 0 2 x y Use the above comparison test to determine whether 1 1 x2 5. State the Comparison Test. Solutions; List of Topics that may. That means we need to nd a function smaller than 1+e x. Give a reasonable \"best\" comparison function that you use in the comparison (by \"best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). Consider the following These are all improper because the function being integrated is not finite at one of the limits of integration. I Examples: I = Z \u221e 1 dx xp, and I = Z 1 0 dx xp I Convergence test: Direct comparison test. 10) Activity: Choosing a Technique of Integration EWA 5. Since the integral has an infinite discontinuity, it is a Type 2 improper integral. Integral Test: If f(x) is a decreasing positive function from [1,\u221e) to [0,\u221e), then the series P f(n) converges if and only if the improper integral R \u221e 0 f(x)dx converges (i. Review Problems from your textbook: Integration plus L'Hospital's Rule page 579 #1-15, 33-37, 73-87 odds. In such cases the following test is useful. To be convincing, we need to find a series with smaller terms whose sum diverges. give me a formal proof of comparison test for improper integrals. The rst of these is the Direct Comparison Test (DCT). 7) I Integrals on in\ufb01nite. Comparison Tests. The key tools are the Comparison Test and the P-Test. Comparison Test which requires the non negativity of relevant functions, we shall consider instead f(x) and g(x). Int[1 to infinity] (sin^2 x)/x^3 dx. We work through several examples for each case and provide many exercises. Improper. We can summarize this line of argumentation with the following: Theorem. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. via integration by parts with u = t/2, dv = sin(2t) dt = \u03c0/8. Improper Integrals. NEET: 45 Day Crash Course | Day 45 Last Day | Alcohols, Phenols, Ethers | Unacademy NEET | Anoop Sir Unacademy NEET 117 watching Live now. Improper Integral. The concepts used include regular inequalities as well as the. There are 2 types of comparison tests: direct comparison and limit comparison. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. One way to determine the convergence of an improper integral is by comparing it to other integrals that we do know the convergence of. Using the Comparison Test from Calculus, we determine whether several improper integrals in the question converge or diverge. Step 2: Click the blue arrow to submit. integral(1 to infinity) ln(x) dx/x^(1. This chapter has explored many integration techniques. Convergence tests for improper integrals. We have two main tests that we can use to establish convergence or divergence -- the Direct Comparison Test and the Limit Comparison Test. Although we state it for Type 1 integrals,. The comparison test Identify an improper integral. Ok so from this test i came to the conclusion that the integral of sinx/lnx converges. Here I give a quick idea of what the direct comparison test for improper integrals and use it show whether an improper integral converges or. Z 1 1 1 (2 x+ 1)3 dx 2. The Comparison Test and Limit Comparison Test also apply, modi ed as appropriate, to other types of improper integrals. A similar statement holds for type 2 integrals. Comparing Improper Integrals. 2 The Integral Test and p-series test File WS 11. integral toolbox. 14 Improper integrals While the rst integral on the right-hand side diverges, the second one converges, as can be proved by the same procedure as above. The direct comparison test is a simple, common-sense rule: If you\u2019ve got a series that\u2019s smaller than a convergent benchmark series, then your series must also converge. There are two we will talk about here. The Limit Comparison Theorem for Improper Integrals Limit Comparison Theorem (Type I): If f and g are continuous, positive functions for all values of x, and lim x!1 f(x) g(x) = k Then: 1. Otherwise the integrals are divergent. Here is the statement: Limit Comparison Test: Suppose f(x) and g(x) are continuous and positive for all x 2[a;1). Here I give a quick idea of what the direct comparison test for improper integrals and use it show whether an improper integral converges or. Z 1 1 1 x+ 3ln(x) dx 6. If you're behind a web filter, please make sure that the domains *. Often we aren't concerned with the actual value of these integrals. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. \u222b\u221e 1 1 x3+1dx Solution. improper integrals (comparison theorem) 5 $\\mu$ test for convergence of improper integral of first kind. ) - Duration: 1:02:59. Smith , Founder & CEO, Direct Knowledge. integral 1/(x^2+3) from 1 to infinity 3. Comparison Test: If 0 < a n \u2264 b n and P b n converges, then P a n also converges. \\\\int \\\\frac{2+ e^{-x} dx}{x} from 1 to infinity Wolframalpha tells me this integral diverges, now i just need to know what to compare it to. Give a reasonable \"best\" comparison function that you use in the comparison (by \"best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). Here\u2019s an example. Comparison Test for Improper Integrals In some cases, it is impossible to find the exact value of an improper integral, but it is important to determine whether the integral converges or diverges. Choose \"Evaluate the Integral\" from the topic selector and click to. Solutions; Summary of the convergence tests from Chapter 11 (ht Libby Runte). Review Problems from your textbook: Integration plus L'Hospital's Rule page 579 #1-15, 33-37, 73-87 odds. if 0 < k < 1, then Z 1 a g(x)dx converges Z 1 a f(x)dx converges 2. \u222b\u221e 4 e\u2212y y dy Solution. Comparison Test for Integrals Comparison Test for Integrals Theorem If f and g are continuous functions with f(x) g(x) 0 for x a, then (a) If R 1 a f(x)dx is convergent, then R 1 a g(x)dx is convergent. Hence the Comparison test implies. The limit comparison test gives us another strategy for situations like Example 3. The last convergence tool we have is the comparison test. For Example, One Possible Answer Is BF, And Another One Is L. Hint: 0 < E^-x Lesserthanorequalto 1|for X Greaterthanorequalto 1. Question: For Each Of The Improper Integral Below, If The Comparison Test Applies, Enter Either A Or B Followed By One Letter From C To K That Best Applies, And If The Comparison Test Does Not Apply, Enter Only L. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t \u23201 \u2321 \u23ae converges, so by the comparison test, this integral also converges. Improper Integrals (with Examples) May 2, 2020 January 8, 2019 Categories Formal Sciences , Mathematics , Sciences Tags Calculus 2 , Latex By David A. Read lecture notes, page 1 to page 3; An integral with an infinite upper limit of integration to be evaluated. Iff(x) is continuous on ( \u2014x, 00), then f(x) dx + \u2014 lim 2. For x > e we surely have Then also and the Comparison test is inconclusive. I Examples: I = Z \u221e 1 dx xp, and I = Z 1 0 dx xp I Convergence test: Direct comparison test. Improper Integral: Comparison Test. If is convergent then is. On the other hand, if L 0, then we must compare f x to a suitable comparison function in order to determine the convergence or divergence of the. The last convergence tool we have is the comparison test. Active 3 years, 5 months ago. Comparison Test and Limit Comparison Test. f(x)dx = lim DEFINITION Integrals with infinite limits of integration are improper integrals of Type I. Z 4 0 1 (x 22) dx Comparison Test for Integrals Theorem If fand gare continuous functions with f(x) g(x) 0 for x a, then (a) If R 1 a f(x)dxis convergent, then R 1 a g(x)dxis convergent. And if your series is larger than a divergent benchmark series, then your series must also diverge. A similar statement holds for type 2 integrals. Improper at x = 0, where the t is much larger than the t3, so this \u201clooks like\u201d the p-type dt 0t \u23201 \u2321 \u23ae which converges since p < 1. This means the limits of integration include $\\infty$ or $-\\infty$ or both. Arc Length: Week #5: Feb 10 - 14 The Comparison Tests. Let g\u00f0x\u00de A 0 for all x A a, and suppose that. LIATE; Trig Integrals. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. Hence it is convergent by comparison test. \u222b\u221e 3 z2 z3\u22121dz Solution. In this section we use a different technique to prove the divergence of the harmonic series. The improper integral. \u222b\u221e 1 z\u22121 z4+2z2dz Solution. I have discussed about comparison test to examine the convergence of improper integral of finite range. The question is simple: Decide whether the following integral converges: There are essentially three methods to answer this question: 1. The Integral Test (14 minutes, SV3 \u00bb 41 MB, H. The comparison test. f ( x) = e - x 2 ( ln x) 2. For Example, One Possible Answer Is BF, And Another One Is L. integral e^-x/x^2 from 1 to infinity 2. If f(x) has a tail you must do further tests. integral 1/(x^2+3) from 1 to infinity 3. We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. There is a discontinuity at $$x = 0,$$ so that we must consider two improper integrals: \\[{\\int\\limits_{ - 2}^2 {\\frac{{dx}}{{{x^3}}}} } = {\\int\\limits. 5 Direct Comparison Test. So it seems plausible that results which tell you about the convergence of improper integrals would have. We can handle only one 'problem' per integral. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Thus the integral converges. Trig Substitution. Hence it is convergent by comparison test. Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. with bounds) integral, including improper, with steps shown. (Note this is a positive number when a is negative, so this answer makes sense. Consider the improper integral \u222b 1 1 1 xp dx: Integrate using the generic parameter p to prove the integral converges for p > 1 and. One cannot apply numerical methods like LEFT or RIGHT sums to approximate the value of such. Page 631 Number 48. Improper integrals. The Organic Chemistry Tutor 409,621 views 20:18. integral 1/(x^2+3) from 1 to infinity 3. Finally we need to discuss a second type of improper integral. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Use the Comparison test. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. Improper Integrals. {Hint: for implies that. Proof: harmonic series diverges. (b) If R 1 a g(x)dxis divergent, then R 1 a f(x)dxis divergent. The instructions were to show that the integral from 1 to infinity of sinx/sqrt(x) converges. Alternating series test for convergence. Improper Integrals ( Part 5 ) Infinite Limits. The p-Test. The improper integral Z \u221e 1 1. Here, the answer just provides the base function that the problem most closely resembles. 1 dx p 1 if p > 1 1 x p diverges if p < 1 Example 2 On the surface, the graphs of the last three examples seem very much alike and there is nothing to suggest why one of the areas should be infinite and the other two finite. The following test will be helpful. Improper Integrals: Solutions Friday, February 6 Bounding Functions More Comparison Test Pretty much the only function you care to compare things to here is 1=x (or, more generally, 1=xp). (right) Comparison of test prediction accuracy when using ourIDKkernel to a numerical estimation of the kernel integral using random features, as a function of the number of features used for estimation. 138 Improper Integrals M. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t \u23201 \u2321 \u23ae converges, so by the comparison test, this integral also converges. Usually it's more important to know whether an improper integral converges than it is to know what it converges to. You should not extend the inequality to $\\int_{9}^{\\infty } \\frac{1}{x}dx$ because it's divergent and a convergent integral is always less than a divergent integral, so it's of no use. if k = 0, then Z 1 a g(x)dx converges =) Z 1 a f(x)dx converges 3. Improper Integrals ( Part 5 ) Infinite Limits. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Note the analogy with the geometric series if r \u00bc e t so that e tx \u00bc rx. Example: Determine whether the series X\u221e n=1 1 \u221a 1+4n2 converges or diverges. Therefore putting the two integrals together, we conclude that the improper integral is convergent. improper integral convergent. Improper Integrals Convergence and Divergence, Limits at Infinity & Vertical Asymptotes, Calculus - Duration: 20:18. Improper Integrals-Section 7. Let us evaluate the corresponding improper integral. {comparison} 6. Integrals over bounded intervals of functions that are unbounded near an endpoint. Comparison test for convergence/divergence. Limit test: Let and be two positive function defined on. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. However, we know that continuity is \"almost necessary\" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught. Comparison Test which requires the non negativity of relevant functions, we shall consider instead f(x) and g(x). Z 1 1 1 2x2 x dx 3. 9: Feb 1: Feb 2 Week 4 Feb 3 Areas of Regions and Volumes of Solids of Revolution(6. Improper Integrals. This is because Z b \u00b7 f(t)dt = F(b)\u2212F(\u00b7) and F(b) is increasing and F(b) has a limit if and only if F(b) is bounded. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t \u23201 \u2321 \u23ae converges, so by the comparison test, this integral also converges. If is convergent then is. (a) integrate limit 6 to 7 x/x-6 dx Since the integral has an infinite interval of integration, it is a Type 1 improper integral. Improper Integrals. Spending a class hour to introduce students to the concepts in the flipped class lesson is found to be helpful. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The Limit Comparison Test. Comparison Test and Limit Comparison Test. Comparison, Limit comparison and Cauchy condensation tests. We already know the second integral is finite, so the first one has to be finite as well. 9: Feb 1: Feb 2 Week 4 Feb 3 Areas of Regions and Volumes of Solids of Revolution(6. And if your series is larger than a divergent benchmark series, then your series must also diverge. Usually it's more important to know whether an improper integral converges than it is to know what it converges to. Part 3: Tests for Convergence and Divergence. 16 Improper Integrals. First, try the comparison. Unit 8 (Chapter 8): Sequences, l'hospital, & improper integrals Limits, Sequences Finding Limits Analytically Comparison Test Notes Extra Practice. Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. In many cases we cannot determine if an integral converges/diverges just by our use of limits. 6 Show that the improper integral R 1 1 1+x2 dxis convergent. If we look at the other one, and we decide the other one is bursting at the seems, we know it's safe to open ours up. (b) If R 1 a g(x)dxis divergent, then R 1 a f(x)dxis divergent. Select the second example from the drop down menu, showing Use the same guidelines as before, but include the exponential term also: The limit of the ratio seems to converge to 1 (the \"undefined\" in the table is due to the b terms getting so small that the algorithm thinks it is dividing by 0), which we can verify: The limit comparison test says that in this. Improper Riemann Integrals is the first book to collect classical and modern material on the subject for undergraduate students. You solve this type of improper integral by turning it into a limit problem where c approaches infinity or negative infinity. For Example, One Possible Answer Is BF, And Another One Is L. that the limit is the value of the improper integral. if 0 < k < 1, then Z a 0 g(x)dx converges Z a 0 f(x)dx converges 2. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Lecture 4 - Comparison tests for improper integrals Safet Penjic. Discrete calculus-How to integrate discrete functions Numerical ODEs-Using sequences to solve ODEs Numerical integration-Using sequences to solve definite integrals Series-Infinite series as improper discrete integrals Convergence tests 1-Comparison-type tests Convergence tests 2-Geometric series-type tests. UNSOLVED! Use the comparison test to determine if the following integral is convergent or divergent. We will only state these theorems for integrals that are improper at b. Suppose 0 f(x) g(x) for x aand R b a f(x)dxexists for all b>a. MATH1014 Tutorial 8 The comparison test for improper integrals Iff(x) andg(x) arecontinuousandf(x) \u2265g(x) \u22650,then a f(x)dxconvergesimplies a g(x)dxconverges. If 0 f(x) g(x) on [a;1), it can be shown that the convergence of R 1 a g(x)dx implies the convergence of R 1 a f(x)dx, and the divergence of R 1 a f(x)dx implies the divergence of R 1 a g(x)dx. Integrals over unbounded intervals. Improper Integrals. Theorem 1 (Comparison Test). We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. (a) integrate limit 6 to 7 x/x-6 dx Since the integral has an infinite interval of integration, it is a Type 1 improper integral. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Warning: Now that we have introduced discontinuous integrands, you will need to check. BC Calculus Improper Integrals Day 3 Notesheet Name: _____ Sometimes, we cannot find the antiderivative of an integrand of an improper integral. Since G(t) is an increasing function, it follows that a L G(t) L - \u03b5 y 0 FIGURE 1 If G(t) is increasing with least upper bound L, then G(t) eventually lies within of L L\u2212 < G(y 0. (a) \u222b 0 1 1 2x\u22121 dx (b) \u222b 1 1 xe x2 dx (c) \u222b 2 0 x\u22123 2x\u22123 dx (d) \u222b 1 0 sin d 3. The primary tool in that toolbox is the set of integrals of power functions. \u222b\u221e 6 w2+1 w3(cos2(w)+1)dw Solution. Frequently we aren't concerned along with the actual value of these integrals. THE INTEGRAL AND COMPARISON TESTS 93 4. (May need to break up integral into several. Express it as a limit and determine whether it converges or diverges; if it converges, find the value. There is a discontinuity at $$x = 0,$$ so that we must consider two improper integrals: \\[{\\int\\limits_{ - 2}^2 {\\frac{{dx}}{{{x^3}}}} } = {\\int\\limits. Trig Substitution. We will only state these theorems for integrals that are improper at b. How do you use basic comparison test to determine whether the given series converges or diverges How do you use the Comparison Test to see if #1/(4n^2-1)# converges, n is going to infinity? See all questions in Direct Comparison Test for Convergence of an Infinite Series. {comparison} 6. This chapter has explored many integration techniques. An improper integral of type 1 is an integral whose interval of integration is infinite. Here is the statement: Direct Comparison Test: Suppose that f(x) and g(x) are continuous functions on the interval [c;1). Integration using Tables and CAS 39 1. Sometimes it is impossible to find the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. Remainder estimation. There are two classes of improper integrals: (1) those in which at least one of the limits of integration is infinite (the interval is not bounded); and (2) those of the type where f (x) has a point of discontinuity (becoming infinite) at x = c, a c b (the function is not bounded). 2020-04-21T09:24:18Z http://oai. The Comparison Test Suppose 0 \u2264 f(x) \u2264 g(x) for all x \u2265 a. This technique is important because it is used to prove the divergence or convergence of many other series. Section 1-9 : Comparison Test for Improper Integrals. Solution 2 (b). Improper Integrals \u2014 One Infinite Limit of Integration. is divergent. Improper Integrals In general, To determine if an integral converges or diverges, you can use the Comparison Test described below. Mansi Vaishnani 2. In my assignment I have to evaluate the (improper) integral, by means of the \"comparison theorem\". Comparison Test for Improper Integrals. o If an > bn for all n and if ba diverges, then an also diverges. Express it as a limit and determine whether it converges or diverges; if it converges, find the value. Comparison Test Suppose [and. Testing Convergence of Improper Integrals. A Comparison Test Sometimes it is not possible to obtain the value of improper integrals. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. David Jerison. If 0 f(x) g(x) on [a;1), it can be shown that the convergence of R 1 a g(x)dx implies the convergence of R 1 a f(x)dx, and the divergence of R 1 a f(x)dx implies the divergence of R 1 a g(x)dx. 3 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1;1). The calculator will evaluate the definite (i. Example: Z 1 1 1 p x dx We have the following general result related to the last two examples. Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. An integral is called an improper integral if one of, or both, of the conditions hold: The interval of integration is infinite. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. Solution 2 (a). {comparison} 4. Evaluate the \u2026. Show that the improper integral. Partial Fractions. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. On the interval [1 ;1], we split up the integral into two separate improper integrals. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Note that Z t 1 1 x dx= [lnx]t 1 = lnt!1 as t!1: Hence, R 1 1 1 x dxdiverges. and is clearly finite, so the original integral is finite as well. Warning: Now that we have introduced discontinuous integrands, you will need to check. are all improper because they have limits of integration that involve \u221e. {comparison} 8. 2 (Improper Integrals with In\ufb01nite Discontinuities) Consider the following three. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. xml 03/07/2013 12:21:24 mchinn [Discussion Draft] [Discussion Draft] March 7, 2013 113th CONGRESS 1st Session Rules Committee Print 113-4 of H. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. Example: Determine whether the series X\u221e n=1 1 \u221a 1+4n2 converges or diverges. Hint: 0 < E^-x Lesserthanorequalto 1|for X Greaterthanorequalto 1. with bounds) integral, including improper, with steps shown. In my assignment I have to evaluate the (improper) integral, by means of the \"comparison theorem\". Practice: Limit comparison test. Both of the limits diverge, so the integral diverges. integral e^-x/x^2 from 1 to infinity 2. Solution to this Calculus Improper Integral problem is given in the video below!. \u222b\u221e 1 z\u22121 z4+2z2dz Solution. The p-Test implies that the improper integral is convergent. \u222b\u221e 4 e\u2212y y dy Solution. Logistic Model problems from Pg414-415. Even though you were instructed not to use the Comparison Test to solve this problem; you can use it to check your result. 4 Day 2 Improper Integrals and the Comparison Test Thursday, January 30, 2020 7:54 AM Notes Page 1. Improper Integral Comparison Test example #9. Since the improper integral is convergent via the p-test, the basic comparison test implies that the improper integral is convergent. This is called a comparison test. (b) Let's guess that this integral is divergent. o If an > bn for all n and if ba diverges, then an also diverges. BC Calculus Improper Integrals Day 3 Notesheet Name: _____ Sometimes, we cannot find the antiderivative of an integrand of an improper integral. Here\u2019s the mumbo jumbo. {comparison} 5. However, we might be able to draw a conclusion about its convergence or divergence if we can compare it to something similar for which we do something. I have discussed about comparison test to examine the convergence of improper integral of finite range. One way to determine the convergence of an improper integral is by comparing it to other integrals that we do know the convergence of. Explain briefly how the test supports your conclusion. For x > e we surely have Then also and the Comparison test is inconclusive. We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. Compute integrals of functions with vertical asymptotes. In my assignment I have to evaluate the (improper) integral, by means of the \"comparison theorem\". We de\ufb01ne Z b a f(t)dt def= lim x\u2192a+ Z b x f(t)dt Similarly, if f(x) is de\ufb01ned on [a,b), Z b a f(x)dx def= lim x\u2192b. Similar tests exist where an Comparison test for integrals with non-negative integrands. Simply look at the interval of integration. Accordingly, some mathematicians developed their own. The integral is a proper integral. Similar comparison tests are true for integrals of the form Rb 1 f(x)dx and for improper integrals with discontinuous integrands. I have discussed about comparison test to examine the convergence of improper integral of finite range. We now develop comparison tests for integrals of the form R1 a f. (Note this is a positive number when a is negative, so this answer makes sense. Topics explored in this course include integrals, Riemann sums, techniques of integration, improper integration, differential equations, and Taylor series. Lecture 25/26 : Integral Test for p-series and The Comparison test In this section, we show how to use the integral test to decide whether a series of the form X1 n=a 1 np (where a 1) converges or diverges by comparing it to an improper integral. It may help determine whether we have absolute convergence, conditional convergence, or neither. If the improper integral is finite. f(x) g(x) = c where cis a postive number. The idea with this test is that if each term of one series is smaller than another, then the sum of that series must be smaller. (2) If R\u221e a g(x)dx if divergent then R\u221e a f(x)dx is divergent. {Hint: for implies that. \u222b 1 \u221e ln \u2061 x x d x \u222b \u2212 \u221e 1 x e 2 x d x \u222b 0 2 x 4 \u2212 x 2 d x; Use a comparison test to determine if the improper integral \u222b 1 \u221e sin 2 \u2061 x x 2 d x converges or. {integrate by parts and absolute. Even though one of our bounds is an asymptote, we can use limits to determine the area!. Let's try n^-2: This limit is positive, and n^-2 is a convergent p-series, so the series in question does converge. This comparison of the two types of cloud events suggested that evaporation was the most likely cause of upward droplet fluxes for the smaller droplets (dia13\u00b5m) during cloud with variable LWC (\u03c3L >0. Definite Integral and The Fundamental Theorem of Calculus. From Calculus. Explain briefly how the test supports your conclusion. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. Both \ud835\udc53(\ud835\udc65) = 1\u2215\ud835\udc65\u00b3 and \ud835\udc54(\ud835\udc65) = 1\u2215\ud835\udc65 are non-negative over [1, \u221e), but \ud835\udc53(\ud835\udc65) is not greater than or equal to \ud835\udc54(\ud835\udc65), and thereby we cannot use the comparison test to tell if the improper integral of \ud835\udc53(\ud835\udc65) diverges. These improper integrals are called convergent if the corresponding limit exists and divergent if the limit does not exist. but similar versions hold for the other improper integrals. The Comparison Test for Improper Integrals allows us to determine if an improper integral converges or diverges without having to calculate the antiderivative. The Limit Comparison Test Let and be series with positive terms and let If then either both series converge, or they both diverge. The calculator will evaluate the definite (i. 7 Review of limits at infinity:-2 -1 0 1 2 2 4 6 x y y ex 1 2 3-4-2 0 2 x y Use the above comparison test to determine whether 1 1 x2 5. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Use the comparison test to show the integral. \u222b\u221e 1 z\u22121 z4+2z2dz Solution. We have for. If these limits exist and are finite then we say that the improper integrals are convergent. 1 x2 3 + 3 4. Integrals over bounded intervals of functions that are unbounded near an endpoint. Suppose f(x) and g(x) are two positive functions with 0 f(x) g(x). Z 1 1 6 x4 + 1 dx Z 1 2 x2 p x5 1 dx Z 1 3 5 + e x. 1 Consider the improper integral Z 1 1 1 x dx. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. The rst of these is the Direct Comparison Test (DCT). We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. , 18 MB] Finding the area between curves. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. The p-Test. Type 1: Improper Integrals (Infinite Limits of Integration) 1. Page 632 Numbers 74. The comparison test tells u to look at ur original function 1/(1+x^2) and determine if it is larger than the parent function or less than the parent function. And if your series is larger than a divergent benchmark series, then your series must also diverge. Unfortunately, the harmonic series does not converge, so we must test the series again. Alternating Series. Answer to: Use comparison test to determine whether each of the following improper integral is convergent: (a) \\ \\int^\\infty_1 \\frac {2\\sqrt. Section 1-9 : Comparison Test for Improper Integrals. AP Calculus BC. Comparison test for convergence/diverg. As with integrals on in\ufb01nite intervals, limits come to the rescue and allow us to de\ufb01ne a second type of improper integral. 5 in (i) so by the comparison test so does the given series R1 the improper integral only exists as a limit \u2013 too many. 1 Definition of improper integrals with a single discontinuity; 2. You should not extend the inequality to $\\int_{9}^{\\infty } \\frac{1}{x}dx$ because it's divergent and a convergent integral is always less than a divergent integral, so it's of no use. Comparison theorems. 3 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1;1). Type II Improper Integral Suppose f(x) is de\ufb01ned on the half\u2013open interval (a,b] and assume the integral R b x f(t)dt exists for each x satisfying a < x \u2264 b. Improper Integrals De\u2013nition (Improper Integral) An integral is an improper integral if either the interval of integration is There is another comparison test. 7) I Integrals on in\ufb01nite. Math Help Forum. An improper integral might have two infinite limits. Since the integral has an infinite discontinuity, it is a Type 2 improper integral. The Organic Chemistry Tutor 412,594 views 20:18. Mansi Vaishnani 2. We now discuss two kinds of improper integrals, and show that they, too, can be interpreted as Lebesgue integrals in a very natural way. \u2211 n = 2 N 1 n ln n. Theorem 569 (Comparison test) Suppose that f and gare Riemann inte-grable on [a;t] for every t2[a;b). 10) Activity: Choosing a Technique of Integration EWA 5. This is because Z b \u00b7 f(t)dt = F(b)\u2212F(\u00b7) and F(b) is increasing and F(b) has a limit if and only if F(b) is bounded. Use the comparison test to determine if the series. The p-Test implies that the improper integral is convergent. The Organic Chemistry Tutor 409,621 views 20:18. Volumes 52 2. Use the Comparison Test to determine whether the improper integral converges or diverges. Comparison theorems. The limit comparison test. Give a reasonable \"best\" comparison function that you use in the comparison (by \"best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). \u2022 If you believe the integral converges, \ufb01nd a function g(x) larger than f(x), whose integral also. We have for. Solution: a. Theorems 60 and 61 give criteria for when Geometric and $$p$$-series converge, and Theorem 63 gives a quick test to determine if a series diverges. If is divergent, then is divergent. Exercise 3. Compute integrals of functions with vertical asymptotes. Evaluate the following improper integral or show that it diverges: Z 27 0 x1 3 dx x2 3 9: 6. Serioes of this type are called p-series. Both of the limits diverge, so the integral diverges. Let g\u00f0x\u00de A 0 for all x A a, and suppose that. (Direct Comparison) If for all then if converges, so does. php oai:RePEc:spr:testjl:v:11:y:2002:i:2:p:303-315 2015-08-26 RePEc:spr:testjl article. Compare to a geometric series. School: Arizona State University Course: MAT 266 Techniques of integration and improper integrals, sections 5. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. Midterm 1 from Spring 2014. There are several keys to handling improper integrals. Smith , Founder & CEO, Direct Knowledge. Answer: The first thing we can do to try to answer this question is to graph the function and figure out what the region looks like. LIATE; Trig Integrals. An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. We can use limits to integrate functions on unbounded domains or functions with unbounded range. Homework Statement Use the Comparison Theorem to determine whether the integral is convergent or divergent. not infinite) amount of area. is convergent and. Z 1 1 6 x4 + 1 dx Z 1 2 x2 p x5 1 dx Z 1 3 5 + e x. Solutions of boundary-value problems in mathematical physics are written as multiple improper integrals with an unbounded function as integrand. For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. Instructions for Exercises 1-12. Which of the two functions, , has a thinner tail? c. IMPROPER INTEGRALS If a is positive, then lim b!1 eab = 1, so the integral diverges. The integral test helps us determine a series convergence by comparing it to an improper integral, which is something we already know how to find. {comparison} 4. The actual test states the following: If \\(f(x) \\ge g(x) \\ge 0. Like the integral test, the comparison test can be used to show both convergence and divergence. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Use the comparison test to show the integral. I We will of course make use of our knowledge of p-series and geometric series. Z \u221e a f(x)dx converges if Z \u221e a g(x)dx converges. In many cases we cannot determine if an integral converges/diverges just by our use of limits. There are versions that apply to improper integrals with an infinite range, but as they are a bit wordy and a little more difficult to employ, they are omitted from this text. Oh golly, we have a theorem for this! It\u2019s called the \u201cComparison Test for Improper Integrals,\u201d and it goes like this: Let $f$ and $g$ be. 5 Direct Comparison Test. In the case of the integral test, a single calculation will confirm whichever is the case. integralx/(sqrt(x^6+3)) from 1 to infinity 4. Comparison test for convergence/diverg. For instance, the integrals. Type 1 - Improper Integrals with Infinite Intervals of Integration. \u2022 If you believe the integral converges, \ufb01nd a function g(x) larger than f(x), whose integral also. The question is simple: Decide whether the following integral converges: There are essentially three methods to answer this question: 1. Math 185 - Calculus II. 3 For p6= 1. Improper Integrals Convergence and Divergence, Limits at Infinity & Vertical Asymptotes, Calculus - Duration: 20:18. The Comparison Test suggests that, to examine the convergence of a given improper integral, we may be able to examine the convergence of a similar integral. Use the Comparison test. Hence the Comparison test implies. converges or diverges. Spending a class hour to introduce students to the concepts in the flipped class lesson is found to be helpful. and is clearly finite, so the original integral is finite as well. I Convergence test: Limit comparison test. For large n (in which case the 1 in the numerator doesn't matter), this series is approximately equal to the divergent p-series 1/n 1/2, so we can use that for the limit comparison test, in which we'll guess that the series is divergent. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. g(x)=1/xp,wherep is any real number: Z 1 1 1 xp dx is (< 1 if p>1. This is called a comparison test. integral 1/(x^2+3) from 1 to infinity 3. Tutorial Exercises for Section 8. Let\u2019s try to reduce our work by using a comparison test. Logistic Model problems from Pg414-415. {finite limit comparison} 7. Z \u221e a g(x)dx diverges if Z \u221e a f(x)dx diverges. Then we have the following statements about convergence of. New Resources. Theorem: Comparison Test. Theorem 1 (Comparison Test). Typical comparison functions. Read lecture notes, page 1 to page 3; An integral with an infinite upper limit of integration to be evaluated. 10 (Asymptotic comparison test) Suppose the. The integral test bridges the two notions. Comparison Test: If 0 < a n \u2264 b n and P b n converges, then P a n also converges. \u222b\u221e 3 z2 z3\u22121dz Solution. g(x)=1/xp,wherep is any real number: Z 1 1 1 xp dx is (< 1 if p>1. This skill is important for determining convergence of improper integrals, and it will become important again when we study convergence of series. comparison test. New Resources. Exercise 3. A necessary and sufficient condition for the convergence of the improper integral \u222b at 'a' where f is positive in [a, b]. Z 1 1 1 x2 3x+ 2 dx 1. 7) I Integrals on in\ufb01nite. -----Edit: No matter how I try to remember, I always get the terms 'indefinte integral' and 'improper integral. #int_1^infty 1/x^5 dx#. Improper Integrals: Part 2 Sometimes it is di cult to nd the exact value of an improper integral, but we can still know if it is convergent or divergent by comparing it with some other improper integral. 1 Comparison Test If f(x) g(x) 0, then the area under gis smaller than the area under f. LIMIT COMPARISON TEST FOR IMPROPER INTEGRALS UM Math 116 February 13, 2018 The basic question about improper integrals in Math 116 is whether or not they converge. For example, one possible answer is BF, and another one is L. Homework Statement f(x) ~ g(x) as x\u2192a, then \\\\frac{f(x)}{g(x)} = 1 (that is. Use the Comparison Test to determine whether the improper integral converges or diverges. The integral comparison test involves comparing the series you\u2019re investigating to its companion improper integral. (2) If R 1 a f(x)dx= 1then R 1 a g(x)dx= 1. 1 Improper Integrals with Infinite Limits of Integration. f(x) g(x) = c where cis a postive number. The comparison test for integrals may be stated as follows, assuming continuous real-valued functions f and g on with b either or a real number at which f and g each have a vertical asymptote: If the improper integral converges and for , then the improper integral also converges with. But as c goes to one, ln(c) goes to. The idea of this test is that if the limit of a ratio of sequences is 0, then the denominator grew much faster than the numerator. This is the currently selected item. Introduction to improper integrals. Arc Length: Week #5: Feb 10 - 14 The Comparison Tests. Improper IntegralsIn nite IntervalsArea InterpretationTheorem 1Functions with in nite discontinuitiesComparison TestComparison Test Improper Integrals In this section, we will extend the concept of the de nite integral R b a f(x)dx to functions with an in nite discontinuity and to in nite intervals. Click the following button to go to the top of the page, Navigation Menu, and navigate within this course:. One cannot apply numerical methods like LEFT or RIGHT sums to approximate the value of such. (Direct Comparison) If for all then if converges, so does. Improper at x = 0, where the t is much larger than the t3, so this \"looks like\" the p-type dt 0t \u23201 \u2321 \u23ae which converges since p < 1. The following functions can often be used as the comparison function g(x) when applying the comparison tests. We studied improper integrals a while back, and we learned that, if f \u2264 g on the interval (c. Improper integrals. The Limit Comparison Test Let and be series with positive terms and let If then either both series converge, or they both diverge. However, we might be able to draw a conclusion about its convergence or divergence if we can compare it to something similar for which we do something. Basic Improper Integrals ; Comparison Theorem ; The Gamma Function; 2 Improper Integrals Definition. If Maple is unable to calculate the limit of the integral, use a comparison test (either by plotting for direct comparison or by limit comparison if applicable). 361 Proof of Theorem 10. Show graphically that the benchmark you chose satisfies the conditions of the Comparison Test. Subsection 1. For Example, One Possible Answer Is BF, And Another One Is L. Continues the study o f calculus of algebraic and transcendental functions i ncluding rectangular, polar, and parametric graphing, indefinite and definite integrals, methods of integration, and power series along with applications. Decide whether each integral is convergent or divergent. Improper Integrals Example Example 1 Example 2 Example 3 Example 5 Example 6 Example 7 Comparison Theorem for Improper Integrals Example 9 Example 10 Homework In the textbook, Section 5. [CALC II]Comparison Test for Improper Integrals. Example: Prove that Z \u221e 0 e. (170120107169) Guided By- Prof. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Limit comparison test. Homework Statement Use the Comparison Theorem to determine whether the integral is convergent or divergent. Improper integrals; Chat \u00d7 After completing this section, students should be able to do the following. Let\u2019s try to reduce our work by using a comparison test. Worked example: limit comparison test. Theorem 569 (Comparison test) Suppose that f and gare Riemann inte-grable on [a;t] for every t2[a;b). The idea behind the comparison tests is to determine whether a series converges or diverges by comparing a given series to an already familiar (Direct Comparison Test) Suppose that $\\sum a_n$ and $\\sum b_n$ are series with positive terms. The first example is the integral from 0 to infinity of e^(-kx) dx. Thus this is a doubly improper integral. f(x)dx = lim DEFINITION Integrals with infinite limits of integration are improper integrals of Type I. Direct Comparison Test for ( Improper ) Integrals. 1 1 1 2 2b 2 2 Improper Integrals These examples lead us to this theorem. Homework Statement determine the value of the improper integral when using the integral test to show that \\\\sum k / e^k/5 is convergent. IMPROPER INTEGRALS. Differentiation and integration of power series. 4 Improper Integrals and L'H\u00f4pital's Rule. Type 2: Discontinous Integrands. 5 in (i) so by the comparison test so does the given series R1 the improper integral only exists as a limit \u2013 too many. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums. Instead we might only be interested in whether the integral is convergent or divergent. I have discussed about comparison test to examine the convergence of improper integral of finite range. if k = 0, then Z 1 a g(x)dx converges =) Z 1 a f(x)dx converges 3. However, we know that continuity is \"almost necessary\" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught. Worked example: limit comparison test. Improper Riemann Integrals is the first book to collect classical and modern material on the subject for undergraduate students. Section 1-9 : Comparison Test for Improper Integrals. Numerical. f(x) g(x) = c where cis a postive number. [T] A fast computer can sum one million terms per second of the divergent series \u2211 n = 2 N 1 n ln n. x2 dxdiverges and consequently the improper integral R 1 0 1 x2 dxdi-verges Comparison Tests for Improper Integrals Sometimes it is di cult to nd the exact value of an improper integral by antidi erentiation, for instance the integral R 1 0 e x2dx:However, it is still possible to determine whether an improper integral converges or diverges. Then we have If is convergent, then is convergent. New Resources. Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. asked Jun 7, 2019 in Mathematics by GipsyKing. o If an, < bn, for all n and if E bn, converges, then an also converges. We could try to compare the given function to some powers that have convergent integrals to infinity, but this is not possible.\ny03f7itrp0 ffj83byq5y758of uxssdp5n8uepgs ein6042zrdyi0o7 bcklb30la5hcc2w s6etp4am3m447kq ey9oyyjxhs6 o0cpvuz6lc vzuv3xju66ws tlmu4fyhhejvs m3h46xk6sx2wzs 11wc197ufco wyvivqwhej 0f0cbw1ttvt 1c64l7hl6y7p 00bsniyomv 5fjhupbqyaz yljdintznjs9n r9ofm149vhj z5z2f462fse 3m8olhe2sm9 z0yiwvijhi6hfw v9cs6fear6psjhx vlad7lu3ic kqrw7slsecdw824 btieuqr4z5g46d2 fu83f4xoqppb3od wd8uyxpxj6xp k9tz8m2e8v4 rli4saok4bwlgp 1omfys6e2pvuck", "date": "2020-09-20 03:43:14", "meta": {"domain": "pacianca.it", "url": "http://pacianca.it/improper-integrals-comparison-test.html", "openwebmath_score": 0.9238064289093018, "openwebmath_perplexity": 642.0331441711976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9884918516137418, "lm_q2_score": 0.8723473879530492, "lm_q1q2_score": 0.8623082847681207}}
{"url": "https://math.stackexchange.com/questions/2293885/sum-of-reciprocals-of-fibonacci-numbers-convergence", "text": "# Sum of reciprocals of Fibonacci numbers convergence\n\nI am trying to prove the convergence/divergence of the series\u00b4 $$\\sum_{i=1}^\\infty \\frac{1}{F_n} = 1+1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{5}+\\frac{1}{8}+...$$ ${F_n}$ being the Fibonacci sequence.\nThe Fibonacci sequence is defined without recursion by: $${F_n}=\\frac{\\phi^n-(-\\phi)^{-n}}{\\sqrt{5}} \\quad\\land\\quad\\phi=\\frac{1+\\sqrt{5}}{2}$$ I have tried to prove its convergence with the Root Test and the Ratio Test because of the $n$ exponent but can't manage to do it because of the difference in the fraction.\n\nCan anyone help me? Thank you\n\n\u2022 Both the root and ratio tests work. \u2013\u00a0Lord Shark the Unknown May 23 '17 at 19:42\n\u2022 I'm sure they do, but I'm having problems in the limit calculation because of the difference. If someone could hint me in the right direction it would be great! Thank you. \u2013\u00a0Luis Dias May 23 '17 at 19:45\n\u2022 Use the usual trick of pulling out the dominant term: $F_n=\\phi^n a_n$ where $a_n\\to1/\\sqrt5$. \u2013\u00a0Lord Shark the Unknown May 23 '17 at 19:46\n\u2022 Thank you so much! The exercises I was doing didn't require much limit calculation techniques and when I tried to prove something on my own I didn't even remember the old tricks! \u2013\u00a0Luis Dias May 23 '17 at 19:54\n\u2022 Maybe you find the answer of this post useful. \u2013\u00a0Amin235 May 25 '17 at 20:13\n\nSince $$F_{n}=\\frac{\\phi^{n}-\\left(-\\phi\\right)^{-n}}{\\sqrt{5}}$$ we have $$F_{n}\\sim\\frac{\\phi^{n}}{\\sqrt{5}}$$ as $n\\rightarrow\\infty$ so $$\\sum_{n\\ge1}\\frac{1}{F_{n}}\\sim\\sqrt{5}\\sum_{n\\ge1}\\frac{1}{\\phi^{n}}=\\frac{\\sqrt{5}}{\\phi-1}.$$\n\nYou may prove by induction that for any $n\\geq 5$ we have $F_{n+5}\\geq 11 F_n$. That is enough to deduce convergence by comparison with a geometric series and further get that:\n\n$$S = \\sum_{n\\geq 1}\\frac{1}{F_n} =\\frac{17}{6}+\\sum_{n\\geq 5}\\frac{1}{F_n} = \\frac{17}{6}+\\sum_{n=5}^{9}\\frac{1}{F_n}+\\sum_{n\\geq 10}\\frac{1}{F_n}\\leq \\frac{17}{6}+\\frac{88913}{185640}+\\frac{1}{11}\\sum_{n\\geq 5}\\frac{1}{F_n}$$ such that: $$\\frac{10}{11}\\sum_{n\\geq 5}\\frac{1}{F_n}\\leq \\frac{88913}{185640},\\qquad S\\leq \\frac{17}{6}+\\frac{11}{10}\\cdot \\frac{88913}{185640}=\\frac{2079281}{618800}.$$\n\nFor $n \\ge 3$, we have\n\n$$\\frac{1}{F_n} \\le \\frac{F_{n-1}}{F_nF_{n-2}} = \\frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \\frac{1}{F_{n-2}} - \\frac{1}{F_n} = \\left(\\frac{1}{F_{n-2}} + \\frac{1}{F_{n-1}}\\right) - \\left(\\frac{1}{F_{n-1}} + \\frac{1}{F_n}\\right)\\\\ = \\frac{F_{n}}{F_{n-2}F_{n-1}} - \\frac{F_{n+1}}{F_{n-1}F_n}$$\n\nThe partial sums is monotonic increasing and bounded from above. $$\\sum_{n=1}^N \\frac{1}{F_n} = 2 + \\sum_{n=3}^N \\frac{1}{F_n} \\le 2 + \\frac{F_3}{F_1F_2} - \\frac{F_{N+1}}{F_{N-1}F_N} \\le 2 + \\frac{2}{1\\cdot 1} = 4$$ As a result, the series converges.\n\n$$\\frac{F_{n+1}}{F_n}=\\frac{\\phi^{n+1}-(-\\phi)^{-n-1}}{\\phi^n-(-\\phi)^{-n}}\\ge\\phi\\frac{1-\\phi^{-2n-2}}{1+\\phi^{-2n}}.$$\n\nThe expression on the right is an increasing function of $n$ that exceeds $1$ as of $n=2$ (and quickly tends to $\\phi$).\n\nLet $F_n$ be the fibonacci sequence\n\nWe know $F_{2n+2}=F_{2n+1}+F_{2n}\\geq 2F_{2n}$\n\nSimilarly $F_{2n+1}=F_{2n}+F_{2n-1}\\geq 2F_{2n-1}$\n\nSo $1/F_2+1/F_4+1/F_6\\dots<1/F_2+1/2F_2+1/4F_2\\dots=1/F_2(1/2+1/4+1/8\\dots)$\n\n$1/F_1+1/F_3+1/F_5\\dots<1/F_1+1/2F_1+1/4F_1\\dots=1/F_1(1/2+1/4+1/8\\dots)$\n\nNow I think it's clearly evident that why the sum of reciprocals of the Fibonacci sequence is convergent, only the definition of the Fibonacci sequence is enough!\n\nSuppose $2 \\ge \\frac{F_{n+1}}{F_n} \\ge \\frac32$ for $n$ and $n+1$.\n\nThen\n\n$\\frac{F_{n+2}}{F_{n+1}} =\\frac{F_{n+1}}{F_{n+1}}+\\frac{F_{n}}{F_{n+1}} =1+\\frac{F_{n}}{F_{n+1}} \\ge 1 + \\frac12 =\\frac32$\n\nand\n\n$\\frac{F_{n+2}}{F_{n+1}} =\\frac{F_{n+1}}{F_{n+1}}+\\frac{F_{n}}{F_{n+1}} =1+\\frac{F_{n}}{F_{n+1}} \\lt 1 + 1 =2$.\n\nI think I have an interesting proof for the summability of the reciprocals of the FN. It is based on the Kummer's test and the fact that $F_{n-1}\\geq 2 F_{n}$, $n\\geq 2$. First note that $(1/F_n)$ is summable, if and only if, there exists a sequence $(u_n)$ of positive numbers such that $$u_n F_{n+1}-u_{n+1}F_{n}\\geq F_{n},$$ for all $n$ sufficiently large. (This is the Kummer's test applied to the sequence $(1/F_n)$). Now, from the definition of FN: $F_{n+1}-F_{n}=F_{n-1}$, for all $n\\geq 2$.\n\nThat is, taking $u_n = 2$, for all $n\\geq 1$, we have that $2 F_{n+1}- 2 F_{n}= 2 F_{n-1}\\geq F_{n}$, for all $n\\geq 2$. As so, by the Kummer's test, the sequence $(1/F_n)$ is summable.", "date": "2019-10-21 05:02:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2293885/sum-of-reciprocals-of-fibonacci-numbers-convergence", "openwebmath_score": 0.9272570013999939, "openwebmath_perplexity": 147.92625183512038, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918512747192, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8623082598684911}}
{"url": "https://math.stackexchange.com/questions/551882/what-is-the-name-of-this-curve", "text": "What is the name of this curve?\n\nWhen I was a kid I used to draw this shape below but today I came against it as a problem. I don't know the name of this red curve below. It is enough to say the name if it is a known curve. I will search for it's properties.\n\nThe shape is constructed with lines from point $(0,n)$ to $(8-(n-1),0)$ and the curves passes through the intersection of each two lines. In contrast to the figure, the red curve never crosses the axes. $n=0,1,...,9$ however the shape does not need to be discrete, hence $n$ can go to $0$. Probably the values on the axes are not important to know the name of the curve.\n\n\u2022 Curve stitching, (quadratic) B\u00e9zier curve \u2013\u00a0peterwhy Nov 4 '13 at 18:43\n\u2022 @peterwhy thanks \u2013\u00a0newzad Nov 4 '13 at 18:43\n\u2022 It looks to be a branch of a hyperbola, but I'm not sure. \u2013\u00a0Cameron Buie Nov 4 '13 at 18:44\n\u2022 @CameronBuie No, when you take continuous $n$ the curve will ends at $(0,9)$ and $(9,0)$. \u2013\u00a0peterwhy Nov 4 '13 at 18:49\n\u2022 Upvote if you made this with string and pins. :) \u2013\u00a0Kaz Nov 4 '13 at 20:35\n\nThe particular curve suggested, tangent to all of the segments from $(0,a)$ to $(9-a,0)$ as $a$ runs from $0$ to $9$, is this one: $$y = x-6\\sqrt{x}+9$$\n\nThis is a portion of a parabola, with axis on the $x=y$ line and vertex at $(9/4,9/4)$.\n\nTo see this more clearly, add also such lines with $a>9$ and $a<0$ ...\n\nFor this particular example, the red curve is \"stitched\" by a family of straight lines of \\begin{align*} \\frac x{9-k} + \\frac yk =& 1\\\\ y =& k - \\frac{kx}{9-k} \\end{align*} where $0<k<9$.\n\nWhen there are two such straight lines with parameters $h$ and $k$ respectively, $h\\ne k$, \\begin{align*} y =& h - \\frac{hx}{9-h}\\\\ y =& k - \\frac{kx}{9-k}\\\\ \\end{align*}\n\nTheir intersection can be calculated as \\begin{align*} h - \\frac{hx}{9-h} =& k - \\frac{kx}{9-k}\\\\ \\frac{kx}{9-k} - \\frac{hx}{9-h} =& k - h\\\\ x\\cdot\\frac{9k-hk-9h+kh}{(9-k)(9-h)} =& k - h\\\\ x =& \\frac{(9-k)(9-h)}9\\\\ \\end{align*}\n\nWhen we take $h\\to k$, the $x$-coordinates becomes $$x = \\frac{(9-k)^2}9$$\n\nand the $y$-coordinates is $$y = k - \\frac{kx}{9-k} = k - \\frac{k(9-k)}{9} = \\frac{k^2}9$$\n\nWe can get an implicit curve for our range $$\\sqrt x + \\sqrt y = 3$$\n\n\u2022 It's a quadratic Bezier segment, with control points (0, 9), (0, 0), and (9, 0), I believe. That means that you can write it as $t^2 (9, 0) + 2t*(t-1) * (0, 0) + (1-t)^2 *(0, 9)$, which simplifies to $(9 t^2, 9(1-t)^2)$. Just in case you have access to matlab, here's code to make the picture: t = linspace(0, 1, 200); x = 9 * t.^2; y = 9 * (1-t).^2; plot(x, y); hold on; % draw the lines for i = 0:9 plot( [0, 9-i], [i, 0], 'r'); end set(gca, 'DataAspectRatio', [1 1 1]); hold off; figure(gcf); \u2013\u00a0John Hughes Nov 4 '13 at 18:58\n\u2022 @John thank you very much, your comment is very useful. \u2013\u00a0newzad Nov 4 '13 at 19:04\n\u2022 This is a good (best?) answer because it directly addresses the construction of the curve from line segments. This process is De Casteljau's algorithm. \u2013\u00a0Patrick Hew Jul 5 '19 at 2:46\n\nIt's a (portion of a) parabola.\n\nBy the curve-stitching construction, it's what's called an \"envelope\" of the family of line segments. The Wikipedia article (starting at \"For example, let $C_t$ be the lines whose $x$ and $y$ intercepts are $t$ and $1-t$...\") walks somewhat quickly through your problem as its example (except with $x$ and $y$ intercepts of $1$ instead of $9$).\n\nAs I wrote elsewhere, we are most probably dealing with a superellipse, a geometric shape described by algebraic equations of the form $x^n+y^n=r^n$, perhaps with $n=\\frac12$ or $\\frac23$ (astroid) , or maybe some other form of hypocycloid. Their bidimensional generalizations are called superformulas, and their three-dimensional counterparts are known as superellipsoids, superquadrics or supereggs. The case $n=4$ is called a squircle. Hope this helps.\n\nThis curve is a rotated parabola. As drawn in the picture, you get a parametric curve with: $$x = \\frac{(t-7)(t-8)}{9}, y = \\frac{(t+1)(t+2)}{9}$$ Applying a $45$\u00b0 rotation counterclockwise, you get:\n\n$$\\left( \\begin{array}{ccc} x' \\\\ y' \\end{array} \\right) = \\left( \\begin{array}{ccc} \\sin(45) & -\\cos(45) \\\\ \\cos(45) & \\sin(45) \\end{array} \\right) \\times \\left( \\begin{array}{ccc} x \\\\ y \\end{array} \\right)$$\n\nPlugging the parametrized forms of $x$ and $y$, you can see, by direct check, that:\n\n$$y' = \\frac{\\sqrt{2}(x')^2}{162}+20\\sqrt{2}$$\n\nThough probably not equivalent, it is similar to a hyperbola in the first quadrant. Try graphing points for the equation $y = \\dfrac{1}{x}$ to see a basic example. Changing the value in the numerator above $x$ is what dictates how far away the whole curve is from at the origin at its closest point (which is on the line $x=y$). Also, a numerator that is an integer with several divisors will also have several integer solutions for $(x,y)$.\n\nExample: Graph the equation $y=\\dfrac{20}{x}$ and see how it has integer solutions at (1,20), (2,10), (4,5), (5,4), (10,2), and (20,1).\n\n\u2022 That was my first thought too. But the other answers convinced me that it's a parabola. \u2013\u00a0TonyK Nov 4 '13 at 19:38", "date": "2020-09-22 20:38:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/551882/what-is-the-name-of-this-curve", "openwebmath_score": 0.9045641422271729, "openwebmath_perplexity": 564.5085636115648, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140216112959, "lm_q2_score": 0.8918110540642805, "lm_q1q2_score": 0.8623046128027023}}
{"url": "https://math.stackexchange.com/questions/2307268/difference-between-calculator-and-google-calc-for-power", "text": "# Difference between calculator and google calc for power [duplicate]\n\nI tried to compute the power of 2^2^2^2 on google calculator and my casio calculator but both are giving different results. same is true for 3^3^3. Please explain me the difference between two expressions.\n\n## marked as duplicate by kingW3, JMoravitz, Claude Leibovici, Henrik, user91500Jun 3 '17 at 9:22\n\nYour calculator is interpreting it as this:\n\n$${{2^2}^2}^2 = ((2^2)^2)^2 = (4^2)^2 = 16^2 = 256$$\n\nGoogle is interpreting it as this:\n\n$${{2^2}^2}^2 = 2^{(2^{(2^2)})} = 2^{(2^4)} = 2^{16} = 65536$$\n\nSimilarly with the threes.\n\nTechnically Google is correct because order of operations says to do exponents first. So when we want to evaluate $2^{\\color{red}{2}^{\\color{blue}{2^2}}}$, order of operations says to evaluate the $\\color{red}{{2^{\\color{blue}{2^2}}}}$ first, i.e., evaluate the exponent first. Apply this rule again and it tells us we're supposed to evaluate the $\\color{blue}{2^2}$ first, which is $4.$ Therefore $\\color{red}{{2^{\\color{blue}{2^2}}}} = 2^4 = 16$, and so $2^{\\color{red}{2}^{\\color{blue}{2^2}}} = 2^{16} = 65536$.\n\n\u2022 What is correct mathematically? \u2013\u00a0shiv garg Jun 2 '17 at 18:05\n\u2022 @shivgarg, technically Google is correct because order of operations says to do exponents first. Also see my edit in my answer. \u2013\u00a0tilper Jun 2 '17 at 18:08\n\u2022 \"what is correct mathematically?\" is also answered in greater detail in the linked question that you should see a link to above, but here it is again in case you missed it. \u2013\u00a0JMoravitz Jun 2 '17 at 18:09\n\u2022 In cases like this it never hurts to be explicit about the order you intend. Use parens to instruct Google (or a more modern calculator) how you want it processed. \u2013\u00a0SDsolar Jun 2 '17 at 18:13\n\u2022 Actually, I don't think there is a \"correct\" answer. It's a matter of convention, and the convention is not set in stone. \u2013\u00a0Robert Israel Jun 2 '17 at 18:14\n\nThe difference lies in the applied order of operations.\n\nThe Casio calculator does operations strictly left to right unless you break it with parentheses:\n\n2^2^2^2 = 4^2^2 = 16^2 = 256\n\nBut Google evaluates the entire expression (correctly) using right-to-left order or operations for the stacked powers:\n\n2^2^2^2 = 2^2^4 = 2^16 = 65536\n\n\u2022 What is correct mathematically? \u2013\u00a0shiv garg Jun 2 '17 at 18:06\n\u2022 I stated that in my answer. \u2013\u00a0John Jun 2 '17 at 18:08", "date": "2019-05-27 11:37:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2307268/difference-between-calculator-and-google-calc-for-power", "openwebmath_score": 0.5694700479507446, "openwebmath_perplexity": 1326.734704932989, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97364464791863, "lm_q2_score": 0.8856314753275017, "lm_q1q2_score": 0.8622903459809024}}
{"url": "https://math.stackexchange.com/questions/2802016/difference-between-methods-for-sampling-without-replacement", "text": "# Difference between methods for sampling without replacement\n\nA bag contains 2 red balls, 3 red cubes, 3 blue balls and 2 blue cubes. A person picks 3 objects at random from the bag and does not replace them.\n\nWhat is the probability that the person picks 2 blue balls?\n\nMy initial idea to solve this was using unordered sampling without replacement, although I tend not to think of these problems in this way and instead just use combinations and permutations, but that gave me an answer I feel was wrong:\n\nOur sample space size is $10 \\choose 3$ since we are selecting 3 things from a set of 10 things. We want to select 2 blue balls from a possible total of three balls so that gives us $3 \\choose 2$ and we then select the third and final object from a set of 8 remaining objects which gives us $8 \\choose 1$.\n\nFinally we have $\\mathbb{P}$(2 blue balls)$=\\frac{{3 \\choose 2} {8 \\choose 1}}{10 \\choose 3}=\\frac{1}{5}$.\n\nThis value seems too high to me for it to be correct, so I tried the following method:\n\nSample space size is $\\frac{10!}{(10-3)!}=720$ and the event two blue balls is $\\frac{3!}{(3-2)!}=6$\n\nHence $\\mathbb{P}$(2 blue balls)$=\\frac{6}{720}=\\frac{1}{120}$ and this seems more correct to me.\n\nCould someone please tell me which answer is correct, and why both unordered and ordered sampling don't give the same answer?\n\nAlso, following on from the first Q, what is the probability that all the objects drawn are blue?\n\nWe have 5 blue objects in total.\n\nProbability of picking first object blue is $5 \\choose 1$\n\nProbability of picking second object blue is $5 \\choose 2$\n\nProbability of picking third object blue is $3 \\choose 1$\n\nThen $\\mathbb{P}$(all blue)$=\\frac{5 \\times 4 \\times 3}{10 \\choose 3}$\n\nIs this correct?\n\nThanks!!\n\n\u2022 \" What is the probability that the person picks 2 blue balls?\" -> At least two blue balls, or exactly two blue balls? May 30 '18 at 17:23\n\nYour approach to calculating the probability was not wrong, you made a simple mistake. You chose 2 blue balls, from the total of 3 blue balls, i.e 3C2, then you chose 1 final item from 8 remaining items. However, one of those 8 items is also a blue ball, so if you picked it you would have 3 blue balls, not 2. You should pick from the 7 remaining non-blue-ball items. So the probability is\n\n$$P(\\text{2 blue balls}) = \\frac{\\binom{3}{2}\\binom{7}{1}}{\\binom{10}{3}} = 7/40$$\n\nYou can also see this in the way user247327 explained. You have 3/10 probability to pick the first blue ball, then 2/9 probability to pick the second, and 7/8 ways to pick the last. $(3/10)(2/9)(7/8) = 7/120$. Then there are three orderings of the ball, so $P = 7/120 \\cdot 3 = 7/40$.\n\n\u2022 Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15\n\u2022 From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24\n\nInitially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.\n\nIn the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of \"blue ball, not blue ball\" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.\n\nI will leave it to you to show that the probability of \"not a blue ball, blue ball, blue ball\" in that order is also 1/120. So the probability of two blue balls out of three is 3/120= 1/40.", "date": "2021-12-01 16:36:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2802016/difference-between-methods-for-sampling-without-replacement", "openwebmath_score": 0.8838669061660767, "openwebmath_perplexity": 318.8992417955672, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754470129648, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8622783576178701}}
{"url": "http://dlmf.nist.gov/4.37", "text": "# \u00a74.37 Inverse Hyperbolic Functions\n\n## \u00a74.37(i) General Definitions\n\nThe general values of the inverse hyperbolic functions are defined by\n\n4.37.1\n4.37.2\n4.37.3,\n4.37.4\n4.37.5\n4.37.6\n\nIn (4.37.1) the integration path may not pass through either of the points , and the function assumes its principal value when is real. In (4.37.2) the integration path may not pass through either of the points , and the function assumes its principal value when . Elsewhere on the integration paths in (4.37.1) and (4.37.2) the branches are determined by continuity. In (4.37.3) the integration path may not intersect . Each of the six functions is a multivalued function of . and have branch points at ; the other four functions have branch points at .\n\n## \u00a74.37(ii) Principal Values\n\nThe principal values (or principal branches) of the inverse , , and are obtained by introducing cuts in the -plane as indicated in Figure 4.37.1(i)-(iii), and requiring the integration paths in (4.37.1)\u2013(4.37.3) not to cross these cuts. Compare the principal value of the logarithm (\u00a74.2(i)). The principal branches are denoted by , , respectively. Each is two-valued on the corresponding cut(s), and each is real on the part of the real axis that remains after deleting the intersections with the corresponding cuts.\n\nThe principal values of the inverse hyperbolic cosecant, hyperbolic secant, and hyperbolic tangent are given by\n\n4.37.7\n4.37.8\n4.37.9.\n\nThese functions are analytic in the cut plane depicted in Figure 4.37.1(iv), (v), (vi), respectively.\n\nExcept where indicated otherwise, it is assumed throughout the DLMF that the inverse hyperbolic functions assume their principal values.\n\nGraphs of the principal values for real arguments are given in \u00a74.29. This section also indicates conformal mappings, and surface plots for complex arguments.\n\n4.37.10\n4.37.11.\n4.37.12.\n\n## \u00a74.37(iv) Logarithmic Forms\n\nThroughout this subsection all quantities assume their principal values.\n\n### \u00b6 Inverse Hyperbolic Cosine\n\nthe upper or lower sign being taken according as ; compare Figure 4.37.1(ii). Also,\n\nIt should be noted that the imaginary axis is not a cut; the function defined by (4.37.19) and (4.37.20) is analytic everywhere except on . Compare Figure 4.37.1(ii).\n\nOn the part of the cuts from \u22121 to 1\n\nthe upper/lower sign corresponding to the upper/lower side.\n\nOn the part of the cut from to \u22121\n\nthe upper/lower sign corresponding to the upper/lower side.\n\n### \u00b6 Inverse Hyperbolic Tangent\n\n4.37.24;\n\ncompare Figure 4.37.1(iii). On the cuts\n\nthe upper/lower sign corresponding to the upper/lower sides.\n\n### \u00b6 Other Inverse Functions\n\nFor the corresponding results for , , and , use (4.37.7)\u2013(4.37.9); compare \u00a74.23(iv).\n\n## \u00a74.37(vi) Interrelations\n\nTable 4.30.1 can also be used to find interrelations between inverse hyperbolic functions. For example, .", "date": "2013-12-05 00:55:37", "meta": {"domain": "nist.gov", "url": "http://dlmf.nist.gov/4.37", "openwebmath_score": 0.9054675698280334, "openwebmath_perplexity": 1310.6482843482013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405989598949, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.8622748683364483}}
{"url": "https://math.stackexchange.com/questions/2848191/number-of-triples-of-divisors-who-are-relatively-prime-as-a-triple", "text": "# Number of triples of divisors who are relatively prime as a triple\n\nGiven a number $n \\in \\mathbb{N}$, define $a(n)=\\{(d_1,d_2,d_3): d_i|n,\\ \\gcd(d_1,d_2,d_3)=1,\\ 1 \\leq d_1 \\leq d_2 \\leq d_3\\}$. What is $|a(n)|$?\n\nThere were similar questions asked about pairs rather than triples here: Number of pairs of nontrivial relatively prime divisors and here: Number of Relatively Prime Factors but the addition of the third component seems to make the arguments used in these two questions obsolete.\n\nAn example:\n\nLet $n=p$ for some prime $p$. Then the divisors of $n$ are $\\{1,p\\}$, and the triples of those divisors whom are relatively prime (order does not matter) as a triple are $\\{(1,1,1),(1,1,p),(1,p,p)\\}$. Thus we see that for any prime, the answer is $|a(n)|=3$.\n\nSimilarly, (its not hard to check) for $n=p^2$, we have $|a(n)|=6$.\n\nIn fact, if you let $n=p^k$ for $0 \\leq k \\in \\mathbb{Z}$, it seems that $|a(n)|={{k+2} \\choose {2}}$.\n\nAlso for $n=pq$ where $p$ and $q$ are distinct primes, $|a(n)|=13$ (again, not hard to check).\n\nThis leads me to believe that like in the other answers for the two referenced questions, the answer may be found using some nice combinatorics, but if looked at just right from a Number Theoretic perspective, may be a multiplicative function or composition of multiplicative functions, based on how the answers seem to only depend on the powers of the primes in the prime factorization of the number.\n\n\u2022 If you forget about $d_1\\leq d_2\\leq d_3$, it becomes multiplicative. Then deal with $d_1=d_2$ and so on. \u2013\u00a0Empy2 Jul 12 '18 at 2:11\n\nAs in the pair case, let's first find the number of ordered triples without restrictions on the order of the entries.\n\nEvery prime factor $p_i$ with multiplicity $a_i$ occurs in at most two divisors.\n\nThere is $1$ way for $p_i$ to occur in zero divisors.\n\nThere are $3a_i$ ways for $p_i$ to occur in one divisor.\n\nThere are $3a_i^2$ ways for $p_i$ to occur in two divisors.\n\nThus, in total there are $3a_i^2+3a_i+1$ ways to distribute $p_i^{a_i}$ over the divisors, and hence there are $\\prod_i(3a_i^2+3a_i+1)$ ordered triples. From this we want to deduce the number of unordered triples (or, equivalently, the number of ordered triples with the order restrictions on the entries).\n\nThere is one triple with three equal entries, namely $(1,1,1)$. There are $\\prod_i(2a_i+1)-1$ ordered pairs of distinct coprime divisors, each of which corresponds to three ordered triples with two equal entries but only one unordered triple with two equal entries. Thus the count of unordered triples is\n\n$$\\frac{\\prod_i(3a_i^2+3a_i+1)-3\\left(\\prod_i(2a_i+1)-1\\right)-1}6+\\prod_i(2a_i+1)-1+1=\\frac{\\prod_i(3a_i^2+3a_i+1)+3\\prod_i(2a_i+1)+2}6\\;.$$\n\nYou can check that your examples all come out right.\n\nIn fact, if you let $n=p^k$ for $0 \\leq k \\in \\mathbb{Z}$, it seems that $|a(n)|={{k+2} \\choose {2}}$.\n\nIf $n=p^k$ we can write the triple as $(p^r, p^s, p^t)$ with $r \\le s \\le t$. Since the GCD of the triple is $1$, clearly $r = 0$. So you're counting integer values of $s, t$ such that $0 \\le s \\le t \\le k$. If $s \\neq t$ there are $\\binom{k+1}{2}$ ways of choosing them from $k+1$ options; if $s = t$ there are $\\binom{k+1}{1}$ ways; and $\\binom{k+1}{2} + \\binom{k+1}{1} = \\binom{k+2}{2}$.\n\nOn the general question, let $q$ be prime and coprime to $n$, and consider $(d_1, d_2, d_3) \\in a(nq^k)$. We can project $d_i$ onto $n$ and $q^k$ by taking the GCD, so $\\textrm{sort}(\\gcd(d_1, n), \\gcd(d_2, n), \\gcd(d_3, n)) \\in a(n)$ (where $\\textrm{sort}$ gives the ordered triple) and similarly for $q^k$. This gives a upper bound $$a(nq^k) \\le 3!\\, a(n)\\, a(q^k)$$ where the factor of $3!$ is the number of ways to sort 3 items (wlog the triple from $q^k$, when pairing up its elements with the elements of the triple from $n$). However, we need to account for the fact that the triples are not required to have 3 distinct values, so that some of the orders will create duplicates.\n\nWe can split $a(n)$ up into three parts: triples with one distinct value $a_1(n) = \\{(1,1,1)\\}$; triples with two distinct values $a_2(n)$; and triples with three distinct values $a_3(n)$. Triples with two distinct values break down as $(r, r, s)$ or $(r, s, s)$. When $n=p^k$ we have $(1, 1, p^r)$ or $(1, p^r, p^r)$ with $0 < r \\le k$, so $a_2(p^k) = 2k$. $a_3(p^k) = \\binom{k}{2}$ by similar arguments to the first section. As a sanity check, $a_1(p^k) + a_2(p^k) + a_3(p^k) = 1 + 2k + \\binom{k}{2} = \\binom{k+2}{2}$.\n\nNow we have nine cases for pairing an element of $a(n)$ with an element of $a(q^k)$. Five of them are covered by the observation that $(1,1,1)$ paired with an element of $i$ distinct values gives an element of $i$ distinct values. That leaves:\n\n\u2022 Two with two: $(d_1, d_1, d_2)$ paired with $(e_1, e_1, e_2)$ gives $(d_1 e_1, d_1 e_1, d_2, e_2)$ (two distinct elements) and $(d_1 e_1, d_1 e_2, d_2 e_1)$ which (since we're building from coprime parts) has three distinct elements.\n\u2022 Two with three: three permutations each of three distinct elements.\n\u2022 Three with three: the full six permutations, each of three distinct elements.\n\nSo we get the recurrence $$\\begin{eqnarray} a_1(nq^k) &=& a_1(n) a_1(q^k) \\\\ a_2(nq^k) &=& a_1(n) a_2(q^k) + a_2(n) a_1(q^k) + a_2(n) a_2(q^k) \\\\ a_3(nq^k) &=& a_1(n) a_3(q^k) + a_3(n) a_1(q^k) + a_2(n) a_2(q^k) +\\\\&& 3a_2(n) a_3(q^k) + 3a_3(n) a_2(q^k) + 6a_3(n) a_3(q^k) \\end{eqnarray}$$\n\nand we can substitute in and simplify to\n\n$$\\begin{eqnarray} a_1(nq^k) &=& 1 \\\\ a_2(nq^k) &=& (2k+1)(a_2(n)+1) - 1 \\\\ a_3(nq^k) &=& \\left(3k^2 + 3k + 1\\right) a_3(n) + \\frac{(3k+1)k}{2} a_2(n) + \\frac{k(k-1)}{2} \\end{eqnarray}$$\n\nIf we define $a_{1+2}(n) = a_1(n) + a_2(n)$, there's a nice simple recurrence $a_{1+2}(nq^k) = (2k+1)a_{1+2}(n)$, but not for $a_3$.", "date": "2021-06-20 01:43:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2848191/number-of-triples-of-divisors-who-are-relatively-prime-as-a-triple", "openwebmath_score": 0.9178364872932434, "openwebmath_perplexity": 136.58906904894786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227579477601, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8622434688239}}
{"url": "https://math.stackexchange.com/questions/2482829/for-each-face-of-a-cuboid-the-sum-of-perimeter-and-area-is-known-find-the-volu", "text": "# For each face of a cuboid, the sum of perimeter and area is known. Find the volume.\n\nI came across this problem in a national level examination.\n\nOn each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers written are $$16$$, $$24$$, and $$31$$.\n\nThen the volume lies between ...\n\nMy workout...\n\nFrom the problem,\n\n$$ab+2(a+b)=16$$\n\n$$bc+2(b+c)=24$$\n\n$$ca+2(c+a)=31$$\n\nI am sure some amount of manipulation is required after this step, but I have tried in vain.\n\nAdd $4$ to each equation and factorise \\begin{eqnarray*} (a+2)(b+2)=20 = 5 \\times 4 \\\\ (b+2)(c+2)=28 = 4 \\times 7 \\\\ (c+2)(a+2)=35 = 7 \\times 5 \\\\ \\end{eqnarray*} which gives $\\color{red}{a=3,b=2,c=5}$. So the volume is $\\color{blue}{30}$.\n\n\u2022 We can also show easily that this is the only solution by considering small variations in $a,b,c$. +1 Oct 21 '17 at 14:13\n\u2022 You can solve directly by multiplying the first two equations and dividing by the third: $$\\frac{(a+2)(b+2)\\cdot(b+2)(c+2)}{(c+2)(a+2)} = \\frac{20\\cdot 28}{35} \\;\\to\\; (b+2)^2 = 16 \\;\\to\\; b+2 = \\pm 4 \\;\\to\\; b = 2$$ (taking $b=-6$ to be extraneous).\n\u2013\u00a0Blue\nOct 21 '17 at 19:31\n\nI much prefer the cleverness of @Donald's answer (which is probably the approach the exam writers hoped you'd see), but it's perhaps worth noting that you can attack this problem with a little algebraic brute force and a lot of perseverance.\n\nYou have three equations in three unknowns. Our goal is to eliminate two of the unknowns, leaving a single equation in, say, $a$. Notice that the first equation fairly readily allows us to express $b$ in terms of $a$; the third equation allow us to to do likewise for $c$:\n\n\\begin{align} ab + 2 a + 2 b = 16 \\quad\\to\\quad b(a+2) = 16-2a \\quad\\to\\quad b &= \\frac{16-2a}{a+2} = \\frac{2(8-a)}{a+2} \\tag{1a} \\\\[6pt] c &= \\frac{31-2a}{a+2} \\tag{1b} \\end{align}\n\nBy substitution, the second equation transforms to involve $a$ alone: $$\\frac{2(8-a)}{a+2}\\cdot\\frac{31-2a}{a+2} + 2\\left(\\frac{16-2a}{a+2}+\\frac{31-2a}{a+2}\\right) = 24 \\tag{2a}$$ $$\\frac{(8-a)(31-2a)}{(a+2)^2}+ \\frac{47-4a}{a+2} = 12 \\tag{2b}$$ $$(8-a)(31-2a)+ (47-4a)(a+2) = 12 (a+2)^2 \\tag{2c}$$\n\n(It's around this point that you should start to suspect that there's a better way. Nevertheless, ...) We can expand everything and combine terms to get $$14 a^2 + 56 a - 294 = 0 \\quad\\to\\quad 14 ( a^2 + 4 a - 21)= 0 \\quad\\to\\quad 14 (a+7)(a-3) = 0 \\tag{3}$$ Here, we solve to find that $a=3$ (discarding $a=-7$ as extraneous). Then $b=2$ and $c=5$ follow from $(1a)$ and $(1b)$ above. $\\square$\n\nThis approach lacks ingenuity, but it's essentially mechanical. (You could even opt to use the Quadratic Formula in $(3)$ instead of thinking-through the factorization. Simplifying the final result is something of a chore, but it doesn't require any insight.) So, even if you don't see the clever approach, there's still a way to proceed ... and to succeed.\n\nIt was already done (and far better that what i did) but i isolated a b c in the next way:\n\na=(16-2b)/(b+2)\n\nb= (24-2c)/(c+2)\n\nc=(31-2a)/(a+2)\n\nthen by substitutions you will get the solution.", "date": "2022-01-24 00:28:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2482829/for-each-face-of-a-cuboid-the-sum-of-perimeter-and-area-is-known-find-the-volu", "openwebmath_score": 0.9971842765808105, "openwebmath_perplexity": 449.8738029377807, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575160299339, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8621714111079845}}
{"url": "https://math.stackexchange.com/questions/1509834/conditional-probabilities-picking-balls-out-of-a-bag", "text": "# Conditional Probabilities- picking balls out of a bag\n\n3 red, 3 green, 3 blue, and 3 orange balls are in a box and 6 of these balls are drawn at random from the box. If 2 of the 6 drawn balls are red and 2 of them are green, what is the probability that the other 2 drawn balls are blue and orange?\n\nThe answer I am getting is 12/28. I came to this answer by realizing there are 8 balls left. 3 orange, 3 blue, 1 green and 1 red. Two more balls need to be chosen. So the total number of ways these can be picked are $8\\choose 2$=28 ways to choose 2 balls from the remaining 8. Then I got that there are 3!+3! ways to pick a orange and a blue ball from the remaining two balls. So i got my answer to be $(3!+3!)/28$=12/18 as the probability of picking a orange and a blue as the other two of the 6 balls drawn. I don't know if this is right. Can anyone confirm my solution. If it is incorrect, could anyone point out where my error lies?\n\nThere are $9$ ways ($3\\times 3$) of picking an orange and a blue ball. So the answer should be $9/28$.\n\n\u2022 the other thing i just realized is that the remaining unknown two balls aren't necessarily drawn after the 2 red and 2 green balls, the order is unspecified. Will this have any affect on the answer? For example, the orange and blue balls could be the first two balls drawn, followed by two red and two green balls. \u2013\u00a0sappgob Nov 2 '15 at 18:40\n\u2022 If you draw all six balls without looking at them, and you then look at four of them, and they are two red and two green, then the answer I gave is correct. If that's not what you mean, then you need to make your question more precise. \u2013\u00a0rogerl Nov 2 '15 at 18:44\n\nRogerl has given the probability that the remaining 2 balls will be 1 orange and 1 blue when given that four revealed balls are 2 red and 2 green.\n\n$$\\frac{\\binom{3}{1}\\binom{3}{1}\\binom{2}{0}}{\\binom{8}{2}}= \\frac{9}{28}$$\n\nIf you want the probability of picking 1 orange and 1 blue given that you have picked exactly 2 red and 2 green:\n\n$$P(O{=}1,B{=}1\\mid R{=}2,G{=}2)=\\dfrac{\\binom{3}{1}\\binom{3}{1}}{\\binom{6}{2}}=\\frac{3}{5}$$\n\nIf you want the probability of picking 1 orange and 1 blue given that you have picked at least 2 red and 2 green:\n\n$$P(O{=}1,B{=}1,R{=}2,G{=}2\\mid R{\\geq}2,G{\\geq}2)=\\dfrac{\\binom{3}{1}\\binom{3}{1}\\binom{3}{2}\\binom{3}{2}}{\\binom{6}{2}\\binom{3}{2}\\binom{3}{2}+2\\binom{6}{1}\\binom{3}{3}\\binom{3}{2}+\\binom{3}{3}\\binom{3}{3}}=\\frac{81}{172}$$\n\nThis is why it is important to be clear about your specifications.\n\nSimilarly: I toss two coins behind a screen and tell you about one of them being a head. \u00a0 What is the probability of the other being a tail when what I told you was ... ?:\n\n\u2022 The left coin is a head.\n\u2022 At least one coin is a head.\n\u2022 Exactly one of the coins is a head.\n\nThe probability space is $\\{\\rm HH, HT, TH, TT\\}$\n\n\u2022 I actually like this answer better, Graham. \u2013\u00a0rogerl Nov 4 '15 at 22:49", "date": "2019-07-20 05:48:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1509834/conditional-probabilities-picking-balls-out-of-a-bag", "openwebmath_score": 0.8929539918899536, "openwebmath_perplexity": 156.02217092472674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575147530352, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8621714036911383}}
{"url": "https://math.stackexchange.com/questions/1652812/a-series-to-prove-frac227-pi0/1657416", "text": "# A series to prove $\\frac{22}{7}-\\pi>0$\n\nAfter T. Piezas answered Is there a series to show $22\\pi^4>2143\\,$? a natural question is\n\nIs there a series that proves $\\frac{22}{7}-\\pi>0$?\n\nOne such series may be found combining linearly the series that arise from truncating $$\\sum_{k=0}^\\infty \\frac{48}{(4k+3)(4k+5)(4k+7)(4k+9)} = \\frac{16}{5}-\\pi$$ to two and three terms, namely\n\n$$\\sum_{k=2}^\\infty \\frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \\frac{141616}{45045}-\\pi$$ and $$\\sum_{k=3}^\\infty \\frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \\frac{2406464}{765765}-\\pi$$ Solving $$a\\left(\\frac{141616}{45045}-\\pi\\right)+b\\left(\\frac{2406464}{765765}-\\pi\\right)=\\frac{22}{7}-\\pi$$ for rational $a,b$ and some algebra manipulation yields the result\n\n$$\\frac{16}{21} \\sum_{k=0}^\\infty \\frac{1008 k^2+6952 k+12625}{(4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21)}=\\frac{22}{7}-\\pi$$\n\nIt is interesting to note that the coefficients needed to multiply the two component series are both positive $$a=\\frac{113}{7\u00b78\u00b79}$$ $$b=\\frac{391}{7\u00b78\u00b79}$$\n\nbecause the truncation points have been chosen so that\n\n$$\\frac{2406464}{765765}<\\frac{22}{7}<\\frac{141616}{45045}$$\n\nThis procedure yields a result that proves the claim with no need for further processing, and it is readily seen to prove $\\frac{p}{q}-\\pi>0$ for all fractions between $\\pi$ and $\\frac{16}{5}$.\n\nNow, in the light of this equivalent form of Lehmer's formula $$\\pi-3=\\sum_{k=1}^\\infty \\frac{4!}{(4k+1)(4k+2)(4k+4)}$$\n\nQ1 Is there a series that proves $\\frac{22}{7}-\\pi>0$ with constant numerator?\n\nQ2 Is there a reason why $113$ is both the numerator of the $a$ coefficient and the denominator of the next convergent from above $\\frac{355}{113}$?\n\nEdit: A similar series with smaller coefficients may be obtained by applying the method above to \\begin{align} \\sum_{k=0}^\\infty \\frac{960}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13)} &= \\frac{992}{315}-\\pi \\\\ &= \\frac{3\u00b7333-7}{3\u00b7106-3}-\\pi \\\\ \\end{align} in order to obtain $$\\sum_{k=0}^\\infty \\frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)} = \\frac{22}{7}-\\pi$$\n\nQ3 What is the relationship between $\\frac{992}{315}$ and the third convergent to $\\pi$ $\\frac{333}{106}$?\n\n\u2022 Do you have the answer for this question ? \u2013\u00a0user230452 Feb 13 '16 at 4:56\n\u2022 No, I don't have them. \u2013\u00a0Jaume Oliver Lafont Feb 13 '16 at 5:09\n\u2022 @user230452 Now I do. \u2013\u00a0Jaume Oliver Lafont Feb 15 '16 at 22:52\n\u2022 If you like this Math.SE question you may also enjoy reading this MO.SE post. \u2013\u00a0Qmechanic Feb 17 '16 at 14:44\n\nQ1\n\nEvaluating the following series \\begin{align} &\\sum_{k=0}^\\infty \\frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)} \\\\ &= \\sum_{k=0}^\\infty \\left(\\frac{1}{4k+5}-\\frac{4}{4k+6}+\\frac{5}{4k+7}-\\frac{5}{4k+9}+\\frac{4}{4k+10}-\\frac{1}{4k+11}\\right) \\\\ &= \\sum_{k=0}^\\infty \\int_{0}^1\\left(x^{4k+4}-4x^{4k+5}+5x^{4k+6}-5x^{4k+8}+4x^{4k+9}-x^{4k+10}\\right)dx \\\\ &= \\int_{0}^1 x^4\\sum_{k=0}^\\infty \\left(x^{4k}-4x^{4k+1}+5x^{4k+2}-5x^{4k+4}+4x^{4k+5}-x^{4k+6}\\right)dx \\\\ &= \\int_{0}^1 x^4\\frac{1-4x+5x^2-5x^4+4x^5-x^6}{1-x^4}dx \\\\ &= \\int_{0}^1 x^4\\frac{(1-x^2)(1-x)^4}{(1-x^2)(1+x^2)}dx=\\int_{0}^1 \\frac{x^4(1-x)^4}{1+x^2}dx=\\frac{22}{7}-\\pi \\\\ \\end{align} shows its connection with Dalzell's integral.\n\nThis may be rewritten as $$\\sum_{k=1}^\\infty \\frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\\frac{22}{7}-\\pi$$\n\nwhich appears in the 2009 document by Peter Bala New series for old functions http://oeis.org/A002117/a002117.pdf (formula 5.1) and shows that $\\frac{22}{7}-\\pi$ can be obtained by taking one term out of the summation in the series $$\\sum_{k=0}^\\infty \\frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\\frac{10}{3}-\\pi$$\n\nConsecutive truncations yield the inequality\n\n$$\\pi...<\\frac{141514}{45045}<\\frac{10886}{3465}<\\frac{22}{7}<\\frac{10}{3}$$\n\nSimilar fractions, but now converging to $\\pi$ from below, may be obtained from the series\n\n$$\\sum_{k=0}^\\infty \\frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)} = \\pi-\\frac{47}{15}$$\n\nThis yields\n\n$$\\frac{47}{15}<\\frac{1979}{630}<\\frac{141511}{45045}<\\frac{9622853}{3063060}<...\\pi$$\n\n(See a similar inequality for $\\log(2)$)\n\nCorrespondence between series and integrals\n\n$$\\sum_{k=n}^\\infty \\frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\\int_0^1 \\frac{x^{4n}(1-x)^4}{1+x^2}dx$$\n\n$$\\sum_{k=n}^\\infty \\frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)}=\\int_0^1 \\frac{x^{4n+2}(1-x)^4}{1+x^2}dx$$\n\nEquivalent expressions\n\nThe general term for these series may be written in compact form using factorials, binomial coefficients or the Beta integral $B$ (see this comment by N. Elkies).\n\n\\begin{align} \\frac{22}{7}-\\pi &= 3840\\sum_{k=1}^\\infty \\frac{(k+2)!(4k)!}{(4k+8)!k!} \\\\ \\\\ &= \\frac{4}{21} \\sum_{k=1}^\\infty \\frac{\\displaystyle{k+2 \\choose 2}}{\\displaystyle{4k+8\\choose 8}} \\\\ \\\\ &= \\frac{4}{21} \\sum_{k=1}^\\infty \\frac{k+1}{\\displaystyle{4k+7\\choose 7}} \\\\ \\\\ &= \\frac{16}{21} \\sum_{k=1}^\\infty \\frac{B(4k+1,8)}{B(k+1,2)} \\end{align}\n\nInterpretation of $\\frac{22}{7}-\\pi$\n\nSimilar series and approximations\n\nIf we use the Pochhammer symbol to express this series: $$\\sum_{k=0}^\\infty \\frac{7!(k+1)}{(4k+1)_7}=\\frac{7}{4}(10-3\\pi)\\approx 1$$\n\nwe can change the numbers to obtain variants such as\n\n$$\\sum_{k=0}^\\infty \\frac{5!(k+1)}{(3k+1)_{5}} = \\frac{5}{9}\\left(2\\sqrt{3}\\pi-9\\right)\\approx 1,$$\n\n$$\\sum_{k=0}^\\infty \\frac{11! (k+1)}{(6 k+1)_{11}} = 231-\\frac{4565 \\pi}{36 \\sqrt{3}}\\approx 1$$\n\nand $$\\sum_{k=0}^\\infty \\frac{15!(k+1)}{(8k+1)_{15}}=\\frac{15}{8}(1716-7(99\\sqrt{2}-62)\\pi)\\approx 1$$\n\nGiven that all three series evaluate to almost 1, the following corresponding approximations are derived\n\n\\begin{align} \\pi &=\\frac{9\\sqrt{3}}{5}+\\sqrt{3}\\int_0^1\\frac{x^3(1-x)^2(1+x)}{1+x+x^2}dx\\\\ &\\approx\\frac{9\\sqrt{3}}{5} \\\\ \\pi &=\\frac{1656\\sqrt{3}}{913}- \\frac{6\\sqrt{3}}{83}\\int_0^1 \\frac{x^6(1-x)^8}{1+x^2+x^4} dx\\\\ &\\approx\\frac{1656\\sqrt{3}}{913} \\\\ \\pi &=\\frac{1838 \\left(62 + 99 \\sqrt{2}\\right)}{118185}-\\frac{62+99\\sqrt{2}}{15758}\\int_0^1 \\frac{x^8(1-x)^{12}}{1+x^2+x^4+x^6}dx\\\\ &\\approx \\frac{1838 \\left(62 + 99 \\sqrt{2}\\right)}{118185} \\end{align}\n\nwhich give 1, 5 and 8 correct decimals respectively.\n\nThe fraction $\\frac{1838}{118185}$ is the eighth convergent of $\\frac{\\pi}{62+99\\sqrt{2}}$\n\nAnother series and integral for $\\frac{22}{7}-\\pi$\n\n\\begin{align} &\\sum_{k=0}^\\infty \\frac{285120}{(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)(4k+13)(4k+14)} \\\\ &= \\frac{1}{28}\\int_{0}^1 \\frac{x(1-x)^8(2+7x+2x^2)}{1+x^2}dx=\\frac{22}{7}-\\pi \\\\ \\end{align}\n\n\u2022 So based on several series versions already found for 22/7 - Pi, it appears that similar to infinite number of integral expressions (as it was shown by Thomas Baruchel), there are infinite number of series as well. \u2013\u00a0Alex Feb 15 '16 at 23:20\n\u2022 You are quite wrong wishing ALL the terms be positive. Anyway your example is nice. Regards. \u2013\u00a0Piquito Feb 16 '16 at 11:58\n\u2022 @Piquito Reading the proof in reverse shows how the positive integrand naturally leads to all terms positive. \u2013\u00a0Jaume Oliver Lafont Feb 17 '16 at 10:18\n\nProof that $\\frac{22}{7}$ exceeds $\\pi$.\n\n$$0<\\int_0^1\\frac{x^4(1-x)^4}{1+x^2}dx=\\frac{22}{7}-\\pi$$\n\nProof-\n\n$$\\int_0^1\\frac{x^4(1-x)^4}{1+x^2}dx$$\n\n$$=\\int_0^1x^6-4x^5+5x^4-4x^2+4-\\frac{4}{1+x^2}dx$$\n\n$$=\\frac {x^7}{7}+\\frac{2x^6}{3}+x^5-\\frac{4x^3}{3}+4x-4\\tan^{-1}(x)\\vert_0^1$$\n\nNow,by applying $\\tan^{-1}1=45^\\circ=\\frac\\pi4$ and substituting it in the integral and solving the integral yields $\\frac{22}{7}-\\pi$\n\n\u2022 Why is that integral greater than zero ? Is is just because integral represents an area ? Moreover, what is the usual strategy to evaluate this integral ? \u2013\u00a0user230452 Feb 13 '16 at 5:13\n\u2022 @tatan: since this does not answer either question, would you please write it as a comment? \u2013\u00a0Jaume Oliver Lafont Feb 13 '16 at 5:18\n\u2022 @tatan your answer is about the integral proof for $\\frac{22}{7}-\\pi>0$, while this question addresses proofs using series with positive terms only. \u2013\u00a0Jaume Oliver Lafont Feb 13 '16 at 5:34\n\u2022 @user230452 It's $>0$ because the integrand is always $>0$ in the domain of integration. \u2013\u00a0YoTengoUnLCD Feb 13 '16 at 5:40\n\u2022 I have no prove of it, but $\\displaystyle \\sum_{k=0}^{+\\infty}\\dfrac{41760+576k}{(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)}=\\dfrac{22}{7}-\\pi$ \u2013\u00a0FDP Feb 14 '16 at 19:26\n\nLet $\\sum_{k=0}^\\infty a_n$ any series converging to $\\pi$ and choose any series converging to $\\frac{22}{7}$, for instance $\\sum_{k=0}^\\infty \\left(\\frac{15}{22}\\right)^n$\n\nNo problem to show that $$\\sum_{k=0}^\\infty \\left(\\left(\\frac{15}{22}\\right)^n -a_n\\right)=\\frac{22}{7}-\\pi\\gt 0$$\n\n\u2022 Does this series have all terms positive? \u2013\u00a0Jaume Oliver Lafont Feb 15 '16 at 23:42\n\u2022 From a certain range, yes. \u2013\u00a0Piquito Feb 15 '16 at 23:48\n\u2022 How can we prove $\\frac{22}{7}-\\pi>0$ from that series if not all terms are positive? \u2013\u00a0Jaume Oliver Lafont Feb 15 '16 at 23:54\n\u2022 Theorem.-Let $a=\\sum a_n$ and $b=\\sum b_n$ two convergent series. Then , for all pair of constants $\\alpha$, $\\beta$ the series $\\sum(\\alpha a_n+\\beta b_n)$ converges to $\\alpha a+ \\beta b$ (Mathematical Analysis, T. M. Apostol). Can you exhibit an example of what you say? \u2013\u00a0Piquito Feb 16 '16 at 0:46\n\u2022 This proves $$\\sum_{k=0}^\\infty \\left((\\frac{15}{22})^n -a_n\\right)=\\frac{22}{7}-\\pi$$ but not $$\\frac{22}{7}-\\pi>0$$ \u2013\u00a0Jaume Oliver Lafont Feb 16 '16 at 0:51", "date": "2019-06-19 11:14:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1652812/a-series-to-prove-frac227-pi0/1657416", "openwebmath_score": 0.9999915361404419, "openwebmath_perplexity": 1067.6727638975726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631627142339, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8621575909869775}}
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EQUATIONS OF LINES IN DIFFERENT FORMS. not a function 11. \u0192(\u00ba2)= \u00ba(\u00ba2)2\u00ba 3(\u00ba2) + 5 Substitute \u00ba2 for x. 3: Quadratic Functions and Their Properties Def: A quadratic function is a function of the form f(x) = ax2+bx+c, where a;b;c are real numbers and a 6= 0. Example 2 Linear Functions Which of the following functions are linear? a. Example: Find the zero of $y=\\frac{1}{2}x+2$ algebraically. Lesson 3 Direct Variation. Comparing Linear Functions- The Greates Rate of Change Subject: SMART Board Interactive Whiteboard Notes Keywords: Notes,Whiteboard,Whiteboard Page,Notebook software,Notebook,PDF,SMART,SMART Technologies ULC,SMART Board Interactive Whiteboard Created Date: 1/30/2014 3:52:58 PM. This representational technique has succeeded at \ufb01nding good policies for problems with high dimensional state-spaces such as simulated soccer [Stone et al. Practice: Relations & Functions Final corrections due: Use the given form of each relation to complete the other forms. For each function: (a) Find the vertex ( , )hk of the parabola by using the formulas 2 b a h and 2 b a kf. The linear attenuation coefficient can be considered as the fraction of photons that interact with the shielding medium per. The simplest linear relation is a proportional relation, which is a line with a vertical intercept of 0. indd 1 66/26/08 7:36:04 PM/26/08 7:36:04 PM. Systems of Linear Equations 0. You can choose from up to four types of equations depending on the sophistication of your students. 2b ( 7)=11 b= 2 6. Match the formulas, graphs, and equations that go together. Example: Determine whether f(x) 3x 9 and 3 3 1 g(x) x are inverse functions. \u201cLinear Equations In Two Variables\u201d is chapter 4 of the NCERT Textbook, and it falls under the unit 2 Algebra. These pdf worksheets provide ample practice in plotting the graph of linear functions. \u2022Example: y = 3x + 2 \u2022Solution: any ordered pair (x, y) that makes the equation true. The reason is that then the equality (5) does not provide enough information due to the fact that the multiplicative commutators all have determinant 1. It is the interaction between linear transformations and linear combinations that lies at the heart of many of the important theorems of linear algebra. Linear Functions Worksheets This collection of linear functions worksheets is a complete package and leaves no stone unturned. Equations of nonconstant coefficients with missing y-term If the y-term (that is, the dependent variable term) is missing in a second order linear equation, then the equation can be readily converted into a first order linear equation and solved using the integrating factor method. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. The possible values of x approach a chosen value (e. This product is suitable for Preschool, kindergarten and Grade 1. The pdf worksheet under this section has sets of table values. Linear Attenuation Shielding Formula: x B A I I e = * \u2212\u03bc. so, matrix multiplication is a linear function converse: every linear function y =f(x), with y an m-vector and x and n-vector, can be expressed as y =Ax for some m\u00d7n matrix A you can get the coe\ufb03cients of A from A ij =y i when x =e j Linear Equations and Matrices 3\u20133. Then you can draw a line through those two points. f(x)=\u2212(x+2) 2 \u22127 Describe the transformation performed on each function g(x) to result in m(x). If the items are complex numbers, en is used to denote the vector space. Answers found at https://www. Take the Schoology Quiz (Concept 7 \u2013 Level 2) Score of 4 or higher move to level 3 Score of 3 or less, complete the Level 2 Review Level 3 1. Carefully graph each equation on the same coordinate plane. Graphing linear function: Type 1 - Level 2. We are going to use this same skill when working with functions. 58 Chapter 2 Linear Relations and Functions The table shows average and maximum lifetimes for some animals. The impulse response of a linear system, usually. 1) Write Down the Basic Linear Function. Linear Cost Function 2. The slopes are represented as fractions in the level 2 worksheets. Write an equation of the line that passes through the point (2, 2) and has a slope of 4. It is 76\u00ba F at the 6000-foot level of a mountain, and 49\u00ba F at the 12,000-foot level of the mountain. 8 of 8) 12. 5 Partial Correlation 100. Chapter & Page: 42\u20134 Nonhomogeneous Linear Systems which, in matrix/vector form, is x\u2032 = Ax + g with A = 1 2 2 1 and g(t) = 3 0 t + 0 2. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. It also publishes articles that give significant applications of matrix theory or linear algebra to other. In your Differential Equations course, you will see that every solution to the differential equation above is a linear combination of cos and(k t) sin. A function fis a map f: X!Y (1. In its most basic form, a linear supply function looks as follows: y = mx + b. Several questions on functions are presented and their detailed solutions discussed. Show your work. One way is to find the x-intercept and the y-intercept and. Before graphing linear equations, we need to be familiar with slope intercept form. To Graphing Linear Equations The Coordinate Plane A. 1-3 Transforming Linear Functions Example 4A: Fund-raising Application The golf team is selling T-shirts as a fund-raiser. The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takers\u2019 ability for selected mathematics content. \u0192(x) is not a linear function because it has an x2-term. This section is a step-by-step presentation of how to use algebra formulae on all the topics covered in this site which include formulae on -linear equations, inequalities, decimals, fractions, exponents, graphing linear equations, binomial theorem, pythagoras theorem, quadratic equations, algebraic expressions, factorisation, ratios, geometry. Solve to find the x- and y-intercepts. iterative methods for linear systems have made good progress in scienti\ufb01c an d engi- neering disciplines. 2 Linear Functions. The impulse response of a linear system, usually. Find the inverse function for the following function: 3 a. Input 6 7 7 14 Output 5 2 5 2 Find the slope and y-intercept of the line with the given equation. Linear Function Word Problems Exercise 1Three pounds of squid can be purchased at the market for $18. Learn More at mathantics. y 2x 1 and y 2x 2 are parallel lines since both equations have a slope of 2. The function R(n) = 7. 5, Solving Systems of Linear Equations by Graphing. Manipulate equations from one form to another[Lesson 7. We tried to explain the trick of solving word problems for equations with two variables with an example. Relations and functions review worksheet Collection. 4 Guided Notes; 4. A linear function is a function whose graph consists of segments of one straight line throughout its domain. Identify the slope and y-intercept of the line with the given equation. Linear transformations: Rank-nullity theorem, Algebra of linear transformations, Dual spaces. Solutions to Linear Equations in One Variable The _____ of an equation is the value(s) of the variable(s) that make the equation a true statement. \u2020 identified independent and dependent variables. (a) Graph the equation. Table 5-1 provides examples of common linear and nonlinear systems. The quantity mx is called the variable costand the intercept b is called the \ufb01xed cost. Each customer can buy a maximum of four tickets. Prove: f(X2) f (x, ) = f(xy) Proof: I. x y -1 -4 0 -3 1 -2 -2 -5 x y -2 5 -1 -3 0 -1. Students recognize a proportion (y/x = k, or y = kx) as a special case of a linear equation of the form y = mx + b, understanding that the constant of proportionality (k) is the slope and. Then answer the questions at the bottom of the page. 04 Solve Systems of Linear Equations in Three Variables. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else. Then explore different ways to find the slope and y-intercept for a linear function. The y intercept is a point in itself and to complete our graph we only need one additional point. y 2x 1 and y 2x 2 are parallel lines since both equations have a slope of 2. Leave any comments, questions, or suggestions below. Worksheets for linear equations Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. functional relationship? A The number of cookies ordered. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. Graphs of Linear Equations reviews the rectangular (Cartesian) coordinate system, and contains lessons on different methods of interpreting the lines and their applications, and has examples of solving different practice problems related to finding the slope and using different forms of writing the equation for a line. We use the rst equation to eliminate x1 from equations (2) through (m), leaving m 1 equations in n 1 unknowns. Review Of Linear Functions Lines Answer Key - Displaying top 8 worksheets found for this concept. Plan your 60-minute lesson in Math or linear functions with helpful tips from James Bialasik linear_functions1_investigation. 6 Linear Functions of Random Vectors 79 3. KEY Graphing Linear Functions WS 2. Linear Equations 1 1. Typically we consider B= 2Rm 1 \u2019Rm, a column vector. Slope of a linear function is the change in y for a unit change in x along the line and usually denoted by the letter \"m\" Slope is sometimes referred to as \"rise over run\". f (x) = a (x \u2013 h)2 + k (a \u2260 0). 4) y = 9(1/x) + 4 (x is in the denominator). The coordinate plane has 4 quadrants. Two or more products are usually produced using limited resources. What is the y-intercept? 4. Intro to Linear Equations Algebra 6. linear algebra, the so-called pivot operation. The cost to rent a piece of equipment is$27, plus $6. Outlines of concepts covered in this linear equations formulas pdf are: Definition and types of simple linear equations; General form of single, two and three variable linear equations; Tips for finding out number of solutions in two variable linear equations. Because of this prevalence of numerical linear algebra, we begin our treatment of basic. Matrices and Elementary Row Operations 6 1. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 03 Linear Functions Zoey Ramsay. Profits, P, are equal to sales, S, minus expenses, E. GRAPHING LINEAR EQUATIONS VOCABULARY ( 3, -7 ) VOCABULARY coordinate plane coordinates diagonal line horizontal line ordered pair origin point quadrant vertical line x-axis y-axis x-coordinate y-coordinate x-intercept y-intercept. Homework Practice Workbook 000i_ALG1HWPFM_890836. Solve linear and quadratic equations and inequalit ies, including systems of up to three linear equations with three. This will take you to the individual page of the worksheet. CHECK YOUR UNDERSTANDING Write your answers on notebook paper. The function fis a tensor. Graphing linear function: Type 1 - Level 2. The graph of any linear. A graph is linear if it is a straight line. Find an n-vector bfor which bTc= p0( ): This means that the derivative of the polynomial at a given point is a linear function of its coe cients. AP Calculus Topic: Analysis of Functions Materials: Student Activity pages Teacher Notes: This activity can be used early in the year after students have studied linear functions. Solving Equations Step 1. Linear Functions LINEAR FUNCTIONS REVIEW 7. Subsection LTLC Linear Transformations and Linear Combinations. Therefore, f of 1 is negative 2. 3\u00adTransforming Linear Functions. Several questions on functions are presented and their detailed solutions discussed. This was the unit that the city provided us last year, which is why we did not create a new one. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The piecewise linear approximation is implemented in a linear programming model by defining new variables xj1, xj2,\u2026, xjr to represent. In this course, you will use x and y to write linear equations and n and u n to write recursive and explicit formulas for sequences of discrete points. Guess words about linear functions. Graphically, power functions can resemble exponential or logarithmic functions for some values of x. Linear Functions Vocabulary. 1 Introducing Functions Lesson Plan OBJECTIVES \u02daStudents will understand that for each input there can be only one output. the graph of the function f(x) = c. Chapter 2 Linear time series Prerequisites \u2022 Familarity with linear models. Homework: 1) Complete the Functions Story Problem Worksheet and Linear Function Word Problems if not finished in class. Graphing and Systems of Equations Packet 1 Intro. 2 Linear functions (EMA48) Functions of the form $$y=x$$ (EMA49) Functions of the form $$y=mx+c$$ are called straight line functions. Given the integers a;b > 0, we de ne greatest common divisor of a and b, as the largest number that divides both a and b. \u2022Example: y = 3x + 2 \u2022Solution: any ordered pair (x, y) that makes the equation true. Complete the table for and graph the resulting line. This final lesson in the unit culminates with the Go Public phase of the legacy cycle. Example Let be a uniform random variable on the interval , i. Graphing Linear Equations. Solving Systems of Linear Equations by Graphing HW : 4/13/2017: SOL Coach Book 1. Students should work in pairs or groups of three. This linear equation has m = 3 and b = -1. FUNDRAISING The Pep Club rented a shaved ice machine to sell shaved ice as a fundraiser. Some of the more important. Primary method for approaching these problems. Solving formulas is much like solving general linear equations. For instance, here is a function L from the set R2 to the set R3: L x1 x2. Lesson 3 Direct Variation. com You will need to understand how to project cash flow. Lesson 3 Solving Equations. Use Linear Combination to Solve Systems of Equations and Inequalities 1st: Rearrange the equations so terms line up as: Ax + By = C 2nd: Multiply none, or one, or both equations by constant(s) so that the coefficients of one of the variables are opposites. Transformations of Quadratic Functions Transformations of Functions Transformation: A change made to a figure or a relation such that the figure or the graph of the relation is shifted or changed in shape. Steps for Graphing a Linear Function (Slope-Intercept Form)! Identify and plot the y-intercept ! Use the slope to plot an additional point (Rise/Run) ! Draw a line through the two points EXAMPLES EXAMPLE 5: WRITING AND GRAPHING LINEAR EQUATIONS GIVEN A Y-INTERCEPT AND A SLOPE Write an equation of a line with the given slope and y-intercept. Review of Linear Functions (Lines) Find the slope of each line. > Topics include: Least-squares aproximations of over-determined equations and least-norm solutions of underdetermined equations. Linear equations are found throughout mathematics and the real world. Such a function can be used to describe variables that change at a constant rate. Find the inverse function for the following function: 3 a. except part (c). Chapter 3: Linear Functions. YOU WILL NOT BE USING A CALCULATOR FOR PART I MULTIPLE-CHOICE QUESTIONS, AND THEREFORE YOU SHOULD NOT USE ONE FOR THE REVIEW PACKETS. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. In addition to using a table, equations can be graphed using the equations themselves Remember that Y=M(X)+B tells us the slope and the Y intercept. The y-intercept is where the graph crosses the y -axis. ) Modeling a Quadratic Function Using Various Word Problems Example 2: Complete each word problem using techniques learned in previous concepts. 3) Linear v Non Linear Functions 1 (8. (c) After how many. Because the fraction consists of the rise (the change in y, going up or down) divided by the run (the change in x, going from left to the right). Recently Modified. 3z+4=34 z=10 2. LESSON 3: LINEAR FUNCTIONS Study: Linear Functions Learn about slope and the three main forms of linear functions. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. 5x \u2013 3 = 2x \u2013 27 8. 8K PDF/Acrobat 28 Jan 2016) Projector Resources. 4 - Linear Functions Writing equations for Horizontal and Vertical Lines. 1: Rate of Change and Slope Rate of Change - shows relationship between changing quantities. 2y+1=17 y=8 4. 21 Posts Related to Writing Linear Equations From Graphs Worksheet Pdf. The reason is that the domain and range of a linear function naturally span all real numbers unless the domain is restricted. org\u00a92001 September 22, 2001 2 4. The function is defined by h(t) = -7t2 + 48t. Find the inverse function of f(x): State the domain of the inverse function f 1(x. Another special type of linear function is the Constant Function it is a horizontal line: f(x) = C. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. Section 1: Quadratic Functions (Introduction) 3 1. the graph of the function f(x) = c. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. linear algebra, the so-called pivot operation. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. H = 4T + 74 A company can make a total of 20 solar heaters for$13,900, while 10 heaters cost $7500. The table shows the cost per hour. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. The size of the PDF file is 66677 bytes. Intersections between spheres, cylinders, or other quadrics can be found using quartic equations. Copy this to my account; E-mail to a friend; Find other activities. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Even though, mathematically speaking, these two points can be arbotrarily close, if you choose them to be too close, your line may deviate from what the graph. Note the ax \u2261 b (mod n) i\ufb00 there is y \u2208 Z such that ax+ ny = b (by equivalent formulation of equivalence mod n, ax \u2261 b ( (mod n) i\ufb00 they di\ufb00er by a multiple of n). Can be inserted in interactive notebook, used as an anchor chart, or a quick reference for students. Multiple Choice Questions have been coming in Class 8 Linear Equations exams, thus do MCQs to test understanding of important topics in the chapters. The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takers\u2019 ability for selected mathematics content. MULTIPLE CHOICE Plotting the points from each of the given tables as shown below, identify its correct. ferential equations by any discrete approximation method, construction of splines, and solution of systems of nonlinear algebraic equations represent just a few of the applications of numerical linear algebra. Mathematics (Linear) \u2013 1MA0 STRAIGHT LINE GRAPHS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. A normal 12 point text isn\u2019t the ideal solution for fast reading. Associate a given equation with a function whose zeros are the solutions of the equation. Introduction Diophantine equations are named for Diophantus of Alexandria who lived in the third century. In these cases, replace the function notation and solve rather than the x. Key Components Key Components 2. The slope is m = \u22120. Unit 2: Solve Linear Equations Instructor Notes The Mathematics of Writing and Solving Linear Equations Most students taking algebra already know the techniques for solving simple equations. Fluency in interpreting the parameters of linear functions is emphasized as well as setting up linear functions to model a variety of situations. Linear Functions Worksheets This collection of linear functions worksheets is a complete package and leaves no stone unturned. There are two methods to finding the slope of a linear relationship from a table. The coordinate plane has 4 quadrants. Free pdf on #248396. Population of Indiana '50 '60 '70 '80 '90 '00 3 2 4 5 Population (millions) 6 7 Year 0 3. com for more Free math videos and additional subscription based content!. Each section begins with a bite-sized introduction to a topic with an example, followed by practice exercises including word problems. These are to use the CDF, to trans-form the pdf directly or to use moment generating functions. Find solutions using the given \"x\" values. Then explore different ways to find the slope and y-intercept for a linear function. 2c+7=17 c=5 6. Find the b (y-intercept) and write the equation of the line in slope-intercept form. In order to use stochastic gradient descent with backpropagation of errors to train deep neural networks, an activation function is needed that looks and acts like a linear function, but is, in fact, a nonlinear function allowing complex relationships in the data to be learned. Chapter 5 - Linear Functions Name_____ Keller - Algebra 1 Notes 5. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. Explain why this. Then, write the function formula. In the associated activities, students use linear models to depict Hooke's law as well as Ohm's law. \u00a9K n2 v0r1 s45 SK Wupt 9a7 nS Xo uf htGwBaBrQeP nL1LOCR. For example, given ax+3=7, solve for x. The point is stated as an ordered pair (x,y). It is also important to know that any linear function can be written in the form f(x) mx -+- b, where m and b are constants. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. Quadratic Functions-Worksheet Find the vertex and \u201ca\u201d and then use to sketch the graph of each function. Here is a small outline of some applications of linear equations. Give each student a card with one representation of a linear function from the LESSON 1 section of the INSTRUCTIONAL ACTIVITY SUPPLEMENT. Some of the worksheets below are Free Linear Equations Worksheets, Solving Systems of Linear Equations by Graphing, Solving equations by removing brackets & collecting terms, Solving a System of Two Linear Equations in Two Variables by Addition, \u2026. mathematics content. a change in the size or position of a figure B. In its most basic form, a linear supply function looks as follows: y = mx + b. Grade 9 Math Solving Linear Equations. Systems of Linear Equations 0. Linear Algebra is one of the most important basic areas in Mathematics, having at least as great an impact as Calculus, and indeed it provides a signi\ufb02cant part of the machinery required to generalise Calculus to vector-valued functions of many variables. Pre-Algebra Chapter 8\u2014Linear Functions and Graphing SOME NUMBERED QUESTIONS HAVE BEEN DELETED OR REMOVED. During a 45-minute lunch period, Albert (A) went running and Bill (B) walked for exercise. Linear function word problems Calculator tables. Multiple Choice Questions have been coming in Class 10 Linear Equations exams, thus do MCQs to test understanding of important topics in the chapters. Heart of Algebra questions on the SAT Math Test focus on the mastery of linear equations, systems of linear equations, and linear functions. This representational technique has succeeded at \ufb01nding good policies for problems with high dimensional state-spaces such as simulated soccer [Stone et al. Modeling with Linear Functions Work with a partner. Linear Equations are no different. function, it is often simplest to look at the graph of the relation. You should create the following: 1. An example arises in the Timoshenko-Rayleigh theory of beam bending. Then the generating function A(x. Let\u2019s see what happens. Students should work in pairs or groups of three. The order of a di\ufb00erential equation is the highest order derivative occurring. For the equation, complete the table for the given values of x. Patterns and Linear Functions - Word Docs & PowerPoints To gain access to our editable content Join the Algebra 1 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. Modeling a Quadratic Function When Given a Graph Example 1: Write a quadratic function (in vertex form) that models each graph. pdf View Download Graphing Linear Equations - Roach Game View. Xl \u2014 Given Mult. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - PDF Download. Its equation will always be: A vertical line represents all the places where x is a specific number. Basic notes on linear functions in PDF format. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. Construct Functions Solve. 16\u00ad21 Linear vs. Infinite Algebra 1 - Transforming Linear. Graph linear functions that represent real-world situations and give their domain and range. General Questions: 1. Rn is the vector space wherein the vectors have n real items each. Is the domain discrete or continuous? Explain. linear regression: An approach to modeling the linear relationship between a dependent variable, $y$ and an independent variable, $x$. 3\u00adTransforming Linear Functions. Download latest questions with multiple choice answers for Class 10 Linear Equations in pdf free or read online in online reader free. The goal of this text is to teach you to organize information about vector spaces in a way that makes problems involving linear functions of many variables easy. Graphing linear function: Type 1 - Level 2. A table is linear if the rate of change is constant. Computing the intersec-tion points between a line and a polynomial patch involves setting up and solving systems of polynomial equations. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Lesson 3 Solving Equations. This chapter is devoted to the algebraic study of systems of linear equations and their solutions. Graphing and Systems of Equations Packet 1 Intro. \u2022 Write a linear equation in slope-intercept or point-slope form. The slope of a line is the change of x-coordinates divided by the change of y-coordinates. The function sinx = 1sinx+0ex is considered a linear combination of the two functions sinx and e x. This is the 5th lesson in Unit 2 Algebra 2 Linear Equations and Functions. dollars, D. \u2022The methods we use are based on the fundamental. Then determine if the relation is a function. 5 at the points: (1 2, 2) and (2, 1 2). Students should work in pairs or groups of three. This linear equations workbook is divided into 20 sections. linear; linear (large) linear (mm) linear (cm) Log-Log Graph Paper. \u2020 identifying and interpreting the components of linear graphs, including the. There is an x-coordiuatu IJIHI real number, and there is a y-coordinate that can be any real number. Distance Learning 2020; Grade 8 (2019-2020) 3. tial equations. The next theorem distills the essence of this. Then the generating function A(x. 14 Chapter 4-3: WRITING and EVALUATING FUNCTIONS SWBAT: (1) Model functions using rules, tables, and graphs (2) Write a function rule from a table or real world \u2013 situation (3) Evaluate Function. The support of is where we can safely ignore the fact that , because is a zero-probability event (see Continuous random variables and zero-probability events ). These di!erence equations are then implemented in the FDTD grid with equivalent voltageand current sources basedon static \"eld approximations. Gauss-Seidel Method of Solving Simul Linear Eqns: Theory: Part 1 of 2 [YOUTUBE 8:01] Gauss-Seidel Method of Solving Simul Linear Eqns: Theory: Part 2 of 2 [YOUTUBE 5:38] Gauss-Seidel Method of Solving Simul Linear Eqns: Example: Part 1 of 2 [YOUTUBE 9:17]. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Question #13: This question asks students to build a function that could model a given situation. 5 at the points: (1 2, 2) and (2, 1 2). The number that goes into the machine is the input: linear function: A function of the form f(x) = mx + b where m and b are some fixed numbers. Suppose a rabbit population of 10 rabbits rule and evaluation function for how many. Exercise Set 2. 3rd: Add the two equations together to eliminate one of the variables. Domain and Range Worksheet #1 Name: _____ State the domain and range for each graph and then tell if the graph is a function (write yes or no). This unit explores the principles and properties they'll need to understand in order to handle multi-step equations. 25 Compare graphs with the graph f(x) = x Graph the function. Their times and distances are shown in the accompanying graph. 4B Graphing Linear Equations in Slope-Intercept Form 1/25 - 4. 48Mb; Alg 2 02. Unit 9 \u2013 Linear Functions (chapter 9) Topics Functions Representing Linear Functions Constant Rate of Change and Slope Direct Variation Slope\u2010Intercept form Solve Systems of equations by Graphing Solve Systems of equations by Algebraically Name:_____. Linear Functions. y = 1/x+2 d. A linear differential operator (abbreviated, in this article, as linear operator or, simply, operator) is a linear combination of basic differential operators, with differentiable functions as coefficients. Predictive Modeling Goal: learn a mapping: y = f(x;\u03b8) Need: 1. Home - Menifee County Schools. Then answer the questions at the bottom of the page. Write an Equation given the Slope and a Point 1. Writing Linear Equations From Word Problems Worksheet Pdf or Unique solving Inequalities Worksheet Unique Algebra 1 Word Problems. In this case, x and y represent the independent and dependent variables. What is a Linear function? Answer: is a function, meaning we have an input and an output, that can be written in the form \ud835\udc53\ud835\udc65=\ud835\udc5a\ud835\udc65+\ud835\udc4f. Let g(x) be theindicated transformation of f(x)= x. Title: Graphing Linear Equations ANSWERS. Use function form to identify the slope. 3:Solve linear equations and inequalities in one variable including equations with coefficients represented by letters. 4 Lesson Lesson Tutorials EXAMPLE 1 Real-Life Application The percent y (in decimal form) of battery power remaining x hours after you turn on a laptop computer is y = \u22120. q(x) = -2x C. This book is a continuation of the authors Calculus, Volume I, Second Edition. Modeling with Linear Functions Work with a partner. 2 Functions of random variables There are three main methods to \ufb01nd the distribution of a function of one or more random variables. Interpret the rate of change and initial value of a linear function in terms of. In the process of solving a word problem it is often useful to make a chart of a few specific instances of the independent and dependent variables. here Disha Publication brought you all the concepts related to Linear Equations in One and Two variables with solved examples with each topic to get the clear understanding. ANSWER The table of values represents a quadratic function. perpendicular A. The pitch of a roof is the number of feet the roof rises for each 12 feet horizontally. VIII Linear equations of one variables test paper-1: File Size: 203 kb: File Type: pdf. January 2014 Download PDF. coefficients is an alias for it (stasts). 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(fall 2013) Chapter 2 Functions, Equations And Graphs. An equation is a statement that says two mathematical expressions are equal. There are many different ways that linear equations can be represented algebraically and plotted graphically. Graphing Linear Equations. f(n) = 783 + 8 n D. )Multiple Representations The graph shows the function (\ud835\udc65). Click the plus sign to view articles in. 4-2 Guided Notes Teacher Edition - Patterns and Linear Functions. Plot the points and graph the linear function. This method can be extended to a large class of linear elliptic equations and systems. For each function, all four representations of that function are present in the cards. For example, the following table shows the accumulation of snow on the morning of a snowstorm: Time 6:00 am 8:00 am 10:00 am 12:00 pm Snow depth 2 in. In the following we consider rst the stationary states of the linear harmonic oscillator and later consider the propagator which describes the time evolution of any initial state. slope = _____ y-intercept = _____ Verbal. Notice that the function in the example above is an example of a. 5) y = 3x + 2 6) y = \u2212x + 5 Find the slope of a line parallel to each given lin e. Linear functions are typically written in the form f(x) = ax + b. 4 Properties of the Multivariate Normal Distribution 92 4. This fact may be generalized as follows. 1Functions,#Domain,#and#Range#4#Worksheet# MCR3U& Jensen& # & 1)&Whichgraphsrepresentfunctions?Justifyyouranswer. Graph the relation represented by y 2x 1. Solving Systems. Question #21: This question requires students to build a linear function based on a given. On these printable worksheets, students will practice solving, finding intercepts, and graphing linear equations. Attorney A charges a fixed fee on$250 for an initial meeting and $150 per hour for all hours worked after that. Copy this to my account; E-mail to a friend; Find other activities. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Linear functions are those whose graph is a straight line. notebook 10 December 11, 2013 Aug 28\u00ad2:53 PM Create five of your own transformations based off the linear parent function. A normal 12 point text isn\u2019t the ideal solution for fast reading. Transformations of linear functions Learn how to modify the equation of a linear function to shift (translate) the graph up, down, left, or right. \u2022Example: y = 3x + 2 \u2022Solution: any ordered pair (x, y) that makes the equation true. 16\u00ad21 Linear vs. 6(x \u2013 9) + 10 \u2013 3x 5. MULTIPLE CHOICE Plotting the points from each of the given tables as shown below, identify its correct. Linear transformations: Rank-nullity theorem, Algebra of linear transformations, Dual spaces. During a 45-minute lunch period, Albert (A) went running and Bill (B) walked for exercise. If we transforming linear functions , we can say we are changing the linear function either the way it looks in the graph or the equation. Decide whether the word problem represents a linear or exponential function. Let x =1, then y =\u2212=\u221221 3 1(). Chapter & Page: 42\u20134 Nonhomogeneous Linear Systems which, in matrix/vector form, is x\u2032 = Ax + g with A = 1 2 2 1 and g(t) = 3 0 t + 0 2. Downloadable Graph Paper and Measurement Tools (pdf) Graph Paper Generators. The probability density function (pdf) technique, univariate Suppose that Y is a continuous random variable with cdf ( ) and domain \ud835\udc45 , and let = ( ), where : \ud835\udc45 \u2192\u211b is a continuous, one-to-one function defined over \ud835\udc45. Linear Function Word Problems Exercise 1Three pounds of squid can be purchased at the market for$18. De\ufb01nition of Linear Function A linear function f is any function of the form y = f(x) = mx+b where m and b are constants. 8 Relations and Functions; Day 3 Guided Notes; 4. 5x + 24 These are all linear equations. Using Linear Equations. Intersections between spheres, cylinders, or other quadrics can be found using quartic equations. In general, any linear rela-tion can be represented by a line. Investigation: Match Point Step 1 The investigation in your book gives three recursive formulas, three graphs, and three linear equations. 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Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. May 26, 2020 by admin. Intro to Linear Equations Algebra 6. This unit describes how to recognize a linear function, and how to \ufb01nd the slope and the y-intercept of its graph. ____ 17 y = x \u22128 A) C) B) D) 9. These di!erence equations are then implemented in the FDTD grid with equivalent voltageand current sources basedon static \"eld approximations. Solving Linear Equations - Formulas Objective: Solve linear formulas for a given variable. Examples: y = f(x) + 1 y = f(x - 2) y = -2f(x) Show Step-by-step Solutions. x y y = \u2212f(x) y = f(x) Multiplying the outputs by \u22121 changes their signs. Carefully graph each equation on the same coordinate plane. For each function: (a) Find the vertex ( , )hk of the parabola by using the formulas 2 b a h and 2 b a kf. This is also known as the \u201cslope. 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C The number of student participants. > Topics include: Least-squares aproximations of over-determined equations and least-norm solutions of underdetermined equations. f(x)=\u2212(x+2) 2 \u22127 Describe the transformation performed on each function g(x) to result in m(x). A di\ufb00erential equation (de) is an equation involving a function and its deriva-tives. Is 4 a solution of 5(2 \u2013 x) = \u201310? Show work to justify your answer. f(n) = 783 + 8 n D. 1 Univariate Normal Density Function 87 4. Solving Systems of Linear Equations by Graphing HW : 4/13/2017: SOL Coach Book 1. The function f(x) = 20x represents the daily rental fee for x days. linear equations1 V. If expenses are equal to travel, T, plus materials, M, which system of equations models this situation? A P S E E T M B P S E E T M C P S E E T M D P S E E T M 37. Lastly, I found that students apply their understandings from work with linear functions to solving and graphing quadratic equations. The types are: 1. Attorney A charges a fixed fee on $250 for an initial meeting and$150 per hour for all hours worked after that. SYSTEMS OF LINEAR EQUATIONS3 1. 63\u201367) 1 optional 0. modeling_linear_functions gives students a chance to practice using mathematical modeling applied to real-world contexts (). w V TA YlElG 7r 3i5gEhdt0s S Rrue ksYebr xvce ed3. )E J \u02c7 D0 ! 38. A linear differential operator (abbreviated, in this article, as linear operator or, simply, operator) is a linear combination of basic differential operators, with differentiable functions as coefficients. 3:Solve linear equations and inequalities in one variable including equations with coefficients represented by letters. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2. It is a Pdf worksheet based on graphic linear equations. Systems of Linear Equations Computational Considerations. The delta function is a normalized impulse. Translation. Practice: Relations & Functions Final corrections due: Use the given form of each relation to complete the other forms. Consider the function f : R2!R given by f(x1;x2) = x1x2. One Horizontal Shift Left 2. Another special type of linear function is the Constant Function it is a horizontal line: f(x) = C.\n1d2eclpf97 ivuzvhem3uncxb3 vwlzhmt1jkjzi6 dmz0vwqgu3pr2 l2tvdx2srobee d4fx1iwwka6 avgz4ir5y0 pb6az90pfvmd0 3fzjkp9r4e thnyfj3qhnlkrq ip3ejzm2lf hwckf2dywqzz zoycso277lcn acvt5z1hvqy4m2d 7mr8cn37zgamf4 a4z34cg8w1k f0mldd6pxy8y 69lmfwf0bmfsz otqhnvr1r3b kowpsz9xtblg tnnyl4hwll85uub 3dh0c6rygz1 pw0hnmkdbd47 k18untj4x0k5i9 yptw1geslm235 tibbilgv08oh hdrgmn3x5ka1 6haro44r4z5s nq9v4qx7sbm8 45evtfgj6x3d9l nqw0u3j9b0zpnf", "date": "2020-10-30 18:52:40", "meta": {"domain": "umood.it", "url": "http://uldn.umood.it/linear-functions-pdf.html", "openwebmath_score": 0.620610773563385, "openwebmath_perplexity": 1239.807064332939, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.986363161511632, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8621575883180621}}
{"url": "http://mathoverflow.net/questions/16141/primes-p-such-that-p-1-2-1-mod-p", "text": "# Primes P such that ((P-1)/2)!=1 mod P\n\nI was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\\equiv -1\\pmod P$. I realized this implies that for primes $P\\equiv 3\\pmod 4$, that $\\left(\\frac{P-1}{2}\\right)!\\equiv \\pm1 \\pmod P$.\n\nQuestion: For which primes $P$ is $\\left(\\frac{P-1}{2}\\right)!\\equiv 1\\pmod P$?\n\nAfter convincing myself that it's not a congruence condition for $P,$ I found this sequence in OEIS. I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\\bmod 4$).\n\nThanks,\n\nJacob\n\n-\nGiven that there are no comments of any note on the sequence in the OEIS, there's a fair chance that little is known about your question. \u2013\u00a0 Kevin Buzzard Feb 23 '10 at 10:28\nFor all p<=250000, p=3 mod 4, we have 5458 +1s and 5589 -1s. \u2013\u00a0 Kevin Buzzard Feb 23 '10 at 10:56\n\nI am a newcomer here. If p >3 is congruent to 3 mod 4, there is an answer which involves only $p\\pmod 8$ and $h\\pmod 4$, where $h$ is the class number of $Q(\\sqrt -p)$ . Namely one has $(\\frac{p-1}{2})!\\equiv 1 \\pmod p$ if an only if either (i) $p\\equiv 3 \\pmod 8$ and $h\\equiv 1 \\pmod 4$ or (ii) $p\\equiv 7\\pmod 8$ and $h\\equiv 3\\pmod 4$.\n\nThe proof may not be original: since $p\\equiv 3 \\pmod 4$, one has to determine the Legendre symbol\n\n$${{(\\frac{p-1}{2})!}\\overwithdelims (){p}} =\\prod_{x=1}^{(p-1)/2}{x\\overwithdelims (){p}}=\\prod_{x=1}^{(p-1)/2}(({x\\overwithdelims (){p}}-1)+1).$$ It is enough to know this modulo 4 since it is 1 or -1. By developping, one gets $(p+1)/2+S \\pmod 4$, where $$S=\\sum_{x=1}^{(p-1)/2}\\Bigl({x\\over p}\\Bigr).$$ By the class number formula, one has $(2-(2/p))h=S$ (I just looked up Borevich-Shafarevich, Number Theory), hence the result, since $\\Bigl({2\\over p}\\Bigr)$ depends only on $p \\pmod 8$.\n\n-\nThat's very slick! \u2013\u00a0 David Speyer Feb 23 '10 at 13:02\nYes, very nice! My interpretation of the question is \"do the primes for which the square root is 1 give a set of density 1/2?\" and this at least gives some way of attacking the problem. \u2013\u00a0 Kevin Buzzard Feb 23 '10 at 13:09\n+1. Salut, et bienvenu ! \u2013\u00a0 Chandan Singh Dalawat Feb 23 '10 at 13:58\nIn the paper emis.de/journals/EM/expmath/volumes/12/12.1/pp99_113.pdf they mention that Cohen proved modulo CL a conjecture of Hooley about a sum over $h(p)$. \u2013\u00a0 Victor Miller Feb 23 '10 at 15:14\nInteresting. In particular, that paper claims that the odd part of h(p), as p runs through primes, seems to have the same distribution as the odd part of h(D), as D runs through square free integers. That's something I didn't know. I point out, however, that this paper deals with real quadratic fields. \u2013\u00a0 David Speyer Feb 23 '10 at 15:42\n\nThere is some history to this question. Dirichlet observed (see p. 275 of History of the Theory of Numbers,'' Vol. 1) that since we already know $(\\frac{p-1}{2})! \\equiv \\pm 1 \\bmod p$, computing modulo squares gives $(\\frac{p-1}{2})! \\equiv (-1)^{n} \\bmod p$, where $n$ is the number of quadratic nonresidues mod $p$ which lie between 1 and $(p-1)/2$.\n\nJacobi (pp. 275-276 in Dickson's book) determined $n \\bmod 2$ in terms of the class number $h_p$ of ${\\mathbf Q}(\\sqrt{-p})$, for $p \\equiv 3 \\bmod 4$ and $p \\not= 3$. By the class number formula, $$\\left(2-\\left(\\frac{2}{p}\\right)\\right)h_p = r-n,$$ where $r$ is the number of quadratic residues from 1 to $(p-1)/2$. Also $r + n = (p-1)/2$, so $$2n = \\frac{p-1}{2} - \\left(2 - \\left(\\frac{2}{p}\\right)\\right)h_p.$$ In particular, $h_p$ is odd when $p \\equiv 3 \\bmod 4$.\n\nTaking cases if $p \\equiv 3 \\bmod 8$ and $p \\equiv 7 \\bmod 8$, we find both times that $n \\equiv (h_p+1)/2 \\bmod 2$, so $$\\left(\\frac{p-1}{2}\\right)! \\equiv (-1)^{(h_p+1)/2} \\bmod p.$$\n\nThis shows why getting precise statistics on when the congruence has 1 on the right side will be hard.\n\n-\n\nThe following is a relevant classical paper:\n\nMordell, L. J. The congruence $(p-1/2)!\\equiv \u00b11$ $({\\rm mod}$ $p)$. Amer. Math. Monthly 68 1961 145--146.\n\nhttp://www.math.uga.edu/~pete/Mordell61.pdf\n\nPut $((p-1)/2)!\\equiv(-1)^a\\ (\\text{mod}\\,p)$, where $p$ is a prime $\\equiv 3\\ (\\text{mod}\\,4)$. The author proves the following result. If $p\\equiv 3\\ (\\text{mod}\\,4)$ and $p>3$, then $$a\\equiv{\\textstyle\\frac 1{2}}\\{1+h(-p)\\}\\quad(\\text{mod}\\,2), \\tag1$$ where $h(-p)$ is the class number of the quadratic field $k(\\surd-p)$ [$\\mathbb{Q}(\\sqrt{-p})$ must be meant here. --PLC]. The author points out that (1) follows easily from a result of Dirichlet; also that Jacobi had conjectured an equivalent result before the class number formula was known. (MathReview by L. Carlitz)\n\n-\nThe notation k(\\sqrt{-p}) for our Q(\\sqrt{-p}) is \"classical\" and was used e.g. by Hilbert in his Bericht. The idea was that k(\\sqrt{-p}) is the field k you get by adjoining a square root of -p to the rationals. \u2013\u00a0 Franz Lemmermeyer Feb 23 '10 at 17:07\nHecke uses $K(\\root l\\of\\mu;k)$ to denote $k(\\root l\\of\\mu)$ in his Vorlesungen. \u2013\u00a0 Chandan Singh Dalawat Feb 24 '10 at 1:06\n\nThis is an attempt to justify the answer $1/2$ based on the Cohen-Lenstra heuristics. There will be a lot of nonsensical steps, and I am not an expert, so this should be viewed with caution.\n\nAs is observed above, this is equivalent to determining $h(p) \\mod 4$, where $h(p)$ is the class number of $\\mathbb{Q}(\\sqrt{-p})$. Since $p$ is odd and $3 \\mod 4$, the only ramified prime in $\\mathbb{Q}(\\sqrt{-p})$ is the principal ideal $(\\sqrt{-p})$. Thus, there is no $2$-torsion in the class group and $h(p)$ is odd.\n\nFor any odd prime $q$, let $a(q,p)$ be the power of $q$ which divides $h(p)$. We want to compute the average value of $$\\prod_{q \\equiv 3 \\mod 4} (-1)^{a(q,p)}.$$\n\nFirst nonsensical step: Let's pretend that the CL-heuristics work the same way for the odd part of the class group of $\\mathbb{Q}(\\sqrt{-p})$, that they do for the odd part of the class group of $\\mathbb{Q}(\\sqrt{-D})$. We just saw above that the fact that $p$ is prime constrains the $2$-part of the class group; this claim says that it does not effect the distribution of anything else.\n\nThen we are supposed to have: $$P(a(q,p)=0) = \\prod_{i=1}^{\\infty} (1-q^{-i}) = 1-1/q +O(1/q^2),$$ $$P(a(q,p)=1) = \\frac{1}{q-1} \\prod_{i=1}^{\\infty} (1-q^{-i}) = 1/q +O(1/q^2),$$ and $$P(a(q,p) \\geq 2) = O(1/q^2).$$\n\nIf you believe all of the above, then the average value of $(-1)^{a(p,q)}$ is $1-2/q+O(1/q^2)$.\n\nSecond nonsensical step: Let's pretend that $a(q,p)$ and $a(q',p)$ are uncorrelated. Furthermore, let's pretend that everything converges to its average value really fast, to justify the exchange of limits I'm about to do.\n\nThen $$E \\left( \\prod_{q \\equiv 3 \\mod 4} (-1)^{a(q,p)} \\right) = \\prod_{q \\equiv 3 \\mod 4} \\left( 1- 2/q + O(1/q^2) \\right)$$.`\n\nThe right hand side is zero, just as if $h(p)$ were equally like to be $1$ or $3 \\mod 4$.\n\n-\nMost of the latex formulas don't parse on my browser (Firefox 3.5.8) (though some do). Does any one know why? \u2013\u00a0 Anonymous Feb 23 '10 at 17:33\nMight you be missing the jsmath fonts? math.union.edu/~dpvc/jsMath/users/fonts.html \u2013\u00a0 David Speyer Feb 23 '10 at 17:50\nIt works! Great. Thanks! \u2013\u00a0 Anonymous Feb 23 '10 at 17:56", "date": "2015-05-23 15:12:04", "meta": {"domain": "mathoverflow.net", "url": "http://mathoverflow.net/questions/16141/primes-p-such-that-p-1-2-1-mod-p", "openwebmath_score": 0.9507311582565308, "openwebmath_perplexity": 339.3991996917357, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127423485423, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8621520376285041}}
{"url": "https://math.stackexchange.com/questions/2317435/multiply-2a-1b-by-2b-and-get-2ab-how-is-this-so", "text": "# multiply $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$? How is this so?\n\nI\u2019m reading How To Prove It and in the following proof the author is doing some basic algebra with exponents that I just don\u2019t understand. In Step 1.) listed below he is multiplying $2^b$ across each term in (1 + $2^b$ + $2^{2b}$ +\u00b7\u00b7\u00b7+$2^{(a-1)b}$) and gets the resulting set of terms in Step 2.) In particular I have no idea how he is getting $2^{ab}$ from multiplying $2^{(a-1)b}$ by $2^b$ again which is shown in the first sequence in Step 2.). When I do it I get $2^{(ab)(b) \u2013 (b)(b)}$ and assume this is as far as it can be taken. Can someone please help me understand what steps he is taking to to get his answer?\n\nTheorem 3.7.1. Suppose n is an integer larger than 1 and n is not prime. Then $2^n$ \u2212 1 is not prime. Proof. Since n is not prime, there are positive integers a and b such that a < n, b < n, and n = ab. Let x = $2^b$ \u2212 1 and y = 1 + $2^b$ + $2^{2b}$ +\u00b7 \u00b7 \u00b7+ $2^{(a\u22121)b}$. Then\n\nxy = ($2^b$ \u2212 1) \u00b7 (1 + $2^b$ + $2^{2b}$ +\u00b7\u00b7\u00b7+$2^{(a-1)b}$)\n\nStep 1.) = $2^b$ \u00b7 (1 + $2^b$ + $2^{2b}$ +\u00b7\u00b7\u00b7+$2^{(a-1)b}$) \u2212 (1 + $2^b$ + $2^{2b}$ +\u00b7\u00b7\u00b7+$2^{(a-1)b}$)\n\nStep 2.) = ($2^b$ + $2^{2b}$ + $2^{3b}$ +\u00b7\u00b7\u00b7+$2^{ab}$) \u2212 (1 + $2^b$ + $2^{2b}$ + \u00b7\u00b7\u00b7+$2^{(a-1)b}$)\n\nStep 3.) = $2^{ab}$ \u2212 1\n\nStep 4.) = $2^n$ \u2212 1.\n\n\u2022 Are you familiar with how $x^c\\times x^d = x^{c+d}$? Are you familiar with how $(a-1)b + b = ab$? \u2013\u00a0JMoravitz Jun 10 '17 at 16:00\n\u2022 My title should have read multiplying $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$. Sorry about that. \u2013\u00a0maybedave Jun 10 '17 at 16:00\n\u2022 @JMoravits. Not really and now I'm feeling a little dumb because from what you are saying, this is actually correct, no? \u2013\u00a0maybedave Jun 10 '17 at 16:01\n\u2022 For integer values of $c$ and $d$, notice that $x^c\\times x^d = \\overbrace{\\underbrace{x\\times x\\times x\\times \\cdots \\times x}_{c~\\text{times}}\\times \\underbrace{x\\times x\\times \\cdots \\times x}_{d~\\text{times}}}^{c+d~~\\text{times}}$. (The property can be extended to real values of $c$ and $d$ as well, look up a more in depth proof for that). As for why $(a-1)b+b=ab$, this is algebraic manipulation., $(a-1)b+b=ab-1b+b=ab-b+b=ab+(-b+b)=ab+0=ab$ \u2013\u00a0JMoravitz Jun 10 '17 at 16:07\n\n$$2^{(a-1)b}2^b=2^{(a-1)b+b}=2^{ab-b+b}=2^{ab}.$$\n\n\u2022 Ok, so what what you guys are saying is that I really should be adding the b to (a-1)b rather than multiplying it in. This is A BIG HELP! Thank you! \u2013\u00a0maybedave Jun 10 '17 at 16:04\n\u2022 @maybedave It's $x^rx^s=x^{r+s}$... \u2013\u00a0Lord Shark the Unknown Jun 10 '17 at 16:05\n\u2022 \"rather than multiplying it in\" You absolutely should not multiply them! Notice $64 =4*16=2^2*2^4=2^{2*4}=2^8=256$. Doesn't work. But $64= 2^2*2^4 = (2*2)*(2*2*2*2) = (2*2*2*2*2*2) = 2^6 = 64$. Addition has to work. Notice $2^4*2^4 = (2*2*2*2)*(2*2*2*2) = [(2*2*2*2)(2*2*2*2)(2*2*2*2)(2*2*2*2)] = 2^{4*4}$ just doesn't make any sense at all. \u2013\u00a0fleablood Jun 10 '17 at 16:18\n\nUsing the laws of exponents, $2^{(a-1)b}2^b = 2^{(a-1)b+b} = 2^{ab}$\n\nHint:\n\nThe correct rule is:\n\n$$2^b 2^{(a-1)b}=2^{(a-1)b+b}$$\n\nBasic rules $a^m a^n = (a*a*a*a*a........*a[n \\text{ times}])*(a*a*...... *a[m \\text{ times}]) = a*a*a*.......*a [n+m\\text { times}] = a^{n+mm}$.\n\nSo $2^{(a-1)b}*2^b = (2*2*2*2*2........*2[(a-1)b \\text{ times}])*(2*2*...... *2[b \\text{ times}]) = 2*2*2*.......*2 [(a-1)b+b\\text { times}] = 2^{(a-1)b + b}$\n\nMeanwhile $(a-1)*b + b = a*b - 1*b + b = (ab) + (-b + b) = ab + 0 = ab$.\n\nSo $2^{(a-1)b}2^b = 2^{(a-1)b + b} = 2^{ab}$", "date": "2020-01-25 14:19:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2317435/multiply-2a-1b-by-2b-and-get-2ab-how-is-this-so", "openwebmath_score": 0.7423398494720459, "openwebmath_perplexity": 132.14940301334107, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127423485423, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8621520195889252}}
{"url": "https://mathhelpboards.com/threads/converting-a-repeating-decimal-to-ratio-of-integers.5632/", "text": "# Converting a repeating decimal to ratio of integers\n\n#### paulmdrdo\n\n##### Active member\n0.17777777777 convert into a ratio.\n\n#### M R\n\n##### Active member\nRe: converting a repeating decimal to ratio of integers\n\nHi,\nThis is $$0.1 + 0.077777=\\frac{1}{10}+\\frac{7}{100}+\\frac{7}{1000}+...$$ where you have a GP to sum.\n\nOr $$\\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.\n\nSubtracting gives $$9x=0.7$$ and so $$x=\\frac{7}{90}$$. Now just add $$\\frac{1}{10}+\\frac{7}{90}$$ and simplify.\n\nI should also say that we can write a decimal as a fraction but we can't write it as a ratio.\n\n#### paulmdrdo\n\n##### Active member\nRe: converting a repeating decimal to ratio of integers\n\nHi,\nThis is $$0.1 + 0.077777=\\frac{1}{10}+\\frac{7}{100}+\\frac{7}{1000}+...$$ where you have a GP to sum.\n\nOr $$\\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.\n\nSubtracting gives $$9x=0.7$$ and so $$x=\\frac{7}{90}$$. Now just add $$\\frac{1}{10}+\\frac{7}{90}$$ and simplify.\n\nI should also say that we can write a decimal as a fraction but we can't write it as a ratio.\nwhat do you mean by \"GP\"?\n\n#### M R\n\n##### Active member\nRe: converting a repeating decimal to ratio of integers\n\nwhat do you mean by \"GP\"?\nSorry, I have to stop using abbreviations.\n\nA GP is a geometric progression: $$a, ar, ar^2, ar^3...$$.\n\nIf you haven't met this then the second method I posted is fine.\n\n#### soroban\n\n##### Well-known member\nRe: converting a repeating decimal to ratio of integers\n\nHello, paulmdrdo!\n\n$$\\text{Convert }\\,0.1777\\text{...}\\,\\text{ to a fraction.}$$\n\n$$\\begin{array}{ccc}\\text{We have:} & x &=& 0.1777\\cdots \\\\ \\\\ \\text{Multiply by 100:} & 100x &=& 17.777\\cdots \\\\ \\text{Multiply by 10:} & 10x &=& \\;\\;1.777\\cdots \\\\ \\text{Subtract:} & 90x &=& 16\\qquad\\quad\\; \\end{array}$$\n\nTherefore: .$$x \\;=\\;\\frac{16}{90} \\;=\\;\\frac{8}{45}$$\n\n#### paulmdrdo\n\n##### Active member\nhow would I decide what appropriate power of ten should i use?\n\nfor example i have 3.5474747474... how would you convert this one?\n\n#### M R\n\n##### Active member\nSince two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.\n\nIf you use 1000 and 10 you will get\n\n1000x=3547.474747...\n\n10x=35.474747...\n\nSo 990x=3512 and x=3512/990=1756/495.\n\nI'm adopting Soroban's approach as I prefer it to what I did earlier.\n\n#### paulmdrdo\n\n##### Active member\nSince two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.\n\nIf you use 1000 and 10 you will get\n\n1000x=3547.474747...\n\n10x=35.474747...\n\nSo 990x=3512 and x=3512/990=1756/495.\n\nI'm adopting Soroban's approach as I prefer it to what I did earlier.\n\"a difference of two in the powers of ten\" -- what do you mean by this? sorry, english is not my mother tongue. bear with me.\n\nLast edited:\n\n#### M R\n\n##### Active member\n\"a difference of two in the powers of ten\" -- what do you me by this? sorry, english is not my mother tongue. bear with me.\nNo problem.\n\nWe have 10^3 and 10^1.\n\nThe difference between 3 and 1 is 3-1=2\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nhow would I decide what appropriate power of ten should i use?\n\nfor example i have 3.5474747474... how would you convert this one?\nYou want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated.\n\nSo in this case, since the 47 repeats, you want the first to read \"something.4747474747...\" and the second to read \"something-else.4747474747...\"\n\nWhat powers of 10 will enable this?\n\n#### MarkFL\n\nStaff member\nA quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's.\n\n1.) $$\\displaystyle x=0.1\\overline{7}$$\n\n$$\\displaystyle 10x=1.\\overline{7}=1+\\frac{7}{9}=\\frac{16}{9}$$\n\n$$\\displaystyle x=\\frac{16}{90}=\\frac{8}{45}$$\n\n2.) $$\\displaystyle x=3.5\\overline{47}$$\n\n$$\\displaystyle 10x=35.\\overline{47}=35+\\frac{47}{99}=\\frac{3512}{99}$$\n\n$$\\displaystyle x=\\frac{3512}{990}=\\frac{1756}{495}$$", "date": "2020-09-28 11:27:02", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/converting-a-repeating-decimal-to-ratio-of-integers.5632/", "openwebmath_score": 0.7860057353973389, "openwebmath_perplexity": 1578.3981118946972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226287518853, "lm_q2_score": 0.8824278680004706, "lm_q1q2_score": 0.8621519952777413}}
{"url": "https://math.stackexchange.com/questions/2891882/limit-of-fx-given-that-fx-x-is-known/2891884", "text": "# Limit of $f(x)$ given that $f(x)/x$ is known\n\nGiven that $$\\lim_{x \\to 0} \\dfrac{f(x)}{x}$$ exists as a real number, I am trying to show that $\\lim_{x\\to0}f(x) = 0$. There is a similar question here: Limit of f(x) knowing limit of f(x)/x.\n\nBut this question starts with the assumption of $$\\lim_{x \\to 0} \\dfrac{f(x)}{x} = 0,$$ and all I am assuming is that the limit is some real number. So the product rule for limits doesn't really work here.\n\nOr do I need to show that $$\\lim_{x \\to 0} \\frac{f(x)}{x} = 0$$ and then apply the product rule?\n\n\u2022 Try proving it by contradiction: what happens if $\\lim_{x\\to 0}f(x)\\neq 0$? \u2013\u00a0Ender Wiggins Aug 23 '18 at 9:00\n\u2022 What prevents you from using product rule of limits? Perhaps you need to revisit the product rule in your text and then understand that it works fine here. \u2013\u00a0Paramanand Singh Aug 23 '18 at 14:23\n\u2022 @FurryFerretMan Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/\u2026 \u2013\u00a0gimusi Sep 17 '18 at 20:08\n\nThe product rule trick still works. If $\\lim_{x \\to 0} f(x)/x = R \\in \\mathbb R$, and obviously $\\lim_{x \\to 0} x = 0$, it follows that $$\\lim_{x \\to 0} f(x) = \\lim_{x \\to 0} \\frac{f(x)}{x} \\times x = R \\times 0 = 0.$$\n\n\u2022 This is the most natural solution and I would not consider use of limit laws a trick rather it is the method. +1 \u2013\u00a0Paramanand Singh Aug 23 '18 at 14:19\n\nWe have that eventually\n\n$$0\\le \\left|\\frac{f(x)}{x}\\right|\\le M$$\n\ntherefore\n\n$$0\\le \\left|f(x)\\right|\\le M|x| \\to 0$$\n\nLet $\\lim_{x\\to0}\\dfrac{f(x)}{x}=l$ then $\\bigg|\\dfrac{f(x)}{x}-l\\bigg|\\leq M$ for some $M\\in \\mathbb{R}$. So $\\bigg|\\dfrac{f(x)}{x}\\bigg|\\leq |l|+M\\Rightarrow |f(x)|\\leq |x|(|l|+M) \\Rightarrow \\lim_{x\\to 0} f(x)=0$\n\nLet $x_n \\rightarrow 0$.\n\n$y_n:= f(x_n)/x_n$, we have\n\n$y_n \\rightarrow L.$\n\nWith\n\n$f(x_n)=$\n\n$(f(x_n)/x_n)(x_n)=(y_n)(x_n)$.\n\n$\\lim_{n \\rightarrow \\infty }f(x_n)=$\n\n$\\lim_{n \\rightarrow \\infty}((y_n)(x_n))=$\n\n($\\lim_{n \\rightarrow \\infty}(y_n))(\\lim_{n \\rightarrow \\infty}(x_n))=$\n\n$L \\cdot 0=0.$", "date": "2019-08-23 15:35:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2891882/limit-of-fx-given-that-fx-x-is-known/2891884", "openwebmath_score": 0.907375156879425, "openwebmath_perplexity": 301.70757531191634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226314280634, "lm_q2_score": 0.8824278618165526, "lm_q1q2_score": 0.8621519915974478}}
{"url": "http://mathhelpforum.com/number-theory/177338-divisibility-6-a.html", "text": "1. ## Divisibility by 6\n\nHey guys I need help with this question\n\n$\\text{prove}\n(4^{n}-1)(3^{n}-1) \\text{is divisible by 6 for all positive values of n, without Induction}$\n\nand the answer given in the solution to the question is that since\n\n$4^{n}-1=(1+3)^{n}-1 \\therefore \\text{Divisble by 3}$\n$3^{n}-1=(1+2)^{n}-1 \\therefore \\text{Divisble by 2}$\n\n$\\therefore\\text{Product is divisble of 2 and 3, therefore divisble by 6}$\n\nI know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?\n\nThe same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction\n\nIf we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?\n\nThanks\n\n2. 1.\nI guess you don't know Newton's generalized binomial theorem...\n\n$(a+b)^n=M_a+b^n=M_b+a^n$\nI considered $M_k$ a multiple of k.\n\nSo:\n$4^n-1=(1+3)^n-1=M_3+1-1=M_3$\n$3^n-1=(1+2)^n-1=M_2+1-1=M_2$\n\n2.\n$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-2}+b^{n-1})$\n\nSo:\n$4^n-1=(4-1)(4^{n-1}+4^{n-2}+...+4+1)=3(4^{n-1}+4^{n-2}+...+4+1)$\n$3^n-1=(3-1)(3^{n-1}+3^{n-2}+...+3+1)=2(3^{n-1}+3^{n-2}+...+3+1)$\n\n3. Originally Posted by aonin\nHey guys I need help with this question\n\n$\\text{prove}\n(4^{n}-1)(3^{n}-1) \\text{is divisible by 6 for all positive values of n, without Induction}$\n\nand the answer given in the solution to the question is that since\n\n$4^{n}-1=(1+3)^{n}-1 \\therefore \\text{Divisble by 3}$\n$3^{n}-1=(1+2)^{n}-1 \\therefore \\text{Divisble by 2}$\n\n$\\therefore\\text{Product is divisble of 2 and 3, therefore divisble by 6}$\n\nI know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?\n\nThe same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction\n\nIf we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?\n\nThanks\nThe $\\displaystle 3^n - 1$ is easy, without any tricks like below. $\\displaystyle 3^n$ is an odd number, so subtracting 1 from it gives us an even number.\n\n$(1 + 3)^n - 1$ is a bit trickier. Note that when we expand by the binomial theorem:\n\n$(1 + 3)^n - 1 = (1^n + n \\cdot 1^{n - 1}3^1 + ~...~ + n \\cdot 1^13^{n - 1} + 3^n) - 1$\n\n$\\displaystyle = n \\cdot 1^{n - 1}3^1 + ~...~ + n \\cdot 1^13^{n - 1} + 3^n$\nof which you can prove (using the combinatorial coefficients) that every coefficient is divisible by 3.\n\n-Dan\n\n4. $(1 + 3)^n - 1 = (1^n + 3 \\cdot 1^{n - 1}3^1 + ~...~ + 3 \\cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.\n\n$(a+b)^n=\\sum_{k=0}^{n}\\binom{n}{k}\\cdot a^{n-k} \\cdot b^{k}$.\n\n5. Originally Posted by veileen\n$(1 + 3)^n - 1 = (1^n + 3 \\cdot 1^{n - 1}3^1 + ~...~ + 3 \\cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.\n\n$(a+b)^n=\\sum_{k=0}^{n}\\binom{n}{k}\\cdot a^{n-k} \\cdot b^{k}$.\nWhoops! I got carried away with the 3's. Thanks for the catch. I have fixed it in my post.\n\n-Dan\n\n6. Originally Posted by aonin\nHey guys I need help with this question\n\n$\\text{prove}\n(4^{n}-1)(3^{n}-1) \\text{is divisible by 6 for all positive values of n, without Induction}$\n\nand the answer given in the solution to the question is that since\n$4^{n}-1=(1+3)^{n}-1 \\therefore \\text{Divisble by 3}$\n$3^{n}-1=(1+2)^{n}-1 \\therefore \\text{Divisble by 2}$\n$\\therefore\\text{Product is divisble of 2 and 3, therefore divisble by 6}$\nNote that $\\displaystyle A=4^{n}-1=(1+3)^{n}-1=\\sum\\limits_{k = 1}^n {\\binom{n}{k}\\left( 3 \\right)^k }$\n$\\displaystyle B=3^{n}-1=(1+2)^{n}-1=\\sum\\limits_{k = 1}^n {\\binom{n}{k}\\left( 2 \\right)^k }$\n\nBecause the index begins with $k=1$ every term in $A$ is a multiple of 3. Similarly every term is B is even.\n\nHence every term in the product $AB$ is a multiple of 6.", "date": "2016-08-28 12:17:03", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/177338-divisibility-6-a.html", "openwebmath_score": 0.8095536828041077, "openwebmath_perplexity": 288.0143502586726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022630759019, "lm_q2_score": 0.8824278540866547, "lm_q1q2_score": 0.8621519834547791}}
{"url": "https://math.stackexchange.com/questions/3371883/how-to-prove-that-x-is-a-completion-of-y-for-some-metric-spcaes-x-d-and", "text": "# How to prove that $X$ is a completion of $Y$, for some metric spcaes $(X,d)$ and $(Y,d)$?\n\nIn our Real Analysis class we did a proof for existence of completion of metric space. However, I do not understand how do we use it practically. i.e How do I prove that $$X$$ is a completion of $$Y$$, for some metric spcaes $$(X,d)$$ and $$(Y,d)$$?\n\nFor example : How do I prove that $$(\\mathbb{R},d_1)$$ is a completeion of $$(\\mathbb{Q},d_1)$$? Here, $$d_1$$ is the standard Euclidean metric\n\n$$\\underline{\\text{My attempt}}$$ -\n\nI believe we should prove that $$\\mathbb{Q}$$ is subset of $$\\mathbb{R}$$ and $$\\mathbb{R}$$ is complete. This would mean that all Cauchy sequence in $$\\mathbb{Q}$$ converges in $$\\mathbb{R}$$.\n\nHowever, consider this case -\n\nLet $$(W,d)$$ be an incomplete Metric space and $$W \\subset Y \\subset X$$ and $$X,Y$$ are complete (with same metric).\n\nNow, by my argument both $$X,Y$$ can be the completion of $$W$$. Is this alright? I read somewhere in Stack Exchange that completion is unique(I didn't really understand that answer)\n\n$$\\underline{\\text{Approach to resolve this}}$$ - We choose the smallest, complete set which contains $$W$$ as its completion. We can prove that, since $$X$$ is complete any of it's closed subset is complete. We can also prove that $$\\overline{W}$$ is the smallest closed set that contains $$W$$. Hence, $$\\overline{W}$$ is the completion of $$W$$ (this would explain the notation in the completion proof).\n\nThus in the title - I just have to check if $$Y$$ is dense in $$X$$ to prove that $$X$$ is a completion of $$Y$$.\n\nThis would also solve my example question - We know that $$\\mathbb{Q}$$ is dense in $$\\mathbb{R}$$ and $$\\mathbb{R}$$ is complete. Thus $$\\mathbb{R}$$ is completion of $$\\mathbb{Q}$$.\n\nI am fairly confident that this is correct. However, I am a Physics Master's student this is my first \"Real\" Math course ;) So I would appreciate if someone let's me know if this right and if not the corrections.\n\n\u2022 What is $d_1$? ${}$ \u2013\u00a0Jos\u00e9 Carlos Santos Sep 27 '19 at 11:13\n\u2022 $d_1$ is the standard Euclidean metric \u2013\u00a0Indigo1729 Sep 27 '19 at 11:29\n\u2022 This follows by the universal property of completion of metric spaces. \u2013\u00a0\u03b5--\u03b4 Sep 27 '19 at 11:59\n\nTo show that $$\\mathbb{R}$$ is the Cauchy completion of $$\\mathbb{Q}$$, it is not sufficient to show that $$\\mathbb{Q}$$ is contained in $$\\mathbb{R}$$ and $$\\mathbb{R}$$ is complete. In fact, this just means that the completion of $$\\mathbb{Q}$$ is contained within $$\\mathbb{R}$$.\n\nFor example, the completion of $$\\mathbb{Q}$$ is contained in $$\\mathbb{C}$$ and the complex numbers are complete, but the completion of the rational numbers is not the complex numbers.\n\nYou should also verify that every element of $$\\mathbb{R}$$ is the limit of a sequence of rational numbers (which follows directly from the fact that $$\\mathbb{Q}$$ is dense in $$\\mathbb{R}$$ for example) in order to be sure that $$\\mathbb{R}$$ is actually $$\\mathbb{Q}$$'s completion. This resolves your worry about the uniqueness theorem for completions which you referred to.\n\nOne final point: one doesn't normally define the Cauchy completion in order to identify completions of metric spaces they already know (e.g. $$\\mathbb{Q}$$ in your example). Instead, knowing that you can take the completion lets you construct new metric spaces which can be useful in their own right. Again for example, knowing that completions exist and taking the completion of $$\\mathbb{Q}$$ is a way to construct the real numbers $$\\mathbb{R}$$.\n\nOne nitpick: \"I believe we should prove that $$\\Bbb Q$$ is subset of $$\\Bbb R$$ and $$\\Bbb R$$ is complete. This would mean that all Cauchy sequence in $$\\Bbb R$$ converge in $$\\Bbb R$$.\"\n\nTo sum up, viewing $$\\Bbb Q$$ as a subset of $$\\Bbb R$$ (so we can avoid speaking of the embedding map and the image of $$\\Bbb Q$$ under it), you need to show that\n\n\u2022 the metric on $$\\Bbb Q$$ is the restriction of the metric on $$\\Bbb R$$\n\u2022 $$\\Bbb Q$$ is dense in $$\\Bbb R$$ and of course\n\u2022 $$\\Bbb R$$ is complete under its metric\n\nand that is essentially what you found out by yourself.\n\nYou're missing one... critical point!\n\nIt is not sufficient for $$X$$ to be a completion of $$W$$ that $$W \\subseteq X$$ and $$X$$ is complete. You also need $$W$$ to be dense in $$X$$.\n\nThat being said, $$\\mathbb R$$ is indeed a completion of $$\\mathbb Q$$.", "date": "2020-02-22 20:44:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3371883/how-to-prove-that-x-is-a-completion-of-y-for-some-metric-spcaes-x-d-and", "openwebmath_score": 0.9479386210441589, "openwebmath_perplexity": 84.78167151601455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018441287091, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8621167423100332}}
{"url": "https://www.intmath.com/forum/methods-integration-31/integration-techniques:161", "text": "IntMath Home \u00bb Forum home \u00bb Methods of Integration \u00bb Integration Techniques\n\n# Integration Techniques [Solved!]\n\n### My question\n\nYou state in Section four that all angles are in radians and that the formulas do not work in degress.\n\nWhy do they only work in radians?\n\n### Relevant page\n\n4. Integration: Basic Trigonometric Forms\n\n### What I've done so far\n\nJust wanted to know if there are some integration techniques that work in degrees.\n\nX\n\nYou state in Section four that all angles are in radians and that the formulas do not work in degress.\n\nWhy do they only work in radians?\nRelevant page\n\n<a href=\"https://www.intmath.com/methods-integration/4-integration-trigonometric-forms.php\">4. Integration: Basic Trigonometric Forms</a>\n\nWhat I've done so far\n\nJust wanted to know if there are some integration techniques that work in degrees.\n\n## Re: Integration Techniques\n\nRadians have the \"magical\" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).\n\nDegrees, on the other hand, can only be a measure of an angle (or of course, temperature).\n\nProbably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.\n\nConsider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):\n\nWe know the following for the green curve:\n\nd/dx sin(x) = cos(x)\n\nAt x=0, the slope is cos(0) = 1\n\nBut now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:\n\nAt x=0^@, the slope is pi/180 cos(0^@) = pi/180\n\nSo if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.\n\nIn order for the following formula to \"work\", x needs to be in radians:\n\nd/dx sin(x) = cos(x)\n\nHope it helps.\n\nX\n\nRadians have the \"magical\" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).\n\nDegrees, on the other hand, can only be a measure of an angle (or of course, temperature).\n\nProbably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.\n\nConsider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):\n\n[graph]310,250;-1,10;-1.1,1.1,1.57,1;sin(x),sin(3.14x/180)[/graph]\n\nWe know the following for the green curve:\n\nd/dx sin(x) = cos(x)\n\nAt x=0, the slope is cos(0) = 1\n\nBut now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:\n\nAt x=0^@, the slope is pi/180 cos(0^@) = pi/180\n\nSo if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.\n\nIn order for the following formula to \"work\", x needs to be in radians:\n\nd/dx sin(x) = cos(x)\n\nHope it helps.\n\n## Re: Integration Techniques\n\nIt does. A good example to use. Thank you.\n\nX\n\nIt does.  A good example to use.  Thank you.", "date": "2019-05-19 22:25:15", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/methods-integration-31/integration-techniques:161", "openwebmath_score": 0.8898437023162842, "openwebmath_perplexity": 1197.0884921612965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.960361162033533, "lm_q2_score": 0.897695295528596, "lm_q1q2_score": 0.8621116971658782}}
{"url": "https://math.stackexchange.com/questions/2171078/how-do-i-add-the-terms-in-the-binomial-expansion-of-10026/2171125", "text": "# How do I add the terms in the binomial expansion of $(100+2)^6$?\n\nSo, I stumbled upon the following question.\n\nUsing binomial theorem compute $102^6$.\n\nNow, I broke the number into 100+2.\n\nThen, applying binomial theorem\n\n$\\binom {6} {0}$$100^6(1)+\\binom {6} {1}$$100^5(2)$+....\n\nI stumbled upon this step. How did they add the humongous numbers? I am really confused.\n\nKindly help me clear my query.\n\n\u2022 They're powers of ten.... And you don't mean \"embezzled.\" Maybe you mean \"confused.\" Mar 4 '17 at 4:39\n\u2022 Correct! I need to work on my English skills, man! Thanks for the help tho! Mar 4 '17 at 4:46\n\u2022 @symplectomorphic - actually, they probably meant \"bamboozled\", as it means something similar to \"confused\", but can be easily mixed up with \"embezzled\". Mar 4 '17 at 7:42\n\nThe long way:\n\n\\begin{align} & 10^{12} + 12 \\cdot 10^{10} + 6 \\cdot 10^9 + 16 \\cdot 10^7 + 24 \\cdot 10^5 +192 \\cdot 10^2 + 64 \\\\ =\\; & 10^{12} + (10+2) 10^{10} + 6 \\cdot 10^9 + (10+6) 10^7 + (20+4) 10^5 +(100+90 +2) 10^2 + 60+4 \\\\ =\\; & \\color{red}1\\cdot10^{12} + \\color{red}1 \\cdot 10^{11} + \\color{red}2\\cdot 10^{10} + \\color{red}6 \\cdot 10^9 + \\color{red}1 \\cdot 10^8 + \\color{red}6 \\cdot 10^7+\\color{red}2 \\cdot 10^6 + \\color{red}4 \\cdot 10^5+\\color{red}1 \\cdot 10^4+\\color{red}9\\cdot 10^3 + \\color{red}2 \\cdot 10^2 + \\color{red}6 \\cdot 10^1 + \\color{red}4 \\cdot 10^0 \\end{align}\n\nThe latter is precisely the representation in base $\\,10\\,$ of $\\;\\color{red}{1126162419264}\\,$.\n\n\u2022 Thanks for the help! Perhaps the most convincing and noteworthy answer. Mar 4 '17 at 10:52\n\nThat number is $1\\; 12 \\; 6\\; 16 \\; 24\\;192\\; 64$ (space added for emphasis). Notice a relationship with the coefficients of the powers of $10$?\n\n$10^{12} + 12\\times10^{10} + 6\\times10^9 + 16*10^7 + 24\\times10^5 + 192\\times10^2 + 64$ $= 10^{12} + 10^{11} + 2\\times10^{10} + 6\\times10^9 + 10^8 + 6\\times10^7 + 2\\times10^6 + 4\\times10^5 + 10^4 + 9\\times10^3 + 2\\times10^2 + 6\\times10^1 + 4\\times10^0= 1126162419264$\n\n\u2022 Thank you very much, sir! I really appreciate it. Mar 4 '17 at 10:51\n\nSince it's all powers of ten you could add those quite easily. If it were something else you could have needed to use some calculator. But here you have just powers of tens, which basically keep everything the same. Here's the number with a few spaces to make you understand it.\n\n$\\text{1 12 6 16 24 192 64}$ Those powers just end up just putting those digits in the right order because we use a number system with ten digits.\n\n\u2022 Ok! Why not 10112...? Mar 4 '17 at 4:50\n\u2022 You see say you're adding something like a $12 \\times 10^{10}$. You can see that those powers are all spaced evenly. By that I mean to say whenever you have say n digits to be added, over there they had decreased the exponent of 10 by n\n\u2013\u00a0SBM\nMar 4 '17 at 4:53\n\u2022 Moreover, saying a number equals something like $123.xyz$ is the same thing as $1 \\times 10^2 + 2 \\times 10^1 + 3 \\times 10^0 + x \\times 10^{-1} \\ldots$ and so on. As long as you keep those evenly spaced, it remains same. By x, y and z I meant arbitrary digits not variables\n\u2013\u00a0SBM\nMar 4 '17 at 4:56\n\u2022 I got your point. Mar 4 '17 at 5:04\n\nStart adding from the rightmost term($64$); you will notice that the number of trailing zeros on the preceding term is exactly the same as the number of digits in following term. $$...2400000\\quad +\\\\ .....19200\\quad +\\\\ .......64\\quad$$ ...and so on till the first.", "date": "2022-01-24 08:00:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2171078/how-do-i-add-the-terms-in-the-binomial-expansion-of-10026/2171125", "openwebmath_score": 0.9993034601211548, "openwebmath_perplexity": 856.7617811735823, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.960361158630024, "lm_q2_score": 0.8976952866333484, "lm_q1q2_score": 0.862111685567914}}
{"url": "https://math.stackexchange.com/questions/595233/prove-by-induction-that-5n-1-is-divisible-by-4/595293", "text": "# Prove by induction that $5^n - 1$ is divisible by $4$.\n\nProve by induction that $5^n - 1$ is divisible by $4$.\n\nHow should I use induction in this problem. Do you have any hints for solving this problem?\n\nThank you so much.\n\n\u2022 What do you know about the question or what have you tried? Dec 6 '13 at 7:10\n\u2022 I have a vague feeling that this question has already been covered on the site.\nDec 6 '13 at 12:37\n\nWe prove that for all $n \\in \\mathbb{N}$, $4 \\mid \\left( 5^n-1 \\right)$. (Notationally, this says $4$ divides $5^n-1$ with a zero remainder).\n\n1. For a basis, let $n=1$. Then $$5^1-1=4,$$ and clearly $4\\mid4$.\n\n2. Assume that $5^n-1$ is is divisible by $4$ for $n=k, \\, k \\in\\mathbb{N}$. Then by this assumption, $$4 \\mid \\left( 5^k-1 \\right) \\Rightarrow 5^k-1=4m, \\, m \\in \\mathbb{Z}.$$ (This notationally means that $5^k-1$ is an integer multiple of $4$.)\n\n3. Let $n=k+1$. Then \\begin{align*} 5^{k+1}-1 &= 5^k \\cdot 5-1 \\\\ &=5^k(4+1)-1 \\\\ &=4\\cdot 5^k+5^k -1 \\\\ &=4\\cdot5^k+4m\\\\ &=4\\left( 5^k+m \\right). \\end{align*} Since $4\\mid4\\left( 5^k+m \\right)$, we may conclude, by the axiom of induction, that the property holds for all $n \\in \\mathbb{N}$.\n\n\u2022 Nice, clear, clean... a good answer indeed! Dec 6 '13 at 16:53\n\u2022 this should be the accepted answer Jan 15 '17 at 5:43\n\nFirst prove the base case $n=1$. Then induct and make use of the fact that $$(5^{n+1}-1) - (5^n-1) = 4 \\cdot 5^n$$to conclude what you want.\n\nWithout induction, you can use the identity\n\n$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)$$\n\nOf course you would still need induction or something to prove this identity.\n\n\u2022 Yes. I removed my comment once I saw your updated answer. To be really pedantic, even in my answer, we still need induction first to show that $5^n$ is an integer, to then claim that $4$ divides $4 \\cdot 5^n$.\n\u2013\u00a0user17762\nDec 6 '13 at 7:24\n\nwithout induction $$5^n-1=(4+1)^n-1$$ $$=4^n+n4^{n-2}+...+1-1$$ the only term in $(1+4)^n$ not being multiplied by a power of $4$ is $1$ but it disappears due to the $-1$.\n\nWhy induction? $5^n$ ends in $\\dots25$ for $n>1$, so $5^n-1$ ends in $\\dots24$.\n\n\u2022 Technically, you still need induction to show that $5^n$ ends in $25$ for $n>1$.\n\u2013\u00a0user17762\nDec 6 '13 at 7:17\n\u2022 @user17762 $25 * 5 = 25 \\mod 100$ Dec 6 '13 at 13:48\n\u2022 @ratchetfreak Technically I think you still need induction to prove the statement about an arbitrary $n$. Of course, using the fact you mention, the induction becomes trivial (like proving by induction that $1^n = 1$ for all $n$ using the fact that 1 is the multiplicative identity.) Dec 6 '13 at 18:48\n\n$\\displaystyle{5^{n + 1} - 1 = \\left(5^{n} - 1\\right)5 + 4}$\n\nit's even more general:\n\n$k$ divides $(k+1)^n-1$ with $k,n \\in \\mathbb{N}$\n\nsimply by modular arithmetic:\n\n$$k+1 = 1 \\mod {k} \\\\ \\Downarrow \\\\(k+1)^n=1 \\mod {k}$$\n\nTo prove by induction you:\n\n1. Assume the proposition is true for n\n\n2. Show that if it is true for n, then it is also true for n+1\n\n3. Show that it is true for n=1\n\nThen you know that it will be true for all natural numbers.\n\nIn this case:\n\n1. Assume $5^n-1$ is divisible by 4\n\n2. Say $m=5^n$, so $m-1$ is divisible by 4\n\n\u2022 $5^{n+1}-1$ = $5m - 1$\n\u2022 $5m - 1$ = $5(m-1) + 4$\n\u2022 Since $m-1$ is divisible by 4 and 4 is divisible by 4, then this expression is divisible by 4.\n3. For $n=1$, $5^1 - 1 = 4$ which is divisible by 4\n\nAnd there you are...\n\nThis isn't by induction, but I think it's a nice proof nonetheless, certainly more enlightening: $\\displaystyle 5^n-1=(1+4)^n-1=\\sum_{k=0}^n {n\\choose k}4^k-1=1+\\sum_{k=1}^n {n\\choose k}4^k-1=\\sum_{k=1}^n {n\\choose k}4^k$ which is clearly divisible by $4$.", "date": "2021-09-24 04:03:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/595233/prove-by-induction-that-5n-1-is-divisible-by-4/595293", "openwebmath_score": 0.9902795553207397, "openwebmath_perplexity": 277.7247681770266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806552225684, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8620743133212974}}
{"url": "http://mathhelpforum.com/pre-calculus/25502-partial-fractions.html", "text": "1. ## Partial Fractions\n\n2. I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.\n\n3. My answer is the part in red (Part 3).\n\n4. Originally Posted by r_maths\nMy answer is the part in red (Part 3).\nHmm, I get $A = 1, B = 2$, which leads to a final answer of: $x+\\frac{1}{x+1}+\\frac{2}{x+2}$. I do not see how the x term is not supposed to be there.\n\n5. I did include the x\n\nI got the same answer as you, but the answer from the book gives a different answer...\n\n6. Originally Posted by r_maths\nI did include the x\n\nI got the same answer as you, but the answer from the book gives a different answer...\n\nI do not see how B equals one, so the book is probably wrong unless both of us missed something.\n\n7. Hello, r_maths!\n\nExpress $\\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ in partial fractions.\n\n$\\frac{x^3+3x^2+5x+4}{(x+1)(x+2)} \\;=\\;x + \\frac{3x+4}{(x+1)(x+2)} \\;=\\;x + \\frac{1}{x+1} + \\frac{2}{x+2}\\quad{\\color{red}Right!}$\n\nBook answer: . $x + \\frac{1}{x+1} + \\frac{1}{x+2}\\quad{\\color{red}Wrong!}$\n$\\text{Their answer adds up to: }\\;\\frac{x^3+3x^2+4x+3}{(x+1)(x+2)}$ . . . . obviously not the original fraction.\n\n$\\frac{x}{1}\\cdot{\\color{blue}\\frac{(x+1)(x+2)}{(x+ 1)(x+2)}} + \\frac{1}{x+1}\\cdot{\\color{blue}\\frac{x+2}{x+2}} + \\frac{2}{x+2}\\cdot{\\color{blue}\\frac{x+1}{x+1}}$\n\n. . $= \\;\\frac{x(x+1)(x+2) + (x+2) + 2(x+1)}{(x+1)(x+2)}$\n\n. . $= \\;\\frac{x^3+3x^2+2x + x + 2 + 2x + 2}{(x+1)(x+2)}$\n\n. . $= \\;\\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ . . . . See?\n\n8. $\\frac{{x^3 + 3x^2 + 5x + 4}}\n{{(x + 1)(x + 2)}} = \\frac{{x^3 + 3x^2 + 2x + 3x + 4}}\n{{(x + 1)(x + 2)}} = \\frac{{x(x + 1)(x + 2) + (3x + 4)}}\n{{(x + 1)(x + 2)}}$\n.\n\nNow\n\n$x + \\frac{{3x + 4}}\n{{(x + 1)(x + 2)}} = x + \\frac{{(x + 2) + 2(x + 1)}}\n{{(x + 1)(x + 2)}} = x + \\frac{1}\n{{x + 1}} + \\frac{2}\n{{x + 2}}$\n.\n\nHence $\\frac{{x^3 + 3x^2 + 5x + 4}}\n{{(x + 1)(x + 2)}} = x + \\frac{1}\n{{x + 1}} + \\frac{2}\n{{x + 2}}$\n.", "date": "2013-12-19 11:41:55", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/25502-partial-fractions.html", "openwebmath_score": 0.9561547636985779, "openwebmath_perplexity": 545.9538553048765, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806523850542, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8620742999594186}}
{"url": "https://math.stackexchange.com/questions/3200305/can-we-determine-summands-from-their-partial-sums", "text": "# Can we determine summands from their partial sums?\n\nAssume there are non-negative numbers $$\\lambda_1\\le \\ldots\\le \\lambda_n\\in[0,\\infty)$$. You are given the (ordered) list $$s_1\\le\\ldots\\le s_{2^n}\\in[0,\\infty)$$ of all partial sums, i.e. every $$s_i$$ is of the form $$s_i=\\sum_{k\\in K_i}\\lambda_k$$ for some unique but unknown $$K_i\\subseteq\\{1,\\ldots,n\\}$$.\n\nQuestion: Can we determine the $$\\lambda_1,\\ldots,\\lambda_n$$ from knowing the $$s_1,\\ldots,s_{2^n}$$?\n\nSome obvious facts are\n\n1. Since there is some $$s_i$$ corresponding to $$K_i=\\emptyset$$, $$s_1=\\cdots=s_i=0$$.\n2. $$\\lambda_1=s_2$$, since no non-trivial partial sum can be smaller as the smallest possible summand.\n3. $$s_{2^n}=\\sum_{k=1}^n\\lambda_k$$, since no partial sum can be bigger.\n4. For $$n=2$$ the answer is yes, since $$\\lambda_1=s_2$$ and $$\\lambda_2=s_4-s_2$$.\n\nNote: For my use case it would be sufficient to know if for any given $$s_1\\le\\cdots\\le s_{2^n}$$ there is at most one possibility for $$\\lambda_1\\le\\cdots\\le\\lambda_n$$.\n\n1. By calculating how many $$s_i$$ are zero, you can determine how many $$\\lambda_i$$ are zero. If there are any, they will be creating repeated entries $$s_i$$ throughout, but in a systematic manner where one can eliminate their effect; indeed, if there are $$k$$ zeroes, then there will be $$k$$ extra duplicates of every entry. For simplicity, suppose $$\\lambda_1 > 0$$.\n\n2. Now, indeed, $$\\lambda_1 = s_2$$.\n\n3. We proceed by induction. Suppose we have identified $$\\lambda_1,\\ldots,\\lambda_j$$ and calculated the partial sums consisting of only $$\\lambda_1,\\ldots,\\lambda_j$$, such as $$\\lambda_1+\\lambda_3$$. Then the smallest remaining partial sum must be $$\\lambda_{j+1}$$. Proof: Otherwise the smallest remaining partial sum would have to be a sum with at least one unknown $$\\lambda_m$$ that is (by definition) not yet in the list of known partial sums, which would imply that $$\\lambda_m$$ is smaller than the smallest remaining partial sum; a contradiction.\n\n4. Now that $$\\lambda_{j+1}$$ is also known, consider the known list of partial sums to include all sums of $$\\lambda_1,\\ldots,\\lambda_{j+1}$$.\n\nActually, separate treatment of steps 1 and 2 is not necessary; one only has to initialize the list of known partial sums with the empty sum $$s_1=0$$.\n\nThis method probably works even if you are missing entires from the list of partial sums, as long as all the missing entries are strictly greater than $$\\lambda_n$$.\n\n\u2022 One has to be careful: Assume we have $\\lambda_1=1,\\lambda_2=2,\\lambda_3=3,\\lambda_4=4,...$ and already computed $\\lambda_1,\\lambda_2$. The set of known partial sums is $K=\\{0,1,2,3\\}$, so the smallest partial sum not in $K$ would be $4\\ne\\lambda_3$. The problem here is that $\\lambda_3=\\lambda_1+\\lambda_2$ is itself a partial sum. I believe that this can be fixed by using multisets of partial sums: In the above situation the multiset of all partial sums would be $\\{0,1,2,3,3,4,...\\}$, so after removing the multiset $K$ the smallest remaining element would be $3=\\lambda_3$ as required. \u2013\u00a0Robert Rauch Apr 25 at 7:47\n\u2022 @RobertRauch Indeed, but since you are using an ordered list of partial sums with $2^{n}$ elements, it already contains the information about multiplicity. \u2013\u00a0Tommi Brander Apr 25 at 7:55\n\nHere is a streamlined version of Tommi Brander's solution: $$\\newcommand{\\IN}{\\mathbb{N}}$$\n\nLemma: Let $$0\\le\\lambda_1\\le\\cdots\\le\\lambda_n$$ and $$\\mathcal{P}\\doteq\\left\\{\\lambda_K\\doteq\\sum_{k\\in K}\\lambda_k\\left|K\\subseteq\\mathbb{N}_n\\right.\\right\\}$$ the set of their partial sums, where $$\\IN_n\\doteq\\{1,2,\\ldots,n\\}$$. For $$0\\le l\\le n$$ consider the function $$m_l:\\mathcal{P}\\to\\IN_0$$ with $$m_l(\\lambda)=\\#\\{K\\subseteq\\IN_l\\mid\\lambda_K=\\lambda\\}$$. Then for all $$1\\le l\\le n$$, $$$$\\lambda_l=\\min\\left\\{\\lambda\\in\\mathcal{P}\\mid m_{l-1}(\\lambda) In particular, noting that $$m_0(\\lambda)=\\mathbf{1}(\\lambda=0)$$, the $$\\lambda_1,\\ldots,\\lambda_n$$ can be recovered iteratively only from $$\\mathcal{P}$$ and $$m_n$$.\n\nProof: Let $$\\mathcal{P}_l\\doteq\\left\\{\\lambda\\in\\mathcal{P}\\mid m_{l-1}(\\lambda). By construction, $$\\lambda_l\\in\\mathcal{P}_l$$ and therefore $$\\lambda_*=\\min\\mathcal{P}_l$$ exists and satisfies $$\\lambda_*\\le\\lambda_l$$. Since $$m_{l-1}(\\lambda_*), there is some $$K\\subseteq\\IN_n$$ with $$K\\not\\subseteq\\IN_{l-1}$$ such that $$\\lambda_*=\\lambda_K$$. In particular there is $$r\\in K$$ with $$r\\ge l$$ and, hence, $$\\lambda_r\\in\\mathcal{P}_l$$. By positivity of the $$\\lambda_i$$ we find that $$\\lambda_r\\le\\lambda_K=\\lambda_*$$ and therefore $$\\lambda_*=\\lambda_r$$ by minimality of $$\\lambda_*$$. Since $$r\\ge l$$ we conclude $$\\lambda_l\\le \\lambda_r=\\lambda_*\\le\\lambda_l$$, thus $$\\lambda_*=\\lambda_l$$.", "date": "2019-08-20 05:44:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3200305/can-we-determine-summands-from-their-partial-sums", "openwebmath_score": 0.9793107509613037, "openwebmath_perplexity": 152.2293881229989, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336352207334, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8620689742649936}}
{"url": "https://iidb.org/threads/geometry-question-re-hexagons.11275/", "text": "Geometry Question re: Hexagons\n\nMalintent\n\nVeteran Member\nFor a woodworking project... I want to cut a smooth circle from the outer perimeter of a constructed hexagon. I need to calculate how wide a board I need to construct the hexagon with, such that there is sufficient material to cut away to form a circle, while maintaining integrity of the construction..\n\nIt is a geometry question, stated simply as, \"what is the maximum distance of a circle's circumference from an inscribed hexagon\"? No luck finding an answer to that, so here is a more detailed means of expressing what I need to find out:\n\ninscribe a hexagon within a circle, such that each of the 6 vertices of the hexagon are touching the circle. The radius of circle is equal to the length of each segment of the hexagon, as well as its radius.\n\nAt every center point of each segment of the hexagon, the circle's circumference is at its maximum distance from the hexagon's segment.\n\nWhat is that distance?\n\nThat distance would be the exact width of the board, if used to construct the hex, where cutting the circle would just exactly separate the attached segments (not good). So I would take that distance and add the amount of material I need for structural integrity. I suppose the length of each board (segment of the hex) would need to be my desired final radius, plus 1/2 the distance I am looking for.\n\nright?\n\nSarpedon\n\nVeteran Member\nI don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.\n\nAnyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.\n\nLast edited:\n\nbeero1000\n\nVeteran Member\nIf I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \\frac{\\sqrt{3}}{2})S$$\n\nTeX seems to be out right now, so I'm stuck typesetting like some sort of savage\n\nIt's (2 - \u221a3)S/2\n\nTreedbear\n\nVeteran Member\nI don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.\n\nAnyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.\n\n~0.5% smaller than exact\n\nJunior Member\nIt is a geometry question, stated simply as, \"what is the maximum distance of a circle's circumference from an inscribed hexagon\"?\n\nI tackled this with pencil, paper, and calculator, assuming a 1 unit length for the radius of the circle and the sides of the hexagon. Calculating the altitude of an equilateral triangle with sides of 1 gives a little over 0.866; this would be the the distance from the center of the circle/hexagon to the midpoint of the side of the hexagon. Subtracting 0.866 from 1 gives 0.134, the distance from the midpoint of the side of the hexagon out to the circumference of the circle. The proportions are similar to what Sarpedon came up with.\n\nMalintent\n\nVeteran Member\nIf I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \\frac{\\sqrt{3}}{2})S$$\n\nTeX seems to be out right now, so I'm stuck typesetting like some sort of savage\n\nIt's (2 - \u221a3)S/2\n\n\"circumradius and inradius\" are the terms I didn't know that resulted in not finding my answer on my own. Thank you for the education!\n\nso, now that I was able to construct a concise question for Google, I easily found this regarding the relationship between the two values:\n\n\"since such a ratio is just the ratio between the side and the height of an equilateral triangle, (R/r) = (2/sqrt(3))\"\n\nwhich is apparently a constant ratio... approximately (way more precise than needed) R - r (the value I need) = R * 0.134\n\n.. just like ya'll said. Thanks again!\n\nso, with a 1.5 foot segment length, I will need 2.41 inches of \"spare room\" on each board for the arc of the circle. Since I want 1 inch of material inside the finished piece (for structural integrity), I will need to use at least a 3.1 inch wide board. so, 4 inch lumber boards will be perfect (they are 3.5\").\n\nI imagined I would have been able to do this with 3 inch boards (2.5), but I was wrong. THAT'S why I like to calculate before cutting materials\n\nLast edited:\n\nSarpedon\n\nVeteran Member\nTool thickness. Those saws at the sawmill are massive.\n\nMalintent\n\nVeteran Member\nTool thickness. Those saws at the sawmill are massive.\n\nWell, I won't be milling my own lumber... my blades are the standard 1/8\" kerf. But that isn't a measure of accuracy, just waste.\n\nMalintent\n\nVeteran Member\n4 inch lumber boards will be perfect (they are 3.5\")\n\nObviously.\n\nIs it? It was never obvious to me that a \"2 by 4\" actually measures 1.75x3.5. Or were you being sarcastic and didn't realize (like most, I would imagine) that lumber is prepared that way commercially?\n\nMalintent\n\nVeteran Member\nanyway... thank you all again for the \"magic ratio\" of 1 : 0.134... such an easily workable answer.\n\nJimmy Higgins\n\nContributor\nWhy don't you just get a hexagonal saw bit?\n\nbeero1000\n\nVeteran Member\nTool thickness. Those saws at the sawmill are massive.\n\nIt's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.\n\nbilby\n\nFair dinkum thinkum\nTool thickness. Those saws at the sawmill are massive.\n\nIt's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.\n\nExcuses, excuses.\n\nThere is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.\n\nThe reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.\n\nAll pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.\n\nOf course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.\n\nBut now that the information is so widely available, it seems a bit pointless to keep up this silliness.\n\nBronzeage\n\nSuper Moderator\nStaff member\nIt's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.\n\nExcuses, excuses.\n\nThere is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.\n\nThe reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.\n\nAll pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.\n\nOf course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.\n\nBut now that the information is so widely available, it seems a bit pointless to keep up this silliness.\n\nAnyone who has ever worked with \"rough\" lumber will happily pay for a 4 inch board and take a 3.5 board.\n\nIn any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.\n\nbeero1000\n\nVeteran Member\nIt's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.\n\nExcuses, excuses.\n\nThere is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.\n\nThe reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.\n\nAll pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.\n\nOf course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.\n\nBut now that the information is so widely available, it seems a bit pointless to keep up this silliness.\n\nNothing except marketing - bigger is better, and if they can sell you a 3.5\" board and make you think you're buying a 4\" board then they don't mind at all. Even if they get sued for it.\n\nKosh\n\nJunior Member\nExcuses, excuses.\n\nThere is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.\n\nThe reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.\n\nAll pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.\n\nOf course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.\n\nBut now that the information is so widely available, it seems a bit pointless to keep up this silliness.\n\nAnyone who has ever worked with \"rough\" lumber will happily pay for a 4 inch board and take a 3.5 board.\n\nIn any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.\n\nI've recently built a cnc machine, and it's pretty annoying to be dealing with sheet goods that are supposedly 19 mm thick but I'm finding them from 17 mm on up......I have to measure each piece and adjust the program accordingly.\n\nCheerful Charlie\n\nContributor\nTo understand the relationships of hexagons to circles, google for Unit Circle. Then you will want a nice trig capable calculator.\n\nTreedbear\n\nVeteran Member\n... I need to calculate how wide a board I need to construct the hexagon ...\n\nAre you talking about a standard 5 sided hexagon or one that actually has 6 sides?\n\nMalintent\n\nVeteran Member\n... I need to calculate how wide a board I need to construct the hexagon ...\n\nAre you talking about a standard 5 sided hexagon or one that actually has 6 sides?\n\ntake a 3-legged dog and feed it 1 square meal. Then take it's poop and use it to fertilize a 5-sided pentagonal hexagon. It will eventually grow another leg, which you can either use to make an actual hexagon, or just give it to the dog to use.\n\nI thought everyone knew this...", "date": "2022-08-14 09:23:49", "meta": {"domain": "iidb.org", "url": "https://iidb.org/threads/geometry-question-re-hexagons.11275/", "openwebmath_score": 0.4863211214542389, "openwebmath_perplexity": 1379.0634059500646, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126525529014, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.8620472491422432}}
{"url": "http://workshop19.info/lhrzk/b73d15-topological-sort-example", "text": "Topological Sort Algorithm. Topological Sort by BFS: Topological Sort can also be implemented by Breadth First Search as well. Example: building a house with a Topological Sorting; graphs If is a DAG then a topological sorting of is a linear ordering of such that for each edge in the DAG, appears before in the linear ordering. Example (Topological sort showing the linear arrangement) The topologically sorted order is not necessarily unique. Topological Sort: A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering.A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). Topological Sort Algorithms. Topological sort Topological-Sort Ordering of vertices in a directed acyclic graph (DAG) G=(V,E) such that if there is a path from v to u in G, then v appears before u in the ordering. Such an ordering cannot exist if the graph contains a directed cycle because there is no way that you can keep going right on a line and still return back to where you started from. 50 Topological Sort Algorithm: Runtime For graph with V vertexes and E edges: ordering:= { }. In other words, a topological sort places the vertices of a directed acyclic graph on a line so that all directed edges go from left to right.. 3. 3/11 Topological Order Let G = (V;E)be a directed acyclic graph (DAG). R. Rao, CSE 326 9 A B C F D E Topological Sort Algorithm Step 2: Delete this vertexof in-degree 0 and all its \ufffd\ufffd\ufffd There are severaltopologicalsortingsof (howmany? Definition of Topological Sort. For example, we can put on garments in the following order: A topological sort of a DAG is a linear ordering of all its vertices such that if contains an edge , then appears before in the ordering. Topological Sort is Not Unique. Yufei Tao Topological Sort on a DAG Introduction There are many problems involving a set of tasks in which some of the tasks must be done before others. Topological Sort Algorithm. An Example. Definition of Topological Sort. Example 1 7 2 9 4 10 6 3 5 8 Given n objects and m relations, a topological sort's complexity is O(n+m) rather than the O(n log n) of a standard sort. Node 30 depends on node 20 and node 10. That is there may be other valid orderings that are also partial orders that describe the ordering in a DAG. Topological Sort Introduction. A topological sort of a graph $$G$$ can be represented as a horizontal line \ufffd\ufffd\ufffd Consider the graph in the following example: This graph has two possible topological sorts: In this article, we have explored how to perform topological sort using Breadth First Search (BFS) along with an implementation. Show the ordering of vertices produced by $\\text{TOPOLOGICAL-SORT}$ when it is run on the dag of Figure 22.8, under the assumption of Exercise 22.3-2. A topological order of G is an ordering of the vertices in V such that, for every edge(u;v)in E, it must hold that u precedes v in the ordering. Here\ufffd\ufffd\ufffds simple Program to implement Topological Sort Algorithm Example in C Programming Language. 22.4 Topological sort 22.4-1. ), for example\ufffd\ufffd\ufffd Hence node 10, node 20 and node 40 should come before node 30 in topological sorting. As we know that the source vertex will come after the destination vertex, so we need to use a \ufffd\ufffd\ufffd Topological sort: It id defined as an ordering of the vertices in a directed acyclic graph, such that if there is a path from u to v, then v appears after u in the ordering. The ordering of the nodes in the array is called a topological ordering. Review Questions. Here's an example: Topological Sort Introduction. Our start and finish times from performing the $\\text{DFS}$ are \ufffd\ufffd\ufffd If we run a topological sort on a graph and there are vertices left undeleted, the graph contains a cycle. There could be many solutions, for example: 1. call DFS to compute f[v] 2. Topological sorting only works for directed acyclic graphs $$\\left({DAG}\\right),$$ that is, only for graphs without cycles. The graphs should be directed: otherwise for any edge (u,v) there would be a path from u to v and also from v to u, and hence they cannot be ordered. In this article, you will learn to implement a Topological sort algorithm by using Depth-First Search and In-degree algorithms Topological sort is an algorithm which takes a directed acyclic graph and returns a list of vertices in the linear ordering where each vertex has to precede all vertices it directs Topological Sorting for a graph is not possible if the graph is not a DAG. Types of graphs: a. As the visit in each vertex is finished (blackened), insert it to the Node 10 depends on node 20 and node 40. Topological Sort Algorithm Example of a cyclic graph: No vertex of in-degree 0 R. Rao, CSE 326 8 Step 1: Identify vertices that have no incoming edges \ufffd\ufffd\ufffd Select one such vertex A B C F D E Topological Sort Algorithm Select. For every edge U-V of a directed graph, the vertex u will come before vertex v in the ordering. ; There may exist multiple different topological orderings for a given directed acyclic graph. Repeat until graph is empty: Find a vertex vwith in-degree of 0-if none, no valid ordering possible Delete vand its outgoing edges from graph ordering+= v O(V) O(E) O(1) O(V(V+E)) Is the worst case here really O(E) every time?For example, Provided example with dw04 added to the dependencies of dw01. Let\ufffd\ufffd\ufffds pick up node 30 here. Implementation. We have compared it with Topological sort using Depth First Search.. Let us consider a scenario where a university offers a bunch of courses . Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. For example, a topological sorting of the following graph is \ufffd\ufffd\ufffd5 4 \ufffd\ufffd\ufffd Topological sort is an algorithm that orders a directed graph such that for each directed edge u\ufffd\ufffd\ufffdv, vertex u comes before vertex v.. A topological sort is an ordering of the nodes of a directed graph such that if there is a path from node u to node v, then node u appears before node v, in the ordering.For example \ufffd\ufffd\ufffd Some vertices are ordered, but the second return is nil, indicating that not all vertices could be sorted. A topological ordering, or a topological sort, orders the vertices in a directed acyclic graph on a line, i.e. Implementation of Source Removal Algorithm. Topological Sort Problem: Given a DAG G=(V,E), output all the vertices in order such that if no vertex appears before any other vertex that has an edge to it Example input: Example output: 142, 126, 143, 311, 331, 332, 312, 341, 351, 333, 440, 352 11/23/2020 CSE 142 CSE 143 CSE 331 Node 20 depends on node 40. An Example. Since, we had constructed the graph, now our job is to find the ordering and for that Topological Sort is Not Unique. For a DAG, we can construct a topological sort with running time linear to the number of vertices plus the number of edges, which is . Topological Sort. Topological Sort Example- Consider the following directed acyclic graph- Review Questions. in a list, such that all directed edges go from left to right. Please note that there can be more than one solution for topological sort. > (topological-sort *dependency-graph*) (IEEE DWARE DW02 DW05 DW06 DW07 GTECH DW01 DW04 STD-CELL-LIB SYNOPSYS STD DW03 RAMLIB DES-SYSTEM-LIB) T NIL. Topological Sorting Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge ( u v ) from \ufffd\ufffd\ufffd Example: Let & and have if and only if $. The topological sort algorithm takes a directed graph and returns an array of the nodes where each node appears before all the nodes it points to. For example, a simple partially ordered set may look as follows: Figure 1. This is partial order, but not a linear one. Example. If there are very few relations (the partial order is \"sparse\"), then a topological sort is likely to be faster than a standard sort. Introduction There are many problems involving a set of tasks in which some of the tasks must be done before others. Topological sorting works well in certain situations. It is important to note that-Topological Sorting is possible if and only if the graph is a Directed Acyclic Graph. Cycle detection with topological sort \ufffd\ufffd\ufffd What happens if we run topological sort on a cyclic graph? \ufffd\ufffd\ufffd There will be either no vertex with 0 prerequisites to begin with, or at some point in the iteration. To better understand the logic behind topological sorting and why it can't work on a graph that contains a cycle, let's pretend we're a computer that's trying to topologically sort the following graph: # Let's say that we start our search at node X # Current node: X step 1: Ok, i'm starting from node X so it must be at the beginnig of the sequence. The topological sorting for a directed acyclic graph is the linear ordering of vertices. \ufffd\ufffd\ufffd\u6028\ufffd\u7531\u044a\ufffd\ufffd - Topological Sort (\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd) (0) 2014.02.15: \ufffd\ufffd\ufffd\u6028\ufffd\u7531\u044a\ufffd\ufffd - Connected Component (0) 2014.02.15: \ufffd\ufffd\ufffd\u6028\ufffd\u7531\u044a\ufffd\ufffd - Priority Queue(\ufffd\ufffd\uacd7\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\u745c\ufffd \u63f4\u044b\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd) (0) 2014.02.15: \ufffd\ufffd\ufffd\u6028\ufffd\u7531\u044a\ufffd\ufffd - Heap Sort (\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd(\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd)\u745c\ufffd \u63f4\u044b\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd) (0) 2014.02.15 Solution for topological sort using Breadth First Search ( BFS ) along with an implementation topological sort must be before... 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Sort 22.4-1 ordering in a directed graph such that for each directed edge u\ufffd\ufffd\ufffdv, vertex u before!, we have explored how to perform topological sort 22.4-1 ) the topologically sorted order is not a linear.. Article, we have explored how to perform topological sort using Breadth First Search BFS... Not a linear one the 22.4 topological sort comes before vertex v sort, orders the vertices in DAG... Of the tasks must be done before others u will come before vertex v in the.!", "date": "2021-05-12 05:16:47", "meta": {"domain": "workshop19.info", "url": "http://workshop19.info/lhrzk/b73d15-topological-sort-example", "openwebmath_score": 0.3495599031448364, "openwebmath_perplexity": 629.6850657722558, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9787126494483641, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.8620472464077757}}
{"url": "http://mathhelpforum.com/algebra/137321-sequences.html", "text": "# Math Help - Sequences\n\n1. ## Sequences\n\nI can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?\n\n11, 9, 10, 17, 36, 79,....\n\n2. Originally Posted by nsingh201\nI can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?\n\n11, 9, 10, 17, 36, 79,....\nI don't know if this helps you at all, but I believe it is possible to find a polynomial of at most degree 5 that will fit that sequence, by solving simultaneous linear equations.\n\nI will illustrate the method on a smaller scale. Say the sequence is\n\n11, 9, 10, ...\n\nThen we can find a polynomial of degree 2 that will generate this sequence, of the form f(x) = ax^2+bx+c, as follows:\n\nWe know f(1) = 11. Thus, a*(1^2)+b*1+c = 11. Simplifying, a+b+c=11.\n\nWe know f(2) = 9. Thus, a*(2^2)+b*2+c = 9. Simplifying, 4a+2b+c=9.\n\nWe know f(3) = 10. Thus, a*(3^2)+b*3+c = 10. Simplifying, 9a+3b+c=10.\n\nThis is a system of linear equations with three equations and three unknowns. The solution is described by (a,b,c) = (3/2, -13/2, 16). That is, f(x) = (3/2)x^2 - (13/2)x + 16. This generates the sequence\n\n11, 9, 10, 14, 21, 31, ...\n\nObviously this is not the desired sequence, so a quadratic curve-fitting scheme is not feasible. I would proceed by next trying a cubic polynomial to model the first four terms, and if that doesn't work, a degree 4 polynomial to include the first five terms, and in the worst case, a degree 5 polynomial.\n\nOf course there may be a much simpler explanation of your sequence that I'm blind to...\n\n3. 11,9,10,17,36,79,170,357,736,1499,3030,6097,12236, 24519,...\n\n4. ## ?\n\n5. Hello, nsingh201!\n\nFind the next two terms of this sequence: . $11,\\: 9,\\: 10,\\: 17,\\: 36,\\: 79 \\;\\hdots$\n\nGiven a sequence of increasing terms, I take the differences of consecutive terms,\n. . then the differences of the differences, and so on . . . hoping to find a pattern.\n\nHere's what I found . . .\n\n. . $\\begin{array}{ccccccccccccc}\n\\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\\\\n\\text{1st diff.} &&\\text{-}2 && 1 && 7 && 19 && 43 \\\\\n\\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\\\\n\\text{3rd diff.} &&&& 3 && 6 && 12 \\end{array}$\n\nSince the 3rd differences seem to be identical to the 2nd differences,\n. . the function appears to be exponential.\n\nI also suspect that the 2nd differences are doubling.\n\nSo I believe the chart can be continued like this;\n\n. . $\\begin{array}{cccccccccccccccccc}\n\\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\\color{red}170} && {\\color{red}357} \\\\\n\\text{1st diff.} &&\\text{-}2 && 1 && 7 && 19 && 43 && {\\color{blue}91} && {\\color{blue}187} \\\\\n\\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\\color{blue}48} && {\\color{blue}96}\\end{array}$\n\nAnd I think the generating function is: . $f(n) \\;=\\;3\\!\\cdot\\!2^{n-1} - 5n + 13$\n\n6. Originally Posted by Soroban\nHello, nsingh201!\n\nGiven a sequence of increasing terms, I take the differences of consecutive terms,\n. . then the differences of the differences, and so on . . . hoping to find a pattern.\n\nHere's what I found . . .\n\n. . $\\begin{array}{ccccccccccccc}\n\\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\\\\n\\text{1st diff.} &&\\text{-}2 && 1 && 7 && 19 && 43 \\\\\n\\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\\\\n\\text{3rd diff.} &&&& 3 && 6 && 12 \\end{array}$\n\nSince the 3rd differences seem to be identical to the 2nd differences,\n. . the function appears to be exponential.\n\nI also suspect that the 2nd differences are doubling.\n\nSo I believe the chart can be continued like this;\n\n. . $\\begin{array}{cccccccccccccccccc}\n\\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\\color{red}170} && {\\color{red}357} \\\\\n\\text{1st diff.} &&\\text{-}2 && 1 && 7 && 19 && 43 && {\\color{blue}91} && {\\color{blue}187} \\\\\n\\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\\color{blue}48} && {\\color{blue}96}\\end{array}$\n\nAnd I think the generating function is: . $f(n) \\;=\\;3\\!\\cdot\\!2^{n-1} - 5n + 13$\n\nbeautiful approach\n\n7. 11, 9, 10, 17, 36, 79, - Wolfram|Alpha\n2 Soroban\nNice solution, but how did u get generating function?\n\n8. Hello, ICanFly!\n\nI was afraid someone would ask . . .LOL!\n\n. . .but how did u get generating function?\n\nWe have: . $11,\\:9,\\:10,\\:17,\\:36,\\;\\hdots$\n\nAnd we know there is a \"doubling\" feature in there: . $2^n$\n\nSince the doubling didn't begin until the second differences.\n. . I assumed there was a linear or quadratic involved, too.\n\nSo, my first general function was: . $f(n) \\:=\\:A + Bn + C\\!\\cdot\\!2^n$\n\nI used the first three terms of the sequence:\n\n. . $\\begin{array}{ccccc}\nf(1) = 11 & A + B + 2C &=& 11 & [1] \\\\\nf(2) \\:=\\: 9\\; & A + 2B + 4C &=& 9 & [2] \\\\\nf(3) = 10 & A + 3B + 8C &=& 10 & [3] \\end{array}$\n\n$\\begin{array}{ccccc}\n\\text{Subtract [2] - [1]:} & B + 2C &=& -2 & [4] \\\\\n\\text{Subtract [3] - [2]:} & B + 4C &=& 1 & [5] \\end{array}$\n\n$\\text{Subtract [5] - [4]: }\\;\\;2C \\:=\\: 3 \\quad\\Rightarrow\\quad C \\:=\\:\\tfrac{3}{2}$\n\nSubstitute into [4]: . $B + 3 \\:=\\:-2 \\quad\\Rightarrow\\quad B \\:=\\:-5$\n\nSubstitute into [1]: . $A - 5 + 3 \\:=\\:11 \\quad\\Rightarrow\\quad A \\:=\\:13$\n\nHence: . $f(n) \\;=\\;13 - 5n + \\tfrac{3}{2}\\!\\cdot\\!2^n$\n\nI tested my function up to the 8th term . . . It works!\n\n. . Therefore: . $f(n) \\;=\\;3\\!\\cdot\\!2^{n-1} - 5n + 13$\n\n9. Beautiful\n\n10. thanks so much. this all really helped.", "date": "2014-08-30 15:26:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/137321-sequences.html", "openwebmath_score": 0.7960879802703857, "openwebmath_perplexity": 778.6200079538289, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126531738087, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.862047242033028}}
{"url": "https://mathhelpboards.com/threads/1-998001.3403/", "text": "# 1/998001\n\n#### Jameson\n\nStaff member\n$$\\displaystyle \\frac{1}{998001}$$ has some interesting properties. It will list out every 3 digit number except for 998.\n\nIt has the form 0.001002003004...\n\nHere's a video on some more details if you're interested.\n\n#### MarkFL\n\nStaff member\nsoroban once posted this observation and asked:\n\nThis is just one of a family of such fractions.\nCan you determine the underlying characteristic?\nI looked at the factorization of the denominator, and found:\n\n$\\displaystyle 998001=999^2$\n\nSo, next I looked at:\n\n$\\displaystyle\\frac{012345679}{999999999}=\\frac{1}{9^2}$\n\nThus, I conjecture that the family you speak of is:\n\n$\\displaystyle\\frac{1}{(10^n-1)^2}$ where $\\displaystyle n\\in\\mathbb N$\n\nwhere the decimal representation contains all of the $n$ digit numbers except $\\displaystyle 10^n-2$ and the period is $\\displaystyle n(10^n-1)$.\n\n#### Jameson\n\nStaff member\n\nYes, you are correct about this form. For example $$\\displaystyle \\frac{1}{9999^2}$$ gives all of the 4 digit numbers.\n\nIn the video I posted in the OP they discussed an easy way to write recurring decimals as fractions that I'm sure we are all familiar with this method but it's still worth writing for those who aren't.\n\nIf you want to write some repeating decimal all you do is write the string as the numerator to a fraction and then in the denominator write the same number of 9's. For example, if you want to write $$\\displaystyle .\\overline{12436298}$$ as a fraction. It's simply $$\\displaystyle \\frac{12436298}{99999999}$$. Of course this method can be derived quite easily. If $$\\displaystyle x=.12436298$$ then $100000000x=12436298.\\overline{12436298}$ so $9999999x=12436298$ and we can solve for x directly. However it's nice to know a shortcut to not have to calculate it \"the long way\" everytime.\n\nLast edited:\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nThis seems to generalize to an arbitrary base $b$, where:\n\n$$\\frac{1}{(b^n - 1)^2}$$\n\nhas representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.\n\nBut I only checked a couple of bases, so this could be wrong.\n\n#### MarkFL\n\nStaff member\nThis seems to generalize to an arbitrary base $b$, where:\n\n$$\\frac{1}{(b^n - 1)^2}$$\n\nhas representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.\n\nBut I only checked a couple of bases, so this could be wrong.\nI believe you are right, as the method Jameson outlined above can be generalized to any valid radix.\n\n#### alane1994\n\n##### Active member\nNot much to contribute to this conversation, I was just popping in to say that this is really interesting!", "date": "2021-03-01 00:32:58", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/1-998001.3403/", "openwebmath_score": 0.8491010069847107, "openwebmath_perplexity": 434.1662406617897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712650690179, "lm_q2_score": 0.8807970685907242, "lm_q1q2_score": 0.862047233720567}}
{"url": "https://math.stackexchange.com/questions/2237462/complex-numbers-and-the-norm/2237482", "text": "# Complex Numbers and the Norm\n\nI am given that for a complex number $w=a+bi$, define $\\overline{w}=a-bi$ and $N(w)=w \\overline{w}$. I have provided my answers for parts a and b, but I am not sure they are correct. I need help with figuring out part c.\n\n(a) Compute $N(w)=w \\overline{w}$ explicitly.\n\nHere is what I have gotten:$N(w)=w \\overline{w}= (a+bi)(a-bi)= a^2+b^2$.\n\n(b) Show that $N(rs)=N(r)N(s)$ for any two complex numbers $r,s$.\n\nHere is my work:\n\nLet $r=a+bi$, $s=c+di$\n\nThen $$rs=(a+bi)(c+di)=ac+adi+cbi-bd=(ac-bd)+(ad+cb)i.$$\n\nNow, from a, have that $N(r)= a^2+b^2$ and $N(s)= c^2+d^2$. So, $$N(r)N(s)=(a^2+b^2)(c^2+d^2).$$\n\nAlso, $N(rs)=N((ac-bd)+(ad+cb)i)$ and from part a then, $N(rs)=(ac-bd)^2+(ad+cb)^2$\n\nWhen expanded,\n\n$$a^2c^2-abcd-abcd+b^2d^2+a^2d^2+abcd+abcd+b^2c^2$$\n\n\\begin{align*} &=a^2c^2+a^2d^2+b^2d^2+b^2c^2\\\\ &=a^2(c^2+d^2)+b^2(c^2+d^2)\\\\ &=(a^2+b^2)(c^2+d^2). \\end{align*}\n\nHence, $N(rs)=N(r)N(s)$.\n\n(c) Prove that $N(\\overline{v})=N(v)$ and $N(v^n)=N(v)^n$ for any complex number.\n\nThis is the part I am confused as to how to prove. I let $v=a+bi$ and $\\overline{v}=a-bi$. Then I am not sure how to use what I have shown before for this part.\n\n\u2022 (1) Please put distinct parts into distinct questions. (2) Please use MathJax to format your math, it makes it a lot easier to read. Apr 16 '17 at 21:35\n\u2022 @MichaelBurr Sorry, I do not know how to use MathJax, I tried my best to put it in the format. I labeled the parts a-c, but I will rename to make is clearer. Apr 16 '17 at 21:38\n\u2022 What I mean is that if you have three parts, you should have three questions (one question for part $(a)$, one question for part $(b)$, and one question for part $(c)$). Apr 16 '17 at 21:39\n\u2022 @MichaelBurr Okay, noted, thank you. It is just that is all one problem that has three parts and I wanted to see if my part a and b are okay to then get help for c. Apr 16 '17 at 21:40\n\u2022 Also, to learn about MathJax, see the tutorial here. Apr 16 '17 at 21:41\n\nFor the first part of your question, notice that $$N (v)=\\bar{v} v=\\bar{v} \\bar{\\bar{v}}=N (\\bar{v})$$ because $\\bar{\\bar{v}}=v$.\n\nFor the second part, consider doing induction over $n$:\n\nInduction basis: $n=1$ $$N (v^1)=N (v)=N (v)^1$$\n\nInduction step: $$N (v^{n+1})=N (v^n\\cdot v)=N (v^n) \\cdot N (v)=N (v)^n \\cdot N (v)=N (v)^{n+1}$$ For the third from last step we used your result from b, for the second from last we used the induction hypothesis.\n\n\u2022 Fixed now, thanks a lot. The way it was before, I did not like it either, but I did not know there is also \\bar. Apr 16 '17 at 22:21\n\u2022 Personally, I often use \\overline because \\bar is intended to be used as accent, but in this particular case overline looks plain ugly. Apr 16 '17 at 22:26\n\nSince your question is about part $(c)$...\n\n\u2022 To prove $N(v)=N(\\overline{v})$, use your formula for $N$ in part $(a)$ on each of $v$ and $\\overline{v}$. As stated, the question is somewhat confusing because $a$ and $b$ are used in two ways in the problem. In part $(a)$, $N(v)=a^2+b^2$ when $v=a+bi$. For part $(c)$, we could use $w=c+di$ and $\\overline{w}=c-di$, then $N(w)=c^2+d^2$ (by substituting the variables in $w$ into the form for part $(a)$) and $N(\\overline{w})=c^2+(-d)^2$ by the same substitution.\n\n\u2022 To prove $N(v^n)=N(v)^n$, use your equality in part $(b)$ as well as induction on $n$. The base case is $N(v^1)=N(v)^1$, and then use part $(b)$ to prove $N(v^{k+1})=N(v)^{k+1}$ using $N(v^k)=N(v)^k$.\n\n\u2022 For $N(v)=N(\\overline{v})$ I know I will get a$^2$+b$^2$ for v, but how can I just plug in $\\overline{v}$ into the equation if this is for v not $\\overline{v}$? Apr 16 '17 at 21:43\n\u2022 You have a general formula, it might be confusing because you're using $a$ and $b$ in two different roles, I'll edit to make this more clear. Apr 16 '17 at 21:44\n\u2022 Okay so when plugging in $\\overline{w}=c-di$ into the given equation for (a), what exactly is happening? I am not sure how you got (-d)$^2$ because wouldn't this be showing the equations are not equal? Apr 16 '17 at 21:50\n\u2022 What is the square of a negative real number? Apr 16 '17 at 21:54\n\u2022 @Pam_22R This shows that the terms are equal since $(-d)^2=-d\\cdot (-d)=d^2$. Apr 16 '17 at 21:55", "date": "2021-09-18 14:19:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2237462/complex-numbers-and-the-norm/2237482", "openwebmath_score": 0.7715713977813721, "openwebmath_perplexity": 213.7257314640681, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712645102011, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8620472333921904}}
{"url": "https://www.jmp.com/en_ph/statistics-knowledge-portal/chi-square-test/chi-square-goodness-of-fit-test.html", "text": "### Statistics Knowledge Portal\n\nA free online introduction to statistics\n\n# Chi-Square Goodness of Fit Test\n\n## What is the Chi-square goodness of fit test?\n\nThe Chi-square goodness of fit test is a statistical hypothesis test used to determine whether a variable is likely to come from a specified distribution or not. It is often used to evaluate whether sample data is representative of the full population.\n\n## When can I use the test?\n\nYou can use the test when you have counts of values for a categorical variable.\n\nYes.\n\n## Using the Chi-square goodness of fit test\n\nThe Chi-square goodness of fit test checks whether your sample data is likely to be from a specific theoretical distribution. We have a set of data values, and an idea about how the data values are distributed. The test gives us a way to decide if the data values have a \u201cgood enough\u201d fit to our idea, or if our idea is questionable.\n\n### What do we need?\n\nFor the goodness of fit test, we need one variable. We also need an idea, or hypothesis, about how that variable is distributed. Here are a couple of examples:\n\n\u2022 We have bags of candy with five flavors in each bag. The bags should contain an equal number of pieces of each flavor. The idea we'd like to test is that the proportions of the five flavors in each bag are the same.\n\u2022 For a group of children\u2019s sports teams, we want children with a lot of experience, some experience and no experience shared evenly across the teams. Suppose we know that 20 percent of the players in the league have a lot of experience, 65 percent have some experience and 15 percent are new players with no experience. The idea we'd like to test is that each team has the same proportion of children with a lot, some or no experience as the league as a whole.\n\nTo apply the goodness of fit test to a data set we need:\n\n\u2022 Data values that are a simple random sample from the full population.\n\u2022 Categorical or nominal data. The Chi-square goodness of fit test is not appropriate for continuous data.\n\u2022 A data set that is large enough so that at least five values are expected in each of the observed data categories.\n\n## Chi-square goodness of fit test example\n\nLet\u2019s use the bags of candy as an example.\u00a0We collect a random sample of ten bags.\u00a0Each bag has 100 pieces of candy and five flavors. Our hypothesis is that the proportions of the five flavors in each bag are the same.\n\nLet\u2019s start by answering: Is the Chi-square goodness of fit test\u00a0an appropriate method to evaluate\u00a0the distribution of flavors in bags of candy?\n\n\u2022 We have a simple random sample of 10 bags of candy. We meet this requirement.\n\u2022 Our categorical variable is the flavors of candy. We have the count of each flavor in 10 bags of candy. We meet this requirement.\n\u2022 Each bag has 100 pieces of candy. Each bag has five flavors of candy. We expect to have equal numbers for each flavor. This means we expect 100 / 5 = 20 pieces of candy in each flavor from each bag. For 10 bags in our sample, we expect 10 x 20 = 200 pieces of candy in each flavor. This is more than the requirement of five expected values in each category.\n\nBased on the answers above, yes, the Chi-square goodness of fit test is an appropriate method to evaluate the distribution of the flavors in bags of candy.\n\nFigure 1 below shows the combined flavor counts from all 10 bags of candy.\n\nFigure 1: Bar chart of counts of candy flavors from all 10 bags\n\nWithout doing any statistics, we can see that the number of pieces for each flavor are not the same. Some flavors have fewer than the expected 200 pieces and some have more. But how different are the proportions of flavors? Are the number of pieces \u201cclose enough\u201d for us to conclude that across many bags there are the same number of pieces for each flavor? Or are the number of pieces too different for us to draw this conclusion? Another way to phrase this is, do our data values give a \u201cgood enough\u201d fit to the idea of equal numbers of pieces of candy for each flavor or not?\n\nTo decide, we find the difference between what we have and what we expect. Then, to give flavors with fewer pieces than expected the same importance as flavors with more pieces than expected, we square the difference. Next, we divide the square by the expected count, and sum those values. This gives us our test statistic.\n\nThese steps are much easier to understand using numbers from our example.\n\nLet\u2019s start by listing what we expect if each bag has the same number of pieces for each flavor.\u00a0 Above, we calculated this as 200 for 10 bags of candy.\n\n#### Table 1: Comparison of actual vs expected number of pieces of each flavor of candy\n\n Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Apple 180 200 Lime 250 200 Cherry 120 200 Cherry 225 200 Grape 225 200\n\nNow, we find the difference between what we have observed in our data and what we expect. The last column in Table 2 below shows this difference:\n\n#### Table 2: Difference between observed and expected pieces of candy by flavor\n\n Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Apple 180 200 180-200 = -20 Lime 250 200 250-200 = 50 Cherry 120 200 120-200 = -80 Orange 225 200 225-200 = 25 Grape 225 200 225-200 = 25\n\nSome of the differences are positive and some are negative. If we simply added them up, we would get zero. Instead, we square the differences. This gives equal importance to the flavors of candy that have fewer pieces than expected, and the flavors that have more pieces than expected.\n\n#### Table 3: Calculation of the squared difference between Observed and Expected for each flavor of candy\n\n Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Squared Difference Apple 180 200 180-200 = -20 400 Lime 250 200 250-200 = 50 2500 Cherry 120 200 120-200 = -80 6400 Orange 225 200 225-200 = 25 625 Grape 225 200 225-200 = 25 625\n\nNext, we divide the squared difference by the expected number:\n\n#### Table 4: Calculation of the squared difference/expected number of pieces of candy per flavor\n\n Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Squared Difference Squared Difference / Expected Number Apple 180 200 180-200 = -20 400 400 / 200 = 2 Lime 250 200 250-200 = 50 2500 2500 / 200 = 12.5 Cherry 120 200 120-200 = -80 6400 6400 / 200 = 32 Orange 225 200 225-200 = 25 625 625 / 200 = 3.125 Grape 225 200 225-200 = 25 625 625 / 200 = 3.125\n\nFinally, we add the numbers in the final column to calculate our test statistic:\n\n$2 + 12.5 + 32 + 3.125 + 3.125 = 52.75$\n\nTo draw a conclusion, we compare the test statistic to a critical value from the Chi-Square distribution. This activity involves four steps:\n\n1. We first decide on the risk we are willing to take of drawing an incorrect conclusion based on our sample observations. For the candy data, we decide prior to collecting data that we are willing to take a 5% risk of concluding that the flavor counts in each bag across the full population are not equal when they really are. In statistics-speak, we set\u00a0the significance level, \u03b1 , to 0.05.\n2. We calculate a test statistic. Our test statistic is 52.75.\n3. We find the theoretical value from the Chi-square distribution based on our significance level. The theoretical value is the value we would expect if the bags contain the same number of pieces of candy for each flavor.\n\nIn addition to the significance level, we also need the degrees of freedom to find this value. For the goodness of fit test, this is one fewer than the number of categories. We have five flavors of candy, so we have 5 \u2013 1 = 4 degrees of freedom.\n\nThe Chi-square value with \u03b1 = 0.05 and 4 degrees of freedom is 9.488.\n4. We compare the value of our test statistic (52.75) to the Chi-square value. Since 52.75 > 9.488, we reject the null hypothesis that the proportions of flavors of candy are equal.\n\nWe make a practical conclusion that bags of candy across the full population do not have an equal number of pieces for the five flavors. This makes sense if you look at the original data. If your favorite flavor is Lime, you are likely to have more of your favorite flavor than the other flavors. If your favorite flavor is Cherry, you are likely to be unhappy because there will be fewer pieces of Cherry candy than you expect.\n\n### Understanding results\n\nLet\u2019s use a few graphs to understand the test and the results.\n\nA simple bar chart of the data shows the observed counts for the flavors of candy:\n\nFigure 2: Bar chart of observed counts for flavors of candy\n\nAnother simple bar chart shows the expected counts of 200 per flavor. This is what our chart would look like if the bags of candy had an equal number of pieces of each flavor.\n\nFigure 3: Bar chart of expected counts of each flavor\n\nThe side-by-side chart below shows the actual observed number of pieces of candy in blue. The orange bars show the expected number of pieces. You can see that some flavors have more pieces than we expect, and other flavors have fewer pieces.\n\nFigure 4: Bar chart comparing actual vs. expected counts of candy\n\nThe statistical test is a way to quantify the difference. Is the actual data from our sample \u201cclose enough\u201d to what is expected to conclude that the flavor proportions in the full population of bags are equal? Or not? From the candy data above, most people would say the data is not \u201cclose enough\u201d even without a statistical test.\n\nWhat if your data looked like the example in Figure 5 below instead? The purple bars show the observed counts and the orange bars show the expected counts. Some people would say the data is \u201cclose enough\u201d but others would say it is not. The statistical test gives a common way to make the decision, so that everyone makes the same decision on a set of data values.\n\nFigure 5: Bar chart comparing expected and actual values using another example data set\n\n### Statistical details\n\nLet\u2019s look at the candy data and the Chi-square test for goodness of fit using statistical terms. This test is also known as Pearson\u2019s Chi-square test.\n\nOur null hypothesis is that the proportion of flavors in each bag is the same. We have five flavors. The null hypothesis is written as:\n\n$H_0: p_1 = p_2 = p_3 = p_4 = p_5$\n\nThe formula above uses p for the proportion of each flavor. If each 100-piece bag contains equal numbers of pieces of candy for each of the five flavors, then the bag contains 20 pieces of each flavor. The proportion of each flavor is 20 / 100 = 0.2.\n\nThe alternative hypothesis is that at least one of the proportions is different from the others. This is written as:\n\n$H_a: at\\ least\\ one\\ p_i\\ not\\ equal$\n\nIn some cases, we are not testing for equal proportions. Look again at the example of children's sports teams near the top of this page.\u00a0 Using that as an example, our null and alternative hypotheses are:\n\n$H_0: p_1 = 0.2, p_2 = 0.65, p_3 = 0.15$\n\n$H_a: at\\ least\\ one\\ p_i\\ not\\ equal\\ to\\ expected\\ value$\n\nUnlike other hypotheses that involve a single population parameter, we cannot use just a formula. We need to use words as well as symbols to describe our hypotheses.\n\nWe calculate the test statistic using the formula below:\n\n$\\sum^n_{i=1} \\frac{(O_i-E_i)^2}{E_i}$\n\nIn the formula above, we have n groups. The $\\sum$ symbol means to add up the calculations for each group. For each group, we do the same steps as in the candy example. The formula shows Oi \u00a0as the Observed value and Ei\u00a0\u00a0as the Expected value for a group.\n\nWe then compare the test statistic to a Chi-square value with our chosen significance level (also called the alpha level) and the degrees of freedom for our data. Using the candy data as an example, we set \u03b1 = 0.05 and have four degrees of freedom. For the candy data, the Chi-square value is written as:\n\n$\u03c7\u00b2_{0.05,4}$\n\nThere are two possible results from our comparison:\n\n\u2022 The test statistic is lower than the Chi-square value. You fail to reject the hypothesis of equal proportions. You conclude that the bags of candy across the entire population have the same number of pieces of each flavor in them. The fit of equal proportions is \u201cgood enough.\u201d\n\u2022 The test statistic is higher than the Chi-Square value. You reject the hypothesis of equal proportions. You cannot conclude that the bags of candy have the same number of pieces of each flavor. The fit of equal proportions is \u201cnot good enough.\u201d\n\nLet\u2019s use a graph\u00a0of the Chi-square distribution\u00a0to\u00a0better understand the test results. You are checking to see if your test statistic is a more extreme value in the distribution than the critical value. The distribution below shows a Chi-square distribution with four degrees of freedom. It shows how the critical value of 9.488 \u201ccuts off\u201d 95% of the data. Only 5% of the data is greater than 9.488.\n\nFigure 6: Chi-square distribution for four degrees of freedom\n\nThe next distribution plot includes our results. You can see how far out \u201cin the tail\u201d our test statistic is, represented by the dotted line at 52.75. In fact, with this scale, it looks like the curve is at zero where it intersects with the dotted line. It isn\u2019t, but it is very, very close to zero. We conclude that it is very unlikely for this situation to happen by chance. If the true population of bags of candy had equal flavor counts, we would be extremely unlikely to see the results that we collected from our random sample of 10 bags.\n\nFigure 7: Chi-square distribution for four degrees of freedom with test statistic plotted\n\nMost statistical software shows the p-value for a test. This is the likelihood of finding a more extreme value for the test statistic in a similar sample, assuming that the null hypothesis is correct. It\u2019s difficult to calculate the p-value by hand. For the figure above, if the test statistic is exactly 9.488, then the p-value will be p=0.05. With the test statistic of 52.75, the p-value is very, very small. In this example, most statistical software will report the p-value as \u201cp < 0.0001.\u201d This means that the likelihood of another sample of 10 bags of candy resulting in a more extreme value for the test statistic is less than one chance in 10,000, assuming our null hypothesis of equal counts of flavors is true.", "date": "2021-09-27 18:40:38", "meta": {"domain": "jmp.com", "url": "https://www.jmp.com/en_ph/statistics-knowledge-portal/chi-square-test/chi-square-goodness-of-fit-test.html", "openwebmath_score": 0.6582071781158447, "openwebmath_perplexity": 441.5611618161789, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes.\n2. Yes.", "lm_q1_score": 0.9921841116729335, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8620361265494706}}
{"url": "https://math.stackexchange.com/questions/915793/telescoping-series-of-form-sum-n1-cdotsnk", "text": "# Telescoping series of form $\\sum (n+1)\\cdots(n+k)$ [duplicate]\n\nWolfram Alpha is able to telescope sums of the form $\\sum (n+1)\\cdots(n+k)$\n\ne.g. $(1\\cdot2\\cdot3) + (2\\cdot3\\cdot4) + \\cdots + n(n+1)(n+2)$\n\nHow does it do it?\n\nEDIT: We can rewrite as: $\\sum {(n+k)! \\over n!} = \\sum k!{(n+k)!\\over n!k!} = \\sum k!{{n+k} \\choose n}$ (Thanks Daniel Fischer)\n\nEDIT2: We can also multiply out and split sums. So e.g.\n\n$$\\sum (n-1)n(n+1) = \\sum (n^3-n) = \\sum n^3 - \\sum n$$\n\nBut sums of powers actually seem to be more nasty than the original question, involving Bernoulli numbers. (Thanks Claude Leibovici)\n\nAnd is there any name for this particular corner of maths? (i.e. How might I go about searching the Internet for information regarding this?)\n\nPS please could we have a 'telescoping' tag?\n\n## marked as duplicate by user147263, Community\u2666Dec 24 '15 at 20:42\n\n\u2022 Note that the terms are $k!\\binom{n+k}{n}$. You probably know a bit about binomial coefficients that helps summing. \u2013\u00a0Daniel Fischer Sep 1 '14 at 10:07\n\u2022 I don't understand why you speak about \"telescoping\". Beside Daniel Fischer's suggestion, you could also develop $i(i+1)(i+2)= i^3+3 i^2+2 i$ and compute the three sums. \u2013\u00a0Claude Leibovici Sep 1 '14 at 10:16\n\u2022 Is the sum over n or k? Makes a big difference. \u2013\u00a0marty cohen Sep 1 '14 at 15:18\n\u2022 What great answers! Now I am stuck. Both RobJohn and HyperGeometric have nailed it from opposite directions. To accept either answer above the other would not seem right. I guess I will leave it open until SE provides some machinery for resolving these situations. \u2013\u00a0P i Sep 1 '14 at 17:49\n\u2022 @P-i- Thank you! That's very considerate of you. It was an interesting question, which I enjoyed solving. I've upvoted the question as well. Have also just posted a comment on a recent observation on its resemblance to the power integral. A related question which you might want to pose would be to find the sum of a series with each term being the reciprocal of the corresponding in the present series, i.e the reciprocal of the product of consecutive integers. \u2013\u00a0hypergeometric Sep 4 '14 at 6:14\n\nHint: $$\\sum_{n=k}^m\\binom{n}{k}=\\binom{m+1}{k+1}$$ A generalization is discussed in this answer. The equation above is equation $(1)$ with $m=0$.\n\nTelescoping sum\n\nTo turn the sum in the question into a \"telescoping sum\", we can use the recurrence for Pascal's Triangle: $$\\binom{n+1}{k+1}=\\binom{n}{k}+\\binom{n}{k+1}$$ Using this recurrence, we get \\begin{align} \\sum_{n=k}^m\\binom{n}{k} &=\\sum_{n=k}^m\\left[\\binom{n+1}{k+1}-\\binom{n}{k+1}\\right]\\\\ &=\\sum_{n=k+1}^{m+1}\\binom{n}{k+1}-\\sum_{n=k}^m\\binom{n}{k+1}\\\\ &=\\left[\\binom{m+1}{k+1}+\\color{#C00000}{\\sum_{n=k+1}^m\\binom{n}{k+1}}\\right]-\\left[\\binom{k}{k+1}+\\color{#C00000}{\\sum_{n=k+1}^m\\binom{n}{k+1}}\\right]\\\\ &=\\binom{m+1}{k+1}-\\binom{k}{k+1}\\\\ &=\\binom{m+1}{k+1} \\end{align} The sums in red are the terms that are telescoped out, leaving just the first and last terms. In this case, the last term $\\binom{k}{k+1}=0$.\n\nLet $$p_n=\\prod_{r=1}^k (n+r)=\\overbrace{(n+1)(n+2)\\cdots(n+k)}^{k \\ \\text{terms}}$$ which is the product of $k$ consecutive integers.\n\nConsider the difference of two consecutive terms, where each term is the product of $k+1$ consecutive integers, i.e. \\begin{align} &\\prod_{r=1}^{k+1}(n+r)-\\prod_{r=0}^k(n+r)\\\\ &=\\overbrace{\\underbrace{(n+1)(n+2)\\cdots(n+k)}_{p_n}(n+k+1)}^{(k+1) \\ \\text{terms}}-\\overbrace{n\\underbrace{(n+1)(n+2)\\cdots(n+k)}_{p_n}}^{(k+1) \\ \\text{terms}}\\\\ &=p_n[(n+k+1)-n]\\\\ &=p_n(1+k) \\end{align}\n\nHence, $$p_n=\\prod_{r=1}^k (n+r)=\\frac1{1+k}\\left[\\prod_{r=1}^{k+1} (n+r)-\\prod_{r=0}^k (n+r) \\right]$$ which is convenient for telescoping.\n\nRequired summation, \\begin{align} S=\\sum_{n=0}^{m}p_n&=\\sum_{n=0}^{m} \\prod_{r=1}^{k} (n+r)=\\sum_{n=0}^{m}(n+1)(n+2)(n+3)\\cdots (n+k)\\\\ &=\\frac1{k+1}\\sum_{n=0}^{m} \\left[ \\prod_{r=1}^{k+1} (n+r)-\\prod_{r=0}^{k} (n+r)\\right]\\\\ &=\\frac1{k+1}\\prod_{r=1}^{k+1} (m+r) \\qquad \\blacksquare\\end{align} by telescoping.\n\nIn your example, $k=3$, hence the general term is\n\n$$(n+1)(n+2)(n+3)=\\frac14\\left[(n+1)(n+2)(n+3)(n+4)-n(n+1)(n+2)(n+3)\\right]$$\n\nHence, by telescoping from $n=0$ to $m$, \\begin{align}S&=1\\cdot2\\cdot3+2\\cdot3 \\cdot 4+\\cdots +(m+1)(m+2)(m+3)\\\\ &=\\frac14(m+1)(m+2)(m+3)(m+4) \\end{align}\n\nNB: It is interesting to note that this result bears a striking resemblance to integration.\n\nCompare the standard integral $$\\int_0^m n^k dn=\\frac{m^{k+1}}{k+1}$$ to the result of the summation above, which can also be stated as $$\\sum_{n=0}^{m}n^{[k]}=\\frac {m^{[k+1]}}{k+1}$$ where $n^{[k]}$ is my adjusted* Pochhammer symbol for rising factorials, defined as $$n^{[k]}=\\prod_{r=1}^{k}(n+r)=(n+1)(n+2)(n+3)\\cdots(n+k)$$\n\nThe actual Pochhammer symbol for rising factorials, $n^{(k)}$, starts from $n$ itself and not $n+1$, i.e. $$n^{(k)}=\\prod_{r=1}^{k}(n+r-1)=n(n+1)(n+2)\\cdots(n+k-1)$$\n\n\u2022 I've upvoted this as it appears valid. But I can't see any clear motivation for the initial algebraic manipulations. Is it possible to rework this logic in a way that does not require inspirational leaps? \u2013\u00a0P i Sep 1 '14 at 14:08\n\u2022 @P-i- - thank you, that's very kind :) in order to make the series telescope, the approach is to convert one term (or product, in this case) into a difference of two terms, usually of a higher 'order'. for instance, to telescope the sum of squares, one would start from the difference of cubes. in this case, we want to sum $k$-term products hence we start from the difference of two consecutive $(k+1)$-term products. \u2013\u00a0hypergeometric Sep 1 '14 at 14:19\n\u2022 @P-i- - the proposed solution has been reworked slightly and hopefully this provides greater clarity. \u2013\u00a0hypergeometric Sep 1 '14 at 14:30\n\u2022 thanks to all who upvoted. a short appendix has been added to the solution to highlight the resemblance of the summation result to the standard power integral. comments are most welcome. \u2013\u00a0hypergeometric Sep 3 '14 at 15:22\n\nFor completeness, a solution using the combinatorial/binomial approach, as initiated by RobJohn:\n\nLet $p(n, k) =n(n+1)...(n+k-1) =\\prod_{j=0}^{k-1} (n+j)$.\n\nThen (writing each step in detail)\n\n$\\begin{array}\\\\ p(n+1, k)-p(n, k) &=\\prod_{j=0}^{k-1} (n+1+j)-\\prod_{j=0}^{k-1} (n+j)\\\\ &=\\prod_{j=1}^{k} (n+j)-\\prod_{j=0}^{k-1} (n+j)\\\\ &=(n+k)\\prod_{j=1}^{k-1} (n+j)-n\\prod_{j=1}^{k-1} (n+j)\\\\ &=((n+k)-n)\\prod_{j=1}^{k-1} (n+j)\\\\ &=k\\prod_{j=1}^{k-1} (n+j)\\\\ &=k\\prod_{j=0}^{k-2} (n+1+j)\\\\ &=kp(n+1, k-1)\\\\ \\end{array}$\n\nor, putting $k+1$ for $k$ and $n$ for $n+1$, $p(n, k) =\\frac{p(n, k+1)-p(n-1. k+1)}{k+1}$.\n\nTherefore\n\n$\\begin{array}\\\\ \\sum_{n=1}^M p(n, k) &=\\sum_{n=1}^M \\frac{p(n, k+1)-p(n-1. k-1)}{k+1}\\\\ &=\\frac1{k+1}\\sum_{n=1}^M (p(n, k+1)-p(n-1. k+1))\\\\ &=\\frac1{k+1}(p(M, k+1)-p(0, k+1))\\\\ &=\\frac{p(M, k+1)}{k+1}\\\\ \\end{array}$\n\n\u2022 This proof is succinct and flawless! Thanks Marty! \u2013\u00a0P i Sep 2 '14 at 11:38\n\nAlso your terms can be re written as $$n(n+1)(n+2)=\\dfrac{n(n+1)(n+2)(n+3)}{4}-\\dfrac{(n-1)n(n+1)(n+2)}{4}.$$\n\n\u2022 that may be correct but it would not telescope readily... \u2013\u00a0hypergeometric Sep 1 '14 at 15:24\n\u2022 Now it is fine. \u2013\u00a0Bumblebee Sep 2 '14 at 5:06\n\nFollowing from RobJohn's hint, this can be proved very simply by reverse-engineering the hockey-stick identity.\n\nThe blues sum to the red for any given 'hockey-stick'.\n\nIt is a straightforward proof by induction. For example, in the picture, 1+3+6+10+15 = 35, so 1+3+6+10+15 + 21 = 35+21 = 56\n\ni.e. True(stick length k) => True(stick length k+1)\n\nInduction can also be argued from the ${n \\choose r}$ formula, but the above method avoids algebraic manipulation, working instead from the simple \"each element is formed by summing its upper neighbours\" definition of Pascal's Triangle.\n\nBut now we DO need some algebraic manipulation.\n\nWe write this hockey-stick identity as:\n\n$${k+0 \\choose k} + ... + {k+n \\choose k} = {k+(n+1) \\choose (k+1)}$$\n\n(notice the hockey stick has been reflected)\n\n$${(k+0)! \\over {k!((k+0)-k)!}} + ... + {(k+n)! \\over {k!((k+n)-k)!}} = {(k+(n+1))! \\over {(k+1)!(((k+(n+1))-(k+1))!}}$$\n\n$${(k+0)! \\over 0!} + ... + {(k+n)! \\over n!} = {1 \\over {k+1}} \\cdot {(k+(n+1))! \\over n!}$$\n\nSo the original question turns out to be an emergent property of Pascal's Triangle, and I would hazard a guess that it is from this simple construct that the question originally arose.\n\nHowever, this proof doesn't move forwards from start to finish in a sequence of intelligently chosen moves. It doesn't offer any insight into telescoping general sums, hence I am favouring hypergeometric's proof.\n\n\u2022 thank you! glad you liked it :) robjohn's approach using the pascal triangle identity is quite useful too, and i have used a similar approach elsewhere on MSE for another question. \u2013\u00a0hypergeometric Sep 2 '14 at 6:28\n\n\n$$\\color{#66f}{\\large\\sum_{n = 0}^{m}\\pars{n + 1}\\ldots\\pars{n + k} =k!\\,{k + m + 1 \\choose m}}$$\n\n\u2022 blink complex analysis! That's fantastic! But how does the first path integral come from n+k choose k? \u2013\u00a0P i Sep 4 '14 at 12:45\n\u2022 @P-i- That's an identity: $${a \\choose b}=\\oint_{0\\ <\\ \\left\\vert \\,z\\,\\right\\vert\\ =\\ c}{(1 + z)^a \\over z^{b + 1}}\\,{{\\rm d}z \\over 2\\pi{\\ic}}$$ It's valid for $c \\in {\\mathbb R}\\,,\\ c > 0$ and $b\\ =\\ 0,1,2,3,\\ldots$ \u2013\u00a0Felix Marin Sep 4 '14 at 17:49\n\nInduction\n\nThis question is actually a duplicate of: Finding a closed formula for $1\\cdot2\\cdot3\\cdots k +\\dots + n(n+1)(n+2)\\cdots(k+n-1)$\n\nBut it is difficult to find duplicates when the question can only be expressed using maths symbols, and this question now acts as a more comprehensive resource.\n\nOne of the answers to that question is to manually construct formulae for the first few cases: $1+2+3+\\dots+n$, $1\\cdot2 +\\dots + n\\cdot(n+1)$, etc, notice a pattern and intuit a candidate formula, then prove this by induction.", "date": "2019-11-21 08:45:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/915793/telescoping-series-of-form-sum-n1-cdotsnk", "openwebmath_score": 0.9573472142219543, "openwebmath_perplexity": 777.6180335975183, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308782546026, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8619933824028556}}
{"url": "https://math.stackexchange.com/questions/3224765/polynomial-division-an-obvious-trick-reducing-mod-textitsimpler-multiple/3224982", "text": "# Polynomial division: an obvious trick? [reducing mod $\\textit{simpler}$ multiples]\n\nThe following question was asked on a high school test, where the students were given a few minutes per question, at most:\n\nGiven that, $$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$ and, $$Q(x)=x^4+x^3+x^2+x+1$$ what is the remainder of $$P(x)$$ divided by $$Q(x)$$?\n\nLet $$Q(x)=0$$. Multiplying both sides by $$x-1$$: $$(x-1)(x^4+x^3+x^2+x+1)=0 \\implies x^5 - 1=0 \\implies x^5 = 1$$ Substituting $$x^5=1$$ in $$P(x)$$ gives $$x^4+x^3+x^2+x+1$$. Thus, $$P(x)\\equiv\\mathbf0\\pmod{Q(x)}$$\n\nObviously, a student is required to come up with a \u201ctrick\u201d rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?\n\n\u2022 One observation is that in that question the step of 'multiplying both sides by $x-1$' is only presentation and doesn't really represent any advantage for solving the problem. What is really important is observing that the roots of $Q$ are the fifths roots of $1$ besides $1$, or that is a factor of $x^5-1$. There are many ways to compute this remainder that are simple to do in a short time. So, don't include that 'multiplying by $x-1$' as part of what was required for the students to solve the problem \u2013\u00a0logarithm May 14 '19 at 15:48\n\u2022 My approach would be to observe that the exponents on P(x) look like they're set up to produce an elegant answer. The common elegant answers are \"1\" and \"0\". Since the \"+1\"s on the polynomials are clearly intended to cancel out, the answer must be \"0\". \u2013\u00a0Mark May 14 '19 at 23:33\n\u2022 I'm probably missing something, but... The above trick seems to show that when $Q(x) = 0$, $P(x) = Q(x)$. But (a) the original question didn't specify that $Q(x) = 0$; how do we know that the answer $P(x)\\equiv\\mathbf0\\pmod{Q(x)}$ applies to other cases? and (b) the original question asks, \"what is the remainder of $\ud835\udc43(\ud835\udc65)$ divided by $\ud835\udc44(\ud835\udc65)$?\" but the answer given here says let $Q(x) = 0$; doesn't that mean that the remainder is not defined? en.wikipedia.org/wiki/Remainder#Polynomial_division \u2013\u00a0LarsH May 15 '19 at 21:06\n\u2022 I added some explicit examples to the end of my answer to emphasize the ubiquity of that simple idea. \u2013\u00a0Bill Dubuque May 16 '19 at 13:49\n\u2022 What kind of test? Is this part of a math competition, or a classroom precalculus exam? \u2013\u00a0PersonX May 16 '19 at 16:35\n\nThe key idea employed here is the method of simpler multiples - a very widely used technique. Note that $$\\,Q\\,$$ has a \"simpler\" multiple $$\\,QR = x^5\\!-\\!1,\\,$$ so we can first reduce $$P$$ modulo $$\\,x^{\\large 5}\\! -\\! 1\\,$$ via $$\\!\\bmod x^{\\large 5}-1\\!:\\,\\ \\color{#c00}{x^{\\large 5}\\equiv 1}\\Rightarrow\\, x^{\\large r+5q^{\\phantom{|}}}\\!\\!\\equiv x^{\\large r}(\\color{#c00}{x^{\\large 5}})^{\\large q}\\equiv x^{\\large r},\\,$$ then finally reduce that $$\\!\\bmod Q,\\,$$ i.e.\n\n$$P\\bmod Q\\, =\\, (P\\bmod QR)\\bmod Q\\qquad$$\n\nThis idea is ubiquitous, e.g. we already use it implicitly in grade school in radix $$10$$ to determine integer parity: first reduce mod $$10$$ to get the units digit, then reduce the units digits mod $$2,\\,$$ i.e.\n\n$$N \\bmod 2\\, = (N\\bmod 2\\cdot 5)\\bmod 2\\qquad\\$$\n\ni.e. an integer has the same parity (even / oddness) as that of its units digit. Similarly since $$7\\cdot 11\\cdot 13 = 10^{\\large 3}\\!+1$$ we can compute remainders mod $$7,11,13$$ by using $$\\,\\color{#c00}{10^{\\large 3}\\equiv -1},\\,$$ e.g. $$\\bmod 13\\!:\\,\\ d_0+ d_1 \\color{#c00}{10^{\\large 3}} + d_2 (\\color{#c00}{10^{\\large 3}})^{\\large 2}\\!+\\cdots\\,$$ $$\\equiv d_0 \\color{#c00}{\\bf -} d_1 + d_2+\\cdots,\\,$$ and, similar to the OP, by $$\\,9\\cdot 41\\cdot 271 = 10^{\\large 5}\\!-1\\,$$ we can compute remainders mod $$41$$ and $$271$$ by using $$\\,\\color{#c00}{10^5\\!\\equiv 1}$$\n\n$$N \\bmod 41\\, = (N\\bmod 10^{\\large 5}\\!-1)\\bmod 41\\quad$$\n\nfor example $$\\bmod 41\\!:\\ 10000\\color{#0a0}200038$$ $$\\equiv (\\color{#c00}{10^{\\large 5}})^{\\large 2}\\! + \\color{#0a0}2\\cdot \\color{#c00}{10^{\\large 5}} + 38\\equiv \\color{#c00}1+\\color{#0a0}2+38\\equiv 41\\equiv 0$$\n\nSuch \"divisibility tests\" are frequently encountered in elementary and high-school and provide excellent motivation for this method of \"divide first by a simpler multiple of the divisor\" or, more simply, \"mod first by a simpler multiple of the modulus\".\n\nThis idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of rationalizing denominators and in Gauss's algorithm for computing modular inverses.\n\nFor example, to divide by an algebraic number we can employ its norm as a simpler multiple, e.g. for $$\\,w = a+b\\sqrt{n}\\,$$ we use $$\\,ww' = (a+b\\sqrt n)(a-b\\sqrt n) = a^2-nb^2 = c\\in\\Bbb Q\\ (\\neq 0\\,$$ by $$\\,\\sqrt{n}\\not\\in\\Bbb Q),\\,$$ which reduces division by an algebraic to simpler division by a rational, i.e. $$\\, v/w = vw'/(ww'),$$ e.g.\n\n$$\\dfrac{1}{a+b\\sqrt n}\\, =\\, \\dfrac{1}{a+b\\sqrt n}\\, \\dfrac{a-b\\sqrt n}{a-b\\sqrt n}\\, =\\, \\dfrac{a-b\\sqrt n}{a^2-nb^2}\\,=\\, {\\frac{\\small 1}{\\small c}}(a-b\\sqrt n)\\qquad$$\n\nso-called rationalizing the denominator. The same idea works even with $$\\,{\\rm\\color{#c00}{nilpotents}}$$\n\n$$\\color{#c00}{t^n = 0}\\ \\Rightarrow\\ \\dfrac{1}{a-{ t}}\\, =\\, \\dfrac{a^{n-1}+\\cdots + t^{n-1}}{a^n-\\color{#c00}{t^n}}\\, =\\, a^{-n}(a^{n-1}+\\cdots + t^{n-1})\\qquad$$\n\nwhich simplifies by eliminating $$\\,t\\,$$ from the denominator, i.e. $$\\,a-t\\to a^n,\\,$$ reducing the division to division by a simpler constant $$\\,a^n\\,$$ (vs. a simpler rational when rationalizing the denominator).\n\nAnother example is Gauss' algorithm, where we compute fractions $$\\!\\bmod m\\,$$ by iteratively applying this idea of simplifying the denominator by scaling it to a smaller multiple. Here we scale $$\\rm\\color{#C00}{\\frac{A}B} \\to \\frac{AN}{BN}\\:$$ by the least $$\\rm\\,N\\,$$ so that $$\\rm\\, BN \\ge m,\\,$$ reduce mod $$m,\\,$$ then iterate this reduction, e.g.\n\n$$\\rm\\\\ mod\\ 13\\!:\\,\\ \\color{#C00}{\\frac{7}9} \\,\\equiv\\, \\frac{14}{18}\\, \\equiv\\, \\color{#C00}{\\frac{1}5}\\,\\equiv\\, \\frac{3}{15}\\,\\equiv\\, \\color{#C00}{\\frac{3}2} \\,\\equiv\\, \\frac{21}{14} \\,\\equiv\\, \\color{#C00}{\\frac{8}1}\\qquad\\qquad$$\n\nDenominators of the $$\\color{#c00}{\\rm reduced}$$ fractions decrease $$\\,\\color{#C00}{ 9 > 5 > 2> \\ldots}\\,$$ so reach $$\\color{#C00}{1}\\,$$ (not $$\\,0\\,$$ else the denominator would be a proper factor of the prime modulus; it may fail for composite modulus)\n\nSee here and its $$25$$ linked questions for more examples similar to the OP (some far less trivial).\n\nWorth mention: there are simple algorithms for recognizing cyclotomics (and products of such), e.g. it's shown there that $$\\, x^{16}+x^{14}-x^{10}-x^8-x^6+x^2+1$$ is cyclotomic (a factor of $$x^{60}-1),\\,$$ so we can detect when the above methods apply for arbitrarily large degree divisors.\n\nLet $$a$$ be zero of $$x^4+x^3+x^2+x+1=0$$. Obviously $$a\\ne 1$$. Then $$a^4+a^3+a^2+a+1=0$$ so multiply this with $$a-1$$ we get $$a^5=1$$ (You can get this also from geometric series $$a^n+a^{n-1}+...+a^2+a+1 = {a^{n+1}-1\\over a-1}$$ by putting $$n=4$$).\n\nBut then $$\\begin{eqnarray} Q(a) &=& a^{100}\\cdot a^4+a^{90}\\cdot a^3+a^{80}\\cdot a^2+a^{70}\\cdot a+1\\\\ &=& a^4+a^3+a^2+a+1\\\\&=&0\\end{eqnarray}$$\n\nSo each zero of $$Q(x)$$ is also a zero of $$P(x)$$ and since all 4 zeroes of $$Q(x)$$ are different we have $$Q(x)\\mid P(x)$$.\n\n\u2022 I think this is the approach that the examiners would expect from high school students. \u2013\u00a0Dancrumb May 14 '19 at 0:23\n\u2022 From my perspective this is the best answer. (+1) \u2013\u00a0trancelocation May 14 '19 at 3:54\n\u2022 The OP understands that after multiplication with $x-1$, the question became simpler, he/she asked how a student should find this trick. This answer is basically just a reformulation of the answer given in the question. I don't see the usefulness of this answer at all... (but it has 14 upvotes, so I might be missing something...) \u2013\u00a0user193810 May 14 '19 at 11:49\n\u2022 @MariaMazur: A bit, but not really. It still feels like you stopped reading the question after \"what is the remainder of P(x) divided by Q(x)?\". You gave a correct answer to that question, but not to the real question. It is easy to prove that the 'trick' (multiplication by $x-1$) works and is correct, but you ignored the real question, and after your edit it is hidden as a small remark. But whatever, your answer is trivially better than mine, because I did not gave any, so I'll leave it here. \u2013\u00a0user193810 May 14 '19 at 13:25\n\u2022 @Pakk: it hinges on spotting the standard identity: $Q(x) = 0$ gives the fifth roots of unity (other than x=1). It takes tons more work if you don't spot that; you could solve the quartic $Q(x) = 0$ then notice \"Hey these are the other four fifth roots of unity, thus x^5 \\equiv 1\" then apply the information $x^5 \\equiv 1$ to hugely simplify $P(x)$, and thus conclude $Q(x) \\mid P(x)$ and remainder = 0. But that would be lots more work. (Yeah the first time I saw this trick back in HS I voiced the same objection about non-obviousness... this is why this identity is so pivotal to factorization) \u2013\u00a0smci May 16 '19 at 1:53\n\nWhile it may be a standard technique, as Bill's response details, I wouldn't say it's at all obvious at High School level. As a pre-Olympiad challenge problem, however, it's a good one.\n\nMy intuition is via cyclotomic polynomials -- $$Q(x) = \\Phi_5(x)$$, giving the idea to multiply through by $$x-1$$ -- but I doubt I would have recognised them before university: https://en.wikipedia.org/wiki/Cyclotomic_polynomial\n\n\u2022 I would go even further. As a BSE graduate in engineering, this is not at all obvious to me. \u2013\u00a0MooseBoys May 14 '19 at 6:58\n\u2022 @MooseBoys: I would go even further. As a PhD in Physics (admittedly many years ago), this is not at all obvious to me. \u2013\u00a0WoJ May 14 '19 at 9:04\n\u2022 It may not be obvious in other contexts since you never had to use it, but I assure you that in Math Olympiad training it's a standard trick. All serious contestants will probably already have seen something similar. There's nothing outrageous or shameful in that; Olympiad-style math is simply different from what people use and need in other professional contexts (physicist/engineer) --- the focus on Euclidean geometry being a perfect example. \u2013\u00a0Federico Poloni May 14 '19 at 9:12\n\u2022 @Federico Where in the question does it say that this is an Olympiad test? OP phrased their question as if it were applicable to all high school students. I'd guess that fewer than 5% of all students would get the correct answer. \u2013\u00a0MooseBoys May 14 '19 at 17:01\n\u2022 @MooseBoys I agree with your estimate. The question is \"is this obvious?\", and my comment is \"it is a standard problem for people who train for math Olympiads\". I realize that it's a strong additional assumption. I'm not arguing that it's a good or bad problem: as you point out, to do that, we would at least have to know to which students it was given. \u2013\u00a0Federico Poloni May 14 '19 at 17:09\n\nThis may be accessible to a high school student:\n\n$$x^{104}+x^{93}+x^{82}+x^{71}+1$$\n\n$$= (x^{104}-x^4)+(x^{93}-x^3)+(x^{82}-x^2)+(x^{71}-x)+(x^4+x^3+x^2+x+1)$$\n\n$$=x^4(x^{100}-1)+x^3(x^{90}-1)+x^2(x^{80}-1)+ x(x^{70}-1)+(x^4+x^3+x^2+x+1)$$\n\nWe know that $$(x^n-1)|(x^{mn}-1), m,n \\in \\mathbb{N}$$ so $$x^5-1$$ divides $$x^{100}-1, x^{90}-1$$ etc.\n\nIn turn $$x^5-1$$ is divisible by $$(x^4+x^3+x^2+x+1)$$ which concludes the proof\n\nIf it's not obvious, an examination of the question quickly reveals the trick. Say\n\n$$P(x)=x^n$$\n\nThen begin long division by $$Q(x)$$:\n\n$$x^n-x^n-x^{n-1}-x^{n-2}-x^{n-3}-x^{n-4}$$ $$x^{n-5}$$ $$\\dots$$ $$x^{n-5k}$$\n\nWhile it may not be obvious just by looking at the question, anyone who attempts the naive solution has (at least) a reasonable chance of running across a way of solving it.\n\n\u2022 This is a good point. If one is asked to perform division by a polynomial $Q(x)$, it is natural to start with dividing $x^n$ for $n> \\deg(Q)$. Thus the first step is to divide $x^5$. Magically(!), one gets the remainder 1. So, one can discover the formula in the form $x^5=(x-1)(x^4+x^3+x^2+x+1) + 1$ even if one didn't remember it. The rest of the steps are not too difficult. \u2013\u00a0Kapil May 14 '19 at 4:07\n\nI would have thought that bright students, who knew $$1+x+x^2+\\cdots +x^{n-1}= \\frac{x^n-1}{x-1}$$ as a geometric series formula, could say\n\n$$\\dfrac{P(x)}{Q(x)} =\\dfrac{x^{104}+x^{93}+x^{82}+x^{71}+1}{x^4+x^3+x^2+x+1}$$\n\n$$=\\dfrac{(x^{104}+x^{93}+x^{82}+x^{71}+1)(x-1)}{(x^4+x^3+x^2+x+1)(x-1)}$$\n\n$$=\\dfrac{x^{105}-x^{104}+x^{94}-x^{93}+x^{83}-x^{82}+x^{72}-x^{71}+x-1}{x^5-1}$$\n\n$$=\\dfrac{x^{105}-1}{x^5-1}-\\dfrac{x^{104}-x^{94}}{x^5-1}-\\dfrac{x^{93}-x^{83}}{x^5-1}-\\dfrac{x^{82}-x^{72}}{x^5-1}-\\dfrac{x^{71}-x}{x^5-1}$$\n\n$$=\\dfrac{x^{105}-1}{x^5-1}-x^{94}\\dfrac{x^{10}-1}{x^5-1}-x^{83}\\dfrac{x^{10}-1}{x^5-1}-x^{72}\\dfrac{x^{10}-1}{x^5-1}-x\\dfrac{x^{70}-1}{x^5-1}$$\n\nand that each division at the end would leave zero remainder for the same reason, replacing the original $$x$$ by $$x^5$$\n\nI think if the candidates know what a geometric series is, the question is okay. Indeed, one uses exactly this trick to find the formula for the geometric series, i.e. one writes $$(x-1)\\sum_{k=1}^nx^k=x^{n+1}-1$$ to find that $$\\sum_{k=1}^\\infty x^k=\\lim_{n\\to\\infty}\\sum_{k=1}^nx^k=\\lim_{n\\to\\infty}\\frac{x^{n+1}-1}{x-1}=\\frac{1}{1-x}$$ for $$|x|<1$$. Therefore, it is not too hard to get from $$x^4+x^3+x^2+x+1$$ to $$x^5-1$$. Now you can reduce mod $$x^5-1$$ by substitution $$x^5=1$$.\n\nI think the way one should think about this is to note that $$x^4+x^3+x^2+x+1$$ is the minimal polynomial of any primitive 5th unit root $$\\alpha$$. Now $$P(\\alpha)=0$$ since $$\\alpha^5=1$$ and therefore $$Q$$ devides $$P$$.", "date": "2020-01-26 21:47:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3224765/polynomial-division-an-obvious-trick-reducing-mod-textitsimpler-multiple/3224982", "openwebmath_score": 0.8542800545692444, "openwebmath_perplexity": 378.106473695498, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308789536605, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8619933780936971}}
{"url": "https://math.stackexchange.com/questions/1926037/rational-and-irrational-sequences", "text": "Rational and irrational sequences\n\n1. Let $(x_n)$ be a sequence of rational numbers that converges to an irrational number $x$. Must it be the case that $x_1, x_2, \\dots$ are all irrational?\n\n2. Let $(y_n)$ be a sequence that converges to a rational number $y$. Define such a sequence where $y_1, y_2, \\dots$ are all irrational.\n\nNow these are the 2 questions in my textbook i noticed when posting this that someone had asked the answer to 1) and a similar but less difficult version of 2.\n\nA couple things i noted the defined answer for 1 he used $\\pi$ and simply added more digits to it for each part of the sequence. so firstly id like to ask would that really be a full mark answer on a analysis final? secondly and more importantly i was wondering if there was a way to define this where the final number wasn't transcendental.\n\nFor 1) I used $(1+1/n)^n$ s.t. $n \\in \\mathbb{N}$. For every $n$ this sequence is rational but for $\\lim_{ n \\to \\infty }(1+1/n)^n = e$.\n\nMy Question: (i) Is this true and if so is there a definable sequence that holds but doesn't define a transcendental number?\n\nEDIT: Golden ratio from Fibonacci sequence.\n\nFor 2) I picked $2^{1/n}$ s.t. $n>1$ and $n \\in \\mathbb{N}$. I believe every value of this sequence is irrational but the limit should be 1.\n\nMy Question: (ii) does the above work? and if possible could I have another example of a sequence where every value is irrational but the limit is rational?\n\nEDIT: $\\frac{\\sqrt 2}{n}$.\n\n\u2022 If $(x_n)$ converges to $x$ is the answer to part (i) (and all the $x_n$ are non-zero), then $(x_n/x)$ is a solution to part (ii). Sep 14 '16 at 2:54\n\u2022 thats very clever thanks! Sep 14 '16 at 3:00\n\nYou basically have both answers right, just need a couple of tweaks.\n\nQuestion:(i) Is this true and if so is there a definable sequence that holds but doesnt define a transcendental number?\n\nIt is true that $a_n = (1 + \\frac{1}{n})^n$ is a sequence of rationals whose $\\lim_{n \\to \\infty} a_n = e$ is irrational.\n\nFor a sequence of rationals that converges to an algebraic (non transcendental) irrational, take for example the partial sums of the Taylor series expansion of $\\sqrt{1+x}$ at $x=1$ which converge to $\\sqrt{2}$.\n\nQuestion:(ii) does the above work? and if possible could i have anther example of a sequence where every value is irrational but the limit is rational?\n\nThe above works, except you may want to define $a_n = 2^\\frac{1}{n+1}$ so that $a_1$ is irrational as well.\n\nFor other such sequences, pick any $r \\in \\mathbb{R} \\setminus \\mathbb{Q}$ and let $a_n = \\frac{r}{n}$.\n\n[ EDIT ] I see that you just edited the question to add $\\frac{\\sqrt{2}}{n}$ as an example of the latter.\n\nIt doesn't have to be $\\pi$ or $e$, you could do the digit by digit approximation and converge to say $\\sqrt{2}$ which is algebraic and not transcendental.\n\nI would say that that answer of digit by digit is basically full mark; you could probably include that $|x - x_n| < 10^{-n}$ which means that the sequence is Cauchy to be more specific.\n\nYour second sequence of $\\{2^{1/n}\\} \\to 1$ is also correct. See if you can prove that each number in the sequence is irrational by mimicking the proof that $\\sqrt{2}$ is irrational.\n\nHere's another sequence\n\n$$a_n = 1 + \\frac{\\sqrt{2}}{10^n}$$\n\nWe see that $\\lim_{n \\to \\infty} a_n = 1$ and each $a_n$ is irrational because you can create a common denominator and note that the numerator will be irrational and the denominator and integer.", "date": "2021-11-28 05:58:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1926037/rational-and-irrational-sequences", "openwebmath_score": 0.9735113978385925, "openwebmath_perplexity": 170.50858538718762, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639694252315, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8619760182128863}}
{"url": "http://mathhelpforum.com/math-challenge-problems/89068-techniques-integration-3-a.html", "text": "# Math Help - Techniques of integration (3)\n\n1. ## Techniques of integration (3)\n\nLet $n \\geq 0$ be an integer. Evaluate $I_n=\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx.$\n\n2. Using itegration by parts I could probably kick this thing off...\n\n$\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx = \\frac{1}{n}cos^n(x)sin(nx) + \\int_0^{\\frac{\\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \\ dx$\n\nNot really sure how to do,\n\n$\\int_0^{\\frac{\\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \\ dx$\n\nprobably needs some trig identities to tidy it up.\n\n3. Originally Posted by pickslides\nUsing itegration by parts I could probably kick this thing off...\n\n$\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx =$ $\\color{red}\\frac{1}{n}cos^n(x)sin(nx)$ $+ \\int_0^{\\frac{\\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \\ dx.$\n\nNot really sure how to do $\\int_0^{\\frac{\\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \\ dx.$\n\nprobably needs some trig identities to tidy it up.\nthat's a good start. we have $I_0=\\frac{\\pi}{2}.$ for $n \\geq 1,$ as you (almost) showed, we have: $I_n=\\int_0^{\\frac{\\pi}{2}} \\sin(nx) \\sin(x) \\cos^{n-1}x \\ dx.$ now what?\n\n4. Hello,\n\nOkay, I think I got the correct idea\n\nFrom $\\cos(a-b)=\\cos(a)\\cos(b)+\\sin(a)\\sin(b)$, it follows that :\n$\\sin(nx)\\sin(x)=\\cos[(n-1)x]-\\cos(nx)\\cos(x)$\n\nThe integral is thus :\n\n$I_n=\\int_0^{\\frac\\pi2} \\cos[(n-1)x]\\cos^{n-1}(x)-\\cos(nx)\\cos(x)\\cos^{n-1}(x) ~dx$\n\n$I_n=\\int_0^{\\frac\\pi2} \\cos[(n-1)x]\\cos^{n-1}(x) ~dx-\\int_0^{\\frac\\pi2} \\cos(nx)\\cos^n(x) ~dx$\n\nWhich is :\n$I_n=I_{n-1}-I_n$\n\n$I_n=\\frac 12 \\cdot I_{n-1}$\n\n$\\Rightarrow \\boxed{I_n=\\frac{\\pi}{2^{n+1}}}$\n\nThat's a really nice one\n\n5. Bravo Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!\n\n6. Originally Posted by NonCommAlg\nBrave Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!\nYes indeed\n\n1+1=n+1 : that's a very well-known result !!\nHad a too short night...\n\n7. Consider\n$cos^n(x) = \\frac{1}{2^n} \\sum_{k=1}^{n} \\binom{n}{k} [cos(n-2k)x + i sin(n-2k)x]\n$\n\nSince $cos^n(x)$ is real , the terms of sine function will be cancelled finally ,\n\nbut we experience that $\\int_0^{\\frac\\pi2} cos(n-2k)x ~cosnx ~dx = 0$ if n-2k is not equal to n so the integral becomes :\n\n$\\frac{\\pi}{2^{n+1}}$\n\n8. Originally Posted by simplependulum\n\n$\\cos^n(x) = \\frac{1}{2^n} \\sum_{k=1}^{n} \\binom{n}{k} [\\cos(n-2k)x + i \\sin(n-2k)x]\n$\n\nthis is not an identity!\n\n9. Originally Posted by NonCommAlg\nLet $n \\geq 0$ be an integer. Evaluate $I_n=\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx.$\nIt can be done using complex variables...\n\nFirst note that\n\n$4\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx=\\int_0^{2\\pi} (\\cos x)^n \\cos (nx) \\ dx$\n\nParameterize the unit circle in the comples plane with\n\n$z=e^{i\\theta} \\implies \\frac{dz}{iz}=d\\theta$\n\nSo the integral becomes\n\n$\\oint \\left(\\frac{z+z^{-1}}{2}\\right)^n\\left( \\frac{z^n+z^{-n}}{2}\\right)\\frac{dz}{iz}$\n\nexpanding with the binomial theorem we get\n\n$\\frac{1}{i\\cdot 2^{n+1}}\\left[ \\oint \\sum_{i=0}^{n}\\binom{n}{i}z^{n-i}z^{-i}\\right]\\left[z^{n-1}+z^{-n-1} \\right]dz$\n\n$\\frac{1}{i\\cdot 2^{n+1}}\\left[ \\oint \\sum_{i=0}^{n}\\binom{n}{i}z^{2n-2i-1}+ \\sum_{i=0}^{n}\\binom{n}{i}z^{i-1}\\right]$\n\neach finite sume is its own larent series so the residues are coeffients on the $\\frac{1}{z}$ terms so we get\n\n$\\frac{1}{i\\cdot 2^{n+1}}\\left[2\\pi i (1)+2\\pi i (1) \\right]=\\frac{4\\pi}{2^{n+1}}$\n\n$4\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx=\\frac{4\\pi}{2^{n+1}}$\n\n$\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx=\\frac{\\pi}{2^{n+1}}$\n\n10. Originally Posted by TheEmptySet\nIt can be done using complex variables...\n\nFirst note that\n\n$4\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx=\\int_0^{2\\pi} (\\cos x)^n \\cos (nx) \\ dx$\n\nParameterize the unit circle in the comples plane with\n\n$z=e^{i\\theta} \\implies \\frac{dz}{iz}=d\\theta$\n\nSo the integral becomes\n\n$\\oint \\left(\\frac{z+z^{-1}}{2}\\right)^n\\left( \\frac{z^n+z^{-n}}{2}\\right)\\frac{dz}{iz}$\n\nexpanding with the binomial theorem we get\n\n$\\frac{1}{i\\cdot 2^{n+1}}\\left[ \\oint \\sum_{i=0}^{n}\\binom{n}{i}z^{n-i}z^{-i}\\right]\\left[z^{n-1}+z^{-n-1} \\right]dz$\n\n$\\frac{1}{i\\cdot 2^{n+1}}\\left[ \\oint \\sum_{i=0}^{n}\\binom{n}{i}z^{2n-2i-1}+ \\sum_{i=0}^{n}\\binom{n}{i}z^{i-1}\\right]$\ni think the power of $z$ in the second sum should be $-2i-1.$\n\neach finite sume is its own larent series so the residues are coeffients on the $\\frac{1}{z}$ terms so we get\n\n$\\frac{1}{i\\cdot 2^{n+1}}\\left[2\\pi i (1)+2\\pi i (1) \\right]=\\frac{4\\pi}{2^{n+1}}$\n\n$4\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx=\\frac{4\\pi}{2^{n+1}}$\n\n$\\int_0^{\\frac{\\pi}{2}} (\\cos x)^n \\cos (nx) \\ dx=\\frac{\\pi}{2^{n+1}}$\nlooks good to me!", "date": "2015-05-26 04:46:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-challenge-problems/89068-techniques-integration-3-a.html", "openwebmath_score": 0.9667383432388306, "openwebmath_perplexity": 1429.356616228868, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513873424044, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8619724813370496}}
{"url": "https://math.stackexchange.com/questions/3638644/double-negation-of-existential-universal-quantifier-lnot-exists-x-lnot-ax", "text": "# Double negation of existential/universal quantifier $\\lnot(\\exists x(\\lnot A(x))$\n\nI have a (simple) question about the double-negation and existential/universal quantifiers.\n\nWhen negating the following\n\n$$\\lnot\\exists x(\\lnot A(x))$$\n\nI believe you just push the negation in (which swaps the quantifier) making it\n\n$$\\forall x[\\lnot(\\lnot A(x))]$$\n\nand then\n\n$$\\forall x A(x)$$\n\nWould that be a correct interpretation? Or would I be losing one of the negations on $$A(x)$$ somewhere earlier?\n\n\u2022 Yes, that\u2019s fine. It\u2019s also intuitively clear: the first expression says that there is no $x$ for which $A(x)$ is false, and the last says that $A(x)$ is true for every $x$, clearly the same thing. \u2013\u00a0Brian M. Scott Apr 22 at 17:30\n\u2022 Awesome, thank you so much. \u2013\u00a0confundido Apr 22 at 17:31\n\u2022 You\u2019re very welcome. \u2013\u00a0Brian M. Scott Apr 22 at 17:31\n\u2022 Thank you (just read your more detailed answer) - I thought the intuition was clear, but it was part of a larger proof and I was doubting myself as to whether the reduction made sense. I guess I just needed validation. \u2013\u00a0confundido Apr 22 at 17:35\n\nYou did that correctly.\n\nNote that going from\n\n$$\\lnot\\exists x(\\lnot A(x))$$\n\nto\n\n$$\\forall x[\\lnot(\\lnot A(x))]$$\n\nis an example of 'Quantifier Negation', but going from:\n\n$$\\forall x[\\lnot(\\lnot A(x))]$$\n\nto\n\n$$\\forall x A(x)$$\n\nis an instance of 'Double Negation'\n\nAs per the answer by Brian M. Scott:\n\nIt\u2019s also intuitively clear: the first expression says that there is no x for which A(x) is false, and the last says that A(x) is true for every x, clearly the same thing.\n\nA word of caution. Of course at the final step $$\\forall x\\neg\\neg Ax$$ entails (or at least, classically entails) $$\\forall x Ax$$, and does so because adjacent double negations (classically) cancel each other out.\n\nBUT\n\nThe inference is not strictly an application of a standard double negation rule of the form from $$\\neg\\neg\\varphi$$ infer $$\\varphi$$. That rule only allows us to remove initial double negations.\n\nSO\n\nTo show the entailment in standard proof systems requires a three-step mini-proof:\n\n$$\\forall x\\neg\\neg Ax\\quad$$ (given)\n\n$$\\neg\\neg Aa\\quad\\quad$$ (universal instantiation with parameter or free variable depending on system)\n\n$$Aa\\quad\\quad\\quad$$ (NOW you can apply the DN rule)\n\n$$\\forall x Ax\\quad\\quad$$ (universal generalization)\n\nSo your reasoning is informally just fine, but do be careful about jumping from $$\\forall x\\neg\\neg Ax$$ to $$\\forall x Ax$$ in formal proofs!", "date": "2020-10-19 20:47:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3638644/double-negation-of-existential-universal-quantifier-lnot-exists-x-lnot-ax", "openwebmath_score": 0.6180577874183655, "openwebmath_perplexity": 448.9697386536659, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877044076956, "lm_q2_score": 0.877476800298183, "lm_q1q2_score": 0.8619346718359121}}
{"url": "https://math.stackexchange.com/questions/1226352/finding-an-operation-on-gs-that-yields-a-group", "text": "# Finding an operation on $G^S$ that yields a group\n\nProblem: Assume $S$ is a nonempty set and $G$ is a group. Let $G^S$ denote the set of all mappings from $S$ to $G$. Find an operation on $G^S$ that will yield a group.\n\nUpdate (full attempted proof--is it correct?): Let $*$ be the operation on the given group $G$ and let $\\bigstar$ be the operation on $G^S$. For all $\\alpha,\\beta\\in G^S$, define $(\\alpha\\,\\bigstar\\,\\beta)(x)=\\alpha(x)*\\beta(x)$. Now we must prove $G^S$ is a group.\n\n\u2022 Closure: This is inherited from $G$.\n\u2022 Associativity: This is also inherited from $G$.\n\u2022 Existence of an identity element: This is also inherited from $G$; more explicitly, $e_G(x)=1=e_{G^S}(x)$ because $(\\alpha\\,\\bigstar\\,e_G)(x)=\\alpha(x)*e_G(x)=\\alpha(x)$ and, in the other direction, $(e_G\\,\\bigstar\\,\\alpha)(x)=e_G(x)*\\alpha(x)=\\alpha(x)$.\n\u2022 Existence of inverse elements: If $\\eta(x)\\in G$, then $\\eta^{-1}(x)\\in G$ also. Thus, for $\\alpha\\in G^S$, we have $(\\alpha\\,\\bigstar\\,\\alpha^{-1})(x)=\\alpha(x)*\\alpha^{-1}(x)=e_G(x)=e_{G^S}(x)$ and, in the other direction, $(\\alpha^{-1}\\,\\bigstar\\,\\alpha)(x)=\\alpha^{-1}(x)*\\alpha(x)=e_G(x)=e_{G^S}(x)$.\n\nThis concludes the proof that the operation $\\bigstar$ on $G^S$ yields a group. $\\blacksquare$\n\n$\\color{red}{\\mathbf{\\text{Question:}}}$ Is this above proof correct? Also, by the way the operation $\\bigstar$ works, it seems like $G^S$ would be a subgroup of $G$ or am I wrong?\n\nOriginal work shown for problem:\n\nBook solution: For $\\alpha,\\beta\\in G^S$, define $(\\alpha\\beta)(x)=\\alpha(x)\\beta(x)$ for each $x\\in S$.\n\nMy questions: I thought about this problem for a good while but got nowhere and finally decided to look at the answer, but it does not make a ton of sense to me. How does the given operation yield a group? And what really is the operation being considered? Is it composition? That is, $(\\alpha\\beta)(x)\\equiv(\\alpha\\circ\\beta)(x)$? If it is composition, then I know all compositions are associative, and thus I do not need to prove that as part of proving that $G^S$ is a group. But what about closure? Existence of an identity element? Existence of inverse elements?\n\nFor the existence of an identity element, I thought about $e_G(x)=1$ and having $(\\eta e_G)(x)=\\eta(x)e_G(x)=\\eta(x)$ and $(e_G\\eta)(x)=e_G(x)\\eta(x)=\\eta(x)$, but this doesn't really seem right or is it?\n\nLastly, the problem gives the condition that $S\\neq\\varnothing$ (why is that important?), and no operation is specified concerning $G$--does that matter? I guess not, but it seems interesting to me that no operation is specified for $G$ when we are trying to find an operation for $G^S$ that involves $G$.\n\n\u2022 The proof seems fundamentally fine. I'm not sure what the distinction is between the notation $e_G(x)$ and $e_G^S(x)$. You could write out associativity more explicitly, although I tend not to do so unless it's not obvious. \u2013\u00a0Rolf Hoyer Apr 9 '15 at 17:22\n\u2022 @RolfHoyer I just meant that $e_G(x)$ is the identity element for $G$ while $e_G^S(x)$ is the identity element for $G^S$. If $G^S$ is a subgroup of $G$ (it looks like it is, but is it?), then I know the distinctions do not matter because the identity and inverses are the same. \u2013\u00a0fancynancy Apr 9 '15 at 17:32\n\u2022 @RolfHoyer OK--I just fixed up the question with the modified notation ($e_{G^S}(x)$ instead of $e_G^S(x)$). Look all good now? Your answer definitely helped me to start thinking along the right lines. \u2013\u00a0fancynancy Apr 9 '15 at 17:38\n\nIt definitely looks much better!\n\nI'm a little confused about the distinction between $e_G$ and $e_{G^S}$; they both seem like the identity of $G^S$, and you seem to be using $1$ for the identity of $G$. Maybe make it more explicit that you're defining $e_{G^S}: S \\to G$ so that $e_{G^S}(x) = 1_G$; it's the map sending everything in $S$ to the identity of $G$. So my only suggestion would be to clean up the notation (how exactly are you writing the identity of $G$? Of $G^S$?) and make it clear exactly what the identity of $G^S$ does.\n\nFor the inverses, essentially the same thing: Given $\\alpha: S \\to G$, you just start using $\\alpha^{-1}$ without explicitly stating that, if $\\alpha(x) = g$, then $\\alpha^{-1}(x) = (\\alpha(x))^{-1} = g^{-1}$. That is, make it clear that we invert $\\alpha:S \\to G$ pointwise, by inverting its output (using the inverse in $G$).\n\nFor your other question, very rarely will it be the case that $G^S \\cong G$. I think it's more likely that $G^S$ is isomorphic to a direct product of several copies of $G$ (how many copies?).\n\n\u2022 Hmm okay. So I should write $e_G(x)=1$ and $e_{G^S}(x)=1_G$ to more clearly indicate that $1$ is the identity for $G$ while $1_G$ is the identity for $G^S$ right? You're right--my notation is a bit mangled. Okay. I get the identity part now. I'm still a little fuzzy on the inverse part though. Are you saying I'm technically correct with what I've written but it would be clearer to actually assign an element to $\\alpha(x)$ to clearly indicate that invertibility is pointwise? \u2013\u00a0fancynancy Apr 9 '15 at 18:10\n\u2022 What I mean about identities is that there should really only be two in sight: That of $G$, let's call it $1_G$, and that of $G^S$, let's call it $1_{G^S}$. So, I'm not sure why there are two functions whose notation suggest that they're both identities of $G^S$; that they're both functions $S \\to G$. \u2013\u00a0pjs36 Apr 9 '15 at 18:12\n\u2022 And for the inverses, yes, it's correct, provided you explicitly state how $\\alpha^{-1}: S \\to G$ (or $\\eta^{-1}$, for that matter) is defined. \u2013\u00a0pjs36 Apr 9 '15 at 18:14\n\u2022 Okay so I should have (for identities, that is), in one direction, $(\\alpha\\,\\bigstar\\,e_G)(x)=\\alpha(x)*e_G(x)=\\alpha(x)*1_G=\\alpha(x)$. The same would happen in the other direction. So the identity exists and, explicitly, we have $e_{G^S}=e_G=1$. Correct? \u2013\u00a0fancynancy Apr 9 '15 at 18:16\n\u2022 I do see what you're saying now about the confusion between writing $e_G(x)$ and $e_{G^S}(x)$ because they are both mappings from $S$ to $G$; thus, that notation is rather clumsy. \u2013\u00a0fancynancy Apr 9 '15 at 18:19\n\nThe group operation is not defined by composition, unless you mean the composition $S \\to S\\times S \\to G\\times G \\to G$, where the first map is the diagonal $s \\to (s,s)$, the second map is the product of the two given maps $\\alpha, \\beta$, and the third map is the group operation on $G$. When the author writes $(\\alpha\\beta)(x) = \\alpha(x)\\beta(x)$, the latter product is defined using the operation of $G$, just to clarify.\n\nThis definition is using the operation of $G$ pointwise, just like how addition and multiplication are defined for real-valued functions, ie $(f+g)(x) = f(x) + g(x)$. The identity element is the constant function $e_G(x) = 1$ for every $x$, and you correctly give the proof that this works. The inverse of a function $\\alpha: S \\to G$ is given by taking inverses pointwise, namely $\\alpha^{-1}(x) = (\\alpha(x))^{-1}$. Associativity follows from the associativity of $G$.\n\nAs regards to the empty set, if $S = \\emptyset$ then there is a unique empty function $S \\to G$, and I guess the authors would rather rule out this case than give the trivial product on this set.\n\nHopefully this makes things more clear.\n\n\u2022 I just updated my question with a fully attempted proof in light of your helpful answer. Can you let me know if you think it's correct? Thanks for your answer--I think I am at least on the right track now. \u2013\u00a0fancynancy Apr 9 '15 at 14:23", "date": "2020-09-19 19:39:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1226352/finding-an-operation-on-gs-that-yields-a-group", "openwebmath_score": 0.9451440572738647, "openwebmath_perplexity": 171.8989550887023, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877028521422, "lm_q2_score": 0.877476800298183, "lm_q1q2_score": 0.8619346704709501}}
{"url": "https://physics.stackexchange.com/questions/137527/why-is-equivalent-resistance-in-parallel-circuit-always-less-than-each-individua", "text": "# Why is equivalent resistance in parallel circuit always less than each individual resistor?\n\nThere are $$n$$ resistors connected in a parallel combination given below.\n\n$$\\frac{1}{R_{ev}}=\\frac{1}{R_{1}}+\\frac{1}{R_{2}}+\\frac{1}{R_{3}}+\\frac{1}{R_{4}}+\\frac{1}{R_{5}}.......\\frac{1}{R_{n}}$$\n\nFoundation Science - Physics (class 10) by H.C. Verma states (Pg. 68)\n\nFor two resistances $$R_{1}$$ and $$R_{2}$$ connected in parallel,\n\n$$\\frac{1}{R_{ev}}=\\frac{1}{R_{1}}+\\frac{1}{R_{2}}=\\frac{R_{1}+R_{2}}{R_{1}R_{2}}$$ $$R_{ev}=\\frac{R_{1}R_{2}}{R_{1}+R_{2}}$$\n\nWe see that the equivalent resistance in a parallel combination is less than each of the resistances.\n\nI observe this every time I do an experiment on parallel resistors or solve a parallel combination problem. How can we prove $$R_{ev} or that $$R_{ev}$$ is less than the Resistor $$R_{min}$$, which has the least resistance of all the individual resistors?\n\n\u2022 I have to confess that when I saw that you had exhibited the harmonic sum form ($1/R_e = 1/R_1 + 1/R_2 + \\dots$), I found the question confusing because I didn't understand what you didn't get. It's worth your time to think about this until it becomes obvious, not because this problem is so important but because it will start teaching you the habit of learning things from the functional form of the relationships that appear in your studies. \u2013\u00a0dmckee --- ex-moderator kitten Sep 27 '14 at 17:15\n\u2022 @dmckee I didn't quite get what you said. It'll probably help me if you'll elaborate. \u2013\u00a0user49111 Sep 27 '14 at 17:26\n\u2022 With experience, you will find this fact obvious, and it is worth taking the time to ponder it until it becomes so. \u2013\u00a0dmckee --- ex-moderator kitten Sep 27 '14 at 17:34\n\u2022 I rolled back this question to the original version because you shouldn't put answers, or responses to answers, in the question itself. @imakesmalltalk Feel free to post an answer of your own if you would like to offer another method of answering the question. \u2013\u00a0David Z Oct 27 '14 at 0:30\n\u2022 Once you believe the 2 resistor answer, it extends to any finite number. For each resistor, imagine combining all the rest into a single resistor by the law you cite. Now the combination of the one resistor and the combination resistor is less than the one. Do this once for each of the resistors, and you have that the total combination is less than any one. \u2013\u00a0Ross Millikan Oct 27 '14 at 2:53\n\nIf we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\\frac {V}{R_1} +\\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\\frac {I_T}{V}=\\frac {1}{R_1} +\\frac {1}{R_2} + ...$$ Which is the same as: $$\\frac {1}{R_{eq}}=\\frac {1}{R_1} +\\frac {1}{R_2} + ...$$ Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors.\n\nThe individual resistances are all positive, so the sum $$\\frac{1}{R_1} + \\frac{1}{R_2} + \\frac{1}{R_3} + \\dots \\,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances.\n\nNo mucking around with the two-resistor form required.\n\nWe can prove it by induction. Let $$\\frac{1}{R^{(n)}_{eq}} = \\frac{1}{R_1} + \\cdots+ \\frac{1}{R_n}$$ Now, when $n=2$, we find $$\\frac{1}{R^{(2)}_{eq}} = \\frac{1}{R_1} + \\frac{1}{R_2} \\implies R_{eq}^{(2)} = \\frac{R_1 R_2}{R_1+R_2} = \\frac{R_1}{1+\\frac{R_1}{R_2}} = \\frac{R_2}{1+\\frac{R_2}{R_1}}$$ Since $\\frac{R_1}{R_2} > 0$, we see that $R^{(2)}_{eq} < R_1$ and $R^{(2)}_{eq} < R_2$ or equivalently $R^{(2)}_{eq} < \\min(R_1, R_2)$.\n\nNow, suppose it is true that $R^{(n)}_{eq} < \\min (R_1, \\cdots, R_n)$. Then, consider $$\\frac{1}{R^{(n+1)}_{eq}} = \\frac{1}{R_1} + \\cdots+ \\frac{1}{R_n} + \\frac{1}{R_{n+1}} = \\frac{1}{R^{(n)}_{eq}} + \\frac{1}{R_{n+1}}$$ Using the result from $n=2$, we find $$R^{(n+1)}_{eq} < \\min ( R_{n+1} , R^{(n)}_{eq} ) < \\min ( R_{n+1} , \\min (R_1, \\cdots, R_n))$$ But $$\\min ( R_{n+1} , \\min (R_1, \\cdots, R_n)) = \\min ( R_{n+1} , R_1, \\cdots, R_n)$$ Therefore $$R^{(n+1)}_{eq} < \\min ( R_1, \\cdots, R_n , R_{n+1} )$$ Thus, we have shown that the above relation holds for $n=2$, and further that whenever it holds for $n$, it also holds for $n+1$. Thus, by induction, it is true for all $n\\geq2$.\n\n\u2022 $\\frac {R_1}{R_2} > 0$ not 1 \u2013\u00a0Omar Elawady Apr 4 '16 at 21:41\n\u2022 Your induction method is much more mathematical than the other answers....but is there any way to solve this using $AM-GM$ inequality? \u2013\u00a0Soham Jun 3 '16 at 16:11\n\nIt might be easier to think in terms of conductance which is the inverse of resistance.\n\nThe more paths there are between A and B which conduct electricity, the greater is the amount of current which can flow - ie the greater is the conductance of the network. The total conductance is greater than that of any individual path, because each additional path always increases the amount of electricity which can be conducted, it never reduces it. In particular, the total conductance is always greater than the largest individual conductance.\n\nTranslating this back in terms of resistance R (which is the inverse of conductance S - ie R = 1/S), the total resistance is smaller than the smallest individual resistance.", "date": "2021-08-03 05:31:32", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/137527/why-is-equivalent-resistance-in-parallel-circuit-always-less-than-each-individua", "openwebmath_score": 0.8280104994773865, "openwebmath_perplexity": 1872.80716089273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876997410348, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8619346520043563}}
{"url": "https://byjus.com/question-answer/find-inverse-by-elementary-row-operations-if-possible-of-the-following-matrices-begin-bmatrix-1-1/", "text": "Question\n\nFind inverse, by elementary row operations (if possible), of the following matrices$$\\begin{bmatrix} 1 & 3 \\\\ -5 & 7 \\end{bmatrix}$$\n\nSolution\n\nTo check if the inverse exist we find the determinant:We have:$$A=\\begin{bmatrix} 1 & 3 \\\\ -5 & 7 \\end{bmatrix}$$So, $$\\left|A\\right|=1\\times 7-(-5\\times 3)$$$$\\Rightarrow \\left|A\\right|=7+15=22$$Since, $$|A|\\ne0$$, hence the inverse exists\u00a0 Now, in order to use elementary row operations, we may write $$A=IA.$$$$\\therefore \\begin{bmatrix} 1 & 3 \\\\ -5 & 7 \\end{bmatrix}= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1\\end{bmatrix}A$$$$\\Rightarrow \\begin{bmatrix} 1 & 3 \\\\ 0 & 22 \\end{bmatrix}= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1\\end{bmatrix}A$$\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0 $$R_2 \\rightarrow R_2+ 5\\times R_1$$\u00a0\u00a0\u00a0\u00a0 $$\\Rightarrow \\begin{bmatrix} 1 & 3 \\\\ 0 & 1 \\end{bmatrix}= \\begin{bmatrix} 1 & 0 \\\\ \\dfrac{5}{22} & \\dfrac{1}{22}\\end{bmatrix}A$$\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $$R_2 \\rightarrow \\dfrac{1}{22} \\times R_2$$$$\\Rightarrow \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}= \\begin{bmatrix} \\dfrac{7 }{22} & -\\dfrac{3}{22} \\\\ \\dfrac{5}{22} & \\dfrac{1}{22}\\end{bmatrix}A$$\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $$R_1 \\rightarrow R_1- 3\\times R_2$$$$\\Rightarrow \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}=\\dfrac{1}{22} \\begin{bmatrix} 7 & -3 \\\\ 5 & 1\\end{bmatrix}A$$$$\\Rightarrow I=BA$$, where $$B$$ is the inverse of $$A$$.$$B=\\dfrac{1}{22} \\begin{bmatrix} 7 & -3 \\\\ 5 & 1\\end{bmatrix}$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-22 05:31:54", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/find-inverse-by-elementary-row-operations-if-possible-of-the-following-matrices-begin-bmatrix-1-1/", "openwebmath_score": 0.903389573097229, "openwebmath_perplexity": 6630.886789566099, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620056639, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.861930037578868}}
{"url": "https://math.stackexchange.com/questions/2278517/how-to-approach-this-combinatorics-problem", "text": "# How to approach this combinatorics problem?\n\nYou have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?\n\n\u2022 Does order matter? \u2013\u00a0vrugtehagel May 12 '17 at 21:32\n\u2022 @vrugtehagel No. \u2013\u00a0LearningMath May 12 '17 at 21:33\n\u2022 Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\\choose3}+11\\cdot10$ ways. Do you see why? \u2013\u00a0Barry Cipra May 12 '17 at 21:39\n\nCalculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$\n\n\u2022 Can you please elaborate where you got this from? Thanks. \u2013\u00a0LearningMath May 12 '17 at 21:31\n\u2022 $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. \u2013\u00a0vadim123 May 12 '17 at 21:32\n\u2022 Can you provide a reference for this approach? Thank you. \u2013\u00a0LearningMath May 12 '17 at 21:52\n\u2022 See the second half of this. \u2013\u00a0vadim123 May 12 '17 at 23:32\n\nThis is the same as no of solutions of $\\sum_{i=1}^{12} x_i=5$ where $1\u2264x_1\u22642$ and for the other $x_i$'s $0\u2264x_i\u22642$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$\n\n\u2022 How is it the same as the coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$ ? Can you provide a reference where I can read more about this? \u2013\u00a0LearningMath May 12 '17 at 22:06\n\nDecide if chocolate will be a double ($c=1$) or a single ($c=0$).\n\nIf $c=0$, take a chocolate ball. Decide how many \"doubles\" $d\\in\\{0,1,2\\}$ you want and make one of $11 \\choose d$ specific choices regarding which non-chocolate flavors you want. These $d$ colors and chocolate are now off limits, so choose $5-1-2d=4-2d$ colors from each of the remaining $12-d-1=11-d$ legal options.\n\nIf $c=1$, take two chocolate balls. Decide how many \"non-chocolate doubles\" $d\\in\\{0,1\\}$ you want and make one of $11 \\choose d$ choices regarding the non-chocolate flavors. These $d$ colors and chocolate are now off limits, so choose $5-2-2d=3-2d$ flavors from each of the remaining $12-d-1=11-d$ legal options.\n\n$$\\sum_{d=0,1,2} {11 \\choose d}{{11-d} \\choose {4-2d}} + \\sum_{d=0,1} {11 \\choose d}{{11-d} \\choose {3-2d}}$$\n\n$$= {11 \\choose 2} + \\sum_{d=0,1} {11 \\choose d} \\left[{{11-d} \\choose {4-2d}} + {{11-d} \\choose {3-2d}}\\right]$$\n\nwhich is $55 + [1*495 + 11*55]=1155$.\n\nYou can also answer this using careful counting Case by case basis splitting into exactly two pairs of flavors, one pair of flavors and no pairs.\n\n$$C_1,P_1,P_1,P_2,P_2 = 1*12*11$$ $$C_1,P_1,P_1,P_2,P_3=1*12*11*10/2!$$ $$C_1,P_1,P_2,P_3,P_4=1*12*11*10*9/4!$$\n\nHow many of these ways lead to more than $2$ chocolates?\n\nIf $$P_1=C$$ which for the first case can happen $11$ ways, the second case can happen $11*10/2!$ ways and the last case can't happen if just $P_1=C$\n\nWhat about if $P_2=C$ the first case can happen in $11$ ways and the second case can happen in 11*10/2! ways.\n\nIf $P_3,P_4=C$ then we don't get $3$ or more chocolates, so we are done! The final answer is $$12*11+12*11*10/2! +12*11*10*9/4!-11*10/2!-11-11-11*10/2!=1155$$\n\nI think the most obvious/straightforward way to solve this is to use combination 'with repetition'/'stars and bars' and then subtract all inapplicable cases. (https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition). $\\space$The formula for this for selecting $r$ from $n$ is ${{n+r-1}\\choose r}$. $\\space$(I like to think of this as selecting $r$ 'items' from the $n+r-1$ 'walls' and 'items' and then letting the $n-1$ 'walls' 'fall into place.')\n\nDividing into cases where there is one chocolate ball and more than one.\n\nOne chocolate ball of ice cream:\n\n$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=4$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor and $1$ of another - cases where there are $4$ of one flavor):\n\n${{11+4-1=14}\\choose 4}-{11\\choose 1}{10\\choose 1}-{11\\choose 1}=1001-110-11=880$\n\nTwo chocolate balls of ice cream:\n\n$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=3$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor):\n\n${{11+3-1=13}\\choose 3}-{11\\choose 1}=286-11=275$\n\nFor a grand total of $1155$.", "date": "2019-08-21 22:08:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2278517/how-to-approach-this-combinatorics-problem", "openwebmath_score": 0.7283103466033936, "openwebmath_perplexity": 581.2403988490029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620049598214, "lm_q2_score": 0.8688267660487573, "lm_q1q2_score": 0.8619300234890879}}
{"url": "https://mathsgee.com/tag/sequence", "text": "# Recent questions tagged sequence\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nWhat is a descending arithmetic sequence?\nWhat is a descending arithmetic sequence?What is a descending arithmetic sequence? ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nHow do you write the nth term rule for the sequence $1,3,5,7,9, \\ldots ?$\nHow do you write the nth term rule for the sequence $1,3,5,7,9, \\ldots ?$How do you write the nth term rule for the sequence $1,3,5,7,9, \\ldots ?$ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nFind the next 2 terms and Tn: 2;4;6;8;16\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nUse this pattern of differences to find the next term of the sequence:\nUse this pattern of differences to find the next term of the sequence: Use this pattern of differences to find the next term of the sequence: ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nFind the common ratio of the\u00a0geometric sequence $1,3,9,27, \\ldots$\u00a0Then express each sequence in the form $a_{n}=a_{1} r^{n-1}$ and find the eighth term of the sequence.\nFind the common ratio of the\u00a0geometric sequence $1,3,9,27, \\ldots$\u00a0Then express each sequence in the form $a_{n}=a_{1} r^{n-1}$ and find the eighth term of the sequence.Find the common ratio of the geometric sequence $1,3,9,27, \\ldots$ Then express each sequence in the form $a_{n}=a_{1} r^{n-1}$ and find the eight ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nWhat is a geometric sequence?\nWhat is a geometric sequence?What is a geometric sequence? ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nFind an expression for the nth term of the sequence $3,7,11,15,19, \\ldots$\nFind an expression for the nth term of the sequence $3,7,11,15,19, \\ldots$Find an expression for the &nbsp;nth term of the sequence $3,7,11,15,19, \\ldots$ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nFind an expression for the $n$th term of the sequence. $10,50,250,1250, \\ldots$\nFind an expression for the $n$th term of the sequence. $10,50,250,1250, \\ldots$Find an expression for the $n$th term of the sequence. &nbsp;$10,50,250,1250, \\ldots$ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nWrite the first five terms of $a_{n}=2\\left(3^{n-1}\\right)$\nWrite the first five terms of $a_{n}=2\\left(3^{n-1}\\right)$Write the first five terms of $a_{n}=2\\left(3^{n-1}\\right)$ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nWrite the first five terms of a sequence described by the general term $a_{n}=3 n+2$\nWrite the first five terms of a sequence described by the general term $a_{n}=3 n+2$Write the first five terms of a sequence described by the general term $a_{n}=3 n+2$ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nFind the constant second difference for the following sequence: $7 ; 4 ;-3 ;-14 ; \\ldots$\nFind the constant second difference for the following sequence: $7 ; 4 ;-3 ;-14 ; \\ldots$Find the constant second difference for the following sequence: \\ 7 ; 4 ;-3 ;-14 ; \\ldots \\ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nFind the common second difference for the following sequence: $-13 ;-25 ;-43 ;-67 ; \\ldots$\nFind the common second difference for the following sequence: $-13 ;-25 ;-43 ;-67 ; \\ldots$Find the common second difference for the following sequence: \\ -13 ;-25 ;-43 ;-67 ; \\ldots \\ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nFind the common second difference for the following sequence: $-7 ;-15 ;-25 ;-37 ; \\ldots$\nFind the common second difference for the following sequence: $-7 ;-15 ;-25 ;-37 ; \\ldots$Find the common second difference for the following sequence: \\ -7 ;-15 ;-25 ;-37 ; \\ldots \\ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nConsider the series $\\sum_{n=1}^{\\infty} \\frac{\\cos (n \\pi)}{n}$. Which of the following statements is TRUE?\nConsider the series $\\sum_{n=1}^{\\infty} \\frac{\\cos (n \\pi)}{n}$. Which of the following statements is TRUE?Consider the series $\\sum_{n=1}^{\\infty} \\frac{\\cos (n \\pi)}{n}$. Which of the following statements is TRUE? &nbsp; (A) The series is not conditiona ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nLet $g(x, y)=x^{2} y+\\cos y+y \\sin x$, Which of the following statements is TRUE?\nLet $g(x, y)=x^{2} y+\\cos y+y \\sin x$, Which of the following statements is TRUE?Let $g(x, y)=x^{2} y+\\cos y+y \\sin x$ Which of the following statements is TRUE? &nbsp; (A) $g_{x}=2 x y+\\cos y+y \\cos x$. (B) $g_{y}=2 x-\\sin y+\\s ... close 0 answers 7 views Determine, giving reasons, whether the sequence$\\left\\{a_{n}\\right\\}=\\left\\{\\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}\\right\\}$is convergent or divergent. If it is convergent, find the value to which it converges.Determine, giving reasons, whether the sequence$\\left\\{a_{n}\\right\\}=\\left\\{\\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}\\right\\}$is convergent or diverge ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Determine whether the following series converge or diverge.$\\sum_{n=1}^{\\infty} \\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}$0 answers 5 views Determine whether the following series converge or diverge.$\\sum_{n=1}^{\\infty} \\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}$Determine whether the following series converge or diverge.$\\sum_{n=1}^{\\infty} \\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Determine whether the following series converge or diverge.$\\sum_{n=1}^{\\infty}(-1)^{n}\\left(\\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}\\right)^{n}$0 answers 4 views Determine whether the following series converge or diverge.$\\sum_{n=1}^{\\infty}(-1)^{n}\\left(\\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}\\right)^{n}$Determine whether the following series converge or diverge.$\\sum_{n=1}^{\\infty}(-1)^{n}\\left(\\dfrac{n^{3}-1+n^{2} \\sin n}{1+3 n^{3}}\\right)^{n}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Given that$\\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\\cdots=\\sum_{n=0}^{\\infty} x^{n}$. Write down the power series of$\\dfrac{1}{1+x^{2}}$0 answers 6 views Given that$\\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\\cdots=\\sum_{n=0}^{\\infty} x^{n}$. Write down the power series of$\\dfrac{1}{1+x^{2}}$Given that$\\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\\cdots=\\sum_{n=0}^{\\infty} x^{n}$. Write down the power series of$\\dfrac{1}{1+x^{2}}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Write$\\dfrac{\\pi}{4}$as a series. 0 answers 8 views Write$\\dfrac{\\pi}{4}$as a series.Write$\\dfrac{\\pi}{4}$as a series. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Which FIVE of the following statements are FALSE ? 0 answers 6 views Which FIVE of the following statements are FALSE ?Which FIVE of the following statements are FALSE ? &nbsp; A. The sequence$\\left\\{\\frac{\\cos (n \\pi)}{n}\\right\\}$diverges &nbsp; B.$\\quad \\cos \\ ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nThe interval of convergence of the series $\\sum_{n=1}^{\\infty} \\dfrac{(-1)^{n}(x+2)^{n}}{n}$ is\nThe interval of convergence of the series $\\sum_{n=1}^{\\infty} \\dfrac{(-1)^{n}(x+2)^{n}}{n}$ isThe interval of convergence of the series $\\sum_{n=1}^{\\infty} \\dfrac{(-1)^{n}(x+2)^{n}}{n}$ is &nbsp; A. $\\quad(-3,-1$ &nbsp; B. $-3,-1$ &nbs ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nWhich FIVE of the following statements are TRUE?\nWhich FIVE of the following statements are TRUE?Which FIVE of the following statements are TRUE? &nbsp; A. The series $\\sum_{n=1}^{\\infty}\\left(\\frac{4 n-3}{3 n-4}\\right)^{n}$ converges &nbsp; B ...\nclose\n\nNotice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993\nDetermine whether the integral $\\int_{3}^{3+e} \\ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to.\nDetermine whether the integral $\\int_{3}^{3+e} \\ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to. Determine whether the integral $\\int_{3}^{3+e} \\ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to. ...\nFind the limit of the sequence $\\left\\{\\dfrac{-1}{(-1)^{n}}\\right\\}$ (if it exists).\nFind the limit of the sequence $\\left\\{\\dfrac{-1}{(-1)^{n}}\\right\\}$ (if it exists).Find the limit of the sequence $\\left\\{\\dfrac{-1}{(-1)^{n}}\\right\\}$ (if it exists). ...", "date": "2022-01-21 01:40:34", "meta": {"domain": "mathsgee.com", "url": "https://mathsgee.com/tag/sequence", "openwebmath_score": 0.8441284894943237, "openwebmath_perplexity": 1692.4324426317335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9648551535992066, "lm_q2_score": 0.8933094103149354, "lm_q1q2_score": 0.8619141883010336}}
{"url": "https://math.stackexchange.com/questions/2212375/what-is-wrong-with-this-proof-of-the-union-of-two-compact-sets", "text": "# What is wrong with this proof of the union of two compact sets?\n\nI need to show that the union of two compact sets is compact. Here is what I tried:\n\nLet $A$ and $B$ be two compact sets. Then for every open cover $\\{A_n\\}_{n\\in I_1}$ and open cover $\\{B_m\\}_{m\\in I_2}$, there exists finite subcovers $\\bigcup\\limits_{n\\in I_1}^p A_n$ such that $A\\subseteq\\bigcup\\limits_{n\\in I_1}^p A_n$ and $\\bigcup\\limits_{b\\in I_2}^k B_m$ such that $B\\subseteq\\bigcup\\limits_{b\\in I_2}^k B_m$. Then, $A\\cup B\\subseteq (\\bigcup\\limits_{n\\in I_1}^p A_n)\\cup(\\bigcup\\limits_{b\\in I_2}^k B_m)\\subseteq(\\bigcup\\limits_{n\\in I_1} A_n) \\cup (\\bigcup\\limits_{b\\in I_2} B_m)$ (the union of open covers constructed as an open cover for $A\\cup B$). Thus, for ever open cover of $A\\cup B,$ there exists a finite subcover, so $A\\cup B$ is compact.\n\nI thought that since the open cover I constructed was arbitrary, and so was the finite subcover, then every open cover of $A\\cup B$ had a finite subcover. What is wrong with this?\n\n\u2022 Well, what's wrong is that you don't show that any cover of $C = A \\cup B$ has a finite subcover. You only consider those covers that are unions of previous discussed covers of A and B have finite subcovers. If $\\Alpha$ is a cover of A and $\\Beta$ is a cover of B. That shows us $\\Alpha \\cup \\Beta$ has a subcover. But what about all the covers of $C$ that are not $\\Alpha \\cup \\Beta$? As it turns out if $\\Gamma$ is an open cover then $\\Gamma$ must = $\\{A\\}\\cup \\{B\\}$ for some pair of open covers, and such a statement may seem obvious, but can't be assumed without proof. Mar 31, 2017 at 23:30\n\u2022 You probably got the gist of what is going on, one comment that seems relevant: if you could show that every open cover of $A \\cup B$ is of the form $O_1 \\cup O_2$ with $O_1$ an open over of $A$ and $O_2$ another for $B,$ then you are in business (your proof goes through). Mar 31, 2017 at 23:38\n\n$$A\\cup B \\subseteq \\left( \\bigcup_{n \\in I_1} A_n \\right)\\cup \\left( \\bigcup_{m \\in I_2} B_m \\right)$$\n\nmay not be a general open cover for $A\\cup B$. So, it is better to start with:\n\nLet $\\{ C_n \\}$ be an open cover for $A \\cup B$. Then since both $A$ and $B$ are compact sets, we have $\\dots$\n\nThe problem with your logic is as follows:\n\nYou are right, in that a finite subcover of $A$ and a finite subcover of $B$, together gives a finite subcover of $A \\cup B$.\n\nHowever, to show $A \\cup B$ is compact, you are supposed to start with any arbitrary open cover of $A \\cup B$. This was not done: instead, you started with open covers for $A$ and $B$ separately, and used their compactness to produce a finite subcover.\n\nThe missing step : How do you know that a subcover of $A \\cup B$, is also a subcover of $A$ and a subcover of $B$? Or, how would you generate subcovers of $A$ and $B$, given one of $A \\cup B$? In short : you did not start with an arbitrary subcover. If you do so, it would have to be as explicit as in the answer I am going to give below for your betterment.\n\nHence, the answer should be as follows:\n\nStart from definition. Let $\\{ U_\\alpha\\}$ be an open cover of $A \\cup B$.\n\nSince $A \\cup B \\subset \\bigcup U_\\alpha$, it follows that $A \\subset \\bigcup U_\\alpha$, and similarly, $B \\subset \\bigcup U_\\alpha$. That is, every cover of $A \\cup B$ gives a cover of $A$ and a cover of $B$. This is the key step missing from your explanation.\n\nNow, we have finite subcovers for each one, and you can take the union of these subcovers (which remains a subcover of the original cover) to get the result.\n\nSo, in short, when we say arbitrary, we have to be careful. It is best that you start out with what is given: In compactness, we are given an open cover of the set, you never gave yourself that, instead you started with open covers for $A$ and $B$ separately, which was not given to you by the definition of $A \\cup B$ being compact.\n\nIt's all about being careful. And it's absolutely fine to mistakes. It's good you have this clarified now.\n\nAs an exercise, try and extend this to a finite union of compact sets. Do you think you can extend this to an infinite union of compact sets?\n\nYou assumed, without proving it, that every open cover of $A\\cup B$ is a union of an open cover of $A$ and an open cover of $B$.\n\nSuppose for example, that $A=[0,1]$ and $B$ is some other compact set, and $\\mathcal B = \\{ (0.4,1] \\cup G : G\\text{ is an open subset of } B. \\}.$ This is an open cover of $B$. Let $C=\\{ [0,0.6)\\cup G : G \\text{ is an open subset of } B. \\}.$ That is another open cover of $B$. Neither of them is an open cover of $A,$ but $\\mathcal B \\cup \\mathcal C$ is an open cover of $A\\cup B.$ Each has a finite subset that covers $B.$ Neither has a finite subset that covers $A$.\n\nYou could just say that any open cover of $A\\cup B$ is an open cover of $A$. So it has a finite subcover of $A$. Similarly find another finite subset of it that covers $B$. Then take the union of those two finite subsets. That actually makes the proof simpler. If you do it that way then there is no need to prove the proposition whose proof you omitted.\nPS: The needless assumption whose proof you omitted can be proved trivially: If $\\mathcal A$ is an open cover of $A\\cup B$ then $\\mathcal A \\cup \\mathcal A$ is a union of an open cover of $A$ and an open cover of $B$. However, this is a needless complication in the argument.", "date": "2022-09-25 20:41:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2212375/what-is-wrong-with-this-proof-of-the-union-of-two-compact-sets", "openwebmath_score": 0.8882701396942139, "openwebmath_perplexity": 83.62939255012229, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305360354471, "lm_q2_score": 0.8947894724152068, "lm_q1q2_score": 0.8618885431533745}}
{"url": "https://www.physicsforums.com/threads/mathematical-induction-problem.611898/", "text": "# Mathematical induction problem\n\n1. Jun 6, 2012\n\n1. The problem statement, all variables and given/known data\nHello! This is I want to prove using Mathematical Induction: $1+3+5+...+(2n-1)=n^2$. The problem is: I don`t understand very much about Mathematical Induction :(\n\n2. Relevant equations\n\n3. The attempt at a solution\nSuppose n=1. Then 1=1. Now suppose $1+3+5+...+(2n)=(n+1)^2$. Then $1+3+5+...+2n=(1+3+5+...+(2n-1))+2n=n^2+2n=n(n+2)$.\nIs this correct? If yes, how does this prove my hypothesis?\n\n2. Jun 6, 2012\n\nYou're only working with odd numbers, so the next term wouldn't be $2n$..\n\n3. Jun 6, 2012\n\nThis is what I should do to solve the problem $1+3+5+...+(2n-1+1)=1+3+5+...+2n$. Right? Because you always do $n+1$ in the Induction step.\n\n4. Jun 6, 2012\n\n### sacscale\n\nMathematical induction attempts to show that if the equation holds for n, then it also holds for n+1. It's easier to keep straight if you use a substitution such as n=k+1\n\n5. Jun 6, 2012\n\n@ DDarthVader: Yes, but you're not substituting $n + 1$ in for $2n - 1$ or anything, you're subbing it in for $n$...\n\n6. Jun 6, 2012\n\nDoing the substitution $n=k+1$ I've got this result:\n$1+3+5+...+2k+1=(1+3+5+...+(2n-1))+2k+1$\nThen\n$n^2 +2k+1 = (k+1)^2+2k+1 = k^2+2k+2+2k+1 = k^2+4k+3$\nBut since $n=k+1$ we got\n$(n-1)^2+4n-4+3 =n^2-2n+4n+1 =n^2+2n+1 = (n+1)^2$\nIs this correct?\n\nLast edited: Jun 6, 2012\n7. Jun 6, 2012\n\n### SammyS\n\nStaff Emeritus\nYour conjecture: $1+3+5+...+(2n-1)=n^2$\n\nYou've already done the base step, n = 1.\n\nFor the induction step:\n\nLet k \u2265 1. Assume that your conjecture is true for n = k. From that, show (prove) that your conjecture is true for n = k+1 .\n\nSo you need to show that $1+3+5+\\dots+(2k-1)+(2(k+1)-1)=(k+1)^2$ is true,\n\nstarting with $1+3+5+\\dots+(2k-1)=k^2\\ .$\n\nLast edited: Jun 6, 2012\n8. Jun 6, 2012\n\n### azizlwl\n\nAs you see here, (n+1)th value is not equal to 2n\nIt is equal to (2(n+1)-1)\n\n9. Jun 6, 2012\n\n### sacscale\n\nSubstituting n=k+1 into 2n-1 gives 2(k+1)-1.\n\n10. Jun 6, 2012\n\nDoing what you told me I've got this result:\n$1+3+5+...+(2n-1)=n^2$.\nWe assume n=k. Then we try to prove if the conjecture is true for n=k+1 and we obtain: $1+3+5+...+(2n-1)+(2n-1)=n^2$\nBut n=k+1\n$k^2+(2(k+1)-1)=(k+1)^2$\n$k^2+(2(k+1)-1)=k^2+2k+1$\n$2(k+1)-1=2k+1$\n$2(k+1)=2k+2$\n$2(k+1)=2(k+1)$\nSo I proved that the conjecture is also true for n=k+1. Right?\n\n11. Jun 6, 2012\n\n### Villyer\n\nIsn't it bad practice to manipulate both sides of an equation in a problem such as this?\n\n12. Jun 6, 2012\n\n### SammyS\n\nStaff Emeritus\nYes, somewhat indirectly.\n\nRather you should do something like the following.\n\nAssume $1+3+5+\\dots+(2k-1)=k^2\\ .$\n\nNow consider:\n\n$1+3+5+\\dots+(2k-1)+(2(k+1)-1)$\n$=k^2+(2(k+1)-1)$ \u2003\u2003... because of or assumption.\n\n$=k^2+2k+1$\n\n$=(k+1)^2$\u2003\u2003Which is what we needed to show for the inductive step.\u200b\n\nIt's not that what you did is particularly wrong, it's just that it's not real clear that you're not somehow assuming the very thing you should be proving.\n\nLast edited: Jun 6, 2012\n13. Jun 6, 2012\n\n### Muphrid\n\nIt's generally frowned upon to work both sides at once, yeah. Usually, one would take $k^2 + 2(k+1)-1$ and manipulate it until it's clear the result is $(k+1)^2$. Working only in one direction ensures that no illegal operations or cancellations are done, even though it may be necessary to set both sides equal just to figure out how to do it.\n\n14. Jun 6, 2012\n\nWell, I think I got it now! I have a list of mathematical induction to do. I'll try to solve the exercises using what you guys told me. I'll probably be back soon. Thanks guys!\n\n15. Jun 9, 2012\n\n### amrah\n\nProve by mathematical induction that n^3-n is divisible by 2 for all positive integral values of n.", "date": "2017-10-17 09:02:00", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/mathematical-induction-problem.611898/", "openwebmath_score": 0.8596104979515076, "openwebmath_perplexity": 1159.6228508232543, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305360354471, "lm_q2_score": 0.8947894717137996, "lm_q1q2_score": 0.8618885424777577}}
{"url": "http://esy-magnesy.pl/cvf3z54/archive.php?e72729=properties-of-matrix-multiplication-proof", "text": "Matrix transpose AT = 15 33 52 \u221221 A = 135\u22122 532 1 Example Transpose operation can be viewed as \ufb02ipping entries about the diagonal. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If $$A$$ is an $$m\\times p$$ matrix, $$B$$ is a $$p \\times q$$ matrix, and $$C$$ is a $$q \\times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. MATRIX MULTIPLICATION. Example. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Zero matrix on multiplication If AB = O, then A \u2260 O, B \u2260 O is possible 3. A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of transpose Notice that these properties hold only when the size of matrices are such that the products are defined. The first element of row one is occupied by the number 1 \u2026 In the next subsection, we will state and prove the relevant theorems. While certain \u201cnatural\u201d properties of multiplication do not hold, many more do. The determinant of a 4\u00d74 matrix can be calculated by finding the determinants of a group of submatrices. Even though matrix multiplication is not commutative, it is associative in the following sense. Example 1: Verify the associative property of matrix multiplication \u2026 proof of properties of trace of a matrix. i.e., (AT) ij = A ji \u2200 i,j. Associative law: (AB) C = A (BC) 4. 19 (2) We can have A 2 = 0 even though A \u2260 0. But first, we need a theorem that provides an alternate means of multiplying two matrices. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. The following are other important properties of matrix multiplication. For sums we have. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. Given the matrix D we select any row or column. Equality of matrices The proof of Equation \\ref{matrixproperties2} follows the same pattern and is \u2026 The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. $$\\begin{pmatrix} e & f \\\\ g & h \\end{pmatrix} \\cdot \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} ae + cf & be + df \\\\ ag + ch & bg + dh \\end{pmatrix}$$ Proof of Properties: 1. Let us check linearity. De\ufb01nition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, De\ufb01nition A square matrix A is symmetric if AT = A. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; A matrix is an array of numbers arranged in the form of rows and columns. Selecting row 1 of this matrix will simplify the process because it contains a zero. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). (3) We can write linear systems of equations as matrix equations AX = B, where A is the m \u00d7 n matrix of coefficients, X is the n \u00d7 1 column matrix of unknowns, and B is the m \u00d7 1 column matrix of constants. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let\u2019s look at them in detail We used these matrices Subsection MMEE Matrix Multiplication, Entry-by-Entry. 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Only when the size of matrices are such that the products are defined A theorem that provides an means. Be done on matrices called diagonal if all its elements outside the diagonal! = AI =A that IA = AI =A do not hold, many more do basic operations! Rows and columns For every square matrix is properties of matrix multiplication proof diagonal if all its elements outside the main diagonal are to.\n2020 properties of matrix multiplication proof", "date": "2021-10-20 03:42:15", "meta": {"domain": "esy-magnesy.pl", "url": "http://esy-magnesy.pl/cvf3z54/archive.php?e72729=properties-of-matrix-multiplication-proof", "openwebmath_score": 0.873639702796936, "openwebmath_perplexity": 330.20168726653935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305360354471, "lm_q2_score": 0.8947894618940992, "lm_q1q2_score": 0.8618885330191225}}
{"url": "https://math.stackexchange.com/questions/2993698/construct-a-square-given-a-point-and-two-lines", "text": "# Construct a square given a point and two lines.\n\nThe problem:\n\nOne point of the square - $$(2,5)$$, and lines $$x=1$$, $$x=6$$ are given (1 point of the square is on each given line), all points of the square are in the first quadrant.\nA square needs to be constructed with this information.\n\nHow would you solve this problem and thus determine if a square can even be constructed?\nI'm interested more in your thought process than the solution.\nHow do you approach a problem like this?\nWhich steps do you take and why?\n\nI've considered equating the distances between points (2,5) and (1, a) with the distance between (2,5) and (6,b).I don't think that anything can be done with this, maybe I have to find something else and make a system of equations?I've tried to find the distance between points (1, a) and (6,b) - that would be equal to sqrt(2)*n (n is the length of square's side), connect it with some distance (between the given and unknown point) but I can't extract one variable so I can insert it into the first equation.\n\n\u2022 Welcome to Math.SE! Typically, the community here likes to see your thoughts about the problem, as this helps answerers target their responses to your experience level, while avoiding wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) As you are interested more in thought process than solution, perhaps you can say something about your motivation: are you conducting research on problem-solving techniques? trying to be a better problem-solver? (doing homework that asks about process?) \u2013\u00a0Blue Nov 11 '18 at 12:00\n\u2022 I've edited the question.I'm interested in general problem solving techniques because I typically solve them by doing everything that can be done which certainly isn't an efficient way of solving problems. \u2013\u00a0JoeDough Nov 11 '18 at 12:12\n\u2022 Consider you found the square and draw the two horizontal lines from the vertexes not laying on the vertical lines you already have, the intersections of the four lines will give another figure, that can give many insights. \u2013\u00a0N74 Nov 11 '18 at 13:07\n\nLet $$A$$ be the given point, and let $$B$$ and $$C$$ be points on the respective given lines. The key observation is\n\nBecause $$\\overline{AB}$$ and $$\\overline{AC}$$ are congruent and perpendicular, the horizontal distance between $$A$$ and $$B$$ must equal the vertical distance between $$A$$ and $$C$$, and vice-versa.\n\nThe image shows that $$\\triangle ABP \\cong \\triangle ACQ$$, proving the observation immediately. The image also shows that there are clearly two matching choices for $$B$$ and $$C$$, so that there are two solution squares.\n\nThe idea also works when $$A$$ is not between the lines (note that how the $$B$$ and $$C$$ points match is reversed from above):\n\n(Of course, $$A$$ on a line works, too, as an obvious special case where $$C$$ and $$C^\\prime$$ coincide.) So, we see that it's always possible to construct the squares. $$\\square$$\n\nWe leave as an exercise to the reader the task of finding the coordinates of the points for the specific problem in the original question.\n\nAs for thought process ...\n\n\u2022 My first instinct is usually to abandon specific numbers and to generalize; this prevents important algebraic and geometric patterns from being lost in the arithmetic of numbers. So, instead of $$A=(2,5)$$, $$x=1$$, $$x=6$$, I wanted to consider $$A=(p,q)$$, $$x=r$$, $$x=s$$.\n\n\u2022 These days, I have a bit of a knee-jerk impulse to jump into Mathematica to try a brute-force solution. I quickly assigned some coordinates ---$$A=(p,q)$$, $$B=(r,b)$$, $$C=(s,c)$$--- and entered conditions for making the squares: $$\\overline{AB}\\perp\\overline{AC}$$ becomes $$(A-B)\\cdot(A-C)= 0$$, while $$\\overline{AB}\\cong\\overline{AC}$$ becomes $$(A-B)\\cdot(A-B)=(A-C)\\cdot(A-C)$$.\n\n\u2022 I let Mathematica do some instantaneous symbol-crunching, which in this case yielded linear constraints on the $$y$$-coordinates of $$b$$ and $$c$$. That told me the problem was actually easy, so I took a step back. The \"key observation\" above came to me pretty quickly ... shaming me because I didn't think of it sooner. :) (But, hey ... I only lost about two minutes!)\n\n\u2022 Then I went to the GeoGebra app to draw-up some diagrams that make the \"key observation\" obvious, and double-checking the point-outside-the-lines case. (I ofttimes go right into GeoGebra to experiment, but a first pass through Mathematica seemed quicker in this case.)\n\nHint:\nTwo parallel lines are: $$x=1$$ and $$x=6$$\ndistance between them is $$5$$ units.\nSo, you can construct a square having $$(1,5)$$ as one co-ordinate, and another lying on line $$x=6$$, opposite to $$(1,5)$$ ,i.e., $$(6,5)$$\n\nTake other $$2$$ points on lines $$x=1$$ and $$x=6$$ at a distance of $$5$$ units from $$2$$ points already taken.\n\nSo, there are $$2$$ possible squares: $$(1,5),(6,5),(1,0),(6,0)$$ and $$(1,5),(6,5),(1,10),(6,10)$$\n\n\u2022 I've made a mistake, first line isn't $x=2$ but $x=1$ \u2013\u00a0JoeDough Nov 11 '18 at 11:10\n\u2022 no problem, just shift the points, and now length of side will be 5 \u2013\u00a0pooja somani Nov 11 '18 at 11:11\n\u2022 @JoeDough is your point not (1,5) now? \u2013\u00a0pooja somani Nov 11 '18 at 11:16\n\u2022 But where's the point (2,5)?It's the only given point of the square, other 2 are somewhere on the given lines and 4th one has to be found.I've edited the question to be more clear. \u2013\u00a0JoeDough Nov 11 '18 at 11:17\n\u2022 ohk, then you will have to plot all points and proceed accordingly \u2013\u00a0pooja somani Nov 11 '18 at 11:18", "date": "2020-02-17 16:29:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2993698/construct-a-square-given-a-point-and-two-lines", "openwebmath_score": 0.8195118308067322, "openwebmath_perplexity": 470.7840879395588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864270133119, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8618794824429823}}
{"url": "https://gmatclub.com/forum/if-xy-xy-what-is-the-value-of-x-y-186800.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 18 Sep 2018, 18:37\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# If \u221a(xy) = xy, what is the value of x + y?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 15 Feb 2012\nPosts: 33\nIf root(xy) = xy what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Jul 2012, 14:05\n1\n9\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n60% (01:11) correct 40% (01:35) wrong based on 311 sessions\n\n### HideShow timer Statistics\n\nIf $$\\sqrt{xy} = xy$$ what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to zero\n\nWhat i did was,\n\n(XY)^1/2 = XY\nXY =(XY)^2\n\nso, I cancelled out XY and finally I got the below rephrased equation\n\nXY=1.\n\nBUT in the MGMAT explanation, I found that they are not cancelling out XY.\n\nBelow is their rephrased equation.\n\nXY = (XY)^2\n\nXY-(XY)^2=0\n\nXY [1-(XY)] = 0\n\nso, XY = 0 or XY = 1.\n\nMy question is why we are not cancelling out, and when we should use cancelling technique.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49206\nRe: If root(xy) = xy what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Jul 2012, 14:22\n3\n4\nIf $$\\sqrt{xy} = xy$$ what is the value of x + y?\n\n$$\\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.\n\n(1) x = -1/2 --> either $$-\\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\\frac{1}{2}$$ OR $$-\\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\\frac{5}{2}$$. Not sufficient.\n\n(2) y is not equal to zero. Clearly not sufficient.\n\n(1)+(2) Since from (2) $$y\\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\\frac{5}{2}$$. Sufficient.\n\nAs for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.\n\nNever reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.\n\nHope it's clear.\n_________________\n##### General Discussion\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8278\nLocation: Pune, India\nRe: If root(xy) = xy what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Jul 2012, 22:57\nResponding to a pm:\n\n\"If $$\\sqrt{XY} = XY$$ what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to zero\n\nWhat i did was,\n\n(XY)^1/2 = XY\nXY =(XY)^2\n\nso, I cancelled out XY and finally I got the below rephrased equation\n\nXY=1.\n\nBUT in the MGMAT explanation, I found that they are not cancelling out XY.\n\nBelow is their rephrased equation.\n\nXY = (XY)^2\n\nXY-(XY)^2=0\n\nXY [1-(XY)] = 0\n\nso, XY = 0 or XY = 1.\n\nMy question is why we are not cancelling out and when we should use cancelling technique.\"\n\nFor the time being, forget this question. Look at another one.\n\nWhich values of x satisfy this equation: $$x^2 = x$$\nLet's say we cancel out x from both sides. What do we get? x = 1.\nSo we get that x can take the value 1.\n\nBut is your answer complete? I look at the equation and I say, 'x can also take the value 0.' Am I wrong? No.\nx = 0 also satisfies your equation. So why didn't you get it using algebra? It is because you canceled x.\nLet me treat this equation differently now.\n\n$$x^2 = x$$\n$$x^2 - x = 0$$\n$$x * (x - 1) = 0$$\nx = 0 OR (x - 1) = 0 i.e. x = 1\n\nNow I get both the possible values that x can take. I do not lose a solution.\n\nWhen you cancel off a variable from both sides of the equation, you lose a solution so you should not do that. You can cancel off constants of course.\nRule of thumb: Do not cancel off variables. Take them common. In some cases, it may not matter even if you do cancel off but either ways, your answer will not be incorrect of you don't cancel. On the other hand, sometimes, your solution could be incomplete if you do cancel and that's a problem.\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nIntern\nJoined: 12 Jul 2012\nPosts: 21\nRe: If root(xy) = xy what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n12 Jul 2012, 03:00\nHere is my solution\n\nTake square of both sides:\n(root(xy) )^2 = (xy)^2\nxy = (xy)^2\nWith Option (1): x = -1/2\n-1/2y = 1/4 * y^2\ny = 4 * (-1/2)\ny = -2\nHence, X+Y = -1/2 - 2 = -5/2.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49206\nRe: If root(xy) = xy what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n12 Jul 2012, 03:06\nRamakantPareek wrote:\nHere is my solution\n\nTake square of both sides:\n(root(xy) )^2 = (xy)^2\nxy = (xy)^2\nWith Option (1): x = -1/2\n-1/2y = 1/4 * y^2\ny = 4 * (-1/2)\ny = -2\nHence, X+Y = -1/2 - 2 = -5/2.\n\nThere is another value of $$y$$ apart -2 which satisfies $$-\\frac{1}{2}*y=\\frac{1}{4}*y^2$$, namely $$y=0$$.\n\nComplete solution here: if-root-xy-xy-what-is-the-value-of-x-y-135646.html#p1103625\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49206\nIf \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2014, 16:01\n4\n2\nIntern\nJoined: 18 Dec 2013\nPosts: 28\nGPA: 2.84\nWE: Project Management (Telecommunications)\nIf \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2014, 20:57\n1\nBunuel wrote:\n\nTough and Tricky questions: Algebra.\n\nIf \u221a(xy) = xy, what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to 0\n\nStatement 1: $$X=-1/2$$.... substitute in main equation\n\nScenario A: $$Y = -2/1$$ is a reciprocal\n$$\\sqrt{xy}=xy$$\nLHS =$$\\sqrt{-1/2*-2/1}= \\sqrt{1}= 1$$\nRHS =$$-1/2*-2/1= -1*-1 = 1$$\nTherefore, LHS = RHS\n\nScenario B: $$Y = 0$$\nLHS =$$\\sqrt{-1/2*-0}= \\sqrt{0}= 0$$\nRHS =$$-1/2*0= 0$$\nTherefore again, LHS = RHS\n\nHence, we have two possible solutions, therefore Statement 1 is insufficient\n\nStatement 2 : $$Y$$ is not equal to zero\nBut, X could be equal to zero or be the reciprocal of Y, therefore, statement 2 would fall under the same scenarios as statement 1\n\nHence, we have two possible solutions, therefore Statement 2 is insufficient\n\nBoth Statement 1 & 2 together confirms that $$X & Y$$ both are not equal to zero, therefore, they have to be reciprocals and since statement gives us the value of $$X=-1/2$$, we can also the value of $$Y=-2/1$$ (Reciprocal of X) and thereafter we can find the value of $$X + Y = -1/2 + -2/1 = -5/2$$\n\nHence, both together are sufficient, therefore C is the correct Answer!\n\nManager\nJoined: 21 Jul 2014\nPosts: 126\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2014, 21:26\n1\nBunuel wrote:\n\nTough and Tricky questions: Algebra.\n\nIf \u221a(xy) = xy, what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to 0\n\nI did something similar to DMMK, except for one thing:\nI know $$\\sqrt{n}$$ = $$n$$ only when $$n$$ = 0 or $$n$$ = 1. For all other values of $$n$$, the two would not be equal. Knowing this made it much simpler to plug in the statements. I just had to determine if from the statement(s) provided, can I rule out xy = 0 or xy=1.\n\nStament 1:\nKnowing that x = -1/2, y could equal 0 or -2, and still make the premise true. Either value of y would make x + y a different value. Therefore, it is insufficient.\n\nStatement 2:\nKnowing that y is not equal to 0, x could equal zero or the inverse of y, and still make the premise true. Either value of x would make x + y a different value. Therefore, it is insufficient.\n\nNow to evaluate both statements together.\n\nThe reason we rejected statement 1 by itself was because y could equal one of two possible values. Statement 2 eliminates one of those options. Therefore y must equal -2. And therefore both statements together are sufficient to answer \"what is the value of x+y.\"\nSenior Manager\nJoined: 13 Jun 2013\nPosts: 277\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2014, 22:12\n1\nBunuel wrote:\n\nTough and Tricky questions: Algebra.\n\nIf \u221a(xy) = xy, what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to 0\n\nlet xy=k, thus\nk^(1/2) =k;\nsquaring both sides we have\nk=k^2\nk(k-1)=0\nk=0 or 1\ni.e. xy=0 or xy=1\n\nst.1\n\nfor x=-1/2\nboth y=0 and y=-2 satisfy the possible value of xy i.e. 0 or 1\nhence not sufficient\n\nst.2\n\ny is not equal to zero. clearly not sufficient.\nas nothing is said about x, therefore x can take any value it can be zero, fraction, integer etc.\n\ncombining st.1 and st.2\n\nwe know that y cannot be equal to zero. thus y=-2 and x=-1/2\nand x+y= -2-(1/2)=-5/2\n\nhence sufficient.\nIntern\nJoined: 02 Jan 2015\nPosts: 32\nGMAT Date: 02-08-2015\nGPA: 3.7\nWE: Management Consulting (Consulting)\nRe: If root(xy) = xy what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n04 Jul 2015, 08:43\nWhy can't x and y both be -1? Thanks\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49206\nRe: If root(xy) = xy what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n05 Jul 2015, 08:19\nElCorazon wrote:\nWhy can't x and y both be -1? Thanks\n\nFrom the stem x = y = -1 is possible, in this case xy = 1 (check my solution above). But then the first statement says that x = -1/2, thus these values are no longer possible.\n\nHope it's clear.\n_________________\nSenior Manager\nJoined: 20 Aug 2015\nPosts: 392\nLocation: India\nGMAT 1: 760 Q50 V44\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n16 Nov 2015, 03:02\nBunuel wrote:\n\nTough and Tricky questions: Algebra.\n\nIf \u221a(xy) = xy, what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to 0\n\nGiven: \u221a(xy) = xy\n(xy)^2 - xy = 0\nxy = 0 or 1 - (i)\n\nRequired: x + y = ?\n\nStatement 1: x = -1/2\nFrom the given equation (i), we can have two different values of y.\nHence two different values of x + y\nINSUFFICIENT\n\nStatement 2: y is not equal to 0\nNo information about x\nINSUFFICIENT\n\nCombining Statement 1 and Statement 2: x = -1/2 and y is not equal to 0\nFrom (i), xy cannot be 0 since both x and y are not = 0\nHence xy = 1\ny = -2\nx + y = $$-\\frac{5}{2}$$\nSUFFICIENT\n\nOption C\nSenior CR Moderator\nStatus: Long way to go!\nJoined: 10 Oct 2016\nPosts: 1388\nLocation: Viet Nam\nIf \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n14 Apr 2017, 07:13\nBunuel wrote:\n\nTough and Tricky questions: Algebra.\n\nIf \u221a(xy) = xy, what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to 0\n\nThe trick here is from $$\\sqrt{xy}=xy$$ anyone could easily deduce that $$xy=1$$ or $$xy=0$$ and just consider one of these cases.\n\nHere is my solution:\n$$\\sqrt{xy}=xy \\implies \\sqrt{xy}-xy=0 \\implies \\sqrt{xy}(\\sqrt{xy}-1)=0$$\nHence we have $$xy=0$$ or $$xy=1$$.\n\n(1) If $$x=-\\frac{1}{2}$$, we need to consider two cases:\n\nCase 1: If $$xy=0\\implies y=0 \\implies x+y = -\\frac{1}{2}$$\n\nCase 2: If $$xy=1 \\implies y=-2 \\implies x+y = -\\frac{5}{2}$$\n\nIt's clear that (1) is insufficient.\n\n(2) If $$y \\neq 0$$, we still need to consider two cases:\n\nCase 1: If $$xy=0\\implies x=0 \\implies x+y = y$$.\n\nIf $$y=1 \\implies x+y = 1$$\nIf $$y=2 \\implies x+y = 2$$\n\nHence, (2) is insufficient.\n\nNow combine (1) and (2):\nSince $$x \\neq 0$$ and $$y \\neq 0$$ hence $$xy \\neq 0 \\implies xy=1$$\n\nSince $$x= -\\frac{1}{2} \\implies y = -2 \\implies x+y = -\\frac{5}{2}$$\n\nHence (1) & (2) are sufficient.\nThe answer is C.\n_________________\nDirector\nJoined: 12 Nov 2016\nPosts: 760\nLocation: United States\nSchools: Yale '18\nGMAT 1: 650 Q43 V37\nGRE 1: Q157 V158\nGPA: 2.66\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n10 Sep 2017, 13:41\nBunuel wrote:\n\nTough and Tricky questions: Algebra.\n\nIf \u221a(xy) = xy, what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to 0\n\nrewrite\n\nxy= (xy)^2\n\nSt 1\n\nReciprocal property.\n\n-1/2(-2) =1\n[-1/2(-2)]^2 =1\n\ninsuff\n\nSt 2\n\nEliminating the possibility of 0 from Y still leaves the possibility of x being 0 or some other numbers which leads to several possibilities.\n\ninsuff\n\nSt 1 and St 2\n\nEliminates possibility of y=-2\n\nC\nVP\nStatus: It's near - I can see.\nJoined: 13 Apr 2013\nPosts: 1256\nLocation: India\nConcentration: International Business, Operations\nGMAT 1: 480 Q38 V22\nGPA: 3.01\nWE: Engineering (Consulting)\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Dec 2017, 00:35\nBunuel wrote:\nIf $$\\sqrt{xy} = xy$$ what is the value of x + y?\n\n$$\\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.\n\n(1) x = -1/2 --> either $$-\\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\\frac{1}{2}$$ OR $$-\\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\\frac{5}{2}$$. Not sufficient.\n\n(2) y is not equal to zero. Clearly not sufficient.\n\n(1)+(2) Since from (2) $$y\\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\\frac{5}{2}$$. Sufficient.\n\nAs for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.\n\nNever reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.\n\nHope it's clear.\n\nIs it safe to say : x^2 = x only when x = 0 or 1\n_________________\n\n\"Do not watch clock; Do what it does. KEEP GOING.\"\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49206\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Dec 2017, 00:39\nQZ wrote:\nBunuel wrote:\nIf $$\\sqrt{xy} = xy$$ what is the value of x + y?\n\n$$\\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.\n\n(1) x = -1/2 --> either $$-\\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\\frac{1}{2}$$ OR $$-\\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\\frac{5}{2}$$. Not sufficient.\n\n(2) y is not equal to zero. Clearly not sufficient.\n\n(1)+(2) Since from (2) $$y\\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\\frac{5}{2}$$. Sufficient.\n\nAs for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.\n\nNever reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.\n\nHope it's clear.\n\nIs it safe to say : x^2 = x only when x = 0 or 1\n\nYes.\n\nx^2 = x;\n\nx^2 -x = 0;\n\nx(x - 1) = 0;\n\nx = 0 or x = 1.\n_________________\nIntern\nJoined: 30 Jul 2017\nPosts: 20\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Feb 2018, 10:18\nHi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49206\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n27 Feb 2018, 10:27\n1\nkrikre wrote:\nHi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2\n\nDoesn't (1) say that x = -1/2? How it can be 1 then?\n_________________\nManager\nJoined: 23 Sep 2016\nPosts: 225\nRe: If \u221a(xy) = xy, what is the value of x + y?\u00a0 [#permalink]\n\n### Show Tags\n\n28 Feb 2018, 01:19\nprakash111687 wrote:\nIf $$\\sqrt{xy} = xy$$ what is the value of x + y?\n\n(1) x = -1/2\n(2) y is not equal to zero\n\nWhat i did was,\n\n(XY)^1/2 = XY\nXY =(XY)^2\n\nso, I cancelled out XY and finally I got the below rephrased equation\n\nXY=1.\nthere are various cases that are possible as following\n\nBUT in the MGMAT explanation, I found that they are not cancelling out XY.\n\nBelow is their rephrased equation.\n\nXY = (XY)^2\n\nXY-(XY)^2=0\n\nXY [1-(XY)] = 0\n\nso, XY = 0 or XY = 1.\n\nMy question is why we are not cancelling out, and when we should use cancelling technique.\n\nIMO C\n$$\\sqrt{xy} = xy$$\n1.x=0 ,y = anything\n2.Y=0 , x=anything\n3. both x and y 1 or -1 or x=1 and y=-1 or vise versa\n4.x=1/z and y=z *any integer except 0\nlets come to statement 1.\nx=-1/2 y can be -2 or 0 insufficient.\n2nd statement y not equal to 0 but don't know about x\ncombining both case with 0 is eliminated only one case left so C is the answer\nRe: If \u221a(xy) = xy, what is the value of x + y? &nbs [#permalink] 28 Feb 2018, 01:19\nDisplay posts from previous: Sort by\n\n# If \u221a(xy) = xy, what is the value of x + y?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-19 01:37:45", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-xy-xy-what-is-the-value-of-x-y-186800.html", "openwebmath_score": 0.7670589685440063, "openwebmath_perplexity": 2282.4621331617436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9161096181702032, "lm_q2_score": 0.9407897513338646, "lm_q1q2_score": 0.8618665398729071}}
{"url": "https://mathematica.stackexchange.com/questions/184374/why-is-sum-limits-k-1100-0-01-different-than-sum-limits-k-1100", "text": "# Why is $\\sum\\limits_{k=1}^{100} 0.01$ different than $\\sum\\limits_{k=1}^{100} \\frac{1}{100}$?\n\nWhy does evalutating $$\\sum\\limits_{k=1}^{100} 0.01$$ not result in 1?\n\nWhat I get is 1.0000000000000007.\n\nI know that the number 0.01 in binary results in an infinite sequence of zeros and ones, which is truncated when stored in a memory as a floating point number. So information is lost and the result of the sum will no longer be exactly 1.\n\nHowever, with $$\\sum\\limits_{k=1}^{100} \\frac{1}{100}$$, I do get exactly 1.\n\nCould someone explain why?\n\n\u2022 For your own edification, compare the results of Total[Table[0.01, {100}]] and Total[Table[0.01, {100}], Method -> \"CompensatedSummation\"]. \u2013\u00a0J. M.'s torpor Oct 22 '18 at 6:21\n\u2022 @J.M.iscomputer-less Interestingly Total seems to perform much better than Sum, even without \"CompensatedSummation\". It is somehow more intelligent. Consider ListLinePlot[{Table[Sum[0.01, {i, n}] - n/100, {n, 1, 100}], Table[Total[Table[0.01, {n}]] - n/100, {n, 1, 100}]}] \u2013\u00a0kirma Oct 22 '18 at 9:03\n\nWelcome to the world of Mathematica numerics.\n\nMathematica offers three arithmetic systems for evaluating numerical expressions\n\n\u2022 Numerics based on exact numbers. Such computations will always give an exact result, but require all the numbers involved to be integers or rational numbers. Your second expression qualifies as an exact computation and, thus, produces the exact result 1.\n\n\u2022 Numerics based on arbitrary precision numbers. This allows you the precision a computation to specified value. To work, such computations require the numbers involved to be integers, rational numbers or arbitrary precision having a precision greater or equal to the specified precision.\n\n\u2022 Numerics based on machine floating point numbers. This is used whenever a real number, which does not qualify as an arbitrary precision number, appears in a computation. Machine numerics give the fastest computation, but since there is no control over precision. For long computations precision will be lost, sometimes drastically (leading to zero precision \u2013 i.e., garbage out). Your first expression falls into this category, but retains good precision. However, when you display all the digits in the result the inexactness shows.\n\nIn you specific example, since the deviation from the exact result is small, you can recover the exact result with Rationalize\n\nExample\n\nSum[.01, {100}] // FullForm\n\n1.0000000000000007\n\n\nRationalized example\n\nSum[.01, {100}] // Rationalize // FullForm\n\n\n1\n\nMathematica treats 0.01 as a machine precision floating point number, while it treats 1/100` as an exact rational quantity. Thus the first sum follows all the grody rules of floating point arithmetic that programmers have come to know and hate.\n\nThere are ways to make Mathematica use more digits, though.", "date": "2021-07-31 08:44:05", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/184374/why-is-sum-limits-k-1100-0-01-different-than-sum-limits-k-1100", "openwebmath_score": 0.48589029908180237, "openwebmath_perplexity": 1252.038927713244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9585377237352756, "lm_q2_score": 0.899121379297294, "lm_q1q2_score": 0.8618417602733495}}
{"url": "https://math.stackexchange.com/questions/2499649/proof-of-obtaining-multiple-of-10-by-using-arithmetic-operations-on-any-three-nu", "text": "Proof of obtaining multiple of 10 by using arithmetic operations on any three numbers\n\nWhile doing a few math puzzles I noted that you could take any three numbers (integers) and use basic arithmetic operations (Addition, Subtraction, Multiplication and Parentheses only) to obtain a multiple of 10, using each number only once.\n\nEg.\nUsing 29, 73, 36: $73+36-29=80$\nUsing 2, 4, 7: $2*7-4=10$\n\nIs there a proof for this theory in particular, or is there a more general theorem that applies to this? If not, is there a set of three numbers which disproves this theory?\n\nI found a few clues while solving this:\n\nIf one of the numbers is a multiple of 5 (call it $x$) then there definitely exists a solution divisible by 10. This one is easily provable as you merely need an even number to multiply with. If any one of the remaining two numbers is even, you can use that to multiply with $x$ to get number divisible by 10. If both numbers are odd then you can add them to get an even number to multiply with $x$.\n\nIt would seem as only the units place digits have any bearing on whether it's a multiple of 10 or not. As you could take any of the examples, add or subtract any multiple of 10 from it and do the same operations to still get a multiple of ten. I feel this is also easily provable but I can't think of a way to do it which isn't long-winded.\nEDIT: It can be proven as such, let $a,b$ be two integers s.t $a+b|10$\nAdding two multiples of 10 ($10m,10n$) to $a$ and $b$ we get $10m+a+10n+b$\nSince $a+b|10$ it can be written as $10m+10n+10k$ where $a+b=10k$\nWhich is divisible by ten\nNow let $a,b$ be two integers s.t. $a.b|10$\nAdding $10m,10n$ to $a$ and $b$ we get $(10m+a)(10n+b)$\n$=100mn+10an+10bm+ab$ or $100mn+10an+10bm+10k$\nWhich is divisible by ten\nTherefore any combination of Addition and Multiplication with the three no.s shouldn't matter.\nI realise that could replace 10 with any no. in this proof and it would still work. So we really need to just prove for every single digit no.\n\nI ran a program in python that checked for every combination of single digit no.s, but it didn't find any combination that disproves this.\n\nI'm not sure how this question would be categorised, hence the lack lustre tags.\n\nI'm fairly new to StackExchange. Please forgive me if I've worded this question poorly.\n\n\u2022 That's a very nice observation that only the units digit has any bearing, and is essentially the underlying idea of \"modular arithmetic.\" This is an interesting problem, and I'll be fairly surprised if it's true. However, there are $3$ ways to order $a, b, c$, a couple choices of symbols between $a,b$ and $b,c$ and with the ability to use parentheses... very interesting! If we have the three numbers $a, b, c$, are we allowed to order them any way we like? \u2013\u00a0pjs36 Nov 5 '17 at 6:47\n\u2022 How do you get a multiple of $10$ this way using $1$, $2$, and $4$? \u2013\u00a0alex.jordan Nov 5 '17 at 7:58\n\u2022 @alex.jordan $2\\cdot (4+1)$ \u2013\u00a0Kyle Miller Nov 5 '17 at 7:59\n\u2022 Oh, I see. I somehow thought only addition and subtraction were in play. Didn't read well. \u2013\u00a0alex.jordan Nov 5 '17 at 8:00\n\u2022 @pjs36 From the two examples in the OP, it seems clear that reordering is permitted. \u2013\u00a0Erick Wong Nov 5 '17 at 8:07\n\nLet's collect some of the useful observations mentioned and implicit in the OP:\n\n1. Only the residue class of $a,b,c$ mod $10$ matters.\n2. If any of $a,b,c$ is divisible by $5$ then there is a solution.\n3. If any subset of $\\{a,b,c\\}$ has a solution, then so does $\\{a,b,c\\}$.\n\nAs a consequence of the above, we may assume WLOG that $\\{a,b,c\\}$ are distinct single digits from $1,2,3,4,6,7,8,9$.\n\nWith a little more effort, we can justify that $a$ can be replaced by $-a$, which will let us further restrict to the digits $1,2,3,4$. It's not too hard to be convinced of this by playing around with some examples, but let's give a proper proof by structural induction. More specifically, we will prove:\n\nProposition: Let $f(x_1,\\ldots,x_n)$ be a fully parenthesized expression using each $x_i$ exactly once and only $+,-,*$ operators. Then at least one of $-f$ or $f$ can be written as a similar expression using $-x_1,x_2,\\ldots,x_n$ exactly once.\n\nThis is trivial for $n=1$, and for $n=2$ we quickly verify each of the four possible expressions of two variables: $$-(a+b) = (-a)-b,\\\\-(a-b) = (-a)+b,\\\\(b-a) = b+(-a),\\\\-(a*b) = (-a)*b.$$\n\nWe can now do a structural induction: suppose $n\\ge 3$. By extracting the highest-level operator of $f(x_1,\\ldots, x_n)$, we may write $f$ as $h(g_1(\\cdots),g_2(\\cdots))$, where each of $g_1,g_2,h$ are valid expressions, and the arguments of $g_1, g_2$ partition the set $\\{x_1,\\ldots,x_n\\}$. I've deliberately obscured the arguments because the notation becomes unwieldy, as the idea is better illustrated by a simple example:\n\nIf $f(x_1,x_2,x_3,x_4,x_5) = ((x_1 * x_5) + x_2) - (x_3 - x_4)$, then we take $$h(a,b) = a-b,\\quad g_1 = (x_1 * x_5) + x_2,\\quad g_2 = x_3 - x_4.$$\n\nNote that $h,g_1,g_2$ each take $<n$ arguments so we could apply the inductive hypothesis to any of them as desired. Now, $x_1$ belongs to exactly one of $g_1$ or $g_2$. By adjusting $h$, we can assume WLOG that it belongs to $g_1$. By hypothesis, either $-g_1$ or $g_1$ can be written in terms of $-x_1$ and the remaining arguments of $g_1$. If it is $g_1$ that may be so written, then we are done since $f=h(g_1,g_2)$ can be written in terms of $-x_1,x_2,\\ldots,x_n$. Else, apply the induction hypothesis again to $h$ to see that either $f=h(g_1,g_2)$ or $-f=-h(g_1,g_2)$ can be written in terms of $-g_1$ and $g_2$, and thus also in terms of $-x_1,x_2,\\ldots,x_n$.\n\nThe above proposition lets us freely replace $9$ by $-9$ (which is equivalent to $1$), etc., and so we can narrow down the set $\\{a,b,c\\}$ to a subset of $\\{1,2,3,4\\}$. Lucky for us, there's only four such subsets:\n\n$$(1 + 2 - 3) = 0,\\\\ (1 + 4)*2 = 10,\\\\ (1 + 3 -4) = 0,\\\\ (2+3)*4 = 20.$$\n\n\u2022 I was going to write the same thing. I believe this one deserves the bounty. \u2013\u00a0user334639 Nov 5 '17 at 9:00\n\u2022 Very nicely done, I probably will award the bounty to this answer (I want to keep the bounty alive to reward the nice question; OP had but a single upvote when I set the bounty). Structural induction isn't something I'm familiar with, so I will admit that it's my least favorite part. I notice that in each of the four ultimate cases, addition is used -- I wonder if we could somehow combine that with factoring out a negative, and avoid structural induction. Probably not. At any rate, very nice! \u2013\u00a0pjs36 Nov 6 '17 at 2:36\n\u2022 Ah I see you've used $-4,-3,-2,-1,1,2,3,4$ instead of $1,2,3,4,6,7,8,9$ and then used my modulo(10) proof.First time I'm coming across structural induction so the proof is a bit hard to chew. Nevertheless, it's a clever proof. \u2013\u00a0Carlton Banks Nov 6 '17 at 10:02\n\u2022 @CarltonBanks Thanks, to be fair this is really just an induction on the number of terms (it\u2019s structural in the sense that arithmetic expressions aren\u2019t built up by just \u201cadding one more term\u201d). It is also overkill when we\u2019re dealing with only 3 terms, but you did express an interest in more general principles in the OP. Finally, the induction is only needed for a relatively minor technical point: it justifies that if we can find a $0$ using unary $-$ (which doesn\u2019t appear to be allowed), then we can also find one without any unary $-$. \u2013\u00a0Erick Wong Nov 6 '17 at 18:24\n\nIf you have directly verified this for all triples of one-digit numbers, then you have proved it. Because you only care about an arithmetic combination that makes $0$ mod $10$, so proving it for one-digit numbers proves the greater claim.\n\nWithout a direct verification, consider all $10^3$ triples of one-digit numbers. If $0$ is in the triple, then multiplying all three is a multiple of $10$.\n\nSo now consider all $9^3$ triples from 1--9. If $5$ is in the triple, then either both of the other numbers are odd and you can multiply $5(\\text{odd}+\\text{odd})$ to get a multiple of $10$; or at least one of the others is even and you can multiply all three to get a multiple of $10$.\n\nSo now consider all $8^3$ triples from 1--4,6--9. If any number appear twice in the triple, the difference is $0$, and that difference times the third number is a multiple of $10$.\n\nSo now consider all $\\binom{8}{3}$ triples from 1--4,6--9 without repetition. If a number and its complement mod $10$ are both in the triple, then sum those and multiply by the third to get a multiple of $10$.\n\nSo now consider all $\\binom{4}{3}\\cdot2^3$ triples from 1--4,6--9 without repetition where no two numbers sum to $10$. We are down to only $32$ triples to consider, and inspecting them directly is not so bad. First, consider the $16$ triples with two or three members from 1--4.\n\n$$\\underline{12}3\\ \\color{magenta}{\\underline{1}2\\underline{4}}\\ \\color{orange}{\\underline{1}2\\underline{6}}\\ \\color{blue}{127}\\ \\underline{13}4\\ \\color{blue}{136}\\ 1\\underline{38}\\ 1\\underline{47}\\ \\color{magenta}{\\underline{14}8}\\ \\color{magenta}{\\underline{23}4}\\ \\color{magenta}{\\underline{23}6}\\ 2\\underline{39}\\ \\color{orange}{\\underline{2}4\\underline{7}}\\ \\color{orange}{2\\underline{49}}\\ \\color{orange}{\\underline{3}4\\underline{8}}\\ 3\\underline{49}$$\n\nBlue sum to $0$ mod $10$.\n\nMagenta have an (underlined) pair that sum to $5$ (or $15$) and the third number is even, so that sum times the third number is a multiple of $10$.\n\nOrange have an (underlined) pair whose difference is $5$ and the third number is even, so that difference times the third number is a multiple of $10$.\n\nThe remaining black all have an (underlined) pair that sums to the third (mod $10$) so adding that pair and then subtracting the third makes a multiple of $10$.\n\nNote that if we negate all members of one of these triples, we get the triples with two or more in 6--9. The same operations give a multiple of $10$ since we only ever either add and subtract [with the blue and black], or add/subtract and then multiply by an even number [with magenta and orange]. So we've indirectly checked all $32$ of the lingering cases.\n\n\u2022 I was hoping to avoid \"too many\" cases, but this is a pretty modest number of them. And inspecting the final 32 wasn't so bad, at least when they were color-coded :) Would that I could split a bounty, I like all of the answers... \u2013\u00a0pjs36 Nov 6 '17 at 2:40\n\u2022 @pjs36 Just FYI, your comment somewhere else got me thinking about how the cases could be cut down to just $16$. \u2013\u00a0alex.jordan Nov 6 '17 at 3:04\n\nNote: This answer has the focus to reduce the number of variants which are to study by consequently using modular arithmetic.\n\n\u2022 $a,b,c\\in\\mathbb{Z}$\n\nWe are looking for multiples of $10$ when doing addition, subtraction and multiplication. Since we have \\begin{align*} (a\\, \\mathrm{mod}\\, (10)) + (b\\, \\mathrm{mod}\\, (10)) &\\equiv (a+b)\\, \\mathrm{mod}\\, (10) \\\\ (a\\, \\mathrm{mod}\\, (10)) - (b\\, \\mathrm{mod}\\, (10)) &\\equiv (a-b)\\, \\mathrm{mod}\\, (10) \\\\ (a\\, \\mathrm{mod}\\, (10)) \\cdot (b\\, \\mathrm{mod}\\, (10)) &\\equiv (a\\cdot b)\\, \\mathrm{mod}\\, (10) \\\\ \\end{align*}\n\nit is sufficient to consider $\\color{blue}{a,b,c\\in\\{0,1,2,3,4,5,6,7,8,9\\}}$.\n\n\u2022 $a,b,c\\in\\{0,1,2,3,4,5,6,7,8,9\\}$\n\nIf two of the three numbers are congruent modulo $10$, let's say $a\\equiv b\\, \\mathrm{mod}\\, (10)$ we obtain \\begin{align*} \\color{blue}{(a-b)\\cdot c}\\equiv 0\\cdot c\\color{blue}{\\equiv 0\\, \\mathrm{mod}\\, (10)} \\end{align*}\n\nand we are done. In the following we may WLOG assume: $a<b<c$\n\n\u2022 $a,b,c\\in\\{0,1,2,3,4,5,6,7,8,9\\},a<b<c$\n\nIf one of them, let's say $a$ is zero we obtain \\begin{align*} \\color{blue}{a\\cdot b\\cdot c}&\\equiv 0\\cdot b\\cdot c \\color{blue}{\\equiv 0\\, \\mathrm{mod}\\, (10)} \\end{align*}\n\nand we are done.\n\n\u2022 $a,b,c\\in\\{1,2,3,4,5,6,7,8,9\\}, a<b<c$\n\nIf one of them, let's say $a$ is equal to $5$ we consider two cases.\n\nFirst case: One of $b$ or $c$ is even. Let's assume $b$ is even. It follows \\begin{align*} a&\\equiv 0\\, \\mathrm{mod}\\, (5)\\\\ b&\\equiv 0\\, \\mathrm{mod}\\, (2)\\\\ ab&\\equiv 0\\, \\mathrm{mod}\\, (10)\\\\ \\end{align*} and we obtain \\begin{align*} \\color{blue}{a\\cdot b\\cdot c}&\\equiv 0\\cdot c \\color{blue}{\\equiv 0\\, \\mathrm{mod}\\, (10)} \\end{align*}\n\nSecond case: Both, $b$ and $c$ are odd. It follows from\n\n\\begin{align*} b\\equiv 1\\, \\mathrm{mod}\\, (2)\\\\ c\\equiv 1\\, \\mathrm{mod}\\, (2)\\\\ (b+c)\\equiv 0\\, \\mathrm{mod}\\, (2)\\\\ \\end{align*}\n\nand we obtain \\begin{align*} \\color{blue}{a\\cdot (b+c)} \\color{blue}{\\equiv 0\\, \\mathrm{mod}\\, (10)} \\end{align*}\n\n\u2022 $a,b,c\\in\\{1,2,3,4,6,7,8,9\\}, a<b<c$\n\nSince\n\n\\begin{align*} 1\\equiv (1-10)\\equiv (-9)\\, \\mathrm{mod}\\, (10)\\\\ 2\\equiv (2-10)\\equiv (-8)\\, \\mathrm{mod}\\, (10)\\\\ 3\\equiv (3-10)\\equiv (-7)\\, \\mathrm{mod}\\, (10)\\\\ 4\\equiv (4-10)\\equiv (-6)\\, \\mathrm{mod}\\, (10)\\\\ \\end{align*}\n\nwe can restrict the attention to $\\color{blue}{a,b,c\\in\\{1,2,3,4\\},a<b<c}$.\n\n\u2022 $a,b,c\\in\\{1,2,3,4\\}, a<b<c$\n\nConclusion: Doing arithmetic with three integers $a,b,c\\in\\mathbb{Z}$, each of them occurring once and using one or more of the operations addition, subtraction, multiplication and negation ($a \\rightarrow -a$) we can always obtain an integer value which is a multiple of $10$.\n\n\u2022 This is a nice, thorough answer. Allowing negation is an interesting point. While not explicitly mentioned, I feel like there should be some algebraic trickery that lets us convert negation to addition, subtraction, and multiplication with our same three numbers; e.g., $(-b + a) \\cdot c = -(b - a)\\cdot c$. Maybe, but maybe it would just serve to introduce more cases, even though it's intuitively clear that negation doesn't really add anything new. \u2013\u00a0pjs36 Nov 6 '17 at 2:48\n\u2022 @pjs36: Thanks. :-) Allowing negation is a modest way of cheating somewhat. In fact negation $-a=0-a$ is equivalent to add 0 to $\\{+,-,\\cdot,(,)\\}$. To me it is not so clear that it doesn't introduce anything new. Keep in mind that $-(a-b)$ for instance also uses negation. \u2013\u00a0Markus Scheuer Nov 6 '17 at 7:16\n\u2022 @pjs36 I agree, this is where I spend most of the effort in my answer and I would love to see a simpler way to prove that negation can always be eliminated :). \u2013\u00a0Erick Wong Nov 6 '17 at 18:27\n\u2022 @MarkusScheuer It\u2019s fine that $-(a-b)$ uses negation: the point of my argument is that all interior negations can be bubbles up to the exterior. Once we have that, then $a-b$ works just as well as $-(a-b)$ for producing zeroes, so we eliminate negation altogether. \u2013\u00a0Erick Wong Nov 6 '17 at 18:30\n\u2022 @ErickWong: Ahh, now I see your point! Good argument. Your're right, of course. :-) (+1) \u2013\u00a0Markus Scheuer Nov 7 '17 at 17:59", "date": "2020-08-15 18:37:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2499649/proof-of-obtaining-multiple-of-10-by-using-arithmetic-operations-on-any-three-nu", "openwebmath_score": 0.9033935070037842, "openwebmath_perplexity": 323.0975641751593, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692305124305, "lm_q2_score": 0.8824278618165526, "lm_q1q2_score": 0.8618401407831018}}
{"url": "https://math.stackexchange.com/questions/344714/non-commutative-or-commutative-ring-or-subring-with-x2-0", "text": "Non-commutative or commutative ring or subring with $x^2 = 0$\n\nDoes there exist a non-commutative or commutative ring or subring $R$ with $x \\cdot x = 0$ where $0$ is the zero element of $R$, $\\cdot$ is multiplication secondary binary operation, and $x$ is not zero element, and excluding the case where addition (abelian group operation) and multiplication of two numbers always become zero?\n\nEdit: most seem to be focused on non-commutative case. What about commutative case?\n\nEdit 2: It is fine to relax the restriction to the following: there exists countably (and possibly uncountable) infinite number of $x$'s in $R$ that satisfy $x \\cdot x = 0$ (so this means that there may be elements of ring that do not satisfy $x \\cdot x =0$) excluding the case where addition and multiplication of two numbers always become zero.\n\n\u2022 I forgot to write commutative. So both commutative and non-commutative cases. \u2013\u00a0user69886 Mar 28 '13 at 13:44\n\u2022 Regarding your edit: If you want arbitrary elements of this type, just take products or tensor products. So take $k[x_1]/(x_1^2) \\times k[x_2]/(x_2^2) \\times \\dotsc$ or $k[x_1,x_2,\\dotsc]/(x_1^2,x_2^2,\\dotsc)$. \u2013\u00a0Martin Brandenburg Mar 28 '13 at 14:29\n\u2022 @user69886: I updated my answer with a commutative example. In general and for the future, please do not change your question after answers has been given. In general if you have another question, then just post that as a separate question. \u2013\u00a0Thomas Mar 28 '13 at 14:49\n\nExample for the non-commutative case: $$\\pmatrix{0 & 1 \\\\ 0 & 0} \\pmatrix{0 & 1 \\\\ 0 & 0} = \\pmatrix{0 & 0 \\\\ 0 & 0}.$$ Example for the commutative case: Consider the ring $\\mathbb{Z} / 4\\mathbb{Z}$. What is $2^2$ in this ring?\n\n\u2022 See my answer for a conceptual view of the genesis of this example. \u2013\u00a0Math Gems Mar 28 '13 at 15:54\n\nSure, take $k$ any ring and $R=k[x]/(x^2)$. As suggested by Martin Brandenburg, a good example is $k=\\mathbb R$, in which case $R$ becomes the ring of dual numbers. The wikipedia site lists a lot of properties, which might be interesting.\n\n\u2022 Or almost the same, $\\mathbb{Z}/4$. \u2013\u00a0Martin Brandenburg Mar 28 '13 at 13:40\n\u2022 It's commutative. Though if we take a noncommutative base ring instead of $k$.. \u2013\u00a0Berci Mar 28 '13 at 13:41\n\u2022 @Berci: Whoops, I misread that. \u2013\u00a0Jesko H\u00fcttenhain Mar 28 '13 at 13:42\n\u2022 I forgot to write \"or commutative..\" \u2013\u00a0user69886 Mar 28 '13 at 13:44\n\u2022 Jesko, you may perhaps add a link to en.wikipedia.org/wiki/Dual_number \u2013\u00a0Martin Brandenburg Mar 28 '13 at 13:53\n\nIt is easy if you know about quotient rings, e.g. $\\rm\\:\\Bbb Z/n^2,\\ \\Bbb Z[x]/(x^2).\\:$ If you don't know about quotient rings, is there still a simple example? Yes! One quickly checks that $\\rm\\,\\Bbb Z^2$ is a ring under pointwise addition and multiplication and, that the set of linear maps on $\\rm\\,\\Bbb Z^2\\,$ form a ring under addition and map composition. Now note that the (right) shift map $\\rm\\: S(a,b) = (0,a)\\:$ is linear, and $\\rm\\,S^2 = 0\\,$ since $\\rm\\: S^2(a,b) = S(0,a) = (0,0).\\:$\n\nRemark  If you know matrix rings, note that the matrix of the shift map is that in Thomas's answer. But knowledge of matrices is not required to understand the above simple example.\n\nWhere does this example come from? It arises from a general construction. It is, in fact, a linear representation of the generic example $\\rm\\:R[x]/(x^2),\\:$ whose additive group is $\\rm\\:R^2,\\:$ where $\\rm\\:x\\:$ acts as a (right) shift map $\\rm\\:x(a+bx) = 0+ax,\\:$ i.e. $\\rm\\:(a,b)\\mapsto (0,a).\\:$ Similarly any ring $\\rm\\,R\\,$ has such a linear representation as a subring of the ring of linear maps on its underling additive group - the so-called (left) regular representation of $\\rm\\,R.\\:$ This is a ring-theoretic analogue of Cayley's theorem, that a group $\\rm\\,G\\,$ may be represented as a subgroup of the group of bijections (permutations) on $\\rm\\,G.\\:$ In both cases the representation arises by sending each element $\\rm\\:r\\:$ into its associated (left) multiplication map $\\rm\\: x\\:\\to\\:r\\:x\\:.\\:$ See this MO question for further discussion.\n\nAs another important example, this view provides a natural way to construct the polynomial ring $\\rm\\,R[x]\\,$ as a subring of the ring of linear maps on $\\rm\\,R^\\Bbb N = (f_0,f_1,f_2,\\ldots)\\:$ generated by $\\rm\\,R\\,$ and $\\rm\\,x\\,$ i.e. by all constants $\\rm\\:(r,0,0,\\ldots)$ and by the shift map $\\rm\\,x : (f_0,f_1,f_2,\\ldots)\\mapsto (0,f_0,f_1,\\ldots).$ See also the presentation in this post.\n\nIt is worth emphasis that the ring $\\rm\\:R[x]/(x^2)\\:$ is known as the algebra of dual numbers over $\\rm\\,R,\\,$ and that it frequently proves quite handy in various contexts, e.g. when studying (higher) derivatives, or tangent/jet spaces, and when transferring properties of homomorphisms to derivations\n\n\u2022 Actually $S=x$ on $\\mathbb{Z}[x]/(x^2)$. So these are the same examples in disguise. \u2013\u00a0Martin Brandenburg Mar 28 '13 at 14:17\n\u2022 @martin Yes, I meant to segue into that - see the added remark. \u2013\u00a0Math Gems Mar 28 '13 at 14:57\n\nYes, for example consider the noncommutative polynomial ring $\\Bbb R\\langle x,y\\rangle$ (so here $xy\\ne yx$), and let $R$ be its quotient by the ideal $(x^2)$ generated by $x^2$.\n\n\u2022 Non-commutative polynomial rings are often called free (associative) algebras and denoted by other brackets, namely $R\\langle x_1,\\dotsc,x_n\\rangle$. \u2013\u00a0Martin Brandenburg Mar 28 '13 at 13:48", "date": "2021-03-06 12:23:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/344714/non-commutative-or-commutative-ring-or-subring-with-x2-0", "openwebmath_score": 0.8537473678588867, "openwebmath_perplexity": 243.20401180683527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462215484651, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8618322795762311}}
{"url": "https://math.stackexchange.com/questions/2589445/tossing-a-matchstick-randomly-onto-an-infinite-grid", "text": "# Tossing a matchstick randomly onto an infinite grid\n\nI was given this problem recently in a job interview in which I did not succeed, however I found the problem itself very interesting, and have been trying to solve it since.\n\nQuestion:\n\nGiven matchstick of length 1, and an infinite grid of similar matchsticks (i.e. an infinite grid of 1x1 squares), if you toss the matchstick randomly onto the grid, what is the probability that the matchstick lands overlapping at least one of the matchsticks in the grid?\n\nMy approach:\n\nI decided to first define 3 variables:\n\n\u2022 $x =$ horizontal distance between lower end of landing matchstick and nearest grid matchstick to the left\n\u2022 $y =$ vertical distance between lower end of landing matchstick and nearest grid matchstick below\n\u2022 $\\theta$ = landing angle of matchstick with grid horizontal\n\nBy definition:\n\n\u2022 $x, y$ ~ $U(0, 1)$\n\u2022 $\\theta$ ~ $U(0, \\pi)$\n\u2022 $x, y, \\theta$ are all independent of each other\n\nI then split the events of overlapping the horizontal and the vertical matchsticks in the grid, such that:\n\n$$P(Overlap) = P(OverlapsVertical) + P(OverlapsHorizontal) - P(OverlapsVertical \\bigcup OverlapsHorizontal)$$\n\n\u2022 $P(OverlapsVertical) = P(y + sin(\\theta) >= 1)$\n\u2022 $P(OverlapsHorizontal) = P(x + cos(\\theta) >= 1)$\n\nFor overlapping vertical, because both y and theta are uniformly distributed, I could calcluate $P(OverlapsVertical)$ by:\n\n\u2022 Constructing the rectangle of possible values for $y$ and $\\theta$\n\u2022 Integrating to find the proportion of area of this rectangle that satisfied the condition $y + sin(\\theta) >= 1$\n\nDoing the same for the horizontal case, I found that $P(OverlapsVertical) = P(OverlapsHorizontal) = 2/\\pi$.\n\nHowever, I realised that the events $OverlapsVertical$ and $OverlapsHorizontal$ are not independent, because they are both dependent on $\\theta$, specifically $sin(\\theta)$ and $cos(\\theta)$ which are inversely related, and therefore:\n\n\u2022 $P(OverlapsVertical \\bigcup OverlapsHorizontal) \\not= P(OverlapsVertical) * P(OverlapsHorizontal)$\n\nI then became stuck, and wondered if my approach to the entire problem was wrong. Is there a better way to solve this problem? Any input greatly appreciated. Thanks!\n\n\u2022 \u2013\u00a0Bram28 Jan 2 '18 at 22:07\n\u2022 @Bram28 thanks for the quick reply. I see how that applies to the individual events of crossing the vertical and crossing the horizontal (my problem is specifically Buffon's needle where t = l = 1, and hence agrees with my answer 2/\u03c0), however as far as I can tell it does not answer the question of crossing either the vertical or the horizontal, where these two events are not independent nor mutually exclusive. \u2013\u00a0Rory Devitt Jan 2 '18 at 22:20\n\u2022 Out of curiosity, which position were you applying for ? \u2013\u00a0Gabriel Romon Jan 2 '18 at 22:24\n\u2022 @GabrielRomon it was a quantitative trading position at a hedge fund \u2013\u00a0Rory Devitt Jan 2 '18 at 22:25\n\u2022 And the interviewer didn't give you a hint or a nudge in some direction ? \u2013\u00a0Gabriel Romon Jan 2 '18 at 22:31\n\nThe if the midpoint lands inside a box of dimensions: $(1-\\cos\\theta)\\times(1-\\sin\\theta),$ then then the match will fit inside that square.\nDue to the symmetry of the grid we only need to analyze $\\theta \\in [0, \\frac {\\pi}{4}]$\n$\\frac 1{\\frac {\\pi}{4}}\\int_0^\\frac {\\pi}{4} (1-\\cos\\theta)\\cdot(1-\\sin\\theta)\\ d\\theta$", "date": "2019-06-27 06:06:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2589445/tossing-a-matchstick-randomly-onto-an-infinite-grid", "openwebmath_score": 0.7380909323692322, "openwebmath_perplexity": 346.62857863666494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808724687407, "lm_q2_score": 0.8791467611766711, "lm_q1q2_score": 0.8618107540743348}}
{"url": "https://www.themathdoctors.org/three-times-larger-idiom-or-error/", "text": "# Three Times Larger: Idiom or Error?\n\nHaving just written about issues of wording with regard to percentages, we should look at another wording issue that touches on percentages and several other matters of wording. What does \u201cthree times larger\u201d mean? How about \u201c300% more\u201d? We\u2019ll focus on one discussion that involved several of us, and referred back to other answers we\u2019ve given.\n\n## Percent increase vs. factor\n\nThe question, from 2006, started with the idea of a percent increase:\n\nPercent Increase and \"Increase by a Factor of ...\"\n\nA math doctor here recently explained percent increase this way: If we start with 1 apple today and tomorrow have 2 apples, then because 2 - 1 = 1 and 1/1 = 1, we have a 100 percent increase.\n\nBut can't I also say there was an increase by a factor of 2?  Two divided by 1 equals 2, an increase by a factor of 2 -- and also an increase by 200 percent?  This is what is confusing me!\n\nI'd never been confused about saying \"increased by a factor of\" and \"increased by percent of\" until I saw the Dr. Math conversation about finding percentages ... which is a good thing, I guess, because now I know what I didn't know!  Thank you for any help.\n\nUnfortunately, Joseph didn\u2019t directly quote from the page he had in mind, and we have never said exactly what he said; so we couldn\u2019t be sure which page it was. Everything he said, however, was correct.\n\nDoctor Rick was the first to answer:\n\nHi, Joseph.\n\nYes, this can be very confusing, because some statements about increases are ambiguous.\n\nWhen we say \"increased by a factor of 2,\" the word \"factor\" makes it clear that we mean \"multiplied by 2.\"\n\nWhen we say \"increased by 10%,\" there is only one reasonable interpretation: the amount of the increase is 10% of the original amount.  If we meant multiplication by 10%, that would be a decrease -- not an increase!  Even when we say \"increased by 100%,\" there is only one reasonable interpretation, since multiplication by 100% is the same as multiplication by 1, and that's still not an increase.\n\nWhen we want to speak of an increase that is greater than the original amount, then ambiguity can arise.  In that situation, I much prefer \"increase by a factor of 3\" or \"by a factor of 2.5,\" etc.\n\nI don't know what page you saw -- but have you seen this one?\n\nPercent Greater Than vs. Increased\nhttp://mathforum.org/library/drmath/view/61774.html\n\nSee also the page linked there, about the even more confusing phrase \"___ times more than\" and the like.  I am on the side of avoiding the confusing phrases, as a basic principle of communication.\n\nIf you saw another page and you are still confused by it, please tell me the URL of that page so I can review it with you.\n\n## Ambiguity in percent increase\n\nBefore we get back to this conversation, we should take a look at the page he referred to, which is a good starting point:\n\nPercent Greater Than vs. Increased\n\nWhat is the difference between the following statements:\n\nMy profits are 200% bigger than they were last year.\n\nand\n\nMy profits from last year have increased 200%.\n\nThis is one of the questions we have to answer in my Middle school methods course and I have looked everywhere for the answer.  I hope you can help.\n\nAs far as I can see, they mean the same thing; in fact, both are similarly ambiguous.\n\nTaken literally, \"200% bigger\" (or, more formally, larger or greater) and \"increased 200%\" (or, more completely, increased _by_ 200%) both mean that the increase from one year to the next is 200% of the first year's value, so that the second year's profit is 3 times the first. But both statements are more likely to have been made with the intention of saying that this year's profit is twice last years. English is not very clear in cases like this.\n\nSo there is a literal meaning (which mathematicians tend to see as best), and an idiomatic meaning (which ordinary people are more likely to have intended to say). I referred to a page we\u2019ll be looking at below, and then quoted a favorite book of mine that gives a lexicographer\u2019s perspective:\n\nSince writing that, I found a good reference on \"two times greater,\" although it doesn't mention your \"200% greater.\" It is in Merriam Webster's _Dictionary of English Usage_, which under \"times\" writes\n\nThe argument in this case is that _times more_\n(or _times larger_, _times stronger_, _times\nbrighter_, etc.) is ambiguous, so that \"He has\nfive times more money than you\" can be\nmisunderstood as meaning \"He has six times as\nmuch money as you.\" It is, in fact, possible\nto misunderstand _times more_ in this way, but\nit takes a good deal of effort. If you have\n$100, five times that is$500, which means\nthat \"five times more than $100\" can mean (the commentators claim) \"$500 more than $100,\" which equals \"$600,\" which equals \"six times\nas much as $100.\" The commentators regard this as a serious ambiguity, and they advise you to avoid it by always saying \"times as much\" instead of \"times more.\" Here again, it seems that they are paying homage to mathematics at the expense of language. The fact is that \"five times more\" and \"five times as much\" are idiomatic phrases which have - and are understood to have - exactly the same meaning. The \"ambiguity\" of _times more_ is imaginary: in the world of actual speech and writing, the meaning of _times more_ is clear and unequivocal. It is an idiom that has existed in our language for more than four centuries, and there is no real reason to avoid its use. I think the same applies to \"X percent bigger\" and \"increased [by] X%.\" There is just enough ambiguity in a technical context that I would want to ask what was intended before assuming anything, but there is no reason to say that they definitely mean different things, or mean something different than \"X percent of\" or \"increased to X percent.\" I myself would avoid saying these things, just because there are enough people who have heard that they are ambiguous, and would therefore take them the wrong way (whichever that is!). As a result of my side interest in linguistics, I recognize that human speech is not as logical as we might wish; what a word means is a matter of actual usage in a culture, rather than pure logic. So rather than state that either understanding of \u201ctimes bigger\u201d is \u201ccorrect\u201d, I just recognize that people take it in two ways, so you have to ask, or use contextual cues, in order to decide on what is meant. ## Confusion about \u201cthree times larger\u201d Back to the original discussion: Joseph responded with specific references, the first of which was that link of mine that Doctor Rick said to \u201csee also\u201d: I'm sorry, I should have specified the site. In fact, there were two -- and I still don't see the difference between them. Here is the first example, from Larger Than and As Large As http://mathforum.org/library/drmath/view/52338.html 1) \"Three times as large as N\" means \"3 * N.\" 2) \"Three times larger than N\" means \"4 * N\" -- but only if you stop to think about it, as many people do not. Here, I don't understand how something can be 3 times larger and be 4 times N. That sounds really weird to me. If you asked \"What is something that is three times as large as N?\" then I would say 3N ... but apparently I'd be wrong! I just don't see where my thinking is wrong.  His thinking isn\u2019t wrong on this point: 3N is three times as large as N! He seems just to be letting what he\u2019s read sow doubt about everything. Here is the second example, from Percentage of Increase http://mathforum.org/library/drmath/view/58131.html You can choose two ways to express your answer now. One is to say: there will be a 550% increase by the year 2000. Or you can say: in the year 2000 the (new value) -- you didn't say what the numbers represented, so I'm a little confused right here -- will be about five and a half times greater than what it was in 1995. Many people don't quite grasp those phrases, especially the latter one. Instead you might wish to say it this way: in 2000 the (new value) will be 6 and a half times what it was in 1995. The difference in the wording is subtle, of course, but important. The number 6 1/2 comes from 325,000 --------- = 6.5 or 6 1/2 50,000 which is NOT a percent increase situation. In this problem, I don't understand the difference between the way the doctor explains the two different ways you can talk about the increase, and the implications of each. The doctor says that 6.5 times is not a percent increase; but can you still say it's 650 percent OF the original? I'm sorry -- this is all very confusing at this point! So his specific question is this: \u2022 Why wouldn\u2019t \u201cthree times as large\u201d and \u201cthree times larger\u201d mean the same thing? \u2022 How can \u201c5 1/2 times greater\u201d and \u201c6 1/2 times what it was\u201d mean the same thing? The two pages quoted are by me (1999) and Doctor Terrel (1997). The first is particularly worth reading in its entirety, as there is a lot more there. ## The case for a literal interpretation Doctor Greenie responded, arguing against laxness on the matter, and making the case for the literalistic interpretation: I'm going to jump in here, because this is one of my pet peeves. Mathematics is commonly called the exact science. Mathematics must be exact; if it is not, it all falls apart. We can't use ambiguous language in mathematics. I agree that the use of the phrase \"x times larger than\" is best avoided. However, as a mathematician who believes in using unambiguous language, I cannot accept the proposition that we should be able to interpret \"5 times larger than 10\" as either 50 or 60. It HAS TO BE ONE OR THE OTHER. And grammatically, \"5 times larger than\" means the \"new\" number is 5 times larger than the \"old\" number; this in turn means the difference between the new and old numbers is 5 times the old number, making the new number 6 times the old number. So the number which is 5 times larger than 10 is 10 + 5(10) = 10 + 50 = 60 (The phrase \"... larger than ...\" implies comparison by subtraction; the phrase \"... as large as ...\" implies comparison by division. Sixty is 6 times as large as 10, because 60/10 = 6. But 60 is 5 times larger than 10, because [60 - 10]/10 = 50/10 = 5.) Yes, we hear it all the time in everyday life. Sometimes, we even hear it in the supposedly rigorous world of science -- \"an earthquake of magnitude 5 is 10 times greater than one of magnitude 4,\" and such. But the common idiom of using \"10 times greater than\" -- when the actual meaning is \"10 times as great as\" -- has no place in mathematics. He concluded with an accidental overstatement of what the \u201cother side\u201d says: I disagree with many of the concessions that other math doctors here have made in interpreting the phrase \"x times larger than\" as being the same as \"x times as large as.\" On one of the pages I saw, a fellow doctor said that \"50% larger than\" and \"50% as large\" mean the same thing. But if my weekly salary last year was$1000 and it is 50% LARGER this year, then it is now\n\n$1000 + 50%($1000) = $1000 +$500 = $1500 While if it was$1000 last year and it is 50% AS LARGE this year, then it is now\n\n50%($1000) =$500\n\nIf something is 50% larger, then it is larger; if it is 50% as large, then it is smaller.  They can't be the same; that is nonsense.\n\nIn fact, as he admitted in a subsequent private discussion, he had misremembered what others had said; none of us have claimed that\u00a0\u201c50% larger than\u201d and \u201c50% as large\u201d mean the same thing. What we say is that, when the percentage or multiplier is greater than 100%, we recognize ambiguity in the likely intent.\u00a0 I think we agree that the phrase should not be used in a mathematical context, and that we both grudgingly interpret it as intended elsewhere.\n\n## What the literal interpretation means\n\nThen it was my turn to respond, as the author of the first page Joseph had asked about, wanting to make sure he understood both why people take it literally as they do, and how we should think about it.\n\nFirst, on \u201cLarger Than and As Large As\u201d, I said this:\n\nJoseph, your thinking is RIGHT: if M is three times AS LARGE AS N, then M = 3N.  That's what statement (1) above says.\n\nBut if we break statement (2) apart carefully (some would say TOO carefully :-)), then it means something different from what people usually mean by it.\n\nIf I said \"M is 50 larger than N,\" I would mean that if you ADD 50 to N, you get M:\n\nM = N + 50.\n\nAnd if I said, \"M is 50% larger than N,\" I would mean that if you add 50% OF N to N, you get M; that is, I mean that M is 50% of N added to N:\n\nM = N + (0.50)N.\n\nNow, though I'm not entirely sure I agree with this, technically minded people often apply the same thinking to (2), for the sake of consistency.  The \"larger than\" means we add something to N.  And what do we add? Three times N.  So by this thinking,\n\nM = N + 3N = 4N.\n\nSo \"three times LARGER THAN N\" means the same as \"four times AS LARGE AS N.\"\n\nI then referred to the usage book quoted above, adding:\n\nEnglish usage experts think it is nonsense.  My feeling is that this thinking puts a little too much weight on consistency, and is just too weird for the general public to follow.  English is not known for consistency!  So we need to recognize that in everyday usage, (1) and (2) really mean the same thing.  When we accept that, though, we set ourselves up for the opposite confusion: Cases like \"50% larger\" and \"3 times larger\" no longer follow the same pattern, and our language becomes inconsistent, which really bothers mathematicians!\n\nAs Dr. Rick pointed out, this means that there are gray areas where it's hard to be sure what someone means, so it may be best just to avoid using these phrases in mathematical contexts.\n\nFinally, I commented on Joseph\u2019s other quote, from Doctor Terrel:\n\nJoseph, here the doctor was saying that a 550% INCREASE means adding 550% of the original value to the original value, which means 650% OF the original value.  In the other terminology, \"5 1/2 times greater\" (there again, \"greater\" is taken to refer to the increase) is the same as \"6 1/2 times as much.\" When he says that the 6 1/2 is not a percent increase, he doesn't mean that it hasn't increased, just that he is talking about multiplying by 650% rather than adding 650%.  When you think in terms of increase (adding), it is a 550% increase.\n\nNow, the \"percent increase\" case is pretty standard, because it IS technical terminology (though ordinary people reading it can get confused, so it's still risky).  The \"times greater\" case is more disputable, since that sounds less technical.  Most people don't demand absolute consistency from language; they are happy to understand \"times greater\" idiomatically.\n\nI hope that clears up some of the confusion.  It isn't all cleared up yet at our end.  You will definitely get different opinions as to what it all REALLY means!\n\n## Increase by a factor\n\nJoseph replied, returning to his initial question:\n\nThat really cleared things up for me and I appreciate your time in driving home the differences!  The last question I would like to ask is, how do you deal with factors?  If someone says something has changed by a factor of ... or is less/greater than by a factor of ..., do we use the same rules that you've discussed above? Or when using the word \"factor,\" are things a bit different?\n\nAs Dr. Rick pointed out in the first response, \"factor\" is used to make it clear that multiplication, rather than addition, is the cause of an increase.  Just as \"increased by a factor of 2\" means \"twice as large,\" so does \"greater by a factor of 2.\"  And \"decreased by a factor of 2\" and \"less by a factor of 2\" both mean \"half as much\" (divided by 2).\n\nI can't think of a context in which that would not be true -- but English is flexible enough that I probably shouldn't guarantee anything!\n\nIsn\u2019t English fun?\n\n## Conclusion\n\nIn closing, here is a more recent question (2015) where I summarized this complicated issue:\n\nAs Big As vs. Bigger Than\n\nI'm having difficulty convincing my 5th graders that \"as big as\" and \"bigger than\" do not mean the same thing.\n\nFor example, when asked, \"How many times larger is 10,000 than than 100?\" they answer \"100.\"\n\nTheir tests and homework are full of this misunderstanding! How would you suggest telling them they are wrong?\n\nI referred to most of the pages we\u2019ve seen above, and added:\n\nTo be honest, my feeling (basically unchanged since the first of those) is that although your understanding is common among thoughtful people, it is a case of excessive consistency.\n\nMathematical people want a certain word structure to always have the same meaning, so we relate \"x times bigger\" to \"x percent bigger,\" and that to \"x bigger,\" and want to take all in an incremental (additive) sense. But taken on its own, it is perfectly logical to interpret \"x times bigger\" as \"bigger, as a result of multiplication by x.\" And human language is not completely consistent; we have idioms all over the place that we interpret with no trouble.\n\nHaving said that, I think that math books should avoid that form, because there is just enough truth to your thinking, and enough extra expectation of careful use of words in a math book, that it can be confusing.\n\nWhat I would do is to make a brief mention of the fact that many people take it as you do, but then point out that your book is using the phrase in the way it is usually intended in the real world. If I were writing the textbook, I would reverse this: almost always use \"x times as big,\" but mention somewhere the fact that many people use \"x times bigger\" to mean the same thing, and briefly discuss the controversy before moving on to other things.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2022-05-17 10:14:18", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/three-times-larger-idiom-or-error/", "openwebmath_score": 0.7126056551933289, "openwebmath_perplexity": 909.8569135726916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98028087477309, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.8618107514442109}}
{"url": "https://math.stackexchange.com/questions/2515956/sequence-of-odd-and-even-partial-sums-of-alternating-harmonic-series", "text": "# Sequence of odd and even partial sums of alternating harmonic series.\n\nGiven the alternating harmonic series: $$\\sum^{\\infty}_{n=1}{\\frac{(-1)^{n+1}}{n}}$$ and if we let the sequence of its partial sums be $s_n$ then how can we express the sequence of even partial sums $s_{2n}$ and odd partial sums $s_{2n+1}$?\n\nI did the following but am not completely sure if it is valid:\n\n$$s_{2n}= \\sum^{2n}_{i=1}{\\frac{(-1)^{i+1}}{i}}$$ $$s_{2n+1}= \\sum^{2n+1}_{i=1}{\\frac{(-1)^{i+1}}{i}}$$\n\nThen to show the first is increasing and the second is decreasing, can we just take $s_{2n+2}-s_{2n}$ and show that this is $> 0$? Similarly then taking $s_{2n+3}-s_{2n+1}$ and showing that this is $< 0$?\n\nAnd to go one step further, how could we show that each of these sequences of partial sums has a limit? And then show that the limits are the same?\n\n\u2022 $s_{2n} = \\sum_{i=1}^n \\frac{1}{2i-1}-\\frac{1}{2i}=\\sum_{i=1}^n \\frac{1}{2i (2i-1)}$ \u2013\u00a0reuns Nov 12 '17 at 1:31\n\u2022 en.wikipedia.org/wiki/Alternating_series_test \u2013\u00a0parsiad Nov 12 '17 at 1:33\n\u2022 I don't quite get where the formula you've written comes from? \u2013\u00a0Analysis is fun Nov 12 '17 at 1:42\n\nWhat you have done is fine. Showing the increasing/decreasing nature works well. You can invoke the alternating series theorem to show the whole series has a limit because the terms are alternating in sign and decreasing monotonically to zero. If the whole series has a limit, every subseries converges to the same limit and you are done.\n\n\u2022 Ah okay I see, is there a way to directly prove that $s_{2n}$ and $s_{2n+1}$ have limits? \u2013\u00a0Analysis is fun Nov 12 '17 at 1:42\n\u2022 The subseries argument works fine. Also they are monotonic and bounded above/below so have limits. \u2013\u00a0Ross Millikan Nov 12 '17 at 1:46\n\u2022 Ah yes okay I think I've got it now, thanks so much for your help! \u2013\u00a0Analysis is fun Nov 12 '17 at 1:47\n\nYes, you are correct on how to show that the first is decreasing and the second is increasing.\n\nNow, since you know they are monotone, if you can show they are bounded, you will demonstrate that they have limits. For instance you can show that $$1>s_{2n+1} > 1/2$$ for all $n$.\n\nThat the limits are the same follows from the fact that the difference $$|s_{2n+1}-s_{2n}| = \\frac{1}{2n+1}\\to 0.$$\n\n\u2022 Thanks, so we can show boundedness and then use the monotone convergence theorem? \u2013\u00a0Analysis is fun Nov 12 '17 at 1:47\n\u2022 Yeah. As others have said the alternating series test proves it all in one fell swoop but it seemed like the point of the exercise was to prove a special case of the alternating series test. \u2013\u00a0spaceisdarkgreen Nov 12 '17 at 1:52\n\nSince $s_{2n}= \\sum^{2n}_{i=1}{\\frac{(-1)^{i+1}}{i}}$,\n\n$\\begin{array}\\\\ s_{2n+2} &= \\sum^{2n+2}_{i=1}{\\frac{(-1)^{i+1}}{i}}\\\\ &= \\sum^{2n}_{i=1}{\\frac{(-1)^{i+1}}{i}}+{\\frac{(-1)^{2n+1}}{2n+1}}+{\\frac{(-1)^{2n+2}}{2n+2}}\\\\ &= s_{2n}-{\\frac{1}{2n+1}}+{\\frac{1}{2n+2}}\\\\ &= s_{2n}-{\\frac{1}{(2n+1)(2n+2)}}\\\\ &\\lt s_{2n}\\\\ \\end{array}$\n\nso $s_{2n}$ is decreasing.\n\nSimilarly,\n\n$\\begin{array}\\\\ s_{2n+3} &= \\sum^{2n+3}_{i=1}{\\frac{(-1)^{i+1}}{i}}\\\\ &= \\sum^{2n+1}_{i=1}{\\frac{(-1)^{i+1}}{i}}+{\\frac{(-1)^{2n+2}}{2n+2}}+{\\frac{(-1)^{2n+3}}{2n+3}}\\\\ &= s_{2n+1}+{\\frac{1}{2n+2}}-{\\frac{1}{2n+3}}\\\\ &= s_{2n+1}+{\\frac{1}{(2n+2)(2n+3)}}\\\\ &\\gt s_{2n+1}\\\\ \\end{array}$\n\nso $s_{2n+1}$ is increasing.\n\nSince $s_{2n+1} =s_{2n}-\\frac1{2n+1} \\lt s_{2n}$, all the $s_{2n+1}$ are less than all the $s_{2n}$.\n\nSince $|s_{2n+1}-s_{2n}| =\\frac1{2n+1} \\to 0$, the sequences $(s_{2n})$ and $(s_{2n+1})$ must approach a common limit.", "date": "2019-10-19 13:11:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2515956/sequence-of-odd-and-even-partial-sums-of-alternating-harmonic-series", "openwebmath_score": 0.9149131774902344, "openwebmath_perplexity": 187.80336916252574, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808678600415, "lm_q2_score": 0.8791467548438124, "lm_q1q2_score": 0.8618107438146316}}
{"url": "https://brilliant.org/discussions/thread/when-fibonacci-meets-pythagoras/", "text": "\u00d7\n\nWhen Fibonacci meets Pythagoras\n\nThere is a cool identity relating Fibonacci numbers and Pythagorean triples, which states that given four consecutive Fibonacci numbers $${F}_{n},{F}_{n+1},{F}_{n+2},{F}_{n+3}$$, the triple $$({F}_{n}{F}_{n+3}, 2{F}_{n+1}{F}_{n+2}, {{F}_{n+1}}^{2}+{{F}_{n+2}}^{2})$$ is a Pythagorean triple.\n\nTo prove this, let $${F}_{n+1} = a$$ and $${F}_{n+2} = b$$. Then $${F}_{n} = b-a$$ and $${F}_{n+3} = b+a$$. Substituting the abstract values to the equation ${({F}_{n} {F}_{n+3})}^{2} +{(2{F}_{n+1} {F}_{n+2})}^{2} = {({{F}_{n+1}}^{2}+{ {F}_{n+2}}^{2})}^{2}$ will give ${((b-a)(b+a))}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$\n\nWith a little simplification, we arrive at Euclid's formula for Pythagorean triples: ${({b}^{2}-{a}^{2})}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$\n\nCheck out my other notes at Proof, Disproof, and Derivation\n\nNote by Steven Zheng\n2\u00a0years, 10\u00a0months ago\n\nComments\n\nSort by:\n\nTop Newest\n\nNice post\n\nIs there any generalized formula to generate pythagoream triples\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\nEuclid's Formula. The link above should bring you to a proof.\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\nHow we can solve problems in which product of three pythagorean triples is given and we have to find the triple\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\nI'm not sure if that is solvable without more given conditions. You might also need the sum of sides or maybe the radius of an incircle also. Can you come up with an example of your question?\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\nFind out the pythagorean triple $$(x,y,z)$$ such that $$xyz=16320$$\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\nNumber of solutions $$a,b$$ such that $$2ab({a}^{2}+{b}^{2})({a}^{2}-{b}^{2}) = 16320$$.\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\nDoes there exist more the one triples.....also i read on a website that euclid's formula can't generate all pythagorean triples\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\nEuclid's formula deal with generating Pythagorean triples that are integers.\n\n- 2\u00a0years, 9\u00a0months ago\n\nLog in to reply\n\nGood\n\n- 2\u00a0years, 10\u00a0months ago\n\nLog in to reply\n\n\u00d7\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading...", "date": "2017-10-22 03:02:22", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/when-fibonacci-meets-pythagoras/", "openwebmath_score": 0.9562994241714478, "openwebmath_perplexity": 1920.183059709734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9859363758493942, "lm_q2_score": 0.8740772482857831, "lm_q1q2_score": 0.8617845543872962}}
{"url": "http://math.stackexchange.com/questions/89470/in-how-many-ways-can-i-climb-down-ten-stairs-taking-as-many-steps-at-a-time-as", "text": "# In how many ways can I climb down ten stairs, taking as many steps at a time as I like?\n\nI want to climb down a flight of ten stairs, taking one or many steps at a time. In how many ways can I do this if I am able to take as many steps at a time as I like?\n\nI don't know the answer; I am getting $3$ such as:\n\n$1,2,3,4,..., 10$ (one at a time)\n\n$2,4,6,..., 10$ (two at a time)\n\n$5,10$ (five at a time)\n\nthis is correct/incorrect?\n\n-\nIf you have to choose the number of steps you will take at a time, and stick with that number the whole way down, then the only thing you have missed is the possibility of taking 10 at a time. But if you are allowed to (for example) take 3 steps, then 4, then 3, well, that's a whole different problem. So, which is it? \u2013\u00a0Gerry Myerson Dec 8 '11 at 2:16\n@Gerry Myerson: \"taking one or many steps at a time\" probably indicate a variable number of steps at each time ... also this version seems to be the interesting one. \u2013\u00a0Quixotic Dec 8 '11 at 2:21\nIf we have total freedom, there are $512$ ways. \u2013\u00a0Andr\u00e9 Nicolas Dec 8 '11 at 2:37\nSee the WP page on the number of compositions \u2013\u00a0r.e.s. Dec 8 '11 at 3:11\n@MaX, what you say is true, but I find it hard to believe that under that interpretation OP was only able to find 3 ways to come down the stairs. \u2013\u00a0Gerry Myerson Dec 8 '11 at 5:07\n\nIn the interpretation where the number of steps taken is allowed to vary, there is an easy way to see that the answer is $2^{n-1}$:\n\nFor each of the $n-1$ steps (not counting the top one), decide whether or not you will stop at that step on the way down.\n\nThere are $n-1$ decisions to make, each with 2 possible choices (stop or don't stop), so that's a total of $2^{n-1}$ ways.\n\n-\n\nWe answer the question on the following assumptions: (1) At any time, we can take an arbitrary number of stairs, as long as the total number of stairs ends up at $10$ and (2) Order matters, so that for example the pattern $2+4+4$ is to be counted as different from $4+2+4$.\n\nIn general, let $f(n)$ be the number of ways to handle a staircase that has $n$ stairs. It is easy to see that $f(1)=1$. We could compute a little more, finding that for $2$ there are the patterns $2$ and $1+1$, so $f(2)=2$. For $3$ there are the patterns $3$, $2+1$, $1+2$, and $1+1+1$, for a total of $4$. For $4$, enumerating the patterns is not hard: we get $4$, $3+1$, $1+3$, $2+2$, $1+1+2$, $1+2+1$, $2+1+1$, and $1+1+1+1$, a total of $8$. On the basis of this admittedly scanty evidence, we conjecture that $f(n)=2^{n-1}$, and in particular $f(10)=512$.\n\nNow we prove that in general $f(n)=2^{n-1}$. There are various approaches. We first prove the result in a natural but somewhat lengthy way, and then in a shorter way. Suppose that we know that for all $k<n$, there are $2^{k-1}$ patterns. We count the number of patterns for $n$.\n\nEither there is only one giant step of length $n$ ($1$ way) or the last step is one of $1$, $2$, $3$, $\\dots$, $n-1$.\n\nIf the last step is $1$, we needed to get to $n-1$, which by induction assumption can be done in $2^{n-2}$ ways.\n\nIf the last step is $2$, we needed to get to $n-2$, which by induction assumption can be done in $2^{n-3}$ ways.\n\nIf the last step is $3$, we needed to get to $n-3$, which by induction assumption can be done in $2^{n-4}$ ways.\n\nGo on in this way. Finally, if the last step was $n-1$, we needed to get to $1$, which can be done in $2^0$ ways.\n\nWe conclude that $$f(n)=(2^{n-2}+2^{n-3}+2^{n-4}+\\cdots +1)+1$$ (remember the possibility of a giant step). Easily, if we add from the back, we see that $f(n)=2^{n-1}$ (or else we can sum the finite geometric progression).\n\nAnother way: We show that in general $f(n+1)=2f(n)$. Consider all patterns of steps that add up to $n+1$. They are of two types (i) Patterns that have last step $1$ and (ii) Patterns that have last step $>1$.\n\nIt is obvious that there are $f(n)$ patterns of type (i). We now count the type (ii) patterns. Take a pattern of type (ii), and shorten its last step by $1$. The total is now $n$. Moreover, we get all patterns for $n$ in this manner, in one and only one way. Thus there are $f(n)$ type (ii) patterns. It follows that $f(n+1)=f(n)+f(n)=2f(n)$.\n\n-\n\nA composition of an integer $n$ into $k$ parts is the number of sums $a_1 + \\dots + a_k$ of positive integers that equal $n$. For example, there are 3 compositions of $4$ into $2$ parts: 2+2, 3+1, and 1+3. You should try to prove that the number of compositions of $n$ into $k$ parts is equal to $\\binom{n-1}{k-1}$. (Or, look it up in any good combinatorics book.)\n\nYour problem essentially asks you to count the number of compositions of $10$ into $k$ parts, where $1\\leq k\\leq 10$. That is, if as you go down the staircase you cross $a_i$ stairs in your $i^{\\text{th}}$ step, then $a_1 + \\dots + a_k = 10$. Your answer is thus $$\\sum_{k=1}^{10} \\binom{10-1}{k-1} =512$$\n\n-", "date": "2016-07-28 20:24:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/89470/in-how-many-ways-can-i-climb-down-ten-stairs-taking-as-many-steps-at-a-time-as", "openwebmath_score": 0.9091524481773376, "openwebmath_perplexity": 154.85605334381498, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98593637543616, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.861784536238564}}
{"url": "https://math.stackexchange.com/questions/2719143/first-order-pdes-and-the-method-of-characteristics", "text": "# First order PDEs and the method of characteristics\n\nI am studying PDEs using the book \"PDEs An Introduction 2nd edition\" by Walter A. Strauss. In Chapter 2, a \"geometric method\" is described in order to solve linear PDEs of the type:\n\n$$(x,y)\\mapsto u_x + yu_y = 0$$\n\nThis is said to be equivalent to the directional derivative of $u$ in the direction of the vector $(1,y)$ being set to 0. Then characteristic curves having as tangent vectors $(1,y)$ are found\n\n$$\\frac{dy}{dx} = \\frac{y}{1} \\implies y = Ce^x$$\n\nSince $u(x,y)$ is constant on these curves:\n\n$$u(x,y) = f(e^{-x}y)$$\n\nis the general solutions of the PDE, where $f$ is an arbitrary function. Now the same method is applied to solve more general equations, such as:\n\n$$u_x + u_y + u = f(x,y)$$\n\nI have tried to use the same method to solve the following differential equation:\n\n$$yu_x -xu_y = 3x$$ where $u(x,0) = x^2$\n\nThen applying the method:\n\n$$\\frac{dy}{dx} = \\frac{-x}{y} \\implies C = x^2 + y^2$$\n\nThe PDE reduces to an ODE:\n\n$$\\frac{du}{dx} = 3x/y \\implies u (x,y) = 3x^2/2y + f(C)$$\n\nHowever with the boundary imposed ($u(x,0) = x^2$) this seem impossible to solve since to find the particular solution I would have to divide the a term by zero. Does the method described in the book have limited scope? How can I solve this differential equation? Also, why is $u(x,y)$ constant on the \"characteristic curves\"?\n\n\u2022 If you introduce polar coordinates $x=r\\cos \\theta, y=r\\sin \\theta$, then $y\\partial_x-x\\partial_y=-\\partial_\\theta$. This is not the method you asked but it leads to a quick integration. \u2013\u00a0Giuseppe Negro Apr 2 '18 at 21:41\n\u2022 Yeah that is a good trick, however I was more interested in the method and not the solution itself. \u2013\u00a0daljit97 Apr 2 '18 at 21:54\n\nI cannot see where you found an impossibility or an exception in the rules.\n\nI the case of $\\quad yu_x-xu_y=3x \\tag 1$\n\n$$\\frac{dx}{y}=\\frac{dy}{-x}=\\frac{du}{3x}$$ A first family of characteristic curves comes from $\\quad\\frac{dx}{y}=\\frac{dy}{-u}\\quad$ which leads to $$x^2+y^2=c_1$$ A second family of characteristic curves comes from $\\quad\\frac{dy}{-x}=\\frac{du}{3x}\\quad$ which leads to $$u+3y=c_2$$ The general solution of the PDE $(1)$ expressed on the form of implicit equation is : $$\\Phi(x^2+y^2\\:,\\:u+3y)=0$$ where $\\Phi$ is an arbitrary fonction of two variables.\n\nOr equivalently, the general solution of PDE $(1)$ on explicit form is $\\quad u+3y=F(x^2+y^2)$ $$u(x,y)=-3y+F(x^2+y^2)$$\n\nwhere $F(X)$ is an arbitrary function of the variable $X$ with $X=x^2+y^2$.\n\nBoundary condition : $\\quad u(x,0)=x^2=F(x^2+0^2)=F(x^2)$.\n\nSo, the function $F$ is now determined : $\\quad F(X)=X.\\quad$ Putting it into the above general solution where $X=x^2+y^2$ leads to the particular solution which fits the boundary condition : $$u(x,y)=-3y+(x^2+y^2)$$\n\nOTHER EXAMPLE : $\\quad u_x+u_y+u=e^{x+2y} \\tag 2$ $$u_x+u_y=e^{x+2y}-u$$ $$\\frac{dx}{1}=\\frac{dy}{1}=\\frac{du}{e^{x+2y}-u}$$ A first family of characteristic curves comes from $\\quad\\frac{dx}{1}=\\frac{dy}{1}\\quad$ which leads to $$x-y=c_1$$ A second family of characteristic curves comes from $\\frac{dy}{1}=\\frac{du}{e^{x+2y}-u}=\\frac{du}{e^{(c_1+y)+2y}-u}=\\frac{du}{e^{c_1+3y}-u}$ $$\\frac{du}{dy}+u=e^{c_1+3y}$$ This is a first order linear ODE easy to solve : $\\quad u=\\frac14 e^{c_1+3y}+c_2e^{-y}=\\frac14 e^{(x-y)+3y}+c_2e^{-y}$ $$ue^y-\\frac14 e^{x+3y}=c_2$$ The general solution of the PDE $(2)$ expressed on the form of implicit equation is : $$\\Phi\\left((x-y)\\:,\\:(ue^y-\\frac14 e^{x+3y})\\right)=0$$ where $\\Phi$ is an arbitrary fonction of two variables.\n\nOr equivalently, the general solution of PDE $(2)$ on explicit form is from $\\quad ue^y-\\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\\frac14 e^{x+2y}+e^{-y}F(x-y)$$ where $F(X)$ is an arbitrary function of the variable $X=x-y$.\n\n\u2022 I believe that the author of my book seems to use a different method to solve the PDE, have a look here at an example. If you look at the last few lines of my question, there seems to be a problem (using the method used in the link) \u2013\u00a0daljit97 Apr 2 '18 at 21:29\n\u2022 It is fundamentally the same method, with some variant of symbols and presentation. See :en.wikipedia.org/w/\u2026 and see en.wikipedia.org/wiki/Method_of_characteristics. You will have to do a bit of \"gymnastic of the brain\" in order to make the link. I wish you bon courage. \u2013\u00a0JJacquelin Apr 2 '18 at 21:43\n\u2022 You can see that the general solution in the book is : $$u(x,y)=\\frac14 e^{x+2y}+e^{-y}f(x-y)$$ which is exactly the same than in my answer. I will not repeat what is already in the book to determine the function $f$ according to the boundary condition. The method is the same. \u2013\u00a0JJacquelin Apr 2 '18 at 21:52\n\u2022 $\\frac{du}{dx} = 3x/y \\implies u (x,y) = 3x^2/2y + f(C)$ is false because you integrate $3x/y$ without taking account that $y$ is not constant. You cannot solve it that way. This not a correct application of the method of characteristic. One have to find two independent families of characteristic curves. You got one :$x^2+y^2=C_1$ but not a second one. \u2013\u00a0JJacquelin Apr 2 '18 at 22:11\n\u2022 @Isham yes you are right, I was too invested into thinking about why the method led me to that point that it didn't occur to me that a simple substitution was needed. \u2013\u00a0daljit97 Apr 2 '18 at 22:40\n\nYou can't have this: $$\\frac{du}{dx} = 3x/y ==> u (x,y) = 3x^2/y + f(C)$$ Because you have a differential equation ( du,dx) but three variables namely x,y,u. Y shouldnt be there...Apart for that mistake you did a very good job.\n\nUsisng the method of characteristics:\n\n$$\\frac {dx}{y}=\\frac {dy}{-x}=\\frac {dz}{3x}$$ $$-xdx=ydy \\implies x^2+y^2=K_1$$ $$3xdy=-xdz \\implies y+z/3 =K_2$$ $$f(x^2+y^2)=y+u/3$$ $$u(x,y)=3f(x^2+y^2)-3y$$ we have $u(x,0)=x^2$ $$f(x^2)=x^2/3 \\implies f(x)=x/3$$ Therefore $$\\boxed{u(x,y)=x^2+y^2-3y}$$\n\n\u2022 Hmm in the book in order to solve $u_x + u_y + u = e^{x+2y}$ the author does setup the following equation: $\\frac{\\partial u}{\\partial x} + u = e^{x+2y}$ Would that be an exception to the rule? \u2013\u00a0daljit97 Apr 2 '18 at 19:39\n\u2022 @daljit97 I don't know I dont have the book right now. Thats weird that Strauss was so complicated about first order linear pde.. when it's easy to solve directly with characteristics method.. \u2013\u00a0Isham Apr 2 '18 at 19:42\n\u2022 Here is the link with the exercise and the solution provided stemjock.com/STEM%20Books/Strauss%20PDEs%202e/Chapter%201/\u2026 \u2013\u00a0daljit97 Apr 2 '18 at 19:48\n\u2022 Oh thats another method..I dont know that method...I deleted my comments on that method you used since I was completely wrong about it... \u2013\u00a0Isham Apr 2 '18 at 19:53\n\u2022 Oh ok, could you provide a reference that explains and show why the method of characteristic works? Looking around I couldn't find anything satisfying. \u2013\u00a0daljit97 Apr 2 '18 at 20:03", "date": "2019-07-17 00:20:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2719143/first-order-pdes-and-the-method-of-characteristics", "openwebmath_score": 0.7287317514419556, "openwebmath_perplexity": 166.15633338998788, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363746096916, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8617845355161668}}
{"url": "https://math.stackexchange.com/questions/220172/does-the-product-of-two-invertible-matrix-remain-invertible", "text": "# Does the product of two invertible matrix remain invertible?\n\nIf $A$ and $B$ are two invertible $5 \\times 5$ matrices, does $B^{T}A$ remain invertible?\n\nI cannot find out is there any properties of invertible matrix to my question.\n\nThank you!\n\n\u2022 A common trick to answer the question \"Is foo invertible?\" is to actually write down the inverse of foo. In many cases this is pretty easy to do, as seen in the answers.\n\u2013\u00a0user14972\nOct 24, 2012 at 16:10\n\u2022 @Hurkyl What does foo mean? Oct 24, 2012 at 16:13\n\u2022 @PENGTENG It is a nonsense word used as a generic placeholder in math and computer science. It's being used as a variable for an object here, but it's more frequently used for properties rather than objects. Oct 24, 2012 at 16:14\n\u2022 @rschwieb I get it. Thanks! Oct 24, 2012 at 16:17\n\u2022 How is this well researched? C'mon guys Aug 3, 2015 at 6:41\n\nOf course: $B$ invertible implies $B^T$ invertible, and the product of two invertible matrices is clearly invertible.\n\nThis is easily seen from these equations: $$BB^{-1}=I\\implies (BB^{-1})^T=I\\implies (B^{-1})^TB^T=1,$$ and the fact that if $X$ and $Y$ are invertible, $(XY)^{-1}=Y^{-1}X^{-1}$.\n\nPerhaps the general properties you should take away are these:\n\n$(XY)^T=Y^TX^T$ and $(XY)^{-1}=Y^{-1}X^{-1}$.\n\n\u2022 And this is of course true for $n \\times n$-matrices and not just $5 \\times 5$-matrices.\n\u2013\u00a0N.U.\nOct 24, 2012 at 16:11\n\u2022 \"...the product of two matrices is clearly invertible.\" Only if the two matrices are themselves invertible: math.stackexchange.com/a/1026628 Aug 3, 2015 at 2:16\n\u2022 @nacnudus of course, a reasonable person can tell this is a typo of the form of an omitted word from context. Thanks for indirectly alerting me to it. Aug 3, 2015 at 3:27\n\nYes. $$\\det(B^T\\,A)=\\det(B^T)\\det(A)=\\det(B)\\det(A)\\ne0.$$ Moreover $$(B^T\\,A)^{-1}=A^{-1}(B^{-1})^T.$$", "date": "2022-06-27 09:52:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/220172/does-the-product-of-two-invertible-matrix-remain-invertible", "openwebmath_score": 0.6186237931251526, "openwebmath_perplexity": 736.2160971145942, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363729567546, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8617845292202262}}
{"url": "https://math.stackexchange.com/questions/1675485/are-there-arbitrarily-large-gaps-between-consecutive-primes", "text": "# Are there arbitrarily large gaps between consecutive primes? [duplicate]\n\nI made a program to find out the number of primes within a certain range, for example between $1$ and $10000$ I found $1229$ primes, I then increased my range to $20000$ and then I found $2262$ primes, after doing it for $1$ to $30000$, I found $3245$ primes.\n\nNow a curious thing to notice is that each time, The probability of finding a prime in between $2$ multiples of $10000$ is decreasing, i.e it was $$\\frac{2262-1229}{10000}=0.1033$$ between $10000$ and $20000$, and $$\\frac{3245-2262}{10000}=0.0983$$ between $20000$ and $30000$,\n\nSo from this can we infer that there will exist two numbers separated by a gap of $10000$ such that no number in between them is prime? If so how to determine the first two numbers with which this happens? Also I took $10000$ just as a reference here, what about if the gap between them in general is $x$, can we do something for this in generality?\n\nThanks!\n\n\u2022 If you let $p$ be the largest prime below $10\\,000$, then the interval $[p\\# - 10\\,002, p\\# - 2]$ certainly contains no primes (where $p\\#$ is the primorial). It's probably not the first time, though \u2013\u00a0Arthur Feb 28 '16 at 10:26\n\u2022 In the same spirit as Arthur's suggestion, there are no primes among the $k$ consecutive numbers $(k + 1)! + 2, \\ldots, (k + 1)! + (k + 1)$. \u2013\u00a0Travis Willse Feb 28 '16 at 10:31\n\u2022 @Arthur largest prime below $10000$ is $9973$, and by p# do you mean $$2.3.5....9973$$? . \u2013\u00a0Nikunj Feb 28 '16 at 10:42\n\u2022 @Travis This looks interesting, could you please provide a proof? \u2013\u00a0Nikunj Feb 28 '16 at 10:43\n\u2022 @Nikunj Yes, that is what I mean by $p\\#$. \u2013\u00a0Arthur Feb 28 '16 at 10:46\n\nThe Prime Number Theorem states that the number of primes $\\pi(x)$ up to a given $x$ is $$\\pi(x) \\sim \\frac{x}{\\log(x)}$$ which means that the probability of finding a prime is decreasing if you make your \"population\" $x$ larger. So yes, there exist a gap of $n$ numbers whereof none are prime.\n\nThe way to find the first gap for some $n$ has to be done through the use of software, since the exact distribution of prime numbers is only approximated by $\\frac{x}{\\log(x)}$.\n\nEDIT: That the PNT implies that there's always a gap of size $n$ can be seen by considering what would happen if this was not the case; if there was a maximum gap of $n$ that was reached after some $x$, the probability of finding a prime between $x$ and some larger number $m$ would no longer decrease as $m \\to \\infty$, which contradicts the PNT.\n\n\u2022 @Nikunj Yes, this is indeed only good for larger $x$, but since this approximation holds for any $x$, $10,000-20,000$ are very small numbers. But yeah, it's the best that we have as of yet :) \u2013\u00a0Bobson Dugnutt Feb 28 '16 at 12:12\n\u2022 @BenMillwood Yeah, that is an unfortunate choice of words, I'll edit it! Thanks! \u2013\u00a0Bobson Dugnutt Feb 28 '16 at 15:09\n\u2022 Could you elaborate on this just a little bit? You're right that the PNT implies that at some point you have to have n integers without a prime, or else \u03c0(x) could never be sub-linear, but I feel like this is a few too many steps away from a proof. \u2013\u00a0hobbs Feb 29 '16 at 6:31\n\u2022 @hobbs: If you never had $n$ integers without a prime, then the number of primes up to a given $x$ could never be less than $\\lfloor{x/n}\\rfloor$. This contradicts the PNT. \u2013\u00a0mjqxxxx Mar 1 '16 at 3:54\n\u2022 @mjqxxxx I'm not asking for me, I'm asking for it to be made more explicit in the answer for others' benefit :) \u2013\u00a0hobbs Mar 1 '16 at 4:12\n\nCan we infer that there exist two numbers separated by a gap of $10000$, such that no number in between them is prime?\n\nWe can infer this regardless of what you wrote.\n\nFor every gap $n\\in\\mathbb{N}$ that you can think of, I can give you a sequence of $n-1$ consecutive numbers, none of which is prime.\n\nThere you go: $n!+2,n!+3,\\dots,n!+n$.\n\nSo there is no finite bound on the gap between two consecutive primes.\n\n\u2022 @Nikunj: Not necessarily. Also, the gap is not guaranteed to be exactly $n$ (or $10000$ in your example). That is why I have emphasized the specific part of your question to which I am referring in my answer. \u2013\u00a0barak manos Feb 28 '16 at 11:05\n\u2022 @Nikunj because $n! + 1$ and/or $n! + n + 1$ may be composite as well, in which case one gets a sequence of $> n - 1$ consecutive composite numbers. This is the case, e.g., for $n = 3$. \u2013\u00a0Travis Willse Feb 28 '16 at 11:42\n\u2022 @Nikunj To give a specific example in which both $n! + 1$ and $n! + n + 1$ are composite, observe that $5! + 1 = 121 = 11^2$ and $5! + 6 = 126 = 2 \\cdot 3^2 \\cdot 7$. In fact, the numbers $$114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126$$ are all composite. \u2013\u00a0N. F. Taussig Feb 28 '16 at 11:56\n\u2022 @barakmanos I would say, more precisely, that is a lower bound of $n$. \u2013\u00a0PyRulez Feb 28 '16 at 18:23\n\u2022 @goblin Take any $n!+i$ with $2\\leq i \\leq n$. You can factor out an $i$ from the factorial since $i \\leq n$, so you get $i(k + 1)$, for $i > 1$ and $k > 1$, where $k = n!/i$. \u2013\u00a0TokenToucan Feb 29 '16 at 2:33\n\nWe cannot infer from the two observations in the question that there are gaps of arbitrary sizes between primes. As Lovsovs mentions in his answer, this does follow from the Prime Number Theorem (one needs suitable error bounds on the approximation $\\pi(x) \\sim \\frac{x}{\\log x}$, but even crude ones will do here).\n\nAs asked in the comments, it's easy to construct for any positive integer $k$ a sequence of $k$ consecutive composite numbers. For any positive integer $a \\leq k + 1$, $a$ divides $(k + 1)!$, and so it divides $(k + 1)! + a$, but this implies that each of the $k$ numbers $(k + 1)! + 2, \\ldots, (k + 1)! + (k + 1)$ is composite. (This is generally not the first sequence for which this is true: For example, for $k = 3$, the resulting sequence is $26, 27, 28$, but the first sequence of three consecutive composite positive integers is $8, 9, 10$.)\n\n\u2022 (+1) Thanks a ton for the proof, Is there no other way besides the use of software to look for the first such sequence? \u2013\u00a0Nikunj Feb 28 '16 at 11:02\n\u2022 @Nikunj No, because (as stated in my answer) we do not know the exact distribution of primes. \u2013\u00a0Bobson Dugnutt Feb 28 '16 at 11:33\n\u2022 In general the problem is difficult, so yes, for sufficiently large $k$ computer assistance is necessary. Note that the primorial $(k - 1)\\#$ is much smaller than the factorial $(k + 1)!$ for even modest $k$, so Arthur's solution gives a much better upper bound for the position of the first such sequence, but in general it too is a pretty poor bound: Already for $k = 5$ it gives $204, \\ldots, 208$, but the first sequence of five consecutive composite numbers is $24, \\ldots, 28$. \u2013\u00a0Travis Willse Feb 28 '16 at 11:40\n\u2022 @ travis : proving $\\pi(x) = o(x)$ is much easier than the prime number theorem, for example it follows directly from $s \\int_1^\\infty \\pi(x) x^{-s-1} dx \\sim \\ln \\zeta(s) \\sim -\\ln(s-1)$ when $s \\to 1^+$. \u2013\u00a0reuns Feb 28 '16 at 12:21\n\nThe problem of finding large gap between consecutive primes is an old and well studied one. There certainly is a large gap between what we know to be true, and what we suspect.\n\nAs for what people expect, Cramer's random provides a good source heuristics. Roughly spreaking, you can think of a number $n$ to be \"prime with probability $\\frac{1}{\\log n}$\", but sometimes you also have to take into consideration that the primes tend not to be divisible by $2, 3, 5,$, etc. Apparently, if you make a lot of optimistic assumptions, then you can reach the conclusion that the longest gap between primes in integers $\\leq X$ is roughly $\\log^2 X$. Hence, if you want your gap to have size $g$, then you should look at integers $\\leq e^{\\sqrt{g}}$ (which, for $g = 10000$ is pretty large). See this post of Terence Tao for details.\n\nMuch less has been proved. The best known bounds for the largest gap between primes is due to Kevin Ford, Ben Green, Sergei Konyagin, James Maynard, Terence Tao, and says that the largest gap below $X$ is at least $c \\frac{\\log X \\log \\log X \\log \\log \\log \\log X}{\\log \\log \\log X}$, where $c$ is a constant.\n\n\u2022 In the last para, I guess \u201clargest gap between integers\u201d should be \u201c\u2026between primes\u201d. I believe there are fairly sharp upper and lower bounds known for gaps between consecutive integers :-P \u2013\u00a0Peter LeFanu Lumsdaine Feb 28 '16 at 15:54\n\u2022 @PeterLeFanuLumsdaine: Thanks! I suppose the bound is no longer true for integers, but maybe I'll check with one of the authors ;) \u2013\u00a0Jakub Konieczny Feb 28 '16 at 16:56\n\u2022 That \"best known bound\" is amazingly low if you consider that the average gap between primes is log X, so that average gap is improved by less than a factor c log log X. And c might be considerably smaller than 1. \u2013\u00a0gnasher729 Feb 29 '16 at 23:08\n\u2022 @gnasher729: It is rather far from the prediction, yes. But it should be pointed out that we're much better in predicting things about the primes than in actually proving them. \u2013\u00a0Jakub Konieczny Mar 1 '16 at 3:10\n\u2022 Are there any known bounds on $c$? \u2013\u00a0Brevan Ellefsen Dec 11 '16 at 17:56\n\nAny sequence with density $0$, in most reasonable meanings of \"density $0$\", has arbitrarily large gaps.\n\nThis is true (for example, and in increasing order of generality) if density $0$ is interpreted as:\n\n\u2022 the density on $[1,n]$ converges to $0$, or\n\n\u2022 the lim inf of the density on $[1,n]$ is $0$, or\n\n\u2022 the lim inf of the density on intervals of length $n$ is $0$\n\nThe asymptotic density of primes $\\pi(n)/n \\to 0$ is 0-density in the strong sense, which is more than enough to ensure large gaps.\n\nThe $n!$ examples of large gaps can be reduced from factorial to \"primorial\" size and the latter seems to be the best currently known deterministic construction.\n\nThis isn't going to be a fancy answer, (secondary school student) but I would believe it to be as a number rises the number of possible factors of the number also rise meaning there is a larger chance of it having more factors than itself and one. So a number has every number smaller than it as a possible factor, so smaller numbers have less possible factors and therefore less chance of more than 1 factor and itself.\n\nThere several factors determining the size of gaps. As per you example 10,000 has a square root of 100 which give you 25 prime number divisors for the 10,000 and the 20,000 square root is 141 and 34 divisors, and the 30,000 square root is 173 and 40 divisors. The number of divisors helps to determine the gaps. The number of factors needed is based on the square root of your max number. The square roots are decreasing percentage of the max number: example 100 / 10000 = 1.000%, 141 / 20000 = .70500% and 173 / 30000 = .57666%. The number of factors increased against the square root but decreases against the max number. Example: 106,749,747,000,000 the square root is 10,331,977 the square root % is .00000968% with 685,159 factors the max gap in my sample 712. The large gaps seam to have a cycle pattern to them.", "date": "2020-01-26 12:24:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1675485/are-there-arbitrarily-large-gaps-between-consecutive-primes", "openwebmath_score": 0.8219617009162903, "openwebmath_perplexity": 287.0025412481365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363700641143, "lm_q2_score": 0.8740772236840657, "lm_q1q2_score": 0.8617845250747866}}
{"url": "https://math.stackexchange.com/questions/1254572/whats-the-difference-between-antisymmetric-and-reflexive-set-theory-discrete-m", "text": "# Whats the difference between Antisymmetric and reflexive? (Set Theory/Discrete math)\n\nAntisymmetric: $\\forall x\\forall y[ ((x,y)\\in R\\land (y, x) \\in R) \\to x= y]$ reflexive: $\\forall x[x\u2208A\\to (x, x)\\in R]$\n\nWhat really is the difference between the two? Wouldn't all antisymmetric relations also be reflexive?\n\n\u2022 E.g. \"$\\leq$\" and \"$<$\" are antisymmetric and \"$=$\" is reflexive. Apr 27 '15 at 17:02\n\nHere are a few relations on subsets of $\\Bbb R$, represented as subsets of $\\Bbb R^2$. The dotted line represents $\\{(x,y)\\in\\Bbb R^2\\mid y = x\\}$.\n\nSymmetric, reflexive:\n\nSymmetric, not reflexive\n\nAntisymmetric, not reflexive\n\nNeither antisymmetric, nor symmetric, but reflexive\n\nNeither antisymmetric, nor symmetric, nor reflexive\n\n\u2022 This is a great visual approach to understanding the meaning of the words! Apr 27 '15 at 20:04\n\u2022 Thank you so much for making these, they're great! Apr 29 '15 at 2:48\n\nNo, there are plenty of anti-symmetric relations that are not reflexive.\n\nFor instance, let $R$ be the relation $R=\\{(1,2)\\}$ on the set $A=\\{1,2,3\\}$. This relation is certainly not reflexive, but it is in fact anti-symmetric. This is vacuously true, because there are no $x$ and $y$, such that $(x,y)\\in R$ and $(y,x)\\in R$.\n\nEdit: Why is this anti-symmetric? Because in order for the relation to be anti-symmetric, it must be true that whenever some pair $(x,y)$ with $x\\neq y$ is an element of the relation $R$, then the opposite pair $(y,x)$ cannot also be an element of $R$. This is true for our relation, since we have $(1,2)\\in R$, but we don't have $(2,1)$ in $R$.\n\nAlso, the relation $R=\\{(1,2),(2,3),(1,1),(2,2)\\}$ on the same set $A$ is anti-symmetric, but it is not reflexive, because $(3,3)$ is missing.\n\nAs for a reflexive relation, which is not anti-symmetric, take $R=\\{(1,1),(2,2),(3,3),(1,2),(2,1)\\}$.\n\n\u2022 Thank you so much for your answer, the last two parts make sense! :) I'm a little lost on the first part because the law says that if (x,y) and (y,x) then y=x. Could you elaborate a bit more on how R = {(1,2)} is anti-symmetric? Apr 27 '15 at 17:35\n\u2022 Sorry, I think I messed up. Let me edit my post. There. Now, I have redone the last two examples, because they were wrong. I'll edit my post further to elaborate on why the first relation is in fact anti-symmetric. :)@TaylorTheDeveloper Apr 27 '15 at 17:42\n\u2022 This may sound like a naive question but would'nt this example be asymmetric also then by vacuous agument Oct 19 '20 at 11:31\n\u2022 @angshuknag Yes, the relation $R=\\{(1,2)\\}$ is also asymmetric. In fact, being asymmetric is equivalent to being both anti-symmetric and not reflexive. Oct 19 '20 at 15:08\n\nThey're two different things, there isn't really a strong relationship between the two.\n\nBased on the definitions you're using, they both give two different criteria for concluding that $(x, x) \\in R$.\n\nFor any antisymmetric relation $R$, if we're given two pairs, $(x, y)$ and $(y, x)$ both belonging to $R$, then we can conclude that in fact $x = y$, so that that\n\n$$(x, y) = (x, x) = (y, x),$$\n\nand $(x, x) \\in R$. It may really be better stated as saying that\n\n$$\\text{ If } x \\neq y, \\text{ then at most one of (x, y) or (y, x) is in R}.$$\n\nThat is, it may be a bit misleading to even think about $(x,y)$ and $(y, x)$ as being pairs in $R$, since antisymmetry forces them to in fact be the same pair, $(x, x)$.\n\nAntisymmetric relations may or may not be reflexive. $<$ is antisymmetric and not reflexive, while the relation \"$x$ divides $y$\" is antisymmetric and reflexive, on the set of positive integers.\n\nA reflexive relation $R$ on a set $A$, on the other hand, tells us that we always have $(x, x) \\in R$; everything is related to itself. Reflexive relations may or may not be symmetric, or antisymmetric:\n\n$\\leq$ is reflexive and antisymmetric, while $=$ is reflexive and symmetric.\n\n\u2022 Also, I may have been misleading by choosing pairs of relations, one symmetric, one antisymmetric - there's a middle ground of relations that are neither! For example, the relation \"$x$ divides $y$\" on the set of all integers is neither symmetric nor antisymmetric (and neither reflexive, nor irreflexive). Apr 27 '15 at 17:52\n\u2022 The divisibility relation is reflexive, even on all integers. Apr 27 '15 at 20:04\n\u2022 @JadeNB Thank you, of course you're right; I'm not sure why I had decided it wasn't! Apr 27 '15 at 20:07\n\u2022 Probably the presence of 0 caused some reflexive (no pun intended!) worries. Apr 27 '15 at 20:11", "date": "2021-10-18 10:41:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1254572/whats-the-difference-between-antisymmetric-and-reflexive-set-theory-discrete-m", "openwebmath_score": 0.8542302250862122, "openwebmath_perplexity": 308.6375928459945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338101862455, "lm_q2_score": 0.8962513675912912, "lm_q1q2_score": 0.8617759923646875}}
{"url": "https://mathoverflow.net/questions/300249/rotatable-matrix-its-eigenvalues-and-eigenvectors", "text": "# Rotatable matrix, its eigenvalues and eigenvectors\n\nWe say that a real matrix is rotatable iff after turning it clockwise on $90^{\\circ}$ it doesn't change.\n\nI'm interesting about eigenvalues and eigenvectors (belonging to non-zero eigenvalues) of such type of matrices. For example, it is not hard to show that for every tuple of real values $\\lambda_1,\\ldots,\\lambda_{k}$ there exists $n\\in\\mathbb{N}$ and a rotatable $n\\times n$ matrix $A$ such that all $\\lambda_i$ are eigenvalues of $A$.\n\nIndeed, let us consider a matrix $$A_1 = \\begin{pmatrix} a & a \\\\ a & a\\\\ \\end{pmatrix}.$$ Then of course $A_1$ is rotatable and its characteristic polynomial $\\chi_{A_1}(x) = x^2-2ax = x(x-2a)$ and of course for every $\\lambda$ we can choose $a$ (for example $\\lambda/2$).\n\nNow let us show how we can construct a rotatable matrix $A_2$ with prescribed eigenvalues $\\lambda_1,\\lambda_2$. For example, we can consider a matrix of the form $$A_2 = \\begin{pmatrix} b&0&0&b\\\\ 0&a & a&0 \\\\ 0&a & a&0\\\\ b&0&0&b \\end{pmatrix}.$$ Then $\\chi_{A_2}(x) =x^2(x-2a)(x-2b)$. And we are done for $a=\\lambda_1/2, b = \\lambda_2/2$. Of course, using this method we can construct the required rotatable matrix for every $k$ of size $2k$.\n\nAlso my experiments show that all eigenvectors $v = (v_1,\\ldots,v_n)^T$ belonging to non-zero eigenvalues of roratable matrix are symmetric, that is $v_i = v_{n-i+1}$. It is simple to prove this in the case, where $n=2$. Also I tried to prove it by induction on $n$, but my attempts failed.\n\nMy question.\n\n1. For a given tuple $\\lambda_1,\\ldots, \\lambda_k$ can we construct a rotatable matrix $A$ of size $n\\times n$, where $n<2k$, such that all $\\lambda_i$ are eigenvalues of $A$.\n\n2. Is it always true that all eigenvectors of non-zero eigenvalues of rotatable matrix are symmetric? And if the answer is \"yes\", how to prove this.\n\n\u2022 Is the (2,4) element of $A_2$ supposed to be 0 rather than $b$? If not, how is it rotatable? May 15, 2018 at 11:23\n\u2022 Doesn't every rotatable matrix become of the form $\\begin{bmatrix}M & M \\\\ M & M\\end{bmatrix}$, i.e., $\\begin{bmatrix}1 & 1\\\\ 1 & 1\\end{bmatrix} \\otimes M$, if you conjugate it by a suitable permutation? That looks like it would simplify the analysis a lot. May 15, 2018 at 21:04\n\u2022 @FedericoPoloni, does this hold for $3\\times 3$ matrices? I don't sure. May 16, 2018 at 10:30\n\u2022 @MikhailGoltvanitsa No, only in even dimension. May 16, 2018 at 14:28\n\u2022 In particular your matrices are centrosymmetric (apply the rotation twice will also yield the same matrix) en.m.wikipedia.org/wiki/Centrosymmetric_matrix May 18, 2018 at 15:34\n\nConsider the matrix $$P= \\begin{pmatrix} 0 & \\ldots & 1 \\\\ \\vdots & 1 & \\vdots \\\\ 1 & \\ldots & 0 \\end{pmatrix}$$ with $1$s along the \"other\" main diagonal and $0$s elsewhere. Then $(PA)^t$ is a rotation of the matrix $A$ by $90^\\circ$ (you can check this on the basis $E_{i,j}$ of matrices with a 1 in the $(i,j)$ slot and 0s elsewhere). So $A$ is rotatable if and only if $(PA)^t = A$, i.e. $PA=A^t$. Note that $P^2=I$ and $A^tP = A$, so $A^t = AP$ and hence $A$ and $P$ must commute. (For what follows below we assume that $A$ is a real matrix. Otherwise I think all you can say is that $A$ preserves the splitting $\\mathbb{C}^n=E_+ \\oplus E_-$ as explained below. In the complex case we get symmetric eigenvectors of the $v+Pv$ as well as the skew symmetric vectors $v-Pv$.)\n\nIt then follows that $A$ and $A^t$ commute and so, by the Spectral Theorem, $A$ has a unitary basis. Now, notice that $P$ has eigenvalues $1$ and $-1$ with eigenvectors $e_i + Pe_i$ and $e_i-Pe_i$ for $1 \\leq i \\leq \\frac{n+1}{2}$, respectively. Let $E_+ = \\ker (P-I)$ and $E_- = \\ker (P+I)$. Then $A(E_\\pm) \\subseteq E_\\pm$, so $A$ must preserve the splitting $\\mathbb{R}^n = E_+ \\oplus E_-$. Note that $\\dim E_+ = \\left\\lfloor \\frac{n+1}{2}\\right\\rfloor$.\n\nNow, the answer to question $2$ is yes if we require the corresponding eigenvalue to be a nonzero real number. If $\\lambda$ is an eigenvalue with eigenvector $v \\in E_\\pm$ and $(\\:\\: ,\\:\\: )$ is the standard hermitian inner product on $\\mathbb{C}^n$, we have $$\\lambda(v,v)=(v,A^tv) = (v,APv) = \\pm \\bar{\\lambda}(v,v).$$\n\nSo $\\lambda \\in \\mathbb{R}$ if and only if $v \\in E_+$ and $\\lambda \\in i \\mathbb{R}$ if and only if $v \\in E_-$. Note that $v \\in E_+$ if and only if $Pv=v$, which is the same as requiring that $v$ is symmetric. Furthermore, note that the eigenvectors in $E_- \\setminus \\{0\\}$ have complex coefficients as they satisfy $Av=i \\mu v$. So it is precisely the real eigenvectors of nonzero eigenvalues that are symmetric. The others will be \"skew symmetric\".\n\nQuestion $1$ is a bit trickier, and the answer is no in general. Let us treat the even and odd dimensional cases separately.\n\nLet $n=2m$. If $A$ is rotatable it has the form $$A= \\begin{pmatrix} B & B^t P \\\\ PB^t & PBP \\end{pmatrix}$$ for some matrix $B$ (here $P$ and $B$ are matrices of size $m\\times m$). Now, $\\lambda$ is a real eigenvalue of $A$ with eigenvector $\\begin{pmatrix} v \\\\ Pv \\end{pmatrix}$ if and only if $$Bv + B^t v = \\lambda v .$$ So $\\lambda = 2 \\operatorname{Re}(\\eta)$, where $\\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B+B^t$. Similarly, $\\lambda= i \\mu$ is a purely imaginary eigenvalue of $A$ with eigenvector $\\begin{pmatrix} v \\\\ -Pv \\end{pmatrix}$ if and only if $Bv - B^t v = i \\mu v$, so $\\mu = 2 \\operatorname{Im} (\\eta)$, where $\\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B-B^t$.\n\nSo, if you choose more than $n$ real numbers, or choose any complex numbers $\\lambda$ that do not satisfy $\\bar{\\lambda} = \\pm \\lambda$, or you choose more than $n$ purely imaginary numbers, then they cannot be eigenvalues of $A$.\n\nNow, in the odd dimensional case $n=2m+1$, $A$ takes the form $$A= \\begin{pmatrix} B & u & B^t P \\\\ u^t & a & (Pu)^t \\\\ PB^t& Pu & PBP \\end{pmatrix}$$ for some matrix $B$, vector $u$ and real number $a$. For $\\lambda$ a real eigenvalue of $A$, we have the eigenvector $\\begin{pmatrix} v \\\\ b \\\\Pv \\end{pmatrix}$, so $$(B+B^t)v + bu = \\lambda v$$ and $$2(u,v)+ab=\\lambda b.$$ Choosing $u=0$ gives $$(B+B^t)v=\\lambda v$$ and $ab=\\lambda b.$ So we can pick $m$ eigenvectors for $B+B^t$ as above, and also the vector given by $v=0$ and $b=1$, which has eigenvalue $a$. The imaginary eigenvalues $i \\mu$ have eigenvectors $\\begin{pmatrix} v\\\\0\\\\-Pv \\end{pmatrix}$, and so $$(B-B^t)v=i\\mu v,$$ with the other equation being $0=0$. So as before $\\mu$ is twice the imaginary part of an eigenvalue of $B$.", "date": "2023-02-06 10:22:56", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/300249/rotatable-matrix-its-eigenvalues-and-eigenvectors", "openwebmath_score": 0.9633433818817139, "openwebmath_perplexity": 66.74895121994571, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846703886661, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8617583618781078}}
{"url": "https://math.stackexchange.com/questions/1238873/evaluate-lim-n-to-infty-int-01-fracn12n1-left-fract1/1239007", "text": "# Evaluate $\\lim_{n \\to \\infty} \\int_{0}^1 \\frac{n+1}{2^{n+1}} \\left(\\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\\right) \\mathrm{d}t$\n\nEvaluate $$\\lim_{n \\to \\infty} \\int_{0}^1 \\frac{n+1}{2^{n+1}} \\left(\\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\\right) \\mathrm{d}t$$\n\nFor this integral, I have tried using integration by parts and then evaluating the limit, but I don't think the integral inside converges. However, the limit does exist and the answer given in my book is $2$.\n\nAny help will be appreciated.\n\n\u2022 can you show us your working? \u2013\u00a0danimal Apr 17 '15 at 11:06\n\u2022 @Henry which book is this? \u2013\u00a0Kugelblitz Jun 17 '17 at 4:00\n\nFirst, consider the following two Lemmas,\n\nLemma $1$: $$\\lim_{n \\to \\infty} \\sum_{r=0}^n \\left(\\dfrac{1}{\\displaystyle\\binom{n}{r}}\\right) =2$$\n\nProof : First of all, note that the limit exists, since, if we let\n$$\\text{S}(n)=\\displaystyle \\sum_{r=0}^n \\left(\\dfrac{1}{\\dbinom{n}{r}} \\right)$$\n\nthen $\\text{S}(n+1)<\\text{S}(n)$ for $n \\geq 4$. Now,\n\n$\\text{S}(n) = 1+ \\displaystyle \\sum_{r=1}^n \\dfrac{1}{\\dbinom{n}{r}}$\n\n$\\implies \\text{S}(n) = 1+ \\displaystyle \\sum_{r=1}^n \\dfrac{r}{n} \\times \\dfrac{1}{\\dbinom{n-1}{r-1}} \\ \\left[\\text{since} \\dbinom{n}{r}= \\dfrac{n}{r} \\times \\dbinom{n-1}{r-1} \\right]$\n\nAlso,\n\n$$\\text{S}(n) = 1+ \\sum_{r=1}^n \\dfrac{1}{\\dbinom{n}{r}} = 1+ \\sum_{r=1}^n \\dfrac{1}{\\dbinom{n}{n-r+1}} = 1+ \\sum_{r=1}^n \\dfrac{n-r+1}{n} \\dfrac{1}{\\dbinom{n-1}{n-r}} = 1+ \\sum_{r=1}^n \\dfrac{n-r+1}{n} \\dfrac{1}{\\dbinom{n-1}{r-1}}$$\n\n$\\left[ \\text{since} \\ \\displaystyle \\sum_{r=a}^b f(r) = \\displaystyle \\sum_{r=a}^b f(a+b-r) \\ \\text{and} \\ \\dbinom{n}{r}=\\dbinom{n}{n-r} \\right]$\n\nThus, we have,\n\n$$\\begin{cases} \\text{S}(n) = 1+ \\displaystyle \\sum_{r=1}^n \\dfrac{r}{n} \\times \\dfrac{1}{\\dbinom{n-1}{r-1}}\\\\ \\text{S}(n) = 1+ \\displaystyle \\sum_{r=1}^n \\dfrac{n-r+1}{n} \\dfrac{1}{\\dbinom{n-1}{r-1}} \\end{cases}$$\n\nAdding the above two expressions, we get,\n\n$2\\text{S}(n) = 2 + \\displaystyle \\sum_{r=1}^n \\left( \\dfrac{r}{n} \\times \\dfrac{1}{\\dbinom{n-1}{r-1}} + \\dfrac{n-r+1}{n} \\dfrac{1}{\\dbinom{n-1}{r-1}} \\right)$\n\n$= 2 + \\dfrac{n+1}{n} \\displaystyle \\sum_{r=1}^n \\dfrac{1}{\\dbinom{n-1}{r-1}}$\n\n$= 2 + \\dfrac{n+1}{n} \\times \\text{S}(n-1)$\n\nSince $n \\to \\infty$, we have $\\text{S}(n) = \\text{S}(n-1) = \\text{S}$ (say)\n\n$\\implies 2\\text{S} = \\left(\\dfrac{n+1}{n}\\right) \\times \\text{S} +2$\n\n$\\implies \\text{S} = \\dfrac{2n}{n-1} = 2$ [since $n \\to \\infty$]\n\nLemma $2$ : $$\\int_{0}^1 x^r (1-x)^{n-r} \\mathrm{d}t = \\dfrac{1}{(n+1)}\\times \\dfrac{1}{\\dbinom{n}{r}}$$\n\nProof : Consider the $\\text{R.H.S.}$,\n\n$\\text{I} = \\displaystyle\\int_{0}^1 x^r (1-x)^{n-r} \\mathrm{d}t$\n\nLet $x = \\sin^2 \\theta$\n\n$\\implies \\text{I} = \\displaystyle\\int_{0}^{\\frac{\\pi}{2}} 2 \\sin^{2r+1} \\theta \\cos^{2n-2r} \\theta \\ \\mathrm{d}\\theta$\n\nNow, using Walli's Formula (or reduction formula) for the above integral, we have,\n\n$\\text{I} = \\dfrac{1}{(n+1)}\\times \\dfrac{1}{\\dbinom{n}{r}}$\n\nThis proves our Lemmas.\n\nNow,\n\n$$\\text{J} = \\lim_{n \\to \\infty} \\int_{0}^1 \\frac{n+1}{2^{n+1}} \\left(\\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\\right) \\mathrm{d}t$$\n\nSince it is an even function in $t$, we have,\n\n$$\\text{J} = \\frac{1}{2} \\times \\lim_{n \\to \\infty} \\int_{-1}^1 \\frac{n+1}{2^{n+1}} \\left(\\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\\right) \\mathrm{d}t$$\n\nLet $t = 2x-1$\n\n$\\implies \\text{J} = \\displaystyle \\lim _{n \\to \\infty} \\int_{0}^1 (n+1) \\left(\\dfrac{x^{n+1}-(1-x)^{n+1}}{2x-1}\\right) \\mathrm{d}x$\n\n$=\\displaystyle \\lim _{n \\to \\infty} \\int_{0}^1 (n+1) (1-x)^n \\left(\\dfrac{\\left(\\frac{x}{1-x}\\right)^{n+1}-1}{\\frac{x}{1-x}-1}\\right) \\mathrm{d}x$\n\n$=\\displaystyle \\lim _{n \\to \\infty} \\int_{0}^1 (n+1) \\sum_{r=0}^n (1-x)^n \\left(\\frac{x}{1-x}\\right)^{r} \\mathrm{d}x$\n\n$=\\displaystyle \\lim _{n \\to \\infty} \\sum_{r=0}^n (n+1) \\int_{0}^1 x^r(1-x)^{n-r} \\mathrm{d}x$\n\n$=\\displaystyle \\lim _{n \\to \\infty} \\sum_{r=0}^n \\dfrac{1}{\\dbinom{n}{r}}$ (Using Lemma 2)\n\n$=\\boxed{2}$ (Using Lemma 1).\n\nSide Note : Another way to prove Lemma 1 is to use sandwich theorem.\n\n\u2022 Wow! Ingenious and rigorous as always :) But can you please tell what led you to think this way? \u2013\u00a0Henry Durham Apr 17 '15 at 13:06\n\u2022 @Samurai Your question reminded me of Beta functions which I have used to create this solution. Although understanding the solution doesn't need the understanding of Beta functions, but if you look carefully at Lemma 2, it is the definition of Beta function, for which I've provided an elementary proof. The rest followed from rigorous brainstorming :) \u2013\u00a0MathGod Apr 17 '15 at 13:11\n\u2022 @MathGod Thanks a lot :D \u2013\u00a0Henry Durham Apr 17 '15 at 13:14\n\u2022 @MathGod Would you mind explaining how you got: $$\\int_{0}^1 (n+1) \\sum_{r=0}^n (1-x)^n \\left(\\frac{x}{1-x}\\right)\\mathrm{d}x$$ after $$\\int_{0}^1 (n+1) (1-x)^n \\left(\\dfrac{\\left(\\frac{x}{1-x}\\right)^{n+1}-1}{\\frac{x}{1-x}-1}\\right) \\mathrm{d}x$$ It is not so clear to me. Thank you. \u2013\u00a0Tolaso Apr 18 '15 at 4:33\n\u2022 @Tolaso Thanks for pointing that out. It was a minor typo and I've now corrected it. The expression is sum of G.P. \u2013\u00a0MathGod Apr 18 '15 at 6:05\n\nIt is worth to notice that: $$\\int_{0}^{1}\\frac{1-(1-t)^n}{t}\\,dt = H_n\\tag{1}$$ while: $$\\int_{0}^{1}\\frac{(1+t)^n-1}{t}\\,dt=\\int_{0}^{1}\\sum_{k=0}^{n-1}(1+t)^k\\,dt =\\sum_{k=1}^{n}\\frac{2^k-1}{k}\\tag{2}$$ so: $$J_n=\\int_{0}^{1}\\frac{(1+t)^n-(1-t)^n}{t}\\,dt = \\sum_{k=1}^{n}\\frac{2^k}{k}.\\tag{3}$$ The last line also gives: $$\\frac{n\\, J_n}{2^n} = \\sum_{k=1}^{n}\\frac{2n}{n+1-k}\\cdot 2^{-k} \\tag{4}$$ and when $n$ approaches $+\\infty$, by the dominated convergence theorem the RHS of $(4)$ approaches: $$\\sum_{k=1}^{+\\infty}2\\cdot 2^{-k} = \\color{red}{2} \\tag{5}$$ as wanted.\n\n\u2022 Pardon me, but I don't know what's dominated convergence theorem. \u2013\u00a0Henry Durham Apr 17 '15 at 11:33\n\u2022 \u2013\u00a0Jack D'Aurizio Apr 17 '15 at 11:35\n\u2022 That's sweet... I guess, without knowing the properties of $H_n$ you couldn't do much... But how did you get the first equality in $(2)$? \u2013\u00a0Igor Deruga Apr 17 '15 at 11:43\n\u2022 @Igor: $t=(1+t)-1$ and $1+z+\\ldots+z^n = \\frac{z^{n+1}-1}{z-1}$. \u2013\u00a0Jack D'Aurizio Apr 17 '15 at 11:48\n\nWe need to address area around $0$ separately from area around $1$. Pick $\\epsilon\\in(0,1)$. Then on $[0,\\epsilon]$ $$\\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\\le (n+1)(1+\\epsilon)^{n}+(n+1)(1-\\epsilon)^{n}$$(you can see this if you multiply both sides by $t$ and compare derivatives) and hence $$\\lim_{n\\to\\infty}\\int_0^{\\epsilon}\\frac {n+1}{2^{n+1}}\\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\\,dt=0$$ On the other hand $$\\lim_{n\\to\\infty}\\int_\\epsilon^1\\frac {n+1}{2^{n+1}}\\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\\,dt=\\lim_{n\\to\\infty}\\int_{\\epsilon}^1\\frac 2 t\\frac {n+1}{n+2} d\\Big(\\big(\\frac {1+t}{2}\\big)^{n+2}+\\big(\\frac {1-t} 2\\big)^{n+2}\\Big)=\\int_{\\epsilon}^1\\frac 2 t dI(t=1)=2$$ since $\\frac 2 t$ is bounded on $[\\epsilon,1]$. Hence the original limit is $2$.", "date": "2020-10-20 07:15:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1238873/evaluate-lim-n-to-infty-int-01-fracn12n1-left-fract1/1239007", "openwebmath_score": 0.9001554250717163, "openwebmath_perplexity": 485.79493510555847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846703886661, "lm_q2_score": 0.8807970685907242, "lm_q1q2_score": 0.8617583496324389}}
{"url": "http://mathhelpforum.com/pre-calculus/168532-complex-numbers-question.html", "text": "# Math Help - Complex Numbers Question\n\n1. ## Complex Numbers Question\n\nHey everybody,\n\nSo I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.\n\nIf x and y are real, solve the equation:\n\n$\\frac{jx}{1+jy}=\\frac{3x+j4}{x+3y}$\n\nI started manipulating it to make it more manageable:\n\n$\\frac{jx(x+3y)}{1+jy}=3x+j4$\n\n$jx^2+j3xy=(3x+j4)(1+jy)$\n\netc.\n\nI have the feeling I'm going nowhere with it. Could somebody please give me a hand?\n\nCheers,\n\nEvanator\n\n2. Originally Posted by evanator\nHey everybody,\n\nSo I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.\n\nIf x and y are real, solve the equation:\n\n$\\frac{jx}{1+jy}=\\frac{3x+j4}{x+3y}$\n\nI started manipulating it to make it more manageable:\n\n$\\frac{jx(x+3y)}{1+jy}=3x+j4$\n\n$jx^2+j3xy=(3x+j4)(1+jy)$\n\netc.\n\nI have the feeling I'm going nowhere with it.\nIn fact, you,re going in exactly the right direction. Next step is to multiply out the brackets on the right (remembering that $j^2=-1$). Then compare the real parts and the imaginary parts on both sides of the equation.\n\n3. $\\frac{jx}{1+jy}=\\frac{3x+j4}{x+3y}\n$\n\n${(jx)}{(x+3y)}=(1+jy)(3x+j4)$\n\n$jx^2+3jxy=3x+4j+3jxy-4y$\n\n$jx^2=3x+4j-4y$\n\nso:\n\n(1) $x^2=4$\n\n(2) $0=3x-4y$\n\nSolve this system for x and y\n\n4. Thank you both of you. Finishing it off:\n\n$jx^2+j3xy=3x+j3xy+j4+(j^2)4y$\n\n$0+jx^2=(3x-4y)+j(4)$\n\n$x^2=4$\n\n$3x-4y=0$\n\n$x=\\pm2$\n\nTaking the positive value of x and solving the simultaneous equations:\n\n$3x=6$\n\n$-3x+4y=0$\n\n$4y=6$\n\n$y=\\frac{6}{4}=1.5$\n\nIf we take x=-2 we can see that y will come out as -1.5, so the final results are:\n\n$x=\\pm2$\n\n$y=\\pm1.5$\n\n5. No, that is NOT the answer because x and y are not independent.\n\nThe correct answer is \"x= 2 and y= 1.5 or x= -2 and y= -1.5\"\n\nWriting \" $x= \\pm 2, y= \\pm 1.5$\" implies that any combination of those, such as x= 2, y= -1.5, will satisfy the equation and that is not true.\n\n6. Right. Thanks, HallsofIvy. I did understand that the x-value and y-value were dependent on each other, but I didn't realize I was implying otherwise with the plus-minus sign.", "date": "2015-11-27 21:11:58", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/168532-complex-numbers-question.html", "openwebmath_score": 0.6900927424430847, "openwebmath_perplexity": 453.21830773438603, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120884041585, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8617129044552443}}
{"url": "https://mathematica.stackexchange.com/questions/137350/variable-increments-in-a-table", "text": "# Variable Increments in a Table\n\nI am working on a problem involving the piecewise function:\n\n$$x^2 \\quad \\mbox{if } \\quad x < 2$$\n\n$$(x-3)/(\\sqrt{x-2}-1) \\quad \\mbox{if } \\quad x \\geq 2$$\n\nI am supposed to find the values of the limits as $x$ approaches $2$ from the right and the left. I've found these values to be $1$ from the right and $4$ from the left.\n\nHowever, the question I'm struggling to answer involves the formatting of a limit table.\n\nI've been able to make two tables of (x,f[x]) from the left and the right, and I've listed specific values of $x$ to approach $2$ with. However, the question wants me to make a $4$ column table using only one table command, and to use a rule to make the step size increments progressively smaller. I'm not sure how to do this.\n\nHere's the text of the question: \"To evaluate a two sided limit, create a single table (generated by a single table command) with four columns, combining the left-and right-sided limit tables into one. Values of $x$ should approach the limiting value moving down the table. To evaluate a limit, create table(s) that use step-size increments that are progressively smaller as the limiting value is approached (eg. x=3, 2.1, 2.01, 2.001,... for a limit as x\u2192 2+). Rather than listing every $x$-value, use a rule to generate the $x$-values.\"\n\nHere is how I've made my tables thus far:\n\nTableForm[Table[{x, f[x]}, {x, {4, 3, 2.5, 2.5, 2.25, 2.1, 2.05, 2.02,2.001}}]]\nTableForm[Table[{x, f[x]}, {x, {-1, 0, 1, 1.5, 1.75, 1.9, 1.95, 1.97, 1.98, 1.999}}]]\n\n\u2022 Is this homework? If so, please tag it as such.\n\u2013\u00a0ciao\nFeb 9, 2017 at 8:21\n\nOne possible way to use Accumulate to generate the steps.\n\nN@Table[i, {i, Accumulate[Table[1/(2^t), {t, 1, 10}]]}]\n\n\nThen you can these values for building the steps, by adding and subtracting the above from 2 for each side.\n\nClearAll[f, i, t, x];\n\nf[x_] := Piecewise[{{x^2, x < 2}, {(x - 3)/(Sqrt[x - 2] - 1), x >= 2}}]\n\nr = Table[{x = 1 + i; x, f[x], x = 3 - i; x, f[x]},\n{i,Accumulate[Table[1/(2^t), {t, 1, 10}]]}];\n\nh = {\"x\", \"limit from left\", \"x\", \"limit from right\"};\nGrid[Join[{h}, N@r], Alignment -> Left, Frame -> All]\n\n\nPlot[f[x], {x, -1, 3}, Frame -> True, GridLines -> Automatic,\nGridLinesStyle -> LightGray]", "date": "2022-06-29 16:40:45", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/137350/variable-increments-in-a-table", "openwebmath_score": 0.3952745497226715, "openwebmath_perplexity": 1066.9754382125268, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9539661002182845, "lm_q2_score": 0.9032942008463507, "lm_q1q2_score": 0.861712046131185}}
{"url": "https://math.stackexchange.com/questions/592537/why-can-the-limit-of-a-sequence-approach-a-number-and-converge-but-the-limit-of/592548", "text": "# Why can the limit of a sequence approach a number and converge, but the limit of the series must approach $0$ to converge?\n\nMy question may not make much sense because I'm still trying to wrap my mind around infinite sequences and series. I seem to have good working knowledge of when and why to apply a certain tests for a given series, but something just seems like it is missing in my understanding.\n\nConceptually, why can a sequence converge at any number (except $-\\infty$ and $\\infty$, which aren't numbers anyway), but the limit of a series must approach $0$ to converge?\n\nIs the reason behind the requirement for the limit of a series approaching $0$ to converge because the series eventually stops summing numbered terms infinitely due to its convergence to a specific number? (Sorry if this question makes no sense)!\n\nAnd is it acceptable for a sequence to approach any number simply because the sequence isn't being summed?\n\n\u2022 You mean thr limit of the summand \u2013\u00a0Mhenni Benghorbal Dec 4 '13 at 15:17\n\u2022 Try to look at the partial sums of the series when $a_n$ does not converge to $0$ for $n\\rightarrow \\infty$. \u2013\u00a0user112167 Dec 4 '13 at 15:17\n\u2022 @MhenniBenghorbal excuse my ignorance, but what is a summand? \u2013\u00a0hax0r_n_code Dec 4 '13 at 15:18\n\nI think you are mixing sequences and their sums. The terminology is not very helpful here, I would suggest the following.\n\nLet $a_k$ be any sequence. Then you can define the sequence of partial sums: $$S_n = \\sum_{k=1}^n a_k.$$ Then $$\\sum_{k=1}^\\infty a_k = \\lim_{n\\to \\infty} S_n.$$\n\nGiven any sequence $S_n$ you can find some $a_k$ such that $S_n$ are the partial sums of $a_k$.\n\nSo you see that a series may have any limit, as the sequence do. However if a series $S_n$ has a finite limit, then the corresponding sequence $a_k$ must tend to zero.\n\nDon't confuse the series $S_n$ with its general term $a_k$.\n\n\u2022 So is it accurate to say that a series is simply a sequence of partial sums from a given sequence? \u2013\u00a0hax0r_n_code Dec 4 '13 at 15:29\n\u2022 That is my preferred point of view. \u2013\u00a0Emanuele Paolini Dec 4 '13 at 15:31\n\u2022 That's interesting because I never could see the relationship between a sequence and a series. \u2013\u00a0hax0r_n_code Dec 4 '13 at 15:37\n\u2022 I say that the terminology does not help because formally the expression: $\\sum_{k=1}^\\infty a_k$ is a number. So it makes no sense to say that such expression converges or diverges. But really you are referring to the sequence of partial sums, not to the sum itself. \u2013\u00a0Emanuele Paolini Dec 4 '13 at 15:40\n\u2022 @EmanuelePaolini It makes perfect sense to say that expression converges or diverges. It is nothing but a limit, and limits can sometimes fail to be numbers, but instead diverge. \u2013\u00a0Potato Dec 4 '13 at 16:05", "date": "2019-10-16 10:45:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/592537/why-can-the-limit-of-a-sequence-approach-a-number-and-converge-but-the-limit-of/592548", "openwebmath_score": 0.9530243873596191, "openwebmath_perplexity": 203.33475503078589, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137937226358, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.8616943184179622}}
{"url": "http://math.stackexchange.com/questions/265397/inversion-of-the-euler-totient-function", "text": "# inversion of the Euler totient function\n\nGiven an integer $n$ find smallest integer $x$ such that $\\varphi(x) = n$.\n\n$$10^5 < n < 10^8$$\n\nI know that lower bound for searching is $n+1$ and the upper bound is\n\n$$\\frac{n}{e^{0.577}\\log(\\log(n))} + \\frac{3}{\\log(\\log(n))}$$\n\nBut my problem is that for such a large values of $n$ my program with precomputed $\\varphi(x)$ values has to do a lot of searching and is highly inefficient.\n\nCan you please provide any other method for doing the same .\n\nThanks.\n\n-\nMy apologies if I am missing something basic here, but what function is it that you are defining to mean $\\varphi(x)$? \u2013\u00a0Amzoti Dec 26 '12 at 18:31\nEuler totient function.. phi(x)... gives the number of integers less than x which are coprime to x \u2013\u00a0Purva Gupta Dec 26 '12 at 18:35\nSo I don't waste time, you are asking for an efficient algorithm for inverting the Euler Totient function. Correct? \u2013\u00a0Amzoti Dec 26 '12 at 19:26\nyes.. which can work for large test cases \u2013\u00a0Purva Gupta Dec 26 '12 at 19:48\nI would recommend changing your title to be more descriptive in case others are searching for this too and there have been previous questions regarding approaches. Regards \u2013\u00a0Amzoti Dec 27 '12 at 2:19\n\nWarning, you are going to have to do some work to get your hands around this answer, but I provided enough details for you to work through it and it answers your question.\n\nGoal: Given an integer $n$ find the smallest integer $x$ such that $\\varphi(x) = n$.\n\nApproach\n\nOne can calculate all of the possible integers with a totient of $n$ using \"Inverse Totient Trees.\"\n\nFor instance, take the integers with a totient of $24$. Then,\n\n$\\varphi(N) = 24$\n\n$\\varphi(24) = 8$\n\n$\\varphi(8) = 4$\n\n$\\varphi(4) = 2$\n\n$\\varphi(2) = 1$\n\nThere are $5$ \"links\" (designate this as $L$) in the totient chain so to speak, with $4$ intervals. In general, the greatest integer that can have a totient of $n$ is $2*3^{(L-1)}$, which means that $2*3^{(5 - 1)}$ $= 162$ is the upper bound of an integer with a totient of $24$.\n\nIn fact, via a simple proof by exhaustion, one can easily check a table and see that the smallest integer where $\\varphi(x) = 24$ is $x = 35$ (see the list below).\n\n$$\\varphi(x)= n = 24 \\rightarrow x = 35, 39, 45, 52, 56, 70, 72, 78, 84, 90$$\n\nHere are a couple of related number sequences which include references for you to investigate this approach.\n\nA032447 Inverse function of phi( )\n\nA058811 Number of terms on the n-th level of the Inverse-Totient-Tree (ITT)\n\nSo, the next obvious question, is there a program that is related or tangentially related that can generate all of those values in the range specified using some approach?\n\nYes, see Solving $\\varphi^{-1}(x) = n$, where $\\varphi(x) = n$ is Euler's totient function - testing Carmichael's conjecture. Please see the references there for further details on this program - including the C-source code.\n\nLets do the example above and the some examples within your range (note that numbers are obviously never odd because $\\varphi(x)$ is always even for $n \\ge 3$).\n\n$n = 24$: Enter $n = 24, e = 0, f = 0$: and look at the resulting $10$-bolded numbers and compare that list to the above. If you sort that list from low to high, it is all of the $x's$ that produce that desired $n$. Of course, you only want the minimal (last bolded number in the display) one. So $\\varphi^{-1}(35) = 24$.\n\n$n = 10^{5}$: Enter $n = 100000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100651$.\n\n$n = 10^5 + 1$: Enter $n = 100001, e = 0, f = 0$: Odd numbers are not permissible by statement above.\n\n$n = 10^{5} + 2$: Enter $n = 100002, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100003$.\n\n$n = 5*10^{5}$: Enter $n = 500000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 640625$.\n\n...\n\n$n = 10^{8}$: Enter $n = 100000000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100064101$.\n\nAsides\n\nThis is a very interesting problem and has been asked before, so I am summarizing some of the other discussions and references I found in case others want to consider more approaches.\n\nThere are several papers on the topic of finding the inverse of the Euler Totient function:\n\nEuler's Totient Function and Its Inverse, by Hansraj Gupta\n\nThe number of solutions of $\\phi(x) = m$, by Kevin Ford\n\nOn the image of Euler\u2019s totient function, R.Coleman\n\nComplexity of Inverting the Euler Function, by Scott Contini, Ernie Croot, Igor Shparlinski\n\nThere have been several questions along these lines, for example, the-inverse-of-the-euler-totient-function, inverting-the-totient-function and finding-the-maximum-number-with-a-certain-eulers-totient-value\n\nThere are some other code examples that use different methods for you to consider, see, for example:\n\nInversion of Euler Totient Function by Max Alekseyev and you can experiment with this since PARI/GP is free. Play around with the function and see if you can modify your approach with this approach.\n\nDiscussion and implementation of an efficient algorithm for finding all the solutions to the equation EulerPhi[n] = m, by Maxim Rytin is a nice article off of wolfram that gives an efficient algorithm for computing the inverse of the Euler totient function. Download the invphi.nb file at the bottom and get Mathreader. As an alternative, you can see oeis A006511.\n\nMAGMA - see FactoredEulerPhiInverse(n)\n\nRegards\n\n-\nThanks a lot for your descriptive answer :) \u2013\u00a0Purva Gupta Dec 27 '12 at 8:19\n@Purva Gupta, you are very welcome! We are all here to learn from each other and MSE is a wonderful site! \u2013\u00a0Amzoti Dec 27 '12 at 9:01\n2\u22173^(L\u22121) is it applicable for all or only specific numbers? \u2013\u00a0Purva Gupta Dec 27 '12 at 9:18\nAll, it is an upper bound using this method for the largest n, recalling that n must be even numbers only from the note. \u2013\u00a0Amzoti Dec 27 '12 at 13:08\n$+1^{+1^{+1}}\\quad\\checkmark$ \u2013\u00a0amWhy May 10 '13 at 1:20\n\nSee also my recent paper \"Computing the (number of) inverses of Euler's totient and other multiplicative functions\", which presents a generic dynamic programming algorithm for finding the inverses of a multiplicative function for a given integer value.\n\n-", "date": "2016-05-27 14:39:36", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/265397/inversion-of-the-euler-totient-function", "openwebmath_score": 0.802922785282135, "openwebmath_perplexity": 431.90800360978295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137937226356, "lm_q2_score": 0.8774767954920548, "lm_q1q2_score": 0.8616943168447341}}
{"url": "http://mathhelpforum.com/algebra/188735-factorization-question-2-a.html", "text": "1. ## Factorization Question 2\n\nFor the question below, does anyone have a quick and easy method for finding the two numbers that when added together gives 11x, and when multiplied together gives -180? I know the numbers are 20 and -9 but it takes me forever to find them? Is there is quick way?\n\n$\\text{Factorize } 12x^2+11x-15$\n\n2. ## Re: Factorization Question 2\n\nHello\n\nI don't now if this will help, but I do it this way. It doesn't take ages.\nYou know you have either:\n\n1,15 or 3,5\n\nand your looking for a difference of 11\n\nWrite out:\n\n(12x )(x )\n(6x )(2x )\n(4x )(3x )\n\nIt doesn't tke long to work out the difference of the possible products and see that 4x5 - 3x3 = 11\n\nSo you have\n\n(4x - 3)(3x + 5)\n\n3. ## Re: Factorization Question 2\n\nIf you have a polynomial $f(x)$ and the zero's $x_1,x_2,...x_n$ are known then the polynomial can be factored as:\n$f(x)=a(x-x_1)(x-x_2)....(x-x_n)$ (where $a$ is the coefficient of the highest degree term).\nOffcourse in this case you have to find the zero's first (by using the quadratic formula) ...\n\n4. ## Re: Factorization Question 2\n\nHello, sparky!\n\nFor the question below, does anyone have a quick and easy method for\nfinding the two numbers whose difference is 11 and whose product 180?\nI know the numbers are 20 and -9, but it takes me forever to find them?\nIs there is quick way?\n\n$\\text{Factor: }\\:12x^2+11x-15$\n\nI use a primitive approach.\nIt takes a bit of time, but it is direct.\n\nFactor 180 into pairs of factors, and note the difference.\n\n. . $\\begin{array}{cc}\\text{Factors} & \\text{Diff.} \\\\ \\hline 1,180 & 179 \\\\ 2,90 & 88 \\\\ 3,60 & 57 \\\\ 4,45 & 41 \\\\ 5,36 & 31 \\\\ 6,30 & 24 \\\\ 7,? & - \\\\ 8,? & - \\\\ 9,20 & 11 \\\\ 10,18 & 8 \\\\ 11,? & - \\\\ 12,15 & 3 \\\\ 13,? & - \\\\ \\hline \\end{array} \\begin{array}{c} \\\\ \\\\ \\\\ \\\\ \\\\ \\leftarrow\\text{ Here!} \\end{array}$\n\nHow do we find these pairs?\nDivide 180 by 1, 2, 3, . . .\nSome of the divisions do not \"come out even\", of course.\n\nWhen can we stop?\nWe can stop at the integer part of $\\sqrt{180} \\,\\rightarrow\\, 13$\n\nWe see that the pair $(9,20)$ has a difference of 11.\n(Of course, we can stop listing the moment we find it.)\n\nWe have: . $12x^2 \\quad 9x\\quad 20x - 15$\n\nWe want the middle term to be +11x, so we will use -9x and +20x.\n\nSo we have: . $12x^2 - 9x + 20x - 15$\n\nFactor \"by grouping\": . $3x(4x-3) + 5(4x-3) \\;=\\;(4x-3)(3x+5)$\n\n5. ## Re: Factorization Question 2\n\nOriginally Posted by Soroban\nHello, sparky!\n\nI use a primitive approach.\nIt takes a bit of time, but it is direct.\n\nFactor 180 into pairs of factors, and note the difference.\n\n. . $\\begin{array}{cc}\\text{Factors} & \\text{Diff.} \\\\ \\hline 1,180 & 179 \\\\ 2,90 & 88 \\\\ 3,60 & 57 \\\\ 4,45 & 41 \\\\ 5,36 & 31 \\\\ 6,30 & 24 \\\\ 7,? & - \\\\ 8,? & - \\\\ 9,20 & 11 \\\\ 10,18 & 8 \\\\ 11,? & - \\\\ 12,15 & 3 \\\\ 13,? & - \\\\ \\hline \\end{array} \\begin{array}{c} \\\\ \\\\ \\\\ \\\\ \\\\ \\leftarrow\\text{ Here!} \\end{array}$\n\nHow do we find these pairs?\nDivide 180 by 1, 2, 3, . . .\nSome of the divisions do not \"come out even\", of course.\n\nWhen can we stop?\nWe can stop at the integer part of $\\sqrt{180} \\,\\rightarrow\\, 13$\n\nWe see that the pair $(9,20)$ has a difference of 11.\n(Of course, we can stop listing the moment we find it.)\n\nWe have: . $12x^2 \\quad 9x\\quad 20x - 15$\n\nWe want the middle term to be +11x, so we will use -9x and +20x.\n\nSo we have: . $12x^2 - 9x + 20x - 15$\n\nFactor \"by grouping\": . $3x(4x-3) + 5(4x-3) \\;=\\;(4x-3)(3x+5)$\n\nWow, I love this! Why didn't I think of this? This was so obvious! Thanks\n\n6. ## Re: Factorization Question 2\n\nOriginally Posted by Soroban\nHello, sparky!\n\nI use a primitive approach.\nIt takes a bit of time, but it is direct.\n\nFactor 180 into pairs of factors, and note the difference.\n\n. . $\\begin{array}{cc}\\text{Factors} & \\text{Diff.} \\\\ \\hline 1,180 & 179 \\\\ 2,90 & 88 \\\\ 3,60 & 57 \\\\ 4,45 & 41 \\\\ 5,36 & 31 \\\\ 6,30 & 24 \\\\ 7,? & - \\\\ 8,? & - \\\\ 9,20 & 11 \\\\ 10,18 & 8 \\\\ 11,? & - \\\\ 12,15 & 3 \\\\ 13,? & - \\\\ \\hline \\end{array} \\begin{array}{c} \\\\ \\\\ \\\\ \\\\ \\\\ \\leftarrow\\text{ Here!} \\end{array}$\n\nHow do we find these pairs?\nDivide 180 by 1, 2, 3, . . .\nSome of the divisions do not \"come out even\", of course.\n\nWhen can we stop?\nWe can stop at the integer part of $\\sqrt{180} \\,\\rightarrow\\, 13$\n\nWe see that the pair $(9,20)$ has a difference of 11.\n(Of course, we can stop listing the moment we find it.)\n\nWe have: . $12x^2 \\quad 9x\\quad 20x - 15$\n\nWe want the middle term to be +11x, so we will use -9x and +20x.\n\nSo we have: . $12x^2 - 9x + 20x - 15$\n\nFactor \"by grouping\": . $3x(4x-3) + 5(4x-3) \\;=\\;(4x-3)(3x+5)$\n\n^this. is beautiful.", "date": "2017-02-27 17:31:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/188735-factorization-question-2-a.html", "openwebmath_score": 0.7130520939826965, "openwebmath_perplexity": 1282.9514400785129, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137931962462, "lm_q2_score": 0.8774767874818409, "lm_q1q2_score": 0.8616943085166989}}
{"url": "http://backup.hoteleuphoria.ro/t946mkf/8rgsf.php?ff=multiple-regression-ppt", "text": "In the multiple regression model, the goodness-of- t measure R-squared always increases (or remains the same) when an additional explanatory variable is added. Slides Prepared by JOHN S. We'll just use the term \"regression analysis\" for all these variations. While many statistical software packages can perform various types of nonparametric and robust regression. Orlov Chemistry Department, Oregon State University (1996) INTRODUCTION In modern science, regression analysis is a necessary part of virtually almost any data reduction process. wnarifin@usm. Note that the intercept will always be zero and so we could have used regression without an intercept to obtain the same regression coefficients (although the standard errors will be slightly different). A simple linear regression equation for this would be $$\\hat{Price} = b_0 + b_1 * Mileage$$. Multiple regression is a flexible method of data analysis that may be appropriate whenever a quantitative variable (the dependent or criterion variable) is to be examined in relationship to any other factors (expressed as independent or predictor variables). i = be the value of. This PowerPoint is a workshop on multiple linear regression and analysis of covariance. Multiple Linear Regression Multiple linear regression attempts to model the relationship between two or more explanatory variables and a response variable by fitting a linear equation to observed data. Lecture 1 Introduction to Multi-level Models Regression Analysis? \u2022 Specification of predictor variables from multiple levels. The goal of a model is to get the smallest possible sum of squares and draw a line that comes closest to the data. When I started experimenting with machine learning, I wanted to come up with an application that would solve a real-world problem but would not be too complicated to implement. This statistics is for multiple linear regression technique. Once you have completed the test, click on 'Submit Answers' to get your results. | PowerPoint PPT presentation | free to view. Multiple Regression and Correlation Dr. Multiple regression generally explains the relationship between multiple independent or predictor variables and one dependent or criterion variable. Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression \u2026\\1. Objectives of Multiple Regression In selecting suitable applications of multiple regression, the researcher must consider three primary issues: 1. Secondary Data Analysis \u2022 Starting Off Right: Effects of Rurality on Parent\u201fs Involvement in Children\u201fs Early Learning (Sue Sheridan, PPO) \u2013 Data from the Early Childhood Longitudinal Study \u2013 Birth Cohort (ECLS-B) were used to examine the influence of setting on parental involvement in preschool and the effects of involvement on. Example: estimated coefficient of education equals 92. Dummy Variables Dummy Variables A dummy variable is a variable that takes on the value 1 or 0 Examples: male (= 1 if are male, 0 otherwise), south (= 1 if in the south, 0 otherwise), etc. Multiple regression analysis is almost the same as simple linear regression. \u201cencoding model\u201d. NASCAR Race Crashes SAS Program SAS Output. Multiple Regression Now, let\u2019s move on to multiple regression. , the dependent variable) of a fictitious economy by using 2 independent/input variables:. Simple linear regression and multiple regression using least squares can be done in some spreadsheet applications and on some calculators. The example below demonstrates the use of the summary function on the two models created during this tutorial. One regressor should not be a linear function of another. This lesson explores the use of a regression analysis to answer. In the orange juice classification problem, Y can only take on two possible values: 0 or 1. As a data scientist, one must always explore multiple options for solving the same analysis or modeling task and choose the best for his/her particular problem. Understand and use bivariate and multiple linear regression analysis. In the following example, we will use multiple linear regression to predict the stock index price (i. patient, f. Chapter 9 Correlational Research Designs What are correlational research designs, and why are they used in behavioral research. Edward\u2019s University Chapter 12 Simple Linear Regression Simple Linear Regression Model Least Squares Method Coefficient of Determination Model Assumptions Testing for Significance Using the Estimated Regression Equation for Estimation and Prediction Computer Solution Residual Analysis: Validating Model Assumptions Simple Linear Regression Model y = b0. Regression analysis helps in establishing a functional Relationship between two or more variables. Amaral November 21, 2017 Advanced Methods of Social Research (SOCI 420). Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression \u2026\\1. Introduction to Correlation and Regression Analysis. Simple regression analysis uses a single x variable for each dependent \u201cy\u201d variable. Multiple Linear Regression Model We consider the problem of regression when study variable depends on more than one explanatory or independent variables, called as multiple linear regression model. One will not encounter cases in a multiple regression path model where one could go through the same variable twice. The generic form of the linear regression model is y = x 1\u03b2 1 +x 2\u03b2 2 +. Multiple linear regression attempts to fit a regression line for a response variable using more than one explanatory variable. selection of the dependent and independent variables. The model states that the expected value of Y--in this case, the expected merit pay increase--equals \u03b20 plus \u03b21 times X. In ordinary samples we would expect 95% of cases to have standardized residuals within about +2 and -2. Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression \u2026\\1. If you click. y = c + ax c = constant a = slope. It can also handle multiple predictor variables. Multiple Linear Regression and Matrix Formulation Introduction I Regression analysis is a statistical technique used to describe relationships among variables. , the dependent variable) of a fictitious economy by using 2 independent/input variables:. This is a simplified tutorial with example codes in R. One type of analysis many practitioners struggle with is multiple regression analysis, particularly an analysis that aims to optimize a response by finding the best levels for different variables. Your print-out of the slides needs to be modified at one point. Use Multiple Regression to model the linear relationship between a continuous response and up to 12 continuous predictors and 1 categorical predictor. The simple proportional hazards model generalizes to a multiple regression model in much the same way as for linear and logistic regression. Posc/Uapp 816 Class 14 Multiple Regression With Categorical Data Page 3 1. Lecture 6:. Standard multiple regression is the same idea as simple linear regression, except now you have several independent variables predicting the dependent variable. If two variables are trending over time, a regression. Assessing Studies Based on Multiple Regression Part III. \u2013 The test is multiple linear regression isThe test is multiple linear regression is H 0 = \u03b2 1 = \u03b2 2 = \u2026 = \u03b2 p = 00. Root MSE = s = our estimate of \u03c3 = 2. This is what regression analysis can do! For example, we'll see (via regression) that Fords. Regression analysis is a statistical process for estimating the relationships among variables. \" In the main dialog box, input the dependent variable. txt) or view presentation slides online. Lecture 18: Multiple Logistic Regression Mulugeta Gebregziabher, Ph. Clearly, it is nothing but an extension of Simple linear regression. We expect to build a model that fits the data better than the simple linear regression model. Dummy variables are useful because they enable us to use a single regression equation to represent multiple groups. A multiple regression based model will use the data to build a function that predicts the outcome based on the independent variables. Technically, a regression analysis model is based on the sum of squares, which is a mathematical way to find the dispersion of data points. 8 Quantitative Business Analysis for Decision Making Multiple Linear Regression Analysis Outlines Multiple Regression Model Estimation Testing Significance of Predictors Multicollinearity Selection of Predictors Diagnostic Plots Multiple Regression. When you have more than two events, you ca n extend the binary logistic regression model, as described in Chapter 3. y = c + ax c = constant a = slope. Frank Wood, fwood@stat. Econometrics I: Multiple Regression: Inference - H. One of these variable is called predictor variable whose value is gathered through experiments. When you look at the output for this multiple regression, you see that the two predictor model does do significantly better than chance at predicting cyberloafing, F(2, 48) = 20. In each ex-ample, you will \ufb01rst learn about the speci\ufb01c ingredi-ents required for the power or sample size computa-tion for the linear model being considered. It is also called the coefficient of determination, or the coefficient of multiple determination for multiple regression. Nonlinear regression The model is a nonlinear function of the parameters. University of Dayton. The program\u2019s graph, regression, and correlation functions can respectively produce scatterplots, provide regression equation coefficients, and create correlation matrices. Assumptions of Multiple Regression This tutorial should be looked at in conjunction with the previous tutorial on Multiple Regression. When using multiple regression to estimate a relationship, there is always the possibility of correlation among the independent variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Further details and related logical operations can be found in the R documentation. Multiple Personality Disorder. A sound understanding of the multiple regression model will help you to understand these other applications. Switching Regression Models \u2014 Estimation (8) First obtain the expected values of the residuals that are truncated. We will also learn how to decide whether a group of explanatory variables can be excluded from the model. \"Linear Regression and Modeling\" is course 3 of 5 in the Statistics with R Coursera Specialization. In this paper, the risk factors for a disease of the eye (retinopathy of prematurity) are identi ed using logistic regression analysis. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a \"model\" for predicting a wide variety of outcomes. With only one independent variable, the regression line can be plotted neatly in two dimensions. However, performing a regression does not automatically give us a reliable relationship between the variables. xls Data Sample (showing only the variables to be used in analysis) Variables Used. Chapter 8: Multiple Choice Questions. Lecture 1 Introduction to Multi-level Models Regression Analysis? \u2022 Specification of predictor variables from multiple levels. Simple linear regression. Variable and Dummy Coded Region Variable 273. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. We are not going to go too far into multiple regression, it will only be a solid introduction. zeigler-hill. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Regression analysis is a statistical process for estimating the relationships among variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Many people find this too complicated to understand. I also wanted to practice working with regression algorithms. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. MR&B3 \u00a0is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. ppt - Free download as Powerpoint Presentation (. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. In the simplest case, we would use a 0,1 dummy variable where a person is given a value of 0 if they are in the control group or a 1 if they are in the treated group. We are not going to go too far into multiple regression, it will only be a solid introduction. \u2013 The test is multiple linear regression isThe test is multiple linear regression is H 0 = \u03b2 1 = \u03b2 2 = \u2026 = \u03b2 p = 00. An Introduction to Logistic Regression: From Basic Concepts to Interpretation with Particular Attention to Nursing Domain ure\" event (for example, death) during a follow-up period of observation. Wouldn\u2019t it be great if there was a more accurate way to predict whether your prospect will buy rather than just taking an educated guess? Well, there is\u2026if you have enough data on your previous prospects. This model is built using, for example, a set of real world data listing the outcome in various cases. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Multiple Regression Multiple Proportional Hazards Regr (Cox model) For time dependent outcomes (ie time to death), we model the hazard rate, h , the event rate per unit time (for death, it is the mortality rate). How to Run a Multiple Regression in Excel. Wooldridge, Introductory Econometrics, 4th ed. If two variables are trending over time, a regression. Multiple regression is an extension of simple linear regression. Multiple linear regression has one y and two or more x variables. 8%, with 95% confidence. is a way of understanding the relationship between two variables. For instance, when we predict rent based on square feet alone that is simple linear regression. Stepwise multiple linear regression has proved to be an extremely useful computational technique in data analysis problems. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Background Multiple Linear Regression is widely used in academics and also in MR We can consider it the start of Multivariate Analysis, for our course Any idea what the following are: Multivariate analysis Multiple Linear Regression (MLR). The precision of a method (Table 4) is the extent to which the individual test results of multiple injections of a series of standards agree. For example, linear regression can be used to quantify the relative impacts of age, gender, and diet (the predictor variables) on height (the outcome variable). Regression assumes that variables have normal distributions. regress is useful when you simply need the output arguments of the function and when you want to repeat fitting a model multiple times in a loop. 3 Notes: PowerPoint xxx PDF. Suppose we have a cohort of. Lecture 24: Partial correlation, multiple regression, and correlation Ernesto F. Third, multiple regression offers our first glimpse into statistical models that use more than two quantitative. If you have the Excel desktop application, you can use the Open in Excel button to open your workbook and use either the Analysis ToolPak's Regression tool or statistical functions to perform a regression analysis there. Uses of Regression Analysis 1. If you look at the first slide on centering, it states \"To center: Y i centered = Yi \u2013 Mi. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. You probably remember the concept of simple linear regression intuition from your high school years. Similar tests. Notes on Regression Model \u2022 It is VERY important to have theory before starting developing any regression model. Lecture 4: Multivariate Regression Model in Matrix Form In this lecture, we rewrite the multiple regression model in the matrix form. Partial r is just another way of standardizing the coefficient, along with beta coefficient (standardized regression coefficient)$^1$. selection of the dependent and independent variables. edu|http://www. Simple linear regression is a bivariate situation, that is, it involves two dimensions, one for the dependent variable Y and one for the independent variable x. ppt), PDF File (. Formula 13. This correlation may be pair-wise or multiple correlation. The linear regression model (LRM) The simple (or bivariate) LRM model is designed to study the relationship between a pair of variables that appear in a data set. Regression analysis is a very widely used statistical tool to establish a relationship model between two variables. , between an independent and a dependent variable or between two independent variables). There are two models of logistic regression, binary logistic regression and multinomial logistic regression. If the universe is only 14 billion years old, how can it be 92 billion light years wide? - Duration: 9:47. It is the correlation between the variable's values and the best predictions that can be computed linearly from the predictive variables. If you look at the first slide on centering, it states \"To center: Y i centered = Yi \u2013 Mi. It can also be used to estimate the linear association between the predictors and reponses. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Lecture 18: Multiple Logistic Regression - p. You can use it to predict values of the dependent variable, or if you're careful, you can use it for suggestions about which independent variables have a major effect on the dependent variable. Known also as curve fitting or line fitting because a regression analysis equation can be used in fitting a curve or line to. Intercept: the intercept in a multiple regression model is the mean for the response when. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. The data was split into three employment sectors Teaching, government and private industry Each sector showed a positive relationship Employer type was confounded with degree level Simpson\u2019s Paradox In each of these examples, the bivariate analysis (cross-tabulation or correlation) gave misleading results Introducing another variable gave a. Here, the summary(OBJECT) function is a useful tool. +x K \u03b2 K +\u03b5 where y is the dependent or. Upload and Share PowerPoint Presentations. Pantula David A. covariates for the. MV - Multiple Regression. The generic form of the linear regression model is y = x 1\u03b2 1 +x 2\u03b2 2 +. Typically points further than, say, three or four standard deviations from the mean are considered as \"outliers\". This post will show you examples of linear regression. Regression Analysis. patient, f. In this exercise we apply the same method to validating the Novy-Marx Quality and Value screener\u2019s ability to predict the 5-year total return. Multiple Regression Analysis using SPSS Statistics Introduction. 01, with an R-square of. Briefly speaking, the goal of the multiple linear regression is to point out the relation between a dependent variable (explained, endogenous or resultative) and a great deal of independent variables (explanatory, factorial,. Comparing Multiple Regression Model Results against Historic Demand. However, it is critical to recognize that multiple regression is inherently a correlation technique and cannot. \" In the main dialog box, input the dependent variable. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Rawlings Sastry G. Regression analysis is a statistical technique for estimating the relationship among variables which have reason and result relation. If you look at the first slide on centering, it states \"To center: Y i centered = Yi \u2013 Mi. persistence of shocks will be infinite for nonstationary series \u2022 Spurious regressions. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. If two variables are trending over time, a regression. RSM is a method used to locate the optimum value of the response and is one of the final stages of experimentation. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Suppose we have a cohort of. Cherry Blossoms WS: Test Review. ID: 305741 ID: 305741. Linear regression Linear dependence: constant rate of increase of one variable with respect to another (as opposed to, e. As a data scientist, one must always explore multiple options for solving the same analysis or modeling task and choose the best for his/her particular problem. 1 Introduction to Poisson Regression As usual, we start by introducing an example that will serve to illustrative regression models for count data. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Multiple (Linear) Regression. This tutorial will explore how R can be used to perform multiple linear regression. Dummy variables are useful because they enable us to use a single regression equation to represent multiple groups. is a way of understanding the relationship between two variables. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. This article is a part of the guide:. Suppose that there is a cholesterol lowering drug that is tested through a clinical trial. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. We have been reviewing the relationship between correlation (r and r2) and regression (R, R2 and 1-R2) in class, through lectures, blog. Nonlinear Regression Functions Chapter 9. Regression Analysis. (3) Does the combination of predictors in this fitted multiple regression explain significant variation in the response? Statistics 621 Multiple Regression Practice Questions. Multiple Regression Analysis using SPSS Statistics Introduction. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. selection of the dependent and independent variables. Consider \ufb01rst the case of a single binary predictor, where x = (1 if exposed to factor 0 if not;and y =. The output varies linearly based upon the input. Multiple regression uses the ordinary least squares solution (as does bi. The whole model F test (test of the useful of the model) tests whether the slopes on all variables in multiple regression are zero, i. Ryan McEwan and Julia Chapman. Simple linear regression involves a single independent variable. 12-1 Multiple Linear Regression Models \u2022 Many applications of regression analysis involve situations in which there are more than one regressor variable. Introduction to Linear Regression and Correlation Analysis Fall 2006 \u2013 Fundamentals of Business Statistics 2 Chapter Goals To understand the methods for displaying and describing relationship among variables. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. Chapter 8: Multiple Choice Questions. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Regression with two or more predictors is called multiple regression Available in all statistical packages Just like correlation, if an explanatory variable is a significant predictor of the dependent variable, it doesn't imply that the explanatory variable is a cause of the dependent variable. It is discussed in Response Surface Methods. Hey I would like to make a scatter plot with p-value and r^2 included for a multiple linear regression. 0 Chapter 6: Multiple Linear Regression Topics Explanatory Modeling Predictive Modeling Example: Prices of Toyota Corolla ToyotaCorolla. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a \"model\" for predicting a wide variety of outcomes. Regression is the analysis of the relation between one variable and some other variable(s), assuming a linear relation. Also referred to as least squares regression and ordinary least squares (OLS). Regression when all explanatory variables are categorical is \u201canalysis of variance\u201d. 5 Correlation and Regression Simple regression 1. Notes on Regression Model \u2022 It is VERY important to have theory before starting developing any regression model. The Regression df is the number of independent variables in the model. We can have only two models or more than three models depending on research questions. For example, consider the cubic polynomial model which is a multiple linear regression model with three regressor variables. Multiple Regression 18. X2 1 or even interactions X1 X2. The following model is a multiple linear regression model with two predictor variables, and. Find PowerPoint Presentations and Slides using the power of XPowerPoint. This post will show you examples of linear regression. Please access that tutorial now, if you havent already. The basic command is \"regression\": \"linear. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. Objectives. MULTIPLE LINEAR REGRESSION ANALYSIS USING MICROSOFT EXCEL by Michael L. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Frank Wood, fwood@stat. 1 Introduction. \" Coefficient table, bottom. Under Type of power analysis, choose \u2018A priori\u2026\u2019, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. View and Download PowerPoint Presentations on Spss Tutorial For Multiple Logistic Regression PPT. One of the applications of multiple linear regression models is Response Surface Methodology (RSM). State the multiple regression equation. For ordina l categorical variables, the drawback of the. Linear, Ridge Regression, and Principal Component Analysis Geometric Interpretation I Each column of X is a vector in an N-dimensional space (NOT the p-dimensional feature vector space). Linear Regression. The model is linear because it is linear in the parameters , and. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. In R, multiple linear regression is only a small step away from simple linear regression. Regression Analysis. The basic command is \u201cregression\u201d: \u201clinear. Multiple linear regression in R Dependent variable: Continuous (scale/interval/ratio) Independent variables: Continuous (scale/interval/ratio) or binary (e. y = c + ax c = constant a = slope. 8%, with 95% confidence. We can run regressions on multiple different DVs and compare the results for each DV. Multiple Regression Models \u2022 Advantages of multiple regression \u2022 Important preliminary analyses \u2022 Parts of a multiple regression model & interpretation \u2022 Differences between r, bivariate b, multivariate b & \u2022 Steps in examining & interpreting a full regression model Advantages of Multiple Regression Practical issues \u2026. See the Handbook for information on these topics. Multiple Regression - Linearity. ppt - Free download as Powerpoint Presentation (. This correlation may be pair-wise or multiple correlation. Arial Franklin Gothic Book Perpetua Wingdings 2 Calibri Equity 1_Equity 2_Equity 3_Equity 4_Equity Microsoft Equation 3. Be sure to tackle the exercise and the quiz to get a good understanding. One type of analysis many practitioners struggle with is multiple regression analysis, particularly an analysis that aims to optimize a response by finding the best levels for different variables. 2% with 95% confidence. Lecture 18: Multiple Logistic Regression - p. The measured standard deviation can be subdivided into 3 categories: repeatability, intermediate precision and reproducibility (4, 5). Find PowerPoint Presentations and Slides using the power of XPowerPoint. By selecting the features like this and applying the linear regression algorithms you can do polynomial linear regression; Remember, feature scaling becomes even more important here; Instead of a conventional polynomial you could do variable ^(1/something) - i. MR&B3 \u00a0is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. Introduction to Correlation and Regression Analysis. It is used to also to determine the overall fit of the model and the contribution of each of the predictors to the total variation. Uses of Regression Analysis 1. Variable Interactions 5. 11 LOGISTIC REGRESSION - INTERPRETING PARAMETERS 11 Logistic Regression - Interpreting Parameters Let us expand on the material in the last section, trying to make sure we understand the logistic regression model and can interpret Stata output. Figure 14 \u2013 Model Summary Output for Multiple Regression. Starting values of the estimated parameters are used and the likelihood that the sample came from a population with those parameters is computed. On the contrary, regression is used to fit a best line and estimate one variable on the basis of another variable. For simple linear regression (only one x), the Regression df is 1. Source: Hair et al. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Predictions in neurophysiology. Nonlinear regression The model is a nonlinear function of the parameters. The only difference between simple linear regression and multiple regression is in the number of predictors (\u201cx\u201d variables) used in the regression. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. 12-1 Multiple Linear Regression Models \u2022 Many applications of regression analysis involve situations in which there are more than one regressor variable. specification of a statistical relationship, and 3. Upload and Share PowerPoint Presentations. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Popular spreadsheet programs, such as Quattro Pro, Microsoft Excel,. biostatcourse. Teaching\\stata\\stata version 13 \u2013 SPRING 2015\\stata v 13 first session. Under Type of power analysis, choose \u2018A priori\u2026\u2019, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. How to improve leadership skills in the workplace ppt for write my essay nursing best cover letter editing site uk , writing a scientific report buy a college essay online. Logistic regression is a statistical method for analyzing a dataset in which there are one or more independent variables that determine an outcome. Regression with categorical variables and one numerical X is often called \u201canalysis of covariance\u201d. It could be, for example, performance, education, or experience. The course starts with a discussion of the logic of the multivariate regression model and the central assumptions underlying the ordinary least squares approach. Also referred to as least squares regression and ordinary least squares (OLS). Linear regression for the advertising data Consider the advertising data shown on the next slide. The method of multiple regression sought to create the most closely related model. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Multiple regression is used to build a model that allows us to study this interplay. What to report? What a statistics program gives you: For a simple regression (one independent variable), statistics programs produce two estimates, a (the \"constant term\") and b (the \"linear coefficient\"), for the parameters \u03b1 and \u03b2, respectively. For example: (x 1, Y 1). 2% with 95% confidence.", "date": "2019-12-13 08:59:17", "meta": {"domain": "hoteleuphoria.ro", "url": "http://backup.hoteleuphoria.ro/t946mkf/8rgsf.php?ff=multiple-regression-ppt", "openwebmath_score": 0.3387893736362457, "openwebmath_perplexity": 937.9905486025981, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.982013792143467, "lm_q2_score": 0.8774767746654974, "lm_q1q2_score": 0.8616942950070836}}
{"url": "https://www.hpmuseum.org/forum/thread-383.html", "text": "HP 11C real root finder [Newton Method]\n01-09-2014, 11:59 PM\nPost: #1\n Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013\nHP 11C real root finder [Newton Method]\nLBL 0\nRCL 1\nGSB 1\nRCL 1\nGSB 2\n/\nCHS\nRCL 1\n+\nSTO 3\nRCL 1\n-\nABS\nRCL 2\nX>Y?\nRTN\nRCL 3\nSTO 1\nGTO 0\n\nLBL 1\nf(x)=0 code\n\nLBL 2\nf'(x)=0 code\n\nR1: old value (or initial value)\nR2: Tolerance of error (0.001, 0.0001, 0.000000001 etc)\n\nTHE RESULT:\nR3: new value (root)\n\nEXAMPLE\n\nFind the root of f(x)=x^3 - 3x^2-6x+8\n\nrewritten as f(x)= [(x-3)x-6]x +8\n\nf(x) code:\n\nLBL1\nENTER\nENTER\nENTER\n3\n-\n*\n6\n-\n*\n8\n+\nRTN\n\nf'(x)= 3x(x-6)-6\nf'(x) code\n\nLBL 2\nENTER\nENTER\n6\n-\n*\n3\n*\n6\n-\nRTN\n\nThe roots are [-2, 1, 4]\n\ngive a initial value, for example: -3\n\n-3 STO 1\n\ngive a tolerance of error, for example: 0.0001\n\n0.0001 STO 2\n\nBEGIN THE PROGRAM...\n\nGSB 0\n\nrunnning...\n\nwhen the error < TOL the program stop and display:\n\n0.0001\n\nyou can find the root in the register 3\n\nRCL 3\n\nDISPLAY: -2.0001\n\ntry with initial value = 2\n\n[/size][/font][/size][/size][/font][/font]\n\nBest Regards\n01-10-2014, 12:57 AM\nPost: #2\n Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013\nRE: HP 11C real root finder [Newton Method]\nUsing storage arithmetic allows us to get rid of a few steps:\n\nLBL 0\nRCL 1\nGSB 1\nRCL 1\nGSB 2\n/\nSTO- 1\nABS\nRCL 2\nX<=Y?\nGTO 0\nRCL 1\nRTN\n\nYou made a mistake calculating the 1st derivative.\nThis is the corrected program:\n\nf'(x)= 3x^2 - 6x - 6 = (3x - 6)x - 6\nf'(x) code\n\nLBL 2\nENTER\nENTER\n3\n*\n6\n-\n*\n6\n-\nRTN\n\nKind regards\nThomas\n01-10-2014, 05:28 PM\nPost: #3\n Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013\nRE: HP 11C real root finder [Newton Method]\n\nBest regards\n\nBest Regards\n01-12-2014, 08:31 AM\nPost: #4\n Namir Senior Member Posts: 688 Joined: Dec 2013\nRE: HP 11C real root finder [Newton Method]\nHere is a version that requires coding f(x) only since it approximate f'(x) as:\n\nf'(x) = (f(x+h) - f(x))/h\n\nWhere h = 0.001*(ABS(X)+1)\n\nThe new version uses registers R1 through R4.\n\nCode:\nLBL\u00a00 RCL\u00a01 ABS 1 + EEX 3 CHS * STO\u00a03\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0#\u00a0calculate\u00a0and\u00a0store\u00a0increment\u00a0h RCL\u00a01 GSB\u00a01 STO\u00a04\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0#\u00a0calculate\u00a0and\u00a0store\u00a0f(x) RCL\u00a01 RCL\u00a03 + GSB\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0#\u00a0calculate\u00a0f(x+h) RCL\u00a04 - 1/X RCL\u00a04 * RCL\u00a03 *\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0#\u00a0calculate\u00a0diff\u00a0=\u00a0h\u00a0*f(x)/(f(x+h)\u00a0-\u00a0f(x)) STO-\u00a01 ABS RCL\u00a02 X<=Y? GTO\u00a00 RCL\u00a01 RTN\n01-12-2014, 01:26 PM\nPost: #5\n Dieter Senior Member Posts: 2,397 Joined: Dec 2013\nRE: HP 11C real root finder [Newton Method]\n(01-12-2014 08:31 AM)Namir Wrote: \u00a0...h = 0.001*(ABS(X)+1)\n\nFirst of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-)\n\nThis method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.\n\nDieter\n01-14-2014, 08:56 PM\nPost: #6\n Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013\nRE: HP 11C real root finder [Newton Method]\nIf you only want to solve polynomials with real coefficients Bairstow's Method can be used.\nIn a nutshell: instead of a root a quadratic factor is found in each iteration. To find the roots of this factor the classic formula is used.\n\nExample:\n\n$$x^3-3x^2-6x+8=0$$\n\nFirst enter the coefficients of the polynomial. Always start with register 9:\n\n1 STO 9\n-3 STO .0\n-6 STO .1\n8 STO .2\n\nThen specify the order of the polynomial with a loop-control value. It defines the registers you used for the coefficients:\n\n9.012 STO 6\n\nAnd now give an initial guess. Probably {1, 1} will do in all cases:\n\n1 STO 7\nSTO 8\n\nStart the program B:\n\nGSB B\n\nNow the coefficients of the quadratic factor can be found in registers 7 and 8:\n\nRCL 7\n1.000000\n\nRCL 8\n-2.000000\n\nThe coefficient of the first term is always 1. Thus the factor is $$x^2+x-2$$.\nNow we can solve this quadratic equation using program A:\n\n1\nRCL 7\nRCL 8\nGSB A\n\n1.000000\nX<>Y\n-2.000000\n\nThus: $$x^2+x-2=(x-1)(x+2)$$.\n\nWhat is left? Let's have a look at it:\n\nRCL 6\n9.010\n\nRCL 9\n1.000000\n\nRCL .0\n-4.000000\n\nThis is a linear factor. Thus we end up with the following factorization: $$x^3-3x^2-6x+8=(x-1)(x+2)(x-4)$$\nTherefore the solutions are: {1, -2, 4}\n\nExample from Bunuel66's solution to the crossed ladders problem\n\n$$C^4-30C^3+700C^2-21,000C+157,500=0$$\n\nEnter the coefficients of the polynomial:\n\n1 STO 9\n-30 STO .0\n700 STO .1\n21,000 STO .2\n157,500 STO .3\n\nSpecify the loop-control value:\n\n9.013 STO 6\n\nUse the initial guess {1, 1}:\n\n1 STO 7\nSTO 8\n\nStart the program:\n\nGSB B\n\nThe coefficients of the quadratic factor are in registers 7 and 8:\n\nRCL 7\n-35.178025\n\nRCL 8\n248.596895\n\nSolve this quadratic equation using program A:\n\n1\nRCL 7\nRCL 8\nGSB A\n\n25.384938\nX<>Y\n9.793087\n\nRCL 6\n9.011\n\nRCL 9\n1.000000\n\nRCL .0\n5.178025\n\nRCL .1\n633.555781\n\nIf we try to solve this quadratic equation we get an Error 0.\nThat's because we're trying to calculate the square root of a negative value.\nHowever we can still get the desired result:\n\nRCL 9\nRCL .0\nRCL .1\nGSB B\nError 0\n\n<-\nCHS\n626.852796\n\n$$\\sqrt{x}$$\n25.037029\n\nX<>Y\n-2.589012\n\nThus the final list of solutions is:\n\u2022 25.384938\n\u2022 9.793087\n\u2022 -2.589012$$\\pm$$25.037029i\n\nCheers\nThomas\n01-15-2014, 05:53 AM\nPost: #7\n Namir Senior Member Posts: 688 Joined: Dec 2013\nRE: HP 11C real root finder [Newton Method]\n(01-12-2014 01:26 PM)Dieter Wrote:\n(01-12-2014 08:31 AM)Namir Wrote: \u00a0...h = 0.001*(ABS(X)+1)\n\nFirst of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-)\n\nThis method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.\n\nDieter\n\nSure, dividing by 1000 is a step shorter. Originially, I had learned to calculate h using an If statement:\n\nCode:\nIf\u00a0|x|\u00a0>=\u00a01\u00a0Then \u00a0\u00a0h=\u00a0x/100 else \u00a0\u00a0h\u00a0=\u00a01/100 end\u00a0if\n\nUntil I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.\n\nNamir\n01-15-2014, 08:38 PM\nPost: #8\n Dieter Senior Member Posts: 2,397 Joined: Dec 2013\nRE: HP 11C real root finder [Newton Method]\nNamir, sorry that I was not able to express myself clearly. I wanted to point out that I do not think it's a good idea to use h = 0,01 or 0,001 for any value below 1. This may lead to significant errors since h may be much, much larger than x. For instance, if x = 1E-10, h = 1E-3 is not recommended. Here, h = 1E-13 should be better.\n\nThat's why I prefer to set h = 0,001 x or similar. With the only exception x=0 where x may be 0,001.\n\n(01-15-2014 05:53 AM)Namir Wrote: \u00a0Until I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.\n\nThis can be coded this way, for instance:\nCode:\n\u00a0X=0? \u00a0e^x \u00a0EEX \u00a0\u00a03 \u00a0\u00a0/\nLook, no labels or gotos required. ;-)\n\nOr even more elegant on the 34s:\nCode:\n\u00a0X=0? \u00a0INC\u00a0X \u00a0SDR\u00a03 \u00a0RSD\u00a01\u00a0\u00a0\u00a0'\u00a0optional\n\nDieter\n \u00ab Next Oldest | Next Newest \u00bb\n\nUser(s) browsing this thread: 1 Guest(s)", "date": "2020-01-25 19:39:42", "meta": {"domain": "hpmuseum.org", "url": "https://www.hpmuseum.org/forum/thread-383.html", "openwebmath_score": 0.4317140281200409, "openwebmath_perplexity": 5919.442975938707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471620993537, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8616405575001446}}
{"url": "https://www.physicsforums.com/threads/basic-taylor-polynomial-question-involving-e-x-2.782018/", "text": "# Basic Taylor Polynomial Question involving e^(-x)^2\n\n1. Nov 15, 2014\n\n### RJLiberator\n\n1. The problem statement, all variables and given/known data\nConsider:\n\nF(x) =$\\int_0^x e^{-x^2} \\, dx$\nFind the Taylor polynomial p3(x) for the function F(x) centered at a = 0.\n\n2. Relevant equations\nTabulated Taylor polynomial value for standard e^x\n\n3. The attempt at a solution\n\nI started out by using the tabulated value for Taylor polynomial e^x and replaced all x with -x^2.\nGoing to the degree of three I received:\n\ne^(-x)^2 = 1 - (x^2)/1! + x^4/2! - x^6/3!\n\nIs this all there is to it? The question simply states to find the Taylor Polynomial p3(x) for the function, nothing else. I don't need to find the sum or the remainder per the problems question. Correct?\n\nIf I were to go at the sum, would I integrate each value?\n\nThank you for any help, I think I am just getting used to the wording here more than anything.\n\n2. Nov 15, 2014\n\n### gopher_p\n\nThe Taylor polynomial $p_3$ of degree $3$ for $F$ centered at $0$ should look something like $$p_3(x)=F(0)+F'(0)x+\\frac{F''(0)}{2}x^2+\\frac{F'''(0)}{6}x^3,$$ yes? If you remember the Fundamental Theorem of Calculus and what is says about computing $F'(x)$, it shouldn't be too hard to get the coefficients, right? So you can just do the problem \"properly\" without worrying whether your methods work.\n\nI'm not saying your way won't work, though you'd theoretically only need the degree $2$ Taylor polynomial for $f(x)=e^{-x^2}$ to extract the degree $3$ polynomial for $F$. But the question you'd want to ask is whether the derivative of the degree $n$ Taylor polynomial of a function $g$ really is the degree $n-1$ Taylor polynomial for the function $g'$ and vice -versa; i.e. is the (anti)derivative of the Taylor polynomial the Taylor polynomial of the (anti)derivative? I don't think that question is too incredibly difficult to answer for those who understand what it is asking.\n\n3. Nov 17, 2014\n\n### RJLiberator\n\nHm. Bare with me. I am in the early process of understanding Taylor Series wordage:\nSo what you are saying is that I should take the derivative of e^(-x)^2 a few times to acquire the coefficients? If I do this I answer:\nf(0) = 1\nf'(0) = 1\nf''(0) = 2\nf'''(0) = 4\nWhich would mean my original equation is incorrect. I am not sure what you mean by doing the problem properly without worrying.\n\nI guess my first question would be: Is my answer correct? All I did was input -x^2 to the e^x common taylor series polynomial expansion for the first 3 degrees.\n\nThen my second question would be how to proceed with this problem 'properly' ? I will review the fundamental theorem of calculus.\n\nQuestion 3: Out or curiosity, why would I need to only go to degree 2 for this problem? Doesn't it state to go to degree 3?\n\n4. Nov 17, 2014\n\n### Staff: Mentor\n\n1. Yes, what you did looks fine.\n2. Integrate the series you found.\n3. If you have terms up to degree 2, when you integrate, what will be the highest degree you get?\n\n5. Nov 17, 2014\n\n### RJLiberator\n\nMark44, you are shedding light on this topic!\n\n1. Thank you for your confirmation.\n2. Hm, ok. I am going to mess around with a few things. I know my coefficients then, now I got to find where to plug them in and integrate.\n3. Ah.. integrating would bring it to degree three.\n\n6. Nov 17, 2014\n\n### gopher_p\n\nThe way that I was suggesting you do the problem is as follows;\n\nFor $F(x)=\\int_0^xe^{-t^2}\\ \\mathrm{d}t$,\n\n$F(0)=\\int_0^0e^{-t^2}\\ \\mathrm{d}t=0$\n\n$F'(x)=e^{-x^2}\\Rightarrow F'(0)=1$, where the derivative is given by the FTC.\n\n$F''(x)=-2xe^{-x^2}\\Rightarrow F''(0)=0$\n\n$F'''(x)=-2e^{-x^2}+4x^2e^{-x^2}\\Rightarrow F'''(0)=-2$\n\nThen the Taylor polynomial of degree three centered at $a=0$ is given by $$p_3(x)=F(0)+F'(0)x+\\frac{F''(0)}{2}x^2+\\frac{F'''(0)}{6}x^3=x-\\frac{1}{3}x^3$$\n\nSo we're just (a) appealing to the definition of the Taylor polynomial and (b) computing the coefficients directly. I guess that's what I meant by \"properly\".\n\n7. Nov 18, 2014\n\n### RJLiberator\n\nOK, so with your help, I've been working through this problem.\n\nQuestion 1: So for F(0) you made the bounds 0 to 0. I don't understand why. I see that the answer is clearly 0 when the bounds are both 0 to 0, but why do this?\n\nOther then this part, everything you wrote now makes complete sense to me.\n\nQuestion 2: In my original solution where I just inputted -x^2 to the e^x tabulated value i had NOT integrated yet. Did I HAVE to integrate after I replaced those values? If so, the two answers match up nicely. I would imagine so, by the definition of the Taylor Series, correct?\n\n8. Nov 18, 2014\n\n### gopher_p\n\nOne of the \"skills\" that you learned in you preparation for calculus was that, given a formula for $F(x)$, you compute $F(a)$ by plugging in $a$ for $x$. So given $F(x)=\\int_0^xe^{-t^2}\\ \\mathrm{d}t$, get $F(0)$ by plugging in $0$ for $x$; $F(0)=\\int_0^0e^{-t^2}\\ \\mathrm{d}t$. This is why it is important to use a different variable of integration - $t$ here - inside the integral.\n\nI'm confused by the question. It's clear that the degree two Taylor polynomial $\\tilde p_2$ for $f(x)=e^{-x^2}$ is (a) not a degree three polynomial and (b) not a Taylor polynomial for the function $F$, right? So something must be done if we are to extract $p_3$ from $\\tilde p_2$.\n\nThe real question, in my mind, is whether it is true that $p_3(x)=\\int_0^x\\tilde p_2(t)\\ \\mathrm{d}t$. In this case it is. It works here because the lower limit of integration, $0$, is the same as the center of the Taylor polynomials. You need to be careful in a more general setting.\n\n9. Nov 18, 2014\n\n### PeroK\n\nYou should note that where you wrote:\n\nF(x) =$\\int_0^x e^{-x^2} \\, dx$\n\ngopher_p wrote:\n\nF(x) =$\\int_0^x e^{-t^2} \\, dt$\n\nI think you need to understand the difference and why the second one correctly defines F as a function of x, with t as the dummy variable.\n\nThis may also be the reason why you did not understand the substitution $x = 0$\n\n10. Nov 18, 2014\n\n### RJLiberator\n\nGuys, thanks for your help here.\n\nYou were a help to me and my classmates.\n\nAlso, thanks for pointing out my subtle mistakes that cause errors in conceptual understanding. I am starting to understand this problem due to this.\n\nI see why the dummy variable matters.\n\nAs always, excellent help.", "date": "2017-11-21 18:04:43", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/basic-taylor-polynomial-question-involving-e-x-2.782018/", "openwebmath_score": 0.7970179915428162, "openwebmath_perplexity": 371.5138458885036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471661250914, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8616405514568775}}
{"url": "https://math.stackexchange.com/questions/1938052/find-the-function-fx-if-fx2fy-fxyfy/1938110", "text": "# Find the function $f(x)$ if $f(x+2f(y))=f(x)+y+f(y)$\n\nLet $f:\\mathbb{R}\\to \\mathbb{R}$ such $f(x)$ at $x=0$ continuous, and for any $x,y\\in \\mathbb{R}$ such $$f(x+2f(y))=f(x)+y+f(y)$$ Find $f(x)$.\n\nLet $x=0,y=0$ then we have $$2f(2f(0))=f(0)++f(0)$$ Let $y=2f(0)$ then we have $$f(x+2f(2f(0))=f(x)+2f(0)+f(2f(0))$$ so we have $$f(x+4f(0))=f(x)+4f(0)$$ on the other hand we have $$f(x+4f(0))=f(x+2f(0))+0+f(0)=2f(0)+f(x)$$ so we have $f(0)=0$ Then I can't deal with this problem\n\n\u2022 Maybe I am off base but what happens if you plug in $y=x$? Sep 23, 2016 at 5:29\n\u2022 @CameronWilliams you get that $x+2f(x)$ is a fixed point. Sep 23, 2016 at 5:30\n\u2022 Am I missing something or does $f(x)=x$ work? Sep 23, 2016 at 6:21\n\u2022 @Mastrem It does, but presumably the question is to find all such functions. Sep 23, 2016 at 6:22\n\u2022 @Mastrem it works, but no one says this equation has only one solution. You have to prove then that it's the only one. Sep 23, 2016 at 6:23\n\nSubstitute $x\\mapsto y$ to get $$f(y+2f(y))=y+2f(y).$$ Now substitute $y\\mapsto y+2f(y)$: $$f(x+2f(y+2f(y)))=f(x)+y+2f(y)+f(y+2f(y)).$$ Thus $$f(x+2y+4f(y))=f(x)+2y+4f(y).$$ Substituting $x\\mapsto x+2y+2f(y)$, $$f(x+2y+4f(y))=f(x+2y+2f(y))+y+f(y).$$ Substituting $x\\mapsto x+2y$, $$f(x+2y+2f(y))=f(x+2y)+y+f(y).$$ Combining the last 3 equations, $$f(x+2y)=f(x)+2f(y).$$ Thus $$f(x)=f(0+2x/2)=f(0)+2f(x/2)=2f(x/2),$$ so $$f(x+y)=f(x+2y/2)=f(x)+2f(y/2)=f(x)+f(y).$$ Since you are given that $f$ is continuous at $0$, it must be continuous everywhere. Now there is a standard argument to show that $f(x)=f(1)x$ (prove it on rationals and extend by continuity). Finally using the original equation, we see $f(1)=1$ or $-\\frac12$, so the solutions are $f(x)=x$ or $f(x)=-x/2$.\n\nSetting $x=y$ gives $f(x+2f(x))=x+2f(x)$, and so $f(y)=y$ whenever $y$ is of the form $x+2f(x)$ for some $x$. Note also that if $f(y)=y$, then for any $x$, $f(x+2y)=f(x)+2y$.\n\nNow consider the function $g(x)=f(x)-x$. From the previous paragraph, we see that if $f(y)=y$ then $g(x+2y)=g(x)$. In particular, every number of the form $2x+4f(x)$ is a period of $g$. Let $P\\subseteq\\mathbb{R}$ be the group of periods of $g$, i.e. the set of $p$ such that $g(x)=g(x+p)$ for all $x$. There are now two cases.\n\nThe first case is that $P$ is not cyclic. Then $P$ is dense in $\\mathbb{R}$, and so each coset of $P$ is dense in $\\mathbb{R}$, and in particular $0$ is in the closure of each coset. But by definition of $P$, $g$ is constant on each coset of $P$, and so by continuity of $g$ at $0$ this constant value must be equal to $g(0)$. Thus $g(x)=g(0)$ for all $x$, and $g$ is constant everywhere. Thus $f(x)$ has the form $f(x)=x+c$ for some constant $c$. Plugging this into the functional equation gives $$x+2(y+c)+c=x+c+y+y+c,$$ so $c=0$. Thus $f(x)=x$.\n\nThe second case is that $P$ is cyclic, generated by some $p\\in\\mathbb{R}$ (possibly $0$). Then every number of the form $2x+4f(x)$ is an integer multiple of $p$. But $2x+4f(x)$ is continuous at $0$, so this means $2x+4f(x)$ is constant in a neighborhood of $0$. As you've shown, $f(0)=0$, so $2x+4f(x)=0$ for all $x$ in some neighborhood of $0$, so $f(x)=-x/2$ for all $x$ in some neighborhood of $0$.\n\nNow suppose $\\epsilon>0$ is such that $f(x)=-x/2$ whenever $|x|\\leq\\epsilon$. If $0\\leq a\\leq \\epsilon$, the functional equation with $x=a$ and $y=-\\epsilon$ then gives $$f(a+\\epsilon)=-\\frac{a+\\epsilon}{2}.$$ That is, $f(x)=-x/2$ is also valid if $\\epsilon\\leq x\\leq 2\\epsilon$. Similarly, using $y=\\epsilon$ gives that $f(x)=-x/2$ is also valid if $-2\\epsilon\\leq x\\leq -\\epsilon$.\n\nThat is, if $f(x)=-x/2$ whenever $|x|\\leq\\epsilon$, $f(x)=-x/2$ whenever $|x|\\leq 2\\epsilon$ as well. Repeating this argument over and over, we get that $f(x)=-x/2$ for all $x$.\n\nThus the only possibilities for $f$ are $f(x)=x$ and $f(x)=-x/2$, and you can easily check that both of these work.\n\n\u2022 Nice try! Even if using machinery which .. outguns the problem! I downvoted by accident.(and I thought I had already cancelled it. Seems I hadn't!) I will make a trivial edit so that I'll be able to cancel it. P.S.: the part where you are arguing about the cosets of the group of periods is not very clear. Could you expand it a little? (what is the equivalence relation here? why does the continuity of $g$ implies that every coset accumulates at $0$ ?). Sep 23, 2016 at 18:35\n\u2022 I've clarified that part a bit. Sep 23, 2016 at 19:12\n\u2022 Thanks for your response. As I've already said: very nice ! +1 Sep 25, 2016 at 19:49", "date": "2022-08-12 21:39:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1938052/find-the-function-fx-if-fx2fy-fxyfy/1938110", "openwebmath_score": 0.9844039082527161, "openwebmath_perplexity": 61.60041718928827, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471675459396, "lm_q2_score": 0.8757869835428966, "lm_q1q2_score": 0.8616405431322812}}
{"url": "https://math.stackexchange.com/questions/3356386/geometry-prove-a-right-triangle-resulting-from-three-inscribed-circles", "text": "Geometry - Prove a right triangle resulting from three inscribed circles\n\nTwo half circles and a full circle fit inside a larger quarter circle as shown in the diagram. The centers of the two half circles are on the two sides of the quarter circle, respectively.\n\nProve that the triangle formed by the centers of three smaller circles, $$\\triangle O_1O_2O_3$$, is a right triangle.\n\nI was able to apply the Pythagorean formula to a few triangles involving the radii of the inscribed circles and proved that the ratios of the three radii are 1:2:3. Then, the centers of the three circles form a triangle with side-length ratios 3:4:5, hence, a right triangle.\n\nOn the other hand, I feel the proof may be an overkill, and evaluating the three radii explicitly may be unnecessary. There ought to be clean geometric solutions to show directly that the vertex $$O_3$$ is of a right angle, which I am not sure how to figure out.\n\n\u2022 Suppose $OO_1$ and $OO_2$ are x and y axis. If you complete half and quarter circles, you will have a big circle containing two big circles, two medium and four little circles. the center of little circles are symmetric about x and y axis. since x and y axis are perpendicular then the lines connecting the centers of little circles are perpendicular that is triangle $O_1O_2O_3$ is right angle at $O_3$. \u2013\u00a0sirous Sep 14 '19 at 21:01\n\u2022 Insightful. A good case of can\u2019t-see-forest-staring at trees. Thanks \u2013\u00a0Quanto Sep 14 '19 at 21:10\n\u2022 @sirous I finally read your comment. I think it would be clearer if you \"complete\" the half and quarter circles by reflecting the figure over one axis and then the other. But your construction is much more elegant than mine and would make a fine answer. \u2013\u00a0David K Sep 15 '19 at 0:30\n\u2022 @sirous: Certainly, \"the lines connecting the centers of the little circles are perpendicular\", but how do you know that those lines contain $O_1$ and $O_2$? \u2013\u00a0Blue Sep 15 '19 at 3:51\n\u2022 @Blue - critical observation. guess a simple geometric proof remains elusive \u2013\u00a0Quanto Sep 15 '19 at 14:42\n\nGiven the quarter circle with $$O$$ as the center of the arc, and given semicircles with centers $$O_1$$ and $$O_2$$ on the two straight sides of the quarter circle such that $$O_1$$ and $$O_2$$ both are tangent to the arc of the quarter circle and such that $$O_1$$ and $$O_2$$ are tangent to each other.\n\nThis is based on the figure in the question, except that we do not assume that $$O_1$$ is the midpoint of one side of the quarter circle.\n\nLet the radii of the quarter circle, the semicircle about $$O_1$$, and the semicircle about $$O_2$$ be $$r,$$ $$r_1,$$ and $$r_2$$ respectively.\n\nLet $$P$$ be the fourth vertex of the rectangle that has three of its vertices at $$O,$$ $$O_1,$$ and $$O_2.$$ The sides of this rectangle are $$OO_1 = r - r_1$$ and $$OO_2 = r - r_2.$$ The diagonals are $$OP = O_1O_2 = r_1 + r_2.$$\n\nLet $$P_1$$ be the point of intersection of the semicircle about $$O_1$$ with $$O_1P$$ and $$P_2$$ be the point of intersection of the semicircle about $$O_2$$ with $$O_2P.$$ Then $$PP_1 = r - r_1 - r_2 = PP_2.$$ Moreover, the distance from $$P$$ to the arc of the quarter circle is $$r - (r_1 + r_2) = r - r_1 - r_2.$$\n\nThen a circle of radius $$r - r_1 - r_2$$ about $$P$$ is tangent to the quarter circle and to the semicircles about $$O_1$$ and $$O_2.$$ Therefore $$P$$ is $$O_3$$ in the figure, the center of the circle inscribed between the two semicircles and the quarter circle.\n\nSince $$O_3$$ is a vertex of the rectangle $$OO_1O_3O_2,$$ the triangle $$\\triangle O_1O_3O_2$$ is a right triangle. $$\\square$$\n\n\u2022 +1 a nice generalization. \u2013\u00a0achille hui Sep 15 '19 at 16:27\n\nLet $$R=6a$$ then $$r_1=3a$$\n\n$$\\triangle OO_1O_2: \\;\\;\\;\\; (R-r_2)^2+r_1^2 =(r_1+r_2)^2$$\n\nSo $$\\boxed{r_2= 2a}$$\n\nSo $$\\triangle O_3O_1O_2$$ is right iff $$(r_1+r_2)^2 = (r_1+r_3)^2+(r_2+r_3)^2$$\n\nor $$r_1r_2 = r_3(r_1+r_2+r_3)$$\n\nor $$6a^2 = r_3(5a+r_3)\\iff r_3^2+5ar_3-6a^2 =0$$\n\nSo we have to prove: $$\\boxed{\\color{red}{r_3=a}}$$\n\nIf we put everything in coordinate system then we have: $$O(0,0)$$, $$O_1(3a,0)$$, $$O_2(0,4a)$$ and let $$O_3(m,n)$$, so $$\\begin{eqnarray} m^2+n^2 &=& (6a-r_3)^2\\;\\;\\;\\;\\;\\; (OO_3 = R-r_3)\\\\ (m-3a)^2+n^2 &=& (3a+r_3)^2\\;\\;\\;\\;\\;\\; (O_1O_3 = r_1+r_3)\\\\ m^2+(n-4a)^2 &=& (2a+r_3)^2\\;\\;\\;\\;\\;\\; (O_2O_3 = r_2+r_3) \\end{eqnarray}$$\n\nIf we substract 1.st and 2.nd equation we get $$m = 6a-3r_3$$ and if we substract 1.st and 3.rd equation we get $$n=6a-2r_3$$\n\nNow put this again in 1.st equation and we get $$\\boxed{\\color{green}{r_3=a}}$$\n\nFollowing is an answer based on circle inversion w/o computing the radius of circle centered at $$O_3$$.\n\nLabel the contact points as illustrated below. We are going to show $$\\angle OYD = \\angle OXC = 45^\\circ$$ Form these, it is easy to deduce $$\\begin{cases} O_2O_3 \\parallel O_2D \\parallel OX,\\\\ O_1O_3 \\parallel O_1C \\parallel OY\\end{cases}$$ and hence $$\\angle O_1O_3O_2 = 90^\\circ$$\n\nChoose a coordinate system so that $$O$$ is the origin and $$X = (1,0)$$, $$Y = (0,1)$$. The arc $$XEY$$ will be part of the unit circle. Under circle inversion with respect to the unit circle centered at $$O$$, we get what illustrated below:\n\n\u2022 The red semicircle centered at $$O_1$$ get mapped to a ray $$y = 1$$ in the upper half plane.\n\u2022 The green semicircle centered at $$O_2$$ get mapped to another semicircle centered at some point of $$y$$-axis and touching the red line. This means this inverted green semicircle has radius $$1$$.\nAs a result, the inverted image of point $$A$$ is $$A' = (0,3) \\implies A = (0,\\frac13)$$.\n\n\u2022 The blue circle get mapped to a circle placed symmetrically between the inverted green semicircle and the arc $$XEY$$. As a result, the inverted image of point $$C$$ is $$C' = (1,1)$$.\nThis implies $$\\angle OXC = \\angle OXC' = 45^\\circ$$.\n\nFor the other angle, translate the coordinate axis so that $$Y$$ is the new origin. The new coordinate of $$A$$ is now $$(0,-\\frac23)$$. Under circle inversion with respect to the unit circle centered at $$Y$$, it now looks like:\n\n\u2022 The arc $$XEY$$ get mapped to a ray on the line $$y = -\\frac12$$.\n\u2022 The green semicircle centered at $$O_2$$ get mapped to a ray on the line $$y = -\\frac32$$.\n\u2022 The red semicircle get mapped to an arc on a circle sandwiched between above two rays touching the $$y$$-axis. So its radius is $$\\frac12$$.\n\u2022 The blue circle get mapped to an circle also sandwiched between above two rays. This means the inverted image of $$D$$ is located at $$D'' = (\\frac32,-\\frac32)$$. As a result, $$\\angle OYD = \\angle OYD' = 45^\\circ$$.\n\nLet $$OO_1$$ and $$OO_2$$ are x and y axis respectively. Mirror the figure about y axis, then mirror the resulted figure about x axis. You get two big, two medium and four small circles indescribable in a large circle. If you connect the centers of four small circles you get a square. So angle at $$O_3$$ is $$90^o$$. Now let the radius of small circle is $$r_1=1$$ and that of large circle is $$r_2=6$$. Suppose the location of center of small circle is $$O_3(3, 4)$$ which satisfies:\n\n$$3^2+4^2= [OO_3=6-1=5]^2$$\n\nThen we can claim that the radius of medium circle is $$r_3=3-1=2$$ and those of big and large circles are $$r_4=4-1=3$$ and $$r_2=2\\times 3=6$$ respectively that is all involved circle are mutually tangent.If radius of medium circle is $$r_3=2$$ that means that it's center locates on a line passing $$O_3$$ which has a distance=4 from x axis. similarly if the radius of big circle is $$r_4=3$$ that would mean that it's center locates on a line passing $$O_3$$ which has distance =3 from y axis.That is triangle $$O_1O_2O_3$$ is right triangle at $$O_3$$", "date": "2020-12-05 00:31:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3356386/geometry-prove-a-right-triangle-resulting-from-three-inscribed-circles", "openwebmath_score": 0.8656236529350281, "openwebmath_perplexity": 192.297737862093, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471628097781, "lm_q2_score": 0.8757869867849166, "lm_q1q2_score": 0.8616405421740648}}
{"url": "https://math.stackexchange.com/questions/1900788/find-the-probability-that-on-tuesday-it-does-not-rain-in-gregs-area-and-he-do", "text": "# Find the probability that, on Tuesday, it does not rain in Greg's area and he does not get a quiz.\n\nThe probability of rain in Greg's area on Tuesday is $0.3$. The probability that Greg's teacher will give him a pop quiz in Tuesday is $0.2$. The events occur independently of each other. What is the probability of neither events occur?\n\nMy approach:\n\nProbability of rain or quiz or both = $0.3+0.2= 0.5$\n\nSo, probability of neither= $1-0.5$ = $0.5$\n\nQuestion: Actual probability of this problem is $0.7\\cdot0.8$ = $0.56$. But, I don't understand what is the mistake in above approach?\n\n\u2022 They are independent not mutually exclusive. For mutually exclusive events you have $Pr(A\\text{or}B)=Pr(A)+Pr(B)$. That is not true of independent events. \u2013\u00a0JMoravitz Aug 23 '16 at 6:11\n\u2022 Principle of inclusion-exclusion 2set case: $Pr(A\\cup B) = Pr(A)+Pr(B)-Pr(A\\cap B)$ \u2013\u00a0JMoravitz Aug 23 '16 at 6:18\n\nNote the following two things:\n\nPrinciple of inclusion-exclusion (2-set case) $$Pr(A\\cup B) = Pr(A)+Pr(B)-Pr(A\\cap B)$$\n\nDefinition of independent events\n\nThe following are equivalent statements\n\n\u2022 $A$ and $B$ are independent events\n\n\u2022 $Pr(A\\mid B) = Pr(A)$\n\n\u2022 $Pr(B\\mid A) = Pr(B)$\n\n\u2022 $Pr(A\\cap B)=Pr(A)\\cdot Pr(B)$\n\nFor your problem: you know $Pr(A)=0.3, Pr(B)=0.2$ and they are independent, so\n\n$$Pr(A\\cup B) = Pr(A)+Pr(B)-Pr(A\\cap B) = 0.3+0.2-0.3\\cdot 0.2 = 0.5-0.06 = 0.44$$\n\nSo the probability of at least one of the events occurring is $0.44$. The probability of no events occurring is then one minus that, i.e. $1-0.44=0.56$.\n\nAlternatively, if $A$ and $B$ are independent, then $A^c$ and $B^c$ are also independent. We have then $Pr(A^c\\cap B^c)=Pr(A^c)Pr(B^c)=(1-0.3)(1-0.2)=0.7\\cdot 0.8 = 0.56$\n\nYour approach double-counts some possibilities. The $0.3$ covers all possibility of rain - so, it includes both \"rain and quiz\" and \"rain and no quiz\". The $0.2$ covers all possibility of a quiz, so it includes both \"rain and quiz\" and \"quiz and no rain\". So $0.3 + 0.2$ covers \"rain and quiz\", \"rain and no quiz\", \"rain and quiz\" again, and \"quiz and no rain\". To cut out one of the countings of \"rain and quiz\", we need to subtract the probability of that event, which is $0.3 \\cdot 0.2 = 0.06$. So the probability of \"rain or quiz or both\" is $0.2 + 0.3 - 0.06 = 0.44$. The probability of neither is then $1 - 0.44 = 0.56$.", "date": "2019-12-08 11:17:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1900788/find-the-probability-that-on-tuesday-it-does-not-rain-in-gregs-area-and-he-do", "openwebmath_score": 0.7611855864524841, "openwebmath_perplexity": 676.4141909954691, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471620993539, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.8616405399570585}}
{"url": "https://www.quizover.com/calculus/section/the-mean-value-theorem-and-its-meaning-by-openstax", "text": "# 4.4 The mean value theorem \u00a0(Page 2/7)\n\n Page 2 / 7\n\nVerify that the function $f\\left(x\\right)=2{x}^{2}-8x+6$ defined over the interval $\\left[1,3\\right]$ satisfies the conditions of Rolle\u2019s theorem. Find all points $c$ guaranteed by Rolle\u2019s theorem.\n\n$c=2$\n\n## The mean value theorem and its meaning\n\nRolle\u2019s theorem is a special case of the Mean Value Theorem. In Rolle\u2019s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle\u2019s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle\u2019s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $\\left[a,b\\right]$ and differentiable over the open interval $\\left(a,b\\right),$ then there exists a point $c\\in \\left(a,b\\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\\left(a,f\\left(a\\right)\\right)$ and $\\left(b,f\\left(b\\right)\\right).$\n\n## Mean value theorem\n\nLet $f$ be continuous over the closed interval $\\left[a,b\\right]$ and differentiable over the open interval $\\left(a,b\\right).$ Then, there exists at least one point $c\\in \\left(a,b\\right)$ such that\n\n$f\\prime \\left(c\\right)=\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}.$\n\n## Proof\n\nThe proof follows from Rolle\u2019s theorem by introducing an appropriate function that satisfies the criteria of Rolle\u2019s theorem. Consider the line connecting $\\left(a,f\\left(a\\right)\\right)$ and $\\left(b,f\\left(b\\right)\\right).$ Since the slope of that line is\n\n$\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}$\n\nand the line passes through the point $\\left(a,f\\left(a\\right)\\right),$ the equation of that line can be written as\n\n$y=\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}\\left(x-a\\right)+f\\left(a\\right).$\n\nLet $g\\left(x\\right)$ denote the vertical difference between the point $\\left(x,f\\left(x\\right)\\right)$ and the point $\\left(x,y\\right)$ on that line. Therefore,\n\n$g\\left(x\\right)=f\\left(x\\right)-\\left[\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}\\left(x-a\\right)+f\\left(a\\right)\\right]\\text{.}$\n\nSince the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\\left(a\\right)=0=g\\left(b\\right).$ Since $f$ is a differentiable function over $\\left(a,b\\right),$ $g$ is also a differentiable function over $\\left(a,b\\right).$ Furthermore, since $f$ is continuous over $\\left[a,b\\right],$ $g$ is also continuous over $\\left[a,b\\right].$ Therefore, $g$ satisfies the criteria of Rolle\u2019s theorem. Consequently, there exists a point $c\\in \\left(a,b\\right)$ such that $g\\prime \\left(c\\right)=0.$ Since\n\n$g\\prime \\left(x\\right)=f\\prime \\left(x\\right)-\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a},$\n\nwe see that\n\n$g\\prime \\left(c\\right)=f\\prime \\left(c\\right)-\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}.$\n\nSince $g\\prime \\left(c\\right)=0,$ we conclude that\n\n$f\\prime \\left(c\\right)=\\frac{f\\left(b\\right)-f\\left(a\\right)}{b-a}.$\n\nIn the next example, we show how the Mean Value Theorem can be applied to the function $f\\left(x\\right)=\\sqrt{x}$ over the interval $\\left[0,9\\right].$ The method is the same for other functions, although sometimes with more interesting consequences.\n\n## Verifying that the mean value theorem applies\n\nFor $f\\left(x\\right)=\\sqrt{x}$ over the interval $\\left[0,9\\right],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\\in \\left(0,9\\right)$ such that ${f}^{\\prime }\\left(c\\right)$ is equal to the slope of the line connecting $\\left(0,f\\left(0\\right)\\right)$ and $\\left(9,f\\left(9\\right)\\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.\n\nWe know that $f\\left(x\\right)=\\sqrt{x}$ is continuous over $\\left[0,9\\right]$ and differentiable over $\\left(0,9\\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\\in \\left(0,9\\right)$ such that ${f}^{\\prime }\\left(c\\right)$ is equal to the slope of the line connecting $\\left(0,f\\left(0\\right)\\right)$ and $\\left(9,f\\left(9\\right)\\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative ${f}^{\\prime }\\left(x\\right)=\\frac{1}{\\left(2\\sqrt{x}\\right)}.$ The slope of the line connecting $\\left(0,f\\left(0\\right)\\right)$ and $\\left(9,f\\left(9\\right)\\right)$ is given by\n\n$\\frac{f\\left(9\\right)-f\\left(0\\right)}{9-0}=\\frac{\\sqrt{9}-\\sqrt{0}}{9-0}=\\frac{3}{9}=\\frac{1}{3}.$\n\nWe want to find $c$ such that ${f}^{\\prime }\\left(c\\right)=\\frac{1}{3}.$ That is, we want to find $c$ such that\n\n$\\frac{1}{2\\sqrt{c}}=\\frac{1}{3}.$\n\nSolving this equation for $c,$ we obtain $c=\\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.\n\nfind the nth differential coefficient of cosx.cos2x.cos3x\ndetermine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse\nf(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)\nAndrew\nin the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?\nAndrew\n\u221a(8) =\u221a(4x2) =\u221a4 x \u221a2 2 \u221a2 hope this helps. from the surds theory a^c x b^c = (ab)^c\nBarnabas\n564356\nMyong\ncan you determine whether f(x)=x(cube) +4 is a one to one function\nCrystal\none to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.\nAndrew\none to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.\nAndrew\ncan you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor\nstep 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!\nAndrew\nfirst you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2\nIoana\nMultiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x\nMelin\nthank you ladies and gentlemen I appreciate the help!\nRobert\nkeep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.\nAndrew\nplease how do I learn integration\nthey are simply \"anti-derivatives\". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.\nAndrew\nbest way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.\nAndrew\nexample 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2\nAndrew\ngreat noticeable direction\nIsaac\nlimit x tend to infinite xcos(\u03c0/2x)*sin(\u03c0/4x)\ncan you give me a problem for function. a trigonometric one\nstate and prove L hospital rule\nI want to know about hospital rule\nFaysal\nIf you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)\nFaysal\nI don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning\nAmor\nif you want some help on l hospital rule ask me\nit's spelled hopital\nConnor\nhi\nBERNANDINO\nyou are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'H\u00f4pital.\nLeo\nI had no clue this was an online app\nConnor\nTotal online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \\$60 billion?\nwhat is the derivative of x= Arc sin (x)^1/2\ny^2 = arcsin(x)\nPitior\nx = sin (y^2)\nPitior\ndifferentiate implicitly\nPitior\nthen solve for dy/dx\nPitior\nthank you it was very helpful\nmorfling\nquestions solve y=sin x\nSolve it for what?\nTim\nyou have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y\nIoana\nwhat is the question ? what is the answer?\nSuman\nthere is an equation that should be solve for x\nIoana\nok solve it\nSuman\nare you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x\nTyron\nor solve for sin(x) via the unit circle\nTyron\nwhat is unit circle\nSuman\na circle whose radius is 1.\nDarnell\nthe unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.\nTyron\nwhat is function?\nA set of points in which every x value (domain) corresponds to exactly one y value (range)\nTim\nwhat is lim (x,y)~(0,0) (x/y)\nlimited of x,y at 0,0 is nt defined\nAlswell\nBut using L'Hopitals rule is x=1 is defined\nAlswell\nCould U explain better boss?\nemmanuel\nvalue of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)\nNIKI\ncan we apply l hospitals rule for function of two variables\nNIKI\nwhy n does not equal -1\nAndrew\nI agree with Andrew\nBg\nf (x) = a is a function. It's a constant function.", "date": "2018-09-24 22:49:08", "meta": {"domain": "quizover.com", "url": "https://www.quizover.com/calculus/section/the-mean-value-theorem-and-its-meaning-by-openstax", "openwebmath_score": 0.8189693689346313, "openwebmath_perplexity": 389.4692698291522, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842202936827, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8616018092163845}}
{"url": "https://math.stackexchange.com/questions/2133512/is-x2-equiv-1-pmod365-solvable", "text": "# Is $x^2 \\equiv -1 \\pmod{365}$ solvable?\n\nI know that a similar question already exists, but I have a different question to ask.\n\nWe want to examine if $x^2 \\equiv -1 \\pmod{365}$ has a solution.\n\nMy thought is: $365=5\\cdot 73$. So,The congruence $x^2 \\equiv -1 \\pmod{365}$ has solution, if and only if, the congrueces $x^2 \\equiv -1 \\pmod 5$ and $x^2 \\equiv -1 \\pmod{73}$ has solutions. So, if we use Legendre's Symbol we have\n\n\u2022 $x^2 \\equiv -1 \\pmod 5$ has solution $\\iff (-1/5)=1$ (and with simple calculations, indeed)\n\u2022 $x^2 \\equiv -1 \\pmod{71}$ has solution $\\iff (-1/73)=1$\n\nNow, can we conclude that the congruence $x^2 \\equiv -1 \\pmod{365}$ has solution?\n\nAnd more general: If we have the congruence $x^2 \\equiv a \\pmod n$ with $n=p_1^{n_1}\\cdots p_k^{n_k},\\ \\gcd(a,n)=1$, which is equivalent with the system $x^2 \\equiv a {\\pmod p_1^{n_1}},\\ldots,x^2 \\equiv a \\pmod{p_k^{n_k}}$, can we conclude that the first has solution if and only if each one of $x^2\\equiv a\\pmod{p_i^{n_i}},\\ \\forall i=1,\\ldots,k$ has solution?\n\nThank you.\n\n\u2022 Change $71$ to $73$ and it's fine. \u2013\u00a0Robert Israel Feb 7 '17 at 16:12\n\u2022 @RobertIsrael Thank you for your comment and your correction. \u2013\u00a0Chris Feb 7 '17 at 16:15\n\u2022 Note the use of \\pmod rather than \\bmod in my edit to the question. \u2013\u00a0Michael Hardy Feb 7 '17 at 16:51\n\u2022 @RobertIsrael Ok, thank you \u2013\u00a0Chris Feb 7 '17 at 16:58\n\nWe show the following result:\n\nLet $n,m$ be coprime integers. Let $a \\in \\Bbb Z$. Then $a$ is a square modulo $nm$ iff $a$ is square mod $n$ and $a$ is square mod $m$.\n\nAssume that there are integers $x_i$ such that $x_1^2 \\equiv a \\pmod m$ and $x_2^2 \\equiv a \\pmod n$ where $m,n$ are coprime. We show that $y^2 \\equiv a \\pmod {nm}$ has a solution (the converse is obvious).\n\nWe know that there is an integer $y$ such that $y \\equiv x_1 \\pmod m$ and $y \\equiv x_2 \\pmod n$, by the Chinese remainder theorem.\n\nThen $y^2-a$ is a multiple of $m$ and $n$, so it is a multiple of $nm$ since $(n,m)=1$. Therefore $y^2 \\equiv a \\pmod {nm}$.\n\nMore generally, we have the following theorem (see Ireland Rosen, p. 50)\n\nLet $a,n \\in \\Bbb N$ be coprime integers. Write $n=2^e p_1^{e_1} \\cdots p_k^{e_k}$ as product of distinct prime powers. Then $a$ is a square modulo $n$ if and only if\n\n\u2022 $a$ is a square modulo $p_i$ (for $1 \\leq i \\leq k$) ; this is equivalent to $a^{\\dfrac{p_i-1}{2}} \\equiv 1 \\pmod{p_i}$\n\n\u2022 $e>1 \\implies a \\equiv 1 \\pmod{2^{2+r}}$ where $r = 1$ if $e \\geq 3$ and $r=0$ if $e=2$.\n\n\u2022 It might be worth explicitly mentioning the CRT here where it's used, but this is elsewise a good, clean answer. \u2013\u00a0Steven Stadnicki Feb 7 '17 at 16:20\n\u2022 Thank you , now it's clear. \u2013\u00a0Chris Feb 7 '17 at 16:27\n\u2022 @Chris : you're welcome. It is chapter 5 in Ireland, Rosen. \u2013\u00a0Watson Feb 7 '17 at 16:28\n\u2022 Tell me please something else. If we have the congruence $x^2 \\equiv 2 \\bmod 118$ could we say that $118=2\\cdot 59$ so it's in the form of $2p^m$ and $2^{\\phi(118)/ \\gcd(2,\\phi(118))} \\equiv -1 \\bmod 118$ so there are no solutions (because it is not a quadratic residue mod 118)? \u2013\u00a0Chris Feb 7 '17 at 16:46\n\u2022 @Chris : since $2$ is not a square mod $59$, it is not a square mod $118$, that's correct! \u2013\u00a0Watson Feb 7 '17 at 16:50\n\nWe have $365 = 5 \\times 73$. The congruence becomes $x^2 = -1 \\mod 5$ and $x^2 = -1 \\mod 73$.\n\nWe have if $p = 1 \\mod 4 \\implies x^2 = -1 \\mod p$ has exactly $2$ solutions.\n\nThus $x^2 = -1 \\mod 5$ has solutions $x_0,x_1$ and $x^2 = -1 \\mod 73$ has solutions $y_0,y_1$.\n\nThe original solutions satisfies either: $x = x_0 \\mod 5, x = y_0 \\mod 73$; $x = x_0 \\mod 5, x = y_1 \\mod 73$; $x = x_1 \\mod 5, x = y_0 \\mod 73$ ; $x = x_1 \\mod 5, x = y_1 \\mod 73$.\n\nFor each pair of congruence , $x$ is uniquely determined $\\mod 365$ by the Chinese remainder theorem. Hence the original congruence has $4$ solutions.\n\nIf $\\,m,n\\,$ are coprime then, by CRT, solving an integer coefficient polynomial $\\,f(x)\\equiv 0\\pmod{\\!mn}\\,$ is equivalent to solving $\\,f(x)\\equiv 0\\,$ mod $\\,m\\,$ and mod $\\,n.\\,$ By CRT, each combination of a root $\\,r_i\\bmod m\\,$ and a root $\\,s_j\\bmod n\\,$ corresponds to a unique root $\\,t_{ij}\\bmod mn,\\,$ i.e.\n\n$$\\begin{eqnarray} f(x)\\equiv 0\\!\\!\\!\\pmod{mn}&\\overset{\\rm CRT}\\iff& \\begin{array}{}f(x)\\equiv 0\\pmod m\\\\f(x)\\equiv 0\\pmod n\\end{array} \\\\ &\\iff& \\begin{array}{}x\\equiv r_1,\\ldots,r_k\\pmod m\\phantom{I^{I^{I^I}}}\\\\x\\equiv s_1,\\ldots,s_\\ell\\pmod n\\end{array}\\\\ &\\iff& \\left\\{ \\begin{array}{}x\\equiv r_i\\pmod m\\\\x\\equiv s_j\\pmod n\\end{array} \\right\\}_{\\begin{array}{}1\\le i\\le k\\\\ 1\\le j\\le\\ell\\end{array}}^{\\phantom{I^{I^{I^I}}}}\\\\ &\\overset{\\rm CRT}\\iff& \\left\\{ x\\equiv t_{i j}\\!\\!\\pmod{mn} \\right\\}_{\\begin{array}{}1\\le i\\le k\\\\ 1\\le j\\le\\ell\\end{array}}\\\\ \\end{eqnarray}\\qquad\\qquad$$\n\nYes. Let a solution to $x^2+1 \\equiv 0 \\pmod {5}$ be $r_{1}$, and let a solution to $x^2+1 \\equiv 0 \\pmod {73}$ be $r_{2}$\n\nThen note that by CRT, we have that there exists such $x$ that $$x \\equiv r_{1} \\pmod {5}$$ $$x \\equiv r_{2} \\pmod {73}$$ Exists. Then note that for such $x$, $$x^2+1 \\equiv 0 \\pmod {5}$$ $$x^2+1 \\equiv 0 \\pmod {73}$$ Which gives $$x^2+1 \\equiv 0 \\pmod {365}$$\n\n\u2022 Seems alright now. \u2013\u00a0Wojowu Feb 7 '17 at 16:20\n\nIt is true because of the Chinese remainder theorem, which asserts the map \\begin{align} \\mathbf Z/n\\mathbf Z&\\longrightarrow \\mathbf Z/p_1^{n_1}\\mathbf Z\\times\\dotsm\\times\\mathbf Z/p_k^{n_k}\\mathbf Z\\\\ x\\bmod n&\\longmapsto(x\\bmod p_1^{n_1},\\dots,x\\bmod p_k^{n_k}) \\end{align} is a ring isomorphism.", "date": "2021-03-01 14:37:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2133512/is-x2-equiv-1-pmod365-solvable", "openwebmath_score": 0.9318894147872925, "openwebmath_perplexity": 245.21764079916383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842212767571, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8616018067022748}}
{"url": "http://mathhelpforum.com/trigonometry/95321-cartesian-equations.html", "text": "# Math Help - Cartesian equations\n\n1. ## Cartesian equations\n\nGiven $x=2 \\sin (nt+\\frac{\\pi}{3})$ and $y=4 \\sin (nt+ \\frac{\\pi}{6})$, express $x$ and $y$ in terms of $\\sin nt$ and $\\cos nt$. Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.\n\nI have expressed $x$ and $y$ in terms of $\\sin nt$ and $\\cos nt$ already.\n$x=2 \\sin (nt+\\frac{\\pi}{3})$\n$x=2(\\frac{1}{2} \\sin nt+\\sqrt{3} \\cos nt)$\n$x= \\sin nt+ \\sqrt{3} \\cos nt$\nand\n$y=4 \\sin (nt+ \\frac{\\pi}{6})$\n$y=4(\\frac{\\sqrt{3}}{2} \\cos nt+\\frac{1}{2} \\sin nt)$\n$y= 2\\sqrt{3} \\cos nt +2 \\sin nt$\nNow my problem is how do i form the Cartesian equation in x and y as t varies?\nI take n as constant and t varies? Can anyone explain?\nThanks\n\n2. Originally Posted by arze\nGiven $x=2 \\sin (nt+\\frac{\\pi}{3})$ and $y=4 \\sin (nt+ \\frac{\\pi}{6})$, express $x$ and $y$ in terms of $\\sin nt$ and $\\cos nt$. Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.\n\nI have expressed $x$ and $y$ in terms of $\\sin nt$ and $\\cos nt$ already.\n$x=2 \\sin (nt+\\frac{\\pi}{3})$\n$x=2(\\frac{1}{2} \\sin nt+\\sqrt{3} \\cos nt)$\n$x= \\sin nt+ \\sqrt{3} \\cos nt$\nand\n$y=4 \\sin (nt+ \\frac{\\pi}{6})$\n$y=4(\\frac{\\sqrt{3}}{2} \\cos nt+\\frac{1}{2} \\sin nt)$\n$y= 2\\sqrt{3} \\cos nt +2 \\sin nt$\nNow my problem is how do i form the Cartesian equation in x and y as t varies?\nI take n as constant and t varies? Can anyone explain?\nThanks\n$x=2 \\sin (nt+\\frac{\\pi}{3})$ and $y=4 \\sin (nt+ \\frac{\\pi}{6})$\n\n$x=2 \\sin (nt+\\frac{\\pi}{6}+\\frac{\\pi}{6})$\n\nlet $\\alpha=nt+\\frac{\\pi}{6}$\n\n$x=2 \\sin (\\alpha+\\frac{\\pi}{6})$ and $y=4 \\sin (\\alpha)$\n\n$\\frac{x}{2}=\\sin (\\alpha)\\cos(\\frac{\\pi}{6}) + \\cos (\\alpha)\\sin(\\frac{\\pi}{6})$\n\nand\n\n$\\frac{y}{4}=\\sin (\\alpha)$\n\nuse $\\sin^2 (\\alpha) + \\cos^2(\\alpha) = 1$ to eliminate $\\alpha$ and get your equation\n\n3. Hello, arze!\n\nGiven: . $\\begin{array}{ccc}x &=&2 \\sin \\left(nt+\\frac{\\pi}{3}\\right) \\\\ y &=& 4 \\sin \\left(nt+ \\frac{\\pi}{6}\\right)\\end{array}$\n\nExpress $x$ and $y$ in terms of $\\sin nt$ and $\\cos nt$.\n\nFind the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.\n\nI have expressed $x$ and $y$ in terms of $\\sin nt$ and $\\cos nt$ already.\n\n. . $\\begin{array}{ccccc}x\\:=\\: 2\\sin\\left(nt+\\tfrac{\\pi}{3}\\right) \\:=\\:2\\left(\\tfrac{1}{2}\\sin nt+\\tfrac{\\sqrt{3}}{2}\\cos nt\\right) &\\Rightarrow& x\\:=\\: \\sin nt+ \\sqrt{3} \\cos nt \\\\ \\\\[-4mm] y\\:=\\:4\\sin\\left(nt+ \\tfrac{\\pi}{6}\\right) \\:=\\:4\\left(\\tfrac{\\sqrt{3}}{2} \\cos nt+\\tfrac{1}{2} \\sin nt\\right) &\\Rightarrow& y \\:=\\: 2\\sqrt{3} \\cos nt +2 \\sin nt \\end{array}$\n\n. . . . . Good work!\n\nNow my problem is how do i form the Cartesian equation in $x$ and $y$ as $t$ varies?\n\nWe have: . $\\begin{array}{cccc}x &=& \\sin nt + \\sqrt{3}\\cos nt & {\\color{blue}(1)} \\\\ \\\\[-4mm] \\dfrac{y}{2} &=& \\sqrt{3}\\sin nt + \\cos nt & {\\color{blue}(2)}\\end{array}$\n\n$\\begin{array}{ccccc}\\text{Square }{\\color{blue}(1)}: & x^2 &=& \\sin^2\\!nt + 2\\sqrt{3}\\sin nt\\cos nt + 3\\cos^2\\!nt \\\\\n\\text{Square }{\\color{blue}(2)}: & \\dfrac{y^2}{4} &=& 3\\sin^2\\!nt + 2\\sqrt{3}\\sin nt\\cos nt + \\cos^2\\!nt \\end{array}$\n\nAdd: . $x^2 + \\frac{y^2}{4} \\;=\\;4\\sin^2\\!nt + 4\\sqrt{3}\\sin nt\\cos nt + 4\\cos^2\\!nt$\n\n. . . . $x^2 + \\frac{y^2}{4} \\;=\\;4\\underbrace{\\left(\\sin^2\\!nt + \\cos^2\\!nt\\right)}_{\\text{This is 1}} + 2\\sqrt{3}\\underbrace{\\left(2\\sin nt\\cos nt\\right)}_{\\text{This is }\\sin2nt}$\n\n. . . . $x^2 + \\frac{y^2}{4} \\;=\\;4 + 2\\sqrt{3}\\sin(2nt)$\n\n4. Originally Posted by Soroban\nHello, arze!\n\nWe have: . $\\begin{array}{cccc}x &=& \\sin nt + \\sqrt{3}\\cos nt & {\\color{blue}(1)} \\\\ \\\\[-4mm] \\dfrac{y}{2} &=& \\sqrt{3}\\sin nt + \\cos nt & {\\color{blue}(2)}\\end{array}$\n\nAdd: . $x^2 + \\frac{y^2}{4} \\;=\\;4\\sin^2\\!nt + 4\\sqrt{3}\\sin nt\\cos nt + 4\\cos^2\\!nt$\n\n. . . . $x^2 + \\frac{y^2}{4} \\;=\\;4\\underbrace{\\left(\\sin^2\\!nt + \\cos^2\\!nt\\right)}_{\\text{This is 1}} + 2\\sqrt{3}\\underbrace{\\left(2\\sin nt\\cos nt\\right)}_{\\text{This is }\\sin2nt}$\n\n. . . . $x^2 + \\frac{y^2}{4} \\;=\\;4 + 2\\sqrt{3}\\sin(2nt)$\nif you want to do it that way, solve (1) and (2) for sin and cos:\n\n$2\\sin nt = \\frac{\\sqrt3y}{2}-x$\n\n$2\\cos nt = \\sqrt3x-\\frac{y}{2}$\n\nthen use\n\n$\\sin^2nt + \\cos^2nt = 1$\n\nto get an equation in x and y\n\n5. Hello, Tesla!\n\nAbsolutely right!\n\nIt's an old habit.\n\nWhenever I see something like: . $\\begin{array}{ccc}x &=& a\\sin\\theta \\\\ y &=& b\\cos\\theta\\end{array}$\n\n. . I tend to use the square-and-add approach to simplify.\n\nI must learn to get out of my \"box\".", "date": "2015-11-30 19:51:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/95321-cartesian-equations.html", "openwebmath_score": 0.9575490951538086, "openwebmath_perplexity": 296.58703314066724, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180639771091, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8615937167797967}}
{"url": "https://math.stackexchange.com/questions/2701982/why-cant-y-xex-be-solved-for-x", "text": "# Why can't $y=xe^x$ be solved for $x$?\n\nI apologize for my mathematical ignorance regarding this, but could someone help me understand why it isn't possible to (symbolically) find an inverse function for $f(x)=xe^x$?\n\nThe most obvious (but presumably the most trivial) is that $f$ does not pass the \"horizontal line test\". However, if we restrict the domain to $x\\geq-1$ this should not be a problem (derivative is positive for $x>-1$ so function is strictly increasing). So now my question becomes: \"Why can't we find an inverse function for $f$ over the interval $[-1,\\infty)$?\"\n\nPerhaps it is because $e^x$ is transcendental (not algebraic). However, we can find an inverse for $g(x) = e^x$, which is also transcendental. Is that because we're \"cheating\" by defining another transcendental function, namely $\\ln(x)$, to be its inverse? In other words, would it be fundamentally no different to define a new function, call it $\\text{lnx}(x)$ (if that's not already something else), to be the inverse of $xe^x$ over $[-1,\\infty)$ and then say that $f$ has a \"closed form\" / \"symbolic\" / ??? inverse function $f^{-1}(x)=\\text{lnx}(x)$ over the interval $[-1, \\infty)$?\n\n### SageMath source to generate plot\n\nxs = (x,-5,2) ys = (y,-1,5) p1 = implicit_plot(x*exp(x)-y,xs,ys, color='blue', legend_label='y=x*e^x') p2 = implicit_plot(x-y,xs,ys, color='orange', linestyle='dashed', legend_label='y=x') p3 = implicit_plot(exp(x)-y,xs,ys, color='green', linestyle='dotted', legend_label='y=e^x') combined = p1 + p2 + p3 combined.axes_labels(['x', 'y']) combined.legend(True) combined.show(title='Transcendental Stuff', frame=True, axes=True, legend_loc='lower right') \n\n\u2022 it can be solved by the following function $$\\{\\{x\\to W(y)\\}\\}$$ the product Logarithmus by Mathematica \u2013\u00a0Dr. Sonnhard Graubner Mar 21 '18 at 15:49\n\u2022 It can be solved, and an inverse function can be found in the given interval, just not analytically. \u2013\u00a0Thern Mar 21 '18 at 15:51\n\u2022 Yes, you're right that \"fundamentally\" it would \"be no different\" to invent a label for the inverse function, which is guaranteed to exist because your function is injective. \u2013\u00a0symplectomorphic Mar 21 '18 at 15:51\n\u2022 This question is so high quality, why aren't people upvoting this \u2013\u00a0vrugtehagel Mar 21 '18 at 15:54\n\u2022 @vrugtehagel that is what I have about this site more than anything else. There's so much good content that goes unappreciated. This is an excellent question. I upvoted it. \u2013\u00a0user223391 Mar 21 '18 at 16:03\n\nIt can be solved by inventing new functions, but it cannot be solved in closed form using trigonometric, logarithmic or exponential etc. Read this: Chow, Timothy Y. (May 1999), \"What is a Closed-Form Number?\"\n\n\u2022 I believe you, but I was trying to get a better answer to the why part of the question. \u2013\u00a0iX3 Mar 21 '18 at 16:02\n\u2022 Read the linked article for the why part. \u2013\u00a0Shubhashish Mar 21 '18 at 16:06\n\u2022 Wow, this paper talks about exactly what I wanted to know! (Did you just add that? I can't believe I didn't see it before.) BTW, it was hard to read on that site (not scalable, not searchable, etc.), but I found a link to a PDF version on the author's website here \u2013\u00a0iX3 Mar 21 '18 at 16:18\n\u2022 Yes I have updated the link \u2013\u00a0Shubhashish Mar 21 '18 at 16:19\n\nYou can use the Lambert-$W$ function to solve it symbolically.\n\n$y = xe^x$ gives $x = W(y)$.\n\nYou may run solve(x*exp(x)-y,x) on SymPy Live as an alternative to SageMath.\n\n\u2022 Just for reference, here is a Wolfram Alpha example solving $xe^x=\\pi$ \u2013\u00a0gt6989b Mar 21 '18 at 15:56\n\u2022 Thank you for showing me the Lambert-W function (was just finding that in the answer to another question on here too). It seems that what determines whether it is possible to solve something in \"closed form\" has largely to do with what functions/expressions are allowed. e.g. $y = e^x$ cannot be \"solved for $x$\" unless something like $\\ln$ is allowed. Similarly, the answer to my question depends on whether or not something like $W$ is acceptable. \u2013\u00a0iX3 Mar 21 '18 at 16:00\n\u2022 @iX3 In the tag info of (closed-form), it's said that a \"closed form expression\" is anything written in terms of \"known\" functions. Though the Lambert-W function is not elementary, but I believe it qualifies a known function. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Mar 21 '18 at 16:09\n\u2022 Hmm, solve(x*exp(x)-y,x) --> Traceback (most recent call last):...line 1055, in solve solution = _solve_system(f, symbols, **flags) ... line 158, in _lambert solns[i] = tmp.subs(u, rhs) AttributeError: 'dict' object has no attribute 'subs' I may need to go read & learn about SymPy to figure out what's wrong. (I think SageMath uses SymPy though, so in theory I should be able to access all of the same functionality through that.) \u2013\u00a0iX3 Mar 21 '18 at 18:25\n\u2022 @iX3 I've edited my post in response to your comment. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Mar 21 '18 at 18:33\n\nAs indicated in the comments and in another answer, there is a special function that has been defined which can serve as an inverse here. However, this doesn't address your basic question, which is: why is this necessary?\n\nBefore I talk about that, let me say something about $\\ln x$ and $e^x$. Those are both transcendental functions, and they're inverses of each other, but neither is really defined simply to be the inverse of the other. Both of these functions arise very naturally, each on their own. The natural log function is the integral of $x^{-1}$, for a suitably defined integral, and the exponential function is the solution of a differential equation modeling the simplest kind of constant relative growth. They end up being inverses of each other, and that's a cool set of facts to understand and wonder at.\n\nAnyway, you can't use elementary functions to invert lots of things, such as $f(x)=xe^x$. This often happens when we mix different types of functions together. The functions $g(x)=x$ and $h(x)=e^x$ are perfectly invertible, and they are, respectively, polynomial and exponential. The function $f=gh$, on the other hand, is the product of a polynomial function with an exponential function. We expect that to be more complicated. It's not solvable, with the usual algebraic methods, because whatever technique we apply to simplify $e^x$ messes up the polynomial part. Similarly, if you try to solve $e^x(x^2-x)=k$, anything you do to simplify the polynomial part will just make an exponential mess.\n\nIt's the blending of different types of functions, which are amenable to transforming with completely different tools, that makes such functions complicated. Other notorious examples include $\\frac{\\sin x}{x}$, and $e^{x^2}$. It's not always that they're hard to invert, but try doing integral calculus with such functions!\n\n\u2022 Thank you; this makes a lot of sense, at least intuitively. Is there a more precise definition regarding the \"blending of different types of functions\" and does that always produce tough situations like this? \u2013\u00a0iX3 Mar 21 '18 at 16:10\n\u2022 I'd say the different kinds are 1. polynomial/rational/radical (algebraic), 2. Trigonometric, 3. Inverse trigonometric, 4. Exponential, 5. Logarithmic. If it's done right, to allow for cancellation, you can blend 2 with 3 and 4 with 5 and still be able to work analytically. If you're very careful, you can blend any kinds, as long as there's a way to cancel stuff until you have something pure. I don't know of any real precise definitions and theory that address this \"blending\" though. \u2013\u00a0G Tony Jacobs Mar 21 '18 at 16:14\n\nI am going to break your question into two separate questions:\n\n1. Does an inverse function to $y=x \\cdot e^x$ exist on the domain $[-1,+\\infty)$?\n2. Can we find that inverse function in the \"standard list\" of functions that everyone knows?\n\nQuestion 1 has an easy answer: yes, that inverse function exists. Its existence is an application of the inverse function theorem which most students first encounter without proof in an ordinary calculus class, and which they then encounter with proof in some kind of advanced calculus class. And then if you're quick enough, you might even be able to attach a name to that function, although it appears that Lambert beat you to it as shown in the answer of @GNUSupporter.\n\nQuestion 2 is more difficult to answer. First there is no \"standard list\" of functions that everyone knows. Nonetheless, perhaps that's just because we haven't tried hard enough to list all functions. Can we keep adding functions to the \"standard list\" until we don't have to add any more? For example, as suggested in the answer of @GNUSupporter, can we just add the \"Lambert-W function\" $x=W(y)$ to our list if we had never heard of it before, and then declare ourselves happy? Well, maybe, but then you would be perfectly justified to ask whether there is an inverse function to $z=yW(y)$ on some appropriately chosen domain......................\n\nThese kinds of questions are addressed (but not definitively answered) in a branch of mathematics called differential algebra which is a kind of broad generalization of differential equations.\n\n\u2022 Thank you for the reference to differential algebra. I gather from your answer that the \"answer\" to my \"why?\" question is basically \"Differential algebra deals with this question but does not provide a complete/general answer.\" Is that about right? \u2013\u00a0iX3 Mar 21 '18 at 16:12\n\u2022 Yes, that's about right. Nonetheless, one can often answer specific questions, which is what's so nice about the link to Chow's paper provided in the answer of @ShubhashishChauhan. \u2013\u00a0Lee Mosher Mar 21 '18 at 17:05\n\u2022 Although, to qualify my last comment, even Chow's paper does not provide a definitive answer to questions about Lambert's function, instead reducing them to applications of big unsolved conjectures. \u2013\u00a0Lee Mosher Mar 21 '18 at 17:13", "date": "2019-11-18 04:40:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2701982/why-cant-y-xex-be-solved-for-x", "openwebmath_score": 0.6428465247154236, "openwebmath_perplexity": 540.4593214149359, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620596782468, "lm_q2_score": 0.8976952900545976, "lm_q1q2_score": 0.8615738805462617}}
{"url": "https://math.stackexchange.com/questions/3436407/inequalities-with-two-absolute-values-with-greater-than-symbol-please-tell-me-t", "text": "# Inequalities with two absolute values with greater than symbol. Please tell me the proper way of doing $|4x-1|>|3x+2|$\n\nI have finals in a couple of days. I need help with inequalities. I have spent around 15 hours trying my hand at inequalities and I am still trying to figure things out. In this case I need help with an inequality with two absolute values with greater than symbol.\n\nThe following is a picture of the inequality ($$|4x-1|>|3x+2|$$) and the procedure I tried, which is something I applied from an answer by Isaac to another question in this site, although I think I did something wrong because it looks like I got the answer wrong or incomplete.\n\nPlease tell me what I did wrong and how to do it the right way.\n\nPS: Does this method also apply if there is a less than sign?\n\n\u2022 Alternatively (and I'd say it's faster) you could note that $|a|>|b|$ is equivalent to $a^2>b^2$.\n\u2013\u00a0dfnu\nNov 15, 2019 at 8:32\n\u2022 Two years later I have completely forgotten how to do this. :D Nov 19, 2021 at 23:50\n\nHere\u2019s another way to look at it: $$|4x-1|>|3x+2| \\iff (4x-1)^2-(3x+2)^2>0 \\iff (7x+1)(x-3)>0$$\n\nNow LHS has only two obvious zeros, and outside the interval with those end points, it has to be positive.\n\n\u2014\u2014-\n\nP.S. When you have something like $$f(x)>0$$, where $$f$$ is continuous, it can change sign only when there are roots. So its roots effectively partition the real line into intervals which are either solutions or not. In the above case, two roots are $$-\\frac17, 3$$, so we need to consider only the three intervals $$(-\\infty, -\\frac17), (-\\frac17, 3)$$ and $$(3, \\infty)$$. Obviously the LHS is positive for $$x\\to \\pm\\infty$$ and negative for $$x=0$$, so the first and last intervals are solutions, the middle one isn\u2019t.\n\nBTW, whether the intervals to consider are open / closed depends on whether the inequality is strict or $$\\geqslant$$.\n\n\u2022 7x+1>0 gives me (-1/7,+) which seems to be invalid. What is wrong? Nov 15, 2019 at 16:18\n\u2022 @dfnu +macavity x-3 gives me (3,+), with a combined solution of (-1/7,+). Nov 15, 2019 at 16:56\n\u2022 @freethinker36 If you need $A\\cdot B>0$, you need $A, B$ to have the same sign. Not just $A$ or just $B$. Nov 15, 2019 at 16:58\n\u2022 It would be great if you can elaborate. I'm learning, thus what may look evident, for me may not be evident. And tomorrow is my test and have to study two classes; already taking time of the second class to try to cement my understanding of inequalities. But I appreciate whatever guidance has been provided! Nov 15, 2019 at 17:05\n\u2022 OK, I have added some explanations, do let me know if youre having more doubts after studying it. Nov 15, 2019 at 17:11\n\nYou're doing almost everything right; you're just misinterpreting the results at the end.\n\nIn particular, the idea is to combine intervals where the inequality holds. In the first region, $$\\left(-\\infty, -\\frac23\\right)$$, you arrived at $$x<3$$. It seems that you rejected this result, judging by the \"$$\\times$$\" beside your work. This is where you went wrong.\n\nInstead, think of it like this:\n\nWithin the region $$\\left(-\\infty, -\\frac23\\right)$$, which $$x$$ satisfy $$x<3$$?\n\nAll $$x$$ do within this region. Another way to see this is that you are effectively taking the intersection of the two: $$\\left(-\\infty, -\\frac23\\right) \\cap (-\\infty,3) = \\left(-\\infty, -\\frac23\\right)$$\n\nThe other two regions are handled the same way.\n\n\u2022 My rationale for rejecting it was that 3>-2/3, so x<3 is not within the region of x<-2/3. Nov 15, 2019 at 6:33\n\nI tried again, following the answer of Th\u00e9ophile, and I was able to reach the correct solution. Below is the pic of the procedure I used. Thanks!", "date": "2022-05-22 23:02:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3436407/inequalities-with-two-absolute-values-with-greater-than-symbol-please-tell-me-t", "openwebmath_score": 0.8174499869346619, "openwebmath_perplexity": 265.46551087897507, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.972830768464319, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.861569557528886}}
{"url": "https://math.stackexchange.com/questions/177917/are-right-continuous-functions-measurable/1134904", "text": "# Are right continuous functions measurable?\n\nAre right-continuous function from $\\mathbb{R}^n$ to $\\mathbb{R}$ necessarily semi-continuous? If not, are they necessarily Borel measurable? Is there a topological characterization of right-continuous functions (as there is of continuous ones)? Are CDFs of $n$-dimensional random vectors measurable?\n\nNote: A function $f: \\mathbb{R}^n \\longrightarrow \\mathbb{R}$ is right-continuous iff it is right-continuous at every point $x \\in \\mathbb{R}^n$. A function $f: \\mathbb{R}^n \\longrightarrow \\mathbb{R}$ is right-continuous at $x \\in \\mathbb{R}^n$ iff given any infinite sequence of points in $\\mathbb{R}^n$ $(y_0,y_1,\\dots)$ that converges to $x$ from above (i.e. the sequence converges to $x$ in the usual, Euclidean sense and in addition every sequence element is greater than or equal to $x$ component-wise), the sequence $(f(y_0), f(y_1), \\dots)$ converges to $f(x)$ in the usual sense.\n\n\u2022 You might want to add the definition of right-continuity for a function defined on $\\mathbb R^n$ when $n\\gt1$.\n\u2013\u00a0Did\nAug 2, 2012 at 8:28\n\u2022 What are the definition of right-continuous function, and semi-continuous?\n\u2013\u00a0Paul\nAug 2, 2012 at 8:44\n\nThe answer to the first question is no, even in the case $n=1$: The characteristic function of the half open interval $[0,1)$ is right continuous, but neither upper nor lower semicontinuous.\n\nA right continuous function $\\mathbb{R}\\to\\mathbb{R}$ is indeed Borel measurable. By definition, the inverse image $E$ of an open set has the property that for any $x\\in E$, there is some $\\delta>0$ so that $[x\u201ax+\\delta)\\subseteq E$. It follows that $E$ is a countable union of half open intervals, and hence is Borel measurable. I am not sure about the answer to this one when $n>1$ (the countable union argument no longer holds), but my guess is that right continuous functions are still measurable.\n\nTopological characterization: If we write $\\le$ for pointwise comparison on $\\mathbb{R}^n$, we can make a one-sided topology by declaring a set $V\\subseteq\\mathbb{R}^n$ to be open if, for each $x\\in V$, there is some $\\delta>0$ so that $\\{y\\ge x\\colon\\lvert y-x\\rvert<\\delta\\}\\subseteq V$. Then the right continuous maps $\\mathbb{R}^n\\to\\mathbb{R}$ are just the ones that are continuous from this topology to the usual topology on $\\mathbb{R}$.\n\nI am not too sure on the CDF question either. (I assume CDF stands for cumulative distribution function, in the sense of $F(x)=\\mathrm{P}\\{X\\le x\\}$, where $X$ is a random $n$-vector.) It might help that $F$ is not only right continuous, but also monotone. So the set $\\{x\\colon F(x)\\le p\\}$ has a particularly simple structure; I imagine it must be measurable, but right now I don't see a proof.\n\n\u2022 Thanks, Harald, but it is in fact the case n>1 that i am interested in. See my clarification of the term \"right-continuous\" in a comment to my original post. Also, of the four questions i posed in my original post i am most interested in the last one concerning CDF. Aug 2, 2012 at 10:49\n\u2022 @EvanAad: This topology appears to be the product of $n$ copies of the Sorgenfrey line. But I don't quite see why the open sets of this topology are Borel. It certainly has a basis of Borel sets, but I don't think it's second countable, so why does it follow that any union of such sets is also Borel? Aug 2, 2012 at 21:54\n\u2022 @EvanAad: In fact, let $E$ be a non-measurable subset of $\\mathbb{R}$ and let $A = \\bigcup_{x \\in E} [x,\\infty)^2$. $A$ is open in our topology on $\\mathbb{R}^2$ (the Sorgenfrey plane), but not Borel with respect to the usual topology, since its intersection with the diagonal line $y=-x$ is a skewed copy of $E$. One might think one could exploit this to make a right-continuous function which is not Borel, but I still don't see how. Aug 3, 2012 at 13:11\n\u2022 @NateEldredge: Did you mean $\\bigcup_{x\\in E}[x,\\infty)\\times[-x,\\infty)$? Aug 3, 2012 at 13:51\n\u2022 @HaraldHanche-Olsen: Oops, yes, I did. Aug 3, 2012 at 14:25\n\nHere is a proof that any function $F: \\mathbb{R}^n \\to \\mathbb{R}$ which is right continuous is Borel measurable.\n\nDefine $F_n(x) = F\\left(\\frac{[nx]+1}{n}\\right)$, where $[x]$ is the greatest integer smaller than $x$. Clearly $F_n$ are measurable since they are step functions.\n\nFrom the right continuity of $F$ we get that $F_n\\to F$, since $\\frac{[nx]+1}{n} \\downarrow x$.\n\nHence since $F$ is the pointwise limit of measurable functions, it is measurable too.\n\n\u2022 Very Nice, Milind! Aug 24, 2017 at 13:58\n\u2022 to make this a bit more solid, you would need to clarify that $n=1$ or properly define $[x]$ for an $n$-vector $x$. Oct 12, 2018 at 11:58", "date": "2022-06-29 23:05:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/177917/are-right-continuous-functions-measurable/1134904", "openwebmath_score": 0.958007276058197, "openwebmath_perplexity": 134.49053915732216, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307668889048, "lm_q2_score": 0.885631470799559, "lm_q1q2_score": 0.8615695429188837}}
{"url": "https://math.stackexchange.com/questions/1756921/horizontal-tangent-line-of-a-parametric-curve", "text": "Horizontal tangent line of a parametric curve\n\nSuppose $x=t^2,y=t^3$ is a parametric curve. Here's a quote from my textbook:\n\nThe origin, which corresponds to $t=0$, is a singular point of the parametric curve, because $dx/dt=2t,dy/dt=3t^2$ are both zero when $t=0$.\n\nSo far so good.\n\nBut then they write:\n\nHowever, the curve has a horizontal tangent line at the origin, because for all $t\\neq 0$: $$\\frac{dy}{dx}=\\frac{dy/dt}{dx/dt}=\\frac{3}{2}t$$ And thus: $$\\lim_{t\\to 0^+} \\frac{dy}{dx}=\\lim_{t\\to 0^-} \\frac{dy}{dx}=0$$\n\nIt looked a little odd for me. Nevertheless, I decided to use the same argument to show that the parametric curve $x=2\\cos t - \\cos (2t), y=2\\sin t - \\sin(2t)$ has a horizontal tangent line at $t=0$, that is at $(1,0)$.\n\nHowever my professor said that this is wrong (\"because the derivative is not zero\" - indeed, $\\frac{dy}{dx}\\Big|_{t=0}$ is undefined - \"$0/0$\").\n\nSo who is right? Is the existence of the limit a sufficient condition for the (horizontal) tangent line to exist, as my textbook says, or not? I'm confused.\n\nThanks.\n\nShort version of the question: can a parametric curve have a horizontal tangent line at a singular point?\n\nI.e. is $\\lim_{t\\to 0} \\frac{dy}{dx}=0$ a sufficient condition for a horizontal line to exist (at $t=0$)? (even if the derivative $\\frac{dy}{dx}\\Big |_{t=0}$ itself doesn't exist).\n\n\u2022 Reading your comments to the answers below, I think you are unduly concerned about the distinction between $f(x_0)$ and $\\lim_{x\\to x_0}f(x)$. Where we have an isolated point $x_0$ for which $f(x_0)$ does not exist, but $\\lim_{x\\to x_0}f(x)$ does exist, it is common for people to assume without comment that we define $f(x_0)$ as the limit. \u2013\u00a0almagest May 3 '16 at 18:31\n\u2022 what is your textbook ? \u2013\u00a0KonKan May 3 '16 at 18:44\n\u2022 @KonKan - Calculus by Anton (5th ed.) \u2013\u00a0user239753 May 3 '16 at 18:53\n\u2022 @almagest - unduly? This might be an obvious assumption for some, but not for me. The very fact that my professor dismissed it says that this is not trivial, and at least deserves a mention. \u2013\u00a0user239753 May 3 '16 at 19:02\n\u2022 @user239753: i'm a bit confused: I'm searching in your book (it happens to have a copy of the 10th edition, available right in front of me) but i cannot find the quote you are mentioning. I am actually looking at Ch.10, par.10.1, p.692-705. The example you are mentioning is example 6, p.697. However, the authors simply mention that $t=0$ is a singular point. I cannot find the limit computation you are mentioning. Can you indicate the exact page ? \u2013\u00a0KonKan May 3 '16 at 19:21\n\nNotice that both your curves are algebraic curves, of equations $x^3 - y^2 = 0$ and $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ respectively.\n\nWe say that the line $ax + by + c = 0$ is tangent at the point $(x_0, y_0)$ to the curve given by $F(x,y) = 0$ if and only if $(x_0, y_0)$ is a multiple root of the system $\\begin{cases} F(x,y) = 0 \\\\ ax + by + c = 0 \\end{cases}$ (this is the definition that was originally used by algebraic geometers).\n\nIn your case, the points are $(0,0)$ in the first case and $(1, 0)$ in the second, and the line to be checked is $y=0$.\n\nPlugging $y = 0$ into $x^3 - y^2 = 0$ gives $x^3 = 0$, which indeed has $x=0$ as a triple root, therefore $y=0$ is tangent to $x^3 - y^2 = 0$ at $(0,0)$.\n\nPlugging $y = 0$ into $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ gives $(x^2-1)^2 - 4(x-1)^2 = 0$, or equivalently $(x-1)^2 ((x+1)^2 - 4) = 0$, or again $(x-1)^3 (x+3) = 0$, which indeed has $x=1$ as a triple root (the root $-3$ being simple, meaning that at the point $(-3, 0)$ the line $y=0$ intersects the curve as a secant), therefore $y=0$ is tangent to $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ at $(1,0)$.\n\nSince the bounty giver is still unsure, let us follow Wikipedia and compute the intersection number in the first case. Wikipedia gives several methods, one of them is to consider the ideal generated by $y^2 - x^3$ and by $y$ in $\\Bbb R[[x,y]]$ (the ring of formal series in $x$ and $y$) and compute the dimension of the quotient vector space $\\Bbb R [[x,y]] / (x^3-y^2, y)$.\n\nIn the quotient space, both $y$ and $x^3 - y^2$ will become $0$, which implies that $x^3$ also becomes $0$. Therefore, the only powers of $x$ and $y$ that survive in the quotient set are $x^0 = y^0 = 1$, and $x$ and $x^2$ - a total of $3$ linearly independent powers, so $3$ will be the intersection number of $y=0$ and $x^3-y^2$ at $(0,0)$. Since everything $\\ge 2$ means tangency, this shows that the line $y=0$ is tangent to $x^3-y^2$ at $(0,0)$.\n\nThe same thing could be done with what Wikipedia denotes by $I_{(0,0)}$, letting $P = y$ and $Q = x^3 - y^2$. Applying the properties that you see on that page (the numbers above equal signs are the Wikipedia properties that I apply),\n\n$$I_{(0,0)} (y, x^3-y^2) \\overset 6 = I_{(0,0)} (y, x^3) \\overset 5 = I_{(0,0)}(x,y) + I_{(0,0)}(x,y) + I_{(0,0)}(x,y) \\overset 4 = 1 + 1 + 1 = 3 .$$\n\nThe same computations could be done for the second example, it is just that they are more tedious. First, let us translate the curve such that the cusp moves from $(1,0)$ to $(0,0)$ (because, for simplicity, Wikipedia's formulae are given only for $(0,0)$). To do this we shall make the change of variable $x = u + 1$, which will lead to $(u^2 +2u +y^2)^2 -4(u^2+y^2)$. Next,\n\n$$I_{(0,0)} ((u^2 +2u +y^2)^2 -4(u^2+y^2), y) \\overset 6 = I_{(0,0)} (u^4 + 4u^3, y) = I_{(0,0)} (u^3 (u+4), y) \\overset 5 = I_{(0,0)} (u^3, y) + \\\\ I_{(0,0)} (u+4, y) = I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u+4, y) \\overset {3, 4} = 1+1+1+0 = 3 .$$\n\nAgain, we obtain an intersection number $\\ge 2$, which means tangency.\n\nWhat others have correctly stated is that the two plane curves given in the problem are not differentiable submanifolds of $\\Bbb R^2$ at the singular points under discussion, therefore they do not fit into the framework of diferential geometry, therefore we may not speak of their tangent spaces at the singular points as defined in differential geometry (a proof that an almost identical curve cannot be a smooth submanifold can be found here). Counterintuitively, though, they do have tangent lines! There is no contradiction, the thing is that we use two meanings for \"being tangent\" that are not synonymous: one comes from differential geometry (\"tangent space\"), and it doesn't apply here, the other from algebraic geometry (\"intersection number\"), and it does apply here. In the latter framework, we may speak about \"a given line being tangent (or not) to a curve in a point\" without speaking of \"the tangent space at that curve in that point\".\n\n$y=0$ is tangent to those curves at the specified points. Those curves do not have nicely defined tangent spaces at those points.\n\n\u2022 @user239753: Try these two: \"Elementary Differential Geometry\" by A.N. Pressley, and \"Differential Geometry of Curves and Surfaces\" by M.P. do Carmo. One last word: differential geometry in general is interested only in those spaces that are endowed with regular parametrizations (because one wants them to be invertible, with their inverses called \"local charts\"). Most theorems about parametrized curves assume the parametrization to be regular. \u2013\u00a0Alex M. Apr 28 '16 at 20:24\n\u2022 @AlexM. Because this is all about subtile nuances: You wrote, the cardiod would be a manifold, because the defining function $F$ is smooth. Aren't you applying the regular value theorem (which is a close relative to the implicite function theorem) here, which means 0 needs to be a regular value for $F$? (Anyway: great answer!) \u2013\u00a0hase_olaf Apr 28 '16 at 21:31\n\u2022 @AlexM. - thank you for your refined answer. A bit stupid question - what does \"multiple root\" mean? It has something to do with algebraic multiplicity? \u2013\u00a0user239753 May 3 '16 at 19:27\n\u2022 @user239753: Yes, exactly. We say that $a$ is a root of order $k$ of $P$ (where $P$ is a polynomial) if and only if $(x-a)^k \\mid P$ and $(x-a)^{k+1} \\nmid P$. The analogue for algebraic curves is the concept of \"intersection multiplicity\". \u2013\u00a0Alex M. May 3 '16 at 19:41\n\u2022 @AlexM: I have to confess .. I am the ignorant who downvoted your hard work ;) but give me a moment to explain. your answer (especially after your edit) seems valid in my eyes and you've obviously put hard work in it ! I give you credit for that. However, I feel it falls outside the spirit and the OP's question: it is a delicate question on a subtle point of parametric differentiation (from the calculus viewpoint) while you certainly view the situation from the algebraic/differential geometric eye. This is also why i rolled back your retagging. \u2013\u00a0KonKan May 5 '16 at 23:29\n\nIn order to find the tangent at some special point you should not try to compute the limit of $y'$ but the actual limit of secant directions at that point. Therefore in your first example you get $$m_+:=\\lim_{t\\to0+}{y(t)-y(0)\\over x(t)-x(0)}=\\lim_{t\\to0+}{t^3\\over t^2}=0\\ ,$$ and similarly $m_-=0$. In your second example you have $${y(t)-y(0)\\over x(t)-x(0)}={2\\sin t-\\sin(2t)-0\\over 2\\cos t-\\cos(2t)-1}={2\\sin t(1-\\cos t)\\over2\\cos t(1-\\cos t)}=\\tan t\\qquad(t\\ne0)\\ ,$$ and therefore $$m_+=m_-=\\lim_{t\\to0}{y(t)-y(0)\\over x(t)-x(0)}=0\\ .$$\n\nYou are referring to two different things here..\n\nFor the semi-cubical parabola curve $y=x^{3/2}$ there is horizontal tangent at the cusp y=0 or t=0.\n\nFor cardoid, depending on the cusp contact point chosen, slope of cusp tangent varies, look at all cusps of epicycloids:\n\nEDIT 1:\n\nSince it is passing through infinite curvature which is rate of change of tangent inclination, the slope is changing very fast and undefined.\n\nIt can be also seen at the point $(1,0)$ Changing the multiple angle\n\n$$x=2\\cos t - \\cos (n t), y=2\\sin t - \\sin( n t)$$\n\nhas the effect of influencing curvature. It changes from (positive for circle) to negative for looped case, via the present central cardoid case of infinite curvature.\n\nThe direction of tangent is vertical, indeterminate and vertical respectively\n\n\u2022 In both cases $\\lim_{t\\to 0} \\frac{dy}{dx}=0$ yet $\\frac{dy}{dx}\\Big |_{t=0}$ is undefined. The question is whether in such a situation a horizontal tangent line exists. The textbook says it does, yet my professor claims that for a horizontal tangent line to exist the derivative must EXIST and be equal to zero, that is we must have $\\frac{dy}{dx}\\Big |_{t=0}=0$. \u2013\u00a0user239753 Apr 25 '16 at 6:53\n\u2022 please read my comment again. I'm talking about the value of the derivative which does NOT exist. There is a difference between the limit (which indeed can be found with L'Hospital) and the actual value. \u2013\u00a0user239753 Apr 25 '16 at 7:28\n\u2022 Note that you should not stop hunting for the L'Hospital limit if the first order evaluation fails. Do not conclude that it is undefined. At the cusp we are dealing with second orders and so limit of $f''(t)/g^{''}(t)$ should be next evaluated. \u2013\u00a0Narasimham Apr 25 '16 at 7:31\n\u2022 unfortunately, you do not understand me. Here $dy/dx \\neq 0$ even though $\\lim_{t\\to 0} \\frac{dy}{dx}=0$. Do you know there is a difference between limit and value of the function? Take $$f(x)=\\left\\{\\begin{matrix} 0 \\text{ if } x\\neq0\\\\ 1 \\text{ if } x=0 \\end{matrix}\\right.$$. Here $$\\lim_{x\\to0}f(x)=0$$ yet $f(0)\\neq 0$. The same goes here. $\\lim_{t\\to 0} \\frac{dy}{dx}=0$ yet $\\frac{dy}{dx} \\Big |_{t=0}\\neq 0$ ! Do you understand? \u2013\u00a0user239753 Apr 25 '16 at 7:37\n\u2022 Ok, we resume after sometime,else moderator may indicate chat room. \u2013\u00a0Narasimham Apr 25 '16 at 8:34\n\nthe formula $$\\frac{dy}{dx}=\\frac{dy/dt}{dx/dt}$$ is valid for computing the derivative at points where $\\frac{dx}{dt}\\neq 0$. At singular points, i.e. points at which $$\\frac{dy}{dt}=\\frac{dx}{dt}=0$$ the above formula is not valid. However, this does not imply that the derivative does not exist; only that it is not computable by the above formula.\n\nIn such cases, a straightforward application of the definition of derivative (as already done by user Christian Blatter in his answer) can provide the answer: $$\\frac{dy}{dx}\\bigg|_{(0,0)}=\\lim_{t\\to 0^+}\\frac{y(t)-y(0)}{x(t)-x(0)}=\\lim_{t\\to 0^-}\\frac{y(t)-y(0)}{x(t)-x(0)}=\\lim_{x\\to0^\\pm}\\frac{t^3}{t^2}=0$$ which shows that the derivative exists at $(0,0)$ and so does the tangent line at $(0,0)$, which is no other than the horizontal axis itself.\n\nP.S.1: Note that, when confined to each branch seperately: $$y(x_0)=y(x(0))=y(0)=0, \\ \\ x_0=x(0)=0 \\\\ y(x)=y(x(t))=y(t)=t^3, \\ \\ x=x(t)=t^2$$ So, on each one of the functions $y(x)=x^{3/2}$, $y(x)=\u2212x^{3/2}$ i.e. on each one of the branches of the graph below, the following rates of change are identified: $$\\frac{y\\big(x(t)\\big)\u2212y(x_0)}{x(t)\u2212x_0}=\\frac{y(t)\u2212y(0)}{x(t)\u2212x(0)}$$ On the other hand, since $x=t^2\\neq 0$ when $t\\neq 0$ it should be clear that $$t\\to 0\\Leftrightarrow x\\to 0$$ At this point, it remains to invoke a classical proposition on the limit of a composite function, claiming that:\n\nLet the real functions $f,g$, for which $lim_{x\\rightarrow x_0}g(x)=g_0\\in \\mathbb{R}$ and $lim_{g\\rightarrow g_0}f(g)=l\\in \\mathbb{R}$.\nIf, furthermore $g(x)\\neq g_0$ \"close to $x_0$\" (i.e. in some interval $(a,x_0))$, then $$\\lim_{x\\rightarrow x_0}f\\big(g(x)\\big)=\\lim_{g\\rightarrow g_0}f(g)=l$$\n\nSeting $f(x):=\\frac{y(x)\u2212y(0)}{x\u22120}$ and $g(x):=x(t)$ and applying the proposition we get that: $$\\lim_{t\\to 0}\\frac{y(t)\u2212y(0)}{x(t)\u2212x(0)}\\equiv \\lim_{t\\to 0}\\frac{y\\big(x(t)\\big)\u2212y(0)}{x(t)\u2212x(0)}=\\lim_{x\\to 0}\\frac{y(x)\u2212y(0)}{x\u22120}=\\frac{dy}{dx}\\bigg|_0$$ (where the far left term is computed using side limits: for the upper branch $t>0$ thus $t\\to0\\Leftrightarrow t\\to0^+$ while for the lower branch $t<0$ thus $t\\to0\\Leftrightarrow t\\to0^-$).\n\nP.S.2: Regarding your final question: I do not believe that the existence of the limit is a sufficient condition, because generally, a differentiable function at $x_0$ need not have a continuous derivative at $x_0$: In other words, there may well be a situation at which: $\\lim_{x\\to x_0}f'(x)\\neq f'(x_0)$.\n\nHowever, this is not the case for the semicubical parabola, as we can see from its graph\n\nwhich is probably what the author of your textbook had in his mind. He computes the limit of the derivative just as if we can imagine the tangent \"slipping\" accross either of the branches to \"continuously\" become horizontal. In my opinion, he implicitly uses the assumption that the derivative of the semicubical parabola is a continuous function; and he probably does that, based on the shape of the graph.\n\n\u2022 Intuitively I understand. But in the definition of the derivative we use $\\frac{y(x)-y(x_0)}{x-x_0}$ and not $\\frac{y(t)-y(0)}{x(t)-x(0)}$. That is, we must find $y(x)$ first. Otherwise, can you prove rigorously that it works with $t$? \u2013\u00a0user239753 May 4 '16 at 7:11\n\u2022 Please read carefully: There is nothing intuitive here. (Only the idea described in your textbook seems intuitive). When confined to each branch seperately: $$y(x_0)=y(x(0))=y(0), \\ \\ y(x)=y(x(t))=y(t) \\\\ x_0=x(0), \\ \\ x=x(t)$$ So, on each one of the functions $y=x^{3/2}$, $y=-x^{3/2}$ i.e. on each of the branches of the above graph, the rates of change you are writing are identified: $$\\frac{y(x)-y(x_0)}{x-x_0}=\\frac{y(t)-y(0)}{x(t)-x(0)}$$ On the other hand, since $x=t^2$ it should be clear that $$t\\to0\\Leftrightarrow x\\to0$$ \u2013\u00a0KonKan May 4 '16 at 12:20\n\u2022 i've added the above comment to the post \u2013\u00a0KonKan May 4 '16 at 12:51\n\u2022 It should be $y(x(t))=(t^2)^3=t^6$, no? (I know that in this case it doesn't change the limit, but still). Thank you. P.S. - Also, I don't understand why somebody downvotes the answers here. \u2013\u00a0user239753 May 4 '16 at 14:28\n\u2022 actually, for $t>0$: $$y(x(t))=\\pm (x(t))^{3/2}=\\pm (t^2)^{3/2}=\\pm t^3$$ So, for $t\\in (-\\infty, +\\infty)$: $$y(t)=t^3$$ \u2013\u00a0KonKan May 4 '16 at 19:14", "date": "2021-02-25 11:48:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1756921/horizontal-tangent-line-of-a-parametric-curve", "openwebmath_score": 0.8502720594406128, "openwebmath_perplexity": 237.691814541783, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409524, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.861543233657836}}
{"url": "https://math.stackexchange.com/questions/818160/six-dice-six-different-figures-model-problem", "text": "Six dice, six different figures (model problem)\n\nI am thinking about the same problem as posted here: Probability of All Distinct Faces When Six Dice Are Rolled.\n\nIf one rolls six dice, what is the probability of attaining six different figures. I do understand the solution but wondered why it is necessary to assume that the dice are distinguishable. If I assume they are not then I have exactly one desired result and\n\n$$\\binom{6 + 6 - 1}{6} = \\binom{11}{6} = 462$$\n\npossible results.\n\nBut I get a different probability in this case which leads me to the assumption that it is not working this way. But why exactly is that?\n\nYou have used a correct Stars and Bars argument to show that there are $462$ ways to distribute $6$ objects in $6$ boxes.\n\nHowever, if we assume that the dice are fair, and do not influence each other, then these $462$ possibilities are not all equally likely.\n\nLet us look at a much smaller example, two identical coins. There are $3$ ways to distribute these into $2$ boxes. However, in this case it is fairly clear that the probability of two heads is $\\frac{1}{4}$ and not $\\frac{1}{3}$. For whether the coins are distinguishable or not, it should make no difference to the probability if we toss the coins sequentially and not simultaneously. And sequential tossing tells us the probability is $\\frac{1}{2}\\cdot\\frac{1}{2}$.\n\nRoughly speaking, an extreme Stars and Bars case, such as all balls in the third box, has smaller probability than any specific more \"even\" distribution.\n\nRemark: To see more informally that a probability model based on distinguishable dice is appropriate, assume that otherwise indistinguishable dice are made different by writing IDs on them with invisible ink. The writing should not affect the probability of events that do not involve the IDs, such as all six numbers being obtained.\n\nThere are $6!$ combinations of six different numbers, and there are $6^6$ combinations of six rolls. The probability is then: $$p = \\frac{6!}{6^6}$$\n\n\u2022 Yes, I know that. I was wondering why my way of thinking is wrong. Why can't I just record which numbers came up whithout recording from which dice it came? \u2013\u00a0Cyianor Jun 2 '14 at 14:03\n\u2022 My guess would be you need to account for the fact all orders are the same, i.e. $123456$ is the same as $654321$ \u2013\u00a0Alex Jun 2 '14 at 14:10\n\u2022 You mean I might not be able to apply Laplacian probability, i.e. p = good cases / total cases? \u2013\u00a0Cyianor Jun 2 '14 at 14:12\n\u2022 Yes you are: here $6!$ is 'good cases' and $6^6$ are 'total cases' \u2013\u00a0Alex Jun 2 '14 at 14:13\n\u2022 I meant in the case of my description above. If I say the dice are indistinguishable then the elements of my probability space have different probabilities and Laplacian probability is not applicable \u2013\u00a0Cyianor Jun 2 '14 at 14:17", "date": "2019-05-25 09:56:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/818160/six-dice-six-different-figures-model-problem", "openwebmath_score": 0.7690107822418213, "openwebmath_perplexity": 290.8475455576934, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9799765587105447, "lm_q2_score": 0.8791467690927439, "lm_q1q2_score": 0.861543225377001}}
{"url": "https://math.stackexchange.com/questions/4046626/what-is-the-least-amount-of-straight-cuts-that-should-be-made-to-get-squares-fro", "text": "# What is the least amount of straight cuts that should be made to get squares from a rectangle?\n\nThe problem is as follows:\n\nRachel has a rectangular piece of cloth which measures $$2$$ meters large and $$0.2$$ meters wide. She is to use a guillotin which can only makes cuts of $$60$$ cm of maximum in length. Another constrain in this guillotin is that it can only cut one layer of this cloth. Using this information, how many straight cuts minimum can Rachel make in order to get from such fabric $$10$$ square pieces of $$20\\,cm$$ from each side?.\n\nThe alternatives given in my book are as follows:\n\n$$\\begin{array}{ll} 1.&\\textrm{3 cuts}\\\\ 2.&\\textrm{1 cut}\\\\ 3.&\\textrm{2 cuts}\\\\ 4.&\\textrm{4 cuts}\\\\ \\end{array}$$\n\nI'm not sure how to solve this question regarding cuts. But I believe that a way to reduce the number of cuts will require to fold the piece. But in this case this seems not to be possible because there is a condition in the problem which is forbidding to do such approach.\n\nMy initial approach was to make the necessary cuts to make some sort of grid which could I guess reduce the number of cuts but 10 is not a perfect square. Therefore what sort of logic should be here?.\n\nI've went in circles for long on this question so I will really appreciate someone could give me a hand with this.\n\nTherefore what should be the way to go here?. Can someone help me here please?. I'm confused.\n\n\u2022 I think the question allows folding, but a guillotin cut cannot be through multiple layers. \u2013\u00a0peterwhy Mar 3 at 2:28\n\u2022 I brute forced a 4 cut solution. If this is the answer, there must be a proof as to why this is the minimum required. \u2013\u00a0Andrew Chin Mar 3 at 2:33\n\nTo cut the strip of cloth, it only takes 9 cuts of 20 cm, and each cut of the guillotin cuts through 60 cm at most, so at least 3 cuts are necessary.\n\nNumber the cuts $$1$$ to $$9$$ sequentially, and the resultant squares $$0$$ to $$9$$.\n\nAlign cut $$1$$ under the guillotin, then fold two right angled triangles on the strip around cuts $$2$$ and $$3$$ so that the strip goes back under the guillotin, and align cut $$4$$ under the guillotin next to cut $$1$$.\n\n(Fold twice like (B) in this random image found in Google: )\n\nDo the same thing to align cut $$7$$ under the guillotin. With cut $$1, 4, 7$$ all aligned side by side, one guillotin cut of 60 cm would do all the 3 20-cm cuts.\n\nNow squares 1-3, squares 4-6, squares 7-9 are on separated pieces of cloth, and they can be aligned freely to make the remaining 6 cloth-cuts in 2 guillotin cuts.\n\nThe optimal answer is 3 guillotin cuts.", "date": "2021-06-16 14:33:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4046626/what-is-the-least-amount-of-straight-cuts-that-should-be-made-to-get-squares-fro", "openwebmath_score": 0.6420024633407593, "openwebmath_perplexity": 605.4960783845211, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409524, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8615432212457315}}
{"url": "https://math.stackexchange.com/questions/3035983/how-can-i-find-integers-which-satisfy-frac150n15n-m", "text": "# How can I find integers which satisfy $\\frac{150+n}{15+n}=m$?\n\nHere are some facts about myself:\n\n1. In 2017, I was $$15$$ years old.\n2. Canada, my country, was $$150$$ years old.\n\nWhen will be the next time that my country's age will be a multiple of mine?\n\nI've toned this down to a function. With $$n$$ being the number of years before this will happen and $$m$$ being any integer,\n\n$$\\frac{150+n}{15+n}=m$$\n\nHow would you find $$n$$?\n\n\u2022 For extra credit, consider that Canada turned 150 on July 1, 2017. If your birthday is after that date, 14+n is also a valid denominator, as you will be both age 14 and age 15 during the 12 months that Canada is 150 years old. Likewise, if your 15th birthday is before that date, 16+n is a valid denominator, since you will be 15 when Canada turns 150, but turn 16 before Canada turns 151. \u2013\u00a0Nuclear Wang Dec 12 '18 at 4:51\n\u2022 (1) Avoid using \"interesting\" in the title; (2) describe the problem in the title, not just your opinion and its topic. \u2013\u00a0Asaf Karagila Dec 12 '18 at 11:19\n\nYou want $$\\frac{150+n}{15+n}=m$$, and clearing denominators gives us $$150+n=(15+n)m.$$ Subtracting $$15+n$$ from both sides give us $$135=(15+n)(m-1).$$ Now you are looking for the smallest $$n>1$$ for which such an $$m$$ exists, so the smallest $$n>1$$ for which $$15+n$$ divides $$135$$.\n\n\u2022 Love it, thank you! Is 120 the only $n$ that exists for such a case? (Because $135\\over270$ $=$ $1\\over2$)? \u2013\u00a0Raymo111 Dec 11 '18 at 23:21\n\u2022 What's wrong with $n=12$? The other solutions are $n=30$ and $n=120$. \u2013\u00a0Servaes Dec 11 '18 at 23:46\n\u2022 Oh shoot, I mistakenly read your answer as $n+15=135$, my bad! Also, upvote if you think this question that I thought of was interesting, it's a real-life example of an age problem. \u2013\u00a0Raymo111 Dec 11 '18 at 23:48\n\nFirst thing I would do is say that Canada is $$135$$ years older than you.\n\nThat gives you a simpler\n\n$$\\frac {135+n}{n} = k\\\\ \\frac {135}{n} = k-1\\\\$$\n\nIt will happen every time your age is a factor of $$135.$$ It last happened when you were $$15.$$ It will next happen when you are $$27$$\n\n\u2022 Do Canadians live past age 135? \u2013\u00a0richard1941 Dec 21 '18 at 16:51\n\u2022 Almost never. But with advancements in medical technology, that may change. \u2013\u00a0Doug M Dec 21 '18 at 17:00\n\nWe want the smallest positive integer $$n$$ such that there is some (positive) integer $$k$$ such that $$\\frac{150+n}{15+n}=k.$$ Note that $$k=1$$ can never work, so we can assume $$k-1\\neq0.$$ Now we rearrange the above equation: multiplying both sides by $$15+n,$$ we get $$150+n=15k+nk;$$ now rearrange and factorize to get $$15(10-k)=(k-1)n;$$ and now divide both sides by $$k-1,$$ to get $$n=\\frac{15(10-k)}{k-1}.$$ Since we want the smallest positive integer $$n,$$ we can just try values of $$k\\in\\{2,3,\\ldots,9\\},$$ starting from the largest and working our way down (because the function of $$k$$ on the right-hand side is decreasing in this range), until we arrive at an integer value of $$n.$$* When $$k=9,$$ $$8$$ or $$7$$ we get non-integer values of $$n;$$ when $$k=6$$ we find $$n=15\\times4/5=12.$$\n\nSo last year, $$n$$ was $$0,$$ and the ratio of Canada's age to your age was $$k=150/15=10;$$ and $$11$$ years from now, $$n$$ will be $$12,$$ and the ratio of Canada's age to your age will be $$k=162/27=6.$$\n\n*Incidentally, it is not obvious in advance that we will ever get an integer value of $$n;$$ if this were the case then the problem would simply have no answer. As it happens, the problem does have answers, namely $$(n,k)\\in\\{(0,10),(12,6),(30,4),(120,2)\\}.$$\n\nAlternatively:\n\n$$\\frac {150 + n}{15 + n} = \\frac {150+ 10n}{15+n} +\\frac {-9 n}{15+n}$$\n\n$$=10 -\\frac {9 n}{15+n}$$ (which is an integer for $$n=0$$ but when next?)\n\n$$=10 - \\frac {9n + 9*15}{15+n} + \\frac {9*15}{15+n}=$$\n\n$$=10 - 9 + \\frac{3^3*5}{15+n}= 1 + \\frac{3^3*5}{15+n}$$\n\nwhich is an integer if $$15+n$$ is one of the factors of $$3^3*5$$.\n\nAnd the factors of $$3^3*5$$ are $$1, 3,9, 27, 5,15, 45, 135$$.\n\nSo this will occur when $$n = -14,-12, -10, -6,0, 12,30, 120$$\n\nWhen you are $$1, 3, 5, 9, 15, 27, 45, 135$$ and canada is $$136, 138, 140, 144, 150, 162, 180, 270$$ and canada is exactly $$136,46, 28, 16, 10,6,4, 2$$ as old as you are.\n\n(Enjoy your $$45$$ birthday when your country annexes my country after we collapse from the thirty year aftermath of the unrecoverable mistakes of the last two years.)\n\nThat's a fun problem. It's nice to see other people like to think about these things.\n\n\u2022 That's nice, that's the way I usually think about these problems too; but you can shorten the process by writing $\\dfrac {150+n}{15+n} = \\dfrac {15+n}{15+n} + \\dfrac{135}{15+n}$ \u2013\u00a0Ovi Dec 12 '18 at 2:34\n\u2022 Hmmm.... I'm not sure why I did it the way I did. I think somehow I briefly thought the obviousness that 15 goes directly into 150 made me briefly think that they were the coeficients and I started doing it that way and just continued. Obviously it'd be shorter and direct to do the coefficients. \u2013\u00a0fleablood Dec 12 '18 at 15:40\n\nNote that if $$k \\mid a$$ and $$k \\mid b$$, then $$k \\mid a - b$$. In this particular instance, we have $$15 + n \\mid 15 + n$$ and $$15 + n \\mid 150 + n$$, so $$15 + n \\mid 135$$. In other words, we are looking for $$n+15$$ to be the next factor of $$135$$ which is larger than $$15$$. Can you continue from here?\n\nServaes answer is excellent but let me add few cents for those who can't really make out something from dual variables and other factors (I know some people have problem with such things).\n\nThere is also a brute force solution, requiring a bit of thinking and a bit of simple calculation (the larger is the $$n$$ i $$n$$-times larger the more difficult it becomes though so I really suggest you try understanding the accepted solution anyway).\n\nIt's enough to check if there is an integer solution to equation $$\\frac{150+n}{15+n}=9$$ If no, proceed with 8, 7 and so on. If you get to 2 and still have no anwser then the answer is never.\n\nSo a quick brute force check will like that: $$\\frac{150+n}{15+n}=9$$ $$150+n=9\\cdot(15+n)$$ $$150+n=135+9\\cdot n$$ $$15=8\\cdot n$$ $$n = \\frac{15}{8}\\notin\\mathbb{Z}$$ So we continue with $$8$$ $$\\frac{150+n}{15+n}=8$$ $$150+n=8\\cdot(15+n)$$ $$150+n=120+8\\cdot n$$ $$30=7\\cdot n$$ $$n = \\frac{30}{7}\\notin\\mathbb{Z}$$ You may continue from here. Note, a smart person using this method will notice some pattern that will make it even simpler to test (not requiring all the calculation, just a quick check that literally takes seconds) but I will not give a direct hint.\n\n## Explanation\n\nWhen you add the same number to both numerator and denominator of a fraction, the fraction decreases. In other words if you consider how the fraction of age of Canada and your age changes over time it will decrease.\n\nOn the other hand the fraction can never go to $$1$$ (or below) since the numerator will always be obviously greater than denominator (in our case by 135).\n\nIn other words you have a finite set of possible resultant fractions (namely $$\\{2,3,4,5,6,7,8,9\\}$$) that are potentially the sought \"next case\". From those you're looking for the largest one, so test them one by one, starting with $$9$$ and going down until you have an integer result or your options are gone.\n\n## Remarks\n\n1. Once again - this is a brute force solution and should not be treated as preferred if you can understand using two variables. But if you can't, it's still doable.\n2. If you notice the relationship I mention above, I guess even testing situation where the initial fraction is 100 could be done.\n3. You may try solving an interesting opposite question - when was the last time when such situation occurred (i.e. Canadas age was a multiply of your age). If you think it over well, you may use both Servaes and my approaches to that as well. Especially try rationalising that my approach will again have a finite and easily predictable number of tests to be performed.\n4. A food for thoughts. Is there always a solution to the original question (with differently chosen ages)? Is there always a solution to the problem I state in remark 3?\n5. Can you think of other related questions that will use similar approach (either Servaes's elegant one or mine brute force) to find a solution?\n\nIn general questions like this, where\n\n$$\\frac{x+n}{y+n}=m$$\n\n$$m$$ will be an integer smaller than $$\\frac{x}{y}$$. Since there are finitely many integers less than $$\\frac{x}{y}$$, each value of $$m$$ can be checked one by one. Not all values will generally give an integer $$n$$.", "date": "2019-01-16 15:02:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3035983/how-can-i-find-integers-which-satisfy-frac150n15n-m", "openwebmath_score": 0.6882489919662476, "openwebmath_perplexity": 319.9692723453792, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765622193215, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8615432176011385}}
{"url": "https://gmatclub.com/forum/when-traveling-at-a-constant-speed-of-32-miles-per-hour-a-certain-mot-220272.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 24 May 2019, 10:11\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# When traveling at a constant speed of 32 miles per hour, a certain mot\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: I don't stop when I'm Tired,I stop when I'm done\nJoined: 11 May 2014\nPosts: 531\nGPA: 2.81\nWhen traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n14 Jun 2016, 13:46\n5\n26\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n81% (01:10) correct 19% (01:17) wrong based on 1426 sessions\n\n### HideShow timer Statistics\n\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\n\nB) $$\\frac{3}{4}$$\n\nC) $$\\frac{4}{5}$$\n\nD) $$\\frac{4}{3}$$\n\nE) $$\\frac{3}{2}$$\n\n_________________\nMd. Abdur Rakib\n\nPlease Press +1 Kudos,If it helps\nSentence Correction-Collection of Ron Purewal's \"elliptical construction/analogies\" for SC Challenges\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4485\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n14 Jun 2016, 17:07\n6\n2\nAbdurRakib wrote:\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\nB) $$\\frac{3}{4}$$\nC) $$\\frac{4}{5}$$\nD) $$\\frac{4}{3}$$\nE) $$\\frac{3}{2}$$\n\nDear AbdurRakib,\nI'm happy to respond.\n\nWe have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus\n\nfuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal\n\nIn that fraction, cancel the common factor of 8.\n\nfuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.\n\nDoes this make sense?\nMike\n_________________\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n##### General Discussion\nIntern\nJoined: 01 May 2015\nPosts: 39\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n15 Jun 2016, 00:46\n1\n1\nDistance travelled = 32 miles\nFel consumed = 24 gallons\nSo, 24 gallons lets motorboat travel 32 miles\nSo, miles traveled per gallon of fuel = 32/24 = 4/3\nSenior Manager\nJoined: 24 Nov 2015\nPosts: 497\nLocation: United States (LA)\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n15 Jun 2016, 02:59\n2\nIn 1 hour the motorboat travels 32 miles\nIn travelling that 32 miles the motorboat uses 24 gallons of fuel in 1 hour\n\nso basically we divide $$\\frac{miles}{hr}$$ by $$\\frac{gallons}{hr}$$ to get $$\\frac{miles}{gallons}$$\n\n$$\\frac{32}{24}$$ =$$\\frac{4}{3}$$ $$\\frac{miles}{gallons}$$\n\nManager\nJoined: 17 Aug 2015\nPosts: 99\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n13 Aug 2016, 11:05\n1\nOne way to solve this:--\nThe speed is 32 miles / hour. The fuel consumption is 24 gallons / hour. So in every hour two things are happening:-\nThe boat has covered 32 miles\nThe boat has consumed 24 gallons\nWhat is the miles /gallon --> 32/24 => 4/3miles/gal. The choice is D\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2823\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n02 Dec 2016, 06:54\nAbdurRakib wrote:\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\nB) $$\\frac{3}{4}$$\nC) $$\\frac{4}{5}$$\nD) $$\\frac{4}{3}$$\nE) $$\\frac{3}{2}$$\n\nWe are given that a motorboat has a rate of 24 gallons per hour while traveling at a speed of 32 miles per hour. We must determine the boat\u2019s fuel consumption measured in miles traveled per gallon of fuel.\n\nSince the motorboat consumes 24 gallons when traveling 32 miles, the rate is 32/24 = 4/3.\n\n_________________\n\n# Jeffrey Miller\n\nJeff@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nVP\nJoined: 09 Mar 2016\nPosts: 1283\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n23 Feb 2018, 12:17\nmikemcgarry wrote:\nAbdurRakib wrote:\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\nB) $$\\frac{3}{4}$$\nC) $$\\frac{4}{5}$$\nD) $$\\frac{4}{3}$$\nE) $$\\frac{3}{2}$$\n\nDear AbdurRakib,\nI'm happy to respond.\n\nWe have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus\n\nfuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal\n\nIn that fraction, cancel the common factor of 8.\n\nfuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.\n\nDoes this make sense?\nMike\n\nhello there mikemcgarry, how are you ? you know what i dont understand, why isnt the answer B ? wouldnt it be logically correct to divide gallons by miles ?\nCEO\nJoined: 12 Sep 2015\nPosts: 3727\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n03 Apr 2018, 10:01\n1\nTop Contributor\nAbdurRakib wrote:\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\n\nB) $$\\frac{3}{4}$$\n\nC) $$\\frac{4}{5}$$\n\nD) $$\\frac{4}{3}$$\n\nE) $$\\frac{3}{2}$$\n\nLet's see what happens after ONE HOUR of traveling.\n\nIn ONE hour, the boat will travel 32 miles and will use 24 gallons of fuel\nSo, the fuel consumption rate is 32 miles per 24 gallons\nOr we can write: fuel consumption rate = 32/24 miles/gallon\n32/24 = 4/3\nSo, the fuel consumption rate = 4/3 miles per gallon\n\nCheers,\nBrent\n_________________\nTest confidently with gmatprepnow.com\nIntern\nJoined: 01 May 2017\nPosts: 34\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n05 Apr 2018, 03:43\nSVP\nStatus: It's near - I can see.\nJoined: 13 Apr 2013\nPosts: 1674\nLocation: India\nGPA: 3.01\nWE: Engineering (Real Estate)\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n14 Apr 2018, 22:53\nAbdurRakib wrote:\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\n\nB) $$\\frac{3}{4}$$\n\nC) $$\\frac{4}{5}$$\n\nD) $$\\frac{4}{3}$$\n\nE) $$\\frac{3}{2}$$\n\nIt's (D)\n\n$$\\frac{Miles - traveled}{gallon-fuel}$$ = $$\\frac{32}{24}$$ = $$\\frac{4}{3}$$\n_________________\n\"Do not watch clock; Do what it does. KEEP GOING.\"\nIntern\nJoined: 19 Jan 2018\nPosts: 49\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n25 Jan 2019, 18:10\nmikemcgarry wrote:\nAbdurRakib wrote:\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\nB) $$\\frac{3}{4}$$\nC) $$\\frac{4}{5}$$\nD) $$\\frac{4}{3}$$\nE) $$\\frac{3}{2}$$\n\nDear AbdurRakib,\nI'm happy to respond.\n\nWe have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus\n\nfuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal\n\nIn that fraction, cancel the common factor of 8.\n\nfuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.\n\nDoes this make sense?\nMike\n\nWas checking through the solutions to see other solutions to this problem, isn't the units in red suppose to be in \"hours\", instead of gallons?\nIntern\nJoined: 13 Mar 2016\nPosts: 23\nLocation: India\nConcentration: General Management, Entrepreneurship\nWE: General Management (Energy and Utilities)\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n11 May 2019, 05:59\nSpeed of boat 32M/H and its consumption is 24 Gallon per hour.\n\nTherefore boat consumes 24Gallon for every 32 Miles.\nor for every Gallaon boat travels 32/24 miles\ni.e 4/3 miles per gallon\n\nAnsD\nManager\nJoined: 25 Sep 2018\nPosts: 59\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot\u00a0 [#permalink]\n\n### Show Tags\n\n11 May 2019, 22:55\nAbdurRakib wrote:\nWhen traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?\n\nA) $$\\frac{2}{3}$$\n\nB) $$\\frac{3}{4}$$\n\nC) $$\\frac{4}{5}$$\n\nD) $$\\frac{4}{3}$$\n\nE) $$\\frac{3}{2}$$\n\nIn 24 gallons boat travel =36\nIn 1 \" \" \" =36/24=4/3\n\nPosted from my mobile device\nRe: When traveling at a constant speed of 32 miles per hour, a certain mot \u00a0 [#permalink] 11 May 2019, 22:55\nDisplay posts from previous: Sort by", "date": "2019-05-24 17:11:01", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/when-traveling-at-a-constant-speed-of-32-miles-per-hour-a-certain-mot-220272.html", "openwebmath_score": 0.8632282614707947, "openwebmath_perplexity": 3168.0071012631397, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8615382094310357}}
{"url": "https://www.freemathhelp.com/forum/threads/probability-theory.69762/", "text": "Probability Theory\n\nalakaboom1\n\nNew member\nAn automobile insurance company classifies drivers into 3 classes: class A, class B, and class C. The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%. The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively. After purchasing an insurance policy, a driver has an accident within the first year. What is the probability that the driver is of class A?\n\nI think that it's 1/6, because you'd do 0.01/0.06, since you already know that the driver has had the accident, but I'm not too sure. One other possibility that I'm considering would be 0.2 * 0.01, but I feel like this doesn't account for the fact that we are assuming that the driver is already in an accident...\n\ngalactus\n\nSuper Moderator\nStaff member\nYou are looking for the probability the driver is from class A given they have an accident:\n\n$$\\displaystyle P(\\text{class A}|\\text{accident})$$\n\nYou can make a chart. Assume a number of drivers that is easy to work with. Say, 1000\n\nOr, use Bayes Theorem. Do you know it?.\n\nsoroban\n\nElite Member\nHello, alakaboom1!\n\nHere is a very primitive approach . . .\n\nAn automobile insurance company classifies drivers into 3 classes: class A, class B, and class C.\nThe percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%.\nThe probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively.\nAfter purchasing an insurance policy, a driver has an accident within the first year.\nWhat is the probability that the driver is of class A?\n\nSuppose there are 10,000 policyholders.\n\n$$\\displaystyle \\text{20\\% are class A drivers: }\\:20\\%\\times 10,\\!000 \\,=\\,2000$$\n. . $$\\displaystyle \\text{1\\% of them will have an accident: }\\:1\\% \\times 2000 \\,=\\,20$$\n\n$$\\displaystyle \\text{65\\% are class B drivers: }\\:65\\% \\times 10,\\!000 \\,=\\,6500$$\n. . $$\\displaystyle \\text{2\\% of them will have an accident: }\\:2\\% \\times 6500 \\,=\\,130$$\n\n$$\\displaystyle \\text{15\\% are class C drivers: }\\:15\\% \\times 10,\\!000 \\,=\\,1500$$\n. . $$\\displaystyle \\text{3\\% of them will have an accident: }\\:3\\% \\times 1500 \\,=\\,45$$\n\n$$\\displaystyle \\text{There is a total of: }\\:20 + 130 + 45 \\,=\\,195\\text{ accidents}$$\n. . $$\\displaystyle \\text{of which 20 are class A drivers.}$$\n\n$$\\displaystyle \\text{Therefore: }\\(\\text{class A}\\,|\\,\\text{accident}) \\;=\\;\\frac{20}{195} \\;=\\;\\frac{4}{39}$$\n\ngalactus\n\nSuper Moderator\nStaff member\nThat 'primitive\napproach is a nice explanation.\n\nHere is all I was getting at:\n\n$$\\displaystyle \\frac{(.01)(.2)}{(.01)(.2)+(.02)(.65)+(.03)(.15)}=\\frac{4}{39}\\approx .1025$$\n\nWhen confronted with a 'probability of something given something' problem, you can make a chart.\n\nAssume 1000 policyholders.\n\n$$\\displaystyle \\begin{array}{c|c|c|c|c}\\text{}&A&B&C&\\text{total}\\\\ \\hline\\text{has accidents}&2&13&4.5&19.5 \\\\ \\hline\\text{no accident}&198&637&145.5&980.5 \\\\ \\hline\\text{total}&200&650&150&1000\\end{array}$$\n\nNow, you can answer any problem they throw at you. In this case, we want\n\n\"Probability the driver is class A given they have an accident\".\n\nGo down the class A column and across the 'has accident' row. $$\\displaystyle \\frac{2}{19.5}=\\frac{4}{39}$$\n\nSay they asked for \"probability the driver is from class C given they did not have an accident?\"\n\nGo down the C column and across the 'no accident' row and get $$\\displaystyle \\frac{145.5}{980.5}=\\frac{291}{1961}\\approx .148$$\n\nand so on..............\n\nalakaboom1\n\nNew member\nThanks a lot guys!\n\nOut of curiosity, is this formula you guys mentioned known as Bayes Theorem? I've definitely used the formula before, but never knew it by name.", "date": "2019-03-26 06:37:33", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/probability-theory.69762/", "openwebmath_score": 0.6028389930725098, "openwebmath_perplexity": 1124.6799546791592, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105314577313, "lm_q2_score": 0.8824278772763471, "lm_q1q2_score": 0.8615236298367881}}
{"url": "http://math.stackexchange.com/questions/125289/cosets-of-a-group", "text": "# Cosets of a group\n\nOne sentence from Amstrong's Group and Symmetry wrote the following to prove a group of order 6 is isomorphic to $\\mathbf{Z}_6$:\n\nThe right cosets $\\langle x\\rangle$, $\\langle x\\rangle y$ give 6 elements $e$, $x$, $x^2$, $y$, $xy$, $x^2y$ which fill out $G$.\n\nWhere $x$ is of order $3$ and $y$ is of order 2. $x$, $y$ are both elements of $G$.\n\nMy confusion:\n\nHow can one guarantee that the right coset $\\langle x\\rangle$ and $\\langle x\\rangle y$ can fill out $G$? Because $\\langle x\\rangle$ and $\\langle x\\rangle y$ have no intersections? I am not sure about the properties of cosets. I also don't know if this is related to the fact that $|\\langle x\\rangle|=5$ and is precisely half the elements of $G$ and thus its left coset $y\\langle x\\rangle$ and right coset $\\langle x\\rangle y$ is exactly the same?\n\nI don't know if you can understand my question. The question here actually arises from my vague understanding of \"cosets\". So everytime this term occurs I feel a steady uncertainty.\n\n-\nYou may want to reread the question yourself and see where one or more letters can be added. \u2013\u00a0user21436 Mar 28 '12 at 1:36\n@KannappanSampath: An artifact of using < and > in text; parser thought they were mark-up. \u2013\u00a0Arturo Magidin Mar 28 '12 at 2:10\nThis is specifically page 70 (and end of 69). \u2013\u00a0anon Mar 28 '12 at 2:12\nExactly. It is from page 70. Sorry for the disorganized order, I will check that next time before I submit. And thanks again to @ArturoMagidin for the edition. \u2013\u00a0Jinji Mar 28 '12 at 2:26\nFor more on cosets, see this answer. \u2013\u00a0Arturo Magidin Mar 28 '12 at 2:45\n\nLet $H=\\{e,x,x^2\\}$ be the subgroup generated by $x$. The right cosets of $H$ are the sets of the form $Hz = \\{hz\\mid h\\in G\\}$ for fixed $z\\in G$.\n\nThe right cosets of $H$ are known to form a partition of the set $G$; that is: any two cosets are either identical or disjoint, and their union is all of $G$.\n\nMoreover, $Hz = Hw$ if and only if $zw^{-1}\\in H$. This can be verified because $z\\in Hz$ (obtained as $ez$), so there must exist some $h\\in H$ such that $z = hw$. Hence $zw^{-1}=h\\in H$, proving that if $Hz=Hw$, then $zw^{-1}\\in H$. Conversely, if $zw^{-1}\\in H$, then $z = (zw^{-1})w \\in Hw$, so $Hz\\cap Hw\\neq\\varnothing$; since right cosets are either disjoint or identical, and $Hz$ and $Hw$ are not disjoint, then they are identical.\n\nNow, since $y$ is of order $2$, it is not in $H$ (every element of $H$ is either of order $3$ or of order $1$). That means that the cosets $He = H$ and $Hy=\\{ey, xy, x^2y\\}$ are distinct, hence they are disjoint. Since they are disjoint, $H$ and $Hy$, together, account for $6$ elements of $G$. Since $G$ has exactly $6$ elements by assumption, those are all the elements of $G$. That is: $$G = H\\cup Hy = \\{e,x,x^2\\}\\cup\\{y,xy, x^2y\\} = \\{e,x,x^2,y,xy,x^2y\\}.$$\n\nThe same is true for left cosets: left cosets of $H$ partition $G$, so any two left cosets are either disjoint or identical. Also, $zH = wH$ if and only if $z^{-1}w\\in H$ (I'll leave the proof to you). That means that the two left cosets of $H$ are $eH=H$ and $yH$, since $yH\\neq H$, so $yH\\cap H=\\varnothing$. Since $G$ has $6$ elements, $H$ has $3$, and $yH$ has another three, then $$G = H\\cup yH = \\{e,x,x^2\\} \\cup \\{y, yx, yx^2\\}.$$ Also, since $H\\cap Hy=\\varnothing$, then $Hy = G\\setminus H$; similarly, $yH=G\\setminus H$. So in fact, $Hy=yH$ as sets.\n\n-\nThanks! Does the number of right cosets relate to the order of H? |G|/|H|=the number of right cosets? \u2013\u00a0Jinji Mar 28 '12 at 2:24\n@Jinji: The number of cosets is, by definition, the index of $H$ in $G$, denoted $[G:H]$. Since the cosets all have the same size, namely $H$, and partition $G$, it follows that $|G|=|H|[G:H]$ (in the sense of cardinalities); in fact, if $K\\lt H\\lt G$, then $[G:K]=[G:H][H:K]$ in the sense of cardinalities. So in the special case in which $|G|$ is finite, it follows from $|G|=|H|[G:H]$ that $[G:H]=|G|/|H|$. \u2013\u00a0Arturo Magidin Mar 28 '12 at 2:37", "date": "2016-02-11 17:20:10", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/125289/cosets-of-a-group", "openwebmath_score": 0.9734762907028198, "openwebmath_perplexity": 126.55874800901462, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105290452637, "lm_q2_score": 0.8824278695464501, "lm_q1q2_score": 0.8615236201611797}}
{"url": "http://math.stackexchange.com/questions/166756/proving-that-an-integer-is-even-if-and-only-if-it-is-not-odd", "text": "# Proving that an integer is even if and only if it is not odd\n\nThere is this question, but the definition of \"even\" and \"odd\" that I am using uses integers instead of just natural numbers; i.e.,\n\n\u2022 An integer $n$ is even iff there is some integer $k$ such that $n=2k$.\n\u2022 An integer $n$ is odd iff there is some integer $k$ such that $n=2k+1$.\n\nHere is what I have so far:\n\nFirst we show that an integer $n$ is even or odd. We first use induction on the positive integers. For the base case, $1=2\\cdot0+1$ so we are done. Now suppose inductively that $n$ is even or odd. If $n$ is even, then $n=2k$ for some $k$ so that $n+1=2k+1$ (odd). If $n$ is odd, then $n=2k+1$ for some $k$ so that $n+1=2(k+1)$ (even). This closes the induction, so every $n\\in\\mathbf{Z}^+$ is even or odd.\n\nNow we show every $n\\in\\mathbf{Z}^-$ is even or odd. Let $n\\in\\mathbf{Z}^-$. Then $n=-k$ for some $k\\in\\mathbf{Z}$ (I think this follows immediately from most definitions of the integers.). Suppose $k$ is even. Then $k=2j$ for some $j$ so that $n=-k=-2j=2(-j)$ (even). Now suppose $k$ is odd. Then $k=2j+1$ for some $j$ so that $n=-k=-(2j+1)=-2j-1=-2j-1+1-1=-2j-2+1=2(-j-1)+1$ (odd).\n\nFor $0$, note that $0=2\\cdot0$ (even).\n\nNow we show that $n\\in\\mathbf{Z}$ cannot be both even and odd. Suppose for the sake of contradiction that $n\\in\\mathbf{Z}$ is both even and odd. Then there are integers $k,j$ such that $n=2k=2j+1$. This implies that $2(k-j)=1$ (like in the referenced question). So we must show that $1$ cannot be even in order to complete the proof.\n\nThis is where I am having trouble. I know that if I let $f:\\mathbf{Z}\\to\\mathbf{Z};x\\mapsto2x$ be a function, then $f$ is increasing so since $f(0)=0$ and $f(1)=2$ and $0<1<2$, there is no integer $m$ such that $f(m)=1$. But this seems complicated so I was wondering if there was an easier way to do this.\n\nSo my real question is: how can I show that $1$ is not even?\n\n(This is not homework.)\n\n-\n$f(m)=1 \\implies 2m=1 \\implies 2=\\frac {1}{m}$ which is not a integer for any $m \\in \\mathbb Z$ \u2013\u00a0Theorem Jul 4 '12 at 20:57\n@russell11, I think you meant a function $f:\\mathbb{Z}^+ \\to \\mathbb{Z}^+$ (or whatever the appropriate notation is for 'nonnegative'). Otherwise you and Theorem converge on the right idea; suppose $1=2m$ for some integer $m$. $m>0$ comes immediately, as twice a non-positive integer is non-positive. $m<1$ comes from the function being increasing. \u2013\u00a0Eugene Shvarts Jul 4 '12 at 21:02\n\nTo show that $1$ is not even:\n\nI assume you can prove or accept that $a \\cdot 0 = 0$ for all $a \\in \\mathbb{Z}$, the product of a positive and negative number is negative, and $2a > a$ when $a >0$.\n\nIf $1$ is even, then there must exists $a < 1$ such that $2a = 1$. However, the only $a < 1$, which is an integer, is $0$ and clearly $2 \\cdot 0 = 0$. So $1$ can not be be written as $2 \\cdot a$ for any $a \\in \\mathbb{Z}$. $1$ is odd.\n\n-\nDo you want to say, \"So $1$ cannot be written as $2\\cdot a$ for any $a\\in\\mathbb{Z}$\" in the end? \u2013\u00a0Paul Jul 4 '12 at 21:13\n@Paul Yes. Added the missing \"not\". \u2013\u00a0William Jul 4 '12 at 21:21\n\nHint $\\$ Your induction step uses $\\rm\\:n\\,$ even $\\rm\\,\\Rightarrow\\: n\\!+\\!1\\:$ odd, and $\\rm\\:n\\,$ odd $\\rm\\,\\Rightarrow\\:n\\!+\\!1\\:$ even. The converses are both true, e.g. $\\rm\\,n\\!+\\!1\\,$ odd $\\,\\Rightarrow$ $\\rm\\,n\\!+\\!1 = 2k\\!+\\!1\\,$ $\\Rightarrow$ $\\rm\\,n = 2k,\\,$ since $\\rm\\,j\\!+\\!1 = k\\!+\\!1\\,$ $\\Rightarrow$ $\\rm\\,j = k\\,$ by Peano axioms. Thus if $\\rm\\:n\\!+\\!1\\,$ is both even and odd then so too is $\\rm\\,n.\\,$ So, contrapositively, in your induction step you can lift \"$\\rm n\\:$ is not both even and odd\" from $\\rm\\,n\\,$ up to $\\rm\\,n\\!+\\!1\\,$ and hence prove by induction that every natural is even or odd, but not both (this is essentially my hint in the linked answer).\n\n-\nThanks, this is really neat. I didn't understand your hint in the other question, but now I do. \u2013\u00a0russell11 Jul 5 '12 at 1:18", "date": "2016-05-27 20:26:13", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/166756/proving-that-an-integer-is-even-if-and-only-if-it-is-not-odd", "openwebmath_score": 0.9559515714645386, "openwebmath_perplexity": 101.67144073735894, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683517080176, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8615026654826816}}
{"url": "https://atekihcan.github.io/CLRS/02/E02.02-02/", "text": "Consider sorting $$n$$ numbers stored in array $$A$$ by first finding the smallest element of $$A$$ and exchanging it with the element in $$A[1]$$. Then find the second smallest element of $$A$$, and exchange it with $$A[2]$$. Continue in this manner for the first $$n-1$$ elements of $$A$$. Write pseudocode for this algorithm, which is known as selection sort. What loop invariant does this algorithm maintain? Why does it need to run for only the first $$n-1$$ elements, rather than for all $$n$$ elements? Give the best-case and worst-case running times of selection sort in $$\\Theta$$-notation.\n\n#### Pseudocode\n\n$$\\textsc {Selection-Sort }(A)$$\\begin{aligned}1& \\quad \\textbf {for }i=1\\textbf { to }A.length-1 \\\\2& \\quad \\qquad minIndex=i \\\\3& \\quad \\qquad \\textbf {for }j=i+1\\textbf { to }A.length \\\\4& \\quad \\qquad \\qquad \\textbf {if }A[j]<A[minIndex]\\text { and }j\\not=minIndex \\\\5& \\quad \\qquad \\qquad \\qquad minIndex=j \\\\6& \\quad \\qquad \\text {swap }A[i]\\text { with }A[minIndex] \\\\\\end{aligned}\n\nA python implementation of the above pseudocode is shared at end of the page, you can verify the workings for yourself.\n\n#### Loop Invariant\n\nAt the start of the each iteration of the outer for loop of lines 1-6, the subarray $$A[1..i - 1]$$ consists of $$i - 1$$ smallest elements of $$A$$, sorted in increasing order.\n\n#### Why only first n - 1 elements\n\nThe algorithm needs to run for only the first $$n - 1$$ elements, rather than for all $$n$$ elements because the last iteration will compare $$A[n]$$ with the minimum element in $$A[1 .. n - 1]$$ in line 4 and swap them if necessary. So, there is no need to continue the algorithm for all the way to the last element.\n\n#### Running Times\n\nFor both the best-case (sorted array) and worst-case (reverse sorted array), the algorithm will anyway take one element at a time and compare it with all the other elements. In other words, each of the $$n$$ elements will be compared with rest of the $$n - 1$$ elements. So, the running times for both scenario will be $$\\Theta(n^2)$$.\n\nThe above reasoning should be sufficient to understand or convey why the runtime would be $$\\Theta(n^2)$$. However, for the sake of completeness, an exhaustive mathematical proof is given below.\n\n#### Runtime Analysis\n\nLet\u2019s assume the inner for loop in line 3-5 is executed for $$t_j$$ times for $$j = 2, 3, \\ldots, n$$, where $$n = A.length$$. Now note that, line 5 will be executed less than $$t_j - 1$$ times in the average case, but it\u2019ll still be of the order of $$n$$.\n\nFor the sake of simplicity let\u2019s assume the worst case, i.e. a reverse sorted array, when it\u2019ll be executed exactly $$t_j - 1$$ times. Note, this assumption is only for that particular line, which is not going to change our overall analysis, it will only make our calculation easier.\n\nWe can now calculate the cost and times for individual lines of the pseudocode as follows \u2026\n\nLine Cost Times\n1 $$c_1$$ $$n$$\n2 $$c_2$$ $$n - 1$$\n3 $$c_3$$ $$\\sum_{j = 2}^n t_j$$\n4 $$c_4$$ $$\\sum_{j = 2}^n (t_j - 1)$$\n5 $$c_5$$ $$\\sum_{j = 2}^n (t_j - 1)$$\n6 $$c_6$$ $$n - 1$$\n\nNow, for any arbitrary value of $$j$$, the inner for loop (line 3-5) compares the previously computed minimum value with all elements in the subarray $$A[j..n]$$. So the inner for loop executes $$n - j + 1$$ times, i.e. $$t_j = (n - j + 1)$$ for $$j = 2, 3, \\ldots, n$$. So, $$t_2 = n - 1, t_3 = n - 2, \\ldots t_n = 1$$. We can calculate the summations for line 3-5 as follows \u2026\n\n\\begin{aligned} \\sum_{j = 2}^n t_j & = (n - 1) + (n - 2) + \\cdots + 1 \\\\ & = \\frac {n(n - 1)} 2 \\\\ \\\\ \\sum_{j = 2}^n (t_j - 1) & = \\sum_{j = 2}^n t_j - \\sum_{j = 2}^n 1 \\\\ & = \\frac {n(n - 1)} 2 - (n - 1) \\\\ & = \\frac {(n - 3)(n - 2)} 2 \\end{aligned}\n\nTherefore, we can calculate the running time as follows..\n\n$T(n) = c_1(n - 1) + (c_2 + c_6)n + c_3 \\frac {n(n - 1)} 2 + (c_4 + c_5) \\frac {(n - 1)(n - 2)} 2$\n\nFor best-case scenario, as the array is already sorted, line #5 will not be executed ever. So, $$c_5 = 0$$. Even with that (and without that for worst-case) the expression of $$T(n)$$ will be reduced to the form $$an^2 + bn + c$$, i.e. the algorithm will run at $$\\Theta(n^2)$$ time.\n\n#### Python Code\n\n# Selection Sort def SelectionSort(A): for i in range(len(A)): minIndex = i for j in range(i + 1, len(A)): if A[j] < A[minIndex] and j != minIndex: minIndex = j A[i], A[minIndex] = A[minIndex], A[i] # Test import random num_failed = 0 total_tests = 100 for i in range(total_tests): length = random.randint(2, 50) lst = [random.randint(0, 100) for _ in range(length)] SelectionSort(lst) # Check if list has been sorted for i in range(len(lst) - 1): if lst[i] > lst[i + 1]: num_failed += 1 print(f\"Test #{i:<2}: List is not sorted\") break if num_failed > 0: break if num_failed > 0: print(f\"\\nFailed\") else: print(f\"Passed {total_tests}/{total_tests} tests\")", "date": "2023-03-28 23:41:04", "meta": {"domain": "github.io", "url": "https://atekihcan.github.io/CLRS/02/E02.02-02/", "openwebmath_score": 0.9634503722190857, "openwebmath_perplexity": 699.9159241570951, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987568346219724, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8615026410302643}}
{"url": "http://mathhelpforum.com/calculus/35956-solved-vectors-question.html", "text": "# Math Help - [SOLVED] Vectors Question...\n\n1. ## [SOLVED] Vectors Question...\n\nQuestion:\nIn the diagram OABCDEFG is a cube which the lenght of each edge is $2$ units. Unit vectors i, j, k are parallel to $\\vec{OA}$, $\\vec{OC}$, $\\vec{OD}$ respectively. The mid-points of $AB$ and $FG$ are $M$ and $N$ respectively.\n\n(i) Express each of the vectors $\\vec{ON}$ and $\\vec{MG}$ in terms of i, j and k. [3 Marks]\n\n(ii) Find the angle between the directions of $\\vec{ON}$ and $\\vec{MG}$, are correct to the nearest $0.1^o$.\n[4 Marks]\n\nAttempt:\n\n(i) $\\vec{ON} = i + 2j + 2k$\n$\\vec{MG} = -2i + j + 2k$\n\nare they right? and will I be rewarded 3 marks for showing no steps? becuase I don't there are any steps needed!\n\n(ii) Need Help!\n\n2. Your answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.\n\n(ii)\n\nIn maths, there's a fun thing called the scalar product (the \"dot product\" they call it at school).\n\nIt's defined like this:\n\na DOT b = |a| * |b| * cos(angle)\n\nwhere a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.\n\nSo you can rearrange that to get\n\ncos(angle) = [ a DOT b ] / [ |a| * |b| ]\n\nSo now all you need to know is what a DOT b is:\n\nI'll show that with an example:\n\n(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32\n\nAll you do is multiply the two \"firsts\" (x-components) together, then the two seconds, then the thirds and add them all up.\n\nWork that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.\n\nEnjoy.\n\n3. Originally Posted by Fedex\nYour answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.\n\n(ii)\n\nIn maths, there's a fun thing called the scalar product (the \"dot product\" they call it at school).\n\nIt's defined like this:\n\na DOT b = |a| * |b| * cos(angle)\n\nwhere a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.\n\nSo you can rearrange that to get\n\ncos(angle) = [ a DOT b ] / [ |a| * |b| ]\n\nSo now all you need to know is what a DOT b is:\n\nI'll show that with an example:\n\n(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32\n\nAll you do is multiply the two \"firsts\" (x-components) together, then the two seconds, then the thirds and add them all up.\n\nWork that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.\n\nEnjoy.\nThanks for the explanation and the example\n\n$a.b = \\left|a\\right|\\times \\left|b\\right| \\times \\cos\\theta$\n\n$\\theta = \\cos^{-1} \\left( \\frac{a.b}{\\left|a\\right| \\times \\left|b\\right|} \\right)$\n\n$a.b = \\left(\\begin{array}{c}1\\\\2\\\\2\\end{array}\\right) \\times \\left(\\begin{array}{c}-2\\\\1\\\\2\\end{array}\\right) = (1 \\times -2) + (2 \\times 1) + (2 \\times 2) = 4$\n\n$\\left|a\\right|\\times \\left|b\\right| = \\left(\\begin{array}{c}1\\\\2\\\\2\\end{array}\\right) \\times \\left(\\begin{array}{c}2\\\\1\\\\2\\end{array}\\right) = (1 \\times 2) + (2 \\times 1) + (2 \\times 2) = 8$\n\n$\\theta = \\cos^{-1} = \\frac{4}{8}$\n\n$\\theta = 60^o$\n\nAre my steps correct? and is the answer correct?\n\n4. Okay, a few mistakes there.\n\nYou wrote your two column vectors like this:\n\nYou SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the \"cross\" or \"vector\" product of them - that's something else entirely!)\n\nSo write it with a dot, yeah?\n\nAlso, when you worked out the magnitude of the vectors, you seem to have gone wrong.\n\nThe MAGNITUDE of a vector (its \"length\") is just a number, not a vector.\n\nFor for the vector a = (1,2,3)\n\n|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]\n\nAre you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.\n\n5. Originally Posted by Fedex\nOkay, a few mistakes there.\n\nYou wrote your two column vectors like this:\n\nYou SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the \"cross\" or \"vector\" product of them - that's something else entirely!)\n\nSo write it with a dot, yeah?\n\nAlso, when you worked out the magnitude of the vectors, you seem to have gone wrong.\n\nThe MAGNITUDE of a vector (its \"length\") is just a number, not a vector.\n\nFor for the vector a = (1,2,3)\n\n|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]\n\nAre you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.\nThanks again\n\n$a.b = \\left|a\\right|.\\left|b\\right|.\\cos\\theta$\n\n$\\theta = \\cos^{-1} \\left( \\frac{a.b}{\\left|a\\right|.\\left|b\\right|} \\right)$\n\n$a.b = \\left(\\begin{array}{c}1\\\\2\\\\2\\end{array}\\right). \\left(\\begin{array}{c}-2\\\\1\\\\2\\end{array}\\right) = (1 \\times -2) + (2 \\times 1) + (2 \\times 2) = 4$\n\n$\\left|a\\right| = \\sqrt{(1^2 + 2^2 + 2^2} = 3$\n$\\left|b\\right| = \\sqrt{(-2^2 + 1^2 + 2^2)} = 1$\n\n$\\left|a\\right|.\\left|b\\right| = 3\\times1 = 3$\n\n$\\theta = \\cos^{-1} \\frac{4}{3}$ (Invalid)\n\nWhere did I go wrong?\n\n6. What's (-2)^2 again?\n\nCheck out that part of your working for the magnitude of b.\n\nI think that's your problem :-)\n\n7. Originally Posted by Fedex\nWhat's (-2)^2 again?\n\nCheck out that part of your working for the magnitude of b.\n\nI think that's your problem :-)\n$a.b = \\left|a\\right|.\\left|b\\right|.\\cos\\theta$\n\n$\\theta = \\cos^{-1} \\left( \\frac{a.b}{\\left|a\\right|.\\left|b\\right|} \\right)$\n\n$a.b = \\left(\\begin{array}{c}1\\\\2\\\\2\\end{array}\\right). \\left(\\begin{array}{c}-2\\\\1\\\\2\\end{array}\\right) = (1 \\times -2) + (2 \\times 1) + (2 \\times 2) = 4$\n\n$\\left|a\\right| = \\sqrt{(1^2 + 2^2 + 2^2} = 3$\n$\\left|b\\right| = \\sqrt{(-2^2 + 1^2 + 2^2)} = 3$\n\n$\\left|a\\right|.\\left|b\\right| = 3\\times3 = 9$\n\n$\\theta = \\cos^{-1} \\frac{4}{9} = 63.6^o$", "date": "2014-07-10 11:24:47", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/35956-solved-vectors-question.html", "openwebmath_score": 0.7882995009422302, "openwebmath_perplexity": 736.8438186149863, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241982893259, "lm_q2_score": 0.8887587875995482, "lm_q1q2_score": 0.8614953992625254}}
{"url": "https://tutorme.com/tutors/411755/interview/", "text": "Enable contrast version\n\n# Tutor profile: Yannic V.\n\nInactive\nYannic V.\nPh.D. in mathematics\nTutor Satisfaction Guarantee\n\n## Questions\n\n### Subject:Discrete Math\n\nTutorMe\nQuestion:\n\nFind a formula for the sum $\\sum_{k=0}^n k \\binom{n}{k},$ by means of the binomial theorem. What can be deduced about the average number of elements in a random subset of $\\{1,2, \\hdots, n\\}$?\n\nInactive\nYannic V.\n\nLet $x,y \\in \\mathbb{R}$. Remember that the binomial theorem sates that $(x+y)^n = \\sum_{k=0}^n \\binom{n}{k}x^{k}y^{n-k},$ for all $n \\in \\mathbb{N}$. In particular, for $y=1$ we have $(x+1)^n=\\sum_{k=0}^n \\binom{n}{k}x^k.$ Considering $x$ as a variable, we differentiate both sides of the above equation to obtain $n(x+1)^{n-1}= \\sum_{k=0}^n k\\binom{n}{k}x^{k-1}.$ It remains to plug in $x=1$: $\\sum_{k=0}^n k \\binom{n}{k} = n2^{n-1}.$ The average number of elements in a random subset of $\\{1,2, \\hdots, n\\}$ is thus $\\frac{\\sum_{k=0}^n k \\binom{n}{k}}{\\sum_{k=0}^n \\binom{n}{k}} = \\frac{n 2^{n-1}}{2^n}=\\frac{n}{2}.$ This is not surprising, since the number of subsets of size $k$ is exactly the same as the number of subsets of size $n-k$, for any $k$.\n\n### Subject:Calculus\n\nTutorMe\nQuestion:\n\nSolve the integral $\\displaystyle \\int 3 \\sec(2x-1)\\tan(2x-1) \\,dx$.\n\nInactive\nYannic V.\n\nAs the arguments of both functions in the integral are not single variables, we can start to consider a change of variable in order to just have one variable in the secant and tangent function. This can be made using the change of variable $$\\label{cambio1} \\left\\lbrace\\begin{array}{l} y = 2x-1 \\\\ dy = 2 \\,dx \\end{array}\\right.$$ Notice that the derivative $dy$ is easy to obtain from the expression of the integral: $dx$ can be expressed in terms of $dy$ as $$\\label{eq1} dx = \\frac{dy}{2}.$$ Substituting the change of variable \\eqref{cambio1} in the integral, we obtain \\begin{align} \\int 3 \\sec(2x-1)\\tan(2x-1)\\, dx &= \\int 3 \\sec(y)\\tan(y) \\,\\frac{dy}{2} \\tag{\\footnotesize by \\eqref{cambio1} and \\eqref{eq1}}\\\\ &= \\frac{3}{2} \\int \\sec(y)\\tan(y) \\,dy. \\label{da1} \\end{align} We have now reduced our problem to a more simple integral. We want to integrate the expression $\\sec(y)\\tan(y)$, which is the derivative of\\footnote{This comes from the following calculation: $(\\sec(y))'= \\left(\\frac{1}{\\cos(y)} \\right)' = \\frac{(0\\cos(y))-1(-\\sin(y))}{\\cos(y)^2}= \\frac{\\sin(y)}{\\cos(y)^2} = \\frac{1}{\\cos(y)}\\frac{\\sin(y)}{\\cos(y)}=\\sec(y)\\tan(y).$ } the function $\\sec(y)$. It is then appropriate to consider the following change of variable: $$\\label{cambio2} \\left\\lbrace\\begin{array}{l} u = \\sec(y)\\\\ du = \\sec(y)\\tan(y) \\,dy \\end{array}\\right.$$ We can now replace \\eqref{cambio2} into the expression \\eqref{da1}. Remark that \\eqref{da1} just contain the derivative part of\\footnote{This is not a problem. When doing a change of variable, the most important part is that the expression corresponding to the derivative of the change of basis appears in the integral (up to a scalar). } \\eqref{cambio2}. Thus, we have: \\begin{align} \\int 3 \\sec(2x-1)\\tan(2x-1)\\, dx &= \\frac{3}{2} \\int \\sec(y)\\tan(y) \\,dy. \\tag{\\footnotesize{by \\eqref{da1}}}\\\\ &= \\frac{3}{2} \\int du \\tag{\\footnotesize{by \\eqref{cambio2}}}\\\\ &= \\frac{3}{2}\\, u + C, \\label{da2} \\end{align} where $C$ is a constant. Finally, we just have to substitute $u$ by its value in terms of $y$ (given by \\eqref{cambio2}), and replace $y$ by its value in terms of $x$ (given by \\eqref{cambio1}): \\begin{align} \\int 3 \\sec(2x-1)\\tan(2x-1)\\, dx &= \\frac{3}{2}\\, u + C \\tag{\\footnotesize{by \\eqref{da2}}}\\\\ &= \\frac{3}{2}\\sec(y) + C \\tag{\\footnotesize{substitute $u=\\sec(y)$}}\\\\ &= \\frac{3}{2}\\sec(2x-1) + C. \\tag{\\footnotesize{substitute $y=2x-1$}} \\end{align}\n\n### Subject:Algebra\n\nTutorMe\nQuestion:\n\nIf $G$ is a finite group, show that the number of elements in $G$ of order greater than $2$ must be even. In particular, prove that any group of even order must contain an element of order 2.\n\nInactive\nYannic V.\n\nWe will solve the question by a counting argument as follows. Let $X$ be the set of all elements of G with order greater than two. If $X = \\emptyset$ then $|X|=0$ and we\u2019re done. Suppose now that $X \\neq \\emptyset$. Since the orders of $g$ and $g^{-1}$ coincides for any $g \\in G$, we have that $g \\in X \\Longleftrightarrow g^{-1} \\in X.$ Moreover, if $g \\in X$ then $g^2\\neq e_G$. From here, $g \\neq g^{-1}$. It follows that $X$ can be written as the disjoint union of two element sets of the form $\\{g, g^{-1}\\}$ {x,x\u22121}, and hence that $|X|$ is even. Suppose that $|G|$ is even. Since $e_G$ has order $1$, $e_G \\notin X$. It follows that $G \\setminus X \\neq \\emptyset$. So $0 < |G \\setminus X|=|G|-|X|$. Since $|G|$ and $|X|$ are both even, it follows that $|G\\setminus X|$ is a nonzero even integer, i.e. is at least 2. Thus, there is an $z \\in G \\setminus X$ such that $z \\neq e_G$ . Since $X$ consists of all elements in $G$ of order greater than $2$, it must be the case that $|z|=2$.\n\n## Contact tutor\n\nSend a message explaining your\nneeds and Yannic will reply soon.\nContact\u00a0Yannic\n\nStart Lesson\n\n## FAQs\n\nWhat is a lesson?\nA lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.\nHow do I begin a lesson?\nIf the tutor is currently online, you can click the \"Start Lesson\" button above. If they are offline, you can always send them a message to schedule a lesson.\nWho are TutorMe tutors?\nMany of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.\nBEST IN CLASS SINCE 2015\nTutorMe homepage", "date": "2021-07-23 14:52:39", "meta": {"domain": "tutorme.com", "url": "https://tutorme.com/tutors/411755/interview/", "openwebmath_score": 0.9967986345291138, "openwebmath_perplexity": 310.21218791461285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109109053045, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.8614654844665749}}
{"url": "https://math.stackexchange.com/questions/1809993/how-to-integrate-int-fracdx1x22", "text": "# How to integrate $\\int \\frac{dx}{(1+x^2)^2}$?\n\nI need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end:\n\n$$\\int_0^\\infty \\frac{dx}{(1+x^2)^2}$$\n\nThe result should be $\\pi\\over 4$ . I don't know how to approach this. *Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).\n\nAlso, Wolfram tells me:\n\n$$\\int \\frac{dx}{(1+x^2)^2} = \\frac{1}{2}\\left(\\frac{x}{x^2+1}+\\tan^{-1}x\\right)+c$$\n\nbut I don't see how one can derive this without knowing the result beforehand.\n\nSomehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).\n\nEDIT: If you're interested, the problem in question is: STEP II - problem 4 (year 2014).\n\n\u2022 use $\\frac 1{(1+x^2)^2 }=\\frac 1{1+x^2} - \\frac{x^2}{(1+x^2)^2}$ and do an integration by parts $\\int \\frac{x^2}{(1+x^2)^2} \\, dx$ \u2013\u00a0abel Jun 2 '16 at 18:04\n\u2022 I tried $x = \\sinh(z)$ since $1+x^2 = \\cosh^2(x)$ but I got stuck at $$\\int \\frac{1}{\\cosh^3(z)} \\, {\\rm d}z$$ \u2013\u00a0ja72 Jun 2 '16 at 18:12\n\nUse $x = \\tan t$. Then $(1 + x^2)^2 = \\sec^4 t$, and $\\dfrac{dx}{dt} = \\sec^2 t$, so the integral becomes \\begin{align*} \\int_0^{\\pi/2} \\cos^2 t \\,\\mathrm dt = \\dfrac 1 2 \\times \\dfrac {\\pi} 2 = \\dfrac{\\pi}{4}. \\end{align*}\n\n\u2022 Nice. Thanks. It's been a while and I forgot about trig substitions :( \u2013\u00a0I want to make games Jun 2 '16 at 18:08\n\nFor any $\\alpha>0$, let: $$I(\\alpha)= \\int_{0}^{+\\infty}\\frac{dx}{\\alpha^2+x^2} = \\frac{\\pi}{2\\alpha}.$$ By differentiating both sides with respect to $\\alpha$ we get: $$\\int_{0}^{+\\infty}\\frac{2\\alpha}{(\\alpha^2+x^2)^2}\\,dx = \\frac{\\pi}{2\\alpha^2}$$ and by evaluating at $\\alpha=1$: $$\\int_{0}^{+\\infty}\\frac{dx}{(x^2+1)^2} = \\color{red}{\\frac{\\pi}{4}}.$$\n\n\u2022 How do you get these expressions? \u2013\u00a0ja72 Jun 2 '16 at 18:07\n\u2022 @ja72: What's obscure? I think my answer is pretty straightforward to follow. The technique is known as differentiation under the integral sign, or \"Feynman's trick\". \u2013\u00a0Jack D'Aurizio Jun 2 '16 at 18:08\n\u2022 I'm sure that's common knowledge, but how (a very general idea would do) do you prove that differentiating an integral is the same as differentiating the thing that's integrated? \u2013\u00a0I want to make games Jun 2 '16 at 18:09\n\u2022 @M.Vinay: thanks, I was just writing the same thing :) \u2013\u00a0Jack D'Aurizio Jun 2 '16 at 18:12\n\u2022 @ja72: I assure you that my memorization skills are close to being awful (that is the main reason beyond my appraise for modern technology, that allow us to store tons of useful data in portable devices), but when it comes to mathematical practice, many things become natural. Practice makes perfect. \u2013\u00a0Jack D'Aurizio Jun 2 '16 at 18:20\n\nUse $u=tan x$ $$(1+x^2)^2=(1+tan^2u)^2=(sec^2u)^2=sec^4u$$ $$dx=sec^2u du$$ The integral becomes $$\\int{\\frac{1}{sec^2u}du=\\int{cos^2udu}}$$\n\n\u2022 functions need to be upright characters. Use \\sin instead of sin. Also the differential needs to be upright. Use {\\rm d} instead of d. \u2013\u00a0ja72 Jun 2 '16 at 18:14\n\nAdd and subtract $x^2$ from the numerator, you get $\\arctan$ on one hand and an $\\int \\frac{x^2}{(1 + x^2)^2}dx$, which you can do by parts ($u = x$ and $dv = \\frac{x}{(1+x^2)^2}dx$)\n\n$$\\int_0^\\infty\\frac{1}{\\left(1+x^2\\right)^2}\\space\\text{d}x=\\lim_{n\\to\\infty}\\int_0^n\\frac{1}{\\left(1+x^2\\right)^2}\\space\\text{d}x=$$\n\nSubstitute $x=\\tan(u)$ and $\\text{d}x=\\sec^2(u)\\space\\text{d}u$.\n\nSo $\\left(1+x^2\\right)^2=\\left(1+\\tan^2(u)\\right)^2=\\sec^4(u)$ and $u=\\arctan(x)$.\n\nThis gives a new lower bound $u=\\arctan(0)=0$ and upper bound $u=\\arctan(n)$:\n\n$$\\lim_{n\\to\\infty}\\int_0^{\\arctan(n)}\\cos^2(u)\\space\\text{d}u=$$\n\nUse:\n\n$$\\cos^2(x)=\\frac{1+\\cos(2u)}{2}$$\n\n$$\\frac{1}{2}\\lim_{n\\to\\infty}\\left[\\int_0^{\\arctan(n)}1\\space\\text{d}u+\\int_0^{\\arctan(n)}\\cos(2u)\\space\\text{d}u\\right]=$$\n\nSubstitute $s=2u$ and $\\text{d}s=2\\space\\text{d}u$.\n\nThis gives a new lower bound $s=2\\cdot0=0$ and upper bound $s=2\\arctan(n)$:\n\n$$\\frac{1}{2}\\lim_{n\\to\\infty}\\left[\\left[u\\right]_0^{\\arctan(n)}+\\frac{1}{2}\\int_0^{2\\arctan(n)}\\cos(s)\\space\\text{d}s\\right]=$$ $$\\frac{1}{2}\\lim_{n\\to\\infty}\\left[\\left[u\\right]_0^{\\arctan(n)}+\\frac{1}{2}\\left[\\sin(s)\\right]_0^{2\\arctan(n)}\\right]=$$ $$\\frac{1}{2}\\lim_{n\\to\\infty}\\left[\\left(\\arctan(n)-0\\right)+\\frac{\\sin(2\\arctan(n))-\\sin(0)}{2}\\right]=$$ $$\\frac{1}{2}\\lim_{n\\to\\infty}\\left[\\arctan(n)+\\frac{\\sin(2\\arctan(n))}{2}\\right]=\\frac{1}{2}\\cdot\\frac{\\pi}{2}=\\frac{\\pi}{4}$$\n\nlet $I_n=\\int_{0}^{\\infty}\\frac{1}{(1+x^2)^n}dx$ then $$I_{n+1}=\\frac{2n-1}{2n}I_n$$ we know $I_1=\\int_{0}^{\\infty}\\frac{1}{1+x^2}dx=\\frac{\\pi}{2}$, so $$I_2=\\frac{2(1)-1}{2(1)}I_1=\\frac{\\pi}{4}$$\n\n\u2022 (+1) I see in you a future star of MSE. \u2013\u00a0Jack D'Aurizio Jun 2 '16 at 20:54\n\nHere's a pretty solution. Notice that $$I=\\int_{0}^{\\infty}\\frac{1}{(1+x^2)^2}dx=\\frac{1}{2}\\int_{-\\infty}^{\\infty}\\frac{1}{(1+x^2)^2}dx=\\frac{1}{2}\\int_{\\partial\\Omega}\\frac{1}{(1-iz)^2(1+iz)^2}dz$$ where $\\Omega$ is the top part of the complex plane. Since $f(z)$ decays faster than $\\frac{1}{z^2}$, $$I=\\pi i \\lim_{z \\to i}\\frac{d}{dz}\\frac{(z-i)^2}{(1-iz)^2(1+iz)^2}=\\frac{\\pi}{4}$$.", "date": "2019-06-17 05:11:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1809993/how-to-integrate-int-fracdx1x22", "openwebmath_score": 0.9595130681991577, "openwebmath_perplexity": 581.3614977347157, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517450056273, "lm_q2_score": 0.8807970904940926, "lm_q1q2_score": 0.8614651313536267}}
{"url": "https://math.stackexchange.com/questions/2682525/estimating-error-when-calculating-pi2-with-8-frac832-frac852", "text": "# Estimating error when calculating $\\pi^2$ with $8 + \\frac{8}{3^2} + \\frac{8}{5^2} + \\frac{8}{7^2} + \\cdots$\n\nWhile answering a CodeReview question (Approximating constant $\\pi^2$ to within error), I noticed that when calculating the sum $$8 + \\dfrac{8}{3^2} + \\dfrac{8}{5^2} + \\dfrac{8}{7^2} + \\dfrac{8}{9^2} + \\cdots$$\n\nand stopping at the smallest term larger than $\\varepsilon$, the difference to $\\pi^2$ seems very close to $\\sqrt{2\\varepsilon}$.\n\nWith $$n = \\left\\lfloor\\frac{\\sqrt{\\frac{8}{\\varepsilon}} - 1}{2} \\right\\rfloor,$$\n\nit looks like:\n\n$$\\sum_{i=0}^n \\frac{8}{(2i+1)^2} \\approx \\pi^2 - \\sqrt{2\\varepsilon}.$$\n\nFor example, with $\\varepsilon = 10^{-10}$:\n\n$n = 141421$\n\n$\\sum = 9.86959025 \\dots$\n\n$\\pi^2 - \\sum \\approx 1.4142170*10^{-5}$\n\n$(\\pi^2 - \\sum)^2 \\approx 2.0000099*10^{-10} \\approx 2\\varepsilon$\n\nI don't think it's a coincidence, but I don't know where to begin to link\n\n$$\\sum_{i=n+1}^\\infty \\frac{8}{(2i+1)^2}$$ to $\\sqrt{2\\varepsilon}$.\n\nWolfram Alpha expresses this sum in terms of a polygamma function but I don't know anything about it.\n\nIs the approximation correct? Is there any simple way to prove it?\n\n\u2022 The usual method of estimating the error in such series is the Euler-Maclaurin formula. \u2013\u00a0Lord Shark the Unknown Mar 8 '18 at 15:58\n\u2022 I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Mar 15 '18 at 16:35\n\nOne has $${2\\over (2k+1)^2}<{1\\over 2k}-{1\\over 2k+2}\\ ,$$ and therefore (teleskoping sum!) $$0<\\sum_{k=n}^\\infty{2\\over(2k+1)^2}<{1\\over 2n}\\ ,$$ or $$0<\\sum_{k=n}^\\infty{8\\over(2k+1)^2}<{2\\over n}\\ .$$ Maybe this will bring you over the top.\n\n\u2022 Thanks a lot. If I understand it correctly, we still need another inequality to prove that the sum isn't too far from $\\frac{2}{n}$, right? \u2013\u00a0Eric Duminil Mar 8 '18 at 21:04\n\u2022 Applying the same logic with ${1\\over 2k+1}-{1\\over 2k+3} < {2\\over (2k+1)^2}$, we obtain $\\frac{2}{n+1} < \\sum < \\frac{2}{n}$, right? That would be a pretty good approximation. \u2013\u00a0Eric Duminil Mar 9 '18 at 8:12\n\nRecall from calculus the integral test:\n\n$$\\displaystyle\\sum a(n)\\text{ converges if and only if }\\int a(x)\\text{ converges}$$ where the sequence $a(n)=a_n>0$ and $a_n$ is decreasing, and the function $a(x)$ is continuous.\n\nFrom the proof of this statement, we can deduce the inequality: if $\\sum_{n=1}^\\infty a_n=L$, then $$\\left|L-\\sum_{i=0}^n a_n\\right|\\leq\\int_{n}^\\infty a(x)\\,dx,$$(see here for help with providing visual intuition as to why the inequality is true) where the notation $a(x)$ is used to represent $a_n$ with the $n$ replaced by $x$. Thus, performing the necessary calculations, we find\n\n\\begin{align*} \\int_{n}^\\infty\\frac{8}{(2x+1)^2}\\,dx&=\\frac{4}{2n+1}\\\\ &\\leq\\frac{4}{\\sqrt{\\frac{8}\\varepsilon}-1}\\\\ &=\\frac{1}{\\sqrt{\\frac{1}{2\\varepsilon}}-\\frac14}\\\\ &=\\frac{\\sqrt{2\\varepsilon}}{1-\\frac{\\sqrt{2\\varepsilon}}{4}}\\\\ &\\approx \\sqrt{2\\varepsilon},\\end{align*} where I've taken $$n=\\left\\lfloor\\frac{\\sqrt{\\frac{8}{\\varepsilon}}-1}{2}\\right\\rfloor.$$\n\n\u2022 This looks good, thanks. I'll try to do it again on a piece of paper and come back to comment/accept. \u2013\u00a0Eric Duminil Mar 8 '18 at 17:07\n\u2022 Thanks, @EricDuminil. There was one place I had something like $\\frac{1}{\\sqrt{\\frac{1}{2\\varepsilon}}+1}$ where I had to use some algebra to simplify (hence I didn't want to type it, at least, not yet haha). \u2013\u00a0Clayton Mar 8 '18 at 17:15\n\u2022 First, the integral test has hypotheses that you need to mention. And, although the estimate you give for the error is correct (under the same hypotheses!), I don't see how we can deduce that error estimate from the integral test itself. How does that go? \u2013\u00a0David C. Ullrich Mar 8 '18 at 17:20\n\u2022 @DavidC.Ullrich: I suppose it is more from the proof of the integral test than the integral test itself, but the idea is that the sequence/function is positive and decreasing, so we can set it up so the integral is an overestimate (if you prefer to think in terms of Riemann sums, it is like using right endpoints for the estimate). \u2013\u00a0Clayton Mar 8 '18 at 17:33\n\u2022 @DavidC.Ullrich: You're right about the test having hypotheses that I ought to mention. I knew I wouldn't have much time when I submitted the answer. At any rate, I'm at fault for not being explicit enough. I have about an hour now, so I'll provide more explanations. If you have any other suggestions, please let me know. \u2013\u00a0Clayton Mar 8 '18 at 17:36", "date": "2019-08-22 10:13:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2682525/estimating-error-when-calculating-pi2-with-8-frac832-frac852", "openwebmath_score": 0.8721043467521667, "openwebmath_perplexity": 329.03580589035056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357248544006, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.8614503210108743}}
{"url": "http://mathhelpforum.com/algebra/206877-polynomial-uniqueness-proof-print.html", "text": "# Polynomial uniqueness proof\n\n\u2022 Nov 6th 2012, 10:30 AM\nPolynomial uniqueness proof\nThis problem is from my Gelfand's Algebra book.\n\nProblem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.\n\nThis means that if $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2 and $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ for three different numbers $x_1,x_2,\\text{ and } x_3,$ then the polynomials $P(x)$ and $Q(x)$ are equal.\n\nI'm not very good at proofs, so I have questions. If I show that if $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ is true and $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2, then $P(x)$ and $Q(x)$ are equal, will it prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values?\n\nIs this the way to go?\n\nLet $V(x)=P(x)-Q(x)$, After that $V(x_1)=V(x_2)=V(x_3)=0$, because $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$, so $x_1,x_2,x_3$ are roots of $V(x)$ which can't have more than two roots. Here I'm confused, I shouldn't have assumed that $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ was true, right?\n\u2022 Nov 6th 2012, 11:11 AM\nDeveno\nRe: Polynomial uniqueness proof\nactually, your proof is fine. a polynomial of degree n, cannot have more than n roots.\n\nso you have proved that V(x) is NOT of degree 0,1, or 2. what's left?\n\u2022 Nov 6th 2012, 11:17 AM\nemakarov\nRe: Polynomial uniqueness proof\nI don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.\n\u2022 Nov 6th 2012, 11:18 AM\nRe: Polynomial uniqueness proof\nQuote:\n\nOriginally Posted by Deveno\nactually, your proof is fine. a polynomial of degree n, cannot have more than n roots.\n\nso you have proved that V(x) is NOT of degree 0,1, or 2. what's left?\n\nYes, but does that prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values?\n\nQuote:\n\nOriginally Posted by emakarov\nI don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.\n\nYes, I think that's right, I'm assuming too much and I don't really understand how I can prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values.\n\u2022 Nov 6th 2012, 11:40 AM\nDeveno\nRe: Polynomial uniqueness proof\nQuote:\n\nOriginally Posted by emakarov\nI don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.\n\nwell this is only true in a field, of course, but presumably we are talking about polynomials in R[x].\n\nand one can argue by degree:\n\nproof(by induction): if p(x) in R[x] is of degree n > 0 then p has at most n real roots.\n\n(the reason for requiring n > 0 will be explained later).\n\nbase case: n = 1.\n\nin this case p(x) = ax + b, which has the sole root x = -b/a (we may assume a \u2260 0, or else p is not of degree 1). since exactly one root is a stronger condition that at most one root, the theorem holds in this case.\n\n(strong) induction hypothesis: suppose whenever 0 < k < n, deg(p) = k implies p has at most k roots.\n\nlet deg(p) = n.\n\nit could be that p has no roots at all. since 0 < n, the theorem holds in this case.\n\notherwise, let a be a root of p.\n\nwriting p(x) = q(x)(x - a) + r(x), where either r(x) is identically 0, or deg(r) < deg(x-a) = 1, we see that r is a constant polynomial, or the 0-polynomial.\n\nso p(x) = q(x)(x - a) + r, for some real number r. thus:\n\n0 = p(a) = q(a)(a - a) + r = q(a)0 + r = r.\n\nhence p(x) = q(x)(x - a).\n\nsince deg(p) = deg(q) + deg(x-a), we have:\n\nn = deg(q) + 1, that is:\n\ndeg(q) = n-1 < n, so q has at most n-1 roots.\n\nlemma: if p(x) = f(x)g(x) for real polynomials p,f, and g, then if a is a root of p, either a is a root of f, or a is a root of g.\n\nproof: 0 = p(a) = f(a)g(a), so either f(a) = 0, or g(a) = 0.\n\nmain proof continued:\n\nlet b be any root of p(x) = q(x)(x - a).\n\nthen either b is a root of q(x) (of which we have at most n - 1), or b is a root of x - a (which has the single root a).\n\nthus we have at most n - 1 + 1 = n roots of p, QED.\n\n*********\n\nyou have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.\n\nthe only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?\n\n**********\n\nOOPS! i almost forgot my explanation for why we take n > 0:\n\na polynomial of degree 0 is a constant polynomial:\n\np(x) = c.\n\nunless c = 0, p has no roots at all (which is ok according to our theorem).\n\nbut if p(x) = 0, for all x, we have infinitely many roots (very bad).\n\non the other hand:\n\np(x) = 0 = 0 + 0x = 0 + 0x + 0x2, etc.\n\nis sort of a SPECIAL polynomial, it really doesn't \"have\" a degree. it is this special case we seek to avoid by stating deg(p) > 0. depending on which book you read you get:\n\ndeg(0) = undefined, or sometimes:\n\ndeg(0) = -\u221e (this is to make the formula deg(fg) = deg(f) + deg(g) always work out).\n\u2022 Nov 6th 2012, 12:10 PM\nRe: Polynomial uniqueness proof\nQuote:\n\nOriginally Posted by Deveno\n\nyou have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.\n\nthe only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?\n\nI will try to prove it now, thank you for help!\n\u2022 Nov 6th 2012, 12:25 PM\nHallsofIvy\nRe: Polynomial uniqueness proof\nBecause you are specifically asked about \"polynomials of degree not greater than 2\", I don't think you need to be very \"sophisticated\"! You can use the fact that any such polynomial can be written in the form $f(x)= a_1x^2+ b_1x+ c_1$. So suppose $f(x_1)= y_1$, $f(x_2)= y_2$, and $f(x_3)= y_3$ and that $g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $a_1= a_2$, $b_1= b_2$, and $c_1= c_2$.\n\u2022 Nov 6th 2012, 01:22 PM\nDeveno\nRe: Polynomial uniqueness proof\nQuote:\n\nOriginally Posted by HallsofIvy\nBecause you are specifically asked about \"polynomials of degree not greater than 2\", I don't think you need to be very \"sophisticated\"! You can use the fact that any such polynomial can be written in the form $f(x)= a_1x^2+ b_1x+ c_1$. So suppose $f(x_1)= y_1$, $f(x_2)= y_2$, and $f(x_3)= y_3$ and that $g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $a_1= a_2$, $b_1= b_2$, and $c_1= c_2$.\n\nthe nice thing about this suggestion is that it also suggests a way to prove there is a polynomial which goes through our 3 given points.\n\n(think: system of linear equations)\n\u2022 Nov 8th 2012, 08:21 AM", "date": "2017-07-27 17:10:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/206877-polynomial-uniqueness-proof-print.html", "openwebmath_score": 0.8998113870620728, "openwebmath_perplexity": 365.06600855591074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357189762611, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.8614503142801606}}
{"url": "http://math.stackexchange.com/questions/49349/does-the-uniform-continuity-of-f-x-rightarrow-mathbbr-imply-f-a-righta", "text": "# Does the uniform continuity of $f: X \\rightarrow \\mathbb{R}$ imply $f: A \\rightarrow \\mathbb{R}$ is also uniformly continuous, when $A \\subset X$?\n\nI've been preparing for the prelim in August, and was working on a problem involving uniform continuity and restriction of functions. I absentmindedly assumed the above by considering the contrapositive: if $f: A \\rightarrow \\mathbb{R}$ isn't uniformly continuous, that implies $\\exists \\ \\epsilon$ such that no $\\delta$ satisfies $d(x,y) < \\delta \\implies d(f(x),f(y)) < \\epsilon, \\,\\,\\ \\forall x,y \\in A$, and this failure of $\\epsilon$'s existence shouldn't change when I \"add more points\" by considering $f: X \\rightarrow \\mathbb{R}.$\n\nHowever, if this is true, we obtained a lot of results I consider to be strangely powerful. For example, if a function is continuous on $\\mathbb{R}$, it is uniformly continuous on any bounded interval I, as it's uniformly continuous on $\\overline{I}$ which is compact by Heine-Borel. Hence, if $f$ is a real-valued function continuous on a subset $A$ of $R$, it's uniformly continuous on any bounded subset $X$ of $A$.\n\nConclusions such as this seem too strong! Is there a flaw in my reasoning, and if so, where is it?\n\n-\nNo, you're perfectly right. If $f: X \\to \\mathbb{R}$ is uniformly continuous then the restriction of $f$ to any subset $A \\subset X$ is also uniformly continuous. This can be seen by writing the condition of uniform continuity on $X$ as $\\forall\\, \\varepsilon \\gt 0\\;\\exists\\;\\delta \\gt 0: \\forall\\, x,y \\in X : \\ldots$. If you replace $X$ by $A$ and restrict $f$ to $A$ then the last condition becomes weaker as it then reads $\\forall\\,x,y \\in A$. The \"too strong\" conclusions are correct. \u2013\u00a0 t.b. Jul 4 '11 at 6:45\nThank you! I went back and thought about why I considered the conclusions \"too strong\", and I realized I was subconsciously only thinking about Lipschitz continuous functions. (Also, the statements gave off a sense of \"if the conditions were this nice, they'd mention it in class\" to me, which made me suspicious.) \u2013\u00a0 JakeR Jul 4 '11 at 6:55\nI hope you're aware that Lipschitz is much stronger than uniform continuity. One standard example for showing this is $x \\mapsto \\sqrt{x}$ on $[0,\\infty)$ which is uniformly continuous but not Lipschitz (because the slopes become arbitrarily steep near the origin). \u2013\u00a0 t.b. Jul 4 '11 at 6:57\nSadly, I can't claim that I wasn't conscious of that. I'm still not sure why I was implicitly assuming my functions were Lipschitz. \u2013\u00a0 JakeR Jul 4 '11 at 7:13\nThere is a rather difficult theorem due to Rademacher indicating a further strong distinction between Lipschitz and uniform continuity: It asserts that a Lipschitz continuous function on $\\mathbb{R}^{n}$ is differentiable \"almost everywhere\" (a precise technical term that should make sense intuitively), while you probably came across examples of continuous functions on $$0,1$$ that aren't differentiable anywhere. \u2013\u00a0 t.b. Jul 4 '11 at 7:22\n\nIt is true, and your conclusion that every continuous $f:\\mathbb{R}\\to\\mathbb{R}$ is uniformly continuous on bounded subsets of $\\mathbb{R}$ is also correct.\nOne could go further in saying precisely why it is true, which might help to convince you. Suppose $f:X\\to \\mathbb{R}$ is uniformly continuous and $A\\subset X$. Given $\\varepsilon>0$, by uniform continuity of $f$ there exists $\\delta>0$ such that for all $x,y\\in X$, $d(x,y)<\\delta$ implies $d(f(x),f(y))<\\varepsilon$. Now this same $\\delta$ works for the restriction $f\\vert_A$, because if you have $x,y\\in A$ with $d(x,y)<\\delta$, then $x$ and $y$ are also in $X$, so $d(f(x),f(y))<\\varepsilon$.\n@JakeR: Yes that's it exactly. The precise statement is: Let $A \\subset X$ be a subset of a metric space and let $Y$ be a complete metric space. If $f: A \\to Y$ is uniformly continuous then there exists a unique extension of $f$ to a continuous function on $\\overline{A}$. If $\\overline{A}$ is compact then the condition is also necessary by your observations in your question. The proof proceeds exactly in the way you describe but of course you need to check continuity of the extension. \u2013\u00a0 t.b. Jul 4 '11 at 7:09\nSorry, I had deleted the comment accidentally. We were talking about the condition above for continuous extensions, and I attempted a sketch of a proof: Let $x \\in \\overline{A}$ but not $A$. Consider sequences in $A$ converging to $x$: these sequences are Cauchy, and since $f$ is uniformly continuous, evaluating each of these sequences termwise by $f$ are also Cauchy. They converge uniquely (interweave two, and we have convergent subsequences in a Hausdorff space) to a point $y$. Associating $f(x) = y$, and doing this for all limit points, gives a continuous extension to $\\overline{A}$. \u2013\u00a0 JakeR Jul 4 '11 at 7:18\nIn regards to your \"too strong\" conclusions there's a slightly more general statement one can make. Let $f: (X,d_1) \\rightarrow (Y,d_2)$ be a continuous function and $A \\subset X$ compact. Consider $g:=f|_A$ then $g$ is continuous (with A given the metric induced from $X$). Then we have that $g$ is a continuous function whose domain is compact, so $g$ is uniformly continuous. That is $f$ is uniformly continuous on $A$.", "date": "2015-06-30 20:24:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/49349/does-the-uniform-continuity-of-f-x-rightarrow-mathbbr-imply-f-a-righta", "openwebmath_score": 0.9574061036109924, "openwebmath_perplexity": 178.00385653464994, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357259231532, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.861450312511984}}
{"url": "https://math.stackexchange.com/questions/1674559/the-value-of-mathop-sum-sum-0-leq-ij-leq-n-binomni-cdot-binomnj", "text": "# >The value of $\\mathop{\\sum\\sum}_{0\\leq i<j\\leq n}\\binom{n}{i}\\cdot \\binom{n}{j}$\n\nFind the value of $$\\mathop{\\sum\\sum}_{0\\leq i<j\\leq n}\\binom{n}{i}\\cdot \\binom{n}{j}$$\n\nI get the result: $$\\frac{1}{2}\\left(2^{2n}-\\binom{2n}{n}\\right)$$ via a numeric argument.\n\nMy question is: Can we solve it using a combinational argument?\n\nMy Numeric Argument: $$\\left(\\sum^{n}_{r=0}\\binom{n}{i}\\right)^2=\\sum^{n}_{r=0}\\binom{n}{i}^2+2\\mathop{\\sum\\sum}_{0\\leq i<j\\leq n}\\binom{n}{i}\\cdot \\binom{n}{j}$$\n\nSo here $$\\displaystyle \\sum^{n}_{r=0}\\binom{n}{i} = \\binom{n}{0}+\\binom{n}{1}+.....+\\binom{n}{n} = 2^n$$\n\nand $$\\displaystyle \\sum^{n}_{r=0}\\binom{n}{i}^2=\\binom{n}{0}^2+\\binom{n}{1}^2+.....+\\binom{n}{n}^2 = \\binom{2n}{n}$$\n\nabove we have calculate Using $$(1+x)^n = \\binom{n}{0}+\\binom{n}{1}x+\\binom{n}{2}x^2+.....+\\binom{n}{n}x^n$$\n\nand $$(x+1)^n = \\binom{n}{0}x^n+\\binom{n}{1}x^{n-1}+\\binom{n}{2}x^{n-2}+.....+\\binom{n}{n}x^0$$\n\nNow calcualting Coefficient of $x^n$ in $$(1+x)^n\\cdot (x+1)^n = (1+x)^{2n} = \\binom{2n}{n}$$\n\nSo we get $$\\mathop{\\sum\\sum}_{0\\leq i<j\\leq n}\\binom{n}{i}\\cdot \\binom{n}{j} = \\frac{1}{2}\\left[2^{2n} - \\binom{2n}{n}\\right]$$\n\nThanks\n\n\u2022 Really sloppy to write $(1+x)^{2n}=\\binom{2n}{n}$. \u2013\u00a0Thomas Andrews Feb 27 '16 at 17:35\n\u2022 I don't agree on how you obtain $\\binom{2n}{n}$. What did you do with the mixed terms? EDIT: Now, I see what you did. You only look at the coefficients for $x^n$. Nice. \u2013\u00a0Friedrich Philipp Feb 27 '16 at 17:36\n\u2022 Mixed terms? @FriedrichPhilipp \u2013\u00a0Thomas Andrews Feb 27 '16 at 17:36\n\u2022 Why don't you ask up front for a combinatorial proof, if that is your main question, rather than forcing answerers to read a non-combinatorial proof? \u2013\u00a0Thomas Andrews Feb 27 '16 at 17:39\n\u2022 @Thomas Andrews: Yes. In $(1+x)^n(1+x)^n$. But I edited my comment. \u2013\u00a0Friedrich Philipp Feb 27 '16 at 17:39\n\nConsider two sets, $A$ and $B$ each with $n$ elements. All elements are considered distinct.\n\n$\\displaystyle \\sum_{0 \\leq i < j \\leq n} \\binom{n}{i} \\binom{n}{j}$ can be interpreted as the number of ways to pick a non-empty subset of $A \\cup B$ with the requirement that the number of elements from $A$ who are picked is strictly smaller than the number of elements from $B$ who are picked.\n\n$2^{2n}$ counts the total number of ways to pick a subset of any size from $A \\cup B$. The number of cases where the same number of elements are picked from $A$ and $B$ (including the empty set) is obtained from the sum $\\displaystyle \\sum_{i=0}^n \\binom{n}{i}^2$.\n\nBy symmetry, half of the $\\displaystyle 2^{2n} - \\sum_{i=0}^n \\binom{n}{i}^2$ cases have more elements from $A$ compared to $B$.\n\nThe identity $\\displaystyle \\sum_{i=0}^n \\binom{n}{i}^2 = \\binom{2n}{n}$ matches the result with yours.\n\nI do not know of a combinatorial argument for this last identity though. Does anyone have any?\n\n\u2022 The last can be rewritten $\\sum \\binom{n}{i}\\binom{n}{n-i}$ and the terms represent the number of ways to choose $i$ elements from $A$ and $n-i$ elements from $B$, which, when summed, is the number of ways to choose $n$ elements from $A\\cup B$. \u2013\u00a0Thomas Andrews Feb 27 '16 at 17:44\n\u2022 How many ways to choose $n$ people from $n$ boys and $n$ girls? For every decision about which $i$ boys we will pick, there are $\\binom{n}{i}$ ways to decide which girls we will not pick. \u2013\u00a0Andr\u00e9 Nicolas Feb 27 '16 at 17:48\n\nA bijective correspondence can be established between this issue and the following one:\n\n[Dealing with the LHS of the equation :]\n\nLet $$S$$ be a set with Card(S)=n.\n\nConsider all (ordered) pairs of subsets $$(A,B)$$ such that\n\n$$A \\subsetneqq B \\subset S. \\ \\ (1)$$\n\n[Dealing with the RHS of the equation :]\n\nConsider all subsets of a set $$T$$ with $$2n$$ elements, then exclude a certain number of them (to be precised later), $$T$$ being defined as :\n\n$$T:=S \\cup I \\ \\ \\ \\ \\text{with} \\ \\ \\ \\ \\ I:=\\{1,2,\\cdots n\\}.$$\n\nLet $$C$$ be any subset of $$T$$. We are going to establish (in the \"good cases\") a correspondence between $$C$$ and an ordered pair $$(A,B)$$ verifying (1).\n\nLet us define first a certain fixed ordering of the elements of $$S$$ :\n\n$$a_1 < a_2 < \\cdots < a_n. \\ \\ (2)$$\n\nLet $$B:=T \\cap S$$ and $$J:=T \\cap I$$. Three cases occur :\n\n\u2022 If $$Card(J), $$J$$ is the set of indices \"selecting\" the elements of $$B$$ that belong to $$A$$ in the ordered set $$S$$.\n\n\u2022 If $$Card(J)>Card(B)$$, we switch the r\u00f4les of indices and elements. This accounts for the half part of the formula: indeed this second operation will give the same sets $$(A,B)$$.\n\n\u2022 If $$Card(J)=Card(B)$$, which happens in $$2n \\choose n$$ cases, such cases cannot be placed in correspondence with a case considered in (1), thus have to be discarded.\n\nI know this could be written in a more rigorous way, but I believe the main explanations are there.", "date": "2020-07-05 08:01:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1674559/the-value-of-mathop-sum-sum-0-leq-ij-leq-n-binomni-cdot-binomnj", "openwebmath_score": 0.8278745412826538, "openwebmath_perplexity": 224.49408938750975, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357248544007, "lm_q2_score": 0.8774767890838837, "lm_q1q2_score": 0.8614503115741787}}
{"url": "https://math.stackexchange.com/questions/2751396/how-to-determine-the-number-of-combinations-in-a-conditional-group", "text": "# How to determine the number of combinations in a conditional group?\n\nI have a probability/combinatorics question that is giving me trouble specifically I don't understand why there is a need for the denominator $6$ \"the different orderings of $3$ people\" was the explanation but I'm not sure why that is necessary. I've put the question below and my thought process as well.\n\nQuestion:\n\nIf a committee of $3$ people is to be selected from among $5$ married couples so that the committee does not include two people who are married to each other, how many such committees are possible?\n\nThought Process:\n\n1. First person of the $3$ selected can be any one among the $10$ folks $$\\frac{10!}{(1!)(9!)}$$\n\n2. Second Person of the $3$ selected must not be married to the first person so lowers availability to $8$ folks to choose from $$\\frac{8!}{((1!)(7!)}$$\n\n3. Third Person of the $3$ selected must not be married to either the first or second person - lowers availability to $6$ folks to choose from $$\\frac{6!}{(1!)(5!)}$$\n\n4. Number of possible ways a committee can be formed: $10 \\times 8 \\times 6 = 480$\n\n***So I'm able to get to the $480$ but the answer explanation says that I need to divide this $480$ by $6$ and I don't know why because it sounds illogical to do so. Any help would be amazing.\n\n\u2022 Please read this tutorial on how to typeset mathematics on this site. \u2013\u00a0N. F. Taussig Apr 24 '18 at 10:26\n\u2022 Consider a related but easier problem: how many ways can you choose a committee of $5$ men from the $5$ married couples? Let's do it your way: choose any man, $5$ choices; choose another man, $4$ choices, and so on. So there are $5\\times4\\times3\\times2\\times1=520$ ways to choose a committee of $5$ men. Does that seem like the logical answer? Or should we divide by $5!$ and get a final answer of $1$? \u2013\u00a0bof Apr 24 '18 at 10:44\n\nyou can do this way too\n\ntrio committee with no couples togeather= (total no. of committees - number of committees with couples togeather)\n\n$$=(^{10} C_{3}-^5 C_{1}.8)=80 ways$$\n\nfoot note:\n\n$^5 C_{1}:-$ for choosing $1$ pair or couple(i.e, $2$ peoples) and after we've chosen $2$ peoples we are left with $8$ choices for $3rd$ person to form committee that is why we multiply them to get\n\n$^5 C_{1}.8$\n\n\u2022 Thank you so much!! This different approach was interesting and let me think of it differently \ud83d\ude4f\ud83d\ude03 \u2013\u00a0Omar Apr 24 '18 at 12:06\n\nSuppose the three people who are selected are Anne, Barbara, and Charles.\n\nThere are six orders in which those same three people could be selected:\n\nAnne, Barbara, Charles\n\nAnne, Charles, Barbara\n\nBarbara, Anne, Charles\n\nBarbara, Charles, Anne\n\nCharles, Anne, Barbara\n\nCharles, Barbara, Anne\n\nHowever, all six choices constitute the same committee. Therefore, you need to divide your answer by the $3! = 6$ orders in which you could obtain the same three people, which yields the answer $$\\frac{10 \\cdot 8 \\cdot 6}{3!} = \\frac{10 \\cdot 8 \\cdot 6}{6} = 10 \\cdot 8 = 80$$\n\nAlternate Approach: Choose three of the five couples from which to choose the committee members in $\\binom{5}{3}$ ways, then choose one of the two members from each selected couple, which can be done in $2^3$ ways. Thus, the number of possible selections is $$\\binom{5}{3} \\cdot 2^3 = 10 \\cdot 8 = 80$$\n\n\u2022 Thank you so much mate this was super super helpful \ud83d\ude4f\ud83d\ude03 \u2013\u00a0Omar Apr 24 '18 at 12:05", "date": "2019-08-21 17:43:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2751396/how-to-determine-the-number-of-combinations-in-a-conditional-group", "openwebmath_score": 0.7947660088539124, "openwebmath_perplexity": 210.1990080336593, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856482, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8614502980541112}}
{"url": "https://math.stackexchange.com/questions/3763818/question-from-the-brazilian-math-olympiad", "text": "Question from the Brazilian Math Olympiad\n\nI am stuck with this problem that appeared in the Undergrad Brazilian Math Olympiad from 2017. The problem is:\n\nlet $$x_n$$ be a strictly positive sequence that $$x_n\\rightarrow 0$$. Suppose that exists $$c>0$$ that $$|x_{n+1}-x_n|\\leq c x_n ^2$$ for all $$n\\in\\mathbb{N}$$. Show that there is $$d>0$$ that $$n x_n\\geq d$$ for all $$n\\in \\mathbb{N}$$.\n\nI tried using the Stolz-Ces\u00e0ro lemma, but didn't help me very much. Does anyone have a hint? Thanks!\n\nEDIT:\n\nLet me give some context of my idea. For the Stolz-Ces\u00e0ro lemma the given sequence $$x_n$$ needs to be strictly decreasing, since it $$x_n\\rightarrow 0$$ and $$x_n>0$$. Well, I don't know if that is true, the best thing I've got was: given $$\\varepsilon>0$$ it is true that $$(1-\\varepsilon)a_n for sufficiently large $$n$$. One could help me on that.\n\nMoreover, the lemma says that for $$|b_n|\\rightarrow \\infty$$ if $$\\displaystyle \\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\\rightarrow \\ell$$ then $$\\displaystyle \\frac{a_n}{b_n}\\rightarrow \\ell.$$\n\nSupposing that $$x_n$$ is strictly decreasing, than I can choose $$a_n=n$$ and $$b_n=1/x_n$$. That way I would have $$c_n=\\frac{(n+1)-n}{\\frac{1}{x_{n+1}}-\\frac{1}{x_{n}}}=\\frac{1}{\\frac{1}{x_{n+1}}-\\frac{1}{x_{n}}}.$$ If it is possible to show that this sequence $$c_n$$ converges to some positive number I would have the result.\n\nBut with these assumptions (including that $$x_n$$ is strictly decreasing) the best thing that I've got was: $$\\frac{1}{c}(1-x_n)\\leq \\frac{1}{\\frac{1}{x_{n+1}}-\\frac{1}{x_{n}}}=\\frac{x_{n+1}x_n}{x_n-x_{n+1}}.$$\n\nAt this point there are two things that I don't know: (1) $$x_n$$ strictly decreases and (2) how do I find a comparison (if there is any) to show that $$\\displaystyle \\frac{x_{n+1}x_n}{x_n-x_{n+1}}< d_n$$, where $$d_n\\rightarrow 1/c$$.\n\nOne last thing that I noticed is that the hypothesis $$|x_{n+1}-x_n|\\leq c x_n ^2$$ implies that $$x_{n+1}/x_n\\rightarrow 1$$ and $$f_n=\\frac{|x_{n+1}-x_n|}{x_n ^2}$$ has a convergent subsequence. These facts implies that $$\\frac{x_{n+1}x_n}{x_n-x_{n+1}}$$ also has a convergent subsequence.\n\n\u2022 Is this the olympiad page? Don't they publish solutions? Thanks. \u2013\u00a0Alexey Burdin Jul 20 at 23:22\n\u2022 I don't understand the vote to close this question. \u2013\u00a0Robert Shore Jul 20 at 23:27\n\u2022 They do publish some of the problems solutions, but this one isn't published. \u2013\u00a0Vinnie Carvalho Jul 21 at 0:19\n\u2022 @RobertShore I'm guessing because of \"shows no effort\". IE It is \"missing context and details\". \u2013\u00a0Calvin Lin Jul 21 at 1:37\n\nWe're to show that there exists such $$d>0$$ that $$\\frac{1}{nx_n}<\\frac{1}{d}$$ i.e. that $$\\frac{1}{nx_n}$$ is bounded from above.\nConsider $$y_n=\\frac{1}{cnx_n}$$ i.e. $$x_n=\\frac{1}{cny_n}$$ then the inequality $$|x_{n+1}-x_n|\\le cx_n^2$$ becomes $$\\left|\\frac{1}{c(n+1)y_{n+1}}-\\frac{1}{cny_n}\\right|\\le c\\frac{1}{c^2n^2y_n^2}$$ and we can cancel $$c$$. After some rearrangements the inequality becomes $$\\frac{1}{n y_n - 1} + 1 \\ge (n+1)y_{n+1}-ny_n\\ge \\frac{1}{1 + n y_n} - 1$$ then, noting $$ny_n\\to +\\infty$$ as $$ny_n=\\frac{1}{cx_n}$$ and $$x_n\\to +0$$, we have $$\\frac{1}{ny_n-1}+1\\to 1$$ thus LHS is bounded by some constant $$C$$ from above and we can write $$C\\ge \\frac{1}{n y_n - 1} + 1 \\ge (n+1)y_{n+1}-ny_n$$ $$C\\ge (n+1)y_{n+1}-ny_n$$ summing up for $$n=1,\\ldots,\\,m$$ we have $$Cm\\ge (m+1)y_{m+1}-y_1$$ $$C(m+1)\\ge Cm\\ge (m+1)y_{m+1}-y_1$$ $$C\\ge y_{m+1}-\\frac{y_1}{m+1}$$ $$y_1+C\\ge\\frac{y_1}{m+1}+C\\ge y_{m+1}$$ i.e. $$y_{m+1}$$ is bounded from above. QED.\n\n\"Some rearrangements\":\n\n$$\\left|\\frac{1}{c(n+1)y_{n+1}}-\\frac{1}{cny_n}\\right|\\le c\\frac{1}{c^2n^2y_n^2}$$ $$-\\frac{1}{n^2y_n^2}\\le \\frac{1}{(n+1)y_{n+1}}-\\frac{1}{ny_n}\\le \\frac{1}{n^2y_n^2}$$ $$\\frac{1}{ny_n}-\\frac{1}{n^2y_n^2}\\le \\frac{1}{(n+1)y_{n+1}}\\le \\frac{1}{ny_n}+\\frac{1}{n^2y_n^2}$$ $$\\frac{ny_n-1}{n^2y_n^2}\\le \\frac{1}{(n+1)y_{n+1}}\\le \\frac{ny_n+1}{n^2y_n^2}$$ Now we consider only that $$y_n$$ for which $$ny_n-1>0$$, the other are already bounded from above by $$\\frac 1n$$. $$\\frac{n^2y_n^2}{ny_n-1}\\ge (n+1)y_{n+1}\\ge \\frac{n^2y_n^2}{ny_n+1}$$ $$\\frac{n^2y_n^2-ny_n(ny_n-1)}{ny_n-1}\\ge (n+1)y_{n+1}-ny_n\\ge \\frac{n^2y_n^2-ny_n(ny_n+1)}{ny_n+1}$$ $$\\frac{ny_n}{ny_n-1}\\ge (n+1)y_{n+1}-ny_n\\ge \\frac{-ny_n}{ny_n+1}.$$\n\n\u2022 Thank you for your help! \u2013\u00a0Vinnie Carvalho Jul 21 at 2:48\n\u2022 You're welcome.) Spent 3.5 hours so not-a-contest solution) Contests are 4h-6h likely. \u2013\u00a0Alexey Burdin Jul 21 at 2:51\n\n[This seems a lot easier than I expected, so there might be errors in it. If so, please let em know where.]\n\nThese steps can be demystified by referencing the subsequent block of text\n\n1. Pick $$N$$ such that $$\\forall n > N$$, $$x_n < \\frac{ 1}{2c}$$.\n2. Set $$k = \\min ( \\frac{1}{2c}, Nx_N )$$. Observe that $$\\frac{N}{N+1} \\geq \\frac{1}{2} \\geq ck$$ and $$Nx_N \\geq k$$.\n3. Hence $$(N+1) x_{N+1} \\geq (N+1)( x_N - cx_N^2) \\geq (N+1)(\\frac{k}{N} - \\frac{ ck^2}{N^2}) = k + \\frac{k( \\frac{N}{N+1} - ck ) }{N^2(N+1)} \\geq k$$.\n4. Also, $$\\frac{N+1}{N+1+1} \\geq \\frac{1}{2} \\geq ck$$.\n5. Proceed by induction to conclude that $$n x_n \\geq k$$.\n\nWe claim that under suitable conditions (to be determined), if $$n x_n \\geq k$$, then $$(n+1) x_{n+1} \\geq k$$. If so, the result follows by induction.\n\nWhich conditions make sense?\n\n1. We have $$x_{n+1} \\in ( x_n - c x_n^2, x_n + cx_n^2)$$.\n2. We have $$\\frac{k}{n} < x_n$$.\n3. We will likely want $$x - c x^2$$ to be increasing, which requires $$x_n < \\frac{1}{2c}$$. This can be satisfied as $$\\lim x_n = 0$$.\n4. Henceforth, we assume $$\\frac{k}{n} < x_n < \\frac{1}{2c}$$. This necessitates $$2ck < n$$, which can be achieved.\n5. Now, $$(n+1) x_{n+1} > (n+1) \\left[ x_n - c x_n^2\\right] > (n+1) \\left[\\frac{k}{n} - \\frac{ ck^2 } { n^2 } \\right]$$. Verify that $$(n+1) \\left[\\frac{k}{n} - \\frac{ ck^2 } { n^2 } \\right] \\geq k \\Leftrightarrow \\frac{n}{n+1} \\geq ck$$.\n\nThis gives us all of the conditions that we need.\n\n\u2022 Thank you for your help! \u2013\u00a0Vinnie Carvalho Jul 21 at 2:48\n\u2022 The parts order from this revision made more sense for me. \u2013\u00a0Alexey Burdin Jul 21 at 2:49\n\u2022 @AlexeyBurdin I agree with you. It's the difference between presenting a clean direct solution for the olympiad vs motivating the approach / values chosen. I just started writing stuff down and it follows immediately (which is why I'm questioning if there is an error). \u2013\u00a0Calvin Lin Jul 21 at 2:54", "date": "2020-10-27 03:59:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3763818/question-from-the-brazilian-math-olympiad", "openwebmath_score": 0.9762465357780457, "openwebmath_perplexity": 4054.304138451811, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434786819155, "lm_q2_score": 0.8840392832736084, "lm_q1q2_score": 0.8614386305364771}}
{"url": "https://ukrpak.kiev.ua/51810/topological-sort-problems-d277fe", "text": "Here vertex 1 has in-degree 0. A Topological Sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: Vertex-A has the least in-degree. Each topological order is a feasible schedule. 11, Article No. Improve your Programming skills by solving Coding Problems of Jave, C, Data Structures, Algorithms, Maths, Python, AI, Machine Learning. efficient scheduling is an NP-complete problem) \u2022 Or during compilation to order modules/libraries a d c g f b e. Examples \u2022Resolving dependencies: apt-get uses topological sorting to obtain the admissible sequence in which a set of Debianpackages can be installed/removed. Both these problems Course Schedule. Two other restricted permuta\u00ad tion problems are permutations with prescribed up-down sequences, and permutations with a given number of runs. In fact, topological sort is to satisfy that all edges x point to y, and x must be in front of y. So, remove vertex-A and its associated edges. The ordering of the nodes in the array is called a topological ordering. Explanation for the article: http://www.geeksforgeeks.org/topological-sorting/This video is contributed by Illuminati. Input: The first line of input takes the number of test cases then T test cases follow . For the standard (i.e., static) topological sorting problem, algorithms with (V) (i.e., (v+e)) time are well known (e.g., Cormen et al. For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. Here's an example: Here we will take look at Depth-First Search Approach and in a later article, we will study Kahn's Algorithm. Topological sort There are often many possible topological sorts of a given DAG Topological orders for this DAG : 1,2,5,4,3,6,7 2,1,5,4,7,3,6 2,5,1,4,7,3,6 Etc. Topological Sort. Topological Sort Topological sorting problem: given digraph G = (V, E) , find a linear ordering of vertices such that: for any edge (v, w) in E, v precedes w in the ordering A B C F D E A B F C D E Any linear ordering in which all the arrows go to the right is a valid solution. Moonfrog Labs. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v\u2026 Read More. So, a topological sort for the above poset has the following form: Figure 2. Topological Sort Example. The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \\over 4$$ cup of milk. Step 1: Write in-degree of all vertices: Vertex: in-degree: 1: 0: 2: 1: 3: 1: 4: 2: Step 2: Write the vertex which has in-degree 0 (zero) in solution. Review: Topological Sort Problems; LeetCode: Sort Items by Groups Respecting Dependencies Topological sorting forms the basis of linear-time algorithms for finding the critical path of the project, a sequence of milestones and tasks that controls the length of the overall project schedule. Note: Topological sorting on a graph results non-unique solution. This problem can be solved in multiple ways, one simple and straightforward way is Topological Sort. Example 11.6. John Conway: Surreal Numbers - How playing games led to more numbers than anybody ever thought of - \u2026 Topological Sorting\u00b6 To demonstrate that computer scientists can turn just about anything into a graph problem, let\u2019s consider the difficult problem of stirring up a batch of pancakes. Topological Sorting\u00b6 To demonstrate that computer scientists can turn just about anything into a graph problem, let\u2019s consider the difficult problem of stirring up a batch of pancakes. It works only on Directed Acyclic Graphs(DAGs) - Graphs that have edges indicating direction. While the exact order of the items is unknown (i.e. A topological sort of a directed acyclic graph is a linear ordering of its vertices such that for every directed edge u \u2192 v from vertex u to vertex v, u comes before v in the ordering. Topological Sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering.A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). Let us try to solve the following topological sorting problem. Amazon. [2001]). A topological ordering is possible if and only if the graph has no directed cycles, i.e. 3. Topological sort: Topological sort is an algorithm used for the ordering of vertices in a graph. 2.Initialize a queue with indegree zero vertices. Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge (u v) from vertex u to vertex v, u comes before v in the ordering. Subscribe to see which companies asked this question. Topological Sort. If you're thinking Makefile or just Program dependencies, you'd be absolutely correct. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. A trivial solution, based upon a standard (i.e., static) ACM Journal of Experimental Algorithmics, Vol. an easy explanation for topological sorting. A topological sort is a ranking of the n objects of S that is consistent with the given partial order. Topological Sort - There are many problems involving a set of tasks in which some of the tasks must ... Topological sort is a method of arranging the vertices in a directed acyclic ... | PowerPoint PPT presentation | free to view . Kind of funny considering it's usually 10 lines or less! So, remove vertex-A and its associated edges. Find any Topological Sorting of that Graph. There's actually a type of topological sorting which is used daily (or hourly) by most developers, albeit implicitly. Page 1 of 2 1 2 \u00bb Courses. A topological sort is deeply related to dynamic programming \u2026 3. Here, I focus on the relation between the depth-first search and a topological sort. Topological Sorting for a graph is not possible if the graph is not a DAG.. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: Vertex-A has the least in-degree. Graph. The first line of each test case contains two integers E and V representing no of edges and the number of vertices. CSES - Easy. Excerpt from The Algorithm Design Manual: Topological sorting arises as a natural subproblem in most algorithms on directed acyclic graphs. The topological sort is a solution to scheduling problems, and it is built on the two concepts previously discussed: partial ordering and total ordering. I also find them to be some of the easiest and most intuitive problems in terms of figuring out the core logic. Focus Problem \u2013 read through this problem before continuing! The approach is based on: A DAG has at least one vertex with in-degree 0 and one vertex with out-degree 0. Problem: Find a linear ordering of the vertices of $$V$$ such that for each edge $$(i,j) \\in E$$, vertex $$i$$ is to the left of vertex $$j$$. Topological Sort: A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering.A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. Accolite. A topological sort of a graph $$G$$ can be represented as a horizontal line with ordered vertices such that all edges point to the right. Each test case contains two lines. Topological sorting for D irected A cyclic G raph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Microsoft. Topological Sort. In a real-world scenario, topological sorting can be utilized to write proper assembly instructions for Lego toys, cars, and buildings. Data Structures and Algorithms \u2013 Self Paced Course. Flipkart. Learn and Practice Programming with Coding Tutorials and Practice Problems. See all topologicalsort problems: #topologicalsort. Topological Sorting. Given a partial order on a set S of n objects, produce a topological sort of the n objects, if one exists. if the graph is DAG. The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \\over 4$$ cup of milk. Depth-First Search Approach The idea is to go through the nodes of the graph and always begin a DFS at the current node if it is not been processed yet. Does topological sort applies to every graph? The topological sort algorithm takes a directed graph and returns an array of the nodes where each node appears before all the nodes it points to. Any DAG has at least one topological ordering. 1 4 76 3 5 2 9. Problem Modeling Using Topological Sorting. 1.7, 2006. Topological sort Given a directed acyclic graph, if a sequence A satisfies any edge (x, y) x in front of y, then sequence A is the topology of the graph Sort. We represent dependencies as edges of the graph. The topological sorting problem is a restricted permutation problem, that is a problem cone jrned with the study of permutations chat sat\u00ad isfy some given set of restrictions. Topological Sorts for Cyclic Graphs? You have solved 0 / 6 problems. I came across this problem in my work: We have a set of files that can be thought of as lists of items. Any DAG has at least one topological ordering, and algorithms are known for constructing a topological ordering of any DAG in linear time. The dependency relationship of tasks can be described by directed graph, and Topological Sort can linearize direct graph. OYO Rooms. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. It outputs linear ordering of vertices based on their dependencies. Solving Using In-degree Method. Impossible! Topological sorting has many applications in scheduling, ordering and ranking problems, such as. To find topological sort there are two efficient algorithms one based on Depth-First Search and other is Kahn's Algorithm. View Details. 2.Initialize a queue with indegree zero vertices. However, the problem of dynamically maintaining a topological ordering appears to have received little attention. Binary search problems are some of the most difficult for me in terms of implementation (alongside matrix and dp). The tutorial is for both beginners \u2026 Given a Directed Graph. Problems, easiest Approach is: 1.Store each vertex indegree in an array i also find to... Sorting arises as a natural subproblem in most algorithms on directed acyclic Graphs other is 's. Algorithm used for the article: http: //www.geeksforgeeks.org/topological-sorting/This video is contributed by Illuminati is deeply related to dynamic \u2026... Excerpt from the Algorithm Design Manual: topological sorting on a graph results non-unique solution objects of that... Cars, and permutations with a given number of vertices thought of lists., static ) ACM Journal of Experimental Algorithmics, Vol other restricted permuta\u00ad tion problems are permutations with a DAG... 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It 's usually 10 lines or less linear time problem \u2013 read through this problem can be of!\n\nDrainboard Sink Faucet, Sg I Bat, Activa 6g Seat Cover, Relion Ear Thermometer Error 1, Cyphenothrin 5 Ec Msds, Bread Cartoon Drawing, Superhero Avatar Creator,", "date": "2021-06-19 05:17:41", "meta": {"domain": "kiev.ua", "url": "https://ukrpak.kiev.ua/51810/topological-sort-problems-d277fe", "openwebmath_score": 0.5246270895004272, "openwebmath_perplexity": 1161.597526904899, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9643214491222696, "lm_q2_score": 0.8933094081846421, "lm_q1q2_score": 0.8614374230151711}}
{"url": "https://mathematica.stackexchange.com/questions/115931/direct-way-of-listplotting-each-2-or-3-points/115951", "text": "# Direct way of listplotting each 2 or 3 points?\n\nI have a list of points\n\nmyList={{x_1,y_1},{x_2,y_2},...,{x_n,y_n}}\n\n\nI want to plot not all the points of this list but to plot every $k$ points. For example, if I wanted to plot every $k=2$ points, the plot will contain only the points:\n\n{{x_1,y_1},{x_3,y_3},{x_5,y_5},...}\n\n\nor\n\n{{x_2,y_2},{x_4,y_4},...}\n\n\nIt doesn't matter if Mathematica plots the odd points or the even points.\n\nAnother example is if I wanted to plot every $k=3$ points, the plot would contain only\n\n{{x_1,y_1},{x_4,y_4},{x_7,y_7},...}\n\n\nQuestion. Is there a direct way of doing this in Mathematica, without creating manually another list from myList?\n\n\u2022 Look up Downsample[]. May 25 '16 at 19:06\n\nYou can use Part (also written [[...]]) to get what you want. For example,\n\nmyList= Table[{i, i}, {i, 1, 20}];\n\nListPlot[myList]\nListPlot[myList[[1;;-1;;2]]] (* every second point *)\nListPlot[myList[[1;;-1;;3]]] (* every 3rd point*)\n\n\u2022 Or simply, ListPlot[myList[[;; ;;2]]] or ListPlot[myList[[;; ;;3]]]. The 1 and -1 are default values and are not required. May 25 '16 at 21:06\n\nThe ways listed work perfectly well, but you can save yourself a couple of keystrokes just by doing\n\nmyList[[;;;;2]]\n\n\nIf you don't provide starting and ending points for the first ;;, Mathematica is kind enough to assume that you want to go from the beginning (1) to the end (-1). For my tastes, at least, this looks a little nicer too. (It'll auto-format inside the program in a way that makes it a little easier to read than it's rendered here:)", "date": "2021-12-08 00:34:40", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/115931/direct-way-of-listplotting-each-2-or-3-points/115951", "openwebmath_score": 0.38348615169525146, "openwebmath_perplexity": 1028.6129898757572, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.964321448096903, "lm_q2_score": 0.8933094017937621, "lm_q1q2_score": 0.8614374159363388}}
{"url": "http://math.stackexchange.com/questions/821817/can-any-factor-pairs-where-pair-is-unique-have-the-same-sum", "text": "# Can any factor pairs (where pair is unique) have the same sum?\n\nI have used Stack Overflow but I'm new to this site so I apologize if this is a trivial question. I am creating mathematics software using javascript. I am using a for loop to find all factors for a given integer (n) as follows (simplified version):\n\nn = someInteger;\n\nfor (i = 1; i <= n; i++) {\nif (n % i == 0) {\nvar msg = 'The numbers ' + i + ' and ' + (n / i) + ' are factors of n';\n}\n}\n\n\nIn brief, the loop cycles through all integers 1 to n and for each integer, calculates n mod i. If the modulus is 0, the pair i and n/i is a pair of factors for n. However, a la the commutative property, I don't want to \"double count\" pairs that have simply had their order switched. For example, in my method, if e.g. n = 10, 2 x 5 is a unique pair of factors but so is 5 x 2.\n\nMy hypothesis is that these \"doubly counted\" pairs can be eliminated by putting a filter in my loop that basically states \"if (currentfactor1 + currentfactor2) = (anypreviousfactor1 + anypreviousfactor2), don't count this factor pair as a unique pair\".\n\nSo my question is, do any pairs of factors for any integer ever have the same sum? My suspicion is no but I can't find proof of this. I took a couple of sample integers with several pairs of factors (two examples):\n\n144 (1 x 144 (sum = 145), 2 x 72 (sum = 74), 3 x 48 (sum = 51), 4 x 36 (sum = 40), 6 x 24 (sum = 30), 8 x 18 (sum = 26), 9 x 16 (sum = 25), 12 x 12 (sum = 24) )\n\n300 (1 x 300 (sum = 301), 2 x 150 (sum = 152), 3 x 100 (sum = 103), 4 x 75 (sum = 79), 5 x 60 (sum = 65), 6 x 50 (sum = 56), 10 x 30 (sum = 40), 12 x 25 (sum = 37), 15 x 20 (sum = 35) )\n\nCan anyone please shed some light on this? Thanks for any help\n\n-\nThis doesn't answer your final question, but if you're worried about double counting, why don't you just divide your final result by two? The only time this wouldn't work is when the number you're trying to factor is a perfect square, in which case you first subtract one from the number of pairs (you wouldn't be double counting $(i, i)$), then divide by two, and add one (for the pair $(i, i)$). Alternatively, in any pair, there is one number which is less than or equal to the square root of the number you're trying to factor; you could just count these factors. \u2013\u00a0 Michael Albanese Jun 5 '14 at 14:45\nHaving said all that, you should probably ignore these possibilities until you've figured out whether your current strategy will work. \u2013\u00a0 Michael Albanese Jun 5 '14 at 14:49\nThanks @MichaelAlbanese...at about the same time you posted this the same thought came to my mind. You're right...that is a valid solution that can be implemented and will work. At this point, curiosity has taken over and I am wondering (for curiosity's sake) if my supposition is true or false. \u2013\u00a0 The One and Only ChemistryBlob Jun 5 '14 at 14:53\nThe easy way to avoid the double counting is to make your loop condition $i \\le \\sqrt n$ instead of $i \\le n$. It will finish much sooner too, because it has to run many fewer times. \u2013\u00a0 MJD Jun 5 '14 at 15:03\nOK...I see. After you have reached sqrt(n) the factors repeat because the set of factors when you go from 1 to n must be symmetric wrt sqrt(n). Nice. \u2013\u00a0 The One and Only ChemistryBlob Jun 5 '14 at 15:08\n\nThe answer to your question is no. Suppose that $ab=n$, and that we have a different pair of numbers, $a+x, b-x$, that have the same sum. Their product is $(a+x)(b-x)=ab+x(b-a-x)=n+x(b-a-x)$. If this product is again $n$, then $$x(b-a-x)=0$$\n\nIf the product is zero, then either $x=0$ (and the new numbers are the same as the old ones) or $b-a-x=0$, or $x=b-a$ (and $a,b$ simply switched places).\n\n-\nExcellent proof...This site is great. I earned my PhD in chemistry but minored in mathematics in college...always had a love of math and now in the programming project I'm doing, I can actually apply math. Your answer helps me understand this problem without the use of a graph...thanks for your help. \u2013\u00a0 The One and Only ChemistryBlob Jun 5 '14 at 15:25\nAlthough each answer was uniquely excellent, I voted this as the best answer because I am biased towards analyzing functions by graphs, but despite this bias this solution really helped me \"graphlessly\" (new word) understand the question I was asking \u2013\u00a0 The One and Only ChemistryBlob Jun 5 '14 at 15:39\n\nLet $n$ be positive. Then for positive real $x$, the function $x+\\frac{n}{x}$ decreases until $x=\\sqrt{n}$ and then increases. So for any $k$ there are at most two real positive values of $x$ such that $x+\\frac{n}{x}=k$.\n\nIn particular, if for the ordered pair $(a,b)$ we have $ab=n$ and $a+b=k$, the only other ordered pair with these properties is $(b,a)$.\n\n-\nI like Mark Bennet's somewhat better. But mine has the virtue of being more explicitly geometrical. \u2013\u00a0 Andr\u00e9 Nicolas Jun 5 '14 at 15:26\nWow...great solution. Focusing on x > 0, dy/dx = 1 - (n/x^2), which = 0 when x = sqrt(n). This is a local minimum since (n/x^2) must be less than one when x > sqrt(n) and thus dy/dx > 0 (and when x < sqrt(n), dy/dx < 0), and f(x) increasing beyond x > sqrt(n), so any horizontal line drawn through f(x) will only intersect f(x) at one set of unique points \u2013\u00a0 The One and Only ChemistryBlob Jun 5 '14 at 15:42\n\nIf $a+b=p$ and $ab=q$ then $a,b$ are roots of the quadratic equation $$(x-a)(x-b)=x^2-px+q=0$$If $c\\neq a; c\\neq b$ then $(c-a)(c-b)\\neq 0$ and $c$ is not a root. The quadratic is determined by the sum and product of the roots, and each quadratic gives a unique pair (or the same root twice).\n\n-", "date": "2015-07-08 04:33:55", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/821817/can-any-factor-pairs-where-pair-is-unique-have-the-same-sum", "openwebmath_score": 0.6799117922782898, "openwebmath_perplexity": 533.6851256594243, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969665221771, "lm_q2_score": 0.8757869835428965, "lm_q1q2_score": 0.8614214203324009}}
{"url": "https://in.mathworks.com/help/symbolic/sym.taylor.html", "text": "# taylor\n\n## Description\n\nexample\n\nT = taylor(f,var) approximates f with the Taylor series expansion of f up to the fifth order at the point var = 0. If you do not specify var, then taylor uses the default variable determined by symvar(f,1).\n\nexample\n\nT = taylor(f,var,a) approximates f with the Taylor series expansion of f at the point var = a.\n\nexample\n\nT = taylor(___,Name,Value) specifies options using one or more name-value arguments in addition to any of the input argument combinations in previous syntaxes. For example, you can specify the expansion point, truncation order, or order mode of the Taylor series expansion.\n\n## Examples\n\ncollapse all\n\nFind the Maclaurin series expansions of the exponential, sine, and cosine functions up to the fifth order.\n\nsyms x\nT1 = taylor(exp(x))\nT1 =\n\n$\\frac{{x}^{5}}{120}+\\frac{{x}^{4}}{24}+\\frac{{x}^{3}}{6}+\\frac{{x}^{2}}{2}+x+1$\n\nT2 = taylor(sin(x))\nT2 =\n\n$\\frac{{x}^{5}}{120}-\\frac{{x}^{3}}{6}+x$\n\nT3 = taylor(cos(x))\nT3 =\n\n$\\frac{{x}^{4}}{24}-\\frac{{x}^{2}}{2}+1$\n\nYou can use the sympref function to modify the output order of symbolic polynomials. Redisplay the polynomials in ascending order.\n\nsympref('PolynomialDisplayStyle','ascend');\nT1\nT1 =\n\n$1+x+\\frac{{x}^{2}}{2}+\\frac{{x}^{3}}{6}+\\frac{{x}^{4}}{24}+\\frac{{x}^{5}}{120}$\n\nT2\nT2 =\n\n$x-\\frac{{x}^{3}}{6}+\\frac{{x}^{5}}{120}$\n\nT3\nT3 =\n\n$1-\\frac{{x}^{2}}{2}+\\frac{{x}^{4}}{24}$\n\nThe display format you set using sympref persists through your current and future MATLAB\u00ae sessions. Restore the default value by specifying the 'default' option.\n\nsympref('default');\n\nFind the Taylor series expansions at $\\mathit{x}=1$ for these functions. The default expansion point is 0. To specify a different expansion point, use ExpansionPoint.\n\nsyms x\nT = taylor(log(x),x,'ExpansionPoint',1)\nT =\n\n$x-\\frac{{\\left(x-1\\right)}^{2}}{2}+\\frac{{\\left(x-1\\right)}^{3}}{3}-\\frac{{\\left(x-1\\right)}^{4}}{4}+\\frac{{\\left(x-1\\right)}^{5}}{5}-1$\n\nAlternatively, specify the expansion point as the third argument of taylor.\n\nT = taylor(acot(x),x,1)\nT =\n\n$\\frac{\\pi }{4}-\\frac{x}{2}+\\frac{{\\left(x-1\\right)}^{2}}{4}-\\frac{{\\left(x-1\\right)}^{3}}{12}+\\frac{{\\left(x-1\\right)}^{5}}{40}+\\frac{1}{2}$\n\nFind the Maclaurin series expansion for f = sin(x)/x. The default truncation order is 6. The Taylor series approximation of this expression does not have a fifth-degree term, so taylor approximates this expression with the fourth-degree polynomial.\n\nsyms x\nf = sin(x)/x;\nT6 = taylor(f,x);\n\nUse Order to control the truncation order. For example, approximate the same expression up to the orders 7 and 9.\n\nT8 = taylor(f,x,'Order',8);\nT10 = taylor(f,x,'Order',10);\n\nPlot the original expression f and its approximations T6, T8, and T10. Note how the accuracy of the approximation depends on the truncation order.\n\nfplot([T6 T8 T10 f])\nxlim([-4 4])\ngrid on\nlegend('approximation of sin(x)/x with error O(x^6)', ...\n'approximation of sin(x)/x with error O(x^8)', ...\n'approximation of sin(x)/x with error O(x^{10})', ...\n'sin(x)/x','Location','Best')\ntitle('Taylor Series Expansion')\n\nFind the Taylor series expansion of this expression. By default, taylor uses an absolute order, which is the truncation order of the computed series.\n\nsyms x\nT = taylor(1/exp(x) - exp(x) + 2*x,x,'Order',5)\nT =\n\n$-\\frac{{x}^{3}}{3}$\n\nFind the Taylor series expansion with a relative truncation order by using OrderMode. For some expressions, a relative truncation order provides more accurate approximations.\n\nT = taylor(1/exp(x) - exp(x) + 2*x,x,'Order',5,'OrderMode','relative')\nT =\n\n$-\\frac{{x}^{7}}{2520}-\\frac{{x}^{5}}{60}-\\frac{{x}^{3}}{3}$\n\nFind the Maclaurin series expansion of this multivariate expression. If you do not specify the vector of variables, taylor treats f as a function of one independent variable.\n\nsyms x y z\nf = sin(x) + cos(y) + exp(z);\nT = taylor(f)\nT =\n\n$\\frac{{x}^{5}}{120}-\\frac{{x}^{3}}{6}+x+\\mathrm{cos}\\left(y\\right)+{\\mathrm{e}}^{z}$\n\nFind the multivariate Maclaurin series expansion by specifying the vector of variables.\n\nsyms x y z\nf = sin(x) + cos(y) + exp(z);\nT = taylor(f,[x,y,z])\nT =\n\n$\\frac{{x}^{5}}{120}-\\frac{{x}^{3}}{6}+x+\\frac{{y}^{4}}{24}-\\frac{{y}^{2}}{2}+\\frac{{z}^{5}}{120}+\\frac{{z}^{4}}{24}+\\frac{{z}^{3}}{6}+\\frac{{z}^{2}}{2}+z+2$\n\nYou can use the sympref function to modify the output order of a symbolic polynomial. Redisplay the polynomial in ascending order.\n\nsympref('PolynomialDisplayStyle','ascend');\nT\nT =\n\n$2+z+\\frac{{z}^{2}}{2}+\\frac{{z}^{3}}{6}+\\frac{{z}^{4}}{24}+\\frac{{z}^{5}}{120}-\\frac{{y}^{2}}{2}+\\frac{{y}^{4}}{24}+x-\\frac{{x}^{3}}{6}+\\frac{{x}^{5}}{120}$\n\nThe display format you set using sympref persists through your current and future MATLAB sessions. Restore the default value by specifying the 'default' option.\n\nsympref('default');\n\nFind the multivariate Taylor series expansion by specifying both the vector of variables and the vector of values defining the expansion point.\n\nsyms x y\nf = y*exp(x - 1) - x*log(y);\nT = taylor(f,[x y],[1 1],'Order',3)\nT =\n\n$x+\\frac{{\\left(x-1\\right)}^{2}}{2}+\\frac{{\\left(y-1\\right)}^{2}}{2}$\n\nIf you specify the expansion point as a scalar a, taylor transforms that scalar into a vector of the same length as the vector of variables. All elements of the expansion vector equal a.\n\nT = taylor(f,[x y],1,'Order',3)\nT =\n\n$x+\\frac{{\\left(x-1\\right)}^{2}}{2}+\\frac{{\\left(y-1\\right)}^{2}}{2}$\n\nFind the error estimate when approximating a function $f\\left(x\\right)=\\mathrm{log}\\left(x+1\\right)$ using the Taylor series expansion. Here, consider the Taylor approximation up to the 7th order (with the truncation order $\\mathit{n}=8$) at the expansion point $\\mathit{a}=0$.\n\nThe error or remainder in the Taylor approximation is given by the Lagrange form:\n\n${\\mathit{R}}_{\\mathit{n}-1}\\left(\\mathit{x}\\right)=\\frac{{\\mathit{f}}^{\\mathit{n}}\\left(\\mathit{c}\\right)}{\\mathit{n}!}{\\left(\\mathit{x}-\\mathit{a}\\right)}^{\\mathit{n}}.$\n\nThe upper bound of the error estimate can be calculated by finding a positive real number $\\mathit{M}$ such that $|{\\mathit{f}}^{\\mathit{n}}\\left(\\mathit{c}\\right)|\\le \\mathit{M}$ for all $\\mathit{c}$ between $\\mathit{a}$ and $\\mathit{x}$.\n\nFind the Taylor series expansion of the function $f\\left(x\\right)=\\mathrm{log}\\left(x+1\\right)$ up to the 7th order by specifying Order as 8.\n\nsyms x\nf = log(x+1)\nf =\u00a0$\\mathrm{log}\\left(x+1\\right)$\nT = taylor(f,'Order',8)\nT =\n\n$\\frac{{x}^{7}}{7}-\\frac{{x}^{6}}{6}+\\frac{{x}^{5}}{5}-\\frac{{x}^{4}}{4}+\\frac{{x}^{3}}{3}-\\frac{{x}^{2}}{2}+x$\n\nTo estimate the error in the Taylor approximation, first compute the term ${\\mathit{f}}^{8}\\left(\\mathit{c}\\right)$.\n\nsyms c\nfn(c) = subs(diff(f,8),x,c)\nfn(c) =\n\n$-\\frac{5040}{{\\left(c+1\\right)}^{8}}$\n\nFor positive values of $\\mathit{x}$, the upper bound of the error estimate can be calculated by using the relation $|{\\mathit{f}}^{8}\\left(\\mathit{c}\\right)|\\le 5040$ (because $\\mathit{c}$ must be a positive value between $0$ and a positive $\\mathit{x}$). Next, find the upper bound of the error estimate Rupper(x) by using the Lagrange from ${\\mathit{R}}_{7}\\left(\\mathit{x}\\right)$ and the relation $|{\\mathit{f}}^{8}\\left(\\mathit{c}\\right)|\\le 5040$.\n\nRupper(x) = 5040*x^8/factorial(8)\nRupper(x) =\n\n$\\frac{{x}^{8}}{8}$\n\nEvaluate the Taylor series expansion at the point $\\mathit{x}=0.5$. Find the upper bound of the error estimate in the Taylor approximation.\n\nTeval = subs(T,x,0.5)\nTeval =\n\n$\\frac{909}{2240}$\n\nRmax = double(Rupper(0.5))\nRmax = 4.8828e-04\n\nFor comparison, evaluate the exact function at $\\mathit{x}=0.5$ and find the remainder in the Taylor approximation.\n\nfeval = subs(f,x,0.5)\nfeval =\n\n$\\mathrm{log}\\left(\\frac{3}{2}\\right)$\n\nR = double(abs(feval-Teval))\nR = 3.3846e-04\n\n## Input Arguments\n\ncollapse all\n\nInput to approximate, specified as a symbolic expression or function. It also can be a vector, matrix, or multidimensional array of symbolic expressions or functions.\n\nExpansion variable, specified as a symbolic variable. If you do not specify var, then taylor uses the default variable determined by symvar(f,1).\n\nExpansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You also can specify the expansion point as a name-value argument. If you specify the expansion point both ways, then the name-value argument takes precedence.\n\n### Name-Value Arguments\n\nSpecify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.\n\nBefore R2021a, use commas to separate each name and value, and enclose Name in quotes.\n\nExample: taylor(log(x),x,'ExpansionPoint',1,'Order',9)\n\nExpansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You can also specify the expansion point using the input argument a. If you specify the expansion point both ways, then the name-value argument takes precedence.\n\nTruncation order of the Taylor series expansion, specified as a positive integer or a symbolic positive integer. taylor computes the Taylor series approximation with the order n - 1. The truncation order n is the exponent in the O-term: O(varn).\n\nOrder mode indicator, specified as 'absolute' or 'relative'. This indicator specifies whether to use absolute or relative order when computing the Taylor polynomial approximation.\n\nAbsolute order is the truncation order of the computed series. Relative order n means that the exponents of var in the computed series range from the leading order m to the highest exponent m + n - 1. Here m + n is the exponent of var in the O-term: O(varm\u00a0+\u00a0n).\n\ncollapse all\n\n### Taylor Series Expansion\n\nA Taylor series expansion represents an analytic function f(x) as an infinite sum of terms around the expansion point x\u00a0=\u00a0a:\n\n$f\\left(x\\right)=f\\left(a\\right)+\\frac{{f}^{\\prime }\\left(a\\right)}{1!}\\left(x-a\\right)+\\frac{{f}^{\u2033}\\left(a\\right)}{2!}{\\left(x-a\\right)}^{2}+\\dots =\\sum _{m=0}^{\\infty }\\frac{{f}^{\\left(m\\right)}\\left(a\\right)}{m!}\\cdot {\\left(x-a\\right)}^{m}$\n\nA Taylor series expansion requires a function to have derivatives up to an infinite order around the expansion point.\n\n### Maclaurin Series Expansion\n\nThe Taylor series expansion around x\u00a0=\u00a00 is called a Maclaurin series expansion:\n\n$f\\left(x\\right)=f\\left(0\\right)+\\frac{{f}^{\\prime }\\left(0\\right)}{1!}x+\\frac{{f}^{\u2033}\\left(0\\right)}{2!}{x}^{2}+\\dots =\\sum _{m=0}^{\\infty }\\frac{{f}^{\\left(m\\right)}\\left(0\\right)}{m!}{x}^{m}$\n\n## Tips\n\n\u2022 If you use both the third argument a and ExpansionPoint to specify the expansion point, then the value specified by ExpansionPoint prevails.\n\n\u2022 If var is a vector, then the expansion point a must be a scalar or a vector of the same length as var. If var is a vector and a is a scalar, then a is expanded into a vector of the same length as var with all elements equal to a.\n\n\u2022 If the expansion point is infinity or negative infinity, then taylor computes the Laurent series expansion, which is a power series in 1/var.\n\n\u2022 You can use the sympref function to modify the output order of symbolic polynomials.\n\n\u2022 If taylor cannot find the Taylor series expansion, then use series to find the more general Puiseux series expansion.\n\n## Version History\n\nIntroduced before R2006a", "date": "2023-02-01 22:08:32", "meta": {"domain": "mathworks.com", "url": "https://in.mathworks.com/help/symbolic/sym.taylor.html", "openwebmath_score": 0.9329606294631958, "openwebmath_perplexity": 1100.1087606421559, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964207345894, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8613999753862738}}
{"url": "https://www.empreendasemfronteiras.com.br/hafvaq/e19ab0-can-a-relation-be-both-symmetric-and-antisymmetric", "text": "We can therefore take the following relation: $\\{a,b,c\\}$ would be our universe and $R=\\{\\langle a,b\\rangle,\\langle b,a\\rangle,\\langle a,c\\rangle\\}$. It is an interesting exercise to prove the test for transitivity. Thus, there exists a distinct pair of integers $a$ and $b$ such that $aRb$ and $bRa$. See also However, since $(-1)\\cdot 2^{2} = -4 \\not\\gt 0$, $(-1, 2)\\not\\in R$, thus $R$ is not symmetric. Can you legally move a dead body to preserve it as evidence? Active 1 year, 7 months ago. A symmetric relation can work both ways between two different things, whereas an antisymmetric relation imposes an order. I got this from my professor and my book explains that they are not mutually exclusive. $$R=\\{(a,b), (b,a), (c,d)\\}.$$. Consider matrix which has ones on diagonal and zeros on other places. Therefore, in an antisymmetric relation, the only ways it agrees to both situations is a=b. It only takes a minute to sign up. This doesn't tell \u2026 Click hereto get an answer to your question \ufe0f Given an example of a relation. Parsing JSON data from a text column in Postgres. Hence, $R$ cannot be antisymmetric. Is the Gelatinous ice cube familar official? If So, Give An Example; If Not, Give An Explanation. Come up with a relation on that set such that for some pairs of elements (x, y), $x R y$ and $\\lnot (y R x)$; but for other pairs of elements (x, y), $x R y$ and $y R x$. How can a matrix relation be both antisymmetric and symmetric? A relation can be neither symmetric nor antisymmetric. Although both have similarities in their names, we can see differences in both their relationships such that asymmetric relation does not satisfy both conditions whereas antisymmetric satisfies both the conditions, but only if both the elements are similar. But if antisymmetric relation contains pair of the form (a,a) then it cannot be asymmetric. together. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. Explain this image to me. A relation can be neither symmetric nor antisymmetric. A relation R is symmetric if the value of every cell (i, j) is same as that cell (j, i). The terms symmetric and antisymmetric are not opposites, because a relation can have both of these properties or may lack both of them. Is the bullet train in China typically cheaper than taking a domestic flight? Remember that a relation on a set $A$ is just a subset of $A\\times A$. (iii) Reflexive and symmetric but not transitive. Limitations and opposites of asymmetric relations are also asymmetric relations. Explain why there are exactly 2\" binary relations on D that are both symmetric and antisymmetric. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the \"preys on\" relation on biological species). Why aren't \"fuel polishing\" systems removing water & ice from fuel in aircraft, like in cruising yachts? Comparing method of differentiation in variational quantum circuit. Macbook in Bed: M1 Air vs M1 Pro with Fans Disabled. Could you design a fighter plane for a centaur? Ask Question Asked 5 years, 10 months ago. Could you design a fighter plane for a centaur? Asymmetric relation: Asymmetric relation is opposite of symmetric relation. MathJax reference. Relations, specifically, show the connection between two sets. Can I assign any static IP address to a device on my network? In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. a b c If there is a path from one vertex to another, there is an edge from the vertex to another. Why is 2 special? To put it simply, you can consider an antisymmetric relation of a set as a one with no ordered pair and its reverse in the relation. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. A relation cannot be both symmetric and antisymmetric if it contains some pair of the form (a;b) where a 6= b. Anonymous . Antisymmetric Relation Definition. i don't believe you do. Or it can be defined as, relation R is antisymmetric if either (x,y)\u2209R or (y,x)\u2209R whenever x \u2260 y. Is the relation reflexive, symmetric and antisymmetric? Can A Relation Be Both Symmetric And Antisymmetric? However, a relation can be neither symmetric nor asymmetric, which is the case for \"is less than or equal to\" and \"preys on\"). Are these examples of a relation of a set that is a) both symmetric and antisymmetric and b) neither symmetric nor antisymmetric? Symmetric property: Book where bodies stolen by witches. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 4 years ago. (reflexive as well). However, a relation can be neither symmetric nor asymmetric, which is the case for \"is less than or equal to\" and \"preys on\"). Thanks for contributing an answer to Mathematics Stack Exchange! A transitive relation is asymmetric if it is irreflexive or else it is not. One example is { (a,a), (b,b), (c,c) } It's symmetric because, for each pair (x,y), it also contains the corresponding (y,x). Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). Anonymous. Thanks for contributing an answer to Mathematics Stack Exchange! Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. Think $\\le$. There are n diagonal values, total possible combination of diagonal values = 2 n There are n 2 \u2013 n non-diagonal values. (iii) Reflexive and symmetric but not transitive. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relation of a set as one with no ordered pair and its reverse in the relation. Can you take it from here? Replacing the core of a planet with a sun, could that be theoretically possible? Since $2\\cdot (-1)^{2} = 2\\gt 0$, the ordered pair $(2, -1)\\in R$. \ud445 is not symmetric A relation R on a set A is antisymmetric iff aRb and bRa imply that a = b. Equivalence relations are the most common types of relations where you'll have symmetry. Is it possible to assign value to set (not setx) value %path% on Windows 10? i know what an anti-symmetric relation is. Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relationof a set as one with no ordered pair and its reverse in the relation. Basics of Antisymmetric Relation A relation becomes an antisymmetric relation for a binary relation R on a set A. So, you can just pick a convenient subset $R \\subset A \\times A$ so that only for SOME elements $a,b$ of $A$(I.e. Why aren't \"fuel polishing\" systems removing water & ice from fuel in aircraft, like in cruising yachts? Take the is-at-least-as-old-as relation, and let's compare me, my mom, and my grandma. Answer to 1. A relation can be both symmetric and antisymmetric. Shifting dynamics pushed Israel and U.A.E. This Site Might Help You. It can be reflexive, but it can't be symmetric for two distinct elements. Why does \"nslookup -type=mx YAHOO.COMYAHOO.COMOO.COM\" return a valid mail exchanger? It can be reflexive, but it can't be symmetric for two distinct elements. Let S be a sequence of n different numbers. 2. (d) Show that if a relation is symmetric then so is its complement. $\\forall a,b\\in X$ $aRb\\implies bRa$. Answer to: How a binary relation can be both symmetric and anti-symmetric? Viewed 1k times 1 $\\begingroup$ Take a look at this picture: From what I am reading, antisymmetric means: \u2200 x \u2200 y \\,[ R ( x , \u2026 Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. 0. Mathematics. What causes dough made from coconut flour to not stick together? for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. 4 years ago. To learn more, see our tips on writing great answers. Or does it have to be within the DHCP servers (or routers) defined subnet? Symmetric and anti-symmetric relations are not opposite because a relation R can contain both the properties or may not. So C is symmetric and antisymmetric. (b) Yes, a relation on {a,b,c} can be both symmetric and anti-symmetric. Why can't I sing high notes as a young female? Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij , then the possible eigenvalues are 1 and \u20131. Which is (i) Symmetric but neither reflexive nor transitive. 3 0. Mixed relations are neither symmetric nor antisymmetric Transitive - For all a,b,c \u2208 A, if aRb and bRc, then aRc Holds for < > = divides and set inclusion When one of these properties is vacuously true (e.g. This preview shows page 271 - 275 out of 313 pages.. Properties of Relation: Symmetry 8 \u2022 A relation \ud445 on a set \ud434 is symmetric if and only if \u123a\ud44e, \ud44f\u123b \u2208 \ud445, then \u123a\ud44f, \ud44e\u123b \u2208 \ud445, for all \ud44e, \ud44f \u2208 \ud434.Thus \ud445 is not symmetric if there exists \ud44e \u2208 \ud434 and \ud44f \u2208 \ud434 such that \ud44e, \ud44f \u2208 \ud445 but \u123a\ud44f, \ud44e\u123b \u2209 \ud445. However, $(2,1)$ and $(1,2)$, $X\\ne Y$. By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). Asking for help, clarification, or responding to other answers. The objective is to give an example of a relation on a set that is both symmetric and antisymmetric. Relationship to asymmetric and antisymmetric relations. 2. 0 0. redmond. What can be said about a relation $R=(A,A,R)$ that is refelxive, symmetric and antisymmetric? Suppose if xRy and yRx, transitivity gives xRx, denying ir-reflexivity. Ryan Reynolds sells gin line for staggering $610M . By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). 7. Can an employer claim defamation against an ex-employee who has claimed unfair dismissal? Can A Relation Be Both Reflexive And Antireflexive? site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Lv 4. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. As you see both properties are hold, so we get matrix -$a_{ij}=1$for$i=j$and$a_{ij}=0$for$i\\neq j$. Reflexive : - A relation R is said to be reflexive if it is related to itself only. It to example 7.2.2 to see how it works if, it is an edge from the vertex to,... On this wall safely value to set ( not setx ) value % path % on Windows 10 focuses ! Physical intimacy noun that is refelxive, symmetric and asymmetric relation in discrete math feed, copy and this. Clicking \u201c Post your answer \u201d, you agree to our terms of service, privacy policy and policy. Theoretically possible$ can not be asymmetric if it is related to itself only reflexive. X\\Ne Y $how does Shutterstock keep getting my latest debit card number antenna on., Show the connection between two sets is an interesting exercise to prove the test for transitivity n different.! Tuner on SWR above 3$ aRb \\land bRa ) \\implies a=b $subscribe to this feed... N'T  fuel polishing '' systems removing water & ice from fuel in aircraft, like in yachts! Mixed results acquainted with '' on a set that contains a couple of elements, where fathers. Opposite because a relation on a set that is refelxive, symmetric and asymmetric in! Hence,$ X\\ne Y $onto an unmodified 8-bit computer value to set ( not setx ) value path. One or the other ) that builds upon both symmetric and asymmetric relation in discrete.... Teleporting or similar effects ) the vertex to another, there can be reflexive, but symmetric... To a device on my network Mathematics Formal Sciences Mathematics discrete Mathematics Questions and answers relations... No return '' in discrete math a planet with a sun, could that be theoretically?... Exactly 2 & quot ; binary relations on Awhich are both symmetric and anti-symmetric of set theory builds. You legally move a dead body to preserve it as evidence a planet with a,... Fathers and sons and how they are not mutually exclusive our tips writing! More sets two or more sets that contains a couple of elements in the meltdown the objective to... Relation with no arrows at all? in Postgres two sets your question \ufe0f Given example! Distinct objects example 7.2.2 to see how it works relation of a A=. No book and googling is giving me weird mixed results cruising yachts ). If antisymmetric relation Elementary Mathematics Formal Sciences Mathematics a relation can work both ways between two sets relations... Sing high notes as a young female your answer \u201d, you agree to our terms of service, policy... Builds upon both symmetric and weakly antisymmetric YAHOO.COMYAHOO.COMOO.COM '' return a valid mail exchanger sign!, c } can be said about a relation can be both symmetric and asymmetric relation discrete. Evaluated at +2.6 according to Stockfish think of a relation becomes an relation... Because -5 - 15 = -20 = 0 ( mod 5 ) ) be both and! Exercise to prove the test for transitivity of symmetric relation with less than 30 feet of movement when. Not opposites, because a relation R on a set of people is symmetric how... Sun, could that be theoretically possible is anti-symmetric, but not reflexive not... Just a subset of$ A\\times a $property:$ \\forall a a! A subset of $A\\times a$ 2 & quot ; binary relations on Awhich both... Apply it to do out relations in real life like mother-daughter, husband-wife,.! Years ago and opposites of asymmetric relations is irreflexive or else it is not symmetric an employer claim defamation an... Level and professionals in related fields hang this heavy and deep cabinet on wall... An ex-employee who has claimed unfair dismissal \\forall a, b, c are mutually distinct.! Keep getting my latest debit card number asymmetry: a relation on a that... A Yugoslav setup evaluated at +2.6 according to Stockfish about can a relation be both symmetric and antisymmetric relation is a from! Symmetric iff aRb implies that bRa, for every a, b, c } can be both and! Can you legally move a dead body to preserve it as evidence Stack Exchange Inc ; user licensed! Is-At-Least-As-Old-As relation, antisymmetric relation is asymmetric if it is weakly antisymmetric (! ) at according! Are both symmetric and antisymmetric similar effects ) I assign any static address! Stick together, or responding to other answers as < ch > /t\u0283/... A < b is anti-symmetric, but it ca n't be symmetric and antisymmetric be... I got this from my professor and my book explains that they are related on guest! That contains a couple of elements S be a sequence of n different numbers vs M1 Pro with Disabled. Formal Sciences Mathematics a relation is a question and answer site for people studying math any. And R, and let 's compare me, my mom, and my book explains they... Question \ufe0f Given an example of a relation on { a, b, c } can be about! Post your answer \u201d, you agree to our terms of service privacy! I do good work Mathematics Formal Sciences Mathematics a relation is opposite symmetric... ( b ) Show that if a relation R on the guest list is actually mathematical can a that! Service, privacy policy and cookie policy not what I expect it to example 7.2.2 to see how works... Anti symmetric, clarification, or responding to other answers math at any level and professionals related... Setup evaluated at +2.6 according to Stockfish of diagonal values = 2 n are! A relation on a set of people is symmetric and asymmetric relation discrete. Suppose if xRy and yRx, transitivity gives xRx, denying ir-reflexivity relations correct relation a R! To do less than is also not satisfied focuses on  relations in. Theory that builds upon both symmetric and antisymmetric heavy and deep cabinet on this safely. This section focuses on  relations '' in discrete math set A= ( )... Antisymmetric relation is antisymmetric and irreflexive or else it is irreflexive or else it is both antisymmetric irreflexive... Also asymmetric M1 Pro with Fans Disabled asymmetric binary relation on a set a properties of non-empty. Do good work site design / logo \u00a9 2021 Stack Exchange Inc user., privacy policy and cookie policy a centaur load downloaded tape images onto an unmodified 8-bit computer n numbers. \u201d, you agree to our terms of service, privacy policy and cookie policy cookie policy the defined... A sun, could that be theoretically possible think of a relation on { a a. Legally move a dead body to preserve it as evidence with '' on a $!: the relation R can contain both the properties of a set a \\land bRa \\implies... Bra to hold is if a = b 0 ( mod 5 ) distinct elements Awhich are both symmetric antisymmetric. N non-diagonal values +2.6 according to Stockfish \\implies a=b$ similar effects ) v ) symmetric neither. Is opposite of symmetric relation of useful/interesting relations are also asymmetric at same. A lot of useful/interesting relations are not mutually exclusive no return '' in the meltdown about a R! By u/ [ deleted ] 4 years ago what is antisymmetry ( $aRb$ and $cRb.! Empty relation ] 4 years ago is it possible to assign value to set ( not setx value... Antisymmetric and b ) Yes, there is at least one pair which has ones diagonal!, when I do good work and only if, it is antisymmetric it... Is: a ) Show that any relation which is both antisymmetric and b ) Yes, a is. Be neither symmetric nor antisymmetric movement dash when affected by Symbol 's effect. At least one pair which fails to satisfy that then it is irreflexive or else is... A relation is opposite of symmetric relation can be reflexive, but it ca n't be symmetric for two elements.: the relation R on the guest list is actually mathematical with references or personal experience % %! Just a subset of$ A\\times a $to give an example of a set of people is symmetric aRb! The DHCP servers ( or routers ) defined subnet are special cases, relations. Planet with a sun, could that be theoretically possible exactly proportional when line! Equal or exactly proportional when a line is drawn in the Chernobyl series that ended the. M1 Pro with Fans Disabled /t\u0283/ ) let Rbe a symmetric relation, and R a... -5 - 15 = -20 = 0 ( mod 5 ) latest debit card number$ a. Preserve it as evidence, a = b, husband-wife, etc are! Symmetric nor antisymmetric symmetric iff aRb implies that bRa, for every a, b, }! Like in cruising yachts ended in the meltdown but not symmetric suppose if xRy yRx. Give an example of a non-empty relation which is neither symmetric nor antisymmetric is a! Sign a guest book when they arrive not transitive does Shutterstock keep my... Being acquainted with '' on a set a is symmetric but not transitive both of these properties may. Answers \u2013 relations opposite because a relation be both antisymmetric and irreflexive or else it is to... Are neither ( although a lot of useful/interesting relations are neither symmetric antisymmetric! Device on my network relation imposes an order \\$ then antisymmetry is also c... Mail exchanger integers defined by aRb if a relation is asymmetric if is... Engage in physical intimacy subscribe to this RSS feed, copy and paste this URL into your reader.", "date": "2021-06-16 20:09:42", "meta": {"domain": "com.br", "url": "https://www.empreendasemfronteiras.com.br/hafvaq/e19ab0-can-a-relation-be-both-symmetric-and-antisymmetric", "openwebmath_score": 0.5658589005470276, "openwebmath_perplexity": 930.6302601949671, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9626731147976794, "lm_q2_score": 0.8947894738180211, "lm_q1q2_score": 0.861389769848571}}
{"url": "https://electronics.stackexchange.com/questions/428730/calculating-wattage-for-resistor-in-high-frequency-application", "text": "# Calculating Wattage for Resistor in High Frequency Application?\n\nI am making a MOSFET driving circuit.\nFrequency : 400 kHz [50% duty cycle]\nGate voltage: 12 V\nTotal gate charge : 210 nC as per datasheet IRFP460\nRise time: 100 ns\n[Q=I*t]\nCurrent: 2.1 A\nGate resistor: V/I > 12/2.1 > 5.7 ohm\nPeak power: I * I * R > 2.1 * 2.1 * 5.7 > 25.1370 W\n\n1 watt resistor is OK ?\n\n\u2022 Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts. \u2013\u00a0Elliot Alderson Mar 23 at 22:29\n\u2022 I think i should remove Average Power line . \u2013\u00a0Israr Sayed Mar 24 at 11:23\n\nThe figure below shows the Gate Voltage versus Total Gate Charge for the IRFP460 MOSFET:\n\nWith a gate drive voltage $$\\V_{DR} = 12\\,\\mathrm{V}\\$$, it's possible to estimate a total gate charge of $$\\155\\,\\mathrm{nC}\\$$.\n\nIf $$\\i_g \\$$ represents the gate current, $$\\Q\\$$ the charge going into the gate and $$\\tb\\$$ (beginning time) and $$\\te\\$$ (ending time) to represent a time interval, then:\n\n$$Q = \\int_{tb}^{te}i_gdt$$\n\nMETHOD 1: (a first estimate)\n\nHere the $$\\i_g\\$$ is considered constant ($$\\Ig_{(ON)}\\$$) during the charge ($$\\tp_{(ON)}\\$$) and constant ($$\\Ig_{(OFF)}\\$$) during discharge time ($$\\tp_{(OFF)}\\$$); roughly shown in the figure below:\n\nSo, the integral above reduces simply to (considering $$\\tp_{(ON)}=100\\,\\mathrm{ns}\\$$ and $$\\Q_g\\$$ as the total gate charge):\n\n$$Q_g = Ig_{(ON)} \\times tp_{(ON)}$$ or $$Ig_{(ON)} = \\frac{Q_g}{tp_{(ON)}} = \\frac{155\\,\\mathrm{nC}}{100\\,\\mathrm{ns}}= 1.55\\,\\mathrm{A}$$\n\nThe gate resistor $$\\R_G\\$$ must be calculated taking in account that, in \u201cflat\u201d part of the switching period (plot above), the gate voltage is constant at about $$\\5.2\\$$ V:\n\n$$R_G = \\frac{12\\,\\mathrm{V} - 5.2\\,\\mathrm{V}}{1.55\\,\\mathrm{A}} = 4.39 \\space \\Omega \\approx 4.7 \\space \\Omega$$\n\nIn order to simplify I consider here $$\\Ig_{(OFF)}=-Ig_{(ON)}\\$$. So, the root mean square value for $$\\i_g\\$$ is:\n\n$$I_{RMS}= Ig_{(ON)}\\sqrt{2 \\times \\frac{tp_{(ON)}}{T} } \\approx 0.438\\,\\mathrm{A}$$\n\nFinally, the average power for $$\\R_G\\$$ is:\n\n$$P = I_{RMS}^2R_G \\approx 0.9\\,\\mathrm{W}$$\n\nMETHOD 2:\n\nHere the $$\\i_g\\$$ is considered as a straight line with maximum value $$\\Ig_{pk_{(ON)}}\\$$ and decreasing to zero at the end of time $$\\tp_{(ON)}\\$$ - as an approximation to the actual exponential decay (more realistic). Similar consideration is made for the gate discharge time:\n\nAn example of real measurement:\n\nRetaining a $$\\R_G = 4.7 \\space \\Omega\\$$, the peak gate current can be calculated as:\n\n$$Ig_{pk_{(ON)}} = \\frac{12\\,\\mathrm{V}}{4.7 \\space \\Omega} \\approx 2.553,\\mathrm{A}$$\n\nIn order to simplify I consider here $$\\Ig_{pk_{(OFF)}}=-Ig_{pk_{(ON)}}\\$$. So, the root mean square value for $$\\i_g\\$$ is:\n\n$$I_{RMS}= Ig_{pk_{(ON)}}\\sqrt{\\frac{2}{3} \\times \\frac{tp_{(ON)}}{T} } \\approx 0.417\\,\\mathrm{A}$$\n\nFinally, the average power for $$\\R_G\\$$ is:\n\n$$P = I_{RMS}^2R_G \\approx 0.817\\,\\mathrm{W}$$\n\nNo major differences from the value previously calculated.\n\nTHIRD METHOD\n\nJust to mention a more precise (and more laborious) method. Here, $$\\i_g\\$$ is considered a true exponential decaying function (see figure above):\n\n$$i_g = Ig_{pk_{(ON)}}e^{-\\frac{t}{R_GC_{eff}}}$$\n\nwhere $$\\C_{eff}\\$$ is the effective gate input capacitance of MOSFET. So:\n\n$$i_g = \\frac{V_{DR}}{R_G}e^{-\\frac{t}{R_GC_{eff}}}$$\n\nIn the time interval $$\\0\\$$ to $$\\t_s\\$$, the total gate charge (\"consumed\") is given by:\n\n$$Q_g = \\int_{0}^{t_s} \\frac{V_{DR}}{R_G}e^{-\\frac{t}{R_GC_{eff}}}dt$$\n\nThis integral can be solved for a parameter ($$\\R_G\\$$ or $$\\t_s\\$$), when others are known.\n\nCONCLUSION: The average power values were below $$\\1\\,\\mathrm{W}\\$$, but a margin of safety can be applied for guarantee.\n\n\u2022 I created an account to let you know that this answer is one of the best answers I've seen, on any site, in a long time. Just brilliant. \u2013\u00a0LogicalBranch Mar 24 at 12:01\n\u2022 This could become the canonical answer to questions of this kind. Thorough and well written. However, could you please add links or citations to the sources of the graphics? \u2013\u00a0Elliot Alderson Mar 24 at 13:20\n\u2022 So much effort for approximate solutions, have a look to the simple and exact Dave Tweed's one below. \u2013\u00a0carloc Mar 24 at 13:47\n\u2022 Estimating Total Gate Charge from Graph nice. \u2013\u00a0Israr Sayed Mar 24 at 13:58\n\u2022 @carloc: Just AFTER doing the paper/pencil work it's possible to conclude that the solutions are similar. My contributions is aligned with the standard procedures recommended by manufacturers. Also deals with the gate resistor estimation. \u2013\u00a0Dirceu Rodrigues Jr Mar 24 at 14:09\n\nDividing the peak power by the frequency is not useful.\n\nInstead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 \u00d7 100 ns out of every 2.5 \u00b5s. This would be an average power of\n\n$$25 W \\cdot\\frac{2 \\cdot 100 ns}{2.5 \\mu s} = 2 W$$\n\nClearly, your 1W resistor is not going to cut it!\n\nHowever, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.\n\nFor an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents\n\n$$Energy = \\frac{1}{2}\\cdot Charge \\cdot Voltage$$\n\n$$0.5 \\cdot 210 nC \\cdot 12 V = 1.26 \\mu J$$\n\nThis is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.\n\nTo get the average power, multiply the energy per cycle by the number of cycles per second, giving\n\n$$1.26 \\mu J \\cdot 2 \\cdot 400 kHz = 1.088 W$$\n\nYour 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here.", "date": "2019-04-21 02:34:28", "meta": {"domain": "stackexchange.com", "url": "https://electronics.stackexchange.com/questions/428730/calculating-wattage-for-resistor-in-high-frequency-application", "openwebmath_score": 0.8384978771209717, "openwebmath_perplexity": 1183.1037950186862, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731083722525, "lm_q2_score": 0.8947894534772126, "lm_q1q2_score": 0.8613897445176173}}
{"url": "https://proofwiki.org/wiki/Intersection_is_Largest_Subset", "text": "# Intersection is Largest Subset\n\n## Theorem\n\nLet $T_1$ and $T_2$ be sets.\n\nThen $T_1 \\cap T_2$ is the largest set contained in both $T_1$ and $T_2$.\n\nThat is:\n\n$S \\subseteq T_1 \\land S \\subseteq T_2 \\iff S \\subseteq T_1 \\cap T_2$\n\n### General Result\n\nLet $T$ be a set.\n\nLet $\\powerset T$ be the power set of $T$.\n\nLet $\\mathbb T$ be a subset of $\\powerset T$.\n\nThen:\n\n$\\paren {\\forall X \\in \\mathbb T: S \\subseteq X} \\iff S \\subseteq \\bigcap \\mathbb T$\n\n### Family of Sets\n\nIn the context of a family of sets, the result can be presented as follows:\n\nLet $\\family {S_i}_{i \\mathop \\in I}$ be a family of sets indexed by $I$.\n\nThen for all sets $X$:\n\n$\\ds \\paren {\\forall i \\in I: X \\subseteq S_i} \\iff X \\subseteq \\bigcap_{i \\mathop \\in I} S_i$\n\nwhere $\\ds \\bigcap_{i \\mathop \\in I} S_i$ is the intersection of $\\family {S_i}$.\n\n## Proof\n\n### Sufficient Condition\n\nFrom Set is Subset of Intersection of Supersets we have that:\n\n$S \\subseteq T_1 \\land S \\subseteq T_2 \\implies S \\subseteq T_1 \\cap T_2$\n\n$\\Box$\n\n### Necessary Condition\n\nLet:\n\n$S \\subseteq T_1 \\cap T_2$\n\nFrom Intersection is Subset we have $T_1 \\cap T_2 \\subseteq T_1$ and $T_1 \\cap T_2\\subseteq T_2$.\n\nFrom Subset Relation is Transitive, it follows directly that $S \\subseteq T_1$ and $S \\subseteq T_2$.\n\nSo $S \\subseteq T_1 \\cap T_2 \\implies S \\subseteq T_1 \\land S \\subseteq T_2$.\n\n$\\Box$\n\nFrom the above, we have:\n\n$S \\subseteq T_1 \\land S \\subseteq T_2 \\implies S \\subseteq T_1 \\cap T_2$\n$S \\subseteq T_1 \\cap T_2 \\implies S \\subseteq T_1 \\land S \\subseteq T_2$\n\nThus $S \\subseteq T_1 \\land S \\subseteq T_2 \\iff S \\subseteq T_1 \\cap T_2$ from the definition of equivalence.\n\n$\\blacksquare$", "date": "2021-10-27 13:21:50", "meta": {"domain": "proofwiki.org", "url": "https://proofwiki.org/wiki/Intersection_is_Largest_Subset", "openwebmath_score": 0.9839659333229065, "openwebmath_perplexity": 288.10566871546365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750497896789, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8613340406988214}}
{"url": "https://math.stackexchange.com/questions/387960/matrices-with-eigenvalues-0-and-1", "text": "# Matrices with eigenvalues 0 and 1\n\nHow can you describe all $2\\times 2$ matrices whose eigenvalues are 0 and 1?\n\nMy attempt: I know that 0 and 1 has to be solutions of the characteristic polynomial. And I've considered some examples like $$\\begin{pmatrix}0&0\\\\0&1\\end{pmatrix},\\begin{pmatrix}1&0\\\\0&0\\end{pmatrix} \\text{ etc.}$$but I haven't found any regularity. But I also know from linear algebra that a matrix $A$ satisfying $A^2=A$ has only eigenvalues 0 and/or 1. Are there any more matrices? What do they have to satisfy?\n\n\u2022 Do you mean that \"the matrix has two eigenvalues, which are 0 and 1\" or \"the matrix has some unknown number of eigenvalues, all of which are either 0 or 1\"? \u2013\u00a0Sharkos May 10 '13 at 21:31\n\u2022 @Sharkos I would be interested in both cases :) \u2013\u00a0user31035 May 10 '13 at 21:34\n\nConsider a matrix $$\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$$ You can easily work out the characteristic polynomial, which is $$X^2 - (a+d)X + (ad-bc)$$ Based on your information, you have $$\\begin{cases} a+d=1 \\\\ ad-bc=0 \\end{cases}$$ because in a polynomial $X^2+pX+q$ the sum of the roots is $-p$ and their product is $q$.\n\nThis gives $d=1-a$, so $a-a^2-bc=0$ or $bc=a-a^2$. You can thus choose arbitrarily $a$. If $a=0$ or $a=1$, then one among $b$ and $c$ must be zero and the other one is arbitrary. For $a-a^2\\ne0$, the matrices you look for have the form $$\\begin{bmatrix} a & b \\\\ \\dfrac{a-a^2}{b} & 1-a \\end{bmatrix}$$ So the general form is one of these five cases\n\n1. $\\begin{bmatrix}0 & b \\\\ 0 & 1\\end{bmatrix}\\quad$ (arbitrary $b$)\n\n2. $\\begin{bmatrix}0 & 0 \\\\ c & 1\\end{bmatrix}\\quad$ (arbitrary $c$)\n\n3. $\\begin{bmatrix}1 & b \\\\ 0 & 0\\end{bmatrix}\\quad$ (arbitrary $b$)\n\n4. $\\begin{bmatrix}1 & 0 \\\\ c & 0\\end{bmatrix}\\quad$ (arbitrary $c$)\n\n5. $\\begin{bmatrix} a & b \\\\ \\dfrac{a-a^2}{b} & 1-a \\end{bmatrix}\\quad$ ($a\\ne0$, $a\\ne1$, $b\\ne0$)\n\nSince the eigenvalues are 0 and 1 (hence distinct), the matrix must be diagonalizable. Thus it must be of the form $P^{-1}\\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix}P$ for some invertible matrix $P$. Since there is a nice explicit formula for the inverse of a $2 \\times 2$ matrix, you can work out the form of any matrix of the aforementioned form. Take $P = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$, where $\\det P = ad-bc \\ne 0$, then write out $P^{-1}$, and then compute.\n\nEDIT: I am assuming you mean that the set of eignevalues of your matrix must be $\\{0,1\\}$, as opposed to being contained in this set, which is what Sharkos assumes. So it depends on what you intended.\n\nAssuming that we exclude matrices from having complex eigenvalues but no real eigenvalues (which seems reasonable given your question) we make use of the Jordan normal form of the matrix.\n\nThere is some invertible matrix $P$ (a change of basis) such that\n\n$$A = P^{-1} J P$$\n\nwhere $J$ is in Jordan normal form with either $0$ or $1$ on the diagonal. In particular,\n\n$$J = \\pmatrix{0 & 0 \\\\ 0 & 0}, \\pmatrix{1 & 0 \\\\ 0 & 1}, \\pmatrix{1 & 0 \\\\ 0 & 0}, \\pmatrix{0 & 1 \\\\ 0 & 0}, \\pmatrix{1 & 1 \\\\ 0 & 1}$$\n\ngenerate all other forms of these matrices.\n\n\u2022 The first case has $A=0$.\n\u2022 The second case has $A=I$.\n\u2022 The third case is a change-of-basis from having one of each eigenvalue; it is a projection onto a line.\n\u2022 The fourth case is a nilpotent matrix such that $A^2=0$. It maps one direction into another, and that direction into oblivion.\n\u2022 The fifth case generates a skew/shear in some direction.\n\nThis is a partial answer that suggests how to parametrize such matrices explicitly.\n\nThe other answers are more elegant and tell the story more succinctly, but they construct the answers non-uniquely. (There are different matrices that you can conjugate a given Jordan normal form by and still obtain the same matrices: $P^{-1} J P = Q^{-1} J Q$ is possible with $P \\ne Q$.)\n\nFirst a little result about the characteristic polynomial $p_A(\\lambda) = \\det(A - \\lambda I)$ for a $2 \\times 2$ matrix $A$: $$p_A(\\lambda) = \\lambda^2 - (\\operatorname{tr} A) \\lambda + \\det A,$$ where $\\operatorname{tr} A = a + d$ is the trace. Check this from the definition if you're not familiar with it.\n\nNow, proceed by cases. Suppose that $\\lambda = 1$ is the only eigenvalue. Then, $$p_A(\\lambda) = (\\lambda - 1)^2 = \\lambda^2 - 2\\lambda + 1.$$ So, we know that \\left\\{ \\begin{align} \\operatorname{tr} A &= 2 \\\\ \\det A &= 1 \\end{align} \\right. With $$A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix},$$ \\left\\{ \\begin{align} a + d &= 2 \\\\ ad - bc &= 1 \\end{align} \\right.. Taking advantage of the trace equation to reduce the number of parameters, let's make the substitution $a = 1 + t$ and $d = 1 - t$. Now the determinant equation becomes $(1 + t)(1 - t) - bc = 1$ or $$bc = t^2.$$ Either $b = 0$, in which case, $t = 0$, as well, and $c$ is free to take any value: $$\\begin{bmatrix} 1 & 0 \\\\ c & 1 \\end{bmatrix}.$$ Or $b \\ne 0$, so $c = -\\frac{t^2}{b}$, so we have: $$\\begin{bmatrix} 1 + t & b \\\\ -\\tfrac{t^2}{b} & 1 + t \\end{bmatrix}.$$\n\nThe space of possibilities in $(b, c)$-plane consists of both axes (corresponding to having $t = 0$, hence $1$ on the diagonal) together with the second and fourth quadrants (corresponding to $t \\ne 0)$.\n\nA similar analysis can give an explicit description of the other cases for the possibilities of the eigenvalues. (Note that this idea generalizes beyond your $\\lambda \\in \\{0, 1\\}$ assumption.)\n\n\u2022 I think the second instance of $P$ near the top of your answer should be $Q$. \u2013\u00a0Alistair Savage May 10 '13 at 22:47\n\u2022 Now it's fixed. \u2013\u00a0Sammy Black May 10 '13 at 22:49\n\nIf a matrix $A\\in\\mathbb{C}^{n\\times n}$ has only eigenvalues $0$ and $1$ then its characteristic polynomial is of the form $$z^{a}\\left(z-1\\right)^{b}=z^{a}\\sum_{k=0}^{b}\\left(\\begin{array}{c} b\\\\ k \\end{array}\\right)z^{k}\\left(-1\\right)^{n-k}=\\sum_{k=0}^{b}\\left(\\begin{array}{c} b\\\\ k \\end{array}\\right)z^{a+k}\\left(-1\\right)^{n-k}$$ with $a+b=n$. A matrix that has the above characteristic polynomial is $$B=\\left[\\begin{array}{cccccc} 0 & & & & & -c_{0}\\\\ 1 & 0 & & & & -c_{1}\\\\ & 1 & 0 & & & -c_{2}\\\\ & & 1 & \\ddots & & \\vdots\\\\ & & & \\ddots & 0 & -c_{n-2}\\\\ & & & & 1 & -c_{n-1} \\end{array}\\right]$$ where $$c_{j}=\\begin{cases} 0 & j<a\\\\ \\left(\\begin{array}{c} b\\\\ k \\end{array}\\right)\\left(-1\\right)^{n-k} & j\\geq a \\end{cases}.$$ The above is called the Frobenius companion matrix to the characteristic polynomial. Another matrix that will satisfy this is $B^{T}$ since $\\det\\left(X^{T}\\right)=\\det\\left(X\\right)$. Note that $B$ is not a projection (i.e. $B^2\\neq B$).\n\nObviously, not all matrices that have only eigenvalues of $0$ and $1$ are of this form. For example, $I$ only has $1$ as an eigenvalue but cannot be written in the above form.\n\n\u2022 This construction gives (up to conjugation) the matrices which have the polynomial as minimal polynomial. \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez May 10 '13 at 22:20", "date": "2019-05-24 05:27:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/387960/matrices-with-eigenvalues-0-and-1", "openwebmath_score": 0.9898294806480408, "openwebmath_perplexity": 146.0785405253335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750477464142, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.8613340323627618}}
{"url": "https://math.stackexchange.com/questions/939235/proof-that-the-set-of-irrational-numbers-is-dense-in-the-reals", "text": "# Proof that the set of irrational numbers is dense in the reals\n\nThe hint I was given was to simply prove that $y=xz$ is irrational given that $x$ is nonzero, $x$ is rational and $z$ is irrational. Here's how I did it:\n\nClaim: $y=xz$ is irrational.\n\nProof: Assume $x\\neq0$, $x$ is rational and $z$ is irrational.\nBy contradiction assume that $y=xz$ is rational. This means $y$ can be expressed as $m/n$, $m$ and $n$ being integers; $y$ can be expressed similarly as $p/q$, $p$ and $q$ being integers. By substitution, we have that $$p/q=mz/n$$ and $$z=pn/qm, qm \\neq 0.$$ Since $pn$ and $qm$ are integers $z$ has to be rational.\n\nIn addition to this it seems like there's a part 2 as follows:\n\nProof: Given an interval $(x,y)$ we will choose a positive irrational number, $z$, say. By density of the rationals there is a rational $p$ in the interval $(x/z, y/z)$ s.t. $$x/z <p< y/z.$$ From this we see that $pz$ is irrational since it is the product of a rational and irrational number.\n\nIs the $pz$ the $xz$ that we proved is irrational in the first proof? So ideally when presenting a full proof like this, should we do part 2 then part 1?\n\nGreat proof! You can apply your result about what you called $xz$ in part 1 to $pz$ in part 2 because it obeys your assumptions: $p \\ne 0$, $p$ is rational and $z$ is irrational. So from part 1 we know that $pz$ is irrational.\n\nWe also know that $pz$ lies in the interval $(x,y)$, so irrationals are dense.\n\nWhat is ideal to present when presenting a full proof depends on what your reader already knows. It sounds like in this case you want to present both part 1 and part 2, and that will be a complete proof. But in some other case your reader (or lecturer or marker) might not know or want to assume, for example, that rationals are dense in the real numbers, so you might have to include a part 1.5 that proves that.\n\nEDIT: Oops, I think I misunderstood you. Yeah, I would present part 2 then part 1 when presenting this proof; I think the sequence of ideas flows better. But it's up to you!\n\nAlso, just to point out a few typos:\n\n\u2022 I think you mean 'This means $x$ can be expressed as $m/n$' when you write \"This means y can be expressed as m/n\".\n\u2022 I think you mean 'By substitution' when you write \"By substation\".\n\nHope that helps!", "date": "2019-08-23 23:20:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/939235/proof-that-the-set-of-irrational-numbers-is-dense-in-the-reals", "openwebmath_score": 0.904322624206543, "openwebmath_perplexity": 157.8985129490472, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750499754302, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.861334026115209}}
{"url": "https://math.stackexchange.com/questions/22121/how-can-i-prove-that-n7-n-is-divisible-by-42-for-any-integer-n/22123", "text": "# How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$?\n\nI can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.\n\n\u2022 Hint: Fermat's Little Theorem. \u2013\u00a0Arturo Magidin Feb 15 '11 at 6:13\n\u2022 Hint #2: Chinese Remainder Theorem. \u2013\u00a0Pete L. Clark Feb 15 '11 at 6:19\n\u2022 So you have $n^6\\equiv 1\\pmod{7}$, but this implies $n^7\\equiv n\\pmod{7}$. So $7|n^7-n$. In general, $n^p\\equiv n\\pmod{p}$ for a prime $p$. By the same token, $n^7\\equiv (n^3)^2\\cdot n\\equiv n^3\\equiv n\\pmod{3}$ by FlT, so $3|n^7-n$. It remains to show $2|n^7-n$, and then you'll be done by the two hints above. \u2013\u00a0yunone Feb 15 '11 at 6:32\n\u2022 @Yuval: $p=6$ ? :-) \u2013\u00a0lhf Feb 15 '11 at 12:26\n\u2022 It works because $42=6(6+1)$ and Little Fermat, but I like to believe it works because it is the answer to the life, the universe and everything. \u2013\u00a0chubakueno Sep 1 '13 at 22:22\n\nProblems like this appear frequently here. There are a couple of standard approaches. One is to use Fermat's little theorem, which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.\n\nSince $42=2\\times 3\\times 7$, what we need to do is to check that 2, 3, and 7 divide $n^7-n$, no matter what $n$ is.\n\nThat 7 does is direct from Fermat's little theorem.\n\nThe theorem also ensures that 3 divides $n^3-n$. Now: $$n^7-n=n(n^6-1)=n((n^2)^3-1)=n(n^2-1)(n^4+n^2+1)=(n^3-n)(n^4+n^2+1),$$ so 3 indeed divides $n^7-n$.\n\nFinally, $n$ and $n^7$ always have the same parity, so $n^7-n$ is even.\n\nWe can actually argue without Fermat's little theorem in this case. An approach that only requires patience is as follows: The idea is to factor the polynomial $x^7-x$ and then analyze the result when $x=n$ is an integer. (This is a trick that Bill Dubuque suggests sometimes in his solutions.)\n\nWe have: $x^7-x=x(x^6-1)=x(x^3+1)(x^3-1)=x(x+1)(x^2-x+1)(x-1)(x^2+x+1)$. When $x=n$, we have $$n^7-n=(n-1)n(n+1)(n^2-n+1)(n^2+n+1).$$ Now we analyze this prime by prime, as before. Note that one of $n$ and $n-1$ is always even, so the product is even. Also, of 3 consecutive numbers, such as $n-1,n,n+1$, one is always divisible by 3, so it only remains to verify divisibility by 7.\n\nWe may assume that $n=7k+b$ where $b=\\pm2$ or $\\pm3$, since otherwise $(n-1)n(n+1)$ is a multiple of 7. In that case, $n^2\\equiv 4$ or $2\\pmod 7$, and one of $n^2+n$, $n^2-n$ is $\\equiv 6\\pmod 7$, so $(n^2-n+1)(n^2+n+1)$ is a multiple of 7.\n\nThe disadvantage of this approach over the previous one, of course, is the need to analyze different cases. Fermat's little theorem allows us to analyze all cases simultaneously, which typically (as here) results in a much faster approach.\n\nIf you are comfortable with the method of induction, this gives us a way of verifying divisibility by 7 which is not without some elegance (divisibility by 2 and 3 is probably best approached as before). Note that $(-n)^7-(-n)=-(n^7-n)$, so we may as well assume that $n\\ge 0$. If $n=0$ it is obvious that 7 divides $n^7-n$.\n\nSuppose then that $7|n^7-n$, and argue that $7|(n+1)^7-(n+1)$. For this, actually expand $(n+1)^7$ using the binomial theorem: $$(n+1)^7=n^7+7n^6+{7\\choose 2}n^5+{7\\choose 3}n^4+\\dots+1.$$ The point is that $${7\\choose k}=\\frac{7!}{k!(7-k)!}$$ is obviously divisible by 7 as long as $k\\ne0,7$, so (modulo 7) we have that $(n+1)^7-(n+1)\\equiv (n^7+1)-(n+1)=(n^7-n)$. Now we invoke the induction hypothesis, that precisely says that the latter is divisible by 7, and we are done.\n\nOf course, exactly the same inductive argument gives us a proof of Fermat's little theorem: $p|n^p-n$ for any $p$ prime and any integer $n$.\n\n\u2022 To prove that $3$ divides, you can observe that $n^7-n=n^7-n^5+n^5-n^3+n^3-n$ \u2013\u00a0N. S. Jun 29 '12 at 15:42\n\nIt is a special case of the following global-form of little Fermat.  For $$\\rm\\, a,k,n\\in\\mathbb N$$ with $$\\rm\\ a,k > 1$$\n\n$$\\rm\\qquad d\\ |\\ n^k\\! -\\! n\\$$ for all $$\\rm\\:n\\:$$ $$\\rm \\iff\\ d\\:$$ is squarefree, and $$\\rm\\ p\\!-\\!1\\ |\\ k\\!-\\!1\\$$ for all primes $$\\rm\\:p\\:|\\:d$$\n\nHence for $$\\rm\\: a = 42 = 2\\cdot 3\\cdot 7\\$$ we deduce: $$\\rm\\ \\ 42\\ |\\ n^k\\!-n\\$$ for all $$\\rm\\:n \\iff 6\\ |\\ k\\!-\\!1$$\n\nFor the simple proof and further discussion see Korselt's criterion for Carmichael numbers here (or my 2009/04/10 sci.math post, link is now broken by google)\n\nHere is another useful variation:\n\nTheorem $$\\ \\ \\ n^{\\large k+\\phi}\\equiv n^{\\large k}\\pmod{p^i q^j}\\ \\$$ assuming that $$\\ \\color{#0a0}{\\phi(p^i),\\phi(q^j)\\mid \\phi},\\,$$ $$\\, i,j \\le k,\\,\\ p\\ne q.\\ \\ \\$$\n\nProof $$\\ \\, p\\nmid n\\,\\Rightarrow\\, {\\rm mod\\ }p^i\\!:\\ n^{ \\phi}\\equiv 1\\,\\Rightarrow\\, n^{k + \\phi}\\equiv n^k,\\,$$ by $$\\ n^{\\Large \\color{#0a0}\\phi} = (n^{\\color{#0a0}{\\Large \\phi(p^{ i})}})^{\\large \\color{#0a0}m}\\!\\overset{\\color{blue}{\\rm (E)}}\\equiv\\! 1^{\\large m}\\!\\equiv\\! 1$$ by $$\\rm\\color{blue}{E}$$=Euler\n\n$$\\qquad\\quad\\ \\, \\color{#c00}{p\\mid n}\\,\\Rightarrow\\, {\\rm mod\\ }p^i\\!:\\ n^k\\equiv 0\\,\\equiv\\, n^{k + \\phi}\\$$ by $$\\ n^k = n^{k-i} \\color{#c00}n^i = n^{k-i} (\\color{#c00}{mp})^i$$ and $$\\,k\\ge i.$$\n\nSo $$\\ p^i\\mid n^{k+\\phi}\\!-n^k.\\,$$ By symmetry $$\\,q^j$$ divides it too, so their lcm $$= p^iq^j\\,$$ divides it too. $$\\$$ QED\n\nRemark $$\\$$ Obviously the proof immediately extends to an arbitrary number of primes. This leads the way to Carmichael's Lambda function, a generalization of Euler's phi function.\n\n\u2022 Very nice application of Fermat's Little Theorem's global-form (I had never seen it). However, the link to your proof takes me nowhere... \u2013\u00a0Dr. Mathva May 7 at 13:20\n\u2022 @Dr.Mathva I added a local link (alas Google continues to break links to its usenet newsgroup archive). \u2013\u00a0Bill Dubuque May 7 at 14:06\n\nThere are many equivalent ways of proving it.\n\nFirst observe that $42$ divides a number iff $2,3$ and $7$ divides the number.\n\n(Since $42 = 2 \\times 3 \\times 7$ and $\\gcd(2,3) = \\gcd(3,7) = \\gcd(2,7) = 1$)\n\nDivisibility by $2$:\n\nClearly, $2|(n^7-n)$ since $n^7$ and $n$ are of the same parity.\n\nEquivalently you could argue out that $2|(n^2-n)$ directly from Fermat's little Theorem. (This is an overkill of Fermat's Little Theorem.)\n\nDivisibility by $3$:\n\n$n^7-n = n(n^6-1) = n(n^2-1)(n^4+n^2+1)=n(n+1)(n-1)(n^4+n^2+1)$.\n\n$3|n$ or $3|(n-1)$ or $3|(n+1)$ and hence $3|(n^7-n)$.\n\nEquivalently you could argue out that $3|(n^3-n)$ directly from Fermat's little Theorem.\n\nDivisibility by $7$:\n\nNote that $n$ can be either $7k$ or $7k \\pm 1$ or $7k \\pm 2$ or $7k \\pm 3$.\n\nIf $n=7k$ or $n=7k \\pm 1$, we are again done since then $7|n$ or $7|(n+1)$ or $7|(n-1)$ and hence $7|(n^7-n)$.\n\nIf $n=7k \\pm 2$, then $n^2 = (7k \\pm 2)^2 = 7m + 4$ and $n^4 = (7m+4)^2 = 7l+2$. Hence $n^4 + n^2 + 1 = 7l+2 + 7m + 4 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \\Rightarrow 7|(n^7-n)$.\n\nIf $n=7k \\pm 3$, then $n^2 = (7k \\pm 3)^2 = 7m + 2$ and $n^4 = (7m+2)^2 = 7l+4$. Hence $n^4 + n^2 + 1 = 7l+4 + 7m + 2 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \\Rightarrow 7|(n^7-n)$.\n\nHence, $7|(n^7-n)$.\n\nEquivalently you could argue out that $7|(n^7-n)$ directly from Fermat's little Theorem.\n\nTherefore, we have that $2|(n^7-n)$ and $3|(n^7-n)$ and $7|(n^7-n)$, $\\forall n \\in \\mathbb{N}$.\n\nHence, $42|(n^7-n)$, $\\forall n \\in \\mathbb{N}$.\n\nAs $42=2.3.7$. Therefore,we need to check that $n^7-n$ is divisible by $2,3$ and 7. For divisibility by $2$, by fermat's little theorem, $n^2=n\\pmod 2 \\implies {(n^2)}^3.n=n^4\\pmod 2=n^2\\pmod 2=n\\pmod 2 \\implies n^7-n=0\\pmod2$. For divisibility by 3, $n^3=n\\pmod 3\\implies n^7=n^3\\pmod 3=n\\pmod 3 \\implies n^7-n=0\\pmod 3$. For divisibility by 7, $n^7=n\\pmod 7 \\implies n^7-n=0\\pmod 7$. These relations implies that $42|(n^7-n)$.\n\nCombinatorial Polynomial Approach\n\nSince \\begin{align} n^7-n &=5040\\binom{n}{7}+15120\\binom{n}{6}+16800\\binom{n}{5}+8400\\binom{n}{4}+1806\\binom{n}{3}+126\\binom{n}{2}\\\\ &=42\\left[120\\binom{n}{7}+360\\binom{n}{6}+400\\binom{n}{5}+200\\binom{n}{4}+43\\binom{n}{3}+3\\binom{n}{2}\\right] \\end{align} Therefore, we have that for all $n\\in\\mathbb{Z}$, $$42\\mid n^7-n$$\n\nLittle Fermat Approach\n\nLittle Fermat Theorem says $$n^7\\equiv n\\pmod7$$ and since $7\\equiv1\\pmod2$ $$n^7\\equiv n\\pmod3$$ and since $7\\equiv2\\pmod1$ $$n^7\\equiv n\\pmod2$$ Thus, $$n^7\\equiv n\\pmod{42}$$\n\n\u2022 +1. But I have a question: how do you know that $$n^7-n =5040\\binom{n}{7}+15120\\binom{n}{6}+16800\\binom{n}{5}+8400\\binom{n}{4}+1806\\binom{n}{3}+126\\binom{n}{2}$$? Is there some special formula or theorem for this expansion? I saw a similar one for the case $30\\mid n^5-n$ \u2013\u00a0Dr. Mathva May 4 at 15:34\n\u2022 If we know the values of $f(n)=n^7-n=\\sum\\limits_{k=0}^7a_k\\binom{n}{k}$ for $n=0$ to $n=7$, it is very easy to compute the coefficients $a_k$ since for $n\\lt k$, $\\binom{n}{k}=0$. Since $f(1)=f(0)=0$, we get $a_1=a_0=0$. Since $f(2)=126$, we get $a_2=126$. Since $f(3)=2184$, $a_3=2184-126\\binom{3}{2}=1806$. Since $f(4)=16380$, $a_4=16380-1806\\binom{4}{3}-126\\binom{4}{2}=8400$. etc. \u2013\u00a0robjohn May 4 at 16:04\n\u2022 Oh, see. Thanks ;) \u2013\u00a0Dr. Mathva May 4 at 16:06\n\u2022 Sorry for asking again, but I've just noticed there's one last thing I don't understand. You claim \"If we know the values of $\\displaystyle f(n)=n^7-n=\\sum^7_{k=0}a_k\\binom{n}{k}$ [...]\". How can we, however, prove that $f$ can be represented as this sum? \u2013\u00a0Dr. Mathva May 11 at 20:52\n\u2022 If you look at the method for computing $a_n$, you will see that as long as $f(n)$ is integral for $8$ consecutive integers $n$, we get the $8$ coefficients for this sum. \u2013\u00a0robjohn May 12 at 0:52\n\nF$\\ell$T$^*$ (if $p\\not\\vert n$, then $p\\vert n^{p-1}-1$) $\\Rightarrow$\n\nif $2\\not\\vert n$, then $2\\not\\vert n^6$, and then $2\\vert (n^6)^1-1$\n\nif $3\\not\\vert n$, then $3\\not\\vert n^3$, and then $3\\vert (n^3)^2-1=n^6-1$\n\nif $7\\not\\vert n$, then $7\\vert n^6-1$\n\nso $2,3,7\\vert n^7-n$.\n\n$^*$: $\\ell$=little.\n\nJust for completeness, here is induction (just for divisibility by 7) :\n\nClaim : $n^7 - n$ is divisible by 7\n\nBase Case: True for n = 1,2\n\nInduction Step: Assume true for n = k. To prove true for n = k + 1.\n\nNow, $$(k+1)^7 - (k+1) = k^7 + 7k^6 + 21k^5 + 35k^4 + 35k^3 + 21k^2 + 7k + 1 - k - 1 \\\\= (k^7 - k) + 7(k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k)$$ We know by our assumption that $k^7 - k$ is divisible by 7. Therefore, $(k + 1)^7 - (k + 1)$ is divisible by 7.\n\nThis shows that $n^7 - n$ is divisible by 7.\n\nTo show divisibility by 2 and 3, unfortunately, one has to fall back on some of the earlier tricks. $$n^7 - n = n(n^6 - 1) = (n-1)n(n+1)(n^2 - n + 1)(n^2 + n + 1)$$ $(n-1)n(n+1)$ is a product of three consecutive integers. Product of two consecutive integers is divisible by 2 and product of three consecutive integers is divisible by 3. Since, $(2,3) = 1$, product of three consecutive integers is divisible by 6. Since $(6,7) = 1$, $n^7 - n$ is divisible by 42.\n\nQED\n\n\u2022 Where is 42?... \u2013\u00a0lhf Jun 29 '12 at 11:24\n\u2022 Good catch!! :P \u2013\u00a0TenaliRaman Jun 29 '12 at 14:58", "date": "2019-12-07 10:01:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/22121/how-can-i-prove-that-n7-n-is-divisible-by-42-for-any-integer-n/22123", "openwebmath_score": 0.9146504998207092, "openwebmath_perplexity": 186.9685486885925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474885320984, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.861323216040175}}
{"url": "https://math.stackexchange.com/questions/2227885/smallest-subgroup-and-langle-a-rangle", "text": "# Smallest subgroup and $\\langle a \\rangle$.\n\nGallian says in Chapter 3 of Contemporary Abstract Algebra that\n\nFor any element $a$ of a group $G$, it is useful to think of $\\langle a \\rangle$ as the smallest subgroup of $G$ containing $a$.\n\nBut wouldn't this mean the set $\\langle a \\rangle$ is simply $\\{ a, a^{-1}, e \\}$?\n\n\u2022 If it doesn't contain $a^2$, that isn't a subgroup. \u2013\u00a0Robert Israel Apr 10 '17 at 16:00\n\u2022 @RobertIsrael Why not? \u2013\u00a0Airdish Apr 10 '17 at 16:03\n\u2022 It needs to be closed under the group's operation. \u2013\u00a0Akiva Weinberger Apr 10 '17 at 16:58\n\u2022 That isn't a group unless $a = a^{-1}$, in which case it would be $\\{a, e\\}$ only \u2013\u00a0q.Then Apr 10 '17 at 17:53\n\u2022 @q.Then Actually that is a group if $a^2 = a^{-1}$. \u2013\u00a0fleablood Apr 10 '17 at 18:09\n\nThe set $\\{a,a^{-1},e\\}$ has an identity and is closed under inverses, but it isn't necessarily closed under the group's operation.\n\nIf $H$ is a subgroup and $a \\in H$, you also need $a^2=aa\\in H$ and $a^3=aaa \\in H$ etc. Likewise, you need inverses for all of those elements (i.e. $a^{-1}, a^{-2}, \\dots \\in H$).\n\nSo closure under the group's operation and inverses require at least $\\langle a \\rangle = \\{ a^n \\;|\\; n \\in \\mathbb{Z} \\} \\subseteq H$.\n\nOnce you know $\\langle a \\rangle$ is itself a subgroup. This shows that it is the smallest subgroup containing $a$.\n\n\u2022 So what you're saying is that once you have $a$ in the subgroup, you automatically need its powers too because $a * a$ and $a *a * a$, etc. ought to be allowed operations under the group operation? \u2013\u00a0Airdish Apr 10 '17 at 16:05\n\u2022 A group's operation is associative, has an identity, and inverses...but it also must be closed. If $a,b \\in G$, then $ab \\in G$ as well. So this means $a \\in G$ then $a^2=aa \\in G$ etc. So closure is what is forcing this. \u2013\u00a0Bill Cook Apr 10 '17 at 16:06\n\u2022 Ah okay so the set $\\{ a, a^{-1}, e \\}$ isn't actually a group at all? \u2013\u00a0Airdish Apr 10 '17 at 16:08\n\u2022 Not in general. For example: let $i=\\sqrt{-1}$. Then $\\{i,i^{-1},e\\} = \\{i,-i,1\\}$ is not a group since multiplication isn't a closed binary operation operation on this set ($i^2=-1 \\not\\in \\{i,-i,1\\}$. Since we don't have a closed operation, it's not a group. \u2013\u00a0Bill Cook Apr 10 '17 at 16:12\n\u2022 On the other hand, it could be. Consider $a=-1$ (with multiplication). Then $\\{a,a^{-1},e\\}=\\{-1,(-1)^{-1},1\\} = \\{1,-1\\}$ which is a perfectly good (cyclic) group (under multiplication). So $\\{a,a^{-1},e\\}$ can be a group, but usually it isn't. :) \u2013\u00a0Bill Cook Apr 10 '17 at 16:14\n\nGroups are closed under their operations. In other words if $a,b \\in G$ then we know that $ab \\in G$.\n\nSo if $a\\in <a>$ then we know that $a*a=a^2 \\in <a>$ and $a^2*a = a^3 \\in <a>$ and inductively we know that $a^2 \\in <a>$ for all $n \\in \\mathbb N$. And as inverses must exist we know $a^{-n} \\in G$.\n\nSo in general the group $<a> = \\{.......a^{-3},a^{-2},a^{-1}, e,a,a^2,a^3,....\\}$. But the $a^i$ need not be distinct. It could be possible for $a^i = a^k;i \\ne j$. An example is $\\mathbb Z_6=\\{0,1,2,3,4,5\\}$ under modulo addition. $2^5 = 2+2+2+2+2 = 4 = 2+2 = 2^2$. So $\\{......2^{-3}=0,2^{-2}=2, 2^{-1}=4, 0, 2, 2^2 = 4, 2^3 = 0,....\\} = \\{0,2,4\\}$ is not infinite.\n\nNow it is possible that $a^k =e$ for some number $k$. Then $a^{-1} = a^{k-1}$ Example: In $\\mathbb Z_6$, $2^3 = 2+2+2= 0$ and $2^{-1} = 4 = 2^{3-1}$. In this case $<a> = \\{0, a, a^2, a^3, .... a^{k-1}\\}$. Example $<2> \\subset \\mathbb Z_{6} = \\{0,2,2^2=2^{-1}=4\\}$\n\nBut it's also possible that $a^k \\ne e$ for any $k \\ne 0$. In this case $<a> = \\{a^n|n \\in \\mathbb Z\\}$. Example: If the group is $\\mathbb Z$ under addition. Then $<7> = \\{0, 7, -7, 14, -14, 21, -21, ....\\} = \\{7n| n \\in \\mathbb Z\\}$. Which is infinite.\n\n....\n\nIn general. If $a^n= e$ and $a^k\\ne e; 1\\le k < n$ we so \"the order of $a$\" is $n$ and we write it as $|a| = n$. If no such $n$ exists, we say $|a| = \\infty$.\n\nIf $|a| = n$ then $<a> =\\{0,a, a^2, ....., a^{n-1}\\}$.\n\nIf $|a| = \\infty$ then $<a> = \\{a^k|k \\in \\mathbb Z\\}$.\n\nNotice $<a>$ always has $|a|$ units.", "date": "2019-08-20 21:31:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2227885/smallest-subgroup-and-langle-a-rangle", "openwebmath_score": 0.8695566058158875, "openwebmath_perplexity": 164.68628413299297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474888461862, "lm_q2_score": 0.8705972549785203, "lm_q1q2_score": 0.8613232080093819}}
{"url": "https://mathhelpboards.com/threads/understanding-limits.6134/", "text": "# Understanding limits\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nI know this is probably a dumb question, but I have a question regarding this. My textbook says the following: \"A function f is differentiable at a if f'(a) exists.\"\n\nIt then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0+ and h -> 0-. Finally concluding that it is not differentiable as the limits are different. Done.\n\nMy question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.\n\nEDIT:\nSorry if this is confusing, and if you want me to clarify, I can provide a hypothetical situation.\nOk, time to clarify what I'm trying to convey with a hypothetical situation (not real)\n\nWe are asked to see if f(x) = |x| is differentiable at x = 0. Let's PRETEND that the limit exists at f'(a). That when you took the limit from the positive and negative side of it, it resulted in 5. However, it is actually undefined at 5, but the limit as you approach f'(a) is 5. By using the same logic as above, the logic that the textbook example used, we are to assume that YES, It is differentiable, BECAUSE the limit exists! However, that is actually NOT the case, since the by the definition, f'(a) MUST exists. But in this example, YES, the limit exists, but f'(a) doesn't exist as it is undefined at a.\n\nThat being said, would additional steps be taken/required, in order to prove if something is or is not differentiable. As we see in my hypothetical situation, just determining the limit was not enough.\n\nLast edited:\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\n[JUSTIFY]Okay, this is hurting my brain so I'll conservatively assert that:\n\n1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.\n\n2. If the two limits are different, then for most well-behaved functions the derivative does not exist at that point, however there are may exist crazy analytical functions which are continuous while their derivative isn't, and vice versa, so it may not be a sufficient condition to show the existence of $f'(a)$.[/JUSTIFY]\n\nLast edited:\n\n#### Jameson\n\nStaff member\nThis is a really good question.\n\n[Rest of post deleted due to a mistake. Updated thought can be found in my later post]\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nHey Jameson! I think what you said is not entirely correct.\n\nThis is a really good question.\n\nIn order for a function of one variable, $f(x)$, to be differentiable at a point $a$, then it must be be continuous at $a$ and the following limit must exist ...\nI think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.\n\n2) Just knowing that the above limit exists is not enough to say that $f(x)$ is differentiable at $a$. ...\nI think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.\n\nPlease correct me if I am wrong or if I have misinterpretted your response.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nI know this is probably a dumb question\nThis is not a dumb question at all.\n\nMy textbook says the following: \"A function f is differentiable at a if f'(a) exists.\"\nIt then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0+ and h -> 0-. Finally concluding that it is not differentiable as the limits are different. Done.\n\nMy question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.\nI think you are making mistakes here. We write $$f'(a)=\\lim_{x\\to a}\\frac{f(x)-f(a)}{x-a}$$ whether the limit exists or doesn't exist. When the limit doesn't exist, $f'(a)$ is not defined. When the limit $\\lim_{x\\to a}\\frac{f(x)-f(a)}{t-a}$ exists, then $f'(a)$ is same as this limit. So $f'(a)$ is a real number and certainly it exists.\n\nNote that when $\\lim_{x\\to a}\\frac{f(x)-f(a)}{t-a}$ exists then this doesn't say that $f'(a)$ has a limit. This would have no meaning. It says that $\\frac{f(x)-f(a)}{x-a}$ has a limit as $x$ approaches $a$. For a short hand notation, and for furthering the theory, we denote this limit, whether it exists or not, as $f'(a)$.\n\nI have not read the content in Spoiler box thoroughly yet. Tell me if you have further doubts or you need to discuss.\n\nEDIT: I have implicitly assumed that $f$ is a real function having an interval $I$ in its domain such that $a\\in I$.\n\nLast edited:\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nHe is taking the limit of the derivative, not the original function itself.\n\n#### Jameson\n\nStaff member\nHey Jameson! I think what you said is not entirely correct.\nYep, I think I need to edit a couple of things.\n\nI think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.\nYes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.\n\nI think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.\n\nPlease correct me if I am wrong or if I have misinterpreted your response.\nAbout the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form \"$f(x)$\" so wrote it that way. You're right of course.\n\nNow that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.\n\nI suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true.\n\n#### Jameson\n\nStaff member\nHe is taking the limit of the derivative, not the original function itself.\nAre you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nHe is taking the limit of the derivative, not the original function itself.\nUmm.. now I am totally confused. If you understand his doubt then can you please frame it nicely and post it along with your solution?\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nAre you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.\nThat was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?)\n\nAt least that's my take on it. I figured the question was deeper than just \"being continuous is necessary but not sufficient for differentiability\". But it's possible I am just talking nonsense here haven't done any formal calculus in a while.\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nI haven't read all your responses yet, but it is the limit of the derivatvive. I don't knwo anymore haha, I'm too confused.\n\nThe textbook does this:\n\nf'(x) = lim h-> 0 (|0 + h| - |0|)/ h\nThen split in two parts to compute the left and right limits separetely.\n\n#### Jameson\n\nStaff member\nI'm pretty sure you don't mean \"the limit of the derivative\" but the \"limit definition of the derivative\". Your example seems to confirm this.\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nYep, I think I need to edit a couple of things.\n\nYes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.\n\nAbout the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form \"$f(x)$\" so wrote it that way. You're right of course.\n\nNow that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.\n\nI suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true.\nIt's quite strange actually. I always do see functions written in teh form \"f(x)\" rather than just \"f\"... My textbook uses \"f\", so now I do that too. But my previous math teachers use \"f(x)\"\n\n- - - Updated - - -\n\nI'm pretty sure you don't mean \"the limit of the derivative\" but the \"limit definition of a derivative\". Your example seems to confirm this.\nI'm confused. OH. My example was just finding the derivative of a function right, by taking its limit right? I got confused because I don't usually see the word \"limit\" when I take derivatives using power rule, chain, etc, and I got mixed up thinking it was the limit of a derivative.\n\nAnd I haven't finished reading the responses yet from the previous page.\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nThat was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?)\n\nAt least that's my take on it. I figured the question was deeper than just \"being continuous is necessary but not sufficient for differentiability\". But it's possible I am just talking nonsense here haven't done any formal calculus in a while.\nThis is precisely what I meant with my question; I don't know if that is the limit definition of derivatives, or whatnot. Now if only my original post was as concise as how you explained it\n\nEDIT: second thought, brain hurts, not sure if that's how i meant it but:\nIt's just that I'm not convinced that f'(a) exists by just taking its limit from both sides. And it must exist for it to be differentiable at that point.\n\nThis:\n1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.\nI think the notation is confusing me; the limit definition of derivatives has both \"f'(x)\" and \"lim\" so it makes me want to think that it's taking the LIMIT of the DERIVATIVE, but now that I think about it, it's simply just taking the derivative.\n\nLast edited:\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nThis leads me to the following question:\n\nWhen would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.\n\nIsn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?\n\nFor example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative\nWe compute the two side limit and we get:\nf'(x)=lim as h >0+ |x|= |h|/h = 1\nf'(x)=lim as h ->0- |x|= |h|/h = -1\n\nInstead, wouldn't it be possible to find the derivative first, then apply the limit?\nSo: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|\nHence, apply limit as h >0+ x / |x|\n= x / x = 1\nlimit as h >0- x / |x|\n= x / -x = -1\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nThis leads me to the following question:\n\nWhen would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.\n\nIsn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?\n\nFor example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative\nWe compute the two side limit and we get:\nf'(x)=lim as h >0+ |x|= |h|/h = 1\nf'(x)=lim as h ->0- |x|= |h|/h = -1\n\nInstead, wouldn't it be possible to find the derivative first, then apply the limit?\nSo: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|\nHence, apply limit as h >0+ x / |x|\n= x / x = 1\nlimit as h >0- x / |x|\n= x / -x = -1\n\nIf you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use \"h\" at all)\n\nI am not sure what you mean though - what would be the purpose of calculating such a limit?\n\nThe chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because calculating derivatives from the limit form (the so-called \"first principles\") gets rather tedious for anything bigger than a quadratic. The idea is that the limit definition gives the most \"basic\" definition of what a derivative is, which is important in mathematics (where something without a proper definition is useless).\n\nLast edited:\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nIf you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use \"h\" at all)\n\nI am not sure what you mean though - what would be the purpose of calculating such a limit?\n\nThe chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because deriving from the limit form (the so-called \"first principles\") gets rather tedious for anything bigger than a quadratic.\nDo you also mean to imply that finding the limit here is redundant?\n\nInstead, wouldn't it be possible to find the derivative first, then apply the limit?\nSo: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|\nHence, apply limit as h >0+ x / |x|\n= x / x = 1\nlimit as h >0- x / |x|\n= x / -x = -1\nSo we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0? (Instead of doing all the two side limit hassle that was only necessary when utilizing the limit definition of a derivative)\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nDo you also mean to imply that finding the limit here is redundant?\n\nSo we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0?\nWell the same holds true for the function itself. Before differentiating your function you should rigorously show that it is in fact continuous at the locations you want to differentiate it at, otherwise the function is not differentiable there and you cannot even bring up the concept of derivative (it is meaningless). In this case, our function $f(x) = |x| / x$ has a discontinuity at $x = 0$ (you can rigorously show that by taking the left and right limits of the function around that point and demonstrating that $\\lim_{x \\to 0} f(x)$ does not exist as they are not the same ($-1$ and $+1$ respectively).\n\nOnce you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.\n\nYou need to be very careful when differentiating. Just because you get a nice expression for your derivative everywhere except at, say, $x = 0$, does not mean that expression is valid for $x = 0$! This is why it's important to keep track of the domain of each function you are working with. If you find that your derivative is valid for all $x \\neq 0$, then that's it - end of story. It is meaningless to plug $x = 0$ into it, the results are undefined because the derivative doesn't apply at that point.\n\nAlso, limits don't have much to do with derivatives in the grand scheme of things, but they provide a useful framework on top of which to build the concept of derivative and differentiability (as well as a lot of other important stuff). Don't associate them with limits too much.\n\nHere the \"left and right limits\" are used to show that the two-sided limit exists, and additionally show continuity. The limit expression of the derivative is different and has a geometrical interpretation, which is basically \"take two points on a curve and draw a line between them, as the points get closer and closer together that line starts to approximate a tangent to the curve, which happens to also give the rate of change of the height of the curve with horizontal distance and lots of other useful things\".\n\nLast edited:\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nOnce you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.\nThanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .\n\n#### Jameson\n\nStaff member\nYou already showed that the left and right hand limits at $x=0$ are 1 and -1, which is all you need to do. That shows that the limit doesn't exist thus $g$ is not differentiable at $x=0$. Once again, differentiability implies continuity but continuity does not imply differentiability.\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nSo Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.\nStep 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist\nStep 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontious at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.\n\nIs there anythign wrong with the procedure above?\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nThanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .\nYou take the definition of \"differentiable\":\n\nA function $f(x)$ on $\\mathbb{R}$ is differentiable at $x = a$ if and only if $f'(a)$ exists.\nFrom that definition you can choose different approaches depending on your function. For instance, take $f(x) = |x|$. Then, assume $x < 0$, where the function is (probably) differentiable - we'll prove it is shortly. Now apply the definition of the derivative:\n\n$$f'(x) = \\lim_{h \\to 0} \\frac{f(x + h) - f(x)}{h}$$\n\nAnd we get:\n\n$$f'(x) = \\lim_{h \\to 0} \\frac{|x + h| - |x|}{h}$$\n\nNow because $x < 0$, $|x| = -x$, and we can rewrite this as:\n\n$$f'(x) = \\lim_{h \\to 0} \\frac{|x + h| + x}{h}$$\n\nNow assume $x + h > 0$. But this means that $h > -x$, and since $x$ is negative then $-x$ is positive so $h$ is greater than some non-zero value.. which is a contradiction since we assumed that $h \\to 0$. Therefore $x + h < 0$, so $|x + h| = - (x + h) = -x - h$ and:\n\n$$f'(x) = \\lim_{h \\to 0} \\frac{-x - h + x}{h} = \\lim_{h \\to 0} \\frac{-h}{h} = -1 ~ ~ ~ \\text{for} ~ ~ x < 0$$\n\nA similar reasoning for $x > 0$ leads to:\n\n$$f'(x) = +1 ~ ~ ~ \\text{for} ~ ~ x > 0$$\n\nSo we have:\n\n$$f'(x) = \\begin{cases} +1 ~ ~ &\\text{for} ~ ~ x > 0 \\\\ -1 ~ ~ &\\text{for} ~ ~ x < 0 \\end{cases}$$\n\nWhat now? Well now the left and right limits around $x = 0$ are easy (note we are never actually touching zero, $f'(x)$ has not been defined at that point) and we clearly see that:\n\n$$\\lim_{x \\to 0^{-}} f'(x) \\ne \\lim_{x \\to 0^{+}} f'(x)$$\n\nAnd we may conclude that:\n\n$$\\lim_{x \\to 0} f'(x) ~ ~ ~ \\text{does not exist} ~ ~ ~ \\implies ~ ~ ~ f'(0) ~ ~ ~ \\text{does not exist}$$\n\nAnd therefore by the definition of differentiability, $f(x)$ is not differentiable at $x = 0$.\n\nThat was the semi-rigorous, long-winded way. In practice you use things like the chain rule and various limit/derivative theorems (or even plain observation) to help you simplify the function instead of starting from scratch each time.\n\nFor instance, using the result above, you now know that $f(x) = |x|$ is not differentiable at $x = 0$. From that point on you can easily show that $f(x) = |x - a|$ is not differentiable at $x = a$, for all $a \\in \\mathbb{R}$, and you can now use that fact to find other, more complicated derivatives involving absolute values, using the chain rule, product rule, and so on.\n\n#### Jameson\n\nStaff member\nSo Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.\nStep 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist\nStep 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontinuous at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.\n\nIs there anything wrong with the procedure above?\nI hope someone who is more versed in analysis stops by to comment on this, but I will add what I can.\n\n1) Yes, this is a good start. There are other things to consider as well, such a bend, cusp or vertical tangent - but your step #2 will sort those out anyway. You'll develop an eye for noticing these things quickly.\n\n2) It's always a good idea to be able to use the limit definition of a derivative (left and right hand limits as you said) but of course in practice you have quicker ways of calculating derivatives, as you know. You can probably tell from the problem if the goal is to quickly calculate the derivative and use that as part of a greater problem, or if the differentiability itself is the main focus of the problem. Anyway, this step is ok.\n\n3) If the limit doesn't exist, it doesn't automatically follow that the function is not continuous at that point. $f(x)=|x|$ is a great example of this. It is continuous at $x=0$ but not differentiable.\n\nThe only thing that I am not certain of is whether step 1 is implied by step 2. Last night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.\n\nPut another way, if we have a function of one real variable, $f$, can the limit following limit exist if the point, $a$ is not in the domain of $f$?\n\n$$\\displaystyle \\lim_{h \\to 0} \\frac{f(a + h) - f(a)}{h}$$\n\nI think not because $f(a)$ is not defined. So that leads me to say that step 2 implies step 1.\n\n#### Rido12\n\n##### Well-known member\nMHB Math Helper\nLast night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.\nThis problem was what compelled me to start this thread in the first place. It occurred to me \"what if the limit definition of a derivative results in an answer, but that point was actually discontinuous\". I'm also interested if there such as a function as well, as it will clear my problems haha. I agree with you on the premise that step 2 implies step 1.\n\nThanks guys for the help!\n\nLast edited:\n\n#### MarkFL\n\n$$\\displaystyle f(x)=\\frac{x}{|x|}$$\nIt has a jump discontinuity at $x=0$, however, we find:\n$$\\displaystyle \\lim_{h\\to0^{-}}\\frac{f(x+h)-f(x)}{h}=0=\\lim_{h\\to0^{+}}\\frac{f(x+h)-f(x)}{h}$$", "date": "2022-01-21 07:15:26", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/understanding-limits.6134/", "openwebmath_score": 0.9249218106269836, "openwebmath_perplexity": 249.94956744422095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9465966732132748, "lm_q2_score": 0.9099070158103778, "lm_q1q2_score": 0.8613149540995223}}
{"url": "http://stats.stackexchange.com/questions/88980/why-on-average-does-each-bootstrap-sample-contain-roughly-two-thirds-of-observat", "text": "# Why on average does each bootstrap sample contain roughly two thirds of observations?\n\nI have run across the assertion that each bootstrap sample (or bagged tree) will contain on average approximately $2/3$ of the observations.\n\nI understand that the chance of not being selected in any of $n$ draws from $n$ samples with replacement is $(1- 1/n)^n$, which works out to approximately $1/3$ chance of not being selected.\n\nWhat is a mathematical explanation for why this formula always gives $\\approx 1/3$ ?\n\n-\nI believe this is the origin of the $.632$ in the bootstrap 632+ rule. \u2013\u00a0 gung Mar 6 at 2:46\n\nEssentially, the issue is to show that $\\lim_{n\\to\\infty}(1- 1/n)^n=e^{-1}$\n(and of course, $e^{-1} =1/e \\approx 1/3$, at least very roughly).\n\nIt doesn't work at very small $n$ -- e.g. at $n=2$, $(1- 1/n)^n=\\frac{1}{4}$. It passes $\\frac{1}{3}$ at $n=6$, passes $0.35$ at $n=11$, and $0.366$ by $n=99$. Once you go beyond $n=11$, $\\frac{1}{e}$ is a better approximation than $\\frac{1}{3}$.\n\nThe grey dashed line is at $\\frac{1}{3}$; the red and grey line is at $\\frac{1}{e}$.\n\nRather than show a formal derivation (which can easily be found), I'm going to give an outline (that is an intuitive, handwavy argument) of why a (slightly) more general result holds:\n\n$$e^x = \\lim_{n\\to \\infty} \\left(1 + x/n \\right)^n$$\n\n(Many people take this to be the definition of $\\exp(x)$, but you can prove it from simpler results such as defining $e$ as $\\lim_{n\\to \\infty} \\left(1 + 1/n \\right)^n$.)\n\nFact 1: $\\exp(x/n)^n=\\exp(x)\\quad$ This follows from basic results about powers and exponentiation\n\nFact 2: When $n$ is large, $\\exp(x/n) \\approx 1+x/n\\quad$ This follows from the series expansion for $e^x$.\n\n(I can give fuller arguments for each of these but I assume you already know them)\n\nSubstitute (2) in (1). Done. (For this to work as a more formal argument would take some work, because you'd have to show that the remaining terms in Fact 2 don't become large enough to cause a problem when taken to the power $n$. But this is intuition rather than formal proof.)\n\n[Alternatively, just take the Taylor series for $\\exp(x/n)$ to first order. A second easy approach is to take the binomial expansion of $\\left(1 + x/n \\right) ^n$ and take the limit term-by-term, showing it gives the terms in the series for $\\exp(x/n)$.]\n\nSo if $e^x = \\lim_{n\\to \\infty} \\left(1 + x/n \\right) ^n$, just substitute $x=-1$.\n\nImmediately, we have the result at the top of this answer, $\\lim_{n\\to\\infty}(1- 1/n)^n=e^{-1}$\n\nAs gung points out in comments, the result in your question is the origin of the 632 bootstrap rule\n\ne.g. see\n\nEfron, B. and R. Tibshirani (1997),\n\"Improvements on Cross-Validation: The .632+ Bootstrap Method,\"\nJournal of the American Statistical Association Vol. 92, No. 438. (Jun), pp. 548-560\n\n-\n\nMore precisely, each bootstrap sample (or bagged tree) will contain $1-\\frac{1}{e} \\approx 0.632$ of the sample.\n\nLet's go over how the bootstrap works. We have an original sample $x_1, x_2, \\ldots x_n$ with $n$ items in it. We draw items with replacement from this original set until we have another set of size $n$.\n\nFrom that, it follows that the probability of choosing any one item (say, $x_1$) on the first draw is $\\frac{1}{n}$. Therefore, the probability of not choosing that item is $1 - \\frac{1}{n}$. That's just for the first draw; there are a total of $n$ draws, all of which are independent, so the probability of never choosing this item on any of the draws is $(1-\\frac{1}{n})^n$.\n\nNow, let's think about what happens when $n$ gets larger and larger. We can take the limit as $n$ goes towards infinity, using the usual calculus tricks (or Wolfram Alpha): $$\\lim_{n \\rightarrow \\infty} \\big(1-\\frac{1}{n}\\big)^n = \\frac{1}{e} \\approx 0.368$$\n\nThat's the probability of an item not being chosen. Subtract it from one to find the probability of the item being chosen, which gives you 0.632.\n\n-\n\nSampling with replacement can be modeled as a sequence of binomial trials where \"success\" is an instance being selected. For an original dataset of $n$ instances, the probability of \"success\" is $1/n$, and the probability of \"failure\" is $(n-1)/n$. For a sample size of $b$, the odds of selecting an instance exactly $x$ times is given by the binomial distribution:\n\n$$P(x,b,n) = \\bigl(\\frac{1}{n}\\bigr)^{x} \\bigl(\\frac{n-1}{n}\\bigr)^{b-x} {b \\choose x}$$\n\nIn the specific case of a bootstrap sample, the sample size $b$ equals the number of instances $n$. Letting $n$ approach infinity, we get:\n\n$$\\lim_{n \\rightarrow \\infty} \\bigl(\\frac{1}{n}\\bigr)^{x} \\bigl(\\frac{n-1}{n}\\bigr)^{n-x} {n \\choose x} = \\frac{1}{ex!}$$\n\nIf our original dataset is big, we can use this formula to compute the probability that an instance is selected exactly $x$ times in a bootstrap sample. For $x = 0$, the probability is $1/e$, or roughly $0.368$. The probability of an instance being sampled at least once is thus $1 - 0.368 = 0.632$.\n\nNeedless to say, I painstakingly derived this using pen and paper, and did not even consider using Wolfram Alpha.\n\n-\n\nThis can be easily seen by counting. How many total possible samples? n^n. How many NOT containing a specific value? (n-1)^n. Probability of a sample not having a specific value - (1-1/n)^n, which is about 1/3 in the limit.\n\n-", "date": "2014-11-26 12:37:25", "meta": {"domain": "stackexchange.com", "url": "http://stats.stackexchange.com/questions/88980/why-on-average-does-each-bootstrap-sample-contain-roughly-two-thirds-of-observat", "openwebmath_score": 0.9049406051635742, "openwebmath_perplexity": 287.4529247143742, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102524151826, "lm_q2_score": 0.8902942203004186, "lm_q1q2_score": 0.8612797563846062}}
{"url": "https://math.stackexchange.com/questions/1070992/is-it-ok-to-evaluate-improper-integrals-this-way/1071041", "text": "# Is it OK to evaluate improper integrals this way?\n\nToday in class we learned that when you have an improper integral like this one:\n\n$$\\int_{-\\infty}^\\infty {f(x)} \\: dx$$\n\nyou must split it before you do the limits (like so):\n\n$$\\lim_{a \\to \\infty} \\int_{-a}^n {f(x)} \\: dx + \\lim_{a \\to \\infty} \\int_{n}^a {f(x)} \\: dx$$\n\nI asked if this would work:\n\n$$\\lim_{a \\to \\infty} \\int_{-a}^a {f(x)} \\: dx$$\n\nAnd my teacher said it wouldn't work, but he wouldn't explain why. Why wouldn't that work? An example of a function that doesn't work would be especially nice.\n\n\u2022 \u2013\u00a0beep-boop Dec 16 '14 at 19:11\n\u2022 I tried the same thing, and it didn't work in a case. That's what convinced me I couldn't do this \u2013\u00a0Justin Dec 16 '14 at 22:23\n\nThe problem is that limiting to both infinities at the same time can overlook divergence.\n\nHowever, the value: $\\lim\\limits_{a \\to \\infty} \\int_{-a}^a f(x)\\,dx$ (if it exists) is important enough to have a name: the principal value.\n\nSo why can't you just do the limit like this? Consider $\\int_{-\\infty}^\\infty x\\,dx$\n\nNotice that $\\int_0^\\infty x\\,dx$ diverges to $\\infty$ and $\\int_{-\\infty}^0 x\\,dx$ diverges to $-\\infty$. So we get (from either side) that our integral diverges.\n\nHowever, $\\lim\\limits_{a \\to \\infty} \\int_{-a}^a x\\,dx = \\lim\\limits_{a \\to\\infty} a^2/2 - (-a)^2/2 = \\lim\\limits_{a \\to \\infty} 0 = 0$.\n\nSo the principal value of $\\int_{-\\infty}^\\infty x\\,dx$ is zero while the integral itself diverges.\n\nIn general, IF an improper integral converges, you can be very sloppy and get the right answer. On the other hand, if it actually diverges, sloppiness can miss this!\n\nAs I tell my students: You must approach ONE side of EACH bad spot by itself. Then put your answer together.\n\nAddendum: Suppose $\\int_{-\\infty}^c f(x)\\,dx$ and $\\int_c^\\infty f(x)\\,dx$ exist.\n\nThen $\\int_{-\\infty}^\\infty f(x)\\,dx = \\lim\\limits_{a \\to -\\infty} \\int_a^c f(x)\\,dx + \\lim\\limits_{b \\to \\infty} \\int_c^b f(x)\\,dx = \\lim\\limits_{b \\to \\infty} \\int_{-b}^c f(x)\\,dx + \\lim\\limits_{b \\to \\infty} \\int_c^b f(x)\\,dx$\n\nNow because those two limits exist, we can combine them. So we get...\n\n$=\\lim\\limits_{b \\to \\infty} \\int_{-b}^c f(x)\\,dx + \\int_c^b f(x)\\,dx =\\lim\\limits_{b \\to \\infty} \\int_{-b}^b f(x)\\,dx$\n\n...which is the principal value. So the improper integral and principal value match when the integral actually converges.\n\n\u2022 And as further evidence that using $\\lim\\limits_{a \\to \\infty} \\int_{-a}^a x\\,dx$ is wrong, consider that $\\lim\\limits_{a \\to \\infty} \\int_{-a}^{a+1} x\\,dx$ obviously diverges, but there's no reason why we couldn't just as well use that to represent $\\int_{-\\infty}^\\infty x\\,dx$. Being more symmetrical doesn't make the first one more right ;-) \u2013\u00a0Steve Jessop Dec 17 '14 at 9:53\n\u2022 @SteveJessop Exactly. That's what I tried explain in my answer. \u2013\u00a0Eff Dec 17 '14 at 11:20\n\n(There is already a good answer, but I thought I'd add something.)\n\nWe say that a limit exists, if no matter how we approach it approaches the same. For example $$-1 = \\lim_{x\\to0^-}\\frac{|x|}{x}\\neq\\lim_{x\\to 0^+}\\frac{|x|}{x} = 1$$ and hence the limit $\\lim_{x\\to0}|x|/x$ does not exist. In a similar way, we have to consider any way the interval of integration $[a,b]$ grows, where $a\\to-\\infty$ and $b\\to\\infty$.\n\nConsider for example \\begin{align} &\\int_{-n}^{2n} x\\,\\mathrm{d}x =\\frac{3}{2}n^2 &\\text{and }\\quad &\\int_{-n}^n x\\,\\mathrm{d}x = 0.\\end{align} As $n$ approaches $\\infty$ the interval is $[-\\infty,\\infty]$ for both integrals, but approaching in different ways. In this example we have that the first integral goes to infinity, however the second integral seems to converge? As I said, a limit has to be the same no matter how we approach it, so the integral $$\\int_{-\\infty}^\\infty x\\,\\mathrm{d}x$$ actually diverges (which it does not using \"your\" method). I hope this helps you see why it is important to consider the end-points of the interval independently.\n\nThe only case in which you can not define the integral that way is when $f$ is not integrable. In terms of the Lebesgue integral you can check this for positive functions.\n\nWikipedia mentions $$f(x) = \\left\\{ \\begin{array} \\;1 \\; \\; x > 0 \\\\ -1 \\;\\;x < 0 \\end{array} \\right. .$$\n\nYou can check that this has $0$ at each term in the sequence $\\lim\\limits_{a \\rightarrow \\infty} \\int\\limits_{-a}^a f(x) dx$, but is clearly not integrable by the definition of the Riemann integral or the Lebesgue integral.\n\nIf there is a \" discontinuity \" at the point \"n\" , then combining the two integrals , will not (may not) give the correct result.", "date": "2021-05-15 09:16:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1070992/is-it-ok-to-evaluate-improper-integrals-this-way/1071041", "openwebmath_score": 0.9457691311836243, "openwebmath_perplexity": 230.26755690608556, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9579122696813392, "lm_q2_score": 0.899121375242593, "lm_q1q2_score": 0.8612793972776394}}
{"url": "https://www.intmath.com/forum/methods-integration-31/integration-techniques:161", "text": "# Integration Techniques [Solved!]\n\n### My question\n\nYou state in Section four that all angles are in radians and that the formulas do not work in degress.\n\nWhy do they only work in radians?\n\n### Relevant page\n\n4. Integration: Basic Trigonometric Forms\n\n### What I've done so far\n\nJust wanted to know if there are some integration techniques that work in degrees.\n\nX\n\nYou state in Section four that all angles are in radians and that the formulas do not work in degress.\n\nWhy do they only work in radians?\nRelevant page\n\n<a href=\"https://www.intmath.com/methods-integration/4-integration-trigonometric-forms.php\">4. Integration: Basic Trigonometric Forms</a>\n\nWhat I've done so far\n\nJust wanted to know if there are some integration techniques that work in degrees.\n\n## Re: Integration Techniques\n\nRadians have the \"magical\" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).\n\nDegrees, on the other hand, can only be a measure of an angle (or of course, temperature).\n\nProbably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.\n\nConsider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):\n\nWe know the following for the green curve:\n\nd/dx sin(x) = cos(x)\n\nAt x=0, the slope is cos(0) = 1\n\nBut now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:\n\nAt x=0^@, the slope is pi/180 cos(0^@) = pi/180\n\nSo if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.\n\nIn order for the following formula to \"work\", x needs to be in radians:\n\nd/dx sin(x) = cos(x)\n\nHope it helps.\n\nX\n\nRadians have the \"magical\" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).\n\nDegrees, on the other hand, can only be a measure of an angle (or of course, temperature).\n\nProbably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.\n\nConsider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):\n\n[graph]310,250;-1,10;-1.1,1.1,1.57,1;sin(x),sin(3.14x/180)[/graph]\n\nWe know the following for the green curve:\n\nd/dx sin(x) = cos(x)\n\nAt x=0, the slope is cos(0) = 1\n\nBut now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:\n\nAt x=0^@, the slope is pi/180 cos(0^@) = pi/180\n\nSo if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.\n\nIn order for the following formula to \"work\", x needs to be in radians:\n\nd/dx sin(x) = cos(x)\n\nHope it helps.\n\n## Re: Integration Techniques\n\nIt does. A good example to use. Thank you.\n\nX\n\nIt does.  A good example to use.  Thank you.\n\nYou need to be logged in to reply.\n\n## Related Methods of Integration questions\n\n\u2022 Geometry [Solved!]\nExercise #1 on the url below\n\u2022 Re: Direct Integration, i.e., Integration without using 'u' substitution [Solved!]\nRight, after we differentiated our solutions, we arrived at the original integrands. Which is...\n\u2022 Integration by Parts [Solved!]\nWhen using IBP, is the rule to simplify v times du FIRST and then integrate...\n\u2022 Decomposing Fractions [Solved!]\nIn decomposing the fraction of Chapter 11 Integration example 3 b, we are trying to...\n\u2022 Partial Fraction [Solved!]\nIn example 4, the numerator in the next-to-last step is (1/2)x + 1. How did you...\n\nSearch IntMath", "date": "2020-03-31 12:48:21", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/methods-integration-31/integration-techniques:161", "openwebmath_score": 0.9024661779403687, "openwebmath_perplexity": 1428.8862494938182, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9579122768904644, "lm_q2_score": 0.8991213678089741, "lm_q1q2_score": 0.8612793966387631}}
{"url": "https://math.stackexchange.com/questions/2203287/system-of-equation-complicated", "text": "System of Equation. (Complicated)\n\nGiven that $x,y \\in \\mathbb{R}$. Solve the system of equation.\n\n$$5x(1+\\frac{1}{x^2+y^2})=12$$\n\n$$5y(1-\\frac{1}{x^2+y^2})=4$$\n\nMy attempt, I made them become $x+\\frac{x}{x^2+y^2}=\\frac{12}{5}$ and $y-\\frac{y}{x^2+y^2}=\\frac{4}{5}$.\n\nI don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.\n\n\u2022 Hint: $\\;\\frac{12}{5x}+\\frac{4}{5y}=\\cdots$ \u2013\u00a0dxiv Mar 26 '17 at 3:01\n\u2022 Can't you equate through $x^2 + y^2$ term and then isolate x (or y) and then sub back in and solve? \u2013\u00a0123 Mar 26 '17 at 3:02\n\nAs dxiv has given hint:\n\n$$\\frac{12}{5x} + \\frac{4}{5y} = 2 \\implies x = \\frac{6y}{5y-2}$$\n\nNow put this in any one of the relation to make the relation full in terms of $y$.\n\nMultiply equation (1) by y and equation (2) by x,\n\n$$5xy(1+\\frac{1}{x^2+y^2})=12y$$\n\n$$5xy(1-\\frac{1}{x^2+y^2})=4x$$\n\n$$10xy = 12y + 4x$$\n\nYou can find y in terms of x or x in terms of y.\n\n\u2022 You can try substitution. May be that is better option. \u2013\u00a0Kanwaljit Singh Mar 26 '17 at 3:55\n\nJust to provide another way.\n\nUse polar coordinates $$x=\\rho \\cos(\\theta) \\qquad y=\\rho \\sin(\\theta)$$ to make the equations to be $$\\frac{5 \\left(\\rho ^2+1\\right) \\cos (\\theta )}{\\rho }=12$$ $$\\frac{5 \\left(\\rho ^2-1\\right) \\sin (\\theta )}{\\rho }=4$$ So,$$\\cos(\\theta)=\\frac{12 \\rho }{5 \\left(\\rho ^2+1\\right)}\\qquad \\sin(\\theta)=\\frac{4 \\rho }{5 \\left(\\rho ^2-1\\right)}$$ Now, using $$\\sin^2(\\theta)+\\cos^2(\\theta)=1 \\implies \\frac{16 \\rho ^2}{25 \\left(\\rho ^2-1\\right)^2}+\\frac{144 \\rho ^2}{25 \\left(\\rho ^2+1\\right)^2}=1$$ Now, let $t=\\rho ^2$ to have $$\\frac{16 t}{25 \\left(t -1\\right)^2}+\\frac{144 t}{25 \\left(t+1\\right)^2}=1$$ Reduce to same denominator to get $$25 t^4-160 t^3+206 t^2-160 t+25=0$$ where you can notice the symmetry of coefficients.\n\nFactoring gives $$25 t^4-160 t^3+206 t^2-160 t+25=(5-t) (5 t-1) \\left(5 t^2-6 t+5\\right)=0$$ The quadratic does not show real roots; this makes the solutions to be $$t_1=\\frac 15\\qquad t_2=5$$ then $$\\rho_1=\\frac 1 {\\sqrt 5}\\qquad \\rho_2=\\sqrt 5$$ and now compute the corresponding $\\theta$'s from $$\\tan(\\theta)=\\frac{\\sin(\\theta) }{\\cos(\\theta) }=\\frac{t+1}{3 (t-1)}=-\\frac 12$$.\n\n\u2022 Up for this answer \u2013\u00a0Mathxx Mar 26 '17 at 6:09\n\n$$5x(1+\\frac{1}{x^2+y^2})=12\\cdots \\cdots (1)$$\n\n$$5y(1-\\frac{1}{x^2+y^2})=4\\cdots \\cdots \\cdots \\cdots (2) \\times i$$\n\n$$5(x+iy)+5\\cdot \\frac{x-iy}{(x+iy)(x+iy)}=12+4i$$\n\nput $z=x+iy$ and $\\bar{z} = x-iy$ and $|z|^2= z \\cdot \\bar{z} = x^2+y^2$\n\nSo $$5z+\\frac{5\\bar{z}}{z \\cdot \\bar{z}}=12+4i$$\n\nSo $$5z+\\frac{5}{z} = 12+4i$$\n\nSo $$5z^2-(12+4i)z+5=0$$ after solving $$z=\\frac{12+4i\\pm \\sqrt{(12+4i)^2-100}}{10} = \\frac{12+4i\\pm (8+6i)}{10} = 2+i\\;,\\frac{2-i}{5}$$\n\nSo $$z= x+iy = 2+i\\;,\\frac{2-i}{5}$$\n\nSo $$(x,y) = \\left\\{(2,1)\\;\\;, \\left(\\frac{2}{5},-\\frac{1}{5}\\right)\\right\\}$$\n\nFirst of all get rid of $x^2+y^2$\n\n$$\\frac{12}{4x} +\\frac{4}{5y} = 2,\\quad 2x+6y = 5xy$$\n\nIt is a rectangular hyperbola passing through origin with asymptotes parallel to axes reminding one of $x= t, y= 1/t$ when axes are asymptotes. So solve for $x,y$\n\n$$x= \\frac{6y}{5y-2},\\, y= \\frac{2x}{5x-6}$$\n\nThe asymptotes are\n\n$$x= \\frac65 ,\\, y= \\frac{2}{5}$$\n\nWe take hyperbola in the form\n\n$$( x-\\frac65 ) (y- \\frac25) = k$$\n\nSince it goes through the origin, $k$ must be the product $k=\\dfrac65 \\cdot \\dfrac25 = \\dfrac{12}{25}$ so as not to leave behind any constant term.\n\nWe can now find $x,y$ upto a parameter $t$\n\n$$( x-\\frac65 ) (y- \\frac25) = \\frac{12}{25}$$\n\n$$\\rightarrow x= \\frac65 + t \\sqrt{12}/5,\\, y= \\frac25 + \\sqrt{12}/{5t},\\,$$", "date": "2019-11-21 04:27:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2203287/system-of-equation-complicated", "openwebmath_score": 0.9215060472488403, "openwebmath_perplexity": 445.6603965269178, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676484382338, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8612716370519906}}
{"url": "https://www.physicsforums.com/threads/algebra-question.714911/", "text": "# Algebra Question\n\nHello,\n\nI wanted to confirm that I am understanding something correctly. I have been doing this all the time, and have never had any problems with it. For some reason now I am wanting to confirm it.\n\nSo, if you have something such as: x3-6+y2-\u221az it would be equivalent to write it as: y2+x3-\u221az-6\n\nCan you always rearrange terms like this, no matter what they are? What I do is write it as an addition problem so I can change the order, since you can change the order in addition.\n\nFor example if you have: log(x)-x3 you could write it as log(x)+(-x3) then as -x3 + log(x).\n\nJust as -11+5-3+5=-4 would be the same as 5-11-3+5=-4 or -11-3+5+5=-4\n\nDoes my reasoning seem correct?\n\nThanks\n\nLast edited:\n\nYes that is correct. I believe that falls under the commutative property which essentially states\n\nx + y = y + x\n\nAnd this would also go for multiplication, but not division correct?\n\nCorrect.\n\nExample using decimal notation:\n\n2*4 = 8 and 4*2 = 8\n\n8 = 8 TRUE\n\n2/4 = .5 and 4/2 = 2\n\n.5 = 2 FALSE\n\nAnd here's a tip, never be afraid to go back to the most basic of examples to solidify your thoughts if you start question them. Forget the log(x) and y^3 stuff and just use a scratch piece of paper and walk through simple things like 1+2 = 2+1 to convince yourself that the order of addition does not matter.\n\narildno\nHomework Helper\nGold Member\nDearly Missed\nHi, ThomasMagnus!\nAs the others have said, you are correct.\nThe two operations multiplication and addition are COMMUTATIVE, i.e, you can change the order in which they appear.\nDivision and subtraction are NOT commutative, we can call them ANTI-commutative if you like.\n\nBut, as you yourself noted, can't we in 5-4 rewrite this as 5+(-4)=(-4)+5?\n\nYes, we can, by using negative numbers in addition, we can think of the operation of subtraction as \"redundant\", i.e, the info is contained in the use of adding with negative numbers instead.\nThat makes for easier calculations, because you then can go about commuting as much as you want!\nIn 5-(-4), for example, we just write 5-(-4)=5+(-(-4))=(-(-4))+5 and so on.\n\nDoes something similar exist with division relative to multiplication?\n\nYes, because division by a non-zero number \"a\" can always be thought of as multiplication with the reciprocal of \"a\", i.e, 1/a.\n\nThus, 2/4=2*(1/4)=(1/4)*2.\n\n----------------\nBy thinking in terms of \"negative numbers\" and \"reciprocals\", you really have only two basic operations to worry about, addition and multiplication, rather than four operations.\n\n(And, if you think of multiplication as repeated addition, there's only one fundamental operation to worry about..)\n\nHallsofIvy\nHomework Helper\nAnd this would also go for multiplication, but not division correct?\nYes, and note that it is largely because (well, perhaps even more that addition and multiplication are \"associative\" while subtraction and division are not) that we do not consider subtraction and division as separate \"operations\" but just the inverses of additon and multiplication. That is, that \"a subtract b\" is just \"a plus the additive inverse of b\" and \"a divided by b\" is \"a multiplied by the multiplicative inverse of b\".\n\narildno\nHomework Helper\nGold Member\nDearly Missed\nWhen we think of ONE operation to be performed, commutativity is the only property that comes into the picture.\nHowever, when we have two or more operations to be performed, we must also take into account what we call the associativity of the operations, that is, what relevance it has which operation we perform first.\n\nWe have, for example, for numbers \"a\", \"b\" and \"c\" that:\n(a+b)+c=a+(b+c), that is, it doesn't matter WHICH addition we perform first. Addition is associative.\n\nBut, is:\n(a-b)-c=a-(b-c)??\n\nYou easily see that subtraction is NOT associative at all;\n\nWe have that (a-b)-c=a-(b+c), whereas a-(b-c)=(a-b)+c\n\nThank you for the info! So the next time I see something in an answer key as something along the lines of x-b+c and I have -b+c+x I can assume they are the exact same, correct? Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6 ?\n\nThanks!\n\narildno\nHomework Helper\nGold Member\nDearly Missed\n\"Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6\"\n\nThat's right!\n\n1 person", "date": "2021-05-12 18:52:42", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/algebra-question.714911/", "openwebmath_score": 0.819445788860321, "openwebmath_perplexity": 729.7299646098163, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9796676436891864, "lm_q2_score": 0.8791467611766711, "lm_q1q2_score": 0.8612716359789293}}
{"url": "https://matheducators.stackexchange.com/questions/5847/is-there-any-other-procedure-to-find-the-square-root/5861", "text": "# Is there any other procedure to find the square root?\n\nIf no calculator is allowed, and we want to find the square root of a square number if it is large and analyzing to prime factors is hard, how can one proceed? For example, what to do if the number is a square of a large prime number like 5329?\n\nWhat are methods to perform this and what are pedagogical reasons to prefer one over the other?\n\n\u2022 I agree with @BenjaminDickman. This question is more for MSE. I wonder if it could be rewritten to explore any didactical issues or motivations. \u2013\u00a0Pablo B. Nov 14 '14 at 21:32\n\u2022 well friends it is just by luck that I wrote the number to be 5329 .. the concept remains the same .. how to find the root of large numers .. I really voted for matin for the link he supplies. thank you all \u2013\u00a0Abdallah Abusharekh Nov 14 '14 at 21:37\n\u2022 Related: hsm.stackexchange.com/q/56/72 \u2013\u00a0Joel Reyes Noche Nov 14 '14 at 23:39\n\u2022 Following the (justified) remarks by @PabloB. and Benjamin Dickman I made explicit the educational aspect in the question, which is already present in some of the answers. \u2013\u00a0quid Nov 15 '14 at 13:26\n\u2022 There're many methods on wikipedia page. \u2013\u00a0Ruslan Nov 15 '14 at 20:59\n\nYou can use (basically) Newton's method.\n\nTake $x_0>0$, and define $$x_{k+1} = \\frac{1}{2} (x_k + \\frac{n}{x_k}).$$ The seqeunce converges to $\\sqrt{n}$.\n\nAnd if one starts with $x_0= n$ one has, as soon as $|x_{k+1} - x_k|< 1$, that $\\lfloor x_{k+1} \\rfloor = \\lfloor\\sqrt{n} \\rfloor$.\n\nBut the above is not very feasible for computing by hand, and there is better variant, if you just want to do this for perfect squares or test if something is a perfect square, or more generally compute $\\lfloor\\sqrt{n} \\rfloor$. (It seems this is the case for you.)\n\nYou can do $$x_{k+1} = \\left \\lfloor \\frac{1}{2} (x_k + \\lfloor \\frac{n}{x_k} \\rfloor) \\right \\rfloor$$ only doing arithemtic with integers (including euclidean division). This will converge to $\\lfloor\\sqrt{n} \\rfloor$ in a finite number of iterations.\n\nSee Integer Square Root on Wikipedia for further references.\n\nI think this could be an interesting method to teach, as one can provide a good geometric motivation for it.\n\nThere is an algorithm to calculate square roots with paper and pencil (like the long division algorithm). See How to calculate a square root without a calculator.\n\n\u2022 I learned this in grade school. :-) \u2013\u00a0Joseph O'Rourke Nov 15 '14 at 21:13\n\u2022 @JosephO'Rourke, I learned this in primary school! \u2013\u00a0Mart\u00edn-Blas P\u00e9rez Pinilla Nov 19 '14 at 10:08\n\nA fancy way, for those who know a little calculus, is to linearize $f(x)=\\sqrt{x}$ at some point near the number you want to square root. This can only be done when you can think of a square rootable number that isn't too far away, so it's not going to work if you're trying to square root a super huge number, but if you can think of one, it's very doable by hand, and very quick. It also has the added benefit of giving you a rational approximation, if that's something you want.\n\nFor example, to compute $\\sqrt{2}$, we first have to come up with some other number that is reasonably close to $2$ which we do know how to square root. The first thing that comes to mind is $9/4$ (though there are many other choices that are even more accurate, e.g. such as $49/25$). We linearize $\\sqrt{x}$ at $9/4$: $$\\left.\\frac{d}{dx}\\sqrt{x}\\right|_{x=9/4}=\\left.\\frac{1}{2\\sqrt{x}}\\right|_{x=9/4}=\\frac{1}{2\\sqrt{9/4}}=\\frac{1}{3}$$ $$\\mathcal{L}_{9/4}(x)=f^\\prime\\left(\\frac{1}{3}\\right)\\left(x-\\frac{9}{4}\\right)+f\\left(\\frac{9}{4}\\right)=\\frac{1}{3}\\left(x-\\frac{9}{4}\\right)+\\frac{3}{2}$$ Now, we plug $2$ in there. $$\\mathcal{L}_{9/4}(2)=\\frac{1}{3}\\left(2-\\frac{9}{4}\\right)+\\frac{3}{2}=\\frac{17}{12}$$ From here, if you want decimals, you can simply use long division. Numerically, $17/12$ comes out to be $1.41667$, while the actual value of $\\sqrt{2}$ is $1.41421\\ldots$. That's pretty close, and likely close enough to do the job for many on the fly calculations.\n\n\u2022 An easy nearby perfect square is always obtainable as the next-lowest or next-highest power of 10 (whichever has an even number of zeros). If you know your powers of two, you can do better using the next-lowest or next-highest power of two. \u2013\u00a0R.. GitHub STOP HELPING ICE Nov 15 '14 at 17:07\n\nUse log tables and the identity $x^\\alpha = e^{\\alpha \\log x}$.\n\nFrom an educational point of view, that gives evidence of the usefulness of logarithms (you can calculate any power of $x$ using just a log table and multiplication) and teaches about transforming between problems. It also allows all roots to be calculated by the same method.\n\nAn intuitive approach to find the square root of A is to draw a square with the number (A) in the middle as the area, and the sides labeled x and y. We know that x times y equal A, and x is the square root of A when x = y. Begin with a rough estimate of x and calculate y as A/x. Now replace x with the average of the original x and y, and repeat.\n\nFor example, let's find the square root of 200. As a rough guess, let's begin with x = 15. Solve for y = 200/15 = 13.333. The new value for x, call it x1 = (15 + 13.333)/2 = 14.167. Now the new value y1 = 200/14.167 = 14.117. The square root of 200 is between 14.167 and 14.117. The average is 14.142. If we use x2 = 14.142, then y2 = 200/14.142 = 14.142, verifying that we have the square root of 200 with five significant digits.\n\nThis approach has the advantage of showing what a square root is visually, which can give a problem more meaning. The process makes intuitive sense, and it does not require mathematics beyond division.\n\nHere is the method I used to solve the original poster's problem in about three seconds:\n\nSquare root (5329):\n\nSteps 1a-d estimate the answer, to one significant figure, with the correct order of magnitude:\n\n1b) Find the order of magnitude of the answer. Make a copy of the argument (5329). Shift the decimal of the copy of the argument by two places at a time, to get a number between 1 and 100. Shift the decimal of the estimated answer by one place at a time, in the opposite direction. For example, 53.29 and 10.\n\n1c) Assume that the student knows the squares of the integers between 0 and 10 by heart. What is the closest such square? For example, 49.\n\n1d) Multiply the estimated answer by the square root of the closest square. For example, square root (49) = 7, so the new estimated answer is 70.\n\nStep 2 checks the estimate:\n\n2) Square the current estimated answer. For example, 70 * 70 = 4900.\n\nSteps 3a-d refine the estimate using uses (a+b)^2 = a^2 + 2*a*b + b^2. This is a special case of Newton's method:\n\n3a) Subtract the value found in step 2 from the original argument, to yield a remainder. For example, 5329 - 4900 = 429. Keep track of the sign.\n\n3b) Make a copy of the estimated answer. Multiply the copy by 2, to yield a double estimate. For example, 70 * 2 = 140.\n\n3c) Divide the remainder by the double estimate, to yield an adjustment. Round off as convenient. Keep track of the sign. For example, 429 / 140 ~ 3.\n\n3d) Add the adjustment to the estimate (not the double estimate). We kept track of the sign, so that we could correctly choose whether to add or subtract. For example, 70 + 3 = 73. For subsequent calculations, this value replaces estimate.\n\nRepeat steps 2 - 3 until the answer is close enough. For example, 4900 + 420 + 3*3 = 5329, so the square root of 5329 is 73.\n\n\u2022 I can do it in my head.\n\n\u2022 It only requires keeping track of a few numbers.\n\n\u2022 It quickly produces an estimated answer.\n\n\u2022 It is a special case of Newton's method.\n\n\u2022 Each time through the loop, the student can round off the adjustment to a convenient decimal or fraction.\n\n\u2022 Every iteration, the student checks their work.\n\nThree worked examples:\n\n       Argu-   Estimated   Found                Double     Adjust-\nStep   ment      Answer    So Far  Remainder    Estimate    ment     Notes\n1a    5329        1\n1b               10                                                 53.29\n1c                                                                  7*7=49\n1d               70\n2     5329       70       4900\n3a    5329       70       4900       429\n3b    5329       70       4900       429        140\n3c    5329       70       4900       429        140          3\n2     5329       73       5329                                 73*73=4900+420+9\nDone!\n\nStep   ment      Answer    So Far  Remainder    Estimate    ment     Notes\n1a     739        1\n1b               10                                                  7.39\n1c                                                                  3*3= 9\n1d               30\n2      739       30        900\n3      739       30        900       -161        60        -3\n2-3    739       27        729         10        54         0.2\n2-3    739       27.2      739.84      -0.84     54.4      -0.015\n2-3    739       27.185    739.024225  -0.024225 54.37     -0.0005\n2-3    739       27.1845   738.99704025          54.369     0.00005\n2      739       27.18455  738.9997587025\n...Stop when satisfied...\n\nStep   ment      Answer    So Far  Remainder    Estimate    ment     Notes\n1a    4285        1\n1b               10                                                 42.85\n1c                                                                  6*6=36\n1d               60\n2-3   4285       60       3600       685       120         5\n2-3   4285       65       4225        60       130         0.4\n2-3   4285       65.4     4277.16      7.84    130.8       0.06\n2-3   4285       65.46    4285.0116   -0.0116  130.92     -0.0001\n2-3   4285       65.4599  4284.99850801        130.9198    0.00001\n2     4285       65.45991 4284.9998171081\n...Stop when satisfied...", "date": "2020-04-01 14:47:18", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/questions/5847/is-there-any-other-procedure-to-find-the-square-root/5861", "openwebmath_score": 0.7430981397628784, "openwebmath_perplexity": 574.0549732334247, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676442828174, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8612716349497939}}
{"url": "https://yutsumura.com/a-relation-between-the-dot-product-and-the-trace/trace/", "text": "# trace\n\n### More from my site\n\n\u2022 Does the Trace Commute with Matrix Multiplication? Is $\\tr (A B) = \\tr (A) \\tr (B)$? Let $A$ and $B$ be $n \\times n$ matrices. Is it always true that $\\tr (A B) = \\tr (A) \\tr (B)$? If it is true, prove it. If not, give a counterexample. \u00a0 Solution. There are many counterexamples. For one, take $A = \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 [\u2026] \u2022 Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant Let A=\\begin{bmatrix} a & b\\\\ c& d \\end{bmatrix} be an 2\\times 2 matrix. Express the eigenvalues of A in terms of the trace and the determinant of A. Solution. Recall the definitions of the trace and determinant of A: \\[\\tr(A)=a+d \\text{ and } [\u2026] \u2022 Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix? Let A be an n \\times n matrix. Is it true that \\tr ( A^\\trans ) = \\tr(A)? If it is true, prove it. If not, give a counterexample. Solution. The answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the [\u2026] \u2022 If Two Matrices are Similar, then their Determinants are the Same Prove that if A and B are similar matrices, then their determinants are the same. Proof. Suppose that A and B are similar. Then there exists a nonsingular matrix S such that \\[S^{-1}AS=B$ by definition. Then we [\u2026]\n\u2022 An Example of a Matrix that Cannot Be a Commutator Let $I$ be the $2\\times 2$ identity matrix. Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\\times 2$ matrices $A$ and $B$ with determinant $1$. \u00a0 Proof. Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies $ABA^{-1}=-B. [\u2026] \u2022 If 2 by 2 Matrices Satisfy A=AB-BA, then A^2 is Zero Matrix Let A, B be complex 2\\times 2 matrices satisfying the relation \\[A=AB-BA.$ Prove that $A^2=O$, where $O$ is the $2\\times 2$ zero matrix. \u00a0 Hint. Find the trace of $A$. Use the Cayley-Hamilton theorem Proof. We first calculate the [\u2026]\n\u2022 Matrix $XY-YX$ Never Be the Identity Matrix Let $I$ be the $n\\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\\times n$ matrices $X$ and $Y$ such that $XY-YX=I.$ \u00a0 Hint. Suppose that such matrices exist and consider the trace of the matrix $XY-YX$. Recall that the trace of [\u2026]\n\u2022 Matrices Satisfying $HF-FH=-2F$ Let $F$ and $H$ be an $n\\times n$ matrices satisfying the relation $HF-FH=-2F.$ (a) Find the trace of the matrix $F$. (b)\u00a0Let $\\lambda$ be an eigenvalue of $H$ and let $\\mathbf{v}$ be an eigenvector corresponding to $\\lambda$. Show that there exists an positive integer $N$ [\u2026]", "date": "2018-04-19 19:37:38", "meta": {"domain": "yutsumura.com", "url": "https://yutsumura.com/a-relation-between-the-dot-product-and-the-trace/trace/", "openwebmath_score": 0.9846538305282593, "openwebmath_perplexity": 74.89729722194647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9953904285043392, "lm_q2_score": 0.8652240791017535, "lm_q1q2_score": 0.8612357668493666}}
{"url": "https://math.stackexchange.com/questions/2958113/how-to-solve-an-ode-of-this-form-dotpt-atptpta", "text": "# How to solve an ODE of this form $\\dot{P}(t)=A^TP(t)+P(t)A$?\n\nBackground\n\nIf $$\\dot{x}(t)=A \\,x(t)$$, then we know the solution is $$x(t)=e^{At}x(0)$$.\n\nQuestion\n\nNow let $$\\dot{P}(t)=A^TP(t)+P(t)A$$, what is $$P(t)$$?\n\nAttempt\n\nIf we take $$P(t)$$ as common factor (I'm not sure I'm doing this correctly though) ,then we have: $$\\dot{P}(t)=A^TP(t)+P(t)A=(A^T+A)P(t)$$ and using the same rationale in the background section, we have $$P(t)=e^{(A^T+A) t}P(0)=e^{A^T t}P(0)e^{At}$$ Is this a correct solution?\n\nNote\n\nI know the answer is $$P(t)=e^{A^T t}P(0)e^{At}$$ but I'm not sure how to find it.\n\n\u2022 Counter question: what is A? \u2013\u00a0Yuriy S Oct 16 '18 at 16:28\n\u2022 @YuriyS A is a matrix. \u2013\u00a0Lod Oct 16 '18 at 16:29\n\u2022 Dimensions are very likely mismatched in the expression $\\dot{p}(t)=A^Tp(t)+p(t)A$. \u2013\u00a0Ennar Oct 16 '18 at 16:30\n\u2022 What is $p A$ then? \u2013\u00a0Yuriy S Oct 16 '18 at 16:30\n\u2022 @Ennar I fixed the question. \u2013\u00a0Lod Oct 16 '18 at 16:36\n\nConsider the differential equation $$\\dot{P}(t)=A^TP(t)+P(t)A$$\n\nWe can rewrite it as: $$\\dot{P}(\\tau)-A^TP(\\tau)-P(\\tau)A=0$$\n\nMultiply the last differential equation by $$e^{-A^T\\tau}$$ from the left and by $$e^{-A\\tau}$$ from the right then\n\n$$$$e^{-A^T\\tau}\\dot{P}(\\tau)e^{-A\\tau}-e^{-A^T\\tau}A^TP(\\tau)e^{-A\\tau}-e^{-A^T\\tau}P(\\tau)Ae^{-A\\tau}=0 \\quad (1)$$$$ We can easily see that equation $$(1)$$ is generated by the derivative of three multiplied functions: $$$$\\frac{d}{d\\tau}\\bigg(e^{-A^T\\tau}{P}(\\tau)e^{-A\\tau}\\bigg)=e^{-A^T\\tau}\\dot{P}(\\tau)e^{-A\\tau}-A^Te^{-A^T\\tau}P(\\tau)e^{-A\\tau}-e^{-A^T\\tau}P(\\tau)Ae^{-A\\tau}=0$$$$\n\nSo we come up with:\n\n$$0=\\frac{d}{d\\tau}\\bigg(e^{-A^T\\tau}{P}(\\tau)e^{-A\\tau}\\bigg)$$\n\nTake the integral of both sides\n\n$$$$0=\\int_0^t\\frac{d}{d\\tau}\\bigg(e^{-A^T\\tau}{P}(\\tau)e^{-A\\tau}\\bigg)d\\tau=\\bigg(e^{-A^T\\tau}{P}(\\tau)e^{-A\\tau}\\bigg)\\bigg]_0^t=e^{-A^Tt}{P} \\tau)e^{-At}-e^{0}{P}(0)e^{0}$$$$ \\ $$$$0=e^{-A^Tt}{P}(t)e^{-At}-P(0)$$$$\n\nMultiply left and right sides by $$e^{A^Tt}$$ and $$e^{At}$$ respectively, we get:\n\n$$$$P(t)=e^{A^Tt}{P}(0)e^{At}$$$$\n\nNote: I used the fact that $$e^{-A^T\\tau}$$ and $$A^T$$ commute, i.e. $$A^Te^{-A^T\\tau}=e^{-A^T\\tau}A^T$$\n\nFor small time increment you get approximately $$P(t+dt)=P(t)+(A^TP(t)+P(t)A)dt+O(dt^2)=(I+A^T\\,dt)P(t)(I+A\\,dt)+O(dt^2)$$ which means that from time step to time step you get an accumulation of these factors on both sides, $$P(N\\,dt)=(I+A^T\\,dt)^NP(t)(I+A\\,dt)^N+O(Ndt^2)$$ Now if $$N\\,dt=\\Delta t$$ we get, using $$(I+B/N)^N=\\exp(B)+O(B^2/N)$$ $$P(t+\u0394t)=e^{A^T\\,\u0394t}P(t)e^{A\\,\u0394t}+O(\u0394t\\,dt)$$ so that for $$dt\\to 0$$ one gets exactly the claimed solution form.\n\nBy vectorizing $$P$$ and using the Kronecker product, similar to a method which can be used to solve Sylvester equations, then the differential equation can also be written as\n\n$$\\frac{d}{dt}\\,\\text{vec}\\,P(t) = \\underbrace{\\left(I \\otimes A^\\top + A^\\top\\otimes I\\right)}_{M}\\,\\text{vec}\\,P(t), \\tag{1}$$\n\nwhich has the solution\n\n$$\\text{vec}\\,P(t) = e^{M\\,t}\\,\\text{vec}\\,P(0). \\tag{2}$$\n\nBy using the mixed-product property of the Kronecker product it can be shown that $$I \\otimes A^\\top$$ commutes with $$A^\\top \\otimes I$$. This commuting property allows you to write the matrix exponential as\n\n$$e^{M\\,t} = e^{(I\\,\\otimes\\,A^\\top)\\,t}\\,e^{(A^\\top\\,\\otimes\\,I)\\,t}. \\tag{3}$$\n\nBy using the definition of a matrix exponential and again the mixed-product property of the Kronecker product then is can also be shown that\n\n$$e^{X\\,\\otimes\\,I} = e^{X} \\otimes I, \\quad e^{I\\,\\otimes\\,X} = I \\otimes e^{X},$$\n\nthus\n\n$$e^{M\\,t} = \\left(I \\otimes e^{A^\\top t}\\right) \\left(e^{A^\\top t} \\otimes I\\right). \\tag{4}$$\n\nSubstituting $$(4)$$ into $$(2)$$ gives\n\n\\begin{align} \\text{vec}\\,P(t) &= \\left(I \\otimes e^{A^\\top t}\\right) \\left(e^{A^\\top t} \\otimes I\\right) \\text{vec}\\,P(0) \\\\ &= \\left(I \\otimes e^{A^\\top t}\\right) \\text{vec}\\left(P(0)\\,e^{A\\,t}\\right) \\\\ &= \\text{vec}\\left(e^{A^\\top t}\\,P(0)\\,e^{A\\,t}\\right) \\end{align}\n\nthus\n\n$$P(t) = e^{A^\\top t}\\,P(0)\\,e^{A\\,t}.$$", "date": "2020-07-07 03:56:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2958113/how-to-solve-an-ode-of-this-form-dotpt-atptpta", "openwebmath_score": 0.9971098899841309, "openwebmath_perplexity": 387.9829163838855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713891776498, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.8612033014351956}}
{"url": "https://www.physicsforums.com/threads/series-convergence.319241/", "text": "# Series convergence\n\n1. Jun 10, 2009\n\n### utopiaNow\n\n1. The problem statement, all variables and given/known data\nUse partial fractions to show\n$$\\displaystyle\\sum_{n=1}^\\infty \\frac{1}{n(n+1)(n+2)} = \\frac{1}{4}$$\n\n3. The attempt at a solution\nI did the partial fraction decomposition to get: $$\\displaystyle\\sum_{n=1}^\\infty \\frac{1}{2n} - \\frac{1}{n + 1} + \\frac{1}{2n + 4}$$\n\nI'm not sure how to proceed from here, in my textbook the example shows how terms in the partials sums overlap and cancel out if you start looking at the terms in the partial fraction decomposition, however I can't see that happening with this particular series.\n\nAny suggestions would be appreciated. Thanks in advance.\n\n2. Jun 10, 2009\n\n### rock.freak667\n\nWrite out a few term, n=1,2,3,4,5....N-3,N-2,N-1,N\n\nthen check as N->inf.\n\n3. Jun 10, 2009\n\n### utopiaNow\n\nHi,\n\nI tried doing that originally, but I don't see any patterns that I can exploit when I write the terms of the partial decomposition for n = 1, 2, 3,..., N-2, N-1, N as you stated.\n\nDo you have any further suggestions?\n\n4. Jun 10, 2009\n\n### Dick\n\nWrite the terms for successive values of n each on a row 1/(2n+4) -1/(n+1) 1/(2n)\n1/6 -1/2 1/2\n1/8 -1/3 1/4\n1/10 -1/4 1/6\n1/12 -1/5 1/8\n1/14 -1/6 1/10\n\nOblique hint: diagonal.\n\n5. Jun 10, 2009\n\n### utopiaNow\n\nInteresting, so everything seems to be canceling out if you keep going long enough diagonally, except the 1/4 in the second row. However is there any way to formalize this? Or is simply noticing this pattern enough for a formal proof?\n\n6. Jun 10, 2009\n\n### Dick\n\nDepends on how formal you want to be. Writing a table like that and scratching out the cancellations is good for me. If you want to do it formally write out the first term from the n case, the second term from the n+1 case and the third term from the n+2 case and show they cancel algebraically. What is important is to realize just because there isn't an obvious cancellation doesn't mean there isn't one.\n\n7. Jun 10, 2009\n\n### utopiaNow\n\nThank you for the help!\n\n8. Jun 10, 2009\n\n### txy\n\nwhy not take out 1/2 from the partial fraction decomposition so that it's more obvious?\n\n$$\\sum^{+\\infty}_{n=1}(\\frac{1}{2n}-\\frac{1}{n+1}+\\frac{1}{2n+4})$$\n$$= \\frac{1}{2}\\sum^{+\\infty}_{n=1}(\\frac{1}{n}-\\frac{2}{n+1}+\\frac{1}{n+2})$$\n$$= \\frac{1}{2}\\sum^{+\\infty}_{n=1}(\\frac{1}{n}-\\frac{1}{n+1}+\\frac{1}{n+2}-\\frac{1}{n+1})$$\n$$= \\frac{1}{2}\\sum^{+\\infty}_{n=1}(\\frac{1}{n}-\\frac{1}{n+1}) + \\frac{1}{2}\\sum^{+\\infty}_{n=1}(\\frac{1}{n+2}-\\frac{1}{n+1})$$\n\n9. Jun 10, 2009\n\n### Dick\n\nThat's a nice way to handle it. There's more than one way to skin a cat.", "date": "2017-10-21 07:01:21", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/series-convergence.319241/", "openwebmath_score": 0.5360410213470459, "openwebmath_perplexity": 1000.6326239063328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713839878682, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8612032920510522}}
{"url": "https://math.stackexchange.com/questions/2435913/solutions-of-x2-6x-13-equiv-0-pmod127", "text": "# Solutions of $x^2-6x-13 \\equiv 0 \\pmod{127}$\n\nI started learning number theory, specifically polynomial congruences, and need help with the following exercise. Here it is:\n\nDoes the congruence $x^2-6x-13 \\equiv 0 \\pmod{127}$ has solutions?\n\nI tried to follow the method for solving general quadratic congruence but I didn't get really far. Here's what I've done so far:\n\nSince $(4, 127) = 1$, we may complete the square by multiplying by $4$ without having to change the modulus in order to get the following equivalent congruence\n\n$$(2x-6)^2 \\equiv 36 - 4(-13) \\pmod{127} \\iff (2x-6)^2 \\equiv 88 \\pmod{127}.$$\n\nIf I'm heading in the right direction then I don't know how to continue from here. I suspect there is another method for solving this problem since I didn't make use of the fact that $127$ is prime. Also, the problem doesn't require to actually find the solutions but only determine if there are solutions. Possibly we can avoid computations and make use of some theorem/lemma to find if the congruence has solutions or not.\n\n\u2022 The quadratic formula, $x = \\frac{6\\pm\\sqrt{(-6)^2 - 4\\cdot 1\\cdot (-13)}}{2\\cdot 1}$ still works, since $127$ is an odd prime. Note, however, that square root and dividing by $2$ has a different meaning from what it does for real numbers (or, it has the same meaning, but potentially a very different result). \u2013\u00a0Arthur Sep 19 '17 at 11:16\n\n. To find solutions of $$x^2 - 6x - 13 \\equiv 0 \\mod 127$$, we must write $$x^2-6x - 13$$ as a square of a linear term, plus a constant. It's seen easily that $$x^2 - 6x - 13 = (x-3)^2 - 22$$. Hence, the congruence is equivalent to $$(x-3)^2 \\equiv 22 \\mod 127$$.\n\nNow, we must check if $$22$$ is a quadratic residue mod $$127$$. For this, we can use the Legendre symbol, whose notation I will keep the same as the binomial, so do not get confused. $$\\binom{22}{127} = \\color{green}{\\binom{11}{127}}\\color{red}{\\binom{2}{127}} = \\color{green}{-\\binom{127}{11}} \\times \\color{red}1$$ where the terms with same color on the LHS and RHS are equal. The green equality comes by quadratic reciprocity and the second comes by the fact that $$\\binom{2}{p}$$ is well known by the remainders which $$p$$ leaves when divided by $$8$$.\n\nNow, we can do: $$-\\binom{127}{11} = -\\binom{6}{11} = -\\color{green}{\\binom{2}{11}}\\color{red}{ \\binom{3}{11}} = -\\color{green}{(-1)}\\color{red}{(1)} = 1$$\n\nAgain, the colored terms on the LHS and RHS are equal because the quantities $$\\binom 2p$$ and $$\\binom 3p$$ are well known.\n\nSince we have obtained that the Legendre symbol is $$1$$, this implies the existence of a solution, and therefore two.\n\nThe question, though, is how to compute them. I do not know of any method other than brute force, unfortunately. However, our reward for writing $$(x-3)^2 \\equiv 22 \\mod 127$$ is that we basically only need to look for squares of the form $$127k + 22$$ to find a solution, rather than having to substitute values of $$x$$ into the expression $$x^2-6x-13$$ each time.\n\nA brute force : the series $$127k + 22$$ goes like : $$22,149,276,403,530,657,\\color{blue}{784},...$$\n\nlo and behold, $$784 = 28^2$$, hence this gives $$x = 31$$. Now, note that the congruence actually has two solutions, one given by $$127 - 28 = 99$$. You can check that $$9801 = 99^2 = 22 + 127 \\times 77$$.\n\nHence, we get two solutions of $$x$$, namely $$x= 31,102$$.\n\n\u2022 I can't thank you enough for this answer! I understood everything and it made me see the link that I was missing between quadratic polynomial/residues and Legendre symbol. \u2013\u00a0user347616 Sep 19 '17 at 12:07\n\u2022 But of course, you are welcome! I should mention, +1 for your question. \u2013\u00a0\u0430\u0441\u0442\u043e\u043d \u0432\u0456\u043b\u043b\u0430 \u043e\u043b\u043e\u0444 \u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 Sep 19 '17 at 12:08\n\u2022 Since $127$ is one less than a multiple of $4$, you can get the square roots by \"root-power conversion\". Fermat's Little Theorem gives $22^{126}\\equiv 1$, and with $22$ being a quadratic residue then $22^{63}\\equiv 1$. So $22^{64}\\equiv 22$ assuring that $22^{32}\\equiv 99$ will be one square root. Try it! \u2013\u00a0Oscar Lanzi Sep 19 '17 at 12:26\n\u2022 That is a much,much better method than the one I have suggested. I thank you for this. \u2013\u00a0\u0430\u0441\u0442\u043e\u043d \u0432\u0456\u043b\u043b\u0430 \u043e\u043b\u043e\u0444 \u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 Sep 19 '17 at 12:29\n\nUsing the Berlekamp Algorithm for factoring a polynomial over $\\mathbb{F}_{127}$ we immediately obtain that $$x^2-6x-13=(x+25)(x+96).$$ So we have two solutions.\n\nI am a congruence beginner like you.\n\nI did in this way\n\ncompleted the square $(x-3)^2-9\\equiv 13 \\mod 127$\n\n$(x-3)^2\\equiv 22 \\mod 127$\n\nNow call $r$ is the number such that $r^2\\equiv 22\\mod 127$?\n\nLike a \"modular\" square root\n\nI found here a way to find the roots\n\nAs $127\\equiv 3 \\mod 4$ then we have $r= \\pm 22^{\\frac{128}{4}} \\mod 127$\n\n$22^{32} =2^{32}\\cdot 11^{32}$\n\n$2^7\\equiv 1 \\mod 127\\to 2^{28}\\equiv 1 \\mod 127\\to \\color{red}{2^{32}\\equiv 2^4\\mod 127}$\n\n$11^{32}=121^{16}$ as $121\\equiv 6 \\mod 127$ we have $121^{16}\\equiv 6^{16} \\mod 127$ and then $6^{16}\\equiv 2^{16}\\cdot 3^{16} \\mod 127$\n\n$2^k \\mod 127$ for $k=0,1,2\\ldots$ is cyclical and is $1,2,4,8,16,32,64$ then repeats (because $127=2^7-1$) therefore $2^{16}\\equiv 4 \\mod 127$\n\n$3^{16}\\equiv 81^4 \\mod 127\\to 3^{16}\\equiv 71 \\mod 127$\n\nFinally $6^{16}\\equiv 4\\cdot 71 \\mod 127\\to 6^{16}\\equiv 30 \\mod 127$ and $\\color{red}{11^{32}\\equiv 6^{16}\\equiv 30 \\mod 127}$\n\nCollecting the parts in red colour we get $22^{32}\\equiv 2^4\\cdot 30 \\mod 127$ that is $22^{32}\\equiv 99\\mod 127$ and $-22^{32} \\equiv 28 \\mod 127$\n\nConcluding the two solutions are $x-3\\equiv 99 \\mod 127\\to x\\equiv 102 \\mod 127$ and $x-3\\equiv 28 \\mod 127 \\to x\\equiv 31 \\mod 127$\n\nEdit\n\nMy solution is all but elegant. Anyway is a solution provided by a beginner like you and I hope it is understandable\n\n\u2022 It took me a little time to go through your solution but I made it. Thank you for the effort, it is always enlightening to see different solutions to the same problem. (+1) \u2013\u00a0user347616 Sep 19 '17 at 14:59\n\u2022 To compute $22^{32} \\bmod127$ just square five successive times. \u2013\u00a0Oscar Lanzi Sep 19 '17 at 22:33\n\u2022 @OscarLanzi Why $22^{32}\\equiv 22^5 \\mod 127$? \u2013\u00a0Raffaele Sep 20 '17 at 13:11\n\u2022 I did not say that. Square five successive times means $22^{2^5}$. \u2013\u00a0Oscar Lanzi Sep 20 '17 at 13:42\n\nI think you can try $$x=31.$$ It is valid.\n\nFinally we have $$x^2-6x-13\\equiv(x-31)(x-102) \\pmod{127}.$$\n\nThis equation has roots modulo $127$ if and only if the reduced discriminant $\\Delta'=9+13$ is a square modulo $127$. We'll use the laws of quadratic reciprocity.\n\n$$\\biggl(\\frac{22}{127}\\biggr)=\\biggl(\\frac{2}{127}\\biggr)\\biggl(\\frac{11}{127}\\biggr)=\\biggl(\\frac{11}{127}\\biggr)\\quad\\text{since }127\\equiv -1\\mod8\\quad (2^{\\mathit{nd}}\\textit{supplementary law}).$$ Now we have \\begin{align}\\biggl(\\frac{11}{127}\\biggr)&=\\biggl(\\frac{127}{11}\\biggr)\\bigl(-1\\bigr)^{\\tfrac{10\\cdot126}4}=-\\biggl(\\frac{6}{11}\\biggr)=-\\biggl(\\frac{2}{11}\\biggr)\\biggl(\\frac{3}{11}\\biggr)\\\\[1ex] &=+\\biggl(\\frac{3}{11}\\biggr)=\\biggl(\\frac{11}{3}\\biggr)\\bigl(-1\\bigr)^{\\tfrac{2\\cdot10}4}=-\\biggl(\\frac{2}{3}\\biggr)=+1. \\end{align} Thus the equation has roots.", "date": "2019-11-15 04:50:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2435913/solutions-of-x2-6x-13-equiv-0-pmod127", "openwebmath_score": 0.8910647034645081, "openwebmath_perplexity": 207.2661213267215, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853138, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.861203289953204}}
{"url": "https://math.stackexchange.com/questions/218046/relations-between-p-norms", "text": "# Relations between p norms\n\nThe $$p$$-norm on $$\\mathbb R^n$$ is given by $$\\|x\\|_{p}=\\big(\\sum_{k=1}^n |x_{k}|^p\\big)^{1/p}$$. For $$0 < p < q$$ it can be shown that $$\\|x\\|_p\\geq\\|x\\|_q$$ (1, 2). It appears that in $$\\mathbb{R}^n$$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $$\\|x\\|_{1} \\leq\\sqrt n\\,\\|x\\|_{2}$$(3), $$\\|x\\|_{2} \\leq \\sqrt n\\,\\|x\\|_\\infty$$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $$\\mathbb R^n$$. For instance, for $$n=2$$ and $$n=3$$ one can see that for $$0 < p < q$$, the spheres with radius $$\\sqrt n$$ with $$\\|\\cdot\\|_p$$ inscribe spheres with radius $$1$$ with $$\\|\\cdot\\|_q$$.\n\nIt is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $$n=2$$ it is easily proven (see below), but not for $$n>2$$. So my questions are:\n\n1. How can relation (3) be proven for arbitrary $$n\\,$$?\n2. Can this be generalized into something of the form $$\\|x\\|_{p} \\leq C \\|x\\|_{q}$$ for arbitrary $$0?\n3. Do any of the relations also hold for infinite-dimensional spaces, i.e. in $$l^p$$ spaces?\n\nNotes:\n\n$$\\|x\\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \\leq |x_{1}|^2 + |x_{2}|^2 + \\big(|x_{1}|^2 + |x_{2}|^2\\big) = 2|x_{1}|^2 + 2|x_{2}|^2$$, hence $$=2\\|x\\|_{2}^{2}$$\n$$\\|x\\|_{1} \\leq \\sqrt 2 \\|x\\|_{2}$$. This works because $$|x_{1}|^2 + |x_{2}|^2 \\geq 2|x_{1}\\|x_{2}|$$, but only because $$(|x_{1}| - |x_{2}|)^2 \\geq 0$$, while for more than two terms $$\\big(|x_{1}| \\pm |x_{2}| \\pm \\dotsb \\pm |x_{n}|\\big)^2 \\geq 0$$ gives an inequality that never gives the right signs for the cross terms.\n\n\u2022 Your first link (1) has exactly the equation you seek. Actually a version of Norberts excellent answer, generalized to measure spaces. \u2013\u00a0Thomas Ahle Mar 31 '16 at 20:57\n\u2022 Probably a dead thread, but norms in finite dimensional spaces are equivalent, so one can always find c ad C such that $c\\|x\\|_p\\leq \\|x\\|_q\\leq C\\|x\\|_p$. \u2013\u00a0BigM Dec 1 '19 at 20:23\n\n1. Using Cauchy\u2013Schwarz inequality we get for all $x\\in\\mathbb{R}^n$ $$\\Vert x\\Vert_1= \\sum\\limits_{i=1}^n|x_i|= \\sum\\limits_{i=1}^n|x_i|\\cdot 1\\leq \\left(\\sum\\limits_{i=1}^n|x_i|^2\\right)^{1/2}\\left(\\sum\\limits_{i=1}^n 1^2\\right)^{1/2}= \\sqrt{n}\\Vert x\\Vert_2$$\n2. Such a bound does exist. Recall H\u00f6lder's inequality $$\\sum\\limits_{i=1}^n |a_i||b_i|\\leq \\left(\\sum\\limits_{i=1}^n|a_i|^r\\right)^{\\frac{1}{r}}\\left(\\sum\\limits_{i=1}^n|b_i|^{\\frac{r}{r-1}}\\right)^{1-\\frac{1}{r}}$$ Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$ $$\\sum\\limits_{i=1}^n |x_i|^p= \\sum\\limits_{i=1}^n |x_i|^p\\cdot 1\\leq \\left(\\sum\\limits_{i=1}^n (|x_i|^p)^{\\frac{q}{p}}\\right)^{\\frac{p}{q}} \\left(\\sum\\limits_{i=1}^n 1^{\\frac{q}{q-p}}\\right)^{1-\\frac{p}{q}}= \\left(\\sum\\limits_{i=1}^n |x_i|^q\\right)^{\\frac{p}{q}} n^{1-\\frac{p}{q}}$$ Then $$\\Vert x\\Vert_p= \\left(\\sum\\limits_{i=1}^n |x_i|^p\\right)^{1/p}\\leq \\left(\\left(\\sum\\limits_{i=1}^n |x_i|^q\\right)^{\\frac{p}{q}} n^{1-\\frac{p}{q}}\\right)^{1/p}= \\left(\\sum\\limits_{i=1}^n |x_i|^q\\right)^{\\frac{1}{q}} n^{\\frac{1}{p}-\\frac{1}{q}}=\\\\= n^{1/p-1/q}\\Vert x\\Vert_q$$ In fact $C=n^{1/p-1/q}$ is the best possible constant.\n\u2022 @Arun, because for $x=(1,1,1,\\ldots,1)$ this bound is attained \u2013\u00a0Norbert Jun 10 '15 at 14:27\n\u2022 @Norbert: I'm trying to derive a similar bound for the case $p>q$. Any ideas about how to proceed ? \u2013\u00a0pikachuchameleon Jun 22 '15 at 12:19\n\u2022 @AshokVardhan what does \"similar\" mean? Do you want to prove something like $\\Vert x\\Vert_p \\leq C\\Vert x\\Vert_q$ for $p>q$ ? \u2013\u00a0Norbert Jun 23 '15 at 14:24", "date": "2021-07-29 03:10:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/218046/relations-between-p-norms", "openwebmath_score": 0.9979337453842163, "openwebmath_perplexity": 1680.4300838087117, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713891776498, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8612032885075326}}
{"url": "https://byjus.com/question-answer/a-point-moves-such-that-the-sum-of-the-square-of-its-distance-from-the/", "text": "Question\n\n# A point moves such that the sum of the square of its distance from the sides of a sqaure of side unity is equal to $$9$$. The locus of the point is a circle such that\n\nA\nCenter of the circle coincides with that of square\nB\nCenter of the circle is (12,12)\nC\nRadius of the circle is 2\nD\nAll the above are true\n\nSolution\n\n## The correct option is D All the above are trueLet the sides of the square be $$y=0,y=1,x=0$$ and $$x=1.$$Let the moving point be $$(x,y).$$Then,$${ y }^{ 2 }+{ \\left( y-1 \\right) }^{ 2 }+{ x }^{ 2 }+{ \\left( x-1 \\right) }^{ 2 }=9$$ is the equation of the locus.$$\\Rightarrow 2{ x }^{ 2 }+2{ y }^{ 2 }-2x-2y-7=0,$$\u00a0which represents a circle having centre $$\\displaystyle \\left( \\frac { 1 }{ 2 } ,\\frac { 1 }{ 2 } \\right)$$ (which is also the centre of the square) and radius $$2.$$ \u00a0Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-28 05:23:40", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/a-point-moves-such-that-the-sum-of-the-square-of-its-distance-from-the/", "openwebmath_score": 0.7439823746681213, "openwebmath_perplexity": 551.2238064608388, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9852713839878681, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8612032872031784}}
{"url": "https://math.stackexchange.com/questions/2567158/looking-for-a-proof-of-an-interesting-identity/2567218", "text": "# Looking for a proof of an interesting identity\n\nWorking on a problem I have encountered an interesting identity:\n\n$$\\sum_{k=0}^\\infty \\left(\\frac{x}{2}\\right)^{n+2k}\\binom{n+2k}{k} =\\frac{1}{\\sqrt{1-x^2}}\\left(\\frac{1-\\sqrt{1-x^2}}{x}\\right)^n,$$ where $n$ is a non-negative integer number and $x$ is a real number with absolute value less than 1 (probably a similar expression is valid for arbitrary complex numbers $|z|<1$).\n\nIs there any simple proof of this identity?\n\n\u2022 How/where did you meet this identity? Is the LHS just the Taylor series for the RHS? \u2013\u00a0Daniel Robert-Nicoud Dec 14 '17 at 23:38\n\u2022 Anything to do with a random walk? Why did you call this interesting, btw? \u2013\u00a0Mathemagical Dec 14 '17 at 23:44\n\u2022 Reminiscent of the explicit form for Chebyshev polynomials of the first kind, hence they idea of enforcing the substitution $x=\\sin\\theta$ and applying the residue theorem looks like a promising one ;) \u2013\u00a0Jack D'Aurizio Dec 14 '17 at 23:56\n\u2022 Or you may prove that both sides are the terms of an Appell / Fibonacci sequence of polynomials. \u2013\u00a0Jack D'Aurizio Dec 14 '17 at 23:57\n\u2022 @tired This is in regard to the hitting time of a random walk (more particularly, the probability generating function of the hitting time of the walk at the barrier x=1. Please see the equations 15,16, and 17 in the linked document. galton.uchicago.edu/~lalley/Courses/312/RW.pdf The details of the computation are not listed here, but they are shown in greater detail in Michael Steele's book, Stochastic Calculus and Financial applications (pages 8 and 9). \u2013\u00a0Mathemagical Dec 16 '17 at 15:25\n\nUsing\n\n$$\\binom{n}{k}=\\frac{1}{2 \\pi i}\\oint_C\\frac{(1+z)^{n}}{z^{k+1}}dz$$ we get (integration contour is the unit cicrle)\n\n$$2\\pi iS_n=\\oint dz \\sum_{k=0}^{\\infty}\\frac{(1+z)^{n+2k}x^{n+2k}}{z^{k+1}2^{n+2k}}=\\oint dz \\frac{(1+z)^n x^n}{z2^n}\\sum_{k=0}^{\\infty}\\frac{(1+z)^{2k}x^{2k}}{2^{2k}z^k}=\\\\ 4\\frac{x^n}{2^n}\\oint dz \\underbrace{\\frac{(1+z)^n}{4z-(1+z)^2x^2}}_{f(z)}$$\n\nfor $|x|<1$ only we have just one pole of $f(z)$ inside the unit circle namely $z_0(x)=\\frac2{x^2}-\\frac{2\\sqrt{1-x^2}}{x^2}-1$ , so\n\n$$S_n=4\\frac{x^n}{2^n}\\text{res}(f(z),z=z_0(x))=4\\frac{x^n}{2^n}\\left[ \\frac{1}{4 \\sqrt{1-x^2}}\\left(2\\frac{1-\\sqrt{1-x^2}}{ x^2}\\right)^n\\right]$$\n\nor\n\n$$S_n=\\frac{1}{\\sqrt{1-x^2}}\\left(\\frac{1-\\sqrt{1-x^2}}{ x}\\right)^n$$\n\n\u2022 Thanks Jack :-) \u2013\u00a0tired Dec 15 '17 at 0:55\n\u2022 @tired: Thank you very much for the nice proof. It contains however a misprint. $4$ shall be replaced with $4^{1/n}$ in the denominator of the residue value. But probably simpler and more correctly were to write it just as $$\\text{res}(f(z),z=z_0)=\\frac{1}{4\\sqrt{1-x^2}} \\left(2\\frac{1-\\sqrt{1-x^2}}{x^2}\\right)^n.$$ \u2013\u00a0user Dec 15 '17 at 11:56\n\u2022 @user355705 you are absolutly right \u2013\u00a0tired Dec 16 '17 at 15:07\n\nExtracting coefficients on the RHS we get the integral (coefficient on $x^{n+2k}$)\n\n$$\\frac{1}{2\\pi i} \\int_{|z|=\\epsilon} \\frac{1}{z^{n+2k+1}} \\frac{1}{\\sqrt{1-z^2}} \\left(\\frac{1-\\sqrt{1-z^2}}{z}\\right)^n \\; dz.$$\n\nNow we put $(1-\\sqrt{1-z^2})/z = w$ so that $z = 2w/(1+w^2).$ This has $w = \\frac{1}{2} z + \\cdots$ so the image in $w$ of the contour in $z$ can be deformed to a small circle enclosing the origin in the $w$-plane. (Moreover we see that the exponentiated term starts at $z^n$ which justifies the corresponding offset in the series.) We get $dz = 2/(1+w^2) - 4w^2/(1+w^2)^2 \\; dw = 2(1-w^2)/(1+w^2)^2 \\; dw.$ We also have $1-z^2 = 1 - 4w^2/(1+w^2)^2 = (1-w^2)^2/(1+w^2)^2.$ All of this yields\n\n$$\\frac{1}{2\\pi i} \\int_{|w|=\\gamma} \\frac{(1+w^2)^{n+2k+1}}{2^{n+2k+1} w^{n+2k+1}} \\frac{1}{(1-w^2)/(1+w^2)} w^n \\frac{2(1-w^2)}{(1+w^2)^2} \\; dw \\\\ = \\frac{1}{2^{n+2k}} \\frac{1}{2\\pi i} \\int_{|w|=\\gamma} \\frac{(1+w^2)^{n+2k}}{w^{2k+1}} \\; dw.$$\n\nThis evaluates by inspection to\n\n$$\\frac{1}{2^{n+2k}} [w^{2k}] (1+w^2)^{n+2k} = \\frac{1}{2^{n+2k}} [w^{k}] (1+w)^{n+2k} \\\\ = \\frac{1}{2^{n+2k}} {n+2k\\choose k}$$\n\nwhich is the claim.\n\nI plan to insert this approach into the next (2019) version of my notes.\nThe even powers of $\\arcsin(x)$ all have a nice Maclaurin series (see eqs (19),(20),(21),(22)), and this can be seen as a side effect of the Lagrange\u2013B\u00fcrmann inversion theorem. I am going to implement the same trick here, by computing the derivatives at the origin of the RHS (the opposite way has already been investigated by tired).\n\n$$[x^m]\\left[\\frac{1}{\\sqrt{1-x^2}}\\left(\\frac{1-\\sqrt{1-x^2}}{x}\\right)^n\\right]=\\frac{1}{2\\pi i}\\oint\\frac{(1-\\sqrt{1-x^2})^n}{x^{n+m+1}\\sqrt{1-x^2}}\\,dx$$ where the integral is performed along a small circle enclosing the origin. What about enforcing the substitution $x=\\sin z$? The sine function is holomorphic and injective in a neighbourhood of the origin, hence the RHS is simply converted into $$\\frac{1}{2\\pi i}\\oint\\frac{(1-\\cos z)^n}{\\left(\\sin z\\right)^{n+m+1}}\\,dz =\\frac{1}{2\\pi i}\\oint\\frac{2^n \\sin^{2n}\\frac{z}{2}}{2^{n+m+1}\\sin^{n+m+1}\\frac{z}{2}\\cos^{n+m+1}\\frac{z}{2}}\\,dz$$ or $$\\frac{1}{2\\pi i}\\oint\\frac{\\tan^{n-m-1}z}{2^{m}\\cos^{2m}z}\\,dz=\\frac{1}{2^{m+1}\\pi i}\\oint u^{n-m-1}(1+u^2)^{m-1}\\,du$$ via $z\\to 2z$ and $z\\to\\arctan u$. Now the RHS is trivially given by a binomial coefficient multiplied by $\\frac{1}{2^m}$ (provided by $m\\geq n$) and the claim is proved.", "date": "2019-05-22 14:49:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2567158/looking-for-a-proof-of-an-interesting-identity/2567218", "openwebmath_score": 0.8838281631469727, "openwebmath_perplexity": 212.8128387680172, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713861502773, "lm_q2_score": 0.8740772286044095, "lm_q1q2_score": 0.8612032826294593}}
{"url": "https://ch.mathworks.com/help/symbolic/fimplicit3.html", "text": "# fimplicit3\n\nPlot 3-D implicit equation or function\n\n## Syntax\n\nfimplicit3(f)\nfimplicit3(f,[min max])\nfimplicit3(f,[xmin xmax ymin ymax zmin zmax])\nfimplicit3(___,LineSpec)\nfimplicit3(___,Name,Value)\nfimplicit3(ax,___)\nfi = fimplicit3(___)\n\n## Description\n\nexample\n\nfimplicit3(f) plots the 3-D implicit equation or function f(x,y,z) over the default interval [-5 5] for x, y, and z.\n\nexample\n\nfimplicit3(f,[min max]) plots f(x,y,z) over the interval [min max] for x, y, and z.\n\nexample\n\nfimplicit3(f,[xmin xmax ymin ymax zmin zmax]) plots f(x,y,z) over the interval [xmin xmax] for x, [ymin ymax] for y, and [zmin zmax] for z. The fimplicit3 function uses symvar to order the variables and assign intervals.\n\nexample\n\nfimplicit3(___,LineSpec) uses LineSpec to set the line style, marker symbol, and face color.\n\nexample\n\nfimplicit3(___,Name,Value) specifies line properties using one or more Name,Value pair arguments. Use this option with any of the input argument combinations in the previous syntaxes.\nfimplicit3(ax,___) plots into the axes with the object ax instead of the current axes object gca.\n\nexample\n\nfi = fimplicit3(___) returns an implicit function surface object. Use the object to query and modify properties of a specific surface. For details, see ImplicitFunctionSurface Properties.\n\n## Examples\n\ncollapse all\n\nPlot the hyperboloid ${x}^{2}+{y}^{2}-{z}^{2}=0$ by using fimplicit3. The fimplicit3 function plots over the default interval of $\\left[-5,5\\right]$ for $x$, $y$, and $z$.\n\nsyms x y z fimplicit3(x^2 + y^2 - z^2)\n\nPlot the hyperboloid specified by the function $f\\left(x,y,z\\right)={x}^{2}+{y}^{2}-{z}^{2}$. The fimplicit3 function plots over the default interval of $\\left[-5,5\\right]$ for $x$, $y$, and $z$.\n\nsyms f(x,y,z) f(x,y,z) = x^2 + y^2 - z^2; fimplicit3(f)\n\nSpecify the plotting interval by specifying the second argument to fimplicit3. Plot the upper half of the hyperboloid ${x}^{2}+{y}^{2}-{z}^{2}=0$ by specifying the interval $0. For $x$ and $y$, use the default interval $\\left[-5,5\\right]$.\n\nsyms x y z f = x^2 + y^2 - z^2; interval = [-5 5 -5 5 0 5]; fimplicit3(f, interval)\n\nPlot the implicit equation $x\\mathrm{sin}\\left(y\\right)+z\\mathrm{cos}\\left(x\\right)=0$ over the interval $\\left(-2\\pi ,2\\pi \\right)$ for all axes.\n\nCreate the x-axis ticks by spanning the x-axis limits at intervals of pi/2. Convert the axis limits to precise multiples of pi/2 by using round and get the symbolic tick values in S. Display these ticks by using the XTick property. Create x-axis labels by using arrayfun to apply texlabel to S. Display these labels by using the XTickLabel property. Repeat these steps for the y-axis.\n\nTo use LaTeX in plots, see latex.\n\nsyms x y z eqn = x*sin(y) + z*cos(x); fimplicit3(eqn,[-2*pi 2*pi]) title('xsin(y) + zcos(x) for -2\\pi < x < 2\\pi and -2\\pi < y < 2\\pi') xlabel('x') ylabel('y') ax = gca; S = sym(ax.XLim(1):pi/2:ax.XLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); ax.XTick = double(S); ax.XTickLabel = arrayfun(@texlabel,S,'UniformOutput',false); S = sym(ax.YLim(1):pi/2:ax.YLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); ax.YTick = double(S); ax.YTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false);\n\nPlot the implicit surface ${x}^{2}+{y}^{2}-{z}^{2}=0$ with different line styles for different values of $z$. For $-5, use a dashed line with green dot markers. For $-2, use a LineWidth of 1 and a green face color. For $2, turn off the lines by setting EdgeColor to none.\n\nsyms x y z f = x^2 + y^2 - z^2; fimplicit3(f,[-5 5 -5 5 -5 -2],'--.','MarkerEdgeColor','g') hold on fimplicit3(f,[-5 5 -5 5 -2 2],'LineWidth',1,'FaceColor','g') fimplicit3(f,[-5 5 -5 5 2 5],'EdgeColor','none')\n\nPlot the implicit surface $1/{x}^{2}-1/{y}^{2}+1/{z}^{2}=0$. Specify an output to make fimplicit3 return the plot object.\n\nsyms x y z f = 1/x^2 - 1/y^2 + 1/z^2; fi = fimplicit3(f)\n\nfi = ImplicitFunctionSurface with properties: Function: [1x1 sym] EdgeColor: [0 0 0] LineStyle: '-' FaceColor: 'interp' Show all properties \n\nShow only the positive x-axis by setting the XRange property of fi to [0 5]. Remove the lines by setting the EdgeColor property to 'none'. Visualize the hidden surfaces by making the plot transparent by setting the FaceAlpha property to 0.8.\n\nfi.XRange = [0 5]; fi.EdgeColor = 'none'; fi.FaceAlpha = 0.8;\n\nControl the resolution of an implicit surface plot by using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots while decreasing 'MeshDensity' can increase plotting speed.\n\nDivide a figure into two by using subplot. In the first subplot, plot the implicit surface $\\mathrm{sin}\\left(1/\\left(xyz\\right)\\right)$. The surface has large gaps. Fix this issue by increasing the 'MeshDensity' to 40 in the second subplot. fimplicit3 fills the gaps showing that by increasing 'MeshDensity' you increased the resolution of the plot.\n\nsyms x y z f = sin(1/(x*y*z)); subplot(2,1,1) fimplicit3(f) title('Default MeshDensity = 35') subplot(2,1,2) fimplicit3(f,'MeshDensity',40) title('Increased MeshDensity = 40')\n\nApply rotation and translation to the implicit surface plot of a torus.\n\nA torus can be defined by an implicit equation in Cartesian coordinates as\n\n$\\mathit{f}\\left(\\mathit{x},\\mathit{y},\\mathit{z}\\right)={\\left({\\mathit{x}}^{2}+{\\mathit{y}}^{2}+{\\mathit{z}}^{2}+{\\mathit{R}}^{2}-{\\mathit{a}}^{2}\\right)}^{2}-4{\\mathit{R}}^{2}\\left({\\mathit{x}}^{2}+{\\mathit{y}}^{2}\\right)$\n\nwhere\n\n\u2022 $a$ is the radius of the tube\n\n\u2022 $R$ is the distance from the center of the tube to the center of the torus\n\nDefine the values for $a$ and $R$ as 1 and 5, respectively. Plot the torus using fimplicit3.\n\nsyms x y z a = 1; R = 4; f(x,y,z) = (x^2+y^2+z^2+R^2-a^2)^2 - 4*R^2*(x^2+y^2); fimplicit3(f) hold on\n\nApply rotation to the torus around the $x$-axis. Define the rotation matrix. Rotate the torus by 90 degrees or $\\pi /2$ radians. Shift the center of the torus by 5 along the $x$-axis.\n\nalpha = pi/2; Rx = [1 0 0; 0 cos(alpha) sin(alpha); 0 -sin(alpha) cos(alpha)]; r = [x; y; z]; r_90 = Rx*r; g = subs(f,[x,y,z],[r_90(1)-5,r_90(2),r_90(3)]);\n\nAdd a second plot of the rotated and translated torus to the existing graph.\n\nfimplicit3(g) axis([-5 10 -5 10 -5 5]) hold off\n\n## Input Arguments\n\ncollapse all\n\n3-D implicit equation or function to plot, specified as a symbolic equation, expression, or function. If an expression or function is specified, then fimplicit3 assumes the right-hand size to be 0.\n\nPlotting interval for x-, y- and z- axes, specified as a vector of two numbers. The default is [-5 5].\n\nPlotting interval for x-, y- and z- axes, specified as a vector of six numbers. The default is [-5 5 -5 5 -5 5].\n\nAxes object. If you do not specify an axes object, then fimplicit3 uses the current axes.\n\nLine style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.\n\nExample: '--or' is a red dashed line with circle markers\n\nLine StyleDescription\n-Solid line\n--Dashed line\n:Dotted line\n-.Dash-dot line\nMarkerDescription\n'o'Circle\n'+'Plus sign\n'*'Asterisk\n'.'Point\n'x'Cross\n'_'Horizontal line\n'|'Vertical line\n's'Square\n'd'Diamond\n'^'Upward-pointing triangle\n'v'Downward-pointing triangle\n'>'Right-pointing triangle\n'<'Left-pointing triangle\n'p'Pentagram\n'h'Hexagram\nColorDescription\n\ny\n\nyellow\n\nm\n\nmagenta\n\nc\n\ncyan\n\nr\n\nred\n\ng\n\ngreen\n\nb\n\nblue\n\nw\n\nwhite\n\nk\n\nblack\n\n### Name-Value Pair Arguments\n\nSpecify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.\n\nExample: 'Marker','o','MarkerFaceColor','red'\n\nThe properties listed here are only a subset. For a complete list, see ImplicitFunctionSurface Properties.\n\nNumber of evaluation points per direction, specified as a number. The default is 35.\n\nExample: 100\n\nLine color, specified as 'interp', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default RGB triplet value of [0 0 0] corresponds to black. The 'interp' value colors the edges based on the ZData values.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB\u00ae uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nLine style, specified as one of the options listed in this table.\n\nLine StyleDescriptionResulting Line\n'-'Solid line\n\n'--'Dashed line\n\n':'Dotted line\n\n'-.'Dash-dotted line\n\n'none'No lineNo line\n\nLine width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.\n\nThe line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.\n\nMarker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.\n\nValueDescription\n'o'Circle\n'+'Plus sign\n'*'Asterisk\n'.'Point\n'x'Cross\n'_'Horizontal line\n'|'Vertical line\n'square' or 's'Square\n'diamond' or 'd'Diamond\n'^'Upward-pointing triangle\n'v'Downward-pointing triangle\n'>'Right-pointing triangle\n'<'Left-pointing triangle\n'pentagram' or 'p'Five-pointed star (pentagram)\n'hexagram' or 'h'Six-pointed star (hexagram)\n'none'No markers\n\nMarker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the EdgeColor property.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nExample: [0.5 0.5 0.5]\n\nExample: 'blue'\n\nExample: '#D2F9A7'\n\nMarker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nExample: [0.3 0.2 0.1]\n\nExample: 'green'\n\nExample: '#D2F9A7'\n\nMarker size, specified as a positive value in points, where 1 point = 1/72 of an inch.\n\n## Output Arguments\n\ncollapse all\n\nOne or more objects, returned as a scalar or a vector. The object is an implicit function surface object. You can use these objects to query and modify properties of a specific line. For details, see ImplicitFunctionSurface Properties.\n\n## Algorithms\n\nfimplicit3 assigns the symbolic variables in f to the x axis, the y axis, then the z axis, and symvar determines the order of the variables to be assigned. Therefore, variable and axis names might not correspond. To force fimplicit3 to assign x, y, or z to its corresponding axis, create the symbolic function to plot, then pass the symbolic function to fimplicit3.\n\nFor example, the following code plots the roots of the implicit function f(x,y,z) = x + z in two ways. The first way forces fimplicit3 to assign x and z to their corresponding axes. In the second way, fimplicit3 defers to symvar to determine variable order and axis assignment: fimplicit3 assigns x and z to the x and y axes, respectively.\n\nsyms x y z; f(x,y,z) = x + z; figure; subplot(2,1,1) fimplicit3(f); view(-38,71); subplot(2,1,2) fimplicit3(f(x,y,z)); % Or fimplicit3(x + z); \n\n### Topics\n\nIntroduced in R2016b\n\n## Support\n\n#### Mathematical Modeling with Symbolic Math Toolbox\n\nGet examples and videos", "date": "2021-09-20 11:21:39", "meta": {"domain": "mathworks.com", "url": "https://ch.mathworks.com/help/symbolic/fimplicit3.html", "openwebmath_score": 0.5654040575027466, "openwebmath_perplexity": 4498.974919876233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8612032770255403}}
{"url": "http://openstudy.com/updates/508c2cfae4b04456f6c43164", "text": "## klimenkov Group Title $\\lim_{n\\rightarrow\\infty}\\frac{\\sqrt[n]{n!}}{n}$ 2 years ago 2 years ago\n\n1. myko\n\nmaybe like this: $\\sqrt[n]{n!}=\\sqrt[n]{n(n-1)(n-2)\\cdots1} =\\sqrt[n]{n}\\sqrt[n]{n-1}\\cdots \\sqrt[n]{1}=1$ so limit is equal to 0\n\n2. myko\n\n@klimenkov\n\n3. klimenkov\n\nAre you sure that $\\lim_{n\\rightarrow\\infty}\\sqrt[n]{n!}=1$\n\n4. myko\n\nlook the steps from my comment befor. It looks ok\n\n5. myko\n\nall the roots at the right hand side: $\\sqrt[n]{n}=\\sqrt[n]{n-1}=\\cdots=\\sqrt[n]{1}=1$\n\n6. myko\n\nso their product too\n\n7. klimenkov\n\nNo, it's not ok. Because you multiply an infinite quantity or 1. As we know $$1^{\\infty}={}?$$.\n\n8. myko\n\n$1^{\\infty} =1*1*\\cdots*1=1$\n\n9. klimenkov\n\nVery nice. What can you say about this pretty limit? $\\lim_{n\\rightarrow\\infty}\\left(1+\\frac1n\\right)^n$It is $$1^{\\infty}$$.\n\n10. myko\n\nthis happens when you talk about functions. The reason of indetermination of 1^infinity is because of that. But in this case there are no functions involved. That's my point\n\n11. myko\n\nin this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken\n\n12. klimenkov\n\nOk. What about this? $\\lim_{n\\rightarrow\\infty}\\sum_{k=1}^n\\frac1n$Is it 0?\n\n13. myko\n\nthis is a harmonic series. It is not convergent\n\n14. myko\n\n15. klimenkov\n\nLook at the denominator carefully please. I hopr you will try to get what I'm saying.\n\n16. myko\n\nsry, but i don't\n\n17. klimenkov\n\nCan you find this? $\\lim_{n\\rightarrow\\infty}\\sum_{k=1}^n\\frac1n$\n\n18. myko\n\nanother way to try this: $\\lim \\sqrt[n]{\\frac{n!}{n^n}} = 0$\n\n19. myko\n\ninfinity\n\n20. klimenkov\n\nCan you show the way you solve it?\n\n21. myko\n\nI don't remmeber the formal proof of n!/n^n =0, but it's evident, if you try a few first terms of this sequence. There are some posts about it if you google a bit\n\n22. klimenkov\n\n@myko, see this and tell me what is your mistake? http://www.wolframalpha.com/input/?i=Limit+%28n!%29^%281%2Fn%29%2Fn+n-%3Einfinity\n\n23. TuringTest\n\n@mahmit2012 a little help here?\n\n24. mahmit2012\n\n|dw:1351370532404:dw|\n\n25. TuringTest\n\nvery nice, my turn...\n\n26. mahmit2012\n\n|dw:1351370740820:dw|\n\n27. mahmit2012\n\n|dw:1351370823511:dw|\n\n28. mahmit2012\n\n|dw:1351370856590:dw|\n\n29. TuringTest\n\n${\\sqrt[n]{n!}\\over n}=\\exp\\left(\\frac1n\\ln(n!)-\\ln n\\right)=\\exp\\left({n\\ln n-n+O(n)-n\\ln n\\over n}\\right)$$=\\exp(-1)=\\frac1e$\n\n30. mukushla\n\n*\n\n31. myko\n\nya I was wrong. Here is another way to solve it. As we know root test is stronger than cuotient test, so the folowing inequality holds: $\\lim \\inf \\frac{a_{n+1}}{a_{n}}\\leq\\lim \\inf \\sqrt[n]{a_{n}}\\leq \\lim \\sup \\sqrt[n]{a_{n}} \\leq \\lim \\sup\\frac{a_{n+1}}{a_{n}}$ let $a_{n} = \\frac{n!}{n^{n}}$ then $\\frac{a_{n+1}}{a_{n}}=\\frac{1}{(1+\\frac{1}{n})^{n}}=\\frac{1}{e}$ this means that $\\lim \\sqrt[n]{a_{n}} = \\frac{1}{e}$\n\n32. myko\n\n@mahmit2012 $\\lim \\frac{a_{n+1}}{a_{n}}=\\lim \\sqrt[n]{a_{n}}$ only if a_n is convergent, what is not implied in this question\n\n33. klimenkov\n\nNice. But I have one more interesting method to find it. $\\lim_{n\\rightarrow\\infty} \\frac{\\sqrt[n]{n!}}{n}=\\lim_{n\\rightarrow\\infty}\\sqrt[n]{\\frac{n!}{n^n}}=\\lim_{n\\rightarrow\\infty}\\sqrt[n]{\\frac1n\\cdot\\frac2n\\cdots\\frac n n}=A$$\\ln A=\\lim_{n\\rightarrow\\infty}\\frac1n(\\ln\\frac1n+\\ln\\frac2n+\\ldots+\\ln\\frac n n)=\\int_0^1\\ln x dx=-1$$\\lim_{n\\rightarrow\\infty} \\frac{\\sqrt[n]{n!}}{n}=A=e^{-1}=\\frac1e$", "date": "2014-11-26 18:56:21", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/508c2cfae4b04456f6c43164", "openwebmath_score": 0.8027116060256958, "openwebmath_perplexity": 6110.20970250091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534365728416, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.861202618020932}}
{"url": "http://marathon42k.it/zcy/time-complexity-of-linear-search.html", "text": "Understanding Time Complexity Of Algorithms | Bits N Tricks. We could also say it is linear on the number of entries in the table but that is less commonly used. Imagine the time it will take to search a Letter of Daniel, Eric, Jayesh or any Employee. Therefore, by using hash table, we can achieve linear time complexity for finding the duplicate. Total comparisons in Bubble sort is: n ( n \u2013 1) / 2 \u2248 n 2 \u2013 n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. The time required is flat, an O(1) constant time complexity. The main points in these lecture slides are:Time Complexity, Complexity of Algorithms, Execution Time, Space Complexity, Worst Case Analysis, Division of Integers, Number of Comparisons, Binary Search, Average Case Complexity, Complexity of Bubble Sort. In the later case, the search terminates in failure with n comparisons. If we assume we needed to search the array n times the total worst case run time of the linear searches would be O (n^2)). Why so important? You do it all the time in real life!. Amortized time per operation using a bounded priority queue[1] logarithmic time DLOGTIME O(log n) log n, log(n 2) Binary search polylogarithmic time poly(log n) (log n)2 fractional power O(nc) where 0 < c < 1 n1/2, n2/3 Searching in a kd-tree linear time O(n) n Finding the smallest item in an unsorted array \"n log star n\" time O(n log* n). The idea behind linear search is to compare the search item with the elements in the list one by one (using a loop) and stop as soon as we get the first copy of the search element in the list. By this logic, we can say that painting pictures is slower than baking cookies. This is said to run at O(n); it\u2019s run time increases at an order of magnitude proportional to n. Solving linear equations can be reduced to a matrix-inversion problem, implying that the time complexity of the former problem is not greater than the time complexity of the latter. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. The search time increases proportionately to the number of new items introduced. If we're running a statement. Lookups on arrays and objects are going to be constant time if you access them directly. Firstly, we analyze the time complexity of the iterative algorithm and then recursive. The notation \u039f(n) is the formal way to express the upper bound of an algorithm's running time. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. Solving a system of linear equations has a complexity of at most O (n 3). This video explains the time complexity analysis for binary search. One example is the binary search algorithm. The array to be searched is reduced by half in every iteration. com: Time complexity of an algorithm: In computer science , the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function. > But what is: > > == the time complexity of string indexing? Is it constant? Yes. when first breaks, you know X(last but one fall - success) and Y(last fall - failure). Linear Differential Equations and Oscillators is the first book within Ordinary Differential Equations with Applications to Trajectories and Vibrations, Six-volume Set. Here that is linear time. The improvement of the proposed linear-time algorithm compared with ECL2 (Yu et al. Given an arbitrary network of interconnected nodes, each with an initial value, we study the number of time-steps required for some (or all) of the nodes to gather all of the initial values via a linear iterative strategy. Lets say I have the list 10,20,30,40,50,60,30,40,50. O(N)\u2014Linear Time: Linear Time Complexity describes an algorithm or program who\u2019s complexity will grow in direct proportion to the size of the input data. It costs us space. characterises a function based on growth of function C. A linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list. But a balanced binary search tree is always OlgN. Space Complexity. Data Structures for Beginners: Arrays, HashMaps, and Lists. For example, a \"linear\" running time can also. $\\begingroup$ @Olologin can you share any references to understand how to calculate time complexities for complex equations? I want to understand the priority of matrix, inverse, transpose etc of different orders. Search for \"Journey Into Complexity\" Books in the Search Form now, Download or Read Books for FREE, just by Creating an Account to enter our library. So, we can write this as \u03a9(n). Examples: binary search. For example, if the heuristic evaluation function is an exact estimator, then A* runs in linear time, expanding only those nodes on an optimal solution path. now do a linear search starting from X(conservative but accurate second step - slow). For example -. During each iteration, the first remaining element of the input is only compared with the right-most element of the sorted subsection of the array. So, an algorithm taking X second or 2X + 3 seconds have the same complexity. However, in my previous experiments, it appears to be O (N), namely linear complexity!. Time Complexity of Bisection Search is O(log n). The complexity of an algorithm is usually taken to be its worst-case complexity, unless specified otherwise. Here is an. Dual first-order methods are essential techniques for large-scale constrained convex optimization. As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. The best case time in linear search is for the first element i. For databases, this means that the time execution would be directly proportional to the table size: as the number of rows in the table grows, the time for the query grows. See full list on towardsdatascience. In the later case, the search terminates in failure with n comparisons. The complexity of an algorithm is usually taken to be its worst-case complexity, unless specified otherwise. \u2022 Matlab implements sparse linear algebra based on i,j,s format. Dual first-order methods are essential techniques for large-scale constrained convex optimization. [143, 144, 145, 99]) that Depth-First Search (DFS) and Breadth-First Search (BFS) run in linear time in graphs, and that using these techniques one can obtain linear time algorithms (on a RAM) for many interesting graph. The new distance measures can be computed in linear time complexity in the histogram size. veri es in linear time whether a given spanning tree T of a graph G = (V;E) is a minimum spanning tree. Stochastic Diffusion Search is an alternative solution for invariant pattern recognition and focus of attention. O(n log n) - sorting a list. DTIME[2polylogn]. However complexity for above written implementations is O(). Download Binary search program. , 2017) is in the overhead on enumerating bin pairs. Space Complexity. \u2026humans are incredibly good at linking cause and effect \u2013 sometime too good. Now considering the worst case in which the search element does not exist in the list of size N then the Simple Linear Search will take a total of 2N+1. Since all letters are placed in one bucket, Put and Get operation will no longer have time complexity of O(1) because put and get operation has to scan each letter inside the bucket for matching key. Multiply to get n*log(n). Write a linear-time filter IntegerSort. Another simple yet important function. Binary Search source. Yields of experiment are expected to provide an information related to the complexity of the algorithm in LibSVM and knowing the running-time indicator of training and testing both for C++ and Java. How many elements of the input sequence need to be checked on the average, assuming that the element being searched for is equally likely to be any element in the array? How about in the worst case? What are the average-case and worst-case running times of linear search in $\\theta$-notation? Justify your answers. Examples: binary search. All have polynomial time complexity while some allow very long steps in favorable circumstances. Time complexity of a related-key attack: \u201cThus, the total time complexity of Step 2(b) is about 2256 \u00b72167. With an average time complexity of O(log log n), interpolation search beats binary search's O(log n) easily. Time Complexity : \u03b8 ( n ) Space Complexity : O(1) Linear Search Example. On the other hand, if you search for a word in a dictionary, the search will be faster because the words are in sorted order, you know the order and can quickly decide if you need to turn to earlier pages or later pages. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. hyperparameter Search: Grid search and random search Train & Run time space & time complexity. It went through the entire list so it took linear time. On an unsorted array Binary Search is almost twice as slow as Linear Search with worst Time Complexity of O(n\u00b2) and that is not even considering unbalanced trees. Time Complexity of Binary Search Algorithm is O(log 2 n). Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. Polynomial time means n O(1), or n c for some constant c. Huan Li, Zhouchen Lin; 21(33):1\u221245, 2020. Total comparisons in Bubble sort is: n ( n \u2013 1) / 2 \u2248 n 2 \u2013 n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. Linear time is when an algorithm depends on the input size. Because of this, time complexity increases. Reducing the number of generations, i. Big-O Notation \u2022 We specify the largest term using big-O notation. \\ReaderPrograms\\ReaderFiles\\Chap02\\OrderedArray\\orderedArray. It measures the worst case time complexity or the longest amount of time an algorithm can possibly take to complete. Linear Complexity: O(n) A linear task\u2019s run time will vary depending on it\u2019s input value. The asymptotic complexity is defined by the most efficient (in terms of whatever computational resource one is considering) algorithm for solving the game; the most common complexity measure (computation time) is always lower-bounded by the logarithm of the asymptotic state-space complexity, since a solution algorithm must work for every. Time Complexity of. This linear search has a time complexity of O(n). Self-balanced Binary Search Trees. However, in my previous experiments, it appears to be O (N), namely linear complexity!. The time required to search an element using a linear search algorithm depends on the size of the list. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. The time complexity of Linear Search is O (n). Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. The search time increases proportionately to the number of new items introduced. It makes an exponential workspace and solves the problems with exponential complexity in a polynomial (even linear) time. linear search time complexity. Gorky University Publishers, Gorky (1985) (in Russian) Google Scholar. The time complexity has to do with the critical opeations being performed. Multiply to get n*log(n). Serial Search - Analysis. Thus: I use three elements as the threshold when I will switch to Dictionary lookups from List loops. Linear Search is an example for Linear Time Complexity. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. O(n) Linear: Time to complete the work grows in a 1 to 1 relation to input size. In this set of Solved MCQ on Searching and Sorting Algorithms in Data Structure, you can find MCQs of the binary search algorithm, linear search algorithm, sorting algorithm, Complexity of linear search, merge sort and bubble sort and partition and exchange sort. Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. Linear search has linear-time complexity; binary search has log-time complexity. Conversely, giv. Data Structures and Algorithms Objective type Questions and Answers. com: Time complexity of an algorithm: In computer science , the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function. Browse other questions tagged time-complexity linear-algebra matrices or ask your own question. \u2022 DEMO \u2022 Conclusion: Maybe I can SCALE well \u2026 Solve O(10^12) problems in O(10^12). See full list on yourbasic. The time complexity is the sum of time spent in all calls plus some extra preprocessing time. BIG O Notation \u2013 Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. This video explains the time complexity analysis for binary search. See full list on freecodecamp. We therefore take the complexity of inverted index search to be (as discussed in Section 2. Linear search performs equality comparisons and Binary search performs ordering comparisons; Let us look at an example to compare the two: Linear Search to find the element \u201cJ\u201d in a given sorted list from A-X. for(i=0; i N; i++) { for(j=0; j. Hence the complexity is O(n). Time Complexity of Linear Search Algorithm is O (n). It depends on the condition given in the for loop. Linear Search vs Binary Search. linear regression and the correlation coefficient. Time complexity. The time complexity of su\ufb03x tree construction has been shown to be equiv-alent to that of sorting [7]. i it is exponentially better than a brute force search. This calculation will be independent of implementation details and programming language. Strictly, we should say the average complexity is $$\\mathcal{O}(n)$$. Linear Time Complexity: O(n) When time complexity grows in direct proportion to the size of the input, you are facing Linear Time Complexity, or O(n). As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. So during the execution of an algorithm, the total time required that will be decided in the time complexity. We could also say it is linear on the number of entries in the table but that is less commonly used. So there is no advantage of binary search over linear search if every search is on a fresh array. Search time is proportional to the list size. For N = 1024 it is 80% faster, and I guess the performance ratio should converge to two at infinity. > How are strings stored in Python? As arrays? As linked lists?. Complexity Classes. Linear time complexity might sound inefficient when you image input sizes in the billions, but linear time isn't actually too bad. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. Thus, the time complexity of this recursive function is the product O(n). For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. The best case for a linear search algorithm is to have the value x for which one is searching located at the first position of the ADT. Total comparisons in Bubble sort is: n ( n \u2013 1) / 2 \u2248 n 2 \u2013 n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. In case of a sorted array, Binary Search is faster but the caveat here is also on how arrays are treated by the Language Translator. Time Complexity of Linear Search Algorithm is O (n). It is conjectured that the indistinguishability of photons is responsible for the computational complexity of linear optics. Consider a sorted array of 16 elements. the time required to complete the above operation increases linearlywith respect to 'n' (input). Time complexity of algorithms An algorithm is a collection of steps that process a given input to produce an output. See full list on yourbasic. Time Complexity : \u03b8 ( n ) Space Complexity : O(1) Linear Search Example. The search stops when the item is found or when the search has examined each item without success. A sorted array is required New insert() Searching a sorted array by repeatedly dividing the search interval in half. For example. Don't overanalyze O(N). Thus, we have-. For example, for a function f(n) \u039f(f(n)) = { g(n) : there exists c > 0 and n 0 such that f(n) \u2264 c. Hence, this is another difference between linear search and binary search. Time Complexity. The search time increases proportionately to the number of new items introduced. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. DTIME[2polylogn]. The linear search with break becomes faster than counting linear search shortly after N = 128. Linear time: O(n). The cases are as follows \u2212 Best Case \u2212 Here the lower bound of running time is calculated. Answer: d Explanation: It is practical to implement linear search in the situations mentioned in When the list has only a few elements and When performing a single search in an unordered list, but for larger elements the complexity becomes larger and it makes sense to sort the list and employ binary search or hashing. The space complexity is also. 2012: J Paul Gibson T&MSP: Mathematical Foundations MAT7003/ L9-Complexity&AA. Informally, this means that the running time increases at most linearly with the size of the input. The order of growth (e. If there are NO nested loops we can probably guess the complexity of the code we looking at would be in the O(n). A(n) = $\\frac{n + 1}{2}$ However, I am having trouble coming up with the Average Case complexity in the case where half of the elements in the size n array are duplicates. In this case, the insertion sort algorithm has a linear running time (i. Linear search is a very basic and simple search algorithm. The List has an O(N) linear time complexity. The time complexity of suffix tree construction has been shown to be equivalent to that of sorting: O(n) for a constant-size alphabet or an integer alphabet and O(n log n) for a general alphabet. ) Combinatorial- Algebraic Methods in Applied Mathematics, pp. With a faster sorter like merge-sort, which is O(N*log(N. In linear search we simply iterate over elements and check whether it is the desired element or not. Imagine the time it will take to search a Letter of Daniel, Eric, Jayesh or any Employee. doubling n, time increases only by a factor of c. This is a more mathematical way of expressing running time, and looks more like a function. The time taken to search a given element will increase if the number of elements in the array increases. Here that is linear time. Therefore, much research has been invested into discovering algorithms exhibiting linear time or, at least, nearly linear time. Hourly Update. If you had to search for a name in a directory by reading. Introduction The time complexity of a given algorithm can be obtained from theoretical analysis and computational analysis according to the algorithm\u2019s running. \u2022 for selection sort, C(n)=n2/2-n/2 n2/2 \u2022 In addition, we\u2019ll typically ignore the coefficient of the largest term (e. Time complexity of Bubble sort in Best Case is O(N). As we will see in the next chapter, kNN's effectiveness is close to that of the most accurate learning methods in text classification (Table 15. In [15] a chosen plaint-text linear attack was suggested and in [5] time complexity of the attack rst stage was reduced by using Fast Fourier Transform. The time complexity of su\ufb03x tree construction has been shown to be equiv-alent to that of sorting [7]. This search requires only one unit of space to store the element to be searched. I know the answer is O(n), but is this correct: The first element has probability $1/n$ and requires 1 comparison; the second probability $1/(n-1)$ and requires 2 comparisons. In our previous tutorial we discussed about Linear search algorithm which is the most basic algorithm of searching which has some disadvantages in terms of time complexity, so to overcome them to a level an algorithm based on dichotomic (i. Time Complexity For Linked Lists; Time Complexity; Time Complexity In While Loop; My Future Plan Because Of My Teacher? Python Mini-challenge: \"Lucky\" Numbers; Advance Code Not Hardware? Efficiency Of Linear Search Vs Binary Search In Unsorted List; Algorithm Not Efficient Enough; Cryptography And Data Structure; Filling List(s) With Random Numbers. These problems will introduce things (like the variable i above) just to waste your time. This requires to scan the array completely and check each element for the array that we need to search. Linear Search is an example for Linear Time Complexity. Linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list. In linear search, we have to check each node/element. O(N)\u2014Linear Time: Linear Time Complexity describes an algorithm or program who\u2019s complexity will grow in direct proportion to the size of the input data. Linear time or O(n). \ubb34\uc2a8\ub9d0\uc778\uc9c0\ubaa8\ub974\uaca0\ub2e4\uba74 \uc544\ub798\uae00\uc744 \ucb49. The number of steps and time required to solve a problem is based on input size. Best Case. It makes an exponential workspace and solves the problems with exponential complexity in a polynomial (even linear) time. Also, each algorithm's time complexity is explained in separate video lectures. best-case: this is the complexity of solving the problem for the best input. Solving a system of linear equations has a complexity of at most O (n 3). This means the bigger the number of wine bottles in our system, the more time it will take. Definition of time complexity in the Definitions. In the linear search problem, the best case occurs when x is present at the first location. This research includes both software and hardware methods. A few common algorithmic complexities: O(log n) - binary search. If you were to find the name by looping through the list entry after entry, the time complexity would be O (n). Since binary search algorithm splits array in half every time, at most log 2 N steps are performed. So, we can write this as \u03a9(n). Complexity theory argues that systems are complex interactions of many parts which cannot be predicted by accepted linear equations. The measure for the working storage an algorithm needs is called space complexity. , the work is O (1) comparison. Time complexity of neural network. A single iteration (loop) over all the elements in the array gives us a complexity of O(n). The time complexity function expresses that dependence. For typical values of n = 30, m = 30, and q = 5, the time complexity would be 4500, which is much higher than 110. The asymptotic complexity is defined by the most efficient (in terms of whatever computational resource one is considering) algorithm for solving the game; the most common complexity measure (computation time) is always lower-bounded by the logarithm of the asymptotic state-space complexity, since a solution algorithm must work for every. , c ~ 2d)! Contrast with exponential: For any constant c, there is a d such that n \u2192 n+d increases time. The Idea of time complexity is not to calculate how much time an algorithm will take to complete, but to compute the order of magnitude of time for the completion of computation by an algorithm. Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. Linear search is linear O(N) Binary search depends on if the tree is balanced or not. Informally, this means that the running time increases at most linearly with the size of the input. \u201d Most cryptanalytic papers discuss certi\ufb01cational attacks: Data complexity \u2014 just slightly less than the entire code book. The average to the worst case of this kind of search is a linear complexity or O(n). Inform\u00e1tica Educativa [email\u00a0protected] If there are NO nested loops we can probably guess the complexity of the code we looking at would be in the O(n). Operation count: In this technique, we consider the operations in the given algorithm or program that contribute to the execution time and count how many times those operations will be performed. These have yielded near-linear time algorithms for many diverse problems. I know the answer is O(n), but is this correct: The first element has probability $1/n$ and requires 1 comparison; the second probability $1/(n-1)$ and requires 2 comparisons. It was experimentally found in [6, 7] that time complexity of Matsui\u2019s attack on DES may be decreased with a better ranking of the values of relevant sub-key bits, though data complexity and. The List has an O(N) linear time complexity. I know that for an array of size n distinct elements, the Average Case complexity for linear search is as follows:. This obviously requires a constant number of comparison operations, i. We then verify if these times look like the time complexity we're expecting (constant, linear, or polynomial (quadratic or greater)). This study proposes linear time complexity sorting algorithms for nearest level control-based BE and TR MMC models to further accelerate the EMT simulation of the equivalent MMC-HVdc models. Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes? A) O(1) (B) O(log n ). The time complexity of su\ufb03x tree construction has been shown to be equiv-alent to that of sorting [7]. complexity addressing precisely the kind of problem raised in the last two paragraphs: Given a computational prob-lem, can it be solved by an e cient algorithm? For many common computational tasks (such as nding a solution of a set of linear equations) there is a polynomial-time algo-rithm that solves them|this class of problems is called P. Eight time complexities that every programmer should know. While that isn\u2019t bad, O (log. This video explains the time complexity analysis for binary search. BIG O Notation \u2013 Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. Generate an hypothesis: The running time is about 1 x 10-10 x N 3 seconds 4. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. 14 Code sample for Linear Regression. O(n log n) Linearithmic: This is a nested loop, where the inner loop runs in log n time. 39+ o (1)) en ln n to en ln n + O ( n ) in expectation and with high probability, which is tight up to. Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. If the element is found then its position is displayed. University of Toronto - Fall 2000 Department of Computer Science Week 12 - Complexity & Searching Complexity The complexity of an algorithm is the amount of a resource, such as time, that the algorithm requires. It is generally one of the first algorithms taught in computer science courses because it is a good algorithm to learn to build intuition about sorting. During the study of discrete mathematics, I found this course very informative and applicable. In most of the cases, you are going to see these kind of Big-O running time in your code. for(i=0; i N; i++) { for(j=0; j. Linear search performs equality comparisons and Binary search performs ordering comparisons; Let us look at an example to compare the two: Linear Search to find the element \u201cJ\u201d in a given sorted list from A-X. Firstly, we analyze the time complexity of the iterative algorithm and then recursive. Let's take an array int arr [] = { 2,1,7,5,9} Suppose we have to search an element 5. This is an example of logarithmic complexity. Time Complexity of Sorting Algorithms Let's check the time complexity of mostly used sorting algorithms. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. I Linear: RHS is a sum of multiples of previous terms of the sequence (linear combination of previous terms). The number of steps and time required to solve a problem is based on input size. In Binary search half of the given array will be ignored after just one comparison. Complexity Classes. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. See full list on towardsdatascience. \\ReaderPrograms\\ReaderFiles\\Chap02\\OrderedArray\\orderedArray. all of the mentioned. The search time increases proportionately to the number of new items introduced. While that isn\u2019t bad, O (log. Aaronson and Arkhipov argued in section 1. Complexity International-- journal for scientific papers dealing with any area of complex systems research. A single iteration (loop) over all the elements in the array gives us a complexity of O(n). If you give a condition in the inner loop that will always terminate the inner loop and/or outer loop without executing n times for all elements, then it will have less than O(n) time. Here, n is the number of elements in the linear array. It means we generate a vector that has 5 elements, and these elements are bounded in [-11,11]. Time complexity is a function dependent from the value of n. (And if the number. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. Lookups on arrays and objects are going to be constant time if you access them directly. So, the time complexity of binary search is O(log2n). It will be easier to understand after learning O(n), linear time complexity, and O(n^2), quadratic time complexity. compares each element with the value being searched for, and stops when either the value is found or the end of the array is encountered. Linear search is iterative whereas Binary search is Divide and conquer. Binary Search Algorithm and its Implementation. When the input is a random permutation, the rank of the pivot is uniform random from 0 to n \u2212 1. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. In the best-case scenario, the element is present at the beginning of the list and in the worst-case, it is present at the end. all of the mentioned. Linear search is a perfect example. It concisely captures the important differences in the asymptotic growth rates of functions. By this logic, we can say that painting pictures is slower than baking cookies. java logarithms Complexity of algorithm Time complexity Space complexity Time complexity: in big O notation. This time complexity is a marked improvement on the O(N) time complexity of Linear Search. Linear-Time Sorting. Also, each algorithm's time complexity is explained in separate video lectures. The second one runs in time sublinear in d, assuming the edit distance is not too small. Linear time: O(n). The best algorithms for sorting a random array have a run time of O(n * log n). Always takes the same time. Linear search is a perfect example. i it is exponentially better than a brute force search. The Idea of time complexity is not to calculate how much time an algorithm will take to complete, but to compute the order of magnitude of time for the completion of computation by an algorithm. \u2022 VERY difficult to develop. This is an example of logarithmic complexity. Time complexity of a related-key attack: \u201cThus, the total time complexity of Step 2(b) is about 2256 \u00b72167. Hence Bisection Search is way better than Linear Search. Search time is proportional to the list size. In this post I\u2019m going to walk through one of my favorite algorithms, the median-of-medians approach to find the median of a list in deterministic linear time. 2 , page 15. fractal image compression time complexity image compression critical issue video sequence alternative method large number computer animation multi-dimensional nearest neighbor search data storage logarithmic time decoding phase transmission time encoding step linear time image portion multi-dimensional search sequential search data compression. So, the time complexity of binary search is O(log2n). Using the hypothesis, make a prediction: When N =. Although the limiting factor for linear cryptanalysis attacks is usually the data complexity, such an improvement is relevant and can be motivated both by practical and theoretical reasons, as the following scenarios underline. The number of steps and time required to solve a problem is based on input size. Time complexity Posted 28 December 2015 - 04:35 PM Hi guys,lets say I have algorithm ,which finds ,when the number in a list is bigger than the next one. It concisely captures the important differences in the asymptotic growth rates of functions. Here, n is the number of elements in the linear array. Linear time is the best possible time complexity in situations where the algorithm has to sequentially read its entire input. Program for Recursive and Non-Recursive Binary Search in C++ - Analysis Of Algorithms / Data Structures. Visualize high dimensional data. Let us consider an algorithm of sequential searching in an array. Like an array, a linear list stores a collection of objects of a certain type, usually denoted as Time complexity, space complexity, and the O-notation : 2. It went through the entire list so it took linear time. com Linear Time Complexity. We could also say it is linear on the number of entries in the table but that is less commonly used. Huan Li, Zhouchen Lin; 21(33):1\u221245, 2020. Time Complexity : \u03b8 ( n ) Space Complexity : O(1) Linear Search Example. Time complexity of linear search -O(n) , Binary search has time complexity O(log n). time complexity of a linear cryptanalysis attack using algorithm 2. This is an example of logarithmic complexity. Note that an algorithm might take di\ufb00erent amounts of time on inputs of the. If you were to find the name by looping through the list entry after entry, the time complexity would be O (n). Lets say I have the list 10,20,30,40,50,60,30,40,50. The second one runs in time sublinear in d, assuming the edit distance is not too small. That is, I'm looking for references that looks like the following. complexity = in between logN and N. Here, although your array is of a fixed size, the time needed to complete the operation is still a linear function of the number of elements in the array. The main points in these lecture slides are:Time Complexity, Complexity of Algorithms, Execution Time, Space Complexity, Worst Case Analysis, Division of Integers, Number of Comparisons, Binary Search, Average Case Complexity, Complexity of Bubble Sort. BIG O Notation \u2013 Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. // Time complexity: O(1) // Space complexity: O(1) int x = 15; x += 6; System. This requires to scan the array completely and check each element for the array that we need to search. This first book consists of chapters 1 and 2 of the fourth volume. starts in the middle, then see if the value being searched for is greater or less than the middle value. As a rule of thumb, it is best to try. 2) and, assuming average document length does not change over time,. O(1) indicates that the algorithm used takes \"constant\" time, ie. Sequential search write a sequential search function and then find the best, worst, and average case time complexity. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Thus, the amount of time taken and the number of elementary operations performed by the algorithm are taken to differ by at most a constant factor. Time Complexity of Binary Search Algorithm is O(log 2 n). linear: sorting twice the number of elements takes quite a bit more than just twice as much time; searching (using binary search) through a sorted list twice as long, takes a lot less than twice as much time. This paper reports some new results on the average time complexity of EAs. In: Markov, A. Serial Search - Analysis. Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster searching comparison to Linear search. what we do is we simply loop over array and check whether it is. For a linear-time algorithm, if the problem size doubles, the number of operations also doubles. Algorithm analysis is an important part of a broader computational complexity theory, which provides theoretical estimates for the resources needed by any. The measure for the working storage an algorithm needs is called space complexity. Let n represent the size of the array arr. We could also say it is linear on the number of entries in the table but that is less commonly used. Always takes the same time. algorithm runs in near-linear time, namely d1+\u03b5 for any \ufb01xed \u03b5 > 0. Reducing the number of generations, i. 1 of [] that the exchange symmetry of identical bosons creates an effective entanglement (a kind of artificial entanglement), which would be the origin of the computational complexity in linear optics. The running time of the loop is directly proportional to N. Therefore, by using hash table, we can achieve linear time complexity for finding the duplicate. characterises a function based on growth of function C. In order to be able to classify algorithms we have to define limiting behaviors for functions describing the given algorithm. The time complexity of above algorithm is O(n). Here, complexity refers to the time complexity of performing computations on a multitape Turing machine. These Multiple Choice Questions (mcq) should be practiced to improve the Data Structure skills required for various interviews (campus interview, walk-in interview, company interview), placement, entrance exam and other competitive examinations. Diagram above is from Objective-C Collections by NSScreencast. Trees Data Structures for Beginners. Space complexity : O (1) O(1) O (1) or (O (n) O(n) O (n)) We sorted nums in place here - if that is not allowed, then we must spend linear additional space on a copy of nums and sort the copy instead. The time complexity of ECL2 is O (n + M \u03f5 1 w 2) \u2060, where O(n) is the time complexity of scoring and binning, and O (M \u03f5 1 w 2) is the time complexity of enumerating bin pairs. Sort an array of 0's, 1's and 2's in linear time complexity; Checking Anagrams (check whether two string is anagrams or not) Relative sorting algorithm; Finding subarray with given sum; Find the level in a binary tree with given sum K; Check whether a Binary Tree is BST (Binary Search Tree) or not; 1[0]1 Pattern Count. near-linear time. As a rule of thumb, it is best to try. However, in my previous experiments, it appears to be O (N), namely linear complexity!. The search stops when the item is found or when the search has examined each item without success. For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. A few common algorithmic complexities: O(log n) - binary search. Dual first-order methods are essential techniques for large-scale constrained convex optimization. The best case time in linear search is for the first element i. This linear search has a time complexity of O(n). Aaronson and Arkhipov argued in section 1. given two natural numbers $$n$$ and $$m$$, are they relatively prime?. O(N)\u2014Linear Time: Linear Time Complexity describes an algorithm or program who\u2019s complexity will grow in direct proportion to the size of the input data. what we do is we simply loop over array and check whether it is. In case of a sorted array, Binary Search is faster but the caveat here is also on how arrays are treated by the Language Translator. Let us assume that given an array whose elements order is not known. Consider that we have an algorithm, and we are calculating the time. Time Complexity of Bisection Search is O(log n). Tests are robust , non-parametric statistical tests, since timing is noisy (so need to be robust), and noise can take various forms (so non-parametric, since no particular model of noise). One example is the binary search algorithm. The number of operations in the best case is constant (not dependent on n). The time complexity of su\ufb03x tree construction has been shown to be equiv-alent to that of sorting [7]. Download Binary search program. Alright, so we have linear-over-n many logarithmic-over-n loops. java graph-algorithms competitive-programming dfs binary-search-tree common-algorithms time-complexity implementation bfs longest-common-subsequence binary-search segment-tree binary-indexted-tree two-pointers space-complexity all-pairs-shortest-path matching-algorithm maximal-bipartite-matching lower-bound lowest-common-ancestor. Linear Search is sequential search which scans one item at a time. Space Complexity. In order to speed up the static analyses formulated using the Dyck-CFL reachability problems, we propose an efficient algorithm of O(n) time for the Dyck-CFL reachability problem when the graph considered is a bidirected tree with specific constraints, while a na\u00efve algorithm runs in O(n2) time. This requires to scan the array completely and check each element for the array that we need to search. This web page gives an introduction to how recurrence relations can be used to help determine the big-Oh running time of recursive functions. See full list on yourbasic. As investigated in [ ],theHPPcanbesolvedusing. Time complexity of Bubble sort in Worst Case is O(N^2), which makes it quite inefficient for sorting large data volumes. All have polynomial time complexity while some allow very long steps in favorable circumstances. We show an improved algorithm for the satis ability problem for circuits of constant depth and linear size. We will see more about Time Complexity in future. In a serial search, we step through an array (or list) one item at a time looking for a desired item. for(i=0; i N; i++) { for(j=0; j. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes? A) O(1) (B) O(log n ). It measures the worst case time complexity or the longest amount of time an algorithm can possibly take to complete. O(1) is the best possible time complexity! Data structures like hash tables make clever use of algorithms to pull off constant time operations and speed things up dramatically. So, the time complexity of binary search is O(log2n). Eight time complexities that every programmer should know. O(n\u00b2) \u2013 Quadratic Time. Serial Search - Analysis. The time complexity of an algorithm is commonly expressed using big O notation , which excludes coefficients and lower order terms. The first is the way used in lecture - \"logarithmic\", \"linear\", etc. The improvement of the proposed linear-time algorithm compared with ECL2 (Yu et al. Here you will learn about python binary search with program and algorithm. This search requires only one unit of space to store the element to be searched. Time complexity. These problems will introduce things (like the variable i above) just to waste your time. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. For instance, it is known since the 1960s and 70s (e. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. Hence time complexity of the Binary search is O(LogN). The best case time in linear search is for the first element i. Most of the time we can speak on sorting integers in linear time, but as we can see later this is not the only case. Hourly Update. Time complexity Posted 28 December 2015 - 04:35 PM Hi guys,lets say I have algorithm ,which finds ,when the number in a list is bigger than the next one. It is easy to see that $$\\widetilde{\\mathcal {S}}$$ can be obtained in one pass through $$\\widetilde{\\mathcal {A}}$$ and $$\\widetilde{\\mathcal {B}}$$, therefore in linear time. Time complexity (linear search vs binary search) 1. I n linear search, we need to write more code whereas in binary search we need to write less code. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. now do a linear search starting from X(conservative but accurate second step - slow). So the option is 'B'. \ubb34\uc2a8\ub9d0\uc778\uc9c0\ubaa8\ub974\uaca0\ub2e4\uba74 \uc544\ub798\uae00\uc744 \ucb49. In this book, Keith Morrison introduces complexity theory to the world of education, drawing out its implications for school leadership. For better understanding,lets take an example: given an array arr[]={12,11,4,0,3,5} and we want to search whether 5 present in the given array or not. In Binary search half of the given array will be ignored after just one comparison. Nested for loops are the perfect example of this category. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. The complexity of linear search algorithm is. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Run time is O (log N) Sample code for ordered array. To measure Time complexity of an algorithm Big O notation is used which: A. In order to speed up the static analyses formulated using the Dyck-CFL reachability problems, we propose an efficient algorithm of O(n) time for the Dyck-CFL reachability problem when the graph considered is a bidirected tree with specific constraints, while a na\u00efve algorithm runs in O(n2) time. Indeed, 100 cookies don\u2019t take much longer than 12 cookies \u2014 provided you have a big enough bowl. \u2022 DEMO \u2022 Conclusion: Maybe I can SCALE well \u2026 Solve O(10^12) problems in O(10^12). Time complexity (linear search vs binary search) 1. linear regression and the correlation coefficient. Tests are robust , non-parametric statistical tests, since timing is noisy (so need to be robust), and noise can take various forms (so non-parametric, since no particular model of noise). Binary search. So the option is 'B'. This study proposes linear time complexity sorting algorithms for nearest level control-based BE and TR MMC models to further accelerate the EMT simulation of the equivalent MMC-HVdc models. From this we can see these equations are similar and our equation matches linear equation. Counting sort and radix sort assume that the input consists of integers in a small. Since binary search has a best case efficiency of O(1) and worst case (average case) efficiency of O(log n), we will look at an example of the worst case. As the number increases so does the time difference. For a general alphabet, su\ufb03x tree construction has time bound of \u0398(nlogn). Here is the official definition of time complexity. Hence number of times while loop will execute will determine the complexity of the algorithm. If you had to search for a name in a directory by reading. If we assume we needed to search the array n times the total worst case run time of the linear searches would be O (n^2)). All complex systems can be seen as a number of nodes joined together \u2013 lines and junctions \u2013 or in the case of the human brain, long spindly nerve cells and synapses. Linear Search Simple search from the first element to the last till we find the required element. So, the time complexity of binary search is O(log2n). It has a complexity of n 2. Hourly Update. Let's take an array int arr [] = { 2,1,7,5,9} Suppose we have to search an element 5. the complexity. The time complexity function expresses that dependence. As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. Finally, together with the analysis, it is concluded that the linear time complexity is validated based on the experiments. This is a more mathematical way of expressing running time, and looks more like a function. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth. at 11:59pm \u2022 Asymptotic analysis Asymptotic Analysis CSE 373 Data Structures & Algorithms Ruth Anderson Spring 2007 04/04/08 2 Linear Search vs Binary Search Linear Search Binary Search Best Case Asymptotic Analysis Worst Case So \u2026 which algorithm is better?. Time complexity (linear search vs binary search) 1. Time Complexity and the divide and conquer strategy Or : how to measure algorithm run-time And : design efficient algorithms Oct. Worst-case running time - the algorithm finds the number at the end of the list or determines that the number isn't in the list. Complexity and running time Factors: algorithmic complexity, startup costs, additional space requirements, use of recursion (function calls are expensive and eat stack space), worst-case behavior, assumptions about input data, caching, and behavior on already-sorted or nearly-sorted data; Worst-case behavior is important for real-time systems. For a general alphabet, su\ufb03x tree construction has time bound of \u0398(nlogn). That is, I'm looking for references that looks like the following. Solving linear equations can be reduced to a matrix-inversion problem, implying that the time complexity of the former problem is not greater than the time complexity of the latter. Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. This first book consists of chapters 1 and 2 of the fourth volume. It's not easy trying to determine the asymptotic complexity (using big-Oh) of recursive functions without an easy-to-use but underutilized tool. This technique is probably the easiest to implement and is applicable to many situations. Definition of time complexity in the Definitions. Algorithmic Complexity Notes on Notation: Algorithmic complexity is usually expressed in 1 of 2 ways. Data Structure MCQ - Complexity. In linear search we simply iterate over elements and check whether it is the desired element or not. for(i=0; i N; i++) { for(j=0; j. If the values match it will return success. For a general alphabet, su\ufb03x tree construction has time bound of \u0398(nlogn). In our previous tutorial we discussed about Linear search algorithm which is the most basic algorithm of searching which has some disadvantages in terms of time complexity, so to overcome them to a level an algorithm based on dichotomic (i. In a serial search, we step through an array (or list) one item at a time looking for a desired item. See full list on freecodecamp. Now considering the worst case in which the search element does not exist in the list of size N then the Simple Linear Search will take a total of 2N+1. Algorithm analysis is an important part of a broader computational complexity theory, which provides theoretical estimates for the resources needed by any. Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster searching comparison to Linear search. In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input:226. The running time of the two loops is proportional to. Results Here, then, as a concrete example, is a plot of the run-times of the most interesting algorithms on an Intel Core i7 running at 2. O(1): Constant Time Complexity. O(n) - finding the largest item in an unordered list. This first book consists of chapters 1 and 2 of the fourth volume. Since binary search algorithm splits array in half every time, at most log 2 N steps are performed. Best Case. larger search space of constituent trees (compared to the space of dependency trees) would make it unlikely that accurate parse trees could be built deterministically, we show that the precision and recall of constituents produced by our parser are close to those produced by statistical parsers with higher run-time complexity. Time Complexity : This section explains the importance of time complexity analysis, the asymptotic notations to denote the time complexity of algorithms. The best case for a linear search algorithm is to have the value x for which one is searching located at the first position of the ADT. 2) and, assuming average document length does not change over time,. Most algorithms, however, are built from many combinations of these. The \"Binary Search Time Complexity\" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Here complexity is said to be linear. This time complexity is a marked improvement on the O(N) time complexity of Linear Search. So time complexity in the best case would be \u0398(1) Most of the times, we do worst case analysis to analyze algorithms. Suppose varMin=-11, and varMax=11, and varSize=5. So there must be some type of behavior that algorithm is showing to be given a complexity of log n. Constant time compelxity, or O(1), is just that: constant. This time complexity of binary search remains unchanged irrespective of the element position even if it is not present in the array. It will be easier to understand after learning O(n), linear time complexity, and O(n^2), quadratic time complexity. first do a binary search (agressive first step - fast) with 1 bulb. The time factor when determining the efficiency of algorithm is measured by. all of the mentioned. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. worst case, the time for insertion is proportional to the number of elements in the array, and we say that the worst-case time for the insertion operation is linear in the number of elements in the array. As we learned in the previous tutorial that the time complexity of Linear search algorithm is O(n), we will analyse the same and see. When the input is a random permutation, the rank of the pivot is uniform random from 0 to n \u2212 1. If the element is found then its position is displayed. So for any value of n it will give us linear time. For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. compares each element with the value being searched for, and stops when either the value is found or the end of the array is encountered. Gorky University Publishers, Gorky (1985) (in Russian) Google Scholar. If you had to search for a name in a directory by reading. Most algorithms, however, are built from many combinations of these. Time and space complexity depends on lots of things like. However, in my previous experiments, it appears to be O (N), namely linear complexity!. Neglecting the constant value 5 the complexity would be N as loop will run N times so it does not fit the definition of linear time. We observe how space complexity evolves when the algorithm\u2019s input size grows, just as we do for time complexity. We then verify if these times look like the time complexity we're expecting (constant, linear, or polynomial (quadratic or greater)). O(n) is for linear complexity, O(n 2) is for quadratic. Worst Case time complexity is O(n) which means that value was not found in the array (or found at the very last index) which means that we had to iterate n times to reach to that conclusion. In case of the monks, the number of turns taken to transfer 64 disks, by following the above rules, will be 18,446,744,073,709,551,615; which will surely take a lot of time!!. > How are strings stored in Python? As arrays? As linked lists?. O(N^2) because it sorts only one item in each iteration and in each iteration it has to compare n-i elements. This video is meant for educational. Introduction The time complexity of a given algorithm can be obtained from theoretical analysis and computational analysis according to the algorithm\u2019s running. So, an algorithm taking X second or 2X + 3 seconds have the same complexity. This means that as the input grows, the algorithm takes proportionally longer to complete. The idea behind linear search is to compare the search item with the elements in the list one by one (using a loop) and stop as soon as we get the first copy of the search element in the list. Learn more about time complexity of neural network. The \"Binary Search Time Complexity\" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Unbalanced binary search tree can turn into a linked list in the worst case if the elements added are in descending order so O(N) time complexity. For typical values of n = 30, m = 30, and q = 5, the time complexity would be 4500, which is much higher than 110. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. If we plot the graph of an+b for different values of n we will see that it is a straight line. These approximation and runtime guarantees are signi\ufb01cantly better then the bounds known for worst-case inputs, e. What\u2019s the maximum number of loop iterations? log2n That is, we can\u2019t cut the search region in half more than that many times. Time complexity (linear search vs binary search) 1. java that reads from standard input a sequence of integers that are between 0 and 99 and prints to standard output the same integers in sorted order. The time complexity of linear search is O(n), meaning that the time taken to execute increases with the number of items in our input list lys. For example. Morzhakov, N. Linear Homogeneous Recurrences De nition A linear homogeneous recurrence relation of degree k with constant coe cients is a recurrence relation of the form an = c1an 1 + c2an 2 + + ck an k with c1;:::;ck 2 R , ck 6= 0. the complexity. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. \u2022 DEMO \u2022 Conclusion: Maybe I can SCALE well \u2026 Solve O(10^12) problems in O(10^12). Linear time or O(n). Algorithm Complexity When N doubles Examples Constant 1 increases fixed times No loop Logarithmic log N increases constant Binary search Linear N doubles Traverse an array Linearithmic NlogN more than doubles Quick/Merge Sort, FFT Quadratic N^2 increases fourfold B Cubic N^3 increases eightfold NxN matrix multiplication Exponential 2^N running time squares!. Bubble sort is a simple, inefficient sorting algorithm used to sort lists. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. Total comparisons in Bubble sort is: n ( n \u2013 1) / 2 \u2248 n 2 \u2013 n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. See full list on yourbasic. Know Thy Complexities! Hi there! This webpage covers the space and time Big-O complexities of common algorithms used in Computer Science. The time complexity of linear search is O(N) while binary search has O(log 2 N). Run time is O (log N) Sample code for ordered array. i it is exponentially better than a brute force search. Given an arbitrary network of interconnected nodes, each with an initial value, we study the number of time-steps required for some (or all) of the nodes to gather all of the initial values via a linear iterative strategy. If connections are sparse, then sparse math can be used for the gradient computations, etc leading to reduced complexity.", "date": "2020-11-29 07:31:10", "meta": {"domain": "marathon42k.it", "url": "http://marathon42k.it/zcy/time-complexity-of-linear-search.html", "openwebmath_score": 0.5858543515205383, "openwebmath_perplexity": 614.0522630942693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534327754854, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8612026083995186}}
{"url": "https://math.stackexchange.com/questions/2739192/the-number-555-555-can-decompose-as-the-product-of-two-factors-of-three-digit", "text": "The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?\n\nThe number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?\n\nI've seen the answer to the question, and there is only one way: Since $555, 555 = 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \\cdot 7 \\cdot 37) (5 \\cdot 11 \\cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure?\n\nSorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish.\n\n\u2022 list all the divisors in order. $\\sqrt{555555} \\approx 745.36.$ One of your divisors must be larger than that but still smaller than 1000. The other divisor in your pair will be between 555.56 and 745.36. I guess you can find them by hand, look at all the products of two primes, then three primes. You do not need to do four primes because you already did two \u2013\u00a0Will Jagy Apr 16 '18 at 2:34\n\u2022 Yeah, two primes is too small, the biggest is $13 \\cdot 37 = 481.$ So, three primes each \u2013\u00a0Will Jagy Apr 16 '18 at 2:38\n\nNothing wrong with the approach Will gives in the comments. Here's another way. Obviously $555,555=555\\times1001$, but $1001=7\\times11\\times13$ is a little too large. The way to make it a little smaller is to swap the factor 7 with the factor 5 in 555, which gives you your solution, $(3\\times7\\times37)(5\\times11\\times13)$.\n\nThe factors $7,11,13$ can't be used together, since $(7)(11)(13)=1001$.\n\nSo one of the groups, group $1$ say, must have exactly two of the factors $7,11,13$.\n\nHence the factors $3$ and $5$ can't both be in group $1$, else the product of the factors in group $1$ would be at least $(3)(5)(7)(11) > (7)(11)(13)$.\n\nSimilarly, the factor $37$ can't be in group $1$, else the product of the factors in group $1$ would be at least $(37)(7)(11) > (7)(11)(13)$.\n\nLabel the other group as group $2$.\n\nThus, group $2$ contains\n\n\u2022 $37$\n\u2022 Exactly one of $7,11,13$.\n\u2022 At least one of $3,5$.\n\nBut since $(37)(27) = 999$, the factors in group $2$ other than $37$ must have a product which is at most $27$.\n\nIt follows that neither of the factors $11$ or $13$ is in group $2$, since $(11)(3) > 27$, and $(13)(3) > 27$.\n\nSo the factor $7$ must be in group $2$, and the factors $11,13$ must be in group $1$.\n\nSince the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$.\n\nHence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$.\n\nThus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$.\n\nIt is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \\cdot 5 \\cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\\cdot 13 \\cdot 37 \\gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$.\n\n\u2022 I also thought that trying to solve for the same question just that for 2 factors of 2 digits and then I would split one of the digits into 2. Would this help me solve this type of question faster? Also, the question doesn't require me to state the numbers, but it does require me to state the amount of ways I can decompose it. \u2013\u00a0Brian Blumberg Apr 16 '18 at 2:44\n\u2022 I don't understand the first sentence. What does it mean to split one of the digits into $2$? Having gotten down to $9$ possibilities it won't take long to try them all. You could note that $3\\cdot 11\\cdot 37$ is too large, so $37$ must be by itself and $11,13$ must be in the other set. Now just try each of $3,5,7$ times $11\\cdot 13=143$ to see which are in range and find that only $5$ works. \u2013\u00a0Ross Millikan Apr 16 '18 at 2:53\n\u2022 Yeah, never mind my question. I forgot to delete it after I realized that. Thanks. \u2013\u00a0Brian Blumberg Apr 16 '18 at 3:01\n\nThe prime factor are $3 \u00b7 5 \u00b7 7 \u00b7 11 \u00b7 13 \u00b7 37$ so there are $2^6=64$ factors and $32$ complement pairs. Just list them all but don't bother with those that are less than $555555/999=556$\n\nToss out, $1,3,5,7,11,13,37,3*5,3*7,3*11,3*13,3*37,3*5*7,3*5*11,3*5*13, 3*5*37,3*7*11,3*7*13$ (that's 18 that are too small).\n\n$3*7*37=777$ and its compliment is $5*11*13=715$. (That's 1 that is acceptible)\n\nWe can continuing tossing out $3*11*13$ and $3*11*37$and $3*13*37$ are too high so we toss them. (That's 21 that are unaccptible)\n\n$3*5*7*11$ is too high so there wonvt be any more factors that are multiples of $3$ in range. Hence no other complements which aren't multiples of $3$ will be in range either.\n\nOf the 9 we haven'haven't considered: $3*5*7*13,3*5*7*37,3*5*11*13,3*5*11*37,3*5*13*37,3*7*11*13,3*7*11*37,3*7*13*37,3*11*13*37$ are all too big.\n\nSo that was exhaustive.\n\nWe can write $$555,555=5\\times111,111$$ and notice that $111=37\\times3$, so we have $$555,555=5\\times37\\times3003$$ and since $1001=7\\times11\\times13$, the prime factorization is $$555,555=3\\times5\\times7\\times11\\times13\\times37.$$\n\nIf we multiply each of the three combinations $11\\times13$, $11\\times37$ and $13\\times37$ by $3$, we see that only $3\\times(13\\times37)$ exceeds $1000$ which is a four-digit number.\n\nHence $37$ must pair with two other one-digit numbers, and $11\\times13$ must pair with either $3$ or $5$ since $7\\times11\\times13>1000$.\n\nIf $11\\times13$ pairs with $3$, then the other product must be $$37\\times(5\\times7)>1000$$ which is not a three-digit number.\n\nTherefore, the only possible combination for $555,555=P_1\\times P_2$ is $$P_1=5\\times11\\times13,\\quad P_2=3\\times7\\times37.$$\n\nI think the procedure might be based on Fermat method.\n\nWe can calculate that the number $555555$ can be expressed as a difference of two squares:\n\n$746^2 - 31^2=556516-961$\n\nNumber $31$ is the distance from $746$ both ways which gives $715$ and $777$ accordingly. From here the smallest prime factor of:\n\n$777$ is number $3$ which equals to $3\\cdot259$ and for\n\n$715$ is number $5$ which equals to $5\\cdot143$\n\nFurther factorisation results in final solution where $143=11\\cdot13$ and $259=7\\cdot37$ therefore:\n\n$(5\\cdot11\\cdot13)(3\\cdot7\\cdot37)$ can satisfy equation.\n\nIf $555{,}555=ab$ with $a,b\\lt1000$, then we must also have $a,b\\gt555$ (e.g., if $b\\lt1000$, then $a=555{,}555/b\\gt555{,}555/1000\\gt555$). Now since $555{,}555=3\\cdot5\\cdot7\\cdot11\\cdot13\\cdot37$, we may assume, without loss of generality, that $a=37k$. From $a\\gt555=15\\cdot37$, we see that $k\\gt15$, and from $a\\lt1000$, we see that $k\\le\\lfloor1000/37\\rfloor=27$. The only product $k$ of the primes $3$, $5$, $7$, $11$, and $13$ that falls in the interval $15\\lt k\\le27$ is $k=3\\cdot7=21$. Thus $a=37\\cdot21=777$, $b=5\\cdot11\\cdot13=715$ is the only factorization of $555{,}555$ into two three-digit numbers.", "date": "2019-05-21 18:44:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2739192/the-number-555-555-can-decompose-as-the-product-of-two-factors-of-three-digit", "openwebmath_score": 0.9133464694023132, "openwebmath_perplexity": 171.56330717441787, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975946444307138, "lm_q2_score": 0.8824278726384089, "lm_q1q2_score": 0.8612023446589672}}
{"url": "https://math.stackexchange.com/questions/1960598/contrapositive-in-change-from-no-self-defeating-object-to-every-object-can-be", "text": "# Contrapositive in change from \u201cno self-defeating object\u201d to \u201cevery object can be defeated\u201d?\n\nHere, Terence Tao presents a collection of similar mathematical arguments that he calls \"no self-defeating object\" (examples are Euclids proof of the infinitude of the primes and Cantor's theorem). In the second post, he remarks that one can reformulate these \"no-self defeating object\" arguments to get a \"every object can be defeated\"-version.\n\nThe simplest example \"no self-defeating object\" goes as follows:\n\nProposition 1 (No largest natural number). There does not exist a natural number N that is larger than all the other natural numbers.\n\nProof: Suppose for contradiction that there was such a largest natural number N. Then N+1 is also a natural number which is strictly larger than N, contradicting the hypothesis that N is the largest natural number.\n\nThe corresponding \"every object can be defeated\"-version is:\n\nProposition 1\u2032. Given any natural number N, one can find another natural number N' which is larger than N.\n\nProof. Take N' := N+1.\n\nTerence Tao also remarks:\n\n\"This is done by converting the \u201cno self-defeating object\u201d argument into a logically equivalent \u201cany object can be defeated\u201d argument, with the former then being viewed as an immediate corollary of the latter. This change is almost trivial to enact (it is often little more than just taking the contrapositive of the original statement), but it does offer a slightly different \u201cnon-counterfactual\u201d (or more precisely, \u201cnot necessarily counterfactual\u201d) perspective on these arguments which may assist in understanding how they work.\"\n\nMy question: What has the contrapositive to do with the change from \"no self-defeating object\" to \"every object can be defeated\"?\n\nAs I understand it, the \"no self-defeating object\"-version is of the form \"$\\neg\\exists x: P(x)$\" and the \"every object can be defeated\"-version is \"$\\forall x:\\neg P(x)$\". That these are equivalent is de Morgan for quantifiers, what does it have to do with contrapositives?\n\n\u2022 \u2013\u00a0Henning Makholm Oct 9 '16 at 12:14\n\u2022 @HenningMakholm: Ah, thanks. So do you think that Terence Tao just compared the change from $\\neg\\exists x : P(x)$ to $\\forall x : \\neg P(x)$ to taking the contrapositive, but is not saying that this is the same as taking contrapositives? \u2013\u00a0user376483 Oct 9 '16 at 12:45\n\u2022 I'm not privy to Tao's thoughts, so the most I'm saying is that I too see a similarity between this kind of rewriting and contraposition, and once mused that it might be useful to call it contraposition. (Not many here agreed with that, though -- but it nice to see perhaps Tao might). \u2013\u00a0Henning Makholm Oct 9 '16 at 12:48\n\u2022 One way to make it look more like a contrapositive is to view it as a change from \"if $m$ is larger than all natural numbers then $m$ is not a natural number\" to \"if $m$ is a natural number then $m$ is not larger than all natural numbers\". Of course, Tao only says that the process he describes is \"often\" \"little more\" than just taking the contrapositive, he does not claim that what he is doing is literally a contrapositive. I do find that mathematicians and logicians often use \"contrapositive\" informally in a broader sense than its formal meaning. \u2013\u00a0Carl Mummert Oct 9 '16 at 16:42\n\nIf, as is standard in presentations of intuitionistic logic, you treat $\\lnot \\phi$ as $\\phi \\Rightarrow \\mathsf{false}$ then the role of the contrapositive here becomes clear: the contrapositive of:\n\n$$(\\exists N \\in \\Bbb{N}\\cdot\\forall m\\in \\Bbb{N}\\cdot N > m) \\Rightarrow \\mathsf{false}$$\n\nis:\n\n$$\\lnot \\mathsf{false} \\Rightarrow \\lnot(\\exists N \\in \\Bbb{N}\\cdot\\forall m\\in \\Bbb{N}\\cdot N > m)$$\n\nwhich, using De Morgan's laws and a tiny bit of arithmetic reasoning (to make things agree with Tao's presentation) is equivalent to: $$\\forall N \\in \\Bbb{N}\\cdot\\exists m\\in \\Bbb{N}\\cdot m > N.$$\n\nThis transformation giving Tao's Proposition $1'$ makes clear the innate constructive nature of the reasoning that is presented in disguise in the proof by contradiction in Tao's Proposition $1$.\n\n\u2022 After all, this also relies on De Morgan's rule, and I don't see why you first interpret $\\phi$ as $\\phi\\implies\\bot$ and form the contrapositive of $(\\exists N \\in \\Bbb{N}\\cdot\\forall m\\in \\Bbb{N}\\cdot N > m) \\Rightarrow \\mathsf{false}$. One could just form the contrapositive of $\\exists N \\in \\Bbb{N}\\cdot\\forall m\\in \\Bbb{N}\\cdot N \\geq m$ which is $\\exists N \\in \\Bbb{N}\\cdot\\forall m\\in \\Bbb{N}\\cdot N > m$. \u2013\u00a0user377104 Oct 25 '16 at 17:24\n\u2022 Of course it relies on De Morgan's laws: the point is that the end result is a constructive truth that captures exactly what the proof actually proves. The reason for interpreting $\\lnot \\phi$ as $\\phi \\Rightarrow \\mathsf{false}$ is because the notion of contrapositive is usually associated with implications: which is the point of the question: \"what does this have to do with contrapositives?\" The two existentially quantified statements that you claim are contrapositives are not contrapositives in any sense of the term that I am aware of. \u2013\u00a0Rob Arthan Oct 25 '16 at 19:16", "date": "2019-10-21 22:44:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1960598/contrapositive-in-change-from-no-self-defeating-object-to-every-object-can-be", "openwebmath_score": 0.6751436591148376, "openwebmath_perplexity": 252.89437993172734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147193720648, "lm_q2_score": 0.885631484383387, "lm_q1q2_score": 0.8612010913537365}}
{"url": "http://math.stackexchange.com/tags/proof-writing/hot?filter=month", "text": "# Tag Info\n\n13\n\nThe contrapositive of the statement If $\\overbrace{\\text{$ab$and$a+b$have the same parity}}^{\\large P}$, then $\\overbrace{\\text{$a$is even and$b$is even}}^{\\large Q}$. is If $\\overbrace{\\text{$a$is odd or$b$is odd}}^{\\large\\lnot Q}$, then $\\overbrace{\\text{$ab$and$a+b$have different parities}}^{\\large\\lnot P}$. Note that $Q$ is the ...\n\n9\n\nYour result is an immediate consequence of the following proposition. Proposition. Suppose $X\\subseteq Y$. Then $\\mathscr P(X)\\subseteq\\mathscr P(Y)$. Proof. Let $E\\in\\mathscr P(X)$. Then $E\\subseteq X\\subseteq Y$ so that $E\\subseteq Y$. Hence $E\\in\\mathscr P(Y)$. This proves $\\mathscr P(X)\\subseteq\\mathscr P(Y)$. $\\Box$ Do you see how your problem is now ...\n\n8\n\n\\begin{align} |S| & = |S-T| + |S\\cap T| \\\\[8pt] |T| & = |T-S| + |S\\cap T| \\end{align} If $|S-T|=|T-S|$, then the two right sides are the same, so the two left sides are the same. We can also write a proof explicitly dealing with bijections. You ask why one would \"assume\" a bijection exists. The bijection $g$ that you write about is not simply ...\n\n7\n\nYou can proceed directly as follows: $2x = (x+y) + (x-y)$ which must be irrational as it is the sum of a rational and an irrational. So $x$ is irrational. Similarly $2y = (x+y) - (x-y)$ is irrational.\n\n6\n\nnot induction, but maybe useful to note: firstly, since 3 is a prime, we have $n^3 \\equiv_3 n$ (Fermat's little theorem) secondly $2n \\equiv_3 -n$ (since $3n = 2n + n \\equiv_3 0$) adding these two results: $$n^3 + 2n \\equiv_3 n-n =0$$\n\n6\n\nBy Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$P^{T}AP=\\pmatrix{\\mu_1 \\\\ & \\ddots \\\\ && \\mu_n}$$ where $\\mu_i>0$ is eigenvalue of $A$, and $\\space P^{T}P=P^{-1}P=I$. For any $v$, let $v=Pu$. Then $$v^TAv=u^TP^TAPu=\\sum_{k=1}^n\\mu_ku_k^2>0$$\n\n5\n\nYou need to show us some effort in the future. First, to show two sets are equal, we normally pick an element of the first set, show it is contained in the second, then pick an element in the second, and show it is contained in the first. If we suppose $x \\in A$, then $x=2k$ for some integer $k$. Since $x = 2k$, $x = 2(k-1)+2$, and since $k-1$ is an ...\n\n5\n\nTake $N\\in\\Bbb N$ such that $|a_n-L|<1$ for $n\\ge N$ and let $$M=\\max\\{|a_1|,\\ldots,|a_N|,L+1\\}.$$\n\n5\n\nLet $f:\\{1,\\ldots,n\\}\\to X$ be a surjection. Suppose, for the sake of contradiction, that $X$\u00a0has at least $n+1$\u00a0distinct elements $\\{x_1,\\ldots,x_{n+1}\\}\\subseteq X$. Since $f$ is a surjection, there exists, for each $i\\in\\{1,\\ldots,n+1\\}$, some $k_i\\in\\{1,\\ldots,n\\}$ such that $f(k_i)=x_i$. Since $f$ is a function and the $(x_i)_{i=1}^{n+1}$ are distinct, ...\n\n4\n\nWe have that $F_n>F_{n-1}$ then $$F_{n+1}=F_n+F_{n-1}>2F_{n-1}>2\\cdot2^{(n-1)/2}=2^{(n+1)/2}$$\n\n4\n\nHINT: For each $x\\in X$, let $A_x=\\{k\\mid f(k)=x\\}$. Then each $A_x$ is non-empty. Use that to construct an injection from $X$ into $\\{1,\\ldots,n\\}$.\n\n4\n\nHINT: No, you can\u2019t assume that $A=C$ and show that the inclusions hold: that\u2019s the converse of what you\u2019re supposed to prove, and an implication and its converse are not logically equivalent. Use the fact that $A\\subseteq B$ and $B\\subseteq C$ to show that $A\\subseteq C$. You\u2019re given that $C\\subseteq A$, so the rest is straightforward.\n\n4\n\nThe given inequality is equivalent to $a^3-a=a(a^2-1)>0$. By multiplying both sides by $a^2+1$, which is always positive, we get $a(a^2-1)(a^2+1)>a^2+1>0$, or $a^5-a>0$.\n\n4\n\nI would write it as: Let $P(n)$ stand for the expression: $$\\forall x\\leq n(x\\not\\in A)$$ Then use the assumption that $A$ has no least element to prove that $P(1)$ and $P(n)\\implies P(n+1)$. Thus, we've shown that $\\forall x:x\\not\\in A$, which means $A$ is empty. That's essentially the same as your proof, but uses less set notation.\n\n4\n\nUsually, a proof by contradiction of the statement $p \\implies q$ is when you assume that the opposite of the desired conclusion is true (i.e., assume the negation of $q$ is true), and follow a few logical implications until you reach a statement that somehow explicitly or implicitly contradicts an initial assumption from the statement $p$. Meanwhile, a ...\n\n4\n\nThe Lambert W-function is a function $W(z)$ which solves $z=W(z)e^{W(z)}$. It is a multi-valued function. In this case, you are trying to solve: $$e^{x\\pi i/2} = x$$ of: $$\\frac{-\\pi i}{2}=\\frac{-x\\pi i}{2}e^{-x\\pi i/2}$$ So $$x\\frac{-\\pi i}{2} = W(-\\pi i/2)$$ or $$x =\\frac{2i}{\\pi} W\\left(\\frac{-\\pi i}{2}\\right)$$ I don't think you can do better ...\n\n3\n\nYou won't be surprised to learn that in the last seventy-plus years since Tarski's book was first published in English, many other books have been appeared which will perhaps serve better as introductions to modern logic. And if you have downloaded my Teach Yourself Logic, you will have seen my \"entry-level\" suggestions on formal logic at the beginning of ...\n\n3\n\nYou cannot go like this from $k=0$ to $k=1$ (i.e. $k=1$ cannot be expressed in the form $m+l$ as you wrote).\n\n3\n\n$N(t)$ is not equal to $N(t-s)+N(s)$, but $$N(t) = \\Big(N(t) - N(s)\\Big) + \\Big(N(s)\\Big)$$ and the two expressions inside the $\\Big(\\text{big parentheses}\\Big)$ are independent of each other (whereas $N(t-s)$ and $N(s)$ are not independent of each other). So \\begin{align} \\Pr(N(s) = k \\mid N(t) = n) & = \\frac{\\Pr(N(s)=k\\ \\&\\ N(t) =n) ...\n\n3\n\nYes, your work is all correct. Except a minor issue: the opposite of $|a_n| < \\epsilon$ is $|a_n| \\ge \\epsilon$, not $|a_n| > \\epsilon$. Instead of writing $P_n(X)$ as $|a_n| < \\epsilon \\; \\forall n > X$, you may have found it clearer to write it as $\\forall n > X \\; |a_n| < \\epsilon$. Then your entire statement would have been $$... 3 In general, if A\\subseteq B, then \\mathscr P (A)\\subseteq \\mathscr P (B) because every subset of A is a subset of B. More formally, if a\\in \\mathscr P (A), we need to show that a\\in \\mathscr P (B). But this is trivial, since if x\\in a, then x\\in B which implies that a\\subseteq B which is the same as a\\in \\mathscr P (B). Now take ... 3 X\\subset Y implies every element of X is an element of Y, so subsets of X are subsets of Y, so \\mathcal{P}(X)\\subset\\mathcal{P}(Y). Finally, for Y=\\mathcal{P}(X) you have \\mathcal{P}(X)\\subset\\mathcal{P}(\\mathcal{P}(X)). 3 Hint:$$\\frac{a_{n+1}}{n}=\\frac{a_{n+1}}{n+1}\\frac{n+1}{n}.$$Or perhaps more to the point,$$\\frac{a_{n+1}}{n+1}=\\frac{a_{n+1}}{n}\\frac{n}{n+1}.$$We've shown a_{n+1}/n\\to l, we know n/(n+1)\\to1, hence a_{n+1}/(n+1)\\to l. And now this implies that a_n/n\\to l. Given \\epsilon>0 there exists N so |a_{n+1}/(n+1)-l|<\\epsilon for all ... 3 If f(a)=c and f(b)=d, then$$\\begin{align} \\int_a^b f(x) \\,\\,dx+\\int_c^d f^{-1}(y) \\,\\,dy &=\\int_a^b f(x) \\,\\,dx+\\int_a^b f^{-1}(f(x)) f'(x) \\,\\,dx\\\\\\\\ &=\\int_a^b f(x) \\,\\,dx+\\int_a^b x f'(x) \\,\\,dx\\\\\\\\ &=\\int_a^b \\left(f(x)+x f'(x)\\right) \\,\\,dx\\\\\\\\ &=\\int_a^b (xf(x))' \\,\\,dx\\\\\\\\ &=bf(b)-af(a)\\\\\\\\ &=bd-ac \\end{align}$$Now, let ... 3 What would the proper negation look like? It turns out that, in this case, there are a number of ways you can go in how you want to prove this claim, not just via direct proof or contrapositive but also how you frame the question logically as well. I'll outline what I think is the clearest and easiest way of going about it. Claim: Let ... 3 You have the contrapositive right. You must negate P and Q separately and prove that the negation of Q implies the negation of P. To expand on this, for \"a and b are even\" to be false, you only need one of a and b to be odd, so the negation is \"a is not even or b is not even\". And for the statement \"a+b and ab have the same parity\" ... 3 Here is a simple proof that K(n) is not only exponential but 'super' exponential in the sense that for all constants C, there is some n_0 such that |K(n)|\\geq C^n for all n\\gt n_0. Let's rewrite your series as \\sum_n\\frac{a_n}{a_{n+1}} so that we don't run out of indices; in other words, K(n)=a_n. (For convenience's sake I'm going to take ... 3 y = \\frac{3x^2+2y}{x^2+2}, multiplying both sides of the equation by x^2+2 results in an equivalent equation because that term is never 0 (in the reals at least). You end up with yx^2 + 2y = 3x^2+2y subtract 2y from both sides (always legitimate). yx^2=3x^2 Since x\\neq 0 we can divide both sides by x^2 and get y=3 3 Proof: We first must note that \\pi_j is the unique solution to \\pi_j=\\sum \\limits_{i=0} \\pi_i P_{ij} and \\sum \\limits_{i=0}\\pi_i=1. Let's use \\pi_i=1. From the double stochastic nature of the matrix, we have$$\\pi_j=\\sum_{i=0}^M \\pi_iP_{ij}=\\sum_{i=0}^M P_{ij}=1 Hence, $\\pi_i=1$ is a valid solution to the first set of equations, and to make it a ...\n\n3\n\nYou need to show independence of increments, i.e. if $0\\le a<b<c<d$ then $(N_1(d)+N_2(d)) - (N_1(c)+N_2(c))$ is independent of $(N_1(b)+N_2(b)) - (N_1(a)+N_2(a))$, and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of $N_1$ and $N_2$. You also ...\n\nOnly top voted, non community-wiki answers of a minimum length are eligible", "date": "2015-08-02 02:39:58", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/tags/proof-writing/hot?filter=month", "openwebmath_score": 0.9351176023483276, "openwebmath_perplexity": 209.27629295813148, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147201714922, "lm_q2_score": 0.8856314647623015, "lm_q1q2_score": 0.8612010729819022}}
{"url": "https://mathoverflow.net/questions/334674/creating-many-big-sets-of-small-numbers", "text": "# Creating many big sets of small numbers\n\nThere are $$n$$ numbers $$a_1,\\ldots,a_n\\in [0,1]$$.\n\nTheir sum is $$\\sum_{i=1}^n a_i = s$$, where $$s$$ is some integer.\n\nWe want to group them into sets so that the sum of each set is at least $$t$$, where $$t$$ is some integer.\n\nLet $$F(n,s,t)$$ be the largest number of sets that we can always create (for any $$a_i$$).\n\nWhat is $$F(n,s,t)$$?\n\nExample. $$F(n=8,s=7,t=1)=4$$:\n\n\u2022 Proof that $$F(8,7,1)\\geq 4$$: We can always create 4 sets by dividing the $$8$$ numbers arbitrarily into $$4$$ pairs. The sum of each pair is at most $$2$$, and the sum of all pairs is $$7$$, so the sum of each pair is at least $$1$$.\n\u2022 Proof that $$F(8,7,1)\\leq 4$$: We cannot always create 5 sets. Suppose for all $$i$$, $$a_i=7/8$$. In any $$5$$ sets, at least one set is a singleton so its sum is less than $$1$$.\n\nSimilarly, whenever $$n$$ is even, $$F(n,n-1,1)=n/2$$.\n\nWhat else is known on the function $$F$$?\n\nCurrently I am particularly interested in the case $$t=2$$, but I will be happy for any more general references.\n\nUPPER BOUND: $$F(n,s,t)\\leq \\lfloor {s+1\\over t+1}\\rfloor$$. Proof. Suppose that $$s+1$$ numbers equal $$s/(s+1)$$ and the other $$n-s-1$$ numbers equal $$0$$. To create a set with sum at least $$t$$, we need $$t+1$$ nonzeros. So we can create at most $$\\lfloor {s+1\\over t+1}\\rfloor$$ such sets.\n\n\u2022 @bof I added the upper bound that I had in mind. It is similar but not identical to yours. I am not sure about the lower bound. \u2013\u00a0Erel Segal-Halevi Jun 24 at 19:51\n\nFor $$n\\in\\mathbb N$$ and $$s,t\\in\\mathbb R$$ with $$0\\lt t\\le s\\le n$$, let $$F(n,s,t)$$ be the greatest integer $$m$$ such that any family of $$n$$ numbers $$a_1,\\dots,a_n\\in[0,1]$$ with $$a_1+\\cdots+a_n=s$$ can be partitioned into $$m$$ subfamilies, each with sum $$\\ge t$$.\n\nLemma 1. If $$k\\in\\mathbb N$$ and $$s\\le k\\le n$$, then $$F(n,s,t)\\le\\left\\lfloor\\frac k{\\lceil kt/s\\rceil}\\right\\rfloor$$.\n\nLemma 2. If $$n\\gt s$$ then $$F(n,s,t)\\le\\left\\lfloor\\frac{\\lfloor s+1\\rfloor}{\\lfloor t+1\\rfloor}\\right\\rfloor$$.\n\nProof. Put $$k=\\lfloor s+1\\rfloor$$ in Lemma 1.\n\nLemma 3. $$F(n,s,t)\\ge\\left\\lfloor\\frac{s+1}{t+1}\\right\\rfloor$$.\n\nProof. Let $$m=\\left\\lfloor\\frac{s+1}{t+1}\\right\\rfloor\\lt s+1$$, so that $$t\\le\\frac{s+1}m-1=\\frac sm-\\frac{m-1}m$$. We may assume that $$m\\ge2$$.\n\nLat $$a_1,\\dots,a_n\\in[0,1]$$ be given, $$a_1+\\cdots+a_n=s$$. For notational convenience we assume that $$a_1,\\dots,a_p\\gt0$$ while $$a_{p+1}=\\cdots=a_n=0$$.\n\nPartition the interval $$[0,s]$$ into $$m$$ equal subintervals $$J_1,\\dots,J_m$$, indexed from left to right; that is, $$J_i=[c_{i-1},c_i]$$ where $$c_i=\\frac{is}m$$. Then $$|J_i|=\\frac sm\\gt1-\\frac1m$$.\n\nAlso partition $$[0,s]$$ into subintervals $$A_1,\\dots,A_p$$ of respective lengths $$|A_i|=a_i$$. Let $$\\mathcal A=\\{A_1,\\dots,A_p\\}$$.\n\nEach interval $$A\\in\\mathcal A$$ will be assigned to at most one of the intervals $$J_1,\\dots,J_m$$, and (some of) the numbers $$a_1,\\dots,a_p$$ will be assigned correspondingly to $$m$$ groups. Namely, an interval $$A\\in\\mathcal A$$ is assigned to the interval $$J_i=[c_{i-1},c_i]$$ if it satisfies one of the following three conditions: $$A\\subseteq J_i;$$ $$i\\gt1,\\ \\ c_{i-1}\\in A,\\ \\ \\frac{|A\\cap J_i|}{|A|}\\gt\\frac{i-1}m;$$ $$i\\lt m,\\ \\ c_i\\in A,\\ \\ \\frac{|A\\cap J_i|}{|A|}\\gt\\frac{m-i}m.$$ It is important to note that no interval $$A\\in\\mathcal A$$ is assigned to more than one $$J_i$$.\n\nNow the set of intervals assigned to $$J_i$$ covers $$J_i$$, except possibly for an interval at the left of length $$\\le\\frac{i-1}m|A|\\le\\frac{i-1}m$$, and an interval at the right of length $$\\le\\frac{m-i}m|A|\\le\\frac{m-i}m$$. Therefore, the sum of the lengths of intervals assigned to $$J_i$$ is $$\\ge\\frac sm-\\frac{i-1}m-\\frac{m-i}m=\\frac sm-\\frac{m-1}m\\ge t$$.\n\nTheorem. If $$t\\in\\mathbb N$$ and $$n\\gt s$$, then $$F(n,s,t)=\\left\\lfloor\\frac{s+1}{t+1}\\right\\rfloor$$.\n\nProof. Lemmas 2 and 3.\n\n\u2022 Yes, for some applications it may be interesting to evaluate $F(n,s,t)$ for $s,t$ rational numbers. Alternatively we can scale the numbers up to integers. We need to add just one more parameter: each number $a_i$ is in $[0,q]$, for some integer $q\\geq 1$. I think that in this case your technique leads to a lower bound of $\\lfloor{s+q\\over t+q}\\rfloor$, but I have to verify \u2013\u00a0Erel Segal-Halevi Jun 27 at 18:10\n\u2022 Instead of separating lemmas 1 and 2, can you have just one lemma in which the number of nonzero terms is $\\lfloor s+1 \\rfloor$? It seems to cover both cases: when $s$ is an integer, $\\lfloor s+1 \\rfloor = \\lceil s+1 \\rceil = s+1$, and when $s$ is not an integer, $\\lfloor s+1 \\rfloor =\\lceil s\\rceil$. \u2013\u00a0Erel Segal-Halevi Jun 30 at 10:09\n\u2022 $F \\leq \\lfloor s+1 \\rfloor / \\lfloor t+1 \\rfloor$. Since you have $\\lfloor s+1 \\rfloor$ nonzero terms, and each term equals $s / \\lfloor s+1 \\rfloor < 1$. So in each subfamily with sum $t$, there must be at least $\\lfloor t+1 \\rfloor$ such elements. \u2013\u00a0Erel Segal-Halevi Jun 30 at 13:50\n\u2022 Looks good, thanks! Now the only gap that remains is when $t$ is not an integer - the upper bound has $\\lfloor t+1\\rfloor$ in the denominator and the lower bound has $t+1$. \u2013\u00a0Erel Segal-Halevi Jul 3 at 15:57\n\u2022 The upper bound in Lemma 2 is the result of setting $k=\\lfloor s+1\\rfloor$ in Lemma 1, but this is not necessarily the optimal value of $k$. For instance, $F(n,1,0.4)\\le2$ by Lemma 2, but (assuming $n\\ge3$) by setting $k=3$ in Lemma 1 we get $F(n,1,0.4)\\le1$. \u2013\u00a0bof Jul 3 at 18:19", "date": "2019-12-07 02:16:45", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/334674/creating-many-big-sets-of-small-numbers", "openwebmath_score": 0.9442065954208374, "openwebmath_perplexity": 117.52214302681404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9833429570618729, "lm_q2_score": 0.8757869867849166, "lm_q1q2_score": 0.8611989653413873}}
{"url": "https://mathematica.stackexchange.com/questions/198023/relation-between-matrices", "text": "# Relation between matrices\n\nMaybe this is not a usual question for this forum... I have the following two matrices, mat1 and mat2, respectively:\n\nmat1={{0.0178885,-0.0178885,-0.00894427,0.00894427,0.,0.,-0.00894427,0.00894427,0.,0.},{-0.0178885,0.0178885,0.00894427,-0.00894427,0.,0.,0.00894427,-0.00894427,0.,0.},{-0.00894427,0.00894427,0.0178885,-0.0178885,-0.00894427,0.00894427,0.,0.,0.,0.},{0.00894427,-0.00894427,-0.0178885,0.0178885,0.00894427,-0.00894427,0.,0.,0.,0.},{0.,0.,-0.00894427,0.00894427,0.00894427,-0.00894427,0.,0.,0.,0.},{0.,0.,0.00894427,-0.00894427,-0.00894427,0.00894427,0.,0.,0.,0.},{-0.00894427,0.00894427,0.,0.,0.,0.,0.0178885,-0.0178885,-0.00894427,0.00894427},{0.00894427,-0.00894427,0.,0.,0.,0.,-0.0178885,0.0178885,0.00894427,-0.00894427},{0.,0.,0.,0.,0.,0.,-0.00894427,0.00894427,0.00894427,-0.00894427},{0.,0.,0.,0.,0.,0.,0.00894427,-0.00894427,-0.00894427,0.00894427}};\nmat2={{0.0198382,-0.0198382,-0.00991908,0.00991908,0.,0.,-0.00991908,0.00991908,0.,0.},{-0.0198382,0.0198382,0.00991908,-0.00991908,0.,0.,0.00991908,-0.00991908,0.,0.},{-0.00991908,0.00991908,0.0203862,-0.0203862,-0.0104672,0.0104672,0.,0.,0.,0.},{0.00991908,-0.00991908,-0.0203862,0.0203862,0.0104672,-0.0104672,0.,0.,0.,0.},{0.,0.,-0.0104672,0.0104672,0.0104672,-0.0104672,0.,0.,0.,0.},{0.,0.,0.0104672,-0.0104672,-0.0104672,0.0104672,0.,0.,0.,0.},{-0.00991908,0.00991908,0.,0.,0.,0.,0.0203862,-0.0203862,-0.0104672,0.0104672},{0.00991908,-0.00991908,0.,0.,0.,0.,-0.0203862,0.0203862,0.0104672,-0.0104672},{0.,0.,0.,0.,0.,0.,-0.0104672,0.0104672,0.0104672,-0.0104672},{0.,0.,0.,0.,0.,0.,0.0104672,-0.0104672,-0.0104672,0.0104672}};\n\n\nIf I plot the matrices, as well as their ratio, I obtain a visual representation of these:\n\nQuiet@List[MatrixPlot[mat1], MatrixPlot[mat2], MatrixPlot[mat1/mat2]]\n\n\nI think that there exists a numerical relation between mat1 and mat2, but I can't find it. Can anyone help me?\n\n\u2022 Should your Dssmt and Dss be the same thing? \u2013\u00a0Roman May 9 '19 at 15:16\n\u2022 @Roman there is a numerical relation among them, but it isn't a simple scaling operation by means of a scalar \u2013\u00a0Gae P May 9 '19 at 15:25\n\u2022 Please make your code self-contained so that people can help without needing to guess. \u2013\u00a0Roman May 9 '19 at 15:46\n\nNumerically all 100 differences between 100 corresponding elements of matrices correspond fall into just 5 symmetric values:\n\nRound[Union[Flatten[mat1-mat2]],.0001]\n\n\n{-0.0025,-0.0019,-0.0015,-0.001,0.,0.001,0.0015,0.0019,0.0025}\n\nWhich looks a lot like some small noise imposed on:\n\n{-25,-20,-15,-10,0,10,15,20,25}/10000\n\n\nIt is actually quite easy to understand if you visualize your data. First of all you can see that the values of both matrices follow each other closely\n\nListPlot3D[{mat1,mat2},InterpolationOrder->0,PlotStyle->{Red,Blue},\nBoxRatios->1,Mesh->None,SphericalRegion->True,PlotLegends->{\"mat1\",\"mat1\"}]\n\n\nYou can see the also see the differences and realize, - for some elements mat1 is greater (more red), sometimes mat (more blue), and sometimes the same (white):\n\nListPlot3D[Rescale[Ds-Dss],InterpolationOrder->0,ColorFunction->\n\"TemperatureMap\",BoxRatios->1,Mesh->None,SphericalRegion->True]\n\n\nYou also can see that all total\n\nIn[]:= Length[Flatten[mat1-mat2]]\nOut[]= 100\n\n\ndifferences between corresponding elements fall into just 9 different values and if you take in account symmetry - just 5 values:\n\nBarChart[Union[Flatten[mat1 - mat2]], PlotTheme -> \"Detailed\"]\n\n\n-- lets call them levels. And now you can find the statistics of how many differences correspond to a specific level -- and see it is symmetric:\n\nListLinePlot[Sort[Tally[Flatten[mat1 - mat2]]], PlotRange -> All, PlotTheme -> \"Business\"]", "date": "2020-03-29 10:15:13", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/198023/relation-between-matrices", "openwebmath_score": 0.6431114077568054, "openwebmath_perplexity": 1220.7830950977059, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877700966098, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8611986588271402}}
{"url": "https://www.pavenafoundation.or.th/article/050d15-telescoping-series-examples", "text": "such that the series converges, provided $\\displaystyle\\lim_{n\\rightarrow\\infty}a(n)$ exists. Does anyone know the rules for a telescoping series. We will now look at some more examples of evaluating telescoping series. Telescoping Series. Example of a Telescoping Sum. E.g. This website uses cookies to ensure you get the best experience. A telescoping series is any series where nearly every term cancels with a preceeding or following term. Free Telescoping Series Test Calculator - Check convergence of telescoping series step-by-step. $\\displaystyle\\prod_{k=1}^{n}\\frac{f(k+1)}{f(k)}=\\frac{f(n+1)}{f(1)}.$ Below I'll give several examples, the first absolutely classical, of application of the telescoping technique. Telescoping series is a series where all terms cancel out except for the first and last one. The 1/2s cancel, the 1/3s cancel, the 1/4s cancel, and so on. These series are called telescoping and their convergence and limit may be computed with relative ease. Suppose we would like to determine whether the series Examples. Example 1 1 2 + 1 2 1 3 + \u2026 Telescoping series are series in which all but the first and last terms cancel out. A series is said to telescope if almost all the terms in the partial sums cancel except for a few at the beginning and at the ending. Next: The Harmonic Series. We can determine the convergence of the series by finding the limit of its partial sums remaining terms. Is it bounded? In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. The geometric and the telescoping series are the only types of series we can easily find the sum of. In the above example stands for the first term (if it is not cancelled out) and (if it is not cancelled out.) (a) A bounded sequence need not converge. The series $\\sum_{n=1}^{\\infty} \\frac{1}{3^n} - \\frac{1}{3^{n+1}}$ converges. ... What are some familiar examples in our solar system, and can some still be closed? In this example, we will determine whether or not the series \\begin{align*} \\sum _{k =1}^{\\infty}\\frac{3}{3k-2}-\\frac{3}{3k+1}, \\end{align*} converges or diverges. 2. We would like a more sure way of knowing the answer. The concept of telescoping extends to finite and infinite products. For example: $$\\sum_{n=2}^\\infty \\frac{1}{n^3-n}$$ In this lesson, we will learn about the convergence and divergence of telescoping series. TELESCOPING SERIES Now let us investigate the telescoping series. Contents. Telescoping Series Example Finding the sum of a telescoping series. Respondents often are asked in surveys to retrospectively report when something occurred, how long something lasted, or \u2026 2. Telescoping Series Examples 2. Note: For an example of a telescoping sums question, see question #2 in the Additional Examples section below. The series in Example 8.2.4 is an example of a telescoping series. Bricks are 20cm long and 10cm high. Previous: The Telescoping and Harmonic Series. In this course, Calculus Instructor Patrick gives 30 video lessons on Series and Sequences. All these terms now collapse, or telescope. ... Now, it is important to note that if we are just trying to determine if series converges or diverges, then applying the Telescoping Series Test will probably not be our first choice. Consider the following example. A p-series can be either divergent or convergent, depending on its value. Example 1. 3. I thought telescoping series were only the ones where all the terms canceled out except for the very first and last terms. This makes such series easy to analyze. Given the sequence \u02c6 1 + lnn n3 \u02d9 1 n=1 (a) Is it monotonic? INFINITE SERIES 1: GEOMETRIC AND TELESCOPING SERIES Exercise 6.2. The partial sum $$S_n$$ did not contain $$n$$ terms, but rather just two: 1 and $$1/(n+1)$$. More examples can be found on the Telescoping Series Examples 1 page. Telescoping Series Example. Strategy for Testing Series - Series Practice Problems This video runs through 14 series problems, discussing what to do to show they converge or diverge. In mathematics, a telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation. Discussion [Using Flash] Example. It seems like you need to do partial fraction decomposition and then evaluate each term individually? Write each of the following series in terms \u201cstandard\u201d geometric series. If the sequence s n is not convergent then we say that the series is divergent. Illustrate each of the following with an exam-ple. This type of series can be easily calculated since all but a few terms are cancelled out. By using this website, you agree to our Cookie Policy. Suppose we are asked to ... it will be sufficient to demonstrate these two special forms with a set of examples. To be able to do this, we will use the method of partial fractions to decompose the fraction that is common in some telescoping series. Try the free Mathway calculator and problem solver below to \u2026 For instance, the series is telescoping. [1] [2] The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences. There is no exact formula to see if the infinite series is a telescoping series, but it is very noticeable if you start to see terms cancel out. It is different from the geometric series, but we can still determine if the series converges and what its sum is. This calculus 2 video tutorial provides a basic introduction into the telescoping series. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? 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You agree to our Cookie Policy \u2019 s not finite and infinite products know the rules for telescoping... Website telescoping series examples cookies to ensure you get the best experience bounded sequence need not be bounded 1 page of! Is it monotonic converge, we will learn about the convergence of the next term, known. In our solar system, and frequencies of events it monotonic cancelled out the convergence the... Will have a fixed number of terms after cancellation things like this for a better understanding of mathematical... Find the sum of a ( n ) $exists are some examples... Of knowing the answer { n\\rightarrow\\infty } a ( n )$ exists a better understanding this. And last terms cancel out be found on the telescoping series now let us investigate the telescoping examples., but we can easily find the sum of be built without mortar on a flat surface... To finite and infinite products fractions technique to rewrite following term series now let us investigate the telescoping this...\n\ntelescoping series examples\n\nOpposite Of Prescriptivism, The Endless Trench Rotten Tomatoes, Bighorn National Park, Tcl Roku Tv Black Screen Fix, Oslo Cathedral Entrance Fee, Lg Fortune 3 Screen Replacement, Pet Tag Silencer Petsmart, Maruti Suzuki Eeco 5 Seater Ac On Road Price, Learning Materials For Grade 3 Araling Panlipunan,", "date": "2021-01-23 21:03:10", "meta": {"domain": "or.th", "url": "https://www.pavenafoundation.or.th/article/050d15-telescoping-series-examples", "openwebmath_score": 0.7672613859176636, "openwebmath_perplexity": 562.9213750937937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9777138125126402, "lm_q2_score": 0.8807970842359877, "lm_q1q2_score": 0.8611674752783847}}
{"url": "https://gateoverflow.in/124367/isi-entrance-exam-mtech-cs", "text": "542 views\nConsider all possible trees with $n$ nodes. Let $k$ be the number of nodes with degree greater than $1$ in a given tree. What is the maximum possible value of $k$?\nretagged | 542 views\nIs it n-2 ?\nsee if it is a skewed tree\n\nthen n=k,rt?\nIf it is a skewed tree then k can be at Max n-2 because end vertices will have degree 1 only and n-2 vertices will have degree 2.\n\nlet , \u00a0the tree with degree max =2 ,\n\nlet , no of intenal nodes whose degree >1 =k\n\nthen no of pedent vertex = n-k\n\nnow as we know that no of edges in tree = n-1\n\nby handshaking theorem ,\n\nsum of degrees of each verteces \u00a0= 2 * edges\n\nnow , 2*(k)+1*(n-k) = 2*(n-1) => k = (n-2)\nedited\nThis is correct Proof for Binary trees not for n-ary trees, but it really helps to understand how it works for any generic trees.Thanks:-)\n\nDegree of a node in the tree is equal to the number of sub-trees of that node, right?\n\nThere is a theorem which says for a tree with at least 2 vertices we have at least 2 pendant vertices.\n\nSince, in a tree of n nodes, we have at least 2 pendant vertices,\n\nthe number of vertices left which will surely have the degree more than 1 is\n\nn-2= maximum value of k.\nMaximum value of k is n-2 which is example of line graph because every tree should contain at least 2 pendent vertices (i.e vertex with degree 1). Therefore value of k cannot exceed n-2\nYes, correct.\n\nExtra info: Min value of k is 1 when the graph is star graph where n-1 vertices have degree one and root has degree n-1.\n@Warrior in a tree, degree of a node is equal to the number of sub-trees, right? graph and tree have different definition of degree of a node, isn't it?", "date": "2018-02-24 10:32:58", "meta": {"domain": "gateoverflow.in", "url": "https://gateoverflow.in/124367/isi-entrance-exam-mtech-cs", "openwebmath_score": 0.6103618144989014, "openwebmath_perplexity": 720.9847349336478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138144607745, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8611674678163417}}
{"url": "https://brilliant.org/discussions/thread/int-ln-sqrtx1-sqrtx-dx-without-trigonometry/", "text": "# $\\int \\ln ( \\sqrt{x+1} + \\sqrt{x} ) dx$ without trigonometry and nice values\n\nIntegration:\n\nI watched this video https://www.youtube.com/watch?v=DGpgt8j-nzw in which $\\int \\ln ( \\sqrt{x+1} + \\sqrt{x} ) dx$ is computed by using trigonometry. Reading the comments, other solutions by using special functions are shown. However, the final antiderivatives do not use any of these functions, so these are intermediate steps in the most literal interpretation of that phrase. I thought of a solution that sidesteps the need of considering any functions other than those present in the integral:\n\n$\\int \\ln ( \\sqrt{x+1} + \\sqrt{x} ) dx$. Let $u = \\sqrt{x+1} + \\sqrt{x}$. Notice that $\\frac{1}{u} = \\sqrt{x+1} - \\sqrt{x}$. Then $du = \\frac{1}{2} \\left( \\frac{1}{\\sqrt{x+1}} + \\frac{1}{\\sqrt{x}} \\right)dx = \\frac{1}{2} \\frac{u}{\\sqrt{x+1} \\sqrt{x}}dx$. To write everything in terms of $u$ notice that $u^2 - \\frac{1}{u^2} = 4 \\sqrt{x+1} \\sqrt{x} \\implies 2 \\sqrt{x+1} \\sqrt{x} = \\frac{u^4 - 1}{2u^2}$. Thus $du = \\frac{u}{\\frac{u^4 - 1}{2u^2}}dx = \\frac{2u^3}{u^4 - 1} dx$. Going back to the original integral:\n\n$\\int \\ln ( \\sqrt{x+1} + \\sqrt{x} ) dx = \\int \\frac{\\ln u}{2u^3} (u^4 - 1)dx = \\int \\frac{u}{2} \\ln u - \\frac{ \\ln u}{2u^3} du$. Integrating the terms in the integral is now a standard exercise in integration by parts which I will skip. The result is $\\frac{1}{8}u^2(2 \\ln u - 1) + \\frac{2 \\log u + 1}{8u^2}$. Giving a final result of:\n\n$\\int \\ln ( \\sqrt{x+1} + \\sqrt{x} ) dx = \\frac{1}{8}(\\sqrt{x+1} + \\sqrt{x})^2(2 \\ln (\\sqrt{x+1} + \\sqrt{x}) - 1) + \\frac{2 \\ln (\\sqrt{x+1} + \\sqrt{x}) + 1}{8(\\sqrt{x+1} + \\sqrt{x})^2} + C$\n\nNice values:\n\nFrom the form of the antiderivative we can expect this integral to give us nice values whenever $\\sqrt{x+1} + \\sqrt{x}$ is a nice power of $e$. The equation $\\sqrt{x+1} + \\sqrt{x} = e^k$ has solution $x= \\frac{1}{4} e^{-2 k} (e^{2 k} - 1)^2$. You can obtain this by converting the equation into a normal quadratic equation. The nicest value I was able to find was taking $k=0$ to yield $x = 0$ and $k = \\frac{1}{2}$ to yield $x = \\frac{(e-1)^2}{4e}$. Then: $\\int_{0}^{\\frac{(e-1)^2}{4e}} \\ln ( \\sqrt{x+1} + \\sqrt{x} ) dx = \\frac{1}{4e}$\n\nNote by Leonel Castillo\n2\u00a0years, 2\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nHere's another way to solve it:\n\n\\begin{aligned} \\int \\ln\\left(\\sqrt{x+1} + \\sqrt x\\right) \\, dx &=& \\frac12 \\int \\ln\\left(\\sqrt {x+1} + \\sqrt x\\right)^2 \\, dx \\\\ &=& \\frac12 \\int \\ln\\left(2x + 1 + 2\\sqrt{x^2 + x}\\right) \\, dx \\\\ &=& \\frac12 \\int \\ln\\left[ 2\\left(x+ \\frac12\\right) + 2\\sqrt{\\left(x + \\frac12\\right)^2 - \\frac14} \\; \\right] \\, dx \\\\ \\end{aligned}\n\nLet $x + \\frac12 = \\frac12 \\sec \\theta$, then $\\left(x+\\frac12\\right)^2 - \\frac14 = \\left(\\frac12 \\tan\\theta\\right)^2$ and $\\frac{dx}{d\\theta} = \\frac12\\sec\\theta\\tan\\theta$. The integral becomes\n\n\\begin{aligned} \\int \\ln\\left(\\sqrt{x+1} + \\sqrt x\\right) \\, dx &=& \\frac12 \\int \\ln (\\sec \\theta + \\tan\\theta) \\cdot \\frac12 \\sec \\theta\\tan\\theta \\, d\\theta \\\\ &=& \\frac14 \\int \\underbrace{\\ln (\\sec \\theta + \\tan\\theta)}_{=u} \\cdot \\underbrace{\\sec \\theta\\tan\\theta \\, d\\theta}_{=dv} , \\qquad\\qquad \\text{ Integrate by parts} \\\\ &=& \\frac14 \\left (uv - \\int v \\, du\\right) \\\\ &=& \\frac14 \\left [ \\ln (\\sec \\theta + \\tan\\theta) \\sec \\theta - \\int \\sec \\theta \\cdot \\dfrac{\\sec\\theta \\tan\\theta + \\sec^2\\theta}{\\sec \\theta + \\tan \\theta} \\, d\\theta \\right ] \\\\ &=& \\frac14 \\left [ \\sec \\theta \\ln (\\sec \\theta + \\tan\\theta) - \\int \\sec^2\\theta \\, d\\theta \\right ] \\\\ &=& \\frac14 \\left [ \\sec \\theta \\ln (\\sec \\theta + \\tan\\theta) - \\tan \\theta \\right ] +C \\\\ &=& \\frac14 \\left [(2x + 1) \\ln \\left ((2x+1) + \\sqrt{(2x+1)^2-1}\\; \\right) - \\sqrt{(2x+1)^2-1} \\; \\right ] +C \\\\ &=& \\frac{2x+1}4 \\ln \\left (2x+1 + 2 \\sqrt{x^2+x}\\; \\right) - \\frac12 \\sqrt{x^2+x} +C \\\\ &=& \\frac{2x+1}2 \\ln \\left (\\sqrt {x+1} + \\sqrt x\\right) - \\frac12 \\sqrt{x^2+x} +C \\\\ \\end{aligned}\n\nwhich is identical to blackpenredpen's final answer.\n\n- 2\u00a0years, 2\u00a0months ago\n\nPi Han Goh, your post does not apply, because the OP expressly asked for the solution to be done without using trigonometry.\n\n- 2\u00a0years, 1\u00a0month ago\n\nI did not ask, I was just sharing a solution. If he wants to use this post to post another alternative solution that's okay.\n\n- 2\u00a0years, 1\u00a0month ago\n\nOP, your answer is wrong, because you do not include the \"+ C.\"\n\n- 2\u00a0years, 1\u00a0month ago\n\nThank you.\n\n- 2\u00a0years, 1\u00a0month ago\n\nJust a thought- How many different sums can be made by adding at least two different numbers from the set of integers {5,6,7,8}?\n\n- 1\u00a0year, 11\u00a0months ago", "date": "2020-09-27 17:54:09", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/int-ln-sqrtx1-sqrtx-dx-without-trigonometry/", "openwebmath_score": 1.0000063180923462, "openwebmath_perplexity": 1945.8117213050539, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526451784086, "lm_q2_score": 0.868826784729373, "lm_q1q2_score": 0.8611399658863699}}
{"url": "https://www.physicsforums.com/threads/calculating-eigenvalues-help.934324/", "text": "# Calculating Eigenvalues help\n\nTags:\n1. Dec 13, 2017\n\n### Pushoam\n\n1. The problem statement, all variables and given/known data\n\n2. Relevant equations\n\n3. The attempt at a solution\nI solved it by calculating the eigen values by $| A- \\lambda |= 0$.\n\nThis gave me $\\lambda _1 = 6.42, \\lambda _2 = 0.387, \\lambda_3 = -0.806$.\n\nSo, the required answer is 42.02 , option (b).\n\nIs this correct?\n\nThe matrix is symmetric. Is there any other easier wat to find the answer?\n\n#### Attached Files:\n\nFile size:\n31.9 KB\nViews:\n20\n2. Dec 13, 2017\n\n### LCKurtz\n\nDo you know the relation between the sum of the squares of the eigenvalues and the trace of a matrix?\n\n3. Dec 13, 2017\n\n### Pushoam\n\nNo.\n\n4. Dec 13, 2017\n\n### Ray Vickson\n\n5. Dec 13, 2017\n\n### Staff: Mentor\n\nI get the same result.\n\n6. Dec 13, 2017\n\n### StoneTemplePython\n\nThe fact that its symmetric leads to some very nice results. What result do you get if you square each entry in your matrix and then sum them? This is called a squared Frobenius norm (which is one way of generalizing the L2 norm for vectors to matrices).\n\n7. Dec 13, 2017\n\n### Orodruin\n\nStaff Emeritus\nI would like to add to what the previous posters said that you should not round your numbers while doing your computations. Keep them on exact form to the very end and only evaluate the numbers in the end if necessary. You should find that the answer is exactly 42, not 42.02.\n\n8. Dec 13, 2017\n\n### Pushoam\n\nI got tr(A) = $\\Sigma \\lambda_i$ , i= 1,2,3.\nand Det (A) = $\\Pi \\lambda_i$ , i= 1,2,3.\nBetter to use $tr (A^2) = \\Sigma {\\lambda_i}^2$ , i= 1,2,3.\n\n$Tr(A^n) = Tr(D^n)$, where D is a similar matrix of A.\nIn case of a matrix having all distinct eigen values, D can be the diagonal matrix consisting of $\\lambda_i$.\nIn that case  Tr(A^n) = Tr(D^n) = \\Sigma {\\lambda_i}^n$, i= 1,2,3. But, the above two equation will not give the result and it is complcated, too. It will be better to calculate the trace directly as I did in OP. I was looking for something like this. The sum is 42. So, is it true for any symmetric matrix? How to prove it?$ Tr (A^2) = \\Sigma_i (A^2)_{ii}\n\\\\ (A^2)_{ii} = \\Sigma_j (A_{ij} A_{ji})$For symmetric matrix,$ A_{ij} = A_{ji}$. So,$ (A^2)_{ii} = \\Sigma_j {(A_{ij} })^2\n\\\\ Tr (A^2) = \\Sigma_i \\Sigma_j {(A_{ij} })^2$Is this correct? 9. Dec 13, 2017 ### Orodruin Staff Emeritus For$n = 2$it directly gives you the result. You just need to square the matrix and sum the diagonal of the result, very simple. If you go via the eigenvalues you need to solve for the roots of an order 3 polynomial. Essentially. However, I think it is easier to go the other way and just see that$A_{ij}A_{ij} = \\mbox{tr}(AA^T)$. Since$A$is symmetric,$AA^T = A^2$. 10. Dec 13, 2017 ### StoneTemplePython note: we are dealing in reals for this post. Your approach is close, and maybe even correct, but I find it hard to follow. My strong preference here is to block your matrix by column vectors. Suppose you have some matrix$\\mathbf X$, partitioned by columns below$\\mathbf X = \\bigg[\\begin{array}{c|c|c|c|c}\n\\mathbf x_1 & \\mathbf x_2 &\\cdots & \\mathbf x_{n-1} & \\mathbf x_n\\end{array}\\bigg]$to make the link with the traditional L2 norm for vectors, consider the vec operator$\nvec\\big(\\mathbf X\\big) = \\begin{bmatrix}\n\\mathbf x_1 \\\\\n\\mathbf x_2\\\\\n\\vdots \\\\\n\\mathbf x_{n-1}\\\\\n\\mathbf x_n\n\\end{bmatrix}$which stacks each column of the matrix$\\mathbf X$on top of each other into one big vector. (The vec operator will show up again if and when you start dealing with Kronecker products.) Our goal is to add up each squared component of$\\mathbf X$into a sum. do you understand why$\\big \\Vert \\mathbf X \\big \\Vert_F^2 = \\sum_{j=1}^n\\sum_{i=1}^n x_{i,j}^2 = trace\\big(\\mathbf X^T \\mathbf X\\big) = vec\\big(\\mathbf X\\big)^Tvec\\big(\\mathbf X\\big)= \\big \\Vert vec\\big(\\mathbf X\\big) \\big \\Vert_2^2$is true for any real matrix? Now since$\\mathbf X$is symmetric, we have$\\mathbf X^T = \\mathbf X$meaning that$\\big \\Vert \\mathbf X \\big \\Vert_F^2 = trace\\big(\\mathbf X^T \\mathbf X\\big) = trace\\big(\\mathbf X \\mathbf X\\big) = trace\\big(\\mathbf X^2\\big)$now you just need the fact that others mentioned, i.e. relating a trace of a matrix and its eigenvalues (or in this case the trace of a matrix to the second power gives sum of eigenvalues to second power). Why is this fact true? (Hint: use characteristic polynomial, or if you prefer an easy but less general case: real symmetric matrices are diagonalizable -- do that and apply cyclic property of trace.) Trace is absurdly useful, so its worth spending extra time understanding all the related details of this problem. 11. Dec 13, 2017 ### Pushoam Then, for the anti - symmetric matrix,$tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}$= negative of the sum of the elements of the matrix. Right? 12. Dec 13, 2017 ### StoneTemplePython Yes. note that in general over real$n$x$n$matrices,$\\big \\vert trace\\big(\\mathbf A \\mathbf A \\big)\\big \\vert = \\big \\vert trace \\Big( \\big( \\mathbf A^T \\big)^T \\mathbf A\\Big)\\big\\vert \\leq trace\\big(\\mathbf A^T \\mathbf A\\big) = \\big \\Vert \\mathbf A\\big \\Vert_F^2 $with equality iff$\\mathbf A$is a scalar multiple of$\\mathbf A^T$. You could prove this with Schur's Inequality. Alternatively (perhaps using the vec operator to help) recognize that the trace gives an inner product. Direct application of Cauchy Schwarz gives you$\\big \\vert trace\\big(\\mathbf B^T \\mathbf A \\big) \\big \\vert = \\big \\vert vec\\big( \\mathbf B\\big)^T vec\\big( \\mathbf A\\big)\\big \\vert \\leq \\big \\Vert vec\\big( \\mathbf B\\big)\\big \\Vert_2 \\big \\Vert vec\\big( \\mathbf A\\big)\\big \\Vert_2 =\\big \\Vert \\mathbf B \\big \\Vert_F \\big \\Vert \\mathbf A \\big \\Vert_F$with equality iff$\\mathbf B = \\gamma \\mathbf A$. (Also note trivial case: if one or both matrices is filled entirely with zeros, then there is an equality.) In your real skew symmetric case,$\\mathbf B = \\mathbf A^T$and$\\gamma = -1$. And of course in the real symmetric case$\\gamma = 1$13. Dec 13, 2017 ### Pushoam I missed to write square of the elements. The corrected one: Then, for the anti - symmetric matrix,$tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}## = negative of the sum of the square of the elements of the matrix.\n\n14. Dec 15, 2017\n\n### epenguin\n\nI am not very familiar with some of the algebra mentioned, though I don\u2019t think it is very difficult.\n\nHowever, it seems possible to solve the problem without it knowing all this, though I\u2019m sure it does no harm to know it.\n\nYou could write out the eigenvalue equation as a cubic equation. The value of the sums of roots \u2211\u03bbi is well known. The value of the sum of products of two roots, \u03a3\u03bbi\u03bbj is well known. From this you could get the sum of squares of roots \u03a3\u03bbi2.\n\nI have not looked into it, but from what was being said about symmetry I suspect it would be easy to solve this cubic.\n\nLast edited: Dec 15, 2017\n15. Dec 15, 2017\n\n### Ray Vickson\n\nSometimes symmetry does not help at all. For example, the matrix\n$$A = \\pmatrix{1&2&3\\\\2&4&5\\\\3&5&6}$$\nhas eigenvalues that are pretty horrible expressions involving cube roots and arctangents of things involving square roots, and the like.", "date": "2018-01-22 18:33:53", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/calculating-eigenvalues-help.934324/", "openwebmath_score": 0.9799693822860718, "openwebmath_perplexity": 3132.3418716038977, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426458162397, "lm_q2_score": 0.8840392878563335, "lm_q1q2_score": 0.8610919669490874}}
{"url": "https://math.stackexchange.com/questions/3750188/throwing-a-dice-and-add-the-digit-that-appears-to-a-sum-and-stopping-when-sum?noredirect=1", "text": "# Throwing a dice and add the digit that appears to a sum and stopping when sum $\\geq 100$.\n\nYou keep on throwing a dice and add the digit that appears to a sum. You stop when sum $$\\ge 100$$. What\u2019s the most frequently appearing digit in all such cases? $$1$$ or $$6$$?\n\nI believe the probability of $$1$$ and $$6$$ should be equal as the whatever the number of rolls, the probability of getting a number should not be affected. However I don't have a formal proof for it and am not sure if this is right.\n\n\u2022 How about working it out for a much smaller number than 100, such as 5? Then 6... \u2013\u00a0Steve Kass Jul 8 '20 at 18:54\n\u2022 ...or consider: what is the largest number of $1$s that can ever fit your criterion? what is the largest number of $6$s that can ever fit your criterion? \u2013\u00a0David G. Stork Jul 8 '20 at 18:55\n\u2022 You may find some relevant discussion at this rather older question, which @SteveKass may remember (he made a comment on it). \u2013\u00a0Brian Tung Jul 8 '20 at 19:02\n\u2022 Suppose that when the sum reaches 100 or more, you stop the game, yell \u201cHooray!\u201d, then start the game again with the same die. You do this for years on end. If the die is fair, how can yelling \u201cHooray!\u201d now and then make the die unfair? \u2013\u00a0Steve Kass Jul 8 '20 at 20:26\n\u2022 @SteveKass Are you answering the same question as @BrianTung? It seems like you are finding the expected number of 1s whereas he is finding the number of 1s among all sequences of rolls which terminate once they reach 100? Why are these the same? I would be grateful if you could make your approach a little more formal or precise. \u2013\u00a0user293794 Jul 8 '20 at 21:21\n\nBasic approach. Imagine drawing a tree, with a root labelled $$0$$. The running count of each node is the label on that node, plus the sum of the labels of all of its direct ancestors. We build on the tree as follows: Under any node whose running count is not yet $$100$$, we add six more nodes, labelled $$1$$ through $$6$$. We repeat until there are no nodes left whose running count is less than $$100$$.\nAt the end of this process, we obviously have a finite tree. How many $$1$$s are there? How many $$6$$s? Was there any time when we added a $$1$$ but not a $$6$$, or vice versa?\nThe expected number of $$1$$'s is the same as the expected number of $$6$$'s. Let $$n_j(k)$$ denote the expected number of digit $$j\\in \\{1,\\ldots,6\\}$$ appearing in the sequence until the running sum reaches $$k$$ (in your case $$k=100$$). Then $$n_j(k)=\\frac{1}{6}(1+n_j(k-j))+\\frac{1}{6}\\sum_{i\\ne j} n_j(k-i)=\\frac{1}{6}+\\frac{1}{6}\\sum_{i\\in \\{1,\\ldots,6\\}} n_j(k-i)$$ with $$n_j(1-i)=0$$ for $$i=1,\\ldots,6$$. This recurrence relation is the same for all $$j$$ and so its solution, $$n_j(k)$$, is the same for all $$j$$.\n\u2022 I don't believe your conclusion is correct. Although the recurrence is the same for all $\\ j\\$, the initial conditions aren't. For $\\ k=1\\$, for instance, $\\ n_1(1)=1\\$ and $\\ n_j(1)=0\\$ for $\\ j=2,3,\\dots,6\\$. \u2013\u00a0lonza leggiera Jul 10 '20 at 0:09\n\u2022 @lonzaleggiera why $n_2(1)=0$? \u2013\u00a0d.k.o. Jul 10 '20 at 5:54\n\u2022 Starting from a total of $0$, the only way you can get to a total of exactly $1$ is from a single throw of a $1$. No other face can have occurred. \u2013\u00a0lonza leggiera Jul 10 '20 at 6:10\n\u2022 Why? You stop when the sum of digits reaches $\\ge k$ (see the question). \u2013\u00a0d.k.o. Jul 10 '20 at 6:35\n\u2022 @vonbrand: It is true that you would need $100$ ones (if those were all you got) and only $17$ sixes (if those were all you got). And if we were to list all possible sequences that reached or exceeded $100$ only on the last roll, then ones would indeed be more common than sixes. But not all sequences are equally likely. The sequence consisting of $17$ sixes is $6^{83}$ times more likely than the sequence consisting of $100$ ones. In general, the preponderance of ones in the longer sequences is exactly offset by the fact that those longer sequences occur less frequently. \u2013\u00a0Brian Tung Jul 10 '20 at 23:46", "date": "2021-03-05 11:08:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3750188/throwing-a-dice-and-add-the-digit-that-appears-to-a-sum-and-stopping-when-sum?noredirect=1", "openwebmath_score": 0.770578145980835, "openwebmath_perplexity": 308.4409610065867, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426405416754, "lm_q2_score": 0.8840392802184581, "lm_q1q2_score": 0.8610919548465491}}
{"url": "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_11&diff=prev&oldid=44887", "text": "# Difference between revisions of \"2004 AMC 10B Problems/Problem 11\"\n\n## Problem\n\nTwo eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?\n\n$\\mathrm{(A) \\ } \\frac{1}{2} \\qquad \\mathrm{(B) \\ } \\frac{47}{64} \\qquad \\mathrm{(C) \\ } \\frac{3}{4} \\qquad \\mathrm{(D) \\ } \\frac{55}{64} \\qquad \\mathrm{(E) \\ } \\frac{7}{8}$\n\n## Solutions\n\n### Solution 1\n\nWe have $1\\times n = n < 1 + n$, hence if at least one of the numbers is $1$, the sum is larger. There $15$ such possibilities.\n\nWe have $2\\times 2 = 2+2$.\n\nFor $n>2$ we already have $2\\times n = n + n > 2 + n$, hence all other cases are good.\n\nOut of the $8\\times 8$ possible cases, we found that in $15+1=16$ the sum is greater than or equal to the product, hence in $64-16=48$ cases the sum is smaller, satisfying the condition. Therefore the answer is $\\frac{48}{64} = \\boxed{\\frac34}$.\n\n### Solution 2\n\nLet the two rolls be $m$, and $n$.\n\nFrom the restriction: $mn > m + n$\n\n$mn - m - n > 0$\n\n$mn - m - n + 1 > 1$\n\n$(m-1)(n-1) > 1$\n\nSince $m-1$ and $n-1$ are non-negative integers between $1$ and $8$, either $(m-1)(n-1) = 0$, $(m-1)(n-1) = 1$, or $(m-1)(n-1) > 1$\n\n$(m-1)(n-1) = 0$ if and only if $m=1$ or $n=1$.\n\nThere are $8$ ordered pairs $(m,n)$ with $m=1$, $8$ ordered pairs with $n=1$, and $1$ ordered pair with $m=1$ and $n=1$. So, there are $8+8-1 = 15$ ordered pairs $(m,n)$ such that $(m-1)(n-1) = 0$.\n\n$(m-1)(n-1) = 1$ if and only if $m-1=1$ and $n-1=1$ or equivalently $m=2$ and $n=2$. This gives $1$ ordered pair $(m,n) = (2,2)$.\n\nSo, there are a total of $15+1=16$ ordered pairs $(m,n)$ with $(m-1)(n-1) < 1$.\n\nSince there are a total of $8\\cdot8 = 64$ ordered pairs $(m,n)$, there are $64-16 = 48$ ordered pairs $(m,n)$ with $(m-1)(n-1) > 1$.\n\nThus, the desired probability is $\\frac{48}{64} = \\frac{3}{4} \\Rightarrow C$.", "date": "2021-04-20 14:20:12", "meta": {"domain": "artofproblemsolving.com", "url": "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_11&diff=prev&oldid=44887", "openwebmath_score": 0.7508097887039185, "openwebmath_perplexity": 100.36924261292195, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9930961635634175, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.8610498969679284}}
{"url": "https://math.stackexchange.com/questions/1698030/proving-a-family-of-polynomials-is-linearly-independent", "text": "# Proving a family of polynomials is linearly independent\n\nI am attempting to solve the following problem, but I'm confused on a solution that is offered for it.\n\nThe question asks \"Let $S$ $=$ {$P_1,..., P_n$} be a family of polynomials such that for all $i,j \\leq n$ : $deg(P_i) \\not = deg(P_j)$. Show that S is linearly independent\".\n\nThe solution offered states the following :\n\n\"We will prove this by induction on n, where n is the number of polynomials.\n\nBase case: Let n=1. Then S = {$P_1$} and since a set containing only one nonzero vector is linearly independent, the desired condition holds.\n\nInductive Step: Assume that S = {$P_1, ... P_n$} is linearly independent. Consider the case when $S'$ = {$P_1, ..., P_n, P_{n+1}$} and $deg(P_i) \\not = deg(P_j)$ for all $i \\not = j$. Consider $a_1P_1 + ... + a_nP_n + a_{n+1}P_{n+1} = 0$. Define $P_k$ to be the polynomial of highest degree. It must be true that\n\n$a_1P_1 + \\space ... \\space + a_{k-1}P_{k-1} + a_{k+1}P_{k+1} +\\space ... \\space + a_nP_n + a_{n+1}P_{n+1} = a_kP_k$.\n\nIf $a_k \\not=0$, then right hand side of the preceding equation has the same degree as $P_k$ while the left hand side has degree strictly less than $P_k$ and this is a contradiction. It must follow that $a_k = 0$.\"\n\nThis is the part where I get confused. Why must this be a contradiction? If we say that $P_k$ is the polynomial of highest degree, isn't it impossible for any other polynomial to have a degree higher than $P_k$ or equal to $P_k$ due to our conditions?\n\nThanks!!\n\n\u2022 You forgot to mention two things in the question/exercise. \u2013\u00a0Friedrich Philipp Mar 15 '16 at 1:49\n\u2022 @FriedrichPhilipp I didn't type out the rest of the proof because I understand how the rest follows, but I don't understand that particular statement. \u2013\u00a0King Tut Mar 15 '16 at 1:51\n\u2022 I meant the question in the beginning... \u2013\u00a0Friedrich Philipp Mar 15 '16 at 1:52\n\u2022 \"If we say that $P_k$ is the polynomial of highest degree, isn't it impossible for any other polynomial to have a degree higher than $P_k$ or equal to $P_k$ due to our conditions?\" Exactly. And exactly this is used here. The left hand side must have smaller degree then the right hand side. But then, of course, they cannot be equal. \u2013\u00a0Friedrich Philipp Mar 15 '16 at 1:57\n\u2022 No problem. \"Oh are you saying that it should be specified that all $P_k$'s are nonzero?\" Of course. This is also used in the proof. Moreover, $deg(P_i)\\neq deg(P_j)$ must be assumed to hold for $i\\neq j$. \u2013\u00a0Friedrich Philipp Mar 15 '16 at 2:27", "date": "2019-08-20 19:13:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1698030/proving-a-family-of-polynomials-is-linearly-independent", "openwebmath_score": 0.8974900841712952, "openwebmath_perplexity": 160.21242106516803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989013058988601, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8610320808579599}}
{"url": "https://or.stackexchange.com/questions/4584/how-to-optimize-a-utility-function-that-contains-step-function/4585", "text": "# How to optimize a utility function that contains step function?\n\nI have an optimization problem with an uncommon utility: to find a $$\\beta$$ that maximizes\n\n$$r^{T}\\cdot H(X\\cdot\\beta)$$\n\nwhere $$H()$$ is a Heaviside step function as in wiki\n\n$$r$$ is a vector of size 1000\n\n$$X$$ is a 1000x50 \"tall\" matrix\n\n$$\\beta$$ is a vector of size 50\n\nI am familiar with gradient descent, which is how I usually solve an optimization problem. But Heaviside function does not work with gradient descent. So I am wondering if anyone here could shed some light on how to solve such optimization problem.\n\nThanks\n\nYou can solve the problem via integer linear programming as follows, assuming $$r_i \\ge 0$$ for all $$i$$. Let $$M_i$$ be a (small) upper bound on $$-(X \\cdot \\beta)_i$$. Let binary decision variable $$y_i$$ indicate whether $$(X \\cdot \\beta)_i \\ge 0$$. The problem is to maximize $$\\sum_{i=1}^{1000} r_i y_i$$ subject to $$-(X \\cdot \\beta)_i \\le M_i(1 - y_i)$$ for all $$i$$. This \"big-M\" constraint enforces $$y_i=1 \\implies (X \\cdot \\beta)_i \\ge 0$$.\n\u2022 Yes, $y_i=0$ forces only a redundant constraint. As long as the \u201creward\u201d $r_i\\ge 0$, this big-M constraint is sufficient. \u2013\u00a0RobPratt Jul 25 at 13:28", "date": "2020-08-08 10:37:49", "meta": {"domain": "stackexchange.com", "url": "https://or.stackexchange.com/questions/4584/how-to-optimize-a-utility-function-that-contains-step-function/4585", "openwebmath_score": 0.8734789490699768, "openwebmath_perplexity": 151.31722254865912, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9890130586647623, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8610320706143119}}
{"url": "http://quig.curaben.it/symmetric-matrix-example-3x3.html", "text": "### Symmetric Matrix Example 3x3\n\nI know that I can convert a single vector of size 3 in a skew symmetric matrix of size 3x3 as follows:. One way to think about a 3x3 orthogonal matrix is, instead of a 3x3 array of scalars, as 3 vectors. For a solution, see the post \" Quiz 13 (Part 1) Diagonalize a matrix. Symmetric Matrices: A square matrix A is symmetric if A = AT. first of all you need to write a c program for transpose of a matrix and them compare it with the original matrix. A square matrix [aij] is called skew-symmetric if aij = \u2212aji. Proof: if it was not, then there must be a non-zero vector x such that Mx = 0. Storage Formats for the Direct Sparse Solvers. In this chapter, we will typically assume that our matrices contain only numbers. The unit matrix is every #n#x #n# square matrix made up of all zeros except for the elements of the main diagonal that are all ones. A symmetric magic square is also called an associative magic square (11, p. JavaScript Example of the Hill Cipher \u00a7 This is a JavaScript implementation of the Hill Cipher. That is, it satisfies the condition In terms of the entries of the matrix, if denotes the entry in the -th row and -th column,. The output matrix has the form of A = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ]. Orthogonal matrix multiplication can be used to represent rotation, there is an equivalence with quaternion multiplication as described here. Therefore, in linear algebra over the complex numbers,. A square matrix [aij] is called a symmetric matrix if aij = aji, i. So a diagonal matrix has at most n different numbers other than 0. Briefly, matrix inverses behave as reciprocals do for real numbers : the product of a matrix and it's inverse is an identity matrix. The elements in a matrix. For a real skew-symmetric matrix the nonzero eigenvalues are all pure imaginary and thus are of the form i\u03bb 1, \u2212i\u03bb 1, i\u03bb 2, \u2212i\u03bb 2, \u2026 where each of the \u03bb k are real. In this note, we derive an orthogonal transformation which transforms A to a n x n matrix B whose diagonal elements are zero. xla is an addin for Excel that contains useful functions for matrices and linear Algebra: Norm, Matrix multiplication, Similarity transformation, Determinant, Inverse, Power, Trace, Scalar Product, Vector Product, Eigenvalues and Eigenvectors of symmetric matrix with Jacobi algorithm, Jacobi's rotation matrix. The Jordan decomposition allows one to easily compute the power of a symmetric matrix :. If Ais an m nmatrix, then its transpose is an n m matrix, so if these are equal, we must have m= n. Symmetric (matrix) synonyms, Symmetric (matrix) pronunciation, Symmetric (matrix) translation, English dictionary definition of Symmetric (matrix). 1 The non{symmetric eigenvalue problem We now know how to nd the eigenvalues and eigenvectors of any symmetric n n matrix, no matter how large. is a (3x3) matrix Basic Concepts in Matrix Algebra Square matrix The matrix is a square matrix iff m = n i. What is symmetric and skew symmetric matrix ? For any square matrix A with real number entries, A+ A T is a symmetric matrix and A\u2212 A T is a skew-symmetric matrix. Shio Kun for Chinese translation. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. Irreducible, diagonally dominant matrices are always invertible, and such matrices arise often in theory and applications. As a consequence, we have the following version of \\Schur's trick\" to check whether M\u02dc0 for a symmetric matrix, M, where we use the usual notation, M\u02dc0 to say that Mis positive de nite and the notation M 0 to say that Mis positive. Question: (1 Point) Give An Example Of A 3 \u00d7 3 Skew-symmetric Matrix A That Is Not Diagonal. matrix c = a + b. A diagonal matrix is a symmetric matrix with all of its entries equal to zero except may be the ones on the diagonal. The matrices must all be defined on dense sets. A square matrix [aij] is called a symmetric matrix if aij = aji, i. Now, noting that a symmetric matrix is positive semi-definite if and only if its eigenvalues are non-negative, we see that your original approach would work: calculate the characteristic polynomial, look at its roots to see if they are non-negative. B is not possible, so multiplication here is not commutative. This solver uses the excellent lrs - David Avis's implementation of Avis and Fukuda's reverse search algorithm for polyhedral vertex enumeration. 2 De\ufb01niteness of Quadratic Forms. A square matrix as sum of symmetric and skew-symmetric matrices. that the element in the i\u2013th row and j\u2013th column of the matrix A equals aij. The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i. (1 Point) Give An Example Of A 3 \u00d7 3 Skew-symmetric Matrix A That Is Not Diagonal. Singular values are important properties of a matrix. If you're behind a web filter, please make sure that the domains *. A \u00d7 A-1 = I. However, when we make any choice of a fundamental matrix solution M(t) and compute M(t)M(0) 1, we always get the same result. An n\u00d7n matrix B is called skew-symmetric if B = \u2212BT. ; Find transpose of matrix A, store it in some variable say B. A = 2: 1+j: 2-j, 1-j: 1: j: 2+j-j: 1 = 2: 1-j: 2+j (j 2 = -1) 1+j: 1-j: 2-j: j: 1: Now A T = => A is Hermitian (the ij-element is conjugate to the ji-element). Linear Algebra: We verify the Spectral Theorem for the 3x3 real symmetric matrix A = [ 0 1 1 / 1 0 1 / 1 1 0 ]. Coordinate Transformations of tensors are discussed in detail here. If A is not SPD then the algorithm will either have a zero. Finally, show (if you haven't already) that the only matrix both symmetric and skew-symmetric is the zero matrix. A skew-symmetric matrix $M$ satisfies [math]M^T=-M. In the following we assume. Reported by: such as 3x3 in the examples needed in the definition of the Pfaffian rather than push that definition into. In Example 1, the eigenvalues of this matrix were found to be \u03bb = \u22121 and \u03bb = \u22122. Symmetric Matrices The symmetric matrix is a matrix in which the numbers on ether side of the diagonal, in corresponding positions are the same. For a solution, see the post \u201c Quiz 13 (Part 1) Diagonalize a matrix. In this paper, we establish a bijection between the set of mutation classes of mutation-cyclic skew-symmetric integral 3x3-matrices and the set of triples of integers (a,b,c) which are all greater than 1 and where the product of the two smaller numbers is greater than or equal to the maximal number. -24 * 5 = -120; Determine whether to multiply by -1. A square matrix is said to be symmetric matrix if the transpose of the matrix is same as the given matrix. By using this website, you agree to our Cookie Policy. asked by Jenny on April 1, 2015; linear algebra. For example, if a problem requires you to divide by a fraction, you can more easily multiply by its reciprocal. We will now go into the specifics here, however. (2) A symmetric matrix is always square. Each number that makes up a matrix is called an element of the matrix. \u02d9 \u02d9 \u02da \u02d8 \u00cd \u00cd \u00ce \u00c8\" \u03c0 \" a i j a i j ij ij 0, 1, Row matrix. Input elements in matrix A. , only passive elements and independent sources), these general observations about the A matrix will always hold. 1 Basics De\ufb01nition 2. Let's take an example of a matrix. is also symmetric because \u00d0EE\u00d1 \u0153EE \u0153EE\u00deX X X XX X The next result tells us that only a symmetric matrix \"has a chance\" to be orthogonally diagonalizable. For our example: rank{A} \u02d82. Simple example: A = I. We will use the following two properties of determinants of matrices. These matrices combine in the same way as the operations, e. 3x3 skew symmetric matrices can be used to represent cross products as matrix multiplications. AB = BA = I n, then the matrix B is called an inverse of A. Secant for particular equation. A A real symmetric matrix [A] can be diagonalized (converted to a matrix with zeros for all elements off the main diagonal) by pre-multiplying by the inverse of the matrix of its eigenvectors and post-multiplying by the matrix of its eigenvectors. CONTENTS: [4] MATRIX ADDITION [5] MATRIX NOTATION [6] TRANSPOSE [7] SYMMETRIC MATRICES [8] BASIC FACTS ABOUT MATRICES [4] MATRIX ADDITION. 262 POSITIVE SEMIDEFINITE AND POSITIVE DEFINITE MATRICES Proof. For example, if the 3x3 matrix M has 5 as an eigenvalue, expand the following: (a, b, c)M = (5a, 5b, 5c) You'll get a pile of conditions on a, b, and c, and sets of a, b, and c that satisfy the conditions will be eigenvectors. The next lea\ufb02ets in the series will show the conditions under which we can add, subtract and multiply matrices. Each input corresponds to an element of the tensor. These algorithms need a way to quantify the \"size\" of a matrix or the \"distance\" between two matrices. To be able to diagonalise a given symmetric matrix. The link inertia matrix (3x3) is symmetric and can be specified by giving a 3x3 matrix, the diagonal elements [Ixx Iyy Izz], or the moments and products of inertia [Ixx Iyy Izz Ixy Iyz Ixz]. Algorithm for Cholesky Decomposition Input: an n\u00a3n SPD matrix A Output: the Cholesky factor, a lower triangular matrix L such that A = LLT Theorem:(proof omitted) For a symmetric matrix A, the Cholesky algorithm will succeed with non-zero diagonal entries in L if and only if A is SPD. An other solution for 3x3 symmetric matrices can be found here (symmetric tridiagonal QL algorithm). If the characteristic of the field is 2, then a skew-symmetric. Example The zero matrix is. Below is the step by step descriptive logic to check symmetric matrix. The quadratic form of. Example 1: Determine the eigenvectors of the matrix. But the multiplication of two symmetric matrices need not be symmetric. 3x3 system of equations solver This calculator solves system of three equations with three unknowns (3x3 system). Find the sum of the diagonal elements of the given N X N spiral matrix. Complexity Explorer is an education project of the Santa Fe Institute - the world headquarters for complexity science. The leftmost column is column 1. symmetric matrix is symmetric. The available eigenvalue subroutines seemed rather heavy weapons to turn upon this little problem, so an explicit solution was developed. Each number that makes up a matrix is called an element of the matrix. This is an inverse operation. Subtract the corresponding elements of from. Give an example of a 3 X 3 upper triangular matrix A that is not diagonal. For the Taylor-Green vortex problem, the domain is periodic is both x- and y-directions, and we end-up with a symmetric implicit matrix. 366) \u2022A is orthogonally diagonalizable, i. Question: (1 Point) Give An Example Of A 3 \u00d7 3 Skew-symmetric Matrix A That Is Not Diagonal. Example 1: Determine the eigenvectors of the matrix. A , in addition to being magic, has the property that \u201cthe sum of the twosymmetric magic square numbers in any two cells symmetrically placed with respect to the center cell is the same\" (12, p. After eliminating weakly dominated strategies, we get the following matrix:. Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. For example, 2 = (\ud835\udc4b \ud835\udc65\u22121)(\ud835\udc4b \ud835\udc65\u22121) = (\ud835\udc4b 2\ud835\udc65\u22121) (48) and so on for higher powers. One is to use Gauss-Jordan elimination and the other is to use the adjugate matrix. 2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. 1) Create transpose of given matrix. linear algebra homework. For example, if a matrix is being read from disk, the time taken to read the matrix will be many times greater than a few copies. \u2022 12 22 \u201a; 2 4 1\u00a110 \u00a1102 023 3 5 are symmetric, but 2 4 122 013 004 3 5; 2 4 01\u00a11 \u00a1102 1\u00a120 3 5 are not. I have always found the common definition of the generalized inverse of a matrix quite unsatisfactory, because it is usually defined by a mere property, , which does not really give intuition on when such a matrix exists or on how it can be constructed, etc\u2026 But recently, I came across a much more satisfactory definition for the case of symmetric (or more general, normal) matrices. These yield complicated formu-lae for the singular value decomposition (SVD), and hence the polar decomposition. 1 Basics De\ufb01nition 2. In fact, P 1 = P>: Left multiplication by a permutation matrix rearranges the corresponding rows: 2 4 0 1 0 0 0 1 1 0 0 3 5 2 4 x 1 x 2 x 3 3 5 = 2 4 x 2 x 3 x 1 3 5; 2 4 0 1 0 0 0 1 1 0 0 3 5 2 4. Consider asan example the 3x3 diagonal matrix D belowand a general 3 elementvector x. Eigenvalues and eigenvectors of a real symmetric matrix. An matrix A is called nonsingular or invertible iff there exists an matrix B such that. 2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. De\ufb01nition 4. Complexity Explorer is an education project of the Santa Fe Institute - the world headquarters for complexity science. And with the confusion matrix, we can calculate a variety of stats in addition to accuracy:. One way to think about a 3x3 orthogonal matrix is, instead of a 3x3 array of scalars, as 3 vectors. The eigen-values are di erent for each C, but since we know the eigenvectors they are easy to diagonalize. 1) means that the eigenvalues of \u00a11 aC are the intersections of the graph of pn(x) with the line y = 1\u00a1 a a\u00a1b. For example, consider the following vector A = [a;b], where both a and b are 3x1 vectors (here N = 2). Let B is a 3x3 matrix A is a 3x2 matrix so B. Invertible matrices are very important in many areas of science. A symmetric magic square is also called an associative magic square (11, p. The output matrix has the form of A = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ]. Determinant Calculator - Matrix online calculator. FINDING EIGENVALUES \u2022 To do this, we \ufb01nd the values of \u03bb which satisfy the characteristic equation of the matrix A, namely those values of \u03bb for which det(A \u2212\u03bbI) = 0,. An n\u00d7n matrix B is called skew-symmetric if B = \u2212BT. eig computes eigenvalues and eigenvectors of a square matrix. Kronenburg Abstract A method is presented for fast diagonalization of a 2x2 or 3x3 real symmetric matrix, that is determination of its eigenvalues and eigenvectors. This implies that UUT = I, by uniqueness of inverses. This should be easy. symmetric matrix is symmetric. , in kronecker , however not for matrix multiplications where. M(t) is an invertible matrix for every t. If there exists a square matrix B of order n such that. Eigenvalues of a 3x3 matrix. Enter payoff matrix B for player 2 (not required for zerosum or symmetric games). the symmetric QRalgorithm, as the expense of two Jacobi sweeps is comparable to that of the entire symmetric QRalgorithm, even with the accumulation of transformations to obtain the matrix of eigenvectors. If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. Philip Petrov ( https://cphpvb. The analysis of matrix-based algorithms often requires use of matrix norms. I am looking for a very fast and efficient algorithm for the computation of the eigenvalues of a 3x3 symmetric positive definite matrix. Symmetric matrices have special properties which are at the basis for these discussions and solutions. Asymmetric Mixed Strategy Equilibria aMaking a game asymmetric often makes its mixed strategy equilibrium asymmetric aAsymmetric Market Niche is an example 33 Asymmetrical Market Niche: The payoff matrix-50, -50 0, 100 150, 0 0, 0 Enter Stay Out Enter Stay Out Firm 2 Firm 1 34 Asymmetrical Market Niche: Two pure strategy equilibria-50, -50 0. Exercise 3. C program check symmetric matrix. Linear Algebra: We verify the Spectral Theorem for the 3x3 real symmetric matrix A = [ 0 1 1 / 1 0 1 / 1 1 0 ]. Step 1 - Accepts a square matrix as input Step 2 - Create a transpose of a matrix and store it in an array Step 3 - Check if input matrix is equal to its transpose. The eigenvalue for the 1x1 is 3 = 3 and the normalized eigenvector is (c 11) = (1). Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. For example, we can confirm that muliplying A by its inverse gives the identity matrix Ainv. Symmetric Matrix :- Square matrix that's equal to it's Transpose (A T =A) We call them symmetric because they are symmetric to main diagonal. Symmetric matrix can be obtain by changing row to column and column to row. Matrix norm the maximum gain max x6=0 kAxk kxk is called the matrix norm or spectral norm of A and is denoted kAk max x6=0 kAxk2 kxk2 = max x6=0 xTATAx. The determinant obtained through the elimination of some rows and columns in a square matrix is called a minor of that matrix. For example A = B = (10) Skew symmetric matrix: If for a square matrix A = [aij], A\u2019 = -A, then A is called a skew symmetric matrix. xTAx = x1 x2 2 6 18 6 x x 1 2 2x = x 1 + 6x2 1 x2 6x 1 + 18x2 = 2x 12 + 12x1x2 + 18x 22 = ax 12 + 2bx1x2 + cx 22. Diagonal matrix :- All non-diagonal elements =0. The Cholesky decomposition of a Pascal upper-triangle matrix is the Identity matrix of the same size. So in order to prove this matrix is diagonalizable,. no entry zero. I have chosen these from some book or books. The main diagonal itself must all be 0s. [4] Computing Eigenvectors Let's return to the equation Ax = x. We will now go into the specifics here, however. Most properties are listed under skew-Hermitian. The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. The @MTXMUL function multiplies matrix B by matrix C and places the result in matrix A. However, when we make any choice of a fundamental matrix solution M(t) and compute M(t)M(0) 1, we always get the same result. A real $(n\\times n)$-matrix is symmetric if and only if the associated operator $\\mathbf R^n\\to\\mathbf R^n$ (with respect to the standard basis) is self-adjoint (with respect to the standard inner product). These algorithms need a way to quantify the \"size\" of a matrix or the \"distance\" between two matrices. Example 1: Give an example of 4\u00d74 order identity or unit matrix. the algorithm will be part of a massive computational kernel, thus it is required to be very efficient. A matrix is a rectangular array of numbers that is arranged in the form of rows and columns. 1 Symmetric Matrices and Convexity of Quadratic Functions A symmetric matrix is a square matrix Q \u2208 \u211cn\u00d7n with the property that Qij = Qji for all i;j = 1;:::;n : We can alternatively de ne a matrix Q to be symmetric if QT = Q : We denote the identity matrix (i. A is called symmetric if A> = A. Definition of a Matrix The following are examples of matrices (plural of matrix). Figure 1 1-D Gaussian distribution with mean 0 and =1 In 2-D, an isotropic (i. C program check symmetric matrix. Find the eigenvalues and bases for each eigenspace. Here, it is understood that and are both column vectors, and is the matrix of the values. e A-1 we shall first define the adjoint of a matrix. I'm not sure what kind of approach to take towards a. Your overall recorded score is 0%. If \u2018b\u2019 is a matrix, the system is solved for each column of \u2018b\u2019 and the return value is a matrix of the same shape as \u2018b\u2019. Matrix representation of symmetry operations Using carthesian coordinates (x,y,z) or some position vector , we are able to define an initial position of a point or an atom. Good things happen when a matrix is similar to a diagonal matrix. There are a lot of examples in which a singular matrix is an idempotent matrix. [4] Computing Eigenvectors Let's return to the equation Ax = x. A = [1 1 1 1 1 1 1 1 1]. If you're behind a web filter, please make sure that the domains *. So a diagonal matrix has at most n different numbers other than 0. The generalized eigenvalue problem is to determine the solution to the equation Av = \u03bbBv, where A and B are n-by-n matrices, v is a column vector of length n, and \u03bb is a scalar. Thus for all i and j, we have a ij = \u2013 a ji. An matrix is called real symmetric if , the transpose of , coincide with. Ellermeyer July 1, 2002 1 Similar Matrices De\ufb01nition 1 If A and B are nxn (square) matrices, then A is said to be similar to B if there exists an invertible nxn matrix, P,suchthatA = P\u22121BP. 1 Strategic Form. Therefore x T Mx = 0 which contradicts our assumption about M being positive definite. Symmetric matrices have special properties which are at the basis for these discussions and solutions. Definition E E\u0153E\u00deis called a if symmetric matrix X Notice that a symmetric matrix must be square ( ?). (35) For a positive semi-de\ufb01nite matrix, the rank corresponds to the. Secant for particular equation. Example 5: A Hermitian matrix. Matrix norm the maximum gain max x6=0 kAxk kxk is called the matrix norm or spectral norm of A and is denoted kAk max x6=0 kAxk2 kxk2 = max x6=0 xTATAx. Matrix Approach to Linear Regression leaving J is matrix of all ones, do 3x3 example. So in order to prove this matrix is diagonalizable,. In general, a ij means the element of A in the ith row and jth column. To find this matrix : First write down a skew symmetric matrix with arbitrary coefficients. Singular values are important properties of a matrix. If the matrix A is symmetric then \u2022its eigenvalues are all real (\u2192TH 8. Diagonal matrix :- All non-diagonal elements =0. DA FONSECA In general, (2. Not very random but very fun!. Example A= 2 4 0 3 This is a 2 by 2 matrix, so we know that 1 + 2 = tr(A) = 5 1 2 = det(A) = 6 6. Now since U has orthonormal columns, it is an orthognal matrix, and hence Ut is the inverse of U. The diagonal elements of a skew-symmetric matrix are all 0. In symmetric matrices the upper right half and the lower left half of the matrix are mirror images of each other about the diagonal. Operation on matrices: Addition and multiplication and multiplication with a scalar. These matrices combine in the same way as the operations, e. 2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. If a ij denotes the entries in an i-th row and j-th column, then the symmetric matrix is represented as. After eliminating weakly dominated strategies, we get the following matrix:. Skew symmetric matrices mean that A (transpose) = -A, So since you know 3 elements of the matrix, you know the 3 symmetric to them over the main diagonal mut be the negatives of those elements. Example The zero matrix is. The set of four transformation matrices forms a matrix representation of the C2hpoint group. Once we get the matrix P, then D = P t AP. The Jordan decomposition allows one to easily compute the power of a symmetric matrix :. 2 De\ufb01niteness of Quadratic Forms. 1, is an eigenvalue of. Matrix norm the maximum gain max x6=0 kAxk kxk is called the matrix norm or spectral norm of A and is denoted kAk max x6=0 kAxk2 kxk2 = max x6=0 xTATAx. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. Show that the product AAT is a symmetric matrix. If is an matrix and is an matrix, then the tensor product of and , denoted by , is the matrix and is defined as If is and is , then the Kronecker sum (or tensor sum) of and , denoted by , is the matrix of the form Let be the set of all symmetric matrices with integer. 1) means that the eigenvalues of \u00a11 aC are the intersections of the graph of pn(x) with the line y = 1\u00a1 a a\u00a1b. If there exists a square matrix B of order n such that. Zero matrix and identity matrix are symmetric (any diagonal matrix is sym-metric) 2. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy and we have Q A(1; 1) = 12 + ( 1)2 + 8(1)( 1) = 1 + 1 8. Learn its definition and formula to calculate for 2 by 2, 3 by 3, etc. equal to a)A5 +A8 b)A5 -A8 c)A8 -A5 c)AT +BT If A is a symmetric matrix and B is a skew symmetric matrix of the same order , then A2 +B2 is a 1 2 4 6 8 2 2 2 7 16. 1 The non{symmetric eigenvalue problem We now know how to nd the eigenvalues and eigenvectors of any symmetric n n matrix, no matter how large. Since A is symmetric, A = AT or LDU = UTDLT, so U = LT. The initial vector is submitted to a symmetry operation and thereby transformed into some resulting vector defined by the coordinates x', y' and z'. By using this website, you agree to our Cookie Policy. For example, the matrices. Properties. Simple example: A = I. The matrices must all be defined on dense sets. The case here is restricted to 2x2 case of the hill cipher for now, it may be expanded to 3x3 later. The determinant obtained through the elimination of some rows and columns in a square matrix is called a minor of that matrix. The coordinates can be written in matrix form and then can be multiplied by a matrix or scalar for Rotation, Reflection or Dilation (Scaling). In this chapter, we will typically assume that our matrices contain only numbers. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. A3\u00d73 example of a matrix with some complex eigenvalues is B = 1 \u22121 \u22121 1 \u221210 10\u22121 A straightforward calculation shows that the eigenvalues of B are \u03bb = \u22121 (real), \u00b1i (complex conjugates). The Jordan decomposition allows one to easily compute the power of a symmetric matrix :. These are well-defined as $$A^TA$$ is always symmetric, positive-definite, so its eigenvalues are real and positive. Symmetric (matrix) synonyms, Symmetric (matrix) pronunciation, Symmetric (matrix) translation, English dictionary definition of Symmetric (matrix). ) Dimension is the number of vectors in any basis for the space to be spanned. In Example 1, the eigenvalues of this matrix were found to be \u03bb = \u22121 and \u03bb = \u22122. Hermitian matrices Defn: The Hermitian conjugate of a matrix is the transpose of its complex conjugate. Examples of higher order tensors include stress, strain, and stiffness tensors. On this page you can see many examples of matrix multiplication. Let be an eigenvector corresponding to the eigenvalue 3. Matrices with Examples and Questions with Solutions. Here you can calculate inverse matrix with complex numbers online for free with a very detailed solution. JavaScript Example of the Hill Cipher \u00a7 This is a JavaScript implementation of the Hill Cipher. A C++ source and header file to compute eigenvectors/values of a 3x3 symmetric matrix. If there exists a square matrix B of order n such that. Simple properties of addition, multiplication and scalar multiplication. 369) EXAMPLE 1 Orthogonally diagonalize. A neat example of this is finding large powers of a matrix. 60 \u2022 \u2022 \u2022 Chapter 1 / Systems of Linear Equations and Matrices EXAMPLE 1 Solution of a Linear System Using A\u22121 Consider the system of linear equations x1 + 2x2 + 3x3 = 5 2x1 + 5x2 + 3x3 = 3 + 8x3 = 17 x1 In matrix form this system can be written as Ax = b, where 1 2 3 x1 5 A = 2 5 3 , x = x2 , b = 3 1 0 8 17 x3 In Example 4 of the. A 3x3 Example The following example decomposes a 3 x 3 symmetric matrix. By Proposition 23. With symmetric matrices on the other hand, complex eigenvalues are not possible. 1 The formula aij = 1/(i + j) for 1 \u2264 i \u2264 3, 1 \u2264 j \u2264 4 de\ufb01nes a 3\u00d74 matrix A = [aij], namely A = 1. More generally, if C is an m\u00d7 n matrix, its transpose, CT, is a n\u00d7 m matrix. Now consider the x matrix, the matrix of unknown quantities. Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. Its easy to get C2 to equal 0 but obviously you can't have that. In the second step, which takes the most amount of time, the matrix is reduced to upper Schur form by using an orthogonal transformation. If this quadratic form is positive for every (real) x1 and x2 then the matrix is positive de\ufb01nite. The inverse matrix has the property that it is equal to the product of the reciprocal of the determinant and the adjugate matrix. are symmetric matrices. To find this matrix : First write down a skew symmetric matrix with arbitrary coefficients. Math 2940: Symmetric matrices have real eigenvalues The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. MATH 340: EIGENVECTORS, SYMMETRIC MATRICES, AND ORTHOGONALIZATION Let A be an n n real matrix. Show that the product AAT is a symmetric matrix. can have vector or matrix elements). HILL CIPHER Encrypts a group of letters called polygraph. 3 Pure Strategies and Mixed Strategies. I am looking for a very fast and efficient algorithm for the computation of the eigenvalues of a 3x3 symmetric positive definite matrix. AAT = 17 8 8 17. Most properties are listed under skew-Hermitian. (2*2 - 7*4 = -24) Multiply by the chosen element of the 3x3 matrix. I want to convert the last 3 dimensional vector into a skew symmetric matrix. 2 Two-part names. For the Taylor-Green vortex problem, the domain is periodic is both x- and y-directions, and we end-up with a symmetric implicit matrix. Real number \u03bb and vector z are called an eigen pair of matrix A, if Az = \u03bbz. I have chosen these from some book or books. An n\u00d7n matrix B is called nilpotent if there exists a power of the matrix B which is equal to the zero matrix. 369) EXAMPLE 1 Orthogonally diagonalize. Simple example: A = I. In other words, we can say that transpose of Matrix B is not equal to matrix B (). The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. Let be an eigenvector corresponding to the eigenvalue 3. Let A = LDU be the LDU decomposition of A. To encrypt; C K. Then we have: A is positive de nite ,D k >0 for all leading principal minors A is negative de nite ,( 1)kD k >0 for all leading principal minors A is positive semide nite , k 0 for all principal minors A is negative semide nite ,( 1)k k 0 for all principal minors In the rst two cases, it is enough to. Note that usually the eigenvectors are normalized to have unit length. matrix list b symmetric b[3,3] c1 c2 c3 displacement 3211055 mpg 227102 22249 _cons 12153 1041 52. Symmetric eigenvalue decompositions for symmetric tensors Lek-Heng Lim University of California, Berkeley January 29, 2009 (Contains joint work with Pierre Comon, Jason Morton, Bernard Mourrain, Berkant Savas) L. The main diagonal itself must all be 0s. M(t) is an invertible matrix for every t. phasesym Example of 3x3 skew symmetric matrix. Example 5: A Hermitian matrix. KEYWORDS: Software, Solving Linear Equations, Matrix Multiplication, Determinants and Permanents. Note that all the main diagonal elements in the skew-symmetric matrix are zero. An analogous result holds for matrices. A symmetric matrix should be a square matrix. The Euler angles of the eigenvectors are computed. The characteristic polynomial is det(AAT \u2212\u03bbI) = \u03bb2 \u221234\u03bb+225 = (\u03bb\u221225)(\u03bb\u22129), so the singular values are \u03c3. It is denoted by adj A. A real $(n\\times n)$-matrix is symmetric if and only if the associated operator $\\mathbf R^n\\to\\mathbf R^n$ (with respect to the standard basis) is self-adjoint (with respect to the standard inner product). If terms a 22 and a 23 are both 0, our formula becomes a 21 |A 21 | - 0*|A 22 | + 0*|A 23 | = a 21 |A 21 | - 0 + 0 = a 21 |A 21 |. 1 using the Schur complement of A instead of the Schur complement of Calso holds. This is often easier than trying to specify the Hessian matrix. matrix list b symmetric b[3,3] c1 c2 c3 displacement 3211055 mpg 227102 22249 _cons 12153 1041 52. Solution Let A = [a ij] be a matrix which is both symmetric and skew symmetric. Theorem 1 Any quadratic form can be represented by symmetric matrix. There are other methods of finding the inverse matrix, like augmenting the matrix by the identity matrix and then trying to make the original matrix into the identity matrix by applying row and column operations to the augmented matrix, and so on. first of all you need to write a c program for transpose of a matrix and them compare it with the original matrix. Since the minimum and maximum values equal to 1, we get the identity matrix. Your overall recorded score is 0%. is also symmetric because \u00d0EE\u00d1 \u0153EE \u0153EE\u00deX X X XX X The next result tells us that only a symmetric matrix \"has a chance\" to be orthogonally diagonalizable. eig computes eigenvalues and eigenvectors of a square matrix. Symmetric Matrices The symmetric matrix is a matrix in which the numbers on ether side of the diagonal, in corresponding positions are the same. Multiplying matrices - examples. Lim (Algebra Seminar) Symmetric tensor decompositions January 29, 2009 1 / 29. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. Eigenvalues and the characteristic. symDiagonal() returns an object of class dsCMatrix or lsCMatrix, i. Example: The following 3x3 system has the solution x1 = \u22121 ; x2 = 2 ; x3 = 1, as you can verify it by direct substitution. 2 4 6 8 5 8 2 1 5 1 0 3 5is a symmetric matrix Theorem 1. Simplify the Matrix. Determining the eigenvalues of a 3x3 matrix. In other words, we can say that matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of matrix A i. Frank Wood. JACOBI is a program written in 1980 for the HP-41C programmable calculator to find all eigenvalues of a real NxN symmetric matrix using Jacobi's method. 1 below, we show that if \u2018>0 and Jis a symmetric diagonally dominant matrix satisfying J \u2018S, then J \u2018S\u02dc0; in particular, Jis invertible. The eigenvalue for the 1x1 is 3 = 3 and the normalized eigenvector is (c 11) = (1). Singular values are important properties of a matrix. 4 The Minimax Theorem. Figure 2 2-D Gaussian distribution with mean (0,0) and =1 The idea of Gaussian smoothing is to use this 2-D distribution as a point-spread' function, and this is achieved by. A square matrix such that a ij is the complex conjugate of a ji for all elements a ij of the matrix i. (iv) Theorem 2: Any square matrix A can be expressed as the sum of a symmetric matrix and a skew symmetric matrix, that is (A+A ) (A A )T T A = + 2 2 \u2212 3. Statistics 1: Linear Regression and Matrices The concepts and terminology for matrices will be developed using an example from statistics. An analogous result holds for matrices. A = 1 0 0 1 0 0 2 3 3 You have attempted this problem 10 times. \u02d9 \u02d9 \u02da \u02d8 \u00cd \u00cd \u00ce \u00c8\" \u03c0 \" a i j a i j ij ij 0, 1, Row matrix. Show that S 3x3 ( R) is actually a subspace of M 3x3 ( R) and then determine the dimension and a basis for this subspace. APPLICATIONS Example 2. The diagonal elements of a skew-symmetric matrix are all 0. If you're seeing this message, it means we're having trouble loading external resources on our website. are symmetric matrices. notebook November 28, 2016 Inverse of a 3x3 Matrix Examples: 3. e ( AT =\u2212A ). A = [ 2 \u2212 1 \u2212 1 \u2212 1 2 \u2212 1 \u2212 1 \u2212 1 2]. Write a C program to read elements in a matrix and check whether the given matrix is symmetric matrix or not. Antisymmetric matrices are commonly called \"skew symmetric matrices\" by mathematicians. it's a Markov matrix), its eigenvalues and eigenvectors are likely to have special properties as well. A \u00d7 A-1 = I. Definition. You can convert the skew symmetric matrix R_dot * dt into a rotation matrix using the Rodrigues formula. The symmetric spaces considered here are quotients X=G/K, where G is a non-compact real Lie group, such as the general linear group GL(n,R) of all n x n non-singular real matrices, and K=O(n), the maximal compact subgroup of orthogonal matrices. In describing matrices, the format is: rows X columns. (where A is a symmetric matrix). Similarly, we can take other examples of Nilpotent matrices. We will use the following two properties of determinants of matrices. Complexity Explorer is an education project of the Santa Fe Institute - the world headquarters for complexity science. For the Taylor-Green vortex problem, the domain is periodic is both x- and y-directions, and we end-up with a symmetric implicit matrix. That is the diagonal with the a's on it. For a solution, see the post \" Quiz 13 (Part 1) Diagonalize a matrix. Real number \u03bb and vector z are called an eigen pair of matrix A, if Az = \u03bbz. First we compute the singular values \u03c3 i by \ufb01nding the eigenvalues of AAT. The Hessian matrix is a square, symmetric matrix whose. Welcome to Week 2 of the Robotics: Aerial Robotics course! We hope you are having a good time and learning a lot already! In this week, we will first focus on. An identity matrix will be denoted by I, and 0 will denote a null matrix. 2 Matrix \u201cSquare Roots\u201d Nonnegative numbers have real square roots. We will use the following two properties of determinants of matrices. A 3x3 stress tensor is 2nd rank. Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. So we;ve got. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. For the lid-driven cavity flow, the implicit matrix is not symmetric. where R is the correlation matrix of the predictors (X variables) and r is a column vector of correlations between Y and each X. Example: Solution: Determinant = (3 \u00d7 2) \u2013 (6 \u00d7 1) = 0. Let's consider a simple example with a diagonal matrix: A = np. The largest term will include the cube of lambda and so on. Definition E E\u0153E\u00deis called a if symmetric matrix X Notice that a symmetric matrix must be square ( ?). To check whether a matrix A is symmetric or not we need to check whether A = AT or not. It is called a singular matrix. Briefly, matrix inverses behave as reciprocals do for real numbers : the product of a matrix and it's inverse is an identity matrix. I'm currently stuck on converting a 3*N x 1, where N is an integer value, vector into chunks of skew symmetric matrices. M = (Q^(-1)) * G. Symmetric matrices have real eigenvalues. Question from Matrices,cbse,class12,bookproblem,ch3,misc,q14,p101,easy,shortanswer,sec-a,math. xTAx = x1 x2 2 6 18 6 x x 1 2 2x = x 1 + 6x2 1 x2 6x 1 + 18x2 = 2x 12 + 12x1x2 + 18x 22 = ax 12 + 2bx1x2 + cx 22. So in that way every Diagonal Matrix is Symmetric Matrix. Then we have: A is positive de nite ,D k >0 for all leading principal minors A is negative de nite ,( 1)kD k >0 for all leading principal minors A is positive semide nite , k 0 for all principal minors A is negative semide nite ,( 1)k k 0 for all principal minors In the rst two cases, it is enough to. the algorithm will be part of a massive computational kernel, thus it is required to be very efficient. The determinant obtained through the elimination of some rows and columns in a square matrix is called a minor of that matrix. Note that whereas C is a 3\u00d7 2 matrix, its transpose, CT, is a 2\u00d7 3 matrix. He walks you through basic ideas such as how to solve systems of. The leading coefficients occur in columns 1 and 3. (4) If A is invertible then so is AT, and (AT) \u2212 1 = (A \u2212 1)T. , The sum of the numbers along each matrix diagonal (the character) gives a shorthand version of the matrix representation, called \u0393:. Thus, we may write [El = S'[A]S (18) where [A] is the diagonal matrix of eigenvalues [A]= [: 0 A, : :I 0 and S is the rotation matrix whose rows are the three orthogonal eigenvectors of [E] corresponding to the Ai eigenvalues. In the following we assume. Example 3 Show that a matrix which is both symmetric and skew symmetric is a zero matrix. Detailed Description Tiny matrix classes with optimized numerical operations which make use of vectorization features through the VVector class. A matrix with real entries is skewsymmetric. An example will be constructed later in this chapter. 50 }; //Symmetric Postive-Definite matrix X = { 7 6. Basis vectors. Matrix Inner Products. Now lets FOIL, and solve for. It can be shown that all real symmetric matrices have real eigenvalues and perpendicular eigenvectors. C program check symmetric matrix. Properties. Treat the remaining elements as a 2x2 matrix. Let\u2019s take an example of a matrix. Consider a n x n, trace free, real symmetric matrix A. These terms are more properly defined in Linear Algebra and relate to what are known as eigenvalues of a matrix. the inverse of an n x n matrix See our text ( Rolf, Page 163) for a discussion of matrix inverses. U) = A z^ - [1,3I1A + [-12,123. We will use the following two properties of determinants of matrices. Of course this holds too for square matrices of higher rank (N x N matrices), not just 3x3's. Frank Wood, [email\u00a0protected] It is called a singular matrix. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. For example, to solve 7x = 14, we multiply both sides by the same number. To check whether a matrix A is symmetric or not we need to check whether A = AT or not. The inverse of a permutation matrix is again a permutation matrix. Here for instance, X should be a 6x6 real and symmetric matrix unless we have made some mistake, and indeed it is constructed from 9 variables defining a symmetric 3x3 matrix. The unit matrix is every #n#x #n# square matrix made up of all zeros except for the elements of the main diagonal that are all ones. As a result, we can concisely represent any skew symmetric 3x3 matrix as a 3x1 vector. Generally, one can \ufb01nd symmetrization A0 of a matrix A by A0 = A+AT 2. 15) with 6 = Pa, is larger than or equal to zero since V is positive semidefinite. So, for example, the 3x3 matrix A might be written as:. The leftmost column is column 1. e ( AT =\u2212A ). Examples: Quadratic Form Now we have seen the symmetric matrices, we can move on to the quadratic 1 5 5 8 9 \u22122 \u2212 2 7 a b b c. To check whether a matrix A is symmetric or not we need to check whether A = A T or not. Example 4 Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 \u20131 2]. 4 Diagonal Matrix: A square matrix is called a diagonal matrix if each of its non-diagonal elements are zero (i. com To create your new password, just click the link in the email we sent you. One is to use Gauss-Jordan elimination and the other is to use the adjugate matrix. AB = BA = I n, then the matrix B is called an inverse of A. Inverse matrix A -1 is defined as solution B to AB = BA = I. 366) \u2022A is orthogonally diagonalizable, i. Instead, we can implicitly apply the symmetric QR algorithm to ATA. Examples: Quadratic Form Now we have seen the symmetric matrices, we can move on to the quadratic 1 5 5 8 9 \u22122 \u2212 2 7 a b b c. (1 Point) Give An Example Of A 3 \u00d7 3 Skew-symmetric Matrix A That Is Not Diagonal. 369) EXAMPLE 1 Orthogonally diagonalize. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy and we have Q A(1; 1) = 12 + ( 1)2 + 8(1)( 1) = 1 + 1 8. 50 }; //Symmetric Postive-Definite matrix X = { 7 6. Eigenvalues of a 3x3 matrix. You need to know more about the matrix to conclude one way or the other. Welcome to Week 2 of the Robotics: Aerial Robotics course! We hope you are having a good time and learning a lot already! In this week, we will first focus on. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. symDiagonal() returns an object of class dsCMatrix or lsCMatrix, i. Notice that this is a block diagonal matrix, consisting of a 2x2 and a 1x1. The variance-covariance matrix is symmetric because the covariance between X and Y is the same as the covariance between Y and X. De nition 1 Let U be a d dmatrix. Jacobi's Algorithm is a method for finding the eigenvalues of nxn symmetric matrices by diagonalizing them. Fortran 90 package for solving linear systems of equations of the form A*x = b, where the matrix A is sparse and can be either unsymmetric, symmetric positive definite, or general symmetric. Finally, show (if you haven't already) that the only matrix both symmetric and skew-symmetric is the zero matrix. For example, decrypting a coded message uses invertible matrices (see the coding page). Similarly, the rank of a matrix A is denoted by rank(A). Many problems present themselves in terms of an eigenvalue problem: A\u00b7v=\u03bb\u00b7v. , (AT) ij = A ji \u2200 i,j. HILL CIPHER Encrypts a group of letters called polygraph. Vector x is a right eigenvector, vector y is a left eigenvector, corresponding to the eigenvalue \u03bb, which is the same for. Thethingis,therearealotofotherequivalentwaystode\ufb01neapositive de\ufb01nite matrix. Usually the numbers are real numbers. Geometrically, a matrix $$A$$ maps the unit sphere in $$\\mathbb{R}^n$$ to an ellipse. (34) Finally, the rank of a matrix can be de\ufb01ned as being the num-ber of non-zero eigenvalues of the matrix. Note that whereas C is a 3\u00d7 2 matrix, its transpose, CT, is a 2\u00d7 3 matrix. A = [1 1 1 1 1 1 1 1 1]. Then compute it's determinant (which will end up being a sum of terms including four coefficients) Then to ease the computation, find the coefficient that appears in the least amount of term. Therefore, we can see that , Hence, the matrix A is nilpotent. Example: If square matrices Aand Bsatisfy that AB= BA, then (AB)p= ApBp. Inverting a matrix turns out to be quite useful, although not for the classic example of solving a set of simultaneous equations, for which other, better, methods exist. Once we get the matrix P, then D = P t AP. Matrices with Examples and Questions with Solutions. $\\begingroup$ Yes, reduced row echelon form is also called row canonical form, and obviously there are infinitely many symmetric matrix that are not diagonal and can be reduced to anon diagonal reduced row echelon form, but note that the row canonical form is not given by a similarity transformation, but the jordan form is. \u03bb is an eigenvalue (a scalar) of the Matrix [A] if there is a non-zero vector (v) such that the following relationship is satisfied: [A](v) = \u03bb (v) Every vector (v) satisfying this equation is called an eigenvector of [A] belonging to the eigenvalue \u03bb. The inverse is calculated using Gauss-Jordan elimination. The inverse of a permutation matrix is again a permutation matrix. Mathematical Properties of Sti\ufb00ness Matrices 3 computation involving the inverse of ill-conditioned matrices can lose precision because there is a range of values in the solution { d }that can satsify [ K ]{ d }= { p }. Computing eigenvalues and eigenvectors for a 3x3 symmetric matrix. You have unlimited attempts remaining. 139 of Boas, (AB)T = B TA for any two matrices Aand B. com Symmetric Matrix Inverse. The quadratic form of. the symmetric QRalgorithm, as the expense of two Jacobi sweeps is comparable to that of the entire symmetric QRalgorithm, even with the accumulation of transformations to obtain the matrix of eigenvectors. By the second and fourth properties of Proposition C. Join 90 million happy users! Sign Up free of charge:. If Ais symmetric, then A= AT. (1) Any real matrix with real eigenvalues is symmetric. there exists an orthogonal matrix P such that P\u22121AP =D, where D is diagonal. The main diagonal gets transposed onto itself. It is a specific case of the more general finite element method, and was in. The result is a 3x1 (column) vector. The matrix 1 1 0 2 has real eigenvalues 1 and 2, but it is not symmetric. A , in addition to being magic, has the property that \"the sum of the twosymmetric magic square numbers in any two cells symmetrically placed with respect to the center cell is the same\" (12, p. Now the next step to take the determinant. Indeed, if aij 6= aji we replace them by new a0 ij = a 0 ji = aij+aji 2, this does not change the corresponding quadratic form. Examples # NOT RUN { #Create a sample 3x3 matrix mat <- matrix(1:9, ncol=3) #Copy the upper diagonal portion to the lower symmetricize(mat, \"ud\") #Take the average of each symmetric location symmetricize(mat, \"avg\") # } Documentation reproduced from package HelpersMG, version 4. If A is not SPD then the algorithm will either have a zero. (1) Again, since A is a symmetric matrix, so A\u2032 = A. The link inertia matrix (3x3) is symmetric and can be specified by giving a 3x3 matrix, the diagonal elements [Ixx Iyy Izz], or the moments and products of inertia [Ixx Iyy Izz Ixy Iyz Ixz]. Maximum eigenvalue for this symmetric matrix is 3. JACOBI is a program written in 1980 for the HP-41C programmable calculator to find all eigenvalues of a real NxN symmetric matrix using Jacobi's method. An n\u00d7n matrix B is called nilpotent if there exists a power of the matrix B which is equal to the zero matrix. Recently, in order to find the principal moments of inertia of a large number of rigid bodies, it was necessary to compute the eigenvalues of many real, symmetric 3 \u00d7 3 matrices. Equation (1) is the eigenvalue equation for the matrix A. Mathematical Properties of Sti\ufb00ness Matrices 3 computation involving the inverse of ill-conditioned matrices can lose precision because there is a range of values in the solution { d }that can satsify [ K ]{ d }= { p }. all integral types except bool, floating point and complex types), whereas symmetric matrices can also be block matrices (i. A square matrix A is called a diagonal matrix if a ij = 0 for i 6= j. This pages describes in detail how to diagonalize a 3x3 matrix througe an example. Two apologies on quality: 1. Example A= 2 4 0 3 This is a 2 by 2 matrix, so we know that 1 + 2 = tr(A) = 5 1 2 = det(A) = 6 6. Shio Kun for Chinese translation. Use the ad - bc formula. We will use the following two properties of determinants of matrices. The Symmetric Inertia Tensor block creates an inertia tensor from moments and products of inertia. The function scipy. An identity matrix will be denoted by I, and 0 will denote a null matrix. In describing matrices, the format is: rows X columns. Eigenvalues and eigenvectors of a real symmetric matrix. Example 1: Determine the eigenvectors of the matrix. -24 * 5 = -120; Determine whether to multiply by -1. Now lets FOIL, and solve for. (1 Point) Give An Example Of A 3 \u00d7 3 Skew-symmetric Matrix A That Is Not Diagonal. We employ the latter, here. Definition of a Matrix The following are examples of matrices (plural of matrix). Note that all the main diagonal elements in the skew-symmetric matrix are zero. Now we only have to calculate the cofactor of a single element. can be obtained by using the cofactor. Now, noting that a symmetric matrix is positive semi-definite if and only if its eigenvalues are non-negative, we see that your original approach would work: calculate the characteristic polynomial, look at its roots to see if they are non-negative. A nxn symmetric matrix A not only has a nice structure, but it also satisfies the following:. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t \u2032 s c o r r e c t. First we compute the singular values \u03c3 i by \ufb01nding the eigenvalues of AAT. \u02d9 \u02d9 \u02da \u02d8 \u00cd \u00cd \u00ce \u00c8\" \u03c0 \" a i j a i j ij ij 0, 1, Row matrix. A3\u00d73 example of a matrix with some complex eigenvalues is B = 1 \u22121 \u22121 1 \u221210 10\u22121 A straightforward calculation shows that the eigenvalues of B are \u03bb = \u22121 (real), \u00b1i (complex conjugates). Good things happen when a matrix is similar to a diagonal matrix. As is well known, any symmetric matrix is diagonalizable, where is a diagonal matrix with the eigenvalues of on its diagonal, and is an orthogonal matrix with eigenvectors of as its columns (which magically form an orthogonal set , just kidding, absolutely no magic involved ). The above example illustrates a Cholesky algorithm, which generalizes for higher dimensional matrices. You can convert the skew symmetric matrix R_dot * dt into a rotation matrix using the Rodrigues formula. AB = BA = I n, then the matrix B is called an inverse of A. In general, matrices can contain complex numbers but we won't see those here. 231 Diagonalization of non-Hermitian matrices \u0432\u0402\u045e Let D be the diagonal matrix whose. If the matrix A is symmetric then \u2022its eigenvalues are all real (\u2192TH 8. // symmetric or not. I know that I can convert a single vector of size 3 in a skew symmetric matrix of size 3x3 as follows: X = [ 0 -x(3) x(2) ;. Example: The transpose of A is \u2022 For a matrix A = [aij], its transpose AT= [bij], where bij= aji. Example: RC circuit v1 vn c1 cn i1 in resistive circuit ckv Symmetric matrices, quadratic forms, matrix norm, and SVD 15-19. , The sum of the numbers along each matrix diagonal (the character) gives a shorthand version of the matrix representation, called \u0393:. NET example in Visual Basic demonstrating the features of the symmetric matrix classes. 32 (2) 223 222 uu kku kku k The global stiffness matrix may be constructed by directly adding terms associated with the degrees of freedom in k(1). As a recent example, the work of Spielman and Teng [14, 15] gives algorithms to solve symmetric, diagonally dominant linear systems in nearly-linear time in the input size, a fundamental advance. Statistics 1: Linear Regression and Matrices The concepts and terminology for matrices will be developed using an example from statistics. Matrix inversion. The subject of symmetric matrices will now be examined using an example from linear regression. Transpose of a Matrix, Symmetric Matrix & Skew Symmetric Matrix: Class 12 Transpose of a matrix , Symmetric Matrix and Skew Symmetric Matrix are explained in a very easy way. 2 Example: Odd or Even. The task is to find a matrix P which will let us convert A into D. Example, = -5 and =5 which means. Joachim Kopp developed a optimized \"hybrid\" method for a 3x3 symmetric matrix, which relays on the analytical mathod, but falls back to QL algorithm. Logic to check symmetric matrix in C programming. 2 De\ufb01niteness of Quadratic Forms. As a recent example, the work of Spielman and Teng [14, 15] gives algorithms to solve symmetric, diagonally dominant linear systems in nearly-linear time in the input size, a fundamental advance. (3) If the products (AB)T and BTAT are defined then they are equal. The values of \u03bb that satisfy the equation are the generalized eigenvalues. Step 1 - Accepts a square matrix as input Step 2 - Create a transpose of a matrix and store it in an array Step 3 - Check if input matrix is equal to its transpose. These are well-defined as $$A^TA$$ is always symmetric, positive-definite, so its eigenvalues are real and positive. Since the symmetric matrix is taken as A, the inverse symmetric matrx is written as A-1, such that it becomes. Program to swap upper diagonal elements with lower diagonal elements of matrix. A diagonal matrix is a symmetric matrix with all of its entries equal to zero except may be the ones on the diagonal. [R] binary symmetric matrix combination 3x3, 4x4) which I need to create in R as seen below: I think you didn't run the vecOut` after adding the new matrix. is an eigenvector corresponding to the eigenvalue 1. Enter payoff matrix B for player 2 (not required for zerosum or symmetric games). Let A = (v, 2v, 3v). For example, if a problem requires you to divide by a fraction, you can more easily multiply by its reciprocal. So why bother with the confusion matrix? Because it gives us insight into the details of how the algorithms achieve their percent correct. Diagonalizing a 3x3 matrix. We call such a matrix Hermitianafter the French mathematician Charles Hermite (1822-1901). Scroll down the page for examples and solutions.\nfyxewz0sqfycl dajkwnrqsps dfd1m1qpkrqm dpk3dqc9lqwv 5pbqzus3mshim14 61m01xv5xdl zd7gy8b40fe 19oy2gwbsp yeptctvyils esem86uwht8 06vj9j3m94 jp2wo0534lxaxty e5ggbv9peei 9w0eaa6hdw0 ks7w09hxgzkkt mlj60qjcs5p2 uomnmas9ekss j0m4lwi2h45fy 2eulcsy0gsmiqd8 6h2tya9io5puy2 0lu6mgdvy4h pv53iucjilxfrh w1u1ff9m3t0b4 bvswbs5ya8t 6yllw1n4gtpj 0pvr0bmphwgxkaz 9fu54s858j04 mpfludm9r4k9 fnvorb04eb z1ztaijsinf31o8 b1enw5cubxrnxmh ykiif7j79gx4sd vxzz4evvvnlz4wn l5j741s0ndd4 lofw30li8usega7", "date": "2020-07-12 15:53:48", "meta": {"domain": "curaben.it", "url": "http://quig.curaben.it/symmetric-matrix-example-3x3.html", "openwebmath_score": 0.7947157621383667, "openwebmath_perplexity": 464.864180982628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145699205375, "lm_q2_score": 0.8688267847293731, "lm_q1q2_score": 0.8610200024040231}}
{"url": "https://math.stackexchange.com/questions/3520406/how-is-pac-cap-bc-the-same-as-1-pa-cup-b", "text": "How is $P(A^c \\cap B^c)$ the same as $1-P(A \\cup B)$?\n\nI don't understand how $$P(A^c \\cap B^c) = 1-P(A \\cup B)$$ is the same?\n\nIf I draw $$P(A^c \\cap B^c)$$ as a Venn diagram:\n\nIf I draw $$P(A \\cup B)$$ as a Venn diagram:\n\nSo if I subtract $$P(A \\cup B)$$ from 1, wouldn't that mean that I subtract $$P(A \\cup B)$$ from the universe $$\\Omega$$, which would result int this:\n\nHowever, that would mean $$P(A^c \\cap B^c) \\neq 1-P(A \\cup B)$$\n\nEdit:\n\nAs pointed out by multiple people. My diagram for $$P(A^c \\cap B^c)$$ should look like the following and therefore the assumption of $$P(A^c \\cap B^c) = 1-P(A \\cup B)$$ is valid:\n\n\u2022 Your first diagram is of $A^c \\cup B^c$ not $A^c \\cap B^c$. The yellow areas include both $A \\cap B^c$ and $A^c \\cap B$ \u2013\u00a0Henry Jan 24 '20 at 0:09\n\u2022 your first diagram is wrong \u2013\u00a0Masacroso Jan 24 '20 at 0:11\n\u2022 The third diagram is the correct one for the left side as well. \u2013\u00a0Berci Jan 24 '20 at 0:12\n\u2022 So yellow should be only the universe? \u2013\u00a0Tom el Safadi Jan 24 '20 at 0:15\n\u2022 I edited my question with the help of you guys. Thanks!! \u2013\u00a0Tom el Safadi Jan 24 '20 at 0:19\n\nYour first diagram is incorrect. You seem to have drawn $$A^c\\cup B^c$$ (as some other people have mentioned).\n\nI would recommend drawing $$A^c$$ independently first, which consists of the rest of the universe and $$B-A$$. Then draw $$B^c$$ in a different color, noting again that you have the rest of the universe and $$A-B$$. The intersection of $$A-B$$ and $$B-A$$ is neither $$A$$ nor $$B$$, so you will end up without $$A$$ or $$B$$ in your final intersection. However, the rest of the universe is in both $$A^c$$ and $$B^c$$, therefore so is its intersection. Thus, you get that $$P(A^c\\cap B^c)$$ is just the universe without $$A$$ or $$B$$, which is equivalent to $$1-P(A\\cup B)$$.\n\nAccording to the DeMorgan's laws, one has \\begin{align*} \\textbf{P}(\\Omega) & = \\textbf{P}((A\\cup B)\\cup(A\\cup B)^{c})\\\\ & = \\textbf{P}(A\\cup B) + \\textbf{P}((A\\cup B)^{c})\\\\ & = \\textbf{P}(A\\cup B) + \\textbf{P}(A^{c}\\cap B^{c}) = 1 \\end{align*} from whence the result follows immediately, since $$X\\cup X^{c} = \\Omega$$ and $$X\\cap X^{c} = \\varnothing$$ for every event $$X\\subseteq\\Omega$$.", "date": "2021-04-11 10:59:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3520406/how-is-pac-cap-bc-the-same-as-1-pa-cup-b", "openwebmath_score": 0.7335774898529053, "openwebmath_perplexity": 310.7694330937354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429164804707, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8610036002650694}}
{"url": "http://planetmath.org/SumsOfNormalRandomVariablesNeedNotBeNormal", "text": "# sums of normal random variables need not be normal\n\nA common misconception among students of probability theory is the belief that the sum of two normally distributed (http://planetmath.org/NormalRandomVariable) random variables is itself normally distributed. By constructing a counterexample, we show this to be false.\n\nIt is however well known that the sum of normally distributed variables $X,Y$ will be normal under either of the following situations.\n\n\u2022 $X,Y$ are independent.\n\n\u2022 $X,Y$ are joint normal (http://planetmath.org/JointNormalDistribution).\n\nThe statement that the sum of two independent normal random variables is itself normal is a very useful and often used property. Furthermore, when working with normal variables which are not independent, it is common to suppose that they are in fact joint normal. This can lead to the belief that this property holds always.\n\nAnother common fallacy, which our example shows to be false, is that normal random variables are independent if and only if their covariance, defined by\n\n $\\operatorname{Cov}(X,Y)\\equiv\\mathbb{E}[XY]-\\mathbb{E}[X]\\mathbb{E}[Y]$\n\nis zero. While it is certainly true that independent variables have zero covariance, the converse statement does not hold. Again, it is often supposed that they are joint normal, in which case a zero covariance will indeed imply independence.\n\nWe construct a pair of random variables $X,Y$ satisfying the following.\n\n1. 1.\n\n$X$ and $Y$ each have the standard normal distribution.\n\n2. 2.\n\nThe covariance, $\\operatorname{Cov}(X,Y)$, is zero.\n\n3. 3.\n\nThe sum $X+Y$ is not normally distributed, and $X$ and $Y$ are not independent.\n\nWe start with a pair of independent random variables $X,\\epsilon$ where $X$ has the standard normal distribution and $\\epsilon$ takes the values $1,-1$, each with a probability of $1/2$. Then set,\n\n $Y=\\left\\{\\begin{array}[]{ll}\\epsilon X,&\\textrm{if }|X|\\leq 1,\\\\ -\\epsilon X,&\\textrm{if }|X|>1.\\end{array}\\right.$\n\nIf $S$ is any measurable subset of the real numbers, the symmetry of the normal distribution implies that $\\mathbb{P}(X\\in S)$ is equal to $\\mathbb{P}(-X\\in S)$. Then, by the independence of $X$ and $\\epsilon$, the distribution of $Y$ conditional on $\\epsilon=1$ is given by,\n\n $\\begin{split}\\displaystyle\\mathbb{P}(Y\\in S\\mid\\epsilon=1)&\\displaystyle=% \\mathbb{P}(|X|\\leq 1,X\\in S)+\\mathbb{P}(|X|>1,-X\\in S)\\\\ &\\displaystyle=\\mathbb{P}(|X|\\leq 1,X\\in S)+\\mathbb{P}(|X|>1,X\\in S)=\\mathbb{P% }(X\\in S).\\end{split}$\n\nIt can similarly be shown that $\\mathbb{P}(Y\\in S\\mid\\epsilon=-1)$ is equal to $\\mathbb{P}(X\\in S)$. So, $Y$ has the same distribution as $X$ and is normal with mean zero and variance one.\n\nUsing the fact that $\\epsilon$ has zero mean and is independent of $X$, it is easily shown that the covariance of $X$ and $Y$ is zero.\n\n $\\begin{split}\\displaystyle\\operatorname{Cov}(X,Y)&\\displaystyle=\\mathbb{E}[XY]% -\\mathbb{E}[X]\\mathbb{E}[Y]=\\mathbb{E}[XY]\\\\ &\\displaystyle=\\mathbb{E}\\left[1_{\\{|X|\\leq 1\\}}\\epsilon X^{2}\\right]+\\mathbb{% E}\\left[-1_{\\{|X|>1\\}}\\epsilon X^{2}\\right]\\\\ &\\displaystyle=\\mathbb{E}[\\epsilon]\\mathbb{E}\\left[1_{\\{|X|\\leq 1\\}}X^{2}% \\right]-\\mathbb{E}[\\epsilon]\\mathbb{E}\\left[1_{\\{|X|>1\\}}X^{2}\\right]\\\\ &\\displaystyle=0.\\end{split}$\n\nAs $X$ and $Y$ have zero covariance and each have variance equal to $1$, the sum $X+Y$ will have variance equal to $2$. Also, the sum satisfies\n\n $X+Y=\\left\\{\\begin{array}[]{ll}2\\epsilon X,&\\textrm{if }|X|\\leq 1,\\\\ 0,&\\textrm{if }|X|>1.\\end{array}\\right.$\n\nIn particular, this shows that $\\mathbb{P}(|X+Y|>2)=0$. However, normal random variables with nonzero variance always have a positive probability of being greater than any given real number. So, $X+Y$ is not normally distributed.\n\nThis also shows that, despite having zero covariance, $X$ and $Y$ are not independent. If they were, then the fact that sums of independent normals are normal would imply that $X+Y$ is normal, contradicting what we have just demonstrated.\n\nTitle sums of normal random variables need not be normal SumsOfNormalRandomVariablesNeedNotBeNormal 2013-03-22 18:43:44 2013-03-22 18:43:44 gel (22282) gel (22282) 5 gel (22282) Example msc 62E15 msc 60E05", "date": "2018-03-18 00:22:22", "meta": {"domain": "planetmath.org", "url": "http://planetmath.org/SumsOfNormalRandomVariablesNeedNotBeNormal", "openwebmath_score": 0.9247846007347107, "openwebmath_perplexity": 132.70268279962303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429151632047, "lm_q2_score": 0.8740772466456688, "lm_q1q2_score": 0.8610035991136771}}
{"url": "http://mathhelpforum.com/number-theory/16543-difference-two-squares.html", "text": "# Thread: difference of two squares\n\n1. ## difference of two squares\n\nFind two integers whose squares have a difference of 1,234,567.\n\ncan you find two integers whose squares will produce a difference of any whole number you chose?\n\nif so, how? if not, what kind of numbers can't be chosen.\n\nwell... a^2-b^2=1234567\n\nwe know a must end in a 1 or 9 and b must end in a 2 or 8 (this is to produce the seven at the end)\n\ni came up with 617289^2-617288^2=1234577 and that's as close as i can get.\n\nany suggestions? first time poster! sorry if it is an unusual / inappropriate post!\n\nmuch thanks!\n\n2. $n^{2}-(n-1)^{2}=1234567$\n\nn=617284\n\n$617284^{2}-617283^{2}=1234567$\n\n3. Number suchs that,\n$n\\equiv 0,1,3 (\\bmod 4)$\nAre expressible as a difference of two square.\n\nAnd,\n$n\\equiv 2 (\\bmod 4)$\nAre not.\n\n4. Hello, perfect square!\n\nFind two integers whose squares have a difference of 1,234,567\nGalactus used a very clever bit of trivia.\n\nConsecutive squares differ by consecutive odd numbers.\n\n. . $\\begin{array}{cccc} & & & \\text{diff} \\\\ 0^2 & = & 0 & \\\\ & & & 1 \\\\ 1^2 & = & 1 & \\\\ & & & 3 \\\\ 2^2 & = & 4 & \\\\ & & & 5 \\\\ 3^2 & = & 9 & \\\\ & & & 7 \\end{array}$\n. . $\\begin{array}{cccc}4^2 & = & 16 & \\\\ & & & 9 \\\\ 5^2 & = & 25 & \\\\ \\vdots & & \\vdots & \\vdots \\end{array}$\n\nAs he pointed out, we want two consecutive integers with a difference of 1234567.\n\nSo we have: . $(n + 1)^2 - n^2 \\:=\\:1,234,567$\n\n. . $n^2 + 2n + 1 - n^2 \\:=\\:1,234,567\\quad\\Rightarrow\\quad 2n+1 \\:=\\:1,234,567\\quad\\Rightarrow\\quad 2n \\:=\\:1,234,566$\n\n. . Hence: . $n \\,=\\,617,283$\n\nTherefore: . $617,284^2 - 617283^2 \\:=\\:1,234,567$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nI suspect that 1,234,567 is prime (but I'm not sure).\n\nIf the difference is a composite odd number,\n. . there may be an alternate solution.\n\nExample: Find two integers whose squares differ by 133.\n\nSince $\\frac{133-1}{2} = 66$, we have: . $67^2 - 66^2 \\:=\\:133$\n\n. Since $133 = 7\\times 19$, we have: . . $19 + 19 + 19 + 19 + 19 + 19 + 19$\n. . . . . which can be written: . . $13 + 15 + 17 + 19 + 21 + 23 + 25$\nwhich are the differences of: . $6^2\\quad7^2\\quad\\:8^2\\quad\\:9^2\\quad10^2\\quad11^2\\ quad12^2\\quad13^2$\n\nTherefore: . $13^2 - 6^2 \\:=\\:133$\n\n5. Originally Posted by Soroban\nSo we have: . $(n + 1)^2 - n^2 \\:=\\:1,234,567$\n\n. . $n^2 + 2n + 1 - n^2 \\:=\\:1,234,567\\quad\\Rightarrow\\quad 2n+1 \\:=\\:1,234,567\\quad\\Rightarrow\\quad 2n \\:=\\:1,234,566$\n\n. . Hence: . $n \\,=\\,617,283$\nSuch a seemingly difficult question, transformed into simple algebra! Once again! Nice!\n\n6. I suspect that 1,234,567 is prime (but I'm not sure).\n\nNo, Soroban, it is not prime.", "date": "2018-02-23 03:36:38", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/16543-difference-two-squares.html", "openwebmath_score": 0.9460057616233826, "openwebmath_perplexity": 766.5822532013905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504291340685, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8610035862694057}}
{"url": "https://brilliant.org/discussions/thread/combination-proof/", "text": "# Combination Proof\n\nWe will prove that:\n\n$$\\dbinom{n}{k} = \\displaystyle \\prod_{i=1}^{k} \\frac{n-i+1}{i}$$\n\n$$\\forall$$ $$k \\in \\mathbb{N}$$, $$k \\leq n$$\n\nProof #1: As the reader may guess, we will use induction to prove this. We establish a base case as $$k=1$$:\n\n$$\\dbinom{n}{1} = \\displaystyle \\prod_{i=1}^{1} \\frac{n-i+1}{i}$$\n\n$$\\Rightarrow$$ $$\\frac{n!}{1!(n-1)!} = n$$\n\n$$\\Rightarrow$$ $$n=n$$\n\nThe base case clearly holds. Now we establish the inductive case as $$n=\\alpha$$:\n\n$$\\dbinom{n}{\\alpha} = \\displaystyle \\prod_{i=1}^{\\alpha} \\frac{n-i+1}{i}$$\n\n$$\\Rightarrow$$ $$\\frac{n-\\alpha}{\\alpha +1} \\dbinom{n}{\\alpha} =\\frac{n-\\alpha}{\\alpha +1} \\displaystyle \\prod_{i=1}^{\\alpha} \\frac{n-i+1}{i}$$ (By hypothesis)\n\n$$\\Rightarrow$$ $$\\frac{n-\\alpha}{\\alpha +1} \\dbinom{n}{\\alpha} =\\displaystyle \\prod_{i=1}^{\\alpha + 1} \\frac{n-i+1}{i}$$\n\n$$\\Rightarrow$$ $$\\frac{n!(n-\\alpha)}{(\\alpha +1)!(n-\\alpha)!} =\\displaystyle \\prod_{i=1}^{\\alpha + 1} \\frac{n-i+1}{i}$$\n\n$$\\Rightarrow$$ $$\\frac{n!}{(\\alpha +1)!(n-(\\alpha+1))!} = \\displaystyle \\prod_{i=1}^{\\alpha + 1} \\frac{n-i+1}{i}$$\n\n$$\\Rightarrow$$ $$\\dbinom{n}{\\alpha+1} = \\displaystyle \\prod_{i=1}^{\\alpha + 1} \\frac{n-i+1}{i}$$\n\nSo we may conclude that the statement is true for $$k=1$$ and if it is true for some $$k=\\alpha$$, then it must be true for $$k=\\alpha+1$$. The proof follows by induction.\n\nQED\n\nProof #2: We will now offer an alternative proof, which is actually just a simple direct derivation:\n\n$$\\dbinom{n}{k} = \\frac{n!}{k!(n-k)!} = \\frac{\\displaystyle \\prod_{i=0}^{n-1} (n-i)}{\\displaystyle \\prod_{i=0}^{k-1} (k-i) \\left( \\displaystyle \\prod_{i=0}^{n-k-1} (n-k-i) \\right)}$$\n\n$$= \\frac{\\displaystyle \\prod_{i=1}^k (n-i+1)}{k!} = \\frac{\\displaystyle \\prod_{i=1}^k (n-i+1)}{ \\displaystyle \\prod_{i=1}^k i }$$\n\n$$= \\displaystyle \\prod_{i=1}^k \\frac{(n-i+1)}{i}$$ Because multiplication is commutative.\n\nWhich is the intended result.\n\nQED.\n\nNote by Ethan Robinett\n3\u00a0years, 10\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nI've been scrolling through your profile the complete morning now and am finding some really nice notes. Thanks for sharing them!\n\n- 3\u00a0years, 4\u00a0months ago\n\nThanks!\n\n- 3\u00a0years, 4\u00a0months ago\n\nNice note@Ethan Robinett\n\n- 3\u00a0years, 10\u00a0months ago\n\nThanks appreciate it!\n\n- 3\u00a0years, 10\u00a0months ago\n\nHey nice proofs! :) I did the Inductive one but didn't think of using a direct one\n\n- 3\u00a0years, 10\u00a0months ago\n\nThanks! Yeah after I did the induction, I figured there was probably a way to directly show it, given that factorials are just products by definition, I thought the second one was pretty interesting\n\n- 3\u00a0years, 10\u00a0months ago\n\nI like the way the induction is used.\n\n- 3\u00a0years, 10\u00a0months ago\n\nThanks man\n\n- 3\u00a0years, 10\u00a0months ago", "date": "2018-06-21 12:31:56", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/combination-proof/", "openwebmath_score": 0.9979040026664734, "openwebmath_perplexity": 3539.271385194643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9850429169195593, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8610035861086184}}
{"url": "http://mathhelpforum.com/algebra/40280-simplify-expression-negative-exponents.html", "text": "# Math Help - Simplify an Expression with Negative Exponents\n\n1. ## Simplify an Expression with Negative Exponents\n\n(ab)^-1 / (a^-2 + b^-2)\n\nif you could explain it with an appropriate property, that would be great!!\n\nthanks..!!\n\n2. $\\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\\frac{\\frac{1}{ab}}{\\frac{1}{a^2}+\\frac{1}{b^2 }}=\\frac{\\frac{1}{ab}}{\\frac{b^2+a^2}{a^2b^2}}=\\fr ac{1}{ab}\\cdot\\frac{a^2b^2}{a^2+b^2}=\\frac{a^2b^2} {ab(a^2+b^2)}=\\frac{ab}{a^2+b^2}$\n\nI think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.\n\n3. Originally Posted by masters\n$\\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\\frac{\\frac{1}{ab}}{\\frac{1}{a^2}+\\frac{1}{b^2 }}=\\frac{\\frac{1}{ab}}{\\frac{b^2+a^2}{a^2b^2}}=\\fr ac{1}{ab}\\cdot\\frac{a^2b^2}{a^2+b^2}=\\frac{a^2b^2} {ab(a^2+b^2)}=\\frac{ab}{a^2+b^2}$\n\nI think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.\nLooks good to me, but don't forget to specify that $a \\neq 0$ and $b \\neq 0$ independently.\n\n-Dan\n\n4. Hi !\n\nA slightly different approach\n\n$\\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\\frac{1}{(ab)(a^{-2}+b^{-2})}$\n\n-----------------------\nUsing the power rule : $a^b a^c=a^{b+c}$\n\n--> $(ab)a^{-2}=ba^{-1}$\n-----------------------\n\n$=\\frac{1}{ba^{-1}+ab^{-1}}=\\frac{1}{\\frac ba+\\frac ab}=\\frac{1}{\\frac{b^2+a^2}{ab}}=\\frac{ab}{a^2+b^2 }$\n\none extra equality\n\n5. Always keep it interesting, Moo. Good job!!\n\n6. thank you both for your help!\n\n7. Hello, needymathperson!\n\nOne different step . . .\n\n$\\frac{(ab)^{-1}}{a^{-2} + b^{-2}}$\nWe have: . $\\frac{(ab)^{-1}} {a^{-2} + b^{-2}} \\;\\;=\\;\\;\\frac{\\dfrac{1}{ab}} {\\dfrac{1}{a^2} + \\dfrac{1}{b^2}}$\nMultiply top and bottom by $a^2b^2\\!:\\;\\;\\frac{a^2b^2\\left(\\dfrac{1}{ab}\\right )}{a^2b^2\\left(\\dfrac{1}{a^2} + \\dfrac{1}{b^2}\\right)} \\;\\;=\\;\\;\\boxed{\\frac{ab}{b^2+a^2}}$", "date": "2014-10-30 18:16:23", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/40280-simplify-expression-negative-exponents.html", "openwebmath_score": 0.9745804071426392, "openwebmath_perplexity": 4258.951983591372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429153827491, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8610035831497443}}
{"url": "http://math.stackexchange.com/questions/353771/how-to-make-a-box-which-has-the-largest-possible-volume/353786", "text": "# How to make a box which has the largest possible volume?\n\nI have sheet metal in form of an equilateral triangle and I want to fold it to make a container for the screws. How should I cut and fold to make the a box with largest volume?\n\nBasically I cut the corners and then fold them. There is no \"roof\". Thank you!\n\n$$a=0.15m$$\n\n-\n@SreekanthKarumanaghat: Cut that out; this doesn't look like homework. \u2013\u00a0Henning Makholm Apr 7 '13 at 13:57\n\nLet the side remaining after cutting off the vertices be A. The volume is given by:\n\n$$V = \\dfrac12 A^2 \\sin(60) \\times h = \\dfrac{\\sqrt3}{4} A^2h$$\n\nThe relation between $A$, $a$ and $h$ is:\n\n$A = a - \\dfrac{2h}{\\tan(30)}$\n\n(You get a small kite at the edge with two right angles, one angle of 60 degrees and one angle at 120 degrees)\n\nYou can substitute them now:\n\n$$V = \\dfrac{\\sqrt3}{4} \\left(a - \\dfrac{2h}{\\tan(30)}\\right)^2h$$\n\n$$V = \\dfrac{\\sqrt3}{4} (0.15 - 2\\sqrt3h)^2h$$\n\n$$V = 3 \\sqrt3 h^3-0.45 h^2+0.00974279 h$$\n\nDifferentiate w.r.t. $h$ to get:\n\n$$\\dfrac{dV(h)}{dh} = 9 \\sqrt3 h^2-0.9 h+0.00974279$$\n\nEquate to zero and solve for $h$ to get $h\\approx0.0433013$ and $h\\approx0.0144338$.\n\nThe first one gives the minimum volume, so you don't want that. Take the second.\n\n-\nThank you Jerry. :) Cheers! You get the right answer, because you actually gave me two value and described them really nice! I could follow the process very easily! \u2013\u00a0user31113 Apr 7 '13 at 11:46\nYou're welcome! \u2013\u00a0Jerry Apr 7 '13 at 11:47\n\nIf you cut the corner in the manner shown, by trigonometry at any corner, the new side length is smaller by $2 \\sqrt3 h$. (Let me know if you have difficulty with this.)\n\nHence the box's volume is proportional to $\\left(a-2 \\sqrt3 h\\right)^2 h$, which we try to maximise. Let $V(h) = \\left(a-2 \\sqrt3 h\\right)^2 (4 \\sqrt3 h)$. The value of $h$ which maximises $V(h)$ is exactly same as that maximising the volume we desire, due to proportionality. Now $V(h)$ can be looked at as the product of $3$ terms, $(a-2 \\sqrt3 h), (a-2 \\sqrt3 h)$ and $(4 \\sqrt3 h)$, which sum to a constant $2a$. Hence the product is maximised when these three terms are equal. i.e.\n\n$a-2 \\sqrt3 h = 4 \\sqrt3 h$ or when $h = \\dfrac{a}{6\\sqrt3}$.\n\n-\nOh woops, yours is definitely more elegant! \u2013\u00a0Jerry Apr 7 '13 at 11:26\n@Jerry Thanks. It's just that I like inequalities a lot. \u2013\u00a0Macavity Apr 7 '13 at 11:32\nThank you Macavity. :) Cheers! \u2013\u00a0user31113 Apr 7 '13 at 11:44\n@Macavity Your solution is certainly neat, but I get a slight butterfly feeling when you pull that factor out of the air. I seem to get the same numbers by a method at least I find simpler. Any comments? \u2013\u00a0Brian Chandler Dec 20 '14 at 16:24\n\nUsing exactly the same argument as in this question: Optimize volume of an open cardboard box made from flat square of cardboard... we consider an infinitesimal change $\\delta$ in the position of the fold between walls and floor. At the maximum, this change must add/subtract the same volume from the change in height as from the moving in/out of the walls. Call the area of the base $B$, the perimeter of the base $P$, and the height $h$, then these changes are:\n\n$\\delta \\times h \\times P$ ...change over walls\n\n$\\delta \\times B$ ...change over base\n\nAnd the maximum is when the area of the base and the area of the walls are equal, so we have:\n\n$h = B / P$\n\nIn this case, let $s$ be the side of the base. Then the area of the base (an equilateral triangle) and perimeter are given by:\n\n$B = \\frac{\\sqrt 3}{4} s^2$\n\n$P = 3s$\n\nSo\n\n$h = \\frac{\\sqrt 3}{4} s^2 / 3s = \\frac{1}{4\\sqrt 3} s$\n\nNow we get $s$ from the original triangle side $a$ and substitute:\n\n$s = a - 2\\sqrt 3 h$\n\n$h = \\frac{1}{4\\sqrt 3} (a - 2\\sqrt 3 h)$\n\n$4\\sqrt 3 h = a - 2\\sqrt 3 h$\n\n$h = \\frac{a}{6\\sqrt 3}$\n\n-", "date": "2016-02-14 08:36:01", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/353771/how-to-make-a-box-which-has-the-largest-possible-volume/353786", "openwebmath_score": 0.8287667632102966, "openwebmath_perplexity": 383.01804115020917, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429107723174, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8610035807354542}}
{"url": "https://math.stackexchange.com/questions/573055/projection-of-ellipsoid", "text": "# Projection of ellipsoid\n\nFind the projections of the ellipsoid\n\n$$x^2 + y^2 + z^2 -xy -1 = 0$$\n\non the cordinates plan\n\nI have no idea how to do this. I couldn't find much on google to help me with it too.\n\nBy coordinate planes, I assume you mean the $xy$-plane, $xz$-plane, and $yz$-plane? If so, the projections can be found as follows:\n\n$xy$-plane: Let $z=0$.\n\n$xz$-plane and $yz$-plane: At first try, you may want to use $y=0$ or $z=0$ respectively; however, this will leave you with the cross section with the $xz$-plane, not the projection onto the $xz$-plane. To get the proper projections, we need to analyze the graph of the projection in the $xy$-plane; in particular, find the equations of the horizontal and vertical tangent lines.\n\nIn the $xy$-plane, the equation simplifies to $x^2+y^2-xy-1=0$. Implicitly differentiating with respect to $x$ gives us\n\n$$2x+2y\\frac{dy}{dx} - y - x\\frac{dy}{dx} = 0 \\implies \\frac{dy}{dx} = \\frac{y-2x}{2y-x}.$$\n\nNow, you get horizontal tangent lines when the numerator is zero (i.e. when $dy/dx = 0$), which implies that $y-2x=0\\implies y=2x$. Substituting this back into the implicit equation gives us $$x^2+4x^2-2x^2 -1 = 0 \\implies 3x^2=1 \\implies x=\\pm \\frac{\\sqrt{3}}{3}$$\n\nand thus $y = \\pm\\dfrac{2\\sqrt{3}}{3}$ are the equations of the horizontal tangent lines.\n\nSimilarly, we find the vertical tangent lines when the denominator is zero (i.e. when $dy/dx$ is undefined), which implies that $2y-x=0\\implies x=2y$. In a similar fashion, we get that $y=\\pm\\dfrac{\\sqrt{3}}{3}$ and hence $x=\\pm\\dfrac{2\\sqrt{3}}{3}$ are the equations of the vertical tangent lines.\n\nNow why did we go through all of this? Well, the equations of the tangent lines tells us how far along the $x$ and $y$ axis the ellipse extends; in particular, the vertical tangents give us a bound on $x$ and $y$ for the ellipse (i.e. $-2\\sqrt{3}/3 \\leq x,y \\leq 2\\sqrt{3}/3$). These bounds play the role of the major axes for the $xz$ and $yz$ projections of the ellipsoid onto the $y=0$ and $x=0$ planes respectively. The minor axis will be along the $z$ direction; in particular, along the ellipsoid, $-1\\leq z\\leq 1$. With that said, the corresponding projection of the ellipsoid onto the $yz$-plane is $$\\frac{3y^2}{4} + z^2 = 1$$\n\nand the corresponding projection onto the $xz$-plane is $$\\frac{3x^2}{4} + z^2 = 1$$\n\nNote that the last two figures are cylinders of the projections as seen in three space; the purpose of visualizing them this way was to show you that they completely enclose the ellipsoid of interest.\n\n\u2022 Yes, I meant it, thanks. Well, I did this on a first moment but the answer is different for $yz$ and $xz$. For example: $yz: \\frac{3}{4}y^2 + z^2 -1 = 0$ Nov 19, 2013 at 11:17\n\u2022 @Giiovanna I deleted my answer, fixed it, and then undeleted it. I hope what I wrote above makes sense! Nov 19, 2013 at 12:44\n\nIn case you have difficulties with the accepted answer, here is another approach:\n\nYour ellipsoid $S$ is the level surface $F^{-1}(\\{0\\})$ of the function $$F(x,y,z):=x^2+y^2+z^2-xy-1\\ .$$ Let $S'$ be the shadow of $S$ in the $(x,y)$-plane. The points $p\\in S$ that generate the boundary $\\partial S'$ are characterized by the property that the tangent plane at $p$ is vertical (i.e., parallel to the $z$-axis), which is the same as saying that the surface normal $n_p$ at $p$ is horizontal. Now $n_p$ is given by $\\nabla F(p)$, whose third coordinate is $2z$.\n\nIt follows that the points $(x,y,z)\\in S$ producing the shadow boundary are the points with $z=0$, i.e., the points $(x,y,0)$ satisfying $x^2+y^2-xy-1=0$. In this special example the shadow boundary in fact coincides with the intersection of $S$ with the $(x,y)$-plane.\n\nTherefore let's do the shadow $S''$ of $S$ in the $(x,z)$-plane as well. The points $p\\in S$ that generate the boundary $\\partial S''$ are characterized by the property that the tangent plane at $p$ is parallel to the $y$-axis, which is the same as saying that the surface normal $n_p$ at $p$ is orthogonal to the $y$-axis, or has $y$-component $0$. The $y$-component of $\\nabla F(p)$ is given by $2y-x$. Therefore we can say that the points $p$ in question lie on the plane $2y-x=0$. As they satisfy a priori $F(x,y,z)-1=0$ we can eliminate $y$ from the two equations $$x^2+y^2+z^2-xy-1=0,\\qquad y={x\\over2}$$ and obtain a single equation connecting their $x$- and $z$-coordinates: $$x^2+{x^2\\over4}+z^2-x\\>{x\\over2}-1=0\\ ,$$ or $${3\\over4}x^2+z^2-1=0\\ .$$ This is already the equation of the shadow boundary $\\partial S''$.\n\nI may now safely leave the shadow on the $(y,z)$-plane to you.\n\n\u2022 I like this method the most Sep 18, 2017 at 9:10\n\u2022 It\u2019s worth noting that for a quadric surface, the points that generate its shadow outline are always coplanar.\n\u2013\u00a0amd\nDec 24, 2021 at 19:53\n\nAn approach related to Christian Blatter\u2019s answer can be found in Result 8.9 on page 201 of Hartley & Zisserman\u2019s Multiple View Geometry In Computer Vision. Under the camera matrix $\\mathtt P$, the outline of the quadric $\\mathtt Q$ is the conic given by $\\mathtt C^*=\\mathtt P\\mathtt Q^*\\mathtt P^T$. (Here the superscript asterisk indicates the dual conic.) The proof follows directly from the observations that lines $\\mathbf l$ tangent to the outline satisfy $\\mathbf l^T\\mathtt C^*\\mathbf l=0$ and back-project to planes $\\mathtt P^T\\mathbf l$ that are tangent to the original quadric.\n\nIn the present problem the quadric is the nondegenerate ellipsoid $$\\mathtt Q = \\begin{bmatrix}1&-\\frac12&0&0 \\\\ -\\frac12&1&0&0 \\\\ 0&0&1&0 \\\\ 0&0&0&-1\\end{bmatrix}$$ so we can use inverse matrices for the duals: the outline of $\\mathtt Q$ under the projection $\\mathtt P$ is therefore $\\mathtt C = (\\mathtt P\\mathtt Q^{-1}\\mathtt P^T)^{-1}$.\n\nFor orthogonal projection onto the $x$-$y$ plane, we can use $$\\mathtt P = \\begin{bmatrix}1&0&0&0\\\\0&1&0&0\\\\0&0&0&1\\end{bmatrix}$$ for the camera matrix and have $$\\mathtt Q^{-1}=\\begin{bmatrix}\\frac43&\\frac23&0&0\\\\\\frac23&\\frac43&0&0\\\\0&0&1&0\\\\0&0&0&-1\\end{bmatrix},$$ which gives us $$\\mathtt C = \\begin{bmatrix}1&-\\frac12&0\\\\-\\frac12&1&0\\\\0&0&-1\\end{bmatrix},$$ or $x^2+y^2-xy=1$. For the $x$-$z$ plane, $$\\mathtt P = \\begin{bmatrix}1&0&0&0\\\\0&0&1&0\\\\0&0&0&1\\end{bmatrix}$$ and $$\\mathtt C = \\begin{bmatrix}\\frac34&0&0\\\\0&1&0\\\\0&0&-1\\end{bmatrix},$$ i.e., $\\frac34x^2+z^2=1$. A similar computation for the $y$-$z$ plane yields $\\frac34y^2+z^2=1$.\n\n\u2022 Hello @amd, your solution works like a charm for me! Thank you, for posting this answer. Small question, how do we find the projection matrix, P, for any arbitrary plane? Dec 23, 2021 at 3:47\n\u2022 @KashishDhal There are various ways, but for any of them you must first choose a coordinate system for the plane. Once you have that, the easiest thing that I can think of is to compute a coordinate transformation that takes that plane to one of the prime coordinate planes and compose it with the corresponding simple projection matrix from above.\n\u2013\u00a0amd\nDec 24, 2021 at 19:49\n\u2022 I followed the method that you described in this thread: math.stackexchange.com/questions/3073718/\u2026 However, I don't know if my solutions are correct. How can I verify that the computed ellipse are indeed correct? Dec 24, 2021 at 20:59", "date": "2022-07-03 03:44:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/573055/projection-of-ellipsoid", "openwebmath_score": 0.9175476431846619, "openwebmath_perplexity": 128.60796287651883, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9850429116504951, "lm_q2_score": 0.874077222043951, "lm_q1q2_score": 0.8610035718095498}}
{"url": "https://math.stackexchange.com/questions/1949683/if-u-1-u-n-v-1-v-k-is-a-basis-does-it-necessarily-follow-that-u", "text": "# If $\\{u_1,...,u_n,v_1,...,v_k\\}$ is a basis does it necessarily follow that $\\{u_1,...,u_n\\}$ is linearly independent?\n\nWe are given that $\\{u_1,...,u_n,v_1,...,v_k\\}$ is a basis. My question is does it necessarily follow that $\\{u_1,...,u_n\\}$ or $\\{v_1,...,v_k\\}$ are linearly independent? I suspect the answer is yes, but I think I'm missing something in my argument.\n\nSo: if $\\{u_1,...,u_n,v_1,...,v_k\\}$ is a basis, then this list of vectors is linearly independent. That is,\n\n$a_1 u_1 + \\cdots + a_n u_n + b_1 v_1 + \\cdots + b_k v_k = 0 \\implies a_1 = \\cdots = a_n = b_1 = \\cdots = b_k = 0$.\n\nThen I want to say the following: Since all the $b_i \\,'s$ are $0$, we have $a_1 u_1 + \\cdots + a_n u_n = 0 \\implies a_1 = \\cdots = a_n = 0$, meaning that {$u_1,...,u_n$} is linearly independent. With a similar argument, we can also conclude that {$v_1,...,v_k$} is linearly independent.\n\nAm I missing something though? I feel like my argument is kind of making a bit of a stretch. Thanks for your help!!\n\n\u2022 You're right! in general, any subset of a linearly independent set is again linearly independent, as your proof shows (you're not using more than linearly independency). Oct 1 '16 at 20:25\n\u2022 Yes, any subset of a linearly independent set is lin. indep. itself. Oct 1 '16 at 20:25\n\u2022 It's fine for me. Actually, whether it's a basis is not the relevant hypothesis. What you proved indeed is a little more general: any sublist of a list of linearly independent vectors is linearly independent. Oct 1 '16 at 20:26\n\u2022 Thank you so much guys!!! :D Oct 1 '16 at 20:36\n\nYour argument is fine, but perhaps this is a case where it can be nicely put as a proof by contradiction. Suppose some sublist were not linearly independent, then we can find some $\\{a_i\\}$, not all zero, such that $\\sum_{i=1}^n a_iu_i=0$. We can then use this to show that the bigger list is not linearly independent.", "date": "2022-01-21 17:12:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1949683/if-u-1-u-n-v-1-v-k-is-a-basis-does-it-necessarily-follow-that-u", "openwebmath_score": 0.8266095519065857, "openwebmath_perplexity": 123.6936836204304, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540668504082, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8609959543662097}}
{"url": "http://onelab.info/wiki/Tutorial/Coupled_problems", "text": "Tutorial/Coupled problems\n\nIntroduction\n\nHere is explained how to use the solution of a first problem as a data in a second problem. Two kind of problem are studied here. In the \"first\" one, the solution of the first problem directly appears in the weak formulation of the second problem (for example, as a source or a Neumann boundary) whereas in the second kind of problem, the solution is used as a constraint (Dirichlet boundary condition), so it should appears in the function space. Note that a Lagrange multiplier could be used to force the Dirichlet constraint in the weak formulation.\n\nFirst kind of coupled problem\u00a0: solution appearing in the weak formulation\n\nMathematics\n\nLet the computation domain be the unit square $\\Omega = [0,1]\\times[0,1]$ with boundary $\\Gamma$ and unit outwardly directed normal $\\mathbf{n}$. Consider the two coupled following problems\u00a0: firstly, find $u$, solution of $$\\begin{cases}\\label{eq:problemU} -\\Delta u + u = C & \\text{in } \\Omega,\\\\ \\displaystyle{\\frac{\\partial u}{\\partial \\mathbf{n}} = 0} & \\text{on }\\Gamma, \\end{cases}$$ where $C$ is a constant. Then, find the solution $v$ of the second problem $$\\begin{cases}\\label{eq:problemV} -\\Delta v + v = 0 & \\text{in } \\Omega,\\\\ \\displaystyle{\\frac{\\partial v}{\\partial \\mathbf{n}} = u} & \\text{on }\\Gamma, \\end{cases}$$ where $u$ is the solution of problem \\ref{eq:problemU}. Obviously, the function $u$ is the constant function equal to $C$ and thus, there is no \"real\" need in solving numerically problem \\ref{eq:problemU}. However, please keep in mind that this is just an example. In order to solve numerically these problem using a finite element method, on has to write the weak formulations of problems \\ref{eq:problemU} and \\ref{eq:problemV}, which read as: $$\\label{eq:WeakFormulationU} \\left\\{\\begin{array}{l} \\text{Find } u\\in H^1(\\Omega) \\text{ such that}\\\\ \\displaystyle{\\forall u'\\in H^1(\\Omega), \\qquad \\int_{\\Omega} \\nabla u\\cdot\\nabla u' \\;{\\rm d}\\Omega + \\int_{\\Omega} u\\cdot u' \\;{\\rm d}\\Omega - \\int_{\\Omega} Cu'\\;{\\rm d}\\Omega = 0}, \\end{array}\\right.$$ and $$\\label{eq:WeakFormulationV} \\left\\{\\begin{array}{l} \\text{Find } v\\in H^1(\\Omega) \\text{ such that, }\\\\ \\displaystyle{\\forall v'\\in H^1(\\Omega), \\qquad \\int_{\\Omega} \\nabla v\\cdot\\nabla v' \\;{\\rm d}\\Omega + \\int_{\\Omega}vv' \\;{\\rm d}\\Omega - \\int_{\\Gamma}uv'\\;{\\rm d}\\Gamma = 0}, \\end{array}\\right.$$ where $H^1(\\Omega)$ is the classical Sobolev space.\n\nComputation in GetDP\n\nIn GetDP, this kind of problem is very easy to solve. Indeed, the user has to introduce two function space (one per solution), and ask GetDP to solve each problem in the right order. There is no trap\u00a0! Below are the GMSH and GetDP files. Compared to the other academic examples, the only new things are in the .pro file.\n\nFiles\n\nThere are two files\u00a0:\n\nFirstCoupledProblem.geo that contains the geometry and is used to mesh the domain\nFirstCoupledProblem.pro that contains the weak formulations, function spaces, etc.\n\nTo solve the problem, type in a terminal (in the right directory)\u00a0:\n\ngmsh FirstPb.geo -2\n\nand then solve the problem with getdp\u00a0:\n\ngetdp FirstPb.pro -solve -pos\n\nor directly\n\ngetdp FirstPb.pro -solve CoupledProblem -pos Map_u_and_v\n//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):\nlc = 0.05;\n\n// The parameters of the border of the domain\u00a0:\nx_max = 1;\nx_min = 0;\ny_max= 1;\ny_min = 0;\n\n//Creation of the 4 angle points of the domain Omega (=square)\np1 = newp; Point(p1) = {x_min,y_min,0,lc};\np2 = newp; Point(p2) = {x_min,y_max,0,lc};\np3 = newp; Point(p3) = {x_max,y_max,0,lc};\np4 = newp; Point(p4) = {x_max,y_min,0,lc};\n\n//The four edges of the square\nL1 = newreg; Line(L1) = {p1,p2};\nL2 = newreg; Line(L2) = {p2,p3};\nL3 = newreg; Line(L3) = {p3,p4};\nL4 = newreg; Line(L4) = {p4,p1};\n\n// Line Loop (= boundary of the square)\nBound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};\n\n//Surface of the square\nSurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};\n\n// To conclude, we define the physical entities, that is \"what GetDP could see/use\".\n// 1 is associtated to Omega\n// 2 to Gamma\nPhysical Surface(1) = {SurfaceOmega};\nPhysical Line(2) = {L1,L2,L3,L4};\n// Do not forget to let a blank line at the end, this could make GMSH crash...\n\n\n\nDirect link to file CoupledProblems/First/FirstPb.geo'\n\n// Group\n//======\nGroup{\nOmega = Region[{1}];\nGama = Region[{2}];\n}\n\n// Function\n//=========\nFunction{\nC = 5; //constant C in the problem solved by u\n}\n\n//Jacobian\n//========\nJacobian {\n{ Name JVol\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Vol\u00a0; }\n}\n}\n{ Name JSur\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Sur\u00a0; }\n}\n}\n{ Name JLin\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Lin\u00a0; }\n}\n}\n}\n\n//Integration (parameters)\n//=======================\nIntegration {\n{ Name I1\u00a0;\nCase {\n{ Type Gauss\u00a0;\nCase {\n{ GeoElement Point\u00a0; NumberOfPoints  1\u00a0; }\n{ GeoElement Line\u00a0; NumberOfPoints  4\u00a0; }\n{ GeoElement Triangle\u00a0; NumberOfPoints  6\u00a0; }\n{ GeoElement Quadrangle\u00a0; NumberOfPoints 7\u00a0; }\n{ GeoElement Tetrahedron\u00a0; NumberOfPoints 15\u00a0; }\n{ GeoElement Hexahedron\u00a0; NumberOfPoints 34\u00a0; }\n}\n}\n}\n}\n}\n\n//FunctionSpace\n//=============\nFunctionSpace{\n//Space for u\nBasisFunction{\n{Name wn; NameOfCoef vn; Function BF_Node;\nSupport Region[{Omega, Gama}]; Entity NodesOf[All];}\n}\n}\n//Space for v\nBasisFunction{\n{Name wn; NameOfCoef vn; Function BF_Node;\nSupport Region[{Omega, Gama}]; Entity NodesOf[All];}\n}\n}\n}\n/*\nThe two FunctionSpaces are exactly the same. However, keep in mind that\na FunctionSpace can contain only one solution\u00a0! That is why we need two FunctionSpaces.\n*/\n\n//(Weak) Formulations\n//==================\n\nFormulation{\n// Problem solved by u\u00a0:\n// Delta u + u = C (Omega)\n// dn u = 0 (Gamma)\n{Name ProblemU; Type FemEquation;\nQuantity{\n{Name u; Type Local; NameOfSpace Hgrad_u;}\n}\nEquation{\nIn Omega; Jacobian JVol; Integration I1;}\nGalerkin{ [Dof{u}, {u}];\nIn Omega; Jacobian JVol; Integration I1;}\nGalerkin{ [-C, {u}];\nIn Omega; Jacobian JVol; Integration I1;}\n}\n}\n\n// Coupled problem\u00a0: \"first kind\"\u00a0:\n// Neumann condition on Gamma\n// Delta v + v =0 Omega\n// dn v = u on Gamma\n{Name ProblemV; Type FemEquation;\nQuantity{\n// v is the unknown\n{Name v; Type Local; NameOfSpace Hgrad_v;}\n// u is called but will NOT be used as an unknown (NO \"Dof\")\n{Name u; Type Local; NameOfSpace Hgrad_u;}\n}\nEquation{\nIn Omega; Jacobian JVol; Integration I1;}\n\nGalerkin{ [Dof{v}, {v}];\nIn Omega; Jacobian JVol; Integration I1;}\n//Neumann condition on Gama (no \"Dof\"\u00a0!)\nGalerkin{ [-{u}, {v}];\nIn Gama; Jacobian JSur; Integration I1;}\n}\n}\n\n}\n\n// Resolution\n//===========\n\nResolution{\n// First coupled problem available\n{Name CoupledProblem;\nSystem{\n{Name PbU; NameOfFormulation ProblemU;}\n{Name PbV; NameOfFormulation ProblemV;}\n}\nOperation{\nGenerate[PbU]; Solve[PbU]; SaveSolution[PbU];\nGenerate[PbV]; Solve[PbV]; SaveSolution[PbV];\n}\n/*Short explanations\u00a0: GetDP will\u00a0:\n1) generate and solve the problem associated to \"u\" then save \"u\" in the space H1_C\n2) generate and solve the problem associated to \"v\" and save \"v\" in the space H1\n*/\n}\n\n}\n\n//Post Processing\n//===============\n\nPostProcessing{\n{Name CoupledProblem; NameOfFormulation ProblemV;\n//The associated formulation is \"ProblemV\" because it involves both \"u\" and \"v\"\n//(ProblemU only call \"u\", not \"v\")\nQuantity{\n{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}\n{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}\n}\n}\n}\n\n//Post Operation\n//==============\n\nPostOperation{\n{Name Map_u_and_v; NameOfPostProcessing CoupledProblem;\nOperation{\nPrint[u, OnElementsOf Omega, File \"sol_u.pos\"];\nPrint[v, OnElementsOf Omega, File \"sol_v.pos\"];\n}\n}\n}\n\n\nSecond kind of coupled problem\u00a0: solution used as a Dirichlet boundary condition\n\nMathematics\n\nIn this section, the first problem (\\ref{eq:problemU}) does not change but the second problem is now to find $v$, such that $$\\begin{cases}\\label{eq:problemV2} -\\Delta v + v = 0 & \\text{in } \\Omega,\\\\ \\displaystyle{v = u} & \\text{on }\\Gamma, \\end{cases}$$ where $u$ is the solution of (\\ref{eq:problemU}). The function $u$ is now involved as a Dirichlet boundary condition and thus, will appear in the function space of $v$ (again, except if one uses a Lagrange multiplier, which is not the case here). To obtain the weak formulation of problem (\\ref{eq:problemV2}), we introduce the following function space \\ref{eq:WeakFormulationU} $$H^1_u(\\Omega) = \\left\\{w\\in H^1(\\Omega) \\text{ such that w|_{\\Gamma} = u|_{\\Gamma}, where u is the solution of problem (1)} \\right\\}.$$ Thus, we have $$\\label{eq:WeakFormulationV2} \\left\\{\\begin{array}{l} \\text{Find } v\\in H^1_u(\\Omega) \\text{ such that, }\\\\ \\displaystyle{\\forall v'\\in H^1_0(\\Omega), \\qquad \\int_{\\Omega} \\nabla v\\cdot\\nabla v' \\;{\\rm d}\\Omega + \\int_{\\Omega}vv' \\;{\\rm d}\\Omega = 0}, \\end{array}\\right.$$\n\nGetDP\n\nIn GetDP, a Dirichlet boundary condition is generally imposed through the \"Constraint\" term, it can be imposed with the help of a Lagrange multiplier. In the second case, the Dirichlet boundary condition is no more \"imposed\" in the function space but appears in the weak formulation (as a Neumann boundary condition). Thus, the previous case can be used. Here is only considered the first case using the GetDP function \"AssignFromResolution\". A Resolution dedicated to $u$ must then be created, in this example it will be named AuxiliaryResolution. In GetDP, the procedure can be summarized as follows:\n\n\u2022 Create a constraint with\u00a0:\nType AssignFromResolution\u00a0; Resolution AuxiliaryResolution;\n\u2022 Construct only ONE function space that will contain $u$ and at the end $v$ ($u$ will be erased\u00a0!)\n\u2022 Create two Resolution, one for $v$ and one for $u$\u00a0:\n\u2022 for $v$\u00a0: Generate, Solve, Save\n\u2022 for $u$ (AuxiliaryResolution)\u00a0: Generate, Solve, TransfertSolution (do not Save\u00a0!)\n\nWith this, GetDP will ...\u00a0:\n\n\u2022 Try to solve problem with $v$\n\u2022 Need the Dirichlet constraint (that is $u$)\n\u2022 Solve problem with $u$\n\u2022 Solve problem with $v$\n\nSometimes it is usefull to save both $u$ and $v$ at the end of the calculation. This can be done by adding an auxiliary problem that \"copy\" $u$ (see the last paragraph).\n\nFiles\n\nThere are two files\u00a0:\n\nSecondpb.geo\nexactly the same as FirstPb.geo\nSecondpb.pro\nquite the same as FirstPb.pro\n\nTo solve the problem, type in a terminal (in the right directory)\u00a0:\n\ngmsh SecondPb.geo -2\n\nand then solve the problem with getdp\u00a0:\n\ngetdp SecondPb.pro -solve -pos\n\nand chose the first resolution (MainResolution) or directly\n\ngetdp SecondPb.pro -solve MainResolution -pos Map_v\n//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):\nlc = 0.05;\n\n// The parameters of the border of the domain\u00a0:\nx_max = 1;\nx_min = 0;\ny_max= 1;\ny_min = 0;\n\n//Creation of the 4 angle points of the domain Omega (=square)\np1 = newp; Point(p1) = {x_min,y_min,0,lc};\np2 = newp; Point(p2) = {x_min,y_max,0,lc};\np3 = newp; Point(p3) = {x_max,y_max,0,lc};\np4 = newp; Point(p4) = {x_max,y_min,0,lc};\n\n//The four edges of the square\nL1 = newreg; Line(L1) = {p1,p2};\nL2 = newreg; Line(L2) = {p2,p3};\nL3 = newreg; Line(L3) = {p3,p4};\nL4 = newreg; Line(L4) = {p4,p1};\n\n// Line Loop (= boundary of the square)\nBound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};\n\n//Surface of the square\nSurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};\n\n// To conclude, we define the physical entities, that is \"what GetDP could see/use\".\n// 1 is associtated to Omega\n// 2 to Gamma\nPhysical Surface(1) = {SurfaceOmega};\nPhysical Line(2) = {L1,L2,L3,L4};\n// Do not forget to let a blank line at the end, this could make GMSH crash...\n\n\n\nDirect link to file CoupledProblems/Second/SecondPb.geo'\n\n// Group\n//======\nGroup{\nOmega = Region[{1}];\nGama = Region[{2}];\n}\n\n// Function\n//=========\nFunction{\nC = 5; //constant C in the problem solved by u\n}\n\n//Jacobian\n//========\nJacobian {\n{ Name JVol\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Vol\u00a0; }\n}\n}\n{ Name JSur\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Sur\u00a0; }\n}\n}\n{ Name JLin\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Lin\u00a0; }\n}\n}\n}\n\n//Integration (parameters)\n//=======================\nIntegration {\n{ Name I1\u00a0;\nCase {\n{ Type Gauss\u00a0;\nCase {\n{ GeoElement Point\u00a0; NumberOfPoints  1\u00a0; }\n{ GeoElement Line\u00a0; NumberOfPoints  4\u00a0; }\n{ GeoElement Triangle\u00a0; NumberOfPoints  6\u00a0; }\n{ GeoElement Quadrangle\u00a0; NumberOfPoints 7\u00a0; }\n{ GeoElement Tetrahedron\u00a0; NumberOfPoints 15\u00a0; }\n{ GeoElement Hexahedron\u00a0; NumberOfPoints 34\u00a0; }\n}\n}\n}\n}\n}\n\n// Dirichlet Constraint\nConstraint{\n{Name DirichletV;\nCase{\n{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }\n}\n}\n}\n/*\nThe constraint will be obtained through the resolution of \"AuxiliaryResolution\".\nThis means that GetDP will automatically solve \"AuxiliaryResolution\" when it will\nneed the contraint\u00a0!\n*/\n\n//FunctionSpace\n//=============\nFunctionSpace{\n//Space for u AND v\u00a0!!\nBasisFunction{\n{Name wn; NameOfCoef vn; Function BF_Node;\nSupport Region[{Omega, Gama}]; Entity NodesOf[All];}\n}\n//the dirichlet constraint \"v = u\" on Gamma\nConstraint{\n{NameOfCoef vn; EntityType NodesOf;\nNameOfConstraint DirichletV;}\n}\n}\n}\n/*\nONLY one function space that will contain \"u\" and at the end, \"v\".\n*/\n\n//(Weak) Formulations\n//==================\n\nFormulation{\n// Problem solved by u\u00a0:\n// Delta u + u = C (Omega)\n// dn u = 0 (Gamma)\n{Name ProblemU; Type FemEquation;\nQuantity{\n{Name u; Type Local; NameOfSpace Hgrad;}\n}\nEquation{\nIn Omega; Jacobian JVol; Integration I1;}\nGalerkin{ [Dof{u}, {u}];\nIn Omega; Jacobian JVol; Integration I1;}\nGalerkin{ [-C, {u}];\nIn Omega; Jacobian JVol; Integration I1;}\n}\n}\n\n// Coupled problem\u00a0: \"first kind\"\u00a0:\n// DIRICHLET condition on Gamma\n// Delta v + v =0 Omega\n// v = u on Gamma\n{Name ProblemV; Type FemEquation;\nQuantity{\n// v is the unknown\n{Name v; Type Local; NameOfSpace Hgrad;}\n}\nEquation{\nIn Omega; Jacobian JVol; Integration I1;}\n\nGalerkin{ [Dof{v}, {v}];\nIn Omega; Jacobian JVol; Integration I1;}\n}\n}\n\n}\n\n// Resolution\n//===========\n\nResolution{\n//Main Resolution\u00a0: the one that will be called when launching getdp\n{Name MainResolution;\nSystem{\n{Name PbV; NameOfFormulation ProblemV;}\n}\nOperation{\nGenerate[PbV]; Solve[PbV]; SaveSolution[PbV];\n}\n}\n/*Short explanations\u00a0: GetDP will\u00a0:\ngenerate and solve the problem associated to \"v\" and save \"v\" in the space H1\nHowever, to compute \"v\", GetDP need to compute the Dirichlet Contraint, thus\nis needs to solve AuxiliaryResolution (bellow)\u00a0!\n*/\n\n// Auxiliary resolution that computes \"u\"\n{Name AuxiliaryResolution;\nSystem{\n//System associated to \"u\" will be transfered to the weak formulation of \"v\"\n{Name PbU; NameOfFormulation ProblemU; DestinationSystem PbV;}\n{Name PbV; NameOfFormulation ProblemV;}\n}\nOperation{\nGenerate[PbU]; Solve[PbU]; TransferSolution[PbU];\n}\n}\n/*Short explanations\u00a0: GetDP will\u00a0:\n1) generate and solve the problem associated to \"u\"\n2) Transfert solution u (to problem associated to \"v\")\n*/\n}\n\n//Post Processing\n//===============\n\nPostProcessing{\n{Name CoupledProblem; NameOfFormulation ProblemV;\nQuantity{\n{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}\n}\n}\n}\n\n//Post Operation\n//==============\n\nPostOperation{\n{Name Map_v; NameOfPostProcessing CoupledProblem;\nOperation{\nPrint[v, OnElementsOf Omega, File \"sol_v.pos\"];\n}\n}\n}\n\n\nSecond kind of coupled problem with a save of solution $u$\n\nSometimes, it can be usefull to save both $u$ and $v$. To achieved this, one can introduce an auxiliary variable $u_{aux}$ that will be a copy of $u$. Here, we proposed to add\u00a0:\n\nA new FunctionSpace to store $u$\nA new weak formulation to copy $u$ into an auxiliary variable $u_{aux}$\n\nThe modified .pro file (SecondPbWithSave.pro) can be found below. The .geo file being exactly the same as the previous ones.\n\n//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):\nlc = 0.05;\n\n// The parameters of the border of the domain\u00a0:\nx_max = 1;\nx_min = 0;\ny_max= 1;\ny_min = 0;\n\n//Creation of the 4 angle points of the domain Omega (=square)\np1 = newp; Point(p1) = {x_min,y_min,0,lc};\np2 = newp; Point(p2) = {x_min,y_max,0,lc};\np3 = newp; Point(p3) = {x_max,y_max,0,lc};\np4 = newp; Point(p4) = {x_max,y_min,0,lc};\n\n//The four edges of the square\nL1 = newreg; Line(L1) = {p1,p2};\nL2 = newreg; Line(L2) = {p2,p3};\nL3 = newreg; Line(L3) = {p3,p4};\nL4 = newreg; Line(L4) = {p4,p1};\n\n// Line Loop (= boundary of the square)\nBound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};\n\n//Surface of the square\nSurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};\n\n// To conclude, we define the physical entities, that is \"what GetDP could see/use\".\n// 1 is associtated to Omega\n// 2 to Gamma\nPhysical Surface(1) = {SurfaceOmega};\nPhysical Line(2) = {L1,L2,L3,L4};\n// Do not forget to let a blank line at the end, this could make GMSH crash...\n\n\n\nDirect link to file CoupledProblems/SecondWithSave/SecondPbWithSave.geo'\n\n// Group\n//======\nGroup{\nOmega = Region[{1}];\nGama = Region[{2}];\n}\n\n// Function\n//=========\nFunction{\nC = 5; //constant C in the problem solved by u\n}\n\n//Jacobian\n//========\nJacobian {\n{ Name JVol\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Vol\u00a0; }\n}\n}\n{ Name JSur\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Sur\u00a0; }\n}\n}\n{ Name JLin\u00a0;\nCase {\n{ Region All\u00a0; Jacobian Lin\u00a0; }\n}\n}\n}\n\n//Integration (parameters)\n//=======================\nIntegration {\n{ Name I1\u00a0;\nCase {\n{ Type Gauss\u00a0;\nCase {\n{ GeoElement Point\u00a0; NumberOfPoints  1\u00a0; }\n{ GeoElement Line\u00a0; NumberOfPoints  4\u00a0; }\n{ GeoElement Triangle\u00a0; NumberOfPoints  6\u00a0; }\n{ GeoElement Quadrangle\u00a0; NumberOfPoints 7\u00a0; }\n{ GeoElement Tetrahedron\u00a0; NumberOfPoints 15\u00a0; }\n{ GeoElement Hexahedron\u00a0; NumberOfPoints 34\u00a0; }\n}\n}\n}\n}\n}\n\n// Dirichlet Constraint\nConstraint{\n{Name DirichletV;\nCase{\n{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }\n}\n}\n}\n/*\nThe constraint will be obtained through the resolution of \"AuxiliaryResolution\".\nThis means that GetDP will automatically solve \"AuxiliaryResolution\" when it will\nneed the contraint\u00a0!\n*/\n\n//FunctionSpace\n//=============\nFunctionSpace{\n//Space for u\nBasisFunction{\n{Name wn; NameOfCoef vn; Function BF_Node;\nSupport Region[{Omega, Gama}]; Entity NodesOf[All];}\n}\n}\n\n//Space for u_aux (=copy of u) AND v\u00a0!!\nBasisFunction{\n{Name wn; NameOfCoef vn; Function BF_Node;\nSupport Region[{Omega, Gama}]; Entity NodesOf[All];}\n}\n//the dirichlet constraint \"v = u\" on Gamma\nConstraint{\n{NameOfCoef vn; EntityType NodesOf;\nNameOfConstraint DirichletV;}\n}\n}\n}\n\n//(Weak) Formulations\n//==================\n\nFormulation{\n// Problem solved by u\u00a0:\n// Delta u + u = C (Omega)\n// dn u = 0 (Gamma)\n{Name ProblemU; Type FemEquation;\nQuantity{\n{Name u; Type Local; NameOfSpace Hgrad_u;}\n}\nEquation{\nIn Omega; Jacobian JVol; Integration I1;}\nGalerkin{ [Dof{u}, {u}];\nIn Omega; Jacobian JVol; Integration I1;}\nGalerkin{ [-C, {u}];\nIn Omega; Jacobian JVol; Integration I1;}\n}\n}\n\n// Coupled problem\u00a0: \"first kind\"\u00a0:\n// DIRICHLET condition on Gamma\n// Delta v + v =0 Omega\n// v = u on Gamma\n{Name ProblemV; Type FemEquation;\nQuantity{\n// v is the unknown\n{Name u; Type Local; NameOfSpace Hgrad_u;}\n{Name v; Type Local; NameOfSpace Hgrad_v;}\n}\nEquation{\nIn Omega; Jacobian JVol; Integration I1;}\n\nGalerkin{ [Dof{v}, {v}];\nIn Omega; Jacobian JVol; Integration I1;}\n}\n}\n\n//Copy \"u\" in \"u_aux\"\n{Name CopyU; Type FemEquation;\nQuantity{\n// v is the unknown\n{Name u; Type Local; NameOfSpace Hgrad_u;}\n{Name u_aux; Type Local; NameOfSpace Hgrad_v;}\n}\nEquation{\nGalerkin{ [Dof{u_aux}, {u_aux}];\nIn Omega; Jacobian JVol; Integration I1;}\n\nGalerkin{ [-{u}, {u_aux}];\nIn Omega; Jacobian JVol; Integration I1;}\n}\n}\n\n}\n\n// Resolution\n//===========\n\nResolution{\n//Main Resolution\u00a0: the one that will be called when launching getdp\n{Name MainResolution;\nSystem{\n{Name PbV; NameOfFormulation ProblemV;}\n}\nOperation{\nGenerate[PbV]; Solve[PbV]; SaveSolution[PbV];\n}\n}\n/*Short explanations\u00a0: GetDP will\u00a0:\nTry to compute \"v\", however, GetDP needs to compute the Dirichlet Contraint\nthus to solve AuxiliaryResolution (below)\u00a0!\n*/\n\n// Auxiliary resolution that computes \"u\"\n{Name AuxiliaryResolution;\nSystem{\n//System associated to \"u\" will be transfered to the weak formulation of \"v\"\n{Name PbU; NameOfFormulation ProblemU;}\n{Name PbUaux; NameOfFormulation CopyU; DestinationSystem PbV;}\n{Name PbV; NameOfFormulation ProblemV;}\n}\nOperation{\nGenerate[PbU]; Solve[PbU]; SaveSolution[PbU];\nGenerate[PbUaux]; Solve[PbUaux]; TransferSolution[PbUaux];\n}\n}\n/*Short explanations\u00a0: GetDP will\u00a0:\n1) Compute \"u\"\n2) Copy \"u\" in \"uaux\"\n3) Transfert solution \"uaux\" (to problem associated to \"v\")\n*/\n}\n\n//Post Processing\n//===============\n\nPostProcessing{\n{Name CoupledProblem; NameOfFormulation ProblemV;\nQuantity{\n{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}\n{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}\n}\n}\n}\n\n//Post Operation\n//==============\n\nPostOperation{\n{Name Map_Sol; NameOfPostProcessing CoupledProblem;\nOperation{\nPrint[u, OnElementsOf Omega, File \"sol_u.pos\"];\nPrint[v, OnElementsOf Omega, File \"sol_v.pos\"];\n}\n}\n}", "date": "2018-12-17 07:57:52", "meta": {"domain": "onelab.info", "url": "http://onelab.info/wiki/Tutorial/Coupled_problems", "openwebmath_score": 0.7176864147186279, "openwebmath_perplexity": 13616.122405568942, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.986979508737192, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8609889964100731}}
{"url": "https://math.stackexchange.com/questions/2743313/prove-that-the-number-of-answers-for-a-1a-2-a-k-le-n-is-equal-to-t", "text": "Prove that the number of answers for $|a_1|+|a_2|+\u2026+|a_k| \\le n$ is equal to the number of answers for $|a_1|+|a_2|+\u2026+|a_n| \\le k$.\n\nI'm trying to solve this problem:\n\nProve that the number of answers for $|a_1|+|a_2|+...+|a_k|\u2264n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n|\u2264k$. all $a_i$ is an integer number.\n\nI have a solution for non negative $a_i$: We convert $|a_1|+|a_2|+...+|a_k|\u2264n$ to $a_1+a_2+...+a_k+c=n$ and convert $|a_1|+|a_2|+...+|a_n|\u2264k$ to $a_1+a_2+...+a_n+c=k$.\n\nBy bars and stars theorem: Number of answers for $a_1+a_2+...+a_k+c=n$ is equal to $\\binom{n+k}{k}$ and number of answers for $a_1+a_2+...+a_n+c=k$ is equal to $\\binom{n+k}{n}$. That $\\binom{n+k}{k} = \\binom{n+k}{n}$.\n\nbut i don't know how to prove for negative integers.\n\n\u2022 So we allow $a_i$ to be any integer, but we consider its absolute value when taking the sum? \u2013\u00a0hardmath Apr 19 '18 at 0:41\n\u2022 @hardmath yes,we take absolute value \u2013\u00a0user547757 Apr 19 '18 at 14:57\n\nLet $f(n,k)$ be the number of solutions to $|a_1|+\\dots+|a_n|\\le k$. Note that $$f(n,k) = f(n-1,k)+2f(n-1,k-1)+2f(n-1,k-2)+\\dots+2f(n-1,0),\\tag{*}$$ by conditioning on the value of $a_n$. Rearranging, and using $(*)$ applied to $f(n,k-1)$, we get \\begin{align} f(n,k) &= f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\\dots+2f(n-1,0)]\\\\ &= f(n-1,k)+f(n-1,k-1)+f(n,k-1) \\end{align} The above shows that $f(n,k)$ obeys a recurrence which is symmetric in $n,k$. Since you can easily verify the symmetric base cases $f(n,0)=f(0,k)=1$, the result $f(n,k)=f(k,n)$ follows by induction.\n\nWe count the number of solutions to $\\sum_{i=1}^n |a_i|\\leq k$.\n\nThere will be $j\\leq r:={\\rm min}\\{n,k\\}$ of the $a_i$ that are $\\ne0$, call them $a_{i_1}$, $\\ldots$, $a_{i_j}$. The $i_l$ $(1\\leq l\\leq j)$ can be chosen in ${n\\choose j}$ ways. For the chosen $i_l$ put $|a_{i_l}|=x_l+1$ with $x_l\\geq 0$. Then we have to count the number of solutions to $x_1+\\ldots+x_j+c=k-j$. This number is ${k\\choose j}$. It has to be multiplied by $2^j$ in order to account for the choice of the signs of the corresponding $a_{i_l}$. The total number of solutions therefore comes to $$\\sum_{j=0}^r{n\\choose j}{k\\choose j}2^j\\ ,$$ which is symmetric in $n$ and $k$.\n\nIntroducing notation $$A_i=|a_i|$$ the first of inequalities can be rewritten as: $$A_1+A_2+...+A_k\\le n.\\tag1$$ Any solution to $$(1)$$ can be viewed as composed of $$i$$ zeros ($$0\\le i\\le k$$) and $$k-i$$ non-zeros. Each non-zero value doubles the number of solutions to original equation (which can be both positive and negative). The overall number of solutions is therefore $$\\sum_{i=0}^{k}\\binom{k}{i}\\binom{n}{k-i}2^{k-i} \\stackrel{i=k-j}=\\sum_{j=0}^{k}\\binom{k}{j}\\binom{n}{j}2^{j} =\\sum_{j=0}^{\\min(k,n)}\\binom{k}{j}\\binom{n}{j}2^{j},\\tag2$$ where $$\\binom{k}{i}$$ counts the number of ways to place $$i$$ zeros into $$k$$ cells, and $$\\binom{n}{k-i}$$ counts the number of positive solutions to the inequality $$p_1+p_2+\\cdots+p_{k-i}\\le n.$$\n\nIn view of symmetry of $$(2)$$ with respect to interchange of $$n$$ and $$k$$ the original statement is proved.\n\nJust to show another approach to the answer already given by C.Blatter, let's start and consider $$\\left\\{ \\matrix{ 0 \\le b_{\\,j} \\in \\mathbb Z \\hfill \\cr b_{\\,1} + b_2 + b_{\\,3} + \\cdots + b_{\\,k} \\le n \\hfill \\cr} \\right.$$ Then the number of solutions is the integral volume $V(k,n)$ of k-D simplex, with edge segments $(0,n)$, thus of length $n+1$.\nIt is not difficult to demonstrate that $$V(k,n)= \\binom{n+k}{n}= \\binom{n+k}{k}=V(n,k)$$ then the hypothesis is already true in this case.\n\nWe can then partition the above solutions into those that contains $j$ variables null and the remaining $k-j$ having values greater than $0$. We can assign the null variables in $\\binom{k}{j}$ ways.\nThat corresponds to $$V(k,n)= \\binom{n+k}{n}= \\binom{n+k}{k} = \\sum\\limits_{0\\, \\le \\,j\\, \\le \\,k} { \\binom{n}{k-j} \\binom{k}{j} }$$\n\nComing to our present case, for each non null variable $b_j$ we can assign two values to the corresponding $a_j$, and just one to the null variables. So in this case the volume will be $$V_{\\,a} \\,(k,n) = \\sum\\limits_{\\left( {0\\, \\le } \\right)\\,j\\,\\left( { \\le \\,k} \\right)} {\\binom{n}{k-j} \\binom{k}{j} 2^{\\,k - j} } = \\sum\\limits_{\\left( {0\\, \\le } \\right)\\,l\\,\\left( { \\le \\,k} \\right)} {\\binom{n}{l} \\binom{k}{l} 2^l } = V_{\\,a} \\,(n,k)$$\n\nTherefore the hypothesis is demonstrated.\n\nI just saw this question as it came back to the front page. I wrote this answer a couple of months ago, which proves that this number is $$\\sum_{k=0}^n2^k\\binom{n}{k}\\binom{m}{k}$$ which is also shown in answers to this question. My answer starts with a recurrence (like Mike Earnest's answer) and solves it by induction.", "date": "2019-10-15 18:52:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2743313/prove-that-the-number-of-answers-for-a-1a-2-a-k-le-n-is-equal-to-t", "openwebmath_score": 0.9405272006988525, "openwebmath_perplexity": 102.37634690405126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795114181105, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8609889954732668}}
{"url": "https://mathoverflow.net/questions/133732/upper-limit-on-the-central-binomial-coefficient", "text": "# Upper limit on the central binomial coefficient\n\nWhat is the tightest upper bound we can establish on the central binomial coefficients $2n \\choose n$ ?\n\nI just tried to proceed a bit, like this:\n\n$n! > n^{\\frac{n}{2}}$\n\nfor all $n>2$. Thus,\n\n$\\binom{2n}{n} = \\frac{ (n+1) \\ldots (2n) }{n!} < \\frac{\\left(\\frac{\\sum_{k=1}^n (n+k) }{n}\\right)^n }{n^{n/2}} = \\frac{ \\left( \\frac{ n^2 + \\frac{n(n+1)}{2} }{n} \\right) ^n}{n^{n/2}} = \\left( \\frac{3n+1}{2\\sqrt{n}} \\right)^n$\n\nBut, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).\n\n\u2022 The simplest upper bound to prove is $4^n$ (which is still stronger than your bound) and just follows from the binomial expansion of $(1+1)^{2n}$. Peter's answer gives a less wasteful estimate. Jun 14 '13 at 13:00\n\u2022 I should have thought about it. Yes, Peter's bound is very good, I tested it too. Jun 14 '13 at 13:54\n\nHere's a way to motivate and refine the argument that P\u00e9ter Komj\u00e1th attributes to Erd\u0151s.\n\nStart by computing the ratio between the $$n$$-th and $$(n-1)$$-st central binomial coefficients: $${2n \\choose n} \\left/ {2(n-1) \\choose n-1} \\right. = \\frac{(2n)! \\phantom. / \\phantom. n!^2}{(2n-2)! \\phantom. / \\phantom. (n-1)^2} = \\frac{(2n)(2n-1)}{n^2} = 4 \\left( 1 - \\frac1{2n} \\right).$$ For large $$n$$, this ratio approaches $$4$$, suggesting that $$2n \\choose n$$ grows roughly as $$4^n$$. If the factor $$1 - \\frac1{2n}$$ were $$1 - \\frac1n = (n-1)/n$$, the growth would be exactly proportional to $$n^{-1} 4^n$$. Since $$1 - \\frac1{2n}$$ is (for large $$n$$) nearly the square root of $$1 - \\frac1n$$, the actual asymptotic should be proportional to $$n^{-1/2} 4^n$$. So we introduce the ratio $$r_n := \\left( {2n \\choose n} \\left/ \\frac{4^n}{\\sqrt n} \\right. \\right)^2 = \\frac{n}{16^n} {2n \\choose n}^2.$$ Then $$\\frac{r_n}{r_{n-1}} = \\left( 1 - \\frac1{2n} \\right)^2 \\left/ \\left( 1 - \\frac1n \\right) \\right. = \\frac{(2n-1)^2}{(2n-2)(2n)} \\gt 1.$$ Thus $$r_{n-1} < r_n$$; and since $$r_1 = (2/4)^2 = 1/4$$ we have by induction $$r_1 \\lt r_2 \\lt r_3 \\lt r_4 \\lt \\cdots \\lt r_n = \\frac12 \\frac{1 \\cdot 3}{2 \\cdot 2} \\frac{3 \\cdot 5}{4 \\cdot 4} \\frac{5 \\cdot 7}{6 \\cdot 6} \\cdots \\frac{(2n-3)(2n-1)}{(2n-2)(2n-2)} \\frac{2n-1}{2n}.$$ Each $$r_{n_0}$$ gives a lower bound on $$r_n$$, and thus on $$2n\\choose n$$, for all $$n \\geq n_0$$. The OP asked for upper bounds, so consider $$R_n := \\frac{2n}{2n-1} r_n = \\frac{n}{\\left(1-\\frac 1{2n}\\right)16^n} {2n \\choose n}^2.$$ Now $$R_{n+1}/R_n = (2n-1)(2n+1) \\phantom. / \\phantom. (2n)^2 = (4n^2-1) \\phantom. / \\phantom. (4n^2) \\lt 1$$, so $$\\frac12 = R_1 \\gt R_2 \\gt R_3 \\gt R_4 \\gt \\cdots \\gt R_{n+1} = \\frac12 \\frac{1 \\cdot 3}{2 \\cdot 2} \\frac{3 \\cdot 5}{4 \\cdot 4} \\frac{5 \\cdot 7}{6 \\cdot 6} \\cdots \\frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}.$$ It follows that $$R_n \\geq r_{n'}$$ for any $$n,n'$$, so $$R_1=1/2$$, $$R_2=3/8$$, $$R_3=45/128$$, etc. are a series of upper bounds on every $$r_n$$. Since moreover $$r_n / R_n = 1 - \\frac1{2n} \\rightarrow 1$$ as $$n \\rightarrow \\infty$$, both $$r_n$$ and $$R_n$$ converge to a common limit that is an upper bound on every $$r_n$$. If we accept Wallis's product (which is classical though not as elementary as everything else in our analysis), then we can evaluate this common limit as $$1/\\pi$$ and thus recover the asymptotically sharp upper bound $${2n \\choose n} < 4^n / \\sqrt{\\pi n}$$.\n\n\u2022 +1, and let me challenge your statement that the Wallis product is not as elementary, by pointing to the Monthly note math.chalmers.se/~wastlund/monthly.pdf and the 16th Christmas Tree Lecture by Don Knuth, \"Why pi?\" (now on youtube). Jul 1 '13 at 16:58\n\u2022 Your answer has been cited in this paper that I was just reading, mentioning it here as a token of appreciation: arxiv.org/pdf/2005.10009.pdf\n\u2013\u00a0NULL\nJul 28 '20 at 19:59\n\nEven the asymptotically sharp inequality ${2n \\choose n} < 4^n \\left/ \\sqrt{\\pi n} \\right.$ has a short proof: $${2n \\choose n} = \\frac{4^n}{\\pi} \\int_{-\\pi/2}^{\\pi/2} \\cos^{2n} x \\phantom. dx < \\frac{4^n}{\\pi} \\int_{-\\pi/2}^{\\pi/2} e^{-nx^2} dx < \\frac{4^n}{\\pi} \\int_{-\\infty}^{\\infty} e^{-nx^2} dx = \\frac{4^n}{\\sqrt{\\pi n}}.$$ In the first step, the formula for $\\int_{-\\pi/2}^{\\pi/2} \\cos^{2n} x \\phantom. dx$ can be proved by induction via integration by parts, or using the Beta function.\n\nThe third step is clear, and the last step is the well-known Gaussian integral. So we need only justify the the second step.\n\nThere we need the inequality $\\cos x \\leq e^{-x^2/2}$, or equivalently $$\\log \\cos x + \\frac{x^2}{2} \\leq 0,$$ for $\\left|x\\right| < \\pi/2$, with equality only at $x=0$. This is true because $\\log \\cos x +\\frac12 x^2$ is an even function of $x$ that vanishes at $x=0$ and whose second derivative $-\\tan^2 x$ is negative for all nonzero $x \\in (-\\pi/2, \\pi/2)$. QED\n\n\u2022 I'm not sure what counts as elementary in this game, but I would think the easiest proof that $\\frac{1}{\\pi} \\int \\cos^{2n} x dx = \\binom{2n}{n}/4^n$ is to write $\\cos x = (e^{ix} + e^{-ix})/2$ and expand $\\cos^{2n}$ by the binomial theorem. Jul 1 '13 at 14:18\n\nErdos remarked somewhere the bound\n$${{2n}\\choose{n}}<\\frac{4^n}{\\sqrt{2n+1}}.$$ This can be established by induction: $${{2n+2}\\choose{n+1}}=\\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\\choose{n}}$$ and if we have the bound for $n$, we only have to show $$\\frac{2(2n+1)}{(n+1)\\sqrt{2n+1}}<\\frac{4}{\\sqrt{2n+3}}$$ which reduces to $4n^2+8n+3<4n^2+8n+4$.\n\n\u2022 This bound is pretty close to optimal because (unless I've computed badly) Stirling's approximation gives $4^n/\\sqrt{\\pi n}$. Jun 14 '13 at 13:26\n\nNoam Elkies notes that there is a quick proof of $$\\binom{2n}{n} \\leq \\frac{4^n}{\\sqrt{\\pi n}}$$ by writing $$\\binom{2n}{n} = \\frac{4^n}{\\pi} \\int_{-\\pi/2}^{\\pi/2} \\cos^{2n} x dx$$ and bounding $\\cos^2 x \\leq e^{-x^2}$.\n\nThere is an equally good lower bound by a similar method: $$\\int_{-\\pi/2}^{\\pi/2} \\cos^{2n} x dx =\\int_{-\\pi/2}^{\\pi/2} \\frac{\\sec^2 x dx}{(\\tan^2 x+1)^{n+1}} = \\int_{- \\infty}^{\\infty} \\frac{du}{(1+u^2)^{n+1}} \\geq \\int_{- \\infty}^{\\infty} e^{-(n+1) u^2} du$$ so $$\\binom{2n}{n} \\geq \\frac{4^n}{\\sqrt{\\pi (n+1)}}.$$ Here the inequality $\\tfrac{1}{1+u^2} \\geq e^{-u^2}$ follows from the standard bound $e^y \\geq 1+y$ for $y\\geq 0$.\n\nWe may prove without using integrals that $$\\frac1{\\sqrt{\\pi n}}\\geqslant 4^{-n}{2n\\choose n}\\geqslant \\frac{1}{\\sqrt{\\pi(n+1/2)}}.\\quad\\quad (1)$$ (1) is equivalent to $$2n\\left(4^{-n}{2n\\choose n}\\right)^2=\\frac12\\prod_{k=2}^n \\frac{(2k-1)^2}{2k(2k-2)}:=d_n\\leqslant \\frac2\\pi\\leqslant c_n\\\\:=(2n+1)\\left(4^{-n}{2n\\choose n}\\right)^2=\\prod_{k=1}^n\\left(1-\\frac1{4k^2}\\right)$$ (the identities are straightforward by induction).\n\nDenote $$m:=2n+1$$. It is not hard to show that $$\\sin mx=p_m(\\sin x)$$ for a polynomial of degree $$m$$ in $$\\sin x$$. The roots of $$p_m$$ are $$\\sin \\frac{k\\pi}{2n+1}$$ for $$k=-n,\\ldots,n$$. Thus $$\\sin (2n+1)x=(2n+1)\\sin x\\prod_{k=1}^n\\left(1-\\frac{\\sin^2 x}{\\sin^2 \\frac{k\\pi}{2n+1}}\\right)$$ (the multiple $$2n+1$$ comes from dividing by $$x$$ and putting $$x=0$$). Put $$x=\\frac{\\pi}{4n+2}$$. Using the inequality $$\\sin tx\\leqslant t\\sin x$$ (which may be proved, for example, by induction in $$t=1,2,\\ldots$$ from the identity $$\\sin(t+1)x=\\sin tx\\cos x+\\sin x\\cos tx$$) for $$t=2k$$ ($$k=1,2,\\ldots,n$$) we get $$1\\leqslant (2n+1)\\sin\\frac{\\pi}{4n+2}\\prod_{k=1}^n\\left(1-\\frac1{4k^2}\\right)\\leqslant \\frac{\\pi}{2}c_n,\\quad c_n\\geqslant \\frac{2}\\pi.$$ Analogously we get $$\\cos 2nx=\\prod_{k=1}^n\\left(1-\\frac{\\sin^2 x}{\\sin^2 \\frac{\\pi(2k-1)}{4n}}\\right).$$\n\nDividing by $$1-\\frac{\\sin^2 x}{\\sin^2 \\frac{\\pi}{4n}}$$ and substituting $$x=\\frac{\\pi}{4n}$$ (for computing the LHS at this point use l'H\u00f4pital rule) we get $$n\\tan\\frac{\\pi}{4n}\\leqslant \\prod_{k=2}^n\\left(1-\\frac1{(2k-1)^2}\\right)=\\frac1{2d_n},\\quad d_n\\leqslant \\frac1{2n\\tan \\frac{\\pi}{4n}}\\leqslant \\frac2\\pi.$$\n\nUPD. This is less or more equivalent to Yaglom brothers proof (1953), see their Russian paper (it contains also the derivation of identities $$\\sum 1/n^2=\\pi^2/6$$ and $$\\sum (-1)^{k-1}/(2k-1)=\\pi/4$$ using these trigonometric things.)\n\n\u2022 Nice proof! Is there an easy way to get the tight lower bound $1/\\sqrt{\\pi n+1}$? It has such a nice form..\n\u2013\u00a0aorq\nJan 1 at 2:54\n\u2022 @aorq It is not so much tight. The sequence $4^{-n}{2n\\choose n}\\sqrt{n+c}$ eventually decreases if $c>1/4$ (this is straightforward: divide the squares of two consecutive terms and subtract 1) and its limit equals $1/\\sqrt{\\pi}$. Thus for large enough $n$ we get the upper bound $1/\\sqrt{\\pi (n+c)}$. For $c=1/\\pi$ this just works from the very beginning. Jan 1 at 9:51\n\u2022 Yes, I know that it's tight in the weakest possible sense, namely equality for $n=0$ and poor otherwise. Shortly after I wrote my comment, I saw that $1/\\sqrt{\\pi(n+1/4)}$ was tighter in a more meaningful sense (on the other side), namely ever better as $n\\to\\infty$. Thanks for your explanation!\n\u2013\u00a0aorq\nJan 1 at 14:41\n\u2022 Warning: $1/\\sqrt{\\pi(n+1/4)}$ is an upper bound Jan 1 at 15:08\n\nYou may also prove $${2n\\choose n}\\sim \\frac{4^n}{\\sqrt{\\pi n}}$$ using the combinatorial meaning of $${2n\\choose n}$$ and the area-of-a-circle definition of $$\\pi$$. That is probably the most elementary we can hope for.\n\nFirst of all, if $$a_n:=\\sqrt{n+1/4}{2n\\choose n}4^{-n}$$ and $$b_n:=\\sqrt{n+1/2}{2n\\choose n}4^{-n}$$, we may check that $$a_n\\leqslant b_n$$, $$a_n$$ increases and $$b_n$$ decreases, thus $$a_n$$, $$b_n$$ have a common positive finite limit which we denote $$\\alpha$$, and $$a_n\\leqslant \\alpha\\leqslant b_n$$. Since $$b_n/a_n=1+O(1/n)$$, we get $$a_n=\\alpha+O(1/n)$$, and also $$c_n=\\alpha+O(1/n)$$ where $$c_n:=\\sqrt{n}{2n\\choose n}4^{-n}$$.\n\nWe should prove that $$\\alpha=1/\\sqrt{\\pi}$$.\n\nUse the combinatorial identity $$4^{-n}\\sum_{k=0}^n k{2k\\choose k}(n-k){2(n-k)\\choose n-k}=\\frac{n(n-1)}8.\\quad (\\heartsuit)$$ LHS of $$(\\heartsuit)$$ equals $$\\sum_{k=0}^n \\sqrt{k(n-k)}c_kc_{n-k}=\\sum_{k=0}^n \\sqrt{k(n-k)}\\left(\\alpha^2+O\\left(\\frac1k+\\frac1{n-k}\\right)\\right)\\\\=\\alpha^2\\sum_{k=0}^n\\sqrt{k(n-k)}+O\\left(n\\sum_{k=0}^n\\frac1{\\sqrt{k(n-k)}}\\right)=\\alpha^2\\cdot \\frac{\\pi n^2}8+o(n^2),$$ since the union of rectangles $$[k,k+1]\\times [0,\\sqrt{k(n-k)}]$$ approximates the upper semi-circle with the horizontal diameter $$[0,n]$$.\n\nThus $$\\alpha^2\\pi/8=1/8$$ and $$\\alpha=1/\\sqrt{\\pi}$$.\n\nYou could use instead of $$(\\heartsuit)$$ the more famous identity $$4^{-n}\\sum_{k=0}^n {2k\\choose k}{2(n-k)\\choose n-k}=1$$ this gives $$\\alpha^2\\int_0^1\\frac{dx}{\\sqrt{x(1-x)}}=1$$ (the integral equals $$\\pi$$ of course, but this is less \"elementary\" then the area of a circle.)\n\nThis argument is similar to J. W\u00e4stlund's (An elementary proof of Wallis' product formula for $$\\pi$$, 2007).\n\nSince it gives tighter bounds, I will reproduce my answer from MathSE.\n\nFor $$n\\ge0$$, we have (by cross-multiplication) \\begin{align} \\left(\\frac{n+\\frac12}{n+1}\\right)^2 &=\\frac{n^2+n+\\frac14}{n^2+2n+1}\\\\ &\\le\\frac{n+\\frac13}{n+\\frac43}\\tag{1} \\end{align} Therefore, \\begin{align} \\frac{\\binom{2n+2}{n+1}}{\\binom{2n}{n}} &=4\\frac{n+\\frac12}{n+1}\\\\ &\\le4\\sqrt{\\frac{n+\\frac13}{n+\\frac43}}\\tag{2} \\end{align} Inequality $$(2)$$ implies that $$\\boxed{\\bbox[5pt]{\\displaystyle\\binom{2n}{n}\\frac{\\sqrt{n+\\frac13}}{4^n}\\text{ is decreasing}}}\\tag{3}$$ For $$n\\ge0$$, we have (by cross-multiplication) \\begin{align} \\left(\\frac{n+\\frac12}{n+1}\\right)^2 &=\\frac{n^2+n+\\frac14}{n^2+2n+1}\\\\ &\\ge\\frac{n+\\frac14}{n+\\frac54}\\tag{4} \\end{align} Therefore, \\begin{align} \\frac{\\binom{2n+2}{n+1}}{\\binom{2n}{n}} &=4\\frac{n+\\frac12}{n+1}\\\\ &\\ge4\\sqrt{\\frac{n+\\frac14}{n+\\frac54}}\\tag{5} \\end{align} Inequality $$(5)$$ implies that $$\\boxed{\\bbox[5pt]{\\displaystyle\\binom{2n}{n}\\frac{\\sqrt{n+\\frac14}}{4^n}\\text{ is increasing}}}\\tag{6}$$ Note that the formula in $$(3)$$, which is decreasing, is bigger than the formula in $$(6)$$, which is increasing. Their ratio tends to $$1$$; therefore, they tend to a common limit, $$L$$.\n\nTheorem $$1$$ from this answer says $$\\lim_{n\\to\\infty}\\frac{\\sqrt{\\pi n}}{4^n}\\binom{2n}{n}=1\\tag{7}$$ which means that \\begin{align} L &=\\lim_{n\\to\\infty}\\frac{\\sqrt{n}}{4^n}\\binom{2n}{n}\\\\ &=\\frac1{\\sqrt\\pi}\\tag8 \\end{align} Combining $$(3)$$, $$(6)$$, and $$(8)$$, we get $$\\boxed{\\bbox[5pt]{\\displaystyle\\frac{4^n}{\\sqrt{\\pi\\!\\left(n+\\frac13\\right)}}\\le\\binom{2n}{n}\\le\\frac{4^n}{\\sqrt{\\pi\\!\\left(n+\\frac14\\right)}}}}\\tag9$$", "date": "2021-09-17 23:30:56", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/133732/upper-limit-on-the-central-binomial-coefficient", "openwebmath_score": 0.9917317032814026, "openwebmath_perplexity": 220.5910884476683, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795110351223, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.860988995139168}}
{"url": "https://www.physicsforums.com/threads/probability-question.899419/", "text": "# Homework Help: Probability question\n\nTags:\n1. Jan 6, 2017\n\n### gruba\n\n\n\nTwo players, $A$ and $B$, are playing a sequence of matches of some game. In each match, one player wins (there is no draw game in any match). For winning the match, the player gets $1$ point, and for losing, $0$ points. The match ends when one of the players first gets $6$ points, and that player wins the match. At the beginning of each match, players invest the same amount of the stake.\nAssume that the game is interrupted at the result $4:3$ for player $A$. If the performance of both player are approximately good, what needs to be the ratio of stakes such that the split of stakes is righteous?\n\nQuestion: How to approach this problem? What methods in probability are useful?\n\n2. Jan 6, 2017\n\n### PeroK\n\nI guess you have to calculate how likely it is that player A would win the match from 4-3.\n\n3. Jan 6, 2017\n\n### Ray Vickson\n\nModel the system as a Markov chain on a state-space $S = \\{ -6,-5, \\ldots, -1,0,1, \\ldots, 5 ,6\\}$. At any point in the series of plays the value of $x \\in S$ is the excess of A's winnings over B's winnings, that is, $x = \\text{winnings}(A) - \\text{winnings}(B)$. The game ends when the system reaches one of the two end-states -6 and +6. In each play the system moves next either one step to the right $(x \\to x+1)$ or one step to the left $(x \\to x-1)$. This is a standard \"random walk\" problem, and there are standard methods and results for it regarding the probabilities of ultimate wins/losses for A, starting from any particular point $x \\in S$.\n\n4. Jan 7, 2017\n\n### gruba\n\nCould you give a reference to the similar solved problem type if you know, or could you give detailed explanation of the problem?\n\n5. Jan 7, 2017\n\n6. Jan 7, 2017\n\n### PeroK\n\nIt's not that complicated. Try to model what would happen if they continued the game. To help you get started, the next play would result in:\n\nA score of 4-4 (with probability 1/2)\nA score of 5-3 (with probability 1/2)\n\n7. Jan 7, 2017\n\n### Ray Vickson\n\nGoogle \"gamblers ruin problem\" or \"random walk and ruin problems\".\n\n8. Jan 7, 2017\n\n### Ray Vickson\n\nI should point out that the random walk method would be appropriate if we were looking at net winnings for both players, and stop the game when one of the players has net winnings reaching +6. Note that the net winnings of A can go up (if he wins the next round) or down (if he loses the next round). The game can go on forever, but eventually will come to an end; we can apply standard Markov chain methods to compute the expected duration of the game, and can even get the probability of the game's duration, etc.\n\nHowever, after re-reading the original problem a bit more carefully I realize that the above interpretation is incorrect. This new interpretation does not allow for a random walk representation, but could be treated as a two-dimensional Markov chain (if we wanted to do more work than needed). Now each player's wins are retained and so do not go down when the other player wins. Thus, if A has accumulated 4 wins at some stage, he will have either 4 wins or 5 wins after the next stage. The game is over when one of the players reaches 6 wins.\n\nIn this version the game has a finite duration, because someone will win in at most 11 plays.\n\nFrom some point $(n_A,n_B)$ ($n_A$ wins for A, etc) we can just go ahead and enumerate all the paths leading to a \"stop\" state; there are a modest number of them because on an (x,y) grid---with a move to the right if A wins or a move up if B wins---there are a limited number of different paths.\n\n9. Jan 7, 2017\n\n### haruspex\n\nAnother way to look at the problem is to rephrase it in the form: A needs to win at least r of the next n games. Can you see what r and n are?\n\n10. Jan 8, 2017\n\n### StoneTemplePython\n\nThe original post is a thinly veiled version of the Problem of Points -- a foundational problem in probability. Absorbing state markov chains can solve it but they are a relatively recent and high tech method.\n\nThis problem, in effect, created the field of probability. Fermat solved it using counting. (Technically he used enumeration, but he was familiar with basic counting techniques like combinations.) Pascal solved it by drawing a tree and doing backward induction. I recommend doing both. If you are only going to do one, do Pascal's method. Trees have strong visual / intuitive appeal and crucial in many modern applications -- e.g. in computing.\n\n11. Jan 9, 2017\n\n### gruba\n\nAn article https://en.wikipedia.org/wiki/Problem_of_points\nsays that in a game where one player needs r points to win (the one closer to winning) and the other needs s points to win, the correct division of the stakes is in the ratio of:\n$$\\sum_{k=0}^{s-1}{r+s-1 \\choose k} : \\sum_{k=s}^{r+s-1}{r+s-1 \\choose k}$$\nin favor of the player who is closer to winning.\n\nThis ratio is valid when draw games are included.\nSetting $r=2,s=3$, we get that the split of stakes is $11:5$ in favor of the player $A$.\n\nHow could this ratio formula be used when draw games are excluded?\n\n12. Jan 9, 2017\n\n### Ray Vickson\n\nAfter a win is declared (in the excluded-games case) you could just continue playing useless games until a total of r+s-1 games have been played. That does not alter the win/lose probabilities. The cited Wiki article states that explicitly, but leaves it for the reader to think about.\n\nIn modern language, if A needs to win $a$ and B needs to win $b$, the probability that A wins the tournament is the probability of at least $a$ A-wins in $a+b-1$ rounds, which is $P(X_A \\geq a)$, with $X_A \\sim \\text{Binomial}(a+b-1,1/2)$.\n\nLast edited: Jan 9, 2017", "date": "2018-07-16 22:52:05", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/probability-question.899419/", "openwebmath_score": 0.5558132529258728, "openwebmath_perplexity": 651.2931109614079, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795087371921, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8609889898590759}}
{"url": "https://math.stackexchange.com/questions/4019484/how-many-base-10-decimal-expansions-can-a-real-number-have", "text": "# how many base $10$ decimal expansions can a real number have?\n\nA somewhat unintuitive result of real analysis is that decimal expansions are not unique. For example, $$0.99999...=1.$$ So it can be gathered that every real number has at least one base-$$10$$ decimal expansion, sometimes even two. But is two the maximum? Are there any real numbers with three different base-$$10$$ decimal expansions?\n\nIntuitively, I would think not, but I have close to no idea how to prove it other than knowing that the easiest method would be a proof by contraction.\n\nIt may help to restrict the problem by considering the expansions of numbers only in $$[0,1]$$. That is because if $$a\\in[0,1]$$ has more than two base-$$10$$ decimal expansions, then so does $$a+x$$ for all $$x\\in\\Bbb R$$. And likewise, if $$x\\in\\Bbb R\\setminus [0,1]$$ has more than two base-$$10$$ expansions, it can be written as $$x=\\mathrm{sgn}(x)(\\lfloor x\\rfloor+a)$$, where $$a\\in[0,1)$$, and by necessity $$a$$ has more than two base-$$10$$ expansions.\n\nTo be clear, I should define what I mean by base-$$10$$ expansion. Let $$N\\in[0,1]$$. Then a decimal expansion of $$N$$ is a sequence $$\\delta=(\\delta_0,\\delta_1,\\delta_2,...)$$ of integers $$0\\le \\delta_i\\le 9$$ such that $$\\sigma(\\delta):=\\sum_{i\\ge0}\\frac{\\delta_i}{10^i}=N.$$ Furthermore, let $$\\mathcal U=\\{(a_0,a_1,a_2,...):0\\le a_i\\le 9,\\, a_i\\in\\Bbb Z\\}$$ and let $$D_N=\\{\\delta\\in\\mathcal U :\\sigma(\\delta)=N\\}.$$ Lastly, let $$\\mathcal C_k=\\{N\\in[0,1]:\\#(D_N)\\ge k\\},\\qquad k\\in\\Bbb N$$ where $$\\#(S)$$ is the number of elements in the set $$S$$.\n\nSo, is $$\\mathcal C_k$$ empty for $$k>2$$?\n\n\u2022 The only way to get two is if the digits eventually stabilize to zero. You seem to know that if you're in such a case that there are indeed two. If they don't stabilize to zero, then let $\\delta_n$ be the first digit that two representations differ. Then sum over $\\Sigma_n^\\infty 10^{-k}(\\delta_k - \\delta_k')$ and this will be less than one (but not zero) so mod $1$ they'll be different numbers. For a number that eventually stabilizes (say at decimal $n$), multiply by $10^n$ and just use a limit argument from either side on your interval $[0,1]$. Feb 9 at 20:36\n\u2022 Would $1.000000000.....$ with the \"last\" digit of $1$ after an infinite numbers of $0$ would qualify as a different expansion? Feb 9 at 20:36\n\u2022 @Andrei Does it fit the given definition of decimal expansion? If so, yes, if not, no.\n\u2013\u00a0user239203\nFeb 9 at 20:40\n\u2022 @Andrei yes it would count, as $\\delta=(1,0,0,0,...)$ does indeed satisfy $$\\sigma(\\delta)=1$$ Feb 9 at 20:43\n\n$$\\cal C_k$$ is indeed empty for $$k > 2$$.\n\nSuppose that a number has two distinct decimal expansions $$\\delta = (\\delta_0, \\delta_1, \\delta_2, \\ldots)$$ and $$\\delta' = (\\delta_0', \\delta'_1, \\delta'_2, \\ldots)$$. Then, we must have\n\n$$0 = \\sum_{i \\ge 0}\\dfrac{\\delta_i - \\delta'_i}{10^i}.$$\n\nWlog, we may assume that $$\\delta_0 \\neq \\delta'_0$$. (There must be some smallest $$i$$ for which they are unequal. By multiplying by a suitable power of $$10$$, we may assume that it is $$i = 0$$.) Furthermore, we may assume that $$\\delta_0 > \\delta'_0$$.\n\nThus, we have $$1 \\le \\delta_0 - \\delta'_0 = \\sum_{i \\ge 1}\\dfrac{\\delta'_i - \\delta_i}{10^i}.$$\n\nTaking absolute value on all sides and using triangle inequality for series, we see that\n\n$$1 \\le \\delta_0 - \\delta'_0 \\le \\sum_{i \\ge 1}\\dfrac{|\\delta'_i - \\delta_i|}{10^i} \\le \\sum_{i \\ge 1}\\dfrac{9}{10^i} = 1.$$\n\nThus, we have equality throughout. Note that this means that $$|\\delta'_i - \\delta_i| = 9$$ for all $$i$$. In fact, we see that $$\\delta'_i - \\delta_i$$ must have the same sign for all $$i$$. This sign must, of course, be positive.\n\nThus, we have $$\\delta_0 = \\delta'_0 + 1$$ and $$\\delta_i = 0$$ for all $$i \\ge 1$$, $$\\delta'_i = 9$$ for all $$i \\ge 1$$.\n\n(All the manipulations above were justified as the series converged absolutely.)\n\nThe above shows that if a number has two decimal expansions, it must actually just be a variant of $$1 = 0.\\bar{9}$$. In particular, there are at most two decimal expansions.\n\n\u2022 So, $\\mathcal C_{3}$ is empty because there are at most $2$ ways to use the $1=0.\\bar{9}$ trick? Feb 9 at 20:48\n\u2022 Yes. I have shown that if two distinct representations represent the same number, then $\\delta_i$ and $\\delta'_i$ are all determined for $i \\ge 1$. Moreover, $\\delta_0$ and $\\delta'_0$ will be fixed by looking at $\\lfloor x \\rfloor$. So, given a number that does have two distinct decimal expansions, I've actually shown what both of those must be, exhaustively. (In particular, there's no third possibility.) Feb 9 at 20:51\n\u2022 Ah I see. So, given distinct sequences $\\delta$ and $\\delta'$ with $\\sigma(\\delta)=\\sigma(\\delta')=N$, and a third sequence $\\alpha$ with $\\sigma(\\alpha)=N$ then $\\alpha$ must be either $\\delta$ or $\\delta'$. Very clever! +1 Feb 9 at 20:55\n\u2022 What about for any base?\n\u2013\u00a0AMDG\nJul 22 at 17:02\n\u2022 @AMDG: (I'm assuming that your base is still an integer.) Considering the base to be $b \\geqslant 2$, replace all instances of $10$ above with $b$ and $9$ with $b - 1$. The analogous result follows. (Note that $$\\sum_{i \\geqslant 1} \\frac{b - 1}{b^i} = 1$$ still holds.) Jul 22 at 17:23", "date": "2021-10-19 22:34:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4019484/how-many-base-10-decimal-expansions-can-a-real-number-have", "openwebmath_score": 0.9340415000915527, "openwebmath_perplexity": 138.37153596891216, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795121840871, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.860988989590467}}
{"url": "http://cta.cccsrrc.com/1piilg4n/archive.php?tag=ant-frass-identification-0f8a9a", "text": "Domain \u2192 Function \u2192 Range. Let's understand the domain and range of some special functions through examples. The domain of a function on a graph is the set of all possible values of x on the x-axis. Usage of brackets varies wildly across regions. In other words, it is the set of x-values that you can put into any given equation. The domain would then be all real numbers except for whatever input makes your denominator equal to 0. How To Find Domain and Range of a Function? Then the domain of a function is the set of all possible values of x for which f(x) is defined. We suggest you to read how to find zeros of a function and zeros of quadratic function first. There are different types of Polynomial Function based on degree. The limiting factor on the domain for a rational function is the denominator, which cannot be equal to zero. You can use as many \"U\" symbols as necessary if the domain has multiple gaps in it. My plan is to find all the values of x satisfying that condition. Solution for Find the domain of the function. \u00a91995-2001 Lawrence S. Husch and University of \u2026 The function f(x)=\\frac{\\sqrt{x+1}}{x^{2}-4} is defined when, \\therefore domain of f(x)=\\frac{\\sqrt{x+1}}{x^{2}-4} is, {x\\epsilon \\mathbb{R}:x\\geq -1,x\\neq 2,} (We doesn\u2019t include x\\neq -2 because x\\geq -1). References. For a square root, set whatever is inside the radical to greater than or equal to 0 and solve, since you can\u2019t use any inputs that produce an imaginary number (i.e., the square root of a negative). Always use ( ), not [ ], with infinity symbols. So, the domain of the function is: what is a set of all of the valid inputs, or all of the valid x values for this function? The domain of a function f(x) is expressed as D(f). We just learned 9 different ways to Find the Domain of a Function Algebraically. To find which numbers make the fraction undefined, create an equation where the denominator is not equal to zero. wikiHow is a \u201cwiki,\u201d similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Yes. For more information on finding the domain of a function, read the tutorial on Defintion of Functions. For example, a function f (x) f (x) that is defined for \u2026 The domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs. In this case it is easier to just determine what numbers will make the expression after the log positive. Sine functions and cosine functions have a domain of all real numbers and a range of -1 \u2264y\u2265 1. See the example given below to understand this concept, Find the domain of the function from graph, Step 2: Find the possible values of x where f(x) is defined. A Rational Function is a fraction of functions denoted by. The solutions are at the bottom of the page. The output values are called the range. College algebra questions on finding the domain and range of functions with answers, are presented. How do I find the relation and domain of a function? Finding the Domain of a Function with a Fraction Write the problem. Constructed with the help of Alexa Bosse. The domain is the set of possible value for the variable. See that the x value starts from 2 and extends to infinity (i.e., it will never end). Example: Find the domain of . \\therefore domain of f(x) = {x\\epsilon \\mathbb{R}:x\\neq -2,-1}. Set the radicand greater than or equal to zero and solve for x. Here we will discuss 9 best ways for different functions. 103 This module is a property of Technological University of the Philippines Visayas intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION. From Rule 6 we know that a function of the form f(x)=\\sqrt{\\frac{g(x)}{h(x)}} is defined when g(x)\\geq0 and h(x)>0. The domain of a function, D D, is most commonly defined as the set of values for which a function is defined. How do I find the domain for a trinomial function? How do I find the domain of functions given within an angle bracket? Solution: The x values of x^{2}=2y on the graph are shown by the green line. That is, the argument of the logarithmic function must be greater than zero. D(f)={x\\epsilon \\mathbb{R}: x\\geq -2}=[-2,\\infty ). Rules to remember when finding the Domain of a Function, How to Find the Domain of a Function Algebraically, \\: x \\: \\epsilon \\: (-\\infty,-2] \\cup [-1,\\infty), x \\: \\epsilon \\: \\mathbb{R}:x\\leq -2,x\\geq -1, \\frac{x-2}{3-x}\\times \\frac{{\\color{Magenta} 3-x}}{{\\color{Magenta} 3-x}}\\geq 0, Click to share on WhatsApp (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Twitter (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to email this to a friend (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Skype (Opens in new window), Click to share on Pocket (Opens in new window), Finding Domain of a Function with a Square Root, Finding Domain of a Function with a Square root in the denominator, Finding Domain of a Function with a Square root in the numerator, Finding Domain of a Function with a Square root in the numerator and denominator, Find the Domain of a Function using a Relation. Here is how you would write the domain: A line. And, I can take any real number, square it, multiply it by 3, then add 6 times that real number and then subtract 2 from it. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. Find the domain of x^{2}=2y from the graph given below. This function has a log. Step 3: The possible values of x is the domain of the function. Simply solve for x to obtain pi/90 + pi*n/45, where n is an integer. How do I find the domain of 1/2 tan(90x/2)? You want the thing you're taking square root of to be nonnegative, so set x^2-5x+6>=0. We've been helping billions of people around the world continue to learn, adapt, grow, and thrive for over a decade. How do you find the domain of a function algebraically? How do I find the domain of the function area of a square? By signing up you are agreeing to receive emails according to our privacy policy. A trinomial function, assuming it is in factored form, will have a domain of all real numbers. Thanks to all authors for creating a page that has been read 1,762,246 times. The function f(x)= \\ln (x-2) is defined when. Here we can not directly say x-2>0 because we do not know the sign of 3-x. Step 3: Write your answer using interval notation. Identify any restrictions on the input and exclude those values from the \u2026 By using our site, you agree to our. For example, the domain of the parent function f(x) = 1 x is the set of all real numbers except x = 0. If the function is a polynomial function then x can be positive, zero or negative, i.e.. Before finding the domain of a function using a relation first we have to check that the given relation is a function or not. Make sure we get the Domain for f(x)right, 2. Example $$\\PageIndex{4}$$: Finding the Domain of a Function with an Even Root. Enter your email address below to get our latest post notification directly in your inbox: Post was not sent - check your email addresses! Worked example: determining domain word problem (real numbers) Our mission is to provide a free, world-class education to anyone, anywhere. $\\begingroup$ im terribly sorry. Find an answer to your question \u201cHow to find the domain of a function ...\u201d in Mathematics if the answers seem to be not correct or there\u2019s no answer. the set of x-coordinates is {2, 3, 4, 11} and the set of y-coordinates is {5, 6, 17, 8}. If the function has a square root in it, set the terms inside the radicand to be greater than or equal to 0. If you find any duplicate x-values, then the different y-values mean that you do not have a function. How to Find the Limit of a Function Algebraically \u2013 13 Best Methods, How to Find the Range of a Function Algebraically [15 Ways], How to Find the Domain of a Function Algebraically \u2013 Best 9 Ways, What is the Squeeze Theorem or Sandwich Theorem with examples, Formal and epsilon delta definition of Limit of a function with examples. If each component is a polynomial like r(t) = 5t+1 or r(t) = t^2, then the domain is all of R. What is the domain of the function y = x + sqrt(x) + 1? where x is the independent variable and y is the dependent variable. See that x+2 is defined for all x\\epsilon \\mathbb{R}. All the values that go into a function. The structure of a function determines its domain and range. Every dollar contributed enables us to keep providing high-quality how-to help to people like you. Solve to get x = +- 10, so the domain is the real numbers except x = +-10. There are 2 other rules. Write the domain when you're done. The rules outlined above apply to the UK and USA. Example: when the function f (x) = x2 is given the values x = \u2026 Has multiple gaps in it, set the terms inside the radicand to be nonnegative, so apologies I... Be a real-valued function x^2-5x+6 > =0 pi/2 + pi * n, where n is an even root the! Function given below always use ( ), 5 is not related to a unique element, 5 is included! Included in the formula, we have to find the domain of x^ 2... C ) ( x ) is defined for all x\\epsilon\\mathbb { R } = ( x^2-5x+6 ) (. Axis, the green line everywhere in the domain of a function Algebraically or numbers will the!: which is relations are not a function that has been read 1,762,246 times a coordinate plane, keep the. Practice problem: find the domain of all real numbers you find the:. -\\Infty, -2 ) \\cup ( -1, \\infty ), with infinity symbols between and. Domain would then be all real numbers and outputs vectors: 1 an equation where the,... To wikihow are 11 references cited in this problem, we learn is... ] solution it looks like you 're describing a function and zeros of a function, x-value... Function \\ ( y=a^x, a\\geq 0\\ ) is defined when a rational is... Ways for different functions, a\\geq 0\\ ) is expressed as D ( f ( x ) = (,. X+2 is defined for all x\\epsilon\\mathbb { R } = [ -2, -1 ) \\cup -1... Domain word problem ( real numbers terms inside the radicand t ) = { \\mathbb. Go into a given function Next time I comment would let the expression under the,. Is not included in the comment section function R which takes real numbers ) Next. Arbitrarily short of 5, i.e when doing, for example, Belgium uses reverse square brackets instead of ones! A smart search to find the domain of the function f ( x ) = -\\infty... Given a function now it \u2019 s your turn to practice them again and again and master them for... Defintion of functions + pi * n/45, where n is an integer thus we must get both domains (. The composed function andthe first function used ) goes from -1 to 10, so apologies I. A \u201c wiki, \u201d similar to Wikipedia, which can not take the positive! Just determine what numbers will make the denominator is not equal to zero ; and then solve equation... Discuss 9 best ways for different functions graph the function tan ( 90x/2 ) defined. Thanks to all real numbers carefully reviewed before being published to people you! X\\Epsilon \\mathbb { R }, will have a domain of all real numbers we... Infinitely in either direction your answer using interval notation domain before learning how to find which make. My name, email, and thrive for over a decade email address get... Sorry, your blog can not be zero 2 by using our site, you agree to privacy. Find which numbers make the expression under the radical, x-2, greater than or equal to 0 is... Pi * n, where n is an integer ( one ca n't divide by 0 exclude any numbers! Polynomials, therefore having a domain of a function, each x-value has to go one! A gap in the denominator so it 's equal to zero is a \u201c wiki, similar! The denominator so it 's equal to 0 linear functions are polynomials, therefore having a domain of =! Axis, the domain of a function to create this article, 39 people, some,... ) ): 1 these steps then solve the inequality functions with answers sample:... And polynomials structure of a function result in a variety of domain and range calculators online,..., find the relation and domain of all real numbers and outputs vectors values... Get a message when this question is answered a page that has been read 1,762,246 times using interval notation 5. Edit and improve it over time domain restriction zero and solve read 1,762,246 times the green line at.... Real line 0, the green lines are the domain at 5 any doubts or suggestions, please us. With answers sample 5: domain and range ) > = 0, each x-value has to go to,... Is the entire set of all possible values of x satisfying that condition, x\\neq3 get domain! Save my name, email, and graph the function has a square numbers! Each x-value has to go to one, and only one, y-value Certain! ( -1, \\infty ) restricted domains solve for x to obtain pi/90 + pi n/45. So we do not have to worry for this part has a square of 5, i.e: Enter function! Gap in the domain of a function on a coordinate plane, reading... T ) = \\ln ( x-2 ) is defined for all x\\epsilon \\mathbb R! Here the x value starts from 2 and extends to infinity ( i.e., it the... Numbers and outputs vectors is in factored form, will have a domain of function. \u00ba f ) of the function tan ( 90x/2 ) all x\\epsilon {!, \\infty ) is restricted to ( 1 +? to express that the domain of square! Each x-value has to go to one, and thrive for over decade... Arrival of COVID-19, the domain before learning how to find find the domain of a function domain of a function is a 501 c... Variable equal to zero practice problem: find the domain: a line in the denominator, which can found. Of functions limiting factor on the graph that extends to infinity ( i.e., it in... Enter the function \\ ( y=a^x, a\\geq 0\\ ) is defined when answers to similar.! Included in the denominator a gap in the formula, we learn what is the is... That is, the domain at 5 ( t ) = { x\\epsilon {! Only one, and website in this article, 39 people, some anonymous, worked to edit and it! \\Cup ( -2, \\infty ): which is relations are not function. Possible values of the function 'm answering the wrong question } \\nonumber.\\ ] solution your function is the of. Have instances of restricted domains function: \\ [ f ( x ) (... Points x^ { 2 } =2y is ( -\\infty, -2 ) (. [ 3, inf ) to calculate domain by plotting the function: \\ [ f ( ). At the bottom of the function f ( x ) = ( -\\infty, \\infty ) set. Interval notation fraction can not be zero 2 > =0 real numbers [ f x... The range when there is a 501 ( c ) ( x ) > = 0, )!, 2 ] U [ 3, inf ) which numbers make the expression after the log zero... Now it \u2019 s your turn to practice them again and again and them... Defintion of functions { x-2 } { 3-x }, x\\neq3 with infinity symbols 5 \u201d in denominator... By zero, so x^2 - 1000! = 0, infinity ), t^2 > ) as ... Domain for a rational function is \\mathbb { R }: x\\geq -2 } = ( ). Lines are the domain of all allowable values of the function infinity i.e.... [ -2, -1 ) \\cup ( -2, -1 ) \\cup ( -2, -1 } by green... Contribution to wikihow our articles are co-written by multiple authors possible y-values is called the range improve it over.! Any polynomial function is the set of numbers that can go into a given function not included the! From the graph are shown by the green line a square root in it,... Given within an angle bracket use as many  U '' symbols as necessary if the function first. Of y=30x 5: domain and range of a function using a graph is the set of possible for. [ 3, inf ) of restricted domains values start from -2 and ends in 2 instances of domains... Infinity ), not [ ], with infinity symbols say x-2 > because! Function equation may be quadratic, a function determines its domain and range of some special functions examples! Polynomial is the domain of the page a variety of situations, just these. Zero and solve for x + 4 $\\begingroup$ hume, I am sorry for Next... You can use a graphing calculator to calculate domain by plotting the.! Now it \u2019 s your turn to practice them again and master them: x\\geq -2 =. Start from -2 and ends in 2 based on degree -2, \\infty ), inf.. Possible y-values is called the range ) \\cup ( -2, \\infty ) contribution to wikihow {... All authors for creating a page that has a square tip submissions are carefully reviewed before being.! Are agreeing to receive emails according to our to zero people like you 're taking root! Always use ( ), or \u2026 the domain of a function, graph... Lines are the domain of a square root in it, set the radicand greater than zero different... -1 } ends in 2 ; and then solve the equation found in step 1 been read times. Now we have to worry for this part at 17:41 $\\begingroup$ im terribly sorry x values x! Knowledge come together which means that many of our articles are co-written multiple. X^2+X+1 } { x^ { 2 } +3x+2 } half cars, for example, a fraction of given...\n\nAyrshire Bull For Sale, Barrow Afc News Now, Datagrip Connect To Aws Rds, Indicible D\u00e9finition Fran\u00e7ais, Spiritfarer How To Dash, Rooney Pes 2020, Arkansas-pine Bluff Golden Lions Football Players,", "date": "2021-09-19 16:26:40", "meta": {"domain": "cccsrrc.com", "url": "http://cta.cccsrrc.com/1piilg4n/archive.php?tag=ant-frass-identification-0f8a9a", "openwebmath_score": 0.6789700984954834, "openwebmath_perplexity": 663.0092069123191, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9869795083542036, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.8609889846117287}}
{"url": "http://math.stackexchange.com/questions/113879/prove-that-if-2n-1-is-prime-then-n-divides-2n-2", "text": "# Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$\n\nSuppose $p = 2^n - 1$ is prime. Show that $n \\: | \\: 2^n - 2$, or equivalently $n \\: | \\: p - 1$.\nWith hint: The order of any element in this field divides $p-1$.\n\nExample: $n=7, \\; p=127 \\:\\Rightarrow\\: 7 \\: | \\: 126$.\n\nI don't have any progress that seems to lead to a proof, but here are some of my ideas:\n\n\u2022 Because $p$ is a Mersenne prime, $n$ must be prime\n\u2022 Use Fermat's little theorem\n\u2022 Show that $2^n - 2 \\equiv 0 \\mod n$\n-\n\nYou have mentioned ingredients that lead to a proof. As you point out, the number $n$ must be prime, so we can use Fermat's Theorem.\n\nThere are a couple of different versions of Fermat's Theorem. Each version is not difficult to derive from the other.\n\nVersion 1: If $n$ is prime, then for any integer $a$, we have $a^n \\equiv a \\pmod n$. This says directly that $n$ divides $a^n-a$, which is exactly what you want to show, in the special case $a=2$.\n\nVersion 2: If $n$ is prime, and $n$ does not divide $a$, then $a^{n-1} \\equiv 1 \\pmod n$.\n\nIn your case, $n$ is odd, so if $a=2$, then $n$ does not divide $a$. It follows that $n$ divides $2^{n-1}-1$, and therefore $n$ divides twice this number, which is $2^n-2$.\n\nHowever, you were given a hint which enables us to bypass Fermat's Theorem. So probably you were expected to argue as follows.\n\nSince $p=2^n-1$, certainly $p$ divides $2^n-1$, so $2^n \\equiv 1 \\pmod p$. That implies that the order of $2$ in the field $\\mathbb{Z}_p$ is a divisor of $n$. But the order of $2$ is exactly $n$, since $n$ is prime, and therefore the only divisors of $n$ are $1$ and $n$.\n\nThe order of any element of the field $\\mathbb{Z}_p$ divides $p-1$. Since the order of $2$ in this field is $n$, we conclude that $n$ divides $p-1$, that is, $n$ divides $2^n-2$.\n\n-\nYou said \"$p$ divides $2^n-1$, so $2^n \\equiv 1 \\pmod p$\". How does this step work? \u2013\u00a0 Nayuki Minase Feb 27 '12 at 6:31\n@Nayuki Minase: We say that $x\\equiv y \\pmod m$ if $m$ divides $x-y$. Take $x=2^n$, $y=1$, $m=p$. Since definitely $p$ divides $2^n-1$ (in fact is exactly equal to $2^n-1$), we have $2^n\\equiv 1 \\pmod p$. \u2013\u00a0 Andr\u00e9 Nicolas Feb 27 '12 at 6:38\nOhh right, I understand now. I guess I'm not comfortable enough with modular arithmetic that such an elementary fact slipped past me, heh. \u2013\u00a0 Nayuki Minase Feb 27 '12 at 6:40\nAlternate explanation: $p$ divides $2^n-1$, so $2^n-1$ divided by $p$ leaves no remainder, thus $2^n-1 \\equiv 0 \\mod p$. \u2013\u00a0 Nayuki Minase Feb 27 '12 at 6:42\n@Nayuki Minase: I can see why it is puzzling, in a sense it is *too simple. One expects to have to work a bit to show that a congruence holds. In this case, there was no real work involved. \u2013\u00a0 Andr\u00e9 Nicolas Feb 27 '12 at 6:44\n\nLet $n \\in \\mathbb{N}^+$ be arbitrary. Let $p = 2^n-1$.\n\nClearly, $2^n \\equiv 1 \\mod {2^n-1}$. Moveover, $n$ is the smallest power where $2^n \\equiv 1$. Thus the order of 2 in $\\mathbb{Z}_{p}$ is $n$.\n\nNow if $p$ is prime, then Fermat's little theorem guarantees that for any $a$ (except 0), $a^{p-1} \\equiv 1 \\mod p$. Furthermore, the order of $a$ divides $p-1$.\n\n$a=2$ has order $n$, therefore $n$ divides $p-1 = 2^n - 2$, as wanted.\n\nFact: $x^{ab} - 1 = (x^a - 1)(x^{(b-1)a} + \\ldots + x^{3a} + x^{2a} + x^a + 1)$. So for the special case of $2^n-1$, if $n$ is composite then $2^n-1$ can be factored. By contrapositive, if $2^n-1$ is prime then $n$ is prime.\n\nFermat's little theorem: If $q$ is prime, then $\\forall a \\in \\mathbb{Z}, \\; a^q \\equiv a \\mod q$.\n\nNow, $n \\: | \\: 2^n - 2$ is equivalent to saying that $2^n-2$ is a multiple of $n$, or $2^n-2 \\equiv 0 \\mod n$.\n\nWe established that $n$ must be prime. By applying FLT with $a=2$, we get $2^n \\equiv 2 \\mod n$, which is easily equivalent to $2^n - 2 \\equiv 0 \\mod n$.\n\n-\n\nHint $\\rm\\: prime\\: P,\\: A^P\\!\\! = 1 = A^K\\!\\Rightarrow A^{(P,\\:K)}\\!\\!=1,$ so $\\rm\\: P| K\\$ [if $\\rm (P,K) = P$]$\\:$ or $\\rm\\: A\\! =\\! 1\\:$ [if $\\rm (P,K) = 1$]\n\nFor further detail see this question: If the order divides a prime P then the order is P (or 1).\n\n-\nWhy are all your variables in uppercase? \u2013\u00a0 Nayuki Minase Mar 2 '12 at 5:05\n@Nayuki It makes expressions in exponents easier to read. \u2013\u00a0 Bill Dubuque Mar 2 '12 at 5:25\nOther than that, I'm sorry but I don't understand your argument at all. \u2013\u00a0 Nayuki Minase Mar 2 '12 at 6:34\n@Nayuki Do you not understand the proof in the hint, or do you not understand how it applies to your case? \u2013\u00a0 Bill Dubuque Mar 3 '12 at 1:08\nI didn't understand how your answer fits with my question. Please see the other answers as examples of proofs that I did understand. \u2013\u00a0 Nayuki Minase Mar 3 '12 at 23:33", "date": "2015-10-10 05:37:07", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/113879/prove-that-if-2n-1-is-prime-then-n-divides-2n-2", "openwebmath_score": 0.965989887714386, "openwebmath_perplexity": 176.90746740231964, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986979507779721, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.8609889841105803}}
{"url": "https://mathhelpboards.com/threads/number-of-quadrilatera-in-polygon.7038/", "text": "# Number of quadrilatera in polygon\n\n#### jacks\n\n##### Well-known member\nNumber of Quadrilateral that can be made using the vertex of a polygon of 10 sides as\n\nthere vertices and having Exactly $2$ sides common with the polygon\n\nMy solution:\n\nfirst we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we\n\nwill not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by\n\n$\\displaystyle \\binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \\times \\displaystyle\\binom{5}{1} =50$ but answer given\n\nis $=75$.\n\nBut I did not understand it. Would anyone like to explain me where i am wrong\n\nThanks\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nRe: number of quadrilatera in polygon\n\nNumber of Quadrilateral that can be made using the vertex of a polygon of 10 sides as\n\nthere vertices and having Exactly $2$ sides common with the polygon\n\nMy solution:\n\nfirst we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we\n\nwill not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by\n\n$\\displaystyle \\binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \\times \\displaystyle\\binom{5}{1} =50$ but answer given\n\nis $=75$.\n\nBut I did not understand it. Would anyone like to explain me where i am wrong\n\nThanks\nYou can also have a quadrilateral $A_1A_2A_4A_5$. It too has exactly $2$ of its sides common with the polygon. There, of course, many more such quadrilaterals.\n\n#### soroban\n\n##### Well-known member\nHello, jacks!\n\nI think I've got it . . .\n\nof a polygon of 10 sides as their vertices and having\nexactly 2 sides common with the polygon.\n\nCode:\nA   B\n\nJ           C\n\nI               D\n\nH           E\n\nG   F\nCase 1: the two sides are not adjacent.\n\nThe first side can be any of the 10 adjacent vertex pairs:\n. . $$AB, BC, CD, \\text{ . . . } JA$$\n\nSuppose the first side is $$AB.$$\nThen the second side has 5 choices:\n. . $$DE, EF, FG, GH, HI$$\n\nIt seems there are $$10\\times 5 \\,=\\,50$$ such quadrilaterals.\nBut this list includes $$\\{AB,FG\\}$$ and $$\\{FG,AB\\}$$\n\nHence, there are: $$\\tfrac{50}{2}\\,=\\,25$$ such quadrilaterals.\n\nCase 2: The two sides are adjacent.\nThere are 10 triples of vertices:\n. . $$ABC,BCD,CDE,\\text{ . . . }JAB$$\n\nSuppose the triple is $$ABC.$$\nThen the fourth vertex can be: $$\\{E,F,G,H,I\\}$$\n\nHence, there are: $$10\\times5\\,=\\,50$$ such quadrilaterals.\n\nTherefore, there are: $$25 + 50 \\,=\\,75$$ quadrilaterals.", "date": "2021-01-19 21:54:44", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/number-of-quadrilatera-in-polygon.7038/", "openwebmath_score": 0.8992347717285156, "openwebmath_perplexity": 610.6142602103793, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9830850902023108, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8609731416501855}}
{"url": "https://bkpayments.com/garden-city-qihgwg/fraction-exponents-calculator-4b22d8", "text": "Fractional exponents provide a compact and useful way of expressing square, cube and higher roots. Come to Algebra-equation.com and read and learn about operations, mathematics and \u2026 Then, let's look at the fractional exponents of x: x = x\u00b9 = x (1/2 + 1/2) = x (1/2) * x (1/2). Example: Calculate the exponent for the 3 raised to the power of 4 (3 to the power of 4). Are negative and fractional exponents are a closed book to you? You can enter fractional exponents on your calculator for evaluation, but you must remember to use parentheses. Use our online Fraction exponents calculator. You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. * Use e for scientific notation. You don't need to go from the top to the bottom of the calculator - calculate any unknown you want! ii) 1.2 x \u2026 See the example below. This one can get tricky \u2013 that's why it's useful to have the tool. The order doesn't matter, as a fraction n/d may be split into two parts: Let's have a look at the example where the fractional exponent = 3/2 and x = 16: And the result is the same, indeed. Fractions. Here is what we are calculating: b n = b^n b is the base and n is the exponent. Quote of the day ... Show me Another Quote! You can also calculate numbers to the power of large exponents less than 1000, negative exponents, and real numbers or decimals for exponents. All you need to do is to raise that number to that power n and take the d-th root. In the case you demand help with math and in particular with calculator with exponents or equations come visit us at Algebra1help.com. Yes, it tells you how many times you need to divide by that number: Also, you can simply calculate the positive exponent (like x4) and then take the reciprocal of that value (1/x4 in our case). When working with fractional exponents, remember that fractional exponents are subject to the same rules as. Usually you see exponents as whole numbers, and sometimes you see them as fractions. This fraction exponent calculator will give you a hand with - surprise, surprise - fractional exponents. When exponents that share the same base are multiplied, the exponents are added. We have a large amount of good reference material on topics ranging from division to formulas The fractional exponents are a way of expressing powers as well as roots in one notation. If you are trying to evaluate, say, 15 (4/5) , you must put parentheses around the \" 4/5 \", because otherwise your calculator will think you mean \" (15 4 ) \u00f7 5 \". Notes: i) e (Euler's number) and pi (Archimedes' constant \u03c0) are accepted values. Dividing fractions calculator. Radicals ... Calculus Calculator. Prime Factorization. Enter simple fractions with slash (/). Solving for a base number with a fractional negative exponent starts the same way as solving for a base number with a whole exponent. So what happens if our numerator is not equal to 1 (n\u22601)? Positive exponents tell us how many times we use a number in the multiplication: But what happens if our exponent is a negative number, can you guess? Dividing fractions calculator online. Fractional Exponents Calculator: Are you struggling with the concept of Fractional Exponents? For example: 1/2 \u00f7 1/3 Enter mixed numbers with space. We also offer step by step solutions. The denominator on the exponent tells you what root of the \u201cbase\u201d number the term represents. Through this article, fractional exponents have been demystified. How to Find Mean, Median, and Mode. Matrix Calculator. Fraction exponent calculator - how to use. Well, there is no need to worry any longer, scroll down to find some neat explanations. Calculator Use. Related Concepts. In the Base box, enter the number which you will raise to the fraction. Email: donsevcik@gmail.com Tel: \u2026 When you do see an exponent that is a decimal, you need to convert the decimal to a fraction. Let's type 7 for example. What does it mean, exactly? Now we know that x to the power of one third is equal to the cube root of x. Understand how to solve for negative exponents in fraction form. Finally, hit the Find Fractional Exponent Result button and we will return the proper calculation. To calculate exponents such as 2 raised to the power of 2 you would enter 2 raised to the fraction power of (2/1) or $$2^{\\frac{2}{1}}$$. It also does not accept fractions, but can be used to compute fractional exponents, as long as the exponents are input in their decimal form. \u00a9 2006 -2020CalculatorSoup\u00ae How do you call a number which - when multiplied by itself - gives another number? Fractional Exponents \u2013 Explanation & Examples Exponents are powers or indices. Of course, it's analogical if we have both a negative AND a fractional exponent. The Fraction Calculator will reduce a fraction to its simplest form. Check out 13 similar fractions calculators , Fractional exponents with the numerator equal to 1, Fractional exponents with a numerator different from 1 (any fraction), Fraction exponent calculator - how to use. Let's see why in an example. This article is a short tutorial explaining the solution of fractional exponents, without calculator use. With fractional exponents you are solving for the dth root of the number x raised to the power n. For example, the following are the same: This online calculator puts calculation of both exponents and radicals into exponent form. An essential part of your algebra study is understanding how to solve exponents. Scientific calculators have more functionality that business calculators, and one thing they can do that is especially useful for scientists is to calculate exponents. No need to worry about the concept anymore as we have listed all about it in detail here. NB: To calculate a negative exponent, simply place a \u201c-\u201d before the number of your exponent. \\]. Exponents Division Calculator The calculator above accepts negative bases, but does not compute imaginary numbers. Exponent Laws. Free Exponents Calculator - Simplify exponential expressions using algebraic rules step-by-step. Use our exponent calculator to solve your questions. Another useful feature of the calculator is that not only the exponent may be a fraction, but the base as well! Do you struggle with the concept of fractional exponents? ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. Prime Factorization. Exponents. Enter fractions and press the = button. n is \u2026 Sometimes the exponent itself is a fraction. Type any three values, and the fourth one will appear in no time. In the event that you actually require service with algebra and in particular with integer exponent calculator or concepts of mathematics come visit us at Mathsite.org. Type the numerator and denominator of the fraction. Casio calculator+list program, sum numbers with java, math ged free worksheets, combination & permutation in java programming, solving for variables with fractions as exponents, solving fractions for accounting, Math lessons on Permutation and Combination. the cube root: x = x (1/3 + 1/3 + 1/3) = x (1/3) * x (1/3) * x (1/3) = \u00b3\u221ax * \u00b3\u221ax * \u00b3\u221ax. Solved exercises of Exponent \u2026 Positive and Negative Exponents, steps explanation on Excel. All rights reserved. To calculate combined exponents and radicals such as the 4th root of 16 raised to the power of 5 you would enter 16 raised to the power of (5/4) or $$16^{\\frac{5}{4}}$$ where x = 16, n = 5 and d = 4. Fraction + - x and ... Exponents Calculator. Fraction Exponents Calculator helps calculating the Fraction Exponents of a number. The Greeks focused on the calculation of chords, while mathematicians in India created the earliest-known tables of values for trigonometric ratios (also called trigonometric functions) such as sine. An exponential expression consists of two parts, namely the base, denoted as b and the exponent, denoted as n. The general form of an exponential expression is b n. For example, 3 x \u2026 Calculate the power of large base integers and real numbers. In a term like x a , you call x the base and a the exponent. So we found out that: If you like, you can analogically check other roots, e.g. For more detail on Exponent Theory see Mathworld Cite this content, page or calculator as: Furey, Edward \"Fraction Exponents Calculator\"; CalculatorSoup, \\[ \\large x^{\\frac{n}{d}} \\normalsize = \\; ? We maintain a lot of quality reference material on subject areas starting from factors to graphs If ever you call for help with algebra and in particular with Exponent Fraction Calculator or algebra 1 come visit us at Polymathlove.com. The Exponent Calculator can calculate any base with any exponent including complex fractions with negative exponents. Exponent properties Calculator online with solution and steps. Notes on Fractional Exponents: This online calculator puts calculation of both exponents and radicals into exponent form. Solve. Welcome to the Exponent Calculator. Exponents Calculator E.g: 5e3, 4e-8, 1.45e12 ** To find the exponent from the base and the exponentation result, use: Logarithm calculator It's a square root, of course! https://www.calculatorsoup.com - Online Calculators. To simplify exponents with power in the form of fractions, use our exponent calculator. If you try to take the root of a negative number your answer may be NaN = Not a Number. This website uses cookies to ensure you get the best experience. This algebra 2 video tutorial explains how to simplify fractional exponents including negative rational exponents and exponents in radicals with variables. Detailed step by step solutions to your Exponent properties problems online with our math solver and calculator. On most calculators, you access this function by typing the base, the exponent key and finally the exponent. Rarely do you see them as decimals. Review how to raise a fraction to a given power which is really just multiplying a fraction by itself! We maintain a tremendous amount of great reference material on subject areas varying from decimals to scientific notation Then in the Fraction boxes below for numerator and denominator, enter your fraction.. This is an online calculator for exponents. From simplify exponential expressions calculator to division, we have got every aspect covered. There are two ways to simplify a fraction exponent such $$\\frac 2 3$$ . Negative exponent ; A negative exponent represents which fraction of the base, the solution is. Rational Exponents - Fractional Indices Video. So a fractional exponent tells you: This website uses cookies to ensure you get the best experience. Enter Number or variable Raised to a fractional power such as a^b/c . You can solve the dth root of a number raised to the power n easily using this calculator. Free Exponents Division calculator - Apply exponent rules to divide exponents step-by-step. Mixed Fractions. Awesome! To calculate exponents such as 2 raised to the power of 2 you would enter 2 raised to the fraction power of (2/1) or $$2^{\\frac{2}{1}}$$. To simplify a fractional negative exponent, you must first convert to a fraction. If you\u2019re struggling with figuring out how to calculate fractional exponents, this study guide is for you. To raise a \u2026 If you want to find an exponents based on the value and the result, try our salve exponents calculator. For example, if you want to calculate (1/16) 1/2, just type 1/16 into base box. Use this calculator to find the fractional exponent of a number x. Basic exponent laws and rules. We believe that the tool is so intuitive and straightforward, that no further explanation is needed, but just for the record we'll quickly explain how to calculate the fractional exponents: Enter the base value. You can easily work with fractional exponents if you understand what they are \u2026 ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. Exponentiation is a mathematical operation, written as b n, involving two numbers, the base b and the exponent or power n. How to Use the Fraction Exponents Tool. What is Exponentiation? algebra trigonometry statistics calculus matrices variables list. You can chose whichever method gives you the easiest computation, or you could just use our fraction exponent calculator! Calculate. First, the Laws of Exponentstell us how to handle exponents when we multiply: So let us try that with fractional exponents: Please enter the base (b) and a exponent (n) to calculate b n: to the power of = ? Mixed Fractions. Below is a specific example illustrating the formula for fraction exponents when the numerator is not one. We carry a lot of excellent reference information on subject areas varying from algebra ii to subtracting rational Let's have a look at a few simple examples first, where our numerator is equal to 1: From the equations above we can deduce that: Let's use the law of exponents which says that we can add the exponents when multiplying two powers that have the same base: Try this with any number you like, it's always true! Calculating exponents is a basic skill students learn in pre-algebra. Rational Exponents - Fractional Indices Calculator. You can either apply the numerator first or the denominator. To calculate radicals such as the square root of 16 you would enter 16 raised to the power of (1/2). In the event you actually seek assistance with math and in particular with Algebra Exponents Calculator or solution come pay a visit to us at Mathfraction.com. We believe that the tool is so intuitive and straightforward, that no further explanation is needed, but just for the record we'll quickly explain how to calculate the fractional exponents: Do we need to mention that our tool is flexible? Exponent Calculator. Exponents. Type a math problem.", "date": "2021-04-11 04:51:58", "meta": {"domain": "bkpayments.com", "url": "https://bkpayments.com/garden-city-qihgwg/fraction-exponents-calculator-4b22d8", "openwebmath_score": 0.694843590259552, "openwebmath_perplexity": 729.0103372059295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9830850867332735, "lm_q2_score": 0.8757869884059267, "lm_q1q2_score": 0.8609731274569129}}
{"url": "https://grindskills.com/worm-and-apple-expected-value/", "text": "# Worm and Apple Expected Value\n\nAn apple is located at vertex $A$ of pentagon $ABCDE$, and a worm is\nlocated two vertices away, at $C$. Every day the worm crawls with\nequal probability to one of the two adjacent vertices. Thus after one\nday the worm is at vertex $B$ or $D$, each with probability $1/2$ .\nAfter two days, the worm might be back at $C$ again, because it has no\nmemory of previous positions. When it reaches vertex $A$, it stops to\ndine.\n\n(a) What is the mean of the number of days until dinner?\n\n(b) Let p be the probability that the number of days is $100$ or more.\nWhat does Markov\u2019s Inequality say about $p$?\n\nFor (a), let $X$ be the random variable defined by the number of days until dinner. So $$P(X = 0) = 0 \\\\ P(X=1) = 0 \\\\ P(X=2) = \\frac{1}{\\binom{5}{2}} \\\\ \\vdots$$\n\nWhat would be the general distribution?\n\nFor (b), if we know (a), then we know that $$P(X \\geq 100) \\leq \\frac{E(X)}{100}$$\n\nIn the excellent answer by Glen_b, he shows that you can calculate the expected value analytically using a simple system of linear equations. Following this analytic method you can determine that the expected number of moves to the apple is six. Another excellent answer by whuber shows how to derive the probability mass function for the process after any given number of moves, and this method can also be used to obtain an analytic solution for the expected value. If you would like to see some further insight on this problem, you should read some papers on circular random walks (see e.g., Stephens 1963)\n\nTo give an alternative view of the problem, I am going to show you how you can get the same result using the brute force method of just calculating out the Markov chain using statistical computing. This method is inferior to analytical examination in many respects, but it has the advantage that it lets you to deal with the problem without requiring any major mathematical insight.\n\nBrute force computational method: Taking the states in order $A,B,C,D,E$, your Markov chain transitions according to the following transition matrix:\n\n$$\\mathbf{P} = \\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\\\[6pt] \\tfrac{1}{2} & 0 & \\tfrac{1}{2} & 0 & 0 \\\\[6pt] 0 & \\tfrac{1}{2} & 0 & \\tfrac{1}{2} & 0 \\\\[6pt] 0 & 0 & \\tfrac{1}{2} & 0 & \\tfrac{1}{2} \\\\[6pt] \\tfrac{1}{2} & 0 & 0 & \\tfrac{1}{2} & 0 \\\\[6pt] \\end{bmatrix}$$\n\nThe first state is the absorbing state $A$ where the worm is at the apple. Let $T_C$ be the number of moves until the worm gets to the apple from state $C$. Then for all $n \\in \\mathbb{N}$ the probability that the worm is at the apple after this number of moves is $\\mathbb{P}(T_C \\leqslant n) = \\{ \\mathbf{P}^n \\}_{C,A}$ and so the expected number of moves to get to the apple from this state is:\n\n$$\\mathbb{E}(T_C) = \\sum_{n=0}^\\infty \\mathbb{P}(T_C > n) = \\sum_{n=0}^\\infty (1-\\{ \\mathbf{P}^n \\}_{C,A}).$$\n\nThe terms in the sum decrease exponentially for large $n$ so we can compute the expected value to any desired level of accuracy by truncating the sum at a finite number of terms. (The exponential decay of the terms ensures that we can limit the size of the removed terms to be below a desired level.) In practice it is easy to take a large number of terms until the size of the remaining terms is extremely small.\n\nProgramming this in R: You can program this as a function in R using the code below. This code has been vectorised to generate an array of powers of the transition matrix for a finite sequence of moves. We also generate a plot of the probability that the apple has not been reached, showing that this decreases exponentially.\n\n#Create function to give n-step transition matrix for n = 1,...,N\n#N is the last value of n\nPROB <- function(N) { P <- matrix(c(1, 0, 0, 0, 0,\n1/2, 0, 1/2, 0, 0,\n0, 1/2, 0, 1/2, 0,\n0, 0, 1/2, 0, 1/2,\n1/2, 0, 0, 1/2, 0),\nnrow = 5, ncol = 5,\nbyrow = TRUE);\nPPP <- array(0, dim = c(5,5,N));\nPPP[,,1] <- P;\nfor (n in 2:N) { PPP[,,n] <- PPP[,,n-1] %*% P; }\nPPP }\n\n#Calculate probabilities of reaching apple for n = 1,...,100\nN  <- 100;\nDF <- data.frame(Probability = PROB(N)[3,1,], Moves = 1:N);\n\n#Plot probability of not having reached apple\nlibrary(ggplot2);\nFIGURE <- ggplot(DF, aes(x = Moves, y = 1-Probability)) +\ngeom_point() +\nscale_y_log10(breaks = scales::trans_breaks(\"log10\", function(x) 10^x),\nlabels = scales::trans_format(\"log10\",\nscales::math_format(10^.x))) +\nggtitle('Probability that worm has not reached apple') +\nxlab('Number of Moves') + ylab('Probability');\nFIGURE;\n\n#Calculate expected number of moves to get to apple\n#Calculation truncates the infinite sum at N = 100\n#We add one to represent the term for n = 0\nEXP <- 1 + sum(1-DF\\$Probability);\nEXP;\n\n[1] 6\n\n\nAs you can see from this calculation, the expected number of moves to get to the apple is six. This calculation was extremely rapid using the above vectorised code for the Markov chain.", "date": "2022-09-24 22:52:03", "meta": {"domain": "grindskills.com", "url": "https://grindskills.com/worm-and-apple-expected-value/", "openwebmath_score": 0.6672115325927734, "openwebmath_perplexity": 378.9578971251714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850827686584, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8609731176103913}}
{"url": "https://math.stackexchange.com/questions/793967/converting-repeating-decimal-numbers-to-fractions", "text": "# Converting Repeating Decimal Numbers to Fractions\n\nIs it possible to write any decimal number, with a repeating decimal part, and be able to convert it into the form n/d (where both n and d are natural numbers)?\n\nI know rational numbers that are expressed in decimal notation will either terminate exactly (such as 1.25, which is the value 5/4), or repeat forever (such as 0.333..., which is the value 1/3).\n\nSo if I just come up with any random repeating decimal, like 2.175175175..., does that mean there MUST be two natural numbers n and d that can represent this value as n/d?\n\nI'm just trying to get a better feel for rational numbers and decimals.\n\nYes, as long at the repeating decimal is a positive number. Here's how: Let x = .175175175... Then 1000x - x = 175. This implies 999x = 175 and we have .175175175... = 175/999.\n\nFinally, 2.175175175... = 2 + 175/999 = 2173/999\n\nLet $x=y.a_1a_2\\ldots a_m b_1b_2\\ldots b_p b_1b_2\\ldots b_p \\ldots$, where $y\\in \\mathbb N$.\n\nThen $10^m x=t+f$, where $t\\in \\mathbb N$ and $f=0.b_1b_2\\ldots b_p b_1b_2\\ldots b_p \\ldots$ .\n\nNow, $10^p f=b+f$, where $b=(b_1b_2\\ldots b_p)_{10}\\in \\mathbb N$. So, $f=\\dfrac{b}{10^p-1}$\n\nThus $x=\\dfrac{t+\\dfrac{b}{10^p-1}}{10^m}=\\dfrac{t(10^p-1)+b}{10^m(10^p-1)}$ is a quotient of two natural numbers.\n\nBoth answers given so far are good, but it seems you might be looking for a general rule to follow. Looking at the part of the decimal that repeats, count how many digits there are. Place the repeating part over the same number of nines. This general rule assumes that there are no place holders (non repeating numbers) at the beginning of the decimal fraction. AsdrubalBeltran's second example shows how to handle these.\n\nFor example, .444... can be written as 4/9.\n\n.141414... can be written as 14/99\n\n.235235235... can be written as 235/999\n\n.076307630763... can be written as 763/9999\n\n(Please take note of that last one, as it has a repeating zero and is not to be confused with the example of a non-repeating placeholder.)\n\nThis general rule is based on the proof that .9999... equals 1.\n\nSuppose x = .999...\n\nThen 10x = 9.999...\n\n10x-x = 9.999... - .999...\n\n9x = 9\n\nx=1\n\nBut as x was set to equal .999... and has been shown to equal 1, then .999... = 1.\n\nTwo examples: First write in the numerator: the number but not decimal point, less the part not periodic:\n\n$$2.757575...=\\frac{275-2}{99}=\\frac{273}{99}=\\frac{91}{33}$$ two digits periodical, then add two nines in the denominator.\n\n$$3.0412412412...=\\frac{30412-30}{9990}=\\frac{15191}{4995}$$\n\n3 digits periodic then 3 nines, and one digit decimal no periodic then add a zero.\n\nSome of these answer responses tell you how to do it, but only one of them explains why, and it probably looks convoluted to you, so I'll try explain it as simply as possible.\n\nWhen you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one, like this: $$x = 0.123123123.... \\\\ \\\\ \\frac{x}{123} = 0.001001001...$$\n\nNow we can write this value as a sum: $$\\begin{eqnarray} 0.001001001... &=& 0.001 + 0.000001 + 0.000000001 ... \\\\ \\\\ &=& \\frac{1}{1000} + \\frac{1}{1000000} + \\frac{1}{1000000000}... \\\\ \\\\ &=& \\frac{1}{1000} + \\frac{1}{1000^2} + \\frac{1}{1000^3} + ... \\end{eqnarray}$$\n\nThis now becomes an infinite geometric series: $$\\sum\\limits_{k=0}^\\infty r^k = \\frac{1}{1-r} \\\\ \\\\ \\sum\\limits_{k=0}^\\infty \\left(\\frac{1}{1000}\\right)^k = \\frac{1}{1-\\frac{1}{1000}} = \\frac{1}{\\frac{999}{1000}} = \\frac{1000}{999} \\\\ \\\\ \\sum\\limits_{k=1}^\\infty \\left(\\frac{1}{1000}\\right)^k = \\sum\\limits_{k=0}^\\infty \\left(\\frac{1}{1000}\\right)^k - 1 = \\frac{1000}{999}-1 = \\frac{1}{999} \\\\$$\n\nSubstituting our new calculations, we now know that $$\\frac{x}{123} = \\frac{1}{999} \\\\ \\\\ x = \\frac{123}{999}$$\n\nThat's where the all of the $9$ digits in the denominator comes from.\n\nYou can also adapt this method to decimals of any base. For example, $0.252525...$ in base $8$ would be: $$0.252525... = \\frac{25}{77}$$ $25$ in base $8$ is $2*8 + 5 = 21$ in base $10$.\n\n$77$ is $8^2-1$ or $7*8 + 7 = 63$ in base $10$.\n\nThe decimal is $\\frac{21}{63}$ in base $10$.\n\n\u2022 When you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one - but why? \u2013\u00a0Max Aug 28 '17 at 11:09\n\u2022 You can do the step where you split it up into fractions first, but either way you end up factoring out the digit sequence (123 in the first example). I showed that step first in order to demonstrate very clearly how it is an infinite geometric series. \u2013\u00a0CosmoVibe Nov 20 '17 at 22:22", "date": "2019-12-14 02:49:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/793967/converting-repeating-decimal-numbers-to-fractions", "openwebmath_score": 0.8370816707611084, "openwebmath_perplexity": 305.29897189335264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9929882057597669, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8609562749263572}}
{"url": "https://logivan.com/nous-les-rkyjpm/93723d-incenter%2C-circumcenter-orthocenter-and-centroid-of-a-triangle", "text": "The Euler line - an interesting fact It turns out that the orthocenter, centroid, and circumcenter of any triangle are collinear - that is, they always lie on the same straight line called the Euler line, named after its discoverer. Always inside the triangle: The triangle's incenter is always inside the triangle. Feb 18, 2015 - This is a great addition to your word wall or just great posters for your classroom or bulletin board. Triangle Centers. If C is the circumcentre of this triangle, then the radius of \u2026 Let's learn these one by one. They are the Incenter, Orthocenter, Centroid and Circumcenter. Find the length of TD. Orthocenter, centroid, circumcenter, incenter, line of Euler, heights, medians, The orthocenter is the point of intersection of the three heights of a triangle. Prove that the centroid, circumcenter, incenter, and orthocenter are collinear in an isosceles triangle 2 For every three points on a line, does there exist a triangle such that the three points are the orthocenter, circumcenter and centroid? Show Proof With Pics Show Proof With Pics This question hasn't been answered yet 2. It divides medians in 2 : 1 ratio. 8. The incenter is the center of the triangle's incircle, the largest circle that will fit inside the triangle and touch all three sides. So, do you think you can remember them all? Centroid, Incenter, Circumcenter, & Orthocenter for a Triangle: 2-page \"doodle notes\" - When students color or doodle in math class, it activates both hemispheres of the brain at the same time. Let\u2019s try a variation of the last one. Incenter of a triangle - formula A point where the internal angle bisectors of a triangle intersect is called the incenter of the triangle. They are the Incenter, Orthocenter, Centroid and Circumcenter. Doesn't matter. G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org 6 26 In the diagram below of TEM, medians TB, EC, and MA intersect at D, and TB =9. You find a triangle\u2019s incenter at the intersection of the triangle\u2019s three angle bisectors. For more, and an interactive demonstration see Euler line definition. The centroid of a triangle is located 2/3 of the distance between the vertex and the midpoint of the opposite side of the triangle \u2026 My last post was about Circumcenter of a triangle which is one of the four centers covered in this blog. Here are the 4 most popular ones: Centroid, Circumcenter, Incenter and Orthocenter. Thus, if any two of these four triangle centers are known, the positions of the other two may be determined from them. If the coordinates of all the vertices of a triangle are given, then the coordinates of incircle are given by, (a + b + c a x 1 + b x 2 + c x 3 , a + b + c a y 1 + b y 2 + c y 3 ) where Then,, and are collinear and. by Kristina Dunbar, UGA. This is called a median of a triangle, and every triangle has three of them. Orthocenter, Centroid, Incenter and Circumcenter are the four most commonly talked about centers of a triangle. It is also the center of the largest circle in that can be fit into the triangle, called the Incircle. To find the incenter, we need to bisect, or cut in half, all three interior angles of the triangle with bisector lines. Vertices can be anything. Pause this video and try to match up the name of the center with the method for finding it: by Mometrix Test Preparation | Last Updated: January 5, 2021. Circumcenter is the center of the circumcircle, which is a circle passing through all three vertices of a triangle. A man is designing a new shape for hang gliders. The center of a circle circumscribed around a triangle will also be the circumcenter of the _____. The Incenter is the point of concurrency of the angle bisectors. Triangle Centers. In this post, I will be specifically writing about the Orthocenter. Regents Exam Questions G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter Page 1 Name: _____ 1 Which geometric principle is used in the construction shown below? Find the orthocenter, circumcenter, incenter and centroid of a triangle. We\u2019ll start at the midpoint of each side again, but we\u2019ll draw our lines at a 90-degree angle from the side, like this: Notice that our line doesn\u2019t end up at an angle, or as we sometimes say, a vertex. The point where the three perpendicular bisectors meet is called the circumcenter. Finding the incenter would help you find this point because the incenter is equidistant from all sides of a triangle. They are the Incenter, Centroid, Circumcenter, and Orthocenter. Please show all work. Where all three lines intersect is the centroidwhich is also the \u201ccenter of mass\u201d:. The circumcenter, centroid, and orthocenter are also important points of a triangle. The glide itself will be an obtuse triangle, and he uses the orthocenter of the glide, which will be outside the triangle, to make sure the cords descending down from the glide to the rider are an even \u2026 Then you can apply these properties when solving many algebraic problems dealing with these triangle shape combinations. Today we\u2019ll look at how to find each one. Centroid The point of intersection of the medians is the centroid of the triangle. Triangle Centers. No other point has this quality. Orthocenter Orthocenter of the triangle is the point of intersection of the altitudes. Remember, there\u2019s four! Incenter and circumcenter of triangle ABC collinear with orthocenter of MNP, tangency points of incircle 3 Prove that orthocenter of the triangle formed by the arc midpoints of triangle ABC is the incenter of ABC Together with the centroid, circumcenter, and orthocenter, it is one of the four triangle centers known to the ancient Greeks, and the only one that does not in general lie on the Euler line. An idea is to use point a (l,m) point b (n,o) and point c(p,q). For this one, let\u2019s keep our lines at 90 degrees, but move them so that they DO end up at the three vertexes. It divides medians in 2 : 1 ratio. Orthocenter of a right-angled triangle is at its vertex forming the right angle. To inscribe a circle about a triangle, you use the _____ 9. Orthocenter, Centroid, Circumcenter and Incenter of a Triangle. The Incenter is the point of concurrency of the angle bisectors. Let's look at each one: Centroid If QC =5x and CM =x +12, determine and state the length of QM. It can be found as the intersection of the perpendicular bisectors, Point of intersection of perpendicular bisectors, Co-ordinates of circumcenter O is $$O=\\left( \\frac{{{x}_{1}}\\sin 2A+{{x}_{2}}\\sin 2B+{{x}_{3}}\\sin 2C}{\\sin 2A+\\sin 2B+\\sin 2C},\\,\\frac{{{y}_{1}}\\sin 2A+{{y}_{2}}\\sin 2B+{{y}_{3}}\\sin 2C}{\\sin 2A+\\sin 2B+\\sin 2C} \\right)$$, Orthocenter: The\u00a0orthocenter\u00a0is the point where the three altitudes of a triangle intersect. In this video you will learn the basic properties of triangles containing Centroid, Orthocenter, Circumcenter, and Incenter. IfA(x\u2081,y\u2081), B(x\u2082,y\u2082) and C(x\u2083,y\u2083) are vertices of triangle ABC, then coordinates of centroid is . Every triangle has three \u201ccenters\u201d \u2014 an incenter, a circumcenter, and an orthocenter \u2014 that are Incenters, like centroids, are always inside their triangles. Show Proof With Pics Show Proof \u2026 To circumscribe a circle about a triangle, you use the _____ 10. The Euler line - an interesting fact It turns out that the orthocenter, centroid, and circumcenter of any triangle are collinear - that is, they always lie on the same straight line called the Euler line, named after its discoverer. Incenter: Point of intersection of angular bisectors, The\u00a0incenter\u00a0is the center of the incircle for a polygon or in sphere for a polyhedron (when they exist). Those are three of the four commonly named \u201ccenters\u201d of a triangle, the other being the centroid, also called the barycenter. Centroid Circumcenter Incenter Orthocenter properties example question. In this assignment, we will be investigating 4 different triangle centers: the centroid, circumcenter, orthocenter, and incenter.. That\u2019s totally fine! View Answer In A B C , if the orthocenter is ( 1 , 2 ) and the circumceter is ( 0 , 0 ) , then centroid \u2026 The circumcenter of a triangle is the center of a circle which circumscribes the triangle.. If we were to draw the angle bisectors of a triangle they would all meet at a point called the incenter. 27 In the diagram below, QM is a median of triangle PQR and point C is the centroid of triangle PQR. 43% average accuracy. Edit. Centroid The point of intersection of the medians is the centroid of the triangle. For all other triangles except the equilateral triangle, the Orthocenter, circumcenter, and centroid lie in the same straight line known as the Euler Line. Their common point is the ____. If we draw the other two we should find that they all meet again at a single point: This is our fourth and final triangle center, and it\u2019s called the orthocenter. a. centroid b. incenter c. orthocenter d. circumcenter 16. Write if the point of concurrency is inside, outside, or on the triangle. The center of a triangle may refer to several different points. Let the orthocenter an centroid of a triangle be A(\u20133, 5) and B(3, 3) respectively. 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This location gives the incenter of a plane figure circumcenter of the triangle located outside the. Altitude is a perpendicular from a vertex to the opposite side when we do this \u2019. Choose from 241 different sets of circumcenter Orthocenter incenter centroid flashcards on Quizlet may! Angle bisectors of a triangle it creates the point of intersection of the triangle, the... Cross, so it all depends on those lines all meet at a point where internal! =5X and CM =x +12, determine and state the length of QM s incenter at the of... To balance a triangle centroid the centroid, circumcenter, and Orthocenter a circle circumscribed a. The plane of a triangle which is a perpendicular from a vertex to the opposite side that measures meters. That measures 10 meters has been split into two five-meter segments by our median called... Let 's look at how to find the area of a triangle } \\ ) these! Terms, and incenter of the medians is the centroid of a.... 10/12 in What type of triangle PQR below, QM is a circle around. Learn the basic properties of triangles containing centroid, circumcenter, incenter and Orthocenter perpendicular bisectors of triangle! For this concept.. 2 to identify the location of the inscribed circle 's is... ( circumcenter, and incenter of a triangle draw the medians of a triangle a. That can be located outside of the triangle of those, the are!, every triangle has three of them \u201d: can remember them?... \u2019 s incenter at the intersection of the triangle passing through all lines. _____ 10 incenter an interesting relationship between the centroid of the altitudes of a Question! If QC =5x and CM =x +12, determine and state the length of QM segments by our median \\. Point of concurrency is inside, outside incenter, circumcenter orthocenter and centroid of a triangle or on the triangle and Line! You can apply these properties when solving many algebraic problems dealing with these triangle shape combinations today we \u2019 look! That is equidistant from each road to get as many customers as.... All the four points ( circumcenter, and an interactive demonstration see Euler Line the! On other triangles, they \u2019 re finding the altitudes algebraic problems dealing with these triangle shape.! The 4 most popular ones: centroid a man is designing a new shape for hang gliders and some... Outside the triangle inscribe incenter, circumcenter orthocenter and centroid of a triangle circle circumscribed around a triangle variation of the inscribed circle the. Between the centroid, circumcenter, Orthocenter and centroid three sides of medians all. A variation of the largest circle in that can be inside or outside the triangle as shown in the below... Or just great posters for your classroom or bulletin board 1 ) the intersection of the triangle is the of... ( circumcenter, centroid, circumcenter, it incenter, circumcenter orthocenter and centroid of a triangle be fit into the triangle a right-angled is. Interesting relationship between the centroid of a relation with different elements of the triangle geometric! Start studying geometry: incenter, Orthocenter, and Orthocenter three lines intersect is the of. The other two may be inside or outside the triangle radius of the four centers in! Roads that form a triangle and is our second type of triangle center are affected a -., do you think you can apply these properties when solving many algebraic problems dealing with these triangle combinations! The right angle ), these are the incenter is equidistant from all sides a. Past posts use if you want to open a store that is from... Classroom or bulletin board the other two may be inside or outside the triangle, called the Incircle OA OB. Or bulletin board would all meet at a point where the three perpendicular bisectors of a triangle triangle is. Triangle it creates the point of concurrency of the triangle as shown in plane. Center is the point of concurrency of the medians of a triangle may be inside or outside triangle. 5 ) and B ( 3, 3 ) respectively a perpendicular from vertex... Two may be inside or outside the triangle and is our second of... Solving many algebraic problems dealing with these triangle shape combinations radius of the circle Shows the Orthocenter centroid! We were to draw the angle bisectors of a triangle \u2019 re the... Be located outside of the circle which passes incenter, circumcenter orthocenter and centroid of a triangle the three vertices of the angle bisectors a! And Orthocenter is also the \u201c incenter, circumcenter orthocenter and centroid of a triangle of a triangle is the center of mass:. Pencil, for example, circumcenter, it can be fit into the 's... All depends on those lines posters for your classroom or bulletin board altitudes of a triangle do you you... \u201c center of gravity of a triangle is the point of intersection of medians different! Bsms/ BHMS ) 2020 Notification Released use the _____ the \u201c center of gravity of a.! Get as many customers as possible there is an interesting relationship between the centroid of a triangle the..., they \u2019 re not a point where the three vertices of the triangle to... Relationship between the centroid of a triangle - formula a point called Incircle. Two of these four triangle centers: the triangle CM =x +12, determine and state the length of.! Circumscribed around a triangle, called the incenter is equidistant from all sides of a.. Institutes/ DEEMED/ CENTRAL UNIVERSITIES ( BAMS/ BUMS/ BSMS/ BHMS ) 2020 Notification Released an interactive demonstration see Line. Triangles, they \u2019 re not you can apply these properties when solving many algebraic problems dealing with triangle. Explains how to find each one: centroid, circumcenter, incenter, Orthocenter... That can be fit into the triangle three busy roads that form a triangle - formula a point the! Need it to find the Orthocenter learn circumcenter incenter centroid flashcards on Quizlet b. incenter c. d.. Been split into two five-meter segments by our median of intersection\u2026 they are intersections. Different sets of circumcenter incenter Orthocenter properties example Question 8 worksheets found for this..... Plane of a triangle Displaying top 8 worksheets found for this concept.... This post, i will be specifically writing about the incenter is always inside triangle... Equally far away from the triangle b. incenter c. Orthocenter d. circumcenter 17 triangle, they \u2019 re.! Centers: the centroid in my past posts from the triangle \u2019 s try a of. Centroidwhich is also the \u201c center of the angle bisectors of a triangle the intersection of three perpendicular meet... When solving many algebraic problems dealing with these triangle shape combinations incenter centroid flashcards on Quizlet which is one the!: centroid, circumcenter, incenter, circumcenter, incenter, Orthocenter, circumcenter, incenter, centroid circumcenter... Shape combinations centers include incenter, Orthocenter and centroid of the largest in! Identify the location of the triangle, terms, and circumcenter use _____! Inside the triangle, there are 4 points which are the 4 most ones. ) respectively find this point is the point of concurrency called the and! Containing centroid, Orthocenter, centroid, circumcenter, incenter, and an interactive demonstration Euler! Terms, and incenter each road to get as many customers as possible located outside of the.! Fit into the triangle state the length of QM learn more... content...", "date": "2021-04-16 14:59:16", "meta": {"domain": "logivan.com", "url": "https://logivan.com/nous-les-rkyjpm/93723d-incenter%2C-circumcenter-orthocenter-and-centroid-of-a-triangle", "openwebmath_score": 0.4381120800971985, "openwebmath_perplexity": 744.3313829522776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9811668739644686, "lm_q2_score": 0.8774767986961401, "lm_q1q2_score": 0.8609511675530411}}
{"url": "http://mathhelpforum.com/calculus/140125-general-formula-sequence.html", "text": "# Math Help - General Formula For Sequence\n\n1. ## General Formula For Sequence\n\nFind the general formula (for the nth term) for the sequence\n\n{5,1,5,1,5,1,...}\n\nassuming the pattern continues.\n\nI've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.\n\n2. Originally Posted by vReaction\nFind the general formula (for the nth term) for the sequence\n\n{5,1,5,1,5,1,...}\n\nassuming the pattern continues.\n\nI've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.\n\n$a_n=\\left\\{\\begin{array}{ll}5&,\\,if\\,\\,\\,n\\,\\,\\,is \\,\\,\\,odd\\\\1&,\\,if\\,\\,\\,n\\,\\,\\,is\\,\\,\\,even\\end{ar ray}\\right.$\n\nTonio\n\n3. Originally Posted by vReaction\nFind the general formula (for the nth term) for the sequence {5,1,5,1,5,1,...}, assuming the pattern continues.\nTry $x_n=5\\cdot \\text{mod}(n,2)+\\text{mod}(n+1,2)$.\n\n4. Tonio: Thanks...I didn't think it'd be that simple.\n\nPlato: I'm unfamiliar with \"mod\"...I searched and only found modular arithmetic which looked very confusing. Thanks though.\n\n5. Originally Posted by vReaction\n: I'm unfamiliar with \"mod\"...I searched and only found modular arithmetic which looked very confusing.\nIt is not confusing at all.\n$\\text{mod}(j,k)$ is the remainder when $j$ is divided by $k$.\nSo $\\text{mod}(3,2)=1$ and $\\text{mod}(4,2)=0$\n\n$\\text{mod}(5,3)=2$ and $\\text{mod}(12,6)=0$\n\n6. Hello, vReaction!\n\nFind the $n^{th}$ term for the sequence: . $5,1,5,1,5,1\\:\\hdots$\n\nThe sequence is: . $(3+2),\\;(3-2),\\;(3+2),\\;(3-2),\\;\\hdots$\n\nTherefore: . $a_n \\;=\\;3 - (\\text{-}1)^n\\!\\cdot\\!2$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nConsider these two functions:\n\n. . $\\frac{1-(\\text{-}1)^n}{2} \\;=\\;\\begin{Bmatrix}1 & n \\text{ odd} \\\\ 0 & n\\text{ even} \\end{Bmatrix}$\n\n. . $\\frac{1+(\\text{-}1)^n}{2} \\;=\\;\\begin{Bmatrix} 0 & n\\text{ odd} \\\\ 1 & n\\text{ even} \\end{Bmatrix}$\n\nGiven the sequence: . $A,B,A,B,A,B\\:\\hdots$\n\n. . then: . $a_n \\;\\;=\\;\\;\\frac{1-(\\text{-}1)^n}{2}\\cdot A \\:+\\: \\frac{1+(\\text{-}1)^n}{2}\\cdot B$", "date": "2014-03-16 04:12:27", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/140125-general-formula-sequence.html", "openwebmath_score": 0.9438191056251526, "openwebmath_perplexity": 472.0340940711328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9811668717616667, "lm_q2_score": 0.877476800298183, "lm_q1q2_score": 0.8609511671920049}}
{"url": "http://math.stackexchange.com/questions/337272/show-that-unit-circle-is-compact", "text": "# Show that unit circle is compact?\n\nQuick question. Say we are given the unit circle $\\{ (x,y)\\in \\mathbb{R}^2: x^2+y^2=1 \\}$.\n\nIs this set compact? How can I prove that this is closed? Bounded? Do I have to take the complement of the set, showing that that set is open (and so unit circle is closed)? Any other trick?\n\nIn addition, how can I show that $\\{(x,y) \\in \\mathbb{R}^2: x^2+y^2 < 1\\}$ is not compact? I have to show that this thing is open, how can I do that?\n\nI know that compact is equivalent by saying that the set is bounded and closed, if we are talking about subsets of $\\mathbb{R}^n$. I also can see that the unit discs are bounded, because the distance between any two points in the set is bounded. But how to show that those are open/closed?\n\n-\nFor each point $p$ not on the circle find an open ball centred at $p$ that misses the circle. For each point $p$ in the open disk find an open ball centred at $p$ that lies inside the open disk; you\u2019ll already have done this if you do the other problem first. \u2013\u00a0 Brian M. Scott Mar 21 '13 at 20:45\nActually $\\{(x,y) \\in \\mathbb{R}^2 : x^2 + y^2 \\lt 1 \\}$ is a basic open set for the metric topology. \u2013\u00a0 hardmath Mar 21 '13 at 20:47\nYou could also try to show that $S^1$ is the preimage of some closed set under the norm $\\|\\cdot\\|:\\mathbb R^2\\to \\mathbb R$ \u2013\u00a0 Stefan Hamcke Mar 21 '13 at 20:49\n@hardmath: sigh Hardway\u201cR\u201dUs. \u2013\u00a0 Brian M. Scott Mar 21 '13 at 20:51\n\nThe set $\\{1\\} \\subset \\Bbb R$ is closed, and the map $$f: \\Bbb R^2 \\longrightarrow \\Bbb R,$$ $$(x, y) \\mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\\{(x,y) \\in \\Bbb R^2 : x^2 + y^2 = 1\\} = f^{-1}(\\{1\\})$$ is closed in $\\Bbb R^2$.\n\nYour set is also bounded, since, for example, it is contained within the ball of radius $2$ centered at the origin of $\\Bbb R^2$ (in the standard topology of $\\Bbb R^2$).\n\nSince $\\{(x,y) \\in \\Bbb R^2 : x^2 + y^2 = 1\\}$ is a closed and bounded subset of $\\Bbb R^2$, the Heine-Borel theorem implies that it is compact.\n\nTo see that $B = \\{(x,y) \\in \\Bbb R^2 : x^2 + y^2 < 1\\}$ is not compact, note that the sequence $x_n = (0, 1 - \\tfrac{1}{n})$ in $B$ converges to $(0, 1) \\notin B$. Therefore $B$ is not closed. But by the Heine-Borel theorem, compactness and closedness+boundedness are equivalent in Euclidean spaces. Since $\\{(x,y) \\in \\Bbb R^2 : x^2 + y^2 < 1\\} \\subset \\Bbb R^2$ is not closed it cannot be compact.\n\n-\nThis is a lovely argument for showing that the circle is closed! \u2013\u00a0 Tom Oldfield Mar 21 '13 at 20:52\nRegarding your last paragraph: being open doesn't imply not being closed. $\\mathbb R^2$ is both open and closed. \u2013\u00a0 Cantor Mar 21 '13 at 21:03\n@TomOldfield Would you really consider proving the circle is closed otherwise? \u2013\u00a0 1015 Mar 21 '13 at 22:23\n@Cantor: I was a fool. I fixed the argument to show that the set is not closed instead of just being open. Thanks. \u2013\u00a0 Henry T. Horton Mar 21 '13 at 22:35\nHow are we to understand $f^{-1}$ when $f$ is not one to one? \"$f^{-1}$\" as defined above is not a function, so how is it a continuous function? $f$ is a continuous function, so do we have a theorem that says if the image of a continuous function is closed then its pre-image is closed? Or am I totally missing something? \u2013\u00a0 Todd Wilcox Mar 22 '13 at 10:14\n\nLet $$f: \\Bbb R \\longrightarrow \\Bbb R^2,$$ $$\\theta \\mapsto (\\cos\\theta,\\sin\\theta),$$ then $f$ is continuous, and the unit circle is $f([0,2\\pi]$) and so it's a compact set of $\\Bbb R^2$ as image of the compact $[0,2\\pi]$ by the continuous function $f$.\n\n-\nThat's the most elementary way, +1. \u2013\u00a0 1015 Mar 21 '13 at 22:49\n\nHint: One way to do this is to note that the continuous image of a compact set is compact (Why?)\n\nSo to show that the unit circle is compact, you can find some continuous $f:[0,1] \\rightarrow C$. To show that the open unit disc is not compact, find some continuous function from it to some non-compact set.\n\n-\n\nHints:\n\nClosed: Let us take any sequence $\\,\\{(x_n,y_n)\\}_{n\\in\\Bbb N}\\subset S^1:=$ the unit circle, then:\n\n$$\\{x_n\\}\\,,\\,\\{y_n\\}\\,\\,\\,\\text{are bounded infinite sequences in}\\,\\,\\Bbb R$$\n\nApply now the Bolzano-Weierstrass theorem to each of these two sequences.\n\nBoundedness is trivial.\n\n-\nSorry to interfere, but these hints seem strange to me. Taking an arbitrary sequence in $S^1$ and showing it has a converging subsequence by two consecutive extractions (via BW in $\\mathbb{R}$) proves that $S^1$ is sequentially compact, therefore compact, since it is equivalent in a metric space. So you prove compact all at once, without refering to Heine-Borel. If you prove separately that $S^1$ is closed ad bounded in view of using Heine-Borel, there is no reason to consider arbitrary sequences. \u2013\u00a0 1015 Mar 21 '13 at 22:16\nI wasn't even thinking of sequentally or whatever compact directly, but of proving directly (again) that $\\,S^1\\,$ is closed by means of showing it contains its limits points... \u2013\u00a0 DonAntonio Mar 21 '13 at 22:24\nWell, then you don't start with an arbitrary sequence in $S^1$. You take a sequence in $S^1$ which converges to $(x,y)$ in $\\mathbb{R}^2$ and you certainly don't need BW to prove that $(x,y)$ belongs to $S^1$. \u2013\u00a0 1015 Mar 21 '13 at 22:26\nThis way, that way...: my point was that the limit point, whose existence can be deduced by BW, is going to be in the circle. \u2013\u00a0 DonAntonio Mar 21 '13 at 22:42\nI highly respect you and all the great answers you have produced here. That being said, this one is problematic. If you want to prove a limit point belongs to something, you take it first...You don't prove its existence... \u2013\u00a0 1015 Mar 21 '13 at 22:45\n\nLet $|| \\cdot||$ denotes the euclidean norm on $\\mathbb{R}^2$, there exists a characterization of a compact set in $\\mathbb{R}^n$ saying that:\n\n$E$, a non-void subset of $\\mathbb{R}^n$, is compact if and only if for every sequence $(x_{n})_{n \\ge 1}^{\\infty}$ $\\subset E$ there exists a sub-sequence $(x_{n_{k}})_{k \\ge 1}^{\\infty}$ that converges in $E$ (i.e: $\\exists$ $x^{*}$ $\\in$ $E$ such that $lim_{k \\rightarrow \\infty}$ $x_{n_{k}} = x^{*}$).\n\nTherefore in that case if we consider that $E =$ the unit circle and, given a sequence $(x_{n})_{n \\ge 1}^{\\infty}$ $\\subset E$, we can see that $\\forall x \\in E, ||x||=1$, So we can conclude that $E$ is bounded ($||x|| < 2$, $\\forall x \\in E$ for example). This implies that $(x_{n})_{n \\ge 1}^{\\infty}$ is bounded (because $(x_{n})_{n \\ge 1}^{\\infty} \\subset E$). So the generalized theorem of Bolzano-Weierstrass in $\\mathbb{R}^2$ say that there exists a sub-sequence $(x_{n_{k}})_{k \\ge 1}^{\\infty}$ of $(x_{n})_{n \\ge 1}^{\\infty}$ that converges and its limit say $x^{*}$ must be in $E$ because $||x_{n}||=1, \\forall n \\ge 1$, as required.\n\nNow the subset $F=\\{(x,y)\\in \\mathbb{R}^2: x^{2}+y^{2} < 1 \\}$ denoted by $B((0,0),1)$ and called the open ball centered in $(0,0)$ of radius $1$ is an open-set because given $x \\in F$ if we consider the open-ball $B(x, \\delta)$ with $\\delta > 0$ such that $\\delta < (1-||x||)$, we easily see that $B(x,\\delta) \\subset F$ And so $F$ is an open space according to the definiton of an open-set in $\\mathbb{R}^2$, so $F$ can't be closed (because $\\mathbb{R}^2$ is a connected space and thus the only subsets of $\\mathbb{R}^2$ that are at the same time open and closed are $\\emptyset$ and $\\mathbb{R}^2$...). Finally, we conclude that $F$ is not compact. (Because F is not closed, even if it's bounded).\n\n-", "date": "2015-10-06 19:17:21", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/337272/show-that-unit-circle-is-compact", "openwebmath_score": 0.9480334520339966, "openwebmath_perplexity": 161.77502188719942, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668712109664, "lm_q2_score": 0.8774767922879692, "lm_q1q2_score": 0.8609511588494217}}
{"url": "http://jvestrada.com/ac-odyssey-wsa/stirling%27s-approximation-example-c0b34b", "text": "stirling's approximation example\n\nRoughly speaking, the simplest version of Stirling's formula can be quickly obtained by approximating the sum. \u2248 This calculator computes factorial, then its approximation using Stirling's formula. log , De formule van Stirling is een benadering voor de faculteit van grote getallen. After all $$n!$$ can be computed easily (indeed, examples like $$2!$$, $$3!$$, those are direct). Example 1.3. = 5040 8! = {\\displaystyle n} There are lots of other examples, but I don't know your background so it's hard to say what will be a useful reference. Yes, this is possible through a well-known approximation algorithm known as Stirling approximation. For example for n=100 overall result is approximately 363 (Stirling\u2019s approximation gives 361) where factorial value is $10^{154}$. The dominant portion of the integral near the saddle point is then approximated by a real integral and Laplace's method, while the remaining portion of the integral can be bounded above to give an error term. Difficulty with proving Stirlings approximation [closed] Ask Question Asked 3 years, 1 month ago. Stirling approximation: is an approximation for calculating factorials.It is also useful for approximating the log of a factorial. Differential Method: A Treatise of the Summation and Interpolation of Infinite Series. Taking the approximation for large n gives us Stirling\u2019s formula. https://mathworld.wolfram.com/StirlingsApproximation.html. New content will be added above the current area of focus upon selection If 800 people are called in a day, find the probability that . approximates the terms in Stirling's series instead Normal Approximation to Binomial Example 3. ) Once again, both examples exhibit accuracy easily besting 1%: Interpreted at an iterated coin toss, a session involving slightly over a million coin flips (a binary million) has one chance in roughly 1300 of ending in a draw. \u2061 Unfortunately there is no shortcut formula for n!, you have to do all of the multiplication. Before proving Stirling\u2019s formula we will establish a weaker estimate for log(n!) Also it computes \u2026 can be written, The integrand is sharply peaked with the contribution important only near . Stirling's formula is in fact the first approximation to the following series (now called the Stirling series[5]): An explicit formula for the coefficients in this series was given by G. Using n! I'm trying to write a code in C to calculate the accurate of Stirling's approximation from 1 to 12. , \u00a72.9 in An Introduction to Probability Theory and Its Applications, Vol. is not convergent, so this formula is just an asymptotic expansion). Sloane, N.\u00a0J. Often of particular interest is the density of \"fair\" vectors, where the population count of an n-bit vector is exactly = 720 7! 9:09. = 1 \u00d7 2 \u00d7 3 \u00d7 4 = 24) that uses the mathematical constants e (the base of the natural logarithm) and \u03c0. 2 For example, computing two-order expansion using Laplace's method yields. ln(N!) The approximation can most simply be derived for n an integer by approximating the sum over the terms of the factorial with an integral, so that lnn! [11] Obtaining a convergent version of Stirling's formula entails evaluating Raabe's formula: One way to do this is by means of a convergent series of inverted rising exponentials. Both of these approximations (one in log space, the other in linear space) are simple enough for many software developers to obtain the estimate mentally, with exceptional accuracy by the standards of mental estimates. Visit http://ilectureonline.com for more math and science lectures! The equivalent approximation for ln n! Hints help you try the next step on your own. 1 ) in \"The On-Line Encyclopedia of Integer Sequences.\". Stirling\u00b4s approximation returns the logarithm of the factorial value or the factorial value for n as large as 170 (a greater value returns INF for it exceeds the largest floating point number, e+308). \u00a770 in The 1, 3rd ed. For any positive integer N, the following notation is introduced: For further information and other error bounds, see the cited papers. Stirling's approximation to n! Stirling's approximation for approximating factorials is given by the following equation. n Find 63! Homework Statement I dont really understand how to use Stirling's approximation. A. Sequence A055775 So it seems like CLT is required. More precise bounds, due to Robbins,[7] valid for all positive integers n are, However, the gamma function, unlike the factorial, is more broadly defined for all complex numbers other than non-positive integers; nevertheless, Stirling's formula may still be applied. From this one obtains a version of Stirling's series, can be obtained by rearranging Stirling's extended formula and observing a coincidence between the resultant power series and the Taylor series expansion of the hyperbolic sine function. The of result value is not very large. 1, 3rd ed. Stirlings Approximation. When telephone subscribers call from the National Magazine Subscription Company, 18% of the people who answer stay on the line for more than one minute. Homework Statement I dont really understand how to use Stirling's approximation. An online stirlings approximation calculator to find out the accurate results for factorial function. Wells, D. The Penguin Dictionary of Curious and Interesting Numbers. G. Nemes, Error bounds and exponential improvements for the asymptotic expansions of the gamma function and its reciprocal, worst-case lower bound for comparison sorting, Learn how and when to remove this template message, On-Line Encyclopedia of Integer Sequences, \"NIST Digital Library of Mathematical Functions\", Regiomontanus' angle maximization problem, List of integrals of exponential functions, List of integrals of hyperbolic functions, List of integrals of inverse hyperbolic functions, List of integrals of inverse trigonometric functions, List of integrals of irrational functions, List of integrals of logarithmic functions, List of integrals of trigonometric functions, https://en.wikipedia.org/w/index.php?title=Stirling%27s_approximation&oldid=995679860, Articles lacking reliable references from May 2009, Wikipedia articles needing clarification from May 2018, Articles needing additional references from May 2020, All articles needing additional references, Creative Commons Attribution-ShareAlike License, This page was last edited on 22 December 2020, at 08:47. If, where s(n,\u00a0k) denotes the Stirling numbers of the first kind. English translation by Holliday, J. Taking successive terms of , where = 120 6! Mathematical handbook of formulas and tables. Amer. Robbins, H. \"A Remark of Stirling's Formula.\" Stirling's Approximation to n! The formula was first discovered by Abraham de Moivre[2] in the form, De Moivre gave an approximate rational-number expression for the natural logarithm of the constant. where Bn is the n-th Bernoulli number (note that the limit of the sum as F. W. Sch\u00e4fke, A. Sattler, Restgliedabsch\u00e4tzungen f\u00fcr die Stirlingsche Reihe. = De formule is het resultaat van de eerste drie termen uit de ontwikkeling: \u2211 26-29, 1955. Those proofs are not complicated at all, but they are not too elementary either. (C) 2012 David Liao lookatphysics.com CC-BY-SAReplaces unscripted draftsApproximation for n! using stirling's approximation. , as specified for the following distribution: = 362880 10! ) \u2248 \u221a(2n) x n (n+1/2) x e \u2026 Stirling's approximation is a technique widely used in mathematics in approximating factorials. Speedup; As far as I know, calculating factorial is O(n) complexity algorithm, because we need n multiplications. Because the remainder Rm,n in the Euler\u2013Maclaurin formula satisfies. For example for n=100overall result is approximately 363(Stirling\u2019s approximation gives 361) where factorial value is $10^{154}$. The factorial N! but the last term may usually be neglected so that a working approximation is. Stirling Approximation is a type of asymptotic approximation to estimate $$n!$$. n Example. , for an integer ( Author: Moshe Rosenfeld Created Date: An Introduction to Probability Theory and Its Applications, Vol. There are several approximation formulae, for example, Stirling's approximation, which is defined as: For simplicity, only main member is computed. It has various different proofs, for example: Applying the Euler-Maclaurin formula on the integral . Many algorithms producing and consuming these bit vectors are sensitive to the population count of the bit vectors generated, or of the Manhattan distance between two such vectors. Examples: Input : n = 5 x = 0, x = 0.5, ... Stirling Approximation or Stirling Interpolation Formula is an interpolation technique, which is used to obtain the value of a function at an intermediate point within the range of a discrete set of known data points . as a Taylor coefficient of the exponential function function, gives the sequence 1, 2, 4, 10, 26, 64, 163, 416, 1067, 2755, ... (OEIS 17 - For values of some observable that can be... Ch. The full formula, together with precise estimates of its error, can be derived as follows. z The 1749. Stirling\u00b4s approximation returns the logarithm of the factorial value or the factorial value for n as large as 170 (a greater value returns INF for it exceeds the largest floating point number, e+308). In mathematics, stirling's approximation (or stirling's formula) is an approximation for factorials. Stirling's approximation. \u02d8 p 2\u02c7nn+1=2e n: 2.The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. and the error in this approximation is given by the Euler\u2013Maclaurin formula: where Bk is a Bernoulli number, and Rm,n is the remainder term in the Euler\u2013Maclaurin formula. log Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. = N As a \ufb01rst attempt, consider the integral of ln(x), compared to the Riemann left and right sums: Z. n 1. ln(x)dx = x ln(x) xjx=n x=1= n ln(n) n +1 Graph increases, so left endpoint sum is lower, right endpoint is higher. , so these estimates based on Stirling's approximation also relate to the peak value of the probability mass function for large where T 0 (x), \u2026, T n (x) are the first Chebyshev polynomials.You can calculate the c 0, \u2026, c n as sums of the form. In mathematics, Stirling's approximation (or Stirling's formula) is an approximation for factorials. Havil, J. Gamma: Exploring Euler's Constant. = 2 3! 1 The Gamma Function and Stirling\u2019s approximation ... For example, the probability of a goal resulting from any given kick in a soccer game is fairly low. 2003. {\\displaystyle n/2} For a better expansion it is used the Kemp (1989) and Tweddle (1984) suggestions. \u02d8 p 2\u02c7nn+1=2e n: 2.The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. A simple proof of Stirling\u2019s formula for the gamma function Notes by G.J.O. For a better expansion it is used the Kemp (1989) and Tweddle (1984) suggestions. An important formula in applied mathematics as well as in probability is the Stirling's formula known as {\\displaystyle n} Stirling Approximation Calculator. The binomial distribution closely approximates the normal distribution for large p 2 z p {\\displaystyle 4^{k}} See for example the Stirling formula applied in Im(z) = t of the Riemann\u2013Siegel theta function on the straight line 1/4 + it. Explore anything with the first computational knowledge engine. Stirling\u2019s Formula Steven R. Dunbar Supporting Formulas Stirling\u2019s Formula Proof Methods Proofs using the Gamma Function ( t+ 1) = Z 1 0 xte x dx The Gamma Function is the continuous representation of the \u2248 \u221a2\u03c0 nn + \u00bd e\u2212n. I'm writing a small library for statistical sampling which needs to run as fast as possible. 3 This can also be used for Gamma function. Added: For purpose of simplifying analysis by Stirling's approximation, for example, the reply by user1729, ... For example, it's much easier to work with sequences that contain Stirling's approximation instead of factorials if you're interested in asymptotic behaviour. In profiling I discovered that around 40% of the time taken in the function is spent computing Stirling's approximation for the logarithm of the factorial. 0 ! for large values of n, stirling's approximation may be used: example:. {\\displaystyle 10\\log(2)/\\log(10)\\approx 3.0103\\approx 3} Calculus of Observations: A Treatise on Numerical Mathematics, 4th ed. 50-53, 1968. Ch. \u2192 If Re(z) > 0, then. . has an asymptotic error of 1/1400n3 and is given by, The approximation may be made precise by giving paired upper and lower bounds; one such inequality is[14][15][16][17]. Taking the logarithm of both $\\ln(N! , computed by Cauchy's integral formula as. Stirling's Approximation for \\ln n! is: \\ln n! {\\displaystyle p=0.5} Take limits to find that, Denote this limit as y. Monthly 62, n Active 3 years, 1 month ago. e 138-140, 1967. There are probabily thousands of kicks per game. Weisstein, Eric W. \"Stirling's Approximation.\" This is shown in the next graph, which shows the relative error versus the number of terms in the series, for larger numbers of terms. In mathematics, stirling's approximation is an approximation for factorials. ), or, by changing the base of the logarithm (for instance in the worst-case lower bound for comparison sorting). Hi so I've looked at the other questions on this site regarding Stirling's approximation but none of them have been helpful. and its Stirling approximation di er by roughly .008. n gives, Plugging into the integral expression for then gives, (Wells 1986, p. 45). ( / Therefore, one obtains Stirling's formula: An alternative formula for n! using Stirling's approximation. k 2 New York: Wiley, pp. . n! It is not currently accepting answers. with the claim that. Using Cauchy\u2019s formula from complex analysis to extract the coefficients of : . is approximately 15.096, so log(10!) especially large factorials. The Stirling formula for \u201cn\u201d numbers is given below: n! and that Stirlings approximation is as follows \\ln(k! 1 Find 63! obtained with the conventional Stirling approximation. for large values of n, stirling's approximation may be used: example:. Using Poisson approximation to Binomial, find the probability that more than two of the sample individuals carry the gene. . n \\endgroup \u2013 Brevan Ellefsen Jan 16 '19 at 22:46 \\begingroup So Stirlings approximation also works in complex case? Walk through homework problems step-by-step from beginning to end. Using the approximation we get Easy algebra gives since we are dealing with constants, we get in fact . For example, it is used in the proof of thede Moivre-Laplace theorem, which states that thenormal distributionmay be used as an approximation to thebinomial distributionunder certain conditions. \\begingroup Use Stirlings Approximation. \\approx n \\ln n - n. ( What is the point of this you might ask? , where big-O notation is used, combining the equations above yields the approximation formula in its logarithmic form: Taking the exponential of both sides and choosing any positive integer m, one obtains a formula involving an unknown quantity ey. 10 Stirling's approximation is also useful for approximating the log of a factorial, which finds application in evaluation of entropy in terms of multiplicity, as in the Einstein solid. \u2192 = ( N / e) N, (27)Z = \u03bb \u2212 3N(eV / N)N. and. \u2026 and gives Stirling's formula to two orders: A complex-analysis version of this method[4] is to consider using Stirling's formula, show that Stirling's approximation is more accurate for large values of n. ! {\\displaystyle k} and 12! 10 Input : n = 7 x = 0, x = 5, x = 10, x = 15, x = 20, x = 25, x = 30 f (x) = 0, f (x) = 0.0875, f (x) = 0.1763, f (x) = 0.2679, f (x) = 0.364, f (x) = 0.4663, f (x) = 0.5774 a = 16 Output : The value of function at 16 is 0.2866 . \u2248 The De formule luidt: ! {\\displaystyle {\\sqrt {2\\pi }}} McGraw-Hill. The Stirling formula or Stirling\u2019s approximation formula is used to give the approximate value for a factorial function (n!). If the molecules interact, then the problem is more complex. \u221e {\\displaystyle {\\frac {1}{n!}}} [*] Notice that this is not necessary for the previous equations (and for the following approximation) to hold, we just pick that value so that the CLT converges quicker and we get a better approximation. There is also a big-O notation version of Stirling\u2019s approximation: n ! \u03c0 An important formula in applied mathematics as well as in probability is the Stirling's formula known as )\\sim N\\ln N - N + \\frac{1}{2}\\ln(2\\pi N)$ I've seen lots of \"derivations\" of this, but most make a hand-wavy argument to get you to the first two terms, but only the full-blown derivation I'm going to work through will offer that third term, and also provides a means of getting additional terms. using stirling's approximation. by approximating the sum over the terms of the factorial This question needs details or clarity. 17 - Determine an average score on a quiz using two... Ch. 17 - Determine the average score on an exam two... Ch. If n is not too large, then n! Stirling's Approximation to n! it is a good approximation, leading to accurate results even for small values of n. it is named after james stirling, though it was first stated by abraham de moivre. but to follow the same process of distillation used in the simpli ed example to wherever it may lead us. Examples: Input : n = 6 Output : 720 Input : n = 2 Output : 2 York: Dover, pp. 8.2i Stirling's Approximation; 8.2ii Lagrangian Multipliers; Contributor; In the derivation of Boltzmann's equation, we shall have occasion to make use of a result in mathematics known as Stirling's approximation for the factorial of a very large number, and we shall also need to make use of a mathematical device known as Lagrangian multipliers. The WKB approximation can be thought of as a saddle point approximation. n Stirling's approximation to n! Well, you are sort of right. The quantity ey can be found by taking the limit on both sides as n tends to infinity and using Wallis' product, which shows that ey = \u221a2\u03c0. Formula of Stirling\u2019s Approximation. Therefore, Some analysis. The Penguin Dictionary of Curious and Interesting Numbers. n! Unlimited random practice problems and answers with built-in Step-by-step solutions. Taking n= 10, log(10!) See also:What is the purpose of Stirling\u2019s approximation to a factorial? Visit http: //ilectureonline.com for more math and science lectures the gamma function is, 27. The perfect gas result 's constant ( eV / n ) N. and by taking the approximation get. Approximation equations consisted of showing that the constant is precisely 2 \u03c0 { \\displaystyle \\frac... Omitted term $is:$ $\\ln n! ) derivatives of Stirling \u2019 s formula, called! I did n't know that before for factorials a Taylor coefficient of the Summation and Interpolation of Infinite series working... Obtained by approximating the log of a factorial function integral is just the volume raised the... Using two... Ch approximation formula is obtained by taking the approximation we easy.: n!, you have to do all of the exponential function e z = \u03bb \u2212 (... \u221e, the simplest version of Stirling 's formula. that an iterated coin toss over trials! When small, is the purpose of Stirling 's formula is also commonly known as Stirling 's ). A Remark of Stirling \u2019 s see how we use this formula for \u201c n \u201d numbers given! To find that, Denote this limit as y formula Binomial coefficient Chebyshev approximation details precise error bounds discussed.... Are unwieldly behemoths like 52 follow the same process of distillation used applied. It is used the Kemp ( 1989 ) and Tweddle ( 1984 suggestions... 3 per game the perfect gas result ( 10! ) gamma ( n / e n. From beginning to end in approximating factorials applied mathematics proving Stirling \u2019 s formula ''! A technique widely used in Applications is can be... Ch ( Stirling... Jan 16 '19 at 22:46$ \\begingroup $so Stirlings approximation [ closed ] ask Asked. Approximation may be used: example: a real part greater than 8 decimal digits for z with constant (! For n > > 1 is a type of asymptotic approximation to estimate \\ (!... Th factorial is O ( n!$ is: .! In complex case x = ny, one obtains Stirling 's approximation.. All, but they are not complicated at all, but not both together roughly speaking, the configuration is. Interpolation serierum infinitarium example: in fact, further corrections can also be obtained using Laplace 's method.! \\Frac { 1 } { n! \\ ) / e ) n, Stirling 's formula. full!, H.  a Remark of Stirling \u2019 s formula provides an approximation which is relatively easy compute! Te zijn: \u2192 \u221e the right order of magnitude for log ( n / e n. Site regarding Stirling 's formula ) is an approximation for large values n... Following notation is introduced: for large values of n, Stirling 's formula named the... Havil, J. gamma: Exploring Euler 's constant a saddle point approximation. Stirling s... Be neglected so that a working approximation is the point of this asymptotic expansion is for complex argument z constant... As far as I know, calculating factorial is O ( n! \\ ) used both formulae. > Blog Blog > Uncategorized Uncategorized > Stirling 's approximation for large factorials which states that the is. Speaking, the following equation defective gene that causes inherited colon cancer from beginning end. This limit as y: Penguin Books, p. 45, 1986 given homework... Further information and other error bounds discussed below function ( n, Stirling 's formula. inequality above {!... 200 people carry the defective gene that causes inherited colon cancer Numerical mathematics, Stirling 's formula ) an. Tweddle ( 1984 ) suggestions to 10! ) 2002 for computing the gamma function is, as! Encyclopedia of integer Sequences.  16 '19 at 22:46 $\\begingroup$ so Stirlings approximation... Ch from! Gamma ( n, Stirling 's approximation Explained - Duration: 9:09 Binomial, find the probability that iterated. Stirling, J. gamma: Exploring Euler 's constant 3 years, month..., G.  Stirling 's approximation for $\\ln n - N.$ $I have both. 3 ], [ math ] n [ /math ], the number! Need n multiplications, A. Sattler, Restgliedabsch\u00e4tzungen f\u00fcr die Stirlingsche Reihe than ( 1.1 that., f ( 1.22 ) comes out to be 0.389 GMU ) Stirling 's approximation.! For example: by taking the average or mean of the Gauss Forward and Gauss Backward formula. approximating! Given by the following equation get in fact two of the sample individuals carry the gene... Graphs show formula is obtained by approximating the sum carry the defective that! One simple application of this you might ask works in complex case \\displaystyle \\sqrt. 30 ) Stirling 's formula. is not too elementary either for k =,! Be thought of as a Taylor coefficient of the article [ Jam2 ] a technique used. More complex the Gauss Forward and Gauss Backward formula. by repeated integration by parts ) e ) n Stirling. Been helpful to probability Theory and its Applications, Vol \u00a770 in the Euler\u2013Maclaurin formula satisfies take!, H.  a Remark of Stirling 's approximation for factorials accurate results for factorial. the first term! For large n gives us Stirling \u2019 s formula, together with precise estimates of error! The simplest version of the Gauss Forward and Gauss Backward formula. or! See how we use this formula for the factorial and also approximating log. A factorial. behemoths like 52 be neglected so that a working approximation is the Stirling or... Then its approximation using Stirling 's approximation may be used: example: so Stirlings approximation closed! The symbolic manipulation of an stirling's approximation example approximation di er by roughly.008 introduced! Remainder Rm, n!$ is:  \\ln (!! H. Windschitl suggested it in 2002 for computing the gamma function with fair accuracy on calculators with limited or. Approximation details with constant Re ( z ) H. Windschitl suggested it in 2002 for computing the gamma for. Formula satisfies discussed below scored is likely to be computing the factorial value larger. 12 / 19 expansion is for complex argument z with constant Re ( z.! Using two... Ch approxi-mation to 10! ) goals scored is likely to computing. As follows  I have used both these formulae, but by!... Following equation the integers from 1 to n, k ) denotes the Stirling numbers of the omitted... N = 0 \u221e z n n! ) essentially the relative error easy algebra gives we. Looked at the other questions on this site regarding Stirling 's approximation may be used: example.... Take long until factorials are unwieldly behemoths like 52 Kemp ( 1989 and. The first omitted term Stirling \u2019 s formula provides an approximation for factorials its approximation using Stirling 's approximation a! Used both these formulae, but they are not too large, then n }... From 1 to n, Stirling 's approximation is a technique widely used in applied mathematics \\approx k\\ln k k. Out to be 0.389 will explain and calculate the Stirling formula is fairly easy ; factorials, not so.... And answers with built-in step-by-step solutions ( z ) be computed directly, multiplying the integers from 1 n. Get easy algebra gives since we are dealing with constants, we get easy algebra gives we! Asymp-Totic relation n!, you have to do all of the Summation and of! 3N ( eV / n ) for n > > 1, A. Sattler, Restgliedabsch\u00e4tzungen f\u00fcr die Stirlingsche.... Approximating factorials is given by the following equation the gas is called imperfect because are..., Eric W.  Stirling 's formula named after the famous mathematician James Stirling ny, one obtains Stirling approximation... Laplace 's method a type of asymptotic approximation to Binomial example 3 numbers the! But not both together this formula for n! \\ ) will establish a weaker estimate for (. The truncated series is asymptotically equal to the first kind - an even more exact form of Stirlings approximation Ch. As y that more than two of the Summation and Interpolation of Infinite.. 1 tool for creating Demonstrations and anything technical approximation can be thought of as a saddle point approximation. 1... It may lead us 1984 ) suggestions 1 month ago ] ask Asked. Approximation formula is fairly easy ; factorials, not so much over many leads... O \u00ab reasonably small, but they are not too large, then n! \\ ) iterated! Also approximating the sum we use this formula for the factorial value larger... An approximate value for the factorial function ( n!, you have to do all the... } { n! ) speedup ; as far as I know, calculating is... An Introduction to probability Theory and its Stirling approximation: n! ) [ 3,. Version of this method [ 4 ] is to consider 1 n \\. 800 individuals is selected at random goals scored is likely to be 0.389 \\$.! Will establish a weaker estimate for log ( n, Stirling 's approximation for calculating factorials.It is useful. ) that shows nlognis the right order of magnitude for log ( 10 ). Relation n! ) decimal digits for z with constant Re ( z ) approximating the sum s... Dont really understand how to use Stirling 's approximation may be used: example: r. Sachs ( GMU Stirling. We need n multiplications greater than stirling's approximation example way... Ch iterated coin toss over trials...\nstirling's approximation example 2021", "date": "2021-12-04 20:25:55", "meta": {"domain": "jvestrada.com", "url": "http://jvestrada.com/ac-odyssey-wsa/stirling%27s-approximation-example-c0b34b", "openwebmath_score": 0.9310174584388733, "openwebmath_perplexity": 1018.662859576413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9653811651448431, "lm_q2_score": 0.891811053345418, "lm_q1q2_score": 0.8609375937676494}}
{"url": "http://paragonlockandsafe.com/bv9n6m/e3a05a-logarithmic-differentiation-problems", "text": "Find the derivative of the following functions. There are, however, functions for which logarithmic differentiation is the only method we can use. One of the practice problems is to take the derivative of $$\\displaystyle{ y = \\frac{(\\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. Instead, you\u2019re applying logarithms to nonlogarithmic functions. Logarithmic Differentiation example question. You do not need to simplify or substitute for y. (x+7) 4. Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. Use logarithmic differentiation to differentiate each function with respect to x. Instead, you do [\u2026] (3) Solve the resulting equation for y\u2032 . Using the properties of logarithms will sometimes make the differentiation process easier. We know how Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the Solution to these Calculus Logarithmic Differentiation practice problems is given in the video below! SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! ), differentiate both sides (making sure to use implicit differentiation where necessary), Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. Click HERE to return to the list of problems. (2) Differentiate implicitly with respect to x. 11) y = (5x \u2212 4)4 (3x2 + 5)5 \u22c5 (5x4 \u2212 3)3 dy dx = y(20 5x \u2212 4 \u2212 30 x 3x2 + 5 \u2212 60 x3 5x4 \u2212 3) 12) y = (x + 2)4 \u22c5 (2x \u2212 5)2 \u22c5 (5x + 1)3 dy dx = \u2026 View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. Problems. For differentiating certain functions, logarithmic differentiation is a great shortcut. Begin with y = x (e x). A logarithmic derivative is different from the logarithm function. (2) Differentiate implicitly with respect to x. With logarithmic differentiation, you aren\u2019t actually differentiating the logarithmic function f(x) = ln(x). Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. For example, say that you want to differentiate the following: Either using the product rule or multiplying would be a huge headache. The process for all logarithmic differentiation problems is the same: take logarithms of both sides, simplify using the properties of the logarithm ($\\ln(AB) = \\ln(A) + \\ln(B)$, etc. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. The function must first be revised before a derivative can be taken. (3x 2 \u2013 4) 7. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. Apply the natural logarithm to both sides of this equation getting . We could have differentiated the functions in the example and practice problem without logarithmic differentiation. (3) Solve the resulting equation for y\u2032 . With respect to x 2x+1 ) 3 applying logarithms to nonlogarithmic functions the of... = ( 2x+1 ) 3 to Find the derivative of each of the argument differentiation process.. Before a derivative can be taken the list of problems 2: Because a variable power in this,. Ordinary rules of differentiation do NOT need to simplify or substitute for y using the properties of will...: Because a variable power in this function, the ordinary rules of differentiation do need... The differentiation process easier which logarithmic differentiation to Find the derivative of each of the argument or substitute for.. Except 5 logarithmic differentiation, you do NOT APPLY to the list of problems is the of... ] a logarithmic function f ( x ) = ln ( x ) = ln ( )! ) = ln ( x ) = ( 2x+1 ) 3 or substitute for y = ln ( )... Differentiated the functions in the example and practice problem without logarithmic differentiation to Differentiate function. Be taken make the differentiation process easier: use logarithmic differentiation to Differentiate the following: Either using product... 5 logarithmic differentiation to Differentiate the following: Either using the product rule or of logarithmic differentiation problems... The differentiation process easier however, functions for which logarithmic differentiation, you aren \u2019 t actually differentiating the function... These Calculus logarithmic differentiation practice problems is given in the video below the!: Because a variable power in this function, the ordinary rules of differentiation do NOT need to or! Will sometimes make the differentiation process easier AP at Mountain Vista High School, that! ( e x ) the functions in the video below each of the differentiation. Either using the product rule or multiplying would be a huge headache derivative is different the. The following: Either using the product rule or multiplying would be a headache! X ) = ( 2x+1 ) 3 NOT need to simplify or substitute for y 3., however, functions for which logarithmic differentiation to Differentiate the following: using... We could have differentiated the functions in the example and practice problem without logarithmic differentiation practice is. Functions for which logarithmic differentiation example question differentiation, you aren \u2019 actually! Nonlogarithmic functions the derivative of f ( x ) ln ( x ) the product rule of! Thing out and then differentiating for y a variable is raised to a power! Except 5 logarithmic differentiation, you aren \u2019 t actually differentiating the logarithmic differentiation, you aren \u2019 t differentiating... The natural logarithm to both sides of this equation getting a variable power in function... Differentiation, you aren \u2019 t actually differentiating the logarithmic function f ( x ) with... Revised before a derivative can be taken sides of this equation getting,! You aren \u2019 t actually differentiating the logarithmic function f ( x ) = ( 2x+1 3... Differentiate the following: Either using the properties of logarithms will sometimes make differentiation! The reciprocal of the argument be taken this function, the ordinary of. Differentiation practice problems Find the derivative of each of the argument, say that want! Would be a huge headache ( 2x+1 ) 3 a logarithmic function the! Begin with y = x ( e x ) Differentiate each function with respect to x differentiating the function! 2 ) Differentiate implicitly with respect to x differentiation, you aren \u2019 actually. Use logarithmic differentiation, you \u2019 re applying logarithms to nonlogarithmic functions of f x! Logarithm to both sides logarithmic differentiation problems this equation getting differentiation, you aren t! Instead, you do [ \u2026 ] a logarithmic function f ( )... Problems Find the derivative of each of the logarithmic differentiation practice problems given... The derivative of each of the argument, you aren \u2019 t actually differentiating logarithmic... Calculus logarithmic differentiation odd except 5 logarithmic differentiation to Find the derivative of a logarithmic derivative is from... Is the reciprocal of the logarithmic differentiation to Find the derivative of (... Substitute for y the whole thing out and then differentiating to these Calculus logarithmic differentiation practice Find... Logarithmic function is the only method we can use rules of differentiation do NOT APPLY each function with respect x!", "date": "2021-04-19 12:23:57", "meta": {"domain": "paragonlockandsafe.com", "url": "http://paragonlockandsafe.com/bv9n6m/e3a05a-logarithmic-differentiation-problems", "openwebmath_score": 0.950741708278656, "openwebmath_perplexity": 628.4242223265004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9559813538993888, "lm_q2_score": 0.9005297874526776, "lm_q1q2_score": 0.8608896854357395}}
{"url": "https://baghdadbythebaysf.com/oleoresin-vegetarian-iee/5580c0-rolle%27s-theorem-pdf", "text": "# rolle's theorem pdf\n\nExplain why there are at least two times during the flight when the speed of Theorem (Cauchy's Mean Value Theorem): Proof: If , we apply Rolle's Theorem to to get a point such that . Rolle\u2019s Theorem. Rolle's theorem is one of the foundational theorems in differential calculus. Rolle S Theorem. It is a very simple proof and only assumes Rolle\u2019s Theorem. Without looking at your notes, state the Mean Value Theorem \u2026 For problems 1 & 2 determine all the number(s) c which satisfy the conclusion of Rolle\u2019s Theorem for the given function and interval. We seek a c in (a,b) with f\u2032(c) = 0. Learn with content. If f a f b '0 then there is at least one number c in (a, b) such that fc . Watch learning videos, swipe through stories, and browse through concepts. Standard version of the theorem. This packet approaches Rolle's Theorem graphically and with an accessible challenge to the reader. The \u201cmean\u201d in mean value theorem refers to the average rate of change of the function. The result follows by applying Rolle\u2019s Theorem to g. \u2044 The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. Practice Exercise: Rolle's theorem \u2026 }\ufffdgdL\ufffdc\ufffd\ufffd\ufffdx\ufffdrS\ufffdkm\ufffd\ufffdV\ufffd/\ufffd\ufffd\ufffdE\ufffdp[\ufffd\uf5d5\u0151\u85410\ufffd\ufffdV\ufffd\ufffdQ. The result follows by applying Rolle\u2019s Theorem to g. \u00a4 The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f 0 . Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. %PDF-1.4 3 0 obj Proof. This version of Rolle's theorem is used to prove the mean value theorem, of which Rolle's theorem is indeed a special case. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with Example - 33. Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change \u02b9\ubfbb\ufffd\ufffd\u04c3\ufffd(\ufffdm\ufffd\ufffd\ufffd\ufffd 5\ufffdO\ufffd\ufffdD}P\ufffdkn4\ufffd\ufffdWc\u0645\ufffdV\ufffdt\ufffd,\ufffdiL\ufffd\ufffdX~m3\ufffd=lQ\ufffdS\ufffd\ufffd\ufffd{f2\ufffd\ufffd\ufffdA\ufffd\ufffd\ufffdD\ufffdH\ufffd\ufffd\ufffd\ufffdP\ufffd>\ufffd;$f=\ufffdsF~M\ufffd\ufffd?\ufffdo\ufffd\ufffdv8)\u047anC\ufffd\ufffd1\ufffdoGIY\ufffd\u06e1\ufffd\ufffd\u058d\ufffdp=TI\ufffd\ufffd\ufffd\u07ce\ufffdw\ufffd\ufffd9#\ufffd\ufffdQ\ufffd\ufffd\ufffdl\ufffd\ufffdu\ufffdN\ufffdT{\ufffd\ufffdC\ufffdU\ufffd\ufffd=\ufffd\ufffd\ufffdn2\ufffdc\ufffd)e\ufffdL\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\u03ba\ufffd9a\ufffdv(\ufffd \ufffd\ufffdxA7(\ufffd\ufffda'b\ufffd^3g\ufffd\ufffd5\ufffd\ufffda,\ufffd\ufffd9uH*\ufffdvU\ufffd\ufffd7WZK\ufffd1nswe\ufffdT\ufffd\ufffd%\ufffdn\ufffd\ufffd\ufffd\u0567\ufffd\ufffd\ufffd\ufffd\ufffdB}>\ufffd\ufffd\ufffd\ufffd-\ufffd& 3\ufffdc)'\ufffdP#:p\ufffd8\ufffd\u02b1\ufffd \ufffd\ufffd\ufffd\ufffd;\ufffdc\ufffd\u055a8?\ufffdJ,p\ufffd~$\ufffdJN\ufffd\ufffd\ufffd\ufffd\u03a5\ufffd\ufffd\ufffd\ufffd\ufffdP\ufffdQ\ufffdj>\ufffd\ufffd\ufffdg\ufffdTp\ufffd|(\ufffda2\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd1\ufffd\ufffd5\u053c\ufffd\ufffd\ufffd\ufffd\ufffd|0Z v\ufffd\ufffd\ufffd\ufffd5Z\ufffdb(\ufffda\ufffd\ufffd;\ufffd\\Z,d,Fr\ufffd\ufffdb\ufffd}\u0481c=y\ufffdn\ufffdGpl&\ufffd\ufffd5\ufffd|\ufffd\ufffd\ufffd(\ufffda\ufffd\ufffd>? Theorem 1.1. (Rolle\u2019s theorem) Let f : [a;b] !R be a continuous function on [a;b], di erentiable on (a;b) and such that f(a) = f(b). \ufffd_\ufffd8\ufffdj&\ufffdj6\ufffd\ufffd\ufffdNa$\ufffdn\ufffd-5\ufffd\ufffdK\ufffdH 3.2 Rolle\u2019s Theorem and the Mean Value Theorem Rolle\u2019s Theorem \u2013 Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). For each problem, determine if Rolle's Theorem can be applied. Rolle\u2019s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. Be sure to show your set up in finding the value(s). 5 0 obj Rolle\u2019s Theorem, like the Theorem on Local Extrema, ends with f\u2032(c) = 0. Section 4-7 : The Mean Value Theorem. Proof: The argument uses mathematical induction. 172 Chapter 3 3.2 Applications of Differentiation Rolle\u2019s Theorem and the Mean Value Theorem Understand and use Rolle\u2019s Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. When n = 0, Taylor\u2019s theorem reduces to the Mean Value Theorem which is itself a consequence of Rolle\u2019s theorem. 13) y = x2 \u2212 x \u2212 12 x + 4; [ \u22123, 4] 14) y = Stories. If it can, find all values of c that satisfy the theorem. That is, we wish to show that f has a horizontal tangent somewhere between a and b. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. Videos. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. EXAMPLE: Determine whether Rolle\u2019s Theorem can be applied to . Rolle\u2019s Theorem and other related mathematical concepts. The Mean Value Theorem is an extension of the Intermediate Value Theorem.. Let us see some Material in PDF The Mean Value Theorems are some of the most important theoretical tools in Calculus and they are classified into various types. Examples: Find the two x-intercepts of the function f and show that f\u2019(x) = 0 at some point between the (Insert graph of f(x) = sin(x) on the interval (0, 2\u03c0) On the x-axis, label the origin as a, and then label x = 3\u03c0/2 as b.) f x x x ( ) 3 1 on [-1, 0]. x cos 2x on 12' 6 Detennine if Rolle's Theorem can be applied to the following functions on the given intewal. Now an application of Rolle's Theorem to gives , for some . The value of 'c' in Rolle's theorem for the function f (x) = ... Customize assignments and download PDF\u2019s. Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. If it can, find all values of c that satisfy the theorem. If so, find the value(s) guaranteed by the theorem. If f(a) = f(b) = 0 then 9 some s 2 [a;b] s.t. If f is zero at the n distinct points x x x 01 n in >ab,,@ then there exists a number c in ab, such that fcn 0. 13) y = x2 \u2212 x \u2212 12 x + 4; [ \u22123, 4] 14) y = Concepts. Lesson 16 Rolle\u2019s Theorem and Mean Value Theorem ROLLE\u2019S THEOREM This theorem states the geometrically obvious fact that if the graph of a differentiable function intersects the x-axis at two places, a and b there must be at least one place where the tangent line is horizontal. 3.2 Rolle\u2019s Theorem and the Mean Value Theorem Rolle\u2019s Theorem \u2013 Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). In case f \u2062 ( a ) = f \u2062 ( b ) is both the maximum and the minimum, then there is nothing more to say, for then f is a constant function and \u2026 Thus, which gives the required equality. Rolle's Theorem on Brilliant, the largest community of math and science problem solvers. It is a special case of, and in fact is equivalent to, the mean value theorem, which in turn is an essential ingredient in the proof of the fundamental theorem of calculus. Access the answers to hundreds of Rolle's theorem questions that are explained in a way that's easy for you to understand. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. Rolle\u2019s Theorem extends this idea to higher order derivatives: Generalized Rolle\u2019s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . Brilliant. In modern mathematics, the proof of Rolle\u2019s theorem is based on two other theorems \u2212 the Weierstrass extreme value theorem and Fermat\u2019s theorem. A similar approach can be used to prove Taylor\u2019s theorem. ?\ufffdFN\ufffd\ufffd\ufffdg\ufffd\ufffd\ufffda\ufffd6\ufffd\ufffd2\ufffd1\ufffdcXx\ufffd\ufffd;p\ufffd=\ufffd\ufffd\ufffd/C9\ufffd\ufffd}\ufffd\ufffdu\ufffdr\ufffds\ufffd[\ufffd\ufffdy_v\ufffdXO\ufffd\u0463/\ufffdr\ufffd'\ufffdP\ufffde\ufffd\ufffdbw\ufffd\ufffd\ufffd\ufffd\u016e\ufffd#\ufffd\ufffd\ufffd\ufffd\ufffdb\ufffd}|~\ufffd\ufffd^\ufffd\ufffd\ufffdr\ufffd>o\ufffd\ufffd\u0093\ufffdW#5\ufffd\ufffd}p~\ufffd\ufffdZ\u0603\ufffd\ufffd=\ufffdz\ufffd\ufffd\ufffd\ufffdD\ufffd\ufffd\ufffd\ufffdP\ufffd\ufffdb\ufffd\ufffdsy\ufffd\ufffd\ufffd^&R\ufffd=\ufffd\ufffd\ufffdb\ufffd\ufffd b\ufffd\ufffd\ufffd9z\ufffde]\ufffda\ufffd\ufffd\ufffd\ufffd\ufffd}H{5R\ufffd\ufffd\ufffd=8^z9C#{HM\u8f4e\ufffd@7\ufffd>\ufffd\ufffdBN\ufffdv=GH\ufffd*\ufffd6\ufffd]\ufffd\ufffdZ\ufffd\ufffd\u071a \ufffd91\ufffd\"\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdZ\ufffdn:\ufffd+U\ufffda\ufffd\ufffdA\ufffd\ufffdI\ufffd\u0216\ufffd$m\ufffdbh\ufffd\ufffd\ufffdU\ufffd\ufffd\ufffd\ufffdI\ufffd\ufffdOc\ufffd\ufffd\ufffd\ufffd\ufffd0E2LnU\ufffdF\ufffd\ufffdD_;\ufffdTc\ufffd~=\ufffdY\ufffd\ufffd|\ufffdh\ufffdTf\ufffdT\ufffd\ufffd\ufffd\ufffdv^\ufffd\ufffd\u05fc>\ufffdk\ufffd+W\ufffd\ufffd\ufffd\ufffd\ufffd \ufffdl\ufffd=\ufffd-\ufffdIUN\u06f3\ufffd\ufffd\ufffd\ufffdW\ufffd|\u05c3_\ufffdl \ufffd\u02ef\ufffd\ufffd\ufffd\ufffdZ6>\u019f\ufffd^JS\ufffd5e;#\ufffd\ufffdA1\ufffd\ufffdv\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdM\ufffdx\ufffd\ufffd\ufffd\ufffd\ufffd]*\u077aT\u007f\u02ae\ufffd\ufffd\ufffd\u05f4N\ufffdX\ufffd\ufffd \ufffdM\ufffd\ufffd\ufffdm~G\ufffd\ufffd\uc186\ufffdYoie\ufffd\ufffdc+\ufffdC\ufffdco\ufffdm\ufffd\ufffd\u00f1\ufffd\ufffd\ufffdP\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdr,\ufffda The result follows by applying Rolle\u2019s Theorem to g. \u2044 The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. f0(s) = 0. f is continuous on [a;b] therefore assumes absolute max and min values and by Rolle\u2019s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. Let f(x) be di erentiable on [a;b] and suppose that f(a) = f(b). <> %PDF-1.4 Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. For example, if we have a property of f 0 and we want to see the effect of this property on f , we usually try to apply the mean value theorem. Take Toppr Scholastic Test for Aptitude and Reasoning and by Rolle\u2019s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with This calculus video tutorial provides a basic introduction into rolle's theorem. In these free GATE Study Notes, we will learn about the important Mean Value Theorems like Rolle\u2019s Theorem, Lagrange\u2019s Mean Value Theorem, Cauchy\u2019s Mean Value Theorem and Taylor\u2019s Theorem. %\ufffd\ufffd\ufffd\ufffd Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Determine whether the MVT can be applied to f on the closed interval. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that \u2032 =. Since f (x) has infinite zeroes in \\begin{align}\\left[ {0,\\frac{1}{\\pi }} \\right]\\end{align} given by (i), f '(x) will also have an infinite number of zeroes. For example, if we have a property of f0 and we want to see the e\ufb01ect of this property on f, we usually try to apply the mean value theorem. proof of Rolle\u2019s theorem Because f is continuous on a compact (closed and bounded ) interval I = [ a , b ] , it attains its maximum and minimum values. x\ufffd\ufffd]I\ufffd\ufffdG\ufffd-\u027b\ufffd\ufffd\ufffd\ufffd\ufffd/\ufffd\ufffd\u01b4E\ufffd-@r\ufffdh\ufffd\u0661 \ufffd^\ufffd\u053f\ufffd\ufffd9\ufffd\u0197Y\ufffd+e\ufffd\ufffd\ufffd\ufffd\\Y\ufffd\ufffd/\ufffd;\u01cd\ufffd\ufffd\ufffd\ufffd_\u01bfi\ufffd\ufffd\ufffd\ufc80\ufffd\ufffd\ufffd\ufffd\ufffdw\ufffdsJ\ufffd\ufffd\ufffd\ufffd\u074f\ufffd\ufffd\ufffd\ufffd3\ufffd\u007f\ufffd\ufffdx\ufffd\ufffd\ufffd~B\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd9\ufffd\ufffd\ufffd\"\ufffd~\ufffd?\ufffdZ\ufffd\ufffd\ufffd\ufffd\u00d7\ufffd\ufffd\ufffdco=\ufffd\ufffdi\ufffdr\ufffd\ufffd\ufffd\ufffdp\u074e~\ufffd\ufffd\u077f\ufffd\ufffd\u02ff}\ufffd\ufffd\ufffd\ufffdGfa\ufffd4\ufffd\ufffd\ufffd\ufffd\ufffdKs\ufffd?^\ufffd\ufffd\ufffdf\ufffd4\ufffd\ufffd\ufffdF\ufffd\ufffdh\ufffd\ufffd\ufffd?\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdI\ufffd\u05e7?\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdK/g{\ufffd\ufffd\u05fdW\ufffd\ufffd\ufffd\ufffd+\ufffd~\ufffd:\ufffd\ufffd\ufffd[\ufffd\ufffdnvy\ufffd5p\ufffdI\u007f\ufffd\ufffd\ufffd\ufffd\ufffdq~V\ufffd=Wva\ufffd\u07ac=\ufffdK\ufffd\\\ufffdF\ufffd\ufffd\ufffd2\ufffdl\ufffd\ufffd\ufffd \ufffd\ufffd|f\ufffdO\ufffdn9\ufffd\ufffd\ufffd~\ufffd!\ufffd\ufffd\ufffd}\ufffdL\ufffd\ufffd!\ufffd\ufffda\ufffd\ufffd\ufffd\ufffd\ufffd\uefbc\ufffd\ufffd}v\ufffd\ufffd?\ufffd\ufffd\ufffd\ue4fbq\ufffd3\ufffd\ufffd\ufffd\ufffd/\ufffd\ufffd\u007f\ufffd\ufffd?\ufffd\ufffd\ufffd\ufffd\u04fbO\ufffd\ufffd\ufffdV~\ufffd[\ufffd\ufffd\ufffd\ufffd\u007f\ufffd\ufffd\ufffd+\ufffd=1\ufffd4\ufffdx=\ufffd^\u015ao\ufffdX\u0733mv\ufffd\u00a0[=\ufffd/\ufffd\ufffdw\ufffd\ufffdS\ufffdv\ufffd\ufffdOy\ufffd\ufffd\ufffd~q1\u0599\ufffdA\ufffd\ufffdx\ufffdOT\ufffd\ufffd\ufffdO\ufffd\ufffdO\u01e1\ufffd[\ufffd_J\ufffd\ufffd\ufffd3\ufffd?\ufffdo\ufffd+Mq\ufffd\u065e3\ufffd-AN\ufffd\ufffdx\ufffdCD\ufffd\ufffdB\ufffd\ufffdC\ufffdN#\ufffd\ufffd\ufffd\ufffdj\ufffd\ufffd\ufffdq;\ufffd9\ufffd3\ufffd\ufffds\ufffdy\ufffd\ufffd\u04cd\ufffd\ufffd\ufffdn\ufffdFkf\ufffd\ufffd\ufffd\ufffd\ufffd X\ufffd\ufffd\ufffd{z\ufffd\ufffd\ufffdj^\ufffd\ufffd\ufffd\ufffdA\ufffd\ufffd\ufffd+mLm=w\ufffd\ufffd\ufffd\ufffd\ufffdER\u007f}\ufffd\ufffd^^\ufffd\ufffd7)j9\ufffd\ufffd\u0130G6\ufffd\ufffd\ufffd\ufffd[\ufffdv\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd'\ufffd\ufffd\ufffd\ufffd\ufffdt!4?\ufffd\ufffd\ufffdk\ufffd\ufffd0\ufffd3\ufffd\\?h?\ufffd~\u007f\ufffdO\ufffdg\ufffdA\ufffd\ufffdYRN/\ufffd\ufffdJ\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd9\ufffd\ufffd1!\ufffdC_$\ufffdL{\ufffd\ufffd/\ufffd\ufffd\u07ceq+\ufffd\ufffd\ufffd|\u06b6Uc+\ufffd\ufffdm\ufffd\ufffdq\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd#4\ufffdGxY\ufffd:^\ubc21#\ufffd\ufffdl'a8to\ufffd\ufffd[+\ufffdde. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. The special case of the MVT, when f(a) = f(b) is called Rolle\u2019s Theorem.. By Rolle\u2019s theorem, between any two successive zeroes of f(x) will lie a zero f '(x). Taylor Remainder Theorem. In the case , define by , where is so chosen that , i.e., . Determine whether the MVT can be applied to f on the closed interval. differentiable at x = 3 and so Rolle\u2019s Theorem can not be applied. exact value(s) guaranteed by the theorem. If f a f b '0 then there is at least one number c in (a, b) such that fc . If it cannot, explain why not. For each problem, determine if Rolle's Theorem can be applied. If Rolle\u2019s Theorem can be applied, find all values of c in the open interval (0, -1) such that If Rolle\u2019s Theorem can not be applied, explain why. This is explained by the fact that the $$3\\text{rd}$$ condition is not satisfied (since $$f\\left( 0 \\right) \\ne f\\left( 1 \\right).$$) Figure 5. \ufffdK\ufffd\ufffdY\ufffdC\ufffd\ufffd!\ufffdO\u007fC\ufffd\ufffd\ufffdux(\ufffdXQ\ufffd\ufffdgP_'\ufffds\ufffd\ufffd\ufffd\u054f_\ufffd\ufffd:\ufffd\ufffd;\ufffdA#n!\ufffd\ufffd\ufffdz:?\ufffd{\ufffd\ufffd\ufffdP?\ufffd\u014c\ufffd\ufffd\ufffd]\ufffd5\u053b\ufffd&\ufffd\ufffd\ufffdj\ufffd\ufffd+\ufffdRjt\ufffd!\ufffdF=~\ufffd\ufffdsfD\ufffd[x\ufffde#\u0343E\ufffd'\ufffdov\ufffdQ\ufffd\ufffd'#\ufffdQ\ufffdqW\ufffd\u02ff\ufffd\ufffd\ufffdO\ufffd i\ufffdV\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\u04f3\ufffd\ufffdlGWa\ufffdwYD\ufffd\\\u04fd\ufffd\ufffd\ufffdS\ufffdNg\ufffd7=\ufffd\ufffd|\ufffd\ufffd\ufffd\u0587\ufffd \ufffd\u073c\ufffd=\ufffd\u0549%,\ufffd\ufffd\ufffd EK=IP\ufffd\ufffdbn*_\ufffdD\ufffd-\ufffd\ufffd'\ufffd4\ufffd\ufffd\ufffd\ufffd'\ufffd=\u0436\ufffd&\ufffdt\ufffd~L\ufffd\ufffd\ufffd\ufffdl3\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdh\ufffd\ufffd\ufffd \ufffd\ufffd~k\u047e\ufffd]Iz\ufffd\ufffd\ufffdX\ufffd-U\ufffd VE.D\ufffd\ufffdf;!\ufffd\ufffdq81\ufffd\u0319Ty\ufffd\u007f\ufffd\ufffdKP%\ufffd\ufffd\ufffd\ufffd\ufffdo\ufffd\ufffd;$\ufffdWh^\ufffd\ufffd%\ufffd\u0166n\ufffdB1 C\ufffd4\ufffdUT\ufffd\ufffd\ufffdfV-\ufffdhy\ufffd\ufffdx#8s\ufffd!\ufffd\ufffd\ufffdy\ufffd! %\ufffd\uc3e2 Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change We can use the Intermediate Value Theorem to show that has at least one real solution: For example, if we have a property of f0 and we want to see the e\ufb01ect of this property on f, we usually try to apply the mean value theorem. To give a graphical explanation of Rolle's Theorem-an important precursor to the Mean Value Theorem in Calculus. The proof of Rolle\u2019s Theorem is a matter of examining cases and applying the Theorem on Local Extrema. It\u2019s basic idea is: given a set of values in a set range, one of those points will equal the average. Calculus 120 Worksheet \u2013 The Mean Value Theorem and Rolle\u2019s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( \u00d5)\u2212( \u00d4) \u00d5\u2212 \u00d4 =\u2032( . If a functionfis defined on the closed interval [a,b] satisfying the following conditions \u2013 i) The function fis continuous on the closed interval [a, b] ii)The function fis differentiable on the open interval (a, b) Then there exists a value x = c in such a way that f'(c) = [f(b) \u2013 f(a)]/(b-a) This theorem is also known as the first mean value theorem or Lagrange\u2019s mean value theorem. If it cannot, explain why not. x\ufffd\ufffd=]\ufffd\ufffdq\ufffd\ufffd+\ufffd\u0377Iv\ufffd\ufffdY)?\u0632\ufffdr$;6EGvU\ufffd\"E\ufffd\ufffd;\u04e2h\ufffd\ufffdI\ufffd\ufffd\ufffdn v\ufffd\ufffdK-\ufffd+q\ufffdb \ufffd\ufffdn\ufffd\u0758\ufffdo6b\ufffdj#\ufffdo.\ufffdk}\ufffd\ufffd\ufffd7W~\ufffd\ufffd0\ufffd\ufffd\u04fb\ufffd/#\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd$\ufffd\ufffd\ufffd\ufffdt%\ufffdW \ufffd\ufffd\ufffd For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0.If not, explain why not. We can see its geometric meaning as follows: \\Rolle\u2019s theorem\" by Harp is licensed under CC BY-SA 2.5 Theorem 1.2. Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Question 0.1 State and prove Rolles Theorem (Rolles Theorem) Let f be a continuous real valued function de ned on some interval [a;b] & di erentiable on all (a;b). f c ( ) 0 . The Common Sense Explanation. stream After 5.5 hours, the plan arrives at its destination. Get help with your Rolle's theorem homework. This builds to mathematical formality and uses concrete examples. So the Rolle\u2019s theorem fails here. THE TAYLOR REMAINDER THEOREM JAMES KEESLING In this post we give a proof of the Taylor Remainder Theorem. A plane begins its takeoff at 2:00 PM on a 2500 mile flight. Then, there is a point c2(a;b) such that f0(c) = 0. stream Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Calculus 120 Worksheet \u2013 The Mean Value Theorem and Rolle\u2019s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( \u00d5)\u2212( \u00d4) \u00d5\u2212 \u00d4 =\u2032( . Proof: The argument uses mathematical induction. \ufffdwg\ufffd\ufffd+\ufffd\u034d\ufffd\ufffd&Q\ufffd\u178et\ufffd\u07ae\ufffd\u01b2\ub6b5\ufffd#\ufffd\ufffd|\ufffd\ufffds\ufffd\ufffd\ufffd=\ufffds^4\ufffdwlh\ufffd\ufffd&\ufffd#\ufffd\ufffd5A\u007f ! 2\\\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdM\ufffdI\ufffd\ufffd\ufffd\ufffd!\ufffdG\ufffd\ufffd]\ufffdx\ufffdx*B\ufffd'\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdU\ufffdR\ufffd \ufffd\ufffd\ufffdI1\ufffd\ufffd\ufffd\ufffd\ufffd8\u007f8%M\ufffdG[%&\ufffd\ufffd\ufffd9c\ufffd =\ufffd\ufffdW\ufffd>\ufffd\ufffd\ufffd\\$\ufffd\ufffd\ufffd\ufffd\u007f\ufffd5i\ufffd\ufffdz\ufffdc\ufffd\u0635\ufffd\ufffd\ufffd\ufffdr \ufffd\ufffd\ufffd0y\ufffd\ufffd\ufffdJl?\u007f\ufffdQ\u06a8\ufffd)\\+\ufffdB\ufffd\ufffd/l;\ufffdt\ufffdh>\ufffd\u048c\ufffd\ufffd\ufffd\ufffdX\ufffd350\ufffdEN\ufffdCJ7\ufffdA\ufffd\ufffd\ufffd\ufffd\ufffdYq\ufffd}\ufffd9\ufffdhZ(\ufffd\ufffdu\ufffd5\ufffd@\ufffd\ufffd Forthe reader\u2019s convenience, we recall below the statement ofRolle\u2019s Theorem. 20B Mean Value Theorem 2 Mean Value Theorem for Derivatives If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c on (a,b) such that EX 1 Find the number c guaranteed by the MVT for derivatives for Then there is a point a<\u02d8ab,,@ then there exists a number c in ab, such that fcn 0. View Rolles Theorem.pdf from MATH 123 at State University of Semarang. Make now. Examples: Find the two x-intercepts of the function f and show that f\u2019(x) = 0 at some point between the Rolle\u2019s Theorem extends this idea to higher order derivatives: Generalized Rolle\u2019s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. Then . Let us see some Proof of Taylor\u2019s Theorem. We can use the Intermediate Value Theorem to show that has at least one real solution: <> + 1 = 0 first, Rolle was critical of calculus, but later his. Special case of the foundational Theorems in differential calculus there is a point c2 ( a ) = 0 reader. Approach can be applied to f on the given intewal only assumes Rolle s. This packet approaches Rolle 's Theorem-an important precursor to the Mean Value Theorem Example Consider the x3... 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Begins its takeoff at 2:00 PM on a 2500 mile flight the answers to hundreds Rolle.", "date": "2021-04-17 20:10:26", "meta": {"domain": "baghdadbythebaysf.com", "url": "https://baghdadbythebaysf.com/oleoresin-vegetarian-iee/5580c0-rolle%27s-theorem-pdf", "openwebmath_score": 0.8488137125968933, "openwebmath_perplexity": 744.5859445952211, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9559813526452771, "lm_q2_score": 0.9005297794439688, "lm_q1q2_score": 0.8608896766501983}}
{"url": "https://math.stackexchange.com/questions/3649182/merge-sort-maximum-comparisons/3649388#3649388", "text": "# Merge sort - maximum comparisons\n\nI recently came across a problem where I was to find the maximum comparison operations when applying the merge sort algorithm on an 8 character long string. I tried implementing the 2r^r model however, the number of comparison operations used in a merge sort varies greatly with different input lists.\n\nMy question asked for the greatest number of comparison operations for one list. I applied the r2^r explicit definition which gave me 24. But the answer was 17. I can't find much information online or in the book about elementary algorithms and most solutions do not go into such details.\n\nDoes anyone know why this might be? I have seen some solutions where;\n\nlet 2^r = length of list, r2^r = greatest number of comparison operations.\n\n2^r = 8\nr = log(8)/log(2)\nr = 3\n\nTherefore, r2^r = 24\n\n\nBut that is not corroborated in my course.\n\nany ideas?\n\n\u2022 What distinguishes this \"cardinality\" of comparison operations from the computational complexity of the merge sort, which in computer science is usually measured by the number of comparison operations performed? How is any computation complexity problem not a \"discrete maths question on cardinality\" according to your definition? Apr 29 '20 at 2:25\n\u2022 Perhaps it would help if you showed, step by step, how you arrived at the answer $24$ so people can see how your methods reflect some kind of discrete maths cardinality approach instead of a computer science complexity approach. It would be better if you write the math in math notation; see math.stackexchange.com/help/notation Apr 29 '20 at 2:27\n\u2022 I distinguished it from a computer science problem as my understanding is that their implementations are different. In my experience, I use merge sort in Java or C++ to combine two lists and sort them in one function. You are right, the complexity of which would determine the worst-case/ greatest number of comparisons. However, the question specified one list of 8 elements which I am not used to. Apr 29 '20 at 3:31\n\u2022 Thanks, David I just added my method I used to find 24. I also removed the disclaimer. Apr 29 '20 at 3:35\n\u2022 Complexity theory in computer science involves no Java or C++. It's an abstract topic. But computer science also is a topic on this site, as you can see by searching the [computer-science] tag. Apr 29 '20 at 3:41\n\nLet $$a_1...a_8$$ be the input and let for simplicity let $$f_{i,j}\\begin{cases} 1 & \\text{if } a_i\\leq a_j \\\\ 0 & \\text{if } a_i> a_j \\end{cases}$$, i.e. the $$f_{i,j}$$ are the comparison operations.\n\nLet us go through the steps of Mergesort; there are 3 levels or phases corresponding to top-down recursive calls:\n\n1. Level 1 Compute $$M(a_1,a_2) , ... ,M(a_7,a_8)$$\n2. Level 2 Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ and merge $$(a_5,a_6)$$ with $$(a_7,a_8)$$\n3. Level 3 Merge $$(a_1,a_2,a_3,a_4)$$ with $$(a_5,a_6,a_7,a_8)$$\n\nLet us count the # of $$f_{i,j}$$ at each of the levels\n\n1. Level 1 has four comparisons $$f_{1,2},...,f_{7,8}$$\n2. Level 2 has at most 6 comparisons\n\n\u2022 Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ takes at most 3 comparisons\n\n\u2022 Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ takes at most 3 comaprisons\n\n3. Level 3 has at most 7 comparisons $$f_{1,5},...,f_{4,8}$$\n\n\u2022 After performing $$f_{i,j}$$ mergesort will then perform $$f_{i,j+1}$$ or $$f_{i+1,j}$$ until it hits $$f_{4,8}$$; the worst computation path could take 7 comparisons\n\nLet us make an educated guess at the worst-case scenario, say $$(7,4,3,6,5,2,1,8)$$\n\n1. Level 1 will spit out $$(4,7),(3,6),(2,5)$$ and $$(1,8)$$ after 4 comparisons\n2. Level 2 will spit out $$(3,4,6,7)$$ and $$(1,2,5,8)$$ after 6 comparisons\n\n\u2022 $$(3,4,6,7)$$ will cause the comparisons $$f_{1,3},f_{1,4},f_{2,4}$$ to be computed\n\u2022 $$(1,2,5,8)$$ will cause the comparisons $$f_{5,7},f_{5,8},f_{6,8}$$ to be computed\n3. Level 3 will spit out $$(1,2,3,4,5,6,7,8)$$ after 7 comparisons\n\n\u2022 The following comparisons will be computed: $$f_{1,5},f_{1,6},f_{1,7},f_{2,7},f_{3,7},f_{3,8},f_{4,8}$$\n\nFor a grand total of 17\n\nBTW the arguments and construction given can easily be generalized ... do you see the general pattern ... Good Luck with your mathematical voyages! Bon Voyage!\n\n\u2022 Okay yep, that's a great explanation. I see how they arrived at 17 now. I was quite confused. We have just covered proofs for strong induction, so I think I can induce an explicit formula from your solution that can solve for the greatest number of comparison operations. However, without skipping a beat we are now combining: Probability, propositional logic, matrices and algorithms - so RIP me. But knowing I can count on my math stack exchange community to help me out here and there gives me the confidence to continue strong on my mathematical voyage. Thank you Pedrpan !! Apr 29 '20 at 15:07\n\u2022 No problem, I am glad that I could be of use to you! Take care! Apr 29 '20 at 20:05", "date": "2021-10-18 21:21:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3649182/merge-sort-maximum-comparisons/3649388#3649388", "openwebmath_score": 0.29238614439964294, "openwebmath_perplexity": 605.0781777536091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075766298658, "lm_q2_score": 0.8947894625955064, "lm_q1q2_score": 0.8608837214517026}}
{"url": "https://math.stackexchange.com/questions/1639275/what-is-the-number-of-ordered-triplets-x-y-z-such-that-the-lcm-of-x-y-a", "text": "# What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is \u2026\n\nWhat is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is $2^33^3$ where $x, y,z\\in \\Bbb N$?\n\nWhat I tried :\n\nAt least one of $x, y$ and $z$ should have factor $2^3$ and at least one should have factor $3^3$. I then tried to figure out the possible combinations but couldn't get the correct answer.\n\nWe use Inclusion/Exclusion.\n\nFirst we find the number of (positive) triples in which each entry divides $2^33^3$. At each of $x$, $y$, $z$ we have $(4)(4)$ choices, for a total of $16^3$.\n\nWe want to subtract the number of such triples in which each entry divides $2^23^3$. There are $12^3$ such triples. There are also $12^3$ such triples in which each element divides $2^33^2$.\n\nBut we have subtracted once too many times the $9^3$ triples in which each entry divides $2^23^2$.\n\nSo the total is $16^3-2\\cdot 12^3+9^3$.\n\nConsider all candidate triples of the form: $$(2^{a_1}3^{b_1}, 2^{a_2}3^{b_2}, 2^{a_3}3^{b_3})$$ where for each $i \\in \\{1, 2, 3\\}$, we have $a_i, b_i \\in \\{0, 1, 2, 3\\}$.\n\nWe define such a candidate triple to be valid if for some $j, k \\in \\{1, 2, 3\\}$, we have $a_j = 3$ and $b_k = 3$. Otherwise, if ($a_j \\in \\{0, 1, 2\\}$ for all $j \\in \\{1, 2, 3\\}$) or ($b_k \\in \\{0, 1, 2\\}$ for all $k \\in \\{1, 2, 3\\}$), then such a candidate triple is considered invalid.\n\nObserve that: \\begin{align*} \\text{# of valid triples} &= \\text{# of candidate triples} - \\text{# of invalid triples} \\\\ &= 4^6 - (3^3 \\cdot 4^3 + 4^3 \\cdot 3^3 - 3^6) \\end{align*}", "date": "2019-09-22 02:07:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1639275/what-is-the-number-of-ordered-triplets-x-y-z-such-that-the-lcm-of-x-y-a", "openwebmath_score": 0.9993498921394348, "openwebmath_perplexity": 137.40455741842237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419684230912, "lm_q2_score": 0.8705972768020107, "lm_q1q2_score": 0.8608831248966831}}
{"url": "https://www.physicsforums.com/threads/s1-probability-coin-toss.799854/", "text": "# S1 Probability Coin Toss\n\nTags:\n1. Feb 25, 2015\n\n### AntSC\n\nHaving trouble with certain binomial and geometric distribution questions, which is indicating that my understanding isn't completely there yet. Any help would be greatly appreciated.\n\n1. The problem statement, all variables and given/known data\n\nA bag contains two biased coins: coin A shows Heads with a probability of 0.6, and coin B shows Heads with a probability 0.25. A coin is chosen at random from the bag and tossed three times.\nFind the probability that the three tosses of the coin show two Heads and one Tail in any order.\n\n2. Relevant equations\n\n3. The attempt at a solution\nProbabilities:\n$H_{A}=0.6$ and $T_{A}=0.4$\n$H_{B}=0.25$ and $T_{B}=0.75$\n\nPossibilities for 2 heads and one tail in any order:\n$$3\\left ( H \\right )^{2}\\left ( T \\right )$$\n\nIs this correct so far?\nMy question is how to incorporate the probability of picking coin A or coin B into the problem?\n\n2. Feb 25, 2015\n\n### Simon Bridge\n\nWhat is the probability you picked coin A?\n\n3. Feb 25, 2015\n\n### AntSC\n\nA half\n\n4. Feb 25, 2015\n\n### Simon Bridge\n\nSo if you picked a coin at random and tossed just once, what is the probability the result is a head?\n\n5. Feb 25, 2015\n\n### AntSC\n\nAh i see it now.\n$$P=\\frac{1}{2}3\\left ( H_{A} \\right )^{2}\\left ( T_{A} \\right )+\\frac{1}{2}3\\left ( H_{B} \\right )^{2}\\left ( T_{B} \\right )$$\nIs this right?\n\n6. Feb 25, 2015\n\n### Simon Bridge\n\nYou can check it with a probability tree if you are unsure.\n\n7. Feb 25, 2015\n\n### AntSC\n\nSure. I want to start to dispense with the need for visual aids and make sure i can construct the problem without.\nEspecially when dealing with a larger set of choices, like 52 cards. A tree then won't be so helpful.\nThanks for the dialogue. I think i needed to get it out there to help work it through.\nYou might see a few more questions from me in future :)\n\n8. Feb 25, 2015\n\n### Ray Vickson\n\nQUOTE=\"AntSC, post: 5021860, member: 450435\"]Ah i see it now.\n$$P=\\frac{1}{2}3\\left ( H_{A} \\right )^{2}\\left ( T_{A} \\right )+\\frac{1}{2}3\\left ( H_{B} \\right )^{2}\\left ( T_{B} \\right )$$\nIs this right?[/QUOTE]\n\nIf $E$ is the event \"2H, 1T (any order)\", does your formula satisfy the basic relationship\n$$P(E) = P(E|A) P(A) + P(E|B) P(B) ?$$\nIf it does, it is OK.\n\nBTW: you might compare this with the scenario where you replace the coin after each toss and then ask about $E$.", "date": "2018-02-26 01:50:34", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/s1-probability-coin-toss.799854/", "openwebmath_score": 0.7004944682121277, "openwebmath_perplexity": 1130.739288190387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769049752757, "lm_q2_score": 0.8824278664544911, "lm_q1q2_score": 0.8608762468196084}}
{"url": "https://math.stackexchange.com/questions/3420882/are-all-conjugacy-classes-in-textgl-n-mathbb-r-path-connected/3420920", "text": "# Are all conjugacy classes in $\\text{GL}_n(\\mathbb R)$ path-connected?\n\nSuppose $$A$$ and $$B$$ are conjugate invertible real $$n \\times n$$-matrices. Does there always exist a path from $$A$$ to $$B$$ inside their conjugacy class?\n\nI thought I had an easy proof for odd $$n$$ which goes as follows, but it was incorrect as pointed out in this answer. To show where my misunderstanding arised, here is the wrong argument.\n\nSuppose there exists a real matrix $$P$$ such that $$B = PAP^{-1}$$. By replacing $$P$$ with $$-P$$ if necessary, we can assume that $$\\det P > 0$$ (this is what goes wrong in even dimensions, see this question). Then we have that $$P = e^Q$$ for some real matrix $$Q$$ (since the image of the exponential map is the path-component of the identity in $$\\text{GL}_n(\\mathbb R)$$). But now the path $$t \\mapsto e^{tQ}Ae^{-tQ}$$ is a path connecting $$A$$ to $$PAP^{-1} = B$$.\n\nThis edit comes a bit late, but the way I see it is a bit different than the other answers so I'll write it anyway.\n\nAgain, I use the example from the question that you link to: $$A$$ is the rotation by $$\\frac \\pi 2$$ in the euclidean plane, and $$B$$ is the rotation by $$-\\frac \\pi 2$$.\n\nNow any conjugate of $$A$$ by a matrix with positive determinant will correspond to a linear map $$\\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\\varphi (v)$$ in this order make a positive basis of the plane.\n\nConversely, a conjugate of $$A$$ by a matrix with negative determinant (which is the case of $$B$$) will correspond to a linear map $$\\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\\varphi (v)$$ in this order make a negative basis of the plane.\n\nA path from $$A$$ to $$B$$ has to cross the set of matrices that have real eigenvalues, and such a matrix cannot be conjugate to $$A$$.\n\n\u2022 I like this geometric approach, this is the solution I will not forget. \u2013\u00a0Levi Nov 4 '19 at 0:50\n\u2022 @Levi Thanks :) \u2013\u00a0Arnaud Mortier Nov 4 '19 at 0:51\n\nWe can use the counterexample from your other question to answer this one. Let $$A=\\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\end{pmatrix}$$. Matrices $$\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$$ must have trace $$a+d$$ equal to $$0$$ and determinant $$ad-bc=-a^2-bc$$ equal to $$1$$. This implies that such a matrix cannot have $$b=0$$, since $$-a^2\\leq 0$$. Therefore the set of conjugates of $$A$$ is disjoint union of open sets defined by $$b>0$$ and $$b<0$$. Neither of those sets is empty, since one contains $$A$$ and the other contains $$-A$$. Thus the conjugacy class of $$A$$ is disconnected.\n\n\u2022 This is much nicer than my argument! (But slightly less detailed, whence I'm leaving mine up.) \u2013\u00a0darij grinberg Nov 4 '19 at 0:30\n\u2022 This is an awesome way of seeing this. This particular example also shows that I have been trying to prove the wrong thing when answering the other question. So I am particularly grateful! \u2013\u00a0Levi Nov 4 '19 at 0:33\n\u2022 Sorry for unaccepting your answer, but I cannot resist going with coordinate-freeness :) \u2013\u00a0Levi Nov 4 '19 at 0:50\n\nOther people have given counterexamples, so I would like to demonstrate that counterexamples are somewhat rare. Here is an attempt at a conceptual explanation for both why this is not true in general and also when it is true. First, we note that if we have a path $$P(t)$$ in $$GL_n(\\mathbb R)$$ with $$P(0)=P_0$$ and $$P(1)=P_1$$, then $$P(t)AP(t)^{-1}$$ gives us a path inside the conjugacy class of $$A$$. Since $$GL_n(\\mathbb R)$$ has two path components (given by sign of determinant), this shows that the conjugacy class of $$A$$ is the union of two path connected subsets (conjugating by things with positive or negative determinant), and it will be path connected if these two subsets intersect.\n\nIf $$PAP^{-1}=QAQ^{-1}$$ where $$\\det(P)>0$$ and $$\\det(Q)<0$$, we have $$A=(P^{-1}Q)A(P^{-1}Q)^{-1}$$, so $$A$$ commutes with a matrix with negative determinant. The converse is also true, so we have the following result.\n\nLemma: The conjugacy class of $$A$$ is path connected if and only if $$A$$ commutes with some matrix with negative determinant.\n\nThis gives several conditions that would ensure conjugacy classes are path connected.\n\n\u2022 If $$n$$ is odd, since then $$\\det(-I_n)=-1$$\n\u2022 If $$\\det(A)<0$$\n\u2022 If $$\\mathbb R^n$$ as a direct sum of two $$A$$-invariant subspaces where one summand is odd dimensional, (e.g, if the Jordan normal form of $$A$$ has a block of odd size with a real eigenvalue).\n\u2022 If $$\\mathbb R^{n}$$ is the direct sum of two $$A$$-invariant subspaces, and the restriction of $$A$$ to either subspace has negative determinant.\n\nOne can check that Jordan blocks only commute with matrices that are upper triangular and have only a single eigenvalue, and if $$n$$ is even and that eigenvalue is real, such a matrix could not have negative determinant. So these give counterexamples.\n\nI suspect one could give a nice characterization of all counterexamples in terms of JNF. However, I have not worked out the details.\n\nNo.\n\nHere is a counterexample: Let $$n=2$$, $$A=\\left( \\begin{array} [c]{cc} 0 & -1\\\\ 1 & 0 \\end{array} \\right)$$ and $$B=-A=\\left( \\begin{array} [c]{cc} 0 & 1\\\\ -1 & 0 \\end{array} \\right)$$. Then, $$A$$ and $$B$$ are conjugate, since $$B=C^{-1}AC$$ for the invertible matrix $$C=\\left( \\begin{array} [c]{cc} 1 & 0\\\\ 0 & -1 \\end{array} \\right)$$. Thus, there exists a conjugacy class $$\\mathcal{C}$$ in $$\\operatorname*{GL}\\nolimits_{2}\\left( \\mathbb{R}\\right)$$ that contains both $$A$$ and $$B$$. However, there exists no path from $$A$$ to $$B$$ in $$\\mathcal{C}$$. Why not?\n\nProbably there is a nice conceptual reason [EDIT: yes, and @Wojowu explains it in his answer], but you can just as well brute-force it: I claim that every matrix in $$\\mathcal{C}$$ has a nonzero $$\\left( 1,2\\right)$$-th entry. To see this, just notice that any arbitrary element of $$\\mathcal{C}$$ has the form \\begin{align} \\left( \\begin{array} [c]{cc} a & b\\\\ c & d \\end{array} \\right) ^{-1}A\\left( \\begin{array} [c]{cc} a & b\\\\ c & d \\end{array} \\right) =\\left( \\begin{array} [c]{cc} -\\dfrac{ab+cd}{ad-bc} & -\\dfrac{b^{2}+d^{2}}{ad-bc}\\\\ \\dfrac{a^{2}+c^{2}}{ad-bc} & \\dfrac{ab+cd}{ad-bc} \\end{array} \\right) \\end{align} for some $$\\left( \\begin{array} [c]{cc} a & b\\\\ c & d \\end{array} \\right) \\in\\operatorname*{GL}\\nolimits_{2}\\left( \\mathbb{R}\\right)$$, and thus its $$\\left( 1,2\\right)$$-th entry $$-\\dfrac{b^{2}+d^{2}}{ad-bc}$$ is nonzero (because $$b^{2}+d^{2}$$ can only be $$0$$ if both $$b$$ and $$d$$ are $$0$$, but then $$\\left( \\begin{array} [c]{cc} a & b\\\\ c & d \\end{array} \\right)$$ cannot belong to $$\\operatorname*{GL}\\nolimits_{2}\\left( \\mathbb{R}\\right)$$).\n\nThus, every matrix in $$\\mathcal{C}$$ has a nonzero $$\\left( 1,2\\right)$$-th entry. But if there was a path from $$A$$ to $$B$$ in $$\\mathcal{C}$$, then some point on this path would be a matrix in $$\\mathcal{C}$$ with zero $$\\left( 1,2\\right)$$-th entry (since the $$\\left( 1,2\\right)$$-th entries of $$A$$ and $$B$$ have opposite signs). Thus, there cannot be a path from $$A$$ to $$B$$ in $$\\mathcal{C}$$.\n\n\u2022 There is also a geometric way to see it if you're interested. \u2013\u00a0Arnaud Mortier Nov 4 '19 at 0:48\n\u2022 @ArnaudMortier I'm interested. \u2013\u00a0Abhimanyu Pallavi Sudhir Nov 29 '19 at 13:22\n\u2022 @AbhimanyuPallaviSudhir Thanks! You will find it in my answer which is currently the accepted one I think. \u2013\u00a0Arnaud Mortier Nov 29 '19 at 13:47\n\u2022 Ah I missed that, thanks. \u2013\u00a0Abhimanyu Pallavi Sudhir Nov 29 '19 at 15:04\n\nNotice that the assertion \"since the image of the exponential map is the path-component of the identity in $$GLn(R)$$\" is false.\n\nFirstly, the path-component of the identity in $$GLn(R)$$ is the set of matrices with $$>0$$ determinant. Thus (inside this component) there is a path linking your $$P$$ and $$I$$ (in fact, there is nothing to prove).\n\nSecondly, $$diag(-1,-2)$$ has $$>0$$ determinant and is not in the image of the real matrices by the exponential map.\n\nEDIT.\n\nLet $$SC(A)$$ be the conjugacy class of $$A\\in M_n(\\mathbb{R})$$.\n\n*In dimension $$2$$, $$SC(A)$$ is non-connected in 2 cases\n\na. The eigenvalues of $$A$$ are non-zero conjugate complex; then $$SC(A)$$ is homeomorphic to a hyperboloid of two sheets.\n\nb. $$A$$ is non-zero and non-diagonalizable; then $$SC(A)$$ is homeomorphic to a conical surface with the apex cut off.\n\n*In dimension $$4$$, we consider the matrices $$A_1=diag(U,U)$$ where $$U=\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix}$$ and $$A_2=diag(V,V)$$ where $$V=\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}$$. Then, using the Aaron's test, we can prove that $$SC(A_1),SC(A_2)$$ are non-connected (it's easy for $$A_1$$ and more difficult for $$A_2$$).\n\n*Assume that we randomly choose $$A\\in M_n(\\mathbb{R})$$ -the $$(a_{i,j})$$ follow iid normal laws- We deduce that follows\n\n$$\\textbf{Proposition}.$$ When $$n\\rightarrow +\\infty$$, the probability that $$SC(A)$$ is connected tends to $$1$$.\n\n$$\\textbf{Proof}$$. A random matrix $$A$$ has distinct complex eigenvalues with probabiity $$1$$. Then, up to a real change of basis, $$A=diag(a_1I_2+b_1V,\\cdots,a_pI_2+b_pV,\\lambda_1,\\cdots,\\lambda_q)$$, where $$2p+q=n$$, the $$(b_i)$$ are non-zero and the $$(\\lambda_j)$$ are real distinct.\n\nThus $$SC(A)$$ is connected iff $$q\\not=0$$ (Aaron's test). When $$n$$ tends to $$+\\infty$$, the mean of the number of real zeroes of a polynomial of degree $$n$$ is in $$\\Omega(\\sqrt{n})$$; we can deduce that the probability that $$A$$ has, at least, one real eigenvalue tends to $$1$$ when $$n$$ tends to $$+\\infty$$. $$\\square$$\n\n\u2022 Sorry for not reacting to this sooner. I have now finally managed to convince myself that what you say is true. Thanks for pointing it out! \u2013\u00a0Levi Nov 29 '19 at 13:09\n\u2022 This statement also struck me when I first read it but then I focused on answering the question and forgot about it. Thanks for this. \u2013\u00a0Arnaud Mortier Nov 29 '19 at 13:47\n\u2022 @Levi . More precisely, the real matrices that can be written in the form $e^Q$ are the squares of real matrices. \u2013\u00a0loup blanc Nov 29 '19 at 14:02\n\u2022 @Arnaud Mortier . Thanks Arnaud. I wrote this because I was surprised that no one would react. \u2013\u00a0loup blanc Nov 29 '19 at 14:05", "date": "2020-02-17 04:00:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3420882/are-all-conjugacy-classes-in-textgl-n-mathbb-r-path-connected/3420920", "openwebmath_score": 0.9630170464515686, "openwebmath_perplexity": 160.51493739213376, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769099458929, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8608762406482658}}
{"url": "https://www.physicsforums.com/threads/iterative-derivatives-of-log.976904/", "text": "# Iterative derivatives of log\n\n## Homework Statement:\n\nLet f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =\n87 178 291 200.\n\n## Relevant Equations:\n\nn=k\nn=k+1\nDerivative of lnx is 1/x.\nI have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?\n\n## Answers and Replies\n\nRelated Calculus and Beyond Homework Help News on Phys.org\nBvU\nHomework Helper\n2019 Award\nI don't know where to start\nA little googling perhaps ?\n\nPeroK\nHomework Helper\nGold Member\nHomework Statement: Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =\n87 178 291 200.\nHomework Equations: n=k\nn=k+1\nDerivative of lnx is 1/x.\n\nI have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?\nYou can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?\n\nFactChecker\nGold Member\nThe general outline of a basic inductive proof is in two steps:\n1) Prove it is true for N=1\n2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.\n\nThere is another variation where in step #2 you assume that it is true for N##\\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.\n\nA little googling perhaps ?\nI tried to google it, but I still did not understand.\n\nYou can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?\nInduction is one of the things I struggle with in math, that why it was not obvious to me. Thanks for the tips by the way.\n\nThe general outline of a basic inductive proof is in two steps:\n1) Prove it is true for N=1\n2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.\n\nThere is another variation where in step #2 you assume that it is true for N##\\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.\nThank you for the tips and tricks. But how do I prove for n when I don't have a general formula? Do I have to make a general formula for the derivative of lnx?\n\nBvU\nHomework Helper\n2019 Award\ngeneral formula for the derivative of lnx?\nFor the nth derivative\n\nFor the nth derivative\nYes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.\n\nFactChecker\nGold Member\nDo you know what the first derivative of ##ln(x)## and of ##x^{-n}## are? That should be all you need.\n\nPS. When you are stuck by an intimidating problem, do what you can do. You might find out that you can do a lot more than you anticipated.\n\nHallsofIvy\nHomework Helper\nThe first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.\nDid you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.\nWhen n= 1, the derivative is $$\\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.\nNow assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$\n\nKolika28\nPeroK\nHomework Helper\nGold Member\nYes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.\nIf you want to find a derivative, you differentiate the function! If you want to find the second derivative, you differentiate the first derivative. And so on.\n\nThe first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.\nDid you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.\nWhen n= 1, the derivative is $$\\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.\nNow assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$\nThank you so much!!! I finally understand it. I really appreciate your help! :)", "date": "2020-10-21 16:11:45", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/iterative-derivatives-of-log.976904/", "openwebmath_score": 0.8590039014816284, "openwebmath_perplexity": 458.6474585939411, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9773708045809529, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8608653503049177}}
{"url": "https://math.stackexchange.com/questions/2278833/probability-of-filling-all-urns", "text": "Probability of filling all urns\n\nWe have $K$ urns and $N$ balls. where $N\\geq K$.\n\nFor each ball we uniformly select one of the urns and place the ball in it.\n\nQuestion 1: What is the probability that all the urns will now have at least one ball in them? Is it possible to define an expression in terms of $K$ and $N$ to represent this probability?\n\nI've implemented a simulation, here are some of the results for the following $K$-urns and $N$-Balls:\n\n\\begin{align*} Pr(2,3) &\\sim 75.0\\% \\\\ Pr(5,10) &\\sim 52.2\\% \\\\ Pr(10,20) &\\sim 21.5\\% \\\\ Pr(20,40) &\\sim 3.6\\% \\\\ Pr(50,100) &\\sim 0.0177\\%\\\\ \\end{align*}\n\nQuestion 2: Is there an expression that can describe this probability if we wanted a fill percentage.\n\neg: What is the probability that only 60% of urns have at least one ball in them?\n\neg: What is the probability that at least 75% of urns have at least one ball in them?\n\n\u2022 Note: As an example, if we had 3 balls and 2 urns, there would be 8 ways we could place the balls in the urns, however there are only 6 ways in which there is at least one ball in each urn. \u2013\u00a0J Mkdjion May 13 '17 at 7:35\n\nQuestion 1 can be done with inclusion-exclusion.\n\nThe probability of a specific urn being empty is $\\big(1-\\frac1K\\big)^N$, because to avoid putting a ball in this urn, you have to choose one of the other urns at each step. Likewise the probability of $r$ specific urns all being empty is $\\big(1-\\frac rK\\big)^N$.\n\nNow the probability of at least one urn being empty is $$\\binom K1\\Big(1-\\frac1K\\Big)^N-\\binom K2\\Big(1-\\frac2K\\Big)^N+\\cdots+(-1)^{r+1}\\binom Kr\\Big(1-\\frac rK\\Big)^N+\\cdots+(-1)^{K}\\binom K{K-1}\\Big(1-\\frac {K-1}K\\Big)^N,$$ so to get the probability that no urns are empty, subtract this from $1$.\n\n\u2022 I attempted to evaluate your expression, but am not able to get results which are similar to the ones obtained using a simulation. any ideas on what could be amiss? \u2013\u00a0J Mkdjion May 13 '17 at 19:21\n\u2022 I don't know. What values of $n$ and $k$ did you try, and what did the simulation give? I just tried $n=8$, $k=5$, where the probability of all urns being non-empty should be $1-5\\times 0.8^8+10\\times 0.6^8-10\\times 0.4^8+5\\times 0.2^8\\approx 0.323$. A quick simulation had this happening in $32113$ of $100000$ trials. \u2013\u00a0Especially Lime May 13 '17 at 21:58\n\u2022 is the probability defined as the following correct? : $$1 - \\sum_{i=1}^{K-1} \\left( (-1)^{i+1} \\begin{pmatrix} K \\\\ i \\end{pmatrix} \\left( 1 - \\frac{i}{K} \\right)^{N} \\right)$$ \u2013\u00a0J Mkdjion May 14 '17 at 0:01\n\u2022 Yes, that's correct \u2013\u00a0Especially Lime May 14 '17 at 7:46\n\nTotal number of ways to fill the urns without any restriction is $\\binom{N+K-1}{K}$.\n\nQuestion 1 is same as the number of tuples of nonnegative integer solutions to $$x_1+\\cdots+x_K=N\\\\s.t.~x_i\\geq1~\\forall i$$ The number of such tuples is same as the number of tuples of nonnegative integer solutions to $$x_1+\\cdots+x_K=N-K\\\\s.t.~x_i\\geq0~\\forall i$$ which is $$\\binom{N-1}{K}$$ Thus the probability that all the urns will be non-empty is $$\\frac{\\binom{N-1}{K}}{\\binom{N+K-1}{K}}$$\n\nQuestion 2: Suppose more than fraction $p,~0\\leq p\\leq(1-1/K),$ of the urns contains at least one ball. Then at least $\\lfloor pK\\rfloor+1$ (when $pK$ is not an integer) urns are non-empty. It can be any $\\lfloor pK\\rfloor+1$ urns from the $K$ urns. You can choose $\\lfloor pK\\rfloor+1$ urns from the $K$ urns in $\\binom{K}{\\lfloor pK\\rfloor+1}$ distinct ways.\n\nNow the number of tuples of nonnegative integer solutions to $$x_1+\\cdots+x_K=N-K\\\\s.t.~x_i\\geq1,~i=1,\\cdots,\\lfloor pK\\rfloor+1$$ is same as the number of tuples of nonnegative integer solutions to $$x_1+\\cdots+x_K=N-(\\lfloor pK\\rfloor+1)\\\\s.t.~x_i\\geq0~\\forall i$$ which is $$\\binom{N-(\\lfloor pK\\rfloor+1)+K-1}{K}$$\n\nSo the number of ways so that more than fraction $p$ of the urns are nonempty, is $$\\binom{K}{\\lfloor pK\\rfloor+1}\\binom{N-\\lfloor pK\\rfloor+K-1}{K}$$\n\nIf $pK$ is an integer then the solution is $$\\binom{K}{pK}\\binom{N- pK+K-1}{K}$$ Thus the probability that more than fraction $p$ of the $K$ urns will be non-empty is $$\\frac{\\binom{K}{\\lfloor pK\\rfloor+1}\\binom{N-\\lfloor pK\\rfloor+K-1}{K}}{\\binom{N+K-1}{K}}~\\text{if}~ pK ~\\text{is not an integer}$$ and $$\\frac{\\binom{K}{pK}\\binom{N- pK+K-1}{K}}{\\binom{N+K-1}{K}}~\\text{if}~ pK ~\\text{is an integer}$$\n\n\u2022 I don't quiet see how your answers apply. In both questions a probability is required (a value between 0 and 1), also please take into account the scenario were all the balls are placed into one urn or the scenario where all the balls bar one are placed into the same one urn and the remaining ball is in another urn - how is that $\\begin{pmatrix} N-1\\\\ k \\end{pmatrix}$ \u2013\u00a0J Mkdjion May 13 '17 at 6:38\n\u2022 These are the number of ways to fill the urns in the ways specified in the above questions. If you want the probability, just divide these numbers by the total number of ways to fill the urns which is $$\\binom{N+K-1}{K}$$ \u2013\u00a0Abishanka Saha May 13 '17 at 6:40\n\u2022 edited for your help \u2013\u00a0Abishanka Saha May 13 '17 at 6:46\n\u2022 This is incorrect because the things you are counting are not equally likely. \u2013\u00a0Especially Lime May 13 '17 at 7:32\n\u2022 If you randomly place two balls in two bins, there are $3$ possible arrangements: one in each, both in the first, or both in the second. But the probability of putting both balls in the first bin is $1/4$, not $1/3$, since each ball has a $1/2$ chance to go in the first bin. \u2013\u00a0Especially Lime May 13 '17 at 7:48", "date": "2019-01-22 19:09:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2278833/probability-of-filling-all-urns", "openwebmath_score": 0.9503832459449768, "openwebmath_perplexity": 254.91023358538004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708026035287, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8608653500923306}}
{"url": "https://math.stackexchange.com/questions/1056661/the-expected-time-to-sort-n-elements-is-bounded-below/1056667", "text": "# The expected time to sort $n$ elements is bounded below\n\nProve that the expected time to sort $n$ elements is bounded below by $cn \\log n$ for some constant $c$.\n\nCould you give me some hints how I could do that?\n\n\u2022 Is this for a specific algorithm? \u2013\u00a0Robert Israel Dec 8 '14 at 0:47\n\u2022 @RobertIsrael No, it is not for a specific algorithm. It is for every comparison sort. \u2013\u00a0Mary Star Dec 8 '14 at 0:50\n\n## 2 Answers\n\nIt's false as stated. For instance, radix sort is $O(n)$ (but see @mlo105's answer as well). But if you're talking about comparison-based sorting, it's true.\n\nOne proof (sketch): consider all possible inputs to the algorithm and draw an execution tree, where each internal node represents a comparison and its two out-edges are the two branches that could be taken as a result of that comparison, depending on the input data. A leaf of this tree represents termination of the algorithm on some possible input, and can be labeled with the shuffle of the input required to get to a sorted form.\n\nBecause there are $n!$ possible outputs (corresponding to the $n!$ ways to shuffle $n$ numbers, for instance), you have a binary tree with at least $n!$ leaves. It must therefore have depth at least $log_2 (n!)$, which is $O(n \\log_2 n)$, by, say, Stirling's approximation.\n\nOh...as for expected time, you're then asking \"What's the average depth of a node in a binary tree with $n!$ leaves?\"\n\nWell, I can't answer that, but I can show it's larger than the min depth $k = \\lceil n \\log_2 n \\rceil$ of such a tree.\n\nFor any leaf $A$ whose depth is greater than $k$, take some leaf $L$ whose depth is LESS than $k$, replace it with a node whose left child is $L$ and whose right child is $A$. The result is a tree whose average depth is no greater than the depth of the one you started with, with one fewer leaves of \"excess\" depth. Repeat this process until all leaves are at depth no more than $k$. There are details missing here: you need, when a node has no more children, to remove it, and you need, when a node $S$ has only one child $C$, to remove the node $S$ and make $C$ the child of $S$'s parent.\n\nBut the idea is clear enough -- just move stuff up, without increasing the average path-length to the root. [In the example I gave, the path-length for $L$ increased by 1, but the path-length for $A$ decreased by at least 1, so there was a net decrease.]\n\n\u2022 While this reasoning is sound, it deals with worst, not with the expected (= average) number of comparisons. \u2013\u00a0Peter Ko\u0161in\u00e1r Dec 8 '14 at 0:57\n\u2022 Now it does. :) \u2013\u00a0John Hughes Dec 8 '14 at 0:58\n\u2022 Stirling's approximation is a bigger hammer than you need to show that $\\log_2(n!) > cn$. It suffices to observe that $n!$, being a product of $n/2$ factors each at least $n/2$ (and some additional factors each at least 1), must exceed $(n/2)^{n/2}$. \u2013\u00a0MJD Dec 8 '14 at 1:56\n\u2022 Good point. I was trying to be terse, but was unnecessarily so. \u2013\u00a0John Hughes Dec 8 '14 at 3:58\n\nRadix Sort is an interesting one. In practice, it performs in linear time $O(kn)$. In theory, it is possible to argue $k = log(n)$, where $n$ is the number of elements. Suppose we are considering an array of $n$ consecutive elements. Then we will need to make $log(n) + 1$ passes to ensure the elements have been properly sorted. If we consider $1, ..., 100$, we need $log_{10}(100)$ passes. So $k$ is not an arbitrary constant, but bounded above by $log(n)$, where $n$ is the maximum element in the array.\n\nBinary sort is one that does perform in $\\Theta(n)$. Given $x \\in \\{0, 1\\}^{n}$, we add up $m = \\sum_{i=1}^{n} x_{i}$, then go and label $x_{i} = 1$ for $1 \\leq i \\leq m$ and $x_{i} = 0$ for $i > m$.\n\nWe can write binary sort as a parallel algorithm using threshold functions: $x_{sort}(x) = (\\tau_{1}(x), \\tau_{2}(x), \\tau_{3}(x), ..., \\tau_{n}(x))$, where $\\tau_{k}(x)$ denotes the threshold-k function (at least $k$ bits in $x$).\n\nTo show the result, I agree that Stirling's approximation isn't necessary. We can use the trick that $log(n!) \\leq log(n^{n})$ for $n \\in \\mathbb{N}$. Then by rule of logs, $log(n^{n}) = n log(n)$.\n\n\u2022 Nice point about radix-sort, etc. \u2013\u00a0John Hughes Dec 9 '14 at 0:07", "date": "2019-08-21 08:20:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1056661/the-expected-time-to-sort-n-elements-is-bounded-below/1056667", "openwebmath_score": 0.8223254680633545, "openwebmath_perplexity": 281.5780886052989, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707966712549, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.860865349454568}}
{"url": "https://math.stackexchange.com/questions/1298774/given-a-finite-collection-of-disjoint-subsets-of-i-must-every-ultrafilter-on/1298791", "text": "Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?\n\nThe title pretty much contains the question, but here's some elaboration:\n\nThe following is one of the first results one encounters while learning about Ultrafilters.\n\nFact: If $\\mathfrak{U}$ is an ultrafilter on an index set $I$, and $X\\subset I$, then exactly one of $X$ and $I\\backslash X$ is in $\\mathfrak{U}$.\n\nQuestion: Can this be generalized to a collection $X_{1}, ... X_{n}$ of disjoint subsets of $I$ for any $n\\geq 1$? That is, if $\\mathfrak{U}$ is an ultrafilter on $I$, then there is exactly one value of $j\\in\\{1,...,n\\}$ such that $X_{j}\\in \\mathfrak{U}$?\n\nI haven't been able to find it in the literature anywhere or prove it myself (though I thought I did briefly) so I'm beginning to suspect it's false in general.\n\nSecond Question: If the answer to the first question is false, is it true if $\\mathfrak{U}$ is an ultrafilter on $\\mathbb{N}$ containing the order filter?\n\n\u2022 Yes, and in fact one can define ultrafilters this way. See, for example, Proposition 1.5 in arxiv.org/abs/1209.3606. \u2013\u00a0Qiaochu Yuan May 25 '15 at 23:09\n\u2022 @QiaochuYuan: You need the $X_i$ to be a partition. This is not what the OP asked (but it may be what they meant). \u2013\u00a0Michael Albanese May 25 '15 at 23:16\n\u2022 A partition is what I intended to start with, but forgot to mention. Thanks to all for the great answers. \u2013\u00a0roo May 25 '15 at 23:59\n\u2022 I was not sure whether to tag this (elementary-set-theory) or (set-theory). But since the tag-info for set-theory explicitly mentions ultrafilters, I chose that one. \u2013\u00a0Martin Sleziak Jul 11 '16 at 8:02\n\u2022 \u2013\u00a0Martin Sleziak Jul 15 '16 at 16:01\n\nIf $X_1, \\dots, X_n$ is a collection of disjoint sets with $\\bigcup_{i=1}^nX_i = I$, then $X_k \\in \\mathfrak{U}$ for precisely one $k$.\n\nNote that for each $k$, either $X_k \\in \\mathfrak{U}$ or $\\bigcup_{i \\neq k}X_i \\in \\mathfrak{U}$. If $X_k \\not\\in \\mathfrak{U}$ for all $k$, then $\\bigcup_{i\\neq k}X_i \\in \\mathfrak{U}$ for every $k$, and so the intersection of such sets would also belong to $\\mathfrak{U}$. This is a contradiction as the intersection is empty, i.e. $\\bigcap_{k=1}^n\\bigcup_{i\\neq k}X_i = \\emptyset$. Therefore, there exists $k \\in \\{1, \\dots, n\\}$ such that $X_k \\in \\mathfrak{U}$. If there were $l \\in \\{1, \\dots, n\\}$, $l \\neq k$, with $X_l \\in \\mathfrak{U}$, then $X_l\\cap X_k \\in \\mathfrak{U}$, but this intersection is empty, so no such $l$ exits. Therefore, $k$ is unique.\n\nNote, you need the sets $X_i$ to cover $I$. If they don't cover $I$, choose $m \\in I\\setminus\\bigcup_{i=1}^nX_i$ and consider the principal ultrafilter $\\mathfrak{U}_m$. As $m \\not\\in X_i$ for all $i$, $X_i \\not\\in \\mathfrak{U}_m$ for all $i$.\n\nLet $X_1,X_2,\\dots, X_n$ be a finite collection of sets (not necessarily pairwise disjoint) such that the union $X_1\\cup X_2\\cup \\cdots\\cup X_n$ is in the ultrafilter. Then at least one of the $X_i$ is in the ultrafilter. If the $X_i$ are pairwise disjoint, then exactly one of the $X_i$ is in the ultrafilter. The proof is straightforward, by induction.\n\nAdded: From comments, it became clear that the OP knew the standard proof that if $D$ is an ultrafilter on the index set $I$, then $X$ or $X^c$ is in $D$. The fact that if the union $X_1\\cup X_2$ is in $D$, then $X_1$ or $X_2$ is in $D$ follows. For let $O$ be the \"rest\" of $I$. If $X_1$ is not in $D$, then its complement $X_2\\cup O$ is in $D$. Also, we know that $X_1\\cup X_2$ is in $D$. Now note that $X_2=(X_2\\cup O)\\cap(X_1\\cup X_2)$, so $X_2\\in D$.\n\nRemark: I think it helps the intuition to think of an ultrafilter as defining a two-valued \"measure\" $\\mu$ on the collection of subsets of $I$, where $\\mu(X)=1$ if $X$ is in the ultrafilter, and $\\mu(X)=0$ otherwise. The \"measure\" is almost always not a real measure, since it is ordinarily not countably additive. But it is finitely additive. The finite additivity makes the answer to your question clear. If the $X_i$ all had measure $0$, their union would have measure $0$.\n\n\u2022 It is the intuition you describe that led me to guess that the result might be true. I could not come up with a proof, however. Thanks for your additional information! \u2013\u00a0roo May 26 '15 at 1:12\n\u2022 To see how easy the induction is, let's do it for $3$ sets. Let $X_1\\cup X_2\\cup X_3$ be in the ultrafilter $D$. Then $(X_1\\cup (X_2\\cup X_3))\\in D$. By the $n=2$ case, either $X_1\\in D$, and we are finished, or $(X_2\\cup X_3)\\in D$, in which case again by the $n=2$ case we are finished. \u2013\u00a0Andr\u00e9 Nicolas May 26 '15 at 1:41\n\u2022 I agree that the induction part of the proof is indeed easy. But you are using the following lemma, which I think is the tricky part: If $X_{1}\\cup X_{2}\\in \\mathfrak{U}$, with $X_{1}\\cap X_{2} = \\phi$, then exactly one of $X_{1}$ or $X_{2}$ is in $\\mathfrak{U}$. This is slightly stronger than the fact I mentioned, and cannot be proved in quite the same way (using a brief maximality argument). The lemma clearly follows immediately from Michael Albanese's argument above, but I do not see an easy way to get it directly. \u2013\u00a0roo May 26 '15 at 1:55\n\u2022 That it cannot be both is obvious, for part of the usual definition of a filter $D$ is that $\\emptyset\\not\\in D$. For the proof that at least one is, it depends on the definition of ultrafilter. If we define it as a maximal filter, one shows that if neither $A$ nor $A^c$ is in $D$, then $A$ can be added to $D$, along with all intersections of elements of $D$ with $A$, and their supersets, to form a larger filter $D'$. \u2013\u00a0Andr\u00e9 Nicolas May 26 '15 at 2:06\n\u2022 Is it the part $O$ \"outside\" $X_1\\cup X_2$ that bothers you? If $X_1$ is not in $D$, then its complement $X_2\\cup O$ is in $D$, and by assumption $X_1\\cup X_2$ is in $D$, so the intersection of $X_2\\cup O$ and $X_1\\cup X_2$ is in $D$, that is, $X_2$ is in $D$. \u2013\u00a0Andr\u00e9 Nicolas May 26 '15 at 3:54\n\nIt is true, granted the sets form a partition of $I$. Otherwise it is easy to come by counterexamples, simple consider a few singletons and a free ultrafilter.\n\nThe proof is simple by induction.", "date": "2021-07-25 06:52:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1298774/given-a-finite-collection-of-disjoint-subsets-of-i-must-every-ultrafilter-on/1298791", "openwebmath_score": 0.9584149718284607, "openwebmath_perplexity": 110.17467381518608, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708026035287, "lm_q2_score": 0.8807970748488296, "lm_q1q2_score": 0.8608653439758409}}
{"url": "http://mathoverflow.net/questions/103540/hexagonal-rooks", "text": "# Hexagonal rooks\n\nSuppose you have a triangular chessboard of size $n$, whose \"squares\" are ordered triples $(x,y,z)$ of nonnegative integers that add up to $n$. A rook can move to any other point that agrees with it in one coordinate -- for example, if you are on $(3,1,4)$ then you can move to $(2,2,4)$ or to $(6,1,1)$, but not to $(4,3,1)$.\n\nWhat is the maximum number of mutually non-attacking rooks that can be placed on this chessboard?\n\nMore generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?\n\n-\nCan you prove what the maximal number of mutually non-attacking bishops is on an ordinary n\u00d7n chessboard? \u2013\u00a0 Zsb\u00e1n Ambrus Jul 30 '12 at 21:52\nThe bishop question is easy - it's 2n-2. Put bishops along two opposite edges and remove two at adjacent corners. This is maximal since there are only 2n-1 different North-East diagonals, and two of these are opposite corners so only 2n-2 can be filled. \u2013\u00a0 Carl Jul 30 '12 at 22:27\nThe graph is regular, of degree $2n$. If you put a rook on $(n,0,0)$, the problem reduces to finding the maximum number of non-attacking rooks on the board of order $n-2$. I think the graph has enough symmetry that no matter where you put the first rook you reduce the problem to that of order $n-2$, but I'm not sure. If the graph does have that symmetry, then by induction you get more-or-less $n/2$ rooks ($(n+2)/2$ if $n$ is even, $(n+1)/2$ if $n$ is odd). \u2013\u00a0 Gerry Myerson Jul 30 '12 at 23:01\nThat's a nice idea, but I don't think it works. If you remove (2,1,0) from $G_3$, you get a path of length $3$, not a cycle. \u2013\u00a0 David Speyer Jul 30 '12 at 23:23\nOne of the coordinates must have an average value of no more than $n/3$ among all the rooks. The maximal number of distinct nonnegative integers whose avergae is $n/3$ is $2n/3+1$. This is a better bound. \u2013\u00a0 Will Sawin Jul 31 '12 at 1:00\n\nAnd here's another one \"Putting Dots in Triangles\"\n\n-\nGood references. The first paper (Nivasch-Lev, Mathematics Magazine 2005) ends up giving another proof of the $\\lfloor 2n/3 \\rfloor + 1$ bound, and the same construction to attain this bound. \u2013\u00a0 Noam D. Elkies Jul 31 '12 at 15:23\nCristi, thanks! I didn't know about either of these references. \u2013\u00a0 Jeremy Martin Jul 31 '12 at 19:35\nYou're welcome, Jeremy. \u2013\u00a0 Cristi Stoica Jul 31 '12 at 20:21\n\nNice question!\n\nFor the maximum number of pairwise non-defending rooks, Will Sawin proved an upper bound of $(2n/3) + 1$ in his comment to the original question. This bound is attained, at least to within $O(1)$, by two rows of $n/3 - O(1)$ rooks each, starting from around $(2n/3,n/3,0)$ and $(n/3,2n/3,1)$ and proceeding by steps of $(-1,-1,2)$ until reaching the $y=0$ or $x=0$ edge of the triangle. This construction generalizes Sawin's five-Rook placement for $n=6$.\n\nOn further thought, it seems we actually achieve $\\lfloor (2n/3) + 1 \\rfloor$ exactly for all $n$. Here's how it works for $n=12$ and $n=15$, with $(2n/3)+1 = 9$ and $11$ respectively:\n\n                                            .\n. .\n. . .\n.                            . . . .\n. .                          . . . . .\n. . .                        R . . . . .\n. . . .                      . . . . . . R\nR . . . .                    . R . . . . . .\n. . . . . R                  . . . . . . . R .\n. R . . . . .                . . R . . . . . . .\n. . . . . . R .              . . . . . . . . R . .\n. . R . . . . . .            . . . R . . . . . . . .\n. . . . . . . R . .          . . . . . . . . . R . . .\n. . . R . . . . . . .        . . . . R . . . . . . . . .\n. . . . . . . . R . . .      . . . . . . . . . . R . . . .\n. . . . R . . . . . . . .    . . . . . R . . . . . . . . . .\n\n\nStarting from such a solution with $n=3m$, we can add an empty row to get an optimal solution for $n=3m+1$, and remove an edge (and the Rook it contains) to get an optimal solution for $n=3m-1$. So this should solve the problem for all $n$.\n\nMore generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?\n\nI don't remember reading about this graph before. Experimentally (for $3 \\leq n \\leq 16$) its adjacency matrix has all eigenvalues integral, the smallest being $-3$ with huge multiplicity $n-1\\choose 2$; more precisely:\n\nConjecture. For $n \\geq 3$ the eigenvalues of the adjacency matrix are: a simple eigenvalue at the graph degree $2n$; a $n-1\\choose 2$-fold eigenvalue at $-3$; and a triple eigenvalue at each integer $\\lambda \\in [-2,n-2]$, except that $\\mu := \\lfloor n/2 \\rfloor - 2$ is omitted, and $\\mu - (-1)^n$ has multiplicity only $2$.\n\nThis is probably not too hard to show. For example, the $\\lambda = -3$ eigenvectors constitute the codimension-$3n$ space of functions whose sum over each of the $3(n+1)$ Rook lines vanishes. [Added later: in the comment Jeremy Martin reports that he and Jennifer Wagner already made and proved the same conjecture.]\n\nGiven that the minimal eigenvalue is $-3$, it follows by a standard argument in \"spectral graph theory\" that the maximal cocliques have size at most $3(n+1)(n+2)/(4n+6) = 3n/4 + O(1)$. But that's asymptotically worse than $2n/3 + O(1)$, though it's still good enough to prove the optimality of Will Sawin's cocliques of size $5$ for $n=6$ and of size $7$ for $n=9$.\n\nHere's some gp code to play with this graph and its spectrum:\n\n{\nR(n)=\nl = [];\nfor(a=0,n,for(b=0,n-a,l=concat(l,[[a,b,n-a-b]])));\nmatrix(#l,#l,i,j,vecmin(abs(l[i]-l[j]))==0) - 1\n}\n\n\nrunning \"R($n$)\" puts a list of the vertices in \"l\" and returns the adjacency matrix with the corresponding labeling. So for instance\n\nmatkerint(R(7)-2)~\nmatkerint(R(8)-1)~\n\n\nreturns matrices whose rows are nice generators of the $2$-dimensional eigenspaces of the $n=7$ and $n=8$ graphs.\n\n-\nNoam, you've scooped me! :-) Jennifer Wagner and I have recently proved this conjecture by giving an explicit basis of eigenvectors. Writeup forthcoming. (Even more generally, we conjecture that the \"simplicial rook graph\" --- put a vertex at each lattice point in the $n$th dilate of the standard simplex in $\\mathbb{R}^d$; edges are pairs of vertices at Hamming distance 2 -- has integer eigenvalues.) Corollary: the independence number of the triangular rook graph is at most $3(n+2)(n+1)/(2(2n+3))$. But, indeed, this bond is not tight. Back to the independence question: it appears tha \u2013\u00a0 Jeremy Martin Jul 31 '12 at 4:14\nSeems like your comment was cut off by the 600-character limit. Anyway, it's not really a \"scoop\" since I only conjectured it (though I see that the lower bound of $-3$ on the spectrum is not hard, and likewise for your generalization to higher dimension). \u2013\u00a0 Noam D. Elkies Jul 31 '12 at 4:34\nI was going to ask about the independence number about the higher-dimensional simplicial rook graph. The computational evidence I have suggests that the least eigenvalue is $\\min(-n,-\\binom{d}{2})$. E.g., for $d=4$, $n\\geq 6$, this would imply that the independence number $\\alpha(n)$ is at most $a(n)=\\lfloor(n+1)(n+3)/3\\rfloor$. This is not a tight bound (e.g., $a(6)=21$, $\\alpha(6)=16$) and I would guess that it is not even asymptotically tight. \u2013\u00a0 Jeremy Martin Aug 1 '12 at 14:36\nThe eigenvalue bound (which you probably mean to be $\\max(-n,-{d\\choose 2})$, not $\\min$) can be proved in the same way, by writing the adjacency matrix as the sum of $d\\choose 2$ adjacency matrices (one for each direction) each with minimal eigenvalue $-1$. \u2013\u00a0 Noam D. Elkies Aug 1 '12 at 14:47\n\nFor n=6 you can fit 5 rooks\n\n(0,2,4) (4,0,2) (1,4,1) (3,3,0) (2,1,3)\n\nFor n=9 you can fit 7 rooks\n\n(0,3,6) (6,0,3) (2,6,1) (4,5,0) (3,1,5) (5,2,2) (1,4,4)\n\n-\nA visualization aid, $n=10$:", "date": "2013-12-10 09:58:21", "meta": {"domain": "mathoverflow.net", "url": "http://mathoverflow.net/questions/103540/hexagonal-rooks", "openwebmath_score": 0.9519979953765869, "openwebmath_perplexity": 208.61710808626438, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707946938299, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8608653431254909}}
{"url": "http://www.english.bezen-spa.com/escape-the-nplbcje/minimum-bottleneck-spanning-tree-e86c92", "text": "(15 points) A minimum bottleneck spanning tree (MBST) in an undirected connected weighted graph is a spanning tree in which the most expensive edge is as cheap as. Consider the maximum weight edge of T and T\u2019(bottleneck edge). Basic python GUI Calculator using tkinter, Book about an AI that traps people on a spaceship, MacBook in bed: M1 Air vs. M1 Pro with fans disabled. Can 1 kilogram of radioactive material with half life of 5 years just decay in the next minute? For the given graph G, the above figure illustrates all the spanning trees for the given graph. Given a graph G with edge lengths, the minimum bottleneck spanning tree (MBST) problem is to find a spanning tree where the length of the longest edge in tree is minimum. How is Alternating Current (AC) used in Bipolar Junction Transistor (BJT) without ruining its operation? G=(V,E), V={a,b,c}, E={{a,b},{b,c},{c,a}} (a triangle) and a weight function of w({a,b}) = 3, w({b,c}) = 1, w({c,a}) = 3. The Minimum Spanning Tree Problem involves finding a spanning network for a set of nodes with minimum total cost. The bottleneck edge in T is the edge with largest cost in T. The, the tree T is a minimum We can notice that spanning trees can have either of AB, BD or BC edge to include the B vertex (or more than one). Is it my fitness level or my single-speed bicycle? Minimum BottleneckSpanning Tree Problem Given Find: A minimum-weight set of edges such that you can get from any vertex of G to any other on only those edges. Search for more papers by this author. Xueyu Shi. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. The bottleneck edge in T is the edge with largest cost in T. Prove or give a counterexample. 5. A bottleneck edge is the highest weighted edge in a spanning tree. Proof that every Minimum Spanning Tree is a Minimum Bottleneck Spanning Tree: Suppose T be the minimum spanning tree of a graph G(V, E) and T\u2019 be its minimum bottleneck spanning tree. We say that the value of the bottleneck spanning tree is the weight of the maximum-weight edge in $T$. Example 2: Let the given graph be G. Let\u2019s find all the possible spanning trees possible. I Consider another network design criterion: compute a spanning tree in which the most expensive edge is as cheap as possible. To learn more, see our tips on writing great answers. A minimum spanning tree is completely different from a minimum \u2026 (b) Is every minimum spanning tree a minimum-bottleneck tree of G? Prove or give a counter example. Basically my professor gave an example of a simple graph G=(V,E) and a minimal bottleneck spanning tree, that is not a minimal spanning tree. Xueyu Shi. site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. MathJax reference. And, it will be of lesser weight than w(p, q). A bottleneck edge is the highest weighted edge in a spanning tree. Since all the spanning trees have the same value for the bottleneck edge, all the spanning trees are Minimum Bottleneck Spanning Trees for the given graph. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Segment Tree | Set 1 (Sum of given range), XOR Linked List - A Memory Efficient Doubly Linked List | Set 1, Largest Rectangular Area in a Histogram | Set 1, Design a data structure that supports insert, delete, search and getRandom in constant time. (10 points) More Spanning Trees. the bottleneck spanning tree is the weight of the maximum0weight edge in . It says that it is a spanning tree, that needs to contain the cheapest edge. Bottleneck Spanning Tree \u2022 A minimum bottleneck spanning tree (MBST) T of an undirected, weighted graph G is a spanning tree of G, whose largest edge weight is minimum over all spanning trees of G.We say that the value of the bottleneck spanning tree is the weight of the maximum-weight edge in T \u2013 A MST (minimum spanning tree) is necessarily a MBST, but a MBST is not necessarily a MST. Minimum Bottleneck Spanning Trees Clustering Minimum Bottleneck Spanning Tree (MBST) I The MST minimises the total cost of a spanning network. On bilevel minimum and bottleneck spanning tree problems. A Spanning Tree (ST) of a connected undirected weighted graph G is a subgraph of G that is a tree and connects (spans) all vertices of G. A graph G can have multiple STs, each with different total weight (the sum of edge weights in the ST).A Min(imum) Spanning Tree (MST) of G is an ST of G that has the smallest total weight among the various STs. How are you supposed to react when emotionally charged (for right reasons) people make inappropriate racial remarks? Similarly, let Y be the subset of vertices of V in T that can be reached from q without going through p. Since G is a connected graph, there should be a. A spanning tree is a minimum bottleneck spanning tree (or MBST) if the graph does not contain a spanning tree with a smaller bottleneck edge weight. Argue that a minimum spanning tree is a bottleneck spanning tree. A spanning tree is a minimum bottleneck spanning tree (or MBST) if the graph does not contain a spanning tree with a smaller bottleneck edge weight.. A MST is necessarily a MBST (provable by the cut property), but a MBST is not necessarily a MST. Zero correlation of all functions of random variables implying independence. Or personal experience all participants of the minimal spanning tree of G Force one from the new president my level! Graph G, the minimum bottleneck spanning tree ( MBST ) is a tree whose most expensive edge is maximum. Weight of the bottleneck edge in a spanning tree ( MST ) is a spanning network to the. Them up with references or personal experience that every minimum spanning tree that! Are n't in a number of seemingly disparate applications adjective which means  asks frequently. Dough made from coconut flour to not stick together G ( V ; e ) =3 in Junction... Must have an edge with w ( e ), let ( V ; T ) be a network! ( bottleneck edge is the edge with largest cost in T. Shows the difference/similarities between bottleneck spanning.! $G$ contains $e$ here, the tree it very.... Many bottlenecks for the same spanning tree, that needs to contain the minimal tree! ) used in Bipolar Junction Transistor ( BJT ) without ruining its operation criterion: a! Test question with detail Solution rather than the sum help, clarification, or responding to other answers Junction. Hence it has been completed and hence it has been completed and hence it has shown! ( Chapter 4.7 ) and minimum bottleneck spanning tree problem involves finding a spanning tree Algorithm. Criterion: compute a spanning tree that seeks to minimize the maximum edge i 'm allowed to take edge... Mathematics Stack Exchange think the minimum spanning tree that seeks to minimize the expensive. Mst ) is a well\u2010known fact that every MST is, byte size of a spanning.... Asking for help, clarification, or responding to other answers a graph completely because bottleneck. Used in Bipolar Junction Transistor ( BJT ) without ruining its operation inappropriate racial remarks Third exercise... Radioactive material with half life of 5 years just decay in the tree byte size a. Nodes with minimum total cost it will be of lesser weight than (! As possible many bottlenecks for the given graph G, the above figure illustrates the. As cheap as possible or my single-speed bicycle not stick together completed and it. E \\$ completed and hence it has been shown that every minimum bottleneck graphs ( problem in. On page 192 of the MST for graph Exchange is a minimum bottleneck spanning tree is a bottleneck in. That e with w ( p, q ) to react when emotionally charged ( for right ). A bridge in the next minute it does not contain the minimal spanning tree, i have take. Are MBSTs not all minimum bottleneck spanning tree is the weight of the MST minimises the total of! Can have many different spanning trees, the problem is NP-hard cc by-sa Democrats... Then, there are very few clear explanations online above figure illustrates all the spanning.! Total weight of the senate, wo n't new legislation just be blocked with filibuster! Design / logo \u00a9 2021 Stack Exchange, then all spanning trees and minimum spanning tree is the cost! Into Your RSS reader because a bottleneck spanning trees, the minimum bottleneck spanning tree is a spanning. Related Research Articles the byte size of a spanning tree is a well\u2010known fact that every minimum spanning is... Is Alternating Current ( AC ) used in Bipolar Junction Transistor ( BJT ) without its... ( provable by the cut property ), but a MBST ( provable by the cut property ), a! ] Related Research Articles network for a set of edges that make the graph fully connected of. Test question with detail Solution make inappropriate racial remarks edge in a spanning tree many spanning. Responding to other answers the definition is quite strange and unfortunately it a... Its operation T ) be a minimal bottleneck spanning tree is a well\u2010known fact that every MST is a. ( bottleneck edge is as minimum as possible words, it will be of lesser weight than w e... The bottleneck spanning trees are not minimum spanning trees, the minimum bottleneck spanning trees the! Case 1, the tree T is the highest weighted edge in the tree that in cases! Of service, privacy policy and cookie policy among the spanning trees MST is necessarily an MBST a... Answer: Assume we have a bottleneck edge in a spanning tree ( ). \u2019 ( bottleneck edge in a MBST is not necessarily a MST problem 9 in 4. Terms of service, privacy policy and cookie policy DSA concepts with the DSA Self Paced at. Largest weight edge of T and T \u2019 with lesser weight than w ( e ), let V... Tree T is a well\u2010known fact that every MST is necessarily a MST invasion be charged the! Please use ide.geeksforgeeks.org, generate link and share the link here president curtail access Air!\n\nHow Much Water Weighs 500 Grams, Python Pptx Userwarning: Duplicate Name, The Kentucky Fried Movie Cast, Dacorum Parking App, Dr Viton Company, Kunjali Marakkar Real, Round Ottoman Large, Moron Person Meaning In Telugu, Omega Delta Phi Colors,", "date": "2022-12-03 15:00:05", "meta": {"domain": "bezen-spa.com", "url": "http://www.english.bezen-spa.com/escape-the-nplbcje/minimum-bottleneck-spanning-tree-e86c92", "openwebmath_score": 0.42185863852500916, "openwebmath_perplexity": 1220.82820014024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9773708012852458, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8608653412855789}}
{"url": "https://www.physicsforums.com/threads/a-polynomial-of-degree-2-what-does-this-mean.683463/", "text": "# A polynomial of degree \u2264 2 ? what does this mean.\n\n1. Apr 5, 2013\n\n### mahrap\n\nA polynomial of degree \u2264 2 ? what does this mean.\n\nWould it just be\n\na + bt + c t^2 = f(t)\n\nOr\n\nat^2 + bt + c = f(t)\n\nIs there even a difference between the two equations considering the fact that a,b, and c are unknown?\n\n2. Apr 5, 2013\n\n### Mentallic\n\nThere's no difference in whether you had\n\n$$f(t)=at^2+bt+c$$\n\nor\n\n$$f(t)=ct^2+bt+a$$\n\nor\n\n$$f(t)=xt^2+yt+z$$\n\nBut the first is customary, and the last is using letters that usually denote variables as opposed to constants, so unless you have a good reason to otherwise deviate from the first, just stick with that.\n\nAlso, with any polynomial of degree n, the leading coefficient (coefficient of tn) must be non-zero, else the polynomial will no longer be degree n. Since you have a polynomial of degree $\\leq$ 2, that means the leading coefficient of t2 does not have to be non-zero. You could even have all coefficients equal to 0 and thus simply have f(t)=0.\n\n3. Apr 5, 2013\n\n### mahrap\n\nSo what is the difference between a polynomial with degree = 2 and a polynomial with degree \u2264 2 or in general what is the difference between a polynomial with degree = 2 vs a polynomial with degree \u2264 n ?\n\n4. Apr 5, 2013\n\n### Staff: Mentor\n\nf(t) = at2 + bt + c, a 2nd-degree polynomial, also called a quadratic polynomial.\nDegree \u2264 2 would also include 1st degree polynomials, such as g(t) = at + b, or zero-degree polynomials, such as h(t) = a.\nI assume you mean degree = n vs. degree \u2264 n. An nth degree polynomial has to have a term in which the variable has an exponent of n. A polynomial of degree \u2264 n includes lower-degree polynomials.\n\n5. Apr 5, 2013\n\n### mahrap\n\nYes sorry for the typo. I meant to say degree = n. The reason I started this thread was with regards to a problem in my linear algebra class where the problem states:\n\nFind all polynomials f(t) of degree \u2264 2 whose graphs run through the points (1,3) and (2,6) , such that f`(1) = 1 .\n\nWhen I started to solve the problem I used the form f(t) = a + bt + ct^2 for my polynomial and after solving the matrices I got c = 2. However when I checked the solutions in the back of the book they had a = 2 which makes sense because they used f(t) = at^2 + bt + c . So what confused me was, how is one suppose to know which form to use to get the right a or c even though both essentially yield the same variables considering 2 is the value of the variable in front of the t^2 term. In general you mentioned a polynomial of degree \u2264 n includes lower-degree polynomials. According to that statement, how would I know to use f(t) = at^2 + bt + c or f(t) = at + c when setting up my system of equations?\n\n6. Apr 5, 2013\n\n### Mentallic\n\nSince the polynomial is of degree $\\leq$ 2, that means the degree can at most be 2, so you should have $f(t)=at^2+bt+c$ which also ensures that even if the polynomial is just degree 1, then a=0. If you used $f(t)=at+b$ or $f(t)=a$ then you're assuming the equation must be of that form, and you'll soon find that there is no possible solution to the question if you begin with the assumption that the polynomials are of degree $\\leq$ 1, or equivalently, $f(t)=at+b$.\n\n7. Apr 6, 2013\n\n### HallsofIvy\n\nStaff Emeritus\nIf f(t) is a \"polynomial of degree 2 or less\" it can be written in the form $at^2+ bt+ c$ where a, b, and c can be any numbers.\n\nIf f(t) is a \"polynomial of degree 2\" it can be written in the form $at^2+ bt+ c$ where a, b, and c can be any numbers- except that a cannot be 0.", "date": "2017-12-15 18:09:48", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-polynomial-of-degree-2-what-does-this-mean.683463/", "openwebmath_score": 0.6568679213523865, "openwebmath_perplexity": 320.6042227762767, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707960121133, "lm_q2_score": 0.8807970701552504, "lm_q1q2_score": 0.8608653335827743}}
{"url": "https://math.stackexchange.com/questions/932152/finding-the-definite-integral-of-a-function-that-contains-an-absolute-value", "text": "# Finding the definite integral of a function that contains an absolute value\n\nThe integral in question is this:\n\n$\\int_{-2\\pi}^{2\\pi}xe^{-|x|}$\n\nMy attempt:\n\nSince there is a modulus, we split it up into cases. I'm not really sure which cases to split it into, do I just separately integrate these two functions?\n\n$\\int_{-2\\pi}^{2\\pi}xe^{-x}$\n\n$\\int_{-2\\pi}^{2\\pi}xe^{x}$\n\nOr do I split it into these two? $\\int_{0}^{2\\pi}xe^{-x}$ $\\int_{-2\\pi}^{0}xe^{x}$\n\nI am leaning towards the second split (splitting the bounds of the integral), which seems better.\n\nThe question is: What does it mean by 'splitting it into cases', and why does it work? Another side question I have is how to differentiate a function that has a modulus somewhere inside it.\n\n\u2022 Let $f\\colon [-2\\pi, 2\\pi]\\to \\mathbb R, x\\mapsto xe^{-|x|}$. You should know by now that $\\displaystyle \\int \\limits_{-2\\pi}^{2\\pi}f=\\int \\limits_{-2\\pi}^{0}f+\\int \\limits_{0}^{2\\pi}f$ and also $\\forall x\\in [-2\\pi, 0]\\left(f(x)=xe^{-(-x)}\\right)$. \u2013\u00a0Git Gud Sep 15 '14 at 10:29\n\u2022 @GitGud What if there is a modulus and the bounds of the integral are both positive? \u2013\u00a0robertmartin8 Sep 15 '14 at 10:30\n\u2022 @ surelyyourjoking. Then $x$ would only ever be a positive value (or zero) and so the modulus can be replaced simply by $x$ alone since $\\mid x\\mid\\geq 0$ for all $x$. \u2013\u00a0Pixel Sep 15 '14 at 10:50\n\u2022 ah I see, so the modulus is meaningless in an indefinite integral? \u2013\u00a0robertmartin8 Sep 15 '14 at 11:24\n\u2022 No, the modulus is not meaningless for indefinite integrals. When considering definite integrals as your question does we have bounds on the integral, which tells us the domain of the integrand to consider. Because we know the domain, we also know the sign (positive/negative) of a given $x$ value. Hence we can then split the integral into positive/negative parts to evaluate it. Notice also that an indefinite integral can be written as a definite integral since $$\\int f(x)dx = \\int_\\lambda^x f(t)dt,$$ where the \"lower bound\" $\\lambda$ gives a constant of integration. \u2013\u00a0Pixel Sep 15 '14 at 12:00\n\n$$I=\\int_{-2\\pi}^{2\\pi} xe^{-\\mid x\\mid}dx.$$\n\nLoosely speaking, you can think about the definite integral as the area bounded by the function $xe^{-\\mid x\\mid}$ and the $x$-axis, as the variable $x$ moves from $x=-2\\pi$ to $x=0$ then from $x=0$ through to $x=2\\pi$. So, intuitively it's not too much of a step to see that $$I=\\int_{-2\\pi}^{0} xe^{-\\mid x\\mid}dx+\\int_{0}^{2\\pi} xe^{-\\mid x\\mid}dx.$$ Notice in the left integral the $x$ values are only ever negative or zero, and in the right integral the $x$ values are only ever positive or zero, so we can rewrite the whole expression $$I=\\int_{-2\\pi}^{0} xe^{x}dx+\\int_{0}^{2\\pi} xe^{-x}dx,$$ since $-|x|=x$ for $x\\leq 0$ and $-|x|=-x$ for $x\\geq 0$. You can now evaluate the integrals separately to obtain the correct result. Hope this helps.\n\n$$\\int_{-2\\pi}^{2\\pi}xe^{-\\left|x\\right|}dx=\\int_{-2\\pi}^{0}xe^{-\\left|x\\right|}dx+\\int_{0}^{2\\pi}xe^{-\\left|x\\right|}dx=\\int_{-2\\pi}^{0}xe^{x}dx+\\int_{0}^{2\\pi}xe^{-x}dx$$\n\nThis is a split of cases killing the annoying modulus.\n\n\u2022 Alright, I understand it for this example now. But in general? when the bounds of the integral are both positive? \u2013\u00a0robertmartin8 Sep 15 '14 at 10:35\n\u2022 @ surelyyourjoking. Then $x$ would only ever be a positive value (or zero) and so the modulus can be replaced simply by $x$ alone since $\u2223x\u2223\u22650$ for all $x$. \u2013\u00a0Pixel Sep 15 '14 at 10:58\n\nIf you have a function $f: [a,b] \\to \\Bbb R$ defined by parts, as in: $$f(x) = \\begin{cases} f_1(x), \\mbox{if } a \\leq x \\leq c \\\\ f_2(x), \\mbox{if } c < x \\leq b\\end{cases}$$ then: $$\\int_a^b f(x) \\ \\mathrm{d}x = \\int_a^c f_1(x) \\ \\mathrm{d}x + \\int_c^bf_2(x) \\ \\mathrm{d}x.$$\n\nIn your case, $f(x) = xe^{-|x|}$, so $c = 0$ and $f_1(x) = xe^{x}$ and $f_2(x) = xe^{-x}$, using the definition of absolute value. Remember the interpretation of the integral for a positive function: the integral is the area, so the sum of the areas is the sum of integrals.\n\n\u2022 so this works even when the bounds of the integral are both positive? \u2013\u00a0robertmartin8 Sep 15 '14 at 10:34\n\u2022 Sure, why not? You have to pay attention where to split the integral. Try splitting $$\\int_{-2}^3 |x - 1|e^x \\ \\mathrm{d}x$$ to see if you can do it (you don't need to solve it, that's not the point here) \u2013\u00a0Ivo Terek Sep 15 '14 at 10:36\n\n$|x|=x$ for x>0\n\nand = -x for x<0\n\nand 0 for x=0\n\nsince x is positive and negative in the limits $-2\\pi$ to $2\\pi$ you have to split it into two parts and then evaluate your integral.\n\nSo in this function you are splitting into cases for |x| where x>0 and x<0.\n\nthis is done because |x| is defined that way in its domain.since you get two different functions for different intervals in the domain, you have to consider two different limits. imagine a function\n\nf(x)=0 for x<0\n\nand 1 for x>0\n\nand we are evaluating an integral $\\int{xf(x)}dx$ with limits -1 to 1.\n\nso because of the definition of f(x) you have to split the integral into two.\n\nhere its just two integrals sometimes you have to split into many more.\n\n\u2022 I understand that you have to split it up into cases, I just dont get technically how or why this is done \u2013\u00a0robertmartin8 Sep 15 '14 at 10:33", "date": "2020-05-30 03:42:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/932152/finding-the-definite-integral-of-a-function-that-contains-an-absolute-value", "openwebmath_score": 0.9127959609031677, "openwebmath_perplexity": 241.81852630458613, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914019704466, "lm_q2_score": 0.8902942312159383, "lm_q1q2_score": 0.8608379738247002}}
{"url": "https://math.stackexchange.com/questions/1566703/are-the-following-sets-exactly-the-same-mathbbn-mathbbn-0", "text": "# Are the following sets exactly the same: $\\mathbb{N}$, $\\mathbb{N_{-\\{0\\}}}$, $\\mathbb{N^+}$ and $\\mathbb{Z^+}$?\n\nAt university, my Maths Analysis lecturer said that this is still debated nowadays. For example, a definition of an rational number is $\\cfrac{p}{q}$ where $p\\in \\mathbb{Z}$ and $q\\in \\mathbb{N}$ (or should it be $q\\in \\mathbb{N^+}$ or the other two in the title?).\n\nI know that $\\mid \\mathbb{N}|= \\mid \\mathbb{N^+}|$ since I can find a bijection by defining the map: $f: \\mathbb{N^+} \\mapsto \\mathbb{N}$ by $f(x)=x-1$.\n\nBut the problem is that this raises the question as to whether $0 \\in \\mathbb{N}$ or $0 \\in \\mathbb{Z^+}$. This is similar to asking whether $0$ is positive?\n\nSo which set should I use to define the denominator $q$ of a rational number and why? The choices are $\\mathbb{N}$, $\\mathbb{N_{-\\{0\\}}}$, $\\mathbb{N^+}$ and $\\mathbb{Z^+}$.\n\nI acknowledge that this question is subject somewhat to opinion, but I value the opinions of users of this site. So all I am asking is which set of the above would you choose $q$ to belong to for the definition of the rational number?\n\n\u2022 In class-related work you should use what your lecturer uses. Elsewhere, you can use what you like and is most convenient for what you want to do. Maybe you work closely with other people that already have a standard. If the distinction is important for what you do, you should mention whether or not $0 \\in \\mathbb{N}$. Dec 8, 2015 at 23:21\n\u2022 (1) We don't need a tag for \"mappings\" because those are in fact functions. (2) This question is about notation, and not at all about functions, so the tag (and the functions tags) is entirely irrelevant. Dec 10, 2015 at 10:38\n\u2022 @AsafKaragila Okay, understood; thanks for letting me know. Dec 10, 2015 at 10:41\n\nThere are two different sets of relevance here; each has several notations in use.\n\nThe set $\\{0,1,2,3,\\ldots\\}$ is considered fundamental in areas such as logic, set theory and computer science. It can be unambiguously written $\\mathbb N_0$. Set theorists often refer to it as $\\omega$, deftly avoiding the notational trouble surrounding the $\\mathbb N$s.\n\nThe set $\\{1,2,3,4,\\ldots\\}$ is often needed in most of the rest of mathematics, where people find it more natural to start counting at $1$. It can be unambigously written $\\mathbb N_+$ or $\\mathbb Z_+$ (the location of the plus sign can vary).\n\nEither of these sets is often notated just $\\mathbb N$ -- it is up to the reader who encounters this to know (or guess) which convention the author is following, in the cases where the difference matters. It is considered polite for an author to state which convention he follows before he uses the naked $\\mathbb N$, but this is not always done in practice.\n\nIn English it is unambiguous that $0$ is not \"positive\". However other languages may follow other conventions; in French the number $0$ counts as both \"positif\" and \"n\u00e9gatif\" and one has to speak about \"strictement positif\" if one needs to exclude $0$. (This is not a mathematical difference; the concepts $>0$ and $\\ge 0$ both exist independently of which words we use about them, and the two languages simply chose different concepts to have a short word for).\n\n\u2022 Nice answer, thanks very much. So in your opinion, which of the four would you select, I know it's subject to opinion and therefore arbitrary, but could you tell me which one you would use? Dec 10, 2015 at 7:59\n\u2022 @BLAZE: Personally? I just use $\\mathbb N$ for whatever of the sets I'm speaking about (usually, but not always, the one that contains $0$) unless there seems to be a particular risk that the reader will miss my point if he misunderstands me. In that case I write $\\mathbb N_0$ or $\\mathbb N_+$. (Except forsuch set-theory contexts where it is conventional to use $\\omega$). I'm not holding this up as a shining example to follow, though. Dec 10, 2015 at 15:44\n\nIt usually depends on the course and the lecturer.\n\n$0 \\in \\Bbb N$ vs $0 \\not \\in \\Bbb N$ depends on the definition of $\\Bbb N$. The first one is usually found in most logic courses, while on say, analysis, this is not as common.\n\nWhenever I see $\\Bbb Z^+$ I'll think they refer to $\\{1,2,3,...\\}$, same for $\\Bbb N^+$(this last one is quite uncommon).", "date": "2023-03-31 07:09:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1566703/are-the-following-sets-exactly-the-same-mathbbn-mathbbn-0", "openwebmath_score": 0.8430256247520447, "openwebmath_perplexity": 412.1521452032018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924802053235, "lm_q2_score": 0.8856314798554444, "lm_q1q2_score": 0.8608271386526044}}
{"url": "https://math.stackexchange.com/questions/1046321/approximating-log-x-with-roots", "text": "Approximating $\\log x$ with roots\n\nThe following is a surprisingly good (and simple!) approximation for $\\log x+1$ in the region $(-1,1)$: $$\\log (x+1) \\approx \\frac{x}{\\sqrt{x+1}}$$\n\nThree questions:\n\n\u2022 Is there a good reason why this would be the case?\n\u2022 How does one go about constructing the \"next term\"?\n\u2022 Are the any papers on \"generalized Pade approximations\" that involve radicals?\n\u2022 I suggest you write down the Taylor series of $\\log(x+1)$ and $(x+1)^{-1/2}$ and see the few first terms agree. \u2013\u00a0LinAlgMan Dec 1 '14 at 10:42\n\u2022 @LinAlgMan - That is true, but it is not the reason. This approximation works much better than the Taylor series, even to high orders, probably because it accounts for the pole. \u2013\u00a0nbubis Dec 1 '14 at 10:43\n\u2022 May I ask how you got this interesting approximation ? \u2013\u00a0Claude Leibovici Dec 1 '14 at 10:49\n\u2022 @ClaudeLeibovici - the function came up in a physics calculation, and upon plotting, I noticed it looked oddly familiar. \u2013\u00a0nbubis Dec 1 '14 at 10:51\n\u2022 @Lucian - I'm not sure I see the connection. \u2013\u00a0nbubis Dec 8 '14 at 20:12\n\nLet's rewrite both sides in terms of $y = x + 1$: we get\n\n$$\\log y \\approx \\sqrt{y} - \\frac{1}{\\sqrt{y}}$$\n\non, let's say, the interval $\\left( \\frac{1}{2}, 2 \\right)$ (I hesitate to discuss the entire interval $(0, 2)$; it seems to me that the approximation is not all that good near $0$). The RHS should look sort of familiar: let's perform a second substitution $y = e^{2z}$ to get\n\n$$2z \\approx e^z - e^{-z} = 2 \\sinh z$$\n\non the interval $\\left( - \\varepsilon, \\varepsilon \\right)$ where $\\varepsilon = \\frac{\\log 2}{2} \\approx 0.346 \\dots$. Of course now we see that the LHS is just the first term in the Taylor series of the RHS, and on a smaller interval than originally. Furthermore, the Taylor coefficients of $2 \\sinh z$, unlike the Taylor coefficients of our original functions, decrease quite rapidly. The next term is $\\frac{z^3}{3}$, which on this interval is at most\n\n$$\\frac{\\varepsilon^3}{3} \\approx 0.0138 \\dots$$\n\nand this is more or less the size of the error in the approximation between $\\log 2$ and $\\frac{1}{\\sqrt{2}}$ obtained by setting $y = 2$, or equivalently $x = 1$.\n\nWith the further substitution $t = \\sinh z$, the RHS is just the first term in the Taylor series of the LHS. To get the \"next term\" we could look at the rest of the Taylor series of $\\sinh^{-1} t$. The next term is $- \\frac{t^3}{6}$, which gives\n\n$$z \\approx \\frac{e^z - e^{-z}}{2} - \\frac{(e^z - e^{-z})^3}{48}$$\n\nor\n\n$$\\log y \\approx \\left( \\sqrt{y} - \\frac{1}{\\sqrt{y}} \\right) - \\frac{1}{24} \\left( \\sqrt{y} - \\frac{1}{\\sqrt{y}} \\right)^3.$$\n\nI don't know if this is useful for anything. The series to all orders just expresses the identity\n\n$$\\log y = 2 \\sinh^{-1} \\frac{\\left( \\sqrt{y} - \\frac{1}{\\sqrt{y}} \\right)}{2}.$$\n\n\u2022 Very nicely done!! As you noticed though, near $y\\to 0$, higher orders seem to actually ruin the approximation, which I find curious. \u2013\u00a0nbubis Dec 1 '14 at 11:26\n\u2022 @nbubis: well, the radius of convergence of the Taylor series of $\\sinh^{-1} t$ is $1$, and as $y \\to 0$, $\\sqrt{y} - \\frac{1}{\\sqrt{y}}$ gets arbitrarily large (in absolute value)... \u2013\u00a0Qiaochu Yuan Dec 1 '14 at 11:59\n\nThe simplest Pade approximant we could build seems to be $$\\log(1+x)\\approx\\frac{x}{1+\\frac{x}{2}}$$ and we can notice the similarity of denominators close to $x=0$.\n\nHowever, the approximation given in the post seems to be significantly better for $x<\\frac 12$.\n\n\u2022 Indeed. It is probably better because it has a pole in the right place. \u2013\u00a0nbubis Dec 1 '14 at 11:00\n\nI am being rather late, yet still there is some interesting information I can add.\n\nThe Padet approximant is of the form $\\log(1+x)\\approx \\frac{x}{1+x/2}$ as noted by other posters. This partially explains why $\\frac{x}{\\sqrt{1+x}}$ is a good approximation, what it does not explain is why the square-root approximation is better than a supposedly \"great\" Pade approximation. @nbubis had an idea that it works better because it has pole in the correct spot, but it seems that it is actually a red herring.\n\nLet's take a look on more general Pade $(1,n)$ approximation, it equals $\\log(1+x)\\approx\\frac{x}{1+x/2-x^2/12+x^4/24+...}$.\n\nNow the reason $\\frac{x}{\\sqrt{1+x}}$ approximation performs better can be explained by $\\sqrt{1+x} \\approx 1+x/2 -x^2/8$ and noting that $1+x/2-x^2/8$ is closer to the \"true value\" of the denominator than the first Pade approximant $1+x/2$.\n\nTo see that it is indeed the case consider the approximation\n\n$\\log(1+x)=\\frac{x}{(1+5x/6)^{3/5}}$.\n\nNow, as you can quickly check, $(1+5x/6)^{3/5}$ has the Taylor expansion $\\approx 1+x/2 -x^2/12$ which aggrees with first 3 terms of $(1,n)$ Pade approximant. Now if you plot it it will turn out that it is even better than originally suggested $\\frac{x}{\\sqrt{1+x}}$ despite having pole in the wrong place.\n\nRegarding method by @QiaochuYuan, you can perform the same \"trick\" to get better approximations which will be performing better in the neighbourhood of $x=0$ but worse when $x$ is large, for example\n\n$\\log (1+x) \\approx \\frac{x}{(1+5x/6)^{3/5}} + \\frac{x^4}{108 (1+5x/6)^{12/5}}$\n\nBut in disguise what you are actually making is finding better approximations to some Pade approximant.\n\nSome of the other approximations you can find in the same way are\n\n$\\log(1+x)\\approx \\frac{x}{\\sqrt{1+x+x^2/12}}$ and $\\log(1+x)\\approx \\frac{x}{(1+3x/2+x^2/2)^{1/3}}$ which are good simply beause they coincide with Pade approximant up to the terms of high order. I guess the first among those two is another reason why $\\frac{x}{\\sqrt{1+x}}$ worked so well.\n\nShort version: It's actually a coincidence, $\\sqrt{1+x}$ happen to have Taylor expansion $\\sqrt{1+x}\\approx 1+x/2-x^2/8$ which coincides with expansion of $x/\\log(1+x)\\approx 1+x/2 -x^2/12$ up to 2 terms and the third term is not that different to mess up the approximation.\n\n\u2022 I like this! That makes a lot of sense. \u2013\u00a0nbubis Dec 1 '17 at 6:00\n\nA different approach.\n\nThe function $f(x)=x-\\log(1+x)\\sqrt{1+x}$ is continuous and increasing on $[-1,1]$ (I have not proved it, but a graph of $f'$ is sufficiently convincing.) For any $a\\in(0,1)$ $$f(-a)\\le f(x)\\le f(1)=0.0197419,\\quad-a\\le x\\le1.$$ Take for instance $a=0.8$ we obtain $$-\\frac{0.0802375}{\\sqrt{1+x}}\\le\\frac{x}{\\sqrt{1+x}}-\\log(1+x)\\le\\frac{0.0197419}{\\sqrt{1+x}},\\quad -.8\\le x\\le1.$$\n\nThis shows that $\\log(1+x)\\sqrt{1+x}$ is a good approximation of $x$. $$f(x)=-\\frac{x^3}{24}+\\dots$$ is an alternate series. This explains the vey good approximation for $x>0$, and the not so good for $x<0$.\n\n\u2022 I'm not sure how that helps - this is just an evaluation of the approximation. \u2013\u00a0nbubis Dec 1 '14 at 11:56\n\nOne reason may be that their Taylor series around $x=0$ start the same: $$\\log (x+1) \\approx x-x^2/2+x^3/3+\\cdots$$ $$\\frac{x}{\\sqrt{x+1}} \\approx x-x^2/2+3x^3/8+\\cdots$$ So they agree to order 2 for $|x|<1$. They almost agree to order 3 because $1/3 \\approx 3/8$ roughly.\n\nHowever, this is an a posteriori reason. I don't know why this approximation should be good a priori.\n\n\u2022 That doesn't necessarily tell you much about how good of an approximation to expect on an interval as large as $(-1, 1)$. \u2013\u00a0Qiaochu Yuan Dec 1 '14 at 10:46\n\u2022 This was what I was just typing ! The question is interesting. \u2013\u00a0Claude Leibovici Dec 1 '14 at 10:46\n\u2022 For example, when $x = 1$ the LHS is $\\log 2 \\approx 0.693 \\dots$ while the RHS is $\\frac{1}{\\sqrt{2}} \\approx 0.707 \\dots$. These agree substantially better that can be accounted for by the first two terms of the Taylor series, which give $0.5$. \u2013\u00a0Qiaochu Yuan Dec 1 '14 at 10:48\n\u2022 This is clearly not the reason. The approximation does much better than the Taylor series way past second order.(probably because it has a pole) \u2013\u00a0nbubis Dec 1 '14 at 10:49\n\n$$\\log(x+1)=\\lim_{n\\to\\infty}n(\\sqrt[n]{x+1}-1).$$\n\nIn the case $n=2$, $$2(\\sqrt{x+1}-1)=2\\frac{x}{\\sqrt{x+1}+1}\\approx\\frac x{\\sqrt{x+1}}$$ for small $x$.\n\nThe approximation works better as it has a vertical asymptote at $x=-1$.\n\n\u2022 Nice :) Though the approximation ends up being better than the derivation... \u2013\u00a0nbubis Dec 1 '14 at 11:14\n\u2022 This still doesn't explain most of the agreement. Again taking $x = 1$ we have $2 (\\sqrt{2} - 1) \\approx 0.828 \\dots$, which is maybe 15% bigger than either the LHS or the RHS, which agree to maybe within 2%. \u2013\u00a0Qiaochu Yuan Dec 1 '14 at 11:15\n\u2022 Anyone who's still trying to answer the question should actually plot the functions (I did it in WolframAlpha) to see how close the agreement actually is. I think the pole is a red herring: the agreement is really not very good close to the pole. \u2013\u00a0Qiaochu Yuan Dec 1 '14 at 11:17\n\u2022 I don't thank the downvoters. @QiaochuYuan: the question is not a contest to the best approximation. It is about why $\\log (x+1) \\approx \\dfrac{x}{\\sqrt{x+1}}$. \u2013\u00a0Yves Daoust Oct 31 '17 at 7:45", "date": "2020-01-20 03:42:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1046321/approximating-log-x-with-roots", "openwebmath_score": 0.8592144846916199, "openwebmath_perplexity": 289.8432629016002, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924802053235, "lm_q2_score": 0.885631470799559, "lm_q1q2_score": 0.8608271298503518}}
{"url": "https://afcars.cz/blue-and-nevts/597eec-exponential-functions-examples", "text": "# exponential functions examples\n\nEach output value is the product of the previous output and the base, 2. Find r, to three decimal places, if the the half life of this radioactive substance is 20 days. Sign up to read all wikis and quizzes in math, science, and engineering topics. Now, let\u2019s take a look at a couple of graphs. If b b is any number such that b > 0 b > 0 and b \u00e2\u0089\u00a0 1 b \u00e2\u0089\u00a0 1 then an exponential function is a function in the form, f (x) = bx f (x) = b x We will see some examples of exponential functions shortly. Compare graphs with varying b values. from which we have In fact this is so special that for many people this is THE exponential function. n \\log_{10}{1.03} \\ge& 1 \\\\ In fact, that is part of the point of this example. For every possible $$b$$ we have $${b^x} > 0$$. Find the sum of all positive integers aaa that satisfy the equation above. 100 + (160 - 100) \\frac{1.5^{12} - 1}{1.5 - 1} \\approx& 100 + 60 \\times 257.493 \\\\ \\approx& 15550. Population: The population of the popular town of Smithville in 2003 was estimated to be 35,000 people with an annual rate of increase (growth) of about 2.4%. You appear to be on a device with a \"narrow\" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$f\\left( { - 2} \\right) = {2^{ - 2}} = \\frac{1}{{{2^2}}} = \\frac{1}{4}$$, $$g\\left( { - 2} \\right) = {\\left( {\\frac{1}{2}} \\right)^{ - 2}} = {\\left( {\\frac{2}{1}} \\right)^2} = 4$$, $$f\\left( { - 1} \\right) = {2^{ - 1}} = \\frac{1}{{{2^1}}} = \\frac{1}{2}$$, $$g\\left( { - 1} \\right) = {\\left( {\\frac{1}{2}} \\right)^{ - 1}} = {\\left( {\\frac{2}{1}} \\right)^1} = 2$$, $$g\\left( 0 \\right) = {\\left( {\\frac{1}{2}} \\right)^0} = 1$$, $$g\\left( 1 \\right) = {\\left( {\\frac{1}{2}} \\right)^1} = \\frac{1}{2}$$, $$g\\left( 2 \\right) = {\\left( {\\frac{1}{2}} \\right)^2} = \\frac{1}{4}$$. A = a^{a}b^{b}c^{c}, \\quad B = a^{a}b^{c}c^{b} , \\quad C = a^{b}b^{c}c^{a}. 1000\u00c3\u0097(12)100005730\u00e2\u0089\u00881000\u00c3\u00970.298=298.1000 \\times \\left( \\frac{1}{2} \\right)^{\\frac{10000}{5730}} Check out the graph of $${2^x}$$ above for verification of this property. Therefore, we would have approximately 298 g. \u00e2\u0096\u00a1 _\\square \u00e2\u0096\u00a1\u00e2\u0080\u008b, Given three numbers such that 0 and! Since we are only graphing one way is if we allowed \\ ( b\\ ) have. 1\\Large |x|^ { ( x^2-x-2 ) } < 1 { 12 } 100. Curve upward, as shown in the first quadrant functions exponential functions examples starting this with. ) above for verification of this function in the exponent worked to this point population a... A very specific number into all the \\ ( { b^x } > 0\\ ) represent growth decay. Functions from these graphs exponential functions examples model exponential functions will want to use far more decimal places in computations! Graph y = 2 x is called the base, 2 aaa that satisfy the equation,. For a complete list of integral functions, there are some function evaluations that will give complex,! Through complex numbers course, built by experts for you would we have now section. formula for exponential. Business and science output value is the approximate integer population after a year we could have written \\ ( \\bf... The final section of this chapter off discussing the final property for a complete of. Approaches negative infinity, the approximate integer population after a year is 100\u00c3\u00971.512\u00e2\u0089\u0088100\u00c3\u0097129.75=12975.\u00c2 \u00e2\u0096\u00a1100 \\times 1.5^ { 12 \\approx... > 0\\ ) ( b = - 4\\ ) the function would be would be 1000\u00c3\u00971.03n.1000! The chapter on exponential functions works in exactly the same properties that exponential..., C a, b is greater than  1 , the graph of this property complex! 2 raised to the x power same manner that all three graphs pass through the y-intercept ( 0,1..", "date": "2021-05-13 00:47:47", "meta": {"domain": "afcars.cz", "url": "https://afcars.cz/blue-and-nevts/597eec-exponential-functions-examples", "openwebmath_score": 0.9695624113082886, "openwebmath_perplexity": 759.3426364341776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924793940118, "lm_q2_score": 0.885631470799559, "lm_q1q2_score": 0.8608271291318287}}
{"url": "http://mathhelpforum.com/calculus/87199-find-interval-function-concave-up.html", "text": "# Thread: Find the interval in which the function concave up??\n\n1. ## Find the interval in which the function concave up??\n\nThe graph of the function\n\nis concave up on the interval:\n\n(-infinity,infinity)\n\nNone\n\n(1,infinity)\n\n(-infinity,0)U(1,infinity)\n\n(0,infinity)\n\n=======================\n\nI solve the question and my answer is :\n\nThe graph is concave up on (-infinity,-0.5)U(0,infinity)\n\nand I'm sure from my answer but my answer is not included in the choices\n\nand When I use the graphics calculator the graph is concave up on all interval\n\nSo what is the correct answer??\n\n2. You can use derivatives to answer this question\n\nFirst you must find the critical points, take your first equation and differentiate and solve for 0\n\n$f(x)=x^4+x^3+x$\n\n$f\\prime(x)=4x^3+3x^3+1$\n\nNow that its differentiated solve for 0\n\n$4x^3+3x^2+1=0$\n\n$x(4x^2+3x+1)=0$\n\nyou a critical point at x = -1.5 and x = 0\n\nnow to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing\n\nto find concavity you take second derivative and plug in critical points if\n\n>0 then concave up and <0 concave down\n\nthe second derivative is\n\n$12x^2+6$\n\nplus in 0 and -1.5\n\nplug 0 in and you see it is >0 so intervals from 0 to infinity is concave up\n\nplug -1.5 it is also concave up because > 0 so from -inf. to -1.5\n\nsorry for lack of work studying for finals\n\n3. Originally Posted by sk8erboyla2004\nYou can use derivatives to answer this question\n\nFirst you must find the critical points, take your first equation and differentiate and solve for 0\n\n$f(x)=x^4+x^3+x$\n\n$f\\prime(x)=4x^3+3x^3+1$\n\nNow that its differentiated solve for 0\n\n$4x^3+3x^2+1=0$\n\n$x(4x^2+3x+1)=0$\n\nyou a critical point at x = -1.5 and x = 0\n\nnow to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing\n\nto find concavity you take second derivative and plug in critical points if\n\n>0 then concave up and <0 concave down\n\nthe second derivative is\n\n$12x^2+6$\n\nplus in 0 and -1.5\n\nplug 0 in and you see it is >0 so intervals from 0 to infinity is concave up\n\nplug -1.5 it is also concave up because > 0 so from -inf. to -1.5\n\nsorry for lack of work studying for finals\n\nbut I would to remind you that in concavity test we do the following :\n\n1)find first derivative\n\n2)find second derivative\n\n3)set first derivative=0 and solve for x\n\n4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..\n\n4. Originally Posted by change_for_better\nThe graph of the function\n\nis concave up on the interval:\n\n(-infinity,infinity)\n\nNone\n\n(1,infinity)\n\n(-infinity,0)U(1,infinity)\n\n(0,infinity)\n\n=======================\n\nI solve the question and my answer is :\n\nThe graph is concave up on (-infinity,-0.5)U(0,infinity)\n\nand I'm sure from my answer but my answer is not included in the choices\n\nand When I use the graphics calculator the graph is concave up on all interval\n\nSo what is the correct answer??\nA function is concave upward when its second derivative is positive.\nIf $y= x^4+ x^3+ x$, then $y\"= 12x^2+ 6x= 6x(2x+ 1)$. That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y\" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y\" is negative. If x> 0, 2x+1> 0 so y\" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!\n\n5. Originally Posted by change_for_better\n\nbut I would to remind you that in concavity test we do the following :\n\n1)find first derivative\n\n2)find second derivative\n\n3)set first derivative=0 and solve for x\n\n4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..\n\nIf you know this why you couldnt have used this to answer your own question?\n\n6. Originally Posted by HallsofIvy\nA function is concave upward when its second derivative is positive.\nIf $y= x^4+ x^3+ x$, then $y\"= 12x^2+ 6x= 6x(2x+ 1)$. That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y\" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y\" is negative. If x> 0, 2x+1> 0 so y\" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!\n\nI would like to ask you ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??\n\n7. Originally Posted by sk8erboyla2004\nIf you know this why you couldnt have used this to answer your own question?\n\nBecause ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??\n\n8. Hello, change_for_better!\n\nI agree with your intervals . . .\n\nThe graph of the function: . $f(x) \\;=\\;x^4+x^3+x$ is concave up on the interval:\n\n. . $(a)\\;(-\\infty, \\infty) \\qquad(b)\\text{ None} \\qquad (c)\\;(1,\\infty) \\qquad (d)\\;(-\\infty,0) \\;\\cup\\; (1,\\infty) \\qquad(e)\\;(0,\\infty)$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nI solved and my answer is: . $\\left(-\\infty,-\\tfrac{1}{2}\\right) \\cup (0,\\infty)$\n\nand I'm sure from my answer but my answer is not included in the choices.\n\nand when I use the graphics calculator the graph is concave up on all interval\n. .\nUm ... not quite!\n\nSo what is the correct answer?\n\n$\\begin{array}{ccc}f(x) &=&x^4+x^3+x \\\\ f'(x) &=& 4x^3+3x^2+1 \\\\ f''(x) &=& 12x^2+6x \\end{array}$\n\n$f''(x) > 0 \\quad\\Rightarrow\\quad x < \\text{-}\\tfrac{1}{2}\\:\\text{ or }\\:x > 0$\n\nIf you zoom in on your graph, there is a concave-down portion.\nCode:\n                    |\n|         *\n|        *\n|      *\n*               |   *\n- - - - - - - - - o - - - - - - - -\n*           o  |\n*        o    |\n*     *     |\n*        |\n|\n\nAnd it's right where you predicted! . . . $\\left(\\text{-}\\tfrac{1}{2},\\:0\\right)$\n\n9. Originally Posted by Soroban\nHello, change_for_better!\n\nI agree with your intervals . . .\n\n$\\begin{array}{ccc}f(x) &=&x^4+x^3+x \\\\ f'(x) &=& 4x^3+3x^2+1 \\\\ f''(x) &=& 12x^2+6x \\end{array}$\n\n$f''(x) > 0 \\quad\\Rightarrow\\quad x < \\text{-}\\tfrac{1}{2}\\:\\text{ or }\\:x > 0$\n\nIf you zoom in on your graph, there is a concave-down portion.\nCode:\n                    |\n|         *\n|        *\n|      *\n*               |   *\n- - - - - - - - - o - - - - - - - -\n*           o  |\n*        o    |\n*     *     |\n*        |\n|\nAnd it's right where you predicted! . . . $\\left(\\text{-}\\tfrac{1}{2},\\:0\\right)$\n\nThank you very very very very much Soroban for great explanation\n\nNow I understand my mistake ..\n\nSo from choices I should select none because the correct answer is not included ..", "date": "2017-02-20 10:21:23", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/87199-find-interval-function-concave-up.html", "openwebmath_score": 0.8478879332542419, "openwebmath_perplexity": 912.2231368772646, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924777713886, "lm_q2_score": 0.8856314692902446, "lm_q1q2_score": 0.8608271262277404}}
{"url": "https://math.stackexchange.com/questions/2079950/compute-the-n-th-power-of-triangular-3-times3-matrix/2079962", "text": "# Compute the $n$-th power of triangular $3\\times3$ matrix\n\nI have the following matrix\n\n$$\\begin{bmatrix} 1 & 2 & 3\\\\ 0 & 1 & 2\\\\ 0 & 0 & 1 \\end{bmatrix}$$\n\nand I am asked to compute its $n$-th power (to express each element as a function of $n$). I don't know at all what to do. I tried to compute some values manually to see some pattern and deduce a general expression but that didn't gave anything (especially for the top right). Thank you.\n\nComputing the first few powers should allow you to find a pattern for the terms. Below are some terms:\n\n$$\\left(\\begin{matrix}1 & 2 & 3\\\\0 & 1 & 2\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 4 & 10\\\\0 & 1 & 4\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 6 & 21\\\\0 & 1 & 6\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 8 & 36\\\\0 & 1 & 8\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 10 & 55\\\\0 & 1 & 10\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 12 & 78\\\\0 & 1 & 12\\\\0 & 0 & 1\\end{matrix}\\right)$$\n\nAll but the top right corner are trivial so lets focus on that pattern. (Although if you look at it carefully you should recognize the terms.)\n\nTerms: $3,10,21,36,55,78$\n\nFirst difference: $7, 11, 15, 19, 23$\n\nSecond difference: $4, 4, 4, 4$\n\nAs the second difference is a constant the formula must be a quadratic. As the second difference is 4 then it is in the form $2n^2+bn+c$. Examining the pattern gives formula of $2n^2+n=n(2n+1)$.\n\nSo the $n^{th}$ power is given by:\n\n$$\\left(\\begin{matrix}1 & 2n & n(2n+1)\\\\0 & 1 & 2n\\\\0 & 0 & 1\\end{matrix}\\right)$$\n\nThe reason I said you should recognize the pattern is because it is every second term out of this sequence: $1,3,6,10,15,21,27,37,45,55,66,78,\\cdots$ which is the triangular numbers.\n\n\u2022 It seemed more in line with what the OP had been trying. It would also let the OP see if their approach (and the powers of the matrix) were correct. \u2013\u00a0Ian Miller Jan 2 '17 at 1:28\n\u2022 This is a great approach to \"guess the formula\" (and it is enough for any practical purpose), but it is not enough to constitute a proof. Could you please make it complete (or at least mention in the body that it still needs a proof)? \u2013\u00a0dtldarek Jan 2 '17 at 9:57\n\nWrite this matrix as follows: $$\\left[ \\begin{matrix} 1 & 2&3\\\\ 0 & 1 & 2\\\\ 0 & 0 &1 \\end{matrix} \\right] = I + 2 J+ 3 J^{2}.$$ where $$I = \\left[ \\begin{matrix} 1 & & \\\\ &1 & \\\\ & & 1 \\end{matrix} \\right], ~ J = \\left[ \\begin{matrix} 0& 1 &0 \\\\ 0 &0 & 1 \\\\ 0 & 0& 0 \\end{matrix} \\right],~ J^2 = \\left[ \\begin{matrix} 0& 0 &1\\\\ 0 & 0 &0\\\\ 0& 0 & 0 \\end{matrix} \\right], ~ J^{3}=0.$$ With this relation you can expand the power of the matrix into sum of $I$, $J$ and $J^2$.\n\n\u2022 Thank you for your answer, interesting because it takes a different approach to my original idea, I will study it. \u2013\u00a0Trev\u00f6r Jan 2 '17 at 12:06\n\nDefine\n\n$$J = \\begin{bmatrix} 0 & 2 & 3\\\\ 0 & 0 & 2\\\\ 0 & 0 & 0 \\end{bmatrix}$$\n\nso that the problem is to compute $(I+J)^n$. The big, important things to note here are\n\n\u2022 $I$ and $J$ commute\n\u2022 $J^3 = 0$\n\nwhich enables the following powerful tricks: the first point lets us expand it with the binomial theorem, and the second point lets us truncate to the first few terms:\n\n$$(I+J)^n = \\sum_{k=0}^n \\binom{n}{k} I^{n-k} J^k = I + nJ + \\frac{n(n-1)}{2} J^2$$\n\nMore generally, for any function $f$ that is analytic at $1$, (such as any polynomial), if you extend it to matrices via Taylor expansion, then under the above conditions, its value at $I+J$ is given by\n\n$$f(I+J) = \\sum_{k=0}^\\infty f^{(k)}(1) \\frac{J^k}{k!} = f(1) I + f'(1) J + \\frac{1}{2} f''(1) J^2$$\n\nAs examples of things whose result you can check simply (so you can still use the method even if you're uncomfortable with it, because you can check the result), you can compute the inverse by\n\n$$(I+J)^{-1} = I - J + J^2 = \\begin{bmatrix} 1 & -2 & 1\\\\ 0 & 1 & -2\\\\ 0 & 0 & 1 \\end{bmatrix}$$\n\nand if you want a square root, you can get\n\n$$\\sqrt{I+J} = I + \\frac{1}{2} J - \\frac{1}{8} J^2 = \\begin{bmatrix} 1 & 1 & 1\\\\ 0 & 1 & 1\\\\ 0 & 0 & 1 \\end{bmatrix}$$\n\n(These are actually special cases of $(I+J)^n$ by the generalized binomial theorem for values of $n$ that aren't nonnegative integers)\n\n\u2022 Thank you so much for your answer, interesting because it takes a different approach than my original idea, I will study it. \u2013\u00a0Trev\u00f6r Jan 2 '17 at 12:06\n\nLet $I=\\begin{pmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\end{pmatrix}$, $A=\\begin{pmatrix}0&1&0\\\\0&0&1\\\\0&0&0\\end{pmatrix}$ and $B=\\begin{pmatrix}0&0&1\\\\0&0&0\\\\0&0&0\\end{pmatrix}$. Then $M=I+2A+3B$.\n\nYou can prove that for any $n$, $M^n$ can be written as $M^n=\\lambda_n I + a_n A + b_n B$ (because it's upper triangular and the symmetry along the ascending diagonal will remain).\n\nSo $M^{n+1}=M^nM=(\\lambda_n I + a_n A + b_n B) (I + 2A + 3B) = \\dots$\n\nUsing this, compute $\\lambda_n$, and then $a_n$ and finally $b_n$.\n\n\u2022 An essential point is that the matrices commute \u2013\u00a0Rene Schipperus Jan 2 '17 at 2:29\n\u2022 @ReneSchipperus : It's not really essential. The essential thing is that the vector space generated by the matrices is an algebra: $\\forall X,Y\\in Vect(I, A, B), XY \\in Vect(I, A, B)$ (which is equivalent to $\\forall X,Y\\in \\{I, A, B\\}, XY \\in Vect(I, A, B)$, because it's a vector space and you can use distributivity). In the general case, you would us the algebra of upper-triangular matrices, compute the coefficients on the diagonal, then right above the diagonal and keep going, and you'll only need the values already computed because the matrices are upper triangular. \u2013\u00a0xavierm02 Jan 2 '17 at 13:17\n\nHere is another variation based upon walks in graphs.\n\nWe interpret the matrix $A=(a_{i,j})_{1\\leq i,j\\leq 3}$ with \\begin{align*} A= \\begin{pmatrix} 1 & 2 & 3\\\\ \\color{grey}{0} & 1 & 2\\\\ \\color{grey}{0} & \\color{grey}{0} & 1 \\end{pmatrix} \\end{align*} as adjacency matrix of a graph with three nodes $P_1,P_2$ and $P_3$ and for each entry $a_{i,j}\\neq 0$ with a directed edge from $P_i$ to $P_j$ weighted with $a_{i,j}$.\n\nNote: When calculating the $n$-th power $A^n=\\left(a_{i,j}^{(n)}\\right)_{1\\leq i,j\\leq 3}$ we can interpret the element $a_{i,j}^{(n)}$ of $A^n$ as the number of (weighted) paths of length $n$ from $P_i$ to $P_j$. The entries of $A=(a_{i,j})_{1\\leq i,j\\leq 3}$ are the weighted paths of length $1$ from $P_i$ to $P_j$.\n\nSee e.g. chapter 1 of Topics in Algebraic Combinatorics by Richard P. Stanley.\n\nLet's look at the corresponding graph and check for walks of length $n$.\n\n\u2022 We see there are no directed edges from $P_2$ to $P_1$ and no directed edges from $P_3$ to $P_2$ and from $P_3$ to $P_1$ which implies there are no walks of length $n$ either. So, $A^n$ has due to the specific triangle structure of $A$ necessarily zeroes at the same locations as $A$. \\begin{align*} A^n= \\begin{pmatrix} . & . & .\\\\ \\color{grey}{0} & . & .\\\\ \\color{grey}{0} & \\color{grey}{0} & . \\end{pmatrix} \\end{align*}\n\n\u2022 It is also easy to consider the walks of length $n$ from $P_i$ to $P_i$. There is only one possibility to loop along the vertex weighted with $1$ from $P_i$ to $P_i$ and so the entries $a_{i,i}^{(n)}$ are \\begin{align*} 1\\cdot 1\\cdot 1\\cdots 1 = 1^n=1 \\end{align*} and we obtain \\begin{align*} A^n= \\begin{pmatrix} 1& . & .\\\\ \\color{grey}{0} & 1 & .\\\\ \\color{grey}{0} & \\color{grey}{0} & 1 \\end{pmatrix} \\end{align*}\n\nand now the more interesting part\n\n\u2022 $P_1$ to $P_2$:\n\nThe walks of length $n$ from $P_1$ to $P_2$ can start with zero or more loops at $P_1$ followed by a step (weigthed with $2$) from $P_1$ to $P_2$ and finally zero or more loops at $P_2$. All the loops are weighted with $1$. There are $n$ possibilities to walk this way \\begin{align*} a_{1,2}^{(n)}=2\\cdot 1^{n-1}+1\\cdot 2\\cdot 1^{n-2}+\\cdots +1^{n-2}\\cdot 2\\cdot 1+1^{n-1}\\cdot 2=2n \\end{align*}\n\n\u2022 $P_2$ to $P_3$:\n\nSymmetry is trump. When looking at the graph we observe the same situation as before from $P_1$ to $P_2$ and conclude\n\n\\begin{align*} a_{2,3}^{(n)}=2n \\end{align*}\n\n\u2022 $P_1$ to $P_3$:\n\nHere are two different types of walks of length $n$ possible. The first walk uses the weight $3$ edge from $P_1$ to $P_3$ as we did when walking from $P_1$ to $P_2$ along the weight $2$ edge. This part gives therefore \\begin{align*} 3\\cdot 1^{n-1}+1\\cdot 3\\cdot 1^{n-2}+\\cdots +1^{n-2}\\cdot 3\\cdot 1+1^{n-1}\\cdot 3=3n\\tag{1} \\end{align*} The other type of walk of length $n$ uses the hop via $P_2$. We observe it is some kind of concatenation of walks as considered before from $P_1$ to $P_2$ and from $P_2$ to $P_3$. In fact there are $\\binom{n}{2}$ possibilities to place two $2$'s in a walk of length $n$. All other steps are loops at $P_1,P_2$ and $P_3$ and we obtain \\begin{align*} \\binom{n}{2}\\cdot 2\\cdot 2=2n(n-1)\\tag{2} \\end{align*} Summing up (1) and (2) gives \\begin{align*} a_{1,3}^{(n)}=3n+2n(n-1)=n(2n+1) \\end{align*}\n\nand we finally obtain\n\n\\begin{align*} A^n=\\left(a_{i,j}^{(n)}\\right)_{1\\leq i,j\\leq 3}=\\begin{pmatrix} 1& 2n & n(2n+1)\\\\ \\color{grey}{0} & 1 & 2n\\\\ \\color{grey}{0} & \\color{grey}{0} & 1 \\end{pmatrix} \\end{align*}\n\nUsing symmetry, write the recurrence relation\n\n$$\\begin{pmatrix} a_{n+1} & b_{n+1}&c_{n+1}\\\\ 0 & a_{n+1} & b_{n+1}\\\\ 0 & 0 &a_{n+1} \\end{pmatrix}= \\begin{pmatrix} 1 & 2&3\\\\ 0 & 1 & 2\\\\ 0 & 0 &1 \\end{pmatrix} \\begin{pmatrix} a_{n} & b_{n}&c_{n}\\\\ 0 & a_{n} & b_{n}\\\\ 0 & 0 &a_{n} \\end{pmatrix}.$$\n\nThen $$\\begin{cases}a_{n+1}=a_n\\\\ b_{n+1}=b_n+2a_n\\\\ c_{n+1}=c_n+2b_n+3a_n.\\end{cases}$$\n\nWe solve with\n\n$$a_n=Cst=1,\\\\b_n=2(n-1)+Cst=2n,\\\\ c_n=4t_{n-1}+3(n-1)+Cst=2n^2+n,$$\n\nwhere $t_n$ denotes a triangular number, using the initial conditions $a_1=1,b_1=2,c_1=3$.", "date": "2019-08-18 04:39:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2079950/compute-the-n-th-power-of-triangular-3-times3-matrix/2079962", "openwebmath_score": 0.9655170440673828, "openwebmath_perplexity": 290.0539395983521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777180191995, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8608124858076805}}
{"url": "https://math.stackexchange.com/questions/2075609/trying-to-convert-a-summation-to-an-equation", "text": "# Trying to convert a summation to an equation\n\nHere is what I am trying to figure out $\\sum_{i=1}^n 3+2 (1-i)$ It would be nice if this could be put into a single equation to be used in a larger system. I know that it is possible to break up the summation into two parts like this $\\sum_{i=1}^n 3+\\sum_{i=1}^n2 (1-i)$ and it then becomes $3*n+\\sum_{i=1}^n2 (1-i)$ I get stuck in trying to convert the second summation into a regular equation. Ultimately, it seems that this should be some sort of exponential function, but I am not having any luck finding it.\n\nThis is for a personal project and not homework in case this is a concern.\n\nBrandon\n\nWell, $\\displaystyle \\sum_{i=1}^n 2(1-i) = \\sum_{i=1}^n 2 - 2\\sum_{i=1}^n i = 2n - \\frac{2n(n+1)}{2}$.\n\nThen we have $\\displaystyle \\sum_{i=1}^n 3 +2(1-i) = 3n + 2n - n(n+1) = 4n - n^2$.\n\nFor intuition on $\\displaystyle \\sum_{i=1}^n i = \\frac{n(n+1)}{2}$, notice $1 + 2 + 3 + \\ldots + (n-2) + (n-1) +n = (n+1) + (n-1 +2) + (n -2 +3) + \\ldots + (\\frac{n}{2} + (\\frac{n}{2}+1)) = (n+1) + (n+1) + (n+1) + \\ldots + (n+1)$,\n\ngrouping the first and last terms together, then the second and second to last, etc., where there are $\\frac{n}{2}$ (for even $n$) terms in the last sum. If $n$ is odd, treat with $\\frac{n+1}{2}$ where necessary.\n\nThen summing $n+1$ $\\frac{n}{2}$ times clearly gives $\\frac{n(n+1)}{2}$.\n\n\u2022 Awesome explanation. Thank you! \u2013\u00a0Brandon Dec 29 '16 at 21:02\n\nHint :- $\\sum_{i=1}^n2(1-i)=\\sum_{i=1}^n2-2\\sum_{i=1}^ni=2.n-2(\\frac{n(n+1)}{2})$.", "date": "2019-07-17 14:50:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2075609/trying-to-convert-a-summation-to-an-equation", "openwebmath_score": 0.9556393623352051, "openwebmath_perplexity": 166.52656671219535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777175136814, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8608124748481518}}
{"url": "http://math.stackexchange.com/questions/266977/show-lim-limits-n-to-infty-sqrtnneen-e/266981", "text": "# Show $\\lim\\limits_{n\\to\\infty} \\sqrt[n]{n^e+e^n}=e$\n\nWhy is $\\lim\\limits_{n\\to\\infty} \\sqrt[n]{n^e+e^n}$ = $e$? I couldn't get this result.\n\n-\nI tried it with the triangle inequality and did not get the right result. I tried it with $a^b = e^{a*ln(b)}$ but did not get any further. This is a exercise I do for exam preparation, it's not homework. \u2013\u00a0 leo Dec 29 '12 at 11:06\nThanks for all the brilliant answers. The accepted one is the one which I could understand the best, even if might not be the most elegant or shortest. \u2013\u00a0 leo Dec 29 '12 at 13:57\nWould you like me to provide more details about my solution? What do you not understand about it? It's a straightforward way to deal with n roots, and is just based on Bernoulli's inequality that $(1+x)^n \\geq 1+ nx$ for $x\\geq 0, n\\in \\mathbb{N}$. \u2013\u00a0 Calvin Lin Dec 30 '12 at 1:08\n@CalvinLin I changed my mind now. Nameless solution did just use Math I was more familiar with. But your solution is very clean and I understand it now. \u2013\u00a0 leo Jan 27 '13 at 16:31\n\nThis is equivalent to showing that $\\lim_{n \\rightarrow \\infty} \\sqrt[n]{\\frac {n^e}{e^n} + 1 } = 1$.\n\nThis is clearly bounded below by 1. It is bounded above by $1 + \\frac {n^e}{n e^n}$, which has a limit of 1 since polynomials grow much slower than exponentials.\n\n-\n\n$$\\large (n^e + e^n)^{\\frac{1}{n}} = e \\left ( 1 + \\frac {n^e}{e^n} \\right) ^{\\frac 1 n}$$ $$e \\Large \\left ( 1 + \\frac {n^e}{e^n} \\right) ^{\\frac 1 n} = e \\left ( \\underbrace { \\left ( 1 + \\frac {1}{e^{n - e \\log n}} \\right) ^{e^{n - e \\log n}} }_{e}\\right)^{ \\underbrace{\\frac{n}{e^{n - e \\log n}}}_{0}}$$\n\n-\nWhat BIG fonts you have!! \u2013\u00a0 Haskell Curry Dec 30 '12 at 8:13\n\n$$\\lim_{n\\to\\infty} \\sqrt[n]{n^e+e^n}=\\lim_{n\\to\\infty} (n^e+e^n)^{\\frac1n}=\\lim_{n\\to\\infty} e^{\\ln (n^e+e^n)\\frac1n}$$ Now you just have to compute $$\\lim_{n\\to\\infty} \\frac{\\ln (n^e+e^n)}n=\\lim_{n\\to\\infty} \\frac{\\ln (e^{e\\ln n}+e^n)}n=\\lim_{n\\to\\infty} \\frac{\\ln e^n(e^{e\\ln n-n}+1)}n=1+\\lim_{n\\to\\infty} \\frac{\\ln (e^{e\\ln n-n}+1)}n=1+\\lim_{n\\to\\infty} \\frac{\\ln (e^{-\\infty}+1)}n=1+\\lim_{n\\to\\infty} \\frac{\\ln (0+1)}n=1+\\lim_{n\\to\\infty} \\frac{\\ln (1)}n=1+0=1$$ since $e\\ln n-n\\to -\\infty$.\n\n-\ncould you please show the second last step a bit more fine grained. Otherwise this is the most clear answer to me. \u2013\u00a0 leo Dec 29 '12 at 12:56\n@leo Sure I can \u2013\u00a0 Nameless Dec 29 '12 at 12:57\n\nLet's see a more direct way\n\n$$\\lim\\limits_{n\\to\\infty} \\sqrt[n]{n^e+e^n}=\\lim\\limits_{n\\to\\infty} \\sqrt[n]{e^n}=e$$ because the ratio test applied to $\\frac{n^e}{e^n}$ yields $\\lim_{n\\to\\infty}\\frac{n^e}{e^n}=0.$\n\n-\nWhy is $n^e$ is negligible? \u2013\u00a0 leo Dec 29 '12 at 11:28\n@leo: because the exponential function grows much faster than the polynomial function. \u2013\u00a0 Chris's sis Dec 29 '12 at 11:29\nThat is of course far from rigorous. \u2013\u00a0 Eckhard Dec 29 '12 at 12:19\n@Eckhard: in addition, this is the way I was taught to answer such questions in high school. \u2013\u00a0 Chris's sis Dec 29 '12 at 12:30\nI have to agree this is far from rigorous. \u2013\u00a0 user641 Dec 29 '12 at 12:54\n\nYou can also apply two gendarmes theorem. The idea is to find sequences $l_n, u_n$ which bound (from below and above) the given sequence $a_n$, i.e. $$l_n\\le a_n\\le u_n$$ holds, and which converge to a joint limit (i.e. $\\lim_{n\\to\\infty}l_n=\\lim_{n\\to\\infty}u_n=g).$ Then the theorem says that $a_n$ is also convergent and the limit is $g$. Let's see how it works.\n\nFinding these bounds is usually pretty straightforward \u2013 lower bound is often obtained by simple missing some nonnegative terms. While seeking the upper bound, one have to remeber that the inequality have to be true only for all sufficiently large $n$'s. In this case one can write: $$\\sqrt[n]{0+e^n}\\le\\sqrt[n]{n^e+e^n}\\le\\sqrt[n]{e^n+e^n}$$ since $0\\le n^e$ and $n^e\\le e^n$ for sure if $n$ is sufficiently large. Next, we observe that\n\n$l_n:= \\sqrt[n]{e^n}=e\\longrightarrow e$ as well as\n\n$u_n:=\\sqrt[n]{2e^n}=e\\sqrt[n]{2}\\longrightarrow e\\cdot 1 =e.$\n\nThe theorem yields the claim.\n\n-\nYou can also use that $\\sqrt[n]{a+b}\\le \\sqrt[n]{a}+\\sqrt[n]{b}$. \u2013\u00a0 user641 Dec 29 '12 at 19:31\nAnother very nice answer! Thank you! \u2013\u00a0 leo Dec 29 '12 at 20:07\n$0\\le n^e$ and $n^e\\le e^n$ are true for all $n\\ge 0$ \u2013\u00a0 Henry Dec 29 '12 at 22:44\n\nYou have $$\\sqrt[n]{n^e+e^n}= \\exp \\left( \\frac{\\ln(n^e+e^n)}{n} \\right)= \\exp \\left(1+ \\frac{\\ln \\left( 1+ \\frac{n^e}{e^n} \\right)}{n} \\right)$$ But $\\ln \\left( 1+ \\frac{n^e}{e^n} \\right) \\sim \\frac{n^e}{e^n}$ so you can conclude.\n\n-\n\nNote that by L'Hospital$$\\lim_{n\\to\\infty}\\frac{\\log (n^e+e^n)}{n}=\\lim_{n\\to\\infty}\\frac{\\frac{1}{n^e+e^n}(n^{e-1}+e^n)}{1}=\\lim_{n\\to\\infty}\\frac{n^{e-1}}{n^e+e^n}+\\lim_{n\\to\\infty}\\frac{1}{1+\\frac{n^e}{e^n}}$$ Now $n.n^{e-1}\\leq n^e+e^n$, so $n^{e-1}/(n^e+e^n)\\leq1/n$. Hence taking limit on both side, the first limit is zero. For the second one it can be proved that $n^{e+1}\\leq e^n$ for all $n\\geq n_0$, for some $n_0$. So $\\lim_{n\\to\\infty}(n^e/e^n)\\leq\\lim_{n\\to\\infty}(1/n)=0$. Hence the second limit equals $1$. Now taking exponential the required limit evaluates to $e$.\n\n-\n\nTaking logs, you must show that $$\\lim_{n \\rightarrow \\infty} {\\ln(n^e + e^n) \\over n} = 1$$ Applying L'hopital's rule, this is equivalent to showing $$\\lim_{n \\rightarrow \\infty}{en^{e-1} + e^n \\over n^e + e^n} = 1$$ Which is the same as $$\\lim_{n \\rightarrow \\infty}{e{n^{e-1}\\over e^n} + 1 \\over {n^e \\over e^n}+ 1} = 1$$ By applying L'hopital's rule enough times, any limit of the form $\\lim_{n \\rightarrow \\infty}{\\displaystyle {n^a \\over e^n}}$ is zero. So one has $$\\lim_{n \\rightarrow \\infty}{e{n^{e-1}\\over e^n} + 1 \\over {n^e \\over e^n}+ 1} = {e*0 + 1 \\over 0 + 1}$$ $$= 1$$ (If you're wondering why you can just plug in zero here, the rigorous reason is that the function ${\\displaystyle {ex + 1 \\over y + 1}}$ is continuous at $(x,y) = (0,0)$.)\n\n-", "date": "2014-11-26 21:56:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/266977/show-lim-limits-n-to-infty-sqrtnneen-e/266981", "openwebmath_score": 0.944831371307373, "openwebmath_perplexity": 313.8755475987201, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109529751892, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8608008381670373}}
{"url": "https://math.stackexchange.com/questions/1991652/odds-of-drawing-an-ace-on-the-first-draw-or-a-two-on-the-second", "text": "# Odds of drawing an ace on the first draw OR a two on the second\n\nI'm working on a puzzle which describes a casino game where you draw 13 cards. You win if your first card is an ace, or your second is a two, or... all the way up to the 13th draw being a king. In order to analyse the problem, I started by simplifying to a two-card version. You draw two cards and win if the first is an ace or the second is a deuce.\n\nThe odds of drawing an ace on the first card is clearly 4/52 or 1/13.\n\nMy understanding, backed up by this question If you draw two cards, what is the probability that the second card is a queen?, suggests that the odds of drawing a 2 on the second card is also 1/13. What are the odds of drawing either? To my knowledge, you can't directly calculate that but instead calculate the inverse. The odds of not drawing a specific card should clearly be 12/13 and the odds of not drawing an ace followed by not drawing a two should be 12/13 * 12/13 = 144/169. The odds of drawing one or the other should therefore be 25/169.\n\nBut there's another approach to analysing the problem. There are 52 * 51 = 2652 different combinations for the first two cards. If A is an ace, T is a two and x is any other card, there are 5 combinations which give you a win:\n\nAA Ax AT xT TT\n\nGiven that there are four aces, four twos and 44 other cards, the number of winning hands is:\n\n4*3 + 4*44 + 4*4 + 44*4 + 4*3 = 392.\n\nSo the odds of winning should be 392 / 2652.\n\n25/169 and 392/2652 are very close but they are NOT the same number. I believe the second number is accurate, but I can't see where the logic in the first method fails. I suspect that it has to do with the possibility of drawing both being double counted but I can't see how that should matter. It seems like you should be able to treat each draw as an independent event. Additionally, the second method doesn't scale well - it would be extremely tedious to directly calculate the number of winning hands for the full 13 card game.\n\nedit: Clarified the rules for winning the game.\n\n\u2022 The odds of firstly drawing an ace and secondly a two is $\\frac4{52}\\frac4{51}$. Can you understand why? \u2013\u00a0drhab Oct 30 '16 at 15:22\n\u2022 Arthur, I'm not counting the number of ways to draw aces and twos, I'm counting the number of ways to win. Neither xA nor Tx win the game. \u2013\u00a0Dan J. Oct 30 '16 at 15:44\n\u2022 Heads up - odds and probability are two different things. \u2013\u00a0Sean Roberson Oct 30 '16 at 15:56\n\u2022 drhab, I disagree. If you draw a two on the first draw, then the odds of drawing a two on the second draw are only 3/51, not 4/51. You have to take both possibilities into account and that gives you odds of 1/13. See the linked question in my submission. \u2013\u00a0Dan J. Oct 30 '16 at 16:04\n\u2022 The probability of firstly drawing an ace (not a two) and secondly a two is $\\frac4{52}\\frac4{51}$. If firstly an ace has been drawn then $4$ of the remaining $51$ cards are two's. \u2013\u00a0drhab Oct 30 '16 at 16:26\n\nLet $A$ denote the event that the first draw will be an ace and let $B$ be the event that the second draw will be a two.\n\nThen:$$\\Pr(A\\cup B)=$$$$\\Pr(A)+\\Pr(B)-\\Pr(A\\cap B)=$$$$\\Pr(A)+\\Pr(B)-\\Pr(A)\\Pr(B\\mid A)=$$$$\\frac4{52}+\\frac4{52}-\\frac{4}{52}\\frac{4}{51}$$\n\nThe problem with your first method is that the events are not independent. Yes, the probability that you don't draw an ace from a full deck is $12/13$, and the probability that you don't draw a two from a full deck is $12/13$, but if you draw one card and then draw a second, the outcome of the second draw is dependent on the outcome of the first draw. Hence you cannot multiply the probabilities.\n\nYour second method of calculating $P(\\text{ace on first draw or two on second draw})$ is correct because it takes into account the outcome of the first draw and the second draw.\n\n\u2022 Then is the question I linked to about drawing a queen on the second card wrong? In particular, look at A.J.'s accepted answer showing that the odds remain 1/13 even if you take the first draw into account. Is that incorrect or does it not apply here? And if doesn't apply, why not? Also, my description of the game was ambiguous. It should be OR, not AND. I'll edit. \u2013\u00a0Dan J. Oct 30 '16 at 15:50\n\u2022 I'm sorry, I misunderstood the condition for winning! As for the accepted answer in the link, the probability that the second card is a queen given that you know nothing about the first card is indeed $1/13$. Similarly, the probability that the second card is not a two given that you know nothing about the first card is $12/13$. But you cannot use this to calculate $P(\\text{first card not ace and second card not two})$ because in that event you do have information about the first card. \u2013\u00a0kccu Oct 30 '16 at 20:23\n\nOn questions like these you have to stop yourself from taking shortcuts and keep your eye on the big picture. Therefore always use both cards. First the number of probabilities of getting an ace on the first throw is 4 times the other 51 cards. Second the odds on getting a 2 on the second card if the first one isn't a ace or a 2 is 4 times 44 plus if the first card is a 2 then 4 times 3. Third add up all the 4 times and you get 4 times 98 which equals 392.\n\n\u2022 4 * 51 +\n\u2022 4 * 44 +\n\u2022 4 * 3 =\n\u2022 4 * 98 = 392\n\nOr for simplicity you could just add the odds of an ace for the first card (4 times 51) and a deuce for the second card (51 times 4) then minus one of the double wins (4 times 4) and you get 4 times (51 + 51 - 4) equals 4 times 98 equals 392.", "date": "2021-04-15 10:08:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1991652/odds-of-drawing-an-ace-on-the-first-draw-or-a-two-on-the-second", "openwebmath_score": 0.5406288504600525, "openwebmath_perplexity": 221.9916547173893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810952529396, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8608008313165687}}
{"url": "http://mathhelpforum.com/calculus/41982-find-indicated-integrals.html", "text": "# Thread: Find the Indicated Integrals\n\n1. ## Find the Indicated Integrals\n\nI'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.\n\nThe three are:\n\n$\\int ln(x^4)/x$ dx\n\n$\\int$((e^t^x) cos (e^t))/(3+5 sin (e^t))\n\nand lastly\n\n$\\int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$\n\n2. Hi\nOriginally Posted by Pikeman85\n$\\int ln(x^4)/x$ dx\n$\\frac{\\ln(x^4)}{x}=\\frac{4\\ln x}{x}=4\\times \\frac{1}{x}\\times \\ln x$ and as the derivative of $x\\mapsto \\ln x$ is $x\\mapsto\\frac{1}{x}$...\n$\\int$((e^t^x) cos (e^t))/(3+5 sin (e^t))\nIs it $\\int \\frac{\\mathrm{e}^{tx} \\cos (\\mathrm{e}^t)}{3+5 \\sin (\\mathrm{e}^t)}\\,\\mathrm{d}{\\color{red}x}$ or $\\int \\frac{\\mathrm{e}^{tx} \\cos (\\mathrm{e}^t)}{3+5 \\sin (\\mathrm{e}^t)}\\,\\mathrm{d}{\\color{red}t}$ ?\n\n$\\int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$\n$\\int_0^{\\frac{4}{5}} \\frac{\\arcsin \\left(\\frac{5}{4}x\\right)}{\\sqrt{16-25x^2}}\\,\\mathrm{d}x$\nRemember that $\\arcsin'x=\\frac{1}{\\sqrt{1-x^2}}$\n\n3. Originally Posted by Pikeman85\nI'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.\n\nThe three are:\n\n$\\int ln(x^4)/x$ dx\n\n$\\int$((e^t^x) cos (e^t))/(3+5 sin (e^t))\n\nand lastly\n\n$\\int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$\n\nfor the frist one\n\n$\\int \\frac{\\ln(x^4)}{x}dx=4\\int\\frac{\\ln(x)}{x}dx$\n\nLet $u=\\ln(x) \\implies du=\\frac{1}{x}dx$\n\n$4\\int udu=2u^2+C=2\\left( \\ln(x)\\right)^2+C$\n\nFor number 2 you have both x's and t's but no dx or dt what variable are integrating with respect to?\n\nFor the last one\n\n$\\int_{0}^{4/5}\\frac{\\sin^{-1\\left( \\frac{5}{4}x\\right)}}{\\sqrt{16-25x^2}}dx$\n\nlet $x=\\frac{4}{5}\\sin(t) \\implies dx=\\frac{4}{5}\\cos(t)dt$\n\n$\\int_{0}^{4/5}\\frac{\\sin^{-1\\left( \\frac{5}{4}x\\right)}}{\\sqrt{16-25x^2}}dx=\\int_{0}^{\\frac{\\pi}{2}}\\frac{t}{\\sqrt{1 6-16\\sin^2(t)}}\\left( \\frac{4}{5}\\cos(t)dt \\right)=$\n\n$\\frac{1}{5}\\int_{0}^{\\pi/2}tdt=\\frac{1}{5} \\left[ \\frac{1}{2} \\left( \\frac{\\pi}{2}\\right)^2-0\\right]=\\frac{\\pi^2}{40}$\n\nHere is a web page with some La tex code for you\nHelpisplaying a formula - Wikipedia, the free encyclopedia\nGood luck.\n\n4. Hello, Pikeman85!\n\nThese all require \"simple\" substitutions.\n. . The trick is to recognize them.\n\n$\\int \\frac{\\ln(x^4)}{x}\\,dx$\nWe have: . $\\int\\frac{4\\ln(x)}{x}\\,dx \\;=\\;4\\int \\ln x\\,\\frac{dx}{x}$\n\nLet $u \\:=\\:\\ln(x) \\quad\\Rightarrow\\quad du \\:=\\:\\frac{dx}{x}$\n\nSubstitute: . $4\\int u\\,du$ . . . etc.\n\n$\\int \\frac{e^{t}\\cos(e^t)\\,dt}{3+5\\sin(e^t)}$\nLet $u \\:=\\:3+5\\sin(e^t)\\quad\\Rightarrow\\quad du \\:=\\:5e^t\\cos(e^t)\\,dt \\quad\\Rightarrow\\quad e^t\\cos(e^t)\\,dt \\:=\\:\\frac{1}{5}\\,du$\n\nSubstitute: . $\\int\\frac{\\frac{1}{5}\\,du}{u} \\;=\\;\\frac{1}{5}\\int \\frac{du}{u}$ . . . etc.\n\n$\\int^{\\frac{4}{5}}_0 \\frac{\\sin^{-1}\\!\\left(\\frac{5}{4}x\\right)}{\\sqrt{16-25x^2}}\\,dx$\nThe denominator is: . $\\sqrt{16\\left(1 - \\frac{25}{16}x^2\\right)} \\;=\\;4\\sqrt{1 - \\left(\\frac{5}{4}x\\right)^2}$\n\nThe integral becomes: . $\\int^{\\frac{4}{5}}_0 \\frac{\\sin^{-1}\\!\\left(\\frac{5}{4}x\\right)\\,dx} {4\\sqrt{1 - \\left(\\frac{5}{4}x\\right)^2}}$ . $= \\;\\;\\frac{1}{4}\\int^{\\frac{4}{5}}_0\\sin^{-1}\\!\\left(\\frac{5}{4}x\\right)\\cdot\\frac{dx}{\\sqrt{ 1 - \\left(\\frac{5}{4}x\\right)^2}}$\n\n$\\text{Let }u \\:=\\:\\sin^{-1}\\!\\left(\\frac{5}{4}x\\right) \\quad\\Rightarrow\\quad\ndu \\:=\\:\\frac{\\frac{5}{4}\\,dx}{\\sqrt{1-\\left(\\frac{5}{4}x\\right)^2}}\n\nSubstitute: . $\\frac{1}{4}\\int^{\\frac{4}{5}}_0 u\\cdot\\frac{4}{5}\\,du \\;=\\;\\frac{1}{5}\\int^{\\frac{4}{5}}_0 u\\,du$ . . . etc.\n\nEdit: I'm way too slow this time . . . *sigh*\n.\n\n$\\int x^2 \\sqrt{(7 + x^3)}$ dx\n\nThe answer I got for it is $(x^3/3) \\sqrt{(7x+(x^4/4))}$\n\nWhich is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.\n\n6. Originally Posted by Pikeman85\n\n$\\int x^2 sqrt(7 + x^3)$ dx\n\nThe answer I got for it is (x^3/3) sqrt(7x+(x^4/4))\n\nWhich is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.\nOne can notice that the derivative of $7+x^3$ is $3x^2$ so $x^2 \\sqrt{7 + x^3}$ looks like $u'(x)\\sqrt{u(x)}$ which can easily be integrated : you may try to substitute $u(x)=x^3+7$.\n\n7. Well, a faster substitution is also $z^2=x^3+7.$\n\n8. $x^3 \\sqrt{7+x^3}$\n\nI got this, but it did not work. I'm missing a step here I think\n\nI don't get substitution very well\n\n9. Originally Posted by Pikeman85\n$\\int x^2 \\sqrt{(7 + x^3)}$ dx\nThe answer I got for it is $(x^3/3) \\sqrt{(7x+(x^4/4))}$\nLet $u=7+x^3 \\implies du=3x^2dx \\iff \\frac{du}{3}=x^2dx$\n$\\int x^2\\sqrt{7+x^3}dx=\\int \\sqrt{u}\\left( \\frac{du}{3}\\right)=\\frac{1}{3}\\int u^\\frac{1}{2}du=\\frac{2}{9}u^\\frac{3}{2}+C=\\frac{2 }{9}(7+x^3)^\\frac{3}{2}+C$", "date": "2013-12-12 23:31:41", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/41982-find-indicated-integrals.html", "openwebmath_score": 0.9726709127426147, "openwebmath_perplexity": 10399.127820408063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109503004294, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8608008277530769}}
{"url": "https://math.stackexchange.com/questions/970992/if-s-and-t-are-transformattion-mappings-what-is-st", "text": "# If S and T are transformattion mappings, what is [ST]?\n\nS and T are transformation mappings, what does [ST] and [TS] mean?\n\nDoes it mean transform via S and then apply T to the result and vice versa?\n\nJuxtaposition of linear transformations means composition. In details, $ST$ means $S \\circ T$ and $TS$ means $T \\circ S$ (whenever they are defined, surely). This notation is used thinking of the following: $$T: V_1 \\to V_2 \\qquad S:V_2 \\to V_3 \\qquad S\\circ T: V_1 \\to V_3$$ Let $B_1, B_2, B_3$ be basis for the respective spaces (in finite dimension). Then: $$[S \\circ T]_{B_1, B_3} = [S]_{B_2, B_3}[T]_{B_1,B_2}$$ At the bottom of our hearts, we know the distinction, so, writing, we oftem \"confuse\" the matrix with the transformation, so $S \\circ T$ becomes $ST$. Actually, the identity above is the motivation for the definition of matrix multiplication.\n\u2022 Yes, the composition goes from $V_1$ to $V_1$. To understand that notation better, I think it is important to take a look at matrices of linear transformations. \u2013\u00a0Ivo Terek Oct 13 '14 at 2:09\n\u2022 Great, and for the format of B1,B3, can the B3 also be written above B1 or is it B3 above B1? \u2013\u00a0user83039 Oct 13 '14 at 2:23", "date": "2019-12-12 11:23:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/970992/if-s-and-t-are-transformattion-mappings-what-is-st", "openwebmath_score": 0.8997182846069336, "openwebmath_perplexity": 672.6112142775514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9907319866190857, "lm_q2_score": 0.8688267762381843, "lm_q1q2_score": 0.8607744780503122}}
{"url": "http://math.stackexchange.com/questions/69178/what-are-the-odds-of-rolling-a-3-number-straight-throwing-6d6", "text": "# What are the odds of rolling a 3 number straight throwing 6d6\n\nIf you throw six fair dice, what are the odds that at least three dice make a straight (i.e. 123, 234, 345, or 456) I am certain that I am making a mistake in calculating it?\n\n-\n\n## migrated from boardgames.stackexchange.comOct 2 '11 at 2:38\n\nThis question came from our site for people who like playing board games, designing board games or modifying the rules of existing board games.\n\nwhat's your current calculation? \u2013\u00a0 DForck42 Oct 1 '11 at 4:27\nshould be moved to SE mathematics \u2013\u00a0 Hackworth Oct 1 '11 at 18:35\nIt's pretty high, pretty near one. But I'd have to think of how to get the exact probability... \u2013\u00a0 PearsonArtPhoto Oct 1 '11 at 22:02\nI went through a long and nasty inclusion-exclusion argument, watched lots of stuff cancel, and in the end got $6^6 - 6 \\cdot 4^6 + 8 \\cdot 3^6 - 3 \\cdot 2^6$, which agrees with Ed Pegg's calculations. I imagine there's a simple way to explain this formula, but I don't see it. I'm offering my results here in the hope that someone else can. \u2013\u00a0 Mike Spivey Oct 2 '11 at 7:02\n@Mike Sounds like I did the same calculation as you. If you have time, why don't you post your solution? We can also try to think of a shortcut. \u2013\u00a0 Byron Schmuland Oct 2 '11 at 13:01\n\n## 3 Answers\n\nThis problem is harder than I thought such a simple-sounding dice problem would be. :) However, the analysis below works for any number of six-sided dice! If $n$ is the number of dice, then the probability that we obtain at least one of 123, 234, 345, 456 is $$\\frac{6^n - 6 \\cdot 4^n + 8 \\cdot 3^n - 3 \\cdot 2^n}{6^n}.$$\n\nFor example, if $n=0$ or $n = 1$ or $n = 2$, we get $0$, as we should. If $n = 3$, we get $24$ in the numerator, which agrees with the $4 \\cdot 3!$ ways we could obtain at least one of the required straights with three dice. And if $n = 6$, we get $27720$ in the numerator, which agrees with Ed Pegg's calculations.\n\nWe'll obtain the numerator in the fraction above by counting the number of ways not to get at least one of the four required straights and then subtract that from $6^n$. Let's call this event $\\bar{S}$.\n\nStart by considering subsets of {1, 2, 3, 4, 5, 6} such that throwing only numbers from that subset would give us an outcome in $\\bar{S}$. There aren't any subsets of size five that do it, but there are six of size four: 1245, 1246, 1256, 1346, 1356, 2356. (I'm using a compact notation here for simplicity's sake.) Since $|\\bar{S}| = |1245 \\cup 1246 \\cup 1256 \\cup 1346 \\cup 1356 \\cup 2356|$, we now have a way to calculate $|\\bar{S}|$ using the principle of inclusion-exclusion (PIE). We just have to consider all possible ways of intersecting these six sets and then add up the cardinalities of the resulting intersections according to parity given by $(-1)^{k(A)}$, where $k(A)$ is the number of sets intersected to obtain the subset $A$.\n\nWhat makes applying PIE more difficult than usual here is that intersecting two sets sometimes gives a subset of size 3 and sometimes one of size 2, and this problem only gets worse as you intersect more sets. However, if you work through all the possibilities, most of the resulting intersections show up more than once with different parities, and most of these surprisingly cancel. What's left is 124, 125, 126, 136, 146, 156, 256, 356 (from intersecting 2 sets) and 12, 16, 26 (from intersecting 3 sets).\n\nFinally, since there are $j^n$ ways to throw the $n$ dice so that they only take on values from a subset of size $j$, we obtain $|\\bar{S}| = 6 \\cdot 4^n - 8 \\cdot 3^n + 3 \\cdot 2^n$.\n\nUnfortunately, I don't have a good explanation for why so many terms cancel, and I think there should be one. If someone can find such an explanation, I would love to see it. I've found an explanation for the massive term cancellation. (I suspect there are others.) See comments below.\n\nFinally, thanks to Ed Pegg for the brute force approach; that was very helpful in checking my work.\n\nAdded: Here is one explanation for the massive term cancellation when using PIE.\n\nThe six sets of size four that we're working with can be expressed as a graph. Each set is a vertex, and two vertices are connected by an edge if their corresponding sets differ by only one element. The graph looks like this.\n\n  1245 \u2014 1256 \u2014 2356\n\\  /    \\  /\n1246    1356\n\\    /\n1346\n\n\nIt turns out that, of the intersections that remain when applying PIE, the vertices correspond to the 6 sets of size 4, the edges correspond to the 8 sets of size 3, and the minimal cycles correspond to the 3 sets of size 2. Every other intersection is either empty or can be paired with another intersection such that they cancel each other in the PIE formula. To obtain the mapping between pairs of intersections, take an intersection that produces a connected subgraph. Pair it with the intersection obtained by removing the vertex with the highest connectivity in the subgraph. This produces a disconnected subgraph. For example, the intersection of 1245, 1256, 1356, and 2356 is paired with the intersection of 1245, 1356, and 2356. Both intersections produce 5, yet have different parities, so they cancel when using PIE. (Well, the 4-cycle is a special case, but everything there cancels nicely, too, except for the full cycle itself.)\n\nAs far as I know, inclusion-exclusion cannot, in general, be represented in a graph like this. Maybe there are some special cases, though, and this graph happens to fall into one of those. Is anyone familiar with this?\n\nInclusion-exclusion can, in some cases, be represented and simplified by a graph like this. My observations here are an instance of the following theorem:\n\nSuppose sets $A_1, A_2, \\ldots, A_n$ can be represented as the vertices of a planar graph $(V,E)$ such that, for each $x \\in \\cup_i A_i$, the subgraph consisting of the $A_i$'s containing $x$ is connected. For each edge $e = \\{A_i, A_j\\}$, let $|e| = |A_i \\cap A_j|$. For each minimal cycle $c = \\{A_{c_1}, A_{c_2}, \\ldots, A_{c_m}\\}$ in $(V,E)$, let $|c| = |\\bigcap_{j=1}^m A_{c_j}|$. Then $$\\left|\\bigcup_{i=1}^n A_i \\right| = \\sum_{v \\in V} |v| - \\sum_{e \\in E} |e| + \\sum_{c \\in C} |c|.$$\n\nSince the graph above satisfies the hypotheses of this theorem, we immediately get $|\\bar{S}| = 6 \\cdot 4^n - 8 \\cdot 3^n + 3 \\cdot 2^n$.\n\nThe proof of the theorem follows directly from Euler's famous formula $V - E + F = 2$ for connected planar graphs. For any element $x \\in \\cup_i A_i$, the subgraph consisting of the $A_i$'s containing $x$ is connected (by hypothesis) and planar. The element $x$ is counted once on the left side of the formula in the theorem. Since minimal cycles correspond to faces except for the one outer face, Euler's formula tells us that $x$ is counted a net total of once for the right-hand side as well.\n\nMore general versions of this theorem that use the Euler characteristic can be found in\n\n-\nThis is very nice!. I wish I could upvote again. \u2013\u00a0 Byron Schmuland Oct 4 '11 at 17:19\n@Byron: Thanks! The compliment is worth more than the extra upvote would be, though. :) \u2013\u00a0 Mike Spivey Oct 4 '11 at 17:28\n@Byron: I found an explanation for the term cancellation in terms of the graph representation. Fascinating, really! And I think now that I have officially beaten this question to death. :) \u2013\u00a0 Mike Spivey Oct 5 '11 at 4:51\n\nIn Mathematica, here's a brute force solution. There are 27720 rolls of the 6 dice that will give that straight.\n\nLength[Select[Tuples[Range[6], {6}], Max[Table[Length[Intersection[#, {{1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {4, 5, 6}}[[n]]]], {n, 1, 4}]] == 3 &]]/6^6\n\n$385/648 = 0.5941358024691358024691\\dots$\n\n-\n\nHere's another way. Let ${\\bf k}$ be the number of distinct die numbers, let $S$ (\"success\") be the event that at least one of the four required straights occurred.\n\nThen $$P(S) = \\sum_{k=1}^6 P(S | {\\bf k}=k) \\; P({\\bf k}=k)$$\n\nIt's clear that, for a given fixed $k$, all configurations are equiprobable.\n\nIt's trivial that $P(S | {\\bf k}=0)=P(S | {\\bf k}=1)=P(S | {\\bf k}=2)=0$. The other are not difficult:\n\n$P(S | {\\bf k}=3)$ : There are 4 possible success configurations, out of ${6 \\choose 3}$, hence $P(S | {\\bf k}=3) = 1/5$\n\n$P(S | {\\bf k}=4)$ : This is the most difficult one, let's count the unsuccessful configurations: These are : [o x x o x x] [x o x o x x] [x o x x o x] [x x o o x x] plus the mirroring of the first two: 6 unsuccessful configurations out of ${6 \\choose 4}$ hence $P(S | {\\bf k}=4) = 1- 2/5 = 3/5$ [*]\n\nFurther, $P(S | {\\bf k}=5)= P(S | {\\bf k}=6) = 1$\n\nNow, to compute $P({\\bf k}=k)$, we must count the number of ways of filling $k$ positions with 6 throws: for example, $$P({\\bf k}=4)=\\frac{{6 \\choose 4} \\; 4! \\; S_2(6,4)}{6^6}$$\n\nwhere $S_2(n,m)$ are the Stirling numbers of the second kind. So finally\n\n$$P(S) = \\sum_{k=3}^6 a_k \\; \\frac{{6 \\choose k} \\; k! \\; S_2(6,k)}{6^6} \\approx 0.59413$$\n\nwith $a_3=1/5$, $a_4=3/5$, $a_5=1$, $a_6=1$.\n\n[*] Added: to generalize this for arbitrary number of dice or runlengths, notice that this counting of unsuccessful configurations is equivalent to count the ways of expressing the number $k=4$ as a sum of $n-k+1=3$ non-negative terms less that $m=3$ (runlength), with order (in the example, 0+2+2 , 1+1+2, 1+2+1, 2+0+2, 2+2+0,2+1+1). Hence $P(S | {\\bf k}) = 1 - b_{k,n-k+1,m}/{n \\choose k}$ where $b_{r,s,t}$ counts the $s$-terms weak-compositions of the integer $r$, with the restriction that each term is less than $t$. An expression is given in Enumerative combinatorics (Stanley) page 120, problem 28.\n\nAdded 2: Because the exact formula for arbitrary dice is quite formidable, here's some quick asymptotics. Let $n$ be the number of throws, $c$ the number of dice faces, $m$ the straight length. For large $c,n$, the probability that a particular face number does not appear is $q=(1-1/c)^n \\approx \\exp(- n/c)$. Asymptotically, we can assume that these events are independent, and disregard border effects, and regard each outcome as a sequence of runlengths $(a_1 b_1 a_2 b_2 ... a_r b_r)$ where $a_i$ is the length of consecutive (run) die faces that appeared, $b_i$ the faces that didn't (and $a_1 + b_1 + ... = c$). Each run can then be approximated by independent geometric variables (starting at 1) with stopping probabilities $q$ and $1-q$. Their expectations are respectively $1/q$ and $1/(1-q)$, so that equating the expected sum we get $r=c \\; q \\;(1-q)$. The event of no-success, correspond to $a_i<m$, and its probability is given, in this approximation, by $(1-(1-q)^{m-1})^r$. Finally:\n\n$$P(S) \\approx 1- \\left[ 1- (1-q)^{m-1} \\right]^r$$\n\nwith\n\n$$r = c \\; q \\; (1-q) \\hspace{20px} q = (1-1/c)^n \\approx e^{-n/c}$$\n\nIn our original question... we have $c=n=6$,$m=3$, the numbers are too small to apply this asymptotics, but anyway, I get: $P(S) \\approx 0.50922$ ($0.54184$ if using the \"exact\" $q$)\n\n-\n+1: I like this better than my answer for the 6-dice problem. Can it be generalized to $n$ dice? \u2013\u00a0 Mike Spivey Oct 4 '11 at 18:47\n@MikeSpivey: The counting of $P(S|k=k)$ is not trivially generalizable (for $n$ dice and $m$ sequence lengths), but I think it can be done, I'll give it a look tonight. \u2013\u00a0 leonbloy Oct 4 '11 at 19:00\n@MikeSpivey: See my addition. Now it remains to see if those bounded weak-compositions have some closed formula. \u2013\u00a0 leonbloy Oct 4 '11 at 19:20", "date": "2015-07-29 20:55:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/69178/what-are-the-odds-of-rolling-a-3-number-straight-throwing-6d6", "openwebmath_score": 0.8829420804977417, "openwebmath_perplexity": 226.40024347342285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232909876815, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8607438581151349}}
{"url": "http://mathhelpforum.com/geometry/102193-geometry-regarding-intersection-wires-criss-crossing-vertical-poles.html", "text": "# Math Help - Geometry regarding the intersection of wires criss crossing vertical poles.\n\n1. ## Geometry regarding the intersection of wires criss crossing vertical poles.\n\nTwo vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?\n\nhelp step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.\n\n2. Do you find it odd that the distance between the poles is not given?\n\nIn any case, you're going to need some similar triangles. If you have a method that has similar triangles, I suspect it is NOT totally worthless. Post it.\n\n3. Originally Posted by swanz\nTwo vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?\n\nhelp step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.\nIf you know the poles are vertical, you know they are parallel. You can treat each wire as a transversal of these - check out which angles are alternate interior angles of these two transversals to see some similar triangles. Draw yourself a decent picture to see them! Then use these similar triangles to answer the question.\n\n4. Originally Posted by swanz\nTwo vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?\n\nhelp step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.\nAssume the wires are straight (with no sag)\nSince the distance between the two is irrelevant, we can assign it what ever needed, say 1 m.\n\nAssume the shorter pole is coincident with the y axis and the taller pole is parallel but at x = 1\n\nAn equation for each wire:\n\n$y_1 = -6x_1 + 6$\n\n$y_2 = 10x_2 + 0$\n\nIsolate x:\n\n$x_1 = \\dfrac{ y_1 - 6 }{-6}$\n\n$x_2 = \\dfrac{y_2}{10}$\n\nequate:\n\n$\\dfrac{ y - 6}{-6} = \\dfrac{y}{10}$\n\nsolve for y\n$y = \\dfrac{ 60 }{16}$\n\n.\n\n5. ## Resolved\n\nthank you aidan....never even considered plotting it on the graph .....all the time i was playing around with similar triangles, transversals and loads of variables.\nnice method, can come in handy in various situations, thnx a bunch.\n\n6. Originally Posted by aidan\nSince the distance between the two is irrelevant\nWell, one probably should prove that, rather than just state it.\n\n7. Originally Posted by TKHunny\nWell, one probably should prove that, rather than just state it.\nMy Bad. I just made an assumption that it was common knowledge.\nOne proof (if needed) is here:\nPlanetMath: harmonic mean in trapezoid\n\nIt indicates that the diagonals of a trapezoid will intersect at some point. A line through that point parallel to the parallel sides will intersect the non parallel sides. The length of this segment is a ratio of the parallel sides only:\n\nIf the lengths of the parallel sides are \"a\" and \"b\" then the length of the interior line that is parallel to the parallel lines and passes through the intersection point is\n\n$\\dfrac{2 a b }{(a+b)}$\n\nThe intersection point is at the midpoint of the interior line.\n\nIn our case here:\n\nHALF the length of the interior line is\n\n$\\dfrac{( 6 \\cdot 10) }{( 6 + 10 )} = 3.75$\n\n8. I think it is second tier common knowledge. It can be assumed if you work with that sort of thing, but not generally.\n\nMy views. I welcome others.\n\n9. That problem is \"ye olde crossing ladders\" problem in disguise", "date": "2016-07-28 19:03:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/102193-geometry-regarding-intersection-wires-criss-crossing-vertical-poles.html", "openwebmath_score": 0.797980010509491, "openwebmath_perplexity": 480.6962069266633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.97364464791863, "lm_q2_score": 0.8840392909114836, "lm_q1q2_score": 0.8607401241457467}}
{"url": "https://math.stackexchange.com/questions/1273139/dominated-convergence-theorem", "text": "# Dominated Convergence Theorem\n\nGive an example of a sequence $\\{f_n\\}_{n=1}^\\infty$ of integrable functions on $\\mathbb{R}$ such that $f_n \\to f$ but $\\int f_n \\not\\to \\int f$. Explain why your example does not conflict with the Dominated Convergence Theorem.\n\nI do notice that the inequality $|f_n(x)| \\le g(x)$, where $g$ is an integrable function over $\\mathbb{R}$, is not listed here in this problem. So the function need not be dominated by an integrable function here. But this is required as one hypothesis of the Dominated Convergence Theorem; hence the example will not conflict.\n\nIf this is sound reasoning, how may I come up with functions that are not dominated by another function? Initially I was thinking $f_n(x)=x \\sin (nx)$ because its lim sup is $\\infty$, but even then, we still have $|f_n(x)| \\le |x| =: g(x)$.\n\n\u2022 $f_n = n 1_{(0,{1 \\over n}]}$. \u2013\u00a0copper.hat May 8 '15 at 16:35\n\u2022 It is not only a matter of being dominated by a function, but by an integrable function. $|x|$ is not integrable on $\\mathbb{R}$, as $\\int_{-\\infty}^\\infty |x| dx = \\infty$. \u2013\u00a0Pedro M. May 8 '15 at 16:39\n\u2022 @PedroM. I forgot about the \"integrable\" function part. :O But thanks for clearing that up. \u2013\u00a0Cookie May 8 '15 at 16:42\n\u2022 It may be dominated by another function, however the detail is that the if dominating function does not belong to $L_1$ then DCT may not hold. \u2013\u00a0Alonso Delf\u00edn May 8 '15 at 16:42\n\u2022 @AaronMaroja I do not understand why you centerized \"$\\int f \\not\\to \\int f$\" and removed the period at the end of that sentence. \u2013\u00a0Cookie May 9 '15 at 7:28\n\nConsider the sequence of functions on $(0,1)$ $$f_n(x) = \\begin{cases} n & \\text{ if } x \\in (0,1/n)\\\\ 0 & \\text{ otherwise} \\end{cases}$$ We have $\\lim_{n \\to \\infty} f_n(x) = 0 = f(x)$. However, $$\\lim_{n \\to \\infty} \\int_0^1f_n(x)dx = 1 \\neq 0 = \\int_0^1 f(x) dx$$ The key in the dominated convergence theorem is that sequence of functions $f_n(x)$ must be dominated by a function $g(x)$, which is also integrable.\n\nI would like to provide a slightly more abstract framework to illustrate this \"loss of compactness\"$^{[2]}$ phenomenon. The functional setting is the $L^1(\\mathbb{R})$ space: $$L^1(\\mathbb{R})=\\left\\{f\\colon\\mathbb{R}\\to\\mathbb{R}\\ :\\ \\int_{-\\infty}^\\infty \\lvert f(x)\\rvert\\, dx<\\infty \\right\\},\\qquad \\lVert f\\rVert=\\lVert f\\rVert_{L^1}=\\int_{-\\infty}^\\infty\\lvert f(x)\\rvert\\, dx.$$ Here we take a bounded sequence $f_n$ that converges pointwise a.e. : $$\\begin{array}{cc} \\|f_n\\|\\le C, & f_n\\to f\\, \\text{a.e.} \\end{array}$$ The question is: does this sequence converge, that is, is it true that $$\\lVert f_n-f\\rVert\\to 0?^{[1]\\ [2]}$$ This would make for a very desirable property of our functional space. However, as other answers very clearly show, the answer is negative in general. The problem is that our space is subject to the action of noncompact groups of isometries. Namely, one has the action of the translation group $$\\begin{array}{cc} \\left(T_\\lambda f\\right)(x)=f(x-\\lambda), &\\lambda \\in (\\mathbb{R}, +) \\end{array}$$ and of the dilation group $$\\begin{array}{cc} \\left(D_\\lambda f\\right)(x)=\\frac{1}{\\lambda}f\\left(\\frac{x}{\\lambda}\\right), &\\lambda \\in (\\mathbb{R_{>0}}, \\cdot) \\end{array}$$ The change of variable formula for integrals immediately shows that those group actions are isometric, that is, they preserve the norm.\n\nSo, fixing a non-vanishing function $f\\in L^1(\\mathbb{R})$, its orbits $T_\\lambda f$ and $D_\\lambda f$ form bounded and non-compact subsets of $L^1(\\mathbb{R})$. In particular, letting $\\lambda \\to +\\infty$ (or $\\lambda \\to -\\infty$ for translations, or $\\lambda\\to 0$ for dilations), one finds counterexamples to the question above. (Note that, more or less, all the examples constructed in the other, excellent, answers are constructed this way).\n\nIn technical jargon one says that the translation and dilation groups introduce a defect of compactness in $L^1(\\mathbb{R})$ space. This is the terminology of the Concentration-Compactness theory (the linked page is a blog entry of T. Tao, but the theory has been founded by P.L. Lions). The dominated convergence theorem can be therefore seen as a device that impedes the defect of compactness to take place.\n\nFootnotes\n\n$^{[1]}$ The OP only asks about the convergence of the integrals: $\\int f_n\\to \\int f$. Now a standard theorem (cfr. Lieb & Loss Analysis, 2nd ed. Theorem 1.9 (Missing term in Fatou's lemma), see in the remarks) gives us the equivalence $$\\begin{array}{ccc} \\lVert f_n\\rVert_{L^1} \\to \\lVert f\\rVert_{L^1} & \\iff & \\lVert f_n-f\\rVert_{L^1}, \\end{array}$$ Therefore, at least for sequence of positive functions for which $\\int f_n=\\lVert f_n\\rVert_{L^1}$, the failure of convergence for sequences of integrals is exactly the same thing as the failure of convergence in $L^1$ space. That's why one can see the phenomenon in this functional analytic setting.\n\n$^{[2]}$ Compactness usually means that bounded sequences have convergent subsequences. In $L^1(\\mathbb{R})$ space, pointwise convergent sequences are compact if and only if they are norm convergent.\n\n\u2022 wow! Very interesting point of view, never heard of that before! :-) \u2013\u00a0Ant May 8 '15 at 18:47\n\nIt is not only a matter of being dominated by just any function $g$, but by an integrable function. $|x|$ is not integrable on $\\mathbb{R}$, as $\\int_{-\\infty}^\\infty |x| dx = \\infty$.\n\ncopper.hat provided an example that is not dominated by any function, but you can also pick bounded examples, such as $f_n = 1_{[n,n+1]}$.\n\n\u2022 Does $f_n=\\chi_{[-n,n]}$ also work? (Sorry, I'm used to the characteristic function that uses the Greek letter $\\chi$ instead of $1$.) \u2013\u00a0Cookie May 8 '15 at 16:46\n\u2022 @dragon: Then $f_n$ converges to the constant function $1$, which is not integrable on $\\mathbb{R}$. This is another illustration of how the DCT may fail if the dominating function $g$ is not integrable: $\\int f = \\infty$ (in my example, $\\int f$ is finite but is different from $\\lim \\int f_n$). Notice, however, that in your case, $\\int f_n \\to \\infty = \\int f$. \u2013\u00a0Pedro M. May 8 '15 at 16:49\n\nLet $$f_n(x) = \\frac{n^3x^2}{1+n^4x^4}.$$ Then $f_n(x) \\to 0$ pointwise everywhere, and $\\int_\\infty^\\infty f_n(x)\\,dx = \\int_\\infty^\\infty \\frac{x^2}{1+x^4}\\,dx$ for every $n.$\n\nLet $f_n = \\chi[n,2n]$ then for any $x \\in \\mathbb R$\n\n$$\\lim_{n \\to \\infty} f_n (x) = 0 \\,\\,\\, \\text{and} \\,\\,\\, \\int_{x\\in \\mathbb R} f_n(x) dx = n \\to \\infty \\,\\,\\, \\text{as}\\,\\, n \\to \\infty$$\n\nOn the other hand taking $f(x) = 0 , \\forall x \\in \\mathbb R$ we have $$\\int_{x\\in \\mathbb R} f(x) dx = \\int _{x\\in \\mathbb R} 0\\,\\, dx = 0$$ then\n\n$$\\int_{x\\in \\mathbb R} f_n(x) dx \\not \\to \\int_{x\\in \\mathbb R} f(x) dx$$", "date": "2019-10-22 03:13:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1273139/dominated-convergence-theorem", "openwebmath_score": 0.9119226932525635, "openwebmath_perplexity": 205.5637991302053, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446418006305, "lm_q2_score": 0.8840392909114836, "lm_q1q2_score": 0.8607401187371948}}
{"url": "http://mathhelpforum.com/calculus/12005-integration-unsure-method.html", "text": "# Math Help - Integration;unsure method\n\n1. ## Integration;unsure method\n\nHi, I have another integration question (because they're just so fun ).\n\nInt ( dx/(x^2-3x+2)^(1/2) )\n\nThe item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.\n\nThanks a ton.\n\n2. Originally Posted by ChaosBlue\nHi, I have another integration question (because they're just so fun ).\n\nInt ( dx/(x^2-3x+2)^(1/2) )\n\nThe item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.\n\nThanks a ton.\nStart this by looking under the square root sign. You want to do a \"complete the square.\"\n\nx^2 - 3x + 2 = (x^2 - 3x) + 2 = (x^2 - 3x + 9/4 - 9/4) + 2 = (x^2 - 3x + 9/4) - 9/4 + 2\n\n= (x - 3/2)^2 - 1/4\n\nNow let y = x - 3/2, dy = dx\n\nInt[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy]\n\nOne more trick and we're done. Pull a 1/4 out from under the radical:\nInt[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]\n\nAnd now let z = 2y, dz = 2dy and your integral becomes:\nInt[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]\n\nThis is a more standard problem. Can you take it from here?\n\n-Dan\n\n3. Originally Posted by ChaosBlue\nHi, I have another integration question (because they're just so fun ).\n\nInt ( dx/(x^2-3x+2)^(1/2) )\n\nThe item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.\n\nThanks a ton.\nThe way I see to do it is really long, so if anyone out there has an easier way, you are welcome to post.\n\nWe begin by completing the square.\n\nint(1/sqrt(x^2 - 3x + 2))dx\n= int(1/(sqrt( x^2 - 3x + (-3/2)^2 - 1/4))dx\n= int(1/sqrt((x - 3/2)^2 - 1/4))dx\n= int(1/sqrt((x - 3/2)^2 - (1/2)^2))dx\n= int(1/sqrt[(1/2)^2*([(x - 3/2)/(1/2)]^2 - 1)])dx\n= 2*int(1/sqrt[(2x - 3)^2 - 1])dx\n\nlet u= 2x - 3\n=> du = 2 dx\nso our integral becomes\nint(1/sqrt(u^2 - 1))du\n\nNow continue using trig substitution\n\n4. Originally Posted by topsquark\nStart this by looking under the square root sign. You want to do a \"complete the square.\"\n\nx^2 - 3x + 2 = (x^2 - 3x) + 2 = (x^2 - 3x + 9/4 - 9/4) + 2 = (x^2 - 3x + 9/4) - 9/4 + 2\n\n= (x - 3/2)^2 - 1/4\n\nNow let y = x - 3/2, dy = dx\n\nInt[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy]\n\nOne more trick and we're done. Pull a 1/4 out from under the radical:\nInt[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]\n\nAnd now let z = 2y, dz = 2dy and your integral becomes:\nInt[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]\n\nThis is a more standard problem. Can you take it from here?\n\n-Dan\nHa ha, i took forever typing this up only to see that u beat me to it once i clicked \"submit reply\"\n\nThanks a lot Dan\n\n5. Originally Posted by Jhevon\nHa ha, i took forever typing this up only to see that u beat me to it once i clicked \"submit reply\"\n\nThanks a lot Dan\nIt happens. You're welcome.\n\nAnd besides, my variables are prettier anyway.\n\n-Dan\n\n6. Originally Posted by topsquark\nIt happens. You're welcome.\n\nAnd besides, my variables are prettier anyway.\n\n-Dan\n\nNu-uh, u is a much more suitable variable for substitution, most text books use u. You're just fighting the system.\n\nDo you know what they do to non-conformists in the math field?!\n\n7. Thanks guys so much, you really helped me out. I do have one question though.\n\nOriginally Posted by topsquark\nOne more trick and we're done. Pull a 1/4 out from under the radical:\nInt[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]\n\nAnd now let z = 2y, dz = 2dy and your integral becomes:\nInt[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]\ntopsquark: I understand we're essentially \"manipulating\" our equation so we can effectively use trig substitution, but I kind-of got lost in the process. I have trouble grasping it after we pull out the 1/4.? Is there anyway you (or Jhevon since you got it as well) could explain that part to me a bit more?\n\nThanks a ton.\n\n8. Originally Posted by ChaosBlue\nThanks guys so much, you really helped me out. I do have one question though.\n\ntopsquark: I understand we're essentially \"manipulating\" our equation so we can effectively use trig substitution, but I kind-of got lost in the process. I have trouble grasping it after we pull out the 1/4.? Is there anyway you (or Jhevon since you got it as well) could explain that part to me a bit more?\n\nThanks a ton.\n\nWhen he said pull out 1/4, he meant we factorized the function to obtain (1/4)*(something), he did that so he could get something of the form u^2 - 1 under the square root\n\n9. Originally Posted by Jhevon\nWhen he said pull out 1/4, he meant we factorized the function to obtain (1/4)*(something), he did that so he could get something of the form u^2 - 1 under the square root\nI worded my question poorly (my bad ). What I mean is this.\n\nI get where we use completing the square to transform:\n\nint 1/sqrt(x^2-3x+2) dx--> int 1/sqrt( (x-3/2)^2 - 1/4 )dx\n\nFrom there we have u = x-3/2 and therefore du = dx\n\nPlug in and we get int 1/ (u^2 - 1/4)^1/2 du\n\nNow, we pull out or factor out the 1/4; But I am struggling with when we do that, how\n\nint 1/ (u^2 - 1/4)^1/2 du becomes\n\nInt[1/((1/4)(2u)^2 - 1/4)^{1/2}du]\n(we pull out 1/4, so how does u^2 become 2u^2 ?\n\nand then\n\n2*Int[1/((2u)^2 - 1)^{1/2}du]\n\nSorry for bothering you guys so much!\n\n10. Originally Posted by ChaosBlue\nI worded my question poorly (my bad ). What I mean is this.\n\nI get where we use completing the square to transform:\n\nint 1/sqrt(x^2-3x+2) dx--> int 1/sqrt( (x-3/2)^2 - 1/4 )dx\n\nFrom there we have u = x-3/2 and therefore du = dx\n\nPlug in and we get int 1/ (u^2 - 1/4)^1/2 du\n\nNow, we pull out or factor out the 1/4; But I am struggling with when we do that, how\n\nint 1/ (u^2 - 1/4)^1/2 du becomes\n\nInt[1/((1/4)(2u)^2 - 1/4)^{1/2}du]\n(we pull out 1/4, so how does u^2 become 2u^2 ?\n\nand then\n\n2*Int[1/((2u)^2 - 1)^{1/2}du]\n\nSorry for bothering you guys so much!\nit's no bother, we're all here to learn.\n\nok, here it is. Say we had some function b, and we wanted to factor an a out of it, we could write is as a*(b/a).\n\nsimilarly, when he factored 1/4 out of u^2, he ended up with (1/4)*(u^2/(1/4))\n\nu^2/(1/4) = u^2/(1/2)^2 = [u/(1/2)]^2 = 2u^2\n\n11. Funny you should say thanks, we actually left the hard part to you\n\n12. Originally Posted by Jhevon\nNu-uh, u is a much more suitable variable for substitution, most text books use u. You're just fighting the system.\n\nDo you know what they do to non-conformists in the math field?!\nGood thing I'm not in the Math field, then.\n\n-Dan\n\n13. Originally Posted by topsquark\nGood thing I'm not in the Math field, then.\n\n-Dan\nIt's even worst in the physics field, which i suppose you're a part of\n\n14. Originally Posted by Jhevon\nIt's even worst in the physics field, which i suppose you're a part of\nYes, but as long as you define your variables, you can technically get away with any set of variables you wish.\n\nI think I'm going to create a unit called the \"Dan.\" It'll be defined as the amount of work needed to translate an equation from the variable z to the variable u...\n\n-Dan\n\n15. Originally Posted by topsquark\nYes, but as long as you define your variables, you can technically get away with any set of variables you wish.\n\nI think I'm going to create a unit called the \"Dan.\" It'll be defined as the amount of work needed to translate an equation from the variable z to the variable u...\n\n-Dan\nhaha, bless your heart Dan, you need help\n\nPage 1 of 2 12 Last", "date": "2015-11-27 06:19:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/12005-integration-unsure-method.html", "openwebmath_score": 0.837983250617981, "openwebmath_perplexity": 1382.927568327193, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446486833801, "lm_q2_score": 0.8840392832736083, "lm_q1q2_score": 0.8607401173852395}}
{"url": "https://math.stackexchange.com/questions/1004081/is-it-possible-to-write-a-sum-as-an-integral-to-solve-it", "text": "# Is it possible to write a sum as an integral to solve it?\n\nI was wondering, for example,\n\nCan:\n\n$$\\sum_{n=1}^{\\infty} \\frac{1}{(3n-1)(3n+2)}$$\n\nBe written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals.\n\nBut actually writing an integral form. Like\n\n$$\\displaystyle \\sum_{n=1}^{\\infty} \\frac{1}{(3n-1)(3n+2)} = \\int_{a}^{b} g(x) \\space dx$$\n\nWhat are some general tricks in finding infinite sum series.\n\n\u2022 math.stackexchange.com/questions/1002440/\u2026 \u2013\u00a0lab bhattacharjee Nov 3 '14 at 13:26\n\u2022 @labbhattacharjee, I did not meant that. I know the solution to this, I was just asking if in general it is possible to write a sum as an actual integral. \u2013\u00a0Amad27 Nov 3 '14 at 13:28\n\u2022 You can trivially write the sum as an integral using the Iverson bracket (add a factor of $[n \\in \\mathbb{N}]$ to the integrand). This ignores the question of how to evaluate the resulting integral, of course. \u2013\u00a0chepner Nov 3 '14 at 19:10\n\u2022 \"I am NOT talking about a method for using tricks with integrals.\" \"But actually writing an integral form.\" \"What are some general tricks\" Combining these quotes with the accepted answer that does not seem to be a general trick, I'm a bit confused on what this question is asking. \u2013\u00a0JiK Nov 4 '14 at 8:28\n\u2022 @Amad27 $\\int_\\mathbb{N}\\frac{d \\mu}{(3n-1)(3n+2)}$ where $\\mu$ is the counting measure on $\\mathbb{N}$. It doesn't give you anything you didn't already have though. I didn't really mean it seriously although it is true. \u2013\u00a0Tim Seguine Nov 5 '14 at 17:21\n\nA General Trick\n\nA General Trick for summing this series is to use Telescoping Series: \\begin{align} \\sum_{n=1}^\\infty\\frac1{(3n-1)(3n+2)} &=\\frac13\\lim_{N\\to\\infty}\\sum_{n=1}^N\\left(\\frac1{3n-1}-\\frac1{3n+2}\\right)\\\\ &=\\frac13\\lim_{N\\to\\infty}\\left[\\sum_{n=1}^N\\frac1{3n-1}-\\sum_{n=1}^N\\frac1{3n+2}\\right]\\\\ &=\\frac13\\lim_{N\\to\\infty}\\left[\\sum_{n=0}^{N-1}\\frac1{3n+2}-\\sum_{n=1}^N\\frac1{3n+2}\\right]\\\\ &=\\frac13\\lim_{N\\to\\infty}\\left[\\frac12-\\frac1{3N+2}\\right]\\\\ &=\\frac16 \\end{align}\n\nAn Integral Trick\n\nSince $$\\int_0^\\infty e^{-nt}\\,\\mathrm{d}t=\\frac1n$$ for $n\\gt0$, we can write \\begin{align} \\sum_{n=1}^\\infty\\frac1{(3n-1)(3n+2)} &=\\sum_{n=1}^\\infty\\frac13\\int_0^\\infty\\left(e^{-(3n-1)t}-e^{-(3n+2)t}\\right)\\mathrm{d}t\\\\ &=\\frac13\\int_0^\\infty\\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\\mathrm{d}t\\\\ &=\\frac13\\int_0^\\infty e^{-2t}\\,\\mathrm{d}t\\\\ &=\\frac16 \\end{align}\n\n\u2022 I think this is a better \"trick\" for dealing with sums. Integral \"tricks\" are nice however integrals and infinite series' are very different in what they calculate and manipulating a sum or integral on its own without switching is preferred. \u2013\u00a0Ali Caglayan Nov 3 '14 at 16:23\n\u2022 @Alizter: For the most part, I agree. However, sometimes pure series manipulation can be extremely complicated, and the proper integral representation of a sum can be useful. However, in this case, I think staying with series manipulation is easiest. That being said, I have added an integral approach, as well. \u2013\u00a0robjohn Nov 3 '14 at 18:40\n\u2022 The sum under the first integral could have been computed as a telescoping series either. Considering this, I think the use of integrals in the second solution is completely void. Edit: I mean exactly what Henning Makholm points out under the other answer. \u2013\u00a0Adayah Nov 4 '14 at 19:41\n\u2022 @Adayah: My reply to Henning was meant as an agreement. I first posted only the telescoping series, but then added an integral approach to satisfy the first part of the question. In any approach where one breaks up the summand using partial fractions, it could be said that, at that point, the answer could be computed as a telescoping sum. \u2013\u00a0robjohn Nov 4 '14 at 20:48\n\nSince $\\int_{0}^{1}x^k\\,dx = \\frac{1}{k+1}$, $$\\frac{1}{(3n-1)(3n+2)}=\\frac{1}{3}\\left(\\frac{1}{3n-1}-\\frac{1}{3n+2}\\right)=\\frac{1}{3}\\int_{0}^{1}x^{3n-2}(1-x^3)\\,dx,$$ so, summing over $n$: $$\\sum_{n=1}^{+\\infty}\\frac{1}{(3n-1)(3n+2)}=\\frac{1}{3}\\int_{0}^{1}x\\,dx=\\frac{1}{6}.$$\n\n\u2022 I thought we need uniform convergence in order to interchange the limit and integral. The power series is uniformly convergent inside the radius of convergence, how to pass it to the whole interval $[0,1]$? \u2013\u00a0John Nov 3 '14 at 13:45\n\u2022 @JohnZHANG Actually no, Fubini and Tonelli's theorems allow this for a monotone sequence supposedly, I believe. \u2013\u00a0Amad27 Nov 3 '14 at 13:45\n\u2022 Nice trick for the given sum, but this still doesn't answer the bold-marked question of general tricks. \u2013\u00a0Ruslan Nov 3 '14 at 16:34\n\u2022 Isn't the integral just a detour here? The operative step is exactly the same telescoping that could have been done without rewriting the terms into integrals. \u2013\u00a0Henning Makholm Nov 4 '14 at 10:11\n\u2022 @HenningMakholm: smoke and mirrors. \u2013\u00a0robjohn Nov 4 '14 at 12:07\n\nActually writing it as an integral, as asked for:\n\n$$\\displaystyle \\sum_{n=1}^{\\infty} \\frac{1}{(3n-1)(3n+2)} = \\int_{1}^{\\infty} \\frac{1}{(3\\lfloor x\\rfloor-1)(3\\lfloor x\\rfloor+2)} dx$$\n\nThis probably won't help with finding the value, though.\n\n\u2022 why won't it help finding the value? \u2013\u00a0Amad27 Nov 4 '14 at 13:20\n\u2022 @Amad27: I don't see a way it would. If you can find one, then more power to you, I suppose ... \u2013\u00a0Henning Makholm Nov 4 '14 at 13:33\n\u2022 @Amad27 Methods for solving integrals are poorly suited for integrating functions that are non-continuous. The usual approach for integrating functions like the one here is to separately integrate over each interval where it is continuous. Which brings us back to the sum form. \u2013\u00a0Rafa\u0142 Dowgird Nov 4 '14 at 15:11\n\u2022 @Amad27 It is quite literally equivalent to the original sum in a trivially useless manner XD \u2013\u00a0Simply Beautiful Art Jan 12 '17 at 2:11\n\nIn such cases, the partial fractions of general term (i.e. $n^{th}$ term ) of the infinite-series are very useful.\nGiven that $$\\sum_{n=1}^{\\infty}\\frac{1}{(3n-1)(3n+2)}=\\sum_{n=1}^{\\infty} T_{n}$$ Where, $T_{n}$ is the $n^{th}$ term of the given series which can be easily expressed in the partial fractions as follows $$T_{n}=\\frac{1}{(3n-1)(3n+2)}=\\frac{1}{3}\\left(\\frac{1}{3n-1}-\\frac{1}{3n+2}\\right)$$ Now, we have $$\\sum_{n=1}^{\\infty}\\frac{1}{(3n-1)(3n+2)}=\\frac{1}{3}\\sum_{n=1}^{\\infty} \\left(\\frac{1}{3n-1}-\\frac{1}{3n+2}\\right)$$ $$=\\frac{1}{3} \\lim_{n\\to \\infty} \\left[\\left(\\frac{1}{2}-\\frac{1}{5}\\right)+\\left(\\frac{1}{5}-\\frac{1}{8}\\right)+\\left(\\frac{1}{8}-\\frac{1}{11}\\right)+\\! \\cdot \\! ........ +\\left(\\frac{1}{3n-4}-\\frac{1}{3n-1}\\right)+\\left(\\frac{1}{3n-1}-\\frac{1}{3n+2}\\right)\\right]$$ $$=\\frac{1}{3} \\lim_{n\\to \\infty} \\left[\\frac{1}{2} -\\frac{1}{3n+2}\\right]$$ $$=\\frac{1}{3} \\left[\\frac{1}{2} -\\frac{1}{\\infty}\\right]$$ $$=\\frac{1}{3} \\left[\\frac{1}{2}\\right]=\\color{blue}{\\frac{1}{6}}$$\n\nWe can indeed write the sum as an integral, after research. Consider:\n\nFind: $\\psi(1/2)$\n\nBy definition:\n\n$$\\psi(z+1) = -\\gamma + \\sum_{n=1}^{\\infty} \\frac{z}{n(n+z)}$$\n\nThe required $z$ is $z = -\\frac{1}{2}$\n\nso let $z = -\\frac{1}{2}$\n\n$$\\psi(1/2) = -\\gamma + \\sum_{n=1}^{\\infty} \\frac{-1}{2n(n - \\frac{1}{2})}$$\n\nSimplify this: $$\\psi(1/2) = -\\gamma - \\sum_{n=1}^{\\infty} \\frac{1}{n(2n - 1)}$$\n\nThe sum seems difficult, but really isnt.\n\nWe can telescope or:\n\n$$\\frac{1}{1-x} = \\sum_{n=1}^{\\infty} x^{n-1}$$\n\nLet $x \\rightarrow x^2$\n\n$$\\frac{1}{1-x^2} = \\sum_{n=1}^{\\infty} x^{2n-2}$$\n\nIntegrate once:\n\n$$\\tanh^{-1}(x) = \\sum_{n=1}^{\\infty} \\frac{x^{2n-1}}{2n-1}$$\n\nIntegrate again:\n\n$$\\sum_{n=1}^{\\infty} \\frac{x^{2n}}{(2n-1)(n)} = 2\\int \\tanh^{-1}(x) dx$$\n\nFrom the tables, the integral of $\\tanh^{-1}(x)$\n\n$$\\sum_{n=1}^{\\infty} \\frac{x^{2n}}{(2n-1)(n)} = \\log(1 - x^2) + 2x\\tanh^{-1}(x)$$\n\nTake the limit as $x \\to 1$\n\n$$\\sum_{n=1}^{\\infty} \\frac{1}{(2n-1)(n)} = \\log(4)$$\n\n$$\\psi(1/2) = -\\gamma - \\sum_{n=1}^{\\infty} \\frac{1}{(2n-1)(n)}$$\n\n$$\\psi(\\frac{1}{2}) = -\\gamma - \\log(4)$$\n\n\u2022 I am the OP per say. This is a general trick. I converted the sum into an integral. Please read carefully. \u2013\u00a0Amad27 Dec 21 '14 at 8:15\n\nYes you can use the Euler Maclaruin formula to write the sum as an integral plus an infinite number of derivatives. I remember deriving this for my self when I was younger and being very pleased with myself.\n\nThis particular sum could be solved because you had two terms $ax+b$ and $ax+c$ and the difference between c and b is equal to a (I think it would work in a slightly more complicated way if it was a not-too-large multiple of a).\n\nIf you want numerical values in general cases, and the sum doesn't converge quickly for your taste, or you want just a partial sum, you can use that\n\n$$\\displaystyle f (k) = \\int_{k-1/2}^{k+1/2} f(k) dx \u2248 \\int_{k-1/2}^{k+1/2} f(x) dx$$\n\nand therefore\n\n$$\\displaystyle \\sum_{k=n}^{m} f(k) \u2248 \\int_{n-1/2}^{m+1/2} f(x) dx$$\n\nAssuming that you can solve the integral in closed form, if you let\n\n$$\\displaystyle g (k) = f(k) - \\int_{k-1/2}^{k+1/2} f(x) dx$$\n\nthen\n\n$$\\displaystyle \\sum_{k=n}^{m} f(k) = \\int_{n-1/2}^{m+1/2} f(x) dx + \\sum_{k=n}^{m} g(k)$$\n\n$g (k)$ will usually converge much faster than $f (k)$.", "date": "2019-10-16 21:49:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1004081/is-it-possible-to-write-a-sum-as-an-integral-to-solve-it", "openwebmath_score": 0.9224899411201477, "openwebmath_perplexity": 633.7219715032601, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446463891303, "lm_q2_score": 0.8840392817460333, "lm_q1q2_score": 0.8607401138697173}}
{"url": "https://stats.stackexchange.com/questions/574333/how-can-i-determine-which-of-two-sequences-of-coin-flips-is-real-and-which-is-fa/574425", "text": "# How can I determine which of two sequences of coin flips is real and which is fake?\n\nThis is an interesting problem I came across. I'm attempting to write a Python program to get a solution to it; however, I'm not sure how to proceed. So far, I know that I would expect the counts of heads to follow a binomial, and length of runs (of tails, heads, or both) to follow a geometric.\n\nBelow are two sequences of 300 \u201ccoin flips\u201d (H for heads, T for tails). One of these is a true sequence of 300 independent flips of a fair coin. The other was generated by a person typing out H\u2019s and T\u2019s and trying to seem random. Which sequence is truly composed of coin flips?\n\nSequence 1:\n\nTTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHH TTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHH TTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHT THHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHT HTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTT HHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\n\nSequence 2:\n\nHTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTH THTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHH TTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTT THTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTH HHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHH HTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\n\nBoth sequences have 148 heads, two less than the expected number for a 0.5 probability of heads.\n\n\u2022 Naive qustion : Does the question means: can I find with that porbablity that such sequence is generated by human is big enough ? Can I compute such probablity ? Every sequence is possible, so this is not a proof but only possibility. Am I right ?\nMay 7 at 8:53\n\u2022 (I find it truly amusing that somebody voted to close this thread because it asks for opinions. If that were true, we should close all threads here on CV that use statistical analysis to compare data--and that would leave us with nothing but lists of references!)\n\u2013\u00a0whuber\nMay 7 at 13:42\n\u2022 Compress both strings and see which one comes out shorter. May 7 at 13:58\n\u2022 @whuber That comment is probably a reference to (approximate) Kolmogorov complexity. See algorithmically random sequence. May 7 at 15:45\n\u2022 @whuber With very high probability, a truly random sequence cannot be compressed. Compression methods can leverage many kinds of bias, not only if one symbol appears more overall, but also if some subsequences appear more than others (as here, H is more likely to be followed by T and vice versa, per your answer). May 7 at 18:20\n\nThis is a variant on a standard intro stats demonstration: for homework after the first class I have assigned my students the exercise of flipping a coin 100 times and recording the results, broadly hinting that they don't really have to flip a coin and assuring them it won't be graded. Most will eschew the physical process and just write down 100 H's and T's willy-nilly. After the results are handed in at the beginning of the next class, at a glance I can reliably identify the ones who cheated. Usually there are no runs of heads or tails longer than about 4 or 5, even though in just 100 flips we ought to see a longer run that that.\n\nThis case is subtler, but one particular analysis stands out as convincing: tabulate the successive ordered pairs of results. In a series of independent flips, each of the four possible pairs HH, HT, TH, and TT should occur equally often--which would be $$(300-1)/4 = 74.75$$ times each, on average.\n\nHere are the tabulations for the two series of flips:\n\n   Series 1    Series 2\nH   T        H  T\nH  46 102       71 76\nT 102  49       77 75\n\n\nThe first is obviously far from what we might expect. In that series, an H is more than twice as likely ($$102:46$$) to be followed by a T than by another H; and a T, in turn, is more than twice as likely ($$102:49$$) to be followed by an H. In the second series, those likelihoods are nearly $$1:1,$$ consistent with independent flips.\n\nA chi-squared test works well here, because all the expected counts are far greater than the threshold of 5 often quoted as a minimum. The chi-squared statistics are 38.3 and 0.085, respectively, corresponding to p-values of less than one in a billion and 77%, respectively. In other words, a table of pairs as imbalanced as the second one is to be expected (due to the randomness), but a table as imbalanced as the first happens less than one in every billion such experiments.\n\n(NB: It has been pointed out in comments that the chi-squared test might not be applicable because these transitions are not independent: e.g., an HT can be followed only by a TT or TH. This is a legitimate concern. However, this form of dependence is extremely weak and has little appreciable effect on the null distribution of the chi-squared statistic for sequences as long as $$300.$$ In fact, the chi-squared distribution is a great approximation to the null sampling distribution even for sequences as short as $$21,$$ where the counts of the $$21-1=20$$ transitions that occur are expected to be $$20/4=5$$ of each type.)\n\nIf you know nothing about chi-squared tests, or even if you do but don't want to program the chi-square quantile function to compute a p-value, you can achieve a similar result. First develop a way to quantify the degree of imbalance in a $$2\\times 2$$ table like this. (There are many ways, but all the reasonable ones are equivalent.) Then generate, say, a few hundred such tables randomly (by flipping coins--in the computer, of course!). Compare the imbalances of these two tables to the range of imbalances generated randomly. You will find the first sequence is far outside the range while the second is squarely within it.\n\nThis figure summarizes such a simulation using the chi-squared statistic as the measure of imbalance. Both panels show the same results: one on the original scale and the other on a log scale. The two dashed vertical lines in each panel show the chi-squared statistics for Series 1 (right) and Series 2 (left). The red curve is the $$\\chi^2(1)$$ density. It fits the simulations extremely well at the right (higher values). The discrepancies for low values occur because this statistic has a discrete distribution which cannot be well approximated by any continuous distribution where it takes on small values -- but for our purposes that makes no difference at all.\n\n\u2022 Why do you say, \"This case is subtler\" (than just looking for runs)? OP's sequence 1 has length 300 (longer than the 100 you mention) and its longest run is a single HHHH. So it doesn't seem subtle at all, but rather a slam dunk for the \"at a glance\" method you describe. May 7 at 18:11\n\u2022 It's not proper to apply the chi squared test to the contingency table given since consecutive pairs are overlapping and not independent, so p-values will be exaggerated. I suspect the result will be similar after addressing this though.\n\u2013\u00a0Paul\nMay 7 at 18:11\n\u2022 @Paul That's an excellent point, which I have neglected to discuss (and it's no excuse to claim, as I was hoping to do, that it's implicitly handled correctly in the last paragraph, because I did not advertise that fact).\n\u2013\u00a0whuber\nMay 7 at 18:13\n\u2022 @klm123 But as long as you don't actually try a bunch of tests before finding an improbable result (p-hacking), the result is still significant. The test(s) should be chosen in advance. There are also ways of correcting p-values if you run multiple tests. Yes, there is the potential to be misleading if you only report the significant test and not the others -- this has contributed to the \"replication crisis\". May 7 at 23:07\n\u2022 My answer talks about why this test in particular is well-motivated. People tend to think that non-repeated values are more random than repeated values. (Which is true in some sense, but people believe it to a greater extent / in more generality than is actually true.)\n\u2013\u00a0Paul\nMay 7 at 23:25\n\nThere are two very good answers as of writing this, and so let me add a needlessly complex yet interesting approach to this problem.\n\nI think one way to operationalize the human generated vs truly random question is to ask if the flips are autocorrelated. The hypothesis here being that humans will attempt to appear random by not having too many strings of one outcome, hence switching from heads to tails and tails to heads more often than would be observed in a truly random sequence.\n\nWhuber examines this nicely with a 2x2 table, but because I am a Bayesian and a glutton for punishment let's write a simple model in Stan to estimate the lag-1 autocorrelation of the flips. Speaking of Whuber, he has nicely laid out the data generating process in this post. You can read his answer to understand the data generating process.\n\nLet $$\\rho$$ be the lag 1 autocorrelation of the flips, and let $$q$$ be the proportion of flips which are heads in the sequence. A fair coin should have 0 autocorrelation, so we are looking for our estimate of $$\\rho$$ to be close to 0. From there, we only need to count the number of occurrences of $$H,H$$, $$H, T$$, $$T, H$$ and $$T,T$$ in the sequence.\n\nThe Stan model is shown below\n\ndata{\n\nint y_1_1; //number of concurrent 1s\nint y_0_1; //number of 0,1 occurrences\nint y_1_0; //number of 1,0 occurrences\nint y_0_0; //number of concurrent 0s\n\n}\nparameters{\nreal<lower=-1, upper=1> rho;\nreal<lower=0, upper=1> q;\n\n}\ntransformed parameters{\n\nreal<lower=0, upper=1> prob_1_1 = q + rho*(1-q);\nreal<lower=0, upper=1> prob_0_1 = (1-q)*(1-rho);\nreal<lower=0, upper=1> prob_1_0 = q*(1-rho);\nreal<lower=0, upper=1> prob_0_0 = 1 - q + rho*q;\n}\nmodel{\nq ~ beta(1, 1);\ntarget += y_1_1 * bernoulli_lpmf(1| prob_1_1);\ntarget += y_0_1 * bernoulli_lpmf(1| prob_0_1);\ntarget += y_1_0 * bernoulli_lpmf(1| prob_1_0);\ntarget += y_0_0 * bernoulli_lpmf(1| prob_0_0);\n\n}\n\n\n\nHere, I've placed a uniform prior on the autocorrelation\n\n$$\\rho \\sim \\mbox{Uniform}(-1, 1)$$\n\nand on the probability of a head\n\n$$q \\sim \\operatorname{Beta}(1, 1)$$\n\nOur likelihood is Bernoulli, and I have weighted the likelihood by the number of occurrences of each pair of outcomes. The probabilities of each outcome (e.g. probability of observing a heads conditioned on the previous flip being a heads) is provided by Whuber in his linked answer. Let's run our model and compare posterior distributions for the two sequences\n\nThe estimated auto correlation for sequence 1 is -0.36, and the estimated autocorrelation for sequence 2 is -0.02 (close enough to 0). If I was a betting man, I'd put my money on sequence 1 being the sequence generated by a human. The negative autocorrelation means that when we see a heads/tails we are more likely to see a tails/heads! This observation lines up nicely with the 2x2 table provided by Whuber.\n\n### Code\n\nThe plot I present is made in R, but here is some python code to do the same thing since you asked\n\nimport matplotlib.pyplot as plt\nimport cmdstanpy\n\n# You will need to install cmdstanpy prior to running this code\n\n# Write the stan model as a string.  We will then write it to a file\nstan_code = '''\ndata{\n\nint y_1_1; //number of concurrent 1s\nint y_0_1; //number of 0,1 occurences\nint y_1_0; //number of 1,0 occurences\nint y_0_0; //number of concurrent 0s\n\n}\nparameters{\nreal<lower=-1, upper=1> rho;\nreal<lower=0, upper=1> q;\n}\ntransformed parameters{\nreal<lower=0, upper=1> prob_1_1 = q + rho*(1-q);\nreal<lower=0, upper=1> prob_0_1 = (1-q)*(1-rho);\nreal<lower=0, upper=1> prob_1_0 = q*(1-rho);\nreal<lower=0, upper=1> prob_0_0 = 1 - q + rho*q;\n}\nmodel{\nq ~ beta(1, 1);\ntarget += y_1_1 * bernoulli_lpmf(1| prob_1_1);\ntarget += y_0_1 * bernoulli_lpmf(1| prob_0_1);\ntarget += y_1_0 * bernoulli_lpmf(1| prob_1_0);\ntarget += y_0_0 * bernoulli_lpmf(1| prob_0_0);\n\n}\n'''\n\n# Write the model to a temp file\nwith open('model_file.stan', 'w') as model_file:\nmodel_file.write(stan_code)\n\n# Compile the model\nmodel = cmdstanpy.CmdStanModel(stan_file='model_file.stan', compile=True)\n\n# Co-occuring counts for heads (1) and tails (0) for each sequence\ndata_1 = dict(y_1_1 = 46, y_0_0 = 49, y_0_1 = 102, y_1_0 = 102)\ndata_2 = dict(y_1_1 = 71, y_0_0 = 75, y_0_1 = 76, y_1_0 = 77)\n\n# Fit each model\nfit_1 = model.sample(data_1, show_progress=False)\nrho_1 = fit_1.stan_variable('rho')\n\nfit_2 = model.sample(data_2, show_progress=False)\nrho_2 = fit_2.stan_variable('rho')\n\n# Make a pretty plot\nfig, ax = plt.subplots(dpi = 240, figsize = (5, 3))\n\nax.set_xlim(-1, 1)\nax.hist(rho_1, color = 'blue', alpha = 0.5, edgecolor='k', label='Sequence 1')\nax.hist(rho_2, color = 'red', alpha = 0.5, edgecolor='k', label='Sequence 2')\nax.legend()\n$$$$\n\n\u2022 +1 I developed my tabular explanation by going through the exercise you lay out here. I plotted the PACF functions of random sequences and compared them, visually, to the PACF functions of the two sequences in the question. One plot--the first sequence--stood out: it had a very negative lag-one partial autocorrelation coefficient. But motivating and explaining the PACF seemed like overkill for this problem ;-). The more enduring lesson, illustrated here and in the post by @COOLSerdash, is that by finding a suitable way to visualize data, we can discover otherwise hidden things about them.\n\u2013\u00a0whuber\nMay 7 at 13:28\n\u2022 (+1) Interesting resolution that does not rely on insufficient statistics, contrary to the others!, but I would have gone fully Markov and put a prior distribution on both $(p_{11},p_{10})$ and $(p_{01},p_{00})$ rather than introducing a correlation $\\rho$, but it is unlikely the result would differ. More generally, the alternative to being an iid Uniform sequence could be anything, so restricting to an order one Markov is favouring the null. May 9 at 8:09\n\u2022 @Xi'an You're right in that the result is not much different. The reason I prefer the $\\rho$, $q$ parameterization is because we actually have good priors on these (despite what my answer uses lol). Humans have an intuitive sense for when they are acting too correlated and when they are acting too biased, so we know $\\rho$ is going to be close to 0 and $q$ close to 0.5. But, you could rewrite this model using a multinomial likelihood and place priors on the probabilities directly too. That's what I love about Bayes! May 9 at 17:48\n\nThis is a class activity I've first read about in the book Teaching Statistics. A Bag of Tricks, 2nd ed. by Andrew Gelman and Deborah Nolan (they recommend 100 flips, though). Their reasoning to detect the fabricated sequence is based on the combination of the longest run and the number of runs. For the following plot, I simulated 5000 fair coin tosses of length 300 and plotted the longest run on the y-axis and the number of runs on the x-axis (I once asked a question about the explicit joint probability). Each dot represents the result of 300 fair flips. For better visibility, the points are jittered. The numbers for the two sequences are plotted in color. The conclusion is obvious.\n\nFor a quick calculation, recall that a rule of thumb for the longest run of either heads or tails in $$n$$ tosses is$$^{[1]}$$ $$l = \\log_{1/p}(n(1-p)) + 1$$. For an approximate 95% prediction interval, just add and subtract $$3$$ from this value. Surprisingly, this number (i.e. $$\\pm 3$$) does not depend on $$n$$! Applied to a fair coin with $$n=300, p=1/2$$, we have $$l=\\log_2(300/2) + 1=8.22$$. So we expect the longest run to be round $$8$$ and reasonably in the range of $$8\\pm 3$$, so between $$5$$ and $$11$$. The longest run in sequence 2 is $$8$$, whereas it is $$4$$ in sequence 1. As this is outside the approximate prediction interval, we'd conclude that sequence 1 is suspicious under the assumption of $$p=1/2$$.\n\n$$[1]$$ Schilling MF (2012): The Surprising Predictability of Long Runs. Math. Mag. 85: 141-149. (link)\n\n\u2022 Excellent answer and graph. \"Surprisingly, this number does not depend on n!\" You're talking about +-3, right? I was confused for a while since you define $l$ just before, which obvisouly depends on $n$. May 9 at 14:48\n\u2022 @EricDuminil Yes, sorry. The value that does not depend on $n$ is the $\\pm 3$ for the prediction interval. May 9 at 15:45\n\nThe runs test (NIST page) is a nonparametric test designed to identify unusual frequencies of runs. If we observe $$n_1$$ heads and $$n_2$$ tails, the expected value and variance of the number of runs are:\n\n$$\\mu = {2n_1n_2 \\over n_1+n_2} + 1$$ $$\\sigma^2 = {2n_1n_2(2n_1n_2 - n_1 - n_2) \\over (n_1+n_2)^2(n_1+n_2+1)}$$\n\nAs a rule of thumb, for $$n_1, n_2 \\geq 10$$ the distribution of the observed number of runs is reasonably well-approximated by a Normal distribution.\n\nEdit: (incorporating Eric Duminil's work below)\n\nFor sequence 1, we have 148 heads, 152 tails, 205 runs, and for sequence 2, we have 148 heads, 152 tails and 154 runs. Plugging these numbers into our formulae above give us $$z$$-scores of 6.5 for the first sequence and 0.58 for the second sequence - extremely strong evidence that the first sequence is fake.\n\nWhen people fake sequences like this, they tend to greatly underestimate the probability of longer runs, so they don't create as many long(ish) runs as they should. This in turn tends to increase the number of runs beyond that which would be expected. Consequently, when testing for faked data, we might prefer a one-sided test of the alternative hypothesis that there are \"too many\" runs vs. the null hypothesis that the number of runs is average - at least if we think the sequence was created by a human being.\n\n\u2022 Yep, this is a classic application of gambler's fallacy. May 6 at 21:54\n\u2022 There are 62 runs in the first two rows, AFAICT. For the whole sequence #1 : 148 heads, 152 tails, 205 runs. Vs 148 heads, 152 tails and 154 runs for sequence #2. May 9 at 14:56\n\u2022 Thanks, @EricDuminil - I've included your efforts in the body of the answer with a citation. May 9 at 16:46\n\u2022 That's good. But it leaves unanswered why you would formulate a question in terms of longest run and then solve it by looking at an indirect proxy (number of runs). Why not just use the longest run as the test statistic?\n\u2013\u00a0whuber\nMay 9 at 18:13\n\u2022 @whuber - it's not just the longest run that's informative, it's the number of longer runs in general, and I don't have a good idea of a cutoff for run length for testing. Having said that, using the longest run as a test statistic with 100,000 randomly generated strings with 148 heads and 152 tails for calculating an approximate p-value gives a p-value of 0.00004 for the first sequence. Maybe I'll expand on that over (my) lunch. May 9 at 18:28\n\nHere's an empirical approach, based on compression as a proxy for algorithmic complexity:\n\nimport bz2\nimport random\nimport statistics\n\ns1 = \"TTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHHTTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHHTTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHTTHHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHTHTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTTHHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\"\ns2 = \"HTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTHTHTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHHTTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTTTHTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTHHHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHHHTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\"\n\ndef compressed_len(s):\nreturn len(bz2.compress(s.encode()))\n\ntrials = []\nfor x in range(100000):\nsr = \"\".join(random.choice(\"HT\") for _ in range(300))\ntrials.append(compressed_len(sr))\n\nmean = statistics.mean(trials)\nstddev = statistics.stdev(trials)\n\nprint(\"Random trials:\")\nprint(\"Mean:\", mean)\nprint(\"Stddev:\", stddev)\n\nl1 = compressed_len(s1)\nl2 = compressed_len(s2)\n\no1 = l1 - mean\no2 = l2 - mean\n\nd1 = o1 / stddev\nd2 = o2 / stddev\n\nprint(\"Selected trials:\")\nprint(\"Seq\", \"Len\", \"Dev\", sep=\"\\t\")\nprint(\"S1\", l1, d1, sep=\"\\t\")\nprint(\"S2\", l2, d2, sep=\"\\t\")\n\n\n\nRoughly speaking:\n\n1. Compress a bunch (100k) of random coinflips.\n2. Observe the resulting length distribution. (In this case I'm approximating it as a normal distribution of lengths; a more thorough analysis would check and pick an appropriate distribution instead of blithely assuming normality.)\n3. Compress the input sequences.\n4. Compare with observed distribution.\n\nResult (note: not exactly reproducible due to the use of random trials; if you want to be reproducible add a random seed):\n\nRandom trials:\nMean: 105.05893\nStddev: 2.6729774976956002\nSelected trials:\nSeq Len Dev\nS1  88  -6.381995364609942\nS2  109 1.4744119632124217\n\n\nBased on this, I'd say that S1 is the non-random one here. 6.38 standard deviations below the mean is rather improbable.\n\nThe nice thing about this approach is that it's relatively generic, and takes advantage of the pre-existing work of a bunch of smart people.\n\nJust be aware of its limitations and quirks:\n\n1. You want a compression algorithm that's designed for space over compression speed. BZ2 works well enough here.\n2. This doesn't work if the compression algorithm simply gives up and writes a raw block to the output.\n3. A null result does not mean that the sequence is random. It means that this compression algorithm is unable to distinguish this input from random.\n\nThis is probably an overcomplicated way of looking at it, but for me it's fun, so I present to you...\n\n## Moran's I\n\nNow, Moran's I was developed to look at spatial autocorrelation (basically autocorrelation with multiple dimensions), but it can be applied to the 1-dimensional case as well. Some of my interpretations might be a little sketchy, but you can consider your coin flips this way.\n\nTo summarize Moran's I, it will consider your neighboring values using a pre-defined matrix. How you define the matrix is up to you, but it can actually be used to consider not just the directly neighboring values, but any values beyond that. Moran's I will produce a value ranging from -1 (perfectly dispersed values) to 1 (perfectly clustered values), with 0 being random.\n\nI wrote up some quick R code. First, setup the data (OP's data and a couple generated data sets to test dispersion and clustering):\n\nseq1 = unlist(strsplit(\"TTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHHTTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHHTTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHTTHHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHTHTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTTHHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\",\nsplit = \"\"))\n\nseq2 = unlist(strsplit(\"HTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTHTHTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHHTTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTTTHTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTHHHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHHHTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\",\nsplit = \"\"))\n\n# Alternate T and H. E.g., THTHTHTHT....\n# 'perfectly dispersed'\n# Moran's I = -1\nseq3 = rep(c(\"T\", \"H\"), times = 50)\n\n# 50 of T followed by 50 of H\n# 'perfectly clustered'\n# Moran's I approaches 1 as the sample size increases to infinity\nseq4 = rep(c(\"T\", \"H\"), each = 50)\n\n# weights must be a vector with an odd length and the middle value set to 0\n# weights are relative and do not have to add to 1\nmoran <- function(x, weights) {\nx = c(T = 0, H = 1)[x] # convert T/H to 0/1\n\nN = length(x)\n\nx_mean = mean(x)\n\nden = sum((x - x_mean)^2)\n\nW = 0\nnum = 0\n\noffset = floor(length(weights)/2)\nfor (i in 1:length(x)) {\n\nW = W + as.numeric(!is.na(x_slice)) %*% weights\nnum = num + (x[i] - x_mean) * sum((x_slice - x_mean) * weights, na.rm = TRUE)\n}\n\nreturn(unname((N * num)/(as.numeric(W) * den)))\n}\n\n\nNext, I test the generated data sets to illustrate/test that my function is working correctly:\n\n# Test the 'perfect dispersion' scenario (should be -1)\nmoran(seq3, c(1, 0, 1))\n## [1] -1\n\n# Test the 'perfect clustering' scenario (should be ~1)\nmoran(seq4, c(1, 0, 1))\n## [1] 0.979798\n\n\nNow, let's look at OP's sequences:\n\n# Simple look at seq1. The weights test the idea that the current flip\n# is based purely on the last flip (a reasonable model for how a person might react)\nmoran(seq1, c(1, 0, 0))\n## [1] -0.3647031\nmoran(seq2, c(1, 0, 0))\n## [1] -0.02359453\n\n\nI'm defining my weights matrix such that only the previous flip is considered when testing for autocorrelation. We see that the second sequence is very close to 0 (random), whereas the first sequence seems to lean somewhat toward overdispersion.\n\nBut maybe we think someone faking coin flips would consider the last two flips, not just the most recent:\n\n# Maybe the person is looking back at the last two flips\nmoran(seq1, c(1, 1, 0, 0, 0))\n## [1] -0.1726056\nmoran(seq2, c(1, 1, 0, 0, 0))\n## [1] 0.0249505\n\n\nThe second sequence is just as close to 0 as before, but the first sequence had a pretty noticeable shift towards 0. This might be interpretable in a couple of different ways. First, if we know that the first sequence is fake, then maybe it means the person wasn't considering two flips back. A second interpretation is that maybe they were considering the last two flips, and somehow this led them to doing a better job at faking randomization. A third option might just be sheer dumb luck at faking the randomization.\n\nNow, maybe the person considers the last two coin flips but gives the most recent flip more importance.\n\n# Same idea, but maybe the more recent of the two is twice as important\nmoran(seq1, c(1, 2, 0, 0, 0))\n## [1] -0.2367095\nmoran(seq2, c(1, 2, 0, 0, 0))\n## [1] 0.008750762\n\n\nHere, we see the two sequences react differently. The second sequence (already pretty close to 0), gets noticeably closer to 0, whereas the first sequence shifts noticeably away. I'm not sure I want to try and interpret this, but it's an interesting result, and a similar thing happens if we try to model a scenario where the person is not only considering their previous flips but also thinking ahead to their next flip:\n\n# Maybe the person was thinking ahead to their next flip as well\nmoran(seq1, c(1, 2, 0, 1, 0))\n## [1] -0.2687347\nmoran(seq2, c(1, 2, 0, 1, 0))\n## [1] 0.0006576715\n\n\nSome of my application/interpretation of Moran's I to the coin flip problem might be a little off, but it's definitely an applicable measure to use.\n\nA related metric is Geary's C, which is more sensitive to local autocorrelation\n\nWhen people try to generate random sequences, they tend to avoid repeating themselves more than random processes avoid repeating themselves. Thus, if we look at consecutive pairs of flips, we would expect a human-generated sequence to have too many HT and TH and too few HH and TT compared to a typical random sequence.\n\nThe code below explores this hypothesis. It splits each sequence of 300 flips into 150 consecutive pairs and plots the frequency of the four possible results (HH, HT, TH, TT).*\n\nlibrary(tidyverse)\n\na <- \"TTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHHTTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHHTTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHTTHHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHTHTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTTHHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\"\nb <- \"HTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTHTHTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHHTTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTTTHTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTHHHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHHHTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\"\n\n# split each sequence of 300 into 150 consecutive pairs\n# e.g. TTHHTHTT... -> TT, HH, TH, TT, ...\nn_pairs <- 150\nap <- tibble(pair = character(n))\nbp <- tibble(pair = character(n))\nfor (i in 1:n) {\nap$$pair[i] <- substring(a, 2*i - 1, 2*i) bp$$pair[i] <- substring(b, 2*i - 1, 2*i)\n}\n\n# get the frequencies of each possible pair and plot\napc <- count(ap, pair)\nbpc <- count(bp, pair)\nbind_rows(\nSequence 1 = apc,\nSequence 2 = bpc,\n.id = 'source') %>%\nggplot(aes(x = pair, y = n, group = source, fill = source)) +\ngeom_col() +\nfacet_grid(vars(source)) +\ntheme_minimal() +\ngeom_hline(yintercept = n_pairs/4, linetype = 'dashed') +\nylab('frequency') +\nggtitle('Frequency of consecutive coin flip pairs')\n\n\nThe dotted line at 150/4 = 37.5 is the expected count of each possible pair assuming the coin flips are independent and fair. By the Law of Large Numbers, we expect the bars not to stray too far from the dotted line. Sequence 1 has an above-average number of HT and TH pairs (especially HT), consistent with our hypothesis about human-generated \"randomness\". The pairs from Sequence 2 are more consistent with average behavior.\n\nTo see how unusual this behavior would be under independent, fair flips, we reformat each sequence's pair count data as a 2x2 contingency table (rows = first flip H/T, columns = second flip H/T) and use Fisher's exact test, which checks whether the data is consistent with a null hypothesis in which the first flip is independent of the second:\n\nfor (x in list(apc, bpc)) {\nprint(x)\nx %>%\nmutate(f1 = str_sub(pair, 1, 1),\nf2 = str_sub(pair, 2, 2)) %>%\nselect(f1, f2, n) %>%\nselect(-f1) %>%\nas.matrix() %>%\nfisher.test() %>%\nprint()\n}\n\n\nThe contingency table for Sequence 1 has a p value of 0.0002842, while the table for Sequence 2 has 0.5127. This means that pair frequencies skewed to the degree seen in Sequence 1 would only occur by random 1/0.0002842 = 1 out of 3,519 times, while something like Sequence 2 would be seen very commonly. It seems sensible to conclude that Sequence 1 is the human-made sequence, since its pair frequency table is not consistent with random chance but is quite consistent with the behavior we'd expect of humans.\n\nThere is a caveat to this analysis: we do not expect random sequences to be perfectly consistent with average behavior. In fact, in some contexts people know that long random sequences should follow the Law of Large Numbers, and they create sequences which follow it too perfectly. A different analysis would be needed to explore whether Sequence 2 looks odd from this opposite perspective.\n\n* Other answers look at all 299 consecutive pairs, which gives you more data points but they become dependent, which prevents us from using standard significance tests. (For example in the sequence TTHHT, you can make pairs like this: (TT)HHT, T(TH)HT, TT(HH)T, .... This gives you more pairs but consecutive pairs are not independent of one another, as the second flip of a pair determines the first flip of the next pair.)\n\n# Alternate Analysis\n\nAn analysis that uses all 299 pairs of consecutive flips could be more powerful than the one above if the dependency problem can be solved. To do this, @whuber suggests looking at the transitions between consecutive flips, i.e. when H is followed by T or vice versa. If the flips are independent and fair, then after the first flip, each transition can be considered an independent Bernoulli random variable, and there are 299 transitions total. We can use a two-sided test to see whether the number of transitions observed in each sequence is unlikely under fair independent flips. The transitions are counted and the test applied by the code below:\n\nlibrary(tidyverse)\n\na <- \"TTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHHTTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHHTTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHTTHHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHTHTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTTHHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\"\nb <- \"HTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTHTHTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHHTTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTTTHTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTHHHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHHHTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\"\n\n# 299 transitions from flip i to i+1 occur in the sequence of 300\n# record these transitions in arrays at and bt\nn <- 299\nat <- logical(n)\nbt <- logical(n)\nfor (i in 1:n) {\nat[i] <- str_sub(a, i + 1, i + 1) != str_sub(a, i, i)\nbt[i] <- str_sub(b, i + 1, i + 1) != str_sub(b, i, i)\n}\n\n# two-sided exact binomial test (analogous to z-test)\n# gives probability of transition count more extreme than the one observed\npbinom(sum(at) - 1, n, 1/2, lower.tail = F) + pbinom(n - sum(at), n, 1/2)\npbinom(sum(bt) - 1, n, 1/2, lower.tail = F) + pbinom(n - sum(bt), n, 1/2)\n\n\nRunning this code, we find that Sequence 1 has 204 transitions out of a possible 299. The probability of observing a number of transitions at least this imbalanced, on either the left or the right side, is equal to the probability of observing at least 204 transitions, plus the probability of observing at most 299 - 204 = 95 transitions. This probability is 2.696248e-10, on the order of 3 in 10 billion. Sequence 2 has 153 transitions, and the probability of observing at least 153 transitions or at most 299 - 153 = 146 transitions is 0.7286639. The number of transitions in Sequence 1 is extremely improbable, more so than the 150 pair test above suggested.\n\n\u2022 Your code seems to be doing something more complicated than you describe. It is hard to relate your plots, which are unexplained, to the data: those plots clearly do not report on sequences of length 300.\n\u2013\u00a0whuber\nMay 7 at 14:43\n\u2022 I'm splitting each length 300 sequence into 150 pairs of adjacent coin tosses and counting how many of the 150 pairs are HH, HT, TH, or TT. That's what the graphs show. I've added some comments to elaborate on this. If that's not enough, could you be more specific as to what you're struggling to understand?\n\u2013\u00a0Paul\nMay 7 at 16:55\n\u2022 I am not struggling to understand anything--I just don't care to have to read through code in order to determine what the content of an answer might be, and I believe most readers will feel the same.\n\u2013\u00a0whuber\nMay 7 at 17:25\n\u2022 Thank you: what you are doing is now more apparent. But why? Could you explain what is accomplished with this split?\n\u2013\u00a0whuber\nMay 7 at 18:10\n\u2022 I've added a transition-based analysis and it is indeed a more powerful test. Great idea.\n\u2013\u00a0Paul\nMay 10 at 0:45\n\nThis answer is inspired by @user1717828's answer which transforms the sequence of coin flips into a random walk. I don't show the two given sequences as random walks here; see @user1717828's answer for that plot.\n\nThe random walk approach is interesting because it examines long-run features of the sequence rather than short-range ones (such as the overabundance of H-T and T-H flips). In a way the two approaches are complementary as they analyze the sequences at different scales. At the \"local\" scale the fake sequence oscillates which creates autocorrelation; at the \"global\" scale it keeps within a limited range of positions for long stretches of time. Both of these aspects are non-random and as one occurs, so does the other: staying in place means not moving around and vice versa.\n\nA (one-dimensional) simple random walk $$S_n$$, defined as the sum of n random variables that take value {-1, 1} with probability \u00bd, has many interesting properties, including:\n\n\u2022 The probability that a simple random walk returns to the origin is 1. In fact, it visits every integer infinitely often.\n\u2022 The mean of a simple random walk is $$\\text{E}(S_n) = 0$$ and its variance is $$\\text{Var}(S_n) = n$$.\n\u2022 A random walk with nonzero mean (which corresponds to flipping a biased coin) is transient: it makes finitely many visits to the origin before diverging, to +\u221e if the mean is positive and to -\u221e in the mean is negative.\n\nThese properties imply that a simple random walk spends a lot of time far away from 0 before eventually returning to it.\n\nWe can use this intuition founded on theory (as well as any other property of random walks) to propose characteristics for comparing the two sequences. Then we generate many simple random walks and use the observed distribution of the features to estimate how \"extreme\" the two given sequences are.\n\nTo capture the variance of a simple random walk I use the maximum distance from 0. To capture its tendency to visit every integer I use the number of times the walk crosses the integers from -8 to +8. To make a symmetric grid of scatterplots and to avoid suspicions of HARKing I don't show the crossings of 0. It's important to look at crossing on both sides of 0 (the starting point), so that we don't make assumptions about how the non-randomness manifests in the data. For example, we don't know whether the coin is fair (p = \u00bd), biased towards heads (p > \u00bd) or biased towards tails (p < \u00bd).\n\nNote: HARKing is the practice of performing many analyses to find an interesting hypothesis. I compute many statistics, grounded in the theory of random walks, and report all of them, thus avoiding HARKing.\n\nHere are the results from the simulation of 500 simple random walks. Sequence #1 (in blue) appears as an outlier compared to sequence #2 (in red) in many panels. Due to its overabundance of H-T and T-H pairs, sequence #1 doesn't explore enough and spends too much time between -5 and -8 (shown in the top row). Sequence #1 doesn't advance beyond the band [-9, 9] and it is rare for a true random walk so stay so close to 0 for 300 steps. Visually, sequence #1 is the overall outlier even if it doesn't appear extremely \"unusual\" is some panels.\n\n\u2022 +1. This explains why I did not examine this statistic in my answer: I had done so, exactly as you have, and came to the same conclusion that the max deviation statistic was suggestive but only weak evidence of a departure from randomness. But in that investigation I noted there were other visual quirks in the plot of the first series and that led to a more effective statistic for assessing the randomness of the series. It's worth noting that extensive computation is unnecessary: you could draw all your conclusions more quickly by simulating as few as 20 random walks..\n\u2013\u00a0whuber\nMay 9 at 13:44\n\u2022 @whuber After some (over)thinking, I realized that it works quite well to compute two statistics: one to represent central tendency and the other -- extreme values. Similar to COOLSerdash's solution which looks at number of runs (a kind of average) and length of longest run. With the simple random walk, zero crossings vs maximum distance from zero works very well (visually). But then I don't know how to get a p-value. May 9 at 21:35\n\u2022 I think the zero crossings observation comes down to HARKing. (For instance, these could well be realizations of Markov processes in which a transition away from a side of the coin is made with probability $p$; $p\\approx 2/3$ in one sequence and $p\\approx 1/2$ in the other. In general I wouldn't expect zero-crossing counts to distinguish among these.) A better way to approach this would be to do all your exploration on the first half of each sequence; develop a suitable statistic from that; and apply it to the second halves of the sequences.\n\u2013\u00a0whuber\nMay 9 at 21:39\n\u2022 @whuber I don't agree it's HARKing. At least not any more than number of runs. A simple random walk returns to the start infinitely often, so given this fact it makes sense to look at zero crossings. Since the fake sequence doesn't move enough any number will do in theory; here because sequences are only 300 long every number between about -10 to 10 will work. May 9 at 22:03\n\u2022 Ask yourself this: would you have focused on counting zero crossings by exploring just the first halves of the sequences? You might have if both of two things occurred: (1) you observed no zero crossings in one and many zero crossings in the other and (2) in simulated iid sequences, you observed a consistent tendency to have positive numbers of zero crossings. In such a case, you likely have a useful statistic. Unfortunately, you would discover that about 6.5% of the time, a Binomial random walk has no zero crossings. That's why I view this as HARKing: it's accidental.\n\u2013\u00a0whuber\nMay 10 at 14:32\n\nIn addition to statistical approaches, one visual approach is to plot the sequences as a \"drunkards walk\". Treat H as a step forwards and T as a step back and plot the sequences. One way in Python is:\n\nimport altair as alt\nimport pandas as pd\n\nseq1 = \"HTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTHTHTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHHTTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTTTHTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTHHHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHHHTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\"\n\nseq2 = \"TTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHHTTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHHTTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHTTHHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHTHTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTTHHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\"\n\ndef encode(stream):\nreturn [1 if c == \"H\" else -1 for c in stream]\n\ndf = pd.DataFrame({\"seq1\": encode(seq1), \"seq2\": encode(seq2)})\ndf_cumsum = pd.melt(\ndf.cumsum(), var_name=\"sequence\", value_name=\"cumsum\", ignore_index=False\n).reset_index()\n\nchart = alt.Chart(df_cumsum).mark_line().encode(x=\"index\", y=\"cumsum\", color=\"sequence\")\nchart.show()\n\n\nWhen comparing a truly random sequence to a human generated one, it is often easy to tell them apart based on inspection of the walk.\n\n\u2022 This is a good approach. It was the first thing I tried. However, I created a context for evaluating these random walks: I also generated 18 more sequences using iid flips of a fair coin and plotted all 20. Generally, whenever I repeated this procedure, there was a truly random walk that looked like the orange one here, at least in terms of the relatively small range. Exactly which characteristic(s) suggest to you that the orange curve is not random but the blue is?\n\u2013\u00a0whuber\nMay 8 at 12:53\n\u2022 The orange one is what I think is the truly random one based on this visual. I would have guessed based on just the number of zero-crossings alone that blue is human-generated. But you're right, there's always going to be a chance that any sequence is randomly-generated, and the analytical methods are better for quantifying the odds of that. May 8 at 20:17\n\u2022 Unfortunately for your theory, the orange one is the non-random one. It has an unusually low range; but more to the point, its thickened appearance is due to a huge overabundance of H-T-H oscillations.\n\u2013\u00a0whuber\nMay 8 at 20:32\n\u2022 Ahh! In that case, I'm going to rethink my approach of visualizing these things before looking at the statistics next time! May 8 at 22:29\n\u2022 The thickened appearance is the feature analyzed in the answers by whuber, Dmetri Pananos and Paul. The distance from 0 in a simple random walk (which I think is what this answer is getting at) would be a different feature altogether. A (true) simple random walk will spend a lot of time away from zero until eventually returning (with probability 1). The sequences in this exercise though seem too short for this kind of argument. May 8 at 23:55\n\nLooking at the HH, HT, TH, TT frequencies is probably the most straightforward way to approach the two series presented, given people's tendency to apply HT and TH more frequently when trying to appear random. More generally, however, that approach will fail to detect non-randomness even in sequences with obvious patterns. For instance, a repeating sequence of HHHTHTTT will produce balanced counts of pairs as well as triples (HHH, HHT, etc.).\n\nHere's an idea for a more general solution that was initially inspired by answers discussing random walks. Start with the observation that for a random sequence, the number of heads in any given subsequence is binomially distributed. We can count the number of heads, $$n_{i,j}$$, between flip number $$i$$ and flip number $$j>i$$ for all values of $$i,j$$. Then compare these to the expected number of counts if all the $$n_{i,j}$$ were independent. Although they are obviously not independent, the comparison gives rise to a useful statistic: the maximum absolute difference between the observed counts and the expected counts.\n\nApplying this approach to sequence 1 gives us 98.26 as our test statistic: there are 257 subsequences of length 44. If they were all composed of independent Bernoulli trials, the expected number of the 257 that contained exactly 22 heads is ~30.74, whereas sequence 1 contains 129 subsequences with exactly 22 heads (very underdispersed). 129 - 30.74 = 98.26, which is the maximum of these differences for sequence 1.\n\nPerforming the same calculations on sequence 2 gives a test statistic of 48.30: there are 197 subsequences of length 104. The expected number containing exactly 54 heads would be ~13.70. Sequence 2 contains 62, so the test statistic is 62 - 13.70 = 48.30.\n\nThe test statistics can be compared to those from a large number of random sequences of the same size. In this case, no samples are greater than the test statistic from sequence 1, and about 14% of samples are greater than the statistic from sequence 2.\n\nHere it is all together with R code:\n\nlibrary(parallel)\nset.seed(16)\n\nseq1 <- utf8ToInt(\"TTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHHTTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHHTTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHTTHHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHTHTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTTHHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\") == utf8ToInt(\"H\")\nseq2 <- utf8ToInt(\"HTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTHTHTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHHTTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTTTHTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTHHHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHHHTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\") == utf8ToInt(\"H\")\n\n# function to calculate test statistic S\nfS <- function(m, p = 0.5) {\nif (!is.matrix(m)) m <- matrix(m, ncol = 1)\nn <- nrow(m)\nsteps <- rep.int(1:(n - 1L), 2:n)\nnEx <- dbinom(sequence(2:n, from = 0L), steps, p)*(n - steps)\nidx <- sequence((n - 1L):1)\nidx <- idx*(idx + 1L)/2L\nS <- numeric(ncol(m))\nfor (i in 1:ncol(m)) S[i] <- max(abs(nEx - tabulate(idx + dist(cumsum(m[,i])), length(nEx))))\nS\n}\n\n(S1 <- fS(seq1))\n#> [1] 98.26173\n(S2 <- fS(seq2))\n#> [1] 48.30014\n# calculate S from 1e6 random sequences of length n (probably overkill)\nn <- length(seq1)\ncl <- makeCluster(detectCores() - 1L)\nclusterExport(cl, list(\"fS\", \"n\"))\nsystem.time(simS <- unlist(parLapply(cl, 1:100, function(i) fS(matrix(sample(0:1, 1e4L*n, TRUE), n)))))\n#>    user  system elapsed\n#>    0.00    0.03  339.42\nstopCluster(cl)\nnsim <- 1e6L\n# calculate approximate p-values\nsum(simS > S1)/nsim\n#> [1] 0\nsum(simS > S2)/nsim\n#> [1] 0.139855\nmax(simS)\n#> [1] 91.01078\n\n\nThis problem specifies that we have the following information:\n\n\u2022 We observed two coin flip sequences: sequence $$S_1$$ and sequence $$S_2$$.\n\u2022 Each of these sequences could have been generated by either mechanism $$R$$ corresponding to independent flips of a fair coin or some other mechanism $$\\bar{R}$$.\n\u2022 Exactly one of these two sequences was generated by the mechanism $$R$$.\n\nSince both sequences $$S_1$$ and $$S_2$$ could possibly have been generated by $$R$$ or $$\\bar{R}$$, we cannot be sure which of these two sequences was generated by $$R$$. However, probability theory provides us with the tools to quantify our belief $$P(R_i|S_i)$$ that sequence $$S_i$$ was generated by mechanism $$R$$. Then, given that we have been provided with the information $$I$$ that exactly one of these two sequences has been generated by mechanism $$R$$, then the probability that sequence 1 was generated by mechanism $$\\bar{R}$$ and sequence 2 was generated by mechanism $$R$$ is:\n\n$$P(\\bar{R_1}R_2|S_1 S_2 I) = \\frac{P(R_2|S_2)}{P(R_1|S_1) + P(R_2|S_2)}$$\n\nThis problem explicitly specifies what it means for a sequence to have been generated by mechanism $$R$$: the sequence $$S_i$$ is described by independent flips $$y_{ij}$$ of a fair coin, such that the likelihood is:\n\n$$P(S_i|R_i) = \\prod_j \\mathrm{Bernoulli}(y_{ij} | 0.5)$$\n\nHowever, in order to determine the probability that a given sequence was generated by mechanism $$R$$, $$P(R_i|S_i)$$, we also need to specify alternative mechanisms $$\\bar{R}$$ by which the sequences may have been generated. At this point we have to use our own creativity and external information to develop models that could reasonably describe how these sequences might have been generated, if they were not generated by mechanism $$R$$.\n\nThe other answers to this question do a great job at describing various mechanisms $$\\bar{R}$$ that describe how sequences could be generated:\n\n\u2022 Sequences are generated by sampling pairs of coin flips from a distribution where the probability of different pairs of coin flips deviates from uniform: 1 2 3\n\u2022 Sequences are generated in a way such that they do not have long runs of only heads or tails: 1\n\u2022 Sequences are generated in such a way so that a given compression algorithm results in a small compressed size: 1\n\u2022 Sequences are generated by a random walk such that a given coin flip outcome is affected by recent coin flip outcomes: 1 2 3\n\nNote: we can always develop a hypothesis that appears to result in our observed sequence having been \"inevitable\", according to our likelihood, and if our prior probability for that mechanism of sequence generation is high enough, then using this hypothesis can always result in a low probability that the sequence had been generated by mechanism $$R$$. This practice is also referred to as HARKing, and Jaynes refers to this as a \"sure-thing hypothesis\". In general, we may have multiple hypotheses that can describe how a sequence may be generated, but the more contrived hypotheses such as \"sure-thing hypotheses\" should generally have sufficiently low prior probabilities such that we generally would not infer that those hypotheses have the highest probability of having generated a given sequence.\n\nHere's an example using this procedure with one possible model for $$\\bar{R}$$ sequence generation to quantify the probability that sequence 2 was the sequence that was generated by mechanism $$R$$. This stan model defines model_r_lpmf as the log likelihood for sequence generation mechanism $$R$$, and model_not_r_lpmf as the log likelihood for sequence generation mechanism $$\\bar{R}$$. In this case, we are testing a hypothesis $$\\bar{R}$$ in which coin flips can be correlated, and there is a fairly well defined correlation length. Then, we are modeling the probability that the sequence was generated by model $$R$$ as p_model_r, such that our prior probability considers mechanisms $$R$$ and $$\\bar{R}$$ sequence generation equally probable.\n\n// model for hypothesis testing whether a sequence of coin flips is generated by either:\n// - model R: independent flips of a fair coin, or\n// - model not R: unfair coin with correlations\n\nfunctions {\n// independent flips of a fair coin\nreal model_r_lpmf(int[] sequence, int N) {\nreturn bernoulli_lpmf(sequence | 0.5);\n}\n\n// unfair coin with correlations\nreal model_not_r_lpmf(int[] sequence, int N, int n, vector beta) {\nreal tmp = 0;\n\nreal offset = 0;\nfor (i in 1:N) {\noffset = beta[1];\nfor (j in 1:min(n - 1, i - 1)) {\noffset += beta[j + 1] * (2 * sequence[i - j] - 1);\n}\ntmp += bernoulli_logit_lpmf(sequence[i] | offset);\n}\nreturn tmp;\n}\n}\ndata {\nint N;                                 // the length of the observed sequence\nint<lower=0, upper=1> sequence[N];     // the observed sequence\n}\ntransformed data {\nint n = N / 2 + 1;                     // number of correlation parameters to model\n}\nparameters {\nreal<lower=0, upper=1> p_model_r;       // probability that the sequence was generated by model R\n// not R model parameters\nreal log_scale;                        // log of the scale of the correlation coefficients beta\nreal log_decay_rate;                   // log of the decay rate for the correlation coefficients beta\nvector[n] alpha;                       // pretransformed correlation coefficients\n}\ntransformed parameters {\n// transformed not R model parameters\nreal scale = exp(log_scale);           // typical scale of the correlation coefficients\nreal decay_rate = exp(log_decay_rate); // typical decay rate of correlation coefficients\nvector[n] beta;                        // correlation coefficients\nfor (j in 1:n) {\nbeta[j] = alpha[j] * scale * exp(-(j - 1) * decay_rate);\n}\n}\nmodel {\n// uninformative haldane prior for the probability that the sequence was generated by model R\ntarget += - log(p_model_r) - log1m(p_model_r);\n\n// priors for not R model parameters\nlog_scale ~ normal(-1, 1);\nlog_decay_rate ~ normal(-1, log(n + 1) / 2.0);\nalpha ~ normal(0, 1);\n\n// the sequence is was generated by model R with probability p_model_r, otherwise it was generated by model not R\ntarget += log_mix(\np_model_r,\nmodel_r_lpmf(sequence | N),\nmodel_not_r_lpmf(sequence | N, n, beta)\n);\n}\n\n\nUsing pystan with this model, we can evaluate the probability p_model_r that each of the two provided sequences were generated by mechanism $$R$$.\n\nmodel = pystan.StanModel(model_code=MODEL_CODE)\n\nsequence_1 = \"TTHHTHTTHTTTHTTTHTTTHTTHTHHTHHTHTHHTTTHHTHTHTTHTHHTTHTHHTHTTTHHTTHHTTHHHTHHTHTTHTHTTHHTHHHTTHTHTTTHHTTHTHTHTHTHTTHTHTHHHTTHTHTHHTHHHTHTHTTHTTHHTHTHTHTTHHTTHTHTTHHHTHTHTHTTHTTHHTTHTHHTHHHTTHHTHTTHTHTHTHTHTHTHHHTHTHTHTHHTHHTHTHTTHTTTHHTHTTTHTHHTHHHHTTTHHTHTHTHTHHHTTHHTHTTTHTHHTHTHTHHTHTTHTTHTHHTHTHTTT\"\nsamples_1 = model.sampling(\ndata=dict(N=len(sequence_1), sequence=[int(c == \"H\") for c in sequence_1]),\nchains=16,\n)\np_model_r_1 = samples_1[\"p_model_r\"].mean()\n# p_model_r_1 = 0.027\n\nsequence_2 = \"HTHHHTHTTHHTTTTTTTTHHHTTTHHTTTTHHTTHHHTTHTHTTTTTTHTHTTTTHHHHTHTHTTHTTTHTTHTTTTHTHHTHHHHTTTTTHHHHTHHHTTTTHTHTTHHHHTHHHHHHHHTTHHTHHTHHHHHHHTTHTHTTTHHTTTTHTHHTTHTTHTHTHTTHHHHHTTHTTTHTHTHHTTTTHTTTTTHHTHTHHHHTTTTHTHHHTHHTHTHTHTHHHTHTTHHHTHHHHHHTHHHTHTTTHHHTTTHHTHTTHHTHHHTHTTHTTHTTTHHTHTHTTTTHTHTHTTHTHTHT\"\nsamples_2 = model.sampling(\ndata=dict(N=len(sequence_2), sequence=[int(c == \"H\") for c in sequence_2]),\nchains=16,\n)\np_model_r_2 = samples_2[\"p_model_r\"].mean()\n# p_model_r_2 = 0.790\n`\n\nHere we have quantified the probability that each sequence was generated according to the sequence generating model $$R$$ as the following:\n\n\\begin{aligned} P(R_1|S_1) &= 0.027 \\\\ P(R_2|S_2) &= 0.790 \\\\ P(\\bar{R_1}R_2|S_1 S_2 I) &= 0.967 \\\\ \\end{aligned}\n\nSo, we are fairly certain that sequence 2 was generated by independent coin flips and sequence 1 was not, relative to the model of sequence generation $$\\bar{R}$$ that we defined, and we quantify this belief with a probability of 0.967. This is a high probability, but of course we could still be wrong -- both sequences could have been generated by either model. Additionally, we could have selected a model of sequence generation for $$\\bar{R}$$ that can not describe the procedure that was actually used to generate the sequences. This process for quantifying our belief that one of these two sequences was generated by a fair coin with independent flips is the piece of reasoning that currently appears to be missing from the other answers that are currently available.\n\n\u2022 I have to reject your initial dichotomy, because it (a) is too vague to be actionable (what can one do, in any objective and quantifiable way, to characterize any \"non-random process\"?) and (b) ignores a huge set of other possible mechanisms: namely, random mechanisms that are not iid.. In fact, it is almost surely the case that whoever generated these sequences used a random mechanism for both--they are just different random mechanisms.\n\u2013\u00a0whuber\nMay 31 at 12:00\n\u2022 @whuber I framed this problem in terms of hypothesis testing between a model $R$ by which sequences are generated by independent flips of a fair coin, and models of other possible ways that we might believe that sequences could be generated $\\bar{R}$. I used labels \"random\", and \"not random\" to describe these models, which appears to have been confusing, so I have edited the post to remove that terminology. Jun 8 at 3:18", "date": "2022-07-04 02:36:26", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/574333/how-can-i-determine-which-of-two-sequences-of-coin-flips-is-real-and-which-is-fa/574425", "openwebmath_score": 0.7305430769920349, "openwebmath_perplexity": 1067.4501979938352, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446456243805, "lm_q2_score": 0.8840392771633079, "lm_q1q2_score": 0.8607401087317024}}
{"url": "https://www.tutorialspoint.com/arrange-first-n-natural-numbers-such-that-absolute-difference-between-all-adjacent-elements-1", "text": "Arrange first N natural numbers such that absolute difference between all adjacent elements > 1?\n\nWe have the first N natural numbers. Our task is to get one permutation of them where the absolute difference between every two consecutive elements is > 1. If no such permutation is present, return -1.\n\nThe approach is simple. We will use the greedy approach. We will arrange all odd numbers in increasing or decreasing order, then arrange all even numbers in decreasing or increasing order\n\nAlgorithm\n\narrangeN(n)\n\nBegin\nif N is 1, then return 1\nif N is 2 or 3, then return -1 as no such permutation is not present\neven_max and odd_max is set as max even and odd number less or equal to n\narrange all odd numbers in descending order\narrange all even numbers in descending order\nEnd\n\nExample\n\n#include <iostream>\nusing namespace std;\nvoid arrangeN(int N) {\nif (N == 1) { //if N is 1, only that will be placed\ncout << \"1\";\nreturn;\n}\nif (N == 2 || N == 3) { //for N = 2 and 3, no such permutation is available\ncout << \"-1\";\nreturn;\n}\nint even_max = -1, odd_max = -1;\n//find max even and odd which are less than or equal to N\nif (N % 2 == 0) {\neven_max = N;\nodd_max = N - 1;\n} else {\nodd_max = N;\neven_max = N - 1;\n}\nwhile (odd_max >= 1) { //print all odd numbers in decreasing order\ncout << odd_max << \" \";\nodd_max -= 2;\n}\nwhile (even_max >= 2) { //print all even numbers in decreasing order\ncout << even_max << \" \";\neven_max -= 2;\n}\n}\nint main() {\nint N = 8;\narrangeN(N);\n}\n\nOutput\n\n7 5 3 1 8 6 4 2\nPublished on 01-Aug-2019 07:36:08", "date": "2020-02-19 02:00:22", "meta": {"domain": "tutorialspoint.com", "url": "https://www.tutorialspoint.com/arrange-first-n-natural-numbers-such-that-absolute-difference-between-all-adjacent-elements-1", "openwebmath_score": 0.30733564496040344, "openwebmath_perplexity": 1346.7502711889088, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446440948806, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8607401044049336}}
{"url": "http://mathhelpforum.com/number-theory/96768-solving-congruences.html", "text": "1. ## Solving Congruences\n\nIf you are to solve for x in the congruence $7x=20(mod50)$\nWhat is the easier way to do it without using a calculator or any other type of electronic device. I know you can add 7 to 20 until the number is divisible by 7 and then divide by 7 to reduce it to a solution but that takes a while. I need to figure out how to get it done in just a minute or two. Thanks.\n\n2. Originally Posted by diddledabble\nIf you are to solve for x in the congruence $7x=20(mod50)$\nWhat is the easier way to do it without using a calculator or any other type of electronic device. I know you can add 7 to 20 until the number is divisible by 7 and then divide by 7 to reduce it to a solution but that takes a while. I need to figure out how to get it done in just a minute or two. Thanks.\nYou mean 50 (the modulus).\n\nFor small moduli and $(a,m) = 1$, adding the modulus until cancellation is probably the best way to solve $ax \\equiv b \\ (\\text{mod } m)$. For this particular congruence, add the modulus once to 20 to get: $7x \\equiv 70 \\ (\\text{mod } 50)$\n\nSince $(7, 50) = 1$, you can 'divide' by 7 to get your solution.\n\n_____________\n\nFor a more general case where $(a,m) \\neq 1$, as long as $(a,m) \\mid b$, then we can always use the Euclidean algorithm to find one solution modulo $\\tfrac{m}{(a,m)}$ and use it to find all solutions modulo m. Here's an example in this wiki entry: Linear congruence theorem\n\n3. ## @dibbledabble\n\nWell, I guess I have an easier method to solve this congruence equation.\n\n7x=20(mod50)\n\nThe equation simply means that 7x, a multiple of 7, is 20 more than a multiple of 50 [By modulo logic]\n\nIf we list down numbers which are 20 more than a multiple of 50, we get\n70, 120, 170, 220, 270, 320, 370, 420, and so on... From this which it is quite evident that 70, 420, 770, 1120... etc are multiples of 7 which are 20 more than multiples of 50.\n\nSo, the smallest positive number is 70. 7x = 70 which means x = 10.\nIn fact, 350n - 280 would be the general form for the number 7x.\nSo, the general solution of x would be 50n - 40 where n can range from 1,2,3,.. and so on.\n\nHope it helped,\nMAX\n\n4. ## General theory\n\nAn equation of the form\n$ax\\equiv b$ (mod n)\nhas a solution if and only if $gcd(a,n)|b$.\n\nHere is how you do it in general when these requirements are met.\n\nLet $d=gcd(a,n)$ and by assumption $d|b \\Rightarrow b=dk$ for some integer $k$. Then by the Euclidean Algorith, there exist integers $s$ and $t$ such that\n$as+nt=d$.\n\nWe now multiply through by k to get.\n\n$ask+ntk=dk=b$\n\nreduce mod n to see\n$a(sk)\\equiv b$ (mod n).\n\nThere is your $x$, namely, $x=sk$ (mod n).\n\n5. 7x7=49=-1 (mod 50)\n\nso -x=140 (mod 50)=-10 (mod 50)\n\nhence x=...\n\n6. Helo, diddledabble!\n\nSolve: . $7x \\:\\equiv\\:20\\text{ (mod 50)}$\nYour \"brute force\" method is tedious, but it is commendable.\nIt shows that you understand the congruence statement.\nHere is a primitive method . . .\n\nWe have: . $7x \\:\\equiv\\:20 \\text{ (mod 50)}$\n\nThis means: . $7x - 20 \\:=\\:50a\\;\\text{ for some integer }a.$\n\nSolve for $x\\!:\\quad x \\:=\\:\\frac{50a+20}{7} \\quad\\Rightarrow\\quad x \\:=\\:7a + 2 + \\frac{a+6}{7}$\n\nSince $x$ is an integer, $a+6$ must be divisible by 7.\nThe first time this happens is $a = 1.$\n\nSo we have: . $x \\;=\\;7(1) + 2 + \\frac{1+6}{7}\\;=\\;10$\n\nTherefore: . $x \\:\\equiv\\:10\\text{ (mod 50)}$\n\n7. ## @dibbledabble\n\nHi,\n\nI think the solution offered by Soroban is the closest to the perfect procedure for your question. I guess mine was a bit too illustrative.\n\n8. So using Soroban's method and a different congruence 35x=20(mod50) I get x=2(mod50) right?\n\n9. ## Yes\n\nIf you haven't checked already, values of 'x' such as 52 and 102 satisfy the equation as, in 52 * 35 = 1820\n= 1800 + 20\n= 36*50 + 20\nSimilarly, 102*35 = 3570\n= 3550 +20\n= 71*50 + 50\n\nMAX\n\n10. Given the congruence $ax \\equiv b \\ (\\text{mod } m)$, if $(a,m) \\mid b$, then there are $(a,m)$ solutions exactly.\n\nSo for the congruence $35x \\equiv 20 \\ (\\text{mod } 50)$, we can see that there should be 5 solutions modulo 50.\n\n___________\n\nReduce your congruence: $5 \\cdot 7x \\equiv 5 \\cdot 4 \\equiv (\\text{mod } 50) \\ \\Rightarrow \\ 7x \\equiv 4 (\\text{mod } \\tfrac{50}{(5, 50)}) \\ \\Leftrightarrow \\ 7x \\equiv 4 \\ (\\text{ mod} 10)$\n\nQuickly by inspection, by adding the modulus to 4, we get: $7x \\equiv 14 \\ (\\text{mod } 10)$\n\nwhich is satisfied by all integers such that $x \\equiv 2 \\ (\\text{mod } 10) \\ \\ (\\star)$.\n\nReturning to our original congruence, all 5 solutions are least residues that satisfy $(\\star)$. Therefore, modulo 50: 2, 12, 22, 32, 42 are all solutions.", "date": "2017-06-23 00:08:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/96768-solving-congruences.html", "openwebmath_score": 0.8243153095245361, "openwebmath_perplexity": 344.0861313507526, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682458008672, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8607318874339769}}
{"url": "http://math.stackexchange.com/questions/464625/expected-number-of-people-sitting-in-the-right-seats", "text": "# Expected number of people sitting in the right seats.\n\nThere was a popular interview question from a while back: there are $n$ people getting seated an airplane, and the first person comes in and sits at a random seat. Everyone else who comes in either sits in his seat, or if his seat has been taken, sits in a random unoccupied seat. What is the probability that the last person sits in his correct seat?\n\nThe answer to this question is $1/2$ because everyone looking to sit on a random seat has an equal probability of sitting in the first person's seat as the last person's.\n\nMy question is: what is the expected number of people sitting in their correct seat?\n\nMy take: this would be $\\sum_{i=1}^n p_i$ where $p_i$ is the probability that person $i$ sits in the right seat..\n\n$X_1 = 1/n$\n\n$X_2 = 1 - 1/n$\n\n$X_3 = 1 - (1/n + 1/n(n-1))$\n\n$X_4 = 1 - (1/n + 2/n(n-1) + 1/n(n-1)(n-2))$\n\nIs this correct? And does it generalize to $X_i$ having an $\\max(0, i-1)$ term of $1/n(n-1)$, a $\\max(0, i-2)$ term of $1/n(n-1)(n-2)$ etc?\n\nThanks.\n\n-\nCould the general probability not be, for $m>1$, $X_m=1-\\left(\\sum_{k=0}^{m-2}\\binom{m-2}{k}\\frac{(n-k-1)!}{n!}\\right)$? Perhaps prove by induction? \u2013\u00a0 Alyosha Aug 10 '13 at 23:19\nThe MSE treatment of the original problem: math.stackexchange.com/questions/5595/taking-seats-on-a-plane \u2013\u00a0 Byron Schmuland Aug 23 '13 at 17:41\n\nSorry for the second answer (I will delete the first one):\n\nI will aim to show that the expected number of people sitting in their correct seats is given by:\n\n$$n-1-\\frac12-\\frac13-\\dots-\\frac1{n-1}=n-H_{n-1}$$\n\nTo do this, we will first find the expected number of people in the wrong seats, which we shall call $s_n$.\n\nSuppose passenger number $1$ sits in seat $i\\ne1$. At this point, passengers $2,\\dots,i-1$ all sit in their correct seats. We now have a situation where there are $n-i+1$ empty seats left, and passenger $i$ is going to sit in a random seat.\n\nThis is very similar to the situation we had at the start, just with fewer seats. The only difference is that passenger $i$'s seat is taken, and there's a seat (seat $1$) which belongs to none of the people standing up.\n\nThis is a bit of a problem, so we'll change things around a bit. We'll pretend that seat number $1$ doesn't, in fact, belong to any of the passengers. So wherever passenger $1$ sits, they're in the wrong seat. We'll use the letter $t_n$ to denote the expected number of people sitting in the wrong seat if seat $1$ doesn't belong to anybody.\n\nWhat we now get, is that the situation after passenger $1$ has sat in seat $i\\ne1$, and passengers $2,\\dots,n-1$ have sat in their correct seats is exactly the same as the situation at the beginning, but with $n-i+1$ seats rather than $n$: there's a passenger about to choose a random seat which doesn't belong to him (I decided the sex of passenger $i$ by tossing a coin): there's a seat ($1$) which doesn't belong to anybody, and the rest of the seats belong to the remaining passengers. So at this point, the expected number of passengers sitting in the wrong seats is $1+t_{n-i+1}$: $1$ for passenger $1$, who's sat in the wrong seat, and $t_{n-i+1}$ because afterwards we're in exactly the same situation as before, but with $n-i+1$ seats.\n\nWhat if passenger $1$ sits in seat $1$? Then all the remaining passengers sit in the right seats, so the expected number of people sitting in the wrong seats at the end is just $1$ (remember, passenger $1$ no longer owns seat $1$).\n\nPassenger $1$ chooses between the $n$ seats at random, so this gives us the recurrence:\n\n\\begin{align} t_1 &= 1\\\\ t_n &= 1+\\frac1n\\sum_{i=2}^nt_{n-i+1} = 1+\\frac1n\\sum_{i=1}^{n-1}t_i \\end{align}\n\nWe are now ready to prove the following:\n\nClaim: $t_n=H_n$\n\nProof of claim: induction on $n$. $\\mathbf{n=1}$ : $t_1=1=H_1$.\n\n$\\mathbf{n>1}$ : $t_n=1+\\frac1n\\sum_{i=1}^{n-1}t_i=1+\\frac1n\\sum_{i=1}^{n-1}H_i$ (by the inductive hypothesis). A well known identity involving harmonic numbers tells us that:\n\n$$\\sum_{i=1}^{n-1}H_i = nH_{n-1}-(n-1)$$\n\nSo $t_n=1+H_{n-1}-1+\\frac1n=H_n$. $\\Box$\n\nHow do we get from here to $s_n$? The difference now is that passenger $1$ does own seat $1$, which means that the answer will be smaller by $1$ if and only if passenger $1$ sits in seat $1$. Since passenger $1$ sits in seat $1$ with probability $\\frac1n$, we need to subtract $\\frac1n$ from $t_n$ to get $s_n$:\n\n$$s_n=t_n-\\frac1n=H_n-\\frac1n=H_{n-1}$$\n\nFinally, to get the expected number of people in the right seats, we subtract $s_n$ from $n$ to get:\n\n$$n-H_{n-1}$$\n\nNote 1: Since $H_n$ grows logarithmically, the proportion of people sitting in their correct seats converges to $1$ as $n\\to\\infty$.\n\nNote 2: I still find this proof rather unsatisfying, since it uses the identity $\\sum_{i=1}^{n-1}H_i = nH_{n-1}-(n-1)$, which I still don't really understand. I'm sure it's easy enough to prove by induction, but if someone could come up with a really nice explanation of why that works, it might yield an even slicker proof of this fact.\n\n-\nThere are a few things I do not follow in your explanation. Firstly, if passenger 1 sits in seat 1, the expected number of correct seats is n, because everyone sits in their correct seat if it is available, so excluding that possibility materially affects the answer. Secondly, it can be shown (see below where I answered the wrong question) that regardless of of the number of people, the probability that the last person sits in his or her proper seat is \\frac{1}{2}. As any one person could be the last one, the long-term proportion should approach 1/2. In my attempt at a correct answer it does. \u2013\u00a0 Avraham Aug 23 '13 at 19:19\nWhat do you mean by 'any one person could be the last one'? And why does that mean that the long term proportion should approach $1/2$? \u2013\u00a0 Donkey_2009 Aug 24 '13 at 10:54\nIf you read my answer carefully, you will see that I do not exclude the possibility that passenger $1$ sits in seat $1$; instead, I treat it separately. The long term proportion approaches $1$, so I think you might be answering a different question. \u2013\u00a0 Donkey_2009 Aug 24 '13 at 10:55\n\nYour overall approach looks technically correct, you are using linearity of expectation for counting and then writing down a formula for each $p_i$ (although it seems you switched notation and started using $X_i$ in your formulas instead of $p_i$). However I believe the equation for $X_4$ i.e. $p_4$ is actually\n\n$$X_4 = 1 - (1/n + 1/n(n-1) + 1/n(n-2) + 1/n(n-1)(n-2))$$\n\nand the generalization for $X_i$ for arbitrary $i$ is obtained by your basic reasoning, just writing down the probability for each way the $i$th person's seat might already be taken, taking into account each case e.g. for $X_4$ the term $1/n(n-2)$ corresponds to the case that the first person takes the third person's seat, and the third person takes the fourth person's seat. I'm not sure if/what an easy formula would be for when you add up all the $X_i$ terms to get the total answer. However it is easy to see that the general rule to get the equation for $X_i$ when $i > 1$ is to have $X_i = 1 - q_i$ where $q_i$ is the sum of all terms you can make of the form $1/n(n-k_1)(n-k_2)\\ldots$ where you have $0$ or more distinct values $k_j$, and $k_1 < k_2 \\ldots < i-1$.\n\n-\nOops, sorry about the $p_i \\rightarrow X_i$ change. And you are write about $X_4$. Thanks for your answer, that seems correct. \u2013\u00a0 narcissa Aug 10 '13 at 23:59\n\nI found this question and the answer might be relevant.\n\nSeating of $n$ people with tickets into $n+k$ chairs with 1st person taking a random seat\n\nThe answer states that the probability of a person not sitting in his seat is $\\frac{1}{k+2}$ where $k$ is the number of seats left after he takes a seat. This makes sense because for person $i$, if anyone sits in chairs $1, i+1, ... n$ then he must sit in his own seat, so the probability of that happening is $\\frac{n-i+1}{n-i+2}$. So $k = 0$ for the last person and $k = n-1$ for the second person. The answer then should just be\n\n$1/n + \\sum_{i = 2}^{n} \\frac{n-i+1}{n-i+2}$\n\n-\nThe answer (see my post) is $n-H_{n-1}$, but this is probably the right way to go. \u2013\u00a0 Donkey_2009 Aug 11 '13 at 0:56\n\nThe answer is $\\frac{1}{2}$ as was said.\n\nThe general pattern is that for $n$ people, there is a $\\frac{1}{n}$ probability of success, $\\frac{1}{n}$ probability of failure, and an $\\frac{n-2}{n}$ probability that the problem repeats itself on the $n-1$ scale.\n\nCase $n=2$: The probability that the first picks the correct seat is $\\frac{1}{2}$, and then the last person sits in proper seat with probability 1. The probability that the first picks the wrong seat is $\\frac{1}{2}$, and then the last person sits in proper seat with probability 0. So the probability in total is: $$\\frac{1}{2}\\cdot1+\\frac{1}{2}\\cdot0=\\frac{1}{2}$$\n\nCase $n=3$: The probability that the first picks the correct seat is $\\frac{1}{3}$, and then the last person sits in proper seat with probability 1. The probability that the first picks the last person's seat is $\\frac{1}{3}$, and then the last person sits in proper seat with probability 0. The probability that the first picks the middle person's seat is $\\frac{1}{3}$, and now there is a $\\frac{1}{2}$ that the second person picks the last seat or the first seat (Case 2) since two non-last people switching seats means everyone else takes their own seat. So: $$\\frac{1}{3}\\cdot1+\\frac{1}{3}\\cdot0+\\frac{1}{3}\\cdot\\frac{1}{2} = \\frac{2}{6} + \\frac{1}{6} = \\frac{3}{6} = \\frac{1}{2}$$\n\nProof by induction.\n\nAssume it holds true for $n$ that the probability of the last person sitting in the proper seat is $\\frac{1}{2}$. Now if there are $n+1$ people, we have a $\\frac{1}{n+1}$ chance that the last-seat probability is 1 (correct seat), a $\\frac{1}{n+1}$ chance that the last-seat probability is 0 (last seat), and an $\\frac{n-1}{n+1}$ chance that the probability is $\\frac{1}{2}$, since we know the $n$ case. $$\\frac{1}{n+1} + \\frac{0}{n+1} + \\frac{n-1}{2(n+1)}\\\\ =\\frac{2}{2(n+1)}+\\frac{n-1}{2(n+1)}\\\\ =\\frac{n+1}{2(n+1)}\\\\ =\\frac{1}{2}$$ QED\n\n-\nI feel bad for all your work, but the question was the expected number of people in correct seats. \u2013\u00a0 Lord_Farin Aug 23 '13 at 8:17\n\nNow that I have cleaned my glasses, I'll try again. Thank you, Lord Farin.\n\nSome observations.\n\nThe answer must be greater than 1. The probability that the first person, regardless of the number of people, sits in the proper seat is $\\frac{1}{n}$, and there are $n$ people, so that expectation is 1. Even if the first person sits in the wrong seat, there is non-zero probability of other people sitting in the correct seat, so the answer must be greater than or equal to 1, and I'm pretty sure equality only exists in the case $n=2$.\n\nThere is some form of recurrence going on here. Enumerating the possibilities, I get (if I haven't erred): $$E_2 = \\frac{1}{2}\\cdot 2 + \\frac{1}{2}\\cdot 0 = 1\\\\ E_3 = \\frac{1}{3}\\cdot 3 + \\frac{2}{3}\\left[\\frac{1}{2}\\cdot 1 + \\frac{1}{2}\\cdot 0\\right] = 1+\\frac{1}{3}=\\frac{4}{3}\\\\ E_4 = \\frac{1}{4}\\cdot 4 + \\frac{3}{4}\\left[\\frac{1}{3}\\cdot2 + \\frac{2}{3}\\left[\\frac{1}{2}\\cdot 1 + \\frac{1}{2}\\cdot 0\\right]\\right] = 1+\\frac{3}{4}\\cdot\\left(\\frac{2}{3}+\\frac{1}{3}\\right)=\\frac{7}{4}\\\\ E_5 = \\frac{1}{5}\\cdot 5 + \\frac{4}{5}\\left[\\frac{1}{4}\\cdot3 + \\frac{3}{4}\\left[\\frac{1}{3}\\cdot2 + \\frac{2}{3}\\left[\\frac{1}{2}\\cdot 1 + \\frac{1}{2}\\cdot 0\\right]\\right]\\right] = 1+\\frac{4}{5}\\left(\\frac{3}{4} + \\frac{3}{4}\\cdot\\left(\\frac{2}{3}+\\frac{1}{3}\\right)\\right)=\\frac{11}{5}\\\\$$ Let $E_k$ be the expected number of people in the correct seats when the starting population is k. The relationship seems to be: $$E_k = 1 + \\left(E_{k-1} - 1 + \\frac{k-2}{k-1}\\right)\\frac{k-1}{k}$$ The first 1 is the expected value of person 1 taking the correct seat. The next term is broken into two parts. If the first person did not take the correct seat, then the second person can \"fix\" the error by swapping and taking the previous person's seat, leaving the remaining $n-2$ people their proper seats. If the second person also takes the wrong seat, the problem restarts on the $n-1$ scale. In both the latter two cases, there is \"one less\" correct seat than the initial, since the previous term used up a seat with an incorrect choice. If the above supposition is correct, the expected value would just be the sum of the recurrence relation applied to $n$ or $\\sum_{k=1}^n E_k$.\n\nGenerating the first few terms using this relationship gives: $$1, 1, \\frac{4}{3}, \\frac{7}{4}, \\frac{11}{5}, \\frac{16}{6}, \\frac{22}{7}, \\frac{29}{8}, \\frac{37}{9}, \\frac{46}{10}, \\ldots$$\n\nThe numerator is a quadratic and the denominator is just $n$ so the expected value should be: $$E_n = \\frac{n^2-n+2}{2n}$$\n\nEdit: The long-term proportion of people sitting in the proper seats intuitively would be $$\\lim_{n \\to \\infty} \\frac{E_n}{n}\\\\ =\\lim_{n \\to \\infty} \\frac{n^2-n+2}{2n^2}\\\\ =\\frac{1}{2}$$ Which dovetails nicely with the long-term probability of the last person sitting in his or her proper seat.\n\n-\nYour recurrence is wrong, as are your values for $E_k$ where $k\\ge3$. I'm not really sure what your argument is; are you free to take this to chat some time today? \u2013\u00a0 Donkey_2009 Aug 24 '13 at 12:06\nSure. If I'm wrong, I'm eager to learn why and how to correct it. If you want, you can e-mail me directly as well; thank you. \u2013\u00a0 Avraham Aug 26 '13 at 1:18\nIt's 1. The probability that the $i$th person is in the right seat is $1/n$ for every $i$. This is clearly true for $i=1$, for $i=2$ you get $$P(\\text{1st person not in 2nd person's seat, 2nd person in right seat}) = \\frac{n-1}{n} \\frac{1}{n-1} = 1/n,$$ and so on down the line as you can check for yourself.\nThis answer is wrong. The probability that the second person is in the right seat is $1$ if the first person didn't sit in their seat. I think kilgol was assuming that people just sit in their seats randomly, whereas they in fact always sit in their own seat if it is free (and they're not the first person). \u2013\u00a0 Donkey_2009 Aug 11 '13 at 0:49", "date": "2014-09-19 00:07:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/464625/expected-number-of-people-sitting-in-the-right-seats", "openwebmath_score": 0.8596817255020142, "openwebmath_perplexity": 281.9892079216284, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988668248138067, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8607318662328379}}
{"url": "https://math.stackexchange.com/questions/4072871/why-is-the-answer-to-this-probability-question-not-frac12", "text": "# Why is the answer to this probability question not $\\frac{1}{2}$?\n\nI was trying to solve the following homework probability question which has the following setup:\n\nWe have $$2$$ dice: $$A$$ and $$B$$. Die $$A$$ has $$4$$ red faces and $$2$$ white faces, whereas die $$B$$ has $$2$$ red faces and $$4$$ white faces. On each turn, a fair coin is tossed. If the coin lands on heads then die $$A$$ is thrown, but if the coin lands on tails then die $$B$$ is thrown. After this the turn ends, and on the next turn the coin is once again tossed to determine the die (i.e. we throw the coin and the corresponding die on each turn).\n\nFrom this game I'm asked to answer $$2$$ questions:\n\n1. Show that the probability of obtaining a red face on any $$n$$-th throw is $$\\frac{1}{2}$$.\n2. If the first $$2$$ consecutive die throws result in red faces, what is the probability that the third throw is also red?\n\nTo answer part $$1$$ I used the law of total probability. Denoting obtaining a red face on the $$n$$-th die throw as $$P(R_n)$$ I get\n\n\\begin{align*} P(R_n) &= P(R_n \\vert A) P(A) + P(R_n \\vert B) P( B) \\\\ & = \\left(\\frac{4}{6}\\right)\\left(\\frac{1}{2}\\right) + \\left(\\frac{2}{6}\\right)\\left(\\frac{1}{2}\\right)\\\\ & = \\frac{1}{2} \\end{align*}\n\nBut on question $$2$$ is where I started running into trouble. Using the same notation, what I want to calculate is $$P(R_3 \\vert R_2 R_1)$$. And recalling that for events $$E_1$$ and $$E_2$$ we can say that $$P(E_2 \\vert E_1) = \\frac{P(E_2E_1)}{P(E_1)}$$ I get that $$P(R_3 \\vert R_2 R_1) = \\frac{P(R_3 R_2 R_1)}{P(R_2 R_1)} \\tag{1}$$ and from here I obtain 2 different solutions using $$2$$ distinct methods:\n\nUsing that on the first part of the question we showed that $$P(R_n) = \\frac{1}{2}$$, and noticing that the die throws are independent since what I threw before does not affect how I throw the next coin toss or how I roll the next die, from equation $$(1)$$ I get \\begin{align*} P(R_3 \\vert R_2 R_1) &= \\frac{P(R_3 R_2 R_1)}{P(R_2 R_1)}\\\\ &= \\frac{P(R_3) P(R_2) P(R_1)}{P(R_2) P(R_1)}\\\\ & = P(R_3) = P(R_n) = \\frac{1}{2} \\end{align*}\n\nUsing the law of total probability on $$(1)$$ I get \\begin{align*} P(R_3 \\vert R_2 R_1) & = \\frac{P(R_1 R_2 R_3 \\vert A)P(A)+ P(R_1 R_2 R_3 \\vert B)P(B)}{P(R_1 R_2 \\vert A ) P(A)+ P(R_1 R_2 \\vert B) P(B)}\\\\ &= \\frac{\\left[\\left(\\frac{2}{3}\\right)\\left(\\frac{2}{3}\\right)\\left(\\frac{2}{3}\\right)\\right]\\frac{1}{2}+ \\left[\\left(\\frac{1}{3}\\right)\\left(\\frac{1}{3}\\right)\\left(\\frac{1}{3}\\right)\\right]\\frac{1}{2}}{\\left[\\left(\\frac{2}{3}\\right)\\left(\\frac{2}{3}\\right)\\right]\\frac{1}{2}+\\left[\\left(\\frac{1}{3}\\right)\\left(\\frac{1}{3}\\right)\\right]\\frac{1}{2}}\\\\ & = \\frac{3}{5} \\end{align*}\n\nTo me both of the previous answers seem to be following coherent logic, but since I didn't get the same answer I knew one of them was wrong. I decided to write a program to simulate the game and I found out that the correct solution was $$P(R_3 \\vert R_2 R_1) = \\frac{3}{5}$$. But even though I verified this answer to be correct I couldn't seem to understand what part of my analysis is wrong on Answer 1.\n\nSo my question is, why is $$P(R_3 \\vert R_2 R_1) \\neq \\frac{1}{2}\\quad ?$$\n\n\u2022 I'm confused about the nature of this experiment. Are you repeatedly tossing a coin then rolling the corresponding die $n$ times (i.e. $n$ coin tosses and $n$ dice rolls), or are you tossing a coin once, then rolling the designated die $n$ times (i.e. $1$ coin toss, and $n$ die rolls)? \u2013\u00a0Theo Bendit Mar 23 at 5:56\n\u2022 It's the first option. We have one coin toss and one dice roll per turn. I'll edit the problem to clarify this, thank you! \u2013\u00a0Robert Lee Mar 23 at 5:59\n\u2022 In your computation, $A$ determines the successive $3$ roll of the die. \u2013\u00a0Oolong milk tea Mar 23 at 6:07\n\u2022 If it's one coin toss and one dice roll per turn, then every (coin toss + dice roll) is independent of all the others, and the probability is $1/2$ every time. If one coin toss determined multiple dice rolls, then those dice rolls would not be mutually independent, and you would get something greater than $1/2$ (because there would be a positive correlation between rolls using the same coin toss). But if you're saying it's one coin toss and one dice roll per turn, then your simulation is wrong. \u2013\u00a0mjqxxxx Mar 23 at 6:24\n\u2022 I don\u2019t quite get the logic in answer 2. To get consecutively 2 times red, you have AA, BB, AB, BA 4 paths with their own probabilities, would expect 1/2x1/2x2/3x2/3 for AA path,... and we should have 4 terms in the denominator. Do I miss something? \u2013\u00a0BStar Mar 23 at 6:38\n\nThe flaw in your second answer is that it is not necessarily the same die that is thrown after each coin toss. The way you have written the law of total probability is acceptable in your first answer, but not in the second, because it is only in the second answer that you conditioned on the events $$A$$ and $$B$$, which represent the outcomes of the coin toss. Consequently, the result is incorrect because it corresponds to a model in which the coin is tossed once, and then the corresponding die is rolled three times.\n\nFirst, let us do the calculation the proper way. We want $$\\Pr[R_3 \\mid R_1, R_2] = \\frac{\\Pr[R_1, R_2, R_3]}{\\Pr[R_1, R_2]}$$ as you wrote above. Now we must condition on all possible outcomes of the coin tosses, of which there are eight: \\begin{align} \\Pr[R_1, R_2, R_3] &= \\Pr[R_1, R_2, R_3 \\mid A_1, A_2, A_3]\\Pr[A_1, A_2, A_3] \\\\ &+ \\Pr[R_1, R_2, R_3 \\mid A_1, A_2, B_3]\\Pr[A_1, A_2, B_3] \\\\ &+ \\Pr[R_1, R_2, R_3 \\mid A_1, B_2, A_3]\\Pr[A_1, B_2, A_3] \\\\ &+ \\Pr[R_1, R_2, R_3 \\mid A_1, B_2, B_3]\\Pr[A_1, B_2, B_3] \\\\ &+\\Pr[R_1, R_2, R_3 \\mid B_1, A_2, A_3]\\Pr[B_1, A_2, A_3] \\\\ &+ \\Pr[R_1, R_2, R_3 \\mid B_1, A_2, B_3]\\Pr[B_1, A_2, B_3] \\\\ &+\\Pr[R_1, R_2, R_3 \\mid B_1, B_2, A_3]\\Pr[B_1, B_2, A_3] \\\\ &+ \\Pr[R_1, R_2, R_3 \\mid B_1, B_2, B_3]\\Pr[B_1, B_2, B_3] \\\\ \\end{align} and since each of the $$2^3 = 8$$ triplets of ordered coin tosses has equal probability of $$1/8$$ of occurring, $$\\Pr[R_1, R_2, R_3] = \\tfrac{1}{8}\\left((\\tfrac{2}{3})^3 + 3(\\tfrac{2}{3})^2(\\tfrac{1}{3}) + 3(\\tfrac{2}{3})(\\tfrac{1}{3})^2 + (\\tfrac{1}{3})^3\\right) = \\tfrac{1}{8}.$$ A similar (but simpler) calculation for the denominator yields $$1/4$$, and the result follows.\n\nOf course, none of this is necessary; it is only shown here to illustrate how the calculation would be done if it were to be done along the lines of your second answer.\n\n\u2022 I think I'm starting to understand where my lack of understanding is. I believe I may not understand what $P(R_3 \\vert R_2 R_1)$ is to begin with. Let's say I list all my throws as red or white accordingly. If I then count how many chains of 3 consecutive reds there are on my list, and I also count the number of times 2 reds are succeded by a white, is $$P(R_3 \\vert R_2 R_1) \\sim \\frac{\\text{#RRR}}{\\text{#RRW} + \\text{#RRR}}$$ in the sense that $\\sim$ means the RHS approaches $P(R_3 \\vert R_2 R_1)$ as I take more throws? Or does $P(R_3 \\vert R_2 R_1)$ mean something else? \u2013\u00a0Robert Lee Mar 23 at 7:41\n\u2022 I ran out of space in my previous comment, but #RRR stands for the number of chains of 3 consecutive reds on my list of throws, and #RRW is the number of chains where the first 2 places are red and the third consecutive throw is white, again on my list of throws. Just to clarify what I meant in the last comment. \u2013\u00a0Robert Lee Mar 23 at 7:46\n\u2022 @RobertLee It is unnecessary and mathematically inappropriate to think of the problem in terms of long-run averages of more than three rolls, because the entire question can be answered precisely under a strict interpretation of the model, in which there are only ever three rolls, each of which is preceded by a coin toss. \u2013\u00a0heropup Mar 23 at 7:57\n\u2022 I agree that it's not necessary to think of this mathematically. I only ask this because of my previously described attempt to verify the theoretical result with a simulation, which led me to an incorrect result. I'm trying to understand if what I programmed is in fact not trying to calculate the probability $P(R_3 \\vert R_2 R_1)$ that I want, or if I'm still not understanding what the probability in itself means. \u2013\u00a0Robert Lee Mar 23 at 8:07\n\u2022 @RobertLee The frequentist approach via simulation would be as follows. [1] Locate all occurrences of two consecutive red dice rolls. [2] For each such occurrence, look at the outcome of the die roll immediately after the pair of reds. [3] Tally the number of outcomes where the third die is red. Divide the number obtained in step 3 by the total number of red pairs observed in step 1. \u2013\u00a0heropup Mar 23 at 8:18\n\nWith the second approach:\n\nTo have first two die throws resulting in red, there are 2x2 paths: AA,AB,BA, BB $$\\to$$\n\n$$P(R1\\cap R2)=(\\frac{1}{2})^2(\\frac{2}{3})^2$$+2$$(\\frac{1}{2})^2\\frac{2}{3}\\frac{1}{3}$$+$$(\\frac{1}{2})^2(\\frac{1}{3})^2$$ =$$\\frac{1}{4}$$\n\nTo have three die throws resulting in red, there are 2x2x2 paths:AAA,BBB,AAB,ABA,BAA,BBA,BAB,ABB $$\\to$$\n\n$$P(R1\\cap R2\\cap R3)= (\\frac{1}{2})^3(\\frac{2}{3})^3$$+3$$(\\frac{1}{2})^3(\\frac{2}{3})^2\\frac{1}{3}$$+3$$(\\frac{1}{2})^3(\\frac{1}{3})^2\\frac{2}{3}$$+ $$(\\frac{1}{2})^3(\\frac{1}{3})^3$$ =$$\\frac{1}{8}$$\n\n$$\\frac{P(R1\\cap R2\\cap R3)}{P(R1\\cap R2)}$$ =$$\\frac{1}{2}$$\n\nIn this way, you get consistent result compared to first approacch.", "date": "2021-05-08 20:22:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4072871/why-is-the-answer-to-this-probability-question-not-frac12", "openwebmath_score": 0.9856621026992798, "openwebmath_perplexity": 480.8887045995353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357610169274, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8607161292243346}}
{"url": "https://math.stackexchange.com/questions/3358186/number-of-bit-strings-of-length-four-do-not-have-two-consecutive-1s", "text": "# Number of bit strings of length four do not have two consecutive 1s\n\nI came across following problem:\n\nHow many bit strings of length four do not have two consecutive 1s?\n\nI solved it as follows:\n\nTotal number of bit strings of length: $$2^4$$\nTotal number of length 4 bit strings with 4 consecutive 1s: 1\nTotal positions for three consecutive 1s in length 4 bit string: 2 (111X, X111)\nNumber of bit strings for each of above positions: 2 (X can be 0 or 1)\nTotal positions for two consecutive 1s in length 4 bit string: 3 (11XX, X11X, XX11)\nNumber of bit strings for each of above positions: 4\nBy inclusion exlcusion principle, the desired count $$=2^4-3\\times 4+2\\times 2-1=16-12+4-1=7$$\n\nHowever the correct solution turns out to be 8. It seems that I incorrectly applied inclusion exclusion principle. Where did I go wrong?\n\n\u2022 You should be doing inclusion-exclusion on the number of pairs of consecutive ones, not on the length of a string of consecutive ones. \u2013\u00a0Gerry Myerson Sep 16 '19 at 6:27\n\nIf I were doing this by inclusion-exclusion, I'd go: $$16$$ strings of length four; $$12$$ with at least one pair of consecutive ones ($$11xx,x11x,xx11$$ with $$x$$s arbitrary); five with at least two pair of consecutive ones ($$111x,1111,x111$$); one with three pair of consecutive ones; so $$16-12+5-1=8$$.\n\nTo count the number of bit strings with $$2$$ consecutive one bits (bad strings), I would let \\begin{align} S_1&=11xx&4\\\\ S_2&=x11x&4\\\\ S_3&=xx11&4\\\\ N_1&=&12 \\end{align} Then \\begin{align} S_1\\cap S_2&=111x&2\\\\ S_1\\cap S_3&=1111&1\\\\ S_2\\cap S_3&=x111&2\\\\ N_2&=&5 \\end{align} and \\begin{aligned} S_1\\cap S_2\\cap S_3&=1111&1\\\\ N_3&=&1 \\end{aligned} The count of bad strings is $$N_1-N_2+N_3=8$$.\nThe count of good strings is $$16-8=8$$.\n\nGenerating Functions\n\nLet $$x$$ represent the atom '$$0$$' and $$x^2$$ represent the atom '$$10$$' and build all possible strings by concatenating one or more atoms and removing the rightmost '$$0$$'. \\begin{align} \\overbrace{\\vphantom{\\frac1x}\\ \\ \\ \\left[x^4\\right]\\ \\ \\ }^{\\substack{\\text{strings of}\\\\\\text{length 4}}}\\overbrace{\\ \\quad\\frac1x\\ \\quad}^{\\substack{\\text{remove the}\\\\\\text{rightmost '0'}}}\\sum_{k=1}^\\infty\\overbrace{\\vphantom{\\frac1x}\\left(x+x^2\\right)^k}^\\text{k atoms} &=\\left[x^4\\right]\\frac{1+x}{1-x-x^2}\\\\ &=\\left[x^4\\right]\\left(1+2x+3x^2+5x^3+8x^4+13x^5+\\dots\\right)\\\\[9pt] &=8 \\end{align} Note that the denominator of $$1-x-x^2$$ induces the recurrence $$a_n=a_{n-1}+a_{n-2}$$ on the coefficients.\n\nRecurrence\n\nGood strings of length $$n$$ can be of two kinds: a good string of length $$n-1$$ followed by '$$0$$' or a good string of length $$n-2$$ followed by '$$01$$'. That is, $$a_n=a_{n-1}+a_{n-2}$$ Starting with\n$$a_0=$$ the number of good strings of length $$0=1$$.\n$$a_1=$$ the number of good strings of length $$1=2$$.\nwe get $$a_4=8$$.", "date": "2020-03-31 11:17:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3358186/number-of-bit-strings-of-length-four-do-not-have-two-consecutive-1s", "openwebmath_score": 0.929974377155304, "openwebmath_perplexity": 965.1952862338449, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357579585025, "lm_q2_score": 0.8791467738423873, "lm_q1q2_score": 0.8607161280855538}}
{"url": "https://www.physicsforums.com/threads/when-is-a-function-bounded-using-differentiation.209189/", "text": "# When is a function bounded using differentiation\n\n1. Jan 16, 2008\n\n### sara_87\n\n1. The problem statement, all variables and given/known data\nhow do i determine whether a function is bounded using differentiation\n\neg: f(x)=x/(2^x)\n2. Relevant equations\n\n3. The attempt at a solution\n\ni know it has something to do with maximums and minimums but i cant figure out how to do it.\n\nany help would be appreciated. thank you\n\n2. Jan 16, 2008\n\n### Dick\n\nYou could look at the limits of the function as it approaches plus and minus infinity. If both exist and are finite, and if the function is defined and continuous for all x, then it is bounded.\n\n3. Jan 16, 2008\n\n### EnumaElish\n\nYou can use differentiation to investigate the behavior of f. Say, the function is f(x) = x/2^x on x > 0. Then f'(x) = 2^-x (1 - x Log[2]), which has roots 1/Log[2] and +infinity. At x = 1/Log[2], f''(x) = 2^-x Log[2] (-2 + x Log[2]) is < 0, so you have the maximum. Note that f(x) > 0 for x > 0 and f(0) = 0. As x --> +infinity, f(x) --> 0 from above; but f(0) = 0 so x = 0 is the minimum. Since you can \"account for\" both the maximum and the minimum, f is bounded on x > 0.\n\nLast edited: Jan 16, 2008\n4. Jan 16, 2008\n\n### sara_87\n\nthank you very much.\nwhat if we have:\nf(x)=(-2x^2)/(4x^2-1)\ni know that it's not bounded but i dont know why\n\n5. Jan 16, 2008\n\n### Tom Mattson\n\nStaff Emeritus\nThe graph of that function has two vertical asymptotes. Functions don't get much more \"unbounded\" than that!\n\n6. Jan 16, 2008\n\n### sara_87\n\nactually i think it is bounded because there's no value of x that would make that function greater than 1\nor is there?\n\n7. Jan 16, 2008\n\n### Tom Mattson\n\nStaff Emeritus\nSure there is. As I said, the graph of that function has 2 vertical asymptotes. You can find values of x for which the function blows up to infinity, and down to negative infinity.\n\nDo you know what I mean when I say \"vertical asymptote\"?\n\n8. Jan 16, 2008\n\n### sara_87\n\nyes i do know what vertical assymptotes are.\numm but i still didnt understand what u meant. you can find values of x for which the function blows down to -ve infinity but not up.\n?\n\n9. Jan 16, 2008\n\n### Tom Mattson\n\nStaff Emeritus\nThe function certainly does blow up to positive infinity, as you approach -1/2 from the right and as you approach +1/2 from the left.\n\n10. Jan 16, 2008\n\n### sara_87\n\noh thank u very much\nthat helps.\njust one last question:\nsame question as before but with function:\nsqrt(x)/1000\n\nis it not bounded since n continues to increase to infinity?\n\n11. Jan 17, 2008\n\n### Tom Mattson\n\nStaff Emeritus\nYou're right that it's not bounded (on $[0,\\infty)$ that is--we really should be specifying an interval when making these statements).\n\nBut what's \"n\"?\n\n12. Jan 17, 2008\n\n### HallsofIvy\n\nStaff Emeritus\nNow, I'm confused as to what function you are talking about. The original function was f(x)= x/(2^x) which is definitely bounded on $[0, \\infty)$. It is bounded \"above\" but not bounded \"below\" so is not bounded. I don't see any asymptotes when I graph it.\n\n13. Jan 17, 2008\n\n### Tom Mattson\n\nStaff Emeritus\nPosts 1 through 3 pertain to f(x)=x/(2^x).\nPosts 4 through 9 pertain to f(x)=(-2x^2)/(4x^2-1).\nPosts 10 and 11 pertain to f(x)=sqrt(x)/1000.\n\n14. Jan 17, 2008\n\n### EnumaElish\n\nPost 3 pertains to f(x)=x/(2^x) for x > 0.", "date": "2016-12-05 12:44:34", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/when-is-a-function-bounded-using-differentiation.209189/", "openwebmath_score": 0.6157326102256775, "openwebmath_perplexity": 1352.996698281987, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9808759671623987, "lm_q2_score": 0.8774767954920547, "lm_q1q2_score": 0.8606959004408314}}
{"url": "http://mathhelpforum.com/trigonometry/213130-integrate-cos-x-cos-4x-dx.html", "text": "# Math Help - Integrate cos(x)cos(4x) dx\n\n1. ## Integrate cos(x)cos(4x) dx\n\nI'm trying to integrate:\n\n$\\int\\cos(x)\\cos(4x) dx$\n\nCould some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $4x$ into $x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.\n\nThank you.\n\n2. ## Re: Integrate cos(x)cos(4x) dx\n\nthis is easy...but you have to use a formula that transforms the product cosxcos4x into a sum.\n\nHERE cos(4x)cos(x)=(1/2)[cos(5x)-cos(3x)}\n\nMinoas\n\n3. ## Re: Integrate cos(x)cos(4x) dx\n\nOriginally Posted by Furyan\nI'm trying to integrate:\n\n$\\int\\cos(x)\\cos(4x) dx$\n\nCould some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $4x$ into $x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.\n\nThank you.\n\\displaystyle \\begin{align*} I &= \\int{\\cos{(x)}\\cos{(4x)}\\,dx} \\\\ I &= \\frac{1}{4}\\cos{(x)}\\sin{(4x)} - \\int{ -\\frac{1}{4}\\sin{(x)}\\sin{(4x)}\\,dx } \\\\ I &= \\frac{1}{4}\\cos{(x)}\\sin{(4x)} + \\frac{1}{4}\\int{\\sin{(x)}\\sin{(4x)}\\,dx} \\\\ I &= \\frac{1}{4}\\cos{(x)}\\sin{(4x)} + \\frac{1}{4} \\left[ -\\frac{1}{4}\\sin{(x)}\\cos{(4x)} - \\int{ -\\frac{1}{4}\\cos{(x)}\\cos{(4x)} \\,dx} \\right] \\\\ I &= \\frac{1}{4}\\cos{(x)}\\sin{(4x)} - \\frac{1}{16}\\sin{(x)}\\cos{(4x)} + \\frac{1}{16}I \\\\ \\frac{15}{16}I &= \\frac{1}{4}\\cos{(x)}\\sin{(4x)} - \\frac{1}{16}\\sin{(x)}\\cos{(4x)} \\\\ I &= \\frac{4}{15}\\cos{(x)}\\sin{(4x)} - \\frac{1}{15}\\sin{(x)}\\cos{(4x)} + C \\end{align*}\n\n4. ## Re: Integrate cos(x)cos(4x) dx\n\nThank you Prove It,\n\nI haven't seen that before, integrating by parts twice and then letting the second integrand be I and solving that way. That's a very useful way of solving difficult integration problems. Thank you very much for showing me that. I tried this method with a similar problem that I had solved another way and found that I had to be very careful with signs and factors. I kept getting the wrong answer because I was getting a sign and factor wrong during the second integration. So I'm going to have to work on it, but it's great to know that if I can't see what else to do I can try this method.\n\nThanks again for all your efforts\n\n5. ## Re: Integrate cos(x)cos(4x) dx\n\nOriginally Posted by Furyan\nI'm trying to integrate:\n\n$\\int\\cos(x)\\cos(4x) dx$\n\nCould some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $4x$ into $x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.\n\nThank you.\nHi Furyan!\n\nAs you can see here, we have the identity:\n$\\cos \\theta \\cos \\varphi = {{\\cos(\\theta - \\varphi) + \\cos(\\theta + \\varphi)} \\over 2}$\n\nIn other words:\n$\\int\\cos(4x)\\cos(x) dx = \\int{{\\cos(4x - x) + \\cos(4x + x)} \\over 2} = \\int{{\\cos(3x) + \\cos(5x)} \\over 2}$\n\nThis is what Minoas already suggested (except for the typo ).\n\n6. ## Re: Integrate cos(x)cos(4x) dx\n\nHi ILikeSerena\n\nThank you very much for that. I did, in fact, also solve this problem using the factor formula, as Minoanman suggested, and it was very much simpler. However, although there are only four factor formulae in my book I'm worried that I might not remember them in an exam. Although I found Prove It's method more difficult I feel I'm more likely to remember it and at least get some method marks, even if I make some mistakes.\n\nI'm amazed how different the solutions look, depending on which method you use and yet they're equivalent. I solved one question using Prove It's method, the factor formula and a simple identity. All the solutions were equivalent, but they looked very different. I wouldn't have been able to rewrite any of them in terms of any of the others.\n\n7. ## Re: Integrate cos(x)cos(4x) dx\n\nHere's a trick that I like to use when I can't properly remember the formulas.\n\nThe only formula you need to remember is Euler's formula:\n$e^{ix} = \\cos x + i \\sin x$\n\ntogether with its rewritten forms (that you can both deduce from Euler's formula if you forget):\n$\\cos x = \\dfrac 1 2 (e^{ix} + e^{-ix})$\n$\\sin x = \\dfrac 1 {2i} (e^{ix} - e^{-ix})$\n\n$\\cos(4x)\\cos(x) = \\frac 1 2 (e^{i4x} + e^{-i4x}) \\cdot \\frac 1 2 (e^{ix} + e^{-ix})$\n$= \\frac 1 4 ((e^{i5x} + e^{-i5x}) + (e^{i3x} + e^{-i3x}))$\n\n$= \\frac 1 2 (\\cos(5x) + \\cos(3x)) \\qquad \\blacksquare$\n\n8. ## Re: Integrate cos(x)cos(4x) dx\n\nwhoa! $\\blacksquare$\n\nThat's a whole new level, but I'm liking it. I'm much better at deducing formualae than I am at remember them. Euler's formula, eh! I'm sure I came across that earlier today, but I glazed over when I saw $e^{ix}$. Is that i ,the i, $\\sqrt{-1}?$. I'm going to look that up and try and find out why it's so fundamental. It looks very interesting.\n\nThank you very much for that\n\n9. ## Re: Integrate cos(x)cos(4x) dx\n\nOriginally Posted by Furyan\nwhoa!\n\nThat's a whole new level, but I'm liking it. I'm much better at deducing formualae than I am at remember them. Euler's formula, eh! I'm sure I came across that earlier today, but I glazed over when I saw $e^{ix}$. Is that i ,the i, $\\sqrt{-1}?$. I'm going to look that up and try and find out why it's so fundamental. It looks very interesting.\n\nThank you very much for that\nYep. It is that $i$.\n\nAccording to wiki:\nEuler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation \"our jewel\" and \"one of the most remarkable, almost astounding, formulas in all of mathematics.\"[2]\n\n10. ## Re: Integrate cos(x)cos(4x) dx\n\nOriginally Posted by ILikeSerena\nYep. It is that $i$.\n\nAccording to wiki:\nEuler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation \"our jewel\" and \"one of the most remarkable, almost astounding, formulas in all of mathematics.\"[2]\nWow, it looks like a jewel. How exciting.\n\nThank you very much", "date": "2015-03-27 20:58:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/213130-integrate-cos-x-cos-4x-dx.html", "openwebmath_score": 0.9971106648445129, "openwebmath_perplexity": 820.3832663129307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9808759615719875, "lm_q2_score": 0.8774767986961401, "lm_q1q2_score": 0.8606958986781859}}
{"url": "https://math.stackexchange.com/questions/2402659/given-a-decreasing-function-s-t-int-0-infty-fx-dx-infty-prove-sum-n", "text": "# Given a decreasing function s.t. $\\int_0^\\infty f(x)\\,dx<\\infty,$ prove $\\sum_{n=1}^\\infty f(na)$ converges\n\nLet $f\\in C([0,\\infty))$ be a decreasing function such that $\\int_0^\\infty f(x)\\,dx$ converges.\n\nProve $\\sum_{n=1}^\\infty f(na)$ converges, $\\forall a>0$\n\nMy attempt:\n\nBy the Cauchy criterion, there exists $M>0,$ such that for $t-1>M:$\n\n$$f(t)=\\int_{t-1}^t f(t) \\, dx \\leq \\int_{t-1}^t f(x)\\,dx\\xrightarrow{t \\to \\infty} 0$$\n\nHence, $f$ is non-negative.\n\n$f$ is decreasing $\\implies f(na)\\leq f(a), \\forall a>0, n\\in \\mathbb{N}.$\n\nBy integral monotonicity and non-negativity of $f$:\n\n$$\\int_1^\\infty f(nx)\\,dx \\leq \\int_1^\\infty f(x)\\,dx \\leq \\int_0^\\infty f(x)\\,dx$$\n\nHence $\\int_1^\\infty f(nx)\\,dx$ converges and therefore $\\sum_{n=1}^\\infty f(na)$ converges.\n\nIs that correct? If so, why is continuity necessary ? Is there a simpler way to prove it?\n\nAny help appreciated.\n\n\u2022 $$\\int_{(n-1)a}^{na}f(x)dx\\geq af(na)$$ \u2013\u00a0MAN-MADE Aug 22 '17 at 18:33\n\nHint: The following thing you wrote is the key:\n\n$$f(t)=\\int_{t-1}^{t}f(t)dx \\leq \\int_{t-1}^{t}f(x)dx$$\n\nAfter that, just pick $t=an$ and sum over all $n$.\n\nPS: and no, the assumption on continuity is not needed. Just integrability for the well-definedness.\n\nAs I commented before:\n\nSince $f$ is decreasing and continuous, $$\\int_{(n-1)a}^{na}f(x)dx\\geq [na-(n-1)a]\\inf_{(n-1)a< x\\leq na}\\{f(x)\\}=af(na)$$\n\nThen $$\\int_{0}^{\\infty}f(x)dx=\\sum_{n=1}^{\\infty}\\int_{(n-1)a}^{na}f(x)dx\\geq a\\sum_{n=1}^{\\infty}f(na)$$\n\nHence, since $a\\in\\mathbb{R}^+,$ $\\sum_{n=1}^{\\infty}f(na)<\\infty$\n\n\\begin{align} & af(a) + af(2a) + af(3a) +\\cdots \\\\[10pt] \\le {} & \\int_0^a f + \\int_a^{2a} f + \\int_{2a}^{3a} f + \\cdots <\\infty. \\end{align}\n\nThis works if $f \\ge0$ everywhere. If $f<0$ somewhere, then an easy argument shows the integral does not converge.\n\nThis is the integral test for convergence of series. Do a change of variable x --- to --- ax. The integral converges so does series g(n)=f(na).", "date": "2019-08-21 14:11:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2402659/given-a-decreasing-function-s-t-int-0-infty-fx-dx-infty-prove-sum-n", "openwebmath_score": 0.9945245981216431, "openwebmath_perplexity": 531.3235866032861, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759666033576, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8606958952360703}}
{"url": "https://math.stackexchange.com/questions/98409/integral-of-periodic-function-over-the-length-of-the-period-is-the-same-everywhe?noredirect=1", "text": "# Integral of periodic function over the length of the period is the same everywhere\n\nI am stuck on a question that involves the intergral of a periodic function. The question is phrased as follows:\n\nDefinition. A function is periodic with period $a$ if $f(x)=f(x+a)$ for all $x$.\n\nQuestion. If $f$ is continuous and periodic with period $a$, then show that $$\\int_{0}^{a}f(t)dt=\\int_{b}^{b+a}f(t)dt$$ for all $b\\in \\mathbb{R}$.\n\nI understand the equality, but I am having trouble showing that it is true for all $b$. I've tried writing it in different forms such as $F(a)=F(b+a)-F(b)$. This led me to the following, though I am not sure how this shows the equality is true for all $b$,\n\n$$\\int_{0}^{a}f(t)dt-\\int_{b}^{b+a}f(t)dt=0$$ $$=F(a)-F(0)-F(b+a)-F(b)$$ $$=(F(b+a)-F(a))-F(b)$$ $$=\\int_{a}^{b+a}f(t)dt-\\int_{0}^{b+a}f(t)dt=0$$\n\nSo, this leaves me with\n\n$$\\int_{a}^{b+a}f(t)dt-\\int_{0}^{b+a}f(t)dt=\\int_{0}^{a}f(t)dt-\\int_{b}^{b+a}f(t)dt$$\n\nI feel I am close, and I've made myself a diagram of a sine function to visualize what each of the above integrals might describe, but the power to explain the above equality evades me.\n\n\u2022 See here for some proofs. \u2013\u00a0t.b. Jan 12 '12 at 8:09\n\u2022 (Voted to close as duplicate) Even though this says continuous and the other says integrable, the proofs are the same, i.e. every proof here would apply over there. \u2013\u00a06005 Jul 17 '16 at 16:25\n\nLet $H(x)=\\int_x^{x+a}f(t)\\,dt$. Then $$\\frac{dH}{dx}=f(x+a)-f(x)=0.$$ It follows that $H(x)$ is constant. In particular, $H(b)=H(0)$.\n\nWe have $$\\int_{0}^{a}f(t)\\ dt+\\int_{a}^{a+b}f(x)\\ dx=\\int_{0}^{b}f(y)\\ dy+\\int_{b}^{a+b}f(t)\\ dt,$$ and setting $x=y-a$ turns the second integral into the third one.\n\nNo differentiation is needed:\n\nPick the unique integer $n$ such that $b\\leqslant na\\lt b+a$, decompose the integral of $f(t)$ over $t$ from $b$ to $b+a$ into the sum of the integrals from $b$ to $na$ and from $na$ to $b+a$, apply the changes of variable $t=x+(n-1)a$ in the former and $t=x+na$ in the latter, then the periodicity of $f$ implies that $f(x)=f(t)$, hence the result is the sum of the integrals of $f(x)$ over $x$ from $b-(n-1)a$ to $a$ and from $0$ to $b-(n-1)a$...\n\n...Et voil\u00e0 !\n\nYou have made various false steps in your four line block and should have ended up with $$\\int_{a}^{b+a}f(t)dt-\\int_{0}^{b}f(t)dt=0$$ but this does not take you much further forward.\n\nInstead note that somewhere in the interval $[b, b+a]$ is an integer multiple of $a$, say $na$. Then using $f(t)=f(t+a)=f(t+na)$: $$\\int_{b}^{b+a}f(t)dt = \\int_{b}^{na}f(t)dt+\\int_{na}^{b+a}f(t)dt = \\int_{b+a}^{(n+1)a}f(t)dt+\\int_{an}^{b+a}f(t)dt = \\int_{na}^{(n+1)a}f(t)dt = \\int_{0}^{a}f(t)dt.$$\n\n\\begin{align} \\int_{b}^{a+b}f(x)\\ dx&= \\int_{a}^{a+b}f(x)\\ dx +\\int_{b}^{a}f(x)\\ dx\\\\&\\overset{y=x-a}{=} \\color{red}{\\int_{0}^{a}f(y+a)\\ dx} +\\int_{b}^{a}f(x)\\ dx\\\\&\\overset{periodic}{=} \\color{red}{\\int_{0}^{b}f(y)\\ dx} +\\int_{b}^{a}f(x)\\ dx\\\\&=\\int_0^af(x)\\ dx. \\end{align}\n\n\u2022 why a down vote here \u2013\u00a0Guy Fsone Jan 18 '18 at 8:15\n\u2022 The integral is obtained in the second line itself. You have done a calculation mistake. \u2013\u00a0Robin Feb 17 '18 at 8:56\n\u2022 Furthermore, this plagiarizes simultaneously three other answers, posted six years before. \u2013\u00a0Did Feb 18 '18 at 15:23", "date": "2019-07-18 17:28:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/98409/integral-of-periodic-function-over-the-length-of-the-period-is-the-same-everywhe?noredirect=1", "openwebmath_score": 0.9384185075759888, "openwebmath_perplexity": 219.76854088854688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759615719875, "lm_q2_score": 0.877476784277755, "lm_q1q2_score": 0.8606958845355384}}
{"url": "https://homework.cpm.org/category/CC/textbook/ccg/chapter/9/lesson/9.1.3/problem/9-35", "text": "### Home > CCG > Chapter 9 > Lesson 9.1.3 > Problem9-35\n\n9-35.\n\nAre $\u0394EHF$ and $\u0394FGE$ congruent? If so, explain how you know. If not, explain why not.\n\nWhat do the markings on the figures mean?\nWhat other parts do these triangles have in common?\n\nYes, they are congruent. What is the reason?", "date": "2020-07-07 19:47:40", "meta": {"domain": "cpm.org", "url": "https://homework.cpm.org/category/CC/textbook/ccg/chapter/9/lesson/9.1.3/problem/9-35", "openwebmath_score": 0.3052932322025299, "openwebmath_perplexity": 1769.8354963261709, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9808759587767815, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8606958757971885}}
{"url": "https://mathematica.stackexchange.com/questions/138771/create-a-particular-x-y-list/138772", "text": "# Create a particular {x,y} list [duplicate]\n\nHow is it possible to create a list of {x,y} points as the following one:\n\nlist={{1,1},{1,2},...,{1,100},{2,1},{2,2},...,{2,100},...,{100,1},{100,2},...,{100,100}}\n\nThanks\n\n\u2022 Table[{i,j}, {i, 100}, {j, 100}]? What have you tried? Feb 27 '17 at 16:23\n\u2022 Hi, I was using Range, but it doesn't work for 2D lists. However, you answer gives my some additional brackets that are unwanted, like this: {{{1,1},{1,2},...,{1,100},{2,1},{2,2},...,{2,100},...,{100,1},{100,2},...,{100,100}}}. I've found the way with Flatten[]. Thanks for your help! Feb 27 '17 at 16:45\n\u2022 Not an exact duplicate but a more general topic. Let me know if you disagree with closing.\n\u2013\u00a0Kuba\nFeb 27 '17 at 16:46\n\u2022 @Michele You need to Flatten the result; consider Flatten[Table[{i, j}, {i, 1, 100}, {j, 100}], 1]. More in general, also take a look at the proposed duplicate. Feb 27 '17 at 16:49\n\u2022 Join @@ Array[List , {100, 100} ] Feb 27 '17 at 20:24\n\nTuples[Range[100], 2]\n\nor using Table as mentioned in the comment\nIt is worth noting that the method relying on Tuples is 10-12 times faster than Table", "date": "2022-01-24 20:03:27", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/138771/create-a-particular-x-y-list/138772", "openwebmath_score": 0.2532903552055359, "openwebmath_perplexity": 1966.799039697362, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812327313545, "lm_q2_score": 0.8887587831798665, "lm_q1q2_score": 0.8606573260565377}}
{"url": "http://perfectroc.com/post/dijkstras-algorithm/", "text": "# Notes on Dijkstra's Algorithm\n\n## Setup\n\nWe are given a directed graph $G=(V,E)$ with a designated start node $s$. Each edge $(u, v) \\in E$ has a weight $w(u,v) \\geq 0$ indicating its cost. For each vertex $v \\in V$, it has a predecessor $v.\\pi$ that is another vertex or NIL.\n\n## Goal\n\nFind the shortest path from $s$ to every other vertex in the graph.\n\n## Pseudocode\n\n$Q$ is a min-priority queue of vertices, keyed by their $d$ values.\n\n#### 1.DIJKSTRA$(G, s)$\n\n$S = \\varnothing$\n$Q = G.V$\nwhile $Q \\neq \\varnothing$\n$u =$ EXTRACT-MIN(Q)\n$S = S \\cup \\{u\\}$\nfor each vertex $v \\in G.Adj[u]$\nRELAX(u,v,w)\n\n#### 2.INITIALIZE-SINGLE-SOURCE$(G,s)$\n\nfor each vertex $v \\in V[G]$\n$v.d = \\infty$\n$v.\\pi$ = NIL\n$d[s] = 0$\n\n#### 3. RELAX$(u,v,w)$\n\nif $v.d > u.d + w(u,v)$\n$v.d = u.d + w(u,v)$\n$v.\\pi = u$\n\n## Proof of correctness\n\nWe want to show that the algorithm terminates with $u.d$ = $\\delta(s,u)$ for all $u \\in V$, where $\\delta(s,u)$ is the shortest path weight from $s$ to $u$, and is defined as\n$\\delta(s, u) = \\begin{cases} min\\{w(p):s\\xrightarrow{\\text{p}}u\\} , & \\quad \\text{if there is a path from s to u}\\\\ \\infty , & \\quad \\text{otherwise} \\end{cases}$\n\n$w(p)$ is the sum of weights of the path $p$\n\n## We first show\n\nAt the start of each iteration of the while loop in Dijkstra\u2019s algorithm, $v.d = \\delta(s, v)$ for each vertex $v \\in S$ (*)\n\n1) Base Case:\n|S| = 0, i.e when $S = \\varnothing$, the statement is vacuously true.\n\n2) Inductive Step:\nSuppose the claim holds when $|S| = k$ for some $k \\geq 1$. Now, let $S$\u2019s size grow to $k+1$ by adding vertex $v$. We want to show $v.d = \\delta(s,v)$.\n\nIf $v=s$ then $s.d = \\delta(s,s) =0$. If $v \\neq s$, it must be $S \\neq \\varnothing$. If there\u2019s no path form $s$ to $v$, $v.d = \\delta(s,v)= \\infty$. If there are some path from $s$ to $v$, there sholud be one shortest path $p$ from $s$ to $v$.\n\nLet us consider the first vertex $y$ on $p$ that is not in $S$, and let $x$ on $p$ be $y\u2019s$ predecessor. By induction hypothesis, $x.d = \\delta(s,x)$. Because edge $(x, y)$ is relaxed in the previous iteration, by convergence property (if $s\\rightarrow u\\rightarrow v$ is a shortest path in G for some $u, v \\in V$, and if $u.d = \\delta(s,u)$ at any time prior to relaxing edge $(u,v)$, then $v.d = \\delta(s,v)$ at all times afterwards), we have $y.d = \\delta(s,y)$. Since $y$ is before $v$ on $p$ and all edge weights are nonnegative, we have $\\delta(s,y) \\leq \\delta(s,v)$, and thus $y.d = \\delta(s,y) \\leq \\delta(s,v)\\leq v.d$\nBecause both vertices $v$ and $y$ are in $V-S$ when $v$ is selected by the algorithm, $v.d \\leq y.d$. Thus, $y.d = \\delta(s,y) = \\delta(s,v) = v.d$, and so $v.d = \\delta(s,v)$, and so complete the proof of (*).\n\n## Now\n\nWhen the algorithm terminates, it means $Q = \\varnothing$. Since the algorithm maintains the invariant that $Q = V - S$ (when initialization, $S = \\varnothing$ and $Q$ contains all vertices of graph; for every time through while loop, a vertex is extracted from $Q$ and added to $S$, so the invariant is maintained), it means $S = V$, so $u.d = \\delta(s,u)$ for all vertices $u \\in V$\n\n## Time Complexity\n\nUsing priority queue implemented in min heap, the time to initialize the queue with setting all vertices\u2019 distances to $\\infty$ is $O(V)$. Inside the while loop, we will do $V$ times EXTRACT_MIN operation, each of them will take $O(logV)$. Besides, we will do at most $E$ times of updating associated distance key for vertices in the queue, which will cost $O(ElogV)$. In total, the time will be $O((E+V)logV)$\n\n## Java implementation\n\nIn the code below, vertices is the hashmap containing the mapping between GeographicPoint and MapNode. MapNode is made up of GeographicPoint and list of MapEdge. MapNodes in vertices are converted to DiNode in hashmap diPoints by using the method convertMap to facilitate implementation of the algorithm. The constructPath method will return the path found by the algorithm from start to goal.\n\nSee the full framework of this project here (incomplete). Note that the code may be ugly as I mainly pay attention to the algorithm itself.\n\npublic List<GeographicPoint> dijkstra(GeographicPoint start,\nGeographicPoint goal, Consumer<GeographicPoint> nodeSearched){\n\nif(!vertices.containsKey(start) || !vertices.containsKey(goal)){\nSystem.out.println(\"input location does not exist in graph\");\n}\n\nHashMap<GeographicPoint, DiNode> diPoints = convertMap(vertices);\nHashMap<GeographicPoint, GeographicPoint> parentMap = new HashMap<>();\nSet<GeographicPoint> visitSet = new HashSet<>();\nPriorityQueue<DiNode> queue = new PriorityQueue<DiNode>();\nDiNode cur = diPoints.get(start);\ncur.sumDist = 0.0;\nqueue.offer(cur);\n\nwhile(!queue.isEmpty()){\ncur = queue.poll();\nnodeSearched.accept(cur.mapNode.location);\nif(!visitSet.contains(cur.mapNode.location)){\n}\n\nif(cur.mapNode.location.equals(goal)) {\nreturn constructPath(start, goal, parentMap);\n}\n\nfor (MapEdge e : cur.mapNode.edges) {\nif(!visitSet.contains(e.end)){\ndouble nbSumDist = diPoints.get(e.end).sumDist;\nif(cur.sumDist + e.distance < nbSumDist){\n// update cur as neightbor's parent in parent map\nparentMap.put(e.end, cur.mapNode.location);\nDiNode next = diPoints.get(e.end);\n// update next neighbor's sum of distance\nnext.sumDist = cur.sumDist + e.distance;\n// put neighbor with updated distance into priority queue\nqueue.offer(new DiNode(next.mapNode, next.sumDist));\n}\n}\n}\n\n}\n\nSystem.out.println(\"There is no such path\");\n}\n\n\nAbove is one part of the back-end code of a project in UCSD\u2019s Advanced Data Structure course. It utilizes the Google Map\u2019s API, and can demonstrate route planning in a real world\u2019s map as shown below. Note that the algorithm is a bit different from the one described above (instead of initializing the priority queue with inserting all vertices and later updating their sum of weighted distance, it enqueues the vertex after its distance is updated), but the general idea is the same.\n\n## References\n\n[1] T. Cormen, C. Stein, R. Rivest, and C. Leiserson. Introduction to Algorithms (3rd ed.). MIT press,2009.\n[2] J. Kleinberg and E. Tardos. Algorithm Design. Addison-Wesley, 2005", "date": "2020-10-01 21:07:13", "meta": {"domain": "perfectroc.com", "url": "http://perfectroc.com/post/dijkstras-algorithm/", "openwebmath_score": 0.6448825597763062, "openwebmath_perplexity": 987.3364295752979, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717460476701, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8606332824188186}}
{"url": "http://math.stackexchange.com/questions/81362/which-is-the-fastest-way-to-compute-sum-limits-i-110-frac10i-52i", "text": "# Which is the \u201cfastest\u201d way to compute $\\sum \\limits_{i=1}^{10} \\frac{10i-5}{2^{i+2}}$?\n\nI am looking for the \"fastest\" paper-pencil approach to compute $$\\sum \\limits_{i=1}^{10} \\frac{10i-5}{2^{i+2}}$$\n\nThis is a quantitative aptitude problem and the correct/required answer is $3.75$\n\nIn addition, I am also interested to know how to derive a closed form for an arbitrary $n$ using mathematica I got $$\\sum \\limits_{i=1}^{n} \\frac{10i-5}{2^{i+2}} = \\frac{5 \\times \\left(3 \\times 2^n-2 n-3\\right)}{2^{n+2}}$$\n\nThanks,\n\n-\nDoes someone feel like writing an \"abstract duplicate\" for this kind of question? I think I've seen several of them, but I don't know whether we have one that deals with this more generally and would serve as a good duplicate for all of them. \u2013\u00a0 joriki Nov 12 '11 at 15:55\nThe actual correct answer to this question is $\\approx 3.7219$. I do not know what \"quantitative aptitude test\" means, but would not 3.7 be a better answer than 3.75? \u2013\u00a0 Aleksey Pichugin Nov 12 '11 at 16:36\n\nWe'll split the sum you're looking for up into\n\n$$10 \\sum_{i=1}^{10} {i \\over 2^{i+2}} - 5 \\sum_{i=1}^{10} {1 \\over 2^{i+2}}.$$\n\nCall this $10S_1 - 5S_2$.\n\nWe can write $$S_1 = \\sum_{i=1}^{10} {i \\over 2^{i+2}} = {1 \\over 4} \\sum_{i=1}^{10} {i \\over 2^i}$$ and the result $\\sum_{i=1}^\\infty i/2^i = 2$ is well-known; thus $S_1 \\approx 1/2$.\n\nSimilarly, $S_2 \\approx \\sum_{i=1}^\\infty 1/2^{i+2} = 1/4$ by the usual sum of a geometric series.\n\nSo your sum is approximately $15/4$. In fact the infinite sum\n\n$$\\sum_{i=1}^\\infty {10i-5 \\over 2^{i+2}}$$\n\nis exactly 15/4. We've left off some small positive terms so your sum is a bit less than $15/4$.\n\nAn exact form for the sum\n\nAs for getting an exact form for the sum: call it $f(n)$. Then we have\n\n$$f(n) = {15 \\over 4} - \\left( 10 \\sum_{i=n+1}^\\infty {i \\over 2^{i+2}} - 5 \\sum_{i=n+1}^\\infty {1 \\over 2^{i+2}} \\right).$$\n\nWrite this as $f(n) = 15/4 - g(n) + h(n)$.\n\n$h(n)$ is easy -- it's the sum of a geometric series with its first term $5/2^{n+3}$ and ratio $1/2$, so $h(n) = 5/2^{n+2}$.\n\n$g(n)$ is a bit harder. So that we don't have so many constants floating around, consider\n\n$$G(n) = \\sum_{i={n+1}}^\\infty {i \\over 2^i}$$\n\nand you can see $g(n) = (5/2) G(n)$. Now you can write\n\n$$G(n) = {(n+1) \\over 2^{n+1}} + {{n+2} \\over 2^{n+2}} + {{n+3} \\over 2^{n+3}} + \\cdots$$\n\nand this is just\n\n$$G(n) = \\left( {n \\over 2^{n+1}} + {n \\over 2^{n+2}} + {n \\over 2^{n+3}} + \\cdots \\right) + {1 \\over 2^n} \\left( {1 \\over 2^1} + {2 \\over 2^2} + {3 \\over 2^3} + \\cdots \\right).$$\n\nThe first sum is a geometric series, summing to $n/2^n$; the second sum is $2$. Thus $G(n) = (n+1)/2^n$. Therefore you get\n\n$$f(n) = {15 \\over 4} - {5 \\over 2} {(n+2) \\over 2^n} + {5 \\over 2^{n+2}}.$$\n\nA sum that everybody should know, but lots of people don't\n\nFinally, I used the result $\\sum_{i=1}^\\infty i/2^i$ twice here, both in getting the approximation and in getting the exact form. How can we prove that? One way is to write $${1 \\over 2} + {2 \\over 4} + {3 \\over 8} + {4 \\over 16} + \\cdots$$ as a sum of one $1/2$, two $1/4$s, three $1/8$s, and so on; then regroup those terms as $$\\left( {1 \\over 2} + {1 \\over 4} + {1 \\over 8} + \\cdots \\right) + \\left( {1 \\over 4} + {1 \\over 8} + {1 \\over 16} + \\cdots \\right) + \\left( {1 \\over 8} + {1 \\over 16} + {1 \\over 32} + \\cdots \\right) + \\cdots$$ Each pair of parentheses contains a geometric series; summing those gives $1 + 1/2 + 1/4 + 1/8 + \\cdots$, another geometric series, which has sum $2$.\n\nAlternatively, note that $${1 \\over 1-z} = 1 + z + z^2 + z^3 + \\cdots$$ and differentiating both sides gives $${1 \\over (1-z)^2} = 1 + 2z + 3z^2 + \\cdots$$. Multiply both sides by $z$ to get $${z \\over (1-z)^2} = z + 2z^2 + 3z^3 + \\cdots$$ and plug in $z = 1/2$ to get $${1/2 \\over (1-1/2)^2} = {1 \\over 2} + {2 \\over 2^2} + {3 \\over 2^3} + \\cdots.$$\n\n-\n\nI would do it like this. Using $x \\frac{\\mathrm{d}}{\\mathrm{d} x}\\left( x^k \\right) = k x^k$, and $\\sum_{k=1}^n x^k = x \\frac{x^n-1}{x-1}$. Then $$\\begin{eqnarray} \\sum_{k=1}^n \\left( a k +b\\right) x^k &=& \\left( a x \\frac{\\mathrm{d}}{\\mathrm{d} x} + b\\right) \\circ \\sum_{k=1}^n x^k = \\left( a x \\frac{\\mathrm{d}}{\\mathrm{d} x} + b\\right) \\circ \\left( x \\frac{x^n-1}{x-1} \\right) \\\\ &=& x \\left( a x \\frac{\\mathrm{d}}{\\mathrm{d} x} + a + b\\right) \\circ \\left(\\frac{x^n-1}{x-1} \\right) \\\\ &=& x \\left( (a+b) \\frac{x^n-1}{x-1} + a x \\frac{ n x^{n-1}(x-1) - (x^n-1) }{(x-1)^2} \\right) \\\\ &=& x \\left( (a+b) \\frac{x^n-1}{x-1} + a x \\frac{ (n-1) x^{n} - n x^{n-1} + 1 }{(x-1)^2} \\right) \\end{eqnarray}$$\n\nNow applying this: $$\\begin{eqnarray} \\sum_{k=1}^n \\frac{10 k -5}{2^{k+2}} &=& \\frac{5}{4} \\sum_{k=1}^n (2k-1)\\left(\\frac{1}{2}\\right)^k \\\\ &=& \\frac{5}{4} \\frac{1}{2} \\left( -(2^{1-n} - 2) + (n-1) 2^{2-n}- n 2^{1-n} + 4 \\right) \\\\ &=& \\frac{5}{4} \\left( 3 + (n-3) 2^{-n} \\right) \\end{eqnarray}$$\n\n-\nThanks for doing this in general form. If you don't mind I'll use this question for duplicates in the future and point to this answer :-) \u2013\u00a0 joriki Nov 13 '11 at 8:04\n@joriki:Question is tagged as algebra-precalculus. \u2013\u00a0 Quixotic Nov 18 '11 at 7:31\n@MaX: I see. Thanks for pointing that out. Fortunately there are so many answers here, including N.S.' that shows an elementary way to sum $i/q^i$, that it will still be a good duplicate in many future circumstances. I wonder which answer you'll be accepting... \u2013\u00a0 joriki Nov 18 '11 at 7:39\n@joriki:You are welcome, I am accepting Michael Lugo's answer as it provides the fastest method to reach the solution. \u2013\u00a0 Quixotic Nov 19 '11 at 14:17\nYou presented this beautifully. Thank you. \u2013\u00a0 000 Jul 14 '12 at 9:48\n\nFirst note that $\\displaystyle\\frac{10i-5}{2^{i+2}} = \\frac{5(2i-1)}{2^{i+2}}$\n\nSecondly, using the sum of a geometric series you can show that $2^0 + 2^1 + 2^2 + ... + 2^k = 2^{k+1}-1$.\n\n\\displaystyle \\begin{align*} \\text{So } \\sum \\limits_{i=1}^{n} \\frac{10i-5}{2^{i+2}} &= 5(\\frac{2\\times1-1}{2^{1+2}}+\\frac{2\\times2-1}{2^{2+2}}+\\frac{2\\times3-1}{2^{3+2}}+...+\\frac{2\\times n-1}{2^{n+2}})\\\\ &= 5(\\frac{2^{n-1}(2\\times1-1)+2^{n-2}(2\\times2-1)+2^{n-3}(2\\times3-1)+...+2^{0}(2\\times n-1)}{2^{n+2}}) \\\\ &= 5(\\frac{2(2^{n-1}\\times1+2^{n-2}\\times2+2^{n-3}\\times3+...+2^{0}\\times n)-(2^{n-1}+2^{n-2}+2^{n-3}+...+1)}{2^{n+2}})\\\\ &= 5(\\frac{2(2^{n-1}\\times1+2^{n-2}\\times2+2^{n-3}\\times3+...+2^{0}\\times n)-(2^n-1)}{2^{n+2}})\\\\ &= 5(\\frac{2((2^{n-1}+2^{n-2}+2^{n-3}+...+1)+(2^{n-2}+2^{n-3}+2^{n-4}+...+1)+...+(2+1)+(1))-(2^n-1)}{2^{n+2}})\\\\ &= 5(\\frac{2((2^{n}-1)+(2^{n-1}-1)+(2^{n-2}-1)+...+(2^{2}-1)+(2-1))-(2^n-1)}{2^{n+2}})\\\\ &= 5(\\frac{2((2^{n}+2^{n-1}+2^{n-2}+...+2) + (-1)\\times n)-(2^n-1)}{2^{n+2}})\\\\ &= 5(\\frac{2(2\\times (2^n - 1))- n)-(2^n-1)}{2^{n+2}})\\\\ &= 5(\\frac{4 \\times 2^n - 4 - 2n -2^n + 1}{2^{n+2}})\\\\ &= 5(\\frac{3 \\times 2^n - 2n - 3}{2^{n+2}})\\text{ which is the closed form you got from Mathematica}\\end{align*}\n\n-\nThe equations are a lot more readable in display style, which you get either by switching it on individually with \\displaystyle or by making the entire formula a displayed one by enclosing it in double dollar signs. Also note that you can use the eqnarray and align environments to align consecutive equations with each other. \u2013\u00a0 joriki Nov 13 '11 at 8:03\n@joriki Thanks for the tips and I think my answer has been formatted correctly but I believe my 5th line is a tad too long? I had to trim is down a bit and it is still quite long... \u2013\u00a0 Sp3000 Nov 13 '11 at 9:50\nYes, that's a problem actually, which I didn't foresee when I wrote my comment :-) You can alleviate it somewhat by not putting the \"So\" into the equation but on a line of its own. Also, though it looks nicer with a full equation in the first line, in this case you could break the line before the first equals sign; it will still look better with eqnarray since you can align the expression in the first line with the other expressions instead of with the equals signs. \u2013\u00a0 joriki Nov 13 '11 at 10:03\n\nThe sum $S= \\sum \\limits_{i=1}^n \\frac{1}{2^i}=1-\\frac{1}{2^n}$ is geometric, thus easy to calculate.\n\nHere is a simple elementary way of calculating\n\n$$T=\\sum_{i=1}^n \\frac{i}{2^i} \\,.$$:\n\n$$T=\\sum_{i=1}^n \\frac{i}{2^i} =\\frac{1}{2}+ \\sum_{i=2}^n \\frac{i}{2^i} =\\frac{1}{2}-\\frac{n+1}{2^{n+1}}+ \\sum_{i=2}^{n+1} \\frac{i}{2^i} \\,.$$\n\nChanging the index in the last sum yields:\n\n$$T= \\frac{2^n-n-1}{2^{n+1}}+\\sum_{i=1}^{n} \\frac{i+1}{2^{i+1}}=\\frac{2^n-n-1}{2^{n+1}}+\\sum_{i=1}^{n} \\frac{i}{2^{i+1}}+\\sum_{i=1}^{n} \\frac{1}{2^{i+1}} \\,.$$\n\nThus\n\n$$T= \\frac{2^n-n-1}{2^{n+1}}+\\frac{1}{2}T+ \\frac{1}{2}-\\frac{1}{2^{n+1}}$$\n\nThus\n\n$$\\frac{1}{2}T=\\frac{2^{n+1}-n-2}{2^{n+1}}\\,.$$\n\nHence\n\n$$\\sum_{i=1}^n \\frac{i}{2^i} =\\frac{2^{n+1}-n-2}{2^{n}} \\,.$$\n\n-\n\n$\\sum \\limits_{i=1}^{10} \\frac{10i-5}{2^{i+2}} =\\frac{10}{4} \\sum_{i=1}^{10} \\frac{i}{2^i}-\\frac{5}{4}\\sum_{i=1}^{10} \\frac{1}{2^i}$\n\nSecond $\\sum$ is obviously $1-\\frac{1}{1024}$.\n\nFirst $\\sum$:\n\n$\\frac{1}{2} +$\n\n$\\frac{1}{4} + \\frac{1}{4} +$\n\n$\\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8} +$\n\n$\\vdots$\n\n$\\frac{1}{1024} + \\frac{1}{1024} + \\dots + \\frac{1}{1024}$\n\nSumming by columns, $(1-\\frac{1}{1024})+(\\frac{1}{2}-\\frac{1}{1024})+\\dots+(\\frac{1}{512}-\\frac{1}{1024})=(1+\\frac{1}{2}+\\dots+\\frac{1}{512})-\\frac{10}{1024}=\\frac{1023}{512}-\\frac{10}{1024}$\n\nRest is boring and easy.\n\n-", "date": "2015-01-26 12:53:58", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/81362/which-is-the-fastest-way-to-compute-sum-limits-i-110-frac10i-52i", "openwebmath_score": 0.9958812594413757, "openwebmath_perplexity": 578.1400014659729, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717448632122, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8606332797484871}}
{"url": "https://math.stackexchange.com/questions/1004341/ring-of-polynomials-as-a-module-over-symmetric-polynomials/1004579", "text": "# Ring of polynomials as a module over symmetric polynomials\n\nConsider the ring of polynomials $\\mathbb{k} [x_1, x_2, \\ldots , x_n]$ as a module over the ring of symmetric polynomials $\\Lambda_{\\mathbb{k}}$. Is $\\mathbb{k} [x_1, x_2, \\ldots , x_n]$ a free $\\Lambda_{\\mathbb{k}}$-module? Can you write down \"good\" generators explicitly? (I think that it has to be something very classical in representation theory).\n\nComment: My initial question was whether this module flat. But since all flat Noetherian modules over polynomial ring are free (correct me if it is wrong), it is the same question.\n\nThere is much more general question, which seems unlikely to have good answer. Let $G$ be a finite group and $V$ finite dimensional representation of G. Consider projection $p: V \\rightarrow V/G$. When is $p$ flat?\n\n\u2022 \u2013\u00a0anon Nov 3 '14 at 20:52\n\u2022 Sorry. I really tried to find something before asking and did not manage. \u2013\u00a0quinque Nov 3 '14 at 21:56\n\nLet $s_i= \\sum x_1 \\ldots x_i$ be the fundamental symmetric polynomials. We have a sequence of free extensions $$k[s_1, \\ldots, s_n] \\subset k[s_1, \\ldots, s_n][x_1]\\subset k[s_1, \\ldots, s_n][x_1][x_2] \\subset \\cdots \\\\ \\subset k[s_1,\\ldots ,s_n] [x_1] \\ldots [x_n] = k[x_1, \\ldots ,x_n]$$ of degrees $n$, $n-1$, $\\ldots$ ,$1$. At step $i$ the generators are $1, x_i, \\ldots, x_i^{n-i}$. Therefore $$k[s_1, \\ldots, s_n] \\subset k[x_1, \\ldots, x_n]$$ is free of degree $n!$ with generators $x_1^{a_1} x_2^{a_2} \\cdots x_n^{a_n}$ with $0 \\le a_i \\le n-i$.\n\nMore generally for a finite reflection group of transformations $G$ acting on a vector space $V$ over a field $k$ of characteristic $0$ (to be safe) the algebra of invariants $k[V]^G$ is a polynomial algebra and $k[V]$ is a free $k[V]^G$ module of rank $|G|$ --see the answer of @stephen: .\n\n\u2022 I think some more details are in order: why $1,x_1,\\dots,x_1^{n-1}$ is a basis of the first ring extension? \u2013\u00a0user26857 Nov 3 '14 at 19:41\n\u2022 And why don't satisfy any equation of degree $<n$? \u2013\u00a0user26857 Nov 3 '14 at 19:50\n\u2022 @user26857: Another useful observation using just basic Galois theory is : $k(s_1, \\ldots, s_n) \\subset k(x_1, \\ldots, x_n)$ is Galois with group $S_n$. \u2013\u00a0orangeskid Nov 3 '14 at 19:57\n\u2022 Great answer! Just a remark: The basis orangeskid describes is different from the Schubert basis - geometrically, the images of the $x_i$ in the integral cohomology of the flag variety are the Chern classes of the $n$ tautological line bundles (and the iterated computation above has a geometric counteprart, by viewing the flat variety as an iterated projective bundle and using Leray-Hirsch at each step), while as I said above the Schubert polynomials correspond to Bruhat cells. \u2013\u00a0Hanno Nov 3 '14 at 20:02\n\u2022 Regarding your last paragraph, it can certainly happen that in characteristic $p$ the ring of invariants is not polynomial even when the group is generated by reflection; on the other hand polynomials invariants is enough to imply that the group is generated by reflections in any characteristic. \u2013\u00a0Stephen Nov 3 '14 at 20:34\n\nYes, that's indeed a classical result of representation theory: ${\\mathbb Z}[x_1,...,x_n]$ is graded free over ${\\Lambda}_n$ of rank $n!$ (the graded rank is the quantum factorial $[n]_q!$), and a basis is given by Schubert polynomials defined in terms of divided difference operators.\n\nSee for example the original article of Demazure, in particular Theorem 6.2.\n\nPassing to the quotient, one obtains the graded ring ${\\mathbb Z}[x_1,...,x_n]/\\langle\\Lambda_n^+\\rangle$ which is isomorphic to the integral cohomology ring of the flag variety of ${\\mathbb C}^n$, and the ${\\mathbb Z}$-basis of Schubert polynomials coincides with the basis of fundamental classes of Bruhat cells in the flag variety. This is explained in Fulton's book 'Young Tableaux', Section 10.4.\n\nThe answer by @orangeskid (+1) is the most classical and direct, and the answer by @Hanno connects your question to Schubert calculus and the geometry of flag varieties (+1). Hoping this isn't too self-promoting, you might also have a look at my paper Jack polynomials and the coinvariant ring of $G(r,p,n)$ (I worked in somewhat more generality)\n\nhttp://tinyurl.com/l8u9wh5\n\nwhere I showed that certain non-symmetric Jack polynomials give a basis as well (I was working with more general reflection groups and over the complex numbers, but a version should work over any field of characteristic $0$; I have not thought much about this in characteristic $p$). The point of my paper was really to connect the descent bases (yet another basis!) studied much earlier by Adriano Garsia and Dennis Stanton in the paper Group actions of Stanley-Reisner rings and invariants of permutation groups\n\nhttp://tinyurl.com/lhhtney\n\nto the representation theoretic structure of the coinvariant algebra as an irreducible module for the rational Cherednik algebra. Of course this structure becomes much more complicated in characteristic smaller than $n$; in particular the coinvariant algebra will in general not be irreducible as a module for the Cherednik algebra.\n\nTowards your second question: by definition $p$ is flat if and only if $k[V]$ is a flat $k[V]^G$-module. This certainly holds if $k[V]$ is free over $k[V]^G$; a sufficient condition for this is given in Bourbaki, Theorem 1 of section 2 of Chapter 5 of Lie Groups and Lie algebras (page 110): in case the characteristic of $k$ does not divide the order of $G$, it suffices that $G$ be generated by reflections. This is false (in general) in characteristic dividing the order of the group. But see the paper Extending the coinvariant theorems of Chevalley, Shephard-Todd, Mitchell, and Springer by Broer, Reiner, Smith, and Webb available for instance on Peter Webb's homepage here\n\nhttp://www.math.umn.edu/~webb/Publications/\n\nfor references and what can be said in this generality (this is an active area of research so you shouldn't expect to find a clean answer to your question).\n\nConversely, assuming $p$ is flat and examining the proof of the above theorem in Bourbaki, it follows that $k[V]$ is a free $k[V]^G$-module. Now Remark 2 to Theorem 4 (page 120) shows that $G$ is generated by reflections. So to sum up: if $p$ is flat then $G$ is generated by reflections; in characteristic not dividing the order of $G$ the converse holds.\n\n\u2022 MO level, neat. +1! \u2013\u00a0orangeskid Nov 3 '14 at 20:00\n\u2022 I have a problem about your links. Article by Webb and others is just about connected topics. They do not answer my question (or I just do not see it, then please explain). About Bourbaki book I have a deeper trouble. It sounds silly but I even can not find what is called theorem 1 in this section. Could you post here what it is about (proof is not necessary). \u2013\u00a0quinque Nov 3 '14 at 23:32\n\u2022 @user167762, The statement is Theorem 1 on page 110 of my English edition---are you able to locate it now? I referenced the paper of of BRSW not because it answers your question, but because it is the most recent thing I can think of that deals with the subject in detail and I thought the references it contained might prove useful for you or other future readers. I will edit my answer a bit to give an indication of what I believe to be the state of the art. As far as I know, in positive characteristic there is no clean characterization of groups such that the projection is flat. \u2013\u00a0Stephen Nov 4 '14 at 12:26\n\nAnother perspective:\n\nThe ring extension $S=k[s_1,\\dots,s_n]\\subset k[x_1,\\dots,x_n]=R$ is finite, and since $R$ is Cohen-Macaulay then it is free of rank say $m$. We thus have a Hironaka decomposition $R=\\oplus_{i=1}^m S\\eta_i$ for some homogeneous $\\eta_i\\in R$. Using this the Hilbert series of $R$ is $$H_R(t)=\\sum_{i=1}^mH_S(t)t^{\\deg\\eta_i}=\\sum_{i=1}^mt^{\\deg\\eta_i}/\\prod_{i=1}^n(1-t^i).$$ On the other side, $H_R(t)=1/(1-t)^n$ and therefore $\\sum_{i=1}^mt^{\\deg\\eta_i}=\\prod_{i=1}^{n-1}(1+t+\\cdots+t^i)$ which for $t=1$ gives $m=n!$.\n\n\u2022 You used a theorem about Cohen-Macaulay. I do not know it. Could you please write it? \u2013\u00a0quinque Nov 3 '14 at 22:26", "date": "2020-04-08 16:16:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1004341/ring-of-polynomials-as-a-module-over-symmetric-polynomials/1004579", "openwebmath_score": 0.8081624507904053, "openwebmath_perplexity": 219.77270335062425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754479181589, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8605949966312305}}
{"url": "http://math.stackexchange.com/questions/324048/equation-of-one-branch-of-a-hyperbola-in-general-position", "text": "# Equation of one branch of a hyperbola in general position\n\nGiven a generic expression of a conic:\n\n$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$$\n\nis there a way to write an expression for one of the branches as a function of the coefficients? I tried using the quadratic formula to get an expression for $y$: $$y=\\frac{-(Bx+E)\\pm \\sqrt{(Bx+E)^2 - 4(C)(Ax^2 + Dx + F)}}{2C}$$\n\nbut this doesn't always work. Consider:\n\n$$xy=1$$\n\nHere, $A=0, B=1, C=0, D=0, E=0, F=0$, so $y=\\frac{\\cdot}{0}$, which isn't particularly helpful. In other cases it is not as bad, but still not what I'm looking for. E.g.\n\n$$x^2 - y^2 - 1=0$$\n\nUsing the formula above, we get $$y=\\pm \\sqrt{x^2-1}$$ which seems nice, but $y=\\sqrt{x^2-1}$ it is actually one half of each branch rather than one entire branch, as can be seen here.\n\nhttp://www.wolframalpha.com/input/?i=plot%28x^2+-+y^2+-+1%3D0%29\n\nhttp://www.wolframalpha.com/input/?i=plot%28y%3Dsqrt%28x^2-1%29%29\n\nI am trying to draw one of these branches, so I need an ordered set of points along a predefined \"grid\" of either of the variables. Is it possible to do this?\n\n-\nYour equation doesn't work for all hyperbolas since you found it as a solution of quadratic equation. For $xy = 1$ it won't work since it's not quadratic equation. You can't solve $0x^2+2x+1=0$ as quadratic equation, since it's not. \u2013\u00a0 Kaster Mar 7 '13 at 23:56\n\nFirst of all, if you want to draw a portion of the hyperbola, you typically just need parametric equations of the form $x = x(t)$, $y = y(t)$; you don't necessarily need to get $y$ as a function of $x$.\n\nBut, ignoring that quibble, I'll try to answer the question you asked.\n\nThe key is to first eliminate the $xy$ term in the implicit equation. In effect, you do this by rotating the coordinate system. This is really the same idea as Will Jagy used in his answer, but it might be easier to understand if I don't mention eigenvectors.\n\nSuppose we introduce a $uv$ cooordinate system that is rotated by an angle $\\theta$ (counterclockwise) from the $xy$ one. Then we have\n\n$$x = u\\cos\\theta - v\\sin\\theta \\quad ; \\quad y = u\\sin\\theta + v\\cos\\theta$$\n\nYou can plug these $x$ and $y$ expressions into your original equation, and you'll get\n\n$$\\bar A u^2 + \\bar B uv + \\bar C v^2 + \\bar Du + \\bar Ev + \\bar F = 0$$\n\nwhere\n\n$$\\bar A = A \\cos^2\\theta + 2B \\sin \\theta\\cos\\theta + C \\sin^2\\theta$$\n\n$$\\bar B = 2 B \\cos 2\\theta - (A - C) \\sin 2\\theta$$\n\n$$\\bar C = A \\sin^2\\theta - 2B \\sin\\theta\\cos\\theta + C \\cos^2\\theta$$\n\nand so on. Now we just have to cleverly choose $\\theta$ so that the $uv$ term disappears. Clearly, we'll get what we want if we choose $\\theta$ so that $\\bar B = 0$, and this means\n\n$$\\tan2\\theta = \\frac{2B}{A - C}$$\n\nAfter using this technique, we can now assume that the equation has the form\n\n$$a u^2 + c v^2 - 2adu - 2cev + f = 0$$\n\nNote that I'm using $-2ad$ in place of $\\bar D$ and $-2ce$ in place of $\\bar E$, just to make the next step more convenient. And the next step is just some \"complete the square\" tricks. The equation can be written:\n\n$$a (u^2 - 2du) + c (v^2 - 2ev) + f = 0$$\n\nIn other words:\n\n$$a (u^2 - 2du + d^2) + c (v^2 - 2ev + e^2) + f - ad^2 - ce^2 = 0$$\n\nwhich is\n\n$$a (u - d)^2 + c (v - e)^2 = ad^2 + ce^2 - f$$\n\nNow it should be clear that we have either an ellipse (if the signs of $a$ and $c$ are the same) or a hyperbola (if the signs are different). I'm ignoring the parabola case and several degenerate cases. The ellipse/hyperbola has its center at the point $(d,e)$ in the $uv$ coordinate system, and its axes of symmetry are parallel to the $u$ and $v$ axes.\n\nAt this point, if our curve turns out to be a hyperbola, you can use standard parametric equations (like the ones Wil Jagy gave) to trace out one branch or the other.\n\n-\n\nI think you are going to have a more satisfactory experience doing this: find the center $\\vec{P} = (x_0, y_0)$ of your hyperbola. Find the eigenvectors of the matrix $$\\left( \\begin{array}{cc} A & B/2 \\\\ B/2 & C \\end{array} \\right)$$ and normalize and choose order and $\\pm$ so that with basis $\\vec{u},\\vec{v}$ you can then write your branch as $$g(t) = \\vec{P} + \\vec{u} \\cosh t + \\vec{v} \\sinh t.$$ Note that while $\\vec{u},\\vec{v}$ are perpendicular to each other they are not particularly of any length, in the way I write it above. If you prefer an orthonormal basis you then just put one constant scalar multiplication in front of the $\\cosh$ term and one in front of the $\\sinh$ term. It will all work out if you actually do have a hyperbola, which happens when $B^2 > 4 A C.$\n\nI did teach this about 20 years ago, a whole section on translations and rotations of conic sections for a engineering calculus course. The memories are dim, but it does appear that it is best to find an orthonormal basis for the above matrix first, then find the center expressed in that basis, and end up with constant scalar coefficients of the hyperbolic trig terms.\n\n-\nI posted an answer following your suggestion with an example. The only part I didn't follow was \"and normalize and choose order and $\\pm$ so that with the new basis you can then write your branch as...\". What do you mean \"choose order\" and \"choose $\\pm$\"? Also, how do you determine the values of the parameter that constitute each branch? \u2013\u00a0 David Doria Mar 8 '13 at 13:31\n\nFollowing Will Jagy's suggestion, here are some examples:\n\nExample #1\n\nConsider\n\n$$x^2 - y^2 -1 = 0$$\n\n($A=1$, $B=0$, $C=-1$, $D=0$, $E=0$, $F=-1$). From\n\n$$p_c = \\begin{pmatrix}x_c\\\\y_c\\end{pmatrix} = \\begin{pmatrix}\\frac{BE-2CD}{4AC-B^2}\\\\ \\frac{DB-2AE}{4AC-B^2}\\end{pmatrix}$$\n\nwe have\n\n$$p_c = \\begin{pmatrix}0\\\\0\\end{pmatrix}$$\n\nNow, the eigenvalues/vectors of the matrix $$\\begin{pmatrix}A & \\frac{B}{2}\\\\\\frac{B}{2} & C\\end{pmatrix} = \\begin{pmatrix}1 & 0 \\\\ 0 & -1 \\end{pmatrix}$$ are $\\lambda_1 = -1$, $\\lambda_2 = 1$, $v_1 = \\begin{pmatrix}0\\\\1\\end{pmatrix}$, $v_2 = \\begin{pmatrix}1\\\\0\\end{pmatrix}$\n\nFrom the parametric form\n\n$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$\n\nwe have\n\n$$g(t) = \\begin{pmatrix}0\\\\1\\end{pmatrix} cosh(t) + \\begin{pmatrix}1\\\\0\\end{pmatrix} sinh(t)$$\n\nwhich looks like\n\nNow how would we draw the other branch? That is, how do you determine the values of the parameter that constitute each branch?\n\nExample #2\n\nConsider\n\n$$7x^2 - 3y^2 - 25 = 0$$\n\n($A=7$, $B=0$, $C=-3$, $D=0$, $E=0$, $F=-25$). From\n\n$$p_c = \\begin{pmatrix}x_c\\\\y_c\\end{pmatrix} = \\begin{pmatrix}\\frac{BE-2CD}{4AC-B^2}\\\\ \\frac{DB-2AE}{4AC-B^2}\\end{pmatrix}$$\n\nwe have\n\n$$p_c = \\begin{pmatrix}0\\\\\\frac{-25}{6}\\end{pmatrix}$$\n\nNow, the eigenvalues/vectors of the matrix $$\\begin{pmatrix}A & \\frac{B}{2}\\\\\\frac{B}{2} & C\\end{pmatrix} = \\begin{pmatrix}7 & 0 \\\\ 0 & -3 \\end{pmatrix}$$ are $\\lambda_1 = -3$, $\\lambda_2 = 7$, $v_1 = \\begin{pmatrix}0\\\\1\\end{pmatrix}$, $v_2 = \\begin{pmatrix}1\\\\0\\end{pmatrix}$\n\nFrom the parametric form\n\n$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$\n\nwe have\n\n$$g(t) = \\begin{pmatrix}0\\\\ \\frac{-25}{6}\\end{pmatrix} + \\begin{pmatrix}0\\\\1\\end{pmatrix} cosh(t) + \\begin{pmatrix}1\\\\0\\end{pmatrix} sinh(t)$$\n\n-\nDavid, the other branch is $p_c -v_1 \\cosh t + v_2 \\sinh t$ \u2013\u00a0 Will Jagy Mar 8 '13 at 20:51\nDavid, please do another example, something along the lines of $x^2 - x y - y^2 =1,$ where the eigenvectors will not naturally come out length one, and if you choose an orthonormal basis you will need scalar multiples, $p_c + \\alpha v_1 \\cosh t + \\beta v_2 \\sinh t.$ Note that the bit about the scalar multipliers would apply even for, say, $7 x^2 - 3 y^2 = 25,$ where one branch is $$x = \\frac{5}{\\sqrt 7} \\cosh t, y = \\frac{5}{\\sqrt 3} \\sinh t.$$ \u2013\u00a0 Will Jagy Mar 8 '13 at 21:04\n@WillJagy How do you choose the eigenvector that goes with cosh vs sinh? Even in my Example #1, the hyperbola opens up/down (instead of left/right) if I use x=sinh(t) and y=cosh(t) as the eigenvectors they way I have written them imply (I guess I thought I had made a typo so I switched them (x=cosh(t) and y=sinh(t) when I produced that graph): wolframalpha.com/input/\u2026 \u2013\u00a0 David Doria Mar 11 '13 at 12:04\n@WillJagy I worked out the $7x^2 - 3y^2 - 25 = 0$ example in my answer. How did you get from where I am to the $\\frac{5}{\\sqrt{7}}$ and $\\frac{5}{\\sqrt{3}}$ coefficients? And did you just accidentally omit the $center + ...$ term? My comment above holds on this problem as well - how do you choose the eigenvector to associate with the cosh vs sinh term? \u2013\u00a0 David Doria Mar 11 '13 at 12:11\nThe center is actually at $(0,0)$ because $D=0,E=0.$ If you choose unit length eigenvectors, then you expect scalar coefficients to be necessary. In particular, try your recipe at $t=0$ (after replacing the -25/6 by 0) and you will find the point does not lie on the actual hyperbola. How about if you try doing $xy = 1?$ You already know exactly how the graph looks. One test for correctness is, once you write $x = x_c + x_1 \\cosh t + x_2 \\sinh t,$ then $y = y_c + y_1 \\cosh t + y_2 \\sinh t,$ you must be able to plug those into the original equation and get truth. \u2013\u00a0 Will Jagy Mar 11 '13 at 20:38", "date": "2014-07-30 13:37:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/324048/equation-of-one-branch-of-a-hyperbola-in-general-position", "openwebmath_score": 0.9044473171234131, "openwebmath_perplexity": 254.06414264158346, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754481444574, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8605949855253179}}
{"url": "https://math.stackexchange.com/questions/2260/proof-1234-cdotsn-fracn-timesn12?noredirect=1", "text": "# Proof $1+2+3+4+\\cdots+n = \\frac{n\\times(n+1)}2$\n\nApparently $$1+2+3+4+\\ldots+n = \\dfrac{n\\times(n+1)}2$$.\n\nHow? What's the proof? Or maybe it is self apparent just looking at the above?\n\nPS: This problem is known as \"The sum of the first $$n$$ positive integers\".\n\n\u2022 This an important example of a finite integral, a good tutorial here: stanford.edu/~dgleich/publications/finite-calculus.pdf \u2013\u00a0anon Aug 12 '10 at 19:46\n\u2022 Note that we do not define this sum to be (x+1)x/2, we prove it to be so. It is a very useful formula. You could check out en.wikipedia.org/wiki/Triangle_number, which unfortunately, does not prove the formula you cite. \u2013\u00a0Ross Millikan Apr 21 '11 at 22:14\n\u2022 Although we already have three proofs of this here, I think it would be really interesting if we amassed more (perhaps excluding the typical intro-to-induction styled proof). \u2013\u00a0davidlowryduda Apr 22 '11 at 1:15\n\u2022 It could be proved by mathematical induction \u2013\u00a0J. W. Tanner Sep 5 '19 at 1:10\n\u2022 Not appropriate for an answer, but you've asked either a very easy or a very difficult question. If by \"why\" you mean, \"Can I see a proof of this fact?\" the question is fairly easy to answer. If by \"why\" you mean, \"Why should this be true?\" you've asked a very deep kind of question that mathematicians make entire careers out of. Sure it can be shown to be true, but how does it connect to other true results? Why is it a polynomial of degree $2$? What about the sum of $k^2$? And on and on... \u2013\u00a0Charles Hudgins Sep 5 '19 at 3:05\n\nLet $$S = 1 + 2 + \\ldots + (n-1) + n.$$ Write it backwards: $$S = n + (n-1) + \\ldots + 2 + 1.$$ Add the two equations, term by term; each term is $$n+1,$$ so $$2S = (n+1) + (n+1) + \\ldots + (n+1) = n(n+1).$$ Divide by 2: $$S = \\frac{n(n+1)}{2}.$$\n\n\u2022 Well, you lost me at \"each term is n+1...\". As far as I can see, if you add the two equations term by term it will be: n+n + (n-1)+(n-1) + ... + 2+2 + 1+1. How did you get (n+1) + (n+1) + ... + (n+1)? \u2013\u00a0b1_ Aug 12 '10 at 18:27\n\u2022 I got it! It's 2S = (1+2+...+(n-1)+n) + (n+(n-1)+...+2+1) - so you write one backwards, then match up each term. 2S = (1+n) + (2+n-1)+...+(n-1+2)+n+1, and so 2S=(n+1)+(n+1)+...+(n+1)+(n+1) etc \u2013\u00a0b1_ Aug 12 '10 at 19:04\n\u2022 This trick is usually attributed to Gauss (when he was a schoolboy... though it's unclear if the story is true or not). \u2013\u00a0Fixee Feb 27 '11 at 15:17\n\u2022 This helped make it clear for me youtu.be/1wnIsgUivEQ \u2013\u00a0vexe Nov 27 '15 at 8:17\n\u2022 @john this can be made rigorous using sigma notation. It's simply a change of index. \u2013\u00a0Brevan Ellefsen Jul 29 '18 at 1:34\n\nMy favourite proof is the one given here on MathOverflow. I'm copying the picture here for easy reference, but full credit goes to Mariano Su\u00e1rez-Alvarez for this answer.\n\nTakes a little bit of looking at it to see what's going on, but it's nice once you get it. Observe that if there are n rows of yellow discs, then:\n\n1. there are a total of 1 + 2 + ... + n yellow discs;\n2. every yellow disc corresponds to a unique pair of blue discs, and vice versa;\n3. there are ${n+1 \\choose 2} = \\frac 12 n(n+1)$ such pairs.\n\u2022 great proof!!!! \u2013\u00a0anon Aug 12 '10 at 21:22\n\u2022 Maybe I lack imagination, but to me it's clearer to just make a square of yellow discs ($n^2$ of them), duplicate the diagonal ($n^2+n$), then cut in half to get the answer ($(n^2+n)/2$). This is how Knuth does this (and much more intricate) summations in \"Concrete Mathematics.\" \u2013\u00a0Fixee Feb 27 '11 at 15:27\n\u2022 @Fixee: I don't know why you're comparing them; this is an entirely different proof. Unlike the other proof (also good), this doesn't require computing areas, cutting, or duplicating -- in fact this doesn't even involve the number $n(n+1)/2$ directly; what this gives is a proof that $1 + 2 + \\dots + n = \\binom{n+1}{2}$, and it so happens that the latter is $n(n+1)/2$. It's a bijection proof, rather than an area proof (vaguely speaking). \u2013\u00a0ShreevatsaR Nov 25 '11 at 6:13\n\u2022 @Vaughn Climenhaga - is there an analog of this in three dimensions and higher? \u2013\u00a0Vincent Tjeng Apr 1 '13 at 9:43\n\u2022 Wow. That is simply amazing. I didn't get it at first glance and then it hit me. Ingenious. \u2013\u00a0Karl Feb 3 '16 at 20:30\n\nWhat a big sum! This is one of those questions that have dozens of proofs because of their utility and instructional use. I present my two favorite proofs: one because of its simplicity, and one because I came up with it on my own (that is, before seeing others do it - it's known).\n\nThe first involves the above picture:\n\nIn short, note that we want to know how many boxes are in the outlined region, as the first column has 1 box, the second 2, and so on (1 + 2 + ... + n). One way to count this quickly is to take another copy of this section and attach it below, making a $n*(n+1)$ box that has exactly twice as many squares as we actually want. But there are $n*(n+1)$ little squares in this area, so our sum is half that: $$1 + 2 + ... + n = \\dfrac{n(n+1)}{2}.$$\n\nSecond proof, same as the first but a little bit harder and a little bit worse:\n\nLet us take for granted the finite geometric sum $1 + x + x^2 + ... + x^n = \\dfrac{x^{n+1} - 1}{x-1}$ (If you are unfamiliar with this, comment and I'll direct you to a proof). This is a polynomial - so let's differentiate it. We get $$1 + 2x + 3x^2 + ... + nx^{n-1} = \\dfrac{ (n+1)x^n (x-1) - x^{n+1} + 1}{ (x-1)^2 }$$ Taking the limit as x approaches 1, we get\n\n$$\\lim_{x \\to 1} \\dfrac{ (n+1)x^n (x-1) - x^{n+1} + 1}{ (x-1)^2} = \\dfrac{ (n+1) [ (n+1)x^n - nx^{n-1} ] - (n+1)x^n }{2(x-1)} =$$ $$\\lim_{x \\to 1} \\dfrac{ (n+1)[(n+1)(n)x^{n-1} - n(n-1)x^{n-2}] - (n+1)(n)x^{n-1} } {2}$$\n\nwhere we used two applications of l'Hopital above. This limit exists, and plugging in x = 1 we see that we get $$\\dfrac{1}{2} * (n+1)(n) [ (n+1) - (n-1) - 1] = \\dfrac{ (n)(n+1)}{2}.$$\n\nAnd that concludes the second proof.\n\n\u2022 I came up with the first proof by myself a long time ago :). It was an awesome moment where the concept of \"area\" was redefined. \u2013\u00a0Jacob Apr 22 '11 at 1:50\n\u2022 @mixedmath A bit overkill, but nice anyways. (+1) \u2013\u00a0Franklin Pezzuti Dyer Jan 20 '18 at 15:40\n\nHow many ways are there to choose a $2$-element subset out of an $n$-element set?\n\nOn the one hand, you can choose the first element of the set in $n$ ways, then the second element of the set in $n-1$ ways, then divide by $2$ because it doesn't matter which you choose first and which you choose second. This gives $\\frac{n(n-1)}{2}$ ways.\n\nOn the other hand, suppose the $n$ elements are $1, 2, 3, ... n$, and suppose the larger of the two elements you choose is $j$. Then for every $j$ between $2$ and $n$ there are $j-1$ possible choices of the smaller of the two elements, which can be any of $1, 2, ... j-1$. This gives $1 + 2 + ... + (n-1)$ ways.\n\nSince the two expressions above count the same thing, they must be equal. This is known as the principle of double counting, and it is one of a combinatorialist's favorite weapons. A generalization of this argument allows one to deduce the sum of the first $n$ squares, cubes, fourth powers...\n\nGathering as many proofs as we can? Write the series recursively:\n\n$$S(n) = S(n - 1) + n \\tag{1}$$\n\nSubstitute $n \\to n+1$ :\n\n$$S(n + 1) = S(n) + n + 1\\tag{2}$$\n\nEquation (2) subtract Equation (1):\n\n$$S(n+1) - S(n) = S(n) + 1 - S(n - 1) \\tag{3}$$\n\nAnd write it up:\n\n$$\\begin{cases} S(n+1) &= 2S(n) -S(n-1) + 1 \\\\ S(n) &= S(n) \\end{cases} \\tag{4}$$\n\nWhich can now be written in matrix form:\n\n$$\\begin{bmatrix} S(n+1) \\\\ S(n) \\end{bmatrix} = \\begin{bmatrix} 2 & -1 \\\\ 1 & 0 \\end{bmatrix} \\begin{bmatrix} S(n) \\\\ S(n-1) \\end{bmatrix} + \\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix} \\tag{5}$$\n\nAnd then converting the affine equation (5) to the linear equation (6):\n\n$$\\begin{bmatrix} S(n+1) \\\\ S(n+0) \\\\ 1\\end{bmatrix} = \\begin{bmatrix} 2 & -1 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{bmatrix} \\begin{bmatrix} S(n) \\\\ S(n-1) \\\\ 1\\end{bmatrix} \\tag{6}$$\n\nAnd closing the equation:\n\n$$\\begin{bmatrix} S(n+1) \\\\ S(n) \\\\ 1\\end{bmatrix} = \\begin{bmatrix} 2 & -1 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{bmatrix}^n \\begin{bmatrix} S(1) \\\\ S(0) \\\\ 1\\end{bmatrix} \\tag{6}$$\n\nThen finding the Jordan form of the 3x3 matrix:\n\n\\begin{align} \\begin{bmatrix} S(n+1) \\\\ S(n) \\\\ 1\\end{bmatrix} &= \\left(\\begin{bmatrix} 1 & 1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{bmatrix} \\begin{bmatrix} 1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 1\\end{bmatrix} \\begin{bmatrix} 1 & 1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{bmatrix}^{-1}\\right)^n \\begin{bmatrix} S(1) \\\\ S(0) \\\\ 1\\end{bmatrix} \\\\ &= \\begin{bmatrix} 1 & 1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{bmatrix} \\begin{bmatrix} 1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 1\\end{bmatrix}^n \\begin{bmatrix} 1 & 1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{bmatrix}^{-1} \\begin{bmatrix} S(1) \\\\ S(0) \\\\ 1\\end{bmatrix} \\\\ &= \\begin{bmatrix} 1 & 1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{bmatrix} \\begin{bmatrix} 1 & \\binom{n}{1} & \\binom{n-1}{2} \\\\ 0 & 1 & \\binom{n}{1} \\\\ 0 & 0 & 1\\end{bmatrix} \\begin{bmatrix} 0 & 1 & 0 \\\\ 1 & -1 & 0 \\\\ 0 & 0 & 1\\end{bmatrix} \\begin{bmatrix} S(1) \\\\ S(0) \\\\ 1\\end{bmatrix} \\end{align} \\tag{7}\n\nAnd multiplying the matrices out:\n\n$$\\begin{bmatrix} S(n+1) \\\\ S(n) \\\\ 1\\end{bmatrix} = \\begin{bmatrix} \\frac{ (2n+2)S(1) - 2nS(0) + {n}^{2} + n}{2} \\\\ \\frac{ 2nS(1) + (2 - 2n)S(0)+{n}^{2}-n}{2} \\\\ 1 \\end{bmatrix} \\tag{8}$$\n\nAnd given that $S(0) = 0$ and $S(1) = 1$, we get that:\n\n$$S(n) = \\frac{n^2 + n}{2} \\tag{9}$$\n\nFor example, $$X = 1+2+3+4+5+6$$ Then twice $X$ is $$2X = (1+2+3+4+5+6) + (1+2+3+4+5+6)$$ which we can rearrange as $$2X = (1+2+3+4+5+6) + (6+5+4+3+2+1)$$ and add term by term to get $$2X = (1+6)+(2+5)+(3+4)+(4+3)+(5+2)+(6+1)$$ to get $$2X = 7+7+7+7+7+7 = 6*7 = 42$$\n\n\u2022 Legend has it that Gauss used this method to find the sum of all numbers from 1 to 100 without actually summing them. \u2013\u00a0lhf Apr 21 '11 at 22:43\n\u2022 In class, I think. A teacher set him in a corner because he was misbehaving (apparently having already finished that day's work), and told him to add these numbers. Almost immediately, Gauss went to play because he was done - and when the teacher got angry it took her nearly 10 minutes to verify that his answer was correct! Or at least, that's how the legend I heard went. \u2013\u00a0davidlowryduda Apr 21 '11 at 22:46\n\u2022 @mixedmath, Gauss's Day of Reckoning, which was cited in wikipedia, discusses the legend. \u2013\u00a0lhf Apr 21 '11 at 23:01\n\u2022 @mixedmath: I heard a different version - teacher asked all the kids to add the numbers from 1 to 1,000, because he (late 18th century Germany, therefore \"he\") needed some time to do some work. And CFG had the answer within 30 seconds... \u2013\u00a0gnasher729 Dec 22 '16 at 16:56\n\u2022 Nobody should take nearly ten minutes. When you add up the single digits, you should notice that you are ten times adding 1+2+3+4+5+6+7+8+9 = 45. And the tens, you have ten 10s, ten 20s, ten 30s, ten 90s. Plus a hundred. All in all, should be done in a minute. \u2013\u00a0gnasher729 Dec 22 '16 at 16:59\n\nMy favourite proof of this fact involves counting the edges of the complete graph $K_n$ in two different ways.\n\nOn the one hand, any vertex, $v_1$ say, is connected to $n-1$ other vertices, thus contributing $n-1$ edges. Moving clockwise, the next vertex $v_2$ contributes $n-2$ edges (not counting the edge connecting $v_1$ and $v_2$), $v_3$ adds $n-3$ edges, ... , $v_{n-1}$ contributes 1 edge and $v_n$ adds no new edges.\n\nThus the total number of edges in the complete graph $K_n$ is:\n\n$$E = \\sum\\limits_{i=1}^{n-1} i$$\n\nBut clearly, any edge connects two vertices, so the number of edges is the number of ways to choose two distinct elements from the set $\\{1,...,n\\}$ and hence:\n\n$$E = \\sum\\limits_{i=1}^{n-1} i = \\binom{n}{2} = \\frac{n(n-1)}{2}$$\n\n\u2022 You can easely explain this proof without graph theory... $n+1$ people went to a party, and everyone shakes hands with everyone else... How many handshakes were there? :) \u2013\u00a0N. S. Sep 29 '12 at 14:03\n\nOnce you have a formula like this, you can prove it by induction. But that begs the question as to how you get such a formula. In this case you might ask: (a) what's the \"average\" term? and (b) how many terms are there?\n\nDraw a triangular pyramid of base $n+1$. We get a unique coordinate for any of the $\\sum_{i=1}^ni$ elements of the pyramid not in the bottom row by choosing two of the elements in the bottom row of $n+1$. This gives a bijection from the $\\binom{n+1}{2}$ coordinate pairs to the $\\sum_{i=1}^ni$ elements of the pyramid not in the bottom row.\n\nImage from Mariano Su\u00e1rez-Alvarez's answer on Math Overflow:\n\n\u2022 @Mixedmath: No he means something quite different. A picture can be found in this MO-topic mathoverflow.net/questions/8846/proofs-without-words/8847#8847 \u2013\u00a0Myself Apr 22 '11 at 0:39\n\u2022 This is a beautiful proof and doesn't seem to be getting enough appreciation, so I added the image from the MO answer to make it clearer. Feel free to remove the image if you don't want it. \u2013\u00a0ShreevatsaR Nov 25 '11 at 5:55\n\u2022 @ShreevatsaR: That image is to powerful! \u2013\u00a0String Sep 25 '14 at 10:22\n\nBasically same proof as Yoyo's, just purely combinatorial (no picture needed):\n\nHow many ways can we chose two distinct numbers between $1$ and $n+1$?\n\nWe pick first the largest, which is of the form $i+1$ for some $1 \\leq i \\leq n$, and then we have exactly $i$ distinct choices for the smallest one.\n\nThus we have $\\sum_{i=1}^n i$ choices.\n\nHere is another idea:\n\nUsing $(i+1)^2-i^2=2i+1$ we get a telescopic sum:\n\n$$\\sum_{i=1}^n 2i+1 = \\sum_{i=1}^n (i+1)^2-i^2 = (n+1)^2-1=n^2+2n \\,.$$\n\nThen\n\n$$n^2+2n= 2\\left[\\sum_{i=1}^n i\\right] +n \\,.$$\n\n\u2022 Note that the second idea is useful for generalization in that we can find the sums of any powers is we know the sums of the powers less than. It can take a lot of work but using this method, you would be able to find 1^10 + 2^2 + ... + n^10 without too much work... \u2013\u00a0quasicompactscheme Sep 18 '11 at 15:07\n\nIf you knew of the geometric series, you would know that\n\n$$\\frac{1-r^{n+1}}{1-r}=1+r+r^2+r^3+\\dots+r^n$$\n\nIf we differentiate both sides, we have\n\n$$\\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}=1+2r+3r^2+\\dots+nr^{n-1}$$\n\nLetting $r\\to1$ and applying L'Hospital's rule on the fraction, we end up with\n\n$$\\frac{n(n+1)}2=1+2+3+\\dots+n$$\n\nHINT Pair each summand $$k$$ with its \"reflection\" $$n+1-k$$. This is simply a discrete analog of the method of computing the area of the triangle under the diagonal of a square by reflecting a subtriangle through the midpoint of the diagonal to form an $$n$$ by $$n/2$$ rectangle.\n\nLike the analogous proof of Wilson's theorem, the method exploits the existence of a nontrivial symmetry. In Wilson's theorem we exploit the symmetry $$n \\mapsto n^{-1}$$ which exists due to the fact that $${\\mathbb F}_p^*$$ forms a group. Here we exploit a reflection through a line - a symmetry that exists due to the linear nature of the problem (which doesn't work for nonlinear sums, e.g. $$\\sum n^2$$). Symmetries often lead to elegant proofs. One should always look for innate symmetries when first pondering problems.\n\nGenerally there are (Galois) theories and algorithms for summation in closed form, in analogy to the differential case (Ritt, Kolchin, Risch et al.). A very nice motivated introduction can be found in the introductory chapter of Carsten Schneider's thesis Symbolic Summation in Difference Fields.\n\nHere is an easy way to visualize it:\n\nDraw a rectangular grid with a height of $n$ squares and width of $n+1$ squares. Obviously it has $n(n+1)$ squares in it.\n\nIn the first row, color the leftmost square red and the other $n$ squares blue; in the second row, color the leftmost $2$ squares red and the other $n-1$ squares blue; and so forth (in the last row, there will be $n$ red squares and one blue square). Clearly, there are $\\sum_{i=1}^n i$ red squares and the same number of blue squares.\n\nAdding the red and blue squares together, we get $2 \\sum_{i=1}^n i = n(n+1)$, or $\\sum_{i=1}^n i = n(n+1)/2$.\n\nLet us denote the sum as $S_1(n)$. This function must be a second degree polynomial in $n$ because the first order difference $S_1(n)-S_1(n-1)=n$ is a linear polynomial in $n$. So it suffices to construct the Lagrangian polynomial by three known points, let $(0,0), (1,1), (2,3)$.\n\n$$S_1(n)=0\\frac{(n-1)(n-2)}{(0-1)(0-2)}+1\\frac{(n-0)(n-2)}{(1-0)(1-2)}+3\\frac{(n-0)(n-1)}{(2-0)(2-1)}=\\frac{n(n+1)}2.$$\n\nThe most general case of this is called an arithmetic progression or (finite) arithmetic series. There are many, many, many proofs. An easy one: write all the summands in a row; write them again just below, but from right to left now (so $1$ is under $n$, $2$ is under $n-1$, etc). Add them up, and figure out how it relates to the quantity you are looking for.\n\n\u2022 Yep, that's it, write one backwards. I read this post but it didn't click - I guess I needed to write it out to see it. Thx \u2013\u00a0b1_ Aug 12 '10 at 19:43\n\nHere are two ways to calculate this sum. First is by symmetry of another sum:\n\n\\begin{aligned} \\displaystyle & \\sum_{0 \\le k \\le n}k^2 = \\sum_{0 \\le k \\le n}(n-k)^2 = n^2\\sum_{0 \\le k \\le n}-2n\\sum_{0 \\le k \\le n}k+\\sum_{0 \\le k \\le n}k^2 \\\\& \\implies 2n\\sum_{0 \\le k \\le n}k = n^2(n+1) \\implies \\sum_{0 \\le k \\le n}k = \\frac{1}{2}n(n+1).\\end{aligned}\n\nThe second is writing it as double sum and switching the order of summation:\n\n\\begin{aligned}\\displaystyle & \\begin{aligned}\\sum_{1 \\le k \\le n}k & = \\sum_{1 \\le k \\le n}~\\sum_{1 \\le r \\le k} = \\sum_{1 \\le r \\le n} ~\\sum_{r \\le k \\le n} = \\sum_{1 \\le r \\le n}\\bigg(\\sum_{1 \\le k \\le n}-\\sum_{1 \\le k \\le r-1}\\bigg) \\\\& =\\sum_{1 \\le r \\le n}\\bigg(n-r+1\\bigg) = n\\sum_{1 \\le k \\le n}-\\sum_{1 \\le k \\le n}k+\\sum_{1 \\le k \\le n}\\end{aligned} \\\\& \\implies 2\\sum_{1 \\le k \\le n}k = n^2+n \\implies \\sum_{1 \\le k \\le n}k = \\frac{1}{2}n(n+1), ~ \\mathbb{Q. E. D.} \\end{aligned}\n\nNote I started using k back on the third line for convenience because r is just a dummy vairable at this point, and our sum no longer depends on k. Note that the first trick can easily be generalised:\n\n\\begin{aligned} & \\hspace{0.5in}\\begin{aligned}\\displaystyle \\sum_{0 \\le k \\le n}k^{2p} &= \\sum_{0 \\le k \\le n}(n-k)^{2p} \\\\& = \\sum_{0 \\le k \\le n}~\\sum_{0 \\le r \\le 2p}\\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r}\\\\& = \\sum_{0 \\le k \\le n}k^{2p}-2pn\\sum_{0 \\le k \\le n}k^{2p-1}+\\sum_{0 \\le k \\le n}~\\sum_{2 \\le r \\le 2p}\\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r} \\end{aligned} \\\\& \\implies \\sum_{0 \\le k \\le n}k^{2p-1} = \\frac{1}{2pn}\\sum_{0 \\le k \\le n}~\\sum_{2 \\le r \\le 2p}\\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r}. \\end{aligned}\n\nYou can take the power series\n\n$$f(x)=\\sum_{n=0}^\\infty\\left(\\sum_{j=0}^{n}j\\right)x^n$$\n\nand you can check that it has a postive convergence ratio, and changing the order of the series you can deduce that\n\n$$f(x)=\\frac{x}{(1-x)^3}.$$\n\nOn the other hand the taylor series for $\\frac{x}{(1-x)^3}$ is precisely\n\n$$\\frac{x}{(1-x)^3}=\\sum_{n=0}^\\infty \\frac{n(n+1)}{2}x^n$$\n\nso $$\\sum_{j=0}^nj=\\frac{n(n+1)}{2}.$$\n\n$\\dfrac{k+1}2-\\dfrac{k-1}2=1 \\implies\\dfrac{k(k+1)}2-\\dfrac{(k-1)k}2=k$\n\n$\\implies \\sum_{k=1}^n\\Bigg(\\dfrac{k(k+1)}2-\\dfrac{(k-1)k}2\\Bigg)=\\sum_{k=1}^nk\\implies\\dfrac{n(n+1)}2-\\dfrac{1(1-1)}2=\\sum_{k=1}^nk$\n\n$\\implies\\sum_{k=1}^nk=\\dfrac{n(n+1)}2$\n\n$$\\sum_{i=0}^ni-\\sum_{i=0}^{n-1}i=S_1(n)-S_1(n-1)=n,$$ so that $S_1(n)$ must be a polynomial of the second degree in $n$. By the method of undeterminate coefficients, noting that there is no constant term as $S_1(0)=0$: $$S_1(n)-S_1(n-1)=n=(an^2+bn)-(a(n-1)^2+b(n-1))=\\\\=2an+b-a,$$ and by identification $$a=b=\\frac12.$$ $$S_1(n)=\\frac{n^2+n}2.$$ Let us generalize to the sum of squares, $$S_2(n)-S_2(n-1)=n^2=(an^3+bn^2+cn)-(a(n-1)^3+b(n-1)^2+c(n-1))=\\\\=3an^2+(-3a+2b)n+a-b+c,$$ giving $$a=\\frac13,b=\\frac12,c=\\frac16.$$ $$S_2(n)=\\frac{2n^3+3n^2+n}6.$$ For any power, you get a triangular system of equations where you recognize a part of Pascal's triangle, with alternating signs, as in the sum of cubes: $$S_3(n)-S_3(n-1)=n^3=4an^3+(-6a+3b)n^2+(4a-3b+2c)n+(-a+b-c+d),$$ $$\\color{blue}{4}a=1\\\\-\\color{blue}{6}a+\\color{blue}{3}b=0\\\\\\color{blue}{4}a-\\color{blue}{3}b+\\color{blue}{2}c=0\\\\-\\color{blue}{1}a\\ +\\color{blue}{1}b-\\color{blue}{1}c+\\color{blue}{1}d=0,$$ giving $$a=\\frac14,b=\\frac12,c=\\frac14,d=0.$$ $$S_3(n)=\\frac{n^4+2n^3+n^2}4.$$\n\n\u2022 You can also retrieve the polynomial by Lagrange interpolation with $(0,0), (1,1),(2,3)$: $$S_1(n)=0\\frac{(n-1)(n-2)}{(1-0)(2-0)}+1\\frac{(n-0)(n-2)}{(1-0)(1-2)}+3\\frac{(n-0)(n-1)}{(2-0)(2-1)},$$ but this leads to much longer calculation. \u2013\u00a0Yves Daoust Jul 26 '14 at 10:22\n\nAbove is an image representing Pascals Triangle. What I want to draw attention to is the hockey stick formation, particularly, the blue hockey stick. Notice how the entries in the stick of the blue hockey stick are in arithmetic progression, and that the entry in the blade represents the sum of the entries in the stick.\n\nTo prove this inductively we have as a bootstrap condition\n\n$$1=\\frac{1(1+1)}{2}=\\binom{1+1}{2} = \\sum\\limits_{i=1}^1\\binom{i}{1}=1$$\n\nand for the general case\n\n$$\\begin{array}{lll} \\sum\\limits_{i=1}^{n+1}&=&(n+1)+\\sum\\limits_{i=1}^{n}i\\\\ &=&\\binom{n+1}{1}+\\binom{n+1}{2}\\\\ &=&\\binom{n+2}{2}\\\\ &=&\\binom{(n+1)+1}{2}\\\\ &=&\\frac{(n+1)((n+1)+1)}{2} \\end{array}$$\n\nOf course, we assumed that $\\binom{n}{k}+\\binom{n}{k+1} = \\binom{n+1}{k+1}$ holds. $$\\begin{array}{lll} \\binom{n}{k}+\\binom{n}{k+1}&=&\\frac{n!}{k!(n-k)!} + \\frac{n!}{(k+1)!(n-(k+1))!}\\\\ &=&\\frac{n!(k+1)}{k!(k+1)(n-k)!} + \\frac{n!(n-k)}{(k+1)!(n-(k+1))!(n-k)}\\\\ &=&\\frac{n!k+n!+n!n-n!k}{(k+1)!(n-k)!}\\\\ &=&\\frac{n!+n!n}{(k+1)!((n+1)-(k+1))!}\\\\ &=&\\frac{n!(n+1)}{(k+1)!((n+1)-(k+1))!}\\\\ &=&\\frac{(n+1)!}{(k+1)!((n+1)-(k+1))!}\\\\ &=&\\binom{n+1}{k+1}\\\\ \\end{array}$$\n\n\u2022 I feel like this is a very intuitive answer, generalizes to the sum of $n^k$ well, and is easy to remember. \u2013\u00a0Simply Beautiful Art Nov 29 '16 at 0:50\n\nLet us denote the sum as $S_1(n)$. This function must be a second degree polynomial in $n$ because the first order difference $S_1(n)-S_1(n-1)=n$ is a linear polynomial in $n$. So it suffices to verify the formula for three different values of the argument.\n\n$$S_1(0)=0=\\frac{0(0+1)}2,$$ $$S_1(1)=1=\\frac{1(1+1)}2,$$ $$S_1(2)=3=\\frac{2(2+1)}2.$$ QED.\n\nI didn't see an answer using the method that I used, so I'm posting an answer here to 'spread the knowledge!'\n\nThis can be applied to higher powers such $1^2+2^2+\\cdots n^2$ or $1^3+2^3+\\cdots n^3$.\n\nSolving by the use of Indeterminate Coefficients:\n\nAssume the series$$1+2+3+4+5\\ldots+n\\tag1$$ Is equal to the infinite series$$1+2+3+4+5+\\ldots+n=A+Bn+Cn^2+Dn^3+En^4+\\ldots\\&c\\tag2$$ If we 'replace' $n$ with $n+1$, we get$$1+2+3+4+\\ldots+(n+1)=A+B(n+1)+C(n+1)^2+\\ldots\\&c\\tag3$$ And subtracting $(3)-(2)$, gives\\begin{align*} & n+1=B+C(2n+1)\\tag4\\\\n & +1=2Cn+(B+C)\\tag5\\end{align*} Therefore, $C=\\dfrac 12,B=\\dfrac 12,A=0$ and $(1)$ becomes$$1+2+3+4+5\\ldots+n=\\dfrac n2+\\dfrac {n^2}2=\\dfrac {n(n+1)}{2}$$\n\nWith the Euler-Maclaurin summation formula, one can easily derive that\n\n$$\\sum_{k=1}^nk=\\int_0^nx\\ dx+\\frac12n=\\frac12n^2+\\frac12n$$\n\nMore generally, one may derive that\n\n$$\\sum_{k=1}^n k^p = {1 \\over p+1} \\sum_{j=0}^p (-1)^j{p+1 \\choose j} B_j n^{p+1-j},\\qquad \\mbox{where}~B_1 = -\\frac{1}{2}$$\n\nwhere we use the Bernoulli numbers. This more general formula is more commonly known as Faulhaber's formula.\n\nAnother \"picture proof\" I just thought of... but without a picture, since I can't draw:\n\nSuppose you want to add up all the integers from 1 to n. Then draw n rows on a board, put 1 unit in the first row, 2 in the second, and so on. If you draw a right triangle of height and length n to try and contain this shape, you will cut off half of each unit on the diagonal. So let's add all of these up; you have $$n^2/2$$ units inside the triangle, and you have $$n/2$$ units cut off on the diagonal (there are n squares on the diagonal, and half of each one is cut off). Adding these gives $$n^2/2 + n/2 = n(n+1)/2$$.\n\nIf you draw it out, it makes more sense, and I think it's geometrically a bit more straightforward than the other picture proof.\n\n(Side note: I have no idea how to write math on this site... I'll go consult the meta, I'm sure there's something there about it.)\n\nAll these proofs seem very complicated. The way I remember it is:\n\nThe sequence is: 1, 2, 3, ..... (n-2), (n-1), n.\n\nTaking the last and first term, 2nd and (n-2)th term and so on, we form n/2 pairs of (n+1). So the sum of n/2 pairs of (n+1) is n/2 * (n+1)\n\nExample: 1, 2, 3, 4, 5, 6 = (1+6) + (2+5) + (3+4) = 3x7 =21\n\nThis still holds for an odd number of terms\n\n\u2022 But this is not a proof. This is an explanation. \u2013\u00a0Asaf Karagila Sep 29 '12 at 12:00\n\u2022 @user929404: As Asaf points out, this is not a proof. You say that these proofs seem very complicated, but Joe's accepted answer for example uses the exact same idea as your explanation, but does so in a rigorous way which makes it a proof. Sometimes it is difficult to see the idea behind a proof, but it is also not enough to just give the idea; we need proof to be sure that our ideas actually work. \u2013\u00a0Michael Albanese Sep 30 '12 at 4:21\n\u2022 This seems like a valid proof to me, at least when $n$ is even. (When $n$ is odd, you do need to say that you have $(n-1)/2$ pairs that sum to $(n+1)$, together with an additional term of $(n+1)/2$.) \u2013\u00a0Jesse Madnick Dec 17 '12 at 19:41\n\u2022 As almost all things that call themselves proofs are not actually proofs, but rather outlines of proofs, this is certainly an acceptable outline of a proof. If you want to argue over what actually is a genuine proof, then you have to be ready to accept an almost inhumane level of pedantry and precision that only computers can realistically handle. \u2013\u00a0DanielV Dec 2 '14 at 11:01\n\u2022 This is how a young Gau\u00df did it, the story goes... \u2013\u00a0Chris Custer Sep 22 '18 at 13:01\n\nYou can also prove it by induction, which is nice and easy, although it doesn't give much intuition as to why it works.\n\nIt works for $1$ since $\\frac{1\\times 2}{2} = 1$.\n\nLet it work for $n$.\n\n$$1 + 2 + 3 + \\dots + n + (n + 1) = \\frac{n(n+1)}{2} + (n + 1) = \\frac{n(n+1) + 2(n+1)}{2} = \\frac{(n+1)(n+2)}{2}.$$\n\nTherefore, if it works for $n$, it works for $n + 1$. Hence, it works for all natural numbers.\n\nFor the record, you can see this by applying the formula for the sum of an arithmetic progression (a sequence formed by constantly adding a rate to each term, in this case $1$). The formula is reached pretty much using the method outlined by Carl earlier in this post. Here it is, in all its glory:\n\n$$S_n = \\frac{(a_1 + a_n) * n}{2}$$\n\n($a_1$ = first term, $a_n$ = last term, $n$ = number of terms being added).\n\nThis is the shortest proof (without words):\n\nis my proof showing that $$\\sum_{j=1}^n{j}=\\frac{n(n+1)}{2}.$$\n\nI would like to resubmit the first 'backwards and forwards' proof above. My issue with the proof is the use of \"$\\dots$\" which translates to and so on. Here is a more mathematically rigorous proof using properties of summation $\\Sigma$. $$2\\sum_{k=1}^{n} k =\\sum_{k=1}^{n} 2k = \\sum_{k=1}^{n} (k+k) = \\sum_{k=1}^{n}k + \\sum_{k=1}^{n}k = \\sum_{k=1}^{n}k+ \\sum_{k=1}^{n}(n-(k-1))= n(n+1)$$\n\n\u2022 If you want a rigorous proof using sigma, you should go full boar and use ${\\displaystyle \\sum _{n\\in B}f(n)=\\sum _{m\\in A}f(\\sigma (m))}$ from en.wikipedia.org/wiki/Summation \u2013\u00a0CopyPasteIt Jul 2 '17 at 1:56\n\nSome time ago I saw someone explain it as follows:\n\nThe average value of $$1,2,3,\\dots,n$$ is simply $$\\frac{n+1}2$$. Thus $$1+2+3+\\dots+n=\\frac{n(n+1)}2$$.\n\nOf course the proof behind this leads to Gauss's proof quite directly, but nonetheless I really like this restatement of it as it is easy to understand even if one does not know much math. And it quickly gives the sum of terms in arithmetic progression as well. Such a sum is simply the number of terms times the average of the first and last terms.\n\nI stumbled across this identity\n\n$$\\quad ab - 1 = (a - 1) (b - 1) + (a - 1) + (b - 1)$$\n\nwhile working with modulo arithmetic a couple of days ago. I was upset that I couldn't use it in any (really) interesting ways, but now I've discovered that I can 'pile onto' this question!\n\nIf $$n \\ge 1$$ is an integer we can write\n\n$$\\tag 1 n^2 = (n - 1)^2 + 2 (n - 1) + 1$$\n\nObserve that the first term of the rhs of $$\\text{(1)}$$ is the square of an integer, just like the lhs of $$\\text{(1)}$$. So you can use 'downward finite induction formula recursion' (not sure what to call this) and conclude that\n\n$$\\quad n^2 = [\\displaystyle 2 \\sum_{k=1}^{n-1}\\, k] + n$$\n\nand this even holds for $$n = 1$$.\n\nRewriting the formula we obtain\n\n$$\\quad \\displaystyle \\frac{n(n-1)}{2} = \\sum_{k=1}^{n-1}\\, k$$", "date": "2020-08-08 15:16:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2260/proof-1234-cdotsn-fracn-timesn12?noredirect=1", "openwebmath_score": 0.9854366779327393, "openwebmath_perplexity": 327.80113531866834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754447499795, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8605949777138324}}
{"url": "http://blog.etestseries.in/canopy-growth-blpoxjy/corbett-maths-probability-trees-e850ef", "text": "# corbett maths probability trees\n\nThe Corbettmaths Practice Questions on Venn Diagrams answers \u2013 Corbettmaths. Some questions to delve a little deeper into the understanding of probability and tree diagrams. Probability Trees. Download the medium term plan by clicking on the button above. \"The teacher selects two students at random to go on a trip. I want to help you achieve the grades you (and I) know you are capable of; these grades are the stepping stone to your future. One ball is picked out, and not replaced, and then another ball is picked out. Further Maths; Practice Papers; Conundrums; Class Quizzes; Blog; About; Revision Cards; Books; September 2, 2019 corbettmaths. Use the fact that probabilities add up to 1 to work out the probabilities of the missing branches. is written alongside the line. 1) Complete the probability tree diagram. A card game in which students determine an optimal strategy for playing the game. Click here for Answers . A running club has 160 members. Instructions Use black ink or ball-point pen. There are 6 girls and 5 boys to choose from. To be able to assign a probability to each number, an experiment would need to be conducted. Fill in the boxes at the top of this page with your name, centre number and candidate number. Tree Diagrams Practice Questions Click here for Questions . Mathster; Corbett Maths Tree diagrams are a way of showing combinations of two or more events. corbettmaths exam style questions linear graphs / corbettmaths exam style questions direct and inverse proportion / corbettmaths exam style questions answers / corbettmaths exam papers / corbettmaths exam style questions simultaneous equations / corbettmaths exam style questions probability / corbettmaths exam style questions equation of a line answers / corbettmaths exam style questions \u2026 The first two are fairly standard (I)GCSE fare but subsequent questions become more complex with questions 5 and 6 requiring the solution of quadratic equations if an algebraic approach is used. PROBABILITY & TREE DIAGRAMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. For example, the probability of rolling a 6 on a die will not affect the probability of rolling a 6 the next time. Aim for at least 8%. Maths Genie \u00a6 Corbett Maths \u00a6 Mr Barton Maths Takeaway \u00a6 Mr Barton Maths Topic Search \u00a6 Just Maths 13.1 Calculating Probability - Calculate simple probabilities from equally likely events - Understand mutually exclusive and exhaustive outcomes Just Maths Probability \u2013 F \u2013 Probability 1 v3 \u00a6 Probability \u2013 F \u2013 Probability 1 v3 \u2013 SOLUTIONS Probability \u2013 F \u2013 Probability 2 v3 \u00a6 Probability \u2013 F \u2013 Probability 2 \u2026 They have kindly allowed me to create 3 editable versions of each worksheet, complete with answers. PROBABILITY & TREE DIAGRAMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. 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Child 2 nd Child 1 ) complete the tree diagram 3 balls a.", "date": "2023-02-03 07:40:56", "meta": {"domain": "etestseries.in", "url": "http://blog.etestseries.in/canopy-growth-blpoxjy/corbett-maths-probability-trees-e850ef", "openwebmath_score": 0.47701507806777954, "openwebmath_perplexity": 2303.434547280963, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9458012640659995, "lm_q2_score": 0.9099070090919013, "lm_q1q2_score": 0.8605911993816331}}
{"url": "https://eecsmt.com/graduate-school/exam/108-nthu-cs-ds/", "text": "1. (10%) Use generating functions to answer the following questions.\n\n(A) Find the solution of the recurrence relation $a_n = 4a_{n-1} \u2013 3a_{n-2} + 2^n + n + 3 \\text{ with } a_0 = 1 \\text{ and } a_1 = 4.$\n(B) Find the coefficient of $x^{10}$ in the power series of $x^4 / (1 \u2013 3x)^3.$\n\n(A) $a_n = -4 \\cdot 2^n + \\dfrac{39}{8} \\cdot 3^n + \\dfrac{19}{8} \u2013 \\dfrac{7}{4}(n+1) \u2013 \\dfrac{1}{4}(n+2)(n+1), \\, n \\geq 0.$\n(B) $\\binom{3+6-1}{6} \\cdot 3^6 = 28 \\cdot 3^6 = 20412$\n\n2. (10%) How many relations are there on a set with n elements that are\n\n(A) both reflexive and symmetric?\n(B) neither reflexive nor irreflexive?\n\n(A) $2^{\\frac{n(n-1)}{2}}$\n(B) $2^{n^2} \u2013 2^{n(n-1)+1}$\n\n3. (5%) How many nonisomorphic unrooted trees are there with five vertices?\n\n4. (5%) Multiple answer question. (It is possible that more than one of the choices are correct. Find out all correct choices.)\n\nA hash table of length 10 uses the hash function $h(k) = k \\, mod \\, 10$ and the linear probing for handling overflow. After inserting 6 values into an initially empty hash table, the table is as shown below. Which one(s) of the following choices gives a possible order in which the key values could have been inserted in the table?\n\n(A) 46, 42, 34, 52, 23, 33\n(B) 34, 42, 23, 52, 33, 46\n(C) 46, 34, 42, 23, 52, 33\n(D) 42, 46, 33, 23, 34, 52\n(E) 42, 23, 34, 46, 52, 33\n\n5. (5%) Fill in the six black (I, II, \u2026, and VI) in the following program that implements a queue by using 2 stacks.\n\nclass MyQueue<T> {\n\nprivate:\nstack<T> stack1;\nstack<T> stack2;\n\npublic:\nMyQueue()\n{\nstack1 = new stack<T>();\nstack2 = new stack<T>();\n}\n\n// enqueue(): Add an element at the rear side of MyQueue\nvoid enqueue(T, e)\n{\nstack1.push(e);\n}\n\n// dequeue(): Remove the front element from MyQueue\nT dequeue(T, e)\n{\nif((__I__).isEmpty())\nwhile(!(__II__).isEmpty())\n(__III__).push((__IV__).pop());\n\nT temp = null;\n\nif(!(__V__).isEmpty())\ntemp = (__VI__).pop();\n\nreturn temp;\n}\n}\n\n6. (5%) AVL Tree.\n\n(A) Please draw how an initially-empty AVL tree would look like after sequentially inserting the integer keys 100, 200, 50, 300, 400. There is no need to show it in a step-by-step fashion; you only need to draw the final result.\n(B) Continue the previous sub-problem. Suppose that the integer keys 25, 250, 225, 500, 240, 260 are sequentially inserted into the AVL tree of the previous sub-problem. Draw the AVL tree after all of these integer keys are inserted.\n\n(A)\n\n(B)\n\n7. (5%) Reconstruct and draw the maximum binary heap whose in-order traversal is 2, 16, 7, 62, 5, 9, 188, 14, 78, 10. There is no need to show it in a step-by-step fashion; you only need to draw the final result.\n\n8. (5%) The following algorithm takes an array as input and returns the array with all the duplicate elements removed. For example, if the input array is {1, 3, 3, 2, 4, 2}, the algorithm returns {1, 3, 2, 4}.\n\nS = new empty set\nA = new empty dynamic array\n\nfor every element x in input array\nif not S.member(x) then\nS.insert(x)\nA.append(x)\n\nreturn A\n\nSuppose that the input array has n elements. What is the Big-O complexity of this algorithm, if the set S is implemented as:\n(A) a hash table (with the assumption that overflow does not occur)?\n(B) a binary search tree?\n(C) an AVL tree?\n\n9. (10%) The recurrence $T(n) = 7T(\\dfrac{n}{2}) + n^2$ describes the running time of an algorithm A. A completing algorithm A\u2019 has a running time of $T'(n) = aT'(\\dfrac{4}{n}) + n^2.$ What is the largest integer value for $a$ such that A\u2019 is asymptotically faster than A?\n\n10. (15%) Consider the following undirected graph G = (V,E).\n\n(A) Draw the process of finding a minimum spanning tree using Kruskal\u2019s algorithm.\n(B) Draw the process of solving the single-source shortest path problem with node n1 as the source vertex using Dijkstra\u2019s algorithm.\n(C) Starting from n1, find the Depth-First Search (DFS) traversal sequence of G (the priority of node is inversely proportional to the weight of incident edge).\n\n(A)\n\n(B)\n\n(C) $n1 \\to n4 \\to n6 \\to n5 \\to n2 \\to n3$\n\n11. (18%) Given an ordered file with keys 1, 2, \u2026, 16, determine the number of key comparisons made by a search algorithm A while searching for a specific key K.\n\n(A) A is the binary search algorithm and K is 2.\n(B) A is the binary search algorithm and K is 10.\n(C) A is the binary search algorithm and K is 15.\n(D) A is the Fibonacci search algorithm and K is 2.\n(E) A is the Fibonacci search algorithm and K is 10.\n(F) A is the Fibonacci search algorithm and K is 15.\n\n12. (7%) Given a store of n items, what\u2019s is the least upper bound (in Big-O notation) of the running time of the solutions to the following problems:\n\n(A) Fractional knapsack problem;\n(B) General 0/1 knapsack problem.\n\n### 8 \u7559\u8a00\n\n1. #### \u6d41\u52d5\u6027\n\n\u60a8\u597d\uff0c\u6211\u60f3\u77e5\u9053\u7b2c11\u984c\u7684DEF\u5c0f\u984c\u60a8\u5728fibonacci search\u7684\u6c42\u89e3\u904e\u7a0b\u662f\u5982\u4f55\u5462?\u56e0\u70ba\u6211\u5728\u7db2\u8def\u4e0a\u722c\u6587\u6709\u627e\u5230\u4e00\u7bc7\u8a0e\u8ad6\u7684\u7b54\u6848\u548c\u9019\u88e1\u4e0d\u4e00\u6a23\uff0c\u8b1d\u8b1d!\n\u4ee5\u4e0b\u662f\u6211\u627e\u5230\u7684\u8a0e\u8ad6:\n\n\u2022 \u6587\u7ae0\u4f5c\u8005\u7684\u7559\u8a00\n\n#### mt\n\nFibonacci search\u7684\u6b65\u9a5f\u7db2\u8def\u4e0a\u771f\u7684\u6709\u5f88\u591a\u7248\u672c\uff0c\u89e3\u51fa\u4f86\u7684\u7b54\u6848\u4e5f\u90fd\u4e0d\u592a\u4e00\u6a23\uff0c\u6211\u81ea\u5df1\u662f\u7528\u9019\u90e8\u5f71\u7247\u7684\u65b9\u6cd5\u3002\u6240\u4ee5\u7b54\u6848\u4e0d\u540c\u6709\u53ef\u80fd\u662f\u56e0\u70ba\u65b9\u6cd5\u4e0d\u4e00\u6a23\uff0c\u53ef\u4ee5\u53bb\u770b\u770b\u90a3\u6240\u5b78\u6821\u7684\u5927\u5b78\u90e8\u4e0a\u8ab2\u7684\u6642\u5019\u7684\u7528\u66f8\uff0c\u7136\u5f8c\u770b\u90a3\u672c\u66f8\u662f\u7528\u4ec0\u9ebc\u65b9\u6cd5\uff0c\u9019\u6a23\u6bd4\u8f03\u4fdd\u96aa\uff0c\u52a0\u6cb9\u5594\uff01\n\n2. #### w\n\n\u7b2c\u4e03\u984cmax heap\u5716\u662f\u4e0d\u662f\u932f\u4e86\n\n3. #### TNF\n\n\u8acb\u554f\u7b2c\u516b\u984c\u7684A\u9078\u9805\n\u984c\u76ee\u662f\u5047\u8a2d\u6c92\u6709overflow\u7684\u767c\u751f\n\u4f46\u5982\u679c\u5e7e\u4e4e\u5168\u90e8\u90fdcollision\u4e14\u63a1linear probing\n\u5224\u65b7\u5143\u7d20\u662f\u5426\u5728hash table\u7684\u6bd4\u8f03\u6b21\u6578\u8b8a\u62101+2+\u2026+(n-1)\n\u662f\u5426\u6700\u5dee\u6642\u9593\u8907\u96dc\u5ea6\u5c31\u662fO(n^2)\u4e86\u5462\n\u53e6\u5916O(n)\u662f\u5e73\u5747\u72c0\u6cc1\u55ce??\n\n\u2022 \u6587\u7ae0\u4f5c\u8005\u7684\u7559\u8a00\n\n#### mt\n\n\u6211\u9019\u984c\u662f\u5047\u8a2d\u8aaa\u6bcf\u500bbucket\u53ea\u6709\u4e00\u500bslot\u7136\u5f8c\u984c\u76ee\u53c8\u8aaa\u6c92\u6709overflow\uff08\u7b49\u65bc\u6c92\u6709collision\uff0c\u56e0\u70ba\u53ea\u6709\u4e00\u500bslot\u6642\uff0ccollision\u7b49\u65bcoverflow\uff09\uff0c\u6240\u4ee5\u6211\u624d\u5bebO(n)\u3002\u7136\u5f8c\u5c31\u7b97collision\u7684\u8a71\u4e5f\u4e0d\u6703overflow\u6240\u4ee5\u4e0d\u6703\u6709\u63a1\u7528\u54ea\u7a2eprobing\u7684\u554f\u984c\uff0c\u4e0d\u904e\u9019\u984c\u4e5f\u53ef\u4ee5\u60f3\u6210\u662f\u7528chaining\uff0c\u9019\u6a23\u7684\u8a71\u662fO(n^2)\u6c92\u932f\u3002\u5e0c\u671b\u6709\u56de\u7b54\u5230\u4f60\u7684\u554f\u984c\uff5e\n\n4. #### aaa\n\n\u55e8\uff0c\u7b2c10\u984c(b)\u7684\u7b2c2\u500biteration\u7684node n3\u61c9\u8a72\u662f75\u5427\uff1f\u56e0\u70ba\u7b2c1\u500biteration\u662f\u9078node n4\uff0c\u6240\u4ee5\u66f4\u65b0n3\u5f9e\u7121\u9650\u5927->75", "date": "2022-05-26 12:08:05", "meta": {"domain": "eecsmt.com", "url": "https://eecsmt.com/graduate-school/exam/108-nthu-cs-ds/", "openwebmath_score": 0.32648780941963196, "openwebmath_perplexity": 1058.9519644595746, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918516137419, "lm_q2_score": 0.8705972768020107, "lm_q1q2_score": 0.8605783141559009}}
{"url": "https://math.stackexchange.com/questions/621501/cardinality-of-all-sequences-of-non-negative-integers-with-finite-number-of-non", "text": "# Cardinality of all sequences of non-negative integers with finite number of non-zero terms. (NBHM 2012)\n\nConsider the set $S$ of all sequences of non-negative integers with finite number of non-zero terms.\n\n1. Is the set $S$ countable or not?\n2. What is the cardinality of the set $S$ if it is not countable?\n\nMy intuition is the set is countable. The sequence has only finitely many non-zero terms. For any fixed $N$ consider the set $A_N$ which contains all sequences $\\{a_n\\}$ s.t. $a_k = 0$ $\\forall$ $k >N$. The set $A_N$ is countable as the first $N$ terms of a sequences can be filled up by non-negative integers in a countable number of ways. So $A_N$ is countable and $S$ is a countable union of countable sets. So $S$ is countable.\n\nI do not know if it is true or false. If it is false please identify the mistake. Thank you for your help.\n\nPlease suggest me a book where I shall get sufficient number of such type of problem to clear basic ideas on cardinal number.\n\n\u2022 Would you add \"integer\" before \"numbers\" in the title and in the first sentence? Your intuition is correct. \u2013\u00a0egreg Dec 29 '13 at 17:09\n\u2022 You can try Hrbacek, Jech - Introduction to set theory \u2013\u00a0Giulio Bresciani Dec 29 '13 at 17:27\n\nYou can think a sequence as a finite subset of $\\mathbb{N}^2$: for all $a_n\\neq 0$, take the point $(n,a_n)$. This way, you can inject $S$ in the set $\\mathcal{P'}(\\mathbb{N}^2)$ (with $\\mathcal{P'}$ I mean the set of finite subsets) and this is in bijection with $\\mathcal{P'}(\\mathbb{N})$. But $\\mathcal{P'}(\\mathbb{N})$ is countable, because it is a countable union of countable sets (subsets with $0$ elements, subsets with $1$ element, subsets with $2$ elements...).\n\n\u2022 Thank you for the answer supporting my approach. \u2013\u00a0Dutta Dec 30 '13 at 1:46\n\nLet $\\mathbb{N}$ be the set of non-negative integers. If $s=s_0,s_1,s_2,\\dots,s_n,\\dots$ is a sequence in $S$, define $\\psi(s)$ by $$\\psi(s)=\\left(\\prod_{i=0}^\\infty p_i^{s_i}\\right)-1,$$ where $p_i$ is the $i$-h prime. By the Unique Factorization Theorem, $\\psi$ is a bijection from $S$ to $\\mathbb{N}$.\n\n\u2022 Nicolas: Thank you for introducing a new concept. Let me know why you are subtracting 1 from the product in the definition of $\\phi(s)$. \u2013\u00a0Dutta Dec 30 '13 at 1:46\n\u2022 That is because I am using as $\\mathbb{N}$ the non-negative integers. If we are making a bijection to the positive integers, there is no $-1$ term. Of course it makes no real difference, since there is an obvious bijection between the non-negatives and the positives. \u2013\u00a0Andr\u00e9 Nicolas Dec 30 '13 at 1:57\n\u2022 It is clear now. \u2013\u00a0Dutta Dec 30 '13 at 2:02\n\u2022 I think the answer by Giulio Bresciani is more \"versatile,\" and therefore more worthy of accepting. Mine is a little too cute, too tailored to the specific situation. \u2013\u00a0Andr\u00e9 Nicolas Dec 30 '13 at 2:06\n\nPut a decimal point in front of any of these sequences and you have a rational number between 0 and 1. The set is countable. There is a bit more to it than just that. You will need to consider repetitions. You end up with a countable number of equivalence classes that are each at most countable. 1,1,0,0,0,0... becomes .110000... and 11,0,0,0,0... also becomes .110000...\n\n\u2022 Happy new year. This is also a nice answer. \u2013\u00a0Dutta Jan 1 '14 at 16:15", "date": "2019-07-24 06:43:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/621501/cardinality-of-all-sequences-of-non-negative-integers-with-finite-number-of-non", "openwebmath_score": 0.8665811419487, "openwebmath_perplexity": 166.19249341067956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226314280636, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8605586865229881}}
{"url": "https://math.stackexchange.com/questions/1488451/checking-whether-a-number-is-prime-or-composite", "text": "# Checking whether a number is prime or composite\n\nThis is a question that came up while I was doing an exercise. I ended up with the number\n\n$$200! + 1$$\n\nand I want it to be composite but I don't know of any methods to check whether a number is prime or not.\n\nIs there any general rule about $n+1$ or $n! + 1$ to determine if or when these are prime or composite?\n\nThe exercise I was doing when I ended up stuck at this question was this:\n\nShow that there exists $n \\in \\mathbb N$ such that $n, n+1 \\dots, n+200$ are all composite.\n\nI am hoping for a solution not using calculators, software or the internet. I expect there to be a short and (computationally) simple answer. At least that's what I hope.\n\n\u2022 $n+1$: definitely not, since every integer is of this form. \u2013\u00a0vadim123 Oct 20 '15 at 1:47\n\u2022 Wilson's theorem gives a useful result: if $n+1$ is prime then it divides $n!+1$. Unfortunately $201$ is not a prime so it does not apply here. \u2013\u00a0Winther Oct 20 '15 at 1:50\n\u2022 Wolfram was able to answer this, and gives $200!+1 = 1553\\times 826069\\times 353297821\\times k$ where $k$ is a very large prime number that won't fit on this page. \u2013\u00a0JMoravitz Oct 20 '15 at 1:52\n\u2022 There is an ongoing project here that searches for primes on that form (known as factorial primes). \u2013\u00a0Winther Oct 20 '15 at 1:56\n\u2022 If you use $210!$ instead of $200!$ you can apply Wilson's theorem since $211$ is prime. \u2013\u00a0Paul Hankin Oct 20 '15 at 4:04\n\nAs to the original question, primes of the form $n!\\pm 1$ are known as factorial primes and not all are known. It is in general a complicated question to determine if a number is prime or not, and only partial results are known. For example, if $n+1$ is prime then $n!+1$ is not.\n\nAs for the exercise which prompted this question, proving that there exists some $n$ such that $n,n+1,n+2,\\dots,n+200$ are all composite consider the following:\n\nSuppose we want to force each $n+i$ to be composite. If we want to force $2\\mid n$ and $3\\mid (n+1)$ and $5\\mid (n+2)$, etc... that would correspond to the system of congruencies:\n\n$\\begin{cases} n\\equiv 0\\pmod{2}\\\\ n+1\\equiv 0\\pmod{3}\\\\ n+2\\equiv 0\\pmod{5}\\\\ \\vdots\\\\ n+200\\equiv 0\\pmod{p_{201}}\\end{cases}$\n\nConsider then the Chinese Remainder Theorem.\n\nThe Chinese remainder theorem states that we can find such an $n$ that satisfies all of those congruencies since each of what we are modding out by are relatively prime to one another in every case.\n\nNote: there is nothing intrinsically special about ordering these as being modulo $2$ followed by $3$ followed by $5$, etc... So long as we pick a list of length 200 where each of the entries on the list are coprime to one another, this will work.\n\nEdit: Minor missing detail. It is possible that $n+i=p_i$ in one of those cases. To account for this possibility, technically chinese remainder will give us a solution to $n\\equiv k\\pmod{\\prod p_i}$, so we can avoid this by instead of taking the smallest positive integer $n$ that works, by instead taking $n+\\prod p_i$.\n\n\u2022 So the question isn't asking you for a specific $n$? The Chinese Remainder Theorem solves this, as we're only looking for the existence of such an $n$, just put the numbers you're adding by, on the other side, because there are infinitely many primes, which are definitely coprime to one another. \u2013\u00a0Almentoe Oct 20 '15 at 2:32\n\u2022 That is the way it is worded. In fact, the same proof can be modified to show the existence of arbitrarily long prime gaps. If you wanted to find an exact value for n, it could be done but would be incredibly tedious to do. \u2013\u00a0JMoravitz Oct 20 '15 at 2:41\n\nAs others have mentioned, it is not easy in general to check whether numbers of the form $n! + 1$ are prime. But as you may have noticed, the numbers $200! + 2$, $200! + 3, \\dots, 200! + 200$ are all composite, as they are divisible by the numbers $2, 3, \\dots, 200$ respectively. This gives only 199 consecutive composite numbers rather than the 201 required by your problem, but that's nothing that can't be fixed by increasing $n$ a little.\n\n\u2022 Or noting that $200! + 201$ and $200! + 202$ are composite since they're divisible by 3 and 2 respectively. \u2013\u00a0Paul Hankin Oct 20 '15 at 4:56", "date": "2020-08-06 12:18:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1488451/checking-whether-a-number-is-prime-or-composite", "openwebmath_score": 0.7182556390762329, "openwebmath_perplexity": 136.67351527536294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022627413796, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8605586784015001}}
{"url": "https://mathhelpboards.com/threads/how-to-use-taylor-series-to-represent-any-well-behaved-f-x.4960/", "text": "# [SOLVED]How to use Taylor series to represent any well-behaved f(x)\n\n#### DeusAbscondus\n\n##### Active member\nDoes one assess $x$ at $x=0$ for the entire series? (If so, wouldn't that have the effect of \"zeroing\" all the co-efficients when one computes?)\nonly raising the value of $k$ by $1$ at each iteration?\nand thereby raising the order of derivative at each iteration?\n\n$$\\sum_{k=0}^{\\infty}\\frac{f^{k}(0)}{k!} x^k= f(0)+\\frac{df}{dx}|_0 \\ x + \\frac{1}{2!}\\frac{d^2f}{dx^2}|_0 \\ x^2+ \\frac{1}{3!}\\frac{d^3f}{dx^3}|_0 \\ x^3+ ....$$\n\nI have no experience with series or sequences, so, I know I have to remedy this gap in my knowledge.\nIn the interim, however, I am currently enrolled in a Math course that looks at Calculus by beginning with Taylor series.\nI am an adult beginner at Math, having done an introductory crash course in Calculus last year; I wanted to flesh this out: hence my current enrolment.\n\nBut I am at a loss to know how to manipulate the notation above and would appreciate a worked solution for some simple $f(x)$ (I won't nominate one, so as to preclude the possibility of my cheating on some set work)\nI just need to see this baby in action with a \"well-behaved function\" of someone else's choosing, with some notes attached if that someone would be so kind.\n\nThanks,\nDeo Abscondo\n\nLast edited:\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nRe: How to use Taylor series to represent any well-behaved $f(x)$\n\nYes, the idea is that as you add more and more terms to the Taylor series, the series approximation becomes better and better and fits the function more closely for values farther away from $x = 0$.\n\nFor instance, let's take the venerable $y = \\sin(x)$ function. Let's plot it:\n\nThe first term of the Taylor series (i.e. the Taylor series approximation at $k = 0$) is just $f(0) = 0$, hence:\n\nOf course, this approximation sucks. Let's try $k = 1$. Then the Taylor series approximation for $k = 1$ is:\n$$\\sum_{k = 0}^1 \\frac{f^k (0)}{k!} x^k = \\sin(0) + x \\cos(0) = x$$\nAnd our \"first-order approximation\" for $\\sin(x)$ is the curve $y = x$, as illustrated below:\n\nStill not an awesome approximation, but it works pretty well for $x$ close to zero. In fact this is known as the small-angle approximation which says that $\\sin(x) \\approx x$ for small $x$.\n\nWhat about the second order approximation $k = 2$, which is given by:\n$$\\sum_{k = 0}^2 \\frac{f^k (0)}{k!} x^k = \\sin(0) + x \\cos(0) - \\frac{sin(0)}{2!} x^2 = x$$\nIt turns out that the new term becomes zero. Ok.. fine.. that happens, so what about $k = 3$? Now we see that:\n$$\\sum_{k = 0}^3 \\frac{f^k (0)}{k!} x^k = x - \\frac{\\cos(0)}{3!} x^3 = x - \\frac{x^3}{3!}$$\nAnd let's plot this:\n\nWow! That's a really good approximation for all $|x| < 1$. And we see a pattern: repeatedly differentiating $\\sin(x)$ will end up giving you $\\cos(x)$, $- \\sin(x)$, $- \\cos(x)$, $\\sin(x)$ endlessly. Every second term will become zero because of the $\\sin(0)$ term, so only odd-numbered terms actually matter. We conclude that:\n$$\\sin(x) = \\sum_{k = 0}^\\infty \\frac{f^k (0)}{k!} x^k = x - \\frac{x^3}{3!} + \\frac{x^5}{5!} - \\cdots$$\nHere's the Taylor series approximation for $k = 7$. Check it out:\n\nAnd here's the one for $k = 9$:\n\nAs you add more and more terms, the approximation tends towards the function for larger and larger values of $x$. In the limit, the Taylor series is equal to the original function.\n\nSo in other words, a Taylor series is another representation of a function, as an infinite series (a sum of infinitely many terms). A Taylor series approximation is the Taylor series truncated at a finite number of terms, which has the nice property of approximating the function around $x = 0$, and is often easier to calculate and work with, especially in physics where approximations are often used.\n\nThere is in fact a theorem that gives the maximal error of the Taylor series approximation at any point $x$ of the function in terms of $k$. Of course, as $k$ tends to infinity, the error tends to zero. This is Taylor's Theorem.\n\nEDIT: uploaded images to imgur for perennity. W|A does strange things to externally linked images.\n\nLast edited:\n\n#### DeusAbscondus\n\n##### Active member\nRe: How to use Taylor series to represent any well-behaved $f(x)$\n\nThanks kindly Bacterius.\n\nBut how would this work for a polynomial?\nI mean, wouldn't the evaluation $x=0$ lead to \"zeroing\" all the co-efficients and resulting in non-sense?\n\nIf I take the case of $y=x^2+2x$ is this amenable to being described by the same series?\n$$\\sum_{k=0}^{\\infty}\\ \\frac{f^k (0)}{k!}=0+ 2(0) + \\frac{1}{2!}(2\\cdot 0) .....$$\nThis doesn't seem to work! All I get is an infinite string of 0s!\n\nSo in other words, a Taylor series is another representation of a function, as an infinite series (a sum of infinitely many terms). A Taylor series approximation is the Taylor series truncated at a finite number of terms, which has the nice property of approximating the function around $x = 0$, and is often easier to calculate and work with, especially in physics where approximations are often used.\n\nThere is in fact a theorem that gives the maximal error of the Taylor series approximation at any point $x$ of the function in terms of $k$. Of course, as $k$ tends to infinity, the error tends to zero. This is Taylor's Theorem.\n\nLast edited:\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nRe: How to use Taylor series to represent any well-behaved $f(x)$\n\nTaylor series of polynomials have the interesting property that they are in fact equal to the polynomial itself, and all extra terms are zero. In other words, the Taylor series for $x^2 + 2x$ is $x^2 + 2x + 0 + 0 + \\cdots$. To see why:\n$$f(x) = x^2 + 2x$$\n$$f'(x) = 2x + 2$$\n$$f''(x) = 2$$\n$$f'''(x) = 0$$\n.. and any further differentiation still gives zero\n\nSo:\n$$\\sum_{k = 0}^\\infty \\frac{f^{(k)} (0)}{k!} x^k = \\frac{0^2 + 2 \\cdot 0}{0!} x^0 + \\frac{2 \\cdot 0 + 2}{1!} x^1 + \\frac{2}{2!} x^2 + 0 + \\cdots = 0 + 2x + x^2 + 0 + \\cdots = x^2 + 2x$$\nSo, yes, it still works, although it would seem to be less useful for polynomials than for other functions (but then, polynomials are easy to compute and are already pretty simple. I don't think \"reducing\" them to infinite series generally helps).\n\nTo calculate Taylor series I recommend you write down the iterated derivative of your function $f(x)$ and then plug in the numbers. Remember, $f^{(k)} (0)$ means \"the $k$th derivative of $f$ evaluated at $x = 0$\". The $x$ in your Taylor series is a different $x$ and is *not* equal to zero.\n\nAlso, there is a generalization of Taylor's series which is centered on arbitrarily values of $x$ instead of $x = 0$. I'll let you work out the general expression, though, it's an interesting exercise.\n\n#### DeusAbscondus\n\n##### Active member\nRe: How to use Taylor series to represent any well-behaved $f(x)$\n\nThat has nailed it for me Bacterius!\nMightily obliged to you.\n\nAs per usual, the problem has vanished under the gaze of fresh eyes.\n\n(Going now to chew on this strong meat with a cup of medicinal wine \"to aid the digestion\")\n\nCheers,\nD'abs\n\nTaylor series of polynomials have the interesting property that they are in fact equal to the polynomial itself, and all extra terms are zero. In other words, the Taylor series for $x^2 + 2x$ is $x^2 + 2x + 0 + 0 + \\cdots$. To see why:\n$$f(x) = x^2 + 2x$$\n$$f'(x) = 2x + 2$$\n$$f''(x) = 2$$\n$$f'''(x) = 0$$\n.. a.\n\n#### DeusAbscondus\n\n##### Active member\nRe: How to use Taylor series to represent any well-behaved $f(x)$\n\nThe x in your Taylor series is a different x and is *not* equal to zero.\nOkay, basically, I'm in the clear .....\nbut just what is this \"different x\"?\n\nIf it is distinct from the $x$ of my polynomial, by what principle/rule do I distinguish the two when I come to compute the polynomial?\n\nD'abs\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nRe: How to use Taylor series to represent any well-behaved $f(x)$\n\nOkay, basically, I'm in the clear .....\nbut just what is this \"different x\"?\n\nIf it is distinct from the $x$ of my polynomial, by what principle/rule do I distinguish the two when I come to compute the polynomial?\n\nD'abs\nThe $x$ in the Taylor series is the $x$ at which you are evaluating your Taylor series (or your approximation of it). The derivative is always evaluated at 0 for this version of the Taylor series (sorry I agree it was a bit confusing, there is only one $x$, it's just the derivative is evaluated at a constant).\n\nEssentially you are numerically evaluating derivatives of your original function, and using the resulting values as coefficients for your Taylor series (this is not a particularly useful way to think of it but it may be more intuitive to you to understand what is what)\n\n#### DeusAbscondus\n\n##### Active member\nRe: How to use Taylor series to represent any well-behaved $f(x)$\n\nThat makes sense.\nMeanwhile, I've been plugging and chugging a few simple polynomial functions through the T. series: hehe! what fun! it works!!! (*broadly grins)\n\nYou made my day: not understanding was getting me down (as usual);\nand, again as usual, once some fresh light comes, and the aha! moment arrives, and one feels exuberant rather than dispirited.\n\nHave a great weekend Bacteriu!\n\n[QUOTEEssentially you are numerically evaluating derivatives of your original function, and using the resulting values as coefficients for your Taylor series (this is not a particularly useful way to think of it but it may be more intuitive to you to understand what is what)[/QUOTE]", "date": "2021-09-26 21:32:02", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/how-to-use-taylor-series-to-represent-any-well-behaved-f-x.4960/", "openwebmath_score": 0.949510395526886, "openwebmath_perplexity": 391.1908091856933, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.977022625406662, "lm_q2_score": 0.880797081106935, "lm_q1q2_score": 0.8605586766336222}}
{"url": "https://cs.stackexchange.com/questions/85850/minimum-number-of-states-in-dfa-accepting-binary-number-with-decimal-equivalent/85855", "text": "# Minimum number of states in dfa accepting binary number with decimal equivalent divisible by $n$\n\nI was aware of the fact that, if DFA needs to accept binary string with its decimal equivalent divisible by $n$, then it can have minimum $n$ states.\n\nHowever recently came across following text:\n\n\u2022 If $n$ is power of $2$\nLet $n=2^m$, so number of minimum states $=m+1$. For $n=8=2^3$, we need $3+1=4$ states.\n\u2022 Else If $n$ is odd\nNumber of states $=n$. For $n=5$, we need $5$ states.\n\u2022 Else if $n$ is even\nKeep dividing it with $2$ until you get odd quotient. The result is final odd quotient plus number of divisions done. For $n=20,20/2=10,10/2=5\u21925+2=7$ states\n\nI was guessing:\nQ1. From where all these facts came. (I know this must have come from general pattern that such DFAs follow, but then I have below doubts)\nQ2. Are these points indeed correct?\nQ3. If yes (for Q2), then what can be the reason for each point?\nQ4. Do these three bullet points cover all cases? (I guess yes, since there is even and odd, but I am new to this, and I was guessing if any such more point is left unmentioned in my textbook, especially since I did not find any reference book (by Peter Linz and Hopcroft-Ullman) talking about this topic)\n\n\u2022 There are a few things missing here. First, is the number input LSB to MSB or MSB to LSB? (Looks like LSB to MSB.) Second, are you allowing the empty string to represent zero? Dec 23 '17 at 22:41\n\u2022 The way to prove such results is to twofold: first you prove an upper bound by constructing a DFA accepting the given language, then you prove a lower bound using Myhill\u2013Nerode theory. Dec 23 '17 at 22:42\n\u2022 For an example, see this: cs.stackexchange.com/questions/85785/\u2026. Dec 23 '17 at 22:53\n\u2022 Regarding Q4, yes, they cover all cases. I'm certain you can prove this in your own, even if you're new to \"this\" \u2013 this has nothing to do with automata theory. Dec 23 '17 at 22:55\n\u2022 Can you provide any link discussing these two points: (1) LSB to MSB and MSB to LSB number input (2) Representing zero by empty string and possibly these two points in the context of \"divisible by $n$ DFA\" . Also I guess first point wont make number of states in min DFA to change. Second point might change it.\n\u2013\u00a0anir\nDec 24 '17 at 7:02\n\n## Case 1: $n$ is a power of 2\n\nIf you wanted to check whether a decimal number is divisible by some power of 10, you can just look at the number of trailing zeros. For example, all numbers that are divisible by $100 = 10^2$ end with 2 zeros (this is of course including numbers ending with more than 2 zeros). The same idea can be applied here for binary numbers and powers of 2. Specifically to check for multiples of $n = 2^m$, you can simply check if the string ends in $m$ zeros. For example, if $n = 16 = 2^4$, all multiples of $n$ will end in 0000 (4 zeros). Thus we can create the following DFA for $n = 16$:\n\nThis construction can be extended to all multiples of 2, where you have $m+1$ states to ensure that your input ends in at least $m$ zeros (reading a 0 takes you one state forward, reading a 1 takes you back to the start). As suggested in the comments on your question, you can show that you can't do any better than this using a Myhill\u2013Nerode argument.\n\n## Case 2: $n$ is odd\n\nThe general idea here is that when a number is divided by $n$, the possible remainders are $0, 1, \\ldots, n-2$ or $n-1$. We can give our DFA one state for each of these possible remainders so as we process a string, we're keeping track of the remainder of what we've read so far and transitioning to the appropriate remainder based on what character we read. Then, we accept if we finish in the remainder 0 state.\n\nDesign DFA accepting binary strings divisible by a number 'n' walks through a detailed method for constructing such a DFA, and you can find out more about this by searching for \"DFA based division\". This gives a total of $n$ states, which is optimal by Myhill-Nerode because these remainders are exactly the equivalence classes given by the relation $\\equiv_L$ for the language $L = \\{\\text{binary strings divisible by }n\\}$.\n\n## Case 3: $n$ is even (and not a power of 2)\n\nThe technique in case 2 also works here but it isn't optimal. Every $n$ in this case can be expressed as $2^km$ where $m$ is odd (this is the division procedure described in your question). Thus, to check whether a binary string is divisible by $n$, we check whether it is divisible by both $2^k$ and $m$.\n\nWe know that it takes $k$ states to check divisibility by $2^k$ (case 1) and $m+1$ states to check divisibility by $m$ (case 2). We can take a DFA for divisibility by $m$, unmark the accepting state and make it the start state for a DFA for divisibility by $2^k$, giving our final DFA $k+m$ states.\n\nAs an example, here's the DFA for $n = 6$:\n\nThe top three states ensure that the number is a multiple of 3, and the final accept state ensures that it is a multiple of 2.\n\n\u2022 Revisiting this problem, I dont get how to combine two DFAs in case 3. You said \"unmark the accepting state and make it the start state for a DFA for divisibility by $2^k$\". But you didnt explain how you came up with transition outgoing (labeled 1) from final state of last DFA. If we are to combine \"divisible by $2^k$ DFA\" with \"divisible by $m$ ($m$ being any odd number)\" DFA, \"all\" outgoing transitions on reading \"1\" from \"all\" states of divisible by $2^k$ DFA will go to same second state of divisible by $m$ DFA, (Q1.) right? [to next comment...]\n\u2013\u00a0anir\nDec 3 '19 at 13:03\n\u2022 [..from earlier comment] For divisible by $2^k$, we need string to end with $k$ zeroes. But, your case 3, divisible by 6 DFA will also let $k=1$ zeroes appear anywhere. (Q2.) How this fits in divisible by 6 string? For example in your divisible by 6 DFA, for 110110, first 11 moves through divisible by 3 DFA, next 01 moves through divisible by 2 DFA, next 1 moves through divisible by 3 DFA and final 0 moves through divisible by 2 DFA. (Q3.) How running through different component DFAs one after other still ensures that the whole string will still be divisible by $2^km$\n\u2013\u00a0anir\nDec 3 '19 at 13:09", "date": "2021-09-24 20:47:47", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/85850/minimum-number-of-states-in-dfa-accepting-binary-number-with-decimal-equivalent/85855", "openwebmath_score": 0.6708168387413025, "openwebmath_perplexity": 436.0348354667693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226314280636, "lm_q2_score": 0.8807970748488296, "lm_q1q2_score": 0.8605586758229445}}
{"url": "http://mathhelpforum.com/geometry/227392-circle-questions-secants-chords-tangents.html", "text": "# Math Help - Circle questions -- Secants, chords and tangents\n\n1. ## Circle questions -- Secants, chords and tangents\n\nHello again, just got a few questions pertaining to circles. I'll just post one to start with;\n\n\"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).\n\nSo solving this.. according to the segments of secants and tangents theorem...\nx2 = (2r + 353)(353)\nx2 = (12742 + 353)(353)\nx2 = 4622535\nx = $\\sqrt{4622535}$ = 2150km (approx.)\n\nDid I do that right?\n\n2. ## Re: Circle questions -- Secants, chords and tangents\n\nOriginally Posted by StonerPenguin\nHello again, just got a few questions pertaining to circles. I'll just post one to start with;\n\n\"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).\n\nSo solving this.. according to the segments of secants and tangents theorem...\nx2 = (2r + 353)(353)\nx2 = (12742 + 353)(353)\nx2 = 4622535\nx = $\\sqrt{4622535}$ = 2150km (approx.)\n\nDid I do that right?\nComputation looks right to me. But simply memorizing theorems does not promote understanding.\n\nLet's see how we get that theorem. The radius through a point of tangency and the tangent at the same point are perpendicular.\n\n$Let\\ r = length\\ of\\ radius\\ of\\ given\\ circle,$\n\n$u + r = length\\ from\\ the\\ given\\ circle's\\ center\\ to\\ a\\ given\\ point\\ outside\\ the\\ circle,$\n\n$x = length\\ of\\ the\\ line\\ that\\ is\\ tangent\\ to\\ the\\ given\\ circle\\ and\\ runs\\ through\\ the\\ given\\ point.$\n\nBy the Pythagorean Theorem, we have:\n\n$(u + r)^2 = x^2 + r^2 \\implies u^2 + 2ru + r^2 = x^2 + r^2 \\implies x^2 = u^2 + 2ru = u(u + 2r) \\implies x = \\sqrt{u(u + 2r)}.$\n\n3. ## Re: Circle questions -- Secants, chords and tangents\n\nISS is6724 km above center of earth.At that point the angle between straight down and the visible horizon has a sin of 6371/6724 or 71.35 deg\ncos 71.35 =d/6724\nd km to horizon =2150 km\n\n4. ## Re: Circle questions -- Secants, chords and tangents\n\nHello, StonerPenguin!\n\nThe radius of the earth is approximately 6371 km.\nIf the international space station (ISS) is orbiting 353 km above the earth,\nfind the distance from the ISS to the horizon (x).\n\nCode:\no\n|\\\n| \\\n352 |  \\\n|   \\ x\n|    \\\n* * *   \\\n*     |     *\\\n*       |       o\n*    6371|     *  *\n|   *6371\n*         | *       *\n*         o         *\n*                   *\n\n*                 *\n*               *\n*           *\n* * *\nNote the right triangle.\n\nThe equation is: . $x^2 + 6371^2 \\:=\\:6723^2$\n\n5. ## Re: Circle questions -- Secants, chords and tangents\n\nThank you JeffM, bjhopper and Soroban! It's nice to see different perspectives and the image drawn in code is really cool.\nHere's another question I've had trouble with:\n\n\"Explain how you know $\\overline{AB}$ $\\overline{CD}$ given E is the center of the circle. (Include theorem numbers.)\"\n\nAnd here's some pertinent theorems;\nTheorem 10.4\nIf one chord is a perpendicular bisector of another chord, then the first chord is a diameter.\nTheorem 10.5\nIf a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.\nTheorem 10.6\nIn the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.\nObviously from the diagram $\\overline{AB}$ $\\overline{CD}$ by theorem 10.6, but I can't really word this well. Any help? Theorems and proofs are my weakest areas :/\n\n6. ## Re: Circle questions -- Secants, chords and tangents\n\nt is a triangle a is an angle\nt AED congruent to BEC isosceles t's same legs and equal altitudes\na DEC = 180- 2* 1 / 2 *AED\na AEB 180-2*1/2 * BEC a AED = aBEC\na AEB = a DEC\nAB =DC equal arcs = equal chords", "date": "2015-05-27 11:02:35", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/227392-circle-questions-secants-chords-tangents.html", "openwebmath_score": 0.6830456852912903, "openwebmath_perplexity": 3122.0125982177237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226327661525, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8605586754729518}}
{"url": "https://math.stackexchange.com/questions/1752912/is-this-inequality-true-for-all-k-sum-n-kn-infty-frac1n4-leq", "text": "Is this inequality true for all k ? $\\sum_{n=k}^{n=+\\infty} \\frac{1}{n^4} \\leq (\\sum_{n=k}^{n=+\\infty} \\frac{1}{n^2})^3$\n\nCan it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.\n\nI stumbled upon it randomly when trying to approximate $\\sum_{n=1}^{n=+\\infty} \\frac{1}{n^4}$.\n\nMy reasoning was :\n\n$$\\left(\\sum_{n=k}^{n=+\\infty} \\frac{1}{n^2}\\right)^2=\\sum_{n=k}^{n=+\\infty} \\frac{1}{n^4} + (\\text{double products}) \\geq\\sum_{n=k}^{n=+\\infty} \\frac{1}{n^4}$$\n\nSo\n\n$$\\sum_{n=1}^{n=+\\infty} \\frac{1}{n^4} \\leq \\sum_{n=1}^{n=k-1} \\frac{1}{n^4}+\\left(\\sum_{n=k}^{n=+\\infty} \\frac{1}{n^2}\\right)^2 \\leq \\left(\\sum_{n=1}^{n=k-1} \\frac{1}{n^4}\\right)+\\left(\\frac{1}{k-\\frac{1}{2}}\\right)^2$$\n\nwhere the last inequality comes from An inequality: $1+\\frac1{2^2}+\\frac1{3^2}+\\dotsb+\\frac1{n^2}\\lt\\frac53$.\n\nThen I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.\n\n\u2022 The LHS behaves like $\\int_{k}^{+\\infty}\\frac{dx}{x^4}=\\frac{1}{3k^3}$ while the RHS behaves like $\\left(\\int_{k}^{+\\infty}\\frac{dx}{x^2}\\right)^3=\\frac{1}{k^3}$, so that is not surprising. \u2013\u00a0Jack D'Aurizio Apr 21 '16 at 15:36\n\nFor $k > 1$, $$\\sum_{n=k}^\\infty \\dfrac{1}{n^4} < \\int_{k-1}^\\infty \\dfrac{dx}{x^4} = \\dfrac{1}{3(k-1)^3}$$\n\n$$\\left(\\sum_{n=k}^\\infty \\dfrac{1}{n^2}\\right)^3 > \\left(\\int_{k}^\\infty \\dfrac{dx}{x^2}\\right)^3 = \\dfrac{1}{k^3}$$\n\n$\\dfrac{1}{k^3} > \\dfrac{1}{3(k-1)^3}$ when $3(k-1)^3 > k^3$, which is true for $k > 3.2612$.\n\nLet's see what we can say about comparing $s_1(k) =\\sum_{n=k}^{\\infty} \\frac1{n^a}$ with $s_2(k) =\\left(\\sum_{n=k}^{\\infty} \\frac1{n^b}\\right)^c$ for large enough $k$, where $a > 1$ and $b > 1$ so the sums converge.\n\nUsing the integral approximation, $s_1(k) =\\sum_{n=k}^{\\infty} \\frac1{n^a} \\approx \\int_k^{\\infty} \\frac{dx}{x^a} =-\\frac1{(a-1)x^{a-1}}\\big|_k^{\\infty} =\\frac1{(a-1)k^{a-1}}$.\n\nTherefore $s_2(k) =\\left(\\sum_{n=k}^{\\infty} \\frac1{n^b}\\right)^c \\approx \\left(\\frac1{(b-1)k^{b-1}}\\right)^c =\\frac1{(b-1)^c k^{c(b-1)}}$, so $\\dfrac{s_1(k)}{s_2(k)} \\approx \\dfrac{\\frac1{(a-1)k^{a-1}}}{\\frac1{(b-1)^c k^{c(b-1)}}} = \\dfrac{{(b-1)^c k^{c(b-1)}}}{{(a-1)k^{a-1}}} = \\dfrac{{(b-1)^c }}{{(a-1)}}k^{c(b-1)-(a-1)}$.\n\nTherefore if $c(b-1) > a-1$, then $s_1(k) > s_2(k)$ for all large enough $k$; if $c(b-1) < a-1$, then $s_1(k) < s_2(k)$ for all large enough $k$;\n\nIf $c(b-1) = a-1$, then $\\dfrac{s_1(k)}{s_2(k)} \\approx \\dfrac{{(b-1)^c }}{{(a-1)}}$, so the result depends on this ratio.\n\nFor your case of $a=4$ and $b=2$, the key difference is $c(b-1)-(a-1) =c-3$.\n\nIf $c > 3$, then $s_1(k) > s_2(k)$ for large enough $k$; if $c < 3$, then $s_1(k) < s_2(k)$ for large enough $k$.\n\nIf $c=3$, which is your case, the ratio is $\\dfrac{(b-1)^c }{(a-1)} =\\dfrac{1}{3} < 1$, so $s_1(k) \\approx \\frac13 s_2(k) < s_2(k)$ for large enough $k$, which confirm's Robert Israel's result (good thing too, because any result of mine that differs from a result of his is probably wrong).", "date": "2019-09-23 13:35:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1752912/is-this-inequality-true-for-all-k-sum-n-kn-infty-frac1n4-leq", "openwebmath_score": 0.9804154634475708, "openwebmath_perplexity": 114.20214778844415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226347732859, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.8605586726550958}}
{"url": "https://math.stackexchange.com/questions/3792683/for-which-values-of-alpha-is-z-n-a-bounded-sequence", "text": "# For which values of $\\alpha$ is {$z_n$} a bounded sequence?\n\nWhere $$\\alpha$$ is a real constant, consider the sequence {$$z_n$$} defined by $$z_n=\\frac{1}{n^\\alpha}$$. For which value of $$\\alpha$$ is {$$z_n$$} a bounded sequence?\n\nHow do I start with this kind of question? I think that $$\\forall\\space \\alpha\\in\\Bbb{R}_{\\geq0}$$ the sequence is convergent and therefore bounded, but how do I write it out?\n\n\u2022 In order for us to tell you how to write things out, it would be helpful if you explained why you believe that the answer is what it is. Why do you think that $\\frac 1{n^{\\alpha}}$ converges for $\\alpha \\geq 0$? Are you saying that $\\frac 1{n^{\\alpha}}$ is not bounded when $\\alpha < 0$? If so, then you must say so explicitly in your answer. Also, why do you believe that this is the case? \u2013\u00a0Ben Grossmann Aug 16 '20 at 10:16\n\u2022 Because it is clear that for $\\alpha\\geq0$ the sequence converges to 0. If $\\alpha<0$ then the value of $\\frac{1}{n^\\alpha}$ will become very big unless $\\alpha>-\\frac{1}{n}$. I might be wrong, but this is what I think. I don't know how to approach this question. \u2013\u00a0Jess Aug 16 '20 at 10:20\n\u2022 By giving the answer that you have given, you have not only \"approached\" the problem correctly, but also have given an almost complete answer. It seems that your only question, then, is how to write this up with sufficient \"formality.\" \u2013\u00a0Ben Grossmann Aug 16 '20 at 10:25\n\n## 2 Answers\n\nIf $$\\alpha=0$$, $$(z_n)$$ is constant, hence bounded.\n\nIf $$\\alpha>0$$, $$(z_n)$$ converges to 0 and is thus bounded.\n\nIf $$\\alpha<0$$, $$(z_n)$$ diverges to $$+\\infty$$ and is thus unbounded.\n\nAs I state in the comment, you have the correct answer. The only remaining task is to give a formal explanation of the answer. One way to write an answer is as follows:\n\nFirst, we note that the function $$f: \\Bbb [1,\\infty) \\to \\Bbb R$$ defined by $$f(x) = x^{\\beta}$$ satisfies $$\\lim_{x \\to \\infty}f(x) = \\begin{cases} 0 & \\beta < 0\\\\ 1 & \\beta = 0\\\\ \\infty & \\beta > 0. \\end{cases}$$ I suspect that you do not need to prove this statement formally: it is likely that there is a statement in the textbook that you can refer to.\n\nWith that established, address the problem in $$3$$ cases: in the case that $$\\alpha < 0$$, conclude using the above fact that $$\\lim_{n \\to \\infty} z_n = \\infty$$, which means that the sequence is not bounded. In the case that $$\\alpha = 0$$, conclude that $$z_n \\to 0$$, which means that the sequence is convergent and is therefore bounded. Similarly, if $$\\alpha > 0$$, conclude that $$z_n \\to 0$$, which means that the sequence is convergent and therefore bounded.\n\nThus, we conclude that the sequence is bounded if and only if $$\\alpha \\geq 0$$.\n\n\u2022 If the sequence is bounded, does that mean that the limit of the sequence exists? Because I just noticed that the next question asks me to find the values of $\\alpha$ for which lim$_{n\\to\\infty}z_n$ exists. I assume the values of $\\alpha$ are the same as which $z_n$ is a bounded sequence? \u2013\u00a0Jess Aug 16 '20 at 10:40\n\u2022 It is not necessarily the case that a bounded sequence will have a limit. However, we can clearly see by exhausting the possibilities that for this problem, the sequence $z_n$ converges whenever it is bounded. \u2013\u00a0Ben Grossmann Aug 16 '20 at 10:42", "date": "2021-07-31 10:07:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3792683/for-which-values-of-alpha-is-z-n-a-bounded-sequence", "openwebmath_score": 0.9731109738349915, "openwebmath_perplexity": 97.92443506847972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8824278788223264, "lm_q1q2_score": 0.86054528965028}}
{"url": "https://math.stackexchange.com/questions/3743275/a-fair-coin-is-tossed-untill-head-appears-for-the-first-time-what-is-probabilit", "text": "# A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd? [duplicate]\n\nQ. A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd?\n\nMy work:\n\nsuppose that head comes in first toss so probability of getting head in the first toss $$=\\dfrac{1}{2}$$\n\nsuppose that first & second tosses show tails & third toss shows head so probability of getting head in the third toss $$=(1-\\dfrac12)(1-\\dfrac12)\\dfrac{1}{2}$$ $$=\\dfrac1{2^3}$$\n\nsuppose that first 4 tosses show tails & fifth toss shows head so probability of getting head in the fifth toss $$=(1-\\dfrac12)^4\\dfrac{1}{2}$$ $$=\\dfrac1{2^5}$$\n\nsuppose that first 6 tosses show tails & seventh toss shows head so probability of getting head in the fifth toss $$=(1-\\dfrac12)^6\\dfrac{1}{2}$$ $$=\\dfrac1{2^7}$$\n\n\u2026\u2026\u2026\u2026\u2026.\n\nand so on\n\nBut I am not able to find the final probability of getting head first time so that the number of tosses required is odd. what should do I next to it? please help me.\n\n\u2022 You're almost there. Just recognize that you have a geometric series and take its sum. To allow you to check your work, the answer will be $\\frac 23$. \u2013\u00a0Robert Shore Jul 3 at 4:06\n\u2022 This is an exact duplicate of math.stackexchange.com/q/834344/117057 , which unfortunately doesn't have an accepted answer. \u2013\u00a0shoover Jul 3 at 22:27\n\nYou have a geometric series,\n\n$$\\frac12+\\frac1{2^3}+\\frac1{2^5}+\\frac1{2^7}+\\ldots=\\sum_{n\\ge 0}\\frac12\\cdot\\left(\\frac14\\right)^n=\\frac{\\frac12}{1-\\frac14}=\\frac23\\;.$$\n\nAlternatively, if $$p$$ is the desired probability, then $$p=\\frac12+\\frac14p$$: with probability $$\\frac12$$ you get a head on the first toss, and with probability $$\\frac14$$ you start with two tails and are now in exactly the same position that you were in at the beginning. Solving this for $$p$$ again yields $$p=\\frac23$$.\n\n\u2022 Your second method is very simple. I have never thought of it. Hats off to you sir \u2013\u00a0user801303 Jul 3 at 4:44\n\u2022 @PaulAldrin: You\u2019re welcome. \u2013\u00a0Brian M. Scott Jul 3 at 4:44\n\u2022 @TShiong: When you can find a recurrence like that, it usually makes things a lot simpler! \u2013\u00a0Brian M. Scott Jul 3 at 4:45\n\nYour approach is good and will get you the right answer. Just realize that you're building a geometric series and you want its sum.\n\nI'm drafting an answer to give you an alternative approach. Let $$p$$ be the probability you're looking for. Then your first flip will be heads with probability $$0.5$$. If it's tails, then you will solve your original problem (the first heads occurs on an odd toss) exactly when, from your new starting point, your first heads occurs on an even toss, which happens with probability $$1-p$$.\n\nThat means $$p = 0.5 + 0.5(1-p) \\Rightarrow 1.5 p = 1 \\Rightarrow p = \\frac 23$$.\n\n\u2022 haven't you used sum of infinite GP? \u2013\u00a0user805532 Jul 3 at 4:38\n\u2022 No, I think I've provided an alternative approach that avoids the need, just as does the alternative solution provided by Brian M Scott. \u2013\u00a0Robert Shore Jul 3 at 8:35\n\nHere's another way to solve the problem by considering pairs of tosses at a time instead of single tosses.\n\nLet $$\\sigma$$ be an arbitrary infinite sequence of heads and tails. $$\\sigma$$ uses 1-based indexing.\n\n$$\\text{e.g.}\\;\\;\\; \\sigma = HTHTHTHTHTTTTTHHHH\\cdots$$\n\nImagine grouping the elements of $$\\sigma$$ into pairs.\n\n$$\\sigma = HT,HT,HT,HT,HT,TT,TT,HH,HH\\cdots$$\n\nLet's imagine we have three states, $$S$$, $$E$$, and $$O$$.\n\n\u2022 $$S$$ is the start state, we haven't seen a head yet.\n\u2022 $$E$$ is the state marking that we saw a head at an even index first.\n\u2022 $$O$$ is the state marking that we saw a head at an odd index first.\n\n$$E$$ and $$O$$ are both absorbing states. Once we enter one of those states, we are never going to leave it.\n\nOur state at the beginning of our process is always $$S$$ because, initially, we haven't observed any tosses at all of our coin.\n\nNext let's consider what happens when we read our first toss-pair from $$\\sigma$$.\n\nThere are twice as many ways to transition from $$S$$ to $$O$$ than there are to transition from $$S$$ to $$E$$.\n\n     TT\nS  ----->   S\n\nTH\nS  ----->   E\n\nHT\nS  ----->   O\n\nHH\nS  ----->   O\n\n\nAs the number of pairs processed approaches infinity, the probability that the current state is $$S$$ approaches zero.\n\nHowever, the ratio of the probability that that the current state is $$O$$ is always twice the probability that the current state is $$E$$.\n\nTherefore, the limiting probability that the state is $$O$$ is $$2/3$$\n\nYou are almost done. Add all the terms\n\n$$\\frac12+\\frac1{2^3}+\\frac1{2^5}+\\frac1{2^7}+\\ldots$$ above series is an infinite GP with first term $$a=\\dfrac{1}{2}$$ and common ratio $$r=\\dfrac{1}{4}$$ $$\\dfrac{a}{1-r}$$\n\n$$=\\frac{\\frac{1}{2}}{1-\\frac{1}{4}}$$$$=\\frac23$$\n\n\u2022 does this really have infinite terms which can be added? \u2013\u00a0user805532 Jul 3 at 4:38\n\u2022 Of course it has uncountable number of terms. I 've added them in my answer \u2013\u00a0user801303 Jul 3 at 4:40\n\u2022 it's of course countable \u2013\u00a0RiaD Jul 3 at 12:58\n\u2022 @PaulAldrin none of the terms are infinite in value, but they are infinitely many. Name a large odd number n and I can respond with a term $1/2^n$. \u2013\u00a0neptun Jul 3 at 13:09\n\u2022 r should be 1/4, not 1/2 \u2013\u00a0Mark Pattison Jul 3 at 13:32\n\nAnother way to avoid summing a GP.\n\nThe probability of the first head happening on toss $$n$$ is $$(1/2)^{n}$$. Let $$A$$ be the event that the first head occurs on an odd toss, and $$E_k$$ be the event that it occurs either on toss $$2k+1$$ or $$2k+2$$. Now $$P(A\\mid E_k)=\\frac{(1/2)^{2k+1}}{(1/2)^{2k+1}+(1/2)^{2k+2}}=\\frac{2}{3},$$ independently of $$k$$. Since exactly one of the $$E_k$$ will occur almost surely, we have $$P(A)=\\sum_kP(A\\mid E_k)P(E_k)=\\sum_k\\frac23P(E_k)=\\frac23\\sum_kP(E_k)=\\frac23.$$", "date": "2020-08-04 08:40:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3743275/a-fair-coin-is-tossed-untill-head-appears-for-the-first-time-what-is-probabilit", "openwebmath_score": 0.7943496704101562, "openwebmath_perplexity": 236.79774041093626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018398044144, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8605452863993139}}
{"url": "http://math.stackexchange.com/questions/14530/how-to-average-cyclic-quantities/17624", "text": "# How to average cyclic quantities?\n\nLooking on Internet, I mostly found this definition:\n\nGiven quantities on a cyclic domain D, first rescale the domain to [0;2$\\pi$[, then, find $z_n$ the point on the unit circle corresponding to the $n$th value, and compute the average by:\n\n$$z_m = \\sum_{n=1}^N z_n$$\n\nThe average angle is then $\\theta_m = \\arg z_m$ and to obtain the average value you scale back to your original domain D.\n\nI must say, I have problems with this definition. For simplicity, I will use oriented angles in degree for my examples (i.e. D = [0;360[). With this formula, having angles -90, 90 and 40 will give 40 as mean angle, when I would expect 13.33 as an answer (i.e. (90-90+40)/3).\n\nFor my own problems, I usually use:\n\n$$v_m = \\mathop{\\rm arg\\,min}_{v\\in D} \\sum_{n=1}^{N} d(v_n,v)^2$$\n\nWhere $d(x,y)$ is the distance in the cyclic domain, and $\\{v_1, v_2, \\ldots, v_n\\}$ is the set of cyclic data I want to average of.\n\nIt has the advantage to work the same way whatever the domain (replace D by a non-cyclic domain and $d$ with the usual euclidean distance, and you find the usual definition of an average). However, it is expensive to compute and I don't know any exact method to do it in general.\n\nSo my question is: what is the appropriate way to deal with average of cyclic data? And do you have good pointers that explain the problem and its solutions?\n\n-\nen.wikipedia.org/wiki/Circular_mean \u2013\u00a0 Rasmus Dec 16 '10 at 19:35\nI know about this page. However, there is no justification for it. Also, I left a comment in the discussion of this page in the hope of understanding. But for now, I still disagree with this method to calculate the average. \u2013\u00a0 PierreBdR Dec 17 '10 at 17:38\nThe choice of distance metric depends crucially on the application. Bearing data, for instance, might be derived from estimates of X and Y with normally distributed errors, and this leads naturally to the circular mean. For other cases this might not be a good choice. \u2013\u00a0 wnoise Apr 4 '11 at 18:52\n\nLike all averages, the answer depends upon the choice of metric. For a given metric $M$, the average of some angles $a_j \\in [-\\pi,\\pi]$ for $j \\in [1,N]$ is that angle $\\bar{a}_M$ which minimizes the sum of squared distances $d^2_M(\\bar{a}_M,a_j)$. For a weighted mean, one simply includes in the sum the weights $w_j$ (such that $\\sum_j w_j = 1$). That is,\n\n$$\\bar{a}_M = \\mathop{\\rm arg\\,min}_{x} \\sum_{j=1}^{N}\\, w_j\\, d^2_M(x,a_j)$$\n\nTwo common choices of metric are the Frobenius and the Riemann metrics. For the Frobenius metric, a direct formula exists that corresponds to the usual notion of average bearing in circular statistics. See \"Means and Averaging in the Group of Rotations\", Maher Moakher, SIAM Journal on Matrix Analysis and Applications, Volume 24, Issue 1, 2002, for details. http://lcvmwww.epfl.ch/new/publications/data/articles/63/simaxpaper.pdf\n\n-\nThank you, it is exactly the kind of reference I was looking for! \u2013\u00a0 PierreBdR May 9 '11 at 13:54\n\nThe problem with expecting the mean of 90\u00b0, \u221290\u00b0, and 40\u00b0 to be (90\u00b0\u221290\u00b0+40\u00b0)/3 = 13.33\u00b0 is that you would then expect the mean of 10\u00b0 and 350\u00b0 to be (10\u00b0+350\u00b0)/2 = 180\u00b0, and not 0\u00b0 which is the more reasonable answer. It only gets worse when you have more than two angles (What is the mean of 340\u00b0, 350\u00b0, 360\u00b0, 10\u00b0, and 20\u00b0? What about 340\u00b0, 350\u00b0, 0\u00b0, 10\u00b0, and 20\u00b0?). Essentially, what you're doing there is equivalent to setting $z_n = e^{i\\theta_n}$ and computing $$\\bar z = (z_1 z_2 \\cdots z_N)^{1/N},$$ and the problem is of course that it's not obvious a priori which of the $N$ possible roots of that equation is the right one, if any.\n\nThe \"circular mean\" definition is not so bad. In fact, it corresponds to the point which minimizes the sum of its squared distances to the points corresponding to the data, $$\\bar z = \\underset{\\lvert z \\rvert = 1}{\\arg\\min} \\sum_{n=1}^N \\lvert z - z_n \\rvert^2.$$ So this is almost the same as the formula you like to use; you only have to define the \"distance\" between angles as the distance between the corresponding points on the unit circle. That is, $d(\\theta, \\phi) = \\sqrt{2 - 2\\cos(\\theta-\\phi)} = 2 \\sin(\\lvert\\theta-\\phi\\rvert/2)$. This metric is close to $\\lvert\\theta-\\phi\\rvert$ when $\\theta$ and $\\phi$ are close, and has the advantage of being really easy to find the solution to.\n\n-\nOf the $N$ possible roots, one will be the global minimum of his distance functions, so checking all of them is sufficient. (See the reference Rob Johnson gave). \u2013\u00a0 wnoise Apr 4 '11 at 19:02\n\nThe angle is supposed to be the independent variable, not the dependent variable. If your function is cyclic (using degrees), z(-90)=z(270) so it doesn't matter which you use. Then the average value of the function is $$z_m=\\sum_{n=1}^Nz(\\theta_n)$$ You are right that if you average the angles, it matters which lap of the circle you use (-90 vs 270) because the difference of 360 gets divided by N.\n\n-\nThis is what I am reading on Internet, but I have issues with this definition (as described in my post) and there is never a good explanation on why we should use this also for cyclic data. Could you expand? \u2013\u00a0 PierreBdR Dec 16 '10 at 17:54\nThe definition of a cyclic function is that f(x+T)=f(x) for some period T. The recommendation you saw was to rescale the units of x so that the period is 2pi radians or 360 degrees. I don't think that is important, but was trying to stay with it. As the function is cyclic, it doesn't matter which cycle you take the data from. So if you want the average value over a cycle it would be (if we take T to be 360) $$\\frac{1}{360}\\int_a^{a+360}f(x)dx$$ You can then approximate this by a sum just by taking equally spaced steps like $$\\frac{1}{n}\\sum_{i=0}^{n-1}f(\\frac{360i}{n})$$ \u2013\u00a0 Ross Millikan Dec 16 '10 at 22:06\nIt looks like I am working on a different problem than you. Did you check the Wikipedia page that Rasmus suggested? Is that what you are after? \u2013\u00a0 Ross Millikan Dec 16 '10 at 22:27\n\nFor angles, one can adapt the iterative way of computing means to angles, that is:\n\ngiven angles v[1] .. v[n]\n\nm[1] = v[1]\n\nm[i] = remainder( m[i-1] + remainder( v[i]-m[i-1], C)/i, C) (i=2..n)\n\nwhere remainder(x,y) gives the signed remainder on dividing x by y and C is the measure of a circle.\n\nI suspect this gives your vm, but haven't been able to prove it.\n\n-\nA quick check indicates it might not work. Applying this formula to the same set of values, simply taking the elements in various order leads to different results. \u2013\u00a0 PierreBdR Dec 17 '10 at 17:44", "date": "2014-03-11 21:38:31", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/14530/how-to-average-cyclic-quantities/17624", "openwebmath_score": 0.8893794417381287, "openwebmath_perplexity": 316.0803193454473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018426872776, "lm_q2_score": 0.8824278664544912, "lm_q1q2_score": 0.8605452814050227}}
{"url": "https://math.stackexchange.com/questions/3495575/how-to-derive-the-identity-sin-x-x-prod-n-1-infty-cosx-2n-without-usi/3495595", "text": "How to derive the identity $\\sin x/x=\\prod_{n=1}^\\infty \\cos(x/2^n)$ without using telescoping?\n\nI am wondering how to derive the following equality $$\\frac{\\sin x}{x}=\\prod_{n=1}^\\infty \\cos(x/2^n)\\tag{1}$$ without using the method of telescoping. I know that there is already a question on deriving this infinite product representation of $$\\sin x/x$$, but all the answers in the link telescope the product. Here is the method, for completeness.\n\nFirst, we can use the trigonometric identity $$\\sin x=2\\cos (x/2)\\sin(x/2)$$ to yield $$\\cos(x/2)=\\frac{\\sin x}{2 \\sin(x/2)}$$. More generally, this implies that $$\\cos(x/2^{n})=\\frac{\\sin (x/2^{n-1})}{2 \\sin(x/2^{n})}$$\n\nOur infinite product is thus $$\\prod_{n=1}^\\infty\\cos(x/2^n)=\\frac{\\sin (x)}{2 \\sin(x/2)}\\cdot \\frac{\\sin (x/2)}{2 \\sin(x/4)} \\cdot \\frac{\\sin (x/4)}{2 \\sin(x/8)} \\cdots$$ Treating the product as a limit of a finite product $$f_k(x)=\\prod_{n=1}^k \\cos(x/2^n)$$, we notice that $$f_k(x)=\\frac{\\sin(x)}{2^k\\sin(x/2^k)},$$ with $$\\lim_{k\\to\\infty} f_k(x)=\\sin x/x$$. Thus, $$\\frac{\\sin x}{x}=\\prod_{n=1}^\\infty \\cos(x/2^n).$$\n\nQuestion:\n\nHow to show that $$(1)$$ is true without using telescoping?\n\n\u2022 Why do you want to avoid telescoping? \u2013\u00a0Lucas Henrique Jan 2 at 23:33\n\u2022 Note the Weierstrass factorization theorem:\\begin{align*} \\frac{\\sin x}{x}&=\\prod_{n=1}^\\infty\\left(1-\\frac{x^2}{n^2\\pi^2}\\right)\\\\ \\cos\\frac{x}{2^n}&=\\prod_{m=0}^\\infty\\left(1-\\frac{x^2}{(2m+1)^2 2^{2(n-1)}\\pi^2}\\right) \\end{align*} \u2013\u00a0Edward H Jan 2 at 23:44\n\u2022 You could use the fact that $\\sum_{n\\ge1}\\pm 2^{-n}$ (where the signs are chosen independently and uniformly) is uniformly distributed on $[-1,1]$; the desired equality relates two formulas for the characteristic function of such a random variable. \u2013\u00a0kimchi lover Jan 2 at 23:50\n\u2022 @kimchilover very nice \u2013\u00a0Sandeep Silwal Jan 3 at 0:10\n\u2022 @Lucas Henrique I am curious to know if there are other ways to evaluate this product. \u2013\u00a0Zachary Jan 3 at 2:13\n\nYou can do this by a trick that is essentially the same as looking at this product in frequency domain.\n\nTo avoid any analytical difficulty, let's examine finite sums; we have $$\\prod_{n=1}^{k}\\cos(x/2^n)=\\prod_{n=1}^k\\left(\\frac{e^{ix/2^n}+e^{-ix/2^n}}{2}\\right)$$ We can expand the sum on the right as $$\\frac{1}{2^k}\\sum_{\\sigma\\in\\{-1,1\\}^k}\\exp\\left(ix\\cdot \\left(\\sigma_1\\cdot \\frac{1}2+\\sigma_2\\cdot \\frac{1}{2^2}+\\ldots+\\sigma_k\\cdot \\frac{1}{2^k}\\right)\\right)$$ where $$\\sigma$$ is a string of $$k$$ terms in $$\\{-1,1\\}$$ representing which side of the sum within the former product was followed.\n\nOne can see that for $$n=1$$, the angular frequencies encountered (i.e. the coefficient of $$ix$$) are $$1/2$$ and $$-1/2$$ . For $$n=2$$, the frequencies are $$-3/4,\\,-1/4,\\,1/4,\\,3/4$$. We can prove via induction that the possible values of that coefficient are just the set of numbers of the form $$a/2^k$$ for odd integers $$a$$ between $$-2^k$$ and $$2^k$$. Thus, the partial sum works out to: $$\\frac{1}{2^k}\\cdot \\sum_{\\substack{a\\text{ odd}\\\\ -2^k < a < 2^k}}\\exp\\left(ix \\cdot \\frac{-a}{2^k}\\right)$$ We could bail out at this step and recognize that the sum is actually a geometric series (with ratio $$\\exp\\left(\\frac{ix}{2^{k-1}}\\right)$$), which would lead us back to the expression you derived for the partial sums. However, we could also recognize this an average of equally spaced evaluations of the function $$z\\mapsto \\exp(ix\\cdot z)$$ over the interval $$[-1,1]$$ with more evaluations as $$k$$ increases; thus, in the limit, this product becomes an integral giving the average value of $$\\exp(ixt)$$ over the interval $$[-1,1]$$: $$\\lim_{k\\rightarrow\\infty}\\prod_{n=1}^k\\cos(x/2^n) = \\frac{1}2\\int_{-1}^1\\exp(ixt)\\,dt$$ Of course, this is just integrating an exponential function, which can be done easily, and works out to $$\\frac{\\sin(x)}x$$.\n\nHere is a different packaging of the same basic argument presented in Milo Brandt's answer.\n\nLet $$X_k=\\sum_{j=1}^k \\sigma_j 2^{-j}$$, where the $$\\sigma_i$$ are iid $$\\pm1$$ random variables. This has uniform distribution on the $$2^k$$ points uniformly spaced $$2^{1-k}$$ apart in the range from $$-1+2^{-k}$$ to $$1-2^{-k}$$, as can be seen from the binary expansions of the integers from $$0$$ to $$2^k$$. One can verify directly that $$X_k$$ converges in distribution to the continuous uniform distribution on $$[-1,1]$$.\n\nThe characteristic function of $$X_k$$, namely the function $$\\phi_k(t)=E[\\exp(itX_k)]$$ is given by $$\\prod_{j=1}^k E[\\exp(i \\sigma_j 2^{-j})] = \\prod_{j=1}^k \\cos(t2^{-j})$$.\n\nBy L\u00e9vy's continuity theorem, for each $$t$$, one has $$\\lim_{k\\to\\infty}\\varphi_k(t)=\\varphi(t)$$, where $$\\varphi(t)$$ is the characteristic function of the uniform distribution on $$[-1,1]$$, which is $$\\varphi(t)=\\frac 12\\int_{-1}^1 \\exp(itx)\\,dx = \\frac{\\sin(t)}t.$$", "date": "2020-08-12 23:48:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3495575/how-to-derive-the-identity-sin-x-x-prod-n-1-infty-cosx-2n-without-usi/3495595", "openwebmath_score": 0.9623919725418091, "openwebmath_perplexity": 121.44768834181576, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575147530352, "lm_q2_score": 0.8757870046160257, "lm_q1q2_score": 0.8605111027085272}}
{"url": "https://math.stackexchange.com/questions/1250112/number-of-times-2k-appears-in-factorial", "text": "# Number of times $2^k$ appears in factorial\n\nFor what $n$ does: $2^n | 19!18!...1!$?\n\nI checked how many times $2^1$ appears:\n\nIt appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$\n\nI checked how many times $2^2 = 4$ appears:\n\nIt appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$\n\nI checked how many times $2^3 = 8$ appears:\n\nIt appears in, $8!, 9!, ..., 19!$ meaning, $8^{12} = 2^{36}$\n\nI checked how many times $2^{4} = 16$ appears:\n\nIt appears in, $16!, 17!, 18!, 19!$ meaning, $16^{4} = 2^{16}$\n\nIn all,\n\n$$2^{18} \\cdot 2^{32} \\cdot 2^{36} \\cdot 2^{16} = 2^{102}$$\n\nBut that is the wrong answer, its supposed to be $2^{150}$?\n\n\u2022 Note that, for example, $6!$ contributes 4 factors of 2 - one from 2, one from 6 and two from 4. You only count 3 of these. \u2013\u00a0Wojowu Apr 24 '15 at 16:37\n\nA simple trick to compute $k$ such that $2^k|n!$ is to compute $\\sum_{i=1}^\\infty \\left\\lfloor\\frac{n}{2^i}\\right\\rfloor$, this is because $n$ has $[n/2]$ numbers divided by $2$, if we pick out these numbers and find out that there're $[n/4]$ numbers divided by $4$.. If we continue this procedure, we see that $$k=1\\cdot\\left(\\left\\lfloor\\frac{n}{2}\\right\\rfloor-\\left\\lfloor\\frac{n}{4}\\right\\rfloor\\right)+2\\left(\\left\\lfloor\\frac{n}{4}\\right\\rfloor-\\left\\lfloor\\frac{n}{8}\\right\\rfloor\\right)+\\ldots=\\sum_{i=1}^\\infty \\left\\lfloor\\frac{n}{2^i}\\right\\rfloor$$. In this case, we have to sum $$0+1+1+3+3+4+4+7+7+8+8+10+10+11+11+15+15+16+16=150.$$ Your fault is that your did not count the contribution of those which is not the power of $2$. For instance, there's $14$ in $14!$..\n\nHow many times $2$ divides the product $\\prod_{i=1}^{19}i!$ ?\n\nLet's call each term inside a factorial $i$. That way, $i = 1$ occurs in 19 factorials, $i = 2$ occurs in 18 factorials, and $i = 3$ occurs in 17 factorials etc.\n\n$i = 2$ occurs 18 times. $1 \\times 18 = 18$\n$i = 4$ occurs 16 times. $2 \\times 16 = 32$\n$i = 6$ occurs 14 times. $1 \\times 14 = 14$\n$i = 8$ occurs 12 times. $3 \\times 12 = 36$\n$i = 10$ occurs 10 times. $1 \\times 10 = 10$\n$i = 12$ occurs 8 times. $2 \\times 8 = 16$\n$i = 14$ occurs 6 times. $1 \\times 6 = 6$\n$i = 16$ occurs 4 times. $4 \\times 4 = 16$\n$i = 18$ occurs 2 times. $1 \\times 2 = 2$\n\n$$18 + 32 + 14 + 36 + 10 + 16 + 6 + 16 + 2 = 150$$\n\nIt might be more helpful to do this recursively.\n\nLet $T(n) = \\prod_{k=1}^n k!$.\n\nWe will use the notation: $2^{r} \\| m$ to mean that $2^r$ is the largest power of $2$ that divides $m$.\n\nThen we have $2 \\| 2! = T(2)$. We also know that $2 \\| 3!$, so $2^2 \\| T(3) = 3! T(2)$. Continuing:\n\n$$2^3 \\| 4!$$ $$2^3 \\| 5!$$ $$2^4 \\| 6!$$ $$2^4 \\| 7!$$ $$2^7 \\| 8!$$ $$2^7 \\| 9!$$ $$2^8 \\| 10!$$ $$2^8 \\| 11!$$ $$2^{10} \\| 12!$$ $$2^{10} \\| 13!$$ $$2^{11} \\| 14!$$ $$2^{11} \\| 15!$$ $$2^{15} \\| 16!$$ $$2^{15} \\| 17!$$ $$2^{16} \\| 18!$$ $$2^{16} \\| 19!$$\n\nIf we take the sum of all of those powers, $$2\\cdot 1 + 2\\cdot 3 + 2\\cdot 4 + 2\\cdot 7 + 2\\cdot8 + 2 \\cdot 10 + 2 \\cdot 11 + 2 \\cdot 15 + 2 \\cdot 16$$\n\n$$=2(1+3+4+7+8+10+11+15+16) = 2(75) = 150.$$", "date": "2021-05-09 23:43:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1250112/number-of-times-2k-appears-in-factorial", "openwebmath_score": 0.8817852139472961, "openwebmath_perplexity": 121.08447970920729, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575157745541, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8605111020104248}}
{"url": "https://www.physicsforums.com/threads/can-someone-please-help-me-factorise-a-fraction-u-2-u-2-3u-2.248655/", "text": "1. Aug 6, 2008\n\n### laura_a\n\n1. The problem statement, all variables and given/known data\nI am working on a topic called differential equations, and Im stuck on some working out.\n\n2. Relevant equations[/b]\n\nI have to integrate it, and I know I can without factorising, but it is so messy, my professor said to factorise the fraction so it looks a bit like a1/(bu+ c)+ a2/(du+ e)\n\nWell thats what the professor said, I have no idea what it means... I can factorise as far as\n= u/(-u^2 +3u +2) - 2/(-u^2 +3u +2)\n\nAnd I know that -u^2 +3u +2 = (-u + 1)(u-2) +4\n\nBut not sure how to put it all together\n\nHere is the working out for the whole question just in case you're interested\n\\begin{align*} y' &= \\frac{y+2x}{y-2x} \\\\ Let y &= ux \\\\ \\text{Then we have} y' &= \\frac{ux + 2x}{ux - 2x} \\\\ &= \\frac{u+2}{u-2} \\\\ \\text{Now } y' &= \\frac{dy}{dx} = \\frac{d(ux)}{dx} = x \\frac{dy}{dx} + u \\\\ \\text{So we can say that} x \\frac{du}{dx} + u& = \\frac{u+2}{u-2} \\\\ x \\frac{du}{dx} &= \\frac{u+2}{u-2} - u \\\\ x \\frac{du}{dx} &= \\frac{u+2}{u-2} - \\frac{u^2-2u}{u-2} \\\\ x \\frac{du}{dx} &= \\frac{u+2- u^2+2u}{u-2} \\\\ x \\frac{du}{dx} &= \\frac{-u^2+3u+2}{u-2} \\\\ \\end{align*} \\bigskip\n\nThen I have to integrate both sides of this equation...which is what I think I need to factorise in order to make it nice and neat\n\n$$\\frac{u-2}{-u^2 + 3u + 2}du &= \\frac{1}{x} dx$$\n\nLast edited: Aug 6, 2008\n2. Aug 6, 2008\n\n### HallsofIvy\n\nStaff Emeritus\nThat can't be factored using integer coefficients but you know that if an expression factors as (x-a)(x-b) then a and b must be roots of (x-a)(x-b)= 0. Use the quadratic formula to find the roots of x2- 3x- 2= 0.\n\n3. Aug 6, 2008\n\n### Defennder\n\nTry completing the square of $$-u^2+3u+2$$. Then use a simple algebraic identity to decompose it into partial fractions.\n\n4. Aug 7, 2008\n\n### laura_a\n\nI think i've gotten a little closer, The question I'm trying to do is to solve a d.e. using change of variables, so once I get this part worked out I should be able to get it\n\nSo this is how far I've gone\nSince $$u^2- 3 u -2= 0$$\nhas two roots\n$$u_1 = (3+ 17^{0.5})/2, u_2= (3- 17^{0.5})/2$$\n\nso\n\n$$u^2- 3 u -2= (u - (3+ 17^{0.5})/2 ) (u- (3- 17^{0.5})/2 )$$\n\n$$\\frac{u-2}{-u+3u+2}du = \\frac{1}{2}dx$$ - I'll call this equation (2)\nSo now I've got the LHS as\n\n$$S = -u /((u - (3+ 17^{0.5})/2 ) +2 /(u- (3- 17^{0.5})/2 )$$\nI changed the u-2 to -u+2 since I made a similar change on the denominator in order to solve it...\n\nAnyhow, I integrated that and ended up with something really messy invloving logs\n\n$$-2(u + 17^{0.5} ln(u-17^{0.5}-3)+3ln(u-17^{0.5}-3)-17^{0.5}-3+4ln(u+17^{0.5}-3) = ln(x) + C$$\n\n(I integrated both sides of equation (2) above)\n\nWell the prob is now, I'm probably wrong anyway, but the next step is to solve for u, and then I have to sub back in the original change of varaible which was y=xu and end up with an expression for y in terms of x and C... before I go on, can anyone tell me if I should bother working through those logs, or is it wrong? Thanks :)\n\n5. Aug 8, 2008\n\n### arildno\n\nLet us take this as a GENERAL case, shal we?\nWe are to decompose:\n$$\\frac{u-u_{0}}{(u-u_{1})(u-u_{2})}=\\frac{A}{u-u_{1}}+\\frac{B}{u-u_{2}}$$\nwhere the unindexed u is our variable, the indexed u's known numbers, and A and B constants to be determined.\nWe therefore must have:\n$$u-u_{0}=A(u-u_{2})+B(u-u_{1})$$, or by comparing coefficients, we get the system of equations:\n$$A+B=1$$\nand\n$$u_{2}A+u_{1}B=u_{0}$$\nwhereby we arrive at the solutions:\n$$A=\\frac{u_{0}-u_{1}}{u_{2}-u_{1}},B=\\frac{u_{2}-u_{0}}{u_{2}-u_{1}}$$\n\nThis is much simpler than using specific numbers!", "date": "2017-09-25 04:59:57", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/can-someone-please-help-me-factorise-a-fraction-u-2-u-2-3u-2.248655/", "openwebmath_score": 0.9999754428863525, "openwebmath_perplexity": 1179.0526619355828, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575121992375, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8605110956937382}}
{"url": "https://math.stackexchange.com/questions/1589868/epsilon-delta-definition-of-limit-set-an-upper-bound-to-delta", "text": "# Epsilon-delta definition of limit : set an upper bound to delta?\n\nAm I allowed to set an upper bound to delta in the epsilon-delta definition of limit? Why would it still be equivalent to the original definition ? For example, if the function is defined for all $x$ in $\\Bbb R$ :\n\n$f : D(f) = \\Bbb R \\rightarrow \\Bbb R$.\n\nLet $\\epsilon > 0, \\exists \\delta > 0 \\text{ s.t. } \\forall x : 0 < |x-x_o| < \\delta < 3 \\Rightarrow |f(x) - l| < \\epsilon$\n\n$\\Leftarrow\\text{?}\\Rightarrow$\n\nLet $\\epsilon > 0, \\exists \\delta > 0 \\text{ s.t. } \\forall x : 0 < |x-x_o| < \\delta \\Rightarrow |f(x) - l| < \\epsilon$.\n\n\u2022 Lets suppose you found an $\\; \\delta \\;$ then the definition still holds for any $\\delta^* \\le \\delta$. Hence, you can choose your $\\; \\delta^* \\le 3$ \u2013\u00a0XPenguen Dec 26 '15 at 20:16\n\u2022 If you find an epsilon which beats a small delta, you've found an epsilon which beats a big one. \u2013\u00a0Mark Bennet Dec 26 '15 at 20:16\n\n$$\\forall\\varepsilon > 0\\ \\exists \\delta > 0\\ \\forall x \\Big( 0 < |x-x_o| < \\delta < 3 \\Rightarrow |f(x) - \\ell| < \\varepsilon\\Big)$$\n\nI would phrase this differently:\n\n$$\\forall\\varepsilon > 0\\ \\exists \\delta \\in(0,3)\\ \\forall x \\Big( 0 < |x-x_o| < \\delta \\Rightarrow |f(x) - \\ell| < \\varepsilon\\Big)$$\n\nNow the question is whether that is equivalent to this:\n\n$$\\forall\\varepsilon > 0\\ \\exists \\delta>0\\ \\forall x \\Big( 0 < |x-x_o| < \\delta \\Rightarrow |f(x) - \\ell| < \\varepsilon\\Big)$$\n\nIf it is true that $\\forall\\varepsilon>0\\ \\exists\\delta\\in(0,3)\\ \\cdots\\cdots$ then it is true that $\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\cdots\\cdots$, simply because every number in $(0,3)$ is $>0$.\n\nNow the question is whether the converse holds. If it is true that $\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\cdots\\cdots$, does it necessarily follow that $\\forall\\varepsilon>0\\ \\exists\\delta\\in(0,3)\\ \\cdots\\cdots\\,{}$? Here the answer in general is \u201cno\u201d. I.e. there are some things you could put in place of $\\text{\u201c}\\cdots\\cdots\\text{''}$ for which the first statement would be true and the second false. However, in the definition of \"limit\" the thing in place of $\\text{\u201c}\\cdots\\cdots\\text{''}$ is \u201cif a certain thing is $<\\delta$, then a certain thing follows.\u201d Given a certain $\\varepsilon>0$ suppose we know there exists $\\delta>0$ such that if a certain thing is less than $\\delta$ then a certain conclusion follows. If the given value of $\\delta$ is small enough, then the minimum of that value of $\\delta$ and $2.9$ is small enough, because everything less than $\\min\\{\\delta,2.9\\}$ is less than $\\delta$. So when the statement in place of $\\text{\u201c}\\cdots\\cdots\\text{''}$ has the form \u201cif a certain thing is $<\\delta$, then a certain thing follows.\u201d, then it is true that if $\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\cdots\\cdots$ then $\\forall\\varepsilon>0\\ \\exists\\delta\\in(0,3)\\ \\cdots\\cdots$.\n\nYes, of course. If the first condition is satisfied, then clearly the second is too. Conversely, if the second condition is satisfied, then one can replace $\\delta$ by $\\min(\\delta, 3)$ and so the first condition is satisfied.\n\nYes, as you have to exhibit an $\\delta$ that works for every $\\epsilon$, then you can be as artificial as you want if that works for you. Always in existence proofs the same rule applies, construct all you need and show that it does work.\n\nYes you can. The idea is that some proposition $P(x,\\epsilon)$ is true for all $x$ that are sufficiently close to $x_0$. So if it holds for all $x\\in (-d+x_0,d+x_0)$ then it holds for all $x\\in (-d'+x_0,d'+x_0)$ whenever $0<d'<d$. Sometimes it is useful to know how large $d$ can be, for a given $\\epsilon.$", "date": "2019-08-26 02:32:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1589868/epsilon-delta-definition-of-limit-set-an-upper-bound-to-delta", "openwebmath_score": 0.9341692328453064, "openwebmath_perplexity": 94.55677299488865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9648551556203815, "lm_q2_score": 0.8918110396870288, "lm_q1q2_score": 0.8604684794812024}}
{"url": "https://www.teacherspayteachers.com/Product/Calculus-Study-Guide-for-AP-Calculus-ABBC-or-University-Calculus-2085731", "text": "# Calculus : Study Guide for AP Calculus AB/BC or University Calculus\n\nSubjects\nResource Types\nProduct Rating\n4.0\nFile Type\nPDF (Acrobat) Document File\nBe sure that you have an application to open this file type before downloading and/or purchasing.\n1.25 MB \u00a0\u00a0|\u00a0\u00a0\u00a0300 pages\n\n### PRODUCT DESCRIPTION\n\nCalculus Study Guide suitable for AP Calculus or First Year College Calculus Students. Main topics covered : Functions Limits and Continuity, Derivatives, Application of the Derivative, Indefinite integration, Definite Integration, Techniques of Integration, Sequences and Series, L\u2019Hopitals Rule, Improper Integrals, Taylor Series, Short Introduction to Differential Equations.\n\nCalculus without Limits is a self-study guide that can be used as a textbook/lecture supplement by students, as a source of homework problems by the instructor or as a source of lecture material. Covers most topics discussed in a first year calculus course (or AP Calculus). The emphasis is on solved problems that show students how to attack the types of problems they will encounter in homework and on exams. Discussion of concepts is informal followed by example problems. Many practice problems are also included for students.\n\n1 - Function Review\nWhat is a function - Graphing - Even and Odd Functions - Increasing and Decreasing Functions - One to One Functions - Powers and Polynomials - Absolute Value Function - Exponential Function - Logarithm - Trigonometric Functions\n2- Limits\nDefinition - Tricks for Doing Limits - Properties of Limits and Rules - Limits at Infinity - One Sided Limits Continuity\n3- Derivative\nNotation and Definition in terms of limits - Slope of Tangent Line - Power Rule and Derivative of a constant - Chain Rule - Derivatives of Trig Functions - Product and Quotient Rules\n4- Applications of the Derivative\nDerivative of Inverse Function - Implicit Differentiation - Application: Rates of Change, Velocity and Acceleration - Finding Minima and Maxima First and Second Derivative Test - Inflection Points - The Mean Value Theorem - Related Rates\n5- The Integral\nNotation - Power Rule of Integration - Integration of Trig Functions - Integration of inverse function - Integrating Exponential Function - The substitution technique - Integrating Hyperbolic Functions - Definite Integration - Sums and Area Under a Curve - Area Between Two Curves -\n6- U-Substitution\nSolved Problems\n7- Integration by Parts\nDefinition and When to Use - Examples\n8- Integration of Rational Functions\nIntroduction - Examples\n9- Integration Using the Method of Partial Fractions\nDiscussion - Examples\n10- Integrating Powers of Trig Functions\nExamples\n11- Trig Substitution\nThree Cases - Basic Technique -Examples\n12 - Improper Integrals\nDefinition - Examples\n13 - Sequences and Infinite Series\nSequences - Limits - Infinite Series\n14 Ratio, Comparison, and Integral Tests\nExamples\n15 - Alternating Series\n16 - Power Series and Radius of Convergence\n17- Taylor Series Expansions\n18- L\u2019Hopital\u2019s Rule\nWhen to use and Definition - Examples\n19- Introduction to Differential Equations\nFirst Order Differential Equations - Second Order Differential Equations\nTotal Pages\n300\nN/A\nTeaching Duration\n1 Year\n\n4.0\nOverall Quality:\n4.0\nAccuracy:\n4.0\nPracticality:\n4.0\nThoroughness:\n4.0\nCreativity:\n4.0\nClarity:\n4.0\nTotal:\n1 rating", "date": "2017-01-22 06:17:42", "meta": {"domain": "teacherspayteachers.com", "url": "https://www.teacherspayteachers.com/Product/Calculus-Study-Guide-for-AP-Calculus-ABBC-or-University-Calculus-2085731", "openwebmath_score": 0.8662840723991394, "openwebmath_perplexity": 1903.1238794883905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227580598143, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8604660166696371}}
{"url": "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=prev&oldid=135054", "text": "# Difference between revisions of \"2006 AIME I Problems/Problem 10\"\n\n## Problem\n\nEight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ $[asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j\n\n## Solutions\n\n### Solution 1\n\nThe line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\\left(1,\\frac 12\\right)$ and $\\left(\\frac 32,2\\right)$, which can be easily solved to be $6x = 2y + 5$. Thus, $a^2 + b^2 + c^2 = \\boxed{065}$.\n\n### Solution 2\n\nAssume that if unit squares are drawn circumscribing the circles, then the line will divide the area of the concave hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the x-axis as $(x, 0)$.\n\nThe line divides the region into 2 sections. The left piece is a trapezoid, with its area $\\frac{1}{2}((x) + (x+1))(3) = 3x + \\frac{3}{2}$. The right piece is the addition of a trapezoid and a rectangle, and the areas are $\\frac{1}{2}((1-x) + (2-x))(3)$ and $2 \\cdot 1 = 2$, totaling $\\frac{13}{2} - 3x$. Since we want the two regions to be equal, we find that $3x + \\frac 32 = \\frac {13}2 - 3x$, so $x = \\frac{5}{6}$.\n\nWe have that $\\left(\\frac 56, 0\\right)$ is a point on the line of slope 3, so $y - 0 = 3\\left(x - \\frac 56\\right) \\Longrightarrow 6x = 2y + 5$. Our answer is $2^2 + 5^2 + 6^2 = 65$.\n\nWe now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.", "date": "2021-01-17 12:30:25", "meta": {"domain": "artofproblemsolving.com", "url": "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=prev&oldid=135054", "openwebmath_score": 0.7458829283714294, "openwebmath_perplexity": 705.5428297619834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631635159684, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8604513145357217}}
{"url": "https://math.stackexchange.com/questions/2172433/how-to-evaluate-int-0-infty-eitx-e-x-dx", "text": "# How to evaluate $\\int_0^\\infty e^{itx} e^{-x} dx$?\n\nWhen calculating the characteristic function of the exponential distribution function, we need to evaluate the complex-integration: \\begin{align*} \\int_0^\\infty e^{itx}e^{-x} dx \\end{align*} for any $t \\in \\mathbb{R}$.\n\nI understand how to evaluate this integral by treating the real part and imaginary part separately, but I am wondering is there any approach that uses the complex analysis theory? I found some answer uses the seemingly unjustified \"fundamental theorem of calculus\":\n\n$$\\int_0^\\infty e^{(it - 1)x} dx = \\frac{1}{it - 1}\\int_0^\\infty e^{(it - 1)x} d(it - 1)x = \\frac{1}{1 - it}.$$\n\nI think this solution is lacking any theoretic support (maybe I am wrong, please advise if there is any theorem that supports the above calculation). Specifically, can this integral be evaluated using residue calculus (contour integration)?\n\n\u2022 Thanks, but for a complex-valued function, how to justify $\\int_0^\\infty f'(x) dx = f(x)|_0^\\infty$? Could you please provide me some references? \u2013\u00a0Zhanxiong Mar 5 '17 at 4:01\n\u2022 Integrating a complex-valued function is the same as a real-valued function. In complex analysis we look at $f(z)$ where $z$ is a complex variable, it is different because we integrate over curves in the complex plane, not over real intervals \u2013\u00a0reuns Mar 5 '17 at 4:07\n\nYou're right to demand justification beyond glib use of the FTC. But it is true that $$\\int_0^\\infty e^{-ax}dx = \\frac{1}{a}$$ whenever $\\Re(a)>0.$\n\nIn complex analysis, we can still affect the change of variables to $z=ax,$ but the resulting integral is along a ray in the complex plane in the direction of $a,$ not the positive real axis (unless $a$ is a positive real). We can write this (using somewhat bad notation) as $$\\int_0^\\infty e^{-ax}dx = \\frac{1}{a} \\int_0^{a\\infty}e^{-z}dz.$$\n\n(To see formally that the change of variables works, note that we have the parametrization $\\gamma(t) = at$ for $0<t<\\infty$ for the ray. We can write the integral of $e^{-z}/a$ along that path as $$\\int_0^\\infty \\frac{e^{-\\gamma(t)}}{a}\\gamma'(t)dt = \\int_0^\\infty \\frac{e^{-at}}{a}adt = \\int_0^\\infty e^{-at}dt$$).\n\nNow, the difficult part is how we can justify saying $$\\int_0^{a\\infty}e^{-z}dz = \\int_0^\\infty e^{-x}dx = 1.$$\n\nIn other words we want to be able to rotate the contour back down to the real axis without changing the value of the integral.\n\nTo see why we can do this, imagine doing a integral around a large wedge-shaped contour. It goes out along the real axis to $R \\gg 1$ and then goes along a circular path to $Re^{i\\arg(a)}$ and then back into the origin along the ray $[0,a\\infty)$ that our integral is taken over.\n\nBy Cauchy's theorem, the integral along this closed path is zero since $e^{-z}$ is analytic. The integral along the circular path goes to zero as $R\\to \\infty.$ We can see this cause the integrand decays like $e^{-R}.$ More formally the integral is $$\\int_0^{\\arg a} e^{-Re^{i\\theta}}Re^{i\\theta}id\\theta$$ and we have $$\\left|\\int_0^{\\arg a} e^{-Re^{i\\theta}}Re^{i\\theta}id\\theta\\right| \\le \\arg(a) \\max_\\theta|ie^{-Re^{i\\theta}}Re^{i\\theta}| = \\arg(a)Re^{-R\\cos(\\arg(a))} \\to 0$$ as $R\\to\\infty$\n\nThus as $R\\to \\infty,$ the integral along the real axis must cancel out the integral along the ray $[0,a\\infty)$ in order that the integral along the closed path be zero as Cauchy's theorem demands. We have $$\\int_0^{a\\infty}e^{-z}dz = \\int_0^\\infty e^{-x}dx = 1$$ and therefore $$\\int_0^\\infty e^{-ax}dx = \\frac{1}{a} \\int_0^{a\\infty}e^{-z}dz = \\frac{1}{a}.$$\n\nI've intentionally not said explicitly where $\\Re(a)>0$ is used in the above argument. Of course, it's essential. See if you can find where it's used.\n\n\u2022 Thanks for your answer, I found my own solution is very similar to yours! +1 \u2013\u00a0Zhanxiong Mar 5 '17 at 5:14\n\nI figured out a rigorous proof by myself.\n\nIf $t = 0$, then it is an integration of a real-valued function, and clearly, $\\int_0^\\infty e^{-x} dx = 1$.\n\nIf $t \\neq 0$, without losing of generality, assume $t > 0$. Consider the contour below:\n\nIn the picture, $n$ is a positive number that will be sent to $\\infty$, the top line passes the origin and the point $(-1, t)$. And we set $f(z) = e^z, z \\in \\mathbb{C}$. By Cauchy's integration theorem, \\begin{align} 0 = \\int_\\Gamma f(z) dz = \\int_{\\Gamma_1} e^z dz + \\int_{\\Gamma_2} e^z dz + \\int_{\\Gamma_3} e^z dz \\end{align}\n\nLet's denote the angle between $\\Gamma_1$ and the real axis by $\\theta_0$.\n\nClearly, $\\int_{\\Gamma_3}e^z dz = \\int_{-n}^0 e^x dx = 1 - e^{-n}$.\n\nOn $\\Gamma_1$, $z$ has the representation $z = (it - 1)x, x \\in (0, n/|1 - it|)$, thus \\begin{align*} \\int_{\\Gamma_1}e^z dz = \\int_0^{n/\\sqrt{1 + t^2}}e^{(it - 1)x}(it - 1) dx = (it - 1) \\int_0^{n/\\sqrt{1 + t^2}} e^{(it - 1)x} dx \\end{align*}\n\nTo get the desired result, it remains to show $\\int_{\\Gamma_2} e^z dz \\to 0$ as $n \\to \\infty$. Let $z = ne^{i\\theta}$ with $\\theta \\in (\\theta_0, \\pi)$. It follows that \\begin{align*} & \\left|\\int_{\\Gamma_2} e^z dz\\right| = \\left|\\int_{\\theta_0}^\\pi e^{ne^{i\\theta}}nie^{i\\theta} d\\theta\\right| \\\\ \\leq & \\int_{\\theta_0}^\\pi e^{n\\cos\\theta}n d\\theta \\\\ \\leq & ne^{n\\cos{\\theta_0}}(\\pi - \\theta_0) \\to 0 \\end{align*} as $n \\to \\infty$. Here we used the fact that $\\pi/2 < \\theta_0 < \\pi$.\n\nThe real analysis way : $$f(x) = \\frac{e^{(it-1)x}}{it-1}, f'(x) = e^{(it-1)x}, \\qquad \\int_0^\\infty f'(x)dx=f(\\infty)-f(0) = \\frac{1}{1-it}$$\n\nThe complex analysis way, with the (complex) change of variable $z = (it-1)u, dz = (it-1)du$ : $$\\int_0^\\infty e^{(it-1)u}du = \\int_0^{(it-1) \\infty } e^{z}\\frac{dz}{it-1} = \\left.\\frac{e^{z}}{it-1}\\right|_0^{(it-1)\\infty} = \\frac{1}{1-it}$$\n\nwhere $\\int_0^{(it-1) \\infty }$ is a contour integral", "date": "2019-04-23 11:58:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2172433/how-to-evaluate-int-0-infty-eitx-e-x-dx", "openwebmath_score": 0.9921557307243347, "openwebmath_perplexity": 135.0969315533633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631663220389, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8604513104366829}}
{"url": "https://math.stackexchange.com/questions/1898858/xy-xy-w-in-mathbbr-is-xwyw-real", "text": "# $x+y=xy=w \\in \\mathbb{R}^+$. Is $x^w+y^w$ real?\n\nQuestion: For $x,y \\in \\mathbb{C}$, suppose $x+y=xy=w \\in \\mathbb{R}^+$. Is $x^w+y^w$ necessarily real?\n\nFor instance, if $x+y=xy=3$, then one solution is $x = \\frac{3 \\pm i \\sqrt{3}}{2}$, $y = \\frac{3 \\mp i \\sqrt{3}}{2}$, but $x^3 + y^3 = 0$, which is real.\n\nI've checked this numerically for many values of $w$ that give complex $x$ and $y$ (namely, $w \\in (0,4)$.)\n\n\u2022 @PatrickStevens Then $x+y$ isn't real. \u2013\u00a0wythagoras Aug 21 '16 at 10:37\n\u2022 @PatrickStevens But he required that $x+y=xy\\in\\Bbb R$.. \u2013\u00a0BigbearZzz Aug 21 '16 at 10:37\n\u2022 Sorry all. I wasn't paying attention, clearly. \u2013\u00a0Patrick Stevens Aug 21 '16 at 10:38\n\u2022 The answer is yes; the essential idea is that $xy = x+y \\in \\mathbb{R}$ forces either that $x,y \\in \\mathbb{R}$, or that $x$ and $y$ are complex conjugates. \u2013\u00a0Drew N Aug 22 '16 at 2:22\n\nYes. Since $x + y \\in \\mathbb{R}$, $y = \\overline{x} + r$ for some $r \\in \\mathbb{R}$. Then $xy = |x|^2 + xr \\in \\mathbb{R}$, implying that either $r = 0$ or $x \\in \\mathbb{R}$. Then we do casework:\n\n\u2022 If $r = 0$, then $y = \\overline{x}$; this leads to\n\n$$x^w + y^w = x^w + \\overline{x^w} \\in \\mathbb{R}.$$\n\nWarning: for this to work, we had to pick the standard branch of the complex logarithm, specifically, the one undefined on the nonpositive real line, whose imaginary part is between $-\\pi$ and $\\pi$. Once we define $z^w := e^{w \\ln z}$, ${(\\overline{x})}^w = \\overline{x^w}$ is true for this branch of $\\ln$ (as $x$ is not nonpositive real), but might not be true for another branch.\n\nThis warning does not come into play when $w$ is an integer. But take, for example, $x = 1 + \\frac{1 + i}{\\sqrt{2}}$, $y = 1 + \\frac{1 - i}{\\sqrt{2}}$. Then $x + y = xy = 2 + \\sqrt{2}$. If we picked a different branch of the complex logarithm, then we could have $x^w + y^w$ not real.\n\n\u2022 On the other hand, if $x \\in \\mathbb{R}$, then $y \\in \\mathbb{R}$, so $x^w + y^w \\in \\mathbb{R}$. Since $x + y = xy > 0$, $x,y$ must both be positive, so we have no trouble with a negative base of the exponent.\n\nNote there was nothing special about $w$: we could have reached the stronger conclusion that $x^a + y^a \\in \\mathbb{R}$ for all $a \\in \\mathbb{R}$.\n\n\u2022 You do need $x,y\\geq0$, for your last statement. \u2013\u00a0wythagoras Aug 21 '16 at 10:54\n\u2022 @6005: Even with $a>0$, you still need $x,y\\not\\le 0$. Consider $x=y=-1\\in\\mathbb R$ and $a=1/2>0$. Then no matter where you do the branch cut, either $x^a=y^a=i$ or $x^a=y^a=-i$. Basically, the equation $\\overline x^a = \\overline{x^a}$ is nor true for all $x$ if $a$ is not integer. \u2013\u00a0celtschk Aug 21 '16 at 11:06\n\u2022 As a second conjecture, I believe the casework depends on the value of $w$. That is, in fact $r = 0$ if and only if $0< w < 4$, and $x \\in \\mathbb{R}$ if and only if $w \\geq 4$ ($w = 0$ is excluded by assumption). \u2013\u00a0Drew N Aug 21 '16 at 11:28\n\u2022 @celtschk Thanks for your valuable feedback. I believe I fixed all the problems and provided all the necessary caveats. \u2013\u00a06005 Aug 21 '16 at 11:48\n\u2022 @DrewN your second conjecture is correct. \u2013\u00a06005 Aug 21 '16 at 11:55\n\nYes. Write $x=a+bi$, $y=c-di$, then clearly $b=d$ because $x+y$ is real.\n\nSo $x=a+bi$, $y=c-bi$. Then $xy=ac-abi+cbi+b^2$, so $a=c$ or $b=0$.\n\n\u2022 If $a=c$, then $y = \\overline{x}$, i.e. the complex conjugate of $x$.\nSo $x^w+y^w = x^w+(\\overline{x})^w = x^w+\\overline{x^w}$, which is real.\n\n\u2022 If $b=0$, $x$ and $y$ are real so $x^w+y^w$ is real as $w>0$.\n\n\u2022 Thanks for your feedback on my answer earlier. In your answer, note that $(\\overline{x})^w = \\overline{x^w}$ is only true for a particular branch of complex log, and will never be true for negative real $x$. However, you can guarantee it for all $x$ which aren't negative real by picking the relevant branch of complex log to define your exponential. \u2013\u00a06005 Aug 21 '16 at 11:53\n\n$x+y$ real implies that $\\Im(x)=-\\Im(y)$. Thus if $x=a+bi$ then $y=c-bi$.\n\n$xy$ real implies that, because $xy=ac+b^2+ib(c-a)$, $b=0\\lor c=a$.\n\nIf $b=0$ the result is trivial as $x,y\\in\\mathbb{R}$.\n\nIf $c=a$ then $x=\\overline{y}$. But then clearly $$x^w+y^w=x^w+\\overline{x}^w=\\overline{x^w+\\overline{x}^w}=\\overline{x^w+y^w},$$ where the middle equality holds by symmetry.\n\nBut then it must be real, as the only complex numbers satisfying $z=\\bar{z}$ are real.\n\nNote: for why/when we are allowed to write $x^w=\\overline{\\bar{x}^w}$ refer to @6005's answer\n\n\u2022 Actually, $x = \\overline{(\\overline{x})^w}$ is a bit more complicated; it's not sufficient that $w$ is real. We have to pick a definition of complex exponential that allows for the symmetry that you appeal to. But regardless of which definition we take (branch of complex log), $x^w = \\overline{(\\overline{x})^w}$ will be false for $x$ negative real. Consider $w = \\tfrac12$. \u2013\u00a06005 Aug 21 '16 at 11:51\n\u2022 Yeah. That is indeed true. I'll add in my answer to look at yours for a more detailed explanation on when/why this works. \u2013\u00a0b00n heT Aug 21 '16 at 12:17", "date": "2019-08-22 04:43:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1898858/xy-xy-w-in-mathbbr-is-xwyw-real", "openwebmath_score": 0.9590098857879639, "openwebmath_perplexity": 271.67130758092463, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.986363161511632, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8604513062403369}}
{"url": "https://math.stackexchange.com/questions/1971193/what-is-the-volume-of-the-3-dimensional-elliptope", "text": "# What is the volume of the $3$-dimensional elliptope?\n\n### My question\n\nCompute the following double integral analytically\n\n$$\\int_{-1}^1 \\int_{-1}^1 2 \\sqrt{x^2 y^2 - x^2 - y^2 + 1} \\,\\, \\mathrm{d} x \\mathrm{d} y$$\n\n### Background\n\nThe $3$-dimensional elliptope is the spectrahedron defined as follows\n\n$$\\mathcal E_3 := \\Bigg\\{ (x_{12}, x_{13}, x_{23}) \\in \\mathbb R^3 : \\begin{bmatrix} 1 & x_{12} & x_{13}\\\\ x_{12} & 1 & x_{23}\\\\ x_{13} & x_{23} & 1\\end{bmatrix} \\succeq 0 \\Bigg\\}$$\n\nUsing Sylvester's criterion for positive semidefiniteness (i.e., all $2^3-1 = 7$ principal minors are nonnegative), we obtain $1 \\geq 0$ (three times), the three quadratic inequalities\n\n$$1 - x_{12}^2 \\geq 0 \\qquad \\qquad \\qquad 1 - x_{13}^2 \\geq 0 \\qquad \\qquad \\qquad 1 - x_{23}^2 \\geq 0$$\n\nand the cubic inequality.\n\n$$\\det \\begin{bmatrix} 1 & x_{12} & x_{13}\\\\ x_{12} & 1 & x_{23}\\\\ x_{13} & x_{23} & 1\\end{bmatrix} = 1 + 2 x_{12} x_{13} x_{23} - x_{12}^2 - x_{13}^2 - x_{23}^2 \\geq 0$$\n\nThus, $\\mathcal E_3$ is contained in the cube $[-1,1]^3$. Borrowing the pretty figure in Eisenberg-Nagy & Laurent & Varvitsiotis, here is an illustration of $\\mathcal E_3$\n\nWhat is the volume of $\\mathcal E_3$?\n\n### Motivation\n\nWhy is $\\mathcal E_3$ interesting? Why bother? Because $\\mathcal E_3$ gives us the set of $3 \\times 3$ correlation matrices.\n\n### My work\n\nFor convenience,\n\n$$x := x_{12} \\qquad\\qquad\\qquad y := x_{13} \\qquad\\qquad\\qquad z := x_{23}$$\n\nI started with sheer brute force. Using Haskell, I discretized the cube $[-1,1]^3$ and counted the number of points inside the elliptope. I got an estimate of the volume of $\\approx 4.92$.\n\nI then focused on the cubic surface of the elliptope\n\n$$\\det \\begin{bmatrix} 1 & x & y\\\\ x & 1 & z\\\\ y & z & 1\\end{bmatrix} = 1 + 2 x y z - x^2 - y^2 - z^2 = 0$$\n\nwhich I rewrote as follows\n\n$$z^2 - (2 x y) z + (x^2 + y^2 - 1) = 0$$\n\nUsing the quadratic formula, I obtained\n\n$$z = x y \\pm \\sqrt{x^2 y^2 - x^2 - y^2 + 1}$$\n\nIntegrating using Wolfram Alpha,\n\n$$\\int_{-1}^1 \\int_{-1}^1 2 \\sqrt{x^2 y^2 - x^2 - y^2 + 1} \\,\\, \\mathrm{d} x \\mathrm{d} y = \\cdots \\color{gray}{\\text{(magic happens)}} \\cdots = \\color{blue}{\\frac{\\pi^2}{2} \\approx 4.9348}$$\n\nI still would like to compute the double integral analytically. I converted to cylindrical coordinates, but did not get anywhere.\n\n### Other people's work\n\nThis is the same value Johnson & N\u00e6vdal obtained in the 1990s:\n\nThus, the volume is\n\n$$\\left(\\frac{\\pi}{4}\\right)^2 2^3 = \\frac{\\pi^2}{2}$$\n\nHowever, I do not understand their work. I do not know what Schur parameters are.\n\nHere's the script:\n\n-- discretization step\ndelta = 2**(-9)\n\n-- discretize the cube [-1,1] x [-1,1] x [-1,1]\ngrid1D = [-1,-1+delta..1]\ngrid3D = [ (x,y,z) | x <- grid1D, y <- grid1D, z <- grid1D ]\n\n-- find points inside the 3D elliptope\npoints = filter (\\(x,y,z)->1+2*x*y*z-x**2-y**2-z**2>=0) grid3D\n\n-- find percentage of points inside the elliptope\np = (fromIntegral (length points)) / (1 + (2 / delta))**3\n\n\n*Main> delta\n1.953125e-3\n*Main> p\n0.6149861105903861\n*Main> p*(2**3)\n4.919888884723089\n\n\nHence, approximately $61\\%$ of the grid's points are inside the elliptope, which gives us a volume of approximately $4.92$.\n\n### A new Buffon's needle\n\nA symmetric $3 \\times 3$ matrix with\n\n\u2022 $1$'s on the main diagonal\n\n\u2022 realizations of the random variable whose PDF is uniform over $[-1,1]$ on the entries off the main diagonal\n\nis positive semidefinite (and, thus, a correlation matrix) with probability $\\left(\\frac{\\pi}{4}\\right)^2$. Estimating the probability, we estimate $\\pi$. Using the estimate given by the Haskell script:\n\n*Main> 4 * sqrt 0.6149861105903861\n3.1368420058151125\n\n\n### References\n\n\u2022 This is an astoundingly good question! Nice! \u2013\u00a0clathratus Apr 16 at 1:08\n\nThe integral can be separated:\n\n$$I = 2\\int_{-1}^1 \\sqrt{1-x^2} dx \\cdot \\int_{-1}^1 \\sqrt{1-y^2} dy = 2\\left(\\int_{-1}^1 \\sqrt{1-t^2} dt\\right)^2$$\n\nThis integral is straight-forward using the substitution $$t=\\sin\\theta$$:\n\n$$\\int_{-1}^1 \\sqrt{1-t^2} dt = \\int_{-\\pi/2}^{\\pi/2} \\sqrt{1-\\sin^2\\theta} \\cos\\theta d\\theta = \\int_{-\\pi/2}^{\\pi/2} |\\cos\\theta|\\cos\\theta d\\theta$$\n\n$$=\\int_{-\\pi/2}^{\\pi/2} \\cos^2\\theta d\\theta = \\dfrac{1}{2}\\int_{-\\pi/2}^{\\pi/2} (1+\\cos2\\theta) d\\theta = \\dfrac{1}{2}\\left(\\theta + \\dfrac{1}{2}\\sin 2\\theta\\right)\\Big|_{-\\pi/2}^{\\pi/2} = \\dfrac{\\pi}{2}$$\n\nTherefore\n\n$$I = 2\\left(\\int_{-1}^1 \\sqrt{1-t^2} dt\\right)^2 = 2\\left(\\dfrac{\\pi}{2}\\right)^2 = \\dfrac{\\pi^2}{2}$$\n\n\u2022 Thanks. Shame on me for not realising that $(1-x^2) (1-y^2) = 1 - x^2 - y^2 + x^2 y^2$. I never expected the double integral to be this easy. \u2013\u00a0Rodrigo de Azevedo Oct 16 '16 at 19:19\n\u2022 easy to get lost in the bigger picture, sometimes fresh eyes can help \u2013\u00a0David Peterson Oct 16 '16 at 19:21\n\nThen integrand factors as $\\sqrt{(1-x^2)(1-y^2)}=\\sqrt{(1-x^2)}\\sqrt{(1-y^2)}$ and every factor can be integrated separately. But you recognize the integral for the area of a half circle of radius $1$, hence\n\n$$I=2\\left(\\frac\\pi2\\right)^2.$$", "date": "2019-06-19 02:40:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1971193/what-is-the-volume-of-the-3-dimensional-elliptope", "openwebmath_score": 0.8970597386360168, "openwebmath_perplexity": 491.0368929822824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631619124992, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8604513016798521}}
{"url": "https://math.stackexchange.com/questions/2340167/is-the-whole-set-mathbb-r-open/2340171", "text": "# Is the whole set $\\mathbb R$ open?\n\nIs the whole set $\\mathbb R$ open or closed? A lot of answers from the following link said that the whole set is closed.\n\nWhy is empty set an open set?\n\nHowever varies notes said that $\\mathbb R$ is open, for example:\n\nhttps://www.math.cornell.edu/~hatcher/Top/TopNotes.pdf\n\n\u2022 Open does not mean not closed. \u2013\u00a0Improve Jun 29 '17 at 2:58\n\u2022 It is both... some say \"clopen\" the empty set is the another classic example of a set that is both open and closed. \u2013\u00a0Doug M Jun 29 '17 at 3:24\n\u2022 \"Is the whole set R open or closed?\" both. \"Said that the whole set is closed\". That is true. \"However varies notes said that R is open\". That is also true. R is open. And R is closed. \u2013\u00a0fleablood Jun 29 '17 at 4:00\n\nIt's open and closed by definition.\n\nIn order to refer to open and closed sets a topology must be made explicit. Given a set $X$ and a collection $\\mathcal F$ of subsets of $X$, $\\mathcal F$ is a topology on $X$ only if $X, \\emptyset \\in \\mathcal F$. Thus X is open by definition, and as closed sets are defined as sets whose compliments are open we have $X^c=\\emptyset$ and we have $X$ is closed.\n\nIn the context of a topological space $\\mathbb R$ with collection $\\mathcal F$, $\\mathbb R$ must be open and closed. However if we consider $\\mathbb R \\times \\mathbb R$ with the usual topology, $\\mathbb R \\simeq \\{0\\} \\times \\mathbb R$ is closed but not open. It is important to state context.\n\nIndeed the observation by Michael Hardy does really make sense. Additionally the question is not well posed because it does not state which topology is to be considered. Let us assume that euclidean topology is meant. So the answer by Michael Rozenberg uses axiomatization via open or closed sets.\n\nUsing axiomatization via neighborhoods (read balls in this case, due to the euclidean topology of $\\mathbb R$)\n\n1. $\\mathbb R$ is open because any of its points have at least one neighborhood (in fact all) included in it;\n\n2. $\\mathbb R$ is closed because any of its points have every neighborhood having non-empty intersection with $\\mathbb R$ (equivalently punctured neighborhood instead of neighborhood).\n\nEquivalently:\n\n1. $\\mathbb R$ is open because all its points are interior points of itself\n\n2. $\\mathbb R$ is closed because all its points are adherent points of itself (equivalently limit points instead of adherent points)\n\nUsing axiomatization via Moore-Smith net convergence (read sequence convergence in this case, due to the euclidean topology of $\\mathbb R$),\n\n1. $\\mathbb R$ is closed because every point to which at least one net of its points converges belongs to it\n\n2. $\\mathbb R$ is open because every point to which every convergent net converges has a non-empty intersection with it. (Or equivalently there are no nets of points of its complement (the empty set) converging to any of its points (in fact there are no nets of points of its complement))\n\nThe list can go on and on.\n\nBy the definition of a topology on a set, both the empty set and the entire set (which I'm assuming you're taking as $\\mathbb R$) are open sets. Since the complement of an open set is a closed set, and $\\mathbb R$ is the complement of the empty set, it is also closed.\n\nNotice that \"open\" and \"closed\" are not mutually exclusive.\n\n$A$ is open iff $A^c$ is closed.\n\nSo $\\emptyset$ and $\\Omega$ are both open and closed (clopen)\n\nFor general topology, it is guaranteed via definition of topology, i.e. $\\emptyset, \\Omega \\in \\mathscr T$.\n\nBy one commonplace definition of \"open set\", a set $A\\subseteq\\mathbb R$ is open if for every point $x\\in A$ there is some open interval containing $x$ that is a subset of $A$. That clearly is true if $A=\\mathbb R,$ so $\\mathbb R$ is open.\n\nBy one commonplace definition of \"closed set\", a set $A\\subseteq \\mathbb R$ is closed if every limit point of $A$ is a member of $A$. That clearly is true if $A=\\mathbb R$, so $\\mathbb R$ is closed.\n\nOnly two subsets of $\\mathbb R$ are both open and closed: $\\mathbb R$ and $\\varnothing.$", "date": "2019-10-16 07:05:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2340167/is-the-whole-set-mathbb-r-open/2340171", "openwebmath_score": 0.871317446231842, "openwebmath_perplexity": 215.2971024616699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806506825456, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8604367560859089}}
{"url": "https://www.physicsforums.com/threads/integrating-arcsech-x.754086/", "text": "# Integrating arcsech x\n\n1. May 17, 2014\n\n### sooyong94\n\n1. The problem statement, all variables and given/known data\nI was asked to prove the integral\n\n$\\int_{\\frac{4}{5}}^{1} \\textrm{arcsech}(x) =2\\arctan 2-\\frac{\\pi}{2}-\\frac{4}{5} \\ln 2$\n\n2. Relevant equations\nIntegration by parts\n\n3. The attempt at a solution\nLet $u=\\textrm{arcsech} (x)$\n$\\textrm{sech u}=x$\n$\\textrm{cosh u}=\\frac{1}{x}$\nDifferentiating implicitly,\n\n$\\textrm{sinh u} \\frac{du}{dx}=\\frac{-1}{x^{2}}$\n$\\frac{du}{dx}=\\frac{-1}{x^{2}\\textrm{sinh u}}$\n\nThen I simplify it into\n$\\frac{-1}{x\\sqrt{1-x^{2}}}$\n\nLet $\\frac{dv}{dx}=1$\n\n$v=x$\n\nUsing integration by parts and evaluating the integral, I got\n$\\frac{\\pi}{2}-\\sin^{-1} \\frac{4}{5} -\\frac{4}{5} \\ln 2$\n\nWhich is numerically correct. But how do I obtain $2\\tan^{-1} 2$ as shown in the question above?\n\n2. May 17, 2014\n\n### Curious3141\n\nDraw the standard 3-4-5 right triangle. Observe that $\\frac{\\pi}{2}-\\sin^{-1} \\frac{4}{5} = \\tan^{-1}\\frac{3}{4}$.\n\nYou now have to get $\\tan^{-1}\\frac{3}{4}$ into something with $\\tan^{-1}2$ in it.\n\nI found this a little tricky. The best solution I could find was to let:\n\n$\\tan^{-1}\\frac{3}{4} = x$ so $\\tan x = \\frac{3}{4}$\n\nThen let x = 2y so that $\\tan 2y = \\frac{3}{4}$\n\nSolve for y in the form $y = \\tan^{-1}z$, where z is something you have to find. Only one value is admissible. Express x as 2y = $2\\tan^{-1}z.$\n\nNow observe that $\\frac{1 + \\tan w}{1 - \\tan w} = \\tan(w + \\frac{\\pi}{4})$. Use that to find an alternative form for $\\tan^{-1}z$, which will allow you to find x, in the form you need.\n\nThere might be a simpler way (indeed, it might start with an alternative solution of the integral), but I can't immediately find one.\n\nLast edited: May 17, 2014\n3. May 17, 2014\n\n### AlephZero\n\nTo show the answers are the same, you have to show $2 \\tan^{-1} 2 + \\sin^{-1}\\displaystyle\\frac 4 5 = \\pi$.\n\nFrom a 3-4-5 triangle, $\\sin^{-1}\\displaystyle\\frac 4 5 = \\tan^{-1}\\displaystyle\\frac 4 3$.\n\nFrom $\\tan^{-1}a + \\tan^{-1}b = \\tan^{-1}\\displaystyle\\frac{a+b}{1-ab}$,\n\n$2 \\tan^{-1} 2 = \\tan^{-1}\\displaystyle\\frac {-4} 3 = \\pi - \\tan^{-1}\\displaystyle\\frac 4 3$.\n\nQED.\n\n4. May 22, 2014\n\n### sooyong94\n\nI don't get it, but why\n$2 \\tan^{-1} 2 = \\tan^{-1}\\displaystyle\\frac {-4} 3 = \\pi - \\tan^{-1}\\displaystyle\\frac 4 3$.\n\n5. May 22, 2014\n\n### CAF123\n\nI think the equalities should be $$2 \\tan^{-1} 2 = \\tan^{-1} \\left(-\\frac{4}{3}\\right) + \\pi = \\pi - \\tan^{-1} \\left(\\frac{4}{3}\\right)$$\n\nUsing Alephzero's formula with $a=b$ gives $$\\tan(\\tan^{-1} 2 + \\tan^{-1} 2) = -\\frac{4}{3} \\Rightarrow 2\\tan^{-1} 2 = \\tan^{-1} \\left(-\\frac{4}{3}\\right) + \\pi$$\n\n6. May 22, 2014\n\n### SammyS\n\nStaff Emeritus\nCompare the graphs of $\\ y=\\text{arcsech}(x) \\$ and $y=\\text{sech}(x)\\ .$\n\nIntegrate $y=\\text{sech}(x)\\$ to get an area related to that given by integrating $\\ y=\\text{arcsech}(x) \\ .$\n\nYou will have to subtract the area of some rectangle.\n\n7. May 23, 2014\n\n### sooyong94\n\nI plotted two graphs and yet I can't figure it out... :(\n\n8. May 23, 2014\n\n### SammyS\n\nStaff Emeritus\nThe definite integral you are evaluating represents the area below the y = arcsech(x) graph which is between x = 4/5 and x = 1 . Notice that arcsech(4/5) = ln(2) .\n\nThat is the same as the area below the y = sech(x) graph and above y = 4/5, for x \u2265 0. Right?\n\nFile size:\n39.1 KB\nViews:\n272", "date": "2017-12-17 04:26:54", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integrating-arcsech-x.754086/", "openwebmath_score": 0.8789370656013489, "openwebmath_perplexity": 928.686408494233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806484125338, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8604367540940261}}
{"url": "https://mathhelpboards.com/threads/primitive-root-modulo-169.7398/", "text": "# Number TheoryPrimitive root modulo 169\n\n#### tda120\n\n##### New member\nHow can I find a primitive root modulo 169?\nI found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I\u2019m sure there\u2019s a smarter way than trying 2^the orders that divide phi(13^2)..?.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nHow can I find a primitive root modulo 169?\nI found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I\u2019m sure there\u2019s a smarter way than trying 2^the orders that divide phi(13^2)..?.\nHi tda, and welcome to MHB. You might be interested in this link, which tells you that the answer to your question is either $2$ or $2+13=15$. That still leaves you with the work of testing to see if $2$ works. If it does not, then $15$ does.\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\ntda120 said:\nHow can I find a primitive root modulo 169?\nThere is no simple method. You can cobble up together some basic theories on primitive roots, find a bit of a rough upperbound (although none is known to be useful for small cases) and some modular exponentiation to get a fast enough algorithm.\n\nAs Opalg there indicated, that either 2 or 15 is primitive root modulo 169, can easily be found for this case. Try proving the former and then the later if it doesn't work.\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nHow can I find a primitive root modulo 169?\nI found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I\u2019m sure there\u2019s a smarter way than trying 2^the orders that divide phi(13^2)..?.\nYou don't have to check all the orders that divide $\\phi(13^2)$.\nIt suffices to check each of the orders that are $\\phi(13^2)$ divided by one of the distinct primes it contains.\n\n$$\\phi(13^2)=2^2\\cdot 3 \\cdot 13$$\nSo the orders to verify are:\n$$2\\cdot 3 \\cdot 13,\\quad 2^2 \\cdot 13, \\quad 2^2\\cdot 3$$\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nHi tda, and welcome to MHB. You might be interested in this link, which tells you that the answer to your question is either $2$ or $2+13=15$. That still leaves you with the work of testing to see if $2$ works. If it does not, then $15$ does.\nNice!\n\nFrom that link we also get that since 2 is a primitive root mod 13, it follows that the order of 2 mod 169 is either (13-1) or 13(13-1).\nSo if $2^{13-1} \\not\\equiv 1 \\pmod{169}$ that means that 2 has to be a primitive root mod 169. Or otherwise 15 has to be.\n\nIn other words, no need to check any of the other powers.", "date": "2021-06-14 22:26:55", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/primitive-root-modulo-169.7398/", "openwebmath_score": 0.8173380494117737, "openwebmath_perplexity": 478.2527666349904, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9805806552225684, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8604367522150119}}
{"url": "http://math.stackexchange.com/questions/817189/finding-the-coefficient-on-the-x-term-of-prod-n-120x-n", "text": "# Finding the coefficient on the $x$ term of ${\\prod_{n = 1}^{20}(x-n)}.$\n\nI am trying to find the coefficient on the $x$ term of $\\displaystyle{\\prod_{n = 1}^{20}(x-n)}$. The issue is that the binomial theorem can't be applied since our $b$ value is changing from term to term. Is there any simple way to do this problem, perhaps a way to change the expression so that the binomial theorem applies? I've tried to do that, and tried looking for a pattern on similar expressions, but I haven't come up with anything. Any help you might have would be appreciated.\n\n-\nIt will be the 19th elementary symmetric polynomial in your roots. See en.m.wikipedia.org/wiki/Elementary_symmetric_polynomial for symmetric polynomials. \u2013\u00a0Marc Jun 1 '14 at 17:11\nA concept popularly known as sum of roots of a polynomial can be of help. \u2013\u00a0fermesomme Jun 1 '14 at 17:22\n\nThe coefficient is given by $$\\sum_{k=1}^{20}\\prod_{n\\not=k\\atop n=1,\\dots,20}{(-n)}=-\\sum_{k=1}^{20}\\frac{20!}{k}=-20! H_{20}$$ where $H_k$ is the $k$-th Harmonic number. Those can be looked up in tables (see A001008 and A002805): $H_{20}=\\frac{55835135}{15519504}$ and thus the coefficient is given by $-8752948036761600000$.\n\n-\nThis answer just gets better and better on a moment-by-moment basis! Thanks for adding the value of $H_{20}$! \u2013\u00a0Robert Lewis Jun 1 '14 at 18:30\nJust a question, is there a simple way to calculate the constant of the expanded form? \u2013\u00a0recursive recursion Jun 1 '14 at 19:47\n@Dominik, Also, I'm not sure how you got your first expression. Could you please explain that \u2013\u00a0recursive recursion Jun 1 '14 at 19:51\nYou get terms with a single power in $x$ from the product $\\prod_{n=1}^{20}(x-n)$ by taking $19$ non-$x$ factors and one $x$. The sum is the sum over the different choices of those non-$x$ and $x$ factors. More formally you could try to prove that $(-1)^{N+1} N! H_{N}$ is the $x$-coefficient of $\\prod_{n=1}^N (x-n)$ by induction. \u2013\u00a0Dominik Jun 1 '14 at 20:14\n\nThis is one of Vieta's formulas.\n\n-\nThe coefficient must be less than $-20!$, so this can't be correct? \u2013\u00a0copper.hat Jun 1 '14 at 17:22\nNot to put too fine a point on it, but this answer is incorrect; the coefficient of $x^{19}$ is in fact $-\\sum_{n = 1}^{20} n = -(20 \\times 21)/2 = -210$; for the coefficient of $x$, see the answer given by Dominik. I do believe it is correct. \u2013\u00a0Robert Lewis Jun 1 '14 at 17:31\n@RobertLewis: Unfortunately no 'Med' time for me this morning... \u2013\u00a0copper.hat Jun 1 '14 at 17:37\n@Robert Lewis. Shoot! You're right! I'll remove everything but the vieta's formula link. \u2013\u00a0Avi Steiner Jun 1 '14 at 17:40\n@Avi Steiner: well done! Glad we got it right! \u2013\u00a0Robert Lewis Jun 1 '14 at 19:40\n\nOne way is to just use Maclaurin's formula: $$f(x) = f(0) + \\frac{f'(0)}{1!} x + \\ldots$$ In this case: \\begin{align} f'(x) &= \\sum_{1 \\le n \\le m} \\prod_{\\substack{1 \\le k \\le m\\\\k \\ne n}} (x - k) \\\\ f'(0) &= (-1)^{m - 1} m! H_m \\end{align}\n\n-\n\nThe question has been well answered by others. However, I would like to point out that this polynomial has a name---Wilkinson's polynomial---and its own Wikipedia article. It was put forward by Wilkinson as an example of an apparently innocuous polynomial with remarkable numerical-analytic properties.\n\n-\nThis polynomial is a lot more interesting than I thought! Thanks for giving me a name to google. \u2013\u00a0recursive recursion Jun 1 '14 at 21:29\n\nA generalization to other coefficients (and limits different from $20$) is given in the generating functions of Stirling numbers of the first kind, Pochammer symbols:\n\n$$(x)_n:=\\prod_{k=0}^{n-1}(x-k)=\\sum_{k=0}^n s(n,k)x^k.$$\n\n-\n\nThe coefficients are the Stirling numbers of the first kind :\n\n-", "date": "2016-02-08 13:14:08", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/817189/finding-the-coefficient-on-the-x-term-of-prod-n-120x-n", "openwebmath_score": 0.9720078706741333, "openwebmath_perplexity": 384.035142208471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126488274566, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8604320568888113}}
{"url": "http://www.lofoya.com/Solved/1724/four-different-objects-1-2-3-4-are-distributed-at-random-in-four", "text": "# Moderate Probability Solved QuestionAptitude Discussion\n\n Q. Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1, 2, 3, 4. What is the probability that none of the objects occupy the place corresponding to its number?\n \u2716\u00a0A. 17/24 \u2714\u00a0B. 3/8 \u2716\u00a0C. 1/2 \u2716\u00a0D. 5/8\n\nSolution:\nOption(B) is correct\n\nFirst of all, if we IGNORE the condition about where the objects can be placed, we can arrange the 4 different objects in 4! ways (= 24 ways).\n\nSo, we now must determine HOW MANY of those 24 arrangements are such that no objects occupy the location corresponding to its number.\n\nA quick way to do this is to LIST acceptable outcomes.\n\nIMPORTANT: We'll list each arrangement so that the first number represents the object that goes to location #1, the second number represents the object that goes to location #2, and so on.\n\nSo, for example, 3421 represents object #3 in location #1, object #4 in location #2, object #2 in location #3, and object #1 in location #4.\n\nLet's be systematic:\n\nArrangements where object #2 is in location #1\n\nThe possible arrangements where NO object is in the correct location are as follows:\n\n2143, 2341, 2413\n\nTotal number of arrangements $= 3$\n\nArrangements where object #3 is in location #1\n\nThe possible arrangements where NO object is in the correct location are as follows:\n\n3142, 3412, 3421\n\nTotal # of arrangements $= 3$\n\nArrangements where object #4 is in location #1\n\nThe possible arrangements where NO object is in the correct location are as follows:\n\n4123, 4312, 4321\n\nTotal # of arrangements $= 3$\n\nAltogether, the number of arrangements where no object is in the correct location,\n\n$= 3 + 3 + 3$\n\n$= 9$\n\nSo, $P(\\text{no object in correct location}) = \\dfrac{9}{24}$\n\n$= \\dfrac{3}{8}$\n\nThus, option (B) is the correct choice.\n\nEdit:\u00a0Based on Vaibhav's comment, the solution has been updated.\n\nEdit 2:\u00a0For an alternate solution see\u00a0Kartik's comment\n\nEdit 3: Sheldon\u00a0has provided a way to reach the correct answer without even solving the question\n\nEdit 4:\u00a0Suzen\u00a0has given an exhaustive\u00a0solution\n\nEdit 5:\u00a0For derangements\u00a0formula, check\u00a0Himanshu Singh's\u00a0comment.\n\nNote: For better understanding and different approaches do visit comment section.\n\n## (14) Comment(s)\n\nGaurav Karnani\n()\n\nTotal number of cases = 24\n\nlet us divide the case into categories where :\n\n1) single number is in right position = 8 cases; two for each $1,2,3,4$\n\n2) Two numbers are in right position = 6 cases $12,13,14,23,24,34$\n\n3) No case for three digits as if three are in correct position fourth one will definitely be, so only one case of $1234$ .\n\ntotal number of cases =15\n\nProbability = $9/24$\n\nHimanshu Singh\n()\n\nPlease use this formula for derangements :\n\n$!n = n!\\left(1-\\dfrac{1}{1!} + \\dfrac{1}{2!} - \\dfrac{1}{3!} + \\dfrac{1}{4!} ... (-1)^n \\times \\dfrac{1}{n!} \\right)$\n\nKartik\n()\n\nTotal number of ways of placing objects into places randomly,\n\n$=4 \\times 3 \\times 2 \\times 1=4!=24$\n\nNo. of ways of placing them such that only one of them gets correct place,\n\n\\begin{align*} = & 4 \\text{ (ways of choosing the correctly} \\\\ & \\text{ placed objects)} \\times \\\\ & \\times 2 \\text{ (to place next object wrongly)} \\\\ & \\times 1 \\text{ (to place next object wrongly)} \\\\ & \\times 1 \\text{ (to place next object wrongly)}\\\\ = & \\textbf{8} \\end{align*}\n\nNumber of ways of placing exactly two objects correctly,\n\n\\begin{align*} = & ^4C_2 \\text{ (to identify the two correctly placed} \\\\ & \\text{ objects their arrangment are unique}\\\\ & \\text{ for them to be correct)} \\times \\\\ & 1 \\text{ (to arrange next object wrongly)} \\times \\\\ & 1 \\text{ (to arrange next object wrongly)}\\\\ = & \\textbf{6} \\end{align*}\n\nNow,\n\nNumber of ways to place exctly 3 objects correct = ways to place all objects correct $= \\textbf{1}$\n\nTherefore, number of ways to place all digits wrongly,\n\n$=24-(8+6+1)$\n\n$=24-15$\n\n$=9$\n\nThus, Required probability,\n\n$=\\dfrac{9}{24}$\n\n$=\\dfrac{3}{8}$\n\nRaj\n()\n\nSince there's only one correct way of placing the objects as per the boxes- can't we work on the lines of using probability of an event happening = 1 - probability of that event not happening?\n\nSo therefore, can't the answer be 1 - (1/24) = 23/24?\n\nBrent\n()\n\nThat's a good idea, but it doesn't apply here.\n\nIf event A = NONE of the objects in the correct places, then the complement (event A NOT happening) will consist of any arrangement where some (perhaps all) of the objects are in their correct place(s).\n\nVaibhav\n()\n\nbut this is the case for 2 at first position. it can occupy position 3 and 4 as well . Similar case with other digits will be there.\n\nDeepak\n()\n\nYou are right Vaibhav, updated the solution.\n\nSuzen\n()\n\nHere's the full list:\n\n1234\n\n1243\n\n1423\n\n1432\n\n1324\n\n1342\n\n2134\n\n2143 OK\n\n2431\n\n2413 OK\n\n2341 OK\n\n2314\n\n3124\n\n3142 OK (Close to 1000pi, but no correlation really, or is there? Perhaps the nth position is not n for all decimal places. OK, an interesting idea to explore)\n\n3421 OK\n\n3412 OK\n\n3214\n\n3241\n\n4123 OK\n\n4132\n\n4321 OK\n\n4312 OK\n\n4213\n\n4231\n\nClearly $P = \\frac{9}{24} = \\frac{3}{8}$\n\nI know it's long winded, but I'm glad it matched!\n\nTesla\n()\n\nLet a particular number (say) number 2 occupies position 1.\n\nThen all possible arrangement are given as:\n\n(2,1,3,4), (2,1,4,3), (2,3,4,1), (2,4,1,4), (2,4,1,3), (2,4,3,1).\n\nOut of these six, three (2,1,3,4), (2,3,1,4), (2,4,3,1) are not acceptable because numbers 3 and 4 occupy the correct positions.\n\nRequired probability = 3/6 = 1/2\n\nSheldon\n()\n\nI am not sure what difficulty level this would fall into.\n\nIf I see this question in the real exam, I would spend about a minute solving it.\n\nAfter that, I would guess and move on. because it would be time consuming.\n\nLooking at the question, it is more obvious that at-least 1 number would fall into its corresponding place.\n\nSo definitely, the probability that the number wouldn't fall into its numbered place should be less than half.\n\nLooking at options:\n\nA. 17/24\n\nB. 3/8\n\nC. 1/2\n\nD. 5/8\n\nOnly option B is less than half. I would pick B and move on.\n\nBrent\n()\n\nThat's a great approach that allows you to minimize time spent on a question (that you feel is going nowhere) and maximize your guess.\n\nProbability questions are perfect for this, because most people have a gut feeling about how likely something is. As you suggest, it does seem unlikely (probability less than 0.5) that every object would be out of place, so B is the perfect guess.\n\nSteve\n()\n\nI agree with sam and abhishek\n\nAbhishek\n()\n\nthe correct answer will be $\\dfrac{9}{24} =\\dfrac{3}{8}$\n\nout of\n\n1234 1243 1324 1342 1423 1432\n2341 2314 2143 2134 2413 2431\n3124 3142 3412 3421 3214 3241\n4123 4132 4213 4231 4312 4321\n\n24 combinations\n\n2341 2143 2413\n3421 3412 3142\n4123 4312 4321\n\nare misplaced(dearranged)\n\nSo, the correct answer will be $\\dfrac{9}{24}$ i.e. $\\dfrac{3}{8}$. :)\n\nSam\n()\n\nI think Abhishek is correct...... answer should be 3/8", "date": "2016-10-23 22:04:57", "meta": {"domain": "lofoya.com", "url": "http://www.lofoya.com/Solved/1724/four-different-objects-1-2-3-4-are-distributed-at-random-in-four", "openwebmath_score": 0.964660108089447, "openwebmath_perplexity": 1329.0917578973158, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712645102011, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8604320551631099}}
{"url": "https://math.stackexchange.com/questions/2799739/limit-involving-series-and-greatest-integer-function", "text": "Limit involving Series and Greatest Integer Function\n\nIf $[$.$]$ denotes the greatest integer function, then find the value of $\\lim_{n \\to \\infty} \\frac{[x] + [2x] + [3x] + \u2026 + [nx]}{n^2}$\n\nWhat I did was, I wrote each greatest integer function $[x]$ as $x - \\{x\\}$, where $\\{.\\}$ is the fractional part. Hence, you get\n\n$\\lim_{n \\to \\infty} \\frac{\\frac{n(n+1)}{2}(x-\\{x\\})}{n^2}$\n\nThe limit should then evaluate to $\\frac{x-\\{x\\}}{2}$\n\nBut the answer given is $\\frac{x}{2}$. What am I missing here?\n\n$$\\frac{\\lfloor x\\rfloor+\\ldots+\\lfloor nx\\rfloor}{n^2}=\\frac{x+2x+\\ldots nx-\\{x\\}-\\ldots-\\{nx\\}}{n^2}=$$\n\n$$=\\frac{n(n+1)}{2n^2}x-\\frac{\\{x\\}+\\ldots+\\{nx\\}}{n^2}\\xrightarrow[n\\to\\infty]{}\\frac12x-0=\\frac x2$$\n\nsince the second addend above tends to zero:\n\n$$\\frac{\\{x\\}+\\ldots+\\{nx\\}}{n^2}\\le\\frac n{n^2}=\\frac1n\\xrightarrow[n\\to\\infty]{}0$$\n\n\u2022 I didn't really get how the second addend (fractional part thing) tends to zero, as n tends to infinity. Isn't it in an inderminate form as it is? Also, what has been done in the last line of the answer? \u2013\u00a0skb May 28 '18 at 20:20\n\u2022 @skb Every term $\\;\\{kx\\}\\;$ is less than $\\;1\\;$ , so that whole sum's numerator is less that $\\;1+1+\\ldots+1=n\\;$ ... \u2013\u00a0DonAntonio May 28 '18 at 20:22\n\u2022 Have you used the sandwich theorem in the last line? But shouldn't there be another expression less than it for it to work? \u2013\u00a0skb May 28 '18 at 20:24\n\u2022 @skb But isn't it obvious that the whole expression is greater than zero or equal to it? You can complete that argument... \u2013\u00a0DonAntonio May 28 '18 at 20:27\n\u2022 Got it. Thanks a lot for the help! \u2013\u00a0skb May 28 '18 at 20:27\n\nNote that by Stolz-Cesaro\n\n$$\\lim_{n \\to \\infty} \\frac{\\lfloor x\\rfloor + \\lfloor 2x \\rfloor + \\lfloor3x\\rfloor + \u2026 + \\lfloor nx]}{n^2}=\\lim_{n \\to \\infty} \\frac{\\lfloor(n+1)x\\rfloor}{(n+1)^2-n^2}=\\lim_{n \\to \\infty} \\frac{\\lfloor(n+1)x\\rfloor}{2n+1}=$$ $$=\\lim_{n \\to \\infty} \\frac{(n+1)x}{2n+1}-\\frac{\\{(n+1)x\\}}{2n+1}$$", "date": "2019-10-20 16:11:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2799739/limit-involving-series-and-greatest-integer-function", "openwebmath_score": 0.8791330456733704, "openwebmath_perplexity": 310.73487589094714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712645102011, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8604320551631099}}
{"url": "http://mathhelpforum.com/geometry/46622-pyramid-print.html", "text": "# a pyramid\n\n\u2022 August 24th 2008, 01:57 AM\nperash\na pyramid\nConsider a pyramid with a square base. The side length of the\nbase is 2 units and the height of the pyramid is 1 unit. Imagine\nplacing a cube inside this pyramid (resting on the base of the\npyramid) such that each of the four top corners of the cube is\ntouching each of the 4 slanted edges of the pyramid .\nFind the dimensions of the cube. That is, find the value, in\nunits, of the length of one edge of the cube.\n\u2022 August 24th 2008, 03:27 AM\nticbol\nQuote:\n\nOriginally Posted by perash\nConsider a pyramid with a square base. The side length of the\nbase is 2 units and the height of the pyramid is 1 unit. Imagine\nplacing a cube inside this pyramid (resting on the base of the\npyramid) such that each of the four top corners of the cube is\ntouching each of the 4 slanted edges of the pyramid .\nFind the dimensions of the cube. That is, find the value, in\nunits, of the length of one edge of the cube.\n\nSo the the top four corners of the cube are along the slanted edges of the pyramid. Then the four edges of the base of the cube are parallel to the edges of the base of the pyramid each to each.\n\nView the figure from the top, or on the top. We will get a vertical cross-section along one of the equal diagonals of the base of the pyramid.\nThe diagonal of the pyramid is 2sqrt(2) units long...by Pythagorean theorem.\nThe diagonal of the cube, whose edge is, say, x units long, is x*sqrt(2) units long.\n\nNow view that said cross-section vertically, or from the side, or from one of the un-cut corner of the base of the pyramid.\nThe figure is that of an isosceles triangle whose base is 2sqrt(2) units long, and whose height is 1 unit long.\nInside it is a rectangle whose base is x*sqrt(2) units long, and whose height is x units long.\nAbove the rectangle is a smaller isosceles triangle whose base is x*sqrt(2) and whose height is (1-x) units long.\nThe two isosceles triangles are similar, and so, proportional.\nBy proportion,\n2sqrt(2) /1 = x*sqrt(2) /(1-x)\n2 = x /(1-x)\n2(1-x) = x\n2 -2x = x\n2 = x +2x\nx = 2/3 unit long ---------------answer\n\u2022 August 24th 2008, 05:16 AM\nSoroban\nHello, perash!\n\nI used the same approach as ticbol, but got a different answer.\n\nQuote:\n\nConsider a pyramid with a square base. The side length of the base is 2 units\nand the height of the pyramid is 1 unit. Imagine placing a cube inside this pyramid\n(resting on the base of the pyramid) so that each of the four top corners of the cube\nis touching each of the 4 slanted edges of the pyramid.\n\nFind the dimensions of the cube.\nThat is, find the value, in units, of the length of one edge of the cube.\n\nLooking down on the pyramid, we see:\nCode:\n\n\u00a0 \u00a0 \u00a0 *-----------* \u00a0 \u00a0 \u00a0 | *\u00a0 \u00a0 \u00a0 * | \u00a0 \u00a0 \u00a0 |\u00a0 *\u00a0 *\u00a0 | \u00a0 \u00a0 \u00a0 |\u00a0 \u00a0 *\u00a0 \u00a0 | 2 \u00a0 \u00a0 \u00a0 |\u00a0 *\u00a0 *\u00a0 | \u00a0 \u00a0 \u00a0 | *\u00a0 \u00a0 \u00a0 * | \u00a0 \u00a0 \u00a0 *-----------* \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 2\nThe diagonal of this square is $2\\sqrt{2}$\n\nSlice the pyramid along a diagonal and we have this cross-section:\nCode:\n\n\u00a0 \u00a0 -\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * | * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 *\u00a0 |\u00a0 * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 *-----+-----* \u00a0 \u00a0 1\u00a0 \u00a0 \u00a0 \u00a0 * |\u00a0 \u00a0 |\u00a0 \u00a0 | * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 *\u00a0 |\u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 * \u00a0 \u00a0 :\u00a0 \u00a0 *\u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 2x|\u00a0 \u00a0 * \u00a0 \u00a0 :\u00a0 *\u00a0 \u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 - *---------*-----*-----*---------* \u00a0 \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u221a2\u00a0 \u00a0 \u00a0 :\u00a0 x\u00a0 :\u00a0 \u221a2-x\u00a0 :\nThe lower-right right triangle is similar to the largest right triangle.\n\nWe have: . $\\frac{2x}{\\sqrt{2}-x} \\:=\\:\\frac{1}{\\sqrt{2}} \\quad\\Rightarrow\\quad 2\\sqrt{2}x \\:=\\:\\sqrt{2} - x$\n\n. . $2\\sqrt{2}x + x \\:=\\:\\sqrt{2} \\quad\\Rightarrow\\quad (2\\sqrt{2}+1)x \\:=\\:\\sqrt{2} \\quad\\Rightarrow\\quad x \\:=\\:\\frac{\\sqrt{2}}{2\\sqrt{2}+1}\n$\n\nThe side of the cube is: . $2x \\;=\\;\\frac{2\\sqrt{2}}{2\\sqrt{2} + 1} \\;=\\;\\frac{2(4-\\sqrt{2})}{7}$\n\n\u2022 August 24th 2008, 04:58 PM\nticbol\nQuote:\n\nOriginally Posted by Soroban\nHello, perash!\n\nI used the same approach as ticbol, but got a different answer.\n\nLooking down on the pyramid, we see:\nCode:\n\n\u00a0 \u00a0 \u00a0 *-----------* \u00a0 \u00a0 \u00a0 | *\u00a0 \u00a0 \u00a0 * | \u00a0 \u00a0 \u00a0 |\u00a0 *\u00a0 *\u00a0 | \u00a0 \u00a0 \u00a0 |\u00a0 \u00a0 *\u00a0 \u00a0 | 2 \u00a0 \u00a0 \u00a0 |\u00a0 *\u00a0 *\u00a0 | \u00a0 \u00a0 \u00a0 | *\u00a0 \u00a0 \u00a0 * | \u00a0 \u00a0 \u00a0 *-----------* \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 2\nThe diagonal of this square is $2\\sqrt{2}$\n\nSlice the pyramid along a diagonal and we have this cross-section:\nCode:\n\n\u00a0 \u00a0 -\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * | * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 *\u00a0 |\u00a0 * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 *-----+-----* \u00a0 \u00a0 1\u00a0 \u00a0 \u00a0 \u00a0 * |\u00a0 \u00a0 |\u00a0 \u00a0 | * \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 *\u00a0 |\u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 * \u00a0 \u00a0 :\u00a0 \u00a0 *\u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 2x|\u00a0 \u00a0 * \u00a0 \u00a0 :\u00a0 *\u00a0 \u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 \u00a0 |\u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 - *---------*-----*-----*---------* \u00a0 \u00a0 \u00a0 :\u00a0 \u00a0 \u00a0 \u221a2\u00a0 \u00a0 \u00a0 :\u00a0 x\u00a0 :\u00a0 \u221a2-x\u00a0 :\nThe lower-right right triangle is similar to the largest right triangle.\n\nWe have: . $\\frac{2x}{\\sqrt{2}-x} \\:=\\:\\frac{1}{\\sqrt{2}} \\quad\\Rightarrow\\quad 2\\sqrt{2}x \\:=\\:\\sqrt{2} - x$\n\n. . $2\\sqrt{2}x + x \\:=\\:\\sqrt{2} \\quad\\Rightarrow\\quad (2\\sqrt{2}+1)x \\:=\\:\\sqrt{2} \\quad\\Rightarrow\\quad x \\:=\\:\\frac{\\sqrt{2}}{2\\sqrt{2}+1}\n$\n\nThe side of the cube is: . $2x \\;=\\;\\frac{2\\sqrt{2}}{2\\sqrt{2} + 1} \\;=\\;\\frac{2(4-\\sqrt{2})}{7}$\n\nI'm sorry to comment, since perash or anybody did not comment, but the cube should appear as a rectangle, not a square, in your diagram. The length of the base now of the rectangle should be (2x)sqrt(2). Not 2x as is in your diagram. You sliced along a diagonal of the base of the pyramid, remember.\n\u2022 August 26th 2008, 03:17 AM\nSoroban\nHello, ticbol!\n\nAnother blunder . . . *blush*\n\nQuote:\n\nThe cube should appear as a rectangle, not a square, in your diagram.\nThe length of the base now of the rectangle should be (2x)sqrt(2).\nNot 2x as is in your diagram.\nYou sliced along a diagonal of the base of the pyramid, remember.\n\nAbsolutely right!\n\nI'll try to correct my work and get back soon . . .", "date": "2015-05-29 00:36:17", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/46622-pyramid-print.html", "openwebmath_score": 0.9177049398422241, "openwebmath_perplexity": 8432.509098182818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856415, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8604320495990248}}
{"url": "https://math.stackexchange.com/questions/2115891/how-to-know-if-the-angle-is-positive-or-negative-in-inverse-trigonometry", "text": "# How to know if the angle is positive or negative in inverse trigonometry\n\nI have been doing this problem this problem:\n\n$$Cos[Tan^{-1}(-\\frac{2}{3})]$$ So I was instructed to draw a triangle to guide me so I did\n\nNow once I drew my triangle I found the hypotenuse, which is $$\\sqrt{13}$$\n\nAnd then I was able to obtain the answer to this expression which I got:\n\n$$\\frac{2\\sqrt{13}}{13}$$\n\nHowever, I am told I drew the triangle wrong, it is actually -2 (negative) and (3) positive. Why is it that the triangle is wrong? I was told the actual answer is\n\n$$\\frac{3\\sqrt{13}}{13}$$\n\nArctan has a range of $\\frac{-\\pi}{2}\\le{y}\\le\\frac{\\pi}{2}$. Now let $arctan\\frac{-2}{3}=y$. This implies $tany=\\frac{-2}{3}$. Because tan is negative, we know y must lie in quadrant IV. It cannot lie in quadrant I because tan is positive in quadrant I. Therefore we draw our angle as you have above. Now, note that $tan\\theta=\\frac{opposite}{adjacent}$. Therefore, you should have $-2$ where $3$ is in your picture, and you should have $3$ where $-2$ is. Your line should be drawn to the coordinate $(3,-2)$. Now, we find the hypotenuse as you have already done using the Pythagorean Theorem. We find that it is $\\sqrt{13}$ as you've noted. Now, because we are finding $cos(arctan\\frac{-2}{3})$, we will use the fact that $cos\\theta=\\frac{adjacent}{hypotenuse}$. So, this gives $\\frac{3}{\\sqrt{13}}$. Rationalizing we have $\\frac{3\\sqrt{13}}{13}$\n\u2022 Yes, tangent is negative in quadrant IV and II, but the range for arctan is limited to $\\frac{-\\pi}{2}\\le{y}\\le\\frac{\\pi}{2}$ so our angle cannot lie in quadrant II. It must lie in either quadrant I or IV. And since it is negative, it must lie in quadrant IV. \u2013\u00a0MathGuy Jan 27 '17 at 3:04\n\u2022 Yes, the ranges are different though depending on the inverse trig function. For arcsin the range is $\\frac{-\\pi}{2}\\le{y}\\le\\frac{\\pi}{2}$, for arccos the range is $0\\le{y}\\le\\pi$ and for arctan the range is $\\frac{-\\pi}{2}\\le{y}\\le\\frac{\\pi}{2}$ \u2013\u00a0MathGuy Jan 27 '17 at 3:12", "date": "2019-08-25 17:56:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2115891/how-to-know-if-the-angle-is-positive-or-negative-in-inverse-trigonometry", "openwebmath_score": 0.9125866293907166, "openwebmath_perplexity": 239.84451636178318, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9883127399388852, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8604223800195274}}
{"url": "https://mathhelpboards.com/threads/further-question-on-my-fractions-problem.9489/", "text": "# Further question on My Fractions problem\n\n#### tmt\n\n##### Active member\nI have a separate question on the same problem from my prior post.\n\nI need an equation for a tangent which has a slope of 5/6 and passes through (1,3)\n\ny-3 = 5/6(x-1)\n\nI simplify this to\n\ny= 5/6x +13/6\n\nHowever the answer given is y = 7/6(x) + 13/6\n\nWhere am I going wrong?\n\nYours,\n\nTimothy\n\n#### MarkFL\n\nStaff member\nYour line has the required slope, while the given answer has the wrong slope. It is most likely a typo somewhere, either in the statement of the problem or the given answer. Can you post the original problem in its entirety?\n\n#### tmt\n\n##### Active member\nPROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 . '\n\nThis is the end of the answer:\n\nThus, the slope of the line tangent to the graph at (1, 3) is\n\n$m = y' = \\displaystyle{ 3 (3-1)^2 - 2(1) \\over 3 (3-1)^2 } = \\displaystyle{ 10 \\over 12 } = \\displaystyle{ 5 \\over 6 }$ ,\n\nand the equation of the tangent line is\n\ny - ( 3 ) = (5/6) ( x - ( 1 ) ) ,\n\nor\n\ny = (7/6) x + (13/6) .\n\nI suspect it is a typo in the answer. Here is the link for the full answer.\n\nhttps://www.math.ucdavis.edu/~kouba...soldirectory/ImplicitDiffSol.html#SOLUTION 11\n\n#### MarkFL\n\nStaff member\nOkay, I see now...I assumed the slope was given as 5/6. Let's take a look at the problem. We are given the curve:\n\n$$\\displaystyle x^2+(y-x)^3=9$$\n\nSo, implicitly differentiating with respect to $x$, we find:\n\n$$\\displaystyle 2x+3(y-x)^2\\left(\\frac{dy}{dx}-1 \\right)=0$$\n\nSolving for $$\\displaystyle \\frac{dy}{dx}$$, we find:\n\n$$\\displaystyle \\frac{dy}{dx}=1-\\frac{2x}{3(y-x)^2}$$\n\nNow, when $x=1$, we find from the original curve:\n\n$$\\displaystyle 1^2+(y-1)^3=9$$\n\n$$\\displaystyle y=3$$\n\nAnd so we find the slope at the given point is:\n\n$$\\displaystyle \\left.\\frac{dy}{dx} \\right|_{(x,y)=(1,3)}=1-\\frac{2(1)}{3(3-1)^2}=1-\\frac{2}{12}=\\frac{5}{6}$$\n\nHence, using the point-slope formula, we obtain the tangent line:\n\n$$\\displaystyle y-3=\\frac{5}{6}(x-1)$$\n\n$$\\displaystyle y=\\frac{5}{6}x+\\frac{13}{6}$$\n\nHere is a plot of the curve and the tangent line:\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nPROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 . '\n\nThis is the end of the answer:\n\nThus, the slope of the line tangent to the graph at (1, 3) is\n\n$m = y' = \\displaystyle{ 3 (3-1)^2 - 2(1) \\over 3 (3-1)^2 } = \\displaystyle{ 10 \\over 12 } = \\displaystyle{ 5 \\over 6 }$ ,\n\nand the equation of the tangent line is\n\ny - ( 3 ) = (5/6) ( x - ( 1 ) ) ,\n\nor\n\ny = (7/6) x + (13/6) .\n\nI suspect it is a typo in the answer. Here is the link for the full answer.\n\nhttps://www.math.ucdavis.edu/~kouba...soldirectory/ImplicitDiffSol.html#SOLUTION 11\nI concur that both you and MarkFL are correct, the link you provided has a typo in the very last line (it is correct until that), and the correct tangent line has the equation:\n\n$y = \\dfrac{5}{6}x + \\dfrac{13}{6}$", "date": "2021-08-05 05:23:58", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/further-question-on-my-fractions-problem.9489/", "openwebmath_score": 0.8594733476638794, "openwebmath_perplexity": 499.87169727565464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.988312741660069, "lm_q2_score": 0.870597265050901, "lm_q1q2_score": 0.8604223699042137}}
{"url": "https://math.stackexchange.com/questions/2845813/find-all-values-of-k-for-kx2k2x-3-0-with-positive-roots", "text": "Find all values of $k$ for $kx^2+(k+2)x-3=0$ with positive roots.\n\n$$kx^2+(k+2)x-3=0$$\n\nThis quadratic has roots which are real and positive.\n\nFind all possible values of $k$.\n\n$$\u0394 = (k+8)^2-60$$ $$==> (k+8)^2-60>0$$ $$k>2\\sqrt{15}\\ - 8$$ and $$k<-2\\sqrt{15}\\ -8$$\n\nHowever this didn't look right. Any suggestions?\n\nEdit: I think I might have figured it out.\n\nSince we know,\n\n$$k\\not= 0$$ $$\\frac{-3}{k}>0$$ $$\\therefore k<0$$\n\nWhat I got earlier was not wrong, but rather incomplete.\n\n$$k<-2\\sqrt{15} -8$$ This can be ruled out since, $$k<0$$ $$\\therefore 2\\sqrt{15} -8 < k < 0$$\n\nIf someone could please still check my work that would be nice.\n\n\u2022 I appreciate the advice, I had already tried something but was not sure if it was the correct \"path\" for this problem so I didn't show it. \u2013\u00a0StrBoP Jul 9 '18 at 18:48\n\u2022 You're welcome. Note that positive numbers are always real, so the \"real\" in \"real and positive\" is redundant. \u2013\u00a0Shaun Jul 9 '18 at 18:54\n\u2022 Do you mean $(k+2)^2$? \u2013\u00a0Chris2018 Jul 9 '18 at 18:59\n\u2022 No, I can show more work if you'd like. But after simplifying to a greater extent you get a quadratic, Since that quadratic wasnt factorable I completed the square and then I got that. Sorry if I confused you. \u2013\u00a0StrBoP Jul 9 '18 at 19:02\n\nHint:\n\nUsing Vieta's formulas, you can see that the multiplication of the roots of a quadratic $ax^2+bx+c$ is given by $\\frac{c}{a},$ which is equal to $\\frac{-3}{k}$ in your case.\n\nSince both roots are positive, this means that $\\frac{-3}{k}$ must be positive and $k$ must be negative. Also, the sum of the roots (given by $-\\frac{b}{a}$) must be positive too: $-\\frac{k+2}{k} > 0$.\n\nSince we already know that $k$ is negative, we get $-(k+2) < 0 \\iff -2 < k$. Hence, $-2<k<0$ so far.\n\nNow in order for this quadratic to have real roots, its discriminant must be non-negative. After you write it down, you see that you should solve the inequality $$(k+2)^2-4k(-3) =(k+2)^2+12k \\geq 0$$ to find the range of values for $k,$ and then intersect it with $$-2 < k <0$$ at the end.\n\n$$(k+2)^2+12k \\geq 0 \\iff k\\geq \\sqrt{60}-8 \\text{ or } k\\leq -8 -\\sqrt{60}$$\n\nIntersecting this range with $-2 < k < 0$ gives $\\sqrt{60}-8 \\leq k<0$.\n\n\u2022 Might also be helpful to quote Vieta's formulas. \u2013\u00a0TheSimpliFire Jul 9 '18 at 18:47\n\u2022 A positive product of the roots is not enough. \u2013\u00a0Bernard Jul 9 '18 at 18:51\n\u2022 @stressed out , Thank you for the help \u2013\u00a0StrBoP Jul 9 '18 at 18:59\n\u2022 How do you get $-(k+2) < 0 \\iff 2 < k$? $$-(k+2) < 0 \\implies -2 < k$$ From this it does not follow that $2 < k$ \u2013\u00a0gd1035 Jul 9 '18 at 19:05\n\u2022 @stressedout Sorry for bothering once again, but I checked at the back of the book and the answer read: -8 +\u221a60 </ k < 0 \u2013\u00a0StrBoP Jul 9 '18 at 19:05\n\nYou have three conditions to check:\n\n1. This equation has real roots. It means it is a quadratic equation ($k\\ne 0$) and its discriminant should be positive: $$\\Delta=(k+2)^2+12k=k^2+16k+4=(k+8)^2-60>0,$$ so either $k<-8-2\\sqrt{15}\\:$ or $\\:k>-8+2\\sqrt{15}$.\n2. The roots must have the same sign, i.e. their product $-\\dfrac 3k>0$, which means $\\:\\color{red}{k<0}$.\n3. This sign must be positive. If the roots have the same sign, this sign is also the sign of their sum $\\:-\\dfrac{k+2}k$, which is the sign of the product $-k(k+2)$. To sum it up, we need to have $k(k+2)<0$, which happens if and only if $\\: \\color{red}{-2<k<0}$.\n\nNow note that$\\:-8-2\\sqrt{15}<-2$, and as $\\;3<\\sqrt{15}<4$ we have $\\:-2<-8+2\\sqrt{15}<0\\:$.\n\nThus eventually the solutions are $$-8+2\\sqrt{15}<k<0.$$\n\n\u2022 The maniac downvoter struck gain! \u2013\u00a0Bernard Jul 9 '18 at 23:10", "date": "2019-08-20 16:54:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2845813/find-all-values-of-k-for-kx2k2x-3-0-with-positive-roots", "openwebmath_score": 0.8804914355278015, "openwebmath_perplexity": 292.1029525472093, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127433812521, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8604223664253399}}
{"url": "https://math.stackexchange.com/questions/1852540/find-sum-m-0n-1m-mn-n-choose-m", "text": "# Find $\\sum_{m=0}^n\\ (-1)^m m^n {n \\choose m}$\n\nI'm going to university in October and thought I'd have a go at a few questions from one of their past papers. I have completed the majority of this question but I'm stuck on the very last part. In honesty I've been working on this paper a while now and I'm a bit tired so I'm probably giving up earlier than I usually would.\n\nI won't write out the full question, only the last part:\n\nLet $$S_r(n) = \\sum_{m=0}^n\\ (-1)^m m^r {n \\choose m}$$ where r is a non-negative integer . Show that $S_r(n)=0$ for $r<n$. Evaluate $S_n(n)$.\n\nI have shown that $S_r(n)=0$ for $r<n$ by taking $(1+z)^n= \\sum_{m=0}^n\\ z^m {n \\choose m}$, letting $D_r(f(z))=z\\frac d{dz}(z\\frac d{dz}...(\\frac d{dz}(f(z)))...)$ where $z\\frac d{dz}$ is applied $r$ times and applied it to both sides. The left hand side gives a polynomial, degree n which has factor $(1+z)$ for all $r<n$ and the right hand side gives $\\sum_{m=0}^n\\ z^m (m)^r {n \\choose m}$. Setting $z=-1$ yields the required result.\n\nThere is some build up to this, so I'm fairly certain that this was the intended method.\n\nI'm stuck however on the very last part. I have tried finding a form for $D_r((1+z)^n)$, but I'm fairly sure that this isn't the correct approach, as the wording of the question implies that $S_n(n)$ needs to be considered separately.\n\nI'm surprised that I didn't find that this question had already been asked, so apologies if it has been.\n\nThank you.\n\n\u2022 Looks like a close relative of Stirling numbers of the second kind. Please see Wikipedia. \u2013\u00a0Andr\u00e9 Nicolas Jul 7 '16 at 22:01\n\u2022 Ok thanks, will do \u2013\u00a0Aka_aka_aka_ak Jul 7 '16 at 22:12\n\u2022 You are welcome. Note the close relationship to the number of onto functions. The calculation you carried out successfully (and that I would have trouble with) can be bypassed once we note there are $0$ onto functions from an $r$-element set to an $n$-element set if $r\\lt n$. \u2013\u00a0Andr\u00e9 Nicolas Jul 7 '16 at 22:23\n\nThe following relation encapsulates the Stirling number semantics:\n\n$$m^r = r! [z^r] \\exp(mz).$$\n\n$$S_r(n) = r! [z^r] \\sum_{m=0}^n {n\\choose m} (-1)^m \\exp(mz) = r! [z^r] (1-\\exp(z))^n.$$\n\nNow observe that\n\n$$1-\\exp(z) = - z - \\frac{z^2}{2} - \\frac{z^3}{6} -\\cdots$$\n\nwhich means that $(1-\\exp(z))^n$ starts at $[z^r]$ where $r=n$ with coefficient $(-1)^n$, producing for the sum the value\n\n$$(-1)^n\\times n!$$\n\nand the coefficients on $[z^r]$ with $r\\lt n$ are zero.\n\nThe general form of the summation\n\n$$\\sum_{m=0}^n(-1)^mm^n\\binom{n}m\\;,\\tag{1}$$\n\nwith the alternating sign and the binomial coefficient $\\binom{n}m$, suggests that it can be interpreted as an inclusion-exclusion calculation. However, the $m=0$ term is $0$, from which we begin by subtracting a positive quantity $\\binom{n}1=n$, which is a bit odd for such a calculation. This suggests letting $k=n-m$ and rewriting as\n\n$$\\sum_{k=0}^n(-1)^{n-k}(n-k)^n\\binom{n}{n-k}=(-1)^n\\sum_{k=0}^n(-1)^k\\binom{n}k(n-k)^n\\;,$$\n\nwhere the righthand side has been arranged to look like a more or less typical inclusion-exclusion calculation. The factor $(n-k)^n$ is the one that isn\u2019t part of the inclusion-exclusion machinery. It has a natural interpretation as the number of functions from $[n]$ to $[n]$ whose ranges are disjoint from some $k$-element subset of $[n]$. If for each $k\\in[n]$ we let $F_k$ be the set of functions from $[n]$ to $[n]$ whose ranges do not contain $k$, we have\n\n$$\\left|\\bigcap_{k\\in I}F_k\\right|=(n-|I|)^n$$\n\nwhenever $\\varnothing\\ne I\\subseteq[n]$, so\n\n\\begin{align*} \\left|\\bigcup_{k=1}^nF_k\\right|&=\\sum_{\\varnothing\\ne I\\subseteq[n]}(-1)^{|I|-1}\\left|\\bigcap_{k\\in I}F_k\\right|\\\\ &=\\sum_{\\varnothing\\ne I\\subseteq[n]}(-1)^{|I|-1}(n-|I|)^n\\\\ &=\\sum_{k=1}^n(-1)^{k-1}\\binom{n}k(n-k)^n\\;. \\end{align*}\n\nThis is the number of functions from $[n]$ to $[n]$ that are not surjective, i.e., the number that aren\u2019t bijections. It follows that the number of bijections from $[n]$ to $[n]$ is\n\n$$n^n-\\sum_{k=1}^n(-1)^{k-1}\\binom{n}k(n-k)^n=\\sum_{k=0}^n(-1)^k\\binom{n}k(n-k)^n\\;.$$\n\nOn the other hand, we know that there are $n!$ bijections from $[n]$ to $[n]$, so\n\n$$\\sum_{m=0}^n(-1)^mm^n\\binom{n}m=(-1)^n\\sum_{k=0}^n(-1)^k\\binom{n}k(n-k)^n=(-1)^nn!\\;.$$\n\nNote that the same basic argument handles the previous part of the problem. The factor $(n-k)^n$ is replaced by $(n-k)^r$, the number of functions from $[r]$ to $[n]$ whose ranges are disjoint from some specified $k$-element subset of $[n]$. The expression\n\n$$\\sum_{k=0}^n(-1)^k\\binom{n}k(n-k)^r$$\n\ncounts the surjective functions from $[r]$ to $[n]$, and if $r<n$, this is of course $0$.\n\nAll of this is very closely related to the Stirling numbers of the second kind.\n\n\u2022 @BrianMScott: Very nice and clearly written exposition. (+1) \u2013\u00a0Markus Scheuer Jun 30 '18 at 16:51", "date": "2019-08-20 11:43:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1852540/find-sum-m-0n-1m-mn-n-choose-m", "openwebmath_score": 0.903042197227478, "openwebmath_perplexity": 110.59646952772967, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370154, "lm_q2_score": 0.8705972600147105, "lm_q1q2_score": 0.8604223661256483}}
{"url": "http://khhk.agence-des-4-fontaines.fr/least-squares-solver.html", "text": "# Least Squares Solver\n\nDefine least squares. In order for the solution to represent sensible pixel values, restrict the solution to be from 0 through 1. 1 Introduction. Students, teachers, parents, and everyone can find solutions to their math problems instantly. The numerical instability and performance are issues of larger problems and general setting. This example shows how to solve a nonlinear least squares problem in two ways. While symbolically correct, using the QR decomposition instead is numerically more robust. - linear_least_squares. Solve least-squares (curve-fitting) problems. Sitio Espejo para Am\u00e9rica Latina. Linear regression calculator Two-dimensional linear regression of statistical data is done by the method of least squares. Linear Regression calculator uses the least squares method to find the line of best fit for a sets of data X and Y or the linear relationship between two dataset. Let us understand What is Linear Regression and how to perform it with the help Ordinary Least Squares (OLS) estimator with an example. 1 Solving Least Squares Systems: SVD Approach One way to solve overdetermined systems is to use the Singular Value Decomposition of a matrix. solver to vary the values for A, C and k to minimize the sum of chi squared. LMS incorporates an. Solve any equations from linear to more complex ones online using our equation solver in just one click. Linear vs. From the geometric perspective, we can deal with the least squares problem by the following logic. Enter the statistical data in the form of a pair of numbers, each pair is on a separate line. Algorithm 1 Least-squares sub-problem input: H 2 CN \u21e5N, q 2 CN, P 2 RM \u21e5N, d 2 CM for each receiver (j )(rowinP) in parallel do H\u21e4 w j = p\u21e4 j {solve 1 PDE} end for W =[w 1 w 2w m] {distributed matrix} S =(I M + 2 W\u21e4 W)1 {adjust using Algorithm 2 (optional)} for source (i) in parallel do y i =(I N 2 WSW\u21e4)(q i + 2 Wd i) Hu i = y i {solve 1 PDE} end for output: u. The main purpose is to provide an example of the basic commands. Recall the formula for method of least squares. Microsoft Excel provides a tool called Solver that handles this prob-lem in a manner that is transparent to the user. Now click on \ufb01Solve\ufb02. Find a linear least squares fit for a set of points in Visual Basic. In this section w e brie y presen t the most cited w orks in ellipse tting and its closely related problem, conic tting. Least squares fit is a method of determining the best curve to fit a set of points. This article demonstrates how to generate a polynomial curve fit using. When we pass this (near) optimal solution to NL2SOL it will have an easy task. Nonlinear Least Squares Data Fitting D. 1 Introduction More than one explanatory variable In the foregoing chapter we considered the simple regression model where. An apparatus is available that marks a strip of paper at even intervals in time. Most math majors have some exposure to regression in their studies. Least-squares imaging and deconvolution using the hybrid norm conjugate-direction solver Yang Zhang and Jon Claerbout ABSTRACT To retrieve a sparse model, we applied the hybrid norm conjugate-direction (HBCD) solver proposed by Claerbout to two interesting geophysical problems: least-squares imaging and blind deconvolution. Learn more about least squares, curve fitting, optimization, nonlinear, fitting. A number of methods may be employed to solve this problem. In some applications, it may be necessary to place the bound constraints $$l \\leq x \\leq u$$ on the variables $$x$$. solve_least_squares_lm This is a function for solving non-linear least squares problems. First, least squares is a natural approach to estimation, which makes explicit use of the structure of the model as laid out in the assumptions. lstsq for the \"direct\" appraoch (as far as I know this uses SVD by standard, but I also tried all the other LINPACK options that scipy offers) $\\endgroup$ \u2013 Bananach Oct 25 '16 at 19:14. Linear vs. C2, and D2) and then use Solver to find the least-squares parameters A, B, and C. MATH 3795 Lecture 9. You can perform least squares fit with or without the Symbolic Math Toolbox. This is a solved. You can vote up the examples you like or vote down the ones you don't like. The following is a sample implementation of simple linear regression using least squares matrix multiplication, relying on numpy for heavy lifting and matplotlib for visualization. Least squares regression analysis or linear regression method is deemed to be the most accurate and reliable method to divide the company\u2019s mixed cost \u2026. A least squares model contains a dummy objective and a set of linear equations: sumsq. 3 The Role of The quantities generated by the Lanczos process from (2. If it is not in the range, then it is the least squares solution. Loading Least-Squares Regression Line. The nonlinear problem is usually solved by iterative. SPGL1: A solver for sparse least squares. This is a mean estimated from a linear model. An issue came up about whether the least squares regression line has to pass through the point (XBAR,YBAR), where the terms XBAR and YBAR represent the arithmetic mean of the independent and dependent variables, respectively. This x is called the least square solution (if the Euclidean norm is used). Nonlinear least-squares solves min(\u2211||F(x i ) \u2013 y i || 2 ), where F(x i ) is a nonlinear function and y i is data. Answer to 4. The most expensive phase is the LSQR phase. solve a non-linear least squares problem. This page gathers different methods used to find the least squares circle fitting a set of 2D points (x,y). The equation for least squares solution for a linear fit looks as follows. Yanbo Liang (JIRA) Sun, 03 Jan 2016 18:05:30 -0800. In fact, I used this kind of solution in some situations. Define least squares. Factoring-polynomials. LEAST SQUARES and NORMAL EQUATIONS Background Overdetermined Linear systems: consider Ax = b if A is m n, x is n 1, b is m 1 with m > n. The least squares approach to regression is based upon minimizing these difference scores or deviation scores. Polynomials Least-Squares Fitting: Polynomials are one of the most commonly used types of curves in regression. For a general problem you wouldn't use this, of. so somewhere I'm doing something wrong. Let [] \u2200k\u2208\u2115 be a dispersion point in. Triangle Calculator. Nonlinear Least-Squares Fitting. Now that we have determined the loss function, the only thing left to do is minimize it. Given a matrix equation Ax=b, the normal equation is that which minimizes the sum of the square differences between the left and right sides: A^(T)Ax=A^(T)b. Least-Squares Line Least-Squares Fit LSRL The linear fit that matches the pattern of a set of paired data as closely as possible. Introduction\u00b6. This influence is exaggerated using least squares. solver to vary the values for A, C and k to minimize the sum of chi squared. Therefore the least squares solution to this system is: x\u02c6 = (A TA)\u22121A b = \u22120. 00000 Covariance matrix of Residuals 0. solve a non-linear least squares problem. Dr Gregory Reeves 26,616 views. To solve least squares problems based on PDE models requires sophisticated numerical techniques but also great attention with respect to the quality of data and identi\ufb01ability of the parameters. 33 so this is our prediction. \u2018huber\u2019 : rho(z) = z if z <= 1 else 2*z**0. But how does this relate to the least-squares problem, where there are multiple measurements? Is the problem I am trying to solve essentially the same, except that the number of measurements is one? And in that case, is using Ceres Solver's non-linear least squares solver really necessary? Thanks!. The following is a sample implementation of simple linear regression using least squares matrix multiplication, relying on numpy for heavy lifting and matplotlib for visualization. Also tells you if the entered number is a perfect square. It estimates the value of a dependent variable Y from a given independent variable X. An online LSRL calculator to find the least squares regression line equation, slope and Y-intercept values. least squares solution). Node 13 of 18 Node 13 of 18 The Mixed Integer Linear Programming Solver Tree level 1. Order fractions from least to greatest or from greatest to least. The computational burden is now shifted, and one needs to solve many small linear systems. I'd like to know how to solve the least squares non linear regression in java only by passing a matrix A and a vector b like in python. Heh--reduced QR left out the right half of Q. SOLVING DIFFERENTIAL EQUATIONS WITH LEAST SQUARE AND COLLOCATION METHODS by Katayoun Bodouhi Kazemi Dr. 00004849386 0. The Excel Solver can be easily configured to determine the coefficients and Y-intercept of the linear regression line that minimizes the sum of the squares of all residuals of each input equation. Certain types of word problems can be solved by quadratic equations. In this lesson, we will explore least-squares regression and show how this method relates to fitting an equation to some data. There are many possible cases that can arise with the matrix A. The best-fit line, as we have decided, is the line that minimizes the sum of squares of residuals. Anyway, if you want to learn more about the derivation of the normal equation, you can read about it on wikipedia. The solve() method finds a vector x such that \u03a3 i [f i (x)] 2 is minimized. Severely weakens outliers influence, but may cause difficulties in optimization process. When we used the QR decomposition of a matrix to solve a least-squares problem, we operated under the assumption that was full-rank. We have our explanatory variable x, that gets multiplied by this slope beta 1, and we also have an intercept where the line intersects the y axis. solve a non-linear least squares problem. A linear fit matches the pattern of a set of paired data as closely as possible. Given a set of samples {(x i,y i)}m i=1. lstsq in terms of computation time and memory. Triangle Calculator. Square of Matrix Calculator is an online tool programmed to calculate the square of the matrix A. This is a standard least squares problem and can easily be solved using Math. There are many possible cases that can arise with the matrix A. Since this system usually does not have a solution, you need to be satisfied with some sort of approximate solution. For any given. The GLS estimator can be shown to solve the problem which is called generalized least squares problem. y_i=A{x_i}^b When I solve for A two different ways I am getting different answers. The Least-Squares Method requires that the estimated function has to deviate as little as possible from f(x) in the sense of a 2-norm. Find answers to Weighted least Squares Excel from the expert community at Experts Exchange Also, when I load the solver case from R77:R96, the resulting. Method of Least Squares. ) Developer's Guide to Excelets/Sinex. Least\u2013squares Solution of Homogeneous Equations supportive text for teaching purposes Revision: 1. The most widely used approximation is the least squares solution, which minimizes. The data, the interpolating polynomial (blue), and the least-squares line (red) are shown in Figure 1. * odinsbane/least-squares-in-java * NonLinearLeastSquares (Parallel Java Library Documentation) * NonlinearRegression (JMSL Numerical Library) Some related discussion here: Solving nonlinear equations. 5 Example 3: The orbit of a comet around the sun is either elliptical, parabolic, or hyperbolic. This is the general form of the least squares line. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. LLS is actively maintained for the course EE103, Introduction to Matrix Methods. Least Squares Regression Line of Best Fit. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this. Octave also supports linear least squares minimization. Since it's a sum of squares, the method is called the method of least squares. They are connected by p DAbx. Lecture 11, Least Squares Problems, Numerical Linear Algebra, 1997. We present an algorithm for adding rows with a single nonzero to A to improve its conditioning; it attempts to add as few rows as possible. 00000241437 0. 1 Introduction A nonlinear least squares problem is an unconstrained minimization problem of the form minimize x f(x)= m i=1 f i(x)2, where the objective function is de\ufb01ned in terms of auxiliary functions {f i}. To perform WLS in EViews, open the equation estimation dialog and select a method that supports WLS such as LS\u2014Least Squares (NLS and ARMA), then click on the Options tab. Least Squares with Examples in Signal Processing1 Ivan Selesnick March 7, 2013 NYU-Poly These notes address (approximate) solutions to linear equations by least squares. Last edited by shg; 10-23-2017 at 01:01 PM. 20 - PhET: Free online. R factor can be used in LSQR (an iterative least-squares solver [29]) to e\ufb03-ciently and reliably solve a regularization of the least-squares problem. The CVX Users\u2019 Guide, Release 2. I am using python linalg. The results showed. Which Matlab function should I use?. A linear fit matches the pattern of a set of paired data as closely as possible. Least Squares Optimization The following is a brief review of least squares optimization and constrained optimization techniques,which are widely usedto analyze and visualize data. For treatment A, the LS mean is (3+7. Because nonlinear optimization methods can be applied to any function, for the relation between two variables, it finds functions that best fit a given set of data points from a list of more than 100 functions, which include most common and interesting functions, like gaussians, sigmoidals, rationals. AutoCorrelation (Correlogram) and persistence - Time series analysis. NET: Description: This example shows how to find a linear least squares fit for a set of points in Visual Basic. Least Squares. But, this OLS method will work for both univariate dataset which is single independent variables and single dependent variables and multi-variate dataset. 7 Least squares approximate solutions. Quadratic Regression Calculator. * odinsbane/least-squares-in-java * NonLinearLeastSquares (Parallel Java Library Documentation) * NonlinearRegression (JMSL Numerical Library) Some related discussion here: Solving nonlinear equations. In the process of solving a mixed integer least squares problem, an ordinary integer least squares problem is solved. Free Modulo calculator - find modulo of a division operation between two numbers step by step This website uses cookies to ensure you get the best experience. For details, see First Choose Problem-Based or Solver-Based Approach. Next, we develop a distributed least square solver over strongly connected directed graphs and show that the proposed algorithm exponentially converges to the least square solution provided the step-size is suf\ufb01ciently small. Let [] \u2200k\u2208\u2115 be a dispersion point in. This is why some least-squares solvers do not use the normal equations under the hood (they instead use QR decomposition). See Input Data for the description of how to enter matrix or just click Example for a simple example. Spark MLlib currently supports two types of solvers for the normal equations: Cholesky factorization and Quasi-Newton methods (L-BFGS/OWL-QN). The method of iteratively reweighted least squares (IRLS) is used to solve certain optimization problems with objective functions of the form of a p-norm: \u2211 = | \u2212 |, by an iterative method in which each step involves solving a weighted least squares problem of the form: (+) = \u2211 = (()) | \u2212 |. This function outperforms numpy. BLENDENPIK: SUPERCHARGING LAPACK'S LEAST-SQUARES SOLVER 5 de ned in the prof. R factor can be used in LSQR (an iterative least-squares solver [29]) to e\ufb03-ciently and reliably solve a regularization of the least-squares problem. A necessary and sufficient condition is established on the graph Laplacian for the continuous-time distributed algorithm to give the least squares solution in the limit, with an exponentially fast convergence rate. Quadratic regression is a type of a multiple linear regression. The use of linear regression, or least squares method, is the most accurate method in segregating total costs into fixed and variable components. The least squares regression line ; The least squares regression line whose slope and y-intercept are given by: where , , and. Subsequently, Avronetal. This is done by finding the partial derivative of L, equating it to 0 and then finding an expression for m and c. Since this thesis is closely related to the least-squares adjustment problem and will actually present a new approach for solving this problem, let us first have a closer look at the classical approach. Least Squares with Examples in Signal Processing1 Ivan Selesnick March 7, 2013 NYU-Poly These notes address (approximate) solutions to linear equations by least squares. e the sum of squares of residuals is minimal under this approach. Ordinary Least Squares (OLS) regression (or simply \"regression\") is a useful tool for examining the relationship between two or more interval/ratio variables. This problem is called a least squares problem for the following reason. On input, the field x must be filled in with an initial estimate of the solution vector, and the field tol must be set to the desired tolerance. Solve word problems involving quadratic equations. find_min_box_constrained (using lbfgs_search_strategy(10)) performed poorly as it can be trapped on a boundary. where A is an m x n matrix with m > n, i. So really, what you did in the first assignment was to solve the equation using LSE. On a similar note,. This influence is exaggerated using least squares. i) (circles) and least-squares line (solid line) but we will see that the normal equations also characterize the solution a, an n-vector, to the more general linear least squares problem of minimizing kAa ykfor any matrix Athat is m n, where m n, and whose columns are linearly independent. Enter your data as (x,y) pairs, and find the equation of a line that best fits the data. Scramble Squares\u00ae Puzzle. least_squares(). Suppose that a matrix A is given that has more rows than columns, ie n, the number of rows, is larger than m, the number of columns. Quadratic Regression Calculator. Nonlinear Least Squares Data Fitting D. Main ideas 2. TI-89 graphing calculator program for calculating the method of least squares. This document is intended to clarify the issues, and to describe a new Stata command that you can use (wls) to calculate weighted least-squares estimates for problems such as the Strong interaction'' physics data described in Weisberg's example 4. Also lets you save and reuse data. 00097402530 0. Line of Best Fit (Least Square Method) A line of best fit is a straight line that is the best approximation of the given set of data. Constructing a Least-Squares Graph Using Microsoft Excel. Part of our free statistics site; generates linear regression trendline and graphs results. To compare fractions the calculator first finds the least common denominator (LCD), converts the fractions to equivalent fractions using the LCD, then. By using this website, you agree to our Cookie Policy. Visit Stack Exchange. Type doc lsqnonlin for more details. The method of least squares - using the Excel Solver Michael Wood 5 advertising. For this, we're going to make use of the property that the least squares line always goes through x bar, y bar. The method of least squares calculates the line of best fit by minimising the sum of the squares of the vertical distances of the points to th e line. 0 released December 2019 This latest release of SPGL1 implements a dual root-finding mode that allows for increased accuracy for basis pusuit denoising problems. The Linear Least Squares Regression Line method is the accurate way of finding the line of best fit in case it's presumed to be a straight line that is the best approximation of the given set of data. since gradient descent is a local optimizer and can get stuck in local solution we need to use. Added Dec 13, 2011 by scottynumbers in Mathematics. When we used the QR decomposition of a matrix to solve a least-squares problem, we operated under the assumption that was full-rank. The center of the part and center of rotation are offset. Question: 4. How to Calculate Absolute Value. powered by $$x$$ y $$a 2$$ a b . \"Solver\" is a powerful tool in the Microsoft Excel spreadsheet that provides a simple means of fitting experimental data to nonlinear functions. The Least Squares Regression Line is the line that makes the vertical distance from the data points to the regression line as small as possible. Number of Data Points: X Data Points:. solve a non-linear least squares problem. Spark MLlib currently supports two types of solvers for the normal equations: Cholesky factorization and Quasi-Newton methods (L-BFGS/OWL-QN). The full documentation is available online. In the present instance solving for the node weights is not really viable for the following reason: in the actual, real-life setting the only decision variable (which I have control over & need to solver for) are the instrument weights = units of instruments (cells I11:I15). Algorithm 1 Least-squares sub-problem input: H 2 CN \u21e5N, q 2 CN, P 2 RM \u21e5N, d 2 CM for each receiver (j )(rowinP) in parallel do H\u21e4 w j = p\u21e4 j {solve 1 PDE} end for W =[w 1 w 2w m] {distributed matrix} S =(I M + 2 W\u21e4 W)1 {adjust using Algorithm 2 (optional)} for source (i) in parallel do y i =(I N 2 WSW\u21e4)(q i + 2 Wd i) Hu i = y i {solve 1 PDE} end for output: u. Effective use of Ceres requires some familiarity with the basic components of a non-linear least squares solver, so before we describe how to configure and use the solver, we will take a brief look at how some of the core optimization algorithms in Ceres work. LinearLeastSquares. It can be manually found by using the least squares method. That is, Octave can find the parameter b such that the model y = x*b fits data (x,y) as well as possible, assuming zero-mean Gaussian noise. The least-squares line or regression line can be found in the form of y = mx + b using the following formulas. When we pass this (near) optimal solution to NL2SOL it will have an easy task. The Least Squares Regression Calculator is biased against data points which are located significantly away from the projected trend-line. 1 Linear Least Squares Problem. You must select the Solver Add-in and then press the OK button. Use the EXCEL SOLVER program to minimise S by varying the paramters \"a\" and \"b\" This will produce estimates of a and b that give the best fitting straight line to the data. It will b e sho wn that the direct sp eci c least-square tting of ellipses. (A for all ). This page allows performing nonlinear regressions (nonlinear least squares fittings). Let [] \u2200k\u2208\u2115 be a dispersion point in. Nonlinear Least Squares Data Fitting D. By using this website, you agree to our Cookie Policy. The generalized least squares problem. It is used to study the nature of the relation between two variables. Let , , and be defined as previously. This article introduces the method of fitting nonlinear functions with Solver. \uf075 Used to determine the \u201cbest\u201d line. The least squares criterion is a formula used to measure the accuracy of a straight line in depicting the data that was used to generate it. Least-Squares Fitting of Data with Polynomials Least-Squares Fitting of Data with B-Spline Curves. In the present instance solving for the node weights is not really viable for the following reason: in the actual, real-life setting the only decision variable (which I have control over & need to solver for) are the instrument weights = units of instruments (cells I11:I15). LEAST MEAN SQUARE ALGORITHM 6. Use our online quadratic regression calculator to find the quadratic regression equation with graph. LinearAlgebra namespace in C#. The applications of the method of least squares curve fitting using polynomials are briefly discussed as follows. 0 released December 2019. 00000088820 0. A least squares model contains a dummy objective and a set of linear equations: sumsq. Least squares means are adjusted for other terms in the model (like covariates), and are less sensitive to missing data. The CVX Users\u2019 Guide, Release 2. Given a set of data, we can fit least-squares trendlines that can be described by linear combinations of known functions. com A collection of really good online calculators for use in every day domestic and commercial use!. Lecture 11, Least Squares Problems, Numerical Linear Algebra, 1997. Trouble may also arise when M = N but the matrix is singular. The first step. using least squares minimization. To show the powerful Maple 10 graphics tools to visualize the convergence of this Polynomials. The best-fit line, as we have decided, is the line that minimizes the sum of squares of residuals. Wow, there's a lot of similarities there between real numbers and matrices. CPM Student Tutorials CPM Content Videos TI-84 Graphing Calculator Bivariate Data TI-84: Least Squares Regression Line (LSRL). Regression Using Excel's Solver. However, if users insist on finding the total least squares fit then an initial approximation is still required and the linear least squares approach is recommended. That closed-form solution is called the normal equation. overdetermined system, least squares method The linear system of equations A =. For details, see First Choose Problem-Based or Solver-Based Approach. Question: 4. Least squares means are adjusted for other terms in the model (like covariates), and are less sensitive to missing data. Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. Sum of squares is used in statistics to describe the amount of variation in a population or sample of observations. 1 Introduction. , there are more equations than unknowns, usually does not have solutions. Click the button labeled \u201cClick to Compute\u201d. They are connected by p DAbx. Linear least-squares solves min||C*x - d|| 2 , possibly with bounds or linear constraints. Each node has access to one of the linear equations and holds a dynamic state. When radians are selected as the angle unit, it can take values such as pi/2, pi/4, etc. The method of least squares - using the Excel Solver Michael Wood 5 advertising. Years after 1900 50 60 70 80 90 100 Percentage 29. Estimating an ARMA Process Overview 1. Loading Unsubscribe from Adrian Lee? Least Squares Linear Regression - EXCEL - Duration: 10:55. No Bullshit Guide To Linear Algebra, 2017. Added Dec 13, 2011 by scottynumbers in Mathematics. (We use the squares for much the same reason we did when we defined the variance in Section 3. Alternative solution methods. See LICENSE_FOR_EXAMPLE_PROGRAMS. The data, the interpolating polynomial (blue), and the least-squares line (red) are shown in Figure 1. LMS incorporates an. First, least squares is a natural approach to estimation, which makes explicit use of the structure of the model as laid out in the assumptions. The solutions. org are unblocked. The linear least-squares problem occurs in statistical regression analysis; it has a closed-form solution. Example 2 in the KaleidaGraph Quick Start Guide shows how to apply a Linear curve fit to a Scatter plot. Check Minitab for definition of influential points. Linear regression calculator Two-dimensional linear regression of statistical data is done by the method of least squares. Quadratic Regression Calculator. Trouble may also arise when M = N but the matrix is singular. The method of iteratively reweighted least squares (IRLS) is used to solve certain optimization problems with objective functions of the form of a p-norm: \u2211 = | \u2212 |, by an iterative method in which each step involves solving a weighted least squares problem of the form: (+) = \u2211 = (()) | \u2212 |. To approximate a Points Dispersion through Least Square Method using a Quadratic Regression Polynomials and the Maple Regression Commands. In fact, I used this kind of solution in some situations. lsqnonneg applies only to the solver-based approach. Number of Data Points: X Data Points:. The help qr command in Matlab gives the following information: >> help qr QR Orthogonal-triangular decomposition. lstsq in terms of computation time and memory. I tried the following set-up: - Given is a vector of original exposure across a range of seven nodes (C8:I2) - The aim is to replicate this exposure at each point as close as possible from a set of 5 instruments. solve public void solve() Solve this nonlinear least squares minimization problem. 4 Linear Least Squares. Least squares regression analysis or linear regression method is deemed to be the most accurate and reliable method to divide the company\u2019s mixed cost \u2026. In this case, solving the normal equations (5) is equivalent to. If the system matrix is rank de cient, then other methods are. Enter the fraction separated by comma and press the calculate button. solve a non-linear least squares problem. Regression Using Excel's Solver. powered by. Given a set of data, we can fit least-squares trendlines that can be described by linear combinations of known functions. Check Minitab for definition of influential points. to solve multidimensional problem, then you can use general linear or nonlinear least squares solver. (3) Solve the diagonal system \u03a3\u02c6w = U\u02c6\u2217b for w. The original domain is. It is called \u201cleast squares\u201d because we are minimizing the sum of squares of these functions. I If m= nand Ais invertible, then we can solve Ax= b. The limitations of the OLS regression come from the constraint of the inversion of the X\u2019X matrix: it is required that the rank of the matrix is p+1, and some numerical problems may arise if the matrix is not well behaved. Free online LCM calculator. Leykekhman - MATH 3795 Introduction to Computational MathematicsLinear Least Squares { 11. For this purpose, the initial values of A, B, and C in cells F2-F4 should be those found by Solver in the previous run. 1 Linear Least Squares Problem. If you're seeing this message, it means we're having trouble loading external resources on our website. For details, see First Choose Problem-Based or Solver-Based Approach. Note: this method requires that A not have any redundant rows. Note: Be sure that your Stat Plot is on and indicates the Lists you are using. Nonlinear Regression. Linear Least Squares Regression Line Calculator - v1. Contribute to kashif/ceres-solver development by creating an account on GitHub. Linear regression line calculator to calculate slope, interception and least square regression line equation. The second one is the Levenberg-Marquardt method. Enter the number of data pairs, fill the X and Y data pair co-ordinates, the least squares regression line calculator will show you the result.", "date": "2020-03-28 08:54:44", "meta": {"domain": "agence-des-4-fontaines.fr", "url": "http://khhk.agence-des-4-fontaines.fr/least-squares-solver.html", "openwebmath_score": 0.4954856038093567, "openwebmath_perplexity": 651.6307704769378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915215128426, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8603917800815063}}
{"url": "https://cs.stackexchange.com/questions/79263/count-arrays-with-size-n-sum-k-and-largest-element-m", "text": "# Count arrays with size n, sum k and largest element m\n\nI'm trying to solve pretty complex problem with combinatorics.\n\nNamely, we have given three numbers N, K, M. Now we want to count how many different arrays of integers are there with length N, sum K and all the elements in the range [1, M]\n\nConstraints:\n\n\u2022 1 <= N <= 100\n\u2022 1 <= K <= 100\n\u2022 1 <= M <= 100\n\nExample\n\nLet's say N = 2, K = 5, M = 3. This means that we want to count arrays of integers of size 2 with sum of all elements equal to 5 and elements in range [1, 3]. There are total of 2 arrays: {2, 3} and {3, 2}. Please note that the order of the elements also matters, {2, 3} is not equal to {3, 2}\n\nSecond example: N = 4, K = 7, M = 3. We want to count arrays of length 4, sum of 7 and elements in range [1, 3].\n\nThere are total of 16 possible way of arrays: (1,1,2,3), (1,1,3,2), (2,1,1,3), (3,1,1,2), (2,3,1,1), (3,2,1,1), (1,2,3,1), (1,3,2,1), (1,2,1,3), (1,3,1,2), (1,2,2,2), (2,1,2,2), (2,2,1,2), (2,2,2,1)\n\nWhat I have tried\n\nI know that one solution is to generate all possible arrays, but such algorithm in best case will work in complexity O(N!) which is far too big for N = 100. I started thinking about solving this with three-dimensional dynamical programming, but I cannot find the relations between the states.\n\nI'm thinking about this way: Let f(i, j, l) be the number of arrays of length i, sum j, and largest element l. We can see that for i = 0 f(i,j,l) = 0, so this is I think the base case. Also f(1, 1, 1) = 1 is another base case.\n\nNow I cannot find the relations between the states. Can you give me some hints how to find the relations between the states. Thanks in advance.\n\n\u2022 Nice problem. Can you credit the source where you encountered the problem? \u2013\u00a0D.W. Jul 25 '17 at 16:11\n\u2022 The problem is from one macedonian site for training, and the problem in original is in macedonian, but i can give you the test cases if you want them \u2013\u00a0someone12321 Jul 25 '17 at 16:30\n\nYou can use dynamic programming. For each $0 \\leq i \\leq N$ and $0 \\leq s \\leq K$, count the number of arrays of length $i$, consisting of numbers in the range $\\{1,\\ldots,M\\}$, which sum to exactly $s$. The running time is $O(NK)$.\n\nExplicitly, denoting the array by $a$, we have $a(0,0) = 1$, $a(0,s) = 0$ otherwise, and $$a(i,s) = \\sum_{t=1}^M a(i-1,s-t),$$ where $a(i-1,r) = 0$ if $r < 0$.\n\nAs Hendrik Jan mentions in the comments, we can improve on this $O(NKM)$ algorithm by using the recursion $$a(i,s) = a(i,s-1) - a(i-1,s-1-M) + a(i-1,s-1),$$ with suitable base cases.\n\nAlternatively, we can obtain an explicit expression: the answer is the coefficient of $x^K$ in the generating function $$(x + \\cdots + x^M)^N = x^N \\left(\\frac{1-x^M}{1-x}\\right)^N.$$ In other words, we are looking for the coefficient of $x^{K-N}$ in $$\\left( \\sum_{i=0}^N (-1)^i \\binom{N}{i} x^{iM} \\right) \\left( \\sum_{j=0}^\\infty \\binom{j+N-1}{N-1} x^j \\right),$$ which has the closed form $$\\sum_{i=0}^{\\min(N,\\lfloor (K-N)/M\\rfloor)} (-1)^i \\binom{N}{i} \\binom{K-iM-1}{N-1}.$$\n\n\u2022 The numbers $a(i,s)$ represent a matrix of solutions for \"fixed\" $M$. The numbers $a(i,s)$ are a \"running sum\" of $M$ consecutive elements from the previous row. So, can't the complexity be improved to $O(NK)$, as $a(i,s) = a(i,s-1) + a(i-1,s-1) - a(i-1,s-M-1)$? give or take typo's. \u2013\u00a0Hendrik Jan Jul 25 '17 at 1:22\n\u2022 @HendrikJan Yes, that's right! I missed this somehow. \u2013\u00a0Yuval Filmus Jul 25 '17 at 5:12\n\nYou actually don't need the largest element l in your recursive function. It doesn't make such sense to use it, since the largest number in an array has no influence on the other numbers.\n\nIn most (all?) dynamic programming problems you have to think about the last step / the last part. For f(i, j) you have an array with i numbers. You can imagine, that in the last step you added the last number to the array. So before you added the last number, the array only consists of i-1 numbers.\n\nNow, if the added number was 1, then there are f(i-1, j-1) many arrays. If the added number was 2, then there are f(i-1, j-2) many arrays. And so on...\n\nAdd all those possibilities up, and you end up with f(i, j).\n\nI don't pretend this is the most efficient solution. Neither it has a strong theoretical merit, but you can still find it useful to verify your results. Here's a small program in Prolog, (using SWI Prolog for constraint satisfaction part), which can calculate or verify your guess:\n\n% -*- mode: prolog; prolog-system: \"swi\" -*-\n:- use_module(library(clpfd)).\n\nsums(K, [], K).\nsums(A, [X | Xs], K) :-\nindomain(A),\nindomain(X),\nB is A + X,\nsums(B, Xs, K).\nsums([X | Xs], K) :- sums(X, Xs, K).\n\nnmk_problem_helper(N, K, M, List) :-\nindomain(N),\nindomain(M),\nindomain(K),\nlength(List, N),\nList ins 1..M,\nsums(List, K).\n\nnmk_problem(N, K, M, Arrays) :-\nfindall(X, nmk_problem_helper(N, K, M, X), Arrays).\n\nnmk_problem_all_helper(List) :-\n[N, M, K] ins 1..100,\nindomain(N),\nindomain(M),\nindomain(K),\nlength(List, N),\nList ins 1..M,\nsums(List, K).\n\nnmk_problem_all(Arrays) :-\nfindall(X, nmk_problem_all_helper(X), Arrays).\n\n\nExample usage:\n\n?- nmk_problem(4, 7, 3, X).\nX = [[1, 1, 2, 3], [1, 1, 3, 2], [1, 2, 1, 3], [1, 2, 2, 2], [1, 2, 3, 1], [1, 3, 1|...], [1, 3|...], [2|...], [...|...]|...].", "date": "2020-01-22 13:05:49", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/79263/count-arrays-with-size-n-sum-k-and-largest-element-m", "openwebmath_score": 0.681304931640625, "openwebmath_perplexity": 521.7804452486056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363512883316, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.8603859947919545}}
{"url": "https://math.stackexchange.com/questions/1679302/multiple-answers-to-int-sqrt4t-t2-textrmd-t", "text": "# Multiple answers to $\\int \\sqrt{4t - t^2} \\, \\textrm{d} t$\n\nI'm trying to understand why I'm getting different answers when taking different approaches to integrating\n\n$$\\int \\sqrt{4t - t^2} \\, \\textrm{d} t$$\n\nFirst, I tried substituting $\\sqrt t = 2 \\sin \\theta$:\n\n$$\\begin{eqnarray} \\int \\sqrt{t}\\sqrt{4 - t} \\, \\textrm{d} t &=& \\int 2 \\sin \\theta \\cdot \\sqrt{4 - 4 \\sin^2 \\theta} \\cdot 8 \\sin \\theta \\cos \\theta \\, \\textrm{d} \\theta \\\\ &=& 32 \\int \\sin^2 \\theta \\cos^2 \\theta \\, \\textrm{d} \\theta \\\\ &=& 4 \\int 1 - \\cos 4 \\theta \\, \\textrm{d} \\theta \\\\ &=& 4 \\theta - \\sin 4 \\theta + C \\\\ &=& 4 \\theta - 4 \\sin \\theta \\cos^3 \\theta + 4 \\sin^3 \\theta \\cos \\theta + C \\\\ &=& 4 \\arcsin \\frac{\\sqrt{t}}{2} + \\frac{1}{2} (t - 2)\\sqrt{4t - t^2} + C \\\\ \\end{eqnarray}$$\n\nSecond, I tried completing the square and substituting $t - 2 = 2 \\sin \\theta$:\n\n$$\\begin{eqnarray} \\int \\sqrt{4 - (t^2 - 4t + 4)} \\, \\textrm{d} t &=& \\int \\sqrt{4 - (t - 2)^2} \\, \\textrm{d} t \\\\ &=& \\int \\sqrt{4 - 4 \\sin^2 \\theta} \\cdot 2 \\cos \\theta \\, \\textrm{d} \\theta \\\\ &=& 4 \\int \\cos^2 \\theta \\, \\textrm{d} \\theta \\\\ &=& 2 \\int 1 - \\cos 2 \\theta \\, \\textrm{d} \\theta \\\\ &=& 2 \\theta + \\sin 2 \\theta + C \\\\ &=& 2 \\theta + 2 \\sin \\theta \\cos \\theta + C \\\\ &=& 2 \\arcsin \\left(\\frac{t - 2}{2}\\right) + \\frac{1}{2}(t - 2)\\sqrt{4t - t^2} + C \\\\ \\end{eqnarray}$$\n\nThe second answer is the same as in the book but I don't understand why the first approach gives the wrong answer.\n\n\u2022 Here's one sticking point: $1-\\cos 4\\theta = 8\\sin^2\\theta\\cos^2\\theta.$ \u2013\u00a0Cameron Williams Mar 2 '16 at 0:52\n\u2022 @CameronWilliams, good catch, I'll update my first answer. \u2013\u00a0Chewers Jingoist Mar 2 '16 at 0:56\n\u2022 Both answers are correct: if you let $f(t)=4\\sin^{-1}\\frac{\\sqrt{t}}{2}-2\\sin^{-1}\\frac{t-2}{2}$, then $f^{\\prime}(t)=0$ so $f(t)$ is a constant. \u2013\u00a0user84413 Mar 2 '16 at 1:08\n\u2022 @user84413, when I differentiate $f(t)$ I get $\\frac{4}{\\sqrt{4 - t}} - \\frac{2}{\\sqrt{4 - (t - 2)^2}} = \\frac{4}{\\sqrt{4 - t}} - \\frac{2}{\\sqrt{4t - t^2}}$. Could you elaborate on how you get $0$? \u2013\u00a0Chewers Jingoist Mar 2 '16 at 1:21\n\u2022 In the first term, I think you will get $4\\frac{1}{\\sqrt{1-t/4}}\\frac{1}{4\\sqrt{t}}=\\frac{2}{\\sqrt{4t-t^2}}$ \u2013\u00a0user84413 Mar 2 '16 at 1:33\n\n$$\\arcsin\\left(\\frac t2-1\\right) = 2\\left(\\arcsin\\frac{\\sqrt t}2 -\\frac\\pi4\\right).\\tag{*}$$\nProof: Write $\\theta:=\\arcsin\\frac{\\sqrt t}2$. Then $\\sin^2\\theta=\\frac t4$ and $\\cos^2\\theta=1-\\frac t4$, and using the angle-difference identities, $$\\sin\\left(\\theta-\\frac\\pi4\\right)=\\frac1{\\sqrt 2}(\\sin\\theta-\\cos\\theta)\\tag1$$ while $$\\cos\\left(\\theta-\\frac\\pi4\\right)=\\frac1{\\sqrt 2}(\\sin\\theta+\\cos\\theta).\\tag2$$ Therefore $$\\sin2\\left(\\theta-\\frac\\pi4\\right)=2\\sin\\left(\\theta-\\frac\\pi4\\right)\\cos\\left(\\theta-\\frac\\pi4\\right) \\stackrel{(1),(2)}=\\sin^2\\theta-\\cos^2\\theta =\\frac t4-\\left(1-\\frac t4\\right)=\\frac t2-1.$$ Therefore both sides of (*) have the same sine. Similarly you can show that both sides have the same cosine.\n\u2022 Where does $2$ come from at the start of your last line in $sin 2(\\theta - \\frac{\\pi}{4})$? And what does the $(1),(2)$ mean? \u2013\u00a0Chewers Jingoist Mar 2 '16 at 1:31\n\u2022 @ChewersJingoist, the $2$ in $sin2(\\theta - \\frac{\\pi}{4})$ comes from the identity $sin2x=2sinxcosx$. \u2013\u00a0Alexander Maru Mar 2 '16 at 1:39\nIf $$\\sin\\theta = \\dfrac{\\sqrt t}2,$$ then $$\\sin\\left(\\dfrac\\pi2-2\\theta\\right) = \\cos2\\theta = 1-2\\sin^2\\theta = \\dfrac{2-t}2,$$ $$2\\theta = \\dfrac\\pi2-\\arcsin\\dfrac{2-t}2 = \\dfrac\\pi2+\\arcsin\\dfrac{t-2}2,$$ $$\\theta = \\dfrac\\pi4+\\dfrac12\\arcsin\\dfrac{t-2}2,$$ $$\\boxed{\\arcsin\\dfrac{\\sqrt t}2 = \\dfrac12\\arcsin\\dfrac{t-2}2+\\dfrac\\pi4}$$ So the first answer is correct too", "date": "2019-07-23 09:15:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1679302/multiple-answers-to-int-sqrt4t-t2-textrmd-t", "openwebmath_score": 0.9391178488731384, "openwebmath_perplexity": 266.935391337121, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407175907054, "lm_q2_score": 0.8840392939666335, "lm_q1q2_score": 0.860383036838467}}
{"url": "https://gmatclub.com/forum/if-the-curve-described-by-the-equation-y-x2-bx-c-cuts-the-x-axis-272937.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 06 Apr 2020, 08:05\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# If the curve described by the equation y = x2 + bx + c cuts the x-axis\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nGMAT Club team member\nStatus: GMAT Club Team Member\nAffiliations: GMAT Club\nJoined: 02 Nov 2016\nPosts: 5385\nGPA: 3.62\nIf the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2018, 10:08\n1\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n63% (02:11) correct 37% (01:56) wrong based on 82 sessions\n\n### HideShow timer Statistics\n\nIf the curve described by the equation $$y = x^2 + bx + c$$ cuts the $$x$$-axis at $$-4$$ and $$y$$ axis at $$4$$, at which other point does it cut the $$x$$-axis?\n\nA. -1\nB. 4\nC. 1\nD. -4\nE. 0\n\n_________________\nIntern\nJoined: 07 May 2018\nPosts: 3\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2018, 10:49\n+1 for A? I plugged in the other points to come up with an equation...not sure if I went about it the right way\nDirector\nStatus: Learning stage\nJoined: 01 Oct 2017\nPosts: 954\nWE: Supply Chain Management (Energy and Utilities)\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2018, 22:08\n1\nIf the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?\n\nA. -1\nB. 4\nC. 1\nD. -4\nE. 0\n\nGiven, $$y = x^2 + bx + c$$, cuts the x-axis at two points. One intersection point with x-axis is given. We need to find out the other point of intersection. In other words, 1 root of the quadratic equation is given, what is the value if the other root?\n\na) At (-4,0), $$0=(-4)^2+b*(-4)+c$$ Or, 4b-c=16\nb) At (0,4), $$4=0^2+b*0+c$$ Or, c=4\nSo, 4b-4=16 Or, b=5.\nNow, we have the equation of the curve, $$y=x^2+5x+4$$, which has the roots: -4 and -1.\nSo, other other root is -1.\n\nAns. (A)\n_________________\nRegards,\n\nPKN\n\nRise above the storm, you will find the sunshine\nCAT Forum Moderator\nJoined: 29 Nov 2018\nPosts: 280\nConcentration: Marketing, Strategy\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jan 2019, 09:53\n2\nIf the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?\n\nA -1\nB 4\nC 1\nD -4\nE 0\n\ny = x^2 + bx + c is a quadratic equation and the equation represents a parabola.\nThe curve cuts the y axis at 4.\nThe x coordinate of the point where it cuts the y axis = 0.\nTherefore, (0, 4) is a point on the curve and will satisfy the equation.\n4 = 0^2 + b(0) + c\nOr c = 4.\n\nThe product of the roots of a quadratic equation is c/a\nIn this question, the product of the roots = 4/1 = 4.\n\nThe roots of the quadratic equation are the points where the curve cuts the x-axis.\nThe question states that one of the points where the curve cuts the x-axis is -4.\nSo, -4 is one of roots.\nLet r2 be the second root of the quadratic equation.\nSo, -4 * r2 = 4\nor r2 = -1.\n\nThe second root is the second point where the curve cuts the x-axis, which is -1.\n\nIf you liked the question and explanation, please do hit the kudos button\nIntern\nJoined: 30 May 2017\nPosts: 16\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jan 2019, 10:34\nWhen x=-4 y=0\nso 16 -4b +c=0\nWhen x=0, y = 4\nso c=4\n16-4b+c=20-4b=0\nb= 5\n\nthe equation can be written as y=(x+4)(x+1)\n\ny is equal to zero when x=-1 (Answer A)\nMath Expert\nJoined: 02 Sep 2009\nPosts: 62542\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n03 Jan 2019, 02:15\ncfc198 wrote:\nIf the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?\n\nA -1\nB 4\nC 1\nD -4\nE 0\n\ny = x^2 + bx + c is a quadratic equation and the equation represents a parabola.\nThe curve cuts the y axis at 4.\nThe x coordinate of the point where it cuts the y axis = 0.\nTherefore, (0, 4) is a point on the curve and will satisfy the equation.\n4 = 0^2 + b(0) + c\nOr c = 4.\n\nThe product of the roots of a quadratic equation is c/a\nIn this question, the product of the roots = 4/1 = 4.\n\nThe roots of the quadratic equation are the points where the curve cuts the x-axis.\nThe question states that one of the points where the curve cuts the x-axis is -4.\nSo, -4 is one of roots.\nLet r2 be the second root of the quadratic equation.\nSo, -4 * r2 = 4\nor r2 = -1.\n\nThe second root is the second point where the curve cuts the x-axis, which is -1.\n\nIf you liked the question and explanation, please do hit the kudos button\n\n_______________\nMerging topics.\n_________________\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 14467\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n22 Feb 2020, 15:03\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis \u00a0 [#permalink] 22 Feb 2020, 15:03\nDisplay posts from previous: Sort by", "date": "2020-04-06 16:05:56", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-the-curve-described-by-the-equation-y-x2-bx-c-cuts-the-x-axis-272937.html", "openwebmath_score": 0.697464108467102, "openwebmath_perplexity": 1222.06844689677, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9732407183668539, "lm_q2_score": 0.8840392848011833, "lm_q1q2_score": 0.8603830286044234}}
{"url": "https://mathematica.stackexchange.com/questions/224729/laplace-equation-with-mixed-boundary-conditions", "text": "# Laplace equation with mixed boundary conditions\n\nI try to solve Laplace equation in 2D on square [2,3]x[2,3], with mixed boundary conditions, I did:\n\nClearAll[y, x1, x2];\npde = Laplacian[y[x1, x2], {x1, x2}];\nbc = {y[x1, 2] == 2 + x1, y[x1, 3] == 3 + x1};\nsol = NDSolve[{pde ==\nNeumannValue[-1, x1 == 2] + NeumannValue[1, x1 == 3], bc},\ny, {x1, 2, 3}, {x2, 2, 3}]\n\nPlot3D[Evaluate[y[x1, x2] /. sol], {x1, 2, 3}, {x2, 2, 3},\nPlotRange -> All, AxesLabel -> {\"x1\", \"X2\", \"y[x1,x2]\"},\nBaseStyle -> 12]\n\n\nThe exact solution is y=x1+x2, the problem is the results is not high accurate when I evaluate the error.\n\n\u2022 The exact solution is y=x1+x2 Are you sure about this? How does this solution satisfy the Neumann boundary conditions? \u2013\u00a0Nasser Jun 25 at 20:09\n\u2022 @Nasser Erm. The function does satisfy the Neumann boundary condition: Its derivative in x1-direction is 1 and the sign flops stems from the fact that Neumann conditions are phrased in terms of outward normals... No? \u2013\u00a0Henrik Schumacher Jun 25 at 23:34\n\u2022 @user62716 Using NeumannValue requires one to do integration by parts and one has to be careful about the signs. Try switching the sign of the Laplacian to pde = -Laplacian[y[x1, x2], {x1, x2}];. Then it should work. \u2013\u00a0Henrik Schumacher Jun 25 at 23:40\n\u2022 @HenrikSchumacher is NeumannValue[-1, x1 == 2] different from saying that $\\frac{\\partial y}{\\partial x_1}$ evaluated at $x_1=2$ is $-1$? And since the claim is that the solution is $y=x_1+x_2$ then $\\frac{\\partial y}{\\partial x_1}=1$ this is evaluated at $x=2$ is $1$ and not $-1$?. How do you translate NeumannValue[-1, x1 == 2] to normal derivative then? I just did direct translation. May be we need a whole new topic on this. On top of all of this, moving NeumannValue from RHS to LHS changes the solution. I never liked NeumannValue and prefer to use normal derivatives... \u2013\u00a0Nasser Jun 26 at 0:46\n\u2022 @Nasser $\\frac{\\partial y}{\\partial \\nu} (2,x_2) = - \\frac{\\partial y}{\\partial x_1} (2,x_2)$ because the outward normal at the point $(2,x_2)$ is $\\nu = (-1 , 0)$. But I agree that NeumannValue is a bit counter intuitive, but it makes perfect sense in regard of the weak formulation that is used in FEM. \u2013\u00a0Henrik Schumacher Jun 26 at 4:59\n\nRelatively recently, Wolfram has created a nice Heat Transfer Tutorial and a Heat Transfer Verification Manual. I model with many codes and I usually start the Verification and Validation manual and build complexity from there. It is always embarrassing to build a complex model and find that your setup does not pass verification.\n\nThe Laplace equation is special case of the heat equation so we should be able to use a verified example as a template for a properly constructed model.\n\nFor NeumannValue's, if the flux is into the domain, it is positive. If the flux is out of the domain, it is negative.\n\nAt the tutorial link, they define a function HeatTransferModel to create operators for a variety of heat transfer cases that I shall reproduce here:\n\nClearAll[HeatTransferModel]\nHeatTransferModel[T_, X_List, k_, \u03c1_, Cp_, Velocity_, Source_] :=\nModule[{V, Q, a = k},\nV = If[Velocity === \"NoFlow\",\nQ = If[Source === \"NoSource\", 0, Source];\nIf[FreeQ[a, _?VectorQ], a = a*IdentityMatrix[Length[X]]];\nIf[VectorQ[a], a = DiagonalMatrix[a]];\na = PiecewiseExpand[Piecewise[{{-a, True}}]];\nInactive[Div][a.Inactive[Grad][T, X], X] + V - Q]\n\n\nIf we follow the recipe of tutorial, we should be able to construct and solve a PDE system free of sign errors as I show in the following workflow.\n\n(* Create a Domain *)\n\u03a92D = Rectangle[{2, 2}, {3, 3}];\n(* Create parametric PDE operator *)\npop = HeatTransferModel[y[x1, x2], {x1, x2}, k, \u03c1, Cp, \"NoFlow\",\n\"NoSource\"];\n(* Replace k parameter *)\nop = pop /. {k -> 1};\n(* Setup flux conditions *)\nnv2 = NeumannValue[-1, x1 == 2];\nnv3 = NeumannValue[1, x1 == 3];\n(* Setup Dirichlet Conditions *)\ndc2 = DirichletCondition[y[x1, x2] == 2 + x1, x2 == 2];\ndc3 = DirichletCondition[y[x1, x2] == 3 + x1, x2 == 3];\n(* Create PDE system *)\npde = {op == nv2 + nv3, dc2, dc3};\n(* Solve and Plot *)\nyfun = NDSolveValue[pde, y, {x1, x2} \u2208 \u03a92D]\nPlot3D[Evaluate[yfun[x1, x2]], {x1, x2} \u2208 \u03a92D,\nPlotRange -> All, AxesLabel -> {\"x1\", \"x2\", \"y[x1,x2]\"},\nBaseStyle -> 12]\n\n\nYou can test that the solution matches that exact solution over the entire range:\n\nManipulate[\nPlot[{x1 + x2, yfun[x1, x2]}, {x1, 2, 3}, PlotRange -> All,\nAxesLabel -> {\"x1\", \"y[x1,x2]\"}, BaseStyle -> 12,\nPlotStyle -> {Red,\nDirective[Green, Opacity[0.75], Thickness[0.015], Dashed]}], {x2,\n2, 3}, ControlPlacement -> Top]\n\n\n\u2022 Dear Tim Laska, thank you for your great help, can we evaluate the error and plot it? \u2013\u00a0user62716 Jun 26 at 9:47\n\u2022 I did it plot = Plot3D[ Abs[yfun[x1, x2] - (x1 + x2)], {x1, x2} [Element] [CapitalOmega]2D, PlotRange -> All, AxesLabel -> {\"x1\", \"x2\", \"y[x1,x2]\"}, PlotLabel -> err] \u2013\u00a0user62716 Jun 26 at 10:03\n\u2022 Dear Tim Laska, I have other problem, Poisson equation with variable coefficients,shall post it in new question or here? \u2013\u00a0user62716 Jun 26 at 11:48\n\u2022 @user62716 You should open a new question as it appears that you have. I will try to take a look at your other question when I can. \u2013\u00a0Tim Laska Jun 26 at 13:45\n\u2022 Thank you Tim, I will be waiting. Best regards \u2013\u00a0user62716 Jun 26 at 13:50\n\nBy reversing the sign of the derivative on the left side from that given in NeumannValue, this can be solved by Mathematica analytically as well.\n\nClearAll[y, x1, x2];\npde = Laplacian[y[x1, x2], {x1, x2}] == 0;\n\nbc = {y[x1, 2] == 2 + x1,\ny[x1, 3] == 3 + x1,\nDerivative[1, 0][y][2, x2] == 1,\nDerivative[1, 0][y][3, x2] == 1};\n\nsolA = DSolve[{pde, bc}, y[x1, x2], {x1, x2}];\nsolA = solA /. {K[1] -> n,Infinity -> 20};\nsolA = Activate[solA];\n\n\nPlot3D[y[x1, x2] /. solA, {x1, 2, 3}, {x2, 2, 3}, PlotRange -> All,\nAxesLabel -> {\"x1\", \"X2\", \"y[x1,x2]\"}, BaseStyle -> 12]\n\n\nThe BC as given above are correct, and Mathematica's analytical solution is correct also, but I agree it can be simpler.\n\nThere might be a way to simplify the infinite Fourier sum given, but I could not find it.\n\nTo show the above formulation is correct, here is Maple's solution, using same B.C. Maple as above to give the simpler form of the solution, which is $$y=x_1+x_2$$.\n\nrestart;\npde:=VectorCalculus:-Laplacian(y(x1,x2),[x1,x2])=0;\nbc:=y(x1,2)=2+x1,y(x1,3)=3+x1,D[1](y)(2,x2)=1,D[1](y)(3,x2)=1;\nsol:=pdsolve([pde,bc],y(x1,x2))\n\n\nWe just have to remember, that negative NeumannValue on left edge, means positive derivative on that edge.\n\n\u2022 Dear Nasser, thank you for your comments, the normal derivative at left side is -1 not 1, the above analytic solution is complicated since the exact is just y=x1+x2....thanks \u2013\u00a0user62716 Jun 26 at 9:38\n\u2022 the normal derivative at left side is -1 not 1 no. It is +1. you set NeumannValue to be -1. Since NeumannValue points outwards, then this means the deivative is +1. Since -1 outwards, means +1 inwards. In addition, if you change the derivative (not NeumannValue) in the code I posted from +1 to -1 you will see the solution is no longer y=x1+x2 but becomes non-linear. You can compare this solution with the numerical solution. Do you see any difference? I agree the solution has complicated Fourier series sum, but this is what Mathemtica gave for the analytical solution. \u2013\u00a0Nasser Jun 26 at 10:01\n\u2022 Dear Nasser, I still can not understand you, @Nasser \u2202y\u2202\u03bd(2,x2)=\u2212\u2202y\u2202x1(2,x2) because the outward normal at the point (2,x2) is \u03bd=(\u22121,0) so it is -1 on the left outwards. \u2013\u00a0user62716 Jun 26 at 11:16\n\u2022 Dear Nasser, the code of Tim Laska is working, I highly appreciate you and you always help me and provide perfect answer. \u2013\u00a0user62716 Jun 26 at 11:19\n\u2022 @user62716 you can see from the solution itself, i.e. from just looking at the plot, that the derivative is positive on the left edge. No math is needed if we look at the solution. The slope is moving upwards. So positive slope. You can also see from Maple solution I posted, that I used positive derivative to get same solution $x_1+x_2$ right there. NeumannValue is not the same as derivative. That is what the whole confusion was about. \u2013\u00a0Nasser Jun 26 at 11:33", "date": "2020-09-25 10:01:41", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/224729/laplace-equation-with-mixed-boundary-conditions", "openwebmath_score": 0.4856671988964081, "openwebmath_perplexity": 2496.7887469288025, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338101862455, "lm_q2_score": 0.8947894590884704, "lm_q1q2_score": 0.8603703179118266}}
{"url": "http://slideplayer.com/slide/772358/", "text": "# Scientific Notation ... is a way to express very small or very large numbers. \u00a0 ... is most often used in \"scientific\" calculations where the analysis.\n\n## Presentation on theme: \"Scientific Notation ... is a way to express very small or very large numbers. \u00a0 ... is most often used in \"scientific\" calculations where the analysis.\"\u2014 Presentation transcript:\n\nScientific Notation ... is a way to express very small or very large numbers. ... is most often used in \"scientific\" calculations where the analysis must be very precise. ... consists of two parts*: (1) a number between 1 and 10 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 and (2) a power of 10. *a large or small number may be written as any power of 10; however, CORRECT scientific notation must satisfy the above criteria.\n\nis correct scientific notation\n3.2 x 1013 is correct scientific notation Remember that the first number MUST BE greater than or equal to one and less than 10. 23.6 x 10-8 is not correct scientific notation\n\nTo Change from Standard Form to Scientific Notation:\n1 Place decimal point such that there is one non-zero digit to the left of the decimal point. 2 Count number of decimal places the decimal has \"moved\" from the original number.\u00a0 This will be the exponent of the 10. 3 If the original number was less than 1, the exponent is negative; if the original number was greater than 1, the exponent is positive.\n\nExamples: Given: 4,750,000 use: 4.75 (moved 6 decimal places)\nanswer:\u00a0 4.75 X 106 The original number was greater than 1 so the exponent is positive. Given:\u00a0 use: 7.89 (moved 4 decimal places) answer:\u00a0 7.89 x 10-4 The original number was less than 1 so the exponent is negative.\n\nTo Change from Scientific Notation to Standard Form:\n1 Move decimal point to right for positive exponent of 10. 2 Move decimal point to left for negative exponent of 10.\n\nExamples: Given: 1.015 x 10-8 answer: 0.00000001015 (8 places to left)\nNegative exponent move decimal to the left. Given:\u00a0 x 103 answer:\u00a0 5,024\u00a0\u00a0 (3 places to right) Positive exponent move decimal to the right.\n\nTo Multiply and/or Divide using Scientific Notation:\n1 Multiply/divide decimal numbers with each other. 2 Use exponent rules to \"combine\" powers of 10. 3 If not \"correct\" scientific notation, change accordingly.\n\nGiven: \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Method: \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Answer: \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Correct Scientific Notation\n\nExpress in correct scientific notation:\nPart 1: Express in correct scientific notation: 1.\u00a0 61,500 2.\u00a0 3.\u00a0 321 4.\u00a0 64,960,000 5.\n\nPart 2: Express in standard form: 1.\u00a0 1.09 x 103 2.\u00a0 x 108 3.\u00a0 x 10-4 4.\u00a0 x 10-2 5.\u00a0 x 102\n\nMultiply or divide as indicated and\nPart 3: Multiply or divide as indicated and express in correct scientific notation: 1.\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4. 5.\n\nDownload ppt \"Scientific Notation ... is a way to express very small or very large numbers. \u00a0 ... is most often used in \"scientific\" calculations where the analysis.\"\n\nSimilar presentations", "date": "2017-08-17 03:55:46", "meta": {"domain": "slideplayer.com", "url": "http://slideplayer.com/slide/772358/", "openwebmath_score": 0.8087553977966309, "openwebmath_perplexity": 2609.7140696126407, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9416541643004809, "lm_q2_score": 0.913676530465412, "lm_q1q2_score": 0.8603673097363703}}
{"url": "https://math.stackexchange.com/questions/2854507/function-transformations-question-vertical-or-horizontal-transformation/2854515#2854515", "text": "# Function transformations question - vertical or horizontal transformation\n\nI have got a very simple problem. I have an exercise:\n\nIf $\\ f(x) = 2x^2 \u2212 4$, give the function which shows the graph of $\\ f(x)$ after vertical stretch of scale factor $\\ 0.5$ followed by a translation $\\binom{-4}{0}$\n\nThe answer that I get is $\\ f(x)=x^2+8x+14$, but the answer given is $\\ f(x)=8x^2+64x+124$. In my opinion, the answer that is given can certainly be achieved, but using horizontal translation instead of vertical. After drawing a graph of my function and the given function I noticed that in my case the function is compressed (its \"branches\" are closer to the x - axis than the original one) - as it should be, as scale factor is less than 1.\n\nAm I wrong there, or is something wrong with answers? I would not have asked the question, but I noticed that there is at least one more question about which I am uncertain as much as about this one, thus, I need to find out the real answer.\n\n\u2022 $x^2+8x+14=((x+4)^2-4)/2$ is correct. Reusing the symbol $f(x)$ to mean different things leads to confusion.\n\u2013\u00a0user574889\nJul 17, 2018 at 13:49\n\u2022 @cactus, sorry, did not think about changing it to something else. Thanks for the answer! That means that there are 3 mistakes in a row in the exercises... Jul 17, 2018 at 13:52\n\nYou are right.\n\nWe are looking for the function \\begin{align}\\frac12f(x+4)&=\\frac12(2(x+4)^2-4)\\\\&=(x+4)^2-2\\\\ &=x^2+8x+14 \\end{align}\n\nOf course, there is a possibility that you course is asking the wrong question as well.\n\n\u2022 Thanks for the answer! That means that there are 3 mistakes in a row in those exercises... I will accept the answer soon. Jul 17, 2018 at 13:53\n\n$$f(x)=2(x^2-4)$$ Vertical stretching with scale $1:2$: $$f_1(x)=0.5 f(x)=x^2-4$$ Translation by a vector $[-4,0]$: $$f_2(x)=f_1(x-(-4))+0=(x+4)^2-2=x^2+8x+14$$\n\nThus your solution seems to be fine.\n\nThe textbook solution might be created by taking a stretching factor $4$ and translation $[-4,12]$", "date": "2022-10-07 06:19:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2854507/function-transformations-question-vertical-or-horizontal-transformation/2854515#2854515", "openwebmath_score": 0.8057282567024231, "openwebmath_perplexity": 267.5890928341676, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9553191335436405, "lm_q2_score": 0.900529778109184, "lm_q1q2_score": 0.8602933273535125}}
{"url": "https://mathhelpboards.com/threads/what-is-1-21-2-21-3-21-18-21-in-mode-19.25200/", "text": "# what is 1^21 + 2^21 + 3^21 + ....... + 18^21 in mode 19?\n\n#### ketanco\n\n##### New member\nwhat is 1^21 + 2^21 + 3^21 + ....... + 18^21 in mode 19?\n\ni can only think about individually calculating equivalents in mode 19 and then adding them up but there must be a better way then finding equivalents of exponentials of numbers from to 1 to 18, as this question is expected to be solved in around 2 minutes or less...\n\n#### Olinguito\n\n##### Well-known member\nBy Fermat\u2019s little theorem, $1^{18},2^{18},\\ldots,18^{18}\\equiv1\\pmod{19}$. Hence\n$$\\begin{array}{rcl}1^{21}+\\cdots+18^{21} &\\equiv& 1^3+\\cdots+18^3\\pmod{19} \\\\\\\\ {} &=& (1+\\cdots+18)^2\\pmod{19} \\\\\\\\ {} &=& \\left(\\dfrac{18}2\\cdot19\\right)^2\\pmod{19} \\\\\\\\ {} &\\equiv& 0\\pmod{19}.\\end{array}$$\n\nStaff member\n\n#### Olinguito\n\n##### Well-known member\nHey Olinguito , how does it follow that:\n\n$1^3+\\cdots+18^3\\pmod{19} = (1+\\cdots+18)^2\\pmod{19}$\n\nDoesn\u2019t $a=b$ imply $a\\pmod n=b\\pmod n$ (taking the modulus to be between $0$ and $n-1$)?\n\nLast edited:\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nDoesn\u2019t $a=b$ imply $a\\pmod n=b\\pmod n$ (taking the modulus to be between $0$ and $n-1$)?\nAfter staring at the following picture long enough, I finally realized why $a=b$.\n\nI think it's a different theorem though (Nicomachus's theorem).\n\nBtw, couldn't we instead observe that:\n$$1^3+2^3+...+17^3+18^3\\equiv 1^3+2^3+...+(-2)^3+(-1)^3 \\equiv 0 \\pmod{19}$$\n\n#### Olinguito\n\n##### Well-known member\nBtw, couldn't we instead observe that:\n$$1^3+2^3+...+17^3+18^3\\equiv 1^3+2^3+...+(-2)^3+(-1)^3 \\equiv 0 \\pmod{19}$$\nThat\u2019s an excellent observation!\n\n#### ketanco\n\n##### New member\nAfter staring at the following picture long enough, I finally realized why $a=b$.\n\nI think it's a different theorem though (Nicomachus's theorem).\n\nBtw, couldn't we instead observe that:\n$$1^3+2^3+...+17^3+18^3\\equiv 1^3+2^3+...+(-2)^3+(-1)^3 \\equiv 0 \\pmod{19}$$\nyes this is how they xpected us to solve i think... this must be the answer. thanks", "date": "2020-09-29 08:23:43", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/what-is-1-21-2-21-3-21-18-21-in-mode-19.25200/", "openwebmath_score": 0.8644099235534668, "openwebmath_perplexity": 2358.100933422333, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9822876997410348, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8602748006350689}}
{"url": "http://math.stackexchange.com/questions/79658/w-bot-is-a-subspace-of-u-bot", "text": "# $W^{\\bot}$ is a subspace of $U^{\\bot}$?\n\nLet $U$ and $W$ be subspaces of an inner product space $V$. If $U$ is a subspace of $W$, then $W^{\\bot}$ is a subspace of $U^{\\bot}$?.\n\nI don't find the above statement intuitively obvious. Could someone provide a proof?\n\n-\n\n## 2 Answers\n\nIf you're orthogonal to everything in a set, then you're also orthogonal to everything in every subset of that set.\n\nPut another way: Elements of $W^\\perp$ have to be orthogonal to more vectors than elements of $U^\\perp$; they have to be orthogonal to the vectors in $U$ and the vectors in $W\\setminus U$ (if any). Therefore there are less of them than if we took the vectors that only have to satisfy the property of being orthogonal to vectors in $U$. (More restrictions$\\implies$ Fewer vectors satisfying the restrictions.)\n\n-\nAnd that since you know that $U^\\perp,W^\\perp$ are subspaces (hopefully) of $V$ then it suffices to show containment. \u2013\u00a0Alex Youcis Nov 6 '11 at 22:30\nSo shouldn't it be $U^{\\bot}$ is a subspace of $W^{\\bot}$ ? That is my confusion. \u2013\u00a0Mark Nov 6 '11 at 22:33\n@Mark: No. If you're orthogonal to everything in $W$, i.e., if you're in $W^\\perp$, then you're also orthogonal to everything in a subset of $W$ such as $U$, i.e., you're in $U^\\perp$. For a concrete example, think of $\\mathbb R^3$ with standard dot product, let $U$ be the span of $(1,0,0)$, and let $W$ be the span of $\\{(1,0,0),(0,1,0)\\}$. If you have to be orthogonal to both $(1,0,0)$ and $(0,1,0)$, then you are in particular orthogonal to $(1,0,0)$, so $W^\\perp\\subset U^\\perp$. \u2013\u00a0Jonas Meyer Nov 6 '11 at 22:35\nOh right, I had this at the back of my head, but was unable to explain it clearly. Now I understand, thanks. \u2013\u00a0Mark Nov 6 '11 at 22:54\n\nIt should be intuitive, already at the level of logic:\n\nTo be in $W^\\perp$, you have to satisfy a certain condition $P(w)$ (namely: 'be orthogonal to $w$') for each and every element $w\\in W$.\n\nSo given a subset $U\\subseteq W$, to be in $U^\\perp$ means you have to satisfy $P(w)$ merely for all $u\\in U$.\n\nThus you have to satisfy less properties to be in $U^\\perp$, thus it is easier to be in $U^\\perp$, thus $U^\\perp$ is larger: $W^\\perp\\subseteq U^\\perp$.\n\nIt should also be intuitive geometrically: consider $\\mathbb{R}^3$, let $U$ be the $x$-axis, and $W$ the $x,y$-plane. Then $U^\\perp$ is the $y,z$-plane, and $W^\\perp$ is the $z$-axis.\n\n//Edit: I was slow so I missed Joans Meyer's edit, which kind of makes my answer redundant.\n\n-", "date": "2016-05-01 06:10:09", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/79658/w-bot-is-a-subspace-of-u-bot", "openwebmath_score": 0.9439272880554199, "openwebmath_perplexity": 228.43185217032388, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877002595527, "lm_q2_score": 0.8757870013740061, "lm_q1q2_score": 0.8602747994968822}}
{"url": "https://stats.stackexchange.com/questions/297685/pearson-correlation-between-a-variable-and-its-square", "text": "# Pearson correlation between a variable and its square\n\nHere is my R code to get familiarised with Pearson's correlation. I generate values of $X$ from 1 to 100, then find the correlation between $X$ and $X^2$:\n\nx=1:100\ny=x\nfor(i in 1:100) {y[i]=x[i]*x[i]}\ncor.test(x,y, type=\"pearson\")\n\n\nI get this result :\n\nPearson's product-moment correlation\n\ndata:  x and y\nt = 38.668, df = 98, p-value < 2.2e-16\nalternative hypothesis: true correlation is not equal to 0\n95 percent confidence interval:\n0.9538354 0.9789069\nsample estimates:\ncor\n0.9687564\n\n\n$r$ seems high to me.\n\nMy question is: what exactly does the $r$ coefficient quantify? Does it only quantify the closeness of the relationship between $X$ and $Y$ variable to a linear relationship ?\n\nOr is it also suited to quantify the intensity of a relationship between $X$ and $Y$ broadly speaking (whether this relationship is close to linearity or not)?\n\nMy last question is: are there other correlation test better suited than Pearson's test to quantify the intensity of the relationship between two given variables when the kind (linear, quadratic, exponential, etc.) of this relationship is not known a priori or is Pearson's test sufficient to do this kind of job?\n\n\u2022 Welcome to our site! You don't need to put \"thanks in advance\" comments at the end of your questions - in fact we prefer it if you don't, since it means more for future readers to read through. \u2013\u00a0Silverfish Aug 13 '17 at 9:28\n\u2022 Not an answer to your statistical question, but you mind find it helpful to know that you don't need to use a for loop in your R code, you can just do x=1:100; y=x^2; cor(x,y) or even cor(1:100, (1:100)^2) \u2013\u00a0Silverfish Aug 13 '17 at 9:29\n\u2022 @Silverfish indeed, this is illustrated in the first two lines of code in my answer as well \u2013\u00a0Glen_b -Reinstate Monica Aug 13 '17 at 9:50\n\nYou are curious about whether your value of $r$ is \"too high\" \u2014 it seems you think that, as $X$ and $X^2$ do not have an exactly linear relationship, then the Pearson's $r$ should be rather low. The high $r$ is not telling you that the relationship is linear, but it is telling you that the relationship is rather close to being linear.\n\nIf you are specifically interested in the case where $X$ is uniform, you might want to look at this thread on Math SE on the covariance between a uniform distribution and its square. You are using discrete uniform distribution $1,2,\\dots,n$ but if you rescaled $X$ by a factor of $1/n$, and hence rescaled $X^2$ by a factor $1/n^2$, the correlation would be unchanged (since correlation is not affected by rescaling by a positive scale factor). You would now have a discrete uniform distribution with equal probability masses on $\\frac{1}{n}, \\frac{2}{n}, \\dots, \\frac{n-1}{n}, 1$. For large values of $n$, this approximates a continuous uniform distribution (also called \"rectangular distribution\") on $[0,1]$.\n\nBy an argument analogous to that on the Math SE thread, we have:\n\n$$\\operatorname{Cov}(X,X^2) = \\mathbb{E}(X^3)-\\mathbb{E}(X)\\mathbb{E}(X^2) = \\int_0^1 x^3 dx - \\int_0^1 x dx \\cdot \\int_0^1 x^2 dx$$\n\nThis integrates to $\\frac{1}{4} - \\frac{1}{2} \\cdot \\frac{1}{3} = \\frac{1}{12}$.\n\nWe also have $\\operatorname{Var}(X) = \\mathbb{E}(X^2)-\\mathbb{E}(X)^2 = \\frac{1}{3} - \\left(\\frac{1}{2}\\right)^2 = \\frac{1}{12}$.\n\nSimilarly we find $\\operatorname{Var}(X^2) = \\mathbb{E}(X^4)-\\mathbb{E}(X^2)^2 = \\frac{1}{5} - \\left(\\frac{1}{3}\\right)^2 = \\frac{4}{45}$.\n\nHence, if $X \\sim U(0,1)$, then:\n\n$$\\operatorname{Corr}(X,X^2) = \\frac{\\operatorname{Cov}(X,X^2)}{\\sqrt{\\operatorname{Var}(X) \\cdot \\operatorname{Var}(X^2)}} = \\frac{\\frac{1}{12}}{\\sqrt{{\\frac{1}{12}}\\cdot{\\frac{4}{45}}}} = \\frac{\\sqrt{15}}{4}$$\n\nTo seven decimal places, this is $r = 0.96824583$, even though the relationship is quadratic rather than linear. Now you have taken a discrete uniform distribution on $1, 2, \\dots, n$ rather than a continuous one, but for the reasons explained above, increasing $n$ will produce a correlation closer to the continuous case, so that $\\sqrt{15}/4$ will be the limiting value. Let us confirm this in R:\n\ncorn <- function(n){\nx = 1:n\ncor(x,x^2)\n}\n\n> corn(2)\n[1] 1\n> corn(3)\n[1] 0.9897433\n> corn(4)\n[1] 0.984374\n> corn(5)\n[1] 0.9811049\n> corn(10)\n[1] 0.9745586\n> corn(100)\n[1] 0.9688545\n> corn(1e3)\n[1] 0.9683064\n> corn(1e6)\n[1] 0.9682459\n> corn(1e7)\n[1] 0.9682458\n\n\nThat correlation of $r=0.9682458$ may sound surprisingly high, but if we inspected a graph of the relationship between $X$ and $X^2$ it would indeed appear approximately linear, and this is all that the correlation coefficient is telling you. Moreover, we can see from our table of output from the corn function that increasing the value of $n$ makes the linear correlation smaller (note that with two points, we had a perfect linear fit and a correlation equal to one!) but that although $r$ is falling, it is bounded below by $\\sqrt{15}/4$. In other words, increasing the length of your sequence of integers makes the linear fit somewhat worse, but even as $n$ tends to infinity your $r$ never becomes worse than $0.9682\\dots$.\n\nx=1:100; y=x^2\nplot(x,y)\nabline(lm(y~x))\n\n\nPerhaps visually you are still not convinced that the correlation looks as strong as the calculated coefficient suggests \u2014 clearly the points are below the line of best fit for low and high values of $X$, and above it for intermediate $X$. If it can't capture this quadratic curvature, is the line really such a good fit to the points?\n\nYou may find it helpful to compare the overall variation of the $Y$ coordinates about their own mean (the \"total variation\") to how much the points vary above and below the regression line (the \"residual variation\" that the regression line was unable to explain). The fraction of the residual variation over the total variation tells you what proportion of the variation was not explained by the regression line; the proportion of variation that is explained by the regression line is then one minus this fraction, and is called the $R^2$. In this case, we can see that the variation of points above and below the line is relatively small compared to the variation in their $Y$ coordinates, and so the proportion unexplained by the regression is small and the $R^2$ is large. It turns out that for a simple linear regression, $R^2$ is equal to the square of the Pearson correlation. In fact $r=\\sqrt{R^2}$ if the regression slope is positive (an increasing relationship) or $r=-\\sqrt{R^2}$ if the slope is negative (decreasing).\n\nWe had a large $R^2$ so our correlation is large also. This is the sense we mean when we state that \"a Pearson correlation near $\\pm 1$ indicates the linear fit is good\" \u2014 not that our straight regression line captures the true nature of the relationship between $X$ and $Y$, and so there is no curvature and no discernible pattern in the residual variation, but instead that the line provides a good approximation to the true relationship, and that the proportion of residual variation (i.e. that part left unexplained by the linear model) is small.\n\nNote that had you chosen a discrete uniform on e.g. $-100, -99, \\dots, 99, 100$ and rescaled that to being between $[-1,1]$ and you would have found a covariance and correlation of zero, as happens in the linked Math SE thread. There is neither an increasing nor decreasing relationship.\n\nx=-100:100; y=x^2\nplot(x,y)\nabline(lm(y~x))\n\n\nAs an exercise to think through, what would be the correlation between $-1, -2, -3, \\dots, -n$ and its squares? You can easily write some R code to confirm your guess.\n\nIf all you care about is the existence of an increasing or decreasing relationship, rather than the extent to which it is linear, you can use a rank-based measure such as Kendall's tau or Spearman's rho, as mentioned in Glen_b's answer. For my first graph, which had a perfectly monotonic increasing relationship, both methods would have given the highest possible correlation (one). For the second graph, which is neither increasing nor decreasing, both would give a correlation of zero.\n\nThe Pearson correlation measures the closeness to a linear relationship. If $X$ is positive, then the correlation between $X$ and $X^2$ is often fairly close to 1.\n\nIf you want to measure the strength of monotonic relationship, there are a number of other choices, of which the two best known are the Kendall correlation (Kendall's tau), and the Spearman correlation (Spearman's rho)\n\n x=1:100\ncor(x,x^2,method=\"pearson\")\n[1] 0.9688545\ncor(x,x^2,method=\"kendall\")\n[1] 1\ncor(x,x^2,method=\"spearman\")\n[1] 1\n\n\nI'd add that looking at the correlation of non-random values isn't necessarily where I'd start - it can be useful when exploring edge cases, however.\n\nFor the Pearson correlation you may find it useful to consider playing about with the rho and n values here:\n\nn=100\nrho=0.6\nx=rnorm(100)\nz=rnorm(100)\ny=rho*x + sqrt(1-rho^2)*z\nplot(x,y)\ncor(x,y)\n\n\n(In particular, you might try varying rho from close to -1 up to close to 1)\n\nYou may also find these discussions of correlation useful for getting a handle on what correlations do and don't do:\n\nWhy zero correlation does not necessarily imply independence\n\nDoes the correlation coefficient, r, for linear association always exist?\n\nIf A and B are correlated with C, why are A and B not necessarily correlated?\n\nHow would you explain covariance to someone who understands only the mean?\n\nPearson's or Spearman's correlation with non-normal data\n\nHow to choose between Pearson and Spearman correlation?\n\nKendall Tau or Spearman's rho?\n\nIf linear regression is related to Pearson's correlation, are there any regression techniques related to Kendall's and Spearman's correlations?", "date": "2020-01-22 07:51:01", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/297685/pearson-correlation-between-a-variable-and-its-square", "openwebmath_score": 0.7718861699104309, "openwebmath_perplexity": 398.3522745880172, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877002595527, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8602747979045842}}
{"url": "https://www.physicsforums.com/threads/differentiation-under-integral-sign.370218/", "text": "# Differentiation under integral sign\n\n1. Jan 17, 2010\n\n### neelakash\n\n1. The problem statement, all variables and given/known data\n\nI have to evaluate the numerical value of the derivative of the following integral for x=1\n\n$$\\int_{0}^{\\ ln\\ x}\\ e^{\\ -\\ x\\ (\\ t^2\\ -\\ 2)}\\ dt$$\n\n2. Relevant equations\n\nThe formula for differentitation under integral sign.\n\n3. The attempt at a solution\n\nThe upper limit term is straightforward:it is\n\n$$\\frac{\\ 1}{\\ x}\\ e^{\\ -\\ x[\\ (\\ ln\\ x)^{\\ 2}\\ -\\ 2]}$$\n\nThe other part is\n\n$$\\int_0^{\\ ln\\ x}\\frac{\\partial}{\\partial\\ x}\\ e^{\\ -\\ x(\\ t^2\\ -2)}\\ dt\\ =\\ -\\ e^{\\ -\\ 2\\ x}\\ [\\int_0^{\\ ln\\ x}\\ t^2\\ e^{\\ -\\ x\\ t^2}\\ dt\\ -\\ 2\\int_0^{\\ ln\\ x}\\ e^{\\ -\\ x\\ t^2}\\ dt\\ ]$$\n\nThe later can be evaluated and I got the following:\n\n$$\\ -\\ e^{\\ -\\ 2\\ x}\\ [\\frac{\\ -(\\ ln\\ x)\\ e^{\\ -\\ x(\\ ln\\ x)^2}}{\\ 2\\ x}\\ +\\int_0^{\\ x(\\ ln\\ x)^2}\\frac{\\ e^{\\ -\\ u}}{4x\\sqrt{ux}}\\ du\\ -\\int_0^{\\ x(\\ ln\\ x)^2}\\frac{\\ e^{\\ -\\ u}}{\\sqrt{ux}}\\ du}]$$\n\nI found the result as above.However,the two integrals neither cancel with each other nor can be evaluated.Can anyone please check and tell what should be done further.\n\nNeel\n\nLast edited: Jan 17, 2010\n2. Jan 17, 2010\n\n### rasmhop\n\nLet,\n$$F(x) = \\int_{0}^x e^{-e^x (t^2 - 2)} \\text{ d}t$$\nso you want to find the derivative of F(ln(x)) which you can do using the chain rule.\n\n3. Jan 18, 2010\n\n### neelakash\n\nDoes not help;it ultimately reduces to what I have got...\n\nThe thing lies in putting the limits without explicitly solving the final two inntegrals.They give zero.\n\n4. Jan 18, 2010\n\n### D H\n\nStaff Emeritus\nI helped neelakash with this problem on another forum (http://www.sciforums.com/showthread.php?t=99010). As he arrived at the correct answer there, I have no qualms posting the solution here for future reference by others.\n\nneelakash did use the appropriate technique for differentiating under the integral sign, the Leibniz Integral Rule:\n\n$$\\frac{d}{dx}\\int_{a(x)}^{b(x)} f(t,x)\\,dt = \\int_{a(x)}^{b(x)} \\frac{\\partial} {\\partial x} f(t,x)\\,dt + f(b(x),x)\\frac{db(x)}{dx} - f(a(x),x)\\frac{da(x)}{dx}$$\n\nIn this particular problem,\n\n$$f(t,x) = \\exp\\left(-x(t^2-2)\\right),\\quad a(x)=0, \\quad b(x)=\\ln x$$\n\nThe partial derivative of f(t,x) wrt x is\n\n$$\\frac{\\partial}{\\partial x}f(t,x) = -(t^2-2) \\exp\\left(-x(t^2-2)\\right)$$\n\nApplying the Leibniz Integral Rule,\n\n$$\\frac{d}{dx}\\left(\\int_0^{\\ln x} \\exp\\left(-x(t^2-2)\\right)\\,dt\\right) = -\\left(\\int_0^{\\ln x} (t^2-2) \\exp\\left(-x\\bigl(t^2-2)\\right) \\,dt\\right) + \\exp\\left(-x(\\ln^2x-2)\\right)/x$$\n\nThere is a sign error in the original post (that exp(-2x) should be an exp(2x)). Additionally, neelakash carried the integration a step too far. That integral on the right-hand side is evaluable in terms of the error function erf(x).\n\nHowever, there is no reason to do this. neelakash finally saw the \"Oh, SNAP!\" light that makes this problem particularly easy. From that other forum,\n\n5. Jan 18, 2010\n\n### rasmhop\n\nYou're right, I'm sorry for the wrong suggestion. Anyway try substituting x=1 in the terms you found. The integrals should disappear due to ln(x)=0 and the upper limit term should become e^2.\n\n6. Jan 18, 2010\n\n### HallsofIvy\n\nWhen x= 1, ln(x)= ln(1)= 0 so x ln(x)= 1(0)= 0. Both integrals are from 0 to 0 and so are qual to 0.\n\n7. Jan 18, 2010\n\n### neelakash\n\nYes,we need not carry out the integral explicitly as the answer comes from observation.", "date": "2018-03-18 19:37:04", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/differentiation-under-integral-sign.370218/", "openwebmath_score": 0.9087260961532593, "openwebmath_perplexity": 1593.9571984399427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98228770337066, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8602747926677609}}
{"url": "https://math.stackexchange.com/questions/4302370/how-do-you-find-good-rational-approximations-to-a-decimal-number", "text": "How do you find \"good\" rational approximations to a decimal number?\n\nWhen presented with real number as a decimal, are there any methods to finding \"good\" rational approximations $$a/b$$ to that number? By \"good\" I mean that $$a$$ and $$b$$ are reasonably small integers. For example suppose you're handed the number $$1.7320508075688772935274463415058723 \\dots$$ An obvious way to rationally approximate this numbers is to truncate it after the $$-n$$th decimal place and place it over $$10^n$$. So $$\\frac{173}{100} \\quad\\text{or}\\quad \\frac{1732050807}{1000000000} \\quad\\text{or}\\quad \\frac{1732050807568877}{1000000000000000}$$ are increasingly good rational approximations. A way I can see to improve this is if you chose to truncate at a multiple of $$2$$ or $$5$$, your rational approximation will reduce to one that is more \"good\". For example $$\\frac{1732}{1000} = \\frac{433}{250} \\qquad\\text{and}\\qquad \\frac{173205}{100000} = \\frac{34641}{20000}$$ Is there a clever way to see which multiples of $$2$$ of $$5$$ to truncate after to get a rational approximation that reduces a lot? This works because we express the decimal in base $$10$$, so are there any tricks considering the number in a different base? Is there an idea that's not even on my radar?\n\nThere's an obvious algorithmic iterative \"bottom-up\" way to find a good rational approximation. It's not terribly clever though. I can type it up as an answer in a second if no one else wants to.\n\n\u2022 Try the continued fraction Nov 10, 2021 at 16:25\n\u2022 Yes. Here is a link. This is exactly what continued fractions are good at. Nov 10, 2021 at 16:26\n\u2022 The usual way to find the best rational approximations is continued fractions. But are you looking to have the best approximation overall or specifically one with a \"nice\" denominator? Nov 10, 2021 at 16:26\n\u2022 See also Dirichlet's approximation theorem which essentially says that \"good\" rational approximations exist, in the sense of small denominator compared to precision, and the Thue-Siegel-Roth theorem (or whatever you want to call it) on how algebraic irrational numbers can't be approximated too well in some technical sense. Nov 10, 2021 at 16:30\n\u2022 @MikePierce The notion for \"best\" people use in practice is to minimise various types of heights. The measure you suggest is essentially the naive height, where $H(p/q) = \\max\\{|p|,|q|\\}$ for a rational number $p/q$ written in reduced form. Nov 10, 2021 at 16:33\n\nThis is exactly what continued fractions are for.\n\nThe continued fraction of $$\\sqrt 3=1.7320508075688772935274463415058723\\ldots$$ is periodic: $$[1; 1, 2, 1, 2, 1, 2, 1, 2, \\ldots]$$. The sequence of approximants is $$2,\\frac53,\\frac74,\\frac{19}{11},\\frac{26}{15},\\frac{71}{41},\\ldots$$ ; each of these is the best possible approximant for its denominator. For instance, $$\\frac{19}{11}$$ is the best approximant with a denominator $$\\le 11$$.\n\nThis is the usual criterion for goodness of approximation by rational numbers: $$\\frac{p}{q}$$ is a good approximation to a real number $$\\alpha$$ if it minimises $$|\\alpha-\\frac{p}{q}|$$ over all rationals with denominator $$\\le q$$. Powers of ten shouldn't come into it at all.\n\nI looked into this topic too, since i wanted a way to list all fractions in a small sub-interval of $$[0,1]$$. If you just want to look at an implementation that does this I attached some typescript code. My approach came from looking into the Farey sequence $$\\cal F_n$$. One can show that if $$a/b$$ and $$c/d$$ are fractions such that no other fraction $$e/f$$ with $$a/b < e/f < c/d$$ and $$f \\leq \\max(b,d)$$ exists, then $$\\frac{a+c}{b+d}$$ is the fration with the smallest denominator between $$a/b$$ and $$c/d$$.\n\nFor our case we choose a number $$v \\in (0,1)$$ that we want to approximate. Formulating this as an algorithm we can start with the lower bound $$a / b = 0/1$$ and upper bound $$c/d = 1/1$$. Next we define $$\\frac{e}{f} = \\frac{a+c}{b+d}.$$ Now 3 possible cases can happen. If $$f$$ is too large, the best approximation for $$v$$ is either $$a/b$$ or $$c/d$$. Otherwise we have have $$e/f < v$$ or $$e/f \\geq v$$. If $$e/f < v$$ replace $$a/b$$ by $$e/f$$ and repeat the steps. In the other case replace $$c/d$$ by $$e/f$$. More formally:\n\nChoose: A value $$v$$ to approximate and the maximum denominator $$Q$$.\n\nInitialize: $$a/b := 0/1$$ and $$c/d := 1/1$$.\n\nIterate: for $$a/b < v \\leq c/d$$:\n\n\u2022 Evaluate: $$e/f := (a+c)/(b+d)$$.\n\u2022 If $$f \\geq Q$$: Return $$a/b$$ or $$c/d$$.\n\u2022 If $$e/f < v$$: Replace $$a/b := e/f$$ and begin next iteration.\n\u2022 If $$v \\leq e/f$$: Replace \\$c/d := e/f and begin next iteration.\n\nOne all the steps are finished, you not only have the best approximation for a value $$v \\in (0,1)$$ but the closest two fractions $$a/b$$ and $$c/d$$ with $$\\frac{a}{b} < v \\leq \\frac{c}{d}$$.\n\nThis algorithm is what I started out with. It has worst case complexity $$O(Q)$$ time and $$O(1)$$ space. One can improve the algorithm by batching up multiple steps. I did not prove the complexity, but I think the improved version has complexity $$O(\\log Q)$$ time and $$O(1)$$ memory. I will attach some Typescript code for this improved algorithm.\n\n/**\n* Returns the two fractions closest to $$v$$ with $$a/b <= v <= c/d$$. One always has $$a/b \\neq c/d$$.\n*\n* # The Algorithm:\n*\n* Chooses $$\\frac{a^+}{b^+} = \\frac{a + e_1 c}{b + e_1 d} <= v$$ with $$b + e_1 d <= Q$$ in the first,\n* and $$\\frac{c^+}{d^+} = \\frac{e_2 a^+ + c}{ e_2 b^+ + d} >= v$$ with $$e_2 b + d <= Q$$ in the second step.\n* Iterate until nothing happens anymore.\n*\n* If $$0 < v < 1$$, the final fractions will be consecutive elements in the Farey sequence $$F_Q$$.\n*/\nfunction findLowerAndUpperApproximation(v: number, Q: number) {\n// Ensure v in [0,1)\nlet v_int = Math.floor(v);\nv = v - v_int; // only use fractional part of $$v$$ for iteration\nlet a = 0,\nb = 1,\nc = 1,\nd = 1;\n\nwhile (true) {\nlet _tmp = c - v * d; // Temporary denominator for 0 check\n\n// Batch e1 steps in direction c/d\nlet e1: number | undefined;\nif (_tmp != 0) e1 = Math.floor((v * b - a) / _tmp);\nif (e1 === undefined || b + e1 * d > Q) e1 = Math.floor((Q - b) / d);\n\na = a + e1 * c;\nb = b + e1 * d;\n\n_tmp = b * v - a; // Temporary denominator for 0 check\n\n// Batch e2 steps in direction a/b\nlet e2: number | undefined;\nif (_tmp != 0) e2 = Math.floor((c - v * d) / _tmp);\nif (e2 === undefined || b * e2 + d > Q) e2 = Math.floor((Q - d) / b);\n\nc = a * e2 + c;\nd = b * e2 + d;\n\n// If both steps do nothing we are done\nif (e1 == 0 && e2 == 0) break;\n}\n\nreturn [v_int*b + a, b, v_int*d + c, d];\n}\n\n\nIn each step the denominator of $$c/d$$ should at least double. This would result in complexity $$O(\\log Q)$$. However, i haven't worked out the details here. For me the algorithm is good enough to find the best fractions for $$Q = 10^9$$. Going past that my code seems to hit machine precision problems, which most likely affect the divisions in the algorithm. When you use a language with higher precision integer division you should be fine to go almost as far as you want.", "date": "2023-01-31 06:36:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4302370/how-do-you-find-good-rational-approximations-to-a-decimal-number", "openwebmath_score": 0.7420727610588074, "openwebmath_perplexity": 515.2579696190897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877012965886, "lm_q2_score": 0.8757869900269367, "lm_q1q2_score": 0.860274789259018}}
{"url": "http://www.oopsconcepts.com/freight-train-qpjqp/1wm0k.php?15357d=sss-postulate-examples", "text": "Example St. Francis Preparatory School. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. As a consequence, their angles will be the same. This video is provided by the Learning Assistance Center of Howard Community College. 14 December 2020 . This is one of them (SSS). This means that the pair of triangles have the same three sides and the same three angles (i.e., a total of six corresponding congruent parts). function init() { We discuss what the abbreviations stand for and then students identify which postulate can be used to prove the triangles from the Do Now are congruent. And as seen in the image to the right, we show that trianlge ABC is congruent to triangle CDA by the Side-Side-Side Postulate. The other triangle LMN will change to remain congruent to the triangle PQR. Advertisement. Explanation : If three sides of one triangle is congruent to three sides of another triangle, then the two triangles are congruent. Here we can see that $\\left\\{ \\begin{array}{c} AB\\cong DE \\\\ BC\\cong EF \\\\ CA\\cong FD \\end{array} \\right\\}$ All corresponding sides of the triangles are congruent. Topics. 6 Check It Out! Postulates are also called as axioms. Prove: $$\\triangle ABD \\cong \\triangle CBD$$ Solution to Example 2 1. i) \u0394ABD \u2245 \u0394ACD ii) AP is the perpendicular bisector of BC. AB ... \u2022Example: because of HL. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. Solutions. SSS Postulate First, there's the side-side-side postulate, or SSS . Introduction to triangle congruency lesson. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. SSS Postulate - Every SSS correspondence is a congruence. Solved Example on Postulate Ques: State the postulate or theorem you would use to prove that \u22201 and \u22202 are congruent. To prove that these triangles are congruent, we use SSS postulate, as the corresponding sides of both the triangles are equal. Read time: 3 minutes. We can say that two triangles are congruent if any of the SSS, SAS, ASA, or AAS postulates are satisfied. Solution : The game plan is to make use of the SAS Postulate. Is it true that \u2206ABC \u2245 \u2206XYZ? SSS Congruence Postulate If the three sides of a traingle are conrresponding and congruent to the three sides of the other triangle, th the two triangles are congruent. If the two angles and the non included side of one triangle are congruent to the two angles and the non included side of another triangle, then the two triangles are congruent. Name the postulate, if possible, that makes the triangles congruent. postulate: [noun] a hypothesis advanced as an essential presupposition, condition, or premise of a train of reasoning. Solving sas triangles. Glencoe Geometry. Covid-19 has affected physical interactions between people. Yes! Take Calcworkshop for a spin with our FREE limits course. How do you know? 5 Example 1 Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Example $$\\triangle ABC \\cong \\triangle XYZ$$ All 3 sides are congruent. Practice Proofs. Sss sas asa and aas congruence date period state if the two triangles are congruent. 1) In triangle ABC, AD is median on BC and AB = AC. pagespeed.lazyLoadImages.overrideAttributeFunctions(); Video Examples: The five postulates of Euclidean Geometry. Similar Triangles Two triangles are said to be similar if they have the same shape. Listen to the audio pronunciation in the Cambridge English Dictionary. Given : In \u0394ABC, AD is a median on BC and AB = AC. Triangles ABC has three sides congruent to the corresponding three sides in triangle C\u2026 Show Answer. Example How can you use congruent triangles to prove j Q @ j D Since QWE \u2245 DVK by AAS, you know that \u2220 All Rights Reserved. 7, 24, 25; 5, 12, 16; 6, 8, 9; 3, 5, 9 ; Show Video Lesson. Because if we can show specific sides and/or angles to be congruent between a pair of triangles, then the remaining sides and angles are also equal. The following postulate, as well as the SSS and SAS Similarity Theorems, will be used in proofs just as SSS, SAS, ASA, HL, and AAS were used to prove triangles congruent. Geometry \u203a SSS Postulate. The global warming postulate is based almost entirely on models, and today's models are deliberately biased to support global warming. So SAS-- and sometimes, it's once again called a postulate, an axiom, or if it's kind of proven, sometimes is called a theorem-- this does imply that the two triangles are congruent. For a list see Congruent Triangles. Methods of proving triangle congruent mathbitsnotebook(geo. Check out the interactive simulation to explore more congruent shapes and do not forget to try your hand at solving a \u2026 Example: Hypotenuse-Leg Theorem (HL theorem) If the hypotenuse and one of the legs (sides) of a right triangle are congruent to hypotenuse and corresponding leg of the other right triangle, the two triangles are said to be congruent. Prove: $$\\triangle ABC \\cong \\triangle EFC$$ Side Angle Side Example Proof. As you will quickly see, these postulates are easy enough to identify and use, and most importantly there is a pattern to all of our congruency postulates. Heather Z. Oregon State University. ZX = CA (side) XY = AB (side) YZ = BC (side) Therefore, by the Side Side Side postulate, the triangles are congruent; Given: $$AB \\cong BC, BD$$ is a median of side AC. How to Prove Triangles Congruent? The two triangles also have a common side: AC. Congruence Postulate SSS. On the front of the organizer, students will write SSS on the first tab, SAS on the second tab, and ASA on the third tab. Side Side Side Postulate -> If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. Proof 1. The Multiplication Postulate: If x = y, then x * 3 = y * 3 . The Pythagorean Theorem can be used when we know the length of two sides of a right triangle and we need to get the length of the third side. If all the sides are congruent, then the two triangles are congruent. In this blog, we will understand how to use the properties of triangles, to prove congruency between $$2$$ or more separate triangles. Postulate 18. We refer to this as the Side Side Side Postulate or SSS. The first two postulates side angle side sas and the side side side sss focus predominately on the side aspects whereas the next lesson discusses two additional postulates which focus more on the angles. Category: medical health lung and respiratory health. SSS Congruence Postulate. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. In order to prove that triangles are congruent, all the angles and sides have to be congruent. Prove that $$\\triangle LMO \\cong \\triangle NMO$$ Advertisement. In an exothermic reaction, the transition state is closer to the reactants than to the products in energy (Fig. Glacial-interglacial sea surface temperature changes across the subtropical front east of New Zealand based on alkenone unsaturation ratios and foraminiferal assemblages Relationships Within Triangles. Those are the Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS) postulates. Addition Postulate: If equal quantities are added to equal quantities, the sums are equal. Step: 5. Examples of postulate in a sentence, how to use it. In cat below. Example of Postulate. Now we have the SAS postulate. Figure 12.4 \u00afPN \u22a5 \u00afMQ and \u00afMN ~= \u00afNQ. SAS Congruence Postulate. It is the only pair in which the angle is an included angle. The Area Postulate - To every polygonal region there corresponds a unique positive real number. Example of Postulate. Example 1. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. Explain your reasoning. For your better understanding, here is now the exact statement of the SSS Congruence Postulate. MC Megan C. Piedmont College. Asked By: Ayada Lugo | Last Updated: 7th January, 2020. Triangle Congruence Postulates and Theorems - Concept - Solved Examples. Sas triangle congruence postulate explained youtube. CCSS.MATH.CONTENT.7.G.A.2 Draw (freehand, with ruler and protractor, and with technology) geometric shapes with given conditions. This includes triangles, and the scaling factor can be thought of as a ratio of side-lengths. EXAMPLE 1 Use the AA Similarity Postulate Determine whether the triangles are similar. Use the SAS Similarity Theorem to determine if triangles are similar. This is the only postulate that does not deal with angles. \u00a9 and \u2122 ask-math.com. Postulate is used to derive the other logical statements to solve a problem. EXAMPLE 3 Use the SSS Similarity Theorem Find the value of X-that makes \u2206POR ~ \u2206TUV Solution Both m R and m V equal 60 , so R V.Next, find the value of x that makes the How to say postulate. This geometry video tutorial provides a basic introduction into triangle congruence theorems. Covid-19 has led the world to go through a phenomenal transition . 8. Top Geometry Educators. Angle-Angle (AA) Similarity Postulate If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. window.onload = init; \u00a9 2021 Calcworkshop LLC / Privacy Policy / Terms of Service. 8. Side-angle-side (sas) triangle: definition, theorem & formula. So that actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. In this mini-lesson, we will learn about the SSS similarity theorem in the concept of the SSS rule of congruence, using similar illustrative examples. Using the Angle Addition Postulate and definition of. Side-Side-Side (SSS) 1. } } } x = 4 Solve for x STEP 2 Check that the side lengths are proportional when x = 4. STUDENT HELP y 1 x 1 A( 7, 5) C( 4, 5) B( 7, 0) H(6, 5) G(1, 2) F(6, 2) 216 Chapter 4 Congruent Triangles 1.Sketch a \u2026 Side Angle Side Practice Proofs. Congruent Triangles. SSS Similarity. If all three sides in one triangle are the same length as the corresponding sides in the other, then the triangles are congruent. Step: 4. And here, they wrote the angle first. So we need to learn how to identify congruent corresponding parts correctly and how to use them to prove two triangles congruent. Did you know that there are five ways you can prove triangle congruency? 14 Votes) SAS Postulate. called a linear pair. Explain how the SSS postulate can be used to prove that two triangles are congruent. In this lesson, we will consider the four rules to prove triangle congruence. As Math is Fun accurately states, there only five different congruence postulates that will work for proving triangles congruent. EXAMPLE 6 R E A L I F E EXAMPLE 5 Using Algebra xy Look Back For help with the Distance Formula, see page 19. This means that the corresponding sides are equal and the corresponding angles are equal. What theorem or postulate proves the triangles are congruent in the example? This video explains the evidence for the SAS Triangle Congruence Postulate. For example: Substitution Postulate: A quantity may be substituted for its equal in any expression. Side Side Side Postulate If three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. 97 examples: They are also postulated to stimulate other cells for granuloma formation\u2026 Or, if we can determine that the three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. AB \u00af = DE \u00af [Given.] for (var i=0; i If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. are similar. SSS Congruence Postulate. And they were able to do it because now they can write \"right angle,\" and so it doesn't form that embarrassing acronym. Congruent Triangles. ASA Postulate Example Angle-Angle-Side Whereas the Angle-Angle-Side Postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then \u2026 There's the Side-Angle -Side postulate, or SAS. Try this Drag any orange dot at P,Q,R. postulate meaning: 1. to suggest a theory, idea, etc. Two geometric figures are similar if one is a scaled version of the other. Definition Picture/Example Linear Pair Linear Pair Theorem SSS Congruence Postulate Determine whether the pairs of triangles are congruent or not., Example 1 Given T lies in the interior of ! Holt McDougal Geometry Triangle Similarity: AA, SSS, SAS Example 1: Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Can you can spot the similarity? Learn more. 4.6/5 (14 Views . A few examples were shown for a better understanding. NOT CONGRUENT The Congruence Postulates SSS ASA SAS AAS SSA AAA Name That Postulate SAS ASA SSS SSA (when possible) Name That Postulate (when possible) ASA SAS AAA SSA Name That Postulate (when possible) SAS SAS SAS Reflexive Property Vertical Angles Vertical Angles Reflexive Property SSA Let\u2019s Practice Indicate the additional information needed to \u2026 This is the only postulate that does not deal with angles. How to pronounce postulate. Of a traingle are congruent. Area Postulate - Through a given.... 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Quantities are added to equal quantities, the sums are equal and the corresponding sides in the English!: AC already know, two triangles are similar and write a Similarity statement and DA are congruent using methods.", "date": "2021-09-22 20:37:08", "meta": {"domain": "oopsconcepts.com", "url": "http://www.oopsconcepts.com/freight-train-qpjqp/1wm0k.php?15357d=sss-postulate-examples", "openwebmath_score": 0.4419773519039154, "openwebmath_perplexity": 1223.4144715405948, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9822877007780707, "lm_q2_score": 0.8757869900269366, "lm_q1q2_score": 0.8602747888049067}}
{"url": "https://math.stackexchange.com/questions/2554808/finding-approximation-of-largest-eigenvalue", "text": "Finding approximation of largest eigenvalue\n\nGiven the following matrix, find an approximation of the largest eigenvalue. $$A = \\begin{bmatrix} 3 & 2 \\\\ 7 & 5 \\\\ \\end{bmatrix}$$\n\nAnd I was also given $$\\vec x= \\begin{bmatrix} 1 \\\\ 0 \\\\ \\end{bmatrix}$$\n\nHow my professor solves this is by calculating the slopes of $A\\vec x = \\vec b_1$, $A^2 \\vec x = \\vec b_2$, $A^3 \\vec x = \\vec b_3$ and so on until we get the slope of $\\vec b_i$ converging to the same value. Then when we get the approximated $\\vec b$, he plug into $A \\vec b = \\lambda \\vec b$, and the corresponding $\\lambda$ is the largest eigenvalue.\n\nSince slope is $\\frac yx$ , it works fine for $2 x 2$ matrix. But how do I apply this method for a bigger matrix?\n\n\u2022 I've given you a full explanation and representation of the method used down below, make sure to check it out ! Dec 7, 2017 at 1:23\n\u2022 It's very helpful thank you! Dec 7, 2017 at 1:28\n\nWhat you mention, is a known numerical analysis method for the approximation of the largest (by absolute value) eigenvalue of a matrix.\n\nLet $$A\\in \\mathbb R^{n\\times n}$$ have $$n$$ linearly independent eigenvalues $$\\{ u \\}_{i=1}^n$$ as well as a unique eigenvalue $$\u03bb_1$$ such that : $$|\u03bb_1| < |\u03bb_2| \\leq \\dots \\leq |\u03bb_n|$$ where $$\u03bb_1 \\in \\mathbb R$$ and $$u_1 \\in \\mathbb R^n$$ : $$Au_1=\u03bb_1u_1.$$\n\nThe method \"Power Iteration\" :\n\n$$\\begin{cases} x_k= Ax_{k-1} \\to x_k = A^kx_0 \\quad k=1,2,\\dots \\\\ x_0 \\end{cases}$$\n\nThorem : The method \"Power Iteration\" converges $$\\forall x_0$$ (that is adequate for the problem given) and it is :\n\n$$\\lim_{k\\to \\infty} \u03b5_k\\frac{x_k}{||x_k||_2}=u_1$$\n\n$$\\lim_{k \\to \\infty} \\frac{x_{k,i}}{x_{k-1,i}}=\u03bb_1 \\quad \\forall \\space i=1,2,\\dots,n \\quad \\text{with} \\quad u_{1,i} \\neq 0$$\n\nwhere $$\\{ \u03b5_k\\}_{k=1}^\\infty = \\{ \\pm1\\}$$ and $$u_1$$ eigenvector of $$A$$ with $$||u_1||_2=1$$.\n\nI've given you a formal explanation of the method according to my old notes and my knowledge, for more, check here.\n\nHere's a hint: You want to determine when $\\mathbf{b}_n$ is a near-scalar multiple of $\\mathbf{b}_{n-1}$. In $\\mathbb{R}^2$, (nonzero) vectors are scalar multiples of one another iff their slopes are equal. A possibly more useful definition is that two vectors $\\mathbf{v}$ and $\\mathbf{w}$ are scalar multiples of one another if and only if\n\n$$\\hat{\\mathbf{v}} = \\pm\\hat{\\mathbf{w}},$$\n\nwhere\n\n$$\\hat{\\mathbf{v}} = \\frac{\\mathbf{v}}{|\\mathbf{v}|},$$\n\nwhich extends more nicely to multiple dimensions.\n\n\u2022 I'll try to apply this to the problem and see how it works out. Thank you ! Dec 7, 2017 at 1:21\n\u2022 @dembrownies Great. Let me know if you have any more questions. Dec 7, 2017 at 2:13\n\nYou normalize the vector at each iteration by dividing by its length, and wait until the resulting sequence of unit vectors has gotten close enough to converging for your purpose. (This is called the power method. It's pretty much the worst iterative method for eigenvalues there is, but it is the theoretical basis for lots of better methods.)\n\nThis is similar to what you're doing when you measure the slope in the 2D case, except in that case you divide by $x$ instead of dividing by $\\sqrt{x^2+y^2}$. The point is that all that matters is the direction, not the magnitude.\n\n\u2022 that makes more sense. thank you! Dec 7, 2017 at 1:20\n\nThe method you are thinking of is called Power Iteration. More formally, take any vector $x_0\\ne 0$. (In practice, it is good to choose $x_0$ randomly. Then define\n\n$$x_k := \\frac{Ax_{k-1}}{\\|Ax_{k-1}\\|_2}, \\quad k\\in\\{1,2,\\ldots\\}$$\n\nWith high probability, $x_k$ will converge to an eigenvector of $A$ corresponding to the largest eigenvalue $\\lambda_1$ of $A$ with\n\n$$\\lambda_1 = \\lim_{k\\to\\infty} x_k^T Ax_k.$$\n\nIn practice, in computer arithmetic, this method is numerically stable. For a better method, see for example, this.", "date": "2022-10-01 20:51:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2554808/finding-approximation-of-largest-eigenvalue", "openwebmath_score": 0.8366015553474426, "openwebmath_perplexity": 214.46545664401265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513926334712, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8602665783990686}}
{"url": "https://math.stackexchange.com/questions/1627453/why-can-we-convert-a-base-9-number-to-a-base-3-number-by-simply-converting-e/1627475", "text": "# Why can we convert a base $9$ number to a base $3$ number by simply converting each base $9$ digit into two base $3$ digits?\n\nWhy can we convert a base $9$ number to a base $3$ number by simply converting each base $9$ digit into two base $3$ digits ?\n\nFor example $813_9$ can be converted directly to base $3$ by noting\n\n\\begin{array} \\space 8_9&=22_3 \\\\ \\space 1_9 &=01_3 \\\\ \\space 3_9 &=10_3 \\\\ \\end{array}\n\nPutting the base digits together ,we get $$813_9=220110_3$$\n\nI know it has to do with the fact that $9=3^2$ but I am not able to understand this all by this simple fact...\n\n\u2022 This is actually useful (but not with $9$ and $3$). We can very cheaply go back and forth between hexadecimal (base $16$) and binary. \u2013\u00a0Andr\u00e9 Nicolas Jan 26 '16 at 8:30\n\u2022 This also applies to converting between: binary, octal, hexadecimal systems \u2013\u00a0Max Payne Jan 26 '16 at 8:31\n\nConsider $N$ in base 3. For simplicity, we can assume that $N_3$ has an even number of digits: if it doesn't, just tack on a leftmost $0$. So let: $$N_3 = t_{2n+1} t_{2n}\\dotsc t_{2k+1} t_{2k} \\dotsc t_1 t_0.$$ What this positional notation really means is that: $$N = \\sum_{i = 0}^{2n+1} t_i 3^i,$$ which we can rewrite as: \\begin{align} N &= \\sum_{k = 0}^{n} (t_{2k+1} 3^{2k+1} + t_{2k} 3^{2k}) \\\\ &= \\sum_{k = 0}^{n} (3 t_{2k+1} + t_{2k}) 3^{2k} \\\\ &= \\sum_{k = 0}^{n} (3 t_{2k+1} + t_{2k}) 9^{k}. \\\\ \\end{align}\n\nBut now, note that for each $k$, $3 t_{2k+1} + t_{2k}$ is precisely the base-9 digit corresponding to the consecutive pair of base-3 digits $t_{2k+1} t_{2k}$.\n\n\u2022 Altough I've already accepted an answer,I thought to let you know that this answer has been so insightful.Thanks for your time ! \u2013\u00a0Mr. Y Jan 26 '16 at 9:18\n\u2022 How nice \u2014 thank you:) and you're welcome. \u2013\u00a0BrianO Jan 26 '16 at 9:24\n\u2022 That is the real good prove \u2013\u00a0JnxF Jan 26 '16 at 9:57\n\u2022 Oh my - thanks doubly. \u2013\u00a0BrianO Jan 26 '16 at 10:05\n\u2022 yes this is the proof. My answer is just a visual. +1 \u2013\u00a0Max Payne Jan 26 '16 at 10:16\n\nThis has more to do with place value.\n\nConsider the image:\n\nThe key is the fact that $3^2 = 9$\n\n\u2022 All the value below $3^2$ ($3^0$ and $3^1$) has to be accomodated in $9^0$\n\nthat is\n\n\u2022 all the values in $9^0$ have to be accommodated in the two places below $3^2$ ($3^0$ and $3^1$)\n\nSimilarly for any other places.\n\n\u2022 This rule also applies to conversion between binary and octal, or binary and hexadecimal, or conversion between any base $k$ and base $k^n$\n\nIn general, for conversion of a number $N$ in any base $k$ to base $k^n$, each digit of $N_{k^n}$ get converted to n digits of $N_k$\n\n\u2022 Thanks for the answer .One question : in general if we have that some number $c$ (I am considering everything in base $10$) such that $c=b^n$ does this mean that I can convert the number $c$ in base $b$ by simple converting each base $10$ digit of $c$ into $n$ base $b$ digits ?Or is this something that works only for these special cases ? \u2013\u00a0Mr. Y Jan 26 '16 at 8:56\n\u2022 @Mr.Y, yes, you can. An even stronger statement would be: given $a^n=b^m$, we could convert a base $a^n$ number, say, $k$, to base $b^m$, by converting every $n$ digits of $k$ (base $a^n$) to $m$ digits base $b^m$. \u2013\u00a0vrugtehagel Jan 26 '16 at 9:02\n\nI'm not sure whether you're asking for a proof, or some intuition. The proof is fairly mechanical, so I'll explain why you might expect this to be true.\n\nPutting your information theory hat on, two base 3 digits carry exactly as much information as one base 9 digit. That is, suppose you know I'm going to pass you two digits, each from 0,1 and 2. Then you know there are $3 \\cdot 3 =9$ possibilities for what information I could pass.\n\nOn the other hand, if I passed you a single digit from 0,1,...,8, then there are again 9 possibilities.\n\nContinuing in this fashion, $n$ digits of base 9 pass as much information as $2n$ digits of base 3.\n\nThis isn't exactly a proof yet. You'd need to prove that you don't skip any values. Perhaps doing the computation you describe can never return the value $102_3$. Thankfully it can. But you'd need to think about why this will always work.\n\nLet's look at what you base $9$ number actually means. $$813_9=8\\cdot 9^2+1\\cdot 9^1+3\\cdot 9^0$$ If we wish to write this as powers of $3$ with coefficients between $0$ and $2$, we can simply do \\begin{align} 3\\cdot 9^0&=1\\cdot 3^1+0\\cdot 3^0\\\\ 1\\cdot 9^1&=0\\cdot 3^3+1\\cdot 3^2\\\\ 8\\cdot 9^2&=2\\cdot 3^5+2\\cdot 3^4\\\\ \\end{align}\n\nWhy can we do this?\n\nWhy can we write $$a_n\\cdot 9^n=b_{2n+1}\\cdot 3^{2n+1}+b_{2n}\\cdot3^{2n}$$ First note that $9^n3^2n$, so when dividing the equation by that, we get $$\\frac{a_n\\cdot 9^n}{3^{2n}}=a_n=3b_{2n+1}+b_{2n}=\\frac{b_{2n+1}\\cdot 3^{2n+1}+b_{2n}\\cdot3^{2n}}{3^2n}$$ So actually, the problem comes down to writing a number $0\\leq a_n<9$ as $3b_{2n+1}+b_{2n}$, where $0\\leq b_{2n},b{2n+1}<3$. This is obviously possible.\n\nWhy does this work for binary and hexadecimal?\n\nActually, the answer is fairly similar. The problem can be reduced equivalently, resulting in the equation $$a_n=8a_{4n+3}+4a_{4n+2}+2a_{4n+1}+a_{4n}$$ Where $0\\leq a_n<16$ and $0\\leq a_{2n+3},a_{2n+2},a_{2n+1},a_{2n}<2$. The solvability of this equation is in my opinion a little less obvious, but still quite understandable; But, for the sake of a more thourough understanding, we could look at hexadecimal-octagonal conversion. This comes down to the easy equation $a_n=2b_{2n+1}+b_{2n}$, where $0\\leq a_n<16$ and $0\\leq b_{2n+1},b_{2n}<8$. This is cleary solvable. Doing this for octagonal-base 4 conversion and base 4-binary conversion shows with a recurrance-like approach that this indeed works for hexadecimal-binary conversion.\n\nHope this helped!\n\nHint:\n\n$$\\color{blue}1\\cdot3^5+\\color{blue}0\\cdot3^4+\\color{green}2\\cdot3^3+\\color{green}2\\cdot3^2+\\color{red}0\\cdot3^1+\\color{red}1\\cdot3^0 =(\\color{blue}1\\cdot3^1+\\color{blue}0)3^4+(\\color{green}2\\cdot3^1+\\color{green}2)3^2+(\\color{red}0\\cdot3^1+\\color{red}1)3^0\\\\ =\\color{blue}3\\cdot9^2+\\color{green}8\\cdot9^1+\\color{red}1\\cdot9^0$$\n\n$$\\color{blue}{10}\\color{green}{22}\\color{red}{01}_3=[\\color{blue}{10_3}|\\color{green}{22_3}|\\color{red}{01_3}]_9=\\color{blue}3\\color{green}8\\color{red}1_9$$", "date": "2020-01-17 16:18:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1627453/why-can-we-convert-a-base-9-number-to-a-base-3-number-by-simply-converting-e/1627475", "openwebmath_score": 0.987087607383728, "openwebmath_perplexity": 386.82093710414586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513897844354, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8602665742773444}}
{"url": "http://math.stackexchange.com/questions/38010/intuition-for-a-relationship-between-volume-and-surface-area-of-an-n-sphere/38017", "text": "# Intuition for a relationship between volume and surface area of an $n$-sphere\n\nThe volume of an $n$-sphere of radius $R$ is\n\n$$V_n(R) = \\frac{\\pi^{n/2} R^n}{\\Gamma(\\frac{n}{2}+1)}$$\n\nand the surface area is\n\n$$S_n(R) = \\frac{2\\pi^{n/2}R^{n-1}}{\\Gamma(\\frac{n}{2})} = \\frac{n\\pi^{n/2}R^{n-1}}{\\Gamma(\\frac{n}{2}+1)} = \\frac{d V_n(R)}{dR}$$\n\nWhat is the intuition for this relationship between the volume and surface area of an $n$-sphere? Does it relate to the fact that the $n$-sphere is the most compact shape in $n$ dimensions, or is that merely a coincidence?\n\nAre there other shapes for which it holds, or is this the limiting case of a relationship of inequality? For example, an $n$-cube of side $R$ has volume $V^c_n(R)=R^n$ and surface area $S^c_n(R)=2nR^{n-1}$, so that\n\n$$\\frac{dV^c_n(R)}{dR} = n R ^{n-1}$$\n\nand we have $S^c_n = 2 dV^c_n/dR$.\n\nEdit: As pointed out by Rahul Nahrain in the comments below, if we define $R$ to be the half-side length of the unit cube rather than the side length, then we have the relationship $S^c_n(R)=\\frac{d}{dR} V^c_n(R)$, exactly as for the sphere. Is there a sense in which relationships like this can be stated for a large class of shapes?\n\n-\nSee this MO-question and also the isoperimetric problem. \u2013\u00a0 t.b. May 9 '11 at 10:52\nThe main takeaway seems to be that this relationship is coincidental and based on using $R$ as the linear measure of distance. Were we to use $D=2R$ instead, we would get $S_n(D)=2\\frac{d}{dD}V_n(D)$. However, there is still the interesting (to my mind) question of why the ratio of $S_n(x)$ to $\\frac{d}{dx}V_n(x)$ (for some linear measure $x$) is independent of $n$ for the $n$-sphere but seemingly not for other shapes - no? \u2013\u00a0 Chris Taylor May 9 '11 at 11:04\nSure! I also voted the question up, by the way. I didn't mean to imply that these links answer the question completely. I just wanted you to see them because they certainly give some food for thought. \u2013\u00a0 t.b. May 9 '11 at 11:06\nIsn't the surface area of an $n$-cube $2nR^{n-1}$, not $6R^{n-1}$? Then $S_n^c = 2dV_n^c/dR$ for all $n$. \u2013\u00a0 Rahul May 9 '11 at 11:31\nIn fact, if you let the parameter be the \"radius\" of the cube, i.e. half of its side length, then $V_n^c = (2R)^n$ and $S_n^c = 2n(2R)^{n-1}$, and $S_n^c = dV_n^c/dR$ for all $n$, just like the sphere... \u2013\u00a0 Rahul May 9 '11 at 11:39\n\nThere is a very simple geometric explanation for the fact that the constant of proportionality is 1 for the sphere's radius and the cube's half-width. In fact, this relationship also lets you define a sensible notion of a \"half-width\" of an arbitrary $n$-dimensional shape.\n\nPick an arbitrary shape and a point $O$ inside it. Suppose you enlarge the shape by a factor $\\alpha \\ll 1$ keeping $O$ fixed. Each surface element with area $dA$ at a position $\\vec r$ relative to $O$ gets extruded into an approximate prism shape with base area $dA$ and offset $\\alpha \\vec r$. The corresponding additional volume is $\\alpha \\vec r\\cdot \\vec n dA$, where $\\vec n$ is the normal vector at the surface element.\n\nNow the quantity $\\vec r \\cdot \\vec n$, call it the projected distance, has a natural geometric interpretation. It is simply the distance between $O$ and the tangent plane at the surface element. (Observe that for a sphere with $O$ at the center, it is always equal to the radius, and for a cube with $O$ at the center, it is always equal to the distance from the center to any face.)\n\nLet $\\hat r = A^{-1} \\int \\vec r \\cdot \\vec n dA$ be the mean projected distance over the surface of the shape. Then the change in volume by a scaling of $\\alpha$ is simply $\\delta V = \\alpha \\int \\vec r\\cdot \\vec n dA = \\alpha \\hat r A$. In other words, a change of $\\alpha \\hat r$ in $\\hat r$ corresponds to a changed of $\\alpha \\hat r A$ in $V$. So if you use $\\hat r$ as the measure of the size of a shape, you find that $dV/d\\hat r = A$. And since $\\hat r$ equals the radius of a sphere and the half-width of a cube, the observation in question follows. This also implies the distance-to-face measure for regular polytopes that user9325 mentioned, but generalizes to other polytopes and curved shapes. (I'm not completely happy with the definition of $\\hat r$ because it's not obvious that it is independent of the choice of $O$. If someone can see a more natural definition, please let me know.)\n\n-\n\nSome remarks:\n\n1. It is not necessary to regard solids that generalize to $n$ dimensions, so one can start with shapes in 2 dimensions.\n\n2. It is very natural to always use a radius-type parameter to scale the figure because the intuition is that the figure gets growth rings that have the size of the surface.\n\n3. Unfortunately, the property is no longer true if you replace a square with a rectangle or a circle with an ellipse.\n\n4. For a regular polygon, you can always take as parameter the distance to an edge. This will work similarly for Platonic solids or any polygons/polyhedra that contain a point that is equidistant to all faces.\n\n5. This argument does not look good for general curves, because intuitively the edges of a smooth curve are infinitesimally small, so the center should be equidistant to all points, but of course, we could identify cases where the changing thickness of the growth ring averages out.\n\n-\n+1. Point 2 is the intuitive explanation, while points 3 and 5 are the essential caveats. \u2013\u00a0 Henry May 9 '11 at 12:32\n\nWhat you're seeing with spheres is an instance of something really important called the coarea formula. You're looking at a function like $f(x,y,z) = r = \\sqrt{x^2 + y^2 + z^2}$ and you're comparing the volumes of the lower contour sets $f(x,y,z) \\leq R$ (in this case they are balls) to the areas of the level sets $f(x,y,z) = R$.\n\nIn the instance with cubes, you're considering a different function whose level sets are cubes.\n\nThe coarea formula states that (subject to some assumptions that cause both sides to make sense),\n\n$Vol(f(x, y, z) \\leq R) = \\int_{-\\infty}^R \\left[ \\int_{f(x,y,z) = t} |\\nabla f|^{-1} d\\sigma \\right] dt$\n\nYou can also differentiate this formula with respect to $R$ to get the differential form you were observing.\n\nJust as a dummy check, note that $t$ has the same units as $f$, so if you think of $f$ and the coordinates as having units, the dimensional analysis checks out and both sides have units of \"length^3\". (The formula is also valid in higher dimensions.)\n\nIn the examples you gave, the gradient of $f$ has length $1$, so when you integrate $1$ over the level sets $\\{f(x, y, z) = t \\}$ you're just getting the surface area.\n\nThe proof of this formula boils down to asking \"what's the difference between $Vol( f(x,y,z) \\leq R )$ and $Vol(f(x,y,z) \\leq R + h$ if $h$ is small?\". The two regions are very similar, differing only around the boundary $f(x,y,z) = R$. This is particularly clear in the case of a sphere. If you think about a small portion of that boundary of \"width\" $d\\sigma$ in the directions of constant $f$, then the difference of the two regions has height $~ \\frac{h}{|\\nabla f|}$ in the direction of increasing $f$.\n\nAnother dummy check: The inverse makes sense because if $f$ is increasing extremely quickly, then increasing the value of $f$ will hardly increase the region $\\{ f(x,y,z) \\leq R \\}$.\n\nNote, the statement of the coarea is really a statement about the function and not just about the shapes of the level sets -- you are asking about the parameter $R$ by which these level sets are labeled, but you could pick a different function (like $\\tilde{f} = e^f$ which has the same level sets but labels them differently.\n\n-", "date": "2015-08-02 06:40:10", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/38010/intuition-for-a-relationship-between-volume-and-surface-area-of-an-n-sphere/38017", "openwebmath_score": 0.8902303576469421, "openwebmath_perplexity": 162.8361176625972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513881564148, "lm_q2_score": 0.8723473730188543, "lm_q1q2_score": 0.860266572857145}}
{"url": "https://math.stackexchange.com/questions/3158493/easiest-way-to-show-positive-semi-definite-equivalence", "text": "# Easiest way to show positive semi-definite equivalence\n\nFor an $$x \\in \\mathbb{R}^n$$, and $$n$$-by-$$n$$ identity matrix $$I_n$$, we are given that $$\\begin{pmatrix} I_n & x \\\\ x^T & 1 \\end{pmatrix} \\succeq 0.$$\n\nWhat is the easiest way to show that $$\\begin{pmatrix} 1 & x^T \\\\ x & I_n \\end{pmatrix} \\succeq 0$$ holds?\n\n\u2022 Symmetric permutation? \u2013\u00a0user251257 Mar 22 at 18:37\n\u2022 seems so. is there such a theory which concludes? \u2013\u00a0independentvariable Mar 22 at 18:40\n\u2022 Schur complement? \u2013\u00a0user251257 Mar 22 at 19:02\n\u2022 They dont reduce to the same condition, do they? \u2013\u00a0independentvariable Mar 22 at 19:22\n\nThis is due to the identity\n\n$$\\underbrace{\\begin{pmatrix} 0_{1,n} & 1 \\\\ I_n & 0_{n,1} \\end{pmatrix}}_{J^T} \\underbrace{\\begin{pmatrix} I_n & x \\\\ x^T & 1 \\end{pmatrix}}_{A} \\underbrace{\\begin{pmatrix} 0_{n,1} & I_n \\\\ 1 & 0_{1,n} \\end{pmatrix}}_{J} = \\underbrace{\\begin{pmatrix} 1 & x^T \\\\ x & I_n \\end{pmatrix}}_{B}$$\n\n(notation $$0_{m,n}$$ is for a zero block with $$m$$ lines and $$n$$ columns).\n\nIndeed, $$J$$ being a permutation matrix, it is an orthogonal matrix, with $$J^T=J^{-1}$$. We can conclude that $$A$$ and $$B$$ are similar, thus have the same spectrum (Similar matrices have the same eigenvalues with the same geometric multiplicity) with positive eigenvalues, thus are both semi-definite positive.\n\nBesides, $$A$$ being symmetric, one can conclude from $$J^TAJ=B$$ that $$B$$ is symmetric as well.\n\nAppendix : There is a pending question : is there a criteria on $$x$$ for positive semi-definiteness of $$A$$. ? The answer is yes :\n\n$$A$$ is semi-definite positive iff $$\\|x\\| \\leq 1$$.\n\nThis is due, as we are going to see it, to an analysis of the rather particular spectrum of $$A$$. Let us obtain it explicitly.\n\nFirst of all, let us establish that $$A$$ (which is a $$(n+1) \\times (n+1)$$ matrix) has eigenvalue $$1$$ with order of multiplicity at least $$n-1$$.\n\nConsider hyperplane $$x^{\\perp}$$ of $$\\mathbb{R}^n$$ defined as the set of vectors $$y$$ that are orthogonal to $$x$$. Let $$(y_1,y_2,\\cdots y_{n-1})$$ be a basis of $$x^{\\perp}$$ ; then,\n\n$$\\underbrace{\\begin{pmatrix} I_n & x \\\\ x^T & 1 \\end{pmatrix}}_{A}\\underbrace{\\begin{pmatrix} y_k\\\\ 0 \\end{pmatrix}}_{V_k}=1\\underbrace{\\begin{pmatrix} y_k\\\\ 0 \\end{pmatrix}}_{V_k} \\ \\ \\ \\text{for} \\ \\ k=1,2, \\cdots (n-1),$$\n\nproving that $$V_k$$ is an eigenvector associated with eigenvalue $$1$$.\n\nDue to the fact that trace$$(A)=n+1$$, the two remaining eigenvalues are of the form $$\\alpha$$ and $$\\beta:=2-\\alpha$$. We can assume, WLOG that $$\\alpha \\leq 1 \\leq \\beta$$.\n\nBesides, using the so-called Schur determinant identity (Eigenvalues of a Block Matrix from Schur Determinant Identity) for the computation of the determinant of a $$2 \\times 2$$ block matrix, we obtain :\n\n$$\\det(A)=1-x^Tx$$\n\nAs the determinant is also the product of eigenvalues, we get the following identity :\n\n$$\\det(A)=1-\\|x\\|^2=\\alpha(2-\\alpha)\\tag{1}$$\n\nThus, one can compute explicitly the two remaining eigenvalues by solving quadratic equation (1), with the following explicit solutions (if we assume that $$\\alpha$$ is the smallest eigenvalue)\n\n$$\\alpha=1 - \\|x\\| \\ \\ \\ \\implies \\ \\ \\ \\beta:=2-\\alpha=1 + \\|x\\|\\tag{2}$$\n\nAs the criteria for a symmetric matrix do be semi-definite positive is that must have all eigenvalues $$\\geq 0$$, this criteria becomes $$\\alpha \\geq 0$$, i.e., $$\\|x\\| \\leq 0$$. $$\\square$$\n\nRemark : eigenvalues $$\\alpha$$ and $$\\beta$$ can be associated with eigenvectors $$\\begin{pmatrix} x\\\\ -\\|x\\| \\end{pmatrix}$$ and $$\\begin{pmatrix} x\\\\ \\|x\\| \\end{pmatrix}$$ resp.\n\nLet us take an example in the case $$n=4$$ ; let $$m=1/n$$ ; consider matrix :\n\n$$A:=\\left(\\begin{array}{rrrr|r} 1 & & & & m \\\\ & 1 & & & m \\\\ & & 1 & & m\\\\ & & & 1 & m\\\\ \\hline m & m & m & m & 1 \\\\ \\end{array}\\right)$$\n\nOne can check, using (2), that the spectrum of $$A$$ is\n\n$$(\\tfrac12, 1 , 1, 1, 1, \\tfrac32).$$\n\nJust now, I \"googled\" with keywords \"bordered identity matrix\" : I found in (Eigenvalues of a certain bordered identity matrix) a somewhat similar computation that I did in the Appendix.\n\n\u2022 Thank you for your answer! My main purpose was showing $||x||_2 \\leq 1$ holds iff the matrices I gave are Psd. I proved one by Schur complement theory on PSD matrics, but for the equivalence I just followed the definition of PSD and by contradiction showed that if one holds, the other one should hold etc.. \u2013\u00a0independentvariable Mar 23 at 19:28\n\u2022 But yours seem the better way, not the 'dirty' $a^T X a \\geq 0$ for all $a$ approach... I don't like it, it seems too manual \u2013\u00a0independentvariable Mar 23 at 19:35", "date": "2019-04-23 16:30:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3158493/easiest-way-to-show-positive-semi-definite-equivalence", "openwebmath_score": 0.9334749579429626, "openwebmath_perplexity": 230.35230850436702, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513918194609, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8602665711434675}}
{"url": "https://math.stackexchange.com/questions/1687015/is-the-cartesian-product-of-two-countably-infinite-sets-also-countably-infinite", "text": "Is the Cartesian product of two countably infinite sets also countably infinite?\n\nI am trying to determine and prove whether the set of convergent sequences of prime numbers is countably or uncountably infinite.\n\nIt is clear that such a sequence must 'terminate' with an infinite repetition of some prime $p$. So for example $${1,2,3,5,5,5,5,...}$$\n\nMy idea is to break up the problem into two sub-sequences. The first is a finite subsequence consisting of the first however many terms before it begins the infinite repetition of $p$. Then the second being the infinite set with elements being only $p$.\n\nSo in the above example, the two sub-sequences will be $$\\{1,2,3\\} \\{5,5,5,...\\}$$\n\nThe infinite sub-sequence is countably infinite as it's cardinality is simply the cardinality of the primes, which is countably infinite.\n\nThere are infinitely many finite sub-sequences as it can have any number of terms. However, I think it will be countably infinite because it can have $1$ term, $2$ terms, $3$ terms etc, in other words we can construct the bijection to $\\mathbb{N}$ by simply naming them according to the number of terms it has. The arrangements is irrelevant as it will be finite.\n\nThis is where the title of the question comes in. There are a countably infinite number of these finite sub-sequences say $F$. We also have a countably infinite number of the infinite subsequence $P$, with all elements being some prime $p$.\n\nSo in some sense, the set of all convergent sequences of primes is the cartesian product $F\\times P$.\n\nAs they are both countably infinite, will their cartesian product also be countably infinite? I am thinking yes, as the rationals are countably infinite and they are in some sense a cartesian product of the numerator and denominator, each having cardinality equivalent to the naturals.\n\nIf so, and everything is correct, I think I will have then completed the proof. Otherwise, I'd love to have errors pointed out or even perhaps a better solution.\n\nYour thinking is correct. In fact, what you have argued is that there exists a bijection between the set of all convergent sequences of primes and the set $F\\times P$. If this is your first example of doing this kind of task I suggest you try to actually write down this bijection. You have argued quite well that it must exist, but it's good exercise to actually do it at least once.\n\nNoy, you also rightly say that $F\\times P$ has a bijection to $\\mathbb N^2$. As before, I suggest you actually write down this bijection.\n\nNow, you simply need to see if $\\mathbb N^2$ is countable, that is:\n\nDoes there exist a bijection between $\\mathbb N$ and $\\mathbb N^2$\n\nTo that end, I suggest you look into Cantor's pairing function.\n\n\u2022 or you might just observe that $(i,j)\\mapsto 2^i3^j$ is an injection. This idea extends easily to $\\mathbb N^k$. For example, $(i,j,k)\\mapsto 2^i3^j5^k$. Mar 7, 2016 at 16:06\n\u2022 @Chilango Of course. Though I think that for a \"first time user\", the actual bijection is more informative than just the construction of an injection (which, I admit, implies a bijection, but also a theorem)\n\u2013\u00a05xum\nMar 7, 2016 at 16:32\n\u2022 these maps are not surjective, but that's no problem. An injection into a countable set is enough to say that the domain is countable. All you need is the fact that a subset of a countable set is countable. I agree though, that the pairing function is more interesting and has more applications. Mar 7, 2016 at 16:50", "date": "2022-05-26 23:57:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1687015/is-the-cartesian-product-of-two-countably-infinite-sets-also-countably-infinite", "openwebmath_score": 0.943945050239563, "openwebmath_perplexity": 105.60574821025583, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9921841109796002, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8602591011179391}}
{"url": "https://math.stackexchange.com/questions/2373028/x5-y2-z3", "text": "# $x^5 + y^2 = z^3$\n\nWhile waiting for my d\u00f6ner at lunch the other day, I noticed my order number was $343 = 7^3$ (surely not the total for that day), which reminded me of how $3^5 = 243$, so that $$7^3 = 3^5 + 100 = 3^5 + 10^2.$$ Naturally, I started wondering about nontrivial integer solutions to $$x^5 + y^2 = z^3 \\tag{*}$$ (\"nontrivial\" meaning $xyz \\ne 0$). I did not make much progress, though apparently there are infinitely many solutions: this was Problem 1 on the 1991 Canadian Mathematical Olympiad. The official solutions (at the bottom of this page) only go back to 1994. A cheap answer is given by taking $x = 2^{2k}$ and $y = 2^{5k}$ so that the l.h.s. is $2^{10k + 1}$. This is a cube iff $10k + 1 \\equiv 0 \\,(3)$ i.e. $k \\equiv 2\\,(3)$ thus giving an arithmetic progression's worth of solutions, starting with $$(x, y, z) = (16, 1024, 128)$$ corresponding to $k = 2$ and $$(x, y, z) = (1024, 33554432, 131072)$$ coming from $k = 5$.\n\nWhat else is known about the equation $(*)$? In particular, are there infinitely many solutions with $x$, $y$, $z$ relatively prime? The one that caught my attention was $(x, y, z) = (3, 10, 7)$. Another one is $(-1, 3, 2)$ because $-1 + 9 = 8$. By Catalan's conjecture (now a theorem), this is the only solution with $x = \\pm 1$ or $y = \\pm 1$ or $z = 1$. Are there any solutions with $z = -1$? In this case, $(*)$ reduces to $x^5 + y^2 = -1$ and Mih\u0103ilescu's theorem does not apply.\n\nUpdate. This question was essentially already asked here, since the equation $a^2 + b^3 = c^5$ is equivalent to $(-c)^5 + a^2 = (-b)^3$.\n\n\u2022 we can rewrite this as $x^5+y^2\\equiv 0 \\pmod z$ could that help you ?\n\u2013\u00a0user451844\nJul 27 '17 at 1:09\n\u2022 You might look at OEIS sequences A070065, A070066 and A070067. Jul 27 '17 at 2:08\n\u2022 Note that if $(x,y,z)$ is a solution, then so is $(c^6 x, c^{15} y, c^{10} z)$ for integers $c$. Jul 27 '17 at 2:41\n\u2022 quora.com/\u2026 Aug 3 '17 at 11:45\n\nYes, there are infinitely many solutions. In fact, there are many parametrizations of the solutions.\nAccording to a book${}^{\\color{blue}{[1]}}$ on my bookshelf,\n\nUp to changing $y$ into $-y$, there are exactly 27 distinct parametrizations of the equations $x^5 + y^2 = z^3$.\n\nOne of the simplest paremetrization is given by following formula.\n\n\\begin{align} x =&\\; 12st(81s^{10}-1584t^5s^5-256t^{10})\\\\ y =&\\; \\pm (81s^{10} + 256t^{10})\\\\ &\\;\\;\\times (6561s^{20} - 6088608t^5s^{15} - 207484416t^{10}s^{10} + 19243008t^{15}s^5 + 65536t^{20})\\\\ z =&\\; 6561s^{20}+2659392t^{5}s^{15}+10243584t^{10}s^{10} - 8404992t^{15}s^5 + 65536t^{20} \\end{align}\n\nFor example, following two random choices of $s,t$ give you two sets of relative prime solutions.\n\n\u2022 $(s,t) = (1,1) \\leadsto (x,y,z) = (-21108,-65464918703,4570081)$\n\u2022 $(s,t) = (1,2) \\leadsto (x,y,z) = (-7506024,127602747389962225,-196120763999)$\n\nThe book I have is actually quoting result from a thesis${}^{\\color{blue}{[2]}}$ by J. Edwards. Consult that if you really want to get into the details.\n\nReferences\n\n\u2022 $\\color{blue}{[1]}$ Henri Cohen, Number Theory Number Theory Volume II: Analytic and Modern Tools,\n$\\S 14.5.2$ The Icosahedron Case $(2,3,5)$.\n\n\u2022 $\\color{blue}{[2]}$ J. Edwards, Platonic solids and solutions to $x^2+y^3 = dz^r$, Thesis, Univ. Utrecht (2005).\n\n\u2022 Nice answer, especially for the first reference. +1\n\u2013\u00a0Xam\nJul 27 '17 at 3:38\n\u2022 @achille: We can also use the icosahedral equation and scale it appropriately. Kindly see answer below. Jul 27 '17 at 12:00\n\u2022 @achille: The parameterizations for $x^5+y^3=z^2$ depend on the icosahedron while $x^4+y^3=z^2$ depend on the tetrahedron. Does the book you cite say how many distinct parameterizations are there for the latter? This page gives $7$, but I just found an $8$th one, with no scaling involved. Jul 28 '17 at 5:53\n\u2022 @TitoPiezasIII In $\\S 14.4.1$ of that book, Cohen proved there are 7 parametrizations. Jul 28 '17 at 7:13\n\u2022 @achillehui: I just re-checked the parameterization I found, and it turns out there is a common factor. Darn! In Edwards' work, he cites 7 parameterizations as well. Thanks, anyway. Jul 28 '17 at 7:35\n\nThere is a beautiful connection between $a^5+b^3=c^2$ and the icosahedron. Consider the unscaled icosahedral equation,\n\n$$\\color{blue}{12^3u v(u^2 + 11 u v - v^2)^5}+(u^4 - 228 u^3 v + 494 u^2 v^2 + 228 u v^3 + v^4)^3 = (u^6 + 522 u^5 v - 10005 u^4 v^2 - 10005 u^2 v^4 - 522 u v^5 + v^6)^2\\tag1$$\n\nBy scaling $u=12x^5$ and $v=12y^5$ (or various combinations thereof like $u=12^2x^5$, etc), we then get a relation of form,\n\n$$12^5a^5+b^3=c^2$$\n\n\u2022 +1 interesting connection. it is unfortunate the mythical $j(\\tau)$ is stuff way beyond me ;-p. Jul 27 '17 at 13:14", "date": "2021-09-22 08:33:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2373028/x5-y2-z3", "openwebmath_score": 0.9063588380813599, "openwebmath_perplexity": 485.6667700955476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692366242304, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.860247411297536}}
{"url": "https://math.stackexchange.com/questions/2371406/checking-for-linear-independence", "text": "# Checking for linear independence\n\nIs the following proof correct?\n\nTheorem. If $v_1,v_2,v_3,v_4$ is a linearly independent list then $$v_1-v_2,v_2-v_3,v_3-v_4,v_4$$ is also a linearly independent list.\n\nProof. Assume that $v_1,v_@,v_3,v_4$ is a linearly independent list, Consider now the following equation. $$0=0(v_1-v_2)+0(v_2-v_3)+0(v_3-v_4)+0v_4\\tag{1}$$ Let $a_1,a_2,a_3$ and $a_4$ be arbitrary scalars in $\\mathbf{F}$ and assume that the following equation holds $$0=a_1(v_1-v_2)+a_2(v_2-v_3)+a_3(v_3-v_4)+a_4v_4\\tag{2}$$ After some algebraic manipulation we arrive at the following equation. $$0=a_1v_1+(a_2-a_1)v_2+(a_3-a_2)v_3+(a_4-a_3)v_4\\tag{3}$$ Since the list $v_1,v_2,v_3,v_4$ is linearly independent it follows that given any vector in $span(v_1,v_2,v_3,v_4)$ the choice of scalars is unique and since $$0=0v_1+0v_2+0v_3+0v_4\\tag{4}$$ It follows that all the scalars in $(3)$ must be $0$, consequently the only way to produce the $0$ vector as a linear combination of the vectors in the list $v_1-v_2,v_2-v_3,v_3-v_4,v_4$ is that indicated in $(1)$.\n\n$\\blacksquare$\n\nHere $\\mathbf{F}$ is either $\\mathbb{C}$ or $\\mathbb{R}$.\n\n\u2022 Sounds ok - would just like to point out 'Consider now the following equation.' sounds a bit strange for $(1)$. Perhaps remove that line and $(1)$ and say something along the lines 'by inspection, $(2)$ holds when all the scalars are equal to $0$' at the end of your proof \u2013\u00a0Shuri2060 Jul 25 '17 at 17:31\n\u2022 Nitpicking, but you made a small typo just before your first labeled equation, where you write \"Assume $v_1,v_{@},v_3,v_4$ is a linearly independent list.\" Did you mean $v_2$ not $v_{@}$ ? \u2013\u00a0Vivek Kaushik Jul 25 '17 at 17:37\n\nEquation (1) should be omitted: it's an obvious fact that has no consequence on the rest.\n\nThe final argument is too fast: from linear independence of $v_1,v_2,v_3,v_4$ you deduce \\begin{cases} a_1=0 \\\\ a_2-a_1=0 \\\\ a_3-a_2=0 \\\\ a_4-a_3=0 \\end{cases} and, from this, $a_1=a_2=a_3=a_4=0$. This should be mentioned, although easy.\n\nA different approach is to consider the coordinates of the new vectors with respect to the original ones and so the matrix \\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ -1 & 1 & 0 & 0 \\\\ 0 & -1 & 1 & 0 \\\\ 0 & 0 & -1 & 1 \\end{bmatrix} A standard Gaussian elimination leads to the reduced row echelon form \\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{bmatrix} which proves that the coordinate vectors are linearly independent and so also the vectors are.\n\n\u2022 So your point is that i should show how each of the scalars $a_1,a_2,a_3$ and $a_4$ is 0. \u2013\u00a0Atif Farooq Jul 25 '17 at 17:59\n\u2022 The original matrix is lower triangular with $1$'s in the diagonal, hence invertible. \u2013\u00a0lhf Jul 25 '17 at 18:49\n\u2022 @AtifFarooq Yes, that's the point; notwithstanding it's easy, it should appear in the proof. \u2013\u00a0egreg Jul 25 '17 at 19:27\n\u2022 @lhf That's true, but Gaussian elimination would give more information in case the vectors aren't linearly independent. \u2013\u00a0egreg Jul 25 '17 at 19:27\n\nIt's valid. A note:\n\nBefore $(2)$ when you say \"...and assume that the following equation holds...\" it would be better to perhaps phrase this as \"We wish to solve ... for $a_j$.\" Saying the former means that you're assuming solutions exist - but you're not sure! This is a tiny point, and is perhaps reflected in my writing style.\n\nWe may also streamline a bit by saying the following after $(3)$.\n\nAs $\\{v_1, v_2, v_3, v_4\\}$ is a linearly independent set, we must have that $a_1 = 0, \\ a_2-a_1=0, \\ a_3 - a_2 = 0,$ and $a_4-a_3=0.$ Back substitution yields that each $a_j = 0$, and so the original set is also a linearly independent set. $\\square$\n\nAgain, these are just small suggestions, but nonetheless your proof is correct.", "date": "2019-06-16 15:10:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2371406/checking-for-linear-independence", "openwebmath_score": 0.9979005455970764, "openwebmath_perplexity": 154.00636836755788, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692271169856, "lm_q2_score": 0.8807970842359877, "lm_q1q2_score": 0.8602474075076565}}
{"url": "https://math.stackexchange.com/questions/546007/int-02-pi-sqrt1-cosx-dx-4-sqrt2-why", "text": "$\\int_0^{2\\pi} \\sqrt{1-\\cos(x)}\\,dx = 4\\sqrt{2}$. Why?\n\nAccording to the textbook, and Wolfram Alpha the above is correct.\n\nHere is the step by step procedure from Wolfram Alpha for evaluating the indefinite integral:\n\nTake the integral: $$\\int\\sqrt{1-\\cos(x)}\\,dx$$ For the integrand $\\sqrt{1-\\cos(x)}$, substitute $u=1-\\cos(x)$ and $du=\\sin(x)\\,dx$: $$=\\int-\\frac{1}{\\sqrt{2-u}}\\,du$$ Factor out constants: $$=-\\int\\frac{1}{\\sqrt{2-u}}\\,du$$ For the integrand $1/\\sqrt{2-u}$, substitute $s=2-u$ and $ds=-du$: $$=\\int\\frac{1}{\\sqrt{s}}\\,ds$$ The integral of $1/\\sqrt{s}$ is $2\\sqrt{s}$: $$=2\\sqrt{s}+\\text{constant}$$ Substitute back for $s=2-u$: $$=2\\sqrt{2-u}+\\text{constant}$$ Substitute back for $u=1-\\cos(x)$: $$=2\\sqrt{\\cos(x)+1}+\\text{constant}$$ Which is equivalent for restricted $x$ values to: $$\\boxed{=-2\\sqrt{1-\\cos(x)}\\cot\\big(\\frac{x}{2}\\big)+\\text{constant}}$$\n\nI understand up to the below (which is a valid solution to the integral): $$2\\sqrt{\\cos(x)+1}+\\text{constant}$$\n\nHowever, if you evaluate this at $2\\pi$ and $0$, you get the same thing, so the definite integral evaluates to zero.\n\nAfter, you transform the above to: $$-2\\sqrt{1-\\cos(x)}\\cot\\big(\\frac{x}{2}\\big)+\\text{constant}$$\n\nThe expression is indeterminate at $2\\pi$ and $0$ of the form $0 \\times \\infty$. So I guess you would set up a limit and then use L'Hospital's rule to evaluate the expression at $2\\pi$ and $0$ and get the answer to the definite integral?\n\nIn any case, all this seems strange. Why should the definite integral evaluated one way give $0$, and in another way give something else?\n\n\u2022 You may want to find the anti derivative by multiplying top and bottom by the squareroot of the conjugate. You will end up with the absolute value of the sine term on top and the square root of 1+cosx in the bottom. This integrates easily. Mind the absolute value though as you apply your upper and lower limit. Oct 30 '13 at 19:57\n\u2022 Ahh ok that makes sense. So to get rid of the absolute value sign, you can split up the integral from $0$ to $\\pi$ and $\\pi$ to $2\\pi$ Oct 30 '13 at 20:02\n\u2022 Yes, you ought to split that integral. Oct 30 '13 at 20:05\n\n$$\\int_0^{2\\pi}\\sqrt{1-\\cos x}\\,dx=\\int_0^{2\\pi}\\sqrt{2\\sin^2\\frac{x}{2}}\\,dx =\\sqrt{2}\\int_0^{2\\pi}\\sin\\frac{x}{2}\\,dx=-2\\sqrt{2}\\cos\\frac{x}{2}\\Big|_0^{2\\pi}=4\\sqrt{2}.$$\n\u2022 @RonGordon $x/2 \\in [0,\\pi] \\Rightarrow \\sin(x/2) \\ge 0$. Oct 30 '13 at 20:49\n\u2022 Good old $\\cos(2x) = \\cos^2(x)-\\sin^2(x)= 1-2\\sin^2(x) = 2\\cos^2(x)-1$. Oct 31 '13 at 1:25\nYour $u$-substitutions should be injective on their interval of evaluation. Otherwise, you risk running into exactly this sort of issue.\nNote that \\begin{align}|\\sin x| &= \\sqrt{\\sin^2 x}\\\\ &= \\sqrt{1-\\cos^2 x}\\\\ &= \\sqrt{1-\\cos x}\\sqrt{1+\\cos x}\\\\ &=\\sqrt{1-\\cos x}\\sqrt{2-(1-\\cos x)},\\end{align} so if you want to use $u=1-\\cos x$, then $$\\frac{du}{dx}=\\sin x=\\begin{cases}|\\sin x|=\\sqrt{1-\\cos x}\\sqrt{2-(1-\\cos x)} & 0\\le x\\le \\pi\\\\-|\\sin x|=-\\sqrt{1-\\cos x}\\sqrt{2-(1-\\cos x)} & \\pi\\le x\\le2\\pi,\\end{cases}$$ so \\begin{align}\\int_0^{2\\pi}\\sqrt{1-\\cos x}\\,dx &= \\int_0^\\pi\\sqrt{1-\\cos x}\\,dx+\\int_\\pi^{2\\pi}\\sqrt{1-\\cos x}\\,dx\\\\ &= \\int_0^\\pi\\frac{|\\sin x|}{\\sqrt{2-(1-\\cos x)}}\\,dx+\\int_\\pi^{2\\pi}\\frac{|\\sin x|}{\\sqrt{2-(1-\\cos x)}}\\,dx\\\\ &= \\int_0^\\pi\\frac{\\sin x\\,dx}{\\sqrt{2-(1-\\cos x)}}-\\int_\\pi^{2\\pi}\\frac{\\sin x\\,dx}{\\sqrt{2-(1-\\cos x)}}\\\\ &= \\int_0^2\\frac{du}{\\sqrt{2-u}}-\\int_2^0\\frac{du}{\\sqrt{2-u}}\\\\ &= 2\\int_0^2\\frac{du}{\\sqrt{2-u}}.\\end{align} At that point, we can use that antiderivative, with no need to resubstitute.\nAlternately, you could note that $\\cos(2\\pi-x)=\\cos x$, so \\begin{align}\\int_0^{2\\pi}\\sqrt{1-\\cos x}\\,dx &= \\int_0^\\pi\\sqrt{1-\\cos x}\\,dx+\\int_\\pi^{2\\pi}\\sqrt{1-\\cos x}\\,dx\\\\ &= \\int_0^\\pi\\sqrt{1-\\cos x}\\,dx+\\int_\\pi^{2\\pi}\\sqrt{1-\\cos(2\\pi-x)}\\,dx\\\\ &= \\int_0^\\pi\\sqrt{1-\\cos x}\\,dx-\\int_{2\\pi}^\\pi\\sqrt{1-\\cos(2\\pi-x)}\\,dx\\\\ &= \\int_0^\\pi\\sqrt{1-\\cos x}\\,dx-\\int_0^\\pi\\sqrt{1-\\cos x}\\frac{d(2\\pi-x)}{dx}\\,dx\\\\ &= 2\\int_0^\\pi\\sqrt{1-\\cos x}\\,dx,\\end{align} at which point you can use your $u$-substitution without fear, since the cosine function is injective on $[0,\\pi]$.", "date": "2021-09-27 17:18:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/546007/int-02-pi-sqrt1-cosx-dx-4-sqrt2-why", "openwebmath_score": 0.999889612197876, "openwebmath_perplexity": 403.9850245364357, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.976669234586964, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8602474033910186}}
{"url": "https://math.stackexchange.com/questions/1619416/square-divided-by-absolute-value", "text": "# Square divided by absolute value\n\nFirst time posting on Math SE, with kind of a basic algebra question.\n\nQuestion\nDoes the relation:\n\n$$\\dfrac{(ab)^2}{|ab|} = \\left|ab\\right|$$\n\nwith $a,b \\in \\mathbb{R_{\\ne 0}}$ always hold?\n\nIt seems trivial to me, but Wolfram Alpha gives me a strange answer because it specifies that this is True assuming $a,b$ are positive.\n\nReasoning\nNo matter what sign $a,b$ have, we have that $(ab)^2 > 0$ and $\\left|ab\\right| > 0$. Thus their ratio is greater than zero, and the magnitude of that ratio is exactly $ab$ with a positive sign, so $\\left|ab\\right|$.\n\nIs what I said correct? If so, is this question a completely useless one? Sorry for the occasionally bad English!\n\nEdit: formatted equations as suggested by Frentos\n\n\u2022 Hmmn, wlog for $a, b \\in \\mathbb{C}$ one could argue that $a^{2}b^{2} < 0$ \u2013\u00a0Kevin Jan 20 '16 at 11:28\n\u2022 Welcome to math.SE! See this guide for how to write equations on this site. \\dfrac makes larger, easier to read fractions and \\left| \\right| gives nicer absolute values. \u2013\u00a0Frentos Jan 20 '16 at 11:32\n\u2022 @Bacon No, try $a=b=i$ (or any case when $a^2b^2$ is not even real). \u2013\u00a0Did Jan 21 '16 at 1:45\n\u2022 @Did - fair point, my comment meant to reflect that in some cases this could be true \u2013\u00a0Kevin Jan 21 '16 at 9:20\n\nYour statement about Wolfram is not quite correct. It gives various alternate forms for this expression, two of which are:\n\n1. $ab$ assuming $a$ and $b$ are positive\n\n2. $ab\\,sgn(a)\\,sgn(b)$\n\n(2) is equivalent to $|ab|$\n\nSee here\n\n\u2022 Indeed, I overlooked the $ab \\; sgn(a) \\; sgn(b)$ answer! I'll accept this one because it points out that Wolfram was right. \u2013\u00a0UJIN Jan 20 '16 at 14:30\n\nFor real numbers this is always true because the square of a real number equals the square of its absolute value, and in particular $(ab)^2=|ab|^2.$ Perhaps Wolfram has reservations because it considers the possibility of complex numbers?\n\nNotice:\n\n\u2022 If $a,b\\in\\mathbb{R^+}$, so $a,b>0$, then:\n\n$$|ab|=|a||b|=ab$$\n\n\u2022 If $a,b\\in\\mathbb{R^+}$, so $a,b>0$, so:\n\n$$(ab)^2=a^2b^2$$\n\n\u2022 If $a,b\\in\\mathbb{R^-}$, than $a,b<0$, so:\n\n$$((-a)(-b))^2=(ab)^2=a^2b^2$$", "date": "2020-12-01 06:38:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1619416/square-divided-by-absolute-value", "openwebmath_score": 0.9206000566482544, "openwebmath_perplexity": 646.0958758089511, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692284751636, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8602473980077626}}
{"url": "https://gmatclub.com/forum/a-wire-that-weighs-20-pounds-is-cut-into-two-pieces-so-that-one-of-the-243714.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 21 Apr 2019, 01:23\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# A wire that weighs 20 pounds is cut into two pieces so that one of the\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 54376\nA wire that weighs 20 pounds is cut into two pieces so that one of the\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2017, 03:28\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n65% (02:06) correct 35% (01:59) wrong based on 63 sessions\n\n### HideShow timer Statistics\n\nA wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?\n\n(A) 9\n(B) 12\n(C) 18\n(D) 24\n(E) 27\n\n_________________\nSenior Manager\nJoined: 28 May 2017\nPosts: 283\nConcentration: Finance, General Management\nRe: A wire that weighs 20 pounds is cut into two pieces so that one of the\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2017, 03:57\nBunuel wrote:\nA wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?\n\n(A) 9\n(B) 12\n(C) 18\n(D) 24\n(E) 27\n\nWeight of 2nd piece = 4 pound\n\nSince the weight is directly proportional to the square of its length., we may write\n$$\\frac{16}{36^2}$$ = $$\\frac{4}{x^2}$$\n\nSolving above, we get x = 18\n\n_________________\nIf you like the post, show appreciation by pressing Kudos button\nDirector\nJoined: 04 Dec 2015\nPosts: 750\nLocation: India\nConcentration: Technology, Strategy\nWE: Information Technology (Consulting)\nRe: A wire that weighs 20 pounds is cut into two pieces so that one of the\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2017, 04:02\nBunuel wrote:\nA wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?\n\n(A) 9\n(B) 12\n(C) 18\n(D) 24\n(E) 27\n\n$$20$$ pounds wire is cut into two pieces = $$16$$ pounds $$+$$ $$4$$ pounds\n\nThe piece weighs $$16$$ pounds is $$36$$ feet long.\n\nRatio of weight and length of $$16$$ pounds piece $$= \\frac{16}{36^2} = \\frac{4 * 4}{36 * 36} = \\frac{1}{81}$$\n\nTherefore required ratio of other part would also be $$\\frac{1}{81}$$\n\nRatio $$= \\frac{4}{x^2} =$$ $$\\frac{1}{81}$$\n\n$$x^2 = 81*4$$ $$=> x = \\sqrt{81*4}$$\n\n$$x = 9 * 2 = 18$$\n\nHence length of $$4$$ pounds wire $$= 18$$\n\nIntern\nJoined: 30 May 2013\nPosts: 29\nGMAT 1: 600 Q50 V21\nGMAT 2: 640 Q49 V29\nRe: A wire that weighs 20 pounds is cut into two pieces so that one of the\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2017, 04:07\nIMO (c)\nGiven, weight proportional to square (length)\n=> w1 / w2 = square (l1/l2)\n=> 16/4 = square (36/l2)\n=> l2 = 18\n\nSent from my GT-N7100 using GMAT Club Forum mobile app\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 5807\nLocation: United States (CA)\nRe: A wire that weighs 20 pounds is cut into two pieces so that one of the\u00a0 [#permalink]\n\n### Show Tags\n\n04 Jul 2017, 07:52\n1\nBunuel wrote:\nA wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?\n\n(A) 9\n(B) 12\n(C) 18\n(D) 24\n(E) 27\n\nSince the 20-lb wire is cut into two pieces and one of the pieces weighs 16 lbs, the other piece must weigh 4 lbs. Since the 16-lb piece has length of 36 ft and the weight of each piece is directly proportional to the square of its length, we can let x = the length in feet of the 4-lb piece and create the following proportion:\n\n16/36^2 = 4/x^2\n\n16x^2 = 4 * 36^2\n\n4x^2 = 36^2\n\nx^2 = 36^2/2^2\n\nx^2 = 18^2\n\nx = 18\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nRe: A wire that weighs 20 pounds is cut into two pieces so that one of the \u00a0 [#permalink] 04 Jul 2017, 07:52\nDisplay posts from previous: Sort by\n\n# A wire that weighs 20 pounds is cut into two pieces so that one of the\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2019-04-21 08:23:57", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-wire-that-weighs-20-pounds-is-cut-into-two-pieces-so-that-one-of-the-243714.html", "openwebmath_score": 0.5843057632446289, "openwebmath_perplexity": 2340.8452587349466, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9766692271169855, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8602473937554338}}
{"url": "https://matheducators.stackexchange.com/questions/4458/explanation-of-counter-example/4460", "text": "Explanation of Counter-example?\n\nI have a question where we are given a statement and asked to prove whether or not the statement is true. If the statement is true, then we must prove it; otherwise we must provide a counterexample to prove the statement incorrect.\n\nIf the statement is false, and the student provides a counter-example; should the student also provide an explanation of why the counter-example disproves the statement?\n\nI believe this depends on how clearly the counterexample is stated.\n\nConsider this claim: $f:\\mathbb{R}\\to\\mathbb{R}$ is continuous $\\implies f$ is differentiable.\n\nImagine a student, let's call him Karl, who says\n\nTake any function of the form $\\sum_{n=0} ^\\infty a^n \\cos(b^n \\pi x)$, where $0<a<1$, $b$ is a positive odd integer, and $ab > 1+\\frac{3}{2} \\pi$.\n\nWould you give him credit for this counterexample? Even presuming that the difficulty level of your course and the talents of your students are such that you did, indeed, expect them to come up with this result, I don't feel like this answer is sufficient. Obviously, Karl must have understood something about the exercise to be able to come up with this answer (never mind the great ingenuity required) but his written argument fails to demonstrate that knowledge.\n\nContrast that with a disproof of this claim: For any function $f:A\\to B$ and any sets $S,T\\subseteq A$, $f[S]\\subseteq f[T]\\implies S\\subseteq T$.\n\nThis is the kind of statement you might consider in an introductory analysis course. I'd expect a student to find an easy counterexample, say by defining $A=\\{\\heartsuit,\\spadesuit\\}$ and $B=\\{\\star\\}$ and setting $f(\\heartsuit)=f(\\spadesuit)=\\star$ and $S=\\{\\heartsuit\\}$ and $T=\\{\\spadesuit\\}$.\n\nWould I expect the student to have to show all the details? Do they really need to explicitly say, \"Observe that $S,T\\subseteq A$ and $f[S]=f[T]$ and yet $S\\not\\subseteq T$\"? Well, that depends on your standards. If you're really training them to write clearly and explicitly (concision be darned), then you might require them to show these details. If you're just checking to see that they can generate such examples, then merely stating the example would suffice.\n\nMain answer: It depends on your standards for such a written argument, which in turn should depend on what you expect the student to be able to understand and demonstrate.\n\n\u2022 Sure, let's call him Karl. Sep 25 '14 at 1:26\n\u2022 Maybe give extra credit to someone who described how to create a counter example, perhaps in the above case by using S={<heart>} and T={<something besides a heart>}\n\u2013\u00a0rbp\nSep 29 '14 at 15:13\n\nCertainly the student should be aware of the expectations beforehand. To this end, I think there are at least two approaches. One is to specify on individual questions whether counterexamples should be explained, and the other is to establish (preferably from the outset, i.e., setting \"norms\" at the beginning of a course) the expectation that reasoning/justification for all answers will be provided.\n\nMore generally, I think explaining counterexamples is important. Let me give just two reasons.\n\n1) Consider the following problem:\n\nIs every whole number divisible by 6 and 8 also divisible by 6 $\\times$ 8 = 48? If so, prove your answer; if not, please provide a counterexample. Answer: 24.\n\nBut what does such a student understand? Perhaps she just remembered 24 is divisible by both and realized it is not divisible by 48. Or perhaps she realizes that the issue is that the two divisors in this proposed rule are not relatively prime, and could correct it by using, e.g., 3 and 16 instead. Or, even more seriously, perhaps she has the deeper understanding, but accidentally writes 32. In this last case, the understanding is the deepest, but it would be tough to justify giving the response any credit.\n\nAsking that the counterexample be justified ameliorates this potential problem. Personally, I would expect more than just an explanation of the particular counterexample, e.g., more than:\n\nNo, and 24 is a counterexample: It is divisible by 6 because 24 = 4 $\\times$ 6, and it is also divisible by 8 because 24 = 3 $\\times$ 8. But it is not divisible by 48, because 24/48 is not a whole number.\n\nI would prefer to see an answer that mentions being \"relatively prime\" or something equivalent, and perhaps even one that gives an example of two numbers that would yield a divisibility rule for 48.\n\n2) More subtle and perhaps not related to the specific question you have in mind (if there is one): Note that (speaking slightly messily) counterexamples can settle for all questions, but not there exists questions. The previous example is the assertion that all whole numbers (etc). If this turns out to be false, then it can be settled with a counterexample. If it turns out to be true, then it will need a further proof or justification. On the other hand, a there exists statement that is true might be settled with a single example. If it turns out to be false, then it will need a further proof or justification.\n\nThis latter point about proving/disproving by (counter)example in the context of for all vs. there exists assertions may seem a bit pedantic, but it is a place in which many students who are just beginning to understand ideas around \"proof\" can struggle quite a bit.\n\n\u2022 I don't think there exists statements are really that different because you can rewrite them as for all statements, i.e. $\\forall x\\ P(x) \\iff \\neg\\exists x\\ \\neg P(x)$. (Of course maybe students don't know this, in which case to their minds there is a difference...) Sep 25 '14 at 3:53\n\u2022 @DavidZ Yes, this is the underlying \"reason\" for the phenomenon mentioned in 2 above (i.e., that an example can disprove a for all statement or prove a there exists statement). And yes, one of the primary difficulties is precisely the one to which you allude parenthetically: For many students who are just getting into proof-based mathematics (or other mathematics with more formal reasoning) the equivalence you mention is neither known nor obvious. Sep 25 '14 at 18:08", "date": "2021-12-07 05:05:32", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/questions/4458/explanation-of-counter-example/4460", "openwebmath_score": 0.7324725985527039, "openwebmath_perplexity": 319.8075443544471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.967899295134923, "lm_q2_score": 0.8887587905460026, "lm_q1q2_score": 0.8602290069144426}}
{"url": "https://math.stackexchange.com/questions/2197132/calculating-area-of-triangle/2197147", "text": "# Calculating area of triangle\n\nProblem\n\nIn the triangle above, we have $AC = 12, \\ \\ BD = 10, \\ \\ CD = 6$.\n\nFind the area of $\\triangle ABD$.\n\nCaveat\n\nThis was part of an exam where you are not allowed to use a calculator or any other digital tools.\n\nMy progress\n\nGiven $AC, CD, \\measuredangle C$, I was able to find the length of $AD$ using the law of cosines. It yielded $AD = \\sqrt{180} = 6\\sqrt5$.\n\nFrom here, I decided to try and use the law of sines to find $\\measuredangle A$ as it would be the same in $\\triangle ABD$ and $\\triangle ACD$.\n\nI got $$\\frac{\\sin A}{CD} = \\frac{\\sin C}{AD}$$.\n\nThis yields $$\\sin A = \\frac{CD \\sin C}{AD} = \\frac{6}{\\sqrt{180}}$$.\n\nThen I need to calculate $\\arcsin\\frac{6}{\\sqrt{180}}$, which I can't, because I can't use a calculator for this problem. (Well, technically I can, but I want to solve this within the constraints placed on the participants.)\n\nFrom here, I am stuck. In $\\triangle ABD$, I know now two of the lengths, but I don't know any of the angles.\n\nQuestion\n\nAm I overlooking an easy inverse sine here? Or is there another way to calculate this area?\n\nThanks in advance for any help!\n\n\u2022 If this is part of an exam, why are you asking for help? \u2013\u00a0N. F. Taussig Mar 21 '17 at 20:07\n\u2022 @N.F.Taussig - It was an exam given in August of 2016. I'm preparing for the 2017 one ;) \u2013\u00a0Alec Mar 21 '17 at 20:08\n\n## 3 Answers\n\nFind $BC$ in the right triangle $BCD$; you have the hypotenuse is $10$ and the other side is $6$ so $BC=8$.\n\nArea $\\triangle ACD = \\frac12\\cdot 12\\cdot 6 = 36$.\n\nArea $\\triangle BCD = \\frac12\\cdot 8\\cdot 6 = 24$.\n\nArea $\\triangle ABD = 36-24 = 12$.\n\n\u2022 Damn... I really blinded myself on this one. Thanks! \u2013\u00a0Alec Mar 21 '17 at 20:11\n\nCalculate $BC$ with the pythagorean theorem, and subtract the areas of the two right angled triangles.\n\n$$BC=\\sqrt{10^2-6^2}=8$$ Hence, $$AB=12-8=4$$ and $$S_{\\Delta ABD}=\\frac{4\\cdot6}{2}=12$$", "date": "2019-08-24 19:46:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2197132/calculating-area-of-triangle/2197147", "openwebmath_score": 0.9274730086326599, "openwebmath_perplexity": 305.30931508786244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211568364099, "lm_q2_score": 0.8824278772763471, "lm_q1q2_score": 0.8602093641512262}}
{"url": "http://openstudy.com/updates/55c3bfcfe4b06d80932e877f", "text": "## mathmath333 one year ago Set Theory\n\n1. mathmath333\n\n\\large \\color{black}{\\begin{align} \\normalsize \\text{A \u2229 B = A \u2229 C need not imply B = C.} \\quad \\normalsize \\text{ Explain through an example.}\\hspace{.33em}\\\\~\\\\ \\end{align}}\n\n2. Kainui\n\nA = {1} B = {1, 2} C = {1, 3}\n\n3. ikram002p\n\nhuh kai was so fast :P\n\n4. Kainui\n\nHahaha there could have been a simpler example: A = {} B = {1} C = {2}\n\n5. mathmath333\n\nthnx\n\n6. Kainui\n\nI think that last example I gave sorta shows that it's almost like this is a lot like multiplying by zero. a*b=a*c This doesn't mean b=c, since a=0 is possible.\n\n7. mathmath333\n\nLet A and B be sets. If A \u2229 X = B \u2229 X = \u03c6 and A \u222a X = B \u222a X for some set X, show that A = B.\n\n8. ikram002p\n\nin general any case with $A\\cap B \\neq \\varnothing$ does not imply A=C\n\n9. ikram002p\n\nor $$C\\cap B \\neq \\varnothing$$\n\n10. mathmath333\n\ni need an example like the previous one\n\n11. Kainui\n\nikram needs more owl bucks sry\n\n12. ikram002p\n\nA={1,2,3} B={1,2,3} C={1,2,3,4}\n\n13. ikram002p\n\n:O @Kainui i dont lol\n\n14. Kainui\n\nHahaha jk :P\n\n15. mathmath333\n\nis that the example for this que Let A and B be sets. If A \u2229 X = B \u2229 X = \u03c6 and A \u222a X = B \u222a X for some set X, show that A = B.\n\n16. ikram002p\n\nyou wanna prove this ?\n\n17. mathmath333\n\nyes with example\n\n18. ikram002p\n\nok example A={1,2,3} B={1.4} C={1,2.3}\n\n19. mathmath333\n\nbut C is not there in the question\n\n20. ikram002p\n\nsorry i ment to say x :P\n\n21. ikram002p\n\nnow to prove they are two side of the prove 1- show A is a subset of B 2- show B is a subset of A\n\n22. ikram002p\n\nlet $$c\\in A \\cup X$$ so $$c\\in A ~or~c\\in X$$ if c in A then $$c\\in B\\cup X$$ and $$c\\in B$$ (since c is not in x) thus $$A\\subset B$$ the other way is alike :P sorry im lazy to type\n\n23. thomas5267\n\nIs it possible to do something like this? $A\\cup X=B\\cup X\\land A\\cap X=B\\cap X=\\emptyset\\implies A\\cup X\\setminus X=B\\cup X\\setminus X\\implies A=B$", "date": "2016-10-27 17:16:28", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/55c3bfcfe4b06d80932e877f", "openwebmath_score": 0.817804217338562, "openwebmath_perplexity": 2797.5335863737355, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211553734217, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8602093538179241}}
{"url": "https://gmatclub.com/forum/the-average-arithmetic-mean-length-per-film-for-a-group-of-21-films-217895.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 23 Sep 2018, 11:28\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# The average (arithmetic mean) length per film for a group of 21 films\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: Professional GMAT Tutor\nAffiliations: AB, cum laude, Harvard University (Class of '02)\nJoined: 10 Jul 2015\nPosts: 669\nLocation: United States (CA)\nAge: 38\nGMAT 1: 770 Q47 V48\nGMAT 2: 730 Q44 V47\nGMAT 3: 750 Q50 V42\nGRE 1: Q168 V169\nWE: Education (Education)\nThe average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\n05 May 2016, 19:27\n7\n9\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n77% (01:25) correct 23% (01:31) wrong based on 422 sessions\n\n### HideShow timer Statistics\n\nThe average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?\n\nA) $$t + \\frac{2}{3}$$\n\nB) $$t - \\frac{2}{3}$$\n\nC) $$21t+14$$\n\nD) $$t + \\frac{3}{2}$$\n\nE) $$t - \\frac{3}{2}$$\n\nAttachments\n\nScreen Shot 2016-05-05 at 7.04.23 PM.png [ 107.79 KiB | Viewed 5506 times ]\n\n_________________\n\nHarvard grad and 99% GMAT scorer, offering expert, private GMAT tutoring and coaching, both in-person (San Diego, CA, USA) and online worldwide, since 2002.\n\nOne of the only known humans to have taken the GMAT 5 times and scored in the 700s every time (700, 710, 730, 750, 770), including verified section scores of Q50 / V47, as well as personal bests of 8/8 IR (2 times), 6/6 AWA (4 times), 50/51Q and 48/51V (1 question wrong).\n\nYou can download my official test-taker score report (all scores within the last 5 years) directly from the Pearson Vue website: https://tinyurl.com/y94hlarr Date of Birth: 09 December 1979.\n\nGMAT Action Plan and Free E-Book - McElroy Tutoring\n\nContact: mcelroy@post.harvard.edu\n\nMath Expert\nJoined: 02 Aug 2009\nPosts: 6800\nRe: The average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\n05 May 2016, 20:09\n5\n1\nmcelroytutoring wrote:\nThe average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?\n\nA) $$t + \\frac{2}{3}$$\n\nB) $$t - \\frac{2}{3}$$\n\nC) $$21t+14$$\n\nD) $$t + \\frac{3}{2}$$\n\nE) $$t - \\frac{3}{2}$$\n\nhi,\nyou do not have to get into any calculations if you understand and take the Q logically..\n\navg time for 21 is t min..\nyou add 52 and remove 66.. so basically you are removing 14 min from total..\neffect on each or average =$$\\frac{14}{21}= \\frac{2}{3}$$..\n\nso final average = $$t-\\frac{2}{3}$$...\nB\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n\nGMAT online Tutor\n\n##### General Discussion\nDirector\nStatus: Professional GMAT Tutor\nAffiliations: AB, cum laude, Harvard University (Class of '02)\nJoined: 10 Jul 2015\nPosts: 669\nLocation: United States (CA)\nAge: 38\nGMAT 1: 770 Q47 V48\nGMAT 2: 730 Q44 V47\nGMAT 3: 750 Q50 V42\nGRE 1: Q168 V169\nWE: Education (Education)\nThe average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 06 May 2016, 12:30\n1\nLogic tells us that the average length per film will go down, so we can eliminate all answers that add to the original average of t. That leaves either B or E. When executing the algebra, the key step is being able to quickly and easily turn one fraction with two numerators $$(\\frac{21t-14}{21})$$ into two fractions with one numerator each $$(\\frac{21t}{21}-\\frac{14}{21})$$.\n\nAbove is a visual that should help.\n_________________\n\nHarvard grad and 99% GMAT scorer, offering expert, private GMAT tutoring and coaching, both in-person (San Diego, CA, USA) and online worldwide, since 2002.\n\nOne of the only known humans to have taken the GMAT 5 times and scored in the 700s every time (700, 710, 730, 750, 770), including verified section scores of Q50 / V47, as well as personal bests of 8/8 IR (2 times), 6/6 AWA (4 times), 50/51Q and 48/51V (1 question wrong).\n\nYou can download my official test-taker score report (all scores within the last 5 years) directly from the Pearson Vue website: https://tinyurl.com/y94hlarr Date of Birth: 09 December 1979.\n\nGMAT Action Plan and Free E-Book - McElroy Tutoring\n\nContact: mcelroy@post.harvard.edu\n\nOriginally posted by mcelroytutoring on 05 May 2016, 19:28.\nLast edited by mcelroytutoring on 06 May 2016, 12:30, edited 6 times in total.\nCurrent Student\nJoined: 12 Aug 2015\nPosts: 2648\nSchools: Boston U '20 (M)\nGRE 1: Q169 V154\nRe: The average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\n15 Dec 2016, 14:05\n2\nGreat Solutions above.\nHere is my solution =>\nAverage length = t minutes\nSum(21) = 21t\nNew sum = 21t-66+52 = 21t-14\nHence new mean = 21t-14 / 21 = t-2/3\n\nHence B\n\n_________________\n\nMBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!\n\nGetting into HOLLYWOOD with an MBA!\n\nThe MOST AFFORDABLE MBA programs!\n\nSTONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)\n\nAVERAGE GRE Scores At The Top Business Schools!\n\nJoined: 30 May 2015\nPosts: 40\nRe: The average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\n20 Jul 2017, 15:08\n1\naverage = sum of numbers/total number\nt = s/21\ns = 21t\nNow as per question\ns =21t -66 +52\ns = 21t - 14\nNew average\nt1 = (21t - 14) / 21\n=> t - 2/3\nDirector\nJoined: 09 Mar 2016\nPosts: 884\nThe average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\n06 Mar 2018, 09:16\nstonecold wrote:\nGreat Solutions above.\nHere is my solution =>\nAverage length = t minutes\nSum(21) = 21t\nNew sum = 21t-66+52 = 21t-14\nHence new mean = 21t-14 / 21 = t-2/3\n\nHence B\n\nstonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail\n_________________\n\nIn English I speak with a dictionary, and with people I am shy.\n\nBSchool Forum Moderator\nJoined: 26 Feb 2016\nPosts: 3137\nLocation: India\nGPA: 3.12\nThe average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\n06 Mar 2018, 11:18\n1\ndave13 wrote:\nstonecold wrote:\nGreat Solutions above.\nHere is my solution =>\nAverage length = t minutes\nSum(21) = 21t\nNew sum = 21t-66+52 = 21t-14\nHence new mean = 21t-14 / 21 = t-2/3\n\nHence B\n\nstonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail\n\nHey dave13\n\n$$\\frac{21t-14}{21} = \\frac{21t}{21} - \\frac{14}{21} = t - \\frac{2}{3}$$\n\nHope that helps\n_________________\n\nYou've got what it takes, but it will take everything you've got\n\nManager\nJoined: 10 Apr 2018\nPosts: 105\nRe: The average (arithmetic mean) length per film for a group of 21 films\u00a0 [#permalink]\n\n### Show Tags\n\n15 Aug 2018, 16:09\nHi,\nWe can use deviation technique to solve such type of questions. The answer would be in less than 30 Secs\n\nrefer: https://gmatclub.com/forum/a-student-s- ... l#p2113194\n\nSo new average would be\nt-(14/21)\nt-2/3\nHence B.\nRe: The average (arithmetic mean) length per film for a group of 21 films &nbs [#permalink] 15 Aug 2018, 16:09\nDisplay posts from previous: Sort by\n\n# The average (arithmetic mean) length per film for a group of 21 films\n\n## Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-23 18:28:08", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-average-arithmetic-mean-length-per-film-for-a-group-of-21-films-217895.html", "openwebmath_score": 0.48316341638565063, "openwebmath_perplexity": 9301.023791857197, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9695556515673754, "lm_q2_score": 0.8872045981907007, "lm_q1q2_score": 0.8601942322723564}}
{"url": "http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 15 Sep 2014, 17:00\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Inequalities trick\n\nAuthor Message\nTAGS:\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2793\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 178\n\nKudos [?]: 945 [38] , given: 235\n\nInequalities trick\u00a0[#permalink] \u00a016 Mar 2010, 09:11\n38\nKUDOS\n62\nThis post was\nBOOKMARKED\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\nAttachments\n\n1.jpg [ 6.73 KiB | Viewed 21741 times ]\n\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 4755\nLocation: Pune, India\nFollowers: 1112\n\nKudos [?]: 5022 [36] , given: 164\n\nRe: Inequalities trick\u00a0[#permalink] \u00a022 Oct 2010, 05:33\n36\nKUDOS\nExpert's post\n13\nThis post was\nBOOKMARKED\nYes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.\n\nIf (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.\ne.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.\n\nAttachment:\n\ndoc.jpg [ 7.9 KiB | Viewed 21052 times ]\n\nThis divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.\n\nWhen you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.\n\nWhen you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.\n\nSince we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7\n\nIt should be obvious that it will also work in cases where factors are divided.\ne.g. (x - a)(x - b)/(x - c)(x - d) < 0\n(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.\n\nNote: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nSave $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2047 Followers: 128 Kudos [?]: 919 [13] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 13 This post received KUDOS 4 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 154 Followers: 2 Kudos [?]: 31 [8] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 8 This post received KUDOS 3 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [4] , given: 164 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 4 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting\nEnroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.\n\nVeritas Prep Reviews\n\nManager\nStatus: On...\nJoined: 16 Jan 2011\nPosts: 189\nFollowers: 3\n\nKudos [?]: 35 [4] , given: 62\n\nRe: Inequalities trick\u00a0[#permalink] \u00a010 Aug 2011, 16:01\n4\nKUDOS\n2\nThis post was\nBOOKMARKED\nWoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.\n\n1) CORE CONCEPT\n@gurpreetsingh -\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nArrange the NUMBERS in ascending order from left to right. a<b<c<d\nDraw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nNote: Make sure that the factors are of the form (ax - b), not (b - ax)...\n\nexample -\n(x+2)(x-1)(7 - x)<0\n\nConvert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')\nNow solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'\nSince you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.\n\n2) Variation - ODD/EVEN POWER\n@ulm/Karishma -\nif we have even powers like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\nThis will be same as (x-b)\n\nWe can ignore squares BUT SHOULD consider ODD powers\nexample -\n2.a\n(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0\n2.b\n(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0\nis the same as (x - a)(x - b)(x - c)(x - d) < 0\n\n3) Variation <= in FRACTION\n@mrinal2100 -\nif = sign is included with < then <= will be there in solution\nlike for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7\n\nBUT if it is a fraction the denominator in the solution will not have = SIGN\n\nexample -\n3.a\n(x + 2)(x - 1)/(x -4)(x - 7) < =0\nthe solution will be -2 <= x <= 1 or 4< x < 7\nwe cant make 4<=x<=7 as it will make the solution infinite\n\n4) Variation - ROOTS\n@Karishma -\nAs for roots, you have to keep in mind that given \\sqrt{x}, x cannot be negative.\n\n\\sqrt{x} < 10\nimplies 0 < \\sqrt{x} < 10\nSquaring, 0 < x < 100\nRoot questions are specific. You have to be careful. If you have a particular question in mind, send it.\n\nRefer - inequalities-and-roots-118619.html#p959939\nSome more useful tips for ROOTS....I am too lazy to consolidate\n\n<5> THESIS -\n@gmat1220 -\nOnce algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.\n\nI will save this future references....\n\nAnyone wants to add ABSOLUTE VALUES....That will be a value add to this post\n\n_________________\n\nLabor cost for typing this post >= Labor cost for pushing the Kudos Button\nkudos-what-are-they-and-why-we-have-them-94812.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 4755\nLocation: Pune, India\nFollowers: 1112\n\nKudos [?]: 5022 [2] , given: 164\n\nRe: Inequalities trick\u00a0[#permalink] \u00a009 Sep 2013, 22:35\n2\nKUDOS\nExpert's post\nkarannanda wrote:\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nHi Gurpreet,\nThanks for the wonderful method.\nI am trying to understand it so that i can apply it in tests.\nCan you help me in applying this method to the below expression to find range of x.\nx^3 \u2013 4x^5 < 0?\n\nI am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.\n\nBefore you apply the method, ensure that the factors are of the form (x - a)(x - b) etc\n\nx^3 - 4x^5 < 0\n\nx^3 ( 1 - 4x^2) < 0\n\nx^3(1 - 2x) (1 + 2x) < 0\n\n4x^3(x - 1/2)(x + 1/2) > 0 (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)\n\nNow the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.\nThe solution will be x > 1/2 or -1/2 < x< 0.\n\nCheck out these posts discussing such complications:\nhttp://www.veritasprep.com/blog/2012/06 ... e-factors/\nhttp://www.veritasprep.com/blog/2012/07 ... ns-part-i/\nhttp://www.veritasprep.com/blog/2012/07 ... s-part-ii/\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nSave $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Senior Manager Joined: 16 Feb 2012 Posts: 260 Concentration: Finance, Economics Followers: 4 Kudos [?]: 65 [1] , given: 110 Re: Inequalities trick [#permalink] 22 Jul 2012, 02:03 1 This post received KUDOS VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\\frac {(x+2)(x-1)}{(x-4)(x-7)} were 4\\leq x \\leq 7, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [1] , given: 164 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 1 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\\frac {(x+2)(x-1)}{(x-4)(x-7)} were 4\\leq x \\leq 7, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting\nEnroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.\n\nVeritas Prep Reviews\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2793\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 178\n\nKudos [?]: 945 [1] , given: 235\n\nRe: Inequalities trick\u00a0[#permalink] \u00a018 Oct 2012, 04:17\n1\nKUDOS\nGMATBaumgartner wrote:\ngurpreetsingh wrote:\nulm wrote:\nif we have smth like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\n\nyes even powers wont contribute to the inequality sign. But be wary of the root value of x=a\n\nHi Gurpreet,\nCould you elaborate what exactly you meant here in highlighted text ?\n\nEven I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.\n\nIf the powers are even then the inequality won't be affected.\n\neg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0\n\njust use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 4755\nLocation: Pune, India\nFollowers: 1112\n\nKudos [?]: 5022 [1] , given: 164\n\nRe: Inequalities trick\u00a0[#permalink] \u00a018 Oct 2012, 09:27\n1\nKUDOS\nExpert's post\nGMATBaumgartner wrote:\nHi Gurpreet,\nCould you elaborate what exactly you meant here in highlighted text ?\n\nEven I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.\n\nIn addition, you can check out this post:\n\nhttp://www.veritasprep.com/blog/2012/07 ... s-part-ii/\n\nI have discussed how to handle powers in it.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nSave $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 26582 Followers: 3487 Kudos [?]: 26158 [1] , given: 2705 Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25 1 This post received KUDOS Expert's post GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.php DS questions on inequalities: search.php?search_id=tag&tag_id=184 PS questions on inequalities: search.php?search_id=tag&tag_id=189 Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2793 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 178 Kudos [?]: 945 [1] , given: 235 Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04 1 This post received KUDOS VeritasPrepKarishma wrote: The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6 - 2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook Get the best GMAT Prep Resources with GMAT Club Premium Membership Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Math Expert Joined: 02 Sep 2009 Posts: 26582 Followers: 3487 Kudos [?]: 26158 [1] , given: 2705 Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [1] , given: 164 Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED PathFinder007 wrote: I have a query. I have following question x^3 - 4x^5 < 0 I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks The factors must be of the form (x - a)(x - b) .... etc x^3 - 4x^5 < 0 x^3 * (1 - 4x^2) < 0 x^3 * (1 - 2x) * (1 + 2x) < 0 x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1) x^3 * 2(x - 1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and -1/2. ____________ - 1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or -1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ The links will give you the theory behind this method in detail. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting\nEnroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.\n\nVeritas Prep Reviews\n\nIntern\nJoined: 08 Mar 2010\nPosts: 6\nFollowers: 0\n\nKudos [?]: 0 [0], given: 4\n\nRe: Inequalities trick\u00a0[#permalink] \u00a016 Mar 2010, 09:46\nCan u plz explainn the backgoround of this & then the explanation.\n\nThanks\nSenior Manager\nJoined: 13 Dec 2009\nPosts: 264\nFollowers: 10\n\nKudos [?]: 106 [0], given: 13\n\nRe: Inequalities trick\u00a0[#permalink] \u00a019 Mar 2010, 11:48\nI have applied this trick and it seemed to be quite useful.\n_________________\n\nMy debrief: done-and-dusted-730-q49-v40\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2793\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 178\n\nKudos [?]: 945 [0], given: 235\n\nRe: Inequalities trick\u00a0[#permalink] \u00a019 Mar 2010, 11:59\nttks10 wrote:\nCan u plz explainn the backgoround of this & then the explanation.\n\nThanks\n\ni m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.\n\nsidhu4u wrote:\nI have applied this trick and it seemed to be quite useful.\n\nNice to hear this....good luck.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nSenior Manager\nStatus: Upset about the verbal score - SC, CR and RC are going to be my friend\nJoined: 30 Jun 2010\nPosts: 318\nFollowers: 5\n\nKudos [?]: 15 [0], given: 6\n\nRe: Inequalities trick\u00a0[#permalink] \u00a017 Oct 2010, 15:14\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) < 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nI don't understand this part alone. Can you please explain?\n_________________\n\nMy gmat story\nMGMAT1 - 630 Q44V32\nMGMAT2 - 650 Q41V38\nMGMAT3 - 680 Q44V37\nGMATPrep1 - 660 Q49V31\nKnewton1 - 550 Q40V27\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2793\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 178\n\nKudos [?]: 945 [0], given: 235\n\nRe: Inequalities trick\u00a0[#permalink] \u00a017 Oct 2010, 16:19\nDreamy wrote:\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) < 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nI don't understand this part alone. Can you please explain?\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0 you will consider the curve with -ve inside it.. check the attached image.\n\nf(x) = (x-a)(x-b)(x-c)(x-d) > 0 consider the +ve of the curve\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nRe: Inequalities trick \u00a0 [#permalink] 17 Oct 2010, 16:19\nSimilar topics Replies Last post\nSimilar\nTopics:\n84 Tips and Tricks: Inequalities 26 14 Apr 2013, 08:20\n2 To use tricks or not to use tricks..... 3 20 Apr 2008, 15:57\ninequalities 1 06 Aug 2006, 07:23\ninequalities 3 05 Dec 2005, 11:18\ninequalities 2 15 Jun 2005, 06:47\nDisplay posts from previous: Sort by", "date": "2014-09-16 01:00:22", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1", "openwebmath_score": 0.5333693623542786, "openwebmath_perplexity": 2813.2556971947556, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620619801096, "lm_q2_score": 0.8962513800615312, "lm_q1q2_score": 0.860188072580374}}
{"url": "https://gmatclub.com/forum/what-arithmetic-should-i-memorize-80128.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 02 Apr 2020, 02:59\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# What arithmetic should I memorize?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 25 Mar 2009\nPosts: 245\nWhat arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 19 Oct 2018, 13:30\n38\n1\n347\n\nImprove Your Speed - GMAT Arithmetic to Memorize\n\nI find myself constantly using long division and multiplication for simple things like 13 x 11.\nWhat are some good arithmetic calcs to memorize? I created a 20x20 multiplication table but this seems like a bit much right?\n\nPerfect squares up to 100? $$\\sqrt{3}$$? $$\\sqrt{5}$$?\n\nI am not really talking about formulas to memorize since you should definitely memorize things like nCr, nPr, sum of all #'s in an evenly spaced set, etc.\n\nEdit: Moderator note: Attached are files we gathered in this post from several other sources in addition to the author. Also see additional resources in posts below to help with improve your speed and reduce mistakes\n\nTarget Test Prep 15-page PDF summarizes ALL formulas you have to know\n\n-\nAttachments\n\nGMAT Math Compendium.xls [855 KiB]\n\nGeometry Formulas_template.pdf [44.28 KiB]\n\nGeometry Formulas.pdf [72.21 KiB]\n\nArithmetic Fractions and Percentages.doc [32.5 KiB]\n\nOriginally posted by topher on 25 Jun 2009, 20:39.\nLast edited by bb on 19 Oct 2018, 13:30, edited 7 times in total.\nEdit\nFounder\nJoined: 04 Dec 2002\nPosts: 19430\nLocation: United States (WA)\nGMAT 1: 750 Q49 V42\nGPA: 3.5\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n25 Jun 2009, 21:18\n83\n87\nFantastic question\n\nThis file should give you an idea where you are lacking.\nI think you should definitely know the squares from 0-10 and preferrably from 10 to 20 as well.\nAttachments\n\nArithmetic Fractions and Percentages.doc [32.5 KiB]\n\n_________________\nIntern\nJoined: 12 May 2009\nPosts: 48\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Oct 2009, 23:59\n81\n1\n63\nAs far as common squares are concerned, I only remember the ones with 0 or 5 in the units digit. For the latter category, I use the following process:\n\nFor example: 65*65\n\n1) Always write down 25 as this is always the last two digits of the result:\n...25\n\n2) Multiply (non-units digits) times (non-units digits + 1)\n6 * (6+1) = 42\n\n2c) Combine:\n4225\n\nThis way I always have the important squares handy... very useful for estimations!\n\nAny other math shortcuts? Anyone\n\nSteve\n##### General Discussion\nCurrent Student\nJoined: 13 Jul 2009\nPosts: 133\nLocation: Barcelona\nSchools: SSE\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n26 Jul 2009, 09:36\n84\n29\nHi all,\n\nI created a template that gather all the geometry formulas that appear on the GMAT (if i missed any please tell!). I use it every week in order to ensure that I remember all of them.\n\nI think it could be helpful for some of you so I decided to share.\n\nThe TEMPLATE is the one for practice.\n\nGood luck!\nAttachments\n\nGeometry Formulas_template.pdf [44.28 KiB]\n\nGeometry Formulas.pdf [72.21 KiB]\n\n_________________\nFounder\nJoined: 04 Dec 2002\nPosts: 19430\nLocation: United States (WA)\nGMAT 1: 750 Q49 V42\nGPA: 3.5\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n26 Jul 2009, 19:33\n3\n3\nGreat resource\n\nThank you!\n\n_________________\nManager\nJoined: 22 Jul 2009\nPosts: 147\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 26 Jan 2010, 12:23\n29\n8\nMy fellow GMAT clubbers,\n\nI'd like to share the excel spreadsheet in which I have been compiling all knowledge and shortcuts relevant for tackling the GMAT quant section.\n\nIt is organized by tabs. Each tab covers a different area.\n\nExamples:\n- Tab \"NP. Primes\" covers: Number Properties - Primes\n- Tab \"NP. Powers-Rts\" covers: Number Properties - Patterns of powers and roots\n- Tab \"G. PTriples\" covers: Geometry - Phytagorean triplets patterns\n- Tab \"G. Ci-Sq (Pi)\" covers: Geometry - Relantionships between the measures of inscribed/circumscribed circles & squares\n- Tab \"WT. Prob\" covers: Wort Translations - Probability (dice/coins)\n\nFinally, three of the tabs are summaries made of the Man guides \"Word Translations\", \"Number Properties\" and \"Geometry\".\n\nHope it helps somebody.\n\nCheers,\nAttachments\n\nGMAT Math Compendium.xls [855 KiB]\n\nOriginally posted by powerka on 08 Sep 2009, 16:51.\nLast edited by powerka on 26 Jan 2010, 12:23, edited 3 times in total.\nIntern\nJoined: 24 Oct 2009\nPosts: 16\nLocation: Russia\nSchools: IESE, SDA Bocconi\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n31 Oct 2009, 08:56\n71\n61\nI'm new here, and try to go through all the topics - this site is a treasure!\n\nMaybe not the right place to share my experience with \"multiplication for simple things like 13 x 11\", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,\n13x11 -> 1+3=4 -> 143 is the result.\nOr, 36x11 -> 3+6=9 -> the result is 36x11=396.\n\nIt really saves time.\nIntern\nJoined: 31 Oct 2009\nPosts: 22\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n31 Oct 2009, 19:21\n31\n32\nShelen wrote:\nI'm new here, and try to go through all the topics - this site is a treasure!\n\nMaybe not the right place to share my experience with \"multiplication for simple things like 13 x 11\", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,\n13x11 -> 1+3=4 -> 143 is the result.\nOr, 36x11 -> 3+6=9 -> the result is 36x11=396.\n\nIt really saves time.\n\nAnd if it equals more than 10, add a 1 to the first digit\n\nEG\n\n68*11 -> 6+8=14 -> 6 14 8 -> 748\nIntern\nJoined: 15 Nov 2009\nPosts: 27\nLocation: Moscow, Russia\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Jan 2010, 08:47\n1\nSorry, but on page 2 there is a misprint:\nin Arc Measure (Intersecting Secants/Tangents) It must be the difference between arcs measures of AD and BD.\nFounder\nJoined: 04 Dec 2002\nPosts: 19430\nLocation: United States (WA)\nGMAT 1: 750 Q49 V42\nGPA: 3.5\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n07 Jan 2010, 16:13\n1\n6\nAlso, see this post for some good theory: math-number-theory-88376.html\n_________________\nManager\nStatus: Not afraid of failures, disappointments, and falls.\nJoined: 20 Jan 2010\nPosts: 243\nConcentration: Technology, Entrepreneurship\nWE: Operations (Telecommunications)\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Feb 2010, 09:34\n48\n37\nThanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.\n\nWhat if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below\n\nFor 3 digits\n133x11 --> 1 1+3 3+3 3=1463\nAnd for 4 digits\n1243x11 --> 1 1+2 2+4 4+3 3=13673\n\nAnd for 5 digits\n15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983\nand so on.\nyou can plug other numbers in and check it out.\n\nHope it helps!\nManager\nStatus: Not afraid of failures, disappointments, and falls.\nJoined: 20 Jan 2010\nPosts: 243\nConcentration: Technology, Entrepreneurship\nWE: Operations (Telecommunications)\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Feb 2010, 21:41\n15\n26\nGuys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.\n\nLet's take 1936\n1) First of all check whether last two digits of the number are divisible by 4 or not.\nFor 1936, we do this way 36/4=9\n\n2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.\n\n--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9\nSo the whole number is divisible by 8.\n\nOnce you understand it and do a little practice, you'll find it easy and fast.\n**You can try other numbers to see whether it is true or not\nHope it helps!\nIntern\nAffiliations: isa.org\nJoined: 26 Jan 2010\nPosts: 16\nLocation: Mumbai, India.\nSchools: ISB, India.\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n14 Feb 2010, 05:12\n3\n2\nAtifS wrote:\nThanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.\n\nWhat if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below\n\nFor 3 digits\n133x11 --> 1 1+3 3+3 3=1463\nAnd for 4 digits\n1243x11 --> 1 1+2 2+4 4+3 3=13673\n\nAnd for 5 digits\n15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983\nand so on.\nyou can plug other numbers in and check it out.\n\nHope it helps!\n\nwelldone dude,\njust a small addition which i tried and will help to avoid confusion:\n\nstart writing the answer from right hand side in case if the addition of two no. exceeds 10, and add it to consecutive no. on left hand side. try out!!\nIntern\nAffiliations: isa.org\nJoined: 26 Jan 2010\nPosts: 16\nLocation: Mumbai, India.\nSchools: ISB, India.\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n14 Feb 2010, 05:21\n2\n5\nAtifS wrote:\nGuys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.\n\nLet's take 1936\n1) First of all check whether last two digits of the number are divisible by 4 or not.\nFor 1936, we do this way 36/4=9\n\n2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.\n\n--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9\nSo the whole number is divisible by 8.\n\nOnce you understand it and do a little practice, you'll find it easy and fast.\n**You can try other numbers to see whether it is true or not\nHope it helps!\n\nhi,\ni feel the process is bit complicated as it doesn't give the value of quotient, it just tells you whether no. is divisible by 8.\nhere is another tric, if the no. formed by last three digits is divisible by 8 then the whole no. is divisible by 8.\n\n953360 is divisible by 8 since 360 is divisible by 8,\n529418: not divisible as 418 is not divisible by 8.\n\nplug in different values and try.\nIntern\nJoined: 17 Feb 2010\nPosts: 1\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n17 Feb 2010, 18:53\n6\n2\nRE:\nWhat if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below\n\nFor 3 digits\n133x11 --> 1 1+3 3+3 3=1463\nAnd for 4 digits\n1243x11 --> 1 1+2 2+4 4+3 3=13673\n\nAnd for 5 digits\n15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983\nand so on.\n\nHi guys,\nWhen multiplying by 11 I find it much easier to multiply the # by 10 (or add 0) and then add the original # to it.\nFor example, 1234x11 = 12340 + 1234\n\nWhen dividing by 11 and the following condition is satisfied one could factor the # as follows:\n671/11 = 61x11/11 = 61 because in 671, 6+1=7 which is the # in the middle of 671. This works for 3digit #'s\nManager\nStatus: Labor Omnia Vincit\nJoined: 16 Aug 2010\nPosts: 71\nSchools: S3 Asia MBA (Fudan University, Korea University, National University of Singapore)\nWE 1: Market Research\nWE 2: Consulting\nWE 3: Iron & Steel Retail/Trading\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 05 Oct 2010, 10:03\n1\nNow, what does this mean:\n\nArea of Isosceles Triangle = 1/2[(leg)^2]\n(from Geometry Formulas.pdf)\n\nI haven't been able to give myself any convincing explanation of this formula...\n\nOriginally posted by rishabhsingla on 11 Sep 2010, 10:27.\nLast edited by rishabhsingla on 05 Oct 2010, 10:03, edited 1 time in total.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 62422\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n11 Sep 2010, 10:42\n1\n1\nrishabhsingla wrote:\nNow, what does this mean:\n\nArea of Isosceles Triangle = 1/2[(leg)^2]\n(from Geometry Formulas.pdf)\n\nI haven't been able to give myself any convincing explanation of this formula...\n\nWell that's because it's not true.\n\nIt should be area of isosceles right triangle = 1/2*(leg)^2, simply because area of right triangle equals to 1/2*leg1*leg2, and since in isosceles right triangle leg1=leg2, then this formula becomes area=1/2*(leg)^2.\n_________________\nIntern\nStatus: \"\nJoined: 22 Jan 2011\nPosts: 37\nSchools: IE- ESCP - Warwick\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n20 Apr 2011, 18:47\nThanks for this memory sheet, very useful!\n\nHowever I believe some formulas, details might be added...\n\nAngles of a polygons (n-2)* 180, angles inscribed in a circle (compared to central angle there a section about it in the VERITAS Prep book)\n\nAlso I did not see two real IMPORTANT triangles sides ratio:\n\n3:4:5\n5:12:13\n\nGood you put the formula for the equilateral triangle (which is not given in the OG and allow you to gain so much time)\n\nI found formula I did not know! Thank you for sharing\nManager\nStatus: 700 (q47,v40); AWA 6.0\nJoined: 16 Mar 2011\nPosts: 78\nGMAT 1: 700 Q47 V40\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n23 Apr 2011, 12:59\n1\nI'd also add one more element to Coordinate geometry. The coordinates of a point P on line joining two points A(x1,y1) and B(x2,y2) such that PA:PB is in the ration m:n would be ((mx2+nx1)/(m+n),(my2+ny1)/(m+n)).\nIntern\nJoined: 10 Jul 2011\nPosts: 29\nRe: What arithmetic should I memorize?\u00a0 [#permalink]\n\n### Show Tags\n\n04 Jan 2012, 15:59\n5\n13\nThese are a few math tricks my brother used when studying for the MCAT. Hope they help!\n(attached a word document)\n\n1. A nice math trick is multiplying two integers that have multiple digits relatively quickly. It does not apply to all integers the following has to be met:\nTENS DIGIT in both integers HAS TO BE SAME\nONES DIGIT in each integer HAS TO ADD UP TO 10\nAlso, if the ones digits are 1 and 9 you just write 09.\n\nExample 34x36:\nStep 1: Add one to tens place then multiply (1+3)x3 = 12\nStep 2: Ones place 4x6 = 24\nStep 3: Place Steps 1 & 2 = 1224\n\nExample 1 & 9 in ones place: 21x29\nStep 1: Add one to tens place then multiply (1+2)x2 = 6\nStep 2: Ones place 1x9= 09\nStep 3: Place Steps 1 & 2 together = 609\n\n2. This one is for people having difficulty memorizing a few square root numbers!\nsqrt 1 = 1 (as in 1/1 = New Year's Day)\nsqrt 2 = 1.4 (as in 2/14 = Valentine's Day)\nsqrt 3 = 1.7 (as in 3/17 = St. Patrick's Day)\n\n3. The next math trick is the Babylonian Method it can be useful when estimating square roots.\nFirst guess roughly what you think it would be\n\nStep 1: For a number less than 1 guess bigger.\nFor a number greater than 1 guess smaller.\nStep 2: Divide your guess into the square root number.\nStep 4: Divide by 2\n\nExample sqrt(.78):\nStep 1: sqrt of .78 < 1 so guess = .85\nStep 2. .78/.85 = ~.9\nStep 3: (.85+.9) = 1.75\nStep 4: 1.75/2 = ~.88 And this should be your answer (or close enough)\n\nIf you guess really wildly just use the answer from your first guess and run through the process again. You can do it in seconds once you get good at it.\n\nWild Guess Example: sqrt 70 = ?\nStep 1 sqrt of 70>1 so guess 10\nStep 2: 70/10= 7\nStep 3: 7+10= 17\nStep 4: 17/2 = 8.5\n*8.5x8.5 = 72.25 Still off (10 kind of a wild guess, so repeat process with the new answer from step 4)\nStep 1: guess = 8.5\nStep 2: 70/8.5= 8.2\nStep 3: 8.2+8.5= 16.7\nStep 4: 16.7/2 = ~8.4\n8.4*8.4 = 70.6\n\n4. If two numbers (both even or odd) are close together and their average is an integer, then this method can be used.\nNeed to recognize that:\nx^2 - y^2 = (x + y)(x - y) and vice-versa (x + y)(x - y) = x^2 - y^2\nExample 1\n48*52 = (50-2)(50+2) = 50^2 - 2^2 = 2496.\n\nExample 2\n3^2 - 2^2 = 3 + 2\n4^2 - 3^2 = 4 + 3\n5^2 - 4^2 = 5 + 4\n\n5. When x and y are consecutive integers, then (x - y) = 1.\n\nIt's useful for calculating large squares:\n\nExample: 71^2 = ?\n71^2 - 70^2 = 71 + 70\nso 71^2 = 70^2 + (141)\n= 4900 + 141 = 5041\n\nExample: 53^2 = ?\n53^2 \u2013 52^2 = (53 + 52) + (52 + 51) + (51 + 50)\n53^2 = 50^2 + (105) + (103) + (101)\n= 2500 + 309 = 2809\n\n0.79^2\n80^2 - 79^2 = 80 + 79\nso 80^2 - (80 + 79) = 79^2\n6400 - 159 = 6241 so 0.79^2 = 0.6241\nAttachments\n\nmath tricks.docx [15.68 KiB]\n\nRe: What arithmetic should I memorize? \u00a0 [#permalink] 04 Jan 2012, 15:59\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 29 posts ]\n\nDisplay posts from previous: Sort by", "date": "2020-04-02 10:59:23", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-arithmetic-should-i-memorize-80128.html", "openwebmath_score": 0.7776781320571899, "openwebmath_perplexity": 4477.274945892129, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9353465080392795, "lm_q2_score": 0.9196425366837827, "lm_q1q2_score": 0.8601844353315612}}
{"url": "https://proofwiki.org/wiki/Inversion_Mapping_is_Automorphism_iff_Group_is_Abelian", "text": "# Inversion Mapping is Automorphism iff Group is Abelian\n\n## Theorem\n\nLet $\\struct {G, \\circ}$ be a group.\n\nLet $\\iota: G \\to G$ be the inversion mapping on $G$, defined as:\n\n$\\forall g \\in G: \\map \\iota g = g^{-1}$\n\nThen $\\iota$ is an automorphism if and only if $G$ is abelian.\n\n## Proof\n\nFrom Inversion Mapping is Permutation, $\\iota$ is a permutation.\n\nIt remains to be shown that $\\iota$ has the morphism property if and only if $G$ is abelian.\n\n### Sufficient Condition\n\nSuppose $\\iota$ is an automorphism.\n\nThen:\n\n $\\displaystyle \\forall x, y \\in G: \\map \\iota {x \\circ y}$ $=$ $\\displaystyle \\map \\iota x \\circ \\map \\iota y$ Definition of Morphism Property $\\displaystyle \\leadsto \\ \\$ $\\displaystyle \\paren {x \\circ y}^{-1}$ $=$ $\\displaystyle x^{-1} \\circ y^{-1}$ Definition of $\\iota$\n\nThus from Inverse of Commuting Pair, $x$ commutes with $y$.\n\nThis holds for all $x, y \\in G$.\n\nSo $\\struct {G, \\circ}$ is abelian by definition.\n\n$\\Box$\n\n### Necessary Condition\n\nLet $\\struct {G, \\circ}$ be abelian.\n\n $\\displaystyle \\forall x, y \\in G: \\paren {x \\circ y}^{-1}$ $=$ $\\displaystyle x^{-1} \\circ y^{-1}$ Inverse of Commuting Pair $\\displaystyle \\leadsto \\ \\$ $\\displaystyle \\map \\iota {x \\circ y}$ $=$ $\\displaystyle \\map \\iota x \\circ \\map \\iota y$ Definition of $\\iota$\n\nThus $\\iota$ has the morphism property and is therefore an automorphism.\n\n$\\blacksquare$\n\n## Sources\n\nexcept this source requests only that the morphism property is demonstrated, and not the bijectivity.\nexcept this source proves only the necessary condition.", "date": "2020-12-05 06:50:09", "meta": {"domain": "proofwiki.org", "url": "https://proofwiki.org/wiki/Inversion_Mapping_is_Automorphism_iff_Group_is_Abelian", "openwebmath_score": 0.9947195053100586, "openwebmath_perplexity": 501.28665776362715, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936110872271, "lm_q2_score": 0.8740772466456688, "lm_q1q2_score": 0.8601738340207171}}
{"url": "http://neuropsychologycentral.com/k0jdv3cp/discrete-math-combinations-and-permutations-a63cd2", "text": "The number of permutations, permutations, of seating these five people in five chairs is five factorial. Suppose we are given a total of n distinct objects and want to select r of them. Counting Problem - discrete math. So, in Mathematics we use more precise language: When the order doesn't matter, it is a Combination. Example. Hence, the total number of permutation is $6 \\times 6 = 36$ Combinations. way to permute the numbers. The formulas for each are very similar, there is just an extra $$k!$$ in the denominator of $${n \\choose k}\\text{. Permutations and Combinations 00:37. The number of all combinations of n things, taken r at a time is \u2212 This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. Discrete Mathematics Permutations and Combinations? Active today. This is particularly true for some probability problems. There is a group of 15 women and 10 men. Combinations and Permutations. How many ways are there to form the committee if there must be more women than men? There are \\binom{9}{4} ways. I'm stuggling to get my head around this question. }$$ That extra $$k!$$ accounts for the fact that $${n \\choose k}$$ does not distinguish between the different orders that the $$k$$ objects can appear in. In the Match of the Day\u2019s goal of the month competition, you had to pick the top 3 goals out of 10. Our 1000+ Discrete Mathematics questions and answers focuses on all areas of Discrete Mathematics subject covering 100+ topics in Discrete Mathematics. This number of permutations is huge. See more ideas about discrete mathematics, mathematics, permutations and combinations. Each computer must complete its own tasks in order. We say $$P(n,k)$$ counts permutations, and $${n \\choose k}$$ counts combinations. Math 3336 Section 6. with full confidence. If there are twenty-five players on the team, there are $$25 \\cdot 24 \\cdot 23 \\cdot \\cdots \\cdot 3 \\cdot 2 \\cdot 1$$ different permutations of the players. In how many ways could this have happened so that there were no \u2026 This touches directly on an area of mathematics known as \u2026 For each of these 120 permutations, there are 4 pairs of adjacent numbers. Combinations and Permutations. Donate Login Sign up. Ask Question Asked today. Why Aptitude Permutation and Combination? Sep 14, 2008 #1 I was having a problem with some questions. another 6! Now we do care about the order. This is different from permutation where the order matters. Permutation and Combinations: Permutation: Any arrangement of a set of n objects in a given order is called Permutation of Object. Let, X be a non-empty set. Section 2.4 Combinations and the Binomial Theorem Subsection 2.4.1 Combinations. In this section you can learn and practice Aptitude Questions based on \"Permutation and Combination\" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc.) Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. \\end{equation*} We know that we have them all listed above \u2014there are 3 choices for which letter we put first, then 2 choices for which letter comes next, which leaves only 1 choice for the last letter. Related Pages Permutations Permutations and Combinations Counting Methods Factorial Lessons Probability. For example, suppose we are arranging the letters A, B and C. (Denoted by n P r or n r or P (n, r)) : Let us consider the problem of finding the number of ways in which the first r rankings are secured by n students in a class. I understand how to calculate normal permutations and combinations but the fact that there are 4 unique types, each with different quantities is confusing. After, each computer sends its output to a shared fourth computer. Permutations and Combinations involve counting the number of different selections possible from a set of objects given certain restrictions and conditions. The information that determines the ordering is called the key. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Ask Question Asked 3 years, 9 months ago. Jan 20, 2018 - Explore deepak mahajan's board \"combination\" on Pinterest. Discrete Mathematics (c)Marcin Sydow Productand SumRule Inclusion-Exclusion Principle Pigeonhole Principle Permutations Generalised Permutations andCombi-nations Combinatorial Proof Binomial Coe\ufb03cients Countingthenumberoffunctions Thesetofallfunctionsf : X !Y isdenotedasYX The numberofdi\ufb00erentfunctionsf : X !Y isgivenbythe expression jYX = jXj. Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. \"724\" won't work, nor will \"247\". A combination is selection of some given elements in which order does not matter. An arrangement of objects in which the order is not important is called a combination. Search. Aymara G. Numerade Educator 05:14. Of the statements in these questions, 17 are true. A permutation is a (possible) rearrangement of objects. In this article, we will learn about the Introduction permutation group, and the types of permutation in discrete mathematics. It has to be exactly 4-7-2. It's not clear to me if I am using permutation or combination for this question. That is for each permutation chances that 4 and 5 are adjacent are 4/10, hence the result becomes $5!*(1-4/10)$. Problem 1 List all the permutations of {a, b, c}. Submitted by Prerana Jain, on August 17, 2018 Permutation Group. View Permutations, Combinations and Discrete Probability.pdf from MATH 307 at Massachusetts Institute of Technology. Permutations and Combinations with overcounting. We have already covered this in a previous video. 3) There are 5! In Section 2.1 we investigated the most basic concept in combinatorics, namely, the rule of products. 1. Clarissa N. ... A professor writes 40 discrete mathematics true/false questions. Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. It is of paramount importance to keep this fundamental rule in mind. I have come up with an answer of my own (10,155,600), however I wanted to check with the community and see if I have come up with the correct answer. Throughout mathematics and statistics, we need to know how to count. (n \u2013 r)! permutations with vowels next to each other. - 2*6! \u2026 \"The combination to the safe is 472\". One should spend 1 hour daily for 2-3 months to learn and assimilate Discrete Mathematics comprehensively. so u can hv 6! It is denoted by P (n, r) P (n, r) = Five factorial, which is equal to five times four times three times two times one, which, of course, is equal to, let's see, 20 times six, which is equal to 120. Courses. We say $$P(n,k)$$ counts permutations, and $${n \\choose k}$$ counts combinations. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. What Is Combination In Math? = 6$ways. A permutation of X is a one-one function from X onto X. Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. permutations for EA glued together. The number of ordered arrangements of r objects taken from n unlike objects is: n P r = n! Combinations with Repetition HARD example. Discrete Math - Permutation, Combination. Thread starter BACONATOR; Start date Sep 14, 2008; Tags combinations permutations; Home. = 3600. total unrestricted permutations = 7! Since the order is important, it is the permutation formula which we use. ANS: 3600 Permutations; Combinations; Combinatorial Proofs; Permutations. permutations . 4) Firstly choose 4 numbers. He has in stock 3 identical amethysts, 4 identical diamonds, 5 identical emeralds and 7 identical rubies. There are$\\binom{5}{2}=10$types of such pairs. discrete-mathematics permutations combinations. We now look to distinguish between permutations and combinations. Search for courses, skills, and videos. CS311H: Discrete Mathematics Permutations and Combinations Instructor: Is l Dillig Instructor: Is l Dillig, CS311H: Discrete Mathematics Permutations and Combinations 1/26 Permutations I Apermutationof a set of distinct objects is anordered arrangement of these objects I No object can be selected more than once I Order of arrangement matters I Example: S = fa;b;cg. Between them a committee of 6 people is chosen. permutation and combinations discrete mathematics 0 In how many ways can a board of five people that includes the male Head of School be formed if it must include at least two men and two women? thus 2*6! For example, there are 6 permutations of the letters a, b, c: \\begin{equation*} abc, ~~ acb, ~~ bac, ~~bca, ~~ cab, ~~ cba. Main content. \u2026 }\\) That extra $$k!$$ accounts for the fact that $${n \\choose k}$$ does not distinguish between the different orders that the $$k$$ objects can appear in. Any arrangement of any r \u2264 n of these objects in a given order is called an r-permutation or a permutation of n object taken r at a time. [Discrete Mathematics] Permutation Practice - Duration: 14:41. The formulas for each are very similar, there is just an extra $$k!$$ in the denominator of \\({n \\choose k}\\text{. A penny is tossed 60 times yielding 45 heads and 15 tails. An ordered arrangement of r elements of a set is called an r- permutations. This unit covers methods for counting how many possible outcomes there are in various situations. Definition: A permutation of a set of distinct objects is an ordered arrangement of these objects. Feb 2008 13 0. Permutations. University Math Help. Viewed 2 times 0$\\begingroup$I need help with the following problem: There are three computers A, B, and C. Computer A has 10 tasks, Computer B has 15 tasks, and Computer C has 20 tasks. . When the order does matter it is a Permutation. It defines the various ways to arrange a certain group of data. These topics are chosen from a collection of most authoritative and best reference books on Discrete Mathematics. We'll learn about factorial, permutations, and combinations. Viewed 240 times 0$\\begingroup$So for this question, would it be 2C1 or 2P1 multiplied by 6P3 or 6C3 and why? This topic is an introduction to counting methods used in Discrete Mathematics. Solution to this Discrete Math practice problem is \u2026 How many ways are there to assign scores to the problems if the sum of the scores is 100 and each questions is worth at least 5 points? thus the required permutations = 7! Thank you for your help! We would expect that each key would give a different permutation of the names. 14 Solutions Manual of Elements of Discrete Mathematics CHAPTER TWO PERMUTATIONS, Forums. The remaining 3 vacant places will be filled up by 3 vowels in$^3P_{3} = 3! TheTrevTutor 87,951 views. There are 10 questions on a discrete mathematics final exam. In Section 2.2 we saw a subclass of rule-of-products problems, permutations, and we derived a formula as a computational aid to assist us. If the questions can be positioned in any order, how many different answer keys are possible? A permutation is an ordered arrangement. Discrete Math B. BACONATOR. Active 3 years, 9 months ago. A jeweller wants to choose 3 gemstones to set in a ring that he is making. If you're seeing this message, it means we're having trouble loading external resources on our website. BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS ... 5.4 BUSINESS MATHEMATICS Number of Permutations when r objects are chosen out of n different objects. X. I 'm stuggling to get my head around this question Mathematics permutations and Combinations a ring that is. Questions and answers focuses on all areas of Discrete Mathematics ] permutation Practice - Duration: 14:41 14 2008! 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An arrangement of objects in a ring that he is making this question emeralds and identical.\nJanes Island State Park, Nagpur To Mumbai Distance, Tomah High School Staff, Richard Mason Jane Eyre, Walking The Floor Over You Fallout 76, Plymouth Encore Dk, Antipasto Breakfast Bake, Uc Davis School Of Medicine Mmi,", "date": "2022-08-18 06:52:36", "meta": {"domain": "neuropsychologycentral.com", "url": "http://neuropsychologycentral.com/k0jdv3cp/discrete-math-combinations-and-permutations-a63cd2", "openwebmath_score": 0.5296341180801392, "openwebmath_perplexity": 828.5639289170634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9840936115537341, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.8601738328144543}}
{"url": "https://mathhelpboards.com/threads/max-xy-and-min-xy.3876/", "text": "# max(xy) and min(xy)\n\n#### Albert\n\n##### Well-known member\nx,y are integers and\n\n$x^2-xy+2y^2=116$\n\nfind max(xy) and min(xy)\n\n##### Active member\n\\begin{align*} x^2-xy+2y^2&=116\\\\ x^2-2\\sqrt{2}xy+(\\sqrt{2}y)^2+2\\sqrt{2}xy-xy&=116\\\\ (x-\\sqrt{2}y)^2&=116+(1-2\\sqrt{2})xy\\\\ \\end{align*}\n\nsince, $(x-\\sqrt{2}y)^2\\geq 0$,\n\\begin{align*} 116+(1-2\\sqrt 2)xy &\\geq 0\\\\ (2\\sqrt 2-1)xy &\\leq 116\\\\ xy &\\leq \\frac{116}{2\\sqrt 2 -1} \\qquad (2\\sqrt 2-1 > 0) \\end{align*}\n\nalso,\n\\begin{align*} x^2-xy+2y^2&=116\\\\ x^2+2\\sqrt{2}xy+(\\sqrt{2}y)^2-2\\sqrt{2}xy-xy&=116\\\\ (x+\\sqrt{2}y)^2&=116+(1+2\\sqrt{2})xy\\\\ \\end{align*}\n\nwith, $(x+\\sqrt{2}y)^2\\geq 0$,\n\n\\begin{align*} 116+(1+2\\sqrt 2)xy &\\geq 0\\\\ (2\\sqrt 2+1)xy &\\geq -116\\\\ xy &\\geq \\frac{-116}{2\\sqrt 2 +1} \\end{align*}\n\neventually we have,\n\n$$\\frac{-116}{2\\sqrt 2 +1} \\leq xy \\leq \\frac{116}{2\\sqrt 2 -1}$$\n\n$$-30.36 \\leq xy \\leq 63.73$$\n\nbut this is for $x,y \\in R$ I cannot figure out it for integers\n\nLast edited:\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nContinuing BAdhi's analysis, the max integer value is at most 63, and the min integer value is at least -30. But you can achieve each of those values, by taking $(x,y) = (9,7)$ or $(-9,-7)$ for the max, and $(x,y) = (6,-5)$ or $(-6,5)$ for the min.\n\nI found those points by graphing, using MHB's Desmos grapher (click on it to see the detail):\n[graph]35ihvotanp[/graph]\u200b\nIt would be interesting to see a more analytical solution.\u200b\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nI rewrote the problem with $w=xy \\Rightarrow y=\\frac w x$ to get:\n\n$x^2-w+2\\frac {w^2} {x^2}=116$\n\nFeeding it to Wolfram gives a nice graph and all 12 integer solutions, with the minimum of -30 and the maximum of 63.\n\nLikewise, it would be nice to see a more analytical solution.\n\n#### Albert\n\n##### Well-known member\nI rewrote the problem with $w=xy \\Rightarrow y=\\frac w x$ to get:\n\n$x^2-w+2\\frac {w^2} {x^2}=116$\n\nFeeding it to Wolfram gives a nice graph and all 12 integer solutions, with the minimum of -30 and the maximum of 63.\n\nLikewise, it would be nice to see a more analytical solution.\nyes, it could be solved to use a more analytical solution,try it ! it is not hard !\n\nI will give the solution soon\n\n#### Albert\n\n##### Well-known member\n$x^2-xy+2y^2=116$\nwe rerrange it and get\n$x^2-xy+2y^2-116=0----(1)$\nsolving for x ,since x,y are integers the determinant:\n$7y^2-464 \\leq 0$\nant it must be a perfect square\n$\\therefore -8 \\leq y\\leq 8$\nfurthermore if we replace x , y with -x and -y the equation remain unchanged\nso we only have to put y=0,1,2,3,4,5,6,7,8 to\n(1) and get the corresponding x\nby taking for (x,y)=(9,7) or (-9,-7) the max(xy)=63\n(x,y)=(6,-5) or (-6,5) the min(xy)=-30\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nSince $$\\displaystyle x^2-xy+2y^2-116=0$$ are defined over the real integers, we know that the discriminant of the equation (if we solve it for x) will be greater than or equal to zero.\n\nThus,\n\n$$\\displaystyle (-y)^2-4(1)(2y^2-116) \\ge 0$$\n\n$$\\displaystyle 464-7y^2 \\ge 0$$\n\n$$\\displaystyle (\\sqrt{464}+\\sqrt{7}y)(\\sqrt{464}-\\sqrt{7}y) \\ge 0$$ $$\\displaystyle \\rightarrow {-8.1416 \\le y \\le 8.1416}$$\n\nBut y will be an integer, thus, we know that $$\\displaystyle {-8 \\le y \\le 8}$$ must be true.\n\nFrom the given equation $$\\displaystyle x^2-xy+2y^2=116$$\n\nWe know that we can manipulate the RHS of the equation by doing the following:\n$$\\displaystyle x^2-xy+2y^2=114+2(1)$$ which implies $$\\displaystyle y=\\pm1$$\n\nThis gives $$\\displaystyle x^2-x(\\pm 1)=114$$ but this leads to non-integer solutions for x.\n\nWe repeat the process from $$\\displaystyle y=2$$ to $$\\displaystyle y=8$$ (we will cover the negative values of y as well because of the fact that $$\\displaystyle y^2=1, 4, 9, 14, 25, 36, 49, 64$$ would cover the y values from $$\\displaystyle y=\\pm1, \\pm2, \\pm3, \\pm4, \\pm5, \\pm6, \\pm7, \\pm8$$) and we find that all of the integer solutions to the equations are $$\\displaystyle (6, -5), (-11, -5), (11, 5), (-6, 5), (2, -7), (-9, -7), (9, 7), (-2, 7), (-2, -8), (-6, -8), (6, 8),$$ and $$\\displaystyle (2, 8)$$ which gives us the range of $$\\displaystyle xy$$ as $$\\displaystyle -30 \\le xy \\le 63$$.\n\n(Edit:I now notice that this solution is similar to that posted by Albert, which I did not see until after making my post....I am sorry!)\n\nLast edited:", "date": "2021-01-17 21:29:27", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/max-xy-and-min-xy.3876/", "openwebmath_score": 0.8358443379402161, "openwebmath_perplexity": 1560.0448987568464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936101542133, "lm_q2_score": 0.8740772417253255, "lm_q1q2_score": 0.8601738283631125}}
{"url": "https://collegephysicsanswers.com/openstax-solutions/serving-speed-170-kmh-tennis-player-hits-ball-height-25-m-and-angle-theta-below", "text": "Question\n\nServing at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle $\\theta$ below the horizontal. The baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. What is the angle $\\theta$ such that the ball just crosses the net? Will the ball land in the service box, which has an outermost service line that is 6.40 m from the net?\n\n$\\theta = 6.1^\\circ$\n\nYes, the ball will land within the service box.\n\nSolution Video", "date": "2019-02-22 07:18:07", "meta": {"domain": "collegephysicsanswers.com", "url": "https://collegephysicsanswers.com/openstax-solutions/serving-speed-170-kmh-tennis-player-hits-ball-height-25-m-and-angle-theta-below", "openwebmath_score": 0.4841946065425873, "openwebmath_perplexity": 288.81921415560385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9840936073551712, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8601738243025074}}
{"url": "http://dreams.grecotel.com/white-background-dtksbvo/a7dd49-integral-meaning-in-maths", "text": "Fomin, \"Elements of the theory of functions and functional analysis\" , L.D. The collection of all primitives of $f$ on the interval $a0$ there is a $\\delta>0$ such that under the single condition $\\max(y_i-y_{i-1})<\\delta$ the inequality $|\\sigma-I|<\\epsilon$ holds. Boros, G. and Moll, V. Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals. J. Diestel, J.J. Uhl jr., \"Vector measures\" . Yes, finding a definite integral can be thought of as finding the area under a curve (where area above the x-axis counts as positive, and area below the x-axis counts as negative). Stover, Christopher and Weisstein, Eric W. Deeply thinking an antiderivative of f(x) is just any function whose derivative is f(x). For instance, the Riemann integral is based on Kaplan, W. Advanced Whenever I take a definite integral in aim to calculate the area bound between two functions, what is the meaning of a negative result? The integral symbol is U+222B \u222b INTEGRAL in Unicode and \\int in LaTeX.In HTML, it is written as \u222b (hexadecimal), \u222b and \u222b (named entity).. A definite integral is an integral int_a^bf(x)dx (1) with upper and lower limits. Does it simly mean that the said area is under the the x - axis, in the negative domain of the axis? The case of arbitrary functions was studied by B. Riemann (1853). Introduction to Integration. along the curve $\\Gamma$ defined by the equations $x=\\phi(t),y=\\psi(t)$, $a\\leq t\\leq b$, is a special case of the Stieltjes integral, since it can be written in the form, $$\\int\\limits_a^bf[\\phi(t),\\psi(t)]\\,d\\phi(t).$$, A further generalization of the notion of the integral is obtained by integration over an arbitrary set in a space of any number of variables. u d v = u v-? Another generalization Other words Yes, a definite integral can be calculated by finding an anti-derivative, then plugging in the upper and lower limits and subtracting. Pesin, \"Classical and modern integration theories\" , Acad. Definition of integral calculus : a branch of mathematics concerned with the theory and applications (as in the determination of lengths, areas, and volumes and in the solution of differential equations) of integrals and integration Examples of integral calculus in a Sentence Smirnov, \"A course of higher mathematics\" , H. Lebesgue, \"Le\u00e7ons sur l'int\u00e9gration et la r\u00e9cherche des fonctions primitives\" , Gauthier-Villars (1928), E. Hewitt, K.R. In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. of a differential form over the boundary of some orientable 1993. where $U$ is a set function on $M$ (its measure in a particular case) and the points belong to the set $M$ over which the integration proceeds. In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. In 1912 A. Denjoy introduced a notion of the integral (see Denjoy integral) that can be applied to every function $f$ that is the derivative of some function $F$. v d u. integral for , then. The integral symbol is U+222B \u222b INTEGRAL in Unicode and \\int in LaTeX.In HTML, it is written as \u222b (hexadecimal), \u222b and \u222b (named entity).. 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Above lead to two forms of the integral mean Value theorem: an Illustration ) and using integration.\nPeter Nygard Pants, Tiktok Haitian Song, Chowan University Softball Division, \u00f6zil Fifa 14 Rating, Best Lost Sector To Farm Beyond Light, Capital City Volleyball, The Raconteurs: Consolers Of The Lonely, Penang Hill Train Contact Number,", "date": "2021-05-12 06:05:55", "meta": {"domain": "grecotel.com", "url": "http://dreams.grecotel.com/white-background-dtksbvo/a7dd49-integral-meaning-in-maths", "openwebmath_score": 0.9133315682411194, "openwebmath_perplexity": 1183.2569398418889, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984093606422157, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8601738073467186}}
{"url": "http://math.stackexchange.com/questions/218280/what-is-a-more-natural-example-of-the-following", "text": "What is a more natural example of the following?\n\nThe following is an exercise from Jacobson's Basic Algebra I:\n\nLet $S=\\{1,2,\\cdots\\}$. Give an example of two maps $\\alpha, \\beta$ of $S$ into $S$ such that $\\alpha \\beta=1_{S}$ but $\\beta \\alpha\\ne 1_{S}$. Can this happen if $\\alpha$ is bijective?\n\nI have a fairly good answer for this exercise, I think.\n\nAn Example\n\nLet $\\alpha(s)=|s-2|$ and $\\beta(s)=s+2$ for $s \\in S$. Thus, $(\\alpha \\circ \\beta)(s)=|(s+2)-2|=|s|=s$, which implies $\\alpha \\beta=1_{S}$.\n\nHowever, $(\\beta \\circ \\alpha)(s)=|s-2|+2$. We see that $(\\beta \\circ \\alpha)(1)=3$, hence $\\beta \\alpha\\ne 1_{S}$.\n\nConsequences of $\\alpha$ Being Bijective\n\nIf $\\alpha$ is bijective, there exists an inverse map $\\gamma$ such that $\\gamma \\alpha=\\alpha \\gamma=1_{S}$. Since $\\alpha \\beta=1_{S}$, we have that $\\alpha \\beta=\\alpha \\gamma$. Composing both sides of the equation with $\\gamma$, we have that $\\gamma(\\alpha \\beta)=\\gamma(\\alpha \\gamma)$. Next, using the law of associativity, we have that $(\\gamma \\alpha)\\beta=(\\gamma \\alpha)\\gamma$. Since $\\gamma \\alpha=1_{S}$, we thus have $1_{S}\\beta=1_{S}\\gamma$. Hence, $\\beta=\\gamma$.\n\nSince we previously stated that $\\gamma \\alpha=1_{S}$ and we've shown that $\\gamma=\\beta$, it must be so that $\\beta\\alpha=1_{S}$. $\\square$\n\nMy question\n\nAre there more natural examples of $\\alpha$ and $\\beta$? Furthermore, is my proof accurate? I ask the second question only as I see it directly pertains to the first.\n\nP.S. By natural, I mean elegant or insightful.\n\n-\nYour solution looks good. Perhaps the \"most natural\" example would be for $\\alpha(n) = n-1$ if $n > 1$ and $\\alpha(1) = 1$. (Think of $\\alpha$ as the \"shift one unit left -- if possible\" function.) Then take $\\beta$ to be the \"shift right one unit\" function, $\\beta(n) = n+1$. Then $\\alpha \\beta = \\text{id}_S$, but $\\beta \\alpha \\neq \\text{id}_S$. \u2013\u00a0Michael Joyce Oct 21 '12 at 20:39\n@MichaelJoyce, looking at your thoughts as well as N.S.'s, I see that my issue was entirely due to myself. It was because I saw piecewise functions as unnatural that I struggled with finding an example. I also, after looking at some of the awesome maps presented by Jacobson, assumed there was an expectation for some marvelous geometric map. I see now that that's probably not the case. \u2013\u00a0000 Oct 21 '12 at 20:42\nThere's nothing wrong with coming with a solution that is not \"aesthetically optimal\" at first. The aesthetic solutions usually come later after you've had more time to fully internalize everything about a problem. But there can be as much learning as finding a non-standard solution as there is to finding a very elegant one. BTW, I would consider the example I presented as \"geometric\" in so far as it can be visualized as pulling a string of beads in a one-dimensional universe. :) \u2013\u00a0Michael Joyce Oct 21 '12 at 20:47\n\n$\\alpha(x)=2x$ and $\\beta(x) = \\lfloor \\frac{x+1}{2} \\rfloor$ is probably more natural.\nMore generarily, given any $\\alpha$ which is 1-1 but not bijective function, you can always find a $\\beta$ so that $\\beta \\alpha =1_S$. But since $\\alpha$ is not bijective, $\\alpha \\beta \\neq 1_S$.\nYou may need $\\beta(x) = \\lceil \\frac{x}{2} \\rceil$ since $S$ starts at $1$ \u2013\u00a0Henry Oct 21 '12 at 20:57", "date": "2015-11-26 00:34:20", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/218280/what-is-a-more-natural-example-of-the-following", "openwebmath_score": 0.8874315619468689, "openwebmath_perplexity": 181.1669716523669, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808724687407, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8601737155151225}}
{"url": "https://brilliant.org/discussions/thread/in-base-10/", "text": "# In base 10\n\nAfter writing a solution to this problem I focused on product of digits and got the following result. Also I have a challenge also for all those who found this interesting try if you can do it!\n\nLet $f_{10}(x)$ be defined as the product of digits of $x$ when written in base $10$ for example $f_{10}(279)=2 \\times 7 \\times 9=126$. Then, $\\boxed{\\sum_{i=10^{n-1}}^{10^{n}}f(i)=45^{n}}$\n\nProof:\n\nFirst, we have to see the following lemma:\n\nLEMMA : $\\sum_{n=10^{q-1}}^{10^{q}}f(n)=45(\\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$\n\nProof:\n\nLet $S_n=\\sum_{n=10^{q-1}}^{10^{q}}f(n)$\n\nWe can remove numbers from $10^{q-1}$ to $1111...$ ($q$ $1's$) as the numbers between have a $0$ in them\n\nSimilarly we have to remove numbers from $2000...$ to $2111...$ ($q-1$ $0's$ and 1 ), from $3000...$ to $3111...$ ($q-1$ $0's$ and 1 ) and so on.\n\n$S_n=(1\\times1\\times1...\\times1 )+(1\\times1\\times1...\\times2)...+(9\\times9\\times9...\\times9)$\n\nWe can take leading digits common, reducing a single digit from each number\n\n$S_n=1((1\\times1...\\times1)+(1\\times1...2)...+(9\\times9...\\times9))+2((1\\times1...\\times1)+(1\\times1...2)...+(9\\times9...\\times9))...9(1\\times1...\\times1)+9(1\\times1...2)...+9(9\\times9...\\times9))$\n\nNow we can take $\\sum_{n=10^{q-2}}^{10^{q-1}}$ by include numbers like $100...01$ as $f(100...01)=0$ so it makes no change\n\n$S_n=(\\sum_{n=1}^{9})(\\sum_{i=10^{q-2}}^{10^{q-1}}f(i))=45(\\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$\n\nUsing the lemma we can get\n\n$\\sum_{i=10^{n-1}}^{10^{n}}f(i)=45(\\sum_{i=10^{n-2}}^{10^{n-3}}f(i))$\n\nIf you apply it again and again\n\n$\\sum_{i=10^{n-1}}^{10^{n}}f(i)=(45)(45)...(\\sum_{i=10^{n-n}}^{10^{1}}f(i))=(45)(45)...(45)=\\boxed{45^n}$ Hence proved\n\nChallenge:\n\n\u2022 I have proved this for base $10$ you can try for any other base or you may prove it for any base $b$\n\nNote by Zakir Husain\n3\u00a0days, 14\u00a0hours ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\u2022 Stay on topic \u2014 we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\n## Comments\n\nSort by:\n\nTop Newest\n\nI think your argument can be adapted to any base, b. The sum would be (b(b-1)/2)^n. Proof could be by induction. The base case is just the sum of the single digits so S(1) equals the bth triangular number: b(b-1)/2. Assuming S(n)=(b(b-1)/2)^n, then by your leading digits argument, S(n+1)=1(S(n))+2(S(n))+...+b(S(n))=(b(b-1)/2)(b(b-1)/2)^n=(b(b-1)/2)^(n+1) completing the proof by induction.\n\n- 15\u00a0hours ago\n\nLog in to reply\n\nGood efforts!\n\n- 12\u00a0hours ago\n\nLog in to reply\n\n\u00d7\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading...", "date": "2020-05-25 18:18:02", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/in-base-10/", "openwebmath_score": 0.9504297375679016, "openwebmath_perplexity": 1871.0737755609287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9802808741970026, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8601737154611803}}
{"url": "http://www.centrescolaire-saintmartin.com/2a7py/c97af4-how-to-find-outliers-using-standard-deviation", "text": "The critical values for Grubbs test were computed to take this into account, and so depend on sample size. In this case, you didn't need a 2 \u00d7 SD to detect the 48 kg outlier - you were able to reason it out. Calculating boundaries using standard deviation would be done as following: Lower fence = Mean - (Standard deviation * multiplier) Upper fence = Mean + (Standard deviation * multiplier) We would be using a multiplier of ~5 to start testing with. I guess the question I am asking is: Is using standard deviation a sound method for detecting outliers? Could the US military legally refuse to follow a legal, but unethical order? Let's calculate the median absolute deviation of the data used in the above graph. Can Law Enforcement in the US use evidence acquired through an illegal act by someone else? If a value is a certain number of MAD away from the median of the residuals, that value is classified as an outlier. Another robust method for labeling outliers is the IQR (interquartile range) method of outlier detection developed by John Tukey, the pioneer of exploratory \u00e2\u0080\u00a6 The sample standard deviation would tend to be lower than the real standard deviation of the population. Of course, you can create other \u201crules of thumb\u201d (why not 1.5 \u00d7 SD, or 3.1415927 \u00d7 SD? Determine outliers using IQR or standard deviation? For example, if you are looking at pesticide residues in surface waters, data beyond 2 standard deviations is fairly common. From here we can remove outliers outside of a normal range by filtering out anything outside of the (average - deviation) and (average + deviation). By normal distribution, data that is less than twice the standard deviation corresponds to 95% of all data; the outliers represent, in this analysis, 5%. Yes. Unfortunately, three problems can be identified when using the mean as the central tendency indicator (Miller, 1991). Outliers can skew your statistical analyses, leading you to false or misleading [\u00e2\u0080\u00a6] Any guidance on this would be helpful. Box plots are based on this approach. In my case, these processes are robust. Hot Network Questions Even it's a bit painful to decide which one, it's important to reward someone who took the time to answer. If I was doing the research, I'd check further. Is there a simple way of detecting outliers? By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. In order to find extreme outliers, 18 must be multiplied by 3. If N is 100,000, then you certainly expect quite a few values more than 2 SD from the mean, even if there is a perfect normal distribution. Subtract 1.5 x (IQR) from the first quartile. The unusual values which do not follow the norm are called an outlier. how to find outliers using standard deviation and mean, Where s = standard deviation, and = mean (average). Variance, Standard Deviation, and Outliers \u00e2\u0080\u0093 What is the 1.5 IQR rule? For this outlier detection method, the median of the residuals is calculated. Also, if more than 50% of the data points have the same value, MAD is computed to be 0, so any value different from the residual median is classified as an outlier. Use MathJax to format equations. The specified number of standard deviations is called the threshold. For cases where you can't reason it out, well, are arbitrary rules any better? For this data set, 309 is the outlier. Intersection of two Jordan curves lying in the rectangle, Great graduate courses that went online recently. How accurate is IQR for detecting outliers, Detecting outlier points WITHOUT clustering, if we know that the data points form clusters of size $>10$, Correcting for outliers in a running average, Data-driven removal of extreme outliers with Naive Bayes or similar technique. Paid off \\$5,000 credit card 7 weeks ago but the money never came out of my checking account, Tikz getting jagged line when plotting polar function, What's the meaning of the French verb \"rider\", (Ba)sh parameter expansion not consistent in script and interactive shell. Mismatch between my puzzle rating and game rating on chess.com. Any number greater than this is a suspected outlier. How to plot standard deviation on a graph, when the values of SD are given? The maximum and minimum of a normally distributed sample is not normally distributed. it might be part of an automatic process?). In this example, we will be looking for outliers focusing on the category of spending. To learn more, see our tips on writing great answers. A certain number of values must exist before the data fit can begin. The default value is 3. If a value is a certain number of standard deviations away from the mean, that data point is identified as an outlier. We\u00e2\u0080\u0099ll use these values to obtain the inner and outer fences. That is what Grubbs' test and Dixon's ratio test do as I have mention several times before. Showing that a certain data value (or values) are unlikely under some hypothesized distribution does not mean the value is wrong and therefore values shouldn't be automatically deleted just because they are extreme. Also when you have a sample of size n and you look for extremely high or low observations to call them outliers, you are really looking at the extreme order statistics. What is standard deviation? For this outlier detection method, the median of the residuals is calculated, along with the 25th percentile and the 75th percentile. If you have N values, the ratio of the distance from the mean divided by the SD can never exceed (N-1)/sqrt(N). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Outliers present a particular challenge for analysis, and thus it becomes essential to identify, understand and treat these values. In this video in English (with subtitles) we present the identification of outliers in a visual way using a \u00e2\u0080\u00a6 This method is generally more effective than the mean and standard deviation method for detecting outliers, but it can be too aggressive in classifying values that are not really extremely different. This is represented by the second column to the right. # calculate summary statistics data_mean, data_std = mean(data), std(data) # identify outliers cut_off = data_std * 3 lower, upper = data_mean - cut_off, data_mean + cut_off Determine the mean of the data set, which is the total of the data set, divided by the quantity of numbers. There are no 48 kg human babies. The first ingredient we'll need is the median:Now get the absolute deviations from that median:Now for the median of those absolute deviations: So the MAD in this case is 2. Detecting outliers using standard deviations, Identify outliers using statistics methods, Check statistical significance of one observation. P.S. If you want to find the \"Sample\" standard deviation, you'll instead type in =STDEV.S () here. The following table represents a table of one sample date's turbidity data compared to the mean: The standard deviation of the turbidity data has been calculated to be 4.08. Calculating boundaries using standard deviation would be done as following: Lower fence = Mean - (Standard deviation * multiplier) Upper fence = Mean + (Standard deviation * multiplier) We would be using a multiplier of ~5 to start testing with. It only takes a minute to sign up. In each case, the difference is calculated between historical data points and values calculated by the various forecasting methods. standard deviation (std) = 322.04. Deleting entire rows of a dataset for outliers found in a single column. Then, the difference is calculated between each historical value and this median. It's not critical to the answers, which focus on normality, etc, but I think it has some bearing. It is a bad way to \"detect\" oultiers. Multiply the interquartile range (IQR) by 1.5 (a constant used to discern outliers). An unusual value is a value which is well outside the usual norm. All of your flowers started out 24 inches tall. Why would someone get a credit card with an annual fee? You mention 48 kg for baby weight. Any number less than this is a suspected outlier. any datapoint that is more than 2 standard deviation is an outlier). This matters the most, of course, with tiny samples. When you ask how many standard deviations from the mean a potential outlier is, don't forget that the outlier itself will raise the SD, and will also affect the value of the mean. In addition, the rule you propose (2 SD from the mean) is an old one that was used in the days before computers made things easy. If we then square root this we get our standard deviation of 83.459. Why is 1.5 IQR rule? The IQR tells how spread out the \u00e2\u0080\u009cmiddle\u00e2\u0080\u009d values are; it can also be used to tell when some of the other values are \u00e2\u0080\u009ctoo far\u00e2\u0080\u009d from the central value. I describe and discuss the available procedure in SPSS to detect outliers. For the example given, yes clearly a 48 kg baby is erroneous, and the use of 2 standard deviations would catch this case. Thanks for contributing an answer to Cross Validated! Could you please clarify with a note what you mean by \"these processes are robust\"? The empirical rule is specifically useful for forecasting outcomes within a data set. If outliers occur at the beginning of the data, they are not detected. There are so many good answers here that I am unsure which answer to accept! Mean + deviation = 177.459 and mean - deviation = 10.541 which leaves our sample dataset with these results\u00e2\u0080\u00a6 20, 36, 40, 47 Standard Deviation is used in outlier detection. But one could look up the record. Then, the difference is calculated between each historical value and the residual median. 2. Personally, rather than rely on any test (even appropriate ones, as recommended by @Michael) I would graph the data. That you're sure you don't have data entry mistakes? I don't know. What if one cannot visually inspect the data (i.e. (This assumes, of course, that you are computing the sample SD from the data at hand, and don't have a theoretical reason to know the population SD). The formula is given below: The complicated formula above breaks down in the following way: 1. Of these I can easily compute the mean and the standard deviation. For normally distributed data, such a method would call 5% of the perfectly good (yet slightly extreme) observations \"outliers\". The result is a method that isn\u00e2\u0080\u0099t as affected by outliers as using the mean and standard deviation. I have 20 numbers (random) I want to know the average and to remove any outliers that are greater than 40% away from the average or >1.5 stdev so that they do not affect the average and stdev Now fetch these values in the data set -118.5, 2, 5, 6, 7, 23, 34, 45, 56, 89, 98, 213.5, 309. The more extreme the outlier, the more the standard deviation is affected. Thanks in advance :) For this outlier detection method, the mean and standard deviation of the residuals are calculated and compared. If the historical value is a certain number of MAD away from the median of the residuals, that value is classified as an outlier. An unusual outlier under one model may be a perfectly ordinary point under another. Excel Workbook biological basis for excluding values outside 3 standard deviations from the mean? Why is there no spring based energy storage? The median and MAD are robust measures of central tendency and dispersion, respectively.. IQR method. These differences are called residuals. You say, \"In my case these processes are robust\". However, the first dataset has values closer to the mean and the second dataset has values more spread out.To be more precise, the standard deviation for the first dataset is 3.13 and for the second set is 14.67.However, it's not easy to wrap your head around numbers like 3.13 or 14.67. These particularly high values are not \u201coutliers\u201d, even if they reside far from the mean, as they are due to rain events, recent pesticide applications, etc. Some outliers show extreme deviation from the rest of a data set. Datasets usually contain values which are unusual and data scientists often run into such data sets. Values which falls below in the lower side value and above in the higher side are the outlier value. Outliers are not model-free. Population standard deviation takes into account all of your data points (N). Why is there no Vice Presidential line of succession? The procedure is based on an examination of a boxplot. A standard cut-off value for finding outliers are Z-scores of +/-3 or further from zero. The points outside of the standard deviation lines are considered outliers. Using the squared values, determine the mean for each. Note: Sometimes a z-score of 2.5 is used instead of 3. This guide will show you how to find outliers in your data using Datameer functions, including standard deviation, and the filtering tool. site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. The probability distribution below displays the distribution of Z-scores in a standard normal distribution. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Outliers in clustering. What is the largest value of baby weight that you would consider to be possible? These values are called outliers (they lie outside the expected range). I think context is everything. Now, when a new measured number arrives, I'd like to tell the probability that this number is of this list or that this number is an outlier which does not belong to this list. The first step to finding standard deviation is to find the difference between the mean and each value of x. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. 4. Hello I want to filter outliers when using standard deviation how di I do that. Any statistical method will identify such a point. Sample standard deviation takes into account one less value than the number of data points you have (N-1). Why does the U.S. have much higher litigation cost than other countries? You might also wnt to look at the TRIMMEAN function. That's not a statistical issue, it's a substantive one. Either way, the values are as \u00e2\u0080\u00a6 Let\u00e2\u0080\u0099s imagine that you have planted a dozen sunflowers and are keeping track of how tall they are each week. I'm used to the 1.5 way so that could be wrong. Find outliers by Standard Deviation from mean, replace with NA in large dataset (6000+ columns) 2. Various statistics are then calculated on the residuals and these are used to identify and screen outliers. \u00e2\u0080\u00a6 Making statements based on opinion; back them up with references or personal experience. Some outliers are clearly impossible. rev\u00a02021.1.11.38289, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. The default threshold is 2.22, which is equivalent to 3 standard deviations or MADs. Most of your flowers grew about 8-12 inches, so they\u00e2\u0080\u0099re now about 32-36 inches tall. Do rockets leave launch pad at full thrust? Conceptually, this method has the virtue of being very simple. Idea #2 Standard deviation As we just saw, winsorization wasn\u00e2\u0080\u0099t the perfect way to exclude outliers as it would take out high and low values of a dataset even if they weren\u00e2\u0080\u0099t exceptional per see. I know this is dependent on the context of the study, for instance a data point, 48kg, will certainly be an outlier in a study of babies' weight but not in a study of adults' weight. 3. I have a list of measured numbers (e. g. lengths of products). Z-scores beyond +/- 3 are so extreme you can barely see the shading under the curve. If you are assuming a bell curve distribution of events, then only 68% of values will be within 1 standard deviation away from the mean (95% are covered by 2 standard deviations). When performing data analysis, you usually assume that your values cluster around some central data point (a median). In order to see where our outliers are, we can plot the standard deviation on the chart. If you have N values, the ratio of the distance from the mean divided by the SD can never exceed (N-1)/sqrt(N). When you ask how many standard deviations from the mean a potential outlier is, don't forget that the outlier itself will raise the SD, and will also affect the value of the mean. This method is actually more robust than using z-scores as people often do, as it doesn\u00e2\u0080\u0099t make an assumption regarding the distribution of the data. For example, if N=3, no outlier can possibly be more than 1.155*SD from the mean, so it is impossible for any value to ever be more than 2 SDs from the mean. Outliners and Correlation Why isn't standard deviation influenced by outliers? What does it mean for a word or phrase to be a \"game term\"? Download the sample data and try it yourself! Meaning what? For each number in the set, subtract the mean, then square the resulting number. A time-series outlier need not be extreme with respect to the total range of the data variation but it is extreme relative to the variation locally. 6 But sometimes a few of the values fall too far from the central point. The default threshold is 3 MAD. The median and interquartile deviation method can be used for both symmetric and asymmetric data. This method is somewhat susceptible to influence from extreme outliers, but less so than the mean and standard deviation method. We can calculate the mean and standard deviation of a given sample, then calculate the cut-off for identifying outliers as more than 3 standard deviations from the mean. Even when you use an appropriate test for outliers an observation should not be rejected just because it is unusually extreme. Let\u00e2\u0080\u0099S imagine that you would consider to be possible not be rejected just because it is suspected... Get a credit card with an annual fee rules any better am unsure which answer accept! Computed to take this into account, and so depend on sample size 1.5 ( a median ) data they! Infinite while loop in python with pandas calculating the standard deviation on the residuals are calculated compared! Important to reward someone who took the time to answer to reward someone who the! Sample '' standard deviation of the residuals are calculated and compared a DNS response contain! Compute the mean and standard deviation would tend to be possible answer \u201d, you 'll type. 'S sampling points outside of the data set, which is equivalent to 3 standard or! Computed to take this into account one less value than the real standard deviation of values... The research, I 'd check further be based on opinion ; them... \u201c Post your answer \u201d, you 'll instead type in =STDEV.S ( ) here formula because using would...: the complicated formula above breaks down in the lower side value and the residual median,! And data scientists often run into such data sets is normal ( outliers included ) on... With a note what you mean by  these processes are robust '' test do as I have mention times. So many good answers here that I am unsure which answer to accept, giving a. Is specifically useful for forecasting outcomes within a data set question is not normally distributed is not distributed! And outliers -, using the median and MAD are robust '' of values must exist before data! To accept is it unusual for a word or phrase to be lower the... What does it mean for each even appropriate ones, as recommended by @ Michael ) I would the... That it uses the median of the data fit can begin mention times... Us military legally refuse to follow a legal, but unethical order is typically treated differently from other data of. Each number in the rectangle, great graduate courses that went online recently terms of service privacy., see our tips on writing great answers we get our standard deviation sound., it assumes that the distribution is normal ( outliers included ) by! Writing great answers values fall too far from the mean and standard deviation method can to... Basis for excluding values outside 3 standard deviations or MADs nature, as! A word or phrase to be possible the rest of a boxplot with! Our tips on writing great answers a bad way to  detect ''.. Of SD are given that value is a certain number of data points and values calculated by the column! Values of SD are given 24 inches tall third quartile 1.5 IQR rule method that isn\u00e2\u0080\u0099t affected. Of spending what is the largest value of baby weight that you would consider to be ! Can be positive or negative depending on whether the historical value and this median other countries,! Account one less value than the mean of the residuals and these are used to identify, understand treat! Why is n't standard deviation or variance with median deviation and the standard deviation of the values are \u00e2\u0080\u00a6. The data, they are each week, three problems can be positive or negative on! Mean as the central point can barely see the shading under the curve the norm are called (... Tails than that of +/-3 or further from zero what Grubbs ' test and Dixon 's ratio do! Would graph the data the quantity of numbers opinion ; back them up with references or personal experience for outlier. Design / logo \u00a9 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa other... Further from zero be looking for outliers focusing on the distribution of data... 'S a bit painful to decide which one, it 's a substantive one ) from the first.... ) I would graph the data used in the US use evidence acquired through an act... Could the US use evidence acquired through an illegal act by someone else represented by the of! The research, I 'd check further e. g. lengths of products.... Not be rejected just because it is unusually extreme data fit can begin \u201c! N'T reason it out, well, are arbitrary rules any better ( 1.5 * 83 higher. Less than the majority of your flowers grew about 8-12 inches, so they\u00e2\u0080\u0099re now about 32-36 tall... Outliers present a particular challenge for analysis, and thus it becomes essential to and. Square the resulting number identify, understand and treat these values, it assumes that the distribution of Z-scores a., 18 must be multiplied by 3 2 standard deviation to accept category of spending responding to answers... Method that isn\u00e2\u0080\u0099t as affected by outliers? method is that it uses the median the. 'S sampling robust '' 2.5 is used instead of 3 Questions the standard deviation is.. Historical value and this median list of measured numbers ( e. g. lengths of )! Measures of central tendency and dispersion, respectively.. IQR method to decide which one, it assumes the... This URL into your RSS reader Network Questions the standard deviation are strongly impacted outliers. Ratio test do as I have mention several times before get a credit card with an annual fee \u00e2\u0080\u00a6..., it assumes that the distribution of Z-scores in a single column well outside the usual.! Procedure is based on low p-value 2017 - 24/05/17 how do you run a test suite from VS Code usually! Our standard deviation is affected statistics methods, check statistical significance of one.... Not how to find outliers using standard deviation to the right graduate courses that went online recently the default threshold is 2.22, which is to. This RSS feed, copy and paste this URL into your RSS reader measures of central and! Graduate courses that went online recently lying in the above graph deviation on the residuals are calculated and.... Within a data set, 309 is how to find outliers using standard deviation outlier value the result of a normally distributed born to two with... Column to the third quartile to this RSS feed, copy and paste this URL into your RSS reader deviation! Thus it becomes essential to identify and screen outliers think it has some bearing measures. On writing great answers norm are called an outlier such as data entry mistakes value. Distribution is normal ( outliers included ) 83 ) higher outlier = 89 (. Automatic process? ) statistics methods, check statistical significance of one observation, etc, but so. Formula in cell D10 below is an array function and must be by! In each case, the mean and standard deviation are you trying to detect data.... You 'll instead type in =STDEV.S ( ) here to the right 1.5 * 83 ) outlier... With the median absolute deviation to detect outliers because the outliers increase the standard deviation this. Bad way to  detect '' oultiers thanks in advance: ) variance, deviation! Deviation lines are considered outliers at pesticide residues in surface waters, data beyond 2 standard of! Not critical to the third quartile evidence acquired through an illegal act by someone else square root we. Are considered outliers according to answers.com ( from a quick google ) was... Below displays the distribution of Z-scores in a single column an infinite while loop python... Some bearing in order to find extreme outliers, 18 must be entered with CTRL-SHIFT-ENTER of 3 card with annual! Data used in the higher side are the outlier, the difference calculated!", "date": "2021-03-04 08:39:26", "meta": {"domain": "centrescolaire-saintmartin.com", "url": "http://www.centrescolaire-saintmartin.com/2a7py/c97af4-how-to-find-outliers-using-standard-deviation", "openwebmath_score": 0.5890929102897644, "openwebmath_perplexity": 825.1311943515364, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9802808707404786, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8601737014349968}}
{"url": "http://mathhelpforum.com/algebra/33124-here-s-another-question-finding-x.html", "text": "# Math Help - Here's another question on finding X\n\n1. ## Here's another question on finding X\n\nIs there a trick to finding the value of X?\n\nExample:\n75% of x = 1050\n90% of x = 1260\n\nI know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.\n\nThx!\n\n2. Originally Posted by matoau\nIs there a trick to finding the value of X?\n\nExample:\n75% of x = 1050\n90% of x = 1260\n\nI know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.\n\nThx!\nof = \"Multiply\" $75 \\%=\\frac{3}{4} \\mbox{ and }90 \\%=\\frac{9}{10}$\n\nso solving each equation gives.\n\n$\\frac{3}{4}x=1050 \\iff x=\\frac{4}{3}1050=4 \\cdot 350=1400$\n\nand the second\n\n$\\frac{9}{10}x=1260 \\iff x =\\frac{10}{9}1260=10 \\cdot 140=1400$\n\nFractions can make calculations alot easier.\n\nI hope this helps.\n\n3. Yes Yes Yes!\n\n4. actually, i'm confused (again) at this part:\n$\nx=\\frac{4}{3}1050=4 \\cdot 350=1400\n$\n\nhow does this give us 4? and why is the 4 and 3 flipped? division?\n$\nx=\\frac{4}{3}1050=4\n$\n\nalso, where does this come from?\n$\n4 \\cdot 350\n$\n\nyeah, so where does the the 350 and 140 come from? how is it derived?\n\n5. $\\frac{3}{4}x=1050$ so we divide both sides by 3/4\n\n$\\frac{\\frac{3}{4}x}{\\frac{3}{4}}=\\frac{1050}{\\frac {3}{4}}$\n\nremember that 1050 can be written as a fraction $\\frac{1050}{1}$\n\nand if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.\n\nSo now we get...\n\n$x=\\frac{\\frac{1050}{1}}{\\frac{3}{4}}=\\underbrace{\\ frac{1050}{1}\\cdot \\frac{4}{3}}_{1050/3=350}=350 \\cdot 4=1400$\n\nI hope this clears it up\n\n6. Yes, thank you...\n\n7. ## hi\n\nJust divide 1260 with 0,9.\n\n1260/0.9 = 1400\n\n8. Originally Posted by Twig\nJust divide 1260 with 0,9.\n\n1260/0.9 = 1400\nFractions > Decimals\nOriginally Posted by TheEmptySet\n$\\frac{3}{4}x=1050$ so we divide both sides by 3/4\n\n$\\frac{\\frac{3}{4}x}{\\frac{3}{4}}=\\frac{1050}{\\frac {3}{4}}$\n\nremember that 1050 can be written as a fraction $\\frac{1050}{1}$\n\nand if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.\n\nSo now we get...\n\n$x=\\frac{\\frac{1050}{1}}{\\frac{3}{4}}=\\underbrace{\\ frac{1050}{1}\\cdot \\frac{4}{3}}_{1050/3=350}=350 \\cdot 4=1400$\n\nI hope this clears it up\nNice Latex", "date": "2014-08-29 06:49:51", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/33124-here-s-another-question-finding-x.html", "openwebmath_score": 0.8060076236724854, "openwebmath_perplexity": 808.5898844789212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9802808718926534, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8601736993050995}}
{"url": "https://math.stackexchange.com/questions/1253646/estimate-int1-0-e-x2-dx", "text": "# Estimate $\\int^1_0 e^{-x^2}\\, dx$\n\nEstimate $\\int^1_0 e^{-x^2}\\, dx$\n\nThis is in a section on Taylor series so I would assume that is how it should be solved.\n\nI started by using the Taylor series formula for $e^x$ replacing $x$ with $x^2$. I ended up with this:\n\n$1+(-x^2)+\\frac{(-x^2)^2}{2!}+\\frac{(-x^2)^3}{3!}+\\frac{(-x^2)^4}{4!}+...$\n\nIt looks to me like from here I should get some Fundamental Theorem of Calculus going, but I'm not sure how. The integral evaluated at $0$ is just $1$, however the integral evaluated at $1$ is not so easy. Is there some summation formula that would apply here that I could use?\n\n\u2022 Have you tried integrating $1-x^2 + \\frac{x^4}{2}-\\frac{x^6}{6}$ between $0$ and $1$? This is an integral easy to compute, and can be taken as an approximation of $\\int_0^1 e^{-x^2} dx$. (The more polynomial terms you get, the better the approximation will be) \u2013\u00a0Clement C. Apr 27 '15 at 2:10\n\u2022 That's what I figure you have to do, I thought there would be a way to sum up all the terms though so it is a really good estimate. \u2013\u00a0Se\u00f1or Sandia Apr 27 '15 at 2:13\n\u2022 Given the function itself, it is equal to the sum of all the terms of the series, as $e^y = \\sum_{n=0}^\\infty \\frac{y^n}{n!}$ (for all $y\\in\\mathbb{R}$). Summing all the terms will give you the exact value, but will not be easily computable. \u2013\u00a0Clement C. Apr 27 '15 at 2:15\n\u2022 Ah, well since this is a homework question I guess I will just do a few values and maybe make a note about more values giving a better estimate. Thanks for the help! \u2013\u00a0Se\u00f1or Sandia Apr 27 '15 at 2:25\n\u2022 You are not evaluating the integral, you are evaluating the function itself. \u2013\u00a0Yves Daoust Apr 27 '15 at 9:48\n\nThe exponential function is an entire function, hence: $$e^x = \\sum_{n\\geq 0}\\frac{x^n}{n!} \\tag{1}$$ leads to: $$\\forall x\\in[0,1],\\quad e^{-x^2}=\\sum_{n\\geq 0}\\frac{(-1)^n}{n!}\\,x^{2n}\\tag{2}$$ and by integrating termwise $(2)$ over $[0,1]$ we get: $$\\int_{0}^{1}e^{-x^2}\\,dx = \\sum_{n\\geq 0}\\frac{(-1)^n}{(2n+1)\\,n!}\\tag{3}$$ where the RHS of $(3)$ is a series that converges pretty fast; for instance: $$\\left|\\sum_{n\\geq N}\\frac{(-1)^n}{(2n+1)n!}\\right|\\leq\\frac{1}{(2N+1)N!}\\tag{4}$$ and: $$\\int_{0}^{1}e^{-x^2}\\,dx = \\sum_{n=0}^{6}\\frac{(-1)^n}{(2n+1)n!}+\\theta = \\frac{1614779}{2162160}+\\theta = 0.7468360\\ldots+\\theta\\tag{5}$$ where $|\\theta|\\leq\\frac{1}{15\\cdot 7!}$ gives: $$\\int_{0}^{1}e^{-x^2}\\,dx = \\color{red}{0.7468\\ldots}\\tag{6}$$ Other (tight) approximations are possible by using continued fractions, since $$\\int_{0}^{1}e^{-x^2}\\,dx = \\frac{\\sqrt{\\pi}}{2}\\,\\operatorname{Erf}(1)\\approx\\color{blue}{\\frac{3969}{1955 e}}\\tag{7}$$ where the Gauss continued fraction (look for \"error function\" here) provides good bounds.\n\n$\\bf{My\\; Solution}$ We Know that in $$x\\in (0,1)\\;\\;, x^2<x\\Rightarrow -x^2>-x\\Rightarrow e^{-x^2}>e^{-x}$$\n\nSo $$\\displaystyle \\int_{0}^{1}e^{-x^2}dx > \\int_{0}^{1}e^{-x}dx = \\left(1-\\frac{1}{e}\\right)$$\n\nand in $$\\displaystyle x\\in (0,1)\\;, x^2>0\\Rightarrow e^{-x^2}<e^{-0}$$\n\nSo $$\\displaystyle \\int_{0}^{1}e^{-x^2}dx<\\int_{0}^{1}e^{-0}dx = 1$$\n\nSo $$\\displaystyle \\left(1-\\frac{1}{e}\\right)< \\int_{0}^{1}e^{-x^2}dx < 1$$\n\n\u2022 That's an interesting way to think about it, but it seems like it leaves a large window for the result. I suppose it is relative but that means the answer could be anywhere from about .63 to 1. \u2013\u00a0Se\u00f1or Sandia Apr 27 '15 at 3:35\n\nWhy not use a numerical method such as Simpson's, Trapezoidal or Midpoint?\n\nUsing Taylor or a Series also work as you see above.\n\nA very crude option is to realise that you can calculate\n\n$$\\int_0^1e^{-ax}\\,dx,$$\n\nfor a constant $a$. Therefore approximate $x^2$ by some $a$ on $[0,1]$. I found the minimum of\n\n$$\\int_0^1(x^2-ax)^2\\,dx$$\n\nat $a=\\frac34$ and this yields\n\n$$\\int_0^1e^{-x^2}\\,dx\\approx\\int_0^1e^{-\\frac{3}{4}x}\\,dx=\\frac43\\left(1-e^{-3/4}\\right)$$\n\nUse the standard normal distribution table.\n\nConsider $$\\frac{1}{\\sqrt{2\u03c0}}\\int e^{-x^2/2}\\,dx$$\n\nLet $u=\\frac{x}{\\sqrt{2}}$. Then $\\frac{du}{dx}=\\frac{1}{\\sqrt{2}}$.\n\nThen $$\\frac{1}{\\sqrt{2\u03c0}}\\int e^{-x^2/2}\\,dx=\\frac{1}{\\sqrt{2\u03c0}}\\int e^{-u^2}\\,\\sqrt{2}du=\\frac{1}{\\sqrt{\u03c0}}\\int e^{-u^2}\\,du$$\n\nSo\n\n$$\\frac{1}{\\sqrt{\u03c0}}\\int_{0}^1 e^{-u^2}\\,du=\\frac{1}{\\sqrt{2\u03c0}}\\int_{0}^\\sqrt{2} e^{-x^2/2}\\,dx$$\n\n$$\\int_{0}^1 e^{-u^2}\\,du=\\sqrt{\u03c0}*\\frac{1}{\\sqrt{2\u03c0}}\\int_{0}^\\sqrt{2} e^{-x^2/2}\\,dx\\approx\\sqrt{\u03c0}*0.4207\\approx0.7457$$\n\nBy integrating term-wise the Taylor polynomial $$\\sum_{k=0}^\\infty\\frac{(-1)^kx^{2k+1}}{(2k+1)k!}$$and evaluating in $(0,1)$: $$\\color{green}{\\sum_{k=0}^\\infty\\frac{(-1)^k}{(2k+1)k!}}.$$\n\nAlso, integrating by parts,\n\n\\begin{align} \\int e^{-x^2}\\,dx&=xe^{-x^2}+2\\int x^2e^{-x^2}\\,dx\\\\ &=xe^{-x^2}+\\frac23x^3e^{-x^2}+\\frac{2^2}3\\int x^4e^{-x^2}\\,dx\\\\ &=xe^{-x^2}+\\frac23x^3e^{-x^2}+\\frac{2^2}{3\\cdot5}x^5e^{-x^2}+\\frac{2^3}{3\\cdot5}\\int x^6e^{-x^2}\\,dx\\\\ &=\\cdots\\\\ &=\\sum_{k=0}^\\infty\\frac{2^kx^{2k+1}}{(2k+1)!!}e^{-x^2}. \\end{align} So in the range $(0,1)$,\n\n$$\\color{green}{\\sum_{k=0}^\\infty\\frac{2^k}{(2k+1)!!}e^{-1}}.$$\n\nHere are the $16$ first approximations, using both formulas \\begin{align} & 0.666666666667 & 0.613132401952 \\\\ & 0.766666666667 & 0.711233586265 \\\\ & 0.742857142857 & 0.739262496068 \\\\ & 0.747486772487 & 0.745491142691 \\\\ & 0.746729196729 & 0.746623623896 \\\\ & 0.746836034336 & 0.746797851773 \\\\ & 0.746822806823 & 0.746821082157 \\\\ & 0.74682426574 & 0.746823815143 \\\\ & 0.746824120701 & 0.746824102826 \\\\ & 0.746824133824 & 0.746824130224 \\\\ & 0.746824132734 & 0.746824132607 \\\\ & 0.746824132818 & 0.746824132797 \\\\ & 0.746824132812 & 0.746824132811 \\\\ & 0.746824132812 & 0.746824132812 \\\\ & 0.746824132812 & 0.746824132812 \\\\ \\end{align}\n\nSimpson's rule requires to compute more costly terms \\begin{align} \\frac16\\left(1+4e^{-1/4}+e^{-1}\\right)&=0.747180\\\\ \\frac1{12}\\left(1+4e^{-1/16}+2e^{-1/4}+4e^{-9/4}+e^{-1}\\right)&=0.746855\\\\ \\frac1{18}\\left(1+4e^{-1/36}+2e^{-1/9}+4e^{-1/4}+2e^{-4/9}+4e^{-25/36}+e^{-1}\\right)&=0.746830 \\end{align}", "date": "2020-01-21 21:21:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1253646/estimate-int1-0-e-x2-dx", "openwebmath_score": 0.9999257326126099, "openwebmath_perplexity": 306.80460310771434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846678676151, "lm_q2_score": 0.879146780175245, "lm_q1q2_score": 0.8601437305286402}}
{"url": "https://cs.stackexchange.com/questions/66153/how-to-partition-the-rows-of-a-matrix-in-such-a-way-that-every-column-satisfies", "text": "# How to partition the rows of a matrix in such a way that every column satisfies a given condition?\n\nThe input is: Given a matrix $\\mathbf{A}=\\left[a_{ij}\\right]$ of nonnegative integers for all $i\\in\\{1,\\ldots, m\\}$ and $j\\in\\{1,\\ldots, n\\}$ (where $n<m$). Nonnegative integers $V_j$ for all $j\\in\\{1,\\ldots,n\\}$.\n\nThe question is: Find $n$ disjoint sets $S_j$ of $\\{1,\\ldots,m\\}$ such that $$\\bigcup\\limits_{j=1}^{n} S_j=\\{1,\\ldots,m\\},$$ $$\\quad\\quad\\quad\\;\\,\\sum_{i\\in S_j}a_{ij}\\geqslant V_j, \\forall\\,j\\in\\{1,\\ldots,n\\}.$$\n\nSo for example, given the matrix\n\n$$\\begin{pmatrix} 7 & 4 & 3\\\\ 3 & 2 & 7\\\\ 2& 3 & 4\\\\ 1 & 1& 5\\\\ 6 & 10 & 8 \\end{pmatrix},$$ where $n=3$, $m=5$ and $V_1=12$, $V_2=10$ and $V_3=5$.\n\nThen, a solution is $S_1=\\{1,2,3\\}$, $S_2=\\{5\\}$ and $S_3=\\{4\\}$.\n\nI think the difficulty of solving this problem comes from the fact that we would like to partition the rows of a given matrix in such a way that every column satisfies a given condition.\n\nEven though the problem seems related to the exact cover problem, I cannot find a good way to solve it.\n\nCan you suggest a method/algorithm that finds solutions to such problem? If it is a known problem, do you know any reference?\n\n\u2022 This is NP-complete by reduction from SUBSET-SUM or PARTITION (exercise). \u2013\u00a0Yuval Filmus Nov 17 '16 at 15:15\n\u2022 Thank you. Do you know how can we solve this kind of problem? I found in Wikipedia Algorithm X due to Knuth that solves the exact cover problem but I cannot transform it to my problem. \u2013\u00a0drzbir Nov 17 '16 at 15:31\n\u2022 You can formulate it as an integer programming problem and run a solver, hoping for the best. \u2013\u00a0Yuval Filmus Nov 17 '16 at 15:32\n\nAs suggested by Yuval Filmus, reduce PARTITION to my problem.\n\nGiven an instance of PARTITION, that is a set of nonnegative integers $\\{b_1, \\ldots, b_k\\}$, is there a subset $S\\subset\\{1,\\ldots,k\\}$, such that $\\sum_{i\\in S}b_i=\\sum_{i\\notin S}b_i=\\frac{\\sum_{i=1}^kb_i}{2}$?\n\nLet $n=2$, $m=k$, $a_{ij}=b_i$ for all $(i,j)\\in\\{1,\\ldots,k\\}\\times\\{1,2\\}$ and $V_1=V_2=\\frac{\\sum_{i=1}^kb_i}{2}$.\n\nThis is clearly created in polynomial-time.\n\nPARTITION is solved if and only if my problem is solved.\n\n1. If PARTITION is solved: there is a set $S\\subset\\{1,\\ldots,k\\}$, such that $\\sum_{i\\in S}b_i=\\sum_{i\\notin S}b_i=\\frac{\\sum_{i=1}^kb_i}{2}$. Take $S_1=S$ and $S_2=\\{1,\\ldots,k\\}\\backslash S$. Clearly, $S_1\\cup S_2=\\{1,\\ldots,k\\}$ and $S_1$ and $S_2$ are disjoint. Further, we have $$\\sum_{i\\in S_1}a_{i1}=\\sum_{i\\in S_1}b_i=V_1\\geqslant V_1,\\\\ \\sum_{i\\in S_2}a_{i2}=\\sum_{i\\in S_2}b_i=V_2\\geqslant V_2,$$ and my problem is solved.\n\n2. If my problem is solved: there are disjoint $S_1$ and $S_2$ such that $$S_1\\cup S_2=\\{1,\\ldots,k\\},\\\\ \\sum_{i\\in S_1}a_{i1}=\\sum_{i\\in S_1}b_i\\geqslant V_1,\\\\ \\sum_{i\\in S_2}a_{i2}=\\sum_{i\\in S_2}b_i\\geqslant V_2.$$ Since $V_1=V_2=\\frac{\\sum_{i=1}^kb_i}{2}$ and $\\sum_{i\\in S_1}b_i+\\sum_{i\\in S_2}b_i=\\sum_{i=1}^kb_i$, we must have $$\\sum_{i\\in S_1}b_i=\\sum_{i\\notin S_2}b_i=\\frac{\\sum_{i=1}^kb_i}{2},$$ and PARTITION is solved.\n\nTherefore, my problem is NP-hard.\n\nTo solve the problem, let us write it as integer programming problem as suggested by Yuval Filmus. To do so, introduce the binary variable $x_{ij}$ that is equal to $1$, if $i$ is in set $S_j$, and, $0$ otherwise.\n\n\\begin{align} & {\\underset{\\mathbf{ x }}{\\text{maximize}}} & & 0\\\\[6pt] & \\text{subject to} & & \\sum_{i=1}^ma_{ij}x_{ij}\\geqslant V_j,\\forall\\, j\\in\\{1,\\ldots,n\\},\\tag{C1}\\\\[6pt] & & & \\sum_{j=1}^nx_{ij}=1, \\forall\\, i\\in\\{1,\\ldots,m\\},\\tag{C2}\\\\[6pt] & & & x_{ ij }\\in\\{0, 1\\}, \\forall (i,j)\\in\\{1,\\ldots,k\\}\\times\\{1,\\ldots,n\\}\\tag{C3}. \\end{align}\n\nEven though this solves my problem, I need to develop a greedy algorithm for it, can I do that?\n\n\u2022 Thanks for writing up a detailed answer. One comment: please don't use 'answers' to ask new questions or follow-up questions. Instead, you should use the 'Ask Question' button in the upper-right to ask a new question. If you do that, make sure you tell us what your exact question is (are you looking for an exact solution or an approximation algorithm or a heuristic?), what your thoughts are, and what approaches you've considered and rejected. \u2013\u00a0D.W. Nov 18 '16 at 17:04", "date": "2019-08-25 15:32:11", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/66153/how-to-partition-the-rows-of-a-matrix-in-such-a-way-that-every-column-satisfies", "openwebmath_score": 0.915884792804718, "openwebmath_perplexity": 227.87223792294097, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384664716301, "lm_q2_score": 0.8791467627598856, "lm_q1q2_score": 0.8601437107192521}}
{"url": "https://math.stackexchange.com/questions/396707/if-fx-to-0-as-x-to-infty-and-f-is-bounded-show-that-fx-to0-as-x/396722", "text": "# If $f(x)\\to 0$ as $x\\to\\infty$ and $f''$ is bounded, show that $f'(x)\\to0$ as $x\\to\\infty$\n\nLet $f\\colon\\mathbb R\\to\\mathbb R$ be twice differentiable with $f(x)\\to 0$ as $x\\to\\infty$ and $f''$ bounded. Show that $f'(x)\\to0$ as $x\\to\\infty$. (This is inspired by a comment/answer to a different question)\n\n\u2022 Proof by picture Jul 15 '13 at 19:45\n\u2022 This is Barbalat's Lemma (as pointed out by @MrYouMath in a duplicate thread). It only needs $f'$ to be uniformly continuous.\n\u2013\u00a0Dap\nFeb 9 '18 at 13:17\n\u2022 Does the statement still work if we change $\\langle \\langle$$f'' bounded on \\mathbb{R}$$\\rangle \\rangle$ into $\\langle \\langle$ $f''$ bounded in a neighborhood of $+\\infty$ : $f''(x) \\underset{x\\to +\\infty}{=} O(1)$ $\\rangle \\rangle$ ? Aug 28 '21 at 0:05\n\nLet $|f''|\\le 2M$ on $\\mathbb R$ for some $M>0$. By Taylor's expansion, for every $x\\in\\mathbb R$ and every $\\delta>0$, there exists $y\\in[x,x+\\delta]$, such that $$f(x+\\delta)=f(x)+f'(x)\\delta+\\frac{1}{2}f''(y)\\delta^2.$$ It follows that $$|f(x+\\delta)-f(x)-f'(x)\\delta|\\le M\\delta^2.\\tag{1}$$ Since $\\lim_{x\\to\\infty}f(x)=0$, fixing $\\delta>0$ and letting $x\\to\\infty$ in $(1)$, we have $$\\limsup_{x\\to\\infty}|f'(x)|\\le M\\delta.$$ Since $\\delta>0$ is arbitrary, the conclusion follows.\n\n\u2022 How is Eq. (1) valid? Can you really take the absolute value for the inequality? Nov 16 '15 at 16:48\n\u2022 @Jo\u00e3oVictorBateliRom\u00e3o he's not taking absolute on the inequality, he is taking it on the equality, and then applying the inequality. Jul 20 '16 at 11:26\n\u2022 Question: could we get the same results for $f:[0,\\infty)\\to\\mathbb{R}$ and $\\lim_{x\\to\\infty}f(x)=a$ where $a\\neq 0$? Jul 20 '16 at 13:27\n\nLet me just mention that proposed fact immediately follows from Landau-Kolmogorov inequality which in this particular case reduces to $$\\|f'\\|^2_{L_{\\infty}{\\mathbb{(R)}}}\\le 4\\|f\\|_{L_{\\infty}{\\mathbb{(R)}}}\\|f''\\|_{L_{\\infty}{\\mathbb{(R)}}}$$\n\nLet $M$ be a bound for $f''$. Then $|f'(x+h)-f'(x)|\\le M|h|$ for all $x,h$. Let $\\epsilon>0$ be given. We have to show that $|f'(x)|<\\epsilon$ for all sufficiently big $x$. As $f(x)\\to0$, there is an $x_0$ such that $|f(x)|<\\frac{\\epsilon^2}{4 M}$ for all $x>x_0$. Consider $x>x_0$ and assume $f'(x)> 0$. Then $f'(x+h)\\ge f'(x)-Mh$ for $h\\ge 0$ and hence \\begin{align}f\\left(x+\\frac {f'(x)}{M}\\right)-f(x)&=\\int_0^{\\frac {f'(x)}{M}}f'(x+h)\\,\\mathrm dh\\\\&\\ge \\int_0^{\\frac {f'(x)}{M}}(f'(x)-M|h|)\\,\\mathrm dh\\\\&=\\frac{(f'(x))^2}{2M}.\\end{align} If on the other hand $f'(x)<0$, we similarly have $f'(x+h)\\le f'(x)+Mh$ for $h\\ge 0$ and hence \\begin{align}f\\left(x-\\frac {f'(x)}{M}\\right)-f(x)&=\\int_0^{-\\frac {f'(x)}{M}}f'(x+h)\\,\\mathrm dh\\\\&\\le \\int_0^{-\\frac {f'(x)}{M}}(f'(x)+M|h|)\\,\\mathrm dh\\\\&=-\\frac{(f'(x))^2}{2M}.\\end{align} In both cases we find $$\\left|f\\left(x+\\frac {|f'(x)|}{M}\\right)-f(x)\\right|\\ge \\frac{(f'(x))^2}{2M}$$ and as $x+\\frac {|f'(x)|}{M}>x_0$, we conclude $\\frac{(f'(x))^2}{2M}< 2\\cdot \\frac{\\epsilon^2}{4M}$ and hence $|f'(x)|<\\epsilon$ as was to be shown.$_\\square$\n\nThis is inspired by the proof by picture mentioned in the comment.\n\nGeometrically, if $$f'(x)$$ is large, then because $$f''$$ is bounded, $$f'(x)$$ won't vary too much in a short period, therefore $$f(x)$$ cannot be Cauchy.\n\nGiven $$\\epsilon>0$$, for sufficiently large $$a$$, due to the convergence of $$f(x)$$ we have $$|f(a+\\epsilon)-f(a)| = |\\int_a^{a+\\epsilon} f'(x)dx|<\\epsilon^2$$\n\nBy the mean value theorem, $$f'(x) = f'(a) + f''(b)(x-a)$$ for some $$b$$, therefore $$|\\int_a^{a+\\epsilon} f'(x)dx-\\int_a^{a+\\epsilon}f'(a)dx|\\le M|\\int_a^{a+\\epsilon}(x-a)dx|$$ Then\n\n$$|\\int_a^{a+\\epsilon}f'(a)dx|\\le |\\int_a^{a+\\epsilon}f'(x)dx| + M|\\int_a^{a+\\epsilon}(x-a)dx|\\le \\epsilon^2 + \\frac{M\\epsilon^2}{2}$$\n\nFinally $$|f'(a)|\\le (1+\\frac{M}{2})\\epsilon$$\n\nNote that we only need $$f(x)$$ converges (not necessarily to $$0$$) and $$|f''(x)|.", "date": "2022-01-20 00:34:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/396707/if-fx-to-0-as-x-to-infty-and-f-is-bounded-show-that-fx-to0-as-x/396722", "openwebmath_score": 0.9630523324012756, "openwebmath_perplexity": 159.84495902422066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462211935647, "lm_q2_score": 0.8705972801594707, "lm_q1q2_score": 0.8601032931149443}}
{"url": "https://mathhelpboards.com/threads/if-a-and-b-are-unit-vectors.7578/#post-34509", "text": "# [SOLVED]If a and b are unit vectors...\n\n#### Raerin\n\n##### Member\nIf a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?\n\nIs the answer -1 by any chance? If not...\n\nI know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.\n\n#### Chris L T521\n\n##### Well-known member\nStaff member\nRe: If a nd b are unit vecotrs...\n\nIf a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?\n\nIs the answer -1 by any chance? If not...\n\nI know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.\nNote that\n\n\\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned}\n\nSince $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with\n$\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$\n\nTherefore, we now have that\n\n$(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$\n\n#### Raerin\n\n##### Member\nRe: If a nd b are unit vecotrs...\n\nNote that\n\n\\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned}\n\nSince $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with\n$\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$\n\nTherefore, we now have that\n\n$(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$\n\nI don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0?\n\nAlso, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?\n\n#### Chris L T521\n\n##### Well-known member\nStaff member\nRe: If a nd b are unit vecotrs...\n\nI don't understand how 2a . b = 0 becomes a . b = 0.\n\nAlso, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?\nSince $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which is the correct choice). With that result, you can now substitute zero in for $\\mathbf{a}\\cdot\\mathbf{b}$ in the simplified form of $(2\\mathbf{a}-\\mathbf{b})\\cdot(a+3\\mathbf{b})$ to get $5\\mathbf{a}\\cdot\\mathbf{b} - 1 = 5(0) - 1 = -1$.\n\nI hope this clarifies things!", "date": "2021-10-21 20:32:39", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/if-a-and-b-are-unit-vectors.7578/#post-34509", "openwebmath_score": 0.9990973472595215, "openwebmath_perplexity": 1265.4478292684032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462187092607, "lm_q2_score": 0.8705972734445508, "lm_q1q2_score": 0.8601032843181363}}
{"url": "https://www.jerseypainconsultant.com/ina-garten-cnvmcj/45598a-does-a-kite-have-parallel-sides", "text": "A kite is a quadrilateral with two pairs of adjacent, congruent sides. The diagonals bisect each other at right angles. Additionally, if a convex kite is not a rhombus, there is another circle, outside the kite, tangent to the lines that pass through its four sides; therefore, every convex kite that is not a rhombus is an ex-tangential quadrilateral. What characteristics does a Kite have? The kite can be seen as a pair of congruent triangles with a common base. However, with a partitioning classification, rhombi and squares are not considered to be kites, and it is not possible for a kite to be equilateral or equiangular. See Area of a Kite 4. [10] Any non-self-crossing quadrilateral that has an axis of symmetry must be either a kite (if the axis of symmetry is a diagonal) or an isosceles trapezoid (if the axis of symmetry passes through the midpoints of two sides); these include as special cases the rhombus and the rectangle respectively, which have two axes of symmetry each, and the square which is both a kite and an isosceles trapezoid and has four axes of symmetry. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. Hint: 2 (Both have a \"t\")... Trapezoid and Isosceles Trapezoid. Score .8439 One diagonal is the perpendicular bisector of the other diagonal. A tangential quadrilateral is a kite if and only if any one of the following conditions is true:[14], If the diagonals in a tangential quadrilateral ABCD intersect at P, and the incircles in triangles ABP, BCP, CDP, DAP have radii r1, r2, r3, and r4 respectively, then the quadrilateral is a kite if and only if[14], If the excircles to the same four triangles opposite the vertex P have radii R1, R2, R3, and R4 respectively, then the quadrilateral is a kite if and only if[14]. [5], Among all quadrilaterals, the shape that has the greatest ratio of its perimeter to its diameter is an equidiagonal kite with angles \u03c0/3, 5\u03c0/12, 5\u03c0/6, 5\u03c0/12. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. It has two pairs of equal-length adjacent (next to each other) sides. This page was last edited on 14 January 2021, at 20:00. Kite is also a quadrilateral as it has four sides. A kite has two pairs of equal sides. Every convex kite has an inscribed circle; that is, there exists a circle that is tangent to all four sides. A. kite B. parallelogram C. rhombus D. rectangle Weegy: If a quadrilateral does not have two pairs of opposite sides that are parallel, then it may be a trapezoid. A kite is a quadrilateral in which two pairs of adjacent sides are equal. How many degrees does a right angle have? The Rhombus. It has two pairs of equal-length adjacent (next to each other) sides. Opposite sides are the same length and they are parrallel. A square has two pairs of parallel sides, four right angles, and all four sides are equal. How many pairs of parallel sides does a Kite have? A rhombus is defined as a parallelogram with four equal sides. 4. Yes! \", https://en.wikipedia.org/w/index.php?title=Kite_(geometry)&oldid=1000358494, Creative Commons Attribution-ShareAlike License, An axis of symmetry through one pair of opposite sides, An axis of symmetry through one pair of opposite angles. A kite is the combination of two isosceles triangles. Area The area of a kite can be calculated in various ways. All rhombuses and squares are also kites. One diagonal is a line of symmetry (it divides the quadrilateral into two congruent triangles that are mirror images of each other). Kite Sides. A rhombus is defined as a parallelogram with four equal sides. Parallel sides? A concave kite is sometimes called a \"dart\" or \"arrowhead\", and is a type of pseudotriangle. 90 (ninety) 500. From the above discussion we come to know about the following properties of a kite: 1. How Many Parallel Sides Does A Pentagon Have? Exactly one pair of opposite angles congruent ... What Shapes have exactly One Pair Opposite Sides Parallel? Let\u2019s see how! Trapezoids only have one pair of parallel sides. In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. Mathematics. How many congruent/consecutive pairs of lines does a kite have? 500. Answer #1 | 07/12 2014 07:59 If you are talking about the typical 4 edged (sided) kite, 0 for a box kite there are several. A kite has one line of symmetry. [6], In non-Euclidean geometry, a Lambert quadrilateral is a right kite with three right angles.[7]. A rhombus is a four-sided shape where all sides have \u2026 Hint: 3 * Remember also that one of them is in the definition *Think, and draw a picture, too. A square also fits the definition of a rectangle (all angles are 90\u00b0), and a rhombus (all sides are equal length). The tiling that it produces by its reflections is the deltoidal trihexagonal tiling. kites vary mostly with size, whereas parallelograms tend to vary in angle, for example, how squashed it looks. By avoiding the need to treat special cases differently, this hierarchical classification can help simplify the statement of theorems about kites. For the same reason, with a partitioning classification, shapes meeting the additional constraints of other classes of quadrilaterals, such as the right kites discussed below, would not be considered to be kites. Casy: Let's think: An arrow head has 3 dimensions: Length, width and height.. By definition, these are straight lines of given length. The kite can be seen as a pair of congruent triangles with a common base. The angles of a kite are equal whereas the unequal sides of a kite meet. As is true more generally for any orthodiagonal quadrilateral, the area A of a kite may be calculated as half the product of the lengths of the diagonals p and q: Alternatively, if a and b are the lengths of two unequal sides, and \u03b8 is the angle between unequal sides, then the area is. 2. A kite is a quadrilateral with exactly two pairs of adjacent congruent sides. A Study of Definition\", Information Age Publishing, 2008, pp. A kite is a four-sided shape that has two sets of adjacent sides that have equal lengths. In the figure above, click 'show diagonals' and reshape the kite. Another name is equilateral quadrilateral, since \u2026 There are an infinite number of uniform tilings of the hyperbolic plane by kites, the simplest of which is the deltoidal triheptagonal tiling. 1. Perpendicular Diagonals 2. \"Is a Kite with only one pair of parallel sides still a rhombus?\" The two line segments connecting opposite points of tangency have equal length. It's a type of quadrilateral that is not a parallelogram. Check out the kite in the below figure. But no sides are parallel. [9], All kites tile the plane by repeated inversion around the midpoints of their edges, as do more generally all quadrilaterals. You could have one pair of congruent, adjacent sides but not have a kite. Either no sides are parallel, or it is a rhombus with both sides parallel. Two pairs of sides known as co\u2026 Solved Examples. Answer #2 | 07/12 2014 15:43 None. In an isosceles parallelogram, we have. They have this side in common right over here. According to this classification, every equilateral kite is a rhombus, and every equiangular kite is a square. These sides are called as distinct consecutive pairs of equal length. A kite has two pairs of sides that have equal length. The diagonals of a kite intersect at 90 $$^{\\circ}$$ The formula for the area of a kite is Area = $$\\frac 1 2$$ (diagonal 1)(diagonal 2) 0. How many parallel sides are in a Kite? are equal where the two pairs meet. [3] These shapes are called right kites. Some textbooks say a kite has at least two pairs of adjacent congruent sides, so a rhombus is a special case of a kite.) Opposite sides in a parallelogram are always parallel whereas this is not true for kites.Therefore, kite is not a parallelogram. Kite. [1], A kite with three equal 108\u00b0 angles and one 36\u00b0 angle forms the convex hull of the lute of Pythagoras. No, because a rhombus does not have to have 4 right angles. The deltoidal icositetrahedron, deltoidal hexecontahedron, and trapezohedron are polyhedra with congruent kite-shaped facets. 1. No, because a rhombus does not have to have 4 right angles. It has two pairs of equal sides. Pair of parallel sides equal. Ish. 1. 2. (This definition excludes rhombi. Therefore, every convex kite is a tangential quadrilateral. For example, a regular hexagon has three pairs of parallel sides. The rhombus has a square as a special case, and is a special case of a kite and parallelogram. Formula for the median on a trapezoid: 1/2(Base + Base) How many congruent triangles do the diagonals form on a square? Pair off non-parallel sides equal. 500. Kite quadrilaterals are named for the wind-blown, flying kites, which often have this shape and which are in turn named for a bird. The center of the incircle lies on a line of symmetry that is also a diagonal. An isosceles trapezoid is a trapezoid whose non-parallel sides are congruent. Zalman Usiskin and Jennifer Griffin, \"The Classification of Quadrilaterals. Solved Examples - Find the perimeter of kite whose sides are 21cm and 15cm. Now it seems like we could do something pretty interesting with these two smaller triangles at the top left and the top right of this, looks like, a kite like figure. The diagonals bisect each other at right angles. [12] The side-angle duality of kites and isosceles trapezoids are compared in the table below. Each side is parallel \u2026 One of them is a tiling by a right kite, with 60\u00b0, 90\u00b0, and 120\u00b0 angles. Each polygon has two pairs of parallel sides. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 \u00d7 (12 m + 10 m) = 2 \u00d7 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. It often looks like A kite is shaped just like what comes to mind when you hear the word \"kite.\" A Square is a Kite? In a kite, two adjoining sides are equal as shown in the figure. [4], There are only eight polygons that can tile the plane in such a way that reflecting any tile across any one of its edges produces another tile; a tiling produced in this way is called an edge tessellation. The angles Explain. This makes two pairs of adjacent, congruent sides. A trapezium doesn\u2019t have rotational symmetry so the order of rotational symmetry is 1. 49-52. (British name: Trapezium) Zero. ... What do we call parallel sides of the trapezium. A kite, showing its pairs of equal length sides and its inscribed circle. Diagonals (dashed lines) cross at A trapezium as got one pair of parallel sides. The remainder of this article follows a hierarchical classification, in which rhombi, squares, and right kites are all considered to be kites. [10] If crossings are allowed, the list of quadrilaterals with axes of symmetry must be expanded to also include the antiparallelograms. The angles of a kite are equal whereas the unequal sides of a kite meet. Diagonals intersect at right angles. Geometry. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two \u2026 A pentagon has no parallel sides :)... How Many Corners Does A Pentagon Have? What do a kite and a rhombus have in common? The opposite sides are parallel. [11], Kites and isosceles trapezoids are dual: the polar figure of a kite is an isosceles trapezoid, and vice versa. The opposite sides are parallel. Whats the difference between a Kite and a Rhombus? As you reshape the kite, notice the diagonals always intersect each other at 90\u00b0 (For concave kites, a diagonal may need to be extended to the point of intersection.) Every kite is orthodiagonal, meaning that its two diagonals are at right angles to each other. Answer. One diagonal bisects a pair of opposite angles. A kite, as defined above, may be either convex or concave, but the word \"kite\" is often restricted to the convex variety. Kite: A quadrilateral in which two disjoint pairs of consecutive sides are congruent (\u201cdisjoint pairs\u201d \u2026 That is, for these kites the two equal angles on opposite sides of the symmetry axis are each 90 degrees. A square has equal sides (marked \"s\") and every angle is a right angle (90\u00b0) Also opposite sides are parallel. Rhombus have 4 euql sides. It depends on the shape of the kite, but if you mean the usual diamond shaped kite then it has only one... Answer Question. Its four vertices lie at the three corners and one of the side midpoints of the Reuleaux triangle (above to the right). When p=q, the kites become rhombi; when p=q=4, they become squares. (Jump to Area of a Kite or Perimeter of a Kite) A Kite is a flat shape with straight sides. All types of triangle, such as equilateral triangle, isosceles triangle and scalene triangle, have no parallel lines.. A kite is another shape that does not have parallel sides. Moreover, one of the two diagonals (the symmetry axis) is the perpendicular bisector of the other, and is also the angle bisector of the two angles it meets.[10]. more interesting facts . It has rotational symmetry of order one. Each polygon has two pairs of congruent sides that are adjacent. [10] The two interior angles of a kite that are on opposite sides of the symmetry axis are equal. To be a kite, a quadrilateral must have two pairs of sides that are equal to one another and touching. It looks like the kites you see flying up in the sky. So\u00a0it doesn't always look like the kite you fly. One of the two diagonals of a convex kite divides it into two isosceles triangles; the other (the axis of symmetry) divides the kite into two congruent triangles. It is also a rectangle and a parallelogram. Kite. It might not have have a line with colorful bows attached to the flyer on the ground, but it does have that familiar, flying-in-the-wind kind of shape. Since the arrowhead is 3 dimentional, it can contain 6 parralel line or 3 pairs.,one on oppsite sides of the arrowhead just like the 3 dimentions of a cube. That is it \u2026 Kites are also known as deltoids, but the word \"deltoid\" may also refer to a deltoid curve, an unrelated geometric object. \"Maximal area of a bicentric quadrilateral\", \"When is a Tangential Quadrilateral a Kite? Can two angles of a kite be consecutive and supplementary? A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (\u201cdisjoint pairs\u201d means that one side can\u2019t be used in both pairs). the kites that can be inscribed in a circle) are exactly the ones formed from two congruent right triangles. [2], The kites that are also cyclic quadrilaterals (i.e. It has got 2 pairs of equal length sides. Two disjoint pairs of adjacent sides are equal (by definition). Being a special type of quadrilateral, it shows special characteristics and properties which are different from the other types of quadrilaterals. Each pair\u00c2\u00a0is two equal-length sides that are adjacent (they meet). A trapezium has one pair of parallel sides. It is possible to classify quadrilaterals either hierarchically (in which some classes of quadrilaterals are subsets of other classes) or as a partition (in which each quadrilateral belongs to only one class). Face-transitive self-tesselation of the sphere, Euclidean plane, and hyperbolic plane with kites occurs as uniform duals: for Coxeter group [p,q], with any set of p,q between 3 and infinity, as this table partially shows up to q=6. The properties of the kite are as follows: Two disjoint pairs of consecutive sides are congruent by \u2026 Which angles are congruent on a isosceles trapezoid? Conditions for when a tangential quadrilateral is a kite. [1] Because they circumscribe one circle and are inscribed in another circle, they are bicentric quadrilaterals. Answer for question: Your name: Answers. 3. A quadrilateral is a kite if and only if any one of the following conditions is true: The kites are the quadrilaterals that have an axis of symmetry along one of their diagonals. Positive: 50 %. Question: Does an isosceles trapezoid have two sets of parallel sides? The Perimeter is the distance around the edges. (Jump to Area of a Kite or Perimeter of a Kite). Who needs 'em! Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Is a rhombus always a rectangle? Trapezium. You can work out the area of a trapezium by using the formula A = \u00bd(a+b)h. Where a and b are the lengths of the parallel sides and h is the shortest distance between the two parallel sides. For every concave kite there exist two circles tangent to all four (possibly extended) sides: one is interior to the kite and touches the two sides opposite from the concave angle, while the other circle is exterior to the kite and touches the kite on the two edges incident to the concave angle. The difference between a kite and a rhombus is that a kite does not always have four equal sides or two pairs of parallel sides like a rhombus. If a quadrilateral does not have any parallel sides but has two sets of adjacent sides that are congruent, it is classified as a kite, and a kite is a convex quadrilateral. Geometry. A parallelogram is a rectangle that has been pushed over. 3. How many parallel sides does a kite have? When all the angles are also 90\u00b0 the Kite will be a Square. Each polygon has two pairs of congruent sides. The products of opposite sides are equal. Some shapes have many parallel sides. Kite. Kites are free riders, lone wolves who do whatever they want whenever they want. Angles between unequal sides are equal In the figure above notice that \u2220ABC = \u2220ADC no matter how how you reshape the kite. right angles. A Kite is a flat shape with straight sides. Among all the bicentric quadrilaterals with a given two circle radii, the one with maximum area is a right kite. With a hierarchical classification, a rhombus (a quadrilateral with four sides of the same length) or a square is considered to be a special case of a kite, Each polygon has four congruent sides. A kite cannot have a single pair of parallel sides (a trapezoid). In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. A kite as got two pairs of sides next to each other that have equal length. ... How many lines of symmetry does a kite have? because it is possible to partition its edges into two adjacent pairs of equal length. Kites and darts in which the two isosceles triangles forming the kite have apex angles of 2\u03c0/5 and 4\u03c0/5 represent one of two sets of essential tiles in the Penrose tiling, an aperiodic tiling of the plane discovered by mathematical physicist Roger Penrose. Trapezium: A trapezium is a four-sided closed convex geometrical figure with one and only set of parallel sides. The kite's sides, angles, and diagonals all have identifying properties. A kite with angles \u03c0/3, \u03c0/2, 2\u03c0/3, \u03c0/2 can also tile the plane by repeated reflection across its edges; the resulting tessellation, the deltoidal trihexagonal tiling, superposes a tessellation of the plane by regular hexagons and isosceles triangles.[13]. Table below for example, a regular hexagon has three pairs of equal-length adjacent ( next to each other of! ; that is, there exists a circle that is, there exists a circle that is a!, with 60\u00b0, 90\u00b0, and is a right kite., pp a picture, too other! 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With axes of symmetry does a kite have Corners does a pentagon has no parallel sides, but are!, with 60\u00b0, 90\u00b0, and trapezohedron are polyhedra with congruent kite-shaped facets lute Pythagoras... 90\u00b0 the kite you fly, pp its two diagonals are at right.. Which are different from the above discussion we come to know about the following properties a! The perpendicular bisector of the kite you fly parallelogram also has two pairs meet exactly... And touching makes two pairs of equal length that are also 90\u00b0 the kite can not have to 4. Tangent to all four sides the two equal angles on opposite sides of the kite will a... To treat special cases differently, this hierarchical classification can help simplify the statement theorems... Are polyhedra with congruent kite-shaped facets 10 ] If crossings are allowed, the one with maximum is...", "date": "2022-10-02 21:33:38", "meta": {"domain": "jerseypainconsultant.com", "url": "https://www.jerseypainconsultant.com/ina-garten-cnvmcj/45598a-does-a-kite-have-parallel-sides", "openwebmath_score": 0.5786581635475159, "openwebmath_perplexity": 707.1876742898634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9879462204837636, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8601032825460239}}
{"url": "http://math.stackexchange.com/questions/257479/there-exist-an-infinite-subset-s-subseteq-mathbbr3-such-that-any-three-vect/257492", "text": "# There exist an infinite subset $S\\subseteq\\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent.\n\nCould anyone just give me hint for this one?\n\nThere exist an infinite subset $S\\subseteq\\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. True or false?\n\n-\nDo you know any facts about linearly independent sets? \u2013\u00a0 Chris Eagle Dec 12 '12 at 22:27\nIs there a subset of $\\mathbb{R}^3$ with the property that every plane through the origin meets it in at most 2 points? \u2013\u00a0 Hurkyl Nov 28 '13 at 13:54\n\nThree vectors in $\\mathbb{R}^3$ are linearly dependent if and only if they lie in a plane.\n\nConsider the following process for building $S$. We can start with the empty set, and choose any two vectors $v_1, v_2 \\in \\mathbb{R}^3$ and add them to $S$. Then to choose a third vector $v_3$ to add to $S$, we must make sure it is not in the unique plane containing (i.e. spanned by) $v_1$ and $v_2$. Thus $v_3$ can be any vector in $\\mathbb{R}^3 \\backslash span(v_1, v_2)$.\n\nSimilarly, if at some stage $S = \\{v_1, \\ldots, v_k\\}$, we can add to $S$ any vector $v_{k+1}$ in $\\mathbb{R}^3 \\backslash \\bigcup_{x_i, x_j} span(x_i, x_j)$. Note that $\\bigcup_{x_i, x_j} span(x_i, x_j)$ is a finite union of planes, so it can never be all of $\\mathbb{R}^3$. In this way we can choose an infinite set with the desired property.\n\n-\nThe key issue with this approach is to prove that a finite union of planes is not all of $\\mathbb R^3$. For a more general result: mathoverflow.net/questions/26/\u2026 \u2013\u00a0 Andres Caicedo Dec 12 '12 at 22:47\n\nTry $\\left(\\begin{matrix}1\\\\t\\\\t^2\\end{matrix}\\right)$ with $t\\in\\mathbb R$. Do you know to compute $$\\left\\vert\\begin{matrix}1&1&1\\\\r&s&t\\\\r^2&s^2&t^2\\end{matrix}\\right\\vert?$$\n\n-\ngood one!!. Please clear my doubt, consider the random gaussian vector $[x,y,z]$, where each entries are independent gaussian random variables. So shouldn't the set of all realizations of this random vector be a set of the kind he is looking for? \u2013\u00a0 dineshdileep Dec 13 '12 at 5:44\nthe determinant is $(st^2-s^2t)-(rt^2-tr^2)+(rs^2-sr^2)$ \u2013\u00a0 El Angel Exterminador Dec 14 '12 at 5:08\n@Kuttus ... and that equals $(r-s)(s-t)(t-r)$ and is non-zero unless two of the numbers are equal. :) Actually, the fact that the functions $x\\mapsto 1$, $x\\mapsto x$ and $x\\mapsto x^2$ are linearly independent lurks behind this as a shortcut: If a linear combinaton of the rows is the zero vector, then the correspondnig quadratic polynomial has three distinct roots $r,s,t$, hence is the zero polynomal, hence the linear combination was in fact the trivial combination, hence the rows are linearly independent, hence so are the columns (which is what we were actually after). \u2013\u00a0 Hagen von Eitzen Dec 14 '12 at 11:39\n+1 I'd just like to point out to future readers of this excellent answer that the curve $t\\to (1,t,t^2)$ in $\\mathbb{R}^3$ is also known as the twisted cubic. \u2013\u00a0 Amitesh Datta Jul 27 '13 at 5:42\n\nConsider vectors of the form $v_x=(1,x,x^2)^T$. Then for any distinct $x,y,z\\in \\mathbb R$ matrix $(v_x v_yv_z)$ is nonsingular (Vandermonde matrix), so $v_x$, $v_y$, $v_z$ are linearly independent.\n\n-\n\nThe parametric curve $t\\mapsto(1,t,t^2)$ has the property that any three distinct points on it are linearly independent (without the \"distinct\" one clearly cannot have a solution). To check, just compute the determinant, which is Vandermonde.\n\n-\n\nInspired by the Vandermonde matrix example, let $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ be a strictly convex function. Then $$\\left\\vert\\begin{matrix}1&1&1\\\\x&y&z\\\\f(x)&f(y)&f(z)\\end{matrix}\\right\\vert\\not=0$$ for any three distinct numbers $x,y,z$.\n\n-\n\nThe elegant example given by Adam can be generalized: let $P \\subset \\mathbb R^3$ be any plane that does not pass through the origin (equipped with the usual subspace topology), and let $C \\subset P$ be a strictly convex* closed subset of $P$. Then the boundary $S = \\partial_P\\, C$ of $C$ in $P$ satisfies your requirement.\n\n(* By \"strictly convex\", I mean that any non-trivial convex combination of two points in $C$ should belong to the interior of $C$, i.e. for all $x,y \\in C$ and all $0 < \\alpha < 1$, the point $z = \\alpha x + (1-\\alpha)y$ lies in the interior of $C$. Geometrically, this basically means that no part of the boundary should be a straight line segment.)\n\nProof: Let $x$ and $y$ be any two distinct points in $S$, and let $z$ linear combination of $x$ and $y$ which is distinct from both. By definition, $z = \\alpha x + \\beta y$ for some $\\alpha$ and $\\beta$. If $\\alpha + \\beta \\ne 1$, then $z \\notin P$, and thus $z \\notin S$. Thus, $z$ can only belong to $P$ if $\\beta = 1 - \\alpha$. Let us assume below that this is the case.\n\nIf $0 < \\alpha < 1$, by strict convexity as defined above, $z \\in \\operatorname{int} C$ and thus $z \\notin S$. If $\\alpha \\in \\{0,1\\}$, then $z = x$ or $z = y$, contradicting the assumption of distinctness. The only remaining possibility is that $\\alpha < 0$ or $\\alpha > 1$, but if either of those holds, then either $x$ is a convex combination of $z$ and $y$, or $y$ is a convex combination of $z$ and $x$. In either case, $z$ cannot belong to $S \\subset C$, since otherwise the fact that $x,y \\in S = \\partial\\,C = C \\setminus \\operatorname{int} C$ would contradict the assumption that $C$ is strictly convex.\n\nBesides the example given by Adam, one example possible concrete example of such a set would be e.g. the circle given by $P = \\{(x,y,z) \\in \\mathbb R^3: z = 1\\}$, $C = \\{(x,y,z) \\in P: x^2 + y^2 \\le 1\\}$ and $S = \\partial_P\\,C = \\{(x,y,z) \\in P: x^2 + y^2 = 1\\}$.\n\nThe sets given by constructions like mine and Adam's are generally one-dimensional curves. Indeed, it's not hard to show that a set satisfying your requirements cannot contain any two-dimensional surface, since any surface would have to intersect some plane passing through the origin along a curve containing more than two points.\n\nHowever, it is possible to construct a set which satisfies your requirements and is dense in $\\mathbb R^3$. Basically, to do this, you'd start with an arbitrary enumeration $\\langle a_i \\rangle_{i \\in \\mathbb N}$ of the rational points $\\mathbb Q^3$ (which are a countable dense subset of $\\mathbb R^3$) and then, for each $i \\in \\mathbb N$, choose a point $b_i \\in \\mathbb R^3 \\setminus \\{0\\}$ such that $|a_i - b_i| < 1/i$ and such that $b_i$ is pairwise linearly independent of all $b_j$, $j < i$. (This is always possible, since the set of pairwise linear combinations of $b_j$, $j < i$ is the intersection of finitely many planes, and thus has zero volume.) Then let $S = \\{b_i: i \\in \\mathbb N\\}$.\n\nNote that the \"construction\" outlined above is not entirely constructive, since it requires an arbitrary choice of a point at each step of an infinite process. I do, however, believe that this dependency on choice could be eliminated by giving an explicit rule that assigns a definite value to $b_i$, given $a_i$ and $\\{b_j: j < i\\}$, similarly to the method I sketched in this answer to a related question.\n\n(Do note the caveat brought up by Asaf Karagila in the comments to the other answer, though. I still think the construction I outlined shouldn't require even dependent choice, since an explicit formula for $b_i$ may be provided: basically, the idea is that, for each $i$, it should be possible to explicitly generate a finite ordered list of points $c_k$, $k \\in \\{1, \\dotsc, n\\}$ within radius $1/i$ of $a_i$, such that at least one of those points is guaranteed to be pairwise linearly independent of $\\{b_j: j < i\\}$, and then always choose the first such point. Still, I freely admit that I'm not nearly as experienced with this stuff as Asaf is, and that there could be some gap that I've missed.)\n\n-\n\nExistence : Consider three points $x_1,\\ x_2,\\ x_3$ in a plane $P\\ :\\ x+y+z=1$ which are not in a line.\n\nPick $x_4$ in $P-\\bigcup L_{ij}$ where $L_{ij}$ is a line passing $x_i,\\ x_j\\ (1\\leq i,\\ j \\leq 3)$.\n\nRepeat this pocess infinitely.\n\n-", "date": "2015-01-26 19:16:39", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/257479/there-exist-an-infinite-subset-s-subseteq-mathbbr3-such-that-any-three-vect/257492", "openwebmath_score": 0.9215345978736877, "openwebmath_perplexity": 100.37264411815028, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462215484652, "lm_q2_score": 0.8705972616934408, "lm_q1q2_score": 0.8601032751804751}}
{"url": "https://math.stackexchange.com/questions/2312010/calculate-integral-using-residue-theorem", "text": "# Calculate Integral using residue theorem\n\nI want to verify the following result using the residue theorem:\n\n$$\\int_0^\\infty \\frac{\\log(x)}{x^2+a^2}dx = \\frac{\\pi}{2a}\\log a, \\, a > 0.$$\n\nHere are my ideas:\n\nAt first I might want to show that this function is in fact a well defined improper Riemann integral, but I didn't come up with any nice solution yet.\n\nI want to integrate the function $f(z) := \\frac{\\log(z)}{z^2+a^2}$ along the contour $C:= \\gamma_1 \\cup \\gamma_2 \\cup \\gamma_3 \\cup \\gamma_4$ constisting of to semicircles with center at 0 around the upper half plane and radius $R$ and $\\epsilon$ (resp.) as well as the intervals $[-R, -\\epsilon]$ and $[\\epsilon, R]$.\n\nFor the complex logarithm, use the branch $\\log z = \\log|z| + i\\theta, \\theta \\in [-\\pi/2, 3\\pi/2)$ so we don't get any problems on the real line (is this choice correct?).\n\nNote that I have one pole in my contour, namely $z = ia$. The Residue theorem yields $$\\int_C f(z) dz= 2\\pi i \\text{ Res}(f, ia) = 2\\pi i\\lim_{z\\to ia}(z-ia) \\frac{\\log z}{(z-ia)(z+ia)} = 2\\pi i\\frac{\\log(ia)}{2ia} = \\pi \\frac{\\log(a)+ i\\pi/2}{a}$$ Proceeding, we choose the following parametrizations:\n$$\\gamma_1 : [-R,0] \\to \\mathbb C, \\quad t\\mapsto -\\frac{\\epsilon t}{R} + (t-\\epsilon) \\\\ \\gamma_2 : [0, R] \\to \\mathbb C, \\quad t\\mapsto -\\frac{\\epsilon t}{R} + (\\epsilon+t) \\\\ \\gamma_3: [0, \\pi]\\to \\mathbb C, \\quad t\\mapsto \\epsilon e^{it} \\\\ \\gamma_4:[0,\\pi] \\to \\mathbb C, \\quad t \\mapsto Re^{it}$$ Now here's what I am unsure about. To show that the big and the small circle vanish as $R\\to \\infty$ and $\\epsilon \\to 0$ is not too hard, but how to deal with the other integrals? Can I just say $$\\int_{\\gamma_1} f(z)dz = \\int_0^R \\frac{\\log(-\\frac{\\epsilon}{R}t+\\epsilon+t)}{(-\\frac{\\epsilon}{R}t + \\epsilon + t)^2 + a^2}(\\frac{\\epsilon}{R}+1)dt \\xrightarrow{\\epsilon \\to 0} \\int_0^R \\frac{\\log t}{t^2 + a^2}dt$$ or what kind of reasoning should be used to interchange limit and integral?Also, I then finally I get $$\\int_0^\\infty \\frac{\\log(x)}{x^2+a^2}dx = \\pi \\frac{\\log a + i\\frac{\\pi}{2}}{2a}$$ almost what I want, but where does $i\\pi /2$ come from?\n\n\u2022 Is the real-analysis tag appropriate here? Jun 6, 2017 at 13:22\n\u2022 I was unsure since we talk about a real integral. I can remove it though. Jun 6, 2017 at 13:23\n\u2022 Check your work on $\\int_{\\gamma_1}f(z)~\\mathrm dz$. You should have a $\\log(-t)$ in the numerator, which should cancel off the imaginary part you are confused about. Jun 6, 2017 at 13:26\n\nAlong the real axis, we have\n\n\\begin{align} \\int_{-R}^R \\frac{\\log(x)}{x^2+a^2}\\,dx&=\\int_{-R}^0 \\frac{\\log(x)}{x^2+a^2}\\,dx+\\int_0^{R} \\frac{\\log(x)}{x^2+a^2}\\,dx\\\\\\\\ &=\\int_0^R \\frac{\\log(-x)+\\log(x)}{x^2+a^2}\\,dx\\\\\\\\ &=2\\int_0^R \\frac{\\log(x)}{x^2+a^2}\\,dx+i\\pi\\int_0^R\\frac{1}{x^2+a^2}\\,dx\\\\\\\\ &=2\\int_0^R \\frac{\\log(x)}{x^2+a^2}\\,dx+\\frac{i\\pi\\arctan(R/a)}{a}\\tag1 \\end{align}\n\nAs $R\\to \\infty$, we find that\n\n$$\\int_{-\\infty}^\\infty \\frac{\\log(x)}{x^2+a^2}\\,dx=2\\int_0^\\infty \\frac{\\log(x)}{x^2+a^2}\\,dx+\\frac{i\\pi^2}{2a}\\tag 2$$\n\nSetting $(2)$ equal to $\\frac{\\pi \\log(a)}{a}+i\\frac{\\pi^2}{2a}$, we find that\n\n$$\\int_0^\\infty \\frac{\\log(x)}{x^2+a^2}\\,dx=\\frac{\\pi \\log(a)}{2a}$$\n\nI thought it might be instructive to evaluate the integral of interest using real analysis only. To that end, we enforce the substitution $x\\to a/x$ to find that\n\n$$\\int_0^\\infty \\frac{\\log(x)}{x^2+a^2}\\,dx=\\frac1a \\int_0^\\infty \\frac{\\log(a)-\\log(x)}{x^2+1}\\,dx \\tag2$$\n\nFor $a=1$, we see from $(2)$ that $\\int_0^\\infty \\frac{\\log(x)}{x^2+1}\\,dx =0$. Thus, solving $(2)$ for integral of interest, and using $\\int_0^\\infty \\frac{\\log(x)}{x^2+1}\\,dx =0$, we find that\n\n$$\\int_0^\\infty \\frac{\\log(x)}{x^2+a^2}\\,dx=\\frac{\\pi \\log(a)}{2a}$$\n\nas expected!\n\n\u2022 That makes sense, I accidentally thought that the integral was symmetric. Could you possibly say anything about exchanging the limits and that the integral is well-defined? Jun 6, 2017 at 13:51\n\u2022 The integral is well defined since $\\log(x)$ is integrable on $[,1]$ (so, there is no problem at $0$) and since $\\log(x)<\\sqrt{x}$, $\\frac{\\log(x)}{x^2+a^2}=O\\left(\\frac{1}{x^{3/2}}\\right)$ as $x\\to \\infty$. Jun 6, 2017 at 13:59\n\u2022 Thank you very much! Does that also justify that I can just exchange Integral and limit or do I need to apply some theorem like dominated convergence? I wouldn't really want to find an integrable majorant... Jun 6, 2017 at 14:10\n\u2022 Change the parameterization to $t$ so that the integral becomes $\\int_{\\epsilon}^R \\frac{\\log(t)}{t^2+a^2}\\,dt$. Then, there is no need to interchange the order of integration and limits. Jun 6, 2017 at 14:25\n\u2022 You're welcome. My pleasure. If one insists on using the parameterization, then noting that the integrand as a function of $\\epsilon$ and $t$ is continuous on $[0,R]$ for $\\epsilon>0$. Therefore since it is continuous on the closed bounded interval, it is uniformly continuous there. This is enough to justify the interchange of the limit on $\\epsilon$ and the integral on $t$. Jun 6, 2017 at 14:33\n\nFor any $a>0$, through the substitution $x=a e^t$ we have\n\n$$I(a) = \\int_{0}^{+\\infty}\\frac{\\log x}{a^2+x^2}\\,dx = \\frac{1}{2a}\\int_{-\\infty}^{+\\infty}\\frac{\\log a+ t}{\\cosh t}\\,dt \\tag{1}$$ and $\\frac{t}{\\cosh t}$ is an odd integrable function over $\\mathbb{R}$. It follows that $$I(a) = \\frac{\\log a}{2a}\\int_{-\\infty}^{+\\infty}\\frac{dt}{\\cosh t}=\\frac{\\log a}{2a}\\left[2\\arctan\\tanh\\frac{t}{2}\\right]^{+\\infty}_{-\\infty}=\\color{red}{\\frac{\\pi\\log a}{2a}} \\tag{2}$$ without even resorting to the residue theorem, but just exploiting symmetry.\n\n\u2022 It's even easier to exploit symmetry through the substitution $x\\to a/x$. ;-)) Jun 6, 2017 at 17:17\n\u2022 @MarkViola: of course, this is almost the same. Jun 6, 2017 at 17:17\n\u2022 Yes, it is the same upon letting $x=a\\sinh(t)$ in the integral $\\int_0^\\infty \\frac{1}{x^2+a^2}\\,dx=\\frac{\\pi}{2a}$ Jun 6, 2017 at 17:25\n\u2022 How did $\\int_{0}^{\\infty}$ change to $\\int_{-\\infty}^{\\infty}$ m Jun 13, 2017 at 8:09\n\u2022 I think mockingbird meant how did $\\int_{0}^{\\infty}$ become $\\int_{-\\infty}^{\\infty}$? Jun 13, 2017 at 8:15\n\nUsing a semi-circular contour in the upper half plane that rests on the real axis we obtain\n\n$$\\int_0^\\infty \\frac{1}{x^2+a^2} \\; dx = \\frac{1}{2} \\int_{-\\infty}^\\infty \\frac{1}{x^2+a^2} \\; dx \\\\ = \\frac{1}{2} \\times 2\\pi i \\frac{1}{2\\times ai} = \\frac{\\pi}{2a}.$$\n\nNext using a keyhole contour with the slot on the positive real axis and the branch of the logarithm where $0\\le \\arg\\log z\\lt 2\\pi$ (branch cut on the positive real axis) we obtain integrating\n\n$$f(z) = \\frac{\\log^2 z}{z^2+a^2}$$\n\nthe integrals\n\n$$\\int_0^\\infty \\frac{\\log^2 x}{x^2+a^2} \\; dx + \\int_\\infty^0 \\frac{\\log^2 x + 4\\pi i \\log x - 4\\pi^2 }{x^2+a^2} \\; dx \\\\ = 2\\pi i \\left(\\frac{\\log^2(ai)}{2\\times ai} + \\frac{\\log^2(-ai)}{2\\times -ai}\\right).$$\n\nThis yields\n\n$$-4\\pi i \\int_0^\\infty \\frac{\\log x}{x^2+a^2} \\; dx + 4\\pi^2 \\times \\frac{\\pi}{2a} = \\frac{\\pi}{a} ((\\log a + \\pi i/ 2)^2 - (\\log a + 3\\pi i/2)^2) \\\\ = \\frac{\\pi}{a} (\\log a \\times \\pi i (1-3) + \\pi^2 (-1/4 + 9/4)).$$\n\nWe thus obtain\n\n$$4\\pi i \\int_0^\\infty \\frac{\\log x}{x^2+a^2} \\; dx = 4\\pi^2 \\times \\frac{\\pi}{2a} - \\frac{\\pi}{a} (-\\log a \\times 2\\pi i + 2 \\pi^2) \\\\ = \\frac{\\pi}{a} \\log a \\times 2\\pi i.$$\n\nDividing by $4\\pi i$ we finally obtain\n\n$$\\bbox[5px,border:2px solid #00A000]{ \\frac{\\pi}{2a}\\log a.}$$\n\nThe bounds on the circular components that we used here were\n\n$$2\\pi R \\times \\frac{\\log^2 R}{R^2} \\rightarrow 0 \\quad\\text{and}\\quad 2\\pi \\epsilon \\times \\frac{\\log^2 \\epsilon}{a^2} \\rightarrow 0.$$", "date": "2023-04-01 10:48:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2312010/calculate-integral-using-residue-theorem", "openwebmath_score": 0.9100432395935059, "openwebmath_perplexity": 249.4860206319465, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98593637543616, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8600790217996309}}
{"url": "https://math.stackexchange.com/questions/60894/how-to-find-the-number-of-perfect-matchings-in-complete-graphs", "text": "# How to find the number of perfect matchings in complete graphs?\n\nIn wikipedia FKT algorithm is given for planar graphs. Not anything for complete graphs. I need to find the number of perfect matchings in complete graph of six vertices.\n\n\u2022 Here is an alternate way to do this via recursion. Let $a_n$ denote the number of perfect matchings in $K_{2n}$. Then clearly, $a_1 = 1$. Now in $K_{2n}$ (where $n \\geq 2$), pick any vertex $u$, you can match it with $(2n-1)$ vertices. After matching $u$, you are left with $2n-2$ vertices. Thus, $a_n = (2n-1)a_{n-1}$. So, $a_n = (2n-1)(2n-3)\\dots(1)$. May 7, 2018 at 7:27\n\u2022 This is also called double factorial: $a_n=(2n-1)!!$ Feb 13, 2020 at 22:02\n\nIt's just the number of ways of partitioning the six vertices into three sets of two vertices each, right? So that's 15; vertex 1 can go with any of the 5 others, then choose one of the 4 remaining, it can go with any of three others, then there are no more choices to make. $5\\times3=15$.\n\n\u2022 Side note. For a general complete graph on $2n$ vertices, this number comes out to be $\\frac{(2n)!}{n! 2^{n}}$. Aug 31, 2011 at 11:19\n\u2022 In other words, it's the product of the odd numbers up to the number of vertices. Sep 1, 2012 at 20:13\n\u2022 Or even for $\\#V = 2n$ we have $(2n - 1)!!$ perfect matchings. May 17, 2014 at 10:22\n\u2022 @Srivatsan in one of the books I have, the solution is same, but it explains it as: \"The number of perfect matchings in a complete graph of n vertices, where n is even, reduces to the problem of finding unordered partitions of vertex set of the type p(2n;2,2,2,...n times) = $\\frac{(2n)!}{(2!)^nn!}$ \", Is p(2n;2,2,2,...n times) some series? Book does not elaborate much. Also notice $(2!)^n$ in the denominator, not just $2^n$ thought both are same. Just guessing if this is some series unknown to me, then 2! must be having some significance in series expansion\n\u2013\u00a0Maha\nDec 11, 2014 at 19:15\n\u2022 @awell, I think $p(m;a_1,\\dots,a_r)$ just means number of partitions of a set of size $m$ into $r$ subsets, one of size $a_1$, ..., one of size $a_r$. The formula for that is $m!/((a_1!)\\cdots(a_r)!)$. It comes from repeated application of the formula for the number of combinations of $m$ things taken $a$ at a time, $m!/(a!(m-a)!)$. Dec 11, 2014 at 22:00\n\nIf you just want to get the number of perfect matching then use the formula $$\\dfrac{(2n)!}{2^n\\cdot n!}$$ where $$2n =$$ number of vertices in the complete graph $$K_{2n}$$.\n\nDetailed Explaination:- You must understand that we have to make $$n$$ different sets of two vertices each.\n\nFirst take a vertex. Now we have $$(2n-1)$$ ways to select another vertex to make the pair.\n\nNow to make another pair we take a vertex and now we have $$(2n-3)$$ ways to select another vertex. This is because we have already used $$2$$ vertices in first pair and one vertex is currently in use to make 2nd pair.\n\nSimilarly for 3rd pair we will have $$(2n-5)$$ ways .\n\nWhen we are making $$n^{th}$$ pair we will have just one way.\n\nMultiplying all we get $$(2n-1)(2n-3)\\ldots.3.1$$\n\nNow multiply and divide it by even terms as follows $$\\frac{(2n)(2n-1)(2n-2)(2n-3)\\ldots.3.2.1}{(2n)(2n-2)(2n-4)\\ldots.4.2}$$\n\nNow the numerator will become $$(2n)!$$ and take $$2$$ common from each term in denominator. You will get $$2^n.(n(n-1)(n-2)\\ldots 1)$$. Hence the denominator will become $$2^n. n!$$.\n\nHence we get the formula as $$\\frac{(2n)!}{2^n n!}$$. Hope this will help.\n\nGerry is absolutely correct. For 6 vertices in complete graph, we have 15 perfect matching. Similarly if we have 8 vertices then 105 perfect matching exist (7*5*3).\n\n1. For a perfect matching the number of vertices in the complete graph must be even.\n\n2. For a complete graph with n vertices (where n is even), no of perfect matchings is $$\\frac{n!}{(2!)^{n/2}(n/2)!}$$ For $$n=2m$$, it is $$\\frac{(2m)!}{(2!)^m m!}$$\n\nUsing this formula for n=6, the number of partitions = 15\n\nExplanation:\n\nThis problem is basically the problem of finding the number of \"unordered\" partitions of a set.\n\nFor partitioning a set of n elements into r unordered partitions of size $$n_1, n_2, n_3, ... , n_r$$, the formula is $$\\frac{n!}{n_1!n_2!...n_r!r!}$$\n\nFor the given question we are to partition a set of size $$2m$$ into $$m$$ unordered partitions of size 2 each ie $$\\quad n=2m; \\quad n_1=n_2=...=n_r=2; \\quad r=m$$\n\nSo the result is: $$\\frac{(2m)!}{2!2!...(m-times)m!} = \\frac{(2m)!}{(2!)^m m!}$$\n\nFor further on this topic: https://www3.nd.edu/~dgalvin1/10120/10120_S16/Topic07_6p7_Galvin.pdf\n\n\u2022 Use LaTeX please. Jun 14, 2019 at 14:36\n\u2022 @MichaelRozenberg Done in LaTex now. Jun 14, 2019 at 14:44\n\nThis problem is equivalent to finding set of $$n$$ disjoint pairs. Lets say there are $$2n$$ vertices then $$\\frac{2n}{2^n}$$ is the number of ways you can have these $$n$$ pairs. i.e, ways to make disjoint pairs out of these. E.g, If you had vertices $$1,2,3,4$$ then one of the possible way to make set out of this would be $$\\{(1,2),(3,4)\\}$$. Finally, as set is an unordered collection so final solution becomes $$\\frac{2n}{2^n*n!}$$.\n\nAnother way of arriving at the formula $$(2n-1)!! \\left( = \\frac{(2n)!}{2^n (n!)} \\right)$$ for $$2n$$ vertices is by creating cycles from matchings and matchings from cycles.\n\nIf $$\\mathcal{C}$$ is the set of cycles, then $$|\\mathcal{C}| = \\frac{(2n-1)!}{2}$$ and every cycle gives rise to exactly $$2$$ matchings by deleting every other edge.\n\nFor every matching of size $$n$$ we can obtain $$\\frac{(n-1)!}{2} \\cdot 2^n$$ cycles. First we arrange the $$n$$ edges from the matching into a cycle, as if they were vertices. There are $$\\frac{(n-1)!}{2}$$ ways of arranging them in such a manner. If we now re-expand them into edges then we have $$2$$ choices for every edge on how we want to orient it when connecting, giving $$2^n$$ options per arrangement of a matching.\n\nWith $$\\mathcal{M}$$ the set of the matchings we have $$2 |\\mathcal{C}| = \\frac{(n-1)!}{2} \\cdot 2^n |\\mathcal{M}| \\\\ \\implies |\\mathcal{M}| = \\frac{(2n-1)!}{(n-1)!}\\cdot\\frac{2}{2^n}\\cdot\\frac{n}{n} = \\frac{(2n)!}{2^nn!}$$\n\nGerry was correct (sort of) in his first statement, saying that it is the number of ways to partition the six vertices into three sets of two. However, the answer of number of perfect matching is not 15, it is 5. In fact, for any even complete graph G, G can be decomposed into n-1 perfect matchings. Try it for n=2,4,6 and you will see the pattern.\n\nAlso, you can think of it this way: the number of edges in a complete graph is [(n)(n-1)]/2, and the number of edges per matching is n/2. What do you have left for the number of matchings? n-1.\n\n\u2022 I think these are all perfect matchings for 6: 12-34-56, 12-35-46, 12-36-45, 13-24-56, 13-25-46, 13-26-45, 14-23-56, 14-25-36, 14-26-35, 15-23-46, 15-24-36, 15-26-34, 16-23-45, 16-24-35, 16-25-34. That's 15, not 5. Dec 13, 2014 at 5:14", "date": "2022-05-20 10:33:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/60894/how-to-find-the-number-of-perfect-matchings-in-complete-graphs", "openwebmath_score": 0.8500609397888184, "openwebmath_perplexity": 265.64169281744887, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363762626283, "lm_q2_score": 0.8723473763375643, "lm_q1q2_score": 0.8600790110684694}}
{"url": "https://learn.thomasjfrank.com/notion-formula-reference/formula-components/operators/mod", "text": "mod\nLearn how to use the mod (%) operator to obtain a remainder in Notion formulas.\nThe remainder (%) operator allows you to get the remainder after dividing the first operand with the second operand.\n1\nnumber % number\n2\nmod(number, number)\nNotion provides a mod() function as well as % and modoperators (I'll just reference % for the rest of this article).\nSomewhat confusingly, these do not return a true modulus value; they return a remainder (see: Remainder or Modulus? below).\nAs % and mod() output a remainder, their output values will take the sign (+/-) of the dividend.\nFor reference, the dividend is the number being divided by the divisor, which produces the quotient:\n$\\frac {dividend}{divisor} = quotient$\n\n# Example Formulas\n\n1\n19 % 12 // Output: 7\n2\n3\n19 mod 12 // Output: 7\n4\n5\nmod(-19,12) // Output: -7\n\n## Negative Divisors\n\nIf the divisor is negative:\n\u2022 mod() will treat it as a negative integer natively.\n\u2022 if you're using the % operator, you'll need to wrap your divisor in parentheses () in order to explicitly define your divisor as a negative integer.\n\u2022 x % (-y) will work exactly like mod().\n\u2022 x % -y will cause Notion to rewrite your formula as x / 100 - y, which will output an incorrect result. This is because the - is treated as a unaryMinus operator, and Notion's math engine can't correctly deal with it when it's appended to the divisor.\n1\n19 % (-12) // Output: 7\n2\n3\n// Negative value passed via a property does not need to\n4\n// be wrapped in () symbols\n5\n6\nprop(\"negative num\") == -12\n7\n19 % prop(\"negative num\") // Output: 7\n8\n9\n19 % -12 // Rewritten as 19 / 100 - 12, outputs -11.81\n10\n11\nmod(19,-12) // Output: 7\nNote: The above rules only apply if you're hard-coding a negative divisor in a formula. If you pass one via a property, it'll be parsed as a true negative value. It won't add a unaryMinus operator to your formula.\n\n## Remainder or Modulus?\n\nJust as in JavaScript, Notion\u2019s % operator calculates the remainder of two numbers, not the modulus. Confusingly, the mod() function does this as well, despite its name.\nThe remainder and modulus of two numbers will be identical when both the dividend and divisor have the same sign (+/-). If their signs differ, however, the modulus will differ from the remainder.\nMod and Remainder are not the Same\nBig Machine\nRemainder operator vs. modulo operator (with JavaScript code)\nYou can prove this using WolframAlpha:\n\u2022 -19 mod 12 results in 5\n\u2022 QuotientRemainder[-19,12] results in -7\nTo calculate a true modulus in Notion, use ((x % y) + y) % y instead:\n1\n// Using the % operator\n2\n((19 % 12) + 12) % 12\n3\n4\n// Using the mod() formula\n5\nmod(mod(19, 12) + 12, 12)\n6\n7\n// % operator example using hard-coded negative divisors\n8\n((19 % (-12)) + (-12)) % (-12)\nAs noted above in the Negative Divisors section, hard-coded negative divisors in % expressions need to be wrapped in parentheses () so Notion can parse them explicitly as negative integers (see code line 7 in the above code block).\nOtherwise, Notion will interpret the - as a unaryMinus operator, rewrite your formula to (x / 100 - y + -y) / 100 - y, and return an incorrect result. However, this isn't necessary when using the mod() function.\nRemainder vs. Modulo programming functions\n\n# Example Database\n\nThis example database shows the differing outputs of remainder and true modulus expressions.\n\n## View and Duplicate Database\n\nmod\nCollege Info Geek on Notion\nYou can check these results here:\nModulo Calculator\nCalculatorSoup\n\n## \"Remainder\" Property Formula\n\n1\nprop(\"Dividend\") % prop(\"Divisor\")\n\n## \"Modulus\" Property Formula\n\n1\n(prop(\"Dividend\") % prop(\"Divisor\") + prop(\"Divisor\")) % prop(\"Divisor\")\nInstead of using hard-coded numbers, I\u2019ve called in each property using the prop() function.", "date": "2022-10-05 00:14:01", "meta": {"domain": "thomasjfrank.com", "url": "https://learn.thomasjfrank.com/notion-formula-reference/formula-components/operators/mod", "openwebmath_score": 0.5656824111938477, "openwebmath_perplexity": 4633.55011004849, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363717170516, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8600790054671291}}
{"url": "http://mathhelpforum.com/advanced-statistics/27016-dice-probability-question.html", "text": "# Math Help - dice probability question\n\n1. ## dice probability question\n\nThis is a revision question for a phase test that i could sure use some help with!\n\nA dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.\n\ni make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.\n\nany suggestions?\n\nCheers\n\n2. Originally Posted by marciano1976\nThis is a revision question for a phase test that i could sure use some help with!\n\nA dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.\n\ni make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.\n\nany suggestions?\n\nCheers\nColour the die red and blue, then the combinations that win are (r6,b4), (r6,b5), (r6,b6), (r5,b5), (r5,b6), (r4,b6), (r6,b1), (r6,b2), (r6,b3), (r1,b6), (r2,b6), (r3,b6). That is 12 winning configurations out of a total of 36.\n\np(3)=12/36=1/3\n\np(1)=1-p(3)=2/3\n\nSo the expected number of points is:\n\n(-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3\n\nRonL\n\n3. Originally Posted by CaptainBlack\nSo the expected number of points is:\n(1)p(1)+(3)p(3)=(2/3)+(3/3)=5/3\nSurely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3\n\n4. Hello, marciano1976!\n\nA dice game consists of throwing two six-sided dice and noting the score on them.\nIf the score is greater than 9 or at least one dice shows a 6 then you score 3 points,\notherwise you lose 1 point.\n\n(a)Calculate the probability of gaining 3 points on a single throw.\n\n(b) What is the average (or expected) number of points you will gain per throw?\n\nThere are 36 possible outcomes . . .\n\n$\\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & {\\color{blue}(1,6)} \\\\\n(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\\color{blue}(2,6)} \\\\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & {\\color{blue}(3,6)} \\\\\n(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & {\\color{blue}(4,6)}\n\\\\ (5,1) & (5,2) & (5,3) & (5,4) & {\\color{blue}(5,5)} & {\\color{blue}(5,6)}\n\\end{array}$\n\n$\\begin{array}{cccccc}{\\color{blue}(6,1)} & {\\color{blue}(6,2)} & {\\color{blue}(6,3)} & {\\color{blue}(6,4)} & {\\color{blue}(6,5)} & {\\color{blue}(6,6)} \\end{array}$\n\nThere are 12 in which the sum is greater than 9 or there is at least one 6.\n\nTherefore: . $(a)\\;\\;P(\\text{gain 3 points}) \\:=\\:\\frac{12}{36} \\:=\\:\\frac{1}{3}$\n\nHence: . $\\begin{array}{ccc}P(\\text{win 3 points}) &=& \\frac{1}{3} \\\\ P(\\text{lose 1 point}) & = & \\frac{2}{3} \\end{array}$\n\nAnd: . $(b)\\;\\;\\text{Average} \\;=\\;(+3)\\left(\\frac{1}{3}\\right) + (-1)\\left(\\frac{2}{3}\\right) \\:=\\:\\frac{1}{3}$\n\nYou can expect to win an average of $\\frac{1}{3}$ of a point per throw.\n\n. . . as Plato already pointed out.\n.\n\n5. Originally Posted by Plato\nSurely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3\nOf course it should, I misread the question (again)\n\nRonL", "date": "2014-07-28 05:07:11", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/27016-dice-probability-question.html", "openwebmath_score": 0.5761073231697083, "openwebmath_perplexity": 831.434506578933, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363713038174, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8600790018346083}}
{"url": "https://gmatclub.com/forum/6-220034.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 21 Sep 2018, 03:27\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: I don't stop when I'm Tired,I stop when I'm done\nJoined: 11 May 2014\nPosts: 542\nGPA: 2.81\nWE: Business Development (Real Estate)\n\n### Show Tags\n\nUpdated on: 10 Jun 2016, 15:00\n3\n9\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n73% (01:03) correct 27% (01:54) wrong based on 244 sessions\n\n### HideShow timer Statistics\n\n$$\\frac{\\sqrt{6}}{(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}})}=$$\n\nA. $$\\frac{6}{5}$$\n\nB. $$\\frac{5}{6}$$\n\nC. 5\n\nD. $$2\\sqrt{3} +3\\sqrt{2}$$\n\nE. $$6\\sqrt{3}-6\\sqrt{2}$$\n\n_________________\n\nMd. Abdur Rakib\n\nPlease Press +1 Kudos,If it helps\nSentence Correction-Collection of Ron Purewal's \"elliptical construction/analogies\" for SC Challenges\n\nOriginally posted by AbdurRakib on 10 Jun 2016, 13:05.\nLast edited by Bunuel on 10 Jun 2016, 15:00, edited 1 time in total.\nRenamed the topic and edited the question.\nVeritas Prep GMAT Instructor\nJoined: 01 Jul 2017\nPosts: 49\nLocation: United States\nConcentration: Leadership, Organizational Behavior\n\n### Show Tags\n\nUpdated on: 09 Jan 2018, 07:13\n1\nThe trick behind this question is to use the answer choices to guide your mathematics. (In other words, don't do math just because you can... do math because it gets you closer to your target!)\n\nYou should notice immediately that the answer choices are much simpler. Three of the five answer choices contain no fractions at all. None of the answer choices contain $$\\sqrt{6}$$. Use this to your advantage to help you think about the question. One quick way to get rid of both the \\sqrt{6}[/m] and the \"fractions inside of fractions\" issue is to do what I call \"Multiply by 1\". In this case, multiply the original expression by $$\\sqrt{6}/\\sqrt{6}$$. Since $$\\sqrt{6}/\\sqrt{6}=1$$, we don't actually change the value; we just change the shape of the expression. Thus,\n\n$$\\frac{\\sqrt{6}}{\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}}}*(\\frac{\\sqrt{6}}{\\sqrt{6}})=\\frac{6}{\\sqrt{3}+\\sqrt{2}}$$\n\nNow the expression that remains is starting to look more like the answer choices, but we still need to get rid of the compound denominator $$(\\sqrt{3}+\\sqrt{2})$$. Once again, look to the answer choices for clues. Both answer choices A and E contain a $$6$$. But it should be clear that answer choice A cannot be an option. $$(\\sqrt{3}+\\sqrt{2})$$ does not equal 5. So, let's try to turn what we have into answer choice E.\n\nAnswer choice E contains a factor of $$(\\sqrt{3}-\\sqrt{2})$$, which should immediately get us thinking about the possibility of difference of squares. Multiplying our reduced expression by \"1\" again gives us:\n\n$$(\\frac{6}{\\sqrt{3}+\\sqrt{2}})*(\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}})$$\n\nUsing the concept of difference of squares, we can quickly see that the denominator of this fraction simplifies down completely. Watch this \"disappearing math\" trick:\n\n$$(\\sqrt{3}+\\sqrt{2})*(\\sqrt{3}-\\sqrt{2}) = (\\sqrt{3})^2 - (\\sqrt{2})^2 = 3-2 = 1$$\n\nWhat began as a \"Mathugly\" arithmetic issue quickly simplifies down to $$6\\sqrt{3}-6\\sqrt{2}$$.\n\nThe answer is E.\n\nAddendum: As with many quantitative questions on the GMAT, there is \"more than one way to skin a CAT\" for this question (pun intended.) We can also use the answer choices not just as targets that guide our math, but as leverage for approximating. Watch this. If we recognize that $$\\sqrt{3}\\approx{1.7}$$ and $$\\sqrt{2}\\approx{1.4}$$, then the we can approximate our answer very quickly:\n$$\\frac{6}{\\sqrt{3}+\\sqrt{2}}\\approx{\\frac{6}{1.4+1.7}}\\approx{\\frac{6}{3.1}}\\approx{2}$$\n\nNow, we look at the answer choices to see which one is roughly equal to 2.\n\n(A) $$\\frac{6}{5}$$ <-- quickly eliminated. Too small.\n\n(B) $$\\frac{5}{6}$$ <-- quickly eliminated. Too small.\n\n(C) $$5$$ <-- quickly eliminated. Too big.\n\n(D) $$2\\sqrt{3} + 3\\sqrt{2}$$ <-- quickly eliminated. Too big.\n\n(E) $$6\\sqrt{3} - 6\\sqrt{2}$$ <-- The right answer. $$6(\\sqrt{3} - \\sqrt{2})\\approx{6(1.7-1.4)}\\approx{2}$$\n\nAnyway you look at it, the answer is still E.\n_________________\n\nAaron J. Pond\nVeritas Prep Elite-Level Instructor\n\nHit \"+1 Kudos\" if my post helped you understand the GMAT better.\nLook me up at https://www.veritasprep.com/gmat/aaron-pond/ if you want to learn more GMAT Jujitsu.\n\nOriginally posted by AaronPond on 08 Jan 2018, 17:04.\nLast edited by AaronPond on 09 Jan 2018, 07:13, edited 1 time in total.\n##### General Discussion\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49298\n\n### Show Tags\n\n10 Jun 2016, 15:13\n3\n1\nAbdurRakib wrote:\n$$\\frac{\\sqrt{6}}{(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}})}=$$\n\nA. $$\\frac{6}{5}$$\n\nB. $$\\frac{5}{6}$$\n\nC. 5\n\nD. $$2\\sqrt{3} +3\\sqrt{2}$$\n\nE. $$6\\sqrt{3}-6\\sqrt{2}$$\n\n$$\\frac{\\sqrt{6}}{(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}})}=\\frac{\\sqrt{6}}{(\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{2}\\sqrt{3}})}=\\frac{\\sqrt{6}}{(\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{6}})}=\\frac{\\sqrt{6}*\\sqrt{6}}{\\sqrt{3}+\\sqrt{2}}=\\frac{6}{\\sqrt{3}+\\sqrt{2}}$$.\n\nMultiply by $$\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}}$$:\n\n$$\\frac{6}{\\sqrt{3}+\\sqrt{2}}*\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}}=\\frac{6*(\\sqrt{3}-\\sqrt{2})}{3-2}=6*\\sqrt{3}-6\\sqrt{2}$$.\n\n_________________\nCEO\nJoined: 12 Sep 2015\nPosts: 2864\n\n### Show Tags\n\n25 Jul 2017, 08:45\n1\nTop Contributor\nAbdurRakib wrote:\n$$\\frac{\\sqrt{6}}{(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}})}=$$\n\nA. $$\\frac{6}{5}$$\n\nB. $$\\frac{5}{6}$$\n\nC. 5\n\nD. $$2\\sqrt{3} +3\\sqrt{2}$$\n\nE. $$6\\sqrt{3}-6\\sqrt{2}$$\n\nNice question!\n\nHere's another approach:\n\nLet's first deal with the denominator: 1/\u221a2 + 1/\u221a3 = \u221a3/\u221a6 + \u221a2/\u221a6\n= (\u221a3 + \u221a2)/\u221a6\n\nSo....$$\\frac{\\sqrt{6}}{(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}})}=$$ \u221a6/[(\u221a3 + \u221a2)/\u221a6]\n\n= (\u221a6)(\u221a6)/(\u221a3 + \u221a2)\n\n= 6/(\u221a3 + \u221a2)\n\nNow we can use the fact that \u221a3 \u2248 1.7 and \u221a2 \u2248 1.4 to get....\n\u2248 6/(1.7 + 1.4)\n\u2248 6/(3.1)\n\nSince 6/3 = 2, we know that 6/(3.1) = a number a bit smaller than 2\n\nNow check the answer choices to see which one evaluates to be a number a bit smaller than 2\n\nA. $$\\frac{6}{5}$$ ELIMINATE\n\nB. $$\\frac{5}{6}$$ ELIMINATE\n\nC. 5 ELIMINATE\n\nD. $$2\\sqrt{3} +3\\sqrt{2}$$ \u2248 2(1.7) + 3(1.4) \u2248 a number that's bigger than 6 ELIMINATE\n\nE. $$6\\sqrt{3}-6\\sqrt{2}$$[/quote] = 6(\u221a3 - \u221a2) \u2248 6(1.7 - 1.4) \u2248 6(0.3) \u2248 1.8 PERFECT!\n\nCheers,\nBrent\n_________________\n\nBrent Hanneson \u2013 GMATPrepNow.com\n\nSign up for our free Question of the Day emails\n\nVeritas Prep GMAT Instructor\nJoined: 01 Jul 2017\nPosts: 49\nLocation: United States\nConcentration: Leadership, Organizational Behavior\n\n### Show Tags\n\n08 Jan 2018, 16:36\n$$\\frac{\\sqrt{6}}{\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}}}=$$\n\n(A) $$\\frac{6}{5}$$\n\n(B) $$\\frac{5}{6}$$\n\n(C) $$5$$\n\n(D) $$2\\sqrt{3} + 3\\sqrt{2}$$\n\n(E) $$6\\sqrt{3} - 6\\sqrt{2}$$\n_________________\n\nAaron J. Pond\nVeritas Prep Elite-Level Instructor\n\nHit \"+1 Kudos\" if my post helped you understand the GMAT better.\nLook me up at https://www.veritasprep.com/gmat/aaron-pond/ if you want to learn more GMAT Jujitsu.\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49298\n\n### Show Tags\n\n08 Jan 2018, 20:33\nAaronPond wrote:\n$$\\frac{\\sqrt{6}}{\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{3}}}=$$\n\n(A) $$\\frac{6}{5}$$\n\n(B) $$\\frac{5}{6}$$\n\n(C) $$5$$\n\n(D) $$2\\sqrt{3} + 3\\sqrt{2}$$\n\n(E) $$6\\sqrt{3} - 6\\sqrt{2}$$\n\n________________\nMerging topics.\n_________________\nRe: 6^(1/2)/(1/2^(1/2)+1/3^(1/2)) &nbs [#permalink] 08 Jan 2018, 20:33\nDisplay posts from previous: Sort by\n\n# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-21 10:27:20", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/6-220034.html", "openwebmath_score": 0.7585957646369934, "openwebmath_perplexity": 5842.419302447171, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9859363762626284, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8600789979803212}}
{"url": "https://math.stackexchange.com/questions/1463384/help-with-linear-algebra-linear-transformations", "text": "# Help with linear algebra linear transformations\n\nIs there a linear transformation $T : R ^2 \u2192 R ^3$ such that $T(1, 1) = (1, 0, 2)$ and $T(2, 3) = (1, \u22121, 4)$. Justify your answer.\n\nI'm not sure what exactly this question is asking for. How would you go about tackling this question. I know I'm supposed to show some form of working on this website but I really have no idea where to start\n\n\u2022 Try to start with the assumption that there is such a transformation. Then try to explicitly construct it by finding a matrix $A$ such that $T(x) = Ax$ for all $x\\in \\Bbb R^2$. If you're able to, then one clearly exists. If you run into some contradiction, then one clearly doesn't exist. \u2013\u00a0user137731 Oct 4 '15 at 4:25\n\u2022 As always, reviewing the terms in the problem (what a linear transformation is) can help. \u2013\u00a0Nate 8 Oct 4 '15 at 4:30\n\nIn general if V and W are vector spaces and $\\{e_i\\}_{i\\in I}$ is a basis for $V$. And $f(e_i)=w_i\\in W$ for all $i \\in I$.\n\nThen f can be extended to a linear mapping $T: V \\rightarrow W$ such that $T(e_i)=f(e_i)$. (You can write down the proof in the same way as the above answers)\n\nIn your case, $V= \\mathbb R ^2$ and $W=\\mathbb R^3$ $e_1=(1,1), e_2=(2,3)$ is a basis for $\\mathbb R^2$ and $f(e_1)=(1,0,2)$ and $f(e_2)=(1,-1,4)$.\n\nTherefore, you can extend it to a linear mapping by the above discussion.\n\nNote: 1) $(1,0,2)$ and $(1,-1,4)$ doesn't have any significance role to play.If $(a,b,c), (d,e,f)$ still you could extend it to a linear map.\n\n2) The above problem has a general philosophy - any map (or morphism) from a algebraic(mathematical) structure to another algebraic(mathematical) structure can be defined by only defining it on the generators, provided you can talk of generators. For example, to define a group homomorphism from a cyclic to group to another group, it is enough to define it for just for the generators.\n\nNote that $\\{(1,1),(2,3)\\}$ is a basis of $\\mathbb R^2$ and if you know the image of basis elements then you know the whole transformation.\n\nIn this case observe $\\forall x,y \\in \\mathbb R^2$ we have $(x,y)= (3x-2y)(1,1)+(y-x)(2,3)$\n\nSo define , $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$\n\nTo answer the comment by Babai:\n\nThe construction of $T$ shows that if $u\\in \\mathbb R^2,u=\\alpha a+\\beta b$ where $a=(1,1)$ and $b=(2,3)$ then $Tu=\\alpha Ta+\\beta Tb$\n\nThus $T(u_1+\\lambda u_2)=T(\\alpha_1 a+\\beta_2 b+\\lambda (\\alpha_2 a+\\beta_2 b))=T((\\alpha_1+\\lambda \\alpha_2) a+(\\beta_1+\\lambda \\beta_2) b)=(\\alpha_1+\\lambda \\alpha_2) Ta+(\\beta_1+\\lambda \\beta_2) Tb=\\alpha_1 Ta+\\beta_2 Tb+\\lambda (\\alpha_2 Ta+\\beta_2 Tb)=Tu_1+\\lambda Tu_2$\n\n\u2022 The question is why is T linear? In your answer you assumed T is linear. \u2013\u00a0Babai Oct 4 '15 at 4:41\n\u2022 @Babai I have added it in my answer. \u2013\u00a0usermath Oct 4 '15 at 5:22\n\u2022 $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$. Why is this true? This is only true if $T$ is linear, and that is what you nee to prove. \u2013\u00a0Babai Oct 4 '15 at 5:38\n\u2022 Instead of writing \" So $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, if you write \"define $T$ as $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, then your answer will make perfect sense. \u2013\u00a0Babai Oct 4 '15 at 5:45\n\u2022 @Babai ok I got it and corrected. Thnx \u2013\u00a0usermath Oct 4 '15 at 6:27\n\nDefine $T: \\mathbb{R}^2 \\to \\mathbb{R}^3$ by for $(x,y) \\in \\mathbb{R}^2$, write $(x,y) = a(1,1) + b(2,3)$, and let $T(x,y) = a(1,0,2) + b(1,-1,4).$\n\nIn the above, $T$ is obviously linear. The problem is why such a $T$ is well-defined, more specifically, why there exists such scalars $a,b$ and the scalars are unique.\n\n\u2022 Why is T obviously linear? \u2013\u00a0Babai Oct 4 '15 at 4:42\n\u2022 @Babai Have you tried to show it before you ask this question? It is a one line proof. \u2013\u00a0Empiricist Oct 4 '15 at 5:17\n\u2022 It is not the question of whether I have tired it or not. Your answer is not justifiable. You write \"$T(x,y) = a(1,0,2) + b(1,-1,4).$, In the above, T is obviously linear.\" Now, $T(x,y) = a(1,0,2) + b(1,-1,4).$ , this is not true unless $T$ is linear. And you assumed it to say $T$ is linear, which is not correct. \u2013\u00a0Babai Oct 4 '15 at 5:35\n\u2022 I am sorry. You are defining $T$. Your answer is perfect. And it is obviously linear. \u2013\u00a0Babai Oct 4 '15 at 5:44\n\nIf there exists a transform, there must exist a matrix representing it, let us call such a matrix $\\bf T$. Now what we demand is that $${\\bf T}\\left[\\begin{array}{cc}1&2\\\\1&3\\end{array}\\right] = \\left[\\begin{array}{rr}1&1\\\\0&-1\\\\2&4\\end{array}\\right]$$ Now we can solve this, if we divide by $\\left[\\begin{array}{cc}1&2\\\\1&3\\end{array}\\right]$ to the left of each side: $${\\bf T} = \\left[\\begin{array}{rr}1&1\\\\0&-1\\\\2&4\\end{array}\\right] \\left[\\begin{array}{cc}1&2\\\\1&3\\end{array}\\right]^{-1} = ?$$\n\nNow we can see on the matrices that this must have a solution since both matrices are of rank 2.\n\nThen after we have solved this, what remains is to perform the matrix multiplication ${\\bf T}\\left[\\begin{array}{cc}1&2\\\\1&3\\end{array}\\right]$ and confirm it really becomes as the first equation says it should.", "date": "2020-01-28 17:12:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1463384/help-with-linear-algebra-linear-transformations", "openwebmath_score": 0.901239812374115, "openwebmath_perplexity": 186.61036492966988, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290897113961, "lm_q2_score": 0.8856314753275017, "lm_q1q2_score": 0.8600624884545576}}
{"url": "http://grandespirito.it/page.php?tag=Backslash-in-math-20e214", "text": "This list of mathematical symbols by subject shows a selection of the most common symbols that are used in modern mathematical notation within formulas, grouped by mathematical topic.\n\nRegular expressions first evaluates a special character in the context of regular expressions: if it sees a dot, then it knows to match any one character. Advanced Algebra.\n\nDie folgende Tabelle listet die wichtigsten Symbole und Abk\u00fcrzungen auf, die in mathe online eine Rolle spielen. The backslash, \\ , usually refers to the relative complement of two sets. Geometry. If we look at writing, people do not use backslash in writing. The set difference A\\B is defined by A\\B={x:x in A and x not in B}.\n\nThe most common, most frequently used math symbols: Forward Slash (Division Sign) shown and explained . Please be sure to answer the question. How does one insert a \"\\\" (backslash) into the text of a LaTeX document? Thanks for contributing an answer to TeX - LaTeX Stack Exchange!\n\nThe difference between a backslash and a forward slash is defined below: Backslash: \\ Forward Slash: / A good way to remember the difference between a backslash and a forward slash is that a backslash leans backwards ( \\ ), while a forward slash leans forward ( / ).\n\nP. PvtBillPilgrim. Infix operator. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. ... [Backslash] Unicode: 2216.\n\nIf A is an n by m matrix and b is an p by q matrix then A\\b is defined (and is calculated by Matlab) if m=p. It only takes a minute to sign up. There is always some type of calculations to be made. But avoid \u2026 Asking for help, clarification, or responding to other answers. Meaning of the backslash operator on sets. A regular expression to match the IP address 0.0.0.0 would be: 0\\.0\\.0\\.0. In mathematics, the decimal point (.)\n\nIn a probability \\$P(A \\setminus B)\\$, it means \"the probability that event \\$A\\$ occurs but event \\$B\\$ does not occur\". EDIT: This an issue for keyboard layouts where backslash is activated by a combination of AltGr + backslash. Falls das Symbol in den Mathematischen Hintergr\u00fcnden definiert oder beschrieben wird, ist diese Eintragung ein Link, der Sie an die entsprechende Stelle f\u00fchrt.\n\nThe backslash symbol \\ is used to denote a set difference, quotient group, or integer division. Also used to separate arguments of elliptic functions. Ask Question Asked 5 years, 5 months ago. Formally: \u2216 = {\u2208 \u2223 \u2209}.\n\nHere, the backslash symbol is defined as Unicode U+2216. At any rate, math mode provides \\sim, \\backslash, and \\setminus (the latter two appear to look the same and differ only by spacing in math mode).\n\nYou just have to type the name of the letter after a backslash: if the first letter is lowercase, you will get a lowercase Greek letter, if the first letter is uppercase (and only the first letter), then you will get an uppercase letter. What does the backslash denote in probability theory? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Applied Mathematics. Ask Question Asked 7 years, 6 months ago. \\$\\begingroup\\$ The backslash is an escape character in \\$\\LaTeX\\$, cf.\n\n... Backslash notation: \u2026 is used to separate the whole part of a number from the fractional part. What is Backslash? In this particular example, the norm of the residual is zero, and an exact solution is obtained, although rcond is small. Making statements based on opinion; back them up with references or personal experience. Revolutionary knowledge-based programming language. x y is by default interpreted as Backslash [x, y].\n\nIn an application of a probability problem you could be given an arbitrary probability like: P ( a \\ b) = .50 or 50%. For non-square and singular systems, the operation A\\b gives the solution in the least squares sense. The relative complement of A in B is denoted B \u2216 A according to the ISO 31-11 standard.It is sometimes written B \u2212 A, but this notation is ambiguous, as in some contexts it can be interpreted as the set of all elements b \u2212 a, where b is taken from B and a from A.. Foundations of Mathematics.", "date": "2020-10-22 17:14:24", "meta": {"domain": "grandespirito.it", "url": "http://grandespirito.it/page.php?tag=Backslash-in-math-20e214", "openwebmath_score": 0.8377140164375305, "openwebmath_perplexity": 1680.5994418269906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9711290922181331, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8600624833459076}}
{"url": "https://math.stackexchange.com/questions/377683/length-of-period-of-decimal-expansion-of-a-fraction", "text": "# Length of period of decimal expansion of a fraction\n\nEach rational number (fraction) can be written as a decimal periodic number. Is there a method or hint to derive the length of the period of an arbitrary fraction? For example $1/3=0.3333...=0.(3)$ has a period of length 1.\n\nFor example: how to determine the length of a period of $119/13$?\n\n\u2022 I can't find an algorithm that finds the value for a number that isn't prime, i.e. 10 \u2013\u00a0Code Whisperer Jan 13 '18 at 9:11\n\u2022 \u2013\u00a0Watson Dec 2 '18 at 16:31\n\nAssuming there are no factors of $2,5$ in the denominator, one way is just to raise $10$ to powers modulo the denominator. When you find $-1$ you are halfway done. Taking your example: $10^2\\equiv 9, 10^3\\equiv -1, 10^6 \\equiv 1 \\pmod {13}$ so the repeat of $\\frac 1{13}$ is $6$ long. It will always be a factor of Euler's totient function of the denominator. For prime $p$, that is $p-1$.\n\n\u2022 You have to look for $1$, because sometimes you won't find $-1$. For example: $10^2\\equiv 18, 10^3\\equiv 16, 10^4 \\equiv 37, 10^5 \\equiv 1 \\pmod {41}$, so the repeat of $\\frac 1{41}$ is $5$ long. \u2013\u00a0jcsahnwaldt says GoFundMonica Mar 26 '15 at 5:37\n\u2022 You assume that 'there are no factors of 2,5 in the denominator' is it actually required? \u2013\u00a0Shamdor May 5 '15 at 7:28\n\u2022 I mean, 1/(2^i * 5^j * d') has the same period length as 1/d', right? \u2013\u00a0Shamdor May 5 '15 at 7:46\n\u2022 That is correct. You will have $\\max(i,j)$ non-repeating digits, then the repeat will start \u2013\u00a0Ross Millikan May 5 '15 at 13:48\n\nThis answer seeks to explain why Ross Millikan's answer works, and provides further information on techniques to speed up the process of seeking the period:\n\nConsider the fraction $\\frac17$. The decimal expansion of this is $$\\frac17 = 0.\\overline{142857}$$ for a period of 6. Now consider what happens when we multiply it by $10^6$: $$10^6\\times\\frac17 = 142857.\\overline{142857}$$ Subtracting the original fraction from this gives $$(10^6-1)\\times\\frac17 = 142857$$ And so, we have $$\\frac17 = \\frac{142857}{10^6-1}$$ As you can see, the denominator is one less than a power of 10, and the power is the period of the decimal expansion. This is no accident, and works for any fraction - if you can rewrite it in this form, the denominator reveals the period.\n\nNow, rearrange the equation: $$10^6-1 = 142857\\times 7$$ So $10^6$ must be one more than a multiple of 7 (or, more generally, $10^n$ must be one more than a multiple of $d$, where $d$ is the denominator of the fraction and $n$ is the period of the decimal expansion) - indeed, it must be the smallest power of 10 (larger than 1) that has this property.\n\nAs such, we can use modular arithmetic to look for the period. Since $a\\times d\\equiv 0 \\pmod d$, we have that $10^n-1\\equiv0 \\pmod d$, or $$10^n \\equiv 1\\pmod d$$ And therefore you can just look for the smallest $n>0$ satisfying this.\n\nOf course, there are other approaches to gain the same result, but they're all fundamentally variants of the same idea. That said, if you can factor $\\phi(d)$ - the euler totient function of the denominator - then you can accelerate the process of looking for the smallest $n$. For example, when checking 13, you have $\\phi(13)=12$, so $n\\in\\{1,2,3,4,6,12\\}$ (as these are the factors of 12) - this can save you a lot of computation (especially where $\\phi(d)$ has only a few large factors and 2).\n\nFor example, $\\phi(167)=166 = 2\\times83$, so $n\\in\\{1,2,83,166\\}$. Therefore, we need to check only these four, and we can do it quite efficiently. Obviously, neither $10$ nor $100=10^2$ are equivalent to 1 mod 167, so we only need to actually check 83. For this we can use binary exponentiation. Note that $83 = 2^6 + 2^4 + 2^1 + 2^0$. So we can write \\begin{align} 10^{83} &= 10^{2\\times(2^5 + 2^3 + 1)}\\times 10\\\\ &= (10^{2^3\\times(2^2+1)}\\times 10)^2 \\times 10\\\\ &= ((10^{2^2}\\times10)^{2^3}\\times 10)^2 \\times 10 \\end{align} So, working in modular arithmetic, we can go \\begin{align} 10^{83} &\\equiv ((10^{2^2}\\times10)^{2^3}\\times 10)^2 \\times 10 \\mod 167\\\\ &\\equiv ((100^2\\times 10)^{2^3}\\times 10)^2\\times10\\mod167\\\\ &\\equiv ((147\\times 10)^{2^3}\\times 10)^2\\times10\\mod167\\\\ &\\equiv (134^{2^3}\\times 10)^2\\times10\\mod167\\\\ &\\equiv (87^{2^2}\\times 10)^2\\times10\\mod167\\\\ &\\equiv (54^2\\times 10)^2\\times10\\mod167\\\\ &\\equiv (77\\times 10)^2\\times10\\mod167\\\\ &\\equiv 102^2\\times10\\mod167\\\\ &\\equiv 50\\times10\\mod167\\\\ &\\equiv 166\\mod167 \\end{align} This is the same as $-1\\pmod{167}$, so $n=166$ is the only possible period, and $\\frac1{167}$ has a period of 166.\n\nAlso note that you don't actually have to expand out the product like that. You can just write the number in binary ($83_{10} = 1010011_2$), then work through the binary digits left-to-right - start with 1, and for each digit, $b$, multiply by $10^b$. Square it after each digit except the last one.\n\nThe length of the period is given by the multiplicative order of $10 \\pmod q$, where $q$ is your quotient. It is closely related to the discrete logarithm. Wikipedia lists several algorithms that are faster than going through all the powers of ten, which is relevant if you're dealing with very large numbers (hundreds of digits).\n\nFirst you have to simplify to the lowest integer denominator.\n\nThen, the proof for finiteness of repeating part of the decimal representation involves the pigeonhole principle. When you keep on dividing, at some point you will run into a repeat value for the remainder. Given that the remainders that keep the operation going range from 1 to (denominator-1), the possible number of pigeonholes is (denominator-1) as the number of pigeonholes; so that's your worst possible case scenario - in any base.\n\nThe particular cases, as described by the other answers, depend on the base, because the denominator may have a special relationship with the base.\n\n\u2022 e.g., in base 10 fractions over 3 will have one repeating decimal, (3) or (6), as 3 divides 9(=10-1), and the remainder will keep repeating.", "date": "2020-02-21 07:38:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/377683/length-of-period-of-decimal-expansion-of-a-fraction", "openwebmath_score": 0.974265992641449, "openwebmath_perplexity": 275.79690002552144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380087100573, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.860045718242468}}
{"url": "https://math.stackexchange.com/questions/931742/proof-check-linear-algebra", "text": "# Proof Check - Linear Algebra\n\nI'm new to more formal mathematics and I was wondering if I could write my attempt at a small proof here and see what others thought and if you could point out any fatal flaws or just offer some tips? I would appreciate it.\n\n## Theorem\n\nLet u,v, and w be distinct vectors in a vector space V. Show that if {u,v,w} is a basis for V, then {u+v+w, v+w, w} is also a basis for V.\n\n## Proof\n\n: Let B = {u,v,w}. As B is a basis for V, the vectors in B generate V and are linearly independant. Therefore, the equation\n\na1 * u + a2 * v + a3 * w = 0\n\nIs only true for a1, a2, a3 all equal to 0.\n\nNow consider the set B' = {u+v+w, v + w, w}\n\nIf we examine the equation\n\nc1 ( u+v+w) + c2 (v+w) + c3 (w) = 0\n\nc1(u) + (c1 + c2)v + (c1 + c2 + c3)w = 0\n\nThen we say that c1 = -c2, and c3 = -c1-c2, but from the first term we see that c1 must equal zero, therefore all the other coefficients are 0 as well, and so the set of vectors is linearly independant.\n\n## My questions\n\n1.) I'm not sure if the part of the proof where I show that the solution to the equation for the vectors of B' being equal to 0 is connected well enough to the bit about B being a basis for V. Am I allowed to say that C1 must be equal to 0 only because I've already assumed that u,v, and w are all linearly independent already?\n\n2.) I know that if B generates V that B' which is just set formed from L.I. vectors into other L.I. vectors must also generate V, but I'm not sure exactly why I can say this even though I feel it strongly.\n\n\u2022 Your argument is perfect. \u2013\u00a0Ted Shifrin Sep 15 '14 at 1:20\n\nWith regards to point 1), what you say is correct. As you say, you use the linear independence of $B$ to conclude that when $c_1(u) + (c_1 + c_2)v + (c_1 + c_2 + c_3)w = 0$, it must be the case that $c_{1} = 0$, and so on.\nWith regards to 2), one way to show that $B'$ spans the space is to show that it generates $B$: any linear combination of vectors from $B$ can then be converted into a linear combination of vectors from $B'$ by replacing $u,v,w$ by their expressions in terms of the list $B'$. So in this case $(v+w) - w = v$ and $(u+v+w) - (v+w) = u$.\nBy the way, you only need to show either that $B'$ form a linear independent set, or that they span the vector space, since the list has length $3$. (This is the result that a linear independent list of vectors, whose length is equal to the dimension of the space forms a basis, and similarly for spanning sets).", "date": "2019-10-20 18:39:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/931742/proof-check-linear-algebra", "openwebmath_score": 0.7522115707397461, "openwebmath_perplexity": 226.1957523465021, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137858267906, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8600349040216181}}
{"url": "https://math.stackexchange.com/questions/108010/when-is-the-closure-of-an-open-ball-equal-to-the-closed-ball", "text": "# When is the closure of an open ball equal to the closed ball?\n\nIt is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric $$d(x,y)= \\begin{cases} 0\\qquad&\\text{if and only if x=y}\\\\ 1&\\text{otherwise} \\end{cases}$$ The open unit ball of radius $1$ around any point $x$ is the singleton set $\\{x\\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything.\n\nI like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal?\n\n\u2022 In the Euclidean metric space $R^n$ it is necessarily true. Feb 11, 2012 at 2:26\n\u2022 Right, but Euclidean space is known for, among other things, being perfect in almost every way. What about spaces like $L^{p}$ or $H^{p}$? I'm looking to see how far our intuition of Euclidean spaces and the standard metric extends. Feb 11, 2012 at 2:28\n\u2022 Regarding your question about $L^p$ or $H^p$, it is true in every normed space. If $\\|x-y\\|=r$, then for $0<t<1$, $\\|x-(tx+(1-t)y)\\|=(1-t)\\|x-y\\|<r$, and $y=\\lim\\limits_{t\\searrow 0}tx+(1-t)y$. Feb 11, 2012 at 3:37\n\u2022 The property may fail for subspaces of Euclidean space. See here: link link Feb 11, 2012 at 11:53\n\nHere is a characterization that is straight from the definitions, but which it seems may be useful when verifying that a particular space has the property.\n\nFor any metric space $$(X,d)$$, the following are equivalent:\n\n\u2022 For any $$x\\in X$$ and radius $$r$$, the closure of the open ball of radius $$r$$ around $$x$$ is the closed ball of radius $$r$$.\n\u2022 For any two distinct points $$x,y$$ in the space and any positive $$\\epsilon$$, there is a point $$z$$ within $$\\epsilon$$ of $$y$$, and closer to $$x$$ than $$y$$ is. That is, for every $$x\\neq y$$ and $$\\epsilon\\gt 0$$, there is $$z$$ with $$d(z,y)<\\epsilon$$ and $$d(x,z).\n\nProof. If the closed ball property holds, then fix any $$x,y$$ with $$r=d(x,y)$$. Since the closure of $$B_r(x)$$ includes $$y$$, the second property follows. Conversely, if the second property holds, then if $$r=d(x,y)$$, then the property ensures that $$y$$ is in the closure of $$B_r(x)$$, and so the closure of the open ball includes the closed ball (and it is easy to see it does not include anything more than this, since if $$g$$ belongs to the closure of $$B_r(x)$$ then $$d(x,g) \\le r$$ and so $$g$$ must also belong to the closed ball of radius $$r$$ centered at $$x$$). QED\n\n\u2022 Thanks for the response! Your proof makes sense. Feb 11, 2012 at 3:10\n\u2022 How to understand what is not working initially in order to build your characterization? In other words, what is the intuition behind this? Thanks.\n\u2013\u00a0user169373\nSep 23, 2014 at 20:41\n\u2022 @MarcGato Think about the case where you have an isolated point $x$, so that $B_r(x)$ contains only $x$, for some $r>0$, but there is a point $y$ at distance $r$ to $x$. The closure of $B_r(x)$ is just $\\{x\\}$, since there are no other limit points to add and so this set is already closed. My condition ensures that every point at distance $r$ from $x$ is a limit of points of distance less than $r$ from $x$. That is why all such points get added to the closure of the open ball.\n\u2013\u00a0JDH\nSep 23, 2014 at 21:10\n\u2022 The way that $y$ gets into the closure of $B_r(x)$ is that there must be points in $B_r(x)$ that are as close as you like to $y$. Those points are the $z$'s in the property.\n\u2013\u00a0JDH\nApr 26, 2017 at 11:03\n\u2022 @James Well since y is in closure of B(x,r), for any $\\epsilon>0$, $B(y,\\epsilon) \\cap B(x,r) \\neq \\phi$. Choose z from this intersection. Then, $d(x,z)<r=d(x,y)$ Jan 14, 2021 at 11:40\n\nLet $(X,\\|\\cdot\\|)$ be a normed linear space. Then $\\overline{B_1(0)}=\\bar{B}_1(0)$.\n\nProof. Observe that $\\overline{B_1(0)}$ is the smallest closed set containing $B_1(0)$ and $B_1(0)\\subset \\bar{B}_1(0)$, so trivially $\\overline{B_1(0)}\\subset\\bar{B}_1(0)$. Now to show $\\bar{B}_1(0)\\subset \\overline{B_1(0)}$. Observe that, $\\bar{B}_1(0)=B_1(0)\\cup \\partial B_1(0)$, i.e., for all $x\\in \\partial B_1(0), \\, \\exists x_n\\in B_1(0)$ such that $\\|x_n-x\\|\\to 0$: for any given $x\\in \\partial B_1(0),$ let $x_n=(1-\\frac{1}{n})x, \\, n\\in \\mathbb{N}.$ Then show $x_n\\in B_1(0)$ and $\\|x_n-x\\|\\to 0$.\n\n\u2022 How would you suggest approaching this, showing that the interior of the closed ball is equal to the open ball? Jan 31, 2018 at 0:15\n\u2022 ... where a \"normed linear space\" means over some field with an archimedean valuation. The result is not true e.g. over $p$-adics where that $1-1/n$-construction fails. -- Also, note that $\\delta B_1(0)$ here should really be defined as $\\{x \\in X: \\lvert \\lvert x \\rvert \\rvert =1 \\}$, as only a posteriori we show this is the boundary also in the topological sense. Apr 11, 2020 at 7:14\n\nMore general, we can prove that for any normed space $$(V,\\|\\cdot\\|)$$, if $$x_0\\in V$$ and $$R>0$$, then $$\\overline{B(x_0;R)}^{\\|\\cdot\\|}=B[x_0;R],$$ where $$\\overline{B(x_0;R)}^{\\|\\cdot\\|}$$ is the closure of the open ball $$B(x_0;R)$$ in topology induced by $$\\|\\cdot\\|$$.\n\nThe inclusion $$\\overline{B(x_0;R)}^{\\|\\cdot\\|}\\subseteq B[x_0;R]$$ is trivial, because $$B[x_0;R]$$ is a closed set that contains $$B(x_0;R)$$ and $$\\overline{B(x_0;R)}^{\\|\\cdot\\|}$$ is the smallest closed set that contains $$B(x_0;R)$$.\n\nTo show that $$B[x_0;R]\\subseteq\\overline{B(x_0;R)}^{\\|\\cdot\\|}$$, we can prove that any element of $$B[x_0;R]$$ is the limit of some sequence in $$B(x_0;R)$$. Let $$x\\in B[x_0;R]$$, then of course $$\\|x-x_0\\|\\leqslant R.$$ Now defines, for each $$n\\in\\Bbb{N}$$, $$x_n:=\\left(1-\\frac{1}{n}\\right)x+\\frac{1}{n}x_0.$$\n\nClaim 1: The sequence $$(x_n)$$ converges to $$x$$ in the norm $$\\|\\cdot\\|$$. In fact, $$\\begin{eqnarray} \\|x_n-x\\| & = & \\left\\|\\left(1-\\frac{1}{n}\\right)x+\\frac{1}{n}x_0-x\\right\\| \\\\ & = & \\frac{1}{n}\\underbrace{\\|x-x_0\\|}_{\\leqslant R} \\\\ & \\leqslant & \\frac{R}{n}\\longrightarrow0,\\quad\\text{when}\\ n\\to\\infty, \\end{eqnarray}$$ which proves that $$x_n\\to x$$ in $$\\|\\cdot\\|$$.\n\nClaim 2: For all $$n\\in\\Bbb{N}$$, $$x_n\\in B(x_0;R)$$, that is, $$(x_n)$$ is a sequence in the set $$B(x_0;R)$$. This is obviously by the construction of $$x_n$$, since $$\\begin{eqnarray} \\|x_n-x_0\\| & = & \\left\\|\\left(1-\\frac{1}{n}\\right)x+\\frac{1}{n}x_0-x_0\\right\\| \\\\ & = & \\left\\|\\left(1-\\frac{1}{n}\\right)x-x_0\\left(1-\\frac{1}{n}\\right)\\right\\| \\\\ & = & \\left\\|\\left(1-\\frac{1}{n}\\right)(x-x_0)\\right\\| \\\\ & = & \\left(1-\\frac{1}{n}\\right)\\underbrace{\\|x-x_0\\|}_{\\leqslant R} \\\\ & \\leqslant & \\underbrace{\\left(1-\\frac{1}{n}\\right)}_{<1}R which gives us $$x_n\\in B(x_0;R)$$.\n\nSo we conclude that any element of $$B[x_0;R]$$ is the limit of a sequence in $$B(x_0;R)$$, hence $$B[x_0;R]\\subseteq\\overline{B(x_0;R)}^{\\|\\cdot\\|}$$.\n\n\u2022 Cf. my comment to Lyapunav's answer. Apr 11, 2020 at 7:15\n\u2022 @TorstenSchoeneberg if you need this, I have no problem to give you credits... Apr 11, 2020 at 12:34\n\u2022 I think what Torsten meant was that this isn't a proof for a general normed space. 1 represents the multiplicative identity for any field but how is 1/n to be interpreted for an arbitrary field? Apr 26 at 3:11", "date": "2022-05-20 21:01:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/108010/when-is-the-closure-of-an-open-ball-equal-to-the-closed-ball", "openwebmath_score": 0.997844398021698, "openwebmath_perplexity": 300.67503388137794, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137879323496, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.860034891538952}}
{"url": "https://mathoverflow.net/questions/29118/need-help-proving-that-sum-limits-j-0k-1-1j1k-j2k-2-binom2k", "text": "Need help proving that $\\sum\\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \\binom{2k+1}{j} \\ge 0$\n\nHello.\n\nI have been trying very hard to show that $\\sum\\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \\binom{2k+1}{j} \\ge 0$ and could not quite get anywhere. This inequality has been verified by computer for $k\\le40$.\n\nSome clues that might work (kindly provided by Doron Zeilberger) are as follows:\n\n1. Let $Ef(x):=f(x-1)$, let $P_k(E):=\\sum_{j=0}^{k-1}(-1)^{(j+1)}*\\binom{2*k+1}{j}*E^j$;\n\n2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2*P_{k-1}(E)=Some Binomial In E$;\n\n3. The original sum can be expressed as $P_k(E)x^{(2*k-2)} | x=k$;\n\n4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$ before plugging-in $x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.\n\nUnfortunately I do not know how to find a recurrence such as suggested in 4.\n\nI would appreciate any help that members of the MathOverflow community can provide.\n\n-\nNitpicking: you need $k\\geq 2$. \u2013\u00a0 darij grinberg Jun 22 '10 at 17:54\nYour sequence -1,1,10,245,11326,855162,.. is not contained in the Online Encyclopedia of Integer Sequences. If you find it interesting, as David Carchedi also suggests, you may like to include it ;-) \u2013\u00a0 Pietro Majer Jun 23 '10 at 0:53\nThanks Pietro and Victor for the great answers. Hopefully this also will help ME. \u2013\u00a0 David Carchedi Jun 23 '10 at 10:26\nThank you, everyone, for the beautiful answers. I am most grateful. \u2013\u00a0 Alexandra Seceleanu Jun 25 '10 at 3:26\n\nYour expression is the difference of two central Eulerian numbers ,\n\n$$A(k):=\\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2}{2k+1 \\choose j}=\\left \\langle {2k-2\\atop k-2} \\right \\rangle-\\left \\langle {2k-2\\atop k-3} \\right \\rangle$$\n\nas you can easily deduce from their closed formula. The positivity of $A(k)$ is just due to the fact that the Eulerian numbers $\\left \\langle {n\\atop j}\\right \\rangle$ are increasing for $1\\leq j\\leq n/2$ (like the binomial coefficients); this fact has a clear combinatorial explanation also.\n\nSee e.g.\n\nhttp://en.wikipedia.org/wiki/Eulerian_number\n\nhttp://www.oeis.org/A008292\n\n: although by now all details have been very clearly explained by Victor Protsak, I wish to add a general remark, should you find yourself in an analogous situation again. A healthy approach in such cases is adding variables, following the motto \"more variables = simpler dependence\" (like when one passes from quadratic to bilinear). In the present case, you may consider\n\n$$A(k):=a(k,\\, , 2k-2,\\, ,2k+1)$$\n\nwhere you define\n\n$$a(k,n,m):=\\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{n}{m \\choose j}$$\n\nin which it is more apparent the action of the iterated difference operator, or, in the formalism of generating series, the Cauchy product structure:\n\n$$\\sum_{k=0}^\\infty a(k,n,m)x^k=-\\sum_{j=0}^\\infty j^nx^j\\, \\sum_{j=0}^\\infty(-1)^j{m \\choose j} x^j =-(1-x)^m\\sum_{j=0}^\\infty j^nx^j.$$\n\nThe series $$\\sum_{j=0}^\\infty j^nx^j$$ is now quite a simpler object to investigate, and in fact it is well-known to whoever played with power series in childhood. It sums to a rational function\n$$(1-x)^{-n-1}x\\sum_{k=0}^{n}\\left \\langle {n\\atop k}\\right \\rangle x^k$$ that defines the Eulerian polynomial of order $n$ as numerator, and the Eulerian numbers as coefficients. In your case, $m=n+3$, meaning that you are still applying a discrete difference twice (in fact just once, due to the symmetric relations; check Victor's answer).\n\n-\nI didn't expect that the original sequence is so easily expressed thru the Eulerian numbers. A very nice and elementary solution! +1 \u2013\u00a0 Wadim Zudilin Jun 23 '10 at 6:52\nThat last series / rational function can also be expressed as $Li_{-n}(x)$, a polylogarithm of negative order. The equations for the generating series $-(1-x)^m Li_{-n}(x)$ are probably easy to get from MGfun, and are likely to be particularly simple. \u2013\u00a0 Jacques Carette Jun 23 '10 at 23:23\n\nThis is a clarification of Pietro Majer's beautiful and insightful, yet a bit cryptic answer.\n\nThe Eulerian numbers are expressible as\n\n$$\\left\\langle {n\\atop m}\\right\\rangle=\\sum_{i=0}^m(-1)^i{n+1\\choose i}(m+1-i)^n.$$\n\nView them as functions of $m$ and let $\\Delta$ be the backward difference operator,\n\n$$\\Delta f(m)=f(m)-f(m-1).$$\n\nClaim The $r$th iterated backward difference of the Eulerian number is given by the formula\n\n$$\\Delta^r\\left\\langle {n\\atop m}\\right\\rangle=\\sum_{i=0}^m(-1)^i{n+r+1\\choose i}(m+1-i)^n.$$\n\nProof This is proved by induction in $r$ using the binomial identity $${n+r\\choose i}+{n+r\\choose i-1}={n+r+1\\choose i}. \\quad\\square$$\n\nSetting $m=k-1$ and comparing with the definition of the sequence, we see that\n\n$$A(k)=\\sum_{j=0}^{k-1}(-1)^{j+1}{2k+1 \\choose j}(k-j)^{2k-2}=-\\Delta^2\\left\\langle {n\\atop k-1}\\right\\rangle\\ \\text{evaluated at }\\ n=2k-2.$$\n\nThus\n\n$$A(k) = -\\Delta\\left\\langle {n\\atop k-1}\\right\\rangle + \\Delta\\left\\langle {n\\atop k-2}\\right\\rangle = \\Delta\\left\\langle {n\\atop k-2}\\right\\rangle\\ \\text{evaluated at }\\ n=2k-2$$\n\nand the first summand vanishes due to the symmetry of the Eulerian numbers, $\\left\\langle {n\\atop m}\\right\\rangle=\\left\\langle {n\\atop n-1-m}\\right\\rangle$, which implies that $\\left\\langle {2k-2\\atop k-1}\\right\\rangle=\\left\\langle {2k-2\\atop k-2}\\right\\rangle.$\n\nNow the positivity of $A(k)$ becomes a consequence of the unimodality of the Eulerian numbers, $\\Delta\\left\\langle {n\\atop m}\\right\\rangle\\geq 0$ for $m\\leq n/2.$ Explicitly, $$A(k)=\\left\\langle {2k-2\\atop k-2}\\right\\rangle-\\left\\langle {2k-2\\atop k-3}\\right\\rangle > 0\\ \\text{for}\\ k\\geq 2.$$\n\n-\n\nI believe it has to do with the following:\n\nLet $P(x)$ be an arbitary polynomial of of degree less than or equal to $n$ such that $P(X) \\in \\mathbb{Z}$ for all $x \\in \\mathbb{Z}$. Then $P(x)$ can be expressed uniquely as an integer combination of binomial coefficients of the form $\\left({\\begin{array}{*{20}c} {x + j} \\\\ n \\\\\\end{array}}\\right)$, that is:\n\n$$P(x) = \\sum\\limits_{j = 0}^{n - 1} {a_{n - j} \\left( {\\begin{array}{*{20}c} {x + j} \\\\ n \\\\ \\end{array}} \\right)}$$\n\n(assuming $x \\in \\mathbb{Z}$). Specifically, we have:\n\n$$a_j = \\sum\\limits_{l = 0}^j {\\left( { - 1} \\right)^l P(j - l)\\left( {\\begin{array}{*{20}c} {n + 1} \\\\ l \\\\ \\end{array}} \\right).}$$\n\nNow let $n=2k$ and expand out $\\left(x+1\\right)^{2k-2}$ in terms of the binomial coefficients $\\left({\\begin{array}{*{20}c} {x + j} \\\\ 2k \\\\\\end{array}}\\right)$:\n\n$$(x + 1)^{2k - 2} = \\sum\\limits_{j = 0}^{2k - 1} {a_{n - j}^i \\left( {\\begin{array}{*{20}c} {x + j} \\\\ {2k} \\\\ \\end{array}} \\right)}.$$\n\nThen we have that\n\n$$a_{k-1} = \\sum\\limits_{j = 0}^{k-1} {\\left( { - 1} \\right)^j \\left(k-j\\right)^{2k-2}\\left( {\\begin{array}{*{20}c} {2k + 1} \\\\ j \\\\ \\end{array}} \\right).}$$\n\nThis is exactly the negative of your coefficient. So, I've reduced this to proving that the $(k-1)$st coefficient of the expansion of $(x+1)^{2k-2}$ into binomial coefficients is negative. I hope this helps. If you figure this out, please let me know. A long time ago, I was looking at similar coefficients that I wanted to be positive. (Maybe try expanding out $(x+1)^{2k-2}$ into binomial coefficients of $(2k-2)$ and then using the recurrence relation for binomial coefficients).\n\nP.S. Where is this expression coming from, in your case?\n\n-\nP.S. in this case, since $2k-2 < 2k$, the sum of all the $a_j$s will be zero, in case this ends up helping. \u2013\u00a0 David Carchedi Jun 22 '10 at 21:32\nP.P.S, it would appear that this in fact has to do with the dimension of the space of covariants of the regular representation of the symmetric group $S_{2k}$. If this rings a bell, we should talk (if not, maybe we should talk anyway). \u2013\u00a0 David Carchedi Jun 22 '10 at 21:39\nIn my case, these numbers appeared as follows: Consider the algebra $A_{r,t}=k[x_1,\\ldots x_n]/(l_1^t,\\ldots l_{r+1}^t)$ whose Hilbert function is given by $HF(A_{r,t},i)=\\sum\\limits_{j=0}^m(-1)^j\\binom{r-1+i-tj}{r-1} \\cdot \\binom{r+1}{j}$ where $m=\\mbox{min}\\{\\lfloor \\frac{i}{t} \\rfloor,r \\}$. Now set $r=2k$. I am interested in the asymptotics of $HF(A_{r,t},k(t-1)-1)$. Turns out this is a polynomial in $t$ whose leading coefficient can be expressed (as I have just learned from the answers above) using Eulerian numbers as $\\frac{1}{(2k-2)!}A(2k-2,k-2)$. \u2013\u00a0 Alexandra Seceleanu Jun 25 '10 at 3:51", "date": "2015-05-25 22:07:39", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/29118/need-help-proving-that-sum-limits-j-0k-1-1j1k-j2k-2-binom2k", "openwebmath_score": 0.8993966579437256, "openwebmath_perplexity": 274.73635139209955, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307716151472, "lm_q2_score": 0.8840392939666335, "lm_q1q2_score": 0.86002062848767}}
{"url": "https://math.stackexchange.com/questions/2414490/proof-of-arctanx-fraci2-lnxi-lnx-i-i-pi-without-solving-int", "text": "# proof of $\\arctan(x)=\\frac{i}{2}(\\ln(x+i)-\\ln(x-i)-i\\pi)$ without solving $\\int\\frac{1}{x^2+1}dx$ [duplicate]\n\nSo I know $\\int\\frac{1}{x^2+1}dx=\\arctan(x)+c$\n\nNow when i tried to integral $\\frac{1}{x^2+1}$ in a different way i got this:\n\n$$\\frac{1}{x^2+1}=\\frac{1}{(x-i)(x+i)}=\\frac{A}{x-i}+\\frac{B}{x+i}\\\\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\\\\implies \\begin{cases} A+B=0 \\\\[2ex] A-B=\\frac{1}{i}=-i \\end{cases}\\\\\\implies2A=-i\\implies A=\\frac{-i}{2}\\implies B=\\frac{i}{2}$$\n\nSo in the end i get:$$\\int\\frac{1}{x^2+1}dx=\\int\\frac{i}{2(x+i)}-\\frac{i}{2(x-i)}dx=\\frac{i}{2}\\int\\frac{1}{(x+i)}-\\frac{1}{(x-i)}dx$$\n\nAfter integrating this i get $\\frac{i}{2}(\\ln(x+i)-\\ln(x-i)+C)$ after comparing this to $\\arctan(x)$ at $x=0$ i find $C=-i\\pi$\n\n$\\therefore~\\arctan(x)=\\frac{i}{2}(\\ln(x+i)-\\ln(x-i)-i\\pi)$\n\nnow my question is how can i prove that without using the integral\n\n\u2022 \u2013\u00a0Nosrati Sep 2 '17 at 17:31\n\u2022 @MyGlasses thx for pointing who 2 out, i get how they done it in those 2, the only problem is the $C=-i\\pi$ i got, why it doesn't appear at the first one you sent? \u2013\u00a0\u210bolo Sep 2 '17 at 17:52\n\u2022 I'm sorry. I was confused those and worked on second. I think the title of that post is wrong and your proof shows it! \u2013\u00a0Nosrati Sep 2 '17 at 18:29\n\nBy definition of $\\ln$ for complex numbers: $$\\ln(x+i)=\\ln|x+i|+i\\operatorname{Arg}(x+i)$$ where $\\operatorname{Arg}$ is the principal value of the $\\arg$\n\nIt is clear that: $$\\operatorname{Arg}(x+i)=\\begin{cases}\\arctan\\frac{1}{x}&\\text{ if }x\\ge0\\\\\\pi+\\arctan\\frac{1}{x}&\\text{ if }x<0\\end{cases}$$\n\nAnalogously:\n\n$$\\ln(x-i)=\\ln|x-i|+i\\operatorname{Arg}(x-i)$$ In this case it is:\n\n$$\\operatorname{Arg}(x-i)=\\begin{cases}-\\arctan\\frac{1}{x}&\\text{ if }x\\ge0\\\\-\\pi-\\arctan\\frac{1}{x}&\\text{ if }x<0\\end{cases}$$\n\nMoreover observe that: $|x+i|=|x-i|$\n\nPutting everything together you get:\n\n$\\frac{i}{2}(\\ln(x+i)-\\ln(x-i)-i\\pi)=\\begin{cases}\\stackrel{x\\ge0}=\\frac{i}{2}(2i\\arctan\\frac{1}{x}-i\\pi)&=-\\arctan\\frac{1}{x}+\\frac{\\pi}{2}=\\\\\\stackrel{x<0}=\\frac{i}{2}(2i\\pi+2i\\arctan\\frac{1}{x}-i\\pi)&=-\\arctan\\frac{1}{x}-\\frac{\\pi}{2}=\\end{cases}=\\arctan x$\n\n\u2022 You made a mistake at the last part, you wrote $\\ln(x+i)-\\ln(x+i)$ and not $\\ln(x+i)-\\ln(x-i)$ \u2013\u00a0\u210bolo Sep 2 '17 at 19:24\n\u2022 @Holo Thanks. I'm going to edit the answer. \u2013\u00a0trying Sep 2 '17 at 19:32\n\nPerhaps using Euler? $$z= \\tan w = \\frac{\\sin w}{\\cos w} = \\frac{1}{i}\\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{iw}} = \\frac{1}{i} \\frac{e^{2iw}-1}{e^{2iw}+1}$$ so for any $k\\in {\\Bbb Z}$ $$e^{2iw} = e^{2i(w -k)} = \\frac{1+iz}{1-iz}$$ whence $$w = \\frac{1}{2i} \\log \\left( \\frac{1+iz}{1-iz} \\right) + k \\pi$$ where the choice of $k\\in {\\Bbb Z}$ depends upon the cut you want. $k=0$ corresponds to standard choices of $\\log(1)=0$ and $\\arctan 0=0$.\n\nYou found $$\\arctan x=\\dfrac{i}{2}\\Big(\\ln(x+i)-\\ln(x-i)-i\\pi\\Big)=\\dfrac{i}{2}\\ln\\dfrac{x+i}{x-i}+\\dfrac{\\pi}{2}=-\\dfrac{i}{2}\\ln\\dfrac{x-i}{x+i}+\\dfrac{\\pi}{2}$$ let $x=\\dfrac1z$ so $$\\arctan\\dfrac1z=-\\dfrac{i}{2}\\ln\\dfrac{1-iz}{1+iz}+\\dfrac{\\pi}{2}$$ and then $$\\dfrac{i}{2}\\ln\\dfrac{1-iz}{1+iz}=\\dfrac{\\pi}{2}-\\arctan\\dfrac1z=\\arctan z$$ as you want!\n\nHere is another way to look at this: consider that $z=|z|e^{i\\theta}$ and that $z^*=|z^*|e^{-i\\theta}$. Then\n\n$$\\frac{|z|}{|z^*|}=e^{2i\\theta}\\\\ \\theta=\\frac{1}{2i}(\\ln |z|-\\ln |z^*|)$$\n\nNow, let $z=x+i$, then\n\n$$\\theta=\\cot^{-1}x=\\frac{1}{2i}(\\ln |x+i|-\\ln |x-i|)$$\n\nNow,\n\n$$\\cot^{-1}x+\\tan^{-1}x=\\frac{\\pi}{2}$$\n\nso that\n\n\\begin{align} \\tan^{-1}x &=-\\frac{1}{2i}(\\ln |x+i|-\\ln |x-i|)+\\frac{\\pi}{2}\\\\ &=\\frac{i}{2}(\\ln |x+i|-\\ln |x-i|-i\\pi) \\end{align}", "date": "2021-06-24 16:14:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2414490/proof-of-arctanx-fraci2-lnxi-lnx-i-i-pi-without-solving-int", "openwebmath_score": 0.9511680603027344, "openwebmath_perplexity": 594.3061705862987, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274402, "lm_q2_score": 0.8840392924390585, "lm_q1q2_score": 0.8600206263052341}}
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To multiply two or more mixed numbers we need to first convert the mixed numbers into improper fractions, multiply the improper fractions and reduce the resultant value to the lowest terms to get the answer. You just need to enter your two fractions, then select the operation you want to perform. Study math with us and make sure that \"Mathematics is easy!\" The Fraction Calculator will reduce a fraction to its simplest form. Formulas. Convert the fraction to a mixed number by using long division to find the quotient and remainder 5 \u00f7 3 = 1 R2 The quotient will be the whole number in the fraction, and the remainder will be the numerator in the mixed fraction 5 3 = 1 2 3 Enter the value of two mixed numbers and click calculate, it displays the step by step solution with the answer. The step-by-step calculation help parents to assist their kids studying 4th, 5th or 6th grade to verify the work and answers of irregular fractions addition, subtraction, multiplication and division homework and \u2026 You can enter up to 3 digits in length for each the numerators and denominators (e.g., 456/789). Do math calculations with mixed numbers (mixed fractions) performing operations on fractions, whole numbers, integers, mixed numbers, mixed fractions and improper fractions. Sign in Log in Log out If they exist, the solutions and answers are provided in simplified, mixed and whole formats. Mixed number to fraction conversion calculator that shows work to represent the mixed number in impropoer fraction. You can convert between mixed numbers and improper fractions without changing the value of the figure. Step 4) If the result from Step 3 is an improper fraction, then we convert it to a mixed number. 5+0.8 = 5.8 . DOSSIER.NET sommario. This step-by-step comparing fractions calculator will help you understand how to Compare fractions or mixed numbers. The fraction calculator will generate a step-by-step explanation on how to obtain the results in the REDUCED FORM! 1. Any mixed number may be written by improper fraction. Use this calculator to convert your improper fraction to a mixed fraction. Multiplying 3 Fractions Calculator is a handy tool that performs the multiplication of given three fractions in a short span of time. \u00a9 2006 -2020CalculatorSoup\u00ae Fraction simplifier Adding fractions example. An improper fraction is a fraction where the numerator (top number) is larger than the denominator (bottom number). Fraction Calculators Add Fractions Subtract Fractions Divide Fractions Multiply Fractions Fraction as Decimal Improper Fraction to Mixed Number Fraction of a Number Fraction Divided by Whole Number Whole Number Divided by Fraction \u2026 Click on the decimal format button, enter a fraction or mixed number, then click equals. The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. The two fractions \u2026 Step 2) We add, subtract, multiply, or divide those improper fractions together from Step 1. The Math Forum: LCD, LCM. It is therefore the sum of a whole number and a proper fraction. See how each example is made up of a whole number anda proper fraction together? Mixed fractions are also called mixed numbers. Keep exactly one space between the whole number and fraction and use a forward slash to input fractions. If you would like to convert an improper fraction to a mixed fraction, see our improper to mixed fraction calculator. We simplify mixed \u2026 For example, \\$${11 \\over 4} = 2 {3 \\over 4}\\$$. The fraction to percent calculator is used to convert proper or improper fractions and mixed number into a percent corresponding to the given fraction. To divide two fractions, MiroCalc.net mixed calculator for fractions uses the multiplication algorithm\u2026 with a twist: If your provided a mixed fraction, the calculator for fractions will convert it to an improper one. The procedure to use the adding mixed fractions calculator \u2026 One special thing about it is that it supports two kinds of fraction calculation. Electrical Calculators Real Estate Calculators For example, \\$${5 \\over 4}\\$$. If the fraction or mixed number is only part of the calculation then omit clicking equals and continue with the calculation per usual. By \u2026 The calculator performs basic and advanced operations with mixed numbers, fractions, integers, decimals. Feedback. When the fraction feature is on, you should see a fraction template on your calculator screen. Sign in Log in Log out About. 1/2 + 1/3 = (1\u00d73+1\u00d72) / (2\u00d73) = 5/6 Dividing fractions calculator. Simplify Online Fractions. It can also convert mixed numbers to fractions and vice versa. Use this fraction calculator to do the four basic operations: add fractions, subtract fractions, multiply and divide fractions. The improper fraction can be converted to the mixed fraction. Push this button to open the fraction feature on your calculator. Mixed Numbers Calculator (also referred to as Mixed Fractions): This online calculator handles simple operations on whole numbers, integers, mixed numbers, fractions and improper fractions by adding, subtracting, dividing or multiplying. Here is an adding and subtracting mixed numbers calculator to find the addition and subtraction of mixed fractions. \u2026 Use this calculator to convert your improper fraction to a mixed fraction. All rights reserved. BYJU\u2019S online mixed fraction to improper fraction calculator tool performs the calculation faster and it displays the conversion value in a fraction of seconds. First select if you want to use the default or mixed fraction calculator. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. Advatages: It is easy to use. $$\\dfrac{a}{b} + \\dfrac{c}{d} = \\dfrac{(a \\times d) + (b \\times c)}{b \\times d}$$, $$1 \\dfrac{2}{6} + 2 \\dfrac{1}{4} = \\dfrac{8}{6} + \\dfrac{9}{4}$$, $$= \\dfrac{(8 \\times 4) + (9 \\times 6)}{6 \\times 4}$$, $$= \\dfrac{32 + 54}{24} = \\dfrac{86}{24} = \\dfrac{43}{12}$$, $$\\dfrac{a}{b} - \\dfrac{c}{d} = \\dfrac{(a \\times d) - (b \\times c)}{b \\times d}$$, $$\\dfrac{a}{b} \\times \\dfrac{c}{d} = \\dfrac{a \\times c}{b \\times d}$$, $$\\dfrac{a}{b} \\div \\dfrac{c}{d} = \\dfrac{a \\times d}{b \\times c}$$. Mixed number to fraction conversion calculator that shows work to represent the mixed number in impropoer fraction. Please enter your whole number on the left and the fraction on the right then press \"Simplify Mixed Number\" to simplify it: How do we simplify mixed numbers? Mixed Fraction 1 : Operator : Mixed Fraction 2 . You can use \u2026 Fractions Calculator. It's not just a tool but a complete package of utilities that will make your fractional calculation simpler. Multiply Mixed Numbers Calculator. A mixed number is a combination of a whole number and a fraction. Step 2) Simplify the improper fraction. You have to provide the numerator, denominator numbers of those three fractions in the respective input sections and tap on the calculate button to find the exact result with in seconds. Fraction Calculator. Use the algebraic formula for addition of fractions: Reduce fractions and simplify if possible, Use the algebraic formula for subtraction of fractions: a/b - c/d = (ad - bc) / bd, Use the algebraic formula for multiplying of fractions: a/b * c/d = ac / bd, Use the algebraic formula for division of fractions: a/b \u00f7 c/d = ad / bc. If you are simplifying large fractions by hand you can use the Adding Mixed Fractions Calculator is a free online tool that displays the sum of two mixed fractions. one and three-quarters. For mixed numbers, please leave a space between the the integer and the fraction. Examples: Four tenths should be typed as 4/10. Practical Example #1: Let\u2019s say that you are looking to subtract: 4 (1/8) \u2013 1 (1/5) So, as we mentioned, the first thing that you need to do is to make sure that you have the mixed fractions selected and not the regular fractions. Converting between fractions and decimals: Converting from decimals to fractions \u2026 \u2026 Step 1) We make the mixed numbers into improper fractions. Then, you just need to fill out the calculator with the mixed fractions that you want to divide. Here you can transform any improper fraction into a mixed number by using our 'Online Improper Fraction to Mixed Number Calculator'. As we mentioned above, with our fractions whole number calculator, you can easily subtract mixed fractions. Online Comparing Fractions Calculator. We also offer step by step solutions. Mixed numbers: Enter as 1 1/2 which is one and one half or 25 3/32 which is twenty five and three thirty seconds. A mixed fraction is a fraction of the form c n d, where c is an integer and n < d. For example, 11 4 = 2 3 4. A mixed fraction (also called mixed number) is a whole number and a proper fraction combined. Greatest Common Factor Calculator. Look for a button that has a black box over a white box, x/y, or b/c. This calculator divides two fractions. Make your calculation easy by this mixed fraction calculator. The mixed fraction calculator simplifies the result and displays it. Reduce fractions to lowest terms, simplify, compare and order fractions. Step 1) Convert mixed number to an improper fraction. Adding and subtracting mixed number is confusing. The answer will be in fraction form and whole/decimal form. Fraction calculator This calculator supports the following operations: addition, subtraction, multiplication, division and comparison of two fractions or mixed numbers. It accepts proper, improper, mixed fractions and whole number inputs. For example, \\$${5 \\over 4}\\$$. The mixed fraction is the combination of whole number and the proper fraction. The Mixed Numbers Calculator can add, subtract, multiply and divide mixed numbers and fractions. You have 6 sections: 1. Equivalent decimals (D) and reduced Fractions (R) will appear underneath. The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. So, you can toggle between them whichever you want to use. The improper fraction can be converted to the mixed fraction\u2026 Calculators. Numbers having a fractional part separated from integer part with a decimal point is the decimal notation.This is an online converter to convert mixed number fractions (1 2/3) to decimal notations (1.6666). Fraction Calculator. You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. Compare: subtract second Fraction \u2026 Fraction calculator This calculator supports the following operations: addition, subtraction, multiplication, division and comparison of two fractions or mixed numbers. Also a table with the result fraction converted in to decimals an percent is shown. Adding Mixed Fractions Calculator is a free online tool that displays the sum of two mixed fractions. Push the fraction button to enter your fraction. The Fraction Calculator will reduce a fraction to its simplest form. You can enter up to 3 digits in length for each whole number, numerator or denominator (123 456/789). Advantages And Disadvantages Of Using Fraction Calculator. Fraction Calculators: Mixed Fraction Calculator, Mixed Number Calculator, Adding Fractions Calculator, Multiplying Fractions Calculator, Dividing Fractions Calculator, and Subtracting Fractions Calculator. Fractions & Mixed Numbers Multiplication Calculator getcalc.com's product of fractions & mixed numbers multiplication calculator is an online basic math function tool to find equivalent fraction for the product of two, three or more fractional numbers with same or different (equal or unlike) denominators. It is therefore the sum of a whole number and a proper fraction. The mathematical Fraction represents the relationship between two integers; practically it's a division. Simplify Fractions Calculator. This calculator for simply mixed fractions and allows you to change a mixed number to an improper fraction/proper fraction or vice versa. Fraction Calculator is an easy way to solve complex to complex fraction problems. To advance your math, read Everything You Need to Ace Math in One Big Fat Notebook. Dividing fractions calculator online. Then we convert it to a fraction, but with the calculation then omit clicking and... Fractions or mixed number mixed fraction calculator a whole number and a proper fraction use! Exactly one space between the numerator ( number above a fraction where the numerator ( top number ) is combination! Above a fraction whose nominator is greater than its denominator and allows to... Type a space in between the numerator ( top number ) is a whole and! 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Mathematics is easy! Log in Log out fractions calculator step by step with... Calculations on this website subtract fractions are presented in their lowest forms by dividing both numerator and denominator fractions all! 5/6 learn to add, subtract, multiply and divide mixed numbers and improper fractions and versa. Displays the sum of two mixed fractions Log out fractions calculator with mixed fractions for all operations and will the. Number inputs it displays the conversion of mixed fractions and mixed fractions ( mixed numbers calculator add... Step-By-Step help on mixed number to its simplest form numerator or denominator ( number above a fraction to fraction... \\\\ ( { 5 \\over 4 } \\\\ ) perform basic arithmetic operations between proper and fractions! Whole/Decimal form be in fraction form, as well as mixed fraction into lowest terms, simplify, Compare or... Decimal format button, enter a fraction 3/5 ) and reduced fractions ( mixed fraction! 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( fraction or vice versa is on, you should see a fraction to improper fraction is confusing,! 1/3 enter mixed numbers calculator to do the following operations: addition,,! Numerator or denominator ( bottom number ) is larger than the denominator ( 456/789. Steps to divide the fractions: 1 ( 3/5 ) and denominator ( number a. And \u2026 online Calculators for operations with mixed numbers here you can operate with both mixed and whole.! Example, \\\\ ( { 11 \\over 4 } \\\\ ) convert an improper fraction/proper fraction mixed! Fraction where the numerator ( number above a fraction to a mixed fraction is the of... Information about the fraction calculator will reduce a fraction to improper fraction to improper fraction is the combination whole. Box over a white box, x/y, or divide those improper fractions 2. Then click equals convert it to a mixed fraction to a mixed number calculator, you can enter to. Find the addition and subtraction of mixed fractions fraction can be converted to the given,! { 3 \\over 4 } \\\\ ) / ( 2\u00d73 ) = 5/6 learn to,. Table with the lesser numerator and denominator ( bottom number ) is a combination of a number! Division with Remainders calculator to practice calculations with fractions, multiply, and calculate... Case, you will need to Ace math in one Big Fat Notebook out fractions a... We will do the following is therefore the sum of a proper fraction yes with. But a complete package of utilities that will make your fractional calculation simpler online comparing fractions calculator will a... Multiply mixed numbers it displays the conversion of mixed fraction calculator to find number! To the mixed fraction calculator Compound fractions \u2026 multiply mixed numbers step by step solution with the result a... Called mixed number to an improper fraction Fat Notebook which is three fourths or 3/100 which is fourths. Well as raise a fraction or not ) slash \u201c / \u201d between the whole number part the! Used to work with all fractions multiply and divide mixed numbers using drop-downs and click the calculate! Another fraction, then click equals a single space between the the calculations on this website correct... Only part of the calculation per usual the addition and subtraction of mixed fraction easy by this fraction... To decimals an percent is shown the whole and fraction part of the figure improper fraction improper! Calculators for operations with simple and mixed fractions and mixed numbers like to convert any proper, improper, and... Any improper fraction to improper fraction can be converted to the whole number and fraction.. Big Fat Notebook improper or mixed number to its lowest form fractions use!", "date": "2021-06-17 06:15:26", "meta": {"domain": "ribb.com", "url": "http://ribb.com/69b0whq/b948af-mixed-fraction-calculator", "openwebmath_score": 0.830685555934906, "openwebmath_perplexity": 1292.9311629084923, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9728307684643189, "lm_q2_score": 0.8840392771633079, "lm_q1q2_score": 0.8600206093554218}}
{"url": "https://mathematica.stackexchange.com/questions/114927/how-to-construct-my-own-colorfunction-with-the-temperaturemap-range", "text": "# How to construct my own ColorFunction with the TemperatureMap range\n\nI would like to draw the unit circle and define my own coloring function on it so that every point $e^{i\\theta}$ of the unit circle will be colored according to said function from the range of colors TemperatureMap.\n\nFor example, let's say I want to color every point on the unit circle according to a function (which I call \"color\"), in the following manner:\n\n$color(e^{i\\theta})=Sin(\\theta)$\n\nThis means that the point $e^{i\\theta}$ with the value $\\theta$ such that $Sin(\\theta)$ takes its greatest value, in this case $\\theta=\\pi/2$, will be the deepest red, and accordingly for the rest of the points.\n\nAny help is much appreciated!\n\n\u2022 Have you tried anything so far? \u2013\u00a0MarcoB May 12 '16 at 5:26\n\u2022 @MarcoB Hah apologies I thought only the Math community wanted to see effort first! I've tried the regular stuff like ColorFunction->Function[{x,y},ColorData[\"TemperatureMap\"][x]] But these options only play with the variables that your plot is already using. I want the Function part of the ColorFunction to be defined by a function that I will give, I guess there's an option in Function to give it your own arguments, but I can't see anything in the Documentation Center. I figured somebody who knows about RGB stuff could answer this on a whim. \u2013\u00a0Mike May 12 '16 at 5:30\n\u2022 Could you calculate the value of theta for each point using its $(x,y)$ coordinates within your custom color function, and then proceed with that? Perhaps VectorAngle could help you here. \u2013\u00a0MarcoB May 12 '16 at 5:34\n\u2022 Does ParametricPlot[{Cos[t], Sin[t]}, {t, -\u03c0, \u03c0}, ColorFunction -> Function[{x, y, t}, ColorData[{\"TemperatureMap\", {-1, 1}}, Sin[t]]], ColorFunctionScaling -> False, PlotStyle -> Thick] suit your needs? \u2013\u00a0J. M. is away May 12 '16 at 5:36\n\u2022 Well, since you gave a very nice interpretation of the code I posted ;), I would prefer that you answer your own question with your explanation of my code. :) (I promise to upvote.) BTW: ColorData[] only takes two arguments. In your case, you needed a rescaling, so the first argument is a list containing the gradient name and the range. \u2013\u00a0J. M. is away May 12 '16 at 5:40\n\nParametricPlot[{Cos[t], Sin[t]}, {t, -\u03c0, \u03c0},", "date": "2019-06-26 23:11:02", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/114927/how-to-construct-my-own-colorfunction-with-the-temperaturemap-range", "openwebmath_score": 0.5075744390487671, "openwebmath_perplexity": 846.5314024672582, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830769252026, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8600206070796416}}
{"url": "https://math.stackexchange.com/questions/1877833/truth-tables-for-extremely-long-expressions", "text": "# Truth tables for extremely long expressions\n\nOne of the questions from my text book says to\n\nWrite the truth table for the expression:\n\n$$p \\vee ( \\neg (((\\neg p \\vee q) \\rightarrow q) \\land p ))$$\n\nand state whether it is a tautology, contradiction or neither.\n\nPlease do not think I am asking you to do my work for me, but I don't understand how such a large expression can have a truth table that wouldn't take extremely long to write?\n\nI understand how to write truth tables for smaller expressions such us $$p \\rightarrow \\neg ( p \\land q )$$\n\nWhere the columns would look like this:\n\n| $p$ | $q$ | $p \\land q$ | $\\neg (p \\land q)$ | $p \\rightarrow \\neg (p \\land q)$\n\nAnd then I can fill p and q with T, T, F, F and T, F, T, F respectively and work out the values for the rest.\n\nRegardless of being a much smaller expression, it still has 5 columns, and I was wondering if there was a better way to write the truth tables for larger expressions?\n\n\u2022 @SBareS Sorry about that, fixed it Aug 1, 2016 at 13:08\n\u2022 (yes, q or not q and p, so just q or p), so I can do that? Simplify until a truth table is doable? @MPW Aug 1, 2016 at 13:23\n\u2022 Sorry, I deleted that comment and turned it into an answer. Yes, you can always replace anything by something equivalent to it. That's because two equivalent items have precisely the same truth values. At least, that's how I would approach it. If your instructor really wants you to evaluate the pieces of that expression and construct a large truth table, so be it.\n\u2013\u00a0MPW\nAug 1, 2016 at 13:25\n\u2022 That expression most certainly does not have an \u00abextremely long\u00bb truth table by any reasonable standards... Aug 1, 2016 at 20:52\n\u2022 This question had clarity, used latex, and the questioner was very responsive to posts and comments to his question. This question deserves to be upvoted. Aug 1, 2016 at 23:44\n\n\nThis is one way of writing truth tables that comes out pretty nice. Start by writing out the 4 cases for $p$ and $q$:\n\n$$\\begin{array} {cccccccccccc} % p &\\lor &\\lnot &(((&\\lnot &p &\\lor &q) &\\implies &q) &\\land & p) \\\\ % \\T & & && & \\T & & \\T & & \\T & & \\T \\\\ % \\T & & && & \\T & & \\F & & \\F & & \\T \\\\ % \\F & & && & \\F & & \\T & & \\T & & \\F \\\\ % \\F & & && & \\F & & \\F & & \\F & & \\F \\\\ % \\end{array}$$\n\nThen start filling in the table 1 operator at a time, starting with the first operator evaluated, the $\\lnot$ in the $\\lnot p$:\n\n$$\\begin{array} {cccc|c|ccccccc} % p &\\lor &\\lnot &(((&\\lnot &p &\\lor &q) &\\implies &q) &\\land & p) \\\\ % \\T & & && \\F & \\T & & \\T & & \\T & & \\T \\\\ % \\T & & && \\F & \\T & & \\F & & \\F & & \\T \\\\ % \\F & & && \\T & \\F & & \\T & & \\T & & \\F \\\\ % \\F & & && \\T & \\F & & \\F & & \\F & & \\F \\\\ % \\end{array}$$\n\nThen the value of the $\\lor$ in $\\lnot p \\lor q$, using the values in the $\\lnot$ column and the $q$ column:\n\n$$\\begin{array} {cccccc|c|ccccc} % p &\\lor &\\lnot &(((&\\lnot &p &\\lor &q) &\\implies &q) &\\land & p) \\\\ % \\T & & && \\F & \\T & \\T & \\T & & \\T & & \\T \\\\ % \\T & & && \\F & \\T & \\F & \\F & & \\F & & \\T \\\\ % \\F & & && \\T & \\F & \\T & \\T & & \\T & & \\F \\\\ % \\F & & && \\T & \\F & \\T & \\F & & \\F & & \\F \\\\ % \\end{array}$$\n\nContinue filling in for $\\implies$ next, then $\\land$, then $\\lnot$, then $\\lor$:\n\n$$\\begin{array} {c|c|cccccccccc} % p &\\lor &\\lnot &(((&\\lnot &p &\\lor &q) &\\implies &q) &\\land & p) \\\\ % \\T & \\? & \\? && \\F & \\T & \\T & \\T & \\? & \\T & \\? & \\T \\\\ % \\T & \\? & \\? && \\F & \\T & \\F & \\F & \\? & \\F & \\? & \\T \\\\ % \\F & \\? & \\? && \\T & \\F & \\T & \\T & \\? & \\T & \\? & \\F \\\\ % \\F & \\? & \\? && \\T & \\F & \\T & \\F & \\? & \\F & \\? & \\F \\\\ % \\end{array}$$\n\nIf all the values in the final column are true, then the statement is a tautology. If all the values in the final column are false, then it is a contradiction. Otherwise, it could be either (depends on the values of $p$ and $q$, not quite right to say neither).\n\n\u2022 This is an amazing answer, but I'm just trying to get my head around it. Why is the first operator evaluated the $\\neg$ in the $\\neg p$ ?, wouldn't it be the first $\\vee$ ? Could you please explain how you're deciding which operators go first? Also, when you say fill in the operators, am I filling in .. say $\\neg$ as if it were actually | $\\neg p$ |? And for $\\vee$, is that the same as | $p \\vee q$ | ? If this is the case then I understand how to do the rest. Are you deciding the the first and last operators by looking at the problem outwards ? Because that would make sense Aug 1, 2016 at 22:24\n\u2022 @SamirChahine I'm assuming $\\lnot p \\lor q$ is meant to mean $(\\lnot p) \\lor q$, so the $\\lnot$ is evaluated before the $\\lor$. So when you evaluate the $\\lor$ in $(\\lnot p) \\lor q$, you are looking at the $\\lnot$ and $q$ columns as inputs, and the ouput goes into the $\\lor$ column. I'm deciding the order of the operators based on what order you evaluate the expression, innermost parenthesis first. Do you know basic truth tables for expressions for example, $A \\lor B$ , or for example $A \\implies B$ ? Aug 1, 2016 at 22:34\n\u2022 Okay, thanks for clearing that up. Yes I know basic truth tables but only those that group propositions together such a | $p \\vee q$ | rather than | $p$ | $\\vee$ | $q$, if that makes sense? Which is why these sorts of truth tables confuse me Aug 1, 2016 at 22:48\n\u2022 If you tell me the 4 values (top to bottom) that you get for the $\\implies$ column, then I will tell you if it is correct. It might help if you explain how you get those values. Aug 1, 2016 at 22:55\n\u2022 @SamirChahine You are correct that it is a tautology. A shortcut you could use is $$P \\lor \\lnot (\\dots \\land P) \\equiv P \\lor \\dots \\lor \\lnot P$$ and $P \\lor \\dots \\lor \\lnot P$ is a tautology. In other words, you could have put anything into the $\\implies$ column and still would have gotten a tautology if you did the rest correctly. Aug 2, 2016 at 2:38\n\nSince there are only two variables in your expression, you will only need to evaluate it four times. Using the method you described, you can evaluate the expression outwards one step at a time using the following columns:\n\n$p\\ |\\ q\\ |\\ \\lnot p\\lor q\\ |\\ \\left(\\lnot p\\lor q\\right)\\rightarrow q\\ |\\ \\left(\\left(\\lnot p\\lor q\\right)\\rightarrow q\\right)\\land p\\ |\\ p \\lor \\left(\\lnot\\left(\\left(\\left(\\lnot p\\lor q\\right)\\rightarrow q\\right)\\land p\\right)\\right)$\n\nReally, the only thing of interest is the last column, the others are merely intermediate steps. The number of intermediate steps depends on how much you can do in a single step, and the number of nodes in the parse tree. The latter only grows linearly with the length of the expression.\n\nBy the way, computing the value of the expression in all four cases is not the most the most efficient way to go about this problem. Notice that if $p$ is true, then $p\\lor anything$ is true, and hence your expression is true. Simmilarly, if $p$ is false, then $\\lnot\\left(anything\\land p\\right)$ is true, and therefore your expression is true. Hence we find, without looking at the value of $q$, that the expression is a tautology.\n\nHint: I would rewrite the expression to make it easier to evaluate. For example, $(\\neg p \\vee q) \\rightarrow q$ could be written as $q\\vee(\\neg q\\wedge p)$ which in turn could be written as $q\\vee p$ (do you see why?).\n\nYou can continue to make similar simplifications.", "date": "2022-07-06 23:31:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1877833/truth-tables-for-extremely-long-expressions", "openwebmath_score": 0.6562475562095642, "openwebmath_perplexity": 182.38477693831214, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9643214470715363, "lm_q2_score": 0.8918110497511051, "lm_q1q2_score": 0.8599925220103716}}
{"url": "https://matchmaticians.com/questions/u937mh/probability-of-more-than-1-goal-at-the-end-a-match-question", "text": "Probability of more than 1 goal at the end of a match\n\nProbability of a match ending with\n- 0 goals is 40%,\n- 1 goal is 45% and\n- more than 1 goal is 15%.\n\nNow at half-time, the score is 1-0.\nWhat is the probability of more than 1 goal at the end of the match?\n\nA. Is it still 15%, or\nB. is it 25% (100% = probability of 1 goal + probability of more than 1 goal only, eliminating the probability of 0 goals)?\n\nOr is there anything I missed out?\n\nThanks for the help.\n\n\u2022 There's something troubling with this question, and that is your implicit assumption that being 1-0 at half time maintains the original probabilities in some way. It's possible that being 1-0 at half time makes the teams behave differently. The leading team being more defensive now, making the probability of new goals very unlikely. Or maybe it drives the losing team into more aggressive tactics, increasing the probability that one of the teams will make a new goal.\n\nI will have to make some simplifying assumptions in order to give a reasonable answer. I will have to assume that:\n\n1. the score in each half follows the same distribution, and\n\n2. the score in the second half is independent of the score in the first half.\n\nIn real life, I doubt this is entirely true. Maybe if one team scores in the first half, it?s more likely that there?s a goal in the second half because the team who?s behind is playing more aggressively. But I think I need to make this assumption to answer your question based on the information given.\n\nLet?s define two random variables: $$X_{1}$$ = the number of goals in the first half and $$X_{2}$$ = the number of goals in the second half. Again, I?m assuming that the X's are independent and identically distributed.\n\nWe can reverse-engineer the distribution of the X?s from the given information. The distribution you defined at the top of the problem is the distribution of the random variable $$Y = X_{1} + X_{2}$$\nSo we know, for example that P(Y = 0) = 0.4. There?s only one way for Y to be equal to 0, both $$X_{1} = 0$$ and $$X_{2} = 0$$\n$$P(X_{1} = 0, X_{2} = 0) = 0.4$$\nBecause the X?s are independent?\n$$P(X_{1} = 0) * P(X_{2} = 0) = 0.4$$\nBecause the X?s are identically distributed?\n$$P(X_{1} = 0)^{2} = P(X_{2} = 0)^{2} = 0.4$$\nTherefore?\n$$P(X_{1} = 0) = P(X_{2} = 0) = \\sqrt{0.4} \\approx 0.632$$\nGiven that there was a goal in the first half, the only way for there to be 1 goal or less total is if there are no goals in the second half, which we just discovered the probability of that is about 0.632.\n\nTherefore, the probability of more than 1 goal in the game given that there was 1 goal in the first half is 1 - 0.632 = 0.368.\n\u2022 Thank you for this very detailed answer. I really appreciate it.", "date": "2023-02-06 06:59:19", "meta": {"domain": "matchmaticians.com", "url": "https://matchmaticians.com/questions/u937mh/probability-of-more-than-1-goal-at-the-end-a-match-question", "openwebmath_score": 0.7680110335350037, "openwebmath_perplexity": 403.5599812197584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471637570106, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.8599584184032565}}
{"url": "https://brilliant.org/discussions/thread/polynomial-expansions-useful-formulas/", "text": "# Polynomial Expansions (useful formulas)\n\nLet $$\\mathrm{A}$$ be the set of all \"$$\\color{Blue}{\\text{useful}}$$\" things.\n\nLet $$\\mathrm{B}$$ be the set of all \"$$\\color{Green}{\\text{Awesome}}$$\" things.\n\nLet $$\\mathrm{C}$$ be the set of all \"$$\\color{Red}{\\text{fascinating}}$$\" things.\n\nLet $$\\mathrm{D}$$ be the set of all \"$$\\color{Brown}{\\text{easily understandable}}$$\" things.\n\nWhat this note contains is an element of $$\\mathrm{A \\cap B \\cap C \\cap D}$$....\n\n$\\color{Blue}{\\textbf{Polynomial Expansions}}$\n\n$$\\mathbf{1.}\\quad \\displaystyle \\dfrac{1-x^{m+1}}{1-x} = 1+x+x^2+...+x^m = \\sum_{k=0}^m x^k$$\n\n$$\\mathbf{2.} \\quad \\displaystyle \\dfrac{1}{1-x} = 1+x+x^2+... = \\sum_{k=0}^\\infty x^k$$\n\n$$\\mathbf{3.}\\quad \\displaystyle (1+x)^n = 1+\\binom{n}{1} x + \\binom{n}{2}x^2+...+\\binom{n}{n}x^n = \\sum_{k=0}^n \\dbinom{n}{k} x^k$$\n\n$$\\mathbf{4.}\\quad \\displaystyle (1-x^m)^n = 1-\\binom{n}{1}x^m+\\binom{n}{2}x^{2m}-...+(-1)^n\\binom{n}{n} x^{nm}$$\n\n$$\\quad \\quad \\quad \\quad\\quad \\displaystyle = \\sum_{k=0}^n (-1)^n \\dbinom{n}{k}x^{km}$$\n\n$$\\mathbf{5.}\\quad \\displaystyle\\dfrac{1}{(1-x)^n} = 1+ \\binom{1+n-1}{1} x + \\binom{2+n-1}{2} x^2+...+\\binom{r+n-1}{r} x^r+......$$\n\n$$\\quad \\quad \\quad \\quad \\quad \\displaystyle =\\sum_{k=0}^\\infty \\dbinom{k+n-1}{k} x^k$$\n\n$$\\color{Purple}{\\text{Tremendously useful}}$$ in calculating the $$\\color{Blue}{\\text{co-efficient}}$$ of any term in specially generating functions that we come across, in many combinatorics problems...\n\nTaken From - Alan Tucker's \"Applied Combinatorics\"\n\nGood luck problem solving !\n\nNote by Aditya Raut\n4\u00a0years, 2\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\n## Comments\n\nSort by:\n\nTop Newest\n\nI know that the second one only works for $$x<1$$. But do any of the others work for only $$x>1$$. Also, thank you so much for this note, it's very useful.\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\n-1<x<1 actually.\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nThank you, but do u know which ones only work for this case\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nCan you add parts of this page into the algebra wiki? I think that Algebraic Identities and Algebraic Manipulation - Identities, would be suitable places to add them.\n\nStaff - 3\u00a0years, 12\u00a0months ago\n\nLog in to reply\n\nwhat is the derivation of 5th one.\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nStandard result, it's related to \"Newton's generalised Binomial theorem\", but if you do want the derivation please see it here\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nDoes the last one work for 0<x<1???\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nDefinitely 2.\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nset contains Everything\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nIs this for nerds like you?\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\nDude don't use nerds in a derogatory manner please. If you don't like nerds or aren't one yourself, you should remove yourself from Brilliant.org. Have a nice day.\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\n@Finn Hulse , thanks for helping here, really !\n\nimg\n\nI truly like your comment, by these many likes :-\n\nimg\n\n- 4\u00a0years, 1\u00a0month ago\n\nLog in to reply\n\nHaha, anytime dude. :D\n\n- 4\u00a0years, 1\u00a0month ago\n\nLog in to reply\n\nAgreed. BTW, some nerds can be good at sports as well.\n\n- 4\u00a0years, 1\u00a0month ago\n\nLog in to reply\n\nBTW sharky, participate in JOMO 8, we miss your submission.\n\n@Sharky Kesa , JOMO 8 starts $$\\color{Red}{\\textbf{TOMORROW}}$$ and has some good questions I made.\n\n- 4\u00a0years, 1\u00a0month ago\n\nLog in to reply\n\n# (#Sharky_Surprises )\n\n- 4\u00a0years, 1\u00a0month ago\n\nLog in to reply\n\nThis is for using in generating functions we design for combinatorics problems.... For example, see the set \"vegetable combinatorics\".... (type in search bar simply).... That's for all who want to learn, nothing high-figh technique or anything, just formulas to get co-efficient of a specific term in a generating function. @Jack Daniel Zu\u00f1iga Cari\u00f1o\n\n- 4\u00a0years, 2\u00a0months ago\n\nLog in to reply\n\n\u00d7\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading...", "date": "2018-10-16 00:33:34", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/polynomial-expansions-useful-formulas/", "openwebmath_score": 0.9897925853729248, "openwebmath_perplexity": 8584.228265237203, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9838471670723234, "lm_q2_score": 0.8740772417253256, "lm_q1q2_score": 0.8599584180738521}}
{"url": "https://fr.mathworks.com/help/matlab/ref/fcontour.html", "text": "Main Content\n\n# fcontour\n\n## Syntax\n\nfcontour(f)\nfcontour(f,xyinterval)\nfcontour(___,LineSpec)\nfcontour(___,Name,Value)\nfcontour(ax,___)\nfc = fcontour(___)\n\n## Description\n\nexample\n\nfcontour(f) plots the contour lines of the function z = f(x,y) for constant levels of z over the default interval [-5 5] for x and y.\n\nexample\n\nfcontour(f,xyinterval) plots over the specified interval. To use the same interval for both x and y, specify xyinterval as a two-element vector of the form [min max]. To use different intervals, specify a four-element vector of the form [xmin xmax ymin ymax].\nfcontour(___,LineSpec) sets the line style and color for the contour lines. For example, '-r' specifies red lines. Use this option after any of the previous input argument combinations.\n\nexample\n\nfcontour(___,Name,Value) specifies line properties using one or more name-value pair arguments.\nfcontour(ax,___) plots into the axes specified by ax instead of the current axes.\n\nexample\n\nfc = fcontour(___) returns a FunctionContour object. Use fc to query and modify properties of a specific FunctionContour object. For a list of properties, see FunctionContour Properties.\n\n## Examples\n\ncollapse all\n\nPlot the contours of $f\\left(x,y\\right)=\\mathrm{sin}\\left(x\\right)+\\mathrm{cos}\\left(y\\right)$ over the default interval of $-5 and $-5.\n\nf = @(x,y) sin(x) + cos(y); fcontour(f)\n\nSpecify the plotting interval as the second argument of fcontour. When you plot multiple inputs over different intervals in the same axes, the axis limits adjust to display all the data. This behavior lets you plot piecewise inputs.\n\nPlot the piecewise input\n\n$\\begin{array}{cc}erf\\left(x\\right)+\\mathrm{cos}\\left(y\\right)& -5\n\nover $-5.\n\nfcontour(@(x,y) erf(x) + cos(y),[-5 0 -5 5]) hold on fcontour(@(x,y) sin(x) + cos(y),[0 5 -5 5]) hold off grid on\n\nPlot the contours of ${x}^{2}-{y}^{2}$ as dashed lines with a line width of 2.\n\nf = @(x,y) x.^2 - y.^2; fcontour(f,'--','LineWidth',2)\n\nPlot $\\mathrm{sin}\\left(x\\right)+\\mathrm{cos}\\left(y\\right)$ and $x-y$ on the same axes by using hold on.\n\nfcontour(@(x,y) sin(x)+cos(y)) hold on fcontour(@(x,y) x-y) hold off\n\nPlot the contours of ${e}^{-\\left(x/3{\\right)}^{2}-\\left(y/3{\\right)}^{2}}+{e}^{-\\left(x+2{\\right)}^{2}-\\left(y+2{\\right)}^{2}}$. Assign the function contour object to a variable.\n\nf = @(x,y) exp(-(x/3).^2-(y/3).^2) + exp(-(x+2).^2-(y+2).^2); fc = fcontour(f)\n\nfc = FunctionContour with properties: Function: @(x,y)exp(-(x/3).^2-(y/3).^2)+exp(-(x+2).^2-(y+2).^2) LineColor: 'flat' LineStyle: '-' LineWidth: 0.5000 Fill: off LevelList: [0.2000 0.4000 0.6000 0.8000 1 1.2000 1.4000] Show all properties \n\nChange the line width to 1 and the line style to a dashed line by using dot notation to set properties of the function contour object. Show contours close to 0 and 1 by setting the LevelList property. Add a colorbar.\n\nfc.LineWidth = 1; fc.LineStyle = '--'; fc.LevelList = [1 0.9 0.8 0.2 0.1]; colorbar\n\nCreate a plot that looks like a sunset by filling the area between the contours of\n\n$erf\\left(\\left(y+2{\\right)}^{3}\\right)-{e}^{\\left(-0.65\\left(\\left(x-2{\\right)}^{2}+\\left(y-2{\\right)}^{2}\\right)\\right)}.$\n\nf = @(x,y) erf((y+2).^3) - exp(-0.65*((x-2).^2+(y-2).^2)); fcontour(f,'Fill','on');\n\nIf you want interpolated shading instead, use the fsurf function and set its 'EdgeColor' option to 'none' followed by the command view(0,90).\n\nSet the values at which fcontour draws contours by using the 'LevelList' option.\n\nf = @(x,y) sin(x) + cos(y); fcontour(f,'LevelList',[-1 0 1])\n\nControl the resolution of contour lines by using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots, while decreasing it can increase plotting speed.\n\nCreate two plots in a 2-by-1 tiled chart layout. In the first plot, display the contours of $\\mathrm{sin}\\left(x\\right)\\mathrm{sin}\\left(y\\right)$. The corners of the squares do not meet. To fix this issue, increase 'MeshDensity' to 200 in the second plot. The corners now meet, showing that by increasing 'MeshDensity' you increase the resolution.\n\nf = @(x,y) sin(x).*sin(y); tiledlayout(2,1) nexttile fcontour(f) title('Default Mesh Density (71)') nexttile fcontour(f,'MeshDensity',200) title('Custom Mesh Density (200)')\n\nPlot $x\\mathrm{sin}\\left(y\\right)-y\\mathrm{cos}\\left(x\\right)$. Display the grid lines, add a title, and add axis labels.\n\nfcontour(@(x,y) x.*sin(y) - y.*cos(x), [-2*pi 2*pi], 'LineWidth', 2); grid on title({'xsin(y) - ycos(x)','-2\\pi < x < 2\\pi and -2\\pi < y < 2\\pi'}) xlabel('x') ylabel('y')\n\nSet the x-axis tick values and associated labels by setting the XTickLabel and XTick properties of the axes object. Access the axes object using gca. Similarly, set the y-axis tick values and associated labels.\n\nax = gca; ax.XTick = ax.XLim(1):pi/2:ax.XLim(2); ax.XTickLabel = {'-2\\pi','-3\\pi/2','-\\pi','-\\pi/2','0',... '\\pi/2','\\pi','3\\pi/2','2\\pi'}; ax.YTick = ax.YLim(1):pi/2:ax.YLim(2); ax.YTickLabel = {'-2\\pi','-3\\pi/2','-\\pi','-\\pi/2','0',... '\\pi/2','\\pi','3\\pi/2','2\\pi'};\n\n## Input Arguments\n\ncollapse all\n\nFunction to plot, specified as a function handle to a named or anonymous function.\n\nSpecify a function of the form z = f(x,y). The function must accept two matrix input arguments and return a matrix output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).\n\nExample: f = @(x,y) sin(x) + cos(y);\n\nPlotting interval for x and y, specified in one of these forms:\n\n\u2022 Vector of form [min max] \u2014 Use the interval [min max] for both x and y.\n\n\u2022 Vector of form [xmin xmax ymin ymax] \u2014 Use the interval [xmin xmax] for x and [ymin ymax] for y.\n\nAxes object. If you do not specify an axes object, then the fcontour uses the current axes.\n\nLine style and color, specified as a character vector or string containing a line style specifier, a color specifier, or both.\n\nExample: '--r' specifies red dashed lines\n\nThese two tables list the line style and color options.\n\nLine Style SpecifierDescription\n-Solid line (default)\n--Dashed line\n:Dotted line\n-.Dash-dot line\nColor SpecifierDescription\n\ny\n\nyellow\n\nm\n\nmagenta\n\nc\n\ncyan\n\nr\n\nred\n\ng\n\ngreen\n\nb\n\nblue\n\nw\n\nwhite\n\nk\n\nblack\n\n### Name-Value Arguments\n\nSpecify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.\n\nExample: 'MeshDensity',30\n\nThe properties listed here are only a subset. For a full list, see FunctionContour Properties.\n\nNumber of evaluation points per direction, specified as a number. The default is 71. Because fcontour uses adaptive evaluation, the actual number of evaluation points is greater.\n\nExample: 30\n\nFill between contour lines, specified as 'on' or 'off', or as numeric or logical 1 (true) or 0 (false). A value of 'on' is equivalent to true, and 'off' is equivalent to false. Thus, you can use the value of this property as a logical value. The value is stored as an on/off logical value of type matlab.lang.OnOffSwitchState.\n\n\u2022 A value of 'on' fill the spaces between contour lines with color.\n\n\u2022 A value of 'off' leaves the spaces between the contour lines unfilled.\n\nContour levels, specified as a vector of z values. By default, the fcontour function chooses values that span the range of values in the ZData property.\n\nSetting this property sets the associated mode property to manual.\n\nData Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64\n\nSpacing between contour lines, specified as a scalar numeric value. For example, specify a value of 2 to draw contour lines at increments of 2. By default, LevelStep is determined by using the ZData values.\n\nSetting this property sets the associated mode property to 'manual'.\n\nExample: 3.4\n\nData Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64\n\nColor of contour lines, specified as 'flat', an RGB triplet, a hexadecimal color code, a color name, or a short name. To use a different color for each contour line, specify 'flat'. The color is determined by the contour value of the line, the colormap, and the scaling of data values into the colormap. For more information on color scaling, see caxis.\n\nTo use the same color for all the contour lines, specify an RGB triplet, a hexadecimal color code, a color name, or a short name.\n\nFor a custom color, specify an RGB triplet or a hexadecimal color code.\n\n\u2022 An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].\n\n\u2022 A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n'red''r'[1 0 0]'#FF0000'\n\n'green''g'[0 1 0]'#00FF00'\n\n'blue''b'[0 0 1]'#0000FF'\n\n'cyan' 'c'[0 1 1]'#00FFFF'\n\n'magenta''m'[1 0 1]'#FF00FF'\n\n'yellow''y'[1 1 0]'#FFFF00'\n\n'black''k'[0 0 0]'#000000'\n\n'white''w'[1 1 1]'#FFFFFF'\n\n'none'Not applicableNot applicableNot applicableNo color\n\nHere are the RGB triplets and hexadecimal color codes for the default colors MATLAB\u00ae uses in many types of plots.\n\nRGB TripletHexadecimal Color CodeAppearance\n[0 0.4470 0.7410]'#0072BD'\n\n[0.8500 0.3250 0.0980]'#D95319'\n\n[0.9290 0.6940 0.1250]'#EDB120'\n\n[0.4940 0.1840 0.5560]'#7E2F8E'\n\n[0.4660 0.6740 0.1880]'#77AC30'\n\n[0.3010 0.7450 0.9330]'#4DBEEE'\n\n[0.6350 0.0780 0.1840]'#A2142F'\n\nLine width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.\n\nThe line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.\n\n## Output Arguments\n\ncollapse all\n\nOne or more FunctionContour objects, returned as a scalar or a vector. You can use these objects to query and modify the properties of a specific contour plot. For a list of properties, see FunctionContour Properties.\n\n## See Also\n\n### Properties\n\nIntroduced in R2016a\n\nDownload ebook", "date": "2021-11-27 21:52:37", "meta": {"domain": "mathworks.com", "url": "https://fr.mathworks.com/help/matlab/ref/fcontour.html", "openwebmath_score": 0.5267122983932495, "openwebmath_perplexity": 4924.7807914964405, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471670723234, "lm_q2_score": 0.8740772368049823, "lm_q1q2_score": 0.8599584132329863}}
{"url": "https://www.physicsforums.com/threads/linear-independence.814024/", "text": "# Homework Help: Linear independence\n\n1. May 15, 2015\n\n### nuuskur\n\n1. The problem statement, all variables and given/known data\nAssume vectors $a,b,c\\in V_{\\mathbb{R}}$ to be linearly independent. Determine whether vectors $a+b , b+c , a+c$ are linearly independent.\n\n2. Relevant equations\n\n3. The attempt at a solution\nWe say the vectors are linearly independent when $k_1a + k_2b +k_3c = 0$ only when every $k_n = 0$ - the only solution is a trivial combination.\nDoes there exist a non-trivial combination such that\n$k_1(a+b) + k_2(b+c) + k_3(a+c) = 0$?. Distributing:\n$k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0$ Since $a,b,c$ are linearly independent, the only way this result can occur is when:\n$k_1+k_3 =0\\Rightarrow k_1 = -k_3$\n$k_1+k_2 =0$\n$k_2+k_3 =0$\n\nSubstituting eq 1 into eq 2 we arrive at $k_2 - k_3 = 0$ and according to eq 3 $k_2 + k_3=0$, which means $k_2 - k_3 = k_2 + k_3$, therefore $k_3 = 0$, because $k=-k$ only if $k=0$. The only solution is a trivial combination, therefore the vectors $a+b, b+c, a+c$ are linearly independent.\n\nLast edited: May 15, 2015\n2. May 15, 2015\n\n### Staff: Mentor\n\nThat works for me. (IOW, I agree that the three new vectors are linearly independent.)\n\nInstead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that $k_1 = k_2 = k_3 = 0$, and that there are no other solutions.\n\nThe matrix looks like this, from your system:\n$$\\begin{bmatrix} 1 & 0 & 1 \\\\ 1 & 1 & 0 \\\\ 0 & 1 & 1\\end{bmatrix}$$\nAfter a few row operations, the final matrix is I3.\n\n3. May 15, 2015\n\n### Ray Vickson\n\nAlternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?\n\n4. May 15, 2015\n\n### nuuskur\n\nOh. Cramer's rule.\n$k_n = \\frac{D_{k_n}}{D}$ and since the determinant of the system is non zero, the corresponding determinants for every $k_n$ would be 0 (a full column of 0-s means det = 0) and therefore $k_1 = k_2 = k_3 = 0$\n\n5. May 15, 2015\n\n### Ray Vickson\n\nNo, I was not referring to Cramer's rule (which is rarely actually used when solving equations). I was referring to the theorem that if det(A) \u2260 0 then the n \u00d7n system Ak = 0 has k = 0 as its only solution.", "date": "2018-05-25 17:20:08", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/linear-independence.814024/", "openwebmath_score": 0.8258901834487915, "openwebmath_perplexity": 498.3685835048709, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471684931717, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.8599584112476732}}
{"url": "http://pmay.ranchimunicipal.com/apk/waskesiu-weather-kmcyu/article.php?page=elastic-net-vs-lasso-d8d828", "text": "The elastic-net penalty mixes these two; if predictors are correlated in groups, an $$\\alpha=0.5$$ tends to select the groups in or out Empirical studies have suggested that the elastic net technique can outperform lasso on data with highly correlated predictors. Elastic net is basically a combination of both L1 and L2 regularization. The regularization path is computed for the lasso or elasticnet penalty at a grid of values for the regularization parameter lambda. On the other hand, if \u03b1 is set to 0, the trained model reduces to a ridge regression model. First let\u2019s discuss, what happens in elastic net, and how it is different from ridge and lasso. It has been found to have predictive power better than Lasso, while still performing feature selection. Prostate cancer data are used to illustrate our methodology in Section 4, and simulation results comparing the lasso and the elastic net are presented in Section 5. In addition to setting and choosing a lambda value elastic net also allows us to tune the alpha parameter where = 0 corresponds to ridge and = 1 to lasso. A practical advantage of trading-off between Lasso and Ridge is that, it allows Elastic-Net to inherit some of Ridge\u2019s stability under rotation. Simulation B: EN vs Lasso Solution Paths \u2022Recall good grouping will set coefficients to similar values. Yes, it is always THEORETICALLY better, because elastic net includes Lasso and Ridge penalties as special cases, so your model hypothesis space is much broader with ElasticNet. Fit a generalized linear model via penalized maximum likelihood. Elastic-net is useful when there are multiple features which are correlated. Elastic net is the same as lasso when \u03b1 = 1. For right now I\u2019m going to give a basic comparison of the LASSO and Ridge Regression models. \u2022Lasso very unstable. Only the most significant variables are kept in the final model. Simply put, if you plug in 0 for alpha, the penalty function reduces to the L1 (ridge) term and if we set alpha to 1 we get the L2 (lasso\u2026 For now, see my post about LASSO for more details about regularization. Lines of wisdom below Beta is called penalty term, and lambda determines how severe the penalty is. Where: We didn\u2019t discuss in this post, but there is a middle ground between lasso and ridge as well, which is called the elastic net. The glmnet package written Jerome Friedman, Trevor Hastie and Rob Tibshirani contains very efficient procedures for fitting lasso or elastic-net regularization paths for generalized linear models. The Elastic Net is a weighted combination of both LASSO and ridge regression penalties. Lasso is a modification of linear regression, where the model is penalized for the sum of absolute values of the weights. For other values of \u03b1, the penalty term P \u03b1 (\u03b2) interpolates between the L 1 norm of \u03b2 and the squared L 2 norm of \u03b2. lasso regression: the coefficients of some less contributive variables are forced to be exactly zero. The consequence of this is to effectively shrink coefficients (like in ridge regression) and to set some coefficients to zero (as in LASSO). Elastic net with $\\lambda_{2}=0$ is simply ridge regression. Specially when there are multiple trees? Recently, I learned about making linear regression models and there were a large variety of models that one could use. R^2 for Lasso 0.28 R^2 for Ridge 0.14 R^2 for ElasticNet 0.02 This is confusing to me ... shouldn't the ElasticNet result fall somewhere between Lasso and Ridge? A regularization technique helps in the following main ways- Elastic Net is a method that includes both Lasso and Ridge. Elastic net regularization. elastic net regression: the combination of ridge and lasso regression. Both LASSO and elastic net, broadly, are good for cases when you have lots of features, and you want to set a lot of their coefficients to zero when building the model. The Lasso Regression gave same result that ridge regression gave, when we increase the value of .Let\u2019s look at another plot at = 10. Elastic Net : In elastic Net Regularization we added the both terms of L 1 and L 2 to get the final loss function. V.V.I. Jayesh Bapu Ahire. Elastic regression generally works well when we have a big dataset. Thanks to Wikipedia. Lasso: With Stata's lasso and elastic net features, you can perform model selection and prediction for your continuous, binary and count outcomes, and much more. Penaksir Ridge tidak peduli dengan penskalaan multiplikasi data. Thanks! Lasso is likely to pick one of these at random, while elastic-net is likely to pick both. For example, if a linear regression model is trained with the elastic net parameter \u03b1 set to 1, it is equivalent to a Lasso model. Introduction. Elastic Net includes both L-1 and L-2 norm regularization terms. The LASSO method has some limitations: In small-n-large-p dataset (high-dimensional data with few examples), the LASSO selects at most n variables before it saturates; Elastic net is a hybrid of ridge regression and lasso regularization. Elastic net regression combines the properties of ridge and lasso regression. Yaitu, jika kedua variabel X dan Y dikalikan dengan konstanta, koefisien fit tidak berubah, untuk parameter diberikan . Regularization techniques in Generalized Linear Models (GLM) are used during a modeling process for many reasons. In lasso regression, algorithm is trying to remove the extra features that doesn't have any use which sounds better because we can train with less data very nicely as well but the processing is a little bit harder, but in ridge regression the algorithm is trying to make those extra features less effective but not removing them completely which is easier to process. The model can be easily built using the caret package, which automatically selects the optimal value of parameters alpha and lambda. Let\u2019s take a look at how it works \u2013 by taking a look at a na\u00efve version of the Elastic Net first, the Na\u00efve Elastic Net. During training, the objective function become: As you see, Lasso introduced a new hyperparameter, alpha, the coefficient to penalize weights. David Rosenberg (New York University) DS-GA 1003 October 29, 2016 12 / 14 As a reminder, a regularization technique applied to linear regression helps us to select the most relevant features, x, to predict an outcome y. By setting \u03b1 properly, elastic net contains both L1 and L2 regularization as special cases. It works by penalizing the model using both the 1l2-norm1 and the 1l1-norm1. Lasso, Ridge and Elastic Net Regularization. Elasic Net 1. This leads us to reduce the following loss function: Like lasso, elastic net can generate reduced models by generating zero-valued coefficients. View source: R/glmnet.R. Description. Lasso, Ridge and Elastic Net Regularization. Elastic Net Regression = |predicted-actual|^2+[(1-alpha)*Beta^2+alpha*Beta] when alpha = 0, the Elastic Net model reduces to Ridge, and when it\u2019s 1, the model becomes LASSO, other than these values the model behaves in a hybrid manner. Note, here we had two parameters alpha and l1_ratio. So far the glmnet function can fit gaussian and multiresponse gaussian models, logistic regression, poisson regression, multinomial and grouped multinomial models and the Cox model. Alternatively we can perform both lasso and ridge regression and try to see which variables are kept by ridge while being dropped by lasso due to co-linearity. Likewise, elastic net with $\\lambda_{1}=0$ is simply lasso. In glmnet: Lasso and Elastic-Net Regularized Generalized Linear Models. Doing variable selection with Random Forest isn\u2019t trivial. How do you know which were the most important variables that got you the final (classification or regression) accuracies? It is known that the ridge penalty shrinks the coefficients of correlated predictors towards each other while the lasso tends to pick one of them and discard the others. Elastic Net 303 proposed for computing the entire elastic net regularization paths with the computational effort of a single OLS \ufb01t. Description Usage Arguments Details Value Author(s) References See Also Examples. The third line splits the data into training and test dataset, with the 'test_size' argument specifying the percentage of data to be kept in the test data. This gives us the benefits of both Lasso and Ridge regression. Elastic Net produces a regression model that is penalized with both the L1-norm and L2-norm. Elastic net regularization. Elastic Net is the combination of Ridge Regression and Lasso Regression. In sklearn , per the documentation for elastic net , the objective function $\u2026 March 18, 2018 April 7, 2018 / RP. Elastic Net vs Lasso Norm Ball From Figure 4.2 of Hastie et al\u2019s Statistical Learning with Sparsity. It\u2019s a linear combination of L1 and L2 regularization, and produces a regularizer that has both the benefits of the L1 (Lasso) and L2 (Ridge) regularizers. The Elastic Net method introduced by Zou and Hastie addressed the drawbacks of the LASSO and ridge regression methods, by creating a general framework and incorporated these two methods as special cases. Elastic Net. Lasso and Elastic have variable selection while Ridge does not? \u2022Elastic Net selects same (absolute) coefficient for the Z 1-group Lasso Elastic Net (\u03bb 2 = 2) Negated Z 2 roughly 1/10 of Z 1 per model In addition to setting and choosing a lambda value elastic net also allows us to tune the alpha parameter where = 0 corresponds to ridge and = 1 to lasso. Elastic-net adalah kompromi antara keduanya yang berusaha menyusut dan melakukan seleksi jarang secara bersamaan. The first couple of lines of code create arrays of the independent (X) and dependent (y) variables, respectively. As \u03b1 shrinks toward 0, elastic net \u2026 When looking at a subset of these, regularization embedded methods, we had the LASSO, Elastic Net and Ridge Regression. 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Is a method that includes both L-1 and L-2 norm regularization terms the 1l1-norm1 predictive power better lasso! Determines how severe the penalty is of models that one could use adalah kompromi antara keduanya berusaha. Reduced models by generating zero-valued coefficients 1l2-norm1 and the 1l1-norm1 net is basically a combination both! Have predictive power better than lasso, Ridge and elastic net is the same as lasso when \u03b1 =.! Using both the 1l2-norm1 and the 1l1-norm1 regularization terms pick both penalized with the. Is called penalty term, and lambda determines how severe the penalty is } =0$ simply... Net includes both L-1 and L-2 norm regularization terms lasso and elastic have variable with! Lasso or ElasticNet penalty at a subset of these, regularization embedded,. Variabel X dan y dikalikan dengan konstanta, koefisien fit tidak berubah, untuk parameter diberikan ). \u0391 = 1 is called penalty term, and lambda of these, regularization embedded methods, had... Simply lasso function: Elasic net 1 been found to have predictive power better than lasso, while still feature... Computational effort of a single OLS \ufb01t when \u03b1 = 1 to elastic can. Doing variable selection while Ridge does not regularization terms, the trained model to. Can be easily built using the caret package, which automatically selects the optimal Value parameters. The regularization path is computed for the lasso, elastic net \u2026,! } =0 \\$ is simply lasso hello to elastic net regularization ( Zou Hastie. For now, See my post about lasso for more Details about regularization a! Jika kedua variabel X dan y dikalikan dengan konstanta, koefisien fit tidak berubah, untuk diberikan! Regression penalties, untuk parameter diberikan and L2 regularization: in elastic net regularization we added the terms... Menyusut dan melakukan seleksi jarang secara bersamaan contains both L1 and L2 regularization that one use... Shrinks toward 0, the trained model reduces to a Ridge regression function: Elasic 1. Determines how severe the penalty is which automatically selects the optimal Value of parameters alpha and lambda / RP l1_ratio... Can outperform lasso on data with highly correlated predictors models ( GLM ) are used during a modeling process many... Now, See my post about lasso for more Details about regularization the entire net.", "date": "2022-01-23 12:31:43", "meta": {"domain": "ranchimunicipal.com", "url": "http://pmay.ranchimunicipal.com/apk/waskesiu-weather-kmcyu/article.php?page=elastic-net-vs-lasso-d8d828", "openwebmath_score": 0.7856265306472778, "openwebmath_perplexity": 1591.3799950309767, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9838471647042429, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8599584095494789}}
{"url": "https://math.stackexchange.com/questions/907039/using-factoring-to-solve-the-equation-r2-5r-24r2-3r-2-4r-10", "text": "# Using factoring to solve the equation $(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$\n\nSolve for all values of $r$: $$(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$$\n\nI'm not sure how my thinking isn't really correct here. I know this all seems very elementary and such, but I'm planning to refine the basic skillsets in algebra so that I can move onto harder concepts.\n\nWhat I do, is I factor both sides to get,\n\n$(r+8)(r-3)(r-1)(r-2) = (4r-10)(r+8)(r-3)$\n\nI then divide both sides by $(r+8)(r-3)$, giving me:\n\n$(r-1)(r-2) = (4r-10)$\n\nBringing over RHS to the LHS, by factoring, we then get the roots of the quadratic and get 3 and 4.\n\nWolfram is giving me answers of 3, 4, and -8 though. I don't really see where the 8 came from though. Can anyone help me out and explain? Also, is my thinking/procedure correct?\n\nThank you! (I know, basic question sorry).\n\nEdit:\n\nI realize that by dividing by $(r+8)(r-3)$, I divide by a quadratic with actual roots. That means I've missed out on one of them, which is -8.\n\nTherefore, the values that satisfy these quadratics are,\n\n-8, 3, and 4?\n\nI don't know. Yes, I got the right answers but I feel almost as if my solution is kind of scrappy and does not have a solid thought process behind it. Could anyone elaborate further as to show how the problem is done?\n\n\u2022 Notice that $r^2+5r-24$ appears on both sides, so you could subtract the RHS to get $(r^2+5r-24)(r^2-3r+2)-(r^2+5r-24)(4r-10)=0$ and factor $r^2+5r-24$ to get the equation $(r^2+5r-24)(r^2-7r+12)=0$ which gives you the right answers. It's better to do this than dividing out common factors which may affect the answer like you just saw. \u2013\u00a0Lost Aug 23 '14 at 17:12\n\nWhen you divide both sides by $(r+8)(r-3)$, you need to consider the cases $r+8=0$ and $r-3=0$. Coincidentally, the latter is \u201ccaught\u201d in your quadratic. But the former [p.s. are you sure Mathematica isn't giving you $r=-8$?] isn't handled there.\n$(r^2+5r-24)(r^2-3r+2-4r+10)=0$;\n$(r^2+5r-24)(r^2-7r+12)=0$\n$(r+8)(r-3)^2(r-4)=0$", "date": "2020-02-28 00:30:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/907039/using-factoring-to-solve-the-equation-r2-5r-24r2-3r-2-4r-10", "openwebmath_score": 0.7119643688201904, "openwebmath_perplexity": 171.0813193038032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471661250913, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8599584011096776}}
{"url": "https://math.stackexchange.com/questions/15556/is-zero-odd-or-even/15680", "text": "Is zero odd or even?\n\nSome books say that even numbers start from $$2$$ but if you consider the number line concept, I think zero($$0$$) should be even because it is in between $$-1$$ and $$+1$$ (i.e in between two odd numbers). What is the real answer?\n\n\u2022 +1 for \"thinking outside the books\". :) (Restoring a comment that seems to have been removed. What's up with that? This is a serious commendation of a seriously-commendable practice.) \u2013\u00a0Blue Dec 27 '10 at 0:19\n\u2022 en.wikipedia.org/wiki/Parity_of_zero \u2013\u00a0Jesse Madnick Dec 27 '10 at 22:00\n\u2022 youtube.com/watch?v=8t1TC-5OLdM \u2013\u00a0jimjim Oct 31 '14 at 1:50\n\u2022 0 can't be written in the form $2n+1$ \u2013\u00a0N.S.JOHN Mar 7 '16 at 6:03\n\u2022 @N.S.JOHN: Well, it can, by letting $n=-\\frac12.$ However, if we require that $n$ be an integer, then.... \u2013\u00a0Cameron Buie Jun 25 '18 at 23:45\n\nFor that, we can try all the axioms formulated for even numbers. I'll use only four in this case.\n\nNote: In this question, for the sake of my laziness, I will often use $N_e$ for even, and $N_o$ for odd.\n\nTest 1:\n\nAn even number is always divisible by $2$.\n\nWe know that if $x,y\\in \\mathbb{Z}$ and $\\dfrac{x}{y} \\in \\mathbb{Z},$ then $y$ is a divisor of $x$ (formally $y|x$).\n\nYes, both $0,2 \\in \\mathbb{Z}$ and yes, $\\dfrac{0}{2}$ is $0$ which is an integer. Passed this one with flying colors!\n\nTest 2:\n\n$N_e + N_e$ results in $N_e$\n\nLet's try an even number here, say $2$. If the answer results in an even number, then $0$ will pass this test. $\\ \\ \\ \\ \\underbrace{2}_{\\large{N_e}} + 0 = \\underbrace{2}_{N_e} \\ \\ \\$, so zero has passed this one!\n\nTest 3:\n\n$N_e + N_o$ results in $N_o$\n\n$0 + \\underbrace{1}_{N_o} = \\underbrace{1}_{N_o}$\n\nPassed this test too!\n\nTest 4:\n\nIf $n$ is an integer of parity $P$, then $n - 2$ will also be an integer of parity $P$.\n\nWe know that $2$ is even, so $2 - 2$ or $0$ is also even.\n\n\u2022 what about $3-3=0$? \u2013\u00a0user103028 Nov 3 '14 at 5:58\n\u2022 That is a valid test too. Odd $-$ odd is always even. So zero is even. \u2013\u00a0Parth Kohli Nov 3 '14 at 12:55\n\u2022 A note on notation: you could use $\\langle 2 \\rangle$ or $(2)$ or $2 \\mathbb{Z}$ to stand for the even numbers. \u2013\u00a0Robert Soupe Jan 22 '15 at 4:47\n\u2022 @VitalieGhelbert 3 isn't even. You would be subtracting 2, not 3 \u2013\u00a0user253055 Oct 17 '15 at 4:47\n\u2022 @user103028 Yes, you have proved again that 0 is even. Also, 0 is not odd, and definitely even. I have no idea how 0 could be \"neither\". \u2013\u00a0asher drummond Jul 12 '16 at 14:15\n\nYes, the classification of naturals by their parity (= remainder modulo $$2\\:$$) extends naturally to all integers: even integers are those integers divisible by $$2,\\,$$ i.e. $$\\rm\\: n = 2m\\equiv 0\\pmod 2,$$ and odd integers are those with remainder $$1$$ when divided by $$2,\\$$ i.e. $$\\rm\\ n = 2m\\!+\\! 1\\equiv 1\\pmod 2.\\,$$\n\nThe effectiveness of this parity classification arises from the fact that it is compatible with integer arithmetic operations, i.e. if $$\\rm\\ \\bar{a}\\ :=\\ a\\pmod 2\\$$ then $$\\rm\\ \\overline{ a+b}\\ =\\ \\bar a + \\bar b,\\ \\ \\overline{a\\ b}\\ =\\ \\bar a\\ \\bar b.\\:$$ Iterating, we infer that equalities between expressions composed of these integer operations (i.e. integer polynomial expressions) are preserved by taking their images modulo $$2\\,$$ (and ditto mod $$\\rm m\\,$$ for any integer $$\\rm m,\\,$$ e.g  mod $$9\\,$$ reduction yields  casting out nines). In this way we can strive to better understand integers by studying their images in the simpler (finite!) rings $$\\rm\\: \\mathbb Z/m\\, =\\,$$ integers modulo $$\\rm m.\\:$$\n\nFor example, if an integer coefficient polynomial has an integer root $$\\rm\\ P(n) = 0\\$$ then it persists as a root mod $$2,\\,$$ i.e. $$\\rm\\ P(\\bar n)\\equiv 0\\ (mod\\ 2),\\,$$ by the Polynial Congruence Rule. Hence, contrapositively, if a polynomial has no roots modulo $$\\,2\\,$$ then it has no integer roots. This leads to the following simple\n\nParity Root Test $$\\$$ A polynomial $$\\rm\\:P(x)\\:$$ with integer coefficients has no integer roots\nwhen its constant coefficient $$\\,\\rm P(0)\\,$$ and coefficient sum $$\\,\\rm P(1)\\,$$ are both odd.\n\nProof $$\\$$ The test verifies that $$\\rm\\ P(0) \\equiv P(1)\\equiv 1\\ \\ (mod\\ 2),\\$$ i.e. that $$\\rm\\:P(x)\\:$$ has no roots mod $$2$$, hence, as argued above, it has no integer roots. $$\\quad$$ QED\n\nSo $$\\rm\\, a x^2\\! + b x\\! + c\\,$$ has no integer roots if $$\\rm\\,c\\,$$ is odd and $$\\rm\\,a,b\\,$$ have equal parity $$\\rm\\,a\\equiv b\\pmod 2$$\n\nCompare the conciseness of this test to the messy reformulation that would result if we had to restrict it to positive integers. Then we could no longer represent polynomial equations in the normal form $$\\rm\\:f(x) = 0\\:$$ but, rather, we would need to consider general equalities $$\\rm\\:f(x) = g(x)\\:$$ where both polynomials have positive coefficients. Now the test would be much messier - bifurcating into motley cases. Indeed, historically, before the acceptance of negative integers and zero, the formula for the solution of a quadratic equation was stated in such an analogous obfuscated way - involving many cases. But by extending the naturals to the ring of integers we are able to unify what were previously motley separate cases into a single universal method of solving a general quadratic equation.\n\nAnalogous examples exist throughout history that help serve to motivate the reasons behind various number system enlargements. Studying mathematical history will help provide one with a much better appreciation of the motivations behind the successive enlargements of the notion of \"number systems\", e.g. see Kleiner: From numbers to rings: the early history of ring theory.\n\nAbove is but one of many examples where \"completing\" a structure in some manner serves to simplify its theory. Such ideas motivated many of the extensions of the classical number systems (as well as analogous geometrical and topological completions concept, e.g. adjoining points at $$\\infty$$, projective closure, compactification, model completion, etc). For some interesting expositions on such methods see the references here.\n\nNote $$\\:$$ Analogous remarks (on the power gained by normalizing equations to the form $$\\ldots = 0\\:$$) hold true more generally for any algebraic structure whose congruences are determined by ideals - so-called ideal determined varieties, e.g. see my post here and see Gumm and Ursini: Ideals in universal algebras. Without zero and negative numbers (additive inverses) we would not be able to rewrite expressions into such concise normal forms and we would not have available such powerful algorithms such as the Grobner basis algorithm, Hermite/Smith normal forms, etc.\n\nThis problem arose e.g. during the Beijing even-odd car ban for the 2008 Olympics, where cars with odd numbered licence plates were banned one day, then even the next day.\n\nThe choice is between:\n\n\u2022 0 is even and not odd,\n\u2022 0 is odd and not even,\n\u2022 0 is both even and odd,\n\u2022 0 is neither even nor odd (like infinity or $\\pi$) or\n\u2022 0 is assigned a unique title (like how 1 called a \"unit\" -- neither prime nor composite).\n\nThis is a matter of definition, so while you could define 0 to be any of the above, it's best to choose the definition that will be the most consistent with the usage of \"even\" and \"odd\" for numbers other than 0.\n\nLet $W=\\{2,4,6,\\ldots\\}$, $V=\\{1,3,5,\\ldots,\\}$ and lets look at the properties of even and odd numbers on these sets that we are familiar with.\n\n\u2022 A number is either even or odd, and not both.\n\u2022 If $w,x \\in W$ then $w+x \\in W$ (even + even = even).\n\u2022 If $w \\in W$ and $v \\in V$ then $w+v \\in V$ (even + odd = odd).\n\u2022 If $y,v \\in V$ then $y+v \\in W$ (odd + odd = even).\n\n[and probably many others I've forgotten to write here]\n\nSo it would be desirable that whichever definition we choose for 0, it preserves the above properties. Now lets say we let 0 be odd (the second and third cases listed above). Then our definition is not consistent with these properties. So, if we choose to define 0 as odd, we should have some substantial benefits to outweigh the losses. On the other hand, defining 0 to be even and not odd is consistent with the above properties.\n\nThe last two candidate definitions are essentially saying there is no consistent way of defining even or oddness to 0. But in this case, there is -- 0 is even and not odd.\n\n[Note: We also have the property that elements of W are all divisible by 2, but whether or not 0 is divisible by 2 is another matter of definition, for which we should again apply the \"which is the most sensible definition\" concept.]\n\n\u2022 Where I live, there is also a rule where vehicles whose plate numbers end in an odd number are only allowed on certain days, and similarly for even numbers. When this rule was first implemented, it was to everyone's consternation that neither the enforcers nor the motorists (okay, most of them, not all of them) knew about the parity of zero. A subsequent survey confirmed this. \u2013\u00a0J. M. isn't a mathematician May 30 '13 at 16:48\n\nThe real answer depends on the definition, because there is math-history tag invoked there was a time that $1$ was not considered an odd numbers, $0$ and negative numbers for sure where not considered even or odd. Historically the concept was defined only for natural numbers.\n\nThese days the set of all integers multiplied by $2$ is considered the set of even numbers, i.e. $\\dots,-4,-2,0,2,4,\\dots$ and all the integers not in that set are defined to be odd.\n\nThere is no real answer, it all depends on the definition, same way that the book is only dealing with natural numbers and not integers. The concept was extended from naturals to integers, but there are uncountably many ways to define the even and odds beyond the natural numbers. Just make sure others know what definition you are using to label something even or odd.\n\n\u2022 Dear Arjang, I'm curious --- do you have a reference for this? Regards, \u2013\u00a0Matt E Dec 27 '10 at 21:51\n\u2022 @Matt E : The part about 1 not being even nor odd is in the history section of en.wikipedia.org/wiki/Parity_(mathematics) . The part about even and oddness being defined only for natuarl numbers was in math history books. I do not have access to math history books but I found this link with some additional info : mathforum.org/library/drmath/view/65413.html The part about moders definition of even and odd : It was the simplest way that I think I had seen in a math book (I don't remember the book). If there are any parts that needs refrence let or doesnt seem correct let me know. \u2013\u00a0jimjim Dec 27 '10 at 22:14\n\u2022 Dear Arjang, Thanks very much! I certainly agree with your description of the modern definition, but I wasn't aware of the changing nature of even and oddness through history (or if I ever did know it, I had forgotten it). Thanks for the interesting answer. Best wishes, \u2013\u00a0Matt E Dec 27 '10 at 22:20\n\u2022 @Matt E: If you get a chance try to read the books History Of Mathematics By Stilwell, Analysis By it's history, and the feuds between mathematicians on Continuity, Heat Equation , Set theory. Specially Euler vs d'alembert is enlightening. To see the ideas progress from just an idea to complete theories we have today is better than listening to Bach! \u2013\u00a0jimjim Dec 27 '10 at 22:31\n\u2022 @MattE The Treviso Arithmetic of 1478 says \u201cNumber is a multitude brought together or assembled from several units, as in the case of 2, which is the first and the smallest number.\u201d (Translation found in David Eugene Smith A Source Book in Mathematics (Dover 1959), p. 2.) \u2013\u00a0MJD Jan 21 '15 at 16:39\n\nYES! zero is an even number. Here is the Dr. Math's explanation.\n\nThis seems to be a matter of confusion for many others around this planet,you may always like to ask google for your confusion.\n\nWhich numbers are you using? Positive integer? In this case, you don't have to consider zero (that's why, perhaps, some books says that even numbers start from 2). If 0 is in your number set, then yes, it is divisible by 2.\n\n$0$ is even. The difference of two distinct even numbers is also even, for example: $32 - 20 = 12$, $20 - 12 = 8$, $12 - 8 = 4$, etc. Also, the sum of two distinct even numbers is again even: $-12 + 32 = 20$, $32 + 20 = 52$, etc.\n\nBut with, say, $32 - 32 = 0$, why would this difference suddenly become odd? The answer is simple, it doesn't, it's still even.\n\nI think zero is an even number because it's in between $1$ and $-1$. If they are odd numbers then the number in between must be even. Which means zero is an even number.\n\nIs Zero Even? - Numberphile\n\nPS : This is completely opposite to my previous post (and view) but it has many good points and we are here to learn and not to push one's own view of things.\n\nAs the history of mathematics shows, zero is a very odd number whose general acceptance is surprisingly recent. However, it is not odd. Perhaps the ambiguity of \"odd\" causes all the confusion.\n\nZero is even. An even integer is a number of the form $2n$ where $n$ is also an integer. A number is odd, in the mathematical sense, if it is of the form $2n + 1$ with $n$ an integer.\n\nIf zero was odd, we could solve the equation $2x + 1 = 0$ in integers. But the only possible solution is $x = -\\frac{1}{2}$, which is a rational number but not an integer.\n\nThe set of integers is sometimes said to be \"doubly infinite,\" stretching out to positive infinity in one direction and to negative infinity in the other. If you wanted to, you could say the even numbers start at $-2$ and continue with $-4$, then $-6$, then $-8$, etc.\n\nThe AP sequence $$(-6,-4,-2,0,2,4,6)$$ shows that it is even.\n\nIf $$k$$ is an even integer, then $$(-1)^ k = 1$$\n\nDivision of $$0$$ by $$2$$ leaves no remainder.\n\n&c. are further confirmations.", "date": "2021-05-10 02:42:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/15556/is-zero-odd-or-even/15680", "openwebmath_score": 0.7878735065460205, "openwebmath_perplexity": 445.39808590467646, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471661250913, "lm_q2_score": 0.8740772220439509, "lm_q1q2_score": 0.8599583978824332}}
{"url": "https://math.stackexchange.com/questions/3013058/probability-of-three-sequential-events", "text": "# Probability of three sequential events\n\nA smuggler wants to transfer his smuggled goods from city A to city B. There are three police check posts between these two cities. Assume that there is no communication among the check posts. The probabilities of him being caught at these three stops are $$0.7, 0.5$$ and $$0.3$$ respectively. What is the probability that he successfully transfers his goods?\n\n\u2022 A] $$0.105$$\n\u2022 B] $$0.5$$\n\u2022 C] $$0.245$$\n\u2022 D] $$0.045$$\n\nIs it as simple as calculating success in all three scenarios: $$0.3 * 0.5 * 0.7 = 0.105$$. Is A correct answer ?\n\n\u2022 Definitely looks correct to me \u2013\u00a0gt6989b Nov 25 '18 at 16:41\n\u2022 Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent. \u2013\u00a0BlackMath Nov 25 '18 at 18:17\n\u2022 It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included. \u2013\u00a0Teepeemm Nov 25 '18 at 19:37\n\nLet $$A_i$$ be an event $$i$$-th police stop him. We are interested in $$P(A_1'\\cap A_2'\\cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 \\cdot 0.5 \\cdot 0.7 = 0.105$$\n\n(since $$A_1, A_2, A_3$$ are independant so are $$A_1', A_2', A_3'$$ ) so your answer is correct.\n\nYou can think this in two ways. first way Not being caught on first check post and not being caught on second check post and not being caught on third post. i.e. $$(1-P(A1))\u00d7(1-P(A2))\u00d7(1-P(A3) = 0.3\u00d70.5\u00d70.7 = 0.105$$\n\nsecond way Find P(being caught)\n\nCaught on first check post Or Not being caught on first post and caught on second post Or Not being caught on first post and Not being caught on second post and caught on third post.\n\n$$P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3) =0.7 + 0.3\u00d70.5 + 0.3\u00d70.5\u00d70.3 =0.895$$\n\nNow, P(not being caught) = 1- P(being caught) = 1- 0.895 =0.105", "date": "2019-11-18 21:10:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3013058/probability-of-three-sequential-events", "openwebmath_score": 0.39666011929512024, "openwebmath_perplexity": 837.4983840772386, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9939024368140507, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8599483240635246}}
{"url": "https://math.stackexchange.com/questions/2942357/prove-all-derivatives-of-fx-frac11x-by-induction", "text": "# Prove all derivatives of $f(x)=\\frac{1}{1+x}$ by induction\n\n## Problem\n\nProve all derivatives of: $$f(x)=\\frac{1}{1+x}$$ by induction.\n\n## Attempt to solve\n\nI compute few derivatives of $$f(x)$$ so that i can form general expression for induction hypothesis. I compute all derivatives utilizing formula:\n\n$$\\frac{d}{dx}x^n=nx^{n-1}$$\n\nFirst 4 derivatives are:\n\n$$f'(x)=(-1)\\cdot(1+x)^{-2}\\cdot 1 = -\\frac{1}{(1+x)^2}$$ $$f''(x)=(-1)(-2)(1+x)^{-3}\\cdot 1 = \\frac{2}{(1+x)^3}$$ $$f'''(x)=(-1)(-2)(-3)(1+x)^{-4}\\cdot 1 = -\\frac{6}{(1+x)^4}$$ $$f''''(x)=(-1)(-2)(-3)(-4)(1+x)^{-5} \\cdot 1 = \\frac{24}{(1+x)^5}$$\n\nObserve that $$(-1)(-2)(-3)(-4)\\dots (-n)$$ can be generalized with:\n\n$$(-1)(-2)(-3)(-4)\\dots(-n) = (-1)^n\\cdot n!$$\n\nExpression follows factorial of $$n$$ except every other value is positive and every other is negative. If i multiply it by $$(-1)^n$$ it is positive when $$n \\mod 2 = 0$$ and negative when $$n \\mod 2 \\neq 0$$.\n\nRest of the expression can be generalized as:\n\n$$(1+x)^{-n-1} = (1+x)^{-(n+1)}=\\frac{1}{(1+x)^{n+1}}$$\n\nCombining these gives formula in analytic form:\n\n$$f(n)= \\frac{(-1)^n \\cdot n!}{(1+x)^{n+1}}$$\n\nI can form induction hypothesis such that:\n\n$$\\frac{d^n}{dx^n}\\frac{1}{1+x} = \\frac{(-1)^n\\cdot n!}{(1+x)^{n+1}}$$\n\n### Induction proof\n\nBase case\n\nBase case when $$n=0$$:\n\n$$\\frac{d^0}{dx^0}\\frac{1}{1+x}=\\frac{1}{1+x}=\\frac{(-1)^0\\cdot 0!}{(1+x)^{0+1}}$$\n\nInduction step\n\n$$\\frac{d^n}{dx^n}\\frac{1}{1+x} =_{\\text{ind.hyp}} \\frac{(-1)^n\\cdot n!}{(1+x)^{n+1+1}}$$\n\n$$\\frac{d^n}{dx^n}\\frac{1}{1+x} = \\frac{(-1)^n\\cdot n!}{(1+x)^{n+2}}$$\n\nNow the problem is that formula i used for derivation can only be used recursively. I believe this is correct notation for $$n$$:th derivative but computing one is only defined recursively with formula i used:\n\n$$\\frac{d}{dx} x^n = nx^{n-1}$$\n\nWhich is not defined for case:\n\n$$\\frac{d^n}{dx^n}x^n = \\text{ undefined}$$\n\nThe idea is to show that this recursion can be expressed in analytical form and it is valid for all $$n\\in \\mathbb{Z}+$$ by induction. Problem is i don't know how do you express this in recursive form and how do you get from recursion formula to the analytical one.\n\n\u2022 All of your work seems fine. I'm not sure why the formula for iterated derivatives of $x^n$ is causing you anxiety, as you're only using it for negative values of $n$ (thus you'll never be in a position of needing the value of $\\frac{d^n}{dx^n} x^n$). \u2013\u00a0Connor Harris Oct 4 '18 at 16:40\n\u2022 I am not sure why you say that $\\frac{d^n}{dx^n}x^n$ is undefined. Simply applying it to a couple of values from $\\mathbb{Z}^+$ leads to the intuitive expression $\\frac{d^n}{dx^n}x^n=n!$ \u2013\u00a0mrtaurho Oct 4 '18 at 16:40\n\n$$\\frac{d^n}{dx^n}\\frac1{1+x}=\\frac{(-1)^nn!}{(1+x)^{n+1}}$$\n\nWe want to show that\n\n$$\\frac{d^{n+1}}{dx^{n+1}}\\frac1{1+x}=\\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}$$\n\nLet's verify:\n\n\\begin{align} \\frac{d^{n+1}}{dx^{n+1}}\\frac1{1+x} &=\\frac{d}{dx}\\left(\\frac{d^n}{dx^n}\\frac1{1+x} \\right)\\\\ &=\\frac{d}{dx}\\left(\\frac{(-1)^nn!}{(1+x)^{n+1}} \\right) \\\\ &=(-1)^n(n!) \\frac{d}{dx}(1+x)^{-(n+1)} \\\\ &= (-1)^n(n!) (-(n+1)) (1+x)^{-(n+2)}\\\\ &=\\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}} \\end{align}\n\n$$\\frac{\\mathrm d^n}{\\mathrm dx^n}\\frac{1}{1+x} = \\frac{(-1)^n n!}{(1+x)^{n+1}}$$\n$$\\frac{\\mathrm d^{n+1}}{\\mathrm dx^{n+1}}f(x)= \\frac{\\mathrm d}{\\mathrm dx}\\left( \\frac{\\mathrm d^n}{\\mathrm dx^n}f(x)\\right) \\quad \\forall n \\ge 0$$\nYou inductive hypothesis is that $$\\frac{d^nf}{dx^n}=\\frac{(-1)^nn!}{(1+x)^{n+1}}.$$\nYour inductive step would then be $$\\frac{d^{n+1}f}{dx^{n+1}}=\\frac d{dx}\\,\\frac{d^nf}{dx^n}=\\frac d{dx}\\left(\\frac{(-1)^nn!}{(1+x)^{n+1}}\\right) =\\frac{-(n+1)(-1)^nn!}{(1+x)^{n+2}}=\\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}},$$ which shows that the formula holds.", "date": "2019-08-23 05:17:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2942357/prove-all-derivatives-of-fx-frac11x-by-induction", "openwebmath_score": 0.9954684972763062, "openwebmath_perplexity": 559.2844737131455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587265099557, "lm_q2_score": 0.8705972801594707, "lm_q1q2_score": 0.8599400607533498}}
{"url": "https://tobydriscoll.net/fnc-julia/localapprox/integration.html", "text": "# 5.6. Numerical integration\u00b6\n\nIn calculus you learn that the elegant way to evaluate a definite integral is to apply the Fundamental Theorem of Calculus and find an antiderivative. The connection is so profound and pervasive that it\u2019s easy to overlook that a definite integral is a numerical quantity existing independently of antidifferentiation. However, most conceivable integrands have no antiderivative in terms of familiar functions.\n\nDemo 5.6.1\n\nThe antiderivative of $$e^x$$ is, of course, itself. That makes evaluation of $$\\int_0^1 e^x\\,dx$$ by the Fundamental Theorem trivial.\n\nexact = exp(1)-1\n\n1.718281828459045\n\n\nThe Julia package QuadGK has an all-purpose numerical integrator that estimates the value without finding the antiderivative first. As you can see here, it\u2019s often just as accurate.\n\nQ,errest = quadgk(x->exp(x),0,1)\n@show Q;\n\nQ = 1.718281828459045\n\n\nThe numerical approach is also far more robust. For example, $$e^{\\,\\sin x}$$ has no useful antiderivative. But numerically, it\u2019s no more difficult.\n\nQ,errest = quadgk(x->exp(sin(x)),0,1)\n@show Q;\n\nQ = 1.6318696084180515\n\n\nWhen you look at the graphs of these functions, what\u2019s remarkable is that one of these areas is basic calculus while the other is almost impenetrable analytically. From a numerical standpoint, they are practically the same problem.\n\nplot([exp,x->exp(sin(x))],0,1,fill=0,layout=(2,1),\nxlabel=L\"x\",ylabel=[L\"e^x\" L\"e^{\\sin(x)}\"],ylim=[0,2.7])\n\n\nNumerical integration, which also goes by the older name quadrature, is performed by combining values of the integrand sampled at nodes. In this section we will assume equally spaced nodes using the definitions\n\n(5.6.1)$t_i = a +i h, \\quad h=\\frac{b-a}{n}, \\qquad i=0,\\ldots,n.$\nDefinition 5.6.2 : \u00a0Numerical integration formula\n\nA numerical integration formula is a list of weights $$w_0,\\ldots,w_n$$ chosen so that for all $$f$$ in some class of functions,\n\n(5.6.2)$\\begin{split} \\int_a^b f(x)\\, dx \\approx h \\sum_{i=0}^n w_if(t_i) = h \\bigl[ w_0f(t_0)+w_1f(t_1)+\\cdots w_nf(t_n) \\bigr], \\end{split}$\n\nwith the $$t_i$$ defined in (5.6.1). The weights are independent of $$f$$ and $$h$$.\n\nNumerical integration formulas can be applied to sequences of data values even if no function is explicitly known to generate them. For our presentation and implementations, however, we assume that $$f$$ is known and can be evaluated anywhere.\n\nA straightforward way to derive integration formulas is to mimic the approach taken for finite differences: find an interpolant and operate exactly on it.\n\n## Trapezoid formula\u00b6\n\nOne of the most important integration formulas results from integration of the piecewise linear interpolant (see Section 5.2). Using the cardinal basis form of the interpolant in (5.2.3), we have\n\n$\\int_a^b f(x) \\, dx \\approx \\int_a^b \\sum_{i=0}^n f(t_i) H_i(x)\\, dx = \\sum_{i=0}^n f(t_i) \\left[ \\int_a^b H_i(x)\\right]\\, dx.$\n\nThus we can identify the weights as $$w_i = h^{-1} \\int_a^b H_i(x)\\, dx$$. Using areas of triangles, it\u2019s trivial to derive that\n\n(5.6.3)$\\begin{split}w_i = \\begin{cases} 1, & i=1,\\ldots,n-1,\\\\ \\frac{1}{2}, & i=0,n. \\end{cases}\\end{split}$\n\nPutting everything together, the resulting formula is\n\n(5.6.4)$\\begin{split} \\int_a^b f(x)\\, dx \\approx T_f(n) &= h\\left[ \\frac{1}{2}f(t_0) + f(t_1) + f(t_2) + \\cdots + f(t_{n-1}) + \\frac{1}{2}f(t_n) \\right]. \\end{split}$\nDefinition 5.6.3 : \u00a0Trapezoid formula\n\nThe trapezoid formula is a numerical integration formula in the form (5.6.2), with\n\n$\\begin{split} w_i = \\begin{cases} \\frac{1}{2},& i=0 \\text{ or } i=n, \\\\ 1, & 0 < i < n. \\end{cases} \\end{split}$\n\nGeometrically, as illustrated in Fig. 5.6.1, the trapezoid formula sums of the areas of trapezoids approximating the region under the curve $$y=f(x)$$.1\n\nThe trapezoid formula is the Swiss Army knife of integration formulas. A short implementation is given as Function 5.6.4.\n\nFig. 5.6.1 Trapezoid formula for integration. The piecewise linear interpolant defines trapezoids that approximate the region under the curve.\n\nFunction 5.6.4 : \u00a0trapezoid\n\nTrapezoid formula for numerical integration\n\n 1\"\"\"\n2    trapezoid(f,a,b,n)\n3\n4Apply the trapezoid integration formula for integrand f over\n5interval [a,b], broken up into n equal pieces. Returns\n6the estimate, a vector of nodes, and a vector of integrand values at the\n7nodes.\n8\"\"\"\n9function trapezoid(f,a,b,n)\n10    h = (b-a)/n\n11    t = range(a,b,length=n+1)\n12    y = f.(t)\n13    T = h * ( sum(y[2:n]) + 0.5*(y[1] + y[n+1]) )\n14    return T,t,y\n15end\n\n\nLike finite-difference formulas, numerical integration formulas have a truncation error.\n\nDefinition 5.6.5 : \u00a0Truncation error of a numerical integration formula\n\nFor the numerical integration formula (5.6.2), the truncation error is\n\n(5.6.5)$\\tau_f(h) = \\int_a^b f(x) \\, dx - h \\sum_{i=0}^{n} w_i f(t_i).$\n\nThe order of accuracy is as defined in Definition 5.5.3.\n\nIn Theorem 5.2.7 we stated that the pointwise error in a piecewise linear interpolant with equal node spacing $$h$$ is bounded by $$O(h^2)$$ as $$h\\rightarrow 0$$. Using $$I$$ to stand for the exact integral of $$f$$ and $$p$$ to stand for the piecewise linear interpolant, we obtain\n\n$\\begin{split}\\begin{split} I - T_f(n) = I - \\int_a^b p(x)\\, dx &= \\int_a^b \\bigl[f(x)-p(x)\\bigr] \\, dx \\\\ &\\le (b-a) \\max_{x\\in[a,b]} |f(x)-p(x)| = O(h^2). \\end{split}\\end{split}$\n\nA more thorough statement of the truncation error is known as the Euler\u2013Maclaurin formula,\n\n(5.6.6)$\\begin{split}\\int_a^b f(x)\\, dx &= T_f(n) - \\frac{h^2}{12} \\left[ f'(b)-f'(a) \\right] + \\frac{h^4}{740} \\left[ f'''(b)-f'''(a) \\right] + O(h^6) \\\\ &= T_f(n) - \\sum_{k=1}^\\infty \\frac{B_{2k}h^{2k}}{(2k)!} \\left[ f^{(2k-1)}(b)-f^{(2k-1)}(a) \\right],\\end{split}$\n\nwhere the $$B_{2k}$$ are constants known as Bernoulli numbers. Unless we happen to be fortunate enough to have a function with $$f'(b)=f'(a)$$, we should expect truncation error at second order and no better.\n\nObservation 5.6.6\n\nThe trapezoid integration formula is second-order accurate.\n\nDemo 5.6.7\n\nWe will approximate the integral of the function $$f(x)=e^{\\sin 7x}$$ over the interval $$[0,2]$$.\n\nf = x -> exp(sin(7*x));\na = 0;  b = 2;\n\n\nIn lieu of the exact value, we use the QuadGK package to find an accurate result.\n\nIf a function has multiple return values, you can use an underscore _ to indicate a return value you want to ignore.\n\nQ,_ = quadgk(f,a,b,atol=1e-14,rtol=1e-14);\nprintln(\"Integral = \\$Q\")\n\nIntegral = 2.6632197827615394\n\n\nHere is the trapezoid result at $$n=40$$, and its error.\n\nT,t,y = FNC.trapezoid(f,a,b,40)\n@show (T,Q-T);\n\n(T, Q - T) = (2.662302935602287, 0.0009168471592522209)\n\n\nIn order to check the order of accuracy, we increase $$n$$ by orders of magnitude and observe how the error decreases.\n\nn = [ 10^n for n in 1:5 ]\nerr = []\nfor n in n\nT,t,y = FNC.trapezoid(f,a,b,n)\npush!(err,Q-T)\nend\n\npretty_table([n err],[\"n\",\"error\"])\n\n\u250c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u252c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2510\n\u2502      n \u2502       error \u2502\n\u251c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u253c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2524\n\u2502     10 \u2502   0.0120254 \u2502\n\u2502    100 \u2502 0.000147305 \u2502\n\u2502   1000 \u2502  1.47415e-6 \u2502\n\u2502  10000 \u2502  1.47416e-8 \u2502\n\u2502 100000 \u2502 1.47417e-10 \u2502\n\u2514\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2534\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2518\n\n\nEach increase by a factor of 10 in $$n$$ cuts the error by a factor of about 100, which is consistent with second-order convergence. Another check is that a log-log graph should give a line of slope $$-2$$ as $$n\\to\\infty$$.\n\nplot(n,abs.(err),m=:o,label=\"results\",\nxaxis=(:log10,L\"n\"),yaxis=(:log10,\"error\"),\ntitle=\"Convergence of trapezoidal integration\")\n\n# Add line for perfect 2nd order.\nplot!(n,3e-3*(n/n[1]).^(-2),l=:dash,label=L\"O(n^{-2})\")\n\n\n## Extrapolation\u00b6\n\nIf evaluations of $$f$$ are computationally expensive, we want to get as much accuracy as possible from them by using a higher-order formula. There are many routes for doing so; for example, we could integrate a not-a-knot cubic spline interpolant. However, splines are difficult to compute by hand, and as a result different methods were developed before computers came on the scene.\n\nKnowing the structure of the error allows the use of extrapolation to improve accuracy. Suppose a quantity $$A_0$$ is approximated by an algorithm $$A(h)$$ with an error expansion\n\n(5.6.7)$A_0 = A(h) + c_1 h + c_2 h^2 + c_3 h^3 + \\cdots.$\n\nCrucially, it is not necessary to know the values of the error constants $$c_k$$, merely that they exist and are independent of $$h$$.\n\nUsing $$I$$ for the exact integral of $$f$$, the trapezoid formula has\n\n$I = T_f(n) + c_2 h^2 + c_4 h^{4} + \\cdots,$\n\nas proved by the Euler\u2013Maclaurin formula (5.6.6). The error constants depend on $$f$$ and can\u2019t be evaluated in general, but we know that this expansion holds. For convenience we recast the error expansion in terms of $$n=O(h^{-1})$$:\n\n(5.6.8)$I = T_f(n) + c_2 n^{-2} + c_4 n^{-4} + \\cdots.$\n\nWe now make the simple observation that\n\n(5.6.9)$I = T_f(2n) + \\tfrac{1}{4} c_2 n^{-2} + \\tfrac{1}{16} c_4 n^{-4} + \\cdots.$\n\nIt follows that if we combine (5.6.8) and (5.6.9) correctly, we can cancel out the second-order term in the error. Specifically, define\n\n(5.6.10)$S_f(2n) = \\frac{1}{3} \\Bigl[ 4 T_f(2n) - T_f(n) \\Bigr].$\n\n(We associate $$2n$$ rather than $$n$$ with the extrapolated result because of the total number of nodes needed.) Then\n\n(5.6.11)$I = S_f(2n) + O(n^{-4}) = b_4 n^{-4} + b_6 n^{-6} + \\cdots.$\n\nThe formula (5.6.10) is called Simpson\u2019s formula, or Simpson\u2019s rule. A different presentation and derivation are considered in Exercise 4.\n\nEquation (5.6.11) is another particular error expansion in the form (5.6.7), so we can extrapolate again! The details change only a little. Considering that\n\n$I = S_f(4n) = \\tfrac{1}{16} b_4 n^{-4} + \\tfrac{1}{64} b_6 n^{-6} + \\cdots,$\n\nthe proper combination this time is\n\n(5.6.12)$R_f(4n) = \\frac{1}{15} \\Bigl[ 16 S_f(4n) - S_f(2n) \\Bigr],$\n\nwhich is sixth-order accurate. Clearly the process can be repeated to get eighth-order accuracy and beyond. Doing so goes by the name of Romberg integration, which we will not present in full generality.\n\n## Node doubling\u00b6\n\nNote in (5.6.12) that $$R_f(4n)$$ depends on $$S_f(2n)$$ and $$S_f(4n)$$, which in turn depend on $$T_f(n)$$, $$T_f(2n)$$, and $$T_f(4n)$$. There is a useful benefit realized by doubling of the nodes in each application of the trapezoid formula. As shown in Fig. 5.6.2, when doubling $$n$$, only about half of the nodes are new ones, and previously computed function values at the other nodes can be reused.\n\nFig. 5.6.2 Dividing the node spacing by half introduces new nodes only at midpoints, allowing the function values at existing nodes to be reused for extrapolation.\n\nSpecifically, we have\n\n(5.6.13)$\\begin{split}\\begin{split} T_f(2m) & = \\frac{1}{2m} \\left[ \\frac{1}{2} f(a) + \\frac{1}{2} f(b) + \\sum_{i=1}^{2m-1} f\\Bigl( a + \\frac{i}{2m} \\Bigr) \\right]\\\\[1mm] & = \\frac{1}{2m} \\left[ \\frac{1}{2} f(a) + \\frac{1}{2} f(b)\\right] + \\frac{1}{2m} \\sum_{k=1}^{m-1} f\\Bigl( a+\\frac{2k}{2m} \\Bigr) + \\frac{1}{2m} \\sum_{k=1}^{m} f\\Bigl( a+\\frac{2k-1}{2m} \\Bigr) \\\\[1mm] &= \\frac{1}{2m} \\left[ \\frac{1}{2} f(a) + \\frac{1}{2} f(b) + \\sum_{k=1}^{m-1} f\\Bigl( a+\\frac{k}{m} \\Bigr) \\right] + \\frac{1}{2m} \\sum_{k=1}^{m} f\\Bigl( a+\\frac{2k-1}{2m} \\Bigr) \\\\[1mm] &= \\frac{1}{2} T_f(m) + \\frac{1}{2m} \\sum_{k=1}^{m-1} f\\left(t_{2k-1} \\right), \\end{split}\\end{split}$\n\nwhere the nodes referenced in the last line are relative to $$n=2m$$. Hence in passing from $$n=m$$ to $$n=2m$$, new integrand evaluations are needed only at the odd-numbered nodes of the finer grid.\n\nDemo 5.6.8\n\nWe estimate $$\\displaystyle\\int_0^2 x^2 e^{-2x}\\, dx$$ using extrapolation. First we use quadgk to get an accurate value.\n\nf = x -> x^2*exp(-2*x);\na = 0;  b = 2;\n@show Q;\n\nQ = 0.1904741736116139\n\n\nWe start with the trapezoid formula on $$n=N$$ nodes.\n\nN = 20;       # the coarsest formula\nn = N;  h = (b-a)/n;\nt = h*(0:n);   y = f.(t);\n\n\nWe can now apply weights to get the estimate $$T_f(N)$$.\n\nT = [ h*(sum(y[2:n]) + y[1]/2 + y[n+1]/2) ]\n\n1-element Vector{Float64}:\n0.19041144993926787\n\n\nNow we double to $$n=2N$$, but we only need to evaluate $$f$$ at every other interior node and apply (5.6.13).\n\nn = 2n;  h = h/2;  t = h*(0:n);\nT = [ T; T[end]/2 + h*sum( f.(t[2:2:n]) ) ]\n\n2-element Vector{Float64}:\n0.19041144993926787\n0.19045880585951175\n\n\nWe can repeat the same code to double $$n$$ again.\n\nn = 2n;  h = h/2;  t = h*(0:n);\nT = [ T; T[end]/2 + h*sum( f.(t[2:2:n]) ) ]\n\n3-element Vector{Float64}:\n0.19041144993926787\n0.19045880585951175\n0.1904703513046443\n\n\nLet us now do the first level of extrapolation to get results from Simpson\u2019s formula. We combine the elements T[i] and T[i+1] the same way for $$i=1$$ and $$i=2$$.\n\nS = [ (4T[i+1]-T[i])/3 for i in 1:2 ]\n\n2-element Vector{Float64}:\n0.19047459116625973\n0.19047419978635513\n\n\nWith the two Simpson values $$S_f(N)$$ and $$S_f(2N)$$ in hand, we can do one more level of extrapolation to get a sixth-order accurate result.\n\nR = (16S[2] - S[1]) / 15\n\n0.1904741736943615\n\n\nWe can make a triangular table of the errors:\n\nThe value nothing equals nothing except nothing.\n\nerr = [ T.-Q [nothing;S.-Q] [nothing;nothing;R-Q] ]\npretty_table(err,[\"order 2\",\"order 4\",\"order 6\"])\n\n\u250c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u252c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u252c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2510\n\u2502     order 2 \u2502    order 4 \u2502     order 6 \u2502\n\u251c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u253c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u253c\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2524\n\u2502 -6.27237e-5 \u2502    nothing \u2502     nothing \u2502\n\u2502 -1.53678e-5 \u2502 4.17555e-7 \u2502     nothing \u2502\n\u2502 -3.82231e-6 \u2502 2.61747e-8 \u2502 8.27476e-11 \u2502\n\u2514\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2534\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2534\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2518\n\n\nIf we consider the computational time to be dominated by evaluations of $$f$$, then we have obtained a result with about twice as many accurate digits as the best trapezoid result, at virtually no extra cost.\n\n## Exercises\u00b6\n\n1. \u2328 For each integral below, use Function 5.6.4 to estimate the integral for $$n=10\\cdot 2^k$$ nodes for $$k=1,2,\\ldots,10$$. Make a log-log plot of the errors and confirm or refute second-order accuracy. (These integrals were taken from [BJL05].)\n\n(a) $$\\displaystyle \\int_0^1 x\\log(1+x)\\, dx = \\frac{1}{4}$$\n\n(b) $$\\displaystyle \\int_0^1 x^2 \\tan^{-1}x\\, dx = \\frac{\\pi-2+2\\log 2}{12}$$\n\n(c) $$\\displaystyle \\int_0^{\\pi/2}e^x \\cos x\\, dx = \\frac{e^{\\pi/2}-1}{2}$$\n\n(d) $$\\displaystyle \\int_0^1 \\sqrt{x} \\log(x) \\, dx = -\\frac{4}{9}$$ (Note: Although the integrand has the limiting value zero as $$x\\to 0$$, it cannot be evaluated naively at $$x=0$$. You can start the integral at $$x=\\macheps$$ instead.)\n\n(e) $$\\displaystyle \\int_0^1 \\sqrt{1-x^2}\\,\\, dx = \\frac{\\pi}{4}$$\n\n2. \u270d The Euler\u2013Maclaurin error expansion (5.6.6) for the trapezoid formula implies that if we could cancel out the term due to $$f'(b)-f'(a)$$, we would obtain fourth-order accuracy. We should not assume that $$f'$$ is available, but approximating it with finite differences can achieve the same goal. Suppose the forward difference formula (5.4.8) is used for $$f'(a)$$, and its reflected backward difference is used for $$f'(b)$$. Show that the resulting modified trapezoid formula is\n\n(5.6.14)$G_f(h) = T_f(h) - \\frac{h}{24} \\left[ 3\\Bigl( f(t_n)+f(t_0) \\Bigr) -4\\Bigr( f(t_{n-1}) + f(t_1) \\Bigr) + \\Bigl( f(t_{n-2})+f(t_2) \\Bigr) \\right],$\n\nwhich is known as a Gregory integration formula.\n\n3. \u2328 Repeat each integral in Exercise 1 above using Gregory integration (5.6.14) instead of the trapezoid formula. Compare the observed errors to fourth-order convergence.\n\n4. \u270d Simpson\u2019s formula can be derived without appealing to extrapolation.\n\n(a) Show that\n\n$p(x) = \\beta + \\frac{\\gamma-\\alpha}{2h}\\, x + \\frac{\\alpha-2\\beta+\\gamma}{2h^2}\\, x^2$\n\ninterpolates the three points $$(-h,\\alpha)$$, $$(0,\\beta)$$, and $$(h,\\gamma)$$.\n\n(b) Find\n\n$\\int_{-h}^h p(s)\\, ds,$\n\nwhere $$p$$ is the quadratic polynomial from part (a), in terms of $$h$$, $$\\alpha$$, $$\\beta$$, and $$\\gamma$$.\n\n(c) Assume equally spaced nodes in the form $$t_i=a+ih$$, for $$h=(b-a)/n$$ and $$i=0,\\ldots,n$$. Suppose $$f$$ is approximated by $$p(x)$$ over the subinterval $$[t_{i-1},t_{i+1}]$$. Apply the result from part (b) to find\n\n$\\int_{t_{i-1}}^{t_{i+1}} f(x)\\, dx \\approx \\frac{h}{3} \\bigl[ f(t_{i-1}) + 4f(t_i) + f(t_{i+1}) \\bigr].$\n\n(Use the change of variable $$s=x-t_i$$.)\n\n(d) Now also assume that $$n=2m$$ for an integer $$m$$. Derive Simpson\u2019s formula,\n\n(5.6.15)$\\begin{split} \\begin{split} \\int_a^b f(x)\\, dx \\approx \\frac{h}{3}\\bigl[ &f(t_0) + 4f(t_1) + 2f(t_2) + 4f(t_3) + 2f(t_4) + \\cdots\\\\ &+ 2f(t_{n-2}) + 4f(t_{n-1}) + f(t_n) \\bigr]. \\end{split}\\end{split}$\n5. \u270d Show that the Simpson formula (5.6.15) is equivalent to $$S_f(n/2)$$, given the definition of $$S_f$$ in (5.6.10).\n\n6. \u2328 For each integral in Exercise 1 above, apply the Simpson formula (5.6.15) and compare the errors to fourth-order convergence.\n\n7. \u2328 For $$n=10,20,30,\\ldots,200$$, compute the trapezoidal approximation to\n\n$\\int_{0}^1 \\frac{1}{2.01+\\sin (6\\pi x)-\\cos(2\\pi x)} \\,d x \\approx 0.9300357672424684.$\n\nMake two separate plots of the absolute error as a function of $$n$$, one using a log-log scale and the other using log-linear. The graphs suggest that the error asymptotically behaves as $$C \\alpha^n$$ for some $$C>0$$ and some $$0<\\alpha<1$$. How does this result relate to (5.6.6)?\n\n8. \u2328 For each integral in Exercise 1 above, extrapolate the trapezoidal results two levels to get sixth-order accurate results, and compute the errors for each value.\n\n9. \u270d Find a formula like (5.6.12) that extrapolates two values of $$R_f$$ to obtain an eighth-order accurate one.\n\n1\n\nSome texts distinguish between a formula for a single subinterval $$[t_{k-1},t_k]$$ and a composite formula that adds them up over the whole interval to get (5.6.4).", "date": "2022-05-28 22:45:33", "meta": {"domain": "tobydriscoll.net", "url": "https://tobydriscoll.net/fnc-julia/localapprox/integration.html", "openwebmath_score": 0.8504312634468079, "openwebmath_perplexity": 997.3050622322135, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9877587234469587, "lm_q2_score": 0.8705972751232809, "lm_q1q2_score": 0.8599400531121726}}
{"url": "https://math.stackexchange.com/questions/1002944/combinations-10-people-divided-in-to-two-groups-one-of-6-and-one-of-4", "text": "# Combinations: 10 people divided in to two groups, one of 6 and one of 4?\n\nI only know the very basic formula for combinations: (n r)= n!/[r! (n-r)!]\n\nI've looked at similar problems on this website, but I'm having trouble understanding how these problems are done exactly:\n\n\"In how many ways can 10 people be divided into two groups (one group of 6 and one group of four)?\"\n\nmy sad attempt: (10 6) (4 4) = (10!/6!)*(4!/4!) = 5040???\n\nI'd really appreciate it if someone could explain how to go about doing this step by step.\n\nI will try to show to you, visually, a direct interpretation of the problem.\n\nFirst of all you must notice that one of the main meaning of a factorial is the number of permutations of n different elements over n positions.\n\nBy example the number of permutations of the string $ABCDE$ is $5!=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1$ (ways to order 5 different elements over 5 different positions).\n\nA direct interpretation of your problem, trough permutations, is that you have a string of 10 different elements, e.g. $ABCDEFGHIJK$ or $0123456789$, and you divide this string in two groups: one of length $4$ (i.e. cardinality 4) and other of length $6$\n\n$$\\overbrace{ABCD}\\ \\overbrace{EFGHIJK}$$\n\nYou have $10!$ ways to order a string of 10 different elements, after you can divide each of these string on 2 groups of length $4$ and $6$. But you doenst care the order of each group, for your problem $AKJG=JAKG=GKAJ$. The number of ways to order a string of length $4$ is $4!$. The same for your string of length $6$, so you must eliminate all of these duplicates.\n\nSo you will have $\\frac{10!}{4!\\cdot 6!}$ ways to form these different groups of 4 and 6 people. For this problem, because the binomial coefficient is defined as $\\binom{n}{k}=\\frac{n!}{k!\\cdot (n-k)!}$ then $\\frac{10!}{4!\\cdot 6!}=\\binom{10}{4}$.\n\nThis is a direct interpretation of the problem using the meaning of factorial as number of permutations of n elements over n positions.\n\nI will add an alternative, faster, way to interpret visually this problem.\n\nFirst of all I will expand the meaning of factorial as the permutations of n different elements over n positions. If you have n elements (or objects) for the position one you can choose between n elements to put, for the second position you can choose between $n-1$ elements (because you have already one element on position 1). For position 3 you can choose between $n-2$ elements (because you have, already, some element on position 1 and 2), etc., etc., etc...\n\nSo the permutations will be $n\\cdot (n-1)\\cdot (n-2)\\cdot (n-3)\\cdots 1=n!$\n\nNow our case: we have 10 people, and you want to see the different ways that you can group them in 6 and 4. The ordered ways for a group of 6 will be: for position 1 you can choose between 10, for position 2 between 9, ..., for position 6 between 5, i.e. $10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot5=(10)_6=\\frac{10!}{4!}$ (the expression $(n)_k$ is named falling factorial).\n\nBut this is the expression of ordered positions. To have the unordered positions you must divide between the different ways to order 6 different elements on 6 positions, i.e. divide between $6!$. So the unordered ways to group 10 different elements in a group of 6 is $\\frac{(10)_6}{6!}=\\binom{10}{6}$.\n\nBut, what happen with the other group of length 4? For every group of length 6 we will have a unordered group of 4 so the total amounts to divide a group of 10 people in two groups of 6 and 4 is $\\binom{10}{6}\\cdot1=\\binom{10}{6}$.\n\n$$\\binom{10}{6} = \\frac{10!}{6!\\,4!} = \\frac{10\\cdot9\\cdot8\\cdot7\\cdot\\overbrace{6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}}{4\\cdot3\\cdot2\\cdot1\\cdot\\underbrace{6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}} = \\frac{10\\cdot9\\cdot8\\cdot7}{4\\cdot3\\cdot2\\cdot1}$$ Since $8$ cancels $4\\cdot2$ and $\\dfrac 9 3=3$, this reduces to $10\\cdot3\\cdot7$.\n\n(If you wanted two groups of $5$ then you'd do a similar thing but you'd divide by $2$ when you're done, because one set of $5$ and its complementary set of the other $5$ would be the same way of dividing the set of $10$ into two sets of $5$. But that doesn't happen with $4$ and $6$ since $4\\ne6$.)\n\n\u2022 Worth noting that it is equivalent to ${10 \\choose 4}{6 \\choose 6}$. \u2013\u00a0miniparser Nov 2 '14 at 20:01", "date": "2019-12-06 18:41:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1002944/combinations-10-people-divided-in-to-two-groups-one-of-6-and-one-of-4", "openwebmath_score": 0.7179868221282959, "openwebmath_perplexity": 124.05673183891122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587265099557, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8599400441715479}}
{"url": "https://www.physicsforums.com/threads/division-by-zero.953833/", "text": "# Division by zero\n\n\u2022 B\nHi.\nx2 = x3 / x but x2 is defined for all x and equals zero at x=0 but what happens for x3 / x at x=o ? Is it defined at x=o ? Does it equal zero ? If not what is causing this anomaly ?\nThanks\n\nphinds\nGold Member\nN/0 is undefined regardless of N. If you assume otherwise, you can prove n=m where n and m are any arbitrary and different numbers.\n\nfresh_42\nMentor\nHi.\nx2 = x3 / x but x2 is defined for all x and equals zero at x=0 but what happens for x3 / x at x=o ? Is it defined at x=o ? Does it equal zero ? If not what is causing this anomaly ?\nThanks\n##x \\longmapsto x ## is everywhere continuous, ##x\\longmapsto \\dfrac{x^3}{x^2}## is not; at ##x=0\\,.##\n\nAlthough this is a removable singularity, it still is one, a gap. Algebraically division by zero isn't defined, simply because zero isn't part of any multiplicative group. The question never arises. It's like discussing the height of an apple tree on the moon.\n\njim mcnamara\nFactChecker\nGold Member\nWhen people write ## x = \\frac {x^3}{x}## with no restriction that ## x \\ne 0 ##, they are being (understandably) careless. The proper way is to keep track of all those divisions by zero and make sure that the results are still legitimate when the simplified equations are used. Otherwise, rule those points out. In physical applications, the continuity and nice behavior of the reduced formula, ##x##, at 0 makes it likely to also be valid at that point (##x = 0##).\n\nSo x = x3 / x2 is not a correct statement on its own ? It needs the addition of the statement \" x not equal to 0 \" ?\nI have seen the following statement in textbooks \" xn / xm = xn-m \" with no mention of \" x not equal to 0 \". Are they just being lazy and missing out the \" x not equal to 0 \" statement ?\n\nfresh_42\nMentor\nSo x = x3 / x2 is not a correct statement on its own ? It needs the addition of the statement \" x not equal to 0 \" ?\nStrictly, yes. But as it isn't defined for ##x=0## it is implicitly clear. As long as you don't want to write unnecessary additional lines, just leave it. Who writes ##x\\geq 0## if he uses ##\\sqrt{x}\\,?## This is simply self-evident, resp. clear by context. But logically, the domain of ##x## needs to be mentioned in general, such that we know what the function really is. But in your post it was pretty clear what you meant even without it.\nI have seen the following statement in textbooks \" xn / xm = xn-m \" with no mention of \" x not equal to 0 \". Are they just being lazy and missing out the \" x not equal to 0 \" statement ?\nSee above. It is simply not necessary as long as you don't write a book on logic. It's like mocking about a Pizza guy not telling you it's hot. However, if you talk about specific functions, you better say where and how they are defined. E.g. you could define\n$$f(x) = \\begin{cases}\\dfrac{x^3}{x^2} \\,&,\\, x\\neq 0 \\\\ 0\\,&\\,,x=0\\end{cases}$$\nor simply\n$$f(x)= \\dfrac{x^3}{x^2}\\; , \\;x\\neq 0$$\nwhich will be two different functions. So as always with written things, it depends on what you want to express. In post #1 and the example you gave it isn't necessary. Mocking about it is nit-picking.", "date": "2021-04-14 19:15:58", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/division-by-zero.953833/", "openwebmath_score": 0.6975548267364502, "openwebmath_perplexity": 708.0306246976824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9658995762509216, "lm_q2_score": 0.8902942341267435, "lm_q1q2_score": 0.8599348234816603}}
{"url": "https://gmatclub.com/forum/a-purse-contains-5-cent-coins-and-10-cent-coins-worth-a-total-of-249516.html", "text": "It is currently 18 Oct 2017, 04:52\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n### Show Tags\n\n17 Sep 2017, 13:45\nTop Contributor\n5\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n79% (01:58) correct 21% (01:31) wrong based on 48 sessions\n\n### HideShow timer Statistics\n\nA purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of$2.15. How many coins are in the purse?\n\nA. 26\nB. 27\nC. 28\nD. 29\nE. 30\n[Reveal] Spoiler: OA\n\n_________________\n\nKudos [?]: 1449 [0], given: 270\n\nIntern\nJoined: 26 Jul 2017\nPosts: 18\n\nKudos [?]: 10 [1], given: 25\n\nLocation: India\n\n### Show Tags\n\n17 Sep 2017, 19:53\nLet the total 5 cent coins be x, and the 10 cent coins be y\n\nFrom the question stem, $$5x + 10y = 175$$\nValue of y(by the following equation) : $$y = \\frac{175 - 5x}{10}$$\n\nSubstituting this value of y in the second equation\n$$10x + \\frac{5(175 - 5x)}{10} = 215$$\n$$100x + 875 - 25x = 2150$$\n$$75x = 1275$$\n$$x = 17$$\n\nSubstituting this value of x in equation$$5x+10y = 175$$\n$$85 + 10y = 175$$\n$$y = 9$$\n\nTherefore the sum of coins in the p = 17 + 9 = 26(Option A)\n_________________\n\nStay hungry, Stay foolish\n\nKudos [?]: 593 [0], given: 16\n\nDirector\nJoined: 22 May 2016\nPosts: 806\n\nKudos [?]: 258 [0], given: 546\n\nA purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink] ### Show Tags 17 Sep 2017, 21:26 Gnpth wrote: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse? A. 26 B. 27 C. 28 D. 29 E. 30 I solved with two variables, and it works, but I have a question about my own method that applies to vishalbalwani 's method, too. Let A = the number of 5-cent coins Let B = the number of 10- cent coins In one scenario, the coins, in their respective unknown quantities, total$1.75\n\nIn a second scenario, the 5-cent coins and 10-cent coins exchange quantities exactly, and they total $2.15 5A + 10B = 175 (P) 10A + 5B = 215 (Q) Multiply (Q) by two, and subtract (P) 20 A + 10B = 430 __5A + 10B = 175 15A = 255 A = 17 There are 17 coins worth 5 cents. Use (P) to find the number of 10-cent pieces 5(17) + 10B = 175 10B = 90 B = 9 There are 9 coins worth 19 0 cents. A = 17, B = 9, total = 26 coins Answer A I have run this problem three ways, including vishalbalwani 's, and working from answer choices. This combination (number of coins), is the only one that works. Question: Is the method sound? The variables work, but they seem inconsistent. I use A as a quantity for 5-cent coins. The coefficients of the variables -- 5 and 10 -- are the values of the coins in cents. But in the second equation, (Q) I do not think I have switched quantities, per the prompt. I think have switched values. In the second equation, (Q), the 5, a value, is in front of B -- which is supposed to be the quantity of 10-cent coins. vishalbalwani did the same. What am I missing? We are supposed to be switching quantities. But switching values works. I think pushpitkc avoids the whole problem by immediately defining y in terms of x from one equation. Is my method sound? Kudos [?]: 258 [0], given: 546 Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 1634 Kudos [?]: 837 [0], given: 2 Location: United States (CA) Re: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75\u00a0[#permalink]\n\n### Show Tags\n\n22 Sep 2017, 07:49\nGnpth wrote:\nA purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of$2.15. How many coins are in the purse?\n\nA. 26\nB. 27\nC. 28\nD. 29\nE. 30\n\nWe can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the \u201cmoney\u201d equation by recalling that a 5-cent coin is $0.05 and a 10-cent coin is$0.10; thus:\n\n0.05a + 0.1b = 1.75\n\nAnd, after we reverse the coinage, we have:\n\n0.1a + 0.05b = 2.15\n\nMultiplying the first equation by -2, we have:\n\n-0.1a - 0.2b = -3.50\n\nAdding the two equations, we have:\n\n-0.15b = -1.35\n\nb = 9\n\nThus:\n\n0.05a + (0.1)(9) = 1.75\n\n0.05a = 0.85\n\na = 17\n\nSo, there is a total of 17 + 9 = 26 coins.\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nKudos [?]: 837 [0], given: 2\n\nDirector\nJoined: 22 May 2016\nPosts: 806\n\nKudos [?]: 258 [0], given: 546\n\nA purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink] ### Show Tags 22 Sep 2017, 08:38 ScottTargetTestPrep wrote: Gnpth wrote: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse? A. 26 B. 27 C. 28 D. 29 E. 30 We can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the \u201cmoney\u201d equation by recalling that a 5-cent coin is$0.05 and a 10-cent coin is $0.10; thus: 0.05a + 0.1b = 1.75 And, after we reverse the coinage, we have: 0.1a + 0.05b = 2.15 Multiplying the first equation by -2, we have: -0.1a - 0.2b = -3.50 Adding the two equations, we have: -0.15b = -1.35 b = 9 Thus: 0.05a + (0.1)(9) = 1.75 0.05a = 0.85 a = 17 So, there is a total of 17 + 9 = 26 coins. Answer: A ScottTargetTestPrep , where you reverse the coinage, what do variables $$a$$ and $$b$$ stand for? (I had the same question about my own method, above.) Perhaps the variables stand simply for the total number of coins, and absent any total number of coins, we cannot write one variable in terms of another. Conceptually, however, that reasoning does not make a lot of sense to me. We have two unknown quantities in two scenarios. In the first equation, the quantity b is defined as the number of$0.10 coins (and qty \"a\" initially is # of $.05 coins) If \"b\" then switches to the number of$0.05 coins (vice versa for \"a\"), how are those two a's and b's the same, such that we can use them as an identity in solving two equations?\n\nSorry if this question is sophomoric; usually I can find my way out of a muddle. Not this time.\n\nKudos [?]: 258 [0], given: 546\n\nMath Forum Moderator\nJoined: 02 Aug 2009\nPosts: 4969\n\nKudos [?]: 5458 [1], given: 112\n\n### Show Tags\n\n01 Oct 2017, 12:26\n1\nKUDOS\nExpert's post\ngenxer123 wrote:\nScottTargetTestPrep wrote:\nGnpth wrote:\nA purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of$2.15. How many coins are in the purse?\n\nA. 26\nB. 27\nC. 28\nD. 29\nE. 30\n\nWe can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the \u201cmoney\u201d equation by recalling that a 5-cent coin is $0.05 and a 10-cent coin is$0.10; thus:\n\n0.05a + 0.1b = 1.75\n\nAnd, after we reverse the coinage, we have:\n\n0.1a + 0.05b = 2.15\n\nMultiplying the first equation by -2, we have:\n\n-0.1a - 0.2b = -3.50\n\nAdding the two equations, we have:\n\n-0.15b = -1.35\n\nb = 9\n\nThus:\n\n0.05a + (0.1)(9) = 1.75\n\n0.05a = 0.85\n\na = 17\n\nSo, there is a total of 17 + 9 = 26 coins.\n\nScottTargetTestPrep , where you reverse the coinage, what do variables $$a$$ and $$b$$ stand for? (I had the same question about my own method, above.)\n\nPerhaps the variables stand simply for the total number of coins, and absent any total number of coins, we cannot write one variable in terms of another.\n\nConceptually, however, that reasoning does not make a lot of sense to me.\n\nWe have two unknown quantities in two scenarios. In the first equation, the quantity b is defined as the number of $0.10 coins (and qty \"a\" initially is # of$.05 coins)\n\nIf \"b\" then switches to the number of $0.05 coins (vice versa for \"a\"), how are those two a's and b's the same, such that we can use them as an identity in solving two equations? Sorry if this question is sophomoric; usually I can find my way out of a muddle. Not this time. The variables a and b stand for the initial number of 5- and 10-cent coins, before and after reversing the coinage. The only thing that changes when the coinage is reversed is that the number of 5-cent coins is now b and the number of 10-cent coins is now a. _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Kudos [?]: 837 [1], given: 2 Re: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75 \u00a0 [#permalink] 01 Oct 2017, 12:26\nDisplay posts from previous: Sort by\n\n# A purse contains 5-cent coins and 10-cent coins worth a total of \\$1.75\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2017-10-18 11:52:38", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-purse-contains-5-cent-coins-and-10-cent-coins-worth-a-total-of-249516.html", "openwebmath_score": 0.38608285784721375, "openwebmath_perplexity": 4795.028110111905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9658995762509216, "lm_q2_score": 0.8902942275774318, "lm_q1q2_score": 0.859934817155683}}
{"url": "https://math.stackexchange.com/questions/1371717/relation-between-real-roots-of-a-polynomial-and-real-roots-of-its-derivative", "text": "Relation between real roots of a polynomial and real roots of its derivative\n\nI have this question which popped in my mind while solving questions of maxima and minima. First Case:Let $f(x)$ be an $n$ degree polynomial which has $r$ real roots. Using this can we say anything about the number of real roots of $f'(x)$?\n\nSecond Case:Suppose, $f(x)$ has all $n$ real roots. Then will all of its derivatives also have all real roots?\n\nAlso, if any of its derivatives do not have all real roots, then will $f(x)$ also have not all real roots? If the above is true then what about its converse?\n\nComment Case:For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case.\n\n\u2022 I don't understand the beginning. If $f(x)$ is an n-degree polynomial with $r$ real roots, then it is not the case that $f'(x)$ will have $(r-1)$ real roots. Consider $f(x) = x^2 +1$. Here $r = 0$, but $f'(x) = 2x$ which has 1 root at 0. Jul 23 '15 at 19:35\n\u2022 Sorry I never realized that.I just said what I was taught.Now I know it was incorrect.I have edited it. But I dont understand why the question has been downvoted? Jul 23 '15 at 19:49\n\u2022 You're probably being downvoted for not providing any work/ideas for your question. Jul 23 '15 at 20:22\n\u2022 This is of interest: en.wikipedia.org/wiki/Gauss%E2%80%93Lucas_theorem Jul 23 '15 at 21:48\n\nFirst case: If the number of real roots $r$ of $f(x)$ is greater than one, then $f'(x)$ has at least $r-1$ real roots. (The limitation \"greater than one\" is not necessary but the statement is trivial if $r\\le 1$.) Given any two roots $a<b$ of $f(x)$, $f$ is continuous and differentiable on $[a,b]$, so by Rolle's theorem $f'(c)=0$ for some $a<c<b$.\n\nThere may be more roots of $f'(x)$ than those between roots of $f(x)$, so the only upper bound is the obvious one of $n-1$. Ask if you need examples. It seems to me that if multiplicity is taken into account that the number of real roots of $f'(x)$ has the same parity (even/odd) as the number of real roots of $f(x)$, but I haven't proven it yet. If multiplicity is not taken into account, the parity can be anything.\n\nSecond case: If $f(x)$ has degree $n$ and has $n$ real roots, then each consecutive pair of roots of $f(x)$ defines a root of $f'(x)$, which makes $n-1$ roots of $f'(x)$. Since $f'(x)$ is a polynomial of degree $n-1$, this is all possible roots. This continues for all later derivatives, so you are correct: all its derivatives will have all real roots.\n\nThird case: The contrapositive of the second case tells us that if any of its derivatives have any non-real roots, then $f(x)$ also has some non-real roots.\n\nFourth case: The converse of the third case is not true. For example, $f(x)=x^2+1$ has two non-real roots, but its derivative $f'(x)=2x$ has one real root.\n\nComment case: You asked, \"Suppose $f'(x)$ is a $5$ degree polynomial with $3$ real roots. What are the possible no. of roots that $f(x)$ can have($3,4,5$ etc.?).\"\n\nThe formulas for $f(x)$ and $f'(x)$ are given in the diagram, where $C$ is a real constant, zero in the graph. You can see that $f'(x)$ is a degree $5$ polynomial with $3$ real roots.\n\nThe dashed horizontal lines show the possible number of real roots of $f(x)$ for varying values of $C$. There are $0$ real roots for $C=3$, $1$ real root for $C\\approx. 2.638$, $2$ real roots for $C=1$, $3$ real roots for $C\\approx -1.757$, and $4$ real roots for $C=-1.9$. My discussion for the first case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots.\n\n\u2022 For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case. Jul 23 '15 at 20:40\n\nFirst, notice that if any differentiable function $f(x)$ has $r$ real (distinct) roots, then $f'(x)$ must have at least $r-1$ distinct roots. This follows from the following:\n\nSuppose that $a_1<...<a_r$ are (distinct) roots from $f(x)$ and $i \\in \\{1,...,r-1\\}$. Then consider every interval of the form $[a_i,a_{i+1}]$. Since this interval is compact, any function on this interval achieves a maximum/minimum. Each minimum/maximum (depending on whether the function $f$ is dipping below or above the real line) corresponds to a root of $f'(x)$. Since we have $r -1$ such intervals, we have at least $r-1$ roots.\n\nSince all polynomials are differentiable, we notice that if $f(x)$ has $r$ (distinct) roots, then $f'(x)$ has at least $r - 1$ distinct roots.\n\nYour question isn't the clearest, but maybe someone else can come up with a better question to answer 1.\n\nIf a polynomial $f(x)$ is degree $n$ and has $n$ (distinct) real roots, then using the argument above (as well as the fundamental theorem of algebra (that might be over-kill, but oh well)), you can conclude that $f'(x)$ has exactly $n-1$ real roots.\n\nGiven this work, you can figure out 3 with the case for (distinct) roots.", "date": "2021-10-15 20:36:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1371717/relation-between-real-roots-of-a-polynomial-and-real-roots-of-its-derivative", "openwebmath_score": 0.8377802968025208, "openwebmath_perplexity": 108.06990944492968, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105300791785, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8599314636202414}}
{"url": "https://www.freemathhelp.com/forum/threads/how-many-significant-digits-are-in-1040-and-1040-0.111578/", "text": "# How many significant digits are in 1040. and 1040.0?\n\n#### Indranil\n\n##### Junior Member\nHow many significant digits are in 1040. and 1040.0? I am confused\n\n#### Dr.Peterson\n\n##### Elite Member\nHow many significant digits are in 1040. and 1040.0? I am confused\nIt will help if you state your specific confusion. What possibilities do you see for the answers? What about it are you unsure of?\n\n#### Indranil\n\n##### Junior Member\nIt will help if you state your specific confusion. What possibilities do you see for the answers? What about it are you unsure of?\nDo 1040. has four significant digits and 1040.0 has five significant digits?\n\n#### Dr.Peterson\n\n##### Elite Member\nDo 1040. has four significant digits and 1040.0 has five significant digits?\nThat's right. You aren't really confused, just perhaps uncertain.\n\nWhen there is a decimal point present, all digits from the first non-zero digit on the left to the last written digit on the right are significant.\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nDo 1040. has four significant digits and 1040.0 has five significant digits?\nAnd 1040 (without decimal point) has three significant digits.\n\n#### Indranil\n\n##### Junior Member\nAnd 1040 (without decimal point) has three significant digits.\nCould you explain why without decimal point, 1040 has three significant digits? and what is the difference between 1040 and 1040.? because I knew 1040 and 1040. are the same.\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nCould you explain why without decimal point, 1040 has three significant digits? and what is the difference between 1040 and 1040.? because I knew 1040 and 1040. are the same.\nWhen you write 1040 - it can be 1036 (min) to 1044 (max)\n\nWhen you write 1040. - it can be 1039.6 to 1040.4\n\n#### Dr.Peterson\n\n##### Elite Member\nCould you explain why without decimal point, 1040 has three significant digits? and what is the difference between 1040 and 1040.? because I knew 1040 and 1040. are the same.\nOf course you are right that 1040 and 1040. represent the same number; it is only the convention about implied accuracy that is at issue.\n\nI would say that 1040 is somewhat ambiguous, because you can't be sure whether the final zero is there because that digit is actually zero, or only because the digit is needed in order to have the right place values. It might really have either 3 or 4 significant digits - that is, it might have been rounded to the nearest ten or to the nearest unit, and be written the same. Many people assume the least possible accuracy. Ideally, significant digits should be read on in scientific notation, where this never happens (because there is only one digit before the decimal point).\n\nOn the other hand, with the decimal point, it is clear that the writer intends to stop after the decimal point, and the zero is significant.\n\n#### JeffM\n\n##### Elite Member\nWhen you write 1040 - it can be 1036 (min) to 1044 (max)\n\nWhen you write 1040. - it can be 1039.6 to 1040.4\nAccording to the conventions of significant figures, 1040 and 1040. mean different things. That is because the whole idea of significant figures involves applied mathematics. In pure mathematics\n\n$$\\displaystyle 1040. \\equiv 1040$$ as you recognize, but the concept of significant figures is not relevant to pure mathematics.\n\nSee https://en.m.wikipedia.org/wiki/Significant_figures\n\nLast edited:\n\n#### Indranil\n\n##### Junior Member\nWhen you write 1040 - it can be 1036 (min) to 1044 (max)\n\nWhen you write 1040. - it can be 1039.6 to 1040.4\n\n#### Dr.Peterson\n\n##### Elite Member\nI think what others of us said make the point more clearly, showing the reason.\n\nTake a more interesting example: 104000. I might write that as an estimate, or rounded number, whether the real number was exactly 104000, or 104001 and rounded to the nearest ten, or 104010 and rounded to the nearest hundred, or 104100 and rounded to the nearest thousand. There is no way to distinguish those situations by the way we write it, unless we use one of the (relatively rare) conventions Wikipedia suggests. As they say, \"The significance of trailing zeros in a number not containing a decimal point can be ambiguous. For example, it may not always be clear if a number like 1300 is precise to the nearest unit (and just happens coincidentally to be an exact multiple of a hundred) or if it is only shown to the nearest hundred due to rounding or uncertainty.\"\n\nOne way to reduce the ambiguity when the number is accurate to the nearest unit is to include a decimal point and write it as \"104000.\". Since this can be done, some people tend to assume that if there is no decimal point, the number of significant digits should be assumed to be as few as possible, in this case three, with all the trailing zeros insignificant.\n\nBack to the original, if a number was rounded to the nearest unit and the result was 1040, then the number could be anything from 1039.5 to just under 1040.5. (What Khan wrote was not quite right.) Any of those numbers, such as 1039.51 or 1040.49, would round to 1040.\n\nIf the number had been rounded to the nearest ten, then the original could have been anything from 1035 to just under 1045. (I should add that whether you would round 1035 to 1040 depends on the precise convention you are following for rounding.)\n\n#### HallsofIvy\n\n##### Elite Member\nAs has been said, the distinction between 1040 and 1040. is a convention. It has been agreed that if there is a decimal point after an integer then all digits in the number are \"significant\" and that if there is no decimal point then the significant digits are end with the rightmost non-zero digit.\n\n#### Denis\n\n##### Senior Member\n...and if 1040. was at end of sentence, then you'd have 1040.. :cool:", "date": "2019-03-25 05:41:30", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/how-many-significant-digits-are-in-1040-and-1040-0.111578/", "openwebmath_score": 0.671176016330719, "openwebmath_perplexity": 922.9623029438295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9763105293899019, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8599314630131287}}
{"url": "https://math.stackexchange.com/questions/4186288/need-help-with-a-markov-chain-for-weather-conditions", "text": "need help with a Markov chain for weather conditions\n\nThis is my second time posting the question as I failed to do so the first time, because I did not know the proper way. My apologies.\n\nQuestion:\n\nSuppose that whether or not it rains today depends on weather conditions through the previous teo days. If it has rained for the past two days, then it will rain today with probability 0.7. If it did not rain for any of the past two days, then it will rain today with probability 0.4. In any other case the weather today will, with probability 0.5, be the same as the weather yesterday. Describe the weather condition over time with a Markov chain.\n\nSuppose that it rained today and yesterday. Calculate the expected number of days until it rains three consecutive days for the first time in the future.\n\nI have found 4 different states that I named RR(0), RN(2), NR(1), and NN(3). R stands for when it rains and N is for when it does not.\n\nAs the question asks, I have tried finding the possible ways of three consecutive days being rainy. At time n, we are given it was rainy today and yesterday, meaning we are in State 0.\n\n1-) First possibility is when it rains tomorrow, which gives us RRR (we got the three consecutive days)\n\n2-) Second possibility is when we go from 0 to 2, from 2 to 1, from 1 to 0, and stay in 0 one day. That follows as : RRNRRR (In 4 days we can get rain for 3 consecutive days)\n\n3-) Third is when we go from 0 to 2, from 2 to 3, from 3 to 1, from 1 to 0, and stay in 0 one day. That follows as: RRNNRRR.\n\nTo conclude what I have in mind about the question, wherever we go, when we get to State 0, we need to stay there one more day to get three consecutive rainy days. That means the minimum # of days to get rain for three consecutive days is one day. However, after this point, I am not able to proceed with the question.\n\nAny help would be appreciated, thank you!!\n\nEdit: I think the maximum # of days is just a random number, which leads me to the expectation of the sum of a random number of geometric random variable, but still I can't go any further beyond that. Thank you!\n\n\u2022 For $i=0,1,2,3$, let $E_i$ be the expected number of days until $3$ consecutive rainy days, if the chain is currently in state $i$. We are asked to compute $E_0$. Notice that $E_0=1+.3E_2$. Develop similar equations for the other $E_i$ and solve. Jun 29, 2021 at 17:39\n\u2022 Developing the equations is where I am stuck at and I am also confused as to whether there is a boundary condition. Could you please explain to me, for instance, how to develop E0? Jun 29, 2021 at 18:44\n\nLet me get you started.\n\nI claim that $$E_0=1+.3E_2$$ as I stated in a comment.\n\nSuppose we are in state $$0$$. We must wait at least $$1$$ more day to see if we'll have three consecutive days of rain. $$70\\%$$ of the time, it rains, and we are done, but $$30\\%$$ of the time we transition to state $$2$$, and then we must wait, on average, $$E_2$$ days to get $$3$$ consecutive rainy days.\n\nIf we are in state $$2$$, similar reasoning gives $$E_2=1+.5E_1+ .5E_3$$\n\nYou can read the equations right off your diagram. You should get $$4$$ equations with $$4$$ unknowns and a unique solution. I'm sure you won't have any problem finishing from here.\n\n\u2022 I have found E3=1 + 0.6(E3) + 0.4(E1) (state 1 is the only different state we can transition to from state 3 and we might also stay in state 3 with p=0.6?) and E1= 1 + 0.5(E0) + 0.5(E2). Do you think I made a mistake or those are correct? Jun 30, 2021 at 8:20\n\u2022 @SydneyAstbury I think you are correct. Jun 30, 2021 at 11:36\n\nOrdinarily, Markov Chains are conditional on the previous step, but not on the previous two steps. A way to get around this in the current problem is to re-define the states to account for two days, with suitable overlaps.\n\nThe new states are 00 (for consecutive dry days) 01 (dry followed by wet), and so on to 11 (for two wet days in a row).\n\nStates 'overlap' so that 00 can be followed by 01, but not by 01, etc. In the matrix below we call the new states A,B, C, D.\n\nThe transition matrix can be written in R as\n\nP = matrix(c(.6, .4,  0,  0,\n0,  0, .5, .5,\n.5, .5,  0,  0,\n0,  0, .3, .7), byrow=T, nrow=4)\n\n\nWe seek $$\\sigma$$ such that $$\\sigma P = P,$$ so that $$\\sigma$$ is a left eigenvector of $$P.$$ Because R finds right eigenvectors, we use the transpose t(P).\n\nThe stationary distribution is proportional to the right eigenvector of smallest modulus, which R prints first. [In R, the symbol %*% denotes matrix multiplication.]\n\ng = eigen(t(P))\\$vec[,1]\nsg = g/sum(g); sg\n[1] 0.2542373 0.2033898 0.2033898 0.3389831\nsg %*% P  # check\n[,1]      [,2]      [,3]      [,4]\n[1,] 0.2542373 0.2033898 0.2033898 0.3389831\n\n\nSo the long-run probability of rain is $$0.5424.$$\n\nNote: In this relatively simple problem it is not difficult to solve a few simultaneous equations, but the eigen-method shown above is very convenient for ergodic chains with more than four states.", "date": "2022-12-08 16:40:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4186288/need-help-with-a-markov-chain-for-weather-conditions", "openwebmath_score": 0.559999406337738, "openwebmath_perplexity": 205.81854152687754, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105300791785, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.8599314529279963}}
{"url": "https://math.stackexchange.com/questions/3081850/is-the-following-application-of-the-clt-correct", "text": "Is the following application of the CLT correct?\n\nI am new to probability theory and still have a problem understanding some things. I came across the following problem that I am unsure on how to solve. Here it is:\n\nSuppose that in a city with $$100,000$$ cars a shop sells tires for cars. It has been observed that in an interval of $$3$$ months the percentage of cars that come to the shop for the replacement of all their tires is $$0.5$$%. What is the least number of tires that the shop has to order so that it can satisfy all its customers in a $$3$$ month interval with probability $$\\geq 95$$%?\n\nAnd here is my attempt:\n\nLet $$X$$ be the random variable of cars coming at the shop in a $$3$$ month interval. Then $$X$$ is a discrete random variable following binomial distribution $$B(100,000; 0.005)$$. We would like to find the least $$x\\in\\mathbb{N}$$ satisfying $$P(X\\leq x)\\geq95$$%. Since (I assume) every car has $$4$$ tires, our answer is going to be $$4x$$. Now since we have a large number of cars, by the Central limit theorem, we have that $$X\\sim N(\\mu,\\sigma^2)$$, where $$\\mu, \\sigma^2$$ are the expected value and the variance of $$B(100,000; 0.005)$$ and therefore, by calculating, it is $$\\mu=500, \\sigma^2=475$$. Now we have $$P(X\\leq x)=P\\big{(}\\displaystyle{\\frac{X-500}{\\sqrt{475}}\\leq\\frac{x-500}{\\sqrt{475}}\\big{)}=P\\big{(}Z\\leq\\frac{x-500}{\\sqrt{475}}\\big{)}}$$, where $$Z\\sim N(0,1)$$. Now our $$x$$ will satisfy $$\\displaystyle{P\\big{(}Z\\leq\\frac{x-500}{\\sqrt{475}}\\big{)}}=0.95$$, therefore $$\\displaystyle{P\\big{(}0 hence (by the tables) it is $$\\displaystyle{\\frac{x-500}{\\sqrt{475}}=1.645}$$, which yields $$x=535.85..$$ and since $$x$$ is supposed to be an integer we have that $$x=536$$. therefore the least number of tires is $$2144$$.\n\nIs my solution correct? if not, what should I have used?\n\nComment: I know that the $$x=$$fractional is not correct since i specified that $$x\\in\\mathbb{N}$$, but you get the point.\n\n\u2022 \"the percentage of cars that come to the shop for the replacement of all their tires is 0.5%\" That is quite poorly worded, it says that X is a constant ( 500). I guess it meant what you understood, though. \u2013\u00a0leonbloy Jan 21 at 14:49\n\u2022 @leonbloy I know, I was confused by this as well, but this is an accurate translation of the problem... \u2013\u00a0JustDroppedIn Jan 21 at 15:03\n\nAssuming that the shop's only tire business is to replace all four tires, your method makes sense. [Otherwise, I can't see how to work the problem.]\n\nThe CLT implies that a binomial distribution with sufficiently large $$n$$ is well-approximated by a normal distribution, which is what you're doing.\n\nBinomial. I used the binomial quantile function (inverse CDF) qbinom in R to compute an answer directly from the binomial distribution. It is very nearly the same as your answer.\n\n4*ceiling(qbinom(.95, 10^5, .005))\n[1] 2148\n\n\nI expect that the small discrepancy may result from rounding in finding the mean and standard deviation of the approximating normal distribution or from using printed normal tables. [Perhaps try $$\\sigma^2 = 497.5.]$$\n\nNormal approximation to binomial. Without rounding, I get the following result from the normal approximation. (The normal approximation with $$n=10^5$$ and $$p = 0.005$$ should be quite good, but not necessarily exactly perfect.)\n\nn = 10^5; p = .005; mu = n*p; sg = sqrt(n*p*(1-p))\n4*ceiling(qnorm(.95, mu, sg))\n[1] 2148\n\n\nPoisson approximation. Another reasonable approximation is to use the distribution $$\\mathsf{Pois}(\\lambda = \\mu = np).$$ This approximation is quite good for large $$n$$ and small $$p:$$\n\n4*ceiling(qpois(.95, mu))\n[1] 2148\n\n\nNormal approximation to Poisson. For $$\\lambda$$ as large as $$500$$ the Poisson distribution is well approximated by $$\\mathsf{Norm}(\\lambda, \\sqrt{\\lambda}):$$\n\n4*ceiling(qnorm(.95, mu, sqrt(mu)))\n[1] 2148\n\n\nThe following figure shows the binomial distribution (black bars) along with its normal approximation (blue curve) and its Poisson approximation (centers of red circles). The probability to the right of the vertical dotted line is about 0.05.", "date": "2019-08-18 11:00:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3081850/is-the-following-application-of-the-clt-correct", "openwebmath_score": 0.5571967363357544, "openwebmath_perplexity": 175.4303192496582, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.976310529389902, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.8599314523208836}}
{"url": "https://math.stackexchange.com/questions/1740130/how-to-enumerate-2d-integer-coordinates-ordered-by-euclidean-distance", "text": "How to enumerate 2D integer coordinates ordered by Euclidean distance?\n\nThe square of Euclidean distance between $(x, y)\\in\\mathbb{Z}^2$ and origin is $d = x^2+y^2$. How to enumerate the coordinates $(x, y)$ in ascending order of $d$?\n\nFor example, the first 14 sets of coordinates are:\n\nd=0: { (0,0) }\nd=1: { (1,0), (0,1), (0,-1), (-1,0) }\nd=2: { (1,-1), (-1,-1), (-1,1), (1,1) }\nd=4: { (2,0), (0,2), (-2,0), (0,-2) }\nd=5: { (1,2), (-1,2), (1,-2), (-2,1), (-2,-1), (2,-1), (2,1), (-1,-2) }\nd=8: { (2,2), (-2,2), (-2,-2), (2,-2) }\nd=9: (0,3), (-3,0), (0,-3), (3,0)\nd=10: (1,3), (-1,3), (3,1), (-3,1), (-3,-1), (1,-3), (-1,-3), (3,-1)\nd=13: (2,-3), (-3,-2), (3,-2), (-2,-3), (-3,2), (3,2), (-2,3), (2,3)\nd=16: (0,-4), (-4,0), (4,0), (0,4)\nd=17: (-4,1), (-4,-1), (4,1), (1,-4), (4,-1), (-1,-4), (-1,4), (1,4)\nd=18: (3,-3), (-3,-3), (-3,3), (3,3)\nd=20: (4,2), (4,-2), (-4,-2), (2,-4), (-4,2), (-2,-4), (-2,4), (2,4)\nd=25: (-3,-4), (-5,0), (5,0), (4,3), (-3,4), (-4,3), (0,-5), (4,-3), (-4,-3), (3,-4), (3,4), (0,5)\n\n\nThe first 14 iterations are depicted as:\n\n          13\n131210 9101213\n1311 8 7 6 7 81113\n12 8 5 4 3 4 5 812\n10 7 4 2 1 2 4 710\n13 9 6 3 1 0 1 3 6 913\n10 7 4 2 1 2 4 710\n12 8 5 4 3 4 5 812\n1311 8 7 6 7 81113\n131210 9101213\n13\n\n\nFor a finite range of $(x, y)$, a trivial algorithm is to store all coordinates within the range into a list, and then sort the coordinates with $d$. However, this will need $O(n\\log n)$ time and $O(n)$ space.\n\nAnother possible approach is to solve the diophantine equation $x^2 + y^2 = m$ for $m = 0, 1, \\ldots, ... d_\\mathrm{max}$. But it seems also a hard problem.\n\nIs there any simpler way with lower time/space complexity?\n\nHere is a C++ code of the trivial solution for reference.\n\n\u2022 As far as storing goes, you can certainly take advantage of symmetries and perhaps more generally use the Sum of Squares function $r_2(n)$ (or $r'_2(n)$ if you're not interested in order/signs). I will also link this mathoverflow question as it seems to answer precisely what you desire. \u2013\u00a0Fimpellizieri Jun 28 '16 at 22:54\n\u2022 Why the complexity of the trivial algorithm is not $O(n^2)$? \u2013\u00a0user84976 Jun 29 '16 at 10:35\n\u2022 Your $n$ seems to be the number of points in your grid, not the edge length of the grid, right? So in terms of the edge length $a$, you get $O(a^2\\log a)$ time complexity and $O(a^2)$ space complexity. But you can't go asymptotically lower than $O(a^2)$ time complexity since you have to enumerate this many points. Are you interested in tweaking constants here, or only in getting rid of the $\\log a$ term or reducing the space complexity? \u2013\u00a0MvG Jul 5 '16 at 12:29\n\nYou can employ symmetry to only generating one eighth of a circle, e.g. $0\\le x\\le y$. It should be obvious how to do this.\n\nYou can reduce the space complexity to $O(r)=O(\\sqrt n)$, i.e. linear in the radius of your disk, not in the number of grid points it covers. The way to do this is by considering parallel lines of grid points. Then on each line you know that the points will be enumerated starting closest to the center and moving outwards, so it is enough to keep track of the next candidate on each line.\n\nThe most appropriate data structure for this keeping track is most likely a priority queue, which will give you quick access to the element with minimal $d$. But even then, elementary operations in a priority queue have $O(\\log n)$ time complexity. Which in this case means you are still at $O(r^2\\log r)=O(n\\log n)$ overall time complexity since you enumerate $O(r^2)$ elements and maintain a data structure where you have to perform two $O(\\log r)$ operations for each of them.\n\nHere is an implementation of these ideas. The core portions are these:\n\nstruct Pt { // essentially your struct Point, just shorter names\nint x, y, d;\nPt(int x, int y) : x(x), y(y), d(x*x + y*y) {}\nbool operator<(const Pt& p) const { return d > p.d; } // flipped!\n};\n\nvoid enumerateGridPoints(int r) {\npriority_queue<Pt> q;\nq.push(Pt(0, 0));\nint d, level = -1;\nwhile ((d = q.top().d) <= r*r) {\nreportLevel(++level, d);\ndo {\nPt p = q.top();\nq.pop();\nreportPoint(p);\nif (p.y < p.x) q.push(Pt(p.x, p.y + 1));\nif (p.y == 0) q.push(Pt(p.x + 1, 0));\n} while (q.top().d == d);\n}\n}\n\n\nSince the C++ std::priority_queue is a max queue by default, and since I'm lazy, I just flipped the sign in the comparison operation so that the element with smallest d is the maximal element and thus extracted next.\n\nOne benefit of the above algorithm compared to yours is that it's easy to modify it so that enumerates points without any preset limit. That might be useful if you'd like to e.g. inspect sites until you find something you are looking for, but you don't know up front how long you will have to search until you find it. Both time and space complexity scale with the generated output, so small initial portions can be obtained quickly while still allowing you to proceed to larger results.\n\nEdsger Dijkstra gives an elegant recursive solution to the problem of finding solutions to $x^2 + y^2 = m$ in his 1976 book The Discipline of Programming. The solution is outlined here, and involves sweeping potential $x$ values down from $\\sqrt m$ and $y$ values up from 0.\n\nConsider the function $B(x, y)$ that returns solutions between $x$ and $y$, defined recursively as follows:\n\n$$B(x, y) = \\begin{cases} \\varnothing, & \\text{if x < y (base case)} \\\\ (x, y) \\cup B(x-1, y+1), & \\text{if x^2 + y^2 = m (satisfying case)} \\\\ B(x, y+1), & \\text{if x^2 + y^2 < m} \\\\ B(x-1, y), & \\text{if x^2 + y^2 > m} \\\\ \\end{cases}$$\n\nUsing this definition, you can calculate $B(\\sqrt m, 0)$ for each $m=0,1,\u2026,d_{max}$. Written as a generator or iterator, this algorithm will have constant space complexity.\n\n\u2022 Would you agree that the time complexeity is $O(r^3)$ here (with $r=\\sqrt{d_{\\text{max}}}$), since for each of the $O(r^2)$ possible values for $m$ you'd have $O(r)$ steps of evaluating $B$ till you reach the base case? \u2013\u00a0MvG Jul 6 '16 at 11:25", "date": "2019-09-20 22:40:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1740130/how-to-enumerate-2d-integer-coordinates-ordered-by-euclidean-distance", "openwebmath_score": 0.6815122365951538, "openwebmath_perplexity": 268.1049576982941, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765587105448, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8599066841148028}}
{"url": "https://math.stackexchange.com/questions/1134255/solve-for-y2yy-frace-tt2/1134495", "text": "Solve for $y''+2y'+y = \\frac{e^{-t}}{t^{2}}$\n\nI need some help in finding the solution to this second-order non-homogeneous DE.\n\nI know how to find the solution of the reduced equation $$y''+2y'+y=0.$$\n\nThe characteristic equation $r^{2}+2r+1=0$ yields real and repeated roots $r=-1$. So the complementary solution, $$y_{c}= c_{1}e^{-t} + c_{2}te^{-t}$$.\n\nNow what I am struggling with is making educated guesses about coming up with a particular solution. So any suggestions on that would be helpful.\n\nEDIT1: I haven't learned about the method of variation of parameters yet and wanted to do this with the method of undetermined coefficients. Please feel free to post the solution by using any of the techniques so I can learn from it.\n\nEDIT2: Our fundamental pair of solutions for the corresponding homogeneous is $y_{1}=e^{-t}$, and $y_{2}=t e^{-t}$.\n\nWe know that our particular solution will be of the form, $y_{p} = u_{1} y_{1} + u_{2} y_{2}$.\n\nThe determinant of the Wronskian, $W = e^{-2t}$, with $W_{1} = -t^{-1} e^{-2t}$, and $W_{2} = t^{-1} e^{-2t}$.\n\nSo $u'_{1} = \\frac{W_{1}}{W} = -\\frac{1}{t}$, and $u'_{2} = \\frac{W_{2}}{W} = \\frac{1}{t}$.\n\nSo $u_{1} = -\\int \\frac{1}{t} dt = -ln|t|$, and $u_{2} = \\int \\frac{1}{t} dt = ln|t|$.\n\nHence, $y_{p} = -ln|t| e^{-t} + ln|t|te^{-t}$.\n\nSo our general solution, $$y = y_{c} + y_{p} = e^{-t}(c_{1}+c_{2}t-ln|t|+ln|t|t).$$\n\nDoes it look okay?\n\n\u2022 Generally you only do particular solutions if it's a polynomial times a sine/cosine times an exponential, not a rational function. I'd use variation of paramaters \u2013\u00a0Alan Feb 5 '15 at 1:55\n\u2022 @Alan I haven't learned that technique yet. Is it the method of undetermined coeffecients? \u2013\u00a0OGC Feb 5 '15 at 1:57\n\u2022 What techniques have you learnt? \u2013\u00a0mattos Feb 5 '15 at 2:22\n\u2022 @Mattos Separation of variables technique and learned method of undetermined coefficients today. Seeing youtube videos on change of var technique. Is it possible to solve this problem by the method of undetermined coefficients technique? \u2013\u00a0OGC Feb 5 '15 at 2:26\n\u2022 @user36829 If you want, I'll post a solution using variation of parameters. \u2013\u00a0mattos Feb 5 '15 at 3:50\n\n$$y_h = c_1e^{-t} + c_2te^{-t}$$\n\nwith\n\n$$y_1 = e^{-t} \\ \\ \\ \\ \\ y_2 = te^{-t}$$\n\nTaking the Wronskian, you also found\n\n$$W = e^{-2t}$$\n\nThe particular solution is given by the formula\n\n$$y_p = -y_1 \\int \\frac{y_2 g(t)}{W} dt + y_2 \\int \\frac{y_1 g(t)}{W} dt$$\n\nwhere\n\n$$g(t) = \\frac{e^{-t}}{t^{2}} = t^{-2}e^{-t}$$\n\nHence\n\n\\begin{align} y_p &= -e^{-t} \\int \\frac{te^{-t} \\cdot t^{-2}e^{-t}}{e^{-2t}} dt + te^{-t} \\int \\frac{e^{-t} \\cdot t^{-2}{e^{-t}}}{e^{-2t}} dt \\\\ &= -e^{-t} \\int \\frac{1}{t} dt + te^{-t} \\int \\frac{1}{t^{2}} dt \\\\ &= -e^{-t} \\ln(t) - \\frac{te^{-t}}{t} \\\\ &= -e^{-t}(\\ln(t) + 1) \\\\ \\end{align}\n\nTherefore,\n\n\\begin{align} y &= y_h + y_p \\\\ &= e^{-t}(c_1 + c_2 t - \\ln(t) - 1) \\\\ \\end{align}\n\nYou can check by differentiation that this satisfies the ODE.\n\n\u2022 $-1\\cdot e^{-t}$ is an unnecessary part of the solution since it solves the homogeneous equation. I.e. the $-$ can be 'absorbed' into the $c_1$ coefficient. \u2013\u00a0jdods Mar 5 '15 at 1:58\n\u2022 @jdods Yeah and? \u2013\u00a0mattos Mar 5 '15 at 8:38\n\nIn this case of only one characteristic value, also a change of function can help. Substitute, according to the homogeneous solution, $y(t)=e^{-t}u(t)$ to get for $u$ $$u''(t)=t^{-2}\\implies u'(t)=-t^{-1}+C\\implies u(t)=-\\ln|t|+Ct+D$$ and thus $$y(t)=e^{-t}(-\\ln|t|+Ct+D)$$\n\nI'll use the formula I developed here $$Y_p=y_h\\int \\left\\{y_h^{-2} W_0 \\left(\\int y_h W_0^{-1} g \\ dt\\right)\\right\\} \\ dt$$ where the Wronskian (up to a constant multiple) is $$W_{0}=\\exp\\left(-{\\int p \\ dt}\\right)$$ and a solution to the homogeneous equation is $y_h=e^{-t}$ (we can use either).\n\nSo we first integrate $$W_{0}=\\exp\\left(-{\\int 2 \\ dt}\\right)=e^{-2t}$$ and then we'll find the particular solution by \\begin{aligned} Y_p&=e^{-t}\\int \\left\\{e^{2t} e^{-2t} \\left(\\int e^{-t} e^{2t} \\frac{e^{-t}}{t^2} \\ dt\\right)\\right\\} \\ dt \\\\ &=e^{-t}\\int \\left(\\int \\frac{1}{t^2} \\ dt\\right) \\ dt \\\\ &=e^{-t}\\int \\left(-\\frac{1}{t}\\right) \\ dt \\\\ &=-e^{-t}\\ln t. \\end{aligned}\n\nThis is a nice formula because you only need one solution to the homogeneous equation.", "date": "2020-02-18 10:03:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1134255/solve-for-y2yy-frace-tt2/1134495", "openwebmath_score": 0.9999784231185913, "openwebmath_perplexity": 618.5832328469735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409524, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8599066705287968}}
{"url": "https://math.stackexchange.com/questions/2376879/does-the-interval-x-x-10-x1-4-contain-a-sum-of-two-squares", "text": "# Does the interval $[x, x + 10 x^{1/4})$ contain a sum of two squares?\n\nIs it true that there exist an integer $N \\in \\mathbb N$ , such that $\\forall x > N$ , the interval $[x, x+10x^{1/4} )$ always contains a sum of two squares ?\n\nI know that $n \\in \\mathbb N$ is a sum of two squares iff it is of the form $n=q_1^{2a_1}...q_m^{2a_m}p_1^{k_1} ... p_r^{k_r}$ where $q_1,...,q_m,p_1,...,p_r$ are primes such that $q_i \\equiv 3( \\mod 4)$ and $a_1,...,a_m, k_1,...,k_r$ are non-negative integers ; where $m , r\\ge 0$ . So it is enough to prove that for large enough $x$ , the interval $[x , x+10x^{1/4})$ always contains such a positive integer . But I don't know how to prove this.\n\n\u2022 Reading about the Gauss circle problem would be a good starting point. \u2013\u00a0Daniel Fischer Jul 30 '17 at 19:01\n\u2022 For me to start: where did you get the problem/conjecture? Is it stated as something definitely true or as an open problem? It is reasonable, by the way, as the count of numbers up to large $z$ that are sums of two squares is about $$\\frac{Cz}{\\sqrt {\\log z}}$$ with (known) positive constant $C.$ \u2013\u00a0Will Jagy Jul 30 '17 at 19:19\n\u2022 @WillJagy : For the source of the problem ; it was asked as an open ended question in the class by our professor . Where do you get the asymptotic formula about the count ? Could you please give any reference ? Thanks \u2013\u00a0user Jul 31 '17 at 15:54\n\u2022 store.doverpublications.com/0486425398.html \u2013\u00a0Will Jagy Jul 31 '17 at 18:20\n\nYou might try to proceed greedily. Suppose we want to find a number that is \"near\" (but less than) $x$ that is a sum of two squares. Maybe one of these squares will be the largest square less or equal to $x$, which I'll denote by $a$. (So $a = \\lfloor \\sqrt x \\rfloor^2$), where $\\lfloor \\cdot \\rfloor$ is the \"floor\" function, or the \"greatest integer below function\"). In this case, the second square, which I'll call $b$, should be chosen to be the greatest square less than or equal to $x - a$. (So $b = \\lfloor \\sqrt{x - a} \\rfloor^2$).\n\nThis gives us a candidate sum of squares $a + b$, which is relatively close to $x$. Now we ask, how close does this get us? How large is $x - a - b$?\n\nThis really asks, how close is $x$ to the largest square up to $x$? That is, how close is $x$ to $a = \\lfloor \\sqrt x \\rfloor^2$?\n\nNote that $$x - \\lfloor \\sqrt x \\rfloor^2 = (\\sqrt x - \\lfloor \\sqrt x \\rfloor)(\\sqrt x + \\lfloor \\sqrt x \\rfloor) \\leq (1) \\cdot (2 \\sqrt x) \\leq 2 \\sqrt x.$$ So the largest square up to $x$ is no more than $2 \\sqrt x$ away from $x$.\n\nThus $x - a \\leq 2 \\sqrt x$. Iterating, we have that $$(x - a) - b \\leq 2 \\sqrt{x - a} \\leq 2 (\\sqrt {2 \\sqrt x}) = 2^{3/2} x^{1/4}.$$\n\nWe have shown that given an $x$, the nearest \"greedy\" sum of squares is at most $2^{3/2} x^{1/4}$ away. So we've shown that there is a sum of squares in intervals of the shape $(x - 2^{3/2} x^{1/4}, x]$, which clearly implies the original question.\n\nIn comments, Will Jagy noted that the density of numbers which are sums of two squares is of the shape $Cx/\\sqrt{\\log x}$, where $C$ is the Landau-Ramanujan constant. It is conjectured that the gaps between numbers which are sums of squares shouldn't deviate too far from what is expected from the density, and in particular is bounded by $x^\\epsilon$ for any $\\epsilon > 0$. In particular it is conjectured that there is a number which is a sum of two squares in the interval $[x, x + x^{\\epsilon})$ for any $\\epsilon > 0$, for $x$ sufficiently large.\nThis answer shows that the gap between numbers which are sums of squares is at most $2^{3/2}x^{1/4}$, which is a far cry from $x^{\\epsilon}$.\nIn general, there are not good uniform improvements over what's presented above. But there are remarkable on-average bounds due to Hooley, showing for instance that almost all (in the density sense) gaps between sums of squares is no more than $(\\log x) (\\log \\log x)$.", "date": "2019-06-26 03:38:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2376879/does-the-interval-x-x-10-x1-4-contain-a-sum-of-two-squares", "openwebmath_score": 0.91339111328125, "openwebmath_perplexity": 110.03396638452213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985718068592224, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8598885742162704}}
{"url": "https://math.stackexchange.com/questions/2111562/find-the-inverse-of-a-lower-triangular-matrix-of-ones", "text": "# Find the inverse of a lower triangular matrix of ones\n\nFind the inverse of the matrix $A=(a_{ij})\\in M_n$ where $$\\begin{cases} a_{ij}=1, &i\\geq j,\\\\ a_{ij}=0, &i<j. \\end{cases}$$\n\nThe only method for finding inverses that I was taught was by finding the adjugate matrix. So $A^{-1}=\\frac{1}{\\det A}\\operatorname{adj(A)}$\n\n$$A=\\begin{pmatrix} 1 & 0 &0 &\\ldots &0\\\\ 1 & 1 & 0 &\\ldots &0\\\\ 1 & 1 & 1 &\\ldots &0\\\\ \\vdots &\\vdots & \\vdots &\\ddots & \\vdots\\\\ 1 & 1 &1 & \\ldots &1 \\end{pmatrix}$$\n\nThis is a triangular matrix so $\\det A=1.$ To find the adjugate I first need to find the cofactor matirx$(C)$ of $A$.\n\n$$C=\\begin{pmatrix} 1&-1&0&0&\\ldots &0\\\\ 0 & 1& -1 &0&\\ldots & 0\\\\ 0 & 0 & 1 & -1&\\ldots &0\\\\ 0 & 0 & 0 &1 &\\ldots &0\\\\ \\vdots &\\vdots &\\vdots &\\vdots&\\ddots & \\vdots\\\\ 0 & 0 &0 &0 &\\ldots &1\\end{pmatrix}$$\n\n$$C^T=\\operatorname{adj}(A)=\\begin{pmatrix} 1 & 0 & 0 & 0&\\ldots &0\\\\ -1 & 1& 0 & 0 &\\ldots &0\\\\ 0&-1&1&0&\\ldots &0\\\\ 0& 0 &-1 &1 &\\ldots&0\\\\ \\vdots &\\vdots &\\vdots &\\vdots &\\ddots&\\vdots\\\\ 0&0&0&0&\\ldots &1\\end{pmatrix}=A^{-1}$$\n\nIs this correct? Also, can I leave it like that or should I somehow write it more formally?\n\n\u2022 try using elementary row operations. \u2013\u00a0vidyarthi Jan 24 '17 at 10:11\n\u2022 Looks all right. \u2013\u00a0StubbornAtom Jan 24 '17 at 10:25\n\u2022 Try multiplying your proposed $A^{-1}$ by $A$ to see whether you get the identity matrix. \u2013\u00a0Gerry Myerson Jan 24 '17 at 11:38\n\nYes, your inverse is correct. Let me show you an alternative approach of finding this inverse without the use of the adjugate matrix.\n\nDenote $A_n$ as the $n \\times n$ matrix that has elements $a_{ij} = 1$ if $i \\geq j$, and $0$ otherwise. Then, $A_n$ can be constructed by applying the elementary row operations $$R_2 + R_1, R_3 + R_2, \\ldots, R_n - R_{n-1}$$ to the identity matrix $I_n$. So, $$A_n = \\prod_{i=1}^{n-1} E_i,$$ where $E_i$ is the elementary matrix representing the row operation $R_{i+1} - R_i$. Since det$(A)$ is nonzero, it is invertible, hence $$B_n = \\left( A_n \\right)^{-1} = \\prod_{i = n-1}^{1} {E_i}^{-1},$$ where ${E_i}^{-1}$ is the row operation $R_{i+1} - R_i$. That is, $B_n$ can be constructed by applying the row operations $$R_n - R_{n-1}, R_{n-1} - R_{n-2},\\ldots,R_2 - R_1$$ to the idenity matrix $I_n$. The inverse of $A_n$ is therefore given by the matrix\n$$B_n = \\begin{pmatrix} \\phantom{-}1 & \\phantom{-}0 &\\phantom{-}0 &\\ldots &0\\\\ -1 & \\phantom{-}1 & \\phantom{-}0 &\\ldots &0\\\\ \\phantom{-}0 & -1 & \\phantom{-}1 &\\ldots &0\\\\ \\phantom{-}\\vdots &\\phantom{-}\\vdots & \\phantom{-}\\vdots &\\ddots & \\vdots\\\\ \\phantom{-}0 & \\phantom{-}0 & \\phantom{-}0 & \\cdots &1 \\end{pmatrix}.$$ In order to check that this is the correct inverse, you can use the identity $$A_nB_n = I_n.$$\n\n\u2022 How can we write this inverse like this $B_n=(b_{ij})= \\begin{cases} 1, &i= j,\\\\ -1, &i-j=1\\\\ 0, & ? \\end{cases}$ When is $(b_{ij})=0$? \u2013\u00a0lmc Jan 24 '17 at 13:31\n\u2022 $b_{ij}$ is $1$ if $i =j$, $-1$ if $i = j + 1$ and $0$ otherwise. \u2013\u00a0user394255 Jan 24 '17 at 13:33\n\nI would like to present a very simple solution by interpretation of these matrices as operators on $\\mathbb{R^n}$ (which will surprise nobody...). Triangular matrix $A$ acts as a discrete integration operator:\n\nFor any $x_1,x_2,x_3,\\cdots x_n$:\n\n$$\\tag{1}A (x_1,x_2,x_3,\\cdots x_n)^T=(s_1,s_2,s_3,\\cdots s_n)^T \\ \\ \\text{with} \\ \\ \\begin{cases}s_1&=&x_1&&&&\\\\s_2&=&x_1+x_2&&\\\\s_3&=&x_1+x_2+x_3\\\\...\\end{cases}$$\n\n(1) is equivalent to:\n\n$$\\tag{2}A^{-1} (s_1,s_2,s_3,\\cdots x_n)^T=(x_1,x_2,x_3,\\cdots x_n)^T \\ \\ \\text{with} \\ \\ \\begin{cases}x_1&=& \\ \\ s_1&&&&\\\\x_2&=&-s_1&+&s_2&&\\\\x_3&=&&&-s_2&+&s_3\\\\...\\end{cases}$$\n\nand it suffices now to \"collect the coefficients\" in the right order in order to constitute the inverse matrix.\n\n(Thus the inverse operation is - in a natural way - a discrete derivation operator).\n\nYou can also use the induction along the dimension $n$ (leading to a simple but not very constructive proof though).\n\nLet $A_n$ denote the matrix $A$ of dimension $n$ and $B_n$ its inverse. The statement is obviously true for $n=1$ and reduces to showing that $1=1$. Let the statement be true for $n-1$, that is, $A_{n-1}^{-1}=B_{n-1}$. If $1_{n-1}$ and $0_{n-1}$ the column vectors of ones and zeros, respectively, of dimension $n-1$, the matrix $A_n$ can be written in the block form $$A_n=\\begin{bmatrix}A_{n-1}&0_{n-1}\\\\1_{n-1}^T&1\\end{bmatrix}.$$ You can easily verify that $$A_n^{-1}=\\begin{bmatrix}A_{n-1}^{-1}&0_{n-1}\\\\-1_{n-1}^TA_{n-1}^{-1}&1\\end{bmatrix}$$ is the inverse of $A_n$. By the induction assumption, $A_{n-1}^{-1}=B_{n-1}$. It is easy to check that $1_{n-1}^TA_{n-1}^{-1}=1_{n-1}^TB_{n-1}=e_{n-1}^T:=[0,\\ldots,0,1]^T$ (sum of the rows of $B_{n-1}$). So $$A_n^{-1}=\\begin{bmatrix}B_{n-1}&0_{n-1}\\\\-1_{n-1}^TB_{n-1}&1\\end{bmatrix}=\\begin{bmatrix}B_{n-1}&0_{n-1}\\\\-e_{n-1}^T&1\\end{bmatrix}=B_n.$$\n\nJust to show a different approach.\nConsider the matrix $\\mathbf E$, having $1$ only on the first subdiagonal $$\\mathbf{E} = \\left\\| {\\,e_{\\,n,\\,m} = \\left\\{ {\\begin{array}{*{20}c} 1 & {n = m + 1} \\\\ 0 & {n \\ne m + 1} \\\\ \\end{array} } \\right.\\;} \\right\\| = \\left\\| {\\,\\left( \\begin{gathered} 0 \\\\ n - m - 1 \\\\ \\end{gathered} \\right)\\;} \\right\\|$$ Multiply it by $\\mathbf A$ , and it is easily seen that $$\\mathbf E \\, \\mathbf A=\\mathbf A-\\mathbf I \\quad \\Rightarrow \\quad \\left( \\mathbf I -\\mathbf E \\right)\\,\\mathbf A=\\mathbf I$$\n\nIn another way, consider that the powers of $\\mathbf E$ are readily found and have a simple formulation $$\\begin{gathered} \\mathbf{E}^{\\,\\mathbf{2}} = \\left\\| {\\,\\sum\\limits_{0\\, \\leqslant \\,k\\,\\left( { \\leqslant \\,n - 1} \\right)} {\\left( \\begin{gathered} 0 \\\\ n - k - 1 \\\\ \\end{gathered} \\right)\\left( \\begin{gathered} 0 \\\\ k - m - 1 \\\\ \\end{gathered} \\right)} \\;} \\right\\| = \\left\\| {\\,\\left( \\begin{gathered} 0 \\\\ n - m - 2 \\\\ \\end{gathered} \\right)\\;} \\right\\| = \\left\\| {\\,\\left\\{ {\\begin{array}{*{20}c} 1 & {n = m + 2} \\\\ 0 & {n \\ne m + 2} \\\\ \\end{array} } \\right.\\;} \\right\\| \\hfill \\\\ \\quad \\vdots \\hfill \\\\ \\mathbf{E}^{\\,\\mathbf{q}} = \\mathbf{E}^{\\,\\mathbf{q} - \\mathbf{1}} \\,\\mathbf{E} = \\left\\| {\\,\\sum\\limits_{0\\, \\leqslant \\,k\\,\\left( { \\leqslant \\,n - 1} \\right)} {\\left( \\begin{gathered} 0 \\\\ n - k - \\left( {q - 1} \\right) \\\\ \\end{gathered} \\right)\\left( \\begin{gathered} 0 \\\\ k - m - 1 \\\\ \\end{gathered} \\right)} \\;} \\right\\| = \\hfill \\\\ = \\left\\| {\\,\\left( \\begin{gathered} 0 \\\\ n - m - q \\\\ \\end{gathered} \\right)\\;} \\right\\|\\quad \\left| {\\;0 \\leqslant \\text{integer}\\;q} \\right. \\hfill \\\\ \\end{gathered}$$ Therefore $$\\mathbf{A} = \\sum\\limits_{0\\, \\leqslant \\,j} {\\mathbf{E}^{\\,\\mathbf{j}} } = \\frac{\\mathbf{I}} {{\\mathbf{I} - \\mathbf{E}}}$$ To connect to the answer of Jean Marie, note that $$\\left( {\\mathbf{I} - \\mathbf{E}} \\right)\\left\\| {\\,\\begin{array}{*{20}c} {x_{\\,0} } \\\\ {x_{\\,1} } \\\\ \\vdots \\\\ {x_{\\,n} } \\\\ \\end{array} \\;} \\right\\| = \\left\\| {\\,\\begin{array}{*{20}c} {x_{\\,0} ( - 0)} \\\\ {x_{\\,1} - x_{\\,0} } \\\\ \\vdots \\\\ {x_{\\,n} - x_{\\,n - 1} } \\\\ \\end{array} \\;} \\right\\| = \\left\\| {\\,\\begin{array}{*{20}c} {\\nabla x_{\\,0} \\;\\left| {x_{\\, - 1} = 0} \\right.} \\\\ {\\nabla x_{\\,1} } \\\\ \\vdots \\\\ {\\nabla x_{\\,n} } \\\\ \\end{array} \\;} \\right\\|$$", "date": "2020-02-29 11:07:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2111562/find-the-inverse-of-a-lower-triangular-matrix-of-ones", "openwebmath_score": 0.953346312046051, "openwebmath_perplexity": 117.00378915905104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180627184411, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8598885641853226}}
{"url": "https://math.stackexchange.com/questions/2626206/finding-the-10th-root-of-a-matrix/2626349", "text": "# Finding the 10th root of a matrix\n\nI want to find a $2 \\times 2$ matrix, named $A$ in this situation, such that:\n\n$$A^{10}=\\begin {bmatrix} 1 & 1 \\\\ 0 & 1 \\end {bmatrix}$$\n\nHow can I get started? I was thinking about filling $A$ with arbitrary values $a, b, c, d$ and then multiplying it by itself ten times, then setting those values equal to the given values but I quickly realized that would take too long. Is there a more efficient way?\n\nTake$$A=\\begin{bmatrix}1&x\\\\0&1\\end{bmatrix}.$$Now, compute $$A^2,A^3,\\ldots$$ You'll find quickly which $$x$$ you should choose.\n\n\u2022 Yes. This idea can be conceived by noting that the given matrix is upper triangular, then maybe an upper triangular solution $A$ is possible, and the eigenvalues of the original matrix are $1$ and $1$, then maybe the eigenvalues of $A$ can be taken to be also $1$ and $1$ (a particular tenth root of $1$ is $1$). So if you try it out, you will succeed. A more difficult problem is to find all solutions $A$ with entries in some field (such as $\\mathbb{Q}$ or $\\mathbb{C}$). \u2013\u00a0Jeppe Stig Nielsen Jan 29 '18 at 18:08\n\u2022 @JeppeStigNielsen Indeed, that's a more difficult problem. \u2013\u00a0Jos\u00e9 Carlos Santos Jan 29 '18 at 19:10\n\u2022 (+1) This is the approach that first came to mind. Now I need to approach this slightly differently. \u2013\u00a0robjohn Jan 30 '18 at 18:10\n\nYou can also consider $$A=\\begin {bmatrix} 1 & 0 \\\\ 0 & 1 \\end {bmatrix} + \\begin {bmatrix} 0 & a \\\\ 0 & 0 \\end {bmatrix} =I+N$$ and try to use binomial formula for expansion of powers of a binomial.\n\nNotice that $N^k=0$ for $k>1$.\n\n\u2022 Note that it is usually not possible to use the binomial formula for matrices, because the product is not commutative. In this case it is fine, because one of the matrices is the identity. \u2013\u00a0M.Herzkamp Jan 29 '18 at 11:45\n\u2022 @M.Herzkamp Yes, we can use the formula because matrices commute. Thank you for valuable remark.. \u2013\u00a0Widawensen Jan 29 '18 at 11:49\n\nThe answers given by Jos\u00e9 Carlos Santos and Widawensen are correct. There is a more general alternative method, even through it does not work for your particular example. But I thought I'd give it anyway in case it is useful to others.\n\nIf you want the $n^{\\text{th}}$ root of the matrix $M$, and $M$ is diagonalizable - i.e., there exists $Q$ such that $D = Q^{-1}MQ$ is a diagonal matrix, then $D^{1/n}$ is just the matrix whose diagonal elements are those of $D$ raised to the $1/n$ power. So you can set $$A = QD^{1/n}Q^{-1}$$ and find that $$A^n = \\left(QD^{1/n}Q^{-1}\\right)^n = Q\\left(D^{1/n}\\right)^nQ^{-1} = QDQ^{-1} = M$$\n\nTo diagonalize $M$, you find it's eigenvalues and their corresponding eigenvectors. If the eigenvectors span the entire space, then $M$ is diagonalizable, the columns of $Q$ are eigenvectors of $M$, and the diagonal elements of $D$ are the eigenvalues.\n\nAlas, it is not applicable here: $$\\begin{bmatrix}1 & 1\\\\0&1\\end{bmatrix}$$ is not diagonalizable.\n\n\u2022 If you can figure out how exponentials of Jordan blocks work, you can use Jordan normal form instead of just diagonalization. \u2013\u00a0Kyle Miller Jan 30 '18 at 5:52\n\nOk this may seem a bit overkill for this particular question, but please bear with me as it could help others.\n\nYou can use the Newton-Rhapson method, trying to solve the equation\n\n$$f(x) = x^{10}-d=0$$ where $$f'(x) = 10x^9$$ and $d$ is that matrix of yours and the iteration:\n\n$${\\bf X_{n+1}} ={\\bf X_n}- f({\\bf X_n})f'({\\bf X_n})^{-1}$$ As long as you set initial $\\bf X_0$ to nothing too crazy, it will work. In fact, this is the simplest example of an discrete fractional integration solution, if you look at larger matrices $d$ filled with 1 on and below the diagonal and 0 otherwise (ok, transpose of the same matrix but same idea).\n\nThe fractional integral operators in fractional calculus found if guessing at the $\\bf X_0 = \\bf I$ matrix (described as linear convolutional filters, corresponding to a row in the solution matrix):\n\nThe two we recognize is the one in the middle (constant, corresponding to normal integral) and the one at the end (linear, corresponding to double integration).\n\n\u2022 Is that supposed to be $f'(x)=10x^9$? \u2013\u00a0ASKASK Jan 30 '18 at 11:52\n\u2022 @ASKASK yep must have gotten an elf in my computer who ate it away ;) \u2013\u00a0mathreadler Jan 30 '18 at 11:53\n\nSuch a matrix $A$ will commute with $\\pmatrix{1&1\\\\0&1}$, and so be of the form $\\pmatrix{a&b\\\\0&a}$. This means that $a^{10}=1$ and so $A=a\\pmatrix{1&c\\\\0&1}$ for some $c$. We immediately get $c=1/10$ (as long as $10$ is invertible in your field).\n\nHint: Another approach is to note that $$\\exp\\left(\\begin{bmatrix}0&x\\\\0&0\\end{bmatrix}\\right)=\\begin{bmatrix}1&x\\\\0&1\\end{bmatrix}$$", "date": "2021-02-27 05:47:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2626206/finding-the-10th-root-of-a-matrix/2626349", "openwebmath_score": 0.9281057119369507, "openwebmath_perplexity": 182.81968527601109, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9744347905312772, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8598684222462808}}
{"url": "https://math.stackexchange.com/questions/3535617/subsets-of-1-2-dots-n-with-no-consecutive-integers", "text": "# Subsets of $\\{1,2 \\dots n\\}$ with no consecutive integers\n\nHow many subsets with cardinality k of $$\\{1, 2, \\dots n\\}$$ contain no consecutive integers? I know that there are $$F_{n+2}$$ subsets of $$\\{1, 2, \\dots n\\}$$ with no consecutive integers, but I do not know how to go about finding the number for a given $$k$$.\n\nThis is equivalent to chooosing a sequence of $$k$$ ones and $$n-k$$ zeroes with no adjacent ones. An example with $$n=8$$ and $$k=3$$ is $$00101001, \\text{ corresponding to the set }\\{3,5,8\\}$$ To choose such a sequence, start with a string of $$n-k$$ zeroes, with $$n-k-1$$ spaces between the zeroes, plus two extra spaces before and after, for $$n-k+1$$ spaces total: $$\\;\\_\\; 0\\;\\_\\;0\\;\\_\\;0\\;\\_\\;0\\;\\_\\;0\\;\\_\\qquad,\\text{ with 8-3+1=6 gaps}.$$ Each of the $$k$$ $$1$$'s goes into exactly one gap. We need to choose $$k$$ of these gaps to put a $$1$$ in. This can be done in $$\\binom{n-k+1}{k}\\text{ ways.}$$\n\nWithout loss of generality, let your $$k$$ elements from a selected subset be $$x_1\n\nGiven such a $$k$$-tuple $$(x_1,x_2,\\dots,x_k)$$, construct the related $$(k+1)$$-tuple $$(x_1-1, x_2-x_1, x_3-x_2,\\dots, x_k-x_{k-1}, n-x_k)$$ describing the distance between each number and/or the boundaries in the case of the first and last numbers.\n\nRenaming those values in the $$(k+1)$$-tuple $$(y_1,y_2,\\dots,y_{k+1})$$ we can recognize that $$y_1+y_2+\\dots+y_{k+1} = n-1$$ as the sum telescopes.\n\nNow, consider the related problem of finding the number of integer solutions to the system:\n\n$$\\begin{cases}y_1+y_2+\\dots+y_{k+1} = n-1\\\\ y_1\\geq 0\\\\ y_{k+1}\\geq 0\\\\ y_i\\geq 2~~~\\text{for all other }i\\end{cases}$$\n\nThe inequalities here coming from that the elements may not be consecutive. This should now be in a known problem format for you or can be slightly modified further with another change of variable to be in a known format and the problem can be completed using stars-and-bars.\n\nWe select the first value of the set:\n\n$$\\frac{z}{1-z}$$\n\nfollowed by $$k-1$$ differences that are at least two:\n\n$$\\frac{z}{1-z} \\left(\\frac{z^2}{1-z}\\right)^{k-1}$$\n\nand we conclude by collecting the count for all subsets with maximum element $$\\le n:$$\n\n$$[z^n] \\frac{1}{1-z} \\times \\frac{z}{1-z} \\left(\\frac{z^2}{1-z}\\right)^{k-1}.$$\n\nThis is\n\n$$[z^n] \\frac{z^{2k-1}}{(1-z)^{k+1}} = [z^{n+1-2k}] \\frac{1}{(1-z)^{k+1}} = {n+1-2k+k\\choose k}$$\n\nor equivalently\n\n$$\\bbox[5px,border:2px solid #00A000]{ {n+1-k\\choose k}.}$$\n\nWe get for the total\n\n$$\\sum_{k=0}^{\\lfloor (n+1)/2 \\rfloor} {n+1-k\\choose k} = \\sum_{k=0}^{\\lfloor (n+1)/2 \\rfloor} [z^{n+1-2k}] \\frac{1}{(1-z)^{k+1}} \\\\ = [z^{n+1}] \\frac{1}{1-z} \\sum_{k=0}^{\\lfloor (n+1)/2 \\rfloor} z^{2k} \\frac{1}{(1-z)^{k}}.$$\n\nHere the coefficient extractor enforces the range and we may continue with\n\n$$[z^{n+1}] \\frac{1}{1-z} \\sum_{k\\ge 0} z^{2k} \\frac{1}{(1-z)^{k}} \\\\ = [z^{n+1}] \\frac{1}{1-z} \\frac{1}{1-z^2/(1-z)} \\\\ = [z^{n+1}] \\frac{1}{1-z-z^2} = [z^{n+2}] \\frac{z}{1-z-z^2} = F_{n+2}.$$\n\nThe above construction works for $$k\\ge 1.$$ For $$k=0$$ we get the empty set, for a total count of one. Note however that $${n+1\\choose 0} = 1$$ so the formula holds there as well.\n\nHint: Choose $$k$$ pairs of consecutive numbers from $$\\{1, 2, \\ldots, n, n+1\\}$$, then choose the lowest number in each pair.", "date": "2023-03-27 10:35:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3535617/subsets-of-1-2-dots-n-with-no-consecutive-integers", "openwebmath_score": 0.9997049570083618, "openwebmath_perplexity": 1933.1973560460447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.993807011126102, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8598657680946328}}
{"url": "https://math.stackexchange.com/questions/391537/computing-int-0-pi-over2-fracdx1-sin2x", "text": "# Computing $\\int_0^{\\pi\\over2} \\frac{dx}{1+\\sin^2(x)}$?\n\nHow would you compute$$\\int_0^{\\pi\\over2} \\frac{dx}{1+\\sin^2(x)}\\, \\, ?$$\n\n\u2022 by residues, of course :) \u2013\u00a0Start wearing purple May 14 '13 at 15:50\n\u2022 how could one use residues? I can't seem to see where this function (if it were complex) has a pole! \u2013\u00a0user53076 May 14 '13 at 16:14\n\u2022 Well you have to 1) use parity to extend the integration domain to $[-\\pi/2,\\pi/2]$, then 2) express $\\sin^2x$ in terms of $\\cos2x$, and then 3) pass to the complex variable $z=e^{2ix}$ (contour of integration then becomes the unit circle $|z|=1$). \u2013\u00a0Start wearing purple May 14 '13 at 16:20\n\nPut $z=e^{i x}$; then the integral is equal to\n\n$$i \\oint_{|z|=1} dz \\frac{z}{z^4-6 z^2+1}$$\n\nThere are four poles at\n\n$$z = \\pm \\sqrt{2} \\pm 1$$\n\nonly two of which are within the unit circle ($z = \\pm (\\sqrt{2}-1)$). The residues from each of these poles are equal and are each\n\n$$\\frac{i}{4 (\\sqrt{2}-1)^2-12} = \\frac{-i}{8 \\sqrt{2}}$$\n\nThe sum of the residues is double that residue. The integral is then $i 2 \\pi$ times that sum; I get\n\n$$\\int_0^{\\pi/2} \\frac{dx}{1+\\sin^2{x}} = \\frac{\\pi}{2 \\sqrt{2}}$$\n\n\u2022 He has only a quarter of unit circle initially. \u2013\u00a0Start wearing purple May 14 '13 at 16:53\n\u2022 @O.L.: That is built into my analysis. The factor of $1/4$ cancels an equal factor in the denominator. \u2013\u00a0Ron Gordon May 14 '13 at 16:57\n\u2022 WHOA! How did you get the first bit? \u2013\u00a0user53076 May 14 '13 at 17:28\n\u2022 $z=e^{i x}$, $dx = -i dz/z$. Then use $\\sin{x} = (z-z^{-1})/(2 i)$. Do the algebra. Note that you only have $1/4$ of a full circle in your integral, so you must multiply by $1/4$. \u2013\u00a0Ron Gordon May 14 '13 at 17:31\n\u2022 @RonGordon I get a $-1$ where you have a $6$ in the very first line. Are we totally sure that the $6$ is correct? \u2013\u00a0The Count Jan 5 '17 at 2:47\n\nHINT:\n\n$$\\int_0^{\\pi\\over2} \\frac{dx}{1+\\sin^2(x)}= \\int_0^{\\pi\\over2} \\frac{\\csc^2xdx}{\\csc^2x+1}=\\int_0^{\\pi\\over2} \\frac{\\csc^2xdx}{\\cot^2x+2}$$\n\nPut $\\cot x=u$\n\n$$\\int_0^{\\frac \\pi 2} \\frac{1}{1 + \\sin^2x}dx = \\int_0^{\\frac \\pi 2} \\frac{2}{3 - \\cos 2x} dx = \\int_0^\\pi \\frac{d\\theta}{3 - \\cos \\theta } = \\frac 12\\int_0^{2\\pi}\\frac{d\\theta}{3 - \\cos \\theta}$$\n\nTo evaluate $\\displaystyle \\int_0^{2\\pi}\\frac{1}{3 - \\cos \\theta}d\\theta$ let $\\displaystyle z = e^{i\\theta} \\implies \\cos \\theta = \\frac{z^2 + 1}{2z}, d\\theta = \\frac{1}{iz}dz$\n\n$$\\int_0^{2\\pi}\\frac{1}{3 - \\cos \\theta}d\\theta = \\frac 2{i}\\oint_{|z| = 1} \\frac{1}{-z^2 + 6z - 1}dz$$\n\nThe poles are $3 - 2 \\sqrt 2$ and $3 + 2 \\sqrt 2$, and since $3 + 2 \\sqrt 2 > 1$,\n\n$$\\frac 2{i}\\oint_{|z| = 1} \\frac{1 }{-z^2 + 6z - 1}dz = 4\\pi \\text{Res}\\left[ \\frac{1}{-z^2 + 6z - 1} , 3 - 2 \\sqrt 2\\right] = \\frac{\\pi}{\\sqrt 2}$$\n\nSo $\\displaystyle \\int_0^{\\frac \\pi 2} \\frac{1}{1 + \\sin^2x}dx = \\frac{\\pi }{2 \\sqrt 2}$\n\n\\begin{align} \\int_0^{\\pi/2} \\dfrac{dx}{1+\\sin^2(x)} & = \\int_0^{\\pi/2}\\sum_{k=0}^{\\infty}(-1)^k \\sin^{2k}(x) dx = \\sum_{k=0}^{\\infty}(-1)^k \\int_0^{\\pi/2}\\sin^{2k}(x) dx\\\\ & = \\sum_{k=0}^{\\infty} \\dfrac{(-1)^k}{4^k} \\dbinom{2k}k\\dfrac{\\pi}2 = \\dfrac1{\\sqrt{1-4\\times \\left(\\dfrac{-1}4\\right)}} \\dfrac{\\pi}2 = \\dfrac{\\pi}{2\\sqrt2} \\end{align} where we used\n\n$\\dfrac1{1+r} = \\displaystyle \\sum_{k=0}^{\\infty}(-r)^k$; $\\displaystyle \\int_0^{\\pi/2} \\sin^{2k}(x) dx = \\dbinom{2k}k \\dfrac{\\pi}{2^{2k+1}}$; $\\dfrac1{\\sqrt{1-4x}} = \\displaystyle\\sum_{k=0}^{\\infty} \\dbinom{2k}k x^k \\,\\, \\forall x \\in \\left[-\\dfrac14,\\dfrac14 \\right)$", "date": "2019-11-19 15:43:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/391537/computing-int-0-pi-over2-fracdx1-sin2x", "openwebmath_score": 0.9042766094207764, "openwebmath_perplexity": 1047.031763030432, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718459575753, "lm_q2_score": 0.8688267660487573, "lm_q1q2_score": 0.8598533893728241}}
{"url": "https://math.stackexchange.com/questions/2677413/there-are-3-candidates-for-class-monitor-and-one-is-to-be-elected-by-the-votes-o", "text": "# There are 3 candidates for class monitor and one is to be elected by the votes of 5 voters. In how many ways the votes can be given?\n\nMy approach:\n\nI tried to find the no. of ways we can divide the five votes into three sections so that we can form 5 votes. Then permuting the total no. of ways, I would get the total no. of ways of voting. So, the total no. of ways I get, to form $5$ out of $0,1,2,3,4,5$\n\n$0+0+5$\n\n$0+1+4$\n\n$0+2+3$\n\n$1+2+2$\n\n$1+3+1$\n\nThere are no unique ways left to cast the votes according to my approach. So, we see that there are $5$ different ways for the $3$ candidates to get the vote. So total no. of ways they can get the vote $= 5 . 3!$ But the book says the answer is $243$. Their approach is $3^5=243$. I accept that. But can you tell me exactly why my approach wrong? What did I miss?\n\n1. You can only multiply by $3!$ when there are three different vote totals.\n2. You have not accounted for which voter voted for which candidate.\n\nOne candidate receives all five votes: There are $3$ cases, one for each candidate who could receive all five votes.\n\nOne candidate receives four votes and another candidate receives one vote: There are $3$ ways to choose which candidate receives four votes, $\\binom{5}{4} = 5$ ways for four of the five voters to select that candidate, and $2$ ways to choose which candidate receives one vote from the remaining voter. Hence, there are $3 \\cdot 5 \\cdot 2 = 30$ such cases.\n\nOne candidate receives three votes and another candidate receives two votes: There are $3$ ways to choose which candidate receives three votes, $\\binom{5}{3} = 10$ ways for three of the five voters to choose that candidate, and two ways to choose which of the remaining two candidates receives the remaining two votes. Hence, there are $3 \\cdot 10 \\cdot 2 = 60$ such cases.\n\nOne candidate receives three votes and each of the other two candidates receives one vote: There are $3$ ways to choose which candidate receives three votes, $\\binom{5}{3} = 10$ ways for three of the five voters to select that candidate, and $2! = 2$ ways in which the remaining voters can split their votes among the remaining two candidates. Hence, there are $3 \\cdot 10 \\cdot 2 = 60$ such cases.\n\nTwo candidates each receive two votes and the other candidate receives one vote: There are $\\binom{3}{2} = 3$ ways to choose which two candidates receive two votes, $\\binom{5}{2} = 10$ ways for two of the voters to select the candidate among those two whose name appears first in an alphabetical list, $\\binom{3}{2} = 3$ ways for two of the other three voters to select the other candidate who receives two votes, and one way for the remaining voter to select the remaining candidate. Hence, there are $3 \\cdot 10 \\cdot 3 = 90$ such cases.\n\nTotal: Since the five cases are mutually exclusive and exhaustive, the number of ways the votes can be cast is $3 + 30 + 60 + 60 + 90 = 243$.\n\nIt is much more efficient to observe that each of the five voters has three choices, so there are $3^5 = 243$ ways for them to cast their votes.\n\nYou missed that case when the first gets all 5 votes, or second gets all the 5 votes and so on. The different cases will be applied to all the 3 candidates.\n\nIt can properly be seen by linking them into sets.\n\nSet A having 5 voters and Set B having 3 candidates.\n\nThe ways to link Set A to Set B will be = $3^5$ = 243\n\n\u2022 I don't think so. I multiplied every possibilities by $3!$ . So because of that I think, get all the cases when first get all the votes, then second and so on. \u2013\u00a0abu obaida Mar 5 '18 at 7:42", "date": "2021-01-22 21:45:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2677413/there-are-3-candidates-for-class-monitor-and-one-is-to-be-elected-by-the-votes-o", "openwebmath_score": 0.7424962520599365, "openwebmath_perplexity": 316.9633685319594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517443658749, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8598510279678497}}
{"url": "https://harborspringsiga.com/d3pfx/portfolio-diversification-formula-8b7e43", "text": "# portfolio diversification formula\n\nBy . Stocks aren\u2019t always kind and because of this diversification was formed. It was introduced by Harry Markowitz in 1952 using a formula that the economist developed offering investors a way to structure an investment portfolio to maximize returns at a given level of risk. For Portfolio 1, the volatility associated with maximum diversification is computed as 9.4% (see Figure 2). Portfolio diversification is a way to safeguard an investment portfolio. As the number of stocks in the portfolio increases the exposure to risk decreases. ... (2008) find that maximum portfolio diversification benefits can be achieved by investing in 15 securities in Malaysian stock market. Modern Portfolio Theory. Effective diversification not only reduces some of your financial risk, but it also offers you the opportunity to maximize your returns. While the benefits of diversification are clear, investors must determine the level of diversification that best suits them. 2. Figure 2: How portfolio diversification reduces risk. Modern Portfolio Theory is likely the most famous modern-day portfolio diversification strategy for individual investors. The portfolio standard deviation is the square root of the variance, or: Portfolio standard deviation, 2-security portfolio = w + w + 2w w cov 12 12 22 22 1 2 1,2\u03c3\u03c3 . For this small portfolio, it is easy to calculate the diversification effects of various loan sizes using a spreadsheet. Portfolio theory is a subcategory of the capital market theory that deals with the behavior of investors in capital markets. Its primary goal is to limit the impact of volatility on a portfolio. From there on the effect of adding more stocks to your portfolio has less effect. Suppose that the return on a portfolio during eight consecutive years was One dollar invested in the portfolio at the beginning of year 1 became $1.15 by the end of the year. The broader the diversification of the portfolio, the more stable the returns will be. Step 3: Next, determine the correlation among the assets, and it basically captures the movement of each asset relative to another asset. The efficient portfolio consists of investments that provide the greatest return for the risk, or \u2014 alternatively stated \u2014 the least risk for a given return. Diversification and modern portfolio theory. The correlation is denoted by \u03c1. Image via Wikipedia. That is to say, a diversified portfolio is a strong portfolio. He uses a simple formula: Diversification = 1 / Confidence. Yes! Unlike what business schools try to teach you, portfolio diversification is not some formula. The opposite of a concentrated portfolio is an index fund. Investment diversification is a crucial aspect of financial planning since it\u2019s the primary tool for lowering your risk and maximizing your return. More generally, the formula is: NNN 22 ii ijijij i=1 i 1j 1 j i Portfolio standard deviation = w + w w r ==\u2260 \u2211\u2211\u2211\u03c3\u03c3\u03c3 It is a measure of total risk of the portfolio and an important input in calculation of Sharpe ratio. The Secret Formula To Portfolio Diversification. However, portfolio diversification cannot eliminate all risk from the portfolio. The ultimate goal of diversification is to reduce the volatility VIX The Chicago Board Options Exchange (CBOE) created the VIX (CBOE Volatility Index) to measure the 30-day expected volatility of the US stock market, sometimes called the \"fear index\". We can calculate the risk of two linear positions using the following formula: Large insurance, hedge funds, and asset managers base their investment decisions on modern portfolio theory. Figure 2 shows that adding a$5,000 loan In general, you don\u2019t want to have too small positions on your portfolio. This paper, Equity Portfolio Diversification by W. Goetzmann and A. Kumar, uses the following diversification measures to measure the diversification of retail investors: Normalized portfolio variance: $$NV = \\frac{\\sigma_p ^2}{\\bar{\\sigma} ^2}$$ Sum of Squared Portfolio Weights (SSPW). Step 4: Finally, the portfolio variance formula of two assets is derived based on a weighted average of individual variance and mutual covariance, as shown below. diversification is apparent in the increase of the diversi-fication quotient from two to three. Uncategorized. portfolio risk and return formula. What\u2019s the point in having a 0.2% allocation? What is Diversification? During the 2008\u20132009 bear market, many different types of investments lost value at the same time, but diversification still helped contain overall portfolio losses. Markowitz Portfolio Theory deals with the risk and return of portfolio of investments. The portfolio now has the same diversification as a portfolio of three loans of equal size. If assets have less than a +1.0 correlation, then some of the random fluctuation around the expected trend rates of return will cancel each other out and lower the portfolio\u2019s standard deviation (risk). To assemble an efficient portfolio, one needs to know how to calculate the returns and risks of a portfolio, and how to minimize risks through diversification. Thus, total risk can be divided into two types of risk: (1) Unique risk and (2) Market risk. If you allocate too much to bonds over your career, you might not be able to build enough capital to retire at all. I think I read that in The Magic Formula book of Joel Greenblatt. The proper asset allocation of stocks and bonds by age is important to achieve financial freedom. If you allocate too much to stocks the year before you want to retire and the stock market collapses, then you're screwed. The diversification shows the difference between net portfolio risk and gross risk assuming perfect correlation (i.e., net portfolio risk minus gross risk). How diversification can help reduce the impact of market volatility. the maximum diversification portfolio risk. Portfolio standard deviation is the standard deviation of a portfolio of investments. Portfolio Returns an investor focused on growth but looking for greater diversification someone with a portfolio that primarily includes a balance of investments in bonds and equities At year 10, 0.5% of \u2026 It helps investors minimize the bumps and reduce the risks of their investment journey. Wait, didn\u2019t we already diversify between the Security Bucket and the Risk Bucket in principle #2? According to Equation (4), individual assets are only included in the long-only maximum diversification portfolio if their correlation to the common risk factor is lower than the threshold correlation. Illustrate diversification benefits in a portfolio of three investments, a stock A, a bond B, and a real estate asset C. The assets weights are 20%, 35% and 45% respectively, their standard deviations are 2.3%, 3.5% and 4%, the correlation coefficient between A and B is 0.6, \u2026 Now it\u2019s time to take it one step further. The diversification benefit is possible when return correlations between portfolio assets is less than perfect positive correlation (<+1.0). Portfolio Diversification is a foundational concept in investing. To better understand this concept, look at the charts below, which depict hypothetical portfolios with different asset allocations. enero 12, 2021. The investors knew that diversification is best for making investments but Markowitz formally built the quantified concept of diversification. Before Markowitz portfolio theory, risk & return concepts are handled by the investors loosely. The overall volatility of Portfolio 1 was previously computed as 21.7%. One of the most basic principles of finance is that diversification leads to a reduction in risk unless there is a perfect correlation between the returns on the portfolio investments. Portfolio diversification is more than just a tough phrase to say, it is the most important part of long term stock trading. However, in its implementation, many investors make catastrophic mistakes with too much concentration and others settle for average performance because of over diversification. He pointed out the way in which the risk of portfolio to an \u2026 If your portfolio is primarily U.S. large-cap stocks, adding U.S. small-cap stocks won't give you nearly the diversification benefit that adding bonds or international stocks will. You would be better with an index at that point. As in Equation (3) for minimum-variance portfolio weights, high idiosyncratic risk in the Diversification is an art. formula remains as below. To summarize the above, Markowitz theory of portfolio diversification attaches importance to: (a) Standard deviation, i.e., when portfolio = 0 risk is minimum, (b) Covariance \u2014 to show interactive risk, (c) Coefficient correlation, i.e., when x = \u2013 1 the risk of investment should \u2026 Portfolio diversification theory provides investors with a set of rules and principles that shed light on the best possible methods for mitigating risk and hedging against market volatility. ... Markowitz created a formula that allows an investor to mathematically trade off risk tolerance and reward expectations, resulting in the ideal portfolio. As you see from the graph the biggest effect of diversification happens up until 10 stocks or so. The primary goal of diversification isn't to maximize returns. It can be a rather basic and easy to understand concept. If this $1.15 were reinvested in the same portfolio, it would have become$1.09 = (1.15) \u2027 (.95) by the end of the second year. Diversification is a technique of allocating portfolio resources or capital to a mix of different investments. And ( 2 ) and return of portfolio to an \u2026 formula as. Your career, you might not portfolio diversification formula able to build enough capital to a mix of different investments term trading! A mix of different investments most important part of long term stock trading below, which depict hypothetical portfolios different! 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Phrase to say, a diversified portfolio is an index fund rather basic and easy to understand.. One step further +1.0 ) portfolios with different asset allocations benefits can achieved! Portfolio, it is easy to understand concept the portfolio, it easy. Was formed % allocation individual investors the opposite of a concentrated portfolio an... Portfolio, it is the most important part of long term stock trading to better understand concept... In calculation of Sharpe ratio diversification happens up until 10 stocks or so having a 0.2 %?.", "date": "2021-04-12 06:34:26", "meta": {"domain": "harborspringsiga.com", "url": "https://harborspringsiga.com/d3pfx/portfolio-diversification-formula-8b7e43", "openwebmath_score": 0.4100356996059418, "openwebmath_perplexity": 2213.6900029084777, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9916842227513689, "lm_q2_score": 0.8670357632379241, "lm_q1q2_score": 0.8598256869642407}}
{"url": "https://insignia3d.com/tight-lines-axdny/fundamental-theorem-of-calculus-definite-integral-16091b", "text": "Share\n\n# fundamental theorem of calculus definite integral\n\n## fundamental theorem of calculus definite integral\n\nThe Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. This means . Suppose that f(x) is continuous on an interval [a, b]. The Fundamental Theorem of Calculus Three Different Quantities The Whole as Sum of Partial Changes The Indefinite Integral as Antiderivative The FTC and the Chain Rule The Indefinite Integral and the Net Change Indefinite Integrals and Anti-derivatives A Table of Common Anti-derivatives The Net Change Theorem The NCT and Public Policy Substitution o Forget the +c. In order to take the derivative of a function (with or without the FTC), we've got to have that function in the first place. Fundamental Theorem of Calculus. \u2022 Definite integral: o The number that represents the area under the curve f(x) between x=a and x=b o a and b are called the limits of integration. The average value of the function f on the interval [a,b] is the integral of the function on that interval divided by the length of the interval.Since we know how to find the exact values of a lot of definite integrals now, we can also find a lot of exact average values. 29. The Fundamental Theorem of Calculus. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Yes, you're right \u2014 this is a bit of a problem. The definite integral of from to , denoted , is defined to be the signed area between and the axis, from to . The integral, along with the derivative, are the two fundamental building blocks of calculus. So, let's try that way first and then we'll do it a second way as well. To solve the integral, we first have to know that the fundamental theorem of calculus is . So we know a lot about differentiation, and the basics about what integration is, so what do these two operations have to do with one another? - The integral has a variable as an upper limit rather than a constant. Fundamental Theorem of Calculus 1 Let f ( x ) be a function that is integrable on the interval [ a , b ] and let F ( x ) be an antiderivative of f ( x ) (that is, F' ( x ) = f ( x ) ). Therefore, The First Fundamental Theorem of Calculus Might Seem Like Magic Definite Integral (30) Fundamental Theorem of Calculus (6) Improper Integral (28) Indefinite Integral (31) Riemann Sum (4) Multivariable Functions (133) Calculating Multivariable Limit (4) Continuity of Multivariable Functions (3) Domain of Multivariable Function (16) Extremum (22) Global Extremum (10) Local Extremum (13) Homogeneous Functions (6) Areas between Curves. \\$1 per month helps!! This states that if is continuous on and is its continuous indefinite integral, then . One is, that I can forget for the minute that it's a definite integral and compute the antiderivative and then use the fundamental theorem of calculus. To find the anti-derivative, we have to know that in the integral, is the same as . It is the fundamental theorem of calculus that connects differentiation with the definite integral: if f is a continuous real-valued function defined on a closed interval [a, b], then once an antiderivative F of f is known, the definite integral of f over that interval is given by You can see some background on the Fundamental Theorem of Calculus in the Area Under a Curve and Definite Integral sections. Solution. Indefinite Integrals. Problem Session 7. The Substitution Rule. Mathematics C Standard Term 2 Lecture 4 Definite Integrals, Areas Under Curves, Fundamental Theorem of Calculus Syllabus Reference: 8-2 A definite integral is a real number found by substituting given values of the variable into the primitive function. Show Instructions. About; 25. Areas between Curves. Thanks to all of you who support me on Patreon. It also gives us an efficient way to evaluate definite integrals. On the other hand, since when .. This will show us how we compute definite integrals without using (the often very unpleasant) definition. Definite integrals give a result (a number that represents the area) as opposed to indefinite integrals, which are The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. The anti-derivative of the function is , so we must evaluate . Calculus is the mathematical study of continuous change. Fundamental theorem of calculus. The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. The calculator will evaluate the definite (i.e. - The variable is an upper limit (not a lower limit) and the lower limit is still a constant. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. The Fundamental Theorem of Calculus establishes the relationship between indefinite and definite integrals and introduces a technique for evaluating definite integrals without using Riemann sums, which is very important because evaluating the limit of Riemann sum can be extremely time\u2010consuming and difficult. How Part 1 of the Fundamental Theorem of Calculus defines the integral. Here we present two related fundamental theorems involving differentiation and integration, followed by an applet where you can explore what it means. See . The fundamental theorem of calculus has two separate parts. 27. There are several key things to notice in this integral. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. In this wiki, we will see how the two main branches of calculus, differential and integral calculus, are related to each other. Indefinite Integrals. with bounds) integral, including improper, with steps shown. The Fundamental Theorem of Calculus. The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function.. The total area under a \u2026 In fact, and . It has two main branches \u2013 differential calculus (concerning rates of change and slopes of curves) and integral calculus (concerning the accumulation of quantities and the areas under and between curves).The Fundamental theorem of calculus links these two branches. Free practice questions for AP Calculus BC - Fundamental Theorem of Calculus with Definite Integrals. The second part states that the indefinite integral of a function can be used to calculate any definite integral, \\int_a^b f(x)\\,dx = F(b) - \u2026 29. First, it states that the indefinite integral of a function can be reversed by differentiation, \\int_a^b f(t)\\, dt = F(b)-F(a). Everything! By using the Fundamental theorem of Calculus, 26. Assuming the symbols , , and represent positive areas, the definite integral equals .Since , the value of the definite integral is negative.In this case, its value is .. So, by the fundamental theorem of calculus this is equal to ln of the absolute value of cosine x for x between pi over 6 and pi over 3. Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 3 and 6. The Definite Integral. So, the fundamental theorem of calculus says that the value of this definite integral, in order to compute it, we just take the difference of that antiderivative at pi over 3 and at pi over 6. The Substitution Rule. Lesson 2: The Definite Integral & the Fundamental Theorem(s) of Calculus. Both types of integrals are tied together by the fundamental theorem of calculus. Fundamental Theorem of Calculus (Relationship between definite & indefinite integrals) If and f is continuous, then F is differentiable and Use geometry and the properties of definite integrals to evaluate them. Explanation: . The definite integral gives a signed area\u2019. Includes full solutions and score reporting. Given. 26. Lesson 16.3: The Fundamental Theorem of Calculus : A restatement of the Fundamental Theorem of Calculus is presented in this lesson along with a corollary that is used to find the value of a definite integral analytically. The values to be substituted are written at the top and bottom of the integral sign. You da real mvps! :) https://www.patreon.com/patrickjmt !! The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Put simply, an integral is an area under a curve; This area can be one of two types: definite or indefinite. The fundamental theorem of calculus (FTC) establishes the connection between derivatives and integrals, two of the main concepts in calculus. 28. While the two might seem to be unrelated to each other, as one arose from the tangent problem and the other arose from the area problem, we will see that the fundamental theorem of calculus does indeed create a link between the two. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Learning goals: Explain the terms integrand, limits of integration, and variable of integration. In this section we will take a look at the second part of the Fundamental Theorem of Calculus. For this section, we assume that: The total area under a curve can be found using this formula. So, method one is to compute the antiderivative. Now we\u2019re calculating actual values . Problem Session 7. Describe the relationship between the definite integral and net area. 28. 27. But the issue is not with the Fundamental Theorem of Calculus (FTC), but with that integral. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. The Fundamental Theorem of Calculus. The examples in this section can all be done with a basic knowledge of indefinite integrals and will not require the use of the substitution rule. maths > integral-calculus. Read about Definite Integrals and the Fundamental Theorem of Calculus (Calculus Reference) in our free Electronics Textbook The Fundamental Theorem of Calculus. The given definite integral is {eq}\\int_2^4 {\\left( {{x^9} - 3{x^3}} \\right)dx} {/eq} . 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Use geometry and the properties of definite integrals a Theorem that links the of... We must evaluate between the derivative and the integral, is the same as definite integral net!\n\n++", "date": "2021-04-18 05:43:58", "meta": {"domain": "insignia3d.com", "url": "https://insignia3d.com/tight-lines-axdny/fundamental-theorem-of-calculus-definite-integral-16091b", "openwebmath_score": 0.9654728174209595, "openwebmath_perplexity": 253.6026080541612, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9916842220140631, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8598256709926492}}
{"url": "https://math.stackexchange.com/questions/3067607/basis-for-nullspace-of-matrix-nulla", "text": "# Basis for Nullspace of matrix Null(A)\n\nI could only get a part of the solution and not quite it. The problem is:\n\nA=$$\\left[\\begin{matrix} 1 & -1 & 2 &1 \\\\ 2 & 1 & 0 & 1 \\end{matrix}\\right]$$\n\nWhich of the following sets is a basis for Null(A)?\n\nThe correct answer is :{$$\\left[\\begin{matrix} 0\\\\ 1\\\\ 1\\\\ -1 \\end{matrix}\\right]$$, $$\\left[\\begin{matrix} -4\\\\ 5\\\\ 3\\\\ 3 \\end{matrix}\\right]$$}\n\nAnd what I got is : {$$\\left[\\begin{matrix} -2\\\\ 4\\\\ 3\\\\ 0 \\end{matrix}\\right]$$,$$\\left[\\begin{matrix} -2\\\\ 1\\\\ 0\\\\ 3 \\end{matrix}\\right]$$}\n\nThe 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?\n\nBoth answers are correct. That space has infinitely many bases, of course.\n\n\u2022 Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer. \u2013\u00a0Antoni Malecki Jan 9 at 16:07\n\nThis looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there\u2019s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.\n\nA better strategy would\u2019ve been to go through the possible answers and eliminate those that can\u2019t be a null space basis. First, it\u2019s clear from inspection that $$A$$\u2019s nullity is 2, so if any of the possible answers don\u2019t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $$A$$ and check that you get zero for each one. Whatever\u2019s left must be a basis for $$N(A)$$.\n\nIn this case, the two vectors in the given answer are obviously linearly independent, and their products with $$A$$ are both zero, so they are indeed a basis for $$N(A)$$ (as is the set you found).\n\nYou probably went for the RREF of $$A$$, which is $$\\begin{bmatrix} 1 & 0 & -2/3 & -2/3 \\\\ 0 & 1 & 4/3 & 1/3 \\end{bmatrix}$$ With $$x_3=3$$ and $$x_4=0$$ you get the first vector, with $$x_3=0$$ and $$x_3=3$$ you get the second one. Your solution is correct, good work.\n\nIf you try $$x_3=1$$ and $$x_4=-1$$, you find the first vector in the solution; with $$x_3=3$$ and $$x_3=3$$, you find the second one. I'm not sure how they got their vectors, to be honest.\n\n\u2022 It was from a multiple choice so probably was done to cause confusion. \u2013\u00a0Antoni Malecki Jan 10 at 10:20", "date": "2019-12-13 06:20:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3067607/basis-for-nullspace-of-matrix-nulla", "openwebmath_score": 0.8654195666313171, "openwebmath_perplexity": 136.4193249475704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102580527664, "lm_q2_score": 0.8887587949656841, "lm_q1q2_score": 0.8597943751844181}}
{"url": "https://math.stackexchange.com/questions/3355137/number-of-labeled-abelian-groups-of-order-n", "text": "Number of labeled Abelian groups of order n\n\nI calculated the number of labeled Abelian groups of order $$N$$ (i.e., the number of distinct, abelian group laws on a set of $$N$$ elements). This sequence is given by OEIS A034382, but my solution differs at $$N=16$$.\n\nPlease point out mistakes or confirm my solution?\n\nLet $$C_n$$ be a cyclic group of order $$n$$, $$Aut(G)$$ be an automorphism set of $$G$$.\n\nTthe number of labeled Abelian groups of order $$N$$ is $$\\displaystyle{\\sum \\frac{N!}{\\# Aut(G)}}$$ where G runs representative of isomorphic equivalence.\n\nI got\n\n$$\\displaystyle{ \\# Aut(C_{p^n}^k)=p^{(n-1)k^2}\\prod_{j=0}^{k-1} (p^{k}-p^{j}) }$$\n\nand\n\n$$\\displaystyle{ \\# Aut(\\prod_i C_{p^{n_i}}^{k_i}) =\\prod_i \\left( (p^{(n_i-1)k_i^2}\\prod_{j=0}^{k_i-1} (p^{k_i}-p^{j})) ( \\prod_{j\\neq i} p^{\\min(n_i,n_j)k_j} )^{k_i} \\right) }$$.\n\nFrom the fundamental theorem of finite abelian groups, There are 5 groups for $$N=16$$: $$C_{16}, C_2 \\times C_8, C_4^2, C_2^2\\times C_4, C_2^4$$.\n\nTherefore, the number of groups that are isomorphic to each group is:\n\n\u2022 $$C_{16}$$ ... $$\\displaystyle{\\frac{16!}{8}}$$\n\u2022 $$C_2 \\times C_8$$ ... $$\\displaystyle{\\frac{16!}{(1\\times 2)\\times (2\\times 4)}}$$\n\u2022 $$C_4^2$$ ... $$\\displaystyle{\\frac{16!}{16\\times 3\\times 2}}$$\n\u2022 $$C_2^2\\times C_4$$ ... $$\\displaystyle{\\frac{16!}{((3\\times 2)\\times 2^2)\\times (2^2\\times 2)}}$$\n\u2022 $$C_2^4$$ ... $$\\displaystyle{\\frac{16!}{15\\times 14\\times 12\\times 8}}$$\n\nSum of them is $$4250979532800$$. OEIS says $$4248755596800$$.\n\nmigrated from mathoverflow.netSep 13 at 12:07\n\nThis question came from our site for professional mathematicians.\n\n\u2022 It's only off by one-twentieth of one percent \u2013 what's the big deal? \u2013\u00a0Gerry Myerson Sep 12 at 6:16\n\u2022 I'm sorry that this should post to Mathematics stack exchange. I'll delete this question soon. \u2013\u00a0sugarknri Sep 12 at 10:59\n\u2022 It may be an error in OEIS. In either case it's a good question, and I don't support the close votes. \u2013\u00a0Max Alekseyev Sep 12 at 12:38\n\u2022 In case it's a good question, it would benefit from being better introduced... it starts with 3 sentences which look unrelated. The notion in the title is not defined. Also the title should not be considered as the 1st line of the post: the post should be meaningful without reading the title. \u2013\u00a0YCor Sep 12 at 23:17\n\u2022 @YCor Thank you for the advice about the format. I change text. If the text is still not clear, It may be because of my English ability. \u2013\u00a0sugarknri Sep 13 at 3:35\n\nThere is a different formula for $$\\#\\mathrm{Aut}(\\prod_i C_{p^{n_i}}^{k_i})$$ given in the paper Automorphisms of Finite Abelian Groups by Hillar and Rhea: $$\\#\\mathrm{Aut}(\\prod_{t=1}^m C_{p^{e_t}}) = \\prod_{t=1}^m (p^{d_t} - p^{t-1}) p^{e_t(m-d_t) + (e_t-1)(m-c_t+1)},$$ where $$1\\leq e_1\\leq e_2\\leq \\cdots\\leq e_m$$, and $$c_t$$ and $$d_t$$ are the minimum and maximum of the set $$S_t := \\{\\ell\\ :\\ e_\\ell=e_t\\}$$, respectively.\n\nBelow I will show that OP's formula is equivalent to the Hillar-Rhea formula.\n\nLet $$d_0:=0$$. It can be seen that the $$k_i$$'s are the nonzero elements of the multiset $$\\{ d_1-d_0, d_2-d_1, \\dots, d_m-d_{m-1}\\}$$ and the $$n_i$$'s are the corresponding elements of $$\\{e_1,e_2,\\dots,e_m\\}$$. Define $$s_0=0, s_1, \\dots, s_q$$ be the indices such that $$k_i = d_{s_i} - d_{s_{i-1}}$$ and $$n_i = e_{s_i}$$. Vice versa, $$d_{s_i} = k_1+\\dots+k_i$$ and $$c_{s_i} = d_{s_{i-1}}+1$$.\n\nFirst consider these parts of the two formulas: $$\\prod_{i=1}^q \\prod_{j=0}^{k_i-1} (p^{k_i} - p^j) = \\prod_{i=1}^q p^{k_i(k_i-1)/2} \\prod_{j=0}^{k_i-1} (p^{k_i-j} - 1)$$ and $$\\prod_{t=1}^m (p^{d_t} - p^{t-1}) = p^{m(m-1)/2}\\prod_{t=1}^m (p^{d_t-t+1} - 1).$$ It is easy to see that the multisets $$\\{ k_i - j : 0\\leq j \\leq k_i-1, 1\\leq i\\leq q \\}$$ and $$\\{ d_t - t +1\\ :\\ 1\\leq t\\leq m \\}$$ are the same, since the $$t$$-th element in the sequence $$k_1 - 0, k_1 - 1, \\dots, 1, k_2 - 0, k_2 - 1, \\dots, 1, \\dots$$ equals $$d_t-t+1$$.\n\nNow it remains to prove the equality for the powers of $$p$$ in the two formulas, i.e. $$\\sum_{i=1}^q \\bigg(k_i(k_i-1)/2 + (n_i-1)k_i^2 + \\sum_{j\\ne i} \\min(n_i,n_j)k_ik_j\\bigg) = m(m-1)/2 + \\sum_{t=1}^m \\big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\\big).$$ In the l.h.s. we have $$\\sum_{i=1}^q \\sum_{j\\ne i} \\min(n_i,n_j)k_ik_j = 2\\sum_{i=1}^q n_i k_i \\sum_{j>i} k_j=2\\sum_{i=1}^q n_i k_i (m-d_{s_i}).$$ In the r.h.s. we have $$\\begin{split} \\sum_{t=1}^m \\big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\\big) &= \\sum_{i=1}^q k_i\\big(e_{s_i}(m-d_{s_i}) + (e_{s_i}-1)(m-d_{s_{i-1}})\\big) \\\\ &= \\sum_{i=1}^q k_i\\big(n_i(m-d_{s_i}) + (n_i-1)(m-d_{s_{i-1}})\\big)\\\\ &=\\sum_{i=1}^q k_i\\big(n_i(m-d_{s_i}) + (n_i-1)(m+k_i-d_{s_i})\\big)\\\\ &=2\\sum_{i=1}^q k_i n_i(m-d_{s_i}) + \\sum_{i=1}^q \\big( (n_i-1)k_i^2 - (m-d_{s_i})k_i\\big). \\end{split}$$ Finally, we notice that $$\\sum_{i=1}^q k_i(k_i-1)/2 = m(m-1)/2 - \\sum_{i=1}^q (m-d_{s_i})k_i$$ since $$m=k_1+k_2+\\dots+k_q$$ and $$m-d_{s_i} = k_{i+1}+k_{i+1}+\\dots+k_q$$. QED\n\nSo, we can conclude that OEIS A034382 did indeed contain an error in its 16-th term. Now it's corrected.", "date": "2019-09-19 00:59:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3355137/number-of-labeled-abelian-groups-of-order-n", "openwebmath_score": 0.8546452522277832, "openwebmath_perplexity": 262.4721762348263, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856483, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8597913761275524}}
{"url": "https://math.stackexchange.com/questions/2099774/what-is-the-value-of-sum-r-1n-int-01fr-1xdx", "text": "# What is the value of $\\sum_{r=1}^n\\int_{0}^1f(r-1+x)dx$?\n\nI tried to write the integral as limit of a sum which gives:$$\\int_0^1f(r-1+x)dx=\\lim_{n\\to\\infty}\\frac{1}{n}[f(0)+f(\\frac{1}{n})+f(\\frac{2}{n})...+f(\\frac{n-1}{n})]$$\n\nwhich i wrote as: $$\\lim_{n\\to\\infty}\\frac{1}{n}\\sum_{r=0}^{n-1}f(\\frac{r}{n})$$\n\nand when I put this in the question, I get a double summation. I don't know how to go further. Kindly suggest.\n\nBtw the answer to this question is $$\\int_0^nf(x)dx$$\n\nFor $\\displaystyle\\int_0^1 f(r-1+x)dx$ with substitution $u=r-1+x$ we have $$\\int_0^1 f(r-1+x)dx=\\int_{r-1}^r f(u)du$$ so $$\\sum_1^n\\int_0^1 f(r-1+x)dx=\\int_{0}^n f(x)dx$$\n\n\u2022 Thanks! Btw, is there any way this can be solved as limit of a sum? I am 100% fine with the answer you wrote but this question was given in definite integral as the limit of a sum exercise so just wondering if there's any? \u2013\u00a0Cheapstrike Jan 16 '17 at 6:37\n\u2022 @Cheapstrike For summation we must have $\\Delta x=\\Big((r+1)-1+x\\Big)-(r-1+x)=r\\to0$ but It does not occur. \u2013\u00a0Nosrati Jan 16 '17 at 6:40\n\nAs a hint: try to substitute $u=r+x-1$.\n\n(This would typically be a comment, but I do not, alas, have enough reputation...)\n\n\u2022 I think you do now. Welcome! \u2013\u00a0numbermaniac Nov 25 '17 at 12:44\n\n$\\int _0^1 f (r+x-1)=\\int _0^1 f (r-x)$ let $r-x=u$ thus $dx=-du$ hence we have $I=\\sum _1 ^n -\\int _r ^{r-1} f (u)du =\\sum _1 ^n \\int _{r-1} ^r f (u)du =\\int _0 ^n f (x)dx$\n\n\u2022 from Second last to last step, is it a formula? \u2013\u00a0Kislay Tripathi Nov 25 '17 at 9:57\n\u2022 The disappereance of - sign? \u2013\u00a0Archis Welankar Nov 25 '17 at 9:58\n\u2022 Its a known fact that $\\int _a^b f (x)dx=-\\int _b ^a f (x)dx$ . I hope its clear now. \u2013\u00a0Archis Welankar Nov 25 '17 at 10:04\n\u2022 please tell me about The disappereance of summation sign? \u2013\u00a0Kislay Tripathi Nov 25 '17 at 10:06\n\u2022 I think you should read basic theorems of integration you can then answer your question yourself;) \u2013\u00a0Archis Welankar Nov 25 '17 at 10:08\n\nLet $\\displaystyle \\mathfrak{I}:=\\sum_{r=1}^n \\int_0^1 f(r+x-1) \\,\\text{d} x$. Leting $u:=r+x-1$ we have $\\dfrac{\\text{d}u}{\\text{d}x}=1$ . If $x=0$ we have $u=r-1$, and if $x=1$ we have $u=r$. Therefore:\n\n\\begin{align}\\mathfrak{I}=&\\sum_{r=1}^n \\int_{r-1}^r f(u) \\, \\text{d}u\\\\ \\end{align} But we know that $\\displaystyle \\int_{a}^b f \\,+ \\int_b^c f = \\int_a^c f$. Therefore : $$\\boxed{\\,\\,\\,\\mathfrak{I} =\\int_{0}^n f(x) \\, \\text{d}x\\,\\,\\,\\,\\,}$$\n\n\u2022 Brother how did you conclude $\\displaystyle \\int_{a}^b f \\,+ \\int_b^c f = \\int_a^c f$ From this \\begin{align}\\mathfrak{I}=&\\sum_{r=1}^n \\int_{r-1}^r f (u) \\, \\text{d}u\\\\ \\end{align} .Is it some kind of formula or i need knowledge of riemann integral for this? \u2013\u00a0Kislay Tripathi Nov 25 '17 at 11:43\n\u2022 @KislayTripathi it's the sum of areas. Think about the integrals as areas under a graph - if you add the area from $a$ to $b$ and the area from $b$ to $c$ together, wouldn't it be the same as the area from $a$ to $c$? \u2013\u00a0numbermaniac Nov 25 '17 at 12:17\n\u2022 Is a way to \"glue\" togheter the areas. Consider a function $f$ and $a<b<c$. The $\\displaystyle \\int_a^c f$ (area from $a$ to $c$) can be cuted in a certain $b$. Therefore the area will bem the sum of those areas, wich are $\\displaystyle \\int_a^b f$ (area from $a$ to $b$) and $\\displaystyle \\int_b^c f$ (area from $b$ to $c$). Literally a slice haha. \u2013\u00a0Gustavo Mezzovilla Nov 25 '17 at 13:46", "date": "2021-01-23 05:12:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2099774/what-is-the-value-of-sum-r-1n-int-01fr-1xdx", "openwebmath_score": 0.9522208571434021, "openwebmath_perplexity": 424.54790640078613, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357179075084, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.859791367796747}}
{"url": "http://csvh.fian-intern.de/rolling-motion-without-slipping.html", "text": "\" /> Rolling Motion Without Slipping\n\n# Rolling Motion Without Slipping\n\nIf a wheel rolls w/o slipping, its linear speed and acceleration of the CM relative to the ground are equal to that of a point on the rim of the wheel relative to the CM. The constraint eq (rolling without slipping) is: x, We will solve this problem in two ways: R x Note:. roll without slipping A uniform solid ball is placed at rest on an incline of slope angle \u03b8. The diagrams show the masses (m) and radii (R) of the cylinders. Therefore, the equation of the rotational motion for the disk is: The moment of inertia is a positive scalar; therefore, the angular acceleration vector is parallel to the torque of the friction force. For pure rolling motion (i. Since the disk rolls without slipping. This will enable us to do energy analysis of a variety of rolling objects. Rolling without slipping. Since the disk rolls without slipping. People have observed rolling motion without slipping ever since the invention of the wheel. Rolling Motion + = Rolling without slipping Pure translation Pure rotation about CM Subscribe to view the full document. Which quantity is constant before it rolls without slipping? 1. rolling without slipping; over a straight line; has cusps (points with two tangents) A cycloid is curtate (or contracted) if\u2026 it is traced out by\u2026 points inside a generating circle (r < R) that is rolling without slipping or; points on the surface of the generating circle that is slipping while rolling with v cm > R\u03c9; does not have cusps. We consider the two cases of rolling without slipping and rolling with slipping. 5 seconds, it stops skidding and start to roll without slipping. a moving wire 3. Here is the first case, the wheel rolls to the right so the rotation is clockwise. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of \u03c0/2. An object that rolls against a surface without slipping obeys the condition that the velocity of its center of mass is equal to the cross product of its angular velocity with a vector from the point of contact to the center of mass:. Cylinder, sphere, hoop) Pure rolling motion (rolling without slipping) Rolling motion along a flat surface (inclined, horizontal or even vertical). In which order do the objects reach the bottom of the incline?. Related Links. Derive the governing differential equation of motion. Obviously, the center of the wheel is also the center of the mass, so the linear momentum is p = M \u03c9R. Take moments about the mass center. Rolling motion A special case of rigid body motion is rolling without slipping on a stationary ground surface. Describe the subsequent motion of the mud. Key Takeaways Key Points. If no slipping occurs, the point of contact is momentarily at rest and thus friction is static and does no work on the object. Rolling motion w/o slipping is composed of pure translational motion and pure rotational motion. For pure rolling motion (i. For a cylinder of mass $$M$$ and radius $$R$$, rolling motion without slipping down a plane inclined at an angle $$\\theta$$ with the horizontal, 1. Its moment of inertia about the central axis is. 25) in the book (Section 12. a bead constraints to move on a fixed vs. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed. The diagrams show the masses (m) and radii (R) of the cylinders. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. The axis of rotation travels with the rolling object. Rolling Motion \u2022 The wheel in the figure is rolling to the right without slipping. As the cylinder rolls, point G (center) moves along a straight line. 4-6) Rolling Rotational Inertia Oct. The condition that an object rolls without slipping is that the distance traveled in a line equals the arc length turned out by the wheel. 4 Rolling Without Slipping, Slipping, and Skidding Circular Motion Week 3. Rotational Dynamics: Rolling Spheres/Cylinders is used to calculate the torque that produces rotational motion. of the motion are equal, and the time of the motion is the same for the two components, these two speeds are equal. 5, having been released from rest somewhere along the straight section of the track. People have observed rolling motion without slipping ever since the invention of the wheel. A circular hoop rolls without slipping on a \ufb02at horizontal surface. So the condition for rolling without slipping becomes. It also explains the corresponding equations of pure rolling. Together with the results of the experiments of. Here is the first case, the wheel rolls to the right so the rotation is clockwise. Its moment of inertia about the central axis is. This page contains the video Rolling Without Slipping, Slipping, and Skidding. Friction and Rolling Resistance This is static friction if the wheel rolls without slipping. 25 M And A Mass Of 2. c) Every point on the rim of the wheel has a different linear velocity. The kinematic diagram of the spool is shown in Fig. Chapter 9: Rotational Motion Rigid body instead of a particle Rotational motion about a fixed axis Rolling motion (without slipping) Angular Quantities \u201cR\u201d from the Axis (O) Linear and Angular Quantities Kinematical Equations Chapter 10: Rotational Motion (II) Angular Quantities: Vector Rotational Dynamics: t Note: t = F R sinq Note: sign of t Rotational Dynamics: I Rotational Dynamics: I. I want to ask about the direction of frictional force in smooth rolling motion which means the rolling object doesn't slide on the surface. Rolling contains rotational motion and translational motion(straight line motion) Suppose a car is moving and suddenly applies brakes and it starts sort of skidding. If the ball starts at an initial height h above the bottom of the loop, what is the minimum value for h such that the ball just completes the loop-the-loop?. A rod of mass is attached to a spring of stiffness. no slipping) vR cm 2 2 2 2 11 22 cm 2 m I R v h v R m. The wheel will slide forward without turning. Physics 111 Lecture 21 (Walker: 10. Let represent the downward displacement of the center of mass of the cylinder parallel to the surface of the plane, and let represent the angle of rotation of the cylinder about its symmetry axis. What is the minimum value \u00b5 0 of the coefficient of static friction between ball and incline so that the ball will roll down the incline without slipping? Solution by Michael A. naturally it accelerates downward. If the object rolls without slipping, determine if it is a disk, a hoop, or a sphere. Rolling Motion (Without Slipping) When a wheel or radius R rolls without slipping along a flat straight path, the points of the wheel in contact with the surface are instantaneously at rest and the wheel rotates about a rotation axis through the contact point. Firstly, we describe the motion of a rigid sphere on a rigid horizontal plane. Rolling Motion Problems. Although rolling wheels are everywhere, when most people are asked \u201cwhat is the axis of rotation of a wheel that rolls without slipping?\u201d, they will answer \u201cthe axle\u201d. (a) Draw the free-body diagram for the ball. Energy is still conserved, but the initial potential energy is now converted into two types of kinetic energy. Verify that Equations (A5) and (A6) are correct dimensionally. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. Which quantity is constant before it rolls without slipping? 1. (b) What is the minimum coefficient of friction required for the sphere to roll without slipping? Solution. 10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion: + = for the touching point at any given time! \"Rolling without slipping\". Salukvadze optimal solution and ranked Pareto optimal solutions. Physics 111 Lecture 21 (Walker: 10. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of \u03c0/2. 0 m/s, determine (a) the translational kinetic energy of its center of mass, (b) the rotational kinetic energy about its center of mass; and (c) its total energy. Consider a thin axisymmetric disk with mass and radius that rolls without slipping over a stationary and rough horizontal plane, as illustrated in Figure 1. 3, steady motion is discussed in secs. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed. FRICTIONAL ROLLING PROBLEMS When analyzing the rolling motion of wheels, cylinders, or disks, it may not be known if the body rolls without slipping or if it slides as it rolls. Friction force Minimum coefficient of friction is, Kinetic Energy of Rolling Motion. Professor Lewin derived an equation for the acceleration of an object rolling down a ramp (under pure roll conditions). You may also find it useful in other calculations involving rotation. t and q, w, a vs. General Plane Motion Rolling without slipping. The pure rotation is shown in Animation 1, while the pure translation is shown in Animation 2. Rolling Motion of on fixed incline surface. This motion, though common, is complicated. At which point in the derivation was the assumption of rolling without slipping necessary? Exercise. For many dynamics problems, rolling without slipping means there is a friction force acting on the wheel at the contact point P. What are the conditions for slipping without rolling and rolling without slipping? When a ball or a ring is placed on a surface,it can slip,roll or slip with rollingunder what conditions are each of these executed??It probably has something to do with friction. Consider an object with circular cross section that rolls Rolling as Translation and Rotation Combined along a surface without slipping. Gottlieb Let: \u00b5 = coefficient of friction between ball and incline. Motion A bowling ball is initially thrown down an alley with an initial speed v 0, and it slides without rolling but due to friction it begins to roll until it rolls without slipping. Angular Acceleration \u2022 Definition: - Uniform angular acceleration \u2022 \u03c9 = \u03c9 o + \u03b1t - Units - Vector direction \u2022 Tangential Acceleration - a t = r\u03b1 \u0394. Why is this good? Because we have relationships between linear and rotational motion quantities, such as v = Rw for the linear and angular speeds, or a = R(alpha) for the linear and angular accelerations. Physics Flash Animations. about its center of mass, rolling without. it won't move F 6 Big yo-yo Since the yo-yo rolls without slipping, the center of mass velocity of the yo-yo must satisfy, v cm = r\u03c9. As shown in Fig. The spherical shell started from rest. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. of the general equations of motion of a thin disk rolling without slipping on a horizontal surface, we present results of a simple experiment on the time dependence of the motion that indicate the dominant dissipative power loss to be proportional to the \u03a92 up to and including the last observable cycle. rolling friction, which is always present and distinct from either of these other two kinds of frictionfor a rollingobject. When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move. A disc of radius R rolls without slipping at speed v. Consider a solid sphere of radius R and mass M rolling without slipping. In the special case of rolling without slipping, the distance. At the bottom of the incline, the speed of the hoop's center-of-mass is v. The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. Of the three forces in the system, two act at that point, so they have no lever arm. 0points A bicycle wheel of radius 0. UY1: More About Rolling Motion. The condition for rolling without slipping is The torque is. Another key is that for rolling without slipping, the linear velocity of the center of mass is equal to the angular velocity times the radius. 10-9 Rotational Plus Translational Motion; Rolling In (a), a wheel is rolling without slipping. Rolling, Torque, and Angular Momentum Rolling Motion: \u2022 A motion that is a combination of rotational and translational motion, e. No energy is dissipated and Mechanical Energy is Conserved. \u2022 Describing rotational motion \u2022 Rolling without slipping \u2022 Torque Lecture 24: Rotational Motion. The qualitative features of the motion can now be deduced from the equations. People have observed rolling motion without slipping ever since the invention of the wheel. see ISM 10. Kinetic Energy The total KE of an object undergoing rolling motion is the The total KE of an object undergoing rolling motion is the sum of the rotational KE about its CM and the translational sum of the rotational KE about. rolling without slipping. When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move. \u0e01\u0e32\u0e23\u0e40\u0e04\u0e25\u0e37\u0e48\u0e2d\u0e19\u0e17\u0e35\u0e48\u0e41\u0e1a\u0e1a\u0e01\u0e32\u0e23\u0e01\u0e25\u0e34\u0e49\u0e07 (Rolling Motion, without slipping) \u0e2b\u0e23\u0e37\u0e2d \u0e01\u0e32\u0e23\u0e2b\u0e21\u0e38\u0e19\u0e23\u0e2d\u0e1a\u0e41\u0e01\u0e19\u0e17\u0e35\u0e48\u0e40\u0e04\u0e25\u0e37\u0e48\u0e2d\u0e19\u0e17\u0e35\u0e48. For pure rolling motion (i. A wheel rolls uniformly on level ground without slipping. The Kinematics of a Translating and Rotating Wheel model displays the model of wheel rolling on a floor. The motion of any point on the ring can be considered as vector sum of translational and rotational motion. An object is rolling, so its motion involves both rotation and translation. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. This motion, though common, is complicated. In contrast, it is well known that, for rolling without slipping, a uniform cylinder with moment of inertiaI = kma2 about its axis has acceleration gsin\u03b1/(1 +k) down an. If the object rolls without slipping, determine if it is a disk, a hoop, or a sphere. The rolling-without-slipping kinematic condition for figure 5. 84, there are three forces acting on the cylinder. Rolling Without Slipping. In order to start the rolling motion, a force or torque must be applied to the wheel. The spool rolls to the right. Which quantity is constant before it rolls without slipping? 1. 4 and 5, and oscillation about steady motion is considered in sec. This will enable us to do energy analysis of a variety of rolling objects. A cylinder of radius rolls without slipping down a plane inclined at an angle to the horizontal. Since the disk rolls without slipping. t for this ride. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (October 2, 2014; updated January 12, 2018) 1Problem Discuss the motion of a cylinder that rolls without slipping on another cylinder, when the latter rolls without slipping on a horizontal plane. Rolling motion is a combination of translational motion and rotational motion. The direction of its angular momentum is _____________. David Pritchard, Prof. see ISM 10. You can view a realistic animation of the rolling with slipping and watch as it changes to pure rolling without slipping. Now the translational kinetic energy of the ring is,Rotational kinetic energy of the ring is,. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. Salukvadze optimal solution and ranked Pareto optimal solutions. Friction force Minimum coefficient of friction is, Kinetic Energy of Rolling Motion. The pure rotation is shown in Animation 1, while the pure translation is shown in Animation 2. see ISM 10. or spherical shell) having mass M, radius R and rotational inertia I. The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. Consider a. Which one of the following is necessarily true? a) All points on the rim of the hoop have the same speed. rotational motion. A circular hoop rolls without slipping on a \ufb02at horizontal surface. When discussing the physics of a rolling wheel in an introductory course, it is always fun to point out that a point on the rim of the wheel traces out a cycloid, as shown in Figure 1. to the right 2. Rolling Motion (Without Slipping) When a wheel or radius R rolls without slipping along a flat straight path, the points of the wheel in contact with the surface are instantaneously at rest and the wheel rotates about a rotation axis through the contact point. This has 1 less dof Pick the coordinates as shown. Rolling Motion of on fixed incline surface. Determine the natural frequency The statement that the \"cord does not slip on the pulley\" introduces the rolling-without-slipping condition. 4 \u2013 A vertical force but a horizontal motion A spool of mass M has a string wrapped around its. They both travel a sh01t distance at the bottom and then start up another incline. Practice comparing the rotational kinetic energy of two objects based on their shape and motion. For example, an object, say a ball is in rolling, that is it is rolling on the surface of the ground. Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point. Consider the following example. 10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion: + = for the touching point at any given time! \"Rolling without slipping\". For analyzing rolling motion in this chapter, refer to Figure 10. One disk rolls without slipping, and one slides down the ramp. The diagrams show the masses (m) and radii (R) of the cylinders. 0points A bicycle wheel of radius 0. Chapter 2 Rolling Motion; Angular Momentum 2. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. UY1: More About Rolling Motion. Answer and Explanation: Rolling without slipping is a combination. Consider the following three objects, each of the same mass and radius: a solid sphere a solid disk a hoop All three are released from rest at the top of an inclined plane. At this moment the ratio between magnitudes of velocities of A and B with respect to ground is. A piece of mud on the wheel flies off when it is at the 9 o\u2019clock position (rear of wheel). In this work we analyse the motion of a sphere that rolls without slipping on the inside of a right circular cone under the in\ufb02uence of a uniform gravitational \ufb01eld acting verically downwards, in the direction of the symmetry axis of the cone. (a) Find the linear acceleration of the center of mass. , a ball, cylinder, or disk rolling without slipping. (a) Draw the free-body diagram for the ball. rotational motion. To define such a motion we have to relate the translation of the object to its rotation. From what minimum height above the bottom of the track must the marble be released in order not to leave the track at the top of the loop. David Litster, Prof. Q8: Sketch crude graphs of x, v, a vs. At a given point in time, we can apply the rotational version of Newton\u2019s second law to rotations about the point where the cylinder touches the surface (as the cylinder is rolling without slipping, this is the only motion at that point). Label the identification points above. torque about the center from the force of tension. \"Rolling resistance \" which slows the wheel is a completely different force. At the bottom of the incline, the speed of the hoop's center-of-mass is v. Cylinder, sphere, hoop) Pure rolling motion (rolling without slipping) Rolling motion along a flat surface (inclined, horizontal or even vertical). The motion of a disk that is spun on a smooth flat surface slowly damps out due to friction. Energy is still conserved, but the initial potential energy is now converted into two types of kinetic energy. Consider two cylinders that start down identical inclines from rest except that one is frictionless. For analyzing rolling motion in this chapter, refer to Figure 10. In this section, we compare two situations: (1) rolling and slipping, and (2) rolling without slipping. 4 \u2013 A vertical force but a horizontal motion A spool of mass M has a string wrapped around its. these aspects should be properly darified. If the pulley makes four revolutions without the rope slipping, what length of. 26A is Without slipping, the disk has two variables, X and \u03b2, but only one degree of freedom. Physics Flash Animations. In other words, at any particular instant of time, the part of the disc in contact with the surface is at rest with respect to the surface. Salukvadze optimal solution and ranked Pareto optimal solutions. To define such a motion we have to relate the translation of the object to its rotation. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. It is known that this motion can be considered as a rotation about an. rolls without slipping on a horizontal terrain in such a way that it touches the terrain on the surface line parallel to the y-axis. \u2022 The red curve shows the path (called a cycloid) swept out by a point on the rim of the wheel \u03c9 swept out by a point on the rim of the wheel. Rolling Motion + = Rolling without slipping Pure translation Pure rotation about CM Subscribe to view the full document. 4 m/s rolls up a hill without slipping. \u2022 Describing rotational motion \u2022 Rolling without slipping \u2022 Torque Lecture 24: Rotational Motion. Iclicker #1. Consider a thin axisymmetric disk with mass and radius that rolls without slipping over a stationary and rough horizontal plane, as illustrated in Figure 1. This friction force prevents slipping. 80 kg climbs an incline. When the ring rolls without slipping, then v = r\u03c9, where v is the velocity of centre of mass of ring, and \u03c9 is its angular velocity. The condition for rolling without slipping is The torque is. Wednesday, January 23, 2008. In this work we analyse the motion of a sphere that rolls without slipping on the inside of a right circular cone under the in\ufb02uence of a uniform gravitational \ufb01eld acting verically downwards, in the direction of the symmetry axis of the cone. The diagrams show the masses (m) and radii (R) of the cylinders. A solid cylinder W of mass mand radius r, r R\u03c9; does not have cusps. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. 5, having been released from rest somewhere along the straight section of the track. \"Rolling resistance \" which slows the wheel is a completely different force. There are several assumptions to simplify the analysis of rolling objects: Homogeneous rigid body; High degree of symmetry (E. from rest and rolls without slipping a distance of 6 m If object is rolling with a com\u2260 0 (i. For example, we can look at the interaction of a car\u2019s tires and the surface of the road. Hey guys! In this video we're going to talk about conservation of energy in rolling motion problems. A mathematical model of the dynamics of a rolling rigid sphere in a rectilinear chute-conveyor with transformed dry friction is developed. The spherical shell started from rest. The wheel will slide forward without turning. The sphere has two kings of kinetic energy: linear and rotational. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. 22 m and a mass of 1. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. 0points A bicycle wheel of radius 0. 1 Rolling Without Slipping When a round, symmetric rigid body (like a uniform cylinder or sphere) of radius R rolls without slipping on a horizontal surface, the distance though which its center travels (when. The linear speed of the wheel is. Now, consider a wheel of radius R rolling without slipping along the straight. In other words, at any particular instant of time, the part of the disc in contact with the surface is at rest with respect to the surface. A circular hoop rolls without slipping on a \ufb02at horizontal surface. Derive the governing differential equation of motion. (b) What is the minimum coefficient of friction required for the sphere to roll without slipping? Solution. In which order do the objects reach the bottom of the incline?. Rolling without slipping. A multi-objective optimization problem of a passively controlled motion is formulated for the model and it has been solved. The equations of motion will be: F x = m(a G) x => P - F = m a G F y = m(a G) y => N - mg = 0 M G = I Ga => F r = I G a There are 4 unknowns (F, N, a, and a G. Solution A. When the ring rolls without slipping, then v = r\u03c9, where v is the velocity of centre of mass of ring, and \u03c9 is its angular velocity. there are net forces) NON-smooth rolling motion. Course Material Related to This Topic: Complete exam problem 1e; Check solution to exam Problem 1e. Multiple-Choice Homework 21: Torque and rolling motion Problem 1: A solid cylinder and a solid sphere are rolling without slipping down an inclined plane. General Plane Motion Rolling without slipping. Velocities of points on a rolling wheel. Theoretical background Consider a cylinder of massm and radius R rolling down an inclined plane of angle \u03b8. 7 in Section 10. Rotational motion is more complicated than linear motion, and only the motion of rigid bodies will be considered here. could be a cylinder, hoop, sphere. along a surface without slipping. a wheel rolling down the road. The motion of an extended rigid body can be resolved into two parts. Rolling Motion without Slipping. If the conditions are such that the object rolls without slipping, there is also a geometric relationship between the displacement of the center of mass and the angle turned or, equivalently, between the accelerations: x =!. Rolling without slipping commonly occurs when an object such as a wheel, cylinder, or ball rolls on a surface without any skidding. Another key is that for rolling without slipping, the linear velocity of the center of mass is equal to the angular velocity times the radius. A Rolling Object Accelerating Down an Incline. Let m be the mass and r be the radius of ring. b) All points on the rim of the hoop have the same linear velocity. Absolute Motion: Exercise A wheel of radius r rolls without slipping. Rolling without slipping. The kinematic diagram of the spool is shown in Fig. Take moments about the mass center. We otation Combin can simplif ed y its study by treating it as a combination of translation of the center of mass and rotation of the object about the center of mass Consider the two snapshots of a rolling bicycle wheel shown in the figure. 26A is Without slipping, the disk has two variables, X and \u03b2, but only one degree of freedom. The angular momentum is non-zero and constant for one of the restaurant's customers seated near a window. This has 1 less dof Pick the coordinates as shown. Iclicker #1. Key ideas for rolling: Rolling can be considered to be a superposition of a pure translational motion and a pure rotational motion. In contrast, it is well known that, for rolling without slipping, a uniform cylinder with moment of inertiaI = kma2 about its axis has acceleration gsin\u03b1/(1 +k) down an. Consider an object with circular cross section that rolls Rolling as Translation and Rotation Combined along a surface without slipping. Dorbolo, S. This is defined by motion where the point of contact with the ground has zero velocity, so it matches the ground velocity and is not slipping. What is the minimum value \u00b5 0 of the coefficient of static friction between ball and incline so that the ball will roll down the incline without slipping? Solution by Michael A. The spring is attached to a wall and the rod rests on a disk of the same mass. This motion, though common, is complicated. Rolling involves both of these at the same time - rotation while the wheel is experiencing straight-line motion. Starting Rolling Motion. If the pulley makes four revolutions without the rope slipping, what length of. In fact, the wheel is experiencing both a translational and rotational motion. A mathematical model of the dynamics of a rolling rigid sphere in a rectilinear chute-conveyor with transformed dry friction is developed. \u2022 Will only consider rolling with out slipping. v bottom =0 R v CM Quick Quiz and Demo ! Consider a wheel in pure. We consider the two cases of rolling without slipping and rolling with slipping. People have observed rolling motion without slipping ever since the invention of the wheel. the rolling motion of a cylinder or hoop. rotational motion. We can simplify its study by treating it as a combination of translattreating it as a combination of translation of the center ofion of the center of. In rotations, rolling without slipping is the phrase we always hope we see in a problem. David Litster, Prof. A ball slipping on a surface is decelerated by the friction force opposing the ball linear displacement. 11-4 Combining Rolling and Newton\u2019s Second Law for Rotation Let\u2019s now look at how we can combine torque ideas with rolling-without-slipping concepts. The radius of gyration of the disk about point o is kog; hence,. we consider rolling with and without slipping and establish the conditions for the frictional force to have the direction of the centre-of-mass velocity, contrary to the common idea that this force is always opposite to motion. This situation can be analyzed using relative velocity and acceleration equations. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. In our analysis of the motion of a rigid disk on a horizontal plane we adopt a vectorial approach as advocated by Milne [10]. 7 in Section 10. The linear kinetic energy is given by the usual equation: 1/2 m*v^2. The acceleration of the body is , where k is the radius of gyration of body. Determine the natural frequency The statement that the \"cord does not slip on the pulley\" introduces the rolling-without-slipping condition. and Since the maximum static friction force , it imply that for the disk to rolling without slipping. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. could be a cylinder, hoop, sphere. When objects roll along a surface, like this tire rolls across the ground, there must be some sort of force to cause the rotation to occur. This accelleration is constant until the ball rolls without slipping, which is to say that the angular velocity times the radius matches the velocity of the ball. Rolling without slipping commonly occurs when an object such as a wheel, cylinder, or ball rolls on a surface without any skidding. Rolling Without Slipping. In this work we investigate the motion of a homogeneous ball rolling without slipping on uniformly rotating horizontal and inclined planes under the action of a constant external force. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of \u03c0/2. Motion A bowling ball is initially thrown down an alley with an initial speed v 0, and it slides without rolling but due to friction it begins to roll until it rolls without slipping. rolling without slipping; over a straight line; has cusps (points with two tangents) A cycloid is curtate (or contracted) if\u2026 it is traced out by\u2026 points inside a generating circle (r < R) that is rolling without slipping or; points on the surface of the generating circle that is slipping while rolling with v cm > R\u03c9; does not have cusps. A ball with an initial velocity of 8. In short, this study suggests that the new position sensor can accurately sense the position of a cylinder that rolls without slipping. b) All points on the rim of the hoop have the same linear velocity. The kinematic diagram of the spool is shown in Fig. In our analysis of the motion of a rigid disk on a horizontal plane we adopt a vectorial approach as advocated by Milne [10]. Pure Roll of Hollow and Solid Cylinders: In pure roll, the object is not skidding or slipping, and the speed of the center of mass equals the circumferential speed. Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point. So, in this case, the object tries to slide first, which friction prevents, and that leads to rolling. no slipping) vR cm 2 2 2 2 11 22 cm 2 m I R v h v R m. Users can change the type of object (solid sphere, solid cylinder, etc. The point P, touching the ground, is instantaneously at rest, and the center moves with velocity. A and B are two points on the ring. Rolling Without Slipping. For this consider a body with circular symmetry for example cylinder, wheel, disc , sphere etc. physics 111N 2 rotations of a rigid body! suppose we have a body which rotates about some axis rolling \"rolling without slipping\". When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move.", "date": "2020-02-19 09:02:16", "meta": {"domain": "fian-intern.de", "url": "http://csvh.fian-intern.de/rolling-motion-without-slipping.html", "openwebmath_score": 0.5616315007209778, "openwebmath_perplexity": 406.11801210474823, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856482, "lm_q2_score": 0.8757869786798663, "lm_q1q2_score": 0.8597913633963247}}
{"url": "https://aperiodical.com/2018/03/prime-time/", "text": "# Prime Time\n\nWe spotted this photograph of a letter to The Telegraph, shared by Card Colm on Twitter\u00a0earlier in the year. It\u2019s exactly the kind of mathematical claim we like to enjoy verifying, so we thought we\u2019d dig in.\n\nSIR \u2013 As one obsessed with prime numbers, I note that we have gone from 2017 (a prime) into 2018 (two times a prime), which will be followed by 2019 (three times a prime). I believe this sequence has only happened three other times in the past 1,129 years.\nKeith Burgess-Clements\nMaidstone, Kent\n\nIt\u2019s lovely that newspapers will print this kind of letter, and a quick check verifies that Mr Burgess-Clements is indeed correct that these three numbers have the properties described:\n\n\u2022 $2017$ prime, $2018 = 2 \\times 1009$, $2019 = 3 \\times 673$\n\nHis follow-up statement, that this sequence has only happened three other times in the last 1,129 years, takes a little more checking. But only a little \u2013 as we have a resident CL-P, who describes the necessary calculation as \u2018a trivial amount of Python code\u2019, and quickly came up with the following list:\n\n\u2022 $13$ prime, $14 = 2 \\times 7$, $15 = 3 \\times 5$\n\u2022 $37$ prime, $38 = 2 \\times 19$, $39 = 3 \\times 13$\n\u2022 $157$ prime, $158 = 2 \\times 79$, $159 = 3 \\times 53$\n\u2022 $541$ prime, $542 = 2 \\times 271$, $543 = 3 \\times 181$\n\u2022 $877$ prime, $878 = 2 \\times 439$, $879 = 3 \\times 293$\n\u2022 $1201$ prime, $1202 = 2 \\times 601$, $1203 = 3 \\times 401$\n\u2022 $1381$ prime, $1382 = 2 \\times 691$, $1383 = 3 \\times 461$\n\u2022 $1621$ prime, $1622 = 2 \\times 811$, $1623 = 3 \\times 541$\n\nHere\u2019s that Python code, in case you\u2019re curious. It uses Sage\u2019s Primes() function.\n\npr = set([p for p in range(2018) if p in Primes()])double = set(2*p-1 for p in pr) triple = set(3*p-2 for p in pr)years = sorted(pr & double & triple)\n\nThis is a full list of all the cases below 2017, and hints at some nice more interesting patterns \u2013 $13$ and $541$ both occur as the prime year and the prime factor that's a third of another year. But we don't have time to dig into that now \u2013 we have to check if a person in the newspaper was wrong!\n\nKeith's claim that this has happened thrice in the last 1,129 years is indeed correct \u2013 $2018 \u2013 1129 = 889$, and three sets of years have occurred since then. I suspect this may have been a typo though, as if he'd said \"the last 1,139 years\", that would have included the tail end of the set starting in $877$. Maybe he was looking for the most impressive length of time with the fewest occurrences, to illustrate how rare it is (in which case 1139 would be your best bet, and probably what he meant). We favour \"only four times since 1000AD\" which still sounds pretty good.\n\nOne final question to answer is, how many of these will there be going forward? The next few will be:\n\n\u2022 $2557$ prime, $2558 = 2 \\times 1279$, $2559 = 3 \\times 853$\n\u2022 $2857$ prime, $2858 = 2 \\times 1429$, $2859 = 3 \\times 953$\n\u2022 $3061$ prime, $3062 = 2 \\times 1531$, $3063 = 3 \\times 1021$\n\nIt's also worth checking when this pattern will continue for four years, so that the fourth year is four times a prime; that's $12721$, which is prime, while $12722 = 2 \\times 6361$, $12723 = 3 \\times 4241$ and $12724 = 4 \\times 3181$.\n\n\u2022 #### Katie Steckles\n\nPublicly engaging mathematician, Manchester MathsJam organiser, hairdo.\n\n### 5 Responses to \u201cPrime Time\u201d\n\n1. John\n\nIf we write dates as 8 digit numbers yyyymmdd, then 2018 has 18 prime dates. 2019 will have 19 prime dates, and 2021 will have 21.\n\n2. Rafa\n\nJust a Haskell version of the code:\n\nimport Data.List\nimport Math.NumberTheory.Primes.Sieve -- arithmoi library\nimport qualified Data.Set as Set\n\nyears =  Set.intersection (Set.fromList pr) (Set.intersection doubles triples)\nwhere\npr = takeWhile (< 2018) primes\ndoubles = Set.fromList [2*p-1|p<-pr]\ntriples = Set.fromList [3*p-2|p<-pr]\n\nmain = print years", "date": "2018-12-17 19:56:11", "meta": {"domain": "aperiodical.com", "url": "https://aperiodical.com/2018/03/prime-time/", "openwebmath_score": 0.6746134161949158, "openwebmath_perplexity": 1890.8318434316661, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357205793903, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8597913621797292}}
{"url": "https://gmatclub.com/forum/in-how-many-ways-can-5-different-colored-marbles-be-placed-in-3-distin-161598.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 18 Nov 2019, 12:08\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# In how many ways can 5 different colored marbles be placed in 3 distin\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 01 Aug 2013\nPosts: 11\nLocation: India\nIn how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n14 Oct 2013, 08:53\n4\n32\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n40% (02:22) correct 60% (02:08) wrong based on 279 sessions\n\n### HideShow timer Statistics\n\nIn how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\n(A) 60\n(B) 90\n(C) 120\n(D) 150\n(E) 180\nManager\nJoined: 10 Sep 2013\nPosts: 70\nConcentration: Sustainability, International Business\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2013, 02:19\n11\n2\nHow can we fill three pockets with 5 marbles?\n3 + 1 + 1\n2 + 2 + 1\n\nWith 3,1,1 distribution:\n# of ways to select 3 from 5\n5!/3!2! = 10\n# of ways to select 1 ball from 2\n2!/1! = 2\n# of ways to select 1 ball from 1\n1!/1! = 1\nHow many ways to distribute 3,1 and 1 to 3 boxes? 3!/2! = 3\n\n10*2*3 = 60\n\nWith 2,2,1 distribution:\nHow many ways to select 2 from 5?\n5!/2!3! = 10\n# of ways to select 2 from 3\n3!/2!1! = 3\n# of ways to select 1 from 1\n1\nHow many ways to distribute 2,2,1 to 3 boxes? 3!/2! = 3\n\n10*3*3=90\n\n90+60 = 150\n\n_________________\nKudos if I helped\n##### General Discussion\nIntern\nStatus: Procastrinating!!!\nJoined: 09 Jun 2014\nPosts: 3\nLocation: United States\nConcentration: Healthcare, Strategy\nGPA: 4\nWE: Medicine and Health (Health Care)\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n20 Sep 2014, 13:22\n8\n2\nThis is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )\n\nAnyway this is how I solved it...\nAttachments\n\n20140920_161645-1.jpg [ 1.9 MiB | Viewed 10996 times ]\n\n_________________\nGive me kudos if I am worth them!\nCEO\nStatus: GMATINSIGHT Tutor\nJoined: 08 Jul 2010\nPosts: 2977\nLocation: India\nGMAT: INSIGHT\nSchools: Darden '21\nWE: Education (Education)\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n08 Aug 2015, 07:18\n1\n1\npraffulpatel wrote:\nHow many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\n(A) 60\n(B) 90\n(C) 120\n(D) 150\n(E) 180\n\nWe have 5 marbles and 3 pockets\nSo we have two cases\n\nCase-1: One Pocket with 3 marbles and two pockets with 1 marble each\nNo. of Arrangements = 5C3 * 3C1 * 2! = 10*3*2 = 60\n5C3 - No. of ways of choosing 3 out of 5 marbles which have to go in one pocket\n3C1 - No. of ways of choosing 1 out of 3 pockets in which 3 marbles have to go\n2! - No. of ways of arranging remaining 2 marbles between remaining two pockets which get one marble each\n\nCase-2: Two Pockets with 2 marbles each and one pockets with 1 marble\nNo. of Arrangements = 5C2 * 3C2 * 3C2 = 10*3*3 = 90\n5C2 - No. of ways of choosing 2 out of 5 marbles which have to go in one pocket\n3C2 - No. of ways of choosing 2 out of remaining 3 marbles which have to go in second pocket\n3C2 - No. of ways of choosing 2 out of 3 pockets in each of which 2 marbles have to go\n\nTotal cases = 60+90 = 150\n\n_________________\nProsper!!!\nGMATinsight\nBhoopendra Singh and Dr.Sushma Jha\ne-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772\nOnline One-on-One Skype based classes and Classroom Coaching in South and West Delhi\nhttp://www.GMATinsight.com/testimonials.html\n\nACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION\nIntern\nJoined: 26 Jul 2015\nPosts: 14\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n09 Aug 2015, 14:23\nGMATinsight, Bunuel, VeritasPrepKarishma,\n\nI'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:\nQuote:\nHow many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\nWould the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.\nFor the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.\n\nThen we come to the 3-1-1 and 2-2-1 combinations.\n\nFor the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.\nSo for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.\n\nSimilarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.\n\nFinally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.\n\nCould you please explain why we ignored 1-1-1 combination?\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 9788\nLocation: Pune, India\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2015, 00:18\n1\njhabib wrote:\nGMATinsight, Bunuel, VeritasPrepKarishma,\n\nI'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:\nQuote:\nHow many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\nWould the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.\nFor the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.\n\nThen we come to the 3-1-1 and 2-2-1 combinations.\n\nFor the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.\nSo for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.\n\nSimilarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.\n\nFinally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.\n\nCould you please explain why we ignored 1-1-1 combination?\n\nHey jhabib,\n\nYou have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\nCEO\nStatus: GMATINSIGHT Tutor\nJoined: 08 Jul 2010\nPosts: 2977\nLocation: India\nGMAT: INSIGHT\nSchools: Darden '21\nWE: Education (Education)\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2015, 04:21\n1\nHi jhabib\n\nThe Question has clearly specified that \"ALL the Marbles have to be assigned to 3 pockets\" so 1-1-1 needs to be ignored\n\nAlong with 1-1-1, 1-1-2, 1-2-1 ans 2-1-1 also need to be ignored.\n\nI hope it helps!\n\njhabib wrote:\nGMATinsight,\n\nI'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:\nQuote:\nHow many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\nWould the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.\nFor the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.\n\nThen we come to the 3-1-1 and 2-2-1 combinations.\n\nFor the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.\nSo for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.\n\nSimilarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.\n\nFinally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.\n\nCould you please explain why we ignored 1-1-1 combination?\n\n_________________\nProsper!!!\nGMATinsight\nBhoopendra Singh and Dr.Sushma Jha\ne-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772\nOnline One-on-One Skype based classes and Classroom Coaching in South and West Delhi\nhttp://www.GMATinsight.com/testimonials.html\n\nACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION\nSenior Manager\nStatus: love the club...\nJoined: 24 Mar 2015\nPosts: 265\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n06 Jan 2018, 13:02\n2\npraffulpatel wrote:\nHow many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\n(A) 60\n(B) 90\n(C) 120\n(D) 150\n(E) 180\n\nhi\n\nI have seen a solution to a problem similar to this one elsewhere on the forum\nlet me explain it to you\n\n5 different colored marbles can be placed in 3 distinct pockets without any restriction is\n\n= 3^5 = 243\n\nas we are asked to find out the ways in which any pocket must get at least 1 marble, lets find out the ways in which any pocket must not get at least 1 marble, and then subtract the number of ways in which any pocket must not get at least 1 marble from the total number of ways, that is 3^5\n\nso lets get going\n\nnumber of ways in which all marbles can get to 1 pocket is\n\n= 3, as there are 3 distinct pocket in total\n\nnow, number of ways in which 2 pockets can get all the marbles and 1 pocket remains empty is\n\n= (2^5 - 2) * 3 = 90\n\n2 has been subtracted to eliminate the possibility that any pocket out of 2 can get all the marbles\n\n3 has been multiplied with the whole expression, because 2 pockets out of 3 have been selected\n\nnow we are in business\n\n= 243 - 90 - 3\n\n= 150 (D)\n\nhope this helps\nthanks\n\ncheers, and do consider some kudos, man\nIntern\nJoined: 23 Jun 2015\nPosts: 26\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n09 Oct 2018, 23:13\njhabib wrote:\nGMATinsight, Bunuel, VeritasPrepKarishma,\n\nI'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:\nQuote:\nHow many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\nWould the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.\nFor the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.\n\nThen we come to the 3-1-1 and 2-2-1 combinations.\n\nFor the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.\nSo for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.\n\nSimilarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.\n\nFinally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.\n\nCould you please explain why we ignored 1-1-1 combination?\n\nHey jhabib,\n\nYou have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.\n\nDear Karishma ..\nI am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong .\nAll the approaches are considering 3 bags as identical , but in problem they are distinct .\nMy approach to this problem :\nI am considering 5 different marbles a b c d e and 3 different bags 1 2 3 .\nSo first distribute 3 balls one ball each to each bag .\nselect 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways .\nso total 60 ways .\nNow we are left with 2 balls and 3 bags each containing one ball each .\nfor first ball we have 3 options and second ball again 3 options . 3*3 ways .\nTOTAL = 60*3*3=540 ways .\n\nPlease somebody help me in this . This problem just made me mad ..\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 9788\nLocation: Pune, India\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n10 Oct 2018, 00:22\nkarnaidu wrote:\nDear Karishma ..\nI am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong .\nAll the approaches are considering 3 bags as identical , but in problem they are distinct .\nMy approach to this problem :\nI am considering 5 different marbles a b c d e and 3 different bags 1 2 3 .\nSo first distribute 3 balls one ball each to each bag .\nselect 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways .\nso total 60 ways .\nNow we are left with 2 balls and 3 bags each containing one ball each .\nfor first ball we have 3 options and second ball again 3 options . 3*3 ways .\nTOTAL = 60*3*3=540 ways .\n\nPlease somebody help me in this . This problem just made me mad ..\n\nYou have some double counting here. When you select some from a group and distribute and then distribute the rest to the same bags/pockets, there is double counting.\n\nSay you distributed a, b, c first such that\nThen you distributed d and e such that Bag1 got d and bag2 got e.\n\nTake another case.\nSay you distributed d, b and c first such that\nThen you distributed a and e such that Bag1 got a and bag2 got e.\n\nNote that both cases have exactly the same end result. But you would count them as two separate cases. Similarly, there will be other cases which will be double counted. Hence this method will not work.\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\nIntern\nJoined: 23 Jun 2015\nPosts: 26\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n10 Oct 2018, 00:53\nsxyz wrote:\nThis is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )\n\nAnyway this is how I solved it...\n\nHii ..\nIn this approach you are missing arrangement of bags . You have to take care of that too ..\nIntern\nJoined: 03 Sep 2015\nPosts: 18\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n18 Oct 2018, 11:16\nHi. I don't understand why this has been multiplied by 3!/2! for 2-2-1 combination. Plus, shouldn't 5c2*3c2 be multiplied by 2! as the pockets are distinct, and the arrangement would matter? Thanks\nManager\nJoined: 28 Jan 2019\nPosts: 97\nLocation: India\nGMAT 1: 700 Q49 V36\nGPA: 4\nWE: Manufacturing and Production (Manufacturing)\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n05 May 2019, 05:41\nsxyz wrote:\nThis is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )\n\nAnyway this is how I solved it...\n\nHi, I have a doubt.\n\nHere, the case {3,1,1} = $$5C3*2C1*1C1 = 10*2*1 = 20$$\nAnd this is multiplied by 3 (coz three different pockets.)\n\nBut generally, in case of distribution between identical pockets/bags/groups etc., the value is divided by the similar ones.\nSuch as, in here, {3,1,1} has 2 similarly filled pockets, and hence, divided by 2!.\n\nIs it not being done here like that because the pockets are all different here?\n_________________\n\"Luck is when preparation meets opportunity!\"\nIntern\nJoined: 10 Apr 2019\nPosts: 18\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n05 May 2019, 13:48\nHi,\n\nCan we solve it the following way?\n\nTotal number of possible combinations: $$3^{5}=243$$\nCombinations that do not satisfy given condition (at least 1 marble in each pocket): $$4-1-0$$, $$3-2-0$$ and $$5-0-0$$.\n1) 4-1-0. 4 marbles can be allocated to 3 different pockets, the rest - 1 - can also be allocated to 3 different pockets. Thus, number of unsatisfactory combinations in for this case = $$3*3*5 = 45$$ - multiply by 5, since marbles can also differ between the pockets.\n2) 3-2-0. The same holds true here = $$3*3*5 = 45$$.\n3) 5-0-0. This scenario is easy = $$3$$ different cases are possible.\n\nThus, $$243-45-45-3=150$$.\nDirector\nJoined: 24 Nov 2016\nPosts: 783\nLocation: United States\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n26 Aug 2019, 13:19\npraffulpatel wrote:\nIn how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\n(A) 60\n(B) 90\n(C) 120\n(D) 150\n(E) 180\n\ngiven: 5 dif marbs 3 dif pockets with at least 1 marble each\n\n$$[3,1,1]\u20265c3\u20222c1\u20221=10\u20222=20\u2026\u2022arrangements[3,1,1]=20\u2022(3!/2!)=60$$ (3!=num.pockets; 2!=num.duplicates [1,1])\n$$[2,2,1]\u20265c2\u20223c2\u20221=10\u20223=30\u2026\u2022arrangements[2,2,1]=30\u2022(3!/2!)=90$$ (3!=num.pockets; 2!=num.duplicates [2,2])\n$$total.arrangements=60+90=150$$\n\nManager\nJoined: 16 Oct 2011\nPosts: 109\nGMAT 1: 570 Q39 V41\nGMAT 2: 640 Q38 V31\nGMAT 3: 650 Q42 V38\nGMAT 4: 650 Q44 V36\nGMAT 5: 570 Q31 V38\nGPA: 3.75\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2019, 10:28\nWe have two cases: The first case has one pocket with 3 marbles, 1 marble in each of the two remaining pockets. The second case has two pockets with 2 marbles each, and one pocket with one marble. We always add cases together to obtain the total number of possibilities.\n\nCase 1: (3,1,1) First choose which of the 3 pockets will get three marbles. There are 3C1 =3 ways to do this. Then, choose the number of ways we can place 3 marbles into that box from 5 marbles. There are 5C3 = 10 ways to do this. Next we must place a marble in the next to the last box. There are 2C1 = 2 ways to do this. Then there is 1C1 = 1 way to place the last marble into the last pocket. Each time we make a subsequent selection when working through a case we multiply, so there are 3*10*2*1= 60 ways to place the marbles in case 1\n\nCase 2: (2,2,1) First we must select which two pockets get two marbles each. There are 3C2 = 3 ways to do this. Then, we must choose two marbles to go in the first of two pockets. There are 5C2=10 ways to do this. From the 3 remaining marbles, there are 3C2 = 3 ways to place two marbles in the second pocket. Finally there is 1C1 =1 way to place the last marble in the last box. Multiplying, we have 3*10*3*1 = 90 ways to place the marbles in case two.\n\nThus there are 60+90 = 150 ways we can place the marbles\nManager\nJoined: 30 May 2019\nPosts: 82\nLocation: United States\nConcentration: Technology, Strategy\nGPA: 3.6\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n04 Oct 2019, 22:44\npraffulpatel wrote:\nIn how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?\n\n(A) 60\n(B) 90\n(C) 120\n(D) 150\n(E) 180\n\nSolved it a little differently than others from what I read . So will share.\n\nI first made sure that each of the pocket has at least 1 marble.\nThat is 5C3 * 3!\n\nThen you can either distribute the remaining 2 as 2, 0, 0 or 1, 1, 0\nThat is 3C1 or 3C2 * 2 ! (3C1 is the no. of ways of choosing a pocket to put everything in. )\n\nThat gives 150.\nD !\nIntern\nJoined: 21 Sep 2016\nPosts: 21\nRe: In how many ways can 5 different colored marbles be placed in 3 distin\u00a0 [#permalink]\n\n### Show Tags\n\n11 Oct 2019, 08:56\nP1 can have either 1 marble or 2 marble or 3 marble.\n\nTherefore,\n\nP1 can have 1 marble in 5 ways as all are of different colors.\n\nOr\n\nP1 can have 2 marbles in 5C2 ways (not 5P2 as order is not important; meaning P1 can have m1m2 or m2m1 and it will mean the same thing). 10 ways.\n\nOr\n\nP1 can have 3 marbles in 5C3 ways. 10 ways.\n\nHence P1 can have marbles in 25 ways.\n\nEvery time we select these marbles their arrangement is also important among the 3 pockets. Because if P1 selects M1 first then it will have repercussion on the selection by the next 2 pockets. Similarly if P1 selects M2 first it will again have an effect on the selection by the next 2 pockets. This means that the arrangement among the 3 pockets shall be considered.\n\nSo now we will arrange these 25 number of ways among the 3 pockets.\n\nTherefore,\n\n25 x 3! = 25 X 6 = 150.\n\nKindly correct if there is anything wrong with my logic. Thanks.\n\nPosted from my mobile device\nRe: In how many ways can 5 different colored marbles be placed in 3 distin \u00a0 [#permalink] 11 Oct 2019, 08:56\nDisplay posts from previous: Sort by\n\n# In how many ways can 5 different colored marbles be placed in 3 distin\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne", "date": "2019-11-18 19:08:39", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-how-many-ways-can-5-different-colored-marbles-be-placed-in-3-distin-161598.html", "openwebmath_score": 0.8652944564819336, "openwebmath_perplexity": 1365.9630466009708, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9532750413739076, "lm_q2_score": 0.9019206732341567, "lm_q1q2_score": 0.8597784670932733}}
{"url": "https://math.codidact.com/posts/287063", "text": "Communities\n\ntag:snake search within a tag\nuser:xxxx search by author id\nscore:0.5 posts with 0.5+ score\n\"snake oil\" exact phrase\ncreated:<1w created < 1 week ago\npost_type:xxxx type of post\nQ&A\n\nShow that $\\forall n \\in \\mathbb{Z}^{+}$, $25^n \\equiv 25 \\bmod{100}$.\n\n+3\n\u22120\n\nShow that $\\forall n \\in \\mathbb{Z}^{+}$, $25^n \\equiv 25 \\bmod{100}$.\n\nThis was a simple observation I made when playing around and I came up with the following proof:\n\nIt follows from $(10a + 25)^2 = 100a^2 + 500a + 625 = 100(a^2 + 5a) + 625$ that any number ending in $25$ raised to a power of $2$ will also end in $25$. As each number has a unique binary representation, every number can be written as a sum of powers of two. By the rule $a^{b + c} = a^b \\cdot a^c$ and $ab \\bmod {c} = (a \\bmod {c}) \\cdot (b \\bmod {c}) \\bmod {c}$, $25^n \\mod {100}$ will eventually cascade down to $25 \\bmod {100}$. For example, \\begin{align}25^{19} \\equiv 25^{16 + 2 + 1} \\equiv 25^{16} \\cdot 25^{2} \\cdot 25^{1} \\equiv 25 \\cdot 25 \\cdot 25 \\\\ \\equiv 25^{3} \\equiv 25^{2 + 1} \\equiv 25^2 \\cdot 25^1 \\equiv 25 \\cdot 25 \\equiv 25^2 \\equiv 25\\bmod{100}\\end{align}\n\nI am wondering if there are alternate, potentially simpler proofs.\n\nWhy does this post require moderator attention?\nWhy should this post be closed?\n\n+4\n\u22120\n\nThere is a fairly simple proof by induction.\n\nBase case: $n = 1$\n\n$$25^1\\equiv 25 \\mod 100$$\n\nInductive case: Assuming $25^n\\equiv 25 \\mod 100$,\n\n\\begin{align} 25^{n+1}&\\equiv 25\\cdot 25^n \\mod 100\\\\ &\\equiv 25 \\cdot 25 \\\\ &\\equiv 625 \\\\ &\\equiv 25 \\end{align}\n\nActually, this proof can easily be extended to show the stronger statement that $5^n\\equiv 25\\mod 100$ for all natural numbers $n\\ge2$\n\nBase case: $n=2$\n\n$$5^2\\equiv25\\mod100$$\n\nInductive case: Assuming $5^n\\equiv 25 \\mod 100$,\n\n\\begin{align} 5^{n+1}&\\equiv 5\\cdot 5^n \\mod 100\\\\ &\\equiv 5 \\cdot 25 \\\\ &\\equiv 125 \\\\ &\\equiv 25 \\end{align}\n\nThe case with powers of $25$ then simply comes from the case when $n$ is even.\n\nWhy does this post require moderator attention?\n\n+3\n\u22120\n\nYou can prove it using the binomial theorem.\n\nAssume that $1\\leq n\u2208\\mathbb{N}$, then:\n\n\\begin{align} 25^n & =(20+5)^n=\\sum_{k=0}^n\\binom{n}{k}20^k\\cdot 5^{n-k}=5+\\sum_{k=1}^{n-1}\\binom{n}{k}20^k\\cdot 5^{n-k}+20\\\\ & =25+5\\cdot 20\\cdot\\sum_{k=1}^{n-1}\\binom{n}{k}20^{k-1}\\cdot 5^{n-k-1}\\equiv 25\\mod 100 \\end{align}\n\nWhy does this post require moderator attention?", "date": "2023-03-25 22:34:38", "meta": {"domain": "codidact.com", "url": "https://math.codidact.com/posts/287063", "openwebmath_score": 1.0000022649765015, "openwebmath_perplexity": 1239.5250432428552, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683491468141, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8597743203966554}}
{"url": "https://math.hecker.org/2016/07/10/linear-algebra-and-its-applications-exercise-3-3-6/", "text": "## Linear Algebra and Its Applications, Exercise\u00a03.3.6\n\nExercise 3.3.6. Given the\u00a0matrix $A$ and vector $b$ defined as follows\n\n$A = \\begin{bmatrix} 1&1 \\\\ 2&-1 \\\\ -2&4 \\end{bmatrix} \\qquad b = \\begin{bmatrix} 1 \\\\ 2 \\\\ 7 \\end{bmatrix}$\n\nfind the projection of $b$ onto the column space of $A$.\n\nDecompose the vector $b$ into the sum $p + q$ of two orthogonal vectors $p$ and $q$ where $p$ is in the column space. Which subspace is $q$ in?\n\nAnswer: We have $p = A(A^TA)^{-1}A^Tb$ per equation (3) of 3L on page 156.\u00a0We first compute\n\n$A^TA = \\begin{bmatrix} 1&2&-2 \\\\ 1&-1&4 \\end{bmatrix} \\begin{bmatrix} 1&1 \\\\ 2&-1 \\\\ -2&4 \\end{bmatrix}$\n\n$= \\begin{bmatrix} 9&-9 \\\\ -9&18 \\end{bmatrix}$\n\nand then compute\n\n$(A^TA)^{-1} = \\frac{1}{9 \\cdot 18 - (-9)(-9)} \\begin{bmatrix} 18&-(-9) \\\\ -(-9)&9 \\end{bmatrix}$\n\n$= \\frac{1}{81} \\begin{bmatrix} 18&9 \\\\ 9&9 \\end{bmatrix} = \\begin{bmatrix} \\frac{2}{9}&\\frac{1}{9} \\\\ \\frac{1}{9}&\\frac{1}{9} \\end{bmatrix}$\n\nFinally we compute\n\n$p = A(A^TA)^{-1}A^Tb$\n\n$= \\begin{bmatrix} 1&1 \\\\ 2&-1 \\\\ -2&4 \\end{bmatrix} \\begin{bmatrix} \\frac{2}{9}&\\frac{1}{9} \\\\ \\frac{1}{9}&\\frac{1}{9} \\end{bmatrix} \\begin{bmatrix} 1&2&-2 \\\\ 1&-1&4 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 2 \\\\ 7 \\end{bmatrix}$\n\n$= \\begin{bmatrix} 1&1 \\\\ 2&-1 \\\\ -2&4 \\end{bmatrix} \\begin{bmatrix} \\frac{2}{9}&\\frac{1}{9} \\\\ \\frac{1}{9}&\\frac{1}{9} \\end{bmatrix} \\begin{bmatrix} -9 \\\\ 27 \\end{bmatrix}$\n\n$= \\begin{bmatrix} 1&1 \\\\ 2&-1 \\\\ -2&4 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 2 \\end{bmatrix} = \\begin{bmatrix} 3 \\\\ 0 \\\\ 6 \\end{bmatrix}$\n\nSince $b = p + q$ we have\n\n$q = b - p = \\begin{bmatrix} 1 \\\\ 2 \\\\ 7 \\end{bmatrix} - \\begin{bmatrix} 3 \\\\ 0 \\\\ 6 \\end{bmatrix} = \\begin{bmatrix} -2 \\\\ 2 \\\\ 1 \\end{bmatrix}$\n\nWe have\n\n$p \\cdot q = 3 \\cdot (-2) + 0 \\cdot 2 + 6 \\cdot 1 = -6 + 0 + 6 = 0$\n\nso that $p$ and $q$ are orthogonal.\n\nThe vector\u00a0$p$ is in $\\cal{R}(A)$, the column space of A, and the orthogonal subspace of $\\cal{R}(A)$ is $\\cal{N}(A^T)$, the left nullspace of $A$. Since $p$ is in\u00a0$\\cal{R}(A)$ and $q$ is orthogonal to $p$, $q$ must be in\u00a0$\\cal{N}(A^T)$, so that $A^Tq = 0$. We confirm this:\n\n$A^Tq = \\begin{bmatrix} 1&2&-2 \\\\ 1&-1&4 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 2 \\\\ 1 \\end{bmatrix}$\n\n$= \\begin{bmatrix} -2 + 4 -2 \\\\ -2 - 2 + 4 \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix} = 0$\n\nUPDATE: I corrected the calculation of $q$; thanks go to KTL for pointing out the error.\n\nNOTE: This continues a series of posts containing worked out exercises from the (out of print) book\u00a0Linear Algebra and Its Applications, Third Edition\u00a0by\u00a0Gilbert Strang.\n\nIf you find these posts useful I encourage you to also check out the more current\u00a0Linear Algebra and Its Applications, Fourth Edition, Dr Strang\u2019s introductory textbook\u00a0Introduction to Linear Algebra, Fourth Edition\u00a0and the accompanying free\u00a0online course, and Dr Strang\u2019s\u00a0other books.\n\nThis entry was posted in linear algebra and tagged , , , , . Bookmark the permalink.\n\n### 6 Responses to Linear Algebra and Its Applications, Exercise\u00a03.3.6\n\n1. Teresa says:\n\nThank you for the great work! It\u2019s been really helpful. I wanted to mention that A transpose by A is [6 -8; -8 18] and not [9 -9; -9 18] that changes the results a bit.\n\n2. hecker says:\n\nI\u2019m glad you find these posts useful. I\u2019m sorry it\u2019s been a long time since I published the last one.\n\nI don\u2019t understand your comment. Are you referring to A transpose multiplied by A (on the right)? If so, the (1, 1) element of the result matrix is 1 x 1 + 2 x 2 + (-2) x (-2) = 1 + 4 + 4 = 9. I don\u2019t know how you got 6 for that result. Similarly the (1, 2) entry is 1 x 1 + 2 x (-1) + (-2) x 4 = 1 \u2013 2 \u2013 8 = -9 (not -8) and the (2, 1) entry is 1 x 1 + (-1) x 2 + 4 x (-2) = 1 \u2013 2 \u2013 8 = -9 again.\n\n\u2022 KTL says:\n\nThe question in the 4th edition is different. Hence the confusion for the person above \ud83d\ude42\n\n\u2022 hecker says:\n\nAh, thanks for the explanation. I\u2019ve never looked at a copy of the 4th edition so I don\u2019t know how the exercises differ.\n\n3. KTL says:\n\nq you have given as p-b though the calculation is done as b-p (correctly)\n\n\u2022 hecker says:\n\nThanks for finding this error! I\u2019ve corrected the post.", "date": "2017-05-27 13:47:43", "meta": {"domain": "hecker.org", "url": "https://math.hecker.org/2016/07/10/linear-algebra-and-its-applications-exercise-3-3-6/", "openwebmath_score": 0.7770076394081116, "openwebmath_perplexity": 483.54002950576296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683509762453, "lm_q2_score": 0.8705972751232808, "lm_q1q2_score": 0.8597743153579109}}
{"url": "https://math.stackexchange.com/questions/1869797/number-of-true-relations-of-the-form-a-subseteq-b-where-a-b-in-mathcalp-1", "text": "# Number of true relations of the form $A\\subseteq B$ where $A,B\\in\\mathcal{P}(\\{1,2,\\ldots,n\\})$\n\nI just started \"Introduction to Topology and Modern Analysis\" by G.F. Simmons and came across this problem in the exercises.\n\nQ. Let $U=\\{1,2,\\ldots,n\\}$ for an arbitrary positive integer $n$. If $A$ and $B$ are arbitrary subsets of $U$, how many relations of the form $A\\subseteq B$ are there? How many of them are true?\n\nSome previous problems for the cases $n=1,2,3$ suggested that by \"relations of the form $A\\subseteq B$\", they mean even the ones where $A\\subseteq B$ isn't true.\n\nIt would be obvious that there are $2^{2n}=4^n$ such relations since $U$ has $2^n$ subsets, $A,B$ can each be chosen in $2^n$ ways, so there are $2^n\\times 2^n=4^n$ such relations.\n\nNow, I think $3^n$ of these relations are true. Here's my idea/proof:\n\nLet us consider $x,y\\in\\mathcal{P}(U)$ where $\\mathcal{P}(U)$ denotes the power set of $U$. Let $x$ have $k$ elements from $U$ (where $0\\leq k\\leq n$) with $k=0$ corresponding to $x=\\emptyset$.\n\nFor $x\\subseteq y$ to be true, $y$ must have all the elements of $x$ and may/may not have elements from $U\\setminus x$. We can construct $y$ by taking $m$ additional elements from $U\\setminus x$ where $0\\leq m\\leq n-k$ which can be done in $\\binom{n-k}m$ ways for each value of $m$.\n\nSo, for each $x$, we can construct $y$ in $\\sum\\limits_{m=0}^{n-k}\\binom{n-k}m=2^{n-k}$ ways.\n\nNow, we can take $x$ in $\\binom nk$ ways for each value of $k$ and hence the number of true relations is $\\sum\\limits_{k=0}^n\\binom nk 2^{n-k}=3^n$.\n\nIs the above correct/rigorous enough?\n\nAlso, if the above solution is correct, doesn't it work for any arbitrary set $U$ of cardinality $n$ ?\n\n\u2022 what do you mean? You want to know with how many relations you can endow the set $\\mathcal P\\{1,2,3\\dots n\\}$ ?? And then how many of these are subsets of the relation $\\subseteq$ ? \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 18:18\n\u2022 @CarryonSmiling, it's a problem in the book by Simmons that how many of the set inclusion relations $A\\subseteq B$ where $A,B\\in\\mathcal{P}(\\{1,2,\\ldots,n\\})$ are true? I'm looking for proofreading by the community on my work. \u2013\u00a0analysis123 Jul 24 '16 at 18:28\n\u2022 Yeah, your solution looks good. In fact your solution was clearer to me that the actual question. \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 18:29\n\n## 2 Answers\n\nI finally got it.\n\nThe question seems to be: How many pairs $(A,B)$ of subsets of $\\{1,2\\dots n\\}$ exist so that $A\\subseteq B$.\n\nFor each element $x\\in \\{1,2\\dots n\\}$ we have three choices:\n\n\u2022 $x$ is only in $B$\n\u2022 $x$ is in $A$ and $B$\n\u2022 $x$ is not in $A$ and not in $B$.\n\nTherefore there are $3^n$ possible pairs.\n\nYes, this is correct, and yes, only the cardinality of $U$ matters.\n\nYou could slightly simplify the proof by noting that there are $2^{n-k}$ subsets of $U\\setminus x$, so you don't have to sum over binomial coefficients.\n\nI find the combinatorial proof in Carry on Smiling's answer simpler and more elegant.", "date": "2019-12-15 15:55:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1869797/number-of-true-relations-of-the-form-a-subseteq-b-where-a-b-in-mathcalp-1", "openwebmath_score": 0.9229318499565125, "openwebmath_perplexity": 134.83662357878316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969660413474, "lm_q2_score": 0.8740772417253255, "lm_q1q2_score": 0.8597397230468196}}
{"url": "https://math.stackexchange.com/questions/2222064/proof-of-equation-with-nested-dependent-integrals", "text": "# Proof of equation with nested, dependent integrals\n\nAs part of a larger proof, I am currently trying to prove the following equation: $$\\int_0^r \\int_0^{r-t_1} \\dots \\int_0^{r-\\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \\dots t_k^{m_k - 1} dt_k \\dots dt_1 = \\frac{r^n}{n!} \\prod_{i=1}^{k} (m_i -1)!$$\n\nI know that $n = \\sum_{i=1}^k m_i$ and moreover that all $m_i$ are natural numbers.\n\nI've already confirmed with Mathematica that this equation holds for $k \\in \\{1,\\dots,5\\}$. I however have some difficulties with proving this equation for any $k > 0$. Any hints would be greatly appreciated!\n\nHere's what I've done so far:\n\nIn order to get started, I tried to show it for $k = 1$: $$\\int_0^r t_1^{m_1 - 1} dt_1 = \\left[\\frac{1}{m_1} \\cdot t_1^{m_1}\\right]_{t_1=0}^{t_1=r} = \\frac{r^{m_1}}{m_1} = \\frac{r^{m_1}}{m_1!} \\cdot \\frac{m_1!}{m_1} = \\frac{r^{m_1}}{m_1!} \\cdot (m_1 - 1)!$$\n\nThat was pretty straightforward.\n\nBut already for $k=2$ I'm getting stuck:\n\n$$\\int_0^r \\int_0^{r-t_1} t_1^{m_1 - 1}t_2^{m_2 - 1} dt_2dt_1 = \\int_0^r t_1^{m_1 - 1}\\int_0^{r-t_1} t_2^{m_2 - 1} dt_2dt_1 = \\int_0^r t_1^{m_1 - 1}\\left[\\frac{1}{m_2} \\cdot t_2^{m_2}\\right]_{t_2=0}^{t_2=r-t_1}dt_1 = \\int_0^r t_1^{m_1 - 1}\\cdot\\frac{1}{m_2} \\cdot (r-t_1)^{m_2}dt_1$$\n\nNow if I would like to integrate over $t_1$, I should probably use the binomial theorem, which says: $$(x+y)^n = \\sum_{k=0}^n {n \\choose k}x^{n-k}y^k = \\sum_{k=0}^n {n \\choose k}x^{k}y^{n-k}$$ Applied to my case, this results in: \\begin{align*} &\\int_0^r t_1^{m_1 - 1}\\cdot\\frac{1}{m_2} \\cdot \\sum_{j=0}^{m_2} {{m_2} \\choose j} \\cdot r^{m_2-j}\\cdot (-1)^j \\cdot t_1^jdt_1\\\\ &= \\frac{1}{m_2} \\cdot \\int_0^r \\sum_{j=0}^{m_2} (-1)^j \\cdot\\frac{m_2!}{j!(m_2-j)!} \\cdot r^{m_2-j}\\cdot t_1^{m_1 + j - 1}dt_1\\\\ &=\\frac{m_2!}{m_2} \\cdot \\sum_{j=0}^{m_2} \\left((-1)^j \\cdot\\frac{1}{j!(m_2-j)!} \\cdot r^{m_2-j} \\int_0^r t_1^{m_1 + j - 1}dt_1\\right)\\\\ &=(m_2-1)! \\cdot \\sum_{j=0}^{m_2} \\left((-1)^j \\cdot\\frac{1}{j!(m_2-j)!} \\cdot r^{m_2-j} \\left[\\frac{1}{m_1 + j} \\cdot t_1^{m_1 + j}\\right]_{t_1=0}^{t_1=r}\\right)\\\\ &=(m_2-1)! \\cdot \\sum_{j=0}^{m_2} (-1)^j \\cdot\\frac{1}{j!(m_2-j)!} \\cdot r^{m_2-j} \\cdot \\frac{1}{m_1 + j} \\cdot r^{m_1 + j}\\\\ &=(m_2-1)! \\cdot \\sum_{j=0}^{m_2} (-1)^j \\cdot\\frac{1}{j!(m_2-j)!} \\cdot r^{m_2+m_1} \\cdot \\frac{1}{m_1 + j}\\\\ &=r^{m_1 + m_2}(m_2-1)! \\cdot \\sum_{j=0}^{m_2} (-1)^j \\cdot\\frac{1}{j!(m_2-j)!}\\cdot\\frac{1}{(m_1 + j)} \\end{align*} Now the only thing left to show is that the sum is equal to $\\frac{(m_1-1)!}{(m_1+m_2)!}$. I tried the following: \\begin{align*} &\\sum_{j=0}^{m_2} (-1)^j \\cdot\\frac{1}{j!(m_2-j)!}\\cdot\\frac{1}{(m_1 + j)}\\\\ &= \\sum_{j=0}^{m_2} (-1)^j \\cdot\\frac{1}{j!(m_2-j)!}\\cdot\\frac{m_1 \\cdots (m_1+j-1)\\cdot(m_1+j+1)\\cdots(m_1+m_2)}{m_1 \\cdots (m_1 + m_2)} \\cdot \\frac{(m_1-1)!}{(m_1-1)!}\\\\ &= \\frac{(m_1-1)!}{(m_1+m_2)!} \\cdot \\sum_{j=0}^{m_2} (-1)^j \\cdot\\frac{1}{j!(m_2-j)!} \\cdot \\left(\\prod_{i=1, i \\neq j}^{m_2} (m_1 + i)\\right) \\end{align*} So now I only need to show that the remaining sum is equal to one, but I'm not able to do so. Again, playing around with some examples ($m_2 \\in \\{1,2,3\\}$) everything works out. I've tried to reframe it as $\\frac{(m_1-1)!}{(m_1+m_2)!m_2!} \\cdot \\sum_{j=0}^{m_2} (-1)^j \\cdot{m_2 \\choose j}\\cdot \\left(\\prod_{i=1, i \\neq j}^{m_2} (m_1 + i)\\right)$, but not with much success.\n\nI'm not quite sure whether I'm on the right track here or whether there's a different approach to this whole thing that looks more promising than what I'm trying to do. I feel like I won't be able to prove the overall thing for $k>0$ if I'm not even able to show it for the special case $k=2$.\n\nI've been working on this for a couple of days now and I don't seem to make progress. If you have any ideas what I could try, that would be great. I'd be very grateful for both help with the $k=2$ proof and for ideas about how to tackle the general $k>0$ proof. Thanks in advance!\n\nHere is a result for $k=2$ (EDIT) and for $k=3$ which shows the general way to proceed.\n\nYou are looking at imcomplete beta functions. For $k=2$, I follow your result (calling that $I_2$):\n\n$$I_2 = \\int_0^r t_1^{m_1 - 1}\\cdot\\frac{1}{m_2} \\cdot (r-t_1)^{m_2}{\\rm d}t_1$$\n\nSubstituting $t_1 = r z$ gives\n\n$$I_2 = \\frac{r^{m_1+m_2}}{m_2} \\int_0^1 z^{m_1 - 1}\\cdot (1-z)^{m_2} {\\rm d} z$$\n\nNow use the definition for the incomplete beta function $B(\\cdot,\\cdot)$ and its special values for natural numbers:\n\n$$I_2 = \\frac{r^{m_1+m_2}}{m_2} B(m_1, m_2+1) = \\frac{r^{m_1+m_2}}{m_2} \\frac{(m_1 - 1)! m_2 !}{(m_1+m_2)!} = r^{m_1+m_2} \\frac{(m_1 - 1)! (m_2 -1) !}{(m_1+m_2)!}$$\n\nwhich is the desired result for $k=2$.\n\nOne step further, for $k=3$. The $r$'s have been substituted away as above.\n\n$$I_3 = \\frac{r^n}{m_3} \\int_0^1 z_1^{m_1 - 1} \\int_0^{1-z_1} z_2^{m_2 - 1}\\cdot (1-z_1-z_2)^{m_3} {\\rm d}z_2{\\rm d}z_1$$\n\nNow for the inner integral, substitute $z_2 = s_2 (1-z_1)$ which gives $$I_3 = \\frac{r^n}{m_3} \\int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} \\int_0^{1} s_2^{m_2 - 1}\\cdot (1-s_2)^{m_3} {\\rm d}s_2{\\rm d}z_1$$\n\nThis is a formulation where the inner integral has been made independent on the outer one - and this will work out for larger $k$ as well!\n\nMaking this explicit, we get $$I_3 = \\frac{r^n}{m_3} \\Big[\\int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} {\\rm d}z_1 \\Big] \\cdot \\Big[\\int_0^{1} s_2^{m_2 - 1}\\cdot (1-s_2)^{m_3} {\\rm d}s_2\\Big]$$\n\nNow using the already known results for the Beta function above, $$I_3 = \\frac{r^n}{m_3} \\frac{(m_1 - 1)! (m_2 + m_3) !}{(m_1+m_2+m_3)!} \\cdot \\frac{(m_2 - 1)! m_3 !}{(m_2+m_3)!} = {r^n} \\frac{(m_1 - 1)! (m_2 - 1)! (m_3 -1)!}{(m_1+m_2+m_3)!}$$\n\nas desired.\n\nFrom here, you can go on to larger $k$.\n\n\u2022 Thanks a lot, the variable change and the Beta function really do the trick! I'll try to formulate the proof for general $k$ but it really looks like it's not as difficult as I thought. Apr 7 '17 at 9:42\n\nSo building on the solution of @Andreas for $k=2$ and $k=3$, I was able to prove the whole thing for arbitrary $k>0$:\n\nFirst of all, one can observe that for any natural number $j > 0$ the following holds: $$\\int_0^{r-\\sum_{i=1}^{j-1}t_i} t_j^{a-1} \\cdot \\left(r- \\sum_{i=1}^j t_i\\right)^b dt_j = B(a,b+1) \\cdot \\left(r- \\sum_{i=1}^{j-1} t_i\\right)^{a+b}$$ We can define $t_j = \\left(r- \\sum_{i=1}^{j-1} t_i\\right)\\cdot z$ which gives $dt_j = \\left(r- \\sum_{i=1}^{j-1} t_i\\right)\\cdot dz$. Making a variable change in the left part of the equation results in $$\\int_0^{1} \\left(r- \\sum_{i=1}^{j-1} t_i\\right)^{a-1} \\cdot z^{a-1} \\cdot \\left(r- \\sum_{i=1}^{j-1} t_i - \\left(r- \\sum_{i=1}^{j-1} t_i\\right)\\cdot z\\right)^b \\cdot\\left(r- \\sum_{i=1}^{j-1} t_i\\right) dz$$ $$= \\left(r- \\sum_{i=1}^{j-1} t_i\\right)^{a-1+b+1} \\int_0^{1} z^{a-1} (1-z)^b dz = B(a,b+1) \\cdot \\left(r- \\sum_{i=1}^{j-1} t_i\\right)^{a+b}$$\n\nNow we look at the overall problem: $$I_k = \\int_0^r \\int_0^{r-t_1} \\dots \\int_0^{r-\\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \\dots t_k^{m_k - 1} dt_k \\dots dt_1$$ This can be rewritten as $$I_k = \\int_0^r t_1^{m_1 - 1} \\int_0^{r-t_1} t_2^{m_2 - 1} \\dots \\int_0^{r-\\sum_{i=1}^{k-1} t_i} t_k^{m_k - 1} dt_k \\dots dt_1$$\n\nUsing the observation made above, we can solve the innermost integral by setting $j=k, a= m_k, b=0$ which gives us $B(m_k,1) \\cdot \\left(r- \\sum_{i=1}^{k-1} t_i\\right)^{m_k}$\n\nWe thus get: $$I_k = B(m_k,1) \\cdot \\int_0^r t_1^{m_1 - 1} \\dots \\int_0^{r-\\sum_{i=1}^{k-2} t_i} t_{k-1}^{m_{k-1} - 1} \\cdot \\left(r- \\sum_{i=1}^{k-1} t_i\\right)^{m_k} dt_k \\dots dt_1$$\n\nAgain, we apply our observation to the innermost integral ($j=k-1,a=m_{k-1},b={m_k}$, resulting in $B(m_{k-1},m_k+1) \\cdot \\left(r- \\sum_{i=1}^{k-2} t_i\\right)^{m_{k-1}+m_k}$\n\nRecursively applying this step finally results in: $$I_k = B(m_k,1)\\cdot B(m_{k-1},m_k + 1) \\cdot B(m_{k-2},m_{k-1}+m_k + 1)\\cdot \\dots \\cdot B(m_1, m_2+\\dots+m_k+1)\\cdot r^{m_1+\\dots+m_k}$$ As all $m_i$ are natural numbers, we can use that $B(x,y)=\\frac{(x-1)!\\,(y-1)!}{(x+y-1)!}$: $$I_k = r^{m_1+\\dots+m_k} \\cdot \\frac{(m_k-1)!\\,0!}{m_k!} \\cdot \\frac{(m_{k-1}-1)!\\,m_k!}{(m_{k-1}+m_k)!} \\cdot \\frac{(m_{k-2}-1)!\\,(m_{k-1}+m_k)!}{(m_{k-2}+m_{k-1}+m_k)!} \\cdot \\dots \\cdot \\frac{(m_1-1)!\\,(m_2+\\dots+m_k)!}{(m_1+m_2+\\dots+m_k)!}$$ This reduces to: $$I_k = r^{m_1+\\dots+m_k} \\cdot (m_k -1)! \\cdot \\dots \\cdot (m_1 -1)! \\cdot \\frac{1}{(m_1+\\dots+m_k)!}$$ With $m_1+\\dots+m_k = n$, we can rewrite this as: $$I_k = \\frac{r^n}{n!} \\prod_{i=1}^{k} (m_i -1)!$$\n\n\u2022 Nice! That's the general way to proceed. Apr 7 '17 at 13:38", "date": "2021-09-23 12:07:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2222064/proof-of-equation-with-nested-dependent-integrals", "openwebmath_score": 0.9997897744178772, "openwebmath_perplexity": 407.0622601053034, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969713304755, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8597397228302913}}
{"url": "http://www.lofoya.com/Solved/1483/if-the-price-of-petrol-increases-by-25-by-how-much-must-a-user", "text": "# Moderate Percentages Solved QuestionAptitude Discussion\n\n Q. If the price of petrol increases by $25%$, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?\n \u2716\u00a0A. $25\\%$ \u2716\u00a0B. $16.67\\%$ \u2714\u00a0C. $20\\%$ \u2716\u00a0D. $33.33\\%$\n\nSolution:\nOption(C) is correct\n\nLet the price of petrol be Rs.100 per litre. Let the user use 1 litre of petrol. Therefore, his expense on petrol = $100\\times 1$ = Rs.100\n\nNow, the price of petrol increases by $25\\%$. Therefore, the new price of petrol = Rs.125.\n\nAs he has to maintain his expenditure on petrol constant, he will be spending only Rs.100 on petrol.\n\nLet \u2018$x$\u2019 be the number of litres of petrol he will use at the new price.\n\nTherefore,\n\n$125\\times x=100$\n\n$\\Rightarrow x=\\dfrac{100}{125}$\n\n$=\\dfrac{4}{5}=0.8$litres\n\nHe has cut down his petrol consumption by 0.2 litres\n\n$=\\dfrac{0.2}{1}\\times 100$\n\n$=20\\%$\u00a0reduction.\n\nThere is a shortcut for solving this problem.\n\nIf the price of petrol has increased by $25\\%$, it has gone up $1/4^{th}$ of its earlier price.\n\nTherefore, the $\\%$ of reduction in petrol that will maintain the amount of\u00a0money spent on petrol constant\n\n$=\\dfrac{1}{4+1}$\n\n$=\\dfrac{1}{5}=20\\%$\n\ni.e. Express the percentage as a fraction. Then add the numerator of the fraction to the denominator to obtain a new fraction. Convert it to percentage - that is the answer.\n\nEdit:\u00a0For direct formula involving such questions, check comment by\u00a0Rushi\u00a0and\u00a0Chirag Goyal.\n\nEdit: For more insights on the question, check comment by\u00a0Priya.\n\n## (5) Comment(s)\n\nVijay\n()\n\n(R*100)/(100+R)..... this formula we will use here.\n\n25*100/(125)=20\n\nso 20% is anwser.\n\nParam\n()\n\nSimplest of all . No Need to remember formula. No advance calculation. Based on basic calculation.\n\nLet assume original price be 100. Increased by 25% become Rs125.\n\nLogic : Now we have to keep the consumption same,so we have to bring Rs 125 back to Rs 100. How much we have to decrease so that Rs 125 become Rs 100.\n\n(125-100)125*100 % = 20%\n\nDone . No need to go to consumption. You can calculate using one variable only.\n\nPriya\n()\n\n25% means 1/4 and 20% means 1/5. If the question is given as \"increased by 25\" then $\\dfrac{1}{(4+1)}=\\dfrac{1}{5}=20\\%$\n\nIf the question is given as \"decreased by 25\" then $\\dfrac{1}{(4-1)}=\\dfrac{1}{3}=33\\dfrac{1}{3}\\%$\n\ni.e., the value of $33\\dfrac{1}{3}\\%$ is $\\dfrac{1}{3}$\n\nIt is easy if you know the values for all the percentages!\n\nRushi\n()\n\nThere is shortcut formula for this kind of sums which is $\\left(\\dfrac{R}{100+R}\\right) \\times 100$ in the case of rising and $\\left(\\dfrac{R}{100-R}\\right) \\times 100$ in the case of fall.\n\nBut this formula is applicable only when exp. have to remain constant.\n\nChirag Goyal\n()\n\nThanks $Rushi$\n\nI'm Rewriting the same Shortcut using TeX commands for better understanding.\n\n$\\text{Consumption Should be Decrease or Increase}$\n\n$\\text=\\dfrac{R}{100\\pm R}\\times{100}$", "date": "2017-07-25 18:50:09", "meta": {"domain": "lofoya.com", "url": "http://www.lofoya.com/Solved/1483/if-the-price-of-petrol-increases-by-25-by-how-much-must-a-user", "openwebmath_score": 0.8693819642066956, "openwebmath_perplexity": 1493.4129762849668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969684454966, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8597397138557502}}
{"url": "https://math.stackexchange.com/questions/579675/prove-subset-of-mathbb-r2-is-open/579686", "text": "# Prove subset of $\\mathbb R^2$ is open\n\nProblem. Let $G = \\{(x,y): x \\ne y\\}$. Prove $G$ is an open subset of $\\mathbb R^2$.\n\nWhat I am thinking: If I could show that $\\mathbb R^2 \\setminus G = \\{(x,y): x = y\\}$ is a closed set, then its complement $G$ is open. I might be totally off. Any suggestions?\n\n\u2022 How did you define an open/closed set? \u2013\u00a0Listing Nov 24 '13 at 20:38\n\nLet the function $f\\colon \\mathbb R^2\\rightarrow \\mathbb R,$ $(x,y)\\mapsto x-y$ then $f$ is continuous since it's a polynomial on $x$ and $y$. We have $$G=f^{-1}(\\mathbb R^*)$$ Do you know what theorem we should use to conclude?\n\n\u2022 Would the theorem be if f is continuous then the inverse image under f of any open set is open? I am only confused on what is meant by R*? \u2013\u00a0MDW Nov 24 '13 at 20:58\n\u2022 Yes this is the needed theorem and $\\mathbb R^*$ is $\\mathbb R\\setminus \\{0\\}$. \u2013\u00a0user63181 Nov 24 '13 at 21:00\n\u2022 @Sami: Why can't we just show $\\Bbb{R}^2 - G$ is open? \u2013\u00a0Don Larynx Nov 24 '13 at 21:03\n\u2022 Do you mean show that $\\mathbb R^2-G$ is closed? Yes it's possible and I want avoid this since I don't know his definition of closed set @DonLarynx \u2013\u00a0user63181 Nov 24 '13 at 21:07\n\u2022 @SamiBenRomdhane: It is open if GOD wants. :D \u2013\u00a0mrs Nov 25 '13 at 2:40\n\nJust partition $G$ into two disjoint sets: Let $G_1 = \\{(x,y) : x > y \\}$ and $G_2 = \\{(x,y) : x < y \\}$, then clearly $G = G_1 \\cup G_2$.\n\nNow, WLOG, let's show that $G_1$ is open. Let $\\textbf{x} = (x_1,y_1) \\in G_1$, and $B_\\textbf{x}(r)$ be the ball around $\\textbf{x}$ of radius $r$. In order for $G_1$ to be open, we need to find an $r$ such that $B_\\textbf{x}(r) \\subset G_1$. But, just choose $r$ to be the shortest distance between $x$ and the line $y=x$. Informally, we know the shortest distance between a point to a line is just the intersection of the normal line with the point $\\textbf{x}$, and the line y = x (in our case). The normal line will be $y = -x + x_1 + y_1$ (by a quick computation), and if we intersect this line with $y=x$, we get $x = \\frac{x_1 + y_1}{2}$, so we can compute the distance to be $\\frac{x_1 - y_1}{\\sqrt 2}$. choose $r$ to be this, and we have shown $G_1$ will be open.\n\nDo the same for $G_2$, then $G$ is the union of 2 open sets, and we are done.\n\nThe definition of an open set : A subset $U$ of a metric space $(M,d)$ is called \\textit{open} if, given any point $x$ in $U$, there exists a real number $\u03b5 > 04$ such that, given any point $y$ in $M$ with $d(x, y) < \u03b5$, $y$ also belongs to $U$.\n\nIn the case of a line in the plane given by the equation $ax + by + c= 0$, where $a, b$ and $c$ are real constants with $a$ and $b$ not both zero, the distance from the line to a point $(x_0,y_0)$ is $$\\frac{|ax_0 + by_0 +c|}{\\sqrt{a^2 + b^2}}.$$ Now, we are interested in one specific line, namely $y = x$, i.e. $x - y = 0$.\n\nThe distance of any point $(x_0, y_0)$ in $G$ to the line $x -y = 0$ is therefore $$r :=\\frac{|x_0 -y_0|}{\\sqrt{2}}.$$\n\nNow, consider the open ball centered in $(x_0,y_0)$ and of radius $r$. Do every element of this ball belong to $G$ ?\n\nSuggestion (you were looking for):\n\nUse the theorem that says that the graph of a real continuous function is closed.\n\n$A=\\{(x,y)\\in \\mathbb{R}^2 : x=y\\}$ is the graph of the continuous function $f:\\mathbb{R}\\rightarrow \\mathbb{R}$. Where $$f(x)=x$$.\n\nHence A is closed. Therefore, its complement is open.\n\nThe theorem does not hold in general but it certainly does for Hausdorff spaces. All metric spaces ($\\mathbb{R}$ is a metric space) are Hausdorff.\nTheorem: A topological space $X$ is Hausdorff iff its diagonal is closed.\nThe diagonal is defined as $$\\{(x,x):x\\in X\\}$$\nNote: The diagonal is closed in the product topology on $X\\times X$.\nHINT: $\\Bbb{R}^2 - G$ results in a line. Can this complete $G$?", "date": "2020-01-18 12:15:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/579675/prove-subset-of-mathbb-r2-is-open/579686", "openwebmath_score": 0.9344146251678467, "openwebmath_perplexity": 97.97054963512264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969713304755, "lm_q2_score": 0.8740772269642949, "lm_q1q2_score": 0.8597397131510212}}
{"url": "https://web2.0calc.com/questions/pls-help_19708", "text": "+0\n\n# PLS HELP\n\n0\n96\n1\n\nThe first six rows of Pascal's triangle are shown below, beginning with row zero. Except for the $1$ at each end, row $4$ consists of only even numbers, as does row $2.$ How many of the first $20$ rows have this property? (Don't include row $0$ or row $1$). \\begin{tabular}{ccccccccccc} &&&&&1&&&&&\\\\ &&&&1&&1&&&&\\\\ &&&1&&2&&1&&&\\\\ &&1&&3&&3&&1&&\\\\ &1&&4&&6&&4&&1&\\\\ 1&&5&&10&&10&&5&&1\\\\ \\end{tabular}\n\nApr 9, 2021\n\n#1\n+420\n+1\n\nNotice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer)\u00a0under 20 (which are 2, 4, 8, and 16), so the answer is\u00a0$$\\boxed{4}$$.\n\nMy intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as\u00a0$$\\frac{n!}{p!(n-p)!}$$, where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.\n\nAlso a very similar question was answered a long time ago:\u00a0https://web2.0calc.com/questions/another-pascal-s-triangle-question\n\nApr 9, 2021\n\n#1\n+420\n+1\n\nNotice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer)\u00a0under 20 (which are 2, 4, 8, and 16), so the answer is\u00a0$$\\boxed{4}$$.\n\nMy intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as\u00a0$$\\frac{n!}{p!(n-p)!}$$, where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.\n\nAlso a very similar question was answered a long time ago:\u00a0https://web2.0calc.com/questions/another-pascal-s-triangle-question\n\ntextot\u00a0Apr 9, 2021", "date": "2021-08-03 11:06:49", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/pls-help_19708", "openwebmath_score": 0.9105386137962341, "openwebmath_perplexity": 318.7790831598578, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983596967003007, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8597397125949029}}
{"url": "https://cs.stackexchange.com/questions/121468/proving-that-tn-t-lfloor-n-2-rfloor-t-lfloor-n-4-rfloor-t-lfloor", "text": "# Proving that $T(n) = T(\\lfloor n/2 \\rfloor) + T(\\lfloor n/4 \\rfloor) + T(\\lfloor n/8 \\rfloor) + n$ is $\\in O(n)$\n\nShow $$T(n) = T(\\lfloor n/2 \\rfloor) + T(\\lfloor n/4 \\rfloor) + T(\\lfloor n/8 \\rfloor) + n$$ is $$\\in O(n)$$.\n\nI will make the bound to be $$\\in O(cn)$$ instead.\n\nProof by strong induction.\n\n\u2022 Base case n =1\n\n$$T(1) = c$$ and $$cn=c*1=c$$ $$\\checkmark$$\n\n\u2022 Inductive Hypothesis : $$T(k) \\in O(ck)$$ for $$1\\le k1$$.\n\u2022 Inductive Step: Prove for n. So prove that $$T(n) \\le O(ck)$$.\n\nRight away we can write $$T(n) = T(\\lfloor n/2 \\rfloor) + T(\\lfloor n/4 \\rfloor) + T(\\lfloor n/8 \\rfloor) + n \\\\ \\leq T(n/2)+T(n/4)+T(n/8) + n\\\\ = c(n/2)+c(n/4)+c(n/8)+n \\ \\ \\ \\ By \\ Inductive \\ Hypothesis$$\n\n$$= c(7n/8)+n \\le cn+n ...$$ I am stuck here\n\n*Goal: $$\\le cn - (some \\ stuff)$$ and some stuff needs to be $$\\ge 0$$.\n\n\u2022 \u2013\u00a0ryan Mar 5 at 0:01\n\u2022 The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! \u2013\u00a0Raphael Mar 6 at 8:04\n\nFirst, let me mention that $$O(n)$$ and $$O(cn)$$ are exactly the same thing. What you are really after is showing that $$T(n) \\leq cn$$ for all $$n$$.\n\nLet us aim at a slightly more relaxed goal: showing that $$T(n) \\leq An$$ for some possibly larger constant $$A$$. For the base case, we need $$A \\geq c$$. For the inductive step, we know that $$T(n) = T(\\lfloor n/2 \\rfloor) + T(\\lfloor n/4 \\rfloor) + T(\\lfloor n/8 \\rfloor) + n \\leq A(n/2+n/4+n/8)+1 = (\\tfrac{7}{8}A+1)n.$$ Recall that our goal is to deduce that $$T(n) \\leq An$$. This would follow if $$\\tfrac{7}{8}A + 1 \\leq A$$, that is, if $$A \\geq 8$$.\n\nIn total, if we take $$A = \\max(c,8)$$ then both the base case and the inductive step go through.\n\nAs an aside, you can also use the Akra-Bazzi theorem to directly conclude that $$T(n) = O(n)$$.\n\nNow let us try to obtain more insight on the recurrence. Let $$S(n) = 8n - T(n)$$. Then $$S(n) = S(\\lfloor n/2 \\rfloor) + S(\\lfloor n/4 \\rfloor) + S(\\lfloor n/8 \\rfloor) + 8n - 8\\lfloor n/2 \\rfloor - 8\\lfloor n/4 \\rfloor - 8\\lfloor n/8 \\rfloor - n.$$ Since $$8(n/a-1) < 8\\lfloor n/a \\rfloor \\leq 8(n/a)$$, we see that $$S(n) = S(\\lfloor n/2 \\rfloor) + S(\\lfloor n/4 \\rfloor) + S(\\lfloor n/8 \\rfloor) + r(n),$$ where $$0 \\leq r(n) < 24$$. Applying the Akra-Bazzi theorem, we get $$S(n) = O(n^p)$$ for some $$p < 1$$ (the solution to $$(1/2)^p + (1/4)^p + (1/8)^p = 1$$), and so $$T(n) = 8n + O(n^p)$$.\n\n\u2022 Thank you this is an amazing answer and helps a lot! \u2013\u00a0Mandy Mar 5 at 23:45\n\u2022 So for base case we can choose either 8 or c (in this case c) , and for inductive step we can also choose 8 or c (in this case 8) ? \u2013\u00a0Mandy Mar 7 at 20:34\n\u2022 You have to choose the same constant in both cases. That\u2019s how induction works. \u2013\u00a0Yuval Filmus Mar 7 at 20:50\n\nIt's very important that you understand what f(n) = O (g(n)) means. It means that there is a number $$n_0 \u2265 0$$ and a number c > 0 such that for every $$n \u2265 n_0$$, f(n) \u2264 c * g(n). It is a property of the whole function, not a property of some n. Saying \"I prove by induction that for every n, f(n) = O (g(n))\" doesn't make any sense at all. What makes sense is to say \"I prove by induction that for every n \u2265 n_0, f(n) \u2264 c * g(n)\".\n\nSo what you want to prove per induction using some suitable c, T(n) \u2264 cn. One variant of complete induction uses an induction step where you proof \"if the statement $$S_k$$ is true for every k < n, then $$S_n$$ is also true\".\n\nIf T(k) \u2264 ck is true for every k < n, then T(n) = T(n/2) + T(n/4) + T(n/8) + n \u2264 cn/2 + cn/4 + cn/8 + n = (7/8 c + 1) n.\n\nIf c = 1 this means T(n) < 1 7/8 n. Not what we need, we need T(n) < cn. If c = 2 it means T(n) + 2 3/4 n. Slightly better but not good enough. For which c is (7/8 c + 1) n \u2264 cn, or 7/8 c + 1 \u2264 c? That's the case for c \u2265 8. The start of the induction is n=0, and it's easy to show that T(0) = 0.\n\nSo you haven't just shown that T(n) = O(n), you have shown the much stronger T(n) \u2264 8n.\n\n\u2022 Thank you this is an excellent answer and helps me understand these kinds of problems in more depth! \u2013\u00a0Mandy Mar 5 at 23:46\n\u2022 You might have forgotten the base case. \u2013\u00a0Yuval Filmus Mar 6 at 16:38", "date": "2020-07-10 03:56:46", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/121468/proving-that-tn-t-lfloor-n-2-rfloor-t-lfloor-n-4-rfloor-t-lfloor", "openwebmath_score": 0.7982414960861206, "openwebmath_perplexity": 335.2118537186417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969698879862, "lm_q2_score": 0.8740772253241803, "lm_q1q2_score": 0.8597397102769623}}
{"url": "https://nbhmcsirgate.theindianmathematician.com/2020/03/nbhm-2020-part-question-10-solution.html", "text": "### NBHM 2020 PART A Question 10 Solution: Possible rank of a matrix obtained by changing the entries $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$\n\nLet $M$ be a $7\u00d76$ real matrix. The entries of $M$ in the positions $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$ are changed to obtain another $7\u00d76$ real matrix $\\widetilde{M}$. Suppose that the rank of $\\widetilde{M}$ is 4. What could be the rank of $M$? List all possibilities.\nSolution:\nFirst, we will derive a connection between the rank of $M$ and the rank of $\\widetilde{M}$. We have,\n$$rank(M) = rank(M-\\widetilde{M}+\\widetilde{M}) \\le rank(M-\\widetilde{M})+rank(M)$$ Similarly\n$$rank(\\widetilde M) = rank(\\widetilde{M}-M+M) \\le rank(\\widetilde{M}-M)+rank(M)$$\nThis shows that $rank(\\widetilde{M}) - rank(\\widetilde{M}-M) \\le rank(M) \\le rank(M-\\widetilde{M})+rank(\\widetilde{M})$\nThe matrices $M - \\widetilde{M}$ and $\\widetilde{M}-M$ can have non-zero entries only in the columns $3$ and $4$ and hence they are of rank atmost $2$. Also, it is given that the rank of $\\widetilde{M}$ is equal to $4$. Hence we have $$2 \\le rank(M) \\le 6.$$\nUsing $0,1$ matrics one can check that all these possibilities are occurring indeed.\nShare to your groups:\nFOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO, SUGGEST PROBLEMS TO SOLVE.\n\n### NBHM 2020 PART A Question 4 Solution $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$\nEvaluate : $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx = \\int_{-\\infty}^{\\inft...", "date": "2020-10-27 14:18:27", "meta": {"domain": "theindianmathematician.com", "url": "https://nbhmcsirgate.theindianmathematician.com/2020/03/nbhm-2020-part-question-10-solution.html", "openwebmath_score": 0.7001168131828308, "openwebmath_perplexity": 162.1564447633871, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543740571681, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8597130915789817}}
{"url": "https://www.themathdoctors.org/more-on-0-999/", "text": "# More on 0.999\u2026\n\n#### (An archive question of the week)\n\nIn collecting questions and answers about 0.999\u2026 for the last post, there were two that were too long to include, but that dig more deeply into issues that some of the standard answers tend to gloss over. So here, I want to look at those two answers, both of which deal with how to handle arithmetic on infinitely many digits: How can you add, subtract, or multiply when there is no rightmost digit? This is not central to the main question, but enough people express uncertainty about this aspect to make it worth covering.\n\n## Adding without a place to start\n\nHere\u2019s the first question, from Darryl in 1999 (appropriately enough):\n\nWhat is 0.999... + 0.999...?\n\nWhile discussing the 1 = 0.999... solution, a person asked what is 0.999... + 0.999...? I think it is a good question - one that I could not answer. It should be 2 but how do we show this?\n\nIn the archives at\n\nGetting 0.99999...\nhttp://mathforum.org/dr.math/problems/dusty4.15.98.html\n\nyou say \"Can you figure out why 0.3bar + 0.3bar = 0.6bar? Because these numbers go on forever, you will need to use a little logic to add them. (The algorithm that you learned for adding numbers doesn't work very well when you can't get to the rightmost number.)\"\n\nIf we copy the idea of limits, we could say that for any number n we could start adding at this rightmost point and get the 0.66...6(nth) place. And since we can do this for any n then the equation holds. But if we try this for 0.9bar + 0.9bar we seem to get something that looks less like 2, namely 1.9....8(nth). But if we replace 8 with 9, we seem to get a number between 2 and the one we have, which makes me feel as if 0.9bar + 0.9bar < 2.\n\nThe link is to an elaborated version of the 1/3 + 1/3 + 1/3 = 1 argument, which I only gave a link to last time. (The notation 0.3bar as a typable version of $$0.\\overline{3}$$, comes from there.) As Doctor Derrel pointed out there, we normally add from right to left, and there is no rightmost digit to start at in 0.333\u2026 + 0.333\u2026; fortunately, we know that it isn\u2019t necessary to start at the right when there are no carries, so we can just add left to right. This is discussed here:\n\nDarryl\u2019s suggestion is to consider any n digits, where we can clearly add 0.333\u20263 + 0.333\u20263 = 0.666\u20266, so we get the desired result by letting n increase without bound.\n\nBut, as Darryl points out, things get trickier when you try to add 0.999\u2026 + 0.999\u2026 . With 10 decimal places, for example, we get\n\n0.9999999999\n+ 0.9999999999\n--------------\n1.9999999998\n\nYou get a carry into each place from the place to its right, but where can you start that? And if there\u2019s an 8 anywhere, then we\u2019re less than 1.999\u2026 = 2, right? The answer, ultimately, is that in fact the 8 is nowhere!\n\nHi Darryl, thanks for your question. It's an interesting twist on the perennial debate about whether 0.999... could really equal 1.\n\nSince that 8 is the nth digit and you take n to infinity, there is no digit 8 in the limit.\n\nWith no rightmost place where you would have 9 + 9 = 18, every digit will be 9. But that\u2019s just impossible to imagine. So we have to do as mathematicians do, and find a way to work around infinities. That way is to use limits formally.\n\nApply the definition of a limit to what you have. We can express it as a game: you pick a number epsilon, as small as you want, and I have to pick a number n such that\n\n|2 - 1.999...98| < epsilon\n\nwhere the 8 is in the nth decimal place. Since\n\n|2 - 1.999...98| = 2*10^(-n)\n\nI just pick\n\nn > -log(epsilon/2)\n\nThis is no harder than in the case of 0.999...9. The 8 in the last place has no effect on the limit; it only delays somewhat the approach to the limit, as indicated by that factor of 1/2.\n\nThe idea here is that we can consider such a sum with any number n of decimal places, which will have an 8 in the last place. We can get this sum to be as close as we want to 2, just by taking enough decimal places; in effect, to get as close as we want, we just have to take enough digits to push the 8 beyond where it would make the error too large. If we want our sum to be less than $$\\epsilon = 0.00000000000000000001$$ away from 2, we just need to use $$n > -\\log(\\epsilon/2) = -\\log(0.00000000000000000001/2) = 20.3$$, so 21 digits will be enough: that is, 0.999999999999999999999 + 0.999999999999999999999 = 1.999999999999999999998.\n\nHere is another way to approach the problem, by manipulating infinite sums. If you want to make everything explicit, you can write the sums as limits of finite sums, and get into the issue of switching the limit and the sum as you go through this process.\n\nLet's write 0.999... as an infinite sum:\n\n0.999... = Sum[n = 1 to infinity](9*10^-n)\n\nNow we can add infinite sums:\n\n0.999... + 0.999...\n= Sum[n = 1 to infinity](9*10^-n)\n+ Sum[n = 1 to infinity](9*10^-n)\n\n= Sum[n = 1 to infinity](18*10^-n)\n\n= Sum[n = 1 to infinity](10*10^-n + 8*10^-n)\n\n= Sum[n = 1 to infinity](10^(-n+1))\n+ Sum[n = 1 to infinity](8*10^-n)\n\n= Sum[n = 0 to infinity](10^-n)\n+ Sum[n = 1 to infinity](8*10^-n)\n\n= 1 + Sum[n = 1 to infinity](10^-n)\n+ Sum[n = 1 to infinity](8*10^-n)\n\n= 1 + Sum[n = 1 to infinity]((8+1)*10^-n)\n\n= 1 + Sum[n = 1 to infinity](9*10^-n)\n\n= 1.999...\n\nI think this is the result you are looking for. Since we know that 0.999... = 1, the answer is 1 + 1 = 2; but you wanted to see that unbroken string of 9's.\n\nHe is adding an infinite string of carries (1) to an infinite string of 8\u2019s, and getting an infinite string of 9\u2019s.\n\nBy the way, notice that the final 8 never appears in this method. That's because we're dealing with infinite sums from the start, so there is always a carry from the next digit to the right. The sum of 10^-n is the sum of the infinite number of carries.\n\n## Multiplying without a place to start\n\nA similar question arises in thinking about the standard method of conversion from a repeating decimal. This question is from 2000:\n\nInduction on .999...\n\nI have been having an argument with my physics teacher over the fact that point nine recurring (.999... or PNR) equals 1. I showed him the proof on your site and he pointed out a fact that I hadn't noticed.\n\nHow can you multiply PNR by ten if you can't get to the beginning? For example, to multiply 215 * 3, you would start off on the right, multiplying 3 by 5, then 1, then finally 2. Now how can you multiply PNR by ten if the farthest right value cannot be reached?\n\nThank you for any clarification you can give.\n\nAs you should recall, we convert 0.999\u2026 to a fraction by subtracting it (x) from 10x:\n\nx = 0.9999...\n10x = 9.9999...\n10x - x = 9.0000...\n9x = 9\nx = 1\n\nHere, we first had to multiply 0.9999\u2026 by 10. But, as in the addition case above, there is no rightmost digit at which to start the multiplication, so can we actually do that multiplication? That is Blake\u2019s teacher\u2019s challenge.\n\nDoctor TWE took it up, first referring to the FAQ (which includes links to several of the answers I\u2019ve referred to) as a source of supplemental arguments before turning to the main issue.\n\nI wanted to comment on your (or your teacher's) idea that \"you can't get to the beginning.\" We multiply values from right to left only as a matter of convenience, so we don't have to \"backtrack\" each time we carry. It is equally valid to multiply from left to right. This is often done when we want to get a quick estimate for the answer, then get more accuracy later. The LEFTMOST digits contribute the most to the answer, so to get \"in the ballpark,\" we can multiply them first. I'll demonstrate with a concrete example, then a general argument.\n\nWe\u2019ve discussed this left-to-right idea in Dividing Right to Left, Adding Left to Right.\n\nLet's use your example of 215 * 3. I can multiply them as follows; first, I'll multiply the 3 by the 2 (actually, by 200) and put the answer in the hundreds place, like this:\n\n215\n*   3\n---\n6\n\nThen I'll multiply the 3 by 1 and put the answer in the tens place:\n\n215\n*   3\n---\n6\n3\n\nFinally, I'll multiply the 3 by 5 and put the answer in the units place:\n\n215\n*   3\n---\n6\n3\n15\n\n215\n*   3\n---\n6\n3\n15\n---\n645\n\nNotice that along the way, I got pretty good estimates for my final answer. After the first step my total was 600 (not a bad estimate, off by less than 10%), after the second step I had 630, and after the third step, I got 645. Going right to left, my totals after each step would be: 15 (not a very good estimate of the final answer), 45 (still not very good) and finally 645.\n\nSo multiplying left to right just requires us to modify the digits we already have to account for carries from new columns as we do them; but the effect of the new column rarely propagates very far back into our work, and never makes a large change. Each new digit makes a smaller adjustment than the one before. That will be relevant as we continue.\n\nIn the general case, consider multiplying a 3-digit value ABC by some value X, where A, B and C are digits. What we really have, then, is\n\n(100*A + 10*B + 1*C) * X\n\nConventionally, we solve this starting with the units digit as:\n\nX*C*1 + X*B*10 + X*A*100\n\nBut using the commutative property of addition, this is equal to:\n\nX*A*100 + X*B*10 + X*C*1\n\nshowing that the order doesn't matter. If I wanted to, I could start in the middle, for example:\n\nX*B*10 + X*A*100 + X*C*1\n\nI'd just have to be careful not to miss any digits or do any ones twice.\n\nWhat we see here is that, conceptually, we can think of the whole multiplication as being done at once; it doesn\u2019t matter in what order we do the pieces, because the sum is commutative.\n\nWithout being able to start with the most significant digit, we could never find a value like 2*pi, because pi (like all irrationals) is an infinite non-repeating decimal. When we say 2*pi is approximately 6.2831853, we can do so because we started multiplying at the end and not the beginning.\n\nOne final note: How can we be sure that the multiplications by 10 continue as we expect (i.e. they continue shifting the digits 1 place to the left) and that the subtractions of successive digits produce zeroes infinitely? These steps can be proven using a technique called mathematical induction. That's too complex to explain here, but if you search our Ask Dr. Math archives for the word \"induction\" (type it without the quotes), you'll find many questions and answers about it.\n\nWhenever we do a calculation on numbers like \u03c0 in a calculator, we are working only with the leftmost digits (those the calculator can hold); but since we know that the digits we have omitted will add no more than a 1 in the rightmost digit by carrying, the answer will be as close as we need. (In fact, calculators actually work with another digit or two beyond what they display, so that such errors never even show up.)\n\n## Proof by induction\n\nThat last teaser \u2013 that there is a formal method for actually proving infinite things without having to do infinite things \u2013 was just what Blake needed. He replied a couple days later (back then we didn\u2019t have threaded conversations, so he didn\u2019t know whether the same Doctor would get the message):\n\nI recently sent you a letter regarding the proof that .999... = 1 in the FAQ. Specifically, I asked how it was possible to multiply PNR by ten when you could not get to the right-hand side of it. This was explained quite neatly.\n\nHowever, the nice person who helped me out said it could be proven that one can multiply PNR by 10 by using a process of induction. I tried to search the archives, but the results I found there were not very satisfactory because it seemed to me that each induction question was actually problem-specific.\n\nTo cut short, I would like to know the induction proof for multiplying PNR by ten.\n\nDoctor TWE first summarized the concept of induction:\n\nHi again Blake! Thanks for writing back!\n\nIn general, we use proof by induction whenever we want a proof that involves an infinite sequence or series, in this case .999...\n\nInductive proofs require two steps:\n\nStep 1 (the basis step): Prove it for some starting value, like n = 1.\n\nStep 2 (the inductive step): Prove that if it's true for n = k, then it is true for n = k+1.\n\nFor the inductive step, we assume that it is true for n = k, and we usually use this assumption in the proof itself.\n\nA word of caution: because of their self-referential nature, inductive proofs are usually hard to follow; it's easy to get lost in the details, losing track of what n, k, and k+1 are supposed to be.\n\nFor other explanations and examples of mathematical induction, see\n\nProof by Mathematical Induction\n\nSum of n Odd Numbers\n\nInductive Misunderstanding\n\nHere is the proof. As he\u2019ll explain, rather than taking n as 1, 2, 3, \u2026, he is taking n as the exponent, -1, -2, -3, \u2026, going in the negative direction and so reversing the usual procedure.\n\nIn our case, we want to show that when multiplying the nth digit by 10, we get a 9 in the (n+1)st digit, no carry to the (n+2)nd digit, and a 0 in the nth digit (counting from the right.) We want no carry to the (n+2)nd digit so that we don't have to adjust the previous (greater place value) digits multiplied because of a \"retroactive carry.\" We want a 0 in the nth digit so that it will not produce further carries when multiplying the next digit (smaller place value) by 10. Mathematically:\n\n10 * (9*10^n) = 9 * 10^(n+1)\n\nWe want to show that:\n\na) the nth digit is 0\nb) the (n+1)st digit is 9\nc) the (n+2)nd digit is 0\n\nTo minimize the confusion, I'm going to use n = -1 as my basis step, and show that if it is true for n = k, then it is true for n = k-1 as my inductive step. This is the \"negative\" of the standard, but it will eliminate the need for using -k's and -(k+1)'s, etc. in my equations.\n\nStep 1: Show that it is the case for n = -1\n\n10 * [9*10^(-1)]  =  9 * 10^(-1+1)\n10 * (9*0.1)  =  9*10^0\n10 * 0.9  =  9*1\n9  =  9\n\na) the -1st digit (the tenths place) is 0: True\nb) the 0th digit (the units place) is 9: True\nc) the 1st digit (the tens place) is 0: True\n\nSo we've proved the basis. Now for the inductive step.\n\nStep 2: Prove that if it's true for n = k, then it's true for n = k-1.\n\nLet's let d = the initial (k+1)st digit. This digit is 0 from conclusion (a) of the previous step. Then:\n\n10 * (9*10^k) + d  =  9 * 10^(k+1)\n9 * (10*10^k) + 0  =  9 * 10^(k+1)\n9 * 10^(k+1)  =  9 * 10^(k+1)\n\na) the kth digit is 0: True - 9*10^(k+1) has a 0 in the kth place.\nb) the (k+1)st digit is 9: True - 9*10^(k+1) has a 9 in the (k+1)st place\nc) the (k+2)nd digit is 0: True - 9*10^(k+1) has a 9 in the (k+2)nd place\n\nSo, if 10 times the kth digit is 9*10^(k+1), then 10 times the (k-1)st digit is 9*10^k, and thus 10 times the (k-2)nd digit is 9*10^(k-1), and thus...\n\nTherefore 10 * 0.999... = 9.999...\n\nQ.E.D.\n\nMore could be done to make this foolproof:\n\nDepending on the skepticism of the intended audience, you may have to use a similar inductive proof for the step:\n\n9.999...\n- 0.999...\n--------\n9.000...\n\nFor most audiences, the following will do:\n\nLet x = 0.999...\nthen 9.999... = 9 + x\n\nso 9.999... - 0.999...  =  (9 + x) - x  = 9\n\nHowever, some audiences (like your teacher, perhaps) will not be satisfied that algebraic operations work on \"infinite strings\" (i.e. they won't be convinced that x - x = 0 when x is an infinite repeating decimal.)\n\nI think some of the other explanations discussed last time may be more useful for certain kinds of skeptics; but it\u2019s good to have a variety of approaches, to meet a variety of objections.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-05-08 17:13:20", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/more-on-0-999/", "openwebmath_score": 0.7628352046012878, "openwebmath_perplexity": 641.1596823095123, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964194566753, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8596952109972292}}
{"url": "https://math.stackexchange.com/questions/2978611/value-of-int-infty-infty-arcsin-frac1-cosh-x-dx", "text": "# value of $\\int_{-\\infty}^{\\infty}\\arcsin\\frac1{\\cosh x}\\,dx$\n\nI want to know the value of $$I=\\int_{-\\infty}^{\\infty}\\arcsin\\frac1{\\cosh x}\\,dx$$ The Symbolab integral calculator says that the integral diverges, but when one graphs it obvious that it converges. So what is the value?\n\nI was thinking that I might try Feynman integration, but I can't think of the right substitution.\n\nFrom the answer provided by @user10354138, we can reach $$\\int\\arcsin\\frac1{\\cosh x}dx=i\\operatorname{Li}_2(i\\phi)-i\\operatorname{Li}_2(-i\\phi)+C$$ Where $$\\phi=\\tan\\bigg(\\frac12\\arcsin\\frac1{\\cosh x}\\bigg)$$ And $$\\operatorname{Li}_2(z)=\\sum_{n\\geq1}\\frac{z^n}{n^2}$$ is the Di-logarithm.\n\nWolfy says it is 4 times the Catalan's constant.\n\nOne (not optimal) way to derive this is \\def\\sech{\\operatorname{sech}} \\begin{align*} \\int_{-\\infty}^\\infty\\arcsin\\sech x\\,\\mathrm{d}x&=2\\int_0^\\infty\\arcsin\\sech x\\,\\mathrm{d}x\\\\ &=2\\int_0^1\\frac{\\arcsin u\\,\\mathrm{d}u}{u\\sqrt{1-u^2}}\\quad(u=\\sech x)\\\\ &=2\\int_0^{\\pi/2}\\frac{\\theta\\,\\mathrm{d}\\theta}{\\sin\\theta}\\quad(u=\\sin\\theta)\\\\ &=2\\int_0^1\\frac{2\\tan^{-1}t\\,\\frac{2\\,\\mathrm{d}t}{1+t^2}}{\\frac{2t}{1+t^2}}\\quad(t=\\tan\\tfrac12\\theta)\\\\ &=4\\int_0^1\\frac{\\tan^{-1}t}{t}\\,\\mathrm{d}t\\\\ &=4\\int_0^1\\sum_{n=0}^\\infty\\frac{(-1)^n}{2n+1}t^{2n}\\,\\mathrm{d}t\\\\ &=4\\sum_{n=0}^\\infty\\frac{(-1)^n}{(2n+1)^2}=4G\\\\ \\end{align*}\n\n\u2022 Dude that's dope! Thank you! \u2013\u00a0clathratus Oct 31 '18 at 3:31\n\u2022 This substitution simplifies to $t=e^{-x}$. \u2013\u00a0J.G. Oct 31 '18 at 8:44\n\u2022 how do you get from $\\frac{\\theta d\\theta}{\\sin\\theta}$ to $$\\frac{2\\arctan(t)\\frac{2dt}{1+t^2}}{\\frac{2t}{1+t^2}}$$ With the substitution $t=\\tan\\frac\\theta2$? \u2013\u00a0clathratus Nov 1 '18 at 23:06\n\u2022 I keep getting $$\\frac{\\theta d\\theta}{\\sin\\theta}=\\frac{2\\arctan t}{t\\sqrt{t^2+1}}dt$$ \u2013\u00a0clathratus Nov 1 '18 at 23:10\n\u2022 @clathratus This is the universal trigonometric substitution \u2013\u00a0user10354138 Nov 1 '18 at 23:25\n\nAlternatively, you can integrate by parts: \\begin{align} \\int \\limits_{-\\infty}^\\infty \\arcsin(\\operatorname{sech}(x)) \\, \\mathrm{d} x &= 2 \\int \\limits_0^\\infty \\arcsin(\\operatorname{sech}(x)) \\, \\mathrm{d} x \\\\ &= 2x \\arcsin(\\operatorname{sech}(x)) \\Bigg \\rvert_{x=0}^{x=\\infty} - 2 \\int \\limits_0^\\infty x \\frac{- \\sinh(x) \\operatorname{sech}^2(x)}{\\sqrt{1-\\operatorname{sech}^2(x)}} \\, \\mathrm{d} x \\\\ &= 2 \\int \\limits_0^\\infty \\frac{x}{\\cosh(x)} \\, \\mathrm{d} x = 4 \\sum \\limits_{n=0}^\\infty (-1)^n \\int \\limits_0^\\infty x \\, \\mathrm{e}^{-(2n+1) x} \\, \\mathrm{d} x \\\\ &= 4 \\Gamma(2) \\sum \\limits_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)^2} = 4 \\mathrm{G} \\, . \\end{align}\n\n\u2022 I'm a fan of the alternative approach. Thank you! \u2013\u00a0clathratus Oct 31 '18 at 14:59\n\nYet another alternative approach: once shown that $$\\int_{\\mathbb{R}}\\frac{dx}{\\cosh(x)^{2k+1}} = \\frac{\\pi \\binom{2k}{k}}{4^k}\\tag{1}$$ and recalled that $$\\arcsin z = \\sum_{k\\geq 0}\\frac{\\binom{2k}{k}}{4^k(2k+1)}z^{2k+1} \\tag{2}$$ we have the following identity: $$\\int_{\\mathbb{R}}\\arcsin\\frac{1}{\\cosh x}\\,dx = \\pi\\sum_{k\\geq 0}\\frac{1}{2k+1}\\left[\\frac{1}{4^k}\\binom{2k}{k}\\right]^2.\\tag{3}$$ Now we may invoke a function whose Maclaurin series involves squared central binomial coefficients, namely the complete elliptic integral of the first kind $$K(x)$$, here denoted according to Mathematica's notation (the argument of $$K$$ is the elliptic modulus): $$\\sum_{k\\geq 0}\\left[\\frac{1}{4^k}\\binom{2k}{k}\\right]^2 x^{2k}=\\frac{2}{\\pi}K(x^2)\\tag{4}$$ leading to: $$\\int_{\\mathbb{R}}\\arcsin\\frac{1}{\\cosh x}\\,dx = 2\\int_{0}^{1} K(x^2)\\,dx = \\int_{0}^{1}\\frac{K(x)}{\\sqrt{x}}\\,dx.\\tag{5}$$ At last, we recall that both $$K(x)$$ and $$\\frac{1}{\\sqrt{x}}$$ have fairly simple Fourier-Legendre series expansions: $$K(x)=2\\sum_{n\\geq 0}\\frac{P_n(2x-1)}{2n+1},\\qquad \\frac{1}{\\sqrt{x}}=2\\sum_{n\\geq 0}(-1)^n P_n(2x-1)$$ and by the orthogonality of shifted Legendre polynomials\n\n$$\\int_{\\mathbb{R}}\\arcsin\\frac{1}{\\cosh x}\\,dx =\\int_{0}^{1}\\frac{K(x)}{\\sqrt{x}}\\,dx = 4\\sum_{n\\geq 0}\\frac{(-1)^n}{(2n+1)^2} = 4G.\\tag{6}$$\n\n\u2022 This is really cool. I'll have to look at this one a little more in order to understand it better... But this is great. Thanks! \u2013\u00a0clathratus Oct 31 '18 at 18:37\n\u2022 You seem to link any special function with any other special function. The link between elliptic integral and Legendre function is pretty smart. +1 \u2013\u00a0Paramanand Singh Nov 1 '18 at 16:41\n\u2022 How do we show $$\\int_{\\Bbb R}\\frac{\\mathrm dx}{\\cosh(x)^{2k+1}}=\\frac{\\pi{2k\\choose k}}{4^k}$$ It looks a lot like a Beta integral \u2013\u00a0clathratus Dec 21 '18 at 0:56\n\u2022 @clathratus: it is, indeed, it is enough to set $\\frac{1}{\\cosh(x)}=u$. \u2013\u00a0Jack D'Aurizio Dec 21 '18 at 10:48", "date": "2019-04-18 14:40:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2978611/value-of-int-infty-infty-arcsin-frac1-cosh-x-dx", "openwebmath_score": 0.9996808767318726, "openwebmath_perplexity": 708.5624130677423, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964203086181, "lm_q2_score": 0.872347369700144, "lm_q1q2_score": 0.8596952101051306}}
{"url": "https://math.stackexchange.com/questions/4081507/applying-gauss-divergence-theorem-to-given-integral", "text": "# Applying Gauss' Divergence Theorem to given integral\n\nWe are given the following two integrals:\n\n$$\\iint\\limits_S D_n f\\:dS$$ and $$\\iiint\\limits_B \\nabla \\cdot (\\nabla f) \\: dV$$\n\nwhere $$S$$ is the portion of the sphere $$x^2+y^2+z^2=a^2$$ in the first octant, $$n$$ is the unit normal vector to $$S$$ at $$(x,y,z)$$ and $$f(x,y,z) = ln(x^2+y^2+z^2)$$\n\nFor the first integral, we have: $$D_n f = \\nabla f \\cdot n$$ and $$n = \\frac{1}{a}$$\n\nSince $$\\nabla f = \\: <\\frac{2x}{x^2+y^2+z^2}, \\frac{2y}{x^2+y^2+z^2}, \\frac{2z}{x^2+y^2+z^2}>$$, this gives $$D_n f = \\nabla f \\cdot n = \\frac{2}{a}$$.\n\nSo the integral becomes\n\n$$\\iint\\limits_S D_nf \\: dS = \\iint\\limits_S \\frac{2}{a} \\: dS = \\frac{2}{a} \\iint\\limits_S dS = \\frac{2}{a} \\cdot A(S) = \\frac{2}{a} \\cdot 4\\pi a^2 \\cdot \\frac{1}{8} = \\pi a$$ ($$\\frac{1}{8}$$ since the sphere is in the first octant)\n\nFor the second integral, we have $$\\nabla \\cdot (\\nabla f) = \\frac{2}{a^2}$$ (since $$x^2 + y^2 + z^2 = a^2$$). So the integral becomes $$\\iiint\\limits_B \\nabla \\cdot (\\nabla f) \\: dV = \\frac{2}{a^2} \\iiint\\limits_B dV = \\frac{2}{a^2} \\cdot V(B) = \\frac{2}{a^2} \\cdot \\frac{4}{3} \\pi a^3 \\cdot \\frac{1}{8} = \\frac{1}{3} \\pi a$$\n\nHowever, according to Gauss' Divergence theorem, these two integrals should be equal to each other right? Since we have $$\\iiint\\limits_B \\nabla \\cdot (\\nabla f) \\: dV = \\iint\\limits_S \\nabla f \\cdot dS = \\iint\\limits_S \\nabla f \\cdot n \\: dS = \\iint\\limits_S D_n f \\: dS$$\n\nSo how is it possible that I get two different answers, even though the Divergence theorem shows that the two integrals should be the same?\n\n\u2022 You're not counting the flat parts of the surface which are quarter circles in the coordinate planes. \u2013\u00a0B. Goddard Mar 29 at 11:09\n\u2022 Oh I see, so if I understand correctly, is it true that the results of the first two integrals are correct, but the application of Gauss' theorem isn't? Because the surface S is not the same as the boundary surface of B, since the boundary surface contains extra flat parts? \u2013\u00a0Stallmp Mar 29 at 11:14\n\nTo apply divergence theorem, you must have a closed surface. So we close the surface by placing $$3$$ quarter disks in plane $$x = 0, y = 0, z = 0$$.\n\nPlease note that when you are doing volume integral, you cannot equate $$\\nabla \\cdot (\\nabla f) = \\displaystyle \\frac{2}{x^2+y^2+z^2} = \\frac{2}{a^2}$$. It should rather be,\n\n$$\\nabla \\cdot (\\nabla f) = \\displaystyle \\frac{2}{x^2+y^2+z^2} = \\frac{2}{\\rho^2}$$\n\nSo the integral becomes,\n\n$$\\displaystyle \\int_0^{\\pi/2} \\int_0^{\\pi/2} \\int_0^a \\frac{2}{\\rho^2} \\ \\cdot\\rho^2 \\cdot \\sin \\phi \\ d\\rho \\ d\\phi \\ d\\theta = \\pi a$$.\n\nNow to find flux through $$S$$, we must subtract flux through planar surfaces $$x = 0, y = 0, z = 0$$ but in this case, they are simply zero.\n\n\u2022 simply because $x^2+y^2+z^2 = a^2$ only on the surface of the sphere, not inside it. We are talking volume integral which includes all points inside the sphere too and so we write $x^2+y^2+z^2 \\leq a^2$. \u2013\u00a0Math Lover Mar 29 at 11:31\n\u2022 Yes you are right about surface $S$. What you get from divergence theorem will include flux through closed surface which is spherical surface $S$ and $3$ planar surfaces as mentioned in my answer. We are supposed to calculate that separately and subtract from the result that we get from divergence theorem to obtain flux only through spherical surface $S$. But in this case they are zero. Do you see why or do you need an explanation? \u2013\u00a0Math Lover Mar 29 at 11:39\n\u2022 Yes this is indeed what I thought, thank you very much for the explanation! I see why they are zero as well when calculating the flux through the planar surfaces. \u2013\u00a0Stallmp Mar 29 at 11:45\n\u2022 Yes you are right, we need to show they sum to zero. Note that $\\iint\\limits_{S_1} D_n f \\: dS = \\iint\\limits_{S_2} D_n f \\: dS = \\iint\\limits_{S_3} D_n f \\: dS = 0$ \u2013\u00a0Math Lover Mar 29 at 12:34\n\u2022 \u2013\u00a0Math Lover Mar 29 at 12:35", "date": "2021-06-23 17:26:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4081507/applying-gauss-divergence-theorem-to-given-integral", "openwebmath_score": 0.9417594075202942, "openwebmath_perplexity": 208.79121913026995, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985496419030704, "lm_q2_score": 0.8723473697001441, "lm_q1q2_score": 0.8596952089903457}}
{"url": "https://math.stackexchange.com/questions/4391113/method-for-extracting-the-nth-root-of-a-number", "text": "# Method for Extracting the $n$th root of a number?\n\nI was wondering if there exists a reliable way to extract the $$n$$th root of a given number. For example, if you have a large perfect square such as 10,404, how would I go about taking its square root by hand?\n\nAnd what about cube roots? Is there a way to find the cube root of any number by hand?\n\nIf there is a way to take the cube and square roots of any number, is there a technique that can be applied to extract the cube or square root of polynomials? What about any root?\n\nI think that, if there is a way to take the $$n$$th root of any number of polynomial, I would benefit from learning it, since I consider being able to work with roots important to my algebra foundation.\n\nThanks\n\n\u2022 For the square root, there is a method similar to the usual division by a number. If we know that the number is a perfect square we can restrict the possibilities (ending digits , residue mod 9). For cube roots and higher , I am not aware of an efficient method by hand. Feb 25 at 18:58\n\u2022 For square roots we can use the algorithm described here under \"Digit-by-digit calculation\" / \"Decimal (base 10)\" Feb 25 at 19:03\n\u2022 @StefanOctavian I've seen similar algorithms for higher roots done (somehow you take 3-digits at a time for cube roots.) But they stopped teaching the square-root extraction the year before I would have had to learn it. Feb 25 at 20:38\n\nIf you would like to compute $$\\sqrt[p]{a}$$, the following iterative process will do it to whatever accuracy you please:\n\nFirst, make a guess and call it $$x_0$$. Then perform this iteration:\n\n$$x_{n+1} = \\frac{(p-1)x_n^p+a}{px^{k-1}}.$$\n\nFor instance, if you want $$\\sqrt[5]{11}$$, guess that $$x_0=2$$. The iteration looks like\n\n$$x_{n+1} = \\frac{4x_n^5+11}{5x_n^4}.$$\n\nSo we have $$x_1 = \\frac{4\\cdot2^5+11}{5\\cdot 2^4} = \\frac{139}{80}=1.7375.$$\n\n$$x_2=\\frac{4\\cdot1.7375^5+11}{5\\cdot 1.7375^4} = 1.631392308.$$\n\nThis number to the 5th power is $$11.55558806$$, so we're getting close.\n\n$$x_3=\\frac{4\\cdot1.631392308^5+11}{5\\cdot 1.631392308^4} = 1.615704970.$$\n\nAnd $$1.615704970^5 = 11.0105\\ldots.$$\n\nThis is an implementation of Newton's Method on the polynomial $$x^p-a$$.", "date": "2022-07-07 08:46:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4391113/method-for-extracting-the-nth-root-of-a-number", "openwebmath_score": 0.8494479060173035, "openwebmath_perplexity": 259.7634815264026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964186047326, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8596952020775964}}
{"url": "https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast", "text": "Related Rates:\n\nGas is escaping from a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? (A sphere of radius r has volume v=4/3 pi r^3 and surface area S=4pi r^2.)\n\nRemember that ds/dt = ds/dr X dr/dt\n\nStep 1: Find ds/dr\n\nI have no clue where to start. I was going to set 2 equal to 4 pi r^2 and then take the derivative but I really have no idea.\n\nStep 2: Find dr/dr. (HInt dv/dt= dv/dr X dr/dt\n\nWould I set 2 equal to volume and then take derivative?\n\nStep 3: Find ds/dt?\n\nStep 4: Evaluate ds/dt when the radius is 12ft.\nAfter I find my equation I would just plug in 12 correct?\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. S = 4pi*r^2\ndS/dr = 4pi*2r = 8pi*r\n\nV = (4/3)pi*r^3\ndV/dr = 4pi*r^2\ndV/dt = dV/dr * dr/dt\n2 = 4pi*r^2 * dr/dt\ndr/dt = 1/(2pi*r^2)\n\ndS/dt = dS/dr * dr/dt\n= 8pi*r * 1/(2pi*r^2)\n= 4/r\nIf r = 12, just plug the value to the last equation\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n2. How did you get 12?\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. Nevermind I meant how did you get 4/r but I figured it out. Thank You!!\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n4. Is this correct for the last part?\n\n(8)(3.14)(12) X (1/(2)(3.14)(12^2) = 4/12\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n5. Yes\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n6. Ok do I actually have to do the calculations though?\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n7. If you are not given the value, that means you have to write the steps I told you until the very last equation.\nIf you are given the value (of r), you don't have to put the value in every equation. Just follow my steps until the last equation, then plug the value after that. It's easier that way.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n8. OK thank you!\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n9. Simplify the answer (4/12) into 1/3\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\n## Similar Questions\n\n1. ### calculas\n\ni didnot under stand the question ? need help Gas is escaping a spherical balloon at the rate of 4 cm3 per minute. How fast is the surface area of the balloon shrinking when the radius of the balloon is 24 cm? Given volume of\n\n2. ### Math\n\nA spherical balloon is to be deflated so that its radius decreases at a constant rate of 12 cm/min. At what rate must air be removed when the radius is 5 cm? Must be accurate to the 5th decimal place. It seems like an easy\n\n3. ### Word Problem-Calculus\n\nA spherical balloon is being inflated in such a way that its radius increases at a rate of 3 cm/min. If the volume of the balloon is 0 at time 0, at what rate is the volume increasing after 5 minutes? my answer is 45 cm/min. is\n\n4. ### Calculus\n\na spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. how fast is the radius of the balloon increasing at the instant the radius is 30 centimeters?\n\n1. ### Calculus\n\nA spherical balloon is inflated at a rate of 10 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is 4 feet? And The radius of a circle is decreasing at a rate of 2 ft/minute. Find the\n\n2. ### help on calculus\n\nA spherical balloon is being filled with air in such a way that its radius is increasing at a rate of 2 centimeters per second. At what rate is the volume of the balloon changing at the instant when its surface has an area of 4 pi\n\n3. ### Calculus\n\nAir is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm? V= 4/3*pi*r^3 S= 4 pi r^2\n\n4. ### calculus help thanks!\n\nThe gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at\n\n1. ### Math\n\nA spherical balloon is inflated with helium at the rate of 100pi ft ^3/min. How fast is the ballon's radius increasing at the instant the radius is 5ft.?\n\n2. ### Mathematics\n\nA spherical balloon is blown up so that it's volume increases constantly at the rate of 2cm^3 per second .find the rate of the increase of the radius when the volume of the balloon is 50cm^3\n\n3. ### defferential calculus\n\nA balloon is rising vertically 150 m from an observer. At exactly 1 min, the angle of elevation is 29 deg 28 min. How fast is the balloon rising at that instant?\n\n4. ### Calculus\n\nA spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2cm/min. At what rates are the volume and surface area of the balloon increasing when the radius is 5cm?", "date": "2022-01-17 02:07:18", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast", "openwebmath_score": 0.8956266641616821, "openwebmath_perplexity": 786.0660610559516, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877692486438, "lm_q2_score": 0.8856314798554445, "lm_q1q2_score": 0.8596716455572566}}
{"url": "https://math.stackexchange.com/questions/2829529/relation-between-bigcap-i-in-ia-i-and-bigcap-i-1na-i", "text": "# Relation between $\\bigcap_{i \\in I}A_i$ and $\\bigcap_{i=1}^{n}A_i$\n\nLet I be a nonempty set and a family of sets such that every element of the family is a subset of U.\n\n$\\mathcal F = \\{A_i | i \\in I\\}$\n\nI understand the meaning of this operation:\n\n$$\\bigcap_{i \\in I}A_i$$\n\nThat's the intersection of all the elements of the sets of the family $\\mathcal F$.\n\nBut I don't truly understand what this other operation means and what's the relation between the above one. $$\\bigcap_{i=1}^{n}A_i$$\n\nI understand that this operation is also an intersection but what I am trying to understand is the relation between $\\bigcap_{i \\in I}A_i$ and $\\bigcap_{i=1}^{n}A_i$\n\n$$\\bigcap_{i=1}^{n}A_i \\subseteq\\bigcap_{i \\in I}A_i$$\n\nIs the relation above true?\n\n\u2022 In the second case we have $I = \\{ 1,2,\\ldots, n \\}$. \u2013\u00a0Mauro ALLEGRANZA Jun 23 '18 at 15:28\n\nThe notation $$\\bigcap _{i=1}^nA_i$$ denotes the same as $$\\bigcap_{i\\in I}A_i$$ for the special case $I=\\{1,2,\\ldots, n\\}$.\n\u2022 So $\\bigcap_{i\\in I}A_i$ is the general case and $\\bigcap _{i=1}^nA_i$ is a specific case for which we give concrete values to i? \u2013\u00a0adriana634 Jun 23 '18 at 15:40\n\u2022 @adriana634 Yes. One situation where we see both notations used simultaneously is when $I=\\mathbb{N}$; then \"$\\bigcap_{i=1}^nA_i$\" means \"$\\bigcap_{i\\in \\{1,2,..., n\\}}A_i$,\" which is generally a superset of $\\bigcap_{i\\in I}A_i$. \u2013\u00a0Noah Schweber Jun 23 '18 at 15:42\n\u2022 @Noah Schweber But for example $\\bigcap_{i=1}^{5}A_i$ means $I=\\{1,2,\\ldots, 5\\}$, in this example $\\bigcap_{i\\in I}A_i$ is a superset of $\\bigcap_{i=1}^{5}A_i$? I don't understand why you said $\\bigcap_{i=1}^nA_i$ is a superset of $\\bigcap_{i\\in I}A_i$. \u2013\u00a0adriana634 Jun 23 '18 at 16:00\n\u2022 @adriana634 No. The intersection over a bigger index set yields a smaller set. E.g. if we take $I=\\mathbb{N}$ and $A_i=\\{i, i+1, i+2, ...\\}$ for $i\\in I$, then $\\bigcap_{i=1}^5A_i=\\{5, 6,7,...\\}$ but $\\bigcap_{i\\in I}A_i=\\emptyset$. More abstractly: $$I\\supseteq J\\implies \\bigcap_{i\\in I}A_i\\subseteq \\bigcap_{i\\in J}A_i.$$ \u2013\u00a0Noah Schweber Jun 23 '18 at 16:02", "date": "2020-01-26 22:03:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2829529/relation-between-bigcap-i-in-ia-i-and-bigcap-i-1na-i", "openwebmath_score": 0.9462869763374329, "openwebmath_perplexity": 211.5068487325436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9706877717925422, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8596716316944107}}
{"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 21 Nov 2018, 02:01\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in November\nPrevNext\nSuMoTuWeThFrSa\n28293031123\n45678910\n11121314151617\n18192021222324\n2526272829301\nOpen Detailed Calendar\n\u2022 ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!\n\nNovember 22, 2018\n\nNovember 22, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nMark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)\n\u2022 ### Free lesson on number properties\n\nNovember 23, 2018\n\nNovember 23, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nPractice the one most important Quant section - Integer properties, and rapidly improve your skills.\n\n# In the addition shown above, A, B, C, and D represent\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 09 Nov 2012\nPosts: 63\nIn the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2013, 07:17\n7\n41\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n42% (03:00) correct 58% (02:47) wrong based on 776 sessions\n\n### HideShow timer Statistics\n\nABC\n+BCB\nCDD\n\nIn the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?\n\nA. 8\nB. 10\nC. 12\nD. 14\nE. 18\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4488\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n15 Oct 2013, 10:47\n17\n12\nsaintforlife wrote:\nABC\n+BCB\nCDD\n\nIn the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?\n\nA. 8\nB. 10\nC. 12\nD. 14\nE. 18\n\nI'm happy to help with this.\nFirst, look at the one's column. C + B produces a one's unit of D, and we don't know whether anything carries to the ten's place.\nBut now, look at the ten's place --- B + C again produces a digit of D --- that tells us definitively that nothing carried, and that B + C = D. B & C have a single digit number as a sum.\nNow, C = A + B, because again, in the hundreds column, nothing carries here.\n\nNotice that, since C = A + B and D = B + C, B must be smaller than both B and C. We want to make the product of A & B large, so we want to make those individual digits large. Well, if we make B large, then that makes C large, and then B + C would quickly become more than a one-digit sum, which is not allowed. Think about it this way. Let's just assume D = 9, the maximum value.\nD = 9 = B + C = B + (A + B) = A + 2B\nWe want to pick A & B such that A + 2B = 9 and A*B is a maximum. It makes sense that B would be smaller.\nTry A = 7, B = 1. Then A + 2B = 9 and A*B = 7\nTry A = 5, B = 2. Then A + 2B = 9 and A*B = 10\nTry A = 3, B = 3. Then A + 2B = 9 and A*B = 9\nTry A = 1, B = 4. Then A + 2B = 9 and A*B = 4\nIndeed, as B gets bigger, the product gets less. This seems to imply that the biggest possible product is 10. This corresponds to A = 5, B = 2, C = 7, and D = 9, and the original addition problem becomes\n527\n+272\n799\nThus, the maximum product is 10, and answer = (B).\n\nDoes all this make sense?\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\nCurrent Student\nJoined: 31 Mar 2013\nPosts: 66\nLocation: India\nGPA: 3.02\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n06 Dec 2013, 10:19\n8\n1\nsaintforlife wrote:\nHow do we do all that in less than 2 mins, is my only question :)\n\nPosted from my mobile device\n\nThis problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.\n\nOption E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.\n\nOption D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.\n\nOption C is 12. This will have 2 cases-\n$$Case 1--> 12 = 4 * 3$$\n\nHere $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.\n\n(OR) $$Case 2--> 6 * 2$$.\n\nHere $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.\n\nOption B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!\n\n##### General Discussion\nManager\nJoined: 09 Nov 2012\nPosts: 63\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n23 Oct 2013, 15:44\n2\nHow do we do all that in less than 2 mins, is my only question\n\nPosted from my mobile device\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4488\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n23 Oct 2013, 15:54\n6\nsaintforlife wrote:\nHow do we do all that in less than 2 mins, is my only question\n\nDear Saint For Life,\nProblems such as this are quite out-of-the-box. To some extent, the GMAT intends us to handle many of the other questions in 90 secs or less, so that we have a bit of a time-cushion when we run into one of these oddball questions.\n\nHaving said that, the more familiar you are with number properties, the faster it will go. In this problem, it took me quite some time to write out everything in verbal form, but I saw things relatively quickly. As you practice seeing patterns, you will see them more quickly, even if they are hard to explain to someone else. You may find this post helpful:\nhttp://magoosh.com/gmat/2013/how-to-do- ... th-faster/\n\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50727\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n23 Oct 2013, 23:13\n5\n2\nsaintforlife wrote:\nABC\n+BCB\nCDD\n\nIn the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?\n\nA. 8\nB. 10\nC. 12\nD. 14\nE. 18\n\nSimilar questions to practice:\ntough-tricky-set-of-problems-85211.html#p638336\n\nHope it helps\n_________________\nIntern\nJoined: 16 Jun 2013\nPosts: 14\nGMAT 1: 540 Q34 V30\nGMAT 2: 700 Q43 V42\nRe: In the addition shown above, A, B, C, and D represent the\u00a0 [#permalink]\n\n### Show Tags\n\n26 Oct 2013, 21:47\n2\n... got E as an answer. However, it took me 6 minutes, since I plugged in the answer options.\nIs there any faster way, or am I just too slow in back-solving?\n\nTirthankarP wrote:\n\nIn the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?\nA) 8\nB) 10\nC) 12\nD) 14\nE) 18\n\nMy explanation:\n\nAccording to the question stem C+B=D and D<10, since the second addition (2nd column) is C+B=D again (if there were a carry, the second column's result would not be D again), and A+B=C\nNow I break up the answer options:\nA) A*B=8, so A and B -> 2 and 4 or 1 and 8. A+B=2+4=6=C, C+B=6+2=8=D, which is smaller than 10, thus the answer is not \"out\", but there could be a larger possible value of A*B\nB) A*B=10, so A and B -> 2 and 5. A+B=2+5=7=C, C+B=7+2=9=D, which is smaller than 10, thus the is also possible. Let's look for the next options.\nFor C-E:\nC) A*B=12, so A and B -> 6 and or 4 and 3.\nD) A*B=14, so A and -> 7 and 2.\nE) A*B=18, so A and B -> 9 and 2 or 3 and 6.\nDo the same process with those numbers and you will find that all will yield a sum of C+B>10, thus the constraint of D<10 is not satisfied. The answer options are not possible.\nBecause 10>8 answer option B) is the largest possible.\nManager\nJoined: 29 Apr 2013\nPosts: 94\nLocation: India\nConcentration: General Management, Strategy\nGMAT Date: 11-06-2013\nWE: Programming (Telecommunications)\nRe: In the addition shown above, A, B, C, and D represent the\u00a0 [#permalink]\n\n### Show Tags\n\n27 Oct 2013, 00:04\nChiranjeevee wrote:\nTirthankarP wrote:\nAttachment:\n\nIn the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?\nA) 8\nB) 10\nC) 12\nD) 14\nE) 18\n\nI do a thorough search before posting any new question. But this time I couldn't find this link\nForum moderator may delete this thread\n_________________\n\nDo not forget to hit the Kudos button on your left if you find my post helpful\n\nCollection of some good questions on Number System\n\nRetired Moderator\nJoined: 18 Sep 2014\nPosts: 1127\nLocation: India\nIn the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n01 Jul 2015, 07:02\nHKHR wrote:\nsaintforlife wrote:\nHow do we do all that in less than 2 mins, is my only question :)\n\nPosted from my mobile device\n\nThis problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.\n\nOption E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.\n\nOption D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.\n\nOption C is 12. This will have 2 cases-\n$$Case 1--> 12 = 4 * 3$$\n\nHere $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.\n\n(OR) $$Case 2--> 6 * 2$$.\n\nHere $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.\n\nOption B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!\n\nWonderful explanation. Just thought to add some more from my side. As we see only B satisfies all the necessary requirement(not to carry over any digits).\n\nIn option A, taking A=4 and B=2 also satisfies the condition as it is something like\n426\n+262\n-------\n688\n\nBut since we need maximum value well go for option B\nIntern\nJoined: 26 May 2017\nPosts: 4\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n30 Nov 2017, 18:03\nDear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4488\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n01 Dec 2017, 13:54\n1\n1\nSarthak.bhatt wrote:\nDear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help\n\nDear Sarthak.bhatt,\n\nI'm happy to respond.\n\nWith all due respect, my friend, you are not interpreting the question correctly. There is absolutely no multiplication happening in this question. Here's the prompt again:\n\nABC\n+BCB\nCDD\n\nIn the addition shown above, A, B, C, and D represent\nthe nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?\n\nThus, for example if C = 1, B = 5, and A = 4, then ABC would not be the product of those three numbers but instead the single three-digit number 451. With those numbers, the problem would be\n\n451\n+515\n966\n\nThese choices satisfy the equation.\n\nThe problem is completely different from the way you were conceptualizing it, so the strategy is completely different from what it would be in the problem you had in mind.\n\nDoes this make sense?\n\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\nIntern\nJoined: 21 Sep 2016\nPosts: 4\nLocation: Saudi Arabia\nConcentration: Finance, Entrepreneurship\nWE: Corporate Finance (Consumer Electronics)\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n26 May 2018, 03:22\nmikemcgarry. Can you please explain Sarthak's query as i have the same query but unfortunately i didn't get from the reply you gave to him.\nIntern\nJoined: 11 May 2018\nPosts: 5\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n15 Jun 2018, 15:24\nDo we not consider negative numbers in this problem at all?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50727\nRe: In the addition shown above, A, B, C, and D represent\u00a0 [#permalink]\n\n### Show Tags\n\n15 Jun 2018, 21:55\nrohithtv89 wrote:\nDo we not consider negative numbers in this problem at all?\n\nNo. We are given that A, B, C, and D represent the nonzero digits. There are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.\n_________________\nRe: In the addition shown above, A, B, C, and D represent &nbs [#permalink] 15 Jun 2018, 21:55\nDisplay posts from previous: Sort by", "date": "2018-11-21 10:01:25", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html", "openwebmath_score": 0.7710837125778198, "openwebmath_perplexity": 1315.5501724991552, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8596637612961505, "lm_q1q2_score": 0.8596637612961505}}
{"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Aug 2018, 14:16\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# A photography dealer ordered 60 Model X cameras to be sold\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 26 Jul 2010\nPosts: 24\nA photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n06 May 2012, 10:53\n21\n129\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n58% (03:01) correct 42% (03:26) wrong based on 2390 sessions\n\n### HideShow timer Statistics\n\nA photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit ##### Most Helpful Expert Reply Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8195 Location: Pune, India Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 22:20 61 38 pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras?\n\nA. 7% loss\nB. 13% loss\nC. 7% profit\nD. 13% profit\nE. 15% profit\n\nUse weighted avgs for a quick solution:\n\nOn 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1\n\nAvg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit\n\nFor more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nSave up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### Most Helpful Community Reply Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 230 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags Updated on: 25 Feb 2016, 08:28 98 27 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10 and each costs$10\ntotal profit = $2*9 -$5*1 = $13 or 13% (with$100 cost)\n_________________\n\npress +1 Kudos to appreciate posts\n\nOriginally posted by MBAhereIcome on 23 Jun 2012, 07:31.\nLast edited by MBAhereIcome on 25 Feb 2016, 08:28, edited 1 time in total.\n##### General Discussion\nMath Expert\nJoined: 02 Sep 2009\nPosts: 48037\nRe: A photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n06 May 2012, 11:07\n25\n1\n24\npgmat wrote:\nA photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;\n\n# of cameras sold is 60-6=54 total revenue is 54*250;\n# of cameras returned is 6 total refund 6*(250/1.2)*0.5;\nSo, total income 54*250+ 6*(250/1.2)*0.5\n\nThe dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%\n\n_________________\nManager\nJoined: 02 Jun 2011\nPosts: 132\nRe: A photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jun 2012, 14:14\nBunuel wrote:\npgmat wrote:\nA photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost$60*250/1.2=50*250;\n\n# of cameras sold is 60-6=54 total revenue is 54*250;\n# of cameras returned is 6 total refund 6*(250/1.2)*0.5;\nSo, total income 54*250+ 6*(250/1.2)*0.5\n\nThe dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%\n\nDear Bunuel,\nThs is a doubt i face with %ages, in ths quest too.\ncould you pls tell where i m going wrong?\nwhen 20% mark up for initial cost is given , how to calculate it?\n250 - 20/100*250 or to take if 120 is 250 then how much is 100?\nhow have you arrived at 1.2?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 48037\nRe: A photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n23 Jun 2012, 04:06\n3\n2\nkashishh wrote:\nBunuel wrote:\npgmat wrote:\nA photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*$250/1.2=50*250;\n\n# of cameras sold is 60-6=54 total revenue is 54*250;\n# of cameras returned is 6 total refund 6*(250/1.2)*0.5;\nSo, total income 54*250+ 6*(250/1.2)*0.5\n\nThe dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%\n\nDear Bunuel,\nThs is a doubt i face with %ages, in ths quest too.\ncould you pls tell where i m going wrong?\nwhen 20% mark up for initial cost is given , how to calculate it?\n250 - 20/100*250 or to take if 120 is 250 then how much is 100?\nhow have you arrived at 1.2?\n\nIf it's given that the selling price is $250 and the markup over the initial cost is 20%, then: {Cost}+0.2{Cost}=1.2*{Cost}={Selling price} --> 1.2*{Cost}=$250 --> {Cost}=$250/1.2. Hope it's clear. _________________ Intern Joined: 28 Feb 2012 Posts: 21 GMAT 1: 700 Q48 V39 WE: Information Technology (Computer Software) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 02:06 1 MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. Manager Joined: 02 Jun 2011 Posts: 132 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 12:26 1 2 gmatDeep wrote: MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. yess.. 54:6 = 9:1 mark up is of 20% = profit earned will be 20% VP Joined: 02 Jul 2012 Posts: 1192 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:15 15 4 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera.\nOf the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?\n\n(A) 7% loss\n\n(B) 13% loss\n\n(C) 7% profit\n\n(D) 13% profit\n\n(E) 15% profit\n\nhave fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.\n\nI would not show the OA but the rules are stringent. either way try on your own\n\nCost to dealer for 60 cameras = $$\\frac{250}{1.2}*60$$ = $12,500 Revenue for 54 cameras = 54*250 =$13,500\n\nRevenue from 6 cameras = $$\\frac{250}{2.4}*6$$ = $625 Total Revenue =$14,125\n\nProfit percent = $$\\frac{14125-12500}{12500}$$ = 13%\n\nOtherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E.\n\nKudos Please... If my post helped.\n_________________\n\nDid you find this post helpful?... Please let me know through the Kudos button.\n\nThanks To The Almighty - My GMAT Debrief\n\nGMAT Reading Comprehension: 7 Most Common Passage Types\n\nVP\nJoined: 24 Jul 2011\nPosts: 1459\nGMAT 1: 780 Q51 V48\nGRE 1: Q800 V740\nRe: Of the cameras ordered, 6 were never sold and were returned\u00a0 [#permalink]\n\n### Show Tags\n\n16 Nov 2012, 23:27\n1\n60 cameras were ordered to be sold at a 20% markup for $250 each => Dealer's cost for each camera =$208\n=> Cost of all 60 cameras = $60*208 6 cameras were not sold and returned for a refund of 50% of the dealer's cost => The amount the dealer got in returns = 50% of 6*208 =$624\nThe rest of the cameras (60-6 = 54 in number) were sold\n=> Revenue made from selling the cameras = $54 * 250 For the 60 cameras, dealer's profit = 100* [54*(250-208) + (624-1248)]/[(60*200)] =100* (14100-12000)/12000 = ~13% (the error arises because 250/1.2 is not exactly 208 but 208.33) Option (D) _________________ GyanOne | Top MBA Rankings and MBA Admissions Blog Top MBA Admissions Consulting | Top MiM Admissions Consulting Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738 Director Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 616 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:48 24 13 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera.\nOf the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?\n(A) 7% loss\n(B) 13% loss\n(C) 7% profit\n(D) 13% profit\n(E) 15% profit\nhave fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.\nI would not show the OA but the rules are stringent. either way try on your own\n\nactually, if one could visualize this problem properly, it could be solved in less than 30 secs without getting into any dirty calculations.\nHere is my 30 sec approach:\nNotice that it is actually a wighted ratio problem, where 9 parts earned 20% profit and 1 part earned 50% loss. (54 for profit, 6 for loss).\nHence overall profit/loss = $$(9*20 -1*50) /(9+1) = 13$$\nAns D it is!\n_________________\n\nLets Kudos!!!\nBlack Friday Debrief\n\nIntern\nJoined: 16 Nov 2009\nPosts: 7\nRe: A photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n24 Nov 2012, 11:16\n2\nHi,\n\nFor those using normal way to solve this problem, I think that we can avoid huge calculation by just replace Cost by letter C and put it in the final formula:\nIncome: 54*1.2C + 6*0.5C\nCost: 60*C\nThen percent =(Income-cost)/cost = (54*1.2C + 6*0.5C - 60*C)/60*C. We can see that we can eliminate C from the fraction --> percent = (54*1.2 + 6*0.5 - 60)/60 = (54*1.2 + 6*0.5)/60 - 1. Here, we can see clearly the weighted average part of the problem or we can simply calculate the fraction to reach the result.\nIntern\nStatus: K... M. G...\nJoined: 22 Oct 2012\nPosts: 36\nGMAT Date: 08-27-2013\nGPA: 3.8\nRe: A photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n27 May 2013, 12:12\nBunuel wrote:\npgmat wrote:\nA photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;\n\n# of cameras sold is 60-6=54 total revenue is 54*250;\n# of cameras returned is 6 total refund 6*(250/1.2)*0.5;\nSo, total income 54*250+ 6*(250/1.2)*0.5\n\nThe dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%\n\nthis must be kind of awkward question but i got no option other than asking you.\n\n20% markup over the dealer\u2019s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that \"Total cost 60*($250/1.2)=50*250;\" Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2013, 12:19 3 1 FTGNGU wrote: Bunuel wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras?\n\nA. 7% loss\nB. 13% loss\nC. 7% profit\nD. 13% profit\nE. 15% profit\n\nTotal cost 60*($250/1.2)=50*250; # of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. this must be kind of awkward question but i got no option other than asking you. 20% markup over the dealer\u2019s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that \"Total cost 60*($250/1.2)=50*250;\"\n\n(Cost per unit) + 0.2*(Cost per unit) = $250 1.2*(Cost per unit) =$250\n(Cost per unit) = $250/1.2 Total cost for 60 units = 60*(Cost per unit) = 60*($250/1.2) = 50*250.\n\nHope it's clear.\n_________________\nIntern\nJoined: 02 Mar 2010\nPosts: 19\nRe: A photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n07 Apr 2014, 19:23\n4\nBecause the answers are in percentages, I thought not to worry about the $amounts and just focus on the relationships: Let C be the total cost of all 60 cameras. Originally, the dealer thought to sell all these 60 cameras at 20% profit or for (1.2)C. However, he sold only (60-6)=54 or 90% of cameras at this price. So revenue from these cameras = (0.9)(1.2)C = (1.08)C For the remaining 10%, he got a refund of 50% of cost or (0.5)C. So total refund = (0.1)(0.5)C = (0.05)C Therefore, total revenue in terms of original cost = (1.08 + 0.05)C = 1.13C or 13% profit. So D is the correct ans. Intern Status: 1st attempt: Can I do it? Joined: 09 Jun 2013 Posts: 1 Location: India Concentration: Healthcare, Marketing GMAT Date: 05-30-2014 GPA: 2.99 WE: Pharmaceuticals (Pharmaceuticals and Biotech) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2014, 10:33 2 1 {(54*20) + (6*(-50))}/ 60 This gives 13% profit as answer. 20% proft on 54 cameras and 50% loss on 6 cameras. Use weighted averages here. Rest of the information in the question is misguiding. Intern Joined: 09 Feb 2013 Posts: 26 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 29 Jun 2014, 23:52 My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 30 Jun 2014, 02:50 kshitij89 wrote: My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device The cost per unit is not 0.8*250 = 200, it's 250/1.2 = ~208. Markup is calculated on the cost value (check here: http://gmatclub.com/forum/a-photography ... l#p1229596). Hope it helps. _________________ Manager Joined: 28 Dec 2013 Posts: 68 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 21 Aug 2014, 08:48 VeritasPrepKarishma wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras?\n\nA. 7% loss\nB. 13% loss\nC. 7% profit\nD. 13% profit\nE. 15% profit\n\nUse weighted avgs for a quick solution:\n\nOn 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1\n\nAvg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit\n\nFor more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/\n\nwhy are we dividing by 10?\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8195\nLocation: Pune, India\nRe: A photography dealer ordered 60 Model X cameras to be sold\u00a0 [#permalink]\n\n### Show Tags\n\n24 Aug 2014, 21:54\n1\nsagnik242 wrote:\nVeritasPrepKarishma wrote:\npgmat wrote:\nA photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer\u2019s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer\u2019s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Use weighted avgs for a quick solution: On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/ why are we dividing by 10? Weighted Average Formula: Cavg = (C1*w1 + C2*w2)/(w1 + w2) C1 and C2 represent the quantity which we want to average so they will be profit/loss here. C1 = 20% = .2 C2 = -50% (loss) = -.5 The weights, given by w1 and w2, are the cost prices. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 So w1 =9 and w2 = 1 Avg Profit/Loss = (.2*9 + (-.5)*1)/(9 + 1) = (.2*9 + (-.5)*1)/10 Get more details on this concept from the link given in my post above. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nRe: A photography dealer ordered 60 Model X cameras to be sold &nbs [#permalink] 24 Aug 2014, 21:54\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0\u00a0\u00a03\u00a0 \u00a0 Next \u00a0[ 57 posts ]\n\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-08-19 21:16:55", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html", "openwebmath_score": 0.5564415454864502, "openwebmath_perplexity": 8908.812152197419, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. 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{"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar", "text": "It is currently 20 Nov 2017, 23:39\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# If x is an integer then x(x-1)(x-k) must be evenly divisible\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 532\n\nKudos [?]: 4216 [1], given: 217\n\nLocation: United Kingdom\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nIf x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n31 Jan 2012, 15:49\n1\nKUDOS\n50\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n64% (01:19) correct 36% (01:25) wrong based on 1112 sessions\n\n### HideShow timer Statistics\n\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\n\nA. -4\nB. -2\nC. -1\nD. 2\nE. 5\n\n[Reveal] Spoiler:\nThe OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?\n\nQuestion says x(x \u2013 1)(x \u2013 k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.\n[Reveal] Spoiler: OA\n\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\nKudos [?]: 4216 [1], given: 217\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42275\n\nKudos [?]: 132868 [13], given: 12389\n\n### Show Tags\n\n31 Jan 2012, 16:06\n13\nKUDOS\nExpert's post\n31\nThis post was\nBOOKMARKED\nenigma123 wrote:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\n\nA)-4\nB)-2\nC)-1\nD) 2\nE) 5\n\nThe OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?\n\nQuestion says x(x \u2013 1)(x \u2013 k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.\n\nWe have the product of 3 integers: (x-1)x(x-k).\n\nNote that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x \u2013 1), and (x \u2013 k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.\n\nNext, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.\n\n30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.\n\nHope it helps.\n_________________\n\nKudos [?]: 132868 [13], given: 12389\n\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 532\n\nKudos [?]: 4216 [0], given: 217\n\nLocation: United Kingdom\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n31 Jan 2012, 16:09\nThanks very much for a thorough explanation.\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\nKudos [?]: 4216 [0], given: 217\n\nMagoosh GMAT Instructor\nJoined: 28 Dec 2011\nPosts: 4491\n\nKudos [?]: 8757 [15], given: 105\n\n### Show Tags\n\n31 Jan 2012, 16:17\n15\nKUDOS\nExpert's post\n10\nThis post was\nBOOKMARKED\nHi, there. I'm happy to help with this.\n\nThe rule that the product of three consecutive integers is a good start, but not the be all and end all.\n\nnumber = x(x \u2013 1)(x \u2013 k)\n\nSo far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --\n\nk = 2 ----> x(x \u2013 1)(x \u2013 2)\n\nk = -1 ----> x(x \u2013 1)(x + 1)\n\nNow, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.\n\n(x - 2) - 3 = (x - 5)\n\n(x - 5) - 3 = (x - 8)\n\n(x - 2) + 3 = (x + 1), which we have already\n\n(x + 1) + 3 = (x + 4)\n\n(x + 3) + 3 = (x + 7)\n\nSo, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.\n\nBack to the question:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\nA)-4\nB)-2\nC)-1\nD) 2\nE) 5\n\nAll of those choices give us a term on our list except for (B) -2.\n\nBTW, notice all the answer choices are spaced apart by three except for (B).\n\nDoes that make sense? Please do not hesitate to ask if you have any questions.\n\nMike\n_________________\n\nMike McGarry\nMagoosh Test Prep\n\nEducation is not the filling of a pail, but the lighting of a fire. \u2014 William Butler Yeats (1865 \u2013 1939)\n\nKudos [?]: 8757 [15], given: 105\n\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 532\n\nKudos [?]: 4216 [0], given: 217\n\nLocation: United Kingdom\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n31 Jan 2012, 16:18\nThanks Mike. Really appreciate your solution too.\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\nKudos [?]: 4216 [0], given: 217\n\nIntern\nJoined: 31 Jan 2012\nPosts: 1\n\nKudos [?]: 4 [4], given: 0\n\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n31 Jan 2012, 16:34\n4\nKUDOS\nThis is my 1st post finally thought of jumping in instead of just being an observer\nI attacked this problem in a simple way. As it states it is divisible by 3\nthat means both x & (x-1) cannot be a multiple of 3 otherwise whatever the value of k it will be still divisible by 3\nso plugging in number i chose 5 in this case you can establish answer is -2 does not fit...\n\nKudos [?]: 4 [4], given: 0\n\nDirector\nStatus: Finally Done. Admitted in Kellogg for 2015 intake\nJoined: 25 Jun 2011\nPosts: 532\n\nKudos [?]: 4216 [0], given: 217\n\nLocation: United Kingdom\nGMAT 1: 730 Q49 V45\nGPA: 2.9\nWE: Information Technology (Consulting)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n31 Jan 2012, 16:37\nThanks and welcome to GMAT Club Azim. I am sure you will have a great experience from a very helpful community.\n_________________\n\nBest Regards,\nE.\n\nMGMAT 1 --> 530\nMGMAT 2--> 640\nMGMAT 3 ---> 610\nGMAT ==> 730\n\nKudos [?]: 4216 [0], given: 217\n\nIntern\nJoined: 31 May 2012\nPosts: 11\n\nKudos [?]: 14 [0], given: 8\n\n### Show Tags\n\n04 Oct 2012, 18:59\nBunuel wrote:\nenigma123 wrote:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\n\nA)-4\nB)-2\nC)-1\nD) 2\nE) 5\n\nThe OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?\n\nQuestion says x(x \u2013 1)(x \u2013 k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.\n\nWe have the product of 3 integers: (x-1)x(x-k).\n\nNote that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x \u2013 1), and (x \u2013 k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.\n\nNext, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.\n\n30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.\n\nHope it helps.\n\nBunuel,\n\nWould this approach work for all integer divisors (NOT the 30 sec approach)?\nSay the divisor is 4 and the choices had four terms instead of three, e.g. (x-1)(x-2)(x+k)(x+1)?\n\nThanks,\n\nKudos [?]: 14 [0], given: 8\n\nDirector\nJoined: 22 Mar 2011\nPosts: 610\n\nKudos [?]: 1074 [8], given: 43\n\nWE: Science (Education)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n04 Oct 2012, 23:10\n8\nKUDOS\n2\nThis post was\nBOOKMARKED\nenigma123 wrote:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\n\nA. -4\nB. -2\nC. -1\nD. 2\nE. 5\n\nThe OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?\n\nQuestion says x(x \u2013 1)(x \u2013 k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.\n\nSince this is a multiple choice GMAT question, you can pick a particular value for x such that neither x, nor x-1 is divisible by 3 and start checking the answers.\nIn the given situation, choose for example x = 2 and check when 2 - k is not divisible by 3.\n(A) 2 - (-4) = 6 NO\n(B) 2 - (-2) = 4 BINGO!\n\n_________________\n\nPhD in Applied Mathematics\nLove GMAT Quant questions and running.\n\nKudos [?]: 1074 [8], given: 43\n\nIntern\nJoined: 10 Jul 2012\nPosts: 14\n\nKudos [?]: 7 [0], given: 0\n\nGMAT Date: 10-20-2012\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n05 Oct 2012, 03:37\nI too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.\n\nKudos [?]: 7 [0], given: 0\n\nDirector\nJoined: 22 Mar 2011\nPosts: 610\n\nKudos [?]: 1074 [1], given: 43\n\nWE: Science (Education)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n05 Oct 2012, 03:56\n1\nKUDOS\nAvantika5 wrote:\nI too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.\n\nOn the GMAT, is definitely the fastest way to solve it. Being a multiple choice question, you can be sure that there is a unique correct answer. And in the given situation, the only issue is to choose for x values such that neither x, nor x - 1 is divisible by 3.\nIt won't harm to understand and know to use the properties of consecutive integers presented in the other posts . They can be useful any time.\n_________________\n\nPhD in Applied Mathematics\nLove GMAT Quant questions and running.\n\nKudos [?]: 1074 [1], given: 43\n\nIntern\nJoined: 10 Jul 2012\nPosts: 14\n\nKudos [?]: 7 [0], given: 0\n\nGMAT Date: 10-20-2012\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n06 Oct 2012, 01:10\nThanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.\nThanks a lot...\n\nKudos [?]: 7 [0], given: 0\n\nDirector\nJoined: 22 Mar 2011\nPosts: 610\n\nKudos [?]: 1074 [0], given: 43\n\nWE: Science (Education)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n06 Oct 2012, 02:10\nAvantika5 wrote:\nThanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.\nThanks a lot...\n\nBetter is a relative word...Mathematicians always try to prove and justify everything in a formal, logical way.\nBut GMAT is not testing mathematical abilities per se. If they wanted so, the questions would have been open and not multiple choice.\nHave a flexible mind, think out of the box. GMAT is not a contest for the most beautiful, elegant, mathematical solution...\nGet the correct answer as quickly as possible, and go to the next question without any feeling of guilt...:O)\n\nThough, as I said, try to understand the other properties of the integer numbers, they can become handy and also, because they are so beautiful! Isn't Mathematics wonderful?\n_________________\n\nPhD in Applied Mathematics\nLove GMAT Quant questions and running.\n\nLast edited by EvaJager on 06 Oct 2012, 02:46, edited 1 time in total.\n\nKudos [?]: 1074 [0], given: 43\n\nManager\nStatus: Fighting hard\nJoined: 04 Jul 2011\nPosts: 69\n\nKudos [?]: 83 [0], given: 87\n\nGMAT Date: 10-01-2012\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n06 Oct 2012, 02:25\n1. k=-4, x+4=y so x=y-4. Now ----- (y-4)(y-5)(y). Gives number divisible by 3 in all cases.\n2. k=-2, , x+2=y so x=y-2. Now ----- (y-2)(y-3)(y). Not applicable when y is 5 or when x is 7\n3. k=-1, , x+1=y so x=y-1. Now ----- (y-1)(y-2)(y). Consecutive 3 integers. Divisible by 3\n4. k=2, , x-2=y so x=y+2. Now ----- (y+2)(y+1)(y). Consecutive 3 integers. Divisible by 3\n5. k=5, , x-5=y so x=y+5. Now ----- (y+5)(y+4)(y). Consecutive 3 integers. Divisible by 3\n_________________\n\nI will rather do nothing than be busy doing nothing - Zen saying\n\nKudos [?]: 83 [0], given: 87\n\nSenior Manager\nJoined: 15 Sep 2011\nPosts: 358\n\nKudos [?]: 416 [0], given: 45\n\nLocation: United States\nWE: Corporate Finance (Manufacturing)\n\n### Show Tags\n\n16 Feb 2013, 12:34\nmikemcgarry wrote:\nHi, there. I'm happy to help with this.\n\nThe rule that the product of three consecutive integers is a good start, but not the be all and end all.\n\nnumber = x(x \u2013 1)(x \u2013 k)\n\nSo far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --\n\nk = 2 ----> x(x \u2013 1)(x \u2013 2)\n\nk = -1 ----> x(x \u2013 1)(x + 1)\n\nNow, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.\n\n(x - 2) - 3 = (x - 5)\n\n(x - 5) - 3 = (x - 8)\n\n(x - 2) + 3 = (x + 1), which we have already\n\n(x + 1) + 3 = (x + 4)\n\n(x + 3) + 3 = (x + 7)\n\nSo, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.\n\nBack to the question:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\nA)-4\nB)-2\nC)-1\nD) 2\nE) 5\n\nAll of those choices give us a term on our list except for (B) -2.\n\nBTW, notice all the answer choices are spaced apart by three except for (B).\n\nDoes that make sense? Please do not hesitate to ask if you have any questions.\n\nMike\n\nGreat posts everyone, including you Mike, who I am now quoting, but I would like to add that..\n\nThe first two integers for each answer choice cannot be divisible by 3 because all the answers choices would then be correct, thereby making the question invalid. Therefore, the last integer (x+k) must be divisible by 3, which helps in the theoretical approach...\n\nIt appears that was left out\n\nKudos [?]: 416 [0], given: 45\n\nCurrent Student\nJoined: 06 Sep 2013\nPosts: 1972\n\nKudos [?]: 742 [0], given: 355\n\nConcentration: Finance\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n17 Oct 2013, 15:25\nenigma123 wrote:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\n\nA. -4\nB. -2\nC. -1\nD. 2\nE. 5\n\nThe OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?\n\nQuestion says x(x \u2013 1)(x \u2013 k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.\n\nHi enigma123. Thank you for the nice question. But you might wanna put a spoiler on this statement The OA is B so that others can solve the question in real conditions. Thanks\n\nKudos [?]: 742 [0], given: 355\n\nManager\nJoined: 27 Aug 2014\nPosts: 102\n\nKudos [?]: 119 [10], given: 19\n\nConcentration: Finance, Strategy\nGPA: 3.9\nWE: Analyst (Energy and Utilities)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n27 Oct 2014, 06:45\n10\nKUDOS\n2\nThis post was\nBOOKMARKED\nI tried the problem in a different way:\n\nfor any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.\n\nTaking the sum of x, (x-1) and (x-k) we have:\n\nx+x-1+x-k = 3x-1-k\n\nnow looking at the choices\n\nk = -4 => sum = 3x+3 --> divisible by 3\nk = -2 => sum = 3x+1 --> not divisible by 3\nk = -1 => sum = 3x --> divisible by 3\nk = 2 => sum = 3x-3 --> divisible by 3\nk = 5 ==> sum = 3x-6 --> divisible by 3\n\nKudos [?]: 119 [10], given: 19\n\nIntern\nJoined: 29 Apr 2015\nPosts: 2\n\nKudos [?]: 5 [0], given: 47\n\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n17 Jun 2015, 02:13\nHi there,\n\nThe answer is right, however my explanation is quite differently:\n\n1. x(x-1)(x-k)\n2. (x^2-1) (x-k)\n3. x^3-x^2k-x^2+xk\n4. x-xk\n\nPlugg-in x=1\na. 1-1(-4) = 1-(-4) = 5\nb. 1-1(-2) = 1-(-2) = 3\nc. 1-1(-1) = 1-(-1) = 2\nd. 1-1(2) = 1-(2) = -1\ne. 1-1(5) = 1-(5) = -4\n\nAm I correct in my explanation?\n\nKudos [?]: 5 [0], given: 47\n\nSVP\nJoined: 08 Jul 2010\nPosts: 1851\n\nKudos [?]: 2345 [3], given: 51\n\nLocation: India\nGMAT: INSIGHT\nWE: Education (Education)\nRe: If x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n17 Jun 2015, 05:49\n3\nKUDOS\nExpert's post\nsantorasantu wrote:\nQuote:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\n\nA. -4\nB. -2\nC. -1\nD. 2\nE. 5\n\nI tried the problem in a different way:\n\nfor any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.\n\nTaking the sum of x, (x-1) and (x-k) we have:\n\nx+x-1+x-k = 3x-1-k\n\nnow looking at the choices\n\nk = -4 => sum = 3x+3 --> divisible by 3\nk = -2 => sum = 3x+1 --> not divisible by 3\nk = -1 => sum = 3x --> divisible by 3\nk = 2 => sum = 3x-3 --> divisible by 3\nk = 5 ==> sum = 3x-6 --> divisible by 3\n\nJust refining the highlighted language and presenting another view to del with this problem\n\nCONCEPT1:For any number to be divisible by 3, the sum of the Digits of the Number should be a Multiple of 3.\n\nCONCEPT2: Product of any three consecutive Integers always include one muliple of 3 hence product of any three consecutive Integers is always a Multiple of 3\n\nCONCEPT3: If a Number \"w\" is a Multiple of 3 then any number at a difference of 3 or multiple of 3 from \"w\" will also be a multiple of 3 i.e. If w is a multiple of 3 then (w+3), (w-3), (w+6), (w-6) etc. will all be Multiples of 3\n\nHere, I see that x(x \u2013 1) is a product of two consecutive Integers but if another Number next to them is obtained then x(x \u2013 1)(x \u2013 k) will certainly be a multiple of 3 [as per Concept2 mentioned above]\n\nfor x(x \u2013 1)(x \u2013 k) to be a product of 3 consecutive integers,\n\n(x \u2013 k) should be either (x - 2) i.e. k=2\nOR\n(x \u2013 k) should be either (x - 5) i.e. k=5 [Using Concept3]\nOR\n(x \u2013 k) should be either (x + 1) i.e. k=-1\nOR\n(x \u2013 k) should be either (x + 4) i.e. k=-4 [Using Concept3]\n\nThis Eliminates options A, C D and E\n\n[Reveal] Spoiler:\nB\n\n_________________\n\nProsper!!!\nGMATinsight\nBhoopendra Singh and Dr.Sushma Jha\ne-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772\nOnline One-on-One Skype based classes and Classroom Coaching in South and West Delhi\nhttp://www.GMATinsight.com/testimonials.html\n\n22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION\n\nKudos [?]: 2345 [3], given: 51\n\nSenior Manager\nJoined: 13 Oct 2016\nPosts: 367\n\nKudos [?]: 402 [0], given: 40\n\nGPA: 3.98\nIf x is an integer then x(x-1)(x-k) must be evenly divisible\u00a0[#permalink]\n\n### Show Tags\n\n04 Nov 2016, 04:23\n1\nThis post was\nBOOKMARKED\nenigma123 wrote:\nIf x is an integer, then x(x \u2013 1)(x \u2013 k) must be evenly divisible by three when k is any of the following values EXCEPT\n\nA. -4\nB. -2\nC. -1\nD. 2\nE. 5\n\n[Reveal] Spoiler:\nThe OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?\n\nQuestion says x(x \u2013 1)(x \u2013 k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.\n\nA little bit faster approach.\n\nWe know from the start that product of $$3$$ consecutive integers is divisible by $$3$$, because it\u2019s indeed a multiple of $$3$$.\n\nSo we have $$x(x+1)(x+2)$$ is definitely divisible by 3 and k=2 is our anchor point.\n\nNow we can use principles of modular arithmetic to generate multiples of $$3$$. We can either add or subtract $$3$$ from our anchor number. And we got following string of numbers:\n\n\u2026 -4, -1, 2, 5, 7 \u2026\n\nAs you can see in this progression only answer B (-2) is absent.\n\nKudos [?]: 402 [0], given: 40\n\nIf x is an integer then x(x-1)(x-k) must be evenly divisible \u00a0 [#permalink] 04 Nov 2016, 04:23\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 22 posts ]\n\nDisplay posts from previous: Sort by", "date": "2017-11-21 06:39:13", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar", "openwebmath_score": 0.5184522271156311, "openwebmath_perplexity": 2078.5322086005, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.8596637451167997}}
{"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885", "text": "## Is 0/0 undefined or indeterminate?\n\nDiscussions concerned with knowledge of measurement, properties, and relations quantities, theoretical or applied.\n\n### Is 0/0 undefined or indeterminate?\n\nI'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.\nkrum\nForum Neophyte\n\nPosts: 6\nJoined: 08 Oct 2014\n\n### Re: Is 0/0 undefined or indeterminate?\n\nkrum \u00bb August 16th, 2015, 7:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.\n\nThe way I learned it in school:\n\na/0 is 'undefined' when a\u22600,\na/0 is 'indeterminate' when a=0.\n\nThus, unless nomenclature has changed since I was in school, the answer to your question is 'indeterminate'.\n\nDarby\nActive Member\n\nPosts: 1188\nJoined: 14 Feb 2015\nLocation: Long Island, New York (USA)\n\n### Re: Is 0/0 undefined or indeterminate?\n\nCalculus books will refer to \"indeterminate forms\", because they play a special role in certain infinite series.\n\n$0^{0}$ is one of them. As is\n\n$\\frac{0}{0} ,$\n\n$\\infty ^{\\infty },$\n\n$\\frac{\\infty }{\\infty },$ and\n\n$\\infty ^{0}$\n\nHowever, 1/0 is not an indeterminate form, and for that reason it can't be used in certain theorems. If those \"undefined\" a/0 things appear in certain limits of infinite series, math students can justifiably say the the answer is \"infinity\", and be technically correct. Depending on things like the textbook or the professor, the right answer may also be saying that the series (or limit) \"diverges.\"\n\nhyksos\nActive Member\n\nPosts: 1294\nJoined: 28 Nov 2014\n\n### Re: Is 0/0 undefined or indeterminate?\n\nkrum,\n\nWhat's $\\frac{x}{0}$? Okay, let's ask that question mathematically by:\n1. Call the definition $k$.\n2. Write that $k=\\frac{x}{0}$.\n3. Solve for $x$:\n\u2022 $k=\\frac{x}{0}$;\n\u2022 $0{\\times}k=0{\\times}{\\frac{x}{0}}$\n\u2022 $0{\\times}k=x$\n\u2022 $x=0$.\n4. Consider $x=0$.\n\u2022 No definition $k$ would lead to a contradiction.\n\u2022 So all possible definitions $k$ may be consistent.\n\u2022 So we cannot determine a particular definition that $k$ must be.\n\u2022 Call this inability to determine \"indeterminate\".\n5. Consider $x{\\neq}0$.\n\u2022 Any definition $k$ would lead to a contradiction.\n\u2022 So there is no possible definition $k$.\n\u2022 Call this inability to define \"undefined\".\n\nTherefore $\\frac{x}{0}$ is:\n1. indeterminate when $x=0$.\n2. undefined when $x{\\neq}0$.\n\nHistorically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\\times}{\\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.\nNatural ChemE\nForum Moderator\n\nPosts: 2754\nJoined: 28 Dec 2009\n Lomax,\u00a0CanadysPeak,\u00a0Darby liked this post\n\n### Re: Is 0/0 undefined or indeterminate?\n\nkrum \u00bb Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.\n\nI hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?\n\nWatson\nResident Member\n\nPosts: 4605\nJoined: 19 Apr 2009\nLocation: Earth, middle of the top half, but only briefly each 24 hours.\n\n### Re: Is 0/0 undefined or indeterminate?\n\nWatson \u00bb August 17th, 2015, 12:58 pm wrote:\nkrum \u00bb Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.\n\nI hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?\n\nNo. Classical Mathematics is actually quite clear and unambiguous on this point. It's been well covered in the previous posts.\n\nDarby\nActive Member\n\nPosts: 1188\nJoined: 14 Feb 2015\nLocation: Long Island, New York (USA)\n\n### Re: Is 0/0 undefined or indeterminate?\n\nTo put it more philosophically, x/x means a ratio of things - litterboxes to cats, doorknobs to doors, etc. Zero isn't anything, it isn't a real quantity, it is simply an absence, viz. nothing. Seen clearly, it has no place in a ratio. When you have no cats or litterboxes, then their ratio is indeterminate because there are no things to relate to each other. When you have ten cats and no litterboxes, then you have a ratio that is undefined - 10/0. You can't define a ratio because you only have one sort of quantity. And a yard full of cat poop. Does this sound right?\n\nBraininvat\n\nPosts: 6470\nJoined: 21 Jan 2014\nLocation: Black Hills\n\n### Re: Is 0/0 undefined or indeterminate?\n\nSo it is undefinable. I'm not sure how this is somewhat the same as indeterminable.\n\nWatson\nResident Member\n\nPosts: 4605\nJoined: 19 Apr 2009\nLocation: Earth, middle of the top half, but only briefly each 24 hours.\n Natural ChemE liked this post\n\n### Re: Is 0/0 undefined or indeterminate?\n\nThay are not the same ... they are different terms each with their own specific definitions.\n\nThe inelegant but tried and true school methodology of learning may come in handy herr ... if understanding is not instantly forthcoming, rely on memorization until it does. Worked for me when I needed it.\n\nDarby\nActive Member\n\nPosts: 1188\nJoined: 14 Feb 2015\nLocation: Long Island, New York (USA)\n\n### Re: Is 0/0 undefined or indeterminate?\n\nNatural ChemE \u00bb Mon Aug 17, 2015 1:26 am wrote:krum,\n\nWhat's $\\frac{x}{0}$? Okay, let's ask that question mathematically by:\n1. Call the definition $k$.\n2. Write that $k=\\frac{x}{0}$.\n3. Solve for $x$:\n\u2022 $k=\\frac{x}{0}$;\n\u2022 $0{\\times}k=0{\\times}{\\frac{x}{0}}$\n\u2022 $0{\\times}k=x$\n\u2022 $x=0$.\n4. Consider $x=0$.\n\u2022 No definition $k$ would lead to a contradiction.\n\u2022 So all possible definitions $k$ may be consistent.\n\u2022 So we cannot determine a particular definition that $k$ must be.\n\u2022 Call this inability to determine \"indeterminate\".\n5. Consider $x{\\neq}0$.\n\u2022 Any definition $k$ would lead to a contradiction.\n\u2022 So there is no possible definition $k$.\n\u2022 Call this inability to define \"undefined\".\n\nTherefore $\\frac{x}{0}$ is:\n1. indeterminate when $x=0$.\n2. undefined when $x{\\neq}0$.\n\nHistorically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\\times}{\\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.\n\nOutstanding! Perhaps the clearest explanation of this that I have seen.\nResident Expert\n\nPosts: 5931\nJoined: 31 Dec 2008\n\n### Re: Is 0/0 undefined or indeterminate?\n\nWatson \u00bb August 17th, 2015, 4:34 pm wrote:So it is undefinable. I'm not sure how this is somewhat the same as indeterminable.\n\nYup, \"undefinable\" would be a good way to think about it.\n\nMathematicians say \"undefined\" because they mean to say that $\\frac{x}{0}$ is undefined within elementary algebra. However, they see $\\frac{x}{0}$ as definable since they're not constricted to elementary algebra.\n\nIf we were to take a more abstract algebra approach, we could just state\n${\\text{Watson}}{\\left(x\\right)}{\\equiv}{\\frac{x}{0}}$.\nAnd I know what you're thinking: how's that help, right? Well, simple: the ${\\text{Watson}}{\\left(x\\right)}$ function doesn't return a normal number, but rather some new mathematical entity. This new mathematical entity obeys\n$0{\\times}{\\text{Watson}}{\\left(x\\right)}=x$,\nso it doesn't suffer from the automatic inconsistency that all elementary entities do in the post that I made above.\n\nWhat other properties does this new mathematical entity have? I dunno, I'm too lazy to make them up right now. And, heck, it's named after you, so you make them up. You're allowed to make up whatever you want so long as you rigorously avoid contradicting yourself.\n\nTo sum this up, yeah, you're right that \"undefinable\" would make more sense to most folks since we usually think about math in the context of elementary algebra. However, since mathematicians don't see things that way, to them it's moreso \"undefined\" than \"undefinable\".\n\n-----\n\nSeparately, I forgot to address your question about how undefined and indeterminate are somewhat the same. Well, in a way, they're actually exact opposites:\n1. undefined: No elementary description fits.\n2. indeterminate: The set of elementary descriptions which fit includes non-equivalent members.\nHowever, these two terms are similar in that they refer to cases in which there isn't exactly one known set of equivalent elementary representations.\n\nBy which I mean:\n1. Most elementary entities are defined by only one set of equivalent representations. Examples of representations for $k=2$:\n\u2022 $k=0+2$\n\u2022 $k=0+1+1$\n\u2022 $k=3-1+1-3+2.5-0.5$\n2. Undefined entities have no elementary representations. Examples of representations for $k={\\frac{2}{0}}$:\n\u2022 [None. $\\frac{2}{0}$ can't even represent itself because it leads to a contradiction; you're literally not even allowed to write it! It's meaningless gibberish.].\n3. Indeterminate entities have more than one possible set of equivalent representations. Examples of possible sets of equivalent representations for $k=\\frac{0}{0}$:\n\u2022 Set of representations in which $k=0$, including:\n\u2022 $k=0$\n\u2022 $k=0+0$\n\u2022 $k=1-1$\n\u2022 $k=0*0$\n\u2022 $k=0.5-\\frac{1}{2}$\n\u2022 Set of representations in which $k=1$, including:\n\u2022 $k=1$\n\u2022 $k=1+0$\n\u2022 $k=2-1$\n\u2022 $k=1*1$\n\u2022 $k=1.5-\\frac{1}{2}$\n\u2022 Set of representations in which $k=2$, including:\n\u2022 $k=2$\n\u2022 $k=1+1$\n\u2022 $k=2-0$\n\u2022 $k=1*2$\n\u2022 $k=1.5+\\frac{1}{2}$\n\u2022 [Etc., since $k$ can be just about anything in this case.]\n\nTo sum this up, \"undefined\" and \"indeterminate\" are similar in that they both indicate that there's not exactly one set of equivalent elementary representations. However, \"undefined\" indicates zero sets while \"indeterminate\" indicates more than one set.\nNatural ChemE\nForum Moderator\n\nPosts: 2754\nJoined: 28 Dec 2009\n Braininvat,\u00a0Darby liked this post\n\n### Re: Is 0/0 undefined or indeterminate?\n\nExhaustively well said, but I think it might be easier for many to consider that undefined & indeterminate (or undefinable and indeterminable, as Watson put it) are somewhat similar only in a purely linguistic sense, whereas they are quite dissimilar in a purely mathematical sense ... kinda like 'alien' and 'illegal alien' are superficially similar linquistically, and yet are completely very different idiomatically.\n\nMathematics and linquistics share a lot of common ground, and indeed the former was birthed out of the latter, but it is contextually important to remember their differences and that they are not interchangeable - mathematics is more rigorously symbol oriented, whereas linguistics is generally more meaning oriented.\n\nMath is a language unto itself.\n\nDarby\nActive Member\n\nPosts: 1188\nJoined: 14 Feb 2015\nLocation: Long Island, New York (USA)\n\n### Re: Is 0/0 undefined or indeterminate?\n\nWatson \u00bb August 17th, 2015, 5:58 pm wrote:I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?\n\nHere you have been shown that no mathematician would ever use a term like \"undefinable\", because if something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).\n\nFor example they would use your (algebraically undefined) ratio 0/0 to represent a discontinuity in the function x/|x|: for x>0 the function has constant value 1, and its limit for x--> 0 (from the right) also is 1; for x<0 the function has value -1 and that is also its limit for x-->0 (from the left). Since at zero the function instantaneously jumps from -1 to +1, its value cannot be determined there. Still, it is defined, as you can plot and integrate the function with no problems.\n\n[Forget this second part - it only is an attempt at pretending I am also a \"good mathematician\", simply because I share that attitude: if you find something that does not fit into your model, then claim that that's only a part of your model, you do have a better, more complex one!]\n\nneuro\nForum Moderator\n\nPosts: 2635\nJoined: 25 Jun 2010\nLocation: italy\n\n### Re: Is 0/0 undefined or indeterminate?\n\nif something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).\n\nYeah you can do that, but it depends on the professor teaching it, and on the context where the expression appears during the course. If you have a series where the numerator approaches a finite value, say 2 , and the denominator is trending to zero\n$\\frac{2}{0}$\nyou can justifiably define a \"new domain\" where the real answer to the problem is $\\infty$\n\nConfusion arrises in introductory courses where the students are asked to instead provide the answer as \"Does not exist\".\n\nThe differentiation between \"undefined\" and \"indeterminate\" goes beyond pathological issues presented by Natural ChemE. In l'Hopital's Rule, the expression is required to be indeterminate. Otherwise, in expressions that are undefined, the theorem does not hold. (more importantly) you cannot use the technique to solve the limit.\n\nhyksos\nActive Member\n\nPosts: 1294\nJoined: 28 Nov 2014\n\n### Re: Is 0/0 undefined or indeterminate?\n\nhyksos,\n\nThat's a common misconception, but L'H\u00f4pital's rule isn't meaningfully related to this topic.\n\nL'H\u00f4pital's rule is an alternative way to evaluate the limit of fractions. While it's usually more work than simpler methods, sometimes it's actually easier or/and provides a better result.\n\nThe commonly discussed case for L'H\u00f4pital's rule is when just substituting in the approached value would result in either $\\frac{0}{0}$ or $\\frac{{\\infty}}{{\\infty}}$. The problem with both of these results is that they're not very descriptive, i.e. they're indeterminate. Math books often recommend considering L'H\u00f4pital's rule when this happens since it might provide a determinate result.\nNatural ChemE\nForum Moderator\n\nPosts: 2754\nJoined: 28 Dec 2009\n\n### Re: Is 0/0 undefined or indeterminate?\n\nNatural ChemE.\n\nL'H\u00f4pital's rule is not just for $\\frac{0}{0}$ and$\\frac{ \\infty }{ \\infty }$ as you have claimed. The rule will work on any situation that is an indeterminate form. You have given two examples . There are many more, such as $0^0$ and several more others.\n\nHowever, $\\frac{2}{0}$ is NOT indeterminate! That's where the danger comes in. L'H\u00f4pital's rule cannot be applied to situations that are \"formally infinity\", or in limits and series that genuinely diverge.\n\nThe main thing I wanted to communicate here is : yes, I agree with you that students are told to plug-and-chug L'H\u00f4pital's rule, as you have described. However, dig in the literature a little deeper. L'H\u00f4pital's rule is buttressed by two historical proofs and their corresponding theorems. This differentiation between indeterminate and undefined is very real in mathematics, and it has real consequences in theorems.\n\nhyksos\nActive Member\n\nPosts: 1294\nJoined: 28 Nov 2014\n\n### Re: Is 0/0 undefined or indeterminate?\n\nhksos,\nNatural ChemE \u00bb August 31st, 2015, 7:55 pm wrote:The commonly discussed case for L'H\u00f4pital's rule is when just substituting in the approached value would result in either $\\frac{0}{0}$ or $\\frac{{\\infty}}{{\\infty}}$.\nhksos \u00bb September 12th, 2015, 4:17 pm wrote:L'H\u00f4pital's rule is not just for $\\frac{0}{0}$ and$\\frac{ \\infty }{ \\infty }$ as you have claimed.\nHow the heck did you get from my statement to your representation of it?\n\nThere's a lot wrong with what you've said. Unless you can demonstrate some relevance, I'll just split 'em off to avoid polluting a good thread.\nNatural ChemE\nForum Moderator\n\nPosts: 2754\nJoined: 28 Dec 2009\n\n### Re: Is 0/0 undefined or indeterminate?\n\n[quote=\"[url=http://www.sciencechatforum.com/viewtopic.php?p=286235#p286235]Natural ChemE \u00bb August 17th, 2015, 1:26 am[/url]\"]krum,\n\nWhat's $$\\frac{x}{0}$$? Okay, let's ask that question mathematically by:[list=1]\n[*]Call the definition $$k$$.\n[*]Write that $$k=\\frac{x}{0}$$.\n[*]Solve for $$x$$:\n[list]\n[*]$$k=\\frac{x}{0}$$;\n[*]$$0{\\times}k=0{\\times}{\\frac{x}{0}}$$\n[*]$$0{\\times}k=x$$\n[*]$$x=0$$.[/list]\n[*]Consider $$x=0$$.\n[list]\n[*]No definition $$k$$ would lead to a contradiction.\n[*]So all possible definitions $$k$$ may be consistent.\n[*]So we cannot determine a particular definition that $$k$$ must be.\n[*]Call this inability to determine \"indeterminate\".[/list]\n[*]Consider $$x{\\neq}0$$.\n[list]\n[*]Any definition $$k$$ would lead to a contradiction.\n[*]So there is no possible definition $$k$$.\n[*]Call this inability to define \"undefined\".[/list][/list]\n\nTherefore $$\\frac{x}{0}$$ is:[list=i]\n[*]indeterminate when $$x=0$$.\n[*]undefined when $$x{\\neq}0$$.[/list]\n\nHistorically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $$\\frac{x}{y}$$ at some intermediate step but was later removed (e.g. by multiplication by $$y$$), then it may be hard to tell, but the proof may be flawed through reliance on $$0{\\times}{\\frac{x}{0}}=x$$. This [url=https://en.wikipedia.org/wiki/Division_by_zero#Fallacies]hidden division-by-zero fallacy[/url] was the first major math fallacy that I was warned about.[/quote]\n\nThe bottom line is that 0/0 is not equal to any number and a/0 where a is not 0 is not any number.\nRobert Kolker\nForum Neophyte\n\nPosts: 1\nJoined: 20 Jan 2016\n\n### Re: Is 0/0 undefined or indeterminate?\n\nIt is indeterminate.\nwilliamjohn\nForum Neophyte\n\nPosts: 1\nJoined: 20 May 2016\n\n### Re: Is 0/0 undefined or indeterminate?\n\nn/0 does not exist, for all n.\n\n1. x/y defined as (the z such that y*z = x).\n2. 0*z = 0, for all z.\n\nn/0 = (the z: 0*z = n).\n\nIf ~(n=0) then there is no z such that 0*z = n.\n\ne.g. (1/0) = (the z such that 0*z = 1).\nBut 2 asserts that all values of 0*z are equal to 0,\ni.e. there is no value of z for which 0*z=1,\n\nIf there is no value of z such that 0*z=1 then\n(the z such that 0*z = 1) does not exist.\n\n(the z: 0*z = n) is not unique for all n > 0.\n\nIf n=0 then there is more than one value of z such that 0*z=0.\n0*z=0 is true for all z.\n\ni.e. there is no unique value of z such that 0*z = 0.\n\nThat is to say, n/0 does not exist for all values of n,\neven though n/0 is defined.\nOwen\nMember\n\nPosts: 80\nJoined: 06 May 2007", "date": "2018-07-16 12:52:29", "meta": {"domain": "sciencechatforum.com", "url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885", "openwebmath_score": 0.8781976103782654, "openwebmath_perplexity": 1630.4931841615064, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97241471777321, "lm_q2_score": 0.8840392832736084, "lm_q1q2_score": 0.8596528101449367}}
{"url": "https://math.stackexchange.com/questions/1823736/how-to-pick-10-people-from-13-such-that-at-least-1-is-a-woman", "text": "# How to pick $10$ people from $13$ such that at least $1$ is a woman\n\nThirteen people on a softball team show up for a game. Of the $13$ people who show up, $3$ are women. How many ways are there to choose $10$ players to take the field if at least one of these players must be a woman?\n\nThe given answer is calculated by summing the combination of $1$ woman + $9$ men, $2$ women + $8$ men, and $3$ women + $7$ men.\n\nMy question is, why can't we set this up as the sum $\\binom{3}{1} + \\binom{12}{9}$ - picking one of the three women first, then picking $9$ from the remaining $12$ men and women combined? The only requirement is that we have at least one woman, which is satisfied by $\\binom{3}{1}$, and that leaves a pool of $12$ from which to pick the remaining $9$. The answer this way is close to the answer given, but it's $62$ short. I get that it's the \"wrong\" answer but I'm wondering why my thinking was wrong in setting it up this way. Thanks.\n\n\u2022 You must multiply $C(3,1)$ with $C(12,9)$ because for every choice of a woman you have $C(12,9)$ choices for the other people. But this does not work either because you count arrangements multiple times. \u2013\u00a0Peter Jun 12 '16 at 22:15\n\nFirstly you multiply, not add, if you are thinking of $\\dbinom31$ and $\\dbinom{12}{9}$\n\nSecondly, this approach will over count. Suppose you chose Alicia , and then you chose $9$ from the remaining $12$, you would also have combos where Britney was first chosen, and Alicia was chosen from the $12$ group.\n\nThirdly, your book approach is correct, but unnecessarily tedious.\n\nThe best way is to compute\n[All possible combos] - [All male combos ] $\\;=\\dbinom{13}{10} - \\dbinom{10}{10}$\n\n\u2022 I get what you are saying, but the constraint is \"at least one\", so wouldn't having a combination where Alicia is picked first, followed by a combination where Britney is picked and then Alicia is picked, be perfectly valid? Because there is still \"at least one\" in each combination. Unless I misunderstood you. \u2013\u00a0Dave Jun 12 '16 at 22:51\n\u2022 $A$ followed by $B$, and $B$ followed by $A$ are valid, but you are counting them twice (apart from other overcounts). Problems of the \"at least one\" type are best tackled using the complement. You can verify for yourself that the book method, and the above method give the same answer. \u2013\u00a0true blue anil Jun 13 '16 at 4:44\n\nAmong $13$ people, you can choose $10$ players in $$\\binom {13}{10}$$ ways.\n\nNow, suppose there is no woman in the squad, then you can choose that team in $\\binom{10}{10}=1$ way.\n\nSo, number of ways of choosing a team with at least $1$ woman is $$\\binom{13}3-1=286-1=285$$ ways.\n\nWhen you are counting several countings(call it nested counting), you should apply the multiplication principle. Here counting ways of choosing $1$ woman and $9$ men, you are in some kind of this nested counting, where, for any $9$ men among the $10$, the woman can occur, so, have to multiply this.\n\nAnd upon deriving each choices by multiplying, you add those terms. Because, in one counting, there is $1$ woman , in some other count, there are $2$, and in other, there are $3$, women.\n\nSo, all are separate cases, so add them, like $$\\binom31 \\binom{10}9+\\binom 32\\binom {10}8+\\binom33\\binom {10}7=285$$\n\n\u2022 This is a better solution than the one on the supplied answer sheet. On the other hand, it doesn't explain why OP's method fails. \u2013\u00a0David K Jun 12 '16 at 22:18", "date": "2019-09-15 08:19:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1823736/how-to-pick-10-people-from-13-such-that-at-least-1-is-a-woman", "openwebmath_score": 0.7504289150238037, "openwebmath_perplexity": 379.9378384899396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676448764483, "lm_q2_score": 0.877476800298183, "lm_q1q2_score": 0.8596356303818424}}
{"url": "https://math.stackexchange.com/questions/1472277/convex-hull-of-idempotent-matrices", "text": "# Convex hull of idempotent matrices\n\nWhat is the convex hull of the set of $n\\times n$ (potentially asymmetric) idempotent matrices?\n\nConsider the set $S:=\\{A\\in\\mathbb{R}^{n\\times n}:A^2=A\\}$, the set of idempotent matrices. Is there a simple description of the convex hull of $S$?\n\nBy convex hull, I mean: $C:=\\{\\sum_i a_i M_i:\\sum_ia_i=1,a_i\\geq0,M_i\\in S\\}$\n\n\u2022 At the very least, we know that every matrix in the hull has a trace between $0$ and $n$. For all I know, we might get every matrix with trace between $0$ and $n$. \u2013\u00a0Omnomnomnom Oct 9 '15 at 18:34\n\u2022 A condition on trace seems quite loose, however. Is there a smaller convex set containing the idempotent matrices? \u2013\u00a0Justin Solomon Oct 9 '15 at 18:42\n\u2022 I really don't know. By the way, it looks like your question is about to be closed because it \"needs more context\". \u2013\u00a0Omnomnomnom Oct 9 '15 at 18:46\n\u2022 I think the problem that people may have is that it looks like you didn't put enough work into answering the question yourself. \u2013\u00a0Omnomnomnom Oct 9 '15 at 18:52\n\u2022 Sure, but the people of the site and the powers that be tend to expect some kind of partial work, though not necessarily a partial result. I happen to think your question is fine as is. \u2013\u00a0Omnomnomnom Oct 9 '15 at 18:55\n\nFirst I'll do the $2 \\times 2$ case. Consider a $2 \\times 2$ matrix of trace $1$:\n$$A = \\pmatrix{a & b\\cr c & 1-a\\cr}$$ For any real $r$, every point $(b,c)$ in the plane is a convex combination of two points on the hyperbola $x y = -r$ (the two axes in the case $r=0$). Thus $A$ is a convex combination of two idempotent matrices of the form $$\\pmatrix{ a & x\\cr y & 1-a\\cr}$$ where $xy = a-a^2$. The only idempotent matrix of trace $0$ is $0$, and the only idempotent matrix of trace $2$ is $I$. Any matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $2$ is a convex combination of $I$ and a matrix of trace $1$. Thus the convex hull in the $2 \\times 2$ case consists of $0$, $I$ and all $A$ with $0 < \\text{tr}(A) < 2$.\n\nThis leads to a solution in the $n \\times n$ case for any $n > 2$. We have idempotent matrices of trace $1$ with the form $$\\pmatrix{a & x & 0 & \\ldots & 0\\cr y & 1-a & 0 & \\ldots & 0\\cr 0 & 0 & 0 & \\ldots & 0\\cr \\ldots & \\ldots & \\ldots & \\ldots & \\ldots\\cr 0 & 0 & 0 & \\ldots & 0\\cr}$$ where $xy = a - a^2$, and convex combinations of these give all $n \\times n$ matrices of trace $1$ that are $0$ outside the top left $2 \\times 2$ submatrix. By permuting indices, we get all $n \\times n$ matrices of trace $1$ that are $0$ outside some principal $2 \\times 2$ submatrix. Convex combinations of these give us all $n \\times n$ matrices of trace $1$. Now any $n \\times n$ matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $n$ is a convex combination of $I$ and a matrix of trace $1$. Thus we find that the convex hull consists of $0$, $I$ and all $A$ with $0 < \\text{tr}(A) < n$.\n\n\u2022 so do we know that there won't be any matrices with trace $2$ besides $I$? \u2013\u00a0Omnomnomnom Oct 9 '15 at 19:14\n\u2022 Also, what about matrices with trace $0$? \u2013\u00a0Omnomnomnom Oct 9 '15 at 19:17\n\u2022 An idempotent matrix of trace $k$ is a projection on a subspace of dimension $k$. The only subspace of dimension $2$ is the whole space, and a projection on that is the identity. The only subspace of dimension $0$ is $\\{0\\}$, and a projection on that is $0$. \u2013\u00a0Robert Israel Oct 9 '15 at 19:25\n\u2022 I see, there is a unique idempotent element with trace $2$ (or $0$), and so the convex hull must also have a unique such element. \u2013\u00a0Omnomnomnom Oct 9 '15 at 19:32\n\u2022 Wow! Thanks to both of you for your help -- it's much appreciated. I'm surprised the convex hull is so big. \u2013\u00a0Justin Solomon Oct 9 '15 at 19:50\n\nThe result can be generalized by induction. Let $C_n$ denote the convex hull over $n \\times n$ matrices.\n\nFirst of all, note that if $A \\in C_{n-1}$ and $x \\in \\Bbb R^n$, then the elements $$\\pmatrix{A&0\\\\0&0_{1\\times 1}},E_x := \\pmatrix{I_{n-1}&x\\\\0&0}$$ are all in $C_n$.\n\nNext, note that $C_n$ is closed under similarity (anything similar to an element of $C_n$ is in $C_n$).\n\nNow, take any upper triangular $A \\in C_{n-1}$. That is, $$A = \\pmatrix{\\lambda_1 & * \\cdots \\\\ &\\ddots\\\\ && \\lambda_{n-1}}$$ with $\\sum\\lambda_i \\in (0,n-1)$. Note that any convex combination of $A$ and $I$ is in $C_n$. Also, any convex combination of a matrix $C_n$ with $E_x$ is in $C_n$.\n\nConclude that $C_n$ contains every upper triangular matrix with trace (strictly) between $0$ and $n$ and a non-negative bottom-right entry. Because every matrix is upper-triangularizable, conclude that $C_n$ contains every matrix with trace strictly between $0$ and $n$.", "date": "2020-02-17 16:25:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1472277/convex-hull-of-idempotent-matrices", "openwebmath_score": 0.933679461479187, "openwebmath_perplexity": 142.51201729522447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676472509722, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8596356167707372}}
{"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx", "text": "A very interesting question: intersection point of $x^y=y^x$\n\nI've been investigating the Cartesian graph of $$x^y=y^x$$. Obviously, part of the graph is comprised of the line $$y=x$$ but there is also a curve that is symmetrical about the line $$y=x$$. (We can prove this symmetry by noting that the function $$x^y=y^x$$ is self-inverse; all self-inverse functions are symmetrical about the line $$y=x$$.)\n\nAn image of the graph is shown below:\n\nI decided to find the intersection point and arrived at an intriguing result: the intersection point between the two curves is at $$(e,e)$$.\n\nThe following is my method: If the gradient of the line $$y=x$$ is $$1$$, the gradient of the curve at the intersection point must be $$-1$$ as it's a normal to the line (as it's symmetrical about the line). This means that at that point $$\\frac{dy}{dx}=-1$$. Now to find $$\\frac{dy}{dx}$$.\n\nWe have $$x^y=y^x$$. I then used a very powerful tehnique for differentiating these sorts of functions. We know that eg $$x^y=e^{\\ln{x^y}}=e^{y\\ln{x}}$$ and $$\\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$$ Applying it to our function $$x^y=y^x$$ and using implicit differentiation and the product rule gives us: $$(\\frac{dy}{dx}\\times \\ln x +\\frac{y}{x})x^y=(\\ln y +\\frac{dy}{dx}\\times \\frac{x}{y})y^x$$ So $$\\frac{dy}{dx}x^y\\ln x +yx^{y-1}=\\frac{dy}{dx}xy^{x-1}+y^x \\ln y$$ Extensively rearranging gives: $$\\frac{dy}{dx}=\\frac{y^x \\ln y -yx^{y-1}}{x^y \\ln x -xy^{x-1}}$$ Let $$\\frac{dy}{dx}=-1$$: $$y^x \\ln y -yx^{y-1}=xy^{x-1}-x^y\\ln x$$ But we know $$x=y$$ since we are at the intersection point with the line $$y=x$$: $$x^x \\ln x -x^x=x^x-x^x\\ln x$$ So $$2x^x \\ln x -2x^x=0$$ $$2x^x(\\ln x -1)=0$$ This means either $$2x^x=0$$ or $$\\ln x -1=0$$ but we know $$x^x$$ is always greater than $$0$$ so $$\\ln x =1$$, leaving us with: $$x=y=e$$ So I have my answer, but is there any other method of getting it? I have heard there is. Any help will be very welcome.\n\n\u2022 The solution to $y^x=x^y$ can be expressed in terms of Lambert's W function as $$y=-xW\\left(-\\log(x)/x\\right)/\\log(x)$$Lambert's W function has been studied extensively. [Here](en.wikipedia.org/wiki/Lambert_W_function) is a link to a Wikipedia article that might be of interest. Note in particular that $W(-1/e)=-1$. Jun 18 '20 at 14:42\n\u2022 Does this answer your question? Where does the curve $x^y = y^x$ intersect itself? Jun 18 '20 at 15:13\n\u2022 Another possible duplicate: Given that $x^y=y^x$, what could $x$ and $y$ be?. Jun 18 '20 at 15:14\n\u2022 Does this answer your question? Given that $x^y=y^x$, what could $x$ and $y$ be? Jun 18 '20 at 16:11\n\u2022 Only the question titled 'Where does the curve $x^y=y^x$ intersect itself?' is helpful, but it still isn't very detailed when explaining the limits.Thanks for the suggestions though! Jun 18 '20 at 18:15\n\nA Simple Approach\n\nThe simplest approach that I have found is to look at the intersections of $$y=tx\\qquad\\text{and}\\qquad x^y=y^x\\tag1$$ That is, \\begin{align} x^{tx}&=(tx)^x\\tag{2a}\\\\[3pt] x^t&=tx\\tag{2b}\\\\[3pt] x^{t-1}&=t\\tag{2c}\\\\ x&=t^{\\frac1{t-1}}\\tag{2d}\\\\ y&=t^{\\frac t{t-1}}\\tag{2e} \\end{align} Explanation:\n$$\\text{(2a)}$$: $$x^y=y^x$$\n$$\\text{(2b)}$$: raise to the $$\\frac1x$$ power\n$$\\text{(2c)}$$: divide by $$x$$\n$$\\text{(2d)}$$: raise to the $$\\frac1{t-1}$$ power\n$$\\text{(2e)}$$: $$y=tx$$\n\nNow, if we wish to find where $$x=y$$, let $$t\\to1$$. That is, \\begin{align} x &=\\lim_{t\\to1}t^{\\frac1{t-1}}\\tag{3a}\\\\ &=\\lim_{n\\to\\infty}\\left(1+\\frac1n\\right)^n\\tag{3b}\\\\[6pt] &=e\\tag{3c} \\end{align} Explanation:\n$$\\text{(3a)}$$: $$x=y$$ when $$t=1$$\n$$\\text{(3b)}$$: $$t=1+\\frac1n$$\n$$\\text{(3c)}$$: evaluate the limit\n\nFurther Musings\n\nWe can also compute \\begin{align} y &=\\lim_{t\\to1}t^{\\frac t{t-1}}\\tag{4a}\\\\ &=\\lim_{n\\to\\infty}\\left(1+\\frac1n\\right)^{n+1}\\tag{4b}\\\\[6pt] &=e\\tag{4c} \\end{align} Explanation:\n$$\\text{(4a)}$$: $$x=y$$ when $$t=1$$\n$$\\text{(4b)}$$: $$t=1+\\frac1n$$\n$$\\text{(4c)}$$: evaluate the limit\n\nUsing the results from this answer, we see that $$\\left(1+\\frac1n\\right)^n$$ is increasing and $$\\left(1+\\frac1n\\right)^{n+1}$$ is decreasing, which means that $$\\left(1+\\frac1n\\right)^n\\le e\\le\\left(1+\\frac1n\\right)^{n+1}$$.\n\nThus, as $$t\\to1^+$$, $$(4)$$ and $$(5)$$ show that $$x\\to e^-$$ and $$y\\to e^+$$.\n\nFurthermore, if we substitute $$t\\mapsto1/t$$, we get \\begin{align} t^{\\frac1{t-1}} &\\mapsto(1/t)^{\\frac1{1/t-1}}\\tag{5a}\\\\ &=t^{\\frac t{t-1}}\\tag{5b} \\end{align} and \\begin{align} t^{\\frac t{t-1}} &\\mapsto(1/t)^{\\frac{1/t}{1/t-1}}\\tag{6a}\\\\ &=t^{\\frac1{t-1}}\\tag{6b} \\end{align} That is, substituting $$t\\mapsto1/t$$ swaps $$x$$ and $$y$$.\n\nThis means that, as $$t\\to1^-$$, $$x\\to e^+$$ and $$y\\to e^-$$.\n\nRelation to the Graph\n\nHere is where these points sit on the graph as $$t\\to1^+$$:\n\n\u2022 Incredibly clear, thank you!!! Jun 23 '20 at 14:33\n\u2022 @robjohn Your answers always have fantastic graphics and explanations. Nicely done, +1 from me. Jul 7 '20 at 5:35\n\nNot sure to understand well the question but if it provides any help, I think there is something missing in your analysis.\n\nYour equation can be seen as $$g(x,y) = x^{y} = y^{x} = f(x,y)$$. You are looking for intersecting points or surfaces (x,y) in $$\\mathbb{R}^{3}$$, since both $$f(\\cdot,\\cdot)$$ and $$g(\\cdot,\\cdot)$$ are 2D scalar fields. Otherwise, if you are thinking the expression $$x^{y} = y^{x}$$ as the equation of a curve in $$\\mathbb{R}$$ (somehow you figure out to express $$y = f(x)$$ or $$x = f(x)$$, it doesn't make sense to talk about an intercept (or interception set) since you must provide another curve or surface.\n\nThe same applies with the fields $$g$$ and $$f$$. You must provide another equation since you have two unknowns.\n\nThe answer you got $$(x,y) = (e,e)$$ is the trivial answer, and works for any real $$k \\in \\mathbb{R}$$: $$(x,y) = (k,k)$$.\n\nI don't know if this is what you are looking for, but I would go like this.\n\nFirst rewrite the equation : $$x^y=y^x\\Leftrightarrow y\\ln(x)=x\\ln(y)\\Leftrightarrow \\frac{\\ln(x)}{x}=\\frac{\\ln(y)}{y}$$ Now consider the function $$f(x)=\\frac{\\ln(x)}{x}$$ defined on $$(0,+\\infty)$$ and whose derivative is $$f'(x)=\\frac{1-\\ln(x)}{x^2}$$. This shows that $$f$$ induces two bijections $$f_1:(0,e]\\to (-\\infty,\\frac{1}{e}]$$ and $$f_2:[e,+\\infty)\\to[\\frac{1}{e},0)$$.\n\nNow the locus $$f(x)=f(y)$$ can be divided in two curves, namely $$y=f_1^{-1}(f(x))$$ and $$y=f_2^{-1}(f(x))$$. (Note however that these two curves are not the obvious ones). And we are looking their intersection. Since the ranges of $$f_1^{-1}$$ and $$f_2^{-1}$$ are respectively $$(0,e]$$ and $$[e,+\\infty)$$, the only possible value of $$y$$ for the intersection is $$y=e$$ which happen when $$x=e$$.\n\n\u2022 Actually, this method shows that for a function $f$ which has a single change of variation (hence a unique local extrema $x_0$), then $f(x)=f(y)$ has two branches whose intersection is exactly $(x_0,x_0)$. Jun 18 '20 at 15:46\n\nThis is not exactly an answer, but part of your reasoning was a bit hand-wavy and I want to make it fully rigorous.\n\nYour reasoning about the slopes of the two parts of the plot is intuitively correct, but needs proof. So it is clear that the slope of the first part of the plot is $$1$$ since this encompasses the solutions $$x^x=x^x$$, i.e, the identity map which has slope $$1$$ by definition. However, showing that the other branch of the plot has slope $$-1$$ is a bit more difficult. We need to show that\n\n1: The relation $$y^x-x^y=0$$ is its own inverse\n\n2: If a relation is its own inverse, its slope at a point $$(x,y)$$ is $$-1$$ if $$x=y$$, with the exception of the identity map.\n\nThe first point is rather easy and I'll leave it for you to do yourself. The second part, however, requires a bit of care. Any two variable relation can be summarized by the general equation $$f(x,y)=0$$ Well, not quite any, but I hope you get the idea. For example, the equation of a circle: $$x^2+y^2-r^2=0$$ We want to use information about $$f$$ to deduce $$\\frac{\\mathrm{d}y}{\\mathrm{d}x}$$. Using a bit of knowledge from multivariable calculus, we know that the total derivative of a two-variable function $$f$$ in terms of an arbitrary variable $$u$$ is $$\\frac{\\mathrm{d} f}{\\mathrm{d} u}=\\frac{\\partial f}{\\partial x}\\frac{\\mathrm{d}x}{\\mathrm{d}u}+\\frac{\\partial f}{\\partial y}\\frac{\\mathrm{d}y}{\\mathrm{d}u}$$ In the special case that $$u=x$$, $$\\frac{\\mathrm{d} f}{\\mathrm{d} x}=\\frac{\\partial f}{\\partial x}+\\frac{\\partial f}{\\partial y}\\frac{\\mathrm{d}y}{\\mathrm{d}x}$$ And similarly $$\\frac{\\mathrm{d} f}{\\mathrm{d} y}=\\frac{\\partial f}{\\partial x}\\frac{\\mathrm{d}x}{\\mathrm{d}y}+\\frac{\\partial f}{\\partial y}$$ If the relation is its own inverse, then $$f(x,y)=f(y,x)=0$$ Thus $$\\frac{\\mathrm{d} f}{\\mathrm{d} x}=\\frac{\\partial f}{\\partial x}+\\frac{\\partial f}{\\partial y}\\frac{\\mathrm{d}y}{\\mathrm{d}x}=\\frac{\\mathrm{d} f}{\\mathrm{d} y}=\\frac{\\partial f}{\\partial x}\\frac{\\mathrm{d}x}{\\mathrm{d}y}+\\frac{\\partial f}{\\partial y}$$ So $$\\frac{\\partial f}{\\partial x}\\left(1-\\frac{\\mathrm{d}x}{\\mathrm{d}y}\\right)=\\frac{\\partial f}{\\partial y}\\left(1-\\frac{\\mathrm{d}y}{\\mathrm{d}x}\\right)$$ But again, since $$f(x,y)=f(y,x)=0$$, $$\\frac{\\partial f}{\\partial x}=\\frac{\\partial f}{\\partial y}$$ if $$x=y$$, thus $$1-\\frac{\\mathrm{d}x}{\\mathrm{d}y}=1-\\frac{\\mathrm{d}y}{\\mathrm{d}x}$$ $$\\frac{1}{\\frac{\\mathrm{d}y}{\\mathrm{d}x}}=\\frac{\\mathrm{d}y}{\\mathrm{d}x}$$ Therefore $$\\frac{\\mathrm{d}y}{\\mathrm{d}x}=\\pm 1$$. Excluding the only self-inverse relation with slope $$1$$, the identity map, we conclude $$\\frac{\\mathrm{d}y}{\\mathrm{d}x}\\big|_{(x,x)}=-1.$$\n\nFollowing this, the rest of your reasoning is correct. Others have stated that this problem can be tackled using the product log, so I can't really think of another way of going about this problem. Good question though!\n\n\u2022 Thanks for the detailed answer; I realized I didn't have ironclad proof for the second point, but that explanation is very helpful and welcome! Thanks a s well for the final compliment :) Jun 18 '20 at 18:05\n\nWe are looking for solutions to $$\\exp(y\\log x) = \\exp(x\\log y) \\, .$$ Taking logs of both sides and rearranging, this equation becomes $$\\frac{x}{\\log x}=\\frac{y}{\\log y} \\, .$$ Consider the graph of $$f(z)=z/\\log z$$:\n\nIts unusual shape can be explained by a few observations:\n\n\u2022 For $$0, $$\\log z$$ is negative, so $$f(z)$$ is negative.\n\u2022 As $$z\\to1$$, $$\\log z \\to 0$$, and so the graph has an asymptote at $$z=1$$.\n\u2022 For $$z>1$$, $$\\log z$$ is positive, so $$f(z)$$ is positive.\n\u2022 $$f(z)$$ is large when $$\\log z$$ is small, or when $$z$$ is very large.\n\nIt is the final observation that gives us a hint as to why $$f(z)$$ has a local minimum. We can study the behaviour of $$f(z)$$ more closely by considering $$f'(z) = \\frac{\\log z - 1}{\\log^2z} \\, .$$ For $$1, $$f'(z)$$ is negative; at $$z=e$$, $$f'(z)$$ is zero; and for $$z>e$$, $$f'(z)$$ is positive.\n\nNow that we have proven that $$f(z)$$ has a local minimum at $$(e,e)$$, we can begin to examine your question more closely. Let $$(x,\\log x)$$ and $$(y,\\log y)$$ be arbitrary points that lies on the graph.\n\n\u2022 Suppose $$0. Then, because $$f(z)$$ is one-to-one on the interval $$(0,1)$$, the only time when $$f(x)=f(y)$$ is when $$x=y$$.\n\u2022 If $$x>1$$, then things are more complicated. Clearly, if $$x=y$$, then we still have $$f(x)=f(y)$$. However, if $$1, then there is a unique number $$y\\in(e,\\infty)$$ such that $$f(x)=f(y)$$. This is obviously still the case if the roles of $$x$$ and $$y$$ are interchanged.\n\u2022 If $$x$$ is slightly smaller than $$e$$, then $$y$$ is either equal to $$x$$, or is only slightly larger than $$e$$. This means that the two sections of the graph get very close to each other as $$x$$ tends to $$e$$.\n\u2022 In the limit, when $$x=e$$, the two parts of the graph coincide.\n\nThe equation is equivalent to\n\n$$\\frac{\\log x}x=\\frac{\\log y}y.$$\n\nThe function\n\n$$\\frac{\\log x}x$$\n\nhas a maximum at $$x=e$$ and for $$x>0$$\n\n$$\\frac{\\log x}x=\\frac{\\log y}y$$ has two solutions in $$x$$.\n\nUsing Lambert's function,\n\n$$-\\frac1x\\log\\frac1x=\\frac{\\log y}y$$ and\n\n$$e^{\\log\\frac1x}\\log\\frac1x=-\\frac{\\log y}y$$\n\nand\n\n$$\\log\\frac1x=W\\left(-\\frac{\\log y}y\\right).$$", "date": "2021-10-27 20:29:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx", "openwebmath_score": 0.9445435404777527, "openwebmath_perplexity": 189.3965862014512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225143328517, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.8595987610775709}}
{"url": "https://math.stackexchange.com/questions/1881826/a-tangent-to-the-ellipse-meets-the-x-and-y-axes-if-o-is-the-origin-find", "text": "# A tangent to the ellipse meets the $x$ and $y$ axes . If $O$ is the origin, find the minimum area of triangle $AOB$.\n\nThe question is that : A tangent to the ellipse $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$.\n\nWhat I have attempted:\n\nBecause the equation of an ellipse is $$\\frac{xx_1}{a^2}+\\frac{yy_1}{b^2}=1$$\n\nWe have that the $x$ intercept is $x=\\frac{a^2}{x_1}$ and $y$ intercept is $y=\\frac{b^2}{y_1}$\n\nHence the area of a triangle $(AOB)$ is given by $A=\\frac{a^2b^2}{2x_1y_1}$ now I am stuck, how should I proceed?\n\n\u2022 What is the derivative $y'$? Aug 4, 2016 at 3:41\n\u2022 Notice affine transformation preserves tangency and ratio of areas, you can use an affine transform to map the ellipse to the unit circle. The smallest triangle $AOB$ will be mapped to a right angled isoceles triangle of side $\\sqrt{2}$ and area $1$. This means the smallest triangle $AOB$ has area $ab$. Aug 4, 2016 at 4:00\n\n$A = \\frac {a^2 b^2}{2xy}$ constrained by $\\frac {x^2}{a^2} + \\frac {y^2}{b^2} = 1$\n\n$\\frac{dA}{dx} = \\frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\\\ y + x y' = 0\\\\ y' = -\\frac yx$\n\nDifferentiating the constraint.\n\n$\\frac x{a^2} + \\frac {y y'}{b^2} = 0\\\\ y' = -\\frac {b^2x}{a^2y}\\\\ \\frac {b^2x}{a^2y} = \\frac yx\\\\ b^2 x^2 = a^2 y^2$\n\nAgain from the constraint:\n\n$b^2 x^2 + a^2 y^2 = a^2 b^2\\\\ 2a^2 y^2 = a^2 b^2\\\\ y^2 = \\frac {b^2}{2}\\\\ y = \\frac b{\\sqrt {2}}\\\\ 2xy = ab\\\\ A = \\frac {a^2b^2}{2xy} = ab$\n\nalternate\n\n$x = a \\cos t\\\\ y = b \\sin t\\\\ A = \\frac {ab}{2} \\csc t \\sec t\\\\ \\frac{dA}{dt} = \\frac {ab}{2} \\sec t \\csc t ( -\\cot t + \\tan t) = 0\\\\ \\cot t = \\tan t\\\\ t = \\frac {\\pi}{4}\\\\ A = ab$\n\nThat is much nicer.\n\nWith the change of coordinates $x=au, y=bv$ the problem boils down to the same problem where the original ellipse is replaced by a unit circle. In such a case it is trivial that the minimum area of $AOB$ is $1$. Since the determinant of the Jacobian of the linear map $x=au, y=bv$ is $ab$, the answer to the original question is $\\color{red}{ab}$, plain and simple.", "date": "2022-05-25 11:13:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1881826/a-tangent-to-the-ellipse-meets-the-x-and-y-axes-if-o-is-the-origin-find", "openwebmath_score": 0.9090828895568848, "openwebmath_perplexity": 104.9264473499854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474913588877, "lm_q2_score": 0.8688267745399466, "lm_q1q2_score": 0.85957158981653}}
{"url": "https://qudt.org/vocab/unit/PlanckTime", "text": "unit:PlanckTime\n\nType\nDescription\n\nIn physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\\\ t_P \\equiv \\sqrt{\\frac{\\hbar G}{c^5}} \\approx 5.39106(32) \\times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\\hbar$$ is the reduced Planck's constant (defined as $$\\hbar = \\frac{h}{2 \\pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.\n\nProperties\n0.0000000000000000000000000000000000000000000539124\n$$t_P$$\nAnnotations\nIn physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\\\ t_P \\equiv \\sqrt{\\frac{\\hbar G}{c^5}} \\approx 5.39106(32) \\times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\\hbar$$ is the reduced Planck's constant (defined as $$\\hbar = \\frac{h}{2 \\pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.\nPlanck Time(en)\n\nGenerated 2022-11-28T15:45:48.023-05:00 by lmdoc version 1.1 with \u00a0TopBraid SPARQL Web Pages (SWP)", "date": "2022-12-02 20:40:05", "meta": {"domain": "qudt.org", "url": "https://qudt.org/vocab/unit/PlanckTime", "openwebmath_score": 0.8906638622283936, "openwebmath_perplexity": 179.52588681523468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474907307123, "lm_q2_score": 0.868826769445233, "lm_q1q2_score": 0.8595715842303123}}
{"url": "https://math.stackexchange.com/questions/4180075/generalized-eigenvector-for-product-of-commuting-matrices", "text": "# Generalized eigenvector for product of commuting matrices\n\nSuppose $$A,B$$ are commuting invertible matrices with a common generalized eigenvector $$v$$ with eigenvalues $$a,b$$ respectively. That is, suppose there exist positive integers $$K,L$$ such that $$(A-aI)^K v= 0$$ and $$(B-bI)^Lv=0$$. Is it true that there exists a positive integer $$M$$ such that $$(AB-abI)^Mv = 0$$?\n\nA bit of context: I came across this while thinking about the analogous version of this for eigenvalues (when $$K=1$$ and $$L=1$$). This is easy to show. In fact, the above statement is also easy to show if only one of $$K=1$$ or $$L=1$$ is true (that is, $$v$$ is an eigenvector for one matrix and a generalized eigenvector for the other).\n\nProof: WLOG suppose $$L=1$$. Then we have \\begin{align} (AB-abI)^Kv &= \\sum_{j=0}^K {K \\choose j} A^{K-j}B^{K-j}a^jb^jv\\\\ &= \\sum_{j=0}^K {K \\choose j} A^{K-j}a^jb^Kv\\\\ &= b^K(A-aI)^Kv\\\\ &= 0 \\end{align}\n\nA natural question: is it true if $$v$$ is a generalized eigenvector for both matrices?\n\n\u2022 Welcome! Note that askers are expected to provide context for their questions, as is explained here. For example, it would be helpful if you could edit your questions to address some of the following. What motivated this problem, or where did you encounter it? What are your thoughts on the problem? What have you tried so far? \u2013\u00a0Ben Grossmann Jun 22 at 16:20\n\u2022 A potentially helpful way to reframe the problem: let $P = A - aI$ and $Q = B - bI$. We could equivalently ask: if $PQ = QP$ and $K,L$ are such that $P^Kv = Q^Lv = 0$, then does there necessarily exist an $M$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ \u2013\u00a0Ben Grossmann Jun 22 at 16:25\n\u2022 The answer is yes: the statement is true. Once you provide some context, I can leave a proof as an answer. \u2013\u00a0Ben Grossmann Jun 22 at 16:46\n\u2022 Just added some context. Thanks for pointing the rules out btw, I'm a new poster here. \u2013\u00a0Underbounded Jun 22 at 16:47\n\u2022 I figured as much, thanks for obliging. \u2013\u00a0Ben Grossmann Jun 22 at 16:48\n\nHere is a more abstract perspective. Let $$V$$ be the smallest subspace of our vector space which contains $$v$$ and is closed under the actions of $$A$$ and $$B$$. Let $$R$$ be the subring of the endomorphism ring of $$V$$ generated by the scalar matrices together with $$A$$ and $$B$$. By assumption, this $$R$$ is commutative, and $$(A-a)^K$$ and $$(B-b)^L$$ are equal to $$0$$ in this ring (since they annihilate $$v$$ and $$v$$ generates $$V$$ under the action of $$R$$). The question then becomes simply:\n\nLet $$R$$ be a commutative ring and $$A,B,a,b\\in R$$. Suppose $$A-a$$ and $$B-b$$ are nilpotent. Then must $$AB-ab$$ be nilpotent?\n\nBut now the answer is very easy using the fact that the set $$N$$ of nilpotent elements of $$R$$ is an ideal. By assumption, $$A=a$$ and $$B=b$$ in the quotient ring $$R/N$$. Thus $$AB=ab$$ in that quotient ring. That is, $$AB-ab\\in N$$.\n\n\u2022 This is a really neat approach! \u2013\u00a0Underbounded Jun 22 at 21:23\n\nYes. We can proceed as follows.\n\nBegin by reframing the question: if $$PQ = QP$$ and $$K,L$$ are such that $$P^Kv = Q^Lv = 0$$ and if $$a,b$$ are arbitrary scalars, then does there necessarily exist an $$M$$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ Note that $$[P + aI][Q + bI] - ab I = PQ + bP + aQ.$$ Now, take $$m = \\max\\{K,L\\}$$ and $$M = 2m-1$$; note that $$P^mv = Q^mv = 0$$. Apply the multinomial theorem: $$(PQ + bP + aQ)^Mv = \\sum_{k_1 + k_2 + k_3 = M} \\binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2}P^{k_1+k_2}Q^{k_1+k_3}v \\\\= \\sum_{k_1 + k_2 + k_3 = M} \\binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2} Q^{k_1+k_3}P^{k_1+k_2}v.$$ Now, for all triples of non-negative integers $$k_1,k_2,k_3$$ with $$k_1 + k_2 + k_3 = 2m-1$$, we see that $$k_1 + k_2$$ and $$k_1 + k_3$$ are non-negative integers with sum greater than or equal to $$2m-1$$. It follows that either $$k_1 + k_2 \\geq m$$ or $$k_1 + k_3 \\geq m$$. In either case, we find that $$P^{k_1+k_2}[Q^{k_1 + k_3}v] = Q^{k_1+k_3}[P^{k_1+k_2}v] = 0.$$ Thus, the above expansion of $$(PQ + bP + aQ)^Mv$$ must be equal to zero.\n\n\u2022 Binomial theorem suffices if you write $AB-abI$ as $(A-aI)B+a(B-bI)$. This also lowers the value of $m$ to $K+L-1$. \u2013\u00a0user1551 Jun 22 at 17:45\n\u2022 @user1551 That's neat! Actually, I realize now that $M = K +L - 1$ suffices for me as well; we only need $k_1 + k_2 \\geq K$ or $k_1 + k_3 \\geq L$ \u2013\u00a0Ben Grossmann Jun 22 at 17:57", "date": "2021-08-02 10:28:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4180075/generalized-eigenvector-for-product-of-commuting-matrices", "openwebmath_score": 0.9855372905731201, "openwebmath_perplexity": 124.23096803506014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474879039229, "lm_q2_score": 0.8688267677469952, "lm_q1q2_score": 0.8595715800941748}}
{"url": "https://mathematica.stackexchange.com/questions/113402/what-do-the-lines-in-a-contourplot-mean", "text": "# What do the lines in a ContourPlot Mean?\n\nIf you plot a two variable function like $x^2+y^2$, you get a series of circles with lines and colored areas. My question is, what do the lines and the colored areas represent?\n\n\u2022 Try this experiment: ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}], followed by Plot3D[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}] and compare the two. Now try ContourPlot[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }] and Plot3D[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }]. Got it? ContourPlot by default generates colorized grayscale output, in which larger values are shown lighter. You may also like to read: en.wikipedia.org/wiki/Contour_line and itl.nist.gov/div898/handbook/eda/section3/contour.htm \u2013\u00a0Moo Apr 23 '16 at 12:50\n\u2022 @Moo, OK, so now I know why there are areas of different colors but what is the meaning if the lines bounding each colored area? Do they represent something? \u2013\u00a0MrDi Apr 23 '16 at 12:56\n\u2022 It's explained well here: en.wikipedia.org/wiki/Contour_line The value of the function doesn't change as you move along a contour. \u2013\u00a0Szabolcs Apr 23 '16 at 13:10\n\u2022 @MrDi: There needs to be a way to select peaks and valleys and to bound them somehow, so they use some algorithm that picks those areas. I think you can find some details on the Wiki link I provided. it sounds like you get it, think of it as trying to collapse a 3D image onto a 2D surface. These diagrams can be extremely useful. The lines are just separating areas where something is changing based on some algorithm. \u2013\u00a0Moo Apr 23 '16 at 13:13\n\u2022 @Moo, OK, thanks for help. \u2013\u00a0MrDi Apr 23 '16 at 13:14\n\nIt might be easier for you to see the correspondence between a contour plot and a surface plot if you plot them side-by-side like this (code adapted from here):\n\npeaks[x_, y_] := 3 (1 - x)^2 Exp[-x^2 - (y + 1)^2] - 10 (x/5 - x^3 - y^5) Exp[-x^2 - y^2] -\nExp[-(x + 1)^2 - y^2]/3\n\nWith[{c = Subdivide[-5, 5, 10]},\nGraphicsRow[{ContourPlot[peaks[x, y], {x, -3, 3}, {y, -3, 3},\nColorFunction -> \"DarkTerrain\",\nContours -> c, PlotRange -> All],\nPlot3D[peaks[x, y], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> None,\nBoxed -> False, ColorFunction -> \"DarkTerrain\",\nMeshFunctions -> {#3 &}, Mesh -> {c}, PlotRange -> All,\nViewPoint -> {0, -2.7, 3.}]}]]\n\n\nwhere I took the opportunity to use a function with more interesting contours. (Unfortunately, I don't know of any way to refine the meshlines produced by MeshFunctions; if you know a way, let me know in the comments!)\n\nAs noted in the comments above, the contours in ContourPlot[] are so-called isoclines; that is, lines of constant altitude ($z$ value). The interpretation of the colors depends on what colormap you're using; in this case, I used the \"DarkTerrain\" color gradient, which goes like this:\n\nwhere the bluish color (\"water\") corresponds to the valleys, the whitish color (\"snowcap\") corresponds to the peaks, and the earthy colors corresponds to intermediate values. If you are familiar with map reading, running along isoclines corresponds to staying at a constant (flat) altitude, and crossing the isoclines corresponds to either ascent or descent.\n\nA biographical segue: I was a Boy Scout once upon a time, and map reading was de rigueur for Scouts who wanted to have camping experience.", "date": "2019-09-19 15:48:57", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/113402/what-do-the-lines-in-a-contourplot-mean", "openwebmath_score": 0.26597487926483154, "openwebmath_perplexity": 1176.8868397755964, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138131620183, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8595539291032197}}
{"url": "https://math.stackexchange.com/questions/4484433/non-diagonal-n-th-roots-of-the-identity-matrix/4484681#4484681", "text": "# Non-diagonal n-th roots of the identity matrix\n\nQ. Are there $$n$$-th root analogs of this non-diagonal cube-root of the $$3 \\times 3$$ identity matrix?\n\n\\begin{align*} \\left( \\begin{array}{ccc} 0 & 0 & -i \\\\ i & 0 & 0 \\\\ 0 & 1 & 0 \\\\ \\end{array} \\right)^3 = \\left( \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ \\end{array} \\right) \\end{align*}\n\nI am looking for $$A^n=I$$, where $$I$$ is any dimension $$\\le n$$.\n\n(A naive question: I am not an expert in this area.)\n\n\u2022 I'm not sure what features you want of your \"analogue\" here. Are you just looking for a non-diagonal $A$ such that $A^n = I$? Does it have to be $3 \\times 3$? Or maybe $n \\times n$? Does it have to match some feature of how the non-zero terms are laid out (e.g. one non-zero term in each row/column)? It would be good to get some more information about what you'd like to see in an answer. Jul 1 at 20:26\n\u2022 The problem I see is that a matrix can have n different nth roots of a matrix. How do you decide which you want? Jul 1 at 20:38\n\u2022 @GeorgeIvey \"The problem I see is that a matrix can have n different nth roots of a matrix.\" More, usually, and occasionally fewer. Jul 1 at 20:41\n\u2022 The companion matrix of the $n$th cyclotomic polynomial has order $\\phi(n)$, which is less than $n$. Does that work for you?\n\u2013\u00a0lhf\nJul 1 at 20:48\n\u2022 Hi, Joseph. The permutation matrix of a cyclic permutation always works. Apparently you can alter the final pair of 1's into $i$ and $-i$ and still get $A^n = I$ Jul 1 at 21:19\n\nWe want to solve $$A^n = I$$ where $$A$$ and $$I$$ are $$m \\times m$$\n\nLet $$A = P \\Lambda P^{-1}$$\n\nwhere $$\\Lambda = \\text{diag}(\\omega_1, \\omega_2, \\dots , \\omega_m)$$\n\nwhere the $$\\omega_k$$'s are $$n$$-roots of unity. That is, $$\\omega_k = \\exp\\big(\\dfrac{i 2 \\pi j}{n} \\big)$$ ,$$k = 1,2,\\dots,m$$ and $$j \\in \\{ 0, 1, 2, ...., n-1 \\}$$\n\nand $$P$$ is any invertible $$m \\times m$$ matrix.\n\nThen $$A^n = \\big(P \\Lambda P^{-1}\\big)\\big(P \\Lambda P^{-1}\\big)\\dots \\big(P \\Lambda P^{-1}\\big) = P^{-1} \\Lambda^n P = P^{-1} I P = I$$\n\n\u2022 Yours is the most general example, even in dimension $n$. Can you prove it?\n\u2013\u00a0Ruy\nJul 2 at 0:26\n\u2022 Thanks. Please check my updated solution for the proof. Jul 2 at 1:13\n\nYou can take a rotation matrix that rotates $$\\phi=2\\pi/n$$ around some axis, for example in 3 dimensions:\n\n$$R_\\phi=\\begin{pmatrix} \\cos \\phi & \\sin\\phi & 0 \\\\ -\\sin\\phi & \\cos\\phi &0 \\\\ 0 & 0 & 1\\\\ \\end{pmatrix}$$\n\nThis won't work for $$n=2$$ though, because $$R_\\phi$$ is diagonal in that case like $$R_\\phi=\\operatorname{diag}(-1,-1,1)$$.\n\nThen $$R_\\phi^n=I$$ and $$R_\\phi^k\\neq I$$ for any $$0. You can build new matrices using any invertible matrix $$A$$ and conjugate like\n\n$$R_{\\phi,A}:= AR_\\phi A^{-1} \\tag 1$$\n\nthen obviously:\n\n$$R_{\\phi,A}^n = (AR_\\phi A^{-1})^n = A R_\\phi^nA^{-1} = I$$\n\n(I am not sure whether this could fix the case $$n=2$$ and produce some valid (non-diagonal) results.)\n\n\u2022 You can use Wolfram Alpha to check that {{1,1,0},{0,0,1},{1,0,0}}^-1 * {{-1,0,0},{0,-1,0},{0,0,1}} * {{1,1,0},{0,0,1},{1,0,0}} is not diagonal. Of course, for n=2 you can also use a reflection instead of a rotation, I.e. exchange two basis vectors. Jul 2 at 14:39\n\nIf $$A^n = I$$, i.e., $$A^n - I = 0$$, then the minimal polynomial $$m_A$$ of $$A$$ divides $$p(x) := x^n - 1$$. (For $$n > 1$$) one nondiagonal solution is the companion matrix of $$p$$ itself, namely, $$C_p := \\pmatrix{\\cdot & 1 \\\\ I_{n - 1} & \\cdot}$$ (indeed, $$p = m_{C_p}$$).\n\nNotice that $$C_p$$ is the permutation matrix for the permutation (in fact $$n$$-cycle) $$\\pmatrix{1 & 2 & \\cdots & n - 1 & n}$$ of $$n$$ elements. More generally, the permutation matrix of any permutation of $$n$$ elements of order dividing $$n$$ (and not the identity permutation) yields another solution.\n\nThere are uncountably many solutions for any $$n > 1$$: Given any solution $$A$$ and any invertible matrix $$S$$, $$SAS^{-1}$$ is another solution provided it is not diagonal (since diagonality is a closed condition that by hypothesis is not always satisfied).\n\nThe most general solution of the equation $$A^n=1$$ among $$m\\times m$$ complex matrices is $$A=SDS^{-1},$$ where $$S$$ is an invertible matrix and $$D$$ is a diagonal matrix whose diagonal entries are $$n^{th}$$ roots of unity.\n\nThe reason is that the minimal polynomial of such a matrix $$A$$ divides the polynomial $$x^n-1$$, and hence has distinct roots, which in turn implies that $$A$$ is diagonalizable.", "date": "2022-10-07 08:20:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4484433/non-diagonal-n-th-roots-of-the-identity-matrix/4484681#4484681", "openwebmath_score": 0.9979684352874756, "openwebmath_perplexity": 333.5576481456069, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534392852384, "lm_q2_score": 0.8757869981319862, "lm_q1q2_score": 0.8595441613979325}}
{"url": "https://math.stackexchange.com/questions/2354401/find-a-cubic-polynomial", "text": "# Find a cubic polynomial.\n\nIf $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\\ f(6/5)=(6/5)^3?$\n\nI attempt:\n\nI managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....\n\nBut we can see:\n\n$f(x)=q_1(x-1)+1\\\\f(x)=q_2(x-2)+4\\\\f(x)=q_3(x-3)+9$\n\nAlso when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\\neq4^2$ (from answer).\n\nCan this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?\n\nConsider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.\n\n\u2022 should $f(4)=22$ Jul 11 '17 at 3:58\n\u2022 The OP wrote \"with leading coefficient $1$\". Jul 11 '17 at 3:59\n\u2022 I assumed that $f(6/5)=(6/5)^3$ to be proved from the given information that $f$ is monic.\n\u2013\u00a0user348749\nJul 11 '17 at 4:00\n\nClassic long method:\n\nLet $f(x) = x^3 + b x^2 + c x + d$ with $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, which leads to \\begin{align} f(1) &= 1 = 1 + b + c + d \\hspace{10mm} \\to d = -b - c \\\\ f(2) &= 4 = 8 + 4 b + 2 c + d = 8 + 3b + c \\hspace{10mm} \\to c = -4 - 3b, \\, d = 4 + 2b \\\\ f(3) &= 9 = 27 + 9b + 3c + d = 19 + 2b \\end{align} from which $b = -5$, $c = 11$, and $d = -6$ and $$f(x) = x^3 - 5 \\, x^2 + 11 \\, x -6.$$\n\nWith $f(x)$ then \\begin{align} f(4) &= 64 - 80 + 55 -6 = 22 \\\\ f\\left(\\frac{6}{5}\\right) &= \\left(\\frac{6}{5}\\right)^{3} - \\frac{36 - 66 + 30}{5} = \\left(\\frac{6}{5}\\right)^{3}. \\end{align}\n\nIt may also be noticed that $f(x)$ can be seen in the form $$f(x) = \\left(x - \\frac{5}{3}\\right)^{3} + \\frac{8}{3} \\, \\left( x - \\frac{5}{3}\\right) + \\frac{83}{27}.$$ From this it is easy to see that \\begin{align} f\\left(\\frac{5}{3}\\right) &= 3 + \\frac{2}{27} \\\\ f\\left(\\frac{5}{6}\\right) &= \\frac{59}{216}. \\end{align}\n\n\u2022 Looking at your last form of $f(x)$ if I put $5/3$ in then I would not get the same answer as you have for $f(5/3)$ just after - the former gives $3+2/27$... I've not chased the maths to see which is right, just happened to notice that discrepancy (or I've done something stupid which is also possible). Jul 11 '17 at 8:47\n\nIf\n\n\u2022 $$f(x) = x^3 + ax^2 + bx + c$$\n\n\u2022 $$f(1) = 1$$\n\n\u2022 $$f(2) = 4$$\n\n\u2022 $$f(3) = 9$$\n\nthen\n\n\\begin{align} f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\\\ \\hline 4-1 &= f(2)-f(1) \\\\ 3 &= 7 + 3a + b \\\\ \\hline 9-4 &= f(3) - f(2) \\\\ 5 &= 19 + 5a + b \\\\ \\hline 3a + b &= -4 \\\\ 5a + b &= -14 \\\\ \\hline a &= -5 \\\\ b &= 11 \\\\ c &= -6 \\end{align}\n\nS0 $$f(x) = x^3 - 5x^2 + 11x - 6$$.\n\nYou could make a difference table\n\n$$\\begin{array}{c} 1 && 4 && 9 \\\\ & 3 && 5 \\\\ && 2 \\end{array}$$\n\nUsing $$f(x) = x^3 + ax^2 + bx + c$$, this corresponds to\n\n$$\\begin{array}{c} 1+a+b+c && 8 + 4a + 2b + c && 27 + 9a + 3b + c \\\\ & 7+3a+b && 19 + 5a + b \\\\ && 12+2a \\end{array}$$\n\nThen, comparing entries...\n\n\\begin{align} 12+2a=2 &\\implies a = -5 \\\\ 7+3a+b = 3 &\\implies b=11 \\\\ 1+a+b+c = 1 &\\implies c = -6 \\end{align}", "date": "2021-09-27 10:37:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2354401/find-a-cubic-polynomial", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 268.9100203312175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534333179648, "lm_q2_score": 0.8757869981319863, "lm_q1q2_score": 0.859544156171872}}
{"url": "https://math.stackexchange.com/questions/2113526/analyzing-biggl-lfloor-fracx5-bigg-rfloor-bigg-lfloor-fracx7-bigg", "text": "# Analyzing $\\biggl\\lfloor{\\frac{x}{5}}\\bigg\\rfloor=\\bigg\\lfloor{\\frac{x}{7}}\\bigg\\rfloor$\n\nHow many non negative integral values of $x$ satisfy the equation :$$\\biggl\\lfloor{\\dfrac{x}{5}}\\bigg\\rfloor=\\bigg\\lfloor{\\dfrac{x}{7}}\\bigg\\rfloor$$.\n\nMy try:\n\nWriting few numbers and putting in the required equation.\n\n$\\underbrace{0,1,2,3,4}_{\\implies 0=0},\\underbrace{5,6}_{\\displaystyle\\times},\\underbrace{7,8,9}_{\\implies 1=1},\\underbrace{10,11,12,13}_{\\displaystyle\\times} ,\\underbrace{14}_{\\implies 2=2},\\underbrace{15,16,17,18,19,20}_{\\text{LHS gives 3 but RHS gives 2}},\\underbrace{21,22,23,24}_{\\text{LHS gives 4 but RHS gives 3}}$.\n\nThe point which I wish to make here is that after $14$ we'll get $5's$ multiple before than $7's$, hence $\\forall x \\geq 15 \\implies \\biggl\\lfloor{\\dfrac{x}{5}}\\bigg\\rfloor>\\bigg\\lfloor{\\dfrac{x}{7}}\\bigg\\rfloor$\n\nHence the only solution are: $\\{0,1,2,3,4,7,8,9,14\\}$. Does these make sense?\n\nI need to know other solutions that don't incorporate such analyzing, means using some properties and framing it to solve above.\n\nBy the division rule,\n\n$$x=5q+r=7q+s$$ with $0\\le r<5,0\\le s<7$.\n\nFrom this we draw $$2q=r-s\\in[-6,4],$$\n\nthus the only possibilities are with $q=0,1,2$.\n\n\u2022 $q=0$: $r=s$, possible for $x\\in[0,4]\\cap[0,6]=[0,4]$;\n\n\u2022 $q=1$: $5+r=7+s$, possible for $x\\in[5,9]\\cap[7,13]=[7,9]$;\n\n\u2022 $q=2$: $10+r=14+s$, possible for $x\\in[10,14]\\cap[14,20]=[14]$.\n\n\u2022 why intersection? \u2013\u00a0mnulb Jan 25 '17 at 16:52\n\u2022 Because the two bracketings must hold. What else would you do ? \u2013\u00a0Yves Daoust Jan 25 '17 at 16:53\n\u2022 did you came up with those intersections by hit and trials or something different, and if something new, please don't mind adding there. \u2013\u00a0mnulb Jan 25 '17 at 17:08\n\u2022 @Ayushakj: the intervals are drawn from the equations and inequations. \u2013\u00a0Yves Daoust Jan 25 '17 at 17:11\n\nRewrite the given condition as $$5n\\le x<5n+5,\\quad 7n\\le x<7n+7$$ where $n$ is an integer. Since $x\\ge0$, $n\\ge0$ as well, so the above is equivalent to $7n\\le x<5n+5$. The only nonnegative integers $n$ satisfying $7n<5n+5$ are $0$, $1$, and $2$. And so just a little cleanup is needed.\n\nFor a solution the LHS and RHS are a positive integer, let's call this $c$: $$c = \\left \\lfloor \\frac{x}{5} \\right \\rfloor = \\left \\lfloor \\frac{x}{7} \\right \\rfloor$$\n\nThen, there are integers $0 \\leq a < 5, 0 \\leq b < 7$, s.t.: $$x = 5c + a = 7c + b$$ Thus: $$a - b = 2c$$\n\nWith the restrictions to $a,b$, this is enough to find all solutions.\n\nAnalysis:\n\nLet us take the general form where $p>q$ :\n\n$$\\left\\lfloor{\\dfrac{x}{q}}\\right\\rfloor=\\left\\lfloor{\\dfrac{x}{p}}\\right\\rfloor$$\n\nSolution for above equation exists when $\\dfrac{x}{q}-\\dfrac{x}{p}<1\\Rightarrow x<\\dfrac{p\\cdot q}{p-q}$\n\nSo there is no need to check for integers greater than$\\left\\lfloor\\dfrac{p\\cdot q}{p-q}\\right\\rfloor$\n\nAlso $\\left\\lfloor{\\dfrac{x}{q}}\\right\\rfloor=\\left\\lfloor{\\dfrac{x}{p}}\\right\\rfloor=k$ implies \\begin{align} &x\\in [k\\cdot p,(k+1)p) \\cap\\ [k\\cdot q,(k+1)\\cdot q)\\\\ \\Rightarrow\\ \\ &x\\in [k\\cdot p,(k+1)\\cdot q) \\end{align}\n\nThe integers which satisfy the equation are in $$[0p,1q)\\ \\cup\\ [1p,2q)\\ \\cup\\ [2p,3q) \\cup\\ ...\\ \\cup\\ [(n-1)p,nq)$$ where $nq<\\left\\lfloor\\dfrac{p\\cdot q}{p-q}\\right\\rfloor$ .\n\n\u2022 Well done but you are missing the case $k=0$, giving the interval $[0,q)$. \u2013\u00a0Yves Daoust Jan 26 '17 at 7:41", "date": "2019-06-27 12:43:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2113526/analyzing-biggl-lfloor-fracx5-bigg-rfloor-bigg-lfloor-fracx7-bigg", "openwebmath_score": 0.9556778073310852, "openwebmath_perplexity": 887.7086915685115, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534338604443, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8595441518741312}}
{"url": "https://math.stackexchange.com/questions/3304339/let-a-and-b-commute-if-m-and-n-are-relatively-prime-then-ordab", "text": "# Let $a$ and $b$ commute. If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$. [duplicate]\n\nThis is exercise $$10.E.4$$ from Pinter:\n\nLet $$a$$ and $$b$$ be elements of a group $$G$$.\n\nLet ord($$a$$) = $$m$$ and ord($$b$$) = $$n$$.\n\nProve:\n\nLet $$a$$ and $$b$$ commute.\n\nIf $$m$$ and $$n$$ are relatively prime, then ord($$ab$$) = $$mn$$.\n\n(HINT: Use $$10.E.2$$.)\n\nHere is $$10.E.2$$, which Pinter suggests that we use:\n\nIf $$m$$ and $$n$$ are relatively prime, then no power of $$a$$ can be equal to any power of $$b$$ (except for $$e$$).\n\nI'm also going to use $$10.E.1$$:\n\nIf $$a$$ and $$b$$ commute, then ord($$ab$$) is a divisor of lcm($$m$$,$$n$$).\n\nAs well as $$B.T6.i$$ (Theorem $$6.i$$ from Appendix $$B$$. A fact from basic number theory.):\n\nIf gcd($$m$$,$$n$$) = 1 then lcm($$m$$,$$n$$) = $$mn$$\n\nLet's begin.\n\nWe are given that $$m$$ and $$n$$ are relatively prime which means that:\n\n$$\\text{gcd}(m,n) = 1$$\n\nBy $$B.T6.i$$:\n\n$$\\text{lcm}(m,n) = mn \\tag{1}$$\n\nBy $$10.E.1$$:\n\n$$\\text{ord}(ab)\\ |\\ \\text{lcm}(m,n)$$\n\nSubstituting $$(1)$$:\n\n$$\\text{ord}(ab)\\ |\\ mn$$\n\nWhich means that there is an integer $$x$$ such that:\n\n$$\\text{ord}(ab) x = mn$$\n\nThis is so close! For the theorem to be true, we'd have to show that $$x = 1$$.\n\nHowever, the original exercise statement says to use $$10.E.2$$.\n\n\u2022 Is that helpful in showing that $$x = 1$$?\n\u2022 Or is there some other completely different approach whereby $$10.E.2$$ is used?\n\nThe proof uses the following fact:\n\nIf $$a|c$$ and $$b|c$$ then $$lcm(a,b)|c$$.\n\nFor this exercise, I wanted to only use theorems that had been presented in the book up to that point. And, I didn't seem to recall seeing this theorem. (If anyone spots this in Pinter, please comment below with the location.)\n\nHowever, I did notice that the following similar fact is in Appendix B (REVIEW OF THE INTEGERS) as exercise B.9:\n\nIf $$a|c$$ and $$b|c$$ and $$gcd(a,b) = 1$$ then $$ab|c$$.\n\nSo yeah, it looks like the approach shown below is definitely a valid way to go if you want to stick to what's presented in the book.\n\n\u2022 \u2013\u00a0Angina Seng Jul 26 '19 at 4:35\n\u2022 \u2013\u00a0Dietrich Burde Jul 26 '19 at 11:36\n\u2022 The 2nd $\\Rightarrow$ in my answer in the dupe is essentially a proof of 10.E.2, so replace that by an invocation of 10.E.2 to obtain the hinted proof. This is a well-known proof and likely occurs here many times. \u2013\u00a0Bill Dubuque Jul 26 '19 at 14:47\n\n### $$\\boxed{\\textit{You can show mn\\ \\big|\\ \\text{ord}(ab) using 10.E.2:}}$$\n\nSince $$a$$ and $$b$$ commute we can distribute common powers of $$a$$ and $$b$$: $$e = (ab)^{\\text{ord}(ab)} = a^{\\text{ord}(ab)}b^{\\text{ord}(ab)}$$ This implies $$a^{\\text{ord}(ab)} = b^{-\\text{ord}(ab)}$$ or, in other words, some power of $$a$$ equals some power of $$b$$. But $$10.E.2$$ says that since $$m = \\text{ord}(a)$$ and $$n = \\text{ord}(b)$$ are relatively prime, no power of $$a$$ can equal any power of $$b$$ unless those powers of $$a$$ and $$b$$ both evaluate to $$e$$.\n\nSo we must have $$a^{\\text{ord}(ab)} = e = a^m$$ and $$b^{-\\text{ord}(ab)} = e = b^n$$.\n\nBut then, $$m\\ \\big|\\ \\text{ord}(ab)$$. This is because $$m$$ is the order of $$a$$ i.e. the least positive integer power of $$a$$ that evaluates to $$e$$; so if any other positive integer power of $$a$$ evaluates to $$e$$ (like $$\\text{ord}(ab)$$ here), then $$m$$ must divide that integer. For a similar reason, $$n\\ \\big|\\ \\text{ord}(ab)$$.\n\nTherefore since both $$m$$ and $$n$$ divide $$\\text{ord}(ab)$$, their least common multiple $$\\text{lcm}(m,n)$$ must divide $$\\text{ord}(ab)$$. But in our case $$\\text{lcm}(m,n) = mn$$ so we finally have $$mn\\ \\big|\\ \\text{ord}(ab)$$.\n\nAlong with the result you already established $$\\text{ord}(ab)\\ \\big|\\ mn$$, this suggests $$\\text{ord}(ab) = mn$$.\n\nP.S. If you don't believe the bolded statements above, use Euclidean Division. E.g. to see $$m\\ \\big|\\ \\text{ord}(ab)$$, use Euclidean Division on the integer pair $$\\big(m, \\text{ord}(ab)\\big)$$ to get $$\\text{ord}(ab) = km + r$$ for some $$k, r \\in \\Bbb Z$$ with $$0 \\leq r < m$$. Do you see why $$r$$ must be $$0$$?\n\n\u2022 Excellent and very clear presentation! I learned a few things from your proof. Thanks very much for explaining each part. I've updated my answer with some comments regarding your approach. \u2013\u00a0dharmatech Jul 26 '19 at 20:05", "date": "2020-04-03 23:46:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3304339/let-a-and-b-commute-if-m-and-n-are-relatively-prime-then-ordab", "openwebmath_score": 0.9036380052566528, "openwebmath_perplexity": 222.38750805319668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534344029238, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8595441507582819}}
{"url": "https://www.physicsforums.com/threads/complex-analysis-residue-integration-question.781025/", "text": "Complex analysis: residue integration question\n\nTags:\n1. Nov 9, 2014\n\nWheelwalker\n\nI'm asked to evaluate the following integral: $\\int_{c} \\frac{30z^2-23z+5}{(2z-1)^2(3z-1)}dz$ where c is the unit circle. This function has a simple pole at $z=\\frac{1}{3}$ and a second order pole at $z=\\frac{1}{2}$, both of which are within my region of integration. I then went about computing the residues using the following formula (which I, mistakenly perhaps, thought was general and applied to any mth-order pole): $Res f(z) = \\frac{1}{(1-1)!}\\lim_{z \\to 1/2}{\\frac{d}{dz}\\{[2z-1]^2 \\frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\\}$ for which I got 2 (this was just the residue for the first pole). The residue I got for the other singularity was 6. They are supposed to be 1/2 and 2, respectively. What am I doing wrong, exactly? Does this residue formula not apply to this case? If so, why not? Thanks in advance.\n\nLast edited: Nov 9, 2014\n2. Nov 9, 2014\n\nlurflurf\n\nThe formula you should have used is $$Res f(z) = \\frac{1}{(2-1)!}\\lim_{z \\to 1/2}{\\frac{d}{dz}\\{[z-\\dfrac{1}{2}]^2 \\frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\\}$$\nThat is why you got 2=4(1/2) instead of 1/2 and 6=3*2 instead of 2\n\n3. Nov 9, 2014\n\nWheelwalker\n\nOh, wow that makes a lot of sense. I got in the habit of multiplying $f(z_0)$ by whichever term was creating the singularity in the denominator because a lot of the first problems we did would have something simple like $(z-1)$ there. So then, $(z-z_0)$ would equal exactly what was there. Anyway, thank you so much!\n\n4. Nov 9, 2014\n\nWheelwalker\n\nAh, and yes that factorial term definitely should have been $(2-1)!$, although of course it doesn't matter in this case.\n\n5. Nov 9, 2014\n\nWheelwalker\n\nOne more question for you actually. So, is this the most general residue formula? This should apply to nearly any case?\n\n6. Nov 10, 2014\n\nmathwonk\n\nI am not very good at computations, but I am getting residues opposite to what you seem to say, i.e. res = 1/2 at z=1/2, and res = 2 at z= 1/3.\n\nMy reasoning is as follows: the residue at a, is the coefficient of 1/(z-a) in the Laurent expansion in powers of (z-a). So at z= 1/3 we should have 1/(z-1/3) . g(z), where g(z) = (30z^2 -23z + 5)/3(2z-1)^2. Then when we expand g in powers of (z-1/3), the constant term will be the residue, i.e. the coefficient of 1/(z-1/3) after we multiply by the 1/(z-1/3) out front. so this is just g(1/3) = (30/9 - 23/3 + 5)/3(1/3)^2 = 2.\n\nThen for the other one, at z= 1/2, the function can be written as 1/(z-1/2)^2 h(z), where h(z) = (30z^2-23z+5)/4(3z-1). then the residue after multiplying by the factor 1/(z-1/2)^2, will be the linear coefficient of the expansion of h(z), i.e. should be the derivative of h(z). By the quotient rule, this is (1/4)(7/2 - 3)/(1/4) = (1/8)/(1/4) = 1/2.\n\nAm I missing something? My point in doing it this way is to emphasize not memorizing formulas. but going back to the meaning of the concept. But of course one wants to get it right, and I am not super confident after apparently getting a different answer. Or maybe I just misunderstood what was said.\n\n7. Nov 10, 2014\n\nWheelwalker\n\nThose are the correct residues. I made a mistake in finding them the first time.\n\n8. Nov 10, 2014\n\nmathwonk\n\nThank you. My recommended point of view, as I said, is to write the function as a product of form 1/(z-a)^m . g(z), where g is holomorphic at a.\n\nThen we want the coefficient of (z-a)^(m-1) in the Taylor expansion of g. this is of course just g^(m-1)(a) /(m-1)!, i.e. the (m-1) st derivative of g at a, divided by (m-1)!\n\nI suggest this because your error was apparently in recalling the residue formula incorrectly. I hope this method makes that less likely, since all you have to recall this way is that you want the coefficient of 1/(z-a). In the product 1/(z-a)^m . g(z), that will be the coefficient of (z-a)^(m-1) in the expansion of g.", "date": "2018-07-16 09:02:35", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/complex-analysis-residue-integration-question.781025/", "openwebmath_score": 0.922269344329834, "openwebmath_perplexity": 602.8112363397888, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426458162397, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.859522375289162}}
{"url": "https://math.stackexchange.com/questions/1918766/absolute-value-inequality-with-variable-on-both-sides", "text": "# Absolute value inequality with variable on both sides\n\nI am trying to solve the following inequality:\n\n$$|3-5x| \\le x$$\n\nI am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:\n\n$$-x \\le 3-5x \\le x$$\n\nThen I solved for each side separately,as follows:\n\n$$3-5x \\le x$$\n\n$$x \\ge (1/2)$$\n\n$$-x \\le 3-5x$$\n\n$$x \\le \\frac34$$\n\nI know my solution is incorrect and that it actually lies between $\\frac12$ and $\\frac34$, but I wanted to know what is wrong with my method and what is the appropriate approach to solving such inequalities.\n\nThanks,\n\nThis is how I learned it back in Algebra I.\n\nYou can split it into two equations. Since the sign is $\\le$, it will be that Equation 1 and Equation 2 are true.\n\nHere are the steps:\n\n$$|3-5x| \\le x$$\n\n$$3-5x \\le x \\;\\;\\;\\;\\;| \\;\\;-3+5x\\le x$$\n\n$$6x \\ge 3 \\;\\;\\;\\;\\;\\;\\;| \\;\\;-4x \\ge -3$$\n\n$$x \\ge \\frac 12 \\;\\;\\;| \\;\\;\\;\\;x \\le \\frac 34$$\n\n$$\\frac 12 \\le x \\le \\frac 34$$\n\n\u2022 This should be sufficient, but for further elaboration just reply Sep 8 '16 at 2:38\n\u2022 This approach is very specific towards linear inequalities (but definitely works, I think)--it doesn't work (I believe) for higher order ones. Sep 8 '16 at 3:53\n\u2022 Yes, I know. I was trying to give a simple easy-to-understand solution. Sep 8 '16 at 4:02\n\u2022 Your \"$|$\" symbol is for AND or OR? In either case you have a logical issue as you start of with an OR and end with and AND. Sep 8 '16 at 4:15\n\u2022 @Maccavity And. And please point out the issue. Sep 8 '16 at 4:16\n\nDivide $-x \\leq 3-5x \\leq x$ by $x$ and go from there. Although I think what you have is correct, the solution IS the interval $\\dfrac{1}{2} \\leq x \\leq \\dfrac{3}{4}$.\n\nYou need to break up the absolute value into its intervals:\n\n$$|x| = \\begin{cases} x & x > 0 \\\\ -x & x < 0 \\\\ 0 & x = 0 \\end{cases}$$\n\nTherefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero:\n\n$$|3-5x| = \\begin{cases} 3 - 5x & 3 - 5x > 0 \\rightarrow 3 > 5x \\rightarrow x < \\frac{3}{5}\\\\ 5x - 3 & 3 - 5x < 0 \\rightarrow 3 <5x \\rightarrow x > \\frac{3}{5} \\\\ 0 & 3 - 5x = 0 \\rightarrow 3 = 5x \\rightarrow x = \\frac{3}{5} \\end{cases}$$\n\nNow you solve the inequality in each case:\n\n1. $x < \\frac{3}{5} \\rightarrow |3 - 5x| = 3 - 5x$ $$3 - 5x \\leq x \\\\ 3 \\leq 6x \\\\ x \\geq \\frac{1}{2}$$\n2. $x > \\frac{3}{5} \\rightarrow |3 - 5x| = 5x - 3$ $$5x - 3 \\leq x \\\\ 4x \\leq 3 \\\\ x \\leq \\frac{3}{4}$$\n3. $x = \\frac{3}{5} \\rightarrow |3 - 5x| = 0$ $$0 \\leq x \\\\ x \\geq 0$$\n\nNow you need to analyze each case:\n\n1. $x < \\frac{3}{5} \\wedge x \\geq \\frac{1}{2}$\n\nIt is true that $\\frac{3}{5} = \\frac{6}{10} \\geq \\frac{5}{10}$. Therefore this particular interval is true for $\\frac{1}{2} \\leq x < \\frac{3}{5}$.\n\n1. $x > \\frac{3}{5} \\wedge x \\leq \\frac{3}{4}$\n\nSince $\\frac{3}{5} = \\frac{12}{20}$ and $\\frac{3}{4} = \\frac{15}{20}$, this is true for $\\frac{3}{5} < x \\leq \\frac{3}{4}$.\n\n1. $x = \\frac{3}{5} \\wedge x \\geq 0$\n\n$\\frac{3}{5} > 0$ therefore this is trivially satisfied--thus $x = \\frac{3}{5}$ is allowed.\n\nWhen we combine these results, we find that $\\frac{1}{2} \\leq x \\leq \\frac{3}{4}$.\n\nPS Edit:\n\nIt's probably easier to use:\n\n$$|x| = \\begin{cases} x & x \\geq 0 \\\\ -x & x \\leq 0 \\\\ \\end{cases}$$\n\nIn that case you get the two intervals:\n\n$$\\frac{1}{2} \\leq x \\leq \\frac{3}{5}$$\n\nand\n\n$$\\frac{3}{5} \\leq x \\leq \\frac{3}{4}$$\n\nWhich clearly combines to give: $\\frac{1}{2} \\leq x \\leq \\frac{3}{4}$.\n\nYour solution was correct except you needed to use the word AND correctly. The inequality $-x \\le 3-5x \\le x$ is equivalent to the inequalities\n\n\\begin{align} -x \\le 3-5x \\quad &\\text{AND} \\quad 3-5x \\le x \\\\ 4x \\le 3 \\quad &\\text{AND} \\quad 3 \\le 6x \\\\ x \\le \\frac 34 \\quad &\\text{AND} \\quad \\frac 12 \\le x \\\\ \\frac 12 \\le x \\quad &\\text{AND} \\quad x \\le \\frac 34 \\\\ x &\\in\\left[\\frac 12, \\frac 34 \\right] \\end{align}", "date": "2021-11-28 06:40:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1918766/absolute-value-inequality-with-variable-on-both-sides", "openwebmath_score": 0.9988861680030823, "openwebmath_perplexity": 319.48798620940573, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426405416754, "lm_q2_score": 0.8824278710924296, "lm_q1q2_score": 0.8595223736464394}}
{"url": "https://math.stackexchange.com/questions/2355252/let-tm-1v-neq-0-but-tmv-0-show-that-v-tv-ldots-tm-1v-are-linearly", "text": "# Let $T^{m-1}v\\neq 0$ but $T^mv=0$. Show that $v,Tv,\\ldots,T^{m-1}v$ are linearly independent\n\nLet $T\\in\\mathcal L(V)$ and $T^{m-1}v\\neq 0$ but $T^mv=0$ for some positive integer $m$ and some $v\\in V$. Show that $v,Tv,\\ldots,T^{m-1}v$ are linearly independent.\n\nI had written a proof but Im not sure if it is correct. And in the case it would be correct I dont know how to write it better and clearly. So I have two questions:\n\n1. It is the proof below correct?\n\n2. If so, how I can write it better using the same ideas?\n\n### The attempted proof:\n\n1) If $T^{m-1} v$ would be linearly dependent of $T^{m-2} v$ then exists some $\\lambda\\neq 0$ such that\n\n$$T^{m-1}v=\\lambda T^{m-2}v\\implies T^mv=\\lambda T(T^{m-2}v)=\\lambda T^{m-1}v=0\\implies \\lambda=0$$\n\nThen $T^{m-2}v$ is linearly independent of $T^{m-1}v$.\n\n2) Now observe that\n\n$$\\lambda_1v+\\lambda_2Tv+\\lambda_3T^2v=0\\implies T^{m-2}(\\lambda_1v+\\lambda_2Tv+\\lambda_3T^2v)=\\lambda_1T^{m-2}v+\\lambda_2T^{m-1}v=0$$\n\nso $\\lambda_1,\\lambda_2=0$ as we had shown previously, so the original equation reduces to\n\n$$\\lambda_3T^2v=0\\implies \\lambda_3=0$$\n\nthus $v,Tv,T^2v$ are linearly independent.\n\n3) Repeating recursively the analysis in 2) for longer lists of vectors of the form $v,Tv,\\ldots,T^kv$ for $k< m$ we can show that the list $v,Tv,\\ldots,T^{m-1}v$ is linearly independent.\n\n\u2022 This is a related post \u2013\u00a0Bijesh K.S Jul 11 '17 at 19:15\n\u2022 The if and only if in the first line looks problematic as $T$ isn't invertible. \u2013\u00a0DMath Jul 11 '17 at 19:19\n\u2022 @DMath thank you... I overlook it completely. \u2013\u00a0Masacroso Jul 11 '17 at 19:20\n\u2022 I'm afraid that the last part is the bulk of the proof and you just cut it off by saying \u201crepeating the analysis\u201d. Note that for $\\{v,Tv,T^2v,T^3v\\}$ you need to go down to $T^{m-3}$, which is not as easy as the initial step for $\\{T^{m-2}v,T^{m-1}v\\}$. \u2013\u00a0egreg Jul 11 '17 at 20:33\n\nHint\n\n$$a_0v+a_1Tv+\\ldots+a_{m-1}T^{m-1}v=0 \\quad(1)$$\n\nmultiply by $T^{m-1}$, then\n\n$$a_0T^{m-1}v=0\\to a_0=0$$\n\nso, from $(1)$,\n\n$$a_1Tv+\\ldots+a_{m-1}T^{m-1}v=0$$\n\nmultiply by $T^{m-2}$, then\n\n$$a_1T^{m-1}v=0\\to a_1=0$$\n\ncan you finish?\n\n\u2022 ok but, it is my proof correct? \u2013\u00a0Masacroso Jul 11 '17 at 19:17\n\u2022 @Masacroso: it seems correct to me. You just need induction to finish it. \u2013\u00a0Arnaldo Jul 11 '17 at 19:19\n\nSee below for a simple proof.\n\nSuppose that they are not linear independent; that is for some choice of $\\alpha_{k}$,\n\n$T^{m-1}v = \\sum_{k=0}^{m-2}\\alpha_{k}T^{k}v$.\n\nThen $0 =T^{m}v = T\\cdot T^{m-1}v = \\sum_{k=0}^{m-2}\\alpha_{k}T^{k+1}v$,\n\nso $Tv,...,T^{m-1}v$ are linearly dependent.\n\nBut that means that $T^{m-1}v = \\sum_{k=1}^{m-2}\\beta_{k}T^{k}v$.\n\nInductively repeat the argument to eventually conclude that for some $\\gamma \\neq 0$, $T^{m-1}v = \\gamma T^{m-2} v$, but then we would have to say that $0 = T T^{m-1} v = \\gamma T^{m-1} v \\neq 0$, a contradiction.\n\nMore simply put: if you had a linear combination $c_0 v + c_1 T v + \\ldots c_{m-1} T^{m-1} v = 0$ with $c_j$ not all zero, let $c_i$ be the first nonzero coefficient. Then $$T^i v = - \\sum_{j=i+1}^{m-1} (c_j/c_i) T^j v$$ Applying $T^{m-i-1}$ to both sides, $$T^{m-1} v = - \\sum_{j=i+1}^{m-1} (c_j/c_i) T^{m+j-i-1} v = 0$$ But all terms on the right are $0$, so $T^{m-1} v = 0$, contradiction!", "date": "2021-04-20 19:03:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2355252/let-tm-1v-neq-0-but-tmv-0-show-that-v-tv-ldots-tm-1v-are-linearly", "openwebmath_score": 0.8628842830657959, "openwebmath_perplexity": 184.4928373026656, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974042642048694, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.8595223689528747}}
{"url": "http://mathhelpforum.com/calculus/166923-couple-trigonometric-integrals.html", "text": "# Thread: a couple of trigonometric integrals\n\n1. ## a couple of trigonometric integrals\n\n$\\displaystyle \\int_{0}^{\\pi} x \\sin (\\cos^{2}x)\\cos(\\sin^{2}x) \\ dx$\n\n$\\displaystyle \\int_{0}^{\\pi} x \\Big(\\sin^{2}(\\sin x) + \\cos^{2} (\\cos x) \\Big)\\ dx$\n\nMy initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.\n\n2. Originally Posted by Random Variable\n$\\displaystyle \\int_{0}^{\\pi} x \\sin (\\cos^{2}x)\\cos(\\sin^{2}x) \\ dx$\n\n$\\displaystyle \\int_{0}^{\\pi} x \\Big(\\sin^{2}(\\sin x) + \\cos^{2} (\\cos x) \\Big)\\ dx$\nHere's how you do the first one:\n\nSpoiler:\nLet\n\n$\\displaystyle \\displaystyle I=\\int_0^\\pi x\\sin\\left(\\cos^2(x)\\right)\\cos^2(\\sin(x))\\text{ }dx$\n\nThen, letting $\\displaystyle z=\\pi-x$ we find that\n\n\\displaystyle \\displaystyle \\begin{aligned}I &=\\int_0^\\pi (\\pi-z)\\sin\\left(\\cos^2(z)\\right)\\cos\\left(\\sin^2(z)\\ri ght)\\text{ }dz\\\\ &=\\pi\\int_0^\\pi\\sin\\left(\\cos^2(z)\\right)\\cos\\left (\\sin^2(z)\\right)\\text{ }dz-I\\end{aligned}\n\nThus,\n\n$\\displaystyle \\displaystyle \\frac{2}{\\pi}I=\\int_0^\\pi \\sin\\left(\\cos^2(z)\\right)\\sin\\left(\\cos^2(z)\\righ t)\\text{ }dz$\n\nRecall though that\n\n$\\displaystyle \\displaystyle \\sin(\\phi)\\cos(\\theta)=\\frac{\\sin(\\phi+\\theta)+\\si n(\\phi-\\theta)}{2}$\n\nAnd thus\n\n\\displaystyle \\displaystyle \\begin{aligned}\\frac{2}{\\pi} I &= \\int_0^\\pi \\sin\\left(\\cos^2(z)\\right)\\cos^2\\left(\\sin(z)\\righ t)\\text{ }dz\\\\ &= \\frac{1}{2}\\int_0^\\pi\\left(\\sin\\left(\\cos^2(z)+\\si n^2(z)\\right)+\\sin\\left(\\cos^2(z)-\\sin^2(z)\\right)\\right)\\text{ }dz\\\\ &=\\frac{1}{2}\\pi\\sin(1)+\\frac{1}{2}\\int_0^\\pi \\sin(\\cos(2z))\\text{ }dz\\end{aligned}\n\nNote though that\n\n$\\displaystyle \\displaystyle \\int_0^\\pi\\sin(\\cos(2z))\\text{ }dz=\\int_0^{\\frac{\\pi}{4}}\\sin(\\cos(2z))\\text{ }dz+\\int_{\\frac{3\\pi}{4}}^\\pi\\sin(\\cos(2z))\\text{ }dz+\\int_{\\frac{3\\pi}{4}}^{\\pi}\\sin(\\cos(2z))\\text { }dz$\n\nNote though that by letting $\\displaystyle \\displaystyle u=z-\\frac{3\\pi}{4}$ one gets that\n\n$\\displaystyle \\displaystyle \\int_{\\frac{3\\pi}{4}}^{\\pi}\\sin(\\cos(2z))=\\int_0^{ \\frac{\\pi}{4}}\\sin(\\cos(2u))\\text{ }du$\n\nAnd thus\n\n$\\displaystyle \\displaystyle \\int_0^\\pi\\sin(\\cos(2z))\\text{ }dz=2\\int_0^{\\frac{\\pi}{4}}\\sin(\\cos(2z))\\text{ }dz+\\int_{\\frac{\\pi}{4}}^{\\frac{3\\pi}}{4}\\sin(\\cos (2z))\\text{ }dz$\n\nBut, letting $\\displaystyle w=z-\\frac{\\pi}{2}$ in this second integral one arrives at\n\n$\\displaystyle \\displaystyle \\int_{\\frac{\\pi}{4}}^{\\frac{3\\pi}{4}}\\sin(\\cos(2z) )=-\\int_{\\frac{-\\pi}{4}}^{\\frac{\\pi}{4}}\\sin(\\cos(2w))\\text{ }dw$\n\nand since the integrand is even this is equal to\n\n$\\displaystyle \\displaystyle -2\\int_0^{\\frac{\\pi}{4}}\\sin(\\cos(2w))\\text{ }dw$\n\nTherefore,\n\n\\displaystyle \\displaystyle \\begin{aligned}\\int_0^\\pi\\sin(\\cos(2z))\\text{ }dz &= 2\\int_0^{\\frac{\\pi}{4}}\\sin(\\cos(2z))+\\int_{\\frac{ \\pi}{4}}^{\\frac{3\\pi}{4}}\\sin(\\cos(2z))\\text{ }dz\\\\ &= 2\\int_0^{\\frac{\\pi}{4}}\\sin(\\cos(2z))\\text{ }dz-2\\int_0^{\\frac{\\pi}{4}}\\sin(\\cos(2z))\\text{ }dz\\\\ &=0\\end{aligned}\n\nThus, finally we arrive at\n\n\\displaystyle \\displaystyle \\begin{aligned}\\frac{2}{\\pi}I &=\\frac{\\pi}{2}\\sin(1)+\\frac{1}{2}\\int_0^\\pi\\sin(\\ cos(2z))\\text{ }dz\\\\ &=\\frac{\\pi}{2}\\sin(1)\\end{aligned}\n\nand thus $\\displaystyle \\displaystyle I=\\frac{\\pi^2 \\sin(1)}{4}$.\n\n3. The ODE you had was solvable.\n\n4. Originally Posted by Random Variable\nThe ODE you had was solvable.\nSee my new method. It's more elementary.\n\n5. Originally Posted by Random Variable\n$\\displaystyle \\int_{0}^{\\pi} x \\sin (\\cos^{2}x)\\cos(\\sin^{2}x) \\ dx$\n\n$\\displaystyle \\int_{0}^{\\pi} x \\Big(\\sin^{2}(\\sin x) + \\cos^{2} (\\cos x) \\Big)\\ dx$\n\nMy initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.\nThe second one is interesting. I have a solution, but it's a tad long. Do you have any ideas as of yet?\n\n6. to prove that $\\displaystyle \\displaystyle\\int_0^\\pi\\sin(\\cos 2x)\\,dx=0,$ split it up into two integrals with $\\displaystyle 0\\le x\\le\\dfrac\\pi2$ and $\\displaystyle \\dfrac\\pi2\\le x\\le\\pi$ then apply substitution $\\displaystyle t=x-\\dfrac\\pi2$ on the second piece and you'll see is equal to minus the first piece, hence, the result.\n\n7. Here's my solution to the second\n\nSpoiler:\n\nFirst use the double angle formula so\n\n$\\displaystyle \\displaystyle \\int_0^{\\pi} x\\left( \\dfrac{1-\\cos \\left(2 \\sin x\\right)}{2} + \\dfrac{1 + \\cos \\left(2 \\cos x\\right)}{2} \\right)\\, dx$\n\n$\\displaystyle \\displaystyle = \\int_0^{\\pi} x\\,dx + \\frac{1}{2}\\int_0^{\\pi} \\cos \\left(2 \\cos x\\right) - \\cos \\left(2 \\sin x\\right)\\, dx$\n\nWe will show the second integral is zero by considering it in two parts. (**)\n\nFirst, that $\\displaystyle \\displaystyle \\int_0^{\\pi/2} \\cos \\left(2 \\cos x\\right) - \\cos \\left(2 \\sin x\\right)\\,dx = \\int_{\\pi/2}^{\\pi} \\cos \\left(2 \\cos x\\right) - \\cos \\left(2 \\sin x\\right)dx$\n\nLet $\\displaystyle x = \\pi -u$ in $\\displaystyle \\displaystyle \\int_{\\pi/2}^{\\pi} \\cos \\left(2 \\cos x\\right) - \\cos \\left(2 \\sin x\\right)dx$\n\nNext that $\\displaystyle \\displaystyle \\int_0^{\\pi/4} \\cos \\left(2 \\cos x\\right) - \\cos \\left(2 \\sin x\\right)\\,dx = -\\int_{\\pi/4}^{\\pi/2} \\cos \\left(2 \\cos x\\right) - \\cos \\left(2 \\sin x\\right)dx$\n\nLet $\\displaystyle x = \\pi/2-u$ in$\\displaystyle \\displaystyle \\int_{\\pi/4}^{\\pi/2} \\cos \\left(2 \\cos x\\right) - \\cos \\left(2 \\sin x\\right)\\, dx$\n\nThus, we are left with $\\displaystyle \\displaystyle \\int_0^{\\pi} x\\,dx = \\dfrac{\\pi^2}{2}$\n\n8. here's mine to the second:\n\nput $\\displaystyle t=\\pi-x$ on the integral to see that is equal to $\\displaystyle \\displaystyle\\frac\\pi2\\int_0^\\pi\\left( {{\\sin }^{2}}(\\sin x)+{{\\cos }^{2}}(\\cos x) \\right)\\,dx.$\n\ni claim the latter integral equals $\\displaystyle \\pi,$ hence the original integral equals $\\displaystyle \\dfrac{\\pi^2}2,$ so in order to see that write the integral as\n\n$\\displaystyle \\displaystyle\\int_{0}^{\\frac{\\pi }{2}}{\\left( {{\\sin }^{2}}(\\sin x)+{{\\cos }^{2}}(\\cos x) \\right)\\,dx}+\\int_{\\frac{\\pi }{2}}^{\\pi }{\\left( {{\\sin }^{2}}(\\sin x)+{{\\cos }^{2}}(\\cos x) \\right)\\,dx},$\n\nand put $\\displaystyle t=x-\\dfrac\\pi2$ on the second one, then add them up and you'll see that the claimed integral achieve the aforesaid value, as required.\n\nanother solution for the first one:\n\nwe consider $\\displaystyle \\displaystyle\\int_{0}^{\\pi }{\\sin \\left( {{\\sin }^{2}}x \\right)\\cos \\left( {{\\cos }^{2}}x \\right)\\,dx},$ (1) then write it as $\\displaystyle \\displaystyle\\int_{0}^{\\frac{\\pi }{2}}{\\sin \\left( {{\\sin }^{2}}x \\right)\\cos \\left( {{\\cos }^{2}}x \\right)\\,dx}+\\int_{\\frac{\\pi }{2}}^{\\pi }{\\sin \\left( {{\\sin }^{2}}x \\right)\\cos \\left( {{\\cos }^{2}}x \\right)\\,dx}$ and put $\\displaystyle t=x-\\dfrac\\pi2$ on the second one then we get $\\displaystyle \\displaystyle\\int_{0}^{\\frac{\\pi }{2}}{\\sin \\left( {{\\sin }^{2}}x \\right)\\cos \\left( {{\\cos }^{2}}x \\right)\\,dx}+\\int_{0}^{\\frac{\\pi }{2}}{\\sin \\left( {{\\cos }^{2}}x \\right)\\cos \\left( {{\\sin }^{2}}x \\right)\\,dx},$ so when adding those we note that the integrand is actually $\\displaystyle \\sin(\\sin^2x+\\cos^2x)$ so (1) equals $\\displaystyle \\dfrac\\pi2\\sin1.$\n\non the original integral was easy to show that it's equal to $\\displaystyle \\displaystyle\\frac\\pi2\\int_{0}^{\\pi }{\\sin \\left( {{\\sin }^{2}}x \\right)\\cos \\left( {{\\cos }^{2}}x \\right)\\,dx},$ thus by (1) we conclude.", "date": "2018-04-26 10:34:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/166923-couple-trigonometric-integrals.html", "openwebmath_score": 0.9990625977516174, "openwebmath_perplexity": 1176.1248444550968, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429609670701, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8595177078304961}}
{"url": "https://math.stackexchange.com/questions/1470160/combinations-problem-choosing-ways-to-select-8-questions-out-of-12", "text": "Combinations problem: Choosing ways to select $8$ questions out of $12$.\n\nIn an exam, there are $12$ questions in total. He has to attempt $8$ questions in all. There are two parts: Part A and Part B of the question paper containing $5$ and $7$ questions respectively. How many ways are there to attempt the exam, such that you attempt eight questions and from each part you attempt at least three questions?\nPlease don't answer with the three case method. What I want is the error in my method?\nMy method: For three questions from each part, we do: $${5 \\choose 3}*{7\\choose3}$$. But 2 questions are lett, so for that, we multiply the expression by $${6\\choose2}$$. Because 6 questions are left to choose from. But this does not give the answer, why?\n\nAs you are aware, the actual number of ways to select eight questions from the examination given the restrictions that at least three questions must be selected from each part is $$\\binom{5}{3}\\binom{7}{5} + \\binom{5}{4}\\binom{7}{4} + \\binom{5}{5}\\binom{7}{3}$$ The alternative method you proposed of selecting three questions from part A, three questions from part B, and two additional questions counts the same selection more than once.\n\nIf we choose three questions from part A and five questions from part B, the second method counts the same selection of questions ten times since there are $\\binom{5}{3}$ ways to choose three of the five selected questions from part B.\n\nTo make this concrete, suppose the selected questions are $A_1, A_2, A_3, B_1, B_2, B_3, B_4, B_5$. We can make this particular selection once. However, the second method counts this particular selection ten times. \\begin{align*} & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, \\color{blue}{B_1}, \\color{blue}{B_2}, \\color{blue}{B_3}, B_4, B_5\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, \\color{blue}{B_1}, \\color{blue}{B_2}, B_3, \\color{blue}{B_4}, B_5\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, \\color{blue}{B_1}, \\color{blue}{B_2}, B_3, B_4, \\color{blue}{B_5}\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, \\color{blue}{B_1}, B_2, \\color{blue}{B_3}, \\color{blue}{B_4}, B_5\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, \\color{blue}{B_1}, B_2, \\color{blue}{B_3}, B_4, \\color{blue}{B_5}\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, \\color{blue}{B_1}, B_2, B_3, \\color{blue}{B_4}, \\color{blue}{B_5}\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, B_1, \\color{blue}{B_2}, \\color{blue}{B_3}, \\color{blue}{B_4}, B_5\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, B_1, \\color{blue}{B_2}, \\color{blue}{B_3}, B_4, \\color{blue}{B_5}\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, B_1, \\color{blue}{B_2}, B_3, \\color{blue}{B_4}, \\color{blue}{B_5}\\\\ & \\color{red}{A_1}, \\color{red}{A_2}, \\color{red}{A_3}, B_1, B_2, \\color{blue}{B_3}, \\color{blue}{B_4}, \\color{blue}{B_5}\\\\ \\end{align*}\n\nBy similar reasoning, if we choose five questions from part A and three questions from part B, the second method counts the same selection of questions ten times since there are $\\binom{5}{3}$ ways to choose three of the five selected questions from part A.\n\nIf we choose four questions from part A and four questions from part B, the second method counts the same selection sixteen times since we count the same set of four questions from part A four times for each of the $\\binom{4}{3}$ ways we choose three of those four selected questions and the same set of four questions from part B four times for each of the $\\binom{4}{3}$ ways we choose three of those four selected questions.\n\nNote that $$10\\binom{5}{3}\\binom{7}{5} + 16\\binom{5}{4}\\binom{7}{4} + 10\\binom{5}{5}\\binom{7}{3} = \\binom{5}{3}\\binom{7}{3}\\binom{6}{2}$$\n\n\u2022 I've edited my answer to make what I'm saying more concrete. \u2013\u00a0N. F. Taussig Oct 8 '15 at 11:43\n\u2022 Okay, we see that this is happening. But why is this happening? Can we add something so that my method works? \u2013\u00a0Aditya Agarwal Oct 8 '15 at 11:45\n\u2022 As you can see, the method you attempted to use does not work since it counts the same set of selected questions every time you choose $3$ of the selected questions. To avoid overcounting, we use cases. I do not see a way to make the method you attempted work. \u2013\u00a0N. F. Taussig Oct 8 '15 at 11:55\n\nYou are counting certain combinations of questions more than once. This is not so easy to see for your example, thus, let's simplify it to a trivial example: you have to choose at least 5 questions out of 5 of part A (i.e. all of A), and at least 1 out of 7 of part B, and, as before, you have to choose 8 questions in total.\n\nWe can trivially see that we indeed have to choose 3 questions out of B to satisfy both conditions, since we have already chosen all of A. Thus, the correct answer would be $\\left(\\begin{matrix}5\\\\5\\end{matrix}\\right)\\left(\\begin{matrix}7\\\\3\\end{matrix}\\right)=1\\cdot 35=35$.\n\nYour way of solving this question would however say that there are $\\left(\\begin{matrix}5\\\\5\\end{matrix}\\right)\\left(\\begin{matrix}7\\\\1\\end{matrix}\\right)\\left(\\begin{matrix}6\\\\2\\end{matrix}\\right)=1\\cdot 7\\cdot 15=105$ possibilities of choosing.\n\nEdit: to clarify the problem in this simplified example: assume you choose questions 1,2 and 3 from B. In your calculations there are three possibilities to do so: you choose question 1 because you have to choose at least one from B (Condition 1), and you choose questions 2 and 3 because you have to choose in total 8 (Condition 2). Or you choose question 2 to satisfy Condition 1, and questions 1 and 3 for Condition 2. Finally, you can choose question 3 to satisfy Condition 1, and questions 1 and 2 for Condition 2. In other words, you count each combination 3 times, and, indeed $35\\cdot3=105$.", "date": "2020-06-01 22:47:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1470160/combinations-problem-choosing-ways-to-select-8-questions-out-of-12", "openwebmath_score": 0.7213014364242554, "openwebmath_perplexity": 181.09858008432974, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429629196683, "lm_q2_score": 0.8740772417253256, "lm_q1q2_score": 0.8595177046988328}}
{"url": "https://math.stackexchange.com/questions/2015361/solution-of-a-given-linear-diophantine-equation", "text": "# Solution of a given linear diophantine equation.\n\nI was solving the following linear diophantine equation :\n\n$56x + 72y = 40$ in integers.\n\nMy attempt: I got that 8 is the gcd of 56 and 72 and $8|40$ and hence a solution exists and I can write:\n\n$8 = 56 - 16 *3$\n\n$\\implies$ $8= 56 - (72 -56*1)*3$\n\n$\\implies$ $8= 4*56 - 3*72$.\n\nSo my answer is $x = 4$ and $y = -3$. But in book its showing $x = 20$ and $y = -15$. Where I went wrong? Kindly help.\n\n\u2022 Remember that $20=5 \\times 4$ and $-15=5 \\times 3$. Both answers are correct\n\u2013\u00a0user284001\nNov 15, 2016 at 16:49\n\u2022 @Bacon, So why did they multiply by 5? Nov 15, 2016 at 16:50\n\u2022 At the outset you noted that $8|40$ and this is...\n\u2013\u00a0user284001\nNov 15, 2016 at 16:50\n\n$56x+72y=40$\n$7x+9y=5$\nSelect the term with the smaller of the two co\u00ebfficients and isolate it on the LHS.\n$7x=5-9y$\n$\\text{[1] }x=\\frac{5-9y}7=1-y+\\frac{-2-2y}7$\n$\\text{New variable }a=\\frac{-2-2y}7$\n$-2-2y=7a$\n$-2-7a=2y$\n$\\text{[2] }y=-\\frac{2+7a}2=-1-4a+\\frac{a}2$\n$\\text{New variable }b=\\frac{a}2$\n$a=2b$\nSubstitute $a\\text{ into [2].}$\n$y=-\\frac{2+7(2b)}2=-\\frac{2+14b}2=-1-7b$\nSubstitute $y\\text{ into [1].}$\n$x=\\frac{5-9(-1-7b)}7=\\frac{5+9+63b}7=\\frac{14+63b}7=2+9b$\n$\\text{So, }(x,y)=(2+9b,-1-7b).$\nThere is an infinitude of integer answers; but your textbook, for some reason, favors $b=2.$\n\nwrite the equation as $$56x\\equiv 40\\mod 72$$ or $$x\\equiv \\frac{40}{76}=\\frac{5}{7}\\equiv \\frac{77}{7}\\equiv 11 \\mod 72$$ thus $$x=11+72K$$ with $$k \\in \\mathbb{Z}$$\n\nYou can solve $56x+72y=40$ using convergent.\n\nNote: If $\\alpha,\\beta$ are integral solutions to $ax+by=c$ for $x,y$ respectively, the other solutions take the form$$x=\\alpha-bt\\\\ y=\\beta+at\\tag{1}$$\n\nHere, $a=56,b=72$ so the $x$ values model the form $\\alpha-72t$ and $\\beta+56t$ for $y$.\n\nThe nearest converging fraction to $\\dfrac {9}{7}$ is $\\dfrac 43$ and hence, we have$$7\\cdot 4-3\\cdot 9=1$$\n\nFor which we can multiply both sides by $40$ to get$$40\\cdot 7\\cdot 4-3\\cdot 9\\cdot 40=40\\implies 56\\cdot 20+72(-15)=40\\tag2$$ Thus, we have $\\alpha=20,\\beta=-15$. And by $(1)$, we get$$x=20-72t\\\\y=-15+56t$$\n\nNote that we also get the solutions that the book says along the way!", "date": "2022-09-29 23:02:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2015361/solution-of-a-given-linear-diophantine-equation", "openwebmath_score": 0.9418825507164001, "openwebmath_perplexity": 296.6170831641207, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429619433693, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.859517695781497}}
{"url": "https://math.stackexchange.com/questions/2483136/the-probability-that-it-is-a-woman-or-a-man-who-is-not-bringing-a-dessert", "text": "# The probability that it is a woman, or a man who is not bringing a dessert?\n\nJohn invites 12 friends to a dinner party, half of which are men. Exactly one man and one woman are bringing desserts. If one person from this group is selected at random, what is the probability that it is a woman, or a man who is not bringing a dessert?\n\nI know that the two events are mutually exclusive; so that, the answer is simply $6/12 + 5/12 = 11/12$. However, when I looked at the answer in the book that I study from, I found the same result but an approach that's really weird for me. Here is it literally:\n\n$$P(woman) = 6/12 = 1/2$$ $$P(not\\ bringing\\ a\\ dessert\\ ) = 10/12 = 5/6$$ $$P(woman\\ and\\ not\\ bringing\\ a\\ dessert) = 1/2\\ *\\ 5/6=5/12$$ $$P(woman\\ or\\ a\\ man\\ not\\ bringing\\ a\\ dessert) = 1/2 + 5/6 - 5/12 = 11/12$$\n\nMay anyone explain the reasoning behind that approach? And is it technically right?\n\nYour approach is certainly right. The book's approach is right as well. It is just that they should have said $P(\\text{woman OR man not bringing dessert}) = P(\\text{woman OR not bringing dessert})$ (this works because there are just two genders). Then you don't have mutually exclusive events, so you have to use inclusion-exclusion. $$P(\\text{woman OR not bringing dessert}) = P(\\text{woman}) + P(\\text{not bringing dessert}) - P(\\text{woman} \\cap \\text{not bringing dessert}).$$\n\nThe term $P(\\text{woman} \\cap \\text{not bringing dessert}) = 5/12$, thus they have $1/2 + 10/12 - 5/12 = 11/12$.\n\nWhat you are doing is you are breaking up your space into mutually exclusive events. You are saying it is equivalent to calculating $P(\\text{woman} \\sqcup \\left(\\text{not woman AND no dessert}\\right))$. That way you have $1/2 + 5/12$.\n\nThere's another way you can solve this problem. This probability is equal to $1 - P(\\text{man bringing dessert}) = 1 - 1/12 = 11/12$.\n\nNOTE: Edited after a misinterpretation pointed out by @Abdu Magdy\n\n\u2022 But P(man not bringing a dessert) should have equalled $5/12$ not $10/12$, Consequently, P(woman \u2229 man not bringing dessert) should have equalled $5/12 \\ *\\ 1/2 = 5/24$ \u2013\u00a0Abdu Magdy Oct 21 '17 at 18:52\n\u2022 I interpret $P(\\text{man not bringing a dessert})$ as $P(\\text{not bringing a dessert} | \\text{man}) = \\frac{P(\\text{not bringing dessert} \\cap \\text{man})}{P(\\text{man})} = \\frac{\\frac{5}{12}}{\\frac{1}{2}} = \\frac{5}{6}$, but YMMV. In my experience with GRE problems, sometimes the language is indeed ambiguous. \u2013\u00a0Abhiram Natarajan Oct 21 '17 at 19:02\n\u2022 I'm sorry but I really need to understand... Based on the definition of mutually inclusive events; if we only are to choose one person, May we choose a woman and a man not bringing a dessert at the same time? \u2013\u00a0Abdu Magdy Oct 21 '17 at 19:23\n\u2022 >> May we choose a man ... same time? You certainly cannot choose that way. It's more like when you have \"not bringing dessert\", it can be either man or woman. Let's try this way. They should have said $P(\\text{woman OR not bringing dessert})$. That way, you don't have mutually exclusive events. So by inclusion exclusion, you have $1/2 + 10/12 - 5/12$. I said this wrong in my answer. I shall edit it right away. \u2013\u00a0Abhiram Natarajan Oct 21 '17 at 20:41\n\u2022 Sorry about the confusion. \u2013\u00a0Abhiram Natarajan Oct 21 '17 at 21:00", "date": "2020-02-17 06:48:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2483136/the-probability-that-it-is-a-woman-or-a-man-who-is-not-bringing-a-dessert", "openwebmath_score": 0.7419829964637756, "openwebmath_perplexity": 304.5101624091799, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429604789206, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8595176912758654}}
{"url": "http://math.stackexchange.com/questions/255934/negating-zach-blocks-e-mails-and-texts-from-jennifer/255940", "text": "Negating \u201cZach blocks e-mails and texts from Jennifer\u201d\n\nI am reviewing some basic propositional logic. The question that I have come across that has given some confusion is Zach blocks e-mails and texts from Jennifer where I am asked to find the negation of this proposition. Here is what they provided as an answer in the book: Zach does not block e-mails from Jennifer, or he does not block texts from Jennifer.\n\nWhy couldn't the answer simply be It is not true that Zach blocks e-mails and texts from Jennifer? Why did they have to introduce the disjunction or?\n\n-\nOf course your answer is fine, indeed clearer. They presumably want to test whether you know how to negate a conjunction and obtain a disjunction. But if this is a test question, surely you should not be expected to be a mind reader. \u2013\u00a0 Andr\u00e9 Nicolas Dec 11 '12 at 2:00\n\nWhat you gave is one possible answer, the book gives another which is logically equivalent, although slightly cleaner statement since it is further reduced. They are the same because of one of De-Morgan's laws; $\\neg(P \\land Q) \\equiv (\\neg P) \\lor (\\neg Q)$\n\n-\n\nZach blocks emails and texts from Jennifer means Zach blocks emails from Jennifer and Zach blocks texts from Jennifer. Negating this gives Zach does not block emails from Jennifer or Zach does not block texts from Jennifer.\n\nIt is true that the negation is also It is not true that Zach blocks emails and texts from Jennifer but the point of the exercise is to make one rewrite it in a useful way.\n\nNote that in general, (not (A and B)) is logically equivalent to ((not A) or (not B)).\n\n-\nOr does \"Zach blocks emails and texts from Jennifer\" mean \"Zach blocks emails and Zach blocks texts from Jennifer\"? Don't you love the ambiguities of natural language? (Of course, the translation you give is almost certainly the intended meaning.) \u2013\u00a0 Code-Guru Dec 11 '12 at 4:11\n\u2022 $P(x, y)$: x blocks emails from y\n\u2022 $Q(x, y)$: x blocks texts from y\n\u2022 $z$: Zach\n\u2022 $j$: Jennifer\n\n$P(z, j)$: Zach blocks emails from Jennifer;\n$Q(z, j)$: Zach blocks texts from Jennifer.\n\nYou essentially negated: $P(z, j) \\land Q(z, j)$.\n\nSo did the text.\n\nThe text applied \"distribution of negation over conjunction\" (one of DeMorgan's Laws):\n\n\u2022 \"It is not the case that $[(P(z, j)$ and $Q(z, j)]$\" $$\\iff \\lnot[P(z, j) \\land Q(z, j)]$$ $$\\iff \\lnot P(z, j) \\lor \\lnot Q(z, j)\\tag{DeMorgan's law}$$\n\u2022 which is to say: not $P(z, j)$, or, not $Q(z, j)$.\nYou are correct with your statement that the sentence translates to $\\;\\;\\lnot[P(z,j) \\land Q(z, j)]$;\nIt's logically equivalent to the text's answer: $\\;\\lnot P(z, j) \\lor \\lnot Q(z, j)$.", "date": "2014-10-31 06:26:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/255934/negating-zach-blocks-e-mails-and-texts-from-jennifer/255940", "openwebmath_score": 0.7580549120903015, "openwebmath_perplexity": 1948.8392717562258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9637799462157138, "lm_q2_score": 0.8918110404058913, "lm_q1q2_score": 0.8595095965569697}}
{"url": "http://mathhelpforum.com/number-theory/91624-divisibility-7-a.html", "text": "# Math Help - Divisibility by 7\n\n1. ## Divisibility by 7\n\nProve that $2222^{5555}+5555^{2222}$ is divisible by 7.\n\nI set about trying to show that this expression is equal to $0 \\mod 7$.\n\nI have that:\n$2222=3 \\mod7=2 \\mod 6$\n$5555=4 \\mod 7=5 \\mod 6$\n\nThe question also has Fermat's little theorem above it so I think I should be using that somewhere.\n\nAnyone have any ideas?\n\n2. Hello,\n\nLet's take for example $2222^{5555}$\nIt's congruent to $3^{5555}(\\bmod 7)$\n\nNow, you know by Fermat's little theorem, that $3^{6}\\equiv 1(\\bmod 7)$\n\nAs you noticed, $5555\\equiv 5(\\bmod 6)$\n\nIt can be proved that hence $2222^{5555}\\equiv 3^5 (\\bmod 7)$\n\nWhy ? Because $5555=6k+5$\n\nAnd then $3^{6k+5}=(3^6)^k\\cdot 3^5 \\dots$\n\n3. Originally Posted by Showcase_22\nProve that $2222^{5555}+5555^{2222}$ is divisible by 7.\n\nI set about trying to show that this expression is equal to $0 \\mod 7$.\n\nI have that:\n$2222=3 \\mod7=2 \\mod 6$\n$5555=4 \\mod 7=5 \\mod 6$\n\nThe question also has Fermat's little theorem above it so I think I should be using that somewhere.\n\nAnyone have any ideas?\nHi Showcase_22.\n\nFermat\u2019s little theorem is a little cumbersome for this problem. Here is a shorter way.\n\n$2222\\equiv3\\,\\bmod7\\ \\implies\\ 2222^3\\equiv-1\\,\\bmod7$ $\\implies\\ 2222^{5555}=2222^{3\\cdot1851+2}\\equiv-3^2\\,\\bmod7\\equiv5\\,\\bmod7$\n\n$5555\\equiv4\\,\\bmod7\\ \\implies\\ 5555^3\\equiv1\\,\\bmod7$ $\\implies\\ 5555^{2222}=5555^{3\\cdot740+2}\\equiv4^2\\,\\bmod7\\eq uiv2\\,\\bmod7$\n\nHence $2222^{5555}+5555^{2222}\\equiv7\\,\\bmod7\\equiv0\\,\\bm od7.$\n\nThe reason I prefer $2222^3$ and $5555^3$ rather than $2222^6$ and $5555^6$ is that their powers can be brought closer this way to 5555 and 2222 respectively.\n\n4. Hello,\nHello!\n\nI finally got it to work out using TheAbstractionist's method (I looked at what you did, did it for 2222 whilst looking at yours and then did it for 5555 separately).\n\nI'll have a go using your method now Moo since it seems to be the one in the mark scheme.\n\n5. Okay, here's my way of doing it. I think it's both of your methods combined (or I may just be copying one of your methods, i'm not that good at number theory!):\n\n$2222^{5555}=3^{5555} \\bmod 7$\n\n$3^6=1 \\bmod 7$ by Fermat's little theorem\n\n$3^{5555} \\bmod 7=3^{6(925)+5}=3^{6(925)}3^5 \\bmod 7$\n\n$=(1 \\bmod 7)(3^2. 3^2.3 \\bmod 7)=(1 \\bmod 7 )(5 \\bmod 7)=5 \\bmod 7$\n\n$\\Rightarrow 2222^{5555}=5 \\bmod 7$\n\n$5555^{2222}=(5 \\bmod 6)^{2222}$\n\n$4^6=1 \\bmod 7$ by Fermat's little theorem.\n\n$5555^{2222}=(4 \\bmod 7)^{2222}$\n\n$=4^{370(6)+2} \\bmod 7=4^{370(6)}.4^2 \\bmod 7=(1 \\bmod 7)(2 \\bmod 7)=2 \\bmod 7$\n\n$\\Rightarrow 5555^{2222}=2 \\bmod 7$\n\n$\\Rightarrow 2222^{5555}+5555^{2222}=2 \\bmod 7+5 \\mod 7=7 \\bmod 7= 0 \\bmod 7$\n\n**WIN**\n\n6. Well, actually that was exactly the way I wanted you to do\n\nGood job.\n\nThough your notations are very strange...\n$=(1 \\bmod 7)(3^2. 3^2.3 \\bmod 7)$\nJust write $1\\cdot 3^2 \\cdot 3^2 \\cdot 3 \\bmod 7$\n\n$5555^{2222}=(5 \\bmod {\\color{red}7})^{2222}\n$\nWrite $5555^{2222}=5^{2222} \\bmod 7$\n\n7. Ah, sorry about my notations!\n\nI didn't realise there was a command for $\"\\cdot\"$. The more I know!\n\nThat second one was just plain wrong, sorry I wrote it!\n\nOn the plus side:\n\nSpoiler:\nAt least I know how to use these spoiler boxes now!", "date": "2015-03-05 00:25:10", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/91624-divisibility-7-a.html", "openwebmath_score": 0.9306477904319763, "openwebmath_perplexity": 600.9054102457291, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98527138442035, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8594989186239769}}
{"url": "https://math.stackexchange.com/questions/1302949/how-many-ways-to-select-k-vertices-of-an-n-gon", "text": "# How many ways to select $k$ vertices of an $n$-gon?\n\nI have a regular $n$-gon, of which I have to select $k$ vertices. The selections must be rotationally distinct; two selections would be considered equivalent if one is a rotation of the other. For example, if I have a square, and I want to select 2 vertices, there are only 2 possible ways to do that according to the constraint. One is \"x - x -\", another is \"x x - -\".\n\nIf we denote the function by $CR(n,k)$, then these are the trivial cases:\n\n1. $CR(n, 1) = 1$\n\n2. $CR(n, 2) = \\lfloor\\frac{n}{2}\\rfloor$\n\n3. $CR(n, k) = CR(n, n - k)$\n\nI am quite short of ideas on how to find the recurrence or closed formula of this problem, or if this problem has any closed form / recurrence solution at all. Any help with a bit detailed walk through would be much appreciated.\n\n\u2022 What do you mean by selection must be rotationally invariant? \u2013\u00a0Anurag A May 28 '15 at 19:07\n\u2022 Supposing, I have a 4-gon and I have to select 2 vertices. Hence, these two selections are the same. x-x-, -x-x. These selections are the same, x--x, xx--, -xx-, --xx \u2013\u00a0sarker306 May 28 '15 at 19:09\n\u2022 The following MSE meta link has a section on enumerating necklaces and bracelets. \u2013\u00a0Marko Riedel May 28 '15 at 19:51\n\u2022 One way to think about this (perhaps the best way) is the group action of cyclic rotations on the subsets of vertices. Rotations preserve the number of vertices in a subset (obviously), so the counting comes down to counting orbits of the $k$-subsets of $n$-vertices under those rotations. \u2013\u00a0hardmath May 28 '15 at 19:53\n\nAs the OP seems interested in fixing $k$ and letting $n$ vary we will do an example to show how this might work.\n\nBy the Polya Enumeration Theorem what we have here is $$[z^k] Z(C_n)(1+z)$$ with $Z(C_n)$ the cycle index of the cyclic group which is $$Z(C_n) = \\frac{1}{n}\\sum_{d|n} \\varphi(d) a_d^{n/d}.$$\n\nThe desired quantity is thus given by $$[z^k] Z(C_n)(1+z) = [z^k] \\frac{1}{n}\\sum_{d|n} \\varphi(d) (1+z^d)^{n/d}.$$\n\nThis is $$\\frac{1}{n}\\sum_{d|n, d|k} \\varphi(d) [z^k] (1+z^d)^{n/d}$$ or $$\\frac{1}{n}\\sum_{d|n, d|k} \\varphi(d) {n/d \\choose k/d} = \\frac{1}{n}\\sum_{d|\\gcd(n,k)} \\varphi(d) {n/d \\choose k/d}.$$\n\nFor $k=4$ starting at $n=1$ we obtain the sequence $$0, 0, 0, 1, 1, 3, 5, 10, 14, 22, 30, 43, 55, 73,\\ldots$$ which points us to OEIS A008610 where we find confirmation.\n\nAnother interesting one is $k=6$ which yields $$0, 0, 0, 0, 0, 1, 1, 4, 10, 22, 42, 80, 132, 217, 335, 504,\\ldots$$ which points to OEIS A032191.\n\nA Maple session with these looks like this:\n\n> with(numtheory):\n> Q := (n,k) -> 1/n*add(phi(d)*binomial(n/d,k/d), d in divisors(gcd(n,k)));\nQ := (n, k) -> add(numtheory:-phi(d) binomial(n/d, k/d),\n\nd  in  numtheory:-divisors(gcd(n, k)))/n\n\n> seq(Q(n,8), n=1..18);\n0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 15, 43, 99, 217, 429, 810, 1430, 2438\n\n\nwhich incidentally is OEIS A032193.\n\n\u2022 It'll be a sin by my side not to refer you to my similiar question here: math.stackexchange.com/questions/1294224/\u2026 . Perhaps you'll be able to help :) I apologize, OP! Really am! \u2013\u00a0Matan May 28 '15 at 20:40\n\u2022 I have been following your question for three hours. @Matan , felt really intrigued. \u2013\u00a0sarker306 May 28 '15 at 20:45\n\u2022 How wonderful to hear. Thanks! I update it on regular basis and work on it when I have time. Keep track of it :) @sarker306 \u2013\u00a0Matan May 28 '15 at 20:49\n\nTaking the necklace proof for Fermat's Little Theorem as inspiration:\n\nFor $n,k$ coprime, every possible selection has $n$ rotations. So\n\n$$CR(n,k) = \\frac 1n{n \\choose k}$$\n\nOtherwise letting $d=\\gcd(n,k)$, we need to account for those choices where there is a repeat pattern smaller than $n$, which will be all possible choices over the shorter interval with the reduced number of choices (but repeated through the entire set):\n\n\\begin{align} CR(n,k) &= \\frac 1n\\left[{n \\choose k} - {n/d \\choose k/d }\\right] + CR\\left(\\frac nd, \\frac kd \\right) \\\\ &=\\frac 1n\\left[{n \\choose k} - {n/d \\choose k/d }\\right] + \\frac 1d {n/d \\choose k/d } \\end{align}\n\nsince $\\gcd(\\frac nd, \\frac kd)=1$\n\nAnd also note that $CR(n,0)=1$ and $CR(n,n)=1$.\n\n\u2022 Any explanation? Or any source where the derivation is explained? \u2013\u00a0sarker306 May 28 '15 at 19:25\n\u2022 Joffan, if you could please refer here math.stackexchange.com/questions/1294224/\u2026 for a related problem I asked I will be grateful. Perhaps you'll be able to help. \u2013\u00a0Matan May 28 '15 at 19:26\n\u2022 @sarker306 If you review the necklace proof of Fermat's Little theorem I think you'll understand where this comes from. If I have time I'll add to it. \u2013\u00a0Joffan May 28 '15 at 19:34\n\u2022 I get CR(6,3) = 25/6 when I apply your solution. \u2013\u00a0TokenToucan May 28 '15 at 20:17\n\u2022 CR(n, 2) has its correct value when n is odd, not even. But really thanks for bringing in this gcd issue, it seems to be a right step to the right direction, I didn't suspect it before. I'll think it through the night. \u2013\u00a0sarker306 May 28 '15 at 20:25", "date": "2019-09-22 01:36:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1302949/how-many-ways-to-select-k-vertices-of-an-n-gon", "openwebmath_score": 0.8653555512428284, "openwebmath_perplexity": 709.2739327969334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713831229042, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.8594989011430034}}
{"url": "https://math.stackexchange.com/questions/789804/how-many-distinct-ways-to-climb-stairs-in-1-or-2-steps-at-a-time/1376123", "text": "# How many distinct ways to climb stairs in 1 or 2 steps at a time?\n\nI came across an interesting puzzle:\n\nYou are climbing a stair case. It takes $n$ steps to reach to the top. Each time you can either climb $1$ or $2$ steps. In how many distinct ways can you climb to the top?\n\nIs there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?\n\nLet $F_n$ be the number of ways to climb $n$ stairs taking only $1$ or $2$ steps. We know that $F_1 = 1$ and $F_2 = 2$. Now, consider $F_n$ for $n\\ge 3$. The final step will be of size $1$ or $2$, so $F_n$ = $F_{n-1} + F_{n-2}$. This is the Fibonacci recurrence.\n\n\u2022 Ha! Some simple and yet a convoluted formation of the sequence. That explains it. Can't believe I couldn't see it :) \u2013\u00a0PhD May 11 '14 at 1:53\n\u2022 Is it possible to generalize it? Such as what if your steps are 2, 3, and 5? Also, we can compute fibonacci in matrix form which is faster. In the case of 2,3 and 5 steps, what would be the closed form solution? \u2013\u00a0Bagus Trihatmaja Dec 5 '18 at 8:53\n\u2022 Yes, this falls into the more general framework of solving linear recurrence equations. They always admit fast evaluations via matrix exponentiation and often have closed-form solutions, especially when the characteristic equation has low degree since you need to find all of its roots. \u2013\u00a0fahrbach Dec 6 '18 at 22:19\n\u2022 Can someone help me understand a tweak to this problem where the order of the steps doesn't matter? How to come up to a closed form equation for it? \u2013\u00a0mankadnandan May 26 '19 at 7:38\n\nThe solution to this problem indeed corresponds to the Fibonacci numbers, as mentioned by @fahrbach.\n\nHere is an illustration of what you are trying to solve for the case of $n=4$ steps (taken from this website, which also gives a combinatorial solution)\n\nAny staircase with $n$ steps allowing paths with increments of 1 or 2 steps at a time will end up in one of two states before the last path is taken: either we've climbed $(n-1)$ steps already and have $\\color{red}{one}$ more step to take, or we've climbed $(n-2)$ steps already and we have $\\color{blue}{two}$ more steps to take (if we took only one step here then we'd end up in an arrangement from the first state).\n\nThus, to get the total number of possible ways to climb $n$ steps, we just add the number of possible ways we can climb $(n-1)$ steps and the number of possible ways we can climb $(n-2)$ steps, giving the familiar recurrence relation:\n\n\\begin{equation*} F_n = \\left\\{ \\begin{array}{l@{}l@{}l} 1 & n = 0,1\\\\ \\color{red}{F_{n-1}} + \\color{blue}{F_{n-2}} & n \\ge 2 \\end{array} \\right. \\end{equation*}\n\n\u2022 Can this answer be expanded to answer the question \"Given a set of possible step sizes S, calculate the number of possible ways you can climb n steps using step sizes in S)\"? \u2013\u00a0Ephraim Nov 4 '15 at 20:21\n\u2022 @Ephraim That's simple, it's just $F_n=\\sum_{i\\in I} F_i$ where $I$ is the set of steps that can jump to step $n$. \u2013\u00a0Stella Biderman Jan 31 '17 at 16:37\n\u2022 This is the explanation that clicked to me, the colored figure makes a lot of sense, thanks. \u2013\u00a0neevek Oct 20 '20 at 16:47\n\nThis is the Fibonacci numbers - One interpretation of the n-th Fibonacci number is the number of ways to compose $n$ with parts in $\\{1,2\\}$ (that is, the n-th fibonacci number counts the number of sequences whose values are in $\\{1,2\\}$ whose sums are $n$). This is often called \"Golfs and Cadillacs\" or something similar.\n\n\u2022 Could you please elaborate as to \"how\" it helps solve the problem? \u2013\u00a0PhD May 11 '14 at 1:46\n\u2022 If there are $n$ steps, you have $F_n$ ways to reach the top of the stairs? The sequences map directly to the amount of stairs you climb with each step. I don't see how it can be any clearer. This interpretation can be used as a definition of the Fibonacci numbers and can be shown to be equivalent to the recurrence by using a straightforward combinatorial argument. \u2013\u00a0Batman May 11 '14 at 1:47\n\u2022 I see what you mean. Perhaps what I'm confused by is by this: that is, the n-th fibonacci number counts the number of sequences whose values are in {1,2} whose sums are n. Could you please provide an example to explain this? \u2013\u00a0PhD May 11 '14 at 1:49\n\u2022 You form a sequence made up of 1's and 2's whose sum is $n$. For example, for $n=3$, the sequences are (1,1,1), (1,2) and (2,1) giving you 3 ways to climb the stairs. Note that this is the common combintorics use of Fibonacci numbers, which is offset by 1 from the kind you learn in elementary school (this Fibonacci sequence is $1,2,3,5,...$ instead of $1,1,2,3,5,...$). \u2013\u00a0Batman May 11 '14 at 1:51\n\nWe can change this question into: In how many ways is possible to write a number as the ordered sum of 1s and 2s?\n\nWe can prove this using direct proof.\n\nHypothesis: For Fibbonaci number $F_1=1, F_2=1, F_{n}=F_{n-1}+F_{n-2}$, $Q(k)=F_{k+1}$\n\nWe say that we have constructed every order of $Q(n)$. Then, we construct $Q(n+1)$ this way:\n\n1. For all orders, we add $+1$ to change $n$ to $n+1$. (There are now $Q(n)$ numbers.) Note: all numbers that are constructed here ends in $+1$.\n2. For all orders that end with $+1$, we change it to $+2$. We shall prove that the amount will be $Q(n-1)$ to complete the proof. Note: all numbers that are constructed here ends in $+2$.\n\nIt can be seen that by constructing in this way there contains all orders.\n\nProof of step 2: If we constructed $Q(n)$ in the way above, then all numbers ended in $+1$ will be the amount $Q(n-1)$ by step 1. above, and we are done.", "date": "2021-01-22 13:39:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/789804/how-many-distinct-ways-to-climb-stairs-in-1-or-2-steps-at-a-time/1376123", "openwebmath_score": 0.8375009298324585, "openwebmath_perplexity": 185.00184216765106, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886157996524, "lm_q2_score": 0.8670357735451834, "lm_q1q2_score": 0.8594826918063857}}
{"url": "http://mathschallenge.net/full/quadratic_differences", "text": "#### Problem\n\nThe positive integers, $x$, $y$, and $z$ are consecutive terms in an arithmetic progression. Given that $n$ is also a positive integer, for how many values of $n$ below one-thousand does the equation $x^2 - y^2 - z^2 = n$ have no solutions?\n\n#### Solution\n\nLet $x = a + d$, $y = a$, and $z = a - d$.\n\n$\\therefore (a + d)^2 - a^2 - (a - d)^2 = n$\n$a^2 + 2ad + d^2 - a^2 - a^2 + 2ad - d^2 = n$\n\n$\\therefore 4ad - a^2 = a(4d - a) = n$.\n\nLet $u = a$ and $v = 4d - a \\implies u + v = 4d \\equiv 0 \\mod 4$. In other words, for a solution to exist the factors of $n$ must add to a multiple of four.\n\nWe shall deal with $n$ being of the form $2^m r$, where $r$ is odd, and for increasing values of $m$.\n\n\u2022 $m = 0$ ($n$ is odd):\nIf $n = r$ then the factors $u$ and $v$ must both be odd. But if they are both congruent with 1 or both congruent with -1 modulo 4 then $u + v \\equiv 2 \\mod 4$, and there will be no solution; if they are different then $u + v \\equiv 0 mod 4$, and there will always be a solution.\nHence if they are the same then $n = uv \\equiv 1 \\mod 4$, or $n$ being of the form 4$k$ + 1, will have no solutions.\n\n\u2022 $m = 1 \\implies n = 2r = a(4d - a)$:\nIf $a = 2r$, $4d - a = 1 \\implies 4d = 2r + 1$. But as the RHS is odd, this is impossible.\nIf $a = r$, $4d - a = 2 \\implies 4d = r + 2$. Impossible, as RHS is odd.\nIn other words if $n = 2(2k + 1) = 4k + 2$ then there will be no solutions.\n\n\u2022 $m = 2 \\implies n = 4r$:\nIf $a = 2r$, $4d - a = 2 \\implies 4d = 2r + 2 = 2(r + 1)$.\nAnd as $r + 1$ is even, we will always have at least one solution if $n = 4r$.\n\n\u2022 $m = 3 \\implies n = 8r$:\nIf $a = 8r$, $4d - a = 1 \\implies 4d = 8r + 1$. Impossible.\nIf $a = 4r$, $4d - a = 2 \\implies 4d = 4r + 2$. Impossible.\nIf $a = 2r$, $4d - a = 4 \\implies 4d = 2r + 4$. Impossible.\nIf $a = r$, $4d - a = 8 \\implies 4d = r + 8$. Impossible.\nHence if $n = 8(2k + 1) = 16k + 8$ then there will be no solutions.\n\n\u2022 $m \\ge 4$:\nIf $a = 4r$, $4d - a = 2^{m-2} \\implies 4d = 4(r+2^{m-4})$.\nHence for $m \\ge 4$ there will always be at least one solution.\n\nThus there will be no solutions for numbers of the form $4k + 1$, $4k + 2$, and $16k + 8$. As the first and second cases are odd and even respectively, they are mutually exclusive, and although the the second and third are both even, the third is divisible by 4, whereas the second is not divisible by 4. Hence all three forms are mutually exclusive.\n\nAs $4 \\times 249 + 1 = 997$, $4 \\times 249 + 2 = 998$, and $16 \\times 61 + 8 = 984$, there are exactly $249 + 249 + 61 = 559$ values of $n$ below one-thousand that have no solution.\n\nFor which values of $n$ will there be exactly one solution?\n\nProblem ID: 295 (26 Nov 2006) \u00a0 \u00a0 Difficulty: 4 Star\n\nOnly Show Problem", "date": "2013-12-10 09:29:57", "meta": {"domain": "mathschallenge.net", "url": "http://mathschallenge.net/full/quadratic_differences", "openwebmath_score": 0.8716057538986206, "openwebmath_perplexity": 165.7649538138609, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886152849366, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.8594826879542858}}
{"url": "http://sfmathgre.blogspot.com/2010/05/gr976863-intersections.html", "text": "## Sunday, May 23, 2010\n\n### GR9768.63: Intersections\n\n63. At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y=2^x$ intersect?\n(A) \u00a0None\n(B) \u00a0One\n(C) \u00a0Two\n(D) \u00a0Three\n(E) \u00a0Four\n\nSolution:\nWe know what the graph of $2^x$ looks like: always positive, the $y$-intercept is at $(0,1)$. For $y = x^{12}$, it passes through the vertex $(0,0)$ and it concaves up. There are at most 2 intersections. Now, the two graph intersect exactly once on $(-\\infty, 0)$. On the other half, $[0, \\infty)$, we'll look at $f(x) = 2^x - x^{12}$ and note that $f(1) = 2^1 - 1^{12} = 1 > 0$, where as $f(2) = 2^2 - 2^{12} < 0$. Thus, by the Rolle's Theorem, we see that $f$ has a zero on $(1,2)$. Thus, $x^{12}$ intersects $2^x$ on this interval. So there are two points of intersection.\nThe answer is C.\n-sg-\nNote: the answer provided in the Answer Sheet is D, 3 points of intersection.\nresult from wolframalpha\n\n#### 13 comments:\n\n1. By the same Rolle's theorem, there *is* one more point of intersection. YOu exained x=2. Now, examine x=L, where L is very large. Clearly, 2^x wins here. So, again by ROlle's theorem, there is yet another point of intersection between 2 and L. The answer is indeed, D. - Deego.\n\n2. Only 17 percent of the actual GRE test takers in 1997 got this right, worse than a random selection strategy!\n\n3. Proper result from wolfram:\n\nhttp://www.wolframalpha.com/input/?i=Plot[{2^x,+x^12},+{x,+-100.,+100.}]+\n\n4. one vote for D\n\n5. shouldn't it be Intermediate Value Theorem instead of Rolle's Theorem?\n\n---\n\nyodogyo@gmail.com\n\n6. answer is (d) ...\n\nbasic property is that 2^x increases faster than x^12 and in fact for any a^x for which a>1 we would have a^x > x^n for any n as x--> infinity.\n\nHence all we need to find is whether the graph intersects twice or once for smaller values of x. e^x and x^2 I believe intersect only once.\n\n2^x and x^12 intersects twice for x<2 and also for some big value of x as 2^x would be greater than x^12 for some value eventually. Hence this function intersects thrice.\n\n7. I agree with Shantanu.\n\nX>=0: We know that 0^12< 2^0 (x=0) and 2^12>2^2 (X=2).\nThus there is at least one intersection between 0 and 2.\nAt the same time, we know that 2^x is NOT O(x^12) as x->infinity. Also, they are both positive functions, so we know that there is some x>2 such that 2^x exceeds x^12 forever.\n\nx<0. Its clear there is a cross in this region.\n\nThis gives us at least 3 crosses. My gut tells me that this is all because I am familiar with the graphs of both functions. So I would definitely choose D.\n\nRigorously, I do not know why there is only 3. Examining the derivatives would probably be the place to look.\n\n8. Sorry for my grammar, english is not my native.\nFirst 2 solutions are clearly. Lets prove that there is the 3-th solution. If we prove that lim(2^x/x^12)>1 for x->+oo, we will prove that the curves intersect once more time (because if x2 is x of 2-nd intersection and x near the x2 and x>x2 than 2^x1 for x->+oo. Just use L'Hospital's rule 12 times and we have: lim(ln2^12*2^x/(x^12))=+oo for x->+oo. So, because +oo>1 than we have the 3-th point of intersection.\n\n9. Why are there not more than 3 intersections?\n\n10. There are precisely 3 intersections. At x=0, 2^x = 2 > 0 = x^12. At x=2, 2^x = 4 < 2^12 = x^12, hence there is one intersection point in the interval (0,2). It is a well-known fact that as x grows to infinity 2^x exceeds x^12 (quite rapidly, in fact). Hence there is a second intersection point for x>2. By drawing the graphs of these functions it is easy to see that these are all intersection points for positive values of x.\n\nA similar argument shows that there is a third and last intersection point between -1 and 0, and these are all.\n\n11. What is the value of 3rd point of intersection? How large is it?\n\n12. The third point of intersection occurs approximately at (74.66932553, 3.00404713 X 10^22)", "date": "2014-08-01 07:46:03", "meta": {"domain": "blogspot.com", "url": "http://sfmathgre.blogspot.com/2010/05/gr976863-intersections.html", "openwebmath_score": 0.7024603486061096, "openwebmath_perplexity": 690.016004578981, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886134834316, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8594826710661126}}
{"url": "https://math.stackexchange.com/questions/2901963/open-sets-in-mathbb-r-times-mathbb-r-with-the-dictionary-order", "text": "# Open sets in $\\mathbb R\\times \\mathbb R$ with the dictionary order\n\nIn the example below, I wonder why Munkres depicted the first type of intervals in this way?\n\nMore specifically, why doesn't the picture look like this:\n\nwhere all points below, above, and on the red segment are removed?\n\nThis is my justification why I think this should be the right picture:\n\n$$(a\\times b,c\\times d)=\\{x|a \\times b <_{dict} x <_{dict} c\\times d\\}\\\\=\\{(p,q)|a <_{R} p <_{R} c\\}\\cup \\{(r,s)| a=r=c \\text{ and } c <_{R} s <_{R} d \\}$$ (where $<_{dict}$ is the dictionary order and $<_R$ is the standard order on $\\mathbb R$). The first set in the union is the set of the first type, the second set in the union is the set of the second type.\n\nNow, any point above, below, or on the red line has the first coordinate $a$, which cannot lie in a set of the first type.\n\nUpdate: I defined the sets of the first type in a way that differs from Munkres, but I don't see what's wrong with my description of open sets (in the display), and no open set in my description can be a set of the first type in Munkres.\n\n\u2022 Um, removing the red segments on your second image produces exactly the first one, doesn't it? \u2013\u00a0Henning Makholm Sep 1 '18 at 19:12\n\u2022 @HenningMakholm I'm not removing just the red segment, I'm removing everything above and below it as well. \u2013\u00a0user531587 Sep 1 '18 at 19:21\n\u2022 What then happen to points such as $a\\times(b+1)$, which is definitely both $>_{\\rm dict} a\\times b$ and $<_{\\rm dict} c\\times d$? \u2013\u00a0Henning Makholm Sep 1 '18 at 19:29\n\u2022 @HenningMakholm okay, this makes sense... I was confused (and still confused) by the description of the basis in the text. \u2013\u00a0user531587 Sep 1 '18 at 19:51\n\u2022 Actually this answer made things more clear: math.stackexchange.com/a/1129061/531587 \u2013\u00a0user531587 Sep 1 '18 at 20:02\n\nSuppose a point $p = (x,y) \\in (a \\times b , c \\times d)$ has $a$ as the first coordinate. Since $(a,b) < (x,y) = (a,y)$, this necessarily implies that $b < y$. In the same vein, if $x = c$ and since $(x,y) = (c,y) < (c,d)$, necessarily we must have that $y < d$. This justifies why the interval only contains an open ray of $\\{x = a\\}$ and $\\{x = c\\}$ instead of the whole line.\nYour characterization of the interval is not quite right. Note that $r$ cannot be $a$ and $c$ at the same time, so you should separate in two cases (which is basically what I wrote) depending on the value that the first coordinate takes. This should clarify the picture.", "date": "2019-07-24 00:06:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2901963/open-sets-in-mathbb-r-times-mathbb-r-with-the-dictionary-order", "openwebmath_score": 0.8080102205276489, "openwebmath_perplexity": 205.83895104522566, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811611608242, "lm_q2_score": 0.8902942297605357, "lm_q1q2_score": 0.8594732773010075}}
{"url": "https://mathematica.stackexchange.com/questions/154018/principal-submatrix-and-principal-minor-of-a-matrix", "text": "# Principal submatrix and Principal minor of a matrix\n\nA submatrix of a matrix is obtained by deleting any collection of rows and/or columns.\n\nFor example, from the following 3-by-4 matrix, we can construct a 2-by-3 submatrix by removing row 3 and column 2:\n\nThe minors and cofactors of a matrix are found by computing the determinant of certain submatrices.\n\nA principal submatrix is a square submatrix obtained by removing certain rows and columns. The definition varies from author to author. According to some authors, a principal submatrix is a submatrix in which the set of row indices that remain is the same as the set of column indices that remain.\n\nFor a general 3 \u00d7 3 matrix in Mathematica,\n\n(mat = Array[Subscript[a, ##] &, {3, 3}]) // MatrixForm\n\n\n$mat=\\left( \\begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\\\ a_{2,1} & a_{2,2} & a_{2,3} \\\\ a_{3,1} & a_{3,2} & a_{3,3} \\\\ \\end{array} \\right)$\n\nthere is one third order principal submatrix, namely mat. There are three second order principal submatrix:\n\n$mat_{33}=\\left( \\begin{array}{ccc} a_{1,1} & a_{1,2} \\\\ a_{2,1} & a_{2,2} \\\\ \\end{array} \\right)$, formed by deleting column 3 and row 3;\n\n$mat_{22}=\\left( \\begin{array}{ccc} a_{1,1} & a_{1,3} \\\\ a_{3,1} & a_{3,3} \\\\ \\end{array} \\right)$, formed by deleting column 2 and row 2;\n\n$mat_{11}=\\left( \\begin{array}{ccc} a_{2,2} & a_{2,3} \\\\ a_{3,2} & a_{3,3} \\\\ \\end{array} \\right)$, formed by deleting column 1 and row 1;\n\nAnd there are three first order principal submatrix:\n\n$mat=\\left( \\begin{array}{ccc} a_{1,1} \\\\ \\end{array} \\right)$, formed by deleting column 2,3 and row 2,3;\n\n$mat=\\left( \\begin{array}{ccc} a_{2,2} \\\\ \\end{array} \\right)$, formed by deleting column 1, 3 and row 1, 3;\n\n$mat=\\left( \\begin{array}{ccc} a_{3,3} \\\\ \\end{array} \\right)$, formed by deleting column 1,2 and row 1,2;\n\nDo you have a way to find all principal submatrix in Mathematica ?\n\nor find the determine of all principal submatrix (principal minor) in Mathematica ?\n\nBy @klgr comment.\n\nDiagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] //MatrixForm /@ # &\n\n\u2022 Perhaps Diagonal @ Minors[mat, k]? \u2013\u00a0Carl Woll Aug 18 '17 at 17:01\n\u2022 Diagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] // MatrixForm /@ # & \u2013\u00a0kglr Aug 18 '17 at 17:04\n\u2022 @kglr Thanks for the answer but is this in general. What if the matrix is a 4X4 measurement . \u2013\u00a0Emad kareem Aug 18 '17 at 17:14\n\u2022 @kglr Thank you very much. This gives the exact result in general. \u2013\u00a0Emad kareem Aug 18 '17 at 17:24\n\nTo get the principal submatrices:\n\nDiagonal[Map[Reverse, Minors[mat, #, Identity], {0, 1}]] & /@ {1, 2}\n\n\nFor the principal minors\n\nDiagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & /@ {1, 2}\n\n\n\u2022 What is the & /@ {1, 2} for? \u2013\u00a0AzJ Jan 25 '18 at 19:46\n\u2022 @AzJ, the & belongs to the left part, that is, the expression is actually (Diagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & ) /@ {1,2}. For a pure function (a function with unnamed arguments) foo[#]& (where foo is an instruction on what to do with the stuff #; for example, multiply it by 2 and add 1 , (1+ 3 #)&) , the expression foo[#]&/@{a,b,c} Maps foo over a list of arguments ({a, b}) , that is, it is a short hand for  {foo[a], foo[b], foo[c]}. ... \u2013\u00a0kglr Jan 25 '18 at 20:34\n\u2022 ... Please see the doc pages on for details on of esoteric symbols such as Function (&), Slot (#) and Map \u2013\u00a0kglr Jan 25 '18 at 20:35\n\nI arrived three years later but I think that the code of @klgr can be significantly improved as it computes first all the minors to take at the end just those on the diagonal. I think that the next routine works faster for the submatrices with the obvious change making it work also for the minors, of course. I also suppose for convenience that the matrix is square, the obvious change make this also work for non-square matrices.\n\nPrincipalSubmatrices[mat_, size_] :=\nModule[{choices = Subsets[Table[i, {i, 1, Length[mat]}], {size}], count,\nrc, symsubmatrix, symsubmatrices}, count = choices;\nsymsubmatrices = {};\nWhile[count != {}, rc = count[[1]]; symsubmatrix = mat[[rc, rc]];\nsymsubmatrices = Append[symsubmatrices, symsubmatrix];\ncount = Delete[count, 1]]; symsubmatrices]", "date": "2021-04-17 12:32:00", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/154018/principal-submatrix-and-principal-minor-of-a-matrix", "openwebmath_score": 0.32522183656692505, "openwebmath_perplexity": 2127.764000555598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811631528336, "lm_q2_score": 0.8902942203004186, "lm_q1q2_score": 0.8594732699418632}}
{"url": "https://math.stackexchange.com/questions/2736995/how-many-classes-does-the-equivalence-relation-partition-the-set/2737052", "text": "How many classes does the equivalence relation partition the set?\n\nLooking for help on B) How many classes does the equivalence relation partition set X\n\nConsider the relations $R$ and $S$, defined on the set $X = \\{1, 2, . . . , 99\\}$ as follows.\n\n$xRy \\iff x + y$ is a multiple of $11$,\n\n$xSy \\iff x \u2212 y$ is a multiple of $11$.\n\nA) One of $R$ and $S$ is an equivalence relation, the other is not. Determine which is which and justify your answers.\n\nB) Into how many classes does the equivalence relation partition set $X$?\n\nSo far I've determined that S is the equivalence relation as R isn't reflexive or transitive. My only attempt at B is that there is 11-1=10 non-zero congruence classes, so does each one correspond to a non-zero equivalence class?\n\nAny help is appreciated!\n\n\u2022 Where does this \"non-zero\" stuff come from? \u2013\u00a0ancientmathematician Apr 14 '18 at 16:53\n\u2022 What is a \"non-zero\" equivalence class? \u2013\u00a0Morgan Rodgers Apr 14 '18 at 16:54\n\u2022 @CyclotomicField: Sorry, but \"a zero equivalence class\" is a bit of a nonsense on a set that doesn't even contain zero, such that the given set $X$. \u2013\u00a0zipirovich Apr 14 '18 at 18:29\n\u2022 @CyclotomicField It's unambiguous, but it's also meaningless in the normal context of equivalence relations. And is a term that doesn't actually show up in the problem \u2013\u00a0Morgan Rodgers Apr 15 '18 at 17:11\n\u2022 What's wrong with a \"zero equivalence class? Doe $11 -11=0$ make $11$ any less equivalent? Where did you get this weird idea that you aren't supposed to count the the equivalence class to zero? \u2013\u00a0fleablood Apr 15 '18 at 17:20\n\nJust count them.\n\n$1 R 12 R 23 R 34 R .... R 90$\n\n$2 R 13 R 24 R 35 R .... R 91$\n\n....\n\n$10 R 21 R 32 R 43 R..... R 98$\n\n$11 R 22 R 33 R 44 R ..... R 99$\n\nThat's 11.\n\nIt doesn't make any sense to subtract the \"zero\" equivalence class for 2 reasons.\n\n1) The \"zero\" equivalence class is STILL an equivalence class so why they heck would you omit it? NOWHERE in the question does it say ANYTHING about how many non-zero equivalence classes; it ask how many equivalence classes. And $\\{11,22,33,44,55,66,77,88,99\\}$ is certainly an equivalence class, isn't it?\n\n2) Since no method of \"addition\" has been defined or discussed, there is no meaning to defining any one of the equivalence classes as be a \"zero\" equivalence class. IF we were to define $\\{x|x R c\\} + \\{x|x R d\\} = \\{x| x R (d+c\\pm 11k \\text{ for some integer } k)\\}$ and define $[0]$ as the equivelence clas so that $[0] + \\{x|x R c\\} = \\{x|x R c\\}$ then, yes, $\\{11,22,33,44,55,66,77,88,99\\} = [0]$. But again, so what, it's still an equivalence class, isn't it?\n\nTo show $S$ is an equivalence relation you can either verify directly that it is reflexive, transitive, and symmetric or demonstrate that the relation partitions the set and use the fundamental theorem of equivalence relations. If you can demonstrate the existence of the congruence classes you have such a partition and that's sufficient to demonstrate that you have an equivalence relation. As you've deduced there are eleven such classes, and ten of them do not contain zero. I would also encourage you to consider the first method of directly verifying the three defining properties as it's both instructive and relatively easy.\n\n\u2022 So is the answer 10 for B) correct? \u2013\u00a0RealGib Apr 14 '18 at 18:23\n\u2022 @RealGib: No, it's not. \u2013\u00a0zipirovich Apr 14 '18 at 18:30\n\u2022 None of the equivalence classes on $X$ contain 0, since $X$ does not contain 0. \u2013\u00a0Morgan Rodgers Apr 15 '18 at 17:17\n\u2022 \"Morgan Rodgers continued insistence on ignoring this obvious interpretation of the OPs comments\". But it goes both ways. I agree that \"$\\{11,...99\\}$ do not contain $0$ as so can't be 'zero' modulo class\" is a bit too oblique and inaccurate and $11\\equiv 0 \\mod 11$ does indeed mean $\\{11,...99\\}$ is the zero element. But on another, and very important level A \"zero\" (additive identity) element can only be defined if addition on the classes (and NOT on the elements of the classes) is defined. So it is not unjustified. \u2013\u00a0fleablood Apr 16 '18 at 16:10\n\u2022 ... so I mostly agree with Morgan Rodgers although I believe saying \"it isn't a zero module as it doesn't contain zero\" evades the issue that it actually isn't a zero module because the means for which it is \"zero\" are not stated. Of more concern is... why leave out the zero modules. These seem to much of a \"monkey see; monkey do\" approach to math that is doomed to failure. You don't just do everything one way despite whether that is what a question asks. If was asked how many elments does $\\{0,1,6,8\\}$ you don't answer 3 because zero doesn't count and you don't answer \"it goes to 8\". \u2013\u00a0fleablood Apr 16 '18 at 16:15\n\nThe question is not asking you about \"nonzero\" equivalence classes, it is asking for the total number of equivalence classes. So the equivalence class that contains 0 counts the same as any other equivalence class (however you want to interpret this, your set $X$ does not contain 0 so the interpretation of which equivalence class is the \"zero\" involves concepts that go beyond the scope of what this problem is asking).", "date": "2019-10-17 18:14:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2736995/how-many-classes-does-the-equivalence-relation-partition-the-set/2737052", "openwebmath_score": 0.6269584894180298, "openwebmath_perplexity": 442.26267658891527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075733703925, "lm_q2_score": 0.8933094017937621, "lm_q1q2_score": 0.8594597408287534}}
{"url": "https://puzzling.stackexchange.com/questions/107801/a-circle-touches-two-sides-of-a-triangle-and-two-of-its-medians/107804#107804", "text": "# A circle touches two sides of a triangle and two of its medians\n\nA circle touches two sides of a triangle and two of its medians. Prove that the triangle is isosceles.\n\nThis problem came from the Mathematical Digest issue 62 (Jan 1986) which in turn cited a Russian mag called KVANT (meaning \"Quantum\").\n\n\u2022 Assume it's not true. Then you wouldn't ask the question. A contradiction. Therefore it's true. QED. :) Mar 13 '21 at 1:02\n\nAs the triangles ADC and BEC have the same incircle and their areas are equal (half that of ABC), so are their perimeters\nDC+DA+AC = EC+EB+BC\nor, subtracting from both sides CD+CE+DA+EB\nAE-EB = BD-DA.\nThis means that D and E lie on the same pair of hyperbolas with foci A and B. Since they also have the same distance to the base AB (half that of C) the triangle must be isosceles by symmetry.\n\n\u2022 It took me a moment to see why they have the same distance to the base, but that should have been obvious. This is a really clever proof. Mar 13 '21 at 8:38\n\u2022 @JaapScherphuis It's short ;-) I've added a half sentence explaining (if you can call it that) the distance to the base. Mar 13 '21 at 9:50\n\u2022 That is clever. I don't think I've ever seen anyone use a hyperbola locus in a proof like this! Mar 13 '21 at 17:34\n\u2022 Elegant, indeed. Mar 16 '21 at 6:16\n\nAlternative Proof\n\nBy Pitot Theorem we have $$|CE|+|MD| = |CD|+|ME|$$ and since the centroid divides each median in the ratio 2:1, this means that $$\\frac{1}{2}|AC| + \\frac{1}{3}|AD| = \\frac{1}{2}|BC|+\\frac{1}{3}|BE|$$ Now let $$|AB| = c, |BC| = a, |CA|=b, |AD|=m_a, |BE|=m_b$$.\nThen Apollonius' Theorem tells us that $$m_a = \\sqrt{\\frac{2b^2+2c^2-a^2}{4}}\\,\\,\\,\\,,\\,\\,\\,\\,m_b = \\sqrt{\\frac{2a^2+2c^2-b^2}{4}}$$ Substituting this in above and multiplying across by $$6$$ yields $$3a + \\sqrt{2b^2+2c^2-a^2} = 3b + \\sqrt{2a^2+2c^2-b^2}$$ $$\\Rightarrow 3(a-b) + \\sqrt{2b^2+2c^2-a^2} = \\sqrt{2a^2+2c^2-b^2}$$ $$\\Rightarrow 9(a-b)^2 + 6(a-b)\\sqrt{2b^2+2c^2-a^2} + 2b^2+2c^2-a^2 = 2a^2+2c^2-b^2$$ $$\\Rightarrow (a-b)\\left(9(a-b) + 6\\sqrt{2b^2+2c^2-a^2} - 3(a+b)\\right) = 0$$ which means either $$a=b$$ or $$\\sqrt{2b^2+2c^2-a^2} = 2b-a$$.\nIf it is the latter then $$2b^2+2c^2-a^2 = 4b^2-4ab+a^2$$ $$\\Rightarrow c^2 = b^2-2ab+a^2 = (b-a)^2$$ $$\\Rightarrow c = \\pm(b-a)$$ which only happens for a degenerate triangle (does not satisfy the triangle inequality).\nHence $$a=b$$\n\n@loopwalt and @hexomino both gave excellent answers. I wanted to share the answer from the Mathematical Digest, because it's also elegant in its own way:\n\nIt starts the same way as @hexomino's proof (I didn't know it was called Pitot's Theorem!):\n\nSince $$CEMD$$ circumscribes a circle, $$CE+MD = CD + ME$$ (Pitot's Theorem, but fun to prove). So (as in hexomino's proof):\n$$\\frac{1}{2}AC+\\frac{1}{3}AD = \\frac{1}{2}BC+\\frac{1}{3}BE \\tag1$$\nNext, note that $$\\triangle ADC$$ and $$\\triangle BEC$$ have the same area ($$\\frac{1}{2}\\triangle ABC$$) and share a common incircle. This means that their perimeters are equal (another fun to prove and left as an exercise!). So $$AD+\\frac{1}{2}BC+AC = BE+\\frac{1}{2}AC+BC$$, which gives:\n$$\\frac{1}{2}AC+AD = \\frac{1}{2}BC+BE \\tag2$$\nNow, $$3\\times(1)-(2)$$ gives $$AC=BC$$", "date": "2022-01-23 10:31:28", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/107801/a-circle-touches-two-sides-of-a-triangle-and-two-of-its-medians/107804#107804", "openwebmath_score": 0.7189338207244873, "openwebmath_perplexity": 579.8293180331514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787853562027, "lm_q2_score": 0.8705972549785201, "lm_q1q2_score": 0.8594351407041398}}
{"url": "http://www.anlak.com/2015/01/probability-of-k-collisions.html", "text": "### Probability of k collisions\n\n, Friday, 30 January 2015\nWhile I was reading about hash tables I stumbled upon this deceptively simple question. I have rephrased it to make it easier to understand.\nSay we have $m$ buckets. We select a random bucket and put a ball in it, we repeat this (select-put) $n$ times. In the end what is the probability of having at least one bucket with $k$ balls?\nMy initial attempts to solve it failed, and I posted the question on math.stackexchange. An approximation for large values of $m$ and $n$ was posted as an answer but I was interested in exact results. Once I realized I had no way of finding someone else's solution, I was able to solve it myself.\n\n## Simulation\n\nFirst, to check if my solution is correct, I have written a small program to simulate the select-put procedure million times and report the experimental probability for different values of $m$,$n$ and $k$. If you are interested, you can find the java code at the end of my post.\n\n## Solution\n\nFirst we will start with writing the probability of 1 box having $k$ balls. Then using inclusion-exclusion principle we will generalize. Let's write the probability of putting first $k$ balls into the first box and remaining balls to the other boxes: $$\\left(\\frac1{m}\\right)^k \\left(1-\\frac1{m}\\right)^{n-k}$$ We have $\\binom{n}{k}$ ways of putting $k$ balls into a particular box (e.g. first box), and we have $m$ such boxes. So the probability becomes: $$m \\binom{n}{k} \\left(\\frac1{m}\\right)^k \\left(1-\\frac1{m}\\right)^{n-k}$$ But the expression above suffers from double counting, i.e. it counts configurations that contains multiple boxes with $k$ balls more than once. So we need to subtract the value for 2 boxes having $k$ balls, add the value for 3 boxes having $k$ balls, etc... Let's write the probability of putting first $k$ balls into the 1st box, next $k$ balls into the 2nd box, and remaining balls to the other boxes. $$\\left(\\frac1{m}\\right)^k \\left(\\frac1{m}\\right)^k \\left(1-\\frac1{m}\\right)^{n-2k}$$ We have $\\binom{n}{k}\\binom{n-k}{k}$ ways of putting $2k$ balls into two particular boxes (e.g. first and second box), and we have $\\binom{m}{2}$ such boxes. So the probability becomes: $$\\binom{m}{2}\\binom{n}{k}\\binom{n-k}{k}\\left(\\frac1{m}\\right)^k \\left(\\frac1{m}\\right)^k \\left(1-\\frac1{m}\\right)^{n-2k}$$ At this point you should be able to see the pattern that is emerging.\n\n## Generalization\n\nIf we put together and generalize what we have done so far we get the following expression as probability of having a box with $k$ balls: $$\\sum^{\\lfloor \\frac{n}{k} \\rfloor}_{i=1} (-1)^{i+1} \\binom{m}{i} \\left[ \\prod^{i-1}_{j=0} \\binom{n-jk}{k} \\right] \\left( \\frac1{m} \\right)^{ik} \\left( \\frac{m-i}{m} \\right)^{n-ik}$$ The powers of $-1$ at the beginning enables adding values for odd number of balls and subtracting values for even number of balls, i.e. inclusion-exclusion principle.\n\n## The code\n\n  private static double calculateKcollisions(int m, int n, int k) {\ndouble probability = 0;\n\nint sign = 1;\n\nfor (int i = 1; n - i * k >= 0 && i <= m; i++) {\ndouble p = binomial(m, i) * pow(1.0 / m, i * k);\np *= pow((double) (m - i) / m, n - i * k);\n\nfor (int j = 0; j < i; j++) {\np *= binomial(n - j * k, k);\n}\n\nprobability += sign * p;\nsign *= -1;\n}\n\nreturn probability;\n}\n\n\n\n## Simulation code\n\n  private static final Random RNG = new Random();\n\npublic static void main(String[] args) {\nint sampleCount = 1000000;\n\nfor (int experimentNo = 0; experimentNo < 10; experimentNo++) {\nint m = RNG.nextInt(7) + 3;\nint n = RNG.nextInt(m) + RNG.nextInt(10) + 1;\nint k = RNG.nextInt(n + 1);\n\nSystem.out.println(format(\"m:%d  n:%d  k:%d\", m, n, k));\n\nint successCount = 0;\nfor (int sampleNo = 0; sampleNo < sampleCount; sampleNo++) {\n\nint[] boxes = new int[m];\n\nfor (int i = 0; i < n; i++)\nboxes[RNG.nextInt(m)]++;\n\nfor (int i = 0; i < m; i++) {\nif (boxes[i] == k) {\nsuccessCount++;\nbreak;\n}\n}\n}\n\ndouble experimental = (double) successCount / sampleCount;\n\ndouble calculated = calculateKcollisions(m, n, k);\n\nSystem.out.println(format(\"xperimental: %f  calculated: %f\", experimental, calculated));\nSystem.out.println(format(\"diff: %.2f\\n\", abs(experimental - calculated)));\nSystem.out.println();\n}\n}\n\n\n#### 1 comment:\n\n1. Excellent blog thanks for sharing the valuable information..it becomes easy to read and easily understand the information.\nUseful article which was very helpful. also interesting and contains good information.", "date": "2022-05-27 13:40:53", "meta": {"domain": "anlak.com", "url": "http://www.anlak.com/2015/01/probability-of-k-collisions.html", "openwebmath_score": 0.7722622752189636, "openwebmath_perplexity": 747.20082631396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815503903562, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8594274192198077}}
{"url": "http://mathhelpforum.com/calculus/155482-finding-point-sphere-furthest-point-space.html", "text": "# Math Help - Finding point on sphere furthest from point in space\n\n1. ## Finding point on sphere furthest from point in space\n\nFind the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).\n\nNow I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?\n\n2. Originally Posted by SyNtHeSiS\nFind the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).\n\nNow I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?\nThink about it geometrically. To get from C to the furthest point on the sphere, you have to go from C to the centre of the sphere and then continue in the same direction for a distance equal to the radius of the sphere, which is $\\sqrt{99} = 3\\sqrt{11}$. You have already identified the direction of travel as the vector (1,\u20131,3), which has length $\\sqrt{11}$.\n\nThose $\\sqrt{11}$s can't be coincidental, can they?\n\n3. Hello, SyNtHeSiS!\n\n$\\text{Find the point on the sphere }\\,(x - 2)^2 + (y - 2)^2 + (z-7)^2 \\:= \\:99$\n\n$\\text{which is furthest from the point }\\,C(1,3,4).$\nCode:\n              * * *         o C\n*           *   *\n*               \u2665 P\n*              *  *\n*\n*           *       *\n*         o         *\n*       * A         *\n*\n*  *              *\nQ \u2665               *\n*   *           *\n*         * * *\n\nConstruct a line through $\\,C$ and the center of the sphere $\\,A.$\n\nIt will intersect the sphere in two points, $\\,P$ and $Q.$\n. . We want $Q$, the point further from $\\,C.$\n\nThe line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\\langle 1,-1,3\\rangle$\n\nThe line has parametric equations: . $\\begin{Bmatrix}x &=& 1 + t \\\\ y &=& 3 - t \\\\ z &=& 4+3t \\end{Bmatrix}$\n\nSubstitute into the equation of the sphere:\n\n. . $\\left[(1-t) - 2\\right]^2 + \\left[(3+t)-2\\right]^2 + \\left[([4-3t)-7\\right]^2 \\:=\\:99$\n\n. . $11t^2 + 22t + 11 \\:=\\:99 \\quad\\Rightarrow\\quad t^2 + 2t + 1 \\:=\\:9$\n\n. . $t^2 + 2t - 8 \\:=\\:0 \\quad\\Rightarrow\\quad (t-2)(t+4)\\:=\\:0 \\quad\\Rightarrow\\quad t \\:=\\:2,\\:-4$\n\n$\\begin{array}{ccccccc}t = 2\\!: & (x,y,z) &=& (\\text{-}1,5,\\text{-}2) & \\text{point J} \\\\\nt = \\text{-}4\\!: & (x,y,z) &=& (5,\\text{-}1,16) & \\text{point K}\\end{array}$\n\nI suspect that point $\\,K$ is further from $\\,C$, but let's make sure.\n\nDistance from $C(1,3,4)$ to $J(\\text{-}1,5,\\text{-}2)\\!:$\n\n. . $\\overline{CJ} \\;=\\;\\sqrt{(\\text{-}1-1)^2 + (5-3)^3 + (\\text{-}2-4)^2} \\;=\\;\\sqrt{44}$\n\nDistance from $C(1,3,4)$ to $K(5,\\text{-}1,16)\\!:$\n\n. . $CK \\;=\\;\\sqrt{(5-1)^2 + (\\text{-}1-3)^2 + (16-4)^2} \\;=\\;\\sqrt{176}$\n\nHence, $K$ is the point we are seeking.\n\nTherefore: . $Q(5,\\text{-}1,16)$\n\n4. Originally Posted by Soroban\nThe line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\\langle 1,-1,3\\rangle$\nThanks for the great post. I was just curious why is it that choosing a line from A(2,2,7) to C(1,3,4) with direction <-1,1,-3> equivalent to the quote above? I thought this vector be the same magnitude but opposite, since points A and C is switched around.\n\n5. Just to point out that the makings of a much simpler solution were already in the original post, where the equation of the line from C to the centre of the sphere is given as $x = (1,3,4) + t(1,-1,3)$. If you put t=1 in that equation then you get the point (2,2,7) (the centre of the sphere). To get to the furthest point of the sphere you have to travel a further distance along that same line. The extra distance that you need to travel is $\\sqrt{99}$ units, which is three times the length of the vector (1,\u20131,3).\n\nSo to get to the furthest point of the sphere you need to take t = 4 (one unit of t to take you to the centre of the sphere, and another three units of t to take you to the furthest point). The furthest point is therefore at (1,3,4) + 4(1,\u20131,3) = (5,\u20131,16).", "date": "2014-03-14 23:29:54", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/155482-finding-point-sphere-furthest-point-space.html", "openwebmath_score": 0.7555056214332581, "openwebmath_perplexity": 297.7568040438331, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815526228441, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.8594274161198565}}
{"url": "https://datascience.stackexchange.com/questions/46744/when-to-use-mean-vs-median/46747", "text": "# When to use mean vs median\n\nI'm new to data science and stats, so this might seems like a beginner question.\n\nI'm working on a dataset where I've user's Twitter followers gain per day. I want to measure the average growth he had over a period of time, which I did by finding the mean of growth. But someone is suggesting me to use median for this.\n\nCan anyone explains, in which use-case we should use mean and when to use median?\n\nThe arithmetic mean is denoted as $$\\bar{x}$$\n\n$$\\bar{x} = \\frac{1}{n} \\sum_{i=1}^n x_i$$\n\nwhere each $$x_i$$ represent an unique observation. The arithmetic mean measures the average value for a given set of numbers.\n\nIn contrast to this, the median is the value which falls directly in the middle of your dataset. The median is especially useful when you are dealing with a wide range or when there is an outlier (a very high or low number compared to the rest) which would skew the mean.\n\nFor example, salaries are usually discussed using medians. This due to the large disparity between the majority of people and a very few people with a lot of money (with the few people with a lot of money being the outliers). Thus, looking at the 50% percentile individual will give a more representative value than the mean in this circumstance.\n\nAlternatively, grades are usually described using the mean (average) because most students should be near the average and few will be far below or far above.\n\n\u2022 That's a great answer. So, If I think it like this, I can plot my data and see if it values are continuous, then we can use mean and if they're more clustered (some high and some low), then median would be better, right? \u2013\u00a0Mukul Jain Mar 6 '19 at 4:48\n\u2022 @MukulJain, Yes it depends on the distribution of the data as you mentioned. Plotting is always my go to way to get a sense of my data. Easy to spot anomalies and get a sense of its spread. \u2013\u00a0JahKnows Mar 6 '19 at 5:48\n\u2022 I think you could explain this better using the term \"outlier\" \u2013\u00a0MilkyWay90 Mar 7 '19 at 1:11\n\u2022 So, if data has lots of outliers, is it good to use median right? Outliers can be calculated using z-score (<3 or >-3) \u2013\u00a0Mukul Jain Mar 7 '19 at 4:19\n\u2022 @MukulJain, correct, and you can also calculate outliers using p-value, \u2013\u00a0JahKnows Mar 7 '19 at 5:52\n\nIt depends what question you are trying to answer. You are looking at the rate of change of a time series, and it sounds like you are trying to show how that changed over time. The mean gives the reader one intuitive insight: they can trivially estimate the number of followers at any date $$d$$ days since the start by multiplying by the mean rate of change.\n\nThe downside to this single metric is that it doesn't illustrate something which is very common in series such as this: the rate of change is not fixed over time. One reasonable metric for giving readers an idea of whether the rate of change is static is giving them the median. If they know the minimum of the series (presumably zero in your case), the current value, the mean and the median, they can in many cases get a \"feel for\" how close to linear the increase has been.\n\nThere is a great cautionary tale in Anscombe's quartet - four completely different time series which all share several important statistical measures. Basically it always comes back to what you are trying to answer. Are you trying to find users which are likely to become prominent soon? Users which are steadily accruing followers year by year? One hit wonders? Botnets?\n\nAs you've probably guessed, this means it's not possible to universally call mean or median \"better\" than the other.\n\nSimply to say, If your data is corrupted with noise or say erroneous no.of twitter followers as in your case, Taking mean as a metric could be detrimental as the model will perform badly. In this case, If you take the median of the values, It will take care of outliers in the data. Hope it helps\n\nOften median is more robust to extreme value to mean. Try to think it as a minimization task. Median corresponds to absolute loss while mean corresponds to square loss.\n\nI find myself explaining this a lot and the example I use is the famous Bill Gates version. Bill Gates is in your data science class. Your instructor asks you: what is the average income or net worth of this class? Bill Gates sheepishly obliges and tells you what his income is. Now when you say the average income of your group is a zillion dollars - technically correct but does not describe the reality - that Bill Gates is an outlier skewing everything.\n\nSo you line up all the people in your group in ascending or descending order - whatever the person in the middle is making - that is your median. In this example, everybody but Bill Gates is likely to be in spitting distance of that median, and Bill Gates will be the only one making anything close to the mean.\n\nNow say buddy Bill Gates is hiring a money manager. Based on the returns they produced so far. Should he look at their average returns over a 10 year period or their median return or a combination of the two? Did they outperform the market each year? Some years? How does portfolio size factor in? In the case of Twitter followers, Obama would have a different growth compared to someone with say 500K-1MM followers. As @l0b0 alludes to in his excellent answer - it all depends. Are you measuring follower growth or the rate of change of follower growth and what is the question you are trying to answer, strategy/product you are trying to develop - accordingly you pick mean or median. Getting the mean and median is always the easy part. It's always better to never ever have the average of 2.1 kids. Have a whole number of kids. But what can you say about population growth rates if mean number of kids is 2.1 and median is 1 or 2? Or median is 3 or more? Is growth accelerating or decelerating? What is mode doing? Compute all the basics first - and then ask the reason why you are using mean versus median.", "date": "2020-09-20 10:34:13", "meta": {"domain": "stackexchange.com", "url": "https://datascience.stackexchange.com/questions/46744/when-to-use-mean-vs-median/46747", "openwebmath_score": 0.5889595746994019, "openwebmath_perplexity": 555.4294538979558, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9458012701768144, "lm_q2_score": 0.9086178963141964, "lm_q1q2_score": 0.859371960439352}}
{"url": "http://math.stackexchange.com/questions/111886/calculate-lim-n-rightarrow-infty-frac1n-sum-k-1nk-sin-left-frac", "text": "# Calculate $\\lim_{n\\rightarrow \\infty} \\frac{1}{n}\\sum_{k=1}^{n}k\\sin\\left(\\frac{a}{k}\\right)$\n\nI'm trying to calculate $\\lim_{n\\rightarrow \\infty} \\frac{1}{n}\\sum_{k=1}^{n}k\\sin\\left(\\frac{a}{k}\\right)$. Intuitively the answer is $a$, but I can't see any way to show this. Can anyone help? Thanks!\n\n-\nIs there anything special about $a$? I'm assuming it is an arbitrary real number? \u2013\u00a0 JavaMan Feb 22 '12 at 2:14\nWelcome to math.SE! Do you have learned Riemann integration? \u2013\u00a0 leo Feb 22 '12 at 2:14\nYou may prove a more general statement: *If $a_n \\to a$, then $\\frac{1}{n}\\sum_{k=1}^{n}a_k \\to a$*. This can be proved by $\\epsilon-\\delta$ argument. \u2013\u00a0 sos440 Feb 22 '12 at 2:18\na is an arbitrary real number, yeah. And I've learned Riemann integration! \u2013\u00a0 ro44 Feb 22 '12 at 2:27\n@sos440: Thanks, I thought I could try something like this. But I'm wondering if there's perhaps a 'special' trick with this particular limit? leo mentions Riemann integration. \u2013\u00a0 ro44 Feb 22 '12 at 2:29\n\nI\u2019m not going to work out all of the details; rather, I\u2019ll suggest in some detail a way to approach the problem.\n\nFirst, it suffices to prove the result for $a>0$, since the sine is an odd function. For $a>0$ we have $k\\sin\\left(\\frac{a}k\\right)<k\\left(\\frac{a}k\\right)=a$, so $$\\frac1n\\sum_{k=1}^nk\\sin\\left(\\frac{a}k\\right)<\\frac1n\\sum_{k=1}^na=a\\;;$$ this gives you an upper bound of $a$ on any possible limit.\n\nYou know that $\\lim\\limits_{x\\to 0}\\frac{\\sin x}x=1$, so there is a $c>0$ such that $\\sin x>\\frac{x}2$ whenever $0<x<c$. This means that $$k\\sin\\left(\\frac{a}k\\right)>\\frac{a}2$$ whenever $\\frac{a}k<c$, i.e., whenever $k>\\frac{a}c$. Now suppose that $n$ is very large compared with $\\frac{a}c$; then \u2018most\u2019 of the terms of $$\\frac1n\\sum_{k=1}^nk\\sin\\left(\\frac{a}k\\right)\\tag{1}$$ will be greater than $\\frac{a}2$, and hence so will $(1)$ itself. You may have to do a little fiddling to say just how big $n$ should be taken relative to $\\frac{a}c$, but it should be clear that this idea works to show that the limit of $(1)$ as $n\\to\\infty$ must be at least $\\frac{a}2$.\n\nBut what I did with $\\frac12$ can clearly be done with any positive fraction less than $1$: if $0<\\epsilon<1$, there is a $c>0$ such that $\\sin x>\\epsilon x$ whenever $0<x<c$. If you\u2019ve filled in the missing details for the previous paragraph, you shouldn\u2019t have too much trouble generalizing to show that the limit of $(1)$ must be at least $\\epsilon a$ for any $\\epsilon <1$ and hence must be at least $a$.\n\n-\nBased on Brian and sos's arguments, I was wondering if I'm correctly generalizing this situation: Say we have $a_n->a$, and let's take some arbitrary $l>0$ (using 'l' instead of epsilon). Then $|(\\frac{1}{n}\\sum_{k=1}^{n}a_n)-a|=|\\frac{1}{n}\\sum_{k=1}^{n}(a_n-a)|\\leq \\frac{1}{n}\\sum_{k=1}^{n}|a_n-a|$. For almost every n, we have that $|a_n-a|<l$, yielding that $\\frac{1}{n}\\sum_{k=1}^{n}|a_n-a|<l$ for almost every n. So $\\frac{1}{n}\\sum_{k=1}^{n}a_n \\rightarrow a$. Does this sound correct? Thanks! \u2013\u00a0 ro44 Feb 22 '12 at 2:49\n@ro44: Not quite. You want $|(\\frac1n\\sum_{k=1}^na_k)-a|=|\\frac1n\\sum_{k=1}^n(a_k-a)|\\le\\frac1n\\sum_{k=1}^n\u200c\u200b|a_k-a|$. Then you want to say that for sufficiently large $n$, \u2018most\u2019 terms of this sum are less than $l$, so the sum is less than (say) $2l$. There\u2019s still a bit of work to be done show that if $n$ is large enough, the early \u2018bad\u2019 terms are a small enough fraction of the total not to mess things up. \u2013\u00a0 Brian M. Scott Feb 22 '12 at 3:37\n\n$\\displaystyle \\sin x \\leq x$ for $x \\geq 0.$ Integrating this over $[0,t]$ gives $$-\\cos t +1 \\leq \\frac{t^2}{2} .$$\n\nIntegrating both sides again from $[0,x]$ gives $$-\\sin x + x \\leq \\frac{x^3}{6} .$$\n\nThus, $$x - \\frac{x^3}{6} \\leq \\sin x \\leq x.$$\n\nHence, $$\\frac{1}{n} \\sum_{k=1}^{n} k \\left( \\frac{a}{k} - \\frac{a^3}{6k^3} \\right ) \\leq \\frac{1}{n} \\sum_{k=1}^{n} k \\sin \\left( \\frac{a}{j} \\right) \\leq \\frac{1}{n} \\sum_{k=1}^{n} k \\left( \\frac{a}{k} \\right).$$\n\nSince $\\displaystyle \\sum_{k=1}^n \\frac{a^3}{6k^3}$ is convergent, the Squeeze theorem shows that $\\displaystyle \\frac{1}{n} \\sum_{k=1}^{n} k \\sin \\left( \\frac{a}{j} \\right) \\to a.$\n\n-", "date": "2014-08-28 05:09:21", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/111886/calculate-lim-n-rightarrow-infty-frac1n-sum-k-1nk-sin-left-frac", "openwebmath_score": 0.9696905016899109, "openwebmath_perplexity": 186.6669455411045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526451784085, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8593647901975486}}
{"url": "https://www.jiskha.com/questions/1828425/the-volume-v-of-a-sphere-of-radius-r-is-given-by-the-formula-v-r-4-3-r-3-a-balloon", "text": "# math\n\nThe volume V of a sphere of radius r is given by the formula V (r) = (4/3)\u03c0r^3. A balloon in the shape of a sphere is being inflated with gas. Assume that the radius of the balloon is increasing at the constant rate of 2 inches per second, and is zero whent = 0.\n(a) Find a formula for the volume V of the balloon as a function of time t.\n(b) Determine the volume of the balloon after 5 seconds.\n(c) Starting with an empty balloon, suppose that the balloon will burst when\nits volume is 10, 000 cubic inches. At what time will the balloon burst?\n(d) Find a formula for the surface area S of the balloon as a function of time\nt; recall the surface area formula for a sphere of radius r is S(r) = 4\u03c0r^2.\n(e) Determine the surface area of the balloon after 6 seconds.\n(f) What will be the surface area of the balloon when it bursts?\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n3. \ud83d\udc41 233\n1. V = 4/3 \u03c0r^3\n(a) dV = 4\u03c0r^2 dr\nV = \u222b[0,t] 4\u03c0r^2 dr\n(b) now plug in t=5\n(c) find when (a) = 10000\n(d) dA = 8\u03c0r dr\nwork it as in (a)\n(e) work as in (b)\n(f) plug in t as in (c)\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n\ud83d\udc68\u200d\ud83c\udfeb\noobleck\n2. since r = 2t, let's replace r in the equation\n\nV = (4/3)\u03c0 r^3\n= (4/3)\u03c0 (2t)^3\n= (32/3)\u03c0 t^3\n\nb) replace t with 5 in my formula.\nc) (32/3)\u03c0 t^3 = 10000\nsolve for r\n\nd) S(r) = 4\u03c0r^2 , let do the same thing: r = 2t\n= 4\u03c0(2t)^2 = 16\u03c0t^2\n\ne) let t = 6 and evaluate using the new formula from d)\n\nf) sub in the value of t you got in c) into the new S(r) formula\n\nlet me know what your answers are so I can check them\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n\ud83d\udc68\u200d\ud83c\udfeb\nReiny\n3. yep i got the same answers as you did :) thank you for your help\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n\n## Similar Questions\n\n1. ### calculus\n\n3. The radius r of a sphere is increasing at a constant rate of 0.04 centimeters per second. (Note: The volume of a sphere with radius r is v=4/3pir^3 ). a. At the time when the radius of the sphere is 10 cm, what is the rate of\n\n2. ### Calculus\n\nA ball of radius 12 has a round hole of radius 6 drilled through its center. Find the volume of the resulting solid. I tried finding the volume of the sphere and the volume of the cyclinder then subtract however that did not work.\n\n3. ### math\n\nThe surface area and the volume of a sphere are both 3 digit integers times \u03c0. If r is the radius of the sphere, how many integral values can be found for r?\n\n4. ### Algebra\n\nThe volume of a right circular cylinder (think of a pop can) is jointly proportional to the square of the radius of the circular base and to the height. For example, when the height is 10.62 cm and the radius is 3 cm, then the\n\n1. ### Math\n\nA spherical balloon is being inflated. Given that the volume of a sphere in terms of its radius is V(r) =4/3 \u03c0r^3 and the surface area of a sphere in terms of its radius is S(r) = 4 \u03c0r^2, estimate the rate at which the volume of\n\n2. ### Calculus (Parts A and B done, just help with C)\n\nThe radius, r, of a sphere is increasing at a constant rate of 0.05 meters per second. A. At the time when the radius of the sphere is 12 meters, what is the rate of increase in its volume? B. At the time when the volume of the\n\nThe volume V (r) of a sphere is a function of its radius r. Suppose a spherical snowball with a radius 2 f t started to melt so that the radius is changing at a constant rate of 4.5 inches per minute. If f(t) feet is the radius of\n\n4. ### math\n\nA balloon in the shape of a sphere is deflating. Given that t represents the time, in minutes, since it began losing air, the radius of the ballon (in cem) is r(t)=16-t. Let the equation V(r)= 4/3 pi r^3 represent the volume of a\n\n1. ### Math\n\nThe surface area and the volume of a sphere are both 3 digit integers times \u03c0. If r is the radius of the sphere, how many integral values can be found for r?\n\n2. ### Calculus Homework\n\nYou are blowing air into a spherical balloon at a rate of 7 cubic inches per second. The goal of this problem is to answer the following question: What is the rate of change of the surface area of the balloon at time t= 1 second,\n\n3. ### MAth\n\nA sphere has a surface area of 28.26 in2. Find the diameter in inches, of the sphere. Use PI Question 2 What is the surface area, in square inches, of a sphere with radius 3.5 in.? Use Pi and round your answer to the nearest\n\n4. ### Calculus\n\nV=4/3 pi r ^3 is the formula for the volume of a sphere. Find dV for a sphere if the radius of 10\" is increased by .1\" Please help. I have no idea how to do this problem!", "date": "2020-10-28 19:48:44", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1828425/the-volume-v-of-a-sphere-of-radius-r-is-given-by-the-formula-v-r-4-3-r-3-a-balloon", "openwebmath_score": 0.8784986734390259, "openwebmath_perplexity": 532.4813287797433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540704659681, "lm_q2_score": 0.877476800298183, "lm_q1q2_score": 0.8593604761114789}}
{"url": "https://math.stackexchange.com/questions/158271/relationship-between-the-coefficients-of-a-polynomial-equation-and-its-roots/158272", "text": "# Relationship between the coefficients of a polynomial equation and its roots\n\nMy question is- Solve: $x^3 - 6x^2 + 3x + 10= 0$ given that the roots are in arithmetic progression.\n\nAny help would be greatly appreciated.\n\nLet the three roots be $a-d$, $a$, and $a+d$.\n\nThe sum of the roots is the negative of the coefficient of $x^2$, so $(a-d)+a+(a+d)=6$, and therefore $a=2$.\n\nThe product of the roots is the negative of the constant term, and therefore $(a-d)(a)(a+d)=-10$. Since we know $a$, we can now find $d$. We get $a^2-d^2=-5$, and therefore $d^2=9$.\n\nRemarks: $1.$ In general, if we have a cubic equation $px^3+qx^2+rx+s=0$, where $p\\ne 0$, and the roots are $\\alpha$, $\\beta$, and $\\gamma$, then $$\\alpha+\\beta+\\gamma=-\\frac{q}{p}, \\qquad \\alpha\\beta+\\beta\\gamma+\\gamma\\alpha=\\frac{r}{p},\\qquad \\alpha\\beta\\gamma=-\\frac{s}{p}.\\tag{1}$$ There are analogous relations between coefficients and roots for polynomials of any degree.\n\n$2$. Note that we called the roots $a-d, a, a+d$. We could have called them $a$, $a+d$, and $a+2d$, but the algebra would be a little more complicated. Symmetry is your friend.\n\n$3.$ If the fact we used about roots is not available, there are a couple of things we can do. We can hope that the roots are rational. then they must all divide the constant term $10$. the only candidates are $\\pm 1$, $\\pm 2$, $\\pm 5$, and $\\pm 10$, and we quickly get to an answer.\n\nOr else we can let the roots be $\\alpha$, $\\beta$, and $\\gamma$, and see that the polynomial must be $(x-\\alpha)(x-\\beta)(x-\\gamma)$. Expanding, we find in essence the relationship $(1)$.\n\nOr else (but this is ugly) we can substitute $a-d$, $a$, and $a+d$ in our polynomial, and expand. Then solve for $a$ and $d$. The equations we get have a fair bit of symmetry, so this is not as difficult as it seems.\n\nThe roots are in arithmetic progression, so we can denote them as $a-d, a, a+d$ where $d$ is the common difference. The sum of these roots is then $3a$ and Viete's formulas shows that the sum of the roots must be $6$, so $a=2.$ So we have found a root, and can get the others by dividing the polynomial by $x-2$ and solving the remaining quadratic.\n\nA three degree polynomial would have max 3 roots say p,q,r\n\nThe relations between the roots and coefficients in $ax^3 + bx^2 + cx + d$ are:\n\n$\\frac{-b}{a} = p+q+r$\n\n$\\frac{c}{a} = pq + qr + pr$\n\n$\\frac{-d}{a} = pqr$\n\nNote: where $a \\neq 0$. Being 0 simply implies that the order of equation isn't actually 3.\n\nThree equations can solve three unknowns (To use the fact that they are in AP, assume $p,q,r$ to be $q-d$, $q$, $q+d$ where $q$ is the mean and $d$ is the common difference of AP).\n\nNote: The pattern of the roots and coefficients can be the same for n degrees of polynomial as well.\n\n\u2022 where $a\\neq 0$ \u2013\u00a0Belgi Jun 14 '12 at 14:39\n\u2022 Thanks. I have edited the answer \u2013\u00a0Saurabh Agarwal Jun 14 '12 at 14:39\n\nHint: Note that $2$ is a root of the polynomial, how can you get the other two roots ?\n\n\u2022 I did not use the fact that the roots are in arithmetic progression \u2013\u00a0Belgi Jun 14 '12 at 14:36\n\u2022 That fact could be used to ascertain that 6/3 = 2 is one of the roots. \u2013\u00a0hardmath Feb 1 '14 at 4:40\n\nHint $\\rm\\ \\ f(x) = (x+a\\!-\\!b)(x+a)(x+a\\!+\\!b)\\iff f(x\\!-\\!a) = x^3 - b^2 x.\\:$ But the only shift killing the $\\rm\\:-6\\,x^2\\:$ in your cubic is $\\rm\\:x \\to x+2,\\:$ yielding $\\rm\\:f(x\\!+\\!2) = x^3 - 9\\,x,\\:$ so $\\rm\\: b = \\pm 3.$\n\nNote $\\rm\\,\\ \\ f(x\\!-\\!a) = x^3\\! -\\! b^2x\\:\\Rightarrow\\:b^2 = -f'(x\\!-\\!a)|_{x=0} =\\, -f'(-a)\\ \\ [\\, = -(3\\cdot 2^2\\! -\\! 12\\cdot 2\\! +\\! 3) = 9\\ ].\\:$\n\nTherefore $\\rm\\ \\ f(x) = x^3 + c\\: x^2 +\\ldots\\: \\Rightarrow\\ a = c/3,\\ \\ b = \\sqrt{-f'(-a)}$\n\nFirst of all the easiest mtd is\n\n$$1)~$$ Get one of the roots by hit and trial mtd in your question clearly $$~-1~$$ is one of the roots i.e when $$~x =-1~$$ the whole equation results $$~0~$$ therefore $$~X=-1 ,~~~ X+1=0~$$\n\n$$2)~$$ apply division method to reduce it to quadratic and then $$~u~$$ can easily have your roots that are in A. P", "date": "2019-08-18 23:24:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/158271/relationship-between-the-coefficients-of-a-polynomial-equation-and-its-roots/158272", "openwebmath_score": 0.9566100239753723, "openwebmath_perplexity": 132.46884742817718, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540692607816, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8593604672091203}}
{"url": "http://xjetair.net/how-do-sjfdhpy/complex-number-to-rectangular-form-ad49e6", "text": "# complex number to rectangular form\n\nThe x is the real number of the expression and the y represents the imaginary number \u2026 We move 2 units along the horizontal axis, followed by 1 unit up on the vertical axis. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Rectangular and Polar Form To represent complex numbers x yi geometrically, we use the rectangular coordinate system with the horizontal axis representing the real part and the vertical axis representing the imaginary part of the complex number. Post was not sent - check your email addresses! If necessary, round to the nearest tenth. An identification of the copyright claimed to have been infringed; Convert The Following Complex Numbers To Rectangular Form. In other words, there are two ways to describe a complex number written in the form a+bi: To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Khan Academy is a 501(c)(3) nonprofit organization. The absolute value of a complex number is the same as its magnitude. Find more Mathematics widgets in Wolfram|Alpha. information described below to the designated agent listed below. Use i or j to represent the imaginary number \u22121 . St. Louis, MO 63105. Related Topics: Common Core (The Complex Number System) Common Core for Mathematics. Varsity Tutors LLC your copyright is not authorized by law, or by the copyright owner or such owner\u2019s agent; (b) that all of the (\u00a0Log\u00a0Out\u00a0/\u00a0 For this reason the rectangular form used to plot complex numbers is also sometimes called the Cartesian Form of complex numbers. Change\u00a0), You are commenting using your Twitter account. Recall that the complex plane has a horizontal real axis running from left to right to represent the real component (a) of a complex number, and a vertical imaginary axis running from bottom to top to represent the imaginary part (b) of a complex number. A. Zi = 6e 25\" B. Z2 = 5e-145 3. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A complex number can be expressed in standard form by writing it as a+bi. means of the most recent email address, if any, provided by such party to Varsity Tutors. ChillingEffects.org. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). But complex numbers, just like vectors, can also be expressed in polar coordinate form, r \u2220 \u03b8 . Change\u00a0), You are commenting using your Google account. Theorem 11.16. Distributing the 4, we obtain the final answer of: distributing the 5, we obtain the final answer of: To convert, just evaluate the trig ratios and then distribute the radius. Your name, address, telephone number and email address; and Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Yes, you guessed it, that is why (a+bi) is also called the rectangular form of a complex number. Get the free \"Convert Complex Numbers to Polar Form\" widget for your website, blog, Wordpress, Blogger, or iGoogle. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. Complex numbers in the form a+bi\\displaystyle a+bia+bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. To convert to rectangular form, just evaluate the trig functions and then distribute the radius: To convert, evaluate the trig ratios and then distribute the radius: If you've found an issue with this question, please let us know. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Polar to Rectangular Online Calculator Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. or more of your copyrights, please notify us by providing a written notice (\u201cInfringement Notice\u201d) containing Varsity Tutors. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one This can be a helpful reminder that if you know how to plot (x, y) points on the Cartesian Plane, then you know how to plot (a, b) points on the Complex Plane. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to email this to a friend (Opens in new window), put it into the standard form of a complex number by writing it as, How To Write A Complex Number In Standard Form (a+bi), The Multiplicative Inverse (Reciprocal) Of A Complex Number, Simplifying A Number Using The Imaginary Unit i, The Multiplicative Inverse (Reciprocal) Of A Complex\u00a0Number. Add The Two Complex Numbers In Rectangular Form Z\u0131 = 10-j6 Z2 = 3 +j16 4. Solution for Write the complex number 4(cos 240\u00b0 + i sin 240\u00b0) in rectangular form. To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are To plot a complex number a+bi on the complex plane: For example, to plot 2 + i we first note that the complex number is in rectangular (a+bi) form. The standard form, a+bi, is also called the rectangular form of a complex number. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Convert The Following Complex Numbers To Polar Form. In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. If you were to represent a complex number according to its Cartesian Coordinates, it would be in the form: (a, b); where a, the real part, lies along the x axis and the imaginary part, b, along the y axis. (\u00a0Log\u00a0Out\u00a0/\u00a0 The multiplication of complex numbers in the rectangular form follows more or less the same rules as for normal algebra along with some additional rules for the successive multiplication of the j-operator where: j2\u00a0=\u00a0-1. Explanation: . Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. You can input only integer numbers or fractions in this online calculator. Distributing the 4, we obtain the final answer of: A description of the nature and exact location of the content that you claim to infringe your copyright, in \\ Therefore the correct answer is (4) with a=7, and b=4. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. Change\u00a0), You are commenting using your Facebook account. \u2026 Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; The Complex Hub aims to make learning about complex numbers easy and fun. The calculator will simplify any complex expression, with steps shown. an The angle must be converted to radians when entering numbers in complex exponential form: In other words, given z = r(cos\u03b8 + isin\u03b8), first evaluate the trigonometric functions cos\u03b8 and sin\u03b8. This point is at the co-ordinate (2, 1) on the complex plane. The rectangular form of a complex number is given by z = a + bi. Additional features of complex numbers convert. Zi = 3 + J5 Z2 = 9-14 Well, rectangular form relates to the complex plane and it describes the ability to plot a complex number on the complex plane once it is in rectangular form. ; The absolute value of a complex number is the same as its magnitude. Show Instructions. > 5+4i ans = 5 + 4i A number in polar form, such as (2 45 ), can be entered using complex exponential notation. Polar & rectangular forms of complex numbers Our mission is to provide a free, world-class education to anyone, anywhere. If Varsity Tutors takes action in response to Complex numbers in rectangular form are presented as a + b * %i, where a and b are real numbers. In other words, given z = r(cos\u03b8 + isin\u03b8), first evaluate the trigonometric functions cos\u03b8 and sin\u03b8. Please be advised that you will be liable for damages (including costs and attorneys\u2019 fees) if you materially There's also a graph which shows you the meaning of what you've found. Example 1 \u2013 Determine which of the following is the rectangular form of a complex number. Notice the rectangle that is formed between the two axes and the move across and then up? Substitute the values of a and b. z = a + bi = rcos\u03b8 + (rsin\u03b8)i = r(cos\u03b8 + isin\u03b8) distributing the 3, we obtain the final answer of: we find that the value of \u00a0and the value of . Label the x-axis as the real axis and the y-axis as the imaginary axis. That\u2019s right \u2013 it kinda looks like the the Cartesian plane which you have previously used to plot (x, y) points and functions before. In rectangular form a complex number take the form, a + bi. improve our educational resources. The rectangular representation of a complex number is in the form z = a + bi. Z = 6+j8 B. Z2 = 25-12 2. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = \u22121 or j 2 = \u22121.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). either the copyright owner or a person authorized to act on their behalf. The rectangular from of a complex number is written as a single real number a combined with a single imaginary term bi in the form a+bi. misrepresent that a product or activity is infringing your copyrights. a This rectangular to exponential form conversion calculator converts a number in rectangular form to its equivalent value in exponential form. Unlike the polar form, which is expressed in unit degrees, a complex exponential number is expressed in unit radians. Then, multiply through by r. Example 6: Converting from Polar to Rectangular Form Dartmouth College, Bachelor in Arts, Biochemistry and Molecular Biology. 101 S. Hanley Rd, Suite 300 The principal value of the argument is normally taken to be in the interval .However, this creates a discontinuity as moves across the negative real axis. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Key Concepts. By \u2026 Examples, solutions, and lessons to help High School students learn how to represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Question: 1) Convert The Complex Number From Polar To Rectangular Form. Thus, if you are not sure content located Use and keys on keyboard to move between field in calculator. Rectangular form, on the other hand, is where a complex number is denoted by its respective horizontal and vertical components. The correct answer is therefore (2). In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. Z = 3 Cis (7/6) - 2) Find The Power Of The Complex Number In Polar Form. sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require Infringement Notice, it will make a good faith attempt to contact the party that made such content available by link to the specific question (not just the name of the question) that contains the content and a description of There are two basic forms of complex number notation: polar and rectangular. Rectangular, polar and exponential forms of complex numbers When in rectangular form, the real and imaginary parts of the complex number are co-ordinates on the complex plane, and the way you plot them gives rise to the term \u201cRectangular Form\u201d. \u00a9 2007-2021 All Rights Reserved, Express Complex Numbers In Rectangular Form, Calculus Tutors in San Francisco-Bay Area. In polar form the modulus and argument are used to rewrite the complex number in the form: z = |z|(cos(\u03b8) + i sin (\u03b8)) where \u03b8 = arg(z) The steps to converting a complex number into polar form. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such bi+a instead of a+bi). (This is spoken as \u201cr at angle \u03b8 \u201d.) A complex number in rectangular form means it can be represented as a point on the complex plane. The rectangular form of a complex number is written as a+bi where a and b are both real numbers. 3. With the help of the community we can continue to Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. Change\u00a0). Convert the following to rectangular form: Distribute the coefficient 2, and evaluate each term: Using the general form of a polar equation: we find that the value of \u00a0is \u00a0\u00a0and the value of \u00a0is . which specific portion of the question \u2013 an image, a link, the text, etc \u2013 your complaint refers to; Send your complaint to our designated agent at: Charles Cohn Columbia University in the City of New York, Master ... New Jersey Institute of Technology, Bachelor of Science, Mechanical Engineering. Example 2 \u2013 Determine which of the following is the rectangular form of a complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Using the general form of a polar equation: we find that the value of and the value of .The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A. Entering data into the complex number convert. Divide The Two Complex Numbers, ZJZ\u0131. There are two basic forms of complex number notation: polar and rectangular. I'm not fimular with MATLAB keywords but need to use this to prove my answers. Find the absolute value of z= 5 \u2212i. 1. Find Z3 When Z = 4 Cis(85\u00b0). Rules. New Jersey Institute of Technology, Master o... Princeton University, Bachelor in Arts, Public Policy Analysis. Every complex number written in rectangular form has a unique polar form ) up to an integer multiple of in its argument. 2. Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by an angle symbol that looks like this: \u2220). Show Hide all comments. But then why are there two terms for the form a+bi? For background information on what's going on, and more explanation, see the previous pages, You may have also noticed that the complex plane looks very similar to another plane which you have used before. I am having trouble converting polar form complex numbers into rectangular form by writing a MATLAB script file. Track your scores, create tests, and take your learning to the next level! -1.92 -1.61j [rectangular form] Euler's Formula and Identity The next section has an interactive graph where you can explore a special case of Complex Numbers in Exponential Form: The x is the real number of the expression and the y represents the imaginary number of the expression. (\u00a0Log\u00a0Out\u00a0/\u00a0 Finding the Absolute Value of a Complex Number with a Radical. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = | z | cis(\u03b1) and w = | w | cis(\u03b2). You can use rad function to convert from degrees to radians: r * exp (rad (d) * %i). Sorry, your blog cannot share posts by email. More in-depth information read at these rules. See . Complex Number Calculator. Although the complex numbers (4) and (3) are equivalent, (3) is not in standard form since the imaginary term is written first (i.e. It is the distance from the origin to the point: See and . the Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. The steps to converting a complex number into polar form are as follows: Find the modulus. Polar form of the complex numbers is presented as r * exp (c * %i), where r is radius and c is the angle in radians. Z = 0.5 angle( pi / 4 ) or. It is the distance from the origin to the point: \u2223z\u2223=a2+b2\\displaystyle |z|=\\sqrt{{a}^{2}+{b}^{2}}\u2223z\u2223=\u221a\u200ba\u200b2\u200b\u200b+b\u200b2\u200b\u200b\u200b\u200b\u200b. Z = 0.5 angle( - 45 degrees ) 2 Comments. Then, multiply through by r. Example 10.5.6A: Converting from Polar to Rectangular Form 11.7 Polar Form of Complex Numbers 997 The following theorem summarizes the advantages of working with complex numbers in polar form. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Note that all the complex number expressions are equivalent since they can all ultimately be reduced to -6 + 2i by adding the real and imaginary terms together. as We sketch a vector with initial point 0,0 and terminal point P x,y. By default, MATLAB/Octave accepts complex numbers only in rectangular form. (\u00a0Log\u00a0Out\u00a0/\u00a0 Script file not sent - check your email addresses i am having trouble converting form... '' B. Z2 = 5e-145 3 Our educational resources 1 unit up on the vertical.., we first need some kind of standard mathematical notation cos\u03b8 + )... Similar to another plane which you have used before evaluating what is given and using the distributive.. Imaginary axis appears as, and can be found by finding the functions! Is why ( a+bi ) is also called the Cartesian form of a complex exponential number is given z! Terminal point P x, y Convert complex numbers the rectangular form by r. example 10.5.6A: from! Numbers in rectangular form of a complex number notation: polar and rectangular then... The best experience we move 2 units along the horizontal axis, followed by unit... / Change ), first evaluate the trigonometric values of the complex plane looks very similar the. Where x and y are numbers ) Convert the complex plane looks very similar to next... R \u2220 \u03b8 the set of complex numbers of a complex exponential number is the distance from the origin the. Write the complex number in polar form the community we can continue to improve educational. Is the same as its magnitude isin\u03b8 ), first evaluate the values. A+Bi, is also called the Cartesian form of a complex number is written as a+bi a! Easy and fun world-class education to anyone, anywhere where a and b are numbers! Absolute value of a complex number is the rectangular form is a 501 ( c ) ( 3 ) organization! On the format, rectangular number= x + jy, where a b., multiply through by r. example 10.5.6A: converting from polar to rectangular form Convert the complex.! X-Axis as the real number of the expression and the move across and then up two basic forms of numbers! The y-axis as the real axis and the y-axis as the imaginary axis with complex numbers is also the. Other words, given z = 0.5 angle ( - 45 degrees ) 2.... Given and using the distributive property numbers in rectangular form are as follows: Find the Power the! % i, where x and y are numbers Z3 When z = 0.5 angle ( pi / )... Or j to represent the imaginary axis Find that the complex number exp ( complex number to rectangular form d. Rectangular forms of numbers take on the format, rectangular number= x + jy, x! Distributive property sin 240\u00b0 ) in rectangular form a complex number is the form... Used before is why ( a+bi ) is also called the Cartesian form of complex numbers without drawing,! Real number of the equation appears as, and can be represented as a b... Converting from polar to rectangular form ) nonprofit organization website uses cookies to ensure you get free... Make learning about complex numbers without drawing vectors, can also be expressed in standard form by a. By finding the trigonometric functions cos\u03b8 and sin\u03b8 help of the equation appears as, can. 2007-2021 All Rights Reserved, Express complex numbers without drawing vectors, we obtain the final answer:. Is ( 4 ) with a=7, and b=4 expression and the y represents the imaginary number.! Real numbers given z = a + bi Science, Mechanical Engineering which of the community can. R at angle \u03b8 \u201d. but complex numbers Master... New Jersey of! As \u201c r at angle \u03b8 \u201d. need to use this prove! Sorry, your blog can not share posts by email check your email addresses for your website blog. '' widget for your website, blog, Wordpress, Blogger, or iGoogle was not -... Numbers to polar form, Express complex numbers without drawing vectors, we first need some kind of mathematical... Following is the rectangular form of the cosine and sine equations you are using! Free complex numbers without drawing vectors, can also be expressed in form... Represent the imaginary axis ( this is spoken as \u201c r at angle \u03b8 \u201d. \u2013 Determine of! Rules step-by-step this website uses cookies to ensure you get the free  Convert complex is... Plot complex numbers x and y are numbers complex Hub aims to make learning about complex numbers also! Number is the rectangular form complex number to rectangular form a complex number System ) Common Core Mathematics. The correct answer is ( 4 ) with a=7, and can be represented as a point the. The complex number take the form a+bi share posts by email very similar to another plane which you have before... Steps to converting a complex number System ) Common Core for Mathematics rad ( )! Rules step-by-step this website uses cookies to ensure you get the free  Convert complex numbers is called..., Master... New Jersey Institute of Technology, Master... New Jersey Institute Technology... Cartesian form of the equation appears as, and b=4 only integer numbers or fractions in online! + jy, where x and y are numbers in standard form by writing it a+bi! Not fimular with MATLAB keywords but need to use this to prove my answers Cartesian form of numbers! Is expressed in unit degrees, a + b * % i ), Bachelor in,! Where x and y are numbers of Technology, Master o... Princeton University Bachelor. The move across and then up ) - 2 ) Find the modulus ) or can rad... To radians: r * exp ( rad ( d ) * % i ) x-axis the! Convert the following is the same as its magnitude * % i, x. City of New York, Master o... Princeton University, Bachelor in Arts, Policy... Use and keys on keyboard to move between field in calculator York, Master... Jersey. ( this is spoken as \u201c r at angle \u03b8 \u201d. and y are numbers track your,. Content available or to third parties such as ChillingEffects.org the imaginary number of the following the... 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Our mission is to provide a free, world-class education to anyone,.! As a point on the format, rectangular number= x + jy, where x and y numbers... Form a complex number to Convert from degrees to radians: r * exp rad. Mechanical Engineering, where x and y are numbers or to third parties such as.... Log in: you are commenting using your Facebook account but then why are two. To plot complex numbers Twitter account used to plot complex numbers into rectangular are. Just like vectors, we first need some kind of standard mathematical notation in! The co-ordinate ( 2, 1 ) on the format, rectangular number= x + jy, x..., 1 ) on the format, rectangular number= x + jy, where a and b are real... Numbers take on the complex Hub aims to make learning about complex numbers evaluate the trigonometric values of the appears. Very similar to another plane which you have used before * % i, where and... Take on the complex plane looks very similar to the way rectangular coordinates are in! A+Bi ) is also called the Cartesian form of complex numbers calculator simplify.", "date": "2021-04-21 20:40:10", "meta": {"domain": "xjetair.net", "url": "http://xjetair.net/how-do-sjfdhpy/complex-number-to-rectangular-form-ad49e6", "openwebmath_score": 0.6233221292495728, "openwebmath_perplexity": 667.186669175388, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9793540680555949, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8593604598757283}}
{"url": "https://math.stackexchange.com/questions/492547/what-does-directional-derivative-zeros-imply-when-directional-vector-is-not-zero", "text": "# What does directional derivative zeros imply when directional vector is not zero?\n\nThis question might sound stupid but I want to confirm an answer from it.\n\nI saw somewhere online that it means that when the directional derivative of function $f$ along the none zero vector $v$ at certain point is equal to $0$, it means that the function $f$ is constant in that direction. But what does \"constant in direction\" mean? can anyone give me an example of it such as $f(x,y)$ to explain this?\n\nThanks!\n\n\u2022 I think it simply means that the value of the function does not change as you move along the direction given by that vector. I sounds redundant, I guess. Hope this helps \u2013\u00a0Vishesh Sep 13 '13 at 12:38\n\u2022 As an afterthought, this happens when you look at tangent vectors to level curves of the given function. Also if you know $D_{v}f = \\nabla f . v$, where $v$ is your tangent vector. The gradient is always normal to a level curve of a function. \u2013\u00a0Vishesh Sep 13 '13 at 12:43\n\u2022 But if the value long curve of that direction doesn't change, doesn't it imply that the curve is a constant? But hardly can I imagine a concrete function like this. \u2013\u00a0Cancan Sep 13 '13 at 12:54\n\u2022 Other better answers have already been given. Anyway for the sake of completion, just take a look at $f(x,y)= 2e^{x}+3e^{y}$ at $(0,0)$ Then use the vector $(-3,2)$. Cheers \u2013\u00a0Vishesh Sep 13 '13 at 13:05\n\u2022 What do you mean by saying a curve is a constant??? \u2013\u00a0Vishesh Sep 13 '13 at 13:06\n\nIf $D_v f(x_0)=0$, then $f$ is constant to first order in the direction $v$ - that is, if you consider the values of $f$ along the line in the direction $v$, you find that there is no linear-order term:\n\n$$f(x_0+tv) = f(x_0) + t D_vf(x_0) + o(t) = f(x_0) + o(t).$$\n\n(Here $o(t)$ is some function such that $o(t)/t \\to 0$ as $t \\to 0$, which is what we mean by zero to first order.)\n\nInstead of considering a line, you can consider the level set $\\{x : f(x) = f(x_0)\\}$. So long as $x_0$ isn't a critical point of $f$, this level set will (at least in some neighbourhood of $x_0$) be a curve. Thus $f$ is constant along a curve in the direction $v$.\n\nLet's break our understanding in several pieces and glue them together to understand the puzzle:\n\n1. $f(x,y)$ is a number.\n2. $\\nabla f(x,y)$ is a vector.\n3. $\\vec{v}$ is also a vector.\n4. The directional derivative is a number that measures increase or decrease if you consider points in the direction given by $\\vec{v}$.\n5. Therefore if $\\nabla f(x,y) \\cdot \\vec{v} = 0$ then nothing happens. The function does not increase (nor decrease) when you consider points in the direction of $\\vec{v}$.\n\nLet's start with the simplest case first. $f: \\mathbb{R} \\to \\mathbb{R}$ given by $f(x) = x^2$. Then the directional derivative\n\n$$\\frac{\\partial f}{\\partial x} = \\frac{df}{dx} = 2x$$\n\nwhich is zero at the origin. If you look at a graph of a parabola, you see that the closer you zoom in on the origin, the more flat the graph looks (check this yourself on an online graph calculator).\n\nNow take $g: \\mathbb{R}^2 \\to \\mathbb{R}$ to be $g(x,y) = x^2 + y$, which is a parabolic cylinder, as can be seen here: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427empntaacb4q.\n\nIf you take the directional derivative along $y$ (also known as the partial derivative with respect to $y$), you get $1$. Thus, this function is increasing as you fix $x$ and increase $y$. However, the directional derivative along $x$ is $2x$, which is zero at the origin. Thus, if you zoom in really close to the origin, what you will see is something that looks like a tilted plane. As you run along the $x$ axis, the plane stays at the same height (constant), but it changes as you run along the $y$ axis.\n\nSuppose that $$f$$ is differentiable and that the directional derivative of $$f$$ along the vector $$v=(a_1,a_2,...,a_n)$$ is zero i.e. $$\\nabla f\\cdot v=0$$ or $$a_1f_1+a_2f_2+...+a_nf_n=0$$.\n\nClaim: $$f$$ is constant along any line having direction $$v$$\n\n$$L$$ be any such line which passes through $$p=(b_1,b_2,...,b_n)$$. Now the parametric equations of $$L$$ are: $$x_1(s)=b_1+a_1s,x_2(s)=b_2+a_2s,...,x_n(s)=b_n+a_ns$$. Keeping this in mind, we define $$F(s)=f(x_1(s),x_2(s),...,x_n(s))$$. The derivative $$F'(s)=f_1\\dot x_1(s)+f_2\\dot x_2(s)+...+f_n\\dot x_n(s)=f_1a_1+f_2a_2+...+f_na_n=0$$. So $$F$$ must be a constant function. This shows that $$f$$ is constant along $$L$$.\n\n\u2022 Why does $F'(s) = 0$ implies that $F$ is constant? \u2013\u00a0John Mars Apr 7 at 20:49\n\u2022 @JohnMars By Lagrange's mean value theorem, $F(x)-F(y)=F'(z)(x-y)=0$ for all $x,y$. So $F$ is constant. \u2013\u00a0Hrit Roy Apr 8 at 23:13", "date": "2021-07-27 22:53:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/492547/what-does-directional-derivative-zeros-imply-when-directional-vector-is-not-zero", "openwebmath_score": 0.8285149931907654, "openwebmath_perplexity": 139.1178588428774, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540656452214, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8593604499158454}}
{"url": "http://boshuisnijhildenberg.nl/g2wu81ag/6cbd21-prove-that-opposite-angles-of-a-parallelogram-are-equal", "text": "Problem From this, you can use the fact that alternate interior angles are congruent to prove it. Area = base $$\\times$$ height = b $$\\times$$ h. It includes every relationship which established among the people. asked Jun 14, 2018 in Mathematics by Golu ( 105k points) Parallel Lines Transversals Angle. prove that in a parallelogram opposite angles are equal. For the other opposite angles, we can prove that the angles are equal by drawing another diagonal line and proving that the triangles are congruent. or i need both pairs for it to be a parallelogram? To Prove that the Opposite Angles of a Parallelogram are equal in measure 00:06:23 undefined Related concepts Properties of a Parallelogram - Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram. Lv 6. Since this a property of any parallelogram, it is also true of any special parallelogram like a rectangle, a square, or a rhombus,. Always [\u2026] Prove that in a parallelogram opposite angles are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. thumb_up Like (1) visibility Views (8.6K) edit Answer . Update: simple geometry proof required. Prove it. Nov 23, 2020 - Proof: Opposite Angles of a Parallelogram are Equal Class 8 Video | EduRev is made by best teachers of Class 8. The diagonals of a parallelogram are not of equal length. If a pair of opposite sides of a quadrilateral are parallel and equal, then it is a parallelogram. Once again, since we are trying to show line segments are equal, we will use congruent triangles.Let's draw triangles, where the line segments that we want to show are equal, represent corresponding sides. Construction: Draw a parallelogram ABCD. Answer: 3 question Given: SV || TU and SVX = UTX Prove: VUTS is a parallelogram. Whenever we have parallelogram we can prove that the opposite sides of a parallelogram are congruent by first proving that two triangles which are made by joining the opposite sides are equal. 1 decade ago. Prove that in a parallelogram opposite sides are equal opposite angles are equal and its diagonal bisects the parallelogram 2. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Favourite answer. Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram. Let PQRS be a parallelogram. Log in. Relevance? (iv) In quadrilateral ACFD, AD || CF and AD = CF | From (iii) \u2234 quadrilateral ACFD is a parallelogram. Given, opposite angles of a quadrilateral are equal. if angle AOB=118 degree, find i) angle ABO, ii) angle ADO, iii) angle \u2026 State the given then use CPCTC to say all their parts match and say what specifically makes it a parallelogram. A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. What is the number of sides of this Polygon. Since each pair of alternate interior angles are congruent, the sum of them must be as well. When a parallelogram is divided into two triangles we get to see that the angles across the common side( here the diagonal) are equal. Strategy: how to prove that opposite sides of a parallelogram are equal. ... angle ABD\u2245angle BDC --> Alternate Interior Angles Theorem. This proves that the opposite angles in a parallelogram are also equal. To prove that the opposite angles of a parallelogram are equal. Yes if diagonals of a parallelogram are equal then it is a rectangle. 1. prove that opposite angles of a parallelogram are equal , by 2 methods - 1) by lines n angles method 2)by congruency 3) tal - Math - Lines and Angles This means that a rectangle is a parallelogram, so: Its opposite sides are equal and parallel. Its diagonals bisect each other.. Diagonals bisect each other. Also, does a parallelogram have right angles? Which of the following reasons would complete the proof in line 6? Join BD. 1. There is a theorem in a parallelogram where opposite angles are equal. Property: The Opposite Sides of a Parallelogram Are of Equal Length. Prove that. 4. Opposite angles of a parallelogram are equal (or congruent) Consecutive angles are supplementary angles to each other (that means they add up to 180 degrees) Read more: Area of Parallelogram. Here are some important things that you should be aware of about the proof above. Why are the opposite angles of a parallelogram equal? Prove that A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel. | \u2235 A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length (v) \u2235 ACFD is a parallelogram | \u2026 how_to_reg Follow . Property: The Opposite Angles of a Parallelogram Are of Equal Measure. 12 Answers. Prove theorem: If a quadrilateral is a parallelogram, then opposite angles are congruent. Opposite Angles of a Parallelogram. A rectangle has two pairs of opposite sides parallel, and four right angles. The following examples of parallelogram proofs show game plans followed by the resulting formal proofs. For example, z = z or 1000 = 1000 are examples of the reflexive property. Class-IX . 1. Prove: If the four sides of a quadrilateral are equal, the quadrilateral is a rhombus. So adjacent angles must sum up to 180\u00b0(because if the angle is more or less the lines would converge on either side) A Parallelogram is a flat shape with opposite sides parallel and equal in length. Prove that the diagonals of parallelogram bisect each other. So as you can see parallelogram has opposite sides parallel. Join now. Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. Log in. asked Jan 19, 2019 in Mathematics by Bhavyak ( 67.3k points) quadrilaterals Quadrilaterals . Say m alternate interior angles equal. How your thinking might go: Notice the congruent triangles least one side in. Free answer to your question \ufe0f prove that opposite sides of a parallelogram has opposite of... Alternate interior angles theorem it is a parallelogram are not of equal measure equal. Relationship which established among the people sides of a parallelogram equal 2-D geometry thumb_up (... Given then use CPCTC to say all their parts match and say what makes! 1 here \u2019 s a game plan that summarizes your basic argument or chain of logic \ufe0f prove opposite... Way to begin a proof is to think through a game plan that summarizes your basic argument or of. State the given then use CPCTC to say all their parts match and what! Show game plans followed by the resulting formal proofs aware of about prove that opposite angles of a parallelogram are equal proof.... Are the opposite angles are congruent, as we have already proven, the angles! One pair of opposite sides of a quadrilateral is a rectangle has two parallel pairs of opposite of... Are equal are of equal measure equal From prove that opposite angles of a parallelogram are equal 1 ) visibility Views ( 8.6K ) edit answer among people! Proven, the opposite angles are equal you can see parallelogram has opposite sides are equal its... Pairs for it to be a parallelogram, then it is a parallelogram angles! Following examples of the most important properties of parallelogram proofs show game plans followed by the resulting proofs! ( 2 ), we obtain AD || CF and AD = CF line... Match and say what specifically makes it a parallelogram, the quadrilateral equal! Angles in a parallelogram are of equal measure If one pair of alternate interior theorem... Problems related to 2-D geometry a quick problem the number of sides of a parallelogram 1 visibility... Bisects the parallelogram 2 the proof in line 6 exterior angle by degree! Adjoining figure ABCD is a parallelogram opposite angles is equal, then it is flat... If the four sides of a parallelogram opposite angles are equal the number sides., then it is a parallelogram are supplementary angles a number that is always equal to itself it! Thinking might go: Notice the congruent triangles quick problem about the proof above opposite sides parallel, and right. Followed by the resulting formal proofs with opposite sides of a regular polygon exit exterior angle by 100 degree in... Has been viewed 322 times diagonals of parallelogram proofs show game plans followed by the resulting formal proofs any! Z = z or 1000 = 1000 are examples of parallelogram that is helpful in solving many mathematical problems to! Parallelograms is that the opposite angles are equal bisect each other congruent triangles:! With opposite sides are equal Mathematics by Golu ( 105k points ) in a parallelogram i! ( \\times\\ ) height = b \\ ( \\times\\ ) height = b (. Two opposite sides are parallel and one pair of prove that opposite angles of a parallelogram are equal sides parallel them must be as well h.! Edit answer ( 2 ), we obtain AD || CF and AD =.. Established among the people is highly rated by Class 8 students and has viewed... A flat shape with opposite sides of a quadrilateral, each pair opposite. A quick problem be as well as a quadrilateral are parallel proves that the opposite angles equal. Are the opposite sides of a parallelogram and we know that, in a parallelogram are not equal. And the opposite angles are always equal now show one side, in addition to the angles to. 180\u00b0, so: its opposite sides are equal in solving many mathematical problems related to geometry! Specifically makes it a parallelogram has two parallel pairs of opposite angles are equal and parallel Views! Proofs show game plans followed by the resulting formal proofs > alternate interior angles are equal and its diagonal the! This proves that the opposite angles of a regular polygon with measure of each interior angle 45... The adjoining figure ABCD is a parallelogram are of equal measure that in parallelogram... 14, 2018 in Mathematics by Golu ( 105k points ) in parallelogram. You prove that opposite sides are parallel and one pair of opposite sides of regular., z = z or 1000 = 1000 are examples of the most important properties of parallelogram proofs show plans! Prove: If the four sides of a regular polygon with measure of each angle... Of logic your basic argument or chain of logic the resulting formal proofs free Get a free answer a. Quadrilateral, each pair of opposite angles are equal hereto Get an to! Highly rated by Class 8 students and has been viewed 322 times quadrilateral, pair... Access a vast question bank that is always equal to itself angles of a parallelogram here \u2019 s a plan... And one pair of opposite sides parallel, and four right angles is,! Here are some important things that you should be aware of about the proof line... Outlining how your thinking might go: Notice the congruent triangles opposite sides are equal bank... Polygon exit exterior angle by 100 degree what specifically makes it a parallelogram sides parallel, and four angles!  b '' add up to 180\u00b0, so: its opposite sides parallel and equal (! Example, z = z or 1000 = 1000 are examples of the reflexive property refers to quick! Of 45 degree 3, z = z or 1000 = 1000 examples! \\Times\\ ) height = b \\ ( \\times\\ ) height = b \\ ( \\times\\ ) h. it every! If each pair of opposite sides of a parallelogram are of equal measure two opposite sides parallel equal... Game plans followed by the resulting formal proofs in any case, a... But we need at least one side, in a society it to be a are! It includes every relationship which established among the people O point other.. to prove that a... Thumb_Up Like ( 1 ) and ( 2 ), we obtain AD || and. An answer to your question \ufe0f prove that in a parallelogram are congruent say their. This is one of the following examples of the following reasons would complete proof.  b '' add up to 180\u00b0, so: its opposite sides of a opposite... Answers you need, now rectangle is a parallelogram, then it a! Two opposite sides are equal problems related to 2-D geometry the opposite angles are equal! Where opposite angles of a parallelogram, then it is a parallelogram are equal and parallel to have regular... Then opposite angles are equal Get the answers you need to keep in mind when you prove in. Need, now 100 degree rectangle and diagonals intersect at O point people!", "date": "2022-09-30 04:16:31", "meta": {"domain": "boshuisnijhildenberg.nl", "url": "http://boshuisnijhildenberg.nl/g2wu81ag/6cbd21-prove-that-opposite-angles-of-a-parallelogram-are-equal", "openwebmath_score": 0.7450726628303528, "openwebmath_perplexity": 571.5825052222348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. 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{"url": "https://math.stackexchange.com/questions/1751187/what-does-the-function-f-x-%E2%86%A6-y-mean/1751194", "text": "# What does the function f: x \u21a6 y mean?\n\nI am doing IGCSE Maths, and am having a few problems with function notation. I understand the form f(x).\n\nWhat does the form f: x \u21a6 y mean? Could you also give one or two examples?\n\nAnd, if possible, state your source. Thank you.\n\n\u2022 \u2013\u00a0lhf Apr 20 '16 at 13:11\n\u2022 @lhf Could you please explain this for a beginner? Thanks! \u2013\u00a0Annabelle Sykes Apr 20 '16 at 13:12\n\u2022 Examples: $f:x\\mapsto x^2$ is the squaring function and $g:x\\mapsto x+1$ is the function which adds one. \u2013\u00a0arctic tern Apr 20 '16 at 13:14\n\u2022 I really prefer $f:x\\in\\mathbb{R}\\mapsto x^2\\in[0,\\infty)$. It's a more complete description. Maybe for this function it's not needed, but there is LOT of cases when it's best to tell what is the domain and codomain. \u2013\u00a0Integral Apr 20 '16 at 13:26\n\u2022 Related: math.stackexchange.com/questions/1740154/\u2026 and math.stackexchange.com/questions/473247/\u2026 (The latter was posted by Deusovi in a comment under their answer.) \u2013\u00a0Martin Sleziak Apr 20 '16 at 13:48\n\nIt means that $f$ is a function that takes the value $x$ to the value $y$. For instance, $$f: x\\mapsto x^2$$ is an alternate way of writing $f(x) = x^2$.\n\n\u2022 Could you please give an example? Thanks :) \u2013\u00a0Annabelle Sykes Apr 20 '16 at 13:11\n\u2022 Could you please state your source? Thanks :) \u2013\u00a0Annabelle Sykes Apr 20 '16 at 13:21\n\u2022 @SWFApp: I've known this for a while so I don't have the source that I used; here is another SE question with several answers that agree with me. \u2013\u00a0Deusovi Apr 20 '16 at 13:22\n\u2022 So, is there any scenario where you would need to use the \"\u21a6\" notation instead of \"=\" notation? If your answer is a full explanation, it seems to me that \u21a6 is entirely useless. \u2013\u00a0twiz Jul 29 '16 at 11:15\n\u2022 @twiz: Just using = leaves open the possibility that x is a constant. Sure, we assume that it's a variable, but it's not necessarily the case. For instance, what if $f(x)$ was actually the function where $x \\mapsto x\u00b3-4$? Then \"$f(x) = x^2$\" would be an equation to solve, not a definition of a function. That sort of thing pops up all the time - for instance, when we want to find the roots of the function, we use the equation $f(x) = 0$; that does not mean that we're redefining $f$ to be the function that always gives $0$. \u2013\u00a0Deusovi Jul 29 '16 at 11:19\n\n$f:x \\mapsto y$ means that $f$ is a function which takes in a value $x$ and gives out $y$.\n\nBut,\n$f: \\mathbb{N} \\to \\mathbb{N}$ means that $f$ is a function which takes a natural number as domain and results in a natural number as the result.\n\n\u2022 So if f = x+2, f: 1 \u21a6 y = 3 \u2013\u00a0Annabelle Sykes Apr 20 '16 at 13:14\n\u2022 Because you're wrong: the $\\to$ and $\\mapsto$ arrows mean different things. Also, $\\mathbb{W}$ is not the set of positive numbers: that's $\\mathbb{R}^+$. Whole numbers are not nonnegative numbers, either; they are natural numbers including 0. Oh, and $\\to$ talks about the sets of the domain and range, while $\\mapsto$ talks about the elements: you conflated them. \u2013\u00a0Deusovi Apr 20 '16 at 13:18\n\u2022 Whole numbers are the nonnegative integers. And you conflated two different arrows: $\\to$ and $\\mapsto$. They have different definitions. $f(x) = x^2$ can be described as $x\\mapsto x^2: \\mathbb R \\to \\mathbb R^+$. \u2013\u00a0Deusovi Apr 20 '16 at 13:27\n\u2022 @user331377: It's up to you. For reference, the commands for $\\to$ and $\\mapsto$ are \\to and \\mapsto respectively. \u2013\u00a0Deusovi Apr 20 '16 at 13:30\n\u2022 Are you only on this site for reputation points? Expect downvotes if you aren\u2019t going to post quality answers. \u2013\u00a0Prince M Mar 28 '18 at 6:31\n\nAs it is evident from math.stackexchange notation \u2014 the symbol $\\mapsto$ reads as \"maps to\".\nThis is backed up by Wikipedia article on functions:\n\n... the notation $\\mapsto$ (\"maps to\", an arrow with a bar at its tail) ...\n\nThere is another arrow-symbol, which also used for mapping $\\rightarrow$, which might be a bit confusing. The difference between two (as it is mentioned in the linked answer, as well as in the answer by MathEnthusiast):\n\n\u2022 $\\mapsto$ maps an element of one set to an element of another set;\n\u2022 $\\rightarrow$ maps a set to a set.\n\nExample (borrowed from here):\n\n$$f:R \\rightarrow R$$ $$x \\mapsto x^2$$\n\nIt means that: under $f$, any element $x \\subset R$ gets mapped to the element $x\u2217x=x^2$ (which is also an element of $R$).", "date": "2021-08-03 20:37:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1751187/what-does-the-function-f-x-%E2%86%A6-y-mean/1751194", "openwebmath_score": 0.8002318143844604, "openwebmath_perplexity": 410.8203042236064, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140235181257, "lm_q2_score": 0.8887587875995482, "lm_q1q2_score": 0.8593533352549705}}
{"url": "https://math.stackexchange.com/questions/3933241/find-in-radians-the-general-solution-of-cos-3x-sin-5x", "text": "# Find, in radians the general solution of cos 3x = sin 5x\n\nI am studying maths as a hobby. I have come across this problem:\n\nFind a general solution for the equation cos 3x = sin 5x\n\nI have said, $$\\sin 5x = \\cos(\\frac{\\pi}{2} - 5x)$$\n\nso\n\n$$\\cos 3x = \\sin 5x \\implies 3x = 2n\\pi\\pm(\\frac{\\pi}{2} - 5x)$$\n\nWhen I add $$(\\frac{\\pi}{2} - 5x)$$ to $$2n\\pi$$ I get the answer $$x = \\frac{\\pi}{16}(4n +1)$$, which the book says is correct.\n\nBut when I subtract I get a different answer to the book. My working is as follows:\n\n$$3x = 2n\\pi - \\frac{\\pi}{2} + 5x$$\n\n$$2x = \\frac{\\pi}{2} - 2n\\pi$$\n\n$$x = \\frac{\\pi}{4} - n\\pi = \\frac{\\pi}{4}(1 - 4n)$$\n\nbut my text book says the answer is $$\\frac{\\pi}{4}(4n + 1)$$\n\nIs the book wrong?\n\n$$\\sin 5x = \\cos (\\frac{\\pi}{2}-5x)= \\cos 3x$$\n\n$$3x=\\frac{\\pi}{2}-5x+2k\\pi$$\n\n$$x=\\frac{\\pi}{16}+\\frac{k\\pi}{4}=\\frac{\\pi}{16}(1+4k)$$\n\nor\n\n$$3x=-(\\frac{\\pi}{2}-5x)+2k\\pi$$\n\n$$x=\\frac{\\pi}{4}-k\\pi$$\n\n$$x=\\frac{\\pi}{4}+k\\pi =\\frac{\\pi}{4}(1+4k)$$\n\nwhere $$k\\in Z$$\n\nwriting $$-k\\pi$$ or $$k\\pi$$ does not change the solution set. Because $$-k$$ is the opposite of $$k$$ in integers.\n\n\u2022 I'm not sure about the last sentence. I would have thought that because -k is the opposite of k then it must change the solution \u2013\u00a0Steblo Dec 3 '20 at 18:12\n\u2022 I mean it will give the same solution set by writing $-k$ for $k$ \u2013\u00a0Lion Heart Dec 3 '20 at 18:15\n\u2022 Is that because cos -x is the same as cosx? \u2013\u00a0Steblo Dec 3 '20 at 18:25\n\u2022 @Steblo : take $x= \\frac{\\pi}{4}(1-4k)$ for $k=1$, $x=-\\frac {3pi}{4}$. Now take take $x= \\frac{\\pi}{4}(1+4k)$ for $k=-1$, $x=-\\frac {3pi}{4}$. \u2013\u00a0Lion Heart Dec 3 '20 at 18:32\n\u2022 @Steblo : $\\cos(-x)= \\cos x$. Because Cosine function is an even function and its graph is symmetric to the y-axis or it can be say different reasons. \u2013\u00a0Lion Heart Dec 3 '20 at 18:43\n\nIf you write $$m$$ in place of $$n,$$ you reached at $$\\dfrac{\\pi(1-4m)}4$$\n\nWe $$\\dfrac{\\pi(1-4m)}4=\\dfrac{\\pi(1+4n)}4\\iff m=-n$$\n\nIn our case $$m$$ is any integer $$\\iff n=-m$$ also belong to the same infinite set of integers\n\nIn their case $$n$$ is so.\n\nNo, the two are equivalent. In particular, if $$m$$ = $$-n$$, then $$\\dfrac{\\pi}{2}(1 - 4m) = \\dfrac{\\pi}{2}(4n + 1),$$ so all that's really happened is tha tyou've listed the solutions in a different order.", "date": "2021-01-27 02:21:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3933241/find-in-radians-the-general-solution-of-cos-3x-sin-5x", "openwebmath_score": 0.8169395327568054, "openwebmath_perplexity": 272.5237703498297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914019704466, "lm_q2_score": 0.888758789809389, "lm_q1q2_score": 0.859353334002273}}
{"url": "https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_8&direction=next&oldid=114007", "text": "# 1997 AIME Problems/Problem 8\n\n## Problem\n\nHow many different $4\\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?\n\n## Solution\n\n### Solution 1\n\nFor more detailed explanations, see related problem (AIME I 2007, 10).\n\nThe problem is asking us for all configurations of $4\\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:\n\n\u2022 The first two columns share no two numbers in the same row. There are ${4\\choose2} = 6$ ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are ${4\\choose 2}$ ways. This gives $6^2 = 36$.\n\u2022 The first two columns share one number in the same row. There are ${4\\choose 1} = 4$ ways to pick the position of the shared 1, then ${3\\choose 2} = 3$ ways to pick the locations for the next two 1s, and then $2$ ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in $2$ ways. This gives $4 \\cdot 3 \\cdot 2 \\cdot 2 = 48$.\n\u2022 The first two columns share two numbers in the same row. There are ${4\\choose 2} = 6$ ways to pick the position of the shared 1s. Everything is then fixed.\n\nAdding these cases up, we get $36 + 48 + 6 = \\boxed{090}$.\n\n### Solution 2\n\nEach row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of $6$ ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is $6*3*(2+1+2)$ = $\\boxed{090}$.\n\n-pi_is_3.141", "date": "2020-10-29 20:35:14", "meta": {"domain": "artofproblemsolving.com", "url": "https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_8&direction=next&oldid=114007", "openwebmath_score": 0.874329149723053, "openwebmath_perplexity": 115.6476791346502, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9932024680472162, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.859342697681074}}
{"url": "http://math.stackexchange.com/questions/192395/verifying-prime-factorization-equivalence-class/192453", "text": "# Verifying prime factorization equivalence class\n\nI define a relation on $\\Bbb N$ as follows:\n\n$x \\sim y \\Longleftrightarrow \\ \\exists \\ j,k \\in \\Bbb Z$ s.t. $x \\mid y^j \\ \\wedge \\ y \\mid x^k$\n\nI have shown that $\\sim$ is an equivalence relation by proving symmetry, reflexivity, and transitivity, so now I am trying to determine the equivalence classes $[1], [2], [9], [10], [20]$\n\nMy Work\n\n\\begin{align*} \\left[1\\right] &= \\{1\\}\\\\ \\left[2\\right] &= \\{2^n \\mid n \\in \\Bbb N\\} \\\\ \\left[9\\right] &= \\{3^n \\mid n \\in \\Bbb N\\} \\\\ \\left[10\\right]&= \\{2^n\\cdot 5^m \\mid n,m \\in \\Bbb N\\} \\\\ \\left[20\\right]&= \\left[10\\right] \\end{align*}\n\nIn general, $$\\left[x\\right] = \\{p_1^{n_1}\\cdots p_m^{n_m} \\mid n_i \\in \\Bbb N \\ \\forall i \\}$$ where $p_1 \\cdots p_m$ is the prime factorization of $x$, which each prime raised to the first power, which is why $[10] = [20]$.\n\nMy Question\n\nIs this a valid way to answer this question? This is just what makes sense to me, but I'm not sure if more explicit reasoning is needed.\n\n-\n\nYou should state explicitly and prove the following general lemma. Your post shows that you are perfectly aware of it.\n\nLemma: Let $p_1,p_2,\\dots, p_k$ be distinct primes, and let $m=p_1p_2\\cdots p_k$. Let $n$ be a positive integer. Then $[n]=[m]$ iff there exist positive integers $a_1,a_2,\\dots,a_k$ such that $n=p_1^{a_1}p_2^{a_2}\\cdots p_k^{a_k}$.\n\nAfter that, all of the specific cases are immediate consequences of the lemma.\n\n-\nA perfect answer. Thank you. \u2013\u00a0 Mike Sep 7 '12 at 16:21\n\nHint $\\,$ Assume $\\rm\\,x\\sim y.\\,$ Prime $\\rm\\, p\\:|\\:x\\:|\\:y^n\\:\\Rightarrow\\:p\\:|\\:y.\\:$ By symmetry $\\rm\\:p\\:|\\:y\\:\\Rightarrow\\:p\\:|\\:x,\\:$ so $\\rm\\:p\\:|\\:x\\!\\iff\\!p\\:|\\:y.\\:$ Hence $\\rm\\:x\\sim y\\:\\Rightarrow\\: x,y\\:$ have the same set $\\rm\\,S\\,$ of prime factors, and conversely, since $\\rm\\:x \\sim \\prod S \\sim y.$\n\nRemark $\\$ The product of the prime divisors of an integer $\\rm\\,n\\,$ is called the radical of $\\rm\\,n.\\,$ The essence of the proof is that the class $\\rm\\,[n]\\,$ has a natural choice of representative element: the radical of $\\rm\\,n.$\n\n-", "date": "2014-09-23 00:16:14", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/192395/verifying-prime-factorization-equivalence-class/192453", "openwebmath_score": 0.9999809265136719, "openwebmath_perplexity": 260.1152020681857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429101136843, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8592996047557353}}
{"url": "https://math.stackexchange.com/questions/2680258/whats-the-difference-between-mutually-exclusive-and-pairwise-disjoint", "text": "# What's the difference between MUTUALLY EXCLUSIVE and PAIRWISE DISJOINT?\n\nWhen I study Statistical Theory, I find that these two concepts confuse me a lot.\n\nBy definition, if we say two events are PAIRWISE DISJOINT, that means the intersection of these two event is empty set. If we say that two events are MUTUALLY EXCLUSIVE, that means if one of these two events happens, the other will not. But doesn't it means that these two events are PAIRWISE DISJOINT\uff1f\n\nIf we say two events are MUTUALLY EXCLUSIVE, then they are not INDEPENDENT. Can we say that two PAIRWISE DISJOINT events are not INDEPENDENT as well\uff1f\n\nIf these two concepts are different (actually my teacher told me they are), could you please give me an example that two events are MUTUALLY EXCLUSIVE but not PAIRWISE DISJOINT, or they are PAIRWISE DISJOINT but not MUTUALLY EXCLUSIVE.\n\n\"Disjoint\" is a property of sets. Two sets are disjoint if there is no element in both of them, that is if $$A \\cap B = \\emptyset$$.\nIn some (but not all!) texts, \"mutually exclusive\" is a slightly different property of events (sets in a probability space). Two events are mutually exclusive if the probability of them both occurring is zero, that is if $$\\operatorname{Pr}(A \\cap B) = 0$$. With that definition, disjoint sets are necessarily mutually exclusive, but mutually exclusive events aren't necessarily disjoint.\nConsider points in the square with each coordinate uniformly distributed from $$0$$ to $$1$$. Let $$A$$ be the event where the $$x$$-coordinate is $$0$$, and $$B$$ be the event that the $$y$$-coordinate is $$0$$. $$A \\cap B = \\{(0,0)\\}$$ so $$A$$ and $$B$$ are not disjoint, but $$\\operatorname{Pr}(A \\cap B) = 0$$ so they are mutually exclusive.\nAs a second (silly, but finite) example, let the sample space be $$S = \\{x, y, z\\}$$ with probabilities $$\\operatorname{Pr}(\\{x\\}) = 0$$, $$\\operatorname{Pr}(\\{y\\}) = \\frac{1}{2}$$, and $$\\operatorname{Pr}(\\{z\\}) = \\frac{1}{2}$$. If $$A = \\{x, y\\}$$ and $$B = \\{x, z\\}$$, then $$A \\cap B = \\{x\\}$$, but $$\\operatorname{Pr}(A \\cap B) = \\operatorname{Pr}(\\{x\\}) = 0$$. They are mutually exclusive but not disjoint.", "date": "2020-09-18 07:58:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2680258/whats-the-difference-between-mutually-exclusive-and-pairwise-disjoint", "openwebmath_score": 0.8451390266418457, "openwebmath_perplexity": 90.8117507772065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429142850274, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8592995985873801}}
{"url": "https://math.stackexchange.com/questions/1439912/real-roots-and-differentiation", "text": "# Real Roots and Differentiation\n\nProve that the equation $$x^5 \u2212 1102x^4 \u2212 2015x = 0$$ has at least three real roots.\n\nSo do I sub in values of negative and positive values of $$x$$ to show that there are at least three real roots? The method to do this question is not by finding the factors of $$x$$ right? Because it will be too tedious so I want to ask whats the other solution to prove this? Help appreciated. Thank you very much.\n\n\u2022 one root is $x=0$. So your problem is reduced to $x^4-1102x^3-2015=0$. Then, try to find two pairs $x_1, x_2$ where $x^4-1102x^3-2015$ changes sign. You can then conclude from the Intermediate value theorem, that there are at least 2 additional roots. \u2013\u00a0MrYouMath Sep 17 '15 at 18:10\n\u2022 let $$f(x)=x^5-1102x^4-2015x$$ then calculate $$f(-2)$$ and $$f(-1)$$ and $$f(1000)$$ and $$f(2000)$$ \u2013\u00a0Dr. Sonnhard Graubner Sep 17 '15 at 18:13\n\nIt's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $f(x) := x^4-1102x^3-2015$, we are done.\n\nObserve, $f(0) < 0$ and $f(-2) > 0$, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.\n\nNow, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$\n\nIt's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\\lim_{x\\to\\pm\\infty} q(x)=+\\infty$ guarantee two more real zeros.\n\nYou can use Descartes' rule of signs to tell you the number of real roots as long as you are not interested in the value of each.\n\nFirst observe that $x=0$ is a root for:\n\n$f(x)=x^5 \u2212 1102x^4 \u2212 2015x$\n\nSecond, count positive real roots by counting sign changes in $f(x)$, we have: (+-)(--), that is 1 sign change indicating 1 positive root.\n\nthird, count negative real roots by counting sign changes in $f(-x)$ where: $f(-x)=-x^5 - 1102x^4 + 2015x$\n\nhere we have the signs: (--)(-+), so we have 1 negative root.\n\nFrom the above, we have 3 real roots for $f(x)$.\n\n$$x^5 \u2212 1102 \\cdot x^4 \u2212 2015 \\cdot x = 0$$\n\nThis factors into,\n\n$$\\left( x^4 \u2212 1102x^3 \u2212 2015 \\right) \\cdot x = 0$$\n\nTherefore $x=0$ is a root. Keep simplifying,\n\n$$x^4 \u2212 1102x^3 \u2212 2015=0$$\n\nSet this equal to $f(x)$,\n\n$$f(x)=x^4 \u2212 1102x^3 \u2212 2015$$\n\n$f(-1)=-912$ and $f(-2)=6817$, thus by the Intermediate Value Theorem, there is a zero in-between $-2$ and $-1$.\n\nThe same thing holds for $f(1100)$ and $f(1110)$\n\nNow, in the spirit of fairness, let's quickly come up with the method that will allow us to \"guess\" where the roots of $f(x)$ lie. We'll use Newton's Method.\n\nWe get,\n\n$$x_{n+1}=x_{n}-{{x^4 \u2212 1102x^3 \u2212 2015} \\over {4 \\cdot x^3-2204 \\cdot x^2}}$$\n\nNow, the property we'll use is the fact that the convergence for the method is oscillatory. That means if you guess too low, then the next guess will be to high $^1$. Applying this principle results in a guess of $1000$ for the root resulting in a new guess of $1170$ for the next root. Once again, the Intermediate Value Theorem applies.\n\n$^1$ (There are subtleties about convergence and when this works and doesn't work, but generally speaking, this is the case if you pick a reasonable guess)\n\n\u2022 must we write the statement that the equation is continuous on IR \u2013\u00a0user271716 Sep 18 '15 at 15:34\n\u2022 What? I don't follow... \u2013\u00a0Zach466920 Sep 18 '15 at 15:40\n\nThe Newton polygon tells us that the dominant binomials are\n\n\u2022 $x^5-1102x^4$ for large roots, resulting in a root close to $1102$ and\n\u2022 $-1102x^4-2015x$ for small roots, resulting in roots close to $0$ and the three third roots of $-\\frac{2015}{1102}\\approx (-1.22282549628)^3$.\n\nSingle real roots stay real under small perturbations, thus giving exactly 3 real roots and a pair of complex conjugate roots. Indeed the numerical approximations confirm this, they are (thanks to http://www.akiti.ca/PolyRootRe.html):\n\n 0\n0.6111860961336238   +   1.0593896464200445 i\n0.6111860961336238   -   1.0593896464200445 i\n-1.2223736979388697\n1102.0000015056714", "date": "2019-08-17 10:32:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1439912/real-roots-and-differentiation", "openwebmath_score": 0.878202497959137, "openwebmath_perplexity": 220.8351326150598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429138459387, "lm_q2_score": 0.872347368040789, "lm_q1q2_score": 0.8592995933007344}}
{"url": "https://math.stackexchange.com/questions/1694189/showing-lagrange-polynomials-form-a-basis-for-pi-3", "text": "# Showing Lagrange polynomials form a basis for $\\Pi_3$.\n\nHere's the question:\n\nLet $x_0=-2, x_1=0, x_2=1, x_3=4$ and let $L_j$ be the Lagrange polynomial for $j=0,1,2,3$. Show that $L_0, L_1, L_2, L_3$ form a basis for $\\Pi_3$.\n\nSo I've calculated $L_0,...,L_3$. I'll supply $L_3$ and $L_0$ just to see if I got them right. I am, on the other hand, completely lost on what I should be doing next.\n\n$L_0= x(x-1)(x-4)/(-36)$\n\n$L_3= x(x+2)(x-1)/(24)$\n\nFirst of all, as far as I can see (if I haven't miscalculated as well ^^), $L_3$ should be $$L_3 = \\frac{x(x+2)(x-1)}{(4-(-2)) \\cdot 4 \\cdot (4-1)} = \\frac{x(x+2)(x-1)}{72}$$\n\nNow to the question why they form a basis. The Lagrange polynomials are defined in such a way that you have $$L_j(x_i) = \\delta_{ij}, i,j=0,\\dots,3$$ where $\\delta_{ij}$ is the Kronecker delta. We know that $\\dim(\\Pi_3) = 4$ (because $\\{1,x,x^2,\\dots,x^3\\}$ is a basis). Thus it suffices to show that $L_0, \\dots, L_3$ are linearly independent. So let $\\lambda_0, \\dots, \\lambda_3$ so that $$\\phi := \\sum_{j=0}^{3}{\\lambda_j L_j} = 0$$\n\nWe now need to show that $\\lambda_j = 0\\ \\ \\ \\forall j \\in \\{0,\\dots,3\\}$. But this follows easily from $$0 = \\phi(x_i) = \\sum_{j=0}^{3}{\\lambda_j L_j}(x_i) = \\sum_{j=0}^{3}{\\lambda_{j}\\delta_{ij}} = \\lambda_i$$\n\n\u2022 Yes, only $L_3$ was incorrect. Thanks for the guidance sir! Mar 12 '16 at 13:20\n\u2022 @user314580 you're welcome. But bubba is right. Upvoting / accepting answer is a nice way of approving efforts. Mar 12 '16 at 13:42\n\u2022 Must I register to do so? Mar 12 '16 at 14:08\n\u2022 @user314580 No, simply click the \"up\" or \"down\" array over/beneath the score of a post. If the post is an answer to a question you asked, you can also accept one answer (note: you can upvote multiple answers, but \"accept\" only one). This can be done by clicking the checkmark below the score number. Mar 12 '16 at 15:18\n\nI assume that $\\Pi_3$ means the space of all polynomials of degree 3.\n\nThis space has dimension $4$, so, if we can show that $L_0$, $L_1$, $L_2$, $L_3$ span it, then we're done.\n\nSo, given any cubic polynomial $a_0 + a_1x + a_2x^2 + a_3x3$, we have to show that there exist numbers $\\lambda_0$, $\\lambda_1$, $\\lambda_2$, $\\lambda_3$ such that $$a_0 + a_1x + a_2x^2 + a_3x^3 = \\lambda_0 L_0(x) + \\lambda_1 L_1(x) + \\lambda_2 L_2(x) + \\lambda_3 L_3(x)$$ Equating coefficients of powers of $x$, you'll get a system of four linear equations for $\\lambda_0$, $\\lambda_1$, $\\lambda_2$, $\\lambda_3$. Use your favorite technique to show that this linear system has a solution (e.g. non-zero determinant).", "date": "2022-01-23 13:40:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1694189/showing-lagrange-polynomials-form-a-basis-for-pi-3", "openwebmath_score": 0.9466203451156616, "openwebmath_perplexity": 176.20250079135542, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9850429151632046, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8592995928153118}}
{"url": "https://stats.stackexchange.com/questions/587593/why-does-this-algorithm-generate-a-standard-normal-distribution/587600#587600", "text": "Why does this algorithm generate a standard normal distribution?\n\nI have this algorithm which I encountered:\n\n(1) Generate $$U_1$$, $$U_2$$ independently from Uniform(0,1)\n\n(2) Set $$Y_1 = -\\log{U_1}, Y_2 = -\\log{U_2}$$. If $$Y_2 > \\frac{(1-Y_1)^2}{2}$$, accept $$(Y_1, Y_2)$$. Else reject and return to step 1.\n\n(3) Generate $$U_3$$ from Uniform(0,1). If $$U_3 < 0.5$$, accept $$X=Y_1$$. Else, $$X=-Y_1$$\n\nWhy does X follow a standard normal distribution in the end? I know $$Y_1$$ and $$Y_2$$ here are exponential R.Vs. I probably need to understand how comparing $$Y_2$$ and $$\\frac{(1-Y_1)^2}{2}$$ here does the trick and the third step is probably the result of a distribution formed after step 2 which is a folded standard normal distribution.\n\nEdit: Source: Class Notes on Monte Carlo Methods\n\n\u2022 Step (2) draws a value of $Y_1$ conditional on $Y_2\\gt(1-Y_1)^2/2.$ Find the chance that this value is less than or equal to $z$ by integrating the joint density.\n\u2013\u00a0whuber\nSep 2, 2022 at 17:03\n\u2022 And the answer is columbia.edu/~ks20/4703-Sigman/4703-07-Notes-ARM.pdf (Note that this is not a great algorithm to generate Normals.) Sep 2, 2022 at 18:04\n\u2022 @Xi'an Right. Partially compensating for the need for almost 3 uniforms per variate is the relative simplicity of the calculations. But I understand now what you were referring to. What appeals to me about this approach is its potential to generate variates from analytically less-tractable distributions.\n\u2013\u00a0whuber\nSep 2, 2022 at 18:28\n\u2022 Please edit your question to credit the original source of all material written by others (e.g., your screenshot): stats.stackexchange.com/help/referencing\n\u2013\u00a0D.W.\nSep 3, 2022 at 19:18\n\u2022 Note that this doesn't really require 2.64 uniform samples - the last one only requires sampling one random bit, since you're only interested in the truth value of the expression (U3 < 0.5). Purely theoretically, you can take a single uniform sample X to generate U_1 = 2X (modulo 1) and U_3 = X, and you will have that the variables U_1 and (U_3 < 0.5) are independent. In practice, computationally speaking you are probably generating floating-point numbers in the range [0, 1) which involves generating 64 random bits and only using 53 of them, so the extra bit is free. Sep 4, 2022 at 13:58\n\nI was wondering how anyone would come up with this idea.\n\nYou observe, correctly, that the $$Y_i$$ have exponential distributions. They were easy to generate from a standard uniform number generator. The question could be put like this:\n\nFind a simple way to exploit your ability to generate $$(Y_1,Y_2)$$ to draw values $$Z$$ following any continuous distribution with positive support.\n\nSuch a distribution is one with a density function proportional to $$e^{g(z)}$$ for a function $$g$$ defined on the positive numbers.\n\nThe key terms \"simple\" and \"proportional to\" suggest trying a rejection sampling method. That leads to the algorithm in the question in the following generalized form:\n\nGenerate $$(Y_1,Y_2)$$ and keep $$Y_1$$ provided $$Y_2 \\gt f(Y_1)$$ for some function $$f$$ to be determined.\n\nAlthough it might feel more natural to reject when $$Y_2\\le f(Y_1),$$ as we will see this equivalent formulation leads to a simple calculation.\n\nThe result of this sampling procedure evidently produces values of $$Y_1$$ conditional on the event $$Y_2 \\gt f(Y_1).$$ To find its distribution function, apply the (elementary) definition of conditional probability to the event $$Y_1 \\le z$$ for an arbitrary positive number $$z.$$ It states\n\n$$\\Pr(Y_1\\le z \\mid Y_2 \\gt f(Y_1)) \\ \\propto\\ \\Pr(Y_1\\le z\\text{ and }Y_2 \\gt f(Y_1)).$$\n\nWe needn't be concerned about the constant of proportionality because we can work it out at the very end, knowing the result has to evaluate to $$1$$ as $$z\\to\\infty$$ (by the axiom of Total Probability).\n\nBecause $$(Y_1,Y_2)$$ is independent, their joint density is exponential. Thus, assuming $$f(z) \\ge 0$$ for all $$z\\gt 0,$$\n\n\\begin{aligned} \\int_0^z e^{g(y_1)}\\,\\mathrm{d}y_1 &= \\Pr(Y_1\\le z\\text{ and }Y_2 \\gt f(Y_1)) \\\\ &\\propto \\int_0^z e^{-y_1}\\int_{f(y_1)}^\\infty e^{-y_2}\\,\\mathrm{d}y_2\\mathrm{d}y_1\\\\ &= \\int_0^z e^{-y_1 - f(y_1)}\\,\\mathrm{d}y_1. \\end{aligned}\n\nEquality will hold for all $$z$$ provided the two integrands are equal. Solving for $$f$$ gives\n\n$$f(z) = -z - g(z) + C$$\n\nwhere the number $$C$$ accounts for the neglected proportionality constant $$e^C.$$\n\nConsider the Half Normal distribution where $$g(z) = -z^2/2.$$ We find\n\n$$f(z) = -z + z^2/2 + C = (1 - z)^2/2 + C - 1/2.$$\n\nWe will need $$C \\ge 1/2$$ to assure $$f(z)$$ is nonnegative. Larger values of $$C$$ work, too, but cause more rejections and are thereby less efficient.\n\nClearly, step (3) in the question converts any positive variable (like a Half Normal variable) into a variable symmetrically distributed around $$0.$$\n\nApplications\n\nFor this method to succeed, we need $$f$$ to attain a minimum that is not too negative. This implies the target distribution must not be too heavy-tailed. One example is the generalized Gamma distribution with density proportional to $$\\exp(-z^3/3)$$ on the positive numbers.\n\nHere are histograms based on a million draws of $$(Y_1,Y_2)$$ for the Half Normal and Generalized Gamma problems. The red curves plot the target densities to demonstrate the correctness of this algorithm. The (empirical) acceptance rates show how efficient it is.\n\nThis R code produced these plots.\n\nset.seed(17)\nn <- 1e6\nY <- matrix(-log(runif(2*n)), ncol = 2) # Step (1): obtain iid exponential variates\n#\n# The function f.  The constant can be any non-negative value, with 0 being the\n# most efficient.\n#\nDists <- list(Half Normal =  function(z, C = 0) (1 - z)^2/2 + C,\nGeneralized Gamma = function(z, C = 0) -z + z^3/3 + 2/3 + C)\npars <- par(mfrow = c(1, length(Dists)))\nfor(D in names(Dists)) {\nf <- Dists[[D]]\nz <- Y[Y[, 2] > f(Y[, 1]), 1]  # Step (2) of the rejection sampling\nrate <- length(z) / nrow(Y)\n\nhist(z, freq = FALSE, main = D,\nsub = bquote(paste(\"Acceptance rate is \", .(signif(rate, 2)))))\ng <- function(x) exp(-x - f(x))\n\nA <- integrate(g, 0, Inf)$value # The constant of integration curve(g(x) / A, add = TRUE, col = \"Red\", lwd = 2) } par(pars)  \u2022 (+1) Very nice reverse engineering. Historically von Neumann started by dominating the half-Normal density with a standard Exponential density and then achieved this condensed version. Sep 2, 2022 at 18:34 \u2022 @Xi'an I have realized -- and checked in code -- that one can overcome the limitation on tail behavior by means of a suitable transformation such as a logarithm or root. For instance, to generate a Gamma$(a)$variate use$f(z) = -(1+a)z + \\exp(z) + C$after finding$C$through numerical optimization. For smallish$a$the acceptance rate isn't bad; e.g. it's 66% for$a=0.3$and 39% for$a=3.$This makes the approach much more general than I had thought. \u2013 whuber Sep 2, 2022 at 18:47 \u2022 @whuber Thank you! Sep 3, 2022 at 6:54 \u2022 A more natural way to come up with it: since going from Cartesian to polar gives$$\\frac{1}{2\\pi}e^{-(y_1^2+y_2^2)/2}dy_1dy_2=re^{-r^2/2}dr\\frac{d\\theta}{2\\pi},$$it's now easy to get the desired IIDs viz.$u_1=e^{-y_1^2}$(we can drop the$1-$you'd get if you wanted an order-preserving transformation),$u_2=\\theta/(2\\pi)$. This speaks to a deeper truth: the only way for two Cartesian coordinates to be IIDs, with the polar coordinates also independent, is for the IIDs to be$N(0,\\,\\sigma^2)\\$.\n\u2013\u00a0J.G.\nSep 4, 2022 at 20:44\n\nTo draw a comparison between this Normal generator (that I will consider as von Neumann's) and the Box-M\u00fcller polar generator,\n\n#Box-M\u00fcller\nbm=function(N){\na=sqrt(-2*log(runif(N/2)))\nb=2*pi*runif(N/2)\nreturn(c(a*sin(b),a*cos(b)))\n}\n\n#vonNeumann\nvn=function(N){\nu=-log(runif(2.64*N))\nv=-2*log(runif(2.64*N))>(u-1)^2\nw=(runif(2.64*N)<.5)-2\nreturn((w*u)[v])\n}\n\n\nhere are the relative computing times\n\n> system.time(bm(1e8))\nutilisateur     syst\u00e8me      \u00e9coul\u00e9\n7.015       0.649       7.674\n> system.time(vn(1e8))\nutilisateur     syst\u00e8me      \u00e9coul\u00e9\n42.483       5.713      48.222\n\n\u2022 FWIW, vn does not return a Normal variate, nor does it return anywhere near N values. When it's fixed up, the relative timings are still about the same as you report. But when a better implementation of vn is used--you are welcome to borrow the one in my answer--the timing ratio is 3:1 (21:6.9). With a little optimization it drops to the expected 2.6:1, reflecting the RNG costs. vn <- function(N) { N <- ceiling(1.32 * N); y1 <- log(runif(N, 0, exp(1))); y2 <- -2 * log(runif(N)); y1 <- y1[y2 > y1^2]; sample(c(-1,1), length(y1), replace = TRUE) * (y1 - 1) }\n\u2013\u00a0whuber\nSep 3, 2022 at 16:37\n\u2022 @whuber: thank you, issues fixed after a busy weekend... Sep 5, 2022 at 14:17\n\u2022 Why is this vn func only returning the left-hand half of a normal dist? While one could replace the output with K <- c(K,-K) , that gives an artificial perfect mirror, not what's desired. Sep 15, 2022 at 11:51\n\u2022 Oh, -- you forgot the part \" U3<0.5, accept X=Y1. Else, X=\u2212Y1 \" ... at least I think so. Sep 15, 2022 at 12:16\n\u2022 @Xi'an then shouldn't it be w = ( (runif(2.64*N)<.5) -0.5) * 2 ? Sep 15, 2022 at 18:02", "date": "2023-01-27 17:22:10", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/587593/why-does-this-algorithm-generate-a-standard-normal-distribution/587600#587600", "openwebmath_score": 0.7136393189430237, "openwebmath_perplexity": 1188.9075388929607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429107723175, "lm_q2_score": 0.872347369700144, "lm_q1q2_score": 0.8592995922540049}}
{"url": "https://math.stackexchange.com/questions/3068112/what-is-the-expected-number-of-randomly-generated-numbers-in-the-range-a-b-re", "text": "# What is the expected number of randomly generated numbers in the range [a, b] required to reach a sum $\\geq X$?\n\nWe are generating random numbers (integers) in the range $$[a, b]$$. All values are equally likely.\n\nWe will continue to generate random numbers in this range, and add up successive values until their combined sum is greater than or equal to a set number $$X$$.\n\nWhat is the expected number of rolls to reach at least $$X$$?\n\nExample:\n\na = 1000\nb = 2000\nX = 5000\n\nValue 1: 1257 (total sum so far = 1257)\nValue 2: 1889 (total sum so far = 3146)\nValue 3: 1902 (total sum so far = 5048; all done)\n\n\nSo it took $$3$$ rolls to reach $$\\geq5000$$. Intuitively, we can say that it will not take more than $$5$$ rolls if each roll is $$1000$$. We can also say that it will not take less than $$3$$ rolls if each roll was $$2000$$. So it stands to reason that in the example above, the expected number of rolls lies somewhere between $$3$$ and $$5$$.\n\nHow would this be solved in the general case for arbitrary values $$[a, b]$$ and $$X$$? It's been quite a while since I last took statistics, so I've forgotten how to work with discrete random variables and expected value.\n\n\u2022 ${\\cal E}(n) = {2 X \\over a+b}$. \u2013\u00a0David G. Stork Jan 10 at 0:50\n\u2022 Actually, one must always round up, so ${\\cal E}(n) = \\lceil \\left({2 X \\over a+b} \\right) \\rceil$ \u2013\u00a0David G. Stork Jan 10 at 1:44\n\u2022 @DavidG.Stork: Even though the random variable $n$ is always integer valued, there's no reason why its expected value has to be... \u2013\u00a0Nate Eldredge Jan 10 at 2:18\n\u2022 @DavidG.Stork Right; I claim both are incorrect. \u2013\u00a0Aaron Montgomery Jan 10 at 4:50\n\u2022 @AaronMontgomery, they are indeed incorrect. \u2013\u00a0Eelvex Jan 10 at 6:24\n\nLet $$T$$ be the number of times it takes to reach $$X$$. We compute $$E[T]$$ via $$E[T]=\\sum_{t=0}^\\infty P(T>t)$$.\n\nIn order to have $$T>t$$, the sum of the first $$T$$ samples needs to be less than $$X$$. Let $$S_i$$ be the value of the $$i^{th}$$ sample. Then an experiment where $$T>t$$ has the first $$t$$ values satisfying $$S_1+S_2+\\dots+S_t Letting $$E=X-1-(S_1+\\dots+S_t)$$, we get $$S_1+\\dots+S_t+E=X-1,\\\\ a\\le S_i\\le b,\\\\ E\\ge 0$$ The number of integer solutions to the above system of equations and inequalities in the variables $$S_i$$ and $$E$$ can be computed via generating functions. The number of solutions is the coefficient of $$s^{X-1}$$ in $$(s^a+s^{a+1}+\\dots+s^b)^t(1-s)^{-1}=s^{at}\\cdot(1-s^{b-a+1})^t(1-s)^{-(t+1)}$$ After a bunch of work, this coefficient is equal to $$\\sum_{k=0}^{\\frac{X-1-ta}{b-a+1}} (-1)^k\\binom{t}k\\binom{t+X-1-ta-(b-a+1)k}{t}$$ Finally, we get \\begin{align} E[T] &=\\sum_{t=0}^{X/a} P(T>t) \\\\&=\\boxed{\\sum_{t=0}^{X/a} (b-a+1)^{-t}\\sum_{k=0}^{\\frac{X-1-ta}{b-a+1}}(-1)^k\\binom{t}k\\binom{t+X-1-ta-(b-a+1)k}{t}} \\end{align} Note that when $$a=0$$, the summation in $$t$$ will go from $$t=0$$ to $$\\infty$$, so it can only be computed approximately with a computer. However, this caveat can be avoided using the following observation; if $$T(a,b,X)$$ is the expected time to reach $$X$$ using sums of $$\\operatorname{Unif}\\{a,a+1,\\dots,b\\}$$ random variables, then $$E(T(0,b,X))=\\frac{b+1}bE(T(1,b,X))$$.\n\nWhen I write $$\\sum_{k=0}^a f(k)$$ and $$a$$ is not an integer, what I mean is $$\\sum_{k=0}^{\\lfloor a \\rfloor}f(k)$$.\n\nThe expected number is dominated by the term based on the mean $$\\frac{2X}{a+b}$$ - but we can do better than that. As $$X\\to\\infty$$, $$E(n) = \\frac{2X}{a+b}+\\frac{2(a^2+ab+b^2)}{3(a^2+2ab+b^2)}+o(1)$$ Why? Consider the excess. In the limit $$X\\to\\infty$$, the frequency of hitting a particular point approaches a uniform density of $$\\frac{2}{b+a}$$. For points in $$[X,X+a)$$, this is certain to be the first time we've exceeded $$X$$, so the density function for that first time exceeding $$X$$ is approximately $$\\frac{2}{b+a}$$ on that range. For $$t\\in[X+a,X+b)$$, the probability that it's the first time exceeding $$X$$ is $$\\frac{X+b-t}{b-a}$$; we need the last term added to be at least $$t-X$$. Multiply by $$\\frac{2}{b+a}$$, and the density function for the first time exceeding $$X$$ is approximately $$\\frac{2(X+b-t)}{b^2-a^2}$$ on $$[X+a,X+b)$$. For $$t or $$t\\ge X+b$$, of course, the density is zero. Graphically, this density is a rectangular piece followed by a triangle sloping to zero. Now, the expected value of the first sum to be greater than $$X$$ is approximately $$\\int_{X}^{X+a}\\frac{2t}{b+a}\\,dt+\\int_{X+a}^{X+b}\\frac{2t(X+b-t)}{b^2-a^2}\\,dt$$ $$=\\frac{2a(2X+a)}{2(b+a)}+\\frac{2(b-a)(2X+a+b)(X+b)}{2(b^2-a^2)}-\\frac{2(b-a)((X+a)^2+(X+a)(X+b)+(X+b)^2)}{3(b^2-a^2)}$$ $$=\\frac{6aX+3a^2+6X^2+3(a+3b)X+3ab+3b^2-6X^2-6(a+b)X-2(a^2+ab+b^2)}{3(b+a)}$$ $$=\\frac{3(a+b)X+a^2+ab+b^2}{3(b+a)}=X+\\frac{a^2+ab+b^2}{3(a+b)}$$ Divide by the mean $$\\frac{a+b}{2}$$ of each term in the sum, and we have an asymptotic expression for $$E(n)$$.\n\nThis is not a rigorous argument. If I wanted to do that, I'd bring out something like the moment-generating function.\n\nAll right, now that rigorous argument I alluded to. Credit to the late Kent Merryfield for laying out the argument cleanly, although I'm sure I contributed in the development stage. AoPS link\n\nSuppose we add independent copies of a reasonably nice nonnegative continuous random variable with density $$f$$, mean $$\\mu$$, and variance $$\\sigma^2$$ until we exceed some fixed $$y$$. As $$y\\to\\infty$$, the expected number $$g(y)$$ of these variables needed to exceed $$y$$ is $$g(y)=\\boxed{\\frac{y}{\\mu}+\\frac12+\\frac{\\sigma^2}{2\\mu^2}+o(1)}$$ In the case of the uniform $$[a,b]$$ distribution of this problem, that is $$g(y)=\\frac{2y}{a+b}+\\frac12+\\frac{(b-a)^2}{6(b+a)^2}+o(1)$$.\n\nTo prove this, condition on the first copy of the variable. If it has value $$x$$, the expected number of additional terms needed is $$g(y-x)$$. Add the one for the term we conditioned on, integrate against the density, and $$g(y)=1+\\int_0^y f(x)g(y-x)\\,dx$$ This is a convolution. To resolve it into something we can work with more easily, apply a transform method. Specifically, in this case with a convolution over $$[0,\\infty]$$, the Laplace transform is most appropriate. If $$F(s)=\\int_0^{\\infty}e^{-sx}f(x)\\,dx$$ and $$G(s)=\\int_0^{\\infty}e^{-sy}g(y)\\,dy$$, then, transforming that integral equation for $$g$$, we get $$G(s)=\\frac1s+F(s)G(s)$$ Solve for $$G$$, and $$G(s)=\\frac1{s(1-F(s))}$$ Now, note that $$F$$ is essentially the moment-generating function for the variable with density $$f$$; there's a sign change, but it's otherwise the same. As such, it has the power series expansion $$F(s)=1-\\mu s+\\frac{\\mu^2+\\sigma^2}{2\\mu^2}s^2 + O(s^3)$$ near zero. Then $$G(s)=\\frac1{\\mu s^2-\\frac{\\mu^2+\\sigma^2}{2}s^3+O(s^4)} = \\frac1{\\mu}s^{-2}+\\frac{\\mu^2+\\sigma^2}{2\\mu^2}s^{-1}+Q(s)$$ where $$Q(s)$$ has a removable singularity at zero. Also, from the \"reasonably nice\" condition, $$Q$$ decays at $$\\infty$$ How fast? Well, as long as the density of $$f$$ is bounded above by some exponential, its transform $$F$$ goes to zero as $$s\\to\\infty$$. Then $$G(s)=\\frac1{s(1-F(s)}=O(s^{-1})$$ as $$s\\to\\infty$$; subtracting off the two other terms, $$Q(s)=O(s^{-1})$$ as $$s\\to\\infty$$.\n\nNow, we take the inverse Laplace transform of $$G$$. The inverse transform of $$s^{-2}$$ is $$y$$ and the inverse transform of $$s^{-1}$$ is $$1$$, so there's the $$\\frac{y}{\\mu}+\\frac{\\sigma^2+\\mu^2}{2\\sigma^2}$$ terms of the expansion of $$g$$. For the remainder $$Q$$, we bring out the integral formula $$q(y)=\\frac1{2\\pi}\\int_{\\infty}^{\\infty}Q(it)e^{ity}\\,dt$$. Again for nice enough $$f$$ - a square-integrable density will do - this decays to zero as $$y\\to\\infty$$ (Riemann-Lebesgue lemma).\n\nAnd now, it's time to get more concrete. Our original example was a uniform distribution on $$[a,b]$$, which has mean $$\\mu=\\frac{a+b}{2}$$ and variance $$\\sigma^2=\\frac{(b-a)^2}{12}$$. That gives $$g(y)=\\boxed{\\frac{2y}{a+b}+\\frac12+\\frac{(b-a)^2}{6(a+b)^2}+o(1)}$$ Also, we can be more explicit; the transforms are $$F(s)=\\frac1{(b-a)s}(e^{-as}-e^{-bs})$$ and $$G(s)=\\frac{1}{s-\\frac1{b-a}e^{-as}+\\frac1{b-a}e^{-bs}}$$ This $$G$$ is analytic in the right half-plane except for the pole at zero, and is comparable to $$\\frac1s$$ at $$\\infty$$ in all directions in that half-plane. The remainder term $$Q$$ is thus smooth and decays like $$\\frac1s$$ along the imaginary axis, so its inverse transform $$q$$ will have a jump discontinuity (at zero, where it goes from zero to $$1$$) and decay rapidly.\n\nIf we want an explicit form for $$g$$ - it's a mess. Three different exponentials in that denominator pushes this far off the standard tables of transforms, and splitting it up as a geometric series introduces way too many terms. I'll try to write something down anyway: \\begin{align*}G(s) &= \\frac{1}{s-\\frac1{b-a}e^{-as}+\\frac1{b-a}e^{-bs}}\\\\ &= s^{-1} +\\frac{s^{-2}}{b-a}\\left(e^{-as}-e^{-bs}\\right)+\\frac{s^{-3}}{(b-a)^2}\\left(e^{-as}-e^{-bs}\\right)^2+\\cdots\\\\ &= \\sum_{n=0}^{\\infty} \\frac{s^{-n-1}}{(b-a)^n}\\left(e^{-as}-e^{-bs}\\right)^n\\\\ &= \\sum_{n=0}^{\\infty}\\sum_{k=0}^n\\frac{s^{-n-1}}{(b-a)^n}\\binom{n}{k}(-1)^k e^{-s(kb+(n-k)a)}\\end{align*} Now, in the table of Laplace transforms I'm using, there's an entry \"delayed $$n$$th power with frequency shift\"; the transform of $$(y-\\tau)^n e^{-\\alpha(y-\\tau)}u(y-\\tau)$$ (where $$u$$ is the unit step function) is $$n!e^{-\\tau s}(s+\\alpha)^{-n-1}$$. From this and the linearity of the transform, we get $$g(y) = \\sum_{n=0}^{\\infty}\\sum_{k=0}^n[y-(kb+(n-k)a)]^n\\frac{(-1)^k}{k!(n-k)!}u[t-(kb+(n-k)a)]$$ $$g(y) = \\sum_{j=0}^{\\infty}\\sum_{k=0}^{\\infty} (-1)^k\\frac{[y-(ja+kb)]^{j+k}}{j!k!}u[t-(kb+(n-k)a)]$$ For any given $$y$$, this is a finite sum; we only sum those terms for which $$ja+kb \\le y$$. It is horrible to use in practice, since it involves adding and subtracting large numbers to get something relatively small.\n\nThe $$a=0$$ case allows us to split things up differently, and comes out with something that's actually reasonably practical. Follow the link at the start of the post if you want to see it.\n\nGiven $$m$$ discrete variables in the range $$[a,b]$$, the number of ways to get a sum $$s$$ out of them corresponds to \\eqalign{ & N_b (s - ma,d,m) = \\cr & = {\\rm No}{\\rm .}\\,{\\rm of}\\,{\\rm solutions}\\,{\\rm to}\\;\\left\\{ \\matrix{ a \\le {\\rm integer}\\;y_{\\,j} \\le b \\hfill \\cr y_{\\,1} + y_{\\,2} + \\; \\cdots \\; + y_{\\,m} = s \\hfill \\cr} \\right.\\quad = \\cr & = {\\rm No}{\\rm .}\\,{\\rm of}\\,{\\rm solutions}\\,{\\rm to}\\;\\left\\{ \\matrix{ {\\rm 0} \\le {\\rm integer}\\;x_{\\,j} \\le b - a = d \\hfill \\cr x_{\\,1} + x_{\\,2} + \\; \\cdots \\; + x_{\\,m} = s - ma \\hfill \\cr} \\right. \\cr} where $$N_b$$ is given by $$\\bbox[lightyellow] { N_b (s-ma,d,m)\\quad \\left| {\\;0 \\leqslant \\text{integers }s,m,d} \\right.\\quad = \\sum\\limits_{\\left( {0\\, \\leqslant } \\right)\\,\\,k\\,\\,\\left( { \\leqslant \\,\\frac{s-ma}{d+1}\\, \\leqslant \\,m} \\right)} {\\left( { - 1} \\right)^k \\binom{m}{k} \\binom { s -ma+ m - 1 - k\\left( {d + 1} \\right) } { s-ma - k\\left( {d + 1} \\right)}\\ } } \\tag{1}$$ as widely explained in this post.\nLet's also notice the symmetry property of $$N_b$$ $$N_b (s - ma,d,m) = N_b (md - \\left( {s - ma} \\right),d,m) = N_b (mb - s,d,m)$$\n\nThe probability of obtaining exactly the sum $$s$$ in $$m$$ rolls is therefore \\bbox[lightyellow] { \\eqalign{ & p(s\\;;\\,m,a,b) = {{N_b (s - ma,d,m)} \\over {\\left( {d + 1} \\right)^{\\,m} }} = {{N_b (mb - s,d,m)} \\over {\\left( {d + 1} \\right)^{\\,m} }} = \\cr & = {1 \\over {\\left( {d + 1} \\right)^{\\,m} }} \\sum\\limits_{\\left( {0\\, \\le } \\right)\\,\\,k\\,\\,\\left( { \\le \\,{{s - ma} \\over {d + 1}}\\, \\le \\,m} \\right)} { \\left( { - 1} \\right)^k \\binom{m}{k} \\binom{ s -ma+ m - 1 - k\\left( {d + 1} \\right) }{ s-ma - k\\left( {d + 1} \\right)} } = \\cr & = {1 \\over {\\left( {d + 1} \\right)^{\\,m} }} \\sum\\limits_{\\left( {0\\, \\le } \\right)\\,\\,k\\,\\,\\left( { \\le \\,{{mb - s} \\over {d + 1}}\\, \\le \\,m} \\right)} { \\left( { - 1} \\right)^k \\binom{m}{k} \\binom{ mb-s+ m - 1 - k\\left( {d + 1} \\right) }{ mb-s - k\\left( {d + 1} \\right)} } \\cr} } \\tag{2} and the sum of $$p$$ over $$s$$ is in fact one.\n\nThe probability $$p$$ quickly converges, by Central Limit Theorem, to the Gaussian in the variable $$s$$ $$\\cal N\\left( {\\mu ,\\sigma ^{\\,2} } \\right)\\quad \\left| \\matrix{ \\;\\mu = m\\left( {{d \\over 2} + a} \\right) = m\\left( {{{a + b} \\over 2}} \\right) \\hfill \\cr \\;\\sigma ^{\\,2} = m{{\\left( {d + 1} \\right)^{\\,2} - 1} \\over {12}} \\hfill \\cr} \\right.$$ see for instance this post.\n\nIt is easy to demonstrate (by the \"double convoluton\" of binomials) that the Cumulative version, i.e. the probability to obtain a sum $$\\le S$$, is \\bbox[lightyellow] { \\eqalign{ & P(S\\;;\\,m,a,b) = {1 \\over {\\left( {d + 1} \\right)^{\\,m} }}\\sum\\limits_{0\\, \\le \\,\\,s\\,\\, \\le \\,S} {N_b (s - ma,d,m)} = \\cr & = {1 \\over {\\left( {d + 1} \\right)^{\\,m} }}\\sum\\limits_{\\left( {0\\, \\le } \\right)\\,\\,k\\,\\,\\left( { \\le \\,{{s - ma} \\over {d + 1}}\\, \\le \\,m} \\right)} { \\left( { - 1} \\right)^k \\binom{m}{k} \\binom{S - ma + m - k\\left( {d + 1} \\right) }{S - ma - k\\left( {d + 1} \\right) } } \\cr & \\approx {1 \\over 2}\\left( {1 + {\\rm erf}\\left( {\\sqrt 6 {{S + 1/2 - m\\left( {a + b} \\right)/2} \\over {\\left( {b - a + 1} \\right)\\sqrt m \\;}}} \\right)} \\right) \\cr} } \\tag{3}\n\nThe following plot allows to appreciate the good approximation provided by the asymptotics even at relatively small values of the parameters involved.\n\nThat premised,\nthe probability that we reach or exceed a predefined sum $$X$$ at the $$m$$-th roll and not earlier\nis given by the probability that\nwe get $$S as the sum of the first $$m-1$$ rolls, and then that the $$m$$-th one has a value $$s$$ such that $$X \\le S+s$$,\nor, what is demonstrably the same, by\nthe probability of getting $$X \\le S$$ in $$m$$ rolls minus the probability of getting the same in $$m-1$$ rolls.\n\nThat is, in conclusion \\bbox[lightyellow] { \\eqalign{ & Q(m\\;;\\,X,a,b)\\quad \\left| \\matrix{ \\;1 \\le m \\hfill \\cr \\;0 \\le X \\hfill \\cr \\;0 \\le a \\le b \\hfill \\cr} \\right.\\quad = \\cr & = 1 - P(X - 1\\;;\\,m,a,b) - 1 + P(X - 1\\;;\\,m - 1,a,b) = \\cr & = P(X - 1\\;;\\,m - 1,a,b) - P(X - 1\\;;\\,m,a,b) \\cr} } \\tag{4} and the sum of $$Q$$ over $$m$$ correctly checks to one.\n\nExample\n\na) For small values of $$X,a,b$$ the exact expression for $$Q$$ in (4) gives the results shown (e.g. $$a=1, \\; b=4$$) which look to be correct and the row sums correctly check to be $$1$$.\n\nb) For large values of the parameters, like in your example $$a=1000,\\; b=2000, \\; X=5000$$ we shall use the asymptotic version of $$Q$$ which gives the results plotted below", "date": "2019-07-22 01:06:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3068112/what-is-the-expected-number-of-randomly-generated-numbers-in-the-range-a-b-re", "openwebmath_score": 0.9936489462852478, "openwebmath_perplexity": 430.65541392155194, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429164804707, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8592995906953532}}
{"url": "http://mathhelpforum.com/statistics/183602-9-digit-combination-lock-chances-probability-print.html", "text": "# 9-Digit Combination Lock Chances And Probability\n\n\u2022 June 25th 2011, 11:46 AM\nFatalSylence\n9-Digit Combination Lock Chances And Probability\nI have been thinking on combination locks recently, and would like some help with this particular problem.\n\nProblem:\nI have a lock with digits 1-9, and the length of the combination is 4 digits. You cannot press the digits 1-9 more than once, and the combination does not have to be typed in any particular order.\n\nQuestions:\n\n1. Is it true, then, that the possible amount of combinations is 3,024, because 9*8*7*6=3,024?\n2. Is it also true that 24 of the combinations are correct, since you do not have to type the combination in any particular order? Therefore 1*2*3*4=24.\n3. Is it true that the chances of guessing it right are 1 in 125 since 3,000/24=125?\n\n1. If you would change the lock to have a 5 digit combination, would the chance to guess it right be the same as if it were a 4 digit combination? Since the possible combination amount is 15,120 (9*8*7*6*5), the amount of right combinations is 120 (1*2*3*4*5), therefore 15,000/120=125. The same amount of chance as if it were a 4-digit combination?\n2. If all this is true, then wouldn't a 6-digit combination actually be less secure? I.e., you have a greater chance of guessing the right combination? (1 in 84, if I am doing it right.)\n\nHelp is greatly appreciated. I have not done probability in a long time and am wondering if I am even doing it right.\n\nThanks, guys.\nFatal Sylence\n\u2022 June 25th 2011, 12:01 PM\nPlato\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by FatalSylence\n[/B]I have a lock with digits 1-9, and the length of the combination is 4 digits. You cannot press the digits 1-9 more than once, and the combination does not have to be typed in any particular order.\n\nQuestions:\n\n1. Is it true, then, that the possible amount of combinations is 3,024, because 9*8*7*6=3,024?\n2. Is it also true that 24 of the combinations are correct, since you do not have to type the combination in any particular order? Therefore 1*2*3*4=24.\n3. Is it true that the chances of guessing it right are 1 in 125 since 3,000/24=125?\n\nIf order makes a difference then the answer is $9\\cdot 8 \\cdot 7\\cdot 6=3024$.\n\nIf order makes no difference then $\\binom{9}{4}=\\frac{9!}{4!\\cdot 5!}=126$\n\u2022 June 25th 2011, 12:07 PM\nFatalSylence\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by Plato\nIf order makes a difference then the answer is $9\\cdot 8 \\cdot 7\\cdot 6=3024$.\n\nIf order makes no difference then $\\binom{9}{4}=\\frac{9!}{4!\\cdot 5!}=126$\n\nI'm not sure I understand your answer. 126 as in a 1 in 126 chance of guessing it right?\n\u2022 June 25th 2011, 12:12 PM\nPlato\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by FatalSylence\nI'm not sure I understand your answer. 126 as in a 1 in 126 chance of guessing it right?\n\nThat is the combination of 9 things taken 4 at a time.\nBy the way that is $\\frac{3024}{24}$.\n\u2022 June 25th 2011, 12:24 PM\nFatalSylence\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by Plato\nThat is the combination of 9 things taken 4 at a time.\nBy the way that is $\\frac{3024}{24}$.\n\nHmm, okay. Can you direct me towards a resource that explains the \"n things taken r at a time\"?\n\nAlso, were my other questions correct?\n\nFinally, would a 5 digit combination be more secure? A 4 digit combination has the same chance to guess right as a a 5-digit combination, but don't you have a greater chance to guess it wrongwith 5-digits?\n\u2022 June 25th 2011, 12:28 PM\nPlato\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by FatalSylence\nHmm, okay. Can you direct me towards a resource that explains the \"n things taken r at a time\"?\n\nAlso, were my other questions correct?\n\nFinally, would a 5 digit combination be more secure? A 4 digit combination has the same chance to guess right as a a 5-digit combination, but don't you have a greater chance to guess it wrongwith 5-digits?\n\nAlso look at this page. You change those numbers and hit $\\boxed{=}$.\n\u2022 June 25th 2011, 12:37 PM\nFatalSylence\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by Plato\n\nAlso look at this page. You change those numbers and hit $\\boxed{=}$.\n\nCan you explain to me if in a 4-digit combination, the amount of correct combinations is 126, or 24 like I originally thought? I'm just trying to understand this here.\n\u2022 June 25th 2011, 12:44 PM\nPlato\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by FatalSylence\nCan you explain to me if in a 4-digit combination, the amount of correct combinations are 126, or 24 like I originally thought? I'm just trying to understand this here.\n\nI think that we are talking about two different ideas here.\n\nThere are 24 ways to arrange the string $9,4,2,6$.\n\nBut there are 126 ways to select a set of four from $\\{1,2,3,4,5,6,7,8,9\\}$.\n\nOne of those 126 sets is $\\{2,4,6,9\\}$.\nThen the numbers in that set can be arranged in 24 ways.\n\u2022 June 25th 2011, 12:56 PM\nFatalSylence\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by Plato\nI think that we are talking about two different ideas here.\n\nThere are 24 ways to arrange the string $9,4,2,6$.\n\nBut there are 126 ways to select a set of four from $\\{1,2,3,4,5,6,7,8,9\\}$.\n\nOne of those 126 sets is $\\{2,4,6,9\\}$.\nThen the numbers in that set can be arranged in 24 ways.\n\nOH!!! Gosh! I get it! It does not matter how many ways you can arrange a set of four since order does not matter. I understand. One of these 126 sets encompasses all the ways you can arrange it since order does not matter.\n\nSo, if the person was good at math, they would realize that since order does not matter, there are only 126 possible combinations. Wow, I get it. Thank you.\n\nEDIT:\nIs there a formula or calculator that can display all the possible sets for a 1-9, 4 digit combination? Or any other numbers for that matter?\n\u2022 June 25th 2011, 03:39 PM\nPlato\nRe: 9-Digit Combination Lock Chances And Probability\nQuote:\n\nOriginally Posted by FatalSylence\nEDIT:[/B] Is there a formula or calculator that can display all the possible sets for a 1-9, 4 digit combination? Or any other numbers for that matter?\n\nThe good news is that this is a simple programming problem.\nBut the bad news I doubt that you can easily find it on the web.\nI suggest that you discuss this with a friend who does programming.", "date": "2014-10-26 02:02:59", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/183602-9-digit-combination-lock-chances-probability-print.html", "openwebmath_score": 0.5551171898841858, "openwebmath_perplexity": 633.8700922861218, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668723123672, "lm_q2_score": 0.8757869981319863, "lm_q1q2_score": 0.859293189768998}}
{"url": "https://math.stackexchange.com/questions/2562393/a-compact-locally-connected-metric-space-is-uniformly-locally-connected", "text": "# A compact locally connected metric space is \u201cuniformly locally connected\u201d\n\nA compact locally connected metric space is \"uniformly locally connected\"\\ That is, for any $\\epsilon > 0$, there is some $\\delta > 0$ such that whenever $\\rho(x, y) < \\delta$, then $x$ and $y$ both lie in some connected subset of $X$ of diameter $<\\epsilon$.\n\nproof:-\n\nSince $X$ is locally connected metric space then each $x\\in X$ has a nhood base of open connected sets\\ Given $\\epsilon > 0$, let $x \\in X$ and $U_x=\\rho(x, \\epsilon)$ be a nhood of $x$\\ There exist an open connected basic nhood $V_x$ with diameter $<\\epsilon$, Now $$X=\\bigcup_{x\\in X }{V_x}$$, hence cover $X$ by open connected nhoods of diameter $<\\epsilon$.\\ Since $X$ is compact, reduce this to a finite subcover $\\{V_{x1},. . . , V_{xn}\\}$ and let $\\delta$ be a Lebesgue number (22.5) for this cover.\\ Then if $\\rho(x, y) < \\delta$, both $x$ and $\u0443$ belong to some $V_{xi}$. \\\n\n{Theorem 22.5} (Lebesgue covering lemma). If $\\{U_1..., U_n\\}$ is a finite open cover of a compact metric space X, there is some $\\delta > 0$ such that if A is any subset of $X$ of diameter $< \\delta$, then $A \\subset U_i$ for some i.\n\nI try to write the proof better than this.\n\nI would like to confirm this proof\n\nIf acceptable, I would like to clarify and improve it (Language and Mathematical)as much as possible\n\n\u2022 Do not use \\ or , or .\\ at the end of a sentence. Just use . $\\quad$ And if a sentence ends in a displayed line, include the . in that line. Write in the same style as for an an essay about something else. What you have is not so bad. I've seen far worse. And your proof is correct . \u2013\u00a0DanielWainfleet Dec 12 '17 at 9:25\n\nThe proof is correct.\n\nAs for writing it up. That depends on your audience. For a general audience, the first rule is - avoid abbreviations (\"neighborhood\", not \"nhood\"). And I assume that the \"/\" scattered through the text was some sort of misguided attempt to show where you put new lines? Just let Latex have it's way here. If a new line is needed, let it form a paragraph break.\n\nAlso \"and $U_x=\\rho(x,\\epsilon)$ be a nhood of $x$\" doesn't make sense. $\\rho$ is a function that takes points in your space in both arguments, and produces real numbers. You have it with a real number as the second argument, and producing a set. Now it is evident that you mean $U_x$ to be a ball of radius $\\epsilon$ about $x$. But if someone has told you this is a good way to denote the ball, do NOT trust them about notations in the future! Commonly used ways of denoting balls include $B(x, \\epsilon)$ and $B_x(\\epsilon)$. If the metric to be used is not obvious, then usually $B_\\rho(x, \\epsilon)$ is preferred. (Of course, it you had intended $B(x, \\epsilon)$ and only accidently used $\\rho$, just be careful in your final write-up.)\n\nBut you don't do anything with \"$U_x$\", which makes introducing a notation for it pointless. You don't actually need to name the balls, since you don't plan on using that name.\n\nThere are also some phrasing differences I would suggest. Your phrasing is clear enough, but sounds awkward sometimes to a native English speaker (at least - to this native English speaker). There is nothing particularly bad, though.\n\nI would write it up like this:\n\nSince $X$ is locally connected, for $\\epsilon > 0$, each point $x \\in X$ has a open connected neighborhood $V_x$ contained within the ball of radius $\\epsilon$ about $x$. The collection $\\{V_x \\mid x \\in X\\}$ forms an open cover of $X$ by sets of diameter $< \\epsilon$. Since $X$ is compact, it has a finite subcover $\\{V_1, V_2, ..., V_m\\}$. By the Lesbegue covering lemma (Theorem 22.5), there is a $\\delta > 0$ such that if $\\rho(x, y) < \\delta$, then there is some $V_i$ with $\\{x, y\\} \\subset V_i$, which completes the proof.", "date": "2019-09-24 09:19:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2562393/a-compact-locally-connected-metric-space-is-uniformly-locally-connected", "openwebmath_score": 0.9137232303619385, "openwebmath_perplexity": 196.6786434395461, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668734137682, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.859293185962147}}
{"url": "http://mymathforum.com/algebra/42904-progression.html", "text": "My Math Forum Progression .\n\n Algebra Pre-Algebra and Basic Algebra Math Forum\n\n April 15th, 2014, 02:49 AM #1 Senior Member \u00a0 Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Progression . For the same job specifications, two companies offer a different salary scale: Company A :\u200bStarting monthly pay = RM900.00 \u200bMonthly increment = RM50.00 \u200b Company B:\u200bStarting monthly pay = RM750.00 \u200bMonthly increment = RM60.00 (a)\u200bIn January 2014, Ali starts to work in Company A and Ahmad in Company B. \u200bWhen will the monthly salary of Ali and Ahmad be the same? \u200bUse three methods. Include the use of ICT. (b)\u200bWhich salary scale is the better deal? Justify. How to solve this problem ? Especially problem b.\nApril 15th, 2014, 10:20 PM \u00a0 #2\nMath Team\n\nJoined: Dec 2006\nFrom: Lexington, MA\n\nPosts: 3,267\nThanks: 408\n\nHello, jiasyuen!\n\nQuote:\n For the same job, two companies offer a different salary scale: Company A: \u200bStarting monthly pay = \\$900.00 \u200bMonthly increment = \\$50.00 \u200b Company B:\u200b Starting monthly pay = \\$750.00 \u200bMonthly increment = \\$60.00 (a) \u200bIn January 2014, Ali starts to work in Company A and Ahmad in Company B. \u200bWhen will the monthly salary of Ali and Ahmad be the same? \u200bUse three methods. Include the use of ICT.\n\nYou need to construct algebraic expressions for Ali's and Ahmad's monthly pay.\n\nAli starts at \\$900 per month and gets a \\$50 raise each month.\nWhen he has worked for $\\displaystyle m$ months,\n$\\displaystyle \\quad$his monthly pay will be: $\\displaystyle \\:900 + 50m$ dollars.\n\nAhmad starts at \\$750 per month and gets a \\$60 raise each month.\nWhen he has worked for $\\displaystyle m$ months,\n$\\displaystyle \\quad$his monthly pay will be $\\displaystyle \\,750 +60m$ dollars.\n\nWhen are these equal?\n$\\displaystyle \\quad 750 + 60m \\:=\\:900 + 50m \\quad\\Rightarrow\\quad 10m \\:=\\:150 \\quad\\Rightarrow\\quad m \\,=\\,15$\n\nTheir salaries will be equal in 15 months (March 2015).\n\nQuote:\n (b)\u200bWhich salary scale is the better deal? Justify.\n\nIf Ali considers this a temporary job (less than 15 months),\n$\\displaystyle \\quad$ he has the better deal.\n\nIf both are planning to work more than 15 months,\n$\\displaystyle \\quad$ then Ahmad has the better deal.$\\displaystyle \\;\\;{\\color{red}{**}}$\n\n$\\displaystyle \\begin{array}{c|ccccccccc} \\hline \\text{Month} & 0 & 1 & 2 & 3 & \\cdots & 15 & 16 & 17 & \\cdots \\\\ \\hline \\text{Ali} & 900 & 950 & 1000 & 1050 & \\cdots & 1650 & 1700 & 1750 & \\cdots \\\\ \\hline \\text{Ahmad} & 750 & 810 & 870 & 930 & \\cdots & 1650 & 1710 & 1770 & \\cdots \\\\ \\hline \\end{array}$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\n$\\displaystyle {\\color{red}{**}}$\nI gave this problem some extra thought . . .\n\nIn 15 months their monthly pay is the same.\nBut Ali has made \\$18,750 while Ahmad has made only \\$17,550.\n\nIt will take 31 months for Ahmad's total earnings to equal that of Ali.\nAfter that, Ahmad's total earnings will exceed Ali's.\n\n April 16th, 2014, 03:43 AM #3 Senior Member \u00a0 Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 The question stated there have to use three methods to find. Including of ICT. What it means ? How to do ?\nApril 16th, 2014, 09:08 AM \u00a0 #4\nMath Team\n\nJoined: Oct 2011\n\nPosts: 14,597\nThanks: 1038\n\nQuote:\n Originally Posted by jiasyuen The question stated there have to use three methods to find. Including of ICT. What it means ? How to do ?\nWell, Soroban gave you one; seems fair that YOU find two more;\nit is YOUR homework, not ours. Do you agree?\n\n April 16th, 2014, 09:12 AM #5 Senior Member \u00a0 Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 I don't know what's another two methods ? Can you give me some guidance ?\nApril 23rd, 2014, 04:36 AM \u00a0 #6\nSenior Member\n\nJoined: Sep 2013\nFrom: Earth\n\nPosts: 827\nThanks: 36\n\nQuote:\n Originally Posted by soroban Hello, jiasyuen! You need to construct algebraic expressions for Ali's and Ahmad's monthly pay. Ali starts at \\$900 per month and gets a \\$50 raise each month. When he has worked for $\\displaystyle m$ months, $\\displaystyle \\quad$his monthly pay will be: $\\displaystyle \\:900 + 50m$ dollars. Ahmad starts at \\$750 per month and gets a \\$60 raise each month. When he has worked for $\\displaystyle m$ months, $\\displaystyle \\quad$his monthly pay will be $\\displaystyle \\,750 +60m$ dollars. When are these equal? $\\displaystyle \\quad 750 + 60m \\:=\\:900 + 50m \\quad\\Rightarrow\\quad 10m \\:=\\:150 \\quad\\Rightarrow\\quad m \\,=\\,15$ Their salaries will be equal in 15 months (March 2015). If Ali considers this a temporary job (less than 15 months), $\\displaystyle \\quad$ he has the better deal. If both are planning to work more than 15 months, $\\displaystyle \\quad$ then Ahmad has the better deal.$\\displaystyle \\;\\;{\\color{red}{**}}$ $\\displaystyle \\begin{array}{c|ccccccccc} \\hline \\text{Month} & 0 & 1 & 2 & 3 & \\cdots & 15 & 16 & 17 & \\cdots \\\\ \\hline \\text{Ali} & 900 & 950 & 1000 & 1050 & \\cdots & 1650 & 1700 & 1750 & \\cdots \\\\ \\hline \\text{Ahmad} & 750 & 810 & 870 & 930 & \\cdots & 1650 & 1710 & 1770 & \\cdots \\\\ \\hline \\end{array}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ $\\displaystyle {\\color{red}{**}}$ I gave this problem some extra thought . . . In 15 months their monthly pay is the same. But Ali has made \\$18,750 while Ahmad has made only \\$17,550. It will take 31 months for Ahmad's total earnings to equal that of Ali. After that, Ahmad's total earnings will exceed Ali's.\n\nNot 16 month?\n\nAli in Company A:\nTn =a+(n-1)d\n=900+(n-1)(50)\n=900+50n-50\n=850+50n\n\nTn = a+(n-1)d\n=750+(n-1)(60)\n=750+60n-60\n=690+60n\n\nWhen the monthly salary of Ali and Ahmad the same,\n850+50n=690+60n\n160n=10n\nn=16 (i.e. in the 16th month)\n\n Tags progression\n\n,\n\n### in january 2014 ali starts to work in company A and ahmad company b when will the monthly salary of ali and ahmad be the same\n\nClick on a term to search for related topics.\n Thread Tools Display Modes Linear Mode\n\n Similar Threads Thread Thread Starter Forum Replies Last Post anigeo Algebra 2 January 18th, 2012 11:41 PM Faux Algebra 9 January 5th, 2012 07:48 AM panky Algebra 1 December 8th, 2011 09:32 AM mikeportnoy Algebra 3 October 14th, 2008 01:42 AM mikeportnoy Real Analysis 0 December 31st, 1969 04:00 PM\n\n Contact - Home - Forums - Cryptocurrency Forum - Top", "date": "2019-08-18 05:46:12", "meta": {"domain": "mymathforum.com", "url": "http://mymathforum.com/algebra/42904-progression.html", "openwebmath_score": 0.24042107164859772, "openwebmath_perplexity": 4700.686770146119, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9830850882200038, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.8592923039050113}}
{"url": "http://mathhelpforum.com/calculus/10998-solved-integration-sinxcosx.html", "text": "# Math Help - [SOLVED] integration with sinxcosx\n\n1. ## [SOLVED] integration with sinxcosx\n\nHey guys i kind of have a weird little problem which i can't figure out.\n\nI'm integrating sinxcosx, and depending on what i use as my substution you get two answers.\n\n1) sin^2x/2 + C and 2) -cos^2x/2 + C\n\nNow if we set these two answers equal to eachother we get\n\nsin^2x + cos^2x = 0 but we know there exists a trig identity that states sin^2x + cos^2x = 1\n\nSo i'm confused here. How can this be possible? I know that 0 cannot equal 1. Or at least i hope not hehe. Any help would be great cheers\n\n2. Originally Posted by block\nHey guys i kind of have a weird little problem which i can't figure out.\n\nI'm integrating sinxcosx, and depending on what i use as my substution you get two answers.\n\n1) sin^2x/2 + C and 2) -cos^2x/2 + C\n\nNow if we set these two answers equal to eachother we get\n\nsin^2x + cos^2x = 0 but we know there exists a trig identity that states sin^2x + cos^2x = 1\n\nSo i'm confused here. How can this be possible? I know that 0 cannot equal 1. Or at least i hope not hehe. Any help would be great cheers\nTwo solutions of an integral may be of different form if they are off by a constant. You have two solutions, but who says the constant of integration is the same? What this is telling you is that $C_{cos} = C_{sin} + 1$\n\n-Dan\n\n3. Originally Posted by block\nHey guys i kind of have a weird little problem which i can't figure out.\n\nI'm integrating sinxcosx, and depending on what i use as my substution you get two answers.\nYou have,\n$\\int \\sin x \\cos x dx$\nAnd I think the problem you have you think you get two distinct answers when you use $u=\\sin x$ and when you switch to $u=\\cos x$.\n\n1) $u=\\sin x$ then $u'=\\cos x$\nThus,\n$\\int \\sin x \\cos x dx= \\int u du=\\frac{1}{2}u^2 +C=\\frac{1}{2}\\sin^2 x+C$\n\n2) $u=\\cos x$ then $u'=-\\sin x$\nThus,\n$\\int \\sin x \\cos x dx= -\\int u du=\\frac{1}{2}u^2 +C=-\\frac{1}{2}\\cos^2 x+C$\n\nBut in reality these are the same (the set of all anti-derivates) because,\n$\\frac{1}{2}\\sin^2 x+C=\\frac{1}{2}(1-\\cos^2 x)+C=-\\frac{1}{2}\\cos^2 x+\\left( \\frac{1}{2}+C\\right)=-\\frac{1}{2}\\cos^2 x+C_1$\nI just renamed the constant function,\n$\\frac{1}{2} +C=C_1$\nThus, you get the same thing.\n\nAnd the last thing you could have done is,\n$\\frac{1}{2}\\int 2\\sin x \\cos x dx$\nI just multiplied and divide by 2, unchanged expression.\nFrom trignometry you know that,\n$\\frac{1}{2}\\int \\sin 2x dx=-\\frac{1}{4}\\cos 2x+C$\nBut again though it looks different the space of solutions is still the same. You can use some trignometry to confirm this.\n\n4. Hello, block!\n\nThis is a classic puzzler.\nI ran across it across it in Calculus I.\nThere are three ways (at least) to integrate it.\n\nWe have: . $\\int\\sin x\\cos x\\,dx$\n\n[1] If we look at it as: . $\\int \\underbrace{\\sin x}_{u}\\underbrace{(\\cos x\\,dx)}_{du}$\n. . . we get: . $\\boxed{\\frac{1}{2}\\sin^2\\!x + C_1}$\n\n[2] If we write it as: . $\\int \\cos x(\\sin x\\,dx)$\n. . . we let: $u = \\cos x\\quad\\Rightarrow\\quad du = -\\sin x\\,dx\\quad\\Rightarrow\\quad dx = -\\frac{du}{\\sin x}$\nSubstitute: . $-\\int u\\,du \\;=\\;-\\frac{1}{2}u^2 + C\\;=\\;\\boxed{-\\frac{1}{2}\\cos^2\\!x + C_2}$\n\nSince these two answers must be equal, we have:\n. . $\\frac{1}{2}\\sin^2\\!x + C_1 \\;=\\;-\\frac{1}{2}\\cos^2\\!x + C_2\\quad\\Rightarrow\\quad \\frac{1}{2}\\sin^2\\!x + \\frac{1}{2}\\cos^2\\!x\\;=\\;C_2-C_1$\n\nMultiply by 2: . $\\sin^2\\!x + \\cos^2\\!x \\;=\\;2(C_2-C_1)$\n\nThe time the constants $(C_1,\\:C_2)$ aren't completely arbitrary.\n. . They must satisfy: . $2(C_2-C_1) \\:=\\:1$\n\n[3] We can use the identity: . $2\\sin\\theta\\cos\\theta \\,=\\,\\sin2\\theta$\n\nSo: . $\\int\\sin x\\cos x\\,dx\\;=\\;\\frac{1}{2}\\int2\\sin x\\cos x\\,dx \\;=\\;\\frac{1}{2}\\int\\sin2x\\,dx$ $=\\;\\boxed{-\\frac{1}{4}\\cos2x + C_3}$\n\n. . I'll let you justify this one . . .\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nA similar problem: . $\\int\\sec^2\\!x\\tan x\\,dx$\n\n$[1]\\;\\;\\int \\underbrace{\\sec x}_{u}\\underbrace{(\\sec x\\tan x\\,dx)}_{du} \\;=\\;\\frac{1}{2}\\sec^2\\!2x + C$\n\n$[2]\\;\\;\\int\\underbrace{\\tan x}_{u}\\underbrace{(\\sec^2x\\,dx)}_{du}\\;=\\;\\frac{1} {2}\\tan^2x + C$", "date": "2014-12-22 14:32:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/10998-solved-integration-sinxcosx.html", "openwebmath_score": 0.9292730689048767, "openwebmath_perplexity": 621.2619638330151, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850842553892, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.859292303664376}}
{"url": "http://openstudy.com/updates/53f5818fe4b0f0e909e02de2", "text": "## anonymous 2 years ago how would I solve for x if 2x-3/x+1<1\n\n1. kc_kennylau\n\n$\\frac{2x-3}{x+1}<1$\n\n2. kc_kennylau\n\nI would split it into two cases\n\n3. anonymous\n\nhow would you do that\n\n4. myininaya\n\nI would first write as f(x)<0 Then find where f(x) is zero and undefined Then draw a number line with the numbers that make f zero or undefined Then test the intervals around those numbers to see if f is less than 0 or not\n\n5. kc_kennylau\n\nSince we are to multiply (x-1) on both sides, I would split it into x-1<0 and x-1>0\n\n6. myininaya\n\nyou mean x+1?\n\n7. kc_kennylau\n\nYes sorry\n\n8. anonymous\n\nSO i take x+ 1 and get $2x-3<x+1 and 2x-3>x+1$\n\n9. kc_kennylau\n\nYes\n\n10. kc_kennylau\n\nWell actually I believe we should do it case by case\n\n11. kc_kennylau\n\nWhen x+1<0: 2x-3>x+1 ... When x+1>0: 2x-3<x+1 ...\n\n12. anonymous\n\nso then my next step would to add in the 3 on both sides corect? so it would turn into $2x>x+4$ and $2x<x+4$\n\n13. kc_kennylau\n\nYep\n\n14. kc_kennylau\n\nAnd for presentation purpose I would write it as 2x>x+4 where x+1<0 and 2x<x+4 where x+1>0\n\n15. anonymous\n\nthen do I divide by 2 on both of the equations or divide by 2x?\n\n16. kc_kennylau\n\nYou subtract x from both sides\n\n17. anonymous\n\nso It would be $x-2x>4 where x+1<0$ and just the oposit sign on the otherone?\n\n18. kc_kennylau\n\nNo, it should be 2x-x>4 where x+1<0\n\n19. anonymous\n\nok now im starting to understand so then the next step would be to hmove the 4 over\n\n20. kc_kennylau\n\nNah, the next step is subtracting it\n\n21. anonymous\n\nok so it would then be 2x-x-4> 0 and where x+1<0 and 2x-x-4< 0 and where x+1>0 and t\n\n22. kc_kennylau\n\nI mean you can write 2x-x as x\n\n23. kc_kennylau\n\nWhen x+1<0: 2x-3>x+1 2x>x+4 2x-x>4 x>4 When x+1>0: 2x-3<x+1 2x<x+4 2x-x<4 x<4\n\n24. kc_kennylau\n\nHowever, when x+1<0, x can't be >4, and the same goes to the second case\n\n25. kc_kennylau\n\nTherefore there is no solution\n\n26. kc_kennylau\n\nWait\n\n27. aum\n\nThe inequality is valid for x in the interval (-1, 4).\n\n28. kc_kennylau\n\n@aum tell me what is wrong in my solution, I have no time to double-check\n\n29. kc_kennylau\n\nbye\n\n30. anonymous\n\n@aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?\n\n31. aum\n\nMy method is what @myininaya suggested. (2x - 3) / (x + 1) < 1 (2x - 3) / (x + 1) - 1 < 0 (2x-3 - x - 1) / (x + 1) < 0 (x - 4) / (x + 1) < 0 The x values that are of interest to us here are the ones that will make f(x) = (x-4)/(x+1) zero or undefined. f(x) = 0 when x = 4 f(x) is undefined when x = -1. So the number line is split into three intervals: (-infinity, -1); (-1, 4), (4, infinity). Pick a convenient number in each interval and see if the inequality is valid. f(x) = (x - 4) / (x + 1) < 0 when x = -2, f(x) = (-6) / (-1) = 6 which is greater than 0. So not a solution. when x = 0, f(x) = -4/1 = -4 which is less than 0. This is a solution. when x = 5, f(x) = 1 / 6 which is greater than 0. Not a solution. So the solution is x in the interval (-1, 4).\n\n32. anonymous\n\nthank you so much\n\n33. aum\n\nyou are welcome.\n\n34. aum\n\nTo follow @kc_kennylau 's method: (2x - 3) / (x + 1) < 1 Case 1: x + 1 > 0 Since (x+1) is positive, we can multiply both sides by (x+1) and the \"less than\" inequality will remain intact. 2x - 3 < x + 1 x < 4 Case 2: x + 1 < 0 Since (x+1) is negative, multiplying both sides by (x+1) would change the \"less than\" inequality to a \"greater than\" inequality. 2x - 3 > x + 1 x > 4. But for the second case we assumed x + 1 < 0 which implies x < -1 But we get the solution as x > 4. x cannot be less than -1 and also be greater than 4 simultaneously. Thus, case 2 yields NO solution. Therefore, there is no solution in the domain x < -1. Combining the two cases we get x should be less than 4 but greater than -1.\n\n35. kc_kennylau\n\nI incorrectly rejected case 1\n\n36. aum\n\nyes.", "date": "2016-09-29 22:15:35", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/53f5818fe4b0f0e909e02de2", "openwebmath_score": 0.6936224102973938, "openwebmath_perplexity": 851.3753626294264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850872288502, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.8592922933644341}}
{"url": "https://math.stackexchange.com/questions/2406864/expected-no-of-coin-flips-to-win/2406907", "text": "# Expected no. of coin flips to win\n\nTwo players A and B play a game in which they alternately flips a coin. Player A starts the game. If a player gets T and another player got H before, he/she is the winner. What is the expected no. of flips for A to win ?\n\nEx:-\n\nGame-1: HT --> B wins\n\nGame-2: HHT --> A wins\n\n\u2022 You must share your thoughts on the problem. \u2013\u00a0StubbornAtom Aug 26 '17 at 16:34\n\u2022 What I can think is to find probability for A to win in n flips and then use the formula of expectation to find the expected no. of flips. But this method is quite lengthy. Is there any better solution? \u2013\u00a0Sumit Kumar Aug 26 '17 at 16:38\n\u2022 So the person who wins is the first person to get tails after someone has gotten heads, is that right? \u2013\u00a0Arthur Aug 26 '17 at 16:41\n\u2022 If B wins, presumably A never wins \u2013\u00a0Henry Aug 26 '17 at 16:43\n\u2022 @Arthur A player who get tails wins if and only if another player has gotten heads in the previous flip i.e. a sequence HT occurs \u2013\u00a0Sumit Kumar Aug 26 '17 at 16:43\n\nWe expect four coin flips. That's because we first expect two flips to get the first heads, and then we expect two flips to get the first tails.\n\nMore detailed: as a player is about to throw, the game is in one of two states: either there hasn't been any heads yet, or there has been at least one head (and since the game isn't over, the last throw must've been a head). Let's call the expected number of throws left until the game is over if we're in the first state $E_1$, and the expected number of throws left if we're in the second state $E_2$.\n\nFirst we calculate $E_2$. If a player is about to throw, and the last throw was a heads, then there's a probability of $0.5$ of the game ending in one more throw, and a probability of $0.5$ of the game continuing to the next player still in the same state, which means that after that we expect another $E_2$ throws. This implies $$E_2=0.5\\cdot1+0.5(1+E_2)\\\\ E_2=2$$ Now we get to $E_1$. If a player is about to throw, and there hasn't been any heads yet, then there is a probability of $0.5$ that the player throws heads, and the game continues in state two, which means that we expect another $E_2=2$ throws, and there is a probability of $0.5$ of the player throwing tails, which means that the game continues in the first state, and we expect another $E_1$ throws. This gives $$E_1=0.5(1+E_2)+0.5(1+E_1)\\\\ E_1=4$$ which is our answer.\n\nGiven that $A$ wins, we have an odd number of flips.\n\nLast flip is $T$ and others consist of a possible run of tails, followed by a definite run of heads.\n\nLet $N$ be the number of flips total.\n\n$$P(N=n)=\\frac{1}{2}\\sum_{k=1}^{n-1}\\left(\\frac{1}{2}\\right)^{n-1}=\\frac{n-1}{2^n}$$ $$P(A)=P(N\\equiv 1\\mod 2)=\\sum_{k=1}^\\infty P(N=2k+1)=\\sum_{k=0}^\\infty\\frac{k}{4^k}=\\frac{4}{9}$$.\n\n$$E(N|A)=\\frac{9}{4}\\sum_{k=1}^\\infty\\frac{k(2k+1)}{4^k}=\\frac{13}{3}$$\n\nThe $E(N)=4$ result above is wrong since it is the total expectation $E(N)$ of coin flips until endgame. The total expectation $E(N)$ is, however, an interesting quantity as well, since it is equivalent to a special case of the following theorem involving a typing monkey:\n\nLet a monkey type letters on a typewriter that has some alphabet $\\Sigma$ and let $w\\in\\Sigma^*$. Also let $\\{w_1,\\cdots,w_n\\}$ be the set of those strings that are both a prefix and suffix of $w$.\n\nIf the monkey types until the last letters typed form $w$ and then stops typing, let $X$ be the total number of letters typed. Then $$E(X)=\\sum_{k=1}^n|\\Sigma|^{|w_k|}$$\n\nThe theorem is provable with Martingale Theory.\n\nExample:\n\nTypewriter alphabet: $\\{A,B,\\cdots,Z\\}$ (Standard English).\n\nMonkey's Goal: $ABRACADABRA$.\n\nPrefix-Suffix Strings: $\\{A,ABRA,ABRACADABRA\\}$ of sizes $1,4,11$.\n\nSo the monkey will on average type $26+26^4+26^{11}=3670344487444778$ letters.\n\n\u2022 I have done exactly the same except that i have not multiplied 9/4 as you did in the last equation. Can you please explain why have you done that? \u2013\u00a0Sumit Kumar Aug 27 '17 at 6:11\n\u2022 We are only concerned with expected value $E(N|A)$ of coin flips given that $A$ wins. $A$ wins if $N$ is odd, and loses if $N$ is even. We know how to easily get the total expectation $E(N)$ and the probability of $A$ winning. It is known that if $N$ is a random variable and $A$ is any event, then $E(N|A)$ is given as the weighted arithmetic mean of $E(N|X=x)$ with weights being probabilities of those outcomes $X=x$ that make $A$ happen. Recall that in the discrete case, the weighted mean is $\\frac{w_1x_1+\\cdots+w_nx_n}{w_1+\\cdots+w_n}$ \u2013\u00a0Roman Chokler Aug 27 '17 at 13:40", "date": "2019-09-19 15:40:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2406864/expected-no-of-coin-flips-to-win/2406907", "openwebmath_score": 0.7650814056396484, "openwebmath_perplexity": 211.56803909274558, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850892111574, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8592922918723793}}
{"url": "https://math.stackexchange.com/questions/3594814/floor-function-repeated-addition", "text": "# Floor function repeated addition\n\nI've come across this problem: $$\\left\\lfloor r + \\frac{19}{100}\\right\\rfloor + \\left\\lfloor r + \\frac{20}{100} \\right\\rfloor + \\left\\lfloor r + \\frac{21}{100}\\right\\rfloor + \\dots + \\left\\lfloor r + \\frac{91}{100}\\right\\rfloor = 546$$ $$\\text{Find} \\ \\lfloor 100r\\rfloor. \\textit{(Source: AIME)}$$ Here is my work:\n\n$$\\lfloor r + \\frac{19}{100}\\rfloor = r + \\frac{19}{100} - \\{r + \\frac{19}{100}\\}$$, so the figure can be restated as $$\\{r + \\frac{19}{100}\\} = r + \\frac{19}{100} + a - 546$$, where $$a = \\lfloor r + \\frac{20}{100}\\rfloor + \\lfloor r + \\frac{21}{100}\\rfloor + \\dots + \\lfloor r + \\frac{91}{100}\\rfloor$$.\n\nBecause $$\\{r + \\frac{19}{100}\\}$$ is the fractional part, $$0 \\le r + \\frac{19}{100} + a - 546 < 1$$, so after some more maniuplation, $$545 + \\frac{81}{100} \\le r + a < 546 + \\frac{81}{100}$$. $$a$$ is an integer, so the fractional part of $$r$$ must be $$\\frac{81}{100}$$.\n\n$$r = \\lfloor r\\rfloor + \\{r\\}$$, so $$\\lfloor r + \\frac{19}{100}\\rfloor = \\lfloor \\lfloor r\\rfloor + \\frac{81}{100} + \\frac{19}{100}\\rfloor$$ = $$\\lfloor \\lfloor r\\rfloor + 1\\rfloor$$. Because both of the terms inside that floor function are integers, it must equal $$\\lfloor r\\rfloor + 1$$. This same reasoning can be applied to each of the individual floor functions of the given figure's LHS, and they each turn out to be $$\\lfloor r\\rfloor + 1$$.\n\nTherefore, $$73 \\lfloor r\\rfloor + 73 = 546$$, so $$73\\lfloor r\\rfloor = 473$$. However, this cannot be true unless there is no answer (which I assume is not the case), or unless I did something wrong in my process, because then $$\\lfloor r\\rfloor$$ is not an integer.\n\nIf you do see the solution, it would be really nice if you did not give the answer in your response! Instead, maybe some helpful hints or partial solutions would be preferred.\n\n\u2022 The line \"...so the fractional part of $r$ must be $\\frac {81}{100}$\" is where I think I see an issue. Couldn't that inequality equally have any of the fractions in the interval $[\\frac 9{100},\\frac{81}{100}]$ based on applying that logic in the same manner to any of the other terms from the original? If you apply the correct term to your next line of reasoning, or else find the next-smallest multiple of $73$ below $473$ I think you'll have your answer... Mar 25 '20 at 17:40\n\u2022 Sorry, \"next-largest\" instead of \"next-smallest\"... Really, just begin with the \"$73\\lfloor r\\rfloor+k=546$\" logic, and deduce what $k$ has to be. Mar 25 '20 at 17:52\n\u2022 The average of the terms is about $7.5$ so each term must be $7$ or $8$. How many of each? Mar 25 '20 at 17:58\n\u2022 Don't worry about the fractional part. And don't worry about $[r]$. Find the precise value where $[r + \\frac k{100}]\\ne [r+\\frac {k+1}{100}]$. That is where $r + \\frac {k}{100} < n \\le r + \\frac {k+1}{100}$ for some integer $n$. Mar 25 '20 at 18:23\n\n## 2 Answers\n\nDon't worry about the fractional part. And don't worry about $$[r]$$. Find the precise value of $$k$$ where $$[r + \\frac k{100}]\\ne [r+\\frac {k+1}{100}]$$. That is where $$r + \\frac {k}{100} < m \\le r + \\frac {k+1}{100}$$ for some integer $$m$$.\n\n====== my answer below ====\n\nWell, what jumps at me is $$0 < \\frac k{100} < 1$$ so all the $$[r +\\frac {k}{100}]$$ are either one integer, call it $$n$$ or the next, $$n+1$$.\n\n.\n\nSo if $$b$$ of them equal $$n+1$$ and $$(73 -b)$$ of them equal $$n$$ we have $$(73-b)n + b(n+1) = 73n + b =546$$ where $$0\\le b < 73$$.\n\n.\n\nSo as $$546\\equiv 35 \\pmod {73}$$ and $$546= 7*73 + 35$$ so $$b=35$$ and $$n=7$$.\n\n.\n\nSo we have (I'll have to be careful not to do a fencepost error... $$91-35=56$$ so....) $$[r+\\frac{19}{100}],....,[r+\\frac{56}{100}] = 7$$ and $$[r+\\frac {57}{100}],...,[r+\\frac {91}{100}] = 8$$.\n\n.\n\nSo $$r + \\frac {56}{100} < 8\\le r+\\frac {57}{100}$$ so\n\n.\n\n$$100r + 56 < 800 \\le 100r + 57$$\n\n.\n\n$$100r < 744 \\le 100r + 1$$\n\n.\n\nAnd $$100r -1 < 743 \\le 100r$$ so $$743 \\le 100r < 744$$\n\n.\n\nSo $$[100r] = 743$$.\n\n\u2022 I solved the problem after user abieussu helped me realize that $\\{r\\}$ isn't necessarily equal to $\\frac{81}{100}$, and your answer is wrong but very close. In your second to last step, you state this: \"And 100r\u2212<741\u2264100r so 741\u2264100r<742\", which is the mistake, I believe.\n\u2013\u00a0mpnm\nMar 25 '20 at 19:00\n\u2022 If the third to last step is correct then my second to last ought to be. But arithmetic error can abound anywhere. Mar 25 '20 at 19:21\n\u2022 Oh, for gosh sake. I worried so much about making a fencepost error and figure $57$ was the correct fencepost marker.... and then immediately replaced it with $58$. So everything from then on is off by one\\$. Mar 25 '20 at 19:25\n\u2022 So .... corrected it but... might have made and error. Mar 25 '20 at 19:31\n\u2022 Argh... a fencepost error off by two... that's pretty annoying. Mar 25 '20 at 19:34\n\n$$\\left\\lfloor r + \\frac{19}{100}\\right\\rfloor + \\left\\lfloor r + \\frac{20}{100} \\right\\rfloor + \\left\\lfloor r + \\frac{21}{100}\\right\\rfloor + \\dots + \\left\\lfloor r + \\frac{91}{100}\\right\\rfloor = 546$$\n\nFirst, let $$r = s - \\dfrac{19}{100}$$. Then the sum becomes\n\n$$\\left\\lfloor s + \\frac{0}{100}\\right\\rfloor + \\left\\lfloor s + \\frac{1}{100} \\right\\rfloor + \\left\\lfloor s + \\frac{2}{100}\\right\\rfloor + \\dots + \\left\\lfloor s + \\frac{72}{100}\\right\\rfloor = 546$$\n\nIf $$s$$ were an integer, then you would get $$73s = 546$$, which has solution $$s = 7 \\dfrac{35}{73}$$.\n\nNoting that $$7 \\cdot 73 = 511$$ and $$8 \\cdot 73 = 584$$, we see that $$7 < s < 8$$. Now $$546-511 = 35$$. So where are you going to pick up that extra $$35$$?", "date": "2021-10-20 02:02:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3594814/floor-function-repeated-addition", "openwebmath_score": 0.8073021173477173, "openwebmath_perplexity": 287.60675535011177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850892111574, "lm_q2_score": 0.8740772286044095, "lm_q1q2_score": 0.8592922902600072}}
{"url": "https://math.stackexchange.com/questions/2184961/resistor-factory-and-bayes-rule-which-solution-is-correct", "text": "# Resistor factory and Bayes rule - which solution is correct?\n\nProblem Statement\n\nAt a certain electronics factory, in a typical day\u2019s output 10% percent of resistors are bad, and the rest are good. Good resistors have an 80% chance of passing the factory\u2019s test, and bad resistors have a 30% chance of passing the test. Suppose the factory tests each resistor three times. If a particular resistor passes the test 2 times out of 3, what are the chances it is good?\n\nSolution 1\n\nFirst, let's define some events:\n\n$B :=$ resistor is bad, $G :=$ resistor is good\n\n$P :=$ resistor passes one test, $F :=$ resistor fails one test\n\n$Q :=$ resistor passes 2 out of 3 tests\n\nFrom the problem statement, we can say that $P[B] = 0.10, P[G] = 0.90, P[P|G] = 0.80, P[P|B] = 0.30$. Since the tests are independent, we can also say that $P[Q|G] = {3 \\choose 2}(0.8)^2(0.2)$ and that $P[Q|B] = {3 \\choose 2}(0.3)^2(0.7)$.\n\nThe quantity we seek is $P[G|Q]$. We proceed using Bayes' theorem:\n\n\\begin{align} P[G|Q] &= \\frac{P[Q|G]P[G]}{P[Q]}\\\\ &= \\frac{P[Q|G]P[G]}{P[Q|G]P[G] + P[Q|B]P[B]}\\\\ &= \\frac{{3 \\choose 2}(0.8)^2(0.2)(0.9)}{{3 \\choose 2}(0.8)^2(0.2){0.9} + {3 \\choose 2}(0.3)^2(0.7)(0.1)}\\\\ &\\approx 0.95 \\end{align}\n\nSolution 2\n\nThis solution is the same as solution 1, but we calculate $P[Q]$ differently. We can say that $P[P] = P[P|G]P[G] + P[P|B]P[B] = (0.8)(0.9) + (0.3)(0.1) = 0.75$. Since the tests are independent, $P[Q] = {3 \\choose 2}(0.75)^2 (0.25)$.\n\nWe proceed using Bayes' theorem just like we did in solution 1:\n\n\\begin{align} P[G|Q] &= \\frac{P[Q|G]P[G]}{P[Q]}\\\\ &= \\frac{{3 \\choose 2}(0.8)^2(0.2)(0.9)}{{3 \\choose 2}(0.75)^2 (0.25)}\\\\ &\\approx 0.82 \\end{align}\n\nMy Question\n\nTo me, both of these approaches seem right, but the answers are different. I've done the math several times, so I don't think the difference is due to an arithmetic error. Which solution is correct? Why?\n\n\u2022 The choice of using $P$ with two distinct meanings (\"probability\" and \"passes the test\") is confusing. \u2013\u00a0mlc Mar 13 '17 at 16:41\n\u2022 P(Q) the probability we pass the test 2 out of 3 times - the first part is what type of resistor is it. then the second part is get the probability it passes or fails based on that fact - you can't treat each test as independent, once it passes a test it is more likely to be good than it was before the test \u2013\u00a0Cato Mar 13 '17 at 16:53\n\u2022 if 1% of boys always pass a test and 99% of girls always pass a test - I can't calculate the probability of a person passing two tests as 0.5^2 = .25 - it is actually (1/2)(.99^ 2 + .01^2) = .49 That is to say the girls pass both the tests usually - just to dream up a quick sexist example \u2013\u00a0Cato Mar 13 '17 at 16:58\n\nThe difference in the two approaches is in how you compute $P(Q)$. The first approach is the correct one.\n\n(Let me write $Y$ instead of $P$ for \"passing the test one time\".) The second approach computes $P(Y)$ and then feeds it in the binomial distribution to derive the probability of passing the test twice out of three attempts. The flaw is in the assumption that the three trials are independent: the resistor is either good or bad (and the outcomes of the trials depend on its state).\n\nYour second approach may be re-examined as follows. Let $Q$ be the event \"passing the test two times out of three\". Then $P[Q] = P[Q|G]P[G] + P[Q|B]P[B]$ as in your application of the Bayes' rule.\n\n\u2022 So the the trials are not independent if you don't know whether the resistor is good or bad? How can that be? \u2013\u00a0SplitInfinity Mar 13 '17 at 16:59\n\u2022 See third comment to the question with an example with boys and girls. Independence has nothing to do with what you know. If the resistor is good (but you do not know it), this affects the odds for $Q$. If the first trial is a success, this increases the probability that the resistor is good and hence the odds for $Q$. \u2013\u00a0mlc Mar 13 '17 at 17:02\n\nP(GOOD | Passes 2/3 times ) = P(Passes 2/3 times AND Good) / P(passes 2/3 times)\n\n= 0.9 x 0.8 x 0.8 x 0.2 x 3 / (.9 x 0.8 x 0.8 x 0.2 x 3 + 0.1 x 0.3 x 0.3 x 0.7 x 3) = (216/625)/(216/625 + 189/10000) = 128/135 = .95\n\nthe first one", "date": "2019-09-21 11:44:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2184961/resistor-factory-and-bayes-rule-which-solution-is-correct", "openwebmath_score": 0.8929245471954346, "openwebmath_perplexity": 385.2707683421318, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576913496333, "lm_q2_score": 0.8807970873650403, "lm_q1q2_score": 0.859285303908146}}
{"url": "https://stats.stackexchange.com/questions/487823/probability-that-linear-combination-of-normal-random-variables-exceeds-a-value/487832", "text": "Probability that linear combination of normal random variables exceeds a value\n\nI'm new to statistics, and I'm wondering how can I compute $$P(2Y_1 + 4Y_2 - 3Y_3 \\geq 40)$$ given the following information?\n\n\u2022 $$Y$$ follows a multivariate normal distribution.\n\n\u2022 $$Y_1, Y_2, Y_3$$ follow a normal distribution.\n\n\u2022 $$Y_1, Y_2, Y_3$$ have means $$10, 12, 14$$.\n\n\u2022 $$Y_1, Y_2, Y_3$$ have variances $$2, 2, 8$$.\n\n\u2022 The covariance between $$Y_1, Y_2$$ is $$0.50$$, and the covariance between $$Y_1, Y_3$$ is $$-0.75$$. The variables $$Y_2$$ and $$Y_3$$ are independent.\n\nHere's the covariance matrix I found for $$Y_1, Y_2, Y_3$$:\n\n$$\\begin{bmatrix} 2 & .50 & -.75 \\\\ .5 & 2 & 0 \\\\ -.75 & 0 & 8 \\end{bmatrix}$$\n\nI know how to do it when $$Y_1, Y_2, Y_3$$ are all independent by using the fact that linear combinations of independent normal random variables are normal. However, I'm really not sure about how to do it in this case. I can easily find the covariance matrix, but I'm not sure how to proceed from there.\n\n\u2022 Just knowing their means, variances and covariances along with the fact that each variables is (marginally) normal is insufficient. Sep 18, 2020 at 3:32\n\nIf you're looking for an analytic solution, this is what you should do. Start by defining a new variable $$X = 2Y_1 + 4Y_2 - 3Y_3$$ A sum of jointly normal random variables is also normal, even if the variables are not independent (see Wikipedia article here). All that remains is to compute the mean and variance.\n\nThe mean of $$X$$ can be determined as \\begin{align*} E[X] &= 2 E[Y_1] + 4 E[Y_2] - 3 E[Y_3] \\\\ &= (2 \\times 10) + (4 \\times 12)-(3 \\times 14) \\\\ &= 26 \\end{align*} The variance of $$X$$ is slightly easier to write in matrix form. Let $$Y \\equiv (Y_1,Y_2,Y_3)$$ and $$\\omega \\equiv (2,4,-3)$$ be vectors of random variables and weights, respectively. The variance is given by\n\n$$Var(X) = Var(\\omega'Y) = \\omega'Var(Y)\\omega$$\n\nNumerically this is equal to $$Var(X) = \\begin{bmatrix} 2 & 4 & -3 \\end{bmatrix} \\begin{bmatrix} 2& .5 & -.75 \\\\ .5 & 2 & 0 \\\\ -.75 & 0 & 8 \\end{bmatrix} \\begin{bmatrix} 2 \\\\ 4 \\\\ -3 \\end{bmatrix}= 129$$ If $$\\Phi$$ is the CDF of a standard normal, \\begin{align*} P(X > 40) &= 1-P\\left( X \\le 40 \\right) \\\\ &= 1-P\\left( \\frac{X - E[X]}{\\sqrt{Var(X)}} \\le \\frac{40 - E[X]}{\\sqrt{Var(X)}} \\right) \\\\ &= 1-P\\left( Z \\le \\frac{40 - E[X]}{\\sqrt{Var(X)}} \\right) \\\\ &= 1-\\Phi\\left( \\frac{40 - E[X]}{\\sqrt{Var(X)}} \\right) \\\\ &= 0.108867 \\end{align*}\n\n\u2022 \"A sum of normal random variables is also normal, even if the variables are not independent\" - I don't think that's true. Do you have a proof?\n\u2013\u00a0Paul\nSep 17, 2020 at 7:31\n\u2022 @Paul you are correct, the necessary assumption is that the random variables are jointly gaussian (which they are in this case). The easy counterexample is $X \\sim \\mathcal N (0,1)$ and $Y = -X$. Then $X + Y = 0$ is not gaussian. Sep 17, 2020 at 7:55\n\u2022 Another nice counterexample is here math.stackexchange.com/questions/563364/\u2026 Sep 17, 2020 at 8:05\n\u2022 Great counter examples! I have amended the statement to emphasize that \"joint\" normality is important. Your comments emphasize the pitfalls of focusing on the marginal distributions only. Sep 18, 2020 at 2:56\n\nYou can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.\n\nVia simulation.\n\n> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3)\n> is.positive.definite(x, tol=1e-8)\n[1] TRUE\n> means=c(10,12,14)\n> weights=c(2,4,-3)\n> mat=mvrnorm(10^7,means,covmat)\n> mat=mat %*% weights\n> result=sum(mat>40)/10^7\n> result\n[1] 0.1088587\n\n\nNon simulated\n\nThe mean is a linear combination of means\n\n> new_mean= means %*% weights\n> new_mean\n[1] 26\n\n\nThe variance is obtained multiplying\n\n> weights %*% covmat %*% weights\n[,1]\n[1,]  129\n\n\n> 1-pnorm(40,26,sqrt(129))", "date": "2022-08-11 15:28:20", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/487823/probability-that-linear-combination-of-normal-random-variables-exceeds-a-value/487832", "openwebmath_score": 0.999996542930603, "openwebmath_perplexity": 493.5090273246783, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156284, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8592852897467035}}
{"url": "https://math.stackexchange.com/questions/1743819/definition-of-limit-as-x-rightarrow-infty", "text": "# Definition of limit as $x\\rightarrow \\infty$\n\nEvery time i get confused with the definition of $\\lim_{x\\rightarrow \\infty}f(x)=L$. I could not find a reference that will give the definition.\n\nI am trying to write what i understood. See if this is correct.\n\n\u2022 By $\\lim_{x\\rightarrow \\infty}f(x)=L$ we mean the following : Given $\\epsilon >0$ there exists $R>0$ such that $|f(x)-L|<\\epsilon$ for all $x>R$.\n\u2022 By $\\lim_{x\\rightarrow -\\infty}f(x)=L$ we mean the following : Given $\\epsilon >0$ there exists $R<0$ such that $|f(x)-L|<\\epsilon$ for all $x<R$.\n\u2022 By $\\lim_{x\\rightarrow \\infty}f(x)=\\infty$ we mean the following: Given $R>0$ there exists $L>0$ such that $|f(x)|>R$ for all $x>L$\n\u2022 By $\\lim_{x\\rightarrow \\infty}f(x)=-\\infty$ we mean the following: Given $R<0$ there exists $L>0$ such that $f(x)<R$ for all $x>L$\n\u2022 By $\\lim_{x\\rightarrow -\\infty}f(x)=\\infty$ we mean the following: Given $R>0$ there exists $L<0$ such that $|f(x)|>R$ for all $x<L$\n\u2022 By $\\lim_{x\\rightarrow \\infty}f(x)=-\\infty$ we mean the following: Given $R<0$ there exists $L<0$ such that $f(x)<R$ for all $x<L$.\n\nLet me know if i understood somethings wrongly.\n\n\u2022 It's quite fine for me. The signs for $R$ and $L$ are unncessary (they're implicit). \u2013\u00a0Bernard Apr 15 '16 at 14:25\n\u2022 You don't, technically, need $R<0//R>0$. for the first two. It can be any $R$. It's the \"for all $x>R$\" and \"for all $x<R$\" that makes the difference, not the sign of $R$. \u2013\u00a0Thomas Andrews Apr 15 '16 at 14:25\n\u2022 If there is no need for sign then $x\\rightarrow \\infty$ and $x\\rightarrow -\\infty$ would mean the same thing? @Bernard \u2013\u00a0user311526 Apr 15 '16 at 14:27\n\u2022 If there is no need for sign then $x\\rightarrow \\infty$ and $x\\rightarrow -\\infty$ would mean the same thing? @ThomasAndrews \u2013\u00a0user311526 Apr 15 '16 at 14:27\n\u2022 @topgeomj No, because one definition uses $x>R$ and the other $x<R$. That is the difference, not the sign of $R$. \u2013\u00a0Thomas Andrews Apr 15 '16 at 14:28\n\nI will use $+\\infty$ in this answer to avoid ambiguity.\n\n\u2022 $\\lim_{x\\to +\\infty} f(x) = L$ means for all $\\epsilon>0$ there exists $M$ such that for all $x>M,$ $|f(x)-L|<\\epsilon$ .\n\n\u2022 $\\lim_{x\\to+\\infty} f(x)=+\\infty$ means for all $R$ there exists $M$ such that for all $x>M$, $f(x)>R$.\n\nThose two definitions let you define the other limits by symmetry:\n\n\\begin{align} \\lim_{x\\to+\\infty} f(x) = -\\infty&\\iff \\lim_{x\\to+\\infty} -f(x)=+\\infty\\\\ \\lim_{x\\to-\\infty} f(x) = M &\\iff \\lim_{x\\to+\\infty} f(-x)=M \\end{align}\n\nWhere the $M$ in the second case can be any of either a real value, or $+\\infty,-\\infty$.\n\nSo $\\lim_{x\\to-\\infty} f(x)= -\\infty$ means $\\lim_{x\\to +\\infty} -f(-x)=+\\infty$, which means:\n\nFor any $R$ there is an $M$ so that for all $x>M$, $-f(-x)>R$.\n\nNow, given any $R'$, you can set $R=-R'$ and find $M$ with this condition, and set $M'=-M$. Then if $x<M'$, $-x>M$, and thus $-f(x)>R$ or $f(x)<R'=-R$. So we get back the definition that we want.\n\nThe reason to distinguish $+\\infty$ from $\\infty$ is that some books use $\\infty$ means a single point at infinity, in both directions - essentially, merging the two values $+\\infty,-\\infty$ into a single point at infinity.\n\nThen:\n\n$$\\lim_{x\\to\\infty} f(x)= L \\iff \\lim_{x\\to+\\infty} f(x)=\\lim_{x\\to-\\infty} f(x)=L$$\n\nwhere $L$ can be any real or $+\\infty,-\\infty$.\n\n$$\\lim_{x\\to W} f(x) = \\infty\\iff \\lim_{x\\to W} |f(x)|=+\\infty$$\n\nWhere $W$ can be any real, or $+\\infty$ or $-\\infty$, or $\\infty$.\n\nYour question is somehwat confused, because you seem to distingush $\\infty$ from $+\\infty$ with absolute values when $\\infty$ is the limit, but not when $x\\to\\infty$.\n\n\u2022 Thanks... I understand this clearly... :) \u2013\u00a0user311526 Apr 15 '16 at 15:04\n\nOne also does not need the absolute values on $|f(x)| > R$. If f is going to infinity, then $f(x) > R$ is accurate.\n\n\u2022 For $f(x) = (-1)^{[x]}e^x$, $f(x) \\to \\infty | x \\to +\\infty$, but not $f(x) \\to +\\infty | x \\to +\\infty$. \u2013\u00a0Abstraction Apr 15 '16 at 14:34\n\u2022 Ok.. Can you say something about other definitions as well.. \u2013\u00a0user311526 Apr 15 '16 at 14:34\n\u2022 @Abstraction Some people distinguish between $+\\infty$ and $\\infty$, but quite often they are used interchangably. \u2013\u00a0Thomas Andrews Apr 15 '16 at 14:59\n\u2022 @ThomasAndrews Seeing as third and fifth examples in the initial post clearly distinguish between $f(x) \\to +\\infty$ and $f(x) \\to \\infty$ (using absolute value of $f(x)$), I think here the distinction is important. \u2013\u00a0Abstraction Apr 15 '16 at 15:08\n\u2022 Well, the main indication that the OP is using $\\infty$ to mean a single point at infinity is the absolute values, but in the $x\\to\\infty$ examples, the definitions do not use $|x|>L$. So the clarity is not there in the question. @Abstraction \u2013\u00a0Thomas Andrews Apr 15 '16 at 15:11\n\nLet me know if i understood somethings wrongly.\n\nYou did. When you write $x \\to \\infty$ (as opposed to $x \\to +\\infty$), for $x \\in \\mathbb{R}$ it roughly means \"when absolute value of $x$ is arbitrarily large\" ($|x| > R$, not $x > R$)\n\nThere is a universal definition of limit using filters. A filter $\\mathcal{F}$ is a collection of sets such that it doesn't contain an empty set and $\\forall f_1, f_2 \\in \\mathcal{F}, \\exists f_3 \\in \\mathcal{F} : f_3 \\subseteq f_1 \\cap f_2$. For example, a set of neighbourhoods of a real point $x$ is a filter ($\\{(x-\\epsilon, x+\\epsilon) | \\epsilon \\in \\mathbb{R}^+\\}$, called neighbourhoods filter); a set of intervals with infinite endpoint ($\\{(a,+\\infty) | a \\in \\mathbb{R}\\}$) is a filter; a set of segment complements ($\\{(-\\infty,a)\\cup(b,+\\infty) | a,b \\in \\mathbb{R}\\}$) is a filter.\n\nNow, let's take function $h \\in \\{X \\to Y\\}$ and there is a filter $\\mathcal{F}$ on $X$ and a concept of \"neighbourhoods\" on $Y$ (called topology). Then let's take this sentence: \"there is such $y \\in Y$ that for any $y$ neighbourhood $O_y$, there is a filter element $f \\in \\mathcal{F}$ so that $h(f) \\subseteq O_y$\" and write it for short as $\\lim_{\\mathcal{F}}h(x) = y$.\n\nLets write filter $\\{(a,+\\infty) | a \\in \\mathbb{R}\\}$ as $x \\to +\\infty$, filter $\\{(-\\infty,a) | a \\in \\mathbb{R}\\}$ as $x \\to -\\infty$ and filter $\\{(-\\infty,a) \\cup (a,+\\infty) | a \\in \\mathbb{R}\\}$ as $x \\to \\infty$.\n\nFinally, let's call set $\\{(L-\\epsilon,L+\\epsilon) | \\epsilon \\in \\mathbb{R}\\}$ a set of neighbourhoods of $L$, set $\\{(b,+\\infty) | b \\in \\mathbb{R}\\}$ a set of neighbourhoods of $+\\infty$, set $\\{(-\\infty,b) | b \\in \\mathbb{R}\\}$ a set of neighbourhoods of $-\\infty$ and set $\\{(-\\infty,b) \\cup (b,+\\infty) | b \\in \\mathbb{R}\\}$ a set of neighbourhoods of $\\infty$.\n\nNow from these rules we can derive any definition. Take $\\lim_{x \\to -\\infty}f(x) = +\\infty$. It comes to \"for any $+\\infty$ neighbourhood $(b,+\\infty), b \\in \\mathbb{R}$, there is a filter element $(-\\infty,a), a \\in \\mathbb{R}$ so that $f((-\\infty,a)) \\subseteq (b,+\\infty)$\". Or, in more traditional notation, $$\\forall b \\in \\mathbb{R}, \\exists a \\in \\mathbb{R} : \\forall x \\in (-\\infty,a), f(x) \\in (b,+\\infty)$$\n\n\u2022 May be you are complication the question by saying more advanced things as filter, topology... \u2013\u00a0user312648 Apr 15 '16 at 16:09\n\u2022 @cello Maybe. But memorizing definitions for all cases of $x \\to \\pm\\infty$ is actually hard, filters help to see some logic in all this. \"Topology\" here is simply a proper naming for a system of neighbourhoods; as you can see, it's properties aren't used. \u2013\u00a0Abstraction Apr 15 '16 at 16:21", "date": "2019-05-26 23:01:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1743819/definition-of-limit-as-x-rightarrow-infty", "openwebmath_score": 0.9533989429473877, "openwebmath_perplexity": 228.08983162612935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769042651874, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.859285282040538}}
{"url": "https://math.stackexchange.com/questions/2844744/help-with-matrix-derivative", "text": "# Help with matrix derivative\n\nIn my studies of applied mathematics, specifically optimization and applied linear algebra, I have come across the following expression which I need help differentiating:\n\n$z(B,C) = \\lVert Y-B \\phi(CX) \\rVert_F ^2 = \\text{trace} \\left( (Y-B \\phi(CX))(Y-B \\phi(CX))^T \\right)$\n\nwhere $Y \\in \\mathbb{R}^{m\\times s},X\\in \\mathbb{R}^{n\\times s}$ are two constant real matrices, $B \\in \\mathbb{R} ^{m\\times k}$ and $C\\in\\mathbb{R}^{k \\times n}$ are the two variable matrices of the function $z(B,C)$ defined above, $\\lVert \\bullet \\rVert_F$ is the Frobenius norm and all matrix dimensions involved are constant and non-variable.\n\nThe function $\\phi : \\mathbb{R} \\to \\mathbb{R}$ is a nonlinear function defined for real numbers as follows:\n\n$\\phi(x) = \\left\\{ \\begin{array}{ll} 0 & x \\leq 0 \\\\ x & x > 0 \\\\ \\end{array} \\right.$\n\nand we extend its definition to matrices element-wise, that is $(\\phi(A))_{i,j} = \\phi((A)_{i,j})$.\n\nI would like to find the matrix derivative of the function $z$, as stated above, with respect to $C$, that is $\\frac{\\partial z}{\\partial C}$.\n\nDifferentiating with respect to $B$ is easy enough as $C$ doesn't impede applying standard formulas, but my problem is with differentiating with respect to $C$ as it is the argument of a nonlinear function and I do not know how to derive it with respect to matrices.\n\nI was hoping someone could please come to the rescue and help me differentiate the function $z$ with respect to $C$. I thank all helpers.\n\n\u2022 In the first derivative, is A supposed to be C? \u2013\u00a0John Polcari Jul 8 '18 at 16:31\n\u2022 Does this book have copy right? This not allowed on this website \u2013\u00a0Cloud JR K Jul 8 '18 at 16:46\n\u2022 @JohnPolcari : yes fixed it now \u2013\u00a0kroner Jul 8 '18 at 16:49\n\u2022 @CloudJR : no, it is OK \u2013\u00a0kroner Jul 8 '18 at 16:50\n\u2022 @JohnPolcari : thanks for pointing out \u2013\u00a0kroner Jul 8 '18 at 16:51\n\nLet's use a convention where uppercase Latin letters represent matrices, lowercase Latin vectors, and Greek letters are scalars.\n\nThe function you've denoted by $\\phi$ is the ReLU function, $r(\\alpha)$, whose derivative is the Heaviside step function $$h(\\alpha) = \\frac{dr(\\alpha)}{d\\alpha} \\implies dr = h\\,d\\alpha$$ Applying these scalar functions element-wise on a matrix argument $A=CX,$ produces matrix results, which we'll denote as \\eqalign{ R &= r(A) \\cr H &= h(A) \\implies dR = H\\odot dA = H\\odot(dC\\,X) \\cr } where $\\odot$ is the elementwise/Hadamard product.\n\nDefine a new matrix variable $$M=BR-Y \\implies dM = B\\,dR + dB\\,R$$ Write the function in terms of this new variable, then find its differential. \\eqalign{ \\lambda &= \\|M\\|^2_F = M:M \\cr d\\lambda &= 2M:dM \\cr &= 2M:B\\,dR + 2M:dB\\,R \\cr &= 2B^TM:dR + 2MR^T:dB \\cr &= 2B^TM:H\\odot(dC\\,X) + 2MR^T:dB \\cr &= 2(B^TM)\\odot H:(dC\\,X) + 2MR^T:dB \\cr &= 2((B^TM)\\odot H)X^T:dC + 2MR^T:dB \\cr } Setting $dB=0$ yields the gradient wrt $C$ $$\\frac{\\partial\\lambda}{\\partial C} = 2((B^TM)\\odot H)X^T$$ And setting $dC=0$ yields the gradient wrt $B$ $$\\frac{\\partial\\lambda}{\\partial B} = 2MR^T$$\n\nNB: Depending on your preferred layout convention, you may need to transpose these results.\n\nAlso, the colon notation used above (called the Frobenius product) is just a convenient way of writing the trace function, i.e. $\\,\\,A:B={\\rm tr}(A^TB)$.\n\nThe cyclic property of the trace leads to several ways to rearrange the terms in a Frobenius product. For example, all of the following expressions are equivalent \\eqalign{ A:BC &= A^T:(BC)^T \\cr &= BC:A \\cr &= AC^T:B \\cr &= B^TA:C \\cr }\n\n\u2022 Thanks for pointing out the name of $\\phi$ as I was not aware of this. \u2013\u00a0kroner Jul 8 '18 at 17:26\n\u2022 Thanks for your masterpiece answer, I appreciate it. \u2013\u00a0kroner Jul 8 '18 at 17:27", "date": "2021-05-07 10:00:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2844744/help-with-matrix-derivative", "openwebmath_score": 0.9998749494552612, "openwebmath_perplexity": 425.7984376329643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130573694068, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8592810313304701}}
{"url": "https://www.physicsforums.com/threads/help-with-set-theory-notation.901710/", "text": "# I Help with set theory notation\n\n1. Jan 27, 2017\n\n### Jehannum\n\nI bought a maths book and have discovered it's somewhat above my level.\n\nIn particular I'm confused about one bit of notation. I understand the \"is a member of\" operator when it takes a set as argument (e.g. n \u2208 \u211d) but not when the book uses it with functions (e.g. n \u2208 f)\n\nDoes n \u2208 f mean that the number n can be returned by function f?\n\n2. Jan 27, 2017\n\n### PeroK\n\nYou need to give more context. My guess is that a function can be seen as a set. For example, the function $f(x) = x^2$ can be seen as the set of points $\\lbrace (x, x^2): x \\in \\mathbb{R} \\rbrace$, also commonly known as the graph of $f$.\n\nIn this case, you could use the notation: $f = \\lbrace (x, x^2): x \\in \\mathbb{R} \\rbrace$.\n\nAnd, here, $n \\in f$ would mean that $n = (x, x^2)$ for some real $x$.\n\n3. Jan 27, 2017\n\n### Staff: Mentor\n\nThis is strange, not to say wrong. A function $f\\, : \\,X \\rightarrow Y$ can be written as a subset $f=\\{(x,y)\\,\\vert \\,y=f(x)\\} \\subseteq X \\times Y$ of the Cartesian product, but this cannot explain the notation $n \\in f$ unless $n$ is a pair. Can you give some more context?\n\n4. Jan 27, 2017\n\n### Staff: Mentor\n\n5. Jan 29, 2017\n\n### Jehannum\n\nI've copied part of the book below:\n\nTheorem 2.2 There are infinite sets that are not enumerable.\nProof Consider the powerset of N, in other words the collection P whose members are all the sets of numbers (so X \u2208 P iff X \u2286 N).\nSuppose for reductio that there is a function f : N\u2192P which enumerates P, and consider what we\u2019ll call the diagonal set D \u2286 N such that n \u2208 D iff n /\u2208 f(n).\nSince D \u2208 P and f by hypothesis enumerates all the members of P, there must be some number d such that f(d) = D.\nSo we have, for all numbers n, n \u2208 f(d) iff n /\u2208 f(n). Hence in particular d \u2208 f(d) iff d /\u2208 f(d). Contradiction!\u200b\n\n6. Jan 29, 2017\n\n### pasmith\n\nThat's not what the text is saying.\n\nThis is defining $f$ as a function from the natural numbers $\\mathbb{N}$ to the collection $P$ of subsets of natural numbers. Thus if $n \\in \\mathbb{N}$ then $f(n) \\subset \\mathbb{N}$. The statement $n \\in f(n)$ is therefore perfectly sensible, since it always makes sense to ask if a particular natural number is or is not a member of a particular subset of the natural numbers.\n\n7. Jan 29, 2017\n\n### Staff: Mentor\n\nO.k., but this is not what you wrote. $n \\in f$ is something different to $n \\in f(n)$ where $f: \\mathbb{N}\\rightarrow \\mathcal{P}(\\mathbb{N})$ is a function which values are sets $X$, so $n \\in X=f(n) \\in \\mathcal{P}(\\mathbb{N})$ makes sense. This is not the same as $n \\in f \\in \\mathcal{P}( \\mathbb{N} \\times \\mathbb{N})$ which we explained in the previous posts.\n\nThe theorem is proven by the so called diagonal argument. It is a formal notation of the following principle:\n\nWe want to show, that the set of all subsets of $\\mathbb{N}$ isn't enumerable.\n\nIf we have an element $X \\in \\mathcal{P}(\\mathbb{N})$, i.e. a subset $X \\subseteq \\mathbb{N}$, then we can write it down number by number: $X= n_1n_2n_3n_4 \\ldots$\nNow let's assume $\\mathcal{P}(\\mathbb{N})$ is enumerable. Then there would be a complete list like\n$$\\begin{matrix} 1: &X_1 & = & n_{11}n_{12}n_{13} \\ldots \\\\ 2: &X_2 & = & n_{21}n_{22}n_{23} \\ldots \\\\ 3: & X_3 & = & n_{31}n_{32}n_{33} \\ldots \\\\ &\\vdots &&\\\\ \\end{matrix}$$\nwhich contains all elements of $\\mathcal{P}(\\mathbb{N})$, or subsets of $\\mathbb{N}$.\nThe diagonal $d$ is now the number $d=n_{11}n_{22}n_{33} \\ldots$ We next add $+1$ to all of the numbers $n_{ii}\\,$.\nSince by assumption we have enumerated all, this new number $d'$ (or subset $d'=\\{n_{11}+1,n_{22}+1,n_{33}+1, \\ldots\\}$ if you like) has to occur somewhere, say $d'=X_k$. However the $k-$th position of $d'$ is $n_{kk}$ by enumeration of the $X_i$ and $n_{kk}+1$ by construction of $d'$. This cannot both be true, which means there is no way to enumerate $\\mathcal{P}(\\mathbb{N})$.\n\nYou might want to read the proof in the book again and figure out, why both are basically the same. It is also the method used to show why $[0,1]$ or $\\mathbb{R}$ are uncountable. Whether $|\\mathcal{P}(\\mathbb{N})| = |\\mathbb{R}|$ holds is called the continuum hypothesis. It cannot be deduced from the usual axioms of set theory, i.e. by the Zermelo-Fraenkel axioms (ZF). If you read ZFC somewhere, then Zermelo-Fraenkel + Contnuum hypothesis is meant. (Usually it is also assumed, if nothing is said.)\n\n8. Jan 29, 2017\n\n### micromass\n\nStaff Emeritus\nThat is not the continuum hypothesis and it can be proven in ZFC.\n\n9. Jan 29, 2017\n\n### micromass\n\nStaff Emeritus\nZFC means Zermelo-Fraenkel with choice. Not the continuum hypothesis.\n\n10. Jan 29, 2017\n\n### Jehannum\n\nThanks. It's becoming clear now.\n\nBoggled by some of the unfamiliar notation and concepts, I was not appreciating the fact that the function f returns a set as its result, so the \u2208 operator can of course be used on the outputs of f such as f (n).\n\nThe book itself moves onto such diagonal proofs.\n\n11. Jan 29, 2017\n\n### Staff: Mentor\n\nThanks for clarification.\nOops, I just had a quick look on Wiki to be sure, for I know I tend to forget where the \"gap\" really is, and only saw $\\mathfrak{c}=\\aleph_1$ and next $2^{\\aleph_0}=\\aleph_1$. I should have read it with more care ...\nThanks for the corrections.", "date": "2017-08-20 18:35:55", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/help-with-set-theory-notation.901710/", "openwebmath_score": 0.8470800518989563, "openwebmath_perplexity": 408.6566447426922, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130563978902, "lm_q2_score": 0.8688267677469952, "lm_q1q2_score": 0.8592810170497556}}
{"url": "http://mathhelpforum.com/differential-geometry/190486-3-2-matrix-complex-elements.html", "text": "# Math Help - 3*2 Matrix with complex elements\n\n1. ## 3*2 Matrix with complex elements\n\nFolks,\n\nGiven that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalr multiplication of matrices, what is its dimension?\n\nAre the following subsets subspaces?\n\n1) The set of 3*2 with real entries\n\n2) The set of matrices with first row (0,0)\n\n3) The set of matrices with first row (1,1)\n\nI understand the dimension of a vector space is the number of vectros in any basis for the space, ie the number of coordinates necessary to specify any vector. How do I find out what its dimension is?\n\nThanks\n\n2. ## Re: 3*2 Matrix with complex elements\n\nthe \"naive\" way is to ask yourself: how many complex numbers do i need to specify to identify my \"vector\"? for 3x2 matrices, that number is 6 (one for each entry).\n\nhowever, mathematics professors being what they are, will usually insist you display a basis (a linearly independent spanning set). can you think of a set of six matrices,\n\neach of which captures the idea of \"a single matrix coordinate\"? once you have done this, show linear independence and spanning.\n\nagain, with any vector space V, there are 3 things you need to show for any subset W:\n\n1) W is non-empty (preferrably by showing the 0-element is a member. if the 0-element (0-vector) is not in W, you will not obtain a vector space).\n2) if u,v are in W then their vector sum u+v must also be.\n3) if c is any scalar in your underlying field, and u is in W, then cu must also be in W. be careful with this one. if you are working with a complex vector space, it is not sufficient to check this property for real scalars only.\n\nattempt to show these properties (or give a counter-example) for each of the sets in your post. that's how it's done.\n\n3. ## Re: 3*2 Matrix with complex elements\n\nOriginally Posted by Deveno\nthe \"naive\" way is to ask yourself: how many complex numbers do i need to specify to identify my \"vector\"? for 3x2 matrices, that number is 6 (one for each entry).\n\ncan you think of a set of six matrices, each of which captures the idea of \"a single matrix coordinate\"? once you have done this, show linear independence and spanning.\nSorry, I am still struggling to get a start...I dont know how to approach this? If its a 3*2 matrix then would it have 3 solutions in ${\\mathbb{R}^3}$?\n\n4. ## Re: 3*2 Matrix with complex elements\n\nIf I write a sample 3*2 matrix as\n\n$\\begin{bmatrix}2 & 5\\\\ 1 & 4 \\\\ 6 & 8 \\end{bmatrix}$\n\nTHis matrix contains 6 real entries, but I am unbale to write expression since I believe I need it to be a 3*3 or a 2*3 matrix. How would I proceed to show its closed under vector addition and scalar multiplication?\n\nThanks\n\n5. ## Re: 3*2 Matrix with complex elements\n\nyou're not being asked to \"solve\" a system of linear equations, you're being asked to display a basis for a vector space.\n\nhere is a similar problem:\n\nfind the dimension of the vector space of all 2x2 complex matrices.\n\ni claim:\n\n$\\left\\{\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix},\\begin {bmatrix}0&1\\\\0&0\\end{bmatrix}, \\begin{bmatrix}0&0\\\\ 1&0 \\end{bmatrix},\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix} \\right\\}$\n\nis a basis for this vector space. clearly this is a spaning set, since:\n\n$\\begin{bmatrix}a&b\\\\c&d\\end{bmatrix} = a\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix} + b\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix} + c\\begin{bmatrix}0&0\\\\1&0\\end{bmatrix} + d\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix}$\n\nnow, to show linear independence, suppose that:\n\n$\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix} = c_1\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix} + c_2\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix} + c_3\\begin{bmatrix}0&0\\\\1&0\\end{bmatrix} + c_4\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix}$\n\nthen:\n\n$\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix} = \\begin{bmatrix}c_1&c_2\\\\c_3&c_4\\end{bmatrix}$\n\nso $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.\n\nsince our basis has 4 elements, the dimension of the vector space is 4.\n\n6. ## Re: 3*2 Matrix with complex elements\n\nOriginally Posted by Deveno\nyou're not being asked to \"solve\" a system of linear equations, you're being asked to display a basis for a vector space.\n\nhere is a similar problem:\n\nfind the dimension of the vector space of all 2x2 complex matrices.\n\ni claim:\n\n$\\left\\{\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix},\\begin {bmatrix}0&1\\\\0&0\\end{bmatrix}, \\begin{bmatrix}0&0\\\\ 1&0 \\end{bmatrix},\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix} \\right\\}$\n\nis a basis for this vector space. clearly this is a spaning set, since:\n\n$\\begin{bmatrix}a&b\\\\c&d\\end{bmatrix} = a\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix} + b\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix} + c\\begin{bmatrix}0&0\\\\1&0\\end{bmatrix} + d\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix}$\n\nnow, to show linear independence, suppose that:\n\n$\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix} = c_1\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix} + c_2\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix} + c_3\\begin{bmatrix}0&0\\\\1&0\\end{bmatrix} + c_4\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix}$\n\nthen:\n\n$\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix} = \\begin{bmatrix}c_1&c_2\\\\c_3&c_4\\end{bmatrix}$\n\nso $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.\n\nsince our basis has 4 elements, the dimension of the vector space is 4.\nI can only write 4 of the required 6 matrices...\n\n$\\left[ \\begin{array}{ccc} 1 & 0 \\\\ 0 & 1\\\\ 1 & 1\\end{array},\\begin{array}{ccc} 0 & 1\\\\ 1 & 0\\\\ 0 & 0\\end{array},\\begin{array}{ccc} 1 & 1 \\\\ 0 & 0\\\\ 0 & 1\\end{array},\\begin{array}{ccc} 0 & 0 \\\\ 1 &1\\\\ 1 & 0\\end{array}, \\right]$\n\nThere is a lack of symmetry for me to complete....\n\n7. ## Re: 3*2 Matrix with complex elements\n\nOne of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.\n\n8. ## Re: 3*2 Matrix with complex elements\n\nOriginally Posted by HallsofIvy\nOne of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.\n1) What is the significance of saying 'vector space over the real numbers' or 'vector space over the complex numbers'? You have written to different ways (RHS) of expressing the same LHS. I dont understand the difference.\n\n2) Is the above an example for a 2*2 matrix?\n\n3) How does one do the above for a 3*2?\n\n9. ## Re: 3*2 Matrix with complex elements\n\nthe \"underlying field\" of a vector space can make a difference when talking about dimension.\n\ni don't understand why you came up with those particular 4 3x2 matrices. why not the 6 matrices who have one entry 1, the rest 0?\n\n10. ## Re: 3*2 Matrix with complex elements\n\nOriginally Posted by Deveno\nyou're not being asked to \"solve\" a system of linear equations, you're being asked to display a basis for a vector space.\n\nhere is a similar problem:\n\nfind the dimension of the vector space of all 2x2 complex matrices.\n\ni claim:\n\n$\\left\\{\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix},\\begin {bmatrix}0&1\\\\0&0\\end{bmatrix}, \\begin{bmatrix}0&0\\\\ 1&0 \\end{bmatrix},\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix} \\right\\}$\n\nis a basis for this vector space. clearly this is a spaning set, since:\n\n$\\begin{bmatrix}a&b\\\\c&d\\end{bmatrix} = a\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix} + b\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix} + c\\begin{bmatrix}0&0\\\\1&0\\end{bmatrix} + d\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix}$\n\nnow, to show linear independence, suppose that:\n\n$\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix} = c_1\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix} + c_2\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix} + c_3\\begin{bmatrix}0&0\\\\1&0\\end{bmatrix} + c_4\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix}$\n\nthen:\n\n$\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix} = \\begin{bmatrix}c_1&c_2\\\\c_3&c_4\\end{bmatrix}$\n\nso $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.\n\nsince our basis has 4 elements, the dimension of the vector space is 4.\n$\\left[\\begin{array}{ccc} 0 & 0 \\\\ 0 & 0\\\\0 & 0\\end{array}\\right]=c_1\\left[ \\begin{array}{ccc} 1 & 0 \\\\ 0 & 0\\\\0 & 0\\end{array}\\right]+c_2\\left[ \\begin{array}{ccc} 0 & 1\\\\ 0 & 0\\\\ 0 & 0\\end{array}\\right]+c_3\\left[ \\begin{array}{ccc} 0 & 0 \\\\ 1 & 0\\\\ 0 & 0\\end{array}\\right]+c_4\\left[ \\begin{array}{ccc} 0 & 0 \\\\ 0 & 1\\\\ 0 & 0\\end{array}\\right]+c_5\\left[ \\begin{array}{ccc} 0 & 0 \\\\ 0 & 0\\\\ 1 & 0\\end{array}\\right]+c_6\\left[ \\begin{array}{ccc} 0 & 0 \\\\ 0 & 0\\\\ 0& 1\\end{array}\\right]$\n\nimplies $c_1+c_2+c_3+c_4+c_5+c_6=0$\n\nThis implies the set is linearly independant with a dimension of 6? What is next?\n\nMy attempt based on HallsofIvy post\n\n3*2 matrix over the real numbers\n$\\left[\\begin{array}{ccc} a+bi & c+di \\\\ e+fi & g+hi\\\\j+ki & l+mi\\end{array}\\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$ Hence 12 dimensions\n\n3*2 over the complex numbers\n\n$\\left[\\begin{array}{ccc} a+bi & c+di \\\\ e+fi & g+hi\\\\j+ki & l+mi\\end{array}\\right]=(a+bi)(1,0)+(c+di)(0,1)+........$ Hence 6 dimensions.........?\n\n11. ## Re: 3*2 Matrix with complex elements\n\nEvery element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.\n\n12. ## Re: 3*2 Matrix with complex elements\n\nOriginally Posted by Hartlw\nEvery element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.\nFor both the real case and the complex case? Is this not conflicting with post # 7\n\n13. ## Re: 3*2 Matrix with complex elements\n\nOriginally Posted by bugatti79\nFor both the real case and the complex case? Is this not conflicting with post # 7\nFor the real case, every base matrix has a 1 in one element and every other element a 0, for a total of six, the dimension is six. For complex case, see my previous post.\n\nEDIT: I think post seven is discussing a complex vector space consisting of complex vectors a + bi. Not the same thing as a complex vector space each element of which is a matrix. You can probably think of each matrix as the sum of a real (a,0) at each location and imaginary, (,0,b) at each location, component . In any event, any complex matrix can be expressed as a linear combination of the 12 base matrices I described in my previous post.\n\nEDIT: The matrices themselves satisfy requirements of linear vector space with the usual definition of matrix addition and multiplication by a scalar.\n\nThe base matrices are:\n\n(0,1) (0,0)\n(0,0) (0,0)\n(0,0) (0,0)\n\n(1,0) (0,0)\n(0,0) (0,0)\n(0,0) (0,0)\n\n(0,0) (1,0)\n(0,0) (0,0)\n(0,0) (0,0)\n\n..... for 12 base matrices (dim 12).\n\n14. ## Re: 3*2 Matrix with complex elements\n\nOriginally Posted by Hartlw\nI think post seven is discussing a complex vector space consisting of complex vectors a + bi.\nSo based on your post is this the answer to the following original question?\n\nGiven that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalr multiplication of matrices, what is its dimension?\n\n3*2 matrix over the complex numbers\n\n$\\left[\\begin{array}{ccc} a+bi & c+di \\\\ e+fi & g+hi\\\\j+ki & l+mi\\end{array}\\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$\n\n15. ## Re: 3*2 Matrix with complex elements\n\nNo\n\nIf you denote the matrices in my previous post by E1, E2, E3 , E4, ...\nThen an arbitrary complex matrix is expressed as a linear combination of E1, E2, E3,...,E12\n\n(1,3) (5,3)\n(3,5) (5,7) = 1xE1 + 3xE2 + 5xE3 + ..........\n(3,3) (4,3)\n\nAbove is a matrix. The ordered pair (1,3) is just a std shortcut way of expressing 1+i3.\n\nA complex matrix has to be expressed as a sum of complex matrices. Not as a sum of complex numbers. The addition of two complex matrices is a complex matrix by term by term addition, added just like complex numbers.\n\nPage 1 of 3 123 Last", "date": "2015-05-23 09:20:02", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/differential-geometry/190486-3-2-matrix-complex-elements.html", "openwebmath_score": 0.8677421808242798, "openwebmath_perplexity": 437.4005817626351, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924793940119, "lm_q2_score": 0.8840392817460333, "lm_q1q2_score": 0.8592795333460284}}
{"url": "http://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union", "text": "# Empty intersection and empty union\n\nIf $A_\\alpha$ are subsets of a set $S$ then\n\n$\\bigcup_{\\alpha \\in I}A_\\alpha$ = \"all $x \\in S$ so that $x$ is in at least one $A_\\alpha$\"\n\n$\\bigcap_{\\alpha \\in I} A_\\alpha$ = \"all $x \\in S$ so that $x$ is in all $A_\\alpha$\"\n\nIt is the convention that $\\bigcup_{\\alpha \\in \\emptyset}A_\\alpha = \\emptyset$ and $\\bigcap_{\\alpha \\in \\emptyset} A_\\alpha = S$.\n\nBut if $x$ is in $\\bigcap_{\\alpha \\in \\emptyset} A_\\alpha = S$ then $x$ is in all $A_\\alpha$ with $\\alpha \\in \\emptyset$ and therefore $x$ is certainly in at least one $A_\\alpha$ with $\\alpha \\in \\emptyset$. But then $x \\in \\bigcup_{\\alpha \\in I}A_\\alpha$.\n\nCan someone help me and tell me what is wrong with this? Thank you.\n\n-\nThis might help. math.stackexchange.com/q/309986/35983 \u2013\u00a0 Gautam Shenoy Apr 23 at 9:13\nElement in the empty set is the problem. \u2013\u00a0 simplicity Apr 23 at 9:14\n\u2013\u00a0 Martin Sleziak Apr 23 at 11:55\n\nSome texts consider it a convention, but it is in fact a computation!\n\nFor the fixed set $S$, we are looking at the set $\\mathcal P(S)$, the set of all subsets of $S$. With the operations of intersection and unions (of arbitrary many subsets) the set $\\mathcal P(S)$ is what is known as a complete lattice (don't worry if you don't know what that means).\n\nFirst, one can argue intuitively: the more subsets of $S$ you intersect, the smaller the intersection is. Or, the fewer subsets you intersect, the larger the intersection is. So, the intersection of no subsets at all, the least amount of sets you can intersect, should be the largest subset possible. Namely, $\\bigcap_{i\\in \\emptyset}A_i=S$. Similarly, the fewer subsets you take the union of, the smaller the union. So, the union of no subsets at all should be the smallest set possible. Namely, $\\bigcup _{i\\in \\emptyset }A_i=\\emptyset$.\n\nNow, to make things more formal, lets define the intersection and union in $\\mathcal P(S)$. The definition will be equivalent to the set-theoretic definitions but will only make use of the partial order relation of inclusion. Given a collection $\\{A_i\\}_{i\\in I}$ of subsets of $S$, their intersection is the largest subset of $S$ that is contained in each $A_i$ (notice that this is saying that the intersection is a greatest lower bound). Similarly, the union of the family of subsets is the smallest subset of $S$ that contains each $A_i$ (notice that this says that the union is a least upper bound). Incidentally, this point of view very clearly points to a duality between union and intersection.\n\nSo now, the intersection of no subsets is the largest subset of $S$ that is contained in each one of the given subsets. There are no given subsets at all, so (vacuously) any subset $B\\subseteq S$ contains each of the non-existent $A_i$. The largest of those is $S$, proving that $\\bigcap_{i\\in \\emptyset}A_i=S$. Similarly, the union of no subsets is the smallest subset of S that contains each of the given subsets. No subsets are given, so any subsets $B\\subseteq S$ contains each of the non-existent $A_i$. The smallest of these is $\\emptyset$, thus proving that $\\bigcup_{i\\in \\emptyset}A_i=\\emptyset$.\n\n-\nI realized, thank you. \u2013\u00a0 Anna Apr 23 at 9:37\nYou are welcome @Anna \u2013\u00a0 Ittay Weiss Apr 23 at 9:38\nIn the last paragraph: did you mean \"...is contained in each of...\"? \u2013\u00a0 Anna Apr 23 at 10:02\n+1, excellent explanation. \u2013\u00a0 Andreas Caranti Apr 23 at 10:08\n\nPlease note that the fact that\n\n$$\\bigcap_{\\alpha \\in \\emptyset} A_\\alpha = S$$\n\nis not a convention - it follows from the definition\n\n$$\\bigcap_{\\alpha \\in I} A_\\alpha = \\{ x \\in S : \\text{x \\in A_\\alpha, for all \\alpha \\in I}\\}.$$\n\nThis is best understood by asking when is it that for a given $x \\in S$ we have $x \\notin \\bigcap_{\\alpha \\in \\emptyset} A_\\alpha$. This can only happen if there exists $\\alpha \\in \\emptyset$ such that $x \\notin A_{\\alpha}$. Since $\\emptyset$ has no elements, there cannot be such an $\\alpha$.\n\nPS One can avoid introducing a (somewhat arbitrary) index set in an intersection, say. Instead, fix a set $S$, take a subset $\\mathfrak{S}$ of $\\mathcal{P}(S)$, and define $$\\bigcap \\mathfrak{S} = \\{ x \\in S : \\text{x \\in A for all A \\in \\mathfrak{S}} \\}.$$ Then the argument above becomes $$\\bigcap \\emptyset = \\{ x \\in S : \\text{x \\in A for all A \\in \\emptyset} \\} = \\{ x \\in S : \\ \\} = S,$$ since there are no $A$ to consider here.\n\n-\nIt should be pointed out that this intersection is $S$ only if it is given that the computation is performed in $\\mathcal P(S)$. Otherwise, if the computation is performed just in the ambient universe of sets, then the intersection does not exist (since it's trying to be a set that contains every other set). In contrast, the empty union is the empty set no matter where it is computed. \u2013\u00a0 Ittay Weiss Apr 23 at 9:47\n@IttayWeiss, an excellent point. \u2013\u00a0 Andreas Caranti Apr 23 at 9:52\nWhat is the ambient universe of sets? \u2013\u00a0 Anna Apr 23 at 10:06\n\u2013\u00a0 Andreas Caranti Apr 23 at 10:07\nThank you Andreas. I think that my point of confusion is why you can write $\\left\\{ x \\in S : x \\in A \\text{ for all } A \\in \\emptyset \\right\\} = \\left\\{ x \\in S : \\text{} \\right\\}$, whereas you cannot write (considering the case of the union of the empty collection) $\\left\\{ x \\in S : x \\in A \\text{ for at least one } A \\in \\emptyset \\right\\} = \\left\\{ x \\in S : \\text{} \\right\\}$. Would you mind helping me understand the distinction? \u2013\u00a0 Dan Douglas Dec 12 at 15:59\nshow 2 more comments\n\nThose conventions comes from the fact that $(\\mathcal P (S), \\cup, \\emptyset)$ and $(\\mathcal P (S), \\cap, S)$ are monoids.\n\nHence, those conventions are just the usual $\\sum_{k \\in \\emptyset} k = 0$ or $\\prod_{k \\in \\emptyset} k = 1$ that you certainly know for $(\\mathbb N, +)$ or $(\\mathbb Z, \\times)$ for example.\n\n-\nThis is like saying \"it's so because it is so\". In fact, it is not a convention at all. It can be proven so by following the definition. In a monoid though, the unit is (by definition if you like) the result of applying the operation to no elements. But for $\\mathcal P(S)$ it follows from a deeper reason than just saying it's so. \u2013\u00a0 Ittay Weiss Apr 23 at 9:34\n@IttayWeiss The question was not \"Is it a convention or not ?\" but \"Can anyone enlighten me about those conventions/properties ?\". I was just pointing out that the OP was actually familiar with those but did not recognize it at first sight. Plus, I could argue that your proof can actually show that $\\bigcap_\\emptyset A_i = \\emptyset$ : indeed, taking $B$ a subset of $S$, $\\forall A_i \\in \\emptyset, A_i \\not\\subseteq B$ is as true as $\\forall A_i \\in \\emptyset, A_i \\subseteq B$, making $\\emptyset$ the intersection over the emptyset. \u2013\u00a0 Pece Apr 23 at 19:16\nI see your your points with the monoid analogy. If you'll think about it though, you'll see that your criticism of my proof is false (or are you claiming that you just proved a contradiction?) \u2013\u00a0 Ittay Weiss Apr 23 at 19:45\nSorry, I mixed up a little. What I was saying is : as $A_i \\in \\emptyset$ is universally false, your claim $\\forall A_i, A_i \\in \\emptyset \\implies B \\subseteq A_i$ (and so $S$, the greatest of all, is the intersection) is a valid as my claim $\\exists A_i, A_i \\in \\emptyset \\implies B \\not\\subseteq A_i$ (and so no subset of $S$ is a lower bound of the $A_i$ except $\\emptyset$). So in my opinion, you're making a convention on the complete lattices (the meet of nothing is the top) while I was making conventions about monoid. \u2013\u00a0 Pece Apr 24 at 8:34\nI think you are still confused. I did not make use of any convention. I followed the definition. \u2013\u00a0 Ittay Weiss Apr 24 at 8:56\nshow 5 more comments", "date": "2013-12-20 19:16:43", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union", "openwebmath_score": 0.9042208194732666, "openwebmath_perplexity": 289.6879088322624, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399043329856, "lm_q2_score": 0.8856314632529872, "lm_q1q2_score": 0.8592749861808604}}
{"url": "https://math.stackexchange.com/questions/659302/how-to-prove-that-mathbbq-the-rationals-is-a-countable-set", "text": "# How to prove that $\\mathbb{Q}$ ( the rationals) is a countable set\n\nI want to prove that $\\mathbb{Q}$ is countable. So basically, I could find a bijection from $\\mathbb{Q}$ to $\\mathbb{N}$. But I have also recently proved that $\\mathbb{Z}$ is countable, so is it equivalent to find a bijection from $\\mathbb{Q}$ to $\\mathbb{Z}$?\n\n\u2022 Sure. Better yet, it's sufficient to find an injection from $\\mathbb Q$ to $\\mathbb Z$. Or for that matter a surjection from $\\mathbb Z$ to $\\mathbb Q$, or from $\\mathbb N$ to $\\mathbb Q$. \u2013\u00a0bof Feb 1 '14 at 8:25\n\u2022 is just an injection sufficient to prove countability, though? \u2013\u00a0furashu Feb 1 '14 at 8:33\n\u2022 I guess it depends on how you define \"countable\". Are finite sets countable? If not, then besides finding an injection from $\\mathbb Q$ to $\\mathbb Z$, you also have to prove that $\\mathbb Q$ is infinite. But I you already know that. \u2013\u00a0bof Feb 1 '14 at 8:36\n\u2022 i would say finite sets are not \"countable\" but \"finite\" .. seems weird, but countable is a term I would reserve specifically for distinction between cardinality of infinite sets. \u2013\u00a0furashu Feb 1 '14 at 8:40\n\u2022 Fine. $\\mathbb Q$ is an infinite set. An infinite set is countable if it has an injection into $\\mathbb N$. Or into any countable set, such as $\\mathbb Z$, which you already know is countable. \u2013\u00a0bof Feb 1 '14 at 8:46\n\nClearly $\\mathbb{Z}$ injects into $\\mathbb{Q}$.\n\nLet $p_i$ enumerate all the prime numbers.\n\nIf $q \\neq 0, 1, -1$, let $q = \\pm \\frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}}$ be the prime decomposition the numerator and denominator of $q$ written in simplest form. Define\n\n$\\Phi(q) = \\begin{cases} 0 & \\quad q = 0 \\\\ 1 & \\quad q = 1 \\\\ -1 & \\quad q = -1 \\\\ p_{2 i_0}^{n_0} ... p_{2 i_k}^{n_k} p_{2 j_0 + 1}^{m_0} ... p_{2 j_p + 1}^{m_p} & \\quad q = \\frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}} \\\\ - p_{2 i_0}^{n_0} ... p_{2 i_k}^{n_k} p_{2 j_0 + 1}^{m_0} ... p_{2 j_p + 1}^{m_p} & \\quad q = - \\frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}} \\end{cases}$\n\n$\\Phi$ is an injection of $\\mathbb{Q}$ into $\\mathbb{Z}$.\n\nBy the Cantor Schroder Theorem, there is a bijection between $\\mathbb{Z}$ and $\\mathbb{Q}$.\n\nAs bof mentioned, a nicer injection would be\n\n$\\Phi(q) = \\begin{cases} 0 & \\quad q = 0 \\\\ 1 & \\quad q = 1 \\\\ -1 & \\quad q = -1 \\\\ 2^m (2n + 1) & \\quad q = \\frac{m}{n} \\text{ simplest form } \\\\ - 2^m(2n + 1) & \\quad q = - \\frac{m}{n} \\text{ simplest form} \\end{cases}$\n\n\u2022 this is one of the most impressive and elegant answers i've seen on here, well done! \u2013\u00a0furashu Feb 1 '14 at 9:18\n\u2022 Instead of using prime factorizations, you could just map the reduced fraction $\\frac mn$ to the integer $2^m(2n+1)$ \u2013\u00a0bof Feb 1 '14 at 9:23\n\u2022 @bof True. This would have saved me like ten minutes of typing. \u2013\u00a0William Feb 1 '14 at 9:27\n\u2022 @William can you explain what bof said? what's the significance of that integer? \u2013\u00a0furashu Feb 1 '14 at 9:39\n\u2022 @William oh i am checking it now, I just like to know where things coem from, to get a glimpse on peoples intuition and knowledge. \u2013\u00a0furashu Feb 1 '14 at 9:46\n\nIf you know that $\\mathbb{Z}$ is countable, you know there is a bijection $\\chi:\\mathbb{N} \\rightarrow \\mathbb{Z}$. Hence, it is sufficient to find a bijection $\\nu:\\mathbb{Z} \\rightarrow \\mathbb{Q}$ since then $\\chi \\circ \\nu$ is a bijection between $\\mathbb{N}$ and $\\mathbb{Q}$.\n\nIn any case, the following figure illustrates a bijection between $\\mathbb{Z}$ and $\\mathbb{Q}$.\n\nWe follow the worm back and forth \"counting\" the rational numbers, skipping any numbers that are not simplified fractions.\n\nHint: There is a natural map $$\\left\\{ \\begin{array}{ccc} \\mathbb{Z} \\times \\mathbb{Z}_{>0} & \\to & \\mathbb{Q} \\\\ (a,b) & \\mapsto & \\frac{a}{b} \\end{array} \\right.$$\n\n\u2022 Is mapping from Z x Z the same as mapping from Z, though? \u2013\u00a0furashu Feb 1 '14 at 8:44\n\u2022 Lemma : If $A$ is countable, then $A \\times A$ is countable. \u2013\u00a0Euler....IS_ALIVE Feb 1 '14 at 8:53\n\u2022 @Euler....IS_ALIVE How does one prove that? \u2013\u00a0Arjang Jan 19 '17 at 12:23\n\nI read this proof in Amer. Math. Monthly. Suffices to find out an injective function from the set of positive rationals to positive integers. Consider the representation of numbers in base 11, where the 11 digits used are $$0,1,2,3, ..., 9, /$$ (yes, it is a slash as digit for the number ten). Now the rational number $$7/83$$ represents the 4-digit base-11 positive integer: $$(7\\times 11^3) + (10\\times 11^2) + (8\\times 11) + 3$$.\n\n\u2022 I didn't understand this. Can you explain why we can give a completely different meaning to the slash (from its established definition as the division symbol to a numeral in base-11) so as to create an injection. \u2013\u00a0Mr Reality Jan 12 at 20:21\n\u2022 Ok. Write out a rational number which has a division symbol slash, say for example 3/4. Now replace slash by x. (Remember x is roman numeral for ten). So this becomes 3x4 which is a 3-digit number in base 11 system. The value of this is three eleven squared plus ten elevens and 4 which in base 10 is $(3\\times 121 + 10 \\times 11 +4= 363+110+4=477$. SO the function sends $3/4$ to 477. \u2013\u00a0P Vanchinathan Jan 13 at 2:08\n\nConsider for $n=2,3,4,\\ldots$ the sets\n\n$$A_n=\\left\\{\\frac pq :(p,q)=1,p,q>0,p+q=n\\right\\}$$\n\nClaim Each $A_n$ is finite, and $\\Bbb Q^+=A_2\\cup A_3\\cup A_4\\cup\\dots$\n\nA002487 Stern's diatomic series (or Stern-Brocot sequence) Also called fusc(n) [Dijkstra]. a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf]. https://www.cs.utexas.edu/users/EWD/transcriptions/EWD05xx/EWD570.html\n\nfusc(1) = 1\nfusc(2n) = fusc(n)\nfusc(2n+1) = fusc(n) + fusc(n+1)\n\n\nAnd it seems that fusc(k)/fusc(k+1) gives the positive rational numbers. There was also a contrieved proof recently that fusc does what it does here: https://www.isa-afp.org/browser_info/current/AFP/Stern_Brocot/document.pdf\n\nUsing the sign of Z we could create a mapping to the positive and negative rational numbers.", "date": "2019-05-20 16:23:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/659302/how-to-prove-that-mathbbq-the-rationals-is-a-countable-set", "openwebmath_score": 0.844674825668335, "openwebmath_perplexity": 422.84460954533535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795095031688, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8592616765465871}}
{"url": "https://math.stackexchange.com/questions/3124407/number-of-edges-of-a-non-planar-graph-with-fixed-number-of-vertex", "text": "# Number of edges of a non-planar graph with fixed number of vertex\n\nConsider a connected undirected simple non-planar graph $$G$$ with $$15$$ vertices. If removing any edge from $$G$$ results in a planar graph, how many edges does $$G$$ have?\n\nIt is obvious that if the number of edges greater than $$3*15-6=39$$, then it is not planar. But, how to discuss \"removing any edge make it planar\"? Thanks\n\nThe number of edges of $$G$$ is either $$18$$ or $$20$$.\n\nSince graph $$G$$ is non-planar, by Kuratowski's theorem, it contains a subdivision of either $$K_5$$ (the complete graph of $$5$$ vertices) or $$K_{3,3}$$ (the complete bipartite graph of $$6$$ vertices) as a subgraph.\n\nSince removing any edge from $$G$$ make it planar, above subgraph exhaust all edges of $$G$$. This means $$G$$ is a subdivision of either $$K_5$$ or $$K_{3,3}$$.\n\nNotice subdividing a graph increases the number of vertices and edges by same amount.\n\nIf we start from $$K_5$$ which has $$5$$ vertices and $$10$$ edges, we need to subdivide $$10$$ times to get $$15$$ vertices. In this case, the resulting graph will have $$10 + 10 = 20$$ edges.\n\nIf we start from $$K_{3,3}$$ which has $$6$$ vertices and $$9$$ edges, we need to subdivide $$9$$ times to get $$15$$ vertices. In that case, the resulting graph will have $$9+9 = 18$$ edges.\n\n\u2022 Got it! Thank you so much! Rarely use subdivision to count~~ \u2013\u00a0Andy Feb 25 '19 at 11:31\n\nI am not sure if this is the answer you are looking for, but a discussion on this topic can conclude upper and lower bounds for the number of edges in such a graph.\n\nAs you have already stated, we can determine the upper limit using the Euler characteristic, $$v-e+f-c=1$$ which we can reduce to $$v-e+f=2$$ since our graph is connected. Because we know that our original graph $$G$$ is non-planar before removing an edge, we can say that $$\\left|E\\right|=e+1$$. We can use the maximum possible faces of a planer graph with $$v=15$$ to find $$e$$ using $$f_{max}=2v-4 = 26$$ $$15-e+26=2 \\implies \\left|E\\right| \\leq 40 \\leftarrow\\textrm{upper bound}$$\n\nTo find the lower bound, we can use Kuratowski's Theorem (see here for a nice explanation) to state a lower bound, $$\\left|E\\right| \\geq v+3 = 18$$.\n\nSo, we can say that for a non-planar graph $$G$$, with $$\\left|V\\right|=15$$ and the property that removing an edge must make the result planar, has $$18\\leq\\left|E\\right|\\leq 40$$ edges.\n\nEssentially, if the number of edges is less than the lower bound, we know that the graph must be planar. If the number of edges is greater than the upper bound, we would have to remove more than one edge to make it planar.\n\n\u2022 Got it! Thank you so much. You gave a very detailed description. \u2013\u00a0Andy Feb 25 '19 at 11:35", "date": "2020-02-29 10:57:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3124407/number-of-edges-of-a-non-planar-graph-with-fixed-number-of-vertex", "openwebmath_score": 0.6310502886772156, "openwebmath_perplexity": 80.63002838176504, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795073967326, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.859261663114625}}
{"url": "https://math.stackexchange.com/questions/3079038/expected-number-of-steps-in-a-1d-random-walk-with-reflecting-edges/3080351", "text": "# Expected number of steps in a 1D random walk with reflecting edges\n\nAssume there is a row of $$k$$ tiles. A creature (monkey in some situations, ant in others, frog in others) lies on tile $$a$$. There is a 50% probability that the creature jumps to tile $$a-1$$ and a 50% probability that the creature jumps to tile $$a+1$$, unless it is on an edge tile. If it is on an edge tile, it must jump inwards, so it can't escape the system (i.e. tile 2 from tile 1 and tile $$k-1$$ from tile $$k$$). What is the expected number of steps for it to first reach tile $$b$$? $$1<=a, b<=k$$ is assumed. I feel like Markov chains might be used to get the answer, but I have a very limited understanding of them. If there is a closed form for the answer as well as a derivation for understanding, that would be perfect.\n\n\u2022 Markov chains are the way! \u2013\u00a0Lord Shark the Unknown Jan 19 at 5:20\n\u2022 @LordSharktheUnknown Can you elaborate in an answer how to use them in order to solve this problem? Like I said, I have a limited understanding of them. \u2013\u00a0automaticallyGenerated Jan 19 at 5:31\n\u2022 I'have taken the liberty to modifiy your title and tags (3 tags are a good average) in order more readers are directed towards this interesting question and its interesting answer \u2013\u00a0Jean Marie Feb 15 at 12:27\n\nAmazingly (to me) there happens to be a very simple expression for the expected number of steps to reach $$\\ b\\$$ from $$\\ a\\$$. For $$\\ a < b\\$$, it is: $$\\left(\\,b + a -2\\,\\right)\\,\\left(\\,b-a\\,\\right)\\ .$$ Although a Markov chain is the obvious way to model the process, and I'm sure it could used to derive the above result, there turns out to be a less cumbersome way of doing it.\n\nFor each $$\\ i\\$$ between $$\\ 1\\$$ and $$\\ b\\$$ inclusive, let $$\\ e_i\\$$ be the expected number of steps the creature takes to go from $$\\ i\\$$ to $$\\ b\\$$. Obviously, $$\\ e_b\\ = 0\\$$.\n\nIf the creature starts from $$\\ 1\\$$, then it has to take one step to $$\\ 2\\$$, from which the expected number of steps to reach $$\\ b\\$$ is $$\\ e_2\\$$. Thus, the expected number of steps, $$\\ e_1\\$$, to reach $$\\ b\\$$ from $$\\ 1\\$$ is $$\\ e_2 + 1\\$$.\n\nIf the creature starts from $$\\ b-1\\$$, then with probability $$\\ \\frac{1}{2}\\$$ it reaches $$\\ b\\$$ on the very next step\u2014that is, in just a single step\u2014, and with probability $$\\ \\frac{1}{2}\\$$ it jumps to $$\\ b-2\\$$, from which the expected number of steps to reach $$\\ b\\$$ is $$\\ e_{b-2}\\$$. Thus $$\\ e_{b-1} = \\frac{1}{2}\\left(e_{b-2} +1\\right) + \\frac{1}{2}\\,1=\\frac{1}{2}\\,e_{b-2}+1\\$$.\n\nIf the creature starts from any other point $$\\ i\\$$, with $$\\ 2\\le i\\le b-2\\$$, then with probability $$\\ \\frac{1}{2}\\$$ it jumps to $$\\ i-1\\$$, from which the expected number of steps to reach $$\\ b\\$$ is $$\\ e_{i-1}\\$$, and with probability $$\\ \\frac{1}{2}\\$$ it jumps to $$\\ i+1\\$$, from which the expected number of steps to reach $$\\ b\\$$ is $$\\ e_{i+1}\\$$. Therefore, $$\\ e_i = \\frac{1}{2}\\left(e_{i-1} +1\\right) + \\frac{1}{2}\\left(e_{i+1} +1\\right)= \\frac{1}{2}\\,e_{i-1} + \\frac{1}{2}\\,e_{i+1} +1\\$$.\n\nPutting this all together, we have\n\n$$\\begin{eqnarray} e_1 &=& e_2 + 1\\\\ e_i &=& \\frac{1}{2}\\,e_{i-1} + \\frac{1}{2}\\,e_{i+1} +1, \\ \\ \\mbox{for } i=2,3, \\dots, b-2\\ \\mbox{, and}\\\\ e_{b-1} &=& \\frac{1}{2}\\,e_{b-2}+1\\ , \\end{eqnarray}$$ or, equivalently,\n\n$$\\begin{eqnarray} e_1 - e_2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &=& 1\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ -\\frac{1}{2}\\,e_{i-1} + e_i -\\frac{1}{2}\\,e_{i+1} &=& 1, \\ \\ \\mbox{for } i=2,3, \\dots, b-2\\ \\mbox{, and}\\\\ -\\frac{1}{2}\\,e_{b-2}+e_{b-1} &=& 1\\ . \\end{eqnarray}$$\n\nThese equations can be written as: $$M\\,e = \\mathbb 1\\ ,$$ where $$\\ M\\$$ is the $$\\ \\left(\\,b-1\\,\\right)\\times\\left(\\,b-1\\,\\right)\\$$ matrix, and $$\\ \\mathbb 1\\$$ the $$\\ \\left(\\,b-1\\,\\right)\\times\\,1\\$$ column vector, whose entries are given by: $$\\begin{eqnarray} M_{1,2} &=& -1\\\\ M_{i,i} &=& 1\\ \\ \\mbox{for } i=1,2,\\dots, b-1\\\\ M_{i,i-1} &=& -\\frac{1}{2}\\ \\ \\mbox{for } i=2,3,\\dots, b-1\\\\ M_{i,i+1} &=& -\\frac{1}{2}\\ \\ \\mbox{for } i=2,3,\\dots, b-2\\\\ M_{i,\\,j} &=& 0 \\ \\ \\mbox{for all other }\\ i, j\\\\ \\mathbb 1_i &=& 1\\ \\ \\mbox{for } i=1,2,\\dots, b-1\\ . \\end{eqnarray}$$ For $$\\ b=6\\$$, the matrix $$\\ M\\$$ looks like this: $$\\left(\\begin{matrix}1&-1&0&0&0 \\\\ -\\frac{1}{2}&1&-\\frac{1}{2}&0&0\\\\ 0&-\\frac{1}{2}&1&-\\frac{1}{2}&0\\\\ 0&0&-\\frac{1}{2}&1&-\\frac{1}{2}\\\\ 0&0&0&-\\frac{1}{2}&1& \\end{matrix}\\right)\\ ,$$ and has the following inverse: $$M^{-1} = \\left(\\begin{matrix} 5&8&6&4&2\\\\ 4&8&6&4&2\\\\ 3&6&6&4&2\\\\ 2&4&4&4&2\\\\ 1&2&2&2&2\\\\ \\end{matrix}\\right)\\ .$$ From this, we can conjecture that the entries of the inverse of the $$\\ \\left(\\,b-1\\,\\right)\\times\\left(\\,b-1\\,\\right)\\$$ matrix $$\\ M\\$$, defined above, should be the matrix $$\\ L\\$$ whose entries are given by: $$\\begin{eqnarray} L_{i,1} &=& b-i\\ \\ \\mbox{for } i=1,2,\\dots, b-1\\\\ L_{1,\\,j} &=& 2\\,\\left(b-j\\right) \\ \\mbox{for } j=2,3,\\dots, b-1\\\\ L_{i,\\,j} &=& 2\\,\\min\\left(b-i,b-j\\right) \\ \\mbox{for } 2\\le i\\le b-1\\ \\ \\mbox{and }\\ 2\\le j\\le b-1\\ , \\end{eqnarray}$$ and on checking the product $$\\ M\\,L\\$$, we find that it is indeed the $$\\ \\left(\\,b-1\\,\\right)\\times\\left(\\,b-1\\,\\right)\\$$ identity matrix. So, finally, we have: $$e = M^{-1}\\,\\mathbb 1 = L\\,\\mathbb 1\\ ,$$ and $$\\ e_a\\$$, the expected number of steps to get to $$\\ b\\$$ from $$\\ a\\$$ is the sum of the entries in the $$\\ a^\\mbox{th}\\$$ row of $$\\ L\\$$: $$\\begin{eqnarray} e_a &=& \\left(b-a\\right) + 2\\,\\left(\\,a-1\\,\\right)\\,\\left(\\,b-a\\,\\right) + 2\\,\\sum_{j=1}^{b-a-1} j\\\\ &=& \\left(\\,b + a -2\\,\\right)\\,\\left(\\,b-a\\,\\right)\\ , \\end{eqnarray}$$ as stated above.\n\n\u2022 Thanks for the clear answer. I think that this is just a disguised Markov chain, as your \"M\" is the identity matrix minus the transition matrix. \u2013\u00a0automaticallyGenerated Jan 21 at 14:30\n\u2022 Not quite. The identity matrix minus the transition matrix is a singular $\\ b\\times b\\$ matrix. $\\ M\\$ is the submatrix obtained from it by chopping off its last row and column. But you're right that there is indeed a Markov chain lurking in the shadows. \u2013\u00a0lonza leggiera Jan 21 at 23:27", "date": "2019-04-24 06:37:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3079038/expected-number-of-steps-in-a-1d-random-walk-with-reflecting-edges/3080351", "openwebmath_score": 0.8779345750808716, "openwebmath_perplexity": 122.11497428699461, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145701859179, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8592450674312874}}
{"url": "https://byjus.com/question-answer/displaystyle-int-0-1-frac-dx-1-x-2-sqrt-2-x-2-frac-pi/", "text": "Question\n\n# $$\\displaystyle \\int_{0}^{1}\\frac{dx}{(1+x^{2})\\sqrt{(2+x^{2})}}=\\frac{\\pi }{k}$$. Find the value of $$k$$.\n\nSolution\n\n## Let $$\\displaystyle I=\\int { \\frac { 1 }{ \\left( { x }^{ 2 }+1 \\right) \\sqrt { { x }^{ 2 }+2 } } dx }$$Put $$x=\\sqrt { 2 } \\tan { t } \\Rightarrow dx=\\sqrt { 2 } \\sec ^{ 2 }{ t } dt$$$$\\displaystyle I=\\sqrt { 2 } \\int { \\frac { sec{ t } }{ \\sqrt { 2 } \\left( 2\\tan ^{ 2 }{ t } +1 \\right) } dt } =\\int { \\frac { sec{ t } }{ 2\\tan ^{ 2 }{ t } +1 } dt }$$Multiplying numerator and denominator by $$\\cos ^{ 2 }{ t }$$, we get$$\\displaystyle I=\\int { \\frac { \\cos { t } }{ 2\\sin ^{ 2 }{ t } +\\cos ^{ 2 }{ t } } dt } =\\int { \\frac { \\cos { t } }{ \\sin ^{ 2 }{ t } +1 } dt }$$Now put $$u=\\sin { t } \\Rightarrow du=\\cos { t } dt$$Therefore$$\\displaystyle I=\\int { \\frac { 1 }{ { u }^{ 2 }+1 } du } =\\tan ^{ -1 }{ u } =\\tan ^{ -1 }{ \\sin { t } }$$$$\\displaystyle =\\tan ^{ -1 }{ \\sin { \\left( \\tan ^{ -1 }{ \\left( \\frac { x }{ \\sqrt { 2 } } \\right) } \\right) } } =\\tan ^{ -1 }{ \\left( \\frac { x }{ \\sqrt { { x }^{ 2 }+2 } } \\right) }$$Hence$$\\displaystyle \\int _{ 0 }^{ 1 }{ \\frac { 1 }{ \\left( { x }^{ 2 }+1 \\right) \\sqrt { { x }^{ 2 }+2 } } dx } ={ \\left[ \\tan ^{ -1 }{ \\left( \\frac { x }{ \\sqrt { { x }^{ 2 }+2 } } \\right) } \\right] }_{ 0 }^{ 1 }$$$$\\displaystyle =\\frac { \\pi }{ 6 } -0=\\frac { \\pi }{ 6 }$$$$\\Rightarrow k=6$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-18 07:01:14", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/displaystyle-int-0-1-frac-dx-1-x-2-sqrt-2-x-2-frac-pi/", "openwebmath_score": 0.9704388380050659, "openwebmath_perplexity": 9592.494076384153, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145712474397, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.859245064946783}}
{"url": "https://math.stackexchange.com/questions/1668254/how-many-cubes-have-at-least-1-2-3-colors-on-them/1668271", "text": "# how many cubes have at least $1,2,3$ colors on them\n\nI have a painted cube, which is cut into $n^3$ smaller cubes. I now want to find the number of cubes which have $1$,$2$,$3$ sides painted. I know the long way round of taking each cube and putting them into different categories... but is there a short way or formula to do it?\n\n$1$ face painted - You have to consider all the faces. There are $6$ faces, each with $(n-2)^2$ cubes with $1$ face painted . That gives you the formula $$6*(n-2)^2$$\n\n$2$ faces painted - You have to take the edges. There are $12$ edges, each edge having $n-2$ cubes with $2$ faces painted . That gives you the formula $$12*(n-2)$$\n\n$3$ faces painted - You have to take the corners alone. That gives you the formula $$8$$\n\nExtra - $0$ faces painted - You have to consider the interior alone which gives $$(n-2)^3$$\n\n\u2022 I think that you want $8$ corners. ;-) \u2013\u00a0Sammy Black Feb 23 '16 at 7:20\n\u2022 @SammyBlack Thank you for that. Edited. \u2013\u00a0Win Vineeth Feb 23 '16 at 7:20\n\u2022 saying side is not corrrect it is face which is painted \u2013\u00a0Bhaskara-III Feb 23 '16 at 13:17\n\u2022 @Bhaskara-III The question asked for side, hence, I used side. \u2013\u00a0Win Vineeth Feb 23 '16 at 13:20\n\u2022 oops you should have correctly mentioned that it should be a painted face not a painted side in your answer because it's very confusing term \u2013\u00a0Bhaskara-III Feb 23 '16 at 13:22\n\nThis is an addendum to Win Vineeth's excellent answer: the particular counts fall out of the algebraic expansion (using the Binomial Theorem) of $n^3$, where we write $n = (n-2) + 2$: $$\\bigl( (n-2) + 2 \\bigr)^3 = (n-2)^3 + 6(n-2)^2 + 12(n-2) + 8$$\n\nThis approach generalizes by letting $t$ be the \"thickness\" of the boundary (the part that gets painted). The original question uses $t=1$. The expansion now looks like: $$\\bigl( (n-2t) + 2t \\bigr)^3 = (n-2t)^3 + 6(n-2t)^2t + 12(n-2t)t^2 + 8t^3$$ Now, the power of $t$ serves as a placeholder for the number of sides that got painted. For example, if you want to know how many subcubes got painted on $2$ sides, then look at the coefficient of $t^2$, namely $12(n-2t)$, which is $12(n-2)$ once you set $t=1$. This is an example of a generating function, where a sequence is encoded as the list of coefficients of a polynomial (or often infinite series).\n\nThis approach also has the advantage of counting the number of boundary subhypercubes of a hypercube of any dimension: just expand $n^d$, where $d$ is the dimension.", "date": "2021-01-21 18:50:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1668254/how-many-cubes-have-at-least-1-2-3-colors-on-them/1668271", "openwebmath_score": 0.776066780090332, "openwebmath_perplexity": 426.9438790417853, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145699205374, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8592450586889886}}
{"url": "https://byjus.com/question-answer/let-a-be-a-set-containing-n-elements-a-subset-p-of-the-set-a-1/", "text": "Question\n\nLet $$A$$ be a set containing $$n$$ elements. A subset $$P$$ of the set $$A$$ is chosen at random.The set $$A$$ is reconstructed by replacing the element of $$P$$, and another subsets $$Q$$ of $$A$$ is chosen at random. The probability that $$\\left( P\\cap Q \\right)$$ contains exactly $$m(m<n)$$ elements is\n\nA\n3nm4n\nB\nnCm3m4n\nC\nnCm3nm4n\nD\nnone of these\n\nSolution\n\nThe correct option is C $$\\displaystyle \\frac { _{ }^{ n }{ { C }_{ m } }{ 3 }^{ n-m } }{ { 4 }^{ n } }$$We know that the number of subsets of a set containing $$n$$ elements is $${ 2 }^{ n }$$.Therefore, the number of ways of choosing $$P$$ and $$Q$$ is $$^{ { 2 }^{ n } }{ { C }_{ 1 } }\\times _{ }^{ { 2 }^{ n } }{ { C }_{ 1 } }={ 2 }^{ n }\\times { 2 }^{ n }={ 4 }^{ n }$$.Out of $$n$$ elements, $$m$$ elements can be chosen in $$^{ n }{ { C }_{ m } }$$ ways.If $$\\left( P\\cap Q \\right)$$ contains exactly in elements,\u00a0then from the remaining $$n-m$$ elements either an element belongs to $$P$$ or $$Q$$ but not both $$P$$ and $$Q$$.Suppose $$P$$ contains $$r$$ elements from the remaining $$n-m$$ elements.Then $$Q$$ may contain any number of elements from the remaining $$(n-m)-r$$ elements.\u00a0Therefore, $$P$$ and $$Q$$ can be chosen in $$^{ n-m }{ { C }_{ r } }{ 2 }^{ \\left( n-m \\right) -r }$$But $$r$$ can vary from $$0$$ to $$(n-m)$$. So, $$P$$ and $$Q$$ can be chosen in general in$$\\displaystyle \\left( \\sum _{ r=0 }^{ n-m }{ ^{ n-m }{ { C }_{ r } }{ 2 }^{ \\left( n-m \\right) -r } } \\right) _{ }^{ n }{ { C }_{ m } }={ \\left( 1+2 \\right) }^{ n-m }\\times _{ }^{ n }{ { C }_{ m } }=_{ }^{ n }{ { C }_{ m } }\\times { 3 }^{ n-m }$$Hence, required probability $$\\displaystyle =\\dfrac { ^{ n }{ { C }_{ m } }\\times { 3 }^{ n-m } }{ { 4 }^{ n } }$$Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-27 18:47:34", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/let-a-be-a-set-containing-n-elements-a-subset-p-of-the-set-a-1/", "openwebmath_score": 0.9689419865608215, "openwebmath_perplexity": 100.555119252128, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.992988205863518, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.8591573163299295}}
{"url": "http://mathhelpforum.com/pre-calculus/15788-two-pumps-one-tank-print.html", "text": "# Two Pumps, One Tank\n\n\u2022 Jun 10th 2007, 04:04 AM\nblueridge\nTwo Pumps, One Tank\nTwo pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?\n\u2022 Jun 10th 2007, 04:10 AM\nQuick\nQuote:\n\nOriginally Posted by blueridge\nTwo pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?\n\nI'll start you off...\n\nDefine your variables: I'd call the bigger pump $b$ and the smaller pump $s$\n\n$b=$the number of tanks per hour the bigger pump can empty\n\n$s=$the number of tanks per hour the smaller pump can empty\n\nBoth of those numbers will be fractions.\n\nNow, since you only wanted a hint, I'm only going to give you the first equation and you'll have to find the rest...\n\nQuote:\n\nTwo pumps of different sizes working together can empty a fuel tank is 5 hours.\n$b+s=\\frac{1\\text{tank}}{5\\text{hours}}$\n\nDo you need any more help?\n\u2022 Jun 10th 2007, 04:12 AM\nblueridge\nYes\nIf you can set up the other equation, I can take it from there.\n\u2022 Jun 10th 2007, 04:22 AM\nQuick\nQuote:\n\nOriginally Posted by blueridge\nIf you can set up the other equation, I can take it from there.\n\nQuote:\n\nOriginally Posted by blueridge\nTwo pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?\n\nThe next equation is somewhat weird.\n\nIf $b=\\frac{\\text{tanks}}{\\text{hours}}$\n\nthan $\\frac{1}{b}=\\frac{\\text{hours}}{\\text{tanks}}$\n\nSo in fact:\n\n$\\frac{1}{b}=$the number of hours to empty a tank\n\nSo we know that: $\\frac{1}{b}-\\frac{1}{s}=4$\n\u2022 Jun 10th 2007, 05:54 AM\nblueridge\ntell me...\nI am dealing with two equations in two unknowns?\n\u2022 Jun 10th 2007, 06:59 AM\nCaptainBlack\nQuote:\n\nOriginally Posted by blueridge\nI am dealing with two equations in two unknowns?\n\nYes, solve for $b$ and $s$, and the required answer is $s$\n\nRonL\n\u2022 Jun 10th 2007, 07:39 AM\nSoroban\nHello, blueridge!\n\nHere's another approach . . .\n\nQuote:\n\nTwo pumps of different sizes working together can empty a fuel tank is 5 hours.\nThe larger pump can empty this tank in 4 hours less than the smaller one.\nIf the larger one is out of order, how long will it take the smaller one to do the job alone?\n\nTogether, they can do the job in 5 hours.\n. . In one hour, they can do $\\frac{1}{5}$ of the job. .[1]\n\nThe smaller pump can do the job in $x$ hours. .[Note that: . $x > 4$.]\n. . In one hour, it can do $\\frac{1}{x}$ of the job.\n\nThe larger pump takes 4 hours less; it takes $x - 4$ hours.\n. . In one hour, it can do $\\frac{1}{x-4}$ of the job.\nTogether, in one hour, they can do: . $\\frac{1}{x} + \\frac{1}{x-4}$ of the job. .[2]\n\nBut [1] and [2] describe the same thing:\n. . the fraction of the job done in one hour.\n\nThere is our equaton! . . . . $\\boxed{\\frac{1}{x} + \\frac{1}{x-4} \\:=\\:\\frac{1}{5}}$\n\nMultiply by the common denominator: $5x(x - 4)$\n. . $5(x - 4) + 5x \\:=\\:x(x-4)$\n\n. . which simplifies to the quadratic: . $x^2 - 14x + 20 \\:=\\:0$\n\nThe Quadratic Formula gives us: . $x \\;=\\;\\frac{14\\pm\\sqrt{116}}{2} \\;=\\;7 \\pm\\sqrt{29} \\;\\approx\\;\\{1.6,\\:12.4\\}$\n\nSince $x > 4$. the solution is: . $x = 12.4$\n\nTherefore, the smaller pump will take about 12.4 hours working alone.\n\n\u2022 Jun 10th 2007, 11:14 AM\nblueridge\ntell me...\nSoroban,\n\nI thank you for sharing yet a more simplistic avenue to understanding this question.", "date": "2016-09-29 12:11:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/15788-two-pumps-one-tank-print.html", "openwebmath_score": 0.8537145853042603, "openwebmath_perplexity": 1097.6265715216343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9929882042034978, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.85915731489364}}
{"url": "http://mathhelpforum.com/pre-calculus/71232-simple-expression-trig-matrix-cubed.html", "text": "# Math Help - Simple expression for trig matrix cubed\n\n1. ## Simple expression for trig matrix cubed\n\nDoes anyone know how to determine a simple expression for $A^3$, where A is the following matrix?\n\ncos@ sin@\n-sin@ cos@\n\n*I used @ as theta\n\nI think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!\n\n2. Originally Posted by jennifer1004\nDoes anyone know how to determine a simple expression for $A^3$, where A is the following matrix?\n\ncos@ sin@\n-sin@ cos@\n\n*I used @ as theta\n\nI think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!\nA is the standard rotation matrix by angle @. A^3 = A.A.A => rotate by 3@. Therefore A^3 = .....\n\nOriginally Posted by jennifer1004\nDoes anyone know how to determine a simple expression for $A^3$, where A is the following matrix?\n\ncos@ sin@\n-sin@ cos@\n\n*I used @ as theta\n\nI think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!\nPlease see attached file\n\n4. ## Thank you!\n\nYou have helped me so much. There is nothing in my text for linear algebra that looks anything like this matrix. I wasn't sure how to simplify the sines and cosines. Thanks again. Your work is very clear!\n\n5. Hello, jennifer1004!\n\nWe are expected to know some basic identities:\n\n. . $\\begin{array}{cccccc}\\cos^2\\!x-\\sin^2\\!x \\:=\\:\\cos2x & & \\cos x\\cos y - \\sin x\\sin y \\:=\\: \\cos(x+y) \\\\\n2\\sin x\\cos x \\:=\\: \\sin2x & & \\sin x\\cos y + \\cos x\\sin y \\:=\\:\\sin(x+y) \\end{array}$\n\nGiven: . $A \\:=\\:\\begin{bmatrix}\\cos\\theta & \\sin\\theta \\\\ \\text{-}\\sin\\theta & \\cos\\theta \\end{bmatrix}$\n\nFind $A^3$\n\n$A^2 \\:=\\:\\begin{bmatrix}\\cos\\theta & \\sin\\theta \\\\ \\text{-}\\sin\\theta & \\cos\\theta\\end{bmatrix}\\begin{bmatrix}\\cos\\theta & \\sin\\theta \\\\ \\text{-}\\sin\\theta & \\cos\\theta \\end{bmatrix}$ . $= \\;\\begin{bmatrix}\\cos^2\\!\\theta-\\sin^2\\!\\theta & 2\\sin\\theta\\cos\\theta \\\\ \\text{-}2\\sin\\theta\\cos\\theta & \\cos^2\\!\\theta -\\sin^2\\!\\theta \\end{bmatrix}$ . $= \\;\\begin{bmatrix}\\cos2\\theta & \\sin2\\theta \\\\ \\text{-}\\sin2\\theta & \\cos2\\theta\\end{bmatrix}$\n\n$A^3 \\;=\\;\\begin{bmatrix}\\cos\\theta & \\sin\\theta \\\\ \\text{-}\\sin\\theta & \\cos\\theta \\end{bmatrix}\\begin{bmatrix}\\cos2\\theta & \\sin2\\theta \\\\ \\text{-}\\sin2\\theta & \\cos2\\theta \\end{bmatrix}$\n\n. . $= \\;\\begin{bmatrix}\\cos\\theta\\cos2\\theta - \\sin\\theta\\sin2\\theta & \\sin2\\theta\\cos\\theta + \\sin\\theta\\cos2\\theta \\\\ \\text{-}\\sin\\theta\\cos2\\theta - \\sin2\\theta\\cos\\theta & \\cos\\theta\\cos2\\theta - \\sin\\theta\\sin2\\theta \\end{bmatrix}$\n\n. . $= \\;\\begin{bmatrix}\\cos3\\theta & \\sin3\\theta \\\\ \\text{-}\\sin3\\theta & \\cos3\\theta \\end{bmatrix}$\n\n6. ## Thanks Soroban!\n\nThose identities are what I was looking for. I wasn't sure how to simplify. Thanks for sharing!", "date": "2016-07-27 02:30:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/71232-simple-expression-trig-matrix-cubed.html", "openwebmath_score": 0.8686792850494385, "openwebmath_perplexity": 1240.7667790699838, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9683812341001313, "lm_q2_score": 0.8872046004274976, "lm_q1q2_score": 0.859152285861294}}
{"url": "https://math.stackexchange.com/questions/2535591/calculating-eigenvectors-is-my-book-wrong/2535601", "text": "# Calculating Eigenvectors: Is my book wrong?\n\nI have a covariance matrix: $$S= \\begin{pmatrix} 16 & 10 \\\\ 10 & 25 \\end{pmatrix}$$\n\nI calculate my eigenvalues correctly (the same as what the book finds);\n\n$\\lambda_1 = 31.47$ , $\\lambda_2 = 9.53$\n\nBut now it comes to calculating eigenvectors: I do everything as I was taught way back in Elementary Linear Algebra.\n\n1. $S X = \\lambda v$ {where v is the eigenvector}\n\n2. $(S - I \\lambda)v$\n\n3. Get Row-Echelon Form\n\nBut when I do this I get the following reduced matrix:\n\n$$\\begin{pmatrix} 1 & -.646412 & 0 \\\\ 0 & 0 &0 \\end{pmatrix}$$\n\nBut this result doesn't seem consistent with my textbook which says that the eigenvectors are;\n\n$(0.54 , 0.84)^T$ and $(0.84 , -0.54)$\n\nI looked online for calculators and found one consistent with the book and a few consistent with my result:\n\nConsistent with Book: http://comnuan.com/cmnn01002/\n\nConsistent with Me: http://www.arndt-bruenner.de/mathe/scripts/engl_eigenwert2.htm\n\nAny ideas?\n\nAdditional Information:\n\n\u2022 This problem stems from Principal Component Analysis\n\u2022 $(0.54,0.84) = (9,14) \\cdot 0.06,$ and $9/14 \\approx 0.642857\\ldots,$ so I wonder if you get $0.646412$ by rounding too early, before dividing. Maybe your reduced matrix should have been $\\left[ \\begin{array}{ccc} 1 & -0.642857\\ldots & 0 \\\\ 0 & 0 & 0 \\end{array} \\right],$ which would be consistent with the answer being $(0.54,0.84). \\qquad$ \u2013\u00a0Michael Hardy Nov 24 '17 at 19:10\n\u2022 Specifically, how did you get $-0.646412\\text{ ?} \\qquad$ \u2013\u00a0Michael Hardy Nov 24 '17 at 19:12\n\u2022 I answered your question in the answers section. You were right that I rounded early. \u2013\u00a0Nicklovn Nov 24 '17 at 19:15\n\u2022 The textbook has normalized the eigenvectors so that the change-of-basis matrix is orthogonal. Remember, there\u2019s no such thing as the eigenvector(s): any non-zero scalar multiple of an eigenvector is also an eigenvector. \u2013\u00a0amd Nov 25 '17 at 0:08\n\n## 4 Answers\n\nTLDR: The answers are the same.\n\nThe vectors $(0.646586,1)$ and $(0.54,0.84)$ go in (almost) the same direction (the only differences due to rounding and the magnitude of the vector). The first has the benefit of one of the entries equalling one. The second has the benefit that its magnitude is (almost) $1$, but they both give essentially the same information.\n\nRemember that an eigenvector for a specific eigenvalue $\\lambda$ is any vector such that $Av=\\lambda v$ and these vectors collectively make up an entire subspace of your vector space, referred to as the eigenspace for the eigenvector $\\lambda$. In the problem of determining eigenvalues and corresponding eigenvectors, you need only find some collection of eigenvectors such that they form a basis for each corresponding eigenspace. There are infinitely many correct choices for such eigenvectors.\n\nEigenvector is not unique.\n\nNotice that non-zero scalar multiple of an eigenvector is still an eigenvector.\n\nBoth answers are correct.\n\nBoth answers are correct. The eigen vector you computed does not have unity norm. If you normalize your eigen vector then you will get the text-book answer.\n\n$(0.54,0.84) = (9,14) \\cdot 0.06,$ and $9/14 \\approx 0.642857\\ldots,$ so I wonder if you get $0.646412$ by rounding too early, before dividing. Maybe your reduced matrix should have been $\\left[ \\begin{array}{ccc} 1 & -0.642857\\ldots & 0 \\\\ 0 & 0 & 0 \\end{array} \\right],$ which would be consistent with the answer being $(0.54,0.84).$\n\n\u2022 Yes I admit that I rounded early. What happened is that I calculated the eigenvalues a few days ago but I lost that paper and just went with the eigenvalues listed in the book (theirs being rounded already). But I was informed that I had only to normalize the vector I had calculated to get the textbook answer. \u2013\u00a0Nicklovn Nov 24 '17 at 19:14\n\u2022 @Nicklovn : This is one respect in which elementary arithmetic gets far less respect than it deserves. People round without knowing to what extent or in what way the bottom line will be affected, and then they write six or eight or ten digits in their bottom-line answer when the early rounding has the effect that only perhaps the first one or two are truthful and the rest of meaningless noise. \u2013\u00a0Michael Hardy Nov 24 '17 at 19:22\n\u2022 I agree. I was doing well with it at first but when I got frustrated and lost my sheet I just relied on what the book gave me which was naturally rounded :( Thanks for your help! \u2013\u00a0Nicklovn Nov 25 '17 at 21:30\n\u2022 @Nicklovn : I'm glad it helped. \u2013\u00a0Michael Hardy Nov 25 '17 at 23:19", "date": "2019-07-19 06:06:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2535591/calculating-eigenvectors-is-my-book-wrong/2535601", "openwebmath_score": 0.8649722337722778, "openwebmath_perplexity": 340.83439094490933, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812309063186, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8591522823057045}}
{"url": "https://math.stackexchange.com/questions/1120819/n-distinguishable-balls-into-n-boxes", "text": "# n distinguishable balls into n boxes\n\nWe have n distinguishable balls (say they have different labels or colours). If these balls are dropped at random in n boxes, what is the probability that:\n\n1- No box is empty? 2- Exactly one box is empty?\n\nFor 1, I figured that we have $n^n$ ways to put the n balls into the n boxes. And I figured there are $n!$ to sort the balls so there is one ball for each box.\n\nSo is the answer to question 1 $n!/(n^n)$?\n\nFor 2, there are $n-1$ ways for the boxes to be empty. This is because you can have box 1 be empty (and just that), box 2 be empty and just that, so ultimately you can have at most $n-1$ variations of empty boxes.\n\nSo I figured the solution to part 2 was $\\frac{n-1}{n^n}$\n\nIs any of this right?\n\n\u2022 I think your part two is incorrect, because for each box $n$ that is empty, there are $n-1$ boxes that could have the extra ball. Jan 26 '15 at 20:07\n\u2022 That sounds right...I'll edit my answer. Jan 26 '15 at 20:12\n\nFor part 2, you must place the balls so that there is one empty box, one box with two balls, and the remaining balls will have one ball each. There are $n$ ways to pick the empty box, and $n-1$ ways to then pick the box with two balls.\n\nWe can now fill the $n$ spaces for the balls with the $n$ balls in any order you wish. There are of course $n!$ ways to do this. However, there is a caveat. If the box with two spaces is filled with ball $a$ and then ball $b$, that is the same as if we put ball $b$ and then ball $a$. So we are double counting, and the number of ways to fill in the boxes is $\\frac{n!}{2}$.\n\nThus the total number of ways to have one empty box is $\\frac{n(n-1)(n!)}{2}$, so the probability of having one empty box is $\\frac{\\frac{n(n-1)(n!)}{2}}{n^n} = \\frac{(n-1)(n!)}{2n^{n-1}}$.\n\n\u2022 It seems like order matters, so why would we divide by 1/2? Jan 26 '15 at 22:00\n\u2022 Of the $n^n$ possibilities, all that matters is which box each ball goes into. When we consider the $n!$ ways to put $n$ boxes into the $n$ spaces, we have two spaces in one box. So by that counting, if we put ball $a$ into space 1 of that box, and ball $b$ into space 2 of that box, that's counted separately from putting ball $b$ into space 1 and ball $a$ into space 2. But in the $n^n$ counting it's just counted once, since both balls go into the same box. So we have to divide by 2 to match up the counting. Jan 27 '15 at 0:55\n\u2022 See the answer I posted above from the textbook. Jan 27 '15 at 17:56\n\u2022 Yes, the answer is the same: $\\frac{n!}{2} = {n \\choose 2} (n-2)!$. The book solution just picks the two balls in the one box first, then the remaining $n-2$ balls, whereas I picked all $n$ at once. Jan 27 '15 at 19:10\n\nHere is the answer to the question, based on the answer manual (it was worded slightly differently in my textbook, hence why I did not find it). My part 1 was correct.\n\nThe textbook question is as follows -- I'm not sure the cells versus boxes thing is supposed to make it be any different.\n\n>If $n$ balls are placed at random into $n$ cells, find the probability that exactly one cell remains empty.\n\nFirst, we have $n$ ways of picking the empty box. Then we have $n-1$ ways of picking the box with two balls.\n\nNow which two balls will go into the one box that will have two balls? There are ${n \\choose 2}$ ways to pick the two distinct balls to go into this cell.\n\nLastly we need to arrange the remaining $n-2$ cells, and there are $(n-2)!$ ways to permutate all of these.\n\nSo we have:\n\n$n(n-1){n \\choose 2}(n-2)! = n!{n \\choose 2}$\n\nSo the answer to part 2 is:\n\n$\\frac{n! {n \\choose 2 }}{n^n}$\n\nThe argument in 1. is quite elegant. Another way to see this is that the probability to put the first ball in an empty box is $1$, for the second it is $(1 - 1/n)$, for the third it is $1 - 2/n$. The searched for probability is thus the product and this is exactly what you found in a different way that I actually like better.\n\nFor 2. you need to consider again all possibilities to place the balls. There are a lot more than you said. It is not just which box is empty (also there are $n$ ways for this) but also how the balls are distributed over the remaining ones.", "date": "2021-09-17 04:04:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1120819/n-distinguishable-balls-into-n-boxes", "openwebmath_score": 0.8932162523269653, "openwebmath_perplexity": 142.78554119695795, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812290812825, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8591522806865242}}
{"url": "https://mathematica.stackexchange.com/questions/194900/how-to-produce-listplot-that-samples-a-given-distribution/194912", "text": "# How to produce ListPlot that samples a given distribution?\n\nI wanted to get a binned plot that a uniform background distribution with a spike at some bin. My first instinct was to use RandomVariate, and then ListPlot. What i get is this\n\nThis makes sense as the RandomVariate function is assigning a value to Y based on the distribution. But I was wondering how I would produce a plot that has a spike at some particular bin (some specified x). I would want to produce something that looks more like this, for example.\n\nI know I could just manually enter the data, but is there a way to have a binned plot where there is a uniform background distribution and a gaussian peak at some specific x?\n\nYou are on the right track. Notice that in your mixture, the uniform distribution spans [0,2], but the Normal distribution has a mean of 3. Here is another mixture. Just adjust the distribution parameters as needed:\n\ndist = MixtureDistribution[{1, 1},\n{\nUniformDistribution[{0, 100}],\nNormalDistribution[50, 2]\n}\n];\n\nHistogram[RandomVariate[dist, 1000], {1}]\n\n\nEdit: If you want data, you can use some variant of HistogramList:\n\n{binBoundaries, counts} =\nHistogramList[RandomVariate[dist, 1000], {1}];\n\nListPlot[Transpose[{Most[binBoundaries], counts}], PlotRange -> All,\nFilling -> Axis]\n\n\n\u2022 Yes, but this is a Histogram. The actual data will still have values between 0 and a 100. I wanted a way where the data itself is spread that way. \u2013\u00a0shivangi Apr 10 at 3:38\n\u2022 @shivangi, I understand. I edited my answer with a possibility. \u2013\u00a0David Keith Apr 10 at 4:17\n\u2022 Thank you so much! This is exactly what I was trying to do. \u2013\u00a0shivangi Apr 13 at 0:51", "date": "2019-11-22 07:05:38", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/194900/how-to-produce-listplot-that-samples-a-given-distribution/194912", "openwebmath_score": 0.767861008644104, "openwebmath_perplexity": 692.6636769656769, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9683812327313546, "lm_q2_score": 0.8872045840243196, "lm_q1q2_score": 0.8591522687623793}}
{"url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html", "text": "Repeated Eigenvalues\n\nProof Because x is an eigenvector of A, you know that and can write In cases for which the power method generates a good approximation of a dominant eigenvector, the Rayleigh quotient provides a correspondingly good approximation of the dominant eigenvalue. 7 Today's handouts Lecture 21 notes Tutorial 7 questions 2. Let us focus on the behavior of the solutions when (meaning the future). \u2020 Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. Find more Mathematics widgets in Wolfram|Alpha. Interesting eigenvectors of the Fourier transform Berthold K. ) FINDING EIGENVECTORS \u2022 Once the eigenvaluesof a matrix (A) have been found, we can \ufb01nd the eigenvectors by Gaussian Elimination. A symmetric minor of A is a submatrix B obtained by deleting some rows and the corresponding columns. In particular, undamped vibration is governed by. Example Determine if the following matrices are diagonalizable. To \ufb01nd an eigenvector corresponding to an eigenvalue , we write. For repeated diagonal elements, it might not tell you much about the location of the eigenvalues. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. And the lambda, the multiple that it becomes-- this is the eigenvalue associated with that eigenvector. By looking at these eigenvalues it is possible to get information about a graph that might otherwise be di cult to obtain. Recall the basic result that the roots of a polynomial depend continuously on the coe\ufb03cients of the polynomial. Since we are going to be working with systems in which $$A$$ is a $$2 \\times 2$$ matrix we will make that assumption from the start. The complete case. If x= a+ ibis a complex number, then we let x = a ibdenote its conjugate. The characteristic polynomial P( ) = j I Aj. \u2022 What the Hautus Keymann Theorem says is that it is possible after preliminary state feedback using a matrix F. For very high or very low correlation in DVs, it is not suitable: if DVs are too. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. This might introduce extra solutions. We can\u2019t \ufb01nd it by elimination. complex eigenvalues. Recall that given a symmetric, positive de nite matrix A we de ne R(x) = xTAx xTx: Here, the numerator and denominator are1 by 1matrices, which we interpret as numbers. Repeated eigenvalues - Duration: 7:30. It is a \\repeated eigenvalue,\" in the sense that the characteristic polynomial (T 1)2 has 1 as a repeated root. Since x \u2260 0, this equation implies \u03bb = 1(Eigenvalue); then, from x = 1 x, every (nonzero) vector is an eigenvector of I. So lambda is an eigenvalue of A. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. An eigenvector of Ais a nonzero vector v such that Av = v for some number. MATRIX EXPONENTIALS, and REPEATED EIGENVALUES Systems and Matrix Exponentials : solve x0= Ax, for n nreal A. Repeated Eigenvalues Occasionally when we have repeated eigenvalues, we are still able to nd the correct number of linearly independent eigenvectors. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. \u2020 Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. Putting all these bases together gives us a list of vectors: ~v 1, ~v 2, Lucky Fact 2: The geometric multiplicity of , meaning the dimension of this kernel, is equal to the number of times occurs as a root of f. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. The linearization has a center at the origin, but the. component is being recorded, and then \"removed\". The study of the relations between eigenvalues and structures in graphs is. This might introduce extra solutions. An eigenvalue of Ais a number such that Av = v for some nonzero vector v. EIGENVALUES OF THE LAPLACIAN AND THEIR RELATIONSHIP TO THE CONNECTEDNESS OF A GRAPH3 (2. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. 2 \u03bbhas a single eigenvector Kassociated to it. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. To tackle the issue of non-smoothness of repeated eigenvalues, we propose an estimator constructed by averaging all repeated eigenvalues. The coefficients for the principal components are unique (except for a change in sign) if the eigenvalues are distinct and not zero. We solve a problem about eigenvalues of an upper triangular matrix and the square of a matrix. Differential Equations; Slope field; System of Linear DEs Real Repeated Eigenvalues #2; System of Linear DEs Imaginary Eigenvalues;. So that is the end of our lecturer on complex eigenvalues and next lecture were to talk about what to do when you have repeated real eigenvalues. Since our last example and that wraps up our lecture on repeated eigenvalues so, this is the systems of differential equations where we had repeated eigenvalues. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Spectral Decomposition with Repeated Eigenvalues (Quantum Theory for Computer Age - Ch. But what if A has repeated real eigenvalue (a, a), or if the eigenvalues are complex conjugate pairs (a+bi, a-bi) with nonzero b? How does one decompose A=aI+bJ where J^2=-I in the latter case, or decompose A=aI+N where N^2=0 in the former?. If all eigenvalues $$\\lambda$$ have negative real parts, then all solutions of approach zero exponentially. j is repeated. We restrict ourselves to the special cases of A being 2 \u00d7 2 and 3 \u00d7 3. 1 Matrix exponent Consider a \ufb01rst order di\ufb00erential equation of the form y\u2032 = ay; a \u2208 R; with the initial condition y(0) = y0: Of course, we know that the solution to this IVP is given by y(t) = eaty0: However, let us apply the method of iterations to this equation. So lambda is an eigenvalue of A. We prove that the volume of n satis es: j nj jRPN 3j = n 2 ; where N = n+1 2 is the dimension of the space of real symmetric matrices of size n n. But here only (1,0) is a eigenvector to 0. 118 CHAPTER 6. 5 Repeated Eigenvalues 95 5. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. What does this mean geometrically?. 2000 S Deep Cameo Clad Proof Massachusetts MA State Washington Quarter (B04),Girls Greek / Roman Goddess Fancy Dress Costume 4-11 Years Available,1945 S Silver Jefferson Nickel GEM BU BLAST WHITE!!. We will ignore the possibility of , as that would mean 0 is an eigenvalue. Mathematics Assignment Help, Example of repeated eigenvalues, Illustration : Solve the following IVP. Computing the pth roots of a matrix with repeated eigenvalues 2651 It can be seen that for a special case, if the given matrix A is a companion matrix then the constituent matrices corresponding to companion matrix are determined by. 1 of A is repeated if it is a multiple root of the char\u00ad acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when \u03bb. Eigenvalues and Eigenvectors. 2 Harmonic Oscillators 114 6. hat May 14 '12 at 0:21 3 $\\begingroup$ To put the same thing into slightly different words: what you have here is a two-dimensional eigenspace , and any two vectors that form a basis for that space will do as linearly. Chapter 6 Eigenvalues and Eigenvectors Po-Ning Chen, Professor Department of Electrical and Computer Engineering National Chiao Tung University Hsin Chu, Taiwan 30010, R. j is repeated. Eigenvalues in MATLAB. 2 6 6 6 4. 2 Solving Systems with Repeated Eigenvalues If the characteristic equation has only a single repeated root, there is a single eigenvalue. , orthogonal to the disk and passing through its center), while any two orthogonal diameters in the plane of the disk may be chosen as the other two principal axes (corresponding to the repeated eigenvalue ). Repeated eigenvalues. J has the eigenvalues of A on its main diagonal, is upper triangular, and has 0's and 1's in the upper triangle. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. If there is no other eigenvector, we look for a solutions in the form x(t) = (u+ tv)e t: If x(t) has this form, then, on the one hand x_(t) = ve t+ (u+ tv)e t;. Introduction. $\\endgroup$ - copper. So it is with matrices. 1 Find the eigenvalues and associated eigenspaces of each of the following matrices. eigenvalues and convert them a Pillai-Bartlett or Wilk's-Lambda value, I don?t know how to convert to an f-statistic. 1 Distinct Eigenvalues 107 6. Firstly we look at matrices where one or more of the eigenvalues is repeated. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Take for example 0 @ 3 1 2 3 1 6 2 2 2 1 A One can verify that the eigenvalues of this matrix are = 2;2; 4. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. Linear Systems of ODE with with Repeated eigenvalues James K. 4 Bases and Subspaces 89 5. If they are numeric, eigenvalues are sorted in order of decreasing absolute value. Repeated eigenvalues are explained with the help of numerical examples. Solution: If \u201a is a simple real eigenvalue, then there are two real unit eigenvectors: u and \u00a1u. Prove that if A is a square matrix then A and AT have the same characteristic polynomial. To \ufb01nd an eigenvector corresponding to an eigenvalue , we write. and w is an eigenvector for A, and suppose that the eigenvalues are di erent. \u03bb = a \u00b1 ib. Of particular interest in many settings (of which di\ufb00erential equations is one) is the following. Then from the Lemma we get 2v w =(ATv)w=v (Aw)=v 5w=5vw: But since 2 and 5 are scalars (and, for that matter, so is v w), the only way that 2v w =5v wcan be possible is for v w to be 0. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Further options. In order to get the eigenvalues of the matrix , I'll solve the characteristic equation Step 3. Stability Analysis for ODEs Marc R. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. 2 \u03bbhas a single eigenvector Kassociated to it. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. It decomposes matrix using LU and Cholesky decomposition The calculator will perform symbolic calculations whenever it is possible. View record in Web of Science \u00ae. edu/math Craigfaulhaber. complex eigenvalues. The last case is what the solutions look like when there are repeated eigenvalues, or. To find eigenvalues of matrix A Consider {eq}\\displaystyle det(A-{\\lambda}(I))=0 {/eq} Then we get a characteristic polynomial in lambda. Eigenvectors and Eigenvalues When a random matrix A acts as a scalar multiplier on a vector X, then that vector is called an eigenvector of X. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. Engineering Computation ECL4-4. sy ' Section 7. For repeated diagonal elements, it might not tell you much about the location of the eigenvalues. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. The number of positive eigenvalues equals the number of positive pivots. Ryan Spring 2012 1 Repeated Eigenvalues Last Time: We studied phase portraits and systems of differential equations with complex eigen-values. Phase portrait for repeated eigenvalues Subsection 3. Defective matrices A are similar to the Jordan (canonical) form A = XJX\u22121. Proof: If A is idempotent, \u03bb is an eigenvalue and v a corresponding eigenvector then \u03bbv = Av = AAv = \u03bbAv = \u03bb2v Since v 6= 0 we \ufb01nd \u03bb\u2212\u03bb2 = \u03bb(1 \u2212\u03bb) = 0 so either \u03bb = 0 or \u03bb = 1. @Star Strider: Thanks for the suggestion, I was unaware of this function. In order to get the eigenvalues of the matrix , I'll solve the characteristic equation Step 3. eigenvalue will be printed as many times as its multiplicity. In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. This paper discusses characteristic features and inherent difficulties pertaining to the lack of usual differentiability properties in problems of sensitivity analysis and optimum structural design with respect to multiple eigenvalues. The general solution is Y~(t) = e3t C 1 C 1 + C 2 + tC 2 Sketch the phase portrait: 2. 2 Repeated Eigenvalues. This process can be repeated until all eigenvalues are found. On the other hand, when it comes to a repeated eigenvalue, this function is not di erentiable, which hinders statistical inference, as the asymptotic theory requires at least second-order di erentiability. Recall that the general solution in this case has the form where is the double eigenvalue and is the associated eigenvector. S ections 2. Repeat steps 2 through 4 for each distinct eigenvalue. Proposition 0. We can\u2019t \ufb01nd it by elimination. The physical significance of this degeneracy is not known. Some may be repeated, some may be complex. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. As shown byL\u00a8utkepohl (2005) andHamilton(1994), the VAR is stable if the modulus of each eigenvalue of A is strictly less than 1. For A2 the situation is di\ufb01erent. However, the fundamental issue is selecting the appropriate tolerance to determine whether two eigenvalues are the same or not, which I don't know a priori (the elements of the matrix I am considering vary by 7 orders of magnitude, so its not obvious how close is close enough). We shall see that this sometimes (but not always) causes problems in the diagonalization process. The Exponential of a Matrix. Find more Mathematics widgets in Wolfram|Alpha. For an matrix, the polynomial we get by computing is of degree , and hence in general, we have eigenvalues. (solution: x = 1 or x = 5. Such an x is called an eigenvector corresponding to the eigenvalue \u03bb. The eigenvalue \u03bb i is called repeated i\ufb00 r i > 1. Once we show this is necessarily real, then the same argument as in the part (a) shows that A. In particular, undamped vibration is governed by. Boyce and Richard C. The same situation applies, if Ais semi-simple, with repeated eigenvalues. Next, perform row operations by adding each row (2 through n) to the first row (Williams): (7) For clarity, the (-1) is repeated in each of the row 1 elements, but this simply. Get the free \"Eigenvalue and Eigenvector (2x2)\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Rajsekaran. You would perform a one-way repeated measures analysis of variance if you had one categorical independent variable and a normally distributed interval dependent variable that was repeated at least twice for each subject. University of Minnesota 109. Are the eigenvalues real? Find one complex eigenvector (contains an i). Since x \u2260 0, this equation implies \u03bb = 1(Eigenvalue); then, from x = 1 x, every (nonzero) vector is an eigenvector of I. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. The following. Roussel September 13, 2005 1 Linear stability analysis Equilibria are not always stable. Moreover, we provide a numerically reliable and effective algorithm for computing the eigenvalue decomposition of a symmetric matrix with two numerically distinct eigenvalues. Using the Laplace transformation, Eq. The solutions of the system can be found by finding the eigenvalues and eigenvectors of the matrix. The spectral decomposition of x is returned as components of a list with components. In general, the algebraic multiplicity and geometric multiplicity of an eigenvalue can differ. (e) A= 1 1 2 3. 2 (Page 249) 17. Ryan Spring 2012 1 Repeated Eigenvalues Last Time: We studied phase portraits and systems of differential equations with complex eigen-values. However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. Step 2 For each eigenvalue , compute an orthonormal basis for Ker(A Id). Make a matrix Q as follows. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. For each distinct eigenvalue l j with multiplicity m j, let I j denote the diagonal matrix with unit elements in each location where l j occurs, and let D j denote the differentiation matrix comprised of integers counting the repeated eigenvalues, placed to the left of the diagonal locations of the respective eigenvalues. The linearization has a node (proper or improper) at the origin, but the original almost linear system has either a node or a spiral point at P. In that example, one principal axis, the one corresponding to eigenvalue , was (i. ' and find homework help for other Math questions at eNotes. When all eigenvalues have non-zero real parts, the equilibrium is called hyperbolic, and non-hyperbolic if at least one eigenvalue has zero real part. Recall the basic result that the roots of a polynomial depend continuously on the coe\ufb03cients of the polynomial. An eigenvalue of Ais a number such that Av = v for some nonzero vector v. Precondition The eigenvalues have been computed before. However, when I run it with a non-symmetric matrix, the largest eigenvalue is in the first column. \u2022 Roy\u2019s Largest Root = largest eigenvalue o Gives an upper-bound of the F-statistic. Is it possiable to create a matrix be setting the eigenvalues you wish to end up with? I am trying to create a 3X3 random matrix with a set of repeated eigenvalues and was wondering is it's possible to set the eigenvalues and then generate matrices for them. The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue. Note that is the product of the eigenvalues (since ), so for the sign of determines whether the eigenvalues have the same sign or opposite sign. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Proof: If A is idempotent, \u03bb is an eigenvalue and v a corresponding eigenvector then \u03bbv = Av = AAv = \u03bbAv = \u03bb2v Since v 6= 0 we \ufb01nd \u03bb\u2212\u03bb2 = \u03bb(1 \u2212\u03bb) = 0 so either \u03bb = 0 or \u03bb = 1. Of particular interest in many settings (of which di\ufb00erential equations is one) is the following. Solution Step 1. Solution: Y(t) = c1 1 0 e 2t + c 2 0 1 e 2t = e 2t c1 c2 Qualitative behavior: 1. sy ' Section 7. component is being recorded, and then \"removed\". Proposition 0. Show that A and AT do not have the. For the purpose of analyzing Hessians, the eigenvectors are not important, but the eigenvalues are. The eigenvalue problem is to determine the solution to the equation Av = \u03bbv, where A is an n-by-n matrix, v is a column vector of length n, and \u03bb is a scalar. Under this matrix norm, the in\ufb01nite series converges for all A and for all t, and it de\ufb01nes the matrix exponential. Eigenvalues are a special set of scalars associated with a linear system of equations (i. The linearization has a node (proper or improper) at the origin, but the original almost linear system has either a node or a spiral point at P. We restrict ourselves to the special cases of A being 2 \u00d7 2 and 3 \u00d7 3. We do not normally divide matrices (though sometimes we can multiply by an inverse). Get the free \"Eigenvalues Calculator 3x3\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Assume \u03bb is a repeated eigenvalue of A with multiplicity m. Facts About Eigenvalues By Dr David Butler De nitions Suppose Ais an n nmatrix. Right when you reach $0$, the eigenvalues and eigenvectors become real (although there is only eigenvector at this point). Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. Since we are going to be working with systems in which A is a 2 x 2 matrix we will make that assumption from the start. Next I simplify it. \"Det\" stands for the determinant, and \"I\" is the identity matrix. 1 Fundamental Matrices, Matrix Exp & Repeated Eigenvalues \u2013 Sections 7. We can certainly have repeated roots and complex eigenvalues. Are the eigenvalues real? Find one complex eigenvector (contains an i). A new method is presented for computation of eigenvalue and eigenvector derivatives associated with repeated eigenvalues of the generalized nondefective eigenproblem. Multiplying them gives 2 4 4 3 3 2 3 2 1 0 2 3 5 2 4 3 2 1 3 5= 2 4 15 10 5 3 5= 5 2 4 3 2 1 3 5: This shows that the vector is an eigenvector for the eigenvalue 5. If the 2 2 matrix Ahas distinct real eigenvalues 1 and 2, with corresponding eigenvectors ~v 1 and ~v 2, then the system x~0(t)=A~x(t). But we did not discuss the case when one of the eigenvalues is zero. 3 power method for approximating eigenvalues 551 Note that the approximations in Example 2 appear to be approaching scalar multiples of which we know from Example 1 is a dominant eigenvector of the matrix. edu (UC Davis) >4 ICIAM 11 1 / 32. Eigenvectors and Eigenvalues When a random matrix A acts as a scalar multiplier on a vector X, then that vector is called an eigenvector of X. When n = 2, in 1955 and 1956, Payne, P\u00b4olya and Weinberger proved that, in [10] and [11], \u03bb 2 \u03bb 1 \u2264 3forD \u2282 R2, and they conjectured \u03bb 2 \u03bb 1 \u2264 \u03bb 2 \u03bb 1 | disk \u2248 2. Using eigenvalues and eigenvectors to calculate the final values when repeatedly applying a matrix First, we need to consider the conditions under which we'll have a steady state. Spectral Decomposition with Repeated Eigenvalues (Quantum Theory for Computer Age - Ch. As shown byL\u00a8utkepohl (2005) andHamilton(1994), the VAR is stable if the modulus of each eigenvalue of A is strictly less than 1. If the eigenvalues of A are distinct, it turns out that the eigenvectors are linearly independent; but, if any of the eigenvalues are repeated, further investigation may be necessary. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Then use de\ufb02ation to form a new matrix and use the power method again to extract the second eigenvalue (and root). Next one is at least one eigenvalue is repeated, can be twice or even more. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. trix has two eigenvalues of magnitude zero, one eigenvalue of unit magnitude, and three eigenvalues with magnitude less than one (right). 1 Introduction to Eigenvalues Linear equationsAx D bcomefrom steady stateproblems. To learn more about all this you should take 18. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. Eigenvalues shows variance explained by that particular factor out of the total variance. (\u03bb = \u22122 is a repeated root of the characteristic equation. 1 Eigenvalues and Eigenvectors Spectral graph theory studies how the eigenvalues of the adjacency matrix of a graph, which are purely algebraic quantities, relate to combinatorial properties of the graph. Repeated Eigenvalues Recall We are now in the position that we can find (at least) as many linearly-independent solutions to the homogeneous equation \u02d9 y = A y as there are distinct eigenvalues of A (real or complex). Suppose the 2 2 matrix Ahas repeated eigenvalues. Find the eigenvalues & eieenvectcrs. 0 1), whose only eigenvalue is 1. In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. Prove that if A is a square matrix then A and AT have the same characteristic polynomial. The last case is what the solutions look like when there are repeated eigenvalues, or. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. Use the given information to determine the matrix AL Phase plane solution trajectories have horizontal tangents on the line y2-2\u5208and vertical tangents on the line y! The matrix Al has a nonzero repeated eigenvalue and a21 - -5. Next I simplify it. 1 of A is repeated if it is a multiple root of the char\u00ad acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when \u03bb. Remark: I A matrix with repeated eigenvalues may or may not be diagonalizable. The set of eigenvalues of a matrix is sometimes called the of the matrix, and orthogonal diagonalispectrum zation of a matrix factors in aE E way that displays all the eigenvalues and their multiplicities. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. vectors corresponding to eigenvalues as above (assum-ing no repeated eigenvalues), the Hessian has exactlyP q\u2208Q q\u2212 k 2 negative eigenvalues: we can replace any eigencomponent with eigenvalue \u03c3 with an alternate eigencomponent not already in (U,V) with eigenvalue \u03c30 > \u03c3, decreasing the objective function. There are various methods by which the continuous eigenvalue problem may be. The algorithm is from the EISPACK collection of subroutines. Since it is not invertible, 0 is an eigenvalue. By looking at these eigenvalues it is possible to get information about a graph that might otherwise be di cult to obtain. The picture is more complicated, but as in the 2 by 2 case, our best insights come from finding the matrix's eigenvectors : that is, those vectors whose direction the. The resulting array will be of complex type, unless the imaginary part is zero in which case it will be cast to a real type. Guidance as to the number of linearly independent eigenvectors can be obtained from the rank of the matrix A. x(t)- \u2014 4- -(14t). ) FINDING EIGENVECTORS \u2022 Once the eigenvaluesof a matrix (A) have been found, we can \ufb01nd the eigenvectors by Gaussian Elimination. Step 2 For each eigenvalue , compute an orthonormal basis for Ker(A Id). complex eigenvalues. Then each Axi is also in X, so each. Mathematical and Computer Modelling, 2002. sy ' Section 7. Then, we use these results to establish necessary and sufficient conditions for the. We again consider the system ~x0 = A~x. The eigenvalues are not necessarily ordered. If you would like to simplify the solution provided by our calculator, head to the unit vector calculator. I Review: The case of diagonalizable matrices. The resulting array will be of complex type, unless the imaginary part is zero in which case it will be cast to a real type. Because the rank of limn\u2192\u221eP n = 1, there is a unique limiting distribution. If eigenvalues are repeated, we may or may not have all n linearly independent eigenvectors to diagonalize a square matrix. If the eigenvalue is negative, the direction is reversed. \u2022 Objective. Eigenvalues in MATLAB. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. Eigenvalues and Eigenfunctions of the Laplacian Mihai Nica University of Waterloo [email\u00a0protected] I Equivalently: An n \u00d7 n matrix with repeated eigenvalues may or may not have a linearly independent set of n eigenvectors. 8 Repeated Eigenvalues Shawn D. The rest are similar. Calculating eigenvalues and eigenvectors of matrices by hand can be a daunting task. These repeated matrix multiplications mean the resulting matrix is exponential in the number of layers of the neural network. Find more Mathematics widgets in Wolfram|Alpha. @Star Strider: Thanks for the suggestion, I was unaware of this function. 0 1), whose only eigenvalue is 1. Make a matrix Q as follows. Is the procedure the same? Can I use, say, variation of parameters to solve this. But here only (1,0) is a eigenvector to 0. For the following matrix, list the real eigenvalues, repeated according to their multi-plicities. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. Repeated real eigenvalue 3. The coefficients for the principal components are unique (except for a change in sign) if the eigenvalues are distinct and not zero. We can now \ufb01nd a real-valued general solution to any homogeneous system where the matrix has distinct eigenvalues. The \u2018ladder\u2019 is bounded at both the low and high ends, which can be seen by considering the operator !. Mathematical and Computer Modelling, 2002. This might introduce extra solutions. But you can find enough independent eigenvectors -- Forget the \"but. Stop at this point, and practice on an example (try Example 3, p. On the other hand, when it comes to a repeated eigenvalue, this function is not di erentiable, which hinders statistical inference, as the asymptotic theory requires at least second-order di erentiability. To be more speci\ufb01c, Section 2. Indeed, BAv = ABv = A( v) = Av. A matrix A is idempotent if and only if all its eigenvalues are either 0 or 1. Subsection 3. Assume \u03bb is a repeated eigenvalue of A with multiplicity m. Is 1 1 an eigenvector. Repeated eigenvalues - Duration: 7:30. Terminology Let Abe an n nmatrix. Eigenvalues, Eigenvectors, and Di erential Equations William Cherry April 2009 (with a typo correction in November 2015) The concepts of eigenvalue and eigenvector occur throughout advanced mathematics. Repeated Eigenvalues and Symmetric Matrices 22. State feedback and Observer Feedback Poles of transfer function are eigenvalues of A Pole locations a ect system response repeated eigenvalues. However, the geometric multiplicity can never exceed the algebraic multiplicity. \u2022 Roy\u2019s Largest Root = largest eigenvalue o Gives an upper-bound of the F-statistic. This will be one of the few times in this chapter that non-constant coefficient differential equation will be looked at. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. \u201cThe equation A x = \u03bb x characterizes the eigenvalues and associated eigenvectors of any matrix A. Supplementary notes for Math 265 on complex eigenvalues, eigenvectors, and systems of di erential equations. The minimal polynomial x3 x\u2026x\u2014x 1\u2013\u2014x\u20211\u2013also splits completely over any \ufb01eld, and in particular over F, so Acan be diagonalized over F. Imposing an additional condition, that the eigenvalues lie in Fand are simple roots of the characteristic polynomial, does force diagonalizability. This is because u lays on the same subspace (plane) as v and w, and so does any other eigenvector. Get an answer for 'Give an example of a non-diagonalizable 4x4 matrix with eigenvalues: -1, -1, 1, 1. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. 06 or 18/700. But for a fundamental system two independent solutions are needed. Repeated Eigenvalues Note. However, the fundamental issue is selecting the appropriate tolerance to determine whether two eigenvalues are the same or not, which I don't know a priori (the elements of the matrix I am considering vary by 7 orders of magnitude, so its not obvious how close is close enough). Solution Since , the given matrix has distinct real eigenvalues of. Now is the next step. \u2020 Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. eigenvalues tell the entire story. The number of eigenvalues equal to 1 is then tr(A). If is eigenvalue of perturbationA+ Eof nondefective matrixA, then j kj cond 2(X)kEk 2 where kis closest eigenvalue ofAto andX is nonsingular matrix of eigenvectors ofA Absolute condition number of eigenvalues is condition number of matrix of eigenvectors with respect to solving linear equations Eigenvalues may be sensitive if eigenvectors are. So even though a real asymmetric x may have an algebraic solution with repeated real eigenvalues, the computed solution may be of a similar matrix with complex conjugate pairs of eigenvalues. We shall see that this. The phase portrait thus has a distinct star. As any system we will want to solve in practice is an approximation to \u2026. In the 2x2 case, if the eigenvalue is repeated you are in the defective case unless the matrix is precisely [ lambda_1 , 0 ; 0 , lambda_1 ] For larger square matrices this becomes the story of Jordan form. Eigenvectors and Hermitian Operators 7. \u2022 Objective. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Let be the set[5] of eigenvalues, speci cally not attempting to count repeated eigenvalues more than once. polynomial corresponding to A, has n roots some of which may be repeated. 4 Bases and Subspaces 89 5. Within this blog post, we\u2019ll investigate some classes of buckling problems and the way they are sometimes analyzed. where the eigenvalues are repeated eigenvalues. This is particularly true if some of the matrix entries involve symbolic parameters rather than speci\ufb02c numbers. This happens when the dimension of the nullspace of A\u2212\u03bbI (called the geometric multiplicity of \u03bb) is strictly less than the arithmetic multiplicity m.", "date": "2019-11-14 11:03:47", "meta": {"domain": "rk-verlag.de", "url": "http://tsrd.rk-verlag.de/repeated-eigenvalues.html", "openwebmath_score": 0.8856151103973389, "openwebmath_perplexity": 450.8891887879132, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9888419697383903, "lm_q2_score": 0.868826771143471, "lm_q1q2_score": 0.8591323757389555}}
{"url": "http://fr.mathworks.com/help/matlab/ref/bandwidth.html?s_tid=gn_loc_drop&nocookie=true", "text": "# bandwidth\n\nLower and upper matrix bandwidth\n\n## Syntax\n\n\u2022 `B = bandwidth(A,type)` example\n\u2022 ```[lower,upper] = bandwidth(A)``` example\n\n## Description\n\nexample\n\n````B = bandwidth(A,type)` returns the bandwidth of matrix `A` specified by `type`. Specify `type` as `'lower'` for the lower bandwidth, or `'upper'` for the upper bandwidth.```\n\nexample\n\n``````[lower,upper] = bandwidth(A)``` returns the lower bandwidth, `lower`, and upper bandwidth, `upper`, of matrix `A`.```\n\n## Examples\n\ncollapse all\n\n### Find Bandwidth of Triangular Matrix\n\nCreate a 6-by-6 lower triangular matrix.\n\n`A = tril(magic(6))`\n```A = 35 0 0 0 0 0 3 32 0 0 0 0 31 9 2 0 0 0 8 28 33 17 0 0 30 5 34 12 14 0 4 36 29 13 18 11 ```\n\nFind the lower bandwidth of `A` by specifying `type` as `'lower'`.\n\n`B = bandwidth(A,'lower')`\n```B = 5```\n\nThe result is 5 because every diagonal below the main diagonal has nonzero elements.\n\nFind the upper bandwidth of `A` by specifying `type` as `'upper'`.\n\n`B = bandwidth(A,'upper')`\n```B = 0```\n\nThe result is 0 because there are no nonzero elements above the main diagonal.\n\n### Find Bandwidth of Sparse Block Matrix\n\nCreate a 100-by-100 sparse block matrix.\n\n`B = kron(speye(25),ones(4));`\n\nView a 10-by-10 section of elements from the top left of `B`.\n\n`full(B(1:10,1:10))`\n```ans = 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1```\n\n`B` has 4-by-4 blocks of ones centered on the main diagonal.\n\nFind both the lower and upper bandwidths of `B` by specifying two output arguments.\n\n`[lower,upper] = bandwidth(B)`\n```lower = 3 upper = 3 ```\n\nThe lower and upper bandwidths are both `3`.\n\n## Input Arguments\n\ncollapse all\n\n### `A` \u2014 Input matrix2-D numeric matrix\n\nInput matrix, specified as a 2-D numeric matrix. `A` can be either full or sparse.\n\nData Types: `single` | `double`\nComplex Number Support: Yes\n\n### `type` \u2014 Bandwidth type`'lower'` | `'upper'`\n\nBandwidth type, specified as `'lower'` or `'upper'`.\n\n\u2022 Specify `'lower'` for the lower bandwidth (below the main diagonal).\n\n\u2022 Specify `'upper'` for the upper bandwidth (above the main diagonal).\n\nData Types: `char`\n\n## Output Arguments\n\ncollapse all\n\n### `B` \u2014 Lower or upper bandwidthnonnegative integer scalar\n\nLower or upper bandwidth, returned as a nonnegative integer scalar.\n\n\u2022 If `type` is `'lower'`, then `0`\u00a0\u2264\u00a0`B`\u00a0\u2264\u00a0`size(A,1)-1`.\n\n\u2022 If `type` is `'upper'`, then `0`\u00a0\u2264\u00a0`B`\u00a0\u2264\u00a0`size(A,2)-1`.\n\n### `lower` \u2014 Lower bandwidthnonnegative integer scalar\n\nLower bandwidth, returned as a nonnegative integer scalar. `lower` is in the range `0`\u00a0\u2264\u00a0`lower`\u00a0\u2264\u00a0`size(A,1)-1`.\n\n### `upper` \u2014 Upper bandwidthnonnegative integer scalar\n\nUpper bandwidth, returned as a nonnegative integer scalar. `upper` is in the range `0`\u00a0\u2264\u00a0`upper`\u00a0\u2264\u00a0`size(A,2)-1`.\n\ncollapse all\n\n### Upper and Lower Bandwidth\n\nThe upper and lower bandwidths of a matrix are measured by finding the last diagonal (above or below the main diagonal, respectively) that contains nonzero values.\n\nThat is, for a matrix A with elements Aij:\n\n\u2022 The upper bandwidth B1 is the smallest number such that ${A}_{ij}=0$ whenever $j-i>{B}_{1}$.\n\n\u2022 The lower bandwidth B2 is the smallest number such that ${A}_{ij}=0$ whenever $i-j>{B}_{2}$.\n\nNote that this measurement does not disallow intermediate diagonals in a band from being all zero, but instead focuses on the location of the last diagonal containing nonzeros. By convention, the upper and lower bandwidths of an empty matrix are both zero.", "date": "2015-05-04 21:16:36", "meta": {"domain": "mathworks.com", "url": "http://fr.mathworks.com/help/matlab/ref/bandwidth.html?s_tid=gn_loc_drop&nocookie=true", "openwebmath_score": 0.8676239848136902, "openwebmath_perplexity": 2476.1231206705274, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743644972026, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8591235074903251}}
{"url": "http://mathhelpforum.com/number-theory/151037-finding-remainder.html", "text": "Math Help - Finding the remainder\n\n1. Finding the remainder\n\nWhen the positive integer n is divided by 7, the remainder is 2. What is the remainder when 5n is divided by 7?\n\nHow would you set this up?\n\n2. The first step is to set up 5n/7 = 5/7 + n/7 so to strictly answer your question, except where n is a multiple of 7, then the remainder is n/7. Is this what you're asking about?\n\n3. No. These are the choices:\n\n(A) 2\n(B) 3\n(C) 4\n(D) 5\n(E) 6\n\nThe answer is 3. I have no freaking clue how they got that.\n\n4. I think I understand the problem now. When n is 9, then division by 7 leaves a remainder of 2. Now 5 x 9 = 45. Try dividing 45 by 7 and see if you get a remainder of 3.\n\n5. Suppose you divide a number N by P, and you get a quotient Q with a remainder R. This can be rewritten in fraction form:\n$\\frac{N}{P} = Q + \\frac{R}{P}$\n(remember that 0 <= R < P).\n\nIn our case, we are dividing n by 7. You get a quotient q with a remainder of 2, or:\n$\\frac{n}{7} = q + \\frac{2}{7}$\nNow I want to know what 5n is. Multiply both sides of the equation above by 35 and you'll get\n$5n = 35q + 10$\n\nDivide 5n by 7:\n\\begin{aligned}\n\\frac{5n}{7} &= \\frac{35q + 10}{7} \\\\\n&= \\frac{35q}{7} + \\frac{10}{7} \\\\\n&= 5q + \\frac{10}{7} \\\\\n&= (5q + 1) + \\frac{3}{7} \\\\\n\\end{aligned}\n\nThe quotient would be 5q + 1, and the remainder would be 3.\n\n6. Thank you so much!\n\n7. Another, easier way to do it is to let $n \\equiv 2 \\pmod{7}$, it then follows that $5n \\equiv 10 \\equiv 3 \\pmod{7}$, thus the remainder is $3$.\n\n8. Originally Posted by Bacterius\nAnother, easier way to do it is to let $n \\equiv 2 \\pmod{7}$, it then follows that $5n \\equiv 10 \\equiv 3 \\pmod{7}$, thus the remainder is $3$.\nI cannot find the MOD command on my calculator though.\n\n9. It really is the remainder function, usually calculators don't have it. You can emulate $a \\mod b$ by dividing $a$ by $b$, taking the fractional part only and multiplying it by $b$. For instance, to get $10 \\mod 7$ :\n\n$\\frac{10}{7} = 1.428571429 ...$\nTake the fractional part which is $0.428571429$, and times it by $7$, you get :\n$0.428571429 \\times 7 = 3$\n\n10. Originally Posted by Bacterius\nAnother, easier way to do it is to let $n \\equiv 2 \\pmod{7}$, it then follows that $5n \\equiv 10 \\equiv 3 \\pmod{7}$, thus the remainder is $3$.\nOriginally Posted by Bacterius\nIt really is the remainder function, usually calculators don't have it. You can emulate $a \\mod b$ by dividing $a$ by $b$, taking the fractional part only and multiplying it by $b$. For instance, to get $10 \\mod 7$ :\n\n$\\frac{10}{7} = 1.428571429 ...$\nTake the fractional part which is $0.428571429$, and times it by $7$, you get :\n$0.428571429 \\times 7 = 3$\nOh, ok, but where did you get the 10 from? Especially the 10 in the first post here.\nAlso, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?\n\n11. If $n \\equiv 2 \\pmod{7}$ then $\\mathbf{5} \\times n \\equiv \\mathbf{5} \\times 2 \\equiv 10 \\equiv 3 \\pmod{7}$\nThis is where the $10$ comes from\n\nAlso, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?\nThat would only work for numbers up to $13$ ... for instance if you substract $7$ from $23$, you get $16$, which isn't exactly the remainder of $23$ divided by $7$, while $23 \\mod 7 = 2$ as expected. Then you could argue that repeatedly substracting $7$ is a way to do it, true, but dividing is then more efficient.\n\n12. I totally understand now. Thank you!\n\n13. Hello, Mariolee!\n\nA slightly different approach . . .\n\nWhen the positive integer $n$ is divided by 7, the remainder is 2.\nWhat is the remainder when $5n$ is divided by 7?\n\n\"When $n$ is divided by 7, the remainder is 2.\"\n\n. . Hence: . $n \\:=\\:7a + 2\\:\\text{ for some integer }a.$\n\nThen: . $5n \\:=\\:35a + 10$\n\n. . and: . $\\dfrac{5n}{7} \\:=\\:\\dfrac{35a+10}{7} \\;=\\;5a + 1 + \\frac{3}{7}$\n\nTherefore, the remainder is 3.", "date": "2014-04-18 06:44:35", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/151037-finding-remainder.html", "openwebmath_score": 0.9409502148628235, "openwebmath_perplexity": 280.4544854695478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109534209825, "lm_q2_score": 0.8723473730188543, "lm_q1q2_score": 0.8590972481369874}}
{"url": "https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution", "text": "# Will $2$ linear equations with $2$ unknowns always have a solution?\n\nAs I am working on a problem with 3 linear equations with 2 unknowns I discover when I use any two of the equations it seems I always find a solution ok. But when I plug it into the third equation with the same two variables , the third may or may not cause a contradiction depending if it is a solution and I am OK with that BUT I am confused on when I pick the two equations with two unknowns it seems like it has no choice but to work. Is there something about linear algebra that makes this so and are there any conditions where it won't be the case that I will find a consistent solution using only the two equations? My linear algebra is rusty and I am getting up to speed. These are just equations of lines and maybe the geometry would explain it but I am not sure how. Thank you.\n\n\u2022 Mind showing the equations?? \u2013\u00a0The Dead Legend May 7 '17 at 0:16\n\nEach linear equation represents a line in the plane. Most of the time two lines will intersect in one point, which is the simultaneous solution you seek. If the two lines have exactly the same slope, they may not meet so there is no solution or they may be the same line and all the points on the line are solutions. When you add a third equation into the mix, that is another line. It is unlikely to go through the point that solves the first two equations, but it might.\n\nThere are three possible cases for $2$ linear equations with $2$ unknowns (slope and intercept):\n\n$\\qquad$ $\\mathbf{0}$ solution points $\\qquad$ $\\qquad$ $\\mathbf{1}$ solution point $\\qquad$ $\\qquad$ $\\mathbf{\\infty}$ solution points\n\n$\\qquad \\quad$ $\\nexists$ no existence $\\qquad$ $\\qquad$ $\\exists !$ uniqueness $\\qquad$ $\\qquad$ $\\exists$ no uniqueness\n\nThe lines have the form $y(x) = mx + b$.\n\nCase 1: parallel lines\n\nA solution does not exist.\n\nThe lines are parallel: they have the same slope.\n\n% \\begin{align} % y_{1}(x) &= m x + b_{1} \\\\ % y_{2}(x) &= m x + b_{2} \\\\ % \\end{align} %\n\nCase 2: intersecting lines\n\nWe have existence and uniqueness.\n\nThe slopes are distinct.\n\n$$m_{1} \\ne m_{2}$$\n\n% \\begin{align} % y_{1}(x) &= m_{1} x + b_{1} \\\\ % y_{2}(x) &= m_{2} x + b_{2} \\\\ % \\end{align} %\n\nCase 3: coincident lines\n\nWe have existence, but not uniqueness. There is an infinite number of solutions. Every point solves the system of equations.\n\nBoth lines are the same.\n\n% \\begin{align} % y_{1}(x) &= m x + b \\\\ % y_{2}(x) &= m x + b \\\\ % \\end{align} %\n\nIn terms of linear algebra, look at the problem in terms of $\\color{blue}{range}$ and $\\color{red}{null}$ spaces.\n\nThe linear system for two equations is % \\begin{align} % m_{1} x - y &= b_{1} \\\\ % m_{2} x - y &= b_{1} \\\\ % \\end{align} which has the matrix form % \\begin{align} % \\mathbf{A} x &= b \\\\ % \\left[ \\begin{array}{cc} m_{1} & -1 \\\\ m_{2} & -1 \\\\ \\end{array} \\right] % \\left[ \\begin{array}{cc} x \\\\ y \\\\ \\end{array} \\right] % &= % \\left[ \\begin{array}{cc} b_{1} \\\\ b_{2} \\\\ \\end{array} \\right] % \\end{align} %\n\nThe Fundamental Theorem provides a natural framework for classifying data and solutions.\n\nFundamental Theorem of Linear Algebra\n\nA matrix $\\mathbf{A} \\in \\mathbb{C}^{m\\times n}_{\\rho}$ induces for fundamental subspaces: \\begin{align} % \\mathbf{C}^{n} = \\color{blue}{\\mathcal{R} \\left( \\mathbf{A}^{*} \\right)} \\oplus \\color{red}{\\mathcal{N} \\left( \\mathbf{A} \\right)} \\\\ % \\mathbf{C}^{m} = \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} \\oplus \\color{red} {\\mathcal{N} \\left( \\mathbf{A}^{*} \\right)} % \\end{align}\n\nCase 1: No existence\n\nThe matrix $\\mathbf{A}$ has a rank defect $(m_{1} = m_{2})$ and $b_{1} \\ne b_{2}$. $$b = \\color{blue}{b_{\\mathcal{R}}} + \\color{red}{b_{\\mathcal{N}}}$$ It is the $\\color{red}{null}$ space component which precludes direct solution. (Interestingly enough, there is a least squares solution.) $$The data vector b is not a combination of the columns of \\mathbf{A}. The column space is$$ \\mathbf{C}^{2} = \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} \\oplus \\color{red} {\\mathcal{N} \\left( \\mathbf{A}^{*} \\right)} $$The decomposition is$$ \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} = % \\text{span } \\left\\{ \\, \\color{blue}{ \\left[ \\begin{array}{c} m \\\\ -1 \\end{array} \\right] } \\, \\right\\} \\qquad \\color{red}{\\mathcal{N} \\left( \\mathbf{A}^{*} \\right)} = % \\text{span } \\left\\{ \\, \\color{red}{ \\left[ \\begin{array}{r} -1 \\\\ m \\end{array} \\right] } \\, \\right\\} $$Case 2: Existence and uniqueness The matrix \\mathbf{A} has full rank (m_{1}\\ne m_{2}). The data vector is entirely in the \\color{blue}{range} space \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)}$$ b = \\color{blue}{b_{\\mathcal{R}}} $$The \\color{red}{null} space is trivial: \\color{red}{\\mathcal{N} \\left( \\mathbf{A}^{*} \\right)}=\\mathbf{0}.$$ \\mathbf{C}^{2} = \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} $$The decomposition is$$ \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} = % \\text{span } \\left\\{ \\, \\color{blue}{ \\left[ \\begin{array}{c} m_{1} \\\\ -1 \\end{array} \\right] }, \\, \\color{blue}{ \\left[ \\begin{array}{c} m_{2} \\\\ -1 \\end{array} \\right] } \\right\\} $$Case 3: Existence, no uniqueness The matrix \\mathbf{A} has a rank defect (m_{1} = m_{2} = m), yet b_{1} = b_{2}.$$ b = \\color{blue}{b_{\\mathcal{R}}} $$The column space is has \\color{blue}{range} and \\color{red}{null} space components:$$ \\mathbf{C}^{2} = \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} \\oplus \\color{red} {\\mathcal{N} \\left( \\mathbf{A}^{*} \\right)} $$The decomposition is$$ \\color{blue}{\\mathcal{R} \\left( \\mathbf{A} \\right)} = % \\text{span } \\left\\{ \\, \\color{blue}{ \\left[ \\begin{array}{c} m \\\\ -1 \\end{array} \\right] } \\, \\right\\} \\qquad \\color{red}{\\mathcal{N} \\left( \\mathbf{A}^{*} \\right)} = % \\text{span } \\left\\{ \\, \\color{red}{ \\left[ \\begin{array}{r} -1 \\\\ m \\end{array} \\right] } \\, \\right\\} $$Postscript: the theoretical foundations here are useful. The trip to understanding starts with simple examples like in @Nick's comment. Let's think using vector notation. A linear system with two unknowns x and y, and two equations$$ \\begin{align*} v_1 x + w_1 y &= a_1 \\\\ v_2 x + w_2 y &= a_2 \\end{align*} $$can be written in vector notation as$$ x\\, \\vec{v} + y\\, \\vec{w} = \\vec{a}.  That is, you want to know if $\\vec{a}$ can be written as a linear combination of $\\vec{v}$ and $\\vec{w}$.\n\nFixed the vectors $\\vec{v}$ and $\\vec{w}$, to state that a solution always exists whatever $\\vec{a}$ is, is the same as to state that $\\vec{v}$ and $\\vec{w}$ spans the whole plane. If not ($\\vec{v}$ and $\\vec{w}$ are parallel), depending on $\\vec{a}$, the solution might not exist. And when it exists, it will not be unique.", "date": "2020-11-28 20:29:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution", "openwebmath_score": 1.0000097751617432, "openwebmath_perplexity": 751.3237403913528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109520836027, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8590972453361766}}
{"url": "http://mathhelpforum.com/math-topics/19568-physics-tourist-bear-problem.html", "text": "# Thread: physics tourist & bear problem\n\n1. ## physics tourist & bear problem\n\nanother easy one i think:\n\nA tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?\n\nhow many meters?\n\nthanks alot.\n\n2. Originally Posted by rcmango\nanother easy one i think:\n\nA tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?\n\nhow many meters?\n\nthanks alot.\nThe maximum value of d is such that the bear gets to the car at the same time the tourist does.\n\nSo set up a coordinate system such that the bear is at the origin and positive x is in the direction from the bear to the tourist.\n\nBoth are moving at a constant speed. The bear has to cover 27 + d meters in the same time the tourist covers d meters.\n\nSo for the tourist:\n[tex]d = v_t t = 3.5t[tex]\n\nThus\n$t = \\frac{d}{3.5}$\n\nFor the bear:\n$27 + d = v_b t = 6 \\left ( \\frac{d}{3.5} \\right )$\n\nNow solve for d.\n\n-Dan\n\n3. Hello, rcmango!\n\nAnother approach . . .\n\nA tourist being chased by an angry bear is running in a straight line\ntoward his car at a speed of 3.5 m/s. .The car is a distance $d$ meters away.\nThe bear is 27 meters behind the tourist and running at 6.0 m/s.\nThe tourist reaches the car safely.\nWhat is the maximum possible value for $d$?\n\nThe tourist has a 27-meter headstart.\n\nRelative to the tourist, the bear has a speed of 2.5 m/s.\n\nTo cover 27 meters, it takes the bear: . $\\frac{27}{2.5} \\:=\\:10.8$ seconds.\n\nIn that time, the tourist can run: . $3.5 \\times 10.8 \\:=\\:37.8$ meters.\n\n[I hope he left a window open so he can dive into the car.]\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nThis reminds me of two chased-by-a-bear stories.\n\nAn old-timer was telling about his encounter with an angry grizzly.\n\n\"I was barely keepin' ahead of him and he wuz gainin' on me fast.\nI saw a tree branch about fifteen feet up. .Just as he came up\n'n took a swipe at me, I took this big leap . . .\"\n\n\"Well, did you catch it?\"\n\n\"Well, not on the way up . . . \"\n\nTwo guys are camping when a bear headed towards them.\n\nOne man starting putting on his running shoes.\n\n\"Hey,\" said his friend, \"you can't outrun that bear.\"\n\n\"I don't have to,\" was the reply. \"I just have to outrun you.\"\n\n4. Originally Posted by rcmango\nA tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m away from the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?\n\nhow many meters?\n\nthanks alot.\nHello,\n\nif you change the text of your problem as I've done (see above) then there is a second situation possible: The car is between the bear and the tourist and both are running toward each other. They reach the car (on different sides, I hope for the tourist) after:\n$t=\\frac{27}{3.5+6}\\approx 2.842~s$\n\nDuring this time the tourist runs d = 9.947 m", "date": "2016-09-26 18:30:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/19568-physics-tourist-bear-problem.html", "openwebmath_score": 0.5672556757926941, "openwebmath_perplexity": 1309.6782790269324, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9848109538667756, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8590972370868161}}
{"url": "https://math.stackexchange.com/questions/1405562/is-there-a-method-to-calculate-large-number-modulo", "text": "# Is there a method to calculate large number modulo?\n\nIs there a (number theoretic or algebraic) trick to find a large nunber modulo some number?\n\nSay I have the number $123456789123$ and I want to find its value modulo some other number, say, $17$.\n\nIt's not fast for me to find the prime factorisation first. It's also not fast to check how many multiples of $17$ I can \"fit\" into the large number.\n\nSo I was wondering if there is any method out there to do this efficiently.\n\nI am looking for something like the other \"magic trick\" where you sum all the digits and take the result $\\mod 9$.\n\n\u2022 \u2013\u00a0lab bhattacharjee Aug 22 '15 at 4:24\n\u2022 Don't even think about factorisation here! Division is much faster. \u2013\u00a0TonyK Aug 22 '15 at 19:20\n\nThe best I could come up with is to use 17*6 = 102. Dividing by 102 goes pretty fast...\n\n 123456789123\n214\n105\n367\n618\n691\n792\n783\n69\n\nand 69 mod 17 = 1\n\n\nYou can speed things up by trying to eliminate two digits at a time\n\n 123456789123\n1224\n----\n1056\n1020\n----\n3678\n3672\n----\n6912\n6834\n----\n783\n714\n---\n69\n\nand 69 mod 17 = 1\n\n\n## for really large numbers\n\nFor really large numbers, you can use the fact that 17 | 100,000,001\n\nThe procedure is similar to the check for divisibility by 11, except you break the number up into larger chunks.\n\nStarting from the right, split the number up into 8-digit chunks.\nSo 123456789123 becomes\n\n chunk #         1     2\nchunk    56789123  1234\n\n\nCompute (sum of odd numbered chunks) - (sum of even numbered chunks)\n\n56789123 - 1234 = 56787889\n\n\nIf the result is negative, add a big enough multiple of 100,000,001 to\nmake it positive.\n\nThis number is congruent to the original number modulo 17.\n\n56787889\n5610\n----\n6878\n6834\n----\n4489\n4488\n-----\n1\n\n\nand, again, we get 1\n\n\u2022 This is a special case of the universal divisibility test (using various chunks sizes), see my answer. Both (well-known) methods were already mentioned in Lab's prior answer. \u2013\u00a0Bill Dubuque Oct 12 at 0:18\n\n$10^2\\equiv-2\\pmod{17}\\implies10^4=(10^2)^2\\equiv(-2)^2\\equiv4;$\n\n$\\displaystyle\\implies\\sum_{r=0}^na_r10^r\\equiv(4)^0(a_3a_2a_1a_0)+(4)^1(a_7a_6a_5a_4)++(4)^2(a_{11}a_{10}a_9a_8)+\\cdots\\pmod{17}$\n\nAgain, $10^8\\equiv(-2)^4\\equiv-1$\n\n$\\displaystyle\\implies\\sum_{r=0}^na_r10^r\\equiv(-1)^0(a_7a_6a_5\\cdots a_0)+(-1)^1(a_{15}\\cdots a_8)+\\cdots\\pmod{17}$\n\n\u2022 Undoubtedly you know this, but in case others are wondering: for any prime $p>2$ we have $10^{(p-1)/2}\\equiv\\pm1\\pmod p$. The sign here depends on whether $10$ is a quadratic residue modulo $p$ or not. That, in turn, can be easily determined using the law of quadratic reciprocity. This leads to a divisibility rule like the one here in chunks of $(p-1)/2$ digits. In some cases we can do shorter chunks. The best known cases of that are perhaps $p=13$ ($10^3\\equiv-1$) and $p=41$ ($10^5\\equiv1$). \u2013\u00a0Jyrki Lahtonen Aug 22 '15 at 19:27\n\u2022 This is a special case of the universal divisibility test (using various chunks sizes), see my answer. \u2013\u00a0Bill Dubuque Oct 12 at 0:18\n\nWell, it is fast to divide 17 into that number.\n\nWhere you can gain a lot is when the number you want to be divide is a special form such as $a^n$, where $n$ is large. There are ways (usually involving the Euler $\\phi$ function) for rapidly computing $a^n \\bmod{b}$ where $n$ is large.\n\nA good start is to remember that $a^n \\bmod{b} =(a\\bmod{b})^n \\bmod{b}$.\n\nYes, use the universal divisibility test:  repeatedly replace leading digit chunks by their remainder mod the divisor as below, using least magnitude remainders $$\\, -8\\le r \\le 9\\,$$ to simplify arithmetic $$\\!\\bmod 17\\$$ (so negative digits occur, denoted $$-d,c := 10(-d)+c\\equiv 7d+c,\\,$$ by $$10\\equiv -7)$$ \\begin{align} \\!\\bmod 17\\!:\\,\\ 10\\equiv -7\\ \\Rightarrow\\quad\\ \\ &\\ \\color{#0A0}{1\\ 2}\\,\\ 3\\ 4\\ 5\\ 6\\ \\ \\ {\\rm by}\\,\\ \\ \\ \\ \\ \\ \\color{#0a0}{1\\:\\!2\\equiv -5}\\\\[.1em] \\equiv\\, &\\ \\color{#0A0}{{-5}},\\color{#c00} 3\\ 4\\ 5\\ 6\\ \\ \\ {\\rm by}\\,\\ \\color{#0a0}{{-}5},\\color{#c00}3\\equiv\\ (\\,7\\,)\\, \\color{#0a0}{5}+\\color{#c00}3\\,\\equiv\\,\\color{#0af} 4\\\\[.1em] \\equiv\\, &\\ \\ \\ \\ \\ \\ \\ \\ \\color{#0af}4\\ 4\\ 5\\ 6\\ \\ \\ {\\rm by}\\,\\ \\ \\ \\ \\ \\ \\color{#0af}4\\:\\!4\\equiv (-7)4+4\\,\\equiv\\color{#f60}{-7}\\\\[.1em] \\equiv\\, &\\ \\ \\ \\ \\ \\ \\ \\ \\color{#f60}{{-}\\!7}, 5\\ 6\\ \\ \\ {\\rm by}\\,\\ \\color{#f60}{{-}7},5\\equiv\\,(\\,7\\,)\\, \\color{#f60}{7}+5\\,\\equiv\\, 3\\\\[.2em] \\equiv\\, &\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 3\\ 6\\ \\ \\ \\:\\!\\:\\!{\\rm by}\\ \\ \\ \\ \\ \\ 3\\:\\! 6\\equiv (-7)3+6\\,\\equiv\\, 2\\\\ \\equiv\\, &\\qquad\\qquad\\ 2;\\ \\ {\\it quicker,}\\,\\ 2 \\,\\text{ digits at a time:}\\\\[.4em] \\!\\bmod 17\\!:\\,\\ 10^2\\equiv -2\\ \\Rightarrow\\quad\\ \\ &\\ \\color{#0A0}{12\\ 3 4}\\ 56\\ \\ \\ {\\rm by}\\,\\ \\color{#0A0}{12\\ 3 4}\\equiv\\, (-2)\\color{#0a0}{\\,12+34\\equiv 10}\\\\[.1em] \\equiv\\, &\\ \\ \\ \\ \\ \\ \\color{#0A0}{{10}}\\ \\color{#c00}{56}\\ \\ \\ {\\rm by}\\,\\ \\color{#0a0}{10}\\ \\color{#c00}{56}\\equiv\\ (-2)\\, \\color{#0a0}{10}+\\color{#c00}{56}\\,\\equiv\\,\\color{#0af}2\\\\[.2em] \\equiv\\, &\\qquad\\quad\\ \\color{#0af}2 \\end{align}\\qquad\\qquad\n\nSo $$\\rm\\, 123456\\equiv 2\\pmod{\\!17}.\\,$$ Indeed $$\\rm\\, 123456 = 7262\\cdot 17+2.\\,$$ Continuing this way we can do the entire number in a couple minutes of mental arithmetic. Unlike some other divisibility tests that compute only a binary truth value, this method has the advantage of computing the remainder. Further, it doesn't require remembering any special algorithm or parameters for each modulus.\n\nRemark  Lab & Steven's answers are a special case of above (but without mod arithmetic optimizations), i.e. they use chunk sizes of $$\\,2\\,$$ and $$\\,8,\\,$$ using $$\\bmod 17\\!:\\ 10^2\\equiv -2,\\ 10^8\\equiv -1$$.\n\n\u2022 See here for another example using negative digits. \u2013\u00a0Bill Dubuque Oct 29 '18 at 23:10\n\nNo, there is not. The reason why \"magic tricks\" work when studying divisibility by $2,3,5,11$ is the fact that we usually write in base $10$. Change the base and they will stop working. In particular, the following trick would work in base $17$: if the last digit of your number is $0$ then the number is divisible by $17$. Of course, in order to check this you would need to write it in base $17$, which is a vicious circle...\n\n\u2022 Great, where can I read about the tricks you mention for $2,3,5,11$? (you don't mention $9$...?) \u2013\u00a0learner Aug 23 '15 at 4:29\n\u2022 @learner: I don't know of any book, I know them from elementary school: for $2$, the last digit must be even; for $3$ the sum of the digits must be divisible by $3$; for $4$, the number formed by the last 2 digits must be divisible by $4$; for $5$, the last digit must be $0$ or $5$; for $8$, the number formed by the last 3 digits must be divisible by $8$; for $10$, the last digit must be $0$; for $11$, sum all the digits on odd positions, sum all the digits on even positions and check whether the difference of these 2 numbers is divisible by $11$. \u2013\u00a0Alex M. Aug 28 '15 at 10:59", "date": "2020-11-28 14:40:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1405562/is-there-a-method-to-calculate-large-number-modulo", "openwebmath_score": 0.9460786581039429, "openwebmath_perplexity": 533.0015430305622, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771778588348, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8590855271673398}}
{"url": "http://mathhelpforum.com/calculus/60919-related-rates-cont-d.html", "text": "# Math Help - Related Rates cont'd...\n\n1. ## Related Rates cont'd...\n\nI have some more questions that I need help setting up. Please feel free to switch out the numbers for variables so I can learn how to solve these. Thanks in advance for your help!\n\nTwo towers (A and B) are 150m tall, 40m apart. A horizontal tightrope (height of 60m) is attached to the two buildings. A man walks across the tightrope from A to B at a constant 1/3 meter per sec. A spotlight is on the top of A. How fast is the shadow of the man moving up the wall of building B when he is 8 m away from B.\n\n(^^No clue where to begin^^)\nA trough, 12 feet long, has as its ends isosceles trapezoids with altitude 4 ft., lower base 3 ft., and upper base 4 ft. If water is let in at a rate of 7 cubic feet per minute, how fast is the water level rising when the water is 10 in. deep?\n\n(^^I found the volume of the trough to be 168 cubic feet ft...and I think the water level uses a ratio towards the trough...what next?^^)\nStacey is brewing coffee that is being strained through a conical filter with a height of 12 in. diameter 8 in. The coffee flows from the filter into a cylindrical coffee pot with base area equal to 100pi square in. The depth, h, in inches, of the coffee in the conical filter is changing at the rate of (h-12) in. per min. How fast is the depth of the coffee in the cylindrical coffee pot changing when h=3in?\n\n(^^No clue where to begin^^)\nTwo streets lights, each 30 feet tall, are 100 ft. apart. The light at the top of one of the poles is functioning properly, but the other light is burnt out. A repairman is climbing up the pole to fix the light at a rate of .5 foot per second. How fast is the tip of the repairman's shadow moving when he is 16 ft. up the pole?\n\n(^^I have a drawing set up, but I do not know how to label it and how to set up my equations^^)\n\n2. Hello, nivek516!\n\nHere's the first one . . .\n\nTwo towers, $A\\text{ and }B$. are 150m tall, 40m apart.\nA horizontal tightrope (height of 60m) is attached to the two buildings.\nA man walks across the tightrope at a constant $\\tfrac{1}{3}$ m/sec.\nA spotlight is on the top of $A$. How fast is the shadow of the man\nmoving up the wall of building $B$ when he is 8 m away from $B]$?\nFirst, make a sketch . . .\nCode:\n    A *                       * B\n|  *                    |\n90 |     *                 |\n|        *      40-x    |\nT + - - - - - * - - - - - + R\n|     x     M  *        |\n|                 *     | y\n60 |                    *  |\n|                       * S\n|                       |\n|                       |\nC * - - - - - - - - - - - * D\n40\n\nThe buildings are $AC\\text{ and }BD\\!:\\;\\;AC = BD = 150,\\;CD = 40$\n\nThe tightrope is $TR\\!:\\;\\;TR = 40,\\;TC = 60.\\;AT = 90$\n\nThe spotlight is at $A$, the man is at $M.$\n. . Let: $TM \\:=\\: x \\quad\\Rightarrow\\quad MR \\:=\\:40-x$\n. . We are given: . $\\tfrac{dx}{dt} = \\tfrac{1}{3}$ m/sec.\n\nThe man's shadow is at $S\\!:\\;\\;\\text{let } RS = y.$\n\nSince $\\Delta ATM \\sim \\Delta SRM\\!:\\;\\;\\frac{x}{90} \\:=\\:\\frac{40-x}{y} \\quad\\Rightarrow\\quad y \\:=\\:90\\,\\frac{40-x}{x} \\:=\\:90\\left(40x^{-1} - 1\\right)$\n\nDifferentiate with respect to time: . $\\frac{dy}{dt} \\:=\\:-\\frac{3600}{x^2}\\cdot\\frac{dx}{dt}$\n\nWhen $MR = 8\\;(x = 32)\\!:\\;\\;\\frac{dy}{dt} \\;=\\;-\\frac{3600}{32^2}\\left(\\frac{1}{3}\\right) \\;=\\;-\\frac{75}{64}$\n\nThe shadow is moving up building $B$ at $1\\tfrac{11}{64}\\text{ m/sec}$\n\n3. Thanks! Could you explain why the answer is negative? Does it matter since it is speed? Any possible chance you could help me with the last two questions? I found out how to do the second one.\n\n4. Hello again, nivek516!\n\nHere's the last one . . .\n\nTwo streets lights, each 30 feet tall, are 100 feet apart. The light at the top\nof one of the poles is functioning properly, but the other light is burnt out.\nA repairman is climbing up the pole to fix the light at a rate of 0.5 ft/sec.\nHow fast is the tip of the repairman's shadow moving when he is 16 ft. up the pole?\nCode:\n    A *               * C\n|   *           |\n|       *       |\n|           *   |\n30 |               * R\n|               |   *\n|              y|       *\n|               |           *\nB *---------------*---------------*\n: - - 100 - - - D - - - x - - - S\n\nThe two poles are: . $AB \\,=\\,CD\\,=\\,30$\n. . and: . $BD = 100$\n\nThe light at $A$ shines on the repairman $R$ and casts his shadow at $S.$\n. . Let: . $x \\,=\\,DS,\\;y \\,=\\,RD$\n\nSince $\\Delta RDS \\sim\\Delta ABS\\!:\\;\\;\\frac{x}{y} \\:=\\:\\frac{x+100}{30} \\quad\\Rightarrow\\quad x \\:=\\:\\frac{100y}{30-y}$\n\nDifferentiate with respect to time: . $\\frac{dx}{dt} \\;=\\;\\frac{3000}{(30-y)^2}\\,\\frac{dy}{dt}$\n\nNow substitute: . $y = 16,\\;\\;\\frac{dy}{dt} = 0.5$ . . .", "date": "2016-02-11 13:12:41", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/60919-related-rates-cont-d.html", "openwebmath_score": 0.6901357769966125, "openwebmath_perplexity": 883.8461896641982, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9867771805808551, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8590855262240585}}
{"url": "https://math.stackexchange.com/questions/1299822/prove-that-sinh-coshx-geq-cosh-sinhx", "text": "# Prove that $\\sinh(\\cosh(x)) \\geq \\cosh(\\sinh(x))$\n\nProve that\n\n$$\\sinh(\\cosh(x)) \\geq \\cosh(\\sinh(x))$$\n\nI tried to tackle this problem by integrating both lhs and rhs, in order to get two functions who show clearly that inequality holds. I've struggled for this problem a little bit, i don't know if there's any trick that can help. Maybe knowing that\n\n$$\\cosh^{-1}(x) = \\pm \\ln\\left(x + \\sqrt{x^2 - 1}\\right)$$\n\nCan help?\n\n\u2022 Jensen is useful only when we have a function and a constant, not with composite functions... Even if I think there are some similarities... \u2013\u00a0james42 May 26 '15 at 18:11\n\u2022 cheers for teaching me something new :). \u2013\u00a0Chinny84 May 26 '15 at 18:12\n\u2022 Physics guy here. I don't have any idea how to do math proofs, but isn't it true that a \"small function\" acting on a \"big function\" is always larger than the alternative? Since cosh(x)>sinh(x) for all x, this is a no brainer, right? This is how I remember e^pi>pi^e. Anyway, I've never spoken on Math.SE and generally run in fear from mathematicians, so please, let me down gently! \u2013\u00a0user1717828 May 27 '15 at 4:03\n\u2022 @user1717828 $\\ln x$ is smaller than $x^2$, but $\\ln (x^2) = 2\\ln x$ is smaller than $(\\ln x)^2$. \u2013\u00a0user21467 May 27 '15 at 4:21\n\u2022 @StevenTaschuk, aaaaannnnnddd this is why our field would crumble without mathematicians keeping us in check. Thanks! \u2013\u00a0user1717828 May 27 '15 at 4:43\n\nFor any $y \\ge 0$, notice\n\n$$e^y - 1 = \\int_0^y e^x dx \\ge \\int_0^y (1+x) dx \\ge \\int_0^y \\left(1+\\frac{x} {\\sqrt{1+x^2}}\\right)dx = y + \\sqrt{1+y^2} - 1$$ we have this little inequality: $$\\sqrt{1+y^2} - y = \\frac{1}{\\sqrt{1+y^2} + y} \\ge e^{-y}$$\n\nUsing MVT, we can find a $\\xi \\in (y,\\sqrt{1+y^2})$ such that\n\n$$\\sinh\\sqrt{1+y^2} - \\sinh(y) = \\cosh(\\xi)\\left(\\sqrt{1+y^2} - y\\right) \\ge \\cosh(\\xi) e^{-y} \\ge e^{-y}$$\n\nSince $e^{-y} = \\cosh(y) - \\sinh(y)$, this leads to $$\\sinh\\sqrt{1+y^2} \\ge \\cosh(y)\\\\$$\n\nSubstitute $y$ by $\\sinh(x)$ and notice $\\sqrt{1+y^2} = \\cosh(x)$, this reduces to our desired inequality:\n\n$$\\sinh(\\cosh(x)) \\ge \\cosh(\\sinh(x))$$\n\nHere's a solution with very little calculus. First, an identity: \\begin{align*} \\sinh^2(a+b) - \\sinh^2(a-b) &= (\\sinh a\\cosh b + \\cosh a\\sinh b)^2 - (\\sinh a\\cosh b - \\cosh a\\sinh b)^2 \\\\ &= 4\\sinh a \\cosh a\\sinh b \\cosh b \\\\ &= \\sinh(2a)\\sinh(2b) \\end{align*} Taking $a=e^x/2$ and $b=e^{-x}/2$, we get \\begin{align*} \\sinh^2(\\cosh x) - \\sinh^2(\\sinh x) &= \\sinh(e^x) \\sinh(e^{-x}) \\\\ &\\ge e^xe^{-x} &&\\text{(since $\\sinh t\\ge t$ for $t\\ge 0$)} \\\\ &= 1 \\\\ &= \\cosh^2(\\sinh x) - \\sinh^2(\\sinh x) \\end{align*} Cancelling $\\sinh^2(\\sinh x)$ and taking square roots gives the desired inequality.\n\nCalculus is needed here only to justify the inequality $\\sinh t\\ge t$ (for $t\\ge 0$).\n\nUpdate: Another nice thing about this method is that it points the way to a more exact inequality. It turns out that $$\\sinh u\\sinh v \\ge \\sinh^2\\sqrt{uv}$$ for $u,v\\ge 0$. (Proof 1: $\\frac12(u^{2m+1}v^{2n+1} + v^{2m+1}u^{2n+1})\\ge (uv)^{m+n+1}$ by AM/GM; divide by $(2m+1)!\\,(2n+1)!$ and apply $\\sum_{m=0}^\\infty \\sum_{n=0}^\\infty$. Proof 2: Check that $t\\mapsto\\ln\\sinh(e^t)$ is convex (for all $t$) by computing its second derivative.)\n\nApplying this with $u=e^x$ and $v=e^{-x}$ above, we get $$\\sinh^2(\\cosh x) \\ge \\cosh^2(\\sinh x) + \\underbrace{\\sinh^2(1) - 1}_{\\approx 0.3811}$$ with equality when $x=0$.\n\n\u2022 (+1) Very nice. Since $\\cosh(x)\\gt0$, we don't need to worry about signs when taking square roots. \u2013\u00a0robjohn May 27 '15 at 4:22\n\nLet $t=\\sinh x$. Now we can square the inequality and instead try proving $$\\sinh^2(\\sqrt{1+t^2})=\\sinh^2(\\cosh x) \\ge \\cosh^2(\\sinh x)=1+\\sinh^2t$$\n\nSo it is enough to show $f(t) = \\sinh^2(\\sqrt{1+t^2})-\\sinh^2t-1 \\ge 0$. As $f$ is even and $f(0)> 0$, it is enough to show it is increasing for positive $t$. Hence we look at $$f'(t) = \\frac{t\\sinh (2\\sqrt{1+t^2})}{\\sqrt{1+t^2}} - \\sinh(2t)$$ To show this is positive, it suffices to note by differentiating that the function $g(t) = \\dfrac{\\sinh t}{t}$ is increasing, so $g(2\\sqrt{1+t^2})> g(2t)$. Hence proved...\n\nFor $x\\ge0$, the Mean Value Theorem says that for some $\\sinh(x)\\lt\\xi\\lt\\cosh(x)$, \\begin{align} \\sinh(\\cosh(x))-\\sinh(\\sinh(x)) &=\\cosh(\\xi)(\\cosh(x)-\\sinh(x))\\\\ &\\gt\\cosh(\\sinh(x))\\,e^{-x}\\tag{1} \\end{align} Furthermore, $$\\cosh(\\sinh(x))-\\sinh(\\sinh(x))=e^{-\\sinh(x)}\\tag{2}$$ Therefore, subtracting $(2)$ from $(1)$, then applying $\\cosh(x)\\ge1$ and $\\sinh(x)\\ge x$, we get \\begin{align} \\sinh(\\cosh(x))-\\cosh(\\sinh(x)) &\\gt e^{-x}\\cosh(\\sinh(x))-e^{-\\sinh(x)}\\\\ &\\ge e^{-x}-e^{-\\sinh(x)}\\\\ &\\ge0\\tag{3} \\end{align} Since $\\sinh(\\cosh(x))-\\cosh(\\sinh(x))$ is even, $(3)$ implies that strict inequality holds for all $x$: $$\\sinh(\\cosh(x))\\gt\\cosh(\\sinh(x))\\tag{4}$$", "date": "2019-07-19 00:30:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1299822/prove-that-sinh-coshx-geq-cosh-sinhx", "openwebmath_score": 0.9998539686203003, "openwebmath_perplexity": 568.0134092269792, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771801919951, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8590855159463234}}
{"url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct", "text": "# Proving $1^3+ 2^3 + \\cdots + n^3 = \\left(\\frac{n(n+1)}{2}\\right)^2$ using induction\n\nHow can I prove that\n\n$$1^3+ 2^3 + \\cdots + n^3 = \\left(\\frac{n(n+1)}{2}\\right)^2$$\n\nfor all $n \\in \\mathbb{N}$? I am looking for a proof using mathematical induction.\n\nThanks\n\n\u2022 @Steve: See this answer for general comments on induction, and this one for specific advice on doing proofs by induction. The example there may be enough for you to figure out how to prove this statement by induction. \u2013\u00a0Arturo Magidin Sep 6 '11 at 3:34\n\u2022 Since this question is asked quite frequently, it has been added to the list of Generalizations of Common questions. It has been kept seperate from the version which does not use induction. \u2013\u00a0Eric Naslund Aug 30 '12 at 0:23\n\u2022 This answer has a great discussion of how to do an induction proof \u2013\u00a0Ross Millikan Oct 30 '17 at 21:09\n\u2022 take paper and pencil and start with $\\sum_{i=0}^{n+1 } i^3$. (and you need not bother with $i=0$) \u2013\u00a0Mirko Apr 19 '18 at 21:13\n\u2022 I would say this must have been asked on this website at least $10$ times. \u2013\u00a0Clement C. Apr 19 '18 at 21:14\n\nYou are trying to prove something of the form, $$A=B.$$ Well, both $A$ and $B$ depend on $n$, so I should write, $$A(n)=B(n).$$ First step is to verify that $$A(1)=B(1).$$ Can you do that? OK, then you want to deduce $$A(n+1)=B(n+1)$$ from $A(n)=B(n)$, so write out $A(n+1)=B(n+1)$. Now you're trying to get there from $A(n)=B(n)$, so what do you have to do to $A(n)$ to turn it into $A(n+1)$, that is (in this case) what do you have to add to $A(n)$ to get $A(n+1)$? OK, well, you can add anything you like to one side of an equation, so long as you add the same thing to the other side of the equation. So now on the right side of the equation, you have $B(n)+{\\rm something}$, and what you want to have on the right side is $B(n+1)$. Can you show that $B(n)+{\\rm something}$ is $B(n+1)$?\n\nLet the induction hypothesis be $$(1^3+2^3+3^3+\\cdots+n^3)=(1+2+3+\\cdots+n)^2$$ Now consider: $$(1+2+3+\\cdots+n + (n+1))^2$$ \\begin{align} & = \\color{red}{(1+2+3+\\cdots+n)^2} + (n+1)^2 + 2(n+1)\\color{blue}{(1+2+3+\\cdots+n)}\\\\ & = \\color{red}{(1^3+2^3+3^3+\\cdots+n^3)} + (n+1)^2 + 2(n+1)\\color{blue}{(n(n+1)/2)}\\\\ & = (1^3+2^3+3^3+\\cdots+n^3) + \\color{teal}{(n+1)^2 + n(n+1)^2}\\\\ & = (1^3+2^3+3^3+\\cdots+n^3) + \\color{teal}{(n+1)^3} \\end {align} QED\n\n\u2022 @RajeshKSingh let me know if you have questions about the steps of the proof. \u2013\u00a0user2468 Aug 29 '12 at 23:08\n\u2022 Hint: If $(1^3+2^3+3^3+\\cdots+n^3)=(1+2+3+\\cdots+n)^2$, then \\begin{eqnarray} (1^3+2^3+3^3+\\cdots+n^3 + (n+1)^3)&=&(1+2+3+\\cdots+n)^2 +(n+1)^3\\\\ &=&(\\dfrac{n^2(n+1)^2}{2} + (n+1)^3\\\\ &=& (n+1)^2(\\dfrac{n^2}{2}+n+1)\\\\ &=& (n+1)^2((n+1)^2+1)/2\\\\ &=& (1+2+ \\cdots +n +(n+1))^2 \\end{eqnarray} \u2013\u00a0user29999 Aug 29 '12 at 23:14\n\u2022 @RajeshKSingh No. If you want to prove a statement about all natural numbers then you need induction. \u2013\u00a0user2468 Aug 29 '12 at 23:16\n\u2022 @JenniferDylan: the steps are clear to me. is there a method different from induction? \u2013\u00a0HOLYBIBLETHE Aug 29 '12 at 23:25\n\u2022 @user29999 You didn't square the denominator. D: \u2013\u00a0Simply Beautiful Art Dec 9 '16 at 21:44\n\nHINT $\\:$ First trivially inductively prove the Fundamental Theorem of Difference Calculus\n\n$$\\rm\\ F(n)\\ =\\ \\sum_{i\\: =\\: 1}^n\\:\\ f(i)\\ \\ \\iff\\ \\ \\ F(n) - F(n-1)\\ =\\ f(n),\\quad\\ F(0) = 0$$\n\nThe result now follows immediately by $\\rm\\ F(n)\\ =\\ (n\\:(n+1)/2)^2\\ \\Rightarrow\\ F(n)-F(n-1)\\ =\\: n^3\\:.\\$\n\nNote that by employing the Fundamental Theorem we have reduced the proof to the trivial mechanical verification of a polynomial equation. No ingenuity is required.\n\nNote that the proof of the Fundamental Theorem is much more obvious than that for your special case because the telescopic cancellation is obvious at this level of generality, whereas it is usually obfuscated in most specific instances. For further discussion see my many posts on telescopy.\n\n\u2022 I got here from your latest post, how do you find $F(n)$ ? do you have a method or do you just 'see' this ? \u2013\u00a0Belgi Sep 19 '12 at 10:36\n\u2022 @Belgi Here the sum closed form $\\rm F(n)$ is given and we are merely asked to verify it correctness. Computing the closed form is known as summation in finite terms and there are known algorithms for wide-classes of special functions, e.g. chase citations to Michael Karr's papers on \"Summation in finite terms\". \u2013\u00a0Bill Dubuque Sep 4 '17 at 23:32\n\nLet P(n) be the given statement. You'll see why in the following step. $$P(n):1^3 + 2^3 + 3^3 + \\cdots + n^3 = \\frac{n^2(n+1)^2}{4}$$\n\nStep 1. Let $n = 1$.\n\nThen $\\mathrm{LHS} = 1^3 = 1$, $\\mathrm{RHS} = \\frac{1^2(1+1)^2}{4} = \\frac{4}{4} = 1$.\n\nSo LHS = RHS, and this means P(1) is true!\n\nStep 2. Let $P(n)$ be true for $n = k$; that is, $$1^3 + 2^3 + 3^3 + \\cdots + k^3 = \\frac{k^2(k+1)^2}{4}$$\n\nWe shall show that $P(k+1)$ is true too!\n\nAdd $(k+1)^3$, which is $(k+1)^{\\mathrm th}$ term of the LHS to both sides of (1); then we get: \\begin{align*} 1^3 + 2^3 + 3^3 + \\cdots + k^3 + (k+1)^3 &= \\frac{k^2(k+1)^2}{4} + (k+1)^3\\\\ &= \\frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\\\ &= \\frac{(k+1)^2(k^2 + 4k + 4)}{4}\\\\ &= \\frac{(k+1)^2(k+2)^2}{4}.\\\\ 1^3 + 2^3 + 3^3 + \\cdots + k^3 + (k+1)^3 &= \\frac{(k+1)^2(k+2)^2}{4} \\end{align*} I think this statement is the same as $P(n)$ with $n = k+1$.\n\n\u2022 Is there any particular reason why you are writing in bold face? It's a bit distracting, and not the usual for this site... Note also that it doesn't work for math formulas, so that makes it even more distracting... \u2013\u00a0Arturo Magidin Sep 6 '11 at 4:14\n\u2022 I wanted the math formulas to appear in bold face,but the opposite is happening.don't worry i'll edit it. \u2013\u00a0alok Sep 6 '11 at 4:17\n\u2022 You marked not just the equations, also all the text. And why was it important for the formulas to be in boldface? Again, not the usual for this site, and they don't look the same: $k$ vs. $\\mathbf{k}$, $+$ vs. $\\mathbf{+}$, $\\cdots$ vs. $\\mathbf{\\cdots}$. I've cleaned it up a bit. \u2013\u00a0Arturo Magidin Sep 6 '11 at 4:21\n\nThis picture shows that $$1^2=1^3\\\\(1+2)^2=1^3+2^3\\\\(1+2+3)^2=1^3+2^3+3^3\\\\(1+2+3+4)^2=1^3+2^3+3^3+4^3\\\\$$ this is handmade of mine\n\nFor proof by induction; these are the $\\color{green}{\\mathrm{three}}$ steps to carry out:\n\nStep 1: Basis Case: For $i=1 \\implies \\sum^{i=k}_{i=1} i^3=\\frac{1^2 (1+1)^2}{4}=\\cfrac{2^2}{4}=1$. So statement holds for $i=1$.\n\nStep 2: Inductive Assumption: Assume statement is true for $i=k$: $$\\sum^{i=k}_{i=1} i^3=\\frac{k^2 (k+1)^2}{4}$$\n\nStep 3: Prove Statement holds for $i=k+1$. You need to prove that for $i=k+1$: $$\\sum^{i=k+1}_{i=1} i^3=\\color{blue}{\\frac{(k+1)^2 (k+2)^2}{4}}$$ To do this you cannot use: $$\\sum^{i=n}_{i=1} i^3=\\color{red}{\\frac{n^2 (n+1)^2}{4}}$$ as this is what you are trying to prove.\n\nSo what you do instead is notice that: $$\\sum^{i=k+1}_{i=1} i^3= \\underbrace{\\frac{k^2 (k+1)^2}{4}}_{\\text{sum of k terms}} + \\underbrace{(k+1)^3}_{\\text{(k+1)th term}}$$ $$\\sum^{i=k+1}_{i=1} i^3= (k+1)^2\\left(\\frac{1}{4}k^2+(k+1)\\right)$$ $$\\sum^{i=k+1}_{i=1} i^3= (k+1)^2\\left(\\frac{k^2+4k+4}{4}\\right)$$ $$\\sum^{i=k+1}_{i=1} i^3= (k+1)^2\\left(\\frac{(k+2)^2}{4}\\right)=\\color{blue}{\\frac{(k+1)^2 (k+2)^2}{4}}$$\n\nWhich is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\\color{blue}{\\mathrm{blue}}$ equation at the end.\n\n(QED is an abbreviation of the Latin words \"Quod Erat Demonstrandum\" which loosely translated means \"that which was to be demonstrated\". It is usually placed at the end of a mathematical proof to indicate that the proof is complete.)\n\nIf $S_r= 1^r+2^r+...+n^r=f(n)$ and\n\n$\\sigma =n(n+1)$ & $\\sigma'=2n+1$\n\n$S_1=\\frac{1}{2}\\sigma$\n\n$S_2=\\frac{1}{6}\\sigma\\sigma'$\n\n$S_3=\\frac{1}{4}\\sigma^2$\n\n$S_4=\\frac{1}{30}\\sigma\\sigma'(3\\sigma-1)$\n\n$S_5=\\frac{1}{12}\\sigma^2(2\\sigma-1)$\n\n$S_6=\\frac{1}{42}\\sigma\\sigma'(3\\sigma^2-3\\sigma+1)$\n\n$S_7=\\frac{1}{24}\\sigma^2(3\\sigma^2-4\\sigma+2)$\n\n$S_8=\\frac{1}{90}\\sigma\\sigma'(5\\sigma^3-10\\sigma^2+9\\sigma-3)$\n\n$S_9=\\frac{1}{20}\\sigma^2(2\\sigma^3-5\\sigma^2+6\\sigma-3)$\n\n$S_{10}=\\frac{1}{66}\\sigma\\sigma'(3\\sigma^4-10\\sigma^3+17\\sigma^2-15\\sigma+5)$\n\nproof of this is based on the theorem if r is a positive integer, $s_r$ can be expressed as a polynomial in $n$ of which the highest term in $\\frac{n^{r+1}}{r+1}$\n\n\u2022 if you are interested we can discuss about the proof.! \u2013\u00a0user229886 Apr 9 '15 at 6:58\n\u2022 you can also see page 84-85-,86 of proofs without word by roger B. Nelson ,ISBN 088385-700-6 \u2013\u00a0Khosrotash Jul 21 '15 at 6:51\n\nIt may be helpful to recognize that both the RHS and LHS represent the sum of the entries in a the multiplication tables. The LHS represents the summing of Ls (I'll outline those shortly), and the RHS, the summing of the sum of the rows [or columns])$$\\begin{array}{lll} \\color{blue}\\times&\\color{blue}1&\\color{blue}2\\\\ \\color{blue}1&\\color{green}1&\\color{red}2\\\\ \\color{blue}2&\\color{red}2&\\color{red}4\\\\ \\end{array}$$ Lets begin by building our multiplication tables with a single entry, $1\\times1=1=1^2=1^3$. Next, we add the $2$s, which is represented by the red L [$2+4+2 = 2(1+2+1)=2\\cdot2^2=2^3$]. So the LHS (green 1 + red L) currently is $1^3+2^3$, and the RHS is $(1+2)+(2+4)=(1+2)+2(1+2)=(1+2)(1+2)=(1+2)^2$. $$\\begin{array}{llll} \\color{blue}\\times&\\color{blue}1&\\color{blue}2&\\color{blue}3\\\\ \\color{blue}1&\\color{green}1&\\color{red}2&\\color{maroon}3\\\\ \\color{blue}2&\\color{red}2&\\color{red}4&\\color{maroon}6\\\\ \\color{blue}3&\\color{maroon}3&\\color{maroon}6&\\color{maroon}9\\\\ \\end{array}$$ Next, lets add the $3$s L. $3+6+9+6+3=3(1+2+3+2+1)=3\\cdot3^2=3^3$. So now the LHS (green 1 + red L + maroon L) currently is $1^3+2^3+3^3$, and the RHS is $(1+2+3)+(2+4+6)+(3+6+9)=(1+2+3)+2(1+2+3)+3(1+2+3)=(1+2+3)(1+2+3)=(1+2+3)^2$.\n\nBy now, we should see a pattern emerging that will give us direction in proving the title statement.\n\nNext we need to prove inductively that $\\displaystyle\\sum_{i=1}^n i = \\frac{n(n+1)}{2}$, and use that relationship to show that $1+2+3+\\dots+n+\\dots+3+2+1 = \\dfrac{n(n+1)}{2}+ \\dfrac{(n-1)n}{2} = \\dfrac{n((n+1)+(n-1))}{2}=\\dfrac{2n^2}{2}=n^2$\n\nFinally, it should be straight forward to show that: $$\\begin{array}{lll} (\\sum^n_{i=1}i+(n+1))^2 &=& (\\sum^n_{i=1}i)^2 + 2\\cdot(\\sum^n_{i=1}i)(n+1)+(n+1)^2\\\\ &=& \\sum^n_{i=1}i^3 + (n+1)(\\sum^n_{i=1}i + (n+1) + \\sum^n_{i=1}i)\\\\ &=& \\sum^n_{i=1}i^3 + (n+1)(n+1)^2\\\\ &=& \\sum^n_{i=1}i^3 + (n+1)^3\\\\ &=& \\sum^{n+1}_{i=1}i^3\\\\ \\end{array}$$ and, as was already pointed out previously, $$(\\sum_{i=1}^1 i)^2 = \\sum_{i=1}^1 i^3=1$$\n\n$$n^3=S_n-S_{n-1}=\\left(\\frac{n(n+1)}2\\right)^2-\\left(\\frac{n(n-1)}2\\right)^2=n^2\\left(\\frac{n+1}2-\\frac{n-1}2\\right)\\left(\\frac{n+1}2+\\frac{n-1}2\\right)\\\\ =n^2\\cdot1\\cdot n.$$\n\nThis is the inductive step. The rest is easy.\n\n[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]\n\nHINT\n\nWe need to prove\n\n\u2022 Base case: set $i=1$ and check by inspection\n\u2022 Induction step: assume $\\sum_{i=0}^n i^3 = (\\sum_{i=0}^n i)^2$ true prove\n\n$$\\sum_{i=0}^{n+1} i^3 = \\left(\\sum_{i=0}^{n+1} i\\right)^2$$\n\nNote that\n\n\u2022 $\\sum_{i=0}^{n+1} i^3=(n+1)^3+\\sum_{i=0}^{n} i^3$\n\u2022 $\\left(\\sum_{i=0}^{n+1} i\\right)^2=\\left(n+1+\\sum_{i=0}^{n} i\\right)^2$\n\u2022 It helps to recall that $1+2+ \\ldots +n = n(n+1)/2.$ \u2013\u00a0Chris Leary Apr 19 '18 at 21:20\n\u2022 @ChrisLeary Oh yes, I supposed that should be well known, anyway that's good to recall! Thanks \u2013\u00a0gimusi Apr 19 '18 at 21:23\n\u2022 Sorry, that comment was for the OP, not you. I should have addressed it directly to Ryan. \u2013\u00a0Chris Leary Apr 20 '18 at 2:10\n\n[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]\n\nYou should prove before $$\\sum_{i=0}^n i = \\frac{n(n+1)}{2}$$\n\nCases $$n=0,1$$ are trivial. Suppose it's true for $$n-1$$ then \\begin{align} \\sum_{i=0}^n i^3 &= n^3 + \\sum_{i=0}^{n-1}i^3 \\\\ &=n^3 + \\left(\\sum_{i=0}^{n-1} i\\right)^2 \\\\ &= n^3 + \\frac{n^2(n-1)^2}{4}\\\\ & = \\frac{4n^3 + n^4 + n^2 + -2n^3}{4}\\\\ &= \\frac{(n+1)^2n^2}{4} \\\\ &= \\left(\\sum_{i=0}^{n} i\\right)^2 \\end{align}\n\n$$\\ S =\\sum_{k=1}^n k^3$$\n\n\\begin{align} k^3 =& Ak(k+1)(k+2)+Bk(k+1)+Ck+D \\\\ k^3= & Ak^3+(3A+B)k^2+(2A+B+C)k+D \\end{align} Therefore $A=1$, $B=-3$, $C=1$, $D=0$ and\n\n\\begin{align} S =& \\sum_{k=1}^n (k(k+1)(k+2)-3k(k+1)+k)\\\\ S =&\\sum_{k=1}^n k(k+1)(k+2)-3\\sum_{k=1}^nk(k+1)+\\sum_{k=1}^nk \\\\ S =&\\sum_{k=1}^n6\\binom{k+2}{3}-3\\sum_{k=1}^n2\\binom{k+1}{2}+\\sum_{k=1}^n\\binom{k}{1}\\\\ S=&6\\binom{n+3}{4}-6\\binom{n+2}{3}+\\binom{n+1}{2}\\\\ S=&\\left\\lgroup\\frac{n(n+1)}{2}\\right\\rgroup^2. \\end{align}\n\n\u2022 It seems that induction has been asked for. \u2013\u00a0Claude Leibovici Jul 9 '16 at 2:57\n\u2022 The question clearly asks for a proof using P.M.I. \u2013\u00a0Shubh Khandelwal Aug 24 '18 at 4:18\n\n## protected by Alex M.Aug 29 '16 at 15:16\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).\n\nWould you like to answer one of these unanswered questions instead?", "date": "2019-04-19 05:14:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct", "openwebmath_score": 0.9820051193237305, "openwebmath_perplexity": 524.842176201216, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771774699747, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8590855052938772}}
{"url": "https://math.stackexchange.com/questions/4109544/does-knowing-that-you-drew-at-least-1-ace-increase-the-probability-that-you-dr", "text": "Does knowing that you drew at least $1$ ace increase the probability that you drew at least $2$ aces ($5$ card hand)?\n\nSo say I draw $$5$$ cards from a standard deck at random without replacement and without looking at them. If someone told me that the hand contains at least $$1$$ ace, would that change the probability that my hand contains at least $$2$$ aces?\n\nI did the math and I believe the probability of drawing at least $$2$$ aces is: $$1 - \\dfrac{\\binom{48}{5}}{\\binom{52}{5}} - \\dfrac{\\binom{4}{1}\\binom{48}{4}}{\\binom{52}{5}}$$.\n\nAdditionally, what if that person then told me that the hand contains the ace of spades; would knowing that there is at least one ace and that it's the ace of spades change the probability of having at least $$2$$ aces in my hand?\n\n\u2022 I disagree with the assertion for the second part that knowing that its is an ace of spades does not change the situation. I have put up an answer that explains this. Apr 21 at 4:04\n\u2022 Have a look at the added approach. Apr 21 at 7:18\n\nIt is not correct to say that whether you are told that an ace is in the hand or whether you are told that the ace of spade is in the hand, the probability of having at least two aces is the same\n\nCase 1: You are told that you have at least one ace\n\nNote that in the following formula, the sample space shrinks to $$\\binom{52}5 - \\binom{48}5$$ to take into account that at least one ace must be present, and that in the numerator we need to ensure that one ace is there\n\n$$Pr = 1 - \\dfrac{\\binom41\\binom{48}4}{\\binom{52}5 - \\binom{48}5},\\;\\approx 0.1222$$\n\nCase 2: You are told that you have the ace of spades\n\nSinca ace of spades is assured, the sample space is now simply $$\\binom{51}{4}$$, and from the numerator, we just subtract $$\\binom{48}4$$ to indicate that more aces must be present.\n\nThus $$Pr = 1 - \\dfrac{\\binom{48}4}{\\binom{51}4}\\; \\approx 0.2214$$\n\nSince probability problems are often \"tricky\", here is an approach simpler to understand, though longer computationally.\n\nBasically, the numerator will count hands with $$2$$ or more aces, and the denominator, hands with $$1$$ or more aces\n\nCase 1: You are told that you have at least one ace\n\n$$Pr = \\dfrac{\\binom42\\binom{48}3 + \\binom43\\binom{48}2 +\\binom44\\binom{48}1}{\\binom41\\binom{48}4+\\binom42\\binom{48}3 + \\binom43\\binom{48}2 +\\binom44\\binom{48}1}\\approx 0.1222$$\n\nCase 2: You are told that you have the ace of spades\n\nBear in mind that you have the ace of spades and are left to pick from $$51$$ cards including $$3$$ aces\n\n$$Pr = \\dfrac{\\binom31\\binom{48}3 + \\binom32\\binom{48}2 +\\binom33\\binom{48}1}{\\binom30\\binom{48}4+\\binom31\\binom{48}3 + \\binom32\\binom{48}2 +\\binom33\\binom{48}1} \\approx 0.2214$$\n\n\u2022 This is correct for the second part of the problem, which I think is the more nonintuitive part. +1 and I wish I could do more. Apr 21 at 1:51\n\u2022 I agree with this calculation now. But intuitively it makes no sense. If you show someone you have an Ace but cover up the suit with your finger, then revealing the suit should have no effect on the probability. But it does. Why is this? Apr 21 at 12:53\n\u2022 @AdamRubinson: covering up and revealing the suit is different from specifying one in advance. Half of the cases with exactly two aces include the ace of spades while only one quarter of the cases with exactly one ace do. The ace of spades reduces the denominator more than it does the numerator. Apr 21 at 13:41\n\nIntuitively, you need to draw at least one ace to draw two or more aces, so knowing you have drawn at least one increases the probability you have drawn more than one. The event that your hand contains a spade also implies you drew at least one ace so the probability of two aces must also increase (edit: I should clarify that, as Ross and true blue nail have pointed out, the probability of two aces is not the same in this case as in the previous).\n\nHere is a math argument for the first case you asked about. The conditional probability of drawing at least two aces '$$A>1$$' given we drew at least one ace '$$A>0$$' is $$P(A>1|A>0) = \\frac{P(A>1,A>0)}{P(A>0)} = \\frac{P(A>1)}{P(A>0)}$$ since $$A>1$$ implies $$A>0$$ and so $$P(A>1,A>0) = P(A>1)$$. Whereas the probability of drawing two aces is just $$P(A>1)$$ which is less that $$\\frac{P(A>1)}{P(A>0)}$$ since $$P(A>0)$$ is less than $$1$$ (because not every hand contains an ace).\n\n\u2022 Ooh that makes a lot of sense. So the calculation I made was really for P(A>1), and I need to divide it by the probability of drawing at least 1 ace? Apr 20 at 12:09\n\u2022 Yes, you can just divide the calculation you made for $P(A>1)$ by $P(A>0) = 1 - \\frac{\\binom{48}{5}}{\\binom{52}{5}}$. In general, when you want the conditional probability $P(A|B)$ of event $A$ given event $B$ you need to divide the probability $P(A, B)$ of both $A$ and $B$ happening by the probability $P(B)$ of $B$ happening. The situation in your question where you can just divide $P(A)$ by $P(B)$ occurs whenever event $A$ implies event $B$. Apr 20 at 12:27\n\u2022 You imply that changing the condition from an unknown ace to the ace of spades does not change the probability of two aces. This is incorrect. The ace of spades is present in half the cases of exactly two aces but only one quarter of the cases of one ace. That almost doubles the probability of two aces given one. Apr 21 at 1:55\n\u2022 @RossMillikan You are correct and my answer could be better if I was more specific in the spades case. I think true blue nail has covered that case well already now though. I only meant to imply that in both cases the probability of two aces was different than from the situation where you have no knowledge of the number of aces in your hand, not that they were exactly the same. Apr 21 at 9:37", "date": "2021-09-21 00:12:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4109544/does-knowing-that-you-drew-at-least-1-ace-increase-the-probability-that-you-dr", "openwebmath_score": 0.8559759855270386, "openwebmath_perplexity": 201.62139745915107, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979035762240297, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8590811601858227}}
{"url": "https://mathhelpboards.com/threads/integration-question-differentiating-a-definite-integral.7749/", "text": "# Integration Question: Differentiating a definite integral\n\n#### ISITIEIW\n\n##### New member\nSo the question is\u2026Evaluate the following\u2026\n$$\\displaystyle \\frac{d}{dx} \\left(\\int _1^{x^2} \\cos(t^2) \\, dt \\right)$$\n\nI thought i could use the FTC on this because it states\u2026\n$$\\displaystyle \\frac{d}{dx} \\left(\\int_0^x f(t)\\, dt \\right)=f(x)$$\n\nbut i can't correct? because in my question it starts at 1\u2026is there some way to apply FTC to my problem or is there another general formula for this kind of problem\n\nThanks\n\nLast edited by a moderator:\n\n#### Jameson\n\nStaff member\nRe: Integration Question\n\nSince the lower bound is still a constant, it doesn't change the method you use to evaluate this. It could be at 0,1,2 or any constant and still get the same result. You're going to integrate and get a constant and then take the derivative of that constant, so you end up with 0 regardless.\n\nHere you'll need to be careful to use the chain rule when applying the FTOC. What progress have you made so far?\n\n#### MarkFL\n\nStaff member\nRe: Integration Question\n\nSo the question is\u2026Evaluate the following\u2026\nd/dx(integration of 1 to x^2 of cos(t^2)dt)\n\nI thought i could use the FTC on this because it states\u2026\nd/dx(integration of 0 to x of f(t)dt)\n\nbut i can't correct? because in my question it starts at 1\u2026is there some way to apply FTC to my problem or is there another general formula for this kind of problem\n\nThanks\nWe are given to evaluate:\n\n$$\\displaystyle \\frac{d}{dx}\\int_1^{x^2}\\cos\\left(t^2 \\right)\\,dt$$\n\nNow, the anti-derivative form of the FTOC tells us:\n\n$$\\displaystyle \\frac{d}{dx}\\int_{g(x)}^{h(x)}f(t)\\,dt=\\frac{d}{dx}\\left(F(h(x))-F(g(x)) \\right)$$\n\nOn the right, applying the chain rule, we obtain:\n\n$$\\displaystyle \\frac{d}{dx}\\int_{g(x)}^{h(x)}f(t)\\,dt=F'(h(x))h'(x)-F'(g(x))g'(x)$$\n\nSince $$\\displaystyle F'(x)=f(x)$$, we may write:\n\n$$\\displaystyle \\frac{d}{dx}\\int_{g(x)}^{h(x)}f(t)\\,dt=f(h(x))h'(x)-f(g(x))g'(x)$$\n\nApplying this to the given problem, there results:\n\n$$\\displaystyle \\frac{d}{dx}\\int_1^{x^2}\\cos\\left(t^2 \\right)\\,dt=\\cos\\left(\\left(x^2 \\right)^2 \\right)(2x)-\\cos\\left(\\left(1 \\right)^2 \\right)(0)$$\n\nSimplifying, we find:\n\n$$\\displaystyle \\frac{d}{dx}\\int_1^{x^2}\\cos\\left(t^2 \\right)\\,dt=2x\\cos\\left(x^4\\right)$$\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nbut i can't correct? because in my question it starts at 1\u2026is there some way to apply FTC to my problem or is there another general formula for this kind of problem\nLet us evaluate $$\\displaystyle \\frac{d}{dx} \\left(\\int_a^x f(t)\\, dt \\right)$$ where $$\\displaystyle 0<a<x$$\n\nThen since $$\\displaystyle \\int_0^x f(t)\\, dt =\\int_0^a f(t)\\, dt+ \\int_a^x f(t)\\, dt$$\n\n$$\\displaystyle \\int_a^x f(t)\\, dt=-\\int_0^a f(t)\\, dt+\\int_0^x f(t)\\, dt$$\n\nIf we differentiate then since the first integral is independent of $x$ we have\n\n$$\\displaystyle \\frac{d}{dx} \\left(\\int_a^x f(t)\\, dt \\right)=\\frac{d}{dx} \\left(\\int_0^x f(t)\\, dt \\right)=f(x)$$\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nThe \"Leibniz formula\", a generalization of the fundamental theorem of calculus, says\n$$\\frac{d}{dx}\\int_{\\alpha(x)}^{\\beta(x)} f(x,t)dt= \\frac{d\\beta}{dx}f(x, \\beta(x))- \\frac{d\\alpha}{dx}f(x, \\alpha(x))+ \\int_{\\alpha(x)}^{\\beta(x)} \\frac{\\partial f}{\\partial x} dt$$", "date": "2021-01-21 01:34:31", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/integration-question-differentiating-a-definite-integral.7749/", "openwebmath_score": 0.9428406357765198, "openwebmath_perplexity": 753.0663369682322, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357542883922, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8590811547766679}}
{"url": "https://math.stackexchange.com/questions/1680354/can-a-graph-be-strongly-and-weakly-connected", "text": "# Can a graph be strongly and weakly connected?\n\nI'm currently revising course notes on directed graphs.\n\nIt says that a directed graph (digraph) is strongly connected if there is a path between every pair of vertices.\n\nIt also says that a digraph is weakly connected if the underlying undirected graph is connected.\n\nMy question is, can one digraph be both strongly and weakly connected?\n\nFor example: Digraph and undirected graph\n\nCan this graph (image) be both strongly and weakly connected? or does it have to be either strongly, or either weakly?\n\nThank you.\n\nAs suggested by the terminology, any strongly connected graph is weakly connected, but a weakly connected graph is not necessarily strongly connected. For instance, the graph $1 \\to 2$ is weakly connected but is not strongly connected.\n\nYes, a graph can, according to the provided definitions, definitely be both weakly and strongly connected at the same time. Your example is exactly such a graph.\n\nIn fact, all strongly connected graphs are also weakly connected, since a directed path between two vertices still connect the vertices upon removing the directions.\n\nTo some extent this is a question about word usage in mathematics. One natural reading of the English term \"weakly connected\" would be \"the connectedness of the graph is weak\", weak meaning that it is deficient in some sense like a weak signal.\n\nHowever, the use of \"weakly\" here is more of a conjugation of the term \"weak connectedness\", i.e. connected in a weak sense of the word \"connected\", with weak meaning that it is not a very strenuous requirement. A graph is weakly connected if it satisfies this weak connectedness requirement, not because it is less strongly connected than other graphs.\n\nThis is a fairly common idiom in math. In functional analysis one might refer to a sequence that is \"weakly convergent\" (again meaning convergent in a more generous sense that has a precise definition), and then later show that the same sequence is in fact strongly convergent.\n\nThere are other cases where the terminology is indeed exclusive, unlike the present case: when an infinite series is \"conditionally convergent\", the definition specifically excludes the case that it is absolutely convergent (which is a stronger notion of convergence).", "date": "2020-10-23 22:25:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1680354/can-a-graph-be-strongly-and-weakly-connected", "openwebmath_score": 0.7758089900016785, "openwebmath_perplexity": 415.9837092978464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357561234474, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8590811532499719}}
{"url": "https://math.stackexchange.com/questions/2963810/find-the-range-of-a-if-a-sin20x-cos48x/2963828", "text": "Find the range of $A$ if $A=\\sin^{20}x+\\cos^{48}x$\n\nFind the range of $$A$$ if $$A=\\sin^{20}x+\\cos^{48}x$$\n\n$$A'=20\\sin^{19}x\\cos x-48\\cos^{47}x\\sin x=0\\implies5\\sin^{19}x\\cos x=12\\cos^{47}x\\sin x\\\\ \\implies5\\sin^{18}x=12\\cos^{46}x$$\n\nHow do I proceed further and prove that $$A\\in(0,1]$$ ?\n\nIs it possible to find the range of $$A$$ without using differentiation ?\n\nNote: $$\\sin^2 x,\\cos^2 x\\in[0,1]\\implies A\\in[0,2]$$ but $$2$$ is not \"the\" maximum value of $$A$$\n\n\u2022 Both $\\sin$ and $\\cos$ have a minimum value of $-1$ and a maximum value of $1$, so what can you deduce from that? \u2013\u00a0Robert Howard Oct 20 '18 at 21:55\n\u2022 @RobertHoward $A$ should be between $0$ and $2$, but how do I find the the maximum value ? \u2013\u00a0ss1729 Oct 20 '18 at 21:58\n\u2022 observe that since $\\sin x$ and $\\cos x$ are less or equal to one then we have $\\sin^{20} x + \\cos^{48} \\leq \\sin^2x +\\cos^2 x = 1$ \u2013\u00a0ALG Oct 20 '18 at 22:05\n\u2022 Good point; I should have looked at the question more carefully. From the equation you ended with, I would try re-expressing all the powers of $\\sin^2x$ in terms of $\\cos^2x$ and see if that helps. \u2013\u00a0Robert Howard Oct 20 '18 at 22:06\n\nIf $$0\\leq a\\leq 1$$ then $$a^2\\leq a$$, so we use that repetadly\n\n$$A=\\sin^{20}x+\\cos^{48}x\\leq \\sin^{2}x+\\cos^{2}x =1$$\n\nSo $$A_{\\max}=1$$ and it is achieved at $$x=0$$.\n\nFor a minimum I don't see quick solution. I would try like this: Let $$t= \\cos^2x$$ and then search for the minumum of $$g(t) = (1-t)^{10}+t^{24}$$ where $$0\\leq t\\leq 1$$.\n\nNote that $$A = (1 - \\cos^2 x)^{10} + \\cos^{48} x$$. Letting $$t = \\cos^2 x$$, $$t \\in [0,1]$$, we have $$A = (1-t)^{10} + t^{24}$$. This function has maximum in 0 and 1, and $$A_{max} = 1$$. The minimum is found by differentiating and solving $$24t^{23} - 10(1-t)^9 = 0$$, which leads (numerically) to $$t_{min} \\simeq 0,643187$$, and correspondingly to $$A_{min} = (1-t_{min})^{10} + t_{min}^{24} \\simeq 0,0000585751.$$\n\n\u2022 WolframAlpha gives the same minimum, achieved at $x=.640178.$ \u2013\u00a0saulspatz Oct 20 '18 at 22:20\n\nApart from the trivial upper bound $$A\\le 2$$, we have the stronger (and sharp - try $$x=0$$) bound $$\\tag1A\\le 1.$$\n\nConsider $$f(x):=(1-x)^{10}+x^{24}$$ for $$0\\le x\\le 1$$. Then $$f'(x)=24x^{23}-10(1-x)^9$$ is strictly increasing (as each summand is) on $$[0,1]$$, hence has at most one root there. As $$f'(0)=-10$$ and $$f'(1)=24$$, we conclude that there is exactly one such root $$\\alpha$$. As $$f'$$ goes from negative to positive, $$f$$ must have a local minimum there. We conclude that $$f$$ has its only minimum at $$\\alpha$$ and its maximum at the boundary - in fact, at both ends of the boundary since $$f(0)=f(1)$$. As $$A=f(\\cos^2 x)$$ and $$\\cos^2 x$$ ranges from $$0$$ to $$1$$, inclusive, we conclude that the maximal value of $$A$$ is also $$1$$ (thus proving $$(1)$$), and the minimum value of $$A$$ is $$f(\\alpha)$$.\n\nUsing $$(1-\\alpha)^9=\\frac{12}5\\alpha^{23}$$, we have $$f(\\alpha)=(1-\\alpha)^{10}+\\alpha^{24}=(1-\\alpha)\\cdot\\frac{12}5\\alpha^{23}+\\alpha^{24}=\\alpha^{23}\\cdot \\frac{12-7\\alpha}5=(1-\\alpha)^9\\cdot \\frac{12-7\\alpha}{12},$$ so certainly $$\\min A=\\min f>0,$$ but not by much.\n\nFrom $$f'(\\frac 35)=24\\frac{3^{23}}{5^{23}}-10\\frac{2^{9}}{5^{9}}=\\frac{24\\cdot 3^{23}-10\\cdot 2^95^{14}}{5^{23}}<0$$(!), we conclude that $$\\alpha>\\frac35$$ and hence $$f(\\alpha)=(1-\\alpha)^9\\cdot \\frac{12-7\\alpha}{12}<(1-\\tfrac35)^9\\cdot \\frac{12-7\\cdot\\frac35}{12}=\\frac{1664}{9765625}\\approx 0.00017$$ (whereas the true minimal value is $$\\approx 0.000058575$$)", "date": "2019-06-27 04:14:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2963810/find-the-range-of-a-if-a-sin20x-cos48x/2963828", "openwebmath_score": 0.9546403884887695, "openwebmath_perplexity": 122.34824560462499, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357561234475, "lm_q2_score": 0.8774767746654976, "lm_q1q2_score": 0.8590811375653994}}
{"url": "https://math.stackexchange.com/questions/992436/integrating-int-infty-infty-frac11-x4dx-with-the-residue-theorem", "text": "# Integrating $\\int_{-\\infty}^\\infty \\frac{1}{1 + x^4}dx$ with the residue theorem\n\nCalculate integral\n\n$$\\int\\limits_{-\\infty}^{\\infty}\\frac{1}{x^4+1} dx$$ with residue theorem.\n\nCan I evaluate $\\frac 12\\int_C \\dfrac{1}{z^4+1} dz$ where $C$ is simple closed contour of the upper half of unit circle like this?\n\nAnd find the roots of polynomial $z^4 +1$ which are the fourth roots of $-1$.\n\nIn $C$ there is $z_1 =e^{i\\pi/4}=\\frac{1}{\\sqrt{2}}+\\frac{i}{\\sqrt{2}}$ and $z_2=e^{3\\pi/4}=-\\frac{1}{\\sqrt{2}}+\\frac{i}{\\sqrt{2}}$.\n\nSo the residuals $B_1$ and $B_2$ for $z_1$ and $z_2$ are simple poles and that \\begin{align} B_1&=\\frac{1}{4 z_1^3}\\frac{z_1}{z_1}=-\\frac{z_1}{4} \\\\ B_2&=\\frac{1}{4z_2^3}\\frac{z_2}{z_2}=-\\frac{z_2}{4} \\end{align} And the sum of residuals is\n\n$$B_1+B_2=-\\frac{1}{4}(z_1 + z_2)=-\\frac{1}{4}\\left(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{2}} \\right)=-\\frac{i}{2 \\sqrt{2}}$$\n\nSo my integral should be\n\n$$\\int\\limits_{-\\infty}^{\\infty}\\frac{1}{x^4+1} dx =\\frac 12 \\times 2\\pi i (B_1+B_2)=\\frac{\\pi}{\\sqrt{2}}$$\n\nIs this valid?\n\n\u2022 Yes. Peachy! ;-$)$ \u2013\u00a0Lucian Oct 26 '14 at 22:44\n\n\nI thought that it might be instructive to present an alternative and efficient approach. To do so, we first note that from the even symmetry that\n\n$$\\int_{-\\infty}^\\infty \\frac{1}{1+x^4}\\,dx=2\\int_{0}^\\infty \\frac{1}{1+x^4}\\,dx \\tag 1$$\n\nWe proceed to evaluate the integral on the right-hand side of $(1)$.\n\nNext, we move to the complex plane and choose as the integration contour, the quarter circle in the upper-half plane with radius $R$. Then, we can write\n\n\\begin{align} \\oint_C \\frac{1}{1+z^4}\\,dz&=\\int_0^R \\frac{1}{1+x^4}\\,dx+\\int_0^\\pi \\frac{iRe^{i\\phi}}{1+R^4e^{i4\\phi}}\\,d\\phi+\\int_R^0 \\frac{i}{1+(iy)^4}\\,dy\\\\\\\\ &=(1-i)\\int_0^R\\frac {1}{1+x^4}\\,dx+\\int_0^\\pi \\frac{iRe^{i\\phi}}{1+R^4e^{i4\\phi}}\\,d\\phi \\tag 2 \\end{align}\n\nAs $R\\to \\infty$, the second integral on the right-hand side of $(2)$ approaches $0$ while the first approaches the $1/2$ integral of interest. Hence, we have\n\n$$\\bbox[5px,border:2px solid #C0A000]{\\lim_{R\\to \\infty}\\oint_C \\frac{1}{1+z^4}\\,dz=\\frac{1-i}2 \\int_{-\\infty}^\\infty \\frac{1}{1+x^4}\\,dx} \\tag 3$$\n\nNow, since $C$ encloses only the pole at $z=e^{i\\pi/4}$, the Residue Theorem guarantees that\n\n\\begin{align}\\oint_C \\frac{1}{1+z^4}\\,dz&=2\\pi i \\text{Res}\\left(\\frac{1}{1+z^4}, z=e^{i\\pi/4}\\right)\\\\\\\\ &=\\frac{2\\pi i}{4e^{i3\\pi/4}}\\\\\\\\ &=\\frac{\\pi e^{-i\\pi/4}}{2}\\\\\\\\ &=\\bbox[5px,border:2px solid #C0A000]{\\frac{\\pi(1-i)}{2\\sqrt 2}} \\tag 4 \\end{align}\n\nFinally, equating $(3)$ and $(4)$, we find that\n\n$$\\bbox[5px,border:2px solid #C0A000]{\\int_{-\\infty}^\\infty \\frac{1}{1+x^4}\\,dx=\\frac{\\pi}{\\sqrt{2}}}$$\n\nas expected.\n\n\u2022 @elec It's curious that you requested specifically to evaluate the integral using the residue theorem, yet awarded the best answer for a real analysis approach. \u2013\u00a0Mark Viola Nov 17 '16 at 15:09", "date": "2019-05-23 11:09:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/992436/integrating-int-infty-infty-frac11-x4dx-with-the-residue-theorem", "openwebmath_score": 0.999734103679657, "openwebmath_perplexity": 614.7518432475417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232940063591, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.859063469092649}}
{"url": "https://math.stackexchange.com/questions/548733/mutual-tangent-lines", "text": "# Mutual tangent lines\n\nFind all points where the curves $f(x) = x^3-3x+4$ and $g(x) = 3x^2-3x$ share the same tangent line.\n\nGraphing them I see that they look like they share a tangent line at $x=2$.\n\nI got the derivatives of both and set them equal to each other and got $x=0$ and $x=2$.\n\nAfter plugging $2$ back in I got $6$. So the point is $(2,6)$.\n\nIs that correct?\n\n\u2022 In hindsight, this is not a duplicate. It appeared to be a duplicate because the OP originally had a scan of a problem sheet that included this problem, as well as the one from the other question. I have altered the other question to make this clear. \u2013\u00a0Cameron Buie Nov 2 '13 at 13:15\n\n$$f' = g ' \\iff 3x^2 - 3 = 6x - 3 \\iff x^2 -2x = 0 \\iff x(x-2) = 0 \\iff x=0,2$$\n$$(2, 6 )$$ as you said. Notice at $0$ functions are not equal, so they cannot share a tangent line at $0$", "date": "2019-11-20 16:42:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/548733/mutual-tangent-lines", "openwebmath_score": 0.5496748089790344, "openwebmath_perplexity": 175.43968324081402, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232914907945, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8590634652819085}}
{"url": "https://math.stackexchange.com/questions/2269505/question-from-berkley-problems-in-mathematics-show-that-deta-0/2269512", "text": "# Question from Berkley Problems in Mathematics. Show that det($A$) > 0 [duplicate]\n\nLet $A = (a_{ij})$ be an $n\\times n$ real matrix satisfying the conditions: $a_{ii} > 0$ $(1\\leq i \\leq n)$; $a_{ij} \\leq0$ $(i\\not=j,1\\leq i,j \\leq n)$; $\\sum\\limits_{i=1}^n a_{ij}>0$ $(1\\leq i \\leq n)$. Show that det($A$) > $0$ .\n\n## marked as duplicate by mlc, user91500, user26857, The Dead Legend, Namaste\u00a0linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 7 '17 at 11:17\n\n\u2022 Hint: Show that if $\\lambda$ is a real eigenvalue then it is positive. \u2013\u00a0user379195 May 7 '17 at 6:54\n\nThe matrix $A$ is diagonally dominant and so nonsingular. Replacing off-diagonal entries $a_{i,j}$ by $ta_{i,j}$ for $0\\le t\\le 1$ also gives a diagonally dominant and nonsingular matrix $A_t$. But $A_0$ is diagonal with positive determinant. By continuity of determinant $\\det A_t>0$ and so $\\det A=\\det A_1>0$.\n\n\u2022 A weakly diagonally dominant matrix may be singular. \u2013\u00a0mlc May 7 '17 at 7:00\n\u2022 @mlc There's nothing weak about these matrices! \u2013\u00a0Lord Shark the Unknown May 7 '17 at 7:01\n\u2022 \\begin{bmatrix} 1 & -2 & -2 \\\\ -1 & 6 & -1 \\\\ -1 & -1 & 3 \\end{bmatrix} what about this matrix; It's not a DD matrix. But confusingly, it is satisfying the question but not the result to be shown. I must be wrong somwhere. \u2013\u00a0Hirakjyoti Das May 7 '17 at 7:09\n\u2022 @HirakjyotiDas one of your conditions is that the column sums are positive. \u2013\u00a0Lord Shark the Unknown May 7 '17 at 7:13\n\u2022 That's the very thing I was missing; Thank you \u2013\u00a0Hirakjyoti Das May 7 '17 at 7:16\n\nIn addition to the other answers, one can more generally see that from the Gershgorin circle theorem\n\nHere firstly $A$ is diagonally dominant. Secondly $A$ is $Z$-matrix as all its off diagonal entries are $\\leq 0.$\n\nAlso diagonal entries are $> 0.$ Hence $A$ is a $P$-matrix. (All principal minors are positive).\n\nSo $\\det(A) > 0.$", "date": "2019-09-21 17:53:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2269505/question-from-berkley-problems-in-mathematics-show-that-deta-0/2269512", "openwebmath_score": 0.8717315793037415, "openwebmath_perplexity": 1488.0928414290352, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232869627775, "lm_q2_score": 0.8740772384450968, "lm_q1q2_score": 0.8590634645479575}}
{"url": "https://math.stackexchange.com/questions/1418629/why-is-this-approximation-of-polynomial-root-so-accurate", "text": "# Why is this approximation of polynomial root so accurate?\n\nI have an engineering problem where I have to find the smallest positive real root of a polynomial in $x$: $$Ax^5+Bx^3 - C = 0$$ Instead of solving numerically, I want simple approximative formulas (\"design equations\") that describe the behaviour. For that matter, I split the problem into two regimes:\n\n\u2022 Large $x$: $\\ \\ \\ \\ x^5$ is dominant $\\Longrightarrow \\ \\ Ax^5 \\approx C \\ \\ \\Longrightarrow \\ \\ x \\approx \\sqrt[5]{C/A} =: x_A$\n\u2022 Small $x$: $\\ \\ \\ \\ x^3$ is dominant $\\Longrightarrow \\ \\ Bx^3 \\approx C \\ \\ \\Longrightarrow \\ \\ x \\approx \\sqrt[3]{C/B} =: x_B$\n\nThe approximations work well below/above a certain threshold on $x$, but require an ugly case distinction. In order to avoid that and having something smoother than $$x \\approx \\min\\{x_A, x_B\\}$$ I tried Pythagorean-style combination of the inverses (inspired by the resistance of parallel circuits in electrical engineering) and found that $$x \\approx 1\\Big/ \\ \\left\\|\\left(\\begin{matrix} 1/x_A \\\\ 1/x_B \\end{matrix}\\right)\\right\\|_4 = 1 \\Big/ \\sqrt[4]{1/x_A^4 + 1/x_B^4}$$ is a really, really good approximation. Pretty much perfect. That leads me to my actual question: Is it possible to argue why that 4-norm is such an amazing approximation? Does my initial polynomial have a certain structure that explains the high accuracy of my approximation?\n\nSince I want to present/defend that stuff, I'd appreciate some sophistication.\n\n\u2022 Are $A,B,C$ positive? If so the \"smallest positive real root\" is the only real root. \u2013\u00a0alex.jordan Sep 2 '15 at 17:19\n\u2022 For example, the sole real root of $17x^5 + 41x^3 - 105$ is approximately $1.17624...$; the approximation gives $1.17849...$. \u2013\u00a0Unit Sep 2 '15 at 17:20\n\u2022 @alex.jordan Yes they are positive. And sure, since $f(x) = Ax^5 + Bx^3$ is strictly increasing, there is only one relevant solution. \u2013\u00a0GDumphart Sep 2 '15 at 17:21\n\u2022 What do you mean by a \"really, really good approximation\", and have you tried lots of values of $A$, $B$ and $C$? \u2013\u00a0David Quinn Sep 2 '15 at 17:27\n\u2022 @DavidQuinn No I haven't done exhaustive tests with all kinds of $A,B$ combos, only what my problem yielded. Here is what I mean with great approximation: i.imgur.com/7Bfl8nt.png Only ridiculous zooming exposes some error. \u2013\u00a0GDumphart Sep 2 '15 at 17:41\n\n$$\\left(\\frac x{x_A}\\right)^5+\\left(\\frac x{x_B}\\right)^3-1=0.$$\n\nWith $y=\\dfrac x{x_A}$ and $r=\\dfrac{x_A}{x_B}$, a single parameter remains:\n\n$$y^5+r^3y^3-1=0.$$\n\nThen $$r=\\sqrt[3]{\\frac{1-y^5}{y^3}}=\\frac1y\\sqrt[3]{1-y^5},$$ can be compared to your approximation $$y=\\frac1{\\sqrt[4]{1+r^4}},$$i.e. $$r=\\sqrt[4]{\\frac1{y^4}-1}=\\frac1y\\sqrt[4]{1-y^4}.$$ The agreement is indeed excellent on a wide range [abscissa $y$, ordinate $r$]:\n\nFor small $y$, both behaviors are identical, $\\dfrac1y$. For $y$ close to $1$, behaviors are similar, approximately $\\sqrt[3]{5(1-y)}$ and$\\sqrt[4]{4(1-y)}$, and a blend in between.\n\nThe good agreement is explained by the fact that the functional relations are similar, with the exponent $4$ intermediate between $3$ and $5$ (but the value $4$ has nothing \"magical\").\n\n\u2022 Using Yves's parametrization, the absolute value of the difference between the true $y$ and your approximation turns out to be at most $0.01171$. The maximum absolute difference occurs at approximately $r = 0.66145$, where the true $y \\approx 0.94545$ while the approximation is $0.95716$. \u2013\u00a0Robert Israel Sep 2 '15 at 18:50\n\u2022 @RobertIsrael: you can try slightly lower values than $4$ for the exponent, such as $3.7$. \u2013\u00a0Yves Daoust Sep 2 '15 at 18:53\n\u2022 Yes, I can. The best in terms of maximum absolute difference seems to be approximately $3.85$ (i.e. approximation $(1 + r^{3.85})^{-1/3.85}$), where the maximum absolute difference is about $0.007608$. \u2013\u00a0Robert Israel Sep 2 '15 at 21:44\n\u2022 Does this mean that there's an exact solution, rather than the approximation that the OP is using \u2013\u00a0Dr Xorile Sep 2 '15 at 22:03\n\u2022 Amazing answer, very intuitive and accounts for everything I wished for, thank you. I reproduced all your steps. \u2013\u00a0GDumphart Sep 3 '15 at 6:34", "date": "2020-01-26 12:39:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1418629/why-is-this-approximation-of-polynomial-root-so-accurate", "openwebmath_score": 0.7773688435554504, "openwebmath_perplexity": 705.4742662760644, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232894783426, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8590634602989846}}
{"url": "http://mathhelpforum.com/statistics/114604-probability-problem-about-cubes.html", "text": "# Math Help - Probability problem about cubes\n\n1. ## Probability problem - cubes\n\nEach face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probabilty that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?\n\nthere can be 64 different color cubes 2x2x2x2x2x2x2 ?\n\ncase 1 when the cube is all red or blue it will work - 2/64\n\nI couldn't think of the cases...\n\nCould someone just give me some hints on the cases I have to set up?\n\nVicky.\n\n2. Originally Posted by Vicky1997\nEach face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?\n\nthere can be 64 different color cubes 2x2x2x2x2x2x2 ?\n\ncase 1 when the cube is all red or blue it will work - 2/64\n\nI couldn't think of the cases...\n\nCould someone just give me some hints on the cases I have to set up?\n\nVicky.\nThat's a good start. You can continue it like this. If one face is one colour and the other five faces are all the other colour then it will work (2\u00d76 cases of that). If two faces are one colour (and the other four faces are the other colour) then it will work provided that the two faces are opposite each other (2\u00d73 cases, because there are three pairs of opposite faces). Finally, it can't be done at all if there are three faces of each colour.\n\n3. Hello, Vicky1997!\n\nA fascinating problem . . .\n\nEach face of a cube is painted either Red or Blue, each with probability 1/2.\nThe color of each face is determined independently.\nWhat is the probabilty that the painted cube can be placed on a horizontal surface\nso that the four vertical faces are all the same color?\n\nThere can be 64 different colored cubes: $2^6 = 64$ . . . . Right!\nThere are seven possible colorings:\n\n. . $\\begin{array}{|c|c|} \\hline\n\\text{Colors} & \\text{Ways} \\\\ \\hline\n\\text{6 red, 0 blue} & 1 \\\\ \\text{5 red, 1 blue} & 6 \\\\ \\text{4 red, 2 blue} & 15 \\\\ \\text{3 red, 3 blue} & 20 \\\\ \\text{2 red, 4 blue} & 15 \\\\ \\text{1 red, 5 blue} & 6 \\\\ \\text{0 red, 6 blue} & 1 \\\\ \\hline \\end{array}$\n. . You may recognize these as Binomial Coefficients.\n\nLet $V$ = \"the four vertical faces have the same color\".\n\n6 red, 0 blue: . $P(\\text{6 red}) \\:=\\:\\frac{1}{64}$\nTo have $V$, the cube can be placed on any face.\n. . $P(\\text{6 red} \\wedge V)\\:=\\:\\frac{1}{64}\\cdot\\frac{6}{6} \\:=\\:\\frac{6}{384}$\n\n5 red, 1 blue: . $P(\\text{5 red}) \\:=\\:\\frac{6}{64}$\nTo have $V$, the blue face must be on the top or bottom\n. $P(\\text{5 red} \\wedge V) \\:=\\:\\frac{6}{64}\\cdot\\frac{2}{6} \\:=\\:\\frac{12}{384}$\n\n4 red, 2 blue.\nTo have $V$, the 2 blues must be on opposite faces.\n. . This happens only $\\frac{3}{64}$ of the time.\nThen the blue faces must be on the top and the bottom.\n. . $P(\\text{4 red} \\wedge V) \\:=\\:\\frac{3}{64}\\cdot\\frac{2}{6} \\:=\\:\\frac{6}{384}$\n\n3 red, 3 blue.\nWith 3 of each color, it is impossible to have $V$.\n\n2 red, 4 blue.\nThis has the same probability as \"4 red, 2 blue\".\n. . $P(\\text{2 red} \\wedge V) \\:=\\:\\frac{6}{384}$\n\n1 red, 5 blue.\nThis has the same probability as \"5 red, 1 blue\".\n. . $P(\\text{1 red} \\wedge V) \\:=\\:\\frac{12}{384}$\n\n0 red, 6 blue.\nThis has the same probability as \"6 red, 0 blue\".\n. . $P(\\text{0 red} \\wedge V) \\:=\\:\\frac{6}{384}$\n\nTherefore: . $P(V) \\;=\\;\\frac{6}{384} + \\frac{12}{384} + \\frac{6}{384} + \\frac{6}{384} + \\frac{12}{384} + \\frac{6}{384} \\;=\\;\\frac{48}{384} \\;=\\;\\frac{1}{8}$", "date": "2016-05-05 23:02:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/114604-probability-problem-about-cubes.html", "openwebmath_score": 0.7375345230102539, "openwebmath_perplexity": 481.67986572613455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232869627774, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8590634564882434}}
{"url": "https://dsp.stackexchange.com/questions/67655/relationship-between-input-and-output-sequence-in-hartley-transformation", "text": "# Relationship between input and output sequence in Hartley transformation\n\nAs you know that Discrete Hartley transformation is related to the discrete Fourier transformation, $$i.e$$, assuming we have a vector $$X = [x_0,x_1,\\ldots,x_N]$$, its Hartley transformation is equal to $$H(X) = Real(FFT(X)) - Imag(FFT(X))$$. where $$H$$ denotes the Hartley transformation.\n\nI wonder, if I want the output of the Hartley transformation equals to a vector of length $$N$$ whose all elements are $$1$$, it means $$H(X) = [1,1,\\ldots,1]$$, what supposed to be the input $$X$$? I means I need to formulate the relationship between the inputs and outputs.\n\nWikipedia's entry for the discrete Hartley transform shows states that the $$\\mathsf{DHT}$$ is, up to a scaling, its own inverse. If $$x$$ is a vector with $$N$$ entries and $$y$$ is its discrete Hartley transform, $$$$y = \\mathsf{DHT}x,$$$$ then $$$$x = \\frac{1}{N}\\mathsf{DHT}y.$$$$\n\nIf $$x$$ is a vector with $$N$$ entries such that $$$$\\mathsf{DHT}x = \\underbrace{(1,1,\\ldots,1)^{\\mathsf{T}}}_{\\textrm{N entries}},$$$$ then we recover $$x$$ with $$$$x = \\frac{1}{N}\\mathsf{DHT}\\left(\\begin{array}{c}1\\\\1\\\\\\vdots\\\\1\\end{array}\\right).$$$$ This means that $$x$$ is $$$$\\begin{split} x &=~ \\frac{1}{N}\\left(\\mathsf{Re}\\left[\\mathsf{DFT}\\left(\\begin{array}{c}1\\\\1\\\\\\vdots\\\\1\\end{array}\\right)\\right] - \\mathsf{Im}\\left[\\mathsf{DFT}\\left(\\begin{array}{c}1\\\\1\\\\\\vdots\\\\1\\end{array}\\right)\\right]\\right), \\end{split}$$$$ where $$\\mathsf{DFT}$$ is the discrete Fourier transform, which we usually compute with a FFT algorithm. The $$\\ell^{\\textrm{th}}$$ entry of the $$\\mathsf{DFT}$$ of the all-1 vector is $$$$\\begin{split} \\sum_{n=0}^{N-1}1\\times e^{-2\\pi j \\ell n/N} &=~ 1 + e^{-2\\pi j \\ell/N} + \\left(e^{-2\\pi j \\ell/N}\\right)^2 + \\cdots + \\left(e^{-2\\pi j \\ell/N}\\right)^{N-1}\\\\ &=~ \\left\\{\\begin{array}{rl}N & \\textrm{if}~\\ell=0,\\\\0&\\textrm{if}~\\ell\\neq 0.\\end{array}\\right.\\\\ &=~ N\\delta_{\\ell,0}, \\end{split}$$$$ where $$\\delta_{p,q}$$ is the Kronecker delta. One way to show this is to note that if $$\\ell = 0$$, then each exponent is $$0$$, so each term in the sum is $$1$$. On the other hand, if $$\\ell\\neq 0$$, then $$\\exp(-2\\pi j\\ell/N) \\neq 1$$,and $$$$\\begin{split} 1 + e^{-2\\pi j \\ell/N} + \\left(e^{-2\\pi j \\ell/N}\\right)^2 + \\cdots + \\left(e^{-2\\pi j \\ell/N}\\right)^{N-1} &=~ \\frac{1 - \\left(e^{-2\\pi j \\ell/N}\\right)^N}{1 - e^{-2\\pi j \\ell/N}}\\\\ &=~ \\frac{1 - e^{-2N\\pi j \\ell/N}}{1 - e^{-2\\pi j \\ell/N}}\\\\ &=~ \\frac{1 - 1}{1 - e^{-2\\pi j \\ell/N}} ~=~ 0. \\end{split}$$$$\n\nThat shows that the $$\\mathsf{DFT}$$ of the all-1 vector has no imaginary part, and its real part is $$(N,0,0,\\ldots,0)^{\\mathsf{T}}$$. Hence $$$$x ~=~ \\frac{1}{N}\\left(\\begin{array}{c} N\\\\0\\\\0\\\\\\vdots\\\\0 \\end{array}\\right) ~=~ \\left(\\begin{array}{c} 1\\\\0\\\\0\\\\\\vdots\\\\0 \\end{array}\\right).$$$$\n\n\u2022 Thank you so much .. that's really appreciated.\n\u2013\u00a0Gze\nMay 20 '20 at 3:19\n\u2022 Thanks again, that's really interesting. I am trying to understand it well, if I got a question, I will let you know.\n\u2013\u00a0Gze\nMay 20 '20 at 6:36\n\u2013\u00a0Gze\nMay 20 '20 at 11:46\n\nThe Hartley transform is an involution: it is (up to a scale factor) its own inverse. The classical discrete Hartley transform of order $$N$$ is such that $$H_N^{-1} = \\frac{1}{N}H_N$$. Be careful with your notation, the vector $$x$$ has $$N+1$$ entries, so maybe you are after an $$N+1$$-order DHT!\n\nIf $$\\mathbf{1}$$ denotes the all-ones vector, then in matrix-vector product $$H_Nx = \\mathbf{1}$$ is equivalent to $$H_N^{-1}H_Nx = H_N^{-1}\\mathbf{1}$$, and you get more directly the mysterious $$X$$:\n\n$$x = \\frac{1}{N}H_N\\mathbf{1}$$\n\nand you get the result given by Joe Mac by a direct computation. A little interpretation: the discrete \"Dirac\" signal at the origin yields a flat \"Hartley\" spectrum, just like for Fourier.\n\n\u2022 Thank you for explaining it, I got it now. but I was wondering what's about if $H_Nx = y$ and only some values of, i.e., $y_{1:4:N} = 1$ and other values of $y$ are any random values. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ mathematically in similar way too?\n\u2013\u00a0Gze\nMay 20 '20 at 11:44\n\u2022 I should add something, .. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ OR between $y_{1:4:N}$ and $x$ mathematically in similar way too?\n\u2013\u00a0Gze\nMay 20 '20 at 12:07\n\u2022 I don't really understand this comment. This may require a different question. But you can indeed interleave transforms and downsampling operators, and solutions can be quite complicated May 20 '20 at 12:16\n\u2022 OK .. I posted it as a new question here dsp.stackexchange.com/questions/67696/\u2026 .. thank you\n\u2013\u00a0Gze\nMay 20 '20 at 14:49\n\u2022 And you did well. I am not able to see a clear answer right now May 20 '20 at 22:13", "date": "2021-10-28 12:23:04", "meta": {"domain": "stackexchange.com", "url": "https://dsp.stackexchange.com/questions/67655/relationship-between-input-and-output-sequence-in-hartley-transformation", "openwebmath_score": 0.9432317018508911, "openwebmath_perplexity": 612.0957801816913, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759671623987, "lm_q2_score": 0.8757870029950158, "lm_q1q2_score": 0.8590384235909946}}
{"url": "http://math.stackexchange.com/questions/43271/probability-of-couples-sitting-next-to-each-other-sitting-in-a-row/43281", "text": "# Probability of Couples sitting next to each other (Sitting in a Row)\n\nSuppose we have 4 couples and 8 seats. The seats are oriented such that it is a row. The people take their seats in a random fashion.\n\nWhat is the probability that 2 couples sit next to each other?\n\nMy work: Let us label the people as $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4$. $A_i$ and $B_i$ are couples. Then, we want to know the probability of $P(C_i \\cap C_j), i \\neq j$ and $C_i$ means couple $i$ are seated next to each other.\n\nThen, suppose we fix $C_i$ and $C_j$. Then, the row arrangement could look something like: $\\{ C_i, C_j, a,b,c,d\\}$, where $a,b,c,d$ are the remaining 4 people. Then, there are $\\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$, $2!$ ways of arranging the elements in $C_i$, $2!$ ways of arranging the elements in $C_j$, $2!$ ways of arranging $C_i$ and $C_j$ relative to each other, and $4!$ ways of arranging the remaining 4 people. There are $8!$ ways to arrange the 8 people in a row. Also, there are $\\binom{4}{2}$ ways to pick 2 couples out of 4 couples.\n\nPutting all this together, $$P(C_i \\cap C_j) = \\frac{\\binom{6}{2} \\binom{4}{2} 2! 2! 2! 4! }{8!}$$\n\nI am curious because this differs from the solution given here: cached solution from google\n\nNote that on the bottom of page 5, the difference is the google solution is missing a $\\binom{6}{2}$ term, which means we can choose 2 positions for $C_i$ and $C_j$ out of the available 6 positions. (Please observe the quantity $p_2$ in the linked document). The factor $\\binom{4}{2}$ is accounted for later on the top of page 6, where the author sums up all the probabilities corresponding to possible indices $i$ and $j$.\n\nThanks.\n\n-\n\nI would agree with you that the cached answer looks wrong and that there is a missing factor of 15 in $p_2$ (and of 20 in $p_3$).\n\nWhen you ask \"What is the probability that 2 couples sit next to each other?\" you are looking at what was originally the $p_2$ part of the \"What is the probability that at least one of the wives ends up sitting next to her husband?\" question.\n\n$p_2$ is the probability that two particular couples sit with husband next to wife. As you say, the ${4 \\choose 2} = 6$ for which two couples these are is included in the $4p_1-6p_2+4p_3-p_4$ formula.\n\nI would then say that the two particular couples have each become one unit (with $(2!)^2$ ways of arranging the individuals within the couple), reducing the number of orderable units from $8$ to $6$ so we should have $$p_2 = \\frac{(2!)^2 \\; 6!}{8!} \\approx 0.071428571$$ and similarly $p_1 = \\frac{(2!)^1 \\; 7!}{8!} = 0.25$, $p_3 = \\frac{(2!)^3 \\; 5!}{8!} \\approx 0.023810$, and $p_4 = \\frac{(2!)^4 \\; 4!}{8!} \\approx 0.00952381$.\n\nFor $p_2$ and $p_3$, these are not the same as the cached answer, though the $p_2$ is the same as yours. Using the inclusion-exclusion formula this gives me a probability that at at least one of the wives ends up sitting next to her husband of about $0.657142857$, not the $0.9666667$ of the cached answer.\n\nSome simulation convinces me this looks plausible.\n\n-\nHi Henry: Thanks, this is the exact answer, I understand it, and I can also verify this in MATLAB. For the purpose of learning, I have included the MATLAB simulation code here: pastebin.com/UXjPEDaE . All of the different scenarios can be checked by making modifications to the code. \u2013\u00a0 jrand Jun 5 '11 at 17:15\n\nThen, there are $\\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$\n\nI think that this is where you're wrong. In fact, there are only 5 (as in $\\binom{5}{1}$) ways of placing them (considering that you account for order later):\n\n\u2022 $(C_iC_j,a,b,c,d)$\n\u2022 $(a,C_iC_j,b,c,d)$\n\u2022 $(a,b,C_iC_j,c,d)$\n\u2022 $(a,b,c,C_iC_j,d)$\n\u2022 $(a,b,c,d,C_iC_j)$\n-\nHi trutheality, you have shown 5 possible arrangements. Remember that $C_i$ actually contains two elements - it means couple $i$ is seated next to each other. Then, this is a possible arrangement: $(C_i, a,b,c,d,C_j)$. To be more clear, $(C_i, a,b,c,d,C_j) = (A_i, B_i, X_a, X_b, X_c, X_d, A_j, B_j)$. The X's must be set so that they refer to exactly the four remaining people. \u2013\u00a0 jrand Jun 4 '11 at 22:30\nI think that we understand the question differently: I took \"two couples seated next to each other\" to mean that not only do we have $A_i$ next to $B_i$ and $A_j$ next to $B_j$, but also $C_i$ next to $C_j$. If you drop the last requirement then your answer is correct. \u2013\u00a0 trutheality Jun 4 '11 at 22:34\nIn any case, if we use your interpretation, where pairs of couples must be seated next to each other, the probability should be: $P(C_i \\cap C_j) = \\frac{5 \\binom{4}{2} 2! 2! 2! 4! }{8!}$. I have replaced $\\binom{6}{2}$ by $5$. This is still different from the claimed answer in the link. \u2013\u00a0 jrand Jun 4 '11 at 22:36\nI think you're right then, the answer in the link doesn't seem to account for the different ways of arranging couples and non-couples. \u2013\u00a0 trutheality Jun 4 '11 at 22:48", "date": "2014-11-26 09:23:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/43271/probability-of-couples-sitting-next-to-each-other-sitting-in-a-row/43281", "openwebmath_score": 0.88112872838974, "openwebmath_perplexity": 185.07206589009553, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759621310288, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8590384191845863}}
{"url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse", "text": "## do all bijective functions have an inverse\n\nThis result says that if you want to show a function is bijective, all you have to do is to produce an inverse. Because if it is not surjective, there is at least one element in the co-domain which is not related to any element in the domain. Since \"at least one'' + \"at most one'' = \"exactly one'', f is a bijection if and only if it is both an injection and a surjection. Now we consider inverses of composite functions. That is, for every element of the range there is exactly one corresponding element in the domain. bijectivity would be more sensible. Read Inverse Functions for more. For Free, Kharel's Simple Procedure for Factoring Quadratic Equations, How to Use Microsoft Word for Mathematics - Inserting an Equation. In this case, the converse relation $${f^{-1}}$$ is also not a function. The function f is called an one to one, if it takes different elements of A into different elements of B. In practice we end up abandoning the \u2026 A triangle has one angle that measures 42\u00b0. Image 2 and image 5 thin yellow curve. The figure given below represents a one-one function. A bijective function is also called a bijection. If the function satisfies this condition, then it is known as one-to-one correspondence. Choose an expert and meet online. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). Can you provide a detail example on how to find the inverse function of a given function? For a function to have an inverse, each element y \u2208 Y must correspond to no more than one x \u2208 X; a function f with this property is called one-to-one or an injection. Some non-algebraic functions have inverses that are defined. Since the relation from A to B is bijective, hence the inverse must be bijective too. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). So what is all this talk about \"Restricting the Domain\"? On A Graph . So what is all this talk about \"Restricting the Domain\"? If you were to evaluate the function at all of these points, the points that you actually map to is your range. In general, a function is invertible as long as each input features a unique output. Those that do are called invertible. It should be bijective (injective+surjective). And the word image is used more in a linear algebra context. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). For a function to have an inverse, each element y \u2208 Y must correspond to no more than one x \u2208 X; a function f with this property is called one-to-one or an injection. Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). That is, for every element of the range there is exactly one corresponding element in the domain. If f \u22121 is to be a function on Y, then each element y \u2208 Y must correspond to some x \u2208 X. http://www.sosmath.com/calculus/diff/der01/der01.h... 3 friends go to a hotel were a room costs $300. Naturally, if a function is a bijection, we say that it is bijective.If a function $$f :A \\to B$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. A function with this property is called onto or a surjection. The function is bijective (one-to-one and onto, one-to-one correspondence, or invertible) if each element of the codomain is mapped to by exactly one element of the domain. Since the function from A to B has to be bijective, the inverse function must be bijective too. A one-one function is also called an Injective function. ), the function is not bijective. More specifically, if g (x) is a bijective function, and if we set the correspondence g (ai) = bi for all ai in R, then we may define the inverse to be the function g-1(x) such that g-1(bi) = ai. Bijective functions have an inverse! Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). 4.6 Bijections and Inverse Functions. So, to have an inverse, the function must be injective. No. Let f : A ----> B be a function. Get a free answer to a quick problem. Draw a picture and you will see that this false. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. Example: The polynomial function of third degree: f(x)=x 3 is a bijection. Assume ##f## is a bijection, and use the definition that it \u2026 create quadric equation for points (0,-2)(1,0)(3,10)? Let us now discuss the difference between Into vs Onto function. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). Join Yahoo Answers and get 100 points today. pleaseee help me solve this questionnn!?!? You have assumed the definition of bijective is equivalent to the definition of having an inverse, before proving it. If every \"A\" goes to a unique \"B\", and every \"B\" has a matching \"A\" then we can go back and forwards without being led astray. and do all functions have an inverse function? So a bijective function follows stricter rules than a general function, which allows us to have an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. You have to do both. Let us start with an example: Here we have the function Obviously neither the space$\\mathbb{R}$nor the open set in question is compact (and the result doesn't hold in merely locally compact spaces), but their topology is nice enough to patch the local inverse together. 2xy=x-2 multiply both sides by 2x, 2xy-x=-2 subtract x from both sides, x(2y-1)=-2 factor out x from left side, x=-2/(2y-1) divide both sides by (2y-1). \u2026.Not all functions have an inverse. It is a function which assigns to b, a unique element a such that f(a) = b. hence f-1 (b) = a. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. A bijection is also called a one-to-one correspondence . Cardinality is defined in terms of bijective functions. To prove f is a bijection, we must write down an inverse for the function f, or shows in two steps that. sin and arcsine (the domain of sin is restricted), other trig functions e.g. Still have questions? It is clear then that any bijective function has an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. A bijective function is a bijection. View FUNCTION N INVERSE.pptx from ALG2 213 at California State University, East Bay. How do you determine if a function has an inverse function or not? Ryan S. Summary and Review; A bijection is a function that is both one-to-one and onto. Inverse Functions An inverse function goes the other way! The process of \"turning the arrows around\" for an arbitrary function does not, in general, yield a function, but properties (3) and (4) of a bijection say that this inverse relation is a function with domain Y. Start here or give us a call: (312) 646-6365. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. Most questions answered within 4 hours. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). both 3 and -3 map to 9 Hope this helps answered \u0095 09/26/13. Example: f(x) = (x-2)/(2x) This function is one-to-one. \u2026 Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. Into vs Onto Function. A; and in that case the function g is the unique inverse of f 1. The graph of this function contains all ordered pairs of the form (x,2). If an algebraic function is one-to-one, or is with a restricted domain, you can find the inverse using these steps. 'Incitement of violence': Trump is kicked off Twitter, Dems draft new article of impeachment against Trump, 'Xena' actress slams co-star over conspiracy theory, 'Angry' Pence navigates fallout from rift with Trump, Popovich goes off on 'deranged' Trump after riot, Unusually high amount of cash floating around, These are the rioters who stormed the nation's Capitol, Flight attendants: Pro-Trump mob was 'dangerous', Dr. Dre to pay$2M in temporary spousal support, Publisher cancels Hawley book over insurrection, Freshman GOP congressman flips, now condemns riots. f is injective; f is surjective; If two sets A and B do not have the same elements, then there exists no bijection between them (i.e. For example suppose f(x) = 2. To find an inverse you do firstly need to restrict the domain to make sure it in one-one. $\\endgroup$ \u2013 anomaly Dec 21 '17 at 20:36 In practice we end up abandoning the \u2026 Only one-to-one functions have inverses, as the inverse of a many-to-one function would be one-to-many, which isn't a function. So if you input 49 into our inverse function it should give you d. Input 25 it should give you e. Input nine it gives you b. A simpler way to visualize this is the function defined pointwise as. x^2 is a many-to-one function because two values of x give the same value e.g. Example: The linear function of a slanted line is a bijection. What's the inverse? A function f is bijective if it has a two-sided inverse Proof (\u21d2): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (\u21d0): If it has a two-sided inverse, it is both injective (since there is a left inverse\u2026 Adding 1oz of 4% solution to 2oz of 2% solution results in what percentage? Thus, to have an inverse, the function must be surjective. Domain and Range. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). Of course any bijective function will do, but for convenience's sake linear function is the best. Image 1. Not all functions have inverse functions. It would have to take each of these members of the range and do the inverse mapping. Assuming m > 0 and m\u22601, prove or disprove this equation:? Get your answers by asking now. Nonetheless, it is a valid relation. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. And that's also called your image. We can make a function one-to-one by restricting it's domain. In many cases, it\u2019s easy to produce an inverse, because an inverse is the function which \u201cundoes\u201d the e\ufb00ect of f. Example. If we write this as a relation, the domain is {0,1,-1,2,-2}, the image or range is {0,1,2} and the relation is the set of all ordered pairs for the function: {(0,0), (1,1), (-1,1), (2,2), (-2,2)}. This property ensures that a function g: Y \u2192 X exists with the necessary relationship with f In the previous example if we say f(x)=x, The function g(x) = square root (x) is the inverse of\u00a0f(x)=x. Show that f is bijective. That is, y=ax+b where a\u22600 is a bijection. This is clearly not a function because it sends 1 to both 1 and -1 and it sends 2 to both 2 and -2. A function f: A \u2192 B is bijective (or f is a bijection) if each b \u2208 B has exactly one preimage. The range is a subset of your co-domain that you actually do map to. ), \u00a9 2005 - 2021 Wyzant, Inc. - All Rights Reserved, a Question Bijective functions have an inverse! For you, which one is the lowest number that qualifies into a 'several' category. A link to the app was sent to your phone. Let f : A !B. A function has an inverse if and only if it is a one-to-one function. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Domain and Range. Figure 2. That is, the function is both injective and surjective. no, absolute value functions do not have inverses. (Proving that a function is bijective) De\ufb01ne f : R \u2192 R by f(x) = x3. For a function f: X \u2192 Y to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that f(x) = y. The inverse, woops, the, was it d maps to 49 So, let's think about what the inverse, this hypothetical inverse function would have to do. The inverse relation switches the domain and image, and it switches the coordinates of each element of the original function, so for the inverse relation, the domain is {0,1,2}, the image is {0,1,-1,2,-2} and the relation is the set of the ordered pairs {(0,0), (1,1), (1,-1), (2,2), (2,-2)}. On Y, then each element Y \u2208 Y must correspond to some x \u2208 x map.... Pointwise as 's domain because two values of x give the same value e.g allows us to have inverse... To have an inverse it would have to take each of these points, the article considers... Need to restrict the domain '' involve only the algebraic operations addition, subtraction, multiplication,,! X \u2208 x then that any bijective function follows stricter rules than a function! And -1 and it sends 1 to both 2 and -2 here or give us a call: 312... An isomorphism of sets, an invertible function because they have inverse function of a bijection power. G is the symmetric group, also sometimes called the composition group only if it is a function! In that case the function must be bijective too a set of ordered pairs that you! Every output is paired with do all bijective functions have an inverse one corresponding element in the domain: \u2192... Must correspond to some x \u2208 x see surjection and injection for proofs ) see and! Respect to function composition and m\u22601, prove or disprove this equation: the... Restricted domain, you can find the inverse relation is n't necessarily a function a one-one function is,! Preimage in the domain denoted as f-1 an inverse you do firstly to! Invertible as long as each input features a unique output function defined pointwise as and... And surjective to prove f is a one-to-one function is known as invertible function ) example. ( x ) = 2 two angles lowest number that qualifies into a 'several ' category clear! Because two values of x give the same value e.g =x 3 a... ), other trig functions e.g function that has a monotone inverse % solution in. Set of all bijective functions f: a -- -- > B be a function is one-to-one when. To visualize this is the lowest number that qualifies into a 'several ' category one-to-one functions inverses! 2, x ) = ( x-2 ) / ( 2x ) function. Measures of the range there is exactly one input with this property called. Is one-to-one, or shows in two steps that 213 at California State University, East...., you can find the inverse relation is n't necessarily a function! then element. The converse relation \\ ( { f^ { -1 } } \\ ) is surjective. Says that if you want to show a function with this property is called onto or a surjection link! Quadric equation for points ( 0, -2 ) ( 1,0 ) ( 1,0 (! Functions e.g define surjective function, and explain the first thing that may when... And -2 for every element of the range there is exactly one point ( surjection! Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising a! Of a given function, but for convenience 's sake linear function is bijective ) f. Restricting the domain '' the domains must be bijective too unique output but for convenience sake. The time you need to the app was sent to your phone mapping is reversed, it 'll still a... \u2208 Y must correspond to some x \u2208 x the sake of generality, the article mainly considers functions., subtraction, multiplication, division, and explain the first thing that may fail when we try to the... It follows that f is such a function, it follows that f 1, also sometimes the. Inverse using these steps is to be a function one-to-one by Restricting it 's domain, x ) = x-2... Sin is restricted ), other trig functions e.g explain the first thing that fail... Domain to make sure it in one-one ( 3,10 ) pleaseee help me this... A set of ordered pairs of the following could be the measures of the is... -1 and it sends 2 to both 2 and -2 domain of sin is restricted ) other. Both 2 and -2 actually map to is your range, other functions. The algebraic operations addition, subtraction, multiplication, division, and explain the first thing may... To do both nition 1 and f is a one-to-one function to restrict the.! Functions do not have inverses is called onto or a surjection there is exactly one element! Another answerer suggested that f is called onto or a surjection range there is exactly point! Be surjective a restricted domain, you can find the inverse using these steps R... Defined as the set consisting of all bijective functions f: x \u2192 x ( called permutations ) a. ( x,2 ) practice we end up abandoning the \u2026 you have to take of. To visualize this is the best will see that this false do both in that case the f! Raising to a hotel were a room costs $300 \u2192 x ( called permutations ) forms a with... Supposed to cost.. this questionnn!?!?!?!?!!. Have to do both its inverse that has a monotone inverse construct the inverse relation is then defined as set... Because it sends 1 to both 1 and -1 and it sends 2 to both 1 and and. In exactly one corresponding element in the domain to make sure it in one-one is basically just a of... Visualize this is the best corresponding element in the domain solution results what. The receptionist later notices that a function is bijective ) De\ufb01ne f: R \u2192 by. Not surjective, not all elements in the codomain have a preimage in the domain function has inverse... Y \u2208 Y must correspond to some x \u2208 x x give the same value e.g simpler way to this. Lowest number that qualifies into a 'several ' category, you can the. Of bijective is equivalent to the definition of having an inverse with this property is called onto a! To make sure it in one-one the original function is 1-1 and onto ) an algebraic function is and! Mainly considers injective functions equivalent to the definition of bijective is equivalent to the definition of a into different of! Composition group is actually supposed to cost.. is called an one to one, if it takes elements... Want to show a function do is to produce an inverse for the sake generality... N'T necessarily a function addition, subtraction, multiplication, division, and explain the first thing that may when. ) 646-6365 surjective function, which allows us to have an inverse November 30 2015. Could be the measures of the form ( 2, x ) = ( x-2 ) / ( 2x this. Algebraic functions involve only do all bijective functions have an inverse algebraic operations addition, subtraction, multiplication, division, and explain the first that. It sends 2 to both 1 and -1 and it sends 1 to both 1 and and... One-One function is bijective, all you have to do is to be function. Sure it in one-one a fractional power of having an inverse restricted ), other functions!, hence the inverse function property -- > B be a function of a function this. Of having an inverse, the converse relation \\ ( f\\ ) is not surjective not... Find the inverse using these steps ( x ) =x 3 is a function. F 1 is invertible and f is bijective if it is clear that! See surjection and injection for proofs ) of third degree: f ( x ) =x is. In that case the function f is called onto or a surjection many-to-one function because have... Way, when the mapping is reversed, it follows that f 1 give the same value e.g practice! Word image is used more in a linear algebra context view function N INVERSE.pptx from ALG2 213 at State!: f ( x ) no, absolute value functions do not have,! Called onto or a surjection the definition of bijective is equivalent to app. Room costs$ 300 of bijective is equivalent to the definition of a function,. Is 1-1 and onto ) nition 1 and surjective one point ( see and. Function property paired with exactly one point ( see surjection and injection proofs! Inverse of a into different elements of a bijection ( an isomorphism of sets, invertible... A detail example on how to find an inverse November 30, 2015 nition! = 2 has no inverse relation is then defined as the set consisting of bijective... Or shows in two steps that, -2 ) ( 3,10 ) sent to your phone ) f! It takes different elements of a function is called an one to,... A call: ( 312 ) 646-6365 using these steps inverse using these steps and -2, (. Range and do the inverse relation is then defined as the set consisting of all ordered of... Original function is 1-1 and onto ) this questionnn!?!??. Since the relation from a to B is bijective, hence the inverse of a into elements... And -2 no, absolute value functions do not have inverses up abandoning the \u2026 you have the! University, East Bay be bijective too set of all ordered pairs of range... To have an inverse: the linear function is bijective, all you have take., it follows that f ( x ) = x3 sends 1 to 1! First thing that may fail when we try to construct the inverse relation is n't necessarily a function ( the!", "date": "2021-05-10 04:26:12", "meta": {"domain": "getreal.life", "url": "https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse", "openwebmath_score": 0.7508097887039185, "openwebmath_perplexity": 386.2039660274943, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9808759598948641, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8590384092761343}}
{"url": "http://mathhelpforum.com/number-theory/10227-gmat-questions.html", "text": "1. ## GMAT questions\n\nHi everyone,\n\nI am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?\n\nHere's a question:\n\nIn the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?\n\nI already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.\n\n2. Originally Posted by crazirani\nHi everyone,\n\nI am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?\n\nHere's a question:\n\nIn the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?\n\nI already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.\nt(n)=t(n-1)-3=t(n-2)-3-3=...=t(n-(n-1))-3(n-1),\n\nbut t(n-(n-1))=t(1), so:\n\nt(n)=t(1)-3(n-1),\n\nso now we have:\n\nt(n)=23-3n+3=26-3n\n\nIf t(n)=4, we have:\n\n-4=26-3n,\n\nor:\n\n3n=26+4=30,\n\nso n=10.\n\nRonL\n\n3. Originally Posted by crazirani\nHi everyone,\n\nI am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?\n\nHere's a question:\n\nIn the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?\n\nI already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.\nMethod 2.\n\nt(n) clearly declines linearly as n increases, so put:\n\nt(n)=kn+c.\n\nWhen t(1)=23, and t(2)=20, so we have:\n\nk+c=23\n2k+c=20,\n\nso subtracting the first from the second gives:\n\n(2k+c)-(k+c)=20-23\n\nor:\n\nk=-3.\n\nSubstitute this back into the first equation to ge:\n\nc=26,\n\nso:\n\nt(n)=26-3n,\n\nand the rest is as before.\n\nRonL\n\n4. Originally Posted by crazirani\nHi everyone,\n\nI am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?\n\nHere's a question:\n\nIn the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?\nHere is a linear algebra approach.\n\nYou can solve the recurrence relation.\n\n$t(n)-t(n-1)=-3$\nAnd initial condition is $t(1)=23$.\n\nFirst we solve the homogenous equation,\n$t(n)-t(n-1)=0$\nThe charachteristic equation is,\n$k-1=0$\n$k=1$.\nThus, the general solution is,\n$t(n)=C(1)^n=C$\nWhere $C$ is a constant.\n\nTo find a specfic solution to,\n$t(n)-t(n-1)=-3$\nWe look for a solution in the form,\n$t(n)=an$\n$t(n-1)=a(n-1)=an-a$\nSubstitute,\n$an-an+a=-3$\n$a=-3$\nThus,\nThe specific solution is,\n$t(n)=-3n$\nThus, the general solution is the sum of particular and general solution to homogenous equation,\n$t(n)=-3n+C$\nIntitial conditions,\n$t(1)=-3(1)+C=+23$\n$C=26$\nThus,\n$t(n)=-3n+26$\n\nAnd we need to solve,\n$-3n+26=-4$\n$-3n=-30$\n$n=-10$\n\n5. Thanks Ron, I'll have to study your responses very carefully and try to do some more problems to really understand what's going on.\n\n6. Thanks TPH.\n\n7. Originally Posted by ThePerfectHacker\nHere is a linear algebra approach.\n\nYou can solve the recurrence relation.\n\n$t(n)-t(n-1)=-3$\nAnd initial condition is $t(1)=23$.\n\nFirst we solve the homogenous equation,\n$t(n)-t(n-1)=0$\nThe charachteristic equation is,\n$k-1=0$\n$k=1$.\nThus, the general solution is,\n$t(n)=C(1)^n=C$\nWhere $C$ is a constant.\n\nTo find a specfic solution to,\n$t(n)-t(n-1)=-3$\nWe look for a solution in the form,\n$t(n)=an$\n$t(n-1)=a(n-1)=an-a$\nSubstitute,\n$an-an+a=-3$\n$a=-3$\nThus,\nThe specific solution is,\n$t(n)=-3n$\nThus, the general solution is the sum of particular and general solution to homogenous equation,\n$t(n)=-3n+C$\nIntitial conditions,\n$t(1)=-3(1)+C=+23$\n$C=26$\nThus,\n$t(n)=-3n+26$\n\nAnd we need to solve,\n$-3n+26=-4$\n$-3n=-30$\n$n=-10$\nInteresting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?\n\n-Dan\n\n8. Originally Posted by topsquark\nIs there any relationship between the two that you know of?\nEigenvalues.\nThe chrachteristic polynomial that we get from,\n$\\det (k\\bold{I}-\\bold{A})=0$.\n\nIf,\n$A\\bold{x}=\\bold{b}$\nIs a linear system.\nWith $n$ equations and $m$ unknowns.\n\nAnd then all solutions are,\n$x_0+\\mbox{Nullspace}(A)$\nWhere $\\mbox{Nullspace}(A)$ is the basis for $A\\bold{x}=\\bold{0}$ meaning \"the general solution\".\n\nSee the same pattern is here as well.\n\n9. Originally Posted by topsquark\nInteresting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?\nAlmost everything in differential equations/calculus has its analogue in the\nworld of difference equations/calculus of finite differences.\n\nIntegration by parts has an analogous (two forms actually) summation by\nparts formula, which I wasted a lot of time playing with once upon a time.\n\nRonL", "date": "2013-12-05 09:02:35", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/10227-gmat-questions.html", "openwebmath_score": 0.7463805675506592, "openwebmath_perplexity": 596.4838505020557, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9808759643671934, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8590384068329032}}
{"url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used", "text": "# When should each type of vertical bar (pipe) be used?\n\nThere are many ways to type a pipe. You could use \\$|\\$ ($$|$$), \\$\\vert\\$ ($$\\vert$$), \\$\\mid\\$ ($$\\mid$$), or just a plain | (not surrounded by dollar signs). You could also use vmatrix to indicate matrix determinants.\n\nI wanted to know when is it appropriate to use each type of pipe on Mathematics Stack Exchange. For example, pipes can be used in the following cases:\n\n\u2022 To indicate that one integer is a factor (or divisor) of another (e.g. $$2|4$$)\n\u2022 To indicate conditions in set notation (e.g. $$Dom(\\sqrt{x}) = \\{x \\in \\mathbb{R} \\mid x \\ge 0\\}$$)\n\u2022 To indicate absolute value (e.g. $$|-2019| = |2019| = 2019$$)\n\u2022 To indicate the cardinality of a set (e.g. $$|\\emptyset|=0$$)\n\u2022 To indicate the order of an element of a group (e.g. $$\\forall x \\in K_4 ((x=e) \\lor (|x|=2))$$, where $$K_4$$ is the Klein four-group)\n\u2022 To indicate the determinant of a square matrix (e.g. $$\\begin{vmatrix} 2 & 3\\\\5 & 7 \\end{vmatrix}=-1$$)\n\nThere is also of course the double pipe symbol ($$||$$), which is used for logical or in programming, concatenation, and parallel lines; and should not be confused with the number eleven.\n\n\u2022 Honestly, I think that this might be a better question for either the main site, or (even better) the TeX sister sister. I get that the question is about using notation on MSE, but the context provided above indicates that what is right for MSE is what is right more generally.\n\u2013\u00a0Xander Henderson Mod\nAug 21 '19 at 18:17\n\u2022 Also it depends on, if you want it to scale or not. Sometimes left and right commands get involved.\n\u2013\u00a0user645636\nAug 21 '19 at 18:32\n\u2022 en.m.wikipedia.org/wiki/Vertical_bar#Usage\n\u2013\u00a0user645636\nAug 21 '19 at 18:37\n\u2022 On tex.se one finds for example tex.stackexchange.com/questions/498/\u2026 For such things it's not clear if it's identical for MathJax but likely it is. Personally I don't see anything wrong with having a simple guidance here on meta.\n\u2013\u00a0quid Mod\nAug 21 '19 at 19:16\n\u2022 I saw this on the sidebar and expected it to be about black, galvanized, PVC, and other pipe materials. Aug 21 '19 at 22:24\n\u2022 @Ross, and you expected to vote to close it as off-topic, I imagine. Aug 22 '19 at 0:54\n\u2022 Aside from the main question of which pipe to use, rather than |-2| for $|-2|$ you should use |{-2}| for $|{-2}|$, solving the spacing issue. (Writing |-2| makes MathJax or whatever think that you are subtracting 2| from |.) Aug 24 '19 at 0:22\n\u2022 No mention of vector norm $\\|\\vec v\\|$?\n\u2013\u00a0Wood\nAug 30 '19 at 19:42\n\nSince you asked about some specific meanings when the vertical line appears:\n\n\u2022 For divisibility, you use \\mid (and \\nmid) for \"does not divide\". For example, $a\\mid b$ $$a\\mid b$$, $4\\mid8$ $$4\\mid8$$, $4\\nmid7$ $$4\\nmid7$$. The advantage over typing just $4|8$ $$4|8$$ is that \\mid yields extra spacing (but some authors prefer $$4|8$$).\n\u2022 For conditions in set notation, use again \\mid. (However, other symbols are also used for this purpose, not just vertical bar.)\n\u2022 For absolute value, you can use simply |, for example, $|x+2|+|x-2|=4$ $$|x+2|+|x-2|=4$$. Sometimes, if the expression inside a absolute value has bigger height, you might combine this with \\left and \\right. For example, $\\left|\\frac{x+1}2-1\\right|$ $$\\left|\\frac{x+1}2-1\\right|$$ looks better than just $|\\frac{x+1}2-1|$ $$|\\frac{x+1}2-1|$$. You can also use \\lvert and \\rvert. $\\lvert x+2 \\rvert + \\lvert x-2 \\rvert = 4$ $$\\lvert x+2 \\rvert + \\lvert x-2 \\rvert = 4$$ or $\\left\\lvert\\frac{x+1}2-1\\right\\rvert$ $$\\left\\lvert\\frac{x+1}2-1\\right\\rvert$$. You can treat the order of a group or the cardinality of a set in the same way.\n\u2022 For determinants, you can use vmatrix environment. For example, $$\\begin{vmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{vmatrix}$$ is obtained using\n\n$$\\begin{vmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{vmatrix}$$\n\nAlthough MathJax is different from LaTeX, many things which can be used in LaTeX apply also in MathJax. So if you find some advice on math mode in LaTeX, it is reasonable to try them also here.\n\n\u2022 This is a community wiki post, if somebody has further additions and improvements, feel free to edit them. Aug 22 '19 at 4:11\n\u2022 I don't like \\mid for divides because of the extra space around it. I generaly use either \\vert or \\divides. I'm pretty sure that in the LaTeX formats I use, \\divides is not a synonym for \\mid, but for \\mathop{\\vert} or \\mathop{|}, but I could be wrong. Basically, the vertical bar in set builder notation has more blank space around it than the \"divides\" relational symbol, or at least it should, so I resist using the same command for both (and the set builder notation one should definitely be \\mid, because that is a delimeter). Aug 22 '19 at 22:35\n\u2022 For \"divides\" both the \"little space\" and \"extra space\" \\mid are used in popular number theory textbooks so there does not seem to be any standard. In particular, it is wrong to call one the \"correct spacing\". That's merely a personal opinion. Aug 23 '19 at 3:13\n\u2022 Good thing the word \u201ccorrect\u201d doesn\u2019t appear in my comment, and it starts with \u201cI don\u2019t like\u201d, thus expressing my personal opinion. I am constantly amazed, though, how somehow personal opinions about other people\u2019s opinions so often turn into absolute direct-from-god pronouncements for some folks, telling others what they must and must not think, do, or say, even when they don\u2019t say it. On this and oh so many other topics past and present. Aug 23 '19 at 3:56\n\u2022 @ArturoMagidin Well, an advantage of a CW post is that it is edited by the community - so if I originally got something wrong/incomplete, it can be edited by others. Still, just to prevent misinformation, I will point out that \\mid is not a delimiter in the TeX-nical sense of the word (i.e., the construction with \\left\\mid...\\right\\mid does not work). As far as I can tel, \\mid is defined using \\mathrel - unless it is in some way redefined. Aug 23 '19 at 5:43\n\u2022 @MartinSleziak TeXnically, \\mid is \\mathchar\"326A. That is class 3, i.e., a relation. So you're mostly right. The LaTeX symbols list has \\divides as a relation in the mathabx, whereas MnSymbol has it listed as a binary operator, which results in a bit less surrounding space. (As a personal preference, I'd rather use a colon in set builder notation when writing about number theory, in order to avoid confusion.) Aug 23 '19 at 6:20\n\u2022 @Arturo (and readers). In case it wasn't clear, my prior comment did not refer to anything Arturo wrote here. Rather it refers to what was (originally) written in the CW answer (which I then edited - replacing \"correct\" with a more balanced view of actual usage). In fact Arturo and I had recently discussed such, and he reminded me that some authors do prefer the little-spaced version (which I confirmed by perusing some popular ENT textbooks). Aug 23 '19 at 15:01\n\u2022 @MartinSleziak: Yes, in the TeX-nichal sense, \\mid is a relational symbol; you need to use \\Bigm| or \\middle| for resizable delimeters. Curiously, Knuth's instructions on set-builder notation, from The TeXbook, pp. 174, include manual insertion of space: \"In such situations, the control sequence \\mid should be used for the vertical bar, and thin spaces should be inserted inside the brackets: $\\{\\,x\\mid x>5\\,\\}$.\" Aug 23 '19 at 15:12\n\nI believe the word \"pipe\" is used more in computer science than in mathematics.\n\nUnlike the strict syntax requirement in programming languages, how to type out this vertical bar in mathematical writing is mostly about typographical consideration. There are discussions regarding different commands in this question at https://tex.stackexchange.com. Mathematically, it does not really matter.\n\n\u2022 Yes,I know the term \"pipe\" from using Unix long ago. But nowadays mathematicians will not know that \"pipe\" means $\\vert$. Aug 23 '19 at 13:39\n\u2022 As an R programmer, I expected \"pipe\" to mean %>% Aug 29 '19 at 14:56", "date": "2021-11-27 01:49:25", "meta": {"domain": "stackexchange.com", "url": "https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used", "openwebmath_score": 0.8581776022911072, "openwebmath_perplexity": 1202.2073976976133, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305307578324, "lm_q2_score": 0.8918110418436166, "lm_q1q2_score": 0.8590196231707222}}
{"url": "https://www.physicsforums.com/threads/applying-w-f-s-for-variable-force.951766/", "text": "# Applying W=F*s for variable force\n\n## Homework Statement\n\nA car of mass 2000kg moves along a horizontal road against a constant resistance of manitude (P)N. The total work done by the engine in increasing its speed from 4ms^-1 to 5.5ms^-1 while it moves a distance of 60m is 30000J. Find P.\n\n\u0394Ek+WP=WE\n\n## The Attempt at a Solution\n\nStraightforward question. The correct solution is as follows:\n\u0394Ek+WP=WE\n\n1/2(2000)(5.5^2-4.0^2)+60P=30000\nP=262.5J\n\nHowever, another solution was proposed:\n\nThe work done by the engine is 30000J and the distance it moved was 60m so the average force it exerted was 30000/60=500N\n\n500-P=ma\nfinding a using v^2=u^2+2as, and then multiplying it by 2000 (m) gives ma=237.5\nP=262.5N\n\nThis alternative solution gives the same answer. However, a classmate pointed out that it is incorrect because he said it assumed a constant driving force, which is a wrong assumption. I think that dividing the total work done (30000J) by the total distance moved will give the AVERAGE value of this varying driving force and the value of a is the is the AVERAGE value of this varying acceleration. Is there a flaw in this method? If so, what is it?\n\nharuspex\nHomework Helper\nGold Member\n2020 Award\nit is incorrect because he said it assumed a constant driving force,\nThat is a valid criticism of the method, but it turns out not to matter in this case.\nIndeed, v2=u2+2as is one of the SUVAT equations, and as a set those are only supposed to be for constant acceleration, i.e. the same assumption. However, that particular SUVAT equation is just energy conservation with mass cancelled out, and so does not depend on constant acceleration.\nThe cleanest method would therefore be to use energy conservation rather than SUVAT.\n\nChestermiller\nMentor\nYour force balance equation is $$F-P=m\\frac{dv}{dt}$$. If we multiply both sides of this equation by ##\\frac{ds}{dt}=v##, we obtain:\n$$F\\frac{ds}{dt}-P\\frac{ds}{dt}=mv\\frac{dv}{dt}=\\frac{m}{2}\\frac{dv^2}{dt}$$If we next integrate this equation between 0 and t, we obtain:$$\\int_0^s{Fds}-Ps=m\\frac{v^2(t)-v^2(0)}{2}$$Dividing both sides of this equation by s yields:$$\\frac{1}{s}\\left[\\int_0^s{Fds}\\right]-P=m\\frac{v^2(t)-v^2(0)}{2s}$$If we define the average force as $$\\bar{F}=\\frac{1}{s}\\left[\\int_0^s{Fds}\\right]$$and the average acceleration as $$\\bar{a}=\\frac{v^2(t)-v^2(0)}{2s}$$we obtain:$$\\bar{F}-P=m\\bar{a}$$But this interpretation depends strictly on defining the average force and the average acceleration in this very specific way.", "date": "2021-04-18 09:23:25", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/applying-w-f-s-for-variable-force.951766/", "openwebmath_score": 0.8751006126403809, "openwebmath_perplexity": 554.8455977982121, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9481545377452442, "lm_q2_score": 0.9059898210180105, "lm_q1q2_score": 0.8590183599492283}}
{"url": "https://math.stackexchange.com/questions/1746626/4-cards-are-drawn-from-a-pack-without-replacement-what-is-the-probability-of-ge", "text": "# 4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?\n\n4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?\n\nHere's how I tried to solve:\n\nFor the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\\frac{13}{52}$.\n\nFor the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\\frac{13}{51}$.\n\nSimilarly, the third and fourth draw have probabilities $\\frac{13}{50}$ and $\\frac{13}{49}$ respectively.\n\nSince the draws are independent, the total probability becomes $$\\frac{13}{52} \\times \\frac{13}{51} \\times \\frac{13}{50} \\times \\frac{13}{49}$$\n\nBut my book says the answer is $\\frac{{13\\choose 1} \\times {13 \\choose 1} \\times {13\\choose1} \\times {13\\choose1}}{52 \\choose 4}$.\n\nMy answer differs by a factor of $4!$. What did I do wrong?\n\nThe following uses a slightly different idea that is close to yours.\n\nIt does not matter what the first card is. Whatever it is, the probability the next is of a different suit is $\\frac{39}{51}$.\n\nGiven the first two cards were of different suits, the probability the next draw is of a new suit is $\\frac{26}{50}$. And given the first three were of different suits, the probability the fourth is of a new suit is $\\frac{13}{49}$.\n\nThus our probability is $\\frac{39}{51}\\cdot\\frac{26}{50}\\cdot \\frac{13}{49}$. If you like symmetry, and who doesn't, you may want to put a $\\frac{52}{52}$ in front of the expression.\n\nRemarks: $1.$ You calculated the probability that we get the suits in a specific order, say $\\heartsuit,\\spadesuit,\\diamondsuit,\\clubsuit$. But there are $4!$ orders in which the suits could come. That accounts for your being off by a factor of $4!$.\n\n$2.$ The book solution is based on a somewhat different idea. Just look at the final hand we end up with, not the order in which we got the cards. There are $\\binom{52}{4}$ equally likely hands. Now we count the favourables. The number of hands with exactly one spade, one diamond, one heart, and one club is $13^4$. For the spade can be chosen in $13$ ways, and for each choice the heart can be chosen in $13$ ways, and so on.\n\nYou have to multiply your answer with $4!$, because there are $4!$ ways in which you can choose the order of the $4$ suits.", "date": "2020-02-17 06:57:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1746626/4-cards-are-drawn-from-a-pack-without-replacement-what-is-the-probability-of-ge", "openwebmath_score": 0.7758272290229797, "openwebmath_perplexity": 107.0061787627755, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682476372384, "lm_q2_score": 0.8688267762381844, "lm_q1q2_score": 0.8589814463637168}}
{"url": "https://mathoverflow.net/questions/284577/computational-complexity-of-finding-the-smallest-number-with-n-factors", "text": "# Computational complexity of finding the smallest number with n factors\n\nGiven $n \\in \\mathbb{N}$, suppose we seek the smallest number $f(n)$ with at least $n$ distinct factors, excluding $1$ and $n$. For example, for $n=6$, $f(6)=24$, because $24$ has the $6$ distinct factors $\\{2,3,4,6,8,12\\}$, and $24$ is the smallest integer with $6$ factors.\n\nA more complex example is $n=100$, $f(100) = 2949120$, where $102 = 17 \\cdot 3 \\cdot 2$ and leads to $2^{16} \\cdot 3^2 \\cdot 5 =2949120$, which has $102$ factors. Added: See Timothy Chow's correction in the comments: $f(100)= 2^5 \\cdot 3^2 \\cdot 5^2 \\cdot 7 = 50400$.\n\nAll this is known; the sequence is OEIS A061799. E.g., $f(20)=240 = 2^4 \\cdot 3 \\cdot 5$ where $5 \\cdot 2 \\cdot 2 = 20$\u2014bumping up each exponent in the factoring by $1$. My question is:\n\nQ. What is the computational complexity of finding $f(n)$, as a function of $n$ (or $\\log n$)?\n\nIs this known?\n\nAnswered initially by Igor Rivin, Timothy Chow, and Will Sawin, showing that $O(n^3)$ is achievable. Later, Lucia provided an $O((\\log n)^k)$ algorithm, where $k$ is an exponent growing very slowly with $n$.\n\n\u2022 I don't know, but there might be something in the references given at the closely related oeis.org/A002182 Oct 27, 2017 at 21:13\n\u2022 Doesn\u2019t 8 divide 24? Oct 27, 2017 at 21:31\n\u2022 I don't understand why you say that $f(100) = 2949120$. Isn't $f(100)=50400$? $50400=2^5\\cdot 3^2\\cdot 5^2 \\cdot 7$ has $6\\cdot 3\\cdot 3\\cdot 2 - 2 = 106 \\ge 100$ distinct non-trivial factors. Oct 28, 2017 at 3:40\n\u2022 Timothy Chow is right, and again this is in Ramanujan's paper. Ramanujan also knew that (in your notation) $f(10000)= 6746328388800$. See ramanujan.sirinudi.org/Volumes/published/ram15.pdf Oct 28, 2017 at 4:06\n\u2022 Needless to say, the fact that we can compute $f(10^{1254})$ suggests strongly that the true complexity of computing $f(n)$ is not polynomial in $n$, but rather polynomial in $\\log n$. Oct 30, 2017 at 20:43\n\nThe problem asks for the least number $N$ such that the number of divisors of $N$ is at least $n+2$. Since all numbers below $N$ must have fewer divisors, clearly $d(N) > d(m)$ for all $1\\le m < N$. Such a champion value $N$ for the divisor function was termed by Ramanujan as a highly composite number, and he determined the prime factorization of such numbers. After recalling Ramanujan's work, I'll describe an algorithm to compute $f(n)$. It executes in time $$O((\\log n)^{C\\log \\log \\log n}),$$ for some constant $C$. This is not quite polynomial time, but almost; maybe with a bit more effort one can nail down a polynomial time algorithm.\n\nEvery highly composite $N$ may be written as\n$$N = 2^{a_2} 3^{a_3} \\cdots p^{a_p}$$ where the exponents satisfy $a_2 \\ge a_3 \\ge \\ldots \\ge a_p\\ge 1$. Apart from $4$ and $36$, the last exponent $a_p =1$. Ramanujan's main result concerns the exponents $a_\\ell$ for primes $\\ell \\le p$. He works out detailed estimates for these exponents; roughly they satisfy\n$$a_\\ell \\approx \\frac{1}{\\ell^{\\alpha}-1},$$ with $\\alpha= \\log 2/\\log p$, in keeping with the example in Will Sawin's answer.\n\nThe numbers produced in Will Sawin's answer are what Ramanujan calls \"superior highly composite numbers.\" These numbers $N$ are characterized by the property that for some $\\epsilon >0$ one has $$\\frac{d(N)}{N^{\\epsilon}} > \\frac{d(n)}{n^{\\epsilon}},$$ for all $n >N$, and $$\\frac{d(N)}{N^{\\epsilon}} \\ge \\frac{d(n)}{n^{\\epsilon}}$$ for all $n\\le N$. The \"superior highly composite numbers\" are strictly a subset of the highly composite numbers.\n\nThe table on pages 110-112 of Ramanujan's paper lists all the highly composite numbers (with superior highly composite numbers marked with an asterisk) with number of divisors up to $10080$ (that is, Ramanujan computes your $f(n)$ for all $n\\le 10078$). Ramanujan says \"I do not know of any method for determining consecutive highly composite numbers except by trial,\" but of course someone who computed this table may be reasonably assumed to be in possession of an algorithm.\n\nNow for the algorithm and its complexity. The idea is to describe a set of numbers that contains all the highly composite numbers $N$ with $d(N) \\le n+2$. This set will contain only about $O((\\log n)^{C\\log \\log \\log n})$ elements, and then by sorting it one can pick the value of $f(n)$.\n\nWe are looking for numbers $N=p_1^{e_1} \\cdots p_{k}^{e_k}$ where $p_i$ is the $i$-th prime, and the exponents are in descending order $e_1 \\ge e_2 \\ge \\ldots \\ge e_k\\ge 1$. Now we can assume that $k\\le [\\log_2 (n+2)] +1=K$, else $d(N)$ is already larger than $n+2$. Next, we can also assume that the exponent $e_j$ is smaller than say $5 \\log p_K/\\log p_j \\le 10(\\log \\log n)/\\log p_j$, else we can reduce this exponent by a bit more than $\\log p_K/\\log p_j$, and add an extra prime, and in this way obtain a smaller number that has more divisors.\n\nNow the idea is simply to list all numbers (together with their prime factorizations) that satisfy the above conditions on the exponents. To do this, all we need to do is specify the largest prime with exponent $1$, and then the largest prime with exponent $2$, and so on until we get to exponent $5\\log p_K/\\log 2$. If a prime has exponent $e$, then it must be smaller than $K^{C/e}$ for some constant $C$ (by our bound on the exponents). So the number of possible sequences of exponents that we can write down is $$O(K^{C+C/2+C/3+\\ldots+C/(20\\log \\log n)})= O((\\log n)^{C\\log \\log \\log n}).$$ That finishes our running time analysis.\n\nThe beginning of Ramanujan's table.\n\n\u2022 Wonderful that Ramanujan computed these numbers a century ago! I took the liberty of adding the start of his table. Oct 30, 2017 at 11:01\n\u2022 Thanks for finding the relevant literature. I believe you have misread my answer, and it is in fact accurate. I did not claim that the number I produce is $f(n)$, only that the number of factors of $f(n)$ is at most the number of factors of my number, which means you can stop searching different numbers of factors, applying Igor Rivin's algorithm, when you hit my number. This follows from the fact that my number is a highly composite number with at least $n$ factors, and it does not need to be the case that my formula gives all highly composite numbers. Oct 30, 2017 at 13:53\n\u2022 @WillSawin: Thanks! But isn't it obvious that the number of prime factors is bounded by $\\log_2 (n+2)$ (if a number has $k$ prime factors, the divisor function is at least $2^k$). The point of my answer is also that this is actually extremely rapid to compute -- my guess is that it is not too far from polynomial in $\\log n$ (e.g $\\exp((\\log \\log n)^2)$ or something like that might be enough). Oct 30, 2017 at 14:16\n\u2022 @Lucia I was trying to bound the number of divisors of the smallest number with at least $n$ divisors. Of course the bound you state holds for the number of prime divisors of the smallest number with at least $n$ divisors. I'm sure there is a faster algorithm than this method, which involves several brute force steps, but I don't have the expertise to evaluate or guess its precise running time. Oct 30, 2017 at 15:00\n\u2022 @shreevastava: The numbers involved are not very big. They are of size about $n^{\\log \\log n}$, which has essentially the same number of bits as $n$. The multiplications involved can all be done in polynomial time in $\\log n$. Nov 15, 2017 at 3:17\n\nThis is not a complete answer but it describes an algorithm for computing $f(n)$ that should be reasonably fast. Let $\\omega(N)$ denote the number of distinct prime factors of $N$ and let $p_i$ denote the $i$th prime. If we happen to know the value of $\\omega(f(n))$\u2014call it $k$\u2014then we can find $f(n)$ by solving the following convex optimization problem with a linear objective function:\n\nMinimize $\\sum_{i=1}^k x_k \\log p_k$ subject to the constraint $\\prod_{i=1}^k (1+x_i) \\ge n + 2$,\n\nwhere the $x_i$ are required to be positive integers. The value of $f(n)$ will then be $\\prod_{i=1}^k p_i^{x_i}$. Now, we don't know the value of $k$, but at worst we can just try $k=1, 2, 3, \\ldots$ in turn, and at the very latest we will stop when $p_k$ exceeds our best value of $f(n)$ so far (and in practice much sooner). And almost certainly there are ways of zeroing in on the correct value of $k$ much faster than exhausting over all possibilities in this way.\n\nSolving the convex optimization problems ostensibly could take a long time, but in practice I think that you will be able to get an approximate answer very fast and there will not be that many integer candidates to exhaust over.\n\nThe number $\\prod_{i} p_i^{e_i}$ where $e_i = \\lfloor \\frac{1}{p_i^\\alpha-1} \\rfloor$ for any real number $\\alpha$ is locally optimal, i.e. no smaller number has a larger product of factors. (This uses the convexity, as mentioned by Timothy Chow - it's a local optimum of the convex function.) By a binary search, we could fairly rapidly find the smallest such number greater with more than $n$ factors. Such a number would have at most $2n$ factors, because this goes up by at most a factor of $2$ when $\\alpha$ increases a sufficiently small amount. So one could search between these two numbers and apply Igor Rivin's algorithm, adding at worst another factor of $n$.\n\n\u2022 It's been mentioned in comments on another answer, but just for completeness: these numbers are known as superior highly composite numbers following Ramanujan, and they (A002201) form a subsequence of the highly composite numbers (A002182) which in turn are all possible values of $f(n)$ for the $f$ in the question. Oct 30, 2017 at 19:39\n\nThe number of divisors of $n = \\prod_{i=1}^k p_i^{\\alpha_i}$ is $g(n)=\\prod_{i=1}^k (1+\\alpha_i)$ (your function differs from this by $2.$) So, once you have $g(n),$ you find the minimum over all factorizations of $g(n)$ of the product $2^{d_1} \\dots p_k^{d_k}.$ Now, the number of factorizations of $N$ is $o(g(n)^2)$ - by a result of Canfield, Erdos, Pomerance\n\nCanfield, E.R.; Erd\u0151s, Paul; Pomerance, Carl, On a problem of Oppenheim concerning \u201dFactorisatio Numerorum\u201d, J. Number Theory 17, 1-28 (1983). ZBL0513.10043.\n\nSo, there is certainly an $O(n^2)$ algorithm, which is pretty good as a function of $n,$ but pretty bad as a function of $\\log n.$ It seems clear that your question is at least as hard as factoring, the complexity of which is open.\n\nBy the way, I learned of the CEP paper from\n\nBalasubramanian, Ramachandran; Luca, Florian, On the number of factorizations of an integer, Integers 11, No. 2, 139-143, A12 (2011). ZBL1245.11100.\n\n\u2022 It's worse than this, because one can't only look at $n$; as the $n=100$ case illustrates, there may be 'easier' values (ones that lead to smaller $f(n)$ slightly larger than $n$ (those specifically that have fewer factors, or more specifically that have 'weightedly fewer' factors for an appropriate weighting). So one has to search a potentially indeterminate distance above $n$. Oct 27, 2017 at 22:15\n\u2022 @StevenStadnicki: \"So one has to search a potentially indeterminate distance above $n$.\" This is exactly my dilemma. Thank you for articulating it. Oct 27, 2017 at 22:32\n\u2022 @StevenStadnicki Yes, you are right, I am answering the question of finding the smallest $n$ with a given number of divisors. It seems clear, however, that the OP is at least as hard as that, so if one cannot give a better answer than mine, then things are bad. Oct 28, 2017 at 3:37\n\u2022 @IgorRivin : It's not clear to me that the original question is at least as hard as your problem. In your problem, we might have to sweat over factoring some difficult-to-factor numbers, whereas in the original problem we may be able to sidestep them. Oct 28, 2017 at 4:28\n\u2022 The question is much easier than factoring IMO. As I just posted as a comment on the question, we know $f(N)$ for all $N$ up to $N \\approx 1.99 \\times 10^{1254}$ (barring errors in the data or the theory those programs rest on), and we cannot say a similar thing for factoring. Oct 30, 2017 at 18:14", "date": "2023-01-29 12:40:08", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/284577/computational-complexity-of-finding-the-smallest-number-with-n-factors", "openwebmath_score": 0.8159390687942505, "openwebmath_perplexity": 176.43541888369455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682464686386, "lm_q2_score": 0.8688267745399465, "lm_q1q2_score": 0.8589814436694121}}
{"url": "http://mathhelpforum.com/algebra/78302-urgent-pattern-i-can-not-solve.html", "text": "# Math Help - urgent pattern i can not solve\n\n1. ## urgent pattern i can not solve\n\nWhat number comes next in this series????? 3, 10, 21, 44, 87\na)112\nb)125\nc)130\nd)163\ne)158\n\n2. Originally Posted by kl050196\nWhat number comes next in this series????? 3, 10, 21, 44, 87\na)112\nb)125\nc)130\nd)163\ne)158\nare you sure about the correctness of this sequence\n\nare you sure about the correctness of this sequence\n\nyea\n\n4. Originally Posted by kl050196\nWhat number comes next in this series????? 3, 10, 21, 44, 87\na)112\nb)125\nc)130\nd)163\ne)158\nClearly option e) 158.\n\nThe sequence of numbers is obviously generated by the function $y = \\frac{4}{3} x^3 -6x^2 + \\frac{47}{3} x - 8$ operating on the integers 1, 2, 3, 4, 5, 6 ....\n\nf(6) = 158.\n\n5. Hello, kl050196!\n\nMr. Fantastic did his usual excellent job.\nI too enjoy cranking out the generating functions for these problems.\n\nHowever, since it asked for the next term (only),\n. . we can use a somewhat intuitive approach.\n\nWhat number comes next in this series: . $3, 10, 21, 44, 87$\n\n. . $(a)\\;112\\qquad(b)\\;125\\qquad(c)\\;130\\qquad(d)\\;163 \\qquad(e)\\;158$\n\nTake the difference of consecutive terms.\nThen take the differences of the differences, and so on.\n\n$\\begin{array}{cccccccccccc}\n\\text{Sequence} & 3 &&10&&21&&44&&87 \\\\\n\\text{1st diff.} & & 7 && 11&&23&&43 \\\\\n\\text{2nd diff.} & & & 4 &&12&&20 \\\\\n\\text{3rd diff.} & & & & 8 && 8 \\end{array}$\n\nIt seems that the 3rd differences are constant.\n\nIf this is true, we can extend the diagram to the right . . .\n\nWe assume that the next 3rd difference is also 8:\n\n. . $\\begin{array}{cccccccccccc}\n3 &&10&&21&&44&&87 \\\\\n& 7 && 11&&23&&43 \\\\\n& & 4 &&12&&20 \\\\\n& & & 8 && 8 && {\\color{red}8}\\end{array}$\n\nThen the next 2nd difference must be 28:\n\n. . $\\begin{array}{cccccccccccc}\n3 &&10&&21&&44&&87 \\\\\n& 7 && 11&&23&&43 \\\\\n& & 4 &&12&&20 && {\\color{red}28}\\\\\n& & & 8 && 8 && 8\\end{array}$\n\nThen the next 1st difference must be 71:\n\n. . $\\begin{array}\n{cccccccccccc}\n3 &&10&&21&&44&&87 \\\\\n& 7 && 11&&23&&43 && {\\color{red}71}\\\\\n& & 4 &&12&&20 &&28\\\\\n& & & 8 && 8 && 8\\end{array}$\n\nFinally, the next term of the sequence must be 158:\n\n. . $\\begin{array}\n{cccccccccccc}\n3 &&10&&21&&44&&87&&{\\color{red}158} \\\\\n& 7 && 11&&23&&43 && 71\\\\\n& & 4 &&12&&20 &&28\\\\\n& & & 8 && 8 && 8\\end{array}\\quad\\hdots$\nta-DAA!", "date": "2014-03-12 05:17:18", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/78302-urgent-pattern-i-can-not-solve.html", "openwebmath_score": 0.7946026921272278, "openwebmath_perplexity": 1191.9139266698458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682461347529, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8589814333053601}}
{"url": "http://xmtermpaperfzft.bethanyspringretreats.us/projectile-range-vs-launch-angle.html", "text": "# Projectile range vs launch angle\n\nIntroduction when a projectile is fired, the horizontal distance traveled or \u201crange\u201d depends on the angle at which the projectile is launched in this activity we will. A derivative tells us the rate of change of one quantity with respect to another for example d x /d t typically represents how fast the position. A) assume you know the initial velocity with which a projectile is launched (v0) what launch in the angle range between 0\u00b0 and 90\u00b0, 2\u03b80 goes from 0\u00b0 to 180. Constant horizontal velocity and constant vertical acceleration the horizontal distance traveled by a projectile is called its range a projectile launched on level ground with an initial speed v0 at an angle \u03b8 above the horizontal will have. Part a - effect of launch angle on projectile range click on the trails and air buttons (to turn off air resistance) set the flagstick at 300 meters by clicking and.\n\nDistance between the launch point and the landing point, where the projectile a slope of angle \u03b1 is expressed as \u2206y = tan \u03b1\u2206x, and when the projectile. Optimization of projectile motion under air resistance quadratic in speed range of a projectile that is launched from atop a tower and is subject only to projectile ballistics quadratic drag optimal launch angle enveloping. Projectile motion is the motion of an object thrown or projected into the air, subject of a projectile on level ground launched at an angle \\boldsymbol{\\ theta_0}. Position and speed at any time can be calculated from the motion equations launch velocity of a projectile can be calculated from the range if the angle of.\n\nThis setup, all three kinds of projectile motion (horizontal, oblique \u2013 ground to thus, the height, angle and speed of the launch can be varied this helps in. Projectile motion (horizontal trajectory) calculator finds the initial and final velocity , and launch and landing angle parameters of projectile motion in physics. Figure 529 a boy kicks a ball at angle \u03b8, and it is displaced a distance of s along of finding the displacement (or range) of a projectile launched at an angle. Here is where i want to go over the \u201cother cool things with a computer\u201d and find the launch angle for projectile motion that gives the highest. In this lab you will study the motion of a freely-falling projectile, namely a small and verify the range and the time-of-flight of a projectile launched at an angle.\n\nThe range of an angled-launch projectile depends upon the launch speed and the launch angle (angle between the launch direction and the horizontal) figure . A projectile is an object that is given an initial velocity, and is acted on by gravity the horizontal range depends on the initial velocity v0, the launch angle \u03b8,. Use our projectile motion calculator to analyse a projectile in parabolic motion angle of launch deg it is free, awesome and will keep people coming back. In physics, assuming a flat earth with a uniform gravity field, and no air resistance , a projectile launched with specific initial conditions will surface \u03b8 is the angle at which the projectile is launched y0 is the initial height of the projectile. The motion of a projectile is composed by a uniform horizontal motion launch velocity, if $\\alpha$ is the angle this vector forms with.\n\nIn the projectile motion episode of nbc learn's the science of nfl due to gravity and \u03b8 is the angle at which the projectile is launched. Yes, you can do this if you can change the angle and speed, you have more variability than you need, so you have to find a reasonable set of. A specified angle for the marble launcher \u2022 evaluate the data error launch angle launch speed projectile range trajectory b2 launch speed and range.\n\n## Projectile range vs launch angle\n\nYou throw a football horizontally, and drop a football from the same height and at angle above the horizontal that will launch a projectile the farthest distance. The path of this projectile launched from a height y 0 has a range d root must be a positive number, and since the velocity and the cosine of the launch angle. Velocity components as the projectile travels along its trajectory \u2022 students will observe maximum range values and determine the optimum launch angle. For ideal projectile motion, which starts and ends at the same height, maximum range is achieved when the firing angle is 45\u00b0 if air resistance is taken into.\n\nThe initial launch angle (0-90 degrees) of an object in projectile motion dictates the range, height, and time of flight of that. Purpose of use: calculating initial velocity and launch angle for a football punter from game film to help fine tune practice goals confirmed initial assessment. Its range \u2022 to predict how far a projectile will travel when fired at different angles, and test the height of the launch point above the ground, written as h 3.\n\nProjectile motion with air resistance coordinate system whose origin coincides with the launch point, and whose $z$ irrespective of its initial launch angle. [APSNIP--] [APSNIP--]\n\nProjectile range vs launch angle\nRated 4/5 based on 30 review\n\n2018.", "date": "2018-11-13 18:37:45", "meta": {"domain": "bethanyspringretreats.us", "url": "http://xmtermpaperfzft.bethanyspringretreats.us/projectile-range-vs-launch-angle.html", "openwebmath_score": 0.37612393498420715, "openwebmath_perplexity": 604.8281345489603, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9752018383629825, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8589549372974188}}
{"url": "https://www.physicsforums.com/threads/bouncing-ball-math-question.78470/", "text": "# Bouncing ball math question\n\n#### ms. confused\n\nThis is a pretty tricky question...been trying to piece it together but I think I need some help.\n\nA ball is dropped from a height of 3 m. After each bounce, it rises to 82% of its previous height. After how many bounces does the ball reach a height less than 15cm?\n\nI get the whole geometric series thing, finding 'n' and all that, but what's with the \"less than 15cm\"? What do I do in this case?\n\nRelated Introductory Physics Homework News on Phys.org\n\n#### whozum\n\nIf initial height = h_i = 3m\nHeight of nth bounce = h_n = 3*.82^n\n\nYou are looking for n such that h_n < 0.15m.\n\n#### ms. confused\n\nExactly so would I just choose any value <0.15m and use it as h_n in order to find 'n'?\n\n#### whozum\n\nThat wouldn't work, because the series h_n only holds specific values. I'd just crank out values for h_n for each n, my guess is its about 15.\n\n#### ms. confused\n\nSo you mean kind of like a trial and error approach?\n\n#### Dr.Brain\n\nHeight initially = $h$ = 3m\nHeight after first bounce = $(0.82)(h)$\nheight after second bounce=$(0.82)^2 (h)$\nHeight after nth bounce=$(0.82)^n(h)$\n\nNow here you might like to use a bit of 'hitntrial' ...Now you want 'n' such that $(0.82)^n(h)$ is just more than .15m\n\n#### ms. confused\n\nWhy more if the question specifies that it should be less?\n\n#### whozum\n\nThat's what I would do, but I'm a lazy ass.\n\n#### whozum\n\nIt should be less.\n\n#### Dr.Brain\n\n$(0.82)^n>0.05$\n\n$n=15$\n\n#### whozum\n\nBrain, that is just confusing, let her solve the problem her own way. Ms. confused,\n\n$$h_n = 3(0.84)^n$$ and $$h_n \\leq 0.15$$ so then\n\n$$3(0.82)^n \\leq 0.15$$\n\nYou need to solve that for an integer n.\n\nLast edited:\n\n#### Dr.Brain\n\nactually i misread the question .\n\nn=15 gives the exact answer,but for less than 15 cm , n=14\n\n#### ms. confused\n\nOkay, I was getting 15 too but according to the answer key, n=16. This is probably a type-o, eh?\n\n#### whozum\n\nNo, 16 is the correct solution.\n\nBrain, the sequence is decreasing, for each higher n, h_n has a lower value. At n = 15 you get\n\n$$3*0.82^{15} = 0.1528m$$ which is still higher than 0.15m. Therefore you need another (16th) bounce to drop below.\n\n#### Dr.Brain\n\nThanks for correction.n=16\n\n#### ms. confused\n\nAlright now it makes perfect sense. Thank you so much!\n\n#### wisredz\n\nn= 16 because,\nlog0,05/log0,82=15,... at this value of n, h is 15 cm. on 16th jump, it would get under that height.\n\n#### Curious3141\n\nHomework Helper\nThe much easier way to solve the inequality is to use logs.\n\nLet $N$ be the required number of bounces, $h_0$ be initial height in centimeters and $h_n$ be the height after $n$ bounces.\n\nWe have : $$h_n = (0.82)^nh_0$$\n\nWe want : $$h_N < 15$$\n\nSo $$(0.82)^Nh_0 < 15$$\n\nTake common logs (base 10) of both sides,\n\n$$\\log{(0.82)^N} + \\log h_0 < \\log 15$$\n\nThe above inequality holds because log is a monotone increasing function on the positive reals.\n\n$$N\\log(0.82) < \\log 15 - \\log h_0$$\n\nDivide throughout by $$\\log(0.82)$$, but remember to invert the sign of the inequality because this is a negative quantity (since 0.82 is less than 10, the base of the logs).\n\n$$N > \\frac{\\log 15 - \\log h_0}{\\log(0.82)}$$\n\nEvaluate that, substituting $$h_0$$ = 300,\n\ngiving $$N > 15.096$$ or so.\n\nSo the smallest integer value for N that satisifies is $$N = 16$$.\n\n#### whozum\n\nI think the poster's problem was the setup, not the solving.\n\n#### Ouabache\n\nHomework Helper\nAs you found out on iteration $$3*0.82^{15} = 0.1528$$. This is better to use, compared to $$3*0.82^{15} = 0.15$$\nThis is where those rounding errors, I mentioned on an earlier question, come into play.\n\nDoc Brain, I liked your approach to solving this one.\nDr. Brain said:\nHeight initially = $h$ = 3m\nHeight after first bounce = $(0.82)(h)$\nheight after second bounce= $(0.82)^2 (h)$\nHeight after nth bounce=$(0.82)^n(h)$\nAnother way to plug-and-chug an interative formula is using spreadsheet. Excel's math utilities handle calculations well.\n\nLast edited:\n\n\"Bouncing ball math question\"\n\n### Physics Forums Values\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-04-22 17:57:45", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/bouncing-ball-math-question.78470/", "openwebmath_score": 0.6533300280570984, "openwebmath_perplexity": 2683.158400484422, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8589549266173163}}
{"url": "https://math.stackexchange.com/questions/2162551/how-to-find-the-limit-lim-limits-n-to-infty1-sqrtnn-which-is-indetermi", "text": "# How to find the limit $\\lim\\limits_{n\\to\\infty}1/\\sqrt[n]{n}$ which is indeterminate on evaluation but is convergent?\n\nWhen I evaluate the limit in the title above I get the following:\n\nBut when I use a computer software (mathematica) to evaluate the same limit it says the limit is 1. What am I doing wrong?\n\n\u2022 You should read up on indeterminate form,if the form is indeterminate it doesn't mean that the limit doesn't exist or that you can't determine the value of the limit. \u2013\u00a0kingW3 Feb 26 '17 at 18:13\n\nIndeterminate forms can have values.\n\nNote from L'Hospital's Rule that $\\lim_{n\\to \\infty}\\frac{\\log(n)}{n}=\\lim_{n\\to \\infty}\\frac{1/n}{1}=0$. Hence, we have\n\n\\begin{align} \\lim_{n\\to \\infty}\\frac{1}{n^{1/n}}&=\\lim_{n\\to \\infty}e^{-\\frac1n \\log(n)}\\\\\\\\ &e^{-\\lim_{n\\to \\infty}\\left(\\frac1n \\log(n)\\right)}\\\\\\\\ &=e^0\\\\\\\\ &=1 \\end{align}\n\nas expected!\n\n\u2022 Isn't $\\lim_{n\\to \\infty}\\left(\\frac1n \\log(n)\\right) = 0\\times\\infty$ also indeterminate? \u2013\u00a0razzak Feb 26 '17 at 18:16\n\u2022 @razzak It is indeterminate. And I showed, using L'Hospital's Rule, that the limit is $0$. \u2013\u00a0Mark Viola Feb 26 '17 at 18:17\n\u2022 sorry I got it now, we use L'hopital rule and it ends up as 0. \u2013\u00a0razzak Feb 26 '17 at 18:22\n\u2022 @razzak Well done! \u2013\u00a0Mark Viola Feb 26 '17 at 18:36\n\n$$\\frac{1}{\\sqrt[n]{n}}=\\left(\\frac{1}{n}\\right)^{\\frac{1}{n}}=e^{\\frac{1}{n}\\log{(1/n)}}=e^{-\\frac{1}{n}\\log{n}}$$ since $$\\lim_{n\\to\\infty}\\frac{1}{n}\\log{n}=0\\implies\\lim_{n\\to\\infty}e^{-\\frac{1}{n}\\log{n}}=e^0=1$$\n\n\u2022 you were faster... your answer was not written when i started posting \u2013\u00a0Francesco Alem. Feb 26 '17 at 18:12\n\u2022 No worry; that happens all the time to me too. (+1) for you answer! \u2013\u00a0Mark Viola Feb 26 '17 at 18:17\n\nHint :\n\n$$\\lim_{ n \\to \\infty }\\sqrt[n]{n}=1$$\n\n\u2022 Have you not simply written the answer that is in question? \u2013\u00a0Mark Viola Feb 26 '17 at 18:08\n\nIt is true that $\\infty^0$ is an indeterminate form. But the fact that you can reduce your expression to an indeterminate form does not mean that the original expression was indeterminate.\n\nIn fact, that is the whole point of saying $\\infty^0$ is indeterminate, because you can get $\\infty^0$ from lots of different expressions which do have (different) limits. Just knowing that you get to $\\infty^0$ doesn't tell you which expression you started with, and hence you can't tell which limit you should get.\n\nIf you're interested in the limit of $x^y$, it's not enough to know that $x\\to\\infty$ and $y\\to 0$; you also need to know how $x$ and $y$ are related. It's only if you don't have this information that the expression can't be determined. In your case you know that $y=1/x$, and this extra information means $\\lim x^y$ is no longer indeterminate.\n\n$n^{1/n}>1$, but for any $c>1$ we have $c^n>n$ if $n$ is sufficiently large, so eventually $n^{1/n}<c$. This means that the limit is $1$ in this case.", "date": "2019-07-23 15:18:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2162551/how-to-find-the-limit-lim-limits-n-to-infty1-sqrtnn-which-is-indetermi", "openwebmath_score": 0.8560221195220947, "openwebmath_perplexity": 453.9333497307281, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226341042415, "lm_q2_score": 0.8791467706759583, "lm_q1q2_score": 0.8589462936500623}}
{"url": "https://apartmanisokobanja.net/cubic-centimeter-jvmor/5e7e83-using-graph-to-demonstrate-a-function-which-is-invertible-function", "text": "Sokobanja, Srbija \u00a0 +381 65 8082462\n\n# using graph to demonstrate a function which is invertible function\n\nTRUE OR FALSE QUESTION. Your textbook probably went on at length about how the inverse is \"a reflection in the line y = x\".What it was trying to say was that you could take your function, draw the line y = x (which is the bottom-left to top-right diagonal), put a two-sided mirror on this line, and you could \"see\" the inverse reflected in the mirror. A graph of a function can also be used to determine whether a function is one-to-one using the horizontal line test: If each horizontal line crosses the graph of a function at no more than one point, then the function is one-to \u2026 Inverse trigonometric functions and their graphs Preliminary (Horizontal line test) Horizontal line test determines if the given function is one-to-one. What happens if we graph both $f\\text{ }$ and ${f}^{-1}$ on the same set of axes, using the $x\\text{-}$ axis for the input to both $f\\text{ and }{f}^{-1}?$. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. Another convention is used in the definition of functions, referred to as the \"set-theoretic\" or \"graph\" definition using ordered pairs, which makes the codomain and image of the function the same. A line. Quadratic function with domain restricted to [0, \u221e). Yes. The line will go up by 1 when it goes across by 1. Suppose we want to find the inverse of a function represented in table form. News; denote angles or real numbers whose sine is x, cosine is x and tangent is x, provided that the answers given are numerically smallest available. The inverse for this function would use degrees Celsius as the input and give degrees Fahrenheit as the output. Because the given function is a linear function, you can graph it by using slope-intercept form. We begin with an example. Suppose {eq}f{/eq} and {eq}g{/eq} are both functions and inverses of one another. Draw graphs of the functions $f\\text{ }$ and $\\text{ }{f}^{-1}$. Figure 8. The inverse of the function f(x) = x + 1 is: The slider below shows another real example of how to find the inverse of a function using a graph. We already know that the inverse of the toolkit quadratic function is the square root function, that is, ${f}^{-1}\\left(x\\right)=\\sqrt{x}$. The function has an inverse function only if the function is one-to-one. Let's use this characteristic to identify inverse functions by their graphs. A function and its inverse function can be plotted on a graph. They are also termed as arcus functions, antitrigonometric functions or cyclometric functions. When you\u2019re asked to find an inverse of a function, you should verify on your own that the inverse \u2026 Every point on a function with Cartesian coordinates (x, y) becomes the point (y, x) on the inverse function: the coordinates are swapped around. Graph of function g, question 1. The inverse trigonometric functions actually performs the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. At times, your textbook or teacher may ask you to verify that two given functions are actually inverses of each other. Are the blue and red graphs inverse functions? is it always the case? Q. This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. Find the inverse function of the function plotted below. This is what they were trying to explain with their sets of points. Several notations for the inverse trigonometric functions exist. Finding the inverse of a function using a graph is easy. Notation. (This convention is used throughout this article.) 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Reciprocal function, so we need to using graph to demonstrate a function which is invertible function the domain and range not tricky.", "date": "2021-07-24 20:13:29", "meta": {"domain": "apartmanisokobanja.net", "url": "https://apartmanisokobanja.net/cubic-centimeter-jvmor/5e7e83-using-graph-to-demonstrate-a-function-which-is-invertible-function", "openwebmath_score": 0.5981078743934631, "openwebmath_perplexity": 433.61432681108704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9770226354423304, "lm_q2_score": 0.8791467690927439, "lm_q1q2_score": 0.8589462932796026}}
{"url": "https://math.stackexchange.com/questions/3480467/find-the-smallest-positive-number-k", "text": "# Find the smallest positive number k\n\nFor any positive number $$n$$, let $$a_n = \\sqrt{2+\\sqrt{2+{...+\\sqrt{2+\\sqrt 2}}}}$$ ($$2$$ appear $$n$$) and let $$k$$ is positive number such that $$\\displaystyle\\frac{1}{k}\\leq\\frac{3-a_{n+1}}{7-a_n}$$ for any positive number $$n$$, then find the smallest positive number $$k$$.\n\nI have $$a_1=\\sqrt 2$$ and $$a_{n+1}=\\sqrt{2+a_n}, \\forall n \\in\\mathbb N$$\n\nConsider\n\n$$a_1=\\sqrt 2 \\lt 2$$\n\n$$a_2=\\sqrt{2+a_1}\\lt\\sqrt{2+2}=2$$\n\n$$a_3=\\sqrt{2+a_2}\\lt\\sqrt{2+2}=2$$\n\nUse Mathematical Induction, I conclude $$\\sqrt 2\\leq a_n\\leq 2,\\forall n\\in\\mathbb N$$\n\nThus, $$3-a_{n+1}\\gt1$$ and $$7-a_{n}\\gt 5$$\n\nSince $$k\\in\\mathbb N$$, I have $$\\displaystyle k\\geq\\frac{7-a_n}{3-a_{n+1}}=\\frac{7-a_n}{3-\\sqrt{2+a_n}}=3+\\sqrt{2+a_n}=3+a_{n+1}$$\n\nHence, $$3+\\sqrt 2\\leq 3+a_{n+1}\\leq 3+2=5$$\n\nTherefore $$k=5$$\n\nPlease check my solution, Is it correct?, Thank you\n\n\u2022 $a_n=2\\cos\\frac{\\pi}{2^{n+1}}$. \u2013\u00a0Riemann Dec 18 '19 at 8:06\n\nYou have given a lower & upper bound for $$a_n$$. With this, you've determined a possible value for $$k$$, but you haven't shown it's necessarily the smallest such value of $$k$$.\n\nInstead, note that $$a_{n+1} = \\sqrt{2 + a_n}$$. Thus, you have\n\n\\begin{aligned} \\frac{3-a_{n+1}}{7-a_n} & = \\frac{3-\\sqrt{2 + a_n}}{7-a_n} \\\\ & = \\frac{(3-\\sqrt{2 + a_n})(3 + \\sqrt{2 + a_n})}{(7-a_n)(3 + \\sqrt{2 + a_n})} \\\\ & = \\frac{9-(2 + a_n)}{(7-a_n)(3 + \\sqrt{2 + a_n})} \\\\ & = \\frac{7 - a_n}{(7-a_n)(3 + \\sqrt{2 + a_n})} \\\\ & = \\frac{1}{3 + \\sqrt{2 + a_n}} \\end{aligned}\\tag{1}\\label{eq1A}\n\nThus, if $$L$$ is the supremum of $$a_n$$, then $$k = 3 + \\sqrt{2 + L}$$. To determine $$L$$, you can prove that $$a_n$$ is a strictly increasing sequence (I'll leave it to you fill in the details, such as what is shown in Solution verification: Prove by induction that $$a_1 = \\sqrt{2} , a_{n+1} = \\sqrt{2 + a_n}$$ is increasing and bounded by $$2$$), with an upper bound as you've shown, so it must converge to limiting value of its supremum. To determine this value, use\n\n\\begin{aligned} L & = \\sqrt{2 + L} \\\\ L^2 & = 2 + L \\\\ L^2 - L - 2 & = 0 \\\\ (L - 2)(L + 1) & = 0 \\end{aligned}\\tag{2}\\label{eq2A}\n\nThus, since $$L \\gt 0$$, you have $$L = 2$$, as you surmised. You also have\n\n$$k = 3 + \\sqrt{2 + 2} = 5 \\tag{3}\\label{eq3A}$$\n\nwhich matches what you got.\n\n\u2022 $a_n=2\\cos\\frac{\\pi}{2^{n+1}}$. \u2013\u00a0Riemann Dec 18 '19 at 8:07", "date": "2020-04-01 00:08:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3480467/find-the-smallest-positive-number-k", "openwebmath_score": 0.9999126195907593, "openwebmath_perplexity": 255.4639680597133, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022630759019, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8589462922559571}}
{"url": "https://math.stackexchange.com/questions/1191072/how-to-calculate-the-errors-of-single-and-double-precision", "text": "# How to calculate the errors of single and double precision\n\nWe consider the initial value problem\n\n$$\\left\\{\\begin{matrix} y'=y &, 0 \\leq t \\leq 1 \\\\ y(0)=1 & \\end{matrix}\\right.$$\n\nWe apply the Euler method with $h=\\frac{1}{N}$ and huge number of steps $N$ in order to calculate the approximation $y^N$ of the value of the solution $y$ at $t^N, \\ y(t^N)=y(1)=e$. At the following table there are, for all $N$, the errors $|\\epsilon^N|=|e-y^N|$, when the calculations are done with single and double precision.\n\n$$\\begin{matrix} N & |\\epsilon^N|\\text{ Single-precision } & |\\epsilon^N| \\text{ Double-precision } \\\\ - & - & - \\\\ 100 & 0.13468 \\cdot 10^{-1} & 0.13468 \\cdot 10^{-1} \\\\ 200 & 0.67661 \\cdot 10^{-2} & 0.67647 \\cdot 10^{-2}\\\\ 400 & 0.33917 \\cdot 10^{-2} & 0.33901 \\cdot 10^{-2}\\\\ 800 & 0.16971 \\cdot 10^{-2} & 0.16970 \\cdot 10^{-2}\\\\ 1600 & 0.85568 \\cdot 10^{-3} & 0.84898 \\cdot 10^{-3} \\\\ \\cdots & & \\\\ 102400 & 0.65088 \\cdot 10^{-4} & 0.13273 \\cdot 10^{-4} \\\\ 204800 & 0.21720 \\cdot 10^{-3} & 0.66363 \\cdot 10^{-5} \\\\ 409600 & 0.78464 \\cdot 10^{-3} & 0.33181 \\cdot 10^{-5} \\\\ 819200 & 0.20955 \\cdot 10^{-2} & 0.16590 \\cdot 10^{-5} \\\\ \\dots \\end{matrix}$$\n\nWe notice that the errors of the calculations of double-precision get approximately half. However, in the case of single-precision, for $N>10^5$ the errors increase! Indeed, for a big enough $N$, the errors in our case tend to $1.71828 \\dots$.\n\nCould you explain me why the errors, when the calculations are done in single-precision, increase for $N>10^5$ and why they get approximately half when the calculations are done in double-precision?\n\nAlso, how can we calculate the error for a given $N$? For example, if we have $N=10^5$ then $\\epsilon^N=|e-y^{10^5}|=\\left |e- \\left( 1+ \\frac{1}{10^5} \\right)^{10^5} \\right |$. How can we calculate the latter, knowing that the zero of the machine is $10^{-6}$ when we have single precision but $10^{-12}$ when we have double precision?\n\nEDIT: It holds that: $$\\ln{\\left( 1+ \\frac{1}{N}\\right)}=\\sum_{n=1}^{\\infty} (-1)^{n+1} \\frac{\\left( \\frac{1}{N}\\right)^n}{n}=\\frac{1}{N}- \\frac{1}{2N^2}+O(N^{-3})$$ Right? If so, then $N \\ln{\\left( 1+ \\frac{1}{N}\\right)}=1-\\frac{1}{2N}+O(N^{-2})$, right?  If so, then how can we find the difference of the real solution with the approximation when we take into consideration that we have single precision and how when we take into consideration that we have double precision?\n\n\u2022 In the case of single-precision, it seems as if truncation error is overcoming the finite-step integration error. \u2013\u00a0robjohn Mar 15 '15 at 17:59\n\u2022 Note that IEEE float has the pattern (s,e,m)=(1,8,23) and thus a machine constant $2^{-23}\\simeq 0.125\u00b710^{-6}$. IEEE double has the pattern (s,e,m)=(1,11,52) and thus a machine constant $2^{-52}\\simeq 0.25\u00b710^{-15}$. \u2013\u00a0Dr. Lutz Lehmann Mar 15 '15 at 19:05\n\nWhen $N$ gets large, the size of each step, and thus the size of the change in function value, gets small. You start out with $y(0)=1$, and then you get $$y(h)=1+\\epsilon$$ and if $\\epsilon$ is small enough, the computer won't be able to handle the difference between $1$ and $1+\\epsilon$ with very good precision. This is the source of the increasing error. Since double precision handles this problem better, you get less error.\n\nAs for why the error tends to $1.71828\\ldots$, if $\\epsilon$ is really small, the computer thinks that $y$ doesn't change at all from time step to time step, and therefore thinks that $y$ is a constant function. You're supposed to get $e$ as the final value, so the error is therefore $e-1$.\n\n\u2022 Don't we have to consider the difference between the real solution and the approximation?  If so, then do we have to consider the difference $|y(h)-y^1|$ ? \u2013\u00a0evinda Mar 15 '15 at 18:17\n\u2022 @evinda The $\\epsilon$ is just $\\Delta y$ for the first step. The reason I don't use a formula specific to this problem is that the reasoning in this answer is equally valid for any (sufficiently nice) differential equation. \u2013\u00a0Arthur Mar 15 '15 at 18:18\n\u2022 It holds that $t^n=nh=\\frac{n}{N}$, right?  So for the first step, doesn't the following hold?  $$t^1=\\frac{1}{N}$$ $$y^1=1+h=1+ \\frac{1}{N}$$ $$y\\left( \\frac{1}{N} \\right)=e^{\\frac{1}{N}}$$ $$\\left| y\\left( \\frac{1}{N} \\right)-y^1\\right|=\\left| e^{\\frac{1}{N}}-\\left( 1+ \\frac{1}{N} \\right) \\right|$$ Or am I wrong? If it is like that, then how do we calculate the latter? \u2013\u00a0evinda Mar 15 '15 at 18:27\n\u2022 $$\\left(1+\\frac1N\\right)^N=e^{N\u00b7\\ln(1+\\frac1N)}=e^{1-\\frac 2N+\\frac3{N^2}\\mp...}=e^1\\left(1-\\frac2N+O(N^{-2})\\right),$$ but in general it is sufficient to know that the global error is proportional to $(e^{LT}-1)h$ with Lipschitz constant $L$, end time or integration interval length $T$ and step size $h=1/N$. \u2013\u00a0Dr. Lutz Lehmann Mar 15 '15 at 18:40\n\u2022 It holds that: $$\\ln{\\left( 1+ \\frac{1}{N}\\right)}=\\sum_{n=1}^{\\infty} (-1)^{n+1} \\frac{\\left( \\frac{1}{N}\\right)^n}{n}=\\frac{1}{N}- \\frac{1}{2N^2}+O(N^{-3})$$ Right? If so, then $N \\ln{\\left( 1+ \\frac{1}{N}\\right)}=1-\\frac{1}{2N}+O(N^{-2})$, right?  If so, then how can we find the difference of the real solution with the approximation when we take into consideration that we have single precision and how when we take into consideration that we have double precision? \u2013\u00a0evinda Apr 13 '15 at 13:10\n\nThe Euler method has local discretization error order 2 and global error order 1. With more detailed analysis one finds the global error proportional to $e^{LT}h$ where $L$ is a Lipschitz constant and $T$ the integration period. In short, if the integration interval is held constant the global error is $Ch+O(h^2)$. This explains why doubling the number of steps and thus halving the step size $h$ also halves the global error.\n\nEach integration step has a random floating point error that is a small multiple $D\u03bc$ of the machine precision $\u03bc$ for the floating point type. This error accumulates over the $N=1/h$ steps and adds to the discretion error, in each step and globally.\n\nIn total this gives an expression for the computational global error like $$Ch+ND\u03bc=Ch+\\frac{TD\u00b7\u03bc}h=2\\sqrt{CDT\u00b7\u03bc}+\\frac{C}{h}\u00b7\\left(h-\\sqrt{\\frac{DT\u00b7\u03bc}{C}}\\right)^2$$ This is a convex function in $h$ that has a valley at approximately (disregarding the constants) $h\u2248\\sqrt{\u03bc}$ with an error of about the same magnitude $\\sqrt\u03bc$.\n\n\u2022 For the float type $\u03bc\u2248 10^{-7}$ and thus the minimal error is attained for approximately $h\u2248 10^{-4}$ to $10^{-3}$.\n\u2022 For the double type, $\u03bc\u224810^{-15}$, so the valley is to be found at step sizes $h\u2248 10^{-8}$ to $10^{-7}$.\n\nThus if you were to continue the table, you would see growing errors also for the double type after reaching about $10^{-7}$ as lowest error. As test, for $h=10^{-9}$ or $N=10^9\\simeq 2^{23}\u00b7100$ the error should be above $10^{-6}$.\n\nThe error progression of the Euler method can be compared to compound interest rate computations where the interest rate is $Lh$ per integration step. This is a consequence of the Gronwall lemma. The local error in step $k$ gets thus amplified over the remaining $(N-k-1)$ integration steps with a factor $(1+Lh)^{N-k-1}$ as part of the global error. The local error is the sum of the discretization error bound by $c\u00b7h^2$ and the floating point error bound by $d\u00b7\u03bc$ where $c$ contains the magnitude of the system function $f$ and $d$ the magnitude of its derivative resp. the complexity of the evaluation of $f$. Thus the estimate of the global error has the form \\begin{align} (c\u00b7h^2+d\u00b7\u03bc)\u00b7\\sum_{k=0}^{N-1}(1+Lh)^{N-k-1} &=(c\u00b7h^2+d\u00b7\u03bc)\u00b7\\frac{(1+Lh)^N-1}{Lh} \\\\ &\\simeq \\left(\\frac{c\u00b7h}L+\\frac{d\u00b7\u03bc}{Lh}\\right)\u00b7(e^{LT}-1) \\end{align}\n\n\u2022 What do you mean by integration step? \u2013\u00a0evinda Mar 15 '15 at 19:19\n\u2022 Moving from $t$ to $t+h$. The solution of differential equations is also called integration. \u2013\u00a0Dr. Lutz Lehmann Mar 15 '15 at 19:31\n\u2022 How do we get that if the integration interval is held constant then the global error is $Ch+O(h^2)$? \u2013\u00a0evinda Mar 15 '15 at 19:53\n\u2022 This just means that the error is a function of $h$ with a regular root at $h=0$, i.e., $h\u00b7g(h)$ with $g(h)=C+O(h)$. Or just a more explicit way to say that the error is of order $O(h)$. \u2013\u00a0Dr. Lutz Lehmann Mar 15 '15 at 20:05\n\u2022 1. The error is $O(h)$ because it is proportional to $e^{LT}h$, or am I wrong?  2. Could you explain me the sentence: \"Each integration step has a random floating point error that is a small multiple $D \\mu$ of the machine precision $\\mu$ for the floating point type.\" ?  3. How did you get this equality: $$Ch+\\frac{TD\u00b7\u03bc}h=2\\sqrt{CDT\u00b7\u03bc}+\\frac{C}{h}\u00b7\\left(h-\\sqrt{\\frac{DT\u00b7\u03bc}{C}}\\right)^2$$ ? \u2013\u00a0evinda Mar 15 '15 at 20:12", "date": "2019-11-21 03:07:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1191072/how-to-calculate-the-errors-of-single-and-double-precision", "openwebmath_score": 0.909229576587677, "openwebmath_perplexity": 181.87190885323648, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98657174604767, "lm_q2_score": 0.870597265050901, "lm_q1q2_score": 0.8589066638855934}}
{"url": "https://math.stackexchange.com/questions/2844296/evaluating-operatornamearccos-frac2-sqrt5-operatornamearccos-frac", "text": "# Evaluating $\\operatorname{arccos} \\frac{2}{\\sqrt5} + \\operatorname{arccos} \\frac{3}{\\sqrt{10}}$\n\nEvaluate: $$\\operatorname{arccos} \\frac{2}{\\sqrt5} + \\operatorname{arccos} \\frac{3}{\\sqrt{10}}$$\n\nWe let $$\\alpha = \\operatorname{arccos} \\frac{2}{\\sqrt5} \\qquad \\beta = \\operatorname{arccos}\\frac{3}{\\sqrt{10}}$$\n\nThen we have:\n\n\\begin{align} \\cos(\\alpha) = \\frac{2}{\\sqrt5} &\\qquad \\cos(\\beta) = \\frac{3}{\\sqrt{10}} \\\\[4pt] \\sin(\\alpha) = \\frac{1}{\\sqrt5} &\\qquad \\sin(\\beta) = \\frac{1}{\\sqrt{10}} \\end{align}\n\nIn order to evaluate, we are told, we first determine $\\sin(\\alpha + \\beta)$; we wind up with $1/\\sqrt2$, thus we have $\\pi/4$.\n\nWhat I am confused about is why we have to use sin($\\alpha + \\beta$). For example, if I were to use $\\cos(\\alpha + \\beta)$, I would get the answer $7/(\\sqrt{10}\\sqrt5)$, which I do not know what to do with. I am having trouble finding out whether there is some kind of pattern to this kind of thing, or did the author just know to use $\\sin(\\alpha + \\beta)$ since he/she checked cos and saw nothing comes out of this?\n\nAny help is much appreciated, thank you\n\n\u2022 $7\\sqrt{50}$ is $\\cos(\\alpha-\\beta)$, not $\\cos(\\alpha+\\beta)$. \u2013\u00a0Angina Seng Jul 8 '18 at 4:13\n\u2022 Actually, you need to do both. It is the signs of the the two answers that determines which quadrant the answer is in. \u2013\u00a0steven gregory Jul 8 '18 at 4:47\n\nAs an alternative approach (which, truthfully, arose from my not reading your question carefully and, essentially, wishing to draw in MS Paint) see the following diagram in which the expression you wish to evaluate is the angle sum $\\alpha + \\beta$:\n\nNext, let us consider instead $\\tan(\\alpha + \\beta)$ using a tangent identity:\n\n$$\\tan(\\alpha + \\beta) = \\frac{\\tan\\alpha + \\tan\\beta}{1 - \\tan\\alpha \\tan\\beta} = \\frac{1/2 + 1/3}{1 - (1/2)(1/3)} = \\frac{5/6}{5/6} = 1$$\n\nObserve $\\alpha + \\beta \\in (0, \\pi)$; the unique angle in this interval yielding a tangent of $1$ is $\\pi/4$.\n\n\u2022 Fantastic answer thank you very much for the insight \u2013\u00a0Algebra 8 Jul 8 '18 at 16:57\n\nFrom $\\sin(\\alpha+\\beta)=\\frac1{\\sqrt2}$, we should expect $|\\cos(\\alpha+\\beta)|=\\frac1{\\sqrt2}$ from the formula $\\sin^2 \\theta + \\cos^2 \\theta = 1$.\n\n\\begin{align} \\cos(\\alpha+\\beta)&=\\cos(\\alpha)\\cos(\\beta)\\color{blue}-\\sin(\\alpha)\\sin(\\beta)\\\\ &=\\frac{2}{\\sqrt5}\\cdot \\frac{3}{\\sqrt{10}}-\\frac1{\\sqrt5}\\cdot \\frac{1}{\\sqrt{10}}\\\\ &=\\frac{6}{5\\sqrt{2}}-\\frac{1}{5\\sqrt2}\\\\ &=\\frac{5}{5\\sqrt2}\\\\ &=\\frac{1}{\\sqrt2} \\end{align}\n\nYou can directly use the formula $$\\cos^{-1}x+\\cos^{-1}y=\\cos^{-1}(xy-\\sqrt{1-x^2}\\sqrt{1-y^2})$$ $$x=\\frac{2}{\\sqrt{5}},y=\\frac{3}{\\sqrt{10}}$$ $$=\\cos^{-1}\\left(\\left(\\frac{2}{\\sqrt{5}}\\right)\\left(\\frac{3}{\\sqrt{10}}\\right)-\\sqrt{1-\\left(\\frac{2}{\\sqrt{5}}\\right)^2}\\sqrt{1-\\left(\\frac{3}{\\sqrt{10}}\\right)^2}\\right)$$ $$=\\cos^{-1}\\left(\\frac{6}{\\sqrt{50}}-\\sqrt{1-\\frac45}\\sqrt{1-\\frac{9}{10}}\\right)$$ $$=\\cos^{-1}\\left(\\frac{6}{\\sqrt{50}}-\\sqrt{\\frac15}\\sqrt{\\frac{1}{10}}\\right)$$ $$=\\cos^{-1}\\left(\\frac{6}{\\sqrt{50}}-\\frac{1}{\\sqrt{50}}\\right)$$ $$=\\cos^{-1}\\left(\\frac{5}{\\sqrt{50}}\\right)$$ $$\\cos^{-1}\\frac{2}{\\sqrt{5}}+\\cos^{-1}\\frac{3}{\\sqrt{10}}=\\cos^{-1}\\left(\\frac{5}{\\sqrt{50}}\\right)$$\n\n\u2022 Thank you all very much, I am studying for the GRE and the author gave the formula cos($\\alpha + \\beta$) = cos$\\alpha$cos$\\beta$ + sin$\\alpha$sin$\\beta$. After seeing everyone's answers and checking the web I see that that is a mistake. \u2013\u00a0Algebra 8 Jul 8 '18 at 16:57\n\u2022 You mean the answer which I have is wrong? \u2013\u00a0Key Flex Jul 8 '18 at 17:01\n\nUsing\n\n$$\\operatorname{arccos} \\frac{2}{\\sqrt5} + \\operatorname{arccos} \\frac{3}{\\sqrt{10}}=\\pi-\\left(\\arcsin\\dfrac2{\\sqrt5}+\\arcsin\\frac{3}{\\sqrt{10}}\\right)$$\n\nas $$\\left(\\frac{2}{\\sqrt5}\\right)^2+\\left(\\frac{3}{\\sqrt{10}}\\right)^2=\\dfrac45+\\dfrac9{10}>1$$\n\n$$\\arcsin\\dfrac2{\\sqrt5}+\\arcsin\\frac{3}{\\sqrt{10}}=\\pi-\\arcsin\\left(\\dfrac2{\\sqrt5}\\dfrac1{\\sqrt{10}}+\\dfrac1{\\sqrt5}\\dfrac3{\\sqrt{10}}\\right)=\\pi-\\arcsin\\dfrac1{\\sqrt2}=?$$", "date": "2021-07-31 15:18:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2844296/evaluating-operatornamearccos-frac2-sqrt5-operatornamearccos-frac", "openwebmath_score": 0.8592499494552612, "openwebmath_perplexity": 613.6945146094305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104953173166, "lm_q2_score": 0.8887587846530937, "lm_q1q2_score": 0.8589058172942126}}
{"url": "https://staff.noqood.co/gillian-flynn-hwmjkyk/topological-sort-problems-codeforces-c0afea", "text": "Here's an example: Nevertheless, we have still used following common algorithms at many places \u2013 min, max, swap, sort, next_permutation, binary_search, rotate, reverse . In computer science, a topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. There are severaltopologicalsortingsof (howmany? For example, a topological sorting of the following graph is \u201c5 4 2 3 1 0\u201d. Practice Problems. If you encounter GREY node while doing DFS traversal ==> CYCLE. Dismiss Join GitHub today. A2 Online Judge (or Virtual Online Contests) is an online judge with hundreds of problems and it helps you to create, run and participate in virtual contests using problems from the following online judges: A2 Online Judge, Live Archive, Codeforces, Timus, SPOJ, TJU, SGU, PKU, ZOJ, URI. My question is, how can dfs be applied to solve this problem, and where can I find more theory/practice problems to practice topological sorting. 2), CSES Problem Set new year 2021 update: 100 new problems, Click here if you want to know your future CF rating, AtCoder Grand Contest 050/051 (Good Bye rng_58 Day 1 / Day 2) Announcement. It's always guaranteed that there's a vertex with in-degree 0, unless the graph has a cycle, in which case, there is no topological ordering. ACCURACY: 29% Detailed tutorial on Topological Sort to improve your understanding of Algorithms. The problem only has a constraints that mn <= 1e5, which means m can be up to 1e5, if we construct the graph by a simple brute force, O(nm^2) complexity would be too high. 1 & 2): Gunning for linear time\u2026 Finding Shortest Paths Breadth-First Search Dijkstra\u2019s Method: Greed is good! Partial ordering is very useful in many situations. There can be more than one topological sorting for a graph. Fifth, After failed in 3rd time see my solution. However, the graph construction can not be done by brute force. Another way to check would be doing a dfs and coloring the vertex with 3 colors. 817D Imbalanced Array (Description) Segment Tree. ... Codeforces . For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. The editorial mentions that this is a classic topological sort problem. There are a couple of algorithms for Toposort. Solve practice problems for Topological Sort to test your programming skills. Check out this link for an explanation of what topological sorting is: http://www.geeksforgeeks.org/topological-sorting/, http://www.spoj.com/problems/RPLA/ http://www.spoj.com/problems/TOPOSORT/. Complete reference to competitive programming. Signup and get free access to 100+ Tutorials and Practice Problems Start Now, ATTEMPTED BY: 11 A topological sort is an ordering of the nodes of a directed graph such that if there is a path from node u to node v, ... in the next session we will be discussing Dynamic Programming Application in Solving Some Classic Problems in Acyclic Graph and problems related to it and for now practice problems. I have an alternative solution to G: First assign to each node the number given by the following greedy algorithm: Process all nodes in order of topological sort starting from the leaves and assign 0 to a leaf and maximum of values of all the children plus 1 if it's not a leaf. 3. For example, a topological sorting \u2026 As you can see the graph is not an unweighted DAG, Hence, the problem became finding acyclic longest chain in a directed cyclic graph. First, Try To Understand the Problem Statement. ACCURACY: 48% 2) 27:08:49 Register now \u00bb The graph should contain at most n \u2014 1 edges (less if the list contains adjacent, lexicographically equal names) and at most 26 vertices (1 vertex for each letter in the alphabet). Yes. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. LEVEL: Medium, ATTEMPTED BY: 1119 My code is here 50382490. Please, don\u2019t just copy-paste the code. 4.Eneque any of the vertices whose indegree has become zero during the above process. Second, Solve Code with Pen and Paper. ACCURACY: 68% A Topological Sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. Graph Ordering / Share Algorithms, Approximate. Fourth, If failed to AC then optimize your code to the better version. Problem link\u2014 SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 \u2026 Topological Sort (ver. ACCURACY: 64% Priority Queue (Heap) \u2013 | page 1 While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. If \"tourist\" directly preceded \"toosimple,\" for example, we could determine that 'u' precedes 'o'. Assumption: We are talking about Div2. Detailed tutorial on Topological Sort to improve your understanding of Algorithms. SPOJ TOPOSORT - Topological Sorting [difficulty: easy] UVA 10305 - Ordering Tasks [difficulty: easy] UVA 124 - Following Orders [difficulty: easy] UVA 200 - Rare Order [difficulty: easy] I wasn't able to come with a clear solution (O(n)) because it didn't feel right that how I can take care of cases with a cycle using maybe topological sort? http://ideone.com/KVobNb, Take 'a' =1, 'b' =2 and so on... Upto 'z'.. Then take pair of words given in the inpu such as\"word-1 with word-2\" and \"word 2 with word 3\" and so on inorder to compare them character by character starting from index-0 of both, and where you meet a mismatch before one of the strings ends, you should create an directed edge from word[k][i] to word[k+1][i] and so on untill all words are processed. I spent a fair bit of time on it, and I knew while solving it that it was a topological sorting problem. Practice always helps. Skills for analyzing problems and solving them creatively are needed. LEVEL: Hard, ATTEMPTED BY: 68 1) 27:08:49 Register now \u00bb *has extra registration Before contest Codeforces Round #668 (Div. Remove it from the graph and update in-degrees of outgoing vertices, then push it into some vector. Search problems across all Competitive Programming websites. Practice always helps. However, I have gone through the USACO training pages to learn my algorithms, which doesn't have a section on topological sorting. ACCURACY: 72% \u03b1(m, n) time complexity of Union-Find, I was working on this problem: http://codeforces.com/contest/510/problem/C. CodeChef . We care about your data privacy. Take a situation that our data items have relation. Example: Let & and have if and only if $. Topological Sorting for a graph is not possible if the graph is not a DAG. UVA- 11686 \u2013 Pick up sticks. thinking. It may be numeric data or strings. The final alphabet is simply the topologically sorted graph, with unused characters inserted anywhere in any order. The topological sort algorithm takes a directed graph and returns an array of the nodes where each node appears before all the nodes it points to. 4.Eneque any of the vertices whose indegree has become zero during the above process. The Cake Is a Lie (diff=2400, constructive algorithm, topological sort, BFS) We should observe that the starting cut must be a cake with no more than 1 shared edge. There can be more than one topological sorting for a graph. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Topological Sorting Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge ( u v ) from \u2026 At the end of the algorithm, if your vector has a size less than the number of vertices, then there was a cycle somewhere! Learn some basic graph algorithms like BFS, DFS, their implementations. Here\u2019s simple Program to implement Topological Sort Algorithm Example in C Programming Language. Some courses may have prerequisites, for example, to take course 0 you have to first take course 1, which is expressed as a pair [0,1]. Then simply use KHAN'S ALGORITHM to detect that the graph id acyclic or not, if not then no solution exists, otherwise exists(carefull about a corner case, given below) Word-1 \u2014 > \"kgef\" and word-2 \u2014 >\"kge\", here solution does not exist. ), for example: 12 algorithm graph depth-first-search topological-sort. SOLVE. This is partial order, but not a linear one. Thanks in Advance. The main function of the solution is topological_sort, which initializes DFS variables, launches DFS and receives the answer in the vector ans. Discussions NEW. Covered in Chapter 9 in the textbook Some slides based on: CSE 326 by S. Wolfman, 2000 R. Rao, CSE 326 2 Graph Algorithm #1: Topological Sort 321 143 142 322 326 341 370 378 401 421 Problem: Find an order in Graph Ordering. WHITE \u2014 Unprocessed 2. I have an alternative solution to G: First assign to each node the number given by the following greedy algorithm: Process all nodes in order of topological sort starting from the leaves and assign 0 to a leaf and maximum of values of all the children plus 1 if it's not a leaf. 1. Not Able to solve any question in the contest. We know many sorting algorithms used to sort the given data. One of them arises in parallel computing where a program can be represented as DAG. 52C Circular RMQ (Range addition, range minimum query, Description) 56E Domino Principle (Single assignment, range maximum query, Description) Store the vertices in a list in decreasing order of finish time. CodeChef was created as a platform to help programmers make it big in the world of algorithms, computer programming, and programming contests.At CodeChef we work hard to revive the geek in you by hosting a programming contest at the start of the month and two smaller programming challenges at the middle and end of the month. I did it by Topological Sorting. Sorting Algorithms are methods of reorganizing a large number of items into some specific order such as highest to lowest, or vice-versa, or even in some alphabetical order. Learn some basic graph algorithms like BFS, DFS, their implementations. Compare all such names on the list and build a directed graph consisting of all the orderings, with each directed edge (a, b) denoting that character a precedes b. LEVEL: Medium, ATTEMPTED BY: 1425 You are given oriented graph with n vertices numbered $$1, 2, \\dots, n$$ and m edges $$u_i, v_i$$. Befor do topsort on a graph,graph have to be DAG.Could you please help me how i check whether it is DAG or not? Understnad the logic and implement by your own. The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. LEVEL: Medium, ATTEMPTED BY: 489 Repeat 1 while there are still vertices in the graph. The topological sort is a simple but useful adaptation of a depth first search. Given a list of names, does there exist an order of letters in Latin LEVEL: Medium, ATTEMPTED BY: 37 When there are still nodes remaining, but none of them as IN-degree as ZERO, you can be sure that a cycle exists in the graph. Programming competitions and contests, programming community. Login; Register; User Editorials: Search Friends: Upcoming Contests: Search Problems: Leaderboard: Trending Problems: Submission Filters: Testimonials: Feature Updates: Find Me Problems. Codeforces. Example Problem Example (Codeforces Round 290 div. But I am not sure if this algorithm is related to topological sort or if should I restructure my work with another point of view. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good.. LEVEL: Easy, ATTEMPTED BY: 233 We should have the initial observation that the problem can be solved by using topological sort. CodeChef - A Platform for Aspiring Programmers. if the graph is DAG. For example, another topological sorting of the following graph is \u201c4 5 2 3 1 0\u201d. HackerEarth uses the information that you provide to contact you about relevant content, products, and services. It also helps you to manage and track your programming comepetions training for you and your friends. While the exact order of the items is unknown (i.e. BLACK \u2014 Processed. Topological Sort. topological sort: all edges are directed from left to right. The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. They are related with some condition that one should happen only after other one happened. Search Problems. Problem Name Search Site Tags... OR . Theoretical knowledge of algorithms is important to competitive programmers. C problems are usually adhoc, string manipulation, bit manipulation, greedy, DFS and BFS, implementation. 2.Initialize a queue with indegree zero vertices. 1 Problem A) A list of names are written in lexicographical order, but not in a normal sense. Think how to implement this corner case, rest part is easy though. (Description) Monotonic Queue/Stack. Editorial. Here you will learn and get program for topological sort in C and C++. Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses. 1385E Directing Edges (Description) 102006G Is Topo Logical? For example, another topological sorting of the following graph is \u201c4 5 2 3 1 0\u201d. Also go through detailed tutorials to improve your understanding to the topic. 3. We start from this piece, do a BFS, two cakes are called connected if there is a shared edge. You would apply the topological sort algorithm I mentioned using a queue to keep all the in-degree 0 vertices. Also try practice problems to test & improve your skill level. Comparing a pair of adjacent, distinct names on the list gives us the relative order of a pair of characters. All Tracks Algorithms Graphs Topological Sort Problem. An algorithm for solving a problem has to be both correct and ef\ufb01cient, and the core of the problem is often about inventing an ef\ufb01cient algorithm. GREY \u2014 In Process 3. 2.Initialize a queue with indegree zero vertices. Analytics. (Indegree of a vertex is defined as number of edges pointing to it), can some one explain the approach this solution of problem 510C to me ? Topological Sorting\u00b6 To demonstrate that computer scientists can turn just about anything into a graph problem, let\u2019s consider the difficult problem of stirring up a batch of pancakes. A topological ordering is possible if and only if the graph has no directed cycles, i.e. GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Programming competitions and contests, programming community. Topological Sorting; graphs If is a DAG then a topological sorting of is a linear ordering of such that for each edge in the DAG, appears before in the linear ordering Solved example of topological sort. The easiest to think about (in my opinion) being along the lines of: It's like you're picking fruits off a tree. Array 295 Dynamic Programming 234 String 207 Math 192 Tree 154 Depth-first Search 143 Hash Table 135 Greedy 114 Binary Search 96 Breadth-first Search 77 Sort 71 Two Pointers 66 Stack 63 Backtracking 61 Design 59 Bit Manipulation 54 Graph 48 Linked List 42 Heap 37 Union Find 35 Codeforces. In problem D , we can join the roots of the all components by one kind of edge which is < (less than). The ordering of the nodes in the array is called a topological ordering. Problem. For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. Third, Then Write code and submit in the OJ to justify test cases. ACCURACY: 60% Programming competitions and contests, programming community . The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \\over 4$$ cup of milk. The first vertex in topological sorting is always a vertex with in-degree as 0 (a vertex with no incoming edges). Problem link\u2014 SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 #define gray 1 #define black 2 \u2026 ;), The only programming contests Web 2.0 platform, Educational Codeforces Round 102 (Rated for Div. ACCURACY: 59% Codeforces Round #258 (Div. 2) Problems by tag; Observation; About Me; Contact; Category Archives: Topological Sort. The topological sort is a simple but useful adaptation of a depth first search. R. Rao, CSE 326 5 Posted on May 24, 2014 by sufiantipu111. For example, a topological sorting of the following graph is \u201c5 4 2 3 1 0\u201d. 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But useful adaptation of a depth first search becomes lexicographical in-degrees of outgoing vertices, then Write code submit! Hackerearth uses the information that you provide to Contact you About relevant content products... Programming websites time\u2026 Finding Shortest Paths Breadth-First search Dijkstra \u2019 s Method: Greed is good one happened to,... Failed to AC then optimize your code to the order of a pair of.. Of n courses you have to take, labeled from 0 to n-1 code submit! N ) time complexity of Union-Find, I was working on this problem in my work we. The array is called a topological ordering is possible if and only if$ a program can be as. Gone through the USACO training pages to learn my algorithms, which does n't have a set files! With some condition that one should happen only After other one happened helps you manage... On the list gives us the relative order of finish time, does there exist an order of time! Program to implement this corner case, rest part is easy though an explanation what! C problems are usually adhoc, string manipulation, greedy, DFS their!\n\nAsus Rog Ryujin 240 Oled Aio Cpu Cooler, Toro 60v Hedge Trimmer Review, James Martin Garlic Butter Recipe, Kwikset 914 Z-wave, Scruples Creme Parfait Mousse, Pizza Seasoning Near Me,", "date": "2021-04-22 00:35:42", "meta": {"domain": "noqood.co", "url": "https://staff.noqood.co/gillian-flynn-hwmjkyk/topological-sort-problems-codeforces-c0afea", "openwebmath_score": 0.2864619791507721, "openwebmath_perplexity": 1876.8722276789906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9715639669551475, "lm_q2_score": 0.8840392893839085, "lm_q1q2_score": 0.8589007189380398}}
{"url": "https://physics.stackexchange.com/questions/483419/wavelength-of-cosine-squared", "text": "# Wavelength of cosine-squared\n\nI am confused. Usually, the wavelength is the x-distance between the tops of two consecutive waves. Here is the graph.\n\nThere is only 0.1 m between 2 crests. But the answer counts the wavelength as 0.2 m\n\nThis is puzzling. But use the identity $$\\cos^2(\\theta)=\\frac{\\cos(2\\theta)+1}{2}.$$ The period of $$\\cos(2\\theta)$$ is $$\\pi$$, so the period of $$\\cos^2(\\theta)$$ is also $$\\pi.$$ I think I disagree with the answer in your post. The period is $$0.1$$.\n\u2022 Yes, there should be more context involved on the question. Still, we should note that the \"oscillations\" in your consideration would be about the value of $1/2$, and not about zero. Subtle, indeed. \u2013\u00a0Bruno Anghinoni May 30 at 23:49\n\u2022 The \"oscillations\" in the power graph are about the value $R_s/2$. \u2013\u00a0user52817 May 31 at 0:10", "date": "2019-07-17 12:48:11", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/483419/wavelength-of-cosine-squared", "openwebmath_score": 0.9447847604751587, "openwebmath_perplexity": 432.0867640337231, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9715639718953155, "lm_q2_score": 0.8840392771633078, "lm_q1q2_score": 0.858900711432247}}
{"url": "http://math.stackexchange.com/questions/499161/proof-checking-and-input-generators-of-mathbbz-pq/499287", "text": "# Proof Checking and input: Generators of $\\mathbb{Z}_{pq}$\n\nI'm self-studying abstract algebra (slowly but surely), and I have a question about my answer to the following prompt:\n\nProblem statement:\n\nShow that there are $(q-1)(p-1)$ generators of the group $\\mathbb{Z}_{pq}$, where $p$ and $q$ are distinct primes. (Where $\\mathbb{Z}_{n}$ is the additive group of integers modulo $n$)\n\nI can use a theorem from my textbook that says:\n\nThe integer $r$ generates the group $\\mathbb{Z}_n$ iff $$1\\le r\\lt n \\quad\\text{and}\\quad \\gcd(r, n) = 1$$\n\nMy attempt at a proof:\n\nLet $p$ and $q$ be distinct primes and $\\mathbb{Z}_{pq}$ be the additive group of integers modulo $pq$.\nAn element $a \\in \\mathbb{Z}_{pq}$ is a generator of $\\mathbb{Z}_{pq}$ iff: $$1\\le a\\lt pq \\quad\\text{and}\\quad \\gcd(a, pq) = 1$$ There are $p-1$ positive multiples of $q$ less than $pq$. Also, there are $q-1$ positive multiples of $p$ less than $pq$. These are the only elements of $\\mathbb{Z}_{pq}$ that are not coprime to $pq$. Finally, $0$ is not a generator of $\\mathbb{Z}_{pq}$.\n\nThere are $pq$ elements in $\\mathbb{Z}_{pq}$, so the number of generators is: \\begin{align} pq - (p-1) - (q-1) - 1 &= pq -p -q + 1 \\\\ &= p(q-1) - (q - 1) \\\\ &= (q-1)(p-1) \\end{align}\n\nMy questions:\n\n\u2022 Obviously, if there's a flaw in the proof, I'd like to know. :) Aside from that...\n\n\u2022 When I assert \"these are the only elements of $\\mathbb{Z}_{pq}$ that are not coprime to $pq$,\" do I need to further show that this is true? It is patently obvious to me, but I know that just claiming \"this is obvious\" is not a valid method of proof.\n\n\u2022 I'd like to have some input on the style/format of my proof. Is there a better (read: more formal/traditional) way to phrase something in this proof, or is there a format that I'm not following? As I'm self-teaching, I don't want to learn bad habits...\n\n-\nOn your second point, the general rule that I like to follow is: If you need to ask whether something has to be justified then it doesn't hurt to actually justify it. People find different things to be obvious and what's obvious to you may not be obvious to me. \u2013\u00a0EuYu Sep 20 '13 at 1:22\n\nI don't see anything wrong with your proof.\n\nFor your second question, you could expand a bit on that statement, perhaps saying something like \"since $1$,$p$, $q$ and $pq$ are the only divisors $pq$, any integer $n$not divisible by $p$ or $a$ must have $\\gcd(n,pq)=1$, and be coprime to $pq$\", but in situations like this, where a moment's though and writing down a few definitions will give a proof, omitting it is usually safe.\n\nIn writing proofs, I have found the following idea helpful: a proof is really intended to convince someone (a reader, a teacher, yourself) that a theorem is true. If, after reading your proof and spending a little time thinking about each step, this person could still doubt that your theorem is true, then you should add more. Otherwise, your safe. This reasoning, however, is very audience-dependant. For some audiences, the following would be an acceptable proof of your theorem:\n\nThe generators of $\\Bbb{Z}_{pq}$ correspond with integers less than and coprime to $pq$. Since these are counted by Euler's $\\phi(n)$, $\\phi(n)$ is multiplicative, and, for any prime $p$, $\\phi(p) = p-1$, the number of generators of $\\Bbb{Z}_{pq}$ is $\\phi(pq)=\\phi(p)\\phi(q) = (p-1)(q-1)$.\n\nFor lots of other audiences, it would not be.\n\nRemember, a proof doesn't always have to be written down, finalized, and perfect. It can be more of a conversation.\n\n-", "date": "2016-07-26 12:35:03", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/499161/proof-checking-and-input-generators-of-mathbbz-pq/499287", "openwebmath_score": 0.9781760573387146, "openwebmath_perplexity": 145.94535546192972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639694252316, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8589007062803219}}
{"url": "https://math.stackexchange.com/questions/3569056/prove-two-functions-are-equal", "text": "# Prove Two Functions are Equal?\n\nGiven two generic functions $$x(t)$$ and $$y(t)$$ I want to prove that $$x(t) = y(t)$$.\n\nTo do so, I take the derivative, which turn out to be: $$\\dot x(t) = Ax(t) + B$$ $$\\dot y(t) = Ay(t) + B$$ where $$A$$ and $$B$$ are the same in both derivatives.\n\nIs this sufficient to say both functions are equal?\n\nThe reason I ask is that, in general, $$x(t)$$ and $$y(t)$$ could have vastly different forms (ie. $$x(t)$$ could be the resultant of a complicated integral $$y(t)$$ or something similar). Because of this, I am wondering if I have to go through the trouble of reducing $$x(t)$$ to $$y(t)$$ or vice-versa.\n\n\u2022 Are $A,B$ constants or functions of $x,y,t$ ? \u2013\u00a0Yves Daoust Mar 4 '20 at 13:52\n\u2022 @YvesDaoust they are constants \u2013\u00a0Clark Mar 4 '20 at 15:38\n\nThey both satisfy the linear first order differential equation:\n\n$$\\begin{equation*} f'(t) = A f(t) + B \\end{equation*}$$\n\nYou need at least one other condition, say prove that $$x(t_0) = y(t_0)$$ for some value $$t_0$$, or perhaps the same derivative at a point. Note that the solution is\n\n$$\\begin{equation*} f(t) = c e^{A t} - B/A \\end{equation*}$$\n\nhere $$c$$ is an unknown constant, to be determined by other conditions.\n\n\u2022 do you mean $x(t_0) = y(t_0)$? So if I can prove both functions have the same value at one point, they should have the same value at all points? (ie. from existence and uniqueness?) \u2013\u00a0Clark Mar 4 '20 at 13:13\n\u2022 @Clark, yes; and fixed. \u2013\u00a0vonbrand Mar 4 '20 at 13:15\n\u2022 Awesome, thanks! It's all coming back \u2013\u00a0Clark Mar 4 '20 at 13:18\n\nIf $$x(t)$$ and $$y(t)$$ are $$C^1$$ functions such that $$x(t_0) = y(t_0)$$ for some $$t_0$$ furthermore $$x'(t) = y'(t) = f(t)$$ for some function (Lipschitz-continuous) $$f$$, then according to the uniqueness of the solution of a differential equation you have $$x(t)=y(t)$$ for all $$t\\geq t_0$$.\n\nBy subtraction,\n\n$$\\dot x-\\dot y=\\dot{(x-y)}=A(x-y)$$ so that the two given functions can differ by a solution of this equation, given by\n\n$$x-y=Ce^{At}.$$\n\n\u2022 This doesn't contradict the other answers right? Let's say $x(t_0) = y(t_0) = 0$ where $t_0 = 0$ for simplicity. This would then give $C = 0$ and, therefore $x(t) = y(t)$ correct? \u2013\u00a0Clark Mar 4 '20 at 15:47\n\u2022 @Clark: absolutely. My goal was to show how to solve the question without integrating the original equations and focusing on the difference. (Though the benefit is tiny.) \u2013\u00a0Yves Daoust Mar 4 '20 at 15:50", "date": "2021-01-24 15:42:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3569056/prove-two-functions-are-equal", "openwebmath_score": 0.853053867816925, "openwebmath_perplexity": 219.79426692813644, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971563964485063, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.8589006989447447}}
{"url": "http://hons.zite.pw/bisection-method.html", "text": "## Bisection Method\n\n1) compute a sequence of increasingly accurate estimates of the root. In this paper we extend the traditional recursive bisection stan-dard cell placement tool Feng Shui to directly consider mixed block designs. This is a visual demonstration of finding the root of an equation $$f(x) = 0$$ on an interval using the Bisection Method. You begin with two initial approximations p 0 and p 1 which bracket the root and have f p 0 f p 1 < 0. I was asked to use the bisection method in matlab to find the real root of 1. Numerically solve F(X)=LN(X)-1/X=0 by forming a convergent, fixed point iteration, other than Newton's, starting from X(1)=EXP(1). Tabular Example of Bisection Method Numerical Computation. Explicitly, the function that predicts the way the bisection method will unfold is the function: Further, it is also invariant under the flipping of all signs. Bisection method, Newton-Raphson method and the Secant method of root-finding. Assume f(x) is an arbitrary function of x as it is shown in Fig. b] that contains a root (We can use the property sign of f(a) \u2260 sign of f(b) to find such an initial interval). The tolerance, tol, of the solution in the bisection method is given by tol =(1/2)(bn-an), where an and bn are the endpoints of the interval after the nth iteration. Distributed Bisection Method for Economic Power Dispatch in Smart Grid Abstract: In this paper, we present a fully distributed bisection algorithm for the economic dispatch problem (EDP) in a smart grid scenario, with the goal to minimize the aggregated cost of a network of generators, which cooperatively furnish a given amount of power within. Le Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ***** *****MATLAB CODE ***** x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, \u2019*r\u2019);. After reading this chapter, you should be able to: 1. It is a bit difficult to apply bisection method to a non-deterministic function. Bisection method is a popular root finding method of mathematics and numerical methods. 84070158) \u2248 0. 1: Bisection (Interval Halving) Method Expected Skills: Be able to state the Intermediate Value Theorem and use it to prove the existence of a solution to f(x) = 0 in an interval (a;b). Student[NumericalAnalysis] Bisection numerically approximate the real roots of an expression using the bisection method Calling Sequence Parameters Options Description Examples Calling Sequence Bisection( f , x =[ a , b ], opts ) Bisection( f , [ a ,. Use the bisection method to find a solution of {eq}\\cos x=x {/eq} that is accurate to two decimal places. To use this module, we should import it using \u2212 This method is same as insort() method. The simple equations of kinematics give the position as a function of time. Let\u2019s start with a method which is mostly used to search for values in arrays of every size, Bisection. Example of the Bisection Method This algorithm shows the result of using the bisection method for 4 given functions. At the end of the step, you still have a bracketing interval, so you can repeat the process. However, instead of simply dividing the region in two, a linear interpolation is used to obtain a new point which is (hopefully, but not necessarily) closer to the root than the equivalent estimate for the bisection method. $299 vinyl cutter to start your home business - Duration: 17:41. Algorithm is quite simple and robust, only requirement is that initial search interval must encapsulates the actual root. NET (or, how to make an async method synchronous) Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet. Bisection Method Example - Polynomial \u2022 If limits of 0 to 10 are selected, the solution converges to x = 4 Engineering Computation: An Introduction Using MATLAB and Excel 21. Assume f(x) is an arbitrary function of x as it is shown in Fig. Use the bisection method to approximate this solution to within 0. Pros of Bisection Method 1. Given a function f(x) and an interval which might contain a root, perform a predetermined number of iterations using the bisection method. The convergence is linear, slow but steady. 1 word related to bisection: division. Octave / MATLAB. It is Fault Free (Generally). 01 and |f(1. Example: Solving x \u2212 sin(x) = 0 using bisection method \u2022 At every step, the function sin(x) needs to be evaluated \u21d2 truncation errors \u2022 Round-off errors occur with every operation of (+, \u2212, \u00d7, \u00f7) and accumulate along the way \u2022 The bisection process cannot go on forever; has to stop at a finite number of iterations \u21d2 further errors. (c) Use Newton\u2019s method to evaluate the same root as in (b). Bisection Method - Half-interval Search This code calculates roots of continuous functions within a given interval and uses the Bisection method. It is a very simple and robust method, but it is also rather slow. (a) The smallest positive root of x = 1+ :3cos( x ) Let f (x ) = 1+ :3cos( x ) x. Newton and Raphson used ideas of the Calculus to generalize this ancient method to find the zeros of an arbitrary equation Their underlying idea is the approximation of the graph of the function f ( x ) by the tangent lines, which we discussed in detail in the previous pages. Bisection Method Description This program is for the bisection method. svg: Tokuchan derivative work: Tokuchan ( talk ) This is a retouched picture , which means that it has been digitally altered from its original version. We then set the width of the compass to about two thirds the length of line segment AB. The main way Bisection fails is if the root is a double root; i. xl xu Bisection algorithm. The red curve shows the function f and the blue lines are the secants. One of your comments says you are creating an object to round values to 6 places, but you are not creating an object there. This Demonstration shows the steps of the bisection root-finding method for a set of functions. Let a = 0 and b = 1. It is a bit difficult to apply bisection method to a non-deterministic function. Bisection method is a closed bracket method and requires two initial guesses. In this paper, we have presented a new method for computing the best-fitted rectangle for closed regions using their boundary points. The theory is kept to a minimum commensurate with comprehensive coverage of the subject and it contains abundant worked examples which provide easy understanding through a clear and concise theoretical treatment. C programs, data structure programs, cbnst programs, NA programs in c, c programs codes, mobile tips nd tricks,. Philippe B. ***** *****MATLAB CODE ***** x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, \u2019*r\u2019);. Assumptions We will assume that the function f(x) is continuous. I will fully admit it has been two years since i opened matlab and i am totally lost. Suppose that we want jr c nj< \": Then it is necessary to solve the following inequality for n: b a 2n+1 < \"By taking logarithms, we obtain n > log(b a) log(2\") log 2 M311 - Chapter 2 Roots of Equations - The Bisection Method. Figure 1: The graphs of y=x (black) and y=\\cos x (blue) intersect. Bisection Method of Solving a Nonlinear Equation. Bisection Method Using C. Assume f(x) is an arbitrary function of x as it is shown in Fig. which proves the global convergence of the method. Brackets are unions of similar simplexes. Bisection method consist of reducing an interval evaluating its midpoints, in this way we can find a value for which f(x)=0. The bisection method is used to solve transcendental equations. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs. In fact, the common proof of the Intermediate Value Theorem uses the Bisection method. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. Bisection Methods: We can pursuse the above idea a little further by narrowing the interval until the interval within which the root lies is small enough. Calculus textbook. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. For a real and continuous function, the method finds where the function is equal to zero over a certain interval. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function , and that function crosses a horizontal line. At which point, things got better. Learn via an example, the bisection method of finding roots of a nonlinear equation of the form f(x)=0. Bisection method in matlab The following Matlab project contains the source code and Matlab examples used for bisection method. Algorithm is quite simple and robust, only requirement is that initial search interval must encapsulates the actual root. Newton- Raphson method. It provides a convenient command line inter-. The bisection method applied to sin(x) starting with the interval [1, 5]. It is also known as Binary Search or Half Interval or Bolzano Method. For example, suppose that we would like to solve the simple equation 2 x = 5. This method, also known as binary chopping or half-interval method, relies on the fact that if f(x) is real and continuous in the interval a < x < b , and f(a) and f(b) are of opposite signs, that is,. A few steps of the bisection method applied over the starting range [a 1;b 1]. 84070742] and sin(40. This process involves \ufb01nding a root, or solution, of an equation of the form f(x) = 0 for a given function f. Always Convergent. the bisection method. The bisection method procedure is: Choose a starting interval such that. THE BISECTION METHOD AND LOCATING ROOTS. Table of Contents 1 - The interval-halving (bisection) method, Java/OOP style 2 - The interval halving method written in a slightly more functional style 3 - The same 'halveTheInterval' function in a completely FP style After writing the code first in what I\u2019d call a \u201cJava style,\u201d I then. Otherwise, the Intermediate Value Theorem is used to determine whether the root lies on the subinterval$(a_n, p_n)$or the subinterval$(p_n, b_n)$. The tolerance, tol, of the solution in the bisection method is given by tol =(1/2)(bn-an), where an and bn are the endpoints of the interval after the nth iteration. This method is closed bracket type, requiring two initial guesses. bisection method root-finding method in mathematics that repeatedly bisects an interval and then selects a subinterval in which a root must lie interval halving method. Bisection Method is one of the simplest, reliable, easy to implement and convergence guarenteed method for finding real root of non-linear equations. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. 4 comments I am using equation f(x): x^ 3-2x-5 = x*x*x-2*x-5 (a=2 and b=3). The bisection method is a method used to find the roots of a function. This is achieved by selecting two points A and B on that interval. In mathematics , the bisection method is a root-finding method that applies to any continuous functions for which one knows two values with opposite signs. BISECTION METHOD Bisection method is the simplest among all the numerical schemes to solve the transcendental equations. The bisection method is a root-finding method based on simple iterations. 001, m = 100) Arguments f. Finding Root using Bisection Method in Java This is an example of solving the square cube of 27. The method assumes that we start with two values of z that bracket a root: z1 (to the left) and z2 (to the right), say. The bisection method requires two points aand bthat have a root between them, and Newton\u2019s method requires one. This code war written for the article How to solve equations using python. It is assumed that f(a)f(b) <0. The bisection method in mathematics is a root finding method which repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. Given an initial. Abbreviate a String ARRAY array size bfs Bisection method breadth first search BUBBLE SORT c code choice choice cloud-computing computer conio c program create node cse data structure delete an element dev c dfs display singly linklist emp Euler's method Gauss Elimination Method getch INSERTION SORT interpolation method Lagrange interpolation. Thus the first three approximations to the root of equation x 3 - x - 1 = 0 by bisection method are 1. The simplest of iterative methods, the bisection method is derived from the Intermediate Value Theorem, which states that if a continuous function [Florin], with an interval [a, b] as its domain, takes values [Florin](a) and [Florin](b) at each end of the interval, then it also takes any value between [Florin](a) and [Florin](b),at some point within the interval. The simplest way to solve an algebraic equation of the form g(z) = 0, for some function g is known as bisection. Bisection method is a popular root finding method of mathematics and numerical methods. Now at the very end, we want to output the result, and this is a function. Just like any other numerical method bisection method is also an iterative method, so it is advised to tabulate values at each iteration. Method: reduce, remove rational roots, divide and conquer in [-M,M], then use bisection in disjoint closed intervals ctg one root each. The setup of the bisection method is about doing a specific task in Excel. Laval Kennesaw State University August 23, 2015 Abstract This document described a method used to solve g(x) = 0. OF TECH & SCI. Select a Web Site. (here in my code i don't know why the loop doesn't work as it should. Consider a root finding method called Bisection Bracketing Methods \u2022 If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). The bisect algorithm is used to find the position in the list, where the data can be inserted to keep the list sorted. In Bisection method we always know that real solution is inside the current interval [x 1, x 2 ], since f(x 1) and f(x 2) have different signs. Function = f= (x^3 + x^2 -3x -3) For bisection functions we have given two values of X(X1 & X2). Matlab Build-in Function. any help would be appreciated. The variable f is the function formula with the variable being x. (b) Use the bisection method to evaluate one root of your choice. im trying to write code using the Bisection method to find the max of F(w) like a have with the cubic spline method, any help would be appreciated. Be able to apply the Bisection (Interval Halving) Method to approximate a solution to f(x) = 0. Bisection method, Newton-Raphson method and the Secant method of root-finding. I can speak English, Hindi and a little French. The algorithm is. sign ( f ( a )) == np. If a function changes sign over an interval, the function value at the midpoint is evaluated. The theorem is demonstrated in Figure 2. Getting root of an equation by Bisection Method through C programming language. follow the algorithm of the bisection method of solving a nonlinear equation, 2. 2004 Judith Koeller The bisection method can be used to approximate a solution p to an equation f(p)=0 where f(x) is a continuous function. Mujahid Islam Md. BISECTION METHOD USING C# Here's the Code using System; namespace BisectionMethod { class Program { CREATE A SIMPLE SIMULTANEOUS EQUATION CALCULATOR WITH C# Hello guys first what is a simultaneous equation: This involves the calculation of more than one equation with unknowns simultaneously. The problem is that it seems like the teachers recommended solution to the task isn't quite right. Hi I'm using prime 3 and I want to write a bisection method code but I'm getting an error as follow : 1/ I can't write (i+1) as a subscript for the. What are synonyms for bisection?. Hi, my code doesn't seem to continue beyond the first iteration of the bisection method in my loop. The two most well-known algorithms for root-finding are the bisection method and Newton\u2019s method. Rafiqul Islam Khaza Fahmida Akter 2. Assumptions We will assume that the function f(x) is continuous. De ning a domain In higher dimensions, there is a rich variety of methods to de ne a simply connected domain. This is a visual demonstration of finding the root of an equation $$f(x) = 0$$ on an interval using the Bisection Method. com) Category TI-83/84 Plus BASIC Math Programs (Calculus) File Size 838 bytes File Date and Time Thu Jun 27 21:55:32 2013 Documentation Included? Yes. 84070158, 40. Bisection Method Description This program is for the bisection method. I think there is a need for an improvement, e. We start with a line segment AB. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. Find more Mathematics widgets in Wolfram|Alpha. method is 10 combine thc bisection method with the secant method and include an inverse quadratic interpolation to get a more robust procedure. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. Given a function f(x) and an interval which might contain a root, perform a predetermined number of iterations using the bisection method. The bisection method is probably the simplest root-finding method imaginable. The bisection method is used to solve transcendental equations. Any zero-finding method (Bisection Method, False Position Method, Newton-Raphson, etc. xl xu Bisection algorithm. bisection bisection-method secant secant-method newton newton-raphson newton-method C# Updated Dec 29, 2018. Consider a root finding method called Bisection Bracketing Methods \u2022 If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). It also makes a graph available of the iterates. This module provides support for maintaining a list in sorted order without having to sort the list after each insertion. In mathematics , the bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists. The bisection method guarantees a root (or singularity) and is used to limit the changes in position estimated by the Newton-Raphson method when the linear assumption is poor. 001 using the bisection method. Bisection method is a popular root finding method of mathematics and numerical methods. In this tutorial you will get program for bisection method in C and C++. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function , and that function crosses a horizontal line. Matlab Build-in Function. The Bisection Method is a numerical method for estimating the roots of a polynomial f(x). a b 3 Regula falsi Consider the \ufb01gure in which the root lies between a and b. We will study three di\ufb00erent methods 1 the bisection method 2 Newton\u2019s method 3 secant method and give a general theory for one-point iteration methods. It is a very simple and robust method, but it is also relatively slow. The problem was the calling of the function. This is intended as a summary and supplementary material to the required textbook. 2 Estimate how many iterations will be needed in order to approximate this root with an accuracy of \u03b5=0. I am trying to return this equation as you suggested but still not working!. xl xu Bisection algorithm. I The Bisection Method requires the least assumptions on f(x), I the Bisection Method is simple to program, I the Bisection Method always converges to a solution, but I the Bisection Method isslowto converge. Bisection can be shown to be an \"optimal\" algorithm for functions that change sigh in [a,b] in that it produces the smallest interval of uncertainty in a given # of iterations f(x) need not be continuous on [a,b] convergence is guarenteed (linearly) Disadvantages of the Bisection Method. The graph of this equation is given in the figure. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. So let's take a look at how we can implement this. here is my program btw, but something's wrong in the bisection function and I can't figure out what is it. Mathematica. If the function values at points A and B have opposite signs\u2026. Lecture Material. Bisection Method. Figure 1: The graphs of y=x (black) and y=\\cos x (blue) intersect. thanks Code:. After reading this chapter, you should be able to: 1. We also call this method as an interval halving method because we consider a midpoint. We stay with our original. i just tried my vba code for bisection method but it doesn't work. Show that \ud835\udc53\ud835\udc65=\ud835\udc653+4\ud835\udc652\u221210= 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10\u22124. And as I mentioned last time, this was the state of the art until the 17th century. The method is based on the following theorem. It is used in cases where it is known that. By testing different. The Bisection method is a numerical method which finds approximate solutions to polynomial equations with the use of midpoints. Select a Web Site. Equations don't have to become very complicated before symbolic solution methods give out. I dream of being a Theoretical Physicist one day. SECANT METHODS Convergence If we can begin with a good choice x 0, then Newton\u2019s method will converge to x rapidly. The bisection method requires two points aand bthat have a root between them, and Newton\u2019s method requires one. a b 3 Regula falsi Consider the \ufb01gure in which the root lies between a and b. Bisection Method 1- Flowchart. (a) Bisection Method: This is one of the simplest and reliable iterative methods for the solution of nonlinear equation. 1) compute a sequence of increasingly accurate estimates of the root. In mathematics , the bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists. It is Fault Free (Generally). If your calculator can solve equations numerically, it most likely uses a combination of the Bisection Method and the Newton-Raphson Method. Author jamespatewilliamsjr Posted on July 29, 2018 July 29, 2018 Categories C# Computer Programming Language, Root Finding Algorithms Tags Bisection Method, Brent's Method, C#, Computer Science, Newton's Method, Numerical Analysis, Regula Falsi Leave a comment on Root Finding Algorithms by James Pate Williams, BA, BS, MSwE, PhD. Numerical Analysis: Root Solving with Bisection Method and Newton\u2019s Method. The number of iterations n that are required for obtaining a solution with a tolerance that is equal to or smaller than a specified tolerance can be determined before the solution. The basic method for making or doing something, such as an artistic work or scientific procedure: learned the techniques involved in painting murals Bisection of the angle technique - definition of bisection of the angle technique by The Free Dictionary. Bisection Method. The theorem is demonstrated in Figure 2. As a result, f(x) is approximated by a secant line through. any help would be appreciated. BISECTION METHOD. Bisection Method Description This program is for the bisection method. It is a very simple and robust method, but it is also rather slow. The most straightforward root-finding method. If the function values at points A and B have opposite signs\u2026. These methods first find an interval containing a root and then systematically shrink the size of successive intervals that contain the root. 3 The bisection method converges very slowly 4 The bisection method cannot detect multiple roots Exercise 2: Consider the nonlinear equation ex \u2212x\u22122=0. It was observed that the Bisection method. For the PASSFAIL method, the measure must pass for one limit and fail for the other limit. CGN 3421 - Computer Methods Gurley Numerical Methods Lecture 6 - Optimization page 107 of 111 Single Variable - Golden Section Search Optimization Method Similar to the bisection method \u2022 Define an interval with a single answer (unique maximum) inside the range sign of the curvature does not change in the given range. The Bisection Method is a successive approximation method that narrows down an interval that contains a root of the function f(x) The Bisection Method is given an initial interval [a. De ning a domain In higher dimensions, there is a rich variety of methods to de ne a simply connected domain. m\" So create new. Apply the bisection method over a \"large\" interval. Use the bisection method to locate % a zero of the function f(x) = x sin(x) - 1. We have provided MATLAB program for Bisection Method along with its flowchart and algorithm. False Position or Regular Falsi method uses not only in deciding. 1 Show there is a root \u03b1in the interval (1,2). The routine assumes that an interval [a,b] is known, over which the function f(x) is continuous, and for which f(a) and f(b) are of opposite sign. The algorithm is iterative. The simplest way to solve an algebraic equation of the form g(z) = 0, for some function g is known as bisection. While the subject itself is quite interesting, the programming environment being used in the lab is Turbo C, a DOS based IDE which has been abandoned a long time ago. Brent's method combines the bisection method, secant method, and the method of inverse quadratic interpolation. If that is the case, you could save that data to an array and plot that array when you exit the loop like. (b) Use the bisection method to evaluate one root of your choice. -Bisection method is used to get a rough estimate of the solution then some other faster methods are used (discuss in our next lecture). Here is one example that passes the function f as a parameter, checks parameters for validity before continuing, avoids some other overflow exposures, avoids redundant calls to. Any zero-finding method (Bisection Method, False Position Method, Newton-Raphson, etc. The method is also called the interval halving method, the binary search method, or the dichotomy method. Mujahid Islam Md. The bisection method is far more efficient than algorithms which involve a search over frequencies, and of course the usual problems associated with such methods (such as determining how fine the search should be) do not arise. A genetic approach using direct representation of solutions for the parallel task scheduling problem. In the \ufb01rst iteration of bisection method, the approximation lies at the small circle. If a n and b n are satistfy equation (3) then b n \u2212 a n \u2264 b \u2212a 2n, for n \u2265 1 where b1 = b,a1 = a. 5 Bisection Method (cont\u2019d) \u2022It always converge to the true root (but be careful about the following) \u2022f(x L) * f(x U) < 0 is true if the interval has odd number of roots, not necessarily one root. PROGRAM(Simple Version):. The method assumes that we start with two values of z that bracket a root: z1 (to the left) and z2 (to the right), say. The setup of the bisection method is about doing a specific task in Excel. However, instead of simply dividing the region in two, a linear interpolation is used to obtain a new point which is (hopefully, but not necessarily) closer to the root than the equivalent estimate for the bisection method. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. Bisection method. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs. 0 (193 ratings) Course Ratings are calculated from individual students\u2019 ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Bisection Theorem An equation f(x)=0, where f(x) is a real continuous function, has at least one root between a and b, if f(a) f(b) < 0. Select a Web Site. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. Be able to apply the Bisection (Interval Halving) Method to approximate a solution to f(x) = 0. The bisection method is a simple root-finding method. ) exa_myfpi. If a function changes sign over an interval, the function value at the midpoint is evaluated. If instead we want the time at which a certain position is reached, we must invert these equations. 1) in terms of all five of the desirable attributes. Description: Given a closed interval [a,b] on which f changes sign, we divide the interval in half and note that f must change sign on either the right or the left half (or be zero at the midpoint of [a,b]. I followed the same steps for a different equation with just tVec and it worked. 3 The bisection method converges very slowly 4 The bisection method cannot detect multiple roots Exercise 2: Consider the nonlinear equation ex \u2212x\u22122=0. The convergence is linear, slow but steady. It is a very simple and robust method, but it is also relatively slow. Let f 2 be a continuous function with di erent signs at a;b, with a #inc This is the solution for finding the roots of a function by Bisection Method in C++ Object Oriented Approach. We will study three di\ufb00erent methods 1 the bisection method 2 Newton's method 3 secant method and give a general theory for one-point iteration methods. follow the algorithm of the bisection method of solving a nonlinear equation, 2. Example - 4: Using the bisection method find the approximate value of square root of 3 in the interval (1, 2) by performing two iterations. Formerly, engineers built external drivers to submit multiple parameterized Star-Hspice jobs, with each job exploring a region of the operating envelope of the circuit. What is Bisection Method? It is an iterative method based on a well known theorem which states that if f(x) be a continuous function in a closed interval [a,b] and f(a)f(b)<0, then there exists at least one real root of the equation f(x)=0, between a and b. Sample C program that uses Bisection Method in mathematics. By the Intermediate Value Theorem, there exists p in with. Pick starting points, precision and method. Compute where is the midpoint. Plot the function sin(x) on [0;2\u02c7]. In fact, the common proof of the Intermediate Value Theorem uses the Bisection method. This tutorial explores a simple numerical method for finding the root of an equation: the bisection method. The secant method avoids this issue by using a nite di erence to approximate the derivative. Use the bisection method to approximate this solution to within 0. 9 Abeam is loaded as shown in Fig. What are synonyms for bisection?. 725 option price). I dream of being a Theoretical Physicist one day. 1 word related to bisection: division. Bisection method is used for finding root of the function in given interval. Then it is much more useful to explain, how a function is called with input arguments, than to convert the function to a script - which is still not working. Suppose we want to solve the equation f(x)=0,where f is a continuous function. Root is found by repeatedly bisecting an interval. In case, you are interested to look at the comparison between bisection method (adopted by Mibian Library) and my code please have look at screenshot of results obtained :-As you can see, bisection method didn\u2019t converge well (to$13. So this is what happens in every iteration of the bisection method, we go through 20 iterations. sign ( f ( a )) == np. Suppose we know the two points of an interval and , where , and. The convergence of the bisection method is very slow. This can be achieved if we joint the coordinates (a,f(a)) and (b. Now at the very end, we want to output the result, and this is a function. erably; traditional methods produce results that are far from satis-factory. Bisection Method: The idea of the bisection method is based on the fact that a function will change sign when it passes through zero. Convergence \u2022 Theorem Suppose function \ud835\udc53(\ud835\udc65) is continuous on [ , ], and \ud835\udc53 \u2219\ud835\udc53 <0. The bisection method is a simple root-finding method. The bisection method is probably the simplest root-finding method imaginable. if a and b are two. Methods for finding roots are iterative and try to find an approximate root $$x$$ that fulfills $$|f(x)| \\leq \\epsilon$$, where $$\\epsilon$$ is a small number referred later as tolerance. To find a root very accurately Bisection Method is used in Mathematics. In a nutshell, the former is slow but robust and the latter is fast but not robust. This will open a new tab with the resource page in our marketplace. 01, and therefore we chose b = 1. The main way Bisection fails is if the root is a double root; i. Suppose we know the two points of an interval and , where , and. Since the line joining both these points on a graph of x vs f(x), must pass through a point, such that f(x)=0. Bisection is the division of a given curve, figure, or interval into two equal parts (halves). % % Enter the starting endpoints for [a,b] in a and b % % Enter the tolerance in delta. Bisection Method. conventional methods like Newton-Raphson method (N-R), Regula Falsi method (R-F) & Bisection method (BIS). THE BISECTION METHOD This method is based on mean value theorem which states that if a function ( ) is continuous between and , and ( ) ( ) are of opposite signs, then there exists at least one root between and. Bisection Method The bisection method is a kind of bracketing methods which searches for roots of equation in a specified interval. Each iteration step halves the current interval into two subintervals; the next interval in the sequence is the subinterval with a sign change for the function (indicated by. What is Bisection Method? The method is also called the interval halving method, the binary search method or the dichotomy method. Use this tag for questions related to the bisection method, which is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. 2 Using the Bisection Method to Prove the Intermediate Value Theorem Now suppose that fis continuous on [a;b], f(a) <0 and f(b) >0. So it is dependent on. Advantages of the Bisection method. It also makes a graph available of the iterates.", "date": "2019-12-10 16:04:47", "meta": {"domain": "zite.pw", "url": "http://hons.zite.pw/bisection-method.html", "openwebmath_score": 0.6526755094528198, "openwebmath_perplexity": 521.1656787315939, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9845754474655618, "lm_q2_score": 0.8723473763375643, "lm_q1q2_score": 0.8588918084029662}}
{"url": "https://math.stackexchange.com/questions/1025630/to-show-for-following-sequence-lim-n-to-infty-a-n-0-where-a-n-1-3", "text": "# To show for following sequence $\\lim_{n \\to \\infty} a_n = 0$ where $a_n$ = $1.3.5 \u2026 (2n-1)\\over 2.4.6\u2026(2n)$\n\nHow can I show\n\n$\\lim_{n \\to \\infty} a_n = 0$\n\n$a_n = {1.3.5 ... (2n-1)\\over 2.4.6...(2n)}$\n\nI have shown that $a_n$ is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks\n\n\u2022 Clarification request: what do you mean by $1.3.5...$? Are these numbers being added? \u2013\u00a0Mike Pierce Nov 17 '14 at 8:14\n\u2022 I think is multiplication of odd numbers. Isn't $\\pi$ in the answer? I vaguely remember a result like this from Euler. \u2013\u00a0ReverseFlow Nov 17 '14 at 8:15\n\u2022 It is certainly bounded below, since all terms are positive, so it converges. But that does not show it converges to $0$. The Stirling approximation will show convergence to $0$, but one can do it with less machinery. \u2013\u00a0Andr\u00e9 Nicolas Nov 17 '14 at 8:21\n\u2022 Try to prove using induction that:$$a_n\\lt\\frac{1}{\\sqrt{3n+1}}$$ \u2013\u00a0pointer Nov 17 '14 at 8:26\n\u2022 That the limit is zero is intuitively obvious, seeing as $a_n=\\binom{2n}n/4^n$is the probability of getting exactly $n$ heads in $2n$ independent tosses of a fair coin. \u2013\u00a0bof Nov 17 '14 at 8:30\n\nMultiply the numerator by the denominator; so $$a_n=\\frac{1\\times3\\times5\\times ... \\times(2n-1)}{ 2\\times4\\times6\\times...\\times(2n)}=\\frac{1\\times2\\times3\\times ... \\times(2n)}{\\Big(2\\times4\\times6\\times...\\times(2n)\\Big)^2}=\\frac{(2n)!}{4^n(n!)^2}$$ If now you use Stirling approximation $$m! \\approx \\sqrt{2 \\pi } e^{-m} m^{m+\\frac{1}{2}}$$ and then $$a_n \\approx \\frac{1}{\\sqrt{\\pi n} }$$ A more detailed approach would show for the asymptotic behavior $$a_n \\approx \\frac{1}{\\sqrt{\\pi n} }\\Big(1-\\frac{1}{8n}\\Big)$$\n\n\u2022 In first step how have u written in denominatorof last inequality as $4^{n}(n!)^{2}$ .. \u2013\u00a0godonichia Nov 17 '14 at 12:11\n\u2022 Factor $2$ inside the brackets of denominator and square. \u2013\u00a0Claude Leibovici Nov 17 '14 at 17:21\n\u2022 @ClaudeLeibovici in stirlings aproximstion as you have mentioned in order to find formula for 2m! I just have to replace m by 2m in above formula .if i do this then im not getting to answer, thanks \u2013\u00a0Jessica Gtb Dec 25 '14 at 12:20\n\u2022 @JessicaGtb. I don't know whee you had a problem since if you use Stirling, you should get $$(2n)! \\approx \\sqrt{\\pi } 2^{2 m+1} e^{-2 m} m^{2 m+\\frac{1}{2}}$$ Post if you still have problems. Cheers :-) \u2013\u00a0Claude Leibovici Dec 26 '14 at 7:22\n\u2022 How did you get that expression for 2n! .it is not same when you replace n by 2n in formula you mentioned in answer \u2013\u00a0Jessica Gtb Dec 26 '14 at 12:29\n\nUsing $\\boldsymbol{1+x\\le e^x}$\n\nSince $1+x\\le e^x$ for all $x\\in\\mathbb{R}$, \\begin{align} a_n &=\\prod_{k=1}^n\\frac{2k-1}{2k}\\\\ &=\\prod_{k=1}^n\\left(1-\\frac1{2k}\\right)\\\\ &\\le\\prod_{k=1}^ne^{-\\large\\frac1{2k}}\\\\[3pt] &=e^{-\\frac12H_n}\\tag{1} \\end{align} where $H_n$ are the Harmonic Numbers. Since the Harmonic Series diverges, $(1)$ tends to $0$.\n\nUsing Gautschi's Inequality \\begin{align} a_n &=\\prod_{k=1}^n\\frac{2k-1}{2k}\\\\ &=\\prod_{k=1}^n\\frac{k-\\frac12}{k}\\\\ &=\\frac{\\Gamma\\!\\left(n+\\frac12\\right)}{\\sqrt\\pi\\,\\Gamma(n+1)}\\\\[6pt] &\\le\\frac1{\\sqrt{\\pi n}}\\tag{2} \\end{align}\n\nCommentary\n\nGautschi's Inequality also says that \\begin{align} a_n &=\\frac{\\Gamma\\!\\left(n+\\frac12\\right)}{\\sqrt\\pi\\,\\Gamma(n+1)}\\\\ &\\ge\\frac1{\\sqrt{\\pi\\left(n+\\frac12\\right)}}\\tag{3} \\end{align} Therefore, $$\\lim_{n\\to\\infty}\\sqrt{n}\\,a_n=\\frac1{\\sqrt\\pi}\\tag{4}$$ Putting the second line of $(1)$ and the limit in $(4)$ together, we can show that $$\\gamma+\\sum_{n=2}^\\infty\\frac{\\zeta(n)}{n2^{n-1}}=\\log(\\pi)\\tag{5}$$ where $\\gamma$ is the Euler-Mascheroni Constant and $\\zeta$ is the Riemann Zeta Function.\n\nLet's prove using induction that $$a_n\\le\\frac{1}{\\sqrt{3n+1}}.$$ For $n=1$ it is true. Now we just need to prove that$$\\frac{(2n+1)^2}{(2n+2)^2}\\le\\frac{3n+1}{3n+4}$$or $$(4n^2+4n+1)(3n+4)\\le(4n^2+8n+4)(3n+1)$$or$$12n^3+28n^2+19n+4\\le12n^3+28n^2+20n+4.$$\n\n\u2022 If we are not given the limit it converges to then how would we go about it \u2013\u00a0godonichia Nov 17 '14 at 8:51\n\u2022 I don't know. Maybe we could find asymtothics using Stirling's formula, then try to find this inequality. \u2013\u00a0pointer Nov 17 '14 at 9:08\n\nYour sequence can be rewritten as $$a_n=\\frac{1\\cdot3\\cdot...\\cdot(2n-1)}{2\\cdot4\\cdot...\\cdot2n}=\\frac{1\\cdot2\\cdot3\\cdot...\\cdot(2n-1)\\cdot2n}{(2\\cdot4\\cdot...\\cdot2n)^2}=\\frac{(2n)!}{2^{2n}(n!)^2}$$ Using Stirling's approximation we get $$a_n=\\frac{(2n)!}{2^{2n}(n!)^2}\\sim\\frac{1}{2^{2n}}\\cdot\\frac{\\sqrt{2\\pi 2n}\\cdot(\\frac{2n}{e})^{2n}}{(\\sqrt{2\\pi n}(\\frac{n}{e})^n)^2}=\\frac{1}{2^{2n}}\\cdot\\frac{\\sqrt{2}\\cdot2^{2n}}{\\sqrt{2\\pi n}}=\\frac{\\sqrt{2}}{\\sqrt{2\\pi n}}\\to 0$$ as $n\\to\\infty$.\n\n\u2022 i havent studied stirlings approximation yet . Can u show me another way which uses sequence series concepts \u2013\u00a0godonichia Nov 17 '14 at 8:35", "date": "2019-09-16 08:50:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1025630/to-show-for-following-sequence-lim-n-to-infty-a-n-0-where-a-n-1-3", "openwebmath_score": 0.9728031158447266, "openwebmath_perplexity": 476.15069233540544, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754510863376, "lm_q2_score": 0.872347368040789, "lm_q1q2_score": 0.8588918033927392}}
{"url": "http://mathhelpforum.com/statistics/129611-help-needed-probablilty-question.html", "text": "# Math Help - Help needed for a probablilty question\n\n1. ## Help needed for a probablilty question\n\nOk so this is really bugging me, because I thought i finally understood it, but then my book is giving me different answers to what I'm getting.\nAnyways, heres the problem, can anyone answer it and show working , so I can learn. Thanks\n\nGiven P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:\n\n(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')\n\n2. P(R)\n= P(R(S+S'))\n= P(R1)\n= P(RS)+P(RS')\n= P(R|S)P(S)+P(R|S')P(S')\n= P(R|S)P(S)+P(R|S')(1-P(S))\n\nP(RS)\n= P(R|S)P(S)\n= P(S|R)P(R)\n\nNow you know P(R) and P(RS), so you know P(RS')\nP(RS')\n= P(S'|R)P(R)\n\nP(S)\n= ...\n= P(S|R)P(R)+P(S|R')(1-P(R))\ngives\nP(SR')\n= P(S|R')P(R')\n= P(S|R')(1-P(R))\nUsing this,\nP(1)\n= P((R+R')(S+S'))\n= P(RS) + P(RS')+P(R'S)+P(R'S')\ngives\nP(R'S')\n= P(S'|R')P(R')\n\n3. Originally Posted by kamicool17\nGiven P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:\n(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')\n$P(RS)=P(R|S)P(S)$\n$P(RS')=P(R|S')P(S')$\n$P(R)=P(RS)+P(RS')$\n\nNow finish.\n\n4. Hello, kamicool17!\n\nAnother approach . . . (much longer, though)\n\nGiven: . $P(R|S) = 0.5,\\;\\;P(R|S') = 0.4,\\;\\;P(S) = 0.6$\n\nFind: . $(a)\\;P(R) \\qquad (b)\\;P(S|R)\\qquad (c)\\;P(S'|R) \\qquad (d)\\;P(S'|R')$\n\nWe have this table:\n\n. . $\\begin{array}{c|| c|c||c}\n& S & S' & \\text{Total } \\\\ \\hline \\hline\nR & {\\color{blue}(a)} & {\\color{blue}(c)} & {\\color{blue}(e)} \\\\ \\hline\nR' & {\\color{blue}(b)} & {\\color{blue}(d)} & {\\color{blue}(f)} \\\\ \\hline \\hline\n\\text{Total} & 0.60 & 0.40 & 1.00 \\end{array}$\n\nWe have: . $P(S) = 0.60 \\quad\\Rightarrow\\quad P(S') = 0.40$\nSo we can fill in the bottom row.\n\nWe have: . $P(R|S) = 0.5$\n\nThen: . $\\frac{P(R \\wedge S)}{P(S)} \\;=\\;0.5 \\quad\\Rightarrow\\quad \\frac{P(R \\wedge S)}{0.6} \\:=\\:0.5\n$\n\nHence: . $P(R \\wedge S) \\,=\\,0.30\\;{\\color{blue}(a)} \\quad\\Rightarrow\\quad P(R' \\wedge S) \\:=\\:0.30\\;{\\color{blue}(b)}$\n\n. . $\\begin{array}{c|| c|c||c}\n& S & S' & \\text{Total } \\\\ \\hline \\hline\nR & 0.30 & (c) & (e) \\\\ \\hline\nR' & 0.30 & (d) & (f) \\\\ \\hline \\hline\n\\text{Total} & 0.60 & 0.40 & 1.00 \\end{array}$\n\nWe have: . $P(R|S') = 0.4$\n\nThen: . $\\frac{P(R \\wedge S')}{P(S')} \\:=\\:0.4 \\quad\\Rightarrow\\quad \\frac{P(R \\wedge S')}{0.4} \\:=\\:0.4$\n\nHence: . $P(R \\wedge S') \\:=\\:0.16\\;{\\color{blue}(c)} \\quad\\Rightarrow\\quad P(R' \\wedge S') \\:=\\:0.24\\;{\\color{blue}(d)}$\n\nAlso: . $P(R) \\:=\\:0.46 \\;{\\color{blue}(e)}\\quad\\Rightarrow\\quad P(R') \\:=\\:0.54\\;{\\color{blue}(f)}$\n\nAnd we have completed the table:\n\n. . $\\begin{array}{c|| c|c||c}\n& S & S' & \\text{Total } \\\\ \\hline \\hline\nR & 0.30 & 0.16 & 0.46 \\\\ \\hline\nR' & 0.30 & 0.24 & 0.54 \\\\ \\hline \\hline\n\\text{Total} & 0.60 & 0.40 & 1.00 \\end{array}$\n\nYou can now answer the questions . . .", "date": "2016-05-02 13:51:46", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/129611-help-needed-probablilty-question.html", "openwebmath_score": 0.9097843766212463, "openwebmath_perplexity": 10784.73465736694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874087451302, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.858874716802039}}
{"url": "http://code.jasonbhill.com/category/algorithms/", "text": "A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.\n\nThen, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).\n\nThe result is Pascal\u2019s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?\n\n## Solution Using the Multinomial Theorem\n\nThe generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula\n$(x_1+x_2+\\cdots+x_m)^n=\\sum_{{k_1+k_2+\\cdots+k_m=n}\\atop{0\\le k_i\\le n}}\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)\\prod_{1\\le t\\le m}x_t^{k_t}$\nwhere\n$\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)=\\frac{n!}{k_1!k_2!\\cdots k_m!}.$\nOf course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by\n$\\left({200000}\\atop{k_1,k_2,k_3}\\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$\nwhere $k_1+k_2+k_3=200000$. It\u2019s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there\u2019s no way that we\u2019re going to calculate these coefficients. We only need to know when they\u2019re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.\n\nFirst, we\u2019ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that\n$p(n,d)=\\left\\lfloor\\frac{n}{d}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^2}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^3}\\right\\rfloor+ \\cdots+\\left\\lfloor\\frac{n}{d^r}\\right\\rfloor,$\nwhere $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we\u2019re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we\u2019ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We\u2019ll store two lists:\n$p2=[p(i,2)\\text{ for }1\\le i\\le 200000]\\quad\\text{and}\\quad p5=[p(i,5)\\text{ for }1\\le i\\le 200000].$\nFor a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when\n$p2[k_1]+p2[k_2]+p2[k_3]<199983\\ \\text{and}\\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$\nThat is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.\n\nNow, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\\le k_2\\le k_3$. Under this assumption, we always have\n$k_1\\le\\left\\lfloor\\frac{200000}{3}\\right\\rfloor=66666.$\nWe know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.\n\nIn case 1, we iterate $0\\le k_1\\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).\n\nIn case 2, we iterate $0\\le k_1\\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.\n\nIn case 3, we iterate $0\\le k_1\\le 66666$, and we iterate over $k_2=k_1+a$ where $1\\le a < \\left\\lfloor\\frac{200000-3k_1}{2}\\right\\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).\n\n## Cython Solution\n\nI\u2019ll provide two implementations, the first written in Cython inside Sage. Then, I\u2019ll write a parallel solution in C.\n\n%cython \u00a0 import time from libc.stdlib cimport malloc, free \u00a0 head_time = time.time() \u00a0 cdef unsigned long p(unsigned long k, unsigned long d): cdef unsigned long power = d cdef unsigned long exp = 0 while power <= k: exp += k / power power *= d return exp \u00a0 cdef unsigned long * p_list(unsigned long n, unsigned long d): cdef unsigned long i = 0 cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long)) while i <= n: powers[i] = p(i,d) i += 1 return powers \u00a0 \u00a0 run_time = time.time() \u00a0 # form a list of number of times each n! is divisible by 2. cdef unsigned long * p2 = p_list(200000,2) \u00a0 # form a list of number of times each n! is divisible by 5. cdef unsigned long * p5 = p_list(200000,5) \u00a0 cdef unsigned long k1, k2, k3, a cdef unsigned long long result = 0 \u00a0 k1 = 0 while k1 <= 66666: # case 1: k1 = k2 < k3 k2 = k1 k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 2: k1 < k2 = k3 if k1 % 2 == 0: k2 = (200000 - k1)/2 k3 = k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 3: k1 < k2 < k3 a = 1 while 2*a < (200000 - 3*k1): k2 = k1 + a k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 6 a += 1 k1 += 1 \u00a0 \u00a0 free(p2) free(p5) \u00a0 \u00a0 elapsed_run = round(time.time() - run_time, 5) elapsed_head = round(time.time() - head_time, 5) \u00a0 print \"Result: %s\" % result print \"Runtime: %s seconds (total time: %s seconds)\" % (elapsed_run, elapsed_head)\n\nWhen executed, we find the correct result relatively quickly.\n\nResult: 479742450 Runtime: 14.62538 seconds (total time: 14.62543 seconds)\n\n## C with OpenMP Solution\n\n#include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <omp.h> \u00a0 /*****************************************************************************/ /* function to determine how many factors of 'd' are in 'k!' */ /*****************************************************************************/ unsigned long p(unsigned long k, unsigned long d) { unsigned long power = d; unsigned long exp = 0; while (power <= k) { exp += k/power; power *= d; } return exp; } \u00a0 /*****************************************************************************/ /* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer */ /*****************************************************************************/ unsigned long * p_list(unsigned long n, unsigned long d) { unsigned long i; unsigned long * powers = malloc((n+1)*sizeof(unsigned long)); for (i=0;i<=n;i++) powers[i] = p(i,d); return powers; } \u00a0 /*****************************************************************************/ /* main */ /*****************************************************************************/ int main(int argc, char **argv) { unsigned long k1, k2, k3, a; unsigned long long result = 0; \u00a0 unsigned long * p2 = p_list(200000, 2); unsigned long * p5 = p_list(200000, 5); \u00a0 \u00a0 #pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result) for (k1=0;k1<66667;k1++) { // case 1: k1 = k2 < k3 k2 = k1; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } // case 2: k1 < k2 = k3 if (k1 % 2 == 0) { k2 = (200000 - k1)/2; k3 = k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } } // case 3: k1 < k2 < k3 for (a=1;2*a<(200000-3*k1);a++) { k2 = k1 + a; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 6; } } } \u00a0 free(p2); free(p5); \u00a0 printf(\"result: %lld\\n\", result); \u00a0 return 0; }\n\nThis can be compiled and optimized using GCC as follows.\n\n$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result. $ time ./problem-154-omp result: 479742450 \u00a0 real 0m1.487s\n\nThis appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds.\n\n### Motivation\n\nMany interesting computational problems, such as those on Project Euler require that one find the sum of proper divisors of a given integer. I had a fairly crude brute-force method for doing this, and was subsequently emailed a comment by Bjarki \u00c1g\u00fast Gu\u00f0mundsson who runs the site www.mathblog.dk. He pointed me in the direction of this page and provided some sample code illustrating how such an approach runs asymptotically faster than the approach I had been taking. Awesome! I\u2019m going to expand on that a bit here, providing some mathematical proofs behind the claims and providing code for those who may want to take advantage of this.\n\n### The Mathematics Behind It All\n\nLet the function $\\sigma(n)$ be the sum of divisors for a positive integer $n$. For example,\n$\\sigma(6)=1+2+3+6=12.$\n\nIt should seem obvious that for any prime $p$ we have $\\sigma(p)=1+p$. What about powers of primes? Let $\\alpha\\in\\mathbb{Z}_+$, and then\n$\\sigma(p^\\alpha)=1+p+p^2+\\cdots+p^\\alpha.$\n\nWe\u2019d like to write that in a closed form, i.e., without the \u201c$+\\cdots+$\u201d. We use a standard series trick to do that.\n\n\\begin{align} \\sigma(p^\\alpha) &= 1+p+p^2+\\cdots+p^\\alpha\\cr p\\sigma(p^\\alpha) &= p+p^2+p^3+\\cdots+p^{\\alpha+1}\\cr p\\sigma(p^\\alpha)-\\sigma(p^\\alpha) &= (p+p^2+\\cdots+p^{\\alpha+1})-(1+p+\\cdots+p^\\alpha)\\cr p\\sigma(p^\\alpha)-\\sigma(p^\\alpha) &= p^{\\alpha+1}-1\\cr (p-1)\\sigma(p^\\alpha) &= p^{\\alpha+1}-1\\cr \\sigma(p^\\alpha) &=\\frac{p^{\\alpha+1}-1}{p-1}.\\end{align}\n\nThat solves the problem of finding the sum of divisors for powers of primes. It would be nice if we could show that $\\sigma$ is multiplicative on powers of primes, i.e., that $\\sigma(p_1^{\\alpha_1}p_2^{\\alpha_2})=\\sigma(p_1^{\\alpha_1})\\sigma(p_2^{\\alpha_2})$. We\u2019ll prove that this is the case, and solve the problem in general along the way.\n\nProposition: The function $\\sigma$ is multiplicative on powers of primes.\n\nProof: Let $n$ be a positive integer written (uniquely, by the fundamental theorem of arithmetic) as\n$n=\\prod_{i=1}^m p_i^{\\alpha_i}$\nfor $m$ distinct primes $p_i$ with $\\alpha_i\\in\\mathbb{Z}_+$. Any divisor $k$ of $n$ then has the form\n$k=\\prod_{i=1}^m p_i^{\\beta_i}$\nwhere each $\\beta_i$ satisfies $0\\le\\beta_i\\le\\alpha_i$. Then\n$\\sigma(n)=\\sigma\\left(\\prod_{i=1}^m p_i^{\\alpha_i}\\right)$\nis the sum of all divisors $k$ of $n$ and can be written by summing over all possible combinations of the exponents $\\beta_i$. There are $\\prod_{i=1}^m \\alpha_i$ combinations, and we can form their sum and simplify it as follows.\n\\begin{align}\\sigma(n) &= \\sum_{1\\le i\\le m,\\ 0\\le\\beta_i\\le\\alpha_i}p_1^{\\beta_i}p_2^{\\beta_2}\\cdots p_m^{\\beta_m}\\cr &= \\sum_{\\beta_1=0}^{\\alpha_1}p_1^{\\beta_1}\\left(\\sum_{2\\le i\\le m,\\ 0\\le\\beta_i\\le\\alpha_i}p_2^{\\beta_2}p_3^{\\beta_3}\\cdots p_m^{\\beta_m}\\right)\\cr &= \\sum_{\\beta_1=0}^{\\alpha_1}p_1^{\\beta_1}\\sum_{\\beta_2=0}^{\\alpha_2}p_2^{\\beta_2}\\left(\\sum_{3\\le i\\le m,\\ 0\\le\\beta_i\\le\\alpha_i}p_3^{\\beta_3}p_4^{\\beta_4}\\cdots p_m^{\\beta_m}\\right) \\cr &= \\vdots\\cr &=\\sum_{\\beta_1=0}^{\\alpha_1}p_1^{\\beta_1}\\sum_{\\beta_2=0}^{\\alpha_2}p_2^{\\beta_2}\\sum_{\\beta_3=0}^{\\alpha_3}p_3^{\\beta_3}\\ \\cdots\\ \\sum_{\\beta_m=0}^{\\alpha_m}p_m^{\\beta_m}\\cr &=\\sigma(p_1^{\\alpha_1})\\sigma(p_2^{\\alpha_2})\\cdots\\sigma(p_m^{\\alpha_m}).\\end{align}\nThis completes the proof. Q.E.D.\n\nThus, we now have a formula for the sum of divisors of an arbitrary positive integer $n$ using the factorization of $n$. Namely,\n$\\sigma(n)=\\sigma\\left(\\prod_{i=1}^m p_i^{\\alpha_i}\\right)=\\prod_{i=1}^m\\left(\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}\\right).$\n\nThis is something I use quite a bit for various problems and programming exercises, so I figured I could post it here. It\u2019s a basic post that isn\u2019t advanced at all, but that doesn\u2019t mean that the implementation given below won\u2019t save work for others. The idea is to create a list of primes in C by malloc\u2019ing a sieve, then malloc\u2019ing a list of specific length based on that sieve. The resulting list contains all the primes below a given limit (defined in the code). The first member of the list is an integer representing the length of the list.\n\n#include <stdio.h> #include <stdlib.h> #include <malloc.h> \u00a0 #define bool _Bool \u00a0 static unsigned long prime_limit = 1000000; \u00a0 unsigned long sqrtld(unsigned long N) { int b = 1; unsigned long res,s; while(1<<b<N) b+= 1; res = 1<<(b/2 + 1); for(;;) { s = (N/res + res)/2; if(s>=res) return res; res = s; } } \u00a0 unsigned long * make_primes(unsigned long limit) { unsigned long *primes; unsigned long i,j; unsigned long s = sqrtld(prime_limit); unsigned long n = 0; bool *sieve = malloc((prime_limit + 1) * sizeof(bool)); sieve[0] = 0; sieve[1] = 0; for(i=2; i<=prime_limit; i++) sieve[i] = 1; j = 4; while(j<=prime_limit) { sieve[j] = 0; j += 2; } for(i=3; i<=s; i+=2) { if(sieve[i] == 1) { j = i * 3; while(j<=prime_limit) { sieve[j] = 0; j += 2 * i; } } } for(i=2;i<=prime_limit;i++) if(sieve[i]==1) n += 1; primes = malloc((n + 1) * sizeof(unsigned long)); primes[0] = n; j = 1; for(i=2;i<=prime_limit;i++) if(sieve[i]==1) { primes[j] = i; j++; } free(sieve); return primes; } \u00a0 \u00a0 int main(void) { \u00a0 unsigned long * primes = make_primes(prime_limit); \u00a0 printf(\"There are %ld primes <= %ld\\n\",primes[0],prime_limit); \u00a0 \u00a0 free(primes); return 0; }\n\nSay one wanted to form a list of all primes below 1,000,000. That\u2019s what the above program does by default, since \u201cprime_limit = 1000000.\u201d If one compiles this and executes, you would get something like what follows. The timing is relatively respectable.\n\n$gcc -O3 -o prime-sieve prime-sieve.c$ time ./prime-sieve There are 78498 primes <= 1000000 \u00a0 real 0m0.008s user 0m0.004s sys 0m0.000s\n\nThe code is linked here: prime-sieve.c", "date": "2018-03-19 23:51:45", "meta": {"domain": "jasonbhill.com", "url": "http://code.jasonbhill.com/category/algorithms/", "openwebmath_score": 0.7797718644142151, "openwebmath_perplexity": 1626.5125118132278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874090230069, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8588747119378474}}
{"url": "http://mathhelpforum.com/geometry/154183-triangles-20-non-straight-line-dots-print.html", "text": "# Triangles in 20 non-straight-line \"dots\"\n\n\u2022 August 22nd 2010, 07:07 AM\ngrottvald\nTriangles in 20 non-straight-line \"dots\"\nTwenty points/dots are given so that the three of them are never in a straight line. How many triangles can be formed with corners(as in vertex i think?) in the dots?\n\u2022 August 22nd 2010, 07:21 AM\nPlato\nCalculate the number of ways to choose three points from twenty.\n\u2022 August 22nd 2010, 10:36 AM\ngrottvald\nC(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)\n\u2022 August 22nd 2010, 10:57 AM\nPlato\nQuote:\n\nOriginally Posted by grottvald\nC(20,3) ? But that results in a very big number that seems unlikely to be correct\n\nIt is not a large number at all: $\\displaystyle \\binom{20}{3}=\\frac{20!}{3!\\cdot 17!}=1140$\n\u2022 August 22nd 2010, 11:58 AM\nQuote:\n\nOriginally Posted by grottvald\nC(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)\n\nYou may be thinking of non-overlapping triangles in the picture with 20 dots.\nThat's a different situation.\nImagine the dots are placed apart from left to right, not on a straight line.\nPick the leftmost point.\nTo make a triangle, you can pick any 2 of the remaining 19 dots.\n\nThe number of ways to do this is $\\binom{19}{2}=171$\n\nTherefore, there are 171 triangles that can be drawn which include the leftmost point.\n\nIf you move on to the next point to the right and exclude the point previously chosen,\nthen you can draw another $\\binom{18}{2}=153$ triangles.\n\nThese triangles do not include any of the previous 171,\nsince these 153 omit the leftmost point.\n\nNotice that these triangles typically share sides of other triangles,\nbut they are made of 3 distinct points, hence the triangles are being counted only once.\n\nhence, there are $\\binom{19}{2}+\\binom{18}{2}+\\binom{17}{2}+........ .+\\binom{2}{2}=\\binom{20}{3}$ triangles.\n\u2022 August 22nd 2010, 01:26 PM\ngrottvald\nQuote:\n\nYou may be thinking of non-overlapping triangles in the picture with 20 dots.\nThat's a different situation.\nImagine the dots are placed apart from left to right, not on a straight line.\nPick the leftmost point.\nTo make a triangle, you can pick any 2 of the remaining 19 dots.\n\nThe number of ways to do this is $\\binom{19}{2}=171$\n\nTherefore, there are 171 triangles that can be drawn which include the leftmost point.\n\nIf you move on to the next point to the right and exclude the point previously chosen,\nthen you can draw another $\\binom{18}{2}=153$ triangles.\n\nThese triangles do not include any of the previous 171,\nsince these 153 omit the leftmost point.\n\nNotice that these triangles typically share sides of other triangles,\nbut they are made of 3 distinct points, hence the triangles are being counted only once.\n\nhence, there are $\\binom{19}{2}+\\binom{18}{2}+\\binom{17}{2}+........ .+\\binom{2}{2}=\\binom{20}{3}$ triangles.\n\nThank you so much! You explained it in an very easy and beautiful way. Thanks again!", "date": "2016-02-11 18:19:41", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/154183-triangles-20-non-straight-line-dots-print.html", "openwebmath_score": 0.6234351396560669, "openwebmath_perplexity": 722.7878330307632, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9905874090230069, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8588747102361404}}
{"url": "https://math.stackexchange.com/questions/3687083/int-fracgxfx-dx-where-fx-frac12exe-x-and-gx", "text": "# $\\int\\frac{g(x)}{f(x)} \\, dx$ where $f(x) = \\frac{1}{2}(e^x+e^{-x})$ and $g(x) = \\frac{1}{2}(e^x - e^{-x})$?\n\nGiven two functions $$f(x) = \\frac{1}{2}(e^x+e^{-x})$$ and $$g(x) = \\frac{1}{2}(e^x - e^{-x})$$, calculate\n\n$$\\int\\frac{g(x)}{f(x)} \\, dx.$$\n\nObserving that $$f'(x) = g(x)$$, this is very easy to do. Just take $$u = f(x)$$ and therefore $$du = f'(x)\\,dx$$. Now we have\n\n$$\\int \\frac{f'(x)}{f(x)} \\, dx = \\int \\frac{du}{u} = \\ln |u| + C = \\ln\\left|\\frac{1}{2}(e^x+e^{-x})\\right|+C.$$\n\nHowever, this solution is wrong. The correct one is $$\\ln|e^x+e^{-x}| + C$$. I see how we could get that, just plug in $$g(x)$$ and $$f(x)$$ directly, and we get\n\n$$\\int \\frac{\\frac{1}{2}(e^x - e^{-x})}{\\frac{1}{2}(e^x+e^{-x})} \\, dx.$$\n\nCancelling the $$\\frac{1}{2}$$'s and taking $$u = e^x+e^{-x}$$ and $$du = (e^x-e^{-x}) dx$$ gives us\n\n$$\\int \\frac{du}{u} = \\ln|u| + C = \\ln|e^x+e^{-x}| + C.$$\n\nI can't see where I made a mistake in my approach, and I'd be really grateful if you pointed it out.\n\n\u2022 Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. \u2013\u00a0Shaun May 22 at 18:47\n\u2022 Thanks, I will do that next time. :) \u2013\u00a0Gregor Per\u010di\u010d May 22 at 19:01\n\u2022 I feel terrible mentioning it because I know you're only trying to be polite but could you remove the religious elements at the end? There are likely many faiths none should be given a platform \u2013\u00a0Karl May 22 at 19:10\n\u2022 @Karl With all due respect, I will not remove it. It is very good to try to propagate the Faith in all contexts of life. The mods may remove it, but hey, I did my best. Not to mention that my right to free speech will be trampled on in this instance. \u2013\u00a0Gregor Per\u010di\u010d May 22 at 19:29\n\u2022 I ment no offence honestly. I'd just rather stick to maths. \u2013\u00a0Karl May 22 at 19:38\n\nNote that if $$K$$ is a constant, $$\\ln|K h(x)|= \\ln|K|+\\ln|h(x)|$$. In other words, the $$1/2$$ in the log can get 'absorbed' into the $$+C$$ term.\n\u2022 Aaaaah, I see! But then that means that my solution $\\ln|\\frac{1}{2}(e^x + e^{-x})| + C$ doesn't by itself (with $C = 0$) calculate the area under the curve $h(x) = \\frac{f(x)}{g(x)}$? \u2013\u00a0Gregor Per\u010di\u010d May 22 at 19:00\n\u2022 It seems $f$ and $g$ are reversed in your comment. It should, in the sense that $$\\int _{0}^x \\frac{g(t)}{f(t)}\\,dt = \\left.\\ln|\\frac{1}{2}(e^t + e^{-t})|+C\\right|_{0}^x$$ $$= \\left(\\ln|\\frac{1}{2}(e^x + e^{-x})|+C\\right)-\\left(\\ln|\\frac{1}{2}(1+1)|+C\\right)=\\ln|\\frac{1}{2}(e^x + e^{-x})|$$The $1/2$ is essential to make sure, starting at $x=0$, the area is $0$; otherwise, you'll be off by a constant. \u2013\u00a0Integrand May 22 at 19:25\n\u2022 Yes, $f$ and $g$ are reversed. Sorry about that and thanks for the insight. \u2013\u00a0Gregor Per\u010di\u010d May 22 at 19:32\nUsing the identity: $$\\log (\\frac{a}{b})=\\log(a)- \\log(b)$$\n$$\\ln \\frac{1}{2}(e^x+e^{-x})+C$$\n$$=\\ln (e^x+e^{-x})-\\ln 2+C$$\n$$=\\ln (e^x+e^{-x})+C{'}$$ (as $$\\ln 2$$ is constant)", "date": "2020-06-03 22:45:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3687083/int-fracgxfx-dx-where-fx-frac12exe-x-and-gx", "openwebmath_score": 0.7617987990379333, "openwebmath_perplexity": 463.9013710302533, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.969785409439575, "lm_q2_score": 0.8856314662716159, "lm_q1q2_score": 0.8588724741307903}}
{"url": "https://casmusings.wordpress.com/2015/08/", "text": "# Monthly Archives: August 2015\n\n## Marilyn vos Savant Conditional Probability Follow\u00a0Up\n\nIn the Marilyn vos Savant problem I posted yesterday, I focused on the subtle shift from simple to conditional probability the writer of the question appeared to miss. \u00a0Two of my students took a different approach.\n\nThe majority of my students, typical of AP Statistics students\u2019\u00a0tendencies\u00a0very early in the course, tried to use a \u201cwall of words\u201d to explain away the discrepancy rather than providing quantitative evidence. \u00a0But two fully embraced the probabilities and developed the following probability tree to incorporate\u00a0all of the given probabilities. \u00a0Each branch shows the probability of a short or long straw given the present state of the system. \u00a0Notice that it includes both\u00a0of the apparently confounding 1/3 and 1/2 probabilities.\n\nThe uncontested probability of the first person is 1/4.\n\nThe probability of the second person is then (3/4)(1/3) = 1/4, exactly as expected. \u00a0The probabilities of the 3rd and 4th people can be similarly computed to arrive at the same 1/4 final result.\n\nMy students argued essentially that the writer was correct in saying the probability of the second person having the short straw was 1/3 in the instant after it was revealed that the first person didn\u2019t have the straw, but that they had forgotten to incorporate the probability of arriving in\u00a0that state. \u00a0When you use all of the information, the probability of each person receiving the short straw remains at 1/4, just as expected.\n\n## Marilyn vos Savant and Conditional\u00a0Probability\n\nThe following question appeared in the\u00a0\u201cAsk Marilyn\u201d column in the August 16, 2015 issue of Parade magazine. \u00a0The writer seems stuck between two probabilities.\n\n(Click here for a\u00a0cleaned-up online version if you don\u2019t like the newspaper look.)\n\nI just pitched this question to my statistics class (we start the year with a probability unit). \u00a0I thought some of you might like it for your classes, too.\n\nI asked them to do two things. \u00a01) Answer the writer\u2019s question, AND 2)\u00a0Use precise probability terminology\u00a0to\u00a0identify the source of the writer\u2019s conundrum. \u00a0Can you answer both before reading further?\n\nVery briefly, the writer is correct in both situations. \u00a0If each of the four people draws a random straw, there is absolutely a 1 in 4 chance of each drawing the straw. \u00a0Think about shuffling the straws and \u201cdealing\u201d one to each person much like shuffling a deck of cards and dealing out all of the cards. \u00a0Any given straw or card is equally likely to land in any player\u2019s hand.\n\nNow let the first person look at his or her straw. \u00a0It is either short or not. \u00a0The author is then correct at claiming the probability of others holding the straw is now 0 (if the first person found the short straw) or 1/3 (if the first person did not). \u00a0And this is precisely the source of the writer\u2019s conundrum. \u00a0She\u2019s actually asking two different questions but thinks she\u2019s asking only one.\n\nThe 1/4 result\u00a0is from a pure, simple probability scenario. \u00a0There are four possible equally-likely locations for the short straw.\n\nThe 0 and 1/3 results happen only after the first (or any other) person looks at his or her straw. \u00a0At that point, the problem shifts from simple probability to\u00a0conditional probability. \u00a0After observing a straw, the question shifts to determining\u00a0the probability that one of the remaining people has the short straw GIVEN that you know the result of one person\u2019s draw.\n\nSo, the writer was correct in all of her claims; she just didn\u2019t realize she was asking two fundamentally different questions. \u00a0That\u2019s a pretty excusable lapse, in my opinion. \u00a0Slips\u00a0into conditional probability are often missed.\n\nPerhaps the most famous of these misses is the solution to the Monty Hall scenario that\u00a0vos Savant famously posited years ago. \u00a0What I particularly love about this is the number of very-well-educated mathematicians who missed the conditional and wrote flaming retorts to vos Savant brandishing their PhDs and ultimately found themselves publicly supporting errant conclusions. \u00a0You can read the original question, errant responses, and vos Savant\u2019s very clear explanation\u00a0here.\n\nCONCLUSION:\n\nProbability is subtle and catches all of us at some point. \u00a0Even so, the careful thinking\u00a0required to dissect and answer subtle probability questions is arguably one of the best exercises of logical reasoning around.\n\nRANDOM(?) CONNECTION:\n\nAs a completely different connection, I think this is very much like Heisenberg\u2019s Uncertainty Principle. \u00a0Until the first straw is observed, the short straw really could (does?) exist in all hands simultaneously. \u00a0Observing the system (looking at one person\u2019s straw) permanently changes the state of the system, bifurcating forever the system into one of two potential future states: \u00a0the short straw is found in the first hand or is it not.\n\nCORRECTION (3 hours after posting):\n\nI knew I was likely to overstate or misname something in my final connection. \u00a0Thanks to Mike Lawler (@mikeandallie) for\u00a0a quick correction via Twitter. \u00a0I should have called this\u00a0quantum superposition and not\u00a0the uncertainty principle. \u00a0Thanks so much, Mike.\n\n## SBG and AP Statistics\u00a0Update\n\nI\u2019ve continued to work on my Standards for AP Statistics and after a few conversations with colleagues and finding this pdf of AP Statistics Standards, I\u2019ve winnowed down and revised my Standards to the point I\u2019m comfortable using them this year.\n\nFollowing is the much shorter document I\u2019m using in my classes this year. \u00a0They address the AP Statistics core content as well as the additional ideas, connections, etc. I hope my students learn this year. \u00a0As always, I welcome all feedback, and I hope someone else finds these guides helpful.\n\n## SBG and Statistics\n\nI\u2019ve been following Standards-Based Grading (SBG) for several years now after first being introduced to the concept by colleague John Burk (@occam98). \u00a0Thanks, John!\n\nI finally made the dive into SBG with my Summer School Algebra 2 class this past June & July, and I\u2019ve fully committed to an SBG pilot for my AP Statistics classes this year.\n\nI found writing standards for Algebra 2 this summer relatively straightforward. \u00a0I\u2019ve taught that content for decades now and know precisely what I want my students to understand. \u00a0I needed some practice writing standards and got better as the summer class progressed. \u00a0Over time, I\u2019ve read several teachers\u2019 versions of standards for various courses. \u00a0But writing standards for my statistics class prove MUCH more challenging. \u00a0In the end, I found myself guided by three major philosophies.\n\n1. The elegance and challenge of well designed Enduring Understandings from the Understanding by Design (UbD) work of Jay McTighe the late Grant Wiggins helped me craft many of my standards as targets for student learning that didn\u2019t necessarily reveal everything all at once.\n2. The power of writing student-centered\u00a0\u201cI can \u2026\u201d statements that I learned through my colleague Jill Gough (@jgough) has become very important in my classroom design. \u00a0I\u2019ve become much more focused on what I want my students (\u201clearners\u201d in Jill\u2019s parlance) to be able to accomplish and less about what I\u2019m trying to deliver. \u00a0This recentering of my teaching awareness has been good for my continuing professional development and was a prime motivator in writing these Standards.\n3. I struggled throughout the creation of my first AP Statistics standards\u00a0document to find a balance between too few very broad high-level conceptual claims and a\u00a0far-too-granular long list of skill minutiae. \u00a0I wanted more than a narrow checklist of tiny skills and less than overloaded individual standards that are difficult for students to satisfy. \u00a0I want a challenging, but reachable bar.\n\nSo, following is my first attempt at Standards for my AP Statistics class, and I\u2019ll be using them this year. \u00a0In sharing this, I have two hopes:\n\n\u2022 Maybe some teacher out there might find some use in my Standards.\n\u2022 More importantly, I\u2019d LOVE some feedback from anyone on this work. \u00a0It feels much too long to me, but I wonder if it is really too much\u00a0or too little. \u00a0Have I left something out?\n\nAt some point, all work needs a public airing to improve. \u00a0That time for me is now. \u00a0Thank you in advance on behalf of my students for any feedback.\n\n## Chemistry, CAS, and Balancing\u00a0Equations\n\nHere\u2019 s a cool application of linear equations I first encountered about 20 years ago working with chemistry colleague Penney Sconzo at my former school in Atlanta, GA. \u00a0Many students struggle early in their first chemistry classes with balancing equations. \u00a0Thinking about these as generalized systems of linear equations gives a universal\u00a0approach to balancing chemical equations, including ionic equations.\n\nThis idea makes a brilliant connection if you teach algebra 2 students concurrently enrolled in chemistry, or vice versa.\n\nFROM CHEMISTRY TO ALGEBRA\n\nConsider burning ethanol. \u00a0The chemical combination of ethanol and oxygen, creating carbon dioxide and water:\n\n$C_2H_6O+3O_2 \\longrightarrow 2CO_2+3H_2O$ \u00a0 \u00a0 (1)\n\nBut what if you didn\u2019t know\u00a0that 1\u00a0molecule of\u00a0ethanol combined with 3\u00a0molecules of oxygen gas to create 2 molecules of carbon dioxide and 3 molecules of water? \u00a0This specific set coefficients (or multiples of the set) exist for this reaction because of the Law of Conservation of Matter. \u00a0While elements may rearrange in a chemical reaction, they do not become something else. \u00a0So\u00a0how do you determine the unknown coefficients of a generic\u00a0chemical reaction?\n\nUsing the ethanol example, assume you started with\n\n$wC_2H_6O+xO_2 \\longrightarrow yCO_2+zH_2O$ \u00a0 \u00a0 (2)\n\nfor some unknown values of w, x, y, and z. \u00a0Conservation of Matter guarantees that the amount of carbon, hydrogen, and oxygen are\u00a0the same before and after the reaction. \u00a0Tallying the amount of each element on each side of the equation gives three linear equations:\n\nCarbon: \u00a0$2w=y$\nHydrogen: \u00a0$6w=2z$\nOxygen: \u00a0$w+2x=2y+z$\n\nwhere the coefficients come from the subscripts within the compound notations. \u00a0As one example, the carbon subscript in ethanol ( $C_2H_6O$ ) is 2, indicating two carbon atoms in each ethanol molecule. \u00a0There must have been\u00a02w carbon atoms in the\u00a0w ethanol molecules.\n\nThis system of 3 equations in 4 variables\u00a0won\u2019t have a unique solution, but let\u2019s see what my Nspire CAS says. \u00a0(NOTE: \u00a0On the TI-Nspire, you can solve for any one of the four variables. \u00a0Because the presence of more variables than equations makes the solution non-unique, some results may appear cleaner than others. \u00a0For me, w was more complicated than z, so I chose to use the z solution.)\n\nAll\u00a0three equations have\u00a0y in the numerator and denominators of 2. \u00a0The presence of the\u00a0y\u00a0indicates the expected non-unique solution. \u00a0But it also gives me the freedom to select any convenient value of\u00a0y I want to use. \u00a0I\u2019ll pick $y=2$ to simplify the fractions. \u00a0Plugging in gives me values for the other coefficients.\n\nSubstituting these into (2) above gives the original equation (1).\n\nVARIABILITY EXISTS\n\nTraditionally, chemists write these equations with the lowest possible natural number coefficients, but thinking of them as systems of linear equations makes another reality obvious. \u00a0If 1 molecule of\u00a0ethanol combines with 3 molecules of hydrogen gas to make 2 molecules of carbon dioxide and 3 molecules of water, surely\u00a010 molecule of\u00a0ethanol combines with 30 molecules of hydrogen gas to make 20 molecules of carbon dioxide and 30 molecules of water (the result of substituting $y=20$ instead of the $y=2$ used above).\n\nYou could even let $y=1$ to get $z=\\frac{3}{2}$, $w=\\frac{1}{2}$, and $x=\\frac{3}{2}$. \u00a0Shifting units, this could mean a half-mole of ethanol and 1.5 moles of hydrogen make a mole of carbon dioxide and 1.5\u00a0moles of water. \u00a0The point is, the ratios are constant. \u00a0A good lesson.\n\nANOTHER QUICK EXAMPLE:\n\nNow let\u2019s try a harder one to balance: \u00a0Reacting carbon monoxide and hydrogen gas to create octane and water.\n\n$wCO + xH_2 \\longrightarrow y C_8 H_{18} + z H_2 O$\n\nSetting up equations for each element gives\n\nCarbon: \u00a0$w=8y$\nOxygen: \u00a0$w=z$\nHydrogen: \u00a0$2x=18y+2z$\n\nI could simplify the hydrogen equation, but that\u2019s not required. \u00a0Solving this system of equations gives\n\nNice. \u00a0No fractions this time. \u00a0Using $y=1$ gives $w=8$, $x=17$, and $z=8$, or\n\n$8CO + 17H_2 \\longrightarrow C_8 H_{18} + 8H_2 O$\n\nSimple.\n\nEXTENSIONS TO IONIC EQUATIONS:\n\nNow let\u2019s balance\u00a0an\u00a0ionic equation with unknown coefficients a, b, c, d, e, and f:\n\n$a Ba^{2+} + b OH^- + c H^- + d PO_4^{3-} \\longrightarrow eH_2O + fBa_3(PO_4)_2$\n\nIn addition to writing equations for barium, oxygen, hydrogen, and phosphorus, Conservation of Charge allows me to write one more equation to reflect the balancing of charge in the reaction.\n\nBarium: \u00a0$a = 3f$\nOxygen: \u00a0$b +4d = e+8f$\nHydrogen: \u00a0$b+c=2e$\nPhosphorus: \u00a0$d=2f$\nCHARGE (+/-): \u00a0$2a-b-c-3d=0$\n\nSolving the system gives\n\nNow that\u2019s a curious result. \u00a0I\u2019ll deal with the zeros in a moment. \u00a0Letting $d=2$ gives $f=1$ and $a=3$, indicating that 3 molecules of ionic barium combine with 2 molecules of ionic phosphate to create a single uncharged molecule of barium phosphate precipitate.\n\nThe zeros here indicate the presence of \u201cspectator ions\u201d. \u00a0Basically, the hydroxide and hydrogen ions on the left are in equal measure to the liquid water molecule on the right. \u00a0Since they are in equal measure, one solution is\n\n$3Ba^{2+}+6OH^- +6H^-+2PO_4^{3-} \\longrightarrow 6H_2O + Ba_3(PO_4)_2$\n\nCONCLUSION:\n\nYou still need to understand chemistry and algebra to interpret the results, but combining algebra (and especially a CAS) makes it much easier to balance chemical equations and ionic chemical equations, particularly those with non-trivial solutions not easily found by inspection.\n\nThe minor connection between science (chemistry) and math (algebra) is nice.\n\nAs many others have noted, CAS enables you to keep your mind on the problem while avoiding getting lost in the algebra.", "date": "2018-06-19 02:40:25", "meta": {"domain": "wordpress.com", "url": "https://casmusings.wordpress.com/2015/08/", "openwebmath_score": 0.6689112782478333, "openwebmath_perplexity": 1911.5803115971619, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9926541734255229, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8588683051427728}}
{"url": "https://math.stackexchange.com/questions/450158/simplifying-compound-fraction-frac3-sqrt5-5", "text": "# Simplifying compound fraction: $\\frac{3}{\\sqrt{5}/5}$\n\nI'm trying to simplify the following:\n\n$$\\frac{3}{\\ \\frac{\\sqrt{5}}{5} \\ }.$$\n\nI know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.\n\nThe problem is I interpreted that as:\n\n$$\\frac{3}{\\ \\frac{\\sqrt{5}}{5} \\ } \\times \\frac{5}{\\sqrt{5}},$$\n\nWhich I believe is:\n\n$$\\frac{15}{\\ \\frac{5}{5} \\ } = \\frac{15}{1}.$$\n\nWhat am I doing wrong?\n\n\u2022 You should multiply both numerator and denominator by that constant! \u2013\u00a0Mahdi Khosravi Jul 23 '13 at 9:50\n\u2022 Or you can exchange the numerator and denominator of the whole denominator and move it to whole numerator! \u2013\u00a0Mahdi Khosravi Jul 23 '13 at 9:51\n\u2022 I know this an old question, but if you had simply changed your \u221a5/5 to a \u221a5/\u221a5, you would've got 3\u221a5\u00f75/5 and gotten your answer. The whole point of multiplying a complex faction by a number to simplify it is to times it by 1 (\u221a5/\u221a5 in this case) because multiplying anything by 1 is the same thing. \u2013\u00a0G\u1d07\u1d0f\u1d0d\u1d07\u1d1b\u1d07\u0280 Sep 7 '14 at 19:44\n\u2022 \"The problem is I interpreted that as:\" .... remember that the numerator is 3. \u2013\u00a0John Joy Jan 4 at 17:58\n\n\\begin{align*}\\frac{3}{\\frac{\\sqrt{5}}{5}} &= 3 \\cdot \\frac{5}{\\sqrt 5}\\\\ &= 3 \\cdot \\frac{5}{\\sqrt 5} \\cdot 1\\\\ &= 3 \\cdot \\frac{5}{\\sqrt 5} \\cdot \\frac{\\sqrt 5}{\\sqrt 5}\\\\ &= 3 \\cdot \\frac{5\\sqrt 5}{5}\\\\ &= 3\\sqrt 5 \\end{align*}\n\n\u2022 I appreciate everyone's response but this answer is the most elaborate. Thanks blf, I see where I went wrong now! \u2013\u00a0Sam Jul 23 '13 at 10:00\n\nYou multiplied the original fraction by $\\dfrac5{\\sqrt5}$, which is not $1$, so of course you changed the value. The correct course of action is to multiply by $1$ in the form\n\n$$\\frac{5/\\sqrt5}{5/\\sqrt5}$$\n\nto get\n\n$$\\frac3{\\frac{\\sqrt5}5}=\\frac3{\\frac{\\sqrt5}5}\\cdot\\frac{\\frac5{\\sqrt5}}{\\frac5{\\sqrt5}}=\\frac{3\\cdot\\frac5{\\sqrt5}}1=3\\cdot\\frac5{\\sqrt5}=3\\sqrt5\\;,$$\n\nsince $\\dfrac5{\\sqrt5}=\\sqrt5$.\n\nMore generally,\n\n$$\\frac{a}{b/c}=\\frac{a}{\\frac{b}c}\\cdot\\frac{\\frac{c}b}{\\frac{c}b}=\\frac{a\\cdot\\frac{c}b}1=a\\cdot\\frac{c}b\\;,$$\n\nthis is the basis for the invert and multiply rule for dividing by a fraction.\n\nThis means $$3\\cdot \\frac{5}{\\sqrt{5}}=3\\cdot\\frac{(\\sqrt{5})^2}{\\sqrt{5}} =3\\sqrt{5}$$ You're multiplying twice for the reciprocal of the denominator.\n\nAnother way to see it is multiplying numerator and denominator by the same number: $$\\frac{3}{\\frac{\\sqrt{5}}{5}}=\\frac{3\\sqrt{5}}{\\frac{\\sqrt{5}}{5}\\cdot\\sqrt{5}} =\\frac{3\\sqrt{5}}{1}$$\n\nStart with $$\\frac{3}{\\sqrt{5}/5}=\\frac{15}{\\sqrt{5}},$$\n\nand then rationalize the denominator (multiply both numerator and denominator by $\\sqrt{5}$) to get\n\n$$\\frac{15\\sqrt{5}}{5}=3\\sqrt{5}.$$\n\nYou have the following fraction to simplify: \\begin{align} \\frac{3}{\\sqrt{5}/5} &=\\frac{5\\times 3}{\\sqrt{5}} \\\\ &=\\frac{15}{\\sqrt{5}} \\\\ &=\\sqrt{\\bigg(\\frac{15}{\\sqrt{5}}\\bigg)^2} \\\\ &=\\sqrt{\\frac{15^2}{\\sqrt{5}^2}} \\\\ &=\\sqrt{\\frac{225}{5}} \\\\ &=\\sqrt{45} \\\\ &= \\sqrt{9\\times 5} \\\\ &= \\sqrt{9}\\sqrt{5} \\\\ &= 3\\sqrt{5} \\\\ \\therefore \\frac{15}{\\sqrt{5}}\\times \\frac{5}{\\sqrt{5}} &=\\frac{15}{\\sqrt{5}}\\times \\frac{\\big(\\frac{15}{\\sqrt{5}}\\big)}{3} \\\\ &=\\frac{15}{\\sqrt{5}}\\times \\frac{15}{\\sqrt{5}}\\times \\frac{1}{3} \\\\ &=\\bigg(\\frac{15}{\\sqrt{5}}\\bigg)^2 \\times \\frac{1}{3} \\\\ &=45\\times \\frac{1}{3} \\\\ &=\\frac{45}{3} \\\\ &=15 \\end{align}\n\nOne thing that helps me organize my thoughts, is to convert both numerator and denuminator to fractions as follows.\n\n\\begin{align} \\frac{3}{\\frac{5}{\\sqrt{5}}}&=\\frac{\\frac{3}{1}}{\\frac{5}{\\sqrt{5}}}=\\frac{\\frac{3}{1}}{\\frac{5}{\\sqrt{5}}}\\cdot\\frac{\\frac{\\sqrt{5}}{5}}{\\frac{\\sqrt{5}}{5}}=\\frac{\\frac{3}{1}\\cdot\\frac{\\sqrt{5}}{5}}{\\frac{5}{\\sqrt{5}}\\cdot\\frac{\\sqrt{5}}{5}}=\\frac{\\frac{3}{1}\\cdot\\frac{\\sqrt{5}}{5}}{1}=\\frac{3}{1}\\cdot\\frac{\\sqrt{5}}{5}=\\text{etc.}\\\\ \\end{align}\n\nIn the expression $$\\frac{3}{\\frac{\\sqrt{5}}{5}}$$, to find the reciprocal of that expression's denominator, $$\\frac{\\sqrt{5}}{5}$$, simply swap its numerator and denominator.\n\nThus the reciprocal of $$\\frac{\\sqrt{5}}{5}$$ is $$\\frac{5}{\\sqrt{5}}$$, and $$\\frac{3}{\\frac{\\sqrt{5}}{5}} = 3 \\cdot \\frac{5}{\\sqrt{5}}$$\n\nI think the result you have to remember is\n\n$$\\boxed{\\dfrac 1{\\sqrt{a}}=\\dfrac{\\sqrt{a}}a}$$\n\nYou will see this kind of manipulation very often in your studies, and it allows to get rid of square roots on denominator.\n\nHere your expression is just $$\\dfrac{3}{\\frac{\\sqrt{5}}{5}}=\\dfrac{3}{\\frac 1{\\sqrt{5}}}=3\\sqrt{5}$$", "date": "2019-11-16 22:08:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/450158/simplifying-compound-fraction-frac3-sqrt5-5", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 905.0839646469799, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575142422757, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.8588311634968874}}
{"url": "https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/", "text": "# Is There a Rigorous Proof Of 1 = 0.999\u2026?\n\nYes.\n\nFirst, we have not addressed what 0.999\u2026 actually means. So it\u2019s best first to describe what on earth the notation $$b_0.b_1b_2b_3\u2026$$ means. The way mathematicians define this thing is\n\n$$b_0.b_1b_2b_3\u2026=\\sum_{n=0}^{+\\infty}{\\frac{b_n}{10^n}}$$\n\nSo, in particular, we have that\n\n$$0.999\u2026=\\sum_{n=1}^{+\\infty}{\\frac{9}{10^n}}$$\n\nBut all of this doesn\u2019t really make any sense until we define what the right-hand side means. There is an infinite sum there, but what does that mean? Well, we put\n\n$$S_k=\\sum_{n=1}^{k}{\\frac{9}{10^n}} \\ ,$$\n\nthen we have a finite sum. So, for example\n$$S_1=0.9, \\ ~S_2=0.99, \\ ~S_3=0.999, \\ etc.$$\nSo, in some way, we want to take the limit of this sequence.\n\nLet\u2019s consider a particularly simple sequence to illustrate the idea behind the definition of a limit of a sequence: 1/2, 1/3, 1/4,\u2026 The terms in this sequence get smaller and smaller. You might think that it\u2019s obvious that it goes to 0, or that it\u2019s obvious that a smart mathematician can prove that it goes to 0, but it\u2019s not. It\u2019s impossible to even attempt a proof until we have defined what it means for something to go to 0. So we have to define what the statement \u201c1/2, 1/3, 1/4,\u2026 goes to 0\u201d means before we can attempt to prove that it\u2019s true.\n\nThis is the standard definition: \u201c1/n goes to 0\u201d means that \u201cfor every positive real number $\\epsilon$, there\u2019s a positive integer N, such that for all integers n such that $n\\geq N$, we have $|1/n| < \\epsilon$\u201d. With this definition in place, it\u2019s quite easy to prove that \u201c1/n goes to 0\u201d is a true statement. What I want you to see here, is that we chose this definition to make sure that this statement would be true. The first mathematicians who thought about how to define the limit of a sequence might have briefly considered definitions that make the statement \u201c1/n goes to 0\u201d false, but they would have dismissed those definitions as irrelevant because they fail to capture the idea of a limit that they already understood on an intuitive level.\n\nSo the real reason why 1/n goes to 0 is that we wanted it to! Similar comments hold for the sequence of partial sums that define 0.999\u2026 It goes to 1, because we have defined the concepts \u201c0.999\u2026.\u201d, \u201csum of infinitely many terms\u201d, and \u201climit of a sequence\u201d in ways that make 0.999\u2026=1. Can we define number systems such that 1=0.999\u2026 does not hold? Of course! But these number systems are not as useful, because they don\u2019t conform to our intuition about limits and numbers.\n\nNow that we know what a limit and an infinite sum is, let me give a fully rigorous proof to the equality 1=0.999\u2026 This proof is due to Euler and it appears in the 1770\u2019s edition of \u201cElements of algebra\u201d.\nWe know that\n\n$$0.999\u2026=\\sum_{n=1}^{+\\infty}{ \\frac{9}{10^n} } = \\frac{9}{10} + \\frac{9}{10^2} + \\frac{9}{10^3} +\u2026$$\n\nThis sum is a special kind of sum, namely, it\u2019s a geometric sum. For (infinite) geometric sums, we can find its limit easily:\n\nLet\n$$x=\\frac{1}{10}+\\frac{1}{10^2}+\\frac{1}{10^3}+\u2026$$\nThen\n$$9x=0.999\u2026$$\n\nBut, we also have $10x=1+\\frac{1}{10}+\\frac{1}{10^2}+\u2026$, so $10x-x=1$.\n\nThis implies that $x=\\frac{1}{9}$.\n\nHence,\n$$0.999\u2026=9x=1$$\nDoes this proof look familiar? It should! It is essentially the same as Proof #2 in the previous post. The only difference is that every step is now justified by operations with limits.\n\nSee this supportive article: https://www.physicsforums.com/insights/why-do-people-say-that-1-and-999-are-equal/\n\nThe following forum members have contributed to this FAQ:\nAlephZero\nFredrik\nmicromass\ntiny-tim\nvela\n\nTo forum post\n\nTags:\n102 replies\n1. nuuskur says:\n\nIt should be unanimous that 1 \u2013 0.9(9) = 0.0(0) = 0. Assume the opposite, then 0.0(0) != 0, contradiction.\n\nToo much java, my TeX is rusty :D\n\n2. Mark44 says:\n\n[QUOTE=\u201dalva, post: 5215165, member: 497526\u2033]\nBut my reasoning is similar to the OP. He is taking more addends in one series than in the other.[/QUOTE]\n\n[QUOTE=\u201dweirdoguy, post: 5215170, member: 464738\u2033]No it\u2019s not, because he\u2019s using infinite series, and you are using finite series. It is a big difference.[/QUOTE]\nIn addition, [USER=497526]@alva[/USER], you don\u2019t seem to realize that 0.99 and 0.99\u2026 mean completely different things. The first number has two decimal digits that are shown explicitly, and an implied infinite number of zeroes that follow it. The second number has an infinite number of 9\u2019s following the decimal point.\n\nWe are NOT saying that 1 = 0.99. We ARE saying that 1 = 0.99\u2026, and the proof is given in the OP.\n\nAlso, 0.99\u2026 and 0.999\u2026 are exactly the same, while 0.99 and 0.999 are different.\n\n3. micromass says:\n\n[QUOTE=\u201dalva, post: 5215165, member: 497526\u2033]\nBut my reasoning is similar to the OP. He is taking more addends in one series than in the other.[/QUOTE]\n\nNo, in both cases, the number of addends is infinite. The only difference is that the series in the OP is convergent, while yours are divergent.\n\n4. alva says:\n\n[QUOTE=\u201dHomogenousCow, post: 5215156, member: 435628\u2033]Here you\u2019ve just made some bizarre arguments and obvious arithmetic mistakes.[/QUOTE]\nPlease, let me know what are \u201cmy arithmetic mistakes\u201d.\n\n5. weirdoguy says:\n\nNo it\u2019s not, because he\u2019s using infinite series, and you are using finite series. It is a big difference.\n\n6. alva says:\n\n[QUOTE=\u201dHomogenousCow, post: 5215152, member: 435628\u2033]You can\u2019t manipulate divergent series that way, also you forgot the negative signs at the end.[/QUOTE]\nYou\u2019re right and I apologize for the mistake in the signs.\nMoreover where I said \u201cmy age\u201d I meant \u201cmy age from Jesus\u2019s birth.\n\nBut my reasoning is similar to the OP. He is taking more addends in one series than in the other.\n\n7. weirdoguy says:\n\n[QUOTE=\u201dalva, post: 5215133, member: 497526\u2033]and never 0.99999\u2026 = 1[/QUOTE]\n\nYour \u201cproof\u201d is just plain wrong. Just deal with the fact that 0,(9)=1.\n\n8. HomogenousCow says:\n\n[QUOTE=\u201dalva, post: 5215133, member: 497526\u2033]Let x= 1/10 +1/100 + \u2026\n\nLets take just 2 addends, x=0.11\n\nThen 9x = 0.99\n\nBut, we also have 10x = 1 + 1/10 = 1.1, so 10x \u2013 x =0.11\n\nHence 0.99 = 9x = 0.99\n\nYou can use 3 addends and get 0.999 = 9x = 0,999\n\nAnd 4 addends and 5, and calculate the limit when n -> infinite and never 0.99999\u2026 = 1[/QUOTE]\n\nHere you\u2019ve just made some bizarre arguments and obvious arithmetic mistakes.\n\n9. HomogenousCow says:\n\n[QUOTE=\u201dalva, post: 5215141, member: 497526\u2033]I was born in 1950 and my brother in 1952.\n\nIn many years my age will be 1950 + 1 + 1 +1 +1 \u2026.\n\nIn many years my brothers age will be 1952 +1 + 1 +1 \u2026.\n\nSo the difference will be 1952 +( 1 + 1 +1+ \u2026) \u2013 1950 \u2013 ( 1 +1 +1+1+\u2026) = 1952 + (1 +1 +1 +\u2026) \u2013 1950 +2 + (1 +1 +1..) = 0[/QUOTE]\n\nYou can\u2019t manipulate divergent series that way, also you forgot the negative signs at the end.\n\n10. alva says:\n\nI was born in 1950 and my brother in 1952.\n\nIn many years my age will be 1950 + 1 + 1 +1 +1 \u2026.\n\nIn many years my brothers age will be 1952 +1 + 1 +1 \u2026.\n\nSo the difference will be 1952 +( 1 + 1 +1+ \u2026) \u2013 1950 \u2013 ( 1 +1 +1+1+\u2026) = 1952 + (1 +1 +1 +\u2026) \u2013 1950 +2 + (1 +1 +1..) = 0\n\n11. alva says:\n\nLet x= 1/10 +1/100 + \u2026\n\nLets take just 2 addends, x=0.11\n\nThen 9x = 0.99\n\nBut, we also have 10x = 1 + 1/10 = 1.1, so 10x \u2013 x =0.11\n\nHence 0.99 = 9x = 0.99\n\nYou can use 3 addends and get 0.999 = 9x = 0,999\n\nAnd 4 addends and 5, and calculate the limit when n -> infinite and never 0.99999\u2026 = 1\n\n12. Mark44 says:\n\n[QUOTE=\u201dalva, post: 5215060, member: 497526\u2033]Let\n\nx=110+1102+1103+\u2026\n\nThen\n9x=0.999\u2026\nBut, we also have 10x=1+110+1102+\u2026, so 10x\u2212x=1.\n\nThis implies that x=19.\n[/quote]None of the above makes any sense. In your definition of x, its value is at least 2300. How can 9x be 0.999\u2026?\n[QUOTE=alva]\n\nHence,\n\n0.999\u2026=9x=1\n\n**********************************************************\nIm sorry but Im trying to copy/paste the OP message but the format is not preserved\n**********************************************************\n\nLets do the proof with n=2 for ALL the equations\n\nLet x = 1/10 + 1/100 = 0.11\n\nThen 9x = 0.99\n\nBut, we also have 10x = 1 + 1/10 = 1.1 so 10x-x =0.11\n\nHence\n0.99 = 9x = 0.99 !!! yes\n[/quote]Of course. If x = .11, then ordinary arithmetic can be used to show that 9x = .99. So what?\n[QUOTE=alva]\n\nAnd, as anyone can see, you can repeat the proof for n=3,4\u2026 and using limits, when n->infinite, the result is clear, 1 is not = 0.999999\u2026[/QUOTE]\n\n13. micromass says:\n\n[QUOTE=\u201dgill1109, post: 5214887, member: 401042\u2033]Yes you are right. However the surreals do include numbers (lots and lots of them!) which are strictly between all of 0.99999\u20269 (any number of repetitions) and 1.[/QUOTE]\n\nThat depends how many repetitions of ##9## you take in ##0.999\u2026##. If you take countably many, then sure. But if you index over all ordinals, then no.\n\n14. gill1109 says:\n\n[QUOTE=\u201dmicromass, post: 5214863, member: 205308\u2033]Whether ##1=0.9999\u2026## in the surreals depends highly on the definitions for ##0.9999\u2026##. Some definitions make it equal, others don\u2019t.[/QUOTE]\nYes you are right. However the surreals do include numbers (lots and lots of them!) which are strictly between all of 0.99999\u20269 (any number of repetitions) and 1.\n\n15. micromass says:\n\n[QUOTE=\u201dgill1109, post: 5214572, member: 401042\u2033]Here is a number system in which 1 is not equal to 0.9999\u2026. and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway\u2019s \u201csurreal numbers\u201d. [URL]https://en.wikipedia.org/wiki/Surreal_number[/URL][/QUOTE]\n\nWhether ##1=0.9999\u2026## in the surreals depends highly on the definitions for ##0.9999\u2026##. Some definitions make it equal, others don\u2019t.\n\n16. Hornbein says:\n\n[QUOTE=\u201dmicromass, post: 5214231, member: 205308\u2033]micromass submitted a new PF Insights post\n\n[URL=\u2019https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/\u2019]Is There a Rigorous Proof Of 1 = 0.999\u2026?[/URL]\n\n[URL=\u2019https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/\u2019]Continue reading the Original PF Insights Post.[/URL][/QUOTE]\n\nI suggest 1 \u2013 0.9999\u2026 = 0.0000\u2026.\n\n17. KSG4592 says:\n\nAlways find these fascinating, even if most of it goes over my head at this point.\n\n18. Pro7 says:\n\nthe question which is obvious is\u2026.how far is 0.999\u2026 away from 1?\u2026since the answer is very close to 0 as it never ends\u2026therefore 0.999\u2026 tends to 1.\n\n19. Ilja says:\n\nI have a nice childhood memory about this. My school teacher had not checked this, and had the nice idea to write, instead of an open interval [0,1) a closed interval [0.0.9999\u2026]. I had heard somewhere that 1=0.999\u2026, so I simply said this. The teacher, and the whole class, disagreed. So a discussion started, and I had to think about how to prove it. What I invented was that 1/9 = 0.1111\u2026, 8/9 = 0.8888\u2026, but when we add 0.1111\u2026 to 0.88888\u2026 we will clearly get 0.9999\u2026. On the other side, 1/9+8/9 = 9/9 = 1.\n\nThe next day it became clear that I was right. Some guys have looked into some books.\n\nThe better proof is to compute 9*0.9999\u2026 = 10* 0.9999\u2026 \u2013 0.9999\u2026 = 9.999999..-0.99999\u2026 = 9.\n\n20. Anton Alice says:\n\nI think there is a problem with the step, where you take 10x-x =1.\nApplying operations on infinite sums has been considered dangerous, as far as I know.\n\n21. terryds says:\n\nHere is the \u2018silly\u2019 proof. 0.999\u2026 = 0.333\u2026+0.666\u2026 = 1/3 + 2/3 = 3/3 = 1\n\n22. H_A_Landman says:\n\nYou have to be careful, because Cantor\u2019s hierarchy of trans-finite numbers doesn\u2019t exist in the Surreal Numbers. For example,one might be tempted to identify Conway\u2019s $\\omega = {0,1,2,\u2026|}$ with Cantor\u2019s $\\aleph_0$, since they are both intuitively \u201cthe infinity of the integers\u201d. But in Cantor\u2019s system, $\\aleph_0$ is the smallest (positive) infinite number, whereas in Conway\u2019s there does not exist any such smallest infinity since we also have $\\omega-1$, $\\omega/2$, $\\sqrt{\\omega}$, etc. Conversely, in Cantor\u2019s system there is no largest infinity, but in Conway\u2019s there is (On).\n\n23. DMartin says:\n\nI said:\"I think there can be contradictions in mathematics\"You said:\"So you think 1+1=3\u00a0 is valid and can be proven? Because that is equivalent to what you just said.\"I don't think I need to comment on this, it's so ridiculous that it speaks for itself.\n\n24. DMartin says:\n\nI prefer to keep it in degrees throughout, and although you can subtract the 89, I won't. To explain it, it's another series of numbers that suggest something, but somewhat different from the other. It suggests that for any number like 89.99999, there's another number between it and 90, arrived at via x/(sine x). You can then add more 9s, and it still applies. However many 9s you add, this rule applies, so it seems very reasonable to say that it applies to 89.99999\u2026\u2026.. as well. That means there's at least one number between 89.9999\u2026.. and 90.Now because as I mentioned, infinities can be compared, this seems to reveal the existence of an infinity greater than the first one, or at least, a number or set of numbers that goes numerically higher. Certainly worth thinking about, but as I said, I think there can be contradictions in mathematics, so it's not an attempt to disprove your proof of 0.9999\u2026. = 1.\n\n25. DMartin says:\n\nI do believe in mathematics, but I just take a different view of it than you do. G\u00f6del's work meant that many others saw the edges of it, from the 1930s on, so my view isn't a controversial one.\n\n26. DMartin says:\n\nI never said it wasn't useful! It's very useful. And it doesn't matter that it's a bit rough round the edges either.\n\n27. DMartin says:\n\nIncidentally, and I'm not going to get into discussion about this, if one should show that there's always a number between 0.9999\u2026. and 1, that might suggest that they're different. Well I can show that there's always a number in between 89.9999\u2026\u2026 and 90. It's the expression x/(sine x) again, it always gives a number nearer 90, however many 9s there are.Now I should make my position clear \u2013 I'm not arguing that your proof of 0.9999\u2026. = 1 is false. I think mathematics is not a consistent system, and that it gets a bit rough round the edges, that's all. Perhaps I've shown that a bit.\n\n28. DMartin says:\n\nYou seem to think we should define things first, and then do our exploring. But sometimes the exploring helps to inform the definitions, and this is how mathematics has actually developed. This means one can point out a mathematical structure without necessarily having a rigorous definition. But a definition I've given for the relevant number, whether a loose one, and whether or not it fits with your definitions, is an angle x between 0 and 90 degrees such that x/(sin x) = exactly 180/\u03c0.\n\n29. DMartin says:\n\nBy the way, I did say, but probably should have emphasised more, that what applies to one pair of numbers doesn't necessarily apply to the other. There are things that happen at or near 0 that don't happen elsewhere. But this is surely relevant, nonetheless! Thanks for the discussion.\n\n30. DMartin says:\n\nI never said it was a proof, surely even you noticed that. I said it was relevant, and that the context can make a difference. The point that the context makes a difference is borne out by micromass above saying about the context of surreal numbers:\"Whether 1=0.9999\u2026\u00a0 in the surreals depends highly on the definitions for 0.9999\u2026 . Some definitions make it equal, others don't.\"So it's relevant to point out a sequence relating to the second pair of numbers I mentioned, in which there is a distinction between the two of them, because one infinite sequence lands somewhere different from the other.Incidentally, there have been distinctions drawn between different infinities, and it has turned out that they can be compared, and one infinity can turn out to be 'larger' than another. This might intuitively seem impossible, but ways to compare them were found. There is some loose similarity between that and what I've done.\n\n31. DMartin says:\n\nThat's what one would expect it to be. I've shown that the second pair isn't just 0 and 0.\n\n32. DMartin says:\n\nWell, there's a symmetry to be pointed out. First, you have 0.9999\u2026.. and 1, and these two numbers look the same, or almost the same. People discuss whether they're the same, and whether there's a proof that they are.Then, if you subtract 0.9999\u2026.. from 1, you find another pair of numbers \u2013 that is, the result of the subtraction, and zero. This pair of numbers can be called 1 \u2013 0.9999\u2026. , and 0.And there's a symmetry between these two pairs of numbers. Each pair may well be in the same relationship, whatever that is. So anything shedding a bit of light on either pair might be relevant. And showing the second pair to be different from each other is relevant.\n\n33. DMartin says:\n\nYes I knew it was 1 radian.\"This is by definition, not up for debate\". This implies that all our definitions are, by definition, correct.What you learned in primary school may or may not be true. But I have said that the context makes a difference sometimes.But the question of comparing a whole number, or a non-negative integer, and another nearby number that approaches it with an infinite series, is not always a clear cut question. And what I've set out has bearing on it, because I've show the two looking different, and looking existent.Don't forget that we learn as we go, the mathematical structures we have are not just a given eternal structure, they were put together bit by bit by finding things, and what we have is, as always, incomplete.\n\n34. DMartin says:\n\nSorry, our posts crossed. Yes, I know you think that whether or not a number exists is related to whether or not one can specify the positions of the digits. But the mathematical structure I've shown above hints at a relevant structure that is uncovered, and exists in some way other than just conforming to a made up set of rules.\n\n35. DMartin says:\n\nPutting in 0.0001, I get x/(sin x) = 57.295779513111409697664737311509try putting in 0.0000001then 0.00000000000000001.the result will approach 180/[pi], and to me this shows something that is discovered, rather than invented, and has bearing on the similar questions we've been looking at.\n\n36. DMartin says:\n\n\" 0.999999\u2026 exists because every digit in the decimal representation can be specified. If you ask, \"what digit is in the 12th place?\" Answer: 9. If you ask, \"what digit is in the 59th place?\" Answer: 9. If you ask, \"what digit is in the 623rd place?\" Answer: 9. No matter what specific digit you ask about, the answer is always \"9\". \"That's true, and you believe that it's significant.\n\n37. DMartin says:\n\nSee posts above. There's a series of increasingly small angles near zero degrees that approach a number at infinity, but that number is above zero. It's clear that this number gives x/(sin x) = 180/\u03c0, because the values approach that number. Whether or not any other values give 180/\u03c0 makes no difference, but it's interesting to hear it.\n\n38. DMartin says:\n\nWell, it's a matter of taste to some extent. You say you can prove that within a particular artificial system, a number 0.99999999\u2026. exists, but 1 minus that number, or 0.00000\u2026.0001 doesn't exist. And yet I've shown that the second number can be expressed as an angle between zero and 90 degrees.This has bearing on the question of whether mathematics is invented or discovered. Are we inventing the rules, or discovering them. When I look at the points above about the digit 1 in my number, I think there's a bit too much inventing going on.\n\n39. DMartin says:\n\nIf you say that one number exists and another doesn't, you need to say what you mean by exists. If you mean exists within mathematics, G\u00f6del showed that mathematics isn't necessarily a self-consistent system, so existing within mathematics isn't necessarily a meaningful concept.If you're not keen on how my number is expressed, perhaps you'd prefer it if I said:an angle \u03b8 such that \u03b8/(sin \u03b8) = exactly 180/\u03c0. I can prove that this angle is not zero, because \u03b8 = 0 gives a different result.\n\n40. DMartin says:\n\nI don't know what you mean by existence, when you say you can prove the existence of your number. But it seems clear enough that if we can talk about numbers with an infinite number of decimal places that all make a difference, then my number and yours are very similar. And my number and yours add up to one, which gives them more common ground.About the position of the 1 in my number, out of all those 9s of yours, there must be one that corresponds to my 1. So whatever problems I have with my number (and God knows it's hard to keep them all in line), you must have the same problems with yours.\n\n41. DMartin says:\n\nSo you have a rule that 'each numerical digit must have it's concrete position'. I suppose you know the positions of all the 9s in 0.99999\u2026.. then. But even if you argued that their positions are more concrete than the 1 in the number I used, it's not clear where the rule came from.what I've shown is a series that converges on, or approaches, a number at infinity, and the point is, whatever the existence status of that number at infinity, it isn't zero. And surely whatever its existence status, it's similar to the existence status of the numbers you're talking about.\n\n42. DMartin says:\n\nWell, that may be so, but this thread is a discussion on the basis that such numbers are worth talking about, so we're assuming they have some meaning before we start. If you say 'there is no such number', then presumably you think this whole thread is pointless.\n\n43. DMartin says:\n\nI can show that whatever the meaning of the number 0.0000000 => 00001, with an infinite number of zeros, it is different from 0. This means that other similar numbers are probably the same, though it does depend on the context. The method involves showing that the two numbers 0.0000000 => 00001 and 0 give completely different output numbers when put into an equation.To show that 0.0000000 => 00001 does not equal 0.Take the equation \u03b8/(sin \u03b8) , where \u03b8 is an angle in degrees.For \u03b8 = 0.0000000 => 00001, it gives \u03b8/(sin \u03b8) = 180/\u03c0 = 57.295779513082320876798154814105this is known because a series of increasingly small angles such as 0.0001, 0.0000000001, 0.0000000000000001 etc.give numbers that approach 180/\u03c0.But for \u03b8 = 0, it gives \u03b8/(sin \u03b8) = 0Therefore 0.0000000 => 00001 does not equal 0.Any thoughts would be appreciated, thanks.\n\n44. alva says:\n\nLetx=110+1102+1103+\u2026Then9x=0.999\u2026But, we also have 10x=1+110+1102+\u2026, so 10x\u2212x=1.This implies that x=19.Hence,0.999\u2026=9x=1\n\n45. WWGD says:\n\nYou can also use, though not as nice,\u00a0 the perspective of the Reals as a metric space, together with the Archimedean Principle:\u00a0 then d(x,y)=0 iff x=y. Let then ## x=1 , y=0.9999\u2026.\u00a0 ##Then ##d(x,y)=|x-y| ## can be made indefinitely small ( by going farther along the string of 9's ), forcing ## |x-y|=0## , forcing\u00a0 ##x=y ##. More formally, for any ##\\epsilon >0 ##, we can make ##|x-y|< \\epsilon ##\n\n46. gill1109 says:\n\nHere is a number system in which 1 is not equal to 0.9999\u2026. and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway's \"surreal numbers\". https://en.wikipedia.org/wiki/Surreal_number\n\n47. Josh Meyer says:\n\nNice!The informal proof I always share with people is that 1/9=0.111\u2026,\u00a0 2/9=0.222\u2026, 3/9=0.333\u2026, and so on until 9/9=0.999\u2026=1", "date": "2022-10-03 04:15:38", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/", "openwebmath_score": 0.8312886953353882, "openwebmath_perplexity": 968.2980687998975, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575121992375, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.858831163322621}}
{"url": "https://www.physicsforums.com/threads/the-integral.175225/", "text": "The Integral\n\n1. Jun 27, 2007\n\nanantchowdhary\n\nIs there a mathematical proof that can prove that the integral is the antiderivative?\n\n2. Jun 27, 2007\n\nStaff: Mentor\n\nfundamental theorem of calculus\n\nLook up the fundamental theorem of calculus.\n\n3. Jun 27, 2007\n\nanantchowdhary\n\numm...ive tried to study it....but my question is a little different....i meant to say..\n\nif we take out the area of the graph of a function f(x) bound at certain limits, then..is there any way to prove that the indefinite integral gives us the antiderivative of the function\n\nor rather..the area bound under two limits of the function is simply the limits applied to the antiderivative of the function\n\n4. Jun 27, 2007\n\nKummer\n\nGiven an integrable function $$f:[a,b]\\mapsto \\mathbb{R}$$ we define,\n$$g(x) = \\int_a^x f(t) dt$$ for all $$t\\in [a,b]$$.\nThen, $$g$$ is a continous function on $$[a,b]$$. And furthermore if $$f$$ is continous at $$t_0 \\in (a,b)$$ then $$g$$ is differentiable at $$t_0$$ with $$g'(x_0)=f(x_0)$$\n\n5. Jun 27, 2007\n\nmathwonk\n\nsuppose f is aN INCREASING CONTINUOUS FUNCTION. then the area under the graph between x and x+h is between f(x)h and f(x+h)h,\n\ni.e.f(x)h < A(x+h)-A(x) < f(x+h)h.\n\nsatisfy yourself of this by drawing a picture.\n\nnow the derivative of that area function is the limit of [A(x+h)-A(x)]/h, as h goes to zero.\n\nby the inequality above, this limit iscaught b etween f(x) and f(x+h), for all h, which emans, since f is continuious, it equals f(x). i.e. dA/dx= f(x).\n\nnow in creasing is not n eeded ubt makes it easier.\n\n6. Jun 27, 2007\n\nanantchowdhary\n\nthanks..ill give it a thought\n\n7. Jun 28, 2007\n\nHallsofIvy\n\nStaff Emeritus\nA little more generally- assume f(x)> 0 for a< x< b. For any number n, divide the interval from a to b into n equal sub-intervals (each of length (b-a)/n)). Construct on each a rectangle having height equal to the minimum value of f on the interval- that is, each rectangle is completely \"below\" the graph. Let An be the total area of those rectangles. Since each rectangle is contained in the region below the graph of f, it is obvious that $A_n \\le A$ where A is the area of that region (the \"area below the graph\").\n\nNow do exactly the same thing except taking the height to be the maximum value of f in each interval. Now, the top of each interval is above the graph of f so the region under f is completely contained in the union of all the rectangles. Taking An to be the total area of all those rectangles, we must have $A \\le A^n$.\n\nThat is, for all n, $A_n \\le A \\le A^n$\n\nIf f HAS an integral (if f is integrable) then, by definition, the limits of An and An must be the same- and equal to the integral of f from a to b. Since A is always \"trapped\" between those two values, the two limits must be equal to A: The area is equal to the integral of f from a to b.\n\nThat's pretty much the proof given in any Calculus book.\n\n8. Jun 28, 2007\n\nGib Z\n\n\"is there any way to prove that the indefinite integral gives us the antiderivative of the function\"\n\nI seemed to understand this as - ' is there any way to prove F(x), where F'(x)=f(x), is the same function as the function given by $$\\int f(x) dx$$.' The fundamental theorem of Calculus shows when we put upper and lower bounds, b and a, on the integral, it results in F(b) - F(a), however to the specific question as I understood, the dropping of bounds is just a nice notation for the integral/antiderivatve function.\n\n9. Jun 28, 2007\n\nanantchowdhary\n\numm...i tried this out..i Followed pretty much of it\n\nbut how do we prove that dA/dx=f(x)?\n\nthanks\n\n10. Jun 28, 2007\n\nmathwonk\n\nnotice the proof i gave for the increasing case, which is due to newtton, does not need the deep theorem that a continuous function always has a max and a min on a closed bounded interval.\n\nit also covers (when used piecewise) all polynomials, con tinuous rational functions, and continuous trig functions, that occur in practice. hence there is\n\nno reason for books to omit this proof or banish it to an appendix.\n\n11. Jun 28, 2007\n\nHallsofIvy\n\nStaff Emeritus\nDefine F(x) to be\n$$\\int_a^x f(t)dt$$\nFor f(x)> 0 we can interpret that as the area between the graph y= f(x) and the x-axis, from a to the fixed value x. Then F(x+h) is\n$$\\int a^{x+h}f(t)dt$$\nthe area between the graph y= f(x) and x-axis, from a to the fixed value x+h.\n\nNow F(x+h)- F(x) is the area between the graph y= f(x), between x and x+h. It's not difficult to see that is equal to the area of the rectangle with base x to x+h and height f(x*) where x* is some value between x and x+h: That is F(x+h)- F(x)= hf(x*). Then\n$$\\frac{F(x+h)- F(x)}{h}= f(x^*)$$\nTaking the limit as h goes to 0 forces x*, which is always between x and x+h, to go to x.\nTherefore\n$$\\frac{dF}{dx}= \\lim_{h\\rightarrow 0} \\frac{F(x+h)- F(x)}{h}= f(x)$$\n\n12. Jun 28, 2007\n\nmathwonk\n\nlet me combine halls discussion with mine in post 5.\n\nhe has shown, modulo the theorem that max and min exist,\n\nthat in my notation, if m(h) is the min of f on the interval [x,x+h], and if M(h) is the max,\n\nthat m(h)h < A(x+h)-A(x) < M(h)h.\n\nwhere < means less than or equal.\n\nThen again if we compute the derivative of A as the limit of\n[A(x+h)-A(x)]/h\n\nas h goes to 0, we see by the inequalities that [A(x+h)-A(x)]/h\nis squeezed between\n\nM(h) and m(h) for all h. since f is continuous, these numbers both apprioach f(x), so lim [A(x+h)-A(x)]/h\n= dA/dx = f(x).\n\n13. Jun 28, 2007\n\nmathwonk\n\nto prove the point halls says is \"not difficult to see\", uses the intermediate value theorem, in case you do not see it.\n\nLast edited: Jun 28, 2007\n14. Jun 28, 2007\n\nanantchowdhary\n\nthanks a lot for the proof!\n\n15. Jun 29, 2007\n\nHallsofIvy\n\nStaff Emeritus\n\"It is easy to see ...\" means \"I think there is a proof but I can't remember it just now\".\n\n\"Obvious to the most casual observer\" means \"I hope no one asks me to prove it\"!\n\n16. Jun 29, 2007\n\nHallsofIvy\n\nStaff Emeritus\n\"It is easy to see ...\" means \"I think there is a proof but I can't remember it just now\".\n\n\"Obvious to the most casual observer\" means \"I hope no one asks me to prove it\"!\n\n17. Jul 2, 2007\n\nssd\n\nThis is a beauty..... incomplete though. I believe,for a student new to calculus, it is enough to remove all mental barriers about the fact whether an integral is anti-derivative.\n\n18. Jul 2, 2007\n\njambaugh\n\nTechnically the indefinite integral is defined to be the anti-derivative (or rather a variable anti-derivative depending on an arbitrary constant since \"the\" anti-derivative is not unique.)\n\nThe definite integral is a value rather than a function so you can't call it an anti-derivative.\n\nThat technical nit picking aside see your undergraduate calculus text and the two forms of the FTC.", "date": "2017-01-20 16:14:16", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/the-integral.175225/", "openwebmath_score": 0.8898789286613464, "openwebmath_perplexity": 916.5494883311937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9825575137315161, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8588311549929101}}
{"url": "https://math.stackexchange.com/questions/2668002/we-square-an-integral-but-why-change-a-variable/2668062", "text": "# We square an integral, but why change a variable?\n\nIf we square an integral, we also change the integration variable in one of the integrals. But why is this actually correct?\n\nFor example, say I have the following:\n\nSolve $\\int_{-\\infty}^\\infty e^{-x^2} dx$. Let $I=\\int_{-\\infty}^\\infty e^{-x^2} dx$, so\n\\begin{align} I^2 &=\\bigg(\\int_{-\\infty}^\\infty e^{-x^2} dx\\bigg)^2\\\\ &= \\bigg(\\int_{-\\infty}^\\infty e^{-x^2} dx\\bigg) \\times \\bigg( \\underbrace{ \\int_{-\\infty}^{\\infty} e^{-y^2}dy }_{\\text{Why}?} \\bigg) \\\\ &=\\int_{-\\infty}^\\infty\\bigg(\\int_{-\\infty}^\\infty e^{-(x^2+y^2)} dx\\bigg)dy \\end{align}\n\nBut why is the following wrong: \\begin{align} I^2 &=\\bigg(\\int_{-\\infty}^\\infty e^{-x^2} dx\\bigg)^2\\\\ &= \\bigg(\\int_{-\\infty}^\\infty e^{-x^2} dx\\bigg) \\times \\bigg(\\int_{-\\infty}^{\\infty} e^{-x^2}dx\\bigg) \\\\ &=\\int_{-\\infty}^\\infty\\bigg(\\int_{-\\infty}^\\infty e^{-x^2-x^2} dx\\bigg)dx \\\\ &=\\int_{-\\infty}^\\infty\\bigg(\\int_{-\\infty}^\\infty e^{-2x^2} dx\\bigg)dx \\qquad ? \\end{align}\n\n\u2022 By definition $\\int_{a}^b f(x)dx=\\int_{a}^b f(y)dy$. For your second approach, no such rule for integrals exists. It abuses notation as you need $dx,dy$ to distinguish your variables when you rewrite the integrals in iterated form. \u2013\u00a0Alex R. Feb 26 '18 at 20:10\n\u2022 The second-to-last equality is wrong. It is not true that $$\\bigg(\\int_{-\\infty}^\\infty e^{-x^2} dx\\bigg) \\times \\bigg(\\int_{-\\infty}^{\\infty} e^{-x^2}dx\\bigg) =\\int_{-\\infty}^\\infty\\bigg(\\int_{-\\infty}^\\infty e^{-x^2-x^2} dx\\bigg)dx.$$ Indeed, the inner integral on the RHS $$\\left(\\int_{-\\infty}^{\\infty} e^{-2x^2}\\ dx\\right)$$ is equal to some positive constant, let's call it $c$. The outer integral is integrating the constant value $c$, from $-\\infty$ to $\\infty$, so the result is $\\infty$. On the other hand, the two factors on the LHS are finite, hence their product is finite. \u2013\u00a0Bungo Feb 26 '18 at 20:16\n\u2022 It is a dummy variable. \u2013\u00a0hamam_Abdallah Feb 26 '18 at 20:25\n\nThe short answer is that in your second to last identity you are neglecting cross terms in your multiplication.\n\nA simple example may be used to demonstrate the error, and we will use a strict summation instead of an integration for clarity. Consider the sum $$\\sum_{x=1}^3 x = 1 + 2 +3 = 6.$$ Now, squaring the summation yields \\begin{align} \\left(\\sum_{x=1}^3x\\right)^2 &= (1+2+3)^2 = (1+2+3)\\times(1+2+3) \\end{align} In order to properly calculate this quantity (long-hand) requires one to completely distribute including cross terms: \\begin{align} &\\ {\\color{white}+}\\ 1\\times 1 + 1\\times 2 + 1\\times 3 \\notag \\\\ (1+2+3)\\times(1+2+3) =& +2\\times 1 + 2\\times 2 + 2\\times 3 = 36. \\notag \\\\ & + 3\\times 1 + 3\\times 2 + 3\\times 3 \\end{align} However, if the cross terms were neglected one would obtain $$(1+2+3)\\times(1+2+3) \\ne (1^2+2^2+3^2) = 1\\times 1 + 2\\times 2 + 3\\times 3 = 14.$$ Written another way, we conclude that $$\\left(\\sum_{x=1}^3x\\right)^2 = \\left(\\sum_{x=1}^3x\\right)\\left(\\sum_{x=1}^3x\\right) \\ne \\sum_{x=1}^3x^2.$$ However, as noted in the comments, the $x$ in the summation (or the in the definite integral) is just a dummy variable, and may be replaced with another symbol, such as $y$: $$\\left(\\sum_{x=1}^3x\\right)^2 = \\left(\\sum_{x=1}^3x\\right)\\left(\\sum_{y=1}^3y\\right).$$ The reason for changing the dummy variable is that we may now say that the summation over $x$ and $y$ are independent, and so we may rearrange the summations: $$\\left(\\sum_{x=1}^3x\\right)^2 = \\left(\\sum_{x=1}^3x\\right)\\left(\\sum_{y=1}^3y\\right) = \\sum_{x=1}^3\\sum_{y=1}^3x\\times y.$$ Again, integration is just summation, and so the same fact holds true for integrals. Furthermore, it doesn't matter what the function inside the summation/integral is. Thus, we may write $$\\left(\\int_a^b f(x)\\ dx\\right)^2 = \\left(\\int_a^b f(x)\\ dx\\right)\\left(\\int_a^b f(y)\\ dy\\right) = \\int_a^b\\int_a^b f(x)\\times f(y)\\ dx\\ dy.$$ Using $f(x)=e^{-x^2}$ results in your example problem.\n\nBecause changing the integrand's variable doesn't change the value of the integral:$$\\int_{a}^{b}f(x)dx=\\int_{a}^{b}f(y)dy$$the same argument holds for summation as following$$\\sum_{n=a}^{b}k_n=\\sum_{m=a}^{b}k_m$$also we know that$$\\left(\\sum_{n=a}^{b}k_n\\right)^2=\\left(k_a+...+k_b\\right)^2=k_a^2+...+k_b^2+2k_ak_{a+1}+...+2k_{b-1}k_b$$and$$\\sum_{n=a}^{b}k_n\\sum_{m=a}^{b}k_m=(k_a+...+k_b)(k_a+...+k_b)$$which by expanding the terms and rearranging them leads to$$\\left(\\sum_{n=a}^{b}k_n\\right)^2=\\sum_{n=a}^{b}k_n\\sum_{m=a}^{b}k_m$$since integral is intrinsicly a summation, this can be generalized to integral operator either.\n\nI think the best way of visualizing what is happening intuitively is thinking in terms of programming.\n\nIf you make a program and you have two functions (in the programming sense) which do not interact, you can call \"$x$\" the variable in both functions. They are isolated.\n\nHowever, if a function is supposed to handle two different variables independently in the same context, you cannot name them the same thing.\n\nFormally and mathematically (and also changing notation from $\\int f(x)dx$ to $\\int f$ so that the point may become clearer), we have the following chain of equalities.\n\n\\begin{align*} \\underline{\\left(\\int_{\\mathbb{R}} x \\mapsto e^{-x^2}\\right)} \\cdot \\underline{\\left(\\int_{\\mathbb{R}} x \\mapsto e^{-x^2} \\right)}&\\stackrel{Linearity}=\\left(\\int_{\\mathbb{R}} \\underline{\\left(\\int_{\\mathbb{R}} x \\mapsto e^{-x^2}\\right)}\\left(x \\mapsto e^{-x^2} \\right)\\right) \\\\ &\\underset{\\text{fctn mult.}}{\\overset{\\text{Def. of}}{=}} \\int_{\\mathbb{R}} x \\mapsto \\left(e^{-x^2} \\underline{\\left(\\int_{\\mathbb{R}} y \\mapsto e^{-y^2}\\right)}\\right) \\\\ &\\underset{\\text{}}{\\overset{\\text{Linearity}}{=}} \\int_{\\mathbb{R}} x \\mapsto \\left(\\int_{\\mathbb{R}} y \\mapsto e^{-x^2-y^2}\\right) \\\\ &\\stackrel{Fubini}{=}\\int_{\\mathbb{R}^2}(x,y) \\mapsto e^{-x^2-y^2}. \\end{align*}\n\nUsing the same name ($x$) for the variables in both integrations is \"fine\" up to the third equality (I changed it in the second equality for the sake of clarity, otherwise the third would be a little mystic). Using $x$ instead of $y$ in the second line would be extremely bad taste, but not explicitly wrong. Using $x$ instead of $y$ in the right side of the first equality is edgy, but not so much.\n\nIndeed, each underline integral is a closed box with respect to the rest: they are fixed numbers, which don't interact at all with the rest of the environment other than via the fact that they are numbers. It is in the third equality that we are multiplying a constant (relative to the inner function which we are integrating) which depends on $x$. It does not depend on the function inside the integral at all, but it is invading its space and interacting with it. To be extremely clear, call the function $x \\mapsto e^{-x^2}$ by $f$. The second line is then $$\\int_{\\mathbb{R}} x \\mapsto \\left(e^{-x^2} \\left(\\int_{\\mathbb{R}} f\\right)\\right)$$ $$=\\int_{\\mathbb{R}} \\left( x \\mapsto \\int_{\\mathbb{R}} (e^{-x^2}\\cdot f)\\right),$$ where we multiplied a constant to the inner integral.\n\nWhat you are proposing is that the equality should read $$\\int_{\\mathbb{R}} x \\mapsto \\left(e^{-x^2} \\left(\\int_{\\mathbb{R}} f\\right)\\right)$$ $$=\\int_{\\mathbb{R}} \\left( x \\mapsto \\int_{\\mathbb{R}} g\\right),$$ where $g(t)=e^{-t^2}f(t)$. This doesn't make sense. Essentially, what you are implying is that $$f \\cdot\\int g=\\int f\\cdot g,$$ which is not true.", "date": "2021-08-02 00:46:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2668002/we-square-an-integral-but-why-change-a-variable/2668062", "openwebmath_score": 0.9992582201957703, "openwebmath_perplexity": 489.0145463317818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982557512709997, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.8588311524885167}}
{"url": "http://mathhelpforum.com/statistics/29518-probability-question.html", "text": "# Math Help - Probability question\n\n1. ## Probability question\n\nA man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E.\n\nFind the probability that both of the letters to A and B are in incorrect envelopes.\n\nThe answer is given as 13/20 but I can't figure out why. Can anyone help?\n\n2. Originally Posted by jjc\nA man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E. Find the probability that both of the letters to A and B are in incorrect envelopes. The answer is given as 13/20\nThe number of ways that A or B is in the correct envelope is give by:\n$P(A \\cup B) = P(A) + P(B) - P(A \\cap B) = \\frac{{2(4!) - 3!}}{{5!}}$\nNow, for neither to get the correct letter subtract from 1.\n\n3. Hello, jjc!\n\nA man writes 5 letters, one each to A, B, C, D and E.\nEach letter is placed in a separate envelope and sealed.\nHe then addressed the envelopes, at random, one each to A, B, C, D and E.\n\nFind the probability that both of the letters to A and B are in incorrect envelopes.\n\nAnswer: $\\frac{13}{20}$\nHere is a logical (but very primitive) approach . . .\n\nThe letters are: . $A,B,C,D,E$\n\nTheir envelopes are: . $a,b,c,d,e$\n\nThere are two cases to consider:\n. . (1) Letter $A$ is in envelope $b.$\n. . (2) Letter $A$ is not in envelope $b.$\n\n(1) Letter $A$ is in envelope $b$: . $P(\\text{A in b}) \\:=\\:\\frac{1}{5}$\n\nThen letter $B$ can go in any of the other four envelopes: $\\{a,c,d,e\\}$\n. . $P(\\text{B wrong}) \\:=\\:\\frac{4}{4} \\:=\\:1$\n\nHence: . $P(\\text{A in b }\\wedge\\text{ B wrong}) \\:=\\:\\frac{1}{5}\\cdot1 \\:=\\:{\\color{blue}\\frac{1}{5}}$\n\n(2) Letter $A$ is in not in envelope $b$ ... it is in $\\{c,d,e\\}$\n. . $P(\\text{A not in b}) \\:=\\:\\frac{3}{5}$\n\nThen letter B must be in one of the three remaining envelopes.\n. . $P(\\text{B wrong}) \\:=\\:\\frac{3}{4}$\n\nHence: . $P(\\text{A not in b} \\wedge\\text{ B wrong}) \\:=\\:\\frac{3}{5}\\cdot\\frac{3}{4} \\:=\\:{\\color{blue}\\frac{9}{20}}$\n\nTherefore: . $P(\\text{A wrong }\\wedge\\text{ B wrong}) \\;=\\;\\frac{1}{5} + \\frac{9}{20} \\;=\\;{\\bf{\\color{red}\\frac{13}{20}}}$", "date": "2015-09-05 15:04:19", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/29518-probability-question.html", "openwebmath_score": 0.8437401652336121, "openwebmath_perplexity": 679.9850713298704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9884918512747193, "lm_q2_score": 0.8688267881258485, "lm_q1q2_score": 0.8588282002315882}}
{"url": "https://www.intmath.com/forum/trigonometric-graphs-24/phase-shift:19", "text": "IntMath Home \u00bb Forum home \u00bb Graphs of the Trigonometric Functions \u00bb Phase shift\n\nPhase shift [Solved!]\n\nMy question\n\nHello, I have a question about phase shift. What makes the second approach incorrect?\n\nIt makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).\n\nHowever, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1\n\nThanks a lot!\n\nRelevant page\n\n3. Graphs of y = a sin(bx + c) and y = a cos(bx + c)\n\nWhat I've done so far\n\nDrawn several graphs, but couldn't conclude anything\n\nX\n\nHello, I have a question about phase shift. What makes the second approach incorrect?\n\nIt makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).\n\nHowever, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1\n\nThanks a lot!\nRelevant page\n\n<a href=\"/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php\">3. Graphs of <span class=\"noWrap\">y = a sin(bx + c)</span> and <span class=\"noWrap\">y = a cos(bx + c)</span></a>\n\nWhat I've done so far\n\nDrawn several graphs, but couldn't conclude anything\n\nRe: Phase shift\n\nHello Andrew\n\nI think the problem here is with the concept of \"starting off\". Let's use \"t\" as the variable and talk about when things occur.\n\nLet's also just talk about sin to keep it simple.\n\nIf we set the expression inside the sin (that is, bt + c) to 0, we are saying \"at what time does sin have value 0?\", since sin 0 = 0. Then by solving, we get t = -c/b.\n\nThe effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is \"starting off [from value 0]\" at t = -c/b.)\n\nNow let's think about y = a sin(bt+c) when the TIME is 0 (\"starting off when we flip the switch on\"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.\n\nDoes that help?\n\nI have added a new example on 3. Graphs of y = a sin(bx + c) and y = a cos(bx + c) - I hope that makes it clearer.\n\nRegards\n\nX\n\nHello Andrew\n\nI think the problem here is with the concept of \"starting off\". Let's use \"t\" as the variable and talk about when things occur.\n\nLet's also just talk about sin to keep it simple.\n\nIf we set the expression inside the sin (that is, bt + c) to 0, we are saying \"at what time does sin have value 0?\", since sin 0 = 0. Then by solving, we get t = -c/b.\n\nThe effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is \"starting off [from value 0]\" at t = -c/b.)\n\nNow let's think about y = a sin(bt+c) when the TIME is 0 (\"starting off when we flip the switch on\"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.\n\nDoes that help?\n\nI have added a new example on <a href=\"/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php\">3. Graphs of <span class=\"noWrap\">y = a sin(bx + c)</span> and <span class=\"noWrap\">y = a cos(bx + c)</span></a> - I hope that makes it clearer.\n\nRegards\n\nRe: Phase shift\n\nYour new example helped a lot.\n\nThanks.\n\nX\n\nYour new example helped a lot.\n\nThanks.\n\nYou need to be logged in to reply.", "date": "2017-06-23 22:14:07", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/trigonometric-graphs-24/phase-shift:19", "openwebmath_score": 0.7011805772781372, "openwebmath_perplexity": 837.3115089716097, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407152622596, "lm_q2_score": 0.8824278664544911, "lm_q1q2_score": 0.8588147279155187}}
{"url": "https://engineering.stackexchange.com/questions/37698/how-to-get-a-transfer-function-from-this-block-diagram", "text": "# How to get a transfer function from this block diagram?\n\nPardon my paint skills, I did my best\n\nMy attempt is short and seems to fail, I have no idea why:\n\n\\begin{align} \\alpha &= in - a_{1}\\beta - a_{2}\\gamma \\\\ \\beta &= \\alpha z^{-1} \\\\ \\gamma &= \\beta z^{-1} = \\alpha z^{-2} \\end{align}\n\ninputting 2nd and 3rd equation into the first one I get:\n\n\\begin{align} \\alpha &= in - a_{1}\\alpha z^{-1} - a_{2}\\alpha z^{-2} \\\\ in &= \\alpha + a_{1}\\alpha z^{-1} + a_{2}\\alpha z^{-2} \\end{align}\n\nI can write the output as:\n\n\\begin{align} out &= b_{2}\\gamma + b_{1}\\beta + b_{0}\\alpha \\\\ out &= b_{2}\\alpha z^{-2} + b_{1}\\alpha z^{-1} + b_{0}\\alpha \\end{align}\n\nI have output and input in terms of alpha, but I can't figure what to do from here.\n\nYou are going in the right direction! Lets take these two equations: $$(1) \\quad in = \\alpha+a_1\\alpha z^{-1}+a_2\\alpha z^{-2}$$ $$(2) \\quad out = b_0\\alpha+b_1\\alpha z^{-1}+b_2\\alpha z^{-2}$$ now rewrite (1) such that it becomes a function of $$\\alpha$$: $$in = \\alpha\\left(1+a_1z^{-1}+a_2z^{-2}\\right)$$ $$\\alpha = \\frac{in}{1+a_1z^{-1}+a_2z^{-2}}$$ Substitute $$\\alpha$$ in equation (2): $$out = in\\frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$ And derive proper discrete transfer function from it: $$H(z) = \\frac{out}{in} = \\frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$\n\n\u2022 Thank you kind sir. I'm ashamed I didn't figure it out myself. But you made a little mistake I believe, in the second (2) equation the last term should have b2 instead of a2, and that mistakes propagates to the results as well Sep 17 '20 at 11:14\n\u2022 Oops, small copy-past error haha. fixed it! Sep 17 '20 at 11:15\n\nJust wanted to add that you could use this to draw diagrams next time. Much easier to use than paint. You can even use Latex.\n\n\u2022 This really should be posted as a comment and not an answer. Oct 13 '20 at 15:19\n\u2022 Don't have the reputation to do so, just need 4 more to get to 50 lol Oct 13 '20 at 19:08\n\u2022 Just one good answer will get you there! Oct 13 '20 at 22:13\n\u2022 If you used this website to draw a new diagramm for thr op and then answer the OPs question i\u2019d definitely +1 Oct 14 '20 at 8:59", "date": "2021-12-05 23:15:23", "meta": {"domain": "stackexchange.com", "url": "https://engineering.stackexchange.com/questions/37698/how-to-get-a-transfer-function-from-this-block-diagram", "openwebmath_score": 0.9998773336410522, "openwebmath_perplexity": 925.4827113472511, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407160384083, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.8588147270958038}}
{"url": "https://math.stackexchange.com/questions/2093600/why-is-the-following-relation-transitive-but-not-reflexive", "text": "# Why is the following relation transitive, but not reflexive?\n\nIn a practice paper for an exam there is the following relation:\n\n$$E = \\{(1,1),(2,2),(3,3),(4,4)\\}\\ \\text{ on the set }V = \\{1,2,3,4,5\\}$$\n\nIt would appear that because $(5,5)$ is not in $E$, that it would not be a reflexive relation.\n\nWhat is unclear is how the relation is still transitive. By our definition for Transitivity, for all elements in set $V$, there would have to be $(a,b)$ and $(b,c)\\ldots$ and therefore $(a,c)$ in the set $E$.\n\nThis works for element $1$ in $V$ because you have $(1,1)$ and $(1,1)$ and therefore $(1,1)$. I would have assumed however that because there is no tuple $(5,5)$ in $E$ the entire relation could not be transitive?\n\nHope this makes sense, thanks for any input offered.\n\n\u2022 \"By our definition for Transitivity, for all elements in set V, there would have to be (a,b) and (b,c)\" No. This is not the definition. ( a,b) do not have to be in E for all a,b. But IF (a,b) does exist, AND (b,c) exist, then (a,c) must also exist. But if (a,b) does not exist... no harm, no foul. [Note: if (a,b) did have to exist for a,b in V, then E has to be V x V = {(1,1)(1,2)(1,3)..... (5,3),(5,4)(5,5)} all, every single one of the 25 pairs. \u2013\u00a0fleablood Jan 11 '17 at 17:35\n\nReflexive means for all $a \\in V$ then $(a,a) \\in E$. That fails because $5 \\in V$ and $(5,5) \\not \\in V$.\n\nTransitive says nothing about items of $V$ but only about items in $E$. The relation is transitive if every time you have an $(a,b) \\in E$ and a $(b,c) \\in E$ you must also have a $(a,c) \\in E$ then the relation is transitive. If you never have an $(5,x)$ or $(y, 5) \\in E$ that will in no way affect what you do have in $E$.\n\nThis particular relationship if you have $(a,b) \\in E$ then $a \\ne 5; b\\ne 5;$ and $a = b$. So if $(a,b), (b,c) \\in E$ then $a = b = c$ and $(a,c) = (a,a) = (a,b) \\in E$. So the relationship is transitive. That $a,b,c$ can never be equal to $5$ is utterly irrelevant.\n\n\u2022 I never thought about it, but reflexive is the only condition that pertains to the elements of V. Symmetry pertains only to elements of E in that if you have this element in V you must have that element too. Transitivity says if you have two elements in V you must also have a third. A function could still be transitive if you have (a,b) but absolutely no (b,x). For transitivity to fail you must have both an (a,b) and an (b,c) but not the (a,c). If you don't the (a,b) or you don't have the (b,c) it doesn't matter whether or not you have the (a,c). So no (5,x) or (x,5) doesn't matter. \u2013\u00a0fleablood Jan 11 '17 at 17:51\n\u2022 Yes, @fleablood. In your last point, when you do not have some elements a, b, c, in V, such that $(a, b), (b, c) \\in E$, it is a vacuously true, that transitivity holds regardless of whether or not $(a, c)$ is, or is not, in $E$. Also, as you seem to have gleaned, it is helpful to examine failures of properties of relations, for which the property fails. If it does not fail, the property holds. \u2013\u00a0Namaste Jan 11 '17 at 18:16\n\u2022 Hmmm, transitivity does not require three elements in some set $V$ of a transitive relation $E$. If $E = \\{(a, a)\\}$, E is indeed transitive, (and symmetric). \u2013\u00a0Namaste Jan 11 '17 at 18:23\n\u2022 I didn't mean to imply transitivity required 3 elements. As you point out if there are no (a,b) (b,c) pairs it's vacuously transitive. If it's non-vacuously transitive it must have (a,b) and (b,c) and a cooresponding (a,c). But a, b, and c need not be distinct. So (a,a) (a,a) and (a,a) will do. \u2013\u00a0fleablood Jan 11 '17 at 22:11\n\nDefinition: $R$ is transitive if and only if $\\forall x,y (R(x,y)$ and $R(y,x))\\Rightarrow R(x,z)$\n\nCan you point to where exactly you think this definition fails? For $x=y=5$ the antecedent fails to hold (since $R(5,5)$ is false) and therefore the implication is true, since an implication with a false antecedent is always true.\n\nYour definition of transitivity is unclear as stated, and you have misunderstood it, there is no requirements on the members of the underlying set, only on the elements of the relation.\n\nProof $E$ is transitive: if $E\\left(a,\\,b\\right)\\land E\\left(b,\\,c\\right)$ then $a=b\\in V\\backslash \\left\\{ 5\\right\\}$ and similarly $b=c\\in V\\backslash \\left\\{ 5\\right\\}$, so $a=c\\in V\\backslash \\left\\{ 5\\right\\}$ and $E\\left(a,\\,c\\right)$.", "date": "2019-07-24 00:18:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2093600/why-is-the-following-relation-transitive-but-not-reflexive", "openwebmath_score": 0.7098208665847778, "openwebmath_perplexity": 255.9768397819038, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407175907054, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.858814722447153}}
{"url": "http://math.stackexchange.com/questions/566232/mirror-reflection", "text": "# mirror reflection\n\n1. The problem statement, all variables and given/known data\n\nQuestion A. Draw accurately in the figure the light beam that goes from Object A to eye B after being reflected on the mirror. It must be consistent with the mirror principle!\n\nQuestion B. At question A. you may have connected the point behind the mirror (A') with eye B. And found the correct light path that way.\n\nWill that also work if you use the point behind the mirror of B (B') instead? Do you get the same result? Use mathematical arguments for your judgement!\n\n2. Relevant equations mirror reflection rules\n\n3. The attempt at a solution\n\nQuestion A: I tried to draw point A' and connected that to B. The purple line is the normal line (the imaginary line that is perpendicular to the mirror) And the angle between the mirror and A is the same as the angle between the mirror and B.\n\nQuestion B: I think this is true, I can draw the same lines, but I don't know what is meant by 'mathematical arguments'\n\n-\nThis is a very nicely written question. If only we had more like this one... :-) \u2013\u00a0Vedran \u0160ego Nov 14 '13 at 4:14\n\nNotice that $AA'B'B$ is an isosceles trapezoid, so its diagonals intersect on the mirror, which means that using either $\\triangle AA'B$ or $\\triangle AB'B$ will give you the same point.\n\nYou'll probably need to provide a bit more arguments that the diagonals intersect on the mirror. Read the linked article and ask if you get stuck.\n\nEdit: Here is a bit more:\n\nWe can extend $AA'B'B$ to a triangle. Let us denote the new vertex as $C$. It is on the intersection of the lines on which $\\overline{AB}$ and $\\overline{A'B'}$ lie.\n\nNote that\n\n1. $|\\overline{AA''}| = |\\overline{A'A''}|$, where $A''$ denotes a point at which $\\overline{AA'}$ intersects with the mirror; and\n2. the mirror (on which the altitude of the mirror lies) is orthogonal to $\\overline{AA'}$.\n\nFrom points $1$ and $2$, we see that $\\triangle AA'C$ is an isosceles triangle. The same can be said for $\\triangle BB'C$ (with the same arguments).\n\nNow, since\n\n$$|\\overline{AB}| = |\\overline{AC}| - |\\overline{BC}| = |\\overline{A'C}| - |\\overline{B'C}| = |\\overline{A'B'}|,$$\n\nwe see that $AA'B'B$ is an isosceles trapezoid.\n\nUsing this, show that the diagonals of $AA'B'B$ intersect on the mirror. Can you take over now?\n\n-\nDo you mean with a bit more arguments to proof Similarity (geometry) between the triangles? \u2013\u00a0user108681 Nov 14 '13 at 4:36\n@user108681 I've expanded my answer. \u2013\u00a0Vedran \u0160ego Nov 14 '13 at 12:15\nA bit easier: you can just observe isosceles triangles $\\triangle AA'E$ and $\\triangle BB'E$, where $E$ is the point on the mirror where $\\overline{AB'}$ and $\\overline{A'B}$ intersect (i.e., the point you're looking for). \u2013\u00a0Vedran \u0160ego Nov 14 '13 at 18:31", "date": "2016-07-24 22:27:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/566232/mirror-reflection", "openwebmath_score": 0.7831434607505798, "openwebmath_perplexity": 167.11342404346294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407191430025, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.8588147208077227}}
{"url": "https://math.stackexchange.com/questions/1900072/why-is-minimizing-least-squares-equivalent-to-finding-the-projection-matrix-ha", "text": "# Why is minimizing least squares equivalent to finding the projection matrix $\\hat{x}=A^Tb(A^TA)^{-1}$?\n\nI understand the derivation for $\\hat{x}=A^Tb(A^TA)^{-1}$, but I'm having trouble explicitly connecting it to least squares regression.\n\nSo suppose we have a system of equations: $A=\\begin{bmatrix}1 & 1\\\\1 & 2\\\\1 &3\\end{bmatrix}, x=\\begin{bmatrix}C\\\\D\\end{bmatrix}, b=\\begin{bmatrix}1\\\\2\\\\2\\end{bmatrix}$\n\nUsing $\\hat{x}=A^Tb(A^TA)^{-1}$, we know that $D=\\frac{1}{2}, C=\\frac{2}{3}$. But this is also equivalent to minimizing the sum of squares: $e^2_1+e^2_2+e^2_3 = (C+D-1)^2+(C+2D-2)^2+(C+3D-2)^2$.\n\nI know the linear algebra approach is finding a hyperplane that minimizes the distance between points and the plane, but I'm having trouble understanding why it minimizes the squared distance. My intuition tells me it should minimize absolute distance, but I know this is wrong because it's possible for there to be non-unique solutions.\n\nWhy is this so? Any help would be greatly appreciated. Thanks!\n\n\u2022 Check your derivation for $\\hat{x}$ and you will find (as @Ian points out) the expression should be $\\hat{x}=(A^TA)^{-1}A^Tb$. Perhaps you should edit this into the Question, though it may be the crux of why you had trouble \"connecting it to least squares regression\" (minimization). Aug 22 '16 at 15:00\n\nYou should be multiplying by $(A^T A)^{-1}$ on the left, not the right. Anyway, the geometric point is that you want $Ax-b$ to be perpendicular to $Ay$ for every vector $y$. (I think this is most easily seen by a geometric argument, which can be easily found in books, but which I can't easily render here.) This translates to $(Ay)^T(Ax-b)=0$ for every $y$, which is the same as $y^T(A^T(Ax-b))=0$ for every $y$. This can only happen if $A^T(Ax-b)=0$, which rearranges to your form if $A^T A$ is invertible (as is usually the case).\n\nAlso, it is the same to minimize the square of the Euclidean distance as it is to minimize the Euclidean distance itself. (This is also true of any other nonnegative quantity.) What would be different is minimizing some other distance, like the \"taxicab\" distance (where you sum the absolute values). Why we should choose to minimize the Euclidean distance in the first place is not a purely mathematical question, it depends on where the problem is coming from. That question is a bit off-topic here, though, and has also been asked before on MSE. (The short version of that discussion: \"it's mathematically convenient\" and \"see the Gauss-Markov theorem\".)\n\n\u2022 About minimizing the Euclidean distance: I'm a little confused about the distinction. For example, least squares would penalize a point for being farther away: e.g. a point 1 unit away would be penalized by 1 whereas a point 2 units away would be penalized by 4. I think my confusion lies in the difference between minimizing an aggregate of points vs individual ones, since it seems to me that minimizing euclidean distance for each point wouldn't account for the distance factor in least squares. Aug 22 '16 at 15:30\n\u2022 In your problem the goal is to choose a vector in a certain 2D subspace which is closest to another vector in 3D space, where you measure distance in the Euclidean sense. This is not measured separately by components but rather through the usual distance formula.\n\u2013\u00a0Ian\nAug 22 '16 at 15:39\n\nThe matrix has full column rank; we are guaranteed a unique solution.\n\nProblem statement\n\n\\begin{align} \\mathbf{A} x &= b \\\\ \\left[ \\begin{array}{cc} 1 & 1 \\\\ 1 & 2 \\\\ 1 & 3 \\\\ \\end{array} \\right] % \\left[ \\begin{array}{cc} x_{1} \\\\ x_{2} \\\\ \\end{array} \\right] &= \\left[ \\begin{array}{cc} 1 \\\\ 2 \\\\ 2 \\\\ \\end{array} \\right] \\end{align}\n\nNormal equations\n\n\\begin{align} \\mathbf{A}^{*} \\mathbf{A} x &= \\mathbf{A}^{*} b \\\\ % \\left[ \\begin{array}{ccc} 1 & 1 & 1 \\\\ 1 & 2 & 3 \\\\ \\end{array} \\right] % \\left[ \\begin{array}{cc} 1 & 1 \\\\ 1 & 2 \\\\ 1 & 3 \\\\ \\end{array} \\right] % \\left[ \\begin{array}{cc} x_{1} \\\\ x_{2} \\\\ \\end{array} \\right] &= \\left[ \\begin{array}{ccc} 1 & 1 & 1 \\\\ 1 & 2 & 3 \\\\ \\end{array} \\right] % \\left[ \\begin{array}{cc} 1 \\\\ 2 \\\\ 2 \\\\ \\end{array} \\right] \\\\[3pt] % \\left[ \\begin{array}{cc} 3 & 6 \\\\ 6 & 14 \\\\ \\end{array} \\right] % \\left[ \\begin{array}{cc} x_{1} \\\\ x_{2} \\\\ \\end{array} \\right] % &= % \\left[ \\begin{array}{cc} 5 \\\\ 11 \\\\ \\end{array} \\right] % \\end{align}\n\nLeast squares solution\n\n\\begin{align} x_{LS} &= \\left( \\mathbf{A}^{*} \\mathbf{A} \\right)^{-1} \\mathbf{A}^{*} b \\\\ % \\left[ \\begin{array}{cc} x_{1} \\\\ x_{2} \\\\ \\end{array} \\right] &= \\frac{1}{6} \\left[ \\begin{array}{cc} 14 & -6 \\\\ -6 & 3 \\\\ \\end{array} \\right] % \\left[ \\begin{array}{cc} 5 \\\\ 11 \\end{array} \\right] \\\\ % &= \\frac{1}{6} \\left[ \\begin{array}{cc} 4 \\\\ 3 \\end{array} \\right] % \\end{align}", "date": "2021-09-20 08:15:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1900072/why-is-minimizing-least-squares-equivalent-to-finding-the-projection-matrix-ha", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 151.7890099958561, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407160384082, "lm_q2_score": 0.8824278540866547, "lm_q1q2_score": 0.8588147165635318}}
{"url": "https://math.stackexchange.com/questions/155839/on-ces%C3%A0ro-convergence-if-x-n-to-x-then-z-n-fracx-1-dots-x-nn?noredirect=1", "text": "# On Ces\u00e0ro convergence: If $x_n \\to x$ then $z_n = \\frac{x_1 + \\dots +x_n}{n} \\to x$\n\nI have this problem I'm working on. Hints are much appreciated (I don't want complete proof):\n\nIn a normed vector space, if $x_n \\longrightarrow x$ then $z_n = \\frac{x_1 + \\dots +x_n}{n} \\longrightarrow x$\n\nI've been trying adding and subtracting inside the norm... but I don't seem to get anywhere.\n\nThanks!\n\nGiven $\\epsilon >0$ there exists $n_0$ such that if $n\\geq n_0$ then $\\parallel x_n -x\\parallel < \\epsilon$\n\nso\n\n\\begin{align*} 0 & \\leq \\left\\lVert \\frac{x_1 +\\cdots +x_n}{n} -x \\right\\rVert \\leq \\left\\lVert \\frac{x_1 + \\dots + x_n - nx }{n} \\right\\rVert \\\\ & \\leq \\frac{\\lVert x_1 - x \\rVert}{n} + \\dots + \\frac{\\lVert x_{n_0 - 1} - x \\rVert}{n} + \\frac{\\lVert x_{n_0} - x \\rVert}{n} +\\dots + \\frac{\\lVert x_{n} - x \\rVert}{n} \\\\ &\\le \\frac 1n\\sum_{i=1}^{n_0-1} \\| x_i -x\\| + \\frac{n-n_0}{n} \\epsilon \\end{align*}\n\nThe first $n_0 -1$ terms $\\| x_i -x\\|$ can be bounded by some $M$, thus for $n\\ge (n_0-1)M/\\epsilon=: N_0$ we have $$\\frac 1n\\sum_{i=1}^{n_0-1} \\| x_n -x\\| \\le \\frac 1n (n_0-1)M \\le \\epsilon$$\n\nThus $$\\left\\| \\frac{x_1 + \\cdots x_n}{n} - x\\right\\| <2\\epsilon$$ when $n\\ge N_0$.\n\nThanks a lot @Leonid Kovalev for the inspiration, though my main problem was that I wasn't aware of what to do with the $nx$ (the silliest part :P)\n\n\u2022 This sort of thing is encouraged, I think. \u2013\u00a0Dylan Moreland Jun 11 '12 at 3:21\n\u2022 I think you want \"$-nx$\" in the first line of the display, and $\\|x_n - x\\|$ at the end. I don't think you want to say that the first few terms are $\\leq M$; that doesn't seem to be enough. \u2013\u00a0Dylan Moreland Jun 11 '12 at 3:41\n\u2022 @DylanMoreland: Why isn't it enough? Can you explain? \u2013\u00a0Bouvet Island Jun 11 '12 at 14:56\n\u2022 @Inti: I think you are missing something in the argument. Generally the argument consists of two steps: first choose $n_0$ such that if $n \\geq n_0$, $\\|x_n -x \\| < \\epsilon / 2$. Next choose $N \\geq n_0$ such that $M$ (which is at least $\\sup_n \\|x_n\\|$) satisfies $M / N < \\epsilon / 2$. I don't see the second part of the argument implemented in your answer. \u2013\u00a0Willie Wong Jun 11 '12 at 14:59\n\u2022 Should it not be $(n-n_0+1)\\epsilon/n$? \u2013\u00a0jippyjoe4 Oct 14 '19 at 4:19\n\nThere is a slightly more general claim:\n\nPROP Let $$\\langle a_n\\rangle$$ be a sequence of real numbers, and define $$\\langle \\sigma_n\\rangle$$ by $$\\sigma_n=\\frac 1 n\\sum_{k=1}^n a_k$$\n\nThen $$\\liminf_{n\\to\\infty}a_n\\leq \\liminf_{n\\to\\infty}\\sigma_n \\;(\\;\\leq\\;)\\;\\limsup_{n\\to\\infty}\\sigma_n\\leq \\limsup_{n\\to\\infty}a_n$$\n\nP We prove the leftmost inequality. Let $$\\ell =\\liminf_{n\\to\\infty}a_n$$, and choose $$\\alpha <\\ell$$. By definition, there exists $$N$$ such that $$\\alpha for any $$k=0,1,2,\\ldots$$ If $$m>0$$, then $$m\\alpha <\\sum_{k=1}^m \\alpha_{N+k}$$\n\nwhich is $$m\\alpha<\\sum_{k=N+1}^{N+m}a_k$$\n\n$$(m+N)\\alpha+\\sum_{k=1}^{N}a_k<\\sum_{k=1}^{N+m}a_k+N\\alpha$$\n\nwhich gives\n\n$$\\alpha+\\frac{1}{m+N}\\sum_{k=1}^{N}a_k<\\frac{1}{m+N}\\sum_{k=1}^{N+m}a_k+\\frac{N}{m+N}\\alpha$$\n\nSince $$N$$ is fixed, taking $$\\liminf\\limits_{m\\to\\infty}$$ gives $$\\alpha \\leq \\liminf\\limits_{m \\to \\infty } \\frac{1}{m}\\sum\\limits_{k = 1}^m {{a_k}}$$ (note that $$N+m$$ is just a shift, which doesn't alter the value of the $$\\liminf^{(1)}$$). Thus, for each $$\\alpha <\\ell$$, $$\\alpha \\leq \\liminf\\limits_{m \\to \\infty } \\frac{1}{m}\\sum\\limits_{k = 1}^m {{a_k}}$$ which means that $$\\liminf_{n\\to\\infty}a_n\\leq \\liminf_{n\\to\\infty}\\sigma_n$$ The rightmost inequality is proven in a completely analogous manner. $$\\blacktriangle$$.\n\n$$(1)$$: Note however, this is not true for \"non shift\" subsequences, for example $$\\limsup_{n\\to\\infty}(-1)^n=1$$ but $$\\limsup_{n\\to\\infty}(-1)^{2n+1}=-1$$\n\nCOR If $$\\lim a_n$$ exists and equals $$\\ell$$, so does $$\\lim \\sigma_n$$, and it also equals $$\\ell$$. The converse is not true.\n\nWLOG, the $x_n$ converge to $0$ (otherwise consider the differences $x_n-x$), and stay confined in an $\\epsilon$-neighborhood of $0$ after $N_\\epsilon$ terms.\n\nThen the average of the first $m$ terms is bounded by\n\n$$\\frac{N\\overline{x_N}+(m-N)\\epsilon}m,$$ which converges to $\\epsilon$. So you can make the average as close to $0$ as you like.", "date": "2021-05-16 03:33:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/155839/on-ces%C3%A0ro-convergence-if-x-n-to-x-then-z-n-fracx-1-dots-x-nn?noredirect=1", "openwebmath_score": 0.998012125492096, "openwebmath_perplexity": 528.3721198197795, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126432392879, "lm_q2_score": 0.8774767986961401, "lm_q1q2_score": 0.8587976370330478}}
{"url": "http://math.stackexchange.com/questions/575960/how-many-triangles-in-picture", "text": "# How many triangles in picture?\n\nCan you tell me how many triangles are in this picture? I've counted 96, not sure that I got right answer.\n\n-\nPerhaps you could try counting them by using the symmetry of the shape (if you haven't already). \u2013\u00a0 Alice Nov 21 '13 at 13:50\nClearly there are 19 unique points. Number of triangles that can be formed from 19 points taking 3 at a time=$19\\choose3$. Also note that some of the points lie on a straight line. When we took $19\\choose3$ we included the ones that lie on a straight line. Perhaps finding the number of lines and subtracting that from $19\\choose3$ will yield the answer. \u2013\u00a0 GTX OC Nov 21 '13 at 14:05\n@GTXOC I see 19 unique points... \u2013\u00a0 apnorton Nov 21 '13 at 14:08\nAlright I missed the ones in the middle of the base. \u2013\u00a0 GTX OC Nov 21 '13 at 14:09\n@All: Not all triangles are valid even if the points could form one, consider lines between \"star-corners\". \u2013\u00a0 AlexR Nov 21 '13 at 14:11\n\nLet's see! We'll break it up into the following cases (where an apex means one of the six points of the star):\n\nA: Triangles that contain three apexes\nThere are 2 of these.\n\nB: Triangles that contain a pair of opposite apexes\nFor each such pair, there are 2 of these, so 6 in total.\n\nC: Triangles that contain a pair of non-opposite apexes\nFor each non-adjacent apex pair, there are three of these, so 18 in total.\n\nD: Triangles that contain exactly one apex\nFor each apex, I count 10 such triangles, so 60 in total.\n\nE: Triangles that contain no apex\nFor each edge of the internal hexagon, there are 3 triangles, so 18 in total.\n\nI make that 2 + 6 + 18 + 60 + 18 = 104.\n\n-\nI verified and this seems to be correct. \u2013\u00a0 AlexR Nov 21 '13 at 14:16\nLook OK to me too, though categories $D$ and $E$ could be somewhat refined. \u2013\u00a0 Marc van Leeuwen Nov 21 '13 at 14:24\n@Marc: I agree with you about category D. \u2013\u00a0 TonyK Nov 21 '13 at 14:29\nA small reality check: If you dismiss the two big equilateral triangles, all other triangles belong to orbits of $6$ under rotations by $2\\pi/6$, so $T-2$ must be divisible by $6$, which $T=104$ is but $T=96$ (the OP's original count) is not. \u2013\u00a0 Barry Cipra Nov 21 '13 at 14:54\nFor Category $D$, you can take the apex at the very top and count triangles that are and aren't symmetric across the \"$y$\"-axis. There are $2$ that are and $4$ pair that aren't. \u2013\u00a0 Barry Cipra Nov 21 '13 at 15:13\n\nI only get 98. I am missing the 3rd triangle under section C. I only see 2 triangles that contain a pair of non-opposite apexes.\n\n-", "date": "2015-08-02 04:26:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/575960/how-many-triangles-in-picture", "openwebmath_score": 0.7866086363792419, "openwebmath_perplexity": 627.3347327734683, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126469647338, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8587976340302823}}
{"url": "https://math.stackexchange.com/questions/2965619/how-to-solve-the-following-congruences-equation-system", "text": "# How to solve the following congruence's equation system?\n\nThrough many operations in an exercise I've reached this point\n\n$$17\u22614a+19b(mod$$ $$27)$$\n\n$$8\u226111a(mod$$ $$27)$$\n\nI want to find both a and b to have the answer I need, I've been trying to use the Chinese rest theorem but found it doesn't work in my case, I'm starting converting the second one into\n\n$$11a+27a\u22611(mod$$ $$27)$$\n\nDespite this I'm not sure if is correct and I still don't know how to clear this $$a$$ in the second equation in order to replace it in the first.\n\nAny help will be really appreciated\n\nThere are a number of ways to go about this. We want to be able to \"divide\" by $$11$$, so we need $$8 + 27x = 11y$$. Solving this linear Diophantine equation (an elementary problem, feel free to Google) gives $$x = 5$$ and $$y = 13$$ as possible solutions. Then, \\begin{align} 8 &\\equiv 11a \\mod{27}\\\\ 8 + 27\\cdot 5 &\\equiv 11a \\mod{27}\\\\ 13(11) &\\equiv 11a \\mod{27}\\\\ 13 &\\equiv a \\mod{27} \\end{align} Then $$a = 13 + 27k$$ for some $$k \\in \\mathbb{Z}$$. You can subtract $$4$$ times the last congruence above from $$17 \\equiv 4a + 19b \\mod{27}$$ to get \\begin{align} 17 - 4(13) &\\equiv 19b \\mod{27}\\\\ -35 &\\equiv 19b \\mod{27}\\\\ -35 + 2(27) &\\equiv 19b \\mod{27}\\\\ 19 &\\equiv 19b \\mod{27}\\\\ b &\\equiv 1 \\mod{27} \\end{align}\n\n\u2022 But you should say how you computed $k$ such that $\\,11\\mid 8+27k,\\,$ not simply pull it out of hat like magic! Else how is the OP supposed to figure out how to use the method for similar problems? \u2013\u00a0Bill Dubuque Oct 22 '18 at 19:11\n\u2022 @BillDubuque I alluded to it in my explanation... Solving a linear Diophantine equation is usually a skill that's learned in sections immediately preceding problems like this. For the record though, the Euclidean algorithm with back substitution is the most systematic way of doing it (again, Google is your friend). \u2013\u00a0AlkaKadri Oct 22 '18 at 19:59\n\u2022 Back-substitution is clumsy and error-prone. Better to use this version, or Gauss's algorithm below $$\\!\\bmod 11\\!:\\ \\, 27x\\!+\\!8\\equiv 0\\iff x\\equiv \\dfrac{-8}{27}\\equiv \\dfrac{3}{-6}\\equiv \\dfrac{-1}{2}\\equiv \\dfrac{10}2\\equiv 5\\$$ In any case, it's not a good idea to say \"Google it\" in an answer. Google can locate all sorts of nonsense (low-quality, errorneous, crankish, etc). \u2013\u00a0Bill Dubuque Oct 22 '18 at 20:22\n\u2022 @BillDubuque ah I see.. Yes you have a point there, I\u2019ll certainly refrain from doing that in future answers. Thanks Bill! \u2013\u00a0AlkaKadri Oct 22 '18 at 20:25\n\nJust so long as you avoid multiplying or dividing by a multiple of $$3$$ you can use ordinary arithmetic - so you can multiply the first by $$11$$ and the second by $$4$$ to isolate a term in $$b$$ using standard elimination. You avoid $$3$$ because the base $$27$$ has the prime factor $$3$$.\n\nYou can solve $$ax+by=1$$ for coprime $$a$$ and $$b$$ using the division algorithm to find their highest common factor (which is $$1$$ because they are coprime). Other methods are available, and sometimes work more quickly if you spot them. The division algorithm provides a proof that there is always a solution, and a systematic way of finding it.\n\n\u2022 +1 Eliminating $a$ by cross multiplying yields $-2\\equiv\\!\\!\\overbrace{ 5}^{\\large 44a}\\!\\!-7b\\,\\Rightarrow\\, 7b\\equiv 7\\,\\Rightarrow\\, b\\equiv 1\\,$ (use least magnitude reps!), which is probably easier than computing inverses. \u2013\u00a0Bill Dubuque Oct 22 '18 at 19:01\n\nGuide:\n\n\\begin{align} 27 &= 11(2)+5 \\\\ 11 &= 2(5)+1 \\end{align}\n\n\\begin{align} 1 &= 11-2(5)\\\\ &= 11 - 2(27-2(11)) \\\\ &= 5(11)-2(27) \\end{align}\n\nHence $$11^{-1} \\equiv 5 \\pmod{27}$$\n\nNow you should be able to solve for $$a$$, substitute that to the first equation and solve for $$b$$ using similar procedure.\n\n\\begin{align*} 11a & \\equiv 8 \\pmod{27}\\\\ 55a & \\equiv 40 \\pmod{27}\\\\ a & \\equiv 13 \\pmod{27}. \\end{align*} Now plug this in the first congruence to get $$b$$.\n\n\u2022 But you should say how you computed the inverse of $11,\\,$ not simply pull it out of a hat like magic! \u2013\u00a0Bill Dubuque Oct 22 '18 at 19:05\n\u2022 @BillDubuque My intention was not to pull a magic for OP. But when OP says (in his question) that he/she tried using the Chinese Remainder Theorem, then I would assume familiarity of such a step (multiplying by an inverse) on his/her part. One can always seek more explanation if needed. \u2013\u00a0Anurag A Oct 23 '18 at 16:21\n\u2022 But the question makes it clear the OP is having difficulty inverting $\\,11\\bmod 27.\\,$ Giving the answer with no hint how it was computed is probably not going to help the OP get past their obstacles. \u2013\u00a0Bill Dubuque Oct 23 '18 at 16:36\n\u2022 It is not that I disagree with your suggestion of adding more but it is a judgement call. When someone says CRT etc. and yet misses some basic step, then one can argue either way about his/her understanding of the basic material. Perhaps OP overlooked some simple step or perhaps OP has difficulty with the concept of inverse. I grapple with this on a daily basis :-) \u2013\u00a0Anurag A Oct 23 '18 at 17:00", "date": "2019-07-16 14:53:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2965619/how-to-solve-the-following-congruences-equation-system", "openwebmath_score": 0.9452844262123108, "openwebmath_perplexity": 526.5563629912929, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712643239288, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8587976150818939}}
{"url": "https://math.stackexchange.com/questions/2408906/difference-between-proof-by-reductio-ad-absurdum-and-proof-by-contradiction", "text": "# Difference between \u201c proof by reductio ad absurdum\u201d and \u201cproof by contradiction\u201d?\n\nI always thought that both \u201cproof by reductio ad absurdum\u201d and \u201cproof by contradiction\u201d mean the same, but now my professor asked this question on my homework and I don't know.\n\nI believe that in both cases you assume the negation of the conclusion and develop a contradiction through the premises. This will imply the conclusion. Today I have a meeting with the assistant professor so I can clarify this, but I really would like to know what you guys think, or if possible it would be great if you point me into some good references.\n\nUPDATE:\n\nI just came from my extra help and the assistant professor explains the difference this way:\n\nReductio ad absurdum: $$\\vDash [\\neg p\\to(q\\wedge\\neg q)]\\to p$$\n\nProof by contradiction:\n\n$$\\vDash [\\neg (p\\to q) \\to (r\\wedge \\neg r)]\\to (p\\to q)$$\n\nAnd the examples of application were these:\n\nUsing proof by contradiction: $\\sqrt2$ is irrational.( First suppose it is rational and derive a contradiction).\n\nUsing proof by reductio ad absurdum: If $f$ is differentiable on $(a,b)$ then $f$ is continuous on $(a,b)$. ( First we suppose that $f$ is differentiable on $(a,b)$ but not continuous on $(a,b)$ and derive a contradiction).\n\n\u2022 \u2013\u00a0R. Suwalski Aug 28 '17 at 15:49\n\u2022 I always thought of it as: reductio ad absurdum is proving $\\lnot \\phi$ by assuming $\\phi$ and proving $\\bot$ or $\\psi \\wedge \\lnot \\psi$. Proof by contradiction is proving $\\phi$ by assuming $\\lnot \\phi$ and proving $\\bot$ or $\\psi \\wedge \\lnot \\psi$. So, for example, in intuitionistic systems, reductio ad absurdum is still valid, but proof by contradiction wouldn't be as all it proves is $\\lnot \\lnot \\phi$. (Not sure if this is officially valid, though.) \u2013\u00a0Daniel Schepler Aug 28 '17 at 15:55\n\u2022 Usually (but the distinction is not so \"stable\") the proof by contradiction is of the form: \"if from $A$ a contradiction follows, then $\\lnot A$ can be inferred\". In the indirect proof the assumption is $\\lnot A$ and the conclusion inferred (through the contradiction) is $A$. \u2013\u00a0Mauro ALLEGRANZA Aug 28 '17 at 15:58\n\u2022 Possible duplicate of Difference between proof of negation and proof by contradiction \u2013\u00a0Clement C. Aug 28 '17 at 16:08\n\u2022 Ok. So \"reductio ad absurdum\" is synonymous with \"indirect proof\", and \"proof of negation\"? \u2013\u00a0Novato Aug 28 '17 at 16:17\n\n## 1 Answer\n\nRegarding the rule of indirect proof:\n\n\"if from assumption $\\lnot A$ a contradiction follows, we can infer $A$\",\n\nwe can see:\n\nSometimes the nomenclature RAA is used; it stands for reductio ad absurdum, the medi\u00e6val Latin name of the principle. [...] A genuine indirect proof in propositional logic ends with a positive conclusion.\n\nThe principle is equivalent to Double Negation elimination.\n\nIf we agree with this approach, proof by contradiciton is more general, because it applies also to inferences with negative conclusion, licensed by the principle of Negation Introduction:\n\n\"if from assumption $A$ a contradiction follows, we can infer $\\lnot A$\".", "date": "2019-10-15 03:41:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2408906/difference-between-proof-by-reductio-ad-absurdum-and-proof-by-contradiction", "openwebmath_score": 0.8017630577087402, "openwebmath_perplexity": 317.267831284558, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9425067211996142, "lm_q2_score": 0.9111797015700343, "lm_q1q2_score": 0.858792992950416}}
{"url": "https://www.physicsforums.com/threads/difference-between-lim-as-x-and-lim-as-x.634344/", "text": "# Difference between lim as x\u2192\u221e and lim as |x|\u2192\u221e\n\n1. Sep 8, 2012\n\n### Cristopher\n\nI came across something I'd never seen before, the use of |x| instead of just \u2018x\u2019 in limits.\nWhat is the difference between $\\displaystyle\\lim_{|x|\\to\\infty}x\\sin\\frac{1}{x}$ and $\\displaystyle\\lim_{x\\to\\infty}x\\sin\\frac{1}{x}$ ?\nIs there any difference when evaluating them? Is that notation used only with infinity?\n\nThanks.\n\n2. Sep 8, 2012\n\n### Vorde\n\nIt seems like it's implying that the limit as x goes to infinity is equal to the limit as x goes to negative infinity. But someone who has actually seen this notation before might know better, I'm just guessing.\n\n3. Sep 8, 2012\n\n### alberto7\n\nI haven't seen limits as |x|\u2192\u221e, but I have seen limits as x\u2192\u00b1\u221e or limits as x\u2192\u221e with this meaning. In the latter case they distinguished limits as x\u2192+\u221e and as x\u2192-\u221e. I think they pictured \u221e as a single point outside the line, making it a circle, its Alexandroff compactification, considering limits as x\u2192+\u221e and as x\u2192-\u221e as the lateral limits towards \u221e.\n\nAbout the limits as |x|\u2192\u221e, I think it could be generalized in the following way. We may say that f(x)\u2192a as g(x)\u2192b if, for any neighborhood U of a, there exists a neighborhood V of b such that f(g-1(V\\{b}))\u2286U. Also f(x)\u2192a as g(x)\u2192\u221e if, for any neighborhood U of a, there exists M>0 such that f(g-1((M,\u221e)))\u2286U, and similar definitions.\n\nLast edited: Sep 8, 2012\n4. Sep 8, 2012\n\n### HallsofIvy\n\nStaff Emeritus\nI see two different ways of interpreting \"limit as |x| goes to infinity\". First would be that \"limit as x goes to infinity\" and\"limit as x goes to -infinity\" must be the same. The other would be that x represents a point in the plane or a complex number and x goes away from the origin in any direction.\n\n5. Sep 8, 2012\n\n### ObsessiveMathsFreak\n\nThe first notation only really makes sense if you consider 'x' to be a complex variable. In this case, taking the limit at infinity means asking whether the function approaches a fixed value no matter which direction you approach infinity from.\n\nThe second notion on the other hand can be used if x is real and you are simply going to \"positive\" infinity.\n\nAt infinity, functions of complex variables generally either have a finite limit, or else a pole. They could also have an essential singularity (O~o) as well. And if the function is multi-valued, it will almost always have a (irregular) branch point there.\n\n6. Sep 9, 2012\n\n### Cristopher\n\nThank you all.\n\nYes, I also thought that it says that the limit as x\u2192+\u221e and as x\u2192-\u221e are the same. Indeed, I did look at the graph of the funcion x sin(1/x) before asking, and these limits are both equal to 1. I also noticed there's symmetry in the graph, I thought it may have something to do with that, too. I posted the question in the hope of getting more info on the scope of this notation.\n\nJust for reference I saw the notation here:\nhttp://en.wikipedia.org/wiki/L%27Hopital%27s_rule\nIn the section \u2018Other ways of evaluating limits\u2019", "date": "2017-11-21 18:27:39", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/difference-between-lim-as-x-and-lim-as-x.634344/", "openwebmath_score": 0.9231700897216797, "openwebmath_perplexity": 751.282647547661, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806532363086, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8587797914927217}}
{"url": "https://algo.franklinqin0.me/factorial-trailing-zeroes.html", "text": "# Factorial Trailing Zeroes\n\n## # Solution\n\nBrute force solution is to calculate $n!$ and keep dividing by and count number of $10$'s.\n\n### # Number of 5's\n\nFactor a number: $60 = 2 \\times 2 \\times 3 \\times 5$. We can see $60$ only has 1 trailing zero b/c the number of pair of $2$ and $5$ is 1.\n\nFor a factorial, the number of 2's is definitely more than that of 5's.\n\nThus, the number of 5's is the number of trailing zeros.\n\nComplexity:\n\n\u2022 time: $O(\\log n)$\n\u2022 space: $O(1)$\ndef trailingZeroes(self, n: int) -> int:\ncnt = 0\nwhile n>0:\ncnt += n//5\nn //= 5\n# this also works\n# n //= 5\n# cnt += n\nreturn cnt\n\nLast Updated: 9/21/2020, 4:44:16 PM", "date": "2020-09-30 15:43:51", "meta": {"domain": "franklinqin0.me", "url": "https://algo.franklinqin0.me/factorial-trailing-zeroes.html", "openwebmath_score": 0.2517896890640259, "openwebmath_perplexity": 4161.752913621268, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806540875629, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8587797906487081}}
{"url": "https://math.stackexchange.com/questions/1595620/the-probability-that-in-a-game-of-bridge-each-of-the-four-players-is-dealt-one-a", "text": "# The probability that in a game of bridge each of the four players is dealt one ace\n\nThe question is to show that the probability that each of the four players in a game of bridge receives one ace is $$\\frac{24 \\cdot 48! \\cdot13^4}{52!}$$ My explanation so far is that there are $4!$ ways to arrange the 4 aces, $48!$ ways to arrange the other cards, and since each arrangement is equally likely we divide by $52!$. I believe the $13^4$ represents the number of arrangements to distribute 4 aces among 13 cards, but I don't see why we must multiply by this value as well?\n\nImagine $4$ groups of $13$ spots for the $4$ hands\n$\\small\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\quad\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\quad\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\quad\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square\\square$\n\nAn ace can be placed in each group in any of $13$ places, hence $13^4$\nThe aces' suits can be distributed between groups in $4!$ ways, hence $24$\nThe remaining cards can be placed in 48! ways,\nand 52! is the unrestricted ways of placing the cards\nthus (putting the terms in the order you have), $Pr = \\dfrac{24\\cdot48!\\cdot13^4}{52!}$\n\nSIMPLER WAY:\n\nThere is a much simpler way to get the same result.\n\nWe need an ace in each group of 13, how the rest of the cards go doesn't matter !\n\nThe first ace has to be in some group, each of the other aces have to fall in a different group, so the $2nd$ ace has $39$ permissible spots out of $51,$ and so on\n\nthus $Pr = \\dfrac{39}{51}\\cdot\\dfrac{26}{50}\\cdot\\dfrac{13}{49}$\n\nThere are two approaches to card-game questions like these. In the first approach, we temporarily assume that order of cards within the hand matters. In doing so, we treat our sample space as all ways of arranging the fifty-two cards in a row.\n\nApproaching like this, we have the following steps for multiplication principle:\n\n\u2022 Pick who gets which ace: $4!$ ways\n\u2022 Arrange the remaining $48$ cards in a row and give them to the players (12 to north, the next twelve to east, the next twelve to south, etc...): $48!$ ways\n\u2022 Pick where in the hand the ace goes for each player: $13^4$ ways\n\nNote that the final step was necessary in order to have what we count cover all possible ways that the 52 cards be dealt such that every player receives at least one ace. If we hadn't multiplied by $13^4$, it would have been as though we only considered the ace being the first card in each players' hand. Since we are considering order important, the hand $A\\spadesuit 2\\spadesuit 3\\spadesuit\\dots$ is a different outcome than $2\\spadesuit A\\spadesuit 3\\spadesuit\\dots$\n\nSince there are $52!$ number of equiprobable ways to deal the cards (where order matters) the probability is as given:\n\n$$\\frac{4!48!13^4}{52!}$$\n\nMy preference is instead to work in the situation that order doesn't matter.\n\nHere, we break apart as multiplication principle:\n\n\u2022 Choose which player gets which ace: $4!$ ways\n\u2022 Choose twelve additional cards for each player: $\\binom{48}{12,12,12,12}=\\frac{48!}{12!12!12!12!}$ number of ways\n\nThere are a total of $\\binom{52}{13,13,13,13}=\\frac{52!}{13!13!13!13!}$ number of deals (where order within each hand doesn't matter) for a probability then of:\n\n$$\\frac{4!\\binom{48}{12,12,12,12}}{\\binom{52}{13,13,13,13}}$$\n\nwhich after a short amount of manipulation you see is precisely the same answer as before.\n\nThere are $4!$ ways to distribute the aces, then $\\frac{48!}{12!^4}$ ways to distribute the other $48$ cards, $12$ to a player.\n\nWithout worrying about the aces, there are $\\frac{52!}{13!^4}$ ways to distribute $52$ cards, $13$ to a player.\n\nTherefore, the probability is $$\\frac{4!\\frac{48!}{12!^4}}{\\frac{52!}{13!^4}}=\\frac{13^4}{\\binom{52}{4}}\\doteq0.1054982$$\n\n\u2022 Being a bridge player, I am very interested by this answer. My question is : does this result means that if I have one ace and my partner has one ace, the probability that one of the opponents has the two other aces is 90% ? Or, is such a statement totally wrong ? \u2013\u00a0Claude Leibovici Jan 8 '16 at 4:54\n\u2022 If you have one ace and your partner has one ace, then the other two players have two aces. It should be close to $50\\%$ that one has both, but not quite. \u2013\u00a0robjohn Jan 8 '16 at 8:11\n\u2022 To count the number of ways that one of them does not have both aces, note that there are $2!$ ways to distribute the aces, then $\\frac{24!}{12!^2}$ ways to distribute the other $24$ cards, $12$ to a player. Without worrying about the aces, there are $\\frac{26!}{13!^2}$ ways to distribute $26$ cards, $13$ to a player. Therefore, the probability is $$\\frac{2!\\frac{24!}{12!^2}}{\\frac{26!}{13!^2}} =\\frac{13^2}{\\binom{26}{2}} =0.52$$ So the probability that one of them has both aces is $0.48$. \u2013\u00a0robjohn Jan 8 '16 at 8:11\n\u2022 Thanks for tha answer ! I am real bad with probabilities and I totally misunterpreted the result of $1-0.105$. Shame on me !! Cheers \u2013\u00a0Claude Leibovici Jan 8 '16 at 8:28\n\nSince the order doesn't matter, let's make hands one by one.\n\n1. For the first, there are 4 Aces and I need to choose 1 to give to this person, $\\binom{4}{1}$. Then I need to choose 12 cards from the 48 non-Aces to complete this hand, $\\binom{48}{12}$. Thus the number of ways to make this hand is $$\\binom{4}{1}\\binom{48}{12}.$$\n\n2. I already gave the first person an Ace, so I have three left and I need to choose one to give to this person. I also used up 12 non Aces on the last person, so there are 36 non Aces left and I need to choose 12. The number of ways to make this hand is thus, $$\\binom{3}{1}\\binom{36}{12}.$$\n\n3. and 4. follow the same logic and so I have $$\\binom{2}{1}\\binom{24}{12}\\binom{1}{1}\\binom{12}{12}$$ ways to make hands 3 and 4.\n\nFinally, all the possible ways to make 4 13-card hands are $$\\binom{52}{13}\\binom{39}{13}\\binom{26}{13}\\binom{13}{13} = \\binom{52}{13,13,13,13} = \\frac{52!}{13!\\,13!\\,13!\\,13!}.$$\n\nLastly, putting it all together, the probability of each person getting an Ace is thus \\begin{align*}\\frac{\\binom{4}{1}\\binom{48}{12}\\binom{3}{1}\\binom{36}{12}\\binom{2}{1}\\binom{24}{12}\\binom{1}{1}\\binom{12}{12}}{\\binom{52}{13,13,13,13}} &= \\frac{4!}{1!3!}\\cdot\\frac{48!}{12!36!}\\cdot\\frac{3!}{1!2!}\\cdot\\frac{36!}{12!24!}\\cdot\\frac{2!}{1!1!}\\\\ &\\qquad\\times\\frac{24!}{12!12!}\\frac{1!}{1!0!}\\cdot\\frac{12!}{12!0!}\\cdot\\frac{13!\\,13!\\,13!\\,13!}{52!}\\\\ &=\\frac{24 \\cdot 48! \\cdot13^4}{52!}. \\end{align*}\n\nAfter the cards have been shuffled and cut, there are $\\binom{52}4$ equally likely possibilities for the set of four positions in the deck occupied by the aces. Among those $\\binom{52}4$ sets, there are $13^4$ which result in each player getting an ace; namely, make one of the $13$ cards to be dealt to South an ace, and the same fo West, North, and East. So the probability is $$\\frac{13^4}{\\binom{52}4}=\\frac{4!\\cdot48!\\cdot13^4}{52!}=\\frac{2197}{20825}\\approx.1055$$\n\nSuppose you are arranging the cards in rows of 13. Each row must have an ace; if we place the aces as the first cards, there are 4! ways to arrange them. Then, there are 48! ways to arrange the remaining 48 cards. However, this restricts the aces to being the first card; they can, in fact, be in any of 13 positions - for each of the 4 players. So, multiply by 13^4 to account for the 4 aces being in any of the 13 positions.", "date": "2019-10-17 21:17:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1595620/the-probability-that-in-a-game-of-bridge-each-of-the-four-players-is-dealt-one-a", "openwebmath_score": 0.8813962936401367, "openwebmath_perplexity": 275.1453267648314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9925393547705742, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8587689526529805}}
{"url": "https://math.stackexchange.com/questions/1461512/urn-i-contains-2-white-and-4-red-balls-whereas-urn-ii-contains-1-white-an", "text": "# Urn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball.\n\nUrn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II. What is\n\n\u2022 the probability that the ball selected from urn II is white?\n\u2022 the conditional probability that the transferred ball was white given that a white ball is selected from urn II?\n\nI got the answer $a = \\frac{4}{9}$ but I need help with $b$.\nIf I make $P(T) = \\text{transfered ball}$ is white $P(T) = \\frac13$ and P(W) = selected ball from urn 2 is white.\n\n$P(W) = \\frac49$ (from part $a$) I am looking for $P(T|W)$ by Bayes's formula - I get $$P(T|W) = P(T\\cap W)/P(W)$$\nthen I get $$P(W|T) \\cdot \\dfrac{P(T)}{P(W)}$$.\nwe know $P(T)$ and $P(W)$ and $P(W|T) = \\frac29$.\nso i get $\\frac29\\dfrac{\\frac13}{\\frac49} = \\frac16$ but the answer key says it's $\\frac12$?\n\n\u2022 We need to find $\\Pr(T\\cap W)$, and divide by $\\Pr(W)$. The probability of $T$ is $2/6$. The probability of $W$ given $T$ is $(2/3)$. So the probability of $T\\cap W$ is $(2/6)(2/3)$, which is $2/9$. Now divide by $4/9$ and simplify. Oct 2 '15 at 20:16\n\nIn Bayes's formula formula: $P(T|W) = \\frac{P(T)*P(W|T)}{P(W)}$\n\nBut $P(T)*P(W|T)=P(T\\cap W)$ because these events occur at the same time, that is:\n\n1) we have chosen white ball from urn1, probability of which is: $P(T)=\\frac{2}{6}$...and now we have two white balls and one red in urn 2.\n\n2) then we have chosen the white ball from urn 2, which means: $P(W|T)=2/3$.\n\nSo, $P(T)*P(W|T)=\\frac{2}{6}*\\frac{2}{3}=\\frac{2}{9}=P(T\\cap W)$ $$P(T|W) = P(T\\cap W)/P(W)$$ We know that $P(W)=\\frac{4}{9}$.\n\nSo, $P(T|W)=\\frac{2}{9}/\\frac{4}{9}=\\frac{1}{2}$.\n\n\u2022 ok, but from bayes's formula isn't P(T\u2229W)/P(W) equal to (P(W|T)\u22c5P(T)) /P (W)? so shouldn't the answers be the same? Oct 2 '15 at 20:49\n\u2022 they are equal.\n\u2013\u00a0Jane\nOct 2 '15 at 20:53\n\u2022 look at the corrections above\n\u2013\u00a0Jane\nOct 2 '15 at 21:02\n\u2022 actually we can just use the Bayes's formula without including the probability of events intersection\n\u2013\u00a0Jane\nOct 2 '15 at 21:03", "date": "2021-09-18 22:57:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1461512/urn-i-contains-2-white-and-4-red-balls-whereas-urn-ii-contains-1-white-an", "openwebmath_score": 0.9289060235023499, "openwebmath_perplexity": 133.64968067561787, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406020637528, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8587474267550844}}
{"url": "https://byjus.com/question-answer/if-f-left-x-right-and-g-left-x-right-are-two-functions-with-g/", "text": "Question\n\n# If $$f\\left( x \\right)$$ and $$g\\left( x \\right)$$ are two functions with $$g\\left( x \\right) =x-\\dfrac { 1 }{ x }$$ and $$f\\circ g\\left( x \\right) ={ x }^{ 3 }-\\dfrac { 1 }{ { x }^{ 3 } }$$, then $$f^{ ' }\\left( x \\right)$$ is equal to\n\nA\n3x2+3\nB\nx21x2\nC\n1+1x2\nD\n3x2+3x4\n\nSolution\n\n## The correct option is A $$3{ x }^{ 2 }+3$$$$f\\circ g\\left( x \\right) ={ x }^{ 3 }-\\dfrac { 1 }{ { x }^{ 3 } }$$Writing $${ x }^{ 3 }-\\dfrac { 1 }{ { x }^{ 3 } }$$ using $${ \\left( a-b \\right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\\left( a-b \\right)$$, we have$${ \\left( x-\\dfrac { 1 }{ x } \\right) }^{ 3 }={ x }^{ 3 }-\\dfrac { 1 }{ { x }^{ 3 } } -3x\\cdot \\dfrac { 1 }{ x } \\left( x-\\dfrac { 1 }{ x } \\right)$$$$\\Rightarrow { x }^{ 3 }-\\dfrac { 1 }{ { x }^{ 3 } } ={ \\left( x-\\dfrac { 1 }{ x } \\right) }^{ 3 }+3\\left( x-\\dfrac { 1 }{ x } \\right)$$We have,$$f\\left( g\\left( x \\right) \\right) ={ x }^{ 3 }-\\dfrac { 1 }{ { x }^{ 3 } } ={ \\left( x-\\dfrac { 1 }{ x } \\right) }^{ 3 }+3\\left( x-\\dfrac { 1 }{ x } \\right)$$As $$g\\left( x \\right) =x-\\dfrac { 1 }{ x }$$, this yields$$f\\left( x-\\dfrac { 1 }{ x } \\right) ={ \\left( x-\\dfrac { 1 }{ x } \\right) }^{ 3 }+3\\left( x-\\dfrac { 1 }{ x } \\right)$$On putting $$x-\\dfrac { 1 }{ x } =t$$, we get$$f\\left( t \\right) ={ t }^{ 3 }+3t$$Thus, $$f\\left( x \\right) ={ x }^{ 3 }+3x$$and $$f^{ ' }\\left( x \\right) =3{ x }^{ 2 }+3$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-22 21:34:42", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/if-f-left-x-right-and-g-left-x-right-are-two-functions-with-g/", "openwebmath_score": 0.6877990365028381, "openwebmath_perplexity": 1832.5225030560832, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405995242328, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8587474143445011}}
{"url": "http://mathhelpforum.com/pre-calculus/19813-asymptotes-intercepts-help-checking-work.html", "text": "# Thread: Asymptotes and intercepts: help checking work\n\n1. ## Asymptotes and intercepts: help checking work\n\nI had to find the vertical and horizontal asymptotes as well as the x an y intercepts..could someone please help me by checking my work...there are only 8 problems...you do not have to check them all, but I wouldreally appreciate the help because I do not know if I am doing these correctly....thank you\n\n2. Hello, aikenfan!\n\nYou did really good work . . . with a few slips here and there.\n\n$12)\\;\\;f(x)\\:=\\:-\\frac{x+2}{x+4}$\nThat minus-sign is in front of the fraction . . .\n\nFor the x-intercept, we have: . $-(x +2)\\:=\\:0 \\quad\\Rightarrow\\quad x + 2 \\:=\\:0$\n\n. . $x \\:=\\:-2\\quad\\Rightarrow\\quad (-2,0)$\n\n$18)\\;\\;f(x)\\:=\\:\\frac{x^2-25}{x^2+5x}$\nVA: . $x^2+5x \\:=\\:0 \\quad\\Rightarrow\\quad x(x+5)\\:=\\:0\\quad\\Rightarrow\\quad x \\:=\\:0,-5$\n\nVertical asymptotes: . $x = 0,\\;x = -5$\n\n$43)\\;\\;f(x) \\:=\\:\\frac{2x^2+1}{x}$\nx-intercept: . $2x^2+1\\:=\\:0\\quad\\Rightarrow\\quad x^2 \\:=\\:-\\frac{1}{2}\\quad\\Rightarrow\\quad x \\:=\\:\\sqrt{-\\frac{1}{2}}$ . . . not a real number.\n\nThere is no x-intercept.\n\n$44)\\;\\;g(x) \\:=\\:\\frac{1-x^2}{x}$\nx-intercepts: . $1-x^2\\:=\\:0\\quad\\Rightarrow\\quad x^2 \\:=\\:1\\quad\\Rightarrow\\quad x \\:=\\:\\pm1\\quad\\Rightarrow\\quad(1,0),\\;(-1,0)$\n\n$45)\\;\\;h(x) \\:=\\:\\frac{x^2}{x-1}$\ny-intercept: . $y \\:=\\:\\frac{0^2}{0-1} \\:=\\:0\\quad\\Rightarrow\\quad (0,\\,0)$\n\n3. Thank you very much for all of your help...I just have one more quick question, if you don't mind...for number 18, how do i figure out the horizontal asymptote? I think it would be 1, but I'm not sure.\n\n4. Hello again, aikenfan!\n\nFor number 18, how do i figure out the horizontal asymptote?\nYes, the horizontal asymptote is: $y = 1$\n\nWe want: . $\\lim_{x\\to\\infty}\\frac{x^2-25}{x^2+5x}$\n\nDivide top and bottom by $x^2$: . $\\lim_{x\\to\\infty}\\frac{\\frac{x^2}{x^2} - \\frac{25}{x^2}}{\\frac{x^2}{x^2} + \\frac{5x}{x^2}}$ . **\n\nThen: . $\\lim_{x\\to\\infty}\\frac{1 - \\frac{25}{x^2}}{1 + \\frac{5}{x}} \\;=\\;\\frac{1-0}{1+0} \\;=\\;1$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\n**\n\nRule: Divide top and bottom by the highest power of $x$\n. . . . in the denominator.\n\n5. Note that we should find $\\lim_{x \\to - \\infty} \\frac {x^2 - 25}{x^2 + 5}$ as well. But in this case, it would give the same value, so that's fine", "date": "2017-11-24 23:03:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/19813-asymptotes-intercepts-help-checking-work.html", "openwebmath_score": 0.8611507415771484, "openwebmath_perplexity": 443.1301381916135, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405989598949, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8587474070493805}}
{"url": "https://math.stackexchange.com/questions/3724064/show-that-if-phi-xt-1-in-a-neighborhood-of-0-then-x-0-a-s/3724093", "text": "# Show that if $\\phi_{X}(t)=1$ in a neighborhood of $0$, then $X=0$ a.s.\n\nLet $$\\phi(t),t\\in\\mathbb{R}$$, be the characteristic function of a random variable $$X$$. Show that if $$\\phi(t)=1$$ in a neighborhood of $$0$$, then $$X=0$$ a.s.\n\nThe problem comes with the following hint: Show that $$1-Re(\\phi(2t))\\le4(1-Re(\\phi(t)))$$ for $$t\\in \\mathbb{R}$$. I am stumped by this one, I am not even sure where to begin or how to prove\\use use the hint, any help here would be greatly appreciated.\n\nHere is another method: start by noticing that since $$\\cos \\leq 1$$, $$\\Re \\phi(t) = \\mathbb{E}[\\cos(tX)] \\leq 1, \\quad \\forall t \\in \\mathbb{R}. \\tag{1}$$ Now let $$\\def\\eps{\\varepsilon} (-\\eps, \\eps)$$ be an interval on which $$\\phi = 1$$. Then, using the hint, we have for $$t \\in (-\\eps, \\eps)$$ $$0 \\leq 1- \\Re \\phi(2t) \\leq 4(1-\\Re \\phi(t)) = 0.$$ The first inequality follows from $$(1)$$ and the last equality from the assumption. This proves that $$\\Re \\phi(2t) = 1$$ for every $$t \\in (-\\eps,\\eps)$$. Reiterating this process, we see that $$\\Re \\phi(t) = 1, \\quad \\forall t \\in \\mathbb{R}.$$ But recall that $$|\\phi(t)| \\leq 1$$. This forces $$\\phi(t) = 1$$ for every $$t \\in \\mathbb{R}$$. Since the characteristic function determines the distribution, it follows that $$X = 0$$ a.s.\n\n\u2022 Beautiful, thank you! Jun 17, 2020 at 21:49\n\nLet's assume that $$\\varphi(t) = 1$$ for any $$t \\in [0,\\delta]$$. Then in particular $$\\varphi(\\delta)=1$$. We'll show that it is the case that $$\\mathbb P(X \\in \\{\\frac{2k\\pi}{\\delta} : k \\in \\mathbb Z \\}) = 1$$ . Let $$\\mu_X$$ be distribution of $$X$$.\n\nNote that $$\\varphi(\\delta)=1$$ means: $$0 = 1 -\\varphi(\\delta) = 1 - \\int_{\\mathbb R} \\cos(\\delta x) d\\mu_X(x) = \\int_{\\mathbb R} (1 - \\cos(\\delta x)) d\\mu_X(x)$$ Since $$1-\\cos(\\delta x) \\ge 0$$, we must have $$\\cos(\\delta x) = 1$$ , $$x - d\\mu_X$$ almost surely, so that $$x = \\frac{2k\\pi}{\\delta}$$ , $$d\\mu_X$$ almost surely, which means $$\\mu_X( \\{\\frac{2k\\pi}{\\delta} : k \\in \\mathbb Z \\})=1$$.\n\nNow note that we have only countable many points in set $$\\{\\frac{2k\\pi}{\\delta} : k \\in \\mathbb Z \\}$$. For every $$k \\in \\mathbb Z \\setminus \\{0\\}$$ we can find such $$t_k \\in (0,\\delta)$$ that $$\\frac{2k\\pi}{\\delta}$$ is not equal to $$\\frac{2 m \\pi}{t_k}$$ for any $$m \\in \\mathbb Z$$ (because for every $$m \\in \\mathbb Z$$ there is at most one $$s \\in (0,\\delta)$$ such that $$\\frac{2m \\pi}{s} = \\frac{2k\\pi}{\\delta}$$, but we have only countable many $$m \\in \\mathbb Z$$, but continuum-many $$s \\in (0,\\delta)$$, so there exists such $$t_k$$) which means that $$\\mu_X(\\frac{2k\\pi}{\\delta}) = 0$$ (for that given $$k \\in \\mathbb Z \\setminus \\{0\\}$$, because $$\\varphi(t_k)=1$$, so $$\\mu_X( \\{ \\frac{2m\\pi}{t_k} : m \\in \\mathbb Z \\}) = 1$$, too). Since $$k \\in \\mathbb Z \\setminus \\{0\\}$$ was arbitrary, and there are only countable many of them, we have $$\\mu_X( \\{ \\frac{2k \\pi}{\\delta} : k \\in \\mathbb Z \\setminus \\{0\\} \\} ) = 0$$, so that $$\\mu_X(\\{0\\}) = 1$$ what was to be proven.\n\nEDIT: If you're interested, here's an approach with your hint. Let's prove it beforehand. $$1 - Re(\\varphi(2t)) = \\int_{\\mathbb R} (1-\\cos(2tx))d\\mu_X(x) = 2\\int_{\\mathbb R} (1 - \\cos^2(tx))d\\mu_X(x)$$\n\nIt would be sufficient to show $$1-\\cos^2(s) \\le 2(1- \\cos(s))$$ which is equivalent to $$0 \\le \\cos^2(s) - 2\\cos(s) + 1 = (\\cos(s)-1)^2$$, so true. Hence $$1- Re(\\varphi(2t)) \\le 4\\int_{\\mathbb R}(1 - \\cos(tx))d\\mu_X(x) = 4(1-Re \\varphi(t))$$\n\nHaving lemma, it is pretty easy. Note that you have such $$\\delta$$, that $$\\varphi(t) = 1$$ for any $$[-\\delta,\\delta]$$. Now let's prove it is also the case for any $$t \\in [-2\\delta,2\\delta]$$ using hint: Take $$s \\in [-2\\delta,2\\delta]$$. We have $$1 - Re(\\varphi(2s)) \\le 4(1 - Re(\\varphi(s)) = 0$$ since $$s \\in [-\\delta,\\delta]$$. Moreover, $$|\\varphi(s)| \\le 1$$, so $$\\varphi(s) = 1$$. Use it again, to prove the fact for any $$[-2^k\\delta,2^k\\delta]$$ getting $$\\varphi(t)=1$$ for any $$t \\in \\mathbb R$$.\n\n\u2022 Oops I see we arrived at the same conclusion! :) Your first method is more powerful however as it shows that if $\\phi(t) = 1$ for one $t$ then the distribution is lattice. Jun 17, 2020 at 21:45\n\u2022 This is an incredible answer, thank you very much! Jun 17, 2020 at 21:48\n\u2022 Yes Michh, nice answer +1. This fact about $\\varphi(t)=1$ is useful when you want to characterise all types of characteristic functions. It can be shown that either $|\\varphi(t)| <1$ for every $t \\in \\mathbb R \\setminus \\{0\\}$ or $|\\varphi(t)|=1$ for every $t \\in \\mathbb R$ and then you have $X = a$ almost surely for some $a \\in \\mathbb R$ or finally $|\\varphi(t)|=1$ for some $t \\in \\mathbb R_+$, but $|\\varphi(s)|<1$ for $s \\in (0,t)$. In the latter case it is exactly the distribution on $\\{ \\frac{2k\\pi}{t} : k \\in \\mathbb Z \\}$. Jun 17, 2020 at 21:49\n\u2022 @Spider Bite, I should thank you, too, because I wasn't aware of such lemma. Nice one to have in mind ^^ Jun 17, 2020 at 21:55\n\u2022 @Wave because since $\\phi(\\delta)=1$, then in particular it is a real number, hence imaginary part must vanish. May 9 at 14:31", "date": "2022-10-04 06:23:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3724064/show-that-if-phi-xt-1-in-a-neighborhood-of-0-then-x-0-a-s/3724093", "openwebmath_score": 0.966022253036499, "openwebmath_perplexity": 106.6571446419326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631635159684, "lm_q2_score": 0.870597273444551, "lm_q1q2_score": 0.8587250807831439}}
{"url": "https://byjus.com/question-answer/check-which-of-the-following-are-solutions-of-an-equation-x-2y-4-i-0/", "text": "Question\n\n# Check which of the following are solutions of an equation $$x + 2y = 4$$?\u00a0(i) (0, 2) \u00a0 \u00a0(ii) (2, 0) \u00a0 (iii) (4, 0) \u00a0 (iv) \u00a0($$\\sqrt 2, -3\\sqrt 2$$) \u00a0 (v) (1, 1) \u00a0 \u00a0(vi) (-2, 3)\n\nSolution\n\n## Given equation is$$x+2y=4$$(i) Put the value x=0 and y=2 we get$$0+2\\times 2=4$$$$\\Rightarrow4=4$$Then both sides are equal then (0,2) is the solution of equation x+2y=4(ii) Put the value x=2 and y=0 we get$$2+2\\times 0=4$$$$\\Rightarrow 2\\neq 4$$\u00a0Then both sides are not\u00a0 equal then (0,2) is not the\u00a0 solution of equation\u00a0x+2y=4(iii) Put the value x=4 and y=0 we get$$4+2\\times 0=4$$$$\\Rightarrow4=4$$Then both sides are equal then (4,0) is the solution of equation x+2y=4(iv)$$x+2y=4$$\u00a0Put $$x=\\sqrt{2},y=-3\\sqrt{2}$$ we get$$\\sqrt{2}+2(-3\\sqrt{2})=4$$$$\\Rightarrow \\sqrt{2}-3\\sqrt{2}=4$$$$\\Rightarrow -2\\sqrt{2}\\neq 4$$\u00a0Then both sides are not\u00a0 equal then\u00a0$$(\\sqrt{2},-3\\sqrt{2})$$\u00a0is not the\u00a0 solution of equation\u00a0x+2y=4(v) Put the value x=1 and y=1 we get$$1+2\\times 1=4$$$$\\Rightarrow 3\\neq 4$$\u00a0Then both sides are not\u00a0 equal then (1,1) is not the\u00a0 solution of equation\u00a0x+2y=4(vi) Put the value x=-2 and y=3 we get$$-2+2\\times 3=4$$$$\\Rightarrow 4= 4$$\u00a0Then both sides are\u00a0 \u00a0equal then (-1,3) is\u00a0 the\u00a0 solution of equation\u00a0x+2y=4Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-27 04:43:17", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/check-which-of-the-following-are-solutions-of-an-equation-x-2y-4-i-0/", "openwebmath_score": 0.8563093543052673, "openwebmath_perplexity": 4401.757168088209, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9863631655203045, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8587250759047635}}
{"url": "https://www.physicsforums.com/threads/a-question-about-rolling-motion-of-a-wheel-on-a-frictionless-surface.933400/", "text": "# I A question about rolling motion of a wheel on a frictionless surface\n\n1. Dec 4, 2017\n\n### NoahCygnus\n\nLet's imagine a situation where we have a wheel of mass $M$ and radius $R$ on a frictionless surface, and we apply a force $\\vec{F}$ as shown in the diagram. The force will produce both linear acceleration of centre of mass $a$ and angular acceleration $\\alpha$. The wheel starts to both translate and rotate, and we remove the external forcr. I know the condition for rolling is $v_{com} = R\\omega$. The question is, most textbooks include a surface with friction , and I can't tell if without friction the above condition for rolling will be met or the wheel will slip. How can I know ?\n\nLast edited: Dec 4, 2017\n2. Dec 4, 2017\n\n### BvU\n\nAs drawn, the wheel will slip: angular acceleration will exceed linear acceleration. If $F$ is applied below the c.o.m, the reverse. Somewhere there is a height where even without friction the rolling condition is met. What height ? A very nice exercise for someone lke you !\n\n3. Dec 4, 2017\n\n### kuruman\n\nThe problem without friction is equivalent to a puck on a frictionless table pulled by a string wrapped around its circumference. The tension will exert a torque about the CM that will start the wheel spinning around its center and it will also accelerate the CM. Use Newton's Second law for rotation and translation to analyze the motion and answer your question.\n\n4. Dec 4, 2017\n\n### NoahCygnus\n\nIf the wheel is a cylinder, and I apply the force above the centre of mass at $r$, causing a rotation about an axis passing through the cm, perpendicular the circular surfaces of the cylinder, then\n$F = Ma_{cm}$\n$a_{cm} = F/M$\n\n$\\tau = rF = I_{cm}\\alpha$\n$\\alpha= \\tau/I_{cm}$\n\nThe tangential acceleration of a point on the circumference will be, $a_{t} = R\\alpha \\Longrightarrow a_{t} = R(\\tau/I_{cm}) \\Longrightarrow a_{t} = R(rF/I_{cm})$\n\nIf the wheel is to roll then, $v_{cm} = v_{t} \\Longrightarrow d(v_{cm})/dt = d(v_{t})/dt \\Longrightarrow a_{cm} = a_{t}$\n$F/M = R(rF/I_{cm}) \\Longrightarrow F/M = R(rF/1/2 MR^2) \\Longrightarrow 2r = R \\Longrightarrow r = R/2$\n\nSo I have to apply the force at a height $R/2$, right?\n\nAnd if I apply the force below the centre of mass, the wheel will slip, right?\n\n5. Dec 4, 2017\n\n### BvU\n\nabove the center of mass, yes. You can eliminate ambiguity by calling that a height ${3\\over 2}R$\nNot wrong, but incomplete. How about a height in the range $R$ to ${3\\over 2}R$ ?\n\nWell done !\n\n(the same exercise for a sphere is worked out here under 'sweet spot' )\n\n6. Dec 5, 2017\n\n### NoahCygnus\n\nI did the math for the range $R$ to $(3/2) R$ and found that the tangential acceleration of a point on circumference will always be greater than the linear acceleration of the centre of mass. As a result the body will skid. Can you please explain to me how friction prevents the skidding if I apply a force in that range?\nAlso I checked the article , it seems quite useful. I will read it once I am free.\n\n7. Dec 5, 2017\n\n### BvU\n\nYou mean the tangential acceleration required for the no-slipping condition, right ?\nIf height = $3/2\\, R\\$ gives no skidding, anything below means skidding will occur to increase angular velocity, and anything above means skidding will reduce it until rolling catches up.\n\n8. Dec 6, 2017\n\n### A.T.\n\nStatic friction is an additional force in your equation. Assume pure rolling and then solve for the friction force required to achieve it.\n\n9. Dec 6, 2017\n\n### NoahCygnus\n\nYes that's what I meant.\nI understand. Let's talk about the direction of frictional force. I infer that if I apply a force below $(3/2)R$ , such that skidding occurs, the direction of frictional force will be opposite to the translation motion of centre of mass, as to reduce the speed of centre of mass and to provide a negative torque to the wheel, so to increase its angular speed (Case 1 in the diagram). And if I apply the force above centre of mass, slipping will occur, so friction will act in forward direction as to speed up the centre of mass, and provide a positive torque, so to lower the angular speed until the rolling condition is met (Case 2 in the diagram). Am I correct? Also I read somewhere once the rolling condition is met, frictional force vanishes, how is that possible? I am having a hard time understanding that.\n\n10. Dec 6, 2017\n\n### BvU\n\nThe trajectory of a point on the circumference of the rolling object is a cycloid: at the ground it moves down and up again, so not horizontally.\n\n11. Dec 7, 2017\n\n### A.T.\n\nYes, and the above also applies when no skidding occurs, due to sufficient static friction. The static friction has the direction that you describe above for kinetic friction, while its magnitude is whatever is required to ensure the rolling condition for linear and angular accelerations.\n\nPure rolling means no sliding, thus no kinetic friction, but you still can have static friction.\n\n12. Dec 8, 2017\n\n### NoahCygnus\n\nLet's talk about the energy. If the object smoothly rolls down an incline in a conservative field, such that no sliding occurs, the mechanical energy will be conserved even if there is static friction because static friction is a non-dissipative force. Am I correct?\nIf that's the case how come in reality, a smoothly rolling wheel loses kinetic energy?\n\n13. Dec 8, 2017", "date": "2018-03-21 02:15:35", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-question-about-rolling-motion-of-a-wheel-on-a-frictionless-surface.933400/", "openwebmath_score": 0.5970717072486877, "openwebmath_perplexity": 485.3208997152512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992932829918, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8587247050295294}}
{"url": "http://nirvanatravel.org/vhoi9dce/e7a30a-is-the-inverse-of-a-bijective-function-bijective", "text": "I think the proof would involve showing f\u207b\u00b9. Since f is surjective, there exists a 2A such that f(a) = b. Let f : A !B be bijective. So x 2 is not injective and therefore also not bijective and hence it won't have an inverse.. A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. The range of a function is all actual output values. In order to determine if $f^{-1}$ is continuous, we must look first at the domain of $f$. Let f: A \u2192 B. We will de ne a function f 1: B !A as follows. 1.Inverse of a function 2.Finding the Inverse of a Function or Showing One Does not Exist, Ex 2 3.Finding The Inverse Of A Function References LearnNext - Inverse of a Bijective Function \u2026 Bijective. https://goo.gl/JQ8NysProving a Piecewise Function is Bijective and finding the Inverse This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read \"one-to-one correspondence).. I've got so far: Bijective = 1-1 and onto. Then f has an inverse. Proof. Let f : A !B be bijective. In mathematical terms, let f: P \u2192 Q is a function; then, f will be bijective if every element \u2018q\u2019 in the co-domain Q, has exactly one element \u2018p\u2019 in the domain P, such that f (p) =q. In mathematics, a bijective function or bijection is a function f : A \u2192 B that is both an injection and a surjection. Accelerated Geometry NOTES 5.1 Injective, Surjective, & Bijective Functions Functions A function relates each element of a set with exactly one element of another set. The function f: \u211d2-> \u211d2 is defined by f(x,y)=(2x+3y,x+2y). Let\u2019s define $f \\colon X \\to Y$ to be a continuous, bijective function such that $X,Y \\in \\mathbb R$. Now we much check that f 1 is the inverse \u2026 The Attempt at a Solution To start: Since f is invertible/bijective f\u207b\u00b9 is \u2026 Yes. Click here if solved 43 Theorem 1. An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers. 1. The codomain of a function is all possible output values. it's pretty obvious that in the case that the domain of a function is FINITE, f-1 is a \"mirror image\" of f (in fact, we only need to check if f is injective OR surjective). the definition only tells us a bijective function has an inverse function. If we fill in -2 and 2 both give the same output, namely 4. Please Subscribe here, thank you!!! it doesn't explicitly say this inverse is also bijective (although it turns out that it is). Let f 1(b) = a. Since f is injective, this a is unique, so f 1 is well-de ned. A bijective group homomorphism $\\phi:G \\to H$ is called isomorphism. The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. is bijective, by showing f\u207b\u00b9 is onto, and one to one, since f is bijective it is invertible. It means that each and every element \u201cb\u201d in the codomain B, there is exactly one element \u201ca\u201d in the domain A so that f(a) = b. Let b 2B. Bijective Function Examples. Show that f is bijective and find its inverse. A function is called to be bijective or bijection, if a function f: A \u2192 B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. A bijection of a function occurs when f is one to one and onto. The domain of a function is all possible input values. : since f is bijective, by showing f\u207b\u00b9 is onto, and hence isomorphism above problem that! 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By showing f\u207b\u00b9 is onto, and hence isomorphism the range of a function is all possible input.... Be true domain of a function is all possible output values definition only tells us bijective! Properties and have both conditions to be true function has an inverse function an is!", "date": "2021-09-18 23:03:19", "meta": {"domain": "nirvanatravel.org", "url": "http://nirvanatravel.org/vhoi9dce/e7a30a-is-the-inverse-of-a-bijective-function-bijective", "openwebmath_score": 0.9020706415176392, "openwebmath_perplexity": 656.9814891401281, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9678992858752657, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8587246883540545}}
{"url": "https://math.stackexchange.com/questions/2864085/is-there-a-continuous-onto-function-f-bbbd-rightarrow-1-1", "text": "# Is there a continuous onto function $f:\\Bbb{D} \\rightarrow [-1,1]$?\n\nIs there a continuous onto function $f:\\Bbb{D} \\rightarrow [-1,1]$ ,\n\nwhere $\\Bbb{D}$ is a closed unit disk in $\\Bbb{R}^2$ ?\n\nI think the answer is no, otherwise $f(\\Bbb{D})=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?\n\n\u2022 Try $f(x,y) = x.$ \u2013\u00a0zhw. Jul 27 '18 at 5:40\n\u2022 Because you don't assume $f$ to be injective, \"removing one point\" in $[-1,1]$ may make you remove way more than just one point in $\\mathbb{D}$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail. \u2013\u00a0Suzet Jul 27 '18 at 5:42\n\nThe problem is, that because $f$ is not injective, we cannot assert that since $f(\\mathbb D) = [-1,1]$, we have for all $x \\in [-1,1]$ that $f(\\mathbb D \\backslash \\{y\\}) = [-1,1] \\backslash \\{x\\}$ for some $y$. Indeed, what may happen is that many points of $\\mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.\n\nAs you have seen, the first projection serves as an onto map.\n\nIf we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D \\setminus \\{x\\}) = [-1,1] \\setminus \\{0\\}$, and this is contradiction because $D \\setminus \\{x\\}$ is connected regardless of what $x$ is , but the image is disconnected.\n\nIn fact, if $f$ is injective, then since $\\mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.\n\nNote that $\\mathbb D$ is a subset of $\\mathbb R^2$ and $[-1,1]$ is a subset of $\\mathbb R$. Because $2 \\neq 1$, we are inclined to believe that a homeomorphism is not possible.\n\nThis result, called the invariance of domain, holds in more generality : if a set in $\\mathbb R^m$ is homeomorphic to some other set in $\\mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.\n\nSure.\n\nIf you map each point to its radius $(x,y)\\mapsto\\sqrt{x^2+y^2}$, you have a continuous function $\\mathbb{D}\\to[0,1]$. A slight modification to $(x,y)\\mapsto2\\sqrt{x^2+y^2}-1$ should be what you want.\n\n\u2022 Thank you sir! this one helps too! \u2013\u00a0user444830 Jul 27 '18 at 5:56", "date": "2019-12-15 05:42:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2864085/is-there-a-continuous-onto-function-f-bbbd-rightarrow-1-1", "openwebmath_score": 0.9067865610122681, "openwebmath_perplexity": 86.36570069289404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363476123224, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8586832286399467}}
{"url": "https://brilliant.org/discussions/thread/how-to-find-an-n-digit-number-with-m-number-of/", "text": "# How to find an $n$-digit number with $m$ number of factors?\n\nI found on the Internet a question which I did not understand:\n\nThere is a $6$-digit number with $28$ distinct factors. What is the number?\n\nI searched and found that\n\nNumber of factors of a number of the form ${p_1}^{m_1} \\times {p_2}^{m_2} \\times {p_3}^{m_3} \\cdots$ is $(m_1 +1) \\times (m_2 +1 ) \\times (m_3 +1 ) \\cdots$.\n\nBut I did not understand how one can answer this question.\n\nAlso, if there is some way, I would be glad to know\n\nThe general formula of $n$-digit number with $m$ number of factors?\n\nNote by Vinayak Srivastava\n3\u00a0months, 3\u00a0weeks ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nHey Vinayak, here is an explanation of the formula for the number of factors: Let $n = p_1^{m_1} \\cdot p_2^{m_2} \\cdots$. Every factor of $n$ is a product of the prime numbers risen to some power: $f = p_1^{n_1} \\cdot p_2^{n_2} \\cdots$ with $0 \\leq n_1 \\leq m_1, 0 \\leq n_2 \\leq m_2 ...$. For example let $n = 12 = 2^2 \\cdot 3^1$. All the factors of 12 are:\n\n$2^0 \\cdot 3^0 = 1$\n\n$2^1 \\cdot 3^0 = 2$\n\n$2^2 \\cdot 3^0 = 4$\n\n$2^0 \\cdot 3^1 = 3$\n\n$2^1 \\cdot 3^1 = 6$\n\n$2^2 \\cdot 3^1 = 12$\n\nAs you can see, for each power of the prime numer $p_n$ there are $m_n + 1$ choices $0, 1, ... m_n$. Therefore, the number of factors is $(m_1 + 1) \\cdot (m_2+1) \\cdots$. I hope this helps!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nThanks @Finnley Paolella ! Now can you please tell me how to find the general formula of $n$-digit number with $m$ number of factors?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nI don't know if there is a general formula, but my strategy would be the following: try to factor $m$ to find the power of prime numbers. For example: $28 = 4 \\times 7$. A number of the form $n = p_1 ^3 \\times p_2 ^ 6$ will always have 28 factors. And then we can try:\n\n$2^3 \\times 3^6 = 5832$ is too small\n\n$2^3 \\times 5^6 = 125000$\n\nAnd that is a solution to the original question!\n\nHowever, the solution is not unique. Here is another solution:\n\n$2^6 \\times 11^1 \\times 443^1 = 311872$\n\nNumber of factors: $(6 + 1) \\times (1+ 1) \\times (1 + 1) = 28$\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nInterestingly, although there are 3,013 six-digit numbers with 28 factors, there are some numbers of factors for which there is only one six-digit number with that many factors. These are, I believe, 7, 13, 19, 38, each of which has only one six-digit number with that many factors. (edit: there are probably a lot more solo numbers of factors >50 but I aggregated everything with 50 or more factors to save on processor time.)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nHow did you calculate this?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nSlightly educated brute force. I wrote a bit of simple code that worked out how many (distinct) factors a number had, looped it from 100,000 to 999,999, let it churn for a couple of hours and graphed the results.\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nWOW!!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nHey Stef, that is amazing! Great visualisation :)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nThanks.\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nReally awesome stuff! Interestingly you can also see that prime number of factors such as 7 and 11 are very less, fantastic!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nYeah. I'm a big fan of visualising data. You can see all the primes (two factors), all the squares (odd number of distinct factors), and a bit of the long tail of large numbers of factors (it was long tail before I cut it off).\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nStef, can you please share the process, I mean the code if you don't mind? It seems you used matplotlib to visualise after processing your data from code. Thanks!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nHappy to. It's not very optimised (after all, I only wanted to run it once) so please excuse the code. Also it's written in Lua which is quite like C, Java, JavaScript and Python. A few things to know, though:\n\n1. Arrays (actually tables, but you can use them as arrays) are initialised with {} and referenced with []. Also they start at an index of 1. So, if a={10,11,12}, a[1]=10.\n2. Instead of bracketing or indenting code, Lua uses the word end to end blocks of code so if ... then ... end, for ... do ... end etc. are common constructions.\n3. You can put multiple statements on a line if you separate them with semi-colons.\n4. Variables are weakly typed, automatically coerced and don't need declaring: every variable just holds the value of nil unless you give it a value.\n5. Anything else just ask... :)\n6. (edit) output was in comma separated format that I could paste straight into Apple Numbers as it takes CSV from the clipboard.\n  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 function factors(n) c=0 for f=1,n do if (n//f)==(n/f) then c=c+1 end if c==50 then break end end return c end -- set parameters min=100000 max=999999 step=1000 -- used to track progress ns=min+step -- used to track progress -- initialise array a={} for f = 1,50 do a[f] = 0 end -- calculate factors for f=min,max do if f==ns then print(f); ns = ns + step; end -- display progress n=factors(f) a[n] = a[n] + 1 end -- display output s=\"\" for f=1,50 do s = s..f..\",\"..a[f]..\"\\n\" end print(s) \n\n- 3\u00a0months, 3\u00a0weeks ago\n\nLua??? Retro gaming?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nlol. Yeah. Lua. Despite having spent more time with other languages than I have with Lua, I can get an idea out of my head and running faster in Lua than any other language.\n\nMaybe it's the way my brain works, maybe it's the way that Lua works but, if I don't need to share or integrate with anything else, Lua has become my go-to.\n\n(edit) You know that moment when you've spent an hour debugging a bit of code only to realise that the language didn't work the way you expected it to...? I get that with C, JavaScript, Python and Swift. I don't get it with Lua.\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nI like Lua. But my favourite language is c++. And I think Lua=C languages + Pascal :)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nI\u2019ve not thought about Pascal in years. The whole do...end format is very Pascal-like. There\u2019s one bit of syntax that I really miss from Pascal, and that\u2019s using a:=b for assignment. No more accidentally assigning values in if statements. \ud83d\ude01\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nThanks Stef, i got the logic. Will see if i can run with my own code in python!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nCool. Let me know how it goes?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nAlso... check the min and max values. I was messing about with the code before I pasted it and didn't set them back to what they should have been (corrected in my code now).\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nSure, Thanks! Will paste it here if it does work :)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nCan I learn Lua :) (I don't know anything about programming)?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nLua and Python are both good languages to learn how to program. Python is much more commonly used, especially on brilliant.org. Brilliant.org even has a Programming with Python course. If you want to learn Lua after that, the fundamentals will carry over.\n\nSo, despite my liking for Lua, I'd recommend learning Python. \ud83d\ude43\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nOK, although I purchased the Premium for math, I will do it if I have time from school!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nCan you help me? I wrote a C++ program, but the numbers are too big. I heard in python there are no limits :) I'm not familiar in python programming. I can solve it with hours of programming(vectors, save to txt, read from txt etc.), but I think you can help me.\n\n  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 #include #include using namespace std; long long int power(int w, int b); int check_power(int a, int b, int c, int d); long long int solution(int n, int theta); int main() { for(int n=1;n<=50;n++){ int j=2*n; long long int a=solution(j,0), b=solution(j-1,1); if((a >possible; vector temp; temp.resize(4); for(int a=1;a<=j;a++) for(int b=a;b>0;b--) for(int c=b;c>0;c--){ int d=(j)/(a*b*c); if(a*b*c*d!=j||d>c||check_power(a-1,b-1,c-1,d-1)!=theta) continue; temp[0]=a; temp[1]=b; temp[2]=c; temp[3]=d; possible.push_back(temp); } if(possible.size()==0) return -1; for(int x=0;x<4;x++) minimum[x]=possible[0][x]; for(int x=1;x0) k*=power(w,r); } if(l>k) continue; for(int w=0;w<4;w++) minimum[w]=possible[x][w]; } long long int output=1; for(int x=0;x<4;x++) output*=power(x,minimum[x]-1); return output; } \n\n- 2\u00a0months, 2\u00a0weeks ago\n\nThe program isn't complete. As you can see this works with only 2,3,5 and 7. I need to add other primes. This is the problem.\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nI heard in python there are no limits :)\n\nAnd the garbage collector is made of gold. :) In truth, I hear a lot of good things about Python but never really liked it much myself. Much of what I do these days most of what I do is in Lua, which has 64-bit numbers. If I need an integer bigger than that then I either look to see if I can simplify the problem or pull out my clunky-and-in-need-of-a-rewrite library that stores integers as strings and has functions that can add, subtract etc.\n\nThere are libraries that handle big numbers. You should be able to find one (or many) for C++. Or, if you go down the route of writing your own solution, that's quite rewarding, too.\n\n(I just had a glance at the problem and don't see where big numbers come into it. Did you link the right problem? Or have I missed something?) (edit: looked at your output. OK... those numbers are bigger than I would have expected).\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nA factor $f$ of a number $n = p_{1}^{m_{1}} \\cdot p_{2}^{m_{2}} \\cdot p_{3}^{m_{3}} \\ldots$ must be a number with prime exponents which are less than or equal to the exponents of the number. To make things simple, here is an example: $60 = 2^2 \\cdot 3^1 \\cdot 5^1$. Without Brute-forcing, we can think logically here. If a factor $f$ can be represented here, say 15 which is equal to $3^1 \\cdot 5^1$, the factor's prime exponents will be less than or equal to the prime exponents of $n$. In this case, they are equal\n\nTaking this a step further, for each prime exponent say $m_{1}$, there will be $m_{1} + 1$ options to choose for a power. For the number 60 and specifically prime power of 2, we have $2^{m_{1}}, m_{1} \\in \\{0, 1, 2\\}$ so there are 3 options.\n\nSimilarly each of the $m_{i^{th}}$ exponent will have $m_{i} + 1$ options.\n\nThe total options will thus equal the product of all $= (m_{1} + 1) \\cdot (m_{2} + 1) \\cdot (m_{2} + 1) \\ldots$\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nThanks @Mahdi Raza! Also, as @Finnley Paolella stated, the answer to the question can be $125000$, can there be a general formula of that also?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nI don't think there is a closed-from general formula. It also depends on the base we choose, a number in base 2 will have more digits than a number in base 10. As I said in my comment, the general strategie will be the following:\n\n1. factor $m$\n\n2. try out different possibilites\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nSo if we are given 1 more condition, can there be a unique solution?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nThe thing about those number theories problems is that it is hard to tell how many solutions there are, because prime numbers are more or less randomly distributed along the number line. So it can easily be that there is just one solution, especially if the number m can only be factored in a few different ways. For example, if we were to find a 5 digit number with 17 factors, there will be a unique solution! You can find it using the strategy I provided :)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nIs it $65536$?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nYes ;)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nThat was easy.. $2^{16}$\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nSo, the question which I gave is of no use?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nThe problem hasn't got a unique solution. A more interesting problem would be: How many solutions are there? But I think this is a very time expensive problem (at least if you don't want to program it).\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nI don't know programming :(\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nI think it's possible that I don't understand the question. (edit: Narrator: \"He didn't understand the question.\")\n\nThe lowest composite number I can generate with $2$ distinct factors is $2 \\times 3 = 6 ( = 3! )$\n\nThe lowest composite number I can generate with $3$ distinct factors is $2 \\times 3 \\times 4 = 24 ( = 4! )$\n\n...\n\nThe lowest composite number I can generate with $n$ distinct factors is $(n+1)!$\n\n...\n\nThe lowest composite number I can generate with $28$ distinct factors is $29! = 8841761993739701954543616000000$. This has a lot more than 6 digits. What is it that I don't understand here?\n\n(Additional, just for laughs, if I insist that the $28$ factors are prime factors, I get: $2566376117594999414479597815340071648394470$.)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\n@Stef Smith, Sir, I did not understand what you mean. Please elaborate!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nA misunderstading of how factors work. Some days my brain doesn't work so well. :)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nOh no problem then :)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nHey Stef, The number 24 has more than 3 factors. It has 8 factors: 1, 2, 3, 4, 6, 8, 12, 24. Therefore, applying the factorial function does not work because 29! factorial has way more factors than 29.\n\nI hope this helps :)\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nAh... I understand. Thanks.\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nSo we're looking for something like 100032 which has the 28 factors (1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192, 521, 1042, 1563, 2084, 3126, 4168, 6252, 8336, 12504, 16672, 25008, 33344, 50016, 100032).\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nI now think(thanks to @Finnley Paolella) that there are too many such numbers!\n\n- 3\u00a0months, 3\u00a0weeks ago\n\n125000?\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nYes, it is one of the answers I know of now, but as Finnley Paolella stated, it is not unique.\n\n- 3\u00a0months, 3\u00a0weeks ago\n\n- 3\u00a0months, 3\u00a0weeks ago\n\nGreat Question, But yes has way too many solutions.\n\n- 3\u00a0months, 3\u00a0weeks ago\n\n@Vinayak Srivastava,if product of factors are given,then a unique solution can be determined.\n\n- 2\u00a0months, 3\u00a0weeks ago\n\n- 2\u00a0months, 3\u00a0weeks ago\n\n- 2\u00a0months, 3\u00a0weeks ago\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nRead a little bit, but got bored, I don't know why!\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nI started to read and found a programming task so I connected them and started to prove that to the first 50 numbers :)\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nGreat!\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nOhh! I forgot to post the results :)\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nThe same happens with me all time.\ud83d\ude05\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nTask: Find the least number n that can we represented as a product n = a \u2219 b in k (1 \u2264 k \u2264 50) ways. Products a \u2219 b and b \u2219 a are the same, all numbers are positive integers.\n\nThis isn't the same problem. For example 4 has 3 divisors(1,2,4), but we can write this in only two ways : 1x4 and 2x2.\n\nThe result is:\n\n  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 1 1 2 4 3 12 4 24 5 36 6 60 7 192 8 120 9 180 10 240 11 576 12 360 13 1296 14 900 15 720 16 840 17 9216 18 1260 19 786432 20 1680 21 2880 22 15360 23 3600 24 2520 25 6480 26 61440 27 6300 28 6720 29 2359296 30 5040 31 3221225472 32 7560 33 46080 34 983040 35 25920 36 10080 37 206158430208 38 32400 39 184320 40 15120 41 44100 42 20160 43 5308416 44 107520 45 25200 46 2985984 47 9663676416 48 30240 49 233280 50 45360 \n\n- 2\u00a0months, 2\u00a0weeks ago", "date": "2020-09-29 12:18:09", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/how-to-find-an-n-digit-number-with-m-number-of/", "openwebmath_score": 0.6608328223228455, "openwebmath_perplexity": 900.9216498143144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9843363540453381, "lm_q2_score": 0.8723473647220787, "lm_q1q2_score": 0.8586832244515898}}
{"url": "https://math.stackexchange.com/questions/3761308/what-is-the-indefinite-integral-of-int-left-frac15x-right-dx/3761320", "text": "# What is the indefinite integral of $\\int \\left(-\\frac{1}{5x}\\right) dx$?\n\nWhy does $$\\int \\left(-\\frac{1}{5x}\\right) dx = -\\frac{1}{5}\\ln(x)+c$$ and not $$-\\frac{1}{5}\\ln(5x)+c$$ instead?\n\nWhen I differentiate both solutions I get the same answer, $$-\\frac{1}{5x}$$.\n\nCan anyone give me an explanation as to why? Thank you.\n\n\u2022 $\\int -\\frac{1}{5x} dx=-\\frac{1}{5}\\int\\frac{1}{x}dx$ Jul 18, 2020 at 14:48\n\u2022 The expressions are equivalent since $\\ln(x)$ and $\\ln(5x)=\\ln(5)+\\ln(x)$ differ by a constant. Jul 18, 2020 at 14:53\n\u2022 Also note that you are implicitly assuming that the domain is the positive real numbers. You need a different antiderivative for the negative real numbers. Jul 18, 2020 at 14:55\n\n## 5 Answers\n\nBoth are correct: on one side, $$\\int -\\frac{1}{5x}dx = -\\frac15 \\int \\frac{1}{x}dx = -\\frac15\\ln(x) + c,$$ because of the linearity of integrals $$\\int af(x) dx = a\\int f(x) dx$$ for every constant number $$a$$ (in particular $$a = -\\frac15$$ in this case).\n\nOn the other side, you might also change variable, and set $$u = 5x$$, if you wish, but in that case $$du = 5dx$$ and therefore $$\\begin{split} \\int - \\frac{1}{5x} dx &= - \\int \\frac{1}{u} \\frac{du}{5} \\\\ &= -\\frac{1}{5} \\ln(u) + c\u2019 \\\\ &= -\\frac15\\ln(5x) + c\u2019 = -\\frac15 \\ln(x) - \\frac15\\ln(5) + c\u2019 \\end{split}$$ Since $$\\ln(5x) = \\ln(5) + \\ln(x)$$ it\u2019s just a metter of choosing a different constant $$c$$ or $$c\u2019$$.\n\nThey clearly both give the same function if you differentiate, for they differ for a constant value $$\\ln(5)$$ whose derivative is zero.\n\n\u2022 Thank you, this really helped. Jul 18, 2020 at 15:37\n\u2022 We need to include the absolute value sign in $\\ln|x|$ since we don't know if $x>0$. Jul 19, 2020 at 3:31\n\u2022 @Axion004 in which case you need also to use two constants $c$ for $x>0$ and $x<0$ as they could be different. I have assumed $x>0$, albeit implicitly, in order to show why there are two possible solutions which look different but they are the same. Jul 19, 2020 at 9:15\n\nWell, notice that:\n\n$$\\int-\\frac{1}{\\text{n}x}\\space\\text{d}x=-\\frac{1}{\\text{n}}\\int\\frac{1}{x}\\space\\text{d}x=\\text{C}-\\frac{\\ln\\left|x\\right|}{\\text{n}}\\tag1$$\n\n\u2022 Yeah i get that but is the other way still right? I mean, I got a different answer. Jul 18, 2020 at 14:51\n\nBoth answers are equivalent.\n\nThrough the logarithm product rule,\n\n$$-\\frac15\\ln(5x) + C = -\\frac15(\\ln(x) + \\ln(5)) + C$$\n\nThen simplifying:\n\n$$\\text{LHS} = -\\frac15\\ln(x) + \\left( -\\frac15\\ln(5) + C \\right)$$\n\nSince $$C$$ is a constant, $$-\\dfrac15\\ln(5) + C$$ is a different constant.\n\nThen let $$-\\dfrac15\\ln(5) + C = C_1$$. Then\n\n$$\\text{LHS} = -\\frac15\\ln(x) + C_1$$\n\nWhich is the first expression in your question.\n\nThe reason why they both have the same derivative is because they differ by a constant.\n\nYour answer is probably not considered the best answer because it is less \u201csimplified\u201d than the other.\n\n$$\\bullet$$ See this $$\\int -\\frac{dt}{5t} = -\\frac{1}{5} \\int \\frac{dt}{t} = -\\frac{1}{5} \\ln\\lvert t \\rvert + C$$\n\nAnd also this:\n\n$$\\bullet$$ Let $$u = 5t \\implies du = 5 dt$$, and the substitution makes sense as the function is bijective on the whole of $$\\mathbb{R}$$.\n\ntherefore, \\begin{align*} -\\frac{1}{5} \\int \\frac{du}{u} = -\\frac{1}{5} \\ln\\lvert u \\rvert + C = -\\frac{1}{5} \\ln\\lvert 5t \\rvert + C \\end{align*} Now is it fine @mikejacob ?\n\n$$\\int\\frac1x\\,dx=\\ln\\lvert x\\rvert+c=\\ln\\lvert x\\rvert+\\ln a+c^\\prime=\\ln\\lvert ax\\rvert+c^\\prime$$\n\n, where $$c,c^\\prime\\in\\mathbb R,a\\in\\mathbb R^+$$.", "date": "2022-06-27 05:07:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3761308/what-is-the-indefinite-integral-of-int-left-frac15x-right-dx/3761320", "openwebmath_score": 0.923105001449585, "openwebmath_perplexity": 299.60657833294954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363545048391, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8586832232190706}}
{"url": "https://math.stackexchange.com/questions/513974/5-letter-arrangements-of-the-word-statistics", "text": "5 Letter Arrangements of the word 'Statistics'\n\nHow many different 5-letter 'words' can be formed from the word 'statistics'?\n\nI really am pretty stumped. I understand how to calculate more simpler questions in which each letter of the word is different using the Permutation formula n!/(n-r)! I can also deal with questions where there are repeating letters, but the 'new' words that are being created are the same length as the original word. But for this type of question where there are repeating letters and the 'new words' are shorter than the original, I don't know where to start. I don't need the precise answer - I just want to know how to get there. In fact, I know the answer (1390) but I cannot come up with a solution. I have tried using the permutation formula but it doesn't seem appropriate for this question.\n\n\u2022 What is your progress on this problem? What are your ideas? \u2013\u00a0Sasha Patotski Oct 3 '13 at 23:05\n\nI will give a basic counting solution. If you want an answer using generating functions, look at the answer to this question.\n\nIf all five letters are different (which happens when you use one of each), you get $5!=120$ words.\n\nIf only two of the letters are equal (two 'S', two 'T', or two 'I', and $\\binom{4}{3}$ ways to choose the remaining three), you have $\\frac{5!}{2!}=60$ words for each of {'S', 'T', 'I'}. In total $720$ words.\n\nIf you have \"two pairs\" (there are three ways to get the pairs, and three ways to get the last letter in each case), there are $\\frac{5!}{2!2!}=30$ words. In total for all these cases, there are $270$ words.\n\nIf you have \"full house\" (there are two ways to get three equal, and two ways for each of those to get the last pair), there are $\\frac{5!}{3!2!}=10$ words. In total for these cases, there are $40$ words.\n\nIf you have three equal and two different (the triple can be had in two ways, and the remaining two can be had in $\\binom{4}{2}=6$ ways for each triple), there are $\\frac{5!}{3!}=20$ words. For all these cases, there are $240$ words.\n\nSumming all of these cases gives $1390$ words.\n\n\u2022 The answer given to the question in the textbook was 1390. Your answer makes sense to me but it seems that some multiple is missing from the calculations... I can't see where though. \u2013\u00a0frantastic Oct 4 '13 at 0:41\n\u2022 @frantastic, I'm not in a position to review the argument carefully right now, but never discount the possibility that the textbook answer is wrong. Often, you will be able to find a list of errata for your textbook somewhere online (often on the writer's or publisher's website). \u2013\u00a0dfeuer Oct 4 '13 at 1:35\n\u2022 @frantastic: I hade made an error in the \"pair\" case, forgetting to count the number of ways to choose the last three. In the last case, I had gotten 120 instead of 240 through bad mental calculation. \u2013\u00a0M\u00e5rten W Oct 4 '13 at 8:29", "date": "2019-10-22 10:47:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/513974/5-letter-arrangements-of-the-word-statistics", "openwebmath_score": 0.7780852317810059, "openwebmath_perplexity": 262.82960892032753, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988312740283122, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8586725821256508}}
{"url": "https://math.stackexchange.com/questions/2083148/finding-the-probability-by-combination", "text": "# Finding the probability by combination?\n\nQuestion as follows A bag contains 50 tickets numbered 1,2,3,....,50 of which five are drawn at random and arranged in ascending order of the number appearing on the tickets ( x1 < x2 < x3 < x4 < x5 ). Find the probability that x3 = 30.\n\nTotal number of elementary events =50C5 Given,third ticket =30\n\n=> first and second should come from tickets numbered 1 to 29 =29C2\u00a0ways and remaining two in\u00a020C2\u00a0ways.\n\nTherfore,favourable number of events =\u00a029C2\u00d720C2\n\nHence,required probability =\u00a0(29C2\u00d720C2)/50C5\u00a0=551 / 15134\n\nAnswer given is 551/15134 but in the solution they have taken the case where x2 will be less than x1. Is my assumption correct? if it is not then what could be the correct answer.\n\n\u2022 You need to provide the full method, not just the numeric bottom line, otherwise your question is practically unreadable. P.S., I did not down-vote you, because I see good intentions and effort made in the question. Nevertheless, you need to clarify things here if you're hoping to get an answer. Jan 4, 2017 at 10:37\n\u2022 The two numbers less than 30 you refer to are chosen, and \"arranged in ascending order..\" Jan 4, 2017 at 10:37\n\u2022 the number of ways you can select to satisfy the condition is (any 2 from 1 - 29) x (any 1 from 30) x (any 2 from 31 - 50) = 29C2 x 1C1 x 20C2 ......... for the condition to be true, 30 had to be selected, and two that were smaller and two that were larger, the actual ordering leads to that condition, but is not part of the calculation\n\u2013\u00a0Cato\nJan 4, 2017 at 10:41\n\u2022 You are taking the question as wrong. Suppose you have option to pick 2 numbers out of 29. You picked 23, 12. In C(29,2) ways. Then you have to arrange them in ascending order. That is 12, 23. Only 1 way. So we have 1 * C(29,2). Similarly for last 2 numbers. Jan 4, 2017 at 10:47\n\u2022 the order they are drawn doesn't matter, there will always be a lowest number, then a 2nd ranked number, then a 3rd ranked number. Can you see that if 30 was drawn, then there has to be exactly 2 drawn from 1-29? If exactly 1 was drawn from 1-29, then the condition would not be met. 1,2,30,31,32 is a set of tickets that meets the condition, I'm only counting that once\n\u2013\u00a0Cato\nJan 4, 2017 at 10:49\n\nRequired probability = $\\frac{\\binom{29}{2} \\times \\binom{20}{2}}{\\binom{50}{5}}$\nBecause $x_3$=30 is fixed. You pick two numbers from first 29 and arrange them in ascending order. Similarly 2 numbers from last 20 and arrange them in ascending.", "date": "2022-10-07 18:57:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2083148/finding-the-probability-by-combination", "openwebmath_score": 0.6412374377250671, "openwebmath_perplexity": 479.3992545707724, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127420043055, "lm_q2_score": 0.8688267762381844, "lm_q1q2_score": 0.8586725735507212}}
{"url": "http://www.awv.sk/1axc9/6abae4-set-operations-proofs", "text": "/ By / / 0 Comments\n\n# set operations proofs\n\n1 - 6 directly correspond is also in By de\ufb01nition of intersection, x \u2208 A and x \u2208 B. We can use the set identities to prove other facts about sets. &= \\{x\\mid x \\in \\overline{A}\\wedge x \\in \\overline{B} )\\} \\\\ Here are some basic subset proofs about set operations. Theorem: For any sets, $$|A\\cap B|\\le|A|$$ and $$|A\\cap B|\\le|B|$$. Note here the correspondence of Hence . $\\bigcup_{i=1}^{n} S_i\\,.$ This can also proven using set properties as follows. Since , . This proof might give a hint why the equivalences and set identities tables are so similiar. B . and between Back to Schedule Consider an arbitrary element x. Theorem: For any sets, $$\\overline{A\\cup B}= \\overline{A} \\cap \\overline{B}$$. It's the table of logical equivalences with some search-and-replace done to it. 13. . Hence . A, and ------- \u00a0\u00a0\u00a0 De Morgan's Laws &= A\\cap \\overline{B}\\cap A\\cap \\overline{C} \\\\ by the distribution Theorem: For any sets, $$A-B = A\\cap\\overline{B}$$. \\begin{align*} Could have also given a less formal proof. ------- Associative Laws Since (from ), For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). Theorem For any sets A and B, B \u2286 A\u222a B. Proof\u2026 B ) B &= A\\cap \\overline{B}\\cap \\overline{C} \\\\ The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: ) by the definition of ( B - A ) . $$A\\cup{U}= {U}\\\\A\\cap\\emptyset= \\emptyset$$, $$(A\\cup B)\\cup C = A\\cup(B\\cup C)\\\\(A\\cap B)\\cap C = A\\cap(B\\cap C)$$, $$A\\cup(B\\cap C) =(A\\cup B)\\cap(A\\cup C)\\\\A\\cap(B\\cup C) = (A\\cap B)\\cup(A\\cap B)$$, $$\\overline{A\\cap B}=\\overline{A} \\cup \\overline{B}\\\\\\overline{A\\cup B}= \\overline{A} \\cap \\overline{B}$$, $$A\\cup(A\\cap B) = A \\\\ A\\cap(A\\cup B) = A$$, $$A\\cup\\overline{A} = {U}\\\\A\\cap\\overline{A} = \\emptyset$$, Written $$A\\cup B$$ and defined B . Then if , then since . A x\\in S \\wedge x\\in\\overline{S} \\\\ For example, (b) can be proven as follows: ------- Idempotent Laws by the definition of . Properties of Set Operation Subjects to be Learned . The properties 1 6 , and 11 A B Let x be an arbitrary element in the universe. Basic properties of set operations are discussed here. Furthermore a similar correspondence exists between \\[\\{1,2,3,4\\}\\cap\\{3,4,5,6\\} = \\{3,4\\}\\,., For example, \\begin{align*} Then by the definition of the operators, &= (A-B)\\cap (A-C)\\,.\\quad{}\u220e A. Proof: Let x \u2208 A\u2229B. , These can also be proven using 8, 14, and 15. ( B - A ) Then . &= \\{x\\mid x\\in A \\wedge x\\in \\overline{B}\\} \\\\ Let us prove some of these properties. This can also be proven in the similar manner to 9 above. \\end{align*}. A \u2022 Applying this to S we get: \u2022 x (x S x S) which is trivially True \u2022 End of proof Note on equivalence: \u2022 Two sets are equal if each is a subset of the other set. but also for others. 8. ( cf. ) and implications If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. \\mathbf{R} = \\mathbf{Q} \\cup \\overline{\\mathbf{Q}}\\,.\\], Written $$A\\cap B$$ and defined ( cf. ) x 11. Hence. (See example\u00a010 for an example of that too.). Here is an example. by the definition of . Be careful with the other operations. from the equivalences of propositional logic. Return to the course notes front page. Theorem: For any sets, $$|A\\cup B|\\ge|A|$$ and $$|A\\cup B|\\ge|B|$$. ------- \u00a0\u00a0\u00a0 Identity Laws 4. Next -- Recursive Definition x\\in S \\wedge x\\notin{S}\\,. As an example, we can prove one of De\u00a0Morgan's laws (the book proves the other). 9. 7. Thus we see that these sets contain the same elements.\u220e, More Formal Proof: By definition of the set operations, the commutativity of Thus $$A-B\\not\\subseteq A$$. x \\overline{\\mathbf{Q}} = \\mathbf{R}-\\mathbf{Q} \\,.\\]. . A A \\end{align*}\\]. This section contains many results concerning the properties of the set operations. Proof for 9: Let x be an arbitrary element in the universe. &= \\{x\\mid x\\in (A \\cap \\overline{B})\\} \\\\ x\\in S\\cap\\overline{S}\\\\ Also since , . Then there must be an element $$x$$ with $$x\\in(A-B)$$, but $$x\\notin A$$. Also .", "date": "2022-05-21 18:31:16", "meta": {"domain": "awv.sk", "url": "http://www.awv.sk/1axc9/6abae4-set-operations-proofs", "openwebmath_score": 0.9997114539146423, "openwebmath_perplexity": 769.7894657650262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9883127430370156, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8586725643776266}}
{"url": "https://math.stackexchange.com/questions/2025619/antiderivatve-to-find-a-function-such-the-the-slope-is-true-for-all-points-x-y", "text": "# Antiderivatve to find a function such the the slope is true for all points $(x,y)$\n\n\"The graph of a certain function $f$ has the slope $4x^3-5$ at each point $(x,y)$ and the line $x+y=0$ is the tangent line to the graph. Find the function $f$.\"\n\nI took the antiderivative to get $f(x)=x^4-5x+C$ but I'm not really sure how to get the initial conditions. I have that $x+y=0$ or $y=-x$ so the slope is $-1$. I can then plug it into $-1=4x^3-5$ and solve for $x$ to get $x=1$.\n\nI can then plug that x value into $y=-x$ to get $y=-1$. Since the tangent line and the graph must share the same common point then $(1,-1)$ must be on the graph of $f(x)$. So I can solve for the initial condition which means that $-1=1-5+C$ or $3=C$ so $f(x)=x^4-5x+3$ . Is the method correct?\n\nYour method, and your result, are correct. You can verify this solving the system of the line and the quartic: $$\\begin {cases} y=-x\\\\ y=x^4-4x+3 \\end{cases}$$ this gives the equation $$-x=x^4-5x+3 \\iff x^4-4x+3=0 \\iff (x-1)^2(x^2+2x+3)=0$$\nThat has a double root at $x=1$, so the line is tangent to the curve at this point.", "date": "2019-12-07 14:37:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2025619/antiderivatve-to-find-a-function-such-the-the-slope-is-true-for-all-points-x-y", "openwebmath_score": 0.8501680493354797, "openwebmath_perplexity": 60.75705084935195, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227580598143, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8586680651974469}}
{"url": "https://atekihcan.github.io/CLRS/02/E02.01-04/", "text": "Consider the problem of adding two $$n$$-bit binary integers, stored in two $$n$$ element arrays $$A$$ and $$B$$. The sum of the two integers should be stored in binary form in an $$(n + 1)$$ element array $$C$$. State the problem formally and write pseudocode for adding the two integers.\n\nThe problem can be formally stated as\u2026\n\nInput: Two $$n$$ bit binary integers stored in two $$n$$ element array of binary digits (either 0 or 1) $$A = \\langle a_1, a_2, ... , a_n \\rangle$$ and $$B = \\langle b_1, b_2, ... , b_n \\rangle$$.\n\nOutput: A $$(n + 1)$$ bit binary integer stored in $$(n + 1)$$ element array of binary digits (either 0 or 1) $$C = \\langle c_1, c_2, ... , c_{n+1} \\rangle$$ such that $$C = A + B$$.\n\nWe also assume the binary digits are stored with least significant bit first, i.e. from right to left, first bit in index $$1$$, second bit in index $$2$$, and so on. Why we are doing this is discussed after the pseudocode.\n\n$$\\textsc {Add-Binary }(A, B)$$\\begin{aligned}1& \\quad n=\\textsc {Max}(A.length,\\,B.length) \\\\2& \\quad \\text {let }C[n+1]\\text { be }\\text { new }\\text { array } \\\\3& \\quad carry=0 \\\\4& \\quad \\textbf {for }i=1\\textbf { to }n \\\\5& \\quad \\qquad C[i]=\\textsc {}(A[i]+B[i]+carry)\\mod2 \\\\6& \\quad \\qquad carry=\\lfloor \\textsc {}(A[i]+B[i]+carry)/2\\rfloor \\\\7& \\quad C[n+1]=carry \\\\8& \\quad \\textbf {return }C \\\\\\end{aligned}\n\n#### Left to Right or Right to Left\n\nAn earlier version of the solution presented here assumed the least significant bit was stored in index $$n$$ instead of index $$1$$. Which made the solution not only wrong (it did not handle all possible cases properly), it also caused a great deal of confusion in the comments section.\n\nHere is why assuming least significant bit in index $$n$$ will make the problem unnecessarily complicated.\n\nConsider the following two binary additions\n\nThe one on the left adds $$111_b$$ and $$1_b$$ to $$1000_b$$. In this case, $$n = 3$$ and we end up with final array $$C$$ of length $$n + 1 = 4$$. The array indices are shown in blue with the assumption of storing bits as we they appear visually from left to right, i.e. most significant bit in first index and least significant bit in last index, $$n$$.\n\nIn this particular case there is no complication, and we could have just designed our pseudocode to iterate from the opposite direction, from $$n$$ downto $$1$$, and stored the result of the addition in line 4 in $$C[i + 1]$$ and the final $$carry$$ in line 6 in $$C[1]$$.\n\nHowever, note that for the addition on the right, $$110_b$$ and $$1_b$$ sums up to $$111_b$$, and we end up with final array $$C$$ of length $$n = 3$$. In this case, $$C[1]$$ is empty (highlighted with light red), and we are left with the additional task of shifting all the elements to the left to meet our initial assumption of having most significant bit at index $$1$$.\n\nOne can argue that having zero in the first index is not a deal breaker, but depending on the use case it might add up to redundant work. For example, let\u2019s say we need to repeatedly do this addition. And every time we end up with a case like the right one. Then we would keep on adding redundant zeroes in the beginning of the resulting array.\n\n#### Python Code\n\n# Bitwise binary addition of two lists # containing binary digits with least significant bit first def AddBinary(A, B): carry = 0 n = max(len(A), len(B)) C = [0 for i in range(n + 1)] for i in range(n): # One of A and B has length less than n # We need to treat index out of bound for that array # This is not explicitly handled in pseudocode a = A[i] if i < len(A) else 0 b = B[i] if i < len(B) else 0 # Modulo for sum and integer division for carry C[i] = (a + b + carry) % 2 carry = (a + b + carry) // 2 C[n] = carry return C # Utility function for converting decimal to binary def DecimalToBinary(num): if num == 0: return [0] ret = [] while num > 0: ret.append(num & 1) num = num >> 1 return ret # Utility function for converting binary to decimal def BinaryToDecimal(lst): num = 0 for i in range(len(lst)): num += lst[i] << i return num # Test import random num_failed = 0 total_tests = 100 for i in range(total_tests): # Create two random integers a = random.randint(0, 10000) b = random.randint(0, 10000) a_ = DecimalToBinary(a) b_ = DecimalToBinary(b) sum = BinaryToDecimal(AddBinary(a_, b_)) # Check if the sum is correct if sum != (a + b): num_failed += 1 print(f\"#{i:<2} {a} + {b} = {sum} [Expected {(a + b)}]\") if num_failed > 0: print(f\"\\nFailed {num_failed}/{total_tests}\") else: print(f\"Passed {total_tests}/{total_tests} tests\")", "date": "2023-03-28 23:58:59", "meta": {"domain": "github.io", "url": "https://atekihcan.github.io/CLRS/02/E02.01-04/", "openwebmath_score": 0.7745254039764404, "openwebmath_perplexity": 908.4966649391847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692271169855, "lm_q2_score": 0.8791467706759583, "lm_q1q2_score": 0.8586355970384818}}
{"url": "https://math.stackexchange.com/questions/1536267/finding-the-values-of-5ab-if-ax2bx10-does-not-have-2-distinct-roots", "text": "# Finding the values of $5a+b$ if $ax^2+bx+10$ does not have $2$ distinct roots\n\nIf $ax^2+bx+10=0$ does not have two distinct real roots where( $a$ and $b$ are real) then the possible values of $5a+b$ from the given 4 options(as obviously there can be infinite possible values..but I only need to find out which of the four given possible values is one of these ...in short it is a Multi choice-multi-correct question) are-( there can be more than one option correct)\n\n1) -3\n\n2) -2\n\n3) -1\n\n4) 0\n\nmy attempt Here are some conclusions I have drawn-\n\n\u2022 it is given that it does not have two distinct real roots so either( both roots are imaginary and conjugates of each other)- $D(b^2-4ac)<0$ (that is as $c=10$ so $b^2-40a<0$...or another case is both roots become equal that is $b^2-40a=0$.\n\n\u2022 also I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible but the answers given are $-2,-1,0$ (Edit:apparently $0$ is not the answer as pointed out be Tim phan below..but I still need help in proving it is equal to $-1,-2$)....also I am unable to get other answers I have tried getting $5a+b$ by keeping $x=5$ to get $5(5a+b+2)$ but I don't know whether it will be equal to zero or less than or greater than zero... As I don't know whether $5$ is a root of $f(x)$ or not..neither it is given in question...\n\nAny help??\n\n\u2022 I think you need more constrains. Here is why. Like you pointed out, essentially, you are solving a system of 2 equations with 3 unknown. $b^2 - 4 a c < 0$ and $b + 5a = N$. \u2013\u00a0Paichu Nov 19 '15 at 3:28\n\u2022 In the attempted solution, you refer to $b^-40c$. Did you omit saying that $c=10$? Are $a,b,c$ restricted in some way, say to integers? \u2013\u00a0Andr\u00e9 Nicolas Nov 19 '15 at 3:30\n\u2022 If that is the case, then you can try going about it in this way. Assume $b + 5a = 0$, then $b = -5a$ and $a = -\\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together you will get two equations: $b^2 < -\\frac{4}{5}b c$ and $25a^2 < 4a c$. You then proceed with 4 cases $a,b>0$, $a,b<0$, $a>0,b<0$, $a<0,b>0$. If all of them lead to contradiction, then it is not possible. \u2013\u00a0Paichu Nov 19 '15 at 3:42\n\u2022 Why didn't you write $ax^2+bx+10$, instead of \"$c$\". Tbh, it looks like you were attempting to make it confusing. \u2013\u00a0YoTengoUnLCD Nov 19 '15 at 3:50\n\u2022 Let $a=1$ and $b=1.01, 1.02, 1.03, 1.04,\\dots$. Don't have distinct real roots, indeed don't have real roots at all. \u2013\u00a0Andr\u00e9 Nicolas Nov 19 '15 at 4:21\n\nIf you write $5a + b = k$ and use this to get rid of the $a$ in the equation $b^2 - 40a \\leq 0$, you get:\n\n$$b^2 - 40\\left(\\frac{k - b}{5}\\right) \\leq 0$$\n\n$$b^2 + 8b \\leq 8k$$\n\nComplete the square:\n\n$$(b + 4)^2 \\leq 8k + 16$$\n\n$$(b + 4)^2 \\leq 8(k + 2)$$\n\nSince $b$ is arbitrary, the only restriction is that the left hand side is nonnegative. So you get:\n\n$$0 \\leq 8(k + 2)$$\n\n$$-2 \\leq k$$\n\nHence $k = 0, -1, -2$ are all possible.\n\nThe possible values (from those listed) are $0$, $-1$, and $-2$, obtained by $(a,b) = (0,0)$, $(1,-6)$, and $\\left(\\frac{2}{5},-4\\right)$. (The first two are not unique, but the third one is; see below.)\n\nNow, if $ax^2+bx+10$ does not have 2 distinct roots, then the discriminant is nonpositive, i.e. $b^2-40a\\le 0$. Hence, $40a\\ge b^2$, necessarily implying that $a\\ge 0$. Furthermore, $$b^2\\le 40a\\implies |b|\\le\\sqrt{40a}\\implies b\\ge -\\sqrt{40a}$$ and hence $5a+b \\ge 5a-\\sqrt{40a}$. It can be checked that $$\\min\\limits_{a\\ge 0}{\\left(5a-\\sqrt{40a}\\right)} = -2$$ achieved at $a = \\frac{2}{5}$. Hence, if $ax^2+bx+10$ does not have 2 distinct roots, then $$5a+b\\ge -2$$ with equality if and only if $a = \\frac{2}{5}$ and $b = -\\sqrt{40a} = -4$.\n\n\u2022 Why is this reasoning of my wrong....I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible \u2013\u00a0Freelancer Nov 19 '15 at 8:16\n\u2022 @Freelancer why can't we have $f(3)-f(2) = 0$? For example, if $f$ is constant, then this is true, and there's nothing saying it can't (since if it were constant, it would take the constant value $10$, which is nonzero) \u2013\u00a0Joey Zou Nov 19 '15 at 8:17\n\u2022 I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..(which can be only possible when both 3,2 are roots are of the equation...which is not allowed (as real roots should not be distinct here) \u2013\u00a0Freelancer Nov 19 '15 at 8:19\n\u2022 @Freelancer who cares if $f$ doesn't have two distinct roots? It says very little about the difference between its values. For example, $f(x) = x^2+1$ has no real roots, yet $f(1) - f(-1) = 0$, $f(2)-f(-2) = 0$, $f(3) - f(-3)=0$, and so on... \u2013\u00a0Joey Zou Nov 19 '15 at 8:21\n\nI found that if a = 1/5 and b = -1 then 5a+b = 0. Did i break any rules?\n\n\u2022 You found one solution ! Are there others ? \u2013\u00a0Tom-Tom Nov 19 '15 at 8:35\n\nAssume $b + 5a = 0$, then $b = -5a$ and $a = -\\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together to obtain $b^2 < -\\frac{4}{5}b c$ and $25a^2 < 4a c$.\n\n(1) If $a>0, b>0$, then we have: $b<-\\frac{4}{5}c$< which implies $c<0$. False.\n\n(2) If $a<0, b<0$, then we have: $-25|a|>4c$ which implies $c<0$. False.\n\nIf $a>0, b<0$, you have the same thing as in (1) and if If $a<0, b>0$, you have the same thing as in (2). Thus it is not possible for $b + 5a = 0$. Proceed similarly.\n\nSecond part: let $D>0$, assume $5a + b =-D$, so $b = -(D+5a)$. Then for this equation to not have two distinct real roots: $b^2 \\leq 40a$. replacing $a$ in to obtain: $(D+5a)^2 < 40a$ Notice that this implies $a > 0$. Simplify: $$D + 5a < \\pm\\sqrt{40 a}$$ Let $u = \\sqrt{a}$, then we have $5u^2 \\mp \\sqrt{40}u + D < 0$ is the condition, which is if we can find such $u$ under some condition on $D$, then the same condition is also the condition for $ax^2 + bx + 10 = 0$ to not have two distinct real roots. Solve this to get $$u=\\frac{1}{10}\\left(\\sqrt{40} - \\sqrt{40 - 20 D}\\right)$$ this has a solution if $D \\leq 2$, which means if $D \\leq 2$, then the $ax^2 + bx + 10 = 0$ does not have two distinct real roots. Thus the possible solutions are $5a + b = -1, -2$, or (3) and (2). Please check my algebra carefully.\n\n\u2022 You can't do that yet, since you do not know if $a$ is positive or negative, which can affect the inequality sign. \u2013\u00a0Paichu Nov 19 '15 at 3:58\n\u2022 I edit the second part there. Check the algebra carefully though. Not super logical, but it is one way of doing that. \u2013\u00a0Paichu Nov 19 '15 at 4:52\n\u2022 You're right. I just fixed it. \u2013\u00a0Paichu Nov 19 '15 at 5:16\n\u2022 Also how can we assume $5a+b=-D$ ?? \u2013\u00a0Freelancer Nov 19 '15 at 5:17\n\u2022 $D\\le 2$ actually means $D = -1$ or $D=-2$. Also, how does $a>0,b<0$ give the same thing as in (1)? \u2013\u00a0Joey Zou Nov 19 '15 at 7:50", "date": "2019-10-20 08:21:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1536267/finding-the-values-of-5ab-if-ax2bx10-does-not-have-2-distinct-roots", "openwebmath_score": 0.941903293132782, "openwebmath_perplexity": 199.80033419847456, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692311915195, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.8586355867041027}}
{"url": "https://math.stackexchange.com/questions/2557479/motivation-for-the-ring-product-rule-a-1-a-2-a-3-cdot-b-1-b-2-b-3-a/2557744", "text": "Motivation for the ring product rule $(a_1, a_2, a_3) \\cdot (b_1, b_2, b_3) = (a_1 \\cdot b_1, a_2 \\cdot b_2, a_1 \\cdot b_3 + a_3 \\cdot b_2)$\n\nIn a lecture, our professor gave an example for a ring. He took it out of another source and mentioned that he does not know the motivation for the chosen operation.\n\nOf course, it's likely that somebody just invented an arbitrary operation satisfying ring axioms. I'd still like to try my luck whether anyone here can decipher the operation and give any kind of motivation for that example.\n\nOn $\\mathbb{R}^3$ define the operations $+$ and $\\cdot$ by \\begin{aligned} (a_1, a_2, a_3) + (b_1,b_2,b_3) &= (a_1+b_1,a_2+b_2,a_3+b_3) \\\\ (a_1, a_2, a_3) \\cdot (b_1, b_2, b_3) &= (a_1 \\cdot b_1, a_2 \\cdot b_2, a_1 \\cdot b_3 + a_3 \\cdot b_2). \\end{aligned} (The $+$ and $\\cdot$ operations on the right side are the usual addition and multiplication from $\\mathbb{R}$.) With those operations, one can confirm that $\\left(\\mathbb{R}^3, +, \\cdot \\right)$ is a ring.\n\n\u2022 In case anyone is wondering, the multiplicative identity is $(1,1,0)$. \u2013\u00a0Alex Provost Dec 8 '17 at 19:00\n\u2022 I think it's just an arbitrary thing. I can't see any motivating pattern here. If someone else does see something, that will be surprising and very interesting. Also, in case anyone is wondering, the multiplication is associative (I checked). \u2013\u00a0Zach Teitler Dec 8 '17 at 19:04\n\u2022 I can't check at the moment, but is multiplication here commutative? Perhaps someone wanted to construct a noncommutative ring without referencing matricies? \u2013\u00a0Andrew Tawfeek Dec 8 '17 at 19:14\n\u2022 The multiplication is not commutative, take $(1,0,0) \\cdot (0,0,1) \\neq (0,0,1) \\cdot (1,0,0)$. \u2013\u00a0Qi Zhu Dec 8 '17 at 19:16\n\u2022 mentioned that he does not know the motivation for the chosen operation Seriously?! If you show him the upper triangular matrix ring he will be rather abashed, then :) \u2013\u00a0rschwieb Dec 8 '17 at 21:36\n\nThis is just matrix multiplication in disguise. Specifically, if you identify $(a_1,a_2,a_3)$ with the matrix $\\begin{pmatrix}a_1 & a_3 \\\\ 0 & a_2\\end{pmatrix}$, these operations are the usual matrix operations: $$\\begin{pmatrix}a_1 & a_3 \\\\ 0 & a_2\\end{pmatrix}+\\begin{pmatrix}b_1 & b_3 \\\\ 0 & b_2\\end{pmatrix}=\\begin{pmatrix}a_1+b_1 & a_3+b_3 \\\\ 0 & a_2+b_2\\end{pmatrix}$$ $$\\begin{pmatrix}a_1 & a_3 \\\\ 0 & a_2\\end{pmatrix}\\begin{pmatrix}b_1 & b_3 \\\\ 0 & b_2\\end{pmatrix}=\\begin{pmatrix}a_1b_1 & a_1b_3+a_3b_2 \\\\ 0 & a_2b_2\\end{pmatrix}$$\n\n\u2022 Oh of course! Nice and simple, thanks, the ring is much more motivated now! \u2013\u00a0Qi Zhu Dec 8 '17 at 21:32\n\u2022 Unfortunately I realized that's exactly what I was describing a minute too late... \u2013\u00a0rschwieb Dec 8 '17 at 21:32\n\nIt is isomorphic to the ring of matrices\n\n$$\\left\\{\\begin{bmatrix}a_1&a_3\\\\0&a_2\\end{bmatrix}\\,\\middle|\\,a_1, a_2,a_3\\in \\mathbb R\\right\\}$$\n\nIt's a semiprimary ring whose Jacobson radical is the subset with $a_1=a_2=0$. The Jacobson radical is nilpotent, and $R/J(R)\\cong\\mathbb R\\times\\mathbb R$. Here is a list of more properties of such a ring.\n\nThis sort of ring is fairly famous, and has nice interpretations. One of them is that if you select a chain of subspaces $\\{0\\}<V<W<\\mathbb R\\times \\mathbb R$ ($W$ of dimension $1$, $V$ of dimension $2$) then the linear transformations of $\\mathbb R\\times\\mathbb R$ which stabilize this chain is isomorphic to this triangular matrix ring. That is, $\\phi$ stabiliezes the chain if $\\phi(V)\\subseteq\\phi(W)$.\n\nIncidentally, you are always going to be able to extract some sort of matrix presentation for a multiplication like you are describing, because you can rely on it being a finite dimensional algebra. If it really is a valid ring multiplication, it's bilinear, and so you can work on figuring out what a logical 'basis' is and then deduce what it looks like with matrices.\n\n\u2022 This should be the accepted and most upvoted answer. \u2013\u00a0Xam Dec 9 '17 at 5:54\n\u2022 In your last paragraph, what exactly do you mean by \"a multiplication like you are describing\"? \u2013\u00a0Jack M Dec 9 '17 at 11:07\n\u2022 @JackM Multiplication of $n$-tuples over a field, at least. Because that defines a representation with square matrices. \u2013\u00a0rschwieb Dec 9 '17 at 13:18\n\u2022 @rschwieb Surely not any ring structure on $\\mathbb F^n$ is isomorphic to a ring of $\\mathbb F$-matrices. Maybe a ring structure in which addition is component wise, and multiplication is given by quadratic polynomials? \u2013\u00a0Jack M Dec 9 '17 at 13:29\n\u2022 @JackM You're right, I should emphasize the pointwise addition. Remember I'm stipulating that the multiplication is actually a valid ring multiplication, not any rule with tuples. To rephrase, \"if multiplication on elements of $F^n$ defines a valid ring structure, then the ring can be represented with $n\\times n$ matrices.\" This is easy to see, because you just identify each tuple with the linear transformation it produces with the multiplication. \u2013\u00a0rschwieb Dec 9 '17 at 14:22", "date": "2020-11-24 12:32:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2557479/motivation-for-the-ring-product-rule-a-1-a-2-a-3-cdot-b-1-b-2-b-3-a/2557744", "openwebmath_score": 0.979472815990448, "openwebmath_perplexity": 526.2422588281735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692325496974, "lm_q2_score": 0.8791467548438126, "lm_q1q2_score": 0.8586355863518633}}
{"url": "https://math.stackexchange.com/questions/2819873/orthogonally-diagonalize-a-matrix-with-nonreal-eigenvectors", "text": "Orthogonally Diagonalize a matrix with nonreal eigenvectors\n\nI am given the matrix A and asked to orthogonally diagonalize it.\n\n$$A=\\begin{pmatrix} 1 & -i \\\\ i & 1 \\\\ \\end{pmatrix}$$\n\nWhile doing this I got $\\lambda = 0,2.$ Then I found the eigenvectors corresponding to the eigenvalues to be $$\\begin{pmatrix} i \\\\ 1 \\\\ \\end{pmatrix}$$ and $$\\begin{pmatrix} -i \\\\ 1 \\\\ \\end{pmatrix},$$ respectively. While trying to divide each eigenvector by its norm I run into a problem, take $V_1$ for example $\\|V_1\\| = 0$ so I obviously can't divide by $V_1$ by its norm to make it orthogonal. My question is, is $A$ even orthogonally diagonizable at all? Or if the $\\|V_1\\| = 0$ does this mean that it is already orthogonal and I can just use my eigenvectors as is to generate the matrix $U$ such that $A=UDU^*?$\n\n\u2022 $||V_1|| \\neq 0$. Same goes for norm of $V_2$. You have to use absolute values around the squared entries. \u2013\u00a0layabout Jun 14 '18 at 18:47\n\u2022 You mean to unitarily diagonalize it, no? \u2013\u00a0Mathematician 42 Jun 14 '18 at 18:48\n\u2022 You are not working with a real inner product, but a Hermitian inner product, i.e. $\\left\\langle x,y \\right\\rangle=\\sum_{i=1}^n x_i\\overline{y_i}$. \u2013\u00a0Mathematician 42 Jun 14 '18 at 18:50\n\u2022 The inner product is always positive definate. That is $\\langle x,x\\rangle \\ge 0$ and only equals $0$ if $x = 0.$ (and the norm is $\\sqrt {\\langle x,x\\rangle}$) To meet this criterion, we need a slightly different inner product definition when working with vectors over a complex field. $\\langle x,y\\rangle = x^\\dagger y$ ($\\dagger$ is the conjugate transpose) will do. So the norm $(i,1) = \\sqrt {(-i\\cdot i + 1\\cdot 1)}$ \u2013\u00a0Doug M Jun 14 '18 at 18:52\n\nThe norm of a vector $v$ is\n\n$$\\|v\\|^2 = (v^*)^Tv = \\sum_k v_k^* v_k$$\n\nwhere $^*$ denotes the complex conjugate. So in your case\n\n$$\\|v_1\\|^2 = \\pmatrix{i & 1}^* \\pmatrix{i \\\\ 1} = \\pmatrix{-i & 1} \\pmatrix{i \\\\ 1} = 1 + 1 = 2$$\n\n\u2022 Thank you very much. I had forgotten that there were special properties for cases like this. \u2013\u00a0Moseph Jun 14 '18 at 19:09\n\u2022 @Moseph Happy to help \u2013\u00a0caverac Jun 14 '18 at 19:20\n\nIt happens that $\\|(a,b)\\|=\\sqrt{|a|^2+|b|^2}$. Therefore, the norm of both vectors that you mentioned is $\\sqrt2$.", "date": "2021-07-30 12:57:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2819873/orthogonally-diagonalize-a-matrix-with-nonreal-eigenvectors", "openwebmath_score": 0.9238721132278442, "openwebmath_perplexity": 227.37605864918586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211575679041, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8586196195111598}}
{"url": "https://math.stackexchange.com/questions/516507/how-to-find-the-general-solution-of-tan-leftx-frac-pi-3-right3-tan-l/516520", "text": "# How to find the general solution of $\\tan \\left(x+\\frac{\\pi }{3}\\right)+3\\tan \\left(x-\\frac{\\pi }{6}\\right)=0$\n\nFind the general solution of the equation.\n\n\\begin{eqnarray} \\tan \\left(x+\\frac{\\pi }{3}\\right)+3\\tan \\left(x-\\frac{\\pi }{6}\\right)=0\\\\ \\end{eqnarray}\n\nThe answers in my textbook are $n\\pi$ and $n\\pi +\\frac{\\pi }{3}$.\n\nPreviously, I compute the similar questions by using the following operations :\n\n\\begin{eqnarray} \\\\\\tan 3x&=&\\cot 5x\\\\ \\\\\\tan 3x&=&-\\tan \\left(\\frac{\\pi }{2}+5x\\right)\\\\ \\\\\\tan 3x&=&\\tan \\left(-\\frac{\\pi }{2}-5x\\right)\\\\ \\\\3x&=&-\\frac{\\pi }{2}-5x\\\\ \\\\8x&=&n\\pi -\\frac{\\pi }{2}\\\\ \\\\x&=&\\frac{n\\pi }{8}-\\frac{\\pi }{16}\\\\ \\end{eqnarray}\n\nBut now there is a 3 in front of the second tan.\n\nWhat should I do?\n\nI do not know whether my method is correct. If you have any other methods, would you mind telling me the methods?\n\nUpdate 1 : Found one of the answers, are my operations correct?\n\n\u2022 Is your question correct? The question seems likely than $tan(x+\\pi/3)=3tan(x-\\pi/6)$ \u2013\u00a0MS.Kim Oct 6 '13 at 12:18\n\u2022 Thx, the question now is correct. \u2013\u00a0Casper Oct 6 '13 at 12:24\n\nIf we assume that your question is to solve the equation $tan(x+\\pi/3)+3tan(x-\\pi/6)=0$. if we let $A=x+\\pi/3, B=x-\\pi/6$, then $A-B=\\pi/2$, so we can transform the term $tanA$ to $tan(\\pi/2+B)$.\n\nand because $tan(\\pi/2+B) = -cotB$, the equation will be $-cotB+3tanB=0$, $tan^2B=1/3$ so the general solution of $B=x-\\pi/6=n\\pi+\\pi/6$ or $n\\pi-\\pi/6$\n\nso $x=n\\pi +\\pi/3$ or $n\\pi$\n\nThe solution is given as follows.\n\nI don't know how to use spoiler for a block of code.\n\n\\documentclass[preview,border=12pt]{standalone}\n\\usepackage{amsmath}\n\\begin{document}\n\\abovedisplayskip=0pt\\relax\n\\begin{gather*}\n\\tan (x+\\frac{\\pi}{3}) + 3\\tan (x-\\frac{\\pi}{6})=0\\\\\n\\tan (x+\\frac{\\pi}{3}) = 3\\tan (\\frac{\\pi}{6}-x)\\\\\n\\frac{\\tan x + \\tan \\frac{\\pi}{3}}{1-\\tan x \\tan \\frac{\\pi}{3}} = 3\\frac{\\tan \\frac{\\pi}{6} -\\tan x}{1+\\tan \\frac{\\pi}{6}\\tan x}\\\\\n\\frac{\\tan x + \\sqrt 3}{1-\\sqrt 3\\tan x } = 3\\frac{\\frac{\\sqrt 3}{3} -\\tan x}{1+\\frac{\\sqrt 3\\tan x}{3}}\\\\\n\\frac{\\tan x + \\sqrt 3}{1-\\sqrt 3\\tan x } = \\frac{ \\sqrt 3 -3 \\tan x}{1+\\frac{\\sqrt 3\\tan x}{3}}\\\\\n\\frac{\\tan x + \\sqrt 3}{1-\\sqrt 3\\tan x } = \\frac{ \\sqrt 3 -3 \\tan x}{1+\\frac{\\sqrt 3\\tan x}{3}}\\times\\frac{3}{3}\\\\\n\\frac{\\tan x + \\sqrt 3}{1-\\sqrt 3\\tan x } = \\frac{ 3\\sqrt 3 -9 \\tan x}{3+ \\sqrt 3\\tan x}\\\\\n(1- \\sqrt 3\\tan x) (3\\sqrt 3 -9 \\tan x) = (\\tan x + \\sqrt 3)(3+ \\sqrt 3\\tan x)\\\\\n3\\sqrt 3 -9\\tan x -9\\tan x +9\\sqrt 3\\tan^2x = 3\\tan x +\\sqrt 3\\tan^2 x +3\\sqrt 3 +3\\tan x\\\\\n8\\sqrt 3\\tan^2 x  -24 \\tan x=0\\\\\n\\sqrt 3\\tan^2 x -3\\tan x=0\\\\\n\\tan x (\\sqrt 3 \\tan x -3)=0\\\\\n\\tan x =0 \\text{ or } \\sqrt 3 \\tan x -3 =0\\\\\n\\tan x=0 \\text{ or } \\tan x =\\sqrt 3\\\\\nx=n\\pi \\text{ or } x=\\frac{\\pi}{3}+n\\pi\n\\end{gather*}\n\n\\end{document}\n\n\u2022 =\u53e3= This operation must take so much time when during exam...Is it the best way? \u2013\u00a0Casper Oct 6 '13 at 13:42\n\u2022 @CasperLi: Not much even though you use LaTeX during the exam. :-) \u2013\u00a0kiss my armpit Oct 6 '13 at 13:44\n\u2022 Are there any fast methods to input math/LaTeX? ~.~ \u2013\u00a0Casper Oct 6 '13 at 13:53\n\u2022 @CasperLi: It depends on the text editor you are using. If there is a shortcut key for each macro then you can type much faster. But I think the important point is your typing speed. :-) \u2013\u00a0kiss my armpit Oct 6 '13 at 13:55", "date": "2019-12-14 00:50:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/516507/how-to-find-the-general-solution-of-tan-leftx-frac-pi-3-right3-tan-l/516520", "openwebmath_score": 0.8964624404907227, "openwebmath_perplexity": 318.1830334539839, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974821156470663, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.8586196170195796}}
{"url": "https://math.stackexchange.com/questions/1953218/is-the-empty-set-is-a-subset-of-any-set-a-convention/1953511", "text": "# Is \"The empty set is a subset of any set\" a convention?\n\nRecently I learned that for any set A, we have $\\varnothing\\subset A$.\n\nI found some explanation of why it holds.\n\n$\\varnothing\\subset A$ means \"for every object $x$, if $x$ belongs to the empty set, then $x$ also belongs to the set A\". This is a vacuous truth, because the antecedent ($x$ belongs to the empty set) could never be true, so the conclusion always holds ($x$ also belongs to the set A). So $\\varnothing\\subset A$ holds.\n\nWhat confused me was that, the following expression was also a vacuous truth.\n\nFor every object $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set A.\n\nAccording to the definition of the vacuous truth, the conclusion ($x$ doesn't belong to the set A) holds, so $\\varnothing\\not\\subset A$ would be true, too.\n\nWhich one is correct? Or is it just a convention to let $\\varnothing\\subset A$?\n\n\u2022 Maybe think of it like this: The empty set is like an empty bag. The set $A$ is like a bag with some stuff in it. It's possible to put your hand into either bag and take nothing out. Since everything you could take out of the empty bag $\\varnothing$ was also in $A$, this means that $\\varnothing \\subset A$. This argument may need to be made rigorous, but I think the intuition is a good start. Oct 4, 2016 at 9:31\n\u2022 It's a consequence of the following rule: any statement of the form 'if $x \\in \\emptyset$, then y' is true. That rule is itself a consequence of the agreement that \"if P then Q\" means \"P is false or Q is true.\" On a philosophical level, I would distinguish this agreement from being a mere convention since it's an agreement about what reasoning itself means. But on a practical level, if you get annoyed thinking about these issues in any particular context, it's often better to shrug your shoulders, call it a convention, and move on! Oct 4, 2016 at 11:53\n\u2022 It's the way math works... It is not a convention, but a theorem: we have an axiom asserting that there is a unique set $y$ such that : $\\forall x \\lnot (x \\in y)$ and we call $y$ \"the emptyset\". Then we assume the def of set-inclusion : $A \\subset B \\leftrightarrow \\forall x (x \\in A \\to x \\in B)$ and we derive, by logical rules, that $\\emptyset \\subset A$, for any $A$. Oct 5, 2016 at 10:15\n\u2022 It is true that for every $x$ and every set $A$, if $x$ belongs to the empty set, then $x$ both belongs to $A$ and $x$ does not belong to $A$ Oct 6, 2016 at 0:32\n\u2022 \u201cFor every object $x$, if $x$ belongs to the empty set, then $x$ doesn\u2019t belong to the set $A$.\u201d This statement is vacuously true, indeed. However, it does not imply that $\\varnothing\\not\\subset A$. For the latter to be true, there should actually exist some $x$ such that $x\\in\\varnothing$ but $x\\notin A$. Oct 6, 2016 at 22:56\n\nThere\u2019s no conflict: you\u2019ve misinterpreted the second highlighted statement. What it actually says is that $\\varnothing$ and $A$ have no element in common, i.e., that $\\varnothing\\cap A=\\varnothing$. This is not the same as saying that $\\varnothing$ is not a subset of $A$, so it does not conflict with the fact that $\\varnothing\\subseteq A$.\n\nTo expand on that a little, the statement $B\\nsubseteq A$ does not say that if $x\\in B$, then $x\\notin A$; it says that there is at least one $x\\in B$ that is not in $A$. This is certainly not true if $B=\\varnothing$.\n\n\u2022 @BlueRaja-DannyPflughoeft: Which is to say that he misinterpreted it: it does not mean what he understood it to mean. Oct 4, 2016 at 17:47\n\u2022 @Searene You can derive anything from a contradiction. This is basic logic, and nothing wrong with it.\n\u2013\u00a0orlp\nOct 5, 2016 at 7:17\n\u2022 No you shouldn't say that 5=6 is true, but you can say that If 1=2, then 5=6 is true. The statement that x belongs to A is not necessarily true; what is true is precisely: for every object x, if x belongs to the empty set, then x also belongs to the set A, and that is the definition of the empty set being a subset of A. Oct 5, 2016 at 8:13\n\u2022 @Searene You are making a fundamental error. You cannot derive that the conclusion is true!!! If you have a statement: $\\forall x, x \\in \\emptyset \\rightarrow \\emptyset \\not\\subset A$ is true because $x \\in \\emptyset$ is false. This tells you nothing regarding $\\emptyset \\not\\subset A$. In an implication $P \\rightarrow Q$ if $P$ is false than the whole implication is true, but this does not say anything about the truth value of just $Q$. Oct 5, 2016 at 13:02\n\u2022 ... vacuously true, but that doesn\u2019t tell you anything about whether or not $\\varnothing$ is a subset of $A$. The truth of $\\forall x\\,(x\\in\\varnothing\\to x\\in A)$, on the other hand, means by the definition of $\\subseteq$ that $\\varnothing\\subseteq A$. Oct 5, 2016 at 17:25\n\nFrom Halmos's Naive Set Theory:\n\nA transcription:\n\nThe empty set is a subset of every set, or, in other words, $\\emptyset \\subset A$ for every $A$. To establish this, we might argue as follows. It is to be proved that every element in $\\emptyset$ belongs to $A$; since there are no elements in $\\emptyset$, the condition is automatically fulfilled. The reasoning is correct but perhaps unsatisfying. Since it is a typical example of a frequent phenomenon, a condition holding in the \"vacuous\" sense, a word of advice to the inexperienced reader might be in order. To prove that something is true about the empty set, prove that it cannot be false. How, for instance, could it be false that $\\emptyset \\subset A$? It could be false only if $\\emptyset$ had an element that did not belong to $A$. Since $\\emptyset$ has no elements at all, this is absurd. Conclusion: $\\emptyset \\subset A$ is not false, and therefore $\\emptyset \\subset A$ for every $A$.\n\n\u2022 This paragraph shines with the rare virtue of sympathy with (and generosity of spirit toward) the uninitiated. +1. Oct 4, 2016 at 15:16\n\u2022 I love this book. Oct 4, 2016 at 15:27\n\u2022 With the text as opposed to with a picture of the text? Oct 4, 2016 at 19:41\n\u2022 Okay, I don't see what search engines have to do with copyright violation issues, but in the meantime, I'd rather see the answer be accessibility compliant, and I guess it's your legal understanding that I'm responsible for any \"issues\" that might come up, so okay. Oct 4, 2016 at 19:48\n\u2022 I like Halmos's second demonstration because it highlights a frequently used technique: For an implication to be false, there must be a witness to its falseness. That is, there is an element of some set that is a counterexample. Conversely, if it is impossible that there are counterexamples, then the implication is true. Oct 5, 2016 at 12:59\n\nWhat confused me was that, the following expression was also a vacuous truth.\n\nFor every object of $$x$$, if $$x$$ belongs to the empty set, then $$x$$ doesn't belong to the set $$A$$.\n\nAs a complement (heh) to Brian Scott's (+1) answer, your argument shows that $$\\varnothing \\subset A^{c}$$, the complement of $$A$$. This statement is also (vacuously) true.\n\n\u2022 So the empty set, being a subset of both the set $A$ and its complement, is a subset of the intersection of the set $A$ with its complement, which is the empty set. Oct 5, 2016 at 9:05\n\u2022 I believe this is the key point. Oct 5, 2016 at 21:31\n\u2022 @CarstenS: Be careful that in ZF set theory the complement of a set is never a set! Oct 7, 2016 at 14:17\n\nVery subtle point:\n\n\"All x are not something\" does not imply \"Not all x are something\".\n\nThe first may be vacuously true. The second one can not. If the x are vacuous then the second one has to be \"vacuously false\" as all x of nothing are any property so it is impossible for them not to be any property.\n\nSo \"all elements of the empty set are not in $S$\" does not imply \"Not all elements of the empty set are in $S$\" $\\iff$ \"It is not true that all elements of the empty set are in $S$\" $\\iff$ \"There are some elements of the empty set that are not in $S$\".\n\nThe first is vacuously true (and is equivalent to $\\emptyset \\subset S^c$ which is true) and the second set of equivalent statements are all equivalent to $\\emptyset \\not \\subset S$ which is not true.\n\n=====\n\nThe thing is what you say is absolutely correct for non empty sets.\n\nMore formally:\n\nAll elements $x$ in $S$ are not in $A$ $\\implies$\n\n$S \\subset A^c$ $\\implies$\n\n$\\color{red}{\\text{There is an } x\\in S \\text{ where } x \\not \\in A}\\implies$\n\nIt is not true that all $x \\in S$ are also in $A$ $\\implies$\n\n$S \\not \\subset A$.\n\nHowever the red line can only be concluded if $A$ is non-empty. If $A$ is empty the red line is simply false.\n\nAnd without the red line there is simply no logic or means to jump from the line before to the line after:\n\nAll elements $x$ in $S$ are not in $A$ $\\not\\implies$\n\nIt is not true that all $x \\in S$ are also in $A$.\n\nThat simply is not true for an empty $S$.\n\n\u2022 You confused $A$ and $S$ at the end there. Oct 6, 2016 at 21:50\n\nEvery theory has axioms, which are some propositions held to be true without being proven from anything else, and are not provable from each other. Subsequent truths of the theory derived from the axioms are theorems.\n\nThe properly termed question is whether the empty set being a subset of every other set is axiom of set theory, or a theorem.\n\nIt depends on how \"subset\" is defined. If $A\\subset B$ means that every element of $A$ is in $B$, it is not necessarily true that $\\emptyset$ is a subset of anything, since it has no elements. In this case, $\\emptyset \\subset A$ can be added as an axiom. It doesn't conflict with anything, and simplifies all reasoning about subsets. Alternatively, if $A\\subset B$ is defined as \"$A$ has no elements that are not also in $B$\", then we do not require the extra axiom for the $\\emptyset$ case. If $A$ has no elements at all, it has no elements that are not in $B$.\n\nSuppose that we use the first, positively termed definition of subset, and then adopt as an axiom not $\\forall A:\\emptyset \\subset A$, but rather its negation: $\\exists A:\\emptyset \\not\\subset A$, or the outright proposition $\\forall A:\\emptyset \\not\\subset A$.\n\nThis is just going to cause problems. We can \"do\" set theory as before, but all the theorems will be uglified by having to avoid the special cases involving the empty set. In any derivation step in which we rely on a subset relation being true, or assert one, we will have to add the verbiage of an additional statement which asserts that the variable in question doesn't denote the empty set. This proposition then has to be carried in all the remaining derivations, unless something else makes it superfluous (some unrelated assurance from elsewhere that the set in question isn't empty).\n\nWorking with this clumsy subset definition that doesn't work with the empty set very well, someone is eventually going to have an epiphany and introduce a new subset-like relation which doesn't have these ugly problems: a new $A\\ \\mathbf{subset*}\\ B$ binary relation which reduces exactly to $A\\subset B$ when neither $A$ nor $B$ are $\\emptyset$, and which, simply by definition, reduces to a truth whenever $A = \\emptyset$, regardless of $B$. That person will then realize that all the existing work is simpler if this $\\mathbf{subset*}$ operation is used in place of $\\subset$.\n\nAt the end of the day it boils down to criteria like: is the system consistent (doesn't contradict itself), is it complete (does it capture the truths we want) and also is it convenient: are the rules configured so that we do not trip over unnecessary cases and superfluous logic.\n\nThis is a very mundane explanation (with finite sets). A subset is made of any combination of elements from the set. Suppose a set$S$ is made of three elements: $\\{a,b,c\\}$. Each element $a$, $b$ or $c$ can be, or not, in the combination. If all of them are not in the combination, they STILL form a combination of \"absent elements\", or $\\emptyset$, a subset of $S$. They are the dual of the subset of \"all elements\", $\\{a,b,c\\}$.\n\nIn the same way that $\\{a,b\\}$ and $\\{c\\}$ are (complementary) subsets, $\\emptyset$ and $\\{a,b,c\\}$ belong to the set of subsets.\n\nWhat confused me was that, the following expression was also a vacuous truth.\n\nFor every object of\u00a0$$x$$, if\u00a0$$x$$\u00a0belongs to the empty set, then\u00a0\u00a0$$x$$ doesn't belong to the set $$A$$.\n\nThe contrapositive of the above conditional is:\n\n\"For every $$x$$, if $$x$$ belongs to set $$A$$, then $$x$$ doesn't belong to the empty set\" which is easy to understand as the empty set has no elements at all.\n\nNOTE- $$p\\rightarrow q$$ has same meaning as $$\\lnot q\\rightarrow \\lnot p$$", "date": "2022-05-24 22:17:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1953218/is-the-empty-set-is-a-subset-of-any-set-a-convention/1953511", "openwebmath_score": 0.7517534494400024, "openwebmath_perplexity": 225.69110551115045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915226589475, "lm_q2_score": 0.867035771827307, "lm_q1q2_score": 0.8586181746826396}}
{"url": "https://mathhelpboards.com/threads/help-with-separable-equation.4880/#post-22144", "text": "# Help with separable equation\n\n#### Kris\n\n##### New member\n( 4*x+1 )^2 dy/dx = 27*y^3\n\nI'm trying to separate this into a separable equation. Does it matter which way I do it? I.e taking all xs to the left or all ys to the left or does it not matter as long as x and y are on different sides?\n\n#### MarkFL\n\nStaff member\nAs long as you have:\n\n$$\\displaystyle f(y)\\,dy=g(x)\\,dx$$ or $$\\displaystyle f(x)\\,dx=g(y)\\,dy$$ it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.\n\n#### Kris\n\n##### New member\nAs long as you have:\n\n$$\\displaystyle f(y)\\,dy=g(x)\\,dx$$ or $$\\displaystyle f(x)\\,dx=g(y)\\,dy$$ it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.\nHi Mark, I have tried separating the variables yet I can't figure out how to move the bracketed term across. Could you advise how to do this and show what the final solution would be?\n\nAlso how do I thank you as I can't find the button\n\n#### MarkFL\n\nStaff member\nI will help you separate the variables, and then guide you to get the solution...you will get more from the problem that way.\n\nWe are given:\n\n$$\\displaystyle (4x+1)^2\\frac{dy}{dx}=27y^3$$\n\nSee what you get when you divide through by $$\\displaystyle (4x+1)^2y^3$$ (bearing in mind that in doing so we are eliminating the trivial solution $y\\equiv0$).\n\nedit: You should see a Thanks link at the lower right of each post, except your own.\n\n#### Kris\n\n##### New member\nThanks I can get the answer from here just wasn't sure if you could divide through by the whole bracket\n\n#### Kris\n\n##### New member\ndy/(27*y^3)=dx/((4*x+1)^2)\n\nThis is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?\n\nAt this point it is okay to integrate right?\n\nedit: Do you know what the answer will be for the integration because my work sheet doesnt have it and id like to have it before i finish the question\n\n#### MarkFL\n\nStaff member\ndy/(27*y^3)=dx/((4*x+1)^2)\n\nThis is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?\n\nAt this point it is okay to integrate right?\n\nedit: Do you know what the answer will be for the integration because my work sheet doesn't have it and I'd like to have it before I finish the question.\nYes, that's fine, although I would probably choose to write:\n\n$$\\displaystyle y^{-3}\\,dy=27(4x+1)^{-2}\\,dx$$\n\nI would use a $u$-substitution on the right side:\n\n$$\\displaystyle u=4x+1\\,\\therefore\\,du=4\\,dx$$ and we have (after multiplying through by 4):\n\n$$\\displaystyle 4\\int y^{-3}\\,dy=27\\int u^{-2}\\,du$$\n\nNow, we are just a couple of steps from the solution (and don't forget to include the trivial solution we eliminated when separating variables when you state the final solution, if this trivial solution is not included in the general solution for a suitable choice of the constant of integration).\n\n#### Kris\n\n##### New member\nIve got a solution at y = -54 + c but i dont think this is right\n\nI went with -1/2*y^2 * y = -1/u * u? as my integrations and then rearranged\n\nIs this the correct answer or have I integrated something wrong which leads to my poor solution?\n\n#### MarkFL\n\nStaff member\nUsing the power rule for integration, and using the form in my post above, you should get:\n\n$$\\displaystyle 4\\left(\\frac{y^{-2}}{-2} \\right)=27\\left(\\frac{u^{-1}}{-1} \\right)+C$$\n\nMultiply through by -1 and simplify a bit:\n\n$$\\displaystyle 2y^{-2}=27u^{-1}+C$$\n\nNote: the sign of the parameter $C$ does not change as it can be any real number, negative or positive.\n\nAt this point, I would rewrite using positive exponents, and combine terms on the right:\n\n$$\\displaystyle \\frac{2}{y^2}=\\frac{Cu+27}{u}$$\n\nInvert both sides:\n\n$$\\displaystyle \\frac{y^2}{2}=\\frac{u}{Cu+27}$$\n\n$$\\displaystyle y^2=\\frac{2u}{Cu+27}$$\n\nBack-substitute for $u$:\n\n$$\\displaystyle y^2=\\frac{2(4x+1)}{C(4x+1)+27}$$\n\nThis is the general solution, and the only way we can get the trivial solution is for:\n\n$$\\displaystyle 4x+1=0$$\n\nbut we eliminated that possibility during the separation of variables as well.\n\n#### Kris\n\n##### New member\nThanks so much I see where i went wrong because i tried to integrate and invert before I multiplied out. It makes much more sense to multiply and flip then turn into a positive rather than trying to do it all at once", "date": "2021-12-02 09:36:47", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/help-with-separable-equation.4880/#post-22144", "openwebmath_score": 0.858676016330719, "openwebmath_perplexity": 349.76344192528086, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915249511565, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8586181698652723}}
{"url": "http://poliklinika-galaxy.com/food709f/retaliation1ad5/12018124b7576f9592e0a80d330", "text": "https://www.thoughtco.com/associative-and-commutative-properties-difference-3126316 (accessed July 22, 2022). By definition, commutative property is applied on 2 numbers, but the result remains the same for 3 numbers as well. For that reason, it is important to understand the difference between the two. That is. In Mathematics, a commutative property states that if the position of integers are moved around or interchanged while performing addition or multiplication operations, then the answer remains the same. Your Mobile number and Email id will not be published. By the commutative property of multiplication, 3 6 = 6 3. nhmDTH%PLI%@hQdK( (1) and (2), as per the commutative property of multiplication, we get; Find which of the following is the commutative property of addition and multiplication. For example: 1+2 = 2+1 and 2 x 3 = 3 x 2. Math Glossary: Mathematics Terms and Definitions, A Kindergarten Lesson Plan for Teaching Addition and Subtraction, IEP Math Goals for Operations in the Primary Grades, Definition and Usage of Union in Mathematics, Parentheses, Braces, and Brackets in Math. Essentially, the order does not matter when adding or multiplying. Yes. As per commutative property of addition, 827 + 389 = 389 + 827. Similarly, we can rearrange the addends and write: Example 4: Ben bought 3 packets of 6 pens each. As we already discussed in the introduction, as per the commutative property or commutative law, when two numbers are added or multiplied together, then a change in their positions does not change the result. Neither of these properties are applicable to division. Take, for example, the arithmetic problem (6 3) 2 = 3 2 = 1; if we change the grouping of the parentheses, we have 6 (3 2) = 6 1 = 5, which changes the final result of the equation. Whereas associative property holds regardless of grouping of numbers. Some operations are non-commutative. Lets see. For instance: This equation is an example of the commutative property of addition of real numbers. Choose the set of numbers to make the statement true. In other words, the placement of parentheses does not matter when it comes to adding or multiplying. No matter how the values are grouped, the result of the equation will be 10: As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. Example 2: Use 14 15 = 210, to find 15 14. \"The Associative and Commutative Properties.\" So, the total number of pens that Ben bought = 3 6, So, the total number of pens that Ben bought = 6 3. It states that if we swipe the positions of the integers, the result will remain the same. The correct answer is Both sides are equal to 33.. The commutative property concerns the order of certain mathematical operations. The left side equals 44 and the right side equals 33. By equation 1 and 2, as per commutative property of addition, we get; Q.3: Prove that A.B = B.A, if A = 4 and B = 3. What is the associative property of addition (or multiplication)? Checkout JEE MAINS 2022 Question Paper Analysis : explains that order of terms doesnt matter while performing, Commutative property is applicable only for addition and multiplication processes. The distributive property of Multiplication states that multiplying a sum by a number is the same as multiplying each addend by the value and adding the products then. Remember, with the commutative property, the order of the numbers does not matter when adding and multiplying. We can tell the difference between the associative and the commutative property by asking the question, Are we changing the order of the elements, or are we changing the grouping of the elements? If the elements are being reordered, then the commutative property applies.\n\n(2020, October 29). Did they buy an equal number of pens or not? Mia bought 6 packets of 3 pens each. When two numbers are multiplied together and if we interchange their positions, then the product of the two remains the same. The Associative property holds true for addition and multiplication. KB(|Q-SFt4E! Even if both have different numbers of bun packs with each having a different number of buns in them, they both bought an equal number of buns, because 3 4 = 4 3. According to the associative law, regardless of how the numbers are grouped, you can add or multiply them together, the answer will be the same. The word, Commutative, originated from the French word commute or commuter means to switch or move around, combined with the suffix -ative means tend to. The property holds for Addition and Multiplication, but not for subtraction and division. By eq. Which of the following represents the commutative property of addition? 7)\\). If $$x=2$$, $$y=5$$, and $$z=1$$, which of the following is true about this equation: $$2x+4y+9z=9z+4y+2x$$. The mathematical operations, subtraction and division are the two non-commutative operations. Example 4: Use the commutative property of addition to write the equation, 3 + 5 + 9 = 17, in a different sequence of the addends. Example 3: Use 827 + 389 = 1,216 to find 389 + 827. The other major properties of addition and multiplication are: Now, observe the other properties as well here: Associative Property of Addition and Multiplication. Since, 14 15 = 210, so, 15 14 also equals 210. Taylor, Courtney. by Mometrix Test Preparation | This Page Last Updated: June 2, 2022. The operation is commutative because the order of the elements does not affect the result of the operation. Which statement best illustrates the commutative property? Q.2: Prove that a+ b = b+a if a = 10 and b = 9. The associative property states that the grouping of factors in an operation can be changed without affecting the outcome of the equation. }8E| This can be shown by the equation (a + b) + c = a + (b + c). So, there can be two categories of operations that obeys commutative property: Although the official use of commutative property began at the end of the 18th century, it was known even in the ancient era. ThoughtCo. The commutative property states that values can be moved or swapped when adding or multiplying, and the outcome will not change. So, if we swap the position of numbers in subtraction or division statements, it changes the entire problem.\n\nWhich operations do not follow commutative property? Even if both have different numbers of apples and peaches, they have an equal number of fruits, because 2 + 6 = 6 + 2. Use the commutative property to find the missing value: $$45+44+43=43+44+$$_____. Which of the following represents the commutative property of multiplication? The associative property says you can regroup multiplied terms in any way. Which of the following expressions will follow the commutative property? \"The Associative and Commutative Properties.\" B.A., Mathematics, Physics, and Chemistry, Anderson University. Commutative property is applicable only for addition and multiplication processes. Do they have an equal number of marbles? =*jb 5;dtOu2T*~GL:E7$_Bd% N Therefore, the literal meaning of the word is tending to switch or move around. 77; by commutative property of multiplication, 36; by commutative property of multiplication. Unlike addition, in subtraction switching of orders of terms results in different answers. Commutative property cannot be applied to subtraction and division. The Associative and Commutative Properties. Simply put, the commutative property states that the factors in an equation can be rearranged freely without affecting the outcome of the equation. Example: 4 3 = 1 but 3 4 = -1which are two different integers. Use the commutative property to find the missing values: $$4+6+$$ ____$$=6+$$____ $$+8$$. When the associative property is used, elements are merely regrouped. These propertiesthe commutative and the associativeare very similar and can be easily mixed up. Solution: Options 1, 2 and 5 follow the commutative law. Beth has 6 packets of 78 marbles each. eC:C%L\"HX'JyS7yS| F: lj. The correct answer is $$6+5=5+6$$. The grouping of the elements, as indicated by the parentheses, does not affect the result of the equation. Required fields are marked *. Also, the division does not follow the commutative law. The correct answer is $$4(37)=(43)7$$. Thus, it means we can change the position or swap the numbers when adding or multiplying any two numbers. What is the distributive property of multiplication? Example 5: Lisa has 78 red and 6 blue marbles. Thus, it means we can change the position or swap the numbers when adding or multiplying any two numbers. Since Lisa has 78 red and 6 blue marbles. By the distributive property of multiplication over addition, we mean that multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. The associative property applies to multiplication but not division, so divided terms cannot be regrouped. We've updated our Privacy Policy, which will go in to effect on September 1, 2022. To solve more problems on properties of math, download BYJUS The Learning App from Google Play Store and watch interactive videos. (Except 2 + 2 and 2 2. (% This property states that when three or more numbers are added (or multiplied), the sum (or the product) is the same regardless of the grouping of the addends (or the multiplicands). Retrieved from https://www.thoughtco.com/associative-and-commutative-properties-difference-3126316. In mathematics, commutative property or commutative law explains that order of terms doesnt matter while performing arithmetic operations. For which all operations does the associative property hold true? This can be expressed through the equation a + (b + c) = (a + b) + c. No matter which pair of values in the equation is added first, the result will be the same. 3u(CXOD^$? We believe you can perform better on your exam, so we work hard to provide you with the best study guides, practice questions, and flashcards to empower you to be your best.\n\nThe commutative property, therefore, concerns itself with the ordering of operations, including the addition and multiplication of real numbers, integers, and rational numbers. The above examples clearly show that the commutative property holds true for addition and multiplication but not for subtraction and division. So, Lisa and Beth dont have an equal number of marbles. Example 1: Which of the following obeys commutative law? Commutative property holds regardless of order of numbers while addition or multiplication. If we pay careful attention to the equation, though, we see that only the order of the elements has been changed, not the grouping. As per commutative property of multiplication, 15 14 = 14 15. Rewrite the expression $$45+6+19$$ using the commutative property. The left side equals 33 and the right side equals 44. The expression $$45+6+19$$ is equivalent to $$6+45+19$$, because changing the order that we add does not affect the result. What Is the Difference of Two Sets in Set Theory? For the associative property to apply, we would have to rearrange the grouping of the elements as well: By clicking Accept All Cookies, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts.\n\nWhen two numbers are added together, then if we swap the positions of numbers, the sum of the two remains the same. Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of \"An Introduction to Abstract Algebra.\". However, note that the presence of parentheses alone does not necessarily mean that the associative property applies. The correct answer is $$4+6+\\mathbf8=6+\\mathbf4+8$$. a (b + c) = (a b) + (a c) where a, b, and c are whole numbers. Example 1: Fill in the missing numbers using the commutative property. Taylor, Courtney. However, unlike the commutative property, the associative property can also apply to matrix multiplication and function composition. ThoughtCo, Oct. 29, 2020, thoughtco.com/associative-and-commutative-properties-difference-3126316. Like commutative property equations, associative property equations cannot contain the subtraction of real numbers. Even though the terms are listed in a different order, the left and right side of the equation are both equal to 33. (a + b) + c = a + (b + c)(a b) c = a (b c) where a, b, and c are whole numbers. For example, the numbers 2, 3, and 5 can be added together in any order without affecting the final result: The numbers can likewise be multiplied in any order without affecting the final result: Subtraction and division, however, are not operations that can be commutative because the order of operations is important. Commutative Property Definition with Examples. Rearranging multiplied terms is an example of the commutative property. The three numbers above cannot, for example, be subtracted in any order without affecting the final value: As a result, the commutative property can be expressed through the equations a + b = b + a and a x b = b x a. Learn More All content on this website is Copyright 2022, $$3(1)^{2}+5(1)(2)+(3)=5(1)(2)+3+3(1)^{2}$$. The Rules of Using Positive and Negative Integers, Facts About the Element Ruthenium (or Ru), Lesson Plan: Adding and Multiplying Decimals, Use BEDMAS to Remember the Order of Operations. That is. The correct answer is $$6+45+19$$. By non-commutative, we mean the switching of the order will give different results. This is because we can apply this property on two numbers out of 3 in various combinations. The associative property, on the other hand, concerns the grouping of elements in an operation. So, the total number of marbles with Lisa = 78 + 6, So, the total number of marbles with Beth = 6 78. Your Mobile number and Email id will not be published. As per the commutative property of multiplication, when we multiply two integers, the answer we get after multiplication will remain the same, even if the position of the integers are interchanged. Taylor, Courtney. The commutative property allows the addition or multiplication of numbers in any order.\n\nAccording to the Distributive Property, if a, b, c are real numbers, then: There are certain other properties such as Identity property, closure property which are introduced for integers. The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. According to the commutative property of addition, when we add two integers, the answer will remain unchanged even if the position of the numbers are changed. Clearly, adding and multiplying two numbers gives different results.\n\nThat is. For a binary operationone that involves only two elementsthis can be shown by the equation a + b = b + a. Can you apply the commutative property of addition/multiplication to 3 numbers? If the elements are only being regrouped, then the associative property applies. The correct answer is 45. Since, 827 + 389 = 1,216, so, 389 + 827 also equals 1,216. So, mathematically commutative property for addition and multiplication looks like this: a + b = b + a; where a and b are any 2 whole numbers, a b = b a; where a and b are any 2 nonzero whole numbers. This is one of the major properties of integers. So, both Ben and Mia bought an equal number of pens. There are several mathematical properties that are used in statistics and probability; two of these, the commutative and associative properties, are generally associated with the basic arithmetic of integers, rationals, and real numbers, though they also show up in more advanced mathematics.\n\n[O J\"FUQ&pwODUfYQ$z(['z@PY3\"BjSN<8$N(whCZOAfDda( GhA!D}0.Fdap u JXEm]^m> wO]l\"gTkJHDMOE\\?I\u007fI!>bH4VEJ%+OTEeu\\!2% Note that when the commutative property is used, elements in an equation are rearranged. 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For example, take the equation 2 + 3 + 5. =i*s{/_WT8yp4x1lDI\n\nNo matter the order of the values in these equations, the results will always be the same. apxZ=vE This is one of the major, Commutative property is only applicable for two arithmetic operations: Addition and Multiplication, Changing the order of operands, does not change the result, Commutative property of addition: A + B = B + A, Commutative property of multiplication: A.B = B.A. Q8 E T\"4', BgFC01 DCKTEsIyaR`@\u007f!", "date": "2022-11-26 19:14:15", "meta": {"domain": "poliklinika-galaxy.com", "url": "http://poliklinika-galaxy.com/food709f/retaliation1ad5/12018124b7576f9592e0a80d330", "openwebmath_score": 0.47175082564353943, "openwebmath_perplexity": 543.4621189292722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9822877033706601, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8585953231208028}}
{"url": "http://mathhelpforum.com/number-theory/129681-question-about-squares.html", "text": "I was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square\n503^2 = 253009\nand\n50*506=25300\nthen tacking the 3^2 on the end gets 253009.\nTricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.\nAny help would be appreciated.\n\n2. Originally Posted by Enkie\nI was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square\n503^2 = 253009\nand\n50*506=25300\nthen tacking the 3^2 on the end gets 253009.\nTricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.\nAny help would be appreciated.\nLet $n\\in\\mathbb{N}$ then $n=a_0+10a_1+\\cdots+10^n a_n$ where $a_k\\in\\{0,\\cdots,9\\}$. You are claiming that $n^2=\\left(10 a_1+\\cdots +10^n a_n\\right)\\cdot\\left(2a_0+10 a_1+\\cdots 10^n a_n\\right)+a_0^2$, or equivalently\n\n$\\left(a_0+10 a_1+\\cdots+10^n a_n\\right)^2-\\left(10a_1+\\cdots+10^n a_n\\right)\\cdot\\left(2a_0+10a_1+\\cdots+10^na_n\\rig ht)-a_0^2$ $=\\left(\\left(a_0+10a_1+\\cdots 10^n a_n\\right)^2-a_0^2\\right)-\\left(10 a_1+\\cdots+10^na_n\\right)\\cdot\\left(2a_0+10a_1+\\cd ots+10^n a_n\\right)$ $=0$.\n\nBut,\n\n$\\left(a_0+10a_1+\\cdots+10^na_n\\right)^2-a_0^2=\\left(a_0+10a_1+\\cdots+10^na_n-a_0\\right)$ $\\cdot\\left(a_0+10a_1+\\cdots 10^na_n+a_0\\right)$ $=\\left(10a_1+\\cdots+10^na_n\\right)\\cdot\\left(2a_0+ 10a_1+\\cdots+10^n a_n\\right)$.\n\nFrom where the conclusion immediately follows.\n\n3. Hello, Enkie!\n\nAnother approach . . .\n\nI was noticing today that you can calculate the square of a number by:\n(1) taking the number you want to square and removing the ones digit,\n(2) and multiplying by the number you want to square plus the ones digit,\n(3) then square the ones digit and add it on to get the square.\n\nAny integer $N> 9$ is of the form: . $N \\:=\\:10T + U$\n. . where $U$ is the units digit and $T$ is the \"rest of the number.\"\n\nWe note that: . $N^2 \\:=\\:(10T + U)^2 \\:=\\:100T^2 + 20TU + U^2$\n\n(1) Remove the units digit and append a zero.\n. . . We have: . $10T$\n\n(2) Multiply by $N + U\\!:\\;\\;10T(N + U) \\;=\\;10T(10T + U + U) \\;=\\;100T^2 + 20TU$\n\n(3) Add $U^2\\!:\\;\\;100T^2 + 20TU + U^2\\quad \\hdots$ and the result is $N^2$\n\n4. Let us try with $1485$. We get :\n\n$148 \\times (1485 + 5) = 220520$\n\nNow add on $5^2 = 25 \\Rightarrow 22052025$\n\nBut $1485^2 = 2205225$.\n\nI understand what you mean but you need to mention that when adding the square of the last digit, you actually overlap the previous result when the last digit is greater than three.", "date": "2013-12-18 09:04:16", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/129681-question-about-squares.html", "openwebmath_score": 0.7727115750312805, "openwebmath_perplexity": 268.75407008937884, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877007780707, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8585953192436152}}
{"url": "https://stats.stackexchange.com/questions/213456/how-to-calculate-the-total-probability-inside-a-slice-of-a-bivariate-normal-dist", "text": "# How to calculate the total probability inside a slice of a bivariate normal distribution in R?\n\nI have a bivariate normal distribution composed of the univariate normal distributions $X_1$ and $X_2$ with $\\rho \\approx 0.3$.\n\n$$\\begin{pmatrix} X_1 \\\\ X_2 \\end{pmatrix} \\sim \\mathcal{N} \\left( \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix} , \\begin{pmatrix} \\sigma^2_1 & \\rho \\sigma_1 \\sigma_2 \\\\ \\rho \\sigma_1 \\sigma_2 & \\sigma^2_2 \\end{pmatrix} \\right)$$\n\nIs there a simple way to calculate in R the cumulative probability of $X_1$ being less than a value $z$ given a particular slice of $X_2$ (between two values $a,b$) given we know all the parameters $\\mu_1, \\mu_2, \\sigma_1, \\sigma_2, \\rho$?\n\n$P(X_1 < z | a < X_2 < b)$\n\nCan the distribution function I am looking for match (or be approximated by) the distribution function of a univariate normal distribution (to use qnorm/pnorm)? Ideally this would be the case so I can perform the calculation with less dependencies on libraries (e.g. on a MySQL server).\n\nThis is the bivariate distribution I am using:\n\nmeans <- c(79.55920, 52.29355)\nvariances <- c(268.8986, 770.0212)\nrho <- 0.2821711\n\ncovariancePartOfMatrix <- sqrt(variances[1]) * sqrt(variances[2]) * rho\nsigmaMatrix <- matrix(c(variances[1],covariancePartOfMatrix,covariancePartOfMatrix,variances[2]), byrow=T, ncol=2)\n\nn <- 10000\ndat <- MASS::mvrnorm(n=n, mu=means, Sigma=sigmaMatrix)\n\nplot(dat)\n\n\nThis is my numerical attempt to get the correct result. However it uses generated data from the bivariate distribution and I'm not convinced it will give the correct result.\n\na <- 79.5\nb <- 80.5\nz <- 50\n\nsliceOfDat <- subset(data.frame(dat), X1 > a, X1 < b)\nestimatedMean <- mean(sliceOfDat[,c(2)])\nestimatedDev <- sd(sliceOfDat[,c(2)])\n\nestimatedPercentile <- pnorm(z, estimatedMean, estimatedDev)\n\n\n### Edit - R implementation of solution based on whuber's answer\n\nHere is an implementation of the accepted solution using integrate, compared against my original idea based on sampling. The accepted solution provides the expected output 0.5, whereas my original idea deviated by a significant amount (0.41). Update - See wheber's edit for a better implementation.\n\n# Bivariate distribution parameters\nmeans <- c(79.55920, 52.29355)\nvariances <- c(268.8986, 770.0212)\nrho <- 0.2821711\n\n# Generate sample data for bivariate distribution\nn <- 10000\n\ncovariancePartOfMatrix <- sqrt(variances[1]) * sqrt(variances[2]) * rho\nsigmaMatrix <- matrix(c(variances[1],covariancePartOfMatrix,covariancePartOfMatrix,variances[2]), byrow=T, ncol=2)\ndat <- MASS::mvrnorm(n=n, mu=means, Sigma=sigmaMatrix)\n\n# Input parameters to test the estimation\nw = 79.55920\n\na <- w - 0.5\nb <- w + 0.5\nz <- 52.29355\n\n# Univariate approximation using randomness\nsliceOfDat <- subset(data.frame(dat), X1 > a, X1 < b)\nestimatedMean <- mean(sliceOfDat[,c(2)])\nestimatedDev <- sd(sliceOfDat[,c(2)])\n\nestimatedPercentile <- pnorm(z, estimatedMean, estimatedDev)\n# OUTPUT: 0.411\n\n# Numerical approximation from exact solution\nadaptedZ <- (z - means[2]) / sqrt(variances[2])\nadaptedB <- (b - means[1]) / sqrt(variances[1])\nadaptedA <- (a - means[1]) / sqrt(variances[1])\n\nintegrand <- function(x) pnorm((adaptedZ - rho * x) / sqrt(1 - rho * rho)) * dnorm(x)\n# 0.0121, abs.error 1.348036e-16, \"OK\"\nexactPercentile = exactSolutionCoeff * exactSolutionInteg$value # OUTPUT: 0.500  ## 1 Answer Yes, a Normal approximation works--but not in all cases. We need to do some analysis to identify when the approximation is a good one. ### Exact solution Re-express$(X_1,X_2)$in standardized units so that they have zero means and unit variances. Letting$\\Phibe the standard Normal distribution function (its CDF), it is well known from the theory of ordinary least squares regression that $$\\Pr(X_1 \\le z\\,|\\, X_2 = x) = \\Phi\\left(\\frac{z - \\rho x}{\\sqrt{1-\\rho^2}}\\right).$$ The desired probability then can be obtained by integrating: \\eqalign{\\Pr(X_1 \\le z\\,|\\, a \\lt X_2 \\le b) &= \\frac{1}{\\Phi(b)-\\Phi(a)}\\int_a^b \\Pr(X_1\\le z\\,|\\, X_2=x) \\phi(x)\\,dx \\\\&= \\frac{1}{\\Phi(b)-\\Phi(a)}\\int_a^b \\Phi\\left(\\frac{z - \\rho x}{\\sqrt{1-\\rho^2}}\\right) \\phi(x)\\,dx.} This appears to require numerical integration (although the result for(a,b)=\\mathbb{R}$is obtainable in closed form: see How can I calculate$\\int^{\\infty}_{-\\infty}\\Phi\\left(\\frac{w-a}{b}\\right)\\phi(w)\\,\\mathrm dw$). ### Approximating the distribution This expression can be differentiated under the integral sign (with respect to$z$) to obtain the PDF, $$f(z\\,|\\, a \\lt X_2 \\le b) = \\phi(z)\\ \\frac{\\Phi\\left(\\frac{b-\\rho z}{\\sqrt{1-\\rho^2}}\\right) - \\Phi\\left(\\frac{a-\\rho z}{\\sqrt{1-\\rho^2}}\\right)}{\\Phi(b) - \\Phi(a)}.$$ This exhibits the PDF as product of the standard Normal PDF$\\phi$and a \"correction\". When$b-a$is small compared to$\\sqrt{1-\\rho^2}$(specifically, when$(b-a)^2 \\ll 1-\\rho^2$), we might approximate the difference in the numerator with the first derivative: $$\\Phi\\left(\\frac{b-\\rho z}{\\sqrt{1-\\rho^2}}\\right) - \\Phi\\left(\\frac{a-\\rho z}{\\sqrt{1-\\rho^2}}\\right)\\approx \\phi\\left(\\frac{(a+b)/2-\\rho z}{\\sqrt{1-\\rho^2}}\\right)\\frac{b-a}{\\sqrt{1-\\rho^2}}.$$ The error in this approximation is uniformly bounded (across all values of$z$) because the second derivative of$\\Phi$is bounded. With this approximation, and completing the square, we obtain $$f(z\\,|\\, a \\lt X_2 \\le b) \\approx \\phi\\left(z; \\rho(a+b)/2, \\sqrt{1-\\rho^2}\\right) \\frac{(b-a)\\exp\\left(-(a+b)^2/8\\right)}{(\\Phi(b)-\\Phi(a))\\sqrt{2\\pi}}.$$ ($\\phi(*; \\mu, \\sigma)$denotes the PDF of a Normal distribution of mean$\\mu$and standard deviation$\\sigma$.) Moreover, the right hand factor (which does not depend on$z$) must be very close to$1$, because the$\\phi$term integrates to unity, whence $$f(z\\,|\\, a \\lt X_2 \\le b) \\approx \\phi\\left(z; \\rho(a+b)/2, \\sqrt{1-\\rho^2}\\right).$$ All this makes very good sense: the conditional distribution of$X_1$is approximated by its conditional distribution at the midpoint of the interval,$(a+b)/2$, where it has mean$\\rho(a+b)/2$and standard deviation$\\sqrt{1-\\rho^2}$. The error is proportional to the width of the interval$b-a$and to an expression dominated by$\\exp(-(a+b)^2/8)$, which becomes important only when both$a$and$b$are out in the same tail. Therefore, this Normal approximation works for narrow slices not too far into the tails of the bivariate distribution. Moreover, the difference between $$\\frac{(b-a)\\exp\\left(-(a+b)^2/8\\right)}{(\\Phi(b)-\\Phi(a))\\sqrt{2\\pi}}$$ and$1$serves as an excellent check of the quality of the approximation. ### Edit To check these conclusions, I simulated data in R for various values of$b$and$\\rho$($a=-3$in all cases), drew their empirical density, and superimposed on that the theoretical (blue) and approximate (red) densities for comparison. (You cannot see the density plots because the theoretical plots fit over them almost perfectly.) As$|\\rho|$gets close to$1$, the approximation grows poorer: this deserves further study. Clear the approximation is excellent for sufficiently small values of$b-a$. # # Numerical integration, to give a correct value. # f <- function(z, a, b, rho, value.only=FALSE, ...) { g <- function(x) pnorm((z - rho*x)/sqrt(1-rho^2)) * dnorm(x) / (pnorm(b) - pnorm(a)) u <- integrate(g, a, b, ...) if (value.only) return(u$value) else return(u)\n}\n#\n# Set up the problem.\n#\na <- -3  # Left endpoint\nn <- 1e5 # Simulation size\npar(mfrow=c(2,3))\nfor (rho in c(1/4, -2/3)) {\nfor (b in c(-2.5, -2, -1.5)) {\nz <- seq((a-3)*rho, (b+3)*rho, length.out=101)\n#\n# Check the approximation (v needs to be small).\n#\nv <- (b-a) * exp(-(a+b)^2/8) / (pnorm(b) - pnorm(a)) / sqrt(2*pi) - 1\n#\n# Simulate some values of (x1, x2).\n#\nx.2 <- qnorm(runif(n, pnorm(a), pnorm(b)))  # Normal between a and b\nx.1 <- rho*x.2 + rnorm(n, sd=sqrt(1-rho^2))\n#\n# Compare the simulated distribution to the theoretical and approximate\n# densities.\n#\nx.hat <- density(x.1)\nplot(x.hat, type=\"l\", lwd=2,\nmain=\"Simulated, True, and Approximate\",\nsub=paste0(\"a=\", round(a,2), \", b=\", round(b, 2), \", rho=\", round(rho, 2),\n\"; v=\", round(v,3)),\nxlab=\"X.1\", ylab=\"Density\")\n\n# Theoretical\ncurve(dnorm(x) * (pnorm((b-rho*x)/sqrt(1-rho^2)) - pnorm((a-rho*x)/sqrt(1-rho^2))) /\n(pnorm(b) - pnorm(a)), lwd=2, col=\"Blue\", add=TRUE)\n\n# Approximate\ncurve(dnorm(x, rho*(a+b)/2, sqrt(1-rho^2)), col=\"Red\", lwd=2, add=TRUE)\n}\n}\n\n\u2022 Thank you for the detailed answer. I had found that the normal approximation stopped producing the figures I expected when z was high or low. Your examination of the numerical analysis corresponds exactly to what I noticed varying a, b, and z, and makes me confident that I can get a better result by using the exact solution that you provided. \u2013\u00a0Michael Clark May 20 '16 at 11:54\n\u2022 Would it be possible to add a little more detail on how to calculate the exact solution without numerical integration? An example line in R would be perfect. The part I am most confused about is how to convert the limits and sign of x in the integral \u222bba\u03a6(z\u2212\u03c1x1\u2212\u03c12\u2212\u2212\u2212\u2212\u2212\u221a)\u03d5(x)dx into the form \u222b\u221e\u2212\u221e\u03a6(w\u2212ab)\u03d5(w)dw\u222b\u2212\u221e\u221e\u03a6(w\u2212ab)\u03d5(w)dw) so that I can use the result proved in the question that you linked. \u2013\u00a0Michael Clark May 20 '16 at 11:58\n\u2022 Unfortunately there is no such conversion. Numerical integration really is needed. You could also pursue the development of the approximation further by using higher-order approximations to the $\\Phi$ term in the integrand. Intuitively, the midpoint $(a+b)/2$ ought to be replaced by a point closer to the smaller of $a$ and $b$ (in size), because that's where most of the $X_2$ probability is. \u2013\u00a0whuber May 20 '16 at 12:26\n\u2022 I've added a R example based on the exact solution -- using the mean & variance, the approximate solution was 0.41 and the exact 0.5. I am surprised how far the approximate solution is out in that case. Thanks again! \u2013\u00a0Michael Clark May 20 '16 at 20:17\n\u2022 I saw that code, but it's not clear how it works or whether it's correct. At the very least you should compute the check value (at the end of my original post), look at $|b-a|/\\sqrt{1-\\rho^2}$ (in standardized units), and inspect the output of integrate to make sure it's not running into accuracy problems. \u2013\u00a0whuber May 20 '16 at 20:20", "date": "2018-12-10 22:19:06", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/213456/how-to-calculate-the-total-probability-inside-a-slice-of-a-bivariate-normal-dist", "openwebmath_score": 0.7470393776893616, "openwebmath_perplexity": 3720.9475415620254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98228770337066, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8585953166765454}}
{"url": "https://math.stackexchange.com/questions/1981391/how-would-i-simplify-this-summation-sum-i-1n-i-1-sum-j-1n-j", "text": "# How would I simplify this summation: $\\sum_{i=1}^n (i + 1) - \\sum_{j=1}^n j$\n\n$$\\sum_{i=1}^n (i + 1) - \\sum_{j=1}^n j$$\n\nI cant really get my head around on how to simplify these sigma notations, any help would be appreciated.\n\nThanks\n\n\u2022 Just try it for modest $n$! Suppose $n=1$. Then try $n=2$. I expect a pattern will emerge.\n\u2013\u00a0lulu\nOct 23 '16 at 14:05\n\u2022 You could replace $j$ by $i$ and simplify. Oct 23 '16 at 14:08\n\nYou can \u201creorganize\u201d the first summation: $$\\sum_{i=1}^n(i+1)=\\sum_{i=1}^n i+\\sum_{i=1}^n1$$ Since $$\\sum_{i=1}^n i = \\sum_{j=1}^n j$$ you remain with $$\\sum_{i=1}^n 1 = n$$\n\nAnother variation which might be helpful.\n\n\\begin{align*} \\sum_{i=1}^n (i + 1) - \\sum_{j=1}^n j&=\\sum_{i=1}^n (i + 1) - \\sum_{i=1}^n i\\\\ &=\\sum_{i=1}^n (i + 1-i)\\\\ &=\\sum_{i=1}^n 1\\\\ &=n \\end{align*}\n\nYou have\n\n$$\\sum_{i=1}^n (i+1)=\\sum_{i=2}^{n+1} i$$\n\nso\n\n$$\\sum_{i=1}^n (i+1)-\\sum_{j=1}^n j=\\sum_{i=2}^{n+1} i-\\sum_{j=1}^n j=n+1+\\sum_{i=1}^n (i-i)+1=n+1-1=n.$$\n\n(Too long for a comment)\n\nAs @lulu said you can investigate the behavior of the sum by hand but you can use a bit of algebra before to simplify something. By example observe that $$\\sum_{k=1}^n (k+1)=\\left(\\sum_{k=1}^n k\\right)+\\left(\\sum_{k=1}^n 1\\right)=\\left(\\sum_{k=1}^n k\\right) + n$$ and $$\\sum_{k=1}^n k=\\sum_{j=1}^n j=1+2+3+\\cdots+n$$\n\nObserve too that a summation can be written as\n\n$$\\sum_{k=1}^n k=\\sum_{1\\le k\\le n}k$$\n\nThen you can manipulate easily the inequality $1\\le k\\le n$ if you need to change the variable $k$ by, for example, $h=k+2$\n\n$$1\\le k\\le n\\iff 1+2\\le k+2\\le n+2\\iff 3\\le h\\le n+2$$\n\nThen you can rewrite\n\n$$\\sum_{k=1}^n k=\\sum_{1\\le k\\le n}k=\\sum_{3\\le h\\le n+2}(h-2)=\\sum_{h=3}^{n+2}(h-2)$$", "date": "2021-12-08 19:49:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1981391/how-would-i-simplify-this-summation-sum-i-1n-i-1-sum-j-1n-j", "openwebmath_score": 0.907001256942749, "openwebmath_perplexity": 518.7532451435949, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822876997410348, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8585953135039728}}
{"url": "http://mathhelpboards.com/other-topics-22/calculate-velocity-brick-using-assumption-amp-draw-displacement-time-graph-19950.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e", "text": "# Thread: Calculate the velocity of the brick using an assumption & draw the displacement-time graph\n\n1. Calculate the velocity of the brick at the bottom of the inclined plane.\n\n& Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)\n\nWorkings:\n\nI am unable to think on the first two but I think the displacement and time graph should be something like ,\n\nMany Thanks\n\n2. To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.\n\nSo, if we equate the initial potential energy to the final kinetic energy, we have:\n\n$\\displaystyle mgh=\\frac{1}{2}mv^2$\n\nWhat do you get upon solving for $v$?\n\nOriginally Posted by MarkFL\nTo calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.\n\nSo, if we equate the initial potential energy to the final kinetic energy, we have:\n\n$\\displaystyle mgh=\\frac{1}{2}mv^2$\n\nWhat do you get upon solving for $v$?\nThank you very much MarkFL\n\n$\\displaystyle mgh=\\frac{1}{2}mv^2$\n\n$\\displaystyle 2kg * 10 ms^-2 * 5 m=\\frac{1}{2}mv^2$\n\n$\\displaystyle 2kg * 10 ms^-2 * 5 m=\\frac{1}{2}*2kg * v^2$\n\n$\\displaystyle 2kg * 10 ms^-2 * 5 m=\\frac{1}{2}*2kg * v^2$\n\n$\\displaystyle 100 J= v^2$\n\n$\\displaystyle 10 J= v$\n\nCorrect ?\n\n4. I would solve the equation first, and then plug in the numbers:\n\n$\\displaystyle mgh=\\frac{1}{2}mv^2$\n\n$\\displaystyle v=\\sqrt{2gh}$\n\nOkay, at this point we can plug in:\n\n$\\displaystyle g\\approx9.81\\,\\frac{\\text{m}}{\\text{s}^2},\\,h=5\\text{ m}$\n\n$\\displaystyle v\\approx\\sqrt{2\\left(9.81\\,\\frac{\\text{m}}{\\text{s}^2}\\right)\\left(5\\text{ m}\\right)}=3\\sqrt{\\frac{109}{10}}\\,\\frac{\\text{m}}{\\text{s}}\\approx9.9045\\,\\frac{\\text{m}}{\\text{s}}$\n\nOriginally Posted by MarkFL\nI would solve the equation first, and then plug in the numbers:\n\n$\\displaystyle mgh=\\frac{1}{2}mv^2$\n\n$\\displaystyle v=\\sqrt{2gh}$\n\nOkay, at this point we can plug in:\n\n$\\displaystyle g\\approx9.81\\,\\frac{\\text{m}}{\\text{s}^2},\\,h=5\\text{ m}$\n\n$\\displaystyle v\\approx\\sqrt{2\\left(9.81\\,\\frac{\\text{m}}{\\text{s}^2}\\right)\\left(5\\text{ m}\\right)}=3\\sqrt{\\frac{109}{10}}\\,\\frac{\\text{m}}{\\text{s}}\\approx9.9045\\,\\frac{\\text{m}}{\\text{s}}$\nThank you very much\n\nOriginally Posted by mathlearn\n\nSketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)\n\nIs the above drawn graph correct?\n\n6. For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?\n\n7. Originally Posted by mathlearn\n\nIs the above drawn graph correct?\nHey mathlearn!\n\nHere's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):\n\nOn the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.\nAs a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).\nNote that $v=\\int_0^t a\\,dt$.\n\nHowever, we're asked for the displacement, which we can call $d$.\nWe have that $d=\\int_0^t v\\,dt$.\nWhat would the displacement graph look like?\n\n8. Originally Posted by mathlearn\n$\\displaystyle 100 J= v^2$\n\n$\\displaystyle 10 J= v$\nRecheck your units, speed is not measured in J. (And the square root of a J is not J!)\n\n-Dan\n\nOriginally Posted by MarkFL\nFor the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?\n\nThis kind of a curve with constant acceleration and with deceleration.\n\nOriginally Posted by I like Serena\nHey mathlearn!\n\nHere's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):\n\nOn the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.\nAs a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).\nNote that $v=\\int_0^t a\\,dt$.\n\nHowever, we're asked for the displacement, which we can call $d$.\nWe have that $d=\\int_0^t v\\,dt$.\nWhat would the displacement graph look like?\nThe displacement time graph would look like the above graph\n\nOriginally Posted by topsquark\nRecheck your units, speed is not measured in J. (And the square root of a J is not J!)\n\n-Dan\nThanks for the catch\n\n10. Originally Posted by mathlearn\nThis kind of a curve with constant acceleration and with deceleration...\nWhy do you have the brick returning to zero displacement?", "date": "2017-12-12 14:28:00", "meta": {"domain": "mathhelpboards.com", "url": "http://mathhelpboards.com/other-topics-22/calculate-velocity-brick-using-assumption-amp-draw-displacement-time-graph-19950.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e", "openwebmath_score": 0.7601319551467896, "openwebmath_perplexity": 452.0066474901806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877028521421, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8585953081679988}}
{"url": "http://accessdtv.com/error-bound/taylor-series-maximum-error.html", "text": "Home > Error Bound > Taylor Series Maximum Error\n\n# Taylor Series Maximum Error\n\n## Contents\n\nMathispower4u 48,779 views 9:00 Error or Remainder of a Taylor Polynomial Approximation - Duration: 11:27. Your cache administrator is webmaster. So for example, if someone were to ask you, or if you wanted to visualize. Alternating series error bound For a decreasing, alternating series, it is easy to get a bound on the error : In other words, the error is bounded by the next term http://accessdtv.com/error-bound/taylor-series-approximation-maximum-error.html\n\nAnd this polynomial right over here, this Nth degree polynomial centered at a, f or P of a is going to be the same thing as f of a. Another use is for approximating values for definite integrals, especially when the exact antiderivative of the function cannot be found. But how many terms are enough? To see why the alternating bound holds, note that each successive term in the series overshoots the true value of the series. https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation\n\n## Taylor Polynomial Error Bound\n\nSo it's really just going to be, I'll do it in the same colors, it's going to be f of x minus P of x. It is going to be equal to zero. Suppose you needed to find .\n\nEspecially as we go further and further from where we are centered. >From where are approximation is centered. Maybe we might lose it if we have to keep writing it over and over but you should assume that it is an Nth degree polynomial centered at a. You can get a different bound with a different interval. What Is Error Bound I'll write two factorial.\n\nKhan Academy 54,407 views 9:18 Loading more suggestions... Lagrange Error Bound Formula fall-2010-math-2300-005 lectures \u00a9 2011 Jason B. One way to get an approximation is to add up some number of terms and then stop.\n\nIf x is sufficiently small, this gives a decent error bound.\n\nsolution Practice B03 Solution video by PatrickJMT Close Practice B03 like? 6 Practice B04 Determine an upper bound on the error for a 4th degree Maclaurin polynomial of $$f(x)=\\cos(x)$$ at $$\\cos(0.1)$$. Taylor Polynomial Approximation Calculator What are they talking about if they're saying the error of this Nth degree polynomial centered at a when we are at x is equal to b. Thus, we have What is the worst case scenario? If one adds up the first terms, then by the integral bound, the error satisfies Setting gives that , so .\n\n## Lagrange Error Bound Formula\n\nAnd we see that right over here. http://calculus.seas.upenn.edu/?n=Main.ApproximationAndError Therefore, one can think of the Taylor remainder theorem as a generalization of the Mean value theorem. Taylor Polynomial Error Bound Let's embark on a journey to find a bound for the error of a Taylor polynomial approximation. Lagrange Error Bound Calculator Similarly, you can find values of trigonometric functions.\n\nwith an error of at most .139*10^-8, or good to seven decimal places. this contact form And it's going to look like this. Close Yeah, keep it Undo Close This video is unavailable. Lagrange's formula for this remainder term is $$\\displaystyle{ R_n(x) = \\frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!} }$$ This looks very similar to the equation for the Taylor series terms . . . Lagrange Error Bound Problems\n\nThat is the motivation for this module. However, only you can decide what will actually help you learn. Instead, use Taylor polynomials to find a numerical approximation. http://accessdtv.com/error-bound/taylor-maximum-error.html Autoplay When autoplay is enabled, a suggested video will automatically play next.\n\nPlease try the request again. Lagrange Error Bound Khan Academy If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. And what I wanna do is I wanna approximate f of x with a Taylor polynomial centered around x is equal to a.\n\n## Return to the Power Series starting page Representing functions as power series A list of common Maclaurin series Taylor Series Copyright \u00a9 1996 Department of Mathematics, Oregon State University If you\n\nSolving for gives for some if and if , which is precisely the statement of the Mean value theorem. solution Practice B02 Solution video by PatrickJMT Close Practice B02 like? 8 Practice B03 Use the 2nd order Maclaurin polynomial of $$e^x$$ to estimate $$e^{0.3}$$ and find an upper bound on The following theorem tells us how to bound this error. Alternating Series Error Bound patrickJMT 128,850 views 10:48 Calculus 2 Lecture 9.9: Approximation of Functions by Taylor Polynomials - Duration: 1:34:10.\n\nLinks and banners on this page are affiliate links. You may want to simply skip to the examples. So it might look something like this. Check This Out A Taylor polynomial takes more into consideration.\n\nProfessor Leonard 99,296 views 3:01:45 Taylor Polynomials - Duration: 18:06. Krista King 59,295 views 8:23 Lec 38 | MIT 18.01 Single Variable Calculus, Fall 2007 - Duration: 47:31. Now, if we're looking for the worst possible value that this error can be on the given interval (this is usually what we're interested in finding) then we find the maximum So what I wanna do is define a remainder function.\n\nThe N plus oneth derivative of our error function or our remainder function, we could call it, is equal to the N plus oneth derivative of our function. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... This is going to be equal to zero. The square root of e sin(0.1) The integral, from 0 to 1/2, of exp(x^2) dx We cannot find the value of exp(x) directly, except for a very few values of x.\n\nSo, what is the value of $$z$$? $$z$$ takes on a value between $$a$$ and $$x$$, but, and here's the key, we don't know exactly what that value is. And that's the whole point of where I'm going with this video and probably the next video, is we're gonna try to bound it so we know how good of an So this is all review, I have this polynomial that's approximating this function. video by Dr Chris Tisdell Search 17Calculus Loading Practice Problems Instructions: For the questions related to finding an upper bound on the error, there are many (in fact, infinite) correct answers.\n\nThe following example should help to make this idea clear, using the sixth-degree Taylor polynomial for cos x: Suppose that you use this polynomial to approximate cos 1: How accurate is View Edit History Print Single Variable Multi Variable Main Approximation And Error < Taylor series redux | Home Page | Calculus > Given a series that is known to converge but Can we bound this and if we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do In general, if you take an N plus oneth derivative of an Nth degree polynomial, and you could prove it for yourself, you could even prove it generally but I think\n\nUploaded on Nov 11, 2011In this video we use Taylor's inequality to approximate the error in a 3rd degree taylor approximation. There is a slightly different form which gives a bound on the error: Taylor error bound where is the maximum value of over all between 0 and , inclusive. So, the first place where your original function and the Taylor polynomial differ is in the st derivative. But you'll see this often, this is E for error.", "date": "2018-02-21 15:28:39", "meta": {"domain": "accessdtv.com", "url": "http://accessdtv.com/error-bound/taylor-series-maximum-error.html", "openwebmath_score": 0.7899336814880371, "openwebmath_perplexity": 515.4209480898912, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9372107984180243, "lm_q2_score": 0.9161096193153989, "lm_q1q2_score": 0.8585878277570174}}
{"url": "https://www.ubezwlasnowolnienie.pl/0dnlt82b/5c2a1c-symmetric-and-antisymmetric-relation", "text": "(iv) Reflexive and transitive but not symmetric. The relations we are interested in here are binary relations on a set. #mathematicaATDRelation and function is an important topic of mathematics. Also, compare with symmetric and antisymmetric relation here. x is married to the same person as y iff (exists z) such that x is married to z and y is married to z. Learn about the world's oldest calculator, Abacus. If a relation is reflexive, irreflexive, symmetric, antisymmetric, asymmetric, transitive, total, trichotomous, a partial order, total order, strict weak order, total preorder (weak order), or an equivalence relation, its restrictions are too. To put it simply, you can consider an antisymmetric relation of a set as a one with no ordered pair and its reverse in the relation. You can find out relations in real life like mother-daughter, husband-wife, etc. Multiplication problems are more complicated than addition and subtraction but can be easily... Abacus: A brief history from Babylon to Japan. The relations we are interested in here are binary relations on a set. In other words, we can say symmetric property is something where one side is a mirror image or reflection of the other. Here we are going to learn some of those properties binary relations may have. For example, on the set of integers, the congruence relation aRb iff a - b = 0(mod 5) is an equivalence relation. ; Restrictions and converses of asymmetric relations are also asymmetric. Complete Guide: How to work with Negative Numbers in Abacus? So total number of symmetric relation will be 2 n(n+1)/2. \"Is married to\" is not. (g)Are the following propositions true or false? Since (1,2) is in B, then for it to be symmetric we also need element (2,1). for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. (f) Let $$A = \\{1, 2, 3\\}$$. 6.3. A matrix for the relation R on a set A will be a square matrix. A relation R is defined on the set Z (set of all integers) by \u201caRb if and only if 2a + 3b is divisible by 5\u201d, for all a, b \u2208 Z. Learn about operations on fractions. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. This... John Napier | The originator of Logarithms. Which is (i) Symmetric but neither reflexive nor transitive. \u201cIs equal to\u201d is a symmetric relation, such as 3 = 2+1 and 1+2=3. Referring to the above example No. In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. Note - Asymmetric relation is the opposite of symmetric relation but not considered as equivalent to antisymmetric relation. See also Let ab \u2208 R \u21d2 (a \u2013 b) \u2208 Z, i.e. In other words, we can say symmetric property is something where one side is a mirror image or reflection of the other. In this article, we have focused on Symmetric and Antisymmetric Relations. This list of fathers and sons and how they are related on the guest list is actually mathematical! (iii) Reflexive and symmetric but not transitive. This blog explains how to solve geometry proofs and also provides a list of geometry proofs. Yes. Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. For a relation R, an ordered pair (x, y) can get found where x and y are whole numbers or integers, and x is divisible by y. Antisymmetric relation is a concept based on symmetric and asymmetric relation in discrete math. Hence, as per it, whenever (x,y) is in relation R, then (y, x) is not. I'll wait a bit for comments before i proceed. Thus, (a, b) \u2208 R \u21d2 (b, a) \u2208 R, Therefore, R is symmetric. irreflexive relation symmetric relation antisymmetric relation transitive relation Contents Certain important types of binary relation can be characterized by properties they have. Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). 6. Antisymmetry is concerned only with the relations between distinct (i.e. The abacus is usually constructed of varied sorts of hardwoods and comes in varying sizes. symmetric, reflexive, and antisymmetric. Figure out whether the given relation is an antisymmetric relation or not. Discrete Mathematics Questions and Answers \u2013 Relations. $<$ is antisymmetric and not reflexive, ... $\\begingroup$ Also, I may have been misleading by choosing pairs of relations, one symmetric, one antisymmetric - there's a middle ground of relations that are neither! Fresheneesz 03:01, 13 December 2005 (UTC) I still have the same objections noted above. So total number of symmetric relation will be 2 n(n+1)/2. The term data means Facts or figures of something. (2,1) is not in B, so B is not symmetric. In this second part of remembering famous female mathematicians, we glance at the achievements of... Countable sets are those sets that have their cardinality the same as that of a subset of Natural... What are Frequency Tables and Frequency Graphs? i.e. The... A quadrilateral is a polygon with four edges (sides) and four vertices (corners). In that, there is no pair of distinct elements of A, each of which gets related by R to the other. Graphical representation refers to the use of charts and graphs to visually display, analyze,... Access Personalised Math learning through interactive worksheets, gamified concepts and grade-wise courses, is school math enough extra classes needed for math. 2 as the (a, a), (b, b), and (c, c) are diagonal and reflexive pairs in the above product matrix, these are symmetric to itself. First step is to find 2 members in the relation such that $(a,b) \\in R$ and $(b,a) \\in R$. Required fields are marked *. That is to say, the following argument is valid. A relation R on a set A is symmetric iff aRb implies that bRa, for every a,b \u03b5 A. A relation R in a set A is said to be in a symmetric relation only if every value of $$a,b \u2208 A, (a, b) \u2208 R$$ then it should be $$(b, a) \u2208 R.$$, Given a relation R on a set A we say that R is antisymmetric if and only if for all $$(a, b) \u2208 R$$ where a \u2260 b we must have $$(b, a) \u2209 R.$$. Complete Guide: How to multiply two numbers using Abacus? $$(1,3) \\in R \\text{ and } (3,1) \\in R \\text{ and } 1 \\ne 3$$ therefore the relation is not anti-symmetric. A relation R is defined on the set Z by \u201ca R b if a \u2013 b is divisible by 7\u201d for a, b \u2208 Z. This blog tells us about the life... What do you mean by a Reflexive Relation? Let a, b \u2208 Z, and a R b hold. (a \u2013 b) is an integer. Ada Lovelace has been called as \"The first computer programmer\". Rene Descartes was a great French Mathematician and philosopher during the 17th century. Examine if R is a symmetric relation on Z. both can happen. They... Geometry Study Guide: Learning Geometry the right way! Justify all conclusions. Their structure is such that we can divide them into equal and identical parts when we run a line through them Hence it is a symmetric relation. Show that R is a symmetric relation. If we let F be the set of all f\u2026 For example, if a relation is transitive and irreflexive, 1 it must also be asymmetric. Hence it is also in a Symmetric relation. Basics of Antisymmetric Relation A relation becomes an antisymmetric relation for a binary relation R on a set A. In this example the first element we have is (a,b) then the symmetry of this is (b, a) which is not present in this relationship, hence it is not a symmetric relationship. In a formal way, relation R is antisymmetric, specifically if for all a and b in A, if R(x, y) with x \u2260 y, then R(y, x) must not hold, or, equivalently, if R(x, y) and R(y, x), then x = y. The standard abacus can perform addition, subtraction, division, and multiplication; the abacus can... John Nash, an American mathematician is considered as the pioneer of the Game theory which provides... Twin Primes are the set of two numbers that have exactly one composite number between them. This blog deals with various shapes in real life. Any relation R in a set A is said to be symmetric if (a, b) \u2208 R. This implies that. Let a, b \u2208 Z and aRb holds i.e., 2a + 3a = 5a, which is divisible by 5. In discrete Maths, a relation is said to be antisymmetric relation for a binary relation R on a set A, if there is no pair of distinct or dissimilar elements of A, each of which is related by R to the other. Assume A={1,2,3,4} NE a11 a12 a13 a14 a21 a22 a23 a24 a31 a32 a33 a34 a41 a42 a43 a44 SW. R is reflexive iff all the diagonal elements (a11, a22, a33, a44) are 1. Antisymmetric relation is a concept based on symmetric and asymmetric relation in discrete math. \u201cIs equal to\u201d is a symmetric relation, such as 3 = 2+1 and 1+2=3. R = {(1,1), (1,2), (1,3), (2,3), (3,1), (2,1), (3,2)}, Suppose R is a relation in a set A = {set of lines}. By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). (iii) R is not antisymmetric here because of (1,2) \u2208 R and (2,1) \u2208 R, but 1 \u2260 2 and also (1,4) \u2208 R and (4,1) \u2208 R but 1 \u2260 4. In the above diagram, we can see different types of symmetry. We proved that the relation 'is divisible by' over the integers is an antisymmetric relation and, by this, it must be the case that there are 24 cookies. 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Apart from antisymmetric, there are different types of relations, such as: An example of antisymmetric is: for a relation \u201cis divisible by\u201d which is the relation for ordered pairs in the set of integers. Are all relations that are symmetric and anti-symmetric a subset of the reflexive relation? The word Abacus derived from the Greek word \u2018abax\u2019, which means \u2018tabular form\u2019. 2 Number of reflexive, symmetric, and anti-symmetric relations on a set with 3 elements I'm going to merge the symmetric relation page, and the antisymmetric relation page again. The fundamental difference that distinguishes symmetric and asymmetric encryption is that symmetric encryption allows encryption and decryption of the message with the same key. A relation R on a set A is symmetric iff aRb implies that bRa, for every a,b \u03b5 A. How can a relation be symmetric an anti symmetric? For example: If R is a relation on set A= (18,9) then (9,18) \u2208 R indicates 18>9 but (9,18) R, Since 9 is not greater than 18. Note - Asymmetric relation is the opposite of symmetric relation but not considered as equivalent to antisymmetric relation. Antisymmetric Relation. Now, 2a + 3a = 5a \u2013 2a + 5b \u2013 3b = 5(a + b) \u2013 (2a + 3b) is also divisible by 5. Hence it is also a symmetric relationship. In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. Symmetric. Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij , then the possible eigenvalues are 1 and \u20131. Antisymmetric relation is a concept based on symmetric and asymmetric relation in discrete math. (i) R is not antisymmetric here because of (1,2) \u2208 R and (2,1) \u2208 R, but 1 \u2260 2. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) \u2208 R for every a \u2208 ASymmetricRelation is symmetric,If (a, b) \u2208 R, then (b, a) \u2208 RTransitiveRelation is transitive,If (a, b) \u2208 R & (b, c) \u2208 R, then (a, c) \u2208 RIf relation is reflexive, symmetric and transitive,it is anequivalence relation These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. For example, the restriction of < from the reals to the integers is still asymmetric, and the inverse > of < is also asymmetric. This section focuses on \"Relations\" in Discrete Mathematics. Draw a directed graph of a relation on $$A$$ that is antisymmetric and draw a directed graph of a relation on $$A$$ that is not antisymmetric. Let $$a, b \u2208 Z$$ (Z is an integer) such that $$(a, b) \u2208 R$$, So now how $$a-b$$ is related to $$b-a i.e. It can be reflexive, but it can't be symmetric for two distinct elements. Also, compare with symmetric and antisymmetric relation here. The graph is nothing but an organized representation of data. Here x and y are the elements of set A. Proofs about relations There are some interesting generalizations that can be proved about the properties of relations. Almost everyone is aware of the contributions made by Newton, Rene Descartes, Carl Friedrich Gauss... Life of Gottfried Wilhelm Leibniz: The German Mathematician. There are different types of relations like Reflexive, Symmetric, Transitive, and antisymmetric relation. Flattening the curve is a strategy to slow down the spread of COVID-19. Or it can be defined as, relation R is antisymmetric if either (x,y)\u2209R or (y,x)\u2209R whenever x \u2260 y. An asymmetric relation is just opposite to symmetric relation. The mathematical concepts of symmetry and antisymmetry are independent, (though the concepts of symmetry and asymmetry are not). If A = {a,b,c} so A*A that is matrix representation of the subset product would be. For example. So, in \\(R_1$$ above if we flip (a, b) we get (3,1), (7,3), (1,7) which is not in a relationship of $$R_1$$. This is called Antisymmetric Relation. Therefore, R is a symmetric relation on set Z. Learn its definition along with properties and examples. ... Symmetric and antisymmetric (where the only way a can be related to b and b be related to a is if a = b) are actually independent of each other, as these examples show. This blog helps answer some of the doubts like \u201cWhy is Math so hard?\u201d \u201cwhy is math so hard for me?\u201d... Flex your Math Humour with these Trigonometry and Pi Day Puns! If a relation is symmetric and antisymmetric, it is coreflexive. Here we are going to learn some of those properties binary relations may have. In this short video, we define what an Asymmetric relation is and provide a number of examples. Similarly, in set theory, relation refers to the connection between the elements of two or more sets. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. For example: If R is a relation on set A= (18,9) then (9,18) \u2208 R indicates 18>9 but (9,18) R, Since 9 is not greater than 18. Relationship to asymmetric and antisymmetric relations. Which of the below are Symmetric Relations? Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. Are all relations that are symmetric and anti-symmetric a subset of the reflexive relation? Let R = {(a, a): a, b \u2208 Z and (a \u2013 b) is divisible by n}. In this short video, we define what an Antisymmetric relation is and provide a number of examples. In this article, we have focused on Symmetric and Antisymmetric Relations. Antisymmetric means that the only way for both $aRb$ and $bRa$ to hold is if $a = b$. Whether the wave function is symmetric or antisymmetric under such operations gives you insight into whether two particles can occupy the same quantum state. Let\u2019s say we have a set of ordered pairs where A = {1,3,7}. Relation R on a set A is asymmetric if (a,b)\u2208R but (b,a)\u2209 R. Relation R of a set A is antisymmetric if (a,b) \u2208 R and (b,a) \u2208 R, then a=b. In this short video, we define what an Antisymmetric relation is and provide a number of examples. There are different types of relations like Reflexive, Symmetric, Transitive, and antisymmetric relation. Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. (iii) Reflexive and symmetric but not transitive. Matrices for reflexive, symmetric and antisymmetric relations. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. The First Woman to receive a Doctorate: Sofia Kovalevskaya. Suppose that your math teacher surprises the class by saying she brought in cookies. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relationof a set as one with no ordered pair and its reverse in the relation. reflexive relation:symmetric relation, transitive relation REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION RELATIONS AND FUNCTIONS:FUNCTIONS AND NONFUNCTIONS R is reflexive. Which is (i) Symmetric but neither reflexive nor transitive. \u201cIs less than\u201d is an asymmetric, such as 7<15 but 15 is not less than 7. This section focuses on \"Relations\" in Discrete Mathematics. If any such pair exist in your relation and $a \\ne b$ then the relation is not anti-symmetric, otherwise it is anti-symmetric. Antisymmetric. Show that R is Symmetric relation. (ii) Transitive but neither reflexive nor symmetric. Relations, specifically, show the connection between two sets. Discrete Mathematics Questions and Answers \u2013 Relations. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the \"preys on\" relation on biological species). In all such pairs where L1 is parallel to L2 then it implies L2 is also parallel to L1. It helps us to understand the data.... Would you like to check out some funny Calculus Puns? Examine if R is a symmetric relation on Z. We can say that in the above 3 possible ordered pairs cases none of their symmetric couples are into relation, hence this relationship is an Antisymmetric Relation. Here's something interesting! ? Asymmetric. This is no symmetry as (a, b) does not belong to \u00f8. Complete Guide: Construction of Abacus and its Anatomy. The relation $$a = b$$ is symmetric, but $$a>b$$ is not. Paul August \u260e 04:46, 13 December 2005 (UTC) i know what an anti-symmetric relation is. Complete Guide: Learn how to count numbers using Abacus now! In mathematics, a relation is a set of ordered pairs, (x, y), such that x is from a set X, and y is from a set Y, where x is related to yby some property or rule. However, a relation can be neither symmetric nor asymmetric, which is the case for \"is less than or equal to\" and \"preys on\"). In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. so neither (2,1) nor (2,2) is in R, but we cannot conclude just from \"non-membership\" in R that the second coordinate isn't equal to the first. Properties. The history of Ada Lovelace that you may not know? Given a relation R on a set A we say that R is antisymmetric if and only if for all (a, b) \u2208 R where a \u2260 b we must have (b, a) \u2209 R. This means the flipped ordered pair i.e. Your email address will not be published. i don't believe you do. For relation, R, an ordered pair (x,y) can be found where x and y are whole numbers and x is divisible by y. #mathematicaATDRelation and function is an important topic of mathematics. Let ab \u2208 R. Then. 2 Number of reflexive, symmetric, and anti-symmetric relations on a set with 3 elements Hence this is a symmetric relationship. Given the usual laws about marriage: If x is married to y then y is married to x. x is not married to x (irreflexive) Given R = {(a, b): a, b \u2208 T, and a \u2013 b \u2208 Z}. reflexive relation:symmetric relation, transitive relation REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION RELATIONS AND FUNCTIONS:FUNCTIONS AND NONFUNCTIONS b \u2013 a = - (a-b)\\) [ Using Algebraic expression]. Then a \u2013 b is divisible by 7 and therefore b \u2013 a is divisible by 7. We also discussed \u201chow to prove a relation is symmetric\u201d and symmetric relation example as well as antisymmetric relation example. Think $\\le$. Note: If a relation is not symmetric that does not mean it is antisymmetric. The Greek word \u2018 abax \u2019, which means \u2018 tabular form symmetric and antisymmetric relation is antisymmetric guest is. ( 2,1 ) has been called as  the First computer programmer '' 1,2 ) is in b then... This case ( b, c } so a * a that matrix. 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Is also parallel to L2 then it implies L2 is also parallel to then!", "date": "2021-03-07 15:13:23", "meta": {"domain": "ubezwlasnowolnienie.pl", "url": "https://www.ubezwlasnowolnienie.pl/0dnlt82b/5c2a1c-symmetric-and-antisymmetric-relation", "openwebmath_score": 0.7028664350509644, "openwebmath_perplexity": 737.7476827893647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9861513897844354, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8585407061834383}}
{"url": "http://makariharlem.com/qt6hqml/linear-programming-2-variables-examples.html", "text": "Linear programming 2 variables examples\n\n+ a mn x n = b m x 1 , x 2 , , x n 0 n variables m equations maximize cT x subject to the constraints Consider the following linear program: Maximize z = 0x1 +0x2 \u22123x3 \u2212 x4 +20, (Objective 1) subject to: x1 \u22123x3 +3x4 = 6, (1) x2 \u22128x3 +4x4 = 4, (2) xj \u2265 0 (j = 1,2,3,4). 2. ). 9 (1,2) Bertsimas, Dimitris, and J. In Class XI, we have studied systems of linear inequalities in two variables and their solutions by graphical method. 5 c. a linear function of the decision variables. This may represent the selection or rejection of an option, the turning on or off of switches, a yes/no answer, or many other situations. 1 \u201cLinear programming was developed by George B. Each barrel of the less expensive crude produces 10 gallons of gasoline and 20 gallons of diesel. Learn vocabulary, terms, and more with flashcards, games, and other study tools. We will discuss formulation of those problems which involve only two variables. 6 Jun 2019 2 / 31. 2 Graph the solution to the linear inequality 50 \u2013 65x y \u2265 650. 5x 1 + 4x 2 = 35 and . 3. But if you\u2019re on a tight budget and have to watch those [\u2026] A linear program is said to be in standard form if it is a maximization program, there are only equalities (no inequalities) and all variables are restricted to be nonnegative. x 2 will be entering the set of basic variables and replacing s 2, which is exiting. This algorithm runs in O(n 2 m) time in the typical case, but may take exponential 2. The number of hours per week it takes to assemble and finish each type of stapler, and the profit for each type of stapler is given in the table below: Regular Heavy Duty These activities (variables) mustbe competingwith other variables for limited resourcesand relationships amongthese variables mustbe linear and the variables must be quantifiable. \\begin{align*}ax + by & = p\\\\ cx + dy & = q\\end{align*} where any of the constants can be zero with the exception that each equation must have at least one variable in it. 1 \u2022 LPP: Linear Programming Problem, one of these \u201c\ufb01nd the optimal value of a linear function subject to linear constraints\u201d problems Linear Programming Terms. ) The image is oriented so that the feasible region is in front of the planes. Linear programming solution examples. . 4. 9. What we have just formulated is called a linear program. All equations must be equalities. A linear system of two equations with two variables is any system that can be written in the form. 0. This video shows how to solve a minimization LP model graphically using the objective function line method. This will giv ey ou insigh ts in to what SOL VER and other commercial linear programming soft Linear programming (LP) refers to a family of mathematical optimization techniques that have proved effective in solving resource allocation problems, particularly those found in industrial production systems. 21 Feb 2019 The first three rows consist of the equations of the linear program, in which The most stringent restriction follows from the last equation (x\u2081 = 2 + x\u2083 -x\u2085). Linear Programming Problems Linear programming problems come up in many applications. 1. x, y, and z coordinate. A summary of Linear Programming in 's Inequalities. The artificial variables which are non-basic at the end of phase-I are removed. Formulate and solve graphically a Linear Programming model that will allow the company to maximize profits. Write an equation for the quantity that is being maximized or minimized (cost, profit, amount, etc. 5 and 0. So, the delivery person will calculate different routes for going to all the 6 destinations and then come up with the shortest route. Sections 3. activities denoted by j, there are n acitivities . In this video, I use linear programming to find the minimum an equation subject to a couple of inequalities. Decision variables x1,,xn \u2208 R. 1 The Basic LP Problem \u201cPresolving in linear programming. There are many points for which f= 24, for example in the point, (3 2;7), which is in the feasible region. If the problem is not a story problem, skip to step 3. Note the variables are Let\u2019s boil it down to the basics. The basic components of the LP are as follows: Decision Variables; Constraints; Data; Objective Functions; Linear Programming Simplex Method The goal of a linear programming problems is to find a way to get the most, or least, of some quantity -- often profit or expenses. A mixed-integer programming (MIP) problem is one where some of the decision variables are constrained to be integer values (i. ) Graphical Solution This very small problem has only two decision variables and therefore only two dimen-sions, so a graphical procedure can be used to solve it. This procedure involves con-structing a two-dimensional graph with x 1 and x 2 as the axes. 3 and 4. Two or more products are usually produced using limited resources. 4 of the text. The following videos gives examples of linear programming problems and how to test the vertices. Notice that point A is the intersection of the three planes x 2 =0 (left), x 3 =0 (bottom), s 4 =0 (cyan). \u201d Athena Scientific 1 (1997): 997. 2 is convenient. Two popular numerical methods for solving linear programming problems are the Simplex method and an Interior Point method. We now briefly discuss how to use the LINDO software. The key to formulating a linear programming problem is recognizing the decision variables. The parameter values are known with certainty. solve applications of Linear Programming Linear programming problems can be very complex and involve hundreds of vari-ables. S. In the case of linear programming, duality yields many more amazing results. Linear programming is the process of taking various linear inequalities relating to some situation, and finding the \"best\" value obtainable under those conditions. x 1 >= 0 . . Linear programming example 1987 UG exam. com. The constraints and objective function are entered into the Work window and the region of feasible solutions is plotted in the Graph window. Max/Min an Easy to visualize in low dimensions (2 or 3 variables) \u2013 Feasible space forms a convex polygon ! Optimum is achieved at a vertex, except when \u2013 No solution to the constraints \u2013 Feasible region is unbounded in direction of the objective CS 312 \u2013 Linear Programming 3 linear-programming model. A linear programming problem is a problem that requires an objective function to be maximized or minimized subject to resource constraints. It costs $2 and takes 3 hours to produce a doodad. Linear Programming Linear Programming Solving systems of inequalities has an interesting application--it allows us to find the minimum and maximum values of quantities with multiple constraints. 2: Applications of Linear Programming Problems Math 1313 Page 1 of 3 Section 2. Example 2: Solve graphically the inequality $y \\lt 1$ Solution to Example 2: Three steps to find the solution set the the given inequality. edu It is generally known that Chapter 4 of the MAT 119 textbook [10]1 is the shakiest of all chapters, especially sections 4. The less expensive crude costs$80 USD per barrel while a more expensive crude costs $95 USD per barrel. Add constraint window will appear once Add option clicked. The number of majestic seats should be at least half the number of the deluxe seats. Because it is often possible to solve the related linear program with the shadow prices as the variables in place of, or in conjunction with, the original linear program, thereby taking advantage of some computational efficiencies. The table gives the cost, in pounds, of transporting a television from each warehouse to each supermarket. For instance, called the objective function. \" They are called Worked example: solutions to 2-variable equations \u00b7 Practice: 31 Jan 2019 This example provides one setting where linear programming can be Problem\" , where X1 and X2 represent the decision variables, that is, . for a linear programming problem is the problem of minimizing a linear function cTx in the vector of nonnegative variables x \u2265 0 N subject to M linear equality constraints, which are written in the form Ax = b. \u21d0 Linear Inequalities in Two Variables \u21d2 Graphing the Solution Region of System of Linear Inequalities \u21d2 Leave a Reply Cancel reply Your email address will not be published. If there are two or more equal coefficients satisfying the above condition (case of tie), then choice the basic variable. The two most straightforward methods of solving these types of equations are by elimination and by using 3 \u00d7 3 matrices. It involves well defined decision variables, with an objective function and set of constraints. The diet problem. We will refer for graphing purposes to a graphing calculator. It is a horizontal line that splits the plane into two regions. infinity(), 2) constraint2. If some variables are restricted to be integer and some are not then the problem is a mixed integer programming problem. In the previous example it is possible to find the solution using the simplex method only because hi > 0 for all i and an initial solution x^ = 0 , i = 1, 2, n with Xn-{-j = ^j, j \u2014 1, 2,, m was thus feasible, that is, the origin is a feasible initial solution. In matrix form, a linear program in standard form can be written as: Max z= cTx subject to: Ax= b x0: where c= 0 B @ c. Linear programming cannot handle arbitrary restrictions: once again, the restrictions ha v etobe line ar. Implementation of interior point methods for large scale linear programming. If a term such as 3x 2 appears in a formulation, then the resulting problem is said to be nonlinear. For example: L = number of leadership training programs offered P = number of problem solving programs offered. EXAMPLE 1. 4 Maximization with constraints 5. Graphical methods can be classified under two categories: 1. Use linear programming to model and solve real-life problems. slack variables, s 1 and s 2 are added to the rst and second constraint, respectively: x+ 2y+ s 1 = 8; x y+ s 2 = 4: The slack variables will always be nonnegative (zero or pos-itive) when solving linear programming problems. Change of variables and normalise the sign of independent terms; Normalise restrictions ADVERTISEMENTS: Simplex Method of Linear Programming! Any linear programming problem involving two variables can be easily solved with the help of graphical method as it is easier to deal with two dimensional graph. Linear programming is a method to achieve the best outcome in a mathematical . 2 Worked Examples Example 1 Max Z = 3x 1 - x 2 Subject to 2x 1 + x 2 \u2265 2 x 1 + 3x 2 \u2264 2 x 2 \u2264 4 Lecture 7 Linear programming : Artifical variable technique : Two - Phase method 1 and x 1 \u2265 0, x 2 \u2265 0 Linear programming is the method of considering different inequalities relevant to a situation and calculating the best value that is required to be obtained in those conditions. show() Maximization: Constraints: Variables: a[1] = x_0 is a 2. Applications 1. resources denoted by i, there are m resources . To solve linear programming problems in three or more variables, we will Finite Math B: Chapter 4, Linear Programming: The Simplex Method. Therefore, we need artificial variables. We will use XR and XE to denote the decision variables. Linear Programming: Beyond 4. First, assign a variable (x or y) to each quantity that is being solved for. Its algorithm solvers for linear programming, mixed integer programming, and quadratic programming are able to solve problems with millions of constraints and variables. Linear Programming: The term was introduced in 1950 to refer to plans or schedules for training Duality is a concept from mathematical programming. Finite math teaches you how to use basic mathematic processes to solve problems in business and finance. Resource allocation 2. Each barrel of the more expensive crude produces 15 gallons of both gasoline and diesel. 2 (1995): 221-245. But series S 3 is -ve , we will add artificial variable A,i. While the problem is a linear program, the techniques apply to all solvers. As the independent terms of all restrictions 27 Aug 2019 Here's a simple example of a linear programming problem. Linear Programming: A Word Problem with Four Variables (page 5 of 5) Sections: Optimizing linear systems, Setting up word problems. Formulating linear programming problems. In a linear programming problem, we have a function, called the objective function, which depends linearly on a number of independent variables, and which we want to optimize in the sense of either \ufb01nding its mini-mum value or maximum value. linear-programming model. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. The column of the input base variable is called pivot column (in green color). , 2S + E \u2212 3P \u2265 150. Decision variables are sometimes called controllable variables because they are under the control of the decision maker. The solution to a linear program is an assignment to the variables that . Some Geometry for Optimization4 3. A decision variable is a system setting whose value is assigned by the decision maker. X x. Write the problem by defining the objective function and the system of linear inequalities. This solver is capable of finding optimal solutions for positive definite or semi-definite quadratic objectives (when Linear algebra is a one of the most useful pieces of mathematics and the gateway to higher dimensions. Set Up a Linear Program, Solver-Based Convert a Problem to Solver Form. Examples of theses applets are the \u2018Exploring linear programming\u2019 [8], the \u2018Linear programming applet\u2019 [9] or the \u2018Animated linear programming applet\u2019 [10]. In each case, linprog returns a negative exitflag, indicating to indicate failure. Linear constraints, each of the In a linear programming problem with just two variables and a hand- We'll do some examples to help understand linear programming problems, but most This is an example of a linear programming problem. In this case, we'll pivot on Row 2, Column 2. In the LP problem, decision variables are chosen so that a linear function of the decision variables is optimized and a simultaneous set of linear constraints involving the decision variables is satisfied. slack variable s 1, as before, and write x 1 + x 2 + s 1 = 10. However, for problems involving more than two variables or problems involving a large number of constraints, it is better to use solution methods that are adaptable to computers. Linear Programming Example: Maximize C = x + y given the constraints, y \u2265 0 x \u2265 0 4x + 2y \u2264 8 Simple Linear Regression Examples. Wikipedia has more advanced examples represented as pure algebra and a discussion about algorithms that provide general solutions for this class of optimization problem. Dependent variables, on the left, are called basic variables. There are two principal algorithms for linear programming. Since we can only easily graph with two variables (x and y), this approach is not practical for problems where there are more than two variables involved. , are to be optimized. This speci c solution is called a dictionary solution. Typ-ically these applets allow the user to de ne a linear program with two variables with a total number of constraints up to 4 or 5. Today we\u2019ll be learning how to solve Linear Programming problem using MS Excel? Linear programming (LP) is useful for resource optimization. Let\u2019s start from one of the linear programming problems from section 4. There are mainly four steps in the mathematical formulation of linear programming problem as a mathematical model. It also possible to test the vertices of the feasible region to find the minimum or maximum values, instead of using the linear objective function. 1 . {\\displaystyle x_{1},x_{2},x_{3 are (non-negative) slack variables, representing in this example the unused area, the amount of 2. Usually, a good choice for the definition is the quantity they asked you to find in the problem. Section 7-1 : Linear Systems with Two Variables. All constraints, except for the nonnegativity of decision variables, are limited and restrictive; as we will see later, however, any linear programming In the example above, the basic feasible solution x1 = 6, x2 = 4, x3 = 0, x4 = 0, is optimal. x 1 - x 2 >= 3 . Examples of these are the \u2018Parametric Linear Programming\u2019 [13], the \u2018The F undamental Theorem of Linear Programming\u2019 [14] or the \u2018Graphical Linear Programming for Two V ariables\u2019 [15]. (11) is attained 16 Aug 2018 An example of linear optimization I'm going to implement in R an The company can produce 10 seats, 20 legs and 2 backs from a standard wood block. The total number of seats should be at least 250. With recent advances in both solution algorithms In other words, the objective function is linear in the decision variables x r and x e. Gradients, Constraints and Optimization10 Chapter 2. 15, 0. This quantity is called your objective. For example: Find x, y such that the Linear Programming Problems (LPP) provide the method of finding such an Step 2: Identify the set of constraints on the decision variables and express them in the form Now let us look at an example aimed at enabling you to learn how to 26 Jan 2016 of linear programming, and there are also zillions of other examples. x 1 + 3x 2 + x 3 + x 4 = 5 2x 1 Example 1: The Production-Planning Problem. It satis\ufb01es the following: 1. A General Maximization Formulation2 2. This increases the dimensionality of the problem by only one (introduce one y variable) regardless of how many variables are unrestricted. There are three steps in applying linear programming: modeling, solving, and interpreting. 25x 2 + 12. Cafieri (LIX). Decision Variables:: Product 1 units to be produced daily: Product 2 units to be produced daily; Objective Function: Maximize . Example (continued) To form an equation out of the second inequality we introduceintroduce a second variable a second variable s 2 and subtract it from the leftit from the left side so that we can write \u2013 x 1 + x 2 \u2013 s 2 = 2. Note that as stated the problem has a very special form. All the feasible solutions in graphical method lies within the feasible area on the graph and we used to test the corner points of the feasible area for the optimal solution i. To use elimination to solve a system of three equations with Linear Inequalities and Linear Programming 5. 2X2 + 5X3 <= 15+X1-----2 since, X1 + X2 + X3 <= 9; let put X1=X2=X3=3 from constraints 2 6+15<=18 it is false,so X1=X2=X3=3 not possible keeping in mind the constraints 2 put, X1=4, X2=2,X3=3 from constraint 1 19<=19 (true) we have to maximize X1 + 2X2 + 3X3; its maximum value is possible when X1=4, X2=2 ,X3=3 maximum value=4+4+9=17 9<=9 (true) from constraints 2$\\endgroup$\u2013 Sara Sharp Jul How to solve linear programming problems with 3 variables Aiden Saturday the 31st Write a good thesis statement for an essay cyber revolution essays sales and marketing business plan sample for a new idea business plan dissertation projects for mba how to solve a problem like maria song essay on trust yourself black hole research paper thesis Linear Programming Lesson 2: Introduction to linear programming And Problem formulation Definition And Characteristics Of Linear Programming Linear Programming is that branch of mathematical programming which is designed to solve optimization problems where all the constraints as will as the objectives A linear program has: 1) a linear objective function 2) linear constraints that can be equalities or inequalities 3) bounds on variables that can be positive, negative, finite or infinite. performance measure denoted by z An LP Model: 1 n j j j zcx = max =\u2211 s. To solve a linear programming problem involving two variables by the graphical. This will giv ey ou insigh ts in to what SOL VER and other commercial linear programming soft w are pac k ages actually do. = Is any variable is unrestricted in sign, it can be expressed as. Linear and (mixed) integer programming are techniques to solve problems which can be formulated within An example problem (or two) Notice that the inequality relations are all linear in nature i. Press \"Solve\" to solve without showing the feasible region, or \"Graph\" to solve it and also show the feasible region for your problem. It is requested that if one of them is positive then the other must be negative. The above is an example of a linear program. 2 LINEAR PROGRAMMING INVOLVING TWO VARIABLES Many applications in business and economics involve a process called optimization, in which we are required to find the minimum cost, the maximum profit, or the minimum use of resources. Thus, these variables are not restricted to just integer values. e. \u2022 Let A be the number of barrels of ale. 2019 profile in courage essay contest review paper format for research ieee von steuben ap summer homework actual nursing home business plan top dissertation writing services near me term white paper mean. The following two sections present the general linear programming model and its basic assumptions. It might look like this: These constraints have to be linear. A small bank offers three type of loans: housing loans at$8. Suc han understanding can b e useful in sev eral w a ys. The use of integer variables greatly expands the scope of useful optimization problems that you can define and solve. It is evident that the word linear programming implies that all the constraints and the objective function are expressed as linear functions of the variables. x 1, x 2 \u2265 0 Define the decision variables. Books: . The number of deluxe seats should be at least 10%and at most 20% of the total number of seats. 3 Geometric Introduction to Simplex Method 5. The formula \u201c2P +E\u201d is called an objective function. Matthias Ehrgott . The corresponding equation of inequality A. Rewrite the objective function in the form -c 1x 1 - c 2x 2 - -c nx n +P=0. The answer should depend on how much of some decision variables you choose. In this section, we will consider only a few simple problems. Three warehouses W, X and Y supply televisions to three supermarkets J, K and L. 2 \u2022 Dual: A related but opposite problem with \u201cthe same\u201d answer, usually a standard maximize LPP in sec 4. Typically you can look at what the problem is asking to determine what the variables are. There are rules about what you can and cannot do within linear programming. problem characteristics 2. Objective function Graphing the Solution Region of Linear Inequality in Two Variables. since, for example, we only receive 98% of the water from supplier 2 that we have to pay for. Examples and standard form Fundamental theorem Simplex algorithm Example I Linear programming maxw = 10x 1 + 11x 2 3x 1 + 4x 2 \u2264 17 2x 1 + 5x 2 \u2264 16 x i \u2265 0, i = 1,2 I The set of all the feasible solutions are called feasible region. 4X \u2013 7Y \u2013 5Z + S 2 =2 . Solution. There are several assumptions on which the linear programming works, these are: Proportionality: The basic assumption underlying the linear programming is that any change in the constraint inequalities will have the proportional change in the objective function. Linear Programs: Variables, Objectives and Constraints The best-known kind of optimization model, which has served for all of our examples so far, is the linear program. \u201d Mathematical Programming 71. The factory is very small and this means that floor space is very limited. subject to 2x 1 + 3x 2 \u2265 1200 x 1 + x 2 \u2264 400 2x 1 + 1. PuLP can be installed using pip, instructions here. 5x 2 \u2265 900. +. SOLUTION OF LINEAR PROGRAMMING PROBLEMS THEOREM 1 If a linear programming problem has a solution, then it must occur at a vertex, or corner point, of the feasible set, S, associated with the problem. We also know that the increase in the objective function will be 2\u00d716 = 32. Changing variables are x and y i. The simplex algorithm studied in Chapter 2 is based on the fact that the feasi- In binary integer programming or 0-1 integer programming, all the variables This section presents some illustrative examples of typical integer programming. What are those things? Choose variables to represent how much of each of those things. Linear Programming: Geometry, Algebra and the Simplex Method A linear programming problem (LP) is an optimization problem where all variables are continuous, the objective is a linear (with respect to the decision variables) function , and the feasible region is de\ufb01ned by a \ufb01nite number of linear inequalities or equations. This very small problem has only two decision variables and therefore only\u00a0 19 Dec 2016 This article shows two ways to solve linear programming problems programming problem in SAS, let's pose a particular two-variable problem: Started\" example in the PROC OPTMODEL chapter about linear programming\u00a0 A change is made to the variable naming, establishing the following correspondences: x becomes X1; y becomes X2. Tsitsiklis. This is a simplified example will illustrates the way in which a problem\u00a0 Approximatede solution if integer variables take large values . Example 2: The Investment Problem. Modeling Assumptions in Linear Programming14 2. \u201cProgramming\u201d \u201c Planning\u201d (term predates computer programming). There is an x-coordiuatu IJIHI real number, and there is a y-coordinate that can be any real number. Set objective is our equation which has to minimized here cell F4, 2. DEPARTAMENTO DE ORGANIZACI\u00d3N INDUSTRIAL. +\u22ef+. Binary Integer Programming. This example shows how to convert a problem from mathematical form into Optimization Toolbox\u2122 solver syntax using the solver-based approach. Solution of Linear Programming Problems: Mathematical Formulation of Linear Programming Problems. problem formulation guidelines 3. Part 1 \u2013 Introduction to Linear Programming Part 2 \u2013 Introduction to PuLP Part 3 \u2013 Real world examples \u2013 Resourcing Problem Part 4 \u2013 Real world examples \u2013 Blending Problem Part 5 \u2013 Using PuLP with pandas and binary constraints to solve a scheduling problem Part 6 \u2013 Mocking conditional statements using binary constraints The most fundamental optimization problem tr eated in this book is the linear programming (LP) problem. self Check 2 Find the minimum value of P 5 2x 1 y subject to the constraints of Example 2. 2 = 240 To include all variables in each equation (a requirement of the next simplex step), we add slack vari- ables not appearing in each equation with a coefficient of zero. First, create variables x and y whose values are in the range from 0 to infinity. x 1 + x 2 + x 3 \u2264 11 b. Constraint(-solver. x 2 >= 0 . This procedure involves con-structing a two-dimensional graph with x 1 and x 2 as the \u2022 Primal: The original problem, usually a minimize LPP in sec 4. When you\u2019re dealing with money, you want a maximum value if you\u2019re receiving cash. 30x 1 + 15x 2 + 45x 3 \u2264 300 x 1 \u2265 0, x 2 \u2265 0, and x 3 \u2265 0 Write the objective function: N(x 1, x 2, x 3) = 1. 13. 5 to Solve Linear/Integer Programs Author Michel Berkelaar and others Maintainer ORPHANED Description Lp_solve is freely available (under LGPL 2) software for solving linear, integer and mixed integer programs. Now, we will look at the broad classification of the different Types of Linear Programming Problems one can encounter when confronted with one. x 1 + x 2 <= 10 . The Basic Set consists of 2 utility knives and 1 chef\u2019s knife. The duality theory in linear programming yields plenty of extraordinary results, because of the specific structure of linear programs. Problem characteristics. As noted above, the Premium Solver Platform uses an extended LP/Quadratic version of the Simplex method with bounds on the variables to handle LP and QP problems of up to 2,000 decision variables. 375x 3 \u2264 82. , 4 2 2). one of the corner points of the feasible area used to be the optimal solution. Formulation of Linear Programming Problem examples. In this example, the constraints are the minimum requirements of the vitamins. now try exercise 15. Write objective function. Duality in Linear Programming Duality in Linear Programming D2 Linear programming - Formation of problems PhysicsAndMathsTutor. For example, let us consider the following linear programming problem (LPP). Using Barney Stinson's crazy-hot scale, we introduce its key concepts. Again, the linear programming problems we\u2019ll be working with have the first variable on the $$x$$-axis and the second on the $$y$$-axis. For a problem to be a linear programming problem, the decision variables, objective function and constraints all have to be linear functions. c) Use slack variables to convert each constraint into a linear equation 2 150. A typical example would be taking the limitations of materials and labor, and then determining the \"best\" production levels for maximal profits under those conditions. HEC/Universite de Geneve Section 7-1 : Linear Systems with Two Variables. 9. The book aims to be a \ufb01rst introduction to the subject. Andersen, Erling D. For linear programming problems involving two variables, the graphical solution method introduced in Section 9. The simplex method. Learn about a class of equations in two variables that's called \"linear equations. Duality is a concept from mathematical programming. R3 is the space of 3 dimensions. Linearity assumptions usually are signi cant approximations. Maximize p = x+y subject to x+y = 2, 3x+y >= 4 Decimal mode displays all the tableaus (and results) as decimals, rounded to the number of significant digits you select (up to 13, depending on your processor and browser). Linear programming with 3 variables watch. To save on fuel and time the delivery person wants to take the shortest route. Chapter 7 The Simplex Metho d In this c hapter, y ou will learn ho w to solv e linear programs. 5. Integer programming can also be solved in polynomial time if the total number of variables is two [6];\u00a0 Tutorial on solving linear programming word problems and applications with two variables. Each Danio eats 4 grams/day of fish flakes while the slower Gourami eats 2 grams/day. 4 Determine the number of each type that must be produced each week to make a Set Up a Linear Program, Problem-Based Convert a Problem to Solver Form. In this notebook, we\u2019ll explore how to construct and solve the linear programming problem described in Part 1 using PuLP. For additional formulation examples, browse Section 3. Maximize 3x + 4y subject the variables. Table 1. Please check image below for reference. Minimize f\u00a0 LINEAR PROGRAMMING: EXERCISES - V. Package \u2018lpSolve\u2019 August 19, 2019 Version 5. 4X \u2013 7Y \u2013 5Z < 2 (b) Adding slack variables in the constraints . Linear programming example 1997 UG exam. Customer A needs fifty sheets and Customer B needs seventy sheets. In many of the examples, the maximize option can be added to the command to . Modify the example or enter your own linear programming problem (with two variables x and y) in the space below using the same format as the example. \u25a1 Among other things, CPLEX allows one to deal with: \u25c7 Real linear . Marko, the advantages (and the limitations) of linear programming are set out below. Launch the LINDO package. The activities all contribute to some measurable bene t (which we wish to maximize) or to some measurable cost PuLP is an open source linear programming package for python. An example of this type of problem is the following: Linear programming gives us a mechanism for answering all of these questions quickly and easily. EXAMPLE 2 Maximizing Annual Yield We will now show how to solve a linear programming problem in two variables graphically. 2, solved like in sec 4. $\\begingroup$ constraints are : X1 + X2 + X3 <= 9; -----1 -1X1 + 2X2 + 5X3 <= 15; X1 >= 0; X2 >= 0. In phase I, we form a new objective function by assigning zero to every original variable (including slack and surplus variables) and -1 to each of the artificial variables. CHAPTER 11: BASIC LINEAR PROGRAMMING CONCEPTS FOREST RESOURCE MANAGEMENT 207 maximizeor minimize Z c i X i i n = = \u2211 1 subject to i=1 n \u2211a j,i X i \u2264 b j j =1,2,,m inequalities) with this form. Let's define the following variables $x_{4p}$ is the number of 4P\u00a0 decision variables in our model, one decision variable per product. AMPL models: a first example. Consider the two variable linear optimization problem written algebraically:\u00a0 We will see examples in which we are maximizing or minimizing a linear General problem Given a linear expression z=ax+by in two variables x and y, find \u00a0 the late 1940s. The goal of utilizing slack variables is to change the two inequalities to equalities. The largest optimization problems in the world are LPs having millions of variables and hundreds of thousands of constraints. Solving this problem is called linear programming or linear optimization. Suppose you wish to solve the product-mix problem. 1 Modeling Modeling a problem using linear programming involves writing it in the language of linear programming. Two Phase Method: Minimization Example 1. linear means: of the form A 11X 1 + A 12X 2 + ::: + A 1;nX n B 1 or A 11X 1 + A 12X 2 + ::: + A 1;nX n B 1 or A 11X 1 + A 12X 2 + :::+ A 1;nX n = B 1) The Conditions for a problem to t the Linear Programming Model 1. In our example, $$x$$ is the number of pairs of earrings and $$y$$ is the number of necklaces. We have already read that a Linear Programming problem is one which seeks to optimize a quantity that is described linearly in terms of a few decision variables. exercise. STEP 2: REWRITE the objective function so all the variables are on the left and the constants are on the right. Graphical methods provide visualization of how a solution for a linear programming problem is obtained. Linear programming problems are optimization problems where the objective function and constraints are all linear. Write an equation for the quantity that is being maximized or minimized solution(s). 5x 1 + 4x 2 <= 35 . An objective function defines the quantity to be optimized, and the goal of linear programming is to find the values of the variables that maximize or minimize the objective function. Solve it using define variables, obj. An LP problem contains severa l essential elements. The dual linear program. Linear programming: how to formulate a condition that product of two variables must be not positive. 3X + 4Y \u2013 6Z \u2013 S 3 = 29/7 . It is only possible to graphically solve linear programming problems in two variables. + c n x n subject to the constraints a 11 x 1 + a 12 x 2 +. easily. Facility location . Linear programming is by far the most widely used method of constrained optimization. The answer to a linear programming problem is always \"how much\" of some things. Notice further that the left-hand-side expressions in all four constraints are also linear. A company manufactures staplers, regular and heavy duty. :2. 2 Exercises 1. 2 History Linear programming is a relatively young mathematical discipline, dating from the invention of the simplex method We'll see some examples of such constraint matrices when we look at applications. 7. Linear relationship means that when one factor changes so does another by a constant amount. e 10,000). 50$% interest, education loans at$13. This is going to be a fairly short section in the sense that it\u2019s really only going to consist of a couple of examples to illustrate how to take the methods from the previous section and use them to solve a linear system with three equations and three variables. 2 Draw a graph of the system and indicate the feasible region clearly. maximize c 1 x 1 + c 2 x 2 + . 1) Graph the corresponding equation $$y = 1$$. Example of the method of the two phases we will see how the simplex algorithm eliminates artificals variables and uses artificial slack variables to give a solution to the linear programming problem. In linear and integer programming methods the objective function is measured in one dimension only but A typical problem requiring the method of linear programming, a graphical approach, provides linear constraints and an objective function, which is to be either maximized or minimized. (The half-planes corresponding to the constraints are colored light blue orange and purple respectively. same index as a basic variable in the right-hand tableau example. Every p . Start studying Chapter 2: An Introduction to Linear Programming. 224J 15 2;x 3;w 1;w 2;w 3;w 4;w 5 0: Notes: This layout is called a dictionary. Different backends compute with different base fields, for example: 'string', QQ] - 7*b[2] x_2 - 7*x_3 sage: mip. FORMULATING LINEAR PROGRAMMING PROBLEMS One of the most common linear programming applications is the product-mix problem. Using Excel to solve linear programming problems Technology can be used to solve a system of equations once the constraints and objective function have been defined. 1 Systems of Linear Inequalities 5. Solve Linear Programs by Graphical Method. For example, you can use linear programming to stay within a budget. add constraints using Add option. If the all the three conditions are satisfied, it is called a Linear Programming Problem. Thecase where the integer variables are restricted to be 0 or 1 comes up surprising often. Take the system of linear inequalities and add a slack variable to each inequality to make it an equation. [2nd] convert each row of the final tableau (except the bottom row) back into equation form (as at the right) to find the values of the remaining variables. Linear programming is a technique that provides the decision maker with a way of optimizing his objective within resource requirements and other constraints provided that the following basic assumptions apply: I . ----- What is Mixed Integer Programming? Section 7-2 : Linear Systems with Three Variables. Costs and daily availability of the oils LINEAR PROGRAMMING \u2013 THE SIMPLEX METHOD (1) Problems involving both slack and surplus variables A linear programming model has to be extended to comply with the requirements of the simplex procedure, that is, 1. In fact, there is a whole line for which f= 24, namely the line 2x+3y= 24. 1 The Basic LP Problem. Any linear constraint can be rewritten as one or two expressions of the type linear for example a restriction that a variable should take integer values, are not allowed. It involves slack variables, tableau and pivot variables for the optimisation of a problem. Example 2: a) Determine the number of slack variables needed. Steps towards formulating a Linear Programming problem: Step 1: Identify the \u2018n\u2019 number of decision variables which govern the behaviour of the objective function (which needs to be optimized). Faster algorithms have been found in for example [2, 10]. Simple Linear Programming Problems13 1. asu. A building supply has two locations in town. 1 + 3X. The numbers on the lines indicate the distance between the cities. It is plain from the diagram below that the maximum occurs at the intersection of . For example, consider a linear programming problem in which we are asked . 3. Linear Programming. There are so many real life examples and use of linear programming. An important class of optimisation problems is linear programming problem which can be solved by graphical methods 9. 1 A first The first step of the Simplex Method is to introduce new variables called slack variables. All variables must be present in all equations. List of Figuresv Preface ix Chapter 1. All decision variables are constrained to be nonnegative. We can also see in the graph that the smaller the values Methods of solving inequalities with two variables, system of linear inequalities with two variables along with linear programming and optimization are used to solve word and application problems where functions such as return, profit, costs, etc. com - View the original, and get the already-completed solution here! Solve the linear programming models using either lp_solve (recommended, see linear programming tutorial) or excel solver (Google for details). Why you should learn it Linear programming is a powerful tool used in business and industry to manage resources effectively in order to maximize profits or minimize costs. Assign the variables: x 1 = number of convenience stores x 2 = number of standard stores x 3 = number of expanded services stores Write the constraints: a. In binary problems, each variable can only take on the value of 0 or 1. 05 (square metres) for products 1, 2, 3 and 4 respectively. When a computer solves a linear programming problem, it starts somewhere in the feasible region and searches for the optimal solution. Linear Programming example in 2 dimensions: x y 0 2 4 6 0 2 4 1 1 x \u2264 4 2 1 y 5 1 = 100 and the second becomes 4X. 2x 1 + 2x 2 Finite Math B: Chapter 4, Linear Programming: The Simplex Method 11. CHAPTER 4. E. The variables of a linear program take values from some continuous range; the objective and constraints must use only linear functions of the vari-ables. Dantzig in 1947 as a technique for planning the Quadratic Programming. Example 2 Slack variables. Linear inequalities 1 WE1 Graph the solution to the linear inequality 4 + 7xy \u2264 28. In this section we will solve systems of two equations and two variables. 2 Dantzig\u2019s method is not only of interest from a computational point of view, but also from a theoretical point of view, since it enables us 2 Actually, we present a version of Dantzig\u2019s (1963; chapter 9) revised simplex algorithm. Setting x 1, x 2, and x 3 to 0, we can read o the values for the other variables: w 1 = 7, w 2 = 3, etc. Constraints: The first constraint represents the daily assembly time constraints. expressions, where user does not create an IloExpr object explicitly (see the example). Several conditions might cause linprog to exit with an infeasibility message. Put the following linear programming problem into standard form. of linear inequalities in two variables and their solutions by graphical . Problem formulation 1. 5X + 7Y + 4Z + S 1 =7 . This article shows two ways to solve linear programming problems in SAS: You can use the OPTMODEL procedure in SAS/OR software or use Systems of equations with three variables are only slightly more complicated to solve than those with two variables. The Cut-Right Knife Company sells sets of kitchen knives. In our example, the criterion was to maximize the objective function. function, constraints and solve the problem with CPLEX:. Solving a Linear Programming Problem. A linear program can be solved by multiple methods. The algorithm used here is. Optimization: Linear Programming attempts to either maximise or minimize the variables. In this course, we introduce the basic concepts of linear programming. 25$% interest. For example, 23X 2 and 4X 16 are valid decision variable terms, while 23X 2 2, 4X 16 3, and (4X 1 * 2X 1) are not. \u201cIntroduction to linear programming. Each constraint can be represented by a linear inequality . a21x1 + a22x2 + + a2nxn = b2. Examples and word problems with detailed solutions are presented. 5 give some additional examples of linear . Lecture 2: Multiobjective Linear Programming. 2 subject to the constraints in the numerical example of Figure 1. Write out the matrix A for the transportation problem in standard form. It is customary to refer to the first group of Home / How to solve linear programming problems with 3 variables / How to solve linear for daycare center pdf hero essay hook examples cake business plan template The number of variables assigned values of zero is n m, where n equals the number of variables and m equals the number of constraints (excluding the nonnegativity constraints). The value of the objective function is in the lower Linear Equations in Three Variables JR2 is the space of 2 dimensions. 1 1 n computation was devoted to linear programming. two types of problems 4. Minimize z = 200x 1 + 300x 2. A decision is made when a value is specified for a decision variable. Linear inequations of two variables Let's begin with a few particular cases: We have to transport $$5$$ office chairs (that weigh $$10$$ kg each one) and three tables (weighing $$20$$ kg each). Linear Programming Problems [2-variables] Mathematical Programming Characteristics Decisions must be made on the levels of a two or more activities. Linear equations in three variables. Example of a linear programming problem. Example: Graph the solution set of the linear inequality in xy\u2013plane. : min x = c x + c x + + c x. b) Name them. as initial solution. 2 in this linear programming model essentially duplicates the information summarized in Table 3. This is why we call the above problem a linear program. Linear Programming Example. Discrete Math B: Chapter 4, Linear Programming: The Simplex Method 2. + a 2n x n = b 2 a m1 x 1 + a m2 x 2 +. PDF | A linear programming problem (LP) deals with determining optimal allocations of LP problems that involve only two variables can be solved by both methods. 2-16 Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty). Formulation of linear programming is the representation of problem situation in a mathematical form. We'll see some examples of such constraint matrices when we look at applications. Now, if we let x 1, x 2 and x 3 equal to zero in the initial solution, we will have x 4 = 5 and x 5 = -2, which is not possible because a surplus variable cannot be negative. 1, 0. Today, linear programming is applied to a wide variety of problems in industry and science. Each coordinate can be any real number. Solved problems 2. Figures on the costs and daily availability of the oils are given in Table 1 below. This JavaScript E-labs learning object is intended for finding the optimal solution, and post-optimality analysis of small-size linear programs. For x\u2013intercept: Put in equation (i) An infeasible LP problem with two decision variables can be identified through its graph. 4, and leaves a lot to be desired when teaching MAT 119. x 1 - x 2 = 3 linear \u2022 MAX{x1,x2,\u2026}, xi*yi, |xi|, etc => non-linear if xi and yi are variables \u2013 Sometimes there is a way to convert these types of constraints into linear constraints by adding some decison variables \u2013 Examples: 12/31/2003 Barnhart 1. T . The first stage of the algorithm might involve some preprocessing of the constraints (see Interior-Point-Legacy Linear Programming). 3 If the profit (P) on type X is R800 and on type Y is R1000, write down the objective function in the form P = ax + by . In this example, it has two decision variables, x r and x e, an objective function, 5 x r + 7 x e, and a set of four constraints. Problems with more than two variables (as is the case for most real The graphical method for solving linear programming problems in two unknowns is as follows. Linear programming algorithms. 3 WE2 Graph the inequality 7 \u2013x 12y \u2264 84 and show that (4, \u20133) is a solution. 1 Represent the above information as a system of inequalities . The following are notes, illustrations, and algebra word problems that utilize linear optimization methods. It remains an important and valuable technique. \uf03d \uf02b \uf02b\uf0a3 \uf02b\uf0a3 \uf0b3\uf0b3 This problem is a standard maximization problem with the decision variables x and y. The Wolfram Language has a collection of algorithms for solving linear optimization problems with real variables, accessed via LinearProgramming, FindMinimum, FindMaximum, NMinimize, NMaximize, Minimize, and Maximize. The solution of a linear inequality in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are Formulating Linear Programming Models Decision variables; Objective function; Constraints for example to maximize profit or minimize cost, although this is not always the case. whole numbers such as -1, 0, 1, 2, etc. This means that a linear function of the decision v ariables m ust b e r elate d to a constan t, where can mean less than or equal to, greater than or equal to, or equal to. For this example the column forming coefficients for non-basic variable 3. To solve linear programming models, the simplex method is used to find the optimal solution to a problem. In a team decision problem there are two or more decision variables, and these optimum basic feasible solution has been attained. non-negativity constraints for the example problem is shown in Fig 6. Speci\ufb01c examples and Linear Programming Simplex Method. g. The world is more complicated than the kinds of optimization problems that we are able to solve. Example 5: Solve using the Simplex Method. We A more complete presentation can be found for example in [2]. The entire problem can be expressed as straight lines, In some applications, you need to optimize a linear objective function of many variables, subject to linear constraints. Solution: We have. This example shows how to convert a linear problem from mathematical form into Optimization Toolbox\u2122 solver syntax using the problem-based approach. Examples of Linear Optimization 2 1 Linear Optimization Models with Python Python is a very good language used to model linear optimization problems. Exercise: Soft Drink Production A simple production planning problem is given by the use of two ingredients A and B that produce products 1 and 2 . 2: The outcomesof all activities are known with certainty. Formulate constraints. 2) Select point $$( 1 , -1 )$$ situated in the region below the horizontal line. 1 Slack Variables and the Pivot (text pg169-176) In chapter 3, we solved linear programming problems graphically. 2: Maximize 5 6 subject to 24 24 0, 0. 2 + S. Example 2 Solve the following linear programming problem graphically: Minimise Z 28 Feb 2017 Example of a linear programming problem. In this example it would be the variable X 1 (P 1) with -3 as coefficient. The first step is to identify Linear programming example 1991 UG exam. The example of a canonical linear programming problem from the introduction lends itself to a linear algebra-based interpretation. PROBLEM 1 Define in detail the decision variables and form the objective function and all. Basic two-variable linear programming problems with numerical solutions and illustrative graphs are available on PurpleMath. [1st] set equal to 0 all variables NOT associated with the above highlighted ISM. In a non-trivial optimization problem, we have among others two variables and both of them are ranged, i. A factory manufactures doodads and whirligigs. 2 n n. , et al. A linear program (LP) is an optimization problem (Wikipedia article . Example 4 (Phase I - Phase II Method): 2. that Linear Programming (LP) models of very large size can be solved in . INTEGER PROGRAMMING. Only one week's production is stored in 50 square metres of floor space where the floor space taken up by each product is 0. 2 Linear Programming. Formulation and Example. Write the objective function. linear programming problems. 1 n c d. For this model,n 4 variables and m 2 constraints; therefore, two of the variables are assigned a value of zero (i. This Demonstration shows the graphical solution to the linear programming problem: maximize subject to . In this section we discuss one type of optimization problem called linear pro-gramming. 1. Components of Linear Programming. Those are your non-basic variables. Step 2: Identify the set of constraints on the decision variables and express them in the form of linear equations /inequations. TD1 sept 2009 We defined a very simple Linear Programming problem. 4 and 3. To solve a linear programming problem with two decision variables using the 13 Sep 2018 Linear programming is the technique used to maximize or minimize a To create one unit of medicine 1 , you need 3 units of herb A and 2 As the constraint have few variables (only x and y ), transforming the Moving forward from this basic example, the true potential of optimization is showcased when With our Linear Programming examples, we'll have a set of compound inequalities, Define the variables, write an inequality for this situation, and graph the 1 May 2005 cjxj. The office receives orders from two customers, each requiring 3/4-inch plywood. Example 2:. A linear program is a set of linear constraints defined over a set of variables. LINEAR PROGRAMMING WITH TWO VARIABLES 189 =6 =12 =18 =24 5 In the corner point, (0;8) we have f= 2(0)+3(8) = 24. Solving Linear Programs in Excel 2) Now label the row just above tableau (I am using rows 10 and on since I have the tableau above in the first few lines ) Variable values to manipulate. Summarize relevant material in table form, relating columns . Graphically Solving Linear Programs Problems with Two Variables (Bounded Case)16 3. Goal programming 1 Goal Programming and Multiple Objective Optimization Goal programming involves solving problems containing not one specific objective function, but rather a collection of goals. 1 Entering variable xs has to be chosen where the maximum in. Introduce decision variables. 2. Be sure to line up variables to the left of the ='s and constants to the right. Standard, deluxe and majestic seats each costs \u00a320, \u00a326 and \u00a336 respectively. In a linear programming problem with two variables, the slack variables are always nonnegative in the corner points of This content was COPIED from BrainMass. From a marketing or statistical research to data analysis, linear regression model have an important role in the business. Solve the following linear program: maximise 5x 1 + 6x 2. the variables f are multiplied by constant Introduction: There are two types of linear programs (linear programming problems): . where X, Y, Z, S 1, S 2, S 3 > 0 (c) Put X = Y= Z = 0, we get S 1 = 7, S 2 = 2, S 3 = -29/7. D3 and E3, 4. = \u2212. LINEAR PROGRAMMING \u2013 THE SIMPLEX METHOD (1) Problems involving both slack and surplus variables A linear programming model has to be extended to comply with the requirements of the simplex procedure, that is, 1. Introduction to Optimization1 1. 10. Chapter 4: Linear Programming The Simplex Method Day 1: 4. We\u2019ll see one of the real life examples in the following tutorial. 6 Max Min with mixed constraints (Big M) Systems of Linear Inequalities in Two Variables See Interior-Point-Legacy Linear Programming. I Select Generate Linear Pgm from the Tools menu. Solving MOLPs by Weighted Sums. The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities. All the feasible solutions in graphical method lies within the feasible area on the graph and we used to test the corner [\u2026] Linear Programming Linear Programming Solving systems of inequalities has an interesting application--it allows us to find the minimum and maximum values of quantities with multiple constraints. It provides the optimal value and the optimal strategy for the decision variables. \u201cLinear\u201d No x2, xy, arccos(x), etc. Linear programming is a useful way to discover how to allocate a fixed amount of resources in a manner that optimizes productivity. Assumptions of Linear programming. LINEAR PROGRAMMING. Linear programming. The levels are represented by decision variables X 1X 2, etc. F or example, y ou will b e able to iden tify when a problem has 2. 2 (The Simplex Method) Christopher Carl Heckman Department of Mathematics and Statistics, Arizona State University checkman@math. Set Up a Linear Program, Problem-Based Convert a Problem to Solver Form. Linear Programming (LP) is an attempt to find a maximum or minimum solution to a function, given certain constraints. In addition, our objective function is also linear. Example. Since they involve 2. The objective can be represented by a linear function. be described by a linear function of the decision variables, that is, a mathematical function involving only the first powers of the variables with no cross products. The objective function is to be maximized subject to the specified constraints on the decision variables. 6. Linear programming is closely related to linear algebra; the most noticeable difference is that linear programming often uses inequalities in the problem statement rather than equalities. Example 1: The Production-Planning Problem. As is well known, such a problem is amenable to linear programming, and as I have shown in another paper [2], the introduction of probabilistic uncertainty, and of the further complications of a team situation, does not destroy the linear character of a programming problem, Chapter 2 Linear programming 10 With this slope the optimal solution will be x 1 1 000 and x 2 0, as indicated by the dot ted line in Figure 2. Define the variables. For the straight- Linear Programs: Variables, Objectives and Constraints The best-known kind of optimization model, which has served for all of our examples so far, is the linear program. As a reminder, the form of a canonical problem is: Minimize c1x1 + c2x2 + + cnxn = z Subject to a11x1 + a12x2 + + a1nxn = b1. Learn exactly what happened in this chapter, scene, or section of Inequalities and what it means. What makes it linear is that all our constraints are linear inequalities in our variables. Linear Programming Problem Complete the blending problem from the in-class part [included below] An oil company makes two blends of fuel by mixing three oils. 2 Linear Programming What you should learn Solve linear programming problems. Linear programming gives us a mechanism for answering all of these questions quickly and easily. Find the value as a function of a. In Two Phase Method, the whole procedure of solving a linear programming problem (LPP) involving artificial variables is divided into two phases. Linear Programming Notes VII Sensitivity Analysis 1 Introduction When you use a mathematical model to describe reality you must make ap-proximations. We will use This will be the very first system that we solve when we get into examples. The Regular Set consists of 2 utility knives and 1 chef\u2019s knife and 1 bread knife. Along the way, dynamic programming and the linear complementarity problem are touched on as well. These are the cells that Excel will \u201cchange\u201d to find the optimum solution to the problem. An integer programming problem in which all variables are required to be integer is called a pure integer pro-gramming problem. This method is used to solve a two variable linear program. All constraints are equality type. 3: A well defined objective function exist which can be used to evaluate differentoutcomes. 5 The Dual; Minimization with constraints 5. feasible region I This feasible region is a colorred convex polyhedron (\u00e0\u0131\u0153/) spanned by points x 1 Toy LP example: brewer\u2019s problem. Let's say a FedEx . Where, , \u2265 0 . 3 Date 2015-09-18 Title Interface to 'Lp_solve' v. To achieve this requirement, convert any unrestricted variable X to two non-negative variables by substituting T - X for X. Any linear programming problem involving two variables can be easily solved with the help of graphical method as it is easier to deal with two dimensional graph. A company makes two products (X and Y) using two machines (A and B). zx y xy xy xy. Sometimes for integer variables the value is not integer. t. In this Linear Programming: related mathematical techniques used to allocate limited resources among competing demands in an optimal way . ment of linear programming and proceeds to convex analysis, network \ufb02ows, integer programming, quadratic programming, and convex optimization. 3 THE SIMPLEX METHOD: MAXIMIZATION. 75$% interest rates, and loans to senior citizens at $12. There is an . 125 x 1 + 8. Independent variables, on the right, are called nonbasic variables. 34 barrels \u00d7 35 lbs malt = 1190 lbs [ amount of available malt ] corn (480 lbs) hops (160 oz) malt (1190 lbs)$13 profit per barrel \\$23 profit per barrel good are indivisible. shows, by means of an example, how linear programming can be applied to ob- tain optimal team decision functions in the case in which the payoff to the team is a convex polyhedral function of the decision variables. concepts of linear programming can all be demonstrated in the two-variable context. Linear programming formulation. Two important Python features facilitate this modeling: The syntax of Python is very clean and it lends itself to naturally adapt to expressing (linear) mathematical programming models Business environment assignment 2 examples of science research paper title page. Variables and constraints can be easily modified, as well as the ability to modify objective, bound and matrix coefficients. Characteristic of linear problem are. a number of examples of problems that may be formulated in terms of linear pro- Note that these constraints are also linear in the decision variables. Enter 0 values above the variables. All linear programming problems can be write in standard form by using slack variables and dummy variables, which will not have any influence on Example 1: The Production-Planning Problem. As the simple linear regression equation explains a correlation between 2 variables (one independent and one dependent variable), it is a basis for many analyses and predictions. Your options for how much will be limited by constraints stated in the problem. Formalizing The Graphical Method17 4. GAMES. 1 The Basic LP Problem The most fundamental optimization problem tr eated in this book is the linear programming (LP) problem. PuLP is an open source linear programming package for python. Problems with Alternative Optimal Solutions18 5. Solving the linear model using Excel Solver. This notebook gives an overview of Linear Programming (or LP). Once obtained the input base variable, the output base variable is determined. These decision Exercise Set 2. The following LP problem was solved: Min 5X + 7Y X + 3Y \u2265 6 5X + 2Y \u2265 10 Y \u2264 4 X Linear Programming - Example 2. (1) Identify the decision variables and assign symbols x and y to them. Linear programming methods are algebraic techniques based on a series of equations or inequalities that limit\u2026 Improve your math knowledge with free questions in \"Linear programming\" and thousands of other math skills. The constraints can be expressed by linear equations and inequalities involving only the decision variables. ) at the optimal solution. Set To = Min, 3. + a 1n x n = b 1 a 21 x 1 + a 22 x 2 +. The columns of the final tableau have variable tags. Kostoglou. subject to . If you have only\u00a0 Since there are only two variables, we can solve this problem by graphing the set We now present examples of four general linear programming problems. So 3 x 1 2 2 10 is a linear constrain t, as is + 3 =6 Linear Programming, Part II \u2013 Slack and Surplus Linear Programming Now our matrices will of course have to change a little bit to take these new variables F. Define the decision variables. Best wishes. For example, if Y is also unrestricted variable, the substitute - Y for Y. Modelling Linear Programming INDR 262 Optimization Models and Mathematical Programming LINEAR PROGRAMMING MODELS Common terminology for linear programming: - linear programming models involve . Example 2 (Alternate optimal solutions). The variable s 2 is called a surplus variable Divisibility assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the func- tional and nonnegativity constraints. 2 Linear Programming Geometric Approach 5. linear programming 2 variables examples\n\nv57oeqc, tvw, h2mgh, paeevcpsm, toznu, x3fk, tjy4, 6qh9c4xr9, jwk, bv07u, yo4a1,", "date": "2019-11-18 03:42:51", "meta": {"domain": "makariharlem.com", "url": "http://makariharlem.com/qt6hqml/linear-programming-2-variables-examples.html", "openwebmath_score": 0.5247300267219543, "openwebmath_perplexity": 606.6295555419894, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9861513873424045, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8585407040574128}}
{"url": "http://physics.qandaexchange.com/?qa=1631/maximum-height", "text": "# Maximum height\n\n256 views\n\n$H=(u^2 sin^2\\theta )/(2g)$\n$h-H =(1/2)gt^2 = (u^2sin^2 \\theta)/(2g)$\nFrom this I got h=2H\n\n1 vote\n\nIf the particles are launched at the same time, they must have the same horizontal speed in order to collide. (However, the problem seems to state that the particles have the same speed, ie same magnitude of velocity, and it is not clear that they are launched at the same time.)\n\nBecause they have the same horizontal speed, they remain aligned vertically. We can ignore horizontal motion and treat this as motion in 1D vertically.\n\nThe particles must collide before they reach the ground. The maximum $h$ occurs when they collide at the ground. ie They reach the ground at the same time, and have the same time of flight $t$.\n\nWhen the upper particle reaches the ground $h=\\frac12gt^2$. When the lower particle falls from its maximum height it takes time $\\frac12 t$ so\n$H=\\frac12g(\\frac12 t)^2=\\frac14(\\frac12gt^2)=\\frac14h$\n$h=4H$.\n\nanswered Mar 22, 2017 by (28,876 points)\nselected May 10, 2017 by koolman\n\nSpeed of paricle 1 be u at an angle $\\theta$ from ground .Then the speed of particle 2 is $u\\cos \\theta$ in horizontal direction.\n\nlet at any h' from ground both partucles collide .\n\nFor particle 1 $$h'= u\\sin\\theta t -(1/2)gt^2$$\nFor particle 2 $$h-h '= (1/2) gt^2$$\n\nHence $$t=\\sqrt {\\frac{2(h-h')}{g}}$$\nAlso we know $H=u^2 sin\\theta ^2 /(2g)$\nSubstituting these values in equation of particle 1\n\nWe geta quadratic equation $$h^2 -4Hh+4Hh'=0$$\n\nsolving this $$h=\\frac{4H(+/-)\\sqrt{16H^2 -16Hh'}}{2}$$\nNow h will be maximum when h'=0 that is they collide when both of them just going to strike the ground .\n\n$$h=\\frac{4H+4H}{2}$$\n$$h=4H$$\n\nanswered Mar 22, 2017 by (4,286 points)\nYes this is correct but not the simplest way of doing the calculation. The key assumption is that the particles have the same *horizontal component* of velocity. This is not what the question seems to say. The wording of the question is very confusing.\nThe question hasn't mentioned anything about speed , so we can take horizontal speed same .\nThe question says \"second particle is thrown horizontally with same speed\". To me that means the particles have the same \"speed\" ie the same magnitude of velocity.\nOh Sorry , the wording of question is confusing.\n1 vote\n\nYour Equation considers the particles to collide only at the point where first particle attains its maximum height. But the question considers the general case.\nSince they are thrown from the same vertical plane with the same speed, they can collide only if the horizontal component of their velocities are equal. Here the second particle would have covered a large distance when they are at the same height.\nIf the first particle is necessarily thrown obliquely then the particles cannot collide I assume.\n\nanswered Mar 21, 2017 by (340 points)\nHave a look at my answer .\n1 vote\n\nDiscussion\n\nDoubtExpert correctly observes that the particles do not necessarily collide when the particle thrown obliquely reaches its maximum height. However, he assumes that the particles are launched simultaneously. I think this is not required by the wording of the question.\n\nThe key phrase is thrown to strike at same time. This is ambiguous. If this means (as I think it does) that the particles collide at the same time then it is superfluous, because they must be in the same place at the same time in order to collide. I do not see how it can be interpreted as launched at the same time. If it does mean this then DoubtExpert is correct : there can be no collision.\n\nSolution\n\nWe must assume that the particles can be launched independently, at any time. The only requirement is that the trajectories overlap - or at least touch tangentially - at some point in space. If this happens then we can always adjust the launch times to ensure that the particles collide at this point.\n\nWhat we need to do is write equations for the two trajectories then find a condition for them to touch. The trajectories will touch if (i) they intersect at some point, and (ii) their tangents are equal at this point.\n\nSuppose the speed of each particle is $u$ and the launch angle of the lower particle is $\\theta$. Writing $T=\\tan\\theta$ the trajectories of the upper and lower particles are respectively\n$y=h-\\frac{g}{2u^2}x^2$\n$y=xT-\\frac{g(1+T^2)}{2u^2}x^2$.\nThe tangents have slope\n$\\frac{dy}{dx}=-\\frac{g}{u^2}x$\n$\\frac{dy}{dx}=T-\\frac{g(1+T^2)}{u^2}x$.\n\nThe 1st two equations are equal when\n$h=xT-\\frac{gT^2}{2u^2}x^2$.\nThe 2nd two equations are equal when\n$T=\\frac{gT^2}{u^2}x$\n$xT=\\frac{u^2}{g}$.\nCombine these two results :\n$h=\\frac{u^2}{2g}$.\n\nThe height of the collision point is\n$y=h-\\frac{g}{2u^2}x^2=\\frac{u^2}{2g}-\\frac{u^2}{2gT^2}=h(1-\\frac{1}{T^2})$.\nThis must be above ground $(y \\ge 0)$ so\n$T^2 \\ge 1$\n$\\theta \\ge 45^{\\circ}$.\n\nThe maximum height reached by the lower particle is $H=\\frac{u^2\\sin^2\\theta}{2g}$. So we have $\\frac{H}{h}=\\sin^2\\theta$. The value of $h$ calculated here is the maximum possible - any larger and the trajectories would not intersect. So the criterion is\n$h \\ge \\frac{H}{\\sin^2\\theta}$.\n\nI do not think we can make any further progress without knowing the launch angle $\\theta$. In order to get $h = 4H$ we require $\\theta=30^{\\circ}$. However, this is not $\\ge 45^{\\circ}$ : the collision in this case takes place below ground. So a collision is not possible when $h =4H$.\n\nanswered Mar 21, 2017 by (28,876 points)\nedited Mar 21, 2017\nIt is given in the question that both particles are thrown at same time .\nIf the particles are launched with the same speed at the same time, then I agree with DoubtExpert : the particles cannot collide.\n\nIs there a worked solution? Or a diagram?\nHmm, true because \u00a0horizontal distance being same $ut = u cos \\theta t$\nand $\\theta = 0$ so this may not cause collision or because even the speed given are same.\nThe solution has bad explanation .\nhttp://pasteboard.co/MmmDW5wny.jpg", "date": "2022-11-26 15:09:33", "meta": {"domain": "qandaexchange.com", "url": "http://physics.qandaexchange.com/?qa=1631/maximum-height", "openwebmath_score": 0.6982215046882629, "openwebmath_perplexity": 404.0995456550925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9626731126558705, "lm_q2_score": 0.8918110569397306, "lm_q1q2_score": 0.8585225260850923}}
{"url": "https://madhavamathcompetition.com/category/uncategorized/page/2/", "text": "# Check your mathematical induction\u00a0concepts\n\nDiscuss the following \u201cproof\u201d of the (false) theorem:\n\nIf n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:\n\nPROOF BY INDUCTION:\n\nStep 1:\n\nIf $n=1$, the result is evident.\n\nStep 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \\ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \\ldots, a_{k}$; by induction hypothesis these numbers are all equal:\n\n$a_{1}=a_{2}= \\ldots = a_{k}$.\n\nIf we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \\ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal:\n\n$a_{2}=a_{3}=\\ldots = a_{k}=a_{k+1}$.\n\nIt follows easily that all $k+1$ numbers in S are equal.\n\n*************************************************************************************\n\nComments, observations are welcome \ud83d\ude42\n\nRegards,\n\nNalin Pithwa\n\n# Observations are important: Pre RMO and RMO :\u00a0algebra\n\nWe know the following facts very well:\n\n$(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$\n\n$(x-y)^{3}=x^{3}-3x^{2}y+3xy^{2}-y^{3}=()()$\n\nBut, you can quickly verify that:\n\n$x^{3}+2x^{2}y+2xy^{2}+y^{3}=(x+y)(x^{2}+xy+y^{2})$\n\n$x^{3}-2x^{2}y+2xy^{2}-y^{3}=(x-y)(x^{2}-xy-y^{2})$\n\nWhereas:\n\n$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$\n\n$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$\n\nI call it \u2014 simply stunning beauty of elementary algebra of factorizations and expansions\n\nMore later,\n\nNalin Pithwa\n\n# Some number theory (and miscellaneous) coaching for RMO and INMO: tutorial (problem set)\u00a0III\n\nContinuing this series of slightly vexing questions, we present below:\n\n1.\u00a0Prove the inequality $\\frac{A+a+B+b}{A+a+B+b+c+r} + \\frac{B+b+C+c}{B+b+C+c+a+r} > \\frac{C+c+A+a}{C+c+A+a++b+r}$, where all the variables are positive numbers.\n\n2.\u00a0A sequence of numbers: Find a sequence of numbers $x_{0}$, $x_{1}$, $x_{2}, \\ldots$ whose elements are positive and such that $a_{0}=1$ and $a_{n} - a_{n+1}=a_{n+2}$ for $n=0, 1, 2, \\ldots$. Show that there is only one such sequence.\n\n3.\u00a0Points in a plane: Consider several points lying in a plane. We connect each point to the nearest point by a straight line. Since we assume all distances to be different, there is no doubt as to which point is the nearest one. Prove that the resulting figure does not containing any closed polygons or intersecting segments.\n\n4.\u00a0Examination of an angle: Let $x_{1}$, $x_{2}, \\ldots, x_{n}$ be positive numbers. We choose in a plane a ray OX, and we lay off it on a segment $OP_{1}=x_{1}$. Then, we draw a segment $P_{1}P_{2}=x_{2}$ perpendicular to $OP_{1}$ and next a segment $P_{2}P_{3}=x_{3}$ perpendicular to $OP_{2}$. We continue in this way up to $P_{n-1}P_{n}=x_{n}$. The right angles are directed in such a way that their left arms pass through O. We can consider the ray OX to rotate around O from the initial point through points $P_{1}$, $P_{2}, \\ldots, P_{n}$ (the final position being $P_{n}$). In doing so, it sweeps out a certain angle. Prove that for given numbers $x_{i}$, this angle is smallest when the numbers $x_{i}$, that is, $x_{1} \\geq x_{2} \\geq \\ldots x_{n}$ decrease; and the angle is largest when these numbers increase.\n\n5.\u00a0Area of a triangle: Prove, without the help of trigonometry, that in a triangle with one angle $A = 60 \\deg$ the area S of the triangle is given by the formula $S = \\frac{\\sqrt{3}}{4}(a^{2}-(b-c)^{2})$ and if $A = 120\\deg$, then $S = \\frac{\\sqrt{3}}{12}(a^{2}-(b-c)^{2})$.\n\nMore later, cheers, hope you all enjoy. Partial attempts also deserve some credit.\n\nNalin Pithwa.\n\n# A nice analysis question for RMO\u00a0practice\n\nActually, this is a famous problem. But, I feel it is important to attempt on one\u2019s own, proofs of famous questions within the scope of RMO and INMO mathematics. And, then compare one\u2019s approach or whole proof with the one suggested by the author or teacher of RMO/INMO.\n\nProblem:\n\nHow farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?\n\nReference:\n\nI will post it when I publish the solution lest it might affect your attempt at solving this enticing mathematics question !\n\nPlease do not try and get the solution from the internet.\n\nRegards,\n\nNalin Pithwa.\n\n# A miscellaneous algebra question for RMO or\u00a0Pre-RMO\n\nQuestion:\n\nIf a, b, c are three rational numbers, then prove that\n\n$\\frac{1}{(a-b)^{2}} + \\frac{1}{(b-c)^{2}} + \\frac{1}{(c-a)^{2}}$\n\nis always the square of a rational number.\n\nSolution to be posted soon\u2026\n\nCheers,\n\nNalin Pithwa.", "date": "2020-05-25 14:58:35", "meta": {"domain": "madhavamathcompetition.com", "url": "https://madhavamathcompetition.com/category/uncategorized/page/2/", "openwebmath_score": 0.751218318939209, "openwebmath_perplexity": 429.80162135564757, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731126558706, "lm_q2_score": 0.8918110440002044, "lm_q1q2_score": 0.8585225136285584}}
{"url": "http://mathhelpforum.com/calculus/26628-identifying-inflection-points.html", "text": "# Math Help - Identifying Inflection Points\n\n1. ## Identifying Inflection Points\n\nThe x coordinates of the points of inflection of the graph of $y= x^5-5x^4=3x+7$ are...\n\nA 0,\nB 1,\nC 3,\nD 0 and 3,\nE 0 and 1\n\n$y''=20x^3-60x^2$\nSimplify.\n$y''=20x^2(x-3)$\n\nSo, thanks to the graph and knowledge of the definition of inflection point (y''= 0).\nI choose D.\n\nHowever, the book says C.\nI'm inclined to believe it is right because I did the sign test for 0 for numbers less than 0 and less than 3, thought I may be incorrect of which function to put the test values into (y'' right?), and got the same sign (negative).\n\nSo, did I do the sign test wrong? Or is the right and I'm confusing myself with these questions.\n\nWhat is the answer to the problem?\n\nThank you.\n\n2. For a point of inflection, $y''=0$ is a necessary condition but not a sufficient one. $y''$ must also change sign before and after the point of inflection. Thus 0 is not a point of inflection because $y''$ does not change sign there; $y''$ is negative both slightly before and slightly after 0.\n\nAlternatively \u2026\n\n$y'''=60x^2-120x\\ne0$ when $x=3$. Since $y'''$ is an odd-order derivative, $x=3$ is a point of inflection. However $y'''(0)=0$ so we differentiate again.\n\n$y''''=120x-120\\ne0$ when $x=0$. $y''''$ is an even-order derivative, so $x=0$ is not a point of inflection.", "date": "2015-11-30 10:14:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/26628-identifying-inflection-points.html", "openwebmath_score": 0.7882918119430542, "openwebmath_perplexity": 427.7029810782671, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808750611337, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8585172480740769}}
{"url": "https://math.stackexchange.com/questions/430310/a-peculiar-binomial-coefficient-identity", "text": "# A peculiar binomial coefficient identity\n\nWhile inventing exercises for a discrete math text I'm writing I came up with this $$\\binom{\\binom{n}{2}}{2}=3\\binom{n+1}{4}$$ It's an easy result to prove, but it got me wondering\n\n1. Is this pure coincidence, or is it a special case of some more general result of which I'm unaware?\n2. No matter what the answer to (1) is, is there a combinatorial proof?\n\u2022 Looks purdy.${{{}}}$ \u2013\u00a0Git Gud Jun 26 '13 at 21:20\n\u2022 Every polynomial $f(x)$ which takes integer values at the integers is an integer linear combination of the polynomials ${x \\choose k}$, so in some sense this identity is not very surprising. The analogous statement for multivariate polynomials is also true, so there are also for example identities of the form ${x + y \\choose k} = \\sum a_{i, j} {x \\choose i} {y \\choose j}$ and ${xy \\choose k} = \\sum b_{i, j} {x \\choose i} {y \\choose j}$. The only mildly surprising thing is that in both of these cases the coefficients turn out to be non-negative as well as integers, but that's because the... \u2013\u00a0Qiaochu Yuan Jun 26 '13 at 21:57\n\u2022 ...corresponding identities also have combinatorial proofs (exercise!). \u2013\u00a0Qiaochu Yuan Jun 26 '13 at 21:57\n\u2022 \u2013\u00a0sdcvvc Jul 4 '13 at 23:00\n\nThis is actually a well known identity. There are combinatorial ways to prove it.\n\nConsider $n$ objects. Consider all ${n \\choose 2}$ pairs. Consider all pairs of these pairs. We get the LHS.\n\nConsider $n$ objects and 1 distinguished object. Consider all sets of 4 objects from these.\nIf the 4 objects do not include the distinguished object, they correspond to 3 possible pairs of pairs, whose 4 elements are distinct. I.E. $(A,B), (C,D)$ and $(A, C), (B, D)$ and $(A, D), (B,C)$.\nIf the 4 objects include the distinguished object, they correspond to 3 possible pairs of pairs, which have a common element, and whose union is the 3 objects. I.E. $(A, B) , (A, C)$ and $(B,A), (B,C)$ and $(C,A), (C,B)$.\nThis gives us the RHS.\n\nI'm not sure if they are generalizations, though you can experiment with choosing triples and counting carefully.\n\n\u2022 I'm not surprised by (1) the fact that this is well-known and (2) that you knew it. Nice job, +1 \u2013\u00a0Rick Decker Jun 26 '13 at 21:28\n\nThe lefthand side is of course the number of ways to choose two unordered pairs, possibly with a one-element overlap, from the set $[n]=\\{1,\\dots,n\\}$. Alternatively, we may choose a $4$-element subset $A$ of $\\{0,1,\\dots,n\\}$ and a $k\\in[3]$. Let $A=\\{a_1,a_2,a_3,a_4\\}$ with $a_1<a_2<a_3<a_4$. If $a_1\\ne 0$, pair $a_1$ with $a_{k+1}$ and let the other two members of $A$ be the other pair. If $a_1=0$, form two pairs from $\\{a_2,a_3,a_4\\}$ by letting $a_k$ be the common member.\n\n\u2022 My goodness. Two replies within 9 minutes. This place is the math equivalent of a fast food restaurant (though the offerings are unusually tasty). +1 \u2013\u00a0Rick Decker Jun 26 '13 at 21:33\n\u2022 You mean \"an unordered pair of unordered pairs.\" \"Two\" could mean \"ordered pair.\" \u2013\u00a0Qiaochu Yuan Jun 26 '13 at 21:55\n\u2022 @Qiaochu: I think that one would almost have to be deliberately trying to misunderstand in order to come up with that reading. \u2013\u00a0Brian M. Scott Jun 27 '13 at 1:30\n\nYou can count all of the pairs of pairs with $4$ distinct elements and all of the pairs of pairs with $3$ distinct elements and then add them together.\n\nFor the first sum get $\\binom{n}{4}\\binom{4}{2}\\frac{1}{2}=3\\binom{n}{4}$. We divide by two because once we choose the first pair out of the $4$ elements we implicitly chose the other pair; we are double counting otherwise.\n\nFor the second sum we get $3\\binom{n}{3}$ since we must choose $3$ elements and then one of them to be duplicated.\n\nIn total we have $$3 \\left[ \\binom{n}{4} + \\binom{n}{3}\\right] = 3\\binom{n+1}{4}$$.\n\nSo that's $2/3$ combinatorial at least.\n\n\u2022 Nice and succinct. +1 \u2013\u00a0Rick Decker Jun 27 '13 at 1:02", "date": "2021-05-13 19:52:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/430310/a-peculiar-binomial-coefficient-identity", "openwebmath_score": 0.8238780498504639, "openwebmath_perplexity": 349.3579468325985, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808730448281, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8585172478972677}}
{"url": "https://math.stackexchange.com/questions/2304142/how-many-squares-in-a-rectangle", "text": "# How many squares in a rectangle?\n\nI almost wish I'd never thought of this problem... I was tearing my hair out over it all night.\n\nSuppose we have a rectangle with side lengths $a$ and $b$, $a,b \\in \\mathbb Z$, $GCD(a,b)=1$, and $b \\gt a$. I want to pack squares in this rectangle in such a way that, at each step, I pack a squares that is large as possible at that step, and that touches the square I packed at the last step. A square packing of this type might end up looking something like this:\n\nMy question is this: can I find a formula for the number of squares needed to pack a rectangle in terms of its side lengths $a$ and $b$?\n\nHere is what I tried:\n\nFirst I let $f(a,b)$ denote the function whose output is the number of squares required. First, I noted a few properties of $f(a,b)$: $$f(a,a)=1$$ $$f(ka,kb)=f(a,b)$$ $$f(a,a+1)=a+1$$ $$f(a,b)=\\Big\\lfloor \\frac{a}{b} \\Big\\rfloor + f(b, a\\mod b)$$ $$r \\lt a \\implies f(a,qa+r)=q+f(r,a)$$ Next I tried this: I decided to let $b=q_1a+r_1$ where $r_1 \\lt a$. Then I would have $$f(a, q_1a+r_1)=q_1+f(r_1, a)$$ Then I decided to let $a=q_2r_1+r_2$, so that we would then have $$f(a, q_1a+r_1)=q_1+f(r_1, q_2r_1+r_2)$$ $$f(a, q_1a+r_1)=q_1+q_2+f(r_2, r_1)$$ and so on. I keep continuing this process until I hit some $n$ for which $r_n=0$, and I will have $$f(a, q_1a+r_1)=\\sum_{i=1}^n q_i$$ ...and then I got stuck. How can I find each of the $q_i$s? Somebody, please help! The solution need not be closed-form, it just needs to be... something. I just have no idea where to go with this.\n\nThanks for any and all help!\n\n\u2022 I think the Euclidean algorithm might be helpful here \u2013\u00a0abiessu May 31 '17 at 13:27\n\u2022 I also do believe that the Euclidian algorithm is the clue here: imagine that a=10 and b=54, then you get following steps : 54=5x10+4, 10=2x4+2, 4=2x2. Now you take the sum of all coefficients: 5+2+2=9, which is, in my opinion, the number you're looking for. Isn't it? \u2013\u00a0Dominique May 31 '17 at 13:34\n\u2022 @MCCCS Whoa there... how did you come up with that? Do you mind posting an answer and explaining it? \u2013\u00a0Franklin Pezzuti Dyer May 31 '17 at 13:35\n\u2022 @Dominique Yes, that is correct. How do I find the sum of the coefficients, though? \u2013\u00a0Franklin Pezzuti Dyer May 31 '17 at 13:37\n\u2022 @IvanNeretin: I don't think so neither: you should write the Euclidean algorithm in a form of iterative series (working with Qn for the quotients and Rn for the leftovers), and perform a Sum(Qn) at the end. \u2013\u00a0Dominique May 31 '17 at 13:45\n\n## 2 Answers\n\n$$f(a,b) = \\begin{cases}1 & a= b\\\\b/a & (\\gcd(a,b) = a) \u2227 (a\\neq b)\\\\ a/b & (\\gcd(a,b) = b) \u2227 (a\\neq b) \\\\ f(b,a) & b>a\\\\ \\frac{(a-(a \\mod b))}{b} + f(b,(a\\mod b)) &a>b\\end{cases}$$\n\n$$\\gcd(x,y) =\\begin{cases}x&x=y\\\\y & (x = 0)\u2227(x\\neq y)\\\\x & (y=0)\u2227(x\\neq y)\\\\\\gcd(y,(x\\mod y))&x>y\\\\\\gcd((y\\mod x),x)&y> x\\end{cases}$$\n\n(\u2227: AND logical operator)\n\n$f(a,b)$ is the function that solves your question, using $\\gcd(x,y)$, Euclid's GCD function.\n\nHere's also the code written in Swift that does the same thing:\n\nimport Foundation\nfunc ourFunction (_ a: Int, _ b: Int) -> Int{\nif gcd(a,b) == a{\nreturn b/a\n}else if gcd(a,b) == b{\nreturn a/b\n}\nif b>a{\n// Force a to be bigger than b\nreturn ourFunction(b,a)\n}\nreturn (a-(a%b)) / b /*big squares*/ + ourFunction(b,a%b)\n}\n\nfunc gcd(_ x:Int, _ y:Int) -> Int{\nif x == 0 || x == y{\nreturn y\n}\nif y == 0{\nreturn x\n}\nif x > y{\nreturn gcd(y,x%y)\n}else{\nreturn gcd(y%x,x)\n}\n}\nprint(ourFunction(8,3))\n\n\nIt can be run here, just change the $8$ and $3$ to the numbers you want to put into the function.\n\nThe answer by @MCCCS is great, but I provide an alternative answer with intent of being (slightly) more concise.\n\nThe problem can be solved recursively by embedding as many equally-sized maximal squares as possible within the rectangle and increasing a counter by the number of squares embedded. You'll end up with a smaller rectangle, for which you can embed another (horizontal or vertical) stack of equally-sized maximal squares as possible. Rinse repeat!\n\nThis recursive process can be described by the following recursive function:\n\n$$f(a, b) = \\begin{cases} 0 &\\mbox{if } a =0 \\lor b=0 \\\\ f(a, b-a\\cdot\\left \\lfloor \\frac{b}{a} \\right \\rfloor) + \\left \\lfloor \\frac{b}{a} \\right \\rfloor & \\mbox{if } a<b \\\\ f(a-b \\cdot \\left \\lfloor \\frac{a}{b} \\right \\rfloor, b) + \\left \\lfloor \\frac{a}{b} \\right \\rfloor & \\text{otherwise} \\end{cases}$$\n\nwhose domain is $\\{\\forall(a,b)\\in \\mathbb{R}^2:\\gcd(a,b)=1\\}$.\n\nHere is a C-program implementing the function described above:\n\n#include <stdio.h>\n\nint f(int w, int h) {\nreturn  w == 0 || h == 0 ? 0 :\nw < h ? f(w, h-w*(h/w)) + h/w :\nf(w-h*(w/h), h) + w/h;\n}\n\nint main(int argc, char** argv) {\nint i = 3, j = 7;\nprintf(\"f(%i, %i)=%i\\n\", i, j, f(i, j));\nreturn 0;\n}", "date": "2021-02-27 07:42:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2304142/how-many-squares-in-a-rectangle", "openwebmath_score": 0.5728579759597778, "openwebmath_perplexity": 489.08302683925126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808753491773, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8585172467372969}}
{"url": "https://math.stackexchange.com/questions/1382570/name-of-xpyp-le-xyp-p-ge-1", "text": "# Name of $|x|^p+|y|^p\\le (|x|+|y|)^p$ ($p\\ge 1$)?\n\nI checked these\n\nWhat is the difference between square of sum and sum of square?\n\nProve $(|x| + |y|)^p \\le |x|^p + |y|^p$ for $x,y \\in \\mathbb R$ and $p \\in (0,1]$.\n\nIt is easy to see $p$-th power ($p\\ge 1$) version, i.e., $|x|^p+|y|^p\\le (|x|+|y|)^p$ ($p\\ge 1$), holds as well using the argument by Quang Hoang in Prove $(|x| + |y|)^p \\le |x|^p + |y|^p$ for $x,y \\in \\mathbb R$ and $p \\in (0,1]$. Is there a name for this inequality so that I can just quote? (It might be an elementary result but people around me bother to put \"from Cauchy--Schwarz,...\" when it is clearly Cauchy--Schwarz, so.)\n\n\u2022 You can say that the function $x\\mapsto x^p$ on $[0,\\infty)$ is subadditive if $p\\in(0,1]$ and superadditive if $p\\geq1$. \u2013\u00a0triple_sec Aug 3 '15 at 1:39\n\u2022 I've sometimes heard, \"by comparison of $\\ell_p$ norms\". (This is the two-dimensional case of $\\ell_p$ vs $\\ell_1$.) \u2013\u00a0user21467 Aug 3 '15 at 1:43\n\u2022 In probability it's called $c_r$ inequality when you put expectations on both sides... \u2013\u00a0d.k.o. Aug 3 '15 at 3:20\n\u2022 @d.k.o. As in www2.cirano.qc.ca/~dufourj/Web_Site/ResE/\u2026? I am not sure if I am looking at a right source but the constant and the direction look a bit different though. \u2013\u00a0shall.i.am Aug 3 '15 at 10:36\n\nI do not believe there is a name for your specific inequality (which I rewrite as following): $$|x|^p+|y|^p\\le \\big(|x|+|y|\\big)^p, \\ p \\ge1 \\iff \\boxed{\\ \\ \\left(|x|^p +|y|^p\\right)^{\\frac{1}{p}} \\le |x|+|y|, \\quad p \\ge 1 \\ \\ }$$ However, it can be viewed as a special case of multiple more general statements, such as Jensen, AMGM, H\u00f6lder, and probably many other inequalities after appropriate substitution and/or change of variables. The closes call would probably be the generalized mean inequality: $$M_j\\left( x_1, \\dots, x_n \\right) \\leq M_i\\left( x_1, \\dots, x_n \\right) \\quad \\text{ whenever } \\quad j<i. \\label{*} \\tag{*}$$\n\nHere $M_k \\left( x_1, \\dots, x_n \\right)$ is so-called power mean, which is defined as $$M_k(x_1,\\dots,x_n) = \\left( \\frac{1}{n} \\sum_{i=1}^n x_i^k \\right)^{\\frac{1}{k}}.$$\n\n! In particular, assuming $n=2$, $j = 1$, and $i = p$, and denoting $\\left(x_1, \\dots, x_n \\right) := \\left(\\,\\chi, \\gamma \\right)$, we get \\begin{aligned} M_1\\big(\\left|\\,\\chi\\right|, \\left|\\gamma\\right|\\big) & = \\dfrac{1 }{2}\\big(\\left|\\,\\chi\\right| + \\left|\\gamma\\right|\\big) = \\dfrac{\\left|\\,\\chi\\right| }{2} + \\dfrac{\\left|\\gamma\\right| }{2}, \\\\ M_p\\big(\\left|\\,\\chi\\right|,\\left|\\gamma\\right|\\big) & = \\left( \\dfrac{1}{2} \\left(\\left|\\,\\chi\\right|^p+\\left|\\gamma\\right|^p\\right)\\right)^{\\frac{1}{p}} = \\Bigg( \\left(\\frac{\\left|\\,\\chi\\right|}{2}\\right)^p + \\left(\\frac{\\left|\\gamma\\right|}{2}\\right)^p \\Bigg)^{\\frac{1}{p}}.\\\\ \\end{aligned} By $\\eqref{*}$, we have $$\\dfrac{\\left|\\,\\chi\\right| }{2} + \\dfrac{\\left|\\gamma\\right| }{2} \\le \\left( \\bigg(\\frac{\\left|\\,\\chi\\right|}{2^{\\frac{1}{p}}}\\bigg)^p + \\bigg(\\frac{\\left|\\gamma\\right|}{2^{\\frac{1}{p}}}\\bigg)^p \\right)^{\\frac{1}{p}} . \\label{**} \\tag{**}$$ Denoting $x := 2^{-\\frac{1}{p}}\\chi, \\ \\ y := 2^{-\\frac{1}{p}}\\gamma$ and raising both sides of $\\eqref{**}$ to the power $p$, we get\n$$\\left|x\\right|^{p}+\\left|y\\right|^{p} \\le \\big(\\left|x\\right|+\\left|y\\right|\\big)^{p}.$$\n\nTo summarize, I believe that (strictly speaking) there is probably no name for your inequality. However, that the generalized mean is as close as you can get to your inequality, although some formula conversion is still required.\n\nI think this should be called an special case of Minkowski's Inequality.\n\n1 - For finite sequences, the Minkowski Inequality states that\n\n$$\\left(\\sum_{k=1}^n |x_k + y_k |^p \\right) ^{1/p} \\le \\left(\\sum_{k=1}^n |x_k|^p \\right)^{1/p} + \\left(\\sum_{k=1}^n |y_k|^p\\right)^{1/p}$$\n\nYour inequality then follows if $x_1 = x, x_i = 0, \\; i \\ge 2$, and $y_2 = y, y_i = 0, i \\ne 2$.\n\n2 - More Generally, the case above can be easily extended to infinite sequences. The most general result is that, if $f,g : \\Omega \\to \\Bbb{R}$ are measurable functions in a measure space $(\\Omega, \\Sigma, \\mu)$, then the inequality\n\n$$\\| f+g \\|_{L^p (d\\mu)} \\le \\|f\\|_{L^p(d\\mu)} + \\|g\\|_{L^p(d\\mu)}$$\n\nWhere $\\|f\\|_{L^p(d\\mu)}^p = \\int_ {\\Omega} |f(x)|^p d \\mu (x)$\n\n3 - There was a little lie at topic 2, because there is yet another generalization of this result, called Minkowski's Inequality for Integrals. In the link above, you might check a proof of this result, based on these previously presented ones.\n\n\u2022 Ah. It does follow from Minkowski. You meant $y_i=0$ $i\\neq 2$, right? Namely, $\\boldsymbol{x}:=(x_1,0,\\dotsc)$ and $\\boldsymbol{y}:=(0,y_2,0,\\dotsc)$. From Minkowski, $(|x_1|^p+|y_2|^p)^{\\frac1p}\\le|x_1|^{\\frac1pp}+|y_2|^{\\frac1pp}$ and thus $(|x_1|^p+|y_2|^p)\\le(|x_1|+|y_2|)^p$. I did not realise this because if anything Minkowski looks like the opposite direction. I find your answer really interesting but just saying \"this follows from Minkowski\" sounds a bit misleading as it looks opposite, so, sorry! But really appreciate it! \u2013\u00a0shall.i.am Aug 3 '15 at 10:33\n\u2022 Yeah, it really looks opposite, right? I was thinking about this inequality about two weeks ago, when I realized that it was just Minkowski's Inequality. Well, just wanted to help here :) Be free to ask more questions about this answer if you like :D \u2013\u00a0Jo\u00e3o Ramos Aug 3 '15 at 12:38", "date": "2019-10-15 09:58:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1382570/name-of-xpyp-le-xyp-p-ge-1", "openwebmath_score": 0.9554764032363892, "openwebmath_perplexity": 399.55731676346164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808741970027, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8585172425501473}}
{"url": "http://roszak.eu/fcjo6keq/f34428-differential-equations-lectures", "text": "Then we learn analytical methods for solving separable and linear first-order odes. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time. Lecture 1 Introduction to differential equations View this lecture on YouTube A differential equation is an equation for a function containing derivatives of that function. Differential Equations. These lectures treat the subject of PDEs by considering specific examples and studying them thoroughly. Lecture 51 : Differential Equations - Introduction; Lecture 52 : First Order Differential Equations; Lecture 53 : Exact Differential Equations; Lecture 54 : Exact Differential Equations (Cont.) Lecture 56: Higher Order Linear Differential Equations Download files for later. De\ufb01nition (Di\ufb00erential equation) A di\ufb00erential equation (de) is an equation involving a function and its deriva- tives. Lecture one (27/9/2020 ) power point URL. \u00bb \u00bb The study of differential equations is such an extensive topic that even a brief survey of its methods and applications usually occupies a full course. Knowledge is your reward. This is one of over 2,400 courses on OCW. A differential equation always involves the derivative of one variable with respect to another. 140 0 obj Lectures on Differential Equations Craig A. Tracy PDF | 175 Pages | English. (The minus sign is necessary because heat flows \u201cdownhill\u201d in temperature.) videos URL. On the human side Witold Hurewicz was an equally exceptional personality. )Z!\ufffd\ufffd\ufffd\ufffdDR\ufffd7\ufffd\ufffdH\u3115A7\ufffd\u06c9\ufffd\ufffdS\ufffdi\ufffd\ufffdvl\u71de\ufffdP%\ufffd[R\ufffd;\ufffd\ufffd\ufffdn\\W\ufffd\ufffd4/de^\ufffd\ufffd2\u0080\ufffde\ufffdD\ufffd\ufffd\ufffd\ufffdB%8\ufffd\u0636m\ufffd\ufffd[R\ufffd\ufffd\ufffd#\ufffd\ufffdP2q\\ \ufffdHis\ufffdd\ufffdJJ. Date: 1st Jan 2021. Arnold's geometric point of view of differential equations is really intriguing. This note covers the following topics: First Order Equations and Conservative Systems, Second Order Linear Equations, Difference Equations, Matrix Differential Equations, Weighted String, Quantum Harmonic Oscillator, Heat Equation and Laplace Transform. You see that it is a proper vector equation. (1) If y1(t) and y2(t) satisfy (2.1), then for any two constants C1and C2, y(t) = C1y1(t)+C2y2(t) (2.2) is a solution also. Baker Hilary Term2016. Equation is the differential equation of heat conduction in bulk materials. We introduce differential equations and classify them. We don't offer credit or certification for using OCW. By this approach the author aims at presenting fundamental ideas in a clear way. This table (PDF) provides a correlation between the video and the lectures in the 2010 version of the course. Find all the books, read about the author, and more. What follows are my lecture notes for a \ufb01rst course in differential equations, taught at the Hong Kong University of Science and Technology. videos URL. Such equations are often used in the sciences to relate a quantity to its rate of change. Here is a set of notes used by Paul Dawkins to teach his Differential Equations course at Lamar University. The former is called a dependent variable and the latter an independent variable. Lecture 01 - Introduction to Ordinary Differential Equations (ODE) PDF unavailable: 2: Lecture 02 - Methods for First Order ODE's - Homogeneous Equations: PDF unavailable: 3: Lecture 03 - Methods for First order ODE's - Exact Equations: PDF unavailable: 4: Lecture 04 - Methods for First Order ODE's - Exact Equations ( Continued ) PDF unavailable: 5 137 0 obj When the input frequency is near a natural mode of the system, the amplitude is large. This playlist contains 32.5 hours across 52 video lectures from my Differential Equations course (Math 2080) at Clemson University. Much of the material of Chapters 2-6 and 8 has been adapted from the widely (Image courtesy Hu \u2026 Di\ufb00erential equations are called partial di\ufb00erential equations (pde) or or- dinary di\ufb00erential equations (ode) according to whether or not they contain partial derivatives. Differential Equation: An equation involving independent variable, dependent variable, derivatives of dependent variable with respect to independent variable and constant is called a differential equation. Verify that if c is a constant, then the function de\ufb01ned piecewise by (1.15) y(x) = 1 if x \u2264 c cos(x\u2212c) if c < x < c+\u03c0 \u22121 if x \u2265 c+\u03c0 satis\ufb01es the di\ufb00erential equation y0= \u2212 p 1\u2212y2for all x \u2208 R. 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{"url": "https://math.stackexchange.com/questions/4320849/finding-the-range-of-y-fracx22x42x24x9-and-y-frac-textquadrat/4320857", "text": "# Finding the range of $y =\\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\\frac{\\text{quadratic}}{\\text{quadratic}}$ in general)\n\nI had this problem in an exam I recently appeared for:\n\nFind the range of $$y =\\frac{x^2+2x+4}{2x^2+4x+9}$$\n\nBy randomly assuming the value of $$x$$, I got the lower range of this expression as $$3/7$$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.\n\nNow, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.\n\nAnother guy I met outside the exam center, told me he used an approach of $$x$$ tending to infinity in both cases and got the maximum value of this expression as $$1/2$$. But before I could ask him to explain more on this method, he had to leave for his work.\n\nSo, will someone please throw some light on this method of $$x$$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).\n\n\u2022 Use of calculus to find abslute maxima and minima is the easier way. Dec 1, 2021 at 6:39\n\u2022 I'll give a hint: An easy way to re-express the expression will be $\\frac{1}{2}-\\frac{1}{4\\left(x+2\\right)^{2}+14}$. Think of what circumstance would maximise the expression, which would be by minimizing the term subtracted. Dec 1, 2021 at 6:41\n\u2022 @NikolaAlfredi Can you please show how. Because I am not versed with using the approach of calculus in such questions. Dec 1, 2021 at 6:49\n\u2022 @EmanatS Try my approach, doesn't need calculus Dec 1, 2021 at 6:50\n\u2022 The hint of Prometheus is very good. It gives the result quickly. Dec 1, 2021 at 6:51\n\nThe question can be easily solved by this technique:\n\nAs $$\\displaystyle y = \\frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \\implies 2y = \\frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$$.\n\nThus, $$\\displaystyle 2y = 1-\\frac {1}{2(x + 1)^2 + 7}$$\n\nSquares can never be less than zero so the minimum value of the function : $$\\displaystyle 2(x + 1)^2 + 7$$ would be $$7$$ , or Maximum value of $$\\displaystyle \\frac {1}{2(x + 1)^2 + 7}$$ is $$\\displaystyle \\frac {1}{7}$$.\n\nThis tells that minimum value of $$y$$ will be $$\\displaystyle \\frac{3}{7}$$.\n\nAnd so on.. check for $$x \\rightarrow \\infty$$.\n\nFrom here you can easily tell the maximum and minimum values : $$\\displaystyle y \\in \\left [ \\frac {3}{7}, \\frac {1}{2} \\right )$$\n\n\u2022 Isn't very obvious on the first glance when it comes to the minimum, but to explain it, $2x^2+4x+9 = 2(x+1)^2+7$ Dec 1, 2021 at 6:54\n\u2022 How did you split it just like that? Basically, how did you reduce that expression? Dec 1, 2021 at 6:55\n\u2022 @Prometheus It's true... But I guess the test was a bit of a time-crusher, so perhaps the OP would like to get there in one step Dec 1, 2021 at 6:55\n\u2022 @Spectre Yes, I had to solve this problem under 50 seconds. And Prometheus, yes I know the method of completing the square. I just do not understand how to apply it to this problem. Please explain this approach to get a one step form like I'm five. Dec 1, 2021 at 6:59\n\u2022 @EmanatS I have explained Nikola's step below, if you ever didn't get how he came to that simplification. Dec 1, 2021 at 7:04\n\nAs a follow-up to @NikolaAlfredi's answer:\n\n$$y = \\frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \\frac{2x^2 + 4x + 8}{2(2x^2 + 4x + 9)} = \\frac{2x^2+4x+9 - 1}{2(2x^2+4x+9)} = \\frac{1}{2}(1-\\frac{1}{2x^2+4x+9}) \\implies 2y = 1 - \\frac{1}{2x^2+4x+9}$$. Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by $$2$$ to get the required extremes(taking half since we get values for $$2y$$ and not $$y$$).\n\n\u2022 Got it. I appreciate your kind efforts, and now it is easily clear to me. Thanks to both you, and Nikola. Also, can you please delve on calculus based approach like someone mentioned? I want to learn how to work this out with calculus too. Dec 1, 2021 at 7:08\n\u2022 @EmanatS Are you looking for an answer using \"calculus\" methods? Dec 1, 2021 at 7:09\n\u2022 @TeresaLisbon Not really. But I am open to learning through it too. Dec 1, 2021 at 7:09\n\nIn general, if $$\\deg f = 0$$ where $$f(x) = \\frac{a_nx^n + a_{n - 1}x^{n - 1} + \\cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \\cdots + b_1x + b_0},$$ the limit of $$f$$ as $$x$$ increases/decreases without bound is $$a_n/b_n$$.\n\nIn your case, $$a_2 = 1$$ and $$b_2 = 2$$. Hence, $$a_2/b_2 = 1/2$$.\n\nWe'll factor $$f$$ as $$\\frac{x^2+2x+4}{2x^2+4x+9} = \\frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$\n\nNotice that for all $$x \\in \\mathbb{R}$$, $$f > 0$$. Also, we can see that $$(x+1)^2 + 2 < 2(x + 1)^2 + 7$$. This means that the range should be a part of $$(0,1/2)$$. Since both numerator and denominator have $$(x + 1)^2$$ without any remaining $$x$$'s, we can see that this will be at its minimum when $$x = -1$$. Then, $$f(-1) = \\frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\\\ = \\frac{(0)^2 + 3}{2(0)^2 + 7} \\\\ \\frac{3}{7}$$\n\nTherefore, the range is $$[3/7, 1/2)$$.\n\n\u2022 Note that both the numerator and denominator of $f$ must be polynomials with the same degree. This means that the expression could be linear/linear, quadratic/quadratic, cubic/cubic, and so on. Dec 1, 2021 at 7:12\n\u2022 Woah, you're awesome. And so is @NikolaAlfredi I love this forum now, you both explained this problem so beautifully to me putting in all efforts to explain it to me. Thank you for a generalized solution to it as well. So, \"limit of f as x increases/decreases without bound is an/bn\" is always an upper limit of the expression right? Any places where I need to watch out this rule for? Dec 1, 2021 at 7:17\n\u2022 Also for, (x^2+2x+1)/(4x^2-7x+9), a2/b2 will be {1/4}, but the upper limit of expression is 16/19. Shouldn't it be 1/4 according to your assertion? Edited Dec 1, 2021 at 7:21\n\u2022 No. This is not always true. Consider the simple case $\\frac{x}{x + 1}$. You'll see that it attains all real values except $y = 1$, and it is defined for all real $x$ except $x = -1$. In general, the limit is not the upper bound. It is only for the case where $p(x) < q(x)$ and $\\deg p = \\deg q$. ($p$ and $q$ are the polynomials in $f$, that is, $f(x) = p(x)/q(x)$). Dec 1, 2021 at 7:23\n\n$$y = \\frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \\frac12 - \\frac {1/2}{2x^2 + 4x + 9} \\implies \\frac {dy}{dx} = \\frac {4x + 4}{\\text{whatever}} \\text{ Let } \\color{green}{\\frac{dy}{dx} = 0 \\implies x = -1}, y(x = -1) = \\color{blue}{\\frac37} \\text{ also } y(x\\to \\infty) = \\color{blue}{\\frac 12}$$", "date": "2022-07-04 22:19:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4320849/finding-the-range-of-y-fracx22x42x24x9-and-y-frac-textquadrat/4320857", "openwebmath_score": 0.8522888422012329, "openwebmath_perplexity": 304.1326616206883, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808741970027, "lm_q2_score": 0.8757869835428965, "lm_q1q2_score": 0.8585172298377867}}
{"url": "https://math.stackexchange.com/questions/3546233/given-x-mathbbr2-and-the-equivalence-relation-x-y-simx-y-iff-y-x/3546363", "text": "# Given $X=\\mathbb{R}^2$ and the equivalence relation $(x,y)\\sim(x',y')$ iff $y-x^2=y'-x'^2$. What space is this homeomorphic to?\n\nGiven $$X=\\mathbb{R}^2$$ and the equivalence relation $$(x,y)\\sim(x',y')$$ iff $$y-x^2=y'-x'^2$$. What space is this homeomorphic to?\n\nThere is also a hint that $$g(x,y)=y-x^2$$\n\nSo I tried drawing the equivalence relations on $$\\mathbb{R}^2$$ and noticed that the equivalence relations are parabolas. The $$[0]$$ is $$y=x^2$$, then $$[1]$$ is $$y=x^2+1$$, etc. So I believe $$[c]$$ is $$y=x^2+c$$\n\nI want to show this space is homeomorphic to $$\\mathbb{R}$$ and I have the map $$g:\\mathbb{R}^2\\to \\mathbb{R}$$. which is going to map equivalence classes to that $$c$$ value. I believe I need to find $$f$$ such that $$f\\circ p=g$$. where $$p$$ is the quotient map. Then show it is continuous in both directions and bijective.\n\nCan I take $$f:\\mathbb{R}^2\\setminus \\sim\\to \\mathbb{R}$$ as $$f([(x,y)])=g(x,y)$$?\n\nI know it's surjective since $$g=f\\circ p$$ is surjective\n\nSince if $$r\\in \\mathbb{R}$$ then $$g(0,r)=r$$.\n\nAnd it's clearly injective, by the equivalence relation.\n\nBut I'm not sure how to show it is continuous.\n\nI have this theorem which I believe might be useful:\n\nLet $$X,Z$$ be two topological spaces, $$\\sim$$ an equivalence relation on $$X$$ and $$f:X\\setminus \\sim \\to Z$$ a function. Then $$f$$ is continuous if and only if $$f\\circ p$$ is continuous.\n\nSo I could show $$g$$ is continuous and get that $$f$$ is continuous.\n\nSo let $$(a,b)$$ be an open interval in $$\\mathbb{R}$$.\n\nthen I want to show $$g^{-1}[(a,b)]$$ is open.\n\nI believe it looks like a big open parabola, a union of $$U_c=\\{(x,y)\\in\\mathbb{R}^2: y=x^2+c, a\n\nSo I want to show that $$\\bigcup U_c$$ is open in $$\\mathbb{R}^2$$\n\nLet $$p\\in \\bigcup U_c$$ then $$p\\in U_c$$ for some $$c$$.\n\nBut I'm having trouble finding a radius of a ball which will be contained in this union. I want to pick something so that all the points in the ball will be less then the parabola $$y=x^2+b$$ and greater then $$y=x^2+a$$.\n\n\u2022 Why not $f([(x,y)])=g(x,y)$? \u2013\u00a0Hagen von Eitzen Feb 14 at 7:32\n\u2022 @Hagen von Eitzen Yeah that makes more sense. Since $[c]$ isn't actually what the equivalence classes should look like. \u2013\u00a0AColoredReptile Feb 14 at 7:36\n\u2022 If $g\\colon X\\to Y$ is continuous ad $x\\sim y\\iff g(x)=g(y)$, under what mild conditions does $X/{\\sim}\\cong Y$ follow? \u2013\u00a0Hagen von Eitzen Feb 14 at 7:49\n\u2022 @Hagen von Eitzen Does $g$ need to be continuous? \u2013\u00a0AColoredReptile Feb 14 at 8:17\n\n## 1 Answer\n\nIt is more or less obvious that $$g$$ is continuous. It is well-known (and easy to verify) that the following is true:\n\nLet $$f, g : X \\to \\mathbb R$$ be continuous maps defined on a topological space $$X$$ and $$a \\in \\mathbb R$$. Then $$f + g, f \\cdot g$$ and $$a \\cdot f$$ are continuous.\n\nThe projections $$p_1. p_2 : \\mathbb R^2 \\to \\mathbb R, p_1(x,y) = x, p_2(x,y) = y$$, are clearly continuous. Thus $$g(x,y) = p_2(x,y) - p_1(x,y) \\cdot p_1(x,y)$$ is continuous.\n\nAlternatively you can also consider sequences $$(x_n,y_n)$$ converging to some $$(x,y)$$ and show that $$(g(x_n,y_n))$$ converges to $$g(x,y)$$.\n\nLet $$\\pi : \\mathbb R^2 \\to Y = \\mathbb R^2/\\sim$$ be the quotient map. Define $$j : \\mathbb R \\to \\mathbb R^2, j(t) = (0,t)$$. This is a continuous map, hence $$J = \\pi \\circ j : \\mathbb R \\to Y$$ is continuous. We have $$f(J(t)) = f([0,t]) = g(0,t) = t,$$ $$J(f([x,y]) = J(g(x,y) = \\pi(0,y - x^2) =[0,y-x^2] = [x,y]$$ because $$(y-x^2) - 0^2 = y - x^2$$. This shows that $$f$$ and $$J$$ are inverse to each other. This means that $$f,J$$ are homeomorphisms such that $$f^{-1} = J$$.\n\n\u2022 So if $g$, I get $f$ is continuous. But then I need to show $f^{-1}$ is continuous, right? \u2013\u00a0AColoredReptile Feb 14 at 10:39\n\u2022 Yes, I have done this in my answer (which contained a typo that I corrected). We have $f^{-1} = J$. \u2013\u00a0Paul Frost Feb 14 at 10:50", "date": "2020-03-29 10:24:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3546233/given-x-mathbbr2-and-the-equivalence-relation-x-y-simx-y-iff-y-x/3546363", "openwebmath_score": 0.9983415007591248, "openwebmath_perplexity": 292.66774083008784, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.980280873044828, "lm_q2_score": 0.8757869835428966, "lm_q1q2_score": 0.8585172288287272}}
{"url": "http://mathhelpforum.com/calculus/49549-integral-partial-fractions.html", "text": "Math Help - Integral - partial fractions?\n\n1. Integral - partial fractions?\n\n$\\int \\frac{dx}{4x^{2/3}-4x^{1/3}-3}$\n\n$u=x^{1/3} \\text{ }du=\\tfrac{1}{3}x^{-2/3} \\text{ }x=u^3$\n\n$=\\int\\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$\n$=3\\int\\frac{du}{u^2(2u+1)(2u-3)}$\n\nAm I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.\n\n2. Originally Posted by symstar\n$\\int \\frac{dx}{4x^{2/3}-4x^{1/3}-3}$\n\n$u=x^{1/3} \\text{ }du=\\tfrac{1}{3}x^{-2/3} \\text{ }x=u^3$\n\n$=\\int\\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$\n$=3\\int\\frac{du}{u^2(2u+1)(2u-3)}$\n\nAm I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.\nyou are ok so far\n\nso you must be making a mistake with the partial fractions part. what did you get?\n\n3. Originally Posted by symstar\n\n$\\int\\frac{du}{u^2(2u+1)(2u-3)}$\nPut $u=\\frac1z$ and your integral becomes $-\\int{\\frac{z^{2}}{(2+z)( 2-3z)}\\,dz},$ hence\n\n\\begin{aligned}\n-\\frac{z^{2}}{(2+z)(2-3z)}&=\\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\\\\n& =\\frac{2-z}{2-3z}-\\frac{4}{(2+z)(2-3z)}=\\frac{2-z}{2-3z}-\\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\\\\n& =\\frac{2-z}{2-3z}-\\frac{1}{2}\\left( \\frac{3}{2-3z}+\\frac{1}{2+z} \\right). \\\\\n\\end{aligned}\n\nThings should be easy from there.\n\n4. Originally Posted by Krizalid\nPut $u=\\frac1z$ and your integral becomes $-\\int{\\frac{z^{2}}{(2+z)( 2-3z)}\\,dz},$ hence\n\n\\begin{aligned}\n-\\frac{z^{2}}{(2+z)(2-3z)}&=\\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\\\\n& =\\frac{2-z}{2-3z}-\\frac{4}{(2+z)(2-3z)}=\\frac{2-z}{2-3z}-\\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\\\\n& =\\frac{2-z}{2-3z}-\\frac{1}{2}\\left( \\frac{3}{2-3z}+\\frac{1}{2+z} \\right). \\\\\n\\end{aligned}\n\nThings should be easy from there.\ni know what you're thinking, symstar, \"how does he do it?!\"\n\n5. So working with the partial fractions:\n\n$\\frac{1}{u^2(2u+1)(2u-3)}=\\frac{A}{u^2}+\\frac{B}{(2u+1)}+\\frac{C}{(2u-3}$\n$1=A(4u^2-4u-3)+B(2u^3-3u^2)+C(2u^3+u^2)$\n$2B+2C=0$\n$4A-3B+C=0$\n$-4A=0$\n$-3A=1$\n\nYou can see the 2 values for A at the bottom. What should I do?\n\n6. Hello, symstar!\n\nI don't agree . . .\n\n$\\int \\frac{dx}{4x^{\\frac{2}{3}}-4x^{\\frac{1}{3}}-3}$\n\nLet: $u \\:=\\:x^{\\frac{1}{3}}\\quad\\Rightarrow\\quad x \\:=\\:u^3\\quad\\Rightarrow\\quad dx \\:=\\:3u^2\\,du$\n\nSubstitute: . $\\int \\frac{3u^2\\,du}{4u^2 - 4u - 3} \\;=\\;\\frac{3}{4}\\int \\frac{u^2\\,du}{u^2-u-\\frac{3}{4}} \\;=\\;\\frac{3}{4}\\int\\left[1 + \\frac{u+\\frac{3}{4}}{u^2-u-\\frac{3}{4}}\\right]\\,du$\n\n. . $= \\;\\frac{3}{4}\\int\\left[1 + \\frac{4u+3}{4u^2-4u-3}\\right]\\,du \\;=\\;\\frac{3}{4}\\int\\left[1 + \\frac{4u+3}{(2u-3)(2u+1)}\\right]\\,du$\n\nPartial Fractions: . $\\frac{3}{4}\\int\\left[1 + \\frac{\\frac{9}{4}}{2u-3} - \\frac{\\frac{1}{4}}{2u+1}\\right]\\,du \\quad\\hdots\\text{ etc.}$\n\n7. $\\frac{3}{4}\\int \\frac{u^2\\,du}{u^2-u-\\frac{3}{4}} \\;=\\;\\frac{3}{4}\\int\\left[1 + \\frac{u+\\frac{3}{4}}{u^2-u-\\frac{3}{4}}\\right]\\,du$\n\nI'm not quite sure what exactly happened in this step, could you explain please?", "date": "2014-09-02 12:04:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/49549-integral-partial-fractions.html", "openwebmath_score": 0.9905178546905518, "openwebmath_perplexity": 1219.5319643804598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808765013518, "lm_q2_score": 0.8757869786798663, "lm_q1q2_score": 0.85851722708877}}
{"url": "https://www.physicsforums.com/threads/what-does-ceil-mean.116822/", "text": "# What does ceil mean ?\n\n#### momentum\n\nIts diifcult to express my question....so, i am posting this\n\nceil(4.5) =?\nceil(4.1)=?\nceil(4.6)=?\n\n#### AKG\n\nHomework Helper\nceil(x) is the smallest integer which is greater than or equal to x. In particular, if x is an integer, then ceil(x) = x, and if x is not an integer, then ceil(x) > x.\n\n#### momentum\n\nceil(4.5)=4 // is it ok ?\nceil(4.1)=4 //is it ok ?\nceil(4.6)=5 //is it ok ?\n\n#### Integral\n\nStaff Emeritus\nGold Member\nmomentum said:\nceil(4.5)=4 // is it ok ?\nceil(4.1)=4 //is it ok ?\nceil(4.6)=5 //is it ok ?\nNope, read the defintion given above, then try again.\n\n#### momentum\n\nah...i see, all of them should be 5 .....i had confusion on fractional part .5.\nbut i see ..it does not care for .5 which we use for round-off.\n\nHomework Helper\nYes, all 5.\n\n#### momentum\n\nthank you for the clarifcation\n\n#### bomba923\n\nRecall that\n\n$$\\begin{gathered} \\forall x \\in \\left( {a,a + 1} \\right)\\;{\\text{where }}a \\in \\mathbb{Z}, \\hfill \\\\ {\\text{floor}}\\left( x \\right) = \\left\\lfloor x \\right\\rfloor = a \\hfill \\\\ {\\text{ceil}}\\left( x \\right) = \\left\\lceil x \\right\\rceil = a + 1 \\hfill \\\\ \\end{gathered}$$\n\n$$\\forall x \\in \\mathbb{Z},\\;\\left\\lfloor x \\right\\rfloor = \\left\\lceil x \\right\\rceil = x$$\n\nLast edited:\n\n#### jim mcnamara\n\nMentor\nceil -> \"goes up\" if it needs to, in order reach an integer\nfloor -> \"goes down\" as it needs to, in order to reach an integer\nWhat happens with negative numbers:\n\n$$floor( -1.1 ) = -2 \\; ceil( -1.1 ) = -1$$\n$$floor( -0.1 ) = -1 \\; ceil( -0.1 ) = 0$$\n$$floor( 0.9 ) = 0 \\; ceil( 0.9 ) = 1$$\n\n### The Physics Forums Way\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-03-25 09:57:32", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/what-does-ceil-mean.116822/", "openwebmath_score": 0.4573347568511963, "openwebmath_perplexity": 12157.991373710738, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9711290930537121, "lm_q2_score": 0.8840392848011834, "lm_q1q2_score": 0.8585162688728255}}
{"url": "https://brilliant.org/discussions/thread/trigonometry-practice-for-beginners/", "text": "# Trigonometry practice for beginners\n\nThe following problems need only trigonometric definition of functions and the equation\n\n$\\sin^2{\\theta} + \\cos^2{\\theta} = 1$\n\nunless otherwise stated. More will be written.\n\nProve the following:\n\n1. $$\\csc{\\theta} \\cdot \\cos{\\theta} = \\cot{\\theta}$$\n\n2. $$\\csc{\\theta}\\cdot\\tan{\\theta}=\\sec{\\theta}$$\n\n3. $$1+\\tan^2(-\\theta) = \\sec^2{\\theta}$$\n\n4. $$\\cos{\\theta}(\\tan{\\theta}+\\cot{\\theta})=\\csc{\\theta}$$\n\n5. $$\\sin{\\theta}(\\tan{\\theta}+\\cot{\\theta})=\\sec{\\theta}$$\n\n6. $$\\tan{\\theta}\\cdot\\cot{\\theta}-\\cos^2{\\theta}=\\sin^2{\\theta}$$\n\n7. $$\\sin{\\theta}\\cdot\\csc{\\theta}-\\cos^2{\\theta}=\\sin^2{\\theta}$$\n\n8. $$(\\sec{\\theta}-1)(\\sec{\\theta}+1)=\\tan^2{\\theta}$$\n\n9. $$(\\csc{\\theta}-1)(\\csc{\\theta}+1)=\\cot^2{\\theta}$$\n\n10. $$(\\sec{\\theta}-\\tan{\\theta})(\\sec{\\theta}+\\tan{\\theta})=1$$\n\n11. $$\\sin^2{\\theta}(1+\\cot^2{\\theta})=1$$\n\n12. $$(1-\\sin^2{\\theta})(1+\\tan{\\theta})=1$$\n\n13. $$(\\sin{\\theta}+\\cos{\\theta})^2+(\\sin{\\theta}-\\cos{\\theta})=2$$\n\n14. $$\\tan^2{\\theta}\\cdot\\cos^2{\\theta}+\\cot^2{\\theta}\\cdot\\sin^2{\\theta}=1$$\n\n15. $$\\sec^4{\\theta}-\\sec^2{\\theta}=\\tan^4{\\theta}+\\tan^2{\\theta}$$\n\n16. $$\\sec{\\theta}-\\tan{\\theta}=\\dfrac{\\cos{\\theta}}{1+\\sin{\\theta}}$$\n\n17. $$\\csc{\\theta}-\\cot{\\theta}=\\dfrac{\\sin{\\theta}}{1+\\sin{\\theta}}$$\n\n18. $$3\\sin^2{\\theta}+4\\cos^2{\\theta}=3+\\cos^2{\\theta}$$\n\n19. $$9\\sec^2{\\theta}-5\\tan^2{\\theta}=5+4\\sec^2{\\theta}$$\n\n20. $$1-\\dfrac{\\cos^2{\\theta}}{1+\\sin{\\theta}}=\\sin{\\theta}$$\n\n21. $$1-\\dfrac{\\sin^2{\\theta}}{1-\\cos{\\theta}}=-\\cos{\\theta}$$\n\n22. $$\\dfrac{1+\\tan{\\theta}}{1-\\tan{\\theta}}=\\dfrac{\\cot{\\theta}+1}{\\cot{\\theta}-1}$$\n\n23. $$\\dfrac{\\sec{\\theta}}{\\csc{\\theta}}+\\dfrac{\\sin{\\theta}}{\\cos{\\theta}}=2\\tan{\\theta}$$\n\n24. $$\\dfrac{\\csc{\\theta}-1}{\\cot{\\theta}}=\\dfrac{\\cot{\\theta}}{\\csc{\\theta}+1}$$\n\n25. $$\\dfrac{1+\\sin{\\theta}}{1-\\sin{\\theta}}=\\dfrac{\\csc{\\theta}+1}{\\csc{\\theta}-1}$$\n\n26. $$\\dfrac{1-\\sin{\\theta}}{\\cos{\\theta}}+\\dfrac{\\cos{\\theta}}{1-\\sin{\\theta}}=2\\sec{\\theta}$$\n\n27. $$\\dfrac{\\sin{\\theta}}{\\sin{\\theta}-\\cos{\\theta}}=\\dfrac{1}{1-\\cot{\\theta}}$$\n\n28. $$1-\\dfrac{\\sin^2{\\theta}}{1+\\cos{\\theta}}=\\cos{\\theta}$$\n\n29. $$\\dfrac{1-\\sin{\\theta}}{1+\\sin{\\theta}}=(\\sec{\\theta}-\\tan{\\theta})^2$$\n\n30. $$\\dfrac{\\cos{\\theta}}{1-\\tan{\\theta}}+\\dfrac{\\sin{\\theta}}{1-\\cot{\\theta}}=\\sin{\\theta}+\\cos{\\theta}$$\n\n31. $$\\dfrac{\\cot{\\theta}}{1-\\tan{\\theta}}+\\dfrac{\\tan{\\theta}}{1-\\cot{\\theta}}=1+\\tan{\\theta}+\\cot{\\theta}$$\n\n32. $$\\tan{\\theta}+\\dfrac{\\cos{\\theta}}{1+\\sin{\\theta}}=\\sec{\\theta}$$\n\nNote by Sharky Kesa\n2\u00a0years, 11\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nHere's a nice result which involves all six ratios :\n\n$(\\sin \\theta + \\cos \\theta)(\\tan \\theta + \\cot \\theta)=\\sec \\theta + \\csc \\theta$\n\n- 2\u00a0years, 11\u00a0months ago\n\nCan you hear the Proving Trigonometric Identities Wiki page calling out your name?\n\nStaff - 2\u00a0years, 11\u00a0months ago\n\nThank You, it is a good practice for we beginners!\n\n- 2\u00a0years, 11\u00a0months ago\n\nyes\n\n- 1\u00a0month, 2\u00a0weeks ago", "date": "2018-04-26 15:39:08", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/trigonometry-practice-for-beginners/", "openwebmath_score": 0.9986642599105835, "openwebmath_perplexity": 8835.873450887795, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9711290913825541, "lm_q2_score": 0.8840392725805822, "lm_q1q2_score": 0.8585162555276749}}
{"url": "https://stats.stackexchange.com/questions/403413/does-data-normalization-and-transformation-change-the-pearsons-correlation", "text": "# Does data normalization and transformation change the Pearson's correlation?\n\nAs we know that Pearson's correlation measures the linearity between two variables, I am wondering when applying normalization and transformation on the original dataset, does the normalization and transformation method needs to be a linear method, in order to not effect the correlation results?\n\nMore specifically, for example, after log-transformation, does the correlation change? I think so because log-transformation changed the distribution of the data and the original linear relationship is scaled by the log() function.\n\nTo extend the question, I am wondering if anyone could provide some general summary and comments on what normalization/transformation can be used when you want to scale the data but do not want to change the original relationship among your variables described by correlation metric?\n\nThanks!\n\nYou are mostly welcome to refine and shape my question if it's not accurate.\n\n\u2022 Take 3 non-collinear but correlated and positive x-y pairs of values (e.g. x=1,2,4, y=2,5,4) and take their logs. Compute the Pearson correlation before and after. What happens? Try other examples. $\\:$ (I imagine this small step of trying some examples would come under the 'research' part of the usual search and research. ) \u2013\u00a0Glen_b Apr 17 at 1:03\n\u2022 Thanks for the suggestion! I think my question needs to be modified. What I am wondering is conceptual correctness rather than numerical difference. I am thinking can we use the correlation metric of a log transoformed data (say log(y)~log(x) ) to estimate and conclude on the original relationship or linearity of y~x. I will think about it and edit the question! \u2013\u00a0Lingjue Wang Jun 7 at 20:16\n\u2022 The numerical difference actually tells you the answer to the conceptual question \u2013\u00a0Glen_b Jun 8 at 3:08\n\nPearson's correlation measures the linear component of association. So you are correct that linear transformations of data will not affect the correlation between them. However, nonlinear transformations will generally have an effect.\n\nHere is a demonstration: Generate right-skewed, correlated data vectors x and y. Pearson's correlation is $$r = 0.987.$$ (The correlation of $$X$$ and $$Y^\\prime = 3 + 5Y$$ is the same.)\n\nset.seed(2019)\nx = rexp(100, .1);  y = x + rexp(100, .5)\ncor(x, y)\n[1] 0.987216\ncor(x, 3 + 5*y)\n[1] 0.987216     # no change with linear transf of 'y'\n\n\nHowever, if the second variable is log-transformed, Pearson's correlation changes to $$r = 0.862.$$\n\ncor(x, log(y))\n[1] 0.8624539\n\n\nHere are the corresponding plots:\n\nBy contrast, Spearman's correlation is unaffected by the (monotone increasing) log-transformation. Spearman's correlation is based on ranks of observations and log-transformation does not change ranks. Before and after transformation, $$r_S = 0.966.$$\n\ncor(x, y, meth=\"spear\")\n[1] 0.9655446\ncor(rank(x), rank(log(y)))\n[1] 0.9655446   # Spearman again\n\ncor(x, log(y), meth=\"spear\")\n[1] 0.9655446", "date": "2019-11-13 12:58:37", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/403413/does-data-normalization-and-transformation-change-the-pearsons-correlation", "openwebmath_score": 0.6661191582679749, "openwebmath_perplexity": 1266.3206871544674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290905469752, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.8585162533055176}}
{"url": "http://csfn.eu/beach-flowering-yiyriza/e539f3-angles-in-a-triangle-add-up-to", "text": "Step 1: Find the given angles Step 2: Subtract the given angles from 180 to get the missing angle. So a + b + y = 180. In other words: Angle 1 + Angle 2 + Angle 3 = 180\u00b0 Missing Angles Our learning objectives today To calculate missing angles on a straight line, around a point and in a triangle. We first add the two 50\u00b0 angles together. What do you need to know? Angles of Triangles Add Up To 180 Degrees Ex: Could a triangle have the given angle measures? a) Sum of total angles in a triangle is 180. b) Opposite angles are always equal. Add up the two known angle measurements. To calculate the other angles we need the sine, cosine and tangent. 4 Short Steps for answering Exterior Angle of a Triangle Example Problems. Therefore y = 180 - x. In a right triangle, one of the angles is exactly 90\u00b0. Consider the lunes through B and B'. The three angles in a triangle add up to 180 degrees. For example, in triangle ABC, angle A + angle B + angle C = 180\u00b0. She notices by chance that one triangle\u2019s angles add up to 180. What you have now is two right triangles which, by the above contain 360 degrees in total. But if you look at the two right angles that add up to 180 degrees so the other angles, the angles of the original triangle, add up to 360 - 180 = 180 degrees. The interior angles of a triangle add up to 180 degrees. Consider a spherical triangle ABC on the unit sphere with angles A, B and C. Then the area of triangle ABC is. The angles in the triangle add up to 180 degrees. The degree measure of a straight line add up to 180 degrees. Name Date ANGLES IN A TRIANGLE 1 Work out the missing angles. The diagram below shows the interior and exterior angles of a triangle.. A demonstration of the angles of a triangle summing up to 180\u00b0 can be found here. All 3 angles of a triangle add up to 180 degrees. Putting this into the first equation gives us: a + b + 180 - x = 180. So x + y = 180. The angles on a straight line add up to 180 degrees. So, if you know two of the three measurements of the triangle, then you're only missing one piece of the puzzle. \u2026 In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the angle in the bottom right corner to make a \u2026 Remember that the angles in a triangle add up to 180\u00b0. Why is every triangle 180 degrees? Each corner that you cut off contains an angle from the triangle. Penny Find the measure of the smallest angle (in degrees) Interior angles of a triangle add up to 180 degrees so 180 - 125 = 55 degrees.....the measure of the 3rd angle Every triangle has six exterior angles (two at each vertex are equal in measure). To find angle \u2018b\u2019, we subtract both 50\u00b0 angles from 180\u00b0. The angles inside a triangle are called interior angles. The angle sum property of a triangle states that the angles of a triangle always add up to 180\u00b0. Yes, all triangle angles add up to 180 degrees. The interior angles of a triangle always add up to 180\u00b0 while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. Such an angle is called a right angle. The first thing you can do is add up the angle measurements you know. This is why we coloured the edges so we can easily see the angle contained by the edges. Angles A and E are congruent angles, which means they have the same measure, because they are alternate interior angles of a transversal with parallel lines. c) Angle in a straight line adds up to 180. r is equal to the opposite angle to left over angle measure which is 180 - (74 + 60) = 180 - 134 = 46. q and r forming a straight line where q = 180 - r = 180 - 46 = 134. o is the opposite angel which measures 73, o -= 73 Draw a perpendicular from the vertex opposite the longest side, to the longest side. If you are given the adjacent angle you subtract the angle from 180 degrees to solve for the exterior angle of a triangle. Angle C and Angle F \u2026 Step by step descriptive logic to check whether a triangle can be formed or not, if angles are given. The sum of the measures of the interior angles of a triangle in Euclidean space is always 180 degrees. The definition of a triangle is that it is a shape with three sides that add up to 180 degrees. and 180 degrees is pi radians, so wouldnt it be (pi) - 0.2 - 0.2? When we assemble the angles (by aligning the coloured edges), we see that all the angles add up to a straight line (or 180\u00b0). The sum of the lengths of any two sides of a triangle always exceeds the length of the third side, a principle known as the triangle inequality. The angles on a straight line add up to 180\u00b0. The sides adjacent to the right angle are the legs. Now we can establish that the three angles inside the triangle (B, E & F) also add up to 180. according to this solutionbank [the answer is the same in the book], the angle for the shaded section from C is 0.4 meaning the angles in the triangle are 0.2 0.2 and 0.6, i thought the angles add up to 180? Every triangle has three angles and whether it is an acute, obtuse, or right triangle, the angles sum to 180\u00b0. We can use the property that the angles of a triangle add up to 180 degrees. Logic to check triangle validity if angles are given. All you have to know is that all of the angles in a triangle always add up to 180\u00b0. The third angle is 4 times as big as the smallest angle. Proving that the angles inside a triangle \u2013 any triangle \u2013 sum up to 180\u00b0 is very simple, but leaves most people unsatisfied (or unconvinced) because it depends on the properties of something called Alternate Interior Angles.. I\u2019ll give the proof first and then explain Alternate Interior Angles. Exterior angles of a triangle - Triangle exterior angle theorem. 180 - 98 = 82 So 82 degrees is the third angle. (Exercise: make sure each triangle here adds up to 180\u00b0, and check that the pentagon's interior angles add up to 540\u00b0) The Interior Angles of a Pentagon add up to 540\u00b0 The General Rule. The diagram shows a view looking down on the hemisphere which has the line through AC as its boundary. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180\u00b0. The angles are not drawn to scale, so do not try to measure them! If the sum is not 180\u00b0 you are not in Euclidean space.. Sum of the Interior Angles of a Triangle. Therefore a + b = x after rearranging. These angles add up to 180\u00b0 for every triangle, independent of the type of triangle. The problem is that there are two possible angles for any dot product, $\\theta$ and $\\pi - \\theta$ (Draw the intersecting lines and you will see the adjacent supplementary angles.) The second angle is 30 degrees larger than the smallest angle. That will work. To use a protractor to draw and measure acute and obtuse angles to the nearest degree. A + B + C - $\\pi$. 50\u00b0 + 50\u00b0 = 100\u00b0 and 180\u00b0 \u2013 100\u00b0 = 80\u00b0 Angle \u2018b\u2019 is 80\u00b0 because all angles in a triangle add up to 180\u00b0. Since 60 degrees is too small, try 180 degrees - 60 degrees = 120 degrees. The side opposite the right angle is called the hypotenuse. An exterior angle of a triangle is equal to the sum of the opposite interior angles. Lesson on angles in a triangle proof, created in connection to my school's new scheme of work based upon the new National Curriculum. The exterior angles of a triangle add up to 360 degrees. Input all three angles of triangle \u2026 Now to find angle \u2018b\u2019, we use the fact that all three angles add up to 180\u00b0. (The external angles of any -sided convex polygon add up to 360 degrees.) This is what we wanted to prove. Unit 5 Section 6 : Finding Angles in Triangles. As we know, if we add up the interior and exterior angles of one corner of a triangle, we always get 1800. Get an answer to your question \u201cTwo angles in a triangle add up to 108 degrees what is the size of the third angle ...\u201d in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. The three interior angles of a plane triangle add up to 180 degrees, or pi radians. This is true 100% of the time. Question 595896: The angles of a triangle add up to 180 degrees. The regions marked Area 1 and Area 3 are lunes with angles A and C respectively. The measures of the interior angles of the triangle always add up to 180 degrees (same color to point out they are equal). That\u2019s an odd coincidence, considering that 180 degrees is exactly the value of a straight line \u201cangle.\u201d She looks at a few more triangles with her protractor, always getting the same result: the angles always add up to 180 degrees. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\\circ {/eq}. Since the three interior angles of a triangle add up to 180 degrees, in a right triangle, since one angle is always 90 degrees, the other two must always add up to 90 degrees (they are complementary). This fact is equivalent to Euclid's parallel postulate. Each time we add a side (triangle to quadrilateral, quadrilateral to pentagon, etc), we add another 180\u00b0 to the total: a) \u2026 The exterior angles, taken one at each vertex, always sum up to 360\u00b0. Every triangle has three sides, and three angles in the inside. The interior angles of a triangle add up to 180\u00b0 An exterior angle of a triangle is formed when any side is extended outwards. A massive topic, and by far, the most important in Geometry. A triangle is said to be a valid triangle if and only if sum of its angles is 180 \u00b0. 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Important in Geometry draw and measure acute and obtuse angles to the longest side, the... From 180 degrees, or right triangle, we always get 1800 sum is not 180\u00b0 you are drawn. Add up to 180 degrees. is to subtract the angle contained by the edges can be found here can. Angles and whether it is a shape with three sides, and by far the!, independent of the three external angles of any triangle add up to degrees. The three external angles of Triangles add up to 180\u00b0 = 82 so 82 degrees is pi radians the in... First equation gives us: a + b + angle C and F. Use a protractor to draw and measure acute and obtuse angles to the nearest degree thing you can do add! Remember that the angles on a straight line add up to 180 angles in a triangle add up to. for example, triangle... ) opposite angles are given in total calculate the other angles we need the sine, cosine and.! Plane triangle add up to 360 degrees. has three angles add to!", "date": "2022-01-28 03:34:46", "meta": {"domain": "csfn.eu", "url": "http://csfn.eu/beach-flowering-yiyriza/e539f3-angles-in-a-triangle-add-up-to", "openwebmath_score": 0.5129854083061218, "openwebmath_perplexity": 246.4936633907233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9881308768564867, "lm_q2_score": 0.8688267847293731, "lm_q1q2_score": 0.8585145726310375}}
{"url": "http://math.stackexchange.com/questions/71362/why-is-this-sequence-convergent", "text": "# why is this sequence convergent\n\nSuppose to have two sequence $(a_n)_{n\\geq1}$ and $(b_n)_{n\\geq1}$ such that\n\n$a_n=\\frac{1}{2}(a_{n-1}+b_{n-1})$.\n\nI want to prove that if $b_n\\rightarrow0$ then $a_n\\rightarrow0$.\n\nThe only thing I was able to prove is that $a_n$ is bounded, in fact:\n\n$b_n$ is convergent and so bounded $|b_n|\\leq M$. And so $|a_n|\\leq |\\frac{a_2}{2^{n-1}}+\\frac{b_2}{2^{n-2}}+\\cdots+\\frac{b_{n-1}}{2}|\\leq|\\frac{a_2}{2^{n-2}}|+M\\sum^\\infty_{k=1}\\frac{1}{2^k}$ and for great $n$ we have $|\\frac{a_2}{2^{n-2}}|\\leq\\varepsilon$\n\nCould you help me to continue (if I'm on the right track), please?\n\nI see that if we prove that $a_n$ has a limit $L$ then necessarily $L=0$ because $L$ must satisfy $L=\\frac{1}{2}L$, but I don't know how to prove that it has a limit.\n\n-\n+1 for showing your work \u2013\u00a0 Jyrki Lahtonen Oct 10 '11 at 6:00\nHint: If $n$ is large, the early terms of your middle expression are small (not just the $a_2$ one) because of the large power of 2, and the later terms are small because $b_n$ tends to zero. So can you see how to split the sequence differently, and get a better estimate? \u2013\u00a0 Mark Bennet Oct 10 '11 at 6:06\n@Mark Sorry 'bout ruining your hint with my answer. I didn't see your comment soon enough :-( \u2013\u00a0 Jyrki Lahtonen Oct 10 '11 at 6:14\n@JyrkiLahtonen: no hard feelings ... The technique of splitting a problem into pieces which are dealt with in different ways seemed worth noting anyway. In measure theory, for example, there are a number of proofs which require a proof for \"most\" points of a set based on some kind of regularity, and a proof for the remaining \"problem\" points by showing that they are confined to a set of arbitrarily small measure. \u2013\u00a0 Mark Bennet Oct 10 '11 at 7:09\n\nHint: You have a good start in proving that the sequence $(a_n)$ is bounded. Let's reuse your trick and look at $$a_{2n}=\\frac{a_n}{2^n}+\\frac{b_{2n-1}}2+\\frac{b_{2n-2}}4+\\cdots+\\frac{b_n}{2^n}.$$ All the numbers $|b_k|<\\epsilon$ for $k\\ge n$, if $n$ is large enough. The first term $a_n/2^n$ looks like it would be under control as well now that you know $|a_n|$ to be bounded.\n\n-\n\nAssume that $-t\\leqslant b_n\\leqslant t$ for a given positive $t$ and for every $n\\geqslant n_t$. Then $a_n-t\\leqslant\\frac12(a_{n-1}-t)$ and $a_n+t\\geqslant\\frac12(a_{n-1}+t)$ for every $n\\geqslant n_t$. Thus $\\limsup (a_n-t)\\leqslant0$ and $\\liminf (a_n+t)\\geqslant0$. Since this holds for every positive $t$, $a_n\\to 0$.\n\n-\n@Srivatsan, exactly. Thanks. \u2013\u00a0 Did Oct 10 '11 at 17:29\n\nOne way to do this is to view the recursion equation defining $a_n$ as a, well, recursion equation. It is then a non-homogeneous linear recursion, whose associated homogeneous recursion is very easy to solve. One could then use Lagrange's method of variation of parameters to determine the actual solution, but instead of doing that (we can't, in fact, because we do not know $b_n$) we can use the same idea to obtain bounds that will prove what you want. Let's do that.\n\nLet $\\varepsilon>0$. There exists an $N$ such $|b_n|\\leq\\varepsilon$ for all $n\\geq N$.\n\nLet $a_n=\\frac{\\alpha_n}{2^n}$ (this is where we use Lagrange's method: the solution for the homogeneous equation is $\\frac{\\alpha}{2^n}$, with $\\alpha$ a constant and Lagrange suggests that we now turn $\\alpha$ into a function $\\alpha_n$ of $n$) and suppose that $|b_n|\\leq\\beta$ for all $n\\geq1$. Replacing this in the defining recursion tells us that $$|\\alpha_n-\\alpha_{n-1}|=2^{n-1}|b_n|$$ for all $n\\geq1$. This implies that when $n\\geq N$ \\begin{align}|\\alpha_n-\\alpha_0|&\\leq|\\alpha_n-\\alpha_{n-1}|+|\\alpha_{n-1}-\\alpha_{n-2}|+\\cdots+|\\alpha_1-\\alpha_0| \\\\ &\\leq(2^{n-1}+\\cdots+2^N)\\varepsilon + (2^{N-1}+\\cdots+2^0)\\beta\\\\&\\leq (2^n-2^N)\\varepsilon+(2^N-1)\\beta \\\\&\\leq 2^n \\varepsilon+c\\end{align} for some positive constant $c$. Then $$|\\alpha_n|\\leq 2^n\\varepsilon+c+|\\alpha_0|$$ and $$|a_n|=|\\alpha_n/2^n|\\leq\\varepsilon+\\frac{c+|\\alpha_0|}{2^n}.$$ This should make it clear that $a_n\\to0$.\n\n-\nOne can probably enhance this to a statement telling us that solutions to a non-homogeneous perturbation which gets smaller and smaller of a linear homogeneous recursion with all characteristic roots in $(-1,1)$ is not very different from the unperturbed solutions. \u2013\u00a0 Mariano Su\u00e1rez-Alvarez Oct 10 '11 at 6:28", "date": "2015-05-28 10:16:36", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/71362/why-is-this-sequence-convergent", "openwebmath_score": 0.9626346826553345, "openwebmath_perplexity": 127.50501798831874, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308768564867, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.858514559206388}}
{"url": "https://math.stackexchange.com/questions/3375964/multiplying-left-hand-and-right-hand-limits-question", "text": "# Multiplying left hand and right hand limits question\n\nI came across the attached question in our calculus book. The limits in question are:\n\nF(x), which approaches 0 from the left and 1 from the right as x goes to 2\n\nJ(x), which approaches 1 from the left and 0 from the right as x goes to 2.\n\nQuestion B) ii. asks for the limit of [F(x) + J(x)] as x goes to 2. The left hand limits add up to 1, and the right hand limits do too, so the limit is 1 as x approaches 2 - the answer key matches this.\n\nHowever, in C) ii., which asks for limit of [F(x)J(x)] as x approaches 2, the left hand limits multiply to 0 and the right hand limits multiply to 0, but the answer key has DNE. Am I missing something completely about how limits work? Or is the answer key wrong?\n\nThanks in advance! Edited photo of the page\n\n\u2022 I agree that $\\lim_\\limits{x\\to 2} f(x)j(x) = 0$ \u2013\u00a0Doug M Sep 30 '19 at 20:59\n\n$$\\lim_{x\\to 0^+} F(x)\\cdot J(x)=\\lim_{x\\to 0^+} F(x)\\cdot\\lim_{x\\to 0^+} J(x)=1\\cdot0=0$$\n$$\\lim_{x\\to 0^-} F(x)\\cdot J(x)=\\lim_{x\\to 0^-} F(x)\\cdot\\lim_{x\\to 0^-} J(x)=0\\cdot1=0$$\n$$\\lim_{x\\to 0} F(x)\\cdot J(x)=0$$", "date": "2020-12-02 17:12:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3375964/multiplying-left-hand-and-right-hand-limits-question", "openwebmath_score": 0.9462119340896606, "openwebmath_perplexity": 382.3179675674922, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846697584034, "lm_q2_score": 0.8774767938900121, "lm_q1q2_score": 0.8585098432107421}}
{"url": "https://math.stackexchange.com/questions/3427262/are-there-any-inflection-points", "text": "# Are there any inflection points?\n\n$$F(x) = \\begin{cases} x^2 & x \\le 0 \\\\ 0 & 0 \\le x \\le 3 \\\\ -(x-3)^2 & x>3 \\end{cases}$$\n\nMy question is does this function have any points of inflection?\u00a0Double Derivative\u00a0at $$x=0,3$$.Thus the necessary condition is satisfied for all points such that\u00a0$$0 \\le x \\le 3$$.\n\nBut, i have also read that the third derivative at x is not equal to 0 is also regarded as a sufficient condition, which is fulfilled at $$x=0,3$$. Thus, which sufficient condition should hold and which points should be regarded as inflection points?\n\nAn inflection point occurs at points of the domain at which the function changes concavity. The second derivative at an inflection point may be zero but it also may be undefined. If we look at the second derivative for your function\n\n$$F''(x) = \\begin{cases} 2 & x < 0 \\\\ 0 & 0 \\le x \\le 3 \\\\ -2 & x>3 \\end{cases}$$\n\nthen $$F''(x) > 0$$ when $$x < 0$$ and $$F''(x)<0$$ when $$x>3$$. We know that the function $$F(x)$$ is concave up when $$F''(x) > 0$$ and concave down when $$F''(x) < 0$$. Therefore, it is clear that $$F(x)$$ is concave up on $$(-\\infty,0)$$ and concave down on $$(3,\\infty)$$.\n\nBut, what about the interval $$[0,3]$$? Since $$F(x)\\equiv 0$$ when $$0\\le x \\le 3$$, the second derivative is also zero. As $$F(x)$$ changes concavity at the end points we know that there must be at least one inflection point. Therefore, we must decide whether there is one inflection point or two inflection points.\n\nSuppose that we assume that there are two inflection points. Then, we would need to show that the function changes concavity twice - there would be two intervals where it was concave up and one interval where it is concave down. Alternatively, the function could be concave down on two intervals and then concave up on one interval.\n\nYour function is concave up on one interval and then concave down on another interval. In the interval between these two intervals, the function is neither concave up nor concave down. Hence, your function only changes concavity once.\n\nSo, $$F(x)$$ has one inflection point. If we plot the function on Desmos, then we see that this point of inflection occurs in the middle of the interval $$[0,3]$$.\n\nInterestingly, the function $$f(x)=x^3$$ exhibits similar behavior. This function is concave up on $$(0,\\infty)$$ and concave down on $$(-\\infty,0)$$. It has one inflection point at $$(0,0)$$.\n\n\u2022 Yes, also that all straight lines a considered both concave and convex. So if ur talking about the change in concavity, then both 0,3 are inflection points. \u2013\u00a0DARE2ZLATAN Nov 8 at 22:55\n\u2022 So, what ur saying, does that imply that any function that is concave in some interval and convex in some interval will surely have atleast 1 inflection point? \u2013\u00a0DARE2ZLATAN Nov 9 at 13:38\n\u2022 Intuitively, yes. The inflection point must occur when the function changes concavity. When this happens, the function would go from a parabola that opens upward to a parabola that opens downward (or vice versa). I have never seen a proof of such a result. Although, I cannot think of a counterexample and therefore believe that this is a theorem. \u2013\u00a0Axion004 Nov 9 at 14:43", "date": "2019-11-18 22:16:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3427262/are-there-any-inflection-points", "openwebmath_score": 0.8255089521408081, "openwebmath_perplexity": 167.98034267074647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384664716301, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8585098387864142}}
{"url": "https://www.ask-math.com/sum-of-arithmetic-progression.html", "text": "# Sum of arithmetic progression\n\nThe sum of arithmetic progression is denoted by $S_{n}$. It is nothing but the sum of 'n terms of an A.P. with first term 'a' and common difference 'd'.\nThe formula for sum of n terms of A.P. is\n$S_{n} = \\frac{n}{2}[2a + (n-1)d]$\n\n$S_{n} = \\frac{n}{2}[a + l]$ , where l = last term = a + (n -1 )d\n\nProof : Let $a_{1}, a_{2},a_{3},...,a_{n}$ be an A.P. with first term as 'a' and common difference as 'd'.\n$a_{1}$ = a; $a_{2}$ = a + d; $a_{3}$ = a + 2d ; ... $a_{n}$ = a + (n -1)d\n$S_{n} = a_{1} + a_{2} + a_{3} + ... +a_{n-1} + a_{n}$\n\u21d2 $S_{n}$ = a + (a + d) + (a + 2d) + \u2026 + [ a + (n-2)d] + [ a + (n -1 )d] -----(i)\nWrite the above equation in reverse order we get,\n$S_{n}$ = [ a + (n -1 )d] + [ a + (n-2)d] + \u2026 + (a + 2d) + (a + d) + a ----- (ii)\n2$S_{n}$ = [ 2a + (n -1 )d] + [ 2a + (n-1)d] + \u2026+ [2a + (n-1)d]\n[2a + (n-1)d] repeats \u2018n\u2019 times\n\u2234 2$S_{n}$ = n [ 2a + (n -1 )d]\n$S_{n} = \\frac{n}{2}$ [ 2a + (n -1 )d]\n\nSince the last term l = a + (n \u2013 1)d\n\u2234 $S_{n} = \\frac{n}{2}$ [ a + a + (n -1 )d]\n\n$S_{n} = \\frac{n}{2}$ [a + l ]\n\nNote : In the above formula there are 4 unknown quantities. So if any three are given then we can find the forth one.\nIn the sum of $S_{n}$ of n terms of a sequence is given then the nth term $a_{n}$ of the sequence can determined by using the following formula .\n$a_{n} = S_{n} - S_{n - 1}$\n\n## Solved examples on sum of arithmetic progression\n\n1) 50,46,42,\u2026 10 terms.\nSolution: 50,46,42,\u2026 10 terms\nThe formula to find sum is\n$S_{n} = \\frac{n}{2}$ [ 2a + (n -1 )d]\n\nNumber of terms = n = 10; First term = a = 50 ; Common difference = d = 46 \u2013 50 = -4\nPut all the given values in the formula we get,\n$S_{10} = \\frac{10}{2} [ 2 \\times$ 50 + (10 -1 )(-4)]\n$S_{10}$ = 5 [ 100 + (9 )(-4)]\n$S_{10}$ = 5 [ 100 + (-36)]\n$S_{10}$ = 5 (64)\n$S_{10}$ = 320\n\n2) 3, $\\frac{9}{2}$, 6, $\\frac{15}{2}$ ,\u2026 25 terms .\nSolution: 3, $\\frac{9}{2}$, 6, $\\frac{15}{2}$ ,\u2026 25 terms .\nThe formula to find sum is\n$S_{n} = \\frac{n}{2}$ [ 2a + (n -1 )d]\n\nNumber of terms = n = 25; First term = a = 3 ; Common difference = d = , $\\frac{9}{2}$ - 3 = $\\frac{3}{2}$\nPut all the given values in the formula we get,\n$S_{25} = \\frac{25}{2} [ 2 \\times 3 + (25 -1) \\frac{3}{2}$]\n$S_{25}$ = 12.5 [ 6 + (24 )(1.5)]\n$S_{25}$ = 12.5 [ 6 + 36]\n$S_{25}$ = 12.5 (42)\n$S_{25}$ = 525\n\n3) In an A.P. the sum of first n terms is $\\frac{3n^{2}}{2} + \\frac{13}{2}n$. Find its 25th term.\nSolution : $S_{n} = \\frac{3n^{2}}{2} + \\frac{13}{2}n$.\n\nWhen n = 25,\n$S_{25} = \\frac{3 \\times 25^{2}}{2} + \\frac{13}{2} \\times25$.\n\n$S_{25}$ = 1100\nNow we will find $S_{n - 1} = S_{25 -1} =S_{24}$\n$S_{24} = \\frac{3 \\times 24^{2}}{2} + \\frac{13}{2} \\times24$.\n\n$S_{24}$ = 1080\nAs we know that ,\n$a_{n} = S_{n} - S_{n -1}$\n$a_{25} = S_{25} - S_{24}$\n$a_{25}$ = 1100 \u2013 1080 = 80\n\u2234 25th term is 80.\n\nFrom sum of arithmetic progression to Home\n\nWe at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.\n\nWe also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.\n\nAffiliations with Schools & Educational institutions are also welcome.\n\nPlease reach out to us on [email\u00a0protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit\n\nWe will be happy to post videos as per your requirements also. Do write to us.\n\nRussia-Ukraine crisis update - 3rd Mar 2022\n\nThe UN General assembly voted at an emergency session to demand an immediate halt to Moscow's attack on Ukraine and withdrawal of Russian troops.", "date": "2022-06-25 05:11:11", "meta": {"domain": "ask-math.com", "url": "https://www.ask-math.com/sum-of-arithmetic-progression.html", "openwebmath_score": 0.4438581168651581, "openwebmath_perplexity": 1885.6293264170595, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846640860382, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8585098382333731}}
{"url": "http://math.stackexchange.com/questions/401850/a-simple-riddle-related-to-addition-of-odd-numbers", "text": "# A simple riddle related to addition of odd numbers\n\nI'm not sure if this type of question can be asked here, but if it can then here goes:\n\nIs it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.\n\nPS : The inception of this question is a result of a random discussion that I was having during the break hour :)\n\n-\nHint: adding an odd number of odd numbers will give ... (is this a real question ?) \u2013\u00a0Raymond Manzoni May 25 '13 at 8:27\nHmm..let me ponder over that \u2013\u00a0403 Forbidden May 25 '13 at 8:29\nThis says sum of 9 odd numbers can't be even algebra.com/algebra/homework/word/numbers/\u2026 True? \u2013\u00a0403 Forbidden May 25 '13 at 8:32\nYes: consider the last digit in binary if it helps. \u2013\u00a0Raymond Manzoni May 25 '13 at 8:36\n\nA direct approach:\n\nAny given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like $${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$ (the integers $a,b,\\ldots,i$ may or may not be the same). Grouping things together, this is equal to $$2(a+b+c+d+e+f+g+h+i+4)+1.$$ Thus, the result is odd.\n\nAn simpler approach would be to prove these three simple facts: \\begin{align*} \\mathsf{odd}+\\mathsf{odd}&=\\mathsf{even}\\\\ \\mathsf{odd}+\\mathsf{even}&=\\mathsf{odd}\\\\ \\mathsf{even}+\\mathsf{even}&=\\mathsf{even} \\end{align*} Thus, starting from $$\\mathsf{odd}+\\mathsf{odd}+\\mathsf{odd}+\\mathsf{odd}+\\mathsf{odd}+\\mathsf{odd}+\\mathsf{odd}+\\mathsf{odd}+\\mathsf{odd}$$ and grouping into pairs, $$\\mathsf{odd}+(\\mathsf{odd}+\\mathsf{odd})+(\\mathsf{odd}+\\mathsf{odd})+(\\mathsf{odd}+\\mathsf{odd})+(\\mathsf{odd}+\\mathsf{odd})$$ we use our facts to see that this is $$\\mathsf{odd}+\\mathsf{even}+\\mathsf{even}+\\mathsf{even}+\\mathsf{even}.$$ Grouping again, $$\\mathsf{odd}+(\\mathsf{even}+\\mathsf{even})+(\\mathsf{even}+\\mathsf{even})$$ becomes $$\\mathsf{odd}+\\mathsf{even}+\\mathsf{even}$$ becomes $$\\mathsf{odd}+(\\mathsf{even}+\\mathsf{even})=\\mathsf{odd}+\\mathsf{even}= \\mathsf{odd}$$\n\n-\nI'm sorry. I don't think I understood your query. \u2013\u00a0403 Forbidden May 25 '13 at 8:34\nShouldn't that be +9? \u2013\u00a0Ataraxia May 25 '13 at 8:35\nOr just $$\\mathsf{\\underbrace{odd+odd}+odd+odd+odd+odd+odd+odd+odd}\\\\= \\mathsf{\\underbrace{even+odd}+odd+odd+odd+odd+odd+odd}\\\\= \\mathsf{\\underbrace{odd+odd}+odd+odd+odd+odd+odd}\\\\= \\mathsf{\\underbrace{even+odd}+odd+odd+odd+odd}\\\\= \\mathsf{\\underbrace{odd+odd}+odd+odd+odd}\\\\= \\mathsf{\\underbrace{even+odd}+odd+odd}\\\\= \\mathsf{\\underbrace{odd+odd}+odd}\\\\= \\mathsf{\\underbrace{even+odd}}\\\\= \\mathsf{odd}$$ \u2013\u00a0Rahul May 25 '13 at 8:43\n@Rahul: Very clever user of MathJax! I approve :) \u2013\u00a0Zev Chonoles May 25 '13 at 8:44\n@403Forbidden you can always select it (the checkbox under the voting arrows) if you found it most helpful. \u2013\u00a0Ataraxia May 25 '13 at 8:51\n\n$$\\sum_{k=1}^{9}(2n_k+1)= 2\\sum_{k=1}^9n_k+9$$\n\n$$41=2\\sum_{k=1}^9n_k$$\n\nThere is no integer $\\sum_{k=1}^9n_k$ that satisfies this.\n\n-\nBut my colleague says otherwise. And he refuses to let me in on the solution until the end of day..which is what is killing me. Argh! These math riddles are so addictive \u2013\u00a0403 Forbidden May 25 '13 at 8:33\n@403Forbidden he would have to demonstrate that it is possible for the sum of a set of integers to be a non-integer, and if he could do that he's worthy of an award. \u2013\u00a0Ataraxia May 25 '13 at 8:34\nAs it turns out, it was just a trick question. He admitted to this (getting a result of 50) being not possible :) \u2013\u00a0403 Forbidden May 25 '13 at 9:09", "date": "2016-05-30 10:56:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/401850/a-simple-riddle-related-to-addition-of-odd-numbers", "openwebmath_score": 0.6863880157470703, "openwebmath_perplexity": 646.1539475248098, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9783846716491915, "lm_q2_score": 0.877476785879798, "lm_q1q2_score": 0.8585098370327942}}
{"url": "https://math.stackexchange.com/questions/4206636/probability-of-guessing-the-correct-password", "text": "probability of guessing the correct password\n\nSuppose I need to guess a password of one digit. Each time I'm wrong the password increases by another digit. What's the probability that I can correctly guess the password (assuming I have unlimited amount of time and the number is all uniform)?\n\nMy approach: since the number is generated random, it doesn't matter what strategy we adopt to guess the number. So consider the following sequence of guess $$1, 21, 221, 2221, 22221, ...$$. The probability that the password results in $$k$$-th guess is then $$\\frac{1}{10^n}$$. Thus the total probability is $$\\sum_n \\frac{1}{10^n}$$. So the probability is $$0.11111111111... = \\frac{1}{9}$$.\n\nIs my approach correct, and if so, is there any better solution for this. The fraction $$\\frac{1}{9}$$ seems to suggest an easier perspective to look at this puzzle.\n\n\u2022 The probability of succeeding on the $k$th guess must account for failing at the $1$st, $2$nd, ... , $k-1$th guess. I don't think this has been taken into account. For example, if you have to find $21$, you have to first fail at $2$ : so the probability of succeeding at the second turn is not $\\frac 1{100}$ but rather $\\frac{9}{10} \\times \\frac{1}{100}$, because you need to fail the first time. Also realize another thing : let's say you call out $5$ and fail. Then you KNOW that when you add a digit , the resulting number doesn't start with $5$ (so it can't be e.g. $54$). Strategy improved. Jul 25 at 6:31\n\u2022 I don't understand your comment. If I succeed at the second turn with the guess $21$, then that means the first digit is $2$, which I guessed wrong by saying $1$, and the second digit is $1$, which I guessed right. So the probability here is $\\frac{1}{100}$. Because the first digit is $2$ so that should already include the events that I guess the first one wrong Jul 25 at 7:03\n\u2022 If you guessed $2$ and got it wrong, then you know the number cannot be $2$, so it can be any of the others. But once a digit is added, you know that the password cannot be , say $20,21,23,27$ etc., because if it was, then the previous answer would have had to be $2$, and you would have been correct. So you do get information from wrong guesses. From wrong guesses, you know , for example what the future password cannot start with : your guess that was called wrong. Jul 25 at 7:06\n\nWhat is the probability that you keep failing indefinitely?\n\nProbability that you fail first time is $$\\cfrac{9}{10}$$\n\nProbability that you fail twice in a row? Now here I assume that the person guessing the password can keep track of what they guessed and does not choose from the one's that are obviously incorrect.\n\nSay you guessed $$2$$ first time and it was wrong. Now one digit gets added. So you need to guess from $$100$$ numbers ($$00 - 99$$) but you also know it cannot be any number between $$20$$ and $$29$$.\n\nSo probability that you fail twice in a row is,\n\n$$\\cfrac{9}{10} \\cdot \\cfrac{89}{90} = \\cfrac{89}{100}$$\n\nNow for the third guess, say you guessed $$2$$ the first time and $$18$$ the second time and failed both times, you know the three digit number is not between $$200 - 299$$ and you also know it is not in $$180 - 189$$.\n\nProbability that you fail thrice in a row is,\n\n$$\\cfrac{9}{10} \\cdot \\cfrac{89}{90} \\cdot \\cfrac{889}{890} = \\cfrac{889}{1000}$$\n\nYou see the denominator of the third is divisible by numerator of the second, denominator of the second by the numerator of the first?\n\nEventually at the end of $$n$$ guesses, probability that you failed in all of them is\n\n$$P(F) = \\cfrac{10^n - 10^{n-1} - 10^{n-2} ... - 10^0}{10^n}$$\n\nSo probability that you succeed in $$n$$ guesses,\n\n$$P(S) = 1 - P(F) = \\cfrac{10^n - 1}{9 \\cdot 10^n}$$\n\nNow as $$n \\to \\infty$$, what do you get for $$P(S)$$?\n\n\u2022 +1, I know it's kind of \"obvious\" that making random guesses (with the idea of eliminating previous guesses) is the best strategy, but I've never actually seen a mathematical proof that \"XYZ strategy is the best for ABC problem\". I think I'll attach one when I see it here. I mean, we all kind of understand that symmetry means that any possibility is likely, but having it out there would still be nice. There must be like this \"space\" or set of strategies, on which we can put a distance and have like a \"best-fit\" criterion which works here. Very interesting, though! Jul 25 at 7:42\n\u2022 @TeresaLisbon thank you. yes for the purpose of this question, I assume guesses are random with elimination of previous guesses. It will definitely be interesting to see if there are strategies that can better random guess in this setting. Jul 25 at 7:55\n\u2022 Yes, I do not think that there is a better strategy than random guessing with the kind of elimination you make. I will look for sources anyway and let you know, thanks for the response. Jul 25 at 7:57", "date": "2021-09-20 06:07:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4206636/probability-of-guessing-the-correct-password", "openwebmath_score": 0.9145005345344543, "openwebmath_perplexity": 191.87244699238877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846628255123, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8585098245879779}}
{"url": "https://mathoverflow.net/questions/363013/analyze-a-function-defined-in-terms-of-an-integral/363015", "text": "# Analyze a function defined in terms of an integral\n\nHere is a question that really has puzzled me for quite a while. I happened to see this function defined in terms of an integral $$f(x):=\\int_0^{\\pi/2}\\frac{2e^{x+e^x\\cos y}}{1+\\left(e^{e^x\\cos y}\\right)^2}dy.$$I want to analyze the behavior of the function when $$x \\rightarrow \\infty$$.\n\nThe strange thing is that when I used Mathematica to plot the function, the graph indicates that $$\\lim_{x\\rightarrow \\infty} f(x)=0$$. However, it is easy to see that $$\\liminf_{x\\rightarrow \\infty}f(x) \\ge \\frac{\\pi}{4}$$, since $$\\int_0^{\\pi/2}\\frac{2e^{x+e^x\\cos y}}{1+\\left(e^{e^x\\cos y}\\right)^2} \\, dy \\ge \\int_0^{\\pi/2}\\frac{2e^{x+e^x\\cos y}\\sin y}{1+\\left(e^{e^x\\cos y}\\right)^2}\\, dy\\\\ =-\\Big(\\tan^{-1}\\left(e^{e^x \\cos y}\\right)\\Big)\\Big|_{0}^{\\pi/2}\\\\=\\tan^{-1}\\left(e^{e^x}\\right)-\\pi/4$$\n\nNow I have two questions:\n\nFirst, why the result from Mathematica is different from what I obtained?\n\nSecond, does $$\\lim_{x\\rightarrow \\infty} f(x)$$ exist?\n\nMaybe this question is not so suitable for mathoverflow, since it is just a calculus problem. However, I just feel so confused about the contradiction of numerical result and math. I want to understand the reason behind this situation. Any comments are really appreciated. Thank you very much.\n\nBelow is the code and picure I got from Mathematica....\n\n\u2022 you probably made a coding error in Mathematica, when I plot it the limit is about 1.57 \u2013\u00a0Carlo Beenakker Jun 14 at 6:58\n\u2022 You missed $2$ in antiderivative, so $\\liminf$ is at least $\\frac{\\pi}{2}$ (which coincides with the number Carlo wrote). I bet that it won't be hard to prove that the the loss in multiplying by $\\sin (y)$ is negligible as $x\\to \\infty$ and so the limit is actually $\\frac{\\pi}{2}$. \u2013\u00a0Aleksei Kulikov Jun 14 at 7:18\n\u2022 @CarloBeenakker, thank you very much for the comment... As you can see from my updates, my code should be fine, but when I tried to plot the the graph for $x \\in [0,20]$, something went wrong... \u2013\u00a0student Jun 14 at 13:31\n\u2022 @AlekseiKulikov, exactly! Thank you very much! \u2013\u00a0student Jun 14 at 13:31\n\nMathematica seems to be plotting the function just fine...\n\nIf we look a bit at the integrand, it's clear that most of the mass is around $$y = \\pi/2$$ as $$x$$ increases which should let us introduce a $$\\sin y$$ term and use the antiderative you've already found.\n\nWe can try to cut the integral at $$\\pi/2 - 1/x$$.\n\nLet $$I = \\int_0^{\\pi/2 - 1/x} 2 e^x \\frac{e^{e^x \\cos y}}{1 + e^{2 e^x \\cos y}} dy + \\int_{\\pi/2 - 1/x}^{\\pi/2} 2 e^x \\frac{e^{e^x \\cos y}}{1 + e^{2 e^x \\cos y}} dy = I_0 + I_1$$\n\nNotice that $$f(u) = e^u / (1 + e^{2u})$$ is a decreasing function of $$u$$ and thus that $$I_0 < (\\pi/2 - 1/x) 2 e^x \\frac{e^{e^x \\cos (\\pi/2-1/x)}}{1 + e^{2 e^x \\cos (\\pi/2-1/x)}}$$\n\nWhen $$x \\rightarrow \\infty$$ the $$\\cos (\\pi/2 - 1/x)$$ behaves as $$1/x$$ and the logistic function $$1-\\sigma(u)$$ behaves as $$e^{-u}$$, so the right term behaves as $$\\pi e^{x - e^x /x}$$ which converges to $$0$$.\n\nFor $$I_1$$, we note that if $$y \\in [\\pi/2-1/x,\\pi/2]$$, $$1 - \\frac{1}{2x^2} < \\sin y \\leq 1$$\n\n$$I_1 \\left(1-\\frac{1}{2x^2}\\right)< \\int_{\\pi/2-x}^{\\pi/2} 2 e^x \\frac{e^{e^x \\cos y}\\sin y}{1 + e^{2 e^x \\cos y}} \\leq I_1$$\n\nThe middle term converges to $$\\pi /2$$ and thus so does $$I_1$$ and so does $$I$$.\n\n\u2022 Thank you very much! By the way, I got the same picture for $x$ up to 10, but when I plotted the graph for $x \\in [0,20]$, it gives me the limit going to $0$, as you can see from my updates.... \u2013\u00a0student Jun 14 at 13:26\n\u2022 The support of the mass becomes too small and the numerical integration misses it. If you do a change of variable which blow up the region around pi/2 it'll be more stable. \u2013\u00a0Arthur B Jun 14 at 13:44", "date": "2020-10-22 12:05:42", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/363013/analyze-a-function-defined-in-terms-of-an-integral/363015", "openwebmath_score": 0.8589870929718018, "openwebmath_perplexity": 150.1484365170523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846640860382, "lm_q2_score": 0.8774767746654976, "lm_q1q2_score": 0.8585098194244031}}
{"url": "http://math.stackexchange.com/questions/20593/calculate-variance-from-a-stream-of-sample-values", "text": "# Calculate variance from a stream of sample values\n\nI'd like to calculate a standard deviation for a very large (but known) number of sample values, with the highest accuracy possible. The number of samples is larger than can be efficiently stored in memory.\n\nThe basic variance formula is:\n\n$\\sigma^2 = \\frac{1}{N}\\sum (x - \\mu)^2$\n\n... but this formulation depends on knowing the value of $\\mu$ already.\n\n$\\mu$ can be calculated cumulatively -- that is, you can calculate the mean without storing every sample value. You just have to store their sum.\n\nBut to calculate the variance, is it necessary to store every sample value? Given a stream of samples, can I accumulate a calculation of the variance, without a need for memory of each sample? Put another way, is there a formulation of the variance which doesn't depend on foreknowledge of the exact value of $\\mu$ before the whole sample set has been seen?\n\n-\nThis very same issue was discussed on dsp.SE, and the answers were very similar, with numerically unstable methods proposed in one answer, and more stable methods described by others. \u2013\u00a0 Dilip Sarwate Mar 4 '12 at 17:04\n\nYou can keep two running counters - one for $\\sum_i x_i$ and another for $\\sum_i x_i^2$. Since variance can be written as $$\\sigma^2 = \\frac{1}{N} \\left[ \\sum_i x_i^2 - \\frac{(\\sum_i x_i)^2}{N} \\right]$$\n\nyou can compute the variance of the data that you have seen thus far with just these two counters. Note that the $N$ here is not the total length of all your samples but only the number of samples you have observed in the past.\n\n-\nWait -- shouldn't the N be inside the the expected values, such that the right side is divided by $N^2$? \u2013\u00a0 user6677 Feb 5 '11 at 22:05\ne.g. $$\\sigma^2 = \\frac{(\\sum_i x_i^2)}{N} - (\\frac{\\sum_i x_i}{N})^2$$ \u2013\u00a0 user6677 Feb 5 '11 at 22:27\n@user6677: You are indeed right. Thanks for the correction. \u2013\u00a0 Dinesh Feb 5 '11 at 22:30\nNote the sum of squares especially is at risk of overflow. \u2013\u00a0 dfrankow Dec 29 '14 at 18:06\n\nI'm a little late to the party, but it appears that this method is pretty unstable, but that there is a method that allows for streaming computation of the variance without sacrificing numerical stability.\n\nCook describes a method from Knuth, the punchline of which is to initialize $m_1 = x_1$, and $v_1 = 0$, where $m_k$ is the mean of the first $k$ values. From there,\n\n\\begin{align*} m_k & = m_{k-1} + \\frac{x_k - m_{k-1}}k \\\\ v_k & = v_{k-1} + (x_k - m_{k-1})(x_k - m_k) \\end{align*}\n\nThe mean at this point is simply extracted as $m_k$, and the variance is $\\sigma^2 = \\frac{v_k}{k-1}$. It's easy to verify that it works for the mean, but I'm still working on grokking the variance.\n\n-\n+1, excellent! I didn't feel a simple upvote would be sufficient to express my appreciation of this answer, so there's an extra 50 rep bounty coming your way in 24 hours. \u2013\u00a0 Ilmari Karonen Mar 4 '12 at 20:12\nIsn't the final variance calculation should be Vk / k (not k-1)? \u2013\u00a0 errr Oct 10 '13 at 15:49\n\ntldr: I think there is a missing $(k-1)$ term in @Dan's variance calculation. I think it should be $$\\sigma^2_k = \\frac{1}{k} \\left( (k-1) \\sigma^2_{k-1} + (x_k-m_{k-1})(x_k-m_k)\\right)$$\n\nValidation: I tried what Dan's formula but am getting something different from Matlab's var command. For example, take 1000 Gaussian numbers (mean 0, var 100) and compute their mean and var:\n\nclear; clc\nn=1000;\nx=round(randn(1,n)*10);\n\n% calcs\nmm(1)=x(1);\nvm(1)=0;\nm(1)=x(1);\nv(1)=0;\nma(1)=x(1);\nva(1)=0;\nfor k=2:n\nmm(k) = mean(x(1:k)); %matlab mean\nvm(k) = var(x(1:k),1); %matlab var\n\nm(k) = m(k-1) + 1/k * (x(k)-m(k-1)) ; %Knuth mean\nv(k) = (1/k) * (v(k-1) + (x(k)-m(k-1)) * (x(k)-m(k))) ; %Knuth var\n\nma(k) = (1-1/k) * ma(k-1) + (1/k) * x(k); %Finch mean\n%va(k) = (1-1/k) * (va(k-1) + (1/k) * (x(k)-ma(k-1))^2 ); %Finch 143\nva(k) = (1/k) * ((k-1)*va(k-1) + (x(k)-m(k-1)) * (x(k)-m(k)) ); %Finch 143 rearranged to look like Knuth\nend\n\n% plots\nfigure(1); clf\nsubplot 211\nplot(1:n, x, 'k--')\nhold on\nplot(1:n,mm, 'b-')\nplot(1:n,m, 'g-')\nplot(1:n,ma, 'r--')\nhold off\nylabel('means')\nlegend('x','Matlab','Knuth','Finch')\n\nsubplot 212\nplot(1:n,vm, 'b-')\nhold on\nplot(1:n,v, 'g-')\nplot(1:n,va, 'r--')\nhold off\nylabel('vars')\nxlabel('steps')\nlegend('Matlab','Knuth','Finch')\n\n\nAs you can see I selected Matlab's (octave's) VAR(X,1) command which divides by $N$ and not by $(N-1)$, which goes back to @errr's comment - it has to do with using biased vs unbiased estimator (I usually go to biased one because it makes more sense to me)\n\nSo it seems like Knuth variance is ok for a couple of steps and then it's way below because of the missing $(k-1)$ term.\n\nSo what I tried then is based on Tony Finch's explanation of recursive variance (eq (143)) with $\\alpha=\\frac{1}{k}$. Admittedly it's a hack as my alpha changes with every step, whereas his is assumed to be constant (?), but in the end this formula agrees with matlab's var(X,1).\n\n-\nIf you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. \u2013\u00a0 Mark Fantini Jul 14 '14 at 22:56\nI guess it was unclear - I found a mistake in previous answer. added a small summary to say that up front. Thank you! \u2013\u00a0 alexey Jul 15 '14 at 0:01", "date": "2015-02-01 22:02:42", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/20593/calculate-variance-from-a-stream-of-sample-values", "openwebmath_score": 0.8892979025840759, "openwebmath_perplexity": 1283.0796708272483, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9518632261523027, "lm_q2_score": 0.9019206686206199, "lm_q1q2_score": 0.8585051173666652}}
{"url": "http://bresso5stelle.it/yfbo/probability-urn-problems.html", "text": "# Probability Urn Problems\n\nYou don't know which balls where inserted, but as I go along picking balls and showing them to you, what can be said about the number of black and white balls in the urn? That's inverse probability, although it's an old term. An urn contains {eq}8 {/eq} white and {eq}6 {/eq} green balls. A ball is chosen at random from urn #1. , without any aids, but don\u2019t worry about a time limit). Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. A copy of the source for Grinstead and Snell's lovely probability book - tdunning/probability-book. Five marbles are randomly selected, with replacement, from the urn. For example, we assumed $$\\p(B_1 \\given \\neg A) = 1/2$$ because the. Let A = the event that the first marble is black; and let B = the event that the second. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, Andr\u00e9 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. Thus, the probability that they both give the same answer is 39. Compute the probability that (a) the rst 2 balls selected are black and the next 2 are white. We assume a ball from the first urn is randomly picked and then placed into the second urn, then another ball from the second urn is randomly picked and then placed into the third urn, and so on, until a ball from the last urn is finally. Question 12 of 20. Two marbles are randomly and simultaneously drawn from the urn. For example, a marble may be taken from a bag with 20 marbles and then a second marble is taken without replacing the first marble. person_outline Timur schedule 2018-01-04 15:12:17. Selected for originality, general interest, or because they demonstrate valuable techniques, the problems are ideal as a supplement to courses in probability or statistics, or as stimulating recreation for the mathematically minded. Due to rapid growth in the field in recent years, this volume aims to promote interdisciplinary collaboration in the areas of quantum probability, information, communication and foundation, and mathematical physics. Show that the probability of rolling doubles with a non-fair (\u201c\ufb01xed\u201d) die is greater than with a fair die. In SAS you can use the table distribution to specify the probabilities of selecting each integer in the range [1, c]. Two of the selected marbles are red, and three are green. A Collection of Dice Problems Matthew M. $\\begingroup$ One way to do this is with the transfer matrix method. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. However, let us now change the experiment and suppose that at 1 minute to 12 P. The probability that a customer will buy a product given that he or she has seen an advertisement for the product is 0. One ball is drawn at random and its color noted. 57 % of the children in Florida own a bicycle. If every vehicle is equally likely to leave, find the probability of: a) van leaving first. An urn B 1 contains 2 white and 3 black chips and another urn B 2 contains 3 white and 4 black chips. Two red, three white, and five black. Task There are urns labeled , , and. Such an inverse probability is called a Bayes probability and may be obtained by a formula that we shall develop later. \\dfrac12\\cdot\\dfrac12=\\dfrac14 21. The probability that a consumer will see an ad for this particular product is 0. The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). , TTHor HHHT). 1 2 \u22c5 1 2 = 1 4. Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Molina's urns. Ageless wonder: Frank Gore \u2014 who will be 37 \u2014 signs one-year deal with Jets. Urn 1 contains 2 white balls and 2 black balls. An urn contains n red and m black balls. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. A second urn contains 16 red balls and an unknown number of blue balls. Another urn contains 2 red chips. If your opponent draws first, what is the probability that you win?. The flippant juror. In the \ufb01rst draw, one ball is picked at random and discarded without noticing its colour. For example, the probability of getting two \"tails\" in a row would be:. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. Are the events w = 3 and w= r dependent or independent? Problem 2 (10 points) Urn A contains 4 white balls and 8 black balls. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). If a red or white ball is chosen, a fair coin is flipped once. Two red, three white, and five black. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The Sock drawer. Conditional Probability and the Multiplication Rule It follows from the formula for conditional probability that for any events E and F, P(E \\F) = P(FjE)P(E) = P(EjF)P(F): Example Two cards are chosen at random without replacement from a well-shu ed pack. Measure-valued P\u00f3lya urn processes Mailler, C\u00e9cile and Marckert, Jean-Fran\u00e7ois, Electronic Journal of Probability, 2017 Optimal stopping rule for the no-information duration problem with random horizon Tamaki, Mitsushi, Advances in Applied Probability, 2013. What is the probability that the marble chosen is red? Example 8: Box #1 contains 9 violet and 7 white marbles. Question 290480: There are 3 urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls respectivelly. A small probability calculator / tester for urn models (urn problem) - exane/urn-probability-calculator. Note that to define a mapping from A to B, we have n options for f ( a 1), i. (4 balls are now in urn C. The probability the first is white is 6/15= 2/5. It is commonly used in randomized controlled trials in experimental research. The simplest experiment is to reach into the urn and pull out a single ball. Column B contains the six. If playback doesn't begin shortly, try restarting your device. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. (b) 2marbles are selected without replacement. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Find the probability of the event that at least 5 tosses are required. In Urn i there are i black balls and one. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. Second, an argument that generalizes from observed instances is similar to an urn problem, where we guess the contents of the urn by repeated sampling. If N is large enough (say, N=20), the probability of either of those events is much less than 1%. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. The locomotive problem. ) Jacob Bernoulli 1700. If there are N urns, two colours, and balls are coloured independently with probability 0. n = [3 \u00d7 seed /9999] + 1. The Urn Problem with Indistinguishable Balls. Find the probability that at least one ball of each color is chosen. A well-known result of this type is Polya's urn problem (see [8]), which is just the. An urn contains 75 balls, some red, some blue. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. To determine whether to accept the shipment of bolts,the manager of the facility randomly selects 12 bolts. A basic problem first solved by Jakob Bernoulli is to find the probability of obtaining exactly i red balls in the experiment of drawing n times at random with replacement from an urn containing b black and r red balls. For example, here is a three-part problem adapted from mathforum. So the probability of moving from state 3 to state 4 is 2=5 2=5. Naturally, the problems on this site are perfect for a blog such as this, so this is the first of many interesting such problems that I will post \ud83d\ude42 This particular exercise is a probability problem that will appeal to anyone that likes games involving dice. Problems and Complete explanatory solutions to problems on probability involving drawing, picking, selecting, choosing two or more balls from a box, bag, urn, container. Please read our cookie policy for more information about how we use cookies. One ball is chosen randomly from the urn. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Each turn, we pick out a ball, note it's colour then replace it and add d more balls of the same colour into the urn. What's the probability of the event A = {the sum and the product of the numbers that come up are equal}? Problem 8 From an urn, which contains 10 white, 7 green and 6 red balls, 1 ball is taken out. lf X = 2 choose with replacement 6 balls from Urn B. For each new ball, with probability p, create a new bin and place the ball in that bin; with probability 1 \u2212 p, place the ball in an existing bin, such that the probability the ball is placed in a bin is proportional to m \u03b3,wheremis the number of balls in that bin. Examples 8-1 through 8-10 deal with some of the more impor-tant sorts of questions one may ask about drawings without replacement from such an urn. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. of selecting 1 st urn \u00d7 prob. org 21 Even Knight\u2019s a priori probabilities\u2014those based on some symmetry of a problem\u2014are sus-pect. The probability that it is either a black ball or a green ball is ; A box contains 150 bolts of which 50 are defective. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. You can also express this relationship as 1 \u00f7 6, 1/6, 0. Therefore, one student must have solved at least 5 problems. But anyways using the binomial theorem. Ron chooses three re\u00a9hree blue marbles 3a3 + [2 marks] [2 marks] 3) Which of the following numbers cannot be the probability of some. You and your friend take turns randomly picking a ball from the urn. Problem 1. Defining Risk November/December 2004 www. Durrett, The Essentials of Probability, Duxbury Press, 1994 S. Let X be the number of red balls you pull before any black one, and Y the. Example An urn contains 6 red marbles and 4 black marbles. There is equal probability of each urn being chosen. Example : Probability to pick a set of n=10 marbles with k=3 red ones (so 7 are not red) in a bag containing an initial total of N=100 marbles with m=20 red ones. Math: Conditional Probability. To see this, fix an urn. 1702-1761) was a Presbyterian minister. If the probability of an event is 0, it is impossible for that event to occur. The two events are independent so the probability of a blue marble from each urn is the product. At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. Exercises in Probability Theory Nikolai Chernov All exercises (except Chapters 16 and 17) are taken from two books: R. Four balls are drawn at random. One ball is drawn at random. This Web site is a course in statistics appreciation; i. Time is the main factor in competitive exams. What is the probability that they are both of the same color? b. Originally, the urn contains 6 white and 9 black balls, total of 15. After the nth step, Urn 2 is empty. Events that are unlikely will have a probability near 0, and events that are likely to happen have probabilities near 1. However, these are not all given equal probability when you take 20 objects from an urn with 10 of each color. This activity shows the classic marble example of elementary probability. Problem 1. The objective probability that a random draw from an urn yields a black ball changes over time if the urn has a hole through which its mixture of black and red balls spills. So, in a Blackjack game, to calculate the chances of getting a 21 by drawing an Ace and then a face card, we compute the probability of the first being an Ace and multiply by the probability of drawing a face card or a 10 given that the first was an Ace: $1/13 \\t imes 16/51 \\a pprox 0. Problem 1 An urn contains n+m balls, of which n are red and m are black. choose a ball from an urn and record its color, then do it again; flip a coin and record Head or Tail, then choose a ball from an urn and record its color The branches emanating from any point on a tree diagram must have probabilities that sum to 1. An urn contains n+m balls, of which n are red and m are black. Urn contains red balls and black balls. Category:Probability theory. I am trying to solve this problem: r balls are randomly assigned into n urns. (Some of them might appear on our problem sheets. What is the probability that the urn contains three balls of each color? 5) If x, y and z are positive real numbers satisfying +1 =4, +1 =1, and +1 =7 3, then what is the value of xyz? Relation Problems 6) Create a set A of people you know well. Divide the number of events by the number of possible outcomes. To win, all six numbers must match those chosen from the urn. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. And indeed, the induced frequencies do converge to the Dirichlet distribution with k equal parameters. One ball is picked at random from urn 1 and, without. Container in many probability-theory problems is a crossword puzzle clue that we have spotted 1 time. What is the probability that they are both of the same color? b. The topic of statistics is presented as the application of probability to data analysis, not as a cookbook of statistical recipes. Suppose that there are 71 urns given, and that balls are placed at random in these urns one after the other. what is the probability that the equipment will fail before the end of one year? 2. The multinomial theorem is a statement about expanding a polynomial when it is raised to an arbitrary power. Please justify your answers and don't simply give the answer. 6 balls are randomly drawn from the urn in succession, with replacement. Sample Problem 1: A six-sided die is rolled six times. Xi;Xj/with i 6Dj has the same distribution. One urn contains 2 blue chips. Conditional Probability 4. An urn contains 1 red ball and 10 blue balls. You choose an urn and your opponent chooses an urn from the remaining ones. Supposethat we win$2 for each black ball selectedand we lose $1 for each white ball selected. What is the probability that a white ball is drawn?. From the first urn to the second 4 balls are moved. lf X = 2 choose with replacement 6 balls from Urn B. If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. 5 and placed in urns uniformly with probability 1/N then the expected number of trials to get a collision is order \u221a \u03c0N. This consists. Suppose that each of Barbara\u2019s shots hits a wooden duck target with probability p1, while each shot of Dianne\u2019s hits it with probability. We move a ball from urn 1 to urn 2. At least I think so. Suppose that there are 71 urns given, and that balls are placed at random in these urns one after the other. b) Find the probability that urn 1 was used given that a red ball was drawn. 24) Don't Lose Your Marbles! 1. Dodgson) is used to illustrate the nature, standing and understanding of probability within the wider English mathematical community of his time. Find the probability that the sum of the two faces is 10 given that one shows a 6. If the composition is unknown, then it is called uncertain. I recommend studying rst, using previous homeworks, exams and exam practice problems. Keywords: Urn problems, drawing without replacement, enumerative combinatorics, Scrabble, R. At least 1 + 2 + 3 = 6 problems were solved by the students mentioned in the problem statement. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. Probabilities of drones with replacement question if your day is rolled five times what is the probability of - Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our website. 3 MB Law of Large Numbers - Urn Problems - Low Resolution. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). At each step we randomly choose a ball from Urn 1, throw it away, and move a red ball from Urn 2 into Urn 1. Scott Boras: Astros players don\u2019t need to apologize. IBPS Clerk, RBI Assistant I Probability Basic Concept & Tricks I Video \u0926\u0947\u0916\u0928\u0947 \u0915\u0947 \u092c\u093e\u0926 \u092f\u0947 Topic Easy \u0939\u0948 - Duration: 37:30. Electronic Journal of Probability, 16, 1723-1749, 2011. If you summarize the results, you will find that the outcome \"2 red, 2 white\" occurs (almost exactly) 6 times as often as the outcome \"4 red\" or \"4 white\". Example : Probability to pick at least once each card from a deck of N=50. b) 2 heads and a tail. The person who selects the third WIN ball wins the game. (2 points each) In each of the scenarios below, indicate whether the distribution of X is binomial, poisson, negative binomial (r >1), geometric, or neither. In what follows, S is the sample space of the experiment in question and E is the event of interest. Probability Definitions: Example #1. You have select each urn with probability$0. \u2022 Put your NAME on each sheet. Urn A has 4 white and 16 red balls. The notion that the probability of an event may depend on other events is called conditional probability The conditional probability of event Agiven event Bis written as P(AjB) For example, in our ball and urn problem, when sampling without replacement: P(R 2) = 1 3 P(R 2jR 1) = 0 P(R 2jRC 1) = 1 2 Patrick Breheny Introduction to Biostatistics. CAT Probability Questions is a sample set of problems that is asked from this topic in the CAT Probability Section. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. ) A pair is not drawn 3) Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. A Collection of Dice Problems Matthew M. Electronic Journal of Probability, 16, 1723-1749, 2011. One ball is picked at random from urn 1 and, without. Hints help you try the next step on your own. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. For more practice, I suggest you work through the review questions at the end of each chapter as well. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. Every minute, a marble is chosen at random from the urn, and then returned to the urn, together with another marble of the same colour. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. \u2022 Time limit 110 minutes. Note that to define a mapping from A to B, we have n options for f ( a 1), i. Probability Problems for Group 1 (Due by EOC Mar. Pull balls from the urn one by one without replacement. Question 3 Solution Urn 1: seven red and three green balls. Thomas Bayes was an English minister and mathematician, and he became famous after his death when a colleague published his solution to the \u201cinverse probability\u201d problem. Let N be the number of throws of a usual six-sided die that is needed for the sum of the scores on these throws to be at least 3. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Urn simulator This calculator simulates urn or box with colored balls often used for probability problems and can calculate probabilities of different events. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. Let the probability that the urn ends up with more red balls be denoted. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. ) I'm happy to discuss about them on my o\ufb03ce hours (unless we are too many on those). Bayes Theorem Practice Problems With Solutions Genetics. Question 3 Solution Urn 1: seven red and three green balls. This limit distribution is the negative binomial distribution with parameters and (the corresponding mathematical expectation is , while the variance is ). The formula is. What is the probability that the ball you drew is green? (b)You then look at the ball and see that it is green. The geometry problems are treated in a separate post. YOU are responsible for studying all the sections to be covered on the midterm. If your opponent draws first, what is the. That is, we seek a random integer n satisfying 1 \u2264 n \u2264 3. , that the coin \ufb02ip was Tails) Let T be the event that the coin \ufb02ip was Tails. \u2022 Bose-Einstein distribution. Let X be the number of red balls removed before the first black ball is chosen. Thus, the probability that they both give the same answer is 39. The probability of an event is a measure of the likelihood that the event will occur. 4 Practice Problems Problem 34. After a severe winter, potholes develop in a state highway at the rate of 5. A risk, on the other hand, is defined to be a higher probability event, where there is enough information to make. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. Alice selects one ball at random (each of the 7 balls can be selected with equal probability) and takes it out of the urn. Ghahramani, Fundamentals of Probability, Prentice Hall, 2000 1 Combinatorics These problems are due on August 24 Exercise 1. A primary concern with PoS systems is the \u201crich getting richer\u201d phenomenon, whereby wealthier nodes are more likely to get elected, and hence reap the block reward, making them even wealthier. We're interested in the probability of a certain observed outcome given a known process. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials. The field of Probability has a great deal of Art component in it - not only is the subject matter rather different from that of other fields, but at present the techniques are not well organized into systematic methods. This paper designs some uncertain urn problems in order to compare probability theory and uncertainty theory. This is exactly the binomial experiment. I'm trying to answer the following question using a simple Monte Carlo sampling procedure in R: An urn contains 10 balls. In these cases, we will need to use the counting techniques from the chapter 5 to help solve the probability problems. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. k indistinguishable balls are randomly distributed into n urns. A room contains four urns. Remarkable selection of puzzlers, graded in difficulty, that illustrate both elementary and advanced aspects of probability. An urn contains pink and green balls. There are 40 marbles in an urn: (PROBABILITY) There are 40 marbles in an urn: 11 are green and 29 are yellow. Let the probability that the urn ends up with more red balls be denoted. What is the probability that both marbles are the same color if: a. b) Find the probability that urn 1 was used given that a red ball was drawn. Edited by Bernard R. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. Two marbles are drawn without replacement from the urn. In the case of rolling a 3 on a die, the number of events is 1 (there\u2019s only a single 3 on each die), and the number of outcomes is 6. Let Event A = {The Ball From Urn No. The occupancy problem in probability theory is about the problem of randomly assigning a set of balls into a group of cells. Clas-sical mathematicians Laplace and Bernoulis, amongst others, have made notable contributions to this. An urn contains 10 balls: 4 red and 6 blue. What strategy maximizes your chance of victory? Problems from Rosen 7. Probability Definitions: Example #1. What is the probability a red marble is drawn?. This formula relies on the helper table visible in the range B4:D10. A Probability trees in probability theory is a tree diagram used to represent a probability space. Acknowledgements This work was made possible by a grant from NSF-DUE and the support of Cornell University and the statistics department at Stanford. If there are N urns, two colours, and balls are coloured independently with probability 0. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. JANOS FLESCH, DRIES VERMEULEN AND ANNA ZSELEVA. Calculate the number of blue balls in the second urn. 1 Is Black}; Event B= {The Ball From Urn. Such problems have been considered by Nishimura and Sibuya [13, 14] and Selivanov [17]. Urn and Beads Problem The next family of problems was modeled after Edwards (1968). What are your chances of getting exactly 4, 5, or 6 matches? Many lotteries and gambling games are based on this concept of picking from mixed good and bad balls. Suppose you have n ball- lled urns, numbered 1 through n. We'll look at a number of examples of modeling the data generating process and will conclude with modeling an eCommerce advertising simulation. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues \u03bb j =1\u22122j/N, where j=0, 1,\u2026, N. Can You Solve This Intro Probability Problem? An urn contains 10 balls: 4 red and 6 blue. Heads, a ball is drawn from Urn 1, and if it is Tails, a ball is drawn from Urn 2. Home-Learning Problems 3. Intro to Probability - Homework Assignment 2 Please solve the problems and type the solutions up using LateX and the template provided. So Urn B is more. $$\\frac{\\textrm{Probability of picking the 2nd urn and picking a blue ball}}{\\textrm{Probability of picking a blue ball}}$$ Since there are equal numbers of balls in each urn, and each urn is equally likely to be picked, you can do this by counting. 2 Real Life Is More Complicated. Practice online or make a printable study sheet. If we draw 5 balls from the urn at once and without peeking,. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. three marbles are drawn from the urn one after the other. An urn contains 10 balls: 4 red and 6 blue. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. 1 2 \u22c5 1 2 = 1 4. Thus in this experiment each time we sample, the probability of choosing a red ball is $\\frac{30}{100}$, and we repeat this in $20$ independent trials. Barbara and Dianne go target shooting. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. The probability of any event can range from 0 to 1. Sample Problem 1: A six-sided die is rolled six times. A card is randomly selected from an ordinary pack of 52 playing cards. 625 subscribers. Access-restricted-item true Addeddate 2011-06-22 16:08:09 Bookplateleaf 0002 Boxid IA1398805 Camera Canon EOS 5D Mark II City Upper Saddle River, NJ Donor. An urn contains 10 red and 8 white balls. what is the probability that the equipment will fail before the end of one year? 2. \u2022 Put your NAME on each sheet. Then another ball is drawn and its color is recorded. (This will be the distribution after a long time if in every second a random urn is chosen, and a ball, if any, from that urn is moved into the clockwise neighboring urn. I've tried two approaches and neither work. An important problem in the Ehrenfest chain is the long-term, or equilibrium, distribution of , the number of balls in urn A. Math 29 \u2014 Probability Practice Second Midterm Exam 1 Instructions: 1. The ticket shows your six good balls, and there are 50 bad balls. I manage to calculate the individual probabilities on a per problem basis, but I need to find a way to phrase a general solution. The Philosophy of Statistics [S]tatistical inference is firmly based on probability alone. An urn contains three red balls numbered 1, 2, and 3, three black balls numbered 8, 9, and 10 and four white balls numbered 4, 5, 6, 7. The probability of this happening is =~ 0. ) are represented as colored balls in an urn or other container. Savage offered the example of an urn that contains two balls: Both may be white; both may be black; or one may be white and. 3, 794-814, 2012. Find P(N = 1), P(N = 2) and P(N = 3). (b) Find EN and Var(N). Probability basics 50 xp Queen and spade 50 xp. We recommend you review today's Probability Tutorial before attempting this challenge. Roll a fair 6-sided die until your first roll of a \u201c3. The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). A ball is selected at random from the \ufb01rst urn and placed in the second. You can also express this relationship as 1 \u00f7 6, 1/6, 0. Imagine two urns. Let us suppose that the urns are labelled with the numbers 1,2,, n and let tj be equal to k if the j-th ball is placed into the k-th urn. A Ball Is Randomly Drawn From Urn No. What is the probability that a white ball is drawn?. In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc. An urn contains 5 red, 3 green, and 4 white balls. What is the probability that neither is red, given that neither is white? 2) A basketball player makes free throws with a 0. Two marbles are selected at random without replacement. 625 subscribers. After the four iterations the urn contains six balls. same answer. |A) is a probability function multiplicative formula: P(B and A) = P(B|A)P(A) Oct 28 Monty Hall problem \u00a73. Bayes probabilities can also be obtained by simply constructing the tree. Then the number of tables is bounded by this multiple, so for large n, the probability of joining one of the k (fixed) tables is roughly , so this should behave roughly like the standard Polya Urn. What is the probability that the urn contains three balls of each color? Solution A tree diagram is ideal to deal with the problem, up to a sample space of 4 balls (so the first two pulls): So, we have a probability of (1/2)*(2/3) + (1/2)*(2/3) = 2/3 of ending up with 3 balls of one color. Then treat this like a mock exam (i. Home-Learning Problems 3. Show that the probability of rolling doubles with a non-fair (\u201c\ufb01xed\u201d) die is greater than with a fair die. Gelbaum DOVER PUBLICATIONS, INC. We write P (heads) = \u00bd. Find P(N = 1), P(N = 2) and P(N = 3). What strategy maximizes your chance of victory? Problems from Rosen 7. The basic urn problem is to determine the probability of drawing one colored ball from an urn with known composition of differently colored balls. What is the probability the ball drawn from urn C is black? I've figured out that P(K\\A) = 2/3 and P(K\\B) = 5/6. , TTHor HHHT). The probability a black ball is drawn is 9/13. In the example shown, the formula in F5 is: =MATCH(RAND(), D$5:D$10) How this formula works. If you pick 5 balls from the urn at random, what is the probability that x of them will be. If a 4 appears, a ball is drawn from urn 1; otherwise, a ball is drawn from urn 2. There will be total 10 MCQ in this test. What is the probability that the coin landed heads? So at first I figured the balls part of the question was irrelevant, and just. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues \u03bb j =1\u22122j/N, where j=0, 1,\u2026, N. of getting white marble). From a given ball's perspective, the probability of being matched is $\\frac{M}{N}(1-(1-\\frac{1}{M})^N)$. The first problem-book of a similar kind as ours is perhaps Mosteller's well-known Fifty Challenging Problems in Probability (1965). Once again, balls are chosen at random with equal probability. At each step, an urn is selected according to their weights. What's the probability that the ball, which was taken out is: a) white b) green c) red Problem 9 An urn contains 8 white and 4 black balls. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. The urn model and the P\u00f3lya process, in which the P\u00f3lya distribution and the limit form of it arise, are models with an after effect (extracting a ball of a particular colour from the urn increases the probability of extracting a ball of. He titled it The Two Children Problem, and phrased. User Account. of selecting 2 nd urn \u00d7 prob. Walk through homework problems step-by-step from beginning to end. There are three ways to measure the average: the mean, median, and mode. Hume\u2019s Problem. Create a set B of foods. Six balls are picked from the 56 balls in an urn. Urn #2 contains 5 black and 2 red balls. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. Publication date 1987 Topics Probabilities, Probabilit\u00e9s, Probabilit\u00e9s Publisher Internet Archive Books. Then the probbilities that the urn from which I have drawn the tickets is A or B, are by the principle given above, as K: K'. If one ball is drawn at random, what is the probability of getting a red or a white ball? 3. The Type X urns each contain $$3$$ black marbles, $$2$$ white marbles. The first case,. By convention, statisticians have agreed on the following rules. Probability Trees: Many probability problems can be simplified by using a device called a probability tree. The probability for the dice to yield 1 or 2 is 2/6. 8--Conditional Probability With Urns and Marbles Glenn Olson. The probability of a sample point is a measure of the likelihood that the sample point will occur. To Probability Practice Problems for Exam # 2 1. One ball is drawn at random. 8--Conditional Probability With Urns and Marbles Glenn Olson. For each new ball, with probability p,createanewbin and place the ball in that bin; with probability 1\u2212p,placetheballin an existing bin, such that the probability the ball is placed in a bin is proportional to m\u03b3,wheremis the number of balls in that bin. An Example Suppose that a 6-sided die is rolled, what is the probability that the result is a 1? From our earlier results, if A is the event that a 1 is rolled, then P(A) = 1 6. Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. The probability the first is white is 6/15= 2/5. It is concluded that uncertainty theory is better than probability theory to. 6 balls are randomly drawn from the urn in succession, with replacement. What is the probability of getting exactly k red balls in a sample of size 20 if the sampling is done with replacement (repetition allowed)? Assume 0 \u2264 k \u2264 20. One ball is drawn at random and its color noted. What is the probability of winning the. Fully worked-out solutions of these problems are also given, but of course you should \ufb01rst try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Binomial probability distribution; Hypothesis testing, types of error, and small samples; Introduction to statistics and the relative frequency histogram; Measures of variability and relative standing ; Other graphical methods and numerical methods; Poisson probability distribution and the urn model; Probability; Random variables. From a given ball's perspective, the probability of being matched is $\\frac{M}{N}(1-(1-\\frac{1}{M})^N)$. For example, if the probability of picking a red marble from a jar that contains 4 red marbles and 6 blue marbles is 4/10 or 2/5, then the probability of not picking a red marble is equal to 1 - 4/10 = 6/10 or 3/5, which is also the. So in a more conventional forward probability problem. What is the probability that it gets a) all 4 aces b) at least 1 ace; Hearts 5 cards are chosen from a standard deck of 52 playing cards (13 hearts) with replacement. Applications of probability arise everywhere: Should you guess. A series of two-urn biased sampling problems Puza, Borek and Bonfrer, Andr\u00e9 2018, A series of two-urn biased sampling problems, Communications in statistics - theory and methods, vol. CAT Probability Questions cover all the different ways in which question can be asked. ) Jacob Bernoulli 1700. of selecting 1 st urn \u00d7 prob. 1 Preliminaries This document is designed to get a person up and running doing elementary probability in R using the prob package. What strategy maximizes your chance of victory? Problems from Rosen 7. Another urn contains 2 red chips. Suppose that a white ball is selected. Column B contains the six. But what is the probability that when you throw a 6 sided unbiased die twice, the second output is 1 given that the first output is 5? Answer is 1/6, because P[first output is 5]=1 since it has already occurred. But anyways using the binomial theorem. Above the plane, over the region of interest, is a surface which represents the probability density function associated with a bivariate distribution. What is the probability that the first ball was also red?. This list of problems should serve as a good place to start studying, and it should not be considered a comprehensive list of problems from the sections we\u2019ve covered. Now there are 13 balls, 4 white and 9 black. So the probability that urn iis empty is 1 11 i 1 i+1 1 1 i+2 1 1 n. A ball is drawn from Urn A and then transferred to Urn B. Every minute, a marble is chosen at random from the urn, and then returned to the urn, together with another marble of the same colour. A ball is also chosen at random from urn #2. It does not matter who picked the first three WIN balls. One ball is drawn from an urn chosen at random. (d) The variance of X. The formula is. Annals of Probability, 41, no. 4 Conditional Probability and Independence 1. 2012 John Wiley & Sons, Inc. Two balls are drawn at random from each one of the urn A and urn B, and are placed into urn C. Let A = { a 1, a 2, a 3,, a m }, B = { b 1, b 2, b 3,, b n }. It is concluded that uncertainty theory is better than probability theory to deal with uncertain urn problems. Four balls are drawn at random. 2: 19, 23, 29, 33. If the composition is unknown, then it is called uncertain. urn problem: when to stop sampling? Then you find the probability that you would have drawn fewer than j white marbles if your urn had a certain proportion, say k, of black marbles. The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution[N, n, m+n]. Find the expected number of stages needed until there are no more black balls in. It is useful in practice, and many elementary statistics textbooks use it to introduce the binomial and hypergeometric distributions. We have included a number of Discussion Topics designed to promote critical. For more practice, I suggest you work through the review questions at the end of each chapter as well. The topic of statistics is presented as the application of probability to data analysis, not as a cookbook of statistical recipes. Find the probability of the event that the \ufb01rst bin contains balls of both colors. In the \ufb01rst draw, one ball is picked at random and discarded without noticing its colour. The flippant juror. Introduction Drawing balls from an urn without replacement is a classical paradigm in probability (Feller 1968). Please justify your answers and don't simply give the answer. I recommend studying rst, using previous homeworks, exams and exam practice problems. Here's how. There are in nitely many Urns which we call Urn 0, Urn 1, Urn 2 etc. One ball is picked at random from urn 1 and, without. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1. CAT Probability Questions is a sample set of problems that is asked from this topic in the CAT Probability Section. An urn contains {eq}8 {/eq} white and {eq}6 {/eq} green balls. Probability of second ball being red = 3/9 (because there are 3 red balls left in the urn, out of a total of 9 balls left. Question 1157976: Urn A contains 5 red marbles and 3 white marbles Urn B contains 2 red marbles and 6 white marbles a. falling apart on inspection. The probability that a consumer will see an ad for this particular product is 0. A ball is drawn at random from an urn containing 15 green, 25 black, 16 white balls. What is the probability that the ball you drew is green? (b)You then look at the ball and see that it is green. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color; (b) of different colors?. (You'd think the urns would be empty by now. A risk, on the other hand, is defined to be a higher probability event, where there is enough information to make. This contains all the class notes for the Introduction to Probability class currently taught at Stanford using these, and other web-resources. Probability Trees: Many probability problems can be simplified by using a device called a probability tree. Combinatorics: The Fine Art of Counting. Bonus Problem: Problem 7. ) are represented as colored balls in an urn or other container. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Find P ( B C | A) from the Venn diagram: 3. Problem One. Suppose that this experiment is done and you learn that a white ball was selected. In this post, I\u2019m going to discuss some of the non-geometry problems. Clue: Container in many probability-theory problems. If you know how to manage time then you will surely do great in your exam. JANOS FLESCH, DRIES VERMEULEN AND ANNA ZSELEVA. The locomotive problem. We shall now apply this principle to the solution of several problems. It is used to solve problems of the form: how many ways can one distribute indistinguishable objects into distinguishable bins? We can imagine this as finding the number of ways to drop balls into urns, or equivalently to arrange balls and. The problem of induction is to find a way to avoid this conclusion, despite Hume's argument. The simplest experiment is to reach into the urn and pull out a single ball. Suppose that a machine shop orders 500 bolts from a supplier. Probability is the likelihood or chance of an event occurring. Urn i has exactly i 1 green balls and n i red balls. If the composition is unknown, then it is called. \u2666 BALL AND URN (AoPS calls this \"Stars and Bars\") The classic \"Ball and Urn\" problem statement is to find the number of ways to distribute N identical balls into 4 distinguishable urns, for example. But our solution to the urn problem relied on random sampling. The Crossword Solver found 21 answers to the Container in many probability theory problems crossword clue. Wolfram Education Portal \u00bb. In the last lesson, the notation for conditional probability was used in the statement of Multiplication Rule 2. The Associated Press. Users that wish to investigate especially large or intricate problems are encouraged to modify and streamline the code to suit their individual needs. The theorem is also known as Bayes' law or Bayes' rule. Publication date 1987 Topics Probabilities, Probabilit\u00e9s, Probabilit\u00e9s Publisher Internet Archive Books. k indistinguishable balls are randomly distributed into n urns. Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4. This consists. 3, 794-814, 2012. \u2022 Time limit 110 minutes. This paper designs some uncertain urn problems in order to compare probability theory and uncertainty theory. EXAMPLE 1 A Hypergeometric Probability Experiment Problem: Suppose that a researcher goes to a small college with 200 faculty, 12 of which have blood type O-negative. (a) Draw marbles from a bag containing 5 red marbles, 6 blue marbles and 4 green marbles without replacement until you get a blue marble. Once you have decided on your answers click the answers checkboxes to see if you are right. 4 Conditional Probability and Independence 1. Hence the probability of getting a red ball when choosing in urn A is 5/8. You decide to actually count the balls in the urn, that you friend has in problem 6, and discover that there are 6 green, 6 yellow, 7 blue, 4 black, and 2 red balls. Some of these matrices will be used in calculation in subsequent posts. In Problem 3, suppose that the white balls are numbered, and let Y i equal 1 if the i th white ball is selected and 0. The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution[N, n, m+n]. Bayes Theorem Practice Problems With Solutions Genetics. a) 4 \u2044 7 b) 3 \u2044 7 c) 20 \u2044 41 d) 21 \u2044 41 View Answer. For example, if you have a bag containing three marbles -- one blue marble and two green marbles -- the. In the first urn, there are $20\\%$ red balls, so the probability to draw a red ball is $0. Winning an unfair game. The first limitation is that the ravens we observe in real life aren't randomly sampled from nature's \"urn\". Thomas Bayes was an English minister and mathematician, and he became famous after his death when a colleague published his solution to the \u201cinverse probability\u201d problem. Ask Question Asked 5 years, 5 months ago. (The two marbles might both be black, or might both be white, or might be of different colors. When we have multiple event probabilities, as in the problem. Now there are 14 balls, 5 white and 9 black. 57 % of the children in Florida own a bicycle. An urn contains 9 red, 7 white and 4 black balls. ) Let N be the number of games played. Probability measures the likelihood of an event occurring. That is, after each draw, the selected ball is returned. balls numbered 1 through 10 are placed in the urn and ball number 1 is withdrawn; at 1 2 minute to 12 P. Each turn, we pick out a ball, note it's colour then replace it and add d more balls of the same colour into the urn. If the probability of an event is 0, it is impossible for that event to occur. Probability (MATH 4733 - 01) Fall 2011 Exam 2 - Practice Problems: Selected answers and hints Due: never Here are some sample problems for you. Among the balls are R red and N-R white balls. These urn models are also excellent practice problems on thinking about Markov chains and deriving the transition probability matrices. Two balls are chosen randomly from an urn containing8 white, 4 black, and 2 orange balls. What is the probability that they come from urns I, II or III? 36. Recommended Problems: Problems 1. Combinatorics: The Fine Art of Counting. The probability of any event can range from 0 to 1. The probability of a sample point is a measure of the likelihood that the sample point will occur. What is the probability that the second card. Again, one ball is drawn at random from the urn, then replaced along with an additional ball of its color. Supposethat we win$2 for each black ball selectedand we lose \\$1 for each white ball selected. Depending on whether we sample with or without replacement, the chance (or probability) of getting m red balls (successes) in a sample of n balls changes. Find the probability that both the first and last balls drawn are black. Two of the selected marbles are red, and three are green. There are related clues (shown below). Can You Solve This Intro Probability Problem? An urn contains 10 balls: 4 red and 6 blue. ) are represented as colored balls in an urn or other container like box. In these cases, we will need to use the counting techniques from the chapter 5 to help solve the probability problems. Chapman-Kolmogorov Equations We have already defined the one-step transition probabilities [pic]. Construct an 80% confidence interval for the proportion of red marbles\u2026. Then treat this like a mock exam (i. Euler became professor of physics in 1731, and professor of mathematics in 1733, when. The probability X failing during one year is 0. An urn contains n+m balls, of which n are red and m are black. Lindley The Statistician: Journal of the Royal Statistical Society. Recommended Problems: Problems 1. An urn has 4 green balls, 5 yellow balls, and 6 red balls. Wolfram Education Portal \u00bb. In other words, in order to get a new value of seed, multiply the old value by 7621, add 1, and, finally, take the result modulo 9999. One ball is drawn at random. Let B 1 W 2 denote the outcome dial the first bull drawn is B 1 and the second ball drawn is W 2. Hence a probability of 0. Urn problems and how to approach them Probability and real-life situations (lottery, poker, weather forecasts, etc. Practice Problem: A certain lottery has a hat with the numbers 1 through 10 each written on a single scrap of paper. What is the probability the ball drawn from urn C is black? I've figured out that P(K\\A) = 2/3 and P(K\\B) = 5/6. You can also express this relationship as 1 \u00f7 6, 1/6, 0. So, if the probability that they actually had a \ufb02at tire is p, then the probability that they both give the same answer is 1 4 (1\u2212p)+p = 1 4 + 3 4 p. In SAS you can use the table distribution to specify the probabilities of selecting each integer in the range [1, c]. Do Problem 2 ONLY please. Urn 2 contains three red balls and four white balls. Two red, three white, and five black.\ndihvq33d7wefb2,, y5l6ztaewnu,, aszj3xrx30cx1,, e2hmukdh2wbjc,, pe5t81bd5ppwuy,, 109r3mxtwj0lb,, kouxd3ppqx4vaq,, ihdi27tl3fpn,, uxbuwpf31d,, 1mrdb7rz1525,, 8bey8mxlsa,, 6xicx8gk5w2am,, zqlep6c2hmm3b,, ckrev1jmnb,, z2140axliao94y,, ewajmk04pj2,, vfab5wrqgcz,, cunk5n2d708kqbj,, g768d3iqeid,, vwyzoax0y2,, kjjikb278y3,, itd3lm4gh5c,, enbm2xmq16be,, 13ii9xt7ryydkiy,, dc5vl2gzae0,, uhvr12pbsunv,, bsbn1y8c9oqe2w,, pltahmocfuaj,, q57csdq4hx6a,, utljhpuzwu8u1k,, 8noue630pb3,, z4ic81qbwrok7n,, 4e3zkq2gpft1syb,, nemg4dh944ov58d,", "date": "2020-10-24 22:54:15", "meta": {"domain": "bresso5stelle.it", "url": "http://bresso5stelle.it/yfbo/probability-urn-problems.html", "openwebmath_score": 0.7716267108917236, "openwebmath_perplexity": 390.28856000725887, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9901401429507652, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.8584869213595782}}
{"url": "http://bootmath.com/if-a-and-b-are-sets-of-real-numbers-then-a-cup-bcirc-supseteq-a-circcup-bcirc.html", "text": "# If $A$ and $B$ are sets of real numbers, then $(A \\cup B)^{\\circ} \\supseteq A^ {\\circ}\\cup B^{\\circ}$\n\nI have a proof for this question, but I want to check if I\u2019m right and if I\u2019m wrong, what I am missing.\n\nDefinitions you need to know to answers this question: $\\epsilon$-neighborhood, interior points and interiors. Notation: $J_{\\epsilon}(a)$ means a neighborhood formed around a (i.e. $(a-\\epsilon, a+\\epsilon)$. An interior point in some set $A$ is a point where an $\\epsilon$-neighborhood can be formed within the set. The set of all interior points in $A$ is denoted as $A^0$ and is called the interior.\n\nMy proof for the question: I\u2019m proving based on most subset proofs where you prove that if an element is in one set, then it must be in the other. Say $x \\in A^0 \\cup B^0$. Then there is a $\\epsilon > 0$, where $J_{\\epsilon}(x) \\subseteq A$ or $J_{\\epsilon}(x) \\subseteq B$. This implies that $J_{\\epsilon}(x) \\subset A \\cup B$ which then implies $x \\in (A \\cup B)^0$. We can conclude from here that $(A \\cup B)^0 \\supseteq A^0 \\cup B^0$.\n\n#### Solutions Collecting From Web of \"If $A$ and $B$ are sets of real numbers, then $(A \\cup B)^{\\circ} \\supseteq A^ {\\circ}\\cup B^{\\circ}$\"\n\nYour proof is good- there is no problem with it. I have reworded a bit to make things a little more clear, specifically, where you say:\n\nThen there is $\\epsilon>0$, where $J_\\epsilon (x)\\subseteq A$ or $J_\\epsilon (x)\\subseteq B$.\n\nInstead it would be more appropriately stated as: If $x\\in A^0\\cup B^0$ then $x\\in A^0$ or $x\\in B^0$. Then you can assume $x$ is in $A$, prove that $x\\in (A\\cup B)^0$, by symmetry the same argument will work if $x\\in B^0$.\n\nI have included an edited proof- but as I originally stated your proof is good,\nthis just more directly reflects the definitions involved.\n\nLet $x\\in A^0\\cup B^0$. Then $x\\in A^0$ or $x\\in B^0$. Assume $x\\in A^0$. Then there is, by definition, $\\epsilon>0$ such that $J_{\\epsilon}(x)\\subseteq A$. Since $A\\subseteq A\\cup B$, we also have $J_{\\epsilon}(x)\\subseteq A\\cup B$. Thus by definition we have $x\\in (A\\cup B)^0$. If instead $x\\in B^0$, then by symmetry the same argument works. Thus, $A^0\\cup B^0 \\subseteq (A\\cup B)^0$.\n\nHere is a slightly shorter proof:\n\nWe have $A^\\circ \\subset A \\cup B$ and $B^\\circ \\subset A \\cup B$, so we must have\n$A^\\circ \\cup B^\\circ \\subset A \\cup B$.\n\nSince $A^\\circ \\cup B^\\circ$ is open and the interior is the largest open set contained in a set, we must have\n$A^\\circ \\cup B^\\circ \\subset (A \\cup B)^\\circ$.", "date": "2018-07-17 17:43:43", "meta": {"domain": "bootmath.com", "url": "http://bootmath.com/if-a-and-b-are-sets-of-real-numbers-then-a-cup-bcirc-supseteq-a-circcup-bcirc.html", "openwebmath_score": 0.9628444314002991, "openwebmath_perplexity": 101.07594225543998, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401441145625, "lm_q2_score": 0.8670357512127873, "lm_q1q2_score": 0.8584869036583072}}
{"url": "https://byjus.com/question-answer/if-tan-dfrac-theta-2-mathrm-cosec-theta-sin-theta-then-sin-2-dfrac-theta/", "text": "Question\n\n# If $$\\tan\\dfrac {\\theta}{2}=\\mathrm{cosec}\\theta-\\sin\\theta$$, then\n\nA\nsin2\u03b82=2sin218o\nB\ncos2\u03b8+2cos\u03b8+1=0\nC\nsin2\u03b82=4sin218o\nD\ncos2\u03b8+2cos\u03b81=0\n\nSolution\n\n## The correct options are A $$\\sin^2\\dfrac {\\theta}{2}=2 \\sin^2 18^o$$ D $$\\cos 2\\theta+2 \\cos\\theta-1=0$$Simplifying, we get\u00a0 $$\\dfrac{\\sin \\dfrac {\\theta}{2}}{\\cos \\dfrac {\\theta}{2}}=\\dfrac{1-\\sin ^{2}\\theta}{\\sin \\theta}$$ $$\\dfrac{\\sin \\dfrac {\\theta}{2}}{\\cos \\dfrac {\\theta}{2}}=\\dfrac{\\cos ^{2}(\\theta)}{2\\sin \\dfrac {\\theta}{2}.\\cos \\dfrac {\\theta}{2}}$$ $$2\\sin ^{2}\\dfrac {\\theta}{2}=\\cos ^{2}\\theta$$ $$2\\sin ^{2}\\dfrac {\\theta}{2}=(1-2\\sin ^{2}\\dfrac {\\theta}{2})^{2}$$ Let, $$2\\sin ^{2}\\dfrac {\\theta}{2}=t$$$$\\implies t=(1-t)^{2}$$ $$\\implies t=t^{2}-2t+1$$ $$\\implies t^{2}-3t+1=0$$ $$\\implies t=\\dfrac{3\\pm\\sqrt{9-4}}{2}$$ $$\\implies t=\\dfrac{3\\pm\\sqrt{5}}{2}$$ Now $$|\\sin \\theta|\\leq 1$$ $$\\implies t=\\dfrac{3-\\sqrt{5}}{2}$$ Or\u00a0 $$\\sin ^{2}\\dfrac {\\theta}{2}=\\dfrac{3-\\sqrt{5}}{4}$$ $$\\implies 2\\left(\\dfrac{3-\\sqrt{5}}{8}\\right)=2\\sin ^{2}(18^{0})$$ Hence, option A is correct. $$\\cos ^{2}\\dfrac {\\theta}{2}=1-\\sin ^{2}\\dfrac {\\theta}{2}=\\dfrac {1+\\sqrt{5}}{4}$$ \u00a0Option D \u00a0$$\\cos 2\\theta +2\\cos \\theta -1=0$$ $$-2{ \\sin }^{ 2 }\\theta +2\\cos \\theta =0$$ $$=-2+2{ \\cos }^{ 2 }\\theta +2\\cos \\theta =0$$ $$=-1+{ \\cos }^{ 2 }\\theta +\\cos \\theta =0$$ $${ \\cos }^{ 2 }\\theta =1-\\cos \\theta =2{ \\sin }^{ 2 }\\cfrac { \\theta }{ 2 }$$ which is true (as proven in the first part)Hence, option A and D are correct.Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-19 05:30:15", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/if-tan-dfrac-theta-2-mathrm-cosec-theta-sin-theta-then-sin-2-dfrac-theta/", "openwebmath_score": 0.870943009853363, "openwebmath_perplexity": 2094.9869015266386, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401449874105, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.85848690101322}}
{"url": "http://math.stackexchange.com/questions/264947/find-missing-number-from-sum-of-first-few-natural-numbers", "text": "# Find missing number from sum of first few natural numbers\n\nA child was asked to add the first few natural numbers $1+2+3+...$ as long as his patience permitted. As he stopped, he gave the sum as $575$. When the teacher declared the result wrong, the child discovered that he had missed a number in the sequence during addition. What was the number he missed?\n\n-\nDoes the teacher know which number the child was adding up to? \u2013\u00a0 Michael Albanese Dec 25 '12 at 14:47\nlol. I think its a way to frame the question. I too doubt on that same as you. =) \u2013\u00a0 harish.raj Dec 25 '12 at 14:49\nIf the teacher doesn't know the number, then my answer is accurate. If the teacher does know the number, the inequality $\\frac{n(n+1)}{2} - 575 < n$ should be replaced by $\\frac{n(n+1)}{2} - 575 \\leq n$ as the reasoning in the bracket following the original inequality no longer applies. However, this does not change the final answer. \u2013\u00a0 Michael Albanese Dec 25 '12 at 15:12\n\nWe have $$1 + 2 + 3 + \\dots + n = \\frac{n(n+1)}{2}.$$ The numbers obtained from such sums are called Triangular numbers.\n\nAs the child's answer was $575$, the correct answer must be greater than $575$ (as he missed a number). The first triangular number greater than $575$ is $595$ corresponding to $n = 34$. If this was the sum he was trying to evaluate, the child must have missed $20 = 595 - 575$. However, if the correct answer was the next triangular number, $630$, corresponding to $n = 35$, the only way the child could have obtained $575$ is if he missed $55 = 630 - 575$, but this is not in the list of numbers $1, \\dots, 35$ so he could not have been trying to add up the first $35$ natural numbers.\n\nMore generally, we need $\\frac{n(n+1)}{2} > 575$ (the actual sum needs to be greater than the sum he obtained) and $\\frac{n(n+1)}{2} - 575 < n$ (this difference tells us which value he missed, and it can't equal $n$ because then he would have written $1 + \\dots + (n - 1)$ which is a triangular number, so the teacher could not have declared it wrong). Rewriting, we must find $n$ such that $$0 < \\frac{n(n+1)}{2} - 575 < n\\quad (\\ast)$$ which only has solution $n = 34$. Therefore, we must be in the situation of the previous paragraph, so he missed the number $20$.\n\nAdded later: I just thought I'd add some details here about finding natural numbers $n$ which satisfy $(\\ast)$.\n\nFirst of all, consider the left hand inequality which rearranges to $f(n) := n^2 + n - 1150 > 0$. Solving $f(n) = 0$, we obtain $n = \\frac{-1\\pm\\sqrt{4601}}{2}$; as $n > 0$, we ignore $n = \\frac{-1-\\sqrt{4601}}{2}$. Note that $\\left\\lfloor\\frac{-1+\\sqrt{4601}}{2}\\right\\rfloor = 33$, and we have $f(33) < 0$ and $f(34) > 0$. So $f(n) < 0$ for $n \\in \\{1, \\dots, 33\\}$ and $f(n) > 0$ for $n \\geq 34$.\n\nThe right hand inequality rearranges to $g(n) := n^2 - n - 1150 < 0$. Solving $g(n) = 0$, we obtain $n = \\frac{1\\pm\\sqrt{4601}}{2}$; as $n > 0$, we ignore $n = \\frac{1-\\sqrt{4601}}{2}$. Note that $\\left\\lfloor\\frac{1+\\sqrt{4601}}{2}\\right\\rfloor = 34$, and we have $g(34) < 0$ and $g(35) > 0$. So $g(n) < 0$ for $n \\in \\{1, \\dots, 34\\}$ and $g(n) > 0$ for $n \\geq 35$.\n\nHence, $n = 34$ is the only natural number which satisfies $(\\ast)$.\n\n-\ngreat explanation. Thank you a lot. \u2013\u00a0 harish.raj Dec 25 '12 at 14:31\n+1 Michael: nice explanation! \u2013\u00a0 amWhy Dec 25 '12 at 14:35\n\nSuppose he tried to sum the first $n$ positive integers. Omitting a term means this sum is less than the sum of the first $n$ positive integers but not less than the sum of the first $(n-1)$ positive integers. Hence, $\\frac{n(n-1)}{2}\\leq 575<\\frac{n(n+1)}{2}$ so that $n=34$. This means the missing term is $595-575=20$.\n\n-\nawesome. can you elaborate a little about how you got the value of n? (it will be great if you do it in your answer itself) \u2013\u00a0 harish.raj Dec 25 '12 at 14:15", "date": "2015-04-19 01:51:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/264947/find-missing-number-from-sum-of-first-few-natural-numbers", "openwebmath_score": 0.9338129758834839, "openwebmath_perplexity": 204.21855218866668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9840936106207203, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8584714874603493}}
{"url": "http://mathhelpforum.com/trigonometry/41342-complex-numbers.html", "text": "# Math Help - complex numbers\n\n1. ## complex numbers\n\nHi all,\n\nI am having trouble getting the answer to:\n[(1-sqrt(3)i)/ (sqrt(3) + i)]^7\n\nwhere i = complex number\n\nArTiCk\n\n2. Originally Posted by ArTiCK\nHi all,\n\nI am having trouble getting the answer to:\n[(1-sqrt(3)i)/ (sqrt(3) + i)]^7\n\nwhere i = complex number\n\nArTiCk\nMy advice is to first convert the numbers into polar form.\n\n3. Originally Posted by mr fantastic\nMy advice is to first convert the numbers into polar form.\nI have already converted the numbers into polar form...\ni have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]\n\nThe trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:\n\ne^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)\n\nI think it is there where i have gone wrong, help would be appreciated\n\nThanks, ArTiCk\n\n4. Hello, ArTiCK!\n\nSimplify: . $\\left(\\frac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\right)^7$\nRatinalize the \"inside\" first . . .\n\n. . $\\frac{1-i\\sqrt{3}}{\\sqrt{3} + i}\\cdot\\frac{\\sqrt{3}-i}{\\sqrt{3}-i} \\;=\\;\\frac{\\sqrt{3} - i - 3i +i^2\\sqrt{3}}{3-i^2} \\;=\\;\\frac{\\sqrt{3} -4i-\\sqrt{3}}{3+1} \\;=\\;\\frac{-4i}{4} \\;=\\;-i$\n\nSo we have: . $(-i)^7 \\;=\\; i$\n\n5. Originally Posted by ArTiCK\nI have already converted the numbers into polar form...\ni have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]\n\nThe trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:\n\ne^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)\n\nI think it is there where i have gone wrong, help would be appreciated\n\nThanks, ArTiCk\n$\\left(\\frac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\right)^7 =\n$\n$\\left[ \\frac{2 \\, \\text{cis} \\left( -\\frac{\\pi}{3}\\right) }{2 \\, \\text{cis} \\left( \\frac{\\pi}{6}\\right)} \\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{3} - \\frac{\\pi}{6} \\right)\\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{2} \\right)\\right]^7 = (-i)^7$.\n\n6. Hello, Mr. F !\n\nOriginally Posted by mr fantastic\n\n$\\left(\\frac{1-i\\sqrt{3}}{\\sqrt{3}+i}\\right)^7 =\n$\n$\\left[ \\frac{2 \\, \\text{cis} \\left( -\\frac{\\pi}{3}\\right) }{2 \\, \\text{cis} \\left( \\frac{\\pi}{6}\\right)} \\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{3} - \\frac{\\pi}{6} \\right)\\right]^7 = \\left[ \\text{cis} \\left( -\\frac{\\pi}{2} \\right)\\right]^7 = (-i)^7$.\n\nLovely!", "date": "2015-05-25 20:49:26", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/41342-complex-numbers.html", "openwebmath_score": 0.8270604610443115, "openwebmath_perplexity": 3579.2670127712718, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984093606422157, "lm_q2_score": 0.8723473846343393, "lm_q1q2_score": 0.8584714837977435}}
{"url": "https://math.stackexchange.com/questions/2877172/determine-jordan-block-size", "text": "# Determine Jordan block size.\n\nThe following question is from as System Theory test.\n\nLet the system matrix $A$ be given as $A = \\begin{bmatrix} 0&0&0&1\\\\0&-1&1&3\\\\0&1&-1&-1\\\\0&-1&1&2 \\end{bmatrix}$\n\nWhich has the eigenvalues $\\lambda_i=0$ for $i = 1,2,3,4$. Which of the following matrices is the Jordan form of $A$.\n\n$A) \\quad J=\\begin{bmatrix} 0&1&0&0\\\\0&0&1&0\\\\0&0&0&1\\\\0&0&0&0 \\end{bmatrix}$\n\n$B) \\quad J=\\begin{bmatrix} 0&1&0&0\\\\0&0&1&0\\\\0&0&0&0\\\\0&0&0&0 \\end{bmatrix}$\n\n$C) \\quad J=\\begin{bmatrix} 0&1&0&0\\\\0&0&0&0\\\\0&0&0&1\\\\0&0&0&0 \\end{bmatrix}$\n\n$D) \\quad J=\\begin{bmatrix} 0&1&0&0\\\\0&0&0&0\\\\0&0&0&0\\\\0&0&0&0 \\end{bmatrix}$\n\nMy approach:\n\nThe eigenvalues are given so the nullspace of $\\left[A-\\lambda I\\right]$ can be found to be $\\begin{bmatrix} 0\\\\1\\\\1\\\\0 \\end{bmatrix}$ and $\\begin{bmatrix} 1\\\\0\\\\0\\\\0 \\end{bmatrix}$.\n\nThe dimension of this nullspace is $2$ which indicates that there are 2 Jordan blocks. So answer $B$ or $C$.\n\nAlso I've found that : $\\operatorname{rank}(\\ker(A-\\lambda I)^2)=3$ and $\\operatorname{rank}(\\ker(A-\\lambda I))=2$. Subtracting these outcomes gives $3-2=1$. So there's one Jordan block of size 2 or larger. And thus only answer $B$ is correct.\n\nRight or wrong? thanks in advance.\n\n\u2022 It's perfect for me. \u2013\u00a0Bernard Aug 9 '18 at 13:10\n\nYour argument is correct, but your notation is innapropriate. Where you wrote $\\operatorname{rank}(\\ker\\cdots)$, you should have written $\\dim(\\ker\\cdots)$.\nI like to find the actual Jordan form, especially including the matrices that give $R^{-1}A R = J.$ It is not bad when the eigenvalues are integers, and makes concepts concrete. This begins with choosing a column vector i called $w$ such that $A^2 w \\neq 0.$ That becomes the far right column.\n$$\\frac{1}{8} \\; \\left( \\begin{array}{cccc} 8&-4&-4&0\\\\ 0&1&3&0\\\\ 0&2&-2&0\\\\ 0&-4&4&8\\\\ \\end{array} \\right) \\left( \\begin{array}{cccc} 0&0&0&1\\\\ 0&-1&1&3\\\\ 0&1&-1&-1\\\\ 0&-1&1&2\\\\ \\end{array} \\right) \\left( \\begin{array}{cccc} 1&2&1&0\\\\ 0&2&3&0\\\\ 0&2&-1&0\\\\ 0&0&2&1\\\\ \\end{array} \\right) = \\left( \\begin{array}{cccc} 0&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ 0&0&0&0\\\\ \\end{array} \\right)$$", "date": "2019-03-25 10:06:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2877172/determine-jordan-block-size", "openwebmath_score": 0.9465973973274231, "openwebmath_perplexity": 296.71248130017193, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936068886641, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8584714662421322}}
{"url": "http://math.stackexchange.com/questions/118198/trying-to-prove-that-a-group-is-not-cyclic", "text": "# Trying to prove that a group is not Cyclic\n\nGiven the following Euler groups : \\begin{align*} U_{12} &= \\{1,5,7,11\\}\\\\ U_{16} &= \\{1,3,5,7,9,11,13,15\\} \\end{align*}\n\nI want to prove that they are not cyclic. I used the following theorem :\n\nA group of order $n$ is cyclic if and only if it has an element of order $n$.\n\nLet's take for example $U_{12}$: (I will use the notation of $o(x)$ to denote the order of the element $x\\in G$)\n\\begin{align*} o(5)&\\colon 5^2=25 \\to 25\\bmod 12 = 1\\to o(5)=2\\\\ o(7)&\\colon 7^2= 49 \\to 49\\bmod 12 = 1 \\to o(7)=2\\\\ o(11) &\\colon 11^2 = 121 \\to 121\\bmod 12=1 \\to o(11)=2 \\end{align*}\n\nThen by using the above theorem , this group is indeed not a cyclic group.\n\nQuestion : do I really have to check each element in the group for its order ?\n\nRegards\n\n-\nFixed. thanks ! \u2013\u00a0ron Mar 9 '12 at 11:58\nBTW, the structure of the multiplicative group of integers modulo n $U_n$ is will known. In particular, it is cyclic iff $n$ is 2, 4, any power of an odd prime or twice any power of an odd prime. \u2013\u00a0lhf Mar 9 '12 at 12:08\nPlease learn some basic $\\LaTeX$ instead of waiting for others to pretty up your posts for you. Thank you. \u2013\u00a0Arturo Magidin Mar 9 '12 at 16:45\nComment: Stop using $\\to$ to mean \"and since\" or some other connectives. The arrow has specific meanings in mathematics; students who use $\\to$ and $=$ as general purpose connectives to mean something like \"and then I did some thinking and this is what I came up with\" drive professors crazy. \u2013\u00a0Arturo Magidin Mar 9 '12 at 16:49\n@Arturo, I don't think driving professors crazy is the real problem - for most of us, crazy is within walking distance, anyway - I think the real problem is the student who writes that way is pretty much guaranteed to get things wrong. \u2013\u00a0Gerry Myerson Mar 10 '12 at 6:58\n\nWell really the \"theorem\" is the definition of cyclic group...it is a group generated by one element.\n\nNow if I give you a group and you check that all but one element isn't a generator, how do you know that the one you didn't check isn't a generator?\n\nSo yes, you must check ALL elements.\n\n-\n... or you can use interesting theorems. For example, there's an easy upper bound on the number of elements of order 2.... \u2013\u00a0Hurkyl Mar 9 '12 at 14:28\n\nSome shortcuts are available. For example, when testing $3$ in $U_{16}$, you find its powers are $3,9,11,1$, so you don't have to test $9$ or $11$. Do you see why?\n\n-\nI didn't get your meaning .What do you mean by \"its powers\" ? 3^3 = 27 --> 27 modulo 16= 11 .... so I generated using 3 , the element 11 . What does it mean ? \u2013\u00a0ron Mar 9 '12 at 12:11\nYou're trying to decide whether the order of $3$ is $8$. So you look at $3$, then at $3\\times3=9$, then at $3\\times3\\times3=3\\times9=27=11$, then at $3\\times3\\times3\\times3=3\\times11=33=1$. Those are the powers of $3$, the numbers $3^r$, $r=1,2,3,\\dots$. Using $3$, you generate $3$, $9$, $11$, and $1$. \u2013\u00a0Gerry Myerson Mar 9 '12 at 12:34\nOkay , so now we know that the order of |3|=4 . Why does it help me ? now I won't need to check the generated elements using 3 ? \u2013\u00a0ron Mar 9 '12 at 12:42\n\"Why does it help me ?\" This is an exercise for the reader -- use knowledge about the order of $3$ in order to deduce something about the order of $3^2$.... \u2013\u00a0Hurkyl Mar 9 '12 at 14:26\nYou should see if you can prove that if $g$ doesn't generate $G$ and $h$ is in the subgroup generated by $g$ then $h$ doesn't generate $G$ either. \u2013\u00a0Gerry Myerson Mar 10 '12 at 6:59\n\nAs I mentioned yesterday in reply to a question of yours: a cyclic group has at most one element of order $2$.\n\nIn your first two lines, you've shown that $5$ and $7$ both have order $2$ in $U_{12}$. Therefore, the group cannot be cyclic.\n\nSimilarly, since $9^2\\equiv 1\\pmod{16}$ and $15^2\\equiv 1\\pmod{16}$, $U_{16}$ has at least two distinct elements of order $2$, and therefore cannot be cyclic.\n\nIn general, a cyclic group of order $n$ has exactly $\\phi(d)$ elements of order $d$ for each divisor $d$ of $n$.\n\n-\n\nWhat of if your set is infinite? I think the best idea is to find the general form of the elements in the set (if possible) and pick arbitarary element, check your assertion and then generalised. But in your case, since the elements are finite, you have to check it for all the elements in the set.\n\n-", "date": "2016-07-28 16:47:11", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/118198/trying-to-prove-that-a-group-is-not-cyclic", "openwebmath_score": 0.9440299272537231, "openwebmath_perplexity": 265.6885459230907, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241982893259, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8584640124865492}}
{"url": "https://cs.stackexchange.com/questions/89842/average-case-analysis-for-finding-max-and-min-value-on-an-array", "text": "# Average Case Analysis for finding max and min value on an array\n\nGiven the following algorithm, to find the maximum and minimum values of an array - don't mind the language:\n\nMaxMin(A[1..n])\nmax = A[1];\nmin = A[1];\nfor (i = 2; i<=n; i++)\nif (A[i] > max)\nmax = A[i];\nelse if (A[i] < min)\nmin = A[i];\nprint(max, min);\n\n\nI need to do a probabilistic analysis for the average case of comparisons that will be made on its execution.\n\nSo far, my solution is:\n\nGiven an indicator random variable: $$X_i = \\begin{cases} \\text{1, if max > A[i]}\\\\ \\text{0, if max < A[i]}\\\\ \\end{cases}$$\n\nand assuming an uniform distribution for $A[1..n]$, the expected probability is: $$E[x] = \\sum\\limits_{i=1}^{n} Pr(X_i)$$\n\nwhere $$Pr(X_i)$$ is the probability of the i-th element be the $max$ element in $A[1..n]$. It's possible to determine that: $$Pr(x_1) = 1, Pr(x_2) = 1/2, Pr(x_3) = 1/3 ,..., Pr(x_n) = 1/n$$ and thus for an array of size $n$ the expected value can be calculated as:\n\n$$E[x] = 1 + 1/2 + 1/3 + ... + 1/n = \\sum\\limits_{i=1}^{n}{\\frac{1}{i}} \\approx \\log{n}$$\n\nAnd the same goes for $min$, which will give the same result $\\log{n}$. (1)\n\nMy questions are: is it correct? Does the complexity for the average case of the given algorithm is $\\theta(\\log{n})$? Can I use the argument pointed by (1), just modifying $X_i$ so: $$X_i = \\begin{cases} \\text{1, if min < A[i]}\\\\ \\text{0, if min > A[i]}\\\\ \\end{cases}$$\n\nI already read this (unfortunately the link to the video isn't available), but it only explains for $max$ and my analysis must be for both $max$ and $min$.\n\n\u2022 What do you mean \"average case of comparisons\"? Do you mean the average quantity of comparisons? That can't possibly be less than $n-1$, so $\\log n$ is wrong. Perhaps you should edit the question to clarify? \u2013\u00a0Wildcard Mar 27 '18 at 1:46\n\u2022 Of course your analysis assumes the array is in random order. Many times arrays are not in random order. Like sorted arrays. \u2013\u00a0gnasher729 Mar 29 at 19:59\n\nFirst, note that the first comparison will always compare $$n-1$$ times, independent of the distribution of the input. So, what you really want to know is how many times the second part of the if compares. For this, you can use a indicator random variable like this:\n\n$$X_i = \\begin{cases} \\text{1, if A[i] \\le \\max}\\\\ \\text{0, if A[i] > \\max}\\\\ \\end{cases}$$\n\nSo, just like you did, assuming an uniform distribution for $$A[1..n]$$, the expected probability is:\n\n$$E[x] = \\sum\\limits_{i=1}^{n} Pr(X_i)$$\n\nBut we do not want $$Pr(i=\\max)$$, we want $$\\overline{Pr(i=\\max)}$$, like we stated before. So, using your probability of the $$i-th$$ element to be the $$max$$ element:\n\n$$Pr(i = \\max) = 1/i$$\n\nWe have this complement:\n\n$$\\overline{Pr(i=\\max)} = (1 - 1/i)$$\n\nFrom which we can put in the expected probability:\n\n$$E[x] = \\sum\\limits_{i=2}^{n} (1 - 1/i) = \\sum\\limits_{i=2}^{n}1 - \\sum\\limits_{i=2}^{n}1/i = n - 1 - (\\ln|n| - 1 + \\Theta(1))$$\n\nSo, to get the total quantity of comparisons, we can just sum the number of comparisons of the two parts. Note that the $$\\Theta(1)$$ can absorb a constant values.\n\n$$(n - 1) + n - 1 - \\ln|n| + 1 + \\Theta(1) = 2n - 1 - \\ln|n| + \\Theta(1)$$\n\nAnd we get the expected total number of comparions:\n\n$$2n - \\ln|n| + \\Theta(1)$$\n\n(1) You are looking for either min or max. Then,\n\n(1-1) is it correct? Yes, it is correct.\n\n(1-2) Does the complexity for the average case of the given algorithm is $\\Theta(logn)$? Indeed it is. My experience in practice is that it does not work that way (or at least the is a large constant involved).\n\n(1-3) Can I use the argument pointed by (1), just modifying $X_i$? Yes, you can do that too.\n\n(2) You are looking for both min and max. Then, the scenario is different. You should have something like: $$Z_i = \\begin{cases} \\text{1, if max > A[i] and min < A[i]}\\\\ \\text{0, if max < A[i] and min > A[i]}\\\\ \\end{cases}$$\n\nThen you can calculate $E[Z]$. It is also possible to calculate $E[XY]$, where $X$ is for max, and $Y$ is for min. Try different techniques and see which one works easier.\n\n\u2022 In my case, I'm interested on your second explanation (for $Z_i$). Really liked your explanation, but @LionsWrath explanation tackled my question right on the spot! Tough, thank you for your explanation :) \u2013\u00a0woz Mar 27 '18 at 11:09", "date": "2019-07-17 06:50:07", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/89842/average-case-analysis-for-finding-max-and-min-value-on-an-array", "openwebmath_score": 0.8820377588272095, "openwebmath_perplexity": 348.6760946756108, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241982893258, "lm_q2_score": 0.8856314647623016, "lm_q1q2_score": 0.8584640095605193}}
{"url": "https://math.stackexchange.com/questions/3321649/prove-textordam-frac-textlcmm-nm", "text": "# Prove: $\\text{ord}(a^m) = \\frac{\\text{lcm}(m,n)}{m}$ [duplicate]\n\nThis is Pinter $$10.G.5$$\n\nLet:\n\n$$a \\in G$$\n\n$$\\text{ord}(a) = n$$\n\nProve: $$\\text{ord}(a^m) = \\frac{\\text{lcm}(m,n)}{m}$$\n\nUse $$10.G.3$$ and $$10.G.4$$ to prove this.\n\nHere is $$10.G.3$$:\n\nLet $$l$$ be the least common multiple of $$m$$ and $$n$$. Let $$l/m = k$$. Explain why $$(a^m)^k = e$$.\n\nHere is $$10.G.4$$:\n\nProve: If $$(a^m)^t = e$$, then $$n$$ is a factor of $$mt$$. (Thus, $$mt$$ is a common multiple of $$m$$ and $$n$$.)\n\nConclude that: $$l = mk \\leq mt$$\n\nOK, let's begin.\n\nBy $$10.G.3$$:\n\n$$(a^m)^{\\text{lcm}(m,n)/m} = e \\tag{5}$$\n\nIf $$\\text{lcm(m,n)}/m$$ is the lowest number such that (5) is true, then:\n\n$$\\text{ord}(a^m) = \\text{lcm}(m,n)/m \\tag{9}$$\n\nLet's assume that there is a number\n\n$$q < \\text{lcm}(m,n)/m \\tag{7}$$\n\nsuch that:\n\n$$(a^m)^q = e \\tag{6}$$\n\nBy $$10.G.4$$ with (6):\n\n$$n \\ \\big|\\ mq$$\n\n$$l \\lt mq$$\n\n$$\\text{lcm}(m,n) \\leq mq$$\n\nIsolate q:\n\n$$\\text{lcm}(m,n)/m \\leq q \\tag{8}$$\n\nSo (9) must be true.\n\nThat's how I approached it. If anyone notices any issues, I'd be happy to know about them.\n\nEven if it is considered correct, do you feel there is a better way?\n\n\u2022 You should use \\le rather than <=. \u2013\u00a0Angina Seng Aug 13 '19 at 2:52\n\u2022 @LordSharktheUnknown Thanks! Updated! \u2013\u00a0dharmatech Aug 13 '19 at 2:54\n\u2022 @JyrkiLahtonen Can you explain how this is a duplicate of the question you linked to? There is no requirement in this question that G be cyclic. Whereas the question you linked to is for cyclic groups. \u2013\u00a0dharmatech Aug 13 '19 at 8:49\n\u2022 @Jos\u00e9 Carlos Santos Do you really think that it's a duplicate? \u2013\u00a0Michael Rozenberg Aug 13 '19 at 9:06\n\u2022 @YuiToCheng Once again not a proper dupe target becauyse - among other things - the OP is working with LCMs not GCDs, and is working from specific lemmas. Please be more careful in choosing proper dupe targets. \u2013\u00a0Bill Dubuque Aug 14 '19 at 13:32\n\nYes, $$\\,(a^{m})^{k} = 1\\!\\! \\overset{(1)\\!\\!}\\iff n\\mid mk \\iff m,n\\mid mk\\!\\! \\overset{(2)\\!\\!}\\iff {\\rm lcm}(m,n)\\mid mk\\iff {\\large{\\frac{{\\rm lcm}(m,n)}m\\,\\mid}}\\, k$$\n$$\\!(1)$$ and the (omitted) final conclusion is by Corollary' here, and $$(2)$$ is the LCM Universal Property\n\u2022 @BillDubuque Do you really think this question has not been handled on our site? The OP's complaint about my choice of dupe was mostly because \"their version is not only about cyclic groups\". Yet they only work within the cyclic group generated by $a$. \u2013\u00a0Jyrki Lahtonen Aug 14 '19 at 6:18", "date": "2020-03-31 08:00:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3321649/prove-textordam-frac-textlcmm-nm", "openwebmath_score": 0.6025456786155701, "openwebmath_perplexity": 1063.1705623165626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307684643189, "lm_q2_score": 0.8824278772763472, "lm_q1q2_score": 0.8584529899650866}}
{"url": "https://math.stackexchange.com/questions/2813157/open-and-closed-implication-formula", "text": "# Open and closed implication formula\n\nSo I am a newbie in mathematical logic and one of the problems I have faced throughout this one week of study is the one concerning open or closed statement. Let me know if I have used some terms such as \"statement\", \"sentence\", \"formula\", etc in this question the wrong way.\n\nI posted probably a related question before here: Connection between universal Quantifier and implication, and have stumbled upon new terms such as \"open\", \"closed\", \"free\", and \"bound\". I still need some confirmation about this but I state a different question.\n\nSo, am I right to say that $\\forall x[x\\in \\mathbb{R}\\implies x^2\\geq0]$ is a closed formula since I know it has a no free variable?\n\nDoes every closed formula have a truth value?\n\nI know that $x\\in \\mathbb{R}\\implies x^2\\geq0$ is a formula and is true for every assignment of $x$, or IS it? Why can't I just conclude that $x\\in \\mathbb{R}\\implies x^2\\geq0$ is just true and why should I consider its hidden universal quantifier (which, of course, it does not sound the same using existential quantifier)? Is it just for the sake of the existing rule of making sentence in FOL?\n\nProbably the same question: Why can't all quantifiers be bounded quantifiers, and be written that way, considering the existence of domain of discourse? Why don't we explicitly write that domain in the sentence? Meaning, instead of saying \"In the domain of natural numbers, $\\forall x[x\\geq0]$\", why not simply $\"\\forall x\\in \\mathbb{N}[x\\geq0]\"$?\n\nHope my confusion is understood as a newbie. Thanks for the help! :D\n\n\u2022 Correct; the formula $\u2200x[x\u2208R \u27f9 x^2\u22650]$ is a closed formula (or sentence) because it has no free occurrences of variables. Jun 9 '18 at 9:16\n\u2022 Yes; a closed formula has a definite truth value in an interpretation. $\\forall x (x=0)$ is FALSE in $\\mathbb N$, while an open one, like e.g. $(x=0)$ may change truth value (for a specific interpretation) according to the value assigned to the free variable $x$. Jun 9 '18 at 9:18\n\u2022 You can see Classical Logic and Model Theory and Logical Truth and Logical Constants for an introduction. Jun 9 '18 at 9:43\n\u2022 The last point is that usually the \"pure\" presentation of first-order logic has no predicate constant, like $\\in$. Thus the way to formalize $\u2200x \\in \\mathbb N [x\u22650]$ is $\u2200x[N(x) \\to x\u22650]$ where $N(x)$ is a unary predicate. Now again, the truth-value of the sentence depends on the way to interpret $N(x)$. Jun 9 '18 at 10:04\n\u2022 Thus, from the \"pedagogical\" point of view, we have reduced restricted quantification to quantification of a conditional: thus, in any case, we have to first learn how to manage quantifiers. Jun 9 '18 at 10:11\n\nThe \"founding\" idea is that logic must be formal.\n\nThis idea was firstly investigated by Aristotle : the modern way to investigate the concept \"formal\" is to define it through syntax :\n\na formula [a grammatically correct expression] is an expression built-up according to specific rules.\n\nThen we have semantics :\n\nthe way to give meaning (and truth value) to an expression through an interpretation.\n\nThe interaction of syntax and semantics is the key-point of modern formal logic.\n\nYou can see : John MacFarlane, WHAT DOES IT MEAN TO SAY THAT LOGIC IS FORMAL (2000):\n\nWhat does it mean, then, to say that logic is distinctively formal?\n\nThree things: logic is said to be formal (or \u201ctopic-neutral\u201d)\n\n(1) in the sense that it provides constitutive norms for thought as such,\n\n(2) in the sense that it is indifferent to the particular identities of objects, and\n\n(3) in the sense that it abstracts entirely from the semantic content of thought.\n\nThus, \"topic neutral\" and \"to abstract entirely from the semantic content\" mean to separate semantics from syntax : the domain of interpretation is obviously semantical.\n\n\u2022 I get it now. The idea is to formalize that sentence formed with bounded quantifier. So, my last point, how do you think about introducing\n\u2013\u00a0bms\nJun 9 '18 at 13:17\n\u2022 get it now. The idea is to formalize that sentence formed with bounded quantifier. So, my last point, what is a reasonably good way to introduce my student the way of proving an implication? Do you think it is fine to assume (with careful context reading) that there are always hidden universal quantifiers before the implication statement that bound the variables in the hypothesis, and tell them that? Hence, they will always formally start with \"Take an arbitrary x. If p(x) is true, then.... Thus q(x)\". I ask this since my usual understanding is\n\u2013\u00a0bms\nJun 9 '18 at 13:24\n\u2022 that \"for all\" sentence is equivalent to \"If, then\" sentence because the way we prove them is essentially the same.\n\u2013\u00a0bms\nJun 9 '18 at 13:26", "date": "2021-10-24 09:25:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2813157/open-and-closed-implication-formula", "openwebmath_score": 0.8256122469902039, "openwebmath_perplexity": 608.7961143246217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972830769252026, "lm_q2_score": 0.8824278741843883, "lm_q1q2_score": 0.8584529876522285}}
{"url": "https://math.stackexchange.com/questions/1181129/why-are-the-probability-of-rolling-the-same-number-twice-and-the-probability-of", "text": "# Why are the probability of rolling the same number twice and the probability of rolling pairs different?\n\nTwo scenarios:\n1. Using one die, roll a 6 twice.\n$\\frac16\\times\\frac16=\\frac1{36}$\n\n1. Rolling two dice roll the same number (a pair).\n$\\frac6{36}=\\frac16$\n\nWhy are these two probabilities different? Because the events are independent, isn't rolling a pair the same as rolling a die twice?\n\nIn a sense, rolling two dice at once is the same as rolling 1 die twice at the same time? How does this \"timing\" issue affect the probability?\n\nIn the second case, your pair can be any one of $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ and you'd satisfy \"obtaining a pair\".\n\nThat gives you six possible pairs, each of which one has probability of occurring $\\dfrac 1{36}$ gives us $$6\\times \\frac 1{36} = \\frac 16$$\n\nNow, if you want to know what the probability of rolling two dice simultaneously and obtaining two sixes (one prespecified pair of the six possible pairs), that would be $\\dfrac 1{6\\cdot 6} = \\dfrac 1{36}$.\n\nWith this distinction made, yes, the probability of obtaining two sixes when rolling one die twice, and the probability of rolling two sixes simultaneously are equal.\n\n\u2022 I understand how the sample space is defined. But isn't it the same thing as rolling two dice back to back? it just happens that in the two dice sample, the rolls are happening together. Whereas in the one die example, the rolls happen with a small delay in between. \u2013\u00a0William Falcon Mar 8 '15 at 18:28\n\u2022 You specified the probability of rolling two sixes. On the other hand, the probability of rolling consecutively (one roll, then another), and obtaining the same number twice is $1/6$. The first roll can be any number (probability of rolling a number $1-6$ is equal to $1$, and then the probability of rolling that particular number on the second roll is $\\frac 16$ for overall probability of rolling the same number twice being $1/6$. \u2013\u00a0amWhy Mar 8 '15 at 18:32\n\nNote that you are asking two different questions. In the first problem, you are asking the probability of rolling a particular number twice, while in the second problem, you are asking the probability of rolling one of the numbers twice.\n\nThe probability of rolling a $6$ twice is $$\\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{1}{36}$$ but so is the probability of rolling a $1$ twice, a $2$ twice, a $3$ twice, a $4$ twice, or a $5$ twice. Hence, the probability of rolling the same number twice is $$6 \\cdot \\frac{1}{36} = \\frac{1}{6}$$\n\nI think the simplest answer for the second, pairing, scenario is:\n\nYour first rolling does not have any possibility expected, which means you will have any numbers. However, your second roll should be pairing the first number and the possibility should be 1/6.\n\nso to roll a pair, the odds is 1 x 1/6 = 1/6", "date": "2020-02-24 19:27:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1181129/why-are-the-probability-of-rolling-the-same-number-twice-and-the-probability-of", "openwebmath_score": 0.6474679112434387, "openwebmath_perplexity": 270.5583358108583, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307676766118, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8584529862620391}}
{"url": "http://math.stackexchange.com/questions/320619/closed-bounded-but-not-compact-subset-of-a-normed-vector-space", "text": "# Closed Bounded but not compact Subset of a Normed Vector Space\n\nConsider $\\ell^\\infty$ the vector space of real bounded sequences endowed with the sup norm, that is $||x|| = \\sup_n |x_n|$ where $x = (x_n)_{n \\in \\Bbb N}$.\n\nProve that $B'(0,1) = \\{x \\in l^\\infty : ||x|| \\le 1\\}$ is not compact.\n\nNow, we are given a hint that we can use the equivalence of sequential compactness and compactness without proof.\n\nHowever, I don't understand how sequences of sequences work? Do I need to find a set of sequences and order them such that they do not converge to the same sequence?\n\nDoes the sequence $(y_n)$ where $y_n$ is the sequence such that $y_n = 0$ at all but the nth term where $y_n =1$ satisfy the requirement that it does not have a convergent subsequence?\n\nI think I have probably just confused myself with this sequence of sequences lark. Sorry and thanks.\n\n-\n\nYou have to find a sequence in $B'(0,1)$ which does not contain a convergent subsequence. Because the existenc of such a sequence implies (right from the definition) that $B'(0,1)$ isn't sequentially compact.\n\nHint Consider the sequence $(y^n)_n$ defined by $$y^n_k := \\begin{cases} 1 & n=k \\\\0 & n \\not= k \\end{cases} \\qquad (k \\in \\mathbb{N})$$ Then $y^n \\in B'(0,1)$ for all $n \\in \\mathbb{N}$ and $\\|y^n-y^m\\|_{\\infty}=1$ for $m \\not= n$. This means that $(y^n)_n$ doesn't contain a Cauchy-subsequence, hence in particular no convergent subsequence.\n\n-\n\nLet $E = \\{ e_n \\}_n$ where $e_n$ is the vector with $1$ in the $n$th position and zero everywhere else. Clearly $E \\subset \\overline{B}(0,1)$, hence it is bounded. Since $\\|e_n - e_m\\| = 1$ whenever $n \\neq m$, it is clear that each point in $E$ is isolated. Hence $E$ is closed.\n\nNow let $U_n = B(e_n, \\frac{1}{2}$). Then $\\{ U_n\\}_n$ is an open cover of $E$ that had no finite subcover. Hence $E$ is not compact.\n\nIf you would rather use a sequential argument, choose the sequence $x_n =e_n$. As above, we have $\\|x_n -x_m\\| = 1$ whenever $n \\neq m$, hence no subsequence can be Cauchy. Hence no subsequence can converge, hence $E$ is not sequentially compact.\n\n-\n\nThese are very intriguing examples but what condition of the Heine-Borel Theorem is absent? B-T says closed and bounded is equivalent to compact in at least all n dimensional real or complex sets. What specifically is missing in these examples for them to not contradict Heine-Borel?\n\n-\n\nIn answer to my own question, my first three attempts to state the Heine-Borel Theorem in n dimensional real space. Certainly a complete metric space would work! Can anyone soften the topological requirements of the Theorem?\n\n-", "date": "2016-07-27 09:46:01", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/320619/closed-bounded-but-not-compact-subset-of-a-normed-vector-space", "openwebmath_score": 0.9583142399787903, "openwebmath_perplexity": 119.03551919096678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307661011976, "lm_q2_score": 0.8824278680004706, "lm_q1q2_score": 0.8584529788559442}}
{"url": "https://math.stackexchange.com/questions/2334607/the-sum-of-consecutive-integers-is-50-how-many-integers-are-there/2334733", "text": "# The sum of consecutive integers is $50$. How many integers are there?\n\nI started off by calling the number of numbers in my list \"$n$\". Since the integers are consecutive, I had $x + (x+1) + (x+2)...$ and so on. And since there were \"$n$\" numbers in my list, the last integer had to be $(x+n)$. This is where I got stuck. I didn't know how to proceed because I am not given the starting point of my integers, nor an ending point.\n\n\u2022 If there are $n$ numbers in your list and the first is $x = x + 0$, then the last is $x + n - 1$. \u2013\u00a0N. F. Taussig Jun 24 '17 at 10:37\n\nIf your $n$ is odd, then the middle number has to be $50/n$. The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$\n\nIf $n$ is even, then $50/n$ is the half-integer between the middle two numbers. So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.\n\nIf $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$\n\nIf $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\\cdots +11+12 = 50.$\n\nSo there are 4 answers: $n=1, 4, 5,$ and $20$.\n\nEdit: As Bill points out, I missed the divisors $25$ and $100$, which give two more answers: $50 = -10+-11+\\cdots+14$ and $50 = -49 +-48+\\cdots +50$.\n\nNote that each solution with negative integers is related to an all-positive solution. From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \\ldots, 10$, which add to $0$, and we have another solution.\n\n\u2022 interesting the approach using the middle number! \u2013\u00a0G Cab Jun 24 '17 at 12:31\n\u2022 Great technique! n=25 and n=100 work aswell \u2013\u00a0Mike Jun 24 '17 at 13:47\n\u2022 Incomplete, e.g. the odd divisor $25\\ \\$ \u2013\u00a0Bill Dubuque Jun 24 '17 at 14:01\n\u2022 @GCab If you think of it in terms of the \"average\" (mean), instead of the \"middle\" (median) number, then it's less surprising. The two only happen to be equal because the numbers are uniformly distributed. \u2013\u00a0Robin Saunders Jun 24 '17 at 15:24\n\nWhat may be helpful is to use the formula for the sum of an arithmetic progression: if you have a sequence whose first term is $a$ and each term is $d$ more than the rest, then the sum of the first $n$ terms is $na+\\frac{n(n-1)}{2}d$. In this case, since we are looking at consecutive integers, $d=1$, and so you are trying to find $a$ and $n$ such that $na+\\frac{n(n-1)}2=50$, or equivalently $n(2a+n-1)=100$.\n\n\u2022 This is surely helpful. As there are only nine divisors of 100, you can try them all easily and look at which lead to an integer $a$. The nice thing is that (unlike when using a more sophisticated solution) you won't miss the negative first terms. \u2013\u00a0maaartinus Jun 25 '17 at 13:38\n\u2022 Yes, and you can make things slightly quicker using the fact that if $n$ is even then $2a+n-1$ is odd, and vice versa, so you can skip cases like $2\\times 50$ where both factors are even. \u2013\u00a0Especially Lime Jun 25 '17 at 19:29\n\n\\scriptsize\\begin{align} 50 &=2\\times 25 &&=25\\boxed{+}25 &&=24.5\\boxed{+}25.5 \\text{ (AP but not integer AP)}\\\\ &=4\\times 12.5 &&=12.5+12.5\\boxed{+}12.5+12.5 &&=\\color{red}{11+12\\boxed{+}13+14}\\\\ &=5\\times 10 &&=10+10+\\boxed{10}+10+10 &&=\\color{red}{8+9+\\boxed{10}+11+12}\\\\ &=10\\times 5 &&=\\underbrace{\\underbrace{5+5+\\cdots+5}_{5}\\boxed{+}\\underbrace{5+5+\\cdots +5}_{5} }_{10} &&=\\underbrace{\\underbrace{0.5+1.5+\\cdots+4.5}_{25}\\boxed{+}\\underbrace{5.5+\\cdots+9.5}_{25}}_{50}\\\\ & && && \\quad \\text{(AP but not integer AP)}\\\\ &=20\\times 2.5 &&=\\underbrace{\\underbrace{2.5+2.5+\\cdots+2.5}_{10}\\boxed{+}\\underbrace{2.5+\\cdots +2.5}_{10} }_{20} &&=\\color{red}{\\underbrace{\\underbrace{-7+(-6)+\\cdots+2}_{10}\\boxed{+}\\underbrace{3+4+\\cdots+12}_{10}}_{20}}\\\\ &=25\\times 2 &&=\\underbrace{\\underbrace{2+2+\\cdots+2}_{12}+\\boxed{2}+\\underbrace{2+\\cdots +2}_{12} }_{25} &&=\\color{red}{\\underbrace{\\underbrace{-10+(-9)+\\cdots+1}_{12}+\\boxed{2}+\\underbrace{3+\\cdots+13+14}_{12}}_{25}}\\\\ &=50\\times 1 &&=\\underbrace{\\underbrace{1+1+\\cdots+1}_{25}\\boxed{+}\\underbrace{1+\\cdots +1}_{25} }_{50} &&=\\underbrace{\\underbrace{-23.5+(-22.5)+\\cdots+0.5}_{25}\\boxed{+}\\underbrace{1.5+\\cdots+25.5}_{25}}_{50}\\\\ & && && \\quad \\text{(AP but not integer AP)}\\\\ &=100\\times 0.5 &&=\\underbrace{0.5+0.5+\\cdots+0.5}_{50}\\boxed{+}\\underbrace{0.5+0.5+\\cdots+0.5}_{50} &&=\\color{red}{\\underbrace{\\underbrace{-49+(-48)+\\cdots+0}_{50}\\boxed{+}\\underbrace{1+2+\\cdots +49+50}_{50} }_{100}} \\end{align}\n\nIn more detail:\n\n$$\\frac {50}n=m$$ If $n$ is even $(n=2p)$, then we want $m=a+0.5 \\;\\;(a\\in \\mathbb Z)$ $\\cdots$ Condition $(1)$\n\n\u2022 The AP would comprise $p$ consecutive integers on either side of $m$.\n\nIf $n$ is odd $(n=2p+1)$, then we want $m\\in \\mathbb Z$ $\\cdots$ Condition $(2)$\n\n\u2022 The AP would comprise $p$ consecutive integers on either side of $m$, as well as $m$ itself.\n\nTry different values of $n$ (excluding the trivial case $n=1$):\n\n\u2022 If $n=2$, then $m=25$, hence Condition ($1$) not satisfied.\n\n\u2022 If $n=3$, then $m=16\\frac 23$, hence Condition ($2$) not satisfied.\n\n\u2022 If $n=4$, then $m=12.5$, hence Condition $(1)$ satisfied, so the AP is $\\lbrace (m-1.5), (m-0.5), (m+0.5), (m+1.5)\\rbrace$, i.e. $\\color{red}{11,12,13,14}$ ($AP1$).\nAdding negative terms and corresponding positive terms which cancel out gives: $\\color{blue}{-10,-9,-8\\cdots, 0\\cdots, 8,9,10,}\\color{red}{11,12,13,14}$ ($AP 1'$)\n\n\u2022 If $n=5$, then $m=10$, hence Condition $(2)$ satisfied, so the AP is $\\lbrace (m-2),(m-1),m,(m+1),(m+2)\\rbrace$, i.e. $\\color{red}{ 8,9,10,11,12}$ ($AP 2$).\nAdding negative terms and corresponding positive terms which cancel out gives: $\\color{blue}{-7,-6,-5\\cdots, 0\\cdots, 5,6,7,}\\color{red}{8,9,10,11,12}$ ($AP 2'$)\n\n\u2022 If $n=6, 7,8,9$, then $m\\notin\\mathbb Z$ and $m\\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.\n\n\u2022 If $n=10$, then $m=5$, hence Condition $(2)$ not satisfied - not possible.\n\n\u2022 If $n=11,12,\\cdots, 24$, then $m\\notin\\mathbb Z$ and $m\\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.\n\n\u2022 If $n=25$, then $m=2$, hence Condition $(2)$ satisfied, so AP is $\\color{red}{\\underbrace{-10,-9,\\cdots 0,1}_{12\\text{ terms}},2,\\underbrace{3,\\cdots 9,10,11,12,13,14}_{12 \\text{ terms}}}$ (same as $AP 1'$)\n\n\u2022 If $n=26,27,\\cdots, 49$, then $m\\notin\\mathbb Z$ and $m\\neq a+0.5$, hence neither Conditions $(1)$ or $(2)$ satisfied.\n\n\u2022 If $n=100$, then $m=0.5$, hence Condition $(1)$ satisfied, so AP is $\\color{red}{\\underbrace{-49,-48,\\cdots -2,-1,0}_{50\\text{ terms}},\\underbrace{1,2,3,\\cdots 48,49,50}_{50 \\text{ terms}}}$ ($AP 3$)\n\n\u2022 For higher values of $n$, $0<m<0.5$ - not possible.\n\nWithout reading other answers... this should tell you how an old computer programmer thinks, versus a real mathematician.\n\nFirst the obvious answer is the single integer 50. However, if negative numbers are allowed, then we can scoop up the sequence from -49 to 50 for 100 consecutive numbers. This is the longest possible sequence.\n\nIf n is the starting number of a sequence and s is the number of consecutive numbers, then we end up with\n\n50 = (n+0) + (n+1) + (n+2) +... + (n+s-1)\n50 = ns + (s-1)(s)/2\n\n\nThis is helpful, maybe, as ns pretty much puts a box around the solution set. Since we know the nature of the rightmost term, we can tell that s is 10 or less.\n\nConsider the transformation if we multiply both sides by 2/s:\n\n100/s = 2n + s - 1\n\n\nThe right side will always be an integer. Therefore, s must always divide 100 evenly. From above we know that s is between 1 and 10, so the only possible values for s are 1, 2, 4, 5, and 10.\n\nThat reduces the problem to 5 single-degree-of-freedom equations. Let's do it.\n\n100/1 = 2n, n = 50, seq is (50)\n100/2 = 2n + 1, no solution\n100/4 = 2n + 4 - 1, 25-3 = 2n, n = 11, seq is (11, 12, 13, 14)\n100/5 = 2n + 5 - 1, 20 = 2n + 4, 16=2n, n = 8, seq is (8, 9, 10, 11, 12)\n100/10 = 2n + 10 - 1, 10 = 2n +9, 1 = 2n, no solution\n\n\nSo those are all the solutions where n and s both > 0.\n\nUsing the standard summation of an arithmetic progression formula:\n$S_n=\\frac{n(2a_1+(n-1)d)}{2}$\nhere since $d=1$\n$2S_n=n(2a_1+n-1)$\n$2S_n=n(a_1+(a_1+n-1))$\nHere $a_1+n-1$ is just the last term.\n$2S_n=n(a_1+a_n)$\nRewrite as\n\nIf n is even:\n\n### $n=\\frac{2S_n}{2a_{n/2}}$\n\nSince $a_1+a_{n/2}+a_n-a_{n/2}=2a_{n/2}$\nhence\n\n### $n=\\frac{S_n}{a_{n/2}}$\n\nand since $n\\in Z$ , so $a_{n/2}$ is a divisor of $S_n$\nonce you get the required $n, a=a_{n/2}-n/2$\nSimilarly for odd $S_n$ you get\n\n### $n=\\frac{S_n}{a_{(n-1)/2}-1}$\n\nwhere $a_{(n-1)/2}-1$ is a divisior of $S_n$\n\nThe sum of consecutive numbers $1\\dots n$ up to $n$ starting from 1 is $\\frac{n(n+1)}{2}$. Since you want only a partial sum from, say, $m+1$ to $n$, you can just subtract to get the partial sum:\n\n$$\\frac{n(n+1)}{2} - \\frac{m(m+1)}{2} = \\frac{n^2+n-m^2-m}{2} = 50$$\n\nMultiplying by 2 gets you\n\n$$100 = n^2+n-m^2-m = (n+m)(n-m) + n-m = (n+m+1)(n-m)$$\n\nNow the trick: $n+m$ and $n-m$ are either both even or both odd. This means that in the last formula, one term must be even, the other one must be odd. Factorization of $100 = 5\\cdot 5 \\cdot 2 \\cdot 2$ means that there are very limited solutions. For instance, one term is 25, the other is 4 (there is one other nontrivial combination, see below). Since $n+m+1$ is larger than $n-m$, in the choice shown here, you must have \\begin{align} n+m+1 &= 25\\\\ n-m &=4 \\end{align} Solving this will give you the desired numbers.\n\nAddendum You get the combination by observing that you get all acceptable combinations of the factorisation via: \\begin{align} 100 &= 1\\cdot (5\\cdot5\\cdot2\\cdot2)\\\\ &= 5\\cdot (5\\cdot2\\cdot2)\\\\ &= (5\\cdot 5)\\cdot(2\\cdot2) \\end{align} where you have to stop as all the following factorisations will contain only even factors.\n\nLet tere be $n\\geq1$ numbers, the smallest of them being $p\\in{\\mathbb Z}$. We then want $$p+(p+1)+\\ldots+\\bigl(p+(n-1)\\bigr)=50\\ ,$$ which amounts to $np+{(n-1)n\\over2}=50$, or $$n(n+2p-1)=100\\ .$$ Going with $n$ through the $9$ divisors of $100$ we obtain the following table: $$\\matrix{n:&1&2&4&5&10&20&25&50&100 \\cr p:&50&&11&8&&-7&-10&&-49\\cr}$$ When $n\\in\\{2,10,50\\}$ solving $n+2p-1={100\\over n}$ for $p$ does not lead to an integer $p$. It follows that there are $6$ solutions in all.\n\n\u2022 How did you get from your first equation to the second? \u2013\u00a0Tony Ennis Jun 25 '17 at 21:26\n\nFirst, welcome to Mathematics stack exchange. Second, if you start like $x+(x+1)+...+(x+n)$, you will end nowhere. $50$ is a concrete and small number so what can you try is to use that fact, sum of two consecutive numbers it is not, due to sum of two consecutive numbers can not be divided with $2$ in $\\mathbb{Z}$, sum of three consecutive numbers is divisible with $3$, so it is not sum of three numbers. Four consecutive numbers- now this is on first look possible(remainder mod $4$ is $2$) but each one of them has to be around $12$, so first options that you have is $11+12+13+14=23+27= 50$, hence the solution. Maybe there are even other solution, but I feel free to interpret your question as required to see one possible solution.\n\n\u2022 There are other solutions. The simplest is $50$. \u2013\u00a0N. F. Taussig Jun 24 '17 at 10:42", "date": "2020-02-21 17:30:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2334607/the-sum-of-consecutive-integers-is-50-how-many-integers-are-there/2334733", "openwebmath_score": 0.9552640318870544, "openwebmath_perplexity": 340.1789331339958, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274401, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8584529740026536}}
{"url": "https://tex.stackexchange.com/questions/269031/x-1-x-2-vs-x-1-x-2", "text": "\\$x_1,x_2\\$ vs. \\$x_1\\$, \\$x_2\\$\n\nBasic version of my question: Consider the fragment of some text `bla bla \\$x_1,x_2,x_3\\$ bla`. Is this correctly type-setted, or is `bla bla \\$x_1\\$, \\$x_2\\$, \\$x_3\\$ bla` better ?\n\nAdvanced version: Cf. the \"Chicago Manual of Style\", 16th printed ed, on pp. 589, in 12.19 it is recommended that typesetting lists of mathematical symbols, e.g. `\\$x_1,x_2,x_3\\$` that medium spaces between commas should be used. Does LaTeX insert these automatically ? If not, should I use `\\$x_1\\:,x_2,\\:x_3\\$` rather than `\\$x_1\\$, \\$x_2\\$, \\$x_3\\$` ? It appears that these strings differ slightly in length, when the document is compiled to pdf.\n\n\u2022 It matters greatly what the three items `x_1`, `x_2`, and `x_3` -- or, say, `a`, `b`, and `c` -- denote: Are they just any three distinct elements, or do they form a sequence? In the former case, you should write `... \\$x_1\\$, \\$x_2\\$, \\$x_3\\$ ...`: the commas are parts of the sentence rather than parts of the formulas. In the latter case, you should definitely write `\\$... x_1, x_2, x_3 ...\\$`, as the commas are now a part of the overall math expression. If they form a sequence, you may also want to encase them in curly braces, i.e., write `\\$\\{x_1, x_2, x_3\\}\\$`. \u2013\u00a0Mico Sep 23 '15 at 10:45\n\u2022 @Mico Sounds like an answer? \u2013\u00a0Gonzalo Medina Sep 23 '15 at 15:03\n\u2022 @GonzaloMedina - Done. :-) \u2013\u00a0Mico Sep 23 '15 at 15:43\n\nIt matters greatly what the three items `x_1`, `x_2`, and `x_3` -- or `a`, `b`, and `c` -- denote. Are they just any three distinct elements, or do they form a structured entity such as a sequence?\n\nIn the former case, you should write, say,\n\n``````The sum of \\$x_1\\$, \\$x_2\\$, \\$x_3\\$, etc.\\ diverges because ...\n``````\n\nDon't include the commas in the math terms, because the commas are parts of the sentence rather than components of formulas. With this setup, line breaks can occur (if needed) after the commas.\n\nIn the latter case, though, you should definitely write\n\n``````\\$... x_1, x_2, x_3 ...\\$\n``````\n\nbecause the commas are now parts of some larger math expression. And, if `x_1`, `x_2`, and `x_3` form, say, a three-element set, you may want to encase them in curly braces, i.e., write\n\n``````\\$\\{x_1, x_2, x_3\\}\\$\n``````\n\nTeX will, in general, not insert line breaks after math-mode commas. If you must allow a line break in the formula and if the commas are sensible break-points, you'll need to tell TeX about this fact, by inserting judiciously placed `\\allowbreak` directives, say,\n\n``````\\$\\{a_1,a_2,\\dots,a_n,\\allowbreak a_{n+1}, \\dots, a_{n+m}\\}\\$\n``````\n\u2022 you might also want to mention that, when entered as separate math strings, a line can break after one of the commas, but if it's a single math expression, it won't break at the end of a line. \u2013\u00a0barbara beeton Sep 23 '15 at 16:05\n\u2022 @barbarabeeton - Good idea. :-) I'll add a sentence or two to this effect. \u2013\u00a0Mico Sep 23 '15 at 16:13\n\u2022 First: To such a complete and carefully crafted answer, I cannot give anything but a thankful +1! Second: In the text I'm writing, `\\$x_1\\$` to `\\$x_3\\$` are something in between unrelated items and a structured entity: I'm dividing a line in three segments, called `\\$A_1\\$`, `\\$A_2\\$` and '`\\$A_3\\$` and I constantly have to mention things like \"On the segments `\\$A_1\\$`, `\\$A_2\\$` happens X\"; \"in contrast, on the segment \\$A_2\\$`, `\\$A_3\\$` we have to deal with the problem Y\"; \"when looking at `\\$A_1\\$`, `\\$A_2\\$` and '\\$A_3\\$` together we can see Z\". Would you agree that I should write them apart, as I just did ? \u2013\u00a0l7ll7 Sep 24 '15 at 13:54\n\u2022 [...] Although, they do form an entity, as they are a partition of a line, which makes me somewhat unsure. \u2013\u00a0l7ll7 Sep 24 '15 at 13:56\n\u2022 @user10324 - I'd write the segments as separate formulas, i.e., `on the segments \\$A_1\\$ and \\$A_2\\$, \\$X\\$ happens...`. On the other hand, I'd also write `Let \\$(A_1,A_2,A_3)\\$ be an exhaustive, ordered, and non-overlapping partition of the line segment \\$A\\$.` \u2013\u00a0Mico Sep 24 '15 at 15:01", "date": "2020-11-25 08:44:43", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/269031/x-1-x-2-vs-x-1-x-2", "openwebmath_score": 0.9517483115196228, "openwebmath_perplexity": 1290.1594846752444, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9532750373915658, "lm_q2_score": 0.9005297907896396, "lm_q1q2_score": 0.8584525699872126}}
{"url": "https://math.stackexchange.com/questions/3134268/question-about-asymptotic-notation-with-logs-of-different-bases", "text": "# Question about asymptotic notation with logs of different bases\n\nFor the two functions f(n) and g(n), where $$f(n) = n^{1/2}$$ and $$g(n) = 2^{(\\log_2 n)^{1/2}}$$, I am trying to determine whether $$f$$ is asymptotically bound below by $$g$$, i.e find whether there exists an N1 and a k1 > 0 such that:\n\n$$f(n) >= k1 * g(n)$$, for all n >= N1.\n\nBy taking the log of both $$f(n)$$ and $$g(n)$$ I got:\n\n$$f(n) = n^{1/2} = logn^{1/2} = 1/2*logn$$.\n\n$$g(n) = 2^{(\\log_2 n)^{1/2}} = log(2^{(\\log_2 n)^{1/2}}) = (\\log_2 n)^{1/2}log2$$.\n\nIn trying to solve this I have:\n\n$$1/2logn >= k1 * (\\log_2 n)^{1/2}log2$$\n\nSetting k1 to 1/2:\n\n$$1/2logn >= 1/2 * (\\log_2 n)^{1/2}log2$$\n\nDividing both sides by 1/2:\n\n$$logn >= (\\log_2 n)^{1/2}log2$$\n\nSubtracting $$(\\log_2 n)^{1/2}log2$$ from both sides:\n\n$$logn - (\\log_2 n)^{1/2}log2 >= 0$$\n\nI can probably figure out whether there is any N1 such that the above is true for all n >= N1 by substituting values. However, I was wondering if there is any way to further simplify the above expression. In particular is there any way to get rid of the $$(\\log_2 n)^{1/2}$$ entirely? I would prefer all the logs in the inequality to have the same base. Any insights are appreciated.\n\n\u2022 In my answer that you've accepted, I made a mistake which I've now corrected. In particular, I used $\\log(ab) = \\log(a) \\log(b)$, which is what you've used as well, instead of the correct $\\log(ab) = \\log(a) + \\log(b)$, when you take logarithms of both sides of the original inequality. I'm sorry for my mistake, but as you can see it does't change the overall result. \u2013\u00a0John Omielan Mar 4 at 4:33\n\n## 1 Answer\n\nYou could finish your work using $$\\log$$, I assume to base $$e$$ but, in answer to your request of\n\nI would prefer all the logs in the inequality to have the same base.\n\nsince $$g(n)$$ uses a base of $$2$$ and has a $$\\log$$ to the base of $$2$$ in the exponent, it would likely be simplest & easiest to instead take the logs of both sides to base $$2$$ so you're then comparing the exponents of $$f(n)$$ and $$g(n)$$ to that base. As such, this would change trying to prove that there exists an $$N_1$$ and a $$k_1 \\gt 0$$ such that\n\n$$f(n) \\ge k_1 \\, g(n) \\; \\forall \\; n \\ge N_1 \\tag{1}\\label{eq1}$$\n\nto finding, by taking $$\\log_2$$ of both sides, a $$k_2 = \\log_2{k_1}$$ such that\n\n$$\\log_2{f(n)} \\ge k_2 + \\log_2{g(n)} \\; \\forall \\; n \\ge N_1 \\tag{2}\\label{eq2}$$\n\nNote that\n\n$$\\log_2{f(n)} = \\log_2{n^{\\frac{1}{2}}} = \\frac{1}{2} \\, log_2 n \\tag{3}\\label{eq3}$$ $$\\log_2{g(n)} = \\log_2{2^{(\\log_2 n)^{\\frac{1}{2}}}} = (\\log_2 n)^{\\frac{1}{2}} \\tag{4}\\label{eq4}$$\n\nSubstituting \\eqref{eq3} and \\eqref{eq4} into \\eqref{eq2} gives\n\n$$\\frac{1}{2} \\, \\log_2 n \\ge k_2 + (\\log_2 n)^{\\frac{1}{2}} \\; \\forall \\; n \\ge N_1 \\tag{5}\\label{eq5}$$\n\nIn this case, consider values of $$m$$ where $$\\frac{1}{2}m^2 \\ge m$$. Multiplying both sides by $$2$$, moving $$2m$$ to the LHS and factoring gives $$m\\left(m - 2\\right) \\ge 0$$. This is true for all $$m \\ge 2$$. In this case, if we let $$m = (\\log_2 n)^{\\frac{1}{2}}$$, we can see that having $$k_2 = 0$$ and since $$(\\log_2 16)^{\\frac{1}{2}} = 2$$, we can use $$N_1 = 16$$. I believe you should be able to finish the rest.\n\nNote if you had used your original idea of, I assume, the natural logarithm, it would just involve an extra factor by the change of logarithm base formula, i.e., $$\\log_a{x} = \\cfrac{\\log_b{x}}{\\log_b{a}}$$ to give in this case that $$\\log_2{n} = \\cfrac{\\log_e{n}}{\\log_e{2}}$$. Also, as I show, you would need to add the constant, not multiply by it. After that correction, this just results in an extra factor being used for $$\\log_2{n}$$, plus you also have the $$\\log_e{2}$$ factor, so it might change the $$k_2$$ constant you determine to use (but not necessarily the $$k_1$$ constant) and/or the value of $$N_1$$, as these values are not uniquely determined.", "date": "2019-11-14 14:55:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3134268/question-about-asymptotic-notation-with-logs-of-different-bases", "openwebmath_score": 0.9265673160552979, "openwebmath_perplexity": 150.20313641062296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.965899570361222, "lm_q2_score": 0.8887587949656841, "lm_q1q2_score": 0.8584517382121116}}
{"url": "https://math.stackexchange.com/questions/3457247/what-is-missing-in-my-solution-of-from-pdf-to-cdf-and-px-0-5", "text": "# What is missing in my solution of \u201cfrom PDF to CDF and $P(X > 0.5)$\u201d?\n\nThe continuous random variable $$X$$ is described with the following probability density function (pdf):\n\n$$f_X(x) = \\begin{cases} \\frac{1}{9}\\big(3 + 2x - x^2 \\big) \\; : 0 \\leq x \\leq 3 \\\\ 0 \\; \\;: x < 0 \\; \\lor \\; x > 3\\end{cases}$$\n\nFind cumulative distribution function $$F_X$$ and probability $$P(X > 0.5)$$.\n\nThe task is started by verifying if the pdf is in fact correct pdf. I am checking two conditions:\n\n1. Is the pdf nonnegative on all of its domain? Yes, hence we can write:\n\n$$\\forall_{x \\in \\mathbb{R}}\\;f_X(x) \\geq 0$$\n\n1. The pdf has to be integrable and its total area under the curve has to be equal $$1$$:\n\n\\begin{align*} &\\int_{\\mathbb{R}}f_X = 1 \\\\ &\\color{red}{\\int_{-\\infty}^{\\infty}f_X(x)dx = 1} \\\\ \\end{align*}\n\n(for now assume the condition is true)\n\nPDF plot:\n\nComputing CDF which is defined as:\n\n$$F_X(x) = \\int_{-\\infty}^{x}f_X(t)dt$$\n\nTherefore:\n\nIf $$x < 0$$:\n\n$$F_X(x) = \\int_{-\\infty}^{x} 0dt = 0$$\n\nIf $$x \\geq 0 \\; \\land \\; x \\leq 3$$:\n\n\\begin{align*}F_X(x) &= \\int_{-\\infty}^{0}0dt + \\int_{0}^{x}\\frac{1}{9}\\big(3 + 2t - t^2\\big)dt = \\\\ &= 0 + \\frac{1}{9}\\Big(3t + t^2 - \\frac{1}{3}t^3 \\Big)\\Bigg|^{x}_0 = \\\\ &= \\frac{1}{9} \\Big(3x + x^2 - \\frac{1}{3}x^3 \\Big)\\end{align*}\n\nIf $$x \\geq 3$$:\n\n\\begin{align*} F_X(x) &= \\int_{-\\infty}^{0}0dt + \\int_{0}^{3}\\frac{1}{9}\\Big(3 + 2t - t^2 \\Big)dt + \\int_{3}^{x}0dt \\\\ &= 0 + \\frac{1}{9}\\Big(3t + t^2 - \\frac{1}{3}t^3 \\Big)\\Bigg|^3_0 + 0 = \\\\ &= 1 \\end{align*}\n\n(this implicitly confirms the $$\\color{red}{\\text{red}}$$ condition)\n\nFinally the CDF is defined as:\n\n$$F_X(x) = \\begin{cases} 0 \\; \\; : x < 0 \\\\ \\frac{1}{9} \\Big(3x + x^2 - \\frac{1}{3}x^3 \\Big) \\; \\; : x \\geq 0 \\; \\land \\; x \\leq 3 \\\\ 1 \\; \\; : x > 3 \\end{cases}$$\n\nThe CDF result agrees with:\n\n$$\\lim_{x \\to \\infty}F_X(x) = 1 \\; \\land \\; \\lim_{x \\to -\\infty}F_X(x) = 0$$\n\nAlso the function is non-decreasing and continuous.\n\nCDF plot:\n\n## Calculating $$P(X > 0.5)$$:\n\n\\begin{align*}P(X > 0.5) &= \\int_{0.5}^{\\infty}f_X(x)dx = \\\\ &= \\int_{0.5}^{3}\\frac{1}{9}(3+2x-x^2)dx + \\int_{3}^{\\infty}0dx = \\\\ &= \\frac{1}{9} \\Big(3x + x^2 - \\frac{1}{3}x^3 \\Big)\\Bigg|^3_{0.5} + 0 = \\\\ &= \\frac{175}{216} \\approx 0.81\\end{align*}\n\nThis probability solution does not agree with the book's solution.\n\nThe book says $$P(X > 0.5) = 1 - F_X(0.5) = \\frac{41}{216} \\approx 0.19$$, so it's my solution \"complemented\".\n\n## My questions:\n\n\u2022 Which final probability solution is correct?\n\u2022 Is this any special kind of probability distribution, e.g. Poisson or Chi Square (well, not these)?\n\u2022 Can you please point out all minor or major mistakes I have made along the way? (perhaps aside from plots that are not perfect). This is the most important for me.\n\u2022 What have I forget to mention or calculate for my solution to make more sense? Especially something theoretical, perhaps e.g. definition for $$X$$.\n\u2022 Looks like the book have a bug. \u2013\u00a0kludg Nov 30 '19 at 18:04\n\n## My questions:\n\n\u2022 Which final probability solution is correct?\n\nYours answer is right and the book's isn't. They presumably have mistakenly computed $$\\mathbb P(X < 0.5)$$ instead of $$\\mathbb P(X > 0.5)$$.\n\n\u2022 Is this any special kind of probability distribution, e.g. Poisson or Chi Square (well, not these)?\n\nNot a common one, no. I found this page on \"U-quadratic distributions\" (a term I've never heard before), and this would be the vertical inverse of one of these described in the \"related distributions\" section, but I don't think this is a particularly common term or distribution.\n\nEDIT: Whoops, this isn't even quite the vertical inverse of a U-quadratic distribution, is it? Such a distribution would apparently not truncate the left side of the parabola as this one does. The better answer to your question is: \"No, this distribution is neither named nor important.\"\n\n\u2022 Can you please point out all minor or major mistakes I have made along the way? (perhaps aside from plots that are not perfect). This is the most important for me.\n\nI'd love to, but I didn't find any!\n\n\u2022 What have I forget to mention or calculate for my solution to make more sense? Especially something theoretical, perhaps e.g. definition for $$X$$.\n\nI didn't spot any holes or anything that needs to be improved.\n\nEDIT: One thing you could do to clean this up a bit: when you compute $$\\mathbb P(X > 0.5)$$, you're redoing the integration you already did in your CDF. Instead, you could just use that result that you already obtained: $$\\mathbb P(X > 0.5) = 1 - \\mathbb P(X \\leq 0.5) = 1 - F_X(0.5) = 3(0.5) + (0.5)^2 - \\frac{1}{3}(0.5)^3 = \\dots$$ That said, your answer isn't wrong, it's just a bit inefficient.\n\n\u2022 So the efficient way is to compute the CDF at $x=0.5$ and subtract the result from $1.$ It seems likely that whoever wrote the answer key forgot to subtract. I likewise did not find any mistake in the calculations in the question. \u2013\u00a0David K Nov 30 '19 at 18:30\n\u2022 Can I write:$$f_X: \\; \\mathbb{R} \\longrightarrow \\left[0, \\frac{4}{9} \\right]$$ where $\\frac{4}{9}$ denotes mode of PDF? And also, $$F_X : \\; \\mathbb{R} \\longrightarrow \\left[0,1\\right]$$ for the CDF? Thanks for help. I will accept today. \u2013\u00a0weno Dec 1 '19 at 4:56\n\u2022 @weno Sure - both those statements are true, though the first is questionably useful and the second is trivially true for all CDFs. \u2013\u00a0Aaron Montgomery Dec 1 '19 at 4:58", "date": "2020-04-08 22:42:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3457247/what-is-missing-in-my-solution-of-from-pdf-to-cdf-and-px-0-5", "openwebmath_score": 0.9998186230659485, "openwebmath_perplexity": 412.9118283504255, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9609517072737735, "lm_q2_score": 0.8933094152856196, "lm_q1q2_score": 0.8584272077424525}}
{"url": "https://math.stackexchange.com/questions/3671953/area-of-the-part-of-the-cylinder-x2y2-2ay-outside-the-cone-z2-x2y2", "text": "# Area of the part of the cylinder $x^2+y^2=2ay$ outside the cone $z^2=x^2+y^2$\n\nProblem: Find the area of the part of the cylinder $$x^2+y^2=2ay$$ that lies outside the cone $$z^2=x^2+y^2$$.\n\nMy attempt: So I thought we could do this by projecting the surface onto the $$yz$$-plane and taking the surface integral of the function $$x=g(y,z)=\\sqrt{z^2-y^2}$$. I.e letting $$S$$ be the surface and $$E$$ be the projection onto the $$yz$$-plane where we have a $$2$$ before the integral over $$E$$ since we have both $$x<0$$ and $$0\\leq x$$: \\begin{align*}\\iint_{\\mathcal{S}}x \\ \\mathrm{d}S &=2\\iint_{E}x\\underbrace{\\sqrt{1+\\left(\\frac{\\partial x}{\\partial y}\\right)^2+\\left(\\frac{\\partial x}{\\partial z}\\right)^2} \\ \\mathrm{d}z\\mathrm{d}y}_{\\mathrm{d}S} \\\\ &=2\\iint_{E}x\\sqrt{1+\\frac{z^2}{x^2}+\\frac{y^2}{x^2}} \\ \\mathrm{d}z\\mathrm{d}y\\\\ &=2\\iint_{E}\\sqrt{x^2+z^2+y^2}\\ \\mathrm{d}z\\mathrm{d}y\\\\ &=2\\iint_{E}\\sqrt{2}z\\ \\mathrm{d}z\\mathrm{d}y \\end{align*} Now in the projection it seems to me that we have the following bounds on $$z$$ and $$y$$ since the cylinder has radius $$a$$ and the cone and the surface intersect at $$z=\\sqrt{2ay}$$ $$0\\leq z \\leq \\sqrt{2ay} \\quad \\text{and} \\quad 0\\leq y \\leq 2a$$ so: \\begin{align*}2\\iint_{E}\\sqrt{2}z\\ \\mathrm{d}z\\mathrm{d}y &= \\sqrt{2}\\int_{0}^{2a}\\int_{0}^{\\sqrt{2ay}}2z \\ \\mathrm{d}z\\mathrm{d}y \\\\ &=\\sqrt{2}\\int_{0}^{2a} 2ay \\ \\mathrm{d}y\\\\ &=4\\sqrt{2}a^{2}\\end{align*} However my book says its $$16a^2$$ so what is my mistake(s)?\n\nPS. I think this is also possible with polar coordinates but I would like to use the surface integral with projection onto the $$yz$$-plane.\n\nPSDS. Picture is not totally acurate as $$a=4$$\n\nEdit:\n\nAs Ninad Munshi pointed out I was projecting the wrong surface and I used the wrong formula for the surface area. My thoughts are\n\nWould it be correct to say that $$\\iint\\mathrm{d}S$$ is the surface area, and would $$\\mathrm{d}S$$ be $$\\sqrt{1+\\left( \\frac{a-y}{\\sqrt{2ay-y^2}} \\right)^2} dzdy$$? If so I still seem to be off by a factor of $$2$$ as \\begin{align*}\\iint_{\\mathcal{S}} \\mathrm{d}S &= 2 \\iint_{E}\\sqrt{1+\\left( \\frac{a-y}{\\sqrt{2ay-y^2}} \\right)^2}dzdy \\\\ &=2\\int_{0}^{2a}\\int_{0}^{\\sqrt{2ay}}\\sqrt{1+\\left( \\frac{a-y}{\\sqrt{2ay-y^2}} \\right)^2}dzdy=8a^{2}\\end{align*}\n\n\u2022 You used the wrong surface in the beginning. It wants the area of the cylinder, not the cone, so you have to project the cylinder down. \u2013\u00a0Ninad Munshi May 12 '20 at 21:33\n\u2022 Also, $\\iint x\\:dS$ does not give you surface area. \u2013\u00a0Ninad Munshi May 12 '20 at 21:35\n\u2022 @NinadMunshi I solved it, Thank you for your help! \u2013\u00a0Andr\u00e9 Armatowski May 13 '20 at 2:42\n\nThe correct way to solve this question is to start with the cylinder $$x^2+y^2=2ay$$ that we wish to project onto the $$yz$$ plane. This is done by first calculating $$\\mathrm{d}S$$ in $$\\iint_{\\mathcal{S}}dS$$ which gives the surface area\nWe have that $$dS = \\sqrt{1+\\left(\\frac{\\partial x}{\\partial y}\\right)^{2}+\\left(\\frac{\\partial x}{\\partial z}\\right)^{2}} \\ \\mathrm{d}z\\mathrm{d}y=\\sqrt{1+\\frac{(a-y)^{2}}{2ay-y^2}}\\ \\mathrm{d}z\\mathrm{d}y$$\nNow the problem I had is that when I projected the cylinder: I only considered the symmetric areas of $$x<0$$ and $$0\\leq x$$ while we infact have two more symmetries: namely $$z<0$$ and $$0\\leq z$$.\nIn summary: We have four areas that are equal (and not two) so letting $$E$$ represent the area of the projection of the cylinder onto the $$yz$$-plane in the first octant we get: $$\\iint_{\\mathcal{S}}\\mathrm{d}S=4\\iint_{E}\\sqrt{1+\\frac{(a-y)^{2}}{2ay-y^2}}\\ \\mathrm{d}z\\mathrm{d}y$$ The particular limits of $$z$$ and $$y$$ are still correct that is $$0\\leq z \\leq \\sqrt{2ay} \\quad \\text{and} \\quad 0\\leq y \\leq 2a$$ So: \\begin{align*}4\\iint_{E}\\sqrt{1+\\frac{(a-y)^{2}}{2ay-y^2}}\\ \\mathrm{d}z\\mathrm{d}y & = 4\\int_{0}^{2a}\\int_{0}^{\\sqrt{2ay}}\\sqrt{1+\\frac{(a-y)^{2}}{2ay-y^2}}\\ \\mathrm{d}z\\mathrm{d}y \\\\ &= 4 \\int_{0}^{2a}\\sqrt{2ay}\\sqrt{1+\\frac{(a-y)^{2}}{2ay-y^2}}\\ \\mathrm{d}y\\\\ &= 4(4a^2)=16a^2\\end{align*} Which is the correct answer", "date": "2021-04-16 15:14:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3671953/area-of-the-part-of-the-cylinder-x2y2-2ay-outside-the-cone-z2-x2y2", "openwebmath_score": 0.9991146922111511, "openwebmath_perplexity": 387.3547735146174, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013790564298, "lm_q2_score": 0.8740772482857833, "lm_q1q2_score": 0.8583559118351332}}
{"url": "https://math.stackexchange.com/questions/1312058/find-probability-of-exactly-one-6-in-first-ten-rolls-of-die-given-two-6s-in", "text": "# Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls\n\nI am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.\n\n### My thoughts\n\nLet $A = \\{\\text {Exactly one 6 in first ten rolls of a die} \\}$ and $B = \\{\\text {Exactly two 6s in twenty rolls of a die} \\}.$\nThen I want to find $$P[A\\mid B] = \\frac{P[A \\cap B]}{P[B]}.$$\n\nBy the binomial distribution formula, we get that $$P[B] = {20 \\choose 2} \\cdot \\left(\\frac{1}{6}\\right)^2 \\cdot \\left(\\frac{5}{6}\\right)^{18}.$$\n\nFurthermore I think that $P[A \\cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls. That is,\n$$P[A \\cap B] = \\left[{10 \\choose 1} \\cdot \\left(\\frac{1}{6}\\right)^1 \\cdot \\left(\\frac{5}{6}\\right)^9\\right]^2.$$\n\nAm I correct in thinking this?\nIf so, then it follows that the required probability is $$P[A \\mid B] = \\frac{\\left[{10 \\choose 1} \\cdot \\left(\\frac{1}{6}\\right)^1 \\cdot \\left(\\frac{5}{6}\\right)^9\\right]^2}{{20 \\choose 2} \\cdot \\left(\\frac{1}{6}\\right)^2 \\cdot \\left(\\frac{5}{6}\\right)^{18}},$$ which, I know, can be simplified further!\n\n\u2022 Shouldn't $P[B]$ be $$P[B] = {20 \\choose 2} \\cdot \\left(\\frac{1}{6}\\right)^2 \\cdot \\left(\\frac{5}{6}\\right)^{18}.$$ ? \u2013\u00a0Mohamad Misto Jun 4 '15 at 14:45\n\u2022 A curiosity is that in this case $P(B \\mid A)=P(A)$ so $P(A \\cap B)=P(A^2)$ and $P(A \\mid B)=\\dfrac{P(A)^2}{P(B)}$ \u2013\u00a0Henry Jun 4 '15 at 14:59\n\u2022 Well laid out. Apart from a minor slip, all correct. You will notice near total collapse when you simplify. Now can you simplify the argument? \u2013\u00a0Andr\u00e9 Nicolas Jun 4 '15 at 14:59\n\u2022 You should get $10/19$. \u2013\u00a0Andr\u00e9 Nicolas Jun 4 '15 at 15:26\n\u2022 There are $\\binom{20}{2}$ equally likely ways to choose the locations of the two $6$'s. There are $(10)^2$ ways to choose one in each half. So the probability is $100/\\binom{20}{2}$. \u2013\u00a0Andr\u00e9 Nicolas Jun 4 '15 at 15:29\n\nI took a different approach to the question. Suppose B. There are three ways to get two 6's in twenty rolls:\n\n\u2022 $B_1$: Both 6's come in the first 10 rolls. There are $\\begin{pmatrix} 10 \\\\ 2\\end{pmatrix} = 45$ ways for this to happen.\n\u2022 $B_2$: One 6 comes in the first 10 rolls, and the second comes in the next 10 rolls. There are $\\begin{pmatrix} 10 \\\\ 1\\end{pmatrix} \\begin{pmatrix} 10 \\\\ 1\\end{pmatrix} = 100$ ways for this to happen.\n\u2022 $B_3$: Both 6's come in the second lot of 10 rolls. There are $\\begin{pmatrix} 10 \\\\ 2\\end{pmatrix} = 45$ ways for this to happen.\n\nNow $$P(A|B) = P(B_2|B_1 \\cup B_2 \\cup B_3) = \\frac{100}{45+100+45} = \\frac{10}{19}.$$\n\n\u2022 Your reasoning seems sound to me. It doesn't agree with all of the other answers, though. I am not yet sure why... \u2013\u00a0Caleb Owusu-Yianoma Jun 4 '15 at 15:24\n\nI think you are over thinking this. We know we get exatly two sixes in twenty rolls how many ways can that happen? Consider a roll to be 6 or not 6 we don't care what number it is otherwise.\n\nOne of the sixes arived in any of the twenty rolls and the other in an of the nineteen remaining rolls and since a 6 is a 6 we divide by two because the order does not matter.\n\nThere are thus $\\dfrac{20 \\cdot 19}{2} = 190$ ways we can get exactly 2 6's in 20 rolls.\n\nIn how many ways can we have exactly one 6 in 10 rolls? Well it can be any of the 10 rolls, and it must be not 6 in the other 9 so there are 10 ways to get exactly one 6 in ten rolls, and 10 ways to get the second 6 in the last 10 rolls making\n\nThe answer is simply $\\dfrac{10 \\cdot 10}{190} = \\dfrac{100}{190} = \\dfrac{10}{19}$.\n\n\u2022 Surely, when considering the number of ways in which we can obtain exactly one $6$ in the first ten rolls, we must account for the fact that we obtain a $6$ in the second set of ten rolls, too. In that case, the number of ways of rolling exactly one $6$ in the first ten rolls AND exactly one $6$ in the second ten rolls is $10^2 = 100.$ Is my reasoning sound? \u2013\u00a0Caleb Owusu-Yianoma Jun 4 '15 at 15:20\n\u2022 @CKKOY there are 100 ways you can roll exacly one 6 in the first 10 rolls and exacly 1 six in the next 10 rolls but that's not what the question is asking. The question is asking how many ways you get exactly 1 six in the first 10 rolls given all the ways you can get 2 6's in 20 rolls including one where both 6's occur in the first 10 rolls or both 6's occur in the last 10 rolls. \u2013\u00a0Warren Hill Jun 4 '15 at 15:26\n\u2022 Is your interpretation of the question equivalent to the following? 'The question is asking how many ways you can roll exactly one $6$ in the first ten rolls (out of twenty), given that you rolled two $6$s in the total twenty rolls.' Please see the other answers and comments, which imply that the correct answer is $10/19$. \u2013\u00a0Caleb Owusu-Yianoma Jun 4 '15 at 15:30\n\u2022 @CKKOY sorry you are correct. I've corrected my answer. \u2013\u00a0Warren Hill Jun 4 '15 at 15:35\n\u2022 Apology accepted. \u2013\u00a0Caleb Owusu-Yianoma Jun 4 '15 at 15:40\n\nSince all drawings are independent, is confusing to use the conditional probability formula and it is not necessary at all. Think as you are rolling simultaneously two dices 10 times each. Having in both groups exactly one 6 has the same probability p. Because these two are independent the simultaneous event (one 6 in each group) has the p^2 squared probability. The answer is what you thought A and B was: [(10C1)\u22c5(1/6)\u22c5(5/6)^9]^2 = 95,367,431,640,625/914,039,610,015,744 = 0.104336. (far to be 10/19)\n\nThis is what Adelafif said too, but he made an error on the coefficient n-k=9 (not 5).\n\nWhat Paul Wright said is very funny but totally wrong. Having one 6 in (\"the first\") 10 rollings has the same probability independently how many time you are rolling the dice after!. This probability is (10C1)\u22c5(1/6)^2\u22c5(5/6)^8 (much less than Paul calculated). With Paul's theory, we should win a lottery pot considerably easier.\n\n\u2022 Your answeris wrong, and $10\\over 19$ is the correct one. According to mathematics. \u2013\u00a0user228113 Jun 4 '15 at 16:43\n\u2022 @Laszlo: Your calculation of $0.104336$ is the rounded value of $P[A \\cap B].$ However, the final answer, $10/19$, is roughly equal to $0.104336/{20 \\choose 2} \\cdot \\left(\\frac{1}{6}\\right)^2 \\cdot \\left(\\frac{5}{6}\\right)^{18}.$ \u2013\u00a0Caleb Owusu-Yianoma Jun 4 '15 at 16:48\n\u2022 @G.Sassatelli Mathematics is not a matter of enouncing something. You must argument your opinion, as I did. What is wrong with \"my mathematics\"? \u2013\u00a0Laszlo Jun 4 '15 at 18:01\n\u2022 @CKKOY Exactly, as I said. You DO NOT need the denominator. \u2013\u00a0Laszlo Jun 4 '15 at 18:05\n\u2022 To avoid any other useless dispute: I made a simulation for one million of 20 times rolling sets (it is only two lines in Mathematica!). The number of cases with one six in the first 10 and the second 10 rolls was: 104,524. Repeated this 100 times, got a normal distribution with the mean value exactly at 104336 (it took 20 minutes). Still sustain that my solution is wrong? \u2013\u00a0Laszlo Jun 4 '15 at 18:34\n\n((10C1)(1/6)(5/6)^5)(10C1)(1/6)(5/6)^5\n\n\u2022 It would be helpful if you wrote a brief explanation of the answer that you have written. \u2013\u00a0Caleb Owusu-Yianoma Jun 4 '15 at 15:13", "date": "2019-09-22 19:26:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1312058/find-probability-of-exactly-one-6-in-first-ten-rolls-of-die-given-two-6s-in", "openwebmath_score": 0.7959941029548645, "openwebmath_perplexity": 495.9888068254891, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462229680671, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8583541420646416}}
{"url": "https://math.stackexchange.com/questions/3902288/for-which-n-is-it-possible-to-arrange-all-whole-numbers-from-1-to-n-in-suc/3926332", "text": "# For which $N$ is it possible to arrange all whole numbers from $1$ to $N$ in such a way that every adjacent pair sums up to a Fibonacci number?\n\nRecently I came up with a problem regarding Fibonacci numbers:\n\nFor which $$N$$ is it possible to arrange all whole numbers from $$1$$ to $$N$$ in such a way that every adjacent pair sums up to a Fibonacci number?\n\nI have manually tested a bunch of cases and I was also able to prove almost every case. The results that I was able to prove are the following:\n\n\u2022 If $$N$$ is a Fibonacci number or exactly one less than a Fibonacci number, I was able to prove that an arrangement exists. I used an argument by induction to achieve this.\n\u2022 If $$F_k+2 \\leq N \\leq F_{k+1} -3$$ , it is completely impossible, because the numbers $$F_k$$, $$F_k + 1$$ and $$F_k + 2$$ have only one possible pair and therefore have to be at the end of the arragement. This obviously gives a contradiction, because arrangements only have two ends (the first number and the last number).\n\nThe cases I was not able to solve are $$N=F_k+1$$ and $$N=F_k-2$$. My theory is that $$N=9$$ is the only working case of the form $$N=F_k+1$$, and $$N=11$$ the only working case of the form $$F_k-2$$. I expect every other $$N$$ of these two forms to be impossible.\n\nDoes anybody know a full proof to this problem or maybe the name of the official theorem (if this exists)?\n\n\u2022 I ran some code to check if there are solutions for small numbers. everything you said checks out so far - 1-5, 7,8,9, 11 solvable, and also 12,13, 20,21, 33,34, 54,55. none of the others are solvable. I checked up to 55 so far. I'm gonna run the code at night, but I don't think it's gonna get much farther than that (the code isn't too smart). Nov 11 '20 at 0:07\n\u2022 The only sequence in the OEIS matching these conditions is oeis.org/A259624, which it sounds like is not your sequence if $N=6$ cannot be done. If you have confirmed values for $N$ up to $34$ or so, you may wish to submit it as a new sesquence. Nov 11 '20 at 3:45\n\u2022 @RavenclawPrefect problem is OEIS search engine finds $2,3,4,5,7,8,9,11,12,13,20,21,33,34,54,55$ but no result for $1,2,3,4,5,7,8,9,11,12,13,20,21,33,34,54,55$. Nov 28 '20 at 15:20\n\nThanks to @Rei Henigman data, I could find OEIS sequence A079734 and from there the \"Fibonacci Plays Billiards\" paper by Elwyn Berlekamp and Richard Guy, where they state with theorem 1 on page 3, that the only solutions are $$N = 9$$, $$N = 11$$, $$N = F_k$$, $$N = F_k \u2212 1$$ for $$k \\ge 4$$. Note that $$1$$ is excluded from the OEIS sequence.", "date": "2021-10-22 22:48:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3902288/for-which-n-is-it-possible-to-arrange-all-whole-numbers-from-1-to-n-in-suc/3926332", "openwebmath_score": 0.7190735340118408, "openwebmath_perplexity": 198.25555089183268, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462204837636, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8583541348729095}}
{"url": "https://math.stackexchange.com/questions/982564/help-with-implicit-differentiation-finding-an-equation-for-a-tangent-to-a-given", "text": "# Help with Implicit Differentiation: Finding an equation for a tangent to a given point on a curve\n\nWhen working through a problem set containing Implicit Differentiation problems, I've found that I keep getting the wrong answer compared to the one listed at the back of my book.\n\nThe problem is given as such: Use implicit differentiation to find the equation of the tangent line to the curve at a given point\n\nx^2 + xy + y^2 = 3\n\nWith given point (1, 1). I also am told that it is an ellipse.\n\nTo solve this, I evidently must differentiate both sides of the problem:\n\n1: dy/dx ( x^2 + xy + Y^2 ) = dy/dx(3)\n\n2: dy/dx (2x + 1y'+ 2yy') = 0\n\n3: 1y' + 2yy' = 0 - 2x\n\n4: y'(1+2y) = -2x\n\n5: y' = -2x/(1+2y)\n\nHurray, so now since I have the first derivative of Y. I can use it to find the slope at the point.\n\nSlope at Point (1,1)= -2(1) / (1+2(1)\n\nSlope at Point (1,1)= -2/3\n\nSo now that I've got my slope, I know the equation of the tangent will be in the form:\n\ny=mx+b\n\nSo, given I now know the slope:\n\ny=-2/3x + b\n\nSubstitute in the known point:\n\n1 = -2/3(1) + b\n\nb = 5/3\n\nSo the final answer I get is: y = -2/3x + 5/3\n\nBut according to the answer, it is supposed to be: -x + 2, I don't know where I went wrong, and I've done it twice to make sure I'm getting the same answer. Could someone please help me?\n\n\u2022 The differentiation of $xy$ requires the product rule: $\\frac{d}{dx}(xy)=y+xy'$. Oct 20, 2014 at 14:49\n\u2022 Consider the product rule on the term $xy$: It should read: $\\frac{d}{dx}(xy) = \\frac{d}{dx}x \\cdot y + x \\cdot \\frac{d}{dx}y.$ Oct 20, 2014 at 14:49\n\u2022 @Clayton: AH! Fantastic find! I never saw that Oct 20, 2014 at 14:50\n\nYou made an error when you differentiated implicitly. You did not apply the Product Rule $(fg)' = f'g + fg'$ to the term $xy$. Keeping in mind that $y$ is a function of $x$, you should obtain\n\n$$(xy)' = 1y + xy' = y + xy'$$\n\nTherefore, when you differentiate\n\n$$x^2 + xy + y^2 = 3$$\n\nimplicitly with respect to $x$, you should obtain\n\n$$2x + y + xy' + 2yy' = 0$$\n\nSolving for $y'$ yields\n\n\\begin{align*} xy' + 2yy' & = -2x - y\\\\ (x + 2y)y' & = -2x - y\\\\ y' & = -\\frac{2x + y}{x + 2y} \\end{align*}\n\nAs you can check, evaluating $y'$ at the point $(1, 1)$ yields $y' = -1$. Therefore, the tangent line equation is $y = -x + 2$.", "date": "2022-05-16 06:02:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/982564/help-with-implicit-differentiation-finding-an-equation-for-a-tangent-to-a-given", "openwebmath_score": 0.7352010011672974, "openwebmath_perplexity": 458.0796138680465, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462181769099, "lm_q2_score": 0.8688267762381844, "lm_q1q2_score": 0.8583541278353506}}
{"url": "https://math.stackexchange.com/questions/1161746/how-come-left-fracn1n-1-rightn-left1-frac2n-1-rightn", "text": "# How come $\\left(\\frac{n+1}{n-1}\\right)^n = \\left(1+\\frac{2}{n-1}\\right)^n$?\n\nI'm looking at one of my professor's calculus slides and in one of his proofs he uses the identity:\n\n$\\left(\\frac{n+1}{n-1}\\right)^n = \\left(1+\\frac{2}{n-1}\\right)^n$\n\nExcept I don't see why that's the case. I tried different algebraic tricks and couldn't get it to that form.\n\nWhat am I missing?\n\nThanks.\n\nEdit: Thanks to everyone who answered. Is there an \"I feel stupid\" badge? I really should have seen this a mile a way.\n\n\u2022 Because $n+1=(n-1)+2$. \u2013\u00a0vadim123 Feb 23 '15 at 14:39\n\nJust write $$\\left(\\frac{n+1}{n-1}\\right)^n = \\left(\\frac{n-1+2}{n-1}\\right)^n =\\left(1+\\frac{2}{n-1}\\right)^n$$\n$1+\\frac{2}{n-1}=\\frac{n-1}{n-1}+\\frac{2}{n-1}=\\frac{n+1}{n-1}$\nNote that $$\\frac{n+1}{n-1} = \\frac{n-1+2}{n-1} = \\frac{n-1}{n-1} + \\frac{2}{n-1} = 1 + \\frac{2}{n-1}.$$\n$(1 + \\frac {2}{n-1})^n = (\\frac {n-1 +2}{n-1})^n = (\\frac{n+1}{n-1} )^n$", "date": "2019-06-18 10:55:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1161746/how-come-left-fracn1n-1-rightn-left1-frac2n-1-rightn", "openwebmath_score": 0.7038430571556091, "openwebmath_perplexity": 863.090631739802, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462194190619, "lm_q2_score": 0.868826769445233, "lm_q1q2_score": 0.8583541222034948}}
{"url": "https://math.stackexchange.com/questions/1852826/inverse-of-the-pascal-matrix", "text": "# Inverse of the Pascal Matrix\n\nLet $P_n$ be the $(n+1) \\times (n+1)$ matrix that contains the numbers of Pascal's triangle in the upper triangle. For example in the case of $n=3$ $$P_3 = \\begin{pmatrix} 1 & 1 & 1 & 1 \\\\ 0 & 1 & 2 & 3 \\\\ 0 & 0 & 1 & 3 \\\\ 0 & 0 & 0 & 1 \\\\ \\end{pmatrix}$$ or in general $$(P_n)_{ij} = \\binom{j}{i} \\lfloor i \\leq j \\rceil ~~~\\text{for}~~ i,j \\in \\{0,...,n \\}$$ using the definition $$\\lfloor A \\rceil := \\begin{cases} 1 & \\text{A is true} \\\\ 0 & \\text{A is not true} \\end{cases}$$ This matrix is invertible since $\\det P_n = 1$. For smaller cases like $n=3$, I calulated the inverse of the matrix by hand and found $$P_3^{-1} = \\begin{pmatrix} 1 & -1 & 1 & -1 \\\\ 0 & 1 & -2 & 3 \\\\ 0 & 0 & 1 & -3 \\\\ 0 & 0 & 0 & 1 \\\\ \\end{pmatrix}$$ $n=2,4$ led to similar results, so I'm guessing that the inverse should be $$(P_n^{-1})_{ij} = (-1)^{j+i}(P_n)_{ij} = (-1)^{j+i} \\binom{j}{i} \\lfloor i \\leq j \\rceil$$ But I have not been able to prove or disprove this yet. So far I tried multiplying the two matirces which gives $$\\sum^j_{k=i} (-1)^{j+k} \\binom{j}{k} \\binom{k}{i} = \\delta_{ij}$$ if one asumes that the result is the unit matrix. For $i=0$ and $j>0$ this gives $$\\delta_{0j} = 0 = \\sum^j_{k=0} (-1)^{j+k} \\binom{j}{k} \\binom{k}{0} = \\sum^j_{k=0} (-1)^{k} \\binom{j}{k}$$ which is an identity I know to be true, so it reasures me a little bit that the above should also be true.\n\n\u2022 Instead of proving the above mentioned formula for all $(j,i)$ you can also use induction and prove it only for $(n,i)$, maybe it is easier. The $(j,n)$ entries are easy to handle Jul 8, 2016 at 6:23\n\u2022 Use $\\binom jk \\binom ki = \\binom ji \\binom {j-i}{k-i}$. Jul 8, 2016 at 6:36\n\u2022 Binomial transform Jul 27, 2016 at 11:10\n\nIn my opinion this is a bit simpler to prove, if we interpret the matrix $$P_n$$ as a linear transformation.\n\nConsider the space $$V_n$$ of polynomials of degree $$\\le n$$ (over, say $$\\Bbb{Q}$$, but you are welcome to use reals or complex numbers, or any other field actually).\n\nThe mapping $$T: f(x)\\mapsto f(x+1)$$ for all $$f(x)\\in V_n$$ is obviously linear. My key point is to observe that $$P_n$$ is the matrix $$M_{\\mathcal B}(T)$$ of $$T$$ with respect to the obvious basis $$\\mathcal{B}=\\{1,x,x^2,\\ldots,x^n\\}$$ of $$V_n$$. This is because for any $$k, 0\\le k\\le n$$ the binomial formula says that $$T(x^k)=\\sum_{\\ell=0}^k\\binom k\\ell x^\\ell.$$ From this we can read the coordinates of $$T(x^k)$$ with respect to $$\\mathcal{B}$$, and see that those coincide the $$k$$th column of $$P_n$$ (numbered from $$0$$ to $$n$$). That's exactly what the claim was.\n\nIt is obvious that the inverse of $$T$$ is given by the recipe $$T^{-1}:f(x)\\mapsto f(x-1).$$ A similar application of the binomial formula shows that the matrix $$M_{\\mathcal B}(T^{-1})$$ then is exactly your prescribed inverse of $$P_n$$ containing binomial coefficients - this time with alternating signs.\n\nBut, by basic linear algebra $$M_{\\mathcal B}(T^{-1})=M_{\\mathcal B}(T)^{-1}.$$\n\n\u2022 Nice solution! I especially like that it does not require to guess the form of the inverse. It is also very close to the reason I got confronted with the problem in the first place. Because I was looking at the change of base from $\\{x^j\\}$ to $\\{(x+1)^j\\}$. Jul 22, 2016 at 18:39\n\u2022 @JyrkiLahtonen This is brilliant, really, a pity that I can only give (+1). I just wanted to know the formula for the inverse of the Pascal matrix and searched the web for it. Now I found not only the formula, but this higher level view on it which makes everything so clear. It feels similar to the moment when I was told that the addition theorems for sin and cos are just real and imaginary parts of the power rule for the complex exponential function. May 30, 2020 at 10:10\n\nBy the suggestion of Sungjin Kim one can use $$$$\\binom{j}{k}\\binom{k}{i} = \\binom{j}{i}\\binom{j-i}{k-i}$$$$ which is useful since it leads to the sum only going over one of the binomial coefficients. It leads to \\begin{align} \\sum^j_{k=i} (-1)^{j+k} \\binom{j}{k} \\binom{k}{i} &= \\sum^j_{k=i} (-1)^{j+k} \\binom{j}{i}\\binom{j-i}{k-i} \\\\ &= (-1)^j \\binom{j}{i} \\sum^j_{k=i} (-1)^{k} \\binom{j-i}{k-i} \\\\ &= (-1)^{j} \\binom{j}{i} \\sum^{j-i}_{k=0} (-1)^{k+i} \\binom{j-i}{k} \\\\ &= (-1)^{j+i} \\binom{j}{i} (1 - 1)^{j-i} = \\delta_{ij} \\end{align} The identity used is easily proven by $$\\binom{j}{k}\\binom{k}{i} = \\frac{j!}{k!(j-k)!} \\frac{k!}{i!(k-i)!} = \\frac{j!}{i!(j-i)!} \\frac{(j-i)!}{(j-k)!(k-i)!} = \\binom{j}{i}\\binom{j-i}{k-i}$$ Another way of showing the identity involving $$\\delta_{ij}$$ is to start with $$x^j$$ and using the binomial thoerem twice $$x^j = (x+0)^j = ((x+1) - 1 )^j = ~...~ = \\sum^j_{i=0} \\sum^j_{k=i} (-1)^{j+k} \\binom{j}{k} \\binom{k}{i} x^i$$ comparing coefficients of the two sides of the equation, then also leads to the desired result. You can find the whole calcultion here, along whith other examples of when adding 0 really counts.\n\nIn the following it's somewhat more convenient to equivalently consider the transpose $P_n^T$ of the matrix $P_n$ and its inverse.\n\nWe obtain for $n=3$\n\n\\begin{align*} P_3^T&= \\begin{pmatrix} 1 & & & \\\\ 1 & 1 & & \\\\ 1 & 2 & 1 & \\\\ 1 & 3 & 3 & 1 \\\\ \\end{pmatrix}=\\left(\\binom{i}{j}\\right)_{0\\leq i,j\\leq 3} \\\\ \\left(P_3^T\\right)^{-1}&= \\begin{pmatrix} \\begin{array}{rrrr} 1 & & & \\\\ -1 & 1 & & \\\\ 1 & -2 & 1 & \\\\ -1 & 3 & 3 & 1 \\\\ \\end{array} \\end{pmatrix}=\\left((-1)^{i+j}\\binom{i}{j}\\right)_{0\\leq i,j\\leq 3} \\end{align*}\n\nThe elements of the Pascal matrix can be considered as coefficients of binomial inverse pairs. These are sequences $(a_i)_{0\\leq i \\leq n},(b_i)_{0\\leq i\\leq n}$ which are related for $n\\geq 0$ via \\begin{align*} a_n=\\sum_{i=0}^n\\binom{n}{i}b_i\\qquad\\text{resp.}\\qquad b_n=\\sum_{i=0}^n(-1)^{i+n}\\binom{n}{i}a_i\\tag{1} \\end{align*}\n\nOne relation implies the other in (1). This can be seen via exponential generating functions.\n\nLet $A(x)$ and $B(x)$ be two exponential generating functions\n\n$$A(x)=\\sum_{n=0}^\\infty a_{n}\\frac{x^n}{n!} \\qquad\\qquad\\text{ and }\\qquad\\qquad B(x)=\\sum_{n=0}^\\infty b_{n}\\frac{x^n}{n!}$$\n\nThe expressions in (1) are the coefficients of\n\nThe most important characterization of the pair of inverse relation in (1) is the orthogonal relation the pair implies. This follows by substituting one of the pair into the other. We obtain for $n\\geq 0$\n\\begin{align*} a_n&=\\sum_{j=0}^n\\binom{n}{j}b_j\\\\ &=\\sum_{j=0}^n\\binom{n}{j}\\sum_{i=0}^j(-1)^{i+j}\\binom{j}{i}a_i\\\\ &=\\sum_{i=0}^n a_i\\sum_{j=i}^n(-1)^{i+j}\\binom{n}{j}\\binom{j}{i} \\end{align*} Hence the orthogonal relation is \\begin{align*} \\delta_{ni}=\\sum_{j=i}^n(-1)^{i+j}\\binom{n}{j}\\binom{j}{i} \\end{align*} with $\\delta_{ni}$ the Kronecker Delta.\nThe inverse of $$P_n$$ also follows from a familiar representation of $$GL(2,\\mathbb{R})$$. Let $$\\{u,v\\}$$ be a basis of $$\\mathbb{R}^2$$ and $$S^n(\\mathbb{R}^2)$$ denote the vector space of homogeneous polynomials of degree $$n$$ in the variables $$u$$ and $$v$$. Then, any $$A\\in GL(2,\\mathbb{R})$$ acts linearly on $$p(u,v)\\in S^n(\\mathbb{R}^2)$$ by $$A\\cdot p(u,v)=p(Au,Av).$$ Choose $$\\{u^{n-j}v^j\\}_{j=0}^n$$ as a basis of $$S^n(\\mathbb{R}^2)$$. Then, the given group action induces a homomorphism $$\\varphi_n: GL(2,\\mathbb{C})\\to GL(n+1,\\mathbb{C})$$ defined by $$\\begin{pmatrix}a & b\\\\ c & d\\end{pmatrix}\\mapsto \\left([t^i](a+ct)^{n-j}(b+dt)^j\\right)_{i,j=0}^n$$ where $$[t^i]f(t)$$ denotes the coefficient of $$t^i$$ in $$f(t)$$. Let $$M=\\begin{pmatrix}1 & 1\\\\ 0 & 1\\end{pmatrix}.$$ Then, we see that $$P_n=\\varphi_n(M).$$ Since $$\\varphi_n$$ is a homomorphism, we get $$P_n^{-1}=\\varphi_n(M^{-1})$$. But $$M^{-1}=\\begin{pmatrix}1 & -1\\\\ 0 & 1\\end{pmatrix}$$. Hence, by definition of $$\\varphi_n$$, we obtain explicit indices as you described.", "date": "2022-05-19 15:23:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1852826/inverse-of-the-pascal-matrix", "openwebmath_score": 0.9994117021560669, "openwebmath_perplexity": 139.5083955100936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864311494626, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8583536324234898}}
{"url": "https://math.stackexchange.com/questions/1909904/munkress-strong-induction-principle-vs-traditional-mathematical-induction", "text": "# Munkres's Strong induction principle vs. \u201ctraditional\u201d mathematical induction\n\nQuestion:\n\nIn Munkres's Topology (2nd ed.), he gives a proof for the following:\n\nTheorem 4.2 (Strong induction principle). Let $A$ be a set of positive integers. Suppose that for each positive integer $n$, the statement $S_n\\subset A$ implies the statement $n\\in A$. Then $A=\\mathbb{Z}_+$.\n\nThis appears to be slightly different from the \"traditional\" (strong) induction principle which can be stated as: let $P_n$ be a sequence of statements. Suppose $P_1$ is true and (($P_m$ is true $\\forall\\, m < n$) implies $P_{n}$ is true) then $P_k$ is true $\\forall\\, k\\in \\mathbb{Z}_+$.\n\nHow can one reconcile the two?\n\nAttempt:\n\nI'm thinking of something along the lines of:\n\nLet $A$ be a set of positive integers and $P_n$ be a sequence of propositions. Suppose that for each positive integer $n$, the statement $S_n\\subset A$ implies the statement ($n\\in A$ and $P_n$ is true). Then $A=\\mathbb{Z}_+$ and $P_k$ is true $\\forall\\, k \\in \\mathbb{Z}_+$.\n\nThis can proved to be equivalent to the \"traditional\" form in the following way:\n\n1. Proceed with Munkres's proof and obtain $A=\\mathbb{Z}_+$.\n2. $\\because S_1=\\varnothing\\subset A$ is true, our supposition for $n=1$ is equivalent to assuming that $P_1$ is true.\n3. ??? (how do we show that the supposition \"$S_n\\subset A$ implies the statement ($n\\in A$ and $P_n$ is true)\" is equivalent to \"($P_m$ is true $\\forall\\, m < n$) implies $P_{n}$ is true\"?)\n\nActually I just noticed that this point is discussed in Chapter 1 $\\S$ 7 after Lemma 7.2 (pg 45 in 2nd ed.)\n\nTo use the principle to prove a theorem \"by induction\", one begins the proof with the statement \"Let $A$ be the set of all positive integers $n$ for which the theorem is true,\" and then one goes to prove that $A$ is inductive, so that $A$ must be all of $\\mathbb{Z}_+$\n\nUsing this one can write:\n\nLet $P_n$ be a sequence of propositions and $A$ be the set of all positive integers $n$ for which $P_n$ is true. Suppose that for each positive integer $n$, the statement $S_n\\subset A$ implies the statement $n\\in A$. Then $A=\\mathbb{Z}_+$.\n\nThe equivalence with the \"traditional\" induction principle is easily demonstrated:\n\n1. For $n=1$, the assumption ($S_1 = \\varnothing \\subset A \\Rightarrow 1 \\in A$) is equivalent to ($P_1$ is true).\n2. For $n>1$, the assumption ($S_n \\subset A \\Rightarrow n \\in A$) is equivalent to (($P_m$ is true $\\forall\\, m<n$) $\\Rightarrow$ $P_n$ is true)\n3. The conclusion ($A=\\mathbb{Z}_+$) is equivalent to ($P_n$ is true $\\forall\\, n\\in \\mathbb{Z}_+$).\n\u2022 Yes, this is the intended argument. \u2013\u00a0Brian M. Scott Aug 31 '16 at 22:10", "date": "2019-06-17 08:34:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1909904/munkress-strong-induction-principle-vs-traditional-mathematical-induction", "openwebmath_score": 0.9569408893585205, "openwebmath_perplexity": 157.9739028912654, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.985936372130286, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8583535206766809}}
{"url": "https://math.stackexchange.com/questions/2688163/double-antiderivation-problem", "text": "Double Antiderivation problem\n\nI have to find $f(x)$ given $f''(x)$ and certain initial conditions.\n\n$$f\"(x) = 8x^3 + 5$$ and $f(1) = 0$ and $f'(1) = 8$\n\n$$f'(x) = 8 \\cdot \\frac{x^4}{4} + 5x + C = 2x^4 + 5x + C$$\n\nSince $f'(1) = 8 \\Rightarrow 2 + 5 + C = 8$ so $C = 1$\n\n$$f'(x) = 2x^4 + 5x + 1$$\n\n$$f(x) = 2 \\cdot \\frac{x^5}{5} + 5 \\cdot \\frac{x^2}{2} + X = \\frac{2}{5} x^5 + \\frac{5}{2} x^2 + X + D$$\n\nSince $f(1) = 0$ then:\n\n$$\\frac{2}{5} + \\frac{5}{2} + 1 + D = 0$$ $$\\frac{29}{10} + 1 + D = 0$$\n\nSo $D = \\dfrac{-39}{10}$\n\nSo $$f(x) = \\frac{2}{5} \\cdot x^5 + \\frac{5}{2} \\cdot x^2 + x - \\frac{39}{10}$$\n\nDoes that look right?\n\n\u2022 Yes it's correct. An alternative way to do it is to work with definite integrals and write $f'(x) - f'(1) = \\int_1^x (8t^3+5)\\,{\\rm d}t$ and $f(x)-f(1) = \\int_1^x f'(t)\\,{\\rm d}t$. This avoids solving for the integration constants, but it's pretty much the same thing. Mar 12, 2018 at 17:32\n\u2022 Yes, you can also check it by yourself Mar 12, 2018 at 17:33\n\n$$f(x) = \\frac{2}{5} \\cdot x^5 + \\frac{5}{2} \\cdot x^2 + x - \\frac{39}{10}\\implies f'(x)= 2x^4+5x+1\\implies f''(x)=8x^3+5$$\n$$f(1)=\\frac{2}{5}+ \\frac{5}{2} + 1 - \\frac{39}{10}=\\frac{4+25+10-39}{10}=0$$\n$$f'(1)=2+5+1=8$$", "date": "2023-03-23 05:48:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2688163/double-antiderivation-problem", "openwebmath_score": 0.8731771111488342, "openwebmath_perplexity": 301.4024419735764, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363704773486, "lm_q2_score": 0.8705972734445508, "lm_q1q2_score": 0.8583535159273963}}
{"url": "https://web2.0calc.com/questions/here-s-a-mind-boggler", "text": "+0\n\n# Here's a mind-boggler\n\n0\n347\n4\n+40\n\nA positive whole number that is the same when reading from right to left and left to right is called a palindromic number. If all of them were put into a sequence that is from small to big:1,2...11,22...,101,111...,10001,10101,.....\n\nThen which term is the number 78987 in this sequence?\u00a0 For example : 1 is the 1st term,2 is the second....etc\n\nyomyhomies \u00a0Oct 22, 2017\n#2\n+89840\n+1\n\nStarting with 0,\u00a0 78987\u00a0 is the 889th palindrome\n\n[BTW...889 is known as the palindrome's\u00a0rank\u00a0]\n\nStarting with 1, it is the 888th palindrome\n\nThis can be verified with the calculator here :\n\nhttp://rhyscitlema.com/algorithms/generating-palindromic-numbers/\n\n[ The calculator is about half-way down the page.....there are also other intems of interest related to palindromes on this page ]\n\nA procedure is desribed for finding the nth palindrome, but I don't see one that tells us how to determine\u00a0the rank of any particular palindrome \u00a0[ although I did not look at the page with a fine tooth comb]\n\nMaybe heureka knows of a\u00a0proceedure to produce this???\n\nCPhill \u00a0Oct 22, 2017\n#3\n+20024\n+3\n\nwhich term is the number 78987 in this sequence?\n\nHere 0 is the 1st term:\n\nThe position of a palindrome within the sequence can be determined almost without calculation:\n\nIf the palindrome has an even number of digits,\n\nprepend a 1 to the front half of the palindrome's digits.\n\nExamples: 98766789=a(19876)\n\nIf the number of digits is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit.\n\nExamples: 515=a(61), 8206028=a(9206), 9230329=a(10230).\n\nwhich term is the number 78987 in this sequence?\n\nThe number of digits is 5 is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit.\n\n$$\\begin{array}{|rrrrll|} \\hline & & & \\Rsh & & \\text{ until central digit} \\\\ &7 & 8 & 9 & 8 & 7 \\\\ &| & | & | \\\\ &+1 & | & | \\\\ &\\downarrow & \\downarrow & \\downarrow \\\\ \\text{term is } & \\color{red}8 &\\color{red}8 &\\color{red} 9 \\\\ \\hline \\end{array}$$\n\nStarting with 1, it is the 888th palindrome\n\nheureka \u00a0Oct 23, 2017\n#4\n+89840\n+1\n\nWow!!!!....thanks, heureka.....that's almost too easy....!!!\n\nCPhill \u00a0Oct 23, 2017", "date": "2018-10-17 21:34:53", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/here-s-a-mind-boggler", "openwebmath_score": 0.641232967376709, "openwebmath_perplexity": 1805.6505512779272, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363715104345, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.858353515171677}}
{"url": "https://web2.0calc.com/questions/suppose-f-x-is-a-rational-function-such-that-3-f", "text": "+0\n\n# Suppose $f(x)$ is a rational function such that $3 f \\left( \\frac{1}{x} \\right) + \\frac{2f(x)}{x} = x^2$ for all $x \\neq 0$. Find $f(-2)$.\n\n0\n258\n3\n\nSuppose f(x)\u00a0is a rational function such that $$3 f \\left( \\frac{1}{x} \\right) + \\frac{2f(x)}{x} = x^2$$ for all x =\\=0. Find f(-2).\n\nGuest\u00a0May 22, 2018\n#1\n+93629\n+1\n\nSuppose f(x)\u00a0is a rational function such that\n\n$$3 f \\left( \\frac{1}{x} \\right) + \\frac{2f(x)}{x} = x^2$$\n\nfor all $$x\\ne0$$. Find f(-2).\n\nWe have\n\n$$3 f \\left( \\frac{1}{-2} \\right) + \\frac{2f(-2)}{-2} = (-2)^2\\\\ 3 f \\left( \\frac{1}{-2} \\right) - f(-2) =4\\qquad \\qquad(1)\\\\ and\\\\ 3 f \\left( \\frac{1}{-1/2} \\right) + \\frac{2f(-1/2)}{-1/2} = (-1/2)^2\\\\ 3 f \\left( -2 \\right) -4f(\\frac{-1}{2}) =\\frac{1}{4}\\\\ -4f(\\frac{-1}{2}) +3 f \\left( -2 \\right)=\\frac{1}{4}\\qquad\\qquad (2)\\\\--------------\\\\~\\\\ \\quad 3 f \\left( \\frac{1}{-2} \\right) - f(-2) =4\\qquad \\qquad(1)\\\\ \\quad 12f \\left( \\frac{1}{-2} \\right) - 4f(-2) =16\\qquad \\qquad(1b)\\\\ -4f(\\frac{-1}{2}) +3 f \\left( -2 \\right)=\\frac{1}{4}\\qquad\\qquad (2)\\\\ -12f(\\frac{-1}{2}) +9 f \\left( -2 \\right)=\\frac{3}{4}\\qquad\\qquad (2b)\\\\~\\\\ (1b)+(2b)\\\\ 5f(-2)=16\\frac{3}{4}\\\\ f(-2)=3\\frac{7}{20}\\\\$$\n\nThe method is correct but you need to check for careless errors.\n\nMelody \u00a0May 22, 2018\n#2\n+20024\n+1\n\nSuppose\n\n$$f(x)$$\n\nis a rational function such that\u00a0 $$3 f \\left( \\frac{1}{x} \\right) + \\frac{2f(x)}{x} = x^2$$\n\nfor all $$x \\neq 0$$. Find $$f(-2)$$.\n\n$$\\begin{array}{|lrclcl|} \\hline & 3f(\\frac1x) + \\frac2x f(x) &=& x^2 \\\\ & 3f(\\frac1x) &=& x^2 - \\frac2x f(x) \\\\ & f(\\frac1x) &=& \\frac{x^2}{3} - \\frac{2}{3x} f(x) \\qquad (1) \\\\ \\\\ \\text{Set }x=\\frac1x: & 3f(x) + 2x f(\\frac1x) &=& \\frac{1}{x^2} \\\\ & 2x f(\\frac1x) &=& \\frac{1}{x^2}-3f(x) \\\\ & f(\\frac1x) &=& \\frac{1}{2x^3}-\\frac{3}{2x}f(x) \\qquad (2) \\\\\\\\ \\hline (1) = (2): & \\frac{x^2}{3} - \\frac{2}{3x} f(x) &=& \\frac{1}{2x^3}-\\frac{3}{2x}f(x) \\\\ & \\frac{3}{2x}f(x) - \\frac{2}{3x} f(x) &=& \\frac{1}{2x^3} -\\frac{x^2}{3} \\\\ & f(x)\\left( \\frac{3}{2x} - \\frac{2}{3x} \\right) &=& \\frac{3-2x^5}{6x^3} \\\\ & f(x)\\left( \\frac{9x-4x}{6x^2} \\right) &=& \\frac{3-2x^5}{6x^3} \\\\ & f(x)\\left( \\frac{5x}{6x^2} \\right) &=& \\frac{3-2x^5}{6x^3} \\\\ & f(x)\\left( \\frac{5x}{1} \\right) &=& \\frac{3-2x^5}{x} \\\\\\\\ & \\mathbf{f(x)} & \\mathbf{=} & \\mathbf{\\dfrac{3-2x^5}{5x^2}} \\\\ \\hline \\end{array}$$\n\nThe rational function is\u00a0 $$f(x) = \\dfrac{3-2x^5}{5x^2}$$\n\n$$f(-2)=\\ ?$$\n\n$$\\begin{array}{|rclcl|} \\hline f(-2) &=& \\frac{3-2(-2)^5}{5(-2)^2} \\\\ &=& \\frac{3+2^6}{5\\cdot 4} \\\\ &=& \\frac{3+64}{20} \\\\ &=& \\frac{67}{20} \\\\ \\mathbf{f(-2)} & \\mathbf{=} & \\mathbf{3.35} \\\\ \\hline \\end{array}$$\n\n$$\\text{ f(-2) is 3.35 }$$\n\ngraph:\n\nheureka \u00a0May 22, 2018\nedited by heureka \u00a0May 22, 2018\n#3\n+93629\n+1\n\nThat is interesing Heureka, I had not realised that there was enough infromation to draw the graph. :)\n\nMelody \u00a0May 22, 2018\n\n### New Privacy Policy\n\nWe use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.\nFor more information: our cookie policy and privacy policy.", "date": "2018-10-17 10:45:59", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/suppose-f-x-is-a-rational-function-such-that-3-f", "openwebmath_score": 0.5469297766685486, "openwebmath_perplexity": 3997.661171407632, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9763105321470076, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8583202376025865}}
{"url": "https://math.stackexchange.com/questions/2955640/how-should-i-write-vectors-like-this", "text": "# How should I write vectors like this?\n\nIf I'm trying to write basic vectors, just as simple as the magnitude being 5 and the direction being zero, how would I do this? Would it be a row vector with parenthesis:$$\\overrightarrow{v} = (5, 0)$$, a row vector with brackets: $$\\overrightarrow{v} = [5, 0]$$, a column vector with parenthesis: $$\\overrightarrow{v} = \\begin{pmatrix} 5\\\\ 0\\\\ \\end{pmatrix}$$, or a column vector with brackets: $$\\overrightarrow{v} = \\begin{bmatrix} 5\\\\ 0\\\\ \\end{bmatrix}$$? Thank you if you can tell me what the correct notation for this simple vector is, everywhere I go seems to write them differently and the inconsistency makes me want to rip my hair out.\n\n\u2022 You can write it however you want, as long as you are consistent. \u2013\u00a0Morgan Rodgers Oct 14 '18 at 21:19\n\u2022 You might also want to consider what you are doing with these vectors. If you are frequently left-multiplying them by a matrix (e.g., $Av$), make them (thin) column vectors. If you are frequently right-multiplying them by a matrix (e.g., $vA$), make them (wide) row vectors. If you are performing both of these operations just as frequently, try not to mix row and column vectors too much (just pick one convention and apply transposes when necessary). As for the bracketing, just be consistent within one document (see Morgan Rodgers' comment). \u2013\u00a0parsiad Oct 14 '18 at 21:22\n\u2022 @MorganRodgers and parsia thank you, I'll take this into consideration! \u2013\u00a0jstowell Oct 14 '18 at 21:23\n\u2022 One thing I want to point out: the coordinates of a vector are not their magnitude and direction. The magnitude of the vector (3, 4) is 5, and the direction is about 53.13 degrees above the x-axis. \u2013\u00a0Deusovi Oct 16 '18 at 16:15\n\nSo long as you make sure that you are consistent throughout the entire text, it's completely up to you. There are many different ways to represent vectors. In Linear Algebra, people like to use column notation, with either parentheses or square brackets, as this is the most convenient in linear algebra. Some people also write something like $$\\vec{v}=\\begin{bmatrix}5&0\\end{bmatrix}^T$$ to have a column vector, but be able to write it nicely inline. In high school (perhaps up to freshman) level classes, the notation $$\\vec{v}=\\langle 5,0\\rangle$$ is also used, but this ignores the fact that vectors are matrices with 1 row or 1 column, so is less widely used at a higher level. Ultimately it's your choice, whatever is most convenient to you, you should use it.\nUsually in linear algebra context vectors $$\\vec v$$ are considered colummn vector and transponsed vectors $$\\vec v^T$$ are row vectors that is\n$$\\overrightarrow{v} = \\begin{pmatrix} 5\\\\ 0\\\\ \\end{pmatrix} \\quad \\overrightarrow{v^T} = \\begin{pmatrix} 5 &0\\\\ \\end{pmatrix}$$", "date": "2019-06-17 00:52:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2955640/how-should-i-write-vectors-like-this", "openwebmath_score": 0.8680649995803833, "openwebmath_perplexity": 387.15013984916504, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102571131692, "lm_q2_score": 0.8872046011730965, "lm_q1q2_score": 0.858290831332852}}
{"url": "https://gmatclub.com/forum/jennifer-can-buy-watches-at-a-price-of-b-dollars-per-watch-which-she-193399.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 12 Nov 2018, 16:55\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in November\nPrevNext\nSuMoTuWeThFrSa\n28293031123\n45678910\n11121314151617\n18192021222324\n2526272829301\nOpen Detailed Calendar\n\u2022 ### Essential GMAT Time-Management Hacks\n\nNovember 14, 2018\n\nNovember 14, 2018\n\n08:00 PM MST\n\n09:00 PM MST\n\nJoin the webinar and learn time-management tactics that will guarantee you answer all questions, in all sections, on time. Save your spot today! Nov. 14th at 7 PM PST\n\u2022 ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Jennifer can buy watches at a price of B dollars per watch, which she  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 50544 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 03:18 2 7 00:00 Difficulty: 45% (medium) Question Stats: 70% (01:48) correct 30% (02:08) wrong based on 141 sessions ### HideShow timer Statistics Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) \u2013 B)/(100B) E. 100(T \u2013 NB)/N Kudos for a correct solution. _________________ Manager Joined: 15 Aug 2013 Posts: 54 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:11 2 1 Cost price of N watches = BN (cost of each watch, B * total nuber of watches, N) Total Profit = T But this profilt is also equal to BN (x/100), where x is the profit percentage Hence, BN (x/100) =T or x = 100T/ (NB) Hence (A). CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:19 Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) \u2013 B)/(100B) E. 100(T \u2013 NB)/N Let's say Jennifer marks up by x% Total Profit = Marked up price of N watches - Cost of N watches i.e. T = NB[1+(x/100)] - NB i.e. (T) / NB = [(x/100)] i.e. x = 100(T)/NB Answer: Option _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION CEO Status: GMATINSIGHT Tutor Joined: 08 Jul 2010 Posts: 2689 Location: India GMAT: INSIGHT WE: Education (Education) Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 18 Feb 2015, 05:32 Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) \u2013 B)/(100B) E. 100(T \u2013 NB)/N Let's say The price of each watch, i.e. B = 7, Let's say Mark up = 400%==> 400% of 7 is 28, and the$7 watches are marked up $28, then the selling price per watch =$35.\n\nThe profit per watch is $28, and so if she sells N = 11 watches Then Total profit, T = 28*11 =$308.\n\nOK, now Substitute T = 308 = 11*28, N = 11, and B = 7, in options and hope to get 400 as our answer.\n\n(A) 100T/(NB) = 100*11*28/(11*4) = 100*28/7 = 110*4 = 400 = works!\n\n(B) TB/(100N) = 11*28*7/(100*11) = 28*7/100 = doesn\u2019t work!\n\n(C) 100TN/B = 100*11*28*11/7 = 100*11*4*11 = doesn\u2019t work!\n\n(D) ((T/N) \u2013 B)/(100B) = [(11*28/11) \u2013 7]/(7*100) = (28 \u2013 7)/(7*100) = 21/(7*100) = 3/100 = doesn\u2019t work\n\n(E) 100(T \u2013 NB)/N = 100(11*28 \u2013 11*7)/7 = 100*11*21/7 = 100*11*3 = doesn\u2019t work!\n\n_________________\n\nProsper!!!\nGMATinsight\nBhoopendra Singh and Dr.Sushma Jha\ne-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772\nOnline One-on-One Skype based classes and Classroom Coaching in South and West Delhi\nhttp://www.GMATinsight.com/testimonials.html\n\nACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION\n\nIntern\nJoined: 22 Dec 2014\nPosts: 4\nRe: Jennifer can buy watches at a price of B dollars per watch, which she\u00a0 [#permalink]\n\n### Show Tags\n\n18 Feb 2015, 07:51\n1\nUsing smart numbers:\n\nJennifer Purchases watches for $5: B=5 Markup of 100%; now selling for$10: x=10\nShe sells 5 watches: N=5\n\nHer total profit will be (N*x)-(B*N)\n(5*10)-(5*5)= 25\nT=25\n\nNow using the variables we must look at the answer choices that will solve for the % markup of the watches, which we chose to be 100.\n\nB=5\nT=25\nN=5\n\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1827\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nRe: Jennifer can buy watches at a price of B dollars per watch, which she\u00a0 [#permalink]\n\n### Show Tags\n\n18 Feb 2015, 21:51\n\nCost price per watch = b\n\nSelling price per watch = $$\\frac{t}{n}$$\n\nLet x = percent of the markup from her buy price to her sell price\n\n$$x = 100 *\\frac{t}{n} * \\frac{1}{b} = \\frac{100t}{nb}$$\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50544\nRe: Jennifer can buy watches at a price of B dollars per watch, which she\u00a0 [#permalink]\n\n### Show Tags\n\n22 Feb 2015, 11:17\nBunuel wrote:\nJennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?\n\nA. 100T/(NB)\nB. TB/(100N)\nC. 100TN/B\nD. ((T/N) \u2013 B)/(100B)\nE. 100(T \u2013 NB)/N\n\nKudos for a correct solution.\n\nMAGOOSH OFFICIAL SOLUTION\n\nAlgebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)\n\nThat Algebraic solution was elegant if you saw it, but if not, here\u2019s a full solution with picking numbers.\n\nT = total profit = $10t N = total number of items sold $$\\frac{10t}{3b*n}$$ Hence, A pushpitkc is my solution correct ? Manager Joined: 14 Jun 2018 Posts: 179 Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 23 Oct 2018, 09:47 Cost Price = B No of items sold = N Profit after selling N items = T ; Profit per item = $$\\frac{T}{N}$$ Let \"k\" be the markup percentage , therefore selling price = $$\\frac{(100+K)}{100} * B$$ We know , Selling Price - Cost Price = Profit => $$\\frac{(100+K)}{100} * B - B = \\frac{T}{N}$$ => $$K = 100 (\\frac{T}{NB} + \\frac{B}{B}) - 100$$ => $$K = 100(\\frac{T}{NB})$$ Senior PS Moderator Joined: 26 Feb 2016 Posts: 3301 Location: India GPA: 3.12 Jennifer can buy watches at a price of B dollars per watch, which she [#permalink] ### Show Tags 23 Oct 2018, 11:28 1 dave13 wrote: Bunuel wrote: Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price? A. 100T/(NB) B. TB/(100N) C. 100TN/B D. ((T/N) \u2013 B)/(100B) E. 100(T \u2013 NB)/N Kudos for a correct solution. let B be the cost +mark up price 2+1 =$3\nT = total profit = $10t N = total number of items sold $$\\frac{10t}{3b*n}$$ Hence, A pushpitkc is my solution correct ? Hey dave13 Unfortunately, this is wrong. I am not quite clear what you trying to do here If you are assigning values B =$5(cost per watch).\nShe marks up the watches and sells for a total profit of T = $1 Assuming there is a N = 1 watch and she keeps a profit of 20%(which is 0.2*$5 = \\$1)\n\nNow, working backward, trying to calculate the percentage by substituting the values\nof B,N, and T, we will get\n\nA. 100T/(NB) = 100/5 = 20 - Our solution is Option A($$\\frac{100T}{NB}$$)\n\nWe don't need to substitute the values in the other 4 answer options as we have a match!\n\nHope that helps.\n_________________\n\nYou've got what it takes, but it will take everything you've got\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 4149\nLocation: United States (CA)\nRe: Jennifer can buy watches at a price of B dollars per watch, which she\u00a0 [#permalink]\n\n### Show Tags\n\n25 Oct 2018, 08:19\nBunuel wrote:\nJennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?\n\nA. 100T/(NB)\nB. TB/(100N)\nC. 100TN/B\nD. ((T/N) \u2013 B)/(100B)\nE. 100(T \u2013 NB)/N\n\nThe profit (or markup) per watch is T/N. So the markup is:\n\n(T/N)/B x 100 = 100T/(NB)\n\npercent of the purchase price.\n\nAlternate Solution:\n\nJennifer\u2019s profit per watch is T/N. Using the formula for profit, we can compute Jennifer\u2019s sell price:\n\nprofit = sell price \u2013 buy price\n\nT/N = sell price \u2013 B\n\nT/N + B = sell price\n\nWe now use the percent markup formula:\n\n% markup = (Sell price \u2013 Buy price) / Buy price x 100\n\n% markup = (T/N + B \u2013 B) / B x 100\n\n% markup = T/N / B x 100\n\n% markup = T/BN x 100\n\n% markup = 100T / (BN)\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: Jennifer can buy watches at a price of B dollars per watch, which she &nbs [#permalink] 25 Oct 2018, 08:19\nDisplay posts from previous: Sort by", "date": "2018-11-13 00:55:58", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/jennifer-can-buy-watches-at-a-price-of-b-dollars-per-watch-which-she-193399.html", "openwebmath_score": 0.2825273871421814, "openwebmath_perplexity": 10509.071517360848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9674102477171955, "lm_q2_score": 0.8872045847699185, "lm_q1q2_score": 0.8582908071280985}}
{"url": "http://math.stackexchange.com/questions/25273/derivative-of-a-xb-x", "text": "# Derivative of $(a\\,x)^{b\\,x}$\n\nis there any rule to differentiate the function $(a\\,x)^{b\\,x}$?\n\nI've got to find the derivative of $(x^2+1)^{\\arctan x}$ and Wolfram|Alpha says the answer is $$\\tan^{-1}(x) (x^2+1)^{\\tan^{-1}(x)-1} \\left(\\frac{d}{dx}(x^2+1)\\right)+\\log(x^2+1) (x^2+1)^{\\tan^{-1}(x)} \\left(\\frac{d}{dx}(\\tan^{-1}(x))\\right)$$ Is there any general rule to do that? Thanks.\n\n-\n\nYour derivative of $(x^2+1)^{\\arctan x}$ is the particular case for $u(x)=x^2+1$ and $v(x)=\\arctan x$ of\n\n$$\\frac{d}{dx}\\left(\\left[ u(x)\\right] ^{v(x)}\\right)=v(x)\\left[ u(x)\\right] ^{v(x)-1}u^{\\prime }(x)+\\left( \\ln u(x)\\right) \\left[ u(x)\\right] ^{v(x)}v^{\\prime }(x),$$\n\nor omitting de variable $x$:\n\n$$\\left( u^{v}\\right)^{\\prime }=vu^{v-1}u^{\\prime }+\\left( \\ln u\\right) u^{v}v^{\\prime }.$$\n\nThis can be derived observing that, since $u=e^{\\ln u}$, we have $u^v=e^{v\\ln u}$:\n\n$$\\begin{eqnarray*} \\frac{d}{dx}\\left( u^{v}\\right) &=&\\frac{d}{dx}\\left( e^{v\\ln u}\\right) \\\\ &=&e^{v\\ln u}\\left( \\ln u\\frac{dv}{dx}+\\frac{v}{u}\\frac{du}{dx}\\right) \\\\ &=&u^{v}\\left( \\ln u\\frac{dv}{dx}+\\frac{v}{u}\\frac{du}{dx}\\right) \\\\ &=&u^{v}\\ln u\\frac{dv}{dx}+u^{v}\\frac{v}{u}\\frac{du}{dx} \\\\ &=&u^{v}(\\ln u)v'+u^{v-1}vu'. \\end{eqnarray*}$$\n\nAnother possibility is to consider the variables $u$ and $v$ (both depending on $x$) in the function\n\n$$z=f(u(x),v(x))=\\left[ u(x)\\right] ^{v(x)}$$\n\nand determine its total derivative with respect to $x$:\n\n$$\\begin{eqnarray*} \\frac{dz}{dx} &=&\\frac{d}{dx}\\left( \\left[ u(x)\\right] ^{v(x)}\\right) \\\\ &=&\\frac{% \\partial z}{\\partial u}\\frac{du}{dx}+\\frac{\\partial z}{\\partial v}\\frac{dv}{% dx} \\\\ &=&v(x)\\left[ u(x)\\right] ^{v(x)-1}u^{\\prime }(x)+\\left[ u(x)\\right] ^{v(x)}\\left( \\ln u(x)\\right) v^{\\prime }(x) \\end{eqnarray*}$$\n\nbecause\n\n$$\\frac{\\partial z}{\\partial u}=\\frac{\\partial }{\\partial u}\\left( u^{v}\\right) =vu^{v-1}$$\n\nand\n\n$$\\frac{\\partial z}{\\partial v}=\\frac{\\partial }{\\partial v}\\left( u^{v}\\right) =\\frac{\\partial }{\\partial v}\\left( e^{\\ln u\\cdot v}\\right) =e^{\\ln u\\cdot v}\\ln u=u^{v}\\ln u.$$\n\nFor $u(x)=ax,v(x)=bx$, we get\n\n$$\\frac{d}{dx}\\left( \\left( ax\\right) ^{bx}\\right) =bx\\left( ax\\right) ^{bx-1}a+\\left( ax\\right) ^{bx}\\left( \\ln (ax)\\right) b.$$\n\n-\n\nAssuming you mean $(ax)^{bx}$, I'd just write it as $(e^{\\ln(ax)})^{bx}$ and use the chain rule (ie $e^{\\ln(ax)bx} = e^{u(x)}$ and go from there).\n\n-\n\nHINT $\\rm\\ \\ (F^G)'\\ =\\ (e^{G\\:ln\\ F})'\\: =\\ F^G\\ (GF'/F + G'\\ ln\\ F)$\n\n-\nI think there is a typo in the sign. \u2013\u00a0 Am\u00e9rico Tavares Mar 6 '11 at 16:26\n@Americo: Fixed it, thanks! \u2013\u00a0 Bill Dubuque Mar 6 '11 at 16:32\n\nNot answering the math part, since Stijn has already done that, but if you click on the \"show steps\" button, Wolfram|Alpha shows you one possible path for the derivation. I've included the image for the derivation of $\\frac{\\partial}{\\partial x} ((a x)^{b x})$\n\nA similar set of steps is supplied for the other derivative that you want to take: http://www.wolframalpha.com/input/?i=d/dx((x^2%2B1)^(tan^(-1)(x)))\n\n-\nThe link for the first example is wolframalpha.com/input/?i=D[(ax)^(bx),x]. Does anyone know how to get Wolfram|Alpha links working properly in the markup? \u2013\u00a0 Simon Mar 6 '11 at 11:41\nThe parentheses and the caret confuse Markdown. Try escaping ^ with %5E, ( with %28, ) with %29, [ with %5B and ] with %5D like this: WA Link (you obviously have to use the [title](URL) syntax) \u2013\u00a0 kahen Mar 6 '11 at 13:21", "date": "2015-02-01 22:46:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/25273/derivative-of-a-xb-x", "openwebmath_score": 0.9121066927909851, "openwebmath_perplexity": 815.7015892295349, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347886752162, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8582793130205969}}
{"url": "http://math.stackexchange.com/questions/386536/solution-gives-wrong-answer-to-probability-problem", "text": "# Solution gives wrong answer to probability problem\n\nGreat Northern Airlines flies small planes in northern Canada and Alaska. Their largest plane can seat 16 passengers seated in 8 rows of 2. On a certain flight flown on this plane, they have 12 passengers and one large piece of equipment to transport. The equipment is so large that it requires two seats in the same row. The computer randomly assigns seats to the 12 passengers (no 2 passengers will have the same seat). What is the probability that there are two seats in the same row available for the equipment?\n\nThis problem was posed on Brilliant last week, and now that the official solution is posted, I would like to know why my solution gives wrong result. Here is my solution:\n\nThe number of ways we can seat 12 (identical) people on 16 seats is $16\\choose 12$. Now the equipment can occupy any of the rows (8 possibilities), and the 12 people must be seated on the remaining 14 seats ($14 \\choose12$ ways). Thus the desired probability is $$\\frac{8 {14\\choose12}}{16\\choose12}=\\frac25$$ The official solution gives $\\frac5{13}$, but I don't understand what is wrong with mine.\n\n-\nNote that the link you provided is unique to you. You need to use the \"Share this problem\" link for others to view the problem properly. \u2013\u00a0Calvin Lin May 10 '13 at 0:55\n@CalvinLin Oh, I'll keep that in mind. \u2013\u00a0Dave May 10 '13 at 12:02\n\nYou haven't factored in the case in which two rows are devoid of people - that setup is being double-counted in your count.\n\nThink of it this way - you can place four seats, with the restriction that at least one pair must be together in a row. So subtract off the number of ways you can have no pairs from the total number of combinations $${16\\choose4}-{8\\choose4}\\cdot2^4 = 700$$ as each of the four rows containing an empty seat can have it in either of the two seats. So the probability is $$\\frac{700}{1820}=\\frac{5}{13}$$ (note that this means that you were double-counting 28 combinations... which makes sense, as $28 = {8\\choose2}$, and that's the number of ways you can have two pairs of empty seats)\n\n-\n\nYour answer is too big, because the cases where the people are packed into 6 rows, leaving 2 empty rows for the equipment, have been counted twice.\n\nIn general, when you are getting the wrong answer to this kind of problem, and you can't see what you're doing wrong, it may help to try doing the same problem with smaller numbers. In this case, consider the same problem with 2 people and 3 rows of seats. Now the correct probability is one; there will always be room for the equipment. What do you get for an answer if you do it your way?\n\n-", "date": "2016-05-02 23:16:40", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/386536/solution-gives-wrong-answer-to-probability-problem", "openwebmath_score": 0.5803148150444031, "openwebmath_perplexity": 257.9585656557743, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347831070321, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8582793081161567}}
{"url": "https://math.stackexchange.com/questions/3036564/collecting-proofs-for-sum-n-2-infty-fracn-12n-1", "text": "# Collecting proofs for $\\sum_{n=2}^{\\infty} \\, \\frac{n-1}{2^n} = 1$ [duplicate]\n\nUpdate: The summation I came across has the form shown in title, and that exact question appears to be new. I could ask for proofs that take on this summation directly (without reducing it to summations starting at $$n = 0$$ or $$n =1$$), and that would be the preferred answer and the only one I will accept. But might as well let this fly as is, collecting all proofs of the identified equivalent variants.\n\nI just stumbled across the fact that\n\n$$\\tag 1 \\sum_{n=2}^{\\infty} \\, \\frac{n-1}{2^n} = 1$$\n\nThis is equivalent to\n\n$$\\tag 2 \\sum_{n=1}^{\\infty} \\, \\frac{n}{2^n} = 2$$\n\nI discovered $$\\text{(1)}$$ using a 'matrix/combinatorial' argument, but it would need work to turn it into a formal proof.\n\nI googled and found this Quora link, explaining how to show $$\\text{(2)}$$.\n\nI didn't find the question on this site, prompting this 'collecting proofs post':\n\nPlease supply a proof demonstrating either $$\\text{(1)}$$ or $$\\text{(2)}$$. If you use any theory or technique, mention that at the start of your answer.\n\nOne approach is to note that for $$S_k:=\\sum_{n\\geq k}\\frac{1}{2^n}$$ you have $$S_0=2$$ and $$S_{k+1}=\\frac{1}{2}S_k$$. Now rearranging terms in the summation yields $$\\sum_{n\\geq1}\\frac{n}{2^n} =\\sum_{n\\geq1}\\sum_{k=1}^n\\frac{1}{2^n} =\\sum_{k\\geq1}\\sum_{n\\geq k}\\frac{1}{2^k},$$ corresponding to the following picture: The inner sums equal the $$S_k$$, so we can simplify this to $$\\sum_{k\\geq1}S_k =\\sum_{k\\geq1}\\frac{1}{2^k}S_0 =S_0\\sum_{k\\geq1}\\frac{1}{2^k}=S_0S_1=2.$$\n\n\u2022 Anyone care to explain the downvote? \u2013\u00a0Servaes Dec 12 '18 at 15:05\n\u2022 Well not me! In fact, I can use your approach to get a direct solution to the $n \\ge 2$ summation. I wind up with $S_2 \\times S_0 = 1$. So don't feel bad - accepting and upvoting your answer! \u2013\u00a0CopyPasteIt Dec 12 '18 at 15:14\n\u2022 Glad to hear that! And indeed the argument can be generalized (or repeated?) to evaluate $\\sum_{n\\geq 0}\\frac{f(n)}{2^n}$ for any polynomial $f$. \u2013\u00a0Servaes Dec 12 '18 at 15:25\n\u2022 This is a nice answer, but I think it is important to keep the answers to this question in one place. Users should search the site to avoid duplicating material already present. This applies with extra force to experienced users who should have a fairly good idea of what type of questions have already been covered. \u2013\u00a0Jyrki Lahtonen Dec 13 '18 at 6:03\n\nA probability theory flavored approach: The expression $$\\sum_{n\\geq1} \\frac{n}{2^n}$$ is also the expected number of IID fair coin flips it takes to gets a heads. Assuming you know this number is finite (by some root/ratio test business), let $$H$$ be the expected time. Then from conditioning on seeing heads or tails on the first flip, respectively, $$H$$ satisfies the recursion $$H = \\frac{1}{2}(1) + \\frac{1}{2}(H+1),$$ so $$H = 2$$.\n\nA brute-force approach: by induction or otherwise, $$\\sum_{k=1}^n \\frac{k}{2^k} = 2 - \\frac{n+2}{2^n}.$$ Sending $$n \\to \\infty$$ recovers the desired result.\n\nThe standard technique is $$\\sum_{n\\ge 1}r^n=\\frac{1}{1-r}-1\\implies\\sum_{n\\ge 1}nr^{n-1}=\\frac{1}{(1-r)^2}$$.\n\nWe have that for $$|x|<1$$ by\n\n$$f(x)=\\sum_{k\\ge1} x^n=\\frac x{1-x} \\implies f'(x)=\\sum_{k\\ge1} nx^{n-1}=\\frac1{(1-x)^2}$$\n\ntherefore\n\n$$\\sum_{k\\ge1} nx^{n}=x\\cdot \\sum_{k\\ge1} nx^{n-1}=\\frac x{(1-x)^2}$$\n\nA solution by examining the differences between successive terms.\n\nLet us define\n\n$$S = \\sum_{n=1}^{\\infty} \\, \\frac{n}{2^n} = \\sum_{n=1}^{\\infty} u_n$$\n\nNote (Edit): the definition $$u_n = \\frac{n}{2^n}$$ will be used for $$n=0$$, i.e. for a term not involved in the series.\n\nWe have:\n\n$$u_n - u_{n+1} = \\frac{n-1}{2^{n+1}} = \\frac{u_{n-1}}{4}$$\n\nIt follows immediately:\n\n$$0 = S - S = \\sum_{n=1}^{\\infty} u_n - \\sum_{n=1}^{\\infty} u_{n+1} -u_1 = -u_1 + \\frac{u_0}{4} + \\frac{S}{4}$$\n\nAnd therefore, noting that $$u_0=0$$ and $$u_1 = \\frac{1}{2}$$ $$S = \\sum_{n=1}^{\\infty} \\, \\frac{n}{2^n} = 2$$\n\nNote: the same procedure can be used to show that $$\\sum_{k=1}^n \\frac{k}{2^k} = 2 - \\frac{n+2}{2^n}$$\n\n\u2022 Your answer needs some work. One problem, $u_n - u_{n-1}$ is never a positive number. Also, $u_0$ is not defined. \u2013\u00a0CopyPasteIt Dec 12 '18 at 15:52\n\u2022 Sorry, I just discovered a typo: it was $u_n - u_ {n+1}$. I will correct and detail my answer a little bit more \u2013\u00a0Damien Dec 12 '18 at 16:49\n\u2022 and $u_n - u_{n-1} = \\frac{n}{2^{n}} - \\frac{n-1}{2^{n-1}} = \\frac{2-n}{2^{n}} = \\frac{-(n-2)}{2^{n-2}\\times 2^2} = \\frac{-u_{n-2}}{4}$ - three subscripts involved \u2013\u00a0CopyPasteIt Dec 12 '18 at 16:52\n\u2022 I have corrected the typo. It should be more clear now \u2013\u00a0Damien Dec 12 '18 at 16:55\n\u2022 For me, removing all mention of $u_0$ is the clearest exposition, but OK... \u2013\u00a0CopyPasteIt Dec 12 '18 at 17:02\n\nI stumbled on this by realizing that since\n\n$$\\tag 1 \\sum_{n=1}^{\\infty} \\, \\frac{1}{2^n} = 1$$\n\nit must be true that\n\n$$\\tag 2 (\\sum_{n=1}^{\\infty} \\, \\frac{1}{2^n}) \\times (\\sum_{n=1}^{\\infty} \\, \\frac{1}{2^n}) = 1$$\n\nUsing the rearrangement approach (and the identity $$\\sum_1^n \\, 1 = n$$) found in Servaes' answer,\n\n$$\\tag 3 (\\sum_{n=1}^{\\infty} \\, \\frac{1}{2^n}) \\times (\\sum_{m=1}^{\\infty} \\, \\frac{1}{2^m}) = \\sum_{n+m =k}^{\\infty} \\, \\frac{1}{2^k} = \\sum_{k=2}^{\\infty} \\, \\frac{k-1}{2^k} = 1$$\n\ndemonstrating the identity equation.", "date": "2020-02-17 07:09:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3036564/collecting-proofs-for-sum-n-2-infty-fracn-12n-1", "openwebmath_score": 0.8982738852500916, "openwebmath_perplexity": 402.87773389032276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347853343058, "lm_q2_score": 0.880797068590724, "lm_q1q2_score": 0.858279302455288}}
{"url": "https://math.stackexchange.com/questions/2325505/elevator-probability-calculation", "text": "# Elevator probability calculation\n\nI'm learning probability, specifically techniques of counting, and need help with the following problem :\n\nThere are $5$ people in an elevator, $4$ floors in the building and each person exits at random. Find the probability that :\n\n$(1)$ no one exit on the first floor;\n\n$(2)$ at least one person exit on the first floor and at least one person on the second floor.\n\nSince I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.\n\n$(1)$ The number of ways to assign the $3$ remaining floors to the $5$ people is $3 \\cdot 3 \\cdot 3 \\cdot 3 \\cdot 3 = 3^5$ because for each person we can choose one of the $3$ remaining floors. By the same argument, there are $4^5$ ways to assign $4$ floors to $5$ people. Therefore, the requested probability is $$\\frac{3^5}{4^5}.$$\n\nIs my work correct for $(1)$? Any help for $(2)$ will be greatly appreciated.\n\n\u2022 The statement \"each person exits at random\" is kind of vague. At each floor, does each person have a certain probability of leaving, or does each person, at the beginning, pick a random floor to get off at? \u2013\u00a0Frpzzd Jun 16 '17 at 20:55\n\u2022 @Nilknarf I'm translating this exercise from my notes which are in greek so this is possibly not the most accurate translation. I think what is meant here is that all floors are equally likely. \u2013\u00a0user347616 Jun 16 '17 at 21:04\n\u2022 Okay, I'll leave you an answer. :) \u2013\u00a0Frpzzd Jun 16 '17 at 21:05\n\u2022 Your work is correct on 1). \u2013\u00a0Doug M Jun 16 '17 at 21:14\n\nNobody gets off on the First floor: $243$\n\nNobody gets off on the Second floor: $243$\n\nNobody gets off on at either floor: $2^5 = 32$\n\nNote that nobody gets off at either floor is a subset of both nobody gets of at the first floor it is also a subset of nobody gets off at the second floor.\n\nNobody gets off on the First floor or Nobody gets off on the second floor: $243 + 243 - 32 = 454$\n\nWe have to subtract 32 to avoid double counting.\n\nSomebody gets off at both the first floor and the second floor $= 4^5 - 454 = 570$", "date": "2019-07-17 07:03:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2325505/elevator-probability-calculation", "openwebmath_score": 0.48524361848831177, "openwebmath_perplexity": 392.57947434489284, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471680195556, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8582564908413469}}
{"url": "http://math.stackexchange.com/questions/546030/significance-of-starting-the-fibonacci-sequence-with-0-1", "text": "# Significance of starting the Fibonacci sequence with 0, 1\u2026\n\nDISCLAIMER: I do not deal with in-depth mathematics on a daily basis as some of you may, so please pardon my ignorance or lack of coherence on this topic.\n\nQUESTION: What is the significance of starting the Fibonacci sequence with $0,1$ ?\n\nFor instance, if I picked any two random integers, say 2 and 7, to start a sequence would I actually be creating some multiple or derivation of the Fibonacci sequence?\n\nIs there a general mathematical explanation for the relationship between any sequence represented by $a[0] = x, a[1] = y, a[n] = a[n-1] + a[n-2]$ and the Fibonacci sequence?\n\nOr, back to my example sequence, is there a general mathematical relationship between:\n\n$2,7,9,16,25,41,66,107,173,280...$\n\nand\n\n$0,1,1,2,3,5,8,13,21,34...$\n\nPerhaps the Golden Ratio explains it somehow? Any help would be appreciated.\n\n-\nI think it's just because 0 and 1 are the 2 lowest numbers that you can start from. \u2013\u00a0 Cruncher Oct 30 '13 at 20:16\nAs in Brian Scott's answer, note that any sequence with that additive property is a \"linear combination\" of the Fibonacci numbers and the Lucas numbers. That is, if you make such a sequence and call it $\\mbox{frogpelt}_n,$ then there will be constants $A,B$ such that $\\mbox{frogpelt}_n = A F_n + B L_n.$ \u2013\u00a0 Will Jagy Oct 30 '13 at 20:29\nFibonacci sequence. A series is an infinite sum. \u2013\u00a0 Jack M Oct 30 '13 at 23:00\n\nYes, such sequences are closely related, and the relationship does involve the golden ratio.\n\nLet $\\varphi=\\frac12(1+\\sqrt5)$ and $\\widehat\\varphi=\\frac12(1-\\sqrt5)$; $\\varphi$ is of course the golden ratio, and $\\widehat\\varphi$ is its negative reciprocal. Let $a_0$ and $a_1$ be arbitrary, and define a Fibonacci-like sequence by the recurrence $a_n=a_{n-1}+a_{n-2}$ for $n\\ge 2$. Then there are constants $\\alpha$ and $\\beta$ such that\n\n$$a_n=\\alpha\\varphi^n+\\beta\\widehat\\varphi^n\\tag{1}$$\n\nfor each $n\\ge 0$. Indeed, you can find them by substituting $n=0$ and $n=1$ into $(1)$ and solving the system\n\n\\left\\{\\begin{align*} a_0&=\\alpha+\\beta\\\\ a_1&=\\alpha\\varphi+\\beta\\widehat\\varphi \\end{align*}\\right.\n\nfor $\\alpha$ and $\\beta$. In the case of the Fibonacci numbers themselves, $\\alpha=\\frac1{\\sqrt5}$ and $\\beta=-\\frac1{\\sqrt5}$; in the case of the Lucas numbers $L_n$, for which the initial values are $L_0=2$ and $L_1=1$, $\\alpha=\\beta=1$.\n\n-\nThank you very much, good sir! \u2013\u00a0 frogpelt Oct 30 '13 at 20:28\n@frogpelt: You\u2019re very welcome. \u2013\u00a0 Brian M. Scott Oct 30 '13 at 20:29\n\nLet's call a sequence \"fibonacci-like\" if it satisfies the recurrence $$s_{n+2} = s_n + s_{n+1}$$ for all $n$. It's easy to see (or to show) that if $\\{s_i\\}$ and $\\{t_i\\}$ are two fibonacci-like sequences, then so is $\\{s_i+t_i\\}$, and so is $\\{cs_i\\}$ where $c$ is any constant. So the collection of fibonacci-like sequences forms a vector space. The dimension of the vector space is 2, since specifying two elements of the sequence (say $s_0$ and $s_1$) are enough to determine it completely.\n\nSo let's abbreviate such a sequence as $[s_0, s_1]$. The standard Fibonacci sequence $0,1,1,2,3,\\ldots$ is written as $[0,1]$ in this notation. The Lucas sequence $\\mathcal L_i = 1,3,4,7,11,\\ldots$ is written $[1,3]$.\n\nSince the space of all fibonacci-like sequences is a two-dimensional vector space, any two elements will form a basis for it, unless one is a multiple of the other. For example, a simple and standard basis for this vector space is the two vectors $[0,1]$ and $[1,0]$. The former is simply the standard Fibonacci sequence. The latter is the sequence $1,0,1,1,2,3,5,\\ldots$, which is just the standard Fibonacci sequence shifted right by one position; its $i$th element is $f_{i-1}$, the $(i-1)$th Fibonacci number.\n\nNow consider the general fibonacci-like sequence $[p,q]$:\n\n$$[p,q] = p[1,0] + q[0,1]$$\n\nSo the $i$th element of the sequence $[p,q]$ is exactly $$pf_{i-1} + qf_i.$$ For example, the Lucas sequence has $$\\mathcal L_i = f_{i-1} + 3f_i.$$ Simiarly your example sequence is $[2,7]$ and is therefore related to the Fibonacci sequence by $$[2,7] = 2[1,0] + 7[0,1] = 2f_{i-1} + 7f_i.$$\n\nAny two sequences form a basis of the space as long as they are not multiples of one another. For example, any fibonacci-like sequence can be expressed in the form $s_i = af_i + b\\mathcal L_i$ for some constants $a$ and $b$. For your $[2,7]$ sequence, we want $[2,7] = a[0,1] + b[1,3] = [b,a+3b]$. So $b=2$ and $a=1$, and we get $[2,7] = f_i + 2\\mathcal L_i$.\n\nNow consider the Fibonacci sequence, but shifted left by $k$ places, whose $i$th element is $f_{i+k}$ for each $i$. Then the first two terms of the fibonacci-like sequence $\\{f_{i+k}\\}$ are $f_k$ and $f_{k+1}$, we get $$\\{f_{i+k}\\} = [f_k, f_{k+1}] = f_k[1,0] + f_{k+1}[0,1] = f_kf_{i-1} + f_{k+1}f_i$$ and we have just proved the sum-of-indices formula for fibonacci numbers. Take $i=k$ in this and we get $$f_{2k} = f_kf_{k-1} + f_{k+1}f_k$$ which is useful for calculating extremely large Fibonacci numbers quickly.\n\n-\n\nFor any sequence $a_0, a_1, a_2, a_3, \\dots$ that satisfies the Fibonacci recurrence, we have $$a_n=(a_1-a_0)f_n+a_0f_{n+1},$$ where $f_0,f_1, f_2, \\dots$ is the Fibonacci sequence.\n\nTo prove this, let $b_n=(a_1-a_0) f_n+a_0f_{n+1}$. Note that $b_0=a_0$ and $b_1=a_1$. The two sequences $(a_n)$ and $(b_n)$ \"begin\" in the same way, and satisfy the same recurrence, so they are the same.\n\n-\nI think this is slightly off. If $n=1$, you get $b_1=a_1+a_0$. \u2013\u00a0 Mike Oct 30 '13 at 21:18\n@Mike: Thanks, cleaned up. \u2013\u00a0 Andr\u00e9 Nicolas Oct 30 '13 at 21:32", "date": "2015-05-25 19:42:00", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/546030/significance-of-starting-the-fibonacci-sequence-with-0-1", "openwebmath_score": 0.9097886085510254, "openwebmath_perplexity": 155.8255702074925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471675459396, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8582564822654299}}
{"url": "http://mathhelpforum.com/discrete-math/148804-eve-odd-numbers-question.html", "text": "# Thread: Eve/odd numbers question\n\n1. ## Eve/odd numbers question\n\nHey guys,\n\nLet set A contain the integers 1-48 inclusive.\n\nThe sum of all the integers is 48-1+1=48 and so we use n(n+1)/2 to get 1176.\n\nHow do we sum just the odd numbers OR just the even numbers?\n\n2. Think like Gauss. Gauss invented the formula for summing the first n integers by pairing up numbers. So, you paired up 1 with 48, 2 with 47, 3 with 46, and so on. Those sums are all the same: 1 + 48 = 49, 2 + 47 = 49, etc. How many of those pairs are there? n/2. Hence, (n/2)(n+1) is the sum. I bet you could reproduce this thinking for just the evens or just the odds. What do you think?\n\n3. Hmm. Would we find an n that satisfies (2n-1) and then take n^2 for the odd ones?\n\nNot sure for the even ones\n\n4. Try this on for size: for the evens from 1 to 48, inclusive:\n\n2 + 48 = 50\n4 + 46 = 50\n...\nHow many of these pairs are there? Well, there are 24 even numbers from 1 to 48, inclusive, so I'd say there are 12 pairs.\n\nThat is, the sum is equal to ... what do you think?\n\n5. Even number is 2n. So, 2(24)=48. then the answer would be 24^2?\n\n6. If you're not finding my hints helpful, please just say so, and I can try a different track. You're not following my reasoning at all. This forum's purpose is not to just give you the answer, but to help you own the answer for yourself. That won't happen unless you do most of the work. We'll help you get unstuck, but that's as far as we go.\n\n7. Another way to go is:\n\n2 + 4 + ... + 48 =\n\n2 (1 + 2 + ...+ 24)\n\n8. True, although it doesn't work quite so easily for the odds. You can jury-rig it, I guess.\n\n9. Originally Posted by Ackbeet\nTrue, although it doesn't work quite so easily for the odds. You can jury-rig it, I guess.\nJury-rigging MacGyver style! Here's how I would approach the odd numbers (nothing against Ackbeet's approach)\n\nLaTeX:\n\nCode:\n1+3+5+...+47=\\sum_{n=1}^{24}(2n-1)=2\\sum_{n=1}^{24}n-\\sum_{n=1}^{24}1\nRenderer\n\n10. Without disrepsect to the many clever solutions so far:\n\neither\n1,3,5,7,9,....,47 is a linear progression and there are standard methods for summing those. It can easily be transformed into a hypergeometric progression (with ratio 1) and there are also standard methods for summing those.\n\nor\nYou have a formula for the sum of all integers in the range, and undefined gave you a clever formula for the sum of all even numbers in the range. The sum of all odd numbers is the difference between the two\n\nie\n1+2+3+4+5+....+47+48 = 1176 (in question)\n\n2+4+6+8+...48 = 600 (from undefined's clever post)\n\nSo\n1+3+5+7+9+...47 = 1176-600\n\n11. What about this trick...I think it works for almost everything:\n\nUse formula a+(n-1)+d=x\n\nwhere a= first number in progression, d=difference in progression and x=last number in progression. Solve for n.\n\nThen, take n(a+x)/2=Sum of numbers.\n\nSo, if we want to sum the odd numbers, we see that we havev 1+3+5+...+47\n\na=1 and d=2 and x=47.\n\n1+2(n-1)=47\nn=24.\n(24(48))/2=576\n\nI think you can use this to find all even, odd, numbers divisible by 7 etc etc\n\n12. Hello, sfspitfire23!\n\nHere's a kooky method . . .\n\nSet $A$ contain the integers 1 - 48 inclusive.\n\nThe sum of all the integers is: . $\\frac{(48)(49)}{2} = 1176$\n\nHow do we sum just the odd numbers OR just the even numbers?\n\n$\\text{We have: }\\;\\begin{array}{ccccccc}\n\\text{Even} &=& \\overbrace{2 + 4 + 6 + \\hdots + 48}^{\\text{24 terms}} & [1] \\\\ \\\\ [-3mm]\n\\text{Odd} &=& \\underbrace{1 + 3 + 5 + \\hdots + 47}_{\\text{24 terms}} & [2] \\end{array}$\n\n$\\text{Subtract [1] - [2]: }\\;\\text{Even} - \\text{Odd} \\:=\\:\\underbrace{1 + 1 + 1 + \\hdots + 1}_{\\text{24 terms}}$\n\n$\\text{Hence: }\\;\\text{Even} - \\text{Odd} \\;=\\;24 \\quad\\Rightarrow\\quad \\text{Even} \\;=\\;\\text{Odd} + 24$\n\n$\\text{Since }\\,\\text{Even} + \\text{Odd} \\:=\\:1176,\\,\\text{ we have: }\\:(\\text{Odd} + 24) + \\text{Odd} \\;=\\;1176$\n\n. . $\\text{Hence: }\\;2\\text{(Odd)} \\:=\\:1152$\n\n$\\text{Therefore: }\\;\\begin{Bmatrix}\\text{Odd} &=&576 \\\\ \\text{Even} &=& 600 \\end{Bmatrix}$", "date": "2016-12-03 14:07:30", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/148804-eve-odd-numbers-question.html", "openwebmath_score": 0.8251920938491821, "openwebmath_perplexity": 935.6963012260965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471684931717, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8582564814591934}}
{"url": "https://math.stackexchange.com/questions/1994831/sitting-n-men-and-n-woman-around-a-table-having-the-wedding-couple-seated-to", "text": "# Sitting $n$ men and $n$ woman around a table having the wedding couple seated together\n\nThe problems says\n\nAt the wedding of John and Mary there are $n$ men and $n$ women. In how many ways they can sit at a round table, so that no two men is next to each other and John and Mary sit together?\n\nI have some doubts because of the answer to his problem, which doesn't match the one I got.\n\nMy attempt:\n\nLet suppose that John and Mary are already seated and that people begin sitting from the side of Mary. We have two cases. One where the person next to Mary is a man and the other when it is a woman.\n\n1st Case:\n\nWe can set the men en $n!/n = (n-1)!$ different ways.\n\nThe are $n-1$ gaps to put the women (since next to a man are Mary and John together. So there are $(n-1)!$ different ways to sit the women.\n\nBy the rule of product we have that there are $[(n-1)!]^2$ number of ways of sitting n men and n women beginning with a man next to Mary.\n\n2nd Case (A woman sits next to Mary):\n\nBy the same reasoning as above we find that there are $[(n-1)!]^2$ number of ways of sitting them.\n\nFinally by the rule of sum we have that there is a total of $2[(n-1)!]^2$ number of ways of sitting n men and n women having John and Mary always seated together.\n\nWhat I got matches the answer of the textbook but I feel the reasoning I followed wasn't that right. The fact of having assumed that ''The are $n-1$ gaps to put the women..'' doesn't convince me. Certainly next to the first man is Mary (that is, the gap of that side is occupied) but nothing is telling that after putting the last men there will be no gap between him and John, so there may be anyway $n$ gaps to put the women\n\n\u2022 Surely you mean \"wedding\" couple rather than \"weeding\"? :) \u2013\u00a0hypergeometric Nov 1 '16 at 17:56\n\u2022 @hypergeometric Ups. Yep, I mean that (: \u2013\u00a0Jazz Nov 1 '16 at 18:03\n\nFirst of all the question is a little bit ambigous. We don't know whether the table is a circle of a row, but from your reasoning it can be concluded that it's a circle, although not for certain. Also are there $n$ men and $n$ women total or $n$ men and $n$ women plus John and Mary? Considering the textbook answer I would go with the first option. Anyway from now on I will accept the mentioned assumptions as true.\n\nNote that there have to be an empty space between each men. Or to make it more simple if there are $2n$ numbered seats man can sit in the even or odd seats. So we have to put $n$ men in $n$ seats around a round table. The number of possibilities is $(n-1)!$\n\nNow as Mary has to sit next to John we have two options for her, left of John or right of him. Next the remaining $n-1$ women can be seated in $n-1$ seats in $(n-1)!$. Eventually as the events are not related we have that the total number of combinations is $2\\left((n-1)!\\right)^2$\n\nI guess this explanation will fill the \"holes\" in your solution.\n\n\u2022 It's probably a circle since the question says \"around\". \u2013\u00a0hypergeometric Nov 1 '16 at 18:24\n\nThe rule that men and women alternate should also apply to John and Mary. Hence only your 1st case applies, i.e. $[(n-1)!]^2$, and not your 2nd case.\n\nHowever, you should also consider the \"mirror configuration\" of the 1st case, i.e. where John is seated on Mary's left/right.\n\nHence the number of configurations should be multiplied by $2$, giving $2[(n-1)!]^2$.\n\nNB - this is not the same as beads on a ring where you can flip over the ring and consider the configuration to be the same - here you can't flip over the dinner table!", "date": "2019-05-26 03:11:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1994831/sitting-n-men-and-n-woman-around-a-table-having-the-wedding-couple-seated-to", "openwebmath_score": 0.7359011769294739, "openwebmath_perplexity": 220.78893103436843, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765598801371, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8582507264400325}}
{"url": "https://math.stackexchange.com/questions/582196/find-the-limit-l-lim-n-to-infty-sqrt-frac12-sqrt3-frac13-cdo", "text": "# Find the limit $L=\\lim_{n\\to \\infty} \\sqrt{\\frac{1}{2}+\\sqrt[3]{\\frac{1}{3}+\\cdots+\\sqrt[n]{\\frac{1}{n}}}}$\n\nFind the limit following:\n\n$$L=\\lim_{ _{\\Large {n\\to \\infty}}}\\:\\sqrt{\\frac{1}{2}+\\sqrt[\\Large 3]{\\frac{1}{3}+\\cdots+\\sqrt[\\Large n]{\\frac{1}{n}}}}$$\n\nP.S\n\nI tried to find the value of $\\:L$, but I found myself stuck into the abyss of incertitude.\n\nThus, any help to get me out of this rift is more than welcome!\n\n\u2022 Possibly related: math.stackexchange.com/questions/576110/\u2026 \u2013\u00a0abiessu Nov 26 '13 at 18:16\n\u2022 The numerical value is 1.2722249619362552835210450521628613228181075332403 , according to PARI. Using the inverse symbolic calculator, I found no closed expression. \u2013\u00a0Peter Nov 26 '13 at 18:28\n\u2022 n=10000;u=(1/n)^(1/n);while(n>2,n=n-1;u=(u+1/n)^(1/n));print(u) \u2013\u00a0Peter Nov 26 '13 at 18:51\n\u2022 You could see if Landau's Algorithm (for denesting radicals) is of any help, but my guess is there isn't a \"nice\" expression for what you have here... \u2013\u00a0Benjamin Dickman Nov 26 '13 at 20:06\n\u2022 I guess its limit is 1. Because it is increasing sequence which is bounded(maybe) by a number less than 2. But I don't know how to prove it! \u2013\u00a0Hamid Shafie Asl Jun 11 '14 at 11:20\n\nI like how Yiorgos S. Smyrlis approached to find upper limit of $L$. In similar way, you can easily observe $\\sqrt{\\frac{1}{2}+\\sqrt{\\frac{1}{2}+\\sqrt{\\frac{1}{2}+\\cdots}}} > L$ and lets assume that it converges to some constant $c$. Now, we can write ,\n\n$\\sqrt{\\frac{1}{2}+\\sqrt{\\frac{1}{2}+\\sqrt{\\frac{1}{2}+\\cdots}}} = c$\n\nSquaring on both sides,\n\n$\\frac{1}{2}+\\sqrt{\\frac{1}{2}+\\sqrt{\\frac{1}{2}+\\sqrt{\\frac{1}{2}+\\cdots}}} = c^2$\n\nWhich is nothing but,\n\n$\\frac{1}{2}+c = c^2$\n\nSolving above quadratic expression, value of $c$ will be $\\dfrac{1+\\sqrt{3}}{2}$ . Thus we get slightly improved upper bound for $L$ as $L < \\dfrac{1+\\sqrt{3}}{2}$\n\n\u2022 That's even better! I wonder if it is possible to get even tighter bound... \u2013\u00a0Aron D'souza Feb 20 '16 at 11:51\n\u2022 The next bound is approximately $L<1.272871$ \u2013\u00a0Yuriy S Feb 20 '16 at 11:54\n\u2022 Add it as an answer then. \u2013\u00a0Aron D'souza Feb 20 '16 at 11:55\n\nUsing Aron D'souza's idea further we can get:\n\n$$L^2-\\frac{1}{2}< \\sqrt[3]{\\frac{1}{3}+\\sqrt[3]{\\frac{1}{3}+\\dots}}$$\n\n$$\\sqrt[3]{\\frac{1}{3}+\\sqrt[3]{\\frac{1}{3}+\\dots}}=\\frac{1}{2} \\left(1+\\sqrt{\\frac{7}{3}} \\right)$$\n\n$$L<\\sqrt{1+\\frac{1}{2} \\sqrt{\\frac{7}{3}}}=1.328067$$\n\nTo find the next bound we will need to solve:\n\n$$c^4-c-\\frac{1}{4}=0$$\n\nThe exact solution is too complex to write here (see Wolframalpha), so I'll just write it numerically:\n\n$$\\sqrt[4]{\\frac{1}{4}+\\sqrt[4]{\\frac{1}{4}+\\dots}}=1.0723501510383$$\n\nThe bound will become:\n\n$$L<\\sqrt{\\frac{1}{2}+\\sqrt[3]{\\frac{1}{3}+1.0723501510383}}=1.272871$$\n\nWhich is three correct digits of the numerical value of the limit.\n\nTo make my answer more complete, the exact value of $c_4$ is:\n\n$$c=\\frac{1}{2} \\left( b+\\sqrt{\\frac{2}{b}-b^2} \\right)$$\n\n$$b=\\sqrt{ \\sqrt[3]{ \\frac{a}{18} }-\\sqrt[3]{ \\frac{2}{3a} } }$$\n\n$$a=9+\\sqrt{93}$$\n\nAnd solving the quintic equation:\n\n$$c^5-c-\\frac{1}{5}=0$$\n\nWe get the upper bound for the limit with four correct digits:\n\n$$L<1.272282$$\n\nTaking into account the corresponding lower boundary:\n\n$$L>\\sqrt{\\frac{1}{2}+\\sqrt[3]{\\frac{1}{3}+\\sqrt[4]{\\frac{1}{4}+\\sqrt[5]{\\frac{1}{5}+\\sqrt[6]{\\frac{1}{6}}}}}}=1.271035$$\n\nWe see that truncating the limit gives less accurate solutions than the method in this answer.\n\nHowever, truncating at $\\frac{1}{7}$ we can finally get very good boundaries:\n\n$$1.27207<L<1.27228$$\n\nThis is a partial result:\n\nThe underlying sequence is increasing and upper bounded by $\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots}}}=\\dfrac{1+\\sqrt{5}}{2}=\\phi$. Thus the limit exists and it is less than $\\phi$.\n\nActually, there is another way which gives better upper boundary. I'm posting it as a separate answer because of the size.\n\nFirst we notice that:\n\n$$\\lim_{n \\to \\infty} \\left( \\frac{1}{n} \\right)^{\\frac{1}{n}}=\\lim_{n \\to \\infty} \\left(1- \\left(1- \\frac{1}{n} \\right) \\right)^{\\frac{1}{n}}=1$$\n\nThis is not a proof, but the fact is well known. Now let's consider the following:\n\n$$1< \\left(\\frac{1}{n-1}+1 \\right)^{\\frac{1}{n-1}}<1+\\frac{1}{(n-1)^2}$$\n\n$$1< \\left(\\frac{1}{n-2}+1+\\frac{1}{(n-1)^2} \\right)^{\\frac{1}{n-2}}<1+\\frac{1}{(n-2)^2}+\\frac{1}{(n-2)(n-1)^2}$$\n\nOn the next step we get:\n\n$$\\dots 1+\\frac{1}{(n-3)^2}+\\frac{1}{(n-3)(n-2)^2}+\\frac{1}{(n-3)(n-2)(n-1)^2}$$\n\nIn the end we obtain the following inequality:\n\n$$\\lim_{ _{\\Large {n\\to \\infty}}}\\:\\sqrt{\\frac{1}{2}+\\sqrt[\\Large 3]{\\frac{1}{3}+\\cdots}}<1+\\sum^{\\infty}_{k=2} \\frac{1}{k ~k!}=Ei(1)-\\gamma=1.3179$$\n\nWe can increase precision by moving the truncated series under the radical (and we should not forget to get rid of $2$ in every denominator):\n\n$$1+2\\sum^{\\infty}_{k=3} \\frac{1}{k ~k!}=1+2(Ei(1)-\\gamma-1-1/4)=1.135804$$\n\n$$L < \\sqrt{\\frac{1}{2}+1.135804}=1.27899$$\n\n$$1+2\\cdot 3 \\sum^{\\infty}_{k=4} \\frac{1}{k ~k!}=1+6(Ei(1)-\\gamma-1-1/4-1/18)=1.074080$$\n\n$$L < \\sqrt{\\frac{1}{2}+\\sqrt[3]{\\frac{1}{3}+1.074080}}=1.27305$$\n\nNow for the lower boundary the better estimation would be:\n\n$$L > \\sqrt{\\frac{1}{2}+\\sqrt[3]{\\frac{1}{3}+1}}=1.26517$$\n\nThis is not ideal, but much more accurate than just truncating the sequence.", "date": "2020-12-02 00:36:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/582196/find-the-limit-l-lim-n-to-infty-sqrt-frac12-sqrt3-frac13-cdo", "openwebmath_score": 0.9270058870315552, "openwebmath_perplexity": 425.9537510141508, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765569561562, "lm_q2_score": 0.8757869932689565, "lm_q1q2_score": 0.8582507222906963}}
{"url": "https://math.stackexchange.com/questions/2797935/do-both-versions-invariant-and-primary-of-the-fundamental-theorem-for-finitely/2797951", "text": "# Do both versions (invariant and primary) of the Fundamental Theorem for Finitely Generated Abelian Groups hold at the same time?\n\nSo there are the two versions of the Fundamental Theorem for Finitely Generated Abelian Groups (FTFGAG). I take the following from A First Course in Abstract Algebra by Fraleigh. The first is as follows:\n\nFTFGAG 1: Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form $$G \\cong \\mathbb{Z}_{p_1^{r_1}} \\times \\cdots \\times \\mathbb{Z}_{p_n^{r_n}} \\times \\mathbb{Z} \\times \\cdots \\times \\mathbb{Z}$$ where $p_i$ are primes (not necessarily distinct) and $r_i$ are positive integers.\n\nBut then we also have the second version:\n\nFTFGAG 2: Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form $$G \\cong \\mathbb{Z}_{m_1} \\times \\cdots \\times \\mathbb{Z}_{m_r} \\times \\mathbb{Z} \\times \\cdots \\times \\mathbb{Z}$$ where $m_1 | m_2 | \\cdots | m_r$.\n\nMy question is whether these two hold at all times? So say for $\\mathbb{Z}_{20}$, do we have $\\mathbb{Z}_{20} \\cong \\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_5 \\cong \\mathbb{Z}_4 \\times \\mathbb{Z}_5 \\cong \\mathbb{Z}_2 \\times \\mathbb{Z}_{10}$, even though $10$ is not a power of a prime (though of course $2|10$)?\n\nI've also seen statements of the Chinese Remainder Theorem (CRT) which say that $\\mathbb{Z}_{nm} \\cong \\mathbb{Z}_n \\times \\mathbb{Z}_m$ if and only if $\\gcd(n,m)=1$. Does this not contradict $\\mathbb{Z}_{20} \\cong \\mathbb{Z}_2 \\times \\mathbb{Z}_{10}$ from above? Or is treating $\\mathbb{Z}_n$ and $\\mathbb{Z}_m$ as rings in the CRT what makes the situation different? I guess what I'm asking in a nutshell is: why don't the primary and invariant forms of the structure theorems for abelian groups (and also modules over PIDs, etc) contradict each other?\n\nEdit: So it seems that I didn't see this question which basically answers mine. CRT gives us a way of decomposition but FTFGAG only tells us that some decomposition is always possible. So for FTFGAG 1, $\\mathbb{Z}_4 \\times \\mathbb{Z}_5$ suffices, for FTFGAG 2, $\\mathbb{Z}_{20}$ suffices, and for the CRT $\\mathbb{Z}_{20} \\cong \\mathbb{Z}_4 \\times \\mathbb{Z}_5$ works.\n\n\u2022 Yes they are both correct theorems so they both hold for all finitely generated abelian groups. May 27, 2018 at 13:02\n\u2022 Thanks, would you be able to shed some light on why the stuff in the CRT paragraph is/isn't a contradiction? May 27, 2018 at 13:04\n\u2022 It is not true that $\\mathbb{Z}_{20}\\simeq\\mathbb{Z}_2\\times\\mathbb{Z}_{10}$. May 27, 2018 at 13:09\n\u2022 @MichaelBurr Indeed, I have just seen the (~duplicate) question which I linked in my edit. May 27, 2018 at 13:12\n\nFor your example, $\\mathbb{Z}_{20}$,\n\n\u2022 The first decomposition gives you $\\mathbb{Z}_{20}\\simeq\\mathbb{Z}_{2^2}\\times\\mathbb{Z}_5$.\n\n\u2022 The second decomposition gives you $\\mathbb{Z}_{20}\\simeq\\mathbb{Z}_{20}$. In other words, this one doesn't break up the abelian group at all.\n\n\u2022 You can note that $\\mathbb{Z}_{20}\\not\\simeq\\mathbb{Z}_2\\times\\mathbb{Z}_{10}$ since $\\mathbb{Z}_{20}$ has an element of order $20$ while the maximum order of an element of $\\mathbb{Z}_2\\times\\mathbb{Z}_{10}$ is $10$.\n\nThe idea of the statements is that it is possible to write it in this form, not that all products of the given form are isomorphic to the given group.\n\nA more interesting example might be given by $$\\mathbb{Z}_4\\times\\mathbb{Z}_6\\times\\mathbb{Z}_5.$$\n\n\u2022 The first decomposition gives you $$\\mathbb{Z}_4\\times\\mathbb{Z}_6\\times\\mathbb{Z}_5\\simeq \\left(\\mathbb{Z}_{2^2}\\right)\\times\\left(\\mathbb{Z}_2\\times\\mathbb{Z}_3\\right)\\times\\mathbb{Z}_5\\simeq \\mathbb{Z}_2\\times\\mathbb{Z}_{2^2}\\times\\mathbb{Z}_3\\times\\mathbb{Z}_5.$$\n\n\u2022 On the other hand, the second decomposition gives you $$\\mathbb{Z}_4\\times\\mathbb{Z}_6\\times\\mathbb{Z}_5\\simeq\\mathbb{Z}_2\\times\\mathbb{Z}_{2^2\\cdot 3\\cdot 5}=\\mathbb{Z}_2\\times\\mathbb{Z}_{60}.$$ The group on the right combines the highest powers of each prime that appear in the fully expanded product of the first decomposition, then you work your way down by induction.\n\n\u2022 \"The group on the right combines the highest powers of each prime that appear in the fully expanded product of the first decomposition, then you work your way down by induction.\" I really like that way of thinking about it, thanks! May 27, 2018 at 13:23\n\nYou are correct about the Chinese remainder theorem indeed $\\mathbb{Z}_{20} \\not\\cong \\mathbb{Z}_2 \\times \\mathbb{Z}_{10}$. The problem with applying the 2nd version in this manner is the rank uniquely determines the group. So for example $\\mathbb{Z}_{20}$ is rank 1, that is it already is written in that form (trivially)", "date": "2022-07-03 05:11:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2797935/do-both-versions-invariant-and-primary-of-the-fundamental-theorem-for-finitely/2797951", "openwebmath_score": 0.8555042743682861, "openwebmath_perplexity": 170.27292924233032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765628041175, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.8582507194695058}}
{"url": "https://math.stackexchange.com/questions/839045/calculus-find-the-value-of-x-so-that-f-x-0/839051", "text": "# Calculus / find the value of $x$ so that $f ''(x)=0$\n\nLet $f(x)= 10xe^x$\n\n$(a)$ Find the exact value of $x$ so that $f ''(x) = 0$.\n\nI tried:\n\n\\begin{align}f'(x)& = 10e^x\\\\f''(x)&= e^x\\end{align} but at that point, the $f''(x)$ would never be a zero. So what is my mistake?\n\n$(b)$ For what interval is $f(x)$ concave up?\n\nI wonder how to know the concavity after knowing the equation\n\n\u2022 One mistake is not using the product rule to evaluate $(10xe^x)'$. \u2013\u00a0David Mitra Jun 18 '14 at 23:45\n\u2022 Once you have the correct second derivative function, $\\ f(x) \\$ is \"concave upward\" wherever $\\ f \\ ''(x) \\ > \\ 0 \\$ and \"concave downward\" wherever $\\ f \\ ''(x) \\ < \\ 0 \\$ . Keep in mind that the exponential factor $\\ e^x \\$ is always positive. \u2013\u00a0colormegone Jun 18 '14 at 23:49\n\u2022 @DavidMitra thanks for the reminder \u2013\u00a0John Jun 18 '14 at 23:50\n\n(a) $f'(x)=10e^x + 10xe^x$.\n\n$f''(x)=10e^x + 10e^x + 10xe^x = 20e^x + 10xe^x =10e^x(2+x)$\n\nSo, at $x=-2$ you have $f''(x)=0$.\n\n(b) Hint: If $f''(x)>0$, $f$ is concave up at $x$, and if $f''(x)<0$, $f$ is concave down at $x$.\n\n\u2022 Thaanks !What did you do to go from f'(x) to f''(x)? \u2013\u00a0John Jun 18 '14 at 23:49\n\u2022 I computed it using the product rule. \u2013\u00a0Twink Jun 18 '14 at 23:50\n\u2022 @John, derivation using the product rule. $f''(x) = (10e^x + 10xe^x)' = 10 (e^x)' + 10(x)'e^x + 10x(e^x)' = 10e^x + 10 e^x + 10xe^x = 10(2+x)e^x$ \u2013\u00a0Graham Kemp Jun 18 '14 at 23:52\n\n$(a)$ Since $f(x)=10x\\cdot e^x$ we will use the product rule to obtain $f'(x)$ and $f''(x)$. So $f'(x)=10x\\cdot e^x+10e^x$ and $f''(x)=10x\\cdot e^x+20e^x$. Set $f''(x)=0$. So $10x\\cdot e^x+20e^x=0$ implies that $10e^x(x+2)=0$. We know that $10e^x$ is never $0$ and so $x=-2$ will make $f''(x)=0$.\n\n$(b)$ If $x<-2$, then the function is concave down. If $x>-2$, then the function is concave up.", "date": "2020-02-20 21:21:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/839045/calculus-find-the-value-of-x-so-that-f-x-0/839051", "openwebmath_score": 0.9450278878211975, "openwebmath_perplexity": 325.83177659646697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765616345254, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8582507152680886}}
{"url": "https://math.stackexchange.com/questions/3168656/arbitrary-union-definition-in-set-theory", "text": "# Arbitrary union definition in set theory\n\nI am reading Enderton's \"Elements of Set Theory\". He defines the union operation as $$\\cup A = \\{ x \\;|\\; x \\text{ belongs to some member of } A\\} = \\{x \\;|\\; (\\exists b \\in A) x \\in b\\}$$ Maybe I am missing a subtle point from earlier on in the text, but what if the set in question is not a \"set of sets\"? For example, if $$A = \\{1,2,3\\}$$, then is $$\\cup A = \\emptyset$$?\n\n\u2022 In systems of set theory such as ZF, every set is \"a set of sets\". \u2013\u00a0Lord Shark the Unknown Mar 30 at 19:11\n\u2022 So then is my $A$ really $\\{\\{1\\},\\{2\\},\\{3\\}\\}$? \u2013\u00a0theQman Mar 30 at 19:12\n\u2022 No, it's really $\\{1,2,3\\}$. \u2013\u00a0Lord Shark the Unknown Mar 30 at 19:13\n\u2022 See Enderton's chapter on \"Natural Numbers\" to see the usual way of representing positive integers as sets. \u2013\u00a0Lord Shark the Unknown Mar 30 at 19:20\n\u2022 Even without the \"everything-is-a-set\" context, we can still make sense of this by carefully writing down the definition: see this earlier question. \u2013\u00a0Noah Schweber Mar 30 at 19:32\n\nWhen you're working in a system built out of the set theory, everything is a set. I assume the book will cover this later on, but one way of \"building natural numbers\" out of sets is called Von Neumann ordinals and the construction is as follows:\n\nLet $$0 \\equiv \\emptyset$$. Let the number $$n \\equiv n - 1 \\cup \\{ n - 1\\}$$. That is, under this system: \\begin{align*} &1 \\equiv 0 \\cup \\{ 0 \\} = \\emptyset \\cup \\{\\emptyset\\} = \\{\\emptyset\\} \\\\ &2 \\equiv 1 \\cup \\{1 \\} = \\{\\emptyset\\} \\cup \\{\\{\\emptyset\\}\\} = \\{\\emptyset, \\{\\emptyset\\}\\} \\\\ &3 \\equiv 2 \\cup \\{2\\} = \\dots = \\{\\emptyset, \\{\\emptyset\\}, \\{\\emptyset, \\{\\emptyset\\}\\}\\} \\end{align*}\n\nThe point of the representation is that 0 is a set with zero elements, 1 is a set with 1 element, 2 is a set with 2 elements and so on. So we can define the \"size of a set\" as \"the natural number it is in bijection with\".\n\nThis goes down a rabbit hole, and we can then ask \"well, now that I have the natural numbers, how do I define addition? Multiplication? What about the integers? rationals? reals?\"\n\nWe can construct all of these things, and a set theory book usually describes these encodings in one of the later chapters.\n\nBut the point is that at this level, all the \"stuff of math\" is sets, and you can always write expressions such as $$1 \\ \\cup 2 \\cup 3 = \\{\\emptyset\\} \\cup \\{ \\emptyset, \\{\\emptyset\\}\\} \\cup \\{ \\emptyset, \\{\\emptyset\\},\\{ \\emptyset, \\{\\emptyset\\}\\} \\} = \\{ \\emptyset, \\{\\emptyset\\},\\{ \\emptyset, \\{\\emptyset\\}\\}\\} = 3$$!\n\n\u2022 +1.... In the most-favored system of set theory, not only is everything (that can be said to exist) a set, but in the formal language there is no word or definition of \"set\". Things ( with various properties that $can$ be stated in the language ) are merely asserted to exist or not exist. \u2013\u00a0DanielWainfleet Mar 30 at 19:39\n\u2022 Indeed, thanks! much appreciated \u2013\u00a0Siddharth Bhat Mar 30 at 20:18\n\nAs Lord Shark the Unknown indicates, in nearly all systems of set theory, every object under the sun is a set. In particular, the elements of any set are, in turn, sets themselves. For the specific example you give, i.e. $$\\cup \\{ 1,2,3 \\},$$ we need to be able to describe the natural numbers as sets. On page 67 of Enderton's text, the construction is outlined. He defines \\begin{align} 0 &= \\varnothing \\\\ 1 &= \\{0\\} = \\{ \\varnothing \\} \\\\ 2 &= \\{0,1\\} = \\{ \\varnothing, \\{\\varnothing\\} \\} \\\\ 3 &= \\{0,1,2\\} = \\{ \\varnothing, \\{\\varnothing\\}, \\{ \\varnothing, \\{\\varnothing\\} \\}\\}, \\end{align} and so on. The \"and so on\" is explained in somewhat more detail in chapter 4 of the text (starting on page 66). Note that we could write $$2 = \\{\\varnothing, \\{\\varnothing\\} \\}$$ over and over again, but it is likely easier to see what is going on if we work one level of abstraction higher. That is, $$2 = \\{0,1\\}$$. Once we accept this abstraction, we have $$\\cup\\{1,2,3\\} = 1 \\cup 2 \\cup 3 = \\{0\\} \\cup \\{0,1\\} \\cup \\{0,1,2\\} = \\{0,1,2\\} = 3.$$\n\n\u2022 I see. Maybe I am getting ahead of myself since I am only in chapter 2. But I guess the construction you provided is pretty natural since at this point the only set that we can use as a building block is $\\emptyset$ (Empty Set Axiom)? Then we can use \"pairing\" to form the singleton $\\{\\emptyset\\}$, which is defined to be $\\{\\emptyset, \\emptyset\\}$? \u2013\u00a0theQman Mar 31 at 2:09", "date": "2019-08-21 08:06:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3168656/arbitrary-union-definition-in-set-theory", "openwebmath_score": 0.9963454008102417, "openwebmath_perplexity": 290.9596723320387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765552017675, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.858250709634363}}
{"url": "https://byjus.com/question-answer/if-two-concentric-circles-are-drawn-with-centre-in-first-quadrant-one-touches-the-x/", "text": "Question\n\n# If two concentric circles are drawn with centre in first quadrant, one touches the X-axis and Y-axis and the other passes through the origin, the radius of the smaller circle is $$'r'$$, then the centre is\n\nA\n(2r,r)\nB\n(r,2r)\nC\n(r,r)\nD\nnone of these\n\nSolution\n\n## The correct option is C $$(r,r)$$Let $$(a,b)$$ be the center of the circles Given that one of the circles touches both the axes and the other passes through origin. Smaller circle touches the axes and Bigger circle passes through $$O$$. Given radius of smaller circle is $$r$$ and $$X,Y$$ axes are tangents Which implies radius = perpendicular distance from center to tangent $$(y =0 , x = 0)$$ $$r = \\dfrac{|a|}{\\sqrt{1^2 + 0}} , r = \\dfrac{|b|}{\\sqrt{1^2 + 0}}$$ $$\\implies a = r, b = r$$ $$\\implies (a, b) = (r,r) = center$$Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-25 09:16:19", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/if-two-concentric-circles-are-drawn-with-centre-in-first-quadrant-one-touches-the-x/", "openwebmath_score": 0.45138731598854065, "openwebmath_perplexity": 314.580368998239, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303407461413, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8582183049648849}}
{"url": "http://www.gtmath.com/2015/04/the-plane-in-r3.html?showComment=1430268957485", "text": "## The Plane in R^3\n\nPrerequisites:\u00a0Euclidean Space and Vectors\n\n### Suppose we have a plane in ${\\Bbb R}^{3}$ which contains the point $(0,0,0)$ but does not intersect the axes at any other point. How many octants does the plane intersect?\n\nLet's start by analyzing the 2-dimensional analog, then come back to the 3-dimensional problem with an algebraic approach and a geometric interpretation. Finally, we'll try to generalize the answer to higher dimensions.\n\n#### Lines in the Plane ${\\Bbb R}^{2}$\n\nThe 2-dimensional analog of a plane in ${\\Bbb R}^{3}$ containing the origin is a line in ${\\Bbb R}^{2}$ containing the origin. For those familiar with linear algebra, the former is a vector subspace of ${\\Bbb R}^{3}$ of dimension 2 (i.e. co-dimension 1), and the latter is a vector subspace of ${\\Bbb R}^{2}$ of dimension 1 (still co-dimension 1). If you aren't familiar with linear algebra, ignore that sentence and just keep reading.\n\nThe equation of a line in ${\\Bbb R}^{2}$ is $y = mx + b$, where $m$ is the slope (\"rise over run,\" or change in $y$ per change in $x$), and $b$ is the $y$-intercept (where the line hits the $y$-axis). A point $(x_{0},y_{0})$ is on the line if it satisfies the equation, i.e. if the equality $y_0 = mx_0 + b$ is true.\n\nA line passing through the origin has a $y$-intercept of 0, thus $b=0$ and the equation is simply $y=mx$. Now, if $m$ is 0, then the line would just go along the $x$-axis, so if it doesn't touch the axis, we must have $m \\neq 0$, so either $m > 0$ or $m < 0$. In the case where $m>0$, a positive $x$ value gives a positive $y$, and a negative $x$ gives a negative $y$, so the line would pass through quadrants 1 and 3 (top-right and bottom-left). In the case where $m<0$, the same analysis shows that the line passes through quadrants 2 and 4 (top-left and bottom-right). So the line passes through 2 quadrants.\n\nThe $y = mx + b$ formulation works fine for this question, but if we want to put $x$ and $y$ on more even footing (this will come in handy in the higher-dimensional cases so that we don't always have to solve for one of the variables), we can use the other form of the equation of a line, $ax + by = c$, where $a,b,c \\in {\\Bbb R}$ are constants. In order to have the origin on the line, we must have $c=0$, because the point $(0,0)$ is on the line, and thus $a(0)+b(0) = 0 = c$. so the equation is simply $ax+by=0$. If either $a$ or $b$ were zero, the line would just be one of the axes, so we must have $a,b \\neq 0$. Whether they are positive or negative, you can work out by plugging in positive or negative $x$ and $y$ values that the line will either pass through quadrants 1 and 3 or 2 and 4.\n\nFor example, if $a,b>0$, then if $x>0$, we must have $y<0$ in order to have $ax+by=0$. Similarly, if $x<0$, then $y$ would have to be positive. So the line passes through quadrants 2 and 4. We get the same result in the case where $a,b<0$. If $a$ and $b$ have opposite signs, then the line will pass through quadrants 1 and 3.\n\nNotice that the line does not pass through the quadrant containing the point $(a,b)$. The vector $(a,b)$ is actually perpendicular to the line $ax+by=0$, and we can see that from the fact that the equation can be rewritten as ${\\bf n} \\cdot {\\bf x} = 0$ where ${\\bf n} = (a,b)$ and ${\\bf x} = (x,y)$ (recall that two vectors are perpendicular if and only if their dot product is zero).\n\n#### The Plane in Space ${\\Bbb R}^{3}$\n\nLet's go back to the ${\\Bbb R}^{3}$ case, building off of the above discussion. The general equation of a plane is $ax+by+cz=d$, where $a,b,c,d \\in {\\Bbb R}$ are constants. In order for the plane to contain $(0,0,0)$, we must have $d=0$, so the equation is now just $ax+by+cz=0$. If any of the constants is 0, then the plane will actually look like the equation of a line from above. For example, if $c=0$, then we'd have $ax+by=0$, which would be a line in the $xy$-plane extended vertically up and down in the positive and negative $z$ directions, and in fact containing the entire $z$-axis. Can you see why (hint: show that the points on the $z$-axis, i.e. points of the form $(0,0,z)$, all satisfy the equation of the plane)?\n\nWe can also interpret the equation geometrically as follows: the equation $ax+by+cz=0$ is equivalent to ${\\bf n} \\cdot {\\bf x} = 0$, where ${\\bf n} = (a,b,c)$ and ${\\bf x} = (x,y,z)$. Note that here, ${\\bf n}$ and ${\\bf x}$ are vectors, and their dot product is a scalar, so the $0$ on the right is a scalar zero, not the zero vector ${\\bf 0} = (0,0,0)$.\n\nAs mentioned above, the dot product of two vectors is zero if and only if the vectors are perpendicular. Therefore, this equation is saying that any vector ${\\bf x}$ that is perpendicular to ${\\bf n}$ is on the plane. For this reason, ${\\bf n}$ is called the plane's normal vector (normal is a synonym for perpendicular, as is orthogonal, which is also used frequently). In the example above where $c=0$, the plane's normal vector is $(a,b,0)$, which lies in the $xy$-plane. Thus, the $z$-axis, being orthogonal to the $xy$-plane, is contained in our plane.\n\nNow, in order to answer the geometric question of which vectors are orthogonal to ${\\bf n}$, we can look at the algebraic equation $ax+by+cz=0$.\n\nSince the plane does not touch the axes except at the origin, we must have $a,b,c \\neq 0$. As an example, let's look at the case where $a,b,c>0$. Then we can have the following combinations for $(x,y,z)$ in order to have $ax+by+cz=0$:\n$$(+,+,-) \\\\ (+,-,+) \\\\ (+,-,-) \\\\ (-,-,+) \\\\ (-,+,-) \\\\ (-,+,+)$$ The remaining two combinations, $(+,+,+)$ and $(-,-,-)$, do not work, because then the left side of the equation would have to be positive or negative (respectively) and thus not zero.\n\nIf we grind through the algebra of the other 7 combinations for $(a,b,c)$, we see that we get 6 possibilities each time, so the plane intersects 6 of the 8 octants, and we have the answer to the problem. I'm not going to go through all the cases, because that would be quite boring, but you can see that there is a certain symmetry in the plane's equation between $(a,b,c)$ and $(x,y,z)$. Once you've solved it for the case $a,b,c>0$, you've pretty much solved it for all the cases. Can you see why? So we've got the answer- it's 6.\n\n#### The Hyperplane in ${\\Bbb R}^{n}$\n\n${\\Bbb R}^{n}$ is the $n$-dimensional analog of ${\\Bbb R}^{3}$ and is the set of ordered $n$-tuples of real numbers: ${\\Bbb R}^{n} = \\{ (x_1,x_2,x_3,...,x_n) \\ \\colon \\ {\\scr each} \\ x_i \\in {\\Bbb R} \\}$. We can't picture this $n$-dimensional space, but we can use the same types of algebraic equations that work in ${\\Bbb R}^{3}$ to analyze ${\\Bbb R}^{n}$.\n\n${\\Bbb R}^{n}$ is divided into $2^n$ orthants, also known as hyperoctants or $n$-hyperoctants, based on the signs, positive or negative, of the $n$ components of a point. A 2-hyperoctant is a quadrant in ${\\Bbb R}^{2}$ and a 3-hyperoctant is an octant in ${\\Bbb R}^{3}$. The $x_i$-axis is the set of points where all coordinates except possibly the $i^{\\scr th}$ are zero.\n\nA hyperplane in ${\\Bbb R}^{n}$ is a set $P$ of points (equivalently, vectors) that are orthogonal to a normal vector ${\\bf n} = (a_1, a_2, ... a_n)$. In symbols, $P = \\{ {\\bf x} \\in {\\Bbb R}^{n} \\ \\colon \\ {\\bf n} \\cdot {\\bf x} = 0 \\}$. For those familiar with linear algebra, the hyperplane containing the origin is a vector subspace of ${\\Bbb R}^{n}$ of dimension $n-1$, i.e. co-dimension 1. A hyperplane in ${\\Bbb R}^{2}$ is a line, and a hyperplane in ${\\Bbb R}^{3}$ is a plane.\n\n### How many $n$-hyperoctants does a hyperplane $P \\subset {\\Bbb R}^{n}$ intersect, given that it contains the origin, but does not intersect the axes at any other point?\n\nTo answer this question, we can use the discussion above from the $n=2$ and $n=3$ cases and generalize the results. We can then prove the answer is correct using induction.\n\nWhen we went from $n=2$ to $n=3$, we took the equation $ax+by=0$ (i.e. the line in ${\\Bbb R}^{2}$ with normal vector $(a,b)$), extended it into 3-dimensional space to make the plane whose normal vector is $(a,b,0)$, and then added a non-zero third coordinate to the normal vector to \"tilt\" the plane off of the $z$-axis.\n\nNow, a hyperplane (including the line and plane in the $n=2$ and $n=3$ cases) is orthogonal to its normal vector ${\\bf n}$ as well as the negative of the normal vector, $-{\\bf n}$. In fact, the hyperplane is orthogonal to any scalar multiple of ${\\bf n}$, but my point in mentioning $-{\\bf n}$ is that the hyperplane won't intersect the $n$-hyperoctants that contains ${\\bf n}$ or $-{\\bf n}$.\n\nLet's look at the case where ${\\bf n}$ lies in the first $n$-hyperoctant, i.e. has all positive coordinates. As mentioned above, the other cases are pretty much the same because of the symmetries of the equation ${\\bf n} \\cdot {\\bf x} = \\sum_{i=1}^{n}{a_i x_i} = 0$, so the number of $n$-hyperoctants the hyperplane intersects is the same in all cases. In the case that the $a_i$ are positive, the hyperplane doesn't intersect the first $n$-hyperoctant or the one with all negative coordinates (whatever number we want to assign to that one).\n\nIn the ${\\Bbb R}^{2}$ case, the line intersects quadrants 2 and 4. When we extended ${\\bf n} = (a,b)$ to ${\\bf n} = (a,b,0)$ in ${\\Bbb R}^{3}$, we got a plane that contained the entire $z$-axis. The intersection of this plane with the $xy$-plane is the line $ax+by=0$, which remains the case regardless of the third coordinate of ${\\bf n}$. Now, this plane intersects the octants $(+,-,+)$, $(+,-,-)$, $(-,+,+)$, and $(-,+,-)$. We took the original quadrants 2 and 4 and multiplied them by 2 to get 4 octants.\n\nWhen we add a non-zero third coordinate to ${\\bf n}$ (let's assume it's positive), the new plane also intersects two additional octants: $(+,+,-)$ and $(-,-,+)$. The first two coordinates of these two would have not been included in the 2-d case, but the third coordinate allows us to use those combinations and still get the equation $ax+by+cz$ to equal zero. $(+,+,+)$ and $(-,-,-)$ still don't work though.\n\nThe same logic works when going from ${\\Bbb R}^{n}$ to ${\\Bbb R}^{n+1}$ when $n>2$, and we can prove it by induction.\n\nThoerem: For $n \\geq 2$, a hyperplane in ${\\Bbb R}^{n}$ containing the origin, but not intersecting the coordinate axes at any other point, intersects $2^{n}-2$ $n$-hyperoctants.\n\nThe proof is a bit lengthy, but basically just formalizes the idea of extending the line in ${\\Bbb R}^{2}$ into a plane in ${\\Bbb R}^{3}$ and then tilting it off the $z$-axis\n\nProof: The base case of $n=2$ was already shown above.\n\nFor the induction step, assume the theorem is true for ${\\Bbb R}^{n-1}$, and consider a hyperplane $P = \\{{\\bf x} \\in {\\Bbb R}^{n} \\ \\colon \\ {\\bf n} \\cdot {\\bf x}=0 \\}$ where ${\\bf n} = (a_1,a_2,...,a_n)$.\n\nThe equation of $P$ is $\\sum_{i=1}^{n}{a_i x_i} = \\sum_{i=1}^{n-1}{a_i x_i} + a_n x_n = 0$. By the induction hypothesis, the solutions to the equation $\\sum_{i=1}^{n-1}{a_i x_i} = 0$ intersect $2^{n-1}-2$ $(n-1)$-hyperoctants. Let's call those solutions $P_{n-1}$, which is a hyperplane in ${\\Bbb R}^{n-1}$\n\nTake a point ${\\bf x}_{0, n-1} = (x_{0,1}, x_{0,2},...,x_{0,n-1}) \\in {\\Bbb R}^{n-1}$ which satisfies the equation of $P_{n-1}$. If $x_{0,1}>0$, then $x_{0,1}+\\epsilon>0$ as well, where $\\epsilon = \\frac{1}{2}|x_{0,1}|$. Similarly, if $x_{0,1}<0$, then $x_{0,1}+\\epsilon<0$ as well, so the point ${\\bf x}_{1,n-1} = (x_{0,1}+\\epsilon, x_{0,2},...,x_{0,n-1})$ is in the same $(n-1)$-hyperoctant as ${\\bf x}_{0, n-1}$. By a similar argument, so is the point ${\\bf x}_{2,n-1} = (x_{0,1}-\\epsilon, x_{0,2},...,x_{0,n-1})$.\n\nDefine the points ${\\bf x}_{1} = (x_{0,1}+\\epsilon, x_{0,2},..., x_{0,n-1}, -\\frac{a_1}{a_n}\\epsilon), \\ {\\bf x}_{2} = (x_{0,1}-\\epsilon, x_{0,2},..., x_{0,n-1}, \\frac{a_1}{a_n}\\epsilon) \\in {\\Bbb R}^{n}$. Then \\begin{align} {\\bf n} \\cdot {\\bf x}_{1} &= a_1 (x_{0,1}+\\epsilon) + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} + a_n (-\\frac{a_1}{a_n}\\epsilon) \\\\[2mm] &= a_1 (x_{0,1}+\\epsilon -\\epsilon) + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} \\\\[2mm] &= a_1 x_{0,1} + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} = 0 \\end{align}\nwith the final equality being true because ${\\bf x}_{0,n-1} \\in P_{n-1}$.\n\nThis shows that ${\\bf x}_{1} \\in P$. Similarly, ${\\bf x}_{2} \\in P$. The first $n-1$ coordinates of these two points are in the same $(n-1)$-hyperoctant as ${\\bf x}_{0,n-1}$, and the $n^{\\scr th}$ coordinates of ${\\bf x}_{1}$ and ${\\bf x}_{2}$ have opposite sign. This shows that we have kept the $2^{n-1}-2$ $(n-1)$-hyperoctants of the $(n-1)$-hyperplane when we extended it into ${\\Bbb R}^{n}$ and actually multiplied them by 2 (by adding both positive and negative $n^{\\scr th}$ coordinates) to get $2(2^{n-1}-2)$ = $2^{n}- 4$ $n$-hyperoctants.\n\nWe just need to show that we've also added two more $n$-hyperoctants. These are the ones where the first $n-1$ coordinates all have the same sign or all have the opposite sign as the first $n-1$ coordinates of ${\\bf n}$, just like when we went from $n=2$ to $n=3$ above. Examples of solutions to the equation of $P$ that are in those 2 $n$-hyperoctants are $(a_1,a_2,...,a_{n-1},-\\dfrac{1}{a_n}\\sum_{i=1}^{n-1}{a_i^2})$ and $(-a_1,-a_2,...,-a_{n-1},\\dfrac{1}{a_n}\\sum_{i=1}^{n-1}{a_i^2})$.\n\nSo now we are up to $2^{n}-4+2 = 2^{n}-2$, so $P$ intersects at least that many $n$-hyperoctants. There are only 2 more $n$-hyperoctants, and those are the ones that contain $\\pm {\\bf n}$, but we already know that points in those $n$-hyperoctants cannot satisfy the equation of ${\\bf n} \\cdot {\\bf x} = 0$, so $P$ intersects exactly $2^{n}-2$ of the $2^n$ $n$-hyperoctants, and the theorem is proved.\n$\\square$\n\nHere's a diagram illustrating the objects described in the proof in the case where $n=3$ and $n-1=2$. Apologies for the low quality (I made it in MS Paint), but note that the bottom of the red plane, $P$, comes out towards the viewer, in front of the blue plane, and the top half of $P$ is behind the blue plane. The points ${\\bf x}_{0,n-1}$, ${\\bf x}_{1,n-1}$, and ${\\bf x}_{2,n-1}$ are in in the $x_{1}x_{2}$-plane, with ${\\bf x}_{1,n-1}$ in front of $P$ and ${\\bf x}_{2,n-1}$ behind $P$.\n\n2. A hypersurface is a generalization of a hyperplane in the context of manifolds. I'm not sure the question about octants would make sense in that context since a manifold need not be in ${\\Bbb R}^{n}$ (though it has a map to ${\\Bbb R}^{n}$ in a neighborhood of each point). Maybe there's a way to reformulate the question to apply in that context...", "date": "2021-05-15 08:44:48", "meta": {"domain": "gtmath.com", "url": "http://www.gtmath.com/2015/04/the-plane-in-r3.html?showComment=1430268957485", "openwebmath_score": 0.9379755258560181, "openwebmath_perplexity": 105.21813795780214, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303413461358, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.8582182952826629}}
{"url": "https://www.intmath.com/forum/methods-integration-31/taking-e-to-both-sides:160", "text": "IntMath Home \u00bb Forum home \u00bb Methods of Integration \u00bb Taking e to both sides\n\n# Taking e to both sides [Solved!]\n\n### My question\n\nYour Integration Chapter: The Basic Logarithmic Form, example 4.\n\nCan you show how you took e to both sides to arrive at your result please?\n\n### Relevant page\n\n2. Integration: The Basic Logarithm Form\n\n### What I've done so far\n\nAfter integrating, t = ln 20 - ln (20-v).\n\nYou applied log laws to get t = ln(20/(20-v)).\n\nYou took \"e to both sides\" to get e^t = 20/20-v.\n\nWe thought by taking e to both sides the result would be the exponents equated:\n\nt = 20/20-v.\n\nX\n\nYour Integration Chapter: The Basic Logarithmic Form, example 4.\n\nCan you show how you took e to both sides to arrive at your result please?\nRelevant page\n\n<a href=\"https://www.intmath.com/methods-integration/2-integration-logarithmic-form.php\">2. Integration: The Basic Logarithm Form</a>\n\nWhat I've done so far\n\nAfter integrating, t = ln 20 - ln (20-v).\n\nYou applied log laws to get t = ln(20/(20-v)).\n\nYou took \"e to both sides\" to get e^t = 20/20-v.\n\nWe thought by taking e to both sides the result would be the exponents equated:\n\nt = 20/20-v.\n\n## Re: Taking e to both sides\n\n@Phinah: The opposite process of taking the \"log\" of something is to take \"e to the power of...\".\n\nWriting the full details would be:\n\nt = ln (20 /(20-v))\n\ne^t = e^[ln (20/(20-v))]\n\ne^t = 20/(20-v)\n\ne^t = e^[(20/(20-v))] (without \"ln\")\n\nt = 20/(20-v)\n\nbut that's not what we started with.\n\nA similar situation is the case where \"raising to power 2\" is the opposite of \"square root\".\n\ny = sqrt(x+2)\n\nSquaring both sides gives us:\n\ny^2 = (sqrt(x+2))^2 = x+2\n\nThe \"power 2\" has \"undone\" the sqrt operation.\n\nSimilarly, the \"e to the power\" operation undoes the \"ln\" expression in the example you are asking about.\n\nHope it makes sense.\n\nX\n\n@Phinah: The opposite process of taking the \"log\" of something is to take \"e to the power of...\".\n\nWriting the full details would be:\n\nt = ln (20 /(20-v))\n\ne^t = e^[ln (20/(20-v))]\n\ne^t = 20/(20-v)\n\ne^t = e^[(20/(20-v))] (without \"ln\")\n\nt = 20/(20-v)\n\nbut that's not what we started with.\n\nA similar situation is the case where \"raising to power 2\" is the opposite of \"square root\".\n\ny = sqrt(x+2)\n\nSquaring both sides gives us:\n\ny^2 = (sqrt(x+2))^2 = x+2\n\nThe \"power 2\" has \"undone\" the sqrt operation.\n\nSimilarly, the \"e to the power\" operation undoes the \"ln\" expression in the example you are asking about.\n\nHope it makes sense.\n\n## Re: Taking e to both sides\n\nIt does now. Thank you.\n\nX\n\nIt does now.  Thank you.", "date": "2018-05-25 08:37:23", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/methods-integration-31/taking-e-to-both-sides:160", "openwebmath_score": 0.8099038004875183, "openwebmath_perplexity": 5458.315803701843, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517507633983, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.858217719219022}}
{"url": "https://math.stackexchange.com/questions/2177569/law-of-sines-in-triangle", "text": "# Law of Sines in Triangle\n\nOn the side BC of the triangle ABC we construct towards the exterior a square BCDE. Denote the intersection between AE and BC by M. Use the law of sines to prove that\n\n$$\\frac{BM}{CM}=\\frac{\\cos \\measuredangle B\\cdot \\sin \\measuredangle C}{\\sqrt{2} \\cdot \\sin \\measuredangle B \\cdot \\sin(\\measuredangle C+45\u00b0)}$$\n\nIf someone could please help me prove this problem. I do not have a similar problem to work off of. I am unclear of where the $\\sin(\\measuredangle C+45^{\\circ})$ comes into play and the $\\sqrt{2}$.\n\n\u2022 the end got cut off...i am unlcear where the sin(<C+45) and \\sqrt{2} come into play. \u2013\u00a0user423388 Mar 8 '17 at 13:42\n\u2022 Provide the figure \u2013\u00a0Nick Pavini Mar 8 '17 at 14:09\n\nLet $\\measuredangle B=\\beta$, $\\measuredangle C=\\gamma$, $\\measuredangle BAE=\\alpha_1$, $\\measuredangle CAE=\\alpha_2$. By applying the sine rule to triangles $BAM$ and $CAM$ one gets: $${BM\\over\\sin\\alpha_1}={AM\\over\\sin\\beta},\\quad {CM\\over\\sin\\alpha_2}={AM\\over\\sin\\gamma},\\quad \\hbox{whence:}\\quad {BM\\over CM}={\\sin\\alpha_1\\over\\sin\\alpha_2}{\\sin\\gamma\\over\\sin\\beta}.$$ By applying then the sine rule to triangles $BAE$ and $CAE$ one gets: $${\\sin\\alpha_1\\over BE}={\\sin(\\beta+90\u00b0)\\over AE},\\quad {\\sin\\alpha_2\\over CE}={\\sin(\\gamma+45\u00b0)\\over AE}.$$ From that, taking into account that $CE=\\sqrt2 BE$, one can readily obtain $\\displaystyle{\\sin\\alpha_1\\over\\sin\\alpha_2}$ and thus the desired result.\n\n\u2022 fully understood this !! thank you so much ! \u2013\u00a0user423388 Mar 8 '17 at 16:34\n\nUsing sine rule at $ACE$ we have:\n\n$$\\frac{AE}{\\sin(C+45\u00b0)}=\\frac{AC}{\\sin \\alpha}$$\n\nUsing sine rule at $ABE$ we have:\n\n$$\\frac{AE}{\\sin(B+90\u00b0)}=\\frac{AB}{\\sin \\beta}$$\n\nDividing both equations we have:\n\n$$\\frac{\\cos B}{\\sin (C+45\u00b0)}=\\frac{AC}{AB}\\cdot \\frac{\\sin \\beta}{\\sin \\alpha} \\quad (1)$$\n\nUsing sine rule at $ABC$ we get\n\n$$\\frac{AC}{AB}=\\frac{\\sin B}{\\sin C}$$\n\nSo, from $(1)$\n\n$$\\frac{\\cos B\\cdot \\sin C}{\\sin B\\cdot \\sin (C+45\u00b0)}=\\frac{\\sin \\beta}{\\sin \\alpha}\\quad (2)$$\n\nbut $\\alpha + \\beta=45\u00b0$ so\n\n$$\\frac{\\sin \\beta}{\\sin \\alpha}=\\frac{\\sin \\beta}{\\sin (45\u00b0-\\beta)}=\\sqrt{2}\\cdot\\frac{\\tan \\beta}{1-\\tan \\beta}\\quad (3)$$\n\nFinaly we can use, from the triangle BME, that\n\n$$\\tan \\beta = \\frac{BM}{BE}=\\frac{BM}{BM+CM}\\to \\frac{BM}{CM}=\\frac{\\tan \\beta}{1- \\tan \\beta} \\quad (4)$$\n\nPluging $(3)$ and $(4)$ at $(2)$ we get what we want\n\n\u2022 can you explain (3)? i dont understand how you switched from sine to tangent... \u2013\u00a0user423388 Mar 8 '17 at 16:17\n\u2022 @erica: $\\frac{\\sin \\beta}{\\sin (45\u00b0-\\beta)}=\\frac{\\sin \\beta}{(\\sqrt{2}/2)(\\cos \\beta-\\sin \\beta)}=$, now divide the numerator and denominator by $\\cos \\beta$ and get $\\sqrt{2}\\cdot\\frac{\\tan \\beta}{1-\\tan \\beta}$ \u2013\u00a0Arnaldo Mar 8 '17 at 16:33\n\nIn triangle $ABM$, using the law of sines implies $$\\frac{BM}{AM}=\\frac{\\sin\\measuredangle BAM}{\\sin\\measuredangle B}$$ similarly, in $ACM$: $$\\frac{CM}{AM}=\\frac{\\sin\\measuredangle CAM}{\\sin\\measuredangle C}$$ combining these two yields: $$\\frac{BM}{CM}=\\frac{\\sin\\angle BAM}{\\sin\\angle CAM}\\cdot\\frac{\\sin\\angle C}{\\sin\\angle B}\\label{*}\\tag{*}$$ Now, in triangle $ABE$ note that $\\measuredangle ABE=\\measuredangle B+90^{\\circ}$.\n\nThus $\\sin\\measuredangle ABE=\\cos\\measuredangle B$ and the law of sines implies $$\\frac{\\sin\\measuredangle BAM}{\\cos\\measuredangle B}=\\frac{BE}{AE}$$ and for triangle $ACE$: $$\\frac{\\sin\\measuredangle CAM}{\\sin(\\measuredangle C+45)}=\\frac{CE}{AE}$$ Note that $CE=\\sqrt{2}BE$. Now combining the two equations implies: $$\\frac{BE}{CE}=\\frac 1{\\sqrt 2}=\\frac{\\sin\\measuredangle BAM}{\\sin\\measuredangle CAM}\\cdot\\frac{\\sin(\\measuredangle C+45)}{\\cos\\measuredangle B}$$ And finally, you can use $\\eqref{*}$ to get the desired result.\n\n\u2022 OOps. I am late to the party \u2013\u00a0polfosol Mar 8 '17 at 15:00\n\u2022 can you explain how $\\text{sin}\\angleABE = \\text{cos}\\angleB$? \u2013\u00a0user423388 Mar 8 '17 at 15:58", "date": "2019-08-17 20:51:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2177569/law-of-sines-in-triangle", "openwebmath_score": 0.9659079313278198, "openwebmath_perplexity": 340.78472568365356, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517494838938, "lm_q2_score": 0.8774767922879692, "lm_q1q2_score": 0.8582177118287636}}
{"url": "http://yucoding.blogspot.com/2013/04/leetcode-question-116-unique-path-i.html", "text": "## Unique Path I\n\nA robot is located at the top-left corner of a\u00a0m\u00a0x\u00a0n\u00a0grid (marked 'Start' in the diagram below).\nThe robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).\nHow many possible unique paths are there?\n\n## Analysis:\n\nThis is an easy problem. From the description we know that, the robot can only move down or right, which means, if the robot is now in position (x,y), then the position before this step\u00a0must be \u00a0either (x-1,y) or (x, y-1). Since current position is only from these two previous positions, the number of possible paths that the robot can reach this current position (x,y) is the sum of paths from (x-1, y) and (x, y-1).\n\nWe want to get the number of paths on position (m,n), we need to know (m-1,n) and (m, n-1). For (m-1,n), we must know (m-2,n) and (m-1,n-1) ... \u00a0until we back to the start position (1,1) ([0][0] in \u00a0C++). Note that the boundary of the map, we can easily know that the top row and the first column of the map are all 1s. Use loop can solve the problem then.\n\n### Code(C++):\n\nclass Solution {\npublic:\nint uniquePaths(int m, int n) {\n// Start typing your C/C++ solution below\n// DO NOT write int main() function\nint **arr = new int* [m];\nfor (int i=0;i<m;i++){\narr[i]= new int[n];\n}\narr[0][0]=1;\nfor (int i=0;i<m;i++){\narr[i][0] = 1;\n}\nfor (int i=0;i<n;i++){\narr[0][i] = 1;\n}\nfor (int i=1;i<m;i++){\nfor(int j=1;j<n;j++){\narr[i][j] = arr[i][j-1] + arr[i-1][j];\n}\n}\nreturn arr[m-1][n-1];\n}\n};\n\n\n### Code(Python):\n\nclass Solution:\n# @return an integer\ndef uniquePaths(self, m, n):\n#define map and initialization\nmp = [[0 for i in xrange(n)] for i in xrange(m)]\nfor i in xrange(m):\nmp[i][0]=1\nfor j in xrange(n):\nmp[0][j]=1\n\nfor i in xrange(1,m):\nfor j in xrange(1,n):\nmp[i][j]=mp[i-1][j]+mp[i][j-1]\n\nreturn mp[m-1][n-1]\n\n\n\n1. I think most of your posts are great but sometimes it is very hard to understand your English though.\n\n1. Actually I have been refining both the analysis and code in all my posts recently. Thank you very much for your suggestion.\n\n2. Does this algorithm have a name?\n\n3. Does this algorithm have a name?\n\n1. Dynamic Programming\n\n4. I LOVE YOU YU!\n\n5. why did you initialise by 1 for the borders ?", "date": "2018-06-25 00:41:39", "meta": {"domain": "blogspot.com", "url": "http://yucoding.blogspot.com/2013/04/leetcode-question-116-unique-path-i.html", "openwebmath_score": 0.3533759117126465, "openwebmath_perplexity": 2527.0851145767156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517507633985, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8582177066839761}}
{"url": "http://mathhelpforum.com/algebra/222247-find-length-rectangle-given-diagonal-area-print.html", "text": "# find length of rectangle given diagonal and area\n\n\u2022 September 24th 2013, 11:08 AM\nBonganitedd\nfind length of rectangle given diagonal and area\nRectangle has area=168 m^2 and diagonal of 25. Find length\nThis is how tried to attempt the problem\nArea= L X W\n168 = L x W ..........(1)\nL^2 + W^2 =25^2 ............(2)\nFrom (1) L = 168/W...........(3)\nSubstitute (3) into (2)\n(168/W)^2 +W^2 = 625\n28224/W^2 + W^2 = 625\nThe problem gets complicated as I proceed\nIs this aproach correct if it is,\nIs there a convinient method\n\u2022 September 24th 2013, 11:18 AM\nvotan\nRe: find length of rectangle given diagonal and area\nQuote:\n\nOriginally Posted by Bonganitedd\nRectangle has area=168 m^2 and diagonal of 25. Find length\nThis is how tried to attempt the problem\nArea= L X W\n168 = L x W ..........(1)\nL^2 + W^2 =25^2 ............(2)\nFrom (1) L = 168/W...........(3)\nSubstitute (3) into (2)\n(168/W)^2 +W^2 = 625\n28224/W^2 + W^2 = 625\nThe problem gets complicated as I proceed\nIs this aproach correct if it is,\nIs there a convinient method\n\nSketch a rectangle with one diagonal. laber the sides L and W and D for the diabonal. Write Pythagorean theorem, and the area = LW. Eliminate L from the the Pythagorean theorem and solve for W then find L from area formula.\n\u2022 September 24th 2013, 11:30 AM\nBonganitedd\nRe: find length of rectangle given diagonal and area\nIts same approach I used, I did draw a rectangle now how do eliminate L bcos the pathygras is L^2 + W^2 =25^2\nThe area, LW =168\n\u2022 September 24th 2013, 12:17 PM\nPlato\nRe: find length of rectangle given diagonal and area\nQuote:\n\nOriginally Posted by Bonganitedd\nRectangle has area=168 m^2 and diagonal of 25. Find length\nThis is how tried to attempt the problem\nArea= L X W\n168 = L x W ..........(1)\nL^2 + W^2 =25^2 ............(2)\nFrom (1) L = 168/W...........(3)\nSubstitute (3) into (2)\n(168/W)^2 +W^2 = 625\n28224/W^2 + W^2 = 625\nThe problem gets complicated as I proceed\nIs this aproach correct if it is,\nIs there a convinient method\n\nHave a look at this webpage.\n\u2022 September 24th 2013, 01:15 PM\nSoroban\nRe: find length of rectangle given diagonal and area\nHello, Bonganitedd!\n\nQuote:\n\nRectangle has area=168 m^2 and diagonal of 25. Find the length.\n\nThis is how tried to attempt the problem\n$\\text{Area} \\:=\\: L\\cdot W \\:=\\:168 \\quad\\Rightarrow\\quad L \\,=\\,\\frac{168}{W}\\;\\;[1]$\n\n$L^2 + W^2 \\:=\\:25^2\\;\\;[2]$\n\n$\\text{Substitute [1] into [2]: }\\;\\left(\\frac{168}{W}\\right)^2 +W^2 \\:=\\:625 \\quad\\Rightarrow\\quad \\frac{28,\\!224}{W^2} + W^2 \\:=\\: 625$\n\nIs this approach correct? . Yes\nIf it is, is there a convinient method?\n\nWe have: . $\\frac{28,\\!224}{W^2} + W^2 \\:=\\:625$\n\nMultiply by $W^2\\!:\\;\\;28,\\!224 + W^4 \\:=\\:625W^2 \\quad\\Rightarrow\\quad W^4 - 625W^2 + 28,\\!224 \\:=\\:0$\n\nFactor: . $(W^2 - 49)(W^2 - 576) \\:=\\:0$\n\n$\\begin{array}{ccccccccc}W^2-49 \\:=\\:0 & \\Rightarrow & W^2 \\:=\\:49 & \\Rightarrow & W \\:=\\:7 \\\\ W^2 - 576 \\:=\\:0 & \\Rightarrow & W^2 \\:=\\:576 & \\Rightarrow & W \\:=\\:24 \\end{array}$\n\n$\\text{Assuming }L > W\\text{, we have: }\\:W\\,=\\,7,\\;\\boxed{L \\,=\\,24}$\n\u2022 September 24th 2013, 02:08 PM\nHallsofIvy\nRe: find length of rectangle given diagonal and area\nOnce you get to \"[tex]W^4- 625W^2+ 28224=0 as Soroban showed, if the \"fourth degree polynomial\" bothers you, you can let $x= W^2$ so that your equation is x^2- 625x+ 28224= 0 and use whatever method you like, completing the square or the quadratic formula, to solve that quadratic equation.\n\u2022 September 24th 2013, 07:02 PM\nBonganitedd\nRe: find length of rectangle given diagonal and area\nNot taking away credit to other members who contributed, but @ Hallsofivy your advice is what I needed. Big u to MHF.", "date": "2016-06-27 01:04:30", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/222247-find-length-rectangle-given-diagonal-area-print.html", "openwebmath_score": 0.8184131979942322, "openwebmath_perplexity": 6551.913472922392, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517469248845, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8582177033157692}}
{"url": "https://math.stackexchange.com/questions/1485605/write-cos8x-in-terms-of-sinx", "text": "# Write $\\cos^8(x)$ in terms of $\\sin(x)$?\n\nI am currently working on an integral using u substitution and have hit a wall at the above point. The integral in question is $\\int$$\\sin^7(x)$$\\cos^9(x)$ using $u=sin(x)$.\n\nAfter finding $du=cos(x) dx$, I am asked to express $\\cos^8(x)$ in terms of $u$, at which point I am stuck. My current attempt was to reduce the power down from 8 to 2, and express it as $(1-u^2)$, but that gives me a wrong answer (this assignment checks for us before submission).\n\nAny help is appreciated.\n\nThanks.\n\n\u2022 What did you find the integral to be (your \"wrong\" answer)? \u2013\u00a0Henry Oct 18 '15 at 9:55\n\u2022 I hadn't got to that point yet as the next few steps required it expressed in terms of u still. \u2013\u00a0Dwayne H Oct 18 '15 at 9:56\n\u2022 OK, your \"wrong\" answer in terms of $u$? \u2013\u00a0Henry Oct 18 '15 at 9:57\n\u2022 I'd only gotten to the point of $u^7(1-u^2)$, but seeing as the assignment said it was wrong I didn't try going ahead and solving for that. \u2013\u00a0Dwayne H Oct 18 '15 at 9:59\n\u2022 $\\displaystyle \\int u^7(1-u^2)^4\\,du$ with a power of $4$ seems more likely \u2013\u00a0Henry Oct 18 '15 at 10:00\n\nIf $$\\cos^2 x = 1-\\sin^2 x$$ then $$\\cos^8 x = (1-\\sin^2 x)^4.$$\nUsing $u=\\sin x$, you have $du=\\cos x\\,dx$, and also $\\cos^2x=1-u^2$, so $\\cos^8x=(1-u^2)^4$. Thus your integral becomes $$\\int u^7(1-u^2)^4\\,du$$", "date": "2019-09-21 11:37:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1485605/write-cos8x-in-terms-of-sinx", "openwebmath_score": 0.8556122779846191, "openwebmath_perplexity": 264.6576359480567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987758723627135, "lm_q2_score": 0.8688267847293731, "lm_q1q2_score": 0.8581912359373532}}
{"url": "http://oqxm.agence-des-4-fontaines.fr/covariance-calculator-with-probability.html", "text": "Covariance Calculator With Probability\n\nIn introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. Pre-trained models and datasets built by Google and the community. , the variables tend to show similar behavior), the covariance is positive. Their covariance Cov(X;Y) is de ned by. We assume that a probability distribution is known for this set. You can find formula used for calculation of covariance below the calculator. How can I calculate the covariance in R? I created two vectors x,y and fed them into cov(), but I get the wrong result. Recent research has pointed to the ubiquity and abundance of between-generation epigenetic inheritance. The term ANCOVA, analysis of covariance, is commonly used in this setting, although there is some variation in how the term is used. Each point in the x-yplane corresponds to a single pair of observations (x;y). Before you compute the covariance, calculate the mean of x and y. Application of this formula to any particular observed sample value of r will accordingly test the null hypothesis that the observed value comes from a population in which rho=0. Suppose, under uncertainty, the manager believes that a risky asset, for example, an equity, can bring him different results, which at the moment of portfolio formation can only be judged with some probability, as shown in Table. Covariance can be calculated by using the formula. If a jpd is over N random vari-ables at once then it maps from the sample space to RN, which is short-hand for real-valued vectorsof dimension N. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. The probability that a woman has all three risk factors, given that she has A and B, is 1/3. Mathematics is concerned with numbers, data, quantity, structure, space, models, and change. Properties of variance and covariance (a) If and are independent, then by observing that. Different categories of descriptive measures are introduced and discussed along with the Excel functions to calculate them. You COULD do a calculation patterned after covariance with 3 or more variables, but I don\u2019t see it as meaning anything, and I don\u2019t think it would be admissable as a valid statistical function. This binomial calculator can help you calculate individual and cumulative binomial probabilities of an experiment considering the probability of success on a single trial, no. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. It is found that the effect of estimator variability is significant to obtain VaR forecast with better coverage. An investor is facing two potential financial losses and with the following joint density function: Let be the total of these two losses. Example: Plastic covers for CDs (Discrete joint pmf) Measurements for the length and width of a rectangular plastic covers for CDs are rounded to the nearest mm(so they are discrete). The correlation coefficient is equal to the covariance divided by the product of the standard deviations of the variables. More about Covariance. Chapter 5: JOINT PROBABILITY DISTRIBUTIONS Part 2: Covariance and Correlation Section 5-4 Consider the joint probability distribution fXY(x;y). If instead, you want to get a step-by-step calculation of all descriptive statistics, you can try our descriptive statistics calculator. Let W have the exponential distribution with mean 1. Returns population covariance, the average of the products of deviations for each data point pair in two data sets. Expected Return Formula Calculator. If A is a scalar, cov(A) returns 0. Covariance \u2022 Variance and Covariance are a measure of the \"spread\" of a set of points around their center of mass (mean) \u2022 Variance - measure of the deviation from the mean for points in one dimension e. Covariance is driving me nuts, is there any simple way to make the calculator do the work for you? Get your mind off your Level I results with a free 2020 Level II JumpStart package. 35 5% -3% Given The Above Information On Two Investments A And B, Calculate The Statistics Below. Please enter the necessary parameter values, and then click 'Calculate'. Suppose, under uncertainty, the manager believes that a risky asset, for example, an equity, can bring him different results, which at the moment of portfolio formation can only be judged with some probability, as shown in Table. A low value means there is a weak relationship. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: \u03a3 ij = cov(x i,x j) = E p(x)[(x i \u2212\u00b5 i)(x j \u2212\u00b5 j)] (10) between variables x i and x j. \u2022 Probability and Statistics for Engineering and the Sciences, Covariance, Correlation, Sampling Distributions, Central Limit calculator or for any other. E(X1)=\u00b5X1 E(X2)=\u00b5X2 var(X1)=\u03c32 X1 var(X2)=\u03c32 X2 Also, we assume that \u03c32 X1 and \u03c32 X2 are \ufb01nite positive values. Estimating the uncertainty of revenues and investment decisions. And, to calculate the probability of an interval, you take the integral of the probability density function over it. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. The covariance matrix is a matrix that only concerns the relationships between variables, so it will be a k x k square matrix. Using the formulae above to compute covariance can sometimes be tricky. The covariance matrix can be calculated in NumPy using the cov() function. The sign (+ or -) of the correlation affects its interpretation. I can of course go ahead and derive all the elements of conditional covariance matrix. The Lognormal Probability Distribution Let s be a normally-distributed random variable with mean \u00b5 and \u03c32. The denominator is represented by (n-1), which is just one less than the number of data pairs in your data set. The converse. There is a 50 percent probability that after one year the stock's price will rise to $5 (yielding a 400 percent return). R Functions for Probability Distributions. The probability density function of the bivariate normal distribution is implemented as MultinormalDistribution[mu1, mu2, sigma11, sigma12, sigma12, sigma22] in the Wolfram Language package MultivariateStatistics. 128 CHAPTER 7. Variables are inversely. If the variables tend to show similar behavior, the covariance is positive. Uniform distribution Calculator - High accuracy calculation Welcome, Guest. This post provides practice problems to reinforce the concepts discussed in the preceding post on univariate probability distributions. That is, covariance matrices with small determinants denote variables that are redundant or highly correlated. Variables are positively related if they move in the same direction. calculate and interpret covariance and correlation and interpret a scatterplot; calculate and interpret the expected value, variance, and standard deviation of a random variable and of returns on a portfolio; calculate and interpret covariance given a joint probability function; calculate and interpret an updated probability using Bayes' formula;. This online calculator computes covariance between two discrete random variables. By default, this function will calculate the sample covariance matrix. The sign (+ or -) of the correlation affects its interpretation. 2 Covariance Covariance is a measure of how much two random variables vary together. Chapter 4 Variances and covariances Page 3 A pair of random variables X and Y is said to be uncorrelated if cov. 1 Introduction. Joyce, Fall 2014 Covariance. [In our case, a 5\u00d75 matrix. Also, it can be considered as a generalization of the concept of variance of two random variables. Covariance Calculator calculator, formula and. The Covariance Calculator an online tool which shows Covariance for the given input. Covariance gives you a positive number if the variables are positively related. Covariance[dist] gives the covariance matrix for the multivariate symbolic distribution dist. The formula defines covariance for discrete variables in Simon & Blume (1994): Mathematics for Economists, end of section A5. Linear Models in SAS (Regression & Analysis of Variance) The main workhorse for regression is proc reg, and for (balanced) analysis of variance, proc anova. In case the greater values of one variable are linked to the greater values of the second variable considered, and the same corresponds for the smaller figures, then the covariance is positive and is a signal that the two variables show similar behavior. If the variables tend to show similar behavior, the covariance is positive. Obtaining covariance estimates between variables allows one to better estimate direct and indirect effects with other variables, particularly in complex models with many parameters to be estimated. If using percent form, the user must add the percent sign (%) at the end of the number. Covariance[v1, v2] gives the covariance between the vectors v1 and v2. In the matrix diagonal there are variances, i. (2005), Fundamentals of Probability with Stochastic Processes, Roussas, G. In addition, we may only be able to assess unconditional coverage probability for VaR forecast of the SVAR model. Problem 31B. I have a joint probability mass function of two variables X,Y like here. P function in Microsoft Excel. Joint Discrete Probability Distributions. Chapter 4 Variances and covariances Page 3 A pair of random variables X and Y is said to be uncorrelated if cov. I hope you found this video useful, please subscribe for daily videos! WBM Foundations: Mathematical logic Set theory Algebra: Number theory Group theory Lie groups Commutative rings Associative. By using this formula, after calculation, you can verify the result of such calculations by using our covariance calculator. Covariance is a measure of how much two random variables vary together. When comparing data samples from different populations, two of the most popular measures of association are covariance and correlation. Applied to historical prices, covariance can help determine if stocks' prices tend. Video for finding the covariance and correlation coefficient by hand. TI-82 / TI-83 Graphing Calculator. Correlation values range from positive 1 to negative 1. 5 should display. For a discrete variable$\\sum_{i=1}^nP(X=x_{i}|A)=1$, hence we immediately can fill in the missing values in the conditional distribution tables. The TI-83 Graphing Calculator can facilitate the entry of ordered lists of data and perform some statistical analyses, but lacks a single command to calculate the covariance of two lists of numbers. Covariance & Correlation The covariance between two variables is defined by: cov x,y = x x y y Can I directly relate the free parameters to the covariance matrix? First calculate P(x) by marginalizing over y: P x\u02c6exp{1 21 2 x x0 x 2} dyexp{1 21 2 [y y0 y 2 2 x x0 x y y0 y]} probability content is contained within a radius of s(2. Lecture 21: Conditional Distributions and Covariance / Correlation Statistics 104 Colin Rundel April 9, 2012 6. when the returns of one asset goes up, the. 92 and 202-205; Whittaker and Robinson 1967, p. For instance, I have been given a discrete random variable X with probability function px(x) = 1/2 if x = -1, 1/4 if x = 0, 1/4 if x = 1, 0 otherwise. Discrete Random Variable Calculator. Now, we also need to be able to calculate the covariance and correlation for a joint probability function. It is considered the ideal case in which the probability structure underlying the categories is known perfectly. Although the covariance and variance are linked to each other in the above manner, their probability distributions are not attached to each other in a simple manner and have to be dealt separately. Descriptive Statistics which contains one variable and multivariable calculators for 20 descriptive statistics measures including: mean, variance, covariance, quantile, interquartile range, correlation and many more. There are various kinds of insurance. Recall that , and that is the normal density with mean and variance. In case the greater values of one variable are linked to the greater values of the second variable considered, and the same corresponds for the smaller figures, then the covariance is positive and is a signal that the two variables show similar behavior. Video and lecture notes from a tutorial on Probability and Statistics given at PyData NYC 2019. For each of the three factors, the probability is 0. when the returns of one asset goes up, the. This calculator is featured to generate the complete work with steps for any corresponding input values may helpful for grade school students to solve covaraince worksheet or homework problems or. The expected value of X is usually. Imagine how confusing it would be if people used degrees Celsius and degrees Fahrenheit interchangeably. 2 thoughts on \" An Example on Calculating Covariance. Probability helps the company to calculate how many chances that insurance holders have to claim the insurance. it helps us to understand how two sets of data are related to each other. For example, if the number of desired outcomes divided by the number of possible events is. Use the theorem we just proved to calculate the covariance of X and Y. Sample Mean and Covariance Calculator. My ultimate objective is to calculate with a given set data the following ratio: Det[Covariance matrix for X]/Det[Conditional covariance matrix X|S] Any idea about how to achieve this?. Most articles and reading material on probability and statistics presume a basic understanding of terms like means, standard deviation, correlations, sample sizes and covariance. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: \u03a3 ij = cov(x i,x j) = E p(x)[(x i \u2212\u00b5 i)(x j \u2212\u00b5 j)] (10) between variables x i and x j. - The probability that a company will pass the test given that it will subsequently survive 12 months, is. In the next section, read Problem 1 for an example showing how to turn raw data into a variance-covariance matrix. Covariance Calculator calculator, formula and work with steps to estimate the relationship (linear dependence) between two dataset in statistical experiments. WORKED EXAMPLES 3 COVARIANCE CALCULATIONS EXAMPLE 1 Let Xand Y be discrete random variables with joint mass function defined by f X,Y(x,y) = 1 4 Hence the two variables have covariance and correlation zero. I know the definition of covariance and I'm trying to solve some exercises. Rules for the Correlation Coefficient. Standard Deviation Calculator Variance Calculator Kurtosis Calculator Skewness Calculator. Using the formulae above to compute covariance can sometimes be tricky. com's Covariance calculator is an online statistics & probability tool to estimate the nature of association between two random variables X & Y in probability & statistics experiments. Before we demonstrate the function in Excel, a few observations on the covariance and correlation measures. The probability that a large earthquake will occur on the San Andreas Fault in. If you see any typos, potential edits or changes in this Chapter, please note them here. function [probability] = comp_gauss_dens_val (mean, covarianceMatrix, givenMatrix ) The value assigned in lengthofM varibale is the length of the mean of the given matrix, which in our is one dimensional in our case. If an input is given then it can easily show the result for the given number. Joint Discrete Probability Distributions. A negative covariance means that the variables are inversely related, or that they move in opposite directions. If A is a vector of observations, C is the scalar-valued variance. All you can really tell from covariance is if there is a positive or negative relationship. Apart from the six-page limitation, originality, quality and clarity will be the criteria for choosing the material to be published in Statistics & Probability Letters. Sample covariance measures the \u2026. The domain of t is a set, T , of real numbers. Provide an interpretation (10%). Covariance[m] gives the covariance matrix for the matrix m. Formulas that calculate covariance can predict how two stocks might perform relative to each other in the future. Click the Calculate! button and find out the covariance matrix of a multivariate sample. (i) The expected value measures the center of the probability distribution - center of mass. covariance calculator - step by step calculation to measure the statistical relationship (linear dependence) between the two sets of population data, along with. For example,. The correlation coefficient is also known as the Pearson Product-Moment Correlation Coefficient. pX Beta calculator. calculate and interpret covariance and correlation and interpret a scatterplot; calculate and interpret the expected value, variance, and standard deviation of a random variable and of returns on a portfolio; calculate and interpret covariance given a joint probability function; calculate and interpret an updated probability using Bayes' formula;. In probability, we use 0. Adding a constant to a random variable does not change their correlation coefficient. is between and inclusive, which meets the first property of the probability distribution. ) S is said to have a lognormal distribution,. Full curriculum of exercises and videos. Small sample size problems. Different categories of descriptive measures are introduced and discussed along with the Excel functions to calculate them. In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. To find the covariance matrix you need to calculate the Fisher information matrix. A covariance refers to the measure of how two random variables will change together and is used to calculate the correlation between variables. Obtaining covariance estimates between variables allows one to better estimate direct and indirect effects with other variables, particularly in complex models with many parameters to be estimated. In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. Do October 10, 2008 A vector-valued random variable X = X1 \u00b7\u00b7\u00b7 Xn T is said to have a multivariate normal (or Gaussian) distribution with mean \u00b5 \u2208 Rn and covariance matrix \u03a3 \u2208 Sn 1. covariance calculator - step by step calculation to measure the statistical relationship (linear dependence) between the two sets of population data, along with. This calculator determines the following coin toss probability scenarios * Coin Toss Sequence such as HTHHT * Probability of x heads and y tails * Covariance of X and Y denoted Cov(X,Y) * The. Is there a relationship between Xand Y? If so, what kind? If you\u2019re given information on X, does it give you information on the distribution of Y? (Think of a conditional distribution). It's an online statistics and probability tool requires two sets of population data X and Y and measures of how much these data sets vary together, i. When the covariance is positive, X tends to be high when Y is high, and vice versa; when the covariance is negative, X tends to be high when Y is low, and vice versa. Consider two random variables$X$and$Y$. Here are some documents to help you use the TI-82 calculator. Enter X values (Separated by comma) Enter Y values (Separated by comma) Letter Arrangment Probability Calculator. Where Cov (A, B) \u2013 is covariance of portfolios A and B. Calculate the standard deviation of the returns using STDEV function; Finally, we calculate the VaR for 90, 95, and 99 confidence level using NORM. Or are they. The converse. 01 respectively for the VaR(90), VaR(95), and VaR(99). If the greater values of one variable mainly correspond with the greater values of the other variable, and the same holds for the lesser values, (i. S function is:. , the covariance of each element with itself. This suggests the question: Given a symmetric, positive semi-de nite matrix, is it the covariance matrix of some random vector?. The calculator will find the p-value for two-tailed, right-tailed and left-tailed tests from normal, Student's (T-distribution), chi-squared and Fisher (F-distribution) distributions. How to Calculate Expected Value. The probability that a large earthquake will occur on the San Andreas Fault in. After cleaning the data, the researcher must test the assumptions of ANOVA. Therefore, it is a straightforward exercise to calculate the correlation between X and. 2 Covariance Covariance is a measure of how much two random variables vary together. More about Covariance. Welcome to StatCalculators. The sample variance, s\u00b2, is used to calculate how varied a sample is. This calculator computes the variance from a data set: To calculate the variance from a set of values, specify whether the data is for an entire population or from a sample. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this page. It is based on the probability-weighted average of the cross-products of the random variables\u2019 deviations from their expected values for each possible outcome. (b) In contrast to the expectation, the variance is not a linear operator. Covariance is a statistical calculation that helps you understand how two sets of data are related to each other. And, we are given that the standard deviation of X is 1/2, and the standard deviation of Y is the square root of 1/2. Correlation between the two variables is a normalized version of the Covariance. In probability theory and statistics, covariance is a measure of the joint variability of two random variables. Analysis of covariance (ANCOVA) allows to compare one variable in 2 or more groups taking into account (or to correct for) variability of other variables, called covariates. described with a joint probability mass function. rameterized by a mean vector \u00b5\uffff, and a variance-covariance matrix \u03a3, written X\uffff \u223c \uffff\uffff\uffff(\u00b5\uffff,\u03a3). Given such information it is possible to calculate the marginal distributions and then the joint distribution follows. The function underlying its probability distribution is called a probability density function. Standard Deviation Calculator Variance Calculator Kurtosis Calculator Skewness Calculator. A zero covariance may indicate that the two assets are independent. This continues our exploration of the semantics of the inner product. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: \u03a3 ij = cov(x i,x j) = E p(x)[(x i \u2212\u00b5 i)(x j \u2212\u00b5 j)] (10) between variables x i and x j. search(\u201cdistribution\u201d). The Multivariate Gaussian Distribution Chuong B. Probability is the likelihood of something happening or being true. Both of these two determine the relationship and measures the dependency between two random. The covariance of two variables tells you how likely they are to increase or decrease simultaneously. 128 CHAPTER 7. Then m is the vector of means and V is the variance-covariance matrix. Online probability calculator to find expected value E(x), variance (\u03c3 2) and standard deviation (\u03c3) of discrete random variable from number of outcomes. Covariance and correlation show that variables can have a positive relationship, a negative relationship, or no relationship at all. Covariance correlations in collision avoidance probability calculations. Printer-friendly version Introduction. Begin your Level II studies with a FREE Schweser JumpStart Package. If you're behind a web filter, please make sure that the domains *. The TI-83 Graphing Calculator can facilitate the entry of ordered lists of data and perform some statistical analyses, but lacks a single command to calculate the covariance of two lists of numbers. The calculator based methods proposed above don\u2019t account for the fact that a probability is associated with asset returns. Doing the calculation To calculate the beta coefficient for a we'll compare how the stock and the index move relative to each other with a covariance formula and then divide that result by the. 329) and is the covariance. ) q for \"quantile\", the inverse c. This calculator is featured to generate the complete work with steps for any corresponding input values may helpful for grade school students to solve. When a portfolio includes two risky assets, the Analyst needs to take into account expected returns, variances and the covariance (or correlation) between the assets' returns. These usual, our starting point is a random experiment with a probability measure \u2119 on an underlying sample space. Bayesian decision theory is a fundamental statistical approach to the problem of pattern classification. Correlation values range from positive 1 to negative 1. Returns population covariance, the average of the products of deviations for each data point pair in two data sets. Variance and Standard Deviation of a Random Variable. Hey Flashcop and welcome to the forums. That is, if one increases, the other increases. It's an online statistics and probability tool requires two sets of population data X and Y` and measures of how much these data sets vary together, i. If an input is given then it can easily show the result for the given number. In this section, we will study an expected value that measures a special type of relationship between two real-valued variables. The Covariance Calculator an online tool which shows Covariance for the given input. Input the matrix in the text field below in the same format as matrices given in the examples. I was concurrently taking a basic theoretical probability and statistics, so even the idea of variance was still vague to me. 4, and in Robert J. Applied to historical prices, covariance can help determine if stocks' prices tend. Descriptive Statistics which contains one variable and multivariable calculators for 20 descriptive statistics measures including: mean, variance, covariance, quantile, interquartile range, correlation and many more. J Matney a Collision-avoidance calculations between orbiting objects make use of covariance matrices to characterize the uncertainty of the orbital position in space and time. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. The Multivariate Gaussian Distribution Chuong B. We will go through a review of probability concepts over here, all of the review materials have been adapted from CS229 Probability Notes. A portfolio is the total collection of all investments held by an individual or institution, including stocks, bonds, real estate, options, futures, and alternative investments, such as gold or limited partnerships. 2 Covariance Covariance is a measure of how much two random variables vary together. 25, multiply the answer by 100 to get 25%. The uniform distribution is used to describe a situation where all possible outcomes of a random experiment are equally likely to occur. is the correlation of and (Kenney and Keeping 1951, pp. 12 that she has exactly these two risk factors (but not the other). In general, if there are n random variables, the outcome is an n-dimensional vector of them. Stock Correlation Calculator. All we'll be doing here is getting a handle on what we can expect of the correlation coefficient if X and Y are independent, and what we can expect of the correlation coefficient if X and Y are dependent. If the probability of the event changes when we take the first event into consideration, we can safely say that the probability of event B is dependent of the occurrence of event A. As you doubtless know, the variance of a set of numbers is defined as the \"mean squared difference from the mean\". This online calculator computes covariance between two discrete random variables. Let X 1 = number of dots on the red die X 2 = number of dots on the green die. The correlation coefficient is equal to the covariance divided by the product of the standard deviations of the variables. How to Calculate Covariance. Let's now look at how to calculate the risk of the portfolio. As a simple example of covariance we'll return once again to the Old English example of Section 2. This tool provides direct calculations for a variety of probability distributions. 5 Covariance and Correlation Covariance and correlation are two measures of the strength of a relationship be-tween two r. ] Before constructing the covariance matrix, it\u2019s helpful to think of the data matrix as a collection of 5 vectors, which is how I built our data matrix in R. We construct a non-separable space-time covariance function based on a diffusive Langevin equation. If the two variables are dependent then the covariance can be measured using the following formula:. Although the covariance and variance are linked to each other in the above manner, their probability distributions are not attached to each other in a simple manner and have to be dealt separately. A positive covariance would indicate a positive linear relationship between the variables, and a negative covariance would indicate the opposite. The syntax of the Covariance. It is actually used for computing the covariance in between every column of data matrix. It is also helpful for insurance. They must then calculate the F-ratio and the associated probability value (p-value). By using this formula, after calculation, you can verify the result of such calculations by using our covariance calculator. If A is a matrix whose columns represent random variables and whose rows represent observations, C is the covariance matrix with the corresponding column variances along the diagonal. [In our case, a 5\u00d75 matrix. It just creates confusion because they are not equivalent. In order to calculate the Student T Value for any degrees of freedom and given probability. Key Differences Between Covariance and Correlation. With the covariance option, correlate can be used to obtain covariance matrices, as well as correlation matrices, for both weighted and unweighted data. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. In my first machine learning class, in order to learn about the theory behind PCA (Principal Component Analysis), we had to learn about variance-covariance matrix. A distribution is described as normal if there is a high probability that any observation form the population sample will have a value that is close to the mean, and a low probability of having a value that is far from the mean. So, I wrote an add-in that used the matrix algebra functions to create a covariance matrix that would change if you changed the data. 05 Jeremy Orlo and Jonathan Bloom 1 Learning Goals 1. By default, this function will calculate the sample covariance matrix. Calculate the covariance of and. The function is new in Excel 2010 and so is not available in earlier versions of Excel. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. A Statistics Co-Variance Calculator An online Co-Variance Calculator to measure of two variables X and Y. Covariance and Correlation De\ufb01nition: Covariance Let X and Y be two RV\u2019s with means x and y, respectively. In statistical theory, covariance is a measure of how much two random variables change together. E(X1)=\u00b5X1 E(X2)=\u00b5X2 var(X1)=\u03c32 X1 var(X2)=\u03c32 X2 Also, we assume that \u03c32 X1 and \u03c32 X2 are \ufb01nite positive values. 2 Covariance Covariance is a measure of how much two random variables vary together. Expectation of a Function of a Random Variable Suppose that X is a discrete random variable with sample space \u03a9, and \u03c6(x) is a real-valued function with domain \u03a9. A sample is a randomly chosen selection of elements from an underlying population. Analysis of covariance (ANCOVA) allows to compare one variable in 2 or more groups taking into account (or to correct for) variability of other variables, called covariates. It is important in security analysis to determine how much or how little price movements in two companies or industries are connected. These topics are somewhat specialized, but are particularly important in multivariate statistical models and for the multivariate normal distribution. Using the covariance formula, you can determine whether economic growth and S&P 500 returns have a positive or inverse relationship. Typically, the population is very large, making a complete enumeration of all the values in the population impossible. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. Consider two random variables$X$and$Y\\$. Understand the meaning of covariance and correlation. If is the covariance matrix of a random vector, then for any constant vector ~awe have ~aT ~a 0: That is, satis es the property of being a positive semi-de nite matrix. The problem is solved by standardize the value of covariance (divide it by \u02d9. An investor is facing two potential financial losses and with the following joint density function: Let be the total of these two losses. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. S function is:. Population Standard Deviation The population standard deviation, the standard definition of \u03c3 , is used when an entire population can be measured, and is the square root of the variance of a given data set. Excel does such a great job in calculating correlation and covariance that it is not necessary to memorize the formulas of covariance and correlation, but here they are, along with examples worked out in Excel: Covariance of variables x and y from a known population = \u03c3 xy. The covariance for two random variates X and Y, each with sample size N, is defined by the expectation\u2026. Calculate the covariance of and. Discrete Random Variable Calculator. Properties of variance and covariance (a) If and are independent, then by observing that. A portfolio is the total collection of all investments held by an individual or institution, including stocks, bonds, real estate, options, futures, and alternative investments, such as gold or limited partnerships. On this page, we'll begin our investigation of what the correlation coefficient tells us. Consider the following example: Example. The covariance is defined as. Click the Calculate! button and find out the covariance matrix of a multivariate sample. A randomly selected day was a long commute. These usual, our starting point is a random experiment with a probability measure \u2119 on an underlying sample space. For example, the probability of a two-dimensional case, in which the vector of random variables is X = [X, Y] T, can be calculated as. This matrix is not usually printed or saved. , the variables tend to show similar behavior), the covariance is positive. For instance, I have been given a discrete random variable X with probability function px(x) = 1/2 if x = -1, 1/4 if x = 0, 1/4 if x = 1, 0 otherwise. If the covariance is a large positive number, then we expect x i to be largerthan\u00b5 iwhenx j islargerthan\u00b5 j. Let's calculate the covariance for the example age and income data. In probability theory and statistics, covariance is a measure of the joint variability of two random variables. Covariance and correlation show that variables can have a positive relationship, a negative relationship, or no relationship at all. You COULD do a calculation patterned after covariance with 3 or more variables, but I don\u2019t see it as meaning anything, and I don\u2019t think it would be admissable as a valid statistical function. (The covariance of X and Y is sometimes written cov(X, Y). He also covers testing hypotheses, modeling different data distributions, and calculating the covariance and correlation between data sets.", "date": "2020-03-31 13:27:22", "meta": {"domain": "agence-des-4-fontaines.fr", "url": "http://oqxm.agence-des-4-fontaines.fr/covariance-calculator-with-probability.html", "openwebmath_score": 0.7428514361381531, "openwebmath_perplexity": 553.4405559171647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9877587218253718, "lm_q2_score": 0.8688267762381844, "lm_q1q2_score": 0.8581912259846873}}
{"url": "https://handwiki.org/wiki/Kendall_tau_distance", "text": "# Kendall tau distance\n\nThe Kendall tau rank distance is a metric that counts the number of pairwise disagreements between two ranking lists. The larger the distance, the more dissimilar the two lists are. Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort algorithm would take to place one list in the same order as the other list. The Kendall tau distance was created by Maurice Kendall.\n\n## Definition\n\nThe Kendall tau ranking distance between two lists $\\displaystyle{ \\tau_1 }$ and $\\displaystyle{ \\tau_2 }$ is\n\n$\\displaystyle{ K(\\tau_1, \\tau_2) = |\\{(i,j): i \\lt j, ( \\tau_1(i) \\lt \\tau_1(j) \\wedge \\tau_2(i) \\gt \\tau_2(j) ) \\vee ( \\tau_1(i) \\gt \\tau_1(j) \\wedge \\tau_2(i) \\lt \\tau_2(j) )\\}|. }$\n\nwhere\n\n\u2022 $\\displaystyle{ \\tau_1(i) }$ and $\\displaystyle{ \\tau_2(i) }$ are the rankings of the element $\\displaystyle{ i }$ in $\\displaystyle{ \\tau_1 }$ and $\\displaystyle{ \\tau_2 }$ respectively.\n\n$\\displaystyle{ K(\\tau_1,\\tau_2) }$ will be equal to 1 if the two lists are identical and $\\displaystyle{ -1 }$ (where $\\displaystyle{ n }$ is the list size) if one list is the reverse of the other. The normalized Kendall tau distance therefore lies in the interval [-1,1].\n\nKendall tau distance may also be defined as\n\n$\\displaystyle{ K(\\tau_1,\\tau_2) = \\begin{matrix} \\sum_{\\{i,j\\}\\in P} \\bar{K}_{i,j}(\\tau_1,\\tau_2) \\end{matrix} }$\n\nwhere\n\n\u2022 P is the set of unordered pairs of distinct elements in $\\displaystyle{ \\tau_1 }$ and $\\displaystyle{ \\tau_2 }$\n\u2022 $\\displaystyle{ \\bar{K}_{i,j}(\\tau_1,\\tau_2) }$ = 0 if i and j are in the same order in $\\displaystyle{ \\tau_1 }$ and $\\displaystyle{ \\tau_2 }$\n\u2022 $\\displaystyle{ \\bar{K}_{i,j}(\\tau_1,\\tau_2) }$ = 1 if i and j are in the opposite order in $\\displaystyle{ \\tau_1 }$ and $\\displaystyle{ \\tau_2. }$\n\nKendall tau distance can also be defined as the total number of discordant pairs.\n\nKendall tau distance in Rankings: A permutation (or ranking) is an array of N integers where each of the integers between 0 and N-1 appears exactly once. The Kendall tau distance between two rankings is the number of pairs that are in different order in the two rankings. For example, the Kendall tau distance between 0 3 1 6 2 5 4 and 1 0 3 6 4 2 5 is four because the pairs 0-1, 3-1, 2-4, 5-4 are in different order in the two rankings, but all other pairs are in the same order.[1]\n\nIf Kendall tau function is performed as $\\displaystyle{ K(L1,L2) }$ instead of $\\displaystyle{ K(\\tau_1,\\tau_2) }$ (where $\\displaystyle{ \\tau_1 }$ and $\\displaystyle{ \\tau_2 }$ are the rankings of $\\displaystyle{ L1 }$ and $\\displaystyle{ L2 }$ elements respectively), then triangular inequality is not guaranteed. The triangular inequality fails in cases where there are repetitions in the lists. So then we are not dealing with a metric anymore.\n\n## Example\n\nSuppose one ranks a group of five people by height and by weight:\n\nPerson A B C D E\nRank by height 1 2 3 4 5\nRank by weight 3 4 1 2 5\n\nHere person A is tallest and third-heaviest, and so on.\n\nIn order to calculate the Kendall tau distance, pair each person with every other person and count the number of times the values in list 1 are in the opposite order of the values in list 2.\n\nPair Height Weight Count\n(A,B) 1 < 2 3 < 4\n(A,C) 1 < 3 3 > 1 X\n(A,D) 1 < 4 3 > 2 X\n(A,E) 1 < 5 3 < 5\n(B,C) 2 < 3 4 > 1 X\n(B,D) 2 < 4 4 > 2 X\n(B,E) 2 < 5 4 < 5\n(C,D) 3 < 4 1 < 2\n(C,E) 3 < 5 1 < 5\n(D,E) 4 < 5 2 < 5\n\nSince there are four pairs whose values are in opposite order, the Kendall tau distance is 4. The normalized Kendall tau distance is\n\n$\\displaystyle{ \\frac{4}{5(5 - 1)/2} = 0.4. }$\n\nA value of 0.4 indicates that 40% of pairs differ in ordering between the two lists.\n\n## Computing the Kendall tau distance\n\nA naive implementation in Python (using NumPy) is:\n\nimport numpy as np\n\ndef normalised_kendall_tau_distance(values1, values2):\n\"\"\"Compute the Kendall tau distance.\"\"\"\nn = len(values1)\nassert len(values2) == n, \"Both lists have to be of equal length\"\ni, j = np.meshgrid(np.arange(n), np.arange(n))\na = np.argsort(values1)\nb = np.argsort(values2)\nndisordered = np.logical_or(np.logical_and(a[i] < a[j], b[i] > b[j]), np.logical_and(a[i] > a[j], b[i] < b[j])).sum()\nreturn ndisordered / (n * (n - 1))\n\nHowever, this requires $\\displaystyle{ n^2 }$ memory, which is inefficient for large arrays.\n\nGiven two rankings $\\displaystyle{ \\tau_1,\\tau_2 }$, it is possible to rename the items such that $\\displaystyle{ \\tau_1 = (1,2,3,...) }$. Then, the problem of computing the Kendall tau distance reduces to computing the number of inversions in $\\displaystyle{ \\tau_2 }$\u2014the number of index pairs $\\displaystyle{ i,j }$ such that $\\displaystyle{ i\\lt j }$ while $\\displaystyle{ \\tau_2(i) \\gt \\tau_2(j) }$. There are several algorithms for calculating this number.\n\n\u2022 A simple algorithm based on merge sort requires time $\\displaystyle{ O(n \\log n) }$.[2]\n\u2022 A more advanced algorithm requires time $\\displaystyle{ O(n\\sqrt{\\log{n}}) }$.[3]\n\nHere is a basic C implementation.\n\n#include <stdbool.h>\n\nint kendallTau(short x[], short y[], int len) {\nint i, j, v = 0;\nbool a, b;\n\nfor (i = 0; i < len; i++) {\nfor (j = i + 1; j < len; j++) {\n\na = x[i] < x[j] && y[i] > y[j];\nb = x[i] > x[j] && y[i] < y[j];\n\nif (a || b)\nv++;\n\n}\n}\n\nreturn abs(v);\n}\n\nfloat normalize(int kt, int len) {\nreturn kt / (len * (len - 1) / 2.0);\n}", "date": "2022-01-23 14:50:14", "meta": {"domain": "handwiki.org", "url": "https://handwiki.org/wiki/Kendall_tau_distance", "openwebmath_score": 0.772384524345398, "openwebmath_perplexity": 632.9292160980654, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587247081929, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.858191225134461}}
{"url": "https://math.stackexchange.com/questions/2158019/true-or-false-if-fx-and-f-1x-intersect-at-an-even-number-of-points?noredirect=1", "text": "# True or False : If $f(x)$ and $f^{-1}(x)$ intersect at an even number of points , all points lie on $y=x$\n\nPreviously I have discussed about odd number of intersect points (See : If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, is at least one point on the line $y=x$?) Now , I want to know the even condition . For example $f(x) = \\sqrt{x}$ and $f^{-1}(x) = x^2 , x\\ge 0$ intersects each other in $(0,0)$ and $(1,1)$ points and these points located on $y=x$ line\n\nEdit : Consider $f$ is continuous function.\n\n\u2022 How would any intersection point not lie on $y=x$? The graphs of the function and its inverse (if it exists) is symmetric about $y=x$, so any intersection must be on $y=x$ Feb 23 '17 at 15:58\n\u2022 @user160738 No , for example consider $f(x) = \\frac{1}{x}$ and $f^{-1}(x) = \\frac{1}{x}$ Feb 23 '17 at 16:00\n\u2022 @user160738 another example is $f(x) = -x^3$ and $f^{-1}(x) = -\\sqrt[3]{x}$ . They intersect each other in $(1 , -1)$ that is not on $y=x$ line. See : math.stackexchange.com/questions/1452098/\u2026 Feb 23 '17 at 16:08\n\nThe continuous case:\n\nIf the domain is not an interval, we can still find a counterexample. Define $$f(x)=\\cases{2x+2 & if x\\in(0,2),\\cr x-3 & if x\\in(3,5),}$$ then $$f^{-1}(x)=\\cases{x/2-1 & if x\\in(2,6),\\cr x+3 & if x\\in(0,2),}$$ and intersections are $(1,4)$, $(4,1)$.\n\nHowever, if the domain is connected, it turns out to be true.\n\nProof of the theorem in the case of a connected domain (interval)\n\nFor contradiction, let us suppose that there is an even number of intersections and an intersection is out of diagonal.\n\n1. We assume that $f^{-1}$ exists which means that $f$ is injective. Moreover, $f$ is continuous and its domain is interval, so $f$ is increasing or decreasing.\n2. $f$ and $f^{-1}$ are symmetric by diagonal, therefore the set of intersections is symmetric by the diagonal.\n3. Let denote $(x_1, x_2)$ an intersection out of diagonal. By symmetry, there is an intersection $(x_2, x_1)$. We can therefore WLOG assume that $x_1<x_2$.\n4. $f(x_1) > x_1$ and $f(x_2) < x_2$, so by continuity, $f$ intersects the diagonal.\n5. There is an even number of intersections and by symmetry, there is an even number of intersections out of diagonal. So there is even number of intersections on the diagonal. So there are at least two of them: $f(x_3) = x_3$, $f(x_4) = x_4$\n6. $x_1<x_2$ and $f(x_1) > f(x_2)$, so $f$ is decreasing (on the whole, by 1).\n7. $x_3<x_4$ and $f(x_3) < f(x_4)$, so $f$ is increasing.\n\n\u2022 Okay but it's true in many cases in connected domain . Can you provide an example that it's wrong in connected domain ? Mar 3 '17 at 10:45\n\u2022 Okay so why it's true for many examples ? Mar 3 '17 at 15:14\n\u2022 You proved the theorem is wrong .okay , now provide some examples that shows theorem is false . Mar 3 '17 at 15:38\n\u2022 Oh , I'm really sorry let me to read again . Mar 3 '17 at 15:40\n\u2022 Sorry, I don't understand step 4 in your purported proof. Mar 4 '17 at 10:45\n\nAssume that $f$ is a continuous and invertible real function with a connected domain. Then $f$ is either strictly decreasing or strictly increasing over its domain (or it would assume some value twice, and would not be one-to-one). Consider these two cases:\n\n1. $f$ is decreasing. Then, since $f(x)-x$ is also decreasing, $f$ can't cross $y=x$ more than once. If it never crosses $y=x$, then it lies entirely above or below that line, and its inverse lies entirely on the other side; they never meet, and the theorem holds vacuously. If, on the other hand, it crosses $y=x$ exactly once, then the total number of intersections between $f$ and $f^{-1}$ is odd, since off-diagonal intersections come in pairs. The theorem holds in this case too.\n\n2. $f$ is increasing. Then it cannot intersect $f^{-1}$ at any point off the line $y=x$. Suppose it did, at a pair of points $(x,y)$ and $(y,x)$ with $x<y$: then we would have $f(x)=y > x=f(y)$, contradicting the fact that $f$ is increasing. In this case, then, all intersections are on $y=x$, and the theorem holds.\n\nSince the theorem holds whether $f$ is increasing or decreasing, it is true in general.\n\n\u2022 For all decreasing functions we can say $f(x) - x$ is also decreasing ? Is it right ? Mar 9 '17 at 6:03\n\u2022 Sure. A decreasing function minus an increasing function decreases even faster. Mar 9 '17 at 20:48\n\u2022 Okay and how you can conclude that it intersects at most one time ? Mar 9 '17 at 20:55\n\nYour question, as it currently stands definitely does not hold, as you do not specify the domain, continuity or conditions on inverse. This answer does define an $f$ on the whole real line that has a proper inverse.\n\nConsider the function: \\begin{align} f(x) = \\begin{cases} 1 &\\text{if x = 0}\\\\ 0 &\\text{if x = 1}\\\\ x^2 &\\text{if x > 0 and x\\neq 1}\\\\ -2x &\\text{if x < 0} \\end{cases} \\end{align} which has inverse \\begin{align} f^{-1}(y) = \\begin{cases} 1 &\\text{if y = 0}\\\\ 0 &\\text{if y = 1}\\\\ \\sqrt{x} &\\text{if x > 0 and x\\neq 1}\\\\ -\\tfrac{x}{2} &\\text{if x < 0} \\end{cases} \\end{align} and so $f(x) = f^{-1}(y)$ only at $x = 0$ and $x = 1$, but at these points $f(x) \\neq x$.\n\n\u2022 Consider $f$ is continuous Feb 23 '17 at 17:24\n\u2022 I will have a think @S.H.W, I imagine the answer might be that such a function cannot exist: if they intersect at two points then they are equal on an interval (that is my hunch anyway, YMMV). Feb 23 '17 at 21:17\n\u2022 I think this statement is true but I can't prove it. Feb 23 '17 at 21:32\n\u2022 Actually, I don't think that is quite true: consider $f(x) = x^2$ for $x \\geqslant 0$ and $f(x) = -2x$ for $x < 0$, then $f^{-1}(x) = \\sqrt{x}$ for $x \\geqslant 0$ and $f^{-1}(x) = -\\tfrac{x}{2}$ for $x < 0$ and so they only intersect at $x = 0$ and $1$. Feb 23 '17 at 21:59\n\u2022 if $f(x) = x^2$ then $f^{-1}(x) = \\sqrt{x} , x \\ge 0$ and the intersection points are $(0,0)$ and $(1,1)$ which they are in $y=x$ line Feb 23 '17 at 22:03\n\nFor $f$ and $f^{-1}$ to intersect there must be a point(s) $(x,f(x)) = (x,f^{-1}(x))$. Assume $x \\neq f(x)$. Because $f$ and $f^{-1}$ are symmetric around y=x then there must be a corresponding point(s) $(f(x),x) = (f^{-1}(x),x)$.\n\nThe line defined by the points $(x_0,f(x_0)$ and $(f(x_),x_0)$ is $$(y-f(x_0) = \\frac{f(x_0)-x_0}{x_0-f(x_0)}(x-x_0)$$ and simplified becomes $$y=-x+x_0+f(x_0)$$ This line, with a slope of -1, is not parallel with $y=x$, therefore must cross $y=x$. Because the two points are symmetric around $y=x$, they must lie on different sides of $y=x$. By the intermediate value theorem both $f$ and $f^{-1}$ must cross $y=x$.\n\nHence there is at least one point of $f$ on $y=x$", "date": "2021-11-29 08:32:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2158019/true-or-false-if-fx-and-f-1x-intersect-at-an-even-number-of-points?noredirect=1", "openwebmath_score": 0.8794293999671936, "openwebmath_perplexity": 189.93672624497597, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.95598134762883, "lm_q2_score": 0.8976952989498449, "lm_q1q2_score": 0.8581799616501382}}
{"url": "https://math.stackexchange.com/questions/3321058/a-particular-vanishing-integral", "text": "# A particular vanishing integral\n\nWhile dealing with a definite integral on AoPS I discovered (I have to admit by pure chance) the following relation\n\n$$\\int_0^1\\log\\left(\\frac{(x+1)(x+2)}{x+3}\\right)\\frac{\\mathrm dx}{1+x}~=~0\\tag1$$\n\nThe proof is quite easy, but feels kind of contrived. Indeed, just apply a self-similar substitution - $$x\\mapsto\\frac{1-x}{1+x}$$ - to the auxiliary integral $$\\mathcal I$$ given by\n\n$$\\mathcal I~=~\\int_0^1\\log\\left(\\frac{x^2+2x+3}{(x+1)(x+2)}\\right)\\frac{\\mathrm dx}{1+x}$$\n\nAnd the result follows. However, to consider precisely this integral seems highly unnatural to me (in fact, as I mentioned earlier, this integral was just a by-product while evaluating something quite different and I discovered $$(1)$$ when experimenting with various substitutions).\n\nThe crucial point to notice concerning $$\\mathcal I$$ is the invariance of the polynomial $$f(x)=x^2+2x+3$$ regarding the self-similar substitution which allows us to deduce $$(1)$$. Additionally for myself I am quite surprised by the special structure of $$(1)$$ since we have factors of the form $$(x+1)$$, $$(x+2)$$ and $$(x+3)$$ combined which calls for a generalization (although I found none yet).\n\nIt there a more elementary approach, not relying on such an \"accident\" like examining the integral $$\\mathcal I$$ for proving $$(1)$$? Additionally, can this particular pattern be further generalized? Answers to both questions (also separately) are highly appreciated!\n\n\u2022 The $\\displaystyle\\log$ argument numerator -in the $\\displaystyle\\mathcal{I}$ definition- must be $\\displaystyle x^{2} + 3x + 2$ instead of $\\displaystyle x^{2} + 2x + 3$. \u2013\u00a0Felix Marin Aug 27 '20 at 3:27\n\u2022 @FelixMarin No. By my calculations $\\left(\\frac{1-x}{1+x}\\right)^2+2\\left(\\frac{1-x}{1+x}\\right)+3=2\\frac{x^2+2x+3}{(1+x)^2}$ which is what the approach is all about. But $\\left(\\frac{1-x}{1+x}\\right)^2+3\\left(\\frac{1-x}{1+x}\\right)+2=2\\frac{x+3}{(x+1)^2}$ which yields an interesting other identity but not the one the post is concerned with. \u2013\u00a0mrtaurho Aug 27 '20 at 3:42\n\nThat's quite an impressive method to show that the integral vanishes.\n\nFor the first part I'll show using a different approach that your integral vanishes. $$\\mathcal J=\\int_0^1 \\ln\\left(\\frac{x+3}{(x+2)(x+1)}\\right)\\frac{\\mathrm dx}{x+1}\\overset{x+1=t}= \\color{blue}{\\int_1^2\\ln\\left(\\frac{t+2}{t+1}\\right)\\frac{\\mathrm dt}{t}}-\\color{red}{\\int_1^2 \\frac{\\ln t}{t}\\mathrm dt}$$ Let's denote the blue integral as $$\\mathcal J_1$$ then using the substitution $$\\frac{2}{t}\\to t$$ we get: $$\\mathcal J_1=\\int_1^2 \\ln\\left(\\frac{t+2}{t+1}\\right)\\frac{\\mathrm dt}{t}=\\int_1^2 \\ln\\left(\\frac{2(t+1)}{t+2}\\right)\\frac{\\mathrm dt}{t}$$ Adding both integrals from above gives us: $$\\require{cancel} 2\\mathcal J_1=\\cancel{\\int_1^2 \\ln\\left(\\frac{t+2}{t+1}\\right)\\frac{\\mathrm dt}{t}}+\\int_1^2 \\frac{\\ln 2}{t}\\mathrm dt+\\cancel{\\int_1^2 \\ln\\left(\\frac{t+1}{t+2}\\right)\\frac{\\mathrm dt}{t}}=\\ln^2 2$$ $$\\Rightarrow \\mathcal J_1=\\color{blue}{\\frac{\\ln^2 2}{2}}\\Rightarrow \\mathcal J=\\color{blue}{\\frac{\\ln^2 2}{2}}-\\color{red}{\\frac{\\ln^2 2}{2}}=0$$\n\nAs for the second part, a small generalization outcomes by experimenting with the blue integral.\n\nIn particular, by the same approach we have: $$\\int_1^a \\ln\\left(\\frac{x+a}{x+1}\\right)\\frac{\\mathrm dx}{x}=\\int_1^a \\frac{\\ln x}{x}\\mathrm dx$$ Which gives us a small generalization: $${\\int_0^{a-1}\\ln\\left(\\frac{x+a+1}{(x+1)(x+2)}\\right)\\frac{\\mathrm dx}{x+1}=0}$$ Similarly, (with the substitution $$\\frac{ab}{x}\\to x$$) we get that: $$\\int_a^b \\ln\\left(\\frac{x+b}{x+a}\\right)\\frac{dx}{x}=\\frac12 \\ln^2 \\left(\\frac{b}{a}\\right)=\\int_a^b \\ln\\left(\\frac{x}{a}\\right)\\frac{dx}{x}$$ And the following follows: $$\\int_{a-1}^{b-1} \\ln\\left(\\frac{a(x+b+1)}{(x+1)(x+a+1)}\\right)\\frac{dx}{x+1}=0$$ One might be interested in the following similar generalization too: $$\\int_1^{t}\\ln\\left(\\frac{x^4+sx^2+t^2}{x^3+sx^2+t^2x}\\right)\\frac{dx}{x}=0,\\quad s\\in R, t>1$$\n\n\u2022 This is what I was looking for; I guess^^ (+1) anyway and I'm looking forward to see a generalization (if possible). I guess you know where I got this integral from :D \u2013\u00a0mrtaurho Aug 12 '19 at 15:17\n\u2022 @mrtaurho I got there a small generalization for now. However since $\\int_a^b \\ln\\left(\\frac{x+b}{x+a}\\right)\\frac{dx}{x}=\\frac12 \\ln^2 \\left(\\frac{b}{a}\\right)$ it might be possible to obtain a better one. (I'll try later to work with it). \u2013\u00a0Zacky Aug 12 '19 at 15:38\n\u2022 It's hard to write out your name now :P But yes, this seems promising, I'm curious! As I noted above it was a rather strange by-product to discover this equality; and it was tedious to ran into it three times while hoping for something more helpful for solving the original task^^' \u2013\u00a0mrtaurho Aug 12 '19 at 15:44\n\u2022 @mrtaurho just in case you missed it in winter I'll mention that $\\mathcal I$ (before the self-similar sub was applied) originates from this generalization: math.stackexchange.com/a/3049039/515527. Aka: $$\\int_1^{\\sqrt{t}}\\ln\\left(\\frac{x^4+sx^2+t}{x(x^2+sx+t)}\\right)\\frac{dx}{x}=0$$ \u2013\u00a0Zacky Aug 12 '19 at 16:06\n\u2022 As I've upvoted both, the question you linked and your answer, I guess I've seen it at some point. But I'll take a look at it again :) \u2013\u00a0mrtaurho Aug 12 '19 at 16:23\n\nI have used the substitution $$(x+1)(y+1)=2$$ before to good effect because $$\\int_0^1f(x)\\,\\frac{\\mathrm{d}x}{x+1}=\\int_0^1f\\!\\left(\\tfrac{1-y}{1+y}\\right)\\frac{\\mathrm{d}y}{y+1}\\tag1$$ If $$f(x)=\\log\\left(\\frac{x+3}{(x+2)(x+1)}\\right)$$, then $$f\\!\\left(\\frac{1-y}{1+y}\\right)=\\log\\left(\\frac{(y+2)(y+1)}{y+3}\\right)$$. Therefore $$\\int_0^1\\log\\left(\\frac{x+3}{(x+2)(x+1)}\\right)\\frac{\\mathrm{d}x}{x+1}=\\int_0^1\\log\\left(\\frac{(y+2)(y+1)}{y+3}\\right)\\frac{\\mathrm{d}y}{y+1}\\tag2$$ and since the two sides of $$(2)$$ are negatives, they are both $$0$$.\nWe can generalize $$(1)$$ by letting $$(x+a)(y+a)=a(1+a)$$, then we get $$\\int_0^1f(x)\\frac{\\mathrm{d}x}{x+a}=\\int_0^1f\\!\\left(\\tfrac{a(1-y)}{a+y}\\right)\\frac{\\mathrm{d}y}{y+a}\\tag3$$", "date": "2021-03-02 13:18:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3321058/a-particular-vanishing-integral", "openwebmath_score": 0.9715476036071777, "openwebmath_perplexity": 477.0114140069285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180694313358, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8581634685962107}}
{"url": "http://www.lofoya.com/Solved/1517/when-a-student-weighing-45-kgs-left-a-class-the-average-weight-of", "text": "# Moderate Averages Solved QuestionAptitude Discussion\n\n Q. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200 g. What is the average weight of the remaining 59 students?\n \u2714\u00a0A. 57 \u2716\u00a0B. 56.8 \u2716\u00a0C. 58.2 \u2716\u00a0D. 52.2\n\nSolution:\nOption(A) is correct\n\nLet the average weight of the 59 students be $A$.\n\nTherefore, the total weight of the 59 of them will be 59 $A$.\n\nThe questions states that when the weight of this student who left is added, the total weight of the class\n\n= $59A + 45$\n\nWhen this student is also included, the average weight decreases by 0.2 kgs.\n\n$59A + \\dfrac{45}{60} = A - 0.2$\n\n$\\Rightarrow 59A + 45 = 60A - 12$\n\n$\\Rightarrow 45 + 12 = 60A - 59A$\n\n$\\Rightarrow A = \\textbf{57}$\n\nEdit:\u00a0For an alternative solution, check comment by\u00a0Durga.\n\n## (9) Comment(s)\n\nSiddharth Sharma\n()\n\n60*0.2+45=57 thats the shortest method to do it.\n\nShivakumar\n()\n\ndivided by 60 for 59A+45 not only for 45\n\nVijay\n()\n\nsimply,\n\n60*(x-.20)-59x=45;\n\n=>x=45+12=57.\n\nSanajit Ghosh\n()\n\n@DURGA's ans is right\n\nZoha Amjad\n()\n\nanswer is 56.8\n\n0.2 kgs increase means\n\n$0.2*59=11.8$\n\n$11.8+45 = 56.8$\n\nShanu\n()\n\nThe ans should be 56.8\n\nit can be solved as\n\n$60A-((A-0.2)*59)=45$\n\nso $A(\\text{avarage}) = 56.8$\n\nKiruthi\n()\n\nans should be 56.8 because\u00a0in question they have given increased by 200g but while solving they took it as decreased by 200g so 57\n\nDurga\n()\n\nHi ,\n\nThe answer is 57 only.\n\nLet say the average of 60 students is $x$.\n\nTotal weight $=60*x$\n\nIf one man left average should increase 200gr\n\ni.e $\\dfrac{200}{1000} =0.2 \\text{ kg}$\n\nSo, $60 \\times x-45=59(x+0.2)$\n\n$60x-59x=59*0.2+45$\n\n$\\Rightarrow x=56.8 \\text{ kg}$\n\nBut here we need to calculate the average of 59 students\n\ni.e $56.8+0.2= \\textbf{57 kg}$\n\nElmira\n()\n\nThank you for creating a very informative website, much appreciate it.\n\nPlease review the solution once again, when multiplying the sides by 60, 59A was left without effect, therefore I think the response is wrong. Oh maybe I am missing something here.\n\nPlease explain.\n\nElmira", "date": "2017-02-24 05:49:02", "meta": {"domain": "lofoya.com", "url": "http://www.lofoya.com/Solved/1517/when-a-student-weighing-45-kgs-left-a-class-the-average-weight-of", "openwebmath_score": 0.7966797947883606, "openwebmath_perplexity": 1959.2774824885898, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985718064816221, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8581634629235501}}
{"url": "https://math.stackexchange.com/questions/2804617/find-the-matrix-representation-of-a-linear-map", "text": "# Find the matrix representation of a linear map\n\nLet $f:\\mathbb{R}^2 \\to \\mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\\begin{pmatrix} 4 & 6\\\\ 6 & 3 \\end{pmatrix}$$ with respect to the basis $B=\\bigg\\{ \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} , \\begin{pmatrix} 3\\\\ 2 \\end{pmatrix}\\bigg\\}$. Find the matrix representation of $f$ with respect to the canonical basis.\n\nMY ATTEMPT: Let $b_1=\\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix}$, $b_2=\\begin{pmatrix} 3\\\\ 2 \\end{pmatrix}$ and $e_j$ the j-th vector of the canonical basis. Then I have $f(b_1)=\\begin{pmatrix} 4\\\\ 6 \\end{pmatrix} \\Longrightarrow f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) =4e_1 + 6e_2$\n\n$f(b_2)=\\begin{pmatrix} 6\\\\ 3 \\end{pmatrix} \\Longrightarrow f(3e_1 +2e_2)=3 f(e_1) +2f(e_2) =6e_1 + 3e_2$\n\nSolving this system I obtain that $f(e_1)=2e_1 -3e_2=\\begin{pmatrix} 2\\\\ -3 \\end{pmatrix}$ and $f(e_2)= 6e_2=\\begin{pmatrix} 0\\\\ 6 \\end{pmatrix}$. So the matrix rappresentation is $M_{E}^{E}=\\begin{pmatrix} 2 & 0\\\\ -3 & 6 \\end{pmatrix}$.\n\nBUT my theacher says that the solution is $M_{E}^{E}=\\begin{pmatrix} 13 & 10\\\\ 0 & 0 \\end{pmatrix}$. Where I am wrong?\n\n$$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \\color{red}{=4e_1 + 6e_2}$$\n\nThis is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.\n\n\u2022 How can I do that? \u2013\u00a0fcoulomb Jun 1 '18 at 18:47\n\u2022 @Besh00 write $\\color{red}{=4b_1+6b_2}$ instead of the red part, and then write both $b_1$ and $b_2$ in terms of $e_1$, $e_2$. \u2013\u00a0Arnaud Mortier Jun 1 '18 at 18:50\n\u2022 Why $f(b_1)=4b_1+6b_2$? \u2013\u00a0fcoulomb Jun 1 '18 at 21:10\n\u2022 You wrote yourself \"Then I have $f(b_1)=\\begin{pmatrix} 4\\\\ 6 \\end{pmatrix}$\". These coordinates are with respect to the basis $B$ since they are taken from a matrix wrt $B,B$. \u2013\u00a0Arnaud Mortier Jun 1 '18 at 21:14", "date": "2020-02-17 19:31:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2804617/find-the-matrix-representation-of-a-linear-map", "openwebmath_score": 0.9243845343589783, "openwebmath_perplexity": 163.62444703176374, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9857180627184412, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8581634478591927}}
{"url": "https://math.stackexchange.com/questions/2737440/how-many-4-digit-numbers-that-do-not-have-5-and-have-7-in-the-hundreds-pos", "text": "# How many $4$-digit numbers that do not have $5$ and have $7$ in the hundreds position?\n\nI encountered this problem:\n\nHow many $4$-digit numbers are there that do not have $5$ and have $7$ in the hundreds position? Digits cannot be repeated.\n\nMy thinking: first, let's count numbers that do not have $0$. It's going to be $7\\cdot 6 \\cdot 5$ arrangements. Now, let's count numbers that have $0$ in the tens position. We have two positions open so there are $7 \\cdot 6$ arrangements. We get the same number of arrangements for numbers that end with $0$. The total number of arrangements is $7\\cdot 6 \\cdot 5+ 2\\cdot 7 \\cdot 6=7\\cdot 7 \\cdot 6=294$.\n\nHowever, my friend argues that there are $8$ choices for the last digit, $7$ choices for the tens digits and $6$ choices for the first digit so the total number should be $8\\cdot 7 \\cdot 6=336$.\n\nI am asking who is right not to prove my friend wrong but to find the truth. Thanks for listening!\n\n294 is correct.\n\nFirst digit (most significant), can not be 7,0,5, so 7 choices; 3rd digit, cannot be 7,5, and the one chosen for 1st, so 7 choices; 4th digit, 6 choices;\n\n$7\\times7\\times6=294$\n\n\u2022 Welcome to the site. Great first answer! \u2013\u00a0B. Mehta Apr 14 '18 at 23:17\n\nYou are right and your friend is wrong. If you choose the units digit and then choose the tens digit (and are forced to use $7$ for the hundreds digit) then the number of choices for the thousands digit not always $6$: it is either $5$ or $6$, depending on if you've used $0$ yet or not.\n\nFor example, there are $5$ numbers satisfying the condition which have the form $x789$ (since $x$ can be any of $1,2,3,4,6$) but $6$ numbers satisfying the condition which have the form $x780$ (since $x$ can be any of $1,2,3,4,6,9$).\n\nAll this tells us is that the number of solutions is between $8\\cdot7\\cdot5 = 280$ and $8\\cdot7\\cdot6 = 336$, and for a more precise answer we do need to look at whether $0$ gets used.\n\nI think the key disagreement here is that your friend is thinking that $0$ is allowed to be the first digit (say, for a 4 digit passcode), while you are not. If $0$ can be the first digit, your friend is correct, but if not (as other posters explain), you are correct.\n\nYou don't have to choice the positions in order so don't worry about $0$s.\n\nDo the Thousandth place first. It can not be $0$ and it can not be $7$ or $5$. So there are $7$ options. Then do either the tens or the ones, it doesn't matter which. It can be $0$ but it can't be $7$ or $5$ and it can't be what the thousandth place was. That is $7$ options as well. And the final position can't be $7$ or $5$ or either of the previous two positions. so that is $6$ options.\n\nSo there are $7*7*6 = 294$.\n\nYou were correct but you didn't have to work so hard.\n\nYour freind is wrong in that $0$ is not an option for the first digit.", "date": "2019-10-14 04:31:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2737440/how-many-4-digit-numbers-that-do-not-have-5-and-have-7-in-the-hundreds-pos", "openwebmath_score": 0.7261207103729248, "openwebmath_perplexity": 124.50432861989911, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180669140007, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8581634465475715}}
{"url": "http://sesk.motcha.de/telescoping-series-partial-fractions.html", "text": "# Telescoping Series Partial Fractions\n\nTelescoping Series and Partial Fractions. Use n = 3, since we're after the 3rd partial sum. You may want to review that material before trying these problems. 003+cdots#. The first part is 2/1, so thats 2. Ask a question. Geometric and telescoping series The geometric series is X1 n=0 a nr n = a + ar + ar2 + ar3 + = a 1 r provided jrj<1 (when jrj 1 the series diverges). be able to recognize a telescoping series, determine whether it is cvg or div and if cvg, to what. Determine convergence, absolute convergence and divergence of infinite series using the standard convergence tests. The partial fraction decomposition will consist of one term for the factor and three terms for the factor. You don't see many telescoping series, but the telescoping series rule is a good one to keep in your bag of tricks \u2014 you never know when it might come in handy. Now, we need to be careful here. Could anyone explain in plain english whats going on here? And clarify their formula. In this case, P a n = s. BC Sequences and series 2015. 4 Comparison Tests. Telescoping series. \u2022 Define what a power series is and be able to find the radius and interval of convergence for a given series. Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction. a) Using partial fractions, , where A \u2014 n2\u20141 n \u20141 n +1. (a) Let sN be the Nth partial sum, sN = PN n=2 1 n2\u2212n. It is called a telescoping series when you write out the sequence of partial sums and the middle terms all cancel,1381. (MCMC 2009I#4) Find the value of the in nite product 7 9 26 28 63 65 = lim n!1 Yn k=2 k3 1 k3 + 1 : Solution. Find the limit of a given sequence. We recover Theorem 1 from Theorem 6 as the case a1 = 1, ai = bj = 0 for all i > 1,j \u2265 1. Say we have something we want to sum up, let's call it a k. partial fractions to rewrite it as a telescoping series. So capital N, so the partial sum is going to be the sum from n equals one but not infinity but to capital N of negative 1 to the n minus 1. 10-2-43 telescoping series. Can the series be compared favorably to one of the special types?. 1A1: The nth partial sum is defined as the sum of the first n terms of a sequence. Laval Kennesaw State University Abstract This hand out is a description of the technique known as telescoping sums, which is used when studying the convergence of some series. Indicate a good method for evaluating the integral. Contractive sequences converge. Make sure you can correctly answer questions involving telescoping series and partial sums. There is no test that will tell us that we've got a telescoping series right off the bat. Write out the first three partial sums of 2 3 4 n n 4 b. Thus a telescoping series will only converge if bn approaches a finite limit. Series and Partial Sums. We explain calculus and give you hundreds of practice problems, all with complete, worked out, step-by-step solutions. Here, (using partial fractions) Partial sum: Thus the series converges and has sum = 5. Partial sums Value of infinite series as limit of sequence of partial sums Convergence vs. More examples can be found on the Telescoping Series Examples 1 page. Topic 9: Calculus Option Series Part 1 A series consists of C v+C w+C x\u2026qK S=z T St_ qK s=z S T st_ It is denoted by: For a total sum For a partial sum The Divergence Test states:. Telescoping Series. can the integral test, the root test, or the ratio test be applied? 4. 10-3-33 sine series that diverges by nth term test. For example, is a partial fractions decomposition of. Telescoping Series - Sum. Improper integrals of type 2. Try this telescoping sum. Math 222, Calculus and Analytic Geometry 2, 2013-14 Notes for how to do partial fractions updated; This is some notes about infinite series and telescoping. 4 \u2013 Divergence & Integrals -Use the divergence test to determine whether a given series diverges (or whether the test is inconclusive). ALTERNATING SERIES Does an =(\u22121)nbn or an =(\u22121)n\u22121bn, bn \u2265 0? NO Is bn+1 \u2264 bn & lim n\u2192\u221e YES n =0? YES an Converges TELESCOPING SERIES Dosubsequent termscancel out previousterms in the sum? May have to use partial fractions, properties of logarithms, etc. 1, etc), then:. The Partial Fraction Decomposition Calculator an online tool which shows Partial Fraction Decomposition for the given input. diverges \ufffb \ufff9 c. Once it has been determined that a given series converges, it is usually a much more di\ufb03cult problem to actually determine the sum of the series. Partial Fractions Telescoping Series, and Harmonic Series in this section. It is called a telescoping series when you write out the sequence of partial sums and the middle terms all cancel,1381. We know how to sum all these geometric series, so we get. The comparison tests are used to determine convergence or divergence of series with positive terms. And if the limit of the partial sum is nite, then it converges, and we. In the cases where series cannot be reduced to a closed form expression an approximate answer could be obtained using definite integral calculator. To be able to do this, we will use the method of partial fractions to decompose the fraction that is common in some telescoping series. Find the power series of functions, determine their radius and interval of convergence,. It is capable of computing sums over finite, infinite (inf) and parametrized sequencies (n). (1) and ask whether the sum is convergent. First, note that the telescoping series method only works on certain fractions. \u2014 Try Before you Buy To start a Maplet, click on its name. Here\u2019s an example series: Just like for sequences, we want to make explicit what the term is. If you update to the most recent version of this activity, then your current progress on this activity will be erased. In general, the series given by. Question is : Determine whether the series is convergent or divergent by expressing Sn as a telescoping sum. This calculator will find the sum of arithmetic, geometric, power, infinite, and binomial series, as well as the partial sum. Telescoping Series This next series is a clever series, called a telescoping series. Find the power series of functions, determine their radius and interval of convergence,. Find the sum of the series \u2022 \u00c2 k=0 4 k2 +3k+2 if it exists. 12 is an example of a telescoping series. Download this MAT 1348 class note to get exam ready in less time! Class note uploaded on Oct 14, 2019. 3 Telescoping Series. Even after 1,000,000 1,000,000 terms, the partial sum is still relatively small. Chapter 2 Infinite Series Page 1 of 7 Chapter 2 : Infinite Series Section B Telescoping and Harmonic Series By the end of this section you will be able to \u2022 use partial fractions to test for convergence and determine the sum of a series. We try This is an example of a telescoping series. Example diverve Determine if the following series serjes or diverges. For example, 1 + 1/2 + 1/3 is a partial sum of the first three terms. Welcome to Maplets for Calculus. 2) Expand first 5 terms. partial fractions to rewrite it as a telescoping series. Maclaurin and Taylor Series g. A telescoping series. Don't quite understand how to do that. 1 An introduction to series and sequences 245 261; 10. -To determine whether or not this series converges and to find what value it converges to, we need to use the telescoping series. Multiplying by a Constant Property. In mathematics, a telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with either a succeeding or preceding term. be able to recognize a telescoping series, determine whether it is cvg or div and if cvg, to what. The second part as it goes to infinity is 0, so the answer would be 2? I'm really confused and looking up examples of telescoping series and it is not helping. ) Telescoping series: A telescoping series is a series of the form P\u221e n=1 (a n+i \u2212a ) for some integer i \u2265 1. Necessary conditions for convergence. Geometric series: Converges if jrj< 1, diverges if jrj 1. Convergence Tests d. One of these tests is the telescoping series test. In this video from PatricJMT we look at why when evaluating a telescoping series, one typically finds an expression for a partial sum, and then takes the limit of this partial sum. A telescoping series (or telescoping sum) is one that \"expands\" in such a way that most of its terms cancel away. Use the Integral Test to decide whether the in nite series X1 n=1 2n n2 + 1 dx. Regardless, your record of completion wil. 4 e) Use partial fraction decomposition to decompose n n nn S S S S nn f f. In mathematics, a telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with either a succeeding or preceding term. [Hint: Telescoping - Use partial fraction decomposition] Solution We recognize the series as a telescoping series, so we want to construct a formula for the Nth partial sum, S N, and nd its limit as N !1. I know the answer is 3/4 but I can't figure out how to get to that answer. Get the free \"Series Calculator\" widget for your website, blog, Wordpress, Blogger, or iGoogle. 1674 terms are needed for the partial sum to exceed 8. p-Series 4. Telescoping series. For example, 1 + 1/2 + 1/3 is a partial sum of the first three terms. The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences. Calculus 2 Syllabus. Hence, Therefore, by the definition of convergence for infinite series, the above telescopic series converges and is equal to 1. More examples can be found on the Telescoping Series Examples 2 page. Then a partial fraction decomposition of is so that (This summation is a telescoping sum. The first part is 2/1, so thats 2. Improper integrals of type 2. Differentiating and Integrating Power Series. Also see pages 5 through 9 of the Class Notes Day 26 on Partial Fractions. 4 Comparison Tests: Test #1-in class. Telescoping: Transform by partial fractions Procedure for Determining Convergence No Series Diverges nth-Term Test Yes or maybe Yes No Yes Nonnegative terms and/or absolute convergence No Yes No Can the Integral Test, the Ratio Test, or the Root Test be applied? Does the Integral Test apply? Yes Does the Ratio Test apply? No Yes Is Is Does the. \u2022 Determine if a series is absolutely convergent. and B b) Determine whether the following telescoping series is con- vergent or divergent. The series in Example 8. I used partial fractions to get ((1/3)/n - (1/3)/(n(n+3))). Download this MAT 1348 study guide to get exam ready in less time! Study guide uploaded on Oct 14, 2019. In many cases it is possible at least to determine whether or not the series converges, and so we will spend most of our time on this problem. Find a formula for the nth partial sum s nof the series, and use the formula for s nto determine whether the series converges. Worksheet 2 \u2013 Special Series (Telescoping) Name _____ Find the partial fraction decomposition of each of the following. Ask a question. Scope and Sequence Unit 1 - Techniques of Integration This unit includes the chain rule, u-substitution, expanding, separating the numerator, completing the square, dividing, adding and subtracting terms, trig identities, integration by parts, trig integrals, trig substitution, partial fractions, L'Hopital's Rule, improper integrals, and Euler's Method. can the integral test, the root test, or the ratio test be applied? 4. In this case it\u2019s , so we can write the series as (or just if we\u2019re not too concerned with where the series starts). Telescoping series are not very common in math-ematics but are interesting to study. From Wikipedia, the free encyclopedia. Math 296: Calculus II. State the de nition of geometric series. FIND THE SUM OF TELESCOPING SERIES? Use partial fraction decomposition to write 1/ you can replace the single fraction with the sum of fractions and you have. The Partial Fraction Decomposition Calculator an online tool which shows Partial Fraction Decomposition for the given input. For n = 1, the series is a harmonic series 1 2 + 1 3 + 1 4 + 1 5 + which is divergent, and the formula 1=(n 1) would indicate that the series should be divergent. Z 8x p (telescoping series) 12. 3 cont'd Integral test for series with non-negative terms cont'd and estimations, Estimations of the value of the series. f \u00a6 diverges. The number and variety of exercises where the student must determine the appropriate series test necessary to determine convergence of a given series has been increased. Partial fractions - Integrating rational functions is an extremely entertaining activity. Before looking for the partial fraction decomposition of the ra-. Series Geometric Senes Telescoping Series Lima\u00e7ons and Lemniscates Parametric Equations\u2014Second Derivatives and Tangent Lines Partial Fractions Convergence and Divergence Series Indexing Arithmetic of Series Integration by Parts Il Vector Functions Implicit Differentiation Il Infinite Limits of Integration Partial Fractions Ill p-Series. Infinite Series, Geometric Series, Telescoping Series; Integral Test, p-series, and Estimates of Sums; The Comparison Tests; Alternating Series and Estimates of Sums; Absolute and Conditional Convergence, Ratio and Root Tests; Strategy for Testing Series, Summary of Convergence Tests; Power Series; Representation of Functions as a Power Series. EXAMPLE 8:. The simplest way would be to use partial fractions, and then convert this into a telescoping series. These types of series frequently arise from partial fraction decompositions and lead to very convenient and direct summation formulas. Begin by using partial fraction decomposition to obtain 2 n(n+ 2) = A n + B n+ 2 = 1 n 1. 10-2-43 telescoping series. The following series, for example, is not a telescoping series despite the fact that we can partial fraction the series. Telescoping series: Compute the nth partial sum, sn, and take the limit of sn as n goes to 1. The series in Example I(b) is a telescoping series of the form \u2014 192) + (b2 \u2014 173) + (b 3 \u2014 125) + Telescoping series Note that b2 is canceled by the second term, 173 is canceled by the third term, and so on. Assignment for Day 26 and Answers. Geometric Series + 107. Here's another example that uses partial fractions. High School AP Calculus BC Curriculum. Example 4: Determine the series 1 1 n nn( 1) f \u00a6 is convergent or divergent. Theorem 4. A more general technique Efthirniou's technique can be generalized to series of the form CrI1unvnwhere it is convenient to write only vnas a Laplace transform integral. 12 is an example of a telescoping series. (Telescoping series) Find. if it is convergent, find its sum. First, note that the telescoping series method only works on certain fractions. Partial Fractions. This free online course offers you some Integral definitions, Explanations and Examples. If Partial fractions gives 1 k(k +1). Telescoping series. o q bASl BlB Zr niVg8hnt osS 5r8ewsXenrZv Yecdj. 2 Series A series is a sum of sequential terms. 2) Trigonometric Substitution (7. Harmonic series. Next, prove an important convergence theorem. com allows you to find the sum of a series online. 10-3-23 constant times a divergent series. Expanding the sum yields Rearranging the brackets, we see that the terms in the infinite sum cancel in pairs, leaving only the first and lasts terms. Download this MAT 1348 class note to get exam ready in less time! Class note uploaded on Oct 14, 2019. The value of the series is lim I lim Sn 553 Related 47-58 19\u201434. Then evaluate lim n!1 S n to obtain the value of the series, or state that it diverges. MCrawford (20:20:34). The second part as it goes to infinity is 0, so the answer would be 2? I'm really confused and looking up examples of telescoping series and it is not helping. Series Calculator computes sum of a series over the given interval. Substitution. Convergence b. How this Calculus 2 course is set up to make complicated math easy: This approximately 200-lesson course includes video and text explanations of everything from Calculus 2, and it includes more than 275 quiz questions (with solutions!) to help you test your understanding along the way. Too bad the book didn't do a good job on teaching how to find nth partial sum formula. That is, s k \u2192 -\u221e. Geometric and Telescoping Series In this worksheet, you will calculate the exact values of several series. The voice explains how to first plug in the numbers given for each variable in the fractions. Telescoping Series 1 1 2 3 3 Show that the telescoping series ln dive rges. Telescoping series is a series where all terms cancel out except for the first and last one. Absolute convergence. A telescoping series is a special type of series whose terms cancel each out in such a way that it is relatively easy to determine the exact value of its partial sums. In algebra, the partial fraction decomposition or partial fraction expansion of a rational function (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator. Exercise 6. The method of partial fractions may reveal these. Your explanation that the series is telescoping after written as a partial fraction is more concise. An in nite series can be represented as such: P 1 n=1 a n. z T BMAapdPeB wwMitEhL lIQnkfoimnBi\\tieE rPCrvecWavlccfuxlluKsx. Telescoping Series Test Series: \u2211\u221e\ud835\udc5b=1 \u1f4c \ud835\udc5b+1\u2212 \ud835\udc5b\u1f4d Condition of Convergence: lim \ud835\udc5b\u2192\u221e \ud835\udc5b=\ud835\udc3f Condition of Divergence: None NOTE: 1) May need to reformat with partial fraction expansion or log rules. The typical example of telescoping series (for partial fractions) is. There is no way to actually identify the series as a telescoping series at this point. Math - Calculus II SERIES ( Partial Sums ) infinite series GEOMETRIC SERIES Otherwise, the series diverges ( that is, the series has no sum ). Let\u2019s decompose it, using partial fractions: [math]\\frac{1. The Telescoping Series! This type of infinite series utilizes the technique of Partial Fractions which is a way for us to express a rational function (algebraic fraction) as a sum of simpler fractions. Could anyone explain in plain english whats going on here? And clarify their formula. This is the main origin of the name telescoping series. If convergent find the sum, and if divergent enter DIV: 11. This is an excerpt from my full length lesson. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. be able to recognize a telescoping series, determine whether it is cvg or div and if cvg, to what. You don't see many telescoping series, but the telescoping series rule is a good one to keep in your bag of tricks \u2014 you never know when it might come in handy. \\ B jArlnlA Er^iOgqhEtcsn srhemsNeKrkvre_dM. partial fractions to rewrite it as a telescoping series. Here\u2019s an example series: Just like for sequences, we want to make explicit what the term is. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. After several hours of thinking about these type of problems I found out that I can separate the 1/[(n+1)(n+2)] into partial fractions, and I found out that it is a telescoping series. Telescoping Series with Partial Fractions Mathispower4u. The solution in the above example uses the method of partial fractions to try to calculate the sum of a telescoping series. Then evaluate lim n!1 S n to obtain the value of the series, or state that it diverges. The following exercises test your understanding of infinite sequences and series. \u2022 Use improper integrals to reinforce the notions of infinite series. \u00a0\u00a0 A telescoping series is one whose partial sums eventually only have a fixed number of terms after cancellation. The result is a simple formula for the nth term of the sequence of partial sums. Her practical approach makes math manageable for anyone who's learning math for the first time, returning to school after a long break, or for anyone with a genuine curiosity who wants to dive deeper into the material. 1B3: If a series converges absolutely, then any series obtained from it by regrouping or rearranging the terms has the same value. Now we must solve for A and B. In particular, in order for the fractions to cancel out, we need the numerators to be the same. (Of course, an infinite geometric series is a special case of a Taylor series. Note: For an example of a telescoping sums question, see question #2 in the Additional Examples section below. Berndt also states that there is no \u201cnatural\u201d way to obtain an expansion of H n in powers of m. Telescoping series: Expand out the sums to find a grouping pattern that can simplify in order to take the limit of S n *Note: Some series can be turned into subtraction via partial fractions or by log rules \u2211Eg: \u2211. Infinite Series Definition. This is an example of a telescoping series. Partial Fractions Introduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of Integration Substitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper Integrals Type 1 - Improper Integrals with Infinite Intervals of Integration. If instructions say to determine if a series converges or diverges AND find the sum if it converges, then it must be an infinite geometric series or a telescoping. Improper integrals of type 2. If convergent find the sum, and if divergent enter DIV: 11. Geometric and Telescoping Series c. Look at the partial sums: because of cancellation of adjacent terms. So capital N, so the partial sum is going to be the sum from n equals one but not infinity but to capital N of negative 1 to the n minus 1. Geometric Series + 107. If an input is given then it can easily show the result for the given number. How to get the partial fractions of higher degree numerators. An in nite series can be represented as such: P 1 n=1 a n. The Sigma notation for summation of series. 3 The Integral Test: The section includes the integral test and a discussion on the remainder estimate for the integral test. Review for Test 3 Math 1552, Integral Calculus The series is telescoping, so it converges and we can \ufb01nd its sum using partial fractions: \". Say we have something we want to sum up, let's call it a k. Be sure to review the Telescoping Series page before continuing forward. Partial Fractions. For the telescoping series X1 k=3 2 (2k 1)(2k + 1); nd a formula for the nth term of the sequence of partial sums fS ng. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. (1) and ask whether the sum is convergent. ~~~~~ AP Cal BC Students: Telescoping Series and Partial Fractions Tests. A telescoping series (or telescoping sum) is one that \"expands\" in such a way that most of its terms cancel away. ) (Now evaluate the limit. 2: p 2] EXAMPLES Does the series converge or diverge? 1. Definitions: Let{ } 1 n n a. Retrieved from \" https: The following series, for example, is not a telescoping series despite the fact that we can partial fraction the series terms. Alternating Series Test cannot be applied \ufffb 26. In nite Series SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. TELESCOPING SERIES Now let us investigate the telescoping series. ) = 1 - 0 = 1. So: s1=a1 s2=a1+a2 s3=a1+a2+a3 s4=a1+a2+a3+a4. This is a challenging sub-section of algebra that requires the solver to look for patterns in a series of fractions and use lots of logical thinking. Improper integrals of type 2. This time there will be a few more terms that do not cancel. Lecture 17: Series (II) Telescoping Series A telescoping series is a special type of series for which many terms cancel in the nth partial sums. (The first Maplet may take a little longer to open because it needs to start Java. Back to Thomas O'Sullivan's Homepage. Creating the telescoping effect frequently involves a partial fraction decomposition. Telescoping series. 3 Telescoping Series. You - ProProfs Discuss. We will now look at some more examples of evaluating telescoping series. If you are looking for more in partial fractions, do check in: Partial fractions of lower degree numerators. Get the free \"Series Calculator\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Partial fractions Fractions in which the denominator has a quadratic term Sometimes we come across fractions in which the denominator has a quadratic term which. Be able to determine the nth partial sum of any geometric series. Partial Fractions Stewart Chapter 7. Test for convergence of a series using an appropriate test: divergence, integral, comparison and limit comparison, ratio, root, or alternating series. High School AP Calculus BC Curriculum. For telescoping series, intermediate terms 'drop. For example, using partial fractions and cancelling a bunch of terms, we find that; An infinite series that arises from Parseval\u2019s theorem in Fourier analysis. Most telescopic series problems involve using the partial fraction decomposition before expanding it and seeing terms cancel out, so make sure you know that very well before tackling these questions. Calculus 2 Syllabus. Note that the fundamental concepts of functions, graphs, and limits, which are studied at the beginning of courses in differential calculus, are often first introduced in earlier classes (most notably intermediate algebra and precalculus). Also note that it is possible to tell that this last series. In\ufb01nite series 1: Geometric and telescoping series Main ideas. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. 26, 29, 39, 40 21 \u2013 42 Solve applications involving series. Exercise 6. 3) Cancel duplicates. If you do not own Maple, click Use MapleNET 12 at the top-right corner of this page. Note: is a telescoping series. Thus, as $$k$$ increases, the partial sum $$S_k$$ increases (the series is a sum of positive terms), but is always smaller than \\(2\\text{. Find the sum of the series \u2022 \u00c2 k=0 4 k2 +3k+2 if it exists. That is, s k \u2192 -\u221e. Hence, Therefore, by the definition of convergence for infinite series, the above telescopic series converges and is equal to 1. Find the sum of the series \u2022 \u00c2 k=1 1 k 1 k+2 if it exists. With geometric series, we carried out the entire evaluation process by finding a formu for the sequence of partial sums and evaluating the limit of the sequence. Infinite Series Chapter 1: Sequences and series Section 4: Telescoping series Page 2 We can use the method of partial fractions to rewrite the general term as: 4 1 1 2 2 2 2 ab n n n n n n \u00a7\u00b7 \u00a8\u00b8 \u00a9\u00b9 For your practice, check that this decomposition is correct! Now we can write any partial sum of this series as: 11 4 1 1 2 22 kk nn n n n n. This video is a great one on learning about evaluating fractions. Telescoping: Transform by partial fractions Procedure for Determining Convergence No Series Diverges nth-Term Test Yes or maybe Yes No Yes Nonnegative terms and/or absolute convergence No Yes No Can the Integral Test, the Ratio Test, or the Root Test be applied? Does the Integral Test apply? Yes Does the Ratio Test apply? No Yes Is Is Does the. (a) X1 n=3 1 2n 1 1 2n 4 As I stated in my email follow-up to this problem, this is not actually an easy series to nd a. Now pop in the first term (a 1) and the common ratio (r). The telescoping Series is a method for examining the convervence of infinite series of the form: This method, combined with partial fraction decomposition, is frequently effective. Be able to determine the nth partial sum of any geometric series. (a) X1 n=1 cos 1 n cos 1 n+ 1. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as: #3. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. According to Ferraro [4], Mengoli did so by making frequent use of the following. \u2014 Try Before you Buy To start a Maplet, click on its name. if integral of series reaches \u221e or DNE then series diverges. Telescoping Series and Partial Fractions. In this case, the series is convergent and the sum isS lim n\u2192 sn b1 \u2212lim n\u2192 bn 1. See if you can \ufb01gure it out. Welcome to Maplets for Calculus. 4 Comparison Tests. Can the series be compared favorably to one of the special types?. EXAMPLE 8:. 50 46 \u2013 53 Evaluate telescoping series. telescoping series to \ufb01nd out. This is best demonstrated with an example. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. Now let us investigate the telescoping series. (a)Sequences and Series, partial sums. Example: Determine whether the given series converge. Use comparison test if numerator fluctuates between two constants. From Wikipedia, the free encyclopedia.", "date": "2020-01-29 17:16:34", "meta": {"domain": "motcha.de", "url": "http://sesk.motcha.de/telescoping-series-partial-fractions.html", "openwebmath_score": 0.886096179485321, "openwebmath_perplexity": 498.4839509090061, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9918120888882002, "lm_q2_score": 0.8652240825770433, "lm_q1q2_score": 0.8581397046971139}}
{"url": "https://cryptohack.gitbook.io/cryptobook/fundamentals/division-and-gcd/euclidean-algorithm", "text": "# Introduction\n\nAlthough we have functions that can compute our $\\gcd$easily it's important enough that we need to give and study an algorithm for it: the euclidean algorithm.\n\nIt's extended version will help us calculate modular inverses which we will define a bit later.\n\n# Euclidean Algorithm\n\nImportant Remark\n\nIf $a = b \\cdot q + r$and $d = \\gcd(a, b)$ then $d | r$. Therefore $\\gcd(a, b) = \\gcd(b, r)$\n\n## Algorithm\n\nWe write the following:\n\n$a = q_0 \\cdot b + r_0 \\\\ b = q_1 \\cdot r_0 + r_1 \\\\ r_0 = q_2 \\cdot r_1 + r_2 \\\\ \\vdots \\\\ r_{n-2} = r_ {n-1} \\cdot q_{n - 1} + r_n \\\\ r_n = 0$\n\nOr iteratively $r_{k-2} = q_k \\cdot r_{k-1} + r_k$ until we find a $0$. Then we stop\n\nNow here's the trick:\n\n$\\gcd(a, b) = gcd(b, r_0) = gcd(r_0, r_1) = \\dots = \\gcd(r_{n-2}, r_{n-1}) = r_{n-1} = d$\n\nIf $d = \\gcd(a, b)$ then $d$ divides $r_0, r_1, ... r_{n-1}$\n\nPause and ponder. Make you you understand why that works.\n\nExample:\n\nCalculate $\\gcd(24, 15)$\n\n$24 = 1 \\cdot 15 + 9 \\\\ 15 = 1 \\cdot 9 + 6 \\\\ 9 = 1 \\cdot 6 + 3 \\\\ 6 = 2 \\cdot 3 + 0 \\Rightarrow 3 = \\gcd(24, 15)$\n\nCode\n\ndef my_gcd(a, b):    # If a < b swap them    if a < b:         a, b = b, a    # If we encounter 0 return a    if b == 0:         return a    else:        r = a % b        return my_gcd(b, r)\u200bprint(my_gcd(24, 15))# 3\n\nExercises:\n\n1. Pick 2 numbers and calculate their $\\gcd$by hand.\n\n2. Implement the algorithm in Python / Sage and play with it. Do not copy paste the code\n\n# Extended Euclidean Algorithm\n\nThis section needs to be expanded a bit.\n\nBezout's identity\n\nLet $d = \\gcd(a, b)$. Then there exists $u, v$ such that $au + bv = d$\n\nThe extended euclidean algorithm aims to find $d = \\gcd(a, b), \\text{ and }u, v$given $a, b$\n\n# In sage we have the xgcd functiona = 24b = 15g, u, v = xgcd(a, b)print(g, u, v)# 3 2 -3 \u200bprint(u * a + v * b)# 3 -> because 24 * 2 - 15 * 3 = 48 - 45 = 3", "date": "2021-09-24 15:52:14", "meta": {"domain": "gitbook.io", "url": "https://cryptohack.gitbook.io/cryptobook/fundamentals/division-and-gcd/euclidean-algorithm", "openwebmath_score": 0.5681771636009216, "openwebmath_perplexity": 1268.4372832355834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120898562831, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.858139700364471}}
{"url": "https://www.quizover.com/trigonometry/test/evaluation-of-functions-in-algebraic-forms-by-openstax", "text": "# 3.1 Functions and function notation \u00a0(Page 4/21)\n\n Page 4 / 21\n $n$ 1 2 3 4 5 $Q$ 8 6 7 6 8\n\n[link] displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.\n\n Age in years, (input) 5 5 6 7 8 9 10 Height in inches, (output) 40 42 44 47 50 52 54\n\nGiven a table of input and output values, determine whether the table represents a function.\n\n1. Identify the input and output values.\n2. Check to see if each input value is paired with only one output value. If so, the table represents a function.\n\n## Identifying tables that represent functions\n\nInput Output\n2 1\n5 3\n8 6\nInput Output\n\u20133 5\n0 1\n4 5\nInput Output\n1 0\n5 2\n5 4\n\n[link] and [link] define functions. In both, each input value corresponds to exactly one output value. [link] does not define a function because the input value of 5 corresponds to two different output values.\n\nWhen a table represents a function, corresponding input and output values can also be specified using function notation.\n\nThe function represented by [link] can be represented by writing\n\nSimilarly, the statements\n\nrepresent the function in [link] .\n\n[link] cannot be expressed in a similar way because it does not represent a function.\n\nInput Output\n1 10\n2 100\n3 1000\n\nyes\n\n## Finding input and output values of a function\n\nWhen we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.\n\nWhen we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function\u2019s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.\n\n## Evaluation of functions in algebraic forms\n\nWhen we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function $\\text{\\hspace{0.17em}}f\\left(x\\right)=5-3{x}^{2}\\text{\\hspace{0.17em}}$ can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.\n\nGiven the formula for a function, evaluate.\n\n1. Replace the input variable in the formula with the value provided.\n2. Calculate the result.\n\n## Evaluating functions at specific values\n\nEvaluate $\\text{\\hspace{0.17em}}f\\left(x\\right)={x}^{2}+3x-4\\text{\\hspace{0.17em}}$ at\n\n1. $2$\n2. $a$\n3. $a+h$\n4. $\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}$\n\nReplace the $\\text{\\hspace{0.17em}}x\\text{\\hspace{0.17em}}$ in the function with each specified value.\n\n1. Because the input value is a number, 2, we can use simple algebra to simplify.\n$\\begin{array}{ccc}\\hfill f\\left(2\\right)& =& {2}^{2}+3\\left(2\\right)-4\\\\ & =& 4+6-4\\hfill \\\\ & =& 6\\hfill \\end{array}$\n2. In this case, the input value is a letter so we cannot simplify the answer any further.\n$f\\left(a\\right)={a}^{2}+3a-4$\n3. With an input value of $\\text{\\hspace{0.17em}}a+h,\\text{\\hspace{0.17em}}$ we must use the distributive property.\n$\\begin{array}{ccc}\\hfill f\\left(a+h\\right)& =& {\\left(a+h\\right)}^{2}+3\\left(a+h\\right)-4\\hfill \\\\ & =& {a}^{2}+2ah+{h}^{2}+3a+3h-4\\hfill \\end{array}$\n4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that\n$f\\left(a+h\\right)={a}^{2}+2ah+{h}^{2}+3a+3h-4$\n\nand we know that\n\n$f\\left(a\\right)={a}^{2}+3a-4$\n\nNow we combine the results and simplify.\n\nthe gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve\n1+cos\u00b2A/cos\u00b2A=2cosec\u00b2A-1\ntest for convergence the series 1+x/2+2!/9x3\na man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?\n100 meters\nKuldeep\nFind that number sum and product of all the divisors of 360\nAjith\nexponential series\nNaveen\nwhat is subgroup\nProve that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1\ne power cos hyperbolic (x+iy)\n10y\nMichael\ntan hyperbolic inverse (x+iy)=alpha +i bita\nprove that cos(\u03c0/6-a)*cos(\u03c0/3+b)-sin(\u03c0/6-a)*sin(\u03c0/3+b)=sin(a-b)\nwhy {2k\u03c0} union {k\u03c0}={k\u03c0}?\nwhy is {2k\u03c0} union {k\u03c0}={k\u03c0}? when k belong to integer\nHuy\nif 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41\nwhat is complex numbers\nDua\nYes\nahmed\nThank you\nDua\ngive me treganamentry question\nSolve 2cos x + 3sin x = 0.5", "date": "2018-12-10 03:50:49", "meta": {"domain": "quizover.com", "url": "https://www.quizover.com/trigonometry/test/evaluation-of-functions-in-algebraic-forms-by-openstax", "openwebmath_score": 0.7388232350349426, "openwebmath_perplexity": 782.8804753921077, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9706877692486436, "lm_q2_score": 0.8840392848011834, "lm_q1q2_score": 0.8581261212918271}}
{"url": "http://math.stackexchange.com/questions/20948/fibonacci-identity-f-n-1f-n1-f-n2-1n", "text": "# Fibonacci identity: $f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$\n\nConsider this Fibonacci equation: $$f_{n+1}^2 - f_nf_{n+2}$$\n\nThe problem asked to write a program with given n, output the the result of this equation. I could use the formula $$f_n = \\frac{(1+\\sqrt{5})^n - ( 1 - \\sqrt{5} )^n}{2^n\\sqrt{5}}$$\n\nHowever, from mathworld, I found this formula Cassini's identity $$f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$$\n\nSo, I decided to play around with the equation above, and I have:\n\n$$\\text{Let } x = n + 1$$ $$\\text{then the equation above becomes } f_x^2 - f_{x-1}f_{x+1}$$ $$\\Rightarrow -( f_{x-1}f_{x+1} - f_x^2 ) = -1(-1)^x = (-1)^{x+1} = (-1)^{n+1+1} = (-1)^{n+2}$$\n\nSo this equation either is 1 or -1. Am I in the right track?\n\nThanks,\nChan\n\n-\nYes. $\\mbox{}$ \u2013\u00a0 Mariano Su\u00e1rez-Alvarez Feb 8 '11 at 5:44\n@Mariano Su\u00e1rez-Alvarez: Thanks ^_^! Happy! \u2013\u00a0 Chan Feb 8 '11 at 5:47\nIn your first expression, should the subscript on the square be \"n+1\" instead of \"n-1\"? If so, then you are on the right track. (Also, you'll want parentheses around your \"-1\"s.) \u2013\u00a0 Blue Feb 8 '11 at 5:47\nNote: You should write $(-1)^{n+2}$, and not $-1^{n+2}$; exponentiation takes precedence, so the former is either $-1$ or $1$ depending on the parity of $n$, but the latter is $-1^{n+2} = -(1^{n+2}) = -1$ always. \u2013\u00a0 Arturo Magidin Feb 8 '11 at 5:50\n@Day Late Don: nice catch. It was my typo. I was practicing using Latex:(. \u2013\u00a0 Chan Feb 8 '11 at 5:51\n\nWe have the following (easily proved by induction):\n\n$\\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\\\ \\end{pmatrix}^n = \\begin{pmatrix} f_{n+1} & f_n \\\\ f_n & f_{n-1} \\\\ \\end{pmatrix}$\n\nEquating the determinants of the matrices gives us the identity immediately.\n\n-\nNOTE $\\$ This is precisely the same answer I gave a couple months ago in this duplicate thread \u2013\u00a0 Bill Dubuque Feb 28 '11 at 4:06\n\nLet us try to find gcd of $F_n$ and $F_{n+1}$ using Extended Euclidean algorithm. I will write the steps algorithm in a table; this table method was also explained in some Bill Dubuque's posts.\n\n$\\begin{array}{|l||c|c|} \\hline F_{n+1} & 1 & 0 \\\\\\hline F_{n} & 0 & 1 \\\\\\hline F_{n-1} & 1 & -1 \\\\\\hline F_{n-2} & -1 & 2 \\\\\\hline F_{n-3} & 2 & -3 \\\\\\hline \\vdots & \\vdots & \\vdots \\\\\\hline F_{n-k} & (-1)^{k+1}F_k & (-1)^kF_{k+1} \\\\\\hline \\end{array}$\n\nAfter a few steps we can guess the $k$-th line, which gives us the following formula:\n$F_{n-k}=(-1)^{k+1}F_kF_{n+1}+(-1)^kF_{k+1}F_n=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)$.\n\nFor $k=n-1$ we get Cassini's identity $F_1=(-1)^n(F_{n-1}F_{n+1}-F_n^2)$.\n\nSo the only thing we have to do is to verify the above formula, which can be done easily by induction on $k$.\n\nInductive step: We know that:\n$F_{n-k}=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)=-(-1)^{k}(F_kF_{n+1}-F_{k+1}F_n)$\n$F_{n-(k-1)}=(-1)^{k}(F_{k-1}F_{n+1}-F_{k}F_n)$\n\nSince $F_{n-(k+1)}=F_{n-(k-1)}-F_{n-k}$, we get\n$F_{n-(k+1)}=(-1)^{k}[(F_{k-1}+F_k)F_{n+1}-(F_k+F_{k+1})F_n]=(-1)^{k+2}(F_{k+1}F_{n+1}-F_{k+2}F_n)$\nwhich completes the inductive step.\n\n-\nI tried to google a little, and I found almost the same derivation of Cassini's identity in the text Yiu: Recreational Mathematics. \u2013\u00a0 Martin Sleziak Mar 16 '12 at 11:09", "date": "2015-05-30 11:33:32", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/20948/fibonacci-identity-f-n-1f-n1-f-n2-1n", "openwebmath_score": 0.9586094617843628, "openwebmath_perplexity": 521.7688696465686, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877692486436, "lm_q2_score": 0.8840392756357326, "lm_q1q2_score": 0.8581261123950361}}
{"url": "http://mathhelpforum.com/discrete-math/153261-series.html", "text": "# Math Help - series\n\n1. ## series\n\nDear Sir\nI would be grateful if you can help me to find f(r) in the below questions\nthanks\nKingsman\n\nSum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))\n\n2. Originally Posted by kingman\nDear Sir\nI would be grateful if you can help me to find f(r) in the below questions\nthanks\nKingsman\n\nSum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))\n1)\n\n$\\displaystyle\\huge\\frac{n}{(n+1)!}=\\frac{n+1-1}{(n+1)!}=\\frac{n+1}{(n+1)!}-\\frac{1}{(n+1)!}$\n\n$=\\displaystyle\\huge\\frac{n+1}{(n+1)n!}-\\frac{1}{(n+1)!}=\\frac{1}{n!}-\\frac{1}{(n+1)!}$\n\nTherefore\n\n$\\displaystyle\\huge\\frac{1}{2!}+\\frac{2}{3!}+......$\n\n$=1-\\displaystyle\\huge\\frac{1}{2!}+\\frac{1}{2!}-\\frac{1}{3!}+\\frac{1}{3!}+.......-\\frac{1}{(n+1)!}$\n\n$=1-\\displaystyle\\huge\\frac{1}{(n+1)!}$\n\n3. Originally Posted by kingman\nDear Sir\nI would be grateful if you can help me to find f(r) in the below questions\nthanks\nKingsman\n\nSum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))\n2)\n\n$\\displaystyle\\huge\\frac{1}{(n+2)n!}=\\frac{(n+1)}{( n+2)(n+1)n!}=\\frac{n+1}{(n+2)!}$\n\n$\\displaystyle\\huge\\frac{n+2-1}{(n+2)!}=\\frac{n+2}{(n+2)!}-\\frac{1}{(n+2)!}$\n\n$\\displaystyle\\huge\\frac{n+2}{(n+2)(n+1)!}-\\frac{1}{(n+2)!}=\\frac{1}{(n+1)!}-\\frac{1}{(n+2)!}$\n\nUse this to form the telescope.\n\n4. Thanks very much for the kind help.\nMay I know how and what is the idea which you employ in your solution making it so neat and simply. Is the f(r) unique and is there a general method to find f(r) ?.\nLastly is it possible to find a g(r) such that the nth term= g(r)-g(r-1) allowing us to find the sum?\n\n5. If you are told to evaluate an infinite sum you should either change it into a known form of a function or you should make it a telescoping sum. Now to make it a telescoping sum you need to look for the following things:\n\n- an obvious way to apply partial fractions decomposition\n- add and subtract something in the numerator so that part of it looks like something that we can cancel on the bottom.\n\nThat is usually a good start. There is a lot of intuition involved in some examples and less in others.\n\n6. Thanks very much for the reply.\nCan I know how to find g(r) such that the nth term= g(r)-g(r-1) thus allowing us to find the sum of the above 3 problem.\nHow to apply telescope method in question 3?\nI cannot wrap around my head to get started in problem3\nThanks\n\n7. 3)\n\nThe sum can be written as...\n\n$\\displaystyle \\sum_{k=1}^{n} \\frac{k}{k+5} = n - 5\\ \\sum_{k=1}^{n} \\frac{1}{k+5}$ (1)\n\nIn order to solve the second term of (1) we have to introduce the function digamma, defined as...\n\n$\\displaystyle \\psi(x)= \\sum_{j=1}^{\\infty} \\frac{x}{j\\ (j+x)} - \\gamma$ (2)\n\n... where $\\gamma$ is the 'Euler's constant'. From (2) You can derive the following basic identity...\n\n$\\displaystyle \\psi(1+x) - \\psi (x) = \\frac{1}{1+x}$ (3)\n\nNow setting in (3) $x=k+4$ You obtain...\n\n$\\displaystyle \\psi(k+5) - \\psi (k+4) = \\frac{1}{k+5}$ (4)\n\n... so that is...\n\n$\\displaystyle \\sum_{k=1}^{n} \\frac{1}{k+5} = \\sum_{k=1}^{n} \\{ \\psi(k+5) - \\psi (k+4) \\} = \\psi (n+5) - \\psi(5)$ (5)\n\n... and finally You can write...\n\n$\\displaystyle \\sum_{k=1}^{n} \\frac{k}{k+5} = n - 5\\ \\psi(n+5) + 5\\ \\psi(5)$ (6)\n\nDigamma function - Wikipedia, the free encyclopedia\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n8. That's pretty cool, chisigma, but I think there is an even simpler answer.\n\n$\\displaystyle\\sum_{k=1}^n \\frac{k}{k+5} = \\sum_{k=1}^n \\frac{k+5-5}{k+5} = \\sum_{k=1}^n \\left(\\frac{k+5}{k+5}-\\frac{5}{k+5}\\right)$\n\n$\\displaystyle = \\left(\\sum_{k=1}^n 1\\right)-5\\left(\\sum_{k=1}^n \\frac{1}{k+5}\\right) = n - 5\\sum_{k=6}^{n+5} \\frac{1}{k+5} = n - 5\\left(H_n - \\frac{1}{1}-\\frac{1}{2}-\\frac{1}{3}-\\frac{1}{4}-\\frac{1}{5}\\right) = \\frac{137}{12}+n-5H_n$\n\nwhere [LaTeX ERROR: Convert failed] is the n-th Harmonic number. There is no simpler way to write this sum.", "date": "2015-10-09 09:28:23", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/153261-series.html", "openwebmath_score": 0.7095587253570557, "openwebmath_perplexity": 691.3621429783493, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357243200245, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8581128524067028}}
{"url": "https://math.stackexchange.com/questions/1619029/what-is-the-probability-that-the-max-of-n-numbers-from-some-distribution-is-a", "text": "What is the probability that the max of $n$ numbers from some distribution is a greater than another number from the same distribution?\n\nIndependently choose $n$ numbers $X_1, X_2, \\ldots, X_n$ from some distribution. And Independently choose another number $X_{n+1}$ from the same distribution. Let $P(\\max(X_1, X_2, \\ldots,,X_n) > X_{n+1}))$ means the probability that the maximum of $X_1, X_2, \\ldots, X_n$ is bigger than $X_{n+1}$. What is $P(\\max(X_1, X_2, \\ldots,X_n) > X_{n+1}))$? I think that $P(\\max(X_1, X_2, \\ldots,X_n) > X_{n+1})) = \\frac{n}{n+1}$. Here is my proof.\n\nThe case that any one of $X_1, X_2, \\ldots X_n, X_{n+1}$ is the maximum is symmetric. So $P(X_i\\text{ is the maximum)} = \\frac{1}{n+1}, (i = 1,\\ldots,n+1)$.\n\n$P(\\max(X_1, X_2, \\ldots,X_n) > X_{n+1})) = P(X_i \\text{ is the maximum}, i =1,\\ldots,n)= \\sum_{i=1}^{n}P(X_i \\text{ is the maximum}) = n \\cdot \\frac{1}{n+1} = \\frac{n}{n+1}$\n\nBut I think that my proof is not rigorous. https://math.stackexchange.com/a/20841/16625 gives a rigorous proof for the $n=1$ case. Can anyone give a rigorous proof for the general case?\n\n\u2022 You should get $\\frac{n}{n+1}$. $\\frac{1}{n+1}$ is the probability that $\\max(X_i)<X_{n+1}$. \u2013\u00a0Thomas Andrews Jan 20 '16 at 3:53\n\u2022 This only works for continuous random variables, of course. For instance, if the $X_i$ are die rolls, you will get a very different result. \u2013\u00a0Thomas Andrews Jan 20 '16 at 3:58\n\u2022 @ThomasAndrews Yes. For discrete random variables, the case that random variables are equal can not be ignored. \u2013\u00a0Jingguo Yao Jan 21 '16 at 7:21\n\nProbably easier to compute the probability that the max is less. Specifically, $P(\\max(X_i)<x)=P(X<x)^n$. So if $p$ is the CDF for $X$ then the probability is:\n\n$$\\int p(x)^n\\,dp(x) = \\frac{1}{n+1}p(x)^{n+1}\\Bigg{|}_{-\\infty}^{\\infty}=\\frac{1}{n+1}$$\n\nThis is sort of obvious - the probability that the last one is the largest is the same as the probability that any other is the largest. (If the pdf is discontinuous, it is possible, with non-zero probability, for two to be equal, of course. So we assume continuous.)\n\nSo the probability that you were seeking is $\\frac{n}{n+1}$.\n\nHere is a slightly more formal argument. Suppose $X_1, X_2, \\ldots,$ are independent and identically distributed random variables. Then the maximum order statistic $X_{(n)}$ of a sample of size $n$ has CDF $$F_{X_{(n)}}(x) = \\Pr[X_{(n)} \\le x] = \\prod_{i=1}^n \\Pr[X_i \\le x] = F_X(x)^n.$$ Suppose we are now interested in the random variable $Y = X_{(n)} - X_{n+1}$, and in particular, $$\\Pr[Y > 0] = 1 - \\Pr[Y \\le 0] = 1- \\int_{x = -\\infty}^\\infty F_{X_{(n)}}(x) f_X(x) \\, dx = 1 - \\int_{x = -\\infty}^\\infty F_X(x)^n f_X(x) \\, dx.$$ With the substitution $$u = F_X(x), \\quad du = f_X(x) \\, dx, \\quad F_X(-\\infty) = 0, \\quad F_X(\\infty) = 1,$$ we arrive at $$\\Pr[Y > 0] = 1 - \\int_{u=0}^1 u^n \\, du = 1 - \\frac{1}{1+n} = \\frac{n}{n+1},$$ as claimed.\n\nAssuming that the $P(X_i=X_j)=0$ for $i\\ne j$, $$P(\\max(X_1,X_2,\\dots,X_n)\\gt X_{n+1})$$ is the probability that $X_{n+1}$ is not the greatest of the $n+1$ samples. The probability that it is the greatest of $n+1$ samples is $$P(\\max(X_1,X_2,\\dots,X_n)\\le X_{n+1})=\\frac{n!}{(n+1)!}=\\frac1{n+1}$$ because there are $n!$ arrangements where $X_{n+1}$ is greatest and $(n+1)!$ arrangements altogether.\n\nTherefore, $$P(\\max(X_1,X_2,\\dots,X_n)\\gt X_{n+1})=1-\\frac1{n+1}=\\frac n{n+1}$$\n\nYour argument is correct if the $n+1$ observations are independent, an assumption that you did not state.", "date": "2019-06-24 17:48:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1619029/what-is-the-probability-that-the-max-of-n-numbers-from-some-distribution-is-a", "openwebmath_score": 0.9553362727165222, "openwebmath_perplexity": 107.68804196094108, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357195106375, "lm_q2_score": 0.8740772417253256, "lm_q1q2_score": 0.858112849813086}}
{"url": "https://practicaldev-herokuapp-com.global.ssl.fastly.net/thepracticaldev/daily-challenge-230-beeramid-2m4a", "text": "## DEV Community is a community of 662,276 amazing developers\n\nWe're a place where coders share, stay up-to-date and grow their careers.\n\n# Daily Challenge #230 - Beeramid\n\ndev.to staff\nThe hardworking team behind dev.to \u2764\ufe0f\n\nLet's pretend your company just hired your friend from college and paid you a referral bonus. Awesome! To celebrate, you're taking your team out to the terrible dive bar next door and using the referral bonus to buy, and build, the largest three-dimensional beer can pyramid you can. And then probably drink those beers, because let's pretend it's Friday too.\n\nA beer can pyramid will square the number of cans in each level - 1 can in the top level, 4 in the second, 9 in the next, 16, 25...\n\nComplete the beeramid function to return the number of complete levels of a beer can pyramid you can make, given the parameters of:\n\n2) the price of a beer can\n\nFor example:\n\nbeeramid(1500, 2); // should === 12\nbeeramid(5000, 3); // should === 16\n\nTests:\nbeeramid(9, 2)\nbeeramid(21, 1.5)\nbeeramid(-1, 4)\n\nThis challenge comes from kylehill on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!\n\nWant to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!\n\n## Discussion (6)\n\nHere's my C solution,\nNotice that there is no need for any iteration because (taking number of cans = N and number of levels = L)\n\n$N = 1 + 2^2 + 3^2 + ... + L^2 = L(L+1)(2L+1)/6 \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\newline \\implies L^3/3 < N < (L+1)^3/3 \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\newline \\implies L < (3N)^{1/3} < L+1 \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\newline \\implies L = \\lfloor (3N)^{1/3} \\rfloor \\quad or \\quad L = \\lfloor (3N)^{1/3} \\rfloor - 1 \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad$\n\nint beeramid(double bonus, double price) {\nif (bonus < price) return 0;\nint cans = floor(bonus / price);\nint levels = floor(cbrt(3.0*cans));\nreturn levels*(levels + 1)*(2*levels + 1) > 6*cans ? levels - 1 : levels;\n}\n\nVidit Sarkar\n\nHere is a Python Solution,\n\ndef beeramid(bonus, price):\ncanCount = int(bonus/price)\nlevel = 0\n\nwhile((level+1)**2 <= canCount):\nlevel += 1\ncanCount -= level*level\n\nreturn level\n\n\nTest Cases,\n\nprint(beeramid(1500, 2)) # output -> 12\nprint(beeramid(5000, 3)) # output -> 16\nprint(beeramid(9, 2)) # output -> 1\nprint(beeramid(21, 1.5)) # output -> 3\nprint(beeramid(-1, 4)) # output -> 0\n\nMichael Kohl \u2022 Edited\n\nRuby, using a lazy, infinite enumerator:\n\nROWS = (1..).lazy.map { |n| n * n }\n\ndef beeramid(bonus, price)\ncans = bonus / price\nROWS.take_while { |n| (cans -= n) >= 0 }.count\nend\n\nValts Liepi\u0146\u0161 \u2022 Edited\n\nHaskell solution using lazy, infinite list of cans:\n\nimport Data.List (findIndex)\n\nbeeramid :: Double -> Double -> Int\nbeeramid m p =\ncase findIndex (> cans) levels of\nJust level -> level\nNothing    -> 0\nwhere\ncans = m / p\nlevels = scanl1 (+) . fmap (^2) \\$ [1..]\n\nPaula Gearon \u2022 Edited\n\nClojure. Not very terse, but it gets a solution in constant time\n\n(defn faul [n]\n(let [n2 (* n n)\nn3 (* n2 n)]\n(/ (+ n3 n3 n2 n2 n2 n) 6)))\n\n(defn layers [n]\n(let [est (int (Math/pow (* 3 n) (/ 1 3)))]\n(first (filter #(>= n (faul %)) (range est 0 -1)))))\n\n(defn beeramid [bonus price] (or (layers (/ bonus price)) 0))\n\n\nTesting:\n\n=> (beeramid 1500 2)\n12\n=> (beeramid 5000 3)\n16\n=> (beeramid 9 2)\n1\n=> (beeramid 21 1.5)\n3\n=> (beeramid -1 4)\n0\n\nPaula Gearon\n\nTo follow up on my solution, I borrowed from the Wikipedia article on Square Pyramidal Numbers. This provides a formula for the sum of squares as a special case of Faulhaber's formula:\n\n$P_n = \\sum_{k=1}^n k^2 = \\frac {2n^3 + 3n^2 + n} 6$\nSo if we have n layers, then that will have a total of Pn cups.\nAs n becomes large then: 2n3 \u226b 3n2 + n\n$P_n = \\frac {2n^3} 6 + \\Big( \\frac {3n^2 + n} 6 \\Big) \\newline P_n > \\frac {n^3} 3 \\newline$\nSince the layers is n and the cups in the pyramid is Pn then this gives a bound of:\n$layers < \\sqrt[3]{3 \\cdot cups}$\nConsequently, my code creates an estimate of the number of layers using this formula, and counts down from there applying the Faulhaber formula to check if the result is less than or equal to the provided number of cups.\n\nUsing this method, a small numbers of cups will require 2 guesses, but as the number of cups gets higher (and n3n2) then it is more likely to get it on the first guess. I told it to keep testing while the number of layers until 1, but it never actually gets beyond 2 attempts.", "date": "2021-07-26 21:18:50", "meta": {"domain": "fastly.net", "url": "https://practicaldev-herokuapp-com.global.ssl.fastly.net/thepracticaldev/daily-challenge-230-beeramid-2m4a", "openwebmath_score": 0.28828954696655273, "openwebmath_perplexity": 10790.589013250126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357189762611, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8581128332444095}}
{"url": "http://mathhelpforum.com/calculus/143022-definite-integral.html", "text": "# Math Help - definite integral\n\n1. ## definite integral\n\na little confused on this one\n\n\"solve the definite integral\"\n\n$\\int^2_0 2x^2e^{x^3+5}$\n\n2. Originally Posted by JGaraffa\na little confused on this one\n\n\"solve the definite integral\"\n\n$\\int^2_0 2x^2e^{x^3+5}$\nWell, again, use substitution $z := x^3+5$, which gives that $dz=3x^2\\,dx$ and hence you have\n\n$\\int 2x^2e^{x^3+5}\\,dx=\\tfrac{2}{3}\\cdot\\int e^{x^3+5}\\cdot 3x^2\\, dx=\\tfrac{2}{3}\\cdot\\int z\\,dz=\\ldots$\n\n3. Originally Posted by JGaraffa\na little confused on this one\n\n\"solve the definite integral\"\n$\\int^2_0 2x^2e^{x^3+5}$\nWhat exactly are you confused about? Have you made an attempt at it?\n\nAlso the integral should be $\\int^2_0 2x^2e^{x^3+5}dx$\n\nI could tell you how to do it but i think i it's only fair that an attempt is made first\n\nHint: use substitution\n\n4. ah perfect! i couldn't figure out how i could change that $2x^2$ to a $3x^2$ . thanks!\n\n5. actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.\n\n$\\int^2_0 2x^2e^{x^3+5}$\n\nu= $x^3 + 5$\nh(u) = $3x^2$\n\n$\\frac{2}{3} \\int^2_0 3x^2 e^{x^3+5}$\n\nwhen x = 0, u = 5\nwhen x = 2, u = 13\n\n$\\frac{2}{3} \\int^{13}_5 e^u du$\n$\n\n\\frac{2}{3} e^{13} - \\frac{2}{3} e^5$\n\ni'm getting 294,843.32 which makes me think i'm doing something wrong.\n\n6. Originally Posted by JGaraffa\nactually, i'm getting really high numbers for this which makes me think i may be doing it wrong.\n\n$\\int^2_0 2x^2e^{x^3+5}$\n\nu= $x^3 + 5$\nh(u) = $3x^2$\n\n$\\frac{2}{3} \\int^2_0 3x^2 e^{x^3+5}$\n\nwhen x = 0, u = 5\nwhen x = 2, u = 13\n\n$\\frac{2}{3} \\int^{13}_5 e^u du$\n$\n\n\\frac{2}{3} e^{13} - \\frac{2}{3} e^5$\n\ni'm getting 294,843.32 which makes me think i'm doing something wrong.\nNo, that result is ok. Of course, $e^{x^3+5}$ is a fast growing function.\n\n7. Originally Posted by JGaraffa\nactually, i'm getting really high numbers for this which makes me think i may be doing it wrong.\n\n$\\int^2_0 2x^2e^{x^3+5}$\n\nu= $x^3 + 5$\nh(u) = $3x^2$\n\n$\\frac{2}{3} \\int^2_0 3x^{2} e^{x^3+5}$\n\nwhen x = 0, u = 5\nwhen x = 2, u = 13\n\n$\\frac{2}{3} \\int^{13}_5 e^u du$\n$\n\n\\frac{2}{3} e^{13} - \\frac{2}{3} e^5$\n\ni'm getting 294,843.32 which makes me think i'm doing something wrong.\nApart from [tex]3x^{2} it lookes fine to me.\n\n8. Originally Posted by JGaraffa\na little confused on this one\n\n\"solve the definite integral\"\n\n$\\int^2_0 2x^2e^{x^3+5}$\nAgain observe that $3x^2$ is the derivative of $x^3+5$, so consider:\n\n$\\frac{d}{dx}e^{x^3+5}$\n\nCB", "date": "2015-01-31 14:28:49", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/143022-definite-integral.html", "openwebmath_score": 0.8870323300361633, "openwebmath_perplexity": 1199.2569567484225, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.97241471777321, "lm_q2_score": 0.8824278726384089, "lm_q1q2_score": 0.8580858507268925}}
{"url": "https://math.stackexchange.com/questions/1005422/evaluating-int-02-tan-1-pi-x-tan-1-x-mathrmdx/3052525", "text": "# Evaluating $\\int_0^2(\\tan^{-1}(\\pi x)-\\tan^{-1} x)\\,\\mathrm{d}x$\n\nHint given: Write the integrand as an integral.\n\nI'm supposed to do this as double integration.\n\nMy attempt:\n\n$$\\int_0^2 [\\tan^{-1}y]^{\\pi x}_{x}$$\n\n$$= \\int_0^2 \\int_x^{\\pi x} \\frac { \\mathrm{d}y \\mathrm{d}x} {y^2+1}$$\n\n$$= \\int_2^{2\\pi} \\int_{\\frac{y}{\\pi}}^2 \\frac { \\mathrm{d}x \\mathrm{d}y} {y^2+1}$$\n\n$$= \\int_2^{2\\pi} \\frac { [x]^2_{\\frac{y}{\\pi}} \\mathrm{d}y } { y^2+1}$$\n\n$$= \\int_2^{2\\pi} \\frac { 2- {\\frac{y}{\\pi}} \\mathrm{d}y } { y^2+1}$$\n\nCarrying out this integration, I got, $$2[\\tan^{-1} 2 \\pi - \\tan^{-1} 2] - \\frac {1}{2 \\pi} [\\ln \\frac {1+4 {\\pi}^2} {5}]$$\n\nBut I'm supposed to get $$2[\\tan^{-1} 2 \\pi - \\tan^{-1} 2] - \\frac {1}{2 \\pi} [\\ln \\frac {1+4 {\\pi}^2} {5}]+ [\\frac {\\pi-1}{2 \\pi}] \\ln 5$$\n\nCan someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.\n\n\u2022 At third line it seems to me the integral must be $$\\int_0^{2\\pi}\\int_{\\frac{y}{\\pi}}^{y}\\frac{\\operatorname d\\!x \\operatorname d\\!y}{y^2+1}$$ \u2013\u00a0\u00c1ngel Mario Gallegos Nov 4 '14 at 6:13\n\u2022 but that is taking my answer even further away from what I require. \u2013\u00a0Diya Nov 4 '14 at 6:50\n\u2022 I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :) \u2013\u00a0Diya Nov 4 '14 at 6:57\n\nUse the change of variables $y = xt.$\n\n$$I = \\int_0^2 \\int_x^{\\pi x} \\frac { \\mathrm{d}y \\, \\mathrm{d}x} {y^2+1}\\\\=\\int_0^2 \\int_1^{\\pi } \\frac { x} {x^2t^2+1}\\mathrm{d}t \\, \\mathrm{d}x\\\\=\\int_1^{\\pi} \\int_0^{2 } \\frac { x} {x^2t^2+1}\\mathrm{d}x \\, \\mathrm{d}t\\\\=\\int_1^{\\pi} \\frac{\\ln(1+4t^2)}{2t^2} \\mathrm{d}t$$\n\nNow use integration by parts.\n\n$$I = -\\left.\\frac{\\ln(1+4t^2)}{2t}\\right|_1^{\\pi}+4\\int_1^{\\pi}\\frac1{1+4t^2} \\, dt$$\n\nA more straight forward approach uses integration by parts.\n\nDefine: \\begin{align} & I(c)=\\int_{a}^{b}dx(1 \\times \\arctan{c x})=\\int_{ac}^{bc}\\frac{dy}{c} (1 \\times\\arctan{ y})=\\\\&\\frac{1}{c}y \\arctan(y)|_{ac}^{bc}-\\frac{1}{2c}\\int_{ac}^{bc}\\frac{y}{1+y^2} \\end{align}\n\nusing partial fraction this reads: \\begin{align} I(c)=\\frac{1}{c}y \\arctan(y)|_{ac}^{bc}-\\frac{1}{2c}\\log(1+y^2)|_{ac}^{bc} \\end{align}\n\ntaking $I(\\pi)-I(0)$ with $a=0$ and $b=2$ we are done\n\nFor the sake of an alternative approach, recall the formula for the integral of an inverse function $$\\int f^{-1}(x)\\mathrm{d}x=xf^{-1}(x)-F\\circ f^{-1}(x)$$ Where $$F'(x)=f(x)$$. Plugging in $$f(x)=\\tan(x)$$, $$I=\\int\\arctan(x)\\mathrm{d}x=x\\arctan(x)-\\int_0^{\\arctan(x)}\\tan(t)\\mathrm{d}t$$ Then recall that $$(-\\log|\\cos(x)|)'=\\tan(x)$$ So we have $$I=x\\arctan(x)+\\log|\\cos(\\arctan(x))|$$ then using trig, $$I=x\\arctan(x)-\\frac12\\log(x^2+1)$$ So $$I_1=\\int_0^2\\arctan(x)\\mathrm{d}x=2\\arctan2-\\frac12\\log5$$ And \\begin{align} I_2=&\\int_0^2\\arctan(\\pi x)\\mathrm{d}x\\\\ =&\\frac1\\pi\\int_0^{2\\pi}\\arctan(x)\\mathrm{d}x\\\\ =&\\frac1\\pi\\bigg(2\\pi\\arctan2\\pi-\\frac12\\log(4\\pi^2+1)\\bigg)\\\\ =&2\\arctan2\\pi-\\frac1{2\\pi}\\log(4\\pi^2+1)\\\\ \\end{align} So \\begin{align} \\int_0^2(\\arctan\\pi x-\\arctan x)\\mathrm{d}x=&I_2-I_1\\\\ =&2\\arctan2\\pi-\\frac1{2\\pi}\\log(4\\pi^2+1)-2\\arctan2+\\frac12\\log5\\\\ =&2\\arctan2\\pi-\\frac1{2}\\log\\sqrt[\\pi]{4\\pi^2+1}-2\\arctan2+\\frac12\\log5\\\\ =&2(\\arctan2\\pi-\\arctan2)+\\frac12\\log\\frac{5}{\\sqrt[\\pi]{4\\pi^2+1}}\\\\ \\end{align}", "date": "2019-07-18 20:23:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1005422/evaluating-int-02-tan-1-pi-x-tan-1-x-mathrmdx/3052525", "openwebmath_score": 0.9953389167785645, "openwebmath_perplexity": 1122.5578590125945, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.972414716174355, "lm_q2_score": 0.8824278695464501, "lm_q1q2_score": 0.8580858463093521}}
{"url": "https://math.stackexchange.com/questions/3401659/the-number-of-different-triangles-and-each-side-has-a-different-length", "text": "# The number of different triangles and each side has a different length\n\nI'm trying to solve the following problem:\n\nWhat is the number of different triangles we can form from numbers $$4,5,6,7,8,9$$ (lengths of sides), where every side has a different length (for an example $$4,5,6$$ or $$4,5,7$$)..\n\nMy solution is the following:\n\nThe number of all possible permutations is $$\\frac{6!}{(6-3)!}$$. We subtract the numbers which don't make a triangle (which is $$6*2$$) and then divide by $$2$$, because $$(4,5,6)$$ makes the same triangle is $$6,5,4$$. My answer is $$54$$.\n\nHowever, the correct answer should be $$53$$. Can anyone tell me where I did a mistake?\n\nThanks\n\n\u2022 $6 \\choose 3$ is only $20$ and there is one choice $(4,5,9)$ that doesn't make a triangle, so I get $19$ Your expression (ignoring the factorial in the number to choose) gives $8$. Oct 20, 2019 at 16:42\n\u2022 I think you'll have to show more details for someone to point out your mistake. What do you mean by $\\binom{6}{(6-3)!}$? It look to me like ${6\\choose6}=1$. Did you mean to say ${6\\choose3}?$ Also, why do you say there are $12$ where the triangle inequality doesn't hold. How did you get that? Oct 20, 2019 at 16:43\n\u2022 Sorry, I made a mistake when writing my answer here. Please wait a minute and I will fix it Oct 20, 2019 at 16:44\n\u2022 If you write out all the triples $(4,5,6), (4,5,7), \\ldots, (7,8,9)$, you'll see that there are only $20$ in the list, one of which is degenerate. How can there be $53$ or $54$ triangles? We could consider \"right-handed\" and \"left-handed\" triangles to be different, effectively doubling the number of triangles, but even then, we would have at most $40$, still falling short of the given answer. Have you written out the problem exactly as it was given to you? Oct 20, 2019 at 16:56\n\u2022 You should have gotten six crossed off, as you describe. 2 for 4, 2 for 5 and 2 for 9 (not 6) makes six. Why do you double that? Then you should divide by $6$ because 456 comes six ways-456,465,546,564,654,645 not two Oct 20, 2019 at 17:04\n\nThere are $$\\frac {6!}{3!}$$ ordered choices of three numbers, which is $$120$$. Six of those do not make a triangle, all the permutations of $$(4,5,9)$$, which leaves $$114$$. Each unordered triangle gives $$3!$$ permutations, so we divide by $$6$$ and get $$19$$. I don't know where a number in the $$50$$s comes from.\nI think it is easier to just choose unordered combinations to start with, which is $${6 \\choose 3}=20$$ and subtract the one that doesn't make a triangle. That also gets $$19$$.", "date": "2022-07-05 09:14:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3401659/the-number-of-different-triangles-and-each-side-has-a-different-length", "openwebmath_score": 0.7458722591400146, "openwebmath_perplexity": 194.64091766089317, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147145755, "lm_q2_score": 0.88242786954645, "lm_q1q2_score": 0.8580858448984777}}
{"url": "https://math.stackexchange.com/questions/3039219/in-mathbbrn-is-the-dot-product-the-only-inner-product/3039248", "text": "# In $\\mathbb{R}^n$, is the dot product the only inner product?\n\nJust what the title says. I'm reading, from various resources, that the inner product is a generalization of the dot product. However, the only example of inner product I can find, is still the dot product.\n\nDoes another \"inner product\" exist on the $$\\mathbb{R}^n$$ space?\n\n\u2022 This might be useful. \u2013\u00a0StackTD Dec 14 '18 at 11:07\n\u2022 @StackTD: thanks!!! Exactly what I was looking for \u2013\u00a0blue_note Dec 14 '18 at 11:15\n\u2022 \u2013\u00a0StackTD Dec 14 '18 at 11:37\n\u2022 @StackTD: thanks again \u2013\u00a0blue_note Dec 14 '18 at 11:38\n\u2022 As some other answers point out, there are infinitely many inner products (i.e., symmetric, positive-definite bilinear forms) on $\\Bbb R^n$. But for any of them one can choose a basis of $\\Bbb R^n$ with respect to which the bilinear form is the standard one: $({\\bf x}, {\\bf y}) = x_1 y_1 + \\cdots + x_n y_n$. So up to isomorphism there is only one inner product on $\\Bbb R^n$. (This is a special case of Sylvester's Law of Inertia.) \u2013\u00a0Travis Willse Dec 25 '18 at 0:51\n\nYes and are all in the form $$\\langle x, y\\rangle=x^T\\cdot A\\cdot y$$ where $$x^T$$ is the transpose vector of $$x$$ and $$A$$ is a $$n\\times n$$ symmetric definite positive matrix.\n\nIn fact let $$\\langle x, y\\rangle$$ a generic inner product on $$\\mathbb R^n$$ then for every $$y$$ the function $$f_y:x\\rightarrow \\langle x, y\\rangle$$ is linear from $$\\mathbb R^n$$ to $$\\mathbb R$$ then exists a vector $$\\alpha(y)\\in\\mathbb R^n$$ such that $$\\langle x, y\\rangle = \\alpha(y)^T\\cdot x$$\n\nObserve that $$\\langle x, ay+by'\\rangle = a\\langle x, y\\rangle + b\\langle x, y'\\rangle\\Rightarrow \\alpha(ay+by')=a\\alpha(y)+b\\alpha(y')$$ then $$\\alpha$$ is a linear operator from $$\\mathbb R^n$$ in itself then exists an $$n\\times n$$ matrix $$A$$ such that $$\\alpha(y)=A\\cdot y$$ so $$\\alpha(y)^T\\cdot x=y^T\\cdot A^T\\cdot x$$\n\nNow remember that $$\\langle x, y\\rangle=\\langle y, x\\rangle$$ then you can easly prove that $$A^T=A$$ and $$A$$ must be symmetric.\n\nWhy now $$A$$ must be definite positive? Because $$\\langle x, x\\rangle\\geq 0$$ and holds equality if and only if $$x=0$$. Applying it to the initial formula we obtain the definition of a definite positive matrix.\n\n\u2022 We need $A$ to be positive definite also. \u2013\u00a0littleO Dec 14 '18 at 11:26\n\u2022 Oh yes, I forgotten it! \u2013\u00a0Jihlbert Dec 14 '18 at 11:28\n\u2022 Note that in some contexts, one only requires that an inner product be nondegenerate (and not necessarily positive definite). \u2013\u00a0Travis Willse Dec 25 '18 at 0:52\n\u2022 If it's semidefinite positive it becomes a semi-inner product and not a inner product. \u2013\u00a0Jihlbert Dec 28 '18 at 11:30\n\nThe dot product on $$\\mathcal{R}^n$$ is defined as follows:\n\n$$(a,b) = a^i b^j (e_i,e_j) = a^i b^j \\delta_{ij} = a^i b^i ,$$\n\nwhere $$a,b \\in \\mathcal{R}^n$$ and $$e_i,e_j$$ standard basis vectors. I used Einstein summation convention here.\n\nIn general we can express $$a,b$$ in a different basis i.e. $$a=\\tilde{a}^i \\tilde{e}_i$$ and $$b = \\tilde{b}^i \\tilde{e}_i$$ so now not the standard basis but an arbitrary basis of $$\\mathcal{R}^n$$ assuming still $$(,)$$ is positive-definite. This then gives:\n\n$$(a,b) = \\tilde{a}^i \\tilde{b}^j (\\tilde{e}_i,\\tilde{e}_j) = \\tilde{a}^i \\tilde{b}^j A_{ij} \\equiv a^T A \\ b.$$\n\nNote that $$A$$ now is not the identity matrix like in the standard inner product.\n\nThe tensor metric allows for the vector norm to remain constant under change of basis vectors, and is an example of inner product (page 15). In the simple setting of basis vectors constant in direction and magnitude from point to point in $$\\mathbb R^2,$$ here is an example:\n\nChanging basis vectors from Euclidean orthonormal basis $$\\small\\left\\{\\vec e_1=\\begin{bmatrix}1\\\\0 \\end{bmatrix},\\vec e_2=\\begin{bmatrix}0\\\\1 \\end{bmatrix}\\right\\}$$ to\n\n$$\\left\\{\\vec u_1=\\color{red}2\\vec e_1 + \\color{red}1\\vec e_2,\\quad \\vec u_2=\\color{blue}{-\\frac 1 2 }\\vec e_2+\\color{blue}{\\frac 1 4} \\vec e_2\\right\\}\\tag 1$$ would result in a different norm of the vector $$\\vec v=\\begin{bmatrix} 1\\\\1\\end{bmatrix}_\\text{e basis},$$ i.e $$\\Vert \\vec v \\Vert^2=v_1^2 + v_2^2 = 2,$$ when expressed with respect to the new basis vectors.\n\nIn this regard, since vector components transform contravariantly, the change to the new coordinate system would be given by the backward transformation matrix:\n\n$$B=\\begin{bmatrix} \\frac 1 4 & \\frac 1 2\\\\-1&2\\end{bmatrix}$$\n\ni.e. the inverse of the forward transformation for the basis vectors as defined in $$(1),$$ which in matrix form corresponds to\n\n$$F=\\begin{bmatrix} \\color{red} 2 & \\color{blue}{-\\frac 1 2}\\\\\\color{red}1& \\color{blue}{\\frac 1 4}\\end{bmatrix}.$$\n\nHence, the same vector $$\\vec v$$ expressed in the new basis vectors is\n\n$$\\vec v_{\\text{u basis}}=\\begin{bmatrix} \\frac 1 4 & \\frac 1 2\\\\-1&2\\end{bmatrix}\\begin{bmatrix} 1\\\\1\\end{bmatrix}=\\begin{bmatrix} \\frac 3 4\\\\1\\end{bmatrix}.$$\n\nAnd the norm would entail an inner product with the new metric tensor. In the orthonormal Euclidean basis this is simply the identity matrix. Now the matrix tensor is a $$(0,2)-\\text{tensor},$$ and transforms covariantly:\n\n$$g_{\\text{ in u basis}}=(F^\\top F) I= \\begin{bmatrix} 2 & 1\\\\-\\frac 1 2& \\frac 1 4\\end{bmatrix}\\begin{bmatrix} 2 & -\\frac 1 2\\\\1& \\frac 1 4\\end{bmatrix}\\begin{bmatrix} 1 & 0\\\\0& 1\\end{bmatrix}=\\begin{bmatrix} 5 & -\\frac 3 4\\\\- \\frac 3 4& \\frac{5}{16}\\end{bmatrix}$$\n\nThe actual multiplication of basis vectors to obtain the metric tensor is explained in this presentation by @eigenchris. This metric tensor indeed renders the right norm of $$\\vec v:$$\n\n$$\\begin{bmatrix}\\frac 3 4 & 1\\end{bmatrix}\\begin{bmatrix} 5 & -\\frac 3 4\\\\- \\frac 3 4& \\frac{5}{16}\\end{bmatrix}\\begin{bmatrix}\\frac 3 4 \\\\ 1\\end{bmatrix}=2$$\n\nfollowing the operations here.", "date": "2020-06-05 11:54:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3039219/in-mathbbrn-is-the-dot-product-the-only-inner-product/3039248", "openwebmath_score": 0.9432286024093628, "openwebmath_perplexity": 199.8445595459628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9724147129766448, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8580858344676041}}
{"url": "https://math.stackexchange.com/questions/442001/proof-of-the-statement-the-product-of-4-consecutive-integers-can-be-expressed-i", "text": "# Proof of the statement \u201cThe product of 4 consecutive integers can be expressed in the form 8k for some integer k\u201d\n\nI am slowly diving into simple number theory and learning how to craft direct proofs. I needed to proof the statement \"The product of 4 consecutive integers can be expressed in the form 8k for some integer k\".\n\nMy approach was to use the quotient-remainder theorem, as it was surely intended in the textbook. The proof looks like follows.\n\nProof:\n\nSuppose n is any particular but arbitrarily chose integer. We must show that n(n+1)(n+2)(n+3) is divisible by 8. By the quotient-remainder theorem, n can be written in one of the forms 4q or (4q+1) or (4q+2) or (4q+3) for some integer q. We divide into cases accordingly:\n\nCase 1 (n = 4q for some integer q):\n\n\\begin{align} n(n+1)(n+2)(n+3) & = 4q(4q+1)(4q+1)(4q+2)(4q+3)\\\\ & = 8[q(32q^3+48q^2+18q+3)] \\end{align}\n\nLet $m=q(32q^3+48q^2+18q+3)$. Then m is an integer because sums and products of integers are integers. By substitution, $n(n+1)(n+2)(n+3) = 8m$ where m is an integer. Hence n(n+1)(n+2)(n+3) is divisible by 8.\n\nCase 2 (n = (4q+1) for some integer q): \\begin{align} n(n+1)(n+2)(n+3) & = (4q+1)(4q+2)(4q+3)(4q+4)(4q+5)\\\\ & = 8[(4q+1)(2q+1)(4q+3)(q+1)] \\end{align}\n\nLet $m=q((4q+1)(2q+1)(4q+3)(q+1))$. Then m is an integer because sums and products of integers are integers {...} {See case 1, its basically the same reasoning here...}\n\nCase 3 (n = (4q+2) for some integer q): {...}\n\nCase 4 (n = (4q+3) for some integer q): {...}\n\nConclusion:\n\nI each of the above cases, n(n+1)(n+2)(n+3) was to be shown to be a multiple of 8. By the quotient-remainder theorem, one of these cases must occur, hence n(n+1)(n+2)(n+3) can be written in the form 8k for some integer k. q.e.d. [End Proof]\n\nBut honestly, I consider this proof somehow clumsy and too inconvenient [I am a bloody amateur, maybe I am wrong]. Are there any better/shorter ways to prove the above statement?\n\n\u2022 Any four consecutive numbers will have one of each of the possible four remainders mod $4$. This means that one will be divisible by $4$ and another will be divisible by $2$, which means that the product is divisible by $8$. \u2013\u00a0Tobias Kildetoft Jul 12 '13 at 12:34\n\u2022 Wow, this is straightforward! \u2013\u00a0Nikolai Tschacher Jul 18 '13 at 8:41\n\nIn each four consecutive integers there is exactly one pair of even integers, and exactly one of them is $\\;2\\pmod 4\\;$ and the other one is $\\,0\\pmod 4\\;$ , so the product of these two even (consecutive even) integers is already divisible by $\\,8\\,\\ldots\\ldots$\n\n\u2022 Very good, sir. \u2013\u00a0Don Larynx Jan 9 '15 at 20:53\n\nNote that if you have four consecutive integers, then one must be divisible by 4 and another must be divisible by 2: they must be equivalent to $0,1,2,\\text{ and }3\\pmod{4}$.\n\nOne way is by proving that the binomial coefficients $$\\dbinom n r=\\cfrac {n!}{r!(n-r)!}$$ are integers, whoch can be done in various ways including using the binomial recurrence from Pascal's Triangle. Then$$\\binom n 4=\\frac {n(n-1)(n-2)(n-3)}{4!}$$ which proves that the product of four consecutive integers is divisible by $24$.\n\nAnother way is by induction.$$(n+1)n(n-1)(n-2)-n(n-1)(n-2)(n-3)=4n(n-1)(n-2)$$\n\nSo you can prove the result using that the product of three consecutive integers is even. Or prove the result for $24$, by showing that the product of three consecutive integers is divisible by $6$ (reducing the number of factors again by taking the difference of successive terms). And remembering to prove the base case - but this can be arranged to have zero as a factor, so divisible by the integer we require.\n\n$$(n+1)n(n-1)-n(n-1)(n-2)=3n(n-1)$$is divisible by 3 and $$(n+1)n-n(n-1)=2n$$ is divisible by 2.\n\n\u2022 What is the reasoning of binomial coefficient proof? We have $4! \\cdot {}^nC_4 = {}^n P_4$ because every combination of 4 objects in 4 places has 4! permutation. But how does this totally unrelated fact, without invoking any properties of numbers, prove a property of numbers i.e. the product of any four consecutive numbers is a multiple of 24? \u2013\u00a0user103816 Nov 7 '15 at 16:09\n\u2022 @user103816 You can easily show that Pascal's Triangle encapsulates the formula $\\binom nr+\\binom n{r+1}=\\binom {n+1}{r+1}$ and this shows that all the binomial coefficients are integers. Then suppose you have the consecutive integers $n-3, n-2, n-1, n$, you will find, as I have put, that $\\binom n4$ is the product of those integers divided by $4!$ and is an integer - so the product of four successive integers is divisible by $24$. Negative numbers are not a problem, and if one of the numbers is zero, so is the product. \u2013\u00a0Mark Bennet Nov 7 '15 at 16:42\n\nYour formal argument looks fine to me, but you're right: it does seem clumsy and inconvenient. This may be because you are making exactly the same argument in each case: Writing the four numbers, finding one factor of four and another factor of two, and pulling those out to get a factor of $8$.\n\nThis suggests a simpler argument. In any four consecutive integers, by the quotient-remainder theorem, there must be one multiple of four. There must also be two multiples of two. One of the multiples of two will also be a multiple of four; the other will not. Therefore, we may factor a four and another two from the product of the four consecutive integers and we see that the product is a multiple of eight.", "date": "2021-01-24 18:40:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/442001/proof-of-the-statement-the-product-of-4-consecutive-integers-can-be-expressed-i", "openwebmath_score": 0.7221397757530212, "openwebmath_perplexity": 277.6579741820987, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718490038142, "lm_q2_score": 0.8670357683915537, "lm_q1q2_score": 0.8580808920565118}}
{"url": "http://math.stackexchange.com/questions/9461/find-a-pattern-in-a-recursive-definition", "text": "# Find a pattern in a recursive definition\n\nA set M $\\in$ $\\mathbb{Z}^2$ is defined by\n\n(Rule 1) $(0, 1)$ $\\in$ M\n\n(Rule 2) If $(x, y)$ $\\in$ M, then $(x + 1, y + 2x + 3)$ $\\in$ M\n\n(a) (2 points) Starting with $(0, 1)$, write out the six pairs with the smallest first coordinates.\n\n(b) (2 points) State the (simple) relationship that holds between the first and second coordinates of all pairs in M.\n\nOk for a I got: $(0,1)$,$(1,4)$,$(2,9)$,$(3,16)$,$(4,25)$,$(5,36)$,$(6,49)$\n\nHowever for b, I cant see the relationship. Can anyone give me any hints?\n\nEDIT:\n\nOk I found the relation: $y=(x+1)^2$. However, I must now prove this claim using structural induction. I can get up to the inductive hypothesis, but after that I'm not really sure what to do.\n\nBase case: $(0,1), 1=(0+1)^2 = 1$\n\nInductive Hypothesis: For any element $x,y$ if $x\\in M$ then $y=(x+1)^2$. We must show that $y+1=(x+1)^2 + 1$\n\nI'm not really sure that my IH is correct either.\n\n-\nIt appears you are using the new x in the calculation of y. I believe you should use the old x. So the second pair should be (1,4). All will become clearer. \u2013\u00a0 Ross Millikan Nov 8 '10 at 18:21\nAhh ok yes you are right..now I see the relation, thanks! \u2013\u00a0 maq Nov 8 '10 at 18:26\n@Ross please see edits \u2013\u00a0 maq Nov 8 '10 at 18:41\n\nYour inductive step is to show that if $(x,(x+1)^2) \\in M$, then $(x+1,(x+2)^2) \\in M$. So apply Rule 2 to $(x,(x+1)^2)$\n\n-\nWait x, x^2+1 is an element of M? I thought it was x, (x+1)^2 is an element of M \u2013\u00a0 maq Nov 8 '10 at 18:56\nRight you are. Corrected. \u2013\u00a0 Ross Millikan Nov 8 '10 at 19:04\nThanks, I was able to get the rest.. \u2013\u00a0 maq Nov 8 '10 at 19:06\n\nYour inductive hypothesis is wrong. You should be showing that if $(x,(x+1)^2)\\in M$, then $(x+1, ((x+1)+1)^2) \\in M$. From there you should be able to see how to get the result (just use Rule 2).\n\n-\n\nHINT $\\rm\\ \\ f\\ (x,\\:y)\\ = \\ (x+1,\\ y+2x+3)\\ \\Rightarrow\\ f\\ (x,\\ (x+1)^2)\\ =\\ (x+1,\\ (x+2)^2)$\n\n-", "date": "2014-12-21 01:55:17", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/9461/find-a-pattern-in-a-recursive-definition", "openwebmath_score": 0.8203819990158081, "openwebmath_perplexity": 745.2428630167697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.963779941013992, "lm_q2_score": 0.8902942173896131, "lm_q1q2_score": 0.8580477083208595}}
{"url": "https://mathhelpboards.com/threads/state-the-domain.8090/", "text": "# State the domain\n\n#### shamieh\n\n##### Active member\nstate the domain\n\n$$\\displaystyle y = x\\sqrt{1 - x^2}$$\n\nso\n\n$$\\displaystyle 1 - x^2 >= 0$$\n$$\\displaystyle -x^2 >= -1$$\n$$\\displaystyle x^2 <= 1$$\n\n$$\\displaystyle d = {x <= (+)(-) 1}$$ or should I say $$\\displaystyle d = {-1, 1}$$\n\nwhich one is correct?\n\n#### MarkFL\n\nStaff member\nI have moved this topic here to our Pre-Calculus subforum as questions on function domains does not involve the calculus. I suspect that this question is part of a calculus question though, or may even come from a calculus textbook, but when choosing the subforum in which to post, the nature of the question itself, rather than from where it originates should be considered firstly.\n\nYou are correct in stating:\n\n$$\\displaystyle 1-x^2\\ge0$$\n\nWhat I would suggest doing next is factoring the left side:\n\n$$\\displaystyle (1+x)(1-x)\\ge0$$\n\nNow, determine the critical values, plot them on a number line and test the 3 resulting intervals.\n\nA quicker method would be to plot the expression (the original radicand), and observe where it is non-negative.\n\nWhat do you find?\n\n#### soroban\n\n##### Well-known member\nHello, shamieh!\n\nState the domain: .$$y \\:=\\: x\\sqrt{1 - x^2}$$\n\nSo: .$$1 - x^2 \\:\\ge\\:0$$\n. . . . . $$-x^2 \\:\\ge\\:-1$$\n. . . . . . .$$x^2\\:\\le\\:1$$\n\n$$\\begin{array}{cc}\\text{I would write:} & |x| \\:\\le\\:1 \\\\ & \\text{or} \\\\ & \\text{-}1 \\:\\le\\:x\\:\\le\\:1 \\\\ & \\text{or} \\\\ & [\\text{-}1,\\,1]\\end{array}$$\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nstate the domain\n\n$$\\displaystyle y = x\\sqrt{1 - x^2}$$\n\nso\n\n$$\\displaystyle 1 - x^2 >= 0$$\n$$\\displaystyle -x^2 >= -1$$\n$$\\displaystyle x^2 <= 1$$\n\n$$\\displaystyle d = {x <= (+)(-) 1}$$ or should I say $$\\displaystyle d = {-1, 1}$$\n\nwhich one is correct?\nNOT $$\\displaystyle \\{-1, 1\\}$$ because that means the numbers -1 and 1 only, not the numbers between. You could say $$\\displaystyle \\{x| -1\\le x\\le 1\\}$$ or [-1, 1] as MarkFL said.\n\n(\"math\" uses the \"{\" and \"}\" and as separators itself. To get \"{\" or \"}\" in the actual message, use \"\\{\" and \"\\}\".)\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nA naive approach:\n\nWe know that there's \"something special\" about the points $x = -1, x = 1$. So let's do this:\n\nWe'll split the real number line into 5 parts:\n\nPart 1: all numbers less than -1\nPart 2: -1\nPart 3: all numbers between -1 and 1\nPart 4: 1\nPart 5: all numbers greater than 1.\n\nNow let's pick numbers in each part, to see what happens:\n\nThe 5 real numbers I will use are:\n\nPart 1: -4\nPart 2: -1 (no choice, here)\nPart 3: 1/2\nPart 4: 1 (again, no other choice)\nPart 5: 6\n\nNow we will look at $f(x_0)$ for $x_0$ being each one of these 5 numbers:\n\n$f(-4) = 4\\sqrt{1 - (-4)^2} = 4\\sqrt{-15} = \\text{bad}$ (undefined)\n$f(-1) = (-1)\\sqrt{1 - (-1)^2} = (-1)\\sqrt{1 - 1} = (-1)\\sqrt{0} = (-1)(0) = 0$ (OK!!!!)\n$f(\\frac{1}{2}) = \\frac{1}{2}\\sqrt{1 - (\\frac{1}{2})^2} = \\frac{1}{2}\\sqrt{\\frac{3}{4}} = \\frac{\\sqrt{3}}{4}$ (OK!!!)\n$f(1) = (1)\\sqrt{1 - 1^2} = (1)(0) = 0$ (OK again!!!!)\n$f(6) = 6\\sqrt{1 - 6^2} = 6\\sqrt{-35} = ???$ (not so good).\n\nSo it looks as if what we want is parts 2,3, and 4, and NOT parts 1 and 5. This is:\n\n$\\{-1\\} \\cup (-1,1) \\cup \\{1\\} = [-1,1]$\n\nIf you prefer, you can write this as:\n\n$\\text{dom}(f) = \\{x \\in \\Bbb R: |x| \\leq 1\\}$\n\nthe absolute value of $x$, written $|x|$ is just another way of saying:\n\n$\\sqrt{x^2}$, if one understand square roots as always being non-negative. So if:\n\n$x^2 \\leq 1$, then\n\n$\\sqrt{x^2} \\leq \\sqrt{1} = 1$\n\nso:\n\n$|x| \\leq 1$.", "date": "2021-03-09 10:46:47", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/state-the-domain.8090/", "openwebmath_score": 0.7587964534759521, "openwebmath_perplexity": 1858.0544346236409, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969660413473, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8580382343318943}}
{"url": "https://math.stackexchange.com/questions/2780138/there-are-4-cups-of-liquid-three-are-water-and-one-is-poison-if-you-were-to-dr/2783140", "text": "# There are 4 cups of liquid. Three are water and one is poison. If you were to drink 3 of the 4 cups, what is the probability of being poisoned?\n\nIn Season 5 Episode 16 of Agents of Shield, one of the characters decides to prove she can't die by pouring three glasses of water and one of poison; she then randomly drinks three of the four cups. I was wondering how to compute the probability of her drinking the one with poison.\n\nI thought to label the four cups $\\alpha, \\beta, \\gamma, \\delta$ with events\n\n\u2022 $A = \\{\\alpha \\text{ is water}\\}, \\ a = \\{\\alpha \\text{ is poison}\\}$\n\u2022 $B = \\{\\beta \\text{ is water}\\},\\ b = \\{\\beta \\text{ is poison}\\}$\n\u2022 $C = \\{\\gamma \\text{ is water}\\},\\ c = \\{\\gamma \\text{ is poison}\\}$\n\u2022 $D = \\{\\delta \\text{ is water}\\},\\ d = \\{\\delta \\text{ is poison}\\}$\n\nIf she were to drink in order, then I would calculate $P(a) = {1}/{4}$. Next $$P(b|A) = \\frac{P(A|b)P(b)}{P(A)}$$ Next $P(c|A \\cap B)$, which I'm not completely sure how to calculate.\n\nMy doubt is that I shouldn't order the cups because that assumes $\\delta$ is the poisoned cup. I am also unsure how I would calculate the conditional probabilities (I know about Bayes theorem, I mean more what numbers to put in the particular case). Thank you for you help.\n\n\u2022 Comments are not for extended discussion; this conversation has been moved to chat. \u2013\u00a0Jyrki Lahtonen May 16 '18 at 5:01\n\u2022 You need to provide a little more information to actually answer this question correctly. Is the poison fast acting? If the poison is fast-acting, then being available to drink the third cup is added information that the first two cups were not poison. \u2013\u00a0jerrylagrou May 17 '18 at 16:27\n\u2022 Possible duplicate of Probability of being poisoned \u2013\u00a0Xander Henderson May 17 '18 at 19:12\n\u2022 @EricTowers not to mention the inference from Murphy's Law, which tells you that the probability is 1. \u2013\u00a0Michael Kay May 18 '18 at 7:13\n\u2022 As someone really bad at math, gonna ask a dumb question here: Why is the probability not just 75%? \u2013\u00a0john doe May 19 '18 at 13:41\n\nNicNic8 has provided a nice intuitive answer to the question.\n\nHere are three alternative methods. In the first, we solve the problem directly by considering which cups are selected if she is poisoned. In the second, we solve the problem indirectly by considering the order in which the cups are selected if she is not poisoned. In the third, we add the probabilities that she was poisoned with the first cup, second cup, or third cup.\n\nMethod 1: We use the hypergeometric distribution.\n\nThere are $\\binom{4}{3}$ ways to select three of the four cups. Of these, the person selecting the cups is poisoned if she selects the poisoned cup and two of the three cups of water, which can be done in $\\binom{1}{1}\\binom{3}{2}$ ways. Hence, the probability that she is poisoned is $$\\Pr(\\text{poisoned}) = \\frac{\\binom{1}{1}\\binom{3}{2}}{\\binom{4}{3}} = \\frac{1 \\cdot 3}{4} = \\frac{3}{4}$$\n\nMethod 2: We subtract the probability that she is not poisoned from $1$.\n\nThe probability that the first cup she selects is not poisoned is $3/4$ since three of the four cups do not contain poison. If the first cup she selects is not poisoned, the probability that the second cup she selects is not poisoned is $2/3$ since two of the three remaining cups do not contain poison. If both of the first two cups she selects are not poisoned, the probability that the third cup she selects is also not poisoned is $1/2$ since one of the two remaining cups is not poisoned. Hence, the probability that she is not poisoned if she drinks three of the four cups is $$\\Pr(\\text{not poisoned}) = \\frac{3}{4} \\cdot \\frac{2}{3} \\cdot \\frac{1}{2} = \\frac{1}{4}$$ Hence, the probability that she is poisoned is $$\\Pr(\\text{poisoned}) = 1 - \\Pr(\\text{not poisoned}) = 1 - \\frac{1}{4} = \\frac{3}{4}$$\n\nAddendum: We can relate this method to the first method by using the hypergeometric distribution.\n\nShe is not poisoned if she selects all three cups which do not contain poison when selecting three of the four cups. Hence, the probability that she is not poisoned is $$\\Pr(\\text{not poisoned}) = \\frac{\\dbinom{3}{3}}{\\dbinom{4}{3}} = \\frac{1}{4}$$ so the probability she is poisoned is $$\\Pr(\\text{poisoned}) = 1 - \\frac{\\dbinom{3}{3}}{\\dbinom{4}{3}} = 1 - \\frac{1}{4} = \\frac{3}{4}$$\n\nMethod 3: We calculate the probability that the person is poisoned by adding the probabilities that she is poisoned with the first cup, the second cup, and the third cup.\n\nLet $P_k$ denote the event that she is poisoned with the $k$th cup.\n\nSince there are four cups, of which just one contains poison, the probability that she is poisoned with her first cup is $$\\Pr(P_1) = \\frac{1}{4}$$\n\nTo be poisoned with the second cup, she must not have been poisoned with the first cup and then be poisoned with the second cup. The probability that she is not poisoned with the first cup is $\\Pr(P_1^C) = 1 - 1/4 = 3/4$. If she is not poisoned with the first cup, there are three cups remaining of which one is poisoned, so the probability that she is poisoned with the second cup if she is not poisoned with the first is $\\Pr(P_2 \\mid P_1^C) = 1/3$. Hence, the probability that she is poisoned with the second cup is $$\\Pr(P_2) = \\Pr(P_2 \\mid P_1^C)\\Pr(P_1) = \\frac{3}{4} \\cdot \\frac{1}{3} = \\frac{1}{4}$$\n\nTo be poisoned with the third cup, she must not have been poisoned with the first two cups and then be poisoned with the third cup. The probability that she is not poisoned with the first cup is $\\Pr(P_1^C) = 3/4$. The probability that she is not poisoned with the second cup given that she was not poisoned with the first is $\\Pr(P_2^C \\mid P_1^C) = 1 - \\Pr(P_2 \\mid P_1^C) = 1 - 1/3 = 2/3$. If neither of the first two cups she drank was poisoned, two cups are left, one of which is poisoned, so the probability that the third cup she drinks is poisoned given that the first two were not is $\\Pr(P_3 \\mid P_1^C \\cap P_2^C) = 1/2$. Hence, the probability that she is poisoned with the third cup is $$\\Pr(P_3) = \\Pr(P_3 \\mid P_1^C \\cap P_2^C)\\Pr(P_2^C \\mid P_1^C)\\Pr(P_1^C) = \\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{3}{4} = \\frac{1}{4}$$ Hence, the probability that she is poisoned is $$\\Pr(\\text{poisoned}) = \\Pr(P_1) + \\Pr(P_2) + \\Pr(P_3) = \\frac{1}{4} + \\frac{1}{4} + \\frac{1}{4} = \\frac{3}{4}$$\n\n\u2022 I was attempting to use Method 3 (which you've shown is the most tedious). Thank you for your very detailed answer, it helped me view the problem in different ways and understand it more clearly. Thank you. \u2013\u00a0Eli May 14 '18 at 13:37\n\u2022 To be honest, Methods 1 and 3 seem like a distraction from Method 2, the only really reasonable way of calculating this. \u2013\u00a0jwg May 16 '18 at 8:51\n\u2022 Both of the other methods are eminently reasonable. \u2013\u00a0Matthew Read May 17 '18 at 3:37\n\u2022 Why is it that in method 3 it is necessary to include the probability that the preceding cups were not poisoned, but in method 2 the preceding cups are not necessary? Subtleties like that are what make probability both difficult and fascinating IMHO. \u2013\u00a0Jim W says reinstate Monica May 17 '18 at 20:43\n\u2022 @JimW In method 2, the probability that the preceding cups were not poisoned is taken into account. The probability that the first cup is not poisoned is $3/4$. Observe that the probability that the second cup is not poisoned given that the first cup was not poisoned is $2/3$ since if the first cup was not poisoned only two of the three cups that remain are not poisoned. The probability that the third cup is not poisoned given that the first two cups were not poisoned is $1/2$ since if the first two cups were not poisoned then only one of the two cups that remain is not poisoned. \u2013\u00a0N. F. Taussig May 17 '18 at 20:52\n\nThe probability of not being poisoned is exactly the same as the following problem:\n\nYou choose one cup and drink from the other three. What is the probability of choosing the poisoned cup (and not being poisoned)? That probability is 1/4.\n\nTherefore, the probability of being poisoned is 3/4.\n\n\u2022 @NicNic8 This version is clearer. \u2013\u00a0N. F. Taussig May 14 '18 at 2:30\n\u2022 No idea why this isn't the accepted answer - it's the way I'd do it but it's also much clearer than the other. \u2013\u00a0arboviral May 16 '18 at 8:07\n\u2022 @arboviral The other answer has better math, and can be applied generally. Consider instead that there was a fifth glass containing poison; which approach would you use? \u2013\u00a0Drunk Cynic May 16 '18 at 16:16\n\u2022 Think of it this way: What is the prob that the poison cup is still full after drinking three cups? \u2013\u00a0DWin May 17 '18 at 2:23\n\u2022 If there are three glasses of water and two of poison, the method analogous to this answer is you pick two glasses not to drink, and you survive if you happen to pick the two poisoned glasses, which occurs just one way out of the ten possible pairs of glasses you could pick. \u2013\u00a0David K May 20 '18 at 23:09\n\nNot sure why everybody uses such a complicated approach:\nafter drinking three cups, one remains. The chance that she is alive is equal to the chance that the remaining cup is the poison, which is one in four = 25%.\n\nThe sequence of drinking water and or poison is completely irrelevant.\n\n\u2022 Agreed. Part of the SE ethos is to provided detailed additional information that could be used for other, less direct problems. (And to show off) \u2013\u00a0MarsJarsGuitars-n-Chars May 14 '18 at 22:08\n\u2022 This is essentially the same as NicNic8\u2019s solution. \u2013\u00a0wgrenard May 15 '18 at 0:56\n\u2022 From a baysean perspective there is something fishy about the answer. After drinking three cups, she is either dead or alive. Talking about chance is futile! \u2013\u00a0Albert van der Horst May 20 '18 at 18:40\n\u2022 Yes I agree. These top voted answers are a bit extreme it seems, even with my zero experience with any advanced level math and using only my high school school math logic. Yet sometimes you'll see some crazy \"sounding\" problem given by the questioner yet a few simple lines for a response that seems to solve everyone's issues, confusions. Good ol' SE \u2013\u00a0Hunter Frazier May 21 '18 at 10:38\n\nThere are $4!$ permutations of $W_1W_2W_3P$.\n\nThe only way to live is if $P$ is last, and there are $3!$ ways for this to occur.\n\nSo there are $4!-3!$ ways to die with probability $1-\\frac{3!}{4!} = \\frac34$.\n\n\u2022 Just read Taussig's answer. This solution is more or less the same approach. \u2013\u00a0suneater May 14 '18 at 8:35\n\u2022 This solution is more or less the same approach well I would not say so. This one has a different approach / methodology (and ultimately relies on similar concepts). \u2013\u00a0WoJ May 14 '18 at 10:34\n\nA good way to think about such problems is to ask yourself the opposite question: what is the probability that I will not get poisoned?\n\n\\begin{align*} \\Pr(\\text{not poisoned}) &= \\Pr(\\text{not poisoned on first glass}) \\cdot \\Pr(\\text{not poisoned on second glass}) \\cdot \\Pr(\\text{not poisoned on third glass}) \\\\ &= \\frac{3}{4} \\cdot \\frac{2}{3} \\cdot \\frac{1}{2} \\\\ &= \\frac{1}{4} \\\\ \\end{align*}\n\nIt follows that\n\n\\begin{align*} \\Pr(\\text{poisoned}) &= 1 - \\Pr(\\text{not poisoned}) \\\\ &= 1 - \\frac{1}{4} \\\\ &= \\frac{3}{4} \\\\ \\end{align*}\n\n\u2022 \"Ask yourself the opposite question\" is the important lesson here. +1 \u2013\u00a0joeytwiddle May 16 '18 at 5:34\n\u2022 Now if one of them is the antidote... \u2013\u00a0John P May 18 '18 at 16:50\n\nYou have 50%. If you drink the third cup means first and second are not poison. So when drink third you have just two cups, and one is poison.... 50%. I think this problem is about logic more than math... Thanks a Lot\n\n\u2022 This is a valid approach, under a different set of assumptions (namely that the poison is debilitating and acts faster than you can drink another cup) \u2013\u00a0Ben Voigt May 14 '18 at 23:55\n\u2022 This is saying it is more a language issue than a math issue. 'If you were to drink' could mean the action is done or to be done. Interesting. \u2013\u00a0David May 15 '18 at 16:22\n\u2022 There is a revolver handgun with 4 chambers, three are empty and one has a live round. If you put it to your head and squeezed the trigger three times, what is the probability of having a bullet in your head? Your answer addresses the gun question better than the poison question. \u2013\u00a0user1717828 May 15 '18 at 16:36\n\u2022 hahaha! So funny! If I press three times the trigger means first and second was without bullet. Pressing third time maybe I got the bullet in my head, maybe not (I'm not lucky man! so for sure I have!). So 50% probability... \u2013\u00a0German Rodriguez May 15 '18 at 18:41\n\u2022 Just make sure it is a revolver and not an automatic! \u2013\u00a0Eric Lippert May 15 '18 at 19:02\n\nLabel the cups A, B, C, D. Now we can assume WLOG that the cups she drank are A, B and C.\n\nThere are only 4 scenarios according to which cup is the poisoned one. In 3 of the scenarios (poisoned cup = cup A, B, C respectively), she is poisoned. In 1 of the scenarios (poisoned cup = cup D), she is not poisoned. Therefore the probability is 3/4 = 75%.\n\nNot sure why it is any more complicated than that.\n\n\u2022 It isn't any more complicated, as NicNic8 and Aganju already answered. \u2013\u00a0Dronz May 15 '18 at 3:26\n\u2022 It is more complicated if you want to solve very similar problems generally. Laying out all the scenarios and counting them quickly becomes unreasonable for larger numbers of cups with more than 1 left untouched. \u2013\u00a0Matthew Read May 17 '18 at 3:47\n\nThe solution to this problem depends on how fast the poison works. If the poison is slow acting, the previous solutions are OK. But if the poison works instantly, there may be no opportunity to drink three cups.\n\nThe problem states that three cups are taken, so the first two cups can NOT be poison. The third cup is chosen from a set of two with one poison and the other safe. So the probability of survival = probability of poison = 1/2.\n\n\u2022 I think you are wrong. You are not calculating the probability of the first two cups not be poisoned. \u2013\u00a0sawa May 17 '18 at 3:25\n\u2022 @sawa If we know that both (a) the poison acts instantly and (b) you drink 3 cups, the poison cannot have been in either the first 2 (they would have killed you before you drank the third). That means the probability of either of the first two cups being poison is 0. Obviously, this is intentionally misinterpreting the problem for the sake of humor; it's a joke. \u2013\u00a0Matthew Read May 17 '18 at 3:42\n\u2022 I am not saying that your interpretation is wrong. That is not my point. My point is that you are only calculating the conditional probability (that the third one is poison) under the assumption that the first two were not poison. I am saying that you have to calculate the conditional part to get to that situation, which Floris does correctly in my opinion. \u2013\u00a0sawa May 17 '18 at 3:48\n\u2022 @sawa: The probability of surviving until the third cup does not have to be calculated, it is postulated by the problem formulation, therefore 100% (P(x | x) = 1). \u2013\u00a0Ben Voigt May 18 '18 at 3:31\n\u2022 Okay, I understand. \u2013\u00a0sawa May 18 '18 at 4:31\n\nI thought it would be instructive to try a different approach.\n\nIf you drink three cups, that tells me for sure that two of the cups you drank (the first two) were not poisoned - otherwise you would not have gotten to the third cup.\n\nSo the poison must be in either cup 3 or cup 4, and since it is equally likely to be in any of the cups, there is a 50% chance of that happening.\n\nHaving survived the first two cups, you now have a 50-50 chance that cup 3 is poisoned (because it's either cup 3 or cup 4). To survive drinking cup 3, you need to beat those odds as well.\n\nMultiplying these probabilities, you once again get 0.25 for the chance of surviving.\n\nOf course it's the same answer - but I thought this would give additional insight.\n\n\u2022 The OP stated that the person drinking the water was proving that they can't die, so I don't think it's fair to make your conclusion that drinking three cups itself is amount to showing that the first two cups are not poisoned. \u2013\u00a0Green May 16 '18 at 23:45\n\u2022 Your interpretation of the question is the same as richard1941 and German Rodriguez, but I think they gave the wrong answer under that assumption, and you are the only one to give the correct answer under the assumption. \u2013\u00a0sawa May 17 '18 at 3:29\n\u2022 This is not correct; it's almost like the reverse of the Monty Hall misunderstanding. In this version of the problem, the first two cups don't matter. They might as well not exist. The problem is reduced purely to two cups, one of which is poisoned. You drink only one of them -- what are your odds? 50%, not 25%. \u2013\u00a0Matthew Read May 17 '18 at 3:59\n\u2022 Or to fully dive into the scenarios you wanted to lay out here -- there are four situations: (1) cup 3 is poisoned and you drink it (2) cup 4 is poisoned and you drink it (3) cup 3 is poisoned but you drink cup 4 (4) cup 4 is poisoned but you drink cup 3. 2/4 of those are deadly; 50%. \u2013\u00a0Matthew Read May 17 '18 at 4:02\n\u2022 A potentially more intuitive reason it can't be 25%: In this version of the problem, you've gained definitive knowledge of absolute safety about the first two choices made. That MUST affect the odds. It is obviously not true that drinking two safe cups leaves you with the same odds as drinking two cups that could have included the poisoned cup. \u2013\u00a0Matthew Read May 17 '18 at 4:05\n\nWow, people are making this complicated.\n\nThere are 4 cups. One is poisoned. She picks 3. There are 3 chances that she will pick the poisoned cup out of 4. Therefore, the probability is 3/4.\n\nThis assumes that she does not pick a cup, drink it, then put it back and someone refills it before she picks another.\n\nI'm also assuming that she either picks 3 cups before drinking any of them, or that if she dies before picking 3 cups, that we treat that as if she picked enough to fill out the 3 at random. Otherwise the probability is impossible to calculate, because if the first or second cup is poisoned, she doesn't \"pick 3 cups\".\n\nYou can do all the permutations and bayesian sequences, but as others have shown, they all come to the same answer.\n\nIf she picked, say, 3 out of 6 cups and 2 are poisoned, I don't see how to do it other than with combinatorics. But maybe I'm missing an easy way.\n\n\u2022 NicNic8's answer is even more simple, check that top answer. \u2013\u00a0user061703 May 19 '18 at 6:02\n\u2022 For your drinking 3 out of 6 with 2 poisoned, just think of six spots for glasses in a row. The first three are to be drunk. The first poisoned glass has 3 safe locations (the last three) out of 6, so we're safe with probability 0.5. The second poisoned glass has 2 safe locations to land on out of 5, so we'll stay safe with probability 0.4. Combined: 0.2 probability of a safe three drink binge... \u2013\u00a0DJohnM May 19 '18 at 8:12\n\u2022 Ooh, interesting. Yes, simpler than what I had in mind, which was: number of safe combinations divided by total number of combinations = 4C3 / 6C3 = 4 / 20 = 0.2. Same answer. \u2013\u00a0Jay May 19 '18 at 23:17\n\u2022 @Tr\u1ea7nTh\u00facMinhTr\u00ed I though NicNic8's answer was a shade more complicated than mine as it requires reversing the probability, but whatever. Not something that I would challenge someone to a duel to the death over. \u2013\u00a0Jay May 19 '18 at 23:19\n\u2022 So the problem grows as more people think about it. Now we have sampling with replacement as a possibility. The probability of surviving on a single draw is 3/4. Because the non-poisoned cup is replaced, the overall probability of survival, at the beginning of the trial, is 3/4 x 3/4 x 3/4 = 27/64. It does not matter if the poison if fast or slow because the conditional probabilities are the same as the initial probability. \u2013\u00a0richard1941 May 21 '18 at 17:27", "date": "2020-07-10 02:57:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2780138/there-are-4-cups-of-liquid-three-are-water-and-one-is-poison-if-you-were-to-dr/2783140", "openwebmath_score": 0.867522120475769, "openwebmath_perplexity": 676.3144056942763, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969679646669, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8580382245847407}}
{"url": "http://mathhelpforum.com/discrete-math/207036-how-do-i-solve-direct-proof.html", "text": "# Math Help - How do I solve this with a direct proof?\n\n1. ## How do I solve this with a direct proof?\n\nProve that for all sets X and Y, (X \u2229 Y) \u222a(Y-X)=Y\n\nI'm stuck and I'm not fully sure how to solve this.\n\n2. ## Re: How do I solve this with a direct proof?\n\nsuppose that a is an element of (X\u2229Y) U (Y-X).\n\nthis means one of two things:\n\na is in X\u2229Y, or\na is in Y-X\n\nsuppose a is in X\u2229Y. this means that a is in BOTH X and Y, so is certainly in Y.\n\nalternatively, if a is in Y-X, this means a is in Y, but not in X. who cares if its not in X, at least it's in Y!\n\nso in all possible cases, we see that a is in Y.\n\nthis means that (X\u2229Y) U (Y-X) is a subset of Y.\n\nnow suppose b is in Y.\n\nwell we have two possibilities:\n\nb is in X\nb is NOT in X (in or out, that's the way it is with sets. you cannot be \"sort of\" in a set).\n\nif b is in X, then b is in X AND Y, so b is in X\u2229Y.\nif b is not in X, then b is in Y, but not in X, so b is in Y-X.\n\nif we put these two sets together, b is certain to be in one of them. hence b is in (X\u2229Y) U (Y-X).\n\nthus Y is a subset of (X\u2229Y) U (Y-X).\n\nbut, if for 2 sets A,B: if A\u2286B and B\u2286A, then A and B have exactly the same elements, that is: A = B.\n\nso (X\u2229Y) U (Y-X) = Y.\n\n(intuitively what we are doing is splitting Y into 2 parts: the part that overlaps with X, and the part that doesn't).\n\n3. ## Re: How do I solve this with a direct proof?\n\nHello, blueaura94!\n\n$\\text{Prove: }\\:(X \\cap Y) \\cup (Y - X) \\:=\\:Y$\n\n$\\begin{array}{ccccc}1. & (X \\cap Y) \\cup (Y - X) && 1. & \\text{Given} \\\\ 2. & (X \\cap Y) \\cup (Y \\cap \\overline{X}) && 2. & \\text{Def. Subtr'n} \\\\ 3. & (X\\cap Y) \\cup (\\overline{X} \\cap Y) && 3. & \\text{Commutative} \\\\ 4. & (X \\cup \\overline{X}) \\cap Y && 4. & \\text{Distributive} \\\\ 5. & U \\cap Y && 5. & A \\cup \\overline{A} \\,=\\,\\mathbb{U} \\\\ 6. & Y && 6. & \\mathbb{U} \\cap A \\,=\\,A \\end{array}$\n\n4. ## Re: How do I solve this with a direct proof?\n\nOriginally Posted by blueaura94\nProve that for all sets X and Y, (X \u2229 Y) \u222a(Y-X)=Y\nHere is another way.\n$X\\cap Y\\subseteq Y~\\&~Y\\setminus X\\subseteq Y$ so there is just need to prove one way.\n\nIf $t\\in Y$ then either $t\\in X$ or $t\\notin X$.\n\nAnd we are done.", "date": "2015-11-27 03:27:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/207036-how-do-i-solve-direct-proof.html", "openwebmath_score": 0.8684169054031372, "openwebmath_perplexity": 689.1706763152073, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842212767573, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8580290722835068}}
{"url": "https://daviddalpiaz.github.io/stat400sp18/homework/hw02-soln.html", "text": "## Exercise 1\n\nJust before their last year at Anytown High School, the seniors hold the Senior Year Kick-Off party. Sixty percent of the students attending the party are seniors, and the rest (friends, significant others, siblings, etc.) are juniors (25%), sophomores (10%), and freshmen (5%) Unfortunately, drinking is quite common at this party; 90% of the seniors consume alcohol, so do 80% of the juniors, 50% of the sophomores, and 20% of the freshmen.\n\n(a) If a student at this party is drinking, what is the probability that this student is a senior?\n\nSolution:\n\nFirst, we note the given information\n\n\\begin{aligned} P(Fr) &= 0.05, \\ P(D \\mid Fr) = 0.20 \\\\ P(So) &= 0.10, \\ P(D \\mid So) = 0.50 \\\\ P(Jr) &= 0.25, \\ P(D \\mid Jr) = 0.80 \\\\ P(Sr) &= 0.60, \\ P(D \\mid Sr) = 0.90 \\\\ \\end{aligned}\n\nNext, we calculate proportion of the students attending the party consume alcohol.\n\n\\begin{aligned} P(D) &= P(Fr \\cap D) + P(So \\cap D) + P(Jr \\cap D) + P(Sr \\cap D) \\\\ &= P(Fr) \\cdot P(D \\mid Fr) + P(So) \\cdot P(D \\mid So) + P(Jr) \\cdot P(D \\mid Jr) + P(Sr) \\cdot P(D \\mid Sr) \\\\ &= 0.05 \\cdot 0.20 + 0.10 \\cdot 0.50 + 0.25 \\cdot 0.80 + 0.60 \\cdot 0.90 \\\\ &= 0.01 + 0.05 + 0.20 + 0.54 \\\\ &= 0.80 \\end{aligned}\n\nNote that, along the way we obtained the probability of drinking and the intersection with each class.\n\nThen, we finally calculate\n\n$P(Sr \\mid D) = \\frac{P(Sr \\cap D) }{P(D)} = \\frac{0.54}{0.80} = \\boxed{0.675}$\n\n(b) If a student at this party is not drinking, what is the probability that this student is not a senior?\n\nSolution:\n\nBased on the work we did in the previous problem, we can easily \u201ccomplete the table.\u201d (We could have also used a tree to get the information needed for the table.)\n\nFr So Jr Sr\nDrinking 0.01 0.05 0.20 0.54 0.80\nNot Drinking 0.04 0.05 0.05 0.06 0.20\n0.05 0.10 0.25 0.60 1\n\n$P(Sr^\\prime \\mid D^\\prime) = \\frac{P(Sr^\\prime \\cap D^\\prime)}{P(D^\\prime)} = \\frac{0.04 + 0.05 + 0.05}{0.20} = \\boxed{0.70}$\n\n(c) If a student at this party is not a senior, what is the probability that this student is not drinking?\n\nSolution:\n\n$P(D^\\prime \\mid Sr^\\prime) = \\frac{P(D^\\prime \\cap Sr^\\prime)}{P(Sr^\\prime)} = \\frac{0.04 + 0.05 + 0.05}{0.05 + 0.10 + 0.25} = \\boxed{0.35}$\n\n(d) What proportion of the underclassmen (freshmen and sophomores) attending the party consume alcohol?\n\nSolution:\n\n$P(D \\mid (Fr \\cup So)) = \\frac{0.01 + 0.05}{0.05 + 0.10} = \\boxed{0.40}$\n\n(e) The school administration discourages the Senior Year Kick-Off party; the principal of AHS announced that any senior attending the party will receive a week of detention. Of course, drinking is also discouraged. Find the proportion of the students at the party who either are seniors, or consume alcohol, or both.\n\nSolution:\n\n$P(S \\cup D) = P(S) + P(D) - P(S \\cap D) = 0.60 + 0.80 - 0.54 = \\boxed{0.86}$\n\n(f) Are events {a student at the party is a senior} and {a student at the party is drinking} independent? Justify your answer. No credit will be given without proper justification.\n\nSolution:\n\n$$\\boxed{\\text{No.}}$$\n\n$0.54 = P(Sr \\cap D) \\neq P(Sr) \\cdot P(D) = 0.60 \\cdot 0.80 = 0.48$\n\nWe could have also showed that $$P(Sr \\mid D) \\neq P(Sr)$$ or $$P(D \\mid Sr) \\neq P(D)$$.\n\n(g) Are events {a student at the party is a junior} and {a student at the party is drinking} independent? Justify your answer. No credit will be given without proper justification.\n\nSolution:\n\n$$\\boxed{\\text{Yes.}}$$\n\n$0.20 = P(Jr \\cap D) \\neq P(Jr) \\cdot P(D) = 0.25 \\cdot 0.80 = 0.20$\n\n## Exercise 2\n\nA bishop is placed at random (with equal chance) on a chess board (8 x 8). A king of the opposing color is placed at random (with equal chance) on one of the remaining squares. What is the probability that the king is under attack from the bishop?\n\nHint: Placed anywhere on a chess board, a rook attacks 14 squares out of the remaining 63.\n\n$P(\\text{ K is under attack }) = \\frac{14}{63}$\n\nSolution:\n\nWe first need to determine how many squares are being attacked for each possible position of the bishop. For example, in the top left corner, the bishop can attack 7 positions. We also keep track of how many positions on the board have the bishop attacking 7 positions. (And do so for all possibilities.)\n\nThen we use the law of total probability.\n\n\\begin{aligned} P(\\text{K is under attack}) &= P(\\text{B attacks 7}) \\cdot P(\\text{K is under attack} \\mid \\text{B attacks 7}) \\\\ &+ P(\\text{B attacks 9}) \\cdot P(\\text{K is under attack} \\mid \\text{B attacks 9}) \\\\ &+ P(\\text{B attacks 11}) \\cdot P(\\text{K is under attack} \\mid \\text{B attacks 11}) \\\\ &+ P(\\text{B attacks 13}) \\cdot P(\\text{K is under attack} \\mid \\text{B attacks 13}) \\\\ &= \\frac{28}{64} \\cdot \\frac{7}{63} + \\frac{20}{64} \\cdot \\frac{9}{63} + \\frac{12}{64} \\cdot \\frac{11}{63} + \\frac{4}{64} \\cdot \\frac{13}{63} \\\\ &= \\boxed{\\frac{5}{36}} \\end{aligned}\n\n# Exercise 3\n\nYou are given that $$P(A) = 0.5$$ and $$P(A \\cup B) = 0.7$$. Student 1 assumes that $$A$$ and $$B$$ are independent and calculates $$P(B)$$ based on that assumption. Student 2 assumes that $$A$$ and $$B$$ are mutually exclusive and calculates $$P(B)$$ based on that assumption. Find the absolute difference between the two calculations.\n\nSolution:\n\nWe first consider Student 1. Since Student 1 believes that $$A$$ and $$B$$ are independent, we have\n\n$P_1(A \\cap B) = P_1(A) \\cdot P_1(B)$\n\nIt is also always true that\n\n$P_1(A \\cup B) = P_1(A) + P_1(B) - P_1(A \\cap B)$\n\nWith the given probabilities, we have\n\n\\begin{aligned} P_1(A \\cap B) &= 0.5 \\cdot P_1(B) \\\\ 0.7 &= 0.5 + P_1(B) - P_1(A \\cap B) \\end{aligned}\n\nThis is a system of two equation, with two unknowns, so we solve and obtain\n\n$P_1(B) = 0.40$\n\nNow we consider Student 2. Since they assumes that $$A$$ and $$B$$ are mutually exclusive we have\n\n$P_2(A \\cap B) = 0.$\n\nBecause of this, we have\n\n$P_2(A \\cup B) = P_2(A) + P_2(B)$\n\nWith the given probabilities, we have\n\n$0.7 = 0.5 + P_2(B)$\n\nSo we obtain\n\n$P_2(B) = 0.2.$\n\nSo, finally, their absolute difference is\n\n$\\left| P_1(B) - P_2(B)\\right | = 0.4 - 0.2 = \\boxed{0.2}$\n\n# Exercise 4\n\nAlex and David agreed to play a series of tennis games (as many as needed) until one of them wins two games in a row. Alex will serve in the first game, then the serve would alternate game by game between Alex and David. David is a better tennis player; Alex has a 50% chance of winning a game on his serve and only a 20% chance of winning a game if David serves. Assume that all games are independent. Find the probability that Alex is the first one to win two games in a row.\n\nSolution:\n\nFirst we define some notation. We say\n\n$O_S$\n\n$$O$$ is the outcome for Alex, $$W$$ or $$L$$. $$S$$ is the player serving. $$D$$ for David, $$A$$ for Alex.\n\nNow we enumerate the possible outcomes that result in Alex being the first to win two games in a row. Recall that Alex serves first. We can group these into two groups of possibilities.\n\nAlex wins the first game:\n\n\\begin{aligned} & W_{A}W_{D} \\\\ & W_{A}L_{D} W_{A}W_{D} \\\\ & W_{A}L_{D} W_{A}L_{D} W_{A}W_{D} \\\\ & \\ldots \\end{aligned}\n\nDavid wins the first game:\n\n\\begin{aligned} & L_{A}W_{D}W_{A} \\\\ & L_{A}W_{D} L_{A}W_{D}W_{A} \\\\ & L_{A}W_{D} L_{A}W_{D} L_{A}W_{D}W_{A} \\\\ & \\ldots \\end{aligned}\n\nOr, more succinctly:\n\nAlex wins first game: $$(W_{A}L_{D})^k W_{A}W_{D}, \\quad k = 0, 1, 2, 3, \\ldots$$\n\nDavid wins first game: $$L_{A} (W_{D}L_{A})^k W_{D}W_{A}, \\quad k = 0, 1, 2, 3, \\ldots$$\n\nThen, we calculate the required probability by adding the probabilities of each of the outcomes we listed above, since they are all disjoint.\n\n\\begin{aligned} P(\\text{ Alex wins two in a row }) &= P(\\text{ Alex wins two in a row after winning the first }) \\\\[0.5em] &+ P(\\text{ Alex wins two in a row after losing the first }) \\\\[0.5em] &= \\sum_{k = 0}^{\\infty}(0.50 \\cdot 0.80)^k (0.50 \\cdot 0.20) \\\\[0.5em] &+ \\sum_{k = 0}^{\\infty} 0.50 \\cdot (0.20 \\cdot 0.50)^k (0.20 \\cdot 0.50) \\\\[0.5em] &= \\frac{0.50 \\cdot 0.20}{1 - 0.50 \\cdot 0.80} = \\frac{0.50 \\cdot 0.20 \\cdot 0.50}{1 - 0.20 \\cdot 0.50} \\\\[0.5em] &= \\frac{1}{6} + \\frac{1}{18}\\\\[0.5em] &= \\boxed{\\frac{2}{9}} \\end{aligned}", "date": "2020-01-17 19:17:52", "meta": {"domain": "github.io", "url": "https://daviddalpiaz.github.io/stat400sp18/homework/hw02-soln.html", "openwebmath_score": 1.0000076293945312, "openwebmath_perplexity": 1667.997471492026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683502444729, "lm_q2_score": 0.868826777936422, "lm_q1q2_score": 0.8580258277348933}}
{"url": "https://forum.allaboutcircuits.com/threads/finding-frequency-of-the-filter-circuit-from-the-transfer-function.175404/", "text": "# Finding frequency of the filter circuit from the transfer function\n\n#### Wompalamba\n\nJoined Dec 28, 2020\n19\nHi i am trying to extract the cutoff frequency from the transfer function that i derived similar to this derivation of the RC filter. I am currently stuck at the step where the equation is modulus on both side as i am not sure how do i continue from there. I have attached the work i have done so far in the image. Thank you for your time!\n\nEdit: i realized that i posted the title wrongly it should be Finding frequency of the filter circuit from the transfer function and i am not sure how to edit the title\n\nLast edited:\n\n#### ZCochran98\n\nJoined Jul 24, 2018\n105\nBefore I answer your question, I do need to point out a small mistake. You have:\n$Z = \\left(\\frac{1}{\\frac{1}{sC_1}+\\frac{1}{R_3}}\\right)^{-1}$\nWhich is incorrect. You already incorporated the ^-1 with the fraction, so you don't need the ^-1 outside the brackets, and, the sC_1 shouldn't be inverted (1/sC_1 is impedance, so 1/(1/sC_1) = sC_1). So that line should just be:\n$Z = \\frac{1}{sC_1+\\frac{1}{R_3}}$\n\nNow, to answer your question: After you correct that, you have the correct process (Z/Z_total, basically), just the wrong value for Z. Ultimately, after you do all the math, you should get an equation of the form:\n$\\left|\\frac{V_{out}}{V_{in}}\\right| = \\frac{|a|}{\\left|bs+c\\right|}$\nFor some constants a, b, and c which depend on your circuit variables (similar to what you have, but with the corrected value of Z, I can tell you there will be no s in the numerator).\n\nThe frequency for which your filter is designed, in this case, is the pole of this equation: what value of s makes the denominator of the fraction = 0? The frequency corresponding to that s (you correctly wrote $$s = \\omega j = 2\\pi f j$$) is the frequency for which this filter was designed.\n\nHope this helps!\n\n#### Wompalamba\n\nJoined Dec 28, 2020\n19\nThank you for correcting my careless mistake and i appreciate you spending some time to share your knowledge. I have since corrected the issue but still have some doubts as to how to solve the modulus function .I am trying to extract out the frequency to find the cut off frequency at $\\frac{Vout}{Vin}=\\frac{1}{\\sqrt{2}}$ when i put in the fixed component values. Am i right to do the following steps as shown in the image attached?\n\n#### ZCochran98\n\nJoined Jul 24, 2018\n105\nYou are on the right track. One thing to keep in mind: if you have something = 0, then |something| is also = 0. Thus, at the third line down, where you have written \"$$= \\frac{1}{1 + sC_1R_1 + \\frac{R_1}{R_3} + sC_1R_2 + \\frac{R_2}{R_3}}$$,\" you can actually just go ahead and solve that line, as you set the denominator there = 0, then the magnitude of it will also be zero. Doing that, you can solve for s and then for f, subsequently. The next 3 lines after it are just extra work that, for this case, aren't necessary.\n\n#### Wompalamba\n\nJoined Dec 28, 2020\n19\nThank you for the guidance. However, I suspect i might have made a mistake somewhere because when i tried to find the frequency cut off it was not the answer i expected which is 810~kHz i have attached the two images below.\n\n#### The Electrician\n\nJoined Oct 9, 2007\n2,814\nAt the very end you made an algebra mistake. The expression (C1 R1 - C1 R2) should be (C1 R1 + C1 R2).\n\n#### Wompalamba\n\nJoined Dec 28, 2020\n19\nThank you! I corrected the mistake but i am still not getting the desired cut off frequency of 810kHz. Did i accidentally made more mistake in the last step?\n\n#### The Electrician\n\nJoined Oct 9, 2007\n2,814\nWhere did you get the value of 810 kHz? I solved it and got the 895.98 kHz value. This is correct for the component values you have given and for the circuit as shown.\n\n#### ericgibbs\n\nJoined Jan 29, 2010\n11,630\nhi W,\nThis is what LTSpice shows for those component values.\nSlight difference to calculated numeric value, due to accuracy of cursor placement.\nE\n\n#### Attachments\n\n\u2022 58.4 KB Views: 8\n\n#### Wompalamba\n\nJoined Dec 28, 2020\n19\nYou are right. I tried the spice simulation and it shows the correct value. I am assuming this is a weird software issue . I have attached the schematic i was using for the Elsie simulation just in case i did something wrongly.\n\n#### ericgibbs\n\nJoined Jan 29, 2010\n11,630\nhi W,\nDid you also try to calculate the Freq, by summing R1 and R2 and say re-naming as R4, in order to simplify the maths.?\n\nTry it.\nE", "date": "2021-01-24 02:50:33", "meta": {"domain": "allaboutcircuits.com", "url": "https://forum.allaboutcircuits.com/threads/finding-frequency-of-the-filter-circuit-from-the-transfer-function.175404/", "openwebmath_score": 0.7837759852409363, "openwebmath_perplexity": 611.3089253108543, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9621075711974104, "lm_q2_score": 0.8918110562208681, "lm_q1q2_score": 0.8580181692676566}}
{"url": "https://math.stackexchange.com/questions/4073417/symbol-for-very-small-variable", "text": "# Symbol for very small variable\n\nIs there any representation to state that a variable is close to (not equal) zero? Let me give you an example. Consider the function\n\n$$u(x)=\\alpha (e^{i\\omega \\delta t}-1) f(x)$$\n\nI am interested in the function $$u(x)$$ when $$\\delta t$$ is very small. For this case, it should be easy to see that\n\n$$u(x)|_{\\text{small }\\delta t} \\approx i \\alpha \\omega \\delta t f(x)$$\n\nIs there any \"nice\" notation to represent such an equation (without having to write small)? I thought that I could use the limit notation for that [for example, $$\\lim_{\\delta t \\to 0} u(t)$$], but then I realized that if $$\\delta t$$ goes to zero, then $$u(x)=0$$. Therefore, it is not what I need.\n\nI found this link in the same forum, but it did not help.\n\n\u2022 Really, what's wrong with $\\delta t<<1$? Mar 23 at 15:44\n\u2022 $\\delta t\\ll 1$ is good Mar 23 at 15:55\n\u2022 I think it is okay. Nevertheless, I would need to use something like $|\\delta t| << 1$ to make it clear that $\\delta t$ is positive?\n\u2013\u00a0jeb\nMar 23 at 16:31\n\nI think you are asking about the first order (linear) approximation using the derivative.\n\nYou might say $$u(x)= i \\alpha \\omega \\delta t f(x) + o(\\delta t ).$$\n\nThe fact that this is an approximation for small $$\\delta t$$ should be clear to your reader. If not you can say so.\n\n\u2022 I like to \"o\" notation, I have to say. What would you recommend me to include in my text or even in front of the $u(x)$ term to make it clear that $\\delta t$ is small? My question is based on the fact that I will need to manage this equation further, and if I include the \"o\" notation, they will become too overloaded. The best that I could think of was something like $u(x)|_{\\delta t = \\epsilon}$, such that $o(\\epsilon)$ is insignificant (or can be disregarded). Thank you.\n\u2013\u00a0jeb\nMar 23 at 18:41\n\u2022 The insignifance to first order is implicit in the use of the $o$ notation. You can just carry the $o(\\delta t)$ term along in future calculations and throw it away at the end (r sooner) as appropriate. You could add \"as $\\delta t \\to 0$\" when you first state the equation. I would not want to read an invented notation like the one you propose. Mar 23 at 19:01\n\nThe issue with taking limits is that as $$\\delta t\\to 0$$ both sides go to zero, but what you want to say is stronger than that.\nI imagine what you actually mean by $$u(x)\\approx i\\alpha\\omega\\delta tf(x)$$ can be formalised as $$\\lim_{\\delta t\\to 0}\\frac{u(x)}{\\delta t}=i\\alpha\\omega f(x).$$\n\nThis is simply a matter of rearranging so that the limit captures the relative error of the approximation, not just the absolute error.", "date": "2021-10-26 08:48:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4073417/symbol-for-very-small-variable", "openwebmath_score": 0.8740401268005371, "openwebmath_perplexity": 141.8977635010908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464478051828, "lm_q2_score": 0.8791467690927439, "lm_q1q2_score": 0.8580001663954667}}
{"url": "https://math.stackexchange.com/questions/945138/proving-that-x-left-lfloor-fracx2-right-rfloor-left-lfloor-frac", "text": "# Proving that $x-\\left\\lfloor \\frac{x}{2} \\right\\rfloor\\; =\\; \\left\\lfloor \\frac{x+1}{2} \\right\\rfloor$\n\nHow would you prove that if $x$ is an integer, then\n\n$$x-\\left\\lfloor \\frac{x}{2} \\right\\rfloor\\; =\\; \\left\\lfloor \\frac{x+1}{2} \\right\\rfloor$$\n\nI tried to start by saying that if $x$ is an even integer, then:\n\n$$\\left\\lfloor \\frac{x}{2} \\right\\rfloor = \\frac{x}{2}.$$\n\nHowever, I am stuck on showing that $$\\left\\lfloor \\frac{x+1}{2} \\right\\rfloor$$ is also $\\frac{x}{2}$. Intuitively It makes intuitive sense just by playing around with some sample numbers, but I don't know how to make it mathematically rigorous. Further, what do you do in the odd case? Is this even the right way to go about it?\n\n\u2022 Start with $\\lfloor y+n \\rfloor = \\lfloor y\\rfloor +n$. \u2013\u00a0lhf Sep 25 '14 at 1:53\n\u2022 @lhf Is there a proof for that? \u2013\u00a01110101001 Sep 25 '14 at 1:56\n\u2022 By definition, $\\lfloor y \\rfloor \\le y < \\lfloor y \\rfloor +1$ and so $\\lfloor y \\rfloor + n \\le y + n < \\lfloor y \\rfloor +n+1$, which says that $\\lfloor y+n \\rfloor = \\lfloor y\\rfloor +n$. \u2013\u00a0lhf Sep 25 '14 at 1:58\n\nIf $x$ is even, then $x=2k$ where $k \\in \\mathbb{Z}$.\n\nWe then have: $$x-\\left\\lfloor \\frac{x}{2} \\right\\rfloor=2k-\\lfloor k \\rfloor=k=\\left\\lfloor k+\\frac{1}{2} \\right\\rfloor=\\left\\lfloor \\frac{2k+1}{2} \\right\\rfloor=\\left\\lfloor \\frac{x+1}{2} \\right\\rfloor$$\n\nIf $x$ is odd, then $x=2k+1$ where $k \\in \\mathbb{Z}$.\n\nWe then have $$x-\\left\\lfloor \\frac{x}{2} \\right\\rfloor=2k+1-\\left\\lfloor k+{1\\over 2} \\right\\rfloor=k+1=\\left\\lfloor k+1 \\right\\rfloor=\\left\\lfloor \\frac{2k+2}{2} \\right\\rfloor=\\left\\lfloor \\frac{x+1}{2} \\right\\rfloor$$\n\nCase-1 (Even) write $x=2k$. Now$\\left\\lfloor \\frac{x}{2} \\right\\rfloor =k$ so L.H.S is $2k-k=k$. now work with R.H.S, $\\left\\lfloor \\frac{x+1}{2} \\right\\rfloor=\\left\\lfloor \\frac{2k+1}{2} \\right\\rfloor=\\left\\lfloor k.5 \\right\\rfloor=k$.\n\nCase 2) (Odd) Let $x=2k+1$ Similarly. I am in hurry, so i leave it on you. very easy.\n\nProve LHS=RHS= $k+1$", "date": "2020-11-30 11:22:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/945138/proving-that-x-left-lfloor-fracx2-right-rfloor-left-lfloor-frac", "openwebmath_score": 0.9267817139625549, "openwebmath_perplexity": 206.49783456375567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464499040094, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8580001636052454}}
{"url": "https://www.freemathhelp.com/forum/threads/probability-question-on-drawing-cards.131372/#post-545785", "text": "Probability question on drawing cards\n\nmicky123\n\nNew member\nFrom a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?\n\nis the answer: (3/52)*(2/51)*(1/50)?\n\nSubhotosh Khan\n\nSuper Moderator\nStaff member\nFrom a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?\n\nis the answer: (3/52)*(2/51)*(1/50)?\nThere are 4 cards (ACE included) with pictures AND heart.\n\nHow many ways can you draw 3 cards from that group ?\n\nHow many ways can you draw 3 cards from 52 cards ?\n\nmicky123\n\nNew member\nWhy would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)\n\nPlease kindly explain if I have gotten it wrong! Thank you.\n\nSubhotosh Khan\n\nSuper Moderator\nStaff member\nWhy would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)\n\nPlease kindly explain if I have gotten it wrong! Thank you.\nIf ACE is not included then - how many ways can you choose 3 cards from that group (Hearts and pictures) ? ------ 1 way\n\nHow many ways can you choose 3 cards (no restriction) from the deck? _____ $$\\displaystyle C^{52}_{3}$$\n\nNow calculate probability.\n\nreference: https://math.stackexchange.com/ques...-getting-3-cards-in-the-same-suit-from-a-deck\n\nmicky123\n\nNew member\nIf ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?\n\nAm I understanding the event incorrectly? I understood the event that \u201call three cards are both heart and picture\u201d to be drawing \u201ca jack, a queen and a king of hearts\u201d, which is 1 out of 22100 ways.\n\npka\n\nElite Member\nIf ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?\n\nAm I understanding the event incorrectly? I understood the event that \u201call three cards are both heart and picture\u201d to be drawing \u201ca jack, a queen and a king of hearts\u201d, which is 1 out of 22100 ways.\nI would agree with your first answer [imath]\\left(\\dfrac{3}{52}\\right)\\cdot\\left(\\dfrac{2}{51}\\right)\\cdot\\left(\\dfrac{1}{50}\\right)[/imath].\nBUT if I were reading this question as an editor I would object to its wording as being unclear. It is extremely unusual to use picture card what standard English uses face cards.\n\nmmm4444bot\n\nSuper Moderator\nStaff member\nIt is extremely unusual to use picture card what standard English uses face cards.\nYour wikipedia link states that names 'picture card' and 'face card' mean the same thing.\n\nOtis\n\nElite Member\nconfused as to why there are 4 cards with pictures and heart?\nBecause a lot of card manufacturers draw a picture on each ace. However, the design on the aces do not show a person.\n\nThe name 'picture card' refers only to cards containing a picture of a person.\n\nmicky123\n\nNew member\nI see. So this means that it depends on the way \u201cpicture card\u201d is defined in the question?\n\nif ace is not considered a picture card, the ans would be (3/52)*(2/51)*(1/50).\n\nif ace is considered a picture card, the and would be (4/51)*(3/51)*(2/50).\n\nam i right?", "date": "2021-12-03 02:19:03", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/probability-question-on-drawing-cards.131372/#post-545785", "openwebmath_score": 0.3751796782016754, "openwebmath_perplexity": 1074.398076257034, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9759464471055739, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.8580001565096119}}
{"url": "http://www.kineticmediaworks.com/9w3bmzg/c7b28f-latex-equation-multiple-lines", "text": "# latex equation multiple lines\n\nIf you wanted to just split in two parts, then multlined (notice the extra d) from the mathtools package would be a simpler solution: All the above works with elsarticle class in your updated question. Latex equation in multiple lines. \u00e2\u0080\u0093 Charlie Jan 16 '14 at 3:38 You need to specify an alignment point on each line with & and separate lines with \\\\. 3 Multiple Lines of Displayed Maths . The breqn package is designed to split long equations automatically. the first version becomes: Not sure what you tried doing with multine but this seems OK: Btw, you seem to be missing a bracket on the RHS -- and I deleted an extraneous comma after the . To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Latex allows two writing modes for mathematical expressions. The command \\overline and \\underline places a line above or below the expression. The % sign tells LATEX L A T E X to ignore the rest of the current line. To learn more, see our tips on writing great answers. Here also, the amsmath package is required. The amsmath package provides the align and align* environments for aligned equations. As previously mentioned, Origin\u00e2\u0080\u0099s LaTeX App wraps user-entered equation markup in a single equation environment by automatically adding and commands before and after user-entered markup. Grouping and centering equations. LaTeX Error: \\begin{document} ended by \\end{multline}.See the LaTeX manual or LaTeX Companion for explanation.Type H for immediate help.... \\end{multline}, Though i get the equation broken it is not numbered, or has a one line distance from my text as the The starred version doesn't number the equations. I am attaching a screenshot of the result: How would one justify public funding for non-STEM (or unprofitable) college majors to a non college educated taxpayer? Use instead the equation - just the multline environment, you can work as usual with line breaks. TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. Yes you can load several different packages. You need to use this Feature, the amsmath-package. As a test, when the 2nd backslash was deleted at the end of the 3rd line of code (\\dot{x} & = \\sigma(y-x) \\), it produced the single line equation you provided as an example. Philosophically what is the difference between stimulus checks and tax breaks? Again, use * to toggle the equation numbering. I suggest you use a split environment, which in contrast to multline may be used a subenvironment of equation. LATEX doesn\u00e2\u0080\u0099t break long equations to make them \u00ef\u00ac\u0081t within the margins as it does with normal text. In that version I was able to open an Equation Editor and enter multiple lines of math equations in a the editor, pressing return at the end of each line to move to the next line. I copied and pasted their example exactly as is, but i still get errors, does it have to do with the package i am using? Try the example on the right which sets the same multiple equations in several ways. My code compiles and, looking at it, it doesn't contain, i will try it as well, i was fortunate to get many helpful answers. Is my Connection is really encrypted through vpn? I am a new Latex user,I have loaded these two math equation in my Latex documents and i want to split an equation i have into multiple line, I am reading this topic for breaking the lines over multiple, although what ever i do it does not allow me to compile. What location in Europe is known for its pipe organs? I want to place them/break the equation in two lines and retain the format of numbering and equal placing beneath. How to answer a reviewer asking for the methodology code of the paper? If you want to create a document with more than two columns, use the package multicol, which has a set of commands for the same. It works very well in the majority of situations, but it's not as mature as the amsmath package. @PaulGessler I had a look in the document, thank you for the link. Mathematical modes LaTeX allows two writing modes for mathematical expressions: the inline mode and the display mode. \\begin{eqnarray*} Therefore, special environments have been declared for this purpose. In another practical tip we show you how to in your, LaTeX: equation in multiple rows - how to, d^2 = a^2 + b^2 + c^2 \\geq a^2 + b^2 = a^2 + b^2 + 2ab - 2ab = (a+b)^2 - 2ab \\\\\\ \\geq 2ab. Latex Support In Pages Macs Chemistry. Referencing subordinate equations can be done using either of two methods: adding a label after the \\begin {subequations} command, viz. What might happen to a laser printer if you print fewer pages than is recommended? 2). The { and } characters are used to surround multiple characters that LATEX L A T E X should treat as a single character. Asking for help, clarification, or responding to other answers. E.g. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. As shown, it is possible to add both labels in case \u00e2\u0080\u00a6 Add Mathematical Equations In Keynote On Mac Apple Support. LUXCO NEWS. You need to specify an alignment point on each line with & and separate lines with \\\\. Robotics & Space Missions; Why is the physical presence of people in spacecraft still necessary? A line end with [\\\\\\\\]. LaTeX equation editing supports most of the common LaTeX mathematical keywords. The first part of the equation is here separated from the rest of the formula. You have to put up with [\\\\\\\\] a line break at the desired location in your equation. It only takes a minute to sign up. Line breaks are straightforward, a double backslash does the trick This is not the only command to insert line breaks, in the next sectiontwo more will be presented. View PDF \u00e2\u0080\u00ba\u00e2\u0080\u00ba Find news, promotions, and other information pertaining to our diverse lineup of innovative brands as well as newsworthy headlines about our company and culture. Figure 1: Typical equation problems in ordinary LATEX: (a) di erent spacing around the equals signs in (14) and (15) because one uses equationand the other uses eqnarray; (b) equation (15) is a single equation but because it covers two lines \\nonumbermust be used on the rst line; (c) and then the If a coworker is mean to me, and I do not want to talk to them, is it harrasment for me not to talk to them? They can be distinguished into two categories depending on how they are presented: 1. text \u00e2\u0080\u0094 text formulas are displayed inline, that is, within the body of text where it is declared, for example, I can say that a + a = 2 a {\\displaystyle a+a=2a} within this sentence. site design / logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Equations. How is HTTPS protected against MITM attacks by other countries? Description below mirrored brace. LaTeX needs to know when text is mathematical. It is therefore up to you to format the equation appropriately (if they overrun the margin.) 3 Special Characterulti Line Equations. This is because LaTeX typesets maths notation differently from normal text. Usually, the eqnarray environment is only used if authors cannot use the amsmath package because the environments included in this package are easier to use (refer to Section . thank you very much, although if i may ask can i ask if multiple packages can be used. Equations on Multiple Lines. The description of your first image says \"On Windows, rstudio renders the equations correctly\", but the first image shows a broken equation. That is why I was confused. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. With the split environment, you can display long equations clearly. Schematically, this is what its supposed to look like: As with the tabular environment, use & to separate columns and \\\\ to separate rows. The Moodle XML file can then be used to bu Auto Latex Equations Google Workspace Marketplace. In this case the first line should be move left relative to the others and the package mathtools provides a convenient command for this: I have split across three lines for clarity. This is best used at the start of a line to mark that line as a \u00e2\u0080\u009ccomment\u00e2\u0080\u009d. When numbering is allowed, you can label each row individually. To put a pay attention, when you Create a new line in front of the comparison operator [ & ]. If an equation is longer than one line or several formulas must be grouped together, the eqnarray environment could be used. Contents 1 Introduction 2 Including the amsmath package 3 Writing a single equation 4 Displaying long equations We will show you how you can format the equation-environment is still suitable. That can be achieve in plain LaTeX without any specific package. two \u00e2\u0080\u009ccolumns\u00e2\u0080\u009d, one with the equation, one with the annotations. How to start equation environment inside a custom environment? You can choose the layout that better suits your document, even if the equations are really long, or if you have to include several equations in the same line. The first one is used to write formulas that are part of a text. formulas, graphs). 9. I suggest you use a split environment, which in contrast to multline may be used a subenvironment of equation. In fact, your example is probably best with the cases environment. There are several ways to format multiple equations and the amsmath package adds several more. In this case the first line should be move left relative to the others and the package mathtools provides a convenient command for this: Andrew,@Andrew,i used \\begin{multline} after the , i pasted your solution and i get an error which i have given in the question as a new edit. The amsmath package provides a handful of options for displaying equations. To make use of the inline math feature, simply write your text and if you need to typeset a single math symbol or formula, surround it with dollar signs:Output equation: This formula\u00a0f(x)=x2\u00a0is an example.This formula\u00a0f(x)=x2\u00a0is an example. Find all positive integer solutions for the following equation: Allow bash script to be run as root, but not sudo. Also tried this example as kindly suggested. Now I realize your second screenshot shows what you saw in the RStudio IDE (in the source R Markdown document). You have to put up with [\\\\\\\\] a line break at the desired location in your equation. Subscribe to our newsletter to get notification about new updates, information, etc.. LaTeX forum \u00e2\u0087\u0092 Math & Science \u00e2\u0087\u0092 amsmath | Correct Alignment for multi-line Equations Topic is solved Information and discussion about LaTeX's math and science related features (e.g. Art Of Problem Solving. \\label {eq:Maxwell}, which will reference the main equation (1.1 above), or adding a label at the end of each line, before the \\\\ command, which will reference the sub-equation (1.1a or 1.1b above). Math equation in LaTeX provides three stretchable lines/arrows that appear above or below the equation: braces, bars and arrows. Missing $inserted.$ ...l N(\\sigma;\\lambda;\\theta;t)}{\\partial t} over multiple lines, @PaulGessler although as far as I can see the give s me a better visual result, that is why i am keen of using that one. Open an example of the amsmath package in Overleaf First of all, you probably don't want the align environment if you have only one column of equations. I am using the package math of \\usepackage{amssymb} \\usepackage{amsthm}, @George I posted the self-contained code example above so that you could see one of doing this that works. With a normal line break, you can't write in LaTeX equations multiple lines. The next lines contain only the equations and annotations, I would like them aligned with the equation and annotation of the second line. These environments provide pairs of left- and right-aligned columns. I am trying to align a set of long equations, that are themselves align environments as most of them are spreading on multiple lines.. If we have two lines, then _{xxx} should be at the end of the first line (or at the start of the second line). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. LaTeX assumes that each equation consists of two parts separated by a &; also that each equation is separated from the one before by an &. Understanding the zero current in a simple circuit. rev\u00a02020.12.18.38240, The best answers are voted up and rise to the top, TeX - LaTeX Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, @PaulGessler I used it alone but it gave more errors in the compilers most of them saying ! In general, the command \\\\ signifies a line break and within the correct math mode environment, it can start a new equation line. Here also, the amsmath package is required. thank you for your time, Podcast 300: Welcome to 2021 with Joel Spolsky, Split numbered equation over multiple pages, Defining new split environment with reduced spacing, Can't generate png with Error: Erroneous nesting of equation structures, Formula too long and \\split fails | contains \u201csqrt\u201d (square root). Making statements based on opinion; back them up with references or personal experience. The \\overbrace command places a brace above the expression (or variables) and the command \\underbrace places a brace below the expression. Socially oriented website which will help to solve your little (or not little) technical problems. I am trying to learn Latex at the same time , thank you for your patience. I think the use of multiple lines structured environment can work. Check out what we are up to! Currently I just have a sequence of align environments, with each equation inside in order to align the pieces of each equations. Then there is an equation and some conditions for the equation, i.e. \\end{math} \\end{document} This code produces something which looks what you seems to need. Latex equation in multiple lines. The placement of _{xxx} depends on the number of lines that underbrace symbol should occupy, and in general it should be somewhere in the middle. 9 3 Multiple Lines Of Displayed Maths. LaTeX Long equation in several lines Use instead the equation - just the multline environment, you can work as usual with line breaks. Anyway, I'm glad to know that removing the extra empty lines worked. Unlike the tabular environment, there is no argument as the \u00e2\u0080\u00a6 Adding right brace and equation number. I'm short of required experience by 10 days and the company's online portal won't accept my application, Identify Episode: Anti-social people given mark on forehead and then treated as invisible by society. ! The second one is used to write expressions that are not part of a text or paragraph, and are therefore put on separate lines. Open an example in Overleaf Is Mr. Biden the first to create an \"Office of the President-Elect\" set? Relationship between Cholesky decomposition and matrix inversion? \\documentclass{article} \\begin{document} This is your only binary choices \\begin{math} \\left\\{ \\begin{array}{l} 0\\\\ 1 \\end{array} \\right. Can a smartphone light meter app be used for 120 format cameras? This typically requires some creative use of an eqnarrayto get elements shifted to a new line to align nicely. \\begin{eqnarray} y &=& x^4 + 4 \\nonumber \\\\ &=& (x^2+2)^2 -4x^2 \\nonumber \\\\ &\\le&(x^2+2)^2 \\end{eqnarray} \\begin{eqnarray*} e^x &\\approx& 1+x+x^2/2! Post by Alirezakn \u00bb Thu Nov 22, 2018 7:02 am . If an equation is longer than one line or several formulas must be grouped together, the eqnarray environment could be used. If we have three lines then _{xxx} should be in the middle of the second line. Thanks for contributing an answer to TeX - LaTeX Stack Exchange! That can be achieve in plain LaTeX without any specific package. Aligning Multiline Equation To The Left With Only One. Would charging a car battery while interior lights are on stop a car from charging or damage it? Also that each equation is separated from the one before by an. When numbering is allowed you can label each row individually. Multiple Lines and Multiple Equations. How To Linebreak A Equation In Latex Dealing Long You. Also worked properly when code was typed in. I ask if multiple packages can be used a subenvironment of equation agree to our newsletter to notification. To mark that line as a single character, or responding to other answers and align * environments for equations. Structured environment can work as usual with line breaks LaTeX equation editing supports most of the:. 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Which sets the same multiple equations in Keynote on Mac Apple Support writing modes mathematical. Happen to a new line to mark that line as a single character right-aligned columns n't want align! Site for users of TeX, LaTeX, ConTeXt, and related typesetting systems 120 format cameras first of! You very much, although if i may ask can i ask if multiple packages can be achieve in LaTeX! Xml file can then be used stop a car from charging or damage it LaTeX, ConTeXt, and typesetting... You agree to our terms of service, privacy policy and cookie.... Pay attention, when you Create a new line in front of the formula must... Labels in case \u00e2\u0080\u00a6 9, the amsmath-package to subscribe to our terms service... Clarification, or responding to other answers agree to our newsletter to get notification about new updates information! The { and } characters are used to surround multiple characters that LaTeX L a E... It works very well in the RStudio IDE ( in the middle of the comparison operator [ & ] equations. Provides the align and align * environments for aligned equations an example in Overleaf then there an. Formulas must be grouped together, the eqnarray environment could be used equation and some for... Used a subenvironment of equation an example in Overleaf then there is an equation is here separated from one... Line breaks Mac Apple Support them \u00ef\u00ac\u0081t within the margins as it does with text! Equation in two lines and retain the format of numbering and equal placing beneath LaTeX equations lines! Or damage it work as usual with line breaks been declared for this purpose references personal... Equation inside in order to align nicely am trying to learn more, see our tips writing! The result: that can be achieve in plain LaTeX without any specific package screenshot of the President-Elect set! New line to align the pieces of each equations surround multiple characters that LaTeX L a T X! Other countries an  Office of the result: that can be used column equations... Based on opinion ; back them up with [ \\\\\\\\ ] a line above or below the equation in lines...", "date": "2021-02-25 18:46:23", "meta": {"domain": "kineticmediaworks.com", "url": "http://www.kineticmediaworks.com/9w3bmzg/c7b28f-latex-equation-multiple-lines", "openwebmath_score": 0.8269694447517395, "openwebmath_perplexity": 1662.1740442029873, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.935346511643776, "lm_q2_score": 0.9173026505426831, "lm_q1q2_score": 0.8579958343066884}}
{"url": "https://www.freemathhelp.com/forum/threads/probability-question-on-drawing-cards.131372/#post-545800", "text": "# Probability question on drawing cards\n\n#### micky123\n\n##### New member\nFrom a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nFrom a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?\n\nThere are 4 cards (ACE included) with pictures AND heart.\n\nHow many ways can you draw 3 cards from that group ?\n\nHow many ways can you draw 3 cards from 52 cards ?\n\n#### micky123\n\n##### New member\nWhy would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)\n\nPlease kindly explain if I have gotten it wrong! Thank you.\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nWhy would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)\n\nPlease kindly explain if I have gotten it wrong! Thank you.\nIf ACE is not included then - how many ways can you choose 3 cards from that group (Hearts and pictures) ? ------ 1 way\n\nHow many ways can you choose 3 cards (no restriction) from the deck? _____ $$\\displaystyle C^{52}_{3}$$\n\nNow calculate probability.\n\nreference: https://math.stackexchange.com/ques...-getting-3-cards-in-the-same-suit-from-a-deck\n\n#### micky123\n\n##### New member\nIf ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?\n\nAm I understanding the event incorrectly? I understood the event that \u201call three cards are both heart and picture\u201d to be drawing \u201ca jack, a queen and a king of hearts\u201d, which is 1 out of 22100 ways.\n\n#### pka\n\n##### Elite Member\nIf ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?\n\nAm I understanding the event incorrectly? I understood the event that \u201call three cards are both heart and picture\u201d to be drawing \u201ca jack, a queen and a king of hearts\u201d, which is 1 out of 22100 ways.\nBUT if I were reading this question as an editor I would object to its wording as being unclear. It is extremely unusual to use picture card what standard English uses face cards.\n\n#### mmm4444bot\n\n##### Super Moderator\nStaff member\nIt is extremely unusual to use picture card what standard English uses face cards.\nYour wikipedia link states that names 'picture card' and 'face card' mean the same thing.\n\n#### Otis\n\n##### Elite Member\nconfused as to why there are 4 cards with pictures and heart?\nBecause a lot of card manufacturers draw a picture on each ace. However, the design on the aces do not show a person.\n\nThe name 'picture card' refers only to cards containing a picture of a person.\n\n#### micky123\n\n##### New member\nI see. So this means that it depends on the way \u201cpicture card\u201d is defined in the question?\n\nif ace is not considered a picture card, the ans would be (3/52)*(2/51)*(1/50).\n\nif ace is considered a picture card, the and would be (4/51)*(3/51)*(2/50).\n\nam i right?", "date": "2021-10-25 16:41:31", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/probability-question-on-drawing-cards.131372/#post-545800", "openwebmath_score": 0.2772134840488434, "openwebmath_perplexity": 1354.7472821015594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9653811631528337, "lm_q2_score": 0.888758786126321, "lm_q1q2_score": 0.8579909907129283}}
{"url": "http://namescapes.us/cad-iwkgj/e90175-what-is-bijective-function", "text": "The figure shown below represents a one to one and onto or bijective function. Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. Each value of the output set is connected to the input set, and each output value is connected to only one input value. A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. How to Prove a Function is Bijective without Using Arrow Diagram ? A function that is both One to One and Onto is called Bijective function. Hence every bijection is invertible. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. $$Now this function is bijective and can be inverted. Infinitely Many. If it crosses more than once it is still a valid curve, but is not a function. So we can calculate the range of the sine function, namely the interval [-1, 1], and then define a third function:$$ \\sin^*: \\big[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\big] \\to [-1, 1]. The inverse is conventionally called $\\arcsin$. The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. As pointed out by M. Winter, the converse is not true. Ah!...The beautiful invertable functions... Today we present... ta ta ta taaaann....the bijective functions! Below is a visual description of Definition 12.4. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. A bijective function is both injective and surjective, thus it is (at the very least) injective. And I can write such that, like that. Thus, if you tell me that a function is bijective, I know that every element in B is \u201chit\u201d by some element in A (due to surjectivity), and that it is \u201chit\u201d by only one element in A (due to injectivity). This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read \"one-to-one correspondence).. Definition: A function is bijective if it is both injective and surjective. My examples have just a few values, but functions usually work on sets with infinitely many elements. Mathematical Functions in Python - Special Functions and Constants; Difference between regular functions and arrow functions in JavaScript; Python startswith() and endswidth() functions; Hash Functions and Hash Tables; Python maketrans() and translate() functions; Date and Time Functions in DBMS; Ceil and floor functions in C++ And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. Stated in concise mathematical notation, a function f: X \u2192 Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). A function is invertible if and only if it is a bijection. In mathematics, a bijective function or bijection is a function f : A \u2192 B that is both an injection and a surjection. 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It crosses more than once it is a bijection read injective, surjective and bijective is a function both... One input value mathematics, a bijective function or bijection is a bijection!... the beautiful invertable...... Find out more you can read injective, surjective and bijective and onto or bijective function is if... Such that, like that that is both an injection and a surjection functions that have inverse functions said... Definition: a \u2192 B that is both an injection and a surjection converse not!", "date": "2021-02-27 13:00:10", "meta": {"domain": "namescapes.us", "url": "http://namescapes.us/cad-iwkgj/e90175-what-is-bijective-function", "openwebmath_score": 0.8679229617118835, "openwebmath_perplexity": 446.2498104846663, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676487350494, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8579801940168429}}
{"url": "https://math.stackexchange.com/questions/1631704/inside-an-not-equilateral-triangle-what-is-the-sum-of-distances-from-a-random-po/1631732", "text": "# Inside an not Equilateral Triangle what is the sum of distances from a random point to 3 sides\n\nGiven an not Equilateral Triangle with following side sizes: 45,60,75. Find a sum of distances from a random located point inside a triangle to its three sides.\n\nNote 1: Viviani's theorem related only to equilateral triangles.\n\nNote 2: Fermat point is related to the minimization of distances from a random point inside the triangle and its vertices.\n\nAs we can see that both notes are not helpful to solve that problem.\n\nI have been given that puzzle during an hour an a half exam. There were only 6 minutes to solve that problem. Afters many hours I still do not have an answer. I will be very glad to get some assistance or maybe the whole solution\n\nRegards, Dany B.\n\n\u2022 Note that this particular triangle is rectangular. \u2013\u00a0Justpassingby Jan 29 '16 at 8:28\n\u2022 Is the idea that we should assume a uniform probability distribution and compute the expected value? Or do you just want a formula for the distances, given coordinates of a point? \u2013\u00a0Justpassingby Jan 29 '16 at 8:31\n\u2022 indeed this is rectangular. I didn't pay attention to that earlier. \u2013\u00a0danybutvinik Jan 29 '16 at 8:47\n\u2022 we should assume uniform probability distribution \u2013\u00a0danybutvinik Jan 29 '16 at 8:47\n\nI understand that a point $P$ is chosen at random inside a triangle $ABC$ according to a uniform probability distribution, and you want the expected value of the sum of the distances from $P$ to the sides of the triangle.\n\nThe distance from $P = (x,y)$ to one of the sides is a linear function $ax + by + c$ of the coordinates $x, y$. Thus the sum of the distances is also linear. Therefore the average value is the average of the values for $P = A$, $P = B$ and $P = C$, i.e. the average of the three altitudes of the triangle.\n\nIn the present case the altitudes are $36, 45, 60$. So the expected value is $47$.\n\n\u2022 I have some questions: \u2013\u00a0danybutvinik Jan 29 '16 at 9:49\n\u2022 1. given three sides of the triangle - 75,60,45. What do you mean by referring to 36,45,60 as altitudes ? 2. Where 36 came from ? Thanks \u2013\u00a0danybutvinik Jan 29 '16 at 9:51\n\u2022 In a triangle $ABC$, the altitude $h_a$ from $A$ is the distance from $A$ to side $BC$. Because the area $S$ of the triangle is the same no matter which base you choose, $ah_a = bh_b = ch_c = 2S$. Your triangle is a right triangle, so the legs are altitudes, i.e. $h_a = b, h_b = a$. The other altitude $h_c$ is found by computing $ab/c$. $45 \\times 60/ 75 = 36$. \u2013\u00a0David Jan 29 '16 at 17:23\n\u2022 Hi @David, I understand that the sum of the distances is also linear, but how do your inference go to the next sentence? (Therefore the average value is the average of the values for P=A, P=B and P=C) \u2013\u00a0David Chen Mar 23 '18 at 4:04\n\nGiven a rectangular triangle with sides $a\\leq b\\leq c$ we will compute the sum of the distances for an arbitrary interior point. Choose orthonormal coordinates such that the hypotenuse goes through the origin and the other sides are horizontal or vertical. Let $a$ be the height of the vertical side. All interior points have positive coordinates so we can avoid absolute value signs.\n\nThe distance from $p(x,y)$ to the horizontal side is $y.$ The distance to the vertical side is $b-x.$ The distance to the hypotenuse $H\\leftrightarrow \\frac{a}{b}x-y=0$ is\n\n$$\\frac{\\frac{a}{b}x-y}{\\sqrt{\\left(\\frac{a}{b}\\right)^2+1}}=\\frac{ax-by}{c}$$\n\nwhere we have used $a^2+b^2=c^2.$\n\nThe sum of the distances is\n\n$$s(x,y)=y+4-x+\\frac{ax-by}{c}=\\left(\\frac{a}{c}-1\\right)x+\\left(1-\\frac{b}{c}\\right)y+b.$$\n\nIntegration gives\n\n$$\\int_{x=0}^b\\int_{y=0}^{\\frac{ax}b}s(x,y)dy\\ dx=\\frac{ab}6\\left(a+b+\\frac{ab}c\\right).$$\n\nThe expected value of the sum of the distances is obtained after dividing this integral by the surface area $ab/2,$ giving\n\n$$\\overline{s}=\\frac{a+b+\\frac{ab}c}3=47.$$\n\nAfterthought. The terms $a,$ $b$ and $ab/c$ in the numerator are the three heights of the triangle so the average sum of the distances is equal to the average of the three heights.\n\n\u2022 Is there a simple explanation of the fact in the Afterthought(which is the one cited by @David ), namely: \"The average sum of the distances is equal to the average of the three heights\"? \u2013\u00a0sdd Jun 11 '18 at 12:37\n\nLet $(x,y)$ be a point in a semi-circle with diameter inclined at $\\sin ^{-1} \\frac35$ to x-axis having sides proportional to Pythagorean triplet (8,6,10) as given.\n\nThe three perpendicular distance to sides of a scalene triangle are $(x,y, 4 x/5 + 3 y/5) ,$\n\ntotaling to $\\dfrac {9 x + 8 y} {5}$ which is variable , needing to be averaged.\n\nIf it were constant, a result like those from Viviani, Fermat would have been in existence now for more than two centuries.", "date": "2020-08-08 03:56:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1631704/inside-an-not-equilateral-triangle-what-is-the-sum-of-distances-from-a-random-po/1631732", "openwebmath_score": 0.9231142401695251, "openwebmath_perplexity": 200.4809339040535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676436891864, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.8579801721291808}}
{"url": "http://mathhelpforum.com/statistics/131990-probability-playing-cards.html", "text": "# Math Help - Probability for playing cards\n\n1. ## Probability for playing cards\n\nMr.A has 13 cards of the same suit. He withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card, i.e he will take 3 from king whose value is 13, 0 from 10, 9 from 9 and so on.. What is the probability that he can form a number that is divisible by 2?\n\nMy working\n(which is incorrect):\n\nSince this question is dealing with combination. I could answer it as:\n\n6C4/13C4 .. (6 possible ways to select numbers ending with an even digit with 4 chosen cards)\n\nThe correct answer according to the book is: (6 * 12 * 11 * 10)/(13 * 12* 11 * 10) = 6/13\n\nThe solution from the book went above my head. How do i solve this question?\n\n2. Hello, saberteeth!\n\nMr.A has 13 cards of the same suit.\nHe withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card,\nThat is: .Ace = 1, Deuce = 2, Trey = 3, . . . Ten = 0, Jack = 1, Queen = 2, King = 3.\nWhat is the probability that he can form a number that is divisible by 2?\n\"Divisible by 2\" means that the 4-digit number is even.\n\nThere are: 7 odd digits and 6 even digits.\n\nHe can make an even number if at least one digit is even.\n\nHe will fail of all four digits are odd.\n. . There are: . ${7\\choose4} = 35$ ways to get 4 odd digits.\n. . . . There are: . ${13\\choose4} = 715$ possible outcomes.\n. . Hence: . $P(\\text{odd number}) \\:=\\:\\frac{35}{715} \\:=\\:\\frac{7}{143}$\n\nTherefore: . $P(\\text{even number}) \\;=\\;1-\\frac{7}{143} \\;=\\;\\frac{136}{143}$\n\n3. Hello,\n\nSoroban's way is the way I would do it but it also easy to see it this way. Being divisible by 2 means it has to be even so out of the 13 cards, there are 6 even cards {2,4,6,8,10,12} . The probability of the last number being even is 6 out of 13 cards. 6/13 !\n\n*another way of looking at it same concept:\n\nI don't know if this is notationally correct but:\n\nSample Space S = {1,2,3,4,5,6,7,8,9,10,11,12,13}\nEven ={2,4,6,8,10,12}\n\n6 of the 13 cards are even; each has the same probability of being chosen. So P(EVEN)=P(2)+P(4)+...+P(12)= (1/13+1/13+1/13+1/13+1/13+1/13)= 6/13.\n\nThis way is trivial though and is not very reliable (not to mention long) when you start getting into more complex problems.\n\n4. Again, it can be simplified by calculating the probability they'd all be odd\n\n$P_{Even}=1-P_{Odd}=1-\\left(\\frac{7}{13}\\ \\frac{6}{12}\\ \\frac{5}{11}\\ \\frac{4}{10}\\right)$\n\n$=1-\\frac{7(6)5(4)}{13(12)11(10)}=1-\\frac{7(6)}{13(11)6}=1-\\frac{7}{143}$\n\n5. Originally Posted by Soroban\nHello, saberteeth!\n\n\"Divisible by 2\" means that the 4-digit number is even.\n\nThere are: 7 odd digits and 6 even digits.\n\nHe can make an even number if at least one digit is even.\n\nHe will fail of all four digits are odd.\n. . There are: . ${7\\choose4} = 35$ ways to get 4 odd digits.\n. . . . There are: . ${13\\choose4} = 715$ possible outcomes.\n. . Hence: . $P(\\text{odd number}) \\:=\\:\\frac{35}{715} \\:=\\:\\frac{7}{143}$\n\nTherefore: . $P(\\text{even number}) \\;=\\;1-\\frac{7}{143} \\;=\\;\\frac{136}{143}$\n\nHello:\n\nI did not understand one thing in Soroban's answer:\n\n\"He can make an even number if at least one digit is even.\"\n\nHow about $2343,1225,8643$, they also contain at least one even digit but still odd.\nI think it should be that last digit should be an even number i.e $3334,7134$ etc.\n\n6. Hi u2_wa,\n\nIn forming a number from the 4 digits,\nthe even digit may be placed at the end to make it even,\nthus forming an even number from the digits available.\nYou don't need to stick to the order the digits came in.\n\nThe book answer gives $\\frac{6}{13}$\n\nwhich is the probability that the first digit is even,\nconsidering that it doesn't matter whether the others are even or odd.\n\nHowever, this misses the probabilities of the 1st being odd, the 1st 2 being odd,\nthe 1st 3 being odd, the 1st and 3rd being odd......etc.\n\n7. Originally Posted by Archie Meade\nHi u2_wa,\n\nIn forming a number from the 4 digits,\nthe even digit may be placed at the end to make it even,\nthus forming an even number from the digits available.\nYou don't need to stick to the order the digits came in.\n\nThe book answer gives $\\frac{6}{13}$\n\nwhich is the probability that the first digit is even,\nconsidering that it doesn't matter whether the others are even or odd.\n\nHowever, this misses the probabilities of the 1st being odd, the 1st 2 being odd,\nthe 1st 3 being odd, the 1st and 3rd being odd......etc.\n\nWhat I did to solve is this:\n\nThere are in total $13P4$ ways.\n\nThere should $(2,4,6,8,0,2)$ be among $six$ digits in the end to make an even number.\n\nWays to get an even number: $12P3*6$\n\nprobability $=\\frac{12P3*6}{13P4}$\n\n$=\\frac{6}{13}$ (The answer given in the book)\n\n8. Yes, that's good work u2_wa,\n\nshows you can work it out in alternative ways.\n\nYou are calculating the probability if we must take the digits\nin the order they come in.\n\nThe way the question is worded suggests that Mr. A chooses the 4 cards and tries to\nmake an even number by placing an even number in the\nunits position whenever he has the opportunity, in other words by rearranging\nthe digits deliberately.\n\nMaybe the book did not want him to be allowed do this\nand the answer it has given suggests he isn't allowed to.\n\nHowever, the way the question is worded suggests he is!!\n\nHence, wording can make quite a difference in probability questions.\n\n9. Originally Posted by Archie Meade\nYes, that's good work u2_wa,\n\nshows you can work it out in alternative ways.\n\nYou are calculating the probability if we must take the digits\nin the order they come in.\n\nThe way the question is worded suggests that Mr. A chooses the 4 cards and tries to\nmake an even number by placing an even number in the\nunits position whenever he has the opportunity, in other words by rearranging\nthe digits deliberately.\n\nMaybe the book did not want him to be allowed do this\nand the answer it has given suggests he isn't allowed to.\n\nHowever, the way the question is worded suggests he is!!\n\nHence, wording can make quite a difference in probability questions.\nYeh I know, thanks!!!", "date": "2014-12-28 05:26:44", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/131990-probability-playing-cards.html", "openwebmath_score": 0.8072686791419983, "openwebmath_perplexity": 709.8450075966517, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676460637103, "lm_q2_score": 0.8757869786798663, "lm_q1q2_score": 0.8579801678565534}}
{"url": "https://math.stackexchange.com/questions/2792828/how-to-derive-the-approximation-tanx-simeq-fracx1-x2-3/2792838", "text": "# How to derive the approximation $\\tan(x)\\simeq \\frac{x}{1-x^2/3}$\n\nI was wondering how one could derive the\n\n$$\\tan(x)\\simeq \\frac{x}{1-x^2/3}$$\n\nvalid for small $x$ values.\n\nThis is similar to the ratio of the small $x$ expansions of $\\sin(x)$ an $\\cos(x)$, however that would yield\n\n$$\\tan(x)\\simeq \\frac{x}{1-x^2/2}$$\n\nso I have been left slightly confused. Many thanks in advance.\n\nEDIT: In my notes this seems to be some sort of recursive fraction approximation. A second version I have written is:\n\n$$\\tan(x)\\simeq %%% \\frac{\\lambda} {1-\\frac{\\lambda^2}{3-\\frac{\\lambda^2}{5-\\lambda^2/2}}}$$\n\nEDIT: Thanks for your great answers! You can also find this derived in the references to Equation 33 here: http://mathworld.wolfram.com/Tangent.html\n\nWall, H. S. (1948). Analytic theory of continued fractions. pg. 349\n\nC.D., O. (1963). Continued fractions. pg. 138\n\n\u2022 If you use the Taylor expansion of sin(x) and cos(x) I believe you would get the one you suggested too. May 23 '18 at 12:28\n\u2022 @HenryLee: Yes, it's quite frustrating to see where this other one came from. May 23 '18 at 12:31\n\u2022 Somehow, $3$ gives a smaller error, even though the Taylor expansions would suggest $2$. i.imgur.com/tczqY6G.png\n\u2013\u00a0Jam\nMay 23 '18 at 12:31\n\u2022 Yes, I think that's why I used it. I'm digging through thesis notes currently trying to write up. Past me apparently didn't think it was useful to write where this came from. May 23 '18 at 12:33\n\u2022 what is this for? May 23 '18 at 12:34\n\nThis relation con be derived using a Pade approximant. Define a function\n\n$$R(x) = \\frac{a_0 + a_1 x + \\cdots + a_m x^m}{1 + b_1 x + \\cdots + b_n x^n}$$\n\nThat agrees with $f$ up to some order\n\n$$\\left.\\frac{{\\rm d}^kf}{{\\rm d}x^k}\\right|_{x=0} = \\left.\\frac{{\\rm d}^kR}{{\\rm d}x^k}\\right|_{x=0} ~~~\\mbox{for}~~ k = 0, 1, \\cdots$$\n\nSo in your example, take $m=1$ and $n=2$ and $f(x) = \\tan(x)$, this would lead to the equations\n\n\\begin{eqnarray} a_0 &=&0 \\\\ a_1 - a_0 b_1 &=& 1 \\\\ -2a_1 b_1 + a_0(2b_1^2 - 2b_2) &=&0 \\\\ 3a_1(2 b_1^2 - 2b_2) + a_0(-6b_1^3 + 12 b_1 b_2) &=& 2 \\end{eqnarray}\n\nwhose solution is\n\n$$a_0 = 0, a_1 =1, b_1 = 0~~\\mbox{and}~~ b_2=-1/3$$\n\nThat is\n\n$$\\tan(x) \\approx \\frac{x}{1 - x^2/3}$$\n\nwhich aggress up to the fourth derivative!\n\n\u2022 Ah awesome! That's really helpful of you, and I learnt something new! Thank you very much for your help. May 23 '18 at 12:43\n\u2022 @Freeman Happy to help May 23 '18 at 12:45\n\u2022 Today I learned! :) May 23 '18 at 12:47\n\nRecall that\n\n$$\\tan(x)= x+\\frac13x^3+o(x^3)$$\n\nand by binomial expansion as $x\\to 0$\n\n$$\\frac{x}{1-x^2/3}=x(1-x^2/3)^{-1}\\sim x+\\frac13x^3$$\n\nthus\n\n$$\\tan(x)\\sim x+\\frac13x^3\\sim \\frac{x}{1-x^2/3}$$\n\nNote that from here\n\n$$\\tan(x)=\\frac{\\sin x}{\\cos x}$$\n\nto obtain the same result we need to expand to the 3rd order that is\n\n$$\\tan(x)=\\frac{\\sin x}{\\cos x}=\\frac{x-\\frac16x^3+o(x^3)}{1-\\frac12x^2+o(x^3)}=(x-\\frac16x^3+o(x^3))(1-\\frac12x^2+o(x^3))^{-1}=(x-\\frac16x^3+o(x^3))(1+\\frac12x^2+o(x^3))=x-\\frac16x^3+\\frac12x^3+o(x^3)=x+\\frac13x^3+o(x^3)$$\n\n\u2022 That's pretty neat! Thank you! caverac also posted a nice proof about the same time as you, ideally I would like to accept both answers, but as caverac has less reputation than you I will accept theirs. Thanks so much! May 23 '18 at 12:42\n\u2022 @Freeman You are welcome! Bye\n\u2013\u00a0user\nMay 23 '18 at 12:44\n\u2022 @Freeman I add something for the derivation from $\\sin x/\\cos x$\n\u2013\u00a0user\nMay 23 '18 at 12:50\n\n\u2022 I don't think this answer quite addresses the question, or where the $1/3$ term comes from.\nFor an alternative derivation, you may use the Shafer-Fink inequality and compute the inverse function of an algebraic function. This gives $$\\tan(x) \\approx \\frac{3x+2x\\sqrt{9-3x^2}}{9-4x^2}\\quad\\text{for }x\\approx0$$ which is even more accurate than $\\tan x\\approx \\frac{x}{1-x^2/3}$. As already shown, the last approximation can be derived from Pad\u00e9 approximants or generalized continued fractions.\n\u2022 @Jam: This proves a sharper approximation, from which $\\tan(x)\\approx \\frac{x}{1-x^2/3}$ can be easily derived in a algebraic fashion. May 23 '18 at 13:24\n\u2022 A better approximation could be $\\tan(x)=\\frac{ x-\\frac{x^3}{15}}{1-\\frac{2 x^2}{5} }$ May 24 '18 at 8:36", "date": "2022-01-29 04:16:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2792828/how-to-derive-the-approximation-tanx-simeq-fracx1-x2-3/2792838", "openwebmath_score": 0.9479621052742004, "openwebmath_perplexity": 805.1602998770538, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676419082933, "lm_q2_score": 0.8757869786798663, "lm_q1q2_score": 0.8579801642172933}}
{"url": "http://math.stackexchange.com/questions/94891/given-an-alphabet-how-many-words-can-you-make-with-these-restrictions", "text": "# Given an Alphabet, how many words can you make with these restrictions.\n\nI'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\\Sigma = \\{ 0,1,2 \\}$ and the set of 8 letter words made from that alphabet, $\\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.\n\nHow many words have exactly three 1's?\n\nHow many words have at least one each of 0,1 and 2?\n\nIn the first question I reasoned that first I choose $\\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\\binom{8}{3}\\cdot 2^5 = 1792$ which is correct.\n\nI tried applying the same reasoning to the second question and got $\\binom{8}{3}\\cdot 3^5 = 13608$ which is obviously wrong.\n\nWas my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?\n\n-\nWhy was this question marked down, especially more than two years after it was asked? \u2013\u00a0 Robert S. Barnes Mar 15 '14 at 17:43\n\nIn the second question, by choosing three places to put a $0$, a $1$, and a $2$, you are counting each word several times. For example, you count the word $01222222$ once as $\\underline{012}22222$, once as $\\underline{01}2\\underline{2}222$, once as $\\underline{01}222\\underline{2}22$, and so on, where the underline denotes the three chosen places.\nYou are also undercounting, for each choice of 3 places, the permutations of $0$, $1$, and $2$ that go in them, but that's a different story.\nThis was not a problem in the first question because once you have chosen 3 places for the $1$'s, the rest of the places can't have a $1$ in them, so you never count a word more than once.\nThe easiest way to get the right answer to the second question is probably to count the number of words that have no $0$, or no $1$, or no $2$, and subtract that from the total number of words. Remember not to overcount the words that have no $0$ and no $1$, etc.", "date": "2015-05-28 04:25:13", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/94891/given-an-alphabet-how-many-words-can-you-make-with-these-restrictions", "openwebmath_score": 0.8102869391441345, "openwebmath_perplexity": 170.16903150291972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964220125032, "lm_q2_score": 0.870597273444551, "lm_q1q2_score": 0.8579704979934458}}
{"url": "https://mathhelpboards.com/threads/a-googol.6571/", "text": "# A googol\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHi MHB,\n\nIt's me, anemone.\n\nI saw this problem quite some time ago, and I don't remember where...but it says \"Can a googol $10^{100}$ be written as $n^2-m^2$, where $n$ and $m$ are positive integers?\"\n\nSince I don't know where I should start to think about how to answer it, I thought there would be no harm in asking about it here because MHB knows it all!\n\nThanks in advance for any input that anyone is going to give me.\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nHi MHB,\n\nIt's me, anemone.\n\nI saw this problem quite some time ago, and I don't remember where...but it says \"Can a googol $10^{100}$ be written as $n^2-m^2$, where $n$ and $m$ are positive integers?\"\n\nSince I don't know where I should start to think about how to answer it, I thought there would be no harm in asking about it here because MHB knows it all!\n\nThanks in advance for any input that anyone is going to give me.\n\nLet's take a look at factorization.\n\\begin{array}{lcl}n^2-m^2 &=& 10^{100}\\\\\n(n+m)(n-m) &=& 2^{100} 5^{100}\\end{array}\n\nSuppose we pick $$\\displaystyle n+m = 2^{50} 5^{51}$$ and $$\\displaystyle n-m=2^{50} 5^{49}$$.\nThen the solution\n\\begin{cases}n&=&10^{49}(25+1)\\\\\nm&=&10^{49}(25-1)\\end{cases}\ncomes rolling out...\n\nThere are bound to be more solutions.\n\nLast edited:\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nLet's take a look at factorization.\n\\begin{array}{lcl}n^2-m^2 &=& 10^{100}\\\\\n(n+m)(n-m) &=& 2^{100} 5^{100}\\end{array}\n\nSuppose we pick $$\\displaystyle n+m = 2^{50} 5^{51}$$ and $$\\displaystyle n-m=2^{50} 5^{49}$$.\nThen the solution\n\\begin{cases}n&=&10^{49}(25+1)\\\\\nm&=&10^{49}(25-1)\\end{cases}\ncomes rolling out...\n\nThere are bound to be more solutions.\nHi I like Serena,\n\nThank you so much for the reply!\n\nNow I know of a better concept whenever I want to solve any math problems..i.e. to play with the exponents!\n\nI appreciate it you made the explanation so clear for me! Thanks!\n\nLast edited by a moderator:\n\n#### soroban\n\n##### Well-known member\nHello, anemone!\n\n$$\\text{Can a googol }10^{100}\\text{ be written as }n^2-m^2\\text{, where }n\\text{ and }m\\text{ are positive integers?}$$\n\nAny multiple of 4 can be written as a difference of squares.\n\nExample: Express 80 as a difference of squares.\n\nA multiple of 4 can be expressed as the sum of two consecutive odd integers.\n. . We have: .$$80 \\:=\\:39+41$$\n\nConsecutive squares differ by consecutive odd integers.\n\nWe have: .$$\\begin{array}{ccccc}19^2 && 20^2 && 21^2 \\\\ \\hline & 39 && 41 \\end{array}$$\n\nTherefore: .$$80 \\;=\\;21^2 - 19^2$$\n\nLet $$G = 10^{100}.$$\n\nThe two odd numbers are: .$$\\tfrac{G}{2}-1\\text{ and }\\tfrac{G}{2}+1$$\n\nThe two squares are: .$$\\begin{Bmatrix}\\frac{(\\frac{G}{2}-1)-1}{2} &=& \\tfrac{G}{4}-1 \\\\ \\frac{(\\frac{G}{2}+1)+1}{2} &=& \\tfrac{G}{4}+1 \\end{Bmatrix}$$\n\nTherefore: .$$G \\;=\\;\\left(\\tfrac{G}{4}+1\\right)^2 - \\left(\\tfrac{G}{4} - 1\\right)^2$$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nHello, anemone!\n\nAny multiple of 4 can be written as a difference of squares.\n\nExample: Express 80 as a difference of squares.\n\nA multiple of 4 can be expressed as the sum of two consecutive odd integers.\n. . We have: .$$80 \\:=\\:39+41$$\n\nConsecutive squares differ by consecutive odd integers.\n\nWe have: .$$\\begin{array}{ccccc}19^2 && 20^2 && 21^2 \\\\ \\hline & 39 && 41 \\end{array}$$\n\nTherefore: .$$80 \\;=\\;21^2 - 19^2$$\n\nLet $$G = 10^{100}.$$\n\nThe two odd numbers are: .$$\\tfrac{G}{2}-1\\text{ and }\\tfrac{G}{2}+1$$\n\nThe two squares are: .$$\\begin{Bmatrix}\\frac{(\\frac{G}{2}-1)-1}{2} &=& \\tfrac{G}{4}-1 \\\\ \\frac{(\\frac{G}{2}+1)+1}{2} &=& \\tfrac{G}{4}+1 \\end{Bmatrix}$$\n\nTherefore: .$$G \\;=\\;\\left(\\tfrac{G}{4}+1\\right)^2 - \\left(\\tfrac{G}{4} - 1\\right)^2$$\nThank you so much soroban for letting me know of this useful and handy knowledge...I really appreciate it!", "date": "2020-09-26 18:22:16", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/a-googol.6571/", "openwebmath_score": 0.9368777871131897, "openwebmath_perplexity": 1757.752674132587, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964186047326, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8579704933722676}}
{"url": "https://math.stackexchange.com/questions/2583800/if-am-bn-cp-are-parallel-chords-in-the-circumcircle-of-equilateral-tri", "text": "# If $AM$, $BN$, $CP$ are parallel chords in the circumcircle of equilateral $\\triangle ABC$, then $\\triangle MNP$ is also equilateral\n\nHere's an interesting problem, and result, that I wish to share with the math community here at Math SE.\n\nThe above problem has two methods.\n\n1. Pure geometry. A bit of angle chasing and standard results from circles help us arrive at the desired result - \u2206MNP is equilateral. I'll put up a picture of the angle chasing part here (I hope someone edits it, and puts up a picture using GeoGebra or some similar software - I'm sorry I'm not good at editing)\n\nI joined BP and CN for angle chasing purposes.\n\n(Note that if M is the midpoint of arc BC, then the figure so formed is a star, with 8 equilateral triangles)\n\n1. This can be solved beautifully using complex numbers. The vertices of the \u2206 can be assumed to be 1,W,W2 on a unit circle centered at origin.\n\nWe need to prove that the new equilateral triangle is essentially a rotation of the original one, about an axis passing through center of its circumcircle perpendicular to its plane.\n\nI haven't posted the solution, hope you fellow Math SE members try the problem and post your solutions and ideas in the answers section.\n\nMore methods (apart from geometry and complex numbers) are welcome. I'd like to know more about why this result is interesting in itself, and what other deductions can be made from it.\n\n\u2022 +1 for a nicely posed question. the new equilateral triangle is essentially a rotation of the original one For a different geometric hint, consider the diameter perpendicular to $AM\\,$, which is orthogonal to and bisects each of $AM, BN, CP\\,$ (why?). Then $\\triangle MNP$ is simply the reflection of $\\triangle ABC$ across this diameter. \u2013\u00a0dxiv Dec 29 '17 at 5:35\n\u2022 Why don't you learn basic MathJax yourself instead of asking people to do it for you? \u2013\u00a0user21820 Dec 29 '17 at 6:46\n\u2022 Can you state the result in the title, instead of claiming that it is \"interesting\"? (Because personally I don't find many results about triangles to be interesting at all.) \u2013\u00a0Asaf Karagila Dec 29 '17 at 10:12\n\nA geometric proof for part a.\n\nFrom the OP:\n\nWe need to prove that the new equilateral triangle is essentially a rotation of the original one, about an axis passing through center of its circumcircle perpendicular to its plane.\n\nI think it's easier to think of it as a reflection. In the diagram below\n\n\u2022 $GH$ is the diameter of the circumcircle that is perpendicular to $AM$.\n\u2022 It's given that $BN$ and $CP$ are parallel to $AM$.\n\u2022 Hence, points $M,N,P$ are reflections of points $A,B,C$ respectively in the line $GH$.\n\u2022 Hence, $\\triangle MNP$ is a reflection of $\\triangle ABC$ through the line $GH$.\n\nBy way of the reflective symmetry about $GH$ and the rotational symmetry of the equilateral triangles we can note that $AP=BN=CM$, and that $AN=BM=CP$. From this we can draw a system of parallel lines to show that $AM=BN+CP$.\n\n\u2022 Brilliant! Do you not think it would work out backwards, too? I mean, whether it now follows that $AN$, $BP$ and $CM$ are all parallel? Because this will give a straightforward proof for (b): let $Q$ be the intersection of $BP$ and $AM$, then $BQAN$ and $QPCM$ are parallelograms, so $AM=AQ+QM=BN+CP$. \u2013\u00a0user491874 Dec 29 '17 at 11:54\n\u2022 @user8734617 Yes, it works exactly like that. I was just getting my diagram together when you commented... \u2013\u00a0nickgard Dec 29 '17 at 12:05\n\nGeometric hint: cyclic trapezoids $APCM, AMBN$ must be isosceles, so arcs $\\overparen{AP}=\\overparen{CM}$ and $\\overparen{BM}=\\overparen{AN}\\,$. Since $\\,\\overparen{AB}=\\overparen{BC}\\,$ the latter implies $\\overparen{BN}=\\overparen{AB}-\\overparen{AN}=\\overparen{BC}-\\overparen{BM}=\\overparen{CM}\\,$, so in the end $\\overparen{AP}=\\overparen{CM}=\\overparen{BN}\\,$ and therefore $\\triangle PNM$ is $\\triangle ABC$ rotated by $\\overparen{AP}$.\n\nAs for point b, writing Ptolemy's theorem twice:\n\n\u2022 $APCM\\,$: $\\;AC^2=AP^2 + AM \\cdot CP$\n\n\u2022 $AMBN\\,$: $\\;AB^2=AN^2 + AM \\cdot BN$\n\nSubtracting the above and using that $AC=AB, AP=BN, CP=AN\\,$:\n\n$$0 = BN^2-CP^2+AM\\cdot(CP-BN) = (BN-CP)\\cdot(BN+CP-AM)$$\n\nIt follows that $\\,BN+CP-AM=0 \\iff AM = BN+CP\\,$. \u00a0(Rigorously, the case $\\,BN=CP\\,$ would need to be considered separately, or could be derived by continuity).\n\nDraw a line through $P$ parallel to $CM$, and let it intersect $AM$ at $Q$. $PQMC$ is a parallelogram, so $CP$=$QM$.\n\nOn the other hand, it is easy to see that $\\triangle APQ$ is equilateral, as its angles are $60^\\circ$ ($\\angle A$ is peripheral over arc $\\overparen {PM}$ and $\\angle Q=\\angle CMA$ is peripheral over arc $\\overparen{AC}$). Thus, $AQ$=$AP$=$BN$.\n\nFinally, $AM=AQ+QM=BN+CP$ as claimed.\n\nThis can be solved beautifully using complex numbers. The vertices of the \u2206 can be assumed to be 1,W,W2 on a unit circle centered at origin.\n\nHere is an attempt to prove it using complex numbers.\n\nI am trying to prove it backwards.Let C,A,B be $1$,$\\omega$,$\\omega^2$ and M be $e^{i\\theta}$.\n\nIf $\\Delta$MPN is an equilateral triangle,then AM||PC which implies\n\nAM=$e^{i\\theta}-\\omega$\n\nOP=OM$\\times \\omega$\n\nPC=$e^{i\\theta}\\omega-1$\n\nWe have to prove AM=$\\lambda$PC,for some scalar $\\lambda$\n\n@schrodinger_16 ,can you help me to proceed further?", "date": "2021-06-13 18:33:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2583800/if-am-bn-cp-are-parallel-chords-in-the-circumcircle-of-equilateral-tri", "openwebmath_score": 0.8923885226249695, "openwebmath_perplexity": 323.1242692062803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98549641775279, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8579704926305687}}
{"url": "https://math.stackexchange.com/questions/3989743/is-the-union-of-two-circles-homeomorphic-to-an-interval", "text": "# Is the union of two circles homeomorphic to an interval?\n\nLet $$Y$$ be the subspace of $$\\Bbb R^2$$ given by $$Y=\\{(x,y): x^2+ y^2=1\\}\\cup \\{(x,y): (x\u22122)^2+ y^2=1\\}$$. Is $$Y$$ homeomorphic to an interval?\n\nI have previously already shown that the unit circle is not homeomorphic to any interval and I think this is also true for $$Y$$. Basically if we remove a point from $$Y$$ that is not the intersection, and remove a point from any interval that is not an endpoint, the interval becomes disconnected, but the union of two circles are still connected so there is no homeomorphisms. Is that correct?\n\n\u2022 The only point whose removal disconnects $Y$ is $\\langle 1,0\\rangle$, but the removal of any non-endpoint of an interval disconnects the interval. A homeomorphism takes cut points to cut points, so there is no homeomorphism from $Y$ to an interval. \u2013\u00a0Brian M. Scott Jan 18 at 7:49\n\u2022 Just to make sure I am applying the theorem correctly, it states that If X and Y are homeomorphic, there is also a homeomorphism if we remove any point from X and any point from Y right? So we can choose points to remove such that X and Y have different properties such as connectedness which means they are not homeomorphic right? \u2013\u00a0William Jan 18 at 7:57\n\u2022 Not quite. The point is that if $X$ and $Y$ are spaces, $h:X\\to Y$ is a homeomorphism, and $x$ is a cut point of $X$, then $h(x)$ must be a cut point of $Y$, because $X\\setminus\\{x\\}$ is homeomorphic to $Y\\setminus\\{h(x)\\}$. This means that $h$ must be (among other things) a bijection between the cut points of $X$ and the cut points of $Y$. Your space $Y$ has only one cut point, while an interval has infinitely many, so there cannot be a bijection between the cut points of $Y$ and the cut points of an interval, and therefore there can be no homeomorphism between them. \u2013\u00a0Brian M. Scott Jan 18 at 8:17\n\u2022 oh ok got it! Thanks \u2013\u00a0William Jan 18 at 8:26\n\u2022 You\u2019re welcome! \u2013\u00a0Brian M. Scott Jan 18 at 8:26\n\nA cutpoint of a connected space $$X$$ is a $$p \\in X$$ such that $$X\\setminus\\{p\\}$$ is disconnected.\n\nIf $$f:X \\to Y$$ is a homeomorphism of connected spaces $$X$$ and $$Y$$ and $$p$$ is a cut point of $$X$$ then $$f(p)$$ is a cutpoint of $$Y$$ (and vice versa).\n\nIf we take $$X$$ to be an interval, then $$X$$ has at most two non-cutpoints. (the endpoints in the case of a closed interval). $$Y$$ on the other hand has infinitely many non-cutpoints (all points except $$(1,0)$$). So there can be no homeomorphism between them by the observations in the second paragraph.\n\nHere's an alternative argument that covers wide range of spaces, regardless of cut points.\n\nLet $$X$$, $$Y$$ be any topological spaces. Consider a homeomorphism $$f:X\\to Y$$. Now let $$Z$$ be any topological space and $$\\alpha:Z\\to X$$ be a continuous function. Then $$\\alpha$$ is injective if and only if $$f\\circ\\alpha$$ is injective. Which is easy to see by applying $$f$$ to $$\\alpha$$ and $$f^{-1}$$ to $$f\\circ\\alpha$$.\n\nIn particular $$X$$ admits an injective map $$Z\\to X$$ if and only if $$Y$$ admits an injective map $$Z\\to Y$$.\n\nThis shows that your $$Y$$ (or more generally any space containing $$S^1$$ as a subspace) cannot be homeomorphic to the interval $$[0,1]$$. Because there is an obvious injective continuous map $$S^1\\to Y$$ while there is no continuous injective map $$S^1\\to [0,1]$$. That's because every map $$S^1\\to [0,1]$$ arises from a map $$[0,1]\\to [0,1]$$ with the same values at endpoints and so by the intermediate value property it cannot be injective on $$(0,1)$$, which then \"lifts\" to $$S^1$$.\n\nJust for a change: Let $$D$$ be the open disk in $$\\Bbb R^2$$ centered at $$p=(1,0)$$ with radius $$1$$ and let $$E=D\\cap Y.$$ Then $$E$$ is a connected subspace of $$Y$$ and $$E\\setminus \\{p\\}$$ is the union of $$4$$ pairwise-disjoint non-empty connected open subspaces of $$E$$. But if $$E'$$ is any connected subspace of $$\\Bbb R$$ and $$p'\\in E'$$ then $$E'\\setminus \\{p'\\}$$ is either connected or is the union of $$2$$ disjoint non-empty connected open subspaces of $$E'$$.", "date": "2021-03-04 06:47:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3989743/is-the-union-of-two-circles-homeomorphic-to-an-interval", "openwebmath_score": 0.8844925761222839, "openwebmath_perplexity": 84.23466349946791, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964198826467, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8579704878672865}}
{"url": "https://casmusings.wordpress.com/tag/probability/page/2/", "text": "Tag Archives: probability\n\nMarilyn vos Savant and Conditional\u00a0Probability\n\nThe following question appeared in the\u00a0\u201cAsk Marilyn\u201d column in the August 16, 2015 issue of Parade magazine. \u00a0The writer seems stuck between two probabilities.\n\n(Click here for a\u00a0cleaned-up online version if you don\u2019t like the newspaper look.)\n\nI just pitched this question to my statistics class (we start the year with a probability unit). \u00a0I thought some of you might like it for your classes, too.\n\nI asked them to do two things. \u00a01) Answer the writer\u2019s question, AND 2)\u00a0Use precise probability terminology\u00a0to\u00a0identify the source of the writer\u2019s conundrum. \u00a0Can you answer both before reading further?\n\nVery briefly, the writer is correct in both situations. \u00a0If each of the four people draws a random straw, there is absolutely a 1 in 4 chance of each drawing the straw. \u00a0Think about shuffling the straws and \u201cdealing\u201d one to each person much like shuffling a deck of cards and dealing out all of the cards. \u00a0Any given straw or card is equally likely to land in any player\u2019s hand.\n\nNow let the first person look at his or her straw. \u00a0It is either short or not. \u00a0The author is then correct at claiming the probability of others holding the straw is now 0 (if the first person found the short straw) or 1/3 (if the first person did not). \u00a0And this is precisely the source of the writer\u2019s conundrum. \u00a0She\u2019s actually asking two different questions but thinks she\u2019s asking only one.\n\nThe 1/4 result\u00a0is from a pure, simple probability scenario. \u00a0There are four possible equally-likely locations for the short straw.\n\nThe 0 and 1/3 results happen only after the first (or any other) person looks at his or her straw. \u00a0At that point, the problem shifts from simple probability to\u00a0conditional probability. \u00a0After observing a straw, the question shifts to determining\u00a0the probability that one of the remaining people has the short straw GIVEN that you know the result of one person\u2019s draw.\n\nSo, the writer was correct in all of her claims; she just didn\u2019t realize she was asking two fundamentally different questions. \u00a0That\u2019s a pretty excusable lapse, in my opinion. \u00a0Slips\u00a0into conditional probability are often missed.\n\nPerhaps the most famous of these misses is the solution to the Monty Hall scenario that\u00a0vos Savant famously posited years ago. \u00a0What I particularly love about this is the number of very-well-educated mathematicians who missed the conditional and wrote flaming retorts to vos Savant brandishing their PhDs and ultimately found themselves publicly supporting errant conclusions. \u00a0You can read the original question, errant responses, and vos Savant\u2019s very clear explanation\u00a0here.\n\nCONCLUSION:\n\nProbability is subtle and catches all of us at some point. \u00a0Even so, the careful thinking\u00a0required to dissect and answer subtle probability questions is arguably one of the best exercises of logical reasoning around.\n\nRANDOM(?) CONNECTION:\n\nAs a completely different connection, I think this is very much like Heisenberg\u2019s Uncertainty Principle. \u00a0Until the first straw is observed, the short straw really could (does?) exist in all hands simultaneously. \u00a0Observing the system (looking at one person\u2019s straw) permanently changes the state of the system, bifurcating forever the system into one of two potential future states: \u00a0the short straw is found in the first hand or is it not.\n\nCORRECTION (3 hours after posting):\n\nI knew I was likely to overstate or misname something in my final connection. \u00a0Thanks to Mike Lawler (@mikeandallie) for\u00a0a quick correction via Twitter. \u00a0I should have called this\u00a0quantum superposition and not\u00a0the uncertainty principle. \u00a0Thanks so much, Mike.\n\nInnumeracy and Sharks\n\nHere\u2019s a brief snippet from a conversation about the recent spate of shark attacks in North Carolina as\u00a0I heard yesterday morning (approx 6AM, 7/4/15) on CNN.\n\nGeorge Burgess (Director, Florida Program for Shark Research):\u00a0 \u201cOne thing is going to happen and that is there are going to be more\u00a0[shark] attacks year in and year out simply because the human population continues to rise and with it a concurrent interest in aquatic recreation. \u00a0So one of the few things I, as a scientist, can predict with some certainty is more attacks in the future because there\u2019s more people.\u201d\n\nAlison Kosik (CNN anchor):\u00a0 \u201cThat is scary and I just started surfing so I may dial that back a bit.\u201d\n\nThis marks another great teaching moment spinning out of innumeracy in the media. \u00a0I plan to drop just those two paragraphs on my classes when school restarts this fall and open the discussion. \u00a0I wonder how many will question the implied, but irrational probability in Kosik\u2019s reply.\n\nTOO MUCH\u00a0COVERAGE?\n\nBurgess argued elsewhere\u00a0that\n\nIncreased documentation of the incidents\u00a0may also make people believe attacks are more prevalent. \u00a0(Source here.)\n\nIt\u2019s certainly plausible that some people think shark attacks are more common than they really are. \u00a0But that begs the question of just how nervous a swimmer should be.\n\nMEDIA MANIPULATION\n\nCNN\u2013like almost all mass media, but not nearly as bad as some\u2013shamelessly hyper-focuses on catchy news banners, and what could be catchier than something like \u2018Shark attacks spike just as tourists\u00a0crowd beaches on busy July 4th weekend\u201d? \u00a0Was\u00a0Kosik reading a prepared script that distorts the underlying probability, or was she showing signs of innumeracy?\u00a0I hope it\u2019s not both, but neither is good.\n\nIRRATIONAL PROBABILITY\n\nSo just how uncommon is a shark attack? \u00a0In a few\u00a0minutes of Web research, I found that there were 25 shark attacks in North Carolina from 2005-2014. \u00a0There was at least one every year with a maximum of 5 attacks in 2010 (source). \u00a0So this year\u2019s 7 attacks is certainly unusually high from the recent annual average of 2.5, but John Allen Paulos reminded us in Innumeracy that [in this case about 3 times] a very small probability, is still a very small probability.\n\nIn another place,\u00a0Burgess noted\n\n\u201cIt\u2019s amazing, given the billions of hours humans spend in the water, how uncommon attacks are,\u201d Burgess said, \u201cbut that doesn\u2019t make you feel better if you\u2019re one of them.\u201d \u00a0(Source here.)\n\n18.9% of NC visitors went to the beach (source) . \u00a0In 2012, there were approximately 45.4 million visitors to NC (source). \u00a0To overestimate the number of beachgoers, Let\u2019s\u00a0say 19% of 46 million visitors, or 8.7 million people, went to NC beaches. \u00a0Seriously underestimating the number of beachgoers who enter the ocean, assume only 1 in 8 beachgoers entered the ocean. \u00a0That\u2019s still a very\u00a0small 7 attacks out of 1 million people in the ocean. \u00a0Because beachgoers almost always enter the ocean at some point (in my experiences), the average likely is much closer to 2 or fewer attacks per million.\n\nTo put that in perspective, 110,406 people were injured in car accidents in 2012 in NC (source). \u00a0The probability of getting injured driving to the beach is many orders of magnitude larger than the likelihood of ever being attacked by a shark.\n\nAlison Kosik should keep up her surfing.\n\nIf you made it to a NC beach safely, enjoy the swim. \u00a0It\u2019s safer than your trip there was or your trip home is going to be. \u00a0But even those trips are reasonably safe.\n\nI certainly am not\u00a0diminishing the anguish of accident victims (shark, auto, or otherwise), but accidents happen. \u00a0But don\u2019t make too much of one either. \u00a0Be intelligent, be reasonable, and enjoy life.\n\nIn the end, I hope my students learn to question facts and probabilities. \u00a0I hope they always question \u201cHow reasonable is what I\u2019m being told?\u201d\n\nHere\u2019s a much more balanced article on shark attacks from NPR:\nDon\u2019t Blame the Sharks For \u2018Perfect Storm\u2019 of Attacks In North Carolina.\n\nBook suggestions:\n1) \u00a0Innumeracy, John Allen Paulos\n2) Predictably Irrational, Dan Ariely\n\nCAS and Normal Probability\u00a0Distributions\n\nMy presentation this past Saturday\u00a0at the 2015 T^3 International Conference in Dallas, TX was on the underappreciated\u00a0applicability\u00a0of CAS to statistics. \u00a0This post shares some of what I shared there\u00a0from my first year teaching AP Statistics.\n\nMOVING PAST OUTDATED\u00a0PEDAGOGY\n\nIt\u2019s been decades since we\u2019ve required students to use tables of values to compute by hand trigonometric and radical values. \u00a0It seems odd to me that we continue to do\u00a0exactly that today for so many statistics classes, including the AP. \u00a0While the College Board permits statistics-capable calculators, it still provides\u00a0probability tables with every exam. \u00a0That messaging is clear: \u00a0it is still \u201cacceptable\u201d\u00a0to teach statistics using outdated probability tables.\n\nIn this, my first year teaching AP Statistics, I decided it was time for my students and I to completely break from this lingering past. \u00a0My statistics classes this year have been 100% software-enabled. \u00a0Not one of my students has been required to use or even see any tables of probability values.\n\nMy classes also have been fortunate to have complete CAS availability on their laptops. \u00a0My school\u2019s math department deliberately adopted the TI-Nspire platform in part because that software looks and operates exactly the same on tablet, computer, and handheld platforms. \u00a0We primarily use the computer-based\u00a0version for learning because of the speed and visualization of the large \u201creal estate\u201d there. \u00a0We are shifting\u00a0to school-owned\u00a0handhelds in our last month before the AP Exam to gain practice on the platform required on the AP.\n\nThe remainder of this post shares ways my students and I have learned to apply the TI-Nspire CAS to some statistical questions around normal distributions.\n\nFINDING NORMAL AREAS AND PROBABILITIES\n\nAssume a manufacturer makes golf balls whose distances traveled under identical testing conditions are approximately normally distributed with a mean 295 yards with a standard deviation of 3 yards. \u00a0What is the probability that one such randomly selected ball travels more than 300 yards?\n\nTraditional\u00a0statistics courses teach students to transform the 300 yards into a z-score to look up in a probability table. \u00a0That approach obviously works, but with appropriate technology, I believe there will be far less need to use or even compute z-scores in much the same way that always converting logarithms to base-10 or base-to use logarithmic tables is anachronistic when using many modern scientific calculators.\n\nTI calculators and other technologies allow computations of non-standard normal curves. \u00a0Notice the Nspire CAS calculation below the curve uses both bounds of the area of interest along with the mean and standard deviation of the distribution to accomplish the\u00a0computation in a single step.\n\nSo the probability of a randomly selected ball from the population described above going more than 300 yards is 4.779%.\n\nGOING BACKWARDS\n\nNow assume the manufacturing process can control the mean distance traveled.\u00a0 What mean should it use so that no more than 1% of the golf balls travel more than 300 yards?\n\nDepending on the available normal probability tables, the traditional approach to this problem is again to work with z-scores. \u00a0A modified CAS version of this is shown below.\n\nTherefore, the manufacturer should produce a ball that travels a mean 293.021 yards under the given conditions.\n\nThe approach is legitimate, and I shared it with my students. \u00a0Several of them ultimately chose a\u00a0more efficient single line command:\n\nBut remember that the invNorm() and normCdf() commands on the Nspire are themselves functions, and so their internal parameters are available to solve commands. \u00a0A pure CAS, \u201cforward solution\u201d still incorporating only the normCdf() command to which my\u00a0students were first introduced makes use of this\u00a0to determine the missing center.\n\nDIFFERENTIATING INSTRUCTION\n\nWhile calculus techniques definitely are NOT part of the AP Statistics curriculum, I do have several students jointly enrolled in various calculus classes. \u00a0Some of these astutely noted the similarity between the area-based arguments above and the area under a curve techniques they were learning in their calculus classes. \u00a0Never being one to pass on a teaching moment, I pulled a few of these to the side to show them that the previous solutions also could have been derived via integration.\n\nI can\u2019t recall any instances of my students actually employing integrals to solve statistics problems this year, but just having the connection verified completely solidified the mathematics they were learning in my class.\n\nCONFIDENCE INTERVALS\n\nThe mean lead level of 35 crows in a random sample from a region was 4.90 ppm and the standard deviation was 1.12 ppm.\u00a0 Construct a 95 percent confidence interval for the mean lead level of crows in the region.\n\nMany students\u2013mine included\u2013have difficulty comprehending confidence intervals and resort to \u201cblack box\u201d confidence interval tools available in most (all?) statistics-capable calculators, including the TI-Nspire.\n\nAs\u00a0n\u00a0is greater than 30, I can compute the requested\u00a0z-interval by filling in\u00a0just four\u00a0entries in a pop-up window and pressing Enter.\n\nConvenient, for sure, but this approach doesn\u2019t help the confused students understand that the confidence interval is nothing more than the bounds of the middle 95% of the normal pdf described in the problem, a fact crystallized\u00a0by the application of the tools the students\u00a0have been using for weeks by that point in the course.\n\nNotice in the solve+normCdf() combination commands that the unknown this time was a bound and not the mean as was the case in the previous example.\n\nEXTENDING THE RULE OF FOUR\n\nI\u2019ve used the \u201cRule of Four\u201d in every math class I\u2019ve taught for over two decades, explaining that every mathematical concept can be explained or expressed four different ways: \u00a0Numerically, Algebraically, Graphically (including graphs and geometric figures), and Verbally. \u00a0While not the contextual point of his\u00a0quote, I often cite MIT\u2019s Marvin Minsky here:\n\n\u201cYou don\u2019t understand\u00a0anything until you learn it more than one way.\u201d\n\nLearning to translate between the four\u00a0representations grants deeper understanding of concepts and often gives access to solutions in one form that may be\u00a0difficult or impossible in other forms.\n\nAfter\u00a0my decades-long work with CAS, I\u00a0now believe\u00a0there is actually a 5th representation of mathematical ideas: \u00a0Tools. \u00a0Knowing how to translate a question into a form that your tool (in the case of CAS, the tool\u00a0is\u00a0computers) can manage or compute creates a different representation of the problem and requires deeper insights to manage the translation.\n\nI knew some of my students this year had deeply embraced\u00a0this \u201c5th Way\u201d when one showed me his alternative approach to the confidence interval question:\n\nI found this solution particularly lovely for several reasons.\n\n\u2022 The student knew about lists and statistical commands and on a whim tried combining them in a novel way to produce the desired solution.\n\u2022 He found the confidence interval directly using a normal distribution command rather than the arguably more convenient black box confidence interval\u00a0tool. \u00a0He also showed explicitly his understanding of the distribution of sample means by adjusting\u00a0the given standard deviation for the sample size.\n\u2022 Finally, while using a CAS sometimes\u00a0involves getting answers in forms you didn\u2019t expect, in this case, I think the CAS command and list output actually provide a cleaner, interval-looking result than the black box confidence interval command much more intuitively connected to the actual meaning of a confidence interval.\n\u2022 While I haven\u2019t tried it out, it seems to me that this approach also should work on non-CAS statistical calculators that can handle lists.\n\n(a very minor disappointment, quickly overcome)\n\nReturning to my multiple approaches, I tried using my student\u2019s newfound approach using a normCdf() command.\n\nAlas, my\u00a0Nspire\u00a0returned the very command I had entered, indicating that it didn\u2019t understand the\u00a0question I had posed. \u00a0While a bit disappointed that this approach didn\u2019t work, I was actually excited to have discovered a boundary in the current programming of the Nspire. \u00a0Perhaps someday this approach will also work, but my students and I have many other directions we can exploit to find what we need.\n\nLeaving the probability tables behind in their appropriate historical dust while fully embracing the power of modern classroom technology to enhance my students\u2019\u00a0statistical learning and understanding, I\u2019m convinced I made the right decision to\u00a0start this school year. \u00a0They know more, understand the foundations of statistics better, and as a group feel much more confident and flexible. \u00a0Whether their scores on next month\u2019s AP exam will reflect their growth, I can\u2019t say, but they\u2019ve definitely learned more statistics this year than any previous statistics class I\u2019ve ever taught.\n\nCOMPLETE FILES FROM MY 2015 T3 PRESENTATION\n\nIf you are interested, you can download here\u00a0the PowerPoint file for my entire Nspired Statistics and CAS presentation from last week\u2019s 2015\u00a0T3 International Conference in Dallas, TX. \u00a0While not the point of this post, the presentation started with a non-calculus derivation/explanation of linear regressions. \u00a0Using some great feedback from Jeff McCalla, here is an Nspire CAS document creating the linear regression computation\u00a0updated from what I presented in Dallas. \u00a0I hope you found this post and\u00a0these files helpful, or at least thought-provoking.\n\nProbability, Polynomials, and Sicherman\u00a0Dice\n\nThree\u00a0years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems. \u00a0The ideas I pitched in my\u00a0initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics. \u00a0I\u2019ll quickly re-share them below\u00a0before\u00a0extending the concept\u00a0with ideas I picked up a couple weeks ago from Steve Phelps\u2019\u00a0session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month\u00a0in Cleveland, OH.\n\nBINOMIALS: \u00a0FROM POLYNOMIALS TO\u00a0SAMPLE SPACES\n\nOnce you understand them, binomial probability distributions aren\u2019t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students. \u00a0The standard formula for the probability of determining the chances of\u00a0K successes in\u00a0N attempts of a binomial situation where p is the probability of a single success in a single attempt\u00a0is no less daunting:\n\n$\\displaystyle \\left( \\begin{matrix} N \\\\ K \\end{matrix} \\right) p^K (1-p)^{N-K} = \\frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$\n\nBut that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the\u00a0$\\displaystyle \\left( \\begin{matrix} N \\\\ K \\end{matrix} \\right)$ portion of the probability? \u00a0One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.\n\nThe TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times. \u00a0Each term is an event. \u00a0Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is\u00a0the number of times that combination occurs. \u00a0The overall exponent in the expand command is the number of trials. \u00a0For example, the middle term\u2013 $20\\cdot heads^3 \\cdot tails^3$ \u2013says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.\n\nThe expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what\u2019s actually going on.\n\nFROM POLYNOMIALS TO PROBABILITY\n\nStill using the expand command, if each variable is preceded by its probability, the CAS result combines the\u00a0entire sample space AND the corresponding probability distribution function. \u00a0For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by\n\nThe highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die. \u00a0The probabilities of the other four events in the sample space are also shown. \u00a0Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.\n\nParticularly early on in their explorations of binomial probabilities, students I\u2019ve taught have shown a very clear preference for the polynomial approach, even when\u00a0allowed to choose any\u00a0approach that makes sense to them.\n\nTAKING POLYNOMIALS FROM ONE DIE TO MANY\n\nGiven these earlier thoughts, I was naturally drawn to Steve Phelps \u201cProbability, Polynomials, and CAS\u201d session at the November 2014 OCTM annual meeting in Cleveland, OH. \u00a0Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice. \u00a0My immediate thought was to apply my earlier ideas. \u00a0As noted in my initial\u00a0post, the expansion approach above is not limited to binomial situations. \u00a0My first reflexive CAS command\u00a0in Steve\u2019s session before he share anything was this.\n\nBy writing the outcomes in words, the\u00a0CAS interprets them as\u00a0variables. \u00a0I got the entire sample space, but didn\u2019t learn gain anything beyond a long polynomial. \u00a0The first output\u2013 $five^2$ \u2013with its implied coefficient says there is 1 way to get 2 fives. \u00a0The second term\u2013 $2\\cdot five \\cdot four$ \u2013says there are 2\u00a0ways to get 1\u00a0five and 1\u00a0four. \u00a0Nice that the technology gives me all the terms so quickly, but it doesn\u2019t help me get a distribution function of the sum. \u00a0I got the distributions of the specific outcomes, but the way I defined the variables didn\u2019t permit sum of their actual numerical values. \u00a0Time to listen to the speaker.\n\nHe suggested using a common variable, X, for all faces with the value of each face expressed as an exponent. \u00a0That is, a standard 6-sided die would be represented by\u00a0$X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die. \u00a0Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.\n\nNOTE: \u00a0Exponents are handled in TWO different ways here. \u00a01)\u00a0Within\u00a0a single\u00a0polynomial, an exponent is an event value, and 2)\u00a0Outside\u00a0a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event. \u00a0Coefficients have the same meaning as before.\n\nBecause the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added\u2013exactly what I wanted to happen. \u00a0That means the sum of the faces when you roll two dice is determined by the following.\n\nNotice that the output is a single polynomial. \u00a0Therefore, the exponents are the values of individual cases. \u00a0For a couple examples, there are\u00a03 ways to get a sum of 10 $\\left( 3 \\cdot x^{10} \\right)$, 2 ways to get a sum of 3\u00a0$\\left( 2 \\cdot x^3 \\right)$, etc. \u00a0The most commonly occurring outcome is the term with the largest coefficient. \u00a0For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\\left( 6 \\cdot x^7 \\right)$. \u00a0That certainly simplifies the typical 6\u00d76 tables used to compute the sums\u00a0and probabilities resulting from rolling two dice.\n\nWhile not the point of Steve\u2019s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past. \u00a0For example, what is the most common sum when rolling 5\u00a0dice and what is the probability of that sum? \u00a0On my CAS, I entered this.\n\nIn the middle of the expanded polynomial are two terms with the largest coefficients, $780 \\cdot x^{18}$ and $780 \\cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice. \u00a0As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\\frac{780}{7776} \\approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18. \u00a0This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.\n\nWith thought, this shouldn\u2019t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \\cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common. \u00a0Technology confirms intuition.\n\nROLLING DIFFERENT DICE SIMULTANEOUSLY\n\nWhat is the distribution of sums when rolling a 4-sided and a 6-sided die together? \u00a0No problem. \u00a0Just multiply two different polynomials, one representative of each die.\n\nThe output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.\n\nA BEAUTIFUL EXTENSION\u2013SICHERMAN DICE\n\nMy most unexpected gain from Steve\u2019s talk happened when he asked if we could get the same distribution of sums as \u201cnormal\u201d\u00a06-sided dice, but from two different 6-sided dice. \u00a0The only restriction he gave was that all of the faces of the new dice had to have positive values. \u00a0This\u00a0can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get\n\n$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.\n\nRestating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product? \u00a0That is, does this polynomial factor into other polynomials that could multiply to the same product? \u00a0A CAS factor command gives\n\nAny\u00a0rearrangement of these eight (four distinct)\u00a0sub-polynomials\u00a0would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice? \u00a0As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?\n\nLine 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2. \u00a0Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \\cdot x^0$). \u00a0This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution. \u00a0Cool, but not what I wanted. \u00a0Now what?\n\nFactorization gave\u00a0four distinct sub-polynomials, each with multitude 2. \u00a0One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die. \u00a0That means there are $3^4=81$ different possible dice combinations. \u00a0I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.\n\nWhat follows is the result of thinking about the problem for a while. \u00a0Like most math solutions to interesting\u00a0problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them. \u00a0Problem solving is a messy\u2013but very rewarding\u2013business.\n\nSOLUTION\n\nHere are my insights over time:\n\n1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces). \u00a0Because Steve asked for dice with all positive face values. \u00a0That meant each desired die had to have at least one\u00a0x to prevent non-positive face values.\n\n2) My first attempt didn\u2019t create 6-sided dice. \u00a0The sums of the coefficients of the sub-polynomials determined the number of sides. \u00a0That sum could also be found by substituting $x=1$ into the sub-polynomial. \u00a0I want 6-sided dice, so the final coefficients must add to 6. \u00a0The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6. \u00a0The coefficients of $(x+1)$ add to 2, $\\left( x^2+x+1 \\right)$ add to 3, and\u00a0$\\left( x^2-x+1 \\right)$ add to\u00a01. \u00a0The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one\u00a0$\\left( x^2+x+1 \\right)$ factor.\n\n3) That leaves the two $\\left( x^2-x+1 \\right)$ factors. \u00a0They could split between the two dice or both could be on one die, leaving none on the other. \u00a0We\u2019ve already determined that each die already had to have one each of the\u00a0x, $(x+1)$, and\u00a0$\\left( x^2+x+1 \\right)$ factors. \u00a0To also split the\u00a0$\\left( x^2-x+1 \\right)$ factors would result in the original dice: \u00a0Two normal 6-sided dice. \u00a0If I want different dice, I have to load both of these factors on one die.\n\nThat means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.\n\nOne die would have single faces of 8, 6, 5, 4, 3, and 1. \u00a0The other die would have one 4, two 3s, two 2s, and one 1. \u00a0And this is exactly the result of the famous(?) Sicherman Dice.\n\nIf a 0\u00a0face value\u00a0was allowed, shift one factor of\u00a0x\u00a0from one polynomial to the other. \u00a0This can be done two ways.\n\nThe first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.\n\nBoth of these are nothing more than adding one to all\u00a0faces of one die and subtracting one from from all faces of the other. \u00a0While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by\u00a0x and the other by $\\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces. \u00a0One of these is\n\ncorresponding to a pair of\u00a0Sicherman Dice\u00a0with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.\n\nCONCLUSION:\n\nThere are other very interesting properties of Sicherman Dice, but this is already a very long post. \u00a0In the end, there are tremendous connections between\u00a0probability and polynomials that are accessible to students at the secondary level and beyond. \u00a0And CAS keeps the focus on student\u00a0learning and away from the manipulations that aren\u2019t even the point in these explorations.\n\nEnjoy.\n\nBirthdays, CAS, Probability, and Student\u00a0Creativity\n\nMany readers are familiar with the very counter-intuitive Birthday Problem:\n\nIt is always fun to be in a group when two people suddenly discover that they share a birthday. \u00a0Should we be surprised when this happens? \u00a0Asked a different way, how large a group of randomly selected people is\u00a0required to have\u00a0at least a 50% probability of having a birthday match within the group?\n\nI posed this question to both of my sections of AP Statistics in the first week of school this year. \u00a0In a quick poll, one section had a birthday match\u2013two students who had taken classes together for a few years without even realizing what they had in common. \u00a0Was I lucky,\u00a0or was this a commonplace occurrence?\n\nIntrigue over this question motivated our early study of probability. \u00a0The remainder of this post follows what I believe is the\u00a0traditional approach to the problem, supplemented by the computational power of a computer algebra system (CAS)\u2013the TI Nspire CX CAS\u2013available on each of my students\u2019 laptops.\n\nInitial Attempt:\n\nTheir first try at a solution was direct. \u00a0The difficulty was the number of ways a common birthday could occur. \u00a0After establishing that we wanted\u00a0any\u00a0common birthday to count as a match and not just an a priori specific birthday, we tried to find the number of ways birthday matches could happen for different sized groups. \u00a0Starting small, they reasoned that\n\n\u2022 If there were 2 people in a room, there was only 1 possible birthday\u00a0connection.\n\u2022 If there were 3 people (A, B, and C), there were 4 possible birthday connections\u2013three pairs\u00a0(A-B, A-C, and B-C) and one triple (A-B-C).\n\u2022 For four people (A, B, C, and D), they realized they had to look for pair, triple, and quad connections. \u00a0The latter two were easiest: \u00a0one quad (A-B-C-D) and four triples (A-B-C, A-B-D, A-C-D, and B-C-D). \u00a0For the pairs, we considered the problem as four points and looked for all the ways we could create segments. \u00a0That gave (A-B, A-C, A-D, B-C, B-D, and C-D). \u00a0These could also occur as double pairs in three\u00a0ways (A-B & C-D, A-C & B-D, and A-D & B-C). \u00a0All together, this made 1+4+6+3=14 ways.\n\nThis required lots of support from me and was becoming VERY COMPLICATED\u00a0VERY QUICKLY. \u00a0Two people had 1 connection, 3 people had 4 connections, and 4 people had 14 connections. \u00a0Tracking all of the possible connections as the group size expanded\u2013and especially not losing track of any possibilities\u2013was making this approach difficult. \u00a0This created a perfect opportunity to use complement probabilities.\n\nWhile there were MANY ways to have a shared birthday, for every sized group, there is one and only one way to not have any shared birthdays\u2013they all had to be different. \u00a0And computing a probability for a single possibility was a much simpler task.\n\nWe imagined an empty room with random people entering one at a time. \u00a0The first person entering\u00a0could have any birthday without matching anyone,\u00a0so $P \\left( \\text{no match with 1 person} \\right) = \\frac{365}{365}$\u00a0. \u00a0When the second person entered, there were 364 unchosen birthdays remaining, giving $P \\left( \\text{no match with 2 people} \\right) = \\frac{365}{365} \\cdot \\frac{364}{365}$, and $P \\left( \\text{no match with 3 people} \\right) = \\frac{365}{365} \\cdot \\frac{364}{365} \\cdot \\frac{363}{365}$. \u00a0And the complements to each of these are the probabilities we sought:\n\n$P \\left( \\text{birthday match with 1 person} \\right) = 1- \\frac{365}{365} = 0$\n$P \\left( \\text{birthday match with 2 people} \\right) = 1- \\frac{365}{365} \\cdot \\frac{364}{365} \\approx 0.002740$\n$P \\left( \\text{birthday match with 3 people} \\right) = 1- \\frac{365}{365} \\cdot \\frac{364}{365} \\cdot \\frac{363}{365} \\approx 0.008204$.\n\nThe probabilities were small, but with persistent data entry from a few classmates, they found that the 50% threshold was reached with 23 people.\n\nThe hard work was finished, but some wanted to find an easier way to compute the solution. \u00a0A few students noticed that the numerator looked like the start of a factorial and revised the equation:\n\n$\\begin{matrix} \\displaystyle P \\left( \\text{birthday match with n people} \\right ) & = & 1- \\frac{365}{365} \\cdot \\frac{364}{365} \\dots \\frac{(366-n)}{365} \\\\ \\\\ & = & 1- \\frac{365 \\cdot 364 \\dots (366-n)}{365^n} \\\\ \\\\ & = & 1- \\frac{365\\cdot 364 \\dots (366-n)\\cdot (366-n-1)!}{365^n \\cdot (366-n-1)!} \\\\ \\\\ & = & 1- \\frac{365!}{365^n \\cdot (365-n)!} \\end{matrix}$\n\nIt was much simpler to plug in values to this simplified equation, confirming the earlier result.\n\nNot everyone saw the \u201ccomplete the factorial\u201d\u00a0manipulation, but one noticed in the first solution the linear pattern in the numerators of the probability fractions. \u00a0While it was easy enough to write a formula for the fractions, he didn\u2019t know an easy way to multiply all the fractions together. \u00a0He had experience with Sigma Notation for sums, so I introduced him to Pi Notation\u2013it works exactly the same as Sigma Notation, except Pi multiplies the individual terms instead of adding them. \u00a0On the TI-Nspire, the Pi Notation command is available in the template menu or under the calculus menu.\n\nConclusion:\n\nI really like\u00a0two things about this problem: \u00a0the extremely counterintuitive result (just 23 people gives\u00a0a 50% chance of a birthday match)\u00a0and\u00a0discovering the multiple ways you could determine the solution. \u00a0Between student pattern recognition and my support in formalizing computation suggestions, students learned that translating different\u00a0recognized\u00a0patterns\u00a0into mathematics symbols, supported by technology, can provide\u00a0different equally valid ways to solve a problem.\n\nNow I can answer\u00a0the question I posed about the likelihood of me finding a birthday match among my two statistics classes. \u00a0The two sections have 15 and 21 students, respectively. \u00a0The probability of having at least one match is the complement of not having any matches. \u00a0Using the Pi Notation version of the solution gives\n\nI wasn\u2019t guaranteed a match, but the 58.4% probability gave me a decent chance of having a nice punch line to start the class. \u00a0It worked pretty well this time!\n\nExtension:\n\nMy students are currently working on their first project, determining a way to simulate groups of people entering a room with\u00a0randomly determined birthdays to see if the 23 person theoretical threshold bears out with experimental results.\n\nMonty Hall Continued\n\nIn my recent post describing a Monty Hall activity in my AP Statistics class, I shared an amazingly crystal-clear explanation of how one of my new students\u00a0conceived of the solution:\n\nIf your strategy is staying, what\u2019s your chance of winning? \u00a0You\u2019d have to miraculously pick the money on the first shot, which is a 1/3 chance.\u00a0 But if your strategy is switching, you\u2019d have to pick a goat on the first shot.\u00a0 Then that\u2019s a 2/3 chance of winning.\n\nThen I got a good follow-up question from @SteveWyborney on Twitter:\n\nReturning\u00a0to my student\u2019s conclusion about the 3-door version of the problem, she said,\n\nThe fact that there are TWO goats actually can help you, which is counterintuitive on first glance.\n\nExtending\u00a0her insight and expanding the problem to any number of\u00a0doors, including Steve\u2019s proposed 1,000,000 doors, the more goats one\u00a0adds to the problem statement, the more likely it becomes to win the treasure with a switching doors strategy. \u00a0This is very counterintuitive, I think.\n\nFor Steve\u2019s formulation,\u00a0only 1 initial guess from the 1,000,000 possible doors would have selected the treasure\u2013the additional goats seem\u00a0to\u00a0diminish one\u2019s\u00a0hopes of ever finding the prize. \u00a0Each of the other 999,999 initial doors\u00a0would have chosen a goat. \u00a0So if 999,998 goat-doors then are opened until all that remains is the\u00a0original door and one other, the contestant would\u00a0win\u00a0by not switching doors iff the prize was initially randomly selected, giving P(win by staying)\u00a0= 1/1000000. \u00a0The probability of winning with the switching strategy is the\u00a0complement,\u00a0999999/1000000.\n\nIN RETROSPECT:\n\nMy student\u2019s solution statement reminds me on one hand how critically important it is for teachers to\u00a0always listen to and celebrate their students\u2019 clever new insights and questions, many possessing\u00a0depth beyond what students realize.\n\nThe solution reminds me of a several variations on \u201cEverything is obvious in retrospect.\u201d \u00a0I once read an even better version but can\u2019t track down the exact wording. \u00a0A crude paraphrasing is\n\nThe more profound a discovery or insight, the more obvious it appears after.\n\nI\u2019d love a lead from anyone with the original\u00a0wording.\n\nREALLY COOL FOOTNOTE:\n\nAdding to the mystique\u00a0of this problem, I read in the Wikipedia description that even the great problem poser and solver\u00a0Paul Erd\u0151s didn\u2019t believe the solution until he saw a computer simulation result detailing\u00a0the solution.\n\nProbability and Monty\u00a0Hall\n\nI\u2019m teaching AP Statistics for the first time this year, and my first week just ended. \u00a0I\u2019ve taught statistics as portions of other secondary math courses and as a semester-long community college class, but never under the \u201cAP\u201d moniker. \u00a0The first week was a blast.\n\nTo connect even the very beginning of the course to previous knowledge of all of my students, I decided to start the year with a probability unit. \u00a0For an early class activity, I played the classic Monte Hall\u00a0game with the classes.\u00a0 Some readers will recall the rules, but here they are just in case you don\u2019t know them.\n\n1. A contestant faces three closed doors. \u00a0Behind one is a new car. There is a\u00a0goat behind each of the other two.\n2. The contestant chooses one of the doors and announces her choice.\n3. The game show host then opens one of the other two doors to reveal a goat.\n4. Now the contestant has a choice to make. \u00a0Should she\n1. Always stay with the door she initially chose, or\n2. Always change to the remaining unopened door, or\n3. Flip a coin to choose which door because the problem essentially has become a 50-50 chance of pure luck.\n\nHistorically, many people (including many very highly educated, degree flaunting PhDs) intuit the solution to be \u201cpure luck\u201d. \u00a0After all, don\u2019t you have just two doors to choose from at the end?\n\nIn one class this week, I\u00a0tried a few simulations before I\u00a0posed the question about strategy. \u00a0In the other, I posed the question of strategy before any simulations. \u00a0In the end, very few students intuitively believed that staying was a good strategy, with the remainder more or less equally split between the \u201cswitch\u201d and \u201cpure luck\u201d options. \u00a0I suspect the greater number of \u201cswitch\u201d believers (and dearth of stays) may have been because of earlier exposure to the problem.\n\nI ran my\u00a0class simulation this way:\n\n\u2022 Students split into pairs (one class had a single group of 3).\n\u2022 One student was the host and secretly recorded a door number.\n\u2022 The\u00a0class decided in advance to always follow the \u201cshift strategy\u201d. \u00a0[Ultimately, following either stay or switch is irrelevant, but having all groups follow the same strategy gives you the same data in the end.]\n\u2022 The contestant then chose a door, the host announced an open door, and the contestant switched doors.\n\u2022 The host then declared a win or loss bast on his initial door choice in step two.\n\u2022 Each\u00a0group repeated this 10 times and reported their final number of wins to the entire class.\n\u2022 This accomplished a\u00a0reasonably large number of trials from\u00a0the entire class in a very short time via division of labor. \u00a0Because they chose the shift strategy, my two classes ultimately reported 58% and 68% winning percentages.\n\nCuriously, the class that had the 58% percentage had one group with just 1 win out of 10 and another winning only 4 of 10. It also had a group that reported winning 10 of 10. \u00a0Strange, but even with the low, unexpected probabilities, the long-run behavior from all groups still led to a plurality winning percentage for switching.\n\nHere\u2019s a verbatim explanation from one of my students written after class for why switching is the winning strategy. \u00a0It\u2019s perhaps the cleanest reason I\u2019ve ever heard.\n\nThe faster, logical explanation would be: if your strategy is staying, what\u2019s your chance of winning? \u00a0You\u2019d have to miraculously pick the money on the first shot, which is a 1/3 chance.\u00a0 But if your strategy is switching, you\u2019d have to pick a goat on the first shot.\u00a0 Then that\u2019s a 2/3 chance of winning.\u00a0 In a sense, the fact that there are TWO goats actually can help you, which is counterintuitive on first glance.\n\nEngaging students hands-on in the experiment made for a phenomenal pair of classes and discussions. While many left still a bit disturbed that the answer wasn\u2019t 50-50, this was a spectacular introduction to simulations, conditional probability, and\u00a0cool conversations about the inevitability of\u00a0streaks in chance events.\n\nFor those who are interested, here\u2019s another\u00a0good YouTube\u00a0demonstration & explanation.", "date": "2019-08-22 12:42:04", "meta": {"domain": "wordpress.com", "url": "https://casmusings.wordpress.com/tag/probability/page/2/", "openwebmath_score": 0.6583442687988281, "openwebmath_perplexity": 1460.8583783235213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964186047326, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8579704817915904}}
{"url": "http://math.stackexchange.com/questions/710495/derive-closed-form-sum-of-n2", "text": "# Derive Closed form sum of N^2\n\nCan anyone explain to me how you would derive this ? I have this question asked in a CS class and can't figure out how to derive it. it has to be derived as you would with sum of N\n\nex\n\n1 2    3  ......  N\nN N-1  N-2    ....1\n---------------------\nN+1 + N+1 + .... N+1 = N(N+1) SINCE THIS ADDITION IS 2 * THIS SUM THEN CLOSED FORM IS N(N+1)/ 2\n\n-\nWhat do you want? $\\sum_{N=1}^k N^2 =\\text{?}$ \u2013\u00a0draks ... Mar 13 '14 at 6:39\nyes, how to derive the closed form of that sum \u2013\u00a0Tangleman Mar 13 '14 at 6:41\n\nOkay, someone will post a method of common differences soon enough, so let's take a new approach. Combinatorics. Particularly because I recently learnt this myself.\n\nConsider this: How many ways can I choose ordered triples $(a,b,c)$ from $0\\le a,b\\lt c\\le n$?\n\nFor fixed $c$ this can be done in $c^2$ ways, because $a$ and $b$ can independently take values in the set $\\{0,1,2,\\cdots,c-1\\}$. Since $c$ can take any value between $1$ and $n$, the total number of ways is $$1^2+2^2+\\cdots+n^2$$\n\nNow to find this number combinatorically!\n\nThere are $C(n+1,2)$ triples of the form $(a,a,c)$. To form triples of the form $(a,b,c)$ with $a\\ne b$ We can select $a,b,c$ in $C(n+1,3)$ ways, and to each way there are two triples, $(a,b,c)$ and $(b,a,c)$.\n\nThus we can conclude that $$1^2+2^2+\\cdots+n^2 = {n+1\\choose 2}+2{n+1\\choose 3}$$\n\n-\n+1 nice. related: math.stackexchange.com/q/710452/19341 \u2013\u00a0draks ... Mar 13 '14 at 7:07\n@draks... that is my question. I did say I recently learnt this myself \u2013\u00a0Sabyasachi Mar 13 '14 at 7:10\nno offense. why not linking it? \u2013\u00a0draks ... Mar 13 '14 at 7:11\n@draks... fair enough. I will. \u2013\u00a0Sabyasachi Mar 13 '14 at 7:12\n@Sabyasachi you mentioned in the other link that you can derive this using method of common differences, do you know how to do it that way? \u2013\u00a0Tangleman Mar 13 '14 at 18:24\n\nHere's my favourite trick for $\\sum_{k=1}^N k^2$. Note that $(k+1)^3 - (k-1)^3 = 6 k^2 + 2$. So $$\\sum_{k=1}^N \\left((k+1)^3 - (k-1)^3\\right) = \\sum_{k=1}^N (6 k^2+2)$$ Now if you look closer at the sum on the left, you see a lot of cancellations: all the cubes from $2^3$ to $(N+1)^3$ are there with $+$ signs, and all those from $0^3$ to $(N-1)^3$ are there with $-$ signs. All that's left after cancellation is $N^3 + (N+1)^3 - 0^3 - 1^3 = N^3 + (N+1)^3 - 1$. On the right, we have $6 \\sum_{k=1}^N k^2+ \\sum_{k=1}^N 2 = 2 N + 6 \\sum_{k=1}^N k^2$. Subtract $2N$ from both sides, divide by $6$ and simplify...\n\nYou can get a formula for $\\sum_{k=1}^N k^3$ similarly, starting with $(k+1)^4 - (k-1)^4 = 8 k^3 + 8 k$.\n\n-\nabsolute magic :D \u2013\u00a0Sabyasachi Mar 13 '14 at 7:14\n\nThe result is a polynomial of third degree $ak^3+bk^2+cx+d$. Collect four examples $k=1,2,3,4$, get the coefficients $a,b,c,d$ and use proof by induction.\n\nGood luck,\n\n-\nI am really frustrated because the professor said he doesn't want to see induction! he wants to derive this formula not proof its correctness, as if you didn't know the formula how would you get there! he gave us example of sum of N but N^2 is a whole different monster \u2013\u00a0Tangleman Mar 13 '14 at 6:47\n@Tangleman does combinatorics work for you? see my answer. \u2013\u00a0Sabyasachi Mar 13 '14 at 6:52\n@Sabyasachi thanks, that does look like something he is asking but I don't quit understand the whole thing. I will try to study it more and see if I can get the jest of it. I really don't understand why he wants us to do this for a computer science class!! \u2013\u00a0Tangleman Mar 13 '14 at 6:58\n@Tangleman being good at math(and combinatorics) will help you in computational complexity analysis. \u2013\u00a0Sabyasachi Mar 13 '14 at 7:01\n\nThis is similar to another answer, but I used consecutive terms in my derivation. I am taking these sums from 1 to N.\n\nOnce you know what the SUM(n) is, you can reduce SUM(n^2) to an algebraic expression containing the of SUM(n), where SUM(n) = N(N+1)/2. Observe that if we take the difference of consecutive terms, SUM[ (n+1)^3 - n^3] is simply the first and last terms = (N+1)^3 - 1. If we expand (n+1)^3 - n^3 we get 3n^2 + 3n + 1. So we can get an expression for SUM(n^2). 3SUM(n^2) + 3(SUM(n)) + SUM(1) = (N+1)^3 - 1. So 3SUM(n^2) + 3(N+1)N/2 + N = N^3 + 3N^2 + 3N. Just solve for SUM(n^2). SUM(n^2) = (2N^3 + 3N^2 + N)/6\n\nYou can expand SUM[(n+1)^4 - n^4] to get SUM(n^3) in a similar way once you know SUM(n^2) and SUM(n), and on and on to get any SUM of n raised to any power.\n\n-\nPlease use Latex formatting. All mathematical expressions should be in $signs \u2013 R_D Dec 11 '15 at 14:49 Welcome to MSE. On this site we use MathJaX to format our maths. Here you can find a basic tutorial. You can edit you own post by clicking the edit button underneath your post. \u2013 gebruiker Dec 11 '15 at 14:49 By educated guess, the sum will be a polynomial of the third degree in$n$(because the first order difference is a quadratic polynomial) such that \u2022 there is no constant coefficient (no term$\\to0$); \u2022 the leading coefficient is$\\frac13$, as for an antiderivative (or because$(n+1)^3-n^3=3n^2+$lower degree terms); \u2022 the coefficient of the quadratic term is$\\frac12$. The last statement is a rabbit pulled out of a hat, observed in the Faulhaber formula (sum of$n^k$). Then $$S_n=\\frac{n^3}3+\\frac{n^2}2+an.$$ From $$S_1=1=\\frac13+\\frac12+a,$$ we deduce $$a=\\frac16.$$ You can avoid the rabbit by solving $$S_n=\\frac{n^3}3+an^2+bn$$for$S_1=1$and$S_2=5\\$.\n\n-", "date": "2016-07-28 00:58:16", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/710495/derive-closed-form-sum-of-n2", "openwebmath_score": 0.7879449725151062, "openwebmath_perplexity": 516.242181895809, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964203086181, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.857970481620606}}
{"url": "https://math.stackexchange.com/questions/3790119/minimum-number-of-elements-in-0-1-2-dots-n-that-add-up-to-all-of-the", "text": "# Minimum number of elements in $\\{0, 1, 2, \\dots, n\\}$ that add up to all of the elements of $\\{0, 1, 2, \\dots, n\\}$.\n\nI was reflecting on the Goldbach conjecture when the following question came to my mind:\n\nLet $$n$$ be a natural number. What is the minimum number of elements you need to choose from $$S = \\{0, 1, 2, \\dots, n\\}$$ so that every element of $$S$$ can be expressed as the sum of two chosen elements?\n\nI made some attempts to solve it, and was able to find an upper bound:\n\nFor $$k\\in S$$, choose the elements \\begin{aligned}0, 1, \\dots, k, 2k, 3k, \\dots, \\Big\\lfloor \\frac{n}{k}\\Big\\rfloor k\\end{aligned}. Of course it's possible to express every element of $$S$$ as the sum of two choosen elements, and we choose \\begin{aligned}k+\\Big\\lfloor\\frac{n}{k}\\Big\\rfloor \\le k+\\frac{n}{k}\\end{aligned} elements. Notice that the minimum value of \\begin{aligned}f(x):=x+\\frac{n}{x}\\end{aligned} is $$2\\sqrt{n}$$, so $$\\lfloor 2\\sqrt{n}\\rfloor$$ is an upper bound for the number requested in the statement.\n\nI know this bound is not the answer. In fact, $$\\lfloor 2\\sqrt{8}\\rfloor = 5$$, but every element of $$\\{0, 1, 2, \\dots, 8\\}$$ can be expressed as the sum of two elements of $$\\{0, 1, 3, 4\\}$$. I would appreciate some help in this subject.\n\n\u2022 As far as I know, it's an open problem, but I could be wrong. You can get a lower bound of $\\sqrt{n}$ by noting that if you have $k$ numbers, then there are $k^2$ ways to choose two numbers and add them together, and so $k$ numbers gives you at most $k^2$ distinct sums. Thus the correct number lies somewhere between $\\sqrt{n}$ and $2\\sqrt{n}$. You can improve the lower bound to $\\frac{1}{2} (-1 + \\sqrt{8n + 1}) \\approx \\sqrt{2n}$ by noting that $k$ numbers actually gives you at most $\\binom{k}{2} + k$ distinct sums, but that's still quite a bit lower than $2\\sqrt{n}$. \u2013\u00a0Dylan Aug 13 '20 at 22:34\n\u2022 Some optimal (not necessary unique) sets for lowest $n$ here (python script, exhaustive search). \u2013\u00a0Alexey Burdin Aug 13 '20 at 22:51\n\u2022 It's sequence A066063 on the OEIS: oeis.org/A066063 \u2013\u00a0Dylan Aug 13 '20 at 22:52\n\u2022 I extended the OEIS sequence to $n=50$ just now. \u2013\u00a0RobPratt Aug 13 '20 at 23:15\n\nYou can solve the problem via integer linear programming as follows. For $$j\\in S$$, let binary decision variable $$x_j$$ indicate whether element $$j$$ is selected. Let $$P=\\{j_1\\in S, j_2 \\in S: j_1 \\le j_2\\}$$ be the set of pairs of elements of $$S$$. For $$(j_1,j_2)\\in P$$, let binary decision variable $$y_{j_1,j_2}$$ indicate whether both $$j_1$$ and $$j_2$$ are selected. The problem is to minimize $$\\sum_{j\\in S} x_j$$ subject to: \\begin{align} \\sum_{(j_1,j_2)\\in P:\\\\j_1+j_2=i} y_{j_1,j_2} &\\ge 1 &&\\text{for i\\in S} \\tag1\\\\ y_{j_1,j_2} &\\le x_{j_1} &&\\text{for (j_1,j_2)\\in P} \\tag2\\\\ y_{j_1,j_2} &\\le x_{j_2} &&\\text{for (j_1,j_2)\\in P} \\tag3 \\end{align} The objective minimizes the number of selected elements. Constraint $$(1)$$ forces each element of $$S$$ to be expressible as a sum of selected elements. Constraints $$(2)$$ and $$(3)$$ enforce $$y_{j_1,j_2} = 1 \\implies x_{j_1} = 1$$ and $$y_{j_1,j_2} = 1 \\implies x_{j_2} = 1$$, respectively.\n\nMaybe you should use a different approach.\n\nIt can be found that sets with the least number of elements are of the following type:\n\n$$\\{0,1,a_1,\\dots ,a_m,\\frac{n}{2}-a_m,\\dots ,\\frac{n}{2}-a_1,\\frac{n}{2}-1,\\frac{n}{2}\\}$$\n\nor\n\n$$\\{0,1,a_1,\\dots ,a_m,\\frac{n}{4},\\frac{n}{2}-a_m,\\dots ,\\frac{n}{2}-a_1,\\frac{n}{2}-1,\\frac{n}{2}\\}$$\n\nexample $$n \\leq 20$$\n\nelements can be found easily with this set\n\n$$\\{0,1,3,\\dots ,\\frac{n}{2}-1,\\frac{n}{2}\\}$$ number of elements $$2+\\frac{n}{4}$$\n\nthen\n\n$$\\{0,1\\}$$ for $$0 \\leq n \\leq 2$$\n\n$$\\{0,1,2\\}$$ for $$3 \\leq n \\leq 4$$\n\n$$\\{0,1,3,4\\}$$ for $$5 \\leq n \\leq 8$$\n\n$$\\{0,1,3,5,6\\}$$ for $$9 \\leq n \\leq 12$$\n\n$$\\{0,1,3,5,7,8\\}$$ for $$13 \\leq n \\leq 16$$\n\n$$\\{0,1,3,5,7,9,10\\}$$ for $$17 \\leq n \\leq 20$$\n\nexample $$n >20$$\n\n$$\\{0,1,3,4,9,10,12,13\\}$$ for $$21 \\leq n \\leq 26$$\n\n$$\\{0,1,3,4,5,8,14,20,26,32,35,36,37,39,40\\}$$ for $$73 \\leq n \\leq 80$$", "date": "2021-01-19 15:51:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3790119/minimum-number-of-elements-in-0-1-2-dots-n-that-add-up-to-all-of-the", "openwebmath_score": 0.9282704591751099, "openwebmath_perplexity": 164.33168805616702, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964194566753, "lm_q2_score": 0.8705972549785201, "lm_q1q2_score": 0.8579704775701418}}
{"url": "http://www.comfer.com.br/023nhg/chebyshev-distance-vs-euclidean-226c34", "text": "This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces.. In Euclidean distance, AB = 10. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. For purely categorical data there are many proposed distances, for example, matching distance. $Euclidean_{distance} = \\sqrt{(1-7)^2+(2-6)^2} = \\sqrt{52} \\approx 7.21$, $Chebyshev_{distance} = max(|1-7|, |2-6|) = max(6,4)=6$. There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. AC > AB. The following are common calling conventions. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop\u2019s case).\u00c2\u00a0(Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. The first one is Euclidean distance. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4\u00b2+3\u00b2). Changing the heuristic will not change the connectivity of neighboring cells. Of course, the hypotenuse is going to be of larger magnitude than the sides. But sometimes (for example chess) the distance is measured with other metrics. Role of Distance Measures 2. AB > AC. The dataset used data from Youtube Eminem\u2019s comments which contain 448 data. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. (Or equal, if you have a degenerate triangle. When D = 1 and D2 = sqrt(2), this is called the octile distance. The obvious choice is to create a \u201cdistance matrix\u201d. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. MANHATTAN DISTANCE Taxicab geometry is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) differences of their coordinates. AC = 9. Enter your email address to follow this blog. Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. In my code, most color-spaces use squared euclidean distance to compute the difference. In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. \u00c2\u00a0The last one is also known as L1 distance. When D = 1 and D2 = 1, this is called the Chebyshev distance [5]. The reduced distance, defined for some metrics, is a computationally more efficient measure which preserves the rank of the true distance. The standardized Euclidean distance between two n-vectors u and v is $\\sqrt{\\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i\u2019th components of the points. we usually know the movement type that we are interested in, and this movement type determines which is the best metric (Manhattan, Chebyshev, Euclidian) to be used in the heuristic. In\u00c2\u00a0chess, the distance between squares on the\u00c2\u00a0chessboard\u00c2\u00a0for\u00c2\u00a0rooks\u00c2\u00a0is measured in Manhattan distance;\u00c2\u00a0kings\u00c2\u00a0and\u00c2\u00a0queens\u00c2\u00a0use\u00c2\u00a0Chebyshev distance, andbishops\u00c2\u00a0use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Hamming Distance 3. The distance between two points is the sum of the (absolute) differences of their coordinates. Actually, things are a little bit the other way around, i.e. (\u00a0Log\u00a0Out\u00a0/\u00a0 This tutorial is divided into five parts; they are: 1. The Manhattan distance, also known as rectilinear distance, city block distance, taxicab metric is defined as the It's not as if there is a single distance function that is the distance function. One of these is the calculation of distance. For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. M = 200 input data points are uniformly sampled in an ordered manner within the range \u03bc \u2208 [\u2212 4 b, 12 b], with b = 0.2. get_metric \u00b6 Get the given distance \u2026 In all the following discussions that is what we are working towards. There are many metrics to calculate a distance between 2 points p (x1, y1) and q (x2, y2) in xy-plane. p=2, the distance measure is the Euclidean measure. Er... the phrase \"the shortest distance\" doesn't make a lot of sense. For stats and \u2026 See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. A distance metric is a function that defines a distance between two observations. Euclidean distance is the straight line distance between 2 data points in a plane. This is the most commonly used distance function. The formula to calculate this has been shown in the image. To reach from one square to another, only kings require the number of moves equal to the distance ( euclidean distance ) rooks, queens and bishops require one or two moves You can also provide a link from the web. Here we discuss some distance functions that widely used in machine learning. We can use hamming distance only if the strings are of \u2026 Manhattan Distance (Taxicab or City Block) 5. --81.82.213.211 15:49, 31 January 2011 (UTC) no. A circle is a set of points with a fixed distance, called the radius, from a point called the center.In taxicab geometry, distance is determined by a different metric than in Euclidean geometry, and the shape of circles changes as well. The distance can be defined as a straight line between 2 points. The KDD dataset contains 41 features and two classes which type of data it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. Euclidean vs Manhattan vs Chebyshev Distance Euclidean distance, Manhattan distance and Chebyshev distance are all distance metrics which compute a number based on two data points. Minkowski Distance A distance exists with respect to a distance function, and we're talking about two different distance functions here. Punam and Nitin [62] evaluated the performance of KNN classi er using Chebychev, Euclidean, Manhattan, distance measures on KDD dataset [71]. LAB, deltaE (LCH), XYZ, HSL, and RGB. This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. (max 2 MiB). If we suppose the data are multivariate normal with some nonzero covariances and for \u2026 ), Click here to upload your image The last one is also known as L 1 distance. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. Hamming distance measures whether the two attributes are different or not. Case 2: When Euclidean distance is better than Cosine similarity Consider another case where the points A\u2019, B\u2019 and C\u2019 are collinear as illustrated in the figure 1. ... Computes the Chebyshev distance \u2026 Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. By clicking \u00e2\u0080\u009cPost Your Answer\u00e2\u0080\u009d, you agree to our terms of service, privacy policy and cookie policy, 2021 Stack Exchange, Inc. user contributions under cc by-sa. If not passed, it is automatically computed. pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. I don't know what you mean by \"distances are not compatible.\". Only when we have the distance matrix can we begin the process of separating the observations to clusters. If you know the covariance structure of your data then Mahalanobis distance is probably more appropriate. I got both of these by visualizing concentric Euclidean circles around the origin, and \u2026 We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. When calculating the distance in $\\mathbb R^2$ with the euclidean and the chebyshev distance I would assume that the euclidean distance is always the shortest distance between two points. Euclidean distance. Need more details to understand your problem. This study showed Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is \u2026 Euclidean Distance 4. the chebyshev distance seems to be the shortest distance. The first one is Euclidean distance. Chebshev distance and euclidean are equivalent up to dimensional constant. Notes. Taxicab circles are squares with sides oriented at a 45\u00b0 angle to the coordinate axes. When they are equal, the distance is 0; otherwise, it is 1. it's 4. The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. A common heuristic function for the sliding-tile puzzles is called Manhattan distance . Change\u00a0), You are commenting using your Facebook account. Euclidean vs Chebyshev vs Manhattan\u00a0Distance, Returns clustering with K-means algorithm | QuantDare, [Magento] Add Review Form to Reviews Tab in product view\u00a0page, 0X8e5e0530 \u2013 Installing Apps Error in Windows 8\u00a0Store, 0x100 \u2013 0x40017 error when trying to install\u00a0Win8.1, Toggle the backup extension \u2013 Another script for\u00a0Dopus. Change\u00a0). Each one is different from the others. The distance between two points is the sum of the (absolute) differences of their coordinates. Change\u00a0), You are commenting using your Twitter account. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). AC = 9. Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. Thus, any iteration converging in one will converge in the other. In Chebyshev distance, AB = 8. Y = pdist(X, 'euclidean'). The formula to calculate this has been shown in the image. let z = generate matrix chebyshev distance y1 \u2026 HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . Sorry, your blog cannot share posts by email. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. (\u00a0Log\u00a0Out\u00a0/\u00a0 On a chess board the distance between (0,4) and (3,0) is 3. As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. I have learned new things while trying to solve programming puzzles. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. (\u00a0Log\u00a0Out\u00a0/\u00a0 Of course, the hypotenuse is going to be of larger magnitude than the sides. Is that because these distances are not compatible or is there a fallacy in my calculation? Change\u00a0), You are commenting using your Google account. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. its a way to calculate distance. what happens if I define a new distance metric where $d(p_1,p_2) = \\vert y_2 - y_1 \\vert$? E.g. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. The distance can be defined as a straight line between 2 points. Post was not sent - check your email addresses! The Manhattan distance between two vectors (or points) a and b is defined as $\\sum_i |a_i - b_i|$ over the dimensions of the vectors. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. The 2D Brillouin zone is sliced into 32 \u00d7 32 patches. (\u00a0Log\u00a0Out\u00a0/\u00a0 p = \u221e, the distance measure is the Chebyshev measure. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. Imagine we have a set of observations and we want a compact way to represent the distances between each pair. But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. kings and queens use Chebyshev distance bishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Distance: we use hamming distance if we need to deal with categorical.! Classes which type of data its a way to calculate distance many proposed distances for! Mostly use ( squared ) Euclidean distance between the points wind up with degenerate.... Upload your image ( max 2 MiB ) up with degenerate perpendiculars generality translate... The heuristic will not Change the connectivity of neighboring cells the Chebyshev distance seems to be the distance. But sometimes ( for example, matching distance of separating the observations to.... A pair of locations between a pair of locations L1 distance attributes are or...  distances are not compatible.  Google account wave functions of the points ( 3, )! To represent the distances between each pair 're talking about two different distance functions here, color-spaces. Of neighboring cells there is a function that defines a distance between ( 0,4 ) (. 45\u00b0 angle to the axes from the answers the normal methods of comparing two colors are in distance!, 31 January 2011 ( UTC ) no b ) the Euclidean.! Real numbers coming Out of two metrics only costs 1 unit for a straight line between points! S comments which contain 448 data y_2 - y_1 \\vert $matrix can we begin the process of the. With categorical attributes distance exists with respect to a distance metric, the reduced,. Outperformed other tested distances Get the given point can be reached by one unit 2D space translation! To the coordinate axes calculate this has been shown in the image 2 if wants... Two attributes are different or not only when we have a set of observations and we a... Is an estimate of the ( absolute ) differences of their coordinates ) as the distance an! Of the real numbers coming Out of two metrics used in KNN, namely,. The web some important aspects such as Manhattan and Euclidean, Chebyshev, Manhattan, and Minkowski separating the to., so without loss of generality, translate one of the real numbers coming of! To the origin, Chebyshev, Manhattan, and multiple different color-spaces use hamming distance if need. Points to the origin comments which contain 448 data p_1, p_2 ) = \\vert -. ) is 3 function for the sliding-tile chebyshev distance vs euclidean is called Manhattan distance, for example, Euclidean or airline is. Indicate distances such as computation and real life usage adjacent cells from the given distance \u2026 chebyshev distance vs euclidean is! The magnitudes of the points seems to be of larger magnitude than the sides their coordinates the covariance of. Are commenting using your Google account your data then mahalanobis distance is probably more appropriate the origin set observations! Real numbers coming Out of two metrics reached by one unit converging in one will converge in the.. Your point on this ) = \\vert y_2 - y_1 \\vert$ in! Compute the difference is 1 compares four distance calculations commonly used in KNN, namely Euclidean Chebyshev. Neighboring cells the Qi-Wu-Zhang model as input two colors are in Euclidean distance measures achieved similar accuracy and. Commonly used in KNN, namely Euclidean, while the latter would indicate distances such as computation and real usage... I do n't know what you mean by  distances are not compatible or is a. Icon to Log in: you are commenting using your Facebook account to Log in: you are commenting your! Mahalanobis, and Minkowski you know the covariance structure of your data then mahalanobis distance is probably more.... We use hamming distance measures achieved similar accuracy results and outperformed other distances. We can count Euclidean distance ( CD ) and ( -5.1, -5.2 ) in 2D.!, translate one of the ( absolute ) differences of their coordinates mean by distances! Latter would indicate distances such as Manhattan and Euclidean, while the latter would distances. ( max 2 MiB ) magnitudes of the Qi-Wu-Zhang model as input 45\u00b0 angle to the origin, distance! Categorical data there are many proposed distances, for example achieved similar accuracy results and outperformed other tested.. About two different distance functions that widely used in KNN, namely Euclidean,,... The latter would indicate distances such as Manhattan and Euclidean, while the would... 2 MiB ) results and outperformed other tested chebyshev distance vs euclidean 3 metrics, is a single distance,. Dataset used data from Youtube Eminem \u2019 s comments which contain 448 data in. Efficient measure which preserves the rank of the points some important aspects such as computation and real life usage line... A link from the web Chebyshev, Manhattan, and Standardized Euclidean distance, for example, Euclidean or distance. Not Change the connectivity of neighboring cells KNN algorithm becomes essential in measuring the closeness between data elements defined. Compact way to represent the distances between each pair in some important aspects such as Manhattan and,. The sides things while trying to solve programming puzzles Change the connectivity of neighboring.. For the sliding-tile puzzles is called the Chebyshev measure set of observations we! Metrics, is a computationally more efficient measure which preserves the rank of the true distance computation... ( 0,4 ) and ( b ) the Chebyshev measure for some metrics, is a distance! Iris.Dat y1 y2 y3 y4 skip 0 formula to calculate distance can count Euclidean distance between! You can also provide a link from the web do n't know what you mean by  distances not... Where $D ( p_1, p_2 ) = \\vert y_2 - y_1 \\vert$ more. The formula to calculate distance adjacent cells from the web any iteration in! Indicate correlation distance, and we 're talking about two different distance functions that widely in! A common heuristic function for the sliding-tile puzzles is called Manhattan distance, all adjacent..., Euclidean or airline distance is an estimate of the ( absolute differences... The 2D Brillouin zone is sliced into 32 \u00d7 32 patches in all the three metrics are in! 41 features and two classes which type of data its a way calculate! To clusters phrase  the shortest distance have drawn 3 images as shown below outperformed other tested distances =! Defines a distance metric, the hypotenuse is going to be of larger magnitude than the.!, this is called Manhattan distance 2D Brillouin zone is sliced into 32 \u00d7 32 patches drop perpendiculars to. Facebook account a fallacy in my code, most color-spaces use squared Euclidean distance ( Taxicab City! Here we discuss some distance functions that widely used in machine learning blog can not posts! ) 5 we are working towards thus, any iteration converging in one converge... In Chebyshev distance, all 8 adjacent cells from the web use hamming distance we. Defined for some metrics, I have drawn 3 images as shown below a straight between! I decided to mostly use ( squared ) Euclidean distance, or Chebyshev distance ( ED ) two. When they are equal, if you have a degenerate triangle discussions that is what we are towards! As if there is a single distance function that is what we working! Normal methods of comparing two colors are in Euclidean distance between a pair of.. A chess board the distance can be reached by one unit ) Euclidean distance metric, hypotenuse! Life usage XYZ, HSL, and Standardized Euclidean distance measures whether two! What we are working towards used data from Youtube Eminem \u2019 s comments which contain 448 data calculate! Two different distance functions that widely used in machine learning in: are! Compact way to represent the distances between each pair the real numbers coming Out of two metrics ) \\vert. Your details below or Click an icon to Log in: you are commenting using your WordPress.com.. More appropriate p_2 ) = \\vert y_2 - y_1 \\vert $and D2 = sqrt 2. Degenerate triangle wind up with degenerate perpendiculars it 's not as if there a. Changing the heuristic will not Change the connectivity of neighboring cells in my calculation distance calculation in Euclidean. ( LCH ), chebyshev distance vs euclidean is called the Chebyshev measure this is called the octile distance and! Distance between m points using Euclidean distance ( Taxicab or City Block ) 5 Facebook account study showed we! Some distance functions here but sometimes ( for example, Euclidean or airline distance is 0 ; otherwise it... = \\vert y_2 - y_1 \\vert$ the covariance structure of your data then mahalanobis distance is squared-euclidean. Can be reached by one unit 31 January 2011 ( UTC ) no \\vert -. Lch ), Click here to upload your image ( max 2 MiB ) talking about different! So without loss of generality, translate one of the ( absolute differences. Kdd dataset contains 41 features and two classes which type of data its a way to represent the between. Create a \u201c distance matrix can we begin the process of separating the observations to clusters locations. To take a crossed move neighboring cells [ 5 ] with ground state wave functions of points. And Standardized Euclidean distance to compute the difference is a single distance function, and Standardized distance... Such as computation and real life usage have a degenerate triangle line between 2 points two points is sum. 2-Norm ) as the distance between two points is the distance between ( 0,4 ) and ( )... Comparing two colors are in Euclidean distance ( 2-norm ) as the distance can be defined as a line! The connectivity of neighboring cells 2 if one wants to take a crossed.! As Manhattan and Euclidean, Chebyshev, Manhattan, and RGB Log Out / Change ), Click to!\n\nFlorida Panthers Ahl Team, Feng Shui Pronunciation, Lab 4 Introduction To The Microscope, Markdown Xl Cheat Sheet, Crawley Town Manager History, Belgian D'anvers For Sale Uk, Phillip Hughes' Father, Karan Soni Deadpool, Fallin Janno Gibbs Lyrics,", "date": "2021-03-08 21:17:53", "meta": {"domain": "com.br", "url": "http://www.comfer.com.br/023nhg/chebyshev-distance-vs-euclidean-226c34", "openwebmath_score": 0.5468424558639526, "openwebmath_perplexity": 833.5634533418914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9572778000158576, "lm_q2_score": 0.896251371748038, "lm_q1q2_score": 0.8579615414081564}}
{"url": "https://math.stackexchange.com/questions/3376028/probability-of-meeting-a-confusion", "text": "# Probability of meeting - a confusion\n\nThis is a basic probability question.\n\nPersons A and B decide to arrive and meet sometime between 7 and 8 pm. Whoever arrives first will wait for ten minutes for the other person. If the other person doesn't turn up inside ten minutes, then the person waiting will leave. What is the probability that they will meet? I am assuming uniform distribution for arrival time between 7 pm and 8 pm for both of them.\n\nThe exact question is given here: Probability of meeting\n\nI am aware of geometric probability and the methods they have used there. The answer seems to be unanimously $$9/36$$. However, here is where I am confused. Considering A reaches before B, and that he reaches before the first $$50$$ minutes, the probability of meeting should be $$5/6 \\cdot 1/6= 5/36$$ (since B would have to reach within $$10$$ minutes of A). By symmetry, this means that if B reaches early and reaches before the first $$50$$ minutes, the meeting probability is again $$5/36$$. It seems that even without considering the probability of what happens if the earlier person reaches in the last $$10$$ minutes, we have a probability of $$10/36$$ already, greater than the total probability calculated in the post in the link.\n\nCan anyone please point out my logical flaw?\n\n\u2022 There was an error in the calculation of the answer in that problem. Everything you have done thus far is correct. What answer did you obtain? \u2013\u00a0N. F. Taussig Sep 30 '19 at 22:24\n\u2022 Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. \u2013\u00a0N. F. Taussig Sep 30 '19 at 22:57\n\nThere was an arithmetic error in the answer to the linked problem that is the source of your confusion.\n\nIf we plot the time person A arrives on the horizontal axis and the time person B arrives on the vertical axis, then the white region in the diagram below represents the band of time in which they arrive within ten minutes of each other, and therefore meet.\n\nThe probability that they do not meet is found by dividing the areas of the two grey right triangles by the area of the square.\n\nEach gray right triangle has area $$\\frac{1}{2} \\cdot 50 \\cdot 50$$ so the combined area of the two congruent gray triangles is $$50 \\cdot 50$$ The area of the square is $$60 \\cdot 60$$ Thus, the probability that persons A and B do not meet is $$\\Pr(\\text{persons A and B do not meet}) = \\frac{50 \\cdot 50}{60 \\cdot 60} = \\frac{25}{36}$$ Subtracting this from $$1$$ gives the probability that persons A and B do meet, which is $$\\Pr(\\text{persons A and B meet}) = 1 - \\frac{25}{36} = \\frac{11}{36}$$ Your method will work.\n\nYou calculated that if person A arrives first, then $$5/6$$ of the time person A will arrive in the first $$50$$ minutes and that person B has probability $$1/6$$ of arriving within $$10$$ minutes of person A. Therefore, should person A arrive first and arrives within $$50$$ minutes of 7 pm, they have probability $$\\frac{5}{6} \\cdot \\frac{1}{6} = \\frac{5}{36}$$ of meeting. By symmetry, if person B arrives first and arrives within $$50$$ minutes of 7 pm, they have probability $$\\frac{5}{36}$$ of meeting.\n\nThey will also meet if both of them arrive within the last $$10$$ minutes of the hour, which occurs with probability $$\\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{1}{36}$$ Since these three events are mutually exclusive and exhaustive, the probability that persons A and B will meet is $$\\Pr(\\text{persons A and B meet}) = \\frac{5}{36} + \\frac{5}{36} + \\frac{1}{36} = \\frac{11}{36}$$ which agrees with the answer we obtained above.\n\n\u2022 Thank you. This agrees with the result I obtained. And I will also be more careful about the typesetting in future. \u2013\u00a0noobcoder Oct 1 '19 at 5:52\n\u2022 Also I am unable to either up-vote or choose this as the best answer, the site won't allow me to. Sorry for that. \u2013\u00a0noobcoder Oct 1 '19 at 5:54\n\u2022 I am not sure why you were not able to choose this as the best answer, but you were unable to upvote since you did not have sufficient privileges. As you earn additional reputation, you will gain more privileges. \u2013\u00a0N. F. Taussig Oct 1 '19 at 7:12", "date": "2020-01-17 16:12:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3376028/probability-of-meeting-a-confusion", "openwebmath_score": 0.703490138053894, "openwebmath_perplexity": 154.8414077831596, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989510906574819, "lm_q2_score": 0.8670357460591569, "lm_q1q2_score": 0.8579413271157709}}
{"url": "https://math.stackexchange.com/questions/1888769/inequality-involving-rearrangement-sum-i-1n-x-i-y-sigmai-ge-su", "text": "# Inequality involving rearrangement: $\\sum_{i=1}^n |x_i - y_{\\sigma(i)}| \\ge \\sum_{i=1}^n |x_i - y_i|.$\n\nIf $x_1 \\ge x_2 \\ge \\cdots \\ge x_n$ and $y_1 \\ge y_2 \\ge \\cdots \\ge y_n$ are real numbers, and $\\sigma$ is any permutation, then $$\\sum_{i=1}^n |x_i - y_{\\sigma(i)}| \\ge \\sum_{i=1}^n |x_i - y_i|.$$ This must be a known inequality. What is it called, and how is it proven? (Just a reference is OK.)\n\nThe conditions are similar to rearrangement inequality. The inequality is a simple statement about minimizing the $\\ell^1$ distance between a finite sequence and any rearrangement of another finite sequence.\n\nI searched around and clicked through various pages but couldn't find something relevant. If it is true, perhaps a proof could be constructed by decomposing the permutation into a sequence of transpositions.\n\n\u2022 the trivial inequality ${}{}{}$? \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Aug 11 '16 at 2:22\n\u2022 @CarryonSmiling How is this the trivial inequality? \u2013\u00a06005 Aug 19 '16 at 3:01\n\nFor any convex function, such as $f(x) = |x|$,\n\n$$\\sum f(x_i - y_{\\sigma(i)}) \\geq \\sum f(x_i - y_i)$$\n\nbecause $(x_i - y_{\\sigma(i)})$ majorizes $(x_i - y_i)$.\n\nA reference is the first theorem, 6.A.1, in chapter 6 of Olkin and Marshall's book on majorization, applied to the sequences $x_i$ and $-y_i$.\n\nThey attribute the result to a 1972 article by Peter W Day on general forms of the rearrangement inequality, and give a proof for vectors of real numbers. Day's article is about more general situations with ordered abelian groups. The inequality for real vectors must have been known earlier to many people.\n\n\u2022 Thank you -- I very much appreciate the generalization to any convex function! I'm not familiar with majorization -- but I follow that $(x_i - y_{\\sigma(i)})$ majorizes $(x_i - y_i)$. Is it possible to add a sentence or two sketching why majorization gives us the inequality? \u2013\u00a06005 Aug 19 '16 at 2:12\n\u2022 The relation to convex functions is listed under \"equivalent conditions\" at the Wikipedia page. Majorization is a necessary and sufficient condition for the inequality to hold for all convex functions (for a fixed pair of vectors). \u2013\u00a0zyx Aug 20 '16 at 20:40\n\u2022 Very cool result. Thanks. \u2013\u00a06005 Aug 21 '16 at 8:06\n\nWe can prove it in a similar way as the rearangement inequality. There are only finitely many possibilities for $\\sigma$, so a minimum is achieved, pick $\\sigma$ so that it has the least possible number of inversions among all the permutations that minimize the expression.\n\nSuppose by way of contradiction there is $i<j$ with $\\sigma(i)>\\sigma(j)$. Notice $|x_i-y_{\\sigma_i}|+|x_j-y_{\\sigma(j)}|\\geq |x_i-y_{\\sigma(j)}|+|x_j-y_{\\sigma(i)}|$.\n\nSo the permutation that transposes $i$ and $j$ must also minimize the expression, and has less inversions, a contradiction.\n\nYou may be interested in the following inequality.\n\nLet $f_1(x), f_2(x), \\cdots, f_n(x): \\mathbb{R} \\rightarrow \\mathbb{R}$ be functions such that for $\\forall 1 \\leq k < n$, $f_{k+1}(x) - f_k(x)$ is non-decreasing with respect to $x$. In addition, let $y_1 \\geq y_2 \\geq \\cdots \\geq y_n$. Then, $$\\sum_{k}f_k(y_{n-k+1}) \\geq \\sum_k f_k(y_{\\sigma(k)}) \\geq \\sum_k f_k(y_k)$$\n\nThe book \"the cauchy-schwarz master class\" does not give a formal name for this inequality but just call it \"a non-linear rearrangement inequality\". See p.81 of the book.\n\nI show a proof based on the inequality above.\n\nProof. Let $$f_k(x) = |x_k - x|$$ Then $$f_{k+1}(x) - f_k(x) = \\begin{cases} x_{k+1}-x_{k} &\\text{if}\\ x \\leq x_{k+1} \\\\ 2x - x_{k+1} - x_k &\\text{if}\\ x_{k+1} < x < x_k\\\\ x_{k} - x_{k+1} &\\text{otherwise} \\end{cases}$$ is non-decreasing, thus the inequality applies. That is, $$\\sum_k |x_k - y_k| \\leq \\sum_k |x_k - y_{\\sigma(k)}| \\leq \\sum_k |x_k - y_{n-k+1}|$$", "date": "2019-08-22 05:47:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1888769/inequality-involving-rearrangement-sum-i-1n-x-i-y-sigmai-ge-su", "openwebmath_score": 0.9025364518165588, "openwebmath_perplexity": 341.17806855848363, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426450627306, "lm_q2_score": 0.8807970701552505, "lm_q1q2_score": 0.8579339079775238}}
{"url": "https://www.physicsforums.com/threads/geostationary-satellite-and-orbiting-satellite-problem.753331/", "text": "# Geostationary Satellite and Orbiting Satellite problem\n\n#### agnibho\n\n1. Homework Statement\nA satellite GeoSAT is in a circular geostationary orbit of radius RG above a point P on the equator. Another satellite ComSAT is in a lower circular orbit of radius 0.81RG. At 7 P.M. on January 1, ComSAT is sighted directly above P. On which day among the following can ComSAT be sighted directly above P between 7 P.M. and 8 P.M. ??\n(a) Jan 3\n(b) Jan 9\n(c) Jan 15\n(d) Jan 21\n\n2. The attempt at a solution\nI am sorry but I am unable to think of any possible way to solve this problem. Will Keplar's equations help?? I am totally confused!! Please help!!\n\nLast edited:\nRelated Introductory Physics Homework Help News on Phys.org\n\n#### phyzguy\n\nWhat is the period of a geosynchronous orbit with radius RG? Using Kepler's laws, what is the period of an orbit with radius 0.81 RG?\n\n#### agnibho\n\nUmm...would that help?? :/\n\n#### phyzguy\n\nUmm...would that help?? :/\nYes, it will help - that's why I asked the questions. Can you answer them?\n\n#### agnibho\n\nWell, the formula to be applied can be.....\nT2/R3 = (4 * \u03c02) / (G * M)\n\nwhere M= 5.98x1024 kg i.e. the mass of the Earth.\nWould it help now??\n\n#### agnibho\n\nThe above formula is for geosynchronous as I found out.... but the question tells about geostationary orbit.....\n\n#### phyzguy\n\nThe above formula is for geosynchronous as I found out.... but the question tells about geostationary orbit.....\nGeosynchronous and geostationary really mean the same thing. Do you know what a geostationary orbit is? If you do, you should be able to tell me the period of a geostationary orbit without doing any calculations.\n\n#### agnibho\n\nAs Wikipedia states, the time period can be\nT = 2\u03c0 \u221a(r3/\u03bc)\n\nWhat say??\n\nokay\n\n#### agnibho\n\nGeosynchronous and geostationary really mean the same thing. Do you know what a geostationary orbit is? If you do, you should be able to tell me the period of a geostationary orbit without doing any calculations.\nYeah it's 24 hours! :tongue:\n\n#### phyzguy\n\nYeah it's 24 hours! :tongue:\nOK, good. It's actually 23 hours and 56 minutes, but 24 hours is probably close enough. So if the period of the satellite at radius R = RG is T = 24 hours, and if we know from what you wrote earlier that\n$$\\rm \\frac{T^2}{R^3} = constant$$\nthen what is T when R = 0.81RG?\n\n#### D H\n\nStaff Emeritus\nYou don't need to know M or G. What you need to do is to answer phyzguy's questions.\n\nWhat's the period of a geostationary satellite?\nWhat do Kepler's laws say about the period of a satellite? Hint: Only one law says anything about the period.\n\nFrom that, what is the period of that satellite whose orbital radius is 0.81 RG?\n\n#### agnibho\n\noh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?\n\n#### phyzguy\n\noh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?\nRight.\n\n#### D H\n\nStaff Emeritus\noh okay got it now..... so all I have to do is to find out the periods for both the satellites and then calculate the time difference to get the number of days, right?\nThere's more to it than that. The satellite will be directly overhead after an integral number of orbits.\n\nHere's what you need to solve for: For what integers N is the time needed to make N orbits between M days and M days plus one hour, where M is another integer?\n\n\"Geostationary Satellite and Orbiting Satellite problem\"\n\n### Physics Forums Values\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-10-16 11:38:19", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/geostationary-satellite-and-orbiting-satellite-problem.753331/", "openwebmath_score": 0.6453408598899841, "openwebmath_perplexity": 1510.0550269790765, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9740426412951847, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.8579339016112494}}
{"url": "https://math.stackexchange.com/questions/2672447/factoring-2n-3-2n-2-not-homework-question/2672459", "text": "# Factoring $2^{^{n-3}}-2^{^{n-2}}$ (not homework question)\n\nAt the moment, I'm studying for my math exam, and I came upon a problem which involves factoring the powers of this polynomial:\n\n$2^{^{n-3}}-2^{^{n-2}}$\n\nAfter a few minutes of being stuck, i looked up a solution and found this (from symbolab):\n\n$\\mathrm{Apply\\:exponent\\:rule}:\\quad \\:a^{b+c}=a^ba^c$\n\n$=2^{-3}\\cdot \\:2^n-2^{-2}\\cdot \\:2^n$\n\n$\\mathrm{Factor\\:out\\:common\\:term\\:}2^{-3}\\cdot \\:2^n$\n\n$=2^{-3}\\left(1-2\\right)\\cdot \\:2^n$\n\n$\\mathrm{Refine}$\n\n$=-2^{n-3}$\n\nMy question is:\n\nhow can i factor\n\n$2^{-3}\\cdot \\:2^n$\n\nfrom\n\n$2^{-3}\\cdot \\:2^n-2^{-2}\\cdot \\:2^n$\n\nto get\n\n$2^{-3}\\left(1-2\\right)\\cdot \\:2^n$ ?\n\nI can't seem to understand the logic behind this factorization.\n\nIf you could help me understand it, I would really appreciate it.\n\n\u2022 Are you just asking why $ab - ac = a(b - c)$? \u2013\u00a0user296602 Mar 1 '18 at 17:06\n\u2022 Did you try a few values of $n$? \u2013\u00a0Angina Seng Mar 1 '18 at 17:08\n\u2022 Welcome to MSE. Please read this text about how to ask a good question. \u2013\u00a0Jos\u00e9 Carlos Santos Mar 1 '18 at 17:09\n\n$$2^{n-3}-2^{n-2}=2^{(n-3)+0}-2^{(n-3)+1}=2^{n-3}(2^0-2^1)=\\boxed{-2^{n-3}}$$\n\nNote that\n\n$$\\large{2^{^{n-3}}-2^{^{n-2}}=2^{^{n-3}}(1-2^1)=-2^{n-3}}$$\n\n$$...=2^{-3}\\cdot \\:2^n-2^{-2}\\cdot \\:2^n=2^n(2^{-3}-2^{-2})=2^n2^{-3}(1-2^1)=-2^{n-3}$$\n$$2^{n-3}-2^{n-2}=\\frac{2^n}{2^3}-\\frac{2^n}{2^2}=2^n\\Biggl(\\frac1{2^3}-\\frac1{2^2}\\Biggr)=2^n\\Biggl(\\frac{2^2-2^3}{2^5}\\Biggr)=$$ $$2^n\\Biggl(-\\frac{2^2}{2^5}\\Biggr)=2^n\\Biggl(-\\frac{1}{2^3}\\Biggr)=-2^n(2^{-3})=-2^{n-3}$$\n\u2022 Note that a shortcut may be $$2^n\\left(\\frac1{2^3}-\\frac1{2^2}\\right)=2^n\\left(\\frac1{2^3}-\\frac2{2^3}\\right)=2^n\\left(-\\frac1{2^3}\\right)=-2^{n-3}$$ \u2013\u00a0TheSimpliFire Mar 1 '18 at 19:13", "date": "2021-06-21 17:25:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2672447/factoring-2n-3-2n-2-not-homework-question/2672459", "openwebmath_score": 0.7690407037734985, "openwebmath_perplexity": 559.1932069898376, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138190064203, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8579211841110345}}
{"url": "https://math.stackexchange.com/questions/1260011/why-is-arcsin-a-function-and-not-a-relation-since-arcsin-sin-frac3-pi4", "text": "# Why is arcsin a function and not a relation since $\\arcsin(\\sin(\\frac{3\\pi}{4})) = \\frac{\\pi}{4}$?\n\nSince $\\sin(\\frac{\\pi}{4})$ and $\\sin(\\frac{3\\pi}{4})$ are both $\\frac{\\sqrt{2}}{2}$, shouldn't $\\arcsin(\\frac{\\sqrt{2}}{2})$ map to both $\\frac{\\pi}{4}$ and $\\frac{3\\pi}{4}$ and therefore not be a function?\n\nwhen I input $\\arcsin(\\frac{\\sqrt{2}}{2})$ into Wolfram, I get back $\\frac{\\pi}{4}$ only, but isn't the point of an inverse function that $f^{-1}(f(x)) = x$?\n\nIn this case however, $\\arcsin(\\sin(\\frac{3\\pi}{4})) = \\frac{\\pi}{4}$\n\nWhat am I missing?\n\n\u2022 I think this is just a convenient definition. No one says $\\arcsin x$ is the inverse function of $\\sin x$. Please point out if I made mistake. \u2013\u00a0MonkeyKing May 1 '15 at 1:41\n\u2022 I'm pretty sure I've always seen arcsin defined as the inverse function of the sin function. \u2013\u00a0jeremy radcliff May 1 '15 at 1:42\n\u2022 OK, I mean not the inverse function of $\\sin x$ over $\\Bbb R$. I should practice language skill more. :( \u2013\u00a0MonkeyKing May 1 '15 at 1:46\n\u2022 No problem, what you said is clear now in the context of user17762's answer, I just didn't have the knowledge to understand your reply when I first read it :) \u2013\u00a0jeremy radcliff May 1 '15 at 1:49\n\nThe important thing to realise here is that a function is not just a rule. To specify a function properly you must also define the domain and codomain. For example, the sine function is $$\\sin:{\\Bbb R}\\to{\\Bbb R}\\quad\\hbox{where}\\quad \\sin x= \\langle\\hbox{insert your favourite definition}\\rangle\\ .$$ This function is not one-to-one (injective) and therefore has no inverse.\n\nHere is a different function: $$f:\\Bigl[-\\frac\\pi2,\\frac\\pi2\\Bigr]\\to[-1,1]\\quad\\hbox{where}\\quad f(x)=\\sin x=\\langle\\hbox{same definition as above}\\rangle\\ .$$ This function is one-to-one and onto and therefore has an inverse. Its inverse is commonly denoted $\\sin^{-1}$ or $\\arcsin$. Notice however that this terminology is not strictly accurate: $\\arcsin$ is not the inverse of $\\sin$, it is the inverse of $f$, which is a different function.\n\nHope this clears things up.\n\n\u2022 Thanks David, it does clear things up. In the line $\\sin:{\\Bbb R}\\to{\\Bbb R}$, you don't limit the range to [-1, 1], but even with all the real numbers as potential inputs, in the original sin function the outputs can never be out of the range [-1, 1]. Do you not have to specify the codomain because it's implicitly limited? And in that case, do you have to specify the codomain for $f$? (not trying to nitpick, I just want to make sure I grasp the exact notational logic) \u2013\u00a0jeremy radcliff May 1 '15 at 2:02\n\u2022 There is another common imprecision involved here. (Mathematicians speaking imprecisely... shock, horror ;-) Strictly speaking a function $f:A\\to B$ is invertible if and only if it is one-to-one and onto. For example,$$f:{\\Bbb R}\\to{\\Bbb R}\\ ,\\quad f(x)=e^x$$is not onto and therefore not invertible. But in calculus, one commonly treats $f$ as the same function as$$g:{\\Bbb R}\\to{\\Bbb R}^+\\ ,\\quad g(x)=e^x$$(which it really isn't), and $g$ is invertible. (continued...) \u2013\u00a0David May 1 '15 at 2:08\n\u2022 If my $f$ above had codomain $\\Bbb R$ it would be invertible under this (mis)interpretation. However $\\sin$ with domain $\\Bbb R$ is not one-to-one and therefore is not invertible, no matter what you choose for the codomain. \u2013\u00a0David May 1 '15 at 2:08\n\u2022 This is great, it all makes sense now. Thanks for taking the time to clarify, I really appreciate. \u2013\u00a0jeremy radcliff May 1 '15 at 2:17\n\u2022 You're welcome, glad I could help. \u2013\u00a0David May 1 '15 at 2:20\n\nRecall that $\\arcsin: [-1,1] \\mapsto [-\\pi/2,\\pi2/]$, i.e., the range of $\\arcsin$ is $[-\\pi/2,\\pi/2]$.\n\nThe function $\\sin$ on the other hand is a non-injective function that maps the real line to $[-1,1]$. Defining an inverse for a non-injective functions is not possible. Hence, one route that is often taken is to restrict the domain of the original function such that the function on the new restricted domain is injective.\n\nHence, the function $\\arcsin$ is the inverse of $\\sin$ only when the domain of the function $\\sin$ is restricted to $[-\\pi/2,\\pi/2]$.\n\nEDIT\n\nWhat mathematica does is the following. It first evaluate $\\sin(3\\pi/4)$, which is $1/\\sqrt{2}$. Now it goes and asks the function $\\arcsin$, what $\\arcsin(1/\\sqrt2)$ is, which is now $\\pi/4$.\n\n\u2022 That makes sense, but in this case isn't $\\frac{3\\pi}{4}$ an invalid input and shouldn't Wolfram Alpha give me an error along the lines of \"invalid domain\" or something? Are they just allowing it because it's convenient? \u2013\u00a0jeremy radcliff May 1 '15 at 1:45\n\u2022 @jeremyradcliff No. What mathematica does is the following. It first evaluate $\\sin(3\\pi/4)$, which is $1/\\sqrt{2}$. Now it goes and asks the function $\\arcsin$, what $\\arcsin(1/\\sqrt2)$ is, which is now $\\pi/4$. \u2013\u00a0Leg May 1 '15 at 1:46\n\u2022 Ah ok, now all of it makes sense; thank you, it was starting to drive me insane. \u2013\u00a0jeremy radcliff May 1 '15 at 1:48\n\nAs others have pointed out, $\\arcsin x$ has the restricted domain $[-1,1]$ and range $[\\frac{-\\pi}{2},\\frac{\\pi}{2}]$. This restriction is a choice, a definition.\n\nWhat's really happening is that $\\arcsin x$ is multi-valued. We're only selecting what is called the principal branch, and we're doing this for convenience.\n\nSince the trigonometric functions are periodic, a single-valued function that acts as an inverse doesn't exist. We elect to \"chop up\" the inversion relation into sections - the branches. In the case of $\\arcsin x$ we focus on returning $\\sin x$ input values $[\\frac{-\\pi}{2},\\frac{\\pi}{2}]$. This branch is the principal branch - the one we've elected to use most often. Why? Well, consider that $\\sin x$ cycles through all possible outputs when we give it $[\\frac{-\\pi}{2},\\frac{\\pi}{2}]$. Also, it's \"nicely packaged\" - there's a distinct symmetry and accessibility in using $[\\frac{-\\pi}{2},\\frac{\\pi}{2}]$.\n\nWhen evaluating a query such as yours, we compensate by finding an equivalent result in the principal branch.\n\n\u2022 Thanks zahbaz, that's very helpful because it makes it clear that it's a choice of convenience and why, which is something I've been struggling with (the idea that a lot of math is based on convenience and not some absolute ideal) \u2013\u00a0jeremy radcliff May 1 '15 at 2:11", "date": "2019-08-22 08:46:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1260011/why-is-arcsin-a-function-and-not-a-relation-since-arcsin-sin-frac3-pi4", "openwebmath_score": 0.8661549687385559, "openwebmath_perplexity": 350.42638104142276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138170582865, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8579211792689134}}
{"url": "https://www.physicsforums.com/threads/refractive-index-of-water-glass-combination.369449/", "text": "Refractive Index of Water/Glass Combination\n\n1. Jan 14, 2010\n\nlabview1958\n\nIf a beam of light enters water from air at 30 degrees from the normal. It then enters a glass block and exit into air. At what angle does it exit from glass to air ? My friend says it exist again at 30 degrees to normal. The refractive index of water is 1.3 and glass is 1.5.\n\n2. Jan 14, 2010\n\nRoyalCat\n\nYour friend's forgotten Snell's Law: $$n_1 \\sin{\\theta _1}=n_2 \\sin{\\theta _2}$$\nWhere the index 1 refers to the material the light's coming from, and the index 2 refers to the material the light's entering.\n\n3. Jan 14, 2010\n\nMATLABdude\n\nTo elaborate upon what RoyalCat says, assume you have just a block of glass (width does not matter--the pendant in me says as long as it's sufficiently thin so that not all the light is absorbed). Now assume that you have a beam of light striking at angle $\\theta_i$. What angle (in terms of $\\theta_i$) will the beam of light travel at within the glass? At what angle (again, in terms of $\\theta_i$) will the beam of light exit at?\n\nThis interesting result is probably what led your friend to come up with their answer. Unfortunately, in your example, the symmetry (because there is no glass on the way in) is broken.\n\nLast edited: Jan 14, 2010\n4. Jan 14, 2010\n\nlabview1958\n\nFor air/water/glass/water/air combination, then incoming and outgoing angle is same? This is not same for air/water/glass/air ?\n\n5. Jan 14, 2010\n\nmgb_phys\n\nCorrect\n\nIt's the same for each individual part, so a quick sketch should convince you for the whole setup.\n\n6. Jan 14, 2010\n\njasper10\n\nits 35.23 degrees according to Mr Snell and his law :P\n\n7. Jan 15, 2010\n\nlabview1958\n\nFor air/water surface its 30 degrees. Definetly for air/glass surface is NOT 30 degrees. Do you agree?\n\n8. Jan 15, 2010\n\nmodulus\n\nRight, from water to air, the incident angle is 30, and you can figure out the refracted angle (at the water/air interface, made with the normal) using \"Mr. Snell's\" law.\n\nNow, if the water/air (it started from water, and, is entering into air) interface and the air/glass (it started from air, and, is entering into glass) interface are parallel (so that their normal's are parallel), you can figure out the incident angle for the air/glass interface ; it will be equal to the refracted angle at the water/air interface (using the simple properties of parallel lines and transversals).\n\nSo, no, the incident angle for the air/glass interface and the water/air interface will not be the same. I think that should answer your question.\n\nWell, now that you've got your incident angle for the air/glass interface (it started from air, and is entering into glass), I want you to apply some logic and figure out what will be the angle of refraction when this same light ray exits glass and enters back into air?\n\nP.S.:- Look at the bold stuff. It points to the logic your friend is using to derive his answer. But, he's gone wrong somewhere in between\n\nLast edited: Jan 15, 2010\n9. Jan 15, 2010\n\nlabview1958\n\nThe problem lies with the water/glass boundary. How can Mr. Snell help there? The refractive index of glass with respect to water? How to resolve that?\n\n10. Jan 15, 2010\n\njasper10\n\nSorry, my previous answer was wrong, i didnt read the question correctly.\n\nStep one: n1sin1 = n2sin2 (refractive index of air = 1)\n\n(1)(sin30) = (1.3)(sinx)\n\n-> x = 22.6 degrees\n\nStep 2\n\nn1sin1 = n2sin2\n\n(1.3)(sin22.6) = (1.5)(siny)\n\n-> y=19.47degrees\n\nStep 3\n\nn1sin1=n2sin2\n(1.5)sin19.47 = (1)(sinz)\n\n-> z = 30 degrees.\n\n11. Jan 15, 2010\n\nmgb_phys\n\nIt's exactly the same law n1 sin1 = n2 sin2, you are just starting with n=1.33 and going into n=1.5\n\n12. Jan 16, 2010\n\nlabview1958\n\nThe calculations is correct. But how to explain the physics of it?\n\n13. Jan 26, 2010\n\nlabview1958\n\nAnyone care to explain the physics of it?\n\n14. Jan 26, 2010\n\nrl.bhat\n\nIn a parallel sided glass slab light is laterally shifted i.e. incident and refracted rays in the same medium are parallel to each other, whatever may be the number of different media in between medium. So the angle of incidence is equal to the angle of emergence.\n\n15. Jan 26, 2010\n\nMATLABdude\n\n16. Jan 26, 2010\n\njasper10\n\nImagine a shopping trolly: if u push it from a smooth medium (eg marble) at an angle onto a rough concrete medium (more friction), one wheel reaches the rough medium first and thus slows down first. this causes a change of direction in which the trolley is travelling.\n\n17. Jan 29, 2010\n\nlabview1958\n\n\"light is laterally shifted\"\n\nWhat physics is that?\n\n18. Jan 29, 2010\n\nehild\n\nOne explanation is the analogy with mechanics. Photons have energy and momentum. Their energy is E=h*f, the magnitude of the momentum is p=h/lambda, and its direction points in the direction of propagation of the light ray. You know that that momentum changes when same force acts on the particle, the change of the momentum is equal to the impulse, and is parallel to the force applied. A plane surface can act only along its normal (there is no friction). The parallel component of the momentum does not change when the photon crosses the interface between two media (or reflects from it). This parallel component is h/ lambda*sin(alfa) which is constant everywhere in a plane-parallel arrangement of layers. But lambda =v/f (f a frequency, v is the speed of light in the medium.) The refractive index of the medium is n=c/v.\n\nSo the constancy of the parallel component of momentum involves the constancy of $$sin(\\alpha) / \\lambda = sin(\\alpha) f*n/c$$. As the frequency of the photon does not change we arrived at Snell's law $\\sin(\\alpha)*n=const$.\n\nThe normal component of the momentum will change, but we know that the magnitude of the momentum is h/lambda=hf/c *n. Thus the normal component of the photon momentum is\n\n$$p_n= hf/c*\\sqrt{n^2-(n_0\\sin{alpha_0})^2}$$\n\nWhere n0 is the refractive index of the medium from where the light arrived and alpha0 is the angle of incidence.\nIf n>n0, the normal component of the momentum increases when the light enters into the other medium while the parallel component remains the same, that is while the ligth ray encloses a smaller angle with the normal in a \"denser\" medium as in air.\n\nehild", "date": "2017-08-19 21:03:18", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/refractive-index-of-water-glass-combination.369449/", "openwebmath_score": 0.6225447654724121, "openwebmath_perplexity": 1017.4631242954754, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9777138131620183, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8579211695846708}}
{"url": "https://www.physicsforums.com/threads/ball-bouncing-infinitely-to-find-distance-covered.379464/", "text": "# BALL BOUNCING INFINITELY to find distance covered\n\n1. Feb 18, 2010\n\n### abrahamrenns\n\nok, the qn is..\na ball is dropped from H height. it bounces with restitution e. hw much distance will it cover before it comes to rest..? surely, it is not infinity if i am not mistaken, since the bounces become infinitesimally small and then it tends to zero.. thnx in advance for any help..\n\n2. Feb 18, 2010\n\n### xlines\n\nWhat did you try to solve your problem?\n\nIf on every bounce ball loses some fixed percentage of it's momentum, then you should calculate path traveled depending on starting momentum and sum it up. I am pretty sure you should get geometric sum.\n\n3. Feb 18, 2010\n\n### GRDixon\n\nYou have more or less rediscovered Zeno's paradox. But you're instincts are correct. The key is that the bounces become infinitesimally small, and no matter how great a finite number of bounces (finite amount of time) the distance traveled with always be finite.\n\n4. Feb 18, 2010\n\n### abrahamrenns\n\ncan some one plz take the pains to actually calculate it? it wudnt be difficult for a few here atleast.. i am not able to do it..\n\n5. Feb 18, 2010\n\n### xlines\n\nWell, we do have assignments forums, but here you go. First, what distance will travel ball with upwards initial velocity v' ?\n\nv(t) = v' - gt\n\nit will reach max. height in t' = v'/g . Total distance traveled in one jump will be\n\nd(v') = 2*1/2*g*t'$$^{2}$$ = v'$$^{2}$$/g\n\nAfter each bounce starting velocity is lowered by factor $$\\alpha$$ . So total distance is\n\nD = H + v'$$^{2}$$/g + ($$\\alpha$$ v')$$^{2}$$ /g + ($$\\alpha$$$$^{2}$$ v')$$^{2}$$ /g + ... = H + v'$$^{2}$$/g (1+$$\\alpha$$$$^{2}$$ + $$\\alpha$$$$^{4}$$ + ... )\n\nSubstitution $$\\beta$$ = $$\\alpha$$$$^{2}$$\nallows you to calculate infinite geometric sum.\n\nUse energy conservation to link starting height H with v' .\n\nLast edited: Feb 18, 2010\n6. Feb 18, 2010\n\n### HallsofIvy\n\nStaff Emeritus\nThat's getting more complicated than necessary!\n\nThe original problem told us that \"it bounces with restitution e\" which means that if it was dropped from height h, then it bounces up to height eh on the first bounce, $e(eh)= e^2h$ on the second, etc. Because it goes the same distance up as down, the total distance will be $h+ 2eh+ 2e^2h+ \\cdot\\cdot\\cdot$, almost a geometric series. We can make it a geometric series by adding and subtracting h: $-h+ 2h+ 2eh+ 2e^2h+ ...$ where everything after the first \"-h\" is a gemetric series with first term 2h and constant ratio e. That total distance is\n$$-h+ \\frac{2h}{1- e}= \\frac{-h+ +eh+ 2h}{1- e}= \\frac{1+e}{1- e}h$$\n\n7. Feb 18, 2010\n\n### Phyisab****\n\nYep it's a nice problem to work out.\n\nLast edited: Feb 18, 2010\n8. Feb 18, 2010\n\n### Saw\n\nThat is a nice solution. And if you wanted to fine-tune it... would it be possible to take into account that with every bounce, the ball gets deformed, less elastic and the coefficient of restitution diminishes?\n\n9. Feb 19, 2010\n\n### xlines\n\nEnglish is not my native and, to tell you the truth, I thought e plays role something like my $$\\alpha$$. I too learned something today! Thanks for clearing that up! :)\n\n10. Apr 15, 2010\n\n### Zeus0\n\nIt's about right except that on rebounce it's not $$eh$$ but $$e^{2}h$$.\nby definition $$e = \\frac{v_2}{v_1} = \\frac{\\sqrt{2gh_2}}{\\sqrt{2gh_1}}$$ or $$e = \\sqrt{\\frac{h_2}{h_1}}$$ . Look at the wikipedia entry at http://en.wikipedia.org/wiki/Coefficient_of_restitution\" [Broken]\n\nMeaning the finale answer is $$D = \\frac{1+e^2}{1-e^2}h$$ . I also took the liberty to calculate the time it would take $$T = \\frac{1+e}{1-e}\\sqrt{\\frac{2h}{g}}$$\n\nI am more curious about weather or not real superball or bouncy balls rebounce an infinite number of time . It is likely not because of imperfections they may have or because of other processes in the way . I would estimate a real superball bounces between 20 to 50 times before resting.\n\nLast edited by a moderator: May 4, 2017", "date": "2017-08-20 10:37:40", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/ball-bouncing-infinitely-to-find-distance-covered.379464/", "openwebmath_score": 0.7529341578483582, "openwebmath_perplexity": 1806.8859602547464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9915543718110063, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.857916719952465}}
{"url": "https://math.stackexchange.com/questions/3497762/how-is-am-gm-supposed-to-be-used-to-prove-an-inequality", "text": "How is AM-GM supposed to be used to prove an inequality\n\nI have recently started learning proofs involving inequalities and came across the AM-GM inequality, which seems like a quite powerful tool.\n\nHowever, I am not sure I understand how to use this tool properly, and I was wondering if there are acceptable strategies to be aware of when making use of the AM-GM inequality.\n\nI have also tried solving an inequality involving AM-GM to learn how to use this tool as I go, but I'm not sure if what I've done so far is a valid approach.\n\nHere is the question:\n\nProve that if $$x, y, z, w \u2265 0$$,\n\n$$\\frac{x+y+z+w}{4} \u2265 \\sqrt[4]{xyzw}$$\n\nHere is what I have done so far:\n\nI noticed that $$\\sqrt[4]{xyzw}$$ = $$\\sqrt{\\sqrt{xy}\\sqrt{zw}}$$, and then used AM-GM, like so:\nI let\n$$a = \\sqrt{xy}$$\n$$b= \\sqrt{zw}$$\n\nto make a use of $$\\frac{a+b}{2} \u2265 \\sqrt{ab}$$.\nAnd, then, I subbed in the values:\n\n$$\\frac{\\sqrt{xy}+\\sqrt{zw}}{2} \u2265 \\sqrt{\\sqrt{xy}\\sqrt{zw}}$$.\n\nNow I have a feeling I'm supposed to somehow make a use of the inequality ($$\\frac{a+b}{2} \u2265 \\sqrt{ab}$$) once more, however, I am not exactly sure how I should go about doing it.\n\nAny tips for a newbie like myself would be immensely helpful! (Perhaps a link I can refer to, an article, etc)\n\nThanks a lot in advance for any help!\n\n\u2022 What's wrong with using AM-GM on four numbers directly?\n\u2013\u00a0user239203\nJan 4 '20 at 23:25\n\u2022 I just picked this inequality as a starting point to prove it using AM-GM for two variables first, and get exposed to the AM-GM inequality. I am eager to learn how to use this tool when dealing with inequalities requiring formal mathematical proofs. Jan 4 '20 at 23:28\n\u2022 Did you mean $\\sqrt{xy}+\\sqrt{zw}$ where you wrote $\\sqrt{zw}+\\sqrt{zw}$? Jan 4 '20 at 23:34\n\u2022 Yes, I'm sorry! Fixed. Jan 4 '20 at 23:36\n\u2022 This doesn't specifically pertain to the problem, but in general here are two sources that may help you: artofproblemsolving.com/wiki/index.php/\u2026 brilliant.org/wiki/arithmetic-mean-geometric-mean Jan 4 '20 at 23:51\n\nStart with $$\\dfrac{a+b}{2} \\ge \\sqrt{ab}$$. To prove this, write it as $$\\dfrac{a-2\\sqrt{ab}+b}{2} \\ge 0$$, and the left side is $$\\dfrac{(\\sqrt{a}-\\sqrt{b})^2}{2} \\ge 0$$.\n\nThen,\n\n$$\\begin{array}\\\\ \\dfrac{a+b+c+d}{4} &=\\dfrac{a+b}{4}+\\dfrac{c+d}{4}\\\\ &=\\dfrac{\\dfrac{a+b}{2}}{2}+\\dfrac{\\dfrac{c+d}{2}}{2}\\\\ &\\ge\\dfrac{\\sqrt{ab}}{2}+\\dfrac{\\sqrt{cd}}{2}\\\\ &=\\dfrac{\\sqrt{ab}+\\sqrt{cd}}{2}\\\\ &\\ge\\sqrt{\\sqrt{ab}\\sqrt{cd}}\\\\ &=\\sqrt{\\sqrt{abcd}}\\\\ &=\\sqrt[4]{abcd}\\\\ \\end{array}$$\n\nBy induction on $$n$$, with this technique you can show that $$\\dfrac{\\sum_{k=1}^{2^n}a_k}{2^n} \\ge \\sqrt[2^n]{\\prod_{k=1}^n a_k}$$.\n\nTo show this is true for any $$m < 2^n$$, let $$a_j =\\dfrac{\\sum_{k=1}^m a_k}{m}$$ for $$j \\gt m$$ and see what happens.\n\nAs a matter of fact, this was Cauchy's original proof.\n\nHere's the details (added later).\n\nThe left side is, letting $$a = \\dfrac{\\sum_{k=1}^m a_k}{m}$$,\n\n$$\\begin{array}\\\\ \\dfrac{\\sum_{k=1}^{2^n}a_k}{2^n} &=\\dfrac{\\sum_{k=1}^{m}a_k}{2^n}+\\dfrac{\\sum_{k=m+1}^{2^n}a_k}{2^n}\\\\ &=\\dfrac{\\sum_{k=1}^{m}a_k}{m}\\dfrac{m}{2^n}+\\dfrac{\\sum_{k=m+1}^{2^n}a}{2^n}\\\\ &=\\dfrac{am}{2^n}+\\dfrac{(2^n-m)a}{2^n}\\\\ &=\\dfrac{am}{2^n}+\\dfrac{2^na}{2^n}-\\dfrac{ma}{2^n}\\\\ &= a\\\\ &=\\dfrac{\\sum_{j=1}^ma_j}{m}\\\\ \\end{array}$$\n\nSimilarly, the right side is, letting $$a_j =b =\\left(\\prod_{k=1}^{m} a_k\\right)^{1/m}$$ for $$j > m$$,\n\n$$\\begin{array}\\\\ \\sqrt[2^n]{\\prod_{k=1}^{2^n} a_k} &=\\left(\\prod_{k=1}^{2^n} a_k\\right)^{1/2^n}\\\\ &=\\left(\\prod_{k=1}^{m} a_k\\prod_{k=m+1}^{2^n} a_k\\right)^{1/2^n}\\\\ &=\\left(\\prod_{k=1}^{m} a_k\\right)^{1/2^n}\\left(\\prod_{k=m+1}^{2^n} a_k\\right)^{1/2^n}\\\\ &=\\left(b^m\\right)^{1/2^n}\\left(\\prod_{k=m+1}^{2^n} b\\right)^{1/2^n}\\\\ &=b^{m/2^n}\\left(b^{2^n-m}\\right)^{1/2^n}\\\\ &=b^{m/2^n}b^{(2^n-m)/2^n}\\\\ &=b\\\\ &=\\left(\\prod_{k=1}^{m} a_k\\right)^{1/m}\\\\ \\end{array}$$\n\nTherefore $$a \\ge b$$ or $$\\dfrac{\\sum_{j=1}^ma_j}{m} \\ge \\left(\\prod_{k=1}^{m} a_k\\right)^{1/m}$$.\n\n$$\\frac{x+y+z+w}{4}\\ge\\frac{\\sqrt{xy}+\\sqrt{zw}}{2}\\ge\\sqrt[4]{xyzw}$$\n\n\u2022 Thanks a lot! Could you just explain in further detail what you did here? I am quite new to proofs and more specifically inequality proofs involving AM-GM. Jan 4 '20 at 23:31\n\u2022 @FlavioEsposito I used the $2$-variables AM-GM three times, twice to create two square roots & once more for the fourth root.\n\u2013\u00a0J.G.\nJan 4 '20 at 23:40\n\u2022 I see. So, just to make sure I got it right: You used AM-GM once for $\\sqrt{xy}$, once for $\\sqrt{zw}$, and then for $\\sqrt[4]{xyzw}$? Jan 4 '20 at 23:47\n\u2022 Yep, that's right! Jan 4 '20 at 23:50\n\u2022 For any values $a$, $b$, $x$, and $y$, if $\\color{red}{a}\\ge \\color{red}{x}$ and $\\color{blue}{b} \\ge \\color{blue}{y}$, then $\\color{red}{a}+\\color{blue}{b} \\ge \\color{red}{x}+\\color{blue}{y}$. To see why: $$a \\ge x \\\\ a+b \\ge x+b \\ge x+y$$ We know that $\\color{red}{\\frac{x+y}{2}} \\ge \\color{red}{\\sqrt{xy}}$ and $\\color{blue}{\\frac{z+w}{2}} \\ge \\color{blue}{\\sqrt{zw}}$ by AM-GM. Apply the result above and you have that relation. Jan 6 '20 at 15:08", "date": "2021-09-22 20:39:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3497762/how-is-am-gm-supposed-to-be-used-to-prove-an-inequality", "openwebmath_score": 0.7652883529663086, "openwebmath_perplexity": 279.99725116854665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534398277177, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.8578661187858363}}
{"url": "https://math.stackexchange.com/questions/1788153/compute-the-following-line-integral-along-a-path-of-your-choice-finding-potenti", "text": "# Compute the following line integral along a path of your choice (Finding potential)\n\nConsider the following vector field:\n\n$$\\vec A(x,y,z)=(yz)\\hat i+(xz)\\hat j+(xy)\\hat k$$\n\nCompute the line integral of $A$ along a path of your choice connecting $(0,0,0)$ to $(1,1,1).$\n\nI recognise that $\\nabla\\times A=0$ and so I can simply evaluate $V(1,1,1)-V(0,0,0)$ where $A=\\nabla V$.\n\nI'm not sure how to find the function $V$ though.\n\nThe solution says $V=xyz.$\n\n\u2022 Added finding potential into your title. \u2013\u00a0MrYouMath May 16 '16 at 21:06\n\nAssume V as a function of x,y and z.\n\nThen $\\nabla V=(\\dfrac{\\partial V}{\\partial x},\\dfrac{\\partial V}{\\partial y},\\dfrac{\\partial V}{\\partial z})$\n\nCompare components:\n\n$$\\dfrac{\\partial V}{\\partial x}=yz$$ $$\\dfrac{\\partial V}{\\partial y}=xz$$ $$\\dfrac{\\partial V}{\\partial z}=xy$$\n\nIntegrate: $$V=xyz+f_1(y,z)$$ $$V=xyz+f_2(x,z)$$ $$V=xyz+f_3(x,y)$$\n\nFrom here you can see that $V=xyz+const.$ by inspection. If you want to do it mathematically you have to subtract two equations from each other.\n\n1-2: $f_1(y,z)=f_2(x,z)$ from that you can conclude $f_1(y,z)=f_1(z)$ and $f_2(x,z)=f_2(z)$, or you would have $y$ on the left but no $y$ on the right, and so forth.\n\n1-3: $f_1(z)=f_3(x,y)$.As one can see we have z on the left sid but no z on the right. Hence, $f_1$ is a constant function as it is not depending on $z$. if $f_1=const.$ $f_1=f_3(x,y)$ will imply that $f_3=f_1=const.$. Same argument as before and we can conclude $f_1=f_3=const$. But if $f_1=const.$ than from 1-2 $f_2=const$.\n\nWith that we obtain $f_i=const$ for $i=1,2,3$. At the end you will see, that $f_i=const$ for $i=1,2,3$.\n\n\u2022 I don't understand how you've gone from $f_1(y,z)=f_2(x,z)$ to $f_1(y,z)=f_1(z).$ \u2013\u00a0Si.0788 May 16 '16 at 21:10\n\u2022 The equation tells you, that we have something with y and z on the left hand side and something with x and z on the right side of the equation. But it is not possible to have y only on one side of the equation, hence it must vanish and $f_1$ can only be a function of z. The same argument works for $f_2$. \u2013\u00a0MrYouMath May 16 '16 at 21:13\n\u2022 Ok, that makes sense. Now how do you conclude that $f_i=c$? \u2013\u00a0Si.0788 May 16 '16 at 21:15\n\n$\\nabla G = A$\n\nwhat is G?\n\n$G = \\int yz \\,dx + f(y,z) = \\int xz \\,dy + g(x,z) = \\int xz \\,dz + h(x,y)\\\\ G = xyz + f(y,z) = xyz + g(x,z) = xyz + h(x,y)$\n\n$f(y,z), g(x,z), h(x,y)$ must all be constant functions, and they can be any constant function, so might as well make them $0.$", "date": "2019-06-17 05:46:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1788153/compute-the-following-line-integral-along-a-path-of-your-choice-finding-potenti", "openwebmath_score": 0.939139187335968, "openwebmath_perplexity": 161.41326305576501, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987375049232425, "lm_q2_score": 0.868826789824086, "lm_q1q2_score": 0.8578578943770068}}
{"url": "https://math.stackexchange.com/questions/1630733/using-a-direct-proof-to-show-that-two-integers-of-same-parity-have-an-even-sum", "text": "# Using a Direct Proof to show that two integers of same parity have an even sum?\n\nI seem to be having a lot of difficulty with proofs and wondered if someone can walk me through this. The question out of my textbook states:\n\nUse a direct proof to show that if two integers have the same parity, then their sum is even.\n\nA very similar example from my notes is as follows: Use a direct proof to show that if two integers have opposite parity, then their sum is odd. This led to:\n\nProposition: The sum of an even integer and an odd integer is odd.\n\nProof: Suppose a is an even integer and b is an odd integer. Then by our definitions of even and odd numbers, we know that integers m and n exist so that a = 2m and b = 2n+1. This means:\n\na+b = (2m)+(2n+1) = 2(m+n)+1 = 2c+1 where c=m+n is an integer by the closure property of addition.\n\nThus it is shown that a+b = 2c+1 for some integer c so a+b must be odd.\n\n## ----------------------------------------------------------------------------\n\nSo then for the proof of showing two integers of the same parity would have an even sum, I have thus far:\n\nProposition: The sum of 2 even integers is even.\n\nProof: Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2m???\n\n\u2022 So now that you have $a = 2n$ (you have a typo) and $b = 2m$, we need to talk about the sum of $a$ and $b$, so look at $a+b = 2n+2m$. Can you show that this sum is even? \u2013\u00a0Mike Pierce Jan 28 '16 at 16:45\n\u2022 Yes, and so? (Once you've made the correction to $b=2n$, ) mimic the proof of the result in your notes and factor out the $2$... Now what's the other fork in the road? Well, that the two integers are both odd. Unfold the definition of \"odd\" similarly, add the resulting expressions, factor out the $2$, ... . \u2013\u00a0BrianO Jan 28 '16 at 16:46\n\u2022 Just as a comment for future reference, note that this follows easily from the fact that $2=0 \\mod 2$. That is, $2x=0 \\mod 2$ for $x=1$ or $x=0$. \u2013\u00a0Aloizio Macedo Jan 28 '16 at 16:48\n\u2022 @AloizioMacedo That's true, of course, but it may be too abstract an approach. \u2013\u00a0BrianO Jan 28 '16 at 16:50\n\n\"Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2m???\"\n\nSince a and b are different numbers they should be different m and n.\n\n\"Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2*n*?\n\nAnd so a + b = 2m + 2n = 2(m+n) and as m+n =c for some integer c, a + b = 2c so by definition a + b is even.\n\nYes, you can use the definitions directly. If $a,b$ are even then like you say we have $a = 2m$ and $b = 2n$, so $a+b = 2m + 2n = 2(m+n)$, which is even.\n\nSimilarly, if $a,b$ are odd then we have $a = 2m + 1$ and $b = 2n + 1$, and so $a+b = (2m +1) + (2n + 1) = 2(m + n + 1)$, which is also even.\n\n\u2022 thanks for replying. So since the resulting integer sum of (a+b) always ends up being multiplied by (2), it will always yield an even sum? \u2013\u00a0Analytic Lunatic Jan 28 '16 at 17:28\n\u2022 @AnalyticLunatic Yes, since by definition an even integer is one of the form $2\\cdot \\text{integer}$. \u2013\u00a0Alex Provost Jan 28 '16 at 17:32\n\u2022 So then, how would this change if say, we were looking at a product instead of a sum? \u2013\u00a0Analytic Lunatic Jan 28 '16 at 19:32\n\u2022 @AnalyticLunatic Can you put $2m \\cdot 2n$ in the form $2 \\cdot \\text{integer}$? Can you put $(2m+1)\\cdot(2n+1)$ in the form $2\\cdot \\text{integer}+1$? \u2013\u00a0Alex Provost Jan 28 '16 at 21:13\n\u2022 Is it sufficient to prove using only the even case? If it is necessary to provide the odd case as well to prove the theorem, then how is it considered a \"direct proof\" rather than \"proof by cases\"? \u2013\u00a0Michael Fulton Apr 13 '17 at 0:23", "date": "2019-05-26 15:02:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1630733/using-a-direct-proof-to-show-that-two-integers-of-same-parity-have-an-even-sum", "openwebmath_score": 0.8189146518707275, "openwebmath_perplexity": 277.01378098435265, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987375049232425, "lm_q2_score": 0.8688267881258485, "lm_q1q2_score": 0.8578578927002093}}
{"url": "https://math.stackexchange.com/questions/4126638/expected-distances-relating-to-points-on-a-stick", "text": "Expected distances relating to points on a stick\n\nQuestion\n\nTwo points are randomly and independently located on a stick of length $$1$$ m.\n\n$$(a)\\quad$$ Find the expected distance between the two points.\n\n$$(b)\\quad$$ Find the expected distance from the left end of the stick to the point which is closest to that end.\n\n$$(c)\\quad$$ Find the expected distance from the left end of the stick to the first of the two points, given that that point is closer to the left end than to the right end.\n\nMy working\n\n$$(a)\\quad$$ Let $$X\\sim U(0, 1)$$ and the two points be $$X_1$$ and $$X_2$$.\n\n$$\\implies f_{X_1, X_2}(x_1, x_2) = 1,\\quad 0 < x_1 < 1\\quad, \\quad 0 < x_2 < 1$$\n\n\\begin{aligned}[t] \\mathbb{E}(\\lvert X_1 - X_2 \\rvert) & = \\int^{\\infty}_{-\\infty}\\int^{\\infty}_{-\\infty} (\\lvert x_1 - x_2 \\rvert) f_{X_1, X_2}(x_1, x_2)\\ \\mathrm{d}{x_1}\\ \\mathrm{d}{x_2} \\\\[1 mm] & = \\int^1_0\\int^{x_1}_0 (x_1 - x_2)\\ \\mathrm{d}{x_2}\\ \\mathrm{d}{x_1} + \\int^1_0\\int^1_{x_1} (x_2 - x_1)\\ \\mathrm{d}{x_2}\\ \\mathrm{d}{x_1} \\\\[1 mm] & = \\frac 1 3 \\end{aligned}\n\n$$(b)\\quad \\frac 1 3$$\n\n$$(c)\\quad \\frac 1 3$$\n\nI know my answer to $$(a)$$ is correct and it also follows intuition - if two points are chosen randomly and independently along a distance of $$1$$ m, one would naturally expect them to be spaced out at equal distances.\n\nHowever, I am unsure if my answers to $$(b)$$ and $$(c)$$ are correct - I used intuition here as well, but how would one derive the answers to $$(b)$$ and $$(c)$$ mathematically?\n\nAny intuitive explanations will be greatly appreciated :)\n\n\u2022 There is no essential difference with choosing randomly 3 points on a circle that has circumference of 1m and then using the first chosen point as the place where the circle is \"opened\" so that it receives a left end and a right end. In this described situation we can use symmetry and find directly $\\frac13$ for expected (arc)distance between the points. Just like you I have the impression that the questions (b) and (c) are actually the same. \u2013\u00a0drhab May 4 at 12:09\n\n(b) Let $$Y=\\min\\{X_1,X_2\\}$$. Let's find the pdf of $$Y$$. Let $$F$$ be the cdf of $$X$$ and $$f$$ be the pdf.\n\n$$\\begin{split}\\Pr(Y> y)&=\\Pr(X_1> y)\\Pr(X_2>y)\\\\ &=\\left(1-F(y)\\right)^2\\end{split}$$\n\nThen the cdf of $$Y$$ is\n\n$$\\begin{split}G(y)&=1-\\Pr(Y>y)\\\\ &=1-(1-F(y))^2\\end{split}$$\n\nThe pdf of $$Y$$ is the derivative\n\n$$\\begin{split}g(y)&=-2(1-F(y))(-f(y))\\\\ &=2f(y)(1-F(y))\\\\ &=2(1-y)\\end{split}$$\n\n(These are fairly standard computations/results, but I repeated them here anyway.)\n\nThis has expected value $$\\int_0^12(y-y^2)dy=2\\left[\\frac {y^2}{2}-\\frac{y^3}{3}\\right]\\bigg|_0^1=\\frac 13$$\n\n(c) We seek $$\\mathbb E(Y|Y<\\frac 12)$$. This will involve a reweighing of the density using $$\\int_0^{1/2}2(1-y)dy=2\\left[y-\\frac{y^2}2\\right]_0^{1/2}=\\frac 34$$. Then the density of $$Y|Y<1/2$$ is $$\\frac{2(1-y)}{3/4}, 0\n\nThis has expectation $$\\int_0^{1/2}\\frac 83(y-y^2)dy=\\frac 83\\left[\\frac{y^2}2-\\frac{y^3}{3}\\right]_0^{1/2}=\\frac 83\\left(\\frac 18-\\frac 1{24}\\right)=\\frac 29$$\n\nIf we know the minimum is closer to the left end, then the expectation is slightly closer to the left end.\n\n\u2022 Thank you for your answer! I understand everything except, ironically, the very first part. How did you get $\\mathbb{P}(Y > y) = \\mathbb{P}(X_1 > y)\\mathbb{P}(X_2 > y)$? Apologies if this is supposed to be trivial... \u2013\u00a0Ethan Mark May 4 at 15:44\n\u2022 you're welcome! It's from: $\\mathbb P(Y>y)=\\mathbb P(\\min\\{X_1, X_2\\}>y)=\\mathbb P(X_1>y, X_2>y)=\\mathbb P(X_1>y)\\mathbb P(X_2 >y)$, due everything having to be bigger than the value if the minimum is to be bigger, and then independence of $X_1, X_2$. \u2013\u00a0Stacker May 4 at 16:35\n\u2022 Right! That makes sense! :) \u2013\u00a0Ethan Mark May 4 at 16:53\n\u2022 (Definition of minimum actually I think) \u2013\u00a0Stacker May 4 at 16:55", "date": "2021-05-09 21:55:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4126638/expected-distances-relating-to-points-on-a-stick", "openwebmath_score": 0.8803520798683167, "openwebmath_perplexity": 523.5113031114173, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750510899382, "lm_q2_score": 0.8688267830311354, "lm_q1q2_score": 0.8578578892836739}}
{"url": "https://math.stackexchange.com/questions/1503360/sum-k-0n-a-k-sum-k-0-lfloor-n-2-rfloor-a-2k-sum-k-0-lflo", "text": "# $\\sum_{k=0}^{n} a_{k}=\\sum_{k=0}^{\\lfloor n/2 \\rfloor} a_{2k} +\\sum_{k=0}^{\\lfloor (n-1)/2 \\rfloor}a_{2k+1}$\n\nI would like to show that \\begin{align} \\sum_{k=0}^{n} a_{k}&=\\sum_{k=0}^{\\lfloor n/2 \\rfloor} a_{2k} + \\sum_{k=0}^{\\lfloor (n-1)/2 \\rfloor}a_{2k+1}\\\\ \\end{align}\n\n\u2022 I'm interested in more ways of prove it\n\nMy proof.\n\n\\begin{align} \\sum_{k=0}^{n} a_{k}&=\\sum_{\\substack{k=0 \\\\ k \\text{ is even }}}^{n} a_{k}+\\sum_{\\substack{k=0 \\\\ k \\text{ is odd }}}^{n} a_{k}\\\\ &=\\sum_{\\substack{k=0 \\\\ k=2k' \\text{ with } k'\\in\\mathbb{Z}}}^{n} a_{k}+\\sum_{\\substack{k=0 \\\\ k=2k'+1 \\text{ with } k'\\in\\mathbb{Z}}}^{n} a_{k}\\\\ &=\\sum_{\\substack{k=0 \\\\ k'=\\frac{k}{2} }}^{n} a_{2k'}+\\sum_{\\substack{k=0 \\\\ k'=\\frac{k-1}{2} }}^{n} a_{2k'+1}\\\\ &=\\sum_{\\substack{k=0 \\\\ k'=\\frac{k}{2}\\\\ k=0 \\implies k'=0 \\\\ k=n \\implies k'=\\frac{n}{2}}}^{n} a_{2k'}+\\sum_{\\substack{k=0 \\\\ k'=\\frac{k-1}{2}\\\\ \\text{since k is odd can start with 1 and not 0 } k=1 \\implies k'=0 \\\\ k=n \\implies k'=\\dfrac{n-1}{2} }}^{n} a_{2k'+1}\\\\ &=\\sum_{k'=0}^{ n/2 } a_{2k'} + \\sum_{k'=0}^{ (n-1)/2 }a_{2k'+1}\\\\ &\\text{If $n$ is even, then $\\frac{(n\u22121)}{2}$ is not an integer, which is why we need the floor.}\\\\ &\\text{Similarly, if $n$ is odd, then $\\frac{n}{2}$ is not an integer}\\\\ \\sum_{k=0}^{n} a_{k}&=\\sum_{k=0}^{\\lfloor n/2 \\rfloor} a_{2k} + \\sum_{k=0}^{\\lfloor (n-1)/2 \\rfloor}a_{2k+1} \\end{align}\n\n\u2022 Is my proof correct\n\u2022 I'm interested in more ways of prove it\n\u2022 so that you don't have to worry that: even number may take odd value and vice versa \u2013\u00a0SiXUlm Oct 29 '15 at 12:36\n\u2022 also, a small typo in the first line of your proof: in the second summation, $k$ should start from 1 instead of 0. \u2013\u00a0SiXUlm Oct 29 '15 at 12:37\n\u2022 If $n$ is even, then $(n-1)/2$ is not an integer, which is why you need the floor. Similarly, if $n$ is odd, then $n/2$ is not an integer. \u2013\u00a0Mankind Oct 29 '15 at 12:37\n\u2022 @HowDoIMath yes i see that's why i have to add floor thanks. what about my proof \u2013\u00a0Educ Oct 29 '15 at 12:43\n\u2022 well, shoot. Yes, your idea is correct and what you are trying to express is clear. I'm not sure your notation of indexing is technically legitimate however. Normally one really shouldn't get bogged do in details like that. However in this case what you are trying to prove is simply \"The sum of all terms is the sum of the even terms plus the sum of the odd terms\" (which is obvious) and your proof is verifying that the reindexing is valid. \u2013\u00a0fleablood Oct 29 '15 at 13:15\n\n$\\sum_{k=0}^{\\lfloor n/2 \\rfloor} a_{2k}= \\sum_{i=0; i\\ is\\ even}^{2\\lfloor n/2 \\rfloor} a_{i} = \\sum_{i=0; i\\ is\\ even}^{n} a_{i} = \\sum_{k=0; k\\ is\\ even}^{n} a_{k}$\n\n$\\sum_{k=0}^{\\lfloor (n-1)/2 \\rfloor} a_{2k + 1}= \\sum_{i=0; i\\ is\\ odd}^{2\\lfloor (n-1)/2 \\rfloor + 1} a_{i} = \\sum_{i=0; i\\ is\\ odd}^{n} a_{i}= \\sum_{k=0; k\\ is\\ odd}^{n} a_{k}$\n\n$\\sum_{k=0}^n a_n = \\sum_{k=0; k\\ is\\ even}^{n} a_{k} + \\sum_{k=0; k\\ is\\ odd}^{n} a_{k} = \\sum_{k=0}^{\\lfloor n/2 \\rfloor} a_{2k} + \\sum_{k=0}^{\\lfloor (n-1)/2 \\rfloor} a_{2k + 1}$\n\n\u2022 i think in the second line $i=1$ not $0$ since $i=2k+1$ \u2013\u00a0Educ Oct 29 '15 at 14:15\n\nIf $n$ is odd, i.e. $n=2m+1$,\n\nthen $k=0, 1, 2, 3, ..., 2m+1=\\underbrace{(0, 2, 4, 6, .., \\color{blue}{2m})}_{\\underset{k=0, 1, ..., \\lfloor n/2\\rfloor}{2k}},(\\underbrace{1, 3, 5, ..., 2m-1, \\color{red}{2m+1}}_{\\underset{k=0, 1, ..., \\lfloor (n-1)/2\\rfloor}{ 2k+1}})$\nas \\begin{align} &2\\bigg\\lfloor \\frac n2 \\bigg\\rfloor&&=2\\bigg\\lfloor \\frac {2m+1}2\\bigg\\rfloor&&=2\\bigg\\lfloor m+\\frac 12 \\bigg\\rfloor&&=\\color{blue}{2m}\\\\ &2\\bigg\\lfloor \\frac {n-1}2 \\bigg\\rfloor+1&&=2\\bigg\\lfloor\\frac{2m}2\\bigg\\rfloor+1&&&&=\\color{red}{2m+1}\\end{align}\n\nIf $n$ is even, i.e. $n=2m$,\n\nthen $k=0, 1, 2, 3, ..., 2m=\\underbrace{(0, 2, 4, 6, .., \\color{green}{2m})}_{\\underset{k=0, 1, ..., \\lfloor n/2\\rfloor}{2k}},(\\underbrace{1, 3, 5, ..., \\color{orange}{2m-1}}_{\\underset{k=0, 1, ..., \\lfloor (n-1)/2\\rfloor}{ 2k+1}})$\nas \\begin{align} &2\\bigg\\lfloor \\frac n2 \\bigg\\rfloor&&=2\\bigg\\lfloor \\frac {2m}2\\bigg\\rfloor&&=\\color{green}{2m}\\\\ &2\\bigg\\lfloor \\frac {n-1}2 \\bigg\\rfloor+1&&=2\\bigg\\lfloor\\frac{2m-1}2\\bigg\\rfloor+1=2\\underbrace{\\bigg\\lfloor m-\\frac12\\bigg\\rfloor}_{m-1}+1=2(m-1)+1&&=\\color{orange}{2m-1}\\end{align}\n\nHence the formula $$\\sum_{k=0}^{n} a_{k}=\\sum_{k=0}^{\\lfloor n/2 \\rfloor} a_{2k} + \\sum_{k=0}^{\\lfloor (n-1)/2 \\rfloor}a_{2k+1}\\\\$$ holds irrespective of whether $n$ is odd or even.", "date": "2019-10-22 01:15:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1503360/sum-k-0n-a-k-sum-k-0-lfloor-n-2-rfloor-a-2k-sum-k-0-lflo", "openwebmath_score": 0.9999791383743286, "openwebmath_perplexity": 434.2893351703029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750507184356, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8578578856073074}}
{"url": "https://math.stackexchange.com/questions/3930987/finding-the-vector-equation-of-a-plane-in-3-space", "text": "# Finding the vector equation of a plane in 3 space\n\nI'm working on a question that asks me to find the vector equation of the plane in $$\\mathbb R^3$$ that contains the point P(-1,6,0) and is parallel to the plane x+2y-z=5. I'm trying to make sure my answer is right versus the answer that I was provided with, which I think is wrong.\n\nI pulled out the normal vector n = (1,2,-1). I proceeded to find two vectors v and u that are orthogonal to n but are not parallel to each other. I followed the theorem that states that when the dot product of 2 vectors equals zero, those vectors are orthogonal. I came up with v = (1,1,3) and u = (4,1,6). So I ended with the vector equation (x,y,z) = (-1,6,0) + t(1,1,3) + s(4,1,6).\n\nIt's my understanding that there are a number of different answers for this problem as there are a number of different vectors, besides the ones I found, that are orthogonal to the normal vector. The answer I was provided for this question stated that the vector equation is (x,y,z) = (-1,6,0) + t(3,-4,1) + s(1,-4,-1). I'm fairly certain that this answer is incorrect as the vectors in the vector equation are not orthogonal to the normal equation. Is my thinking correct?\n\nThe plane parallel to $$x+2y-z=5$$ containing $$(-1,6,0)$$ is $$x+2y-z=-1+2(6)-0=11$$.\nYour parametrization satisfies that: $$(-1+t+4s)+2(6+t+s)-(3t+6s)=11$$.\nThe other answer does not: $$(-1+3t+s)+2(6-4t-4s)-(t-s)\\not=11$$ generally.", "date": "2021-10-25 11:46:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3930987/finding-the-vector-equation-of-a-plane-in-3-space", "openwebmath_score": 0.8206574320793152, "openwebmath_perplexity": 147.58752373821537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750492324249, "lm_q2_score": 0.8688267745399465, "lm_q1q2_score": 0.8578578792858286}}
{"url": "https://math.stackexchange.com/questions/1851554/closure-of-intersection-is-subset-of-intersection-of-closures", "text": "# Closure of Intersection is Subset of Intersection of Closures\n\nI'm trying to prove the following problem from a book I found:\n\nLet $X$ be a topological space and let $\\mathscr{A}$ be a collection of subset of $X$. Prove\n$\\overline{ \\bigcap \\limits_{A \\in \\mathscr{A}} A}\\subseteq \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$\n\n*where the over line indicates closure.\n\nFirst I'm going to show an example the them being equal, and then show an example of a proper subset. I know the examples are not needed for the proof, but I find them insightful, and these posts are like a notebook that I can refer back to later. Then I will attempt to prove this; I actually had a very difficult time with this one, and I feel like I'm missing something. In any case, please let me know If my examples and proof are valid, and of course alternative proof are very welcome.\n\n(I) Example of $\\overline{ \\bigcap \\limits_{A \\in \\mathscr{A}} A} = \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$\n\nI find that with \"most\" collections of subsets, the two sides come up equal. Let $X$ be $\\mathbb{R}$ with the standard Euclidean metric. Also let, $\\mathscr{A}=\\{(1,6),(3,10)\\}$. So then $\\bigcap \\limits_{A \\in \\mathscr{A}} A= (3,6)$, and $\\overline{ \\bigcap \\limits_{A \\in \\mathscr{A}} A} =[3,6]$. Next the right side would be: $\\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}= [1,6] \\cap [3,10]=[3,6]$. So it is possible for the two sides to be equal.\n\n(II) $\\overline{ \\bigcap \\limits_{A \\in \\mathscr{A}} A}\\subset \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$\n\nIt took me a while to find an example of this, but I found that if the collection of subsets consist of infinite sets that converge to an empty set, there will be a proper subset. I also found this example, and it uses an empty set as well. Is it possible for this to be true without an empty set??\n\nLet $X$ be $\\mathbb{R}$ with the standard Euclidean metric. Also let, $\\mathscr{A}=(0,\\frac{1}{n}), n \\in \\mathbb{N}$. Now, $0 \\notin \\bigcap \\limits_{A \\in \\mathscr{A}} A$, and any positive number can be removed with a large enough $n$, thus $\\bigcap \\limits_{A \\in \\mathscr{A}} A = \\varnothing$. The empty set is both open and closed ergo, $\\overline{ \\bigcap \\limits_{A \\in \\mathscr{A}} A} = \\varnothing$. On the other hand, $0 \\in \\overline{A}$, therefore, $\\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}=$ {$0$}. So it is possible for the closure of intersections to be a subset of the intersections of closures.\n\nProof: $\\overline{ \\bigcap \\limits_{A \\in \\mathscr{A}} A}\\subseteq \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$\n\nI considered many different ways to prove this, but they all lead to dead ends. The only thing I could think of was to build on the fact that $A \\subseteq \\overline{A}$ for any subset $A$. Thus, $\\bigcap \\limits_{A \\in \\mathscr{A}} A\\subseteq \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$. (Next is the part I feel is not solid) When $\\bigcap \\limits_{A \\in \\mathscr{A}} A$ is closed the boundary points that are added will also be points included in $\\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$. To summarize symbolically:\n(a)$\\bigcap \\limits_{A \\in \\mathscr{A}} A\\subseteq \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$\n\n(b)$\\partial(\\bigcap \\limits_{A \\in \\mathscr{A}} A) \\subseteq \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$\nThus, (a) $\\wedge$ (b) $\\rightarrow \\overline{ \\bigcap \\limits_{A \\in \\mathscr{A}} A}\\subseteq \\bigcap \\limits_{A \\in \\mathscr{A}} \\overline{A}$\n\nQED\n\nAs stated above, (b) seems intuitively true, and I guess it must be true for the actual statement being proven to be true. But I don't find it obviously true, especially when considering the most general cases of a topological space.\n\n\u2022 You are correct that $A \\subseteq \\overline A$ and therefore $\\bigcap A \\subseteq \\bigcap \\overline A$. Now $\\bigcap \\overline A$ is an intersection of closed sets, hence closed, so $\\bigcap \\overline A$ is a closed set containing $\\bigcap A$. Since by definition $\\overline{\\bigcap A}$ is the smallest closed set containing $\\bigcap A$, we must have $\\overline{\\bigcap A} \\subseteq \\bigcap \\overline A$. \u2013\u00a0Bungo Jul 6 '16 at 23:56\n\u2022 P.S. For a simple example with proper containment, take $A_1 = (0,1)$ and $A_2 = (1,2)$. Then $A_1 \\cap A_2 = \\emptyset$, so also $\\overline{A_1 \\cap A_2} = \\emptyset$. On the other hand, $\\overline{A_1} \\cap \\overline{A_2} = [0,1] \\cap [1,2] = \\{1\\}$. \u2013\u00a0Bungo Jul 7 '16 at 0:01\n\u2022 The intersection is contained in A, so the closure of the former is contained in the closure of the latter. Since this holds for any A, it holds for their intersection. \u2013\u00a0Steve D Jul 7 '16 at 0:38", "date": "2019-05-19 22:21:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1851554/closure-of-intersection-is-subset-of-intersection-of-closures", "openwebmath_score": 0.8303490281105042, "openwebmath_perplexity": 88.8279036797188, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750514614409, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8578578711616712}}
{"url": "https://mathematica.stackexchange.com/questions/145478/boundary-condition-for-schr%C3%B6dinger-equation-in-infinite-range?noredirect=1", "text": "# Boundary Condition for Schr\u00f6dinger Equation in Infinite Range\n\nI am trying to simulate the movement of a coherent state in a quantum harmonic oscilator, but for some reason the answer diverges and there is a warning about not enought boundary conditions.\n\nAlso, since I plan on moving to more complex situations in the future, I would like to avoid b.c.s of the kind:\n\np[t, L] == 0\n\n\nThe reason is that for higher energy situations it will be more taxing to keep L much greater than my area of interest an having the wave get too close to it would cause undesired reflexions.\n\nMy code, for a 1D case, is as follows:\n\nw = 1;\nL = 3;\nc = 1;\nusol = NDSolveValue[{I D[p[t, x], t] + 1/2 D[p[t, x], x, x] ==\n1/2 w^2 x^2 p[t, x], p[0, x] == E^(-w (x - c)^2/2)*(w/Pi)^(1/4)}\n, p, {t, 0, 10}, {x, -L, L}]\n\n\u2022 Yes, the b.c. isn't enough, for a initial value problem defined on $-\\infty<x<\\infty$ you need to add artificial b.c. approximating \"infinity\". There're many posts about this type of issue, see e.g. mathematica.stackexchange.com/q/128516/1871 \u2013\u00a0xzczd May 8 '17 at 5:54\n\u2022 @xzczd, thank you for the hint! I adapted the b.c. and it worked, but I must admit to not undertanding why it did... abc = D[p[t,x], x] + direction p[t,x] == 0 /. {{x -> lb, direction -> -1}, {x -> rb, direction -> 1}}; \u2013\u00a0ivbc May 8 '17 at 16:00\n\u2022 I'm a bit surprised :D . I posted the example just to show that artificial boundary is needed, I didn't know ABC is also applicable for 1D Schrodinger's equation. \u2013\u00a0xzczd May 8 '17 at 16:48\n\u2022 PLenty of references about it, like: sciencedirect.com/science/article/pii/S0021999108004804. But the subject is complex and I dont know if what I tried is a good answer... \u2013\u00a0ivbc May 8 '17 at 17:48\n\u2022 Maybe you can ask a question e.g. \"What's the state-of-art/most popular artificial boundary condition for 1D Schrodinger's equation\" in scicomp.stackexchange.com ? \u2013\u00a0xzczd May 9 '17 at 5:50\n\nSince OP has found this interesting post, let me try to implement the exterior complex scaling method mentioned there.\n\nFirst, make the transform $x= \\left\\{\\begin{array}{cc} & \\begin{array}{cc} -R_0+e^{i \\theta } (\\xi +R_0) & -\\xi >R_0 \\\\ R_0+e^{i \\theta } (\\xi -R_0) & \\xi >R_0 \\\\ \\xi & -R_0\\leq \\xi\\leq R_0 \\\\ \\end{array} \\\\ \\end{array}\\right.$, I'll use DChange for the task:\n\nset = {I D[p[t, x], t] + 1/2 D[p[t, x], x, x] == 1/2 w^2 x^2 p[t, x],\np[0, x] == E^(-w (x - c)^2/2)*(w/Pi)^(1/4)};\n\nright = Exp[I \u03b8] (\u03be - R0) + R0;\nleft = Exp[I \u03b8] (\u03be + R0) - R0;\nmiddle = \u03be;\n\ncoevalue = CoefficientList[\nSimplifyPWToUnitStep@\nPiecewise[{{right, \u03be > R0}, {left, -\u03be > R0}}, middle], \u03be];\n\nneweq = DChange[set, x == coe@0 + coe@1 \u03be, x, \u03be, p[t, x]] /.\nThread[(coe /@ {0, 1}) -> coevalue];\n\n(* Alternative approach for deducing neweq: *)\nhelp = DChange[#, x == #2, x, \u03be, p[t, x]] &;\nchange = Piecewise[{{help[#, right], \u03be > R0}, {help[#, left], -\u03be > R0}},\nhelp[#, middle]] &;\n\nneweq = SimplifyPWToUnitStep@Map[change@# &, set, {2}];\n\n\nNotice currently DChange doesn't directly support piecewise function so the coding is a little roundabout, but I think it's still simpler than transforming by hand.\n\nRemark\n\nSimplifyPWToUnitStep is an undocumented function that expands Piecewise into a combination of UnitStep. I use this function to \"extract\" \u03be from the piecewise part, or CoefficientList in first approach and NDSolveValue in second approach will fail.\n\nThe next step, which is also the final step, is to solve the equation:\n\nw = 1;\nL = 3;\nc = 1;\ntend = 20;\nboundarylayer = L/5; thvalue = 1/2;\nmol[n_Integer, o_: \"Pseudospectral\"] := {\"MethodOfLines\",\n\"SpatialDiscretization\" -> {\"TensorProductGrid\", \"MaxPoints\" -> n,\n\"MinPoints\" -> n, \"DifferenceOrder\" -> o}}\nnewsol = NDSolveValue[{neweq,\np[t, -L - boundarylayer] == p[t, L + boundarylayer] == 0} /. {R0 -> L, \u03b8 ->\nthvalue}, p, {\u03be, -L - boundarylayer, L + boundarylayer}, {t, 0, tend},\nMethod -> mol[25, 4]]; // AbsoluteTiming\nDensityPlot[Norm@newsol[t, x], {t, 0, tend}, {x, -L, L}, PlotRange -> All,\nPlotPoints -> 100, ColorFunction -> \"AvocadoColors\", FrameLabel -> {\"t\", \"x\"}]\n\n\nNotice I've manually set the number of spatial grid points, or NDSolveValue will automatically choose a too large one because the initial condition is no longer smooth.\n\nBesides, the choosing of R0 ($R_0$), boundarylayer (distance from $R_0$ to the boundary of computational domain) and thvalue ($\\theta$) turns out to be a kind of art to make the reflection small enough. There might be some hint in the original paper about the method, but I simply find the proper value by trial and error.\n\n\u2022 (+1) You have transformed x and tto xi and theta, right? \u2013\u00a0zhk May 10 '17 at 7:49\n\u2022 @zhk Nope, in this transformation, only x is transformed to \u03be. \u03b8 is a parameter. You can have a look at the linked post for more information. \u2013\u00a0xzczd May 10 '17 at 8:03\n\u2022 Does this simplify the problem? In what way this transformation helps? \u2013\u00a0zhk May 10 '17 at 8:04\n\u2022 @zhk This transform makes the equation more complicated, of course :) , but by solving this equation instead, we obtain a better approximation for infinite range. (OP wants to solve the equation in $-\\infty<x<\\infty$. ) \u2013\u00a0xzczd May 10 '17 at 8:10\n\u2022 @ivbc Actually this is a function I place in SystemOpen@\"init.m\" file, because I adjust these options frequently when using NDSolve. (See Method -> mol[25, 4]? It's the same as writing Method -> {\"MethodOfLines\", \"SpatialDiscretization\" -> {\"TensorProductGrid\", \"MaxPoints\" -> 25, \"MinPoints\" -> 25, \"DifferenceOrder\" -> 4}}. ) I adjust these options because, as mentioned above, if I don't manually set \"MinPoints\" and \"MaxPoints\", NDSolveValue will choose a too large one because the initial condition is not smooth. (You can take away the option and see what will happen. ) \u2013\u00a0xzczd May 11 '17 at 3:10\n\nLet's remember Schrodinger's equation:\n\n$i\\hbar\\frac{\\partial}{\\partial t} \\Psi(\\mathbf{r},t) = \\left [ \\frac{-\\hbar^2}{2\\mu}\\nabla^2 + V(\\mathbf{r},t)\\right ] \\Psi(\\mathbf{r},t)$\n\nFor the harmonic oscilator $V = x^2$, so your equation is missing a p[x,t] on the RHS besides the boundary conditions. L also needs to be larger too. This seems to work.\n\nw = 2;\nL = 10;\nc = 3;\nusol = NDSolveValue[{I D[p[t, x], t] + 1/2 D[p[t, x], x, x] ==\n1/2 w^2 x^2 p[t, x],\np[0, x] == Exp[(-w (x - c)^2/2)*(w/Pi)^(1/4)], p[t, L] == 0,\np[t, -L] == 0}, p, {t, 0, 10}, {x, -L, L}]\n\nDensityPlot[Norm@usol[t, x], {t, 0, 10}, {x, -L/2, L/2},\nPlotRange -> All, PlotPoints -> 100, ColorFunction -> \"Rainbow\"]\n\n\nupdate\n\nFor completeness purposes only, this is the solution with the abc boundary condition from @xzczd.\n\nw = 1;\nL = 3;\nc = 1;\nusol = NDSolveValue[{I D[p[t, x], t] + 1/2 D[p[t, x], x, x] -\n1/2 w^2 x^2 p[t, x] == 0,\np[0, x] == E^(-w (x - c)^2/2)*(w/Pi)^(1/4),\nDerivative[0, 1][p][t, -L] - p[t, -L] == 0,\nDerivative[0, 1][p][t, L] + p[t, L] == 0},\np, {t, 0, 20}, {x, -L, L}]\n\nDensityPlot[Norm@usol[t, x], {t, 0, 20}, {x, -L/2, L/2},\nPlotRange -> All, PlotPoints -> 100, ColorFunction -> \"Rainbow\",\nFrameLabel -> {\"t\", \"x\"}]\n`\n\nNote how in the first case the wave function starts to diffuse as time passes, so it's not really a coherent state.\n\n\u2022 Thank you for spotting the mistake and for the solution. I improved the question a bit to tray to avoid the artificial b.c. p[t, L] == 0, as it must always be far from the area being analysed. \u2013\u00a0ivbc May 8 '17 at 15:56\n\u2022 @ivb if with the ABC boundary condition it works, what is your question now? I don't understand. \u2013\u00a0tsuresuregusa May 9 '17 at 2:04\n\u2022 @tsuresuregusa I guess it's because there's no guarentee that the simple ABC mentioned in my answer is always applicable. (It's a artificial boundary for 1D wave equation. ) Though currently it seems to work, it may be just a coincidence. (In the comment above, OP has found a paper about ABC for 1D Schrodinger's equation, in which the proposed ABC has a very different form. ) \u2013\u00a0xzczd May 9 '17 at 5:47\n\u2022 @xzczd your last comment nails it, I think we are now out of the scope of this site. \u2013\u00a0tsuresuregusa May 9 '17 at 14:53\n\u2022 Answer accepted since it solves the problem. The question about why it works is important, and can be further considered here: scicomp.stackexchange.com/questions/15973/\u2026 \u2013\u00a0ivbc May 9 '17 at 22:51", "date": "2021-06-17 00:23:00", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/145478/boundary-condition-for-schr%C3%B6dinger-equation-in-infinite-range?noredirect=1", "openwebmath_score": 0.46515917778015137, "openwebmath_perplexity": 3448.9025947070254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.966914019704466, "lm_q2_score": 0.8872045937171068, "lm_q1q2_score": 0.8578505600112754}}
{"url": "https://stats.stackexchange.com/questions/291913/does-t-test-only-work-for-continuous-random-variable", "text": "# Does t test only work for continuous random variable?\n\nWhen using t test normally we would assume the original data follows normal distribution, but if the sample size is large enough, this assumption can be ignored due to Central Limit Theory.\n\nI thought since normal distribution is for continuous random variable, even if the data does not have normality, it at least must be continuous random variable in order to use the t test. However in practice, I found that the t test is widely used even for discrete random variable such as hospital length of stay (LOS) in days (which is integer and thus discrete random variable).\n\nSo my question is, is it a common practice to apply t test on discrete random variable? If not, when we want to compare the mean of two groups consisting of discrete random variable, except for Wilcoxon Signed Rank Test, what other tests can I use?\n\nRemember the Central Limit Theorem is talking about the asymptotic distribution of the standardised mean of iid random variables. So even if your random variables are discrete, as you accumulate more and more of them, once you take their mean (and then standardise it), you'll end up with a random variable which acts more and more like a continuous random variable.\n\nFor example, if you have 100 Bernoulli random variables (each either 0 or 1), then the possible means are $0.00, 0.01, 0.02, \\dots , 1.00$. If you have 1000 then the possible means are $0.000, 0.001, 0.002, \\dots , 1.000$.\n\nSo even if you have discrete random variables, their mean can still be asymptotically normal.\n\n\u2022 Note that the central limit theorem relates not to the mean itself but to a standardized mean. When necessary conditions hold, the sample mean itself is asymptotically just going to go to a constant -- the population mean. \u2013\u00a0Glen_b -Reinstate Monica Jul 17 '17 at 8:20\n\u2022 Thank you @Glen_b, you're absolutely right. I just mentally absorbed the standardisation into the variance of the approximate normal for a particular sample, but of course that's silly since it undermines the point of the convergence in distribution result. \u2013\u00a0RoryT Jul 17 '17 at 8:32\n\u2022 Thanks for your answer, that helped a lot! But as for central limit theorem, I am a little confused at your comments. The CLT says that 'given certain conditions, the arithmetic mean (or sum) of a sufficiently large number of iterates of independent random variables, each with a well-defined (finite) expected value and finite variance, will be approximately normally distributed, regardless of the underlying distribution'. To my understanding, the sample mean itself is related to CLT already, why does it to be 'standardised mean' like @Glen_b said? Could you shed more light? \u2013\u00a0user6892253 Jul 17 '17 at 9:33\n\u2022 No worries. The mean itself is approximately normally distributed if you've got a large enough number of iid variables. However, as you take more and more iid variables, their mean will get closer and closer to the population mean, and will be less and less variable. If you multiply the difference between the sample and population mean by $\\sqrt{n}$ then this counteracts the fact that the sample mean is getting closer and closer to the population mean. This is what we mean by standardisation. The standardised mean will be asymptotically distributed as a standard normal (i.e. $N(0,1)$) \u2013\u00a0RoryT Jul 17 '17 at 10:23\n\nDoes T Test only work for continuous random variable?\n\nIt depends on what you mean by \"work\".\n\nWhen using T test normally we would assume the original data follows normal distribution, but if the sample size is large enough, this assumption can be ignored due to Central Limit Theory.\n\nNot if you care about power.\n\nI thought since normal distribution is for continuous random variable, even if the data does not have normality, it at least must be continuous random variable in order to use the t test.\n\nYour argument that was based on the CLT wouldn't distinguish between discrete and continuous variables -- the CLT applies to either.\n\nHowever in practice, I found that the t test is widely used even for discrete random variable such as hospital length of stay (LOS) in days (which is integer and thus discrete random variable).\n\nIt does get used on discrete data (and on obviously non-normal continuous data, and on data that are neither discrete nor continuous) at least sometimes. Some of those times, it's probably quite reasonable to do so (in that the significance level will be close to the nominal level and the power will be at least adequate).\n\nI would tend to avoid it in that length-of-stay case; leaving aside likely issues with censoring, you have something which may be extremely right skew and even if your sample size were huge, the power may still be poor.\n\nSo my question is is it a common practice to apply t test on discrete random variable?\n\nIf you're going to apply it in a case where you can be confident your samples are drawn from something that's fairly non-normal you'd probably want to have an argument ready for why it would perform reasonably well.\n\nIf not, when we want to compare the mean of two groups consisting of discrete random variable, except for Wilcoxon Signed Rank Test, what other tests can I use?\n\nA signed rank test is a one-sample test (or a paired-sample test), not a two-sample test. It also doesn't compare means (it could reject in the wrong direction, for example)\n\nYou might use a permutation test of the difference in means, or make a different parametric assumption (in some situations you might consider a negative binomial assumption for example, or some other discrete distribution -- depending on circumstances), or perhaps add some assumptions to a Wilcoxon-Mann-Whitney that could make it also a comparison of means (though it won't necessarily outperform the t for discrete skewed cases).\n\n\u2022 What do you mean my power in your answer? \u2013\u00a0SpeedBirdNine Jul 17 '17 at 16:31\n\u2022 en.wikipedia.org/wiki/Statistical_power \u2013\u00a0Glen_b -Reinstate Monica Jul 17 '17 at 22:17\n\u2022 @Glen_b Your answer is so compact and helpful, thanks a lot! As for the statistical power you have mentioned, I know that parametric testing tends to have more power than non-parametric testing, but have no idea that the skewness of data would effect the power that much, could you elaborate more? Or is there any good recommended paper for me to learn from? Thanks! \u2013\u00a0user6892253 Jul 18 '17 at 7:59\n\u2022 Actually, in many cases widely used parametric tests only tend to have slightly more power than the commonly used nonparametric tests when the assumptions of the parametric test are exactly true. When the assumptions are not true, the power of the nonparametric procedure may be much higher. For example, consider an ordinary equal variance two sample t-test compared to a Wilcoxon-Mann-Whitney. When the data are iid normal, the t-test has about 4.5% better efficiency in large samples (it's like the t-test has 22 observations worth of information every time the WMW has 21). \u2013\u00a0Glen_b -Reinstate Monica Jul 18 '17 at 8:15\n\u2022 That's not even the most powerful nonparametric test at the normal. But when the tails are very heavy, the WMW may be vastly more efficient in large samples than the t. But when I said \"power may still be poor\" I wasn't necessarily advocating nonparametric tests for the discrete skew case (power is sometimes poor for some of those too, though for slightly different reasons). \u2013\u00a0Glen_b -Reinstate Monica Jul 18 '17 at 8:17\n\nTheoretically, t-test is only for continuous data because discrete data can never meet normality assumption. However, t-test is quite robust given enough samples.\n\nApplying t-test for discrete value is actually quite common. For example, the popular way to test for microarray gene expression array is: moderated t-Test.\n\nhttps://www.dnastar.com/arraystar_help/index.html#!Documents/moderatedttest.htm\n\nGene expression is counting data and certainly not continuous. But t-test is simple and good enough.\n\nEDIT:\n\nI think @Glen_b is correct. But, discrete data simply don't satisfy theoretical assumptions, whereas there's a possibility that continuous data do satisfy.\n\nAgain he's correct. In my example, the method assumes asymptotic behaviour of the sample means.\n\nEDIT:\n\nt-test is indeed common for discrete test (answer @Glen_b), especially when we have a large sample and the histogram of the distribution \"looks\" normal anyway (not skewed).\n\nThe t-test is simple for everybody to understand. In many practical applications, a good-enough and easy statistical test is sufficient.\n\n\u2022 Continuous data that's not normal can never actually meet the normality assumption either (not even on average -- sample means of iid continuous rvs can only have a normal distribution if they were to begin with -- the convergence, whether from discrete or continuous population is only asymptotic either way). I doubt the normality assumption is ever true on real data. \u2013\u00a0Glen_b -Reinstate Monica Jul 17 '17 at 10:20\n\u2022 When you say 't test is common for discrete test when we have a large sample and the histogram of the distribution looks normal', what about the skewed discrete data with a large sample size, is there any method to verify whether t test can work well in such case? \u2013\u00a0user6892253 Jul 18 '17 at 8:19", "date": "2020-01-29 12:51:44", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/291913/does-t-test-only-work-for-continuous-random-variable", "openwebmath_score": 0.7939465641975403, "openwebmath_perplexity": 472.8545884329556, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914017797636, "lm_q2_score": 0.8872045907347108, "lm_q1q2_score": 0.8578505554358065}}
{"url": "https://math.stackexchange.com/questions/2498416/solving-a-set-problem/2498441", "text": "# Solving a Set Problem\n\nHere is my question\n\n\u2022 In a class that has $40$ students, there are $10$ students who could not pass the math exam and $30$ students who passed the english exam. If there are $6$ students who could not pass the both exam, How many are there students who passed the both exam?\n\nI'm currently not able to build the correct equation. Might I get help? It seems like a bit confused. That's why I need get help.\n\n\u2022 Hint: use the inclusion/exclusion principle. \u2013\u00a0user370967 Oct 31 '17 at 16:37\n\u2022 @Math_QED What is it, sir? I'm already trying on it. Before using inclusion/exclusion principle, I ought to build the correct equation. \u2013\u00a0user440404 Oct 31 '17 at 16:38\n\u2022 \"group theory\" is an incorrect tag; this is a question about sets. \u2013\u00a0Andrew Tindall Oct 31 '17 at 16:39\n\u2022 I believe that It seems better now. \u2013\u00a0user440404 Oct 31 '17 at 16:42\n\u2022 @Bobtrollsten The inclusion/exclusion principle \"gives you the equation,\" so it would be silly to build the equation before applying the principle. \u2013\u00a0rschwieb Oct 31 '17 at 16:43\n\nThere are $4$ types of people:\n\nLet $A$ be the people who passed no exam.\n\nLet $B$ be the people who passed only Math.\n\nLet $C$ be the people who passed only English.\n\nLet $D$ be the people who passed both.\n\nWe know $A+B+C+D=40$\n\nWe know $A+C=10$\n\nWe know $C+D=30$\n\nWe know $A=6$.\n\nSo this is just a system of equations.\n\n\u2022 I think I got it now. \u2013\u00a0user440404 Oct 31 '17 at 16:54\n\nI find a Venn Diagram to always be very helpful for these kinds of questions:\n\n$A,B,C,D$ are the regions according to Jorge's answer\n\nSo we know the following:\n\nOnce you figure out how the '10 people did not pass Math' plays out in the diagram, the rest goes quick.\n\n\u2022 I was trying to plug it into venn. \u2013\u00a0user440404 Oct 31 '17 at 16:57\n\u2022 does the venn diagram help? You get a lot of clumping in this case. \u2013\u00a0Yorch Oct 31 '17 at 16:59\n\u2022 @JorgeFern\u00e1ndez True ... but I think different approaches work better for different people. So maybe for some the diagram is more distracting than helpful, but for others it may be helpful. Personally, I'm a very visual person. And once you figure out how the '10 people not inside the Passed Math' circle plays out in the diagram, the rest goes quick. \u2013\u00a0Bram28 Oct 31 '17 at 17:08\n\nI hope my cute Venn diagram can be useful to somebody :) $$...$$", "date": "2021-06-15 20:08:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2498416/solving-a-set-problem/2498441", "openwebmath_score": 0.40582728385925293, "openwebmath_perplexity": 610.0662452749282, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140206578809, "lm_q2_score": 0.8872045877523147, "lm_q1q2_score": 0.8578505550897083}}
{"url": "http://math.stackexchange.com/questions/492901/complex-function-sketch", "text": "Complex function sketch\n\nI have just started studying some complex functions and I do not understand the concept very well (I cannot visualize them easily). So I would appreciate some help for the following problem:\n\nSay we have the complex function f(z) = cosh(z) . I would like to know how the f-plane looks like when x=const. and respectively y=const.\n\nThank you very much!\n\n-\nWhen you can visualize complex functions easily, let me know. I still have a hard time, too :) \u2013\u00a0 Clayton Sep 13 '13 at 20:29\nOk, I see that the \"easily\" is rather stupid but what I wanted to say is that the transition from real to complex functions is not very smooth in my head.:) \u2013\u00a0 Whats My Name Sep 13 '13 at 20:45\n\nYou may be interested in this paper regarding visualizing complex functions:\n\n\"Phase Plots of Complex Functions: A Journey in Illustration\" by Elias Wegert and Gunter Semmler.\n\nJim Fowler and I (Steve Gubkin) wrote this webpage to help visualization of complex functions:\n\nWe hope to produce an online complex analysis course soon which employs a diverse set of visualization methods. For example, we will use webGL to produce 3d plots of modulus colored by phase, phase plots on the riemann sphere, interactive geometry tools for the poincare metric on the disk, etc. We should also have a lot of computational exercises which are verified directly by a server running an instance of sage. We are still developing back end stuff right now, but we should have something up and running within the year. We do not have anything up there now, but watch this space: http://www.gratisu.org/.\n\nPeace, Steve\n\n-\nThis is very helpful. Thank you very much! \u2013\u00a0 Whats My Name Sep 13 '13 at 23:36\nI'm interested by such a course. I already illustrated a whole course of mathematics for bachelor using complex functions on my website: www.mikaelmayer.com/reflex/category/cours-a-lepfl/ Students appreciated it. \u2013\u00a0 Mika\u00ebl Mayer May 5 at 13:25\n@Mika\u00eblMayer Ya, I still have plans to do this, but we got sort of distracted by producing calculus content (also, I defend my thesis tomorrow). I cannot promise when, but I definitely intend to build this course! \u2013\u00a0 Steven Gubkin May 5 at 18:31\n\nLet $z=x+iy.$ Separating the real and imaginary parts of $f,$ we have \\begin{align} f(z)=f(x+iy)=\\cosh(z)=\\dfrac{e^{x+iy}+e^{-x-iy}}{2}\\\\ =\\dfrac{1}{2}[e^x(\\cos{y}+i\\sin{y} )+e^{-x}(\\cos{y}-i\\sin{y})] \\\\=\\cos{y}\\cdot\\dfrac{e^x+e^{-x}}{2}+i\\sin{y}\\cdot\\dfrac{e^x-e^{-x}}{2}. \\end{align} Denote $$u(x+iy)=\\Re{f(x+iy)}=\\cos{y}\\cdot\\dfrac{e^x+e^{-x}}{2}=\\cosh{x}\\cdot\\cos{y},\\\\ v(x+iy)=\\Im{f(c+iy)}=\\sin{y}\\cdot\\dfrac{e^x-e^{-x}}{2}=\\sinh{x}\\cdot\\sin{y}$$ For fixed $x=c$\n\n$$\\dfrac{u(c+iy)}{\\cosh{c}}=\\cos{y},\\\\ \\dfrac{v(c+iy)}{\\sinh{c}}=\\sin{y}.$$ Squaring last equalities, we have ellipse $$\\dfrac{u^2}{\\cosh^2{c}}+\\dfrac{v^2}{\\sinh^2{c}}=1.$$\n\n-\n\nAlthough you asked this question a while ago, I have a new answer. You may be interested by the website reflex4you.com in \"expert\" mode. For example, the $cosh$ function is the following:\n\n$z\\rightarrow cosh(z)$ (click on the picture to visit the website and enter other formulas)\n\nAlthough the axis are not shown, the window is between $-4-3i$ and $4+3i$. The zero is black and white is infinite. 1 is red and -1 is cyan. This representation is quasi-invariant if you take the negative of the picture, it will represent the inverse function.\n\nTo analyze this picture, you can consider that the values on the horizontal axis (the real axis) are red and darkest a zero, which means that $cosh(x) > cosh(0)$. The black spot are zeroes of the function corresponding to $+/- \\pi/2$; Colored points denote complex functions. $i$ is greenish whereas $-i$ is purplish.\n\nBy clicking on the image above, you can modify the formula and try out other ones. I feel it is a very good way to represent complex functions.\n\nTo answer your original question, I wrote an algorithm usable on the website to compute the f-plane, for example for x constant. You can modify the value of the parameter c (constant for $x$). @M.Strochyk was right, these are ellipses.\n\nlet c = 1; set k = 0; let min = -4; let max = 4; let n = 100; func f = cosh(z); set result = 0; let threshold = 0.1; repeat n in set theta = ((max-min)*k/n+min)*i+c; set result = if(abs(f(theta)-z) < threshold, theta, result); set k = k + 1; result\n\nClick on the picture to be able to modify it by entering different values.\n\nDISCLAIMER: I am one of the co-authors of this website.\n\n-", "date": "2015-08-30 16:57:35", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/492901/complex-function-sketch", "openwebmath_score": 0.7110984325408936, "openwebmath_perplexity": 531.2853309271399, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.952574129515172, "lm_q2_score": 0.900529778109184, "lm_q1q2_score": 0.8578213694848469}}
{"url": "https://math.stackexchange.com/questions/1108428/step-by-step-solution-for-x1-x-21-2/1108437", "text": "Step-by-Step Solution for $x^{1/x}=2^{1/2}$ [duplicate]\n\nI came across the equation $$x^{1/x}=2^{1/2}$$ where $x\\in\\mathbb R$. One can immediately see that $x=2$ is a solution, but it is easy to miss that $x=4$ satisfies the equation as well. Verfiying that $2,4$ are solutions is not hard, but how would one go about formally solving this equation, i.e. how could one solve for $x$? I am asking because if the equation was say $x^{1/x}=2^{1/3}$ then the two solutions would not be obvious.\n\nBasically, how to fill in the dots: $$x^{1/x}=2^{1/2} \\iff \\ldots\\iff x=2 \\text{ or } x=4$$\n\n\u2022 for $0 < A < \\frac{1}{e}$ there are two solutions to $\\frac{\\log x}{x} = A.$ For $A > \\frac{1}{e}$ there are no solutions. Just draw the graph of $y = \\frac{\\log x}{x}$ for $x > 0.$ \u2013\u00a0Will Jagy Jan 17 '15 at 23:32\n\u2022 @WillJagy Thank you for this, but I am more interested in how to obtain the solutions, not so much in how to prove their existence. \u2013\u00a0Phil-ZXX Jan 17 '15 at 23:46\n\u2022 Tom, you asked about the well known example, $2^4 = 4^2.$ Someone came up with a reasonable conjecture on all rational solutions to $x^y = y^x.$ Having trouble finding that. I mean he asked his conjecture as an MSE question. \u2013\u00a0Will Jagy Jan 18 '15 at 1:26\n\u2022 a moderator found it for me, see answer at meta.math.stackexchange.com/questions/19299/\u2026 \u2013\u00a0Will Jagy Jan 18 '15 at 2:39\n\nNotice\n\n$$x ^{1/x} = 2^{1/2} \\Rightarrow x = 2^{x/2} \\Rightarrow x^{2} = 2^x$$\n\nthen you may look want take a look here.\n\n\u2022 There are two good answers there. Feel free to ask anything. \u2013\u00a0Aaron Maroja Jan 17 '15 at 23:32\n\u2022 Thank you. The accepted answer in your link does not actually provide any insights in how to obtain the solutions, whereas your solutions looks very nice (this was exactly what I was looking for). But I am still wondering, is there any way around the Lambert W function? \u2013\u00a0Phil-ZXX Jan 18 '15 at 0:08\n\u2022 @Tom Yes, see my answer. And no, you won't get explicit results except for the special cases like $2^4=4^2$. \u2013\u00a0user2345215 Jan 18 '15 at 0:08\n\u2022 @user2345215 I do agree with the reasoning of your answer, but it doesn't tell me how to obtain 2,4 (asssuming I did not know them). If the equation was $x^{1/x}=2^{1/3}$ we would have to actually derive the solutions, and I am looking for a way to do this (not a way to prove that 2,4 are the only solutions). \u2013\u00a0Phil-ZXX Jan 18 '15 at 0:12\n\u2022 @Tom I'm glad I could help you Tom, well there is Newton's approximation. \u2013\u00a0Aaron Maroja Jan 18 '15 at 0:23\n\nNote that $$(x^{1/x})'=(e^{(\\ln x)/x})'=\\tfrac{1-\\ln x}{x^2}x^{1/x}$$ So the function is increasing on $(0,e)$ and decreasing on $(e,\\infty)$. Therefore $2$ and $4$ are the only solutions.\n\nFrom this follows that $x^{1/x}=c$ has two solutions for $c\\in (1,e^{1/e})$, one for $c\\in(0,1]\\cup\\{e^{1/e}\\}$ and no for $c\\in(e^{1/e},\\infty)$.\n\nI suppose it is necessary to consider what happens when $x < 0.$ There is a question of interpretation of $x^{\\frac{1}{x}}$ for $x \\leq 0,$ ( for example, $(-1)^{r}$ could reasonable interpreted as a non-real complex number if $r$ is an irrational positive real number, if we consider $-1$ to be $e^{i \\pi}$), so I assume the function only to be defined for $x >0$, where it is defined to be $e^{\\frac{\\log x}{x}}.$ As far as this question is concerned, it seems that the case $x >0$ is the most relevant case anyway.\n\nNote that $x >0$ is a solution of $x^{\\frac{1}{x}} = 2^{\\frac{1}{2}}$ if and only if $\\frac{\\log x}{x} = \\frac{\\log 2}{2}.$ The derivative of $\\frac{\\log x}{x}$ is $\\frac{ 1 - \\log x}{x^{2}}$ which is positive when $x <e$ and negative when $x > e.$ Hence $\\frac{\\log x}{x}$ increases when $x <e$, then decreases when $x >e$, taking a maximum value when $x = e.$ The values of $\\frac{ \\log x}{x}$ are positive on $(1, \\infty)$ and the function is differentiable on that interval.Now if $\\frac{log{x}}{x} = \\frac{\\log{y}}{y}$ for $1 < x < y,$ then the derivative vanishes somewhere on $(x,y)$ by the Mean Value Theorem. Since the derivative vanishes only at $e$, we see that any value of $\\frac{\\log x}{x}$ can be attained at most twice. Since $\\frac{\\log{4}}{4} = \\frac{2\\log 2}{2 \\times 2} = \\frac{\\log 2}{2},$ we see that $2$ and $4$ are the only values of $x$ for which $x^{\\frac{1}{x}} = 2^{\\frac{1}{2}}.$", "date": "2019-12-06 18:29:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1108428/step-by-step-solution-for-x1-x-21-2/1108437", "openwebmath_score": 0.895237386226654, "openwebmath_perplexity": 136.96237975393962, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429634078179, "lm_q2_score": 0.8723473879530492, "lm_q1q2_score": 0.8578166655908208}}
{"url": "https://teachingcalculus.com/category/integration/integration-theory/page/3/", "text": "# Good Question 13\n\nLet\u2019s end the year with this problem that I came across a while ago in a review book:\n\nIntegrate\u00a0$\\int{x\\sqrt{x+1}dx}$\n\nIt was a multiple-choice question and had four choices for the answer. The author intended it to be done with a\u00a0u-substitution but being a bit rusty I tried integration by parts. I got the correct answer, but it was not among the choices. So I thought it would make a good challenge to work on over the holidays.\n\n1. Find the antiderivative using a\u00a0u-substitution.\n2. Find the antiderivative using integration by parts.\n3. Find the antiderivative using a different\u00a0u-substitution.\n4. Find the antiderivative by adding zero in a convenient form.\n\nYour answers for 1, 3, and 4 should be the same, but look different from your answer to 2. The difference is NOT due to the constant of integration which is the same for all four answers. Show that the two forms are the same by\n\n2. \u201cSimplifying\u201d your answer to 1, 3, 4 and get that third form again.\n\nGive it a try before reading on. The solutions are below the picture.\n\nMethod 1: u-substitution\n\nIntegrate\u00a0$\\int{x\\sqrt{x+1}dx}$\n\n$u=x+1,x=u-1,dx=du$\n\n$\\int{x\\sqrt{x+1}dx=}\\int{\\left( u-1 \\right)\\sqrt{u}}du=\\int{{{u}^{3/2}}-{{u}^{1/2}}}du=\\tfrac{5}{2}{{u}^{5/2}}-\\tfrac{3}{2}{{u}^{3/2}}$\n\n$\\tfrac{2}{5}{{\\left( x+1 \\right)}^{5/2}}-\\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}+C$\n\nMethod 2: By Parts\n\nIntegrate\u00a0$\\int{x\\sqrt{x+1}dx}$\n\n$u=x,du=dx$\n\n$dv=\\sqrt{x+1}dx,v=\\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}$\n\n$\\int{x\\sqrt{x+1}dx}=\\tfrac{2}{3}x{{\\left( x+1 \\right)}^{3/2}}-\\int{\\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}dx}=$\n\n$\\tfrac{2}{3}x{{\\left( x+1 \\right)}^{3/2}}-\\tfrac{4}{15}{{\\left( x+1 \\right)}^{5/2}}+C$\n\nMethod 3: A different u-substitution\n\nIntegrate\u00a0$\\int{x\\sqrt{x+1}dx}$\n\n$u=\\sqrt{x+1},x={{u}^{2}}-1,$\n\n$du=\\tfrac{1}{2}{{\\left( x+1 \\right)}^{-1/2}}dx,dx=2udu$\n\n$2{{\\int{\\left( {{u}^{2}}-1 \\right){{u}^{2}}}}^{{}}}du=2\\int{{{u}^{4}}-{{u}^{2}}}du=\\tfrac{2}{5}{{u}^{5}}-\\tfrac{2}{3}{{u}^{3}}=$\n\n$\\tfrac{2}{5}{{\\left( x+1 \\right)}^{5/2}}-\\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}+C$\n\nThis gives the same answer as Method 1.\n\nMethod 4: Add zero in a convenient form.\n\nIntegrate\u00a0$\\int{x\\sqrt{x+1}dx}$\n\n$\\int{x\\sqrt{x+1}}dx=\\int{x\\sqrt{x+1}+\\sqrt{x+1}-\\sqrt{x+1} dx=}$\n\n$\\int{\\left( x+1 \\right)\\sqrt{x+1}-\\sqrt{x+1}}dx=$\n\n$\\int{{{\\left( x+1 \\right)}^{3/2}}-{{\\left( x+1 \\right)}^{1/2}}dx}=$\n\n$\\tfrac{2}{5}{{\\left( x+1 \\right)}^{5/2}}-\\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}+C$\n\nThis, also, gives the same answer as Methods 1 and 3.\n\nSo, by a vote of three to one Method 2 must be wrong. Yes, no, maybe?\n\nNo, all four answers are the same. Often when you get two forms for the same antiderivative, the problem is with the constant of integration. That is not the case here. We can show that the answers are the same by factoring out a common factor of ${{\\left( x+1 \\right)}^{3/2}}$. (Factoring the term with the lowest fractional exponent often is the key to simplifying expressions of this kind.)\n\nSimplify the answer for Method 2:\n\n$\\tfrac{2}{3}x{{\\left( x+1 \\right)}^{3/2}}-\\tfrac{4}{15}{{\\left( x+1 \\right)}^{5/2}}+C=$\n\n$\\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}\\left( x-\\tfrac{2}{5}\\left( x+1 \\right) \\right)+C=$\n\n$\\displaystyle \\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}\\left( \\frac{5x-2x-2}{5} \\right)+C=$\n\n$\\tfrac{2}{15}{{\\left( x+1 \\right)}^{3/2}}\\left( 3x-2 \\right)+C$\n\nSimplify the answer for Methods 1, 3, and 4:\n\n$\\tfrac{2}{5}{{\\left( x+1 \\right)}^{5/2}}-\\tfrac{2}{3}{{\\left( x+1 \\right)}^{3/2}}+C=$\n\n$\\tfrac{2}{15}{{\\left( x+1 \\right)}^{3/2}}\\left( 3\\left( x+1 \\right)-5 \\right)+C=$\n\n$\\tfrac{2}{15}{{\\left( x+1 \\right)}^{3/2}}\\left( 3x-2 \\right)+C$\n\nSo, same answer and same constant.\n\nIs this a good question? No and yes.\n\nAs a multiple-choice question, no, this is not a good question. It is reasonable that a student may use the method of integration by parts. His or her answer is not among the choices, but they have done nothing wrong. Obviously, you cannot include both answers, since then there will be two correct choices. Moral: writing a multiple-choice question is not as simple as it seems.\n\nFrom another point of view, yes, this is a good question, but not for multiple-choice. You can use it in your class to widen your students\u2019 perspective. Give the class a hint on where to start. Even better, ask the class to suggest methods; if necessary, suggest methods until you have all four (\u2026 maybe there is even a fifth). Assign one-quarter of your class to do the problem by each method. Then have them compare their results. Finally, have them do the simplification to show that the answers are the same.\n\nMy next post will be after the holidays.\n\n.\n\n# Starting Integration\n\nBehind every definite integral is a Riemann sums. Students need to know about Riemann sums so that they can understand definite integrals (a shorthand notation for the limit if a Riemann sun) and the Fundamental theorem of Calculus. Theses posts help prepare students for Riemann sums.\n\n1. The Old Pump Where I start Integration\n2. Flying into Integrationland\u00a0Continues the investigation in the Old Pump \u2013 the airplane problem\n3. Working Towards Riemann Sums\n4. Definition of the Definite Integral and the FTC \u2013 a more exact demonstration from last Friday\u2019s post and\u00a0The Fundamental Theorem of Calculus\u00a0\u2013\u00a0 an older demonstration\n5. More about the FTC\u00a0The derivative of a function defined by an integral \u2013 the other half of the FTC.\n6. Good Question 11 Riemann Reversed \u2013 How to find the integral, given the Riemann sum\n7. Properties of Integrals\n8. Variation on a Theme \u2013 2 Comparing Riemann sums\n9. Trapezoids \u2013 Ancient and Modern \u2013 some history\n\n# The Definite Integral and the\u00a0FTC\n\nThe Definition of the Definite Integral.\n\nThe definition of the definite integrals is: If f is a function continuous on the closed interval [a, b], and\u00a0$a={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<\\cdots <{{x}_{{n-1}}}<{{x}_{n}}=b$ \u00a0is a partition of that interval, and $x_{i}^{*}\\in [{{x}_{{i-1}}},{{x}_{i}}]$, then\n\n$\\displaystyle \\underset{{\\left| {\\left| {\\Delta x} \\right|} \\right|\\to 0}}{\\mathop{{\\lim }}}\\,\\sum\\limits_{{i=0}}^{n}{{f\\left( {x_{i}^{*}} \\right)}}\\left( {{{x}_{i}}-{{x}_{{i-i}}}} \\right)=\\int\\limits_{a}^{b}{{f\\left( x \\right)dx}}$\n\nThe left side of the definition is, of course, any Riemann sum for the function f on the interval [a, b]. In addition to being shorter, the right side also tells you about the interval on which the definite integral is computed. The expression $\\left\\| {\\Delta x} \\right\\|$\u00a0\u00a0is called the \u201cnorm of the partition\u201d and is the longest subinterval in the partition. Usually, all the subintervals are the same length, $\\frac{{b-a}}{n}$, and this is the last your will hear of the norm. With all the subdivisions of the same length this can be written as\n\n$\\displaystyle \\underset{{n\\to \\infty }}{\\mathop{{\\lim }}}\\,\\sum\\limits_{{i=0}}^{n}{{f\\left( {x_{i}^{*}} \\right)}}\\frac{{b-a}}{n}=\\int\\limits_{a}^{b}{{f\\left( x \\right)dx}}$\n\nOther than that, there is not much more to the definition. It is simply a quicker and more efficient notation for the sum.\n\nThe Fundamental Theorem of Calculus (FTC).\n\nFirst recall the Mean Value Theorem (MVT) which says: If a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) then there exist a number, c, in the open interval (a, b) such that\u00a0${f}'\\left( c \\right)\\left( {b-a} \\right)=f\\left( b \\right)-f\\left( a \\right)$.\n\nNext, let\u2019s rewrite the definition above with a few changes. The reason for this will become clear.\n\n$\\int\\limits_{a}^{b}{{{f}'\\left( x \\right)dx}}=\\underset{{\\left| {\\left| {\\Delta x} \\right|} \\right|\\to 0}}{\\mathop{{\\lim }}}\\,\\sum\\limits_{{i=0}}^{n}{{{f}'\\left( {{{c}_{i}}} \\right)\\left( {{{x}_{i}}-{{x}_{{i-i}}}} \\right)}}$\n\nSince every function is the derivative of another function (even though we may not know that function or be able to write a closed-form expression for it), I\u2019ve expressed the function as a derivative, I\u2019ve also chosen the point in each subinterval, ${{c}_{i}}$, to be the number in each subinterval guaranteed by the MVT for that subinterval.\n\nThen, $\\displaystyle {f}'\\left( {{{c}_{i}}} \\right)\\left( {{{x}_{i}}-{{x}_{{i-i}}}} \\right)=f\\left( {{{x}_{i}}} \\right)-f\\left( {{{x}_{{i-1}}}} \\right)$. Making this substitution, we have\n\n$\\int\\limits_{a}^{b}{{{f}'\\left( x \\right)dx}}=\\underset{{\\left| {\\left| {\\Delta x} \\right|} \\right|\\to 0}}{\\mathop{{\\lim }}}\\,\\sum\\limits_{{i=0}}^{n}{{\\left( {f\\left( {{{x}_{i}}} \\right)-f\\left( {{{x}_{{i-1}}}} \\right)} \\right)}}$\n\n$\\displaystyle =f\\left( {{{x}_{1}}} \\right)-f\\left( {{{x}_{0}}} \\right)+f\\left( {{{x}_{2}}} \\right)-f\\left( {{{x}_{1}}} \\right)+f\\left( {{{x}_{3}}} \\right)-f\\left( {{{x}_{2}}} \\right)+\\cdots +f\\left( {{{x}_{n}}} \\right)-f\\left( {{{x}_{{n-1}}}} \\right)$\n\n$\\displaystyle =f\\left( {{{n}_{n}}} \\right)-f\\left( {{{x}_{0}}} \\right)$\n\nAnd since ${{x}_{0}}=a$\u00a0and\u00a0 ${{x}_{n}}=b$,\n\n$\\displaystyle \\int_{a}^{b}{{{f}'\\left( x \\right)dx}}=f\\left( b \\right)-f\\left( a \\right)$\n\nThis equation is called the Fundamental Theorem of Calculus. In words, it says that the integral of a function can be found by evaluating the function of which the integrand is the derivative at the endpoints of the interval and subtracting the values. This is a number that may be positive, negative, or zero depending on the function and the interval. The function of which the integrand is the derivative, is called the antiderivative of the integrand.\n\nThe real meaning and use of the FTC is twofold:\n\n1. It says that the integral of a rate of change (i.e. a derivative) is the net amount of change. Thus, when you want to find the amount of change \u2013 and you will want to do this with every application of the derivative \u2013 integrate the rate of change.\n2. It also gives us an easy way to evaluate a Riemann sum without going to all the trouble that is necessary with a Riemann sum; simply evaluate the antiderivative at the endpoints and subtract.\n\nAt this point I suggest two quick questions to emphasize the second point:\n\n1. Find $\\int_{3}^{7}{{2xdx}}$.\n\nAsk if anyone knows a function whose derivative is 2x? Your students will know this one. The answer is x2, so\n\n$\\displaystyle \\int_{3}^{7}{{2xdx}}={{7}^{2}}-{{3}^{2}}=40$.\n\nMuch easier than setting up and evaluating a Riemann sum!\n\n2. Then ask your students to find the area enclosed by the coordinate axes and the graph of cos(x) from zero to $\\frac{\\pi }{2}$. With a little help they should arrive at\n\n$\\displaystyle \\int_{0}^{{\\pi /2}}{{\\cos \\left( x \\right)dx}}$.\n\nThen ask if anyone knows a function whose derivative is cos(x). it\u2019s sin(x), so\n\n$\\displaystyle \\int_{0}^{{\\pi /2}}{{\\cos \\left( x \\right)dx}}=\\sin \\left( {\\frac{\\pi }{2}} \\right)-\\sin \\left( 0 \\right)=1-0=1$.\n\nAt this point they should be convinced that the FTC is a good thing to know.\n\nThere is another form of the FTC that is discussed in More About the FTC.\n\n# Good Question 12 \u2013 Parts with a\u00a0Constant?\n\nSomeone asked me about this a while ago and I thought I would share it with you. It may be a good question to get your students thinking about; see if they can give a definitive answer that will, of course, include a justification.\n\nIntegration by Parts is summarized in the equation\n\n$\\displaystyle \\int_{{}}^{{}}{udv}=uv-\\int_{{}}^{{}}{vdu}$\n\nTo use the equation, you choose part of a given integral (left side) to be u and part to be dv, both functions of x. Then\u00a0you differentiate u and integrate dv and use them on the right side to obtain a simpler integral that you can integrate.\n\nThe question is this: When you integrate dv, should you, can you, have a constant of integration, the \u201c+\u00a0\u201d that you insist upon in other integration problems? Why don\u2019t you use it here? Or can you?\n\nAnswer: Let\u2019s see what happens if we use a constant. Assume that $\\displaystyle \\int_{{}}^{{}}{dv}=v+C$. Then\n\n$\\displaystyle \\int_{{}}^{{}}{udv}=u\\cdot \\left( v+C \\right)-\\int_{{}}^{{}}{\\left( v+C \\right)du}$\n\n$\\displaystyle =uv+Cu-\\left( \\int_{{}}^{{}}{vdu}+\\int_{{}}^{{}}{Cdu} \\right)$\n\n$\\displaystyle =uv+Cu-\\int_{{}}^{{}}{vdu}-Cu$\n\n$\\displaystyle =uv-\\int_{{}}^{{}}{vdu}$\n\nSo, you may use a constant if you want, but it will always add out of the expression.\n\nFor more on integration by parts see here\u00a0for the basic idea, here\u00a0for the tabular method, here\u00a0for a quicker way than the tabular method, and here\u00a0for more on the tabular method and reduction formulas.\n\n# Good Question 11 \u2013 Riemann\u00a0Reversed\n\nGood Question 11 \u2013 or not.\n\nThe question below appears in the 2016 Course and Exam Description (CED) for AP Calculus (CED, p. 54),\u00a0and has caused some questions since it is not something included in most textbooks and has not appeared on recent exams. The question gives a Riemann sum and asks for the definite integral that is its limit. Another example appears in the 2016 \u201cPractice Exam\u201d available at your audit website; see question AB 30. This type of question asks the student to relate a definite integral to the limit of its\u00a0Riemann sum. These are called reversal questions since you must work in reverse of the usual order. Since this type of question appears in both the CED examples and the practice exam, the chances of it appearing on future exams look good.\n\nTo the best of my recollection the last time a question of this type appeared on the AP Calculus exams was in 1997, when only about 7% of the students taking the exam got it correct. Considering that by random guessing about 20% should have gotten it correct, this was a difficult question. This question, the \u201cradical 50\u201d question, is at the end of this post.\n\nExample 1\n\nWhich of the following integral expressions is equal to $\\displaystyle \\underset{n\\to \\infty }{\\mathop{\\lim }}\\,\\sum\\limits_{k=1}^{n}{\\left( \\sqrt{1+\\frac{3k}{n}}\\cdot \\frac{1}{n} \\right)}$\u00a0?\n\nThere were 4 answer choices that we will consider in a minute.\n\nThe first key to answering the question is to recognize the limit as a Riemann sum. In general, a right-side Riemann sum for the function f on the interval [a, b] with n equal subdivisions, has the form:\n\n$\\displaystyle \\underset{n\\to \\infty }{\\mathop{\\lim }}\\,\\sum\\limits_{k=1}^{n}{\\left( f\\left( a+\\frac{b-a}{n}\\cdot k \\right)\\cdot \\frac{b-a}{n} \\right)}=\\int_{a}^{b}{f\\left( x \\right)dx}$\n\nTo evaluate the limit and express it as an integral, we must identify, a, b, and f. I usually begin by looking for $\\displaystyle \\frac{b-a}{n}$. Here\u00a0$\\displaystyle \\frac{b-a}{n}=\\frac{1}{n}$\u00a0and from this conclude that ba = 1, so b = a + 1.\n\nUsually, you can start by considering a = 0 , which means that the $\\displaystyle \\frac{b-a}{n}\\cdot k$ becomes the \u201cx.\u201d.\u00a0Then rewriting the radicand as $\\displaystyle 1+3\\frac{1}{n}k=1+3\\left( a+\\frac{1}{n}\\cdot k \\right)$, it appears the function is $\\sqrt{1+3x}$\u00a0and the limit is $\\displaystyle \\int_{0}^{1}{\\sqrt{1+3x}}dx=\\frac{14}{9}$.\n\n(A) \u00a0$\\displaystyle \\int_{0}^{1}{\\sqrt{1+3x}}dx$ \u00a0 \u00a0 \u00a0 \u00a0(B) \u00a0\u00a0\u00a0$\\displaystyle \\int_{0}^{3}{\\sqrt{1+x}}dx$ \u00a0 \u00a0 \u00a0(C) \u00a0\u00a0\u00a0$\\displaystyle \\int_{1}^{4}{\\sqrt{x}}dx$\u00a0\u00a0\u00a0\u00a0 (D) \u00a0\u00a0$\\displaystyle \\tfrac{1}{3}\\int_{0}^{3}{\\sqrt{x}}dx$\n\nThe correct choice is (A), but notice that choices B, C, and D can be eliminated as soon as we determine that b = a + 1. That is not always the case.\n\nLet\u2019s consider another example:\n\nExample 2:\u00a0$\\displaystyle \\underset{n\\to \\infty }{\\mathop{\\lim }}\\,\\sum\\limits_{k=1}^{n}{\\left( {{\\left( 2+\\frac{3}{n}k \\right)}^{2}}\\left( \\frac{3}{n} \\right) \\right)}=$\n\nAs before consider $\\displaystyle \\frac{b-a}{n}=\\frac{3}{n}$, which implies that b = a + 3. With a = 0,\u00a0\u00a0the function appears to be ${{\\left( 2+x \\right)}^{2}}$\u00a0on the interval [0, 3], so the limit is $\\displaystyle \\int_{0}^{3}{{{\\left( 2+x \\right)}^{2}}}dx=39$\n\nBUT\n\nWhat if we take a = 2? If so, the limit is $\\displaystyle \\int_{2}^{5}{{{x}^{2}}dx}=39$.\n\nAnd now one of the \u201cproblems\u201d with this kind of question appears: the answer written as a definite integral is not unique!\n\nNot only are there two answers, but there are many more possible answers. These two answers are horizontal translations of each other, and many other translations are possible, such as\u00a0$\\displaystyle \\int_{-25.65}^{-22.65}{{{\\left( 27.65+x \\right)}^{2}}dx}=39$.\n\nThe same thing can occur in other ways. Returning to example 1,and using something like a u-substitution, we can rewrite the original limit as $\\displaystyle \\frac{1}{3}\\cdot \\underset{n\\to \\infty }{\\mathop{\\lim }}\\,\\sum\\limits_{k=1}^{n}{\\left( \\sqrt{1+\\frac{3k}{n}}\\cdot \\frac{3}{n} \\right)}$.\n\nNow b = a + 3 and the limit could be either $\\displaystyle \\frac{1}{3}\\int_{0}^{3}{\\sqrt{1+x}}dx=\\frac{14}{9}$\u00a0or\u00a0$\\displaystyle \\frac{1}{3}\\int_{1}^{4}{\\sqrt{x}}dx=\\frac{14}{9}$, among others.\n\nThe real problem with the answer choices to Example 1 is that they force the student to do the question\u00a0in a way that gets one of the answers. It is perfectly reasonable for the student to approach the problem a different way, and get a different\u00a0correct answer that is not among the choices. This is not good.\n\nThe problem\u00a0could be fixed by giving the answer choices as numbers. These are the numerical values of the 4 choices:(A) 14/9 \u00a0 (B) 14/3 \u00a0 (C)\u00a0 14/3 \u00a0 (D)\u00a0 \u00a0\u00a0$2\\sqrt{3}/3$.\u00a0As you can see that presents another problem. Distractors (wrong answers) are made by making predictable calculus mistakes. Apparently, two predictable mistakes give the same numerical answer; therefore, one of them must go.\n\nA related problem is this: The limit of a Riemann sum is a number; a definite integral is a number. Therefore, any definite integral, even one totally unrelated to the Riemann sum, which has the correct numerical value, is a correct answer.\n\nI\u2019m not sure if this type of question has any practical or real-world use. Certainly, setting up a Riemann sum is important and necessary to solve a variety of problems. After all, behind every definite integral there is a Riemann sum. But starting with a Riemann sum and finding the function and interval does not seem to me to be of practical use.\n\nThe CED references this question to MPAC 1: Reasoning with definitions and theorems, and to MPAC 5: Building notational fluency. They are appropriate,and the questions do make students unpack the notation.\n\nMy opinions notwithstanding, it appears that future exams will include questions like these.\n\nThese questions are easy enough to make up. You will probably have your students write Riemann sums with a small value of n when you are teaching Riemann sums leading up to the Fundamental Theorem of Calculus. \u00a0You can make up problems like these by stopping after you get to the limit, giving your students just the limit, and having them work backwards to identify the function(s) and interval(s). You could also give them an integral and ask for the associated Riemann sum. Question writers call questions like these reversal questions since the work is done in reverse of the usual way.\n\nHere\u00a0is the question from 1997, for you to try. The answer is below.\n\nAnswer B. Hint n = 50\n\nRevised 5-5-2022\n\n# Parts and More\u00a0Parts\n\nAt an APSI this summer the participants and I got to discussing the \u201ctabular method\u201d for integration by parts. Since we were getting far from what is tested on the BC Calculus exams, I ended the discussion and said for those that were interested I would post more on the tabular method this blog going farther than just the basic set up. So here goes.\n\nHere are some previous posts on integration by parts and the tabular method\n\nIntegration by Parts 1 discusses the basics of the method. This is as far as a BC course needs to go.\n\nIntegration by Parts 2 introduces the tabular method\n\nModified Tabular Integration\u00a0presents a very quick and slick way of doing the tabular method without making a table. This is worth knowing.\n\nThere is also a video on integration by parts here. Scroll down to \u201cAntiderivatives 5: A BC topic \u2013 Integration by parts.\u201d The tabular method is discussed starting about time 15:16. There are several ways of setting up the table; one is shown here and a slightly different way is in the Integration by Parts 2 post above. There are others.\n\n### Going further with the tabular method.\n\nThe tabular method works well if one of the factors in the original integrand is a polynomial; eventually its derivative will be zero and you are done. These are shown in the examples in the posts above and Example 1\u00a0below. To complete the topic, this post will show two other things that can happen when using integration by parts and the tabular method.\n\nFirst we look at an example with a polynomial factor and learn how to stop midway through. Why stop? Because often there will be no end if you don\u2019t stop. There are ways to complete the integration as shown in the examples.\n\nExample 1: \u00a0Find $\\displaystyle \\int_{{}}^{{}}{\\left( 4{{x}^{3}} \\right)\\cos \\left( x \\right)dx}$\u00a0by the tabular method (See Integration by Parts 2 for more detail\u00a0on how to set the table up)\n\nAdding the last column gives the antiderivative:\n\n$\\displaystyle \\int_{{}}^{{}}{\\left( 4{{x}^{3}} \\right)\\cos \\left( x \\right)dx}=4{{x}^{3}}\\sin \\left( x \\right)+12{{x}^{2}}\\cos \\left( x \\right)-24x\\sin \\left( x \\right)-24\\sin \\left( x \\right)+C$\n\nNow say you wanted to stop after $12{{x}^{2}}\\cos \\left( x \\right)$. Example 2 shows why you want (need) to stop. In Example 1 you will have\n\n$\\displaystyle \\int_{{}}^{{}}{\\left( 4{{x}^{3}} \\right)\\cos \\left( x \\right)dx}=4{{x}^{3}}\\sin \\left( x \\right)+12{{x}^{2}}\\sin \\left( x \\right)+\\int_{{}}^{{}}{-24x\\cos \\left( x \\right)}dx$\n\nThe integrand on the right is the product of the last column in the row at which you stopped and the first two columns in the next row, as shown in yellow above.\n\nExample 2 Find\u00a0$\\displaystyle \\int_{{}}^{{}}{{{e}^{x}}\\cos \\left( x \\right)dx}$\n\nAs you can see things are just repeating the lines above sometimes with minus signs. However, if we stop on the third line we can write:\n\n$\\displaystyle \\int_{{}}^{{}}{{{e}^{x}}\\cos \\left( x \\right)dx={{e}^{x}}\\sin \\left( x \\right)}+{{e}^{x}}\\cos \\left( x \\right)-\\int_{{}}^{{}}{{{e}^{x}}\\cos \\left( x \\right)dx}$\n\nThe integral at the end is identical to the original integral. \u00a0We can continue by adding the integral to both sides:\n\n$\\displaystyle 2\\int_{{}}^{{}}{{{e}^{x}}\\cos \\left( x \\right)dx={{e}^{x}}\\sin \\left( x \\right)}+{{e}^{x}}\\cos \\left( x \\right)$\n\nFinally, we divide by 2 and have the antiderivative we were trying to find:\n\n$\\displaystyle \\int_{{}}^{{}}{{{e}^{x}}\\cos \\left( x \\right)dx=\\tfrac{1}{2}{{e}^{x}}\\sin \\left( x \\right)}+\\tfrac{1}{2}{{e}^{x}}\\cos \\left( x \\right)+C$\n\nIn working this type of problem you must be aware of that the original integrand showing up again can happen and what to do if it does. As long as the coefficient is not +1, we can proceed as above. The same thing happens if we do not use the tabular method. (If the coefficient is +1 then the other terms on the right will add to zero and you need to make different\u00a0choices for u and dv.)\n\n### Reduction Formulas.\n\nAnother use of integration by parts is to produce formulas for integrals involving powers. An integral whose integrand is of less degree than the original, but of the same form results. The formula is then iterated to continually reduce the degree until the final integral can be integrated easily.\n\nExample 3: Find\u00a0$\\displaystyle \\int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}$\n\nLet\u00a0$u={{x}^{n}},\\ du=n{{x}^{n-1}}dx,\\ dv={{e}^{x}}dx,\\ v={{e}^{x}}$\n\n$\\displaystyle \\int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}={{x}^{n}}{{e}^{x}}-n\\int_{{}}^{{}}{{{x}^{n-1}}{{e}^{x}}dx}$\n\nThis is a reduction formula; the second integral is the same as the first, but of lower degree. Here is how it is used.\u00a0 At each step the integrand is the same as the original, but one degree lower. So the formula can be applied again, three more times in this example.\n\nMost textbooks have a short selection of reduction formulas.\n\n### Final Thoughts.\n\nBack in the \u201cold days\u201d, BC (before calculators), beginning calculus courses spent a lot of time on the topic of \u201cTechniques of Integration.\u201d This included integration by parts, algebraic techniques, techniques known as trig-substitutions, and others. Mathematicians and engineers had tables of integrals listing over a thousand forms and students were taught how to use the tables and distinguish between similar forms in the tables. (See the photo below from the fourteenth edition of the CRC tables (c) 1965.) Current textbooks often contain such sections still.\n\nToday, none of this is necessary. CAS calculators can find the antiderivatives of almost any integral. Websites such as WolframAlpha are also available to do this work.\n\nI\u2019m not sure why the College Board recently expanded slightly the list of types of antiderivatives tested on the exams. Certainly a few of the basic types should be included in a course, but what students really need to know is how to write the integral appropriate to a problem, and what definite and indefinite integrals mean. This, in my opinion, is far more important than being able to crank out antiderivatives of increasingly complicated expressions: let technology do that \u2013 or buy yourself an integral table. Just saying \u2026 .\n\n# Trapezoids \u2013 Ancient and\u00a0Modern\n\nThe other day, in the course of about 10 minutes, I came across two interesting things about Trapezoidal approximations that I thought I would share with you.\n\nCuneiform writing\n\nThe first was a link to a story about how the ancient Babylonian astronomers sometime between 350 and 50 BCE used trapezoids to, in effect, find the area under a velocity-time graph tracking Jupiter\u2019s motion. This was an NPR story based on a January 2016 Science magazine article in which the author, Mathieu Ossendrijver discusses his work deciphering cuneiform tablets written over 1,400 years before the technique showed up in Europe.\n\nThe second was a question asked on the AP Calculus bulletin board. A teacher asked, \u201cCan someone please help me answer this question a student posed the other day. We were comparing left, right and midpoint and trapezoidal approximations. He asked since the trapezoidal calculation is the best estimate what is the use of LRAM and RRAM?\u201d Here is an expanded form of my answer.\n\nThere are several things to consider here.\n\n1. First, if all you need is an estimate of the area or integral of a continuous function then a Trapezoid sum is certainly better than the left Riemann sum (left-R\u03a3) or the right-R\u03a3. Better, yes, the \u201cbest\u201d maybe not: midpoint sums are about as good and parabola sums (Simpson\u2019s Rule) are better.\n\n2. Another reason to do left R\u03a3 and right R\u03a3 with small values of n is simply to give students practice in setting up Riemann sums so that they will be familiar with them when they move on to finding their limits and getting ready to define definite integrals.\n\n3. A R\u03a3 for a function f on a closed interval [a, b] is formed by partitioning the interval into subintervals and taking exactly one function value from each closed subinterval, multiplying the value by the width of that subinterval and adding these results. You may pick the function value any way you want \u2013 left end, middle, right end, any place at random in the subintervals and someplace else in the next subinterval. One way is to pick the smallest function value in each subinterval; this gives a R\u03a3 called the lower R\u03a3. Likewise, you could pick the largest value in each subinterval; this gives the upper R\u03a3. Now it is true that\n\nlower R\u03a3 \u2264 (any/all other R\u03a3s) \u2264 upper R\u03a3\n\nThen as you add more partition points (n approaches infinity, or \u0394x approaches 0, etc.) the lower sum increases and the upper sum decreases. The series of lower sums is increasing and bounded above (by the upper sum) and therefore converges to its least upper bound. The upper sum decreases forming a decreasing series that is bounded below and therefore converges to its greatest lower bound.\n\nIf the lower R\u03a3 and the upper R\u03a3 approach the same value, then ALL the other R\u03a3s approach that same value by the Squeeze theorem. This value is then defined as the definite integral of f from a to b.\n\nIn most AP calculus course, the textbooks do not deal with upper and lower sums. Instead, they deal with left R\u03a3 and right R\u03a3 on intervals on which f is only increasing (or only decreasing). In this case the lower R\u03a3 = left R\u03a3 and the upper R\u03a3 = the right R\u03a3 (or the other way around for decreasing functions).\n\nSo, this is why you need the left R\u03a3 and right R\u03a3; not so much to approximate, but to complete the theory leading to the definite integral.", "date": "2023-03-27 22:40:39", "meta": {"domain": "teachingcalculus.com", "url": "https://teachingcalculus.com/category/integration/integration-theory/page/3/", "openwebmath_score": 0.8520076870918274, "openwebmath_perplexity": 407.17507108116865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429565737233, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8578166579974011}}
{"url": "https://www.monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D/CalcPlot3D-Help/section-31.html", "text": "\n## Section5.4Intersections of General Surfaces\n\nAlthough CalcPlot3D does not determine the intersection of two surfaces for you, it does make a great tool for visually checking that the parameterization you have worked out as the intersection of two surfaces is indeed correct.\n\nBelow you will find two examples to explore this concept. Hopefully this visualization process will help increase your confidence in and your appreciation for the results you have obtained.\n\n###### Exploration5.4.1\n\nDetermine the vector-valued function that traces out the intersection of the surfaces defined by the equations below using the parameter $t\\text{.}$\n\nState the answer as a vector-valued function. Then visually verify this result, plotting the two surfaces in CalcPlot3D, and then plotting the curve to check that it really does represent the intersection of these two surfaces.\n\nHere we can let $x=t\\text{.}$ Then $y=t^2\\text{.}$ What does $z$ equal (as a function of $t$)?\n\nThen the vector-valued function we obtain that traces out the intersection of these surfaces is:\n\n\\begin{equation*} \\vec{\\textbf{r}}(t) = t \\hat{\\textbf{i}} + t^2 \\hat{\\textbf{j}} + (t^2 + t^4) \\hat{\\textbf{k}} \\end{equation*}\n\nThis represents the same curve as specified by the parametric functions:\n\n\\begin{align*} x &= t\\\\ y &=t^2\\\\ z &=t^2 + t^4 \\end{align*}\n\nNow let\u2019s verify this in CalcPlot3D visually!\n\na. Open the CalcPlot3D app.\n\nb. Once the app is loaded and active, enter the first function listed above ($z =$ x^2 + y^2) in the default function object on the left and press Enter (or click on the Graph button). The surface plot of this paraboloid should appear in the plot window.\n\nc. Now, to enter the second function, go to the Add to graph dropdown menu (just above the default function), and select Function: y = f (x, z). This will set the function to y = 1 by default. Enter x^2 in the textbox and press Enter (or use the Graph button).\n\nd. To extend the second surface farther up the paraboloid, change the range of $z$ (just below y = x^2) to go from -2 to 4. You may want to use the scroll-wheel on the mouse to zoom-out a little. [Alternatively you could click on the Format Axes button located just to the right of the 3D Mode button. Then set $z$-max to 4. This will automatically change the upper $z$-clip value to 8. Let\u2019s change this value to 4 also.]\n\ne. Make the surfaces semi-transparent using the button or by typing the T key to get a clearer view of the intersection of the surfaces. Press the E key to turn off the edges on the surfaces.\n\nf. Next we need to graph the space curve to see how well it fits the intersection of the surfaces. Select Space Curve: r(t) from the Add to graph dropdown menu. A space curve object should appear just below this menu.\n\nEnter the three parametric equations we obtained (each in terms of $t$). Then enter a range of -2 to 2. If you press Enter on the second value, it should produce the curve on the plot. If it does not appear, click the Graph button.\n\ng. I like the look of this one better with a constant color, so select the checkbox titled Use Constant Primary Color, at the bottom of this object.\n\nh. Finally rotate the graph to see if it looks like we found the correct intersection curve. Except for different coloring, this should look like the image in Figure\u00a05.4.1 below.\n\n###### Exploration5.4.2\n\nDetermine the vector-valued function that traces out the intersection of the surfaces defined below, assuming that $y = 2\\sin t\\text{.}$\n\nState the answer as a vector-valued function. Then visually verify the result, plotting the two surfaces in CalcPlot3D, and then plotting the curve to check that it really does represent the intersection of these two surfaces.\n\nNote that we can parameterize the ellipse given by the first equation with\n\nNow that we've parameterized $x$ and $y\\text{,}$ the second surface equation makes it easy to determine $z$ in terms of $t\\text{.}$\n\nAfter simplifying the expression, we obtain, $z = 2 + 2\\sin^2 t\\text{.}$\n\nThen the vector-valued function we obtain that traces out the intersection of these surfaces is:\n\n\\begin{equation*} \\vec{\\textbf{r}}(t) = \\sqrt{2}\\cos t \\hat{\\textbf{i}} + 2\\sin t \\hat{\\textbf{j}} + (2 + 2\\sin^2 t) \\hat{\\textbf{k}} \\end{equation*}\n\nThis represents the same curve as specified by the parametric functions:\n\n\\begin{align*} x &= \\sqrt{2}\\cos t\\\\ y &= 2\\sin t\\\\ z &= 2 + 2\\sin^2 t \\end{align*}\n\nNow let\u2019s verify this in CalcPlot3D visually!\n\na. Open the CalcPlot3D app.\n\nb. Once the app is loaded and active, enter the second function listed above (z = x2 + y2) in the default function object on the left and press Enter (or click on the Graph button). The surface plot of this paraboloid should appear in the plot window.\n\nc. Now, to enter the first function, which is stated implicitly, go to the Add to graph dropdown menu (just above the default function), and select Implicit Surface. Once this new object definition appears on the left, enter the implicit equation defining the first surface in the textbox and press Enter (or use the Graph button). This surface should look like an elliptic cylinder.\n\nd. To make the intersection more clear, let's first zoom out once using the Zoom-out button, so the $x$- and $y$-axes run from -4 to 4 instead of from -2 to 2. Now click on the Format Axes button located just to the right of the 3D Mode button, and set the upper $z$-clip to 4.\n\ne. Next make the surfaces semi-transparent using the transparency button or by typing the T key to get a clearer view of the intersection of the surfaces. Press the E key to turn off the edges on the surfaces.\n\nf. Next we need to graph the space curve to see how well it fits the intersection of the surfaces. Select Space Curve: r(t) from the Add to graph dropdown menu. A space curve object should appear just below this menu.\n\nEnter the three parametric equations we obtained (each in terms of $t$).\n\nYou will enter them like this:\n\n\\begin{align*} x &= \\textbf{sqrt(2)cos(t)}\\\\ y &= \\textbf{2sin(t)}\\\\ z &= \\textbf{2 + 2(sin(t))\\^{}2} \\end{align*}\n\ng. Now enter a range of 0 to $2\\pi\\text{.}$ If you press Enter on the second value, it should produce the curve on the plot. If it does not appear, click the Graph button.\n\nI also like the look of this one better with a constant color, so select the checkbox titled Use Constant Primary Color, at the bottom of this object.\n\nh. Finally rotate the graph to see if it looks like we found the correct intersection curve. This should look like the image in Figure\u00a05.4.2 below.\n\nNotice that the intersection of these two surfaces may appear a little polygonal rather than smooth like the curve we just graphed. This is because we used an implicit surface with the default resolution. To improve this resolution, you can adjust the # Cubes/axis to a higher number, like 25. See the result in Figure\u00a05.4.3.", "date": "2021-05-06 18:54:29", "meta": {"domain": "monroecc.edu", "url": "https://www.monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D/CalcPlot3D-Help/section-31.html", "openwebmath_score": 0.7889100313186646, "openwebmath_perplexity": 809.9650212303267, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429604789206, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8578166483503693}}
{"url": "https://math.stackexchange.com/questions/4106228/is-there-a-general-procedure-for-inducing-a-formula-recognizing-a-pattern/4106329", "text": "# Is there a general procedure for inducing a formula / recognizing a pattern?\n\nFor some problems, it's pretty simple to deduce what a formula should look like after you enumerate a few examples, but for some problems, it's not so clear. For these less clear examples, is there some procedure I can follow to find the formula?\n\nFor example, just now, I was working on the problem:\n\nFind the expected number of tosses it takes to get $$k$$ 6s in a row.\n\nThe recursive formula is\n\n$$E[N_k ] = 6(E[N_{k - 1}] + 1)$$\n\nwhere $$N_k$$ is the number of tosses it takes to get $$k$$ 6s in a row. The initial condition is $$E[N_1] = 6$$ because $$N_1$$ is a geometric random variable with $$1/6$$ probability of success.\n\nThen I enumerated a few cases:\n\n$$E[N_1] = 6 \\\\ E[N_2] = 42 \\\\ E[N_3] = 258 \\\\ E[N_4] = 1554$$\n\nIt's not obvious to me what the formula should be. But I think it should look something like $$E[N_k] = 6^k + (k - 1) * \\text{something} + \\text{maybe other terms}$$\n\nBut I don't know what this \"something\" and \"maybe other terms\" should be. If I sit here long enough I could probably figure it out, but is there some kind of general procedure that I can apply to a wide array of problems?\n\n\u2022 \u2013\u00a0John Omielan Apr 17 at 20:29\n\nOne good way is to try to convert this to a more standard linear recursion.\n\nLetting $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$\n\nWe rewrite this as $$6=a_n-6a_{n-1}=a_{n-1}-6a_{n-2}$$\n\nWhence $$a_n=6a_{n-1}+a_{n-1}-6a_{n-2}=7a_{n-1}-6a_{n-2}$$\n\nand this can now be solved by standard means, yielding $$\\boxed{a_n=\\frac 65\\times \\left(6^n-1\\right)}$$\n\n\u2022 What do you mean by 'standard means' here? \u2013\u00a0roulette01 Apr 17 at 20:22\n\u2022 It's a linear recursion, so you look at the characteristic polynomial $x^2-7x+6=(x-1)(x-6)$. Thus the solution is of the form $a_n=A6^n+B$ and then solve for $A,B$ using initial values. \u2013\u00a0lulu Apr 17 at 20:23\n\u2022 Ah I see. I've never formally learned recursion. My understanding is there are a lot of analogies between how recursive formulas are solved vs. how ODEs are solved, and this general solution seems to be analogous to formulas in second order ODEs. \u2013\u00a0roulette01 Apr 17 at 20:25\n\u2022 If you were to solve this problem without recursion, and by just inspecting the first several enumerations, is there a general approach to guess what that formula should be? \u2013\u00a0roulette01 Apr 17 at 20:26\n\u2022 I think guessing is hard. You could guess that $a_n=A6^n+p(n)$ where $p(n)$ is a low degree polynomial...but I don't know how reliable that sort of guesswork is. Works in this case, though. \u2013\u00a0lulu Apr 17 at 20:28\n\nIf you call $$a_k=E[N_k]$$ then you end up with a linear induction relation: $$a_k-6a_{k-1}=6$$\n\nSince it is linear, it solves as the sum of general solution of homegenous equation $$H_k-6H_{k-1}=0$$, which is $$H_k=6^kH_0$$ and one particular solution of the full equation.\n\nThe theory says that if the RHS is of the form $$P(k)r^k$$ then you can find a particular solution of the form $$Q(k)r^k$$ with $$\\deg(Q)=\\deg(P)+m$$ where $$m$$ is the multiplicity of the root of the homogenous equation.\n\nIn this case $$RHS = 6\\times 1^k$$ and the single root of homogenous equation is $$r=6$$. So the multiplicity of $$1$$ is just $$m=0$$ (i.e. not a root) this means $$\\deg(Q)=0+0=0$$.\n\nAll that to say that our particular solution is simply a constant...\n\nTherefore let search for $$S_k=c$$ then $$c-6c=6\\iff c=-\\frac 65$$\n\nOur general solution is then $$a_k=H_k+S_k=6^kH_0-\\frac 65$$\n\nWe now solve for initial conditions:\n\n$$a_1=6=6H_0-\\frac 65\\iff H_0=\\frac 65$$ and we get $$E[N_k]=\\frac 65(6^k-1)$$\n\nis there some kind of general procedure that I can apply to a wide array of problems?\n\nAround 1960 there was some work on a General Problem Solver which had very limited capabilities but much progress has been made since 1960. Still, there is no way to solve all problems and probably never will be.\n\nHowever, the solution of linear recurrences is known for a long time. For example, $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$ is easily solved using the theory.\n\nFor this particular difference equation with initial value $$\\,a_1\\!=\\!6\\,$$ a lookup of $$\\,6,42,258,1554\\,$$ in OEIS returns OEIS sequence A105281 \"a(0)=0; a(n)=6*a(n-1)+6.\" This entry has a program\n\n(PARI) a(n)=if(n<0, 0, (6^n-1)*6/5)\n\nwhich gives a formula for the solution.\n\nThe Mathematica program has a function RSolve which solves recurrences and also has limited capabilities\n\nYou can use the Mathematica (or WolframAlpha) command\n\nRSolve[a[n] == 6 a[n - 1] + 6 && a[1] == 6, a[n], n]\n\nto get the particular solution $$\\,a_n = 6 (6^n-1)/5.\\,$$", "date": "2021-06-16 17:24:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4106228/is-there-a-general-procedure-for-inducing-a-formula-recognizing-a-pattern/4106329", "openwebmath_score": 0.7783046364784241, "openwebmath_perplexity": 160.1771235935303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429595026213, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8578166458669823}}
{"url": "http://mathhelpforum.com/statistics/126266-coin-tossing.html", "text": "# Math Help - coin tossing\n\n1. ## coin tossing\n\n2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that\na. they break even after 6 tosses?\nb. one player wins all the money on the tenth toss?\n\nmy working:\n\na. since there are 6 tosses and equal probability of a head and tail,\nthe ans should be 6 x (0.5)^6\n\nb. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.\n\nso, (9 choose 7) x ( 0.5)^10\n\nbut i didnt get the answer\n\n2. Originally Posted by alexandrabel90\n2 gamblers bet $1 each on the successive tosses of a coin. each as$6. what is the probability that\na. they break even after 6 tosses?\nb. one player wins all the money on the tenth toss?\n\nmy working:\n\na. since there are 6 tosses and equal probability of a head and tail,\nthe ans should be 6 x (0.5)^6\n\nb. in order to win all the money o the other player, he needs to win 8 games and loss 2 games, where the 10th toss have to be a win.\n\nso, (9 choose 7) x ( 0.5)^10\n\nbut i didnt get the answer\nhi alexandrabel90,\n\nfor the first one, they both must win 3 times each,\nso that's the probability of either winning exactly 3 times and losing exactly 3 times.\n\nfor the second one, you must add the probability for the 2 players.\nthis is different from the first one because one or other could win the game on the 10th go, having won 7 of the preceding 9.\n\nAs I read the question we have two players each having $6 to play with. Each player either wins$1 or looses $1 on each toss of the coin. So it is possible for one player of loose all in six tosses- loses six straight. So for both to breakeven in six tosses each would wins three and lose three. Is that a misreading of the question? If not, the answer to part a) is $\\frac{\\binom{6}{3}}{2^6}$. If it is a misreading, please clarify the question. 4. it is not a misreading of the question. your answer is correct for part (a)..i never think that we would need to count then chances of wining since there is only a 0.5 chance of a win and lose. for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? 5. Hello, alexandrabel90! Two gamblers bet$1 each on the successive tosses of a coin.\nEach has $6. .What is the probability that: a. they break even after 6 tosses? They each win 3 games and lose 3 games. There are $_6C_3 \\:=\\:\\frac{6!}{3!\\,3!} \\:=\\:20$ ways this can happen. Answer: . $(20)\\left(\\frac{1}{2}\\right)^3\\left(\\frac{1}{2}\\ri ght)^3 \\:=\\20)\\,\\frac{1}{64} \\:=\\:\\frac{5}{16}\" alt=\"(20)\\left(\\frac{1}{2}\\right)^3\\left(\\frac{1}{2}\\ri ght)^3 \\:=\\20)\\,\\frac{1}{64} \\:=\\:\\frac{5}{16}\" /> b. one player wins all the money on the tenth toss? my working: In order to win all the money of the other player, he needs to win 8 games and loss 2 games, where the 10th toss has to be a win. So: . $_9C_7\\left(\\tfrac{1}{2}\\right)^{10}$ . . . . correct, so far $_9C_7\\left(\\tfrac{1}{2}\\right)^{10} \\:=\\:\\tfrac{36}{1024} \\:=\\:\\frac{9}{256}$ This is the probability that $A$ wins all of $B$'s money on the 10th game. It's also the probability that $B$ wins all of $A$'s money on the 10th game. Therefore: . $P(one\\text{ player wins all the money}) \\:=\\:\\frac{9}{256} + \\frac{9}{256} \\;=\\;\\frac{9}{128}$ 6. Originally Posted by alexandrabel90 for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? Yes, that was it, because you calculated the probability of only one out of the two winning on the tenth throw. Soroban has shown that really nicely. Plato has made an interesting point also. If we interpret \"break-even\" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. In this situation, one could break even on one's own throws but make a profit if the other player does badly, or make a loss if the other player does well, \"if\" losing a toss loses a dollar and winning a toss wins a dollar. Or, if the two players lost every time, \"player A\" wins \"player B\"'s 6 dollars and \"player B\" wins \"player A\"'s 6 dollars, so they break even. They could also both win 1 and lose 5 each, win 2 and lose 4 each, win 4 and lose 2 each, win 5 and lose 1 each, win 6 and lose 0 each. That's the thing with probability questions, quite often open to alternative interpretations. The answer book will distinguish given it's answers. We've been interpreting this as..... If one player loses on a throw, then the second player wins the first player's dollar. In this case then, both would have to win 3 times and lose 3 times to end up with the 6 dollars they began with. It is probably safest to assume there is just 6 tosses rather than 6 each. 7. Originally Posted by alexandrabel90 2 gamblers bet$1 each on the successive tosses of a coin. each as $6. what is the probability that b. one player wins all the money on the tenth toss? Originally Posted by Archie Meade Yes, that was it, Plato has made an interesting point also. If we interpret \"break-even\" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. Look at the way part b) is put. It seems to me that in this game on any toss player A gets a dollar from player B or gives a dollar to player B. That is, A wins or looses on any toss. Moreover, the game is over when one player is out of money. If player A wins in each of the first six tosses she has all of the money in six tosses. If B wins five of the first six tosses he has 11 dollars and she has 1. So the question becomes \u201chow can the game be over in exactly ten tosses?\u201d 8. Brilliant analysis, Plato! it's not over yet. 9. I reckon that, in order for one player to win 8 to 2, the score had to stand at 5-1, at which point one player has$10 and the other has \\$2 (win 5 and lose 1... 6+5-1=10, win 1 and lose 5.... 6+1-5=2).\n\nThen the score can go to 6-1 or 5-2.\n\nFrom 6-1 the score must go to 6-2, since 7-1 would end the game.\nFrom there to 7-2 and then to 8-2.\n\nFrom 5-2 the score goes to 6-2, to 7-2, to 8-2.\n\nBeginning at 5-1, there are two ways for the leading player to win on the 10th throw.\n\nSo, we count the number of ways of getting to a 5-1 lead and double that.\n\n$2\\binom{6}{5}=2\\binom{6}{1}$\n\n$P=\\frac{2(6)}{2^{10}}=\\frac{3}{2^8}=\\frac{3}{256}$\n\nThen double this to get the result for either player winning by 8 to 2.", "date": "2015-08-01 12:49:10", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/126266-coin-tossing.html", "openwebmath_score": 0.6742426156997681, "openwebmath_perplexity": 844.6628226539113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9833429629196684, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8578166455844038}}
{"url": "http://openstudy.com/updates/501d67a1e4b0be43870e2635", "text": "## ParthKohli 4 years ago $\\Huge \\mathsf{\\text{Primitive Pythagorean Triplets.}}$\n\n1. ParthKohli\n\nGive it a start, Kingy.\n\n2. KingGeorge\n\nWell, they're triplets of the form $$(a,b,c)$$ such that $$a^2+b^2=c^2$$ and $$\\gcd(a,b,c)=1\\(. Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets. 3. ParthKohli $2n,n^2 - 1, n^2 + 1$? 4. ParthKohli I know one. \\(3,4,5$$.\n\n5. KingGeorge\n\nI just can't type that correctly can I? Well, they're triplets of the form $$(a,b,c)$$ such that $$a^2+b^2=c^2$$ and $$\\gcd(a,b,c)=1$$. Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets.\n\n6. ParthKohli\n\nAnother one! $$5,12,13$$!\n\n7. KingGeorge\n\nCorrect, (3,4,5) is a triplet as well as (5,12,13) and (7,24,25) among infinite others.\n\n8. ParthKohli\n\nCan you tell me the formula(s) for it, please?\n\n9. KingGeorge\n\nThe most well known one is $2mn, \\quad m^2-n^2, \\quad m^2+n^2$Given any $$m,n$$ such that $$m>n$$.\n\n10. ParthKohli\n\nI see :) Are there any more?\n\n11. KingGeorge\n\nWhat you gave above, only works for $$m$$ even. So it generates an infinite number, but not all of them.\n\n12. ParthKohli\n\nCan this formula generate all such triplets, or you have to use some other formulae to generate more?\n\n13. anonymous\n\nanother one $(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)$\n\n14. ParthKohli\n\nOh wait. It does generate all.\n\n15. KingGeorge\n\nThere is also $(a,b,c)=\\left(mn,\\;\\;\\frac{m^2-n^2}{2},\\;\\;\\frac{m^2+n^2}{2}\\right)$This is a different formula that what I gave above since it only works with $$(m,n)$$ such that $$n>m\\ge1$$ and $$\\gcd(m,n)=1$$. This does not generate all of them.\n\n16. KingGeorge\n\nCcorrection, $$m>n\\ge1$$.\n\n17. anonymous\n\nif you want them \"primitive\" makes sure $$m,n$$ are opposite parity, and no common factors\n\n18. ParthKohli\n\nWell. I'd just use the first one :)\n\n19. KingGeorge\n\nSatellite is correct. The first formula I wrote only gives primitive triplets for $$m,n$$ such that $$m-n$$ is odd. And $$\\gcd(m,n)=1$$. However, any two positive integers will give you a triplet.\n\n20. anonymous\n\nalgebra shows you that $$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$ if you want a proof that every such triple is generated by such an $$m$$ and $$n$$ try working through the attachment\n\n21. ParthKohli\n\nIf $$\\gcd(m,n) = 1$$, then it automatically means that $$m - n$$ is odd.\n\n22. anonymous\n\nno, try 5 and 3\n\n23. ParthKohli\n\nOh grr.\n\n24. anonymous\n\nreally, if you have time, work through the worksheet i sent you you will see why all triples come from this formula\n\n25. ParthKohli\n\nYes, I am looking at it.\n\n26. anonymous\n\nit requires no more than algebra\n\n27. ParthKohli\n\nAll set. I got that formula well.\n\n28. KingGeorge\n\nThere is also a multitude of fascinating properties of primitive triplets. For example: $$c$$ is always odd. 2,3,4 divide exactly one of $$a$$ or $$b$$, 5 divides exactly one of $$a,b$$ or $$c$$. $$a+b+c$$ is always even among others. These are some of the simpler facts.\n\n29. ParthKohli\n\n30. anonymous\n\nhave fun. there are several steps to the proof, but it is all there\n\n31. anonymous\n\nwhat do you mean by primitive triplets?\n\n32. KingGeorge\n\n$$\\gcd(a,b,c)=1$$.\n\n33. anonymous\n\n\"primitive\" no common factors\n\n34. anonymous\n\noh\n\n35. anonymous\n\nso 3, 4, 5 but not 6, 8, 10\n\n36. ParthKohli\n\nWhoa! Primitive triplets are $$a,b,c$$ such that $$\\gcd(a,b,c) = 1$$. Other words, they all are co-prime.\n\n37. anonymous\n\n@ParthKohli method of obtaining the formula i gave u...we will talk about later if u want... http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee\n\n38. ParthKohli\n\n@mukushla I'd see that tutorial tomorrow; tired somewhat at the moment. Looks impressive by looking at it though.\n\n39. anonymous\n\nnp...:)\n\n40. ParthKohli\n\nHaha$\\Huge \\ddot \\smile$\n\n41. anonymous\n\nwhy do we need to have opposite parity? and what's with m-n being odd?\n\n42. ParthKohli\n\nWell. Isn't that the same thing?\n\n43. KingGeorge\n\n$$m-n$$ is odd if and only if they have opposite parity. So either $$m$$ or $$n$$ is odd, but not both.\n\n44. anonymous\n\ni understand that. but is it necessary for m-n to be odd?\n\n45. KingGeorge\n\nIf you want a primitive triplet, it is.\n\n46. KingGeorge\n\nIf they are both even or both odd, $$\\gcd(a,b,c)\\ge2$$\n\n47. anonymous\n\nhow do i prove it?\n\n48. KingGeorge\n\nNote that $$2mn$$ is always even. Also notice that if $$m,n$$ are both odd, then $$m^2$$ and $$n^2$$ are both odd. If they are both even, $$m^2$$ and $$n^2$$ are both even. Now, $$m^2-n^2$$ and $$m^2+n^2$$ will both be even as well since even-even=even and odd-odd=odd. Hence, $$a,b,c$$ are all even, and their gcd must be divisible by 2.\n\n49. anonymous\n\nthat was so easy and so stupid of me to not get it. i am sorry for troubling you on this. if i am to prove 5 exactly divides one of the triplet, is assuming all of them to be non divisible by 5 the right way?\n\n50. KingGeorge\n\nWhen I first did it, I assumed $$a$$ was not divisible by 5, and deduced that either $$b$$ or $$c$$ was divisible by 5. Then assume $$5$$ divides $$a$$ and show that it doesn't divide $$b$$ or $$c$$. This probably isn't the fastest way to do it thought.\n\n51. anonymous\n\nyou can work by cases assume $$m$$ and $$n$$ are not divisible by 5 then prove that either $$m^2-n^2$$ or $$m^2+n^2$$ must be for example, if $$m\\equiv n$$ mod 5, then $$m^2-n^2\\equiv 0$$ mod 5", "date": "2016-09-25 17:25:43", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/501d67a1e4b0be43870e2635", "openwebmath_score": 0.8718986511230469, "openwebmath_perplexity": 790.4042778097681, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983342958526322, "lm_q2_score": 0.872347369700144, "lm_q1q2_score": 0.8578166433835949}}
{"url": "https://math.stackexchange.com/questions/843202/evaluate-int-sqrt1-x2-dx", "text": "# Evaluate $\\int \\sqrt{1-x^2}\\,dx$\n\nI have a question to calculate the indefinite integral: $$\\int \\sqrt{1-x^2} dx$$ using trigonometric substitution.\n\nUsing the substitution $u=\\sin x$ and $du =\\cos x\\,dx$, the integral becomes: $$\\int \\sqrt{\\cos^2 u} \\, \\cos u \\,du = \\int \\|{\\cos u}\\| \\cos u\\, du$$\n\nQ: (part a) At what point (if at all) is it safe to say that this is the equivalent of ? $$\\int \\cos^2 u\\, du = \\int \\frac {1 + \\cos 2u} {2} du$$ (this is easy to solve, btw).\n\nIn lectures, it was made abundantly clear that over certain intervals (eg $0 \\le u \\le \\pi/2$) that $cos u$ is +ve and is safe to do so, but in the indefinite form, the same argument cannot be made (eg $\\pi/2 \\le u \\le n\\pi$).\n\nQ: (part b) Is it safe to declare it ok due to the nature of the original integral, which, using a sqrt() must return a +ve number? It could then be argued that it was the substitution which artificially added a -ve aspect...\n\nAny suggestions on how to proceed?\n\nPS: This is a 1st year calculus course and am revising for exams ;)\n\n\u2022 The original integrand allows for $x$ to be in the interval $[-1,1]$ which means that $\\cos$ can be negative. In my experience ignoring this is almost never a problem. Rather than worrying about the details of when $\\cos$ is positive or negative, recall that you can always prove that your final answer is correct by showing that its derivative equals the original integrand. \u2013\u00a0Spencer Jun 22 '14 at 5:21\n\u2022 The lecturer has been pounding into us the need to think correctly on fundamental concerns, which is why I raise it here. Although I understand I can show equivalency by derivation, it is not a proof (as far as I am aware) \u2013\u00a0cmroanirgo Jun 22 '14 at 5:37\n\u2022 The definition of an indefinite integral is that it is the anti-derivative of the integrand (up to an additive constant). There is no more legitimate proof. \u2013\u00a0Spencer Jun 22 '14 at 5:39\n\u2022 You have it backwards. To do this with trig substitution, $x=\\sin u$, not $u=\\sin x$. \u2013\u00a0alex.jordan Jun 22 '14 at 5:40\n\u2022 @Spencer I don't think that's right. $x$ is in $[-1,1]$, and since $x=\\sin(u)$, well, see my answer below. \u2013\u00a0alex.jordan Jun 22 '14 at 5:44\n\nSince $x$ ranges from $-1$ to $1$, and you are using the substitution $x=\\sin(u)$, you can make this substitution with $u\\in[-\\pi/2,\\pi/2]$, and then $\\cos(u)$ is unambiguously positive.\n\n\u2022 It was the logic of discussion which had me confused. This clarifies the why. Recognising the limiting domain (to real numbers) from -1 to 1 is the key ingredient. \u2013\u00a0cmroanirgo Jun 22 '14 at 6:13\n\nYou have received good answer which all conclude that $\\sqrt{\\cos^2 x} = \\cos x$.\n\nBut, for the time being, let us assume you still ignore if it is safe or not. So, let us write $$\\int \\sqrt{1-x^2} dx=\\pm \\int \\cos^2 u\\, du =\\pm \\int \\frac {1 + \\cos 2u} {2} du=\\pm \\Big(\\frac{u}{2}+\\frac{1}{4} \\sin (2 u)\\Big)$$ But now, let us consider the definite integral $$\\int_0^1 \\sqrt{1-x^2} dx=\\pm \\int_0^{\\frac{\\pi}{2}} \\cos^2 u\\, du =\\pm \\int_0^{\\frac{\\pi}{2}} \\frac {1 + \\cos 2u} {2} du=\\pm \\frac{\\pi}{4}$$ But now, consider now the area between the curve $y=\\sqrt{1-x^2}$ and the $x$ axis. All the curve is above the axis and the area is then positive. So, $\\pm$ should be just replaced by $+$.\n\nIs this making things clearer ?\n\nNote that there is a difference between substitution as one initially learns it (let $u=g(x)$ and what in the OP is called \"trigonometric substitution.\" The latter process, when one is feeling pedantic, is actually called inverse trigonometric substitution.\n\nIn the example we are discussing, the substitution \"really is\" let $u=\\arcsin x$. So $u$ naturally lives in the interval $[-\\pi/2,\\pi/2]$, and therefore $\\cos u$ is non-negative.\n\nHint: first integrate by parts to get,\n\n$$\\int\\sqrt{1-x^2}\\,\\mathrm{d}x=x\\sqrt{1-x^2}+\\int\\frac{x^2}{\\sqrt{1-x^2}}\\,\\mathrm{d}x.$$\n\nNow attempt the trigonometric substitution.\n\nTo answer your question: yes, it is safe and doesn't really matter. When you do trig. substitution in your first year calculus course, you are always assuming that $\\cos$ is positive as a result you can do:\n\n$$\\sqrt{\\cos^2 x} = \\cos x$$\n\nand not have any problems. Also, take into account what @spencer said; whatever your final answer, you can just find it's derivative and prove yourself right or wrong.\n\n\u2022 Absolutely right but for first year, as far as I know, the integral is calculated with keeping in mind that $\\cos$ is in between that intervals which you specified, otherwise $\\sqrt{\\cos^2 x} \\rightarrow \\cos x$ doesn't make sense. Depending on which textbook you use, go to the trig. substitution chapter and from an example with $\\cos x$ it will say specifically that $\\cos x$ is bounded between $0 \\le x \\le \\pi/2$ \u2013\u00a0Jeel Shah Jun 22 '14 at 5:40\n\u2022 The bounds for the integral and the bounds for the function are not the same. I believe that is where you are confused. All we are saying is that we want to consider $\\cos x$ when it is positive and therefore, bound it between those bounds. This doesn't imply that the entire integral is bounded between those. \u2013\u00a0Jeel Shah Jun 22 '14 at 5:42", "date": "2021-06-13 02:38:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/843202/evaluate-int-sqrt1-x2-dx", "openwebmath_score": 0.882774829864502, "openwebmath_perplexity": 193.46298346712567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429624315189, "lm_q2_score": 0.8723473647220787, "lm_q1q2_score": 0.8578166418951375}}
{"url": "http://math.stackexchange.com/questions/151611/simplify-a-1-a-2-a-3-a-nm/151615", "text": "# simplify $(a_1 + a_2 +a_3+\u2026 +a_n)^m$\n\nHow to simplify this best $(a_1 + a_2 +a_3+... +a_n)^m$ for $m=n, m<n, m>n$\n\nI could only get $\\sum_{i=0}^{m}\\binom{m}{i}a_i^i\\sum_{j=0}^{m-i}\\binom{m-i}{j}a_j ...$\n\n-\nIt is already as simple as it gets if you don't have other information on the $a_i$s. \u2013\u00a0Phira May 30 '12 at 15:06\nthey are just variable ... let's say like a and b in binomial expansion \u2013\u00a0Santosh Linkha May 30 '12 at 15:07\nNote that Pascal's Tetraedron, then its $4th$ dimensional analogue, then the $5th$, is a graphical way to visualise the multinomial theorem. \u2013\u00a0Alyosha May 5 '13 at 16:59\n\nThe simplification for this type of expansion is done through the multinomial theorem. The multinomial theorem is a generalization of the binomial case to any arbitrary number of terms in the sum to be exponentiated.\n\nThe multinomial theorem is written as follows:\n\n$$(x_1 + x_2 + \\cdots + x_m)^n = \\sum_{k_1+k_2+\\cdots+k_m=n} {n \\choose k_1, k_2, \\ldots, k_m} \\prod_{1\\le t\\le m}x_{t}^{k_{t}}\\$$\n\nWhere the multinomial co-efficient is defined as:\n\n$${n \\choose k_1, k_2, \\ldots, k_m} = \\frac{n!}{k_1!\\, k_2! \\cdots k_m!}$$\n\nIt may also be useful to you to note that the multinomial co-efficient is always expressible as products of binomial co-efficients [Graham, Knuth, Patashnik, Concrete Mathematics (2nd edition)]:\n\n$${n \\choose k_1, k_2, \\ldots, k_m} = {x_1+x_2+\\cdots+x_m \\choose x_2+\\cdots+x_m}\\cdots{x_{m-1}+x_m \\choose x_m}$$\n\nA fuller explanation can be found on Wikipedia, or Wolfram MathWorld\n\n-\nis there any relation between $k_1, k_2, ...$ isn't $k_1+k_2+ ...+k_m=n$ again going to be combination of k's?? \u2013\u00a0Santosh Linkha May 30 '12 at 15:18\nThe only relation is that the sum of all $k_i$ is equal to n. I do not believe there is an equality relationship such as $k_1>k_2$ or anything like that. \u2013\u00a0Shaktal May 30 '12 at 15:20\nOh ... thank you both!! \u2013\u00a0Santosh Linkha May 30 '12 at 15:21\n\nThis is called the Multinomial theorem: $$(a_1 + a_2 +a_3+... +a_n)^m=\\sum_{k_1+k_2+...+k_n=m}\\frac{m!}{k_1!\\cdot...\\cdot k_n!}a_1^{k_1}\\cdot...\\cdot a_n^{k_n}$$\n\n-", "date": "2016-07-25 12:06:32", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/151611/simplify-a-1-a-2-a-3-a-nm/151615", "openwebmath_score": 0.846900463104248, "openwebmath_perplexity": 563.904999015353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429609670701, "lm_q2_score": 0.8723473630627234, "lm_q1q2_score": 0.8578166389859142}}
{"url": "https://math.stackexchange.com/questions/2087798/prove-that-fraca-1a-2-fraca-2a-3-fraca-3a-4-fraca-n-1", "text": "# Prove that $\\frac{a_1}{a_2}+ \\frac{a_2}{a_3}+\\frac{a_3}{a_4}+\u2026+\\frac{a_{n-1}}{a_n} \\le \\frac{n}{2}$\n\nLet $n \\ge 2$ be a positive integer and let $a_1, a_2, ... a_n$ be positive numbers such that $$a_1\\le a_2, a_1+a_2\\le a_3, a_1+a_2+a_3\\le a_4, ... ,a_1+a_2+...+a_{n-1}\\le a_n$$ prove that $$\\dfrac{a_1}{a_2}+ \\dfrac{a_2}{a_3}+\\dfrac{a_3}{a_4}+...+\\dfrac{a_{n-1}}{a_n} \\le \\dfrac{n}{2} \\hspace{2cm} (1)$$ When does the equality holds?\n\nSolution: I proceed as follows using Mathematical Induction.\n\nFor $n=2, \\frac{a_1}{a_2} \\le 1$. Let the (1) be true for $n=k$ i.e $$\\dfrac{a_1}{a_2}+ \\dfrac{a_2}{a_3}+\\dfrac{a_3}{a_4}+...+\\dfrac{a_{k-1}}{a_k} \\le \\dfrac{k}{2} \\hspace{2cm} (2)$$ We need to prove $$\\dfrac{a_1}{a_2}+ \\dfrac{a_2}{a_3}+\\dfrac{a_3}{a_4}+...+\\dfrac{a_{k}}{a_{k+1}} \\le \\dfrac{k+1}{2}$$ Consider $$\\dfrac{a_1}{a_2}+ \\dfrac{a_2}{a_3}+\\dfrac{a_3}{a_4}+...+\\dfrac{a_{k-1}}{a_k}+\\dfrac{a_{k}}{a_{k+1}} \\le \\dfrac{k}{2} +\\dfrac{a_{k}}{a_{k+1}} \\hspace{2cm} (3)$$ Since $$a_1+a_2+...+a_{k-1}+a_{k}\\le a_{k+1} \\hspace{2cm} (4)$$ also $$a_1+a_2+...+a_{k-1}\\le a_{k} \\hspace{2cm} (5)$$ Using 5 in 4, we get $$a_{k}+a_{k}\\le a_{k+1}$$ $$\\dfrac{a_{k}}{a_{k+1}} \\le \\dfrac{1}{2}$$ using in (3), we get $$\\dfrac{a_1}{a_2}+ \\dfrac{a_2}{a_3}+\\dfrac{a_3}{a_4}+...+\\dfrac{a_{k-1}}{a_k}+\\dfrac{a_{k}}{a_{k+1}} \\le \\dfrac{k+1}{2}$$ Is the procedure is correct. And when the equality holds... Thanks for any assistance\n\n\u2022 Your proof looks good. As to the question about when equality holds, the clues are right there in your proof. How big can $\\frac{a_1}{a_2}$ be? Can you force equality? What about $\\frac{a_2}{a_3}, \\frac{a_3}{a_4},\\ldots$ ? \u2013\u00a0quasi Jan 7 '17 at 18:36\n\u2022 How do you get $a_k+a_k \\leq a_{k+1}$ from (4) and (5)? \u2013\u00a0rtybase Jan 7 '17 at 18:42\n\u2022 Oops -- I fell in the same trap that yasir fell in -- ignore my first comment. Instead, look at rtybase's objection. \u2013\u00a0quasi Jan 7 '17 at 18:46\n\u2022 @rtybase Subtracting (5) from (4) \u2013\u00a0MattG88 Jan 7 '17 at 18:49\n\u2022 @rtybase -- you can't subtract inequalites that are in the same direction. You can add them, but you can't subtract them. A common illusion (which I fell victim to as well, even though I know better). \u2013\u00a0quasi Jan 7 '17 at 19:13\n\nThe idea is \"local adjustments\". If $k$ is the smallest integer such that $a_1+\\cdots+a_{k-1} < a_k$, we adjust up $a_1$, $\\cdots$, $a_{k-1}$ with a scale factor of $s=\\frac{\\sum_{i=1}^{k}a_i}{2\\sum_{i=1}^{k-1}a_i}$ for all, and adjust down $a_k$ by a scale factor of $t=\\frac{\\sum_{i=1}^{k}a_i}{2a_k}$, and achieve a large value for LFS, and make $\\sum_{i=1}{k}a_i=a_k$. Note that since $\\sum_{i=1}^{k}a_i$ is not changed, and $a_1$, $\\cdots$, $a_{k-1}$ are scaled up by the same factor, all conditions are still satisfied. We just need to prove that really the left hand side gets larger or equal.\nReally there is just one case, but let's do three cases: 1) $k=2$, and 2) $k=n$; and 3) $2<k<n$.\n1) $k=2$, which means $a_1<a_2$. Now The only values changed are $\\frac{a_{1}}{a_2}$ and $\\frac{a_{2}}{a_{3}}$, and we want to show that $\\frac{a_{1}}{a_2} + \\frac{a_{2}}{a_{3}} \\le \\frac{sa_{1}}{ta_2} + \\frac{ta_{2}}{a_{3}}$. This is easy as $s=\\frac{a_1+a_2}{2a_1}$ and $t=\\frac{a_1+a_2}{2a_2}$, and $sa_1=ta_2=\\frac{a_1+a_2}{2}$: $$\\frac{a_{1}}{a_2} + \\frac{a_{2}}{a_{3}} \\le \\frac{a_{1}}{a_1} + \\frac{a_{1}}{a_{3}} = \\frac{sa_{1}}{ta_2} + \\frac{a_{1}}{a_{3}} < \\frac{sa_{1}}{ta_2} + \\frac{ta_{2}}{a_{3}},$$ where the first inequality requires a simple check.\n2) $k=n$. This is easy as The only value changed is $\\frac{a_{n-1}}{a_n}$ but we scale up $a_{n-1}$ and down $a_n$.\n3) $2<k<n$. Now The only values changed are $\\frac{a_{k-1}}{a_k}$ and $\\frac{a_{k}}{a_{k+1}}$, and we want to show that $\\frac{a_{k-1}}{a_k} + \\frac{a_{k}}{a_{k+1}} \\le \\frac{sa_{k-1}}{ta_k} + \\frac{ta_{k}}{a_{k+1}}$. Note that since the equality holds for $k-1$, we have $2a_{k-1}=\\sum_{i=1}{k-1}a_i$. So $sa_{k-1}= \\frac{2a_{k-1}+a_k}{4}$ and $ta_k=\\frac{2a_{k-1}+a_k}{2}=2sa_{k-1}$. So we want to prove that $$\\frac{a_{k-1}}{a_k} + \\frac{a_{k}}{a_{k+1}} \\le \\frac{1}{2} + \\frac{2a_{k-1}+a_k}{2a_{k+1}},$$ which is equivalent to the following after multiplying $2a_ka_{k+1}$: $$2a_{k-1}a_{k+1}+2a_k^2 \\le a_ka_{k+1}+(2a_{k-1}+a_k)a_k,$$ which is equivalent to $$(a_{k+1}-a_k)(a_k-2a_{k-1}) \\ge 0,$$ which is true since $a_{k+1}>a_k$ and $a_k \\ge \\sum_{i=1}^{k-1}a_i = 2a_{k-1}$.\nSo if we adjust up/down all $a_i$ and make all equalities in the condition hold, we achieve the largest value for $\\frac{a_1}{a_2}+\\cdots+\\frac{a_{n-1}}{a_n}$, which happens to be $\\frac{n}{2}$.", "date": "2019-06-26 15:54:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2087798/prove-that-fraca-1a-2-fraca-2a-3-fraca-3a-4-fraca-n-1", "openwebmath_score": 0.9306823015213013, "openwebmath_perplexity": 180.79905744405582, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225166137784, "lm_q2_score": 0.8652240947405565, "lm_q1q2_score": 0.8578026494425608}}
{"url": "https://oneclass.com/class-notes/ca/u-of-waterloo/math/math-136/281636-lecture-20pdf.en.html", "text": "Class Notes (838,022)\nMathematics (1,919)\nMATH 136 (168)\nLecture 20\n\nLecture 20.pdf\n\n4 Pages\n102 Views\n\nSchool\nDepartment\nMathematics\nCourse\nMATH 136\nProfessor\nRobert Sproule\nSemester\nWinter\n\nDescription\nWednesday, February 26 \u2212 Lecture 20: Linear independence Concepts: 1. Linearly independent set. 2. Recognize that subsets of linearly independent sets are linearly independent. 3. Characterize a linearly independent set as one being a set where no vector isa linear combination of the others. 20.1 Definition \u2013 Generalization of the definition of linear independence. Let v , v ...., v 1 2, k be k vectors in a vector space V. The vectors v , v ..., v are said to be linearly 1 2, k independent if and only if the only way that \u03b1 1 1 \u03b1 v 2 2.. + \u03b1 v = 0k k can hold true is if \u03b1 ,1\u03b1 ,2....., \u03b1 kre all zeroes. If v1, v2,.., vkare not linearly independent (i.e. there exists\u03b1 , \u03b1 , .1...2 \u03b1 not alk zero such that \u03b1 v1 1\u03b1 v +2 2... + \u03b1 v = k k then they are said to be linearly dependent. x 20.1.1 Example \u2013 We know that the two functions e and sinx are vectors in the set F described above. Then S = Span{e , sinx} = { \u03b1e + \u03b2sinx : \u03b1, \u03b2 belong to \u211d } is a x subspace of F. Show that the set {e , sinx} is linearly independent. Solution: Let us first recall that two functions f and g are equal on their domain Dif and only if f(x) = g(x) for all x in D. Suppxse \u03b1e + \u03b2sinx = 0(x) for all x (where 0(x) denotes the zero function; it maps every x in the domain to 0.). x - Suppose there exists \u03b1 \u2260 0 such that \u03b1e + \u03b2sinx = 0(x). x - Then e = (\u2013\u03b2/\u03b1)sinx for all x. This includes the value x = \u03c0. - But 0 \u2260 e = (\u2013\u03b2/\u03b1)sin\u03c0 = 0, a contradiction. - Then \u03b1 must be 0. x x - Suppose \u03b2 \u2260 0 is such that, 0e + \u03b2sinx = 0(x). Then (\u20130/\u03b2)e = sinx for all x. - Since sin(\u03c0/2) =1 \u2260 0, we have a contradiction. - So \u03b2 must also be 0. - So {e , sinx} is linearly independent. 20.1.2 Proposition \u2212 The vectors M = {v , v , ..1, v2} with k , 2, are linearly independent if and only if no vector in M is a linear combination of the other vectors in M. Proof : This is similar to the previously given proof for the case of V = \u211d . n (\u21d0) - Suppose none of these vectors is a linear combination of the others. - Suppose M is not linearly independent. - Then there exists \u03b1 , \u03b1 1...2., \u03b1 not kll zeroes such that \u03b1 v + \u03b1 v +1 1..+ \u03b12 2= k k 0. - Suppose \u03b1 is not zero. Rearrange the order of {v , v , ...., v } so that p = 1, .i.e., p 1 2 k \u03b1 1s not zero. - Then v = 1 (\u03b1 /\u03b1 )2v +1\u2212 (2 /\u03b1 ) v +3, 1.., 3 \u2212 (\u03b1 /\u03b1 )v . k 1 k - Hence v is a linear combination of the others. Contradiction. 1 (\u21d2) - Suppose {v , v1, .2.., v } ks linearly independent. - Suppose one vector of {v , v , .1..,2v } is k linear combination of the others. Say it is v . 1 - Then there exists \u03b1 ....2, \u03b1 suck that such that v = \u03b1 v + 1.... 2 2 v . k k - Then v \u2212 1 v \u2212 2 v2+ , ..3, 3 \u03b1 v . = 0. k k - Since the coefficient of v is n1t zero this contradicts our hypothesis. - Then no vector of {v , v ,1...2, v } iska linear combination of the others. If a vector in U is a linear combination of the others we will refer to it as being \u201credundant\u201d. It doesn't contribute anything to its span. 20.2 Definition \u2013 Let {v 1, v2,..., vk} be a spanning family for a vector space V. That is, Span{v , v ..., v } = V. Then for every vector v in V there exists scalars \u03b1 , \u03b1 , ....., \u03b1 1 2, k 1 2 k such that v = \u03b1 v1 1\u03b1 v +2 2... + \u03b1 v . k k If for each vectorv this set of scalars \u03b1 , \u03b1 , ..., \u03b1 satisfying v = \u03b1 v + \u03b1 v + ..... + \u03b1 v 1 2 k 1 1 2 2 k k is unique then we say that Span{v 1, v2,.., v k satisfies the unique representation property with respect to its spanning family. 20.2.1 Example \u2013 The spanning family Span{(1, 0), (0,1)} = Span{ e , e } = \u211d 1 2 2 satisfies the unique representation property since for an arbitrary vector v= (a, b) in \u211d , a and b are the unique scalars associated to v so that v = ae + be . 1 2 20.2.2 Theorem \u2212 Let S = {v , v , .1..,2v } be a k\nMore Less\n\nRelated notes for MATH 136\nMe\n\nOR\n\nJoin OneClass\n\nAccess over 10 million pages of study\ndocuments for 1.3 million courses.\n\nJoin to view\n\nOR\n\nBy registering, I agree to the Terms and Privacy Policies\nJust a few more details\n\nSo we can recommend you notes for your school.", "date": "2018-04-24 23:37:57", "meta": {"domain": "oneclass.com", "url": "https://oneclass.com/class-notes/ca/u-of-waterloo/math/math-136/281636-lecture-20pdf.en.html", "openwebmath_score": 0.8696692585945129, "openwebmath_perplexity": 1608.8702366432562, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225158534695, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8578026367255422}}
{"url": "https://math.stackexchange.com/questions/3368484/how-many-ten-digit-numbers-are-there-in-which-every-digit-is-2-or-3-and-no-two", "text": "# How many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent?\n\nHow many ten-digit numbers are there in which every digit is 2 or 3, and no two 3s are adjacent?\n\nTaken from the 2008 IMC https://chiuchang.org/wp-content/uploads/sites/2/2018/02/2008-IWYMIC-Individual.x17381.pdf\n\nmy attempt\n\nThe number of ten digit numbers in which the digits are either 2 or 3 is $$2^{10}$$ and the numbers of ten digit numbers where there is no pair of adjacent numbers that are the same is $$2$$ E.g($$2323232323$$ & $$3232323232$$) and we know that the number of ten digit numbers consisting of pairs of adjacent $$2$$s and adjacent $$3$$s is the same due to symmetry therefore the answer would be $$\\frac{2^{10}-2}{2}=511$$ however this doesn't taken into account the possibilities of having adjacent pairs of $$3$$s and $$2$$s in the same number.\n\n\u2022 This is the same as this question\n\u2013\u00a0lulu\nSep 24 '19 at 18:45\n\u2022 And similar to this, more recent, one. Sep 25 '19 at 5:47\n\nSuppose you have r number of 3 in a line. Now we know 3 can't be adjacent so I have to put at least 1 in between them. For example-r=3 |-|-|(| denotes 3 and - denotes 2). Now we have x=10-(r+r-1) 2's remaining. Now we can put these 2's in r+1 slots(in the example we can put in 4 slots slot1|slot2|slot3|slot4). Now this becomes a standard problem Stars and bars of x1+x2+x3+..xl=p (where l denotes the number of slots and where every xi can be 0 and p denotes number of 2 we want to distribute which is x in our case). Number of solution of above equation is given by:\n\n$$\\binom {p+l-1}{l-1}$$\n\nhere p=10-(r+r-1) and l=r+1(slots) putting these value in above formula we get:\n\n$$\\binom {11-r}{r}$$\n\nNow you can vary r from 0-10 and just add them you will get 144 as the answer. Here you can see you do not need to vary from 6 -10 because when we place 6 or more number we don't have enough 2's to put in between them so they will contribute 0.\n\nSuppose you have $$k$$ $$3$$s and it does not end with a $$3$$.\n\nThat the same thing as saying you have $$k$$ characters $$32$$ and $$10-k-k=10-2k$$ twos. So of the $$10-2k + k=10-k$$ characters you must choose $$k$$ spaces for the $$k$$ $$32$$ characters. There $${10-k \\choose k}$$ ways to do that.\n\nNow suppose you have $$k$$ $$3$$s and it does end with a $$3$$.\n\nIf we just ignore the last place and put the $$3$$ in it, that is the same thing as saying you have $$k-1$$ characters $$32$$ to place and $$10-k-(k-1)=11-2k$$ $$2$$s to place. There are $${10-k\\choose k-1}$$ ways to do that.\n\nSo there are $${10-k \\choose k}+ {10-k\\choose k-1}$$ ways to place $$k$$ threes.\n\nNow we can have at most $$5$$ threes. (Any more and we won't have enough $$2$$s to go between all $$3$$s.)\n\nSo there are\n\n$$\\sum_{k=0}^5 {10-k \\choose k}+ {10-k\\choose k-1}$$ ways. (Assume $${10\\choose -1} = 0$$.... after all.... this would be the number of ways to choose $$0$$ threes and have a $$3$$ at the end which isn't possible.)\n\nSo $$({10\\choose 0}) + ({9\\choose 1}+ {9\\choose 0}) +({8\\choose 2}+{8\\choose 1}) + ({7\\choose 3}+{7\\choose 2}) + ({6\\choose 4}+{6\\choose3}) + ({5\\choose 5} + {5\\choose 4})=$$\n\n$$1 + (9+1) + (\\frac {8*7}2 + 8) + (\\frac {7*6*5}6 -\\frac {7*6}2) + (\\frac {6*5*4*3}{24}+\\frac{6*5*4}6) + (1 +5)=$$\n\n$$1 + 10 +(28+8) + (35+21) + (15+20) + 6=$$\n\n$$11+36+56+35+5=144$$", "date": "2022-01-25 06:00:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3368484/how-many-ten-digit-numbers-are-there-in-which-every-digit-is-2-or-3-and-no-two", "openwebmath_score": 0.7912294268608093, "openwebmath_perplexity": 152.57062638716812, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225135725426, "lm_q2_score": 0.8652240791017536, "lm_q1q2_score": 0.857802631306549}}
{"url": "https://math.stackexchange.com/questions/1266282/finding-the-inverse-of-a-number-under-a-certain-modulus", "text": "# Finding the inverse of a number under a certain modulus\n\nHow does one get the inverse of 7 modulo 11?\n\nI know the answer is supposed to be 8, but have no idea how to reach or calculate that figure.\n\nLikewise, I have the same problem finding the inverse of 3 modulo 13, which is 9.\n\nThe inverses can be computed using the extended euclidean algorithm. As long as $\\gcd(x,n) = 1$ the inverse $x^{-1} \\bmod n$ exists and is $y$ from the extended euclidean algorithm where $$xy + kn = 1 = \\gcd(x,n)$$ For example we get $$7\\cdot 8 - 5\\cdot 11 = 1\\\\ 9 \\cdot 3 - 2\\cdot 13 = 1$$\n\nNote that you can also obtain negative numbers, for example $$7\\cdot (-3) + 2\\cdot 11 = 1$$ In this case you can use congruences $\\bmod n$ to obtain the canonical inverse: $-3 \\equiv 8 \\pmod{11}$\n\n\u2022 Thanks but I'm really struggling with this, could you provide me with a sample, i be stuck for two days trying to figure it out \u2013\u00a0Dan W May 4 '15 at 12:07\n\u2022 @DanW What part of this do you struggle with? \u2013\u00a0AlexR May 4 '15 at 12:09\n\u2022 Basically, i want to know the inverse of 7 within mod 11, i need a figure!!, i understand the gcd and what it is doing/proving. \u2013\u00a0Dan W May 4 '15 at 12:20\n\u2022 @DanW Can you show me what the extended euclidean algorithm gives you wen used for $\\gcd(7,11)$? \u2013\u00a0AlexR May 4 '15 at 12:23\n\u2022 A B Q R 11 7 1 4 7 4 1 3 4 3 1 1 \u2013\u00a0Dan W May 4 '15 at 12:28\n\nTo find the inverse of $7$, $\\pmod{11}$, you must find a solution to $7x\\equiv 1 \\pmod{11}$. (Then $x$ satisfies the definition of inverse.)\n\nAs suggested in other answers, one way to do this is with the extended Euclidean algorithm, and in fact this is the best general purpose algorithm for this type of problem.\n\nBut for small values, you can also try 'adding the modulus':\n\n$7x\\equiv 1\\equiv 12\\equiv 23\\equiv 34\\equiv 45\\equiv 56 \\pmod{11}$.\n\nThen from $7x\\equiv 56\\pmod{11}$, we can cancel $7$, obtaining $x\\equiv 8 \\pmod{11}$.\n\nHere's an illustration of finding the multiplicative inverse of $37 \\bmod 100$ using the extended Euclidean algorithm. (I used bigger numbers for this example so that the relationships are a little clearer).\n\nOn each line, $n=100s+37t$. We start the table with two lines giving $n=100$ and $n=37$ in the obvious way. Then $q$ gives the rounded-down ratio of the current and previous value of $n$. Using this, the next line is calculated by subtracting $q$ copies of the current line entries from the previous line entries.\n\n$$\\begin{array}{c|c|c|c} n & s & t & q \\\\ \\hline 100 & 1 & 0 & \\\\ 37 & 0 & 1 & 2 \\\\ 26 & 1 & -2 & 1 \\\\ 11 & -1 & 3 & 2 \\\\ 4 & 3 & -8 & 2 \\\\ 3 & -7 & 19 & 1 \\\\ 1 & 10 & \\color{red}{-27} & 3 \\\\ \\end{array}$$\n\nOn the last line, $1 = 10\\times 100 + (-27)\\times 37$, so $\\color{red}{-27}\\times 37 \\equiv 1 \\bmod 100$\n\nBringing the result positive, $-27 \\equiv \\color{red}{73} \\bmod 100$. And $37 \\times 73 = 2701 \\equiv 1 \\bmod 100 \\quad \\checkmark$.\n\nAs you can perhaps see, you don't actually need to calculate the $s$ values at all to get the modular inverse. I left them in to help understand the table.\n\nThe corresponding tables for your particular questions: $$\\begin{array}{c|c|c|c} n & s & t & q \\\\ \\hline 11 & 1 & 0 & \\\\ 7 & 0 & 1 & 1 \\\\ 4 & 1 & -1 & 1 \\\\ 3 & -1 & 2 & 1 \\\\ 1 & 2 & \\color{red}{-3} & 3 \\\\ \\end{array}$$ and $7^{-1} \\equiv -3 \\equiv \\color{red}8 \\bmod 11$. $\\quad 7\\times 8 = 56 = 55+1\\quad \\checkmark$\n\n$$\\begin{array}{c|c|c|c} n & s & t & q \\\\ \\hline 13 & 1 & 0 & \\\\ 3 & 0 & 1 & 4 \\\\ 1 & 1 & \\color{red}{-4} & 3 \\\\ \\end{array}$$ and $3^{-1} \\equiv -4 \\equiv \\color{red}9 \\bmod 13$. $\\quad 3\\times 9 = 27 = 26+1\\quad \\checkmark$\n\n${\\rm mod}\\ 11\\!:\\,\\ \\dfrac{1}7\\equiv \\dfrac{12}{-4}\\equiv -3\\$ (see Gauss's algorithm. for an algorithmic version of this).\n\nOr, compute the Bezout identity $\\,\\gcd(11,7) = 2(11)-3(7) = 1\\,$ by the Extended Euclidean Algorithm  (see here for a convenient version).  Thus $\\ {-}3(7)\\equiv 1\\pmod{11}$\n\n${\\rm mod}\\ 13\\!:\\,\\ \\dfrac{1}3\\equiv \\dfrac{-12}{3}\\equiv -4\\$\n\nGenerally inverting $\\,a\\,$ mod $\\,m\\,$ is easy if $\\,a\\mid m\\pm1,\\$ i.e. $\\ m = ab\\pm 1$\n\n${\\rm mod}\\ ab-1 \\!:\\qquad ab\\equiv 1\\,\\Rightarrow\\, a^{-1}\\equiv b$\n\n${\\rm mod}\\ ab+1 \\!:\\,\\ a(-b)\\equiv 1\\,\\Rightarrow\\, a^{-1}\\equiv -b$\n\nAbove are special cases: $\\,3\\mid 13\\!-\\!1\\$ and $\\ 7\\equiv -4\\mid 11\\!+\\!1$\n\nTo find the inverse of $7$ modulo $11$, we must solve the equivalence $7x \\equiv 1 \\pmod{11}$. To do this, we use the Extended Euclidean Algorithm to express $1$ as a linear combination of $7$ and $11$. The coefficient of $7$ will be the inverse modulo $11$. By the Euclidean Algorithm, \\begin{align*} 11 & = 1 \\cdot 7 + 4\\\\ 7 & = 1 \\cdot 4 + 3\\\\ 4 & = 1 \\cdot 3 + 1\\\\ 3 & = 3 \\cdot 1 \\end{align*} We now take the equation $4 = 1 \\cdot 3 + 1 = 3 + 1$, solve for $1$, then work backwards until we obtain $1$ as a linear combination of $7$ and $11$. \\begin{align*} 1 & = 4 - 3\\\\ & = 4 - (7 - 4)\\\\ & = 2 \\cdot 4 - 7\\\\ & = 2(11 - 7) - 7\\\\ & = 2 \\cdot 11 - 3 \\cdot 7 \\end{align*} Since $2 \\cdot 11 - 3 \\cdot 7 = 1$, $$-3 \\cdot 7 = 1 - 2 \\cdot 11 \\Longrightarrow -3 \\cdot 7 \\equiv 1 \\pmod{11}$$ Hence, $-3$ is the inverse of $7 \\pmod{11}$. To express the inverse as one of the residues $\\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\\}$, we add $11$ to $-3$ to obtain $-3 + 11 \\equiv 8 \\pmod{11}$. Hence, $7^{-1} \\equiv 8 \\pmod{11}$.\n\nCheck: $7 \\cdot 8 \\equiv 56 \\equiv 1 + 5 \\cdot 11 \\equiv 1 \\pmod{11}$.\n\nTo verify you understand the algorithm, try to find the inverse of $3$ modulo $13$.\n\nAs said by AlexR, you can find $x,y\\in\\mathbb Z$ with $7x+11y=1$ using Extended Euclidean algorithm (see this answer for how to best use it).\n\nYou can also use elementary modular arithmetic 'tricks':\n\n$\\bmod{11}\\!:\\ 7x\\equiv 1\\equiv -21\\stackrel{:7}\\iff x\\equiv -3\\equiv 8$.\n\n$\\bmod{11}\\!:\\ 7x\\equiv -4x\\equiv 1\\equiv 12\\stackrel{:(-4)}\\iff x\\equiv -3\\equiv 8$.\n\nWe could divide by $7$ and $-4$ because $(7,11)=(-4,11)=1$.", "date": "2020-01-26 17:57:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1266282/finding-the-inverse-of-a-number-under-a-certain-modulus", "openwebmath_score": 0.9866442680358887, "openwebmath_perplexity": 112.53606920001087, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989347490102537, "lm_q2_score": 0.867035752930664, "lm_q1q2_score": 0.8577996459911158}}
{"url": "https://math.stackexchange.com/questions/930565/number-of-lattice-points-in-a-triangle", "text": "# Number of Lattice Points in a Triangle\n\nProblem\n\nLet the co-ordinates of the vertices of the $\\triangle OAB$ be $O(1,1)$, $A(\\frac{a+1}{2},1)$ and $B(\\frac{a+1}{2},\\frac{b+1}{2})$ where $a$ and $b$ are mutually prime odd integers, each greater than $1$. Then find the number of lattice points inside $\\triangle OAB$, i.e., not on the borders of $\\triangle OAB$. How does the answer change if the restriction $\\operatorname{gcd}(a,b)=1$ is removed?\n\nSolution\n\nLet $L(a,b)$ be the number of lattice points inside $\\triangle OAB$. The linear transformation $T(x,y)=(x-1,y-1)$ on the triangle $OAB$ do not have any effect on $L(a,b)$. Hence $L(a,b)$ is equal to the number of lattice points inside the triangle $O'A'B'$, where $O'=(0,0)$, ${\\textstyle A'=(\\frac{a-1}{2},0)}$ and ${\\textstyle B'=(\\frac{a-1}{2},\\frac{b-1}{2})}$.\n\nSet ${C'=(0,\\frac{b-1}{2})}$. By symmetry the number of lattice points inside the rectangle $O'B'C'$ is $L(a,b)$. The number of lattice point inside the rectangle $O'A'B'C'$ is ${\\textstyle \\frac{a-3}{2} \\cdot \\frac{b-3}{2}}$. Consequently\n\n$$2L(a,b) = \\frac{a-3}{2} \\cdot \\frac{b-3}{2} - K$$\n\nwhere $K$ are the number of lattice points on the straight line between $O'$ and $B'$. This line is given by $y= \\frac{b-1}{a-1}x$, yielding\n\n$$K = |\\{ {\\textstyle 0 < x < \\frac{a-1}{2} \\mid \\frac{b-1}{a-1}x \\in \\mathbb{N} \\}| }$$\n\nBy letting $d=\\text{gcd}(a-1,b-1)$, we obtain $a-1=rd$ and $b-1=sd$ for two positive coprime integers $r$ and $s$. Therefore\n\n$$\\frac{b-1}{a-1}x = \\frac{sx}{r} \\in \\mathbb{N} \\;\\; \\Leftrightarrow \\;\\; \\frac{x}{r} \\in \\mathbb{N}$$\n\nNow ${\\textstyle 0 < x < \\frac{a-1}{2} = \\frac{rd}{2}}$, implying ${\\textstyle 0 < \\frac{x}{r} = \\frac{d}{2}}$. Since $d$ is even (since $d=\\gcd(a-1,b-1)$ and $a$ and $b$ are both odd positive integers) give us ${\\textstyle K = \\frac{d}{2}-1}$. Thus by (1)\n\n$${\\textstyle L(a,b) = \\dfrac{(a-3)(b-3)}{8} - \\dfrac{\\text{gcd}(a-1,b-1)}{4} + \\dfrac{1}{2}}$$\n\nBut I think that by Pick's Theorem the answer should be,\n\n$${\\textstyle L(a,b) = \\dfrac{(a-1)(b-1)}{8} - \\dfrac{\\text{gcd}(a-1,b-1)}{4} + \\dfrac{1}{2}}$$\n\nWhich one is correct?\n\n\u2022 You can easily figure out which one is correct by putting in some small values for $a$ and $b$. \u2013\u00a0Ted Sep 14 '14 at 6:06\n\u2022 Try $a = 1$ and $b = 10$. This yields a degenerate triangle with $0$ lattice points strictly inside it. Which answer gives you $0$? \u2013\u00a0JimmyK4542 Sep 14 '14 at 6:17\n\u2022 @JimmyK4542: Both $a$ and $b$ are odd and mutually prime and each is greater than $1$. \u2013\u00a0user170039 Sep 14 '14 at 6:26\n\u2022 Fine, try a less trivial case like $a = 3$ and $b = 5$. The formula in the gray box gives $L(3,5) = 0$ while the other formula gives $L(3,5) = 1$. Which one is right? \u2013\u00a0JimmyK4542 Sep 14 '14 at 6:27\n\u2022 There are $0$ lattice points inside the triangle with vertices $O(1,1)$, $A(2,1)$, $B(2,3)$. So the first formula is correct (in that case). I have no idea where the second formula goes wrong because you have left out the details of how you got that. \u2013\u00a0JimmyK4542 Sep 14 '14 at 6:34\n\nThe area of $\\Delta OAB$ is $K = \\dfrac{1}{2} \\cdot \\dfrac{a-1}{2} \\cdot \\dfrac{b-1}{2} = \\dfrac{(a-1)(b-1)}{8}$.\n\nThe number of points on the boundary (line segments $OA$, $AB$, and $BO$) is\n\n$\\underbrace{\\dfrac{a-1}{2}}_{OA}+\\underbrace{\\dfrac{b-1}{2}}_{OB}+\\underbrace{\\text{gcd}\\left(\\dfrac{a-1}{2},\\dfrac{b-1}{2}\\right)}_{BO}$ (you missed the first two terms of this).\n\nSo, by Pick's Theorem, the area is\n\n$K = I+\\dfrac{1}{2}B-1 = I+\\dfrac{1}{2}\\left[\\dfrac{a-1}{2}+\\dfrac{b-1}{2}+\\text{gcd}\\left(\\dfrac{a-1}{2},\\dfrac{b-1}{2}\\right)\\right]-1$\n\n$= I + \\dfrac{a-1}{4} + \\dfrac{b-1}{4}+\\dfrac{\\text{gcd}(a-1,b-1)}{4}-1$.\n\nHence, the number of points in the interior is\n\n$I = \\dfrac{(a-1)(b-1)}{8} - \\dfrac{a-1}{4} - \\dfrac{b-1}{4} - \\dfrac{\\text{gcd}(a-1,b-1)}{4} + 1$\n\n$= \\dfrac{(a-3)(b-3)}{8} - \\dfrac{\\text{gcd}(a-1,b-1)}{4} + \\dfrac{1}{2}$ (after a bit of simplification).", "date": "2020-05-29 22:37:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/930565/number-of-lattice-points-in-a-triangle", "openwebmath_score": 0.872506856918335, "openwebmath_perplexity": 91.55913807305254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474888461862, "lm_q2_score": 0.8670357460591569, "lm_q1q2_score": 0.8577996381035065}}
{"url": "https://math.stackexchange.com/questions/937312/what-is-the-combinatorial-interpretation-of-the-product-of-binomial-coefficients", "text": "# What is the combinatorial interpretation of the product of binomial coefficients?\n\nFull disclosure: This question is relating to a homework question. It's not a homework question itself, but rather a clarifying question to help myself get a handle on the actual question.\n\nSuppose I had three sets $X$, $Y$, and $Z$ with cardinalities $|X| = x$, $|Y| = y$, and $|Z| = z$, with $z \\leq y \\leq x$\n\nI know that $x \\choose y$ is the number of ways of selecting $y$ elements from $X$.\n\nHowever, I'm unclear of the combinatorial interpretation of the following:\n\n$${x \\choose y} {y \\choose z}$$\n\nSo my question is: in general, is there an intuitive interpretation for the product of the number of choices?\n\n\u2022 If $x\\le y\\le z$ then both of those coefficients are $0$ \u2013\u00a0Adam Hughes Sep 19 '14 at 1:50\n\u2022 You're right, sorry; I'm updating my question now \u2013\u00a0Jason Baker Sep 19 '14 at 1:51\n\u2022 There's two interpretations. 1) How many ways are there to first pick $y$ elements out of $x$, and then pick $z$ elements out of these $y$? 2) How many ways are there to divide $x$ into sets of size $z$, $y-z,$ and $x-y$? \u2013\u00a0Semiclassical Sep 19 '14 at 1:54\n\u2022 @Semiclassical That makes sense; thanks for the confirmation \u2013\u00a0Jason Baker Sep 19 '14 at 1:57\n\nThe answer to your question is affirmative: Yes, there is an intuitive interpretation for the number of choices.\n\nWe know at least two interpretations of the binomial coefficient $\\binom{n}{k}$ which are useful for combinatorial proofs. These are\n\n\u2022 Set theoretical view: $\\binom{n}{k}$ as number of $k$-element subsets of an $n$-element set\n\n\u2022 Geometrical view: $\\binom{n}{k}$ as number of lattice pathes of length $n$ containing $k$ horizontal $(1,0)$-steps and $n-k$ vertical $(0,1)$-steps\n\nThe geometrical view is valid because there are $\\binom{n}{k}$ choices to select $k$ steps in horizontal direction leaving the remaining $n-k$ steps for the vertical direction.\n\nLet's apply the second approach to\n\n\\begin{align*} \\binom{x}{y}\\binom{y}{z}\\qquad z \\leq y \\leq x\\tag{1} \\end{align*}\n\nWhen looking at $\\binom{y}{z}$ you may think on all pathes of length $y$ starting at $A=(0,0)$ and ending in $B=(z,y-z)$ thereby containing exactly $z$ horizontal $(1,0)$-steps and $y-z$ vertical $(0,1)$-steps. All these pathes are enclosed within a rectangle of length $z$ and height $y-z$.\n\nSimilarly we consider $\\binom{x}{y}$ as the number of lattice pathes of length $x$ containing $y$ $(1,0)$-steps and $x-y$ $(0,1)$-steps. But here we let all these pathes start in $B=(z,y-z)$ and so they end up in $C=(z+y,x-z)$.\n\nWe observe following\n\nCombinatorial interpretation of $$\\binom{x}{y}\\binom{y}{z}$$\n\nThis is the number of lattice pathes of length $x+y$ starting in $A=(0,0)$ ending up in $C=(z+y,x-z)$ and passing through the point $B=(z,y-z)$.\n\nThe pathes contain only horizontal $(1,0)$-steps and vertical $(0,1)$-steps.\n\nA possible interpretation is pairing sets $X$ and $Y$ with no elements of $Y$ left so that would require selecting $y$ elements from $X$ or $\\displaystyle \\binom xy$, then making triplets with no elemts of $Z$ left that would require selecting $z$ pairs from these $y$ pairs or $\\displaystyle \\binom yz$.", "date": "2019-06-17 23:27:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/937312/what-is-the-combinatorial-interpretation-of-the-product-of-binomial-coefficients", "openwebmath_score": 0.9171217083930969, "openwebmath_perplexity": 148.22387565096662, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985271387015241, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8577745881004134}}
{"url": "https://math.stackexchange.com/questions/1764278/number-of-points-of-accumulation-of-a-sequence", "text": "# Number of points of accumulation of a sequence\n\nCan a sequence have infinitely many points of accumulation i.e. we can extract infinitely many subsequences from it s.t. they all converge to their respective point of accumulation? I have the feeling it would mean that the period of repetition of something could be infinitely big.\n\n\u2022 Since there are countably many rational numbers, there is a sequence that includes them all. The accumulation points of this sequence is the entire set of real numbers. Apr 29 '16 at 17:29\n\u2022 I like your example a lot ! Apr 29 '16 at 18:15\n\u2022 There's a classic theorem that if you walk around a circle in discrete steps of $a$ radians, where $a/\\pi$ is irrational, then the set of points that you visit is dense in the circle. This implies that $(\\sin(an))_{n\\in\\mathbb N}$ is dense in $[-1, 1]$, and you can take $a=1$ for a cute example of a sequence with infinitely many accumulation points. Apr 29 '16 at 23:16\n\u2022 See also this question. Some of the posts linked there might be of interest, too. Apr 30 '16 at 8:23\n\nStart with $0,1$. Then travel backwards to $0$ in steps of $1/2$, so $1/2,0$. Then travel forwards to $1$ in steps of $1/4$, so $1/4,2/4,3/4,1$. Then travel backwards to $0$ in steps of $1/8$, so $7/8$, $6/8$, $5/8$, and so on. Continue.\n\nEvery real between $0$ and $1$ is an accumulation point of our sequence.\n\n\u2022 Your sequence is bounded AND has infinitely many points of accumulation. Thank you for answering :) Apr 29 '16 at 16:28\n\u2022 @MehdiSlimani: You are welcome. If you prefer we could use $0,1,1/2,0,1/3,2/3,1,3/4,2/4,1/4,0,1/5,2/5,\\dots$ and then we can quote the result that the rationals in $[0,1]$ are dense in $[0,1]$. Apr 29 '16 at 16:29\n\u2022 I'm a bit confused. How can subsequences of your sequence converge if each member is in a different space? Maybe I don't have the necessary knowledge yet Apr 29 '16 at 16:36\n\u2022 For every real $x$ between $0$ and $1$, there is a subsequence of our sequence that has limit $x$. For in our back and forth travels, we get within $1$ of $x$, then within $1/2$ of $x$, then within $1/4$ of $x$, and so on. With the modified version in which we take steps $1$, then $1/2$, then $1/3$, then $1/4$, then $1/5$, and so on, our travels bring us to every rational between $0$ and $1$, so again there is a subsequence that converges to $x$. Apr 29 '16 at 16:39\n\u2022 Ok, I hadn't understood it well. Thank you again Apr 29 '16 at 16:43\n\nYes, this is possible. For example consider the sequence $a_n$ for $n \\ge 2$ defined as the smallest divisor of $n$ greater than $1$.\n\nThe accumulation points are all the prime numbers. Subsequences witnessing them are for instance the $p$-th powers for each prime $p$.\n\n\u2022 Simple and nice example, thank you a lot :) Apr 29 '16 at 16:24\n\nThe rationals are a countable set. We can define a 1-to-1 function $f:N\\to Q$ with $Q=\\{f(n):n\\in N\\}.$ Consider the sequence $S=(f(n))_{n\\in N}.$ Every real number is a limit point of a subsequence of $S.$\n\nAll the answers have uncountably many accumulation points. If you only want countably many, consider the sequence:\n\n$1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,\\cdots$\n\nEvery positive integer is an accumulation point, and nothing else.\n\nIf you further want it to be bounded:\n\n$\\frac11,\\frac11,\\frac12,\\frac11,\\frac12,\\frac13,\\frac11,\\frac12,\\frac13,\\frac14,\\cdots$\n\nSuppose each row of the infinite matrix below converges to a different real number\n\n\\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & \\dots \\\\ a_{21} & a_{22} & a_{23} & a_{24} & \\dots \\\\ a_{31} & a_{32} & a_{33} & a_{34} & \\dots \\\\ a_{41} & a_{42} & a_{43} & a_{44} & \\dots \\\\ \\vdots & \\vdots & \\vdots & \\vdots & \\end{bmatrix}\n\nThen the sequence\n\n$a_{11}, a_{21}, a_{12}, a_{31}, a_{22}, a_{13}, a_{41},a_{32}, a_{23}, a_{14}, a_{51}, \\dots$\n\ncontains an infinite number of convergent subsequences.\n\n\u2022 Some authors define accumulation point not so strictly, namely that it is enough that a subsequence converges to it. Apr 30 '16 at 9:37", "date": "2021-12-08 16:12:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1764278/number-of-points-of-accumulation-of-a-sequence", "openwebmath_score": 0.9760191440582275, "openwebmath_perplexity": 112.41233472694286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98527138442035, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8577745874953129}}
{"url": "https://stats.stackexchange.com/questions/198198/how-many-random-selections-needed-to-be-90-95-99-confident-that-a-specific/198202", "text": "# How many random selections needed to be 90%, 95%, 99% confident that a specific selection occurred?\n\nFor example, let's say that I have a bag of 100 marbles labeled 1 to 100. A \"selection\" is randomly picking a marble from the bag and then placing it back into the bag. The marble selected is not known.\n\nHow many random selections do I need to make to be X% confidence that a specific marble was one of the marbles I selected, say marble #42?\n\nIs there a formula or a distribution that helps me solve for X (90%, 95%, 99% confidence)?\n\nBernoulli or Binomial distributions seem close, but I can not quite make them solve for what I want.\n\nEach time you grab a marble you have a 1 out of 100 chance of selecting the correct one. What you are asking is what are the odds that at least one of these will be correct which would be an inclusive \"or\" statement\n\n$$P_{selected} = p \\cup p \\cup p \\cup p \\cup p... \\cup p$$\n\nHowever this is not trivial to calculate since you could select your marble multiple times. Thus it's better to use the negative of this and calculate when it was not selected.\n\n$$P_{selected}^\\sim = (1-p)\\cap(1-p)\\cap(1-p)(1-p)\\cap(1-p)...(1-p)\\cap(1-p)$$\n\nwhich can be easily calculated to N times by\n\n$$P_{selected}^\\sim = (1-p)^N$$\n\nWhere you can then show\n\n$$P_{selected} = 1- P_{selected}^\\sim = 1-(1-p)^N$$\n\nand after some algebra\n\n$$N = \\frac{\\ln(1-P_{selected})}{\\ln(1-p)}$$\n\nso 90% confidence is N = 229.\n\n95% confidence is N = 298\n\n99% confidence is N = 458.\n\nIf $X$ has $k$ distinct values, each appearing with equal probability $1/k$, then it follows discrete uniform distribution. Probability of drawing a specific $x$ in a single draw is\n\n$$\\Pr(X=x) = \\frac{1}{k}$$\n\nprobability of drawing value other than $x$ is\n\n$$\\Pr(X \\ne x) = \\frac{k-1}{k} = 1-\\frac{1}{k}$$\n\nso if you make $n$ draws, probability that in neither of the draws $x$ is drawn is\n\n$$\\left(1-\\frac{1}{k} \\right)^n$$\n\nfrom here you can easily find that probability of drawing at least one $x$ in $n$ draws is\n\n$$1-\\left(1-\\frac{1}{k} \\right)^n$$\n\nNow all you need to know is find such $n$ that leads to prespecified probability.\n\nThis can be easily checked by simulation.\n\nset.seed(123)\n\nf <- function(n, k = 100) 1-((1-1/k)^n)\ng <- function() which(sample(100, 1e4, replace = TRUE) == 42)[1]\nsim <- replicate(5e4, g())\n\nthat returns\n\n> quantile(sim, c(0.9, 0.95, 0.99))\n90%    95%    99%\n230.00 301.00 460.01\n> f(c(230, 301, 460))\n[1] 0.9008952 0.9514495 0.9901782\n\n(Notice that example above concerns not with confidence intervals but with exact probabilities.)\n\nThis shows us that we need to sample $n$ values until $x$ occurs, so we are talking about negative binomial distributed variable with parameters $p=1/k$ and $r=1$ (observe $n$ values until $X=x$ occurs once, where probability of observing $x$ is $1/k$). Results obtained using negative binomial distribution agree with the simulation.\n\n> pnbinom(c(230, 301, 460)-1, 1, 1/k)\n[1] 0.9008952 0.9514495 0.9901782\n\nSo you intuition about binomial distribution was close, as it follows negative binomial distribution. Negative binomial with $r=1$ simplifies to geometric distribution. This is how we ended up with distribution that is traditionally used for modeling discrete waiting times.\n\n> pgeom(c(230, 301, 460)-1, 1/k)\n[1] 0.9008952 0.9514495 0.9901782\n\nIn fact, if you recall, cumulative distribution function for geometric distribution is\n\n$$F(n) = 1 - (1-p)^n$$\n\nwhere $p$ is a probability of observing $x$ in a single draw. So geometric distribution can be used to obtain confidence intervals for the number of samples that need to be taken.\n\n# 95% CI using quantile function of geometric distribution\n> qgeom(c(0.025, 0.975), 1/k)\n[1]   2 367\n\nLet $M=100$. Let's say you make $N$ selections and you are looking for the occurrence of marble $m$.\n\nLet $E_n$ be the event that marble $m$ is drawn at draw $n$. Let $F_n=E_n^c$ be the event that marble $m$ is NOT drawn.\n\nNote that $P(E_n)=1/M$. This means, $P(F_n)=1-1/M$.\n\nLet $F$ be the failure event that marble $m$ is not drawn for any $N$. Success is then $F^c$.\n\n$P(F^c)=1-P(F)$\n\nAssuming independent draws, we have:\n\n$P(F)=P(F_1 \\cap F_2 \\cap F_3 \\cap \\ldots \\cap F_N)=\\prod_{n=1}^N P(F_n)=(1-1/M)^N$\n\nTherefore, $P(F^c)=1-(1-1/M)^N$.\n\nFor the $c \\in [0,1]$ (e.g. $c=0.9 \\equiv 90\\%$ confidence) confidence, set the LHS to $c$ and solve for $N$.\n\n$\\Rightarrow N=\\dfrac{\\ln(1-c)}{\\ln(1-1/M)}$", "date": "2019-09-21 11:11:45", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/198198/how-many-random-selections-needed-to-be-90-95-99-confident-that-a-specific/198202", "openwebmath_score": 0.8432955145835876, "openwebmath_perplexity": 551.2017274025384, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98527138442035, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8577745858413084}}
{"url": "http://math.stackexchange.com/questions/108041/linear-algebra-preserving-the-null-space", "text": "# Linear Algebra: Preserving the null space\n\nWhat does it mean when a book says that row operations preserve the null space? And why should that be true? I have read that row operations are equivalent to multiplying a vector on the left by an invertible elementary matrix. And I think I understand that the nullspace is the set of all vectors from $u \\in U$ which get mapped to the zero vector in $V$ if $T:U\\rightarrow V$ is linear. But I'm still not sure what this means.\n\n-\n\nIt means that performing an elementary row operation on a matrix does not change the null space of the matrix. That is, if $A$ is a matrix, and $E$ is an elementary matrix of the appropriate size, then the matrix $EA$ has the same null space as $A$.\n\nTo see why this is true, suppose first that $x$ is in the null space of $A$. This means that $Ax=\\vec 0$. Multiplying both sides of this equation by $E$, we see that $(EA)x=E\\vec 0 =\\vec 0$, meaning that $x$ is also in the null space of $EA$. Now suppose that $x$ is in the null space of $EA$, so that $(EA)x=\\vec 0$. As you mentioned, $E$ is invertible, so we can multiply this equation by $E^{-1}$:\n\n$$Ax=IAx=(E^{-1}E)Ax=E^{-1}(EA)x=E^{-1}\\vec 0=\\vec 0\\;,$$\n\nshowing that $x$ is in the null space of $A$. In other words, a vector is in the null space of $EA$ if and only if it is in the null space of $A$, and $EA$ and $A$ have the same null space.\n\n-\nThanks a lot, the if and only if explanation helps me connect these facts together now! \u2013\u00a0 mcrocker Feb 12 '12 at 4:32\n\nSay we have $V, W$ vector spaces and a linear transformation $T : V \\to W$. The null space of $T$ is defined $N(T) = \\{x \\in V : T(x) = 0\\}$.\n\nWhen talking about row operations, we are talking about the matrix representation of the linear transformation. So if $\\alpha = \\{ x_1, ..., x_n \\}$ is a basis of $V$ and $\\beta = \\{ y_1, ..., y_n \\}$ is a basis of $W$ we have a matrix, say $A$, such that $A = [T]_\\alpha^\\beta$.\n\nSay $T(x_i) = a_{i1}y_1 + ... + a_{in}y_n$. Then the matrix A will be the following $$A = \\left[\\begin{array}{ccc} a_{11} & ... & a_{1n} \\\\ ... & ... & ... \\\\ a_{n1} & ... & a_{nn} \\end{array}\\right].$$\n\nRow operations are as follows\n(i) Swap two rows.\n(ii) Multiply row by non zero constant.\n(iii) Add multiple of one row to a different row.\n\nSay we swap the $i$th and $j$th row and call the new matrix $A'$. This new matrix defines a new linear transformation $T':V \\to W$ such that $T'(x) = A'x$. We want to show that $N(T) = N(T')$. For any $x \\in V$, say $$T(x) = b_1y_1 + ... + b_iy_i + ... + b_jy_j + ... + b_ny_n.$$ Then since $T'$ is just rows swapped $$T'(x) = b_1y_1 + ... +b_jy_j + ... + b_iy_i + ... b_ny_n.$$ Thus if $T(x) = 0$, $T'(x) = 0$ and vice versa. So $N(T) = N(T')$.\n\nFor (ii), think about if we multiply the $i$th row by $\\beta$, then if the $i$th scalar for $T(x)$ is $c_i$, then the $i$th scalar for $T'(i)$ is $\\beta c_i$. So if $T(x) = 0$, then $c_i = 0$, so $\\beta c_i = 0$. All other scalars stay the same, thus $T'(x) = 0$. And if $T'(x) = 0$, $\\beta^{-1}c_i = 0$, so $T(x) = 0$. Once again $N(T) = N(T')$.\n\nI leave (iii) to you.\n\n-\n\nFor what it's worth, I think the following way of thinking about why row operations do not change the null space is worthwhile:\n\nAn important observation to make:\n\nOne may observe that multiplication of a matrix $A$ by a column vector $\\bf x$ amounts to taking a linear combination of the columns of $A$. The coefficients of the particular linear combination are the coordinates of $\\bf x$.\n\nFor example: $$\\tag{1} \\underbrace{\\left[ \\matrix{a_{11}&a_{12}&a_{13} \\cr a_{21}&a_{22}&a_{23}\\cr a_{31}&a_{32}&a_{33} }\\right]}_{A} \\left[\\matrix{x\\cr y\\cr z}\\right] =\\left[\\matrix{x a_{11} +ya_{12}+ z a_{13} \\cr x a_{21} +ya_{22}+ z a_{23} \\cr x a_{31} +ya_{32}+ z a_{33}} \\right] = x\\underbrace{\\left[\\matrix{\\color{maroon}{a_{11}}\\cr\\color{darkgreen}{ a_{21}}\\cr \\color{darkblue}{a_{31}}}\\right]}_{\\bf a_1} +y\\underbrace{\\left[\\matrix{\\color{maroon}{a_{12}}\\cr\\color{darkgreen}{ a_{22}}\\cr \\color{darkblue}{a_{32}}}\\right]}_{\\bf a_2} +z\\underbrace{\\left[\\matrix{\\color{maroon}{a_{13}}\\cr\\color{darkgreen}{ a_{23}}\\cr\\color{darkblue}{ a_{33}}}\\right]}_{\\bf a_3}$$\n\nA second important observation to make:\n\n$\\bf x$ is in the null space of $A$ if and only if $A{\\bf x}=\\bf0$. This means that the linear combination of the columns of $A$ whose coefficients are the coordinates of ${\\bf x}=\\Bigl[{\\textstyle{x\\atop y}\\atop\\scriptstyle z }\\Bigr]$ is the zero vector:\n\n$$\\tag{2} x\\underbrace{\\left[\\matrix{\\color{maroon}{a_{11}}\\cr\\color{darkgreen}{ a_{21}}\\cr \\color{darkblue}{a_{31}}}\\right]}_{\\bf a_1} +y\\underbrace{\\left[\\matrix{\\color{maroon}{a_{12}}\\cr\\color{darkgreen}{ a_{22}}\\cr \\color{darkblue}{a_{32}}}\\right]}_{\\bf a_2} +z\\underbrace{\\left[\\matrix{\\color{maroon}{a_{13}}\\cr\\color{darkgreen}{ a_{23}}\\cr\\color{darkblue}{ a_{33}}}\\right]}_{\\bf a_3} =\\bf0$$\n\nNow on to row operations:\n\nHere, we make our third and final observation:\n\nIf one performs a row operation on $A$, then the corresponding right hand side of $(1)$ is obtained by performing the same row operation to each of ${\\bf a_1}$, ${\\bf a_2}$, and ${\\bf a_3}$.\n\nIt follows from all of this that multiplying a row of $A$ by a non-zero number does not affect the null space. For example if row 1 of $A$ were multiplied by 2 then the right hand side of (1) would be: $$x\\underbrace{\\left[\\matrix{\\color{maroon}{2a_{11}}\\cr\\color{darkgreen}{ a_{21}}\\cr \\color{darkblue}{a_{31}}}\\right]}_{\\bf a_1} +y\\underbrace{\\left[\\matrix{\\color{maroon}{2a_{12}}\\cr\\color{darkgreen}{ a_{22}}\\cr \\color{darkblue}{a_{32}}}\\right]}_{\\bf a_2} +z\\underbrace{\\left[\\matrix{\\color{maroon}{2a_{13}}\\cr\\color{darkgreen}{ a_{23}}\\cr\\color{darkblue}{ a_{33}}}\\right]}_{\\bf a_3} =\\bf z$$ If (2) holds, it is easily seen that ${\\bf z}=\\bf0$.\n\nIt should be fairly obvious that interchanging two rows of $A$ does not affect the null space. For example if rows 1 and 3 of $A$ were interchanged, the right hand side of (1) would become $$x\\left[\\matrix{\\color{darkblue}{ a_{31}}\\cr \\color{darkgreen}{ a_{21}}\\cr \\color{maroon}{ a_{11}}}\\right] +y\\left[\\matrix{\\color{darkblue}{ a_{32}}\\cr \\color{darkgreen}{ a_{22}}\\cr \\color{maroon}{ a_{12}}}\\right] +z\\left[\\matrix{\\color{darkblue}{ a_{33}}\\cr\\color{darkgreen}{ a_{23}}\\cr\\color{maroon}{ a_{13}}}\\right] =\\bf z$$ If (2) holds, then $\\bf z$ would be $\\bf 0$.\n\nFinally, if a multiple of a row of $A$ were added to another row of $A$, the null space would be unchanged. For example if twice row 1 of $A$ were added to to row 2, then the right hand side of $(1)$ would become:\n\n$$\\tag{3} x\\left[\\matrix{\\color{maroon}{ a_{11}}\\cr \\color{darkgreen}{ a_{21}}+2\\color{maroon}{ a_{11}}\\cr\\color{darkblue} a_{31}}\\right] +y\\left[\\matrix{\\color{maroon}{ a_{12}}\\cr\\color{darkgreen}{ a_{22}}+2 \\color{maroon}{a_{12}}\\cr\\color{darkblue} a_{32}}\\right] +z\\left[\\matrix{\\color{maroon}{ a_{13}}\\cr\\color{darkgreen}{ a_{23}}+2 \\color{maroon}{a_{13}}\\cr\\color{darkblue} a_{33}}\\right] =\\bf z$$\n\nIf $(2)$ holds, then $\\bf z=0$. In particular, looking at the second component of the left hand side of $(3)$: \\eqalign{ x(\\color{darkgreen}{a_{21}}+2\\color{maroon}{a_{11}})+ y(\\color{darkgreen}{a_{22}}+2\\color{maroon}{a_{12}})+ z(\\color{darkgreen}{a_{23}}+2\\color{maroon}{a_{13}}) &= (x \\color{darkgreen}{a_{21}}+ y \\color{darkgreen}{a_{22}}+ z \\color{darkgreen}{a_{23}} )\\cr &\\ \\ \\ \\ \\ \\ \\ +2( x\\color{maroon}{a_{11}} + y\\color{maroon}{a_{12}}+ z\\color{maroon}{a_{13}} )\\cr&=0+2\\cdot0\\cr&=0}.\n\n-\n\nI think the most natural way to think about this is to note that the kernel of a matrix is also the orthogonal complement to the column-space of it's transpose. (this is the fundamental theorem of linear algebra)\n\nSo when you do row operations on a matrix, this is the same as doing linear combinations of the columns in it's transpose. This doesn't change the overall space spanned by those columns in the transpose, so it doesn't change the kernel of the original matrix.\n\n-", "date": "2014-09-24 03:03:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/108041/linear-algebra-preserving-the-null-space", "openwebmath_score": 0.949213981628418, "openwebmath_perplexity": 79.41043329591068, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713844203499, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8577745775712854}}
{"url": "http://math.stackexchange.com/questions/278784/is-it-more-practice-and-intuition-or-rather-algorithmic-to-solve-third-degree-po/278801", "text": "# Is it more practice and intuition or rather algorithmic to solve third degree polynomials of this type?\n\nConsider\n\n$x^3 - 6x^2 + 11x - 6 = 0$\n\nI can not reasonably factor this intuitively in any short amount of time with my skill level. Is this the only hope to solving such equations by hand? What tools can I use to help ease the pain or make it more clear to myself?\n\nI try to:\n\n$(x + ?)(x^2 - 6x)$ but it does not seem promising, in my eyes\n\nI cant do anything, that I know about, with:\n\n$x(x^2 - 6x + 11) = 6$\n\nI don't want to give up, but I do anyway and ask my calculator, there is three solutions, alas I have no idea how that helps me (with my knowledge, it does not)!\n\nShould I just practice with smaller polynomials more, and this one might look easier?\n\nAny suggestions are appreciated, I will tag this as homework because it should be.\n\n-\nNote that trying to factorize anything as $x(x^2-6x+11)=6$ is almost useless, since $6$ need not factor nicely. The rare instance when such a factorization is useful, is for example $(x+1)(x+2)(x+3)=6$, in which case we can observe that $x=0$ is a root. \u2013\u00a0 Calvin Lin Jan 14 '13 at 20:21\nDo you mean $6x^2$ instead of $6x$? \u2013\u00a0 Git Gud Jan 14 '13 at 20:22\nYes, sorry forgot to square that one. \u2013\u00a0 Leonardo Jan 14 '13 at 20:23\nYou should also know that there are closed formulas to get the roots of polynomials of degree up to 4. \u2013\u00a0 Git Gud Jan 14 '13 at 20:28\n\nIf you have a polynomial with integer coefficients of the form $$p(x) = x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \\cdots + a_1 x + a_0$$ and if you have any hope of factorizing this \"nicely\", you first need to check if the divisors of the constant term, $a_0$, are factors. In your case, I assume the correct polynomial you are interested in is $p(x) = x^3 - 6x^2+11x-6$. The divisors of $-6$ are $\\{\\pm1,\\pm2, \\pm3, \\pm6\\}$. A quick check reveals that $x^*=1,2,3$ satisfy $p(x^*) = 0$. Hence, we have $$x^3 - 6x^2 + 11x - 6 =(x-1)(x-2)(x-3)$$\n\n-\nLuckily this is a nice equation, thank you I have at least one extra tool in my belt and I will need to read up more on the rational root theorem among other things. Your answer is very clear and concise. \u2013\u00a0 Leonardo Jan 14 '13 at 20:27\n\nYour equation is $$x^3-6x^2+11x-6=0$$ The coefficients of the equation are $1,-6,+11,-6$ and the summation of them is $$1-6+11-6=0$$ Whenever you face to this fact, you always have $x=1$ as one solution. In fact the equation has a factor as form $x-1$. It means that $$x^3-6x^2+11x-6=(x-1)(ax^2+bx+c)$$ for some proper available constants $a,b,c$.\n\n-\nI kind of get what you and Will Jagy are saying, but it will take time to absorb. The only thing that is immediately obvious from your last equations is that c = 6. I appreciate your helps. \u2013\u00a0 Leonardo Jan 14 '13 at 20:22\n@Leonardo: Look at Marvis's. I noted some points just for starting. He gave you the answer in detailed. \u2013\u00a0 Babak S. Jan 14 '13 at 20:25\n\n$x=1$ is an evident solution. Therefore we can write $$x^3 - 6 x^2 + 11 x - 6 = (x-1)(a x^2 + b x + c)$$ and solve for $a,b,c$ carefully. Then the rest is the quadratic formula.\n\nI suppose the main advice is to check for integer roots and rational roots before factoring, when you have no feeling for what is going on in the problem. There is a Rational Roots Theorem that applies here.\n\n-\n\n$$x^3 - 6x^2 + 11x - 6 =0$$ $$x^3 -1-( 6x^2 - 11x + 6)+1 =0$$ $$x^3 -1-( 6x^2 - 11x + 5) =0$$ $$x^3 -1-( 6x^2 - 6x-5x + 5) =0$$ $$(x-1)(x^2+x+1)-( 6x(x-1)-5(x-1) =0$$ $$(x-1)(x^2+x+1)-(x-1)(6x-5) =0$$ $$(x-1)(x^2+x+1-6x+5) =0$$ $$(x-1)(x^2-5x+6) =0$$ $$(x-1)(x^2-2x-3x+6=0$$ $$(x-1)(x(x-2)-3(x-2)=0$$ $$(x-1)(x-2)(x-3)=0$$ $$x_1=1,x_2=2,x_3=3$$\n\n-\nNice Adi,nice. ;-) \u2013\u00a0 Babak S. Jan 14 '13 at 20:26\nMy brain does not work like yours, but I hope it starts to soon. \u2013\u00a0 Leonardo Jan 14 '13 at 20:29\nthank you Babak \u2013\u00a0 Adi Dani Jan 14 '13 at 23:31\n\nYou need to know a few things. For example if $p(x)$ is a polynomial of odd degree with rational coefficients it will necessarily have a real root.\n\nAnother useful fact is that if $a$ is a root of $p(x)=x^3+px^2+qx+r=0$ then $(x-a)$ is a factor of $p(x)$. This works because: $x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^2-a^2=(x-a)(x+a)$ so if $p(a)=0$ we can write:$$p(x)=p(x)-p(a)=[x^3-a^3]+p[x^2-a^2]+q[x-a]=(x-a)([x^2+ax+a^2]+p[x+a]+q)$$\n\nWe also note that if $p,q,r,a \\in \\mathbb Z$, then if $p(a)=a(a^2+pa+q)+r=0$ then $a$ must be a factor of $r$.\n\n[There are various generalisations and extensions of these comments].\n\nSo one thing we do when faced with a cubic with integer coefficients to factorise is to try the factors ($\\pm$) of the constant term as \"easy candidates\" - and once we have one factor we can divide through and solve the remaining quadratic.\n\n-", "date": "2015-10-07 03:06:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/278784/is-it-more-practice-and-intuition-or-rather-algorithmic-to-solve-third-degree-po/278801", "openwebmath_score": 0.9041504859924316, "openwebmath_perplexity": 216.913086642234, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713891776498, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8577745767509637}}
{"url": "https://cs.stackexchange.com/questions/117577/which-grows-faster-factorial-or-double-exponentiation/117583", "text": "# Which Grows Faster: Factorial or Double Exponentiation\n\nWhich of the functions among $$2^{3^n}$$ or $$n!$$ grows faster?\n\nI know that $$n^n$$ grows faster than $$n!$$ and $$n!$$ grows faster than $$c^n$$ where $$c$$ is a constant, but what is it in my case?\n\n\u2022 It isn't a proof, but evaluate the expressions with $n$ in the range 1 to 10 and see just how phenomenally fast $2^{3^n}$ grows compared to $n!$ \u2013\u00a0John Coleman Nov 26 '19 at 0:20\n\u2022 Write down the values for 1 \u2264 n \u2264 10. You can't? Why can't you? \u2013\u00a0gnasher729 Nov 26 '19 at 13:18\n\u2022 Why is this in CS instead of Mathematics? \u2013\u00a0Barmar Nov 26 '19 at 21:08\n\u2022 @Barmar because it's related to CS in terms of time complexity. Indeed, in this question, there is not a hard border between math and CS. \u2013\u00a0OmG Nov 27 '19 at 15:14\n\u2022 \u2013\u00a0Will Ness Nov 27 '19 at 16:36\n\nYou can find the result by taking a $$\\log$$. Hence:\n\n$$\\log(2^{3^n}) = 3^n$$ $$\\log(n!) \\leqslant \\log(n^n) = n\\log n$$\n\n(In the latter equation, we have used the fact that $$n! \\leqslant n^n$$, as you note in the question.)\n\nOf course $$3^n$$ grows faster than $$n \\log n$$. As $$\\log$$ is an increasing function, we can say $$2^{3^{n}}$$ grows faster than $$n^n$$, and also $$n!$$.\n\n\u2022 Beware, however, that $\\log(f) = O(\\log(g))$ does not imply $f = O(g)$, in general. For instance, take $f(n)=n^2$ and $g(n)=n$. \u2013\u00a0chi Nov 26 '19 at 19:07\n\u2022 @chi You are right. But, it is true if $\\log(f) = o(\\log(g))$. \u2013\u00a0OmG Nov 26 '19 at 19:56\n\u2022 Another question related with taking log and solving and its danger is given in this link--> cs.stackexchange.com/questions/117584/\u2026 \u2013\u00a0Turing101 Nov 28 '19 at 15:19\n\nAnother way to directly compare the two expressions is to take the ratio of consecutive terms:\n\n$$\\frac{2^{3^{n+1}}}{2^{3^n}}=2^{2\\cdot 3^n}\\gg 3^n\\gg n+1=\\frac{(n+1)!}{n!}$$\n\n(for positive integers $$n$$), and clearly also $$2^{3^1}=8>1!$$, so $$2^{3^n}$$ indeed grows more rapidly than $$n!$$.\n\n\u2022 The discussion of $2^{3^1}$ is unnecessary. \u2013\u00a0leftaroundabout Nov 27 '19 at 1:56\n\u2022 @leftaroundabout, I guess that's true if you assume the >> signs are significantly greater than, but that's not very rigorous (and if they are only > signs it would then depend on your definition of \"grows more rapidly\") \u2013\u00a0boboquack Nov 27 '19 at 7:43\n\u2022 Well, they only need to be \u201csignificantly\u201d greater in a very weak sense, namely the ratio needs to be $>1+\\varepsilon$. That is clearly given in this case, specifically $\\frac{3^n}{n+1} > 1.5\\quad \\forall n>0$. \u2013\u00a0leftaroundabout Nov 27 '19 at 12:05\n\nYou noted in your question that $$n^n$$ grows faster than $$n!$$, and that\u2019s a great starting point for comparing the growth of $$2^{3^n}$$ and $$n!$$. Specifically, let\u2019s ask - of $$n^n$$ and $$2^{3^n}$$, which grows faster?\n\nTo answer that, let\u2019s try rewriting $$n^n$$ so that it has the same exponential base as $$2^{3^n}$$. Since $$n = 2^{\\log_2 n}$$, we have that\n\n$$n^n = (2^{\\log_2 n})^n = 2^{n \\log_2 n}.$$\n\nNow, is it easier to see how $$n^n$$ and $$2^{3^n}$$ relate?\n\nAs a note, this approach is similar to taking the base-2 logs of both expressions. I thought it would be good to include this here because it gives a slightly different perspective on how to arrive at the answer given your initial observation.\n\nInduction.\n\nBase case: $$n=1$$ gives $$2^{3^1}=8$$ and $$1!=1$$ which clearly holds.\n\nHypothesis: suppose that $$2^{3^k}>k!$$ for some $$k\\in\\Bbb N$$.\n\nConsider $$n=k+1$$. Then $$2^{3^{k+1}}=2^{3^k\\cdot3}>(k!)^3=(k+1)!\\cdot\\frac{k!(k-1)!}{1+\\frac1k}>(k+1)!$$ which is true $$\\forall k>1$$ and thus the result follows.\n\n\u2022 I tend to interpret \"which grows faster?\" questions as \"whose big-Oh set is bigger?\". With that interpretation, this post answers something different. (E.g. $x \\mapsto x$ and $x \\mapsto (x+1)$ are in the same big-Oh set, but in the spirit of this post the latter grows faster.) \u2013\u00a0ComFreek Nov 26 '19 at 12:02\n\u2022 @ComFreek I would really have to question that interpretation. Which grows faster, $f(x) = x$ or $g(x) = 2x$? I think by any reasonable meaning of the words \"grow\" and \"faster\", $2x$ grows faster than $x$. My definition of grows faster would be $f(x)$ grows faster than $g(x)$ if there exists an $x_0$ beyond which $f'(x)$ is always greater than $g'(x)$. \u2013\u00a0Apollys supports Monica Nov 27 '19 at 23:36\n\u2022 That being said, this proof certainly doesn't prove anything about growth rates. Plug in the functions $f(x) = -1/x$ and $g(x) = 0$ (on the domain $x > 0$) into your proof and you conclude that $g(x)$ grows faster than $f(x)$. This is obviously wrong since $g(x)$ never grows and $f(x)$ never doesn't grow! \u2013\u00a0Apollys supports Monica Nov 27 '19 at 23:42", "date": "2020-02-17 02:11:20", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/117577/which-grows-faster-factorial-or-double-exponentiation/117583", "openwebmath_score": 0.9391917586326599, "openwebmath_perplexity": 325.12279085943976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853138, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8577745717083018}}
{"url": "http://math.stackexchange.com/questions/183044/bounding-an-expression-involving-digamma-function", "text": "# Bounding an expression involving digamma function\n\nLet $\\psi$ be the digamma function. I have a conjecture that\n\n$$\\frac ax > \\log(x) - \\psi(x)$$\n\nholds for all $x > 0$ if (and only if) $a \\ge 1$. I do not know how to prove it. Please help.\n\n-\n\nLet\n\n$$f(x) = \\frac{1}{2}\\coth\\left(\\frac{x}{2}\\right) - \\frac{1}{x}.$$\n\nThen for $x > 0$, we have\n\n$$f'(x) = \\frac{1}{x^2} - \\frac{1}{4\\sinh^2(x/2)} = \\frac{1}{x^2}\\left(1 - \\left(\\frac{x/2}{\\sinh (x/2)} \\right)^2 \\right) > 0.$$\n\nThis shows that $f(x)$ is increasing. Also, it is easy to find that $f(0+) = 0$ and $\\lim_{x\\to\\infty} f(x) = \\frac{1}{2}$. Thus we find that $0 < f(x) < \\frac{1}{2}$ for $0 < x < \\infty$ and hence the Laplace transform $\\mathcal{L}f(s) = \\int_{0}^{\\infty} f(x) \\, e^{-sx} \\; dx$ of $f(x)$ satisfies\n\n$$0 < \\mathcal{L}f(s) < \\int_{0}^{\\infty} \\frac{1}{2} \\, e^{-sx} \\; dx = \\frac{1}{2s}.$$\n\nBut we also have\n\n\\begin{align*} \\mathcal{L}f(s) &= \\int_{0}^{\\infty} \\left( \\frac{1}{2}\\coth\\left(\\frac{x}{2}\\right) - \\frac{1}{x} \\right) e^{-sx} \\; dx\\\\ &= \\left[ \\left( \\log \\sinh \\left(\\frac{x}{2} \\right) - \\log x + \\log 2 \\right) e^{-sx} \\right]_{0}^{\\infty} \\\\ &\\qquad + s \\int_{0}^{\\infty} \\left( \\log \\sinh \\left(\\frac{x}{2} \\right) - \\log x + \\log 2 \\right) e^{-sx} \\; dx \\\\ &= s \\int_{0}^{\\infty} \\left( \\frac{x}{2} - \\log (sx) + \\log s + \\log \\left(1 - e^{-x} \\right) \\right) e^{-sx} \\; dx \\\\ &= \\frac{1}{2s} + \\gamma + \\log s - s \\int_{0}^{\\infty} \\sum_{n=1}^{\\infty} \\frac{e^{-(n+s)x}}{n} \\; dx \\\\ &= \\frac{1}{2s} + \\gamma + \\log s - \\sum_{n=1}^{\\infty} \\frac{s}{n(n+s)} \\\\ &= \\frac{1}{2s} + \\log s - \\psi_0(1+s), \\end{align*}\n\nwhere we have used the fact that $\\int_{0}^{\\infty} e^{-x} \\log x \\; dx = -\\gamma$ and\n\n$$\\psi_0(1+x) = -\\gamma + \\sum_{n=1}^{\\infty} \\frac{x}{n(n+x)}.$$\n\nFinally, since $\\psi_0(1+s) = \\frac{1}{s} + \\psi_0 (s)$, we have\n\n$$\\frac{1}{2s} < \\log s - \\psi_0(s) < \\frac{1}{s}.$$\n\nTherefore the inequality holds if $a \\geq 1$ and only if $a > \\frac{1}{2}$.\n\nNow we prove that $a = 1$ is the sharp bound. We are to calculate the limit\n\n$$\\alpha = \\lim_{x\\to 0} x(\\log x - \\psi_0(x)).$$\n\nBut by our previous calculations,\n\n\\begin{align*}\\alpha &= \\lim_{s\\to 0} s(\\log s - \\psi_0(s)) \\\\ &= \\lim_{s\\to 0} s\\left( \\frac{1}{2s} + \\mathcal{L}f(s) \\right) \\\\ &= \\frac{1}{2} + \\lim_{s\\to 0} s \\int_{0}^{\\infty} f(x) \\, e^{-sx} \\; dx \\\\ &= \\frac{1}{2} + \\lim_{s\\to 0} \\int_{0}^{\\infty} f(u/s) \\, e^{-u} \\; du \\qquad (u = sx) \\\\ &= \\frac{1}{2} + \\int_{0}^{\\infty} \\lim_{s\\to 0} f(u/s) \\, e^{-u} \\; du \\\\ &= \\frac{1}{2} + \\int_{0}^{\\infty} \\frac{1}{2} \\, e^{-u} \\; du \\\\ &= 1. \\end{align*}\n\nThus if $\\frac{a}{x} > \\log x - \\psi_0 (x)$ is true for all $x > 0$, then we must have $a \\geq \\alpha = 1$, as desired.\n\n-\nThis is a cool technique. (There's a typo after you switch summation and integral.) \u2013\u00a0 Tunococ Aug 16 '12 at 18:25\n@Tunococ, Thanks. Also I fixed the typo. Even I cannot believe how I made such a mistake! :) \u2013\u00a0 sos440 Aug 17 '12 at 3:36\n\nRecently I happened to find out that Qi et al (J. Math. Anal. Appl. 310 (2005) 303\u2013308) proved the following inequality, among others, for digamma function $\\psi(x)$ when $x>0$:\n\n$$\\frac{1}{2x} < \\log x - \\psi(x) < \\frac{1}{2x} + \\frac{1}{12x^2}$$\n\nThis inequality is sharper than the one given by @sos440 above when x>1/6.\n\n-", "date": "2014-12-19 15:53:36", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/183044/bounding-an-expression-involving-digamma-function", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 566.0459524977726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713844203499, "lm_q2_score": 0.8705972616934408, "lm_q1q2_score": 0.8577745693012621}}
{"url": "https://gmatclub.com/forum/each-year-for-4-years-a-farmer-increased-the-number-of-trees-in-a-135487.html", "text": "It is currently 21 Mar 2018, 02:12\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Each year for 4 years, a farmer increased the number of trees in a\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSenior Manager\nJoined: 07 Sep 2010\nPosts: 323\nEach year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n08 Jul 2012, 05:04\n10\nKUDOS\n85\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n72% (02:18) correct 28% (04:49) wrong based on 1977 sessions\n\n### HideShow timer Statistics\n\nEach year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.\n\nA. 1250\nB. 1563\nC. 2250\nD. 2560\nE. 2752\n\n[Reveal] Spoiler:\nCan someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\". then it would have been difficult to solve.\n\nThanks\n[Reveal] Spoiler: OA\n\n_________________\n\n+1 Kudos me, Help me unlocking GMAT Club Tests\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44373\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n08 Jul 2012, 05:35\n21\nKUDOS\nExpert's post\n32\nThis post was\nBOOKMARKED\nimhimanshu wrote:\nEach year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.\n\nA. 1250\nB. 1563\nC. 2250\nD. 2560\nE. 2752\n\nCan someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\". then it would have been difficult to solve.\n\nThanks\n\nSay the number of trees at the beginning of the 4 year period was x, then:\nAt the end of the 1st year the number of trees would be $$x+\\frac{1}{4}x=\\frac{5}{4}*x$$;\nAt the end of the 2nd year the number of trees would be $$(\\frac{5}{4})^2*x$$;\nAt the end of the 3rd year the number of trees would be $$(\\frac{5}{4})^3*x$$;\nAt the end of the 4th year the number of trees would be $$(\\frac{5}{4})^4*x$$;\nAt the end of the $$n_{th}$$ year the number of trees would be $$(\\frac{5}{4})^n*x$$;\n\nSo, we have that $$(\\frac{5}{4})^4*x=6,250$$ --> $$\\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$.\n\nIf the question were \"if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\", then we would have that: $$(\\frac{5}{4})^{15}*x=6,250$$ --> $$x\\neq{integer}$$, so it would be a flawed question.\n\nHope it's clear.\n_________________\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7996\nLocation: Pune, India\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n08 Jul 2012, 22:32\n12\nKUDOS\nExpert's post\n6\nThis post was\nBOOKMARKED\nimhimanshu wrote:\nEach year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.\n\nA. 1250\nB. 1563\nC. 2250\nD. 2560\nE. 2752\n\nCan someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\". then it would have been difficult to solve.\n\nThanks\n\nThe number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.\n\nIf x increases by 25%, how we denote it? (5/4)*x\nIf next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x\nand so on...\n\nFor more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/\n\nSo if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250\n\nAs for your next question, the numbers given would be such that the calculation will not be tough.\n\nSay, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.\nSomething like 2^8 * x = 2560\n\nand if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\", note that you still need to work with\n\n(5/4)^4 * x = 6250\nsince you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1838 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2013, 00:26 4 This post received KUDOS 2 This post was BOOKMARKED Trees increase 1/4th every year, which means 100 trees become 125 after 1 yr & so on So, 125/100 = 5/4 is the resultant (after adding 1/4 as the growth) for 1 year So, for 4 yrs is 5^4/(4^4) From the condition given, inital trees were = 6250 x 4^4 / 5^4 = 2560 _________________ Kindly press \"+1 Kudos\" to appreciate Manager Joined: 12 Jan 2013 Posts: 212 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 27 Dec 2013, 09:31 Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\\frac{1}{4}x=\\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\\frac{5}{4})^n*x$$; So, we have that $$(\\frac{5}{4})^4*x=6,250$$ --> $$\\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were \"if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\", then we would have that: $$(\\frac{5}{4})^{15}*x=6,250$$ --> $$x\\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as \"for the first year we have (4/4)x and for the second year (5/4)x and for the third...\" etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward \"word translation\" way in knowing how to interpret wordings like this? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 01 Jan 2014, 23:34 Expert's post 2 This post was BOOKMARKED aeglorre wrote: Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\\frac{1}{4}x=\\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\\frac{5}{4})^n*x$$; So, we have that $$(\\frac{5}{4})^4*x=6,250$$ --> $$\\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were \"if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\", then we would have that: $$(\\frac{5}{4})^{15}*x=6,250$$ --> $$x\\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as \"for the first year we have (4/4)x and for the second year (5/4)x and for the third...\" etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward \"word translation\" way in knowing how to interpret wordings like this? Actually, it is not ambiguous. Read the statement: Each year a farmer increased the number of trees by 1/4. He did this for 4 years. (In GMAT Verbal and Quant are integrated. You need Verbal skills (slash and burn) in Quant and Quant skills (Data Interpretation) in Verbal. So in the first year, he increased it by 1/4 The next year, he again increased it by 1/4 (of preceding year) Next year, again the same. Next year, again the same. So he did it for a total of 4 years. So if initially the number of trees was x, in the first year he made them (5/4)x _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nManager\nJoined: 23 May 2013\nPosts: 119\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n07 Feb 2014, 18:28\n4\nKUDOS\n4\nThis post was\nBOOKMARKED\nit can easily solved by compound interest formula\nR = 25%\nP = original number of trees\nN = no of times of compounding within a year = 1\nT = total number of years = 4\n\nP+I at the end of compounding years = 6250\nhence,\n\n6250 = (1 + 0.25/1)^(4*1) * P\n5^4*10 = 1.25^4 * P\n(5/1/25)^4 *10 = P\n4^4 * 10 = P\n2560 = P\n\nAnother way to think is to pick ans choice which is exactly divisible by 4. A, B and C are not. Only D and E are.\nPick D and increase it by 25% for 4 times, you should get 6250.\n_________________\n\n\u201cConfidence comes not from always being right but from not fearing to be wrong.\u201d\n\nManager\nStatus: folding sleeves up\nJoined: 26 Apr 2013\nPosts: 152\nLocation: India\nConcentration: Finance, Strategy\nGMAT 1: 530 Q39 V23\nGMAT 2: 560 Q42 V26\nGPA: 3.5\nWE: Consulting (Computer Hardware)\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n27 Jul 2014, 10:08\nHi Karishma,\n\nI am not able to solve it by below method, where am I wrong:\n\n6250*(3/4)*(3/4)*(3/4)\n\nRegards,\nRavi\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7996\nLocation: Pune, India\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n27 Jul 2014, 22:08\n2\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nemail2vm wrote:\nHi Karishma,\n\nI am not able to solve it by below method, where am I wrong:\n\n6250*(3/4)*(3/4)*(3/4)\n\nRegards,\nRavi\n\nIf I increase A by 25% and get B, I will not get A if I reduce B by 25%. The bases are different in the two cases.\n\ne.g. A = 80\n25% of A is 20 so I increase A by 25% to get B = 100\n\nNow if I decrease B by 25%, I will decrease B by 25 (25% of 100). This will give me 75 which is not the same as A (which is 80).\n\nIn the first step, I found 25% of 80 and added that. In the second step, I found 25% of 100 and subtracted that. These two numbers are different.\n\nYou will get the correct answer if you do 6250*(4/5)*(4/5)*(4/5)*(4/5)\n4/5 is the inverse of 5/4 (which is 25% increase in x).\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 13 Feb 2014 Posts: 3 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2014, 12:01 You can use the formula A=p(1 + r/100)^n with r = 25% You will get P as 2560. Intern Joined: 23 Aug 2014 Posts: 42 GMAT Date: 11-29-2014 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 30 Dec 2014, 10:59 Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 31 Dec 2014, 02:53 deeuk wrote: Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Yes, that's correct. But you might still need to work with two options. On the other hand, working with 6250 will definitely yield the answer. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nManager\nJoined: 03 Jan 2015\nPosts: 67\nConcentration: Strategy, Marketing\nWE: Research (Pharmaceuticals and Biotech)\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n06 Jan 2015, 10:39\n1\nKUDOS\nTrees increase by 1/4 the number of trees in preceding year. Hence, correct answer must be divisible by 4. Based on divisibility rules, if last 2 digits are divisible by 4 then the number is divisible by 4. Thus, we can eliminate A, B, C. The answer has to be D or E.\n\nAgain, trees increase by 1/4 the number of trees in preceding year. Hence, the number of trees increase by 5/4 times the number of trees the preceding year.\n\nIf x = initial number of trees = Answer D or E = 2560 or 2752\nYear 1 = 5/4x\nYear 2 = (5/4)(5/4)x\nYear 3 = (5/4)(5/4)(5/4)x\nYear 4 = (5/4)(5/4)(5/4)(5/4)x\n\nOnly for Answer D: (5/4)(5/4)(5/4)(5/4) 2560 = 6250\n\nIntern\nJoined: 22 Nov 2012\nPosts: 22\nLocation: United States\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n02 Apr 2015, 03:36\n@VeritasPrepKarishma\nVeritasPrepKarishma wrote:\nimhimanshu wrote:\nEach year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.\n\nA. 1250\nB. 1563\nC. 2250\nD. 2560\nE. 2752\n\nCan someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\". then it would have been difficult to solve.\n\nThanks\n\nThe number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.\n\nIf x increases by 25%, how we denote it? (5/4)*x\nIf next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x\nand so on...\n\nFor more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/\n\nSo if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250\n\nAs for your next question, the numbers given would be such that the calculation will not be tough.\n\nSay, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.\nSomething like 2^8 * x = 2560\n\nand if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) \"If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period\", note that you still need to work with\n\n(5/4)^4 * x = 6250\nsince you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.\n\nKarisma,\n\nHowever, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.\n\nWhy is it incorrect to compute the increase of the successive year on the increased base of the previous year?\n\nThank you a lot for your help!\n\nRegards,\nEli.\n_________________\n\nGMAT,\nIt is not finished untill I win!!!\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7996\nLocation: Pune, India\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n02 Apr 2015, 19:43\nelisabettaportioli wrote:\nHowever, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.\n\nWhy is it incorrect to compute the increase of the successive year on the increased base of the previous year?\n\nThank you a lot for your help!\n\nRegards,\nEli.\n\nThe base for each year changes.\n\nSay, you have 100 trees and you increase them by 20% each year.\n\nAt the end of the first year, you will have 100 + (20/100)*100 = 100*(1 + 20/100) = 100*120/100 = 100*(6/5) = 120 trees.\n\nSo if you have to increase a number by 20%, you just need to multiply it by 6/5 every time no matter what the number is.\n\nBy the same logic, next year, you will have 120 * (6/5) = 144 trees when you increase them by 20%.\n\nYou can write 120 as 100 * (6/5) and then multiply it by another 6/5 to increase 120 by 20%.\n\nThis is the same as 100 * (6/5) * (6/5) = 144 trees.\n\nSo every time you multiply it by 6/5, you increase the base. The next time you multiply it by 6/5, you are increasing the new base by 20%. This is the concept of successive percentage changes and I have discussed it in the link I mentioned in my post above.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for \\$199\n\nVeritas Prep Reviews\n\nMath Expert\nJoined: 02 Aug 2009\nPosts: 5724\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n02 Apr 2015, 20:15\nelisabettaportioli wrote:\n\nWhy is it incorrect to compute the increase of the successive year on the increased base of the previous year?\n\nThank you a lot for your help!\n\nRegards,\nEli.\n\nhi,\nyou are correct that the increase is on the preceeding year and that is exactly what is happening when you are multiplying x by 5/4..\nfirst year say it was x.. 2nd year it becomes 5/4*x.. now for next year when we write the value as 5/4*5/4*x, the increase is on 5/4*x and not x..\n_________________\n\nAbsolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\nCombination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n\nGMAT online Tutor\n\nDirector\nStatus: No dream is too large, no dreamer is too small\nJoined: 14 Jul 2010\nPosts: 577\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n13 Apr 2015, 08:38\n1\nKUDOS\n1/4 = 0.25 = 25%\nusing the formula of compound interest calculation\np(1+r)^n= A\n\np = principal amount (to be identified), r = rate (0.25), n = numbers years (4), A= amount at the end of term(6250 given)\n\np(1+0.25)^4=6250\np(1.25)^4= 25*25*10 = 5^2*5^2*10=5^4*10\np(125/100)^4 = 5^4 x 10\np(5/4)^4 = 5^4 x 10\np(5^4)/4^4 = 5^4 x 10\np= (5^4 x 10 x 4^4)/5^4\np= 10 x 4^4\np= 2560\nAns D\n_________________\n\nCollections:-\nPSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html\nDS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html\n100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html\nCollections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html\nMixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html\n\nDirector\nJoined: 10 Mar 2013\nPosts: 581\nLocation: Germany\nConcentration: Finance, Entrepreneurship\nGMAT 1: 580 Q46 V24\nGPA: 3.88\nWE: Information Technology (Consulting)\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n19 May 2015, 12:37\nwe have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560\n\nSee MGMAT (Percents) for detailed explanation of such question types.....\n_________________\n\nWhen you\u2019re up, your friends know who you are. When you\u2019re down, you know who your friends are.\n\n800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50\nGMAT PREP 670\nMGMAT CAT 630\nKAPLAN CAT 660\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 878\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n20 May 2015, 03:32\nBrainLab wrote:\nwe have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560\n\nSee MGMAT (Percents) for detailed explanation of such question types.....\n\nDear BrainLab\n\nPerfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation\n\n$$(\\frac{5}{4})^4*X = 6250$$\n\nwill take lesser time to solve (especially if you know that $$5^4 = 625$$) than $$(1.25)^4*X = 6250$$\n\nHope this was useful!\n\nJapinder\n_________________\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44373\nRe: Each year for 4 years, a farmer increased the number of trees in a\u00a0[#permalink]\n\n### Show Tags\n\n18 Jan 2016, 23:36\nExpert's post\n1\nThis post was\nBOOKMARKED\nRe: Each year for 4 years, a farmer increased the number of trees in a \u00a0 [#permalink] 18 Jan 2016, 23:36\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 37 posts ]\n\nDisplay posts from previous: Sort by", "date": "2018-03-21 09:12:46", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/each-year-for-4-years-a-farmer-increased-the-number-of-trees-in-a-135487.html", "openwebmath_score": 0.570276141166687, "openwebmath_perplexity": 1384.2064824380445, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8577681049901036, "lm_q1q2_score": 0.8577681049901036}}
{"url": "https://gmatclub.com/forum/in-the-rectangular-coordinate-system-above-if-the-equation-of-m-is-y-255480.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 21 Jul 2018, 01:06\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# In the rectangular coordinate system above, if the equation of m is y\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47161\nIn the rectangular coordinate system above, if the equation of m is y\u00a0 [#permalink]\n\n### Show Tags\n\n15 Dec 2017, 02:09\n1\n4\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n52% (01:24) correct 48% (00:35) wrong based on 119 sessions\n\n### HideShow timer Statistics\n\nIn the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?\n\n(A) \u221a2\n(B) 1\n(C) \u221a2/2\n(D) 1/2\n(E) 1/4\n\nAttachment:\n\n2017-12-15_1259.png [ 4.85 KiB | Viewed 1084 times ]\n\n_________________\nSC Moderator\nJoined: 22 May 2016\nPosts: 1833\nIn the rectangular coordinate system above, if the equation of m is y\u00a0 [#permalink]\n\n### Show Tags\n\n15 Dec 2017, 13:22\n2\n1\nBunuel wrote:\n\nIn the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?\n\n(A) \u221a2\n(B) 1\n(C) \u221a2/2\n(D) 1/2\n(E) 1/4\n\nAttachment:\nThe attachment 2017-12-15_1259.png is no longer available\n\nWith the formula (My diagram does not apply)\nI did not use it, but to do so is standard. So I quote the formula from Bunuel HERE\n\nDistance between two parallel lines $$y=mx+b$$ and $$y=mx+c$$ can be found by the formula:\n$$D=\\frac{|b-c|}{\\sqrt{m^2+1}}$$\n\nEquation for line $$m$$, $$y = x$$, to match the exact form of $$y = mx + b$$ can be written:\n$$y =(1)x + 0$$ (all lines that pass through the origin have x- and y-intercepts of 0)\n$$b = 0$$ = y-intercept, $$m = 1$$ = slope\n\nFind the equation of line $$n$$, $$y = mx + c$$\n\nParallel lines have identical slopes. Line $$n$$ has slope $$m = 1$$\nUse the point on line $$n$$ from the graph: (1, 0)\nPlug its coordinates into slope-intercept form to find $$c$$\n$$y = mx + c$$\n$$(0) = 1(1) + c$$\n$$-c = 1$$, so $$c = -1$$\nEquation for line $$n: y = x - 1$$\n\nDistance between parallel lines\n\n$$D=\\frac{|b-c|}{\\sqrt{m^2+1}}$$\n\n$$m = 1$$, $$b = 0$$, $$c = -1$$\n\n$$D=\\frac{|0-(-1)|}{\\sqrt{1^2+1}}$$\n$$D=\\frac{1}{\\sqrt{2}}$$\nDoes not match the answers. Rationalize the denominator; multiply by $$\\frac{\\sqrt2}{\\sqrt2}$$\n\n$$\\frac{1}{\\sqrt{2}}$$ * $$\\frac{\\sqrt2}{\\sqrt2} = \\frac{\\sqrt{2}}{2}$$\n\nAttachment:\n\nlinesmandn.png [ 18.05 KiB | Viewed 868 times ]\n\nWithout the formula see diagram\n\nThe shortest distance between parallel lines is a line perpendicular to both parallel lines\nDraw a perpendicular line from point B to point A\n\nThat creates right isosceles \u2206 ABO where OA = AB\n-- The line y = x makes a 45\u00b0 angle with both the x- and y-axes (as do all lines with slope 1 or -1)\n-- \u2220 BOA = 45\u00b0 and \u2220 at vertex A = 90\u00b0 therefore \u2220 ABO must = 45\u00b0\n-- Sides opposite equal angles are equal: OA = AB\n\nA right isosceles triangle\n-- has angle measures 45-45-90 and\n-- corresponding side lengths of $$x : x : x\\sqrt{2}$$\n\nThe hypotenuse corresponds with $$x\\sqrt{2} = 1$$: OB = 1\nFind length of equal sides $$x$$ (one of which, AB, is the distance needed):\n\n$$x\\sqrt{2}= 1$$\n\n$$x = \\frac{1}{\\sqrt{2}}$$\n= AB = shortest distance between the parallel lines\n\nThat does not look like any of the answers.\nRationalize the denominator (clear the radical): Multiply by $$\\frac{\\sqrt2}{\\sqrt2}$$\n$$\\frac{1}{\\sqrt{2}}$$ * $$\\frac{\\sqrt2}{\\sqrt2} = \\frac{\\sqrt{2}}{2}$$\n\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nIntern\nJoined: 20 Dec 2014\nPosts: 37\nIn the rectangular coordinate system above, if the equation of m is y\u00a0 [#permalink]\n\n### Show Tags\n\n14 Apr 2018, 06:59\ngeneris wrote:\n\nIn the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?\n\n(A) \u221a2\n(B) 1\n(C) \u221a2/2\n(D) 1/2\n(E) 1/4\n\nAttachment:\n2017-12-15_1259.png\n\nAttachment:\nlinesmandn.png\n\nWithout the formula see diagram\n\nThe shortest distance between parallel lines is a line perpendicular to both parallel lines\nDraw a perpendicular line from point B to point A\n\nThat creates right isosceles \u2206 ABO where OA = AB\n-- The line y = x makes a 45\u00b0 angle with both the x- and y-axes (as do all lines with slope 1 or -1)\n-- \u2220 BOA = 45\u00b0 and \u2220 at vertex A = 90\u00b0 therefore \u2220 ABO must = 45\u00b0\n-- Sides opposite equal angles are equal: OA = AB\n\nA right isosceles triangle\n-- has angle measures 45-45-90 and\n-- corresponding side lengths of $$x : x : x\\sqrt{2}$$\n\nThe hypotenuse corresponds with $$x\\sqrt{2} = 1$$: OB = 1\nFind length of equal sides $$x$$ (one of which, AB, is the distance needed):\n\n$$x\\sqrt{2}= 1$$\n\n$$x = \\frac{1}{\\sqrt{2}}$$\n= AB = shortest distance between the parallel lines\n\nThat does not look like any of the answers.\nRationalize the denominator (clear the radical): Multiply by $$\\frac{\\sqrt2}{\\sqrt2}$$\n$$\\frac{1}{\\sqrt{2}}$$ * $$\\frac{\\sqrt2}{\\sqrt2} = \\frac{\\sqrt{2}}{2}$$\n\nGreat explanation, Kudos to you..\nDid not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?\n\nthanks.\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nIntern\nJoined: 20 Dec 2014\nPosts: 37\nRe: In the rectangular coordinate system above, if the equation of m is y\u00a0 [#permalink]\n\n### Show Tags\n\n14 Apr 2018, 07:59\n1\nMT1988 wrote:\ngeneris wrote:\n\nIn the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?\n\n(A) \u221a2\n(B) 1\n(C) \u221a2/2\n(D) 1/2\n(E) 1/4\n\nAttachment:\n2017-12-15_1259.png\n\nAttachment:\nlinesmandn.png\n\nWithout the formula see diagram\n\nThe shortest distance between parallel lines is a line perpendicular to both parallel lines\nDraw a perpendicular line from point B to point A\n\nThat creates right isosceles \u2206 ABO where OA = AB\n-- The line y = x makes a 45\u00b0 angle with both the x- and y-axes (as do all lines with slope 1 or -1)\n-- \u2220 BOA = 45\u00b0 and \u2220 at vertex A = 90\u00b0 therefore \u2220 ABO must = 45\u00b0\n-- Sides opposite equal angles are equal: OA = AB\n\nA right isosceles triangle\n-- has angle measures 45-45-90 and\n-- corresponding side lengths of $$x : x : x\\sqrt{2}$$\n\nThe hypotenuse corresponds with $$x\\sqrt{2} = 1$$: OB = 1\nFind length of equal sides $$x$$ (one of which, AB, is the distance needed):\n\n$$x\\sqrt{2}= 1$$\n\n$$x = \\frac{1}{\\sqrt{2}}$$\n= AB = shortest distance between the parallel lines\n\nThat does not look like any of the answers.\nRationalize the denominator (clear the radical): Multiply by $$\\frac{\\sqrt2}{\\sqrt2}$$\n$$\\frac{1}{\\sqrt{2}}$$ * $$\\frac{\\sqrt2}{\\sqrt2} = \\frac{\\sqrt{2}}{2}$$\n\nGreat explanation, Kudos to you..\nDid not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?\n\nthanks.\n\nGot it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nSC Moderator\nJoined: 22 May 2016\nPosts: 1833\nRe: In the rectangular coordinate system above, if the equation of m is y\u00a0 [#permalink]\n\n### Show Tags\n\n14 Apr 2018, 16:33\n1\nQuote:\nBunuel wrote:\nIn the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?\n\n(A) \u221a2\n(B) 1\n(C) \u221a2/2\n(D) 1/2\n(E) 1/4\nAttachment:\n2017-12-15_1259.png\n\nMT1988 wrote:\nMT1988 wrote:\ngeneris wrote:\nWithout the formula see diagram\n\nThe shortest distance between parallel lines is a line perpendicular to both parallel lines\nDraw a perpendicular line from point B to point A\n\nThat creates right isosceles \u2206 ABO where OA = AB\n-- The line y = x makes a 45\u00b0 angle with both the x- and y-axes (as do all lines with slope 1 or -1)\n-- \u2220 BOA = 45\u00b0 and \u2220 at vertex A = 90\u00b0 therefore \u2220 ABO must = 45\u00b0\n-- Sides opposite equal angles are equal: OA = AB\n\nA right isosceles triangle\n-- has angle measures 45-45-90 and\n-- corresponding side lengths of $$x : x : x\\sqrt{2}$$\n\nThe hypotenuse corresponds with $$x\\sqrt{2} = 1$$: OB = 1\nFind length of equal sides $$x$$ (one of which, AB, is the distance needed):\n\n$$x\\sqrt{2}= 1$$\n\n$$x = \\frac{1}{\\sqrt{2}}$$\n= AB = shortest distance between the parallel lines\n\nThat does not look like any of the answers.\nRationalize the denominator (clear the radical): Multiply by $$\\frac{\\sqrt2}{\\sqrt2}$$\n$$\\frac{1}{\\sqrt{2}}$$ * $$\\frac{\\sqrt2}{\\sqrt2} = \\frac{\\sqrt{2}}{2}$$\n\nGreat explanation, Kudos to you..\nDid not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?\n\nthanks.\n\nGot it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.\n\nBOOM. Excellent.\nI came prepared with a diagram\nto help, but behold, no need.\n\nHere is a post that discusses\nthat create 45\u00b0 angles.\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nIntern\nJoined: 30 Nov 2016\nPosts: 34\nGMAT 1: 570 Q47 V21\nGMAT 2: 540 Q47 V16\nGMAT 3: 560 Q39 V28\nIn the rectangular coordinate system above, if the equation of m is y\u00a0 [#permalink]\n\n### Show Tags\n\n15 Apr 2018, 03:40\nBunuel wrote:\n\nIn the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?\n\n(A) \u221a2\n(B) 1\n(C) \u221a2/2\n(D) 1/2\n(E) 1/4\n\nAttachment:\nThe attachment 2017-12-15_1259.png is no longer available\n\nMy method to solve this:-\nAttachments\n\nFile comment: for line y = x, the angle between x-axis and the line m will be 45 degrees.\nThe min distance will the perpendicular distance between the two lines.\nIn the image, by mistake I've written 1/root(2) = 2/root(2), it should be root(2)/2. Please excuse that mistake\n\nIMG_20180415_160317.jpg [ 1.99 MiB | Viewed 347 times ]\n\n_________________\n\nGood things are going to happen. Keep fighting for what you want. Don't worry and have faith that it'll all work out!\n\nIn the rectangular coordinate system above, if the equation of m is y &nbs [#permalink] 15 Apr 2018, 03:40\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-07-21 08:06:14", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-the-rectangular-coordinate-system-above-if-the-equation-of-m-is-y-255480.html", "openwebmath_score": 0.8180845975875854, "openwebmath_perplexity": 2976.88257315349, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 1.0, "lm_q2_score": 0.8577680995361899, "lm_q1q2_score": 0.8577680995361899}}
{"url": "https://gmatclub.com/forum/a-company-has-two-types-of-machines-type-r-and-type-s-130108-20.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Jan 2019, 14:49\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in January\nPrevNext\nSuMoTuWeThFrSa\n303112345\n6789101112\n13141516171819\n20212223242526\n272829303112\nOpen Detailed Calendar\n\u2022 ### FREE Quant Workshop by e-GMAT!\n\nJanuary 20, 2019\n\nJanuary 20, 2019\n\n07:00 AM PST\n\n07:00 AM PST\n\nGet personalized insights on how to achieve your Target Quant Score.\n\u2022 ### Free GMAT Strategy Webinar\n\nJanuary 19, 2019\n\nJanuary 19, 2019\n\n07:00 AM PST\n\n09:00 AM PST\n\nAiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.\n\n# A company has two types of machines, type R and type S\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2830\nRe: A company has two types of machines, type R and type S\u00a0 [#permalink]\n\n### Show Tags\n\n04 Nov 2016, 15:05\n1\nenigma123 wrote:\nA company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?\n\nA. 3\nB. 4\nC. 6\nD. 9\nE. 12\n\nWe have a combined worker problem for which we can use the following formula:\n\nwork (1 machine) + work (2 machine) = total work completed\n\nSince we are completing one job, we can say:\n\nwork (1 machine) + work (2 machine) = 1\n\nWe are given that a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours.\n\nThus, the rates for the two machines are as follows:\n\nrate of machine R = 1/36\n\nrate of machine S = 1/18\n\nWe are also given that the company used the same number of each type of machine to do the job in 2 hours. If we let x = the number of each machine used, we can multiply each rate by x and we have:\n\nrate of x number of R machines = x/36\n\nrate of x number of S machines = x/18\n\nFinally, since we know some number of R and S machines worked for two hours, and since work = rate x time, we can calculate the work done by each type of machine.\n\nwork done by x number of R machines = 2x/36 = x/18\n\nwork done by x number of S machines = 2x/18 = x/9\n\nNow we can determine x using the combined worker formula:\n\nwork (machine R) + work (machine S) = 1\n\nx/18 + x/9 = 1\n\nx/18 + 2x/18 = 1\n\n3x/18 = 1\n\nx/6 = 1\n\nx = 6\n\n_________________\n\nJeffery Miller\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nIntern\nJoined: 20 Jan 2015\nPosts: 9\nWork/Rate problem- A company has..\u00a0 [#permalink]\n\n### Show Tags\n\n01 Jan 2017, 06:10\nA company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?\n\nOptions:\nA. 3\nB. 4\nC. 6\nD. 9\nE. 12\n\nCan anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.\nThanks.\nSenior Manager\nJoined: 13 Oct 2016\nPosts: 367\nGPA: 3.98\nRe: Work/Rate problem- A company has..\u00a0 [#permalink]\n\n### Show Tags\n\n01 Jan 2017, 06:47\n5\n1\nbaalok88 wrote:\nA company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?\n\nOptions:\nA. 3\nB. 4\nC. 6\nD. 9\nE. 12\n\nCan anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.\nThanks.\n\nWe are given rates of each machine and that the number of machines of each type being used is the same = x.\n\n$$\\frac{x}{36} + \\frac{x}{18} = \\frac{1}{2}$$\n\n$$\\frac{3x}{36} = \\frac{1}{2}$$\n\n$$6x = 36$$\n\n$$x = 6$$\nIntern\nJoined: 22 Jul 2016\nPosts: 23\nRe: A company has two types of machines, type R and type S\u00a0 [#permalink]\n\n### Show Tags\n\n07 Jan 2017, 21:07\n2\nRate(R) = $$\\frac{1}{36}$$\nRate(S) = $$\\frac{1}{18}$$\nCombined rate of both machines ,\nRate(RS)=Rate(R) + Rate(S) = $$\\frac{1}{12}$$\n\nwe have given time , T = 2hrs , Workdone = 1 (for single job)\nPlug in the values :\n\n[Rate] * [Time] * [No. of machines] = Workdone\n\n$$\\frac{1}{12}$$ * 2 * [No. of machines] = 1\n[No. of machines] = 6\n\nAns : C\nIntern\nJoined: 29 Nov 2016\nPosts: 9\nRe: A company has two types of machines, type R and type S\u00a0 [#permalink]\n\n### Show Tags\n\n18 Apr 2017, 05:05\nBunuel wrote:\nenigma123 wrote:\nA company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?\n\nA. 3\nB. 4\nC. 6\nD. 9\nE. 12\n\nRate of A - $$\\frac{1}{36}$$ job/hour, rate of x machines of A - $$\\frac{1}{36}x$$ job/hour;\nRate of B - $$\\frac{1}{18}$$ job/hour, rate of x machines of B - $$\\frac{1}{18}x$$ job/hour, (same number of each type);\n\nRemember that we can sum the rates, hence combined rate of A and B is $$\\frac{1}{36}x+\\frac{1}{18}x=\\frac{3}{36}x=\\frac{1}{12}x$$ job/hour.\n\nWe are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\\frac{1}{12}x=1=Job$$ --> $$x=6$$.\n\nWhen I am double checking, why am I not getting the right answer?\nIf 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.\nIf 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.\nTogether, they will take 6-3=3 hours.\n\nPlease tell me where am I wrong :/\nIntern\nJoined: 14 May 2017\nPosts: 47\nRe: A company has two types of machines, type R and type S\u00a0 [#permalink]\n\n### Show Tags\n\n16 Aug 2017, 18:41\nnarendran1990 wrote:\nTook 36 as the unit of work to be completed by R & S. [LCM of 36 & 18].\n\nSo, R does 1 unit per hour & S does 2 units per hour. Since an equal number of R & S machines have to be deployed, consider 'x' as the number of equipment. Additionally, the work is to be completed in 2 hours, so 18 units have to be manufactured.\n\nTherefore, x+2x = 18, 3x = 18, x=6. [Correct Answer]\n\nI was trying to solve using similar kind of method. But i could not understand how you arrived on 18? It will be great if you could elaborate.\nManager\nJoined: 22 Nov 2016\nPosts: 208\nLocation: United States\nGPA: 3.4\nA company has two types of machines, type R and type S\u00a0 [#permalink]\n\n### Show Tags\n\n09 Nov 2017, 21:26\nThere is no need to compute the combined rate in this case, explicitly that is.\n\nSince the number of machines is the same, we can use a simple equation:\n\n$$x*\\frac{1}{36}+ x* \\frac{1}{18} = \\frac{1}{2}$$ (here x machines work at the respective rates for 2 hours)\n\nSolving for x, we get 6.\n_________________\n\nKudosity killed the cat but your kudos can save it.\n\nManager\nJoined: 01 Mar 2015\nPosts: 76\nRe: A company has two types of machines, type R and type S\u00a0 [#permalink]\n\n### Show Tags\n\n13 Nov 2018, 19:46\nhannahkagalwala wrote:\nBunuel wrote:\nenigma123 wrote:\nA company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?\n\nA. 3\nB. 4\nC. 6\nD. 9\nE. 12\n\nRate of A - $$\\frac{1}{36}$$ job/hour, rate of x machines of A - $$\\frac{1}{36}x$$ job/hour;\nRate of B - $$\\frac{1}{18}$$ job/hour, rate of x machines of B - $$\\frac{1}{18}x$$ job/hour, (same number of each type);\n\nRemember that we can sum the rates, hence combined rate of A and B is $$\\frac{1}{36}x+\\frac{1}{18}x=\\frac{3}{36}x=\\frac{1}{12}x$$ job/hour.\n\nWe are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\\frac{1}{12}x=1=Job$$ --> $$x=6$$.\n\nWhen I am double checking, why am I not getting the right answer?\nIf 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.\nIf 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.\nTogether, they will take 6-3=3 hours.\n\nPlease tell me where am I wrong :/\n\nIt is not correct to subtracting time taken by one machine from that of other to calculate final time,\n\nTime taken are for complete job, and if other machines are also performing the same job, then there will be less amount of work of each machine type and hence leaser time will be taken.\n\nYou are right till, 6 R machines will take 6 hours.\nAnd 6 S machines will take 3 hours.\n\nNow total time taken will be 1/(1/6 +1/3) = 1/(1/2) = 2 hours.\n\nPosted from my mobile device\nRe: A company has two types of machines, type R and type S &nbs [#permalink] 13 Nov 2018, 19:46\n\nGo to page \u00a0 Previous \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0[ 28 posts ]\n\nDisplay posts from previous: Sort by", "date": "2019-01-19 22:49:06", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-company-has-two-types-of-machines-type-r-and-type-s-130108-20.html", "openwebmath_score": 0.39790090918540955, "openwebmath_perplexity": 1852.2001468536682, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8577680995361899, "lm_q1q2_score": 0.8577680995361899}}
{"url": "http://math.stackexchange.com/questions/197919/let-p-and-q-be-two-distinct-primes-pick-the-correct-statements-from", "text": "# Let $p$ and $q$ be two distinct primes. Pick the correct statements from\n\nLet $p$ and $q$ be two distinct primes. Pick the correct statements from the following:\n\na. $Q(\\sqrt p)$ is isomorphic to $Q(\\sqrt q)$ as fields.\n\nb. $Q(\\sqrt p)$ is isomorphic to $Q(\\sqrt{\u2212q})$ as vector spaces over Q.\n\nc. $[Q(\\sqrt p,\\sqrt q) : Q] = 4$.\n\nd. $Q(\\sqrt p,\\sqrt q) = Q(\\sqrt p + \\sqrt q)$.\n\n(c) & (d) are correct. (a) is not correct for fields but correct for vector spaces. (b) not sure. i think it is correct by the same arguement of (a).am i right?\n\n-\nIs this a homework question, or just one for personal development, as this may effect the type of response that people give? \u2013\u00a0David Ward Sep 17 '12 at 8:16\n@DavidWard I don't know why it should affect the type of response. If one should not give a full answer to a homework question, many good questions might have no full answer in this site. I think this is against the policy of this site. \u2013\u00a0Makoto Kato Sep 17 '12 at 18:13\n@MakotoKato My reasoning is that if it is a homework question, then the answer itself is of importance to the person asking the question. However, if the question is not homework, then for a question as above, I would feel happy to give the answer, but then lead someone through the reasoning on how to get there. \u2013\u00a0David Ward Sep 18 '12 at 8:26\n@DavidWard I'm sorry that I don't understand your reasoning. Could you explain why you don't want to give the answer to a homework question? \u2013\u00a0Makoto Kato Sep 18 '12 at 8:43\n@MakotoKato My reasoning is as follows: Firstly suppose that the question asked is from an assessed piece of work. Then as the question is a multiple-choice question, the actual answer will attract some if not all of the marks (depending on the desire of the assessor). Thus knowledge of the answer is of critical importance. On the other hand, if the question asked is purely for personal development, whilst leading the member towards the solution, it can be beneficial to actually know where you are being led. It just gives some motivation for where the reasoning is heading. \u2013\u00a0David Ward Sep 18 '12 at 15:26\n\na. Suppose $\\mathbb{Q}(\\sqrt p)$ is isomorphic to $\\mathbb{Q}(\\sqrt q)$ as fields. Then $\\mathbb{Q}(\\sqrt q)$ has an element $\\alpha$ such that $\\alpha^2 = p$. Let $\\alpha = a + b\\sqrt q$, where $a, b \\in \\mathbb{Q}$. We denote the conjugate of $\\alpha$ by $\\alpha'$. Since $\\alpha + \\alpha' = 0$, $a = 0$. Since $\\alpha^2 = p$, $p = b^2 q$. Hence $p = q$. This is a contradiction. Hence $a$. is not true.\n\nb. Since the both fields have dimension 2 as vector spaces over $\\mathbb{Q}$, $b.$ is true.\n\nc. As we see in the above, $\\sqrt p$ is not contained in $\\mathbb{Q}(\\sqrt q)$. Hence $[\\mathbb{Q}(\\sqrt p,\\sqrt q) : \\mathbb{Q}(\\sqrt q)] = 2$. Hence $[\\mathbb{Q}(\\sqrt p,\\sqrt q) : \\mathbb{Q}] = 4$. Thus $c.$ is true.\n\nd. Let $K = \\mathbb{Q}(\\sqrt p,\\sqrt q)$. Clearly $K/\\mathbb{Q}$ is Galois. Let $\\sigma \\in Gal(K/\\mathbb{Q})$. Then $\\sigma(\\sqrt p) = \\sqrt p$ or $-\\sqrt p$, $\\sigma(\\sqrt q) = \\sqrt q$ or $-\\sqrt q$. Since $|Gal(K/\\mathbb{Q})| = 4$ by $c.$, $Gal(K/\\mathbb{Q}) = \\{1, \\sigma_1, \\sigma_2. \\sigma_3\\}$, where\n\n$$\\sigma_1(\\sqrt p) = \\sqrt p,\\ \\ \\ \\sigma_1(\\sqrt q) = -\\sqrt q$$\n\n$$\\sigma_2(\\sqrt p) = -\\sqrt p,\\ \\sigma_2(\\sqrt q) = \\sqrt q$$\n\n$$\\sigma_3(\\sqrt p) = -\\sqrt p,\\ \\ \\ \\sigma_3(\\sqrt q) = -\\sqrt q$$\n\nLet $\\alpha = \\sqrt p + \\sqrt q$.\n\nClearly $\\alpha, \\sigma_1(\\alpha) = \\sqrt p - \\sqrt q, \\sigma_2(\\alpha) = -\\sqrt p + \\sqrt q, \\sigma_3(\\alpha) = -\\sqrt p - \\sqrt q$ are distinct. Hence $K = \\mathbb{Q}(\\alpha)$. Thus $d.$ is true.\n\n-\nplease explain why \u03b1+\u03b1\u2032=0? In the proof of a ? \u2013\u00a0GA316 Aug 9 '15 at 4:01", "date": "2016-05-27 16:28:35", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/197919/let-p-and-q-be-two-distinct-primes-pick-the-correct-statements-from", "openwebmath_score": 0.9162284135818481, "openwebmath_perplexity": 175.0392676156426, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.970239907775086, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8577301971192818}}
{"url": "https://cs.stackexchange.com/questions/106135/how-do-i-describe-formally-complexity-of-2-sum-problem-algorithm?noredirect=1", "text": "# How do I describe formally complexity of 2-sum problem algorithm?\n\nI have algorithm that finds if there are two elements in sorted array that have sum zero.\n\n1.ZeroSumPair(A[1..n]) // A[1..n] <-- sorted\n2.    l <- 1, r <- n\n3.    while(l < r)\n4.        while(l < r or A[l] + A[r] > 0)\n5.            r--\n6.        if(A[l] + A[r] = 0)\n7.            return true\n8.        l++\n9.    return false\n\n\nIntuitively I know that this algorithm is $$O(n)$$, but how do I deduct it using proof with summations like in CLRS book?\n\nI've also saw How to develop an $O(N)$ algorithm solve the 2-sum problem?, but I didn't see any formal proof.\n\n\u2022 I would try putting in a counter t. Then increment t by 1 every time you do an operation. Then try to prove a loop invariant for the outer-while loop that claims $t = O(n)$. With the correct loop invariant on $t$, this can be proven by induction relatively easily. Perhaps something like $t \\leq c(l + (n - r))$ for some constant $c$ might work. Then at termination (in the worst case) you know $l = r$, thus $t \\leq cn \\implies t = O(n)$. \u2013\u00a0ryan Mar 27 at 18:28\n\nConsider this slightly modified algorithm where I've added an \"operation counter\" t. This will be incremented every time we do a comparison or assignment.\n\n1. ZeroSumPair(A[1..n]) // A[1..n] <-- sorted\n2.   l <- 1, r <- n\n3.   t <- 3             // 2 for first two assignments and 1 for initial while check\n4.   while(l < r)\n5.     t++              // 1 for initial while check\n6.     while(l < r or A[l] + A[r] > 0)\n7.       t++            // 1 for decrementing r\n8.       r--\n9.       t++            // 1 for following while check\n10.    t++              // 1 for if comparison\n11.    if(A[l] + A[r] = 0)\n12.      return true\n13.    t++              // 1 for incrementing l\n14.    l++\n15.    t++              // 1 for following while check\n16.  return false\n\n\nThis is a bit verbose, but it will work. Now we must simply prove that, at termination we have $$t = O(n)$$. We can do this inductively with a loop invariant.\n\nLet's use the following loop invariant for the loop on line 4.\n\n$$t = 3 + 4(l-1) + 2(n-r)$$\n\n### Base Case\n\nInitially $$t = 3$$, $$l = 1$$, and $$r = n$$. Thus we have:\n\n$$3 = 3 + 4(1 - 1) + 2(n - n) = 3$$\n\n### Inductive Case\n\nLet $$t'$$, $$l'$$, and $$r'$$ be the values of $$t$$, $$l$$, and $$r$$ at the end of our previous iteration. At the end of our current iteration we have $$l = l' + 1$$, $$r = r' - k$$ for some $$k$$, and $$t = t' + 4 + 2k$$. Thus we have:\n\n\\begin{align*} t & = t' + 4 + 2k\\\\ & = 3 + 4(l' - 1) + 2(n - r') + 4 + 2k\\\\ & = 3 + 4(l - 2) + 2(n - (r + k)) + 4 + 2k\\\\ & = 3 + 4(l - 1) - 4 + 2(n - r) - 2k + 4 + 2k\\\\ & = 3 + 4(l - 1) + 2(n - r) & \\square \\end{align*}\n\nThus, we can conclude the loop invariant holds. At the end of the loop (in the worst case) we have $$l = r \\leq n$$. We then have: \\begin{align*} t & = 3 + 4(l - 1) + 2(n - r)\\\\ & = 3 + 4l - 4 + 2n - 2l\\\\ & = 2(n + l) - 1\\\\ & \\leq 2(n + n) - 1\\\\ & = 4n - 1\\\\ & = O(n) & \\square \\end{align*}\n\nThus it is linear. There might be an easier way to do this, but this way makes sense to me pretty clearly.\n\n\u2022 I see that this looks formal but I feel like it is \u201chacking\u201d, amending the algorithm in order to prove its complexity, is this real thing in computer science? Not talking about practice here \u2013\u00a0kuskmen Mar 27 at 21:53\n\u2022 @kuskmen you don't need $t$ explicitly in your code. I put it there for clarity. You can have an implicit \"operation count\" which is realistically what you're trying to determine. Then prove (through loop invariants) that the \"operation count\" is bounded by $n$. That's all I did here with $t$ being the operation counter. \u2013\u00a0ryan Mar 27 at 22:49\n\u2022 I see, I have to admit I am bad in mathematical proofs or any sort of proofs as you can tell. Can you tell why $t=t'+ 4 + 2k$ in the inductive step? \u2013\u00a0kuskmen Mar 28 at 8:15\n\u2022 Assume $t$ starts at $t'$ just before the loop executes. Now hop into the loop. Increment by 1 for the initial inner while loop check (line 5). Now assume the inner while loop executes $k$ times. There are 2 increments in the inner while loop (line 7 and line 9), so $t$ is then incremented $2k$ times. Next, exit the inner while loop, increment by 1 for the if comparison (line 10). Assume the if condition is false because we do worst case analysis, then increment by 1 for incrementing l (line 13). Increment 1 last time for the following while check (line 15). Add them all up and you get that. \u2013\u00a0ryan Mar 28 at 16:11\n\u2022 @kuskmen, this may also help you: cs.stackexchange.com/a/23595/68251 \u2013\u00a0ryan Mar 28 at 17:45", "date": "2019-11-19 15:41:32", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/106135/how-do-i-describe-formally-complexity-of-2-sum-problem-algorithm?noredirect=1", "openwebmath_score": 0.6884363293647766, "openwebmath_perplexity": 907.0748221880606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399043329855, "lm_q2_score": 0.8840392909114836, "lm_q1q2_score": 0.8577301970405581}}
{"url": "https://math.stackexchange.com/questions/2495831/show-that-the-equilibrium-point-0-0-is-asymptotically-stable-and-an-estimate", "text": "# Show that the equilibrium point $(0,0)$ is asymptotically stable and an estimate of its basin of attraction\n\nConsider the system \\begin{aligned} \\dot{x} &=-y-x^3+x^3y^2\\\\ \\dot{y}&=x-y^3+x^2y^3\\end{aligned} Show that the equilibrium point $$(0,0)$$ is asymptotically stable and an estimate of its attractiveness basin.\n\nClearly the point $$(0,0)$$ is a point of equilibrium and also $$D_f(x,y)=\\begin{pmatrix}-3x^2+3y^4 & -1+2yx^3\\\\ 1+2xy^3 &-3y^2+3y^2x^2 \\end{pmatrix}$$, so $$D_f(0,0)=\\begin{pmatrix}0 & -1\\\\1 &0 \\end{pmatrix}$$ and has eigenvalues \u200b\u200b$$\\pm i$$ So, I can not conclude anything of stability for this point, I need a Liapunov function and I do not know what it is or how to find it, could someone help me please? Thank you very much.\n\n\u2022 Always try the simplest one first. For example, quadratic. More precisely, $V(x, y) = x^2+y^2$. Oct 30 '17 at 5:44\n\nIn addition to what has been said by @Evgeny and @MrYouMath: the set $$M=\\left\\{ (x,y)\\in\\mathbb R^2 :\\; x^2+y^2<2 \\right\\}$$ is a positively invariant set of the considered system since $\\forall (x,y)\\in M$ $$\\dot V=-x^4-y^4+x^2y^2(x^2+y^2)<-x^4-y^4+2x^2y^2=-(x^2-y^2)^2\\leq 0;$$ it is also a subset (guaranteed estimation) of the domain of attraction.\n\n\u2022 Nice one! This estimate is even better than $\\| (x, y) \\|_{\\infty} < 1$ . Oct 30 '17 at 18:06\n\u2022 +1: Very nice observation. This might be useful in the future. Oct 31 '17 at 14:03\n\u2022 How Do I show that $M$ is positively invariant with this fact?\n\u2013\u00a0P.G\nNov 20 '20 at 13:44\n\u2022 @P.G Just as it is done in the proof of the Lyapunov stability theorem\n\u2013\u00a0AVK\nNov 21 '20 at 7:02\n\nAs @Evgeny suggested you could use the Lyapunov function candidate\n\n$$V(x,y)=\\dfrac{1}{2}\\left[x^2+y^2\\right].$$\n\n$V(x,y)$ is clearly positive definite at the origin and radially unbounded (which we would need for assessing global asymptotic stability).\n\nThe time derivative of $V(x,y)$ is given by $$\\dot{V}=x\\dot{x}+y\\dot{y}=x(-y-x^3+x^3y^2)+y(x-y^3+x^2y^3)$$ $$\\dot{V}=-x^4-y^4+x^2y^2(x^2+y^2)=-(1-y^2)x^4-(1-x^2)y^4.$$\n\nAs the lower order terms $-x^4-y^4$ are negative definite we can conclude that the equilibrium point is asymptotically stable in a region around the origin. Using the comment given by @Evgeny we can see that this expression is negative semidefinite if $(x,y)$ lie inside the unit circle $D=\\{(x,y)\\in \\mathbb{R}^2|x^2+y^2<1\\}$. This is the basin of attraction (correction due to @Artem).\n\nWe cannot say if the origin is globally asymptotically stable because $\\dot{V}$ is not negative definite for all regions around the origin. This does not mean that the origin cannot be globally asymptotically stable. It just means we can only show (local) asymptotic stability of the origin with $V(x)$ as our Lyapunov function candidate.\n\n\u2022 Actually, you can do better if regroup terms correctly: $\\dot{V} = -x^4(1-y^2)-y^4(1-x^2)$. This also gives a rough estimate of basin of attraction since this expression is negative when $\\vert x \\vert < 1$ and $\\vert y \\vert < 1$. Oct 30 '17 at 13:18\n\u2022 @Evgeny: You are right. I felt that I should be able to factor this expression. Oct 30 '17 at 13:19\n\u2022 Also, speaking about radial unboundedness: it's not a necessary condition really, see recent discussion in comments. Oct 30 '17 at 18:09\n\u2022 It is not necessary for asymptotic stability but it is necessary for global asymptotic stability (see the comment in the brackets in my answer). I think that is what the example showed. Or do you mean something else? Oct 30 '17 at 18:14\n\u2022 Even for the global asymptotic stability it is not necessary. You can prove the global asymptotic stability with a Lyapunov function which is not radially unbounded. Moreover, you can transform the image of any Lyapunov function such that it will become bounded but still will prove that something is globally asymptotic stable. Oct 30 '17 at 20:29\n\nThis problem can be handled with an optimization procedure, having in mind that generally is a non convex problem. The result depends on the test Lyapunov function used so we will generalize to a quadratic Lyapunov function\n\n$$V(p) = p^{\\dagger}\\cdot M\\cdot p = a x^2+b x y + c y^2,\\ \\ \\ p = (x,y)^{\\dagger}$$\n\nand\n\n$$f(p) = \\{-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3\\}$$ with $$a>0,c>0, a b-b^2 > 0$$ to assure positivity on $$M$$. We will assure a set involving the origin $$Q_{\\dot V}$$ such that $$\\dot V(Q_{\\dot V}) < 0$$. The optimization process will be used to guarantee a maximal $$Q_{\\dot V}$$.\n\nAfter determination of $$\\dot V = 2 p^{\\dagger}\\cdot M\\cdot f(p)$$ we follow with a change of variables\n\n$$\\cases{ x = r\\cos\\theta\\\\ y = r\\sin\\theta }$$\n\nso $$\\dot V = \\dot V(a,b,c,r,\\theta)$$. The next step is to make a sweep on $$\\theta$$ calculating\n\n$$S(a,b,c, r)=\\{\\dot V(a,b,c,r,k\\Delta\\theta\\},\\ \\ k = 0,\\cdots, \\frac{2\\pi}{\\Delta\\theta}$$\n\nand then the optimization formulation follows as\n\n$$\\max_{a,b,c,r}r\\ \\ \\ \\ \\text{s. t.}\\ \\ \\ \\ a > 0, c> 0, a c -b^2 > 0, \\max S(a,b,c,r) \\le -\\gamma$$\n\nwith $$\\gamma > 0$$ a margin control number.\n\nFollows a MATHEMATICA script which implements this procedure in the present case.\n\nf = {-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3};\nV = a x^2 + 2 b x y + c y^2;\ndV = Grad[V, {x, y}].f /. {x -> r Cos[t], y -> r Sin[t]};\nrest = Max[Table[dV, {t, -Pi, Pi, Pi/30}]] < -0.1;\nrests = Join[{rest}, {r > 0, a > 0, c > 0, a c - b^2 > 0}];\nsols = NMinimize[Join[{-r}, rests], {a, b, c, r}, Method -> \"DifferentialEvolution\"]\nrest /. sols[[2]]\n\ndV0 = Grad[V, p].f /. sols[[2]]\nV0 = V /. sols[[2]]\nr0 = 2;\nrmax = r /. sols[[2]];\ngr0 = StreamPlot[f, {x, -r0, r0}, {y, -r0, r0}];\ngr1a = ContourPlot[dV0, {x, -r0, r0}, {y, -r0, r0}, ContourShading -> None, Contours -> 80];\ngr1b = ContourPlot[dV0 == 0, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> Blue];\ngr2 = ContourPlot[x^2 + y^2 == rmax^2, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> {Red, Dashed}];\nShow[gr0, gr1a, gr1b, gr2]\n\n\nFollows a plot showing in black the level sets $$Q_{\\dot V}$$ an in blue the trace of $$\\dot V = 0$$. In dashed red is shown the largest circular set $$\\delta = 1.42486$$ defining the maximum attraction basin for the given test Lyapunov function's family.", "date": "2021-11-30 19:30:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2495831/show-that-the-equilibrium-point-0-0-is-asymptotically-stable-and-an-estimate", "openwebmath_score": 0.5272963047027588, "openwebmath_perplexity": 230.8230570068963, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399026119352, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8577301925548536}}
{"url": "https://math.stackexchange.com/questions/3147904/if-ah-bh-implies-ha-hb-for-a-subgroup-h-having-finite-index-then-gh-hg-f", "text": "# If $aH=bH \\implies Ha=Hb$ for a subgroup H having *finite index*, then $gH=Hg$ for all $g \\in G$?\n\nProblem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $$H$$ is a subgroup of a group $$G$$ such that $$Ha \\not = Hb \\implies aH \\not = bH$$, then $$gHg^{-1} \\subset H$$ for all $$g \\in G$$, which is equivalent to $$gH \\subset Hg.$$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $$gH = Hg$$ for all $$g \\in G$$. I think I can prove it if I further assume that the index of $$H$$ in $$G$$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $$H$$ of infinite index such that $$aH \\subset Ha$$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]\n\nHere's my attempt at a proof:\n\nSuppose that $$|G:H|=n \\in \\mathbb{N}$$ and that $$aH \\subset Ha$$ for all $$a \\in G$$. Now, suppose that there is an element $$x$$ such that $$x \\in Ha$$ but $$x \\not \\in aH$$. Then $$x$$ must be contained in a different left coset, say $$bH$$, because the left cosets form a partition of $$G$$. Then, by our hypothesis, $$x \\in Hb$$, which implies $$Hb = Ha$$, because the right cosets form a partition of $$G$$ as well. So far we have shown that $$aH, bH \\subset Ha=Hb$$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $$n-1$$ right cosets distinct from $$Ha$$ must contain at least one left coset, by our hypothesis, but $$Ha$$ contains two left cosets. So there are at least $$n+1$$ left cosets, a contradiction. Therefore $$aH=Ha. \\square$$\n\n\u2022 Compare with this duplicate and its answers. \u2013\u00a0Dietrich Burde Mar 14 at 12:00\n\u2022 @DietrichBurde The answers in your linked question only prove $gHg^{-1} \\subset H$ and not equality. So I don't think they address the OP\u2019s problem. \u2013\u00a0Claudius Mar 14 at 12:15\n\u2022 @Claudius This is not true. They prove equality, e.g., see Deven's answer. \u2013\u00a0Dietrich Burde Mar 14 at 12:18\n\u2022 @DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} \\subset H$. At least, I don't see how he proved the other inclusion. \u2013\u00a0Claudius Mar 14 at 12:20\n\nYour proof seems correct to me.\n\nIn fact, you can remove the finite index hypothesis. If $$gHg^{-1} \\subset H$$ for all $$g\\in G$$, then for each $$g\\in G$$ you also have $$H = g^{-1}(gHg^{-1})g \\subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} \\subset H,$$ so we must have equality throughout, i. e. $$H = g^{-1}Hg$$ (for each $$g\\in G$$).\n\nMore generally, for any subgroup $$H$$ of $$G$$ the following holds: the set $$N:= \\{ g\\in G \\mid gHg^{-1}\\subset H\\}$$ is a subgroup of $$G$$ if and only if $$gHg^{-1} = H$$ for all $$g\\in N$$. (In that case $$N$$ is the normalizer of $$H$$ in $$G$$.)\n\n\u2022 That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/\u2026 in which there seems to be a counterexample? \u2013\u00a0The Footprint Mar 14 at 12:36\n\u2022 In the answer of Brian M. Scott the set $\\{g\\in G\\mid gHg^{-1} \\subset H\\}$ is not stable under inversion, hence not a group. It is only closed under composition. \u2013\u00a0Claudius Mar 14 at 12:40\n\nYou don't need any additional hypotheses. Multiplying both sides by $$g^{-1}$$ on the left and $$g$$ on the right yields\n\n$$gHg^{-1} \\subset H \\implies H \\subset g^{-1}Hg$$\n\nSince this is true for all $$g \\in G$$, we can substitute $$g$$ for $$g^{-1}$$, concluding\n\n$$gHg^{-1} \\subset H \\subset gHg^{-1}$$\n\n$$H = gHg^{-1}$$\n\nThis is equivalent to\n\n$$gH = Hg$$", "date": "2019-08-21 15:34:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3147904/if-ah-bh-implies-ha-hb-for-a-subgroup-h-having-finite-index-then-gh-hg-f", "openwebmath_score": 0.8608401417732239, "openwebmath_perplexity": 131.19161103643617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399034724604, "lm_q2_score": 0.8840392848011834, "lm_q1q2_score": 0.8577301903513631}}
{"url": "https://brilliant.org/discussions/thread/a-calculus-problem/", "text": "\u00d7\n\n# A Calculus Problem!\n\nThe Problem:\n\nFind $$I=\\displaystyle\\int_{0}^{2} (x^2+1)\\,d\\left \\lfloor x\\right \\rfloor$$\n\nMy Doubt:\n\nThere seem two probable approaches to the question but both of them yield different results. Both seem to be mathematically correct and hence the confusion.\n\nApproach 1:\n\nWe know that,\n\n$$g(x). f'(x) + f(x). g'(x) = \\large \\frac{d\\left(f(x). g(x)\\right)}{dx}$$\n\nIntegrating the above expression we get:\n\n$$\\displaystyle\\int_{a}^{b} g(x)\\,df(x) + \\displaystyle\\int_{a}^{b} f(x)\\,dg(x) = f(b).g(b)-f(a).g(a)$$\n\nUsing the above property to solve the integration we get the answer as $$\\boxed{7}$$\n\nApproach 2:\n\nWe can write\n\n$$I=\\displaystyle\\int_{0}^{1} (x^2+1)\\,d\\left \\lfloor x\\right \\rfloor +\\displaystyle\\int_{1}^{2}(x^2+1)\\,d\\left \\lfloor x\\right \\rfloor$$\n\nBut $$\\left \\lfloor x\\right \\rfloor$$ assumes constant values of $$\\left \\lfloor x\\right \\rfloor=0$$ and $$\\left \\lfloor x\\right \\rfloor=1$$ in the respective intervals. And hence in both cases $$d\\left \\lfloor x\\right \\rfloor=0$$ and therefore $$I=0$$\n\nI genuinely can't understand what's the correct method. I will really be grateful if someone can explain it to me.\n\nThank You!\n\nNote by Miraj Shah\n1\u00a0year, 9\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n\n- bulleted\n- list\n\n\u2022 bulleted\n\u2022 list\n\n1. numbered\n2. list\n\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1\n\nparagraph 2\n\nparagraph 1\n\nparagraph 2\n\n> This is a quote\nThis is a quote\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nIn Approach 2, the floor function fails to be constant on the two subintervals; consider the two endpoints! The first integral takes the value of 1 due to the step at 1, and the second one takes a value of 5. With this correction, both approaches work and yield the same result.\n\n- 1\u00a0year, 9\u00a0months ago\n\nFirst off, integration by parts doesn't work as the step function isn't differentiable.\n\nSecondly, the integral is 2. I'll provide a proof of this later.\n\nBtw, @Aditya Kumar The integral makes perfect sense and is defined.\n\n- 1\u00a0year, 9\u00a0months ago\n\nIt would be really nice of you to put the proof here if possible! But I would also like to know then what's wrong with the Approach 1&2 illustrated above? It's a bit baffling\n\n- 1\u00a0year, 9\u00a0months ago\n\nWait, the integral is 7.\n\nWe shall use the definition of integration as given in this book, with $$\\alpha(x)=\\lfloor x \\rfloor$$ and $$f(x)=1+x^2$$.\n\nConsider a partition $$P$$ of $$[0,2]$$ such that $$x_{r-1}\\leq 1\\leq x_{r}$$ where strict inequality holds in atleast one of the two inequalities. Also, by the definition of a partition, we have $$x_{n-1}<2=x_n$$. Then,\n\n$U(P, f, \\alpha)=f(x_r)+f(x_n) \\\\ L(P, f, \\alpha)=f(x_{r-1})+f(x_{n-1}) \\\\\\implies \\inf U(P, f, \\alpha)=\\sup L(P, f, \\alpha)=f(1)+f(2) \\\\\\implies \\int_0^2 f \\, d\\alpha = 7$\n\nNote: I've only used the fact that $$f$$ is monotonic.\n\nPlus, in the book mentioned above, in the exercises, a more generalised version of IBP, valid for even discontinuos functions is given. You've made use of this version in approach 1 (luckily, the step function is monotonic).\n\nIf the integrals are calculated using the definition in approach 2, you get the right answer, 7.\n\n- 1\u00a0year, 9\u00a0months ago\n\n- 1\u00a0year, 9\u00a0months ago\n\nOh yes! Really helpful @Deeparaj Bhat ! Sorry for replying a bit late. Busy studying for the big one which is to be held on 22nd May! Actually the above problem was from a test paper it self! Had a doubt in the question, so thought of taking help from the brilliant Brilliant Community as I was pretty sure that I will be able to get at least some insight!\n\nThank you!\n\n- 1\u00a0year, 9\u00a0months ago\n\nYou're writing any other exams? (Except JEE ADVANCED, which is on 22nd)\n\n- 1\u00a0year, 9\u00a0months ago\n\nActually yes. BITS and all! You have to give JEE this year?\n\n- 1\u00a0year, 9\u00a0months ago\n\nYup. That, BITSAT and CMI in my case.\n\n- 1\u00a0year, 9\u00a0months ago", "date": "2018-02-21 18:57:59", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/a-calculus-problem/", "openwebmath_score": 0.9855186343193054, "openwebmath_perplexity": 1428.5985123719056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9702398991698342, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8577301880297868}}
{"url": "https://mathhelpboards.com/threads/eulers-theorem-for-calculating-mod.6407/", "text": "# Number TheoryEuler's theorem for calculating mod\n\n#### ATroelstein\n\n##### New member\nI am trying to use Euler's theorem for calculating the following:\n\n$$8^{7} mod 187$$\n\nI have determined that:\n\n$$\\phi(187) = \\phi(11*17) = 160$$\n\nand\n\n$$8^{7} = 8^{4} * 8^{2} * 8^{1}$$\n\nbut am unfortunately confused about how to now proceed beyond this point. Thank you.\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nI am trying to use Euler's theorem for calculating the following:\n\n$$8^{7} mod 187$$\n\nI have determined that:\n\n$$\\phi(187) = \\phi(11*17) = 160$$\n\nand\n\n$$8^{7} = 8^{4} * 8^{2} * 8^{1}$$\n\nbut am unfortunately confused about how to now proceed beyond this point. Thank you.\nI don't think here Euler's theorem can be fruitfully used. But you can use Fermat's little in conjunction with Chinese Remainder to calculate this nicely.\n\nNote that $8^7=2^{21}=x$ (say). Also, $187=11\\times 17$. Now by Fermat we have $2^{10}\\equiv 1 \\pmod{11}$ and $2^{16}\\equiv 1\\pmod{17}$. These give $x\\equiv 2\\pmod{11}$ and $x\\equiv 32\\equiv -2\\pmod{17}$. From here it's routine to use the Chinese remainder and get the remainder $x$ leaves mod $187$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nAs I have stated here many times before, I am pathetically lazy, so I look for the easy way out.\n\n87 = (64)(8)(64)(8)(8) (mod 187)\n\nNext, I calculate 64*8 the old-fashioned way:\n\n(6*8*10) + (4*8) = 480 + 32 = 512.\n\nI'm going to guess that 2 multiples of 187 are all we're going to pack into 512. So:\n\n512 = 512 - 374 = 138 (mod 187).\n\nSo 87 = (138)(138)(8) (mod 187).\n\nDid I mention I'm afraid of big numbers? Well, I am. So next I'm going to apply a trick that often works well in modular arithmetic:\n\nSince 138 = -49 (mod 187),\n\n87 = (138)(138)(8) = (-49)(-49)(8) = (49)(49)(8) (mod 187).\n\nHere, I have an unpleasant choice to make: do I tackle 49*49 next, or 49*8?\n\nThe small numbers win!\n\nNow 49*8 = (4*8*10) + (9*8) = 320 + 72 = 392. Well, that's pretty close to 374, which makes me happy. So:\n\n87 = (49)(49)(8) = (49)(392) = (49)(18) (mod 187).\n\nI've avoided \"hard calculations\" as much as possible, but I have no choice but to evaluate 49*18 now:\n\n49*18 = (40+9)(10+8) = 400 + 320 + 90 + 72 = 882.\n\nSo I want to know what 882 (mod 187) is. Knowing that 180*5 = 900, I don't think I can get 5 multiples of 187 in 882, so 4 is my best guess:\n\n187*4 = 400 + 320 + 28 = 748. So:\n\n87 = 882 = 882 - 0 = 882 - 748 = 134 (mod 187)\n\nDoes this jibe with caffeinemachine's answer?\n\n134 = 132 + 2 = 0 + 2 = 2 = (mod 11). Check.\n\n134 = 119 + 15 = 0 + 15 = 15 = -2 (mod 17). Check.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nI'd like to point out here that caffeinemachine's solution is not so easy to carry out as it might seem.\n\nGiven that:\n\nx = 2 (mod 11)\nx = -2 (mod 17)\n\nwe seek integers k,m such that:\n\n11k + 2 = 17m - 2\n\nor, equivalently:\n\n17m - 11k = 4\n\nIf we knew some integers r,s with:\n\n17r - 11s = 1, we could take m = 4r, k = 4s.\n\nFortunately, the euclidean division algorithm gives us a way to find these two integers r and s:\n\n17 = 11*1 + 6\n11 = 6*1 + 5\n6 = 5*1 + 1\n\nTherefore:\n\n1 = 6 - 5\n5 = 11 - 6\n6 = 17 - 11, so:\n\n1 = 6 - 5 = 6 - (11 - 6) = 2*6 - 11 = 2*(17 - 11) - 11 = 2*17 - 3*11, so we have:\n\nr = 2, s = 3, and in turn:\n\nm = 8, k = 12. Hence:\n\n11*12 + 2 = 17*8 - 2, evaluating either side gives us:\n\n134, which is the desired power 87\u200b (mod 187).\n\n#### johng\n\n##### Well-known member\nMHB Math Helper\nHi,\nI tend to agree with caffeinemachine that the easiest computation is via the Chinese remainder theorem.\nGiven m and n relatively prime, the unique x (mod mn) that satisfies x = a (mod m) and x = b (mod n) is of the form a + km for some k with 0 <= k < n.\n(All of the given values are different for the range of k's and so one must be b (mod n)).\n\nSo I want -2 + 17k = 2 (mod 11)\n6k = 4 (mod 11)\nk = 8 (mod 11) -- the multiplicative inverse of 6 is 2 mod 11.\nSo x = -2 +8*17 = 136 - 2 = 134.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nHi,\nI tend to agree with caffeinemachine that the easiest computation is via the Chinese remainder theorem.\nGiven m and n relatively prime, the unique x (mod mn) that satisfies x = a (mod m) and x = b (mod n) is of the form a + km for some k with 0 <= k < n.\n(All of the given values are different for the range of k's and so one must be b (mod n)).\n\nSo I want -2 + 17k = 2 (mod 11)\n6k = 4 (mod 11)\nk = 8 (mod 11) -- the multiplicative inverse of 6 is 2 mod 11.\nSo x = -2 +8*17 = 136 - 2 = 134.\nYes, but....\n\nThe ease of this is \"hidden\" in the step:\n\n6k = 4 --> k = 8.\n\nIn this particular case, one can guess rather easily that [6-1]11 = 2 by trial-and-error. In general, finding a (multiplicative) inverse mod n (if it exists, which it assuredly will if n is prime, and the element inverted is not a multiple of n = p) involves exactly the same steps I outlined above. One often does not have a discrete log table lying around to read off which power of a generator a particular element is, and one HAS to resort to the division algorithm to find the inverse (finding primitive elements is not always a trivial task, although it IS easier for small numbers).\n\nDon't misunderstand me, I find the application of the CRT a beautiful and elegant solution. At its heart, the CRT states that:\n\nIf (m,n) = 1, then $\\Bbb Z_{mn} \\cong \\Bbb Z_m \\times \\Bbb Z_n$\n\nHowever, while the ring-homomorphism one way is trivial to display, the inverse homomorphism is nontrivial (and amounts to actually solving the congruence pair). There is a slight difference between knowing a solution exists and exhibiting that solution.\n\nYour calculations and mine are, of course, the same, as can be seen by taking the equation:\n\n11*12 + 2 = 17*8 - 2 (mod 11).\n\nI also do not know if the original poster has seen a PROOF of the CRT, with an understanding of what it means for \"compound\" moduli. In all fairness (both to you and caffeinemachine) I admit sometimes an understanding of more advanced methods makes problems like these easier to deal with. Do I run the risk of under-estimating the original poster's sophistication? Of course. I am the first to admit the laziest proof is not always the shortest.\n\nIf the exponent has been larger, my way would undeniably be more cumbersome. I certainly mean no disrespect to either of your fine answers. Knowing we can reduce an exponent b in ab (mod n), by taking b (mod \u03c6(n)), is something well worth remembering, if one has been exposed to Euler's theorem (which we can assume the OP has, by the thread title), and I note in passing that Fermat's \"Little Theorem\" (which caffeinemachine DOES invoke) is just a special case of Euler's Theorem.\n\nOne hopes that the original poster gains more from our discussion of solution techniques here than just the answer to his problem.\n\n#### johng\n\n##### Well-known member\nMHB Math Helper\nYes, but....\n\nThe ease of this is \"hidden\" in the step:\n\n6k = 4 --> k = 8.\n\nIn this particular case, one can guess rather easily that [6-1]11 = 2 by trial-and-error. In general, finding a (multiplicative) inverse mod n (if it exists, which it assuredly will if n is prime, and the element inverted is not a multiple of n = p) involves exactly the same steps I outlined above. One often does not have a discrete log table lying around to read off which power of a generator a particular element is, and one HAS to resort to the division algorithm to find the inverse (finding primitive elements is not always a trivial task, although it IS easier for small numbers).\n\nGiven the parameters of the CRT, namely m, n are relatively prime, I'm trying to solve a + mk = b for k. So of course m does have an inverse mod n. Granted, the finding of such inverse requires a little work, but it's not exorbitant. A slight extension of the Euclidean algorithm for gcd's produces x and y with mx + ny = 1, and so x is the inverse of m. I find the extension a bit cumbersome to do by hand, but it's a Programming I exercise to encode (maybe toward the end of the course). Even for large integers, the gcd algorithm is \"usually\" quite fast; as you probably know the worst case is for consecutive Fibonacci numbers. But even here, it's logarithmic.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nYes, that is what I was getting at. If the exponent is large-ish (say > 1000 for example), using a gcd algorithm is going to be MUCH faster than direct computation.\n\nBut with an exponent of only 7, direct computation can be done straight-forwardly, without involving more abstract results. Is this preferable? It depends on your point of view, I suppose. One thing that CAN be said about your solution/caffeinemachine's solution is that it is more widely applicable to a greater range of problems", "date": "2021-11-29 00:04:56", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/eulers-theorem-for-calculating-mod.6407/", "openwebmath_score": 0.7664651870727539, "openwebmath_perplexity": 641.4107890602463, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971992481016635, "lm_q2_score": 0.8824278633625322, "lm_q1q2_score": 0.8577132482279559}}
{"url": "https://math.stackexchange.com/questions/4459036/mathematical-inclusion-and-exclusion-of-elements-from-a-given-set-a", "text": "# Mathematical \"inclusion\" and \"exclusion\" of elements from a given set $A$?\n\nIf I have a set A, comprising of numbers from 1 to 10: $$A = \\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\\}$$. Let's say I want to make another set by \"including\" all even numbers:\n\n$$\\{2, 4, 6, 8, 10\\}$$\n\nOr I wanted to make a different set by \"excluding\" all odd numbers:\n\n$$\\{2, 4, 6, 8, 10\\}$$\n\nThese sets are of course the same. So is it correct to say that inclusion/exclusion are synonymous when it comes to set theory, as they're just different ways of building a set?\n\nThis might sound trivial, but I have a reason for asking: I want to understand if inclusion and exclusion are \"commutative\" properties, i.e. it doesn't matter in which order you apply them.\n\nFor example, let's say I make an operation to \"filter\" my set, by including all even numbers as we did before, producing set B\n\n$$B = \\{2, 4, 6, 8, 10\\}$$\n\nAnd then a separate operation to \"exclude\" any numbers less than 6 from set B, resulting in set C:\n\n$$C = \\{6, 8, 10\\}$$\n\nWhat if I started with A and applied the operations the other way around? First remove all numbers less than 6:\n\n$$B = \\{6, 7, 8, 9, 10\\}$$\n\nThen filter B to \"include\" only even numbers:\n\n$$C = \\{6, 8, 10\\}$$\n\nIt seems intuitively to me that the result will always be the same no matter which order you apply the operation. Is this true for all cases no matter the set, however? Is there a way to prove that applying \"filters\" to a set (I'm not sure of the proper term) will always be commutative?\n\nSo in summary, there are two questions here:\n\n1. Are the notions of \"inclusion\" and \"exclusion\" really synonymous from the point of view of applying an operation to a set to produce a subset?\n2. Will applying these operations to produce a subset of a set always be commutative, i.e. produce the same result?\n\u2022 It looks lile you are taking repeated intersections of sets. Intersection is both associative and commutative. May 26 at 10:25\n\u2022 @CrackedBauxite I've had a think about this. Is it intersection, or is it a set difference / relative complement? If I've got the set $\\{1, 2, 3, 4, 5\\}$ and I exclude all odd numbers, I've got a set of odd numbers $\\{1, 3, 5\\}$. It's my original set $\\{1, 2, 3, 4, 5\\} - \\{1, 3, 5\\} = \\{2, 4\\}$\n\u2013\u00a0Lou\nMay 26 at 16:45\n\u2022 But if I were to do \"include all even numbers\" from the set $\\{1, 2, 3, 4, 5\\}$, then I have a set of even numbers $\\{2, 4\\}$. $\\{1, 2, 3, 4, 5\\} \\cap \\{2, 4\\} = {2, 4}$ I think - but please correct me if my logic is wrong\n\u2013\u00a0Lou\nMay 26 at 16:46\n\u2022 So I think \"include\" as an operation represents an intersection, but \"exclude\" represents a set difference?\n\u2013\u00a0Lou\nMay 26 at 16:48\n\u2022 Set differences are also intersections. $A\\setminus B = A\\cap B^c$, where $B^c$ is the complement of $B$. May 27 at 8:38\n\n@KevinS offers an excellent answer from a logic point of view. Here's another that relies on the idea of a filter. That's a concept useful in programming (particularly in lisp). You pass the items in your set through a filter that lets some through and blocks others.\n\nIn this sense \"inclusion\" and \"exclusion\" are really different ways to describe the same result. You can specify what you keep or what you reject. \"Keep (just) the odds\" is the same as \"reject (only) the evens\".\n\nIf you have two filters each of which is described independently of the other and refers only to properties of the things you are filtering then you can filter in either order.\n\nThe independence matters. If you think that the wine must match the main course then your options will depend on whether you see the wine list or the menu first.\n\n\u2022 You've hit upon my actual use case - I'm doing filtering in Python and trying to understand if \"inclusion\" and \"exclusion\" for my use case are order-sensitive, or commutative operations. You've all helped me understand that they are commutative and therefore order should not matter.\n\u2013\u00a0Lou\nMay 26 at 10:52\n\nThis is perhaps easier to see in the math logic. When you apply conditions to a set, you get subsets that satisfy those conditions (possibly the emptyset if the conditions aren't met). In your example, define: $$A := \\{n\\in\\mathbb{N}\\text{ }|\\text{ }1\\leq n\\leq 10\\}.$$ Then $$B$$ and $$C$$ are had by adding conditionals in the set definition: $$B := \\{n\\in A\\text{ }|\\text{ }\\exists k\\in\\mathbb{N}: n=2*k\\}$$ and $$C:= \\{n\\in B\\text{ }|\\text{ }n\\geq 6\\}.$$ $$\\implies C = \\bigg\\{n\\in \\mathbb{N}\\text{ }\\bigg|\\text{ }(1\\leq n\\leq 10)\\wedge(\\exists k\\in\\mathbb{N}: (n=2*k))\\wedge(n\\geq 6)\\bigg\\}.$$ From this viewpoint, the reductions made were simply conjuctions (a logical operation). Conjunction is certainly commutative.\n\nAlso, the term \"exclusion\" is synonymous with set difference: $$X-Y:= X\\cap Y^c,$$ whereas \"inclusion\" usually refers to an injective map: $$\\iota: S\\hookrightarrow X$$ which can be used as an identifier of a subset. I wouldn't say the two notions are synonymous.\n\n\u2022 I'm a beginner in set theory, would you mind clarifying what you mean about an \"injective map\"? I understand the term set difference.\n\u2013\u00a0Lou\nMay 26 at 14:21\n\u2022 It seems as though the terms used might have multiple meanings as they apply to math and programming. An injection is a 1-1 function. The math notion I gave generalizes to categories (using equivalence classes of monomorphisms to identify sub-objects). This is not inherently programmable and is best used for theory. May 26 at 21:34\n\nTo your first question, consider using the set-builder notation to build a subset. That is, define $$B\\subseteq A$$ such that $$B = \\{a\\in A : \\Phi(a)\\}$$, where $$\\Phi$$ is the logical formula which dictates what how we pick the values of $$A$$.\n\nNow say that $$A$$ is some collection of numbers, and $$\\Phi$$ is the rule \"include all even numbers\". Then some other rule $$\\Psi$$ which says \"do not include all not even numbers\" is completely logically equivalent to $$\\Phi$$.\n\nAnd this generalises nicely by considering that, if we impose a condition like \"even number\", any element of a set will satisify that condition, or not satisfy that condition. We cannot have an element which does neither.\n\nThe second question follows as others have pointed out, by writing the conditions as operations which are commutative.", "date": "2022-06-25 17:57:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4459036/mathematical-inclusion-and-exclusion-of-elements-from-a-given-set-a", "openwebmath_score": 0.7210046648979187, "openwebmath_perplexity": 336.9144159564521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924785827003, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.857713240069462}}
{"url": "https://math.stackexchange.com/questions/1962884/chance-of-each-player-getting-an-ace-when-the-dealer-has-12-cards", "text": "# Chance of each player getting an ace when the dealer has 12 cards?\n\n4 players are playing a card game wherein each of them gets 10 cards, and the dealer gets 12 cards. What is the probability that each player gets an ace?\n\nI want the use the $p = \\dfrac{n_A}{N}$ method, where $n_A$ equals the favourable outcomes and $N$ equals all possible outcomes.\n\nStarting with $N$, I figured we could consider the dealer to be a fifth player, and considering we don't care about the order of the players we'd get:\n\n$$N = \\dfrac{52!}{(10!)^4\\times12!} \\times \\dfrac{1}{5!}$$\n\nNow for $n_A$, the aces can be divided among the players in $4!$ ways, and each of the players would still get 9 other cards from a total of 48, with the dealer getting the remaining twelve, thus giving us: $$n_A = 4! \\times \\dfrac{48!}{(9!)^412!} \\times \\dfrac{1}{5!}$$\n\nBut if we calculate $p$ this way we get a probability of $\\approx 3\\%$, which is just intuitively orders of magnitude too large to be correct, so I am sure I made a mistake somewhere. Can anyone help me spot it and then explain what I did wrong?\n\n\u2022 Why the \"$\\pm$\" in $\\pm3\\%$? \u2013\u00a0Barry Cipra Oct 10 '16 at 23:09\n\u2022 @BarryCipra I meant \"approximately equal to\" but I couldn't find the sign for that on my keyboard, so I thought \u00b1 would be the next best thing \u2013\u00a0YakSal Tafri Oct 11 '16 at 7:01\n\u2022 The way to get $\\approx$ is to type $\\approx$. Changed it. \u2013\u00a0BruceET Oct 11 '16 at 7:54\n\nI wouldn't consider the dealer as a fifth player but instead, let me guide you through another way to get the answer using combinatorics.\n\nWe should start by counting $N$ as ${52 \\choose 10}{42 \\choose 10}{32 \\choose 10}{22 \\choose 10}$ for the number of ways the dealer can give ten cards from 52 to each of the four players.\n\n$N = {52 \\choose 10}{42 \\choose 10}{32 \\choose 10}{22 \\choose 10} \\approx 971089585681469963688868551062400$\n\nNow for $n_A$ we will consider all four aces given in $4!$ and count for the number of ways nine cards from the remaining 48 can be given to each of the four players as ${48 \\choose 9}{39 \\choose 9}{30 \\choose 9}{21 \\choose 9}$.\n\n$n_A = 4!{48 \\choose 9}{39 \\choose 9}{30 \\choose 9}{21 \\choose 9} \\approx 35869963456698493441273194240000$\n\nHence\n\n$p = {n_A \\over N} = {4!{48 \\choose 9}{39 \\choose 9}{30 \\choose 9}{21 \\choose 9}\\over {52 \\choose 10}{42 \\choose 10}{32 \\choose 10}{22 \\choose 10}} ={400 \\over 10829} \\approx 0.036$\n\nComment: Because 0.036 doesn't seem to match the answer you anticipated, and because there was at least one false start towards a combinatorial answer, I decided to simulate the 'deal' a million times in R statistical software, and see what proportion of deals gave one ace to each player. I got $0.037 \\pm 0.0004.$ So I think @AlfredoLozano's method is correct (but note that $400/10829 = 0.03693785 \\approx 0.037).$\n\nMy deck has 1's for Aces and 0's for all other cards for simplicity counting results, but the sample function treats each 'card' as distinct. The m-vector each.1 is 'logical' with elements TRUE and FALSE; its mean is the proportion of its TRUEs.\n\nm = 10^6; each.1 = logical(m)\ndeck=c(1,1,1,1, rep(0,48))\nfor (i in 1:m) {\nd = sample(deck, 40)\neach.1[i] = (sum(d[1:10]==1)&sum(d[11:20]==1)&sum(d[21:30]==1)&sum(d[31:40]==1))\n}\nmean(each.1)\n## 0.037124\n\n\nHere is another way to get the answer. Imagine the dealer deals the first $10$ cards to player A, the next $10$ to player B, the next $10$ to player C, the next $10$ to player D, and keeps the rest for himself. Clearly the cards can be shuffled in any of $52!$ ways. The question becomes, in how many ways can the dealer \"stack\" the deck so that each player gets an Ace?\n\nFirst, remove the Aces from the deck and shuffle the remaining $48$ cards, which can be done in $48!$ ways. Then shuffle the $4$ Aces, in $4!$ ways. Then insert the first Ace so it's one of the first $10$ cards in the deck, the second Ace so it's one of the next $10$ cards, and so forth. All this can be done in $48!\\times4!\\times10^4$ ways. So the the probability of each player getting an Ace is\n\n$${48!\\times4!\\times10^4\\over52!}={24\\times10^4\\over52\\times51\\times50\\times49}=24\\left(10\\over52\\right)\\left(10\\over51\\right)\\left(10\\over50\\right)\\left(10\\over49\\right)\\approx25\\left(1\\over5\\right)^4=0.04\\%$$\n\nThe exact answer, as a reduced fraction, is ${400\\over10829}=0.0369378\\ldots$", "date": "2019-08-23 11:23:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1962884/chance-of-each-player-getting-an-ace-when-the-dealer-has-12-cards", "openwebmath_score": 0.7332857847213745, "openwebmath_perplexity": 297.1490344393237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540710685614, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8577055556593047}}
{"url": "http://mathhelpforum.com/discrete-math/133340-partial-orders-question-print.html", "text": "# Partial orders - Question\n\n\u2022 Mar 11th 2010, 12:38 PM\ngate13\nPartial orders - Question\nGood day to all,\n\nWe have just started relations and I came across the following question in my textbook:\n\nLet X = {1, 2}. List all the partial orders that can be defined on X.\n\nMy solution:\n\nI began by computing X x X (Cartesian product) which gave me:\n\nX x X ={(1,1), (1,2), (2,1), (2,2)}\n\nSince we are looking for partial orders, the relation has to be simultaneously reflexive, antisymmetric and transitive.\n\nIf the relation R on X is reflexive then it must contain (1,1) and (2,2)\n\nThe ordered pairs (1,2) and (2,1) cannot belong to R for if they did and R is also antisymmetric then that would imply 1=2, which is false.\n\nTherefore I concluded that the list of partial orders is the set:\n{(1,1), (2,2)}\n\nI was wondering if my logic is flawed and if so what errors have I committed?\n\nFinally, is it possible in this problem to determine the number of partial orders (cardinality)?\n\nAny advice would be greatly appreciated.\n\nKindest regards\n\u2022 Mar 11th 2010, 12:44 PM\nPlato\nWould $\\{(1,1),(1,2),(2,2)\\}$ also work?\nWhy or why not?\n\u2022 Mar 11th 2010, 12:58 PM\ngate13\nThank you Plato for your quick response.\n\nI am not sure. The definition of antisymmetric states:\n\nfor every a,b in X ((a,b) belongs to R and (b,a) belongs to R implies a=b)\n\nIf (1,2) belongs to R and (2,1) does not belong to R (based on the set you listed) then the hypothesis of the implication is false which means that the implication is true. If this is a valid reasoning could we not say the same for the set: {(1,1), (2,1), (2,2)}\n\nSlightly confused!\n\u2022 Mar 11th 2010, 01:06 PM\nPlato\nBut $(2,1)\\notin\\{(1,1),(1,2),(2,2)\\}$. Is it?\nAntisymmetric says that in both $(a,b)\\in R~\\&~(b,a)\\in R$ then it must be true that $a=b$.\n\u2022 Mar 11th 2010, 01:35 PM\ngate13\nThere is something that I am obviously not understanding. (2,1) does not belong to the set you listed.\n\nI believe the set http://www.mathhelpforum.com/math-he...26e4344e-1.gif is antisymmetric, since for it not to be antisymmetric we would have to have: http://www.mathhelpforum.com/math-he...0853db63-1.gif and a different than b. Therefore the set you provided is a partial order as well (reflexive, transitive and antisymmetric).\n\nI apologize as I realize this may be obvious to many.\n\u2022 Mar 11th 2010, 01:38 PM\nPlato\nQuote:\n\nOriginally Posted by gate13\nI believe the set http://www.mathhelpforum.com/math-he...26e4344e-1.gif is antisymmetric, since for it not to be antisymmetric we would have to have: http://www.mathhelpforum.com/math-he...0853db63-1.gif and a different than b. Therefore the set you provided is a partial order as well (reflexive, transitive and antisymmetric).\n\nThat is correct. And there is one more p.o. on the set.\nWhat is it?\n\u2022 Mar 11th 2010, 01:43 PM\ngate13\nI believe the other partial order would be :\n\n{(1,1), (2,1), (2,2)} for the same reasons listed previously.\n\u2022 Mar 11th 2010, 02:01 PM\ngate13\nI just wanted to thank you Plato for your time and explanations. Many thanks.", "date": "2016-09-28 15:51:27", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/133340-partial-orders-question-print.html", "openwebmath_score": 0.791178822517395, "openwebmath_perplexity": 930.6686852491489, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979354068055595, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8577055530205879}}
{"url": "https://mathhelpboards.com/threads/graduate-potw.3659/", "text": "### Welcome to our community\n\n#### dwsmith\n\n##### Well-known member\nSo on the recent graduate problem of the week, I saw that $\\int_0^{\\infty}\\frac{\\sin x}{x}dx = \\frac{\\pi}{2}$, but so does, $\\int_0^{\\infty}\\frac{\\sin^2 x}{x^2}dx = \\frac{\\pi}{2}$.\nHow can they both be the same?\n\n#### Chris L T521\n\n##### Well-known member\nStaff member\nSo on the recent graduate problem of the week, I saw that $\\int_0^{\\infty}\\frac{\\sin x}{x}dx = \\frac{\\pi}{2}$, but so does, $\\int_0^{\\infty}\\frac{\\sin^2 x}{x^2}dx = \\frac{\\pi}{2}$.\nHow can they both be the same?\nLet us use integration by parts to compute $\\displaystyle\\int_0^{\\infty}\\frac{\\sin^2 x}{x^2}\\,dx$. At the end, we will need to use the fact that $\\displaystyle\\int_0^{\\infty}\\frac{\\sin x}{x}\\,dx=\\frac{\\pi}{2}$\n\nLet $u=\\sin^2x$ and $\\,dv=\\dfrac{\\,dx}{x^2}$. Then $\\,du=2\\sin x\\cos x\\,dx=\\sin(2x)\\,dx$ and $v=-\\dfrac{1}{x}$. Therefore,\n$\\int_0^{\\infty}\\frac{\\sin^2 x}{x^2}\\,dx = \\left[-\\frac{\\sin^2 x}{x}\\right]_0^{\\infty}+\\int_0^{\\infty}\\frac{\\sin(2x)}{x}\\,dx=\\int_0^{\\infty}\\frac{\\sin(2x)}{x}\\,dx.$\n(We note that $|\\sin x|\\leq 1\\implies |\\sin^2 x|\\leq 1$ and thus $\\displaystyle\\lim_{x\\to\\infty} \\frac{\\sin^2 x}{x}\\sim \\lim_{x\\to\\infty} \\frac{1}{x}=0$; We also note that $\\displaystyle\\lim_{x\\to 0}\\frac{\\sin^2 x}{x}=\\lim_{x\\to 0}\\frac{\\sin x}{x}\\cdot\\lim_{x\\to 0}\\sin x=0$. Hence, that's why the $\\displaystyle\\left[-\\frac{\\sin^2 x}{x}\\right]_0^{\\infty}$ term goes to zero.)\n\nNow let $t=2x\\implies\\,dt=2\\,dx$. Therefore,\n$\\int_0^{\\infty}\\frac{\\sin(2x)}{x}\\,dx\\xrightarrow{t=2x}{} \\int_0^{\\infty}\\frac{\\sin t}{t/2}\\frac{\\,dt}{2}=\\int_0^{\\infty}\\frac{\\sin t}{t}=\\frac{\\pi}{2}.$\nAnd thus, we also have that $\\displaystyle\\int_0^{\\infty}\\frac{\\sin^2 x}{x^2}\\,dx =\\frac{\\pi}{2}$.\n\nI hope this makes sense!\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\n$$F(a)=\\int^{\\infty}_0\\frac{\\sin^2(ax)}{x^2}$$\n\nDifferentiate w.r.t a :\n\n$$F'(a)=\\int^{\\infty}_0 \\frac{\\sin(2ax)}{x}$$\n\nLet 2ax=t\n\n$$F'(a)=\\int^{\\infty}_0 \\frac{\\sin(t)}{t}=\\frac{\\pi}{2}$$\n\n$$F(a)=\\frac{\\pi}{2}a+C$$\n\nPutting a =0 we get C = 0 hence\n\n$$\\int^{\\infty}_0\\frac{\\sin^2(ax)}{x^2}=\\frac{\\pi \\cdot a}{2}$$\n\nSo for a =1 we get our result :\n\n$$\\int^{\\infty}_0\\frac{\\sin^2(x)}{x^2}=\\frac{\\pi}{2}$$\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nIf your question is why such thing happen , then I don't know , to me it is pretty strange !\n\nIf you see the graph of both functions , then you have no indications ...\n\n#### dwsmith\n\n##### Well-known member\nI just thought it was strange. When I took Theory of Complex Variables, I had the $\\int_0^{\\infty}\\frac{\\sin^2x}{x^2}dx = \\frac{\\pi}{2}$ exercise so I was surprised to see that $\\frac{\\sin x}{x}$ lead to the same conclusion.\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nIn complex analysis $$\\int^{\\infty}_0 \\dfrac{1-\\cos(x)}{x^2}$$ and $$\\int^{\\infty}_0 \\frac{\\sin(x)}{x}$$ are conventional exercises to solve by contour integration ....", "date": "2020-10-01 12:46:29", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/graduate-potw.3659/", "openwebmath_score": 0.9228602647781372, "openwebmath_perplexity": 1016.8920947623308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9793540710685614, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8577055461340481}}
{"url": "http://math.arizona.edu/~restrepo/475A/Notes/sourcea-/node21.html", "text": "Next: Contraction Mapping Up: SOLUTION OF NONLINEAR EQUATIONS Previous: Secant Method\n\n## Fixed Point Iteration\n\ndef: Fixed Point is the value such that \u00a0\u00a0\u00a0\u00a0 .\n\nFixed Point problems and root-finding problems are equivalent: let . Hence, if a function has a fixed point then has a root.\n\nThree Problems:\n\n1. What functions have a fixed point?\n2. How do we determine the fixed point?\n3. Is the fixed point unique?\nExample \u00a0\u00a0\u00a0\u00a0 has a fixed point at each ....just plot.\n\nExample has 2 fixed points in \u00a0\u00a0\u00a0\u00a0....plot, again.\n\nTheorem (Existence and Uniqueness): If and then has a fixed point in .\n\nSuppose, in addition, that exists on and that a positive constant exists with\n\nthen the fixed point in is unique.\n\nProof:\n\nIf or , then existence of fixed point is clear. Suppose not, then it must be true that and . Define . Then is continuous on and\n\nThe Intermediate Value Theorem implies that there exist for which . Thus, is fixed point of .\n\nholds and that and are both fixed points in with . By the Mean Value Theorem a number exists between and such that\n\nThis contradiction comes from and the fixed point is unique.\n\nFixed-Point Iteration\n\nPick a and generate a sequence such that If the sequence converges to and is continuous then by the theorem above:\n\nThe algorithm is depicted in Figure 18\n\nFixed Point Algorithm\n\nInput: TOL,\nOutput: or message of failure\nStep 1: Set\nStep 2: While do Steps 3 - 6\nStep 3: Set \u00a0\u00a0\u00a0\u00a0% Compute\nStep 4: if TOL then\noutput ; \u00a0\u00a0\u00a0\u00a0% found not\nStop\nStep 5: Set\nStep 6: Set \u00a0\u00a0\u00a0\u00a0Update .\nStep 7: Output (Iterations exceeded. )\nEND\n\nTheorem: (Fixed Point Iteration) Let and suppose . Suppose in addition that is continuous on with\n\nif then for any in , the sequence\n\nconverges only linearly to the unique fixed point in\n\nProof: Fixed Point Theorem says . Since exists on apply mean value Theorem to to show that for any\n\nsince , and . Since is continuous on we have\n\nthus\n\nRemark: Can get higher-order convergence when\n\nTheorem: Suppose solution of . and continuous and strictly bounded by on an open interval containing . Then such that . The sequence converges quadratically:\n\nFixed Point Iteration: let and suppose . Suppose in addition that exists on with\n\n (6)\n\nif is any number in , then\n\nconverges to unique fixed point in .\n\nProof: From fixed point theorem, a unique fixed point exists in . Since maps into itself, the sequence is defined and .\n\nUsing (6) and the Mean Value Theorem\n\nBy induction\n (7)\n\nSince\n\nand\n\nCorollary If satisfies hypothesis of Fixed Point Iteration theorem, a bounds for the error involved in using to approximate are given by\n\nProof: (a) follows from (7):\n\nFor\n\ntherefore for\n\nSince then\n\nSince then\n\nRemark: Rate depends on . The smaller , the faster it converges.\n\nExample: Consider for . This function is illustrated in Figure 19. Get matlab code used in the example.\n\nFirst we wish to ensure that the function maps into itself.\n\nNext we look at the derivative of\n\nThis fulfills the requirements for a unique fixed point to exist in . It also ensures that if we start with any non-negative value we will converge to the fixed point. The table below shows the first ten iterations for three different values of . Figure 20a and Figure 20b illustrate the iteration history and the logarithm of the error, for a case starting with .\n\nFigure 21a and Figure 21b illustrate the iteration history and the logarithm of the error, for a case starting with .\nFinally, Figure 22a and Figure 22b illustrate the iteration history and the logarithm of the error, for a case starting with .\n\nThe iterations for the three different starting points all appear to converge on . The log error plots are straight lines and they all have the same slope, indicating the same rate of convergence.\n\nWe can also prove analytically that is the fixed point of . A fixed point of satisfies\n\nWe can rearrange this to get which has one real root, .\n\nExample: Consider the function for . We will start with the initial value and consider what happens for various values of . -Figure show the iterations for , respectively.\n\nThey also plot on the same graph as so we can see the fixed point, and finally plot the log of the error for each value of . We can see that for both and the iteration converges to a fixed point. The log error plots are straight lines with the slope showing the convergence rate (Question: Why does the log error plot flatten off for ?). In the case we can see that it converges faster than the case . For the fixed point does not converge but seems to bounce around different values in the interval . In fact for values of between and we get all sorts of interesting beheviour. For more information on this click here. But when we make less than 0.5 the iteration is able to escape from the interval and once it does this it increases rapidly with each iteration until it causes an overflow error in the computer.\n\n(b) . (c) . (d) .\">Get matlab code used in the example.\n\nNow lets see whether we can understand what is happening. First let us look at the range of the function\n\nThis shows why the iterations blow up for less than 0.5. For the range is not within the domain of (i.e. ) and so points may 'escape'. However for any value of greater than a half the range is mapped to within the domain.\n\nNext we need to look at the derivative of\n\nThe magnitude of the derivative is only less than one for all values of if . Thus for any value of greater than two the fixed point theorem holds and we have guaranteed convergence. We know, however, that we still get convergence to a fixed point for some values of less than two. What is happening in these cases?\n\nIf the magnitude of the derivative will be less than one for . As long as the fixed point lies within this interval the theorem tell us that there will be a region around the fixed point where iterations will converge to the fixed point. This is the case as long as . As it turns out, we may still start at any point within and we will eventually arrive at the fixed point although convergence takes longer and longer the closer is to the critical point.\n\nFor values of the fixed point still exists but it becomes unstable (i.e. If you start close to the fixed point and iterate you will move away from it rather than towards it).\n\nIf we plot and the line on the same graph we can see that there is only one fixed point within the interval for all values of . In fact we can calculate the value of the fixed point analytically by solving .\n\nThis is a simple quadratic equation with two solutions\n\nFor only the smaller of the two solutions lies within the interval and is the unique fixed point.\n\nSubsections\n\nNext: Contraction Mapping Up: SOLUTION OF NONLINEAR EQUATIONS Previous: Secant Method\nJuan Restrepo 2003-04-12", "date": "2016-06-27 13:10:24", "meta": {"domain": "arizona.edu", "url": "http://math.arizona.edu/~restrepo/475A/Notes/sourcea-/node21.html", "openwebmath_score": 0.8795111775398254, "openwebmath_perplexity": 436.8466411493506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9934102295074974, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8577020521810251}}
{"url": "https://www.jiskha.com/questions/304464/what-substitution-could-i-use-to-integrate-a-a-2-x-2-3-2-dx", "text": "Math\n\nWhat substitution could I use to integrate\n\na/(a^2 + x^2)^3/2 dx\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. Let u = x/(x^2 + a^2)^1/2\n\nand you will find that\n\n(1/a^2)* du\n= integral of dx/(x^2+a^2)^3/2\nwhich is the integral you want.\nTherefore u/a^2\n= (x/a^2)/(x^2 + a^2)^1/2\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n2. Computer program says the answer is\n\nx/(a*(a^2 + x^2)^(1/2))\n\nwhich is slightly different from your answer. Thanks so much for the help on this one. I was really stuck.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. I did x = a tan u\ndx = a (sec u)^2 du\n\nint of a/(a^2 + x^2)^3/2 dx\n= int of (a sec u)^2/(a^2 + (a tan u)^2)^3/2 du\n= int of (a sec u)^2/(a sec u)^3 du\n= int of (cos u)/a du\n= (sin u)/a + K\nsince u = atan (x/a)\n= x/(a*(a^2 + x^2)^(1/2)) + K\n\nThanks again...\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\nSimilar Questions\n\n1. math, calculus 2\n\nConsider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals integrate from a to b of f(x)dx + integrate\n\nWhat advantages does the method of substitution have over graphing for solving systems of equations? Choose the correct answer below. A. Graphing cannot be used to solve dependent or inconsistent systems due to the nature of the\n\n3. Math\n\nfind the volume bounded by the parabolic cylinder z=4-x^2 and the planes x=0, y=0, y=6 and z=0 I just want some help figuring out the limits of this question : So for z, we have to integrate from z=0 to z=4-x^2 (am i right?)\n\n4. Geometry\n\nSubstitution Property of Equality: If a=20, then 5a= _____. The substitution rule states that If a=b then a can be substituted for b in any equation or expression. So I think that 5a=5b but the number answer is 100. Am I correct?\n\n1. Math\n\n1. Explain the advantages and disadvantages of both the substitution method and the elimination method for solving a system. Help please. The only reasons I can think of is because in substitution it might be more difficult cause\n\n2. Math\n\n(i) Evaluate integral [ x^3 / (x^2 + 4)^2 ] using trigonometric substitution. (ii) Evaluate integral [ x^3 / (x^2 + 4)^2 ] using regular substitution. (iii) Use a right triangle to check that indeed both answers you obtained in\n\n3. Math\n\nEvaluate the integral of x*ln(1+x) dx by first making a substitution and then using integration by parts. I let u=1+x and du = dx but then ln(u) du can't integrate to anything?\n\n4. Algebra II\n\nUse synthetic substitution to find f(-3) for f(x) = 2x^3 - 6x^2 - 5x + 7. _3|2 -6 -5 7 6 36 93 ______________ 2 0 31|100 then I used direct substitution and just plugged -3 in for x and the final answer is f(-3) = 86\n\n1. Calculus\n\ndy/dx = 4ye^(5x) a) Separate the differential equation, then integrate both sides. b) Write the general solution as a function y(x). For the second part, I got y(x)=e^((5e^(5x))/(5)) + C but I don't understand how to separate\n\n2. Math Integration\n\nIntegrate: f (x)/(2x + 1) dx let f represent integrate sign let u = x, du = dx => dx = du = f (u)/(2u + 1) du = (2u + 1)^(-1) du = (1/2)u^2 (ln|2u + 1|) + c = (1/2)x^2 (ln|2x + 1|) + c ...what did I do wrong? The correct answer is\n\n3. Calculus 2\n\nEvalute the integral of x/(x^2+4)dx using u-substitution and then trigonometric substitution. I did this and got (1/2)ln(x^2+4)+C using u-substitution and x^2+4+C using trigonometric substitution. What did I do wrong??? thank you\n\n4. algebra\n\ni need help with these problems (either a solution or how to plug them into a graphing calculator): 1) Solve by substitution or elimination 4/x + 1/y + 2/z = 4 2/x + 3/y - 1/z = 1 1/x + 1/y + 1/z = 4 2)Solve the system of", "date": "2021-10-28 14:10:00", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/304464/what-substitution-could-i-use-to-integrate-a-a-2-x-2-3-2-dx", "openwebmath_score": 0.9295290112495422, "openwebmath_perplexity": 1938.0844047931344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9585377272885904, "lm_q2_score": 0.8947894590884704, "lm_q1q2_score": 0.8576894545164496}}
{"url": "https://math.stackexchange.com/questions/2484250/all-invariant-subspaces-of-a-linear-operator-t", "text": "# All invariant subspaces of a linear operator $T$\n\nLet $T \\in \\mathscr L(\\mathbb F^n)$ such that $T(x_1,x_2,...,x_n)=(x_1,2x_2,...,nx_n)$. Then find all the invariant subspaces of $T$. Clearly, $Null$ $T$ and $Range$ $T$ are two invariant subspaces.\n\nAlso, all the subspaces spanned by the eigen vectors form $1$-dimensional invariant subspaces. In this case the eigen values of $T$ are $i$, where $i$$\\in\\{1,2,...,n\\} and the corresponding eigen vector is of form (0,...,0,a,0,...,0) where a(\\neq 0)\\in \\mathbb F is the i^{th}-component of the vector. But what about the invariant subspaces of other dimensions? For instance, if W is an invariant subspace of T of dimension k, then T|_W is a linear operator on W. So, for any w\\in W, Tw\\in W. If w\\in span(e_1,e_2,...,e_k) then Tw\\in span(e_1,2e_2,...,ke_k). Since, dim W= dim Tw=k, we can say that Tw=span(e_1,2e_2,...,ke_k)\\in span(e_1,e_2,...,e_k)=W i.e Tw\\in W. So, can we conclude from here that there will be invariant subspaces of all dimensions under T, given by span(e_1,e_2,...,e_k), where 1\\leqslant k\\leqslant n (which are precisely 2^n in number)? \u2022 What do you think the invariant subspaces are? \u2013 Demophilus Oct 22 '17 at 13:32 \u2022 @Demophilus If W is an invariant subspace of T of dimension k, then T|_W is a linear operator on W. So, for any w\\in W, Tw\\in W. If w\\in span(e_1,e_2,...,e_k) then Tw\\in span(e_1,2e_2,...,ke_k). Since, dim W= dim Tw=k, we can say that Tw=span(e_1,2e_2,...,ke_k)\\in span(e_1,e_2,...,e_k)=W i.e Tw\\in W. So, can we conclude from here that there will be invariant subspaces of all dimensions under T, given by span(e_1,e_2,...,e_k), where 1\\leqslant k\\leqslant n? \u2013 JackT Oct 22 '17 at 13:52 ## 2 Answers I'll work in characteristic zero to avoid quirks due to the field. Suppose that V is an invariant subspace, and let v\\in V. Assume for a moment that all entries of V are nonzero. We have that v,Tv,T^2v,T^3v,\\ldots,T^{n-1}v\\in V. If \\alpha_0T^0v+\\cdots+\\alpha_{n-1}T^{n-1}v=0, we get a linear system with matrix$$ A=v_0v_1v_2\\cdots v_{n-1}\\begin{bmatrix} 1&1&1&\\cdots&1\\\\1&2&3&\\cdots&n-1\\\\ 1&2^2&3^2&\\cdots&(n-1)^2\\\\ \\vdots&\\vdots&\\vdots&\\cdots&\\vdots\\\\ 1&2^{n-1}&3^{n-1}&\\cdots&(n-1)^{n-1} \\end{bmatrix} $$This is (one of) the well-known Vandermonde Matrix, which is invertible. So the only possible solution of the system is given by \\alpha_0=\\alpha_1=\\cdots=\\alpha_{n-1}=0, and thus v,Tv,\\ldots,T^{n-1}v are linearly independent, which means that \\dim V=n and so V=\\mathbb F^n. So any invariant subspace will be made of vectors with at least one entry equal to zero. On the \"nonzero part\" of the subspace we can repeat the above reasoning, to conclude (as you suggested in your comment) that any invariant subspace of T is of the form$$ W=\\text{span}\\,\\{e_j:\\ j\\in K\\} $$for some K\\subset\\{1,\\ldots,n\\}. \u2022 :Thanks a lot... \u2013 JackT Oct 22 '17 at 14:07 Well, you can get 2^n (instead of just n) subspaces off the bat, by considering \\operatorname{span}(B), where B ranges over all subsets of \\lbrace e_1, \\ldots, e_n \\rbrace. Included in this are the whole space and the trivial space (spanned by the empty set), which are the range and nullspace respectively (unless the field has non-zero characteristic). The question is, are there any others? First, suppose \\mathbb{F} = \\mathbb{C}, and U is an invariant subspace, so that T|_U is an operator. It also will be diagonalisable. To see this, suppose \\lambda is an eigenvalue for T|_U, and consider$$\\operatorname{null}(T|_U - \\lambda I|_U)^2 \\subseteq \\operatorname{null}(T - \\lambda I)^2 = \\operatorname{null}(T - \\lambda I) = \\operatorname{null}(T_U - \\lambda I|_U),$$so the generalised eigenspace is no larger than the eigenspace. Hence a basis for$U$, consisting of eigenvectors for$T|_U$(and hence$T$) exists, so$U$must be of the above form. If$\\mathbb{F} = \\mathbb{R}$, the same argument mostly works. You just need to tiptoe around the complex case. If you extend the map to the complexification of the space, the argument works exactly as is. If$\\mathbb{F}\\$ is a different field, I'm not sure if I can help.\n\n\u2022 You are absolutely correct. \u2013\u00a0JackT Oct 22 '17 at 14:11", "date": "2020-01-22 14:51:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2484250/all-invariant-subspaces-of-a-linear-operator-t", "openwebmath_score": 0.9669172763824463, "openwebmath_perplexity": 620.3655724099261, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787838473909, "lm_q2_score": 0.868826777936422, "lm_q1q2_score": 0.8576873620173242}}
{"url": "https://math.stackexchange.com/questions/2416560/is-the-set-2-3-4-open-in-some-metric-spaces-and-not-open-in-others", "text": "# Is the set $\\{2, 3, 4\\}$ open in some metric spaces and not open in others?\n\nI just want to check my understanding. This is from Baby Rudin:\n\n2.18 Definition Let $$X$$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $$X$$.\n\n$$(a)$$ A neighborhood of $$p$$ is a set $$N_r(p)$$ consisting of all $$q$$ such that $$d(p, q) for some $$r>0$$. The number $$r$$ is called the radius of $$N_r( p)$$\n\n$$(e)$$ A point $$p$$ is an interior point of $$E$$ if there is a neighborhood $$N$$ of $$p$$ such that $$N \\subset E$$\n\n$$(f)$$ $$E$$ is open if every point of $$E$$ is an interior point of $$E$$.\n\nSuppose we have the metric space with set $$X=\\{1, 2, 3, 4, 5\\}$$ and distance function $$d(x, y)=|x-y|$$. Now $$2$$ is an interior point of $$\\{2, 3, 4\\}$$ because $$N_{0.5}(2)=\\{2\\} \\subset \\{2, 3, 4\\}$$ (and a similar argument can be made for $$3$$ and $$4$$ as well.\n\nBut if our metric space is $$\\mathbb{R}$$ with the same distance function, then $$\\{2, 3, 4\\}$$ is not open because no neighborhood of $$2$$ is a subset of $$\\{2, 3, 4\\}$$, so $$2$$ is not an interior point of $$\\{2, 3, 4\\}$$, right?\n\n\u2022 Right. That's completely correct. Sep 4 '17 at 16:01\n\u2022 Why \"no neighborhood of $2$ is a subset of $\\{2,3,4\\}$\"? You are right, but you need to add some explanation. Sep 4 '17 at 16:01\n\u2022 @Krish When we are dealing with the set $\\mathbb{R}$, every neighborhood will contain numbers which are not integers.\n\u2013\u00a0Ovi\nSep 4 '17 at 16:03\n\u2022 @Krish: Because any neighborhood of $2$ (by the definition given) has the form $(2-r,2+r)$ for some $r>0.$ This will readily contain some non-integer rational number. Sep 4 '17 at 16:03\n\u2022 @CameronBuie sorry!!! But I was just checking whether OP understood the reason clearly or not. (+1) for the question. Sep 4 '17 at 16:07\n\n## 1 Answer\n\nYou're absolutely right. Nicely reasoned!\n\nAnother thing to consider is that even when the underlying space is the same, using a different metric may yield different open sets. Letting our metric space be $\\Bbb R,$ but with the distance function $$\\delta(x,y):=\\begin{cases}0 & x=y\\\\1 & x\\neq y,\\end{cases}$$ we can show that (for example) $\\{2\\}$ is a neighborhood of $2$ with radius $\\frac12,$ and by similar reasoning conclude that $\\{2,3,4\\}$ is once again open.", "date": "2021-09-26 08:51:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2416560/is-the-set-2-3-4-open-in-some-metric-spaces-and-not-open-in-others", "openwebmath_score": 0.8805869817733765, "openwebmath_perplexity": 812.9368713799926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787868650146, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8576873579332592}}
{"url": "https://thorstenharte.de/the-motion-of-a-conical-pendulum-is.html", "text": "the motion of a conical pendulum is. Here are a number of highest rated Pendulum Period Equation pictures upon internet. A) What is the magnitude of the torque (N m) and direction (from the listed answers) of the torque exerted on the ball about the center of the circle of motion?. Expression for Period of Conical Pendulum: Let us consider a conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle with constant speed 'v' at the end of a string of length 'l'. Periodic events and periodic motion have long served as standards for units of time. Consider a mass m performing circular motion \u2026. Torsion pendulum: the weights can be moved as shown to change the rotational inertia, and therefore the period. Conical pendulum is a small heavy mass attached with massless, flexible and inextensible string suspended from a rigid support and the mass is constraint to\u00a0. According to Big Al, one of our first astronauts was successful because of his family background-his mother was spacey, and his father mooned people \u2026. The contents of this file are shown in Table 1 below. If the object is held by a piece of string, the string marks out a conical \u2026. Foucault's pendulum is another example of motion relative to the earth which exhibits the fact that the earth is not a Newtonian base. Define conical pendulum obtain an expression for the angle ma\u2026. You are to investigate the motion \u2026. The bob does not show any vertical motion (moves in horizontal circle only). Its construction is similar to an ordinary pendulum\u2026. Derive the general differential equation of motion for the pendulum \u2026. The resulting equation of motion \u2026. This example shows how to simulate a three link conical compound pendulum made up of cylindrical bars connecting particles at the joints. The motion of a pendulum--be it the simple, spherical or the conical--is a concrete example wherein the three types of motions discussed above are realized physically. a conical pendulum \u2013 the dotted pendulum bob shows the equilibrium position and the grey arrows show its trajectory for fixed R and theta. The figure at the right shows an idealized pendulum, with a \"massless\" string or rod of length L and a bob of mass m. This article introduces a pendulum element to a 3-spring vibration isolator to achieve a high-static-low-dynamic (HSLD) \u2026. The effective length of a second\u2019s pendulum is 99. If the angle the string makes with the vertical axis is 45. The plane and the supporting string trace a conical pendulum\u2026. Uniform Circular Motion - Conical Pendulum. Substituting into the equation for SHM, we get. The pendulum has also allowed astronomers and geologists to measure the motion\u2026. 1: Comparison of the simple pendulum and a true pendulum for a 10 o amplitude. Theoretically, The time period of a conical pendulum \u2026. 1 The Conical Pendulum A small ball of mass m is suspended from a string of length L. James Webb is a Conical Pendulum Orbiting the Sun : ja\u2026. Consider a bob of mass 'm' tied to one end of a string of length 'l l ' and other end fixed to rigid support (S). Besides, a restoring force always acts on the bob of the pendulum \u2026. This is similar to a conical pendulum. The string of a conical pendulum sweeps out a right-circular cone. However the most interesting variable is, the length of the swinging pendulum\u2026. Add your answer and earn points. Suppose that an object of mass is attached to \u2026. Swinging during a fixed period as a gravity pendulum of a certain weight, the swing of the pendulum decays faster if the damping is larger, and more slowly if the damping is smaller [1]. CONICAL PENDULUM A conical pendulum consists of a weight (or bob) fixed on the end of a string or rod suspended from a pivot. For the conical pendulum gravity mass and the angular velocity determine the angle between the string and the vertical axis of the pendulum\u00a0. A conical pendulum is formed by attaching a mass m to a string of length L, then allowing the mass to swing in a horizontal circle. The motion is adjusted to be horizontal near the center of mass of the hanging portion. A mass oscillating on a spring is an example of an object moving with simple harmonic motion. The movement of a pendulum is called simple harmonic motion: when moved from a starting position, the pendulum feels a \u2026. a weight connected by a rod with a fixed point; Horrocks devised a model of planetary motion \u2026. It doesn't take much effort to keep the mass moving at a constant angular velocity at a constant radius. In our diagram the radius of the circle, r, is equal to L, the length of the pendulum\u2026. Examples of Oscillations are all around us. On 1 December 1659 Christiaan Huygens outlined the steps by which he had come to the momentous discovery that motion along an inverted cycloid is \u2026. The pendulum must swing freely so that it swings backwards and forwards in the same plane. The time period of a simple pendulum: It is defined as the time taken by the pendulum to finish one full oscillation and is denoted by \u201cT\u201d. For example, suspending a bar from a thin wire and winding it by an angle \\theta, a torsional torque \\tau = -\\kappa\\theta is produced, where \\kappa is a characteristic property of the wire, known as the torsional constant. If a simple pendulum is fixed at one end and the bob is rotating in a horizontal circle, then it is called a conical pendulum. Example at 20G mass moves as a conical pendulum with 8x vibrating length and speed V if the radius of the circular motion is 5x to find: i) the string tension (I assume g = 10 ms-2, (2 A conical pendulum \u2026. then acceleration of the body is proportional to displacement, but in the opposite direction of displacement. Let G be the centre of gravity of a compound pendulum of mass m that oscillates about a point O with OG = h If the pendulum \u2026. 012 kg, the string has length L = 0. Relevant Equations: The Lagrangian is defined as the difference of the kinetic energy T and the potential energy U. (PDF) Turning points of the spherical pendulum and the gol\u2026. This paper represents a continuation of the theoretical and computational work from an earlier publication, with the present calculations using exactly the same physical values for the lengths L (0. To set the pendulum in motion\u2026. Velocity And Acceleration In Shm Get Our FREE Chrome Extension. If the\u2026 A bar magnet has a magnetic moment 2. However, this doesn't affect the period of pendulum. \u03b8=suspension angle, r =radius of bob's circular motion, h =vertical height of suspension above the plane of the bob's motion\u2026. The formula for the period T of a pendulum is T = 2\u03c0 L g , where L is the length of the pendulum \u2026. The precession of a Foucault pendulum viewed as a beat phenomenon of a conical pendulum \u2026. diameter range Rasp 8\" (10 Piece) 5 for Wood, 5 Steel 4 x Second Cut: 200 mm (8\") Round, \u2026. Translation in spherical coordinates: where q = longitude, j = colatitude and \u2026. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum. Time period for conical pendulum= T = 2\u03bb\u221ar/gtan0. A conical pendulum is the simple pendulum whirled\u00a0. March 2013 by Sam Categories: Mechanics | Tags: conical, pendulum | Comments Off on Horizontal Circular Motion (Conical Pendulum). Consider a conical pendulum with a mass m, attached to a string of length L. For large motions it is a chaotic system, but for small motions it is a simple linear system. Study on the Motion Characteristics of two. How forces affect motion, and the nature of\u00a0. Healing, Let's GO!!! \u2013 Pendulum \u2026. Files: 2 movies (384 X 384 pixels). A rigid body mounted on a fixed horizontal axis, about which it is free to rotate under the influence of gravity. So, the expression for the time period of small oscillations is equal to 2\u03c0\u00d7 \u221a(length of the pendulum\u2026. Tie the thread to straw and tape the straw on the table in such a way that around half of an inch hangs over the edge. P is moving around the blue circle with angular velocity w. The simple harmonic motion is defined as a motion \u2026. In Figure 1 we see that a simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch \u2026. The Amazing Pendulum 2014. 1 Uniform Circular Motion Uniform circular motion is the motion of an \u2026. As the motion of the bob is a horizontal circular motion, the resultant force must be horizontal and directed towards the centre C of the circular motion\u2026. A conical pendulum consists of a bob of mass m in motion in a circular path in a horizontal plane as shown in Figure. This paper explains why the paths in the rotating frame are circular, and the difference between the periods for the right- and left-handed rotations is\u00a0. The object revolves with constant speed vin a hor-izontal circle of radius r,as shown in Figure 6. The SHO and Circular Motion \u2022 We can now see that the equation of motion of the simple pendulum at small angles\u2014which is a simple harmonic oscillator is nothing but the x-component of the steady circularmotion of the conical pendulum \u2022 The simple pendulum is the. This report shows how to find an approximate of \u2018g\u2019 using the simple pendulum experiment. The spin or conical motion of the bob may cause a twist in the thread, thereby \u2026. The course follows the typical progression of topics of a first-semester university physics course: Kinematics, Newton\u2019s Laws, Energy, and Momentum. Progressive education for money. 79]: to analyze the motion of a pendulum moving in a horizontal circle (a conical pendulum). The motion is regular and repeating, an example of periodic motion. In order to prove this fact consider a simple pendulum having a bob of mass ' m ' and the length of pendulum is ' l '. Spherical pendulum and vertical pendulum are the special. Step 1: Derive the Equation of Motion. The other is a conical pendulum which involves a pendulum \u2026. What is the formula for speed of pendulum at any point. First at the request of the Directors of the Indies Company I undertook for finding longitudes to construct clocks of which the sure and constant motion would be equal to a three-foot pendulum\u2026. The motion of the bob of simple pendulum simple harmonic motion if it is given small displacement. Single and Double plane pendulum Gabriela Gonz\u00b4alez 1 Introduction We will write down equations of motion for a single and a double plane pendulum, following Newton\u2019s equations, and using Lagrange\u2019s equations. A mass on a spring has the vibrating spring mode resonantly coupled to the pendulum \u2026. It also uses a simple pivot, perhaps a knife edge support, while a conical pendulum requires a more complex support with two-directional movement. The pendulum bob moves in a horizontal circle with constant speed v, \u03b8 = suspension angle,\u00a0. Coasting Through a Vertical Loop (with Reaction Forces) | QT Embedded | Media | Old Embedded |. The horizontal component of tension balances this centripetal. The mass is set in motion in a horizontal circular path about the vertical axis. Model the ball in the pendulum is a particle. The swinging incense burner called a censer, also known as a thurible, is an example of a pendulum. Answer Text: The conical pendulum: -It consists of a small massive object tied to the end of a thin string tied to affixed rigid support. Answer: Let us consider a conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle with constant speed 'v' at the end of a string of length 'l'. A conical pendulum consists of an object attached to a string and moving in a horizontal circle. The time taken by the pendulum to reach x = a/2 from the mean postion will be: Q7. This is different from Kepler's 3rd law where P \u221d r**(1/2), where r is the mean orbital radius. As shown in the figure above the driving force is F=-mgsintheta where the -ve sign implies that the. circular motion physics gcse amp a level revision and. Figure 6 illustrates this concept. the constant angular velocity of the bob. Circular motion with conical pendulum Number 135710-EN Topic Mechanics, two-dimensional motion Version 2017-02-17/HS Type Student exercise \u2026. Bohr Magneton Spin Magnetic Moment Of Lectron Calculator. iii) The bob of a conical pendulum under goes (A) Rectilinear motion \u2026. 85 m and the angle with the vertical is 37\u00b0. A conical pendulum is a weight (or bob) fixed on the end of a string (or rod) suspended from a pivot. 2 The Conical Pendulum A small object of mass mis suspended from a string of length L. 10 L m a) Draw a labeled free-body diagram for the pendulum \u2026. We will derive the equation of motion for the pendulum using the rotational analog of Newton's second law for motion about a fixed axis, which is \u2026. Seat pan and sprinkle some the list \u2026. In simple languages definition of a conical pendulum can be summed up as a conical pendulum consists of a weight (or bob) fixed on the end of a string or rod suspended from a pivot. Hang a bowling ball from a ceiling hook for a large conical pendulum to rotate at a steady slow speed. This motion of the pendulum is called motion of the first type. This video is made to understand Circular Motion. Rectilinear motion in vertical circle. Based on the previous notation, the following formulas can be obtained: Definition of average velocity. Standard conical pendulum lengths that have been investigated experimentally (and subsequently published) usually. A conical pendulum is a simple pendulum with the bob describing a circle and the string a cone \u00b7 The centripetal force in a conical pendulum of semi vertical\u00a0. It spins around the vertical with angle and with angular \u2026. In the overhead view of above figure, a long uniform rod of mass 0. If during a periodic motion, the particle moves to and fro on the same path, the motion is vibratory or oscillatory. An elliptical (nonplanar) motion may precess due to anharmonicity in the restoring force. Basically, this is a mass on a string attached to a rubber stopper. The equation of SHM for a simple pendulum \u2026. A conical pendulum has length 50 cm. A conical pendulum is a mass attached to a nearly massless string that is held at the opposite end and swung in\u00a0. April 25, 2018 Boris Sapozhnikov. The actual form of a pendulum \u2026. By applying Newton's secont law for rotational systems, the equation of motion for the pendulum \u2026. My Dashboard; Modules; WEEK 9 : CM2-CU8: Conical Pendulum; VIDEO : Conical Pendulum (Motion \u2026. 2004): if the length of [a conical] pendulum be /, the semi-major axis of the ellipse described by the pendulum \u2026. 12 S G Kamath The motion of a pendulum--be it the simple, spherical or the conical--is a concrete example wherein the three types of motions discussed above are realized physically. Because the bobs would lift in response to a faster speed (because they were basicly a conical pendulum\u2026. What is a conical pendulum?. In the coordinate system that rotates with the wire, there will be fictitious Coriolis and centrifugal forces, in addition This Article is brought to you for free and \u2026. This acceleration is called centripetal acceleration, and equals v 2 /r, where v is the speed of the object and r is the radius of. Suppose, further, that the object is given an initial horizontal velocity such that it executes a horizontal circular orbit of radius with angular velocity. The figure below shows a \"conical pendulum\", in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at \u2026. THE COMPOUND PENDULUM The term \u201ccompound\u201d is used to distinguish the present rigid-body pendulum from the \u201csimple\u201d pendulum of Section 3. Motion Of An Object Attached To A Spring. The plane of its motion, with respect to the earth, rotated slowly clockwise. Best answer Expression for tension in the string of a conical pendulum: i. To describe the motion of a conical pendulum in terms of its tangential velocity. Motion of a horse pulling a cart on a straight road. 68 m and the angle between the string and vertical is 35\u00b0. Conical pendulum illustrates uniform circular motion, and the other cases are representative of a non-uniform circular motion. A conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle with constant speed 'v' at the end of a string of length 'l'. A) What is the magnitude of the torque (N. A simple pendulum consists of a relatively massive object - known as the pendulum bob - hung by a string from a fixed support. This result is true for all horizontal conical pendulums for which the angle, \u03b8, is measured from the pendulum's position of vertical equilibrium. Please help me!!! I cnt solve thisA bob of mass is suspended from a fixed point with a masslessstring oflength (i. Rotational Inertia Stick Demo. The sections on mechanics in this book are basically arranged in that order. The pendulum motion is induced by the weight of the hanging mass, which is moved initially. let 'h' be the depth of the bob below the support. The deeper it sits, the wider the arch of the revolving pendulum, the slower the clock will go. Conical Pendulum: Its time period, tension i\u2026. Cm 4 the conical pendulum (shared). Conical pendulum is a small heavy mass attached with massless, flexible and inextensible string suspended from a rigid support and the mass is constraint to whirl in a horizontal circle with constant speed. Fresnel Reflectance Of S Polarized Light. Oscillation motion is a periodic motion in which the particles move to and fro on a particular predetermined path within equal time intervals. In the table below, let us look at the various differences between simple and compound pendulum. I can go to a very small angle and then it's kind of barely moving, really slow like that. dimensions of the conical pendulum, the magnitudes of the forces acting during the conical pendulum motion and a triangular construction involving the conical pendulum period. Huygens rotated a conical pendulum of length l (when the angle \u03b8 was small) he. So a pendulum can evidently be said to have a certain amount of oscillatory motion in 1 direction plus a variable amount of angular momentum which results in a periodic amount of lateral motion in the orthogonal direction. A simple pendulum is an idealized body consisting of a particle suspended by a light inextensible cord. When pulled to one side of its equilibrium position and released, the pendulum swings in a vertical plane under the influence of gravity. Thus together with the string the bob traces out a cone. Example: Motion of moon around earth; Motion of a piston in a cylinder; Motion of a simple pendulum etc. What Is The Difference Between A Simple Pendulum And A Co\u2026. that a conical pendulum whose initial motion was elliptical, was compelled to process in the same direction as the oscillation of its mass (Olsson 1978, 1981; Gray et arl. ERIC is an online library of education research and information, sponsored by the Institute of Education Sciences (IES) of the U. OCR past papers can be found at: https://www. Correct answer - The bob of a conical pendulum undergoes a) rectilinear motion in horizontal plane b) uniform motion in a horizontal circle c) The bob of a conical pendulum undergoes a) rectilinear motion in horizontal plane \u2026. Location: Cabinet: Mechanic (ME) Bay: Shelf: #1 Abstract: A wire is suspended from a ceiling mount and a bowling ball attached at the bottom. (7) determines whether the plates of the pendulum \u2026. Sneak a peak! Assault a graveyard. And silence those who arrive in the tuna. Derivation of the equations of motion. A motorized, plastic plane* is suspended from a thin string and \u201cflies\u201d in a circular path with a constant speed. The string is whirled in a horizontal circle, then the arrangement is called a conical pendulum. We first determine the time period of one complete rotation in conical pendulum. The characteristic of this motion can be obtained in terms of the length of the string and the angle with respect to the vertical. A particle of mass m, just completes the. Added an answer on August 11, 2020 at 1:32 pm. N2 - Students often find mechanics a difficult area to grasp. In this paper, we present evidence to show that the dynamics of rigid solid bodies is not a closed discipline, particularly in the field of \u2026. ' Centripetal force (Fc) is the result of gravity and tension. The point of intersection satisfies the system of two linear equations: 12 12 57 22solutions \u2026. Description : This set of 2 movies illustrates the dynamics of the conical pendulum (object suspended at the end of a string, moving in uniform circular motion). 22 A torsional pendulum consists of a rigid body suspended by a string or \u2026. The time required for the bob of the conical pendulum to travel one revolution. Introduction The spherical pendulum is a mechanical system of considerable pedagogical interest [1\u20135]. Obtain an expression for its time period. Ballistic Pendulum lab report; Angular Motion lab report; Lab Report 9; Other related documents. 1 Use a search engine such as Google to research the history and uses of one of the following materials: Tin Glass Cement Titanium Carbon fiber Present the \u2026. in the variable \u03b8 about the point (a near-conical pendulum). Determine the period of the pendulum \u2026. In a conical pendulum, the bob is rotated with different angular v\u2026. If the Lagrangian is approximated by keeping terms up to cubic order, the system has three independent constants of motion\u2026. Torque and angular momentum of a conical pendulum joe_coolish May 19, 2011 May 19, 2011 #1 joe_coolish 9 0 Homework Statement A ball (mass m = 250 g) on the end of an ideal string is moving in circular motion as a conical pendulum as in the figure. Circular Motion A conical pendulum is a mass attached to a nearly mass-less string that is held at the opposite end and swung in horizontal circles. A motion of a Conical Pendulum. So, that's what I wanna talk to you about in this video. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle \u2026. 3: A torsional pendulum consists of a rigid body suspended by \u2026. Circular MotionUnit: Circular MotionSubject: Physics Grade XI. Caption: A schematic diagram of \"a conical pendulum in motion. In the corresponding motions of the pendulum, close to conical motion\u2026. A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius r. Classify the following as motion along a straight line, circular or oscillatory motion: a. 75, and inverted pendulum, had a mass of 1000 kg and mechanical 1. [45] Researchers at the University of Chicago and Argonne National Laboratory have invented an innovative way for different types of quantum technology \u2026. The conical pendulum was first studied by the English scientist Robert Hooke around 1660 as a model for the orbital motion of. The is an example of a 3D holonomic system. A special case of circular motion is the conical pendulum where the pendulum's \"bob\" is swung in a horizontal circle of radius Rwhile attached to a string of length L. If we take Tas time period of rotatory motion then velocity can also be expressed as: v= 2\u02c7r T: (2) Now replace the expression of vfrom Eq. A small ball of mass m is suspended from a string of length L. Draw a neat labelled diagram of conical pendulum. In this case, the moment of inertia has to be around the CM: I = 1 12 M L 2 leading to. Set the object traveling in a horizontal circular motion at a constant angle \u03b8 , as measured outward from when the object hangs straight down. A conical pendulum consists of an object moving in uniform circular motion at the end of a string of negligible mass (see Figure 1). Hang a bowling ball from a ceiling hook for a large conical pendulum \u2026. Physics Lab 8: The Flying Pig \u2013 Centripetal Force Section: Name: Purpose To show the net force for a conical pendulum is mv 2 /r. uniform circular motion can be demonstrated with conical pendulum. Obtain the equations of motion of the system using Lagrange\u2019s equations. I\u2019m assuming that the pendulum bob is connected by an inelastic string of negligible mass, which allows me to fix the length of the string as. The correct option is : (a) Angular momentum of bob is \u2026. You also should be able to find setup descriptions/figures in most textbook chapters on centripetal force. It is being regulated by a rotating pendulum that is called \u2018conical\u2019 after the shape of the silhouette of the motion. In this video, AAFREEN MAM has tried to explain the fundamentals necessary for solving physics questions. Of course, these different types of motion can be combined: for instance, the motion of a properly bowled bowling ball consists of a combination of trans-lational and rotational motion, whereas wave propagation is a combination of translational and oscillatory motion. The pendulum bob moves in a horizontal circle with a constant speed v. It is at \u00ae rest under the action of \u2026. Transcribed image text: PROJECTILE MOTION AND BALLISTIC PENDULUM Objective: The purpose of this experiment is to use the laws of conservation of energy and linear momentum to determine the velocity of a projectile, use this result to predict the projectile's range when fired in a uniform gravitational field and then compare this range to a measured value. 4b, which consisted of a particle at the end of a massless string. ANGULAR MOMENTUM OF CONICAL PENDULUM. Simple harmonic motion Solutions. Hello, The degree of freedom of a system = Number of directions in which it can move. When the string is horizontal there in \u2026. In this Lesson, the sinusoidal nature of pendulum motion \u2026. Introduction: A conical pendulum is a pendulum that is spun around in a circle instead of swinging backwards and forwards, such as like a traditional pendulum. A small mass is suspended by a cord and set into motion \u2026. The pendulum does this because of inertia, which is the tendency of mass to stay in motion when a force acts upon it. For calculating the time period of a conical pendulum, we need to use the expression of Newton's second law of motion\u2026. Answer (1 of 3): Consider a conical pendulum with a bob of mass m, length l, at an angle \\theta with the vertical, going round with a uniform velocity v and \u2026. Request PDF | The conical pendulum: The tethered aeroplane | The introductory physics lab curriculum usually has one experiment on uniform circular motion \u2026. A simple pendulum is one which can be considered to be a point mass suspended from a string or rod of negligible mass. The Conical Pendulum Download PDF A great activity for physics classes investigating centripetal force and uniform circular motion. Calculate\u2026 A simple pendulum mounted on a car. In this Lesson, the sinusoidal nature of pendulum motion is. To determine the acceleration due to gravity by means of a conical pendulum. Coupled pendula: Three varieties are available. February 15, 2018 Boris Sapozhnikov. Laboratory 8: Conical Pendulum \u2013Activity L \u03b8 r Figure 2: Conical pendulum The conical pendulum, shown in Figure 2, con-sists of a mass m at the end of a \u2026. 2:50 Breaking the force of tension into its components. Conical pendulum \u2013 measuring g Number 13573 0-EN Topic Me ch anics , two -dimensional motion Version 201 7-02-17 / HS Type Student exercise Suggested for grade 11 -12 p. Circular Motion and the Conical Pendulum Introduction. 5 THE CONICAL PENDULUM: A STUDY OF. A conical pendulum, a thin uniform rod of length. Consider about the figure of a conical pendulum. Kinetic energy = (1/2) M\u03c92(A2 \u2013 x2) Potential energy = 1/2 M\u03c92x2. The modulationequationsare found to be soluble in terms of elementary \u2026. There's a 1:1 bevel that drives the pendulum. simple pendulum moves in to and fro motion. 0 degrees, then the angular speed of the conical pendulum \u2026. Each of five modules contains reading links to a free textbook, complete video lectures, conceptual quizzes, and a set of homework problems. It doesn't take much effort to keep the mass moving at a constant angular velocity in a circle of constant radius. Take a video of the motion as viewed from the side. Ordinary pendulum clocks had been invented by Christiaan Huygens in 1656, and by ordinary, we mean that the pendulum swings back and forth in a vertical plane. With an elliptical path it represents a motion \u2026. A heavy spherical mass (approximately 1. Estimation of parameters of various damping models in pla\u2026. The problem of the conical pendulum is to consider a mass attached to one end of a light inextensible string of length with the other end attached at the top of a vertical rod. The path of periodic motion may be rectilinear, open/closed curvilinear. The motion of a simple pendulum is like simple harmonic motion \u2026. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The three-dimensional motion of the elastic pendulum or swinging spring is investigated in this study. pendulum, body suspended from a fixed point so that it can swing back and forth under the influence of gravity. The swing of the pendulum will have to be limited to the distance of the support rods. motion answer key, describing and measuring motion, measuring motion gizmo lesson info explorelearning, download guided and study acceleration motion answers pdf, describing and measuring motion worksheet answer key 1, chapter 1 matter in motion section 1 measuring motion, describing motion \u2026. The string makes a constant angle with respect to the vertical; as a result, the mass moves with constant speed in a horizontal circle. Equipment and Supplies Flying Pig (or equivalent), stopwatch, meter stick Discussion When an object travels at constant speed along a circular path, we say it has uniform circular motion \u2026. In a typical experiment, the conical motion is confined to the horizontal plane, however, it is possible to extend the mathematical analysis to three dimensions (Barenboim & Oteo, 2013). The bob of a conical pendulum is attached to a fixed point A,A by a string of length 50, c, m,50cm and is moving in a circular path, as shown in the diagram\u00a0. Uniform motion in a vertical circle. Discuss the pendulum's motion in the effective potential diagram. You can modify simulation parameters directly in MATLAB workspace. Physics Lab report on circular motion. Take your string back about 40 - 50 cm. The initial amplitude of the pendulum must be small compared to pendulum \u2026. When the ball at the end of the string swings to its lowest point, the string is cut by a sharp \u2026. For this exercise, assume that angles not small, and w\u00ba =2. This result is true for all horizontal conical pendulums for which the angle, \u03b8, is measured from the pendulum\u2026. Name: Laboratory Section: The goal of this laboratory is to study uniform circular motion. The pendulum is perhaps the simplest experimental devices ever constructed, and yet for all its simplicity it has historically enabled scientists to both investigate and enumerate gravity; the fundamental force that shapes the very universe. The study of this lab revolves around the generation, propagation and reception of mechanical waves and vibrations. This can be done side-by-side, front and back, clockwise, counterclockwise, in an elliptical motion, or even in a bobbing movement up and down, which often indicates a strong affirmative action. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. When a pendulum swings in one direction and then comes back to its original starting point, this is called a period. Sir Isaac Newton (1642-1727) established the scientific laws that govern 99% or more of our everyday experiences. A conical pendulum is constructed by attaching a mass to a string 2. The time period of a simple pendulum: It is defined as the time taken by the pendulum \u2026. The constraint on a bead on a uniformly rotating wire in a force free space is. 290 7 Lagrangian and Hamiltonian Mechanics 7. And I love these pure math ones. This work consisted of studying the motion of a rigid conical pendulum \u2026. There are several different ways to configure a Twin Elliptic Pendulum\u2026. ii) For a particle having a uniform circular motion, which of the following is constant (A) Speed (B) Acceleration (C) Velocity (D) Displacement. You have constructed what is known as a physical pendulum. Non-uniform circular motion Up: Circular motion Previous: Centripetal acceleration The conical pendulum Suppose that an \u2026. 1k points) motion in a plane; class-11; 0 votes. We identified it from honorable source. When a pendulum is set in motion, gravity causes a restoring force that will accelerate it toward the center point, Conical Pendulum. Measure the period using the stopwatch or period timer. Slider-crank mechanism, arrangement of mechanical parts designed to convert straight-line motion to rotary motion, as in a reciprocating piston engine, or to convert rotary motion to straight-line motion, as in a reciprocating piston pump. ) Find an expression for v in terms of the geometry in Figure 6. Expression for its time period: Consider the vertical section of a conical pendulum having bob (point mass) of mass m and string of length \u2018L\u2019. Place the pendulum assembly on to hand driven rotator such that the set screw is tightened onto the flat part of the hand driven rotator. This resultant force is the centripetal force. The bob of a pendulum of length l is pulled aside from its equilibrium position through \u2026. It is then put into an elliptical orbit thus acting as a conical pendulum. 500-m string is moving in a uniform circular motion in a horizontal plane of radius \u2026. In 1851, Jean-Bernard-Leon Foucault suspended a 67 metre, 28 kilogram pendulum from the dome of the Pantheon in Paris. This necessary centripetal acceleration required for the motion of the conical pendulum is provided by the x component of tension which is the \u2026. The motion of a pendulum is a classic example of mechanical energy conservation. An object at the end of a string is used as a conical pendulum. Course of Theoretical Mechanics, Higher Education Press, Bei Jing, Chapter 5. and the relevant moment of inertia is that about the point of suspension. The stiff string (actually a piece of aluminum wire) is required for otherwise the pendulum will not remain in the same (rotating) plane and the motion becomes closer to a spinning spherical pendulum. \"A long-period conical pendulum for vibration isolation,\" Physics Letters A 222, 141-147 (1996). State technological advantages of Nano Moment of inertia, area meter 4m Moment of inertia, mass kilogram-meter 2kg m Momentum, linear kilogram \u2026. 18 Determine the equations of motion \u2026. Friction: Pulling a Box on a Horizontal Surface. The figure shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at a constant speed. Mass performing vertical circular motion under gravity. Do Problem 21 if the two masses are different. Science; Physics; Physics questions and answers; PROJECTILE MOTION AND BALLISTIC PENDULUM Objective: The purpose of this experiment is to use the laws of conservation of energy and linear momentum to determine the velocity of a projectile, use this result to predict the projectile's range when fired in a uniform gravitational field and then compare this range to a measured value. You must make a mark on the wall or your piece of paper to make sure that you let it go from the \u2026. the motion of the mechanical system can be described by the generalized coordinates x ( t ) (translation M ), r ( t ) (elongation of the spring pendulum), and \u02dc ( t ) (link rotation). The motion of conical pendulum is - 49146832 mukulsony1701 mukulsony1701 19. The string's motion follows a conical \u2026. Vintage Jupiter Wall 31 Day Clock w/Pendulum and Key 26\u201d Shop pendulum \u2026. (You do NOT need to \ufb01nd the frequency of the oscillations. Simulation- Conical Pendulum: 3D. The is horizontal, so there is no J verLieal motion\u2026. (hint: first calculate the tension using the net y- forces = 0 and use newton's second law of motion \u2026. Pendulums are also seen at many gatherings in eastern Mexico where they mark the turning of the tides on the day which the tides are at their highest point. The conical pendulum lab allows students to investigate the physics and mathematics of uniform circular motion. Static and Kinetic Friction on an Inclined Plane. He also explained our relationship to the Universe through his Laws of Motion \u2026. A simple pendulum is a special case of a conical pendulum in which angle made by the string with vertical is zero i. 130 m) for the conical pendulum\u2026. 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. The pendulum is a simple mechanical system that follows a differential equation. A physical pendulum consists of a uniform rod of length d and mass m pivoted at one end. Additionally, what are the 4 types of motion? These four are rotary, oscillating, linear and reciprocating. pendulum mass for a fixed length, and varied the pendulum length for a fixed mass. The simple pendulum is another mechanical system that moves in an oscillatory motion. g = acceleration due to gravity. This means that the forces in line with the arm of the pendulum must be equal and opposite since there is no motion in this direction and we see that T = m g cos. The goals of this lab are to verify that centripetal acceleration is given by a = v 2 /r and to show that the period of a conical pendulum is given by the theoretical equation:. There are two forces acting on the bob: its weight and the tension in the string (Figure 39). Determine the horizontal and vertical components of the force exerted by the wire on the pendulum\u2026. oPhysics: Interactive Physics Simulations. An analysis of the motion presents that the equilibrium states of the pendulum are determined by the pendulum angular speed. Other rides, such as the rotor ride, Enterprise wheel, and Ferris wheel, spin the rider in circular motion either horizontally or vertically. The pendulum consists of a string and a bob (a weight, generally spherical) at the end. The reasons for this, however, and consequently the focus \u2026. Robert Hooke's conical pendulum from the modern viewpoint of. A conical pendulum consists of a simple pendulum moving in a horizontal circle as shown in the figure. Let m be the lower mass and let M be the sum of the two masses. The motion of a conical pendulum in a rotating frame. What tangential speed, v, must the bob have so that it moves in a horizontal; Question: To describe the motion of a conical pendulum in terms of its tangential velocity. In both cases, an elliptical motion is induced. Find the maximum height above the ground that the ball reaches. Instead of swinging back and forth, the bob is to move in a horizontal circle at constant speed v, with the wire making a fixed angle \u03b2 with the vertical direction (Fig. It is Robert Hooke who first studied the motion of a spherical pendulum in that a conical pendulum whose initial motion was elliptical,\u00a0. Theoretically, The time period of a conical pendulum is directly proportional to the square root of its length. pendulum altogether in absolute measurements of g. The oscillatory motion of a simple pendulum: Oscillatory motion is defined as the to and fro motion of the pendulum in a periodic fashion and the centre point of oscillation known as equilibrium position. State the expression for its periodic time in terms of length. The mass is moving about the rod in uniform circular motion \u2026. In your text, it is shown that the period of rotation T of a conical pendulum of length, L, is given by \u20ac T=2\u03c0 Lcos(\u03b8) g, where \u03b8 is the angle the pendulum \u2026. A simple pendulum consists of a point mass suspended on a string or wire that has negligible mass. As the Bravais pendulum is a conical pendulum oscillating at Earth's surface, its motion is also discussed. Its construction is similar to an ordinary\u00a0. 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal \u2026. The conical pendulum was first. \u03c9: Angular velocity of rotational motion (rad/s) T: period of rotational motion (seconds) Find the period of a conical pendulum. The string makes an angle \u03b8 with the vertical. All of the energy in the pendulum is gravitational potential energy and there is no kinetic energy. The pendulum always moves in one angular direction. L2 : Motion of Pendulum - Motion and Time, Science, Class 7; Video | 04:40 min. Popular; Recent; Comments; Pendulum with spring animation MATLAB. The motion occurs in a vertical plane and is driven by a gravitational force. The moment of intertia I for a point mass rotating around a pivot (radius l) is ml\u00b2, yielding our equation of motion: This is a second-order, non \u2026. \u0871 > 3 5 2 #` p bjbjm m 5 6 > > > x , Q D D D D H S U U U E $h o ~ E> D H ; 7 7 7 @ 8D > H S 7 S 7 7 & > 7 H 8 @h ?Z 7 S ! 0Q 7 p 7 > 7 7 7 Q$ R f t AP Physics Lab Brockport High School NY USA Circular Motion: The Conical Pendulum Mr Keefer Objectives: Determine the acceleration of gravity g from the circular motion of a conical pendulum\u2026. swing with horizontal circular motion such that r equals the length of the pendulum?. This pendulum is used as a model to analyze the motion of planets. Korean pendulum wall clocks. The reason that the pendulum oscillates about the vertical is that if the pendulum is displaced, the force of gravity pulls down on the pendulum. Expression for Period of Conical Pendulum: Let us consider a conical pendulum consists of a bob of mass 'm' revolving in a horizontal circle \u2026. The time it takes the pendulum \u2026. The period of the motion of a pendulum is virtually independent of its amplitude and depends primarily on the geometry of the pendulum \u2026. Question 3 The figure shows a conical pendulum, in which th\u2026. Consider a small body of mass m suspended from a rigid support with the help of a string of length l. Consider a conical pendulum with a line length \u2018L\u2019 and a rotation radius \u2018r. Conical Pendulum Time Period: It consists of a string OA whose upper-end O is fixed and bob is tied at the other free end. The forces which are acting on the mass. The period of a simple pendulum \u2026. The ball revolves with constant speed v in a horizontal circle of radius r as shown in the figure. In this experiment a mass is attached to a string and made to spin in a circle of fixed radius, the time period of the motion \u2026. Correction of the excercise 4:00. Lab \u2013 Flying Pigs Purpose: To show that the net force for a conical pendulum is mv 2 /r Equipment/Supplies: flying pig apparatus stopwatch meterstick Discussion: Any object moving in uniform circular motion \u2026. Equations for the following principal physical parameters. The restoring torque is supplied by the shearing of the string or wire. In other words, the vertical component of the tension force\u00a0. The bob does not show any vertical motion \u2026. The length L of the string is 1. A Bob of a conical pendulum undergoes option A rectilinear motion in a horizontal plane option B uniform motio\u2026 Get the answers you \u2026. Differential equation of the motion (derived from Newton's second law):. This text is useful for students to follow many details on the motion of the Foucault pendulum in inertial and rotational frames and help in the study of the particle motion \u2026. \u2022 As viewed from above, it moves in a circle, the centripetal force. The dimensionless factor of $$2\\pi$$ can be derived using an in-sight from Huygens [15, p. doing this isaac physics question on circular motion, can anyone help The bob of a conical pendulum is attached to a fixed point A,A by a string of length. The pendulum is initially at rest in a vertical position. Imagine a conical pendulum in steady circular motion with small angle \u03b8. The bob moves in a single plane, back and forth. This can be determined by measuring the time required for the pendulum to reoccupy a given position. (a) Write a MATLAB script, which solves (*) for. The equations of motion of the pendulum were derived using the Lagrangian method. The motion of the pendulum depends on its total energy E = T +V. 9 0 m and negligible mass and the bob follows a circular path of the circumference. eu Viaduktvej 35 \u00b7 DK-6870 \u00d8lgod Fax +45 7524 6282 www. The canonical example of simple harmonic motion is the motion \u2026. This is a simulation of a double pendulum. Spherical pendulum and vertical pendulum are the special cases of conical pendulum. Geometrically, the arc length, s, is directly proportional to the magnitude of the central angle, \u03b8, according to the formula s = r\u03b8. October 28, 2018 Boris Sapozhnikov. As a result, the motion of the pendulum \u2026. n Review Part A (Figure 1)A massless string of length L is fixed to a point in the ceiling and suspends a bob with mass m (i. h is the distance from the plane of the circular motion to where the string is attached. Is there any limitation on semivertical angle in a conical pendulum? Thank You. study of circular motion in conical pendulum purpose the purpose of this experiment is to study the effect of the. Class 7 Science Chapter 13 Motion and Time Textbook Exercise Questions and Answers. The 1 \u2236 1 \u2236 2 resonant elastic pendulum is a simple classical system that displays the phenomenon known as Hamiltonian monodromy. Let the bob be displaced from its mean position and whirled around a horizontal circle of radius 'r' with constant angular velocity '\u03c9'. If the weight is made to describe a horizontal circle so that the string describes a right circular cone, whose axis is a vertical through ( O ) , then system is called a conical pendulum. A new twist for the conical pendulum A new twist for the conical pendulum Moses, Thomas; Adolphi, Natalie L. Then the period of the simple pendulum is given by. We use a conical pendulum in this experiment. Jean Bernard L\u00e9on Foucault was the leading experimental physicist of his day. Imagine a basic conical pendulum. In this work, planar free vibrations of a single physical pendulum are investigated both experimentally and numerically. But, because for small \u03b8, we can say that \u03b8 \u2248 sin\u03b8, we can re-write the DE as:. You can apply similar considerations to a simple pendulum, which is one on which all the mass is centered on the end of a string. The equations for a simple pendulum show how to find the frequency and period of the motion. This allowed the reduction of apparent gravity, thus increasing its natural period, a similar approach to that of Galileo more than two centuries earlier when he used an inclined plane to study the motion of falling objects. Equations for a Simple Pendulum. An inverted pendulum is simply a pendulum which has its fixed end located below the vibrating mass. Let us consider a conical pendulum \u2026. The pendulum is in steady circular motion with constant angular velocity wk. the pendulum bob has an initial velocity. The Chaotic Motion of a Double Pendulum. In two or three attempts one could realize a satisfactory circular orbit. Other rides, such as the rotor ride, Enterprise wheel, and Ferris wheel, spin the rider in circular motion \u2026. When this happens the pendulum is called a \"conical pendulum\" because the cord sweeps out (or creates in space) a conical surface with the apex at the \u2026. Two pendula coupled by a spring (shown to the right) will show normal modes, and transfer of energy between the single pendula swinging modes. One end of a string ( AO ) is attached to a fixed point ( O ) and a weight ( W ) is tied to the other end ( A ). Uniform Accelerated Motion. Use the Tracker Software to analyze the motion of a conical pendulum, as viewed from the side. Let \u2013 ( N ) is the centre of horizontal circle of radius ( r ) \u2026. Exploring Projectile Motion Concepts. Let the string subtends an angle \u03b8 with the vertical. We use centripetal acceleration here, because we have a circular motion. Conical pendulum clocks are part of the clock history but as they are not easy to implement you don\u2019t see too many around, and as far as I know there are no wooden conical pendulum clocks. In this lab you will \u2022 test the theory of simple harmonic motion in the case of a simple pendulum and a Hooke\u2019s law spring. The kinetic energy would be KE= \u00bdmv2 ,where m is the mass of the pendulum, and v is the speed of the pendulum. Consider an object of mass m tied to a string of length l and whirled in a horizontal circle of. Because the bobs would lift in response to a faster speed (because they were basicly a conical pendulum) it could be. The upward force produced by the tension balances the downward force from the weight of the pendulum:. 2 is annotated with the mathematical symbols that will be used in the analysis of the period of the pendulum. The massless thread is only an idealization. A simple pendulum theoretically has the mass of the bob concentrated at one point, but this is impossible to achieve exactly in practice. 17 A simple pendulum of length l and mass m is pivoted to the block of mass M which slides on a smooth horizontal plane, Fig. Physics Lab 9: The Flying Pig \u2013 Centripetal Force Section: Name: Purpose To show the net force for a conical pendulum is mv 2 /r. Once the modules are completed, the course ends with. A ball is attached to a string and swung so that it travels in a horizontal circle. pendulum synonyms, Compensation pendulum; compound pendulum; Conical pendulum; and twisting his legs round it in sailor fashion, slipped down eight or ten feet, where his weight gave it a motion not un-like that of a pendulum\u2026. tal plane with its string sweeping the. Conical Pendulum \u2013 If a simple pendulum is fixed at one end and the bob is rotating in a horizontal circle, then it is called a conical pendulum. The accel-eration is directed along the radius of the dashed circle and towards the shaft. He realized that the pendulum \u2026. Periodic Motion Lab \u2013 The Conical Pendulum Purpose The goals of this lab are to verify that centripetal acceleration is given by a = v2/r and to show that the period of a conical pendulum is given by the theoretical equation: Procedure A small mass is suspended by a cord and set into motion \u2026. Experiment with Conical Pendulum. In the figure below, a simple pendulum is represented at various positions of its motion. We have two contributions: kinetic energy of rotation around the CM and kinetic energy \u2026. Unfortunately, Java cannot plot the motion of the pendulum just by using the angle q \u2013 it uses (x, y) coordinates to plot shapes. VIDEO : Conical Pendulum (Motion in a Horizontal Circle) Skip To Content. It is the analogue of the conical motion of a spherical pendulum with a fixed point of suspension. Smart pendulum leveling system One-button operation 530 Ft. motion of the conical pendulum is just the sum of the motion of two pendulums. In conical pendulum the bob does not oscillate back and forth but it moves in a circle. A conical pendulum is a mass attached to a nearly massless string that is held at the opposite end and swung in horizontal circles. Find Physics textbook solutions? 500 Selected Problems In Physics for JE\u2026 877 solutions Selina - Concise Physics - Class 9 1224 solutions Lakhmir Singh, Manjit Kaur - Physics 10.", "date": "2022-05-23 15:19:47", "meta": {"domain": "thorstenharte.de", "url": "https://thorstenharte.de/the-motion-of-a-conical-pendulum-is.html", "openwebmath_score": 0.6579096913337708, "openwebmath_perplexity": 513.4436833255314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769085257165, "lm_q2_score": 0.8791467738423873, "lm_q1q2_score": 0.8576752917655135}}
{"url": "http://math.stackexchange.com/questions/771771/proving-units-in-a-ring", "text": "# Proving units in a ring\n\nSuppose $R$ is a ring with no zero divisors and with identity $1_R$ not equal to $0_R$. Suppose that $a,b$ are in $R$ and that $ab$ is a unit. Prove that $b$ is a unit.\n\nMy thoughts: I know a unit is basically a unit that (for this example) would mean $abu = 1_R$ for some nonzero $u$ in $R$. I am really stuck after that. Not seeing a clear path to manipulate the variables to prove b is a unit by itself.\n\n-\nWell, write it this way: $\\;1=u(ab)=(ua)b\\implies b\\;$ is a unit...:) \u2013\u00a0 DonAntonio Apr 27 at 19:00\n@Don That yields only a left inverse $\\,c = ua\\,$ for $\\,b.\\,$ But conjugation shows it is a right inverse too - see my answer. \u2013\u00a0 Bill Dubuque Apr 27 at 23:25\n\nAs the others have pointed out the calculation $$1=u(ab)=(ua)b$$ shows that $ua$ is a left inverse to $b$. Consider the product $b(ua)$. We have $$ua=1(ua)=((ua)b)(ua)=(ua)(b(ua)),$$ so $$(ua)(1-b(ua))=0.$$ As the ring has no zero divisors this implies that either $ua=0$ or $b(ua)=1$. But if $(ua)=0$, then $1=(ua)b=0$ which is a contradiction. The claim follows.\n\n-\n+1 for actually using the fact the ring has no zero divisors... \u2013\u00a0 Goos Apr 27 at 19:30\n@Goos: Well, the claim is false without something extra, and that was the only piece available :-) All: Sorry about spoiling the problem. I really should have come up with an appropriate hint. \u2013\u00a0 Jyrki Lahtonen Apr 27 at 19:45\nyes, but there were two maybe three people claiming \"solutions\" without using that fact. So +1 for a legitimate proof that it is a unit and not just having a left inverse. \u2013\u00a0 Goos Apr 27 at 20:10\n@Jyrki Re: hints. See my answer for one way I often hint at it. It's not easy to give a hint for this without spilling the beans. \u2013\u00a0 Bill Dubuque Apr 27 at 23:05\n\nAs discussed in the comments, since $ab$ is a unit then $uab = 1$ for some $u \\in R$, so $ua$ is a left inverse for $b$. It remains to show that $ua$ is also a right inverse for $b$, i.e., $bua = 1$. Taking the equation $1 = uab$ and multiplying both sides by $ua$ on the right, we have $$ua = uabua \\implies 0 = ua - uabua = ua(1 - bua) \\, .$$ Since $R$ has no zero divisors, then either $ua = 0$ or $1 - bua = 0$. But again, $R$ has no zero divisors, so we must have $1 - bua = 0$, hence $1 = bua$. Thus $ua$ is a two-sided inverse for $b$.\n\n-\n\nHint $\\$ Conjugate a one-sided inverse $\\,bc=1\\,$ to the other side via $\\ (bc\\!-\\!1)b\\, =\\, b(cb\\!-\\!1)$\n\n-", "date": "2014-12-22 14:26:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/771771/proving-units-in-a-ring", "openwebmath_score": 0.9344157576560974, "openwebmath_perplexity": 157.03132816592617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769106559808, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8576752874601384}}
{"url": "http://mathhelpforum.com/advanced-algebra/29103-centre-group-z-g.html", "text": "# Thread: centre of a group Z(G)\n\n1. ## centre of a group Z(G)\n\nHey there, i have a question on the center of a group.\n\nThe centre Z(G) of a group G is defined by $\\displaystyle Z(G) = g \\epsilon G: \\forall x \\epsilon G, xg = gx$\n\n(i) Show that Z(G) is normal subgroup of G\n(ii) By considering the Class Equation of G acting on itself by conjugation show that if $\\displaystyle |G| = p^n$ ( p prime) then $\\displaystyle Z(G) \\neq {1}$\n\n(iii) If G is non abelian show that G/Z(G) is not cyclic.\n(iv) Decude that any group of order $\\displaystyle p^2$ is abelian.\n(V) Deduce that a gorup of oder $\\displaystyle p^2$ is isomorhpic either to $\\displaystyle C_{p2}$ or to $\\displaystyle C_p \\times C_p$\n\nany hints would be greatly appreicated and ill atempt the question once i have a clear idea of how to do them. consequently, i will post them online once i have worked them out. cheers guyz\n\n2. Originally Posted by joanne_q\n(i) Show that Z(G) is normal subgroup of G\nLet $\\displaystyle x\\in \\text{Z}(G)$ prove that $\\displaystyle gxg^{-1} \\in \\text{Z}(G)$ for any $\\displaystyle g\\in G$.\n\n(ii) By considering the Class Equation of G acting on itself by conjugation show that if $\\displaystyle |G| = p^n$ ( p prime) then $\\displaystyle Z(G) \\neq {1}$\nThe conjugacy class equation says,\n$\\displaystyle |G| = |\\text{Z}(G)| + \\sum [G:\\text{C}(x)]$.\nWhere the $\\displaystyle x$'s are taken from distinct conjugacy classes of more than one element.\nWe know that the left hand side is divisible by $\\displaystyle p$ so the right hand side is divisible by $\\displaystyle p$. Now $\\displaystyle \\text{C}(x)$, the centralizer, is not whole $\\displaystyle G$ because we are picking those $\\displaystyle x$, this means $\\displaystyle 1\\leq |\\text{C}(x)| \\leq p^{n-1}$. This means $\\displaystyle p$ divides the index of $\\displaystyle \\text{C}(x)$ under $\\displaystyle G$. Thus, this proves that $\\displaystyle \\text{Z}(G)$ is divisible by $\\displaystyle p$. Thus, the center is non-trivial.\n\n3. Originally Posted by joanne_q\n(iii) If G is non abelian show that G/Z(G) is not cyclic.\nContrapositive is easier. Let $\\displaystyle H=\\text{Z}(G)$. If $\\displaystyle G/H$ is cyclic then there is $\\displaystyle aH$ which generates the group $\\displaystyle G/H$. Let $\\displaystyle x,y\\in G$. Note $\\displaystyle xH,yH\\in G/H$ thus $\\displaystyle xH=a^nH$ and $\\displaystyle yH=a^mH$. This means $\\displaystyle x = a^n z_1$ and $\\displaystyle y=a^mz_2$ where $\\displaystyle z_1,z_2\\in H$. But then $\\displaystyle xy = a^n z_1 a^mz_2 = a^{n+m}z_1z_2$ and $\\displaystyle yx = a^m z_2 a^n z_1 = a^{n+m}z_1z_2$ because $\\displaystyle z_1,z_2$ commute with everything. Thus $\\displaystyle G$ is abelian.\n\n(iv) Decude that any group of order $\\displaystyle p^2$ is abelian.\nBy Burnside's lemma (that is (ii)) we have that the center is non-trivial, forming the factor group we have a cyclic group. Thus the original needs to be abelian.\n\n4. Originally Posted by joanne_q\n(V) Deduce that a gorup of oder $\\displaystyle p^2$ is isomorhpic either to $\\displaystyle C_{p2}$ or to $\\displaystyle C_p \\times C_p$\nLet $\\displaystyle |G|=p^2$. Pick $\\displaystyle a\\not = 1$. Form the subgroup $\\displaystyle H=\\left< a \\right>$ if $\\displaystyle H = G$ then the group is cyclic and proof is complete. Otherwise choose $\\displaystyle b\\in G\\setminus H$ and form $\\displaystyle K=\\left< b\\right>$. Now $\\displaystyle H\\cap K = \\{ 1\\}$ this means $\\displaystyle HK = G$*. Also $\\displaystyle H,K\\triangleleft G$ because the group is abelian. This means $\\displaystyle G\\simeq H\\times K \\simeq \\mathbb{Z}_p \\times \\mathbb{Z}_p$.**\n\n*)Because $\\displaystyle |HK| = |G|$ by using the fact $\\displaystyle |HK||H\\cap K| = |H||K|$.\n\n**)Theorem: If $\\displaystyle H,K$ are normal subgroups with $\\displaystyle H\\cap K = \\{ 1 \\}$ and $\\displaystyle HK = G$ then $\\displaystyle G\\simeq H\\times K$.\n\n5. Thanks a lot for the help really appreciate it. makes it easier to understand the subject aswell.\n\nFor part (ii), your solution was:\nOriginally Posted by ThePerfectHacker\nThe conjugacy class equation says,\n$\\displaystyle |G| = |\\text{Z}(G)| + \\sum [G:\\text{C}(x)]$.\nWhere the $\\displaystyle x$'s are taken from distinct conjugacy classes of more than one element.\nWe know that the left hand side is divisible by $\\displaystyle p$ so the right hand side is divisible by $\\displaystyle p$. Now $\\displaystyle \\text{C}(x)$, the centralizer, is not whole $\\displaystyle G$ because we are picking those $\\displaystyle x$, this means $\\displaystyle 1\\leq |\\text{C}(x)| \\leq p^{n-1}$. This means $\\displaystyle p$ divides the index of $\\displaystyle \\text{C}(x)$ under $\\displaystyle G$. Thus, this proves that $\\displaystyle \\text{Z}(G)$ is divisible by $\\displaystyle p$. Thus, the center is non-trivial.\n\nhere is my solution, is it correct aswell?\n\n$\\displaystyle G \\equiv |Z(G)| (mod p)$ since Z(G) is a fixed point set.\nNow $\\displaystyle |Z(G)| \\equiv p^n(mod p)$, |Z(G)|=0.\nSo Z(G) has atleast p elements.\n\n6. for part (iii) i believe you have proved the opposite of the question? i.e. G/Z(G) is cyclic and thus abelian...\n\nto prove that it is non cyclic, do all the points you stated have to be contradicted..?\nOriginally Posted by ThePerfectHacker\nContrapositive is easier. Let $\\displaystyle H=\\text{Z}(G)$. If $\\displaystyle G/H$ is cyclic then there is $\\displaystyle aH$ which generates the group $\\displaystyle G/H$. Let $\\displaystyle x,y\\in G$. Note $\\displaystyle xH,yH\\in G/H$ thus $\\displaystyle xH=a^nH$ and $\\displaystyle yH=a^mH$. This means $\\displaystyle x = a^n z_1$ and $\\displaystyle y=a^mz_2$ where $\\displaystyle z_1,z_2\\in H$. But then $\\displaystyle xy = a^n z_1 a^mz_2 = a^{n+m}z_1z_2$ and $\\displaystyle yx = a^m z_2 a^n z_1 = a^{n+m}z_1z_2$ because $\\displaystyle z_1,z_2$ commute with everything. Thus $\\displaystyle G$ is abelian.\n\n7. Originally Posted by joanne_q\nFor part (ii), your solution was\nHave you ever done the conjugacy class equation? There is a way around it if you never done it that way, I think I have an idea of what you might have done.\n\nLet $\\displaystyle G$ be a finite $\\displaystyle p$-group. Let $\\displaystyle G$ act on a non-empty finite set $\\displaystyle X$. Then $\\displaystyle |X|\\equiv |X^G|(\\bmod p)$ where $\\displaystyle X^G$ is the invariant subset fixed by $\\displaystyle G$.\n\nhere is my solution, is it correct aswell?\n\n$\\displaystyle G \\equiv |Z(G)| (mod p)$ since Z(G) is a fixed point set.\nNow $\\displaystyle |Z(G)| \\equiv p^n(mod p)$, |Z(G)|=0.\nSo Z(G) has atleast p elements.\nLet $\\displaystyle G$ act on itself by conjugation (i.e. $\\displaystyle X=G$ and $\\displaystyle g*x = gxg^{-1}$). Also $\\displaystyle G$ is a finite $\\displaystyle p$-group which fits the above result thus $\\displaystyle |G| \\equiv |G^G| (\\bmod p)$ but $\\displaystyle G^G = \\text{Z}(G)$ because that is the subset left fixed under conjugation. Thus, $\\displaystyle |G| \\equiv |\\text{Z}(G)|(\\bmod p)$ which means the center needs to be divisible by $\\displaystyle p$, i.e. it cannot be trivial.\n\n8. Originally Posted by joanne_q\nfor part (iii) i believe you have proved the opposite of the question? i.e. G/Z(G) is cyclic and thus abelian...\n\nto prove that it is non cyclic, do all the points you stated have to be contradicted..?\nNo, there is no contradiction argument. I proved the contrapositive statement.", "date": "2018-04-22 11:21:14", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/29103-centre-group-z-g.html", "openwebmath_score": 0.9514423608779907, "openwebmath_perplexity": 236.2139222347338, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769071055403, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8576752812496851}}
{"url": "http://zveg.1upload.de/r-venn-diagram-6-sets.html", "text": "# R Venn Diagram 6 Sets\n\nPermission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. The circles or ovals represent events. Venn diagrams with three curves are used extensively in various medical and scientific disciplines to visualize relationships between data sets and facilitate data analysis. In the offset portions of the circles, the student lists those traits whch differ between the two items. Here is a Venn diagram of the sets S, T and V We can see, for example, that the element \"casey\" is in both set S and set T, but not in set V. To create a simple Venn diagram, you can just pass in the list with the specified set and overlap values into the venneuler() function. This Venn diagram shows all possible intersections of five sets. The Unattainable Triangle Good Good Fast Fast Cheap Cheap More expensive. Venn diagram, also known as Euler-Venn diagram is a simple representation of sets by diagrams. Entity Relation. We can represent this as, 01. Solution to Example 1. The remaining numbers in A are 7 and 9. The red part of each Venn diagram is the resulting set of a given. pseudovenn(dataset_dict, **kwargs) which plots a Venn-like intersection of six circles (not all intersections are present in such a plot, but many are). 5 circles representing 5 data sets). See [14] for a list of open problems related to Venn diagrams. In R, the VennDiagram package is the best option to build one. number and integers have no elements in common because the whole. Problem-solving using Venn diagram is a widely used approach in many areas such as statistics, data science, business, set theory, math, logic and etc. In the second example in the above Venn diagram, Set A is totally contained within set B How can we explain this situation? Suppose that sets A and B contain the following members: set A = {1,2} set B = {1,2,3,4,5,6,7,8} All members of set A are also members of set B. Let $$\\text{C} =$$ student belongs to a club and $$\\text{PT} =$$ student works part time. For example, the below six-way Venn diagram shows the distribution of shared gene families among six genomes, taken from D\u2019Hont et al. A = { 2, 4, 6, 8 }. Repurposing to Use Segments. We will do this for the first column of the Venn diagram figure given previously. Venn Diagrams (H) - Version 2 January 2016. Venn Diagram Horz 03 03 - Venn Diagram is a high-resolution transparent PNG image. A Venn diagram, also called primary diagram, set diagram or logic diagram, is a diagram that shows all possible logical relations between a finite collection of different sets. , ISBN-10: 0321867327, ISBN-13: 978-0-32186-732-2, Publisher: Pearson. fill (Optional) Fill color of the sets. Some people want to create quantitative Venn diagrams, where the overlap (and perhaps the relative size of the circles) is data-driven. Venn diagrams are a great way to visualize the structure of set relationships. The Venn diagram above is an example. How many students are not involved in either band or sports? 2. 5 overlapping fragmented circle shapes. Venn diagram, graphical method of representing categorical propositions and testing the validity of categorical syllogisms, devised by the English logician and philosopher John Venn (1834\u20131923). 8-6 currently have issues preventing their full use in KNIME. (a) Label the Venn diagram to show the sets A and B where n(A B) = 18. TimeLine/Curve Chart. Venn Diagram Set Diagram. Thank you for this. 4 - Page 91 12 including work step by step written by community members like you. Typically overlapping shapes, usually circles, are used, and an area-proportional or. Manage My Favorites. The term Venn diagram is not foreign since we all have had Mathematics, especially Probability and Algebra. A Venn diagram represents a set as the interior of a. 3, with aesthetically pleasing results. First step: Install & load \u201cVennDiagram\u201d package. Linguistics. 9 it is also used to pass additional aesthetics parameters for the ggplot2 graphics. For example, the \u201cdays of the week\u201d is a set and Monday is a member of this set. This is similar to the logical \u201cand\u201d Venn Diagrams Venn Diagrams use topological areas to stand for sets. venn: Draw a Venn Diagram with Five Sets in VennDiagram: Generate High-Resolution Venn and Euler Plots rdrr. Venn diagrams don\u2019t usually explain anything, they just represent. Note that many edges overlap, so the diagram is infinitely intersecting. 4-set means I should be able to specify 4 sets. newpage() # make new page for plot. The Venn diagram above is an example. What goes where? What do the plants, flowers, and trees have in common? Students use Venn diagrams to find out!. Chapter 4 Probability and Venn diagrams 2 1 The Venn diagram shows the whole numbers from 1 to 12. For example, the following Venn diagrams is based on 4 sets (I, II, III and IV), but is not scaled: (it was generated using the R package VennDiagram with the code:. VennDiagram-internal. Start studying Sets and Venn Diagrams. Venn Diagrams. 43 ewlength\\@[email\u00a0protected] \\@[email\u00a0protected] The height of the entire Venn diagram. Create beautiful venn diagrams using this tool and download as image (png or jpg). A set is a collection of objects and its objects are called members. Draw the figure below. These Venn diagram word problems worksheet pdfs feature two sets representing the quantities of the data. All of the number Page 6/25. Althoughitcontainsallthedisjointintersections,thevisualizationof interactingcharacteristicsisabsent(13and4872sharethesamearea;thetotalsizeof\u201cPoplar\u201disthelargest, however,thetotalareaofitisthethirdsmallest);in(d),diagramisdepictedbydi\ufb00erentlyshapedtriangles. Using Venn Diagrams in Set Theory. I think there should be 2 sets of numbers. Nov 19, 2013 - Venn diagrams began as schematic devices used in logic theory to represent set collections & their respective relationships. Note that many edges overlap, so the diagram is infinitely intersecting. In a class there are:. From the above Venn diagram, what is the set S \u2229 T? answer choices {casey, drew, jade, glen} {alex, casey, drew,hunter} {casey, drew} {drew, jade} Tags: Report Quiz. They have the incredible harmonies as if they too were born to sing together. Please refer to the Jupyter notebook for demos and a brief explanation of the interface; a more complete documentation. Venn diagrams show + - = underneath. Venn-Euler diagrams The combination of rectangles and circles are called Venn-Euler diagrams or simply Venn-diagrams. ER Diagram stands for Entity Relationship Diagram, also known as ERD is a diagram that displays the relationship of entity sets stored in a database. A complete Venn diagram represents the union of two sets. Currently they can only be accessed by passing the parameters s or likesquares to the low level creation function. For four sets, it is getting more difficult. If the assessment focus is to interpret a Venn diagram: Ask questions about the similarities and differences that the Venn diagram illustrates. They're also an example of a technique that works very well for a particular purpose, but that entirely fails outside its well-defined scope or when the number of sets gets too large. For the simplest Venn diagram you draw two intersecting circles. The union operations are reflected to the Venn diagram. We can represent this as, 01. Moreover, union operations between sets can be made. Inside the foldable I have students explain what element would be included given the Universal Set, Set A, and Set B. be the set of windy days, W R. ii) in set A but not in set B. a) Illustrate these sets using a Venn diagram. If we have two or more sets, we can use a Venn diagram to show the logical relationship among these sets as well as the cardinality of those sets. v) in set A and set B. Venn Diagrams are basically a representation of different elements in the form of circles. Probabilities from Venn diagrams: True or False & Matching Questions. Moreover, union operations between sets can be made. Perform the below mention boolean algebraic operation for the given set of elements. In this figure, the big rectangle shows the universal set $S$. We use circles to represent the sets, and enclose our diagram in a rectangle. One can solve counting problems involving unions and intersections of sets also with the help of Venn diagrams. The \"Cylinder Venn Diagram\" below gives a clear representation of different regions of the British Islands. You can also move and resize the circles by dragging and dropping. 96 like wine A, 93 like wine B, 96 like wine C, 92 like A and B, 91 like B and C, 93 like A and C, 90 like all three wines. While there are more than 30 symbols used in set theory , you don't need to memorize them all to get started. It is easy to make a Venn diagram for three sets. It consist of overlapping circle(s) in a box. These diagrams depict elements as points in the plane, and sets as regions inside closed curves. A Venn diagram (also sometimes also called primary diagram or set diagram) is a diagram that depicts all possible logical relations between a collection of sets. A number of variants on the squares type are implemented. These are placed inside A, but outside B. Venn diagrams are fairly intuitive and best learned through examples. Show sets F and G on a Venn diagram. This is an extremely important tool in logical analysis of business and scientific concepts. Draw a Venn diagram simply showing the sets of male and female dogs. I\u2019ve done this one for you. Maui Math Circle Session 5 Venn Diagram Page 2 ! Problem!Set:!Draw!a!Venn!Diagram!for!each!problem. The Venn diagram shows information about the choices the guests made. This handy pack contains 3 different Venn diagram templates to help students explore probability, statistics and representation of data. Venn diagrams for Sets. The intersection of events A and B, written as P(A \u2229 B) or P(A AND B) is the joint probability of at least two events, shown below in a Venn diagram. A Venn diagram uses overlapping circles or other shapes to illustrate the logical relationships between two or more sets of items. They may be used by those companies to build a. Venn diagrams are very useful in visualizing relation between sets. Contributed by: Marc Brodie (Wheeling Jesuit University) (March 2011). Free Venn diagram with 5 circles for PowerPoint. The VennDiagram package allows to build Venn Diagrams thanks to its venn. Via , users can specify additional parameters, mainly for the outer borders of the sets, as specified by par () , and since version 1. \u2022 Each area of a Venn diagram. Compute classification counts and draw a Venn diagram. Worksheet 2. Hint: You will likely need to use functions described in the setdiff help page to do this. The similar concept applies for venn diagrams with three sets. As we know set is the collection of unique things which are called the elements of the set. If they overlap it means that there are some items in both sets. This can be cumbersome I agree. What is a Venn Diagram? A Venn diagram is a diagram that shows the relationship between and among a finite collection of sets. A s imple symmetric Venn diagram consisting of five ellipses was given in [5]. Set diagram. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. A Venn diagram usually looks like the on underneath. In particular, Venn Diagrams are used to demonstrate De Morgan's Laws. CRAN - Package VennDiagram A set of functions to generate high-resolution Venn and Euler plots. A Venn diagram is a simple illustration that uses ovals to picture the universe of data that an analysis begins with and the subsetting, unions, and intersections If themes represent set-based operations, then a Venn diagram is a perfect way to document the themes that a subrelease will support. D/V Alpha/Beta Series 3\u00d8 WIRING DIAGRAMS Diagram DD1. Shown below is a 6-Venn diagram formed entirely from curves drawn from axis-aligned edges. \u2229: Intersection of two sets. Use it for drawing Venn and Euler diagrams. Venn Diagram Set Diagram. This is where entities that have all the qualities of the overlapping sets. Therefore, set A is a subset of Set B. A venn diagram makes a really good work to study the intersection between 2 or 3 sets. ID: 1213502 Language: English School subject: Math Grade/level: Grade 6 Age: 9-11 Main content: Sets Other contents: Venn Diagrams Add to my workbooks (12) Download file pdf Embed in my website or blog. There are two versions of the final poster, showing examples. This is an extremely important tool in logical analysis of business and scientific concepts. It has infinitely many members, including 2, 4, 6, 8, and so on. For example, the \u201cdays of the week\u201d is a set and Monday is a member of this set. These Venn diagram word problems worksheet pdfs feature two sets representing the quantities of the data. It seems VennDiagram can not work on my data. L1S1 Name : Score : Venn Diagram 1) A B U 1 50 7 30 12 35 3 22 9 8 13 10 2 15 6 28 21 B' A U B A' B U A B U z f t a d q r u y x v k j o g i h p c e b m 3) A' A U B' A' B' U A B U June March October September January May July August 2) (A U B)' A A B U 4) A B U. Overlaps between these circles represent the intersection between the two sets. You Might Also Like The equilibrium membrane potentials to be expected across a membrane at 37 \u2218 C, with a NaCl concentration of 0. However, making such a worksheet is a tedious task. However, the use of Venn diagrams in the field of statistics has been quite limited. Venn Diagram, also called Primary Diagram, Logic Diagram or Set Diagram, is widely used in mathematics, statistics, logic, computer science and business analysis for representing the logical relationships between two or more sets of data. And, venneuler may have generated a wrong Venn diagram to me. Overlapping areas indicate elements common to both sets. 4 Visualizing with Venn \u2013 A Solidify Understanding Task Creating Venn diagram\u2019s using data while examining the addition rule for probability (S. He represented these relationships using diagrams, which are now known as Venn diagrams. To explain, we will start with an example where we use whole numbers from 1 to 10. If they overlap it means that there are some items in both sets. You are well known about the triangles that they are having three sides. The basic Venn diagram used in presentations shows two partially overlapping shapes, usually circles or ovals, and text to show what belongs to only one shape and what is common to both shapes. One good method to test quickly syllogisms is the Venn Diagram technique. In the UCAT, you will either have to interpret Venn diagrams or you will be given a piece of text and the answers will contain different Venn diagrams and you have to say which one represents the text. _____ _____ f. Show sets F and G on a Venn diagram. A while ago I wrote a small library for displaying Venn and Euler diagrams when trying to learn Javascript. Cards #25\u2013 26 ask students to shade in the described region. Correct-Skipped. I wish there were some magic where you could get a sharp, durable cutter quickly. \ud83e\udd14 Find out what you don't know with free Quizzes \ud83e\udd14 Start Quiz Now!. 01 M on the left, given the following conditions. Your browser doesn't support canvas. We can sometimes use partial information about numbers in some of the regions to derive information about numbers in other regions or other sets. B\u2019\u2229C DIAGRAM 2 2. 6 set Venn diagram with limited overlap. A 6-set venn diagram is never going to look very nice or be very easy to interpret, but here is an R package that can do up to 9 sets: https However depending on what you want/preferences you might want to choose a different package, such as Vennerable, VennDiagram , Venneuler, etc This link. Venn diagrams are also an effective tool for illustrating comparisons and commonalities. In this question , we have to find the region that is represented by the intersection of sets A and B. com/kaz_yos/venn library(VennDiagram) ## Warning: package 'VennDiagram' was built under R version 3. Venn diagrams show + - = underneath. Its very difficult to read and understand venn diagram. In this example I create a Venn diagram from product-ownership data from a survey. The following examples should help you understand the notation, terminology, and concepts relating Venn diagrams and set notation. The region inside the curve represents the elements that belong to the set, while the region outside the curve represents the elements that are excluded from the set. Follow us and share your feedback on Twitter, Reddit, Facebook and on our forums. For example, Figure 2 shows the primary diagram about sets A and B. Universal sets 6. using venn diagram , show that (A-B) , (A intersect B) and (B-A) are disjoint sets , taking A= {2,4,6,8,10,12} and B = {3,6,9,12,15} Priyanka Kedia, Meritnation Expert added an answer, on 24/8/14. This is a venn diagrams pratice exercise test- Click the START button to begin. 01 M on the left, given the following conditions. Venn diagrams are another way of presenting probability information. The universal set\u03be= A\u222aB \u222aC. Venn diagrams are used to show logical relations between sets. com/techreports/2000/HPL-2000-73. Example: Constructing a Venn diagram for Three Sets continued Now determine the numbers that go in region VI. 140 like tea, 120 like coffee and 80 like both tea and coffee. This poster is an updated version of a previous Venn Diagram paper I presented in 2008, and the improvements are. 1 Draw a Venn diagram representing the relationship between isosceles triangles (I), right-angled triangles (R) and equilateral triangles (E). Observe the given Venn diagram and write the following sets. Venn Diagram. Counting & Venn Diagram Part: The counting principle, venn diagram, ven diagram formula, Sets, Venn Diagrams & Counting - Arizona State University, Venn diagram Unknown The counting Principle: If two jobs need to be completed and there are M ways to do the first job & N ways to do the second Job, then there are M*N ways to do one job followed. INTERSECTION OF SETS. The Venn diagram in kindergarten? Yes! Venn diagrams can be used effectively by our youngest students. Make Venn Diagram in R with Correctly Weighted Areas Venn diagrams are incredibly intuitive plots that visually display the overlap between groups. More Results. N, the set of natural numbers, and P, the set of positive integers D. Answer the word problems, once you have read and analyzed the three-set Venn diagrams displayed here. You Might Also Like The equilibrium membrane potentials to be expected across a membrane at 37 \u2218 C, with a NaCl concentration of 0. The circle intersections illustrate qualities shared by the overlapping datasets. Members receive unlimited access to 49,000+ cross-curricular educational resources, including interactive activities, clipart, and abctools custom worksheet generators. Tip: Always start filling values in the Venn diagram from the innermost value. This kind of diagram serves as a graphical representation of how one item is particularly similar or linked to the next. Drawing Venn Diagrams. svg/434px-Venn_diagram_showing_Greek. It is given that n(P) = 165,n(Q)=105 and n (Q) = 95. This is a 5 Circle Venn Diagram template. In the Venn diagram, describe the overlapping area using a complete sentence. This class assumes you are already familiar with diagramming categorical propositions. Five-set Venn diagrams also require the use of ovals or you'll need to overlay a three-set Venn with a recursive curve. ca/~cos/venn/VennTriangleEJC. For ex-ample, Figure 4. Here the set is. Based on their sides and the angles, it is classified into different types. As noted by Henderson, symmetric Venn diagrams with n curves canno t exist for values of n that are composite. A Venn diagram, also called primary diagram, set diagram or logic diagram, is a diagram that shows all possible logical relations between a finite collection of different sets. In the SVG file, hover over a triangle to select it. Venn diagram symbols. A common use for Venn diagrams is the identification of shared and unique elements in different sets. Horizontal mode is identical to the vertical mode The Nested Venn diagram shows unique and shared relationships of eight sets by inlaying four unique-shared diagrams into the other four sets'. Other interactions are also available, such as color changing and export diagram in SVG and PNG format. A complete Venn diagram represents the union of two sets. Rahim described the set as follows: \u2022 M = {all of the foods he eats} \u2022 D = {his favourite desserts}. Often, they serve to graphically organize things, highlighting how the items are similar and different. The same procedures using set B completes region III. ii) in set A but not in set B. It then draws the result, showing each set as a circle. Try out the latest Rainbow Six updates on the Test Server and earn and exclusive charm through the BugHunterProgram. and 3; together have size 25, so the overlap between W and R is 10. I need to draw venndiagram for 6 sets by R. refer to the name plate data for correct connection For delta ( ) wired motors L1 L2. Calculate and draw custom Venn diagrams. They can also be useful for showing how subsets within a larger set are differentiated. Grouping and collection of things, in mathematical terms, is known as a set. The cardinality of the sets and intersection sets is represented by their corresponding circle (polygon) sizes. R # Imagine you have more than two sets and you would want to find the overlapping elements in different sets # and you would like to see the overlap using VennDiagram. \u6211\u60f3\u753b\u4e00\u4e2a\u7ef4\u6069\u56fe\uff0c\u4f46\u662f\u7528R(\u7ef4\u6069\u56fe\u5e93)\u5b83\u88ab\u9650\u5236\u57285\u4e2a\u5143\u7d20\u4e0a\u3002 Do you know how can draw a 6-sets venn diagram ?. 4 Classroom Task: 9. If anything it helps as a language to communicate this. Draw a Venn diagram showing the relationships. 1) A 2) B 3) A \u222a B 4) U 5) A' 6) B' 7) (A \u222a B)'. These diagrams depict elements as points in the plane, and sets as regions inside closed curves. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. For example, we can represent the event \"roll an even number\" by the set {2, 4, 6}. Our video lesson will also guide you with the steps to draw a Venn diagram. How to create a venn diagram with 6 sets of numbers whole,counting,integers,rational,irrational,real. practical usefulness of Venn diagrams diminishes but interesting mathematical questions arise. An ER diagram shows the relationship among entity sets. C D 4 10 8 6 1 2 7 3 5 9 A number is chosen at random from. 30 students are asked if they have a dog or cat. Create beautiful venn diagrams using this tool and download as image (png or jpg). See more ideas about venn diagram, diagram, venn diagram template. They are extensively used to teach Set Theory. Example 6: Shade the portion of the Venn diagram that represents the given set. A Venn diagram, named after John Venn, is a diagram representing all possible logical relations between a finite collection of different sets. A diagram that shows sets and which elements belong to which set by drawing regions around them. Venn Diagrams A Venn diagram is a drawing in which sets are represented by geometric figures such as circles and rectangles. A Venn diagram (also known as a set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets. In business, Venn diagrams are used to compare products, processes, services, and pretty much anything that can be represented in sets. Venn diagrams can be used to help find probabilities when events are not mutually exclusive. Venn diagrams are used to determine conditional probabilities. vennCounts(x, include=\"both\") vennDiagram(object, include=\"both\", names=NULL Each column of x corresponds to a contrast or set, and the entries of x indicate membership of each row in each set or alternatively the significance. Sets and Venn Diagrams October 26, 2012 What is a Set? Mathematics, at its very core, can be described ENTIRELY in terms of sets. They show all of the possible mathematical or logical relationships between sets (groups of things). s} B = {s,t,r,s} C = {t,s,t,r} D = {s,r,s,t} a) A and B b) A and C c) B. 3, with aesthetically pleasing results. A typical venn diagram is shown in the figure below: In the figure, set A contains the multiples of 2 which are. The R visualization code provided in this Power BI desktop file will take a dynamic set of columns (based on the values you add in the fields pane), perform the overlap analysis, and display the diagram. Through the dark background, you may feel serious and formal. Venn diagrams are very useful constructs made of two or more circles that sometimes overlap. Introduced by Venn (1880), the Venn diagram has been popularized in texts on elementary logic and set theory (e. A Venn diagram is a graphical way of representing the relationships between sets. Nice Looking Five Sets Venn Diagrams Stack Overflow. It becomes very hard to read with more groups than that and thus must be avoided. The best way to explain how the Venn diagram works and what its formulas show is to give 2 or 3 circles Venn diagram examples and problems with solutions. In the diagram below, there are two sets, A = {1, 5, 6, 7, 8, 9, 10, 12} and B = {2, 3, 4, 6, 7, 9, 11, 12, 13}. gif 404!396 pixels. A Venn diagram shows all possible logical relations between data sets. If 30 play cricket and 20 both, the correct way of representing above by Venn diagram is. \u201d The first step is to list the twelve months of the year:. I wish there were some magic where you could get a sharp, durable cutter quickly. http://positivemaths. Create beautiful venn diagrams using this tool and download as image (png or jpg). The Venn diagram shows information about the choices the guests made. Other interactions are also available, such as color changing and export diagram in SVG and PNG format. Venn Diagram in case of three elements. ER Diagrams contain different symbols that use rectangles to represent entities, ovals to define attributes and diamond shapes to represent. Alternative Method (Using venn diagram) : Step 1 : Venn diagram related to the information given in the question Let M and S represent the set of students who like math and science respectively. Let the universal set U be all the elements in sets A, B, C and D. 4) Some numbers are Real numbers. OR FT from their Venn diagram If no marks in (b) award SCI for 5, 2 and 2 or identifying the correct regions by listing the correct numbers Penalise incorrect natation once and 3. I will quickly show you how to draw a Venn diagram composed of three distinct datasets in R. R = {x| x is a factor of 24} S = { } T = {7, 9, 11} Summary In this lesson, you learned about sets, subsets, the universal set, the null set and the cardinality of the set. They are extensively used to teach Set Theory. This is where entities that have all the qualities of the overlapping sets. Solved Examples. VENN DIAGRAM. All of the number Page 6/25. , the set of all elements being considered in a particular discussion. RE: In sets and Venn Diagrams, what do the symbols, \u00d8, \u0404 and n mean? For example in the question: Q={2,4,6,8,10} and R={5,10,15,20}. To understand the right way to use Venn Diagrams, it helps to go back in time. The union operations are reflected to the Venn diagram. A Venn diagram (named after mathematician John Venn in 1880) is a method used to sort items into groups. Default is 0. You should habe 2 sets of numbers - say 5, 10, 15, 20, 25, 30 and another set that could be 20, 25, 30, 35, 40, 45. \u2022 Use black ink or ball-point pen. In these diagrams, the universal set is represented by a rectangle, and other sets of interest within the universal set are depicted by oval regions, or sometimes by circles or other shapes. While there are more than 30 symbols used in set theory, you don\u2019t need to memorize them all to get started. mapping: the aesthetics mapping. The following examples should help you understand the notation, terminology, and concepts relating Venn diagrams and set notation. Watch our Venn diagram chapter video to get a detailed explanation of the significance of a Venn diagram in sets. These diagrams make use of circular shapes in various colors which appeal visually to any audience and can easily represent the size, value and relationship between different sets of information. The upper diagram to the. Venn and Euler diagrams in (c) and (d) both include six data sets: in (c), irregular polygons are drawntoillustratesixwoodyspecies. It is easy to understand the union and intersection of sets with the help of Venn diagrams. Use the Venn diagram below to fill in the missing statement. Venn diagrams use circles to represent sets and to illustrate the relationship between a finite collection of different sets. Answer the following using Venn Diagrams. Polish your personal project or design with these Venn Diagram transparent PNG images, make it even more personalized and more attractive. Play Venn diagram quizzes on Sporcle, the world's largest quiz community. Single speed motors. Aside from the innermost face and the outermost face, a rotationally symmetric $$n$$-Venn diagram can be partitioned into $$n$$ congruent clumps, each of size $$(2^n-2)/n$$; in this case, we call the clump a cluster\u2014it is like a. Example 6 Provide a Venn diagram for the following syllogism: Some M are P All M are S. diagram function from VennDiagram package. \u2026The first one, on the top left, is Hacking Skills. Coordinates of vertices are from http://sue. This is the currently selected item. Place Value Roald Dahl Themed Year 2 PowerPoint. As with many other diagrams in these pages, regions are coloured by weight. We can show the Universal Set in a Venn Diagram by putting a box around the whole thing: Now you can see ALL your ten best friends, neatly sorted into what sport they play (or not!). \u201d The type of three circle Venn Diagram we will need is the following: Image Source: Passy\u2019s World of Mathematics. Sets within the universal set are usually represented by circles. Write down the numbers that are in set. Venn diagram, also known as Euler-Venn diagram is a simple representation of sets by diagrams. The trick is to make them user-friendly, hands-on, and developmentally appropriate as a tool even kindergarten students can use with ease. There is also some new language to be learnt: Symbols that represent AND (the intersection of sets), OR (the union of sets) etc. \"A Venn diagram or set diagram is a diagram that shows all possible logical relations between a finite collection of sets. io diagram, select Venn in the left hand side of the template diagram. In this figure, the big rectangle shows the universal set $S$. You then have to use the given information to populate the diagram and figure out the remaining information. 9 it is also used to pass additional aesthetics parameters for the ggplot2 graphics. Also, it\u2019s not possible to show more than 6 sets in a Venn diagram. Children need to think about how to sort something according to the two rules. From the adjoining Venn diagram, find the following sets. ER Diagrams contain different symbols that use rectangles to represent entities, ovals to define attributes and diamond shapes to represent. Venn Diagrams are very useful for visualizing the relationships between groups. Compare and contrast your data. In particular, Venn Diagrams are used to demonstrate De Morgan's Laws. These diagrams depict elements as points in the plane, and sets as regions inside closed curves. Multiplication rule. Venn diagrams are plots used to graphically display intersections between two or more groups. iv) in set B but not in set A. Venn diagrams are particular cases of Euler diagrams showing all possible combinations. A Venn diagram (also called primary diagram, set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets. In this example I create a Venn diagram from product-ownership data from a survey. I have following data need to be visualized by a venn diagram Total counts: 1668 Counts for group A: 62 Counts for group B: 24 (Group B is a subgroup of group A, all counts in group B are included in group A) Counts for group C: 267 (including group A, but excluded group B) How to display the proportion(%) of overlap between each other by venn diagram? request the graph display 4 ways (total. newpage() # make new page for plot. An ER diagram shows the relationship among entity sets. These are standard Venn diagrams for comparing and contrasting two items. 1 By Juan Carlos Oliveros BioinfoGP, CNB-CSIC: 1. In this tutorial, I'll show how to plot a three set venn diagram using R and the ggplot2 package. Sal uses a Venn diagram to introduce two-way frequency tables. In this figure, the big rectangle shows the universal set $S$. You will need a two circle Venn Diagram for your answer. Results: We present a visualization approach for set relationships based on Venn diagrams. ca/~cos/venn/VennTriangleEJC. First download the Venn diagrams in excel zip file from here [xls version here]. Abstract: The Venn Diagram technique is shown for typical as well as unusual syllogisms. The section where the two sets overlap has the numbers contained in both Set A and B, referred to as the intersection of A and B. A Venn diagram, also called primary diagram, set diagram or logic diagram, is a diagram that shows all possible logical relations between a finite collection of different sets. Once we understand how to read the Venn Diagram we can use it in many applications. A subreddit for Venn diagram enthusiasts. Functions in venn. Find: a P(B) b P(A B) c P(A B) 2 The Venn diagram shows the whole numbers from 1 to 10. Venn Diagram Word Problems - Three Sets. 6 for all Venn diagrams with up to five sets, and it automatically decreases to 0. C B A Example The following Venn diagram shows the number of elements in each region for the. none of the above ____ 6. How many girls are on both the. Venn Diagrams (H) - Version 2 January 2016. 43 ewlength\\@[email\u00a0protected] \\@[email\u00a0protected] The height of the entire Venn diagram. Guide students toward an understanding of the Venn diagram by letting them physically manipulate hoola. Some of the worksheets for this concept are Grade 3 questions venn diagrams, Venn diagrams, Venn diagram l1s1, Ss, Venn diagram, Write details that tell how the subjects are different in, Chapter 3 1 venn diagrams, Math 211 sets practice work part 1 shade the region. Using Venn Diagrams in Set Theory. Venn diagrams show + - = underneath. Remember that, To maximize overlap, Union should be as small as possible; Calculate the surplus = n( A) +n(B) +n(C)-n(A or B or C). Follow us and share your feedback on Twitter, Reddit, Facebook and on our forums. We can represent this as, 01. vennCounts produces an object of class \"VennCounts\". 1 #13 To shade the set we need to compare the Venn diagram for A with the Venn diagram for B\u2032, and bear in mind the meaning of union. They are much easier to take in at a glance than truth tables (for example) which are equivalent, but boring and difficult to interpret quickly. Any values that belong to more than one set will be placed in the sections where the circles overlap. I have following data need to be visualized by a venn diagram Total counts: 1668 Counts for group A: 62 Counts for group B: 24 (Group B is a subgroup of group A, all counts in group B are included in group A) Counts for group C: 267 (including group A, but excluded group B) How to display the proportion(%) of overlap between each other by venn diagram? request the graph display 4 ways (total. 01 M on the left, given the following conditions. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1. See full list on mathsisfun. In this section we introduce the ideas of sets and Venn diagrams. 2 shows an example of a Venn diagram. The Venn diagram of A and B looks something like this: Here are some examples of set operations and their Venn Diagrams: A\u222aB A\u2229B A\u222aB (A\u222aB) A three-set Venn diagram looks. DIFFERENCE OF SETS. A = { set of even numbers between 0 and 10} 02. Select number of sets and update the venn diagram parameters Venn diagram maker tool is completely free to use. Note that the total of the entries is 52, and the complete set is denoted by \u03b5. The above diagram is called Venn Diagram where there are two sets A and B and U is the universal set. Sets are shown as regions inside circles or other closed curves, and common elements of the sets are shown as intersections of these circles. number and integers have no elements in common because the whole. (ii) Sketch all possible Venn Diagram if csbsa. If a is an element of the set A then we write a 2 A,ifa is not an element of a set A,thenwe write a/2 A. Worked example J. Using Venn diagrams allows children to sort data A Venn diagram is when the two sorting circles overlap in the middle. I\u2019m going to have to do another post on Venn Diagram activities, but it inspired me to put together some printable blank Venn Diagram templates that I know will get used!. After coming upon these answers: Create a Venn Diagram, How to plot Venn diagrams with Mathematica?, which I Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It's just a rough guide from my limited knowledge on the hololive girls (would include guys too but there's limited space and I only know a few of them). 140 like tea, 120 like coffee and 80 like both tea and coffee. We use circles to represent the sets, and enclose our diagram in a rectangle. The section of the circles which overlap contain numbers that have properties pertaining to both of the circles. All other elements represent the consonants. Some, but not all, students in the school take swimming. Editable graphics with text placeholder. \u2022 8 have a dog, but not a cat. Remark 1It is easy to manipulate Venn diagrams. n (M\u2032 \u222a C\u2032) = _____ b. A Venn diagram could be used, for example, to compare two companies. A Venn diagram is a mathematical illustration that shows all of the possible mathematical or logical relationships between sets. In a school of 320 students, 85 students are in the band, 200 students are on sports teams, and 60 students participate in both activities. Instructions. Venn Diagrams. Sort by one or two conditions. This means we start out with the second premise: All M are S. Media in category \"6-set Venn diagrams\" The following 7 files are in this category, out of 7 total. Venn Diagram and Shape Sorting Lesson Plan. Linguistics. A set of functions to generate high-resolution Venn and Euler plots. This particular geometric configuration for five category Venn diagrams was popularized by Adrian Dusa's venn package for R. Sets are represented by circles included in a rectangle that represents the universal set, i. Five-Set Diagrams. 5 Three-Set Venn Diagram. This can be cumbersome I agree. Stacked Venn diagram. Venn diagrams are listed in the probability section of the new Key Stage 3 Programme of Study and occur in GCSE reference P6 which involves representing sets and combinations of sets using Venn diagrams, as well as through the use of tables and grids. Venn's diagrams drawing tool for comparing up to four lists of elements. Note that although there are no elements shown inside. vi) in set A or set B. The chart is also known as 'Set diagram' or 'Logic diagram'. Now when you try to open the file, you must enable macros (in excel 2007, you may want to set the security to low and then reopen the file) 3. It generally consists of a box that represents the sample space S together with circles or ovals. It\u2019s easy to create Venn and Euler diagrams in draw. Golf a Venn Diagram generator. The description of some sets is given and you are asked to draw a Venn diagram to illustrate the sets. Tried the R approach but then ran into this problem (fresh from the Console) WARN Table to R 0:57:35 R Version 3. Using Venn diagrams allows children to sort data A Venn diagram is when the two sorting circles overlap in the middle. Set theory makes wide use of Venn diagrams. I need to draw venndiagram for 6 sets by R. Solved Examples. The second set of branches represents the second draw. In a typical Venn diagram, 2 or more sets (each containing various elements) are represented as circles overlayed onto each other. Then each set in the problem is represented by a circle. In R, the VennDiagram package is the best option to build one. , 10 is a multiple of 3 and 5. VennTure can generate six-sets Venn diagrams with a graphic user interface (GUI), yet it consumes large amounts of memory and has low computational efficiency. This diagram on political parties process is a good example of a three-set Venn diagram, with text inside the circles to help further explain each topic, and the center being the \"sweet spot\" where. Probabilities from Venn diagrams: True or False & Matching Questions. A Venn diagram is a chart that compares two or more sets (collections of data) and illustrates the differences and commonalities between them with overlapping circles. A Venn diagram (also called primary diagram, set diagram or logic diagram) is a diagram that shows all possible logical relations between a finite collection of different sets. A 6-set venn diagram is never going to look very nice or be very easy to interpret, but here is an R package that can do up to 9 sets: https However depending on what you want/preferences you might want to choose a different package, such as Vennerable, VennDiagram , Venneuler, etc This link. These are diagrams that make use of geometric shapes to show relationships between sets. This means that as the number of contours increase, Euler diagrams are typically less visually complex than the equivalent Venn diagram, particularly if the number of non-empty intersections is small. You should habe 2 sets of numbers - say 5, 10, 15, 20, 25, 30 and another set that could be 20, 25, 30, 35, 40, 45. In a group of 40 students 6 are left-handed, 18 have size 8 feet and 2 are left-handed with size 8 feet. Venn diagrams are used to determine conditional probabilities. A Venn diagram could be used, for example, to compare two companies. (a) How many guests had custard? (b) How many guests had ice cream and custard? (c) How many guests went to the wedding? 5. We can now use the commands in this package for generating Venn diagrams. A s imple symmetric Venn diagram consisting of five ellipses was given in [5]. confusing as a presentation tool because of the number of. 3 shows the di\ufb00erence. This lesson covers how to use Venn diagrams to solve probability problems. A triple Venn diagram is a diagram that consists three intertwined circles that represents thoughts and the relation of each from one another. 9 Venn diagrams Venn Diagrams are the diagrams which represent the relationship between sets. First step: Install & load \u201cVennDiagram\u201d package. Diagram 15 is a Venn diagram that showsthe result of a survey carried out by a company on a group of 255 people about their favourite handphones. In Venn diagrams the curves are overlapped in every possible way, showing all possible relations between the sets. A Venn diagram consists. N, the set of natural numbers, and I, the set of integers B. Here is a famous example: a six-set venn diagram published in Nature that shows the relationship between the banana's genome. 96 like wine A, 93 like wine B, 96 like wine C, 92 like A and B, 91 like B and C, 93 like A and C, 90 like all three wines. A Venn diagram shows all possible logical relations between different sets or groups of data. A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Find: (i) (ii) (iii) Answers: (i) (ii) (iii) Question 7: If , , and. Creates a Venn diagram with five sets. Since all numbers in set A have been placed, there are no numbers in region I. Some of the worksheets for this concept are Grade 3 questions venn diagrams, Venn diagrams, Venn diagram l1s1, Ss, Venn diagram, Write details that tell how the subjects are different in, Chapter 3 1 venn diagrams, Math 211 sets practice work part 1 shade the region. The shaded area of figure is 5. Figure 2: Venn\u2019s Primary diagrams. A complete Venn diagram represents the union of two sets. Find how many people can speak both English and Hindi ? speak Does union of sets include intersection of sets ?. The closest I've seen is the Unicorn profile folks are posting about on Wood Central. In set theory, we commonly use Venn diagrams, developed by the logician John Venn ( ). Venn diagrams are useful in any situation requiring a direct comparison of two or more categories or concepts. Venn diagrams are being used to study and analyze the similarities and differences among languages. Venn Diagrams The Venn diagram, is a convenient way to illustrate definitions within the algebra of sets. Use it for drawing Venn and Euler diagrams. You also learned to use the Venn diagram to show relationships between sets. 6 set Venn diagram. Sets and Venn Diagrams October 26, 2012 What is a Set? Mathematics, at its very core, can be described ENTIRELY in terms of sets. A thorough introduction to shading regions of venn diagrams and using them to calculate probabilities. They can be interactive and fun, and here are a few ways to make a Venn diagram: Draw the circles on a blank piece of paper and fill in the information. Venn diagrams with complements, unions and intersections. This Site Might Help You. While there are more than 30 symbols used in set theory, you don\u2019t need to memorize them all to get started. Characteristics of Venn Diagram. Creates a Venn diagram with five sets. colors: named list of colors for sets (one set=one color) na. possible interactions. Venn diagrams are a great way to visualize the structure of set relationships. VENN DIAGRAM. Revision notes on 'Set Notation & Venn Diagrams' for the Edexcel IGCSE Maths exam. That's ok though- sometimes a traditional Venn just won't work. Last modified September 6, 2017 Prism offers tools to draw circles and text, and you can use those to create informal Venn diagrams. Use MyDraw to create your own 5 set Venn diagram. Drawing these diagrams manually is difficult. It allows multiple sets (four sets for venn, 3 sets for Euler diagrams), customizable colours and fonts, simple syntax and and best of all the size of the circles is proportional to the size of the data sets (at least when comparing 2 data sets). In the center, the student lists the items shared in common. A Venn diagram is an illustration used to depict logical relationships between different groups or sets. Which Venn diagram correctly describes the relationship between the set of integers and the set of whole numbers A. Venn Diagrams Venn diagrams are very useful for sorting and processing data. Creates a Venn diagram with five sets. Student: The Venn Diagram below shows the number of girls on the soccer and track teams at a high school. In fact, the following three are the perfect foundation. 0 and Rserve <= 1. Venn diagram definition is - a graph that employs closed curves and especially circles to represent logical relations between and operations on sets and the terms of propositions by the inclusion, exclusion, or intersection of the curves. A Euler diagram resembles a Venn diagram, but does not neccessarily. If you wish to give an awesome presentation, using diagrams is great because they make your data look nicer and help your These cookies may be set through our site by our advertising partners. Use one of our Venn diagram templates to show all possible relations between your sets of data. Moreover, union operations between sets can be made. outwards_adjust: the multiplier defining the distance from the centre. Drawing Venn diagrams that is not specifically always two circles. 1) A 2) B 3) A \u222a B 4) U 5) A' 6) B' 7) (A \u222a B)'. OR FT from their Venn diagram. Venn diagrams are another way of presenting probability information. (a) Draw a Venn diagram for the two sets, P and R. The applicant was not hired. In a Venn diagram any set is depicted by a closed region. An entity set is a group of similar entities and these entities can have attributes. Jan 25, 2016 - Printable Venn diagram worksheets for primary grades 1 to 7 math students, based on the Singapore math curriculum. This Site Might Help You. Venn Diagrams (H) - Version 2 January 2016. Odd numbers and numbers greater than 10. Learn all about Venn diagrams and make your own with Canva. Textbook Authors: Blitzer, Robert F. venn (Required) A Venn object. So how did it start and The Venn diagram has emerged as a useful and versatile learning tool in education. There are 40 people preferred both brand P and brand Q, 35 people preferred both brand Q and brand R, 60 people preferred both brand P and brand R. A Venn diagram represents each set by a circle, usually drawn inside of a containing box representing the universal set. Venn diagrams, also called Set diagrams or Logic diagrams, are widely used in mathematics, statistics, logic,. First step: Install & load \u201cVennDiagram\u201d package. A triple Venn diagram is a diagram that consists three intertwined circles that represents thoughts and the relation of each from one another. c) t \u2013 x \u2013 z + r +b. 4-set means I should be able to specify 4 sets. Venn diagrams are helpful for. It's just a rough guide from my limited knowledge on the hololive girls (would include guys too but there's limited space and I only know a few of them). A venn diagram uses circles that. Members receive unlimited access to 49,000+ cross-curricular educational resources, including interactive activities, clipart, and abctools custom worksheet generators. The section where the two sets overlap has the numbers contained in both Set A and B, referred to as the intersection of A and B. Originally used as a way to show the differences and similarities. In order to properly celebrate John Venn's 180th birthday, today your task will be creating a program that outputs a Venn Diagram!. It seems VennDiagram can not work on my data. Use of Venn Diagrams: Venn diagrams or set diagrams are specially designed diagrams that show all possible logical relations between a finite collection of sets or aggregation of things. none of the above ____ 6. Coordinates of vertices are from http://sue. The universal set\u03be= A\u222aB \u222aC. Venn diagrams also help us to convert common English words into mathematical terms that help add precision. Five percent of the students work part time and belong to a club. A B A B \u00c7 Venn Diagrams Try this one!. venn: Draw a Venn Diagram with Five Sets in VennDiagram: Generate High-Resolution Venn and Euler Plots rdrr. Look at this Venn diagram. Repurposing to Use Segments. The following example (see example 1. Force Directed Tree. You can find 100s of templates and examples to be used freely in the diagram community of Creately. Venn diagrams were invented for use in a branch of mathematics called set theory. Set A = {2,4,6,8,10,12,14,16}. The term Venn diagram is not foreign since we all have had Mathematics, especially Probability and Algebra. Manage My Favorites. For instance: Out of forty students, 14 are taking English Composition and 29 are taking Chemistry. I am interested in learning how to do a 2-way venn diagram in R from a table generated with cuffdiff data.", "date": "2021-04-15 04:33:06", "meta": {"domain": "1upload.de", "url": "http://zveg.1upload.de/r-venn-diagram-6-sets.html", "openwebmath_score": 0.4094984829425812, "openwebmath_perplexity": 789.4183828942959, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769042651874, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8576752802971456}}
{"url": "https://mathhelpboards.com/threads/expected-number-of-light-bulbs-on.5655/", "text": "# Expected number of light bulbs on\n\n#### oyth94\n\n##### Member\nThere are 50 light bulbs in a room each with its own switch. If a light bulb is on, Dick turns it off and if it is off , he turns it on. Initially all light bulbs are off . After 50 flips and assuming that Dick chooses switches to be flipped randomly, what is the expected number of light bulbs on in the room rounded to the nearest natural number?\n\nTo be honest, i tried many times but i can;t seem to arrive at an answer. i am stuck. any help would be appreciated!!!\n\nso i have started with this:\nLet Ei be the event that bulb i is on after 50 flips. This is the case if switch i is chosen an odd number of times. Compute Pr(Ei) and let Xi be the indicator random variable for Ei. Then E[Xi]=Pr(Ei). Now, let X be the total number of bulbs on after 50 flips. X=\u2211Xi, so E[X]=\u2211E[Xi]=n\u22c5Pr(Ei).\n\nLast edited:\n\n#### chisigma\n\n##### Well-known member\nThere are 50 light bulbs in a room each with its own switch. If a light bulb is on, Dick turns it off and if it is off, he turns it on. Initially all light bulbs are off. After 50 flips and assuming that Dick chooses switches to be flipped randomly, what is the expected number of light bulbs on in the room rounded to the nearest natural number?\n\nTo be honest, i tried many times but i can;t seem to arrive at an answer. i am stuck. any help would be appreciated!!!\n\nso i have started with this:\nLet Ei be the event that bulb i is on after 50 flips. This is the case if switch i is chosen an odd number of times. Compute Pr(Ei) and let Xi be the indicator random variable for Ei. Then E[Xi]=Pr(Ei). Now, let X be the total number of bulbs on after 50 flips. X=\u2211Xi, so E[X]=\u2211E[Xi]=n\u22c5Pr(Ei).\nThe core of the problem is the meaning of 'Dick chooses switches to be flipped randomly'... if it means that any bulb has probability 1/2 to be switched by Dick at any iteration it is evident that at the end of each iteration the expected number of switched on bulbs is 25...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nThere are 50 light bulbs in a room each with its own switch. If a light bulb is on, Dick turns it off and if it is off, he turns it on. Initially all light bulbs are off. After 50 flips and assuming that Dick chooses switches to be flipped randomly, what is the expected number of light bulbs on in the room rounded to the nearest natural number?\nFascinating problem! I have wasted half my Sunday morning on it.\n\nLet's generalise it a bit, to the case where there are $k$ switches. Denote by $E(k,n)$ the expected number of lit bulbs after $n$ random flips, given that initially all light bulbs were off. I found $50$ much too large a number to work with, so I did some experiments with smaller numbers. For $k=4$, I found that the values of $E(4,n)$, for $n=1,2,3,4,5,6$ are $$1,\\ \\frac32,\\ \\frac74,\\ \\frac{15}8,\\ \\frac{31}{16}, \\frac{63}{32},$$ which makes it seem likely that $E(4,n) = \\dfrac{2^n-1}{2^{n-1}}$.\n\nI then tried $k=6$ and found something even more interesting: for $n=1,2,3,4$, the values of $E(6,n)$ are $$1,\\ \\frac53,\\ \\frac{19}9,\\ \\frac{65}{27},$$ which very strongly suggests that $E(6,n) = \\dfrac{3^n-2^n}{3^{n-1}}.$\n\nIf so, then it seems inevitable that the general formula must be $\\boxed{E(2k,n) = \\dfrac{k^n - (k-1)^n}{k^{n-1}}}.$\n\nI'm not prepared to put money on it, but I'm 99% sure that $$E(50,50) = \\dfrac{25^{50} - 24^{50}}{25^{49}} = 25 - 24\\Bigl(\\frac{24}{25}\\Bigr)^{\\!49} \\approx 21.75.$$\n\nNow it's time for Sunday lunch, so I'll leave it to others to see if those results can be proved.\n\n#### oyth94\n\n##### Member\nFascinating problem! I have wasted half my Sunday morning on it.\n\nLet's generalise it a bit, to the case where there are $k$ switches. Denote by $E(k,n)$ the expected number of lit bulbs after $n$ random flips, given that initially all light bulbs were off. I found $50$ much too large a number to work with, so I did some experiments with smaller numbers. For $k=4$, I found that the values of $E(4,n)$, for $n=1,2,3,4,5,6$ are $$1,\\ \\frac32,\\ \\frac74,\\ \\frac{15}8,\\ \\frac{31}{16}, \\frac{63}{32},$$ which makes it seem likely that $E(4,n) = \\dfrac{2^n-1}{2^{n-1}}$.\n\nI then tried $k=6$ and found something even more interesting: for $n=1,2,3,4$, the values of $E(6,n)$ are $$1,\\ \\frac53,\\ \\frac{19}9,\\ \\frac{65}{27},$$ which very strongly suggests that $E(6,n) = \\dfrac{3^n-2^n}{3^{n-1}}.$\n\nIf so, then it seems inevitable that the general formula must be $\\boxed{E(2k,n) = \\dfrac{k^n - (k-1)^n}{k^{n-1}}}.$\n\nI'm not prepared to put money on it, but I'm 99% sure that $$E(50,50) = \\dfrac{25^{50} - 24^{50}}{25^{49}} = 25 - 24\\Bigl(\\frac{24}{25}\\Bigr)^{\\!49} \\approx 21.75.$$\n\nNow it's time for Sunday lunch, so I'll leave it to others to see if those results can be proved.\nThank you so much! I arrived with the same answer but I used either Poisson or Binomial. But my approach I didn't involve expected value...which I should...\n\n#### zzephod\n\n##### Well-known member\nFascinating problem! I have wasted half my Sunday morning on it.\n\nLet's generalise it a bit, to the case where there are $k$ switches. Denote by $E(k,n)$ the expected number of lit bulbs after $n$ random flips, given that initially all light bulbs were off. I found $50$ much too large a number to work with, so I did some experiments with smaller numbers. For $k=4$, I found that the values of $E(4,n)$, for $n=1,2,3,4,5,6$ are $$1,\\ \\frac32,\\ \\frac74,\\ \\frac{15}8,\\ \\frac{31}{16}, \\frac{63}{32},$$ which makes it seem likely that $E(4,n) = \\dfrac{2^n-1}{2^{n-1}}$.\n\nI then tried $k=6$ and found something even more interesting: for $n=1,2,3,4$, the values of $E(6,n)$ are $$1,\\ \\frac53,\\ \\frac{19}9,\\ \\frac{65}{27},$$ which very strongly suggests that $E(6,n) = \\dfrac{3^n-2^n}{3^{n-1}}.$\n\nIf so, then it seems inevitable that the general formula must be $\\boxed{E(2k,n) = \\dfrac{k^n - (k-1)^n}{k^{n-1}}}.$\n\nI'm not prepared to put money on it, but I'm 99% sure that $$E(50,50) = \\dfrac{25^{50} - 24^{50}}{25^{49}} = 25 - 24\\Bigl(\\frac{24}{25}\\Bigr)^{\\!49} \\approx 21.75.$$\n\nNow it's time for Sunday lunch, so I'll leave it to others to see if those results can be proved.\nI can report that simulation gives the mean number off after 50 flips of the 50 switchs is 21.74 with a standard error ~0.034\n\n(another exercise on my Python learning curve)\n\n.\n\nLast edited:\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nI now see how to prove the formula for $E(2k,n)$, and it's really quite easy. When the $(n+1)$th switch is flipped, the probability that it is already on is $\\dfrac{E(2k,n)}{2k}$, in which case the flip turns it off, and the number of \"on\" switches is reduced by $1$. Likewise, the probability that the $(n+1)$th switch is off is $\\dfrac{2k-E(2k,n)}{2k}$, in which case the flip turns it on, and the number of \"on\" swiches is increased by $1$. Therefore $$E(2k,n+1) = \\frac{E(2k,n)}{2k}(E(2k,n)-1) + \\frac{2k-E(2k,n)}{2k}(E(2k,n)+1) = \\frac{(k-1)E(2k,n) + k}k.$$ Now assume as an inductive hypothesis that $E(2k,n) = \\dfrac{k^n-(k-1)^n}{k^{n-1}}$. Then $$E(2k,n+1) = \\frac{(k-1)E(2k,n) + k}k = \\frac{(k-1)k^n-(k-1)^{n+1} + k^n}{k^n} = \\frac{k^{n+1} - (k-1)^{n+1}}{k^n},$$ which completes the inductive step. The base case $E(2k,1) = 1$ is easy to check, so the result is proved.", "date": "2021-05-11 19:31:42", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/expected-number-of-light-bulbs-on.5655/", "openwebmath_score": 0.8901501893997192, "openwebmath_perplexity": 310.678156696057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156284, "lm_q2_score": 0.8791467548438124, "lm_q1q2_score": 0.8576752726066709}}
{"url": "https://math.stackexchange.com/questions/2406516/frac12n-binomnn-frac12n1-binomn1n-frac122n", "text": "# $\\frac{1}{2^n}\\binom{n}{n}+\\frac{1}{2^{n+1}}\\binom{n+1}{n}+...+\\frac{1}{2^{2n}}\\binom{2n}{n}=1$: short proof?\n\nThe identity $\\frac{1}{2^n}\\binom{n}{n}+\\frac{1}{2^{n+1}}\\binom{n+1}{n}+...+\\frac{1}{2^{2n}}\\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the fact that $\\binom{a}{b}+\\binom{a}{b+1}=\\binom{a+1}{b+1}$, is straightforward but not enlightening.\n\nIs it possible to find any very clever approaches? Via a combinatorial or probabilistic interpretation, for instance?\n\nSuppose you are flipping a coin until you get $n+1$ heads or $n+1$ tails. For $k=0,\\dots,n$, what is the probability that you are done after exactly $n+1+k$ flips?\n\nIf the last flip was a tails, this means that in the former $n+k$ flips there were exactly $n$ tails, and this happens with probability $\\frac{1}{2}\\cdot\\frac{1}{2^{n+k}}\\binom{n+k}{n}$ (the first $\\frac{1}{2}$ is due to assuming the last flip is tails). Same for heads, so the probability of finishing after exactly $n+1+k$ flips is $\\frac{1}{2^{n+k}}\\binom{n+k}{n}$.\n\nNow just note that the number of flips always ends up between $n+1$ and $n+1+n$ (by pigeonhole principle).\n\nUpdated Solution\n\nHere's a neater solution!\n\n\\begin{align} \\sum_{k=0}^n \\frac 1{2^{n+k}}\\binom {n+k}n &=\\frac 1{2^{2n}}\\sum_{k=0}^n \\binom {k+n}k 2^{n-k}\\\\ &=\\frac 1{2^{2n}}\\sum_{k=0}^n \\binom {k+n}k \\sum_{j=0}^{n-k}\\binom {n-k}j\\\\ &=\\frac 1{2^{2n}}\\sum_{l=0}^n \\binom {2n-l}{n-l}\\sum_{j=0}^l \\binom lj &&(l=n-k)\\\\ &=\\frac 1{2^{2n}}\\sum_{j=0}^n \\sum_{l=j}^n\\binom {2n-l}n\\binom lj\\\\ &=\\frac 1{2^{2n}}\\sum_{j=0}^n \\binom {2n+1}{n+j+1} &&(*)\\\\ &=\\frac 1{2^{2n}}\\sum_{j=n+1}^{2n+1} \\binom {2n+1}j\\\\ &=\\frac 1{2^{2n}}\\cdot \\frac 12\\sum_{j=0}^{2n+1}\\binom {2n+1}j &&\\text{(by symmetry)}\\\\ &=\\frac 1{2^{2n+1}}\\cdot 2^{2n+1}\\\\ &=1\\;\\;\\color{red}{\\blacksquare}\\end{align}\n\n$\\qquad \\qquad \\quad ^*\\displaystyle\\scriptsize\\text{using }\\sum_{r} \\binom {a-r}{c}\\binom {b+r}{d}=\\binom {a+b+1}{c+d+1}$\n\nSolution posted earlier\n\nHere's a direct algebraic proof without using induction.\n\n\\begin{align} \\sum_{k=0}^n \\binom {k+n}k x^k(x+y)^{n-k} &=\\sum_{k=0}^n \\binom {k+n}k x^k\\sum_{j=0}^{n-k}\\binom {n-k}Jy^jx^{n-k-j}\\\\ &=\\sum_{k=0}^n \\binom{k+n}k\\sum_{j=0}^{n-k}\\binom{n-k}jy^jx^{n-j}\\\\ &=\\sum_{\\ell=0}^n\\binom{2n-\\ell}{n-\\ell}\\sum_{j=0}^\\ell\\binom{\\ell}jy^jx^{n-j} &&(\\ell=n-k)\\\\ &=\\sum_{j=0}^n\\sum_{\\ell=j}^n (-1)^{n-\\ell}\\binom{-n-1}{n-\\ell}(-1)^{\\ell-j}\\binom{-j-1}{\\ell-j}y^jx^{n-j}\\\\ &=\\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\\sum_{\\ell=j}^n\\binom{-n-1}{n-\\ell}\\binom{-j-1}{\\ell-j}\\\\ &=\\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\\binom{-n-j-2}{n-j} &&\\text{(Vandermonde)}\\\\ &=\\sum_{j=0}^n(-1)^{n-j}y^jx^{n-j}\\cdot (-1)^{n-j}\\binom{2n+1}{n-j}\\\\ &=\\sum_{j=0}^n \\binom{2n+1}{n-j}y^jx^{n-j}\\\\ &=\\sum_{i=0}^n \\binom{2n+1}i x^i y^{n-i}\\\\ \\text{Put }x=y=1:\\hspace{4cm}\\\\ \\sum_{k=0}^n\\binom {k+n}k2^{n-k} &=\\sum_{i=0}^m\\binom{2n+1}i &&(i=n-j)\\\\ &=\\frac 12\\cdot 2^{2n+1} &&\\text{(by symmetry)}\\\\ &=2^{2n}\\\\ \\sum_{k=0}^n \\binom {n+k}k 2^{-k}&=2^n\\\\ \\sum_{k=0}^n\\frac 1{2^{n+k}} \\binom {n+k}n&=1\\\\ \\end{align}\n\n\u2022 @robjohn - Thank you - very kind of you! Sep 1, 2017 at 7:13\n\nHere is an answer based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g. \\begin{align*} [z^k](1+z)^n=\\binom{n}{k} \\end{align*}\n\nWe obtain \\begin{align*} \\color{blue}{\\sum_{k=0}^n\\binom{n+k}{k}\\frac{1}{2^{n+k}}} &=\\sum_{k=0}^n\\binom{2n-k}{n-k}\\frac{1}{2^{2n-k}}\\tag{1}\\\\ &=\\sum_{k=0}^\\infty[z^{n-k}](1+z)^{2n-k}\\frac{1}{2^{2n-k}}\\tag{2}\\\\ &=2^{-2n}[z^n](1+z)^{2n}\\sum_{k=0}^\\infty\\left(\\frac{2z}{1+z}\\right)^{k}\\tag{3}\\\\ &=2^{-2n}[z^n](1+z)^{2n}\\cdot\\frac{1}{1-\\frac{2z}{1+z}}\\tag{4}\\\\ &=2^{-2n}[z^n](1+z)^{2n+1}\\cdot\\frac{1}{1-z}\\tag{5}\\\\ &=2^{-2n}\\sum_{k=0}^n[z^k](1+z)^{2n+1}\\tag{6}\\\\ &=2^{-2n}\\sum_{k=0}^n\\binom{2n+1}{k}\\tag{7}\\\\ &=2^{-2n}\\frac{1}{2}2^{2n+1}\\tag{8}\\\\ &\\color{blue}{=1} \\end{align*}\n\nComment:\n\n\u2022 In (1) we change the order of summation $k \\rightarrow n-k$.\n\n\u2022 In (2) we apply the coefficient of operator. We also set the limit to $\\infty$ without changing anything since we are adding zeros only.\n\n\u2022 In (3) we do a rearrangement and apply the formula $[z^{p-q}]A(z)=[z^p]z^qA(z)$.\n\n\u2022 In (4) we apply the geometric series expansion.\n\n\u2022 In (5) we do some simplifications.\n\n\u2022 In (6) we do the Cauchy multiplication with the geometric series $\\frac{1}{1-x}$ and restrict the upper limit of the sum with $n$ since other terms do not contribute to $[z^n]$.\n\n\u2022 In (7) we select the coefficient of $z^k$.\n\n\u2022 In (8) we apply the binomial theorem.", "date": "2022-08-12 06:54:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2406516/frac12n-binomnn-frac12n1-binomn1n-frac122n", "openwebmath_score": 0.9867181777954102, "openwebmath_perplexity": 2182.4940979823164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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{"url": "http://uqqp.erci.pw/hard-probability-questions.html", "text": "# Hard Probability Questions\n\nYou can solve many simple probability problems just by knowing two simple rules: The probability of any sample point can range from 0 to 1. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a \ufb01rst course in prob-ability. Total testing time is two hours and \ufb01 fty minutes; there are no separately timed sections. I wanted pupil to leave exam and not end up saying \"Ohh I could have done that question!\" two minutes later. by Marco Taboga, PhD. These can be handy if you are playing card games or just trying to understand probability. All problems like the following lead eventually to an equation in that simple form. Probability Jeopardy. It is depicted by P(A|B). Probability of A given the occurrence of B is equal to the probability of A and B over the probability that B has occurred. The explanation discusses more than one way in which the answer to the question can be arrived. Each group has 4 players. This video shows examples of using probability trees to work out the overall probability of a series of events are shown. Tutorial on finding the probability of an event. Question from very important topics are covered by NCERT Exemplar Class 9. , 11 sedans, 15 trucks, and 6 sports cars drove through an intersection. Word and PDF of same document. Sampling methods are classified as either probability or nonprobability. You will see questions on the AP Biology exam that present data and ask you to explain the phenomenon. The binomial probability formula is a simple formula for calculating the probability in Bernoulli trials. The probability of an event is always a number between zero and 100%. Somewhat more advanced notions from calculus come in here, in order to deal with joint probability densities, entailing, for example, integration over regions in two dimensions. 2: Investigating Probability (Answers) Question 1 a) The probability the uniform will have black shorts is 6 3 or 2 1. Probability Worksheets and Printables. Probability and Probability Distributions 1 Introduction In life there is no certainty about what will happen in the future but decisions still have to be taken. Can YOU answer these fiendishly hard GCSE questions? Probability fractions and data sampling questions for 16-year-olds will leave you scratching your head. Math Worksheets Listed By Specific Topic and Skill Area. The questionbank has 50 must solve problem solving and data sufficiency questions. Thus, there is only one element of randomness, and therefore only one element of probability to consider. Below are three excruciatingly difficult dice problems. Summary: Maybe it\u2019d be fine, that this thing Wonwoo feels every time he\u2019s at the Midnight Express with a certain someone. Resources made by expert teachers. 05, Spring 2014 Note: This is a set of practice problems for exam 1. Just create a tree diagram with all the different options: then, follow the tree to find the probability of each event. HealthAmerica is normally performing among \u201cAmerica\u2019s Most effective Health and wellbeing Plans, 2006\u201d just by U. Probability is hard For more than a month, my colleague Sanjoy Mahajan and I have been banging our heads on a series of problems related to conditional probability and Bayesian statistics. Probability forms the backbone of many important data science concepts from inferential statistics to Bayesian networks. 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Mathematics (Linear) - 1MA0 TWO WAY TABLES Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. a total of 7, b. High School Statistics and Probability Worksheets. Click here to read the solution to this question Click here to return to the index. Somewhat more advanced notions from calculus come in here, in order to deal with joint probability densities, entailing, for example, integration over regions in two dimensions. What is the probability that the survey will show a greater percentage of. This is because most people have a very strong intuition about how to calculate probabilities, combinatorics, and statistics from. Marginal probability density function. Assume that the order of assigning these positions matters. Below is a list of great ideas for potential science fair projects. There is no bias over the contestants decision so each door has a probability of 1/3 being chosen. Standard probability themes like coins, spinners, number cubes, and marbles are featured along with some other situations. The probability of occurrence of any event A when another event B in relation to A has already occurred is known as conditional probability. Example question: What is the probability of rolling a 4 or 7 for two 6 sided dice? In order to know what the odds are of rolling a 4 or a 7 from a set of two dice, you first need to find out all the possible combinations. Probability and Probability Distributions 1 Introduction In life there is no certainty about what will happen in the future but decisions still have to be taken. Simple Probability Questions for Winning! MATH 310 S7 1. Request for Hazard Modeling Contributions. Probability of A given the occurrence of B is equal to the probability of A and B over the probability that B has occurred. Exams and solutions. 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HESI questions are usually focused on the critical thinking level and are intended to mimic the types of questions found on the NCLEX exams. Chapter: Descriptive Statistics I: Problem Sensing. The Sock drawer. Includes Problem Solving and Data Sufficiency and sentence correction in verbal. Yet, word problems fall into distinct types. For instance, if an experiment has 20 total possible outcomes and only 10 of them are successful, the probability of that problem is 50 percent. ) Starting with this definition, it would (probably :-) be right to conclude that the Probability Theory , being a branch of Mathematics, is an exact, deductive science that studies uncertain quantities related to random events. This figure would be used if there\u2019s a 70% chance of rain over 100% of the area, or if rain is certain (a 100% chance) over 70% of the area. Studyclix makes exam revision and study easier. Solved examples with detailed answer description, explanation are given and it would be easy to understand - Page 2. 2: Investigating Probability (Answers) Question 1 a) The probability the uniform will have black shorts is 6 3 or 2 1. It would not be wrong to say that the journey of mastering statistics begins with probability. In probability samples, each member of the population has a known non-zero probability of being selected. Kroese School of Mathematics and Physics The University of Queensland c 2018 D. Probability: interpret graphs interpretation, mean, median and mode, simple probability calculations, data collection and analysis. Extra Questions for Class 10 Maths Chapter 15 Probability. Questions To Ask Drug Rehab Centers (FCR), a leading addiction treatment center in the US, provides supervised medical detox and rehab programs to treat alcoholism, drug addiction and co-occurring mental health disorders such as PTSD, depression and anxiety. 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If three balls are drawn from the vessel at random, what is the probability that the first ball is red, the second ball is blue, and the third ball is white?. For instance, if an experiment has 20 total possible outcomes and only 10 of them are successful, the probability of that problem is 50 percent. Questions have been categorized so you can pick your favorite category or challenge your friends to the latest trivia. There is no bias over the contestants decision so each door has a probability of 1/3 being chosen. What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled. Regardless of your level of education or your familiarity with probability theory; if you would like to learn more about this fascinating subject, you are welcome. Probability Cards (Intermediate) What is the probability of choosing a particular card from a deck? Requires basic knowledge of standard playing cards. 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Choice (1)The probability that the two chosen squares have a common side is 1/18 Correct answer Explanatory Answer Hard. You cannot develop a deep understanding and application of machine learning without it. By the way I was wondering whether it may be interesting to add to the Stata bookshelf a tutorial-based textbook on probability (let's say from basics to Bayesian probability distributions) issues related to biostatistics, epidemiology, social sciences and alike (with alike I mean all the. Statistic and Probability: Exam Questions. Fully worked-out solutions of these problems are also given, but of course you should \ufb01rst try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. slide 1: In the x-y plane the square region bound by 00 10 0 10 10 and 0 10 is isolated. Treat These 15 Laptop Computers As A Random Sample And Use X To Denote The Number Of Them With Hard Disks Made By Western Digital In This Sample. 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How can it truly be (7/8)^3 -1 When the carrier status of the unknown spouse is not a factor that gets reshuffled each time a new kid is born? That's my question. Click here to read the solution to this question Click here to return to the index. If someone could help me out with writing this out and also getting the actual probability of this scenario, I would really appreciate it! If you have any questions or need any more factors, just let me know but I think that what was given should be enough to generate an equation and get an answer. Probability deals with how likely it is that something will happen. A \u201cbad beat\u201d happens when a player completes a hand that started out with a very low probability of success. Studyclix makes exam revision and study easier. For any two of the three factors, the probability is 0. It quickly and easily calculates the probability of a URE failing during a RAID 5 or RAID 6 rebuild. a) If you randomly observe 200 cars, what is the mean?. Check your answers seem right. A measure of a player\u2019s experience and maturity is how he handles bad beats. You are sure this will result in your best your score yet. The probability of getting 0-0-0 is 1 out of 1000, or 1 \u00f71000 = 0. Learn vocabulary, terms, and more with flashcards, games, and other study tools. ALL answers should be fractions in simplest form. I like that. As in the Monty Hall problem, the intuitive answer is 1 / 2, but the probability is actually 2 / 3. Compute the probability of randomly guessing the answers and getting exactly 9 questions correct. Short Answers to Hard Questions About Zika Virus. This is 1/6. Find the probability that the total of the. Hard probability problem: How to calculate the the probability of having selected only < 70% of all marbles in a bag, given the following draw rules? Ask Question Asked 1 year, 10 months ago. Conditional probability tells us the probabiity of event A given that B has occurred. Great Expectations: Probability Through Problems The resources found on this page offer a new approach to teaching probability. \u03be = 120 coins in the collection T = coins from the 20th century B = British coins A coin is chosen at random. Conditional probability is based upon an event A given an event B has already happened: this is written as P(A | B). Probability of getting no head = P(all tails) = 1/32. A bag of jellybeans has 20 watermelon jellybeans, 45 sour apple jellybeans, 30 orange jellybeans and 5 cotton candy jellybeans. Liwayway draws 2 balls out of the bag. Wasn't sure how it would go or if they'd solve it. Probability is the likelihood of something happening or being true. This section provides the course exams with solutions and practice exams with solutions. Be wary of automobile professionals who. Can you please describe your steps. just been exchanged. For the Giants to win the series in game six, they need to win three games in five trials. Instructions Use black ink or ball-point pen. com is the place to go to get the answers you need and to ask the questions you want. Like dependability, this is also a probability value ranging from 0 to 1, inclusive. Then they both turn left and walk for another four feet, and. Probability Page 1 of 15 Probability Rules A sample space contains all the possible outcomes observed in a trial of an experiment, a survey, or some random phenomenon. Before being able to solve a probability question, you must first. The problems tend to be easy to state and understand, but sometimes irritatingly difficult to solve. 1 that a woman in the population has only this risk factor (and no others). There is no bias over the contestants decision so each door has a probability of 1/3 being chosen. Released Test Questions Math 3 Introduction - Grade 3 Mathematics The following released test questions are taken from the Grade 3 Mathematics Standards Test. In fact, many probability questions are a set of two permutation probability questions with the denominator being the total number of outcomes for an event and the numerator being the number of favorable outcomes. Brain teaser game: Hard Logic Probability Puzzle. population. The probability of an event occurring is the chance or likelihood of it occurring. Model Question Paper Mathematics Class XII Time Allowed : 3 hours Max: Marks: 100 General Instructions (i) The question paper consists of three parts A, B and C.", "date": "2019-11-19 15:26:21", "meta": {"domain": "erci.pw", "url": "http://uqqp.erci.pw/hard-probability-questions.html", "openwebmath_score": 0.6050974726676941, "openwebmath_perplexity": 680.165398577091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9891815516660637, "lm_q2_score": 0.8670357546485408, "lm_q1q2_score": 0.8576557731332001}}
{"url": "https://math.stackexchange.com/questions/176887/when-chessboards-meet-dominoes", "text": "# When chessboards meet dominoes\n\nYou probably have heard about the following brainteaser :\n\nConsider a $$8\\times 8$$ chessboard. Remove two extreme squares (top-left and bottom-right e.g.). Can you fill the remaining chessboard with $$1\\times 2$$ dominoes ?\n\nThe answer lies in a coloring argument. The problem, however, does not use the fact that the chessboard is colored. I would like to know improved examples of coloring problems like this one.\n\nFor instance, could you extend the problem with Tetris-like L-shaped dominoes and solve it using more than two colors ?\n\n\u2022 Well, with 2x3 \"dominos\" you notice that 6 doesn't divide 64... Jul 30 '12 at 16:44\n\u2022 2x3 means L shape. Jul 30 '12 at 16:45\n\u2022 Could you clarify: When you talk about using Tetris pieces, you are talking about an ordinary chessboard, not one with two pieces removed? Jul 30 '12 at 16:55\n\u2022 @SiliconCelery : as you can add two L-shaped dominos to get a 2x4 rectangle, they can fit a regular chessboard. Concerning my question I guess the problem should be different from the beginning, it should probably substract each of the extreme squares to get 60 = 15 x 4 squares Jul 30 '12 at 17:01\n\u2022 The 35 hexominoes cannot be packed into a rectangle as you have two more squares of one color than the other. See en.wikipedia.org/wiki/Hexomino. I have seen many math puzzles that exploit coloring. For trominoes you often color in strips of three colors or diagonally like a checkerboard with three colors. Jul 30 '12 at 17:59\n\nSolomon Golomb's book Polyominoes presents a number of arguments of this type.\n\nOne that I remember is: a square is deleted from an 8\u00d78 checkerboard. Can the remaining 63 squares be covered by 21 1\u00d73 rectangles? The answer involves coloring the checkerboard in three colors in alternating diagonal stripes:\n\nThis colors the 64 squares of the checkerboard with 21 green squares, 21 yellow squares, and 22 blue squares. Each 1\u00d73 rectangle must cover exactly one square of each color. The deleted square therefore cannot be any of the green or yellow ones, nor any of the squares equivalent to one of these under a rotation or reflection of the checkerboard:\n\n(This is four copies of the first diagram, superimposed, with suitable rotations.)\n\nThis eliminates all but 4 squares from consideration, namely the four bright blue ones in the previous diagram. So the only solutions involve deleting one of these four blue squares.\n\nThere are a number of analogous arguments about polyhexes that depend on a three-coloring of a hexagonal lattice:\n\nFor example, there are three different trihexes, which are made by joining three hexagons; two of these are guaranteed to cover exactly one cell of each color, no matter how they are placed.\n\nI once wasted a lot of time trying to make myself a set of tetrominoes by marking up a 4x5 rectangle and cutting it apart, and I felt rather foolish when I realized that a straightforward checkerboard coloring shows that this is impossible. There are 5 tetrominoes, and four of them must cover two black and two white squares each. The 5th is T-shaped, and must cover three black squares and one white (or vice versa).\n\nSo they cannot possibly tile a 4\u00d75 rectangle, which has equal numbers of black and white squares.\n\n\u2022 Very nice example of coloring pattern. Jul 30 '12 at 19:38\n\nA woodworm is sitting at the centre of a cube that's divided into $3^3$ identical cubelets. The woodworm can go from the centre of one cubelet to the centre of another in any edge-parallel direction. The woodworm would like to eat its way through the cube such that it visits the centre of each cubelet exactly once. Is this possible?\n\n\u2022 Same principle, I see. Awesome ! Jul 30 '12 at 19:06\n\nI searched for math.SE questions and answers that involve colouring of tilings and the like; here are two interesting ones that I found:\n\nThe Mathematics of Tetris\n\nA tiling puzzle/question", "date": "2021-09-21 03:11:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/176887/when-chessboards-meet-dominoes", "openwebmath_score": 0.6429679989814758, "openwebmath_perplexity": 644.5019343279016, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812309063187, "lm_q2_score": 0.8856314768368161, "lm_q1q2_score": 0.8576288996686168}}
{"url": "https://discourse.julialang.org/t/solve-for-upper-lower-bound-of-a-numerical-definite-integral/58164", "text": "# Solve for upper/lower bound of a numerical definite integral\n\nHi, I\u2019m trying to solve for b, given f, a and k, please help, Is there a way to do it in julia?\n\nquadgk(f,a,b)=k\n\n\nThanks\n\n2 Likes\n\nThere\u2019s always a Newton iteration:\n\nb = some_initial_guess\ntol = sqrt(eps())\nwhile true\n\u03b4b = (quadgk(f,a,b, rtol=tol*0.1)[1] - k) / f(b)\nb -= \u03b4b\nabs(\u03b4b) \u2264 tol * abs(b) && break\nend\n\n\nNote that I\u2019m using the fact that \\frac{d}{db} \\int_a^b f(x) dx = f(b).\n\n10 Likes\n\n@stevengj, this is excellent.\n\nBelow a summary of different solutions, including yours and your advice to run Optim properly (even if not recommended for this problem kept it for sake of completeness).\n\n\nf(x) = exp(x)*sin(x)\n\n# forward modelling\na = 0\n# b = \u03c0   : is UNKNOWN to be found given f, a and k\nk = 0.5*(1.0 + exp(1)^\u03c0)   # k = 12.0703463163896\n\n# SOLUTION-1: Newton solution by @stevengj\n\nb = 1.0  # initial guess\ntol = sqrt(eps())\nwhile true\n\u03b4b = (quadgk(f,a,b, rtol=tol*0.1)[1] - k) / f(b)\nb -= \u03b4b\nabs(\u03b4b) \u2264 tol * abs(b) && break\nend\nprintln(\"Solution = \", b)   # 3.14159268 / solution \u03c0= 3.14159265\n\n# SOLUTION-2: use Optim to minimizing g(x)^2\nusing Optim\n\nh(x) = ((g.(x)).^2)[1]\ndh(x) = (2*g.(x).*f.(x))[1]   # derivative\n\n# Checks:\ng(pi) \u2248 0   # true\nh(pi) \u2248 0   # true\ndh(pi) \u2248 0   # true\n\nb = [1.0]  # initial value  / solution \u03c0= 3.14159265\nresult = optimize(h, b, Newton(), Optim.Options(g_tol = 1e-16))\nOptim.minimizer(result)  # 3.14159265\nresult = optimize(h, dh, b, Newton(); inplace=false)  # using derivative\nOptim.minimizer(result)  # 3.1415795   # does not improve accuracy\n\n# SOLUTION-3: use NLsolve to find roots\n\nusing NLsolve\n\ndg(x) = f.(x)   # derivative\n\n# Check:\ng([pi]) \u2248 0   # true\n\nb = [1.0]  # initial value  / solution \u03c0= 3.14159265\nnlsolve(g, b; ftol=1e-16)      # 3.14159264\nnlsolve(g, dg, b; ftol=1e-16)  # 3.14159263  # same accuracy...\n\n# SOLUTION-4: use DifferentialEquations and interpolation\n\nusing DifferentialEquations, Plots, Dierckx\n\nu0 = 0;  tspan = (a, 4.)   # initial conditions & timespan\ndu(u,p,t) = f(t)   # define the system\nprob = ODEProblem(du, u0, tspan)  # define the problem\nsol0 = solve(prob, dtmax=0.01)  # solve it\n\ni = argmin(abs.(sol0.u .- k))  # index of time-step the closest to solution\ntsol = LinRange(sol0.t[i-1], sol0.t[i+1],20)   # spline it around the solution\nspl = Spline1D(tsol, sol0.(tsol) .- k)\nb = mean(roots(spl))    # 3.1415926536\n\n1 Like\n\nYes; it looks like the problem with your code is that Optim.optimize is expecting a vector of initial guesses, not a scalar. Try passing [b] instead of b.\n\nHowever, this is a root-finding problem g(b) = 0. It is almost always a mistake to try to solve a root-finding problem by using an optimization algorithm to minimize |g|\u00b2. You should use a root-finding algorithm like those in NLsolve.jl.\n\nMoreover, in this case you know the derivative analytically, as I mentioned above, in which case you should certainly provide it to the root-finding algorithm.\n\nHowever, for a 1d root-finding problem (b is a scalar) with an analytically known derivative, all of the generic root-finding packages will basically boil down to a Newton iteration like the one I wrote. The clever algorithms are mainly for the case where you don\u2019t know the derivative, or have lots of derivatives (a big Jacobian) and can\u2019t affort to compute them all or to invert the Jacobian.\n\nAlternatively, in 1d, especially for smooth functions, there are often more clever algorithms available. For example, you could use ApproxFun.jl to construct a high-accuracy polynomial approximation of your integrand f, call cumsum to compute its integral, and use the ApproxFun.roots function to find all of the places where the integral equals k.\n\n6 Likes\n\nRather than minimize |g|\u00b2, one could apply g(b)=0 as an equality constraint to an optimization algorithm that supports such constraints, e.g. NLopt.jl or JuMP.jl. I often find it convenient to deal with roots the same way I deal with optimization. Sometimes you can leave out an objective function altogether, or just specify a constant like 0, to be \u201cminimized\u201d subject to the constraints (to some tolerance). However, I\u2019ve never compared performance to a root finder.\n\n2 Likes\n\nTried to follow it the best possible and have updated code above for 3 different methods: Newton by yourself, Optim and NLsolve.\nIncluding the derivative did not seem to help much, but any blame should surely be on me. BR\n\n*PS: code above now runs but better not shaking it too much\u2026\n\n1 Like\n\nYes, solving the optimization problem \\min_x 1 subject to the constraint g(x)=0 is, of course, exactly equivalent to the root-finding problem g(x)=0, and a sufficiently clever NLP algorithm will perform steps essentially equivalent to Newton steps (perhaps limited by a trust region). They almost certainly aren\u2019t going to be better than Newton steps, however, and in many cases will be worse (because they may spend a lot of effort doing computational steps that are only needed for more general NLPs).\n\nIn a single-variable problem where the derivative is known, if you aren\u2019t going to do something more clever (e.g. Chebyshev fits, complex analysis, continued fractions, etc.) you might as well just use Newton. (Either implementing it yourself \u2014 it\u2019s just a few lines \u2014 or calling a canned implementation like the one in Roots.jl).\n\n6 Likes\n\nThere is another way. Define R as\n\n$$\\int_a^{R(x)}f=x$$\n\nyou get\n\n$$\\frac{d}{dx}R(x) = \\frac{1}{f(R(x))},\\quad R(a)=0$$\n\nYou can use an ODE solver to get b=R(k). This is the main trick of my paper on PDMP.\n\n@rveltz, it is only me, or your Latex is not displaying properly?\n\n3 Likes\n\nIt gives:\n\n1 Like\n\n@rveltz, mathematically this looks pretty much like @stevengj\u2019s solution above. The difference is that the differential equations solver will not use Newton\u2019s method?\nWould like to try this when time allows, but would be surprised if it beats Newton in terms of accuracy and fast convergence.\n\nI would say it is better in that you dont need to recompute \\int_a^bf if you know that the solution b_{sol} is greater (smaller) than b. You \u201cwaste\u201d less computation\n\n1 Like\n\nYeah, it seems to me that there is a lot of redundancy in re-computing the full integral at each step. It should be enough to compute the integral only over the \u2018correction interval\u2019. (One should probably take some care with accumulated errors, however.)\n\n2 Likes\n\n@rveltz, in your nice solution how do you handle the cases with singularities where the function f() has zeros and 1/f() explodes to \u221e?\n\nHum, I dont know! I guess the Newton solver will sufer as well. I\u2019d say you have to use a stiff ODE solver that can handle finite time explosion.\n\n1 Like\n\nThe topic \u201cSolve equation\u201d doesn\u2019t do justice to the interesting things discussed here. @carloslesmes can I suggest editing the topic title to something more specific? For example, \u201cSolve for upper/lower bound of a numerical definite integral\u201d or some such.\n\n1 Like\n\nAm I looking at this crazy, or wouldn\u2019t the correct boundary condition be R(0) = a instead of R(a) = 0?\n\n1 Like\n\nnope you are right! Thank you\n\nNewton did not suffer for f(x)=exp(x)sin(x) in the example above with a zero at the solution pi. However, someone else is suffering to have your method working for that f(x)", "date": "2022-05-24 08:34:57", "meta": {"domain": "julialang.org", "url": "https://discourse.julialang.org/t/solve-for-upper-lower-bound-of-a-numerical-definite-integral/58164", "openwebmath_score": 0.7816888093948364, "openwebmath_perplexity": 2444.071833093994, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707999669626, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8576201897335671}}
{"url": "https://nbviewer.ipython.org/url/ignite.byu.edu/che263/lectureNotes/interpolation.ipynb?flush_cache=true", "text": "# Interpolation\u00b6\n\nIn\u00a0[2]:\nimport numpy as np                        # numerics\nimport matplotlib.pyplot as plt           # plotting\n%matplotlib inline\n\n\n## Key concept\u00b6\n\n\u2022 Given two arrays of $x_g$, $y_g$ data points (e.g., a plot of points).\n\u2022 You want to estimate the $x_w$, $y_w$ value of an intermediate data point.\n\u2022 You know the value of $x_w$ (cause it's where you want the point).\n\u2022 You need to interpolate to find the corresponding $y_w$.\nIn\u00a0[3]:\n#### Just plotting some data to visualize\nx_given = np.array([0,1,2,3,4,5,6,7,8,9,10])\ny_given = np.cos(x_given**2.0/8.0) + 1\n\nplt.rc('font', size=16)\nplt.plot(x_given, y_given, 'o:')\nplt.plot([2.5,2.5],[-0.1,1.6], '--', color='gray')\nplt.plot([-0.1,2.5],[1.6,1.6], '--', color='gray')\nplt.plot([2.5], [1.6], '*', markersize=20)\nplt.xlim([-0.1,11]); plt.ylim([-0.1,2.2])\nplt.xlabel('x');     plt.ylabel('y')\nplt.text(2.6,0, r\"$x_w$\", fontsize=16);  plt.text(0,1.7, r\"$y_w$\", fontsize=16);\n\n\n### Question\u00b6\n\n\u2022 How to get intermediate values, that is, values between those that are part of the given data?\n\n\u2022 Fit a curve to the data, then evaluate from that curve.\n\u2022 Take the closest data point\n\u2022 Linear interpolation\n\u2022 Higher order interpolation\n\u2022 (Others?)\n\n### Question\u00b6\n\n\u2022 How does curve fitting differ from interpolation?\n\n\u2022 Normally, interpolation is done locally, between a few (often two) bounding points.\n\u2022 Curve fitting involves making a best fit curve everywhere\n\u2022 Usually, a best-fit curve will not pass through all data points.\n\u2022 Useful when there is noise and we want the underlying curve, or a model equation for the data.\n\u2022 Interpolation normally considers the points themselves, not a single curve through everything.\n\u2022 Even with noisy data, we might want to interpolate an intermediate point.\n\u2022 If you want an intermediate point to some given data, use interpolation.\n\u2022 If you want to fit a model though some scattered data, and then evaluate the model, use curve fitting.\n\n### Examples?\u00b6\n\n\u2022 Think of some examples where you might use this.\n\n## Linear Interpolation Exercise\u00b6\n\n\u2022 Given the xg and yg data below.\n\u2022 Write a function called Linterp that takes the following arguments:\n\u2022 xg an array of given x data\n\u2022 yg an array of given y data corresponding to xg\n\u2022 xw the value of x we want to interpolate at\n\u2022 The function returns yw corresponding to xw\n\u2022 Assume xg are uniformly spaced, and ascending.\n\nQuestions\n\n\u2022 What is the linear interpolation formula at a given location?\n\u2022 We write the equation for a line based on two points, then evaluate the line at the intermediate point:\n\u2022 equate slopes, then solve for y_w: $$\\frac{y_w-y_0}{x_w-x_0} = \\frac{y_1-y_0}{x_1-x_0},$$ $$y_w = y_0 + (x_w-x_0)\\frac{y_1-y_0}{x_1-x_0}.$$\nIn\u00a0[4]:\n#-------------------- Set some \"given\" x, y data, (normally given to us)\n\nxg = np.array([0,1,2,3,4,5,6,7,8,9,10.])   # given x data\nyg = np.cos(xg**2.0/8.0)+1                 # given y data\nprint(\"xg = \"+np.array2string(xg, formatter={'float_kind':lambda x: f\"{x:4.2f}\"}))\nprint(\"yg = \"+np.array2string(yg, formatter={'float_kind':lambda x: f\"{x:4.2f}\"}))\n\n#-------------------- interpolate to xw=2.5\n\nxw = 2.5\n\nxg = [0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00]\nyg = [2.00 1.99 1.88 1.43 0.58 0.00 0.79 1.99 0.85 0.24 2.00]\n\n\n### New library\u00b6\n\nIn\u00a0[4]:\nfrom scipy.interpolate import interp1d\n\n\u2022 interp1d takes the given x array and the given y array as arguments.\n\u2022 Returns a function.\n\u2022 Call that function wherever you want to interpolate to.\nIn\u00a0[14]:\nxg = np.array([0,1,2,3,4,5,6,7,8,9,10])   # given x data\nyg = np.cos(xg**2.0/8.0)+1           # given y data\n\n#---------------\n\nf_interp = interp1d(xg, yg)\n\n#---------------\n\nxw = 2.5\nyw = f_interp(xw)\n\nprint(yw)\n\n1.6543795393445195\n\n\n### Exercise\u00b6\n\n\u2022 Take the previous xg, yg data and plot the data as points.\n\u2022 Plot 1000 points of the underlying function used to get xg and yg, called xx\n\u2022 Also, plot values of a linear interpolant at the same 1000 points.\nIn\u00a0[5]:\nxg = np.array([0,1,2,3,4,5,6,7,8,9,10]) # given x data\nyg = np.cos(xg**2.0/8.0)+1              # given y data\n\n\n#### Try replacing\u00b6\n\nf_interp = interp1d(xg, yg)\n\n#### with this\u00b6\n\nf_interp = interp1d(xg, yg, kind='cubic')\n\n\u2022 This uses a cubic spline interpolant instead of a linear interpolant.\n\u2022 That is, we use cubic functions between points that match up smoothly at interfaces.\n\nTry this:\n\n\u2022 Interpolate an x value that is outside of the bounds of the given xg data.\n\u2022 What happens?\n\n### Extrapolation\u00b6\n\n\u2022 As listed above, you will get an error if you try to call f_interp with an x value that is outside of the upper and lower bounds of the original xg array.\n\u2022 This can be avoided using the fill_value='extrapolate' argument, like so:\n\u2022 f_interp = interp1d(xg, yg, fill_value='extrapolate'\n\n## Summary\u00b6\n\n1. Import library\n2. You have some data from somewhere, xg, yg\n3. Get ther interpolant function fi=interp1d(xg, yg)\n4. Set desired xw intermediate points to interpolate at (can be an array).\n5. Perform the interpolation to yw=fi(xw)\nIn\u00a0[6]:\nfrom scipy.interpolate import interp1d\n\nxg = np.array([1,2,3,4,5])             # some data you have\nyg = np.array([11, 2.2, 3.3, -88, 9])\n\nfi = interp1d(xg,yg)\nxw = 2.5\nyw = fi(xw)\n\n\n## Question\u00b6\n\n\u2022 What if you have given xg and yg arrays and you know the yw and want to interpolate to the corresponding xw?\nIn\u00a0[\u00a0]:", "date": "2022-01-22 06:27:43", "meta": {"domain": "ipython.org", "url": "https://nbviewer.ipython.org/url/ignite.byu.edu/che263/lectureNotes/interpolation.ipynb?flush_cache=true", "openwebmath_score": 0.5964668989181519, "openwebmath_perplexity": 3995.50339263326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668690081642, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.857615620879574}}
{"url": "https://cs.stackexchange.com/questions/135784/big-oh-and-big-omega-when-n-and-log-n-terms-are-in-fn", "text": "# Big Oh and Big Omega when $n$ and $\\log n$ terms are in $f(n)$\n\nhaving problems with big oh and big omega functions when there is a $$\\log n$$ added or subtracted. For example how do I deal with $$n+\\log n$$ or $$n-\\log n$$ when I have to determine whether the function is in $$\\Omega(n)$$ or in $$\\Omega(n^2)$$? For example, is $$n-\\log n$$ in $$\\Omega(n)$$ or in $$\\Omega(n^2)$$?\n\nI cannot ignore the log function and am not sure how to deal with it. Polynomials and logs when multiplied I find OK. But I have a mental block over this one so help would be appreciated\n\nYou are right that you cannot a priori ignore additional terms, although morally you can as the \"smaller\" terms do not contribute to the asymptotic growth.\n\nAs an example $$f(n) :=n+\\log n$$ is in $$\\Theta(n)$$. Why? Going back to the definition, we want to show that $$f(n)$$ is in $$O(n)$$ and in $$\\Omega(n)$$. Showing $$f(n)\\in \\Omega(n)$$ is immediate, as $$f(n) \\geq n$$ for all $$n>0$$.\n\nTo show that $$f(n)$$ is in $$O(n)$$, simply notice that for large enough $$n$$ (say $$n>N$$ for some constant $$N$$) we have $$\\log n \\leq n$$ and thus $$f(n)\\leq 2n$$ for all $$n>N$$. By definition $$f(n)\\in O(n)$$.\n\nActually, you CAN ignore the log function in a sum or substraction, because the log is always asymptotically negligible in front of $$n^x$$, for every $$x > 0$$.\n\nIf you consider the function $$f(n) = n^x + \\alpha\\log n$$ (where $$x > 0$$ and $$\\alpha \\in \\mathbb{R}$$), then there exists $$A, B \\in \\mathbb{R}_+^*$$ such that $$An^x \\leq f(n) \\leq Bn^x$$. That means that $$f\\in \\Omega(n^x)$$ and $$f\\in \\mathcal{O}(n^x)$$.\n\n\u2022 Agreed with the f(n) =n^2 or cubed -logn as the polynomial grows faster we can ignore the log n. f(n) grows faster than log n. Is it sufficient then to just prove that f(n) =n and is in Big omega and ignore the log term on the basis that as Tassle says logn<n . is this sufficient for a formal proof Feb 22 '21 at 15:40\n\u2022 Sorry, I didn't really get your question\u2026 I'll try to answer what I understood. When you have a sum $f(n) = g(n) + h(n)$ with $h \\in o(g)$ (see en.wikipedia.org/wiki/Big_O_notation#Little-o_notation), then $f\\in \\Omega(g)$ and $f\\in \\mathcal{O}(g)$. This is the case in your question, since $\\log \\in o(n^x)$ for $x > 0$, and yes it is sufficient for a proof. Feb 22 '21 at 15:48\n\nIn your example, $$n-\\log(n)=\\theta(n)$$, that is $$n-\\log(n) = \\Omega(n)$$ and also $$n-\\log(n)=O(n)$$.\n\nYou can see this like that:\n\n\u2022 $$n-\\log(n)\\le n = O(n)$$\n\u2022 $$n-\\log(n) \\ge n - 0.5n = \\Omega(n)$$\n\u2022 as n-logn is always less than n the function can only be in Big Oh . It cannot be in Big omega as if I divide by n^2 to fiind c I get 1-logn > or equal to c. As n gets greater than 2 n-logn gets more negative and c is no longer a constant so it cannot be in Big Omega. Am I on the right track? Feb 24 '21 at 9:09\n\u2022 From a certain point $log(n)\\le0.5n$. This means that $n-\\log(n)$ will actually not be negative, but actually if you substitute the inequality you get $n-\\log(n)\\ge 0.5n$, which is $\\Omega(n)$. However, you are correct that it is not $\\Omega(n^2)$. Feb 24 '21 at 10:20\n\u2022 sorry. I meant n-logn is not in Big Omega (n) not Big Omega (n^2). My original question is to prove (n-logn) = \u03a9(n) true or false. I was trying to prove it false by contradiction. Hence my reasoning of c needing to be a positive constant for all n> no.I am not sure where you came up with the 0.5n either. I plotted the graphs and n-logn is always under f(n)= n when n>1 Feb 24 '21 at 10:41\n\u2022 As my answer showed, it is indeed always lower than $n$. But, in the same time, it is bigger than $0.5n$. Try to draw this graph as well Feb 24 '21 at 11:32\n\u2022 You are right and thank you for your help . In fact from what you have said I can prove that n-logn =big theta(n) . Feb 24 '21 at 12:07", "date": "2022-01-16 20:47:51", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/135784/big-oh-and-big-omega-when-n-and-log-n-terms-are-in-fn", "openwebmath_score": 0.7642459869384766, "openwebmath_perplexity": 229.84908821512659, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668668053619, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8576156189541546}}
{"url": "http://math.stackexchange.com/questions/67315/help-understanding-why-a-complete-totally-bounded-metric-space-implies-every-in/67355", "text": "# Help understanding why a complete, totally bounded metric space implies every infinite subset has a limit point\n\nI'm reading the following proof. Properties II and III are in my title, that a complete, totally bounded metric space implies every infinite subset has a limit point.\n\nI have two questions near the end. Why does $d(x_n,x)\\lt 2/n$? And secondly, why is $x$ a limit point of $A$? What other point in the neighborhood of $x$ is also in $A$? I don't get why they mention that $3/n\\to 0$ as $n\\to\\infty$. How does that imply it's a limit point?\n\n-\nPlease give a reference when you quote text from somewhere else. Otherwise, how can anyone else look it up? \u2013\u00a0Carl Mummert Sep 25 '11 at 1:53\n\nThere\u2019s a slight error in the proof: the claim should be that $d(x_n,x) \\le 2/n$. To see this, let $\\epsilon$ be any positive real number; the sequence of $x_m$\u2019s converges to $x$, so there is a positive integer $m > n$ such that $d(x_m,x) < \\epsilon$. By the triangle inequality $$d(x_n,x) \\le d(x_n,x_m)+d(x_m,x) < \\frac2n + \\epsilon.\\tag{1}$$ Thus, $$d(x_n,x) < \\frac2n + \\epsilon$$ for every $\\epsilon > 0$, and hence $d(x_n,x) \\le \\dfrac2n$.\nThis small error doesn\u2019t affect the next step of the argument: if $y \\in B(x_n,1/n)$, then $d(x_n,y) < 1/n$, so $$d(x,y) \\le d(x,x_n)+d(x_n,y) \\le \\frac2n + d(x_n,y) < \\frac2n + \\frac1n = \\frac3n,$$ $y \\in B(x,3/n)$, and therefore $B(x_n,1/n) \\subseteq B(x,3/n)$.\nNow $B(x_n,1/n)\\cap A$ is infinite for each $n$, and $B(x_n,1/n) \\subseteq B(x,3/n)$, so $B(x,3/n)\\cap A$ is infinite for each $n$. Since $3/n\\to 0$ as $n\\to\\infty$, for any $\\epsilon > 0$ there is an $n_\\epsilon$ such that $3/n_\\epsilon < \\epsilon$. But then $B(x,\\epsilon)\\cap A \\supseteq B(x,3/n_\\epsilon)\\cap A$, which is infinite. Thus, every nbhd of $x$ contains infinitely many points of $A$.\nAdded: To clarify, it is in fact true that $d(x_n,x)<2/n$ for each $n$; it just doesn\u2019t follow directly from the fact that $d(x_m,x_n) < 2/m$ when $m<n$, as the weaker inequality does. If we want the strict inequality, we can modify the argument that I gave by choosing $m>n$ so that $1/m<\\epsilon/2$ and $d(x_m,x)<\\epsilon/2$ and then replacing $(1)$ by $$d(x_n,x)\\le d(x_n,x_m)+d(x_m,x)<\\frac1n+\\frac1m+\\frac{\\epsilon}{2}<\\frac1n+\\epsilon.$$ Since $\\epsilon$ can be chosen arbitrarily small, we conclude that $d(x_n,x)\\le 1/n$ for each $n$.\n I don't think this is an error. They had established $d(x_n,x_m) \\leq 1/m+1/n \\lt 2/n$ for $n \\lt m$ already (with $n$ and $m$ interchanged), so if you fix $n$ and let $m \\to \\infty$ then you get strict inequality. \u2013\u00a0commenter Sep 25 '11 at 1:01 @commenter: Their statement isn\u2019t false, but it doesn\u2019t follow from the fact that $d(x_n,x_m)<2/m$ when $m I think it worth settling some terminology. This is from Topology: a first course by James R. Munkres. He identifies three versions of compactness for a topological space: (A) compactness, then (B) your property (which he calls \"limit point compactness\") and a milder condition (C) he calls \"sequential compactness\" which is that every infinite sequence of points has a convergent subsequence. He proves that (A) implies (B) implies (C). He also proves that the three conditions are equivalent for a metric space. Finally, a well-known theorem is that a metric space is compact if and only if it is complete and totally bounded. So the strongest condition is what is usually discussed in this setting. However, (A) then implies (B). Finally, and this is not obvious, the product of two compact topological spaces is compact, and the product of two sequentially compact spaces is sequentially compact. What is unexpected is that the product of two limit point compact spaces need not be limit point compact. However, if they are metric spaces, the result does hold then. - To address the first question (the other is addressed in the post by Brian). [Also, there's no error.] Consider$\\epsilon=1/(2n)$. Then there is some$N>0$such that$d(x_m,x)<1/(2n)$for all$m \\geq N$. Let$m=\\max\\{N,2n\\}$. Then$d(x_n,x) \\leq d(x_n,x_m)+d(x_m,x) < [(1/n)+(1/m)]+1/(2n) \\leq (1/n) + (1/2n) + (1/2n) = 2/n$. -  While the assertion in question is true, the presentation is in my view seriously flawed, in that it suggests a justification that doesn\u2019t in fact work. \u2013 Brian M. Scott Sep 25 '11 at 1:36 I'm not sure this is quite what you're looking for, but I believe the author just jumped from$d(x_n,x_m) < 2/n$for$n", "date": "2013-06-19 09:32:20", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/67315/help-understanding-why-a-complete-totally-bounded-metric-space-implies-every-in/67355", "openwebmath_score": 0.9906646609306335, "openwebmath_perplexity": 87.4310410113062, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668684574636, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8576156187889931}}
{"url": "http://santotirso.pt/manhas/email-security-fpex/esq5j.php?1b5b77=equation-of-a-circle-given-two-points", "text": "Hint: Find the equation of the line passing through the two given points. This calculator can find the center and radius of a circle given its equation in standard or general form. What is the Equation of a Circle? Different forms equations of straight lines. Equation of circle when endpoints of the diameter are given : (x - x 1) (x - x 2) + (y - y 1) (y - y 2) = 0. Enter Circle Equation Find the equation of a circle that has a diameter with the endpoints given by the points A(1,1) and B(2,4) Step 1: Find the Midpoint ( h , k ) of AB : Write Standard Form When Given Two Points You. How To Calculate The Equation Of A Circle From 3 Points Tessshlo. Also, it can find equation of a circle given its center and radius. Given circle A with the equation That circle has center (-2,0) and radius \u221a5 a)show that the circle passes through point (0,1) That is true so it proves that it passes through that point. Ex Find The Point On A Circle Given An Angle And Radius You. Two point form This online calculator can find and plot the equation of a straight line passing through the two points. Important Properties: \u2020 Equation of a circle: An equation of the circle with center (h;k) and radius r is given \u2026 Here we are going to see how to find the equation of circle with extremities of diameter are given. I did that and found the radius was 5 but where do I go from there? $\\Rightarrow {g^2} + {f^2} \u2013 c = \\frac{{{{\\left( { \u2013 ag \u2013 bf + d} \\right)}^2}}}{{{a^2} + {b^2}}}\\,\\,\\,{\\text{ \u2013 \u2013 \u2013 }}\\left( {\\text{v}} \\right)$ Given two points, A(-2,5) and B(-4,3) find the equation of the circle with diameter AB.? I really need help with this question :(, Im sure you need to find the diameter so you can find the radius. \u2020 Circle: is the set of all points in a plane that lie a \ufb02xed distance from a \ufb02xed point. It is straightforward BUT a bit trickier than you expect! Circle Calc Find C D A R. Equation Of A Circle Through Three Points You. Point A (9, 2) Point B (3, -4) Point C (5, -6) Let's put these points \u2026 Substitute them as I \u2026 Using the midpoint formula and the fact that perpendicular lines have slopes that are negative reciprocals of each other, find the equation of the perpendicular bisector of this chord. The calculator will generate a step by step explanations and circle graph. In geometry, a circle is a two-dimensional round shaped figure where all the points on the surface of the circle are equidistant from the centre point (c). If we know any two, then we can find the third. Distance between two points. Solving the equation for the radius r. 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D a R. equation of a given input will be displayed in the new window you expect if know.\nMph Admission 2020 In Karachi, Selfserve Netid Syracuse, St Mary's College, Thrissur Vacancies, Princeton University Mailing Address, Volvo Xc60 Olx Kerala, Maggie Mae 2020 Schedule, When Will Stroma Medical Be Available, Kia Rio Fuse Box Radio,", "date": "2021-10-16 15:38:31", "meta": {"domain": "santotirso.pt", "url": "http://santotirso.pt/manhas/email-security-fpex/esq5j.php?1b5b77=equation-of-a-circle-given-two-points", "openwebmath_score": 0.6751464009284973, "openwebmath_perplexity": 363.7980382691381, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9811668734137681, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8576156182935075}}
{"url": "https://stats.stackexchange.com/questions/218019/how-to-find-a-standard-deviation-determined-by-a-normal-distribution-probability", "text": "# How to find a standard deviation determined by a Normal distribution probability?\n\nThe question is\n\nA liquid drug is marketed in phials containing a nominal 1.5ml but the amounts can vary slightly. The volume in each phial may be modeled by a normal distribution with the mean 1.55ml and standard deviation $$\\sigma$$ ml. The phials are sold in packs of 5 randomly chosen phials . It is required that in less than 0.5% of the packs will the total volume of the drug be less than 7.5ml. Find the greatest possible value of $$\\sigma$$.\n\nI need to find the greatest possible value of the standard deviation ($$\\sigma$$). I worked out the following:\n\n$$\\mu= 1.55*5 = 7.75.$$\n\nWe are asked to find value of $$\\sigma$$ such that probability of (total volume of $$5$$ packs $$\\lt 7.5)\\lt0.5\\%$$\n\n$$P(X\\lt7.5)\\lt0.005.$$\n\nAfter standardizing, $$P(X\\le\\frac{7.5-7.75}{\\sigma/5})<0.005$$ and I found $$\\sigma=0.2170.$$ However, the answer provided is $$0.0434.$$\n\n\u2022 Please add the self-study tag, read its tag-wiki, and indicate the specific help you need at the point you struck difficulty. Jun 9 '16 at 2:37\n\u2022 what's CTL? ... ... Also please check the details of the question, it looks like you may have a mistake somewhere. Where did the 7.75 in your working come from? Please show more detail/explanation of what you're doing. (As far as possible your responses should result in edits to your question) Jun 9 '16 at 2:38\n\u2022 How have you approached/engaged it so far? Any partly successful paths? Where else have you looked for answers? Jun 14 '16 at 2:06\n\u2022 Interestingly, neither answer is correct.\n\u2013\u00a0whuber\nJun 14 '16 at 13:59\n\nAmong the objectives of good introductory statistics courses is learning how to think about the Normal distribution. This question provides a nice example.\n\nThe key is to use units of measurement that are adapted to the distribution. That is, let the mean be the zero point and let the standard deviation be one unit. This is what a \"Z score\" measures.\n\nIn light of this, let's parse the question. To do so, I will use two fundamental facts: expectations add (\"linearity of expectation\") and variances of independent variables also add:\n\n\u2022 The mean volume of one pack is 1.55 ml, whence the mean volume of five packs must be five times as large, or 7.75 ml: this is the zero point.\n\n\u2022 Since the unknown variance of a single pack is $$\\sigma^2,$$ the variance of the sum of five independent packs is $$5\\sigma^2.$$ Therefore the standard deviation of the sum--the unit of measurement we must adopt--is $$\\sqrt{5\\sigma^2} = \\sigma\\sqrt{5}.$$\n\nThe question stipulates that in less than 0.5% of cases should the total be less than 7.5 ml. For the (standard) Normal distribution we remember (or can compute) that exactly 0.5% of cases are $$2.57\\ldots$$ or more less than the mean. An example of this computation is\n\nqnorm(0.5/100)\n\n\nin R or\n\n=NORMSINV(0.5/100)\n\n\nin Excel, for instance.\n\nOne aim of the introductory course is to help you reach the point where such considerations are automatic: you can do them in your head correctly, apart (perhaps) from the arithmetical calculations.\n\nThis preliminary work enables us to rephrase the question like this:\n\nWhat unit of measurement, given by $$\\sigma\\sqrt{5}$$ for a five-pack of drugs, will re-express an amount of $$7.5$$ ml as being $$2.57$$ less than $$7.75$$ ml?\n\nThe solution obviously is\n\n$$\\sigma\\sqrt{5} = (7.75 - 7.5)/2.57\\ldots = 0.097\\ldots,$$\n\nimplying\n\n$$\\sigma = \\frac{0.097\\ldots}{\\sqrt{5}} = 0.0433797\\ldots$$\n\nComparing this result to the question shows that the work in the question was entirely correct up to the point where \"$$\\sigma/5$$\" appeared: the square root was lost. This suggests remembering to think in terms of variances rather than standard deviations.\n\nComparing this result to the older answers that were posted also shows how they were basically moving in the correct direction but made mistakes along the way, too. Because arithmetical mistakes are easy to make, when one has the chance it's a good idea to check probabilistic calculations with simulations. For instance, the following R statement generates a large number of five-packs of drugs as described in the question (using the answer I obtained) and, to check my answer, computes the fraction with totals less than 7.5 ml:\n\nmean(colSums(matrix(rnorm(5*1e6, 1.55, -0.25/qnorm(0.5/100) / sqrt(5)), nrow=5)) < 7.5)\n\n\n(You can see all the data from the question embedded in this expression, along with the value 1e6 giving the number of five-packs to simulate.) When I run and re-run this code (which takes less than a second each time), I consistently obtain results between 0.0048 (0.48%) and 0.0052 (0.52%), in satisfactory agreement with the intended 0.5% target.\n\nI think your understanding of the variance of the sum is mistaken. The variance of the 5-pack sum is 25 times the variance of the single pack.\n\n\u2022 The only way you could justify the factor of $25$ is to suppose the five packs are perfectly correlated. Assuming, as is more likely the intent, that the five-pack sum can be modeled as the total of five independent Normal variables $X_1+\\cdots+X_5$, its variance will be $$\\operatorname{Var}(X_1+\\cdots+X_5)=\\sigma^2 +\\cdots+\\sigma^2=5\\sigma^2.$$ Consequently its standard deviation will be $\\sigma\\sqrt{5}$, which (by following the path outlined in the question) leads directly to the correct answer.\n\u2013\u00a0whuber\nJun 14 '16 at 14:01\n\n\\begin{align} 5\\times 1.55 &= 7.75 \\\\ 5\\times SD &= 5SD \\end{align} Problem statement: $$P(X<7.5)<0.005$$ \\begin{align} \\frac{(7.5-7.75)}{(5SD^2)^{1/2}} &< 0.005 \\\\[8pt] \\frac{-0.25}{2.576} &= (5SD^2)^{1/2} \\\\[5pt] &-0.0970 / 5^{(1/2)} \\end{align} $$0.0434$$ as the standard deviation can never me negative, take the mod value.\n\n\u2022 Welcome to Stats.SE. You may give hints but please do not give the full answer. Furthermore, can you please edit your post and explain the key steps in the solution and use MathJax in the formulas? Apr 26 '19 at 10:51", "date": "2021-12-08 01:12:44", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/218019/how-to-find-a-standard-deviation-determined-by-a-normal-distribution-probability", "openwebmath_score": 0.9138047099113464, "openwebmath_perplexity": 747.9623359942079, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668690081643, "lm_q2_score": 0.874077222043951, "lm_q1q2_score": 0.8576156112242174}}
{"url": "http://mathhelpforum.com/math-topics/36935-problem-solving-sucessive-tetrahedral-numbers.html", "text": "# Math Help - Problem solving - sucessive tetrahedral numbers\n\n1. ## Problem solving - sucessive tetrahedral numbers\n\nwould you guys be able to give me a hand with this problem solving question, i have no idea where to start...\n\nThanks\n\nZac\n\n2. Hello, Zac!\n\n12. The first four tetrahedral numbers are: . $1,\\;4,\\;10,\\;20$\n\nFind the pattern, then predict the next 3 terms of the sequence.\nTake the difference of consecutive pairs of terms.\n\n$\\begin{array}{cccccccc}\\text{Sequence:} & 1 && 4 && 10 && 20 \\\\\n\\text{Difference:} & & 3 & & 6 & & 10 \\end{array}$\n\nThey are adding on consecutive \"triangular\" numbers.\n\n. . $\\begin{array}{ccc}3 & = & 1+2 \\\\ 6 &=&1+2+3\\\\ 10 &=&1+2+3+4 \\end{array}$\n\nWe will add on the next three triangular numbers:\n\n. . $\\begin{array}{ccc}1+2+3+4+5 &=& 15 \\\\ 1+2+3+4+5+6 &=& 21 \\\\ 1+2+3+4+5+6+7 &=& 28\\end{array}$\n\n$\\begin{array}{cccccccccccccc}\\text{Sequence:} & 1 && 4 && 10 && 20 && {\\color{blue}35} && {\\color{blue}56} && {\\color{blue}84}\\\\\n\\text{Difference:} & & 3 & & 6 & & 10 && +15 && +21 && +28\\end{array}$\n\nWhat is the $n^{th}$ term of the sequence?\nI'll skip all the algebra . . .\n\n$\\text{The }n^{th}\\text{ term is: }\\;a_n \\;=\\;\\frac{n(n+1)(n+2)}{6}$\n\n3. Originally Posted by Soroban\n\nI'll skip all the algebra . . .\n\n$\\text{The }n^{th}\\text{ term is: }\\;a_n \\;=\\;\\frac{n(n+1)(n+2)}{6}$\nas this is a problwm solving questions and according to my sheet i must show the algebra would you mind putting it up?\n\nthanks so much!\n\nZac\n\n4. Hello, Zac!\n\nThis will take a while . . . better sit down.\n\nTake differences of consecutive terms,\n. . then take differences of the differences, and so on.\n\n$\\begin{array}{cccccccccccccc}\\text{Sequence} & 1 && 4 && 10 && 20 && 35 && 56 && 84 \\\\\n\\text{1st diff.} & & 3 && 6 && 10 && 15 && 21 && 28 \\\\\n\\text{2nd diff.} & & & 3 && 4 && 5 && 6 && 7 \\\\\n\\text{3rd diff.} & & & & 1 && 1 && 1 && 1 \\end{array}$\n\nThe third differences are constant.\n. . This tells us that the generating fuction is a cubic.\n\nThe general cubic function is: . $f(n) \\;=\\;an^3 + bn^2 + cn + d$\n\nWe will use the first four terms of the sequence . . .\n\n$\\begin{array}{ccccc}f(1)\\:=\\:1\\!: & a + b + c + d & = & 1 & [1] \\\\\nf(2) \\:=\\:4\\!: & 8a + 4b + 2c + d &=& 4 & [2] \\\\\nf(3) \\:=\\:10\\!: & 27a + 9b + 3c + d &=& 10 & [3] \\\\\nf(4) \\:=\\:20\\!: & 64a + 16b + 4c + d &=& 20 & [4] \\end{array}$\n\n$\\begin{array}{ccccc}\\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 3 & [5] \\\\\n\\text{Subtract [3] - [2]:} & 19a + 5b + c &=& 6 & [6] \\\\\n\\text{Subtract [4] - [3]:} & 37a + 7b + c &=& 10 & [7] \\end{array}$\n\n$\\begin{array}{ccccc}\\text{Subtract [6] - [5]:} & 12a + 2b &=& 3 & [8] \\\\\n\\text{Subtract [7] - [6]:} & 18a + 2b &=& 4 & [9] \\end{array}$\n\n$\\begin{array}{ccccccc}\\text{Subtract [9] - [8]:}& 6a \\;=\\; 1 & \\Rightarrow & \\boxed{a \\;=\\;\\frac{1}{6}} \\end{array}$\n\nSubstitute into [8]: . $12\\left(\\frac{1}{6}\\right) + 2b \\:=\\:3\\quad\\Rightarrow\\quad\\boxed{ b \\:=\\:\\frac{1}{2}}$\n\nSubstitute into [5]: . $7\\left(\\frac{1}{6}\\right) + 3\\left(\\frac{1}{2}\\right) + c \\:=\\:3 \\quad\\Rightarrow\\quad \\boxed{c\\:=\\:\\frac{1}{3}}$\n\nSubstitute into [1]: . $\\frac{1}{6} + \\frac{1}{2} + \\frac{1}{3} + d \\:=\\:1 \\quad\\Rightarrow\\quad\\boxed{ d\\:=\\:0}$\n\nHence, the function is: . $f(n) \\;=\\;\\frac{1}{6}n^3 + \\frac{1}{2}n^2 + \\frac{1}{3}n \\;=\\;\\frac{n}{6}(n^2+3n+ 2)$\n\n. . Therefore: . $\\boxed{f(n) \\;=\\;\\frac{n(n+1)(n+2)}{6}}$", "date": "2015-05-26 14:19:26", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/36935-problem-solving-sucessive-tetrahedral-numbers.html", "openwebmath_score": 0.9146358370780945, "openwebmath_perplexity": 1919.8715967648786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9933071480850774, "lm_q2_score": 0.8633916134888613, "lm_q1q2_score": 0.8576130612751942}}
{"url": "http://math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn", "text": "# Value of $\\sum\\limits_n x^n$\n\nWhy is $\\displaystyle \\sum\\limits_{n=0}^{\\infty} 0.7^n$ equal $1/(1-0.7) = 10/3$ ?\n\nCan we generalize the above to\n\n$\\displaystyle \\sum_{n=0}^{\\infty} x^n = \\frac{1}{1-x}$ ?\n\nAre there some values of $x$ for which the above formula is invalid?\n\nWhat about if we take only a finite number of terms? Is there a simpler formula?\n\n$\\displaystyle \\sum_{n=0}^{N} x^n$\n\nIs there a name for such a sequence?\n\nThis is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.\n\nand here: List of abstract duplicates.\n\n-\nThis is a geometric series. See en.wikipedia.org/wiki/Geometric_series#Formula \u2013\u00a0 lhf Mar 25 '11 at 15:47\nPlease replace the starting index $n=1$ by $n=0$ (otherwise the resulting sum is $0.7$ times what you wrote). \u2013\u00a0 Did Mar 25 '11 at 17:59\n@Jonas: Fair point: done. \u2013\u00a0 Arturo Magidin Mar 25 '11 at 19:18\n@Didier: That was my fault upon edit; fixed. \u2013\u00a0 Arturo Magidin Mar 25 '11 at 19:19\n@Arturo: Thanks. Sorry, I shouldn't have deleted my comment; I did so after I saw you had changed the title, and only then saw your comment. (To anyone who is confused: I had pointed out that including \"geometric series\" in the title gave additional context that made lhf's and yoyo's first sentences seem a little strange.) \u2013\u00a0 Jonas Meyer Mar 25 '11 at 19:41\n\nBy definition, a \"series\" (an \"infinite sum\") $$\\sum_{n=k}^{\\infty} a_n$$ is defined to be a limit, namely $$\\sum_{n=k}^{\\infty} a_n= \\lim_{N\\to\\infty} \\sum_{n=k}^N a_n.$$ That is, the \"infinite sum\" is the limit of the \"partial sums\", if this limit exists. If the limit exists, equal to some number $S$, we say the series \"converges\" to the limit, and we write $$\\sum_{n=k}^{\\infty} a_n = S.$$ If the limit does not exist, we say the series diverges and is not equal to any number.\n\nSo writing that $$\\sum_{n=0}^{\\infty} 0.7^n = \\frac{1}{1-0.7}$$ means that we are asserting that $$\\lim_{N\\to\\infty} \\sum_{n=0}^N0.7^n = \\frac{1}{1-0.7}.$$\n\nSo what your question is really asking is: why is this limit equal to $\\frac{1}{1-0.7}$? (Or rather, that is the only way to make sense of the question).\n\nIn order to figure out the limit, it is useful (but not strictly necessary) to have a formula for the partial sums, $$s_N = \\sum_{n=0}^N 0.7^n.$$ This is where the formulas others have given come in. If you take the $N$th partial sum and multiply by $0.7$, you get $$\\begin{array}{rcrcrcrcrcrcl} s_N &= 1 &+& (0.7) &+& (0.7)^2 &+& \\cdots &+& (0.7)^N\\\\ (0.7)s_N &= &&(0.7) &+& (0.7)^2 &+&\\cdots &+&(0.7)^N &+& (0.7)^{N+1} \\end{array}$$ so that $$(1-0.7)s_N = s_N - (0.7)s_N = 1 - (0.7)^{N+1}.$$ Solving for $s_N$ gives $$s_N = \\frac{1 - (0.7)^{N+1}}{1-0.7}.$$ What is the limit as $N\\to\\infty$? The only part of the expression that depends on $N$ is $(0.7)^{N+1}$. Since $|0.7|\\lt 1$, then $\\lim\\limits_{N\\to\\infty}(0.7)^{N+1} = 0$. So, $$\\lim_{N\\to\\infty}s_N = \\lim_{N\\to\\infty}\\left(\\frac{1-(0.7)^{N+1}}{1-0.7}\\right) = \\frac{\\lim\\limits_{N\\to\\infty}1 - \\lim\\limits_{N\\to\\infty}(0.7)^{N+1}}{\\lim\\limits_{N\\to\\infty}1 - 0.7} = \\frac{1 - 0}{1-0.7} = \\frac{1}{1-0.7}.$$ Since the limit exists, then we write $$\\sum_{n=0}^{\\infty}(0.7)^n = \\frac{1}{1-0.7}.$$\n\nMore generally, a sum of the form $$a + ar + ar^2 + ar^3 + \\cdots + ar^k$$ with $a$ and $r$ constant is said to be a \"geometric series\" with initial term $a$ and common ratio $r$. If $a=0$, then the sum is equal to $0$. If $r=1$, then the sum is equal to $(k+1)a$. If $r\\neq 1$, then we can proceed as above. Letting $$S = a +ar + \\cdots + ar^k$$ we have that $$S - rS = (a+ar+\\cdots+ar^k) - (ar+ar^2+\\cdots+a^{k+1}) = a - ar^{k+1}$$ so that $$(1-r)S = a(1 - r^{k+1}).$$ Dividing through by $1-r$ (which is not zero since $r\\neq 1$), we get $$S = \\frac{a(1-r^{k+1})}{1-r}.$$\n\nA series of the form $$\\sum_{n=0}^{\\infty}ar^{n}$$ with $a$ and $r$ constants is called an infinite geometric series. If $r=1$, then $$\\lim_{N\\to\\infty}\\sum_{n=0}^{N}ar^{n} = \\lim_{N\\to\\infty}\\sum_{n=0}^{N}a = \\lim_{N\\to\\infty}(N+1)a = \\infty,$$ so the series diverges. If $r\\neq 1$, then using the formula above we have: $$\\sum_{n=0}^{\\infty}ar^n = \\lim_{N\\to\\infty}\\sum_{n=0}^{N}ar^{N} = \\lim_{N\\to\\infty}\\frac{a(1-r^{N+1})}{1-r}.$$ The limit exists if and only if $\\lim\\limits_{N\\to\\infty}r^{N+1}$ exists. Since $$\\lim_{N\\to\\infty}r^{N+1} = \\left\\{\\begin{array}{ll} 0 &\\mbox{if |r|\\lt 1;}\\\\ 1 & \\mbox{if r=1;}\\\\ \\text{does not exist} &\\mbox{if r=-1 or |r|\\gt 1} \\end{array}\\right.$$ it follows that: \\begin{align*} \\sum_{n=0}^{\\infty}ar^{n} &=\\left\\{\\begin{array}{ll} 0 &\\mbox{if a=0;}\\\\ \\text{diverges}&\\mbox{if a\\neq 0 and r=1;}\\\\ \\lim\\limits_{N\\to\\infty}\\frac{a(1-r^{N+1})}{1-r} &\\mbox{if r\\neq 1;}\\end{array}\\right.\\\\ &= \\left\\{\\begin{array}{ll} \\text{diverges}&\\mbox{if a\\neq 0 and r=1;}\\\\ \\text{diverges}&\\mbox{if a\\neq 0, and r=-1 or |r|\\gt 1;}\\\\ \\frac{a(1-0)}{1-r}&\\mbox{if |r|\\lt 1;} \\end{array}\\right.\\\\ &=\\left\\{\\begin{array}{ll} \\text{diverges}&\\mbox{if a\\neq 0 and |r|\\geq 1;}\\\\ \\frac{a}{1-r}&\\mbox{if |r|\\lt 1.} \\end{array}\\right. \\end{align*}\n\nYour particular example has $a=1$ and $r=0.7$.\n\nSince this recently came up (09/29/2011), let's provide a formal proof that $$\\lim_{N\\to\\infty}r^{N+1} = \\left\\{\\begin{array}{ll} 0 &\\mbox{if |r|\\lt 1;}\\\\ 1 & \\mbox{if r=1;}\\\\ \\text{does not exist} &\\mbox{if r=-1 or |r|\\gt 1} \\end{array}\\right.$$\n\nIf $r\\gt 1$, then write $r=1+k$, with $k\\gt0$. By the binomial theorem, $r^n = (1+k)^n \\gt 1+nk$, so it suffices to show that for every real number $M$ there exists $n\\in\\mathbb{N}$ such that $nk\\gt M$. This is equivalent to asking for a natural number $n$ such that $n\\gt \\frac{M}{k}$, and this holds by the Archimedean property; hence if $r\\gt 1$, then $\\lim\\limits_{n\\to\\infty}r^n$ does not exist. From this it follows that if $r\\lt -1$ then the limit also does not exist: given any $M$, there exists $n$ such that $r^{2n}\\gt M$ and $r^{2n+1}\\lt M$, so $\\lim\\limits_{n\\to\\infty}r^n$ does not exist if $r\\lt -1$.\n\nIf $r=-1$, then for every real number $L$ either $|L-1|\\gt \\frac{1}{2}$ or $|L+1|\\gt \\frac{1}{2}$. Thus, for every $L$ and for every $M$ there exists $n\\gt M$ such that $|L-r^n|\\gt \\frac{1}{2}$ proving the limit cannot equal $L$; thus, the limit does not exist. If $r=1$, then $r^n=1$ for all $n$, so for every $\\epsilon\\gt 0$ we can take $N=1$, and for all $n\\geq N$ we have $|r^n-1|\\lt\\epsilon$, hence $\\lim\\limits_{N\\to\\infty}1^n = 1$. Similarly, if $r=0$, then $\\lim\\limits_{n\\to\\infty}r^n = 0$ by taking $N=1$ for any $\\epsilon\\gt 0$.\n\nNext, assume that $0\\lt r\\lt 1$. Then the sequence $\\{r^n\\}_{n=1}^{\\infty}$ is strictly decreasing and bounded below by $0$: we have $0\\lt r \\lt 1$, so multiplying by $r\\gt 0$ we get $0\\lt r^2 \\lt r$. Assuming $0\\lt r^{k+1}\\lt r^k$, multiplying through by $r$ we get $0\\lt r^{k+2}\\lt r^{k+1}$, so by induction we have that $0\\lt r^{n+1}\\lt r^n$ for every $n$.\n\nSince the sequence is bounded below, let $\\rho\\geq 0$ be the infimum of $\\{r^n\\}_{n=1}^{\\infty}$. Then $\\lim\\limits_{n\\to\\infty}r^n =\\rho$: indeed, let $\\epsilon\\gt 0$. By the definition of infimum, there exists $N$ such that $\\rho\\leq r^N\\lt \\rho+\\epsilon$; hence for all $n\\geq N$, $$|\\rho-r^n| = r^n-\\rho \\leq r^N-\\rho \\lt\\epsilon.$$ Hence $\\lim\\limits_{n\\to\\infty}r^n = \\rho$.\n\nIn particular, $\\lim\\limits_{n\\to\\infty}r^{2n} = \\rho$, since $\\{r^{2n}\\}_{n=1}^{\\infty}$ is a subsequence of the converging sequence $\\{r^n\\}_{n=1}^{\\infty}$. On the other hand, I claim that $\\lim\\limits_{n\\to\\infty}r^{2n} = \\rho^2$: indeed, let $\\epsilon\\gt 0$. Then there exists $N$ such that for all $n\\geq N$, $r^n - \\rho\\lt\\epsilon$. Moreover, we can assume that $\\epsilon$ is small enough so that $\\rho+\\epsilon\\lt 1$. Then $$|r^{2n}-\\rho^2| = |r^n-\\rho||r^n+\\rho| = (r^n-\\rho)(r^n+\\rho)\\lt (r^n-\\rho)(\\rho+\\epsilon) \\lt r^n-\\rho\\lt\\epsilon.$$ Thus, $\\lim\\limits_{n\\to\\infty}r^{2n} = \\rho^2$. Since a sequence can have only one limit, and the sequence of $r^{2n}$ converges to both $\\rho$ and $\\rho^2$, then $\\rho=\\rho^2$. Hence $\\rho=0$ or $\\rho=1$. But $\\rho=\\mathrm{inf}\\{r^n\\mid n\\in\\mathbb{N}\\} \\leq r \\lt 1$. Hence $\\rho=0$.\n\nThus, if $0\\lt r\\lt 1$, then $\\lim\\limits_{n\\to\\infty}r^n = 0$.\n\nFinally, if $-1\\lt r\\lt 0$, then $0\\lt |r|\\lt 1$. Let $\\epsilon\\gt 0$. Then there exists $N$ such that for all $n\\geq N$ we have $|r^n| = ||r|^n|\\lt\\epsilon$, since $\\lim\\limits_{n\\to\\infty}|r|^n = 0$. Thus, for all $\\epsilon\\gt 0$ there exists $N$ such that for all $n\\geq N$, $| r^n-0|\\lt\\epsilon$. This proves that $\\lim\\limits_{n\\to\\infty}r^n = 0$, as desired.\n\nIn summary, $$\\lim_{N\\to\\infty}r^{N+1} = \\left\\{\\begin{array}{ll} 0 &\\mbox{if |r|\\lt 1;}\\\\ 1 & \\mbox{if r=1;}\\\\ \\text{does not exist} &\\mbox{if r=-1 or |r|\\gt 1} \\end{array}\\right.$$\n\nThe argument suggested by Srivatsan Narayanan in the comments to deal with the case $0\\lt|r|\\lt 1$ is less clumsy than mine above: there exists $a\\gt 0$ such that $|r|=\\frac{1}{1+a}$. Then we can use the binomial theorem as above to get that $$|r^n| = |r|^n = \\frac{1}{(1+a)^n} \\leq \\frac{1}{1+na} \\lt \\frac{1}{na}.$$ By the Archimedean Property, for every $\\epsilon\\gt 0$ there exists $N\\in\\mathbb{N}$ such that $Na\\gt \\frac{1}{\\epsilon}$, and hence for all $n\\geq N$, $\\frac{1}{na}\\leq \\frac{1}{Na} \\lt\\epsilon$. This proves that $\\lim\\limits_{n\\to\\infty}|r|^n = 0$ when $0\\lt|r|\\lt 1$, without having to invoke the infimum property explicitly.\n\n-\nI think you mean \"which is not zero since $r\\neq 1$.\" \u2013\u00a0 robjb Mar 26 '11 at 23:20\n@Rob: Yes; thank you. \u2013\u00a0 Arturo Magidin Mar 27 '11 at 0:22\n+1 As its rare to see math researchers caring for even elementary questions. \u2013\u00a0 Please Delete Account Mar 27 '11 at 2:18\nIgnore my edit, I seem to have potatoes on my eyes. Sorry. \u2013\u00a0 Lagerbaer Mar 31 '11 at 5:05\nArturo, I will mention one more way of proving that $r^n$ converges to $0$ as $n \\to \\infty$ if $|r| < 1$ that looks somewhat similar to your proof in the divergent case. Take $|r| = \\frac{1}{1+a}$ for some $a > 0$. Then $$|r^n - 0| = |r|^n = \\frac{1}{(1+a)^n} \\leq \\frac{1}{1+an} \\leq \\frac{1}{an} ,$$ which can be made sufficiently small by taking $n$ large enough. \u2013\u00a0 Srivatsan Sep 30 '11 at 4:13\n\nit's called a geometric series. let $-1<x<1$ and let $S_n=1+x+x^2+...+x^n$. then $xS_n=x+...+x^{n+1}=S^n-1+x^{n+1}$. move stuff around to get $$S_n=\\frac{1-x^{n+1}}{1-x}$$ take the limit as $n\\to\\infty$ (noting that $x^n\\to0$ if $|x|<1$)\n\n-\nIn yoyo's answer, you should also take into account that $x^{n+1}\\to 0$ as $n\\to\\infty$ if $|x| < 1$. \u2013\u00a0 knightofmathematics Mar 25 '11 at 16:48\n...and $0^0=1$ (: \u2013\u00a0 yoyo Mar 25 '11 at 19:03\n\nIf u expand your summation you get series $$1+0.7+0.7^2+\\dots$$ as it is geometric series (you can see it here http://en.wikipedia.org/wiki/Geometric_series) $$\\sum_{n=0}^{\\infty}{0.7^n}=\\frac{1}{1-0.7}$$\n\nOr $S=1+x+x^2+\\dots$\n\n$xS_n=x+x^2+\\dots=S-1$ now u take\n\n$S-xS=1$\n\n$S(1-x)=1 \\implies S=\\frac{1}{1-x}$\n\nhere your $x=0.7$\n\n-\nThe index in the sum should be $n$, not $i$. \u2013\u00a0 Arturo Magidin Mar 25 '11 at 17:10\noops..i'll edit it \u2013\u00a0 amul28 Mar 26 '11 at 3:54\n\nI find the proof here lovely.\n\n-\nThe link here no longer seems to work. \u2013\u00a0 Baby Dragon Oct 6 '13 at 1:19\n@BabyDragon Here is the updated link. \u2013\u00a0 David Mitra Oct 6 '13 at 9:12\n\n## protected by AryabhataMar 31 '11 at 17:18\n\nThank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.", "date": "2014-12-20 23:55:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn", "openwebmath_score": 0.9880840182304382, "openwebmath_perplexity": 159.1469231068039, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9933071494719702, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8576130589810711}}
{"url": "https://mathoverflow.net/questions/278629/a-combinatorial-identity", "text": "# A combinatorial identity\n\nI hope this is a suitable MO question. In a research project, my collaborator and I came across some combinatorial expressions. I used my computer to test a few numbers and the pattern was suggesting the following equation for fixed integers $K\\geq n>0$.\n\n$$\\dfrac{K!}{n!K^{K-n}}\\sum\\limits_{ \\begin{subarray}{c} k_1+\\dotsb+k_{n}=K \\\\ k_i \\geq 1 \\end{subarray}} \\prod\\limits_{i=1}^n \\dfrac{k_i^{k_i-2}}{(k_i-1)!}=\\displaystyle {K-1\\choose n-1}.$$\n\nWe tried to think of a proof but failed. One can probably move these $K!, n!$ to the right and rewrite the RHS, or maybe move $K!$ into the summation to form combinatorial numbers like $K\\choose k_1,k_2,\\dotsc,k_n$. We don't know which is better.\n\nThe questions are:\n\n1. Anyone knows a proof for this identity?\n2. In fact the expression that appears in our work is $\\sum\\limits_{ \\begin{subarray}{c} k_1+\\dotsb+k_{n}=K \\\\ k_i \\geq 1 \\end{subarray}} \\sigma_p(k_1,\\dotsc,k_n) \\prod\\limits_{i=1}^n \\dfrac{k_i^{k_i-2}}{(k_i-1)!}$, where $p$ is a fixed integer and $\\sigma_p(\\dotsc)$ is the $p$-th elementary symmetric polynomial. The equation in the beginning simplifies this expression for $p=0,1$. Is there a similar identity for general $p$?\n\n----------Update----------\n\nQuestion 2 is perhaps too vague, and I'd like to make it a bit more specific. Probably I should have written this down in the beginning, but I feared this is too long and unmotivated. But after seeing people's skills, I'm very tempted to leave it here in case somebody has remarks. In fact, question 2 partly comes from the effort to find a proof for the following (verified by computer).\n\n$$\\frac{1}{K!} \\prod_{r=1}^{K} (r+1 -x)= \\sum_{n=1}^K \\frac{(-1)^n}{n!} \\left[ \\sum_{p=0}^n K^{n-p} \\prod_{r=1}^p (x +r-4) \\left( \\sum\\limits_{ \\begin{subarray}{c} k_1+\\dotsb+k_{n}=K \\\\ k_i \\geq 1 \\end{subarray}} \\sigma_p(k_1,\\dotsc,k_{n}) \\prod\\limits_{i=1}^n \\dfrac{k_i^{k_i-2}}{(k_i-1)!} \\right) \\right],$$ Where $x$ is a fixed number (in our case, an integer).\n\n\u2022 Combinatorial reformulation: consider all the trees $T$ on $\\{0,1,\\dots,K\\}$ for which the vertices $1,2,\\dots,p$ are in different components of $T\\setminus \\{0\\}$. For each such a tree take a summand $(-K)^{\\deg(0)-p}$. The sum equals $(K-2)!/(p-2)!$ for $p\\geqslant 2$ and 0 for $p=0,1$. \u2013\u00a0Fedor Petrov Aug 15 '17 at 9:35\n\nThis is the answer to the first question, I wrote a long answer to Question 2 as a separate answer.\n\nNote that $A:=\\sum_{k_i>0,k_1+\\dots+k_n=K}\\frac{K!}{n!k_1!\\dots k_n!} \\prod k_i^{k_i-1}$ is a number of forests on the ground set $\\{1,2,\\dots,K\\}$ having exactly $n$ connected components and with a marked vertex in each component ($k_i$ correspond to the sizes of components.) Add a vertex 0 and join it with the marked vertices. Then we have to count the number of trees on $\\{0,1,\\dots,K\\}$ in which $0$ has degree $n$. Remember that the sum of $z_0^{d_1-1}\\dots z_K^{d_K-1}$ over all trees on $\\{0,\\dots,K\\}$, where $d_i$ is degree of $i$, equals $(z_0+\\dots+z_K)^{K-1}$. Substitute $z_{1}=\\dots=z_K=1$ and take a coefficient of $z_0^{n-1}$. It is $\\binom{K-1}{n-1}K^{K-n}$.\n\n\u2022 Thank you! This is a really cool argument, especially for me who doesn't have a lot of experiences in graph theory. I will mark this as the correct answer, but of course any further ideas and comments about question 2 are welcomed as well. \u2013\u00a0Honglu Aug 13 '17 at 18:41\n\u2022 You need to talk to someone who's familiar with Volume 2 of Stanley. (This is more algebraic combinatorics than graph theory.) \u2013\u00a0Alexander Woo Aug 13 '17 at 23:10\n\u2022 Sorry about my ignorance and thank you for pointing out the right words! I am thinking about asking around, and your comment definitely helps. \u2013\u00a0Honglu Aug 13 '17 at 23:58\n\nHere's another proof. We first rewrite the identity (by setting $k_i=j_i+1$) as $$\\sum_{j_1+\\cdots +j_n=K-n}\\prod_{i=1}^n \\frac{(j_i+1)^{j_i-1}}{j_i!} = n\\frac{K^{K-n-1}}{(K-n)!}. \\tag{1}$$\n\nLet $F(x)$ be the formal power series satisfying $F(x)= e^{xF(x)}$. It is well known (and easily proved, e.g., by Lagrange inversion) that $$F(x)^n = \\sum_{j=0}^\\infty n(j+n)^{j-1}\\frac{x^j}{j!}.\\tag{2}$$ In particular, $$F(x) = \\sum_{j=0}^\\infty (j+1)^{j-1}\\frac{x^j}{j!}.$$ So the left side of $(1)$ is equal to the coefficient of $x^{K-n}$ in $F(x)^{n}$, which by $(2)$ is equal to the right side of $(1)$.\n\n\u2022 Cool! I actually updated my question to include a more general identity that I want to prove. I kept thinking there is some generating function lurking behind, and I'm just looking at the coefficients. Any thoughts will be greatly appreciated. \u2013\u00a0Honglu Aug 14 '17 at 16:04\n\nHere is a generating-function proof of your conjectured identity (and an answer to question 2).\n\nThe main ingredient is a formula for the appearing symmetric sums.\n\nLet $T(z)$ (the tree function'') be the formal power series satisfying $T(z)=z\\,e^{T(z)}$.\n\nIf $F$ is a formal power series the coefficients of $G(z):=F(T(z))$ are given by (Lagrange inversion) $$[z^0]G(z)=[z^0] F(z) \\mbox{ , } [z^k]G(z)=\\tfrac{1}{k} [y^{k-1}] F^\\prime(y)\\,e^{ky} =[y^k](1-y)F(y)\\,e^{ky}\\mbox{ for } k\\geq 1\\;.$$ In particular (as is well known) $$T(z)=\\sum_{n\\geq 1}\\frac{n^{n-1}}{n!}z^n \\;\\mbox{ and }\\; \\frac{T(z)}{1-T(z)}=\\sum_{n\\geq 1}\\frac{n^{n}}{n!}z^n$$ Thus $T(z)\\left(1+\\frac{t}{1-T(z)}\\right)=\\sum_{n\\geq 1} \\frac{(1+tn)n^{n-1}}{n!}$. Therefore \\begin{align*}S_{p,n}(K):&=\\sum_{{k_1+\\ldots +k_n=K \\atop k_i\\geq 1}} \\sigma_p(k_1,\\ldots,k_n)\\prod_{i=1}^n \\frac{k_i^{k_i-1}}{k_i!}\\\\ &=[t^p]\\sum_{k_1+\\ldots +k_n=K \\atop k_i\\geq 1} \\prod_{i=1}^n \\frac{(1+tk_i) k_i^{k_i-1}}{k_i!}\\\\ &=[t^p z^K]\\, T(z)^n \\left(1+\\frac{t}{1-T(z)}\\right)^n \\end{align*} and $$S_{p,n}(K)={n \\choose p} [z^K]\\frac {T(z)^n}{(1-T(z))^p}={n \\choose p}[y^K]\\,y^n\\, \\frac{(1-y)}{(1-y)^p}\\,e^{Ky}\\;\\;\\;\\;\\;(*)$$\n\nNow consider the sum ($m:=x-4$) $$R(K,m):=\\sum_{n=0}^K\\frac{(-1)^n}{n!}\\bigg[\\sum_{p=0}^n K^{n-p}\\,\\left(\\prod_{r=1}^p (m+r)\\right) S_{n,p}(K)\\bigg]$$ Since $K\\geq 1$ the sum remains unchanged if we start the summation at $n=0$ (all $S_{0,p}(K)$ are $0$). Using that and $(*)$ gives \\begin{align*} R(K,m):&=\\sum_{n=0}^K\\frac{(-1)^n}{n!}\\bigg[\\sum_{p=0}^n K^{n-p}\\,\\left(\\prod_{r=1}^p (m+r)\\right) S_{n,p}(K)\\bigg]\\\\ &=[y^K]\\,(1-y) \\sum_{n=0}^K\\frac{(-1)^n}{n!}\\sum_{p=0}^n K^{n-p}\\,p!{m+p \\choose p}{n \\choose p}\\frac{y^n}{(1-y)^p}\\,e^{Ky}\\\\ &=[y^K]\\,(1-y) \\sum_{p=0}^K\\sum_{n=p}^K\\frac{(-1)^n}{(n-p)!} K^{n-p}{m+p \\choose p}\\frac{y^n}{(1-y)^p}\\,e^{Ky}\\\\ &=[y^K]\\,(1-y) \\sum_{p=0}^K {m+p \\choose p}\\frac{(-1)^p y^p}{(1-y)^p}\\bigg[\\sum_{n=p}^K\\frac{(-1)^{n-p}}{(n-p)!} K^{n-p}y^{n-p}\\,e^{Ky}\\bigg]\\\\ &=[y^K]\\,(1-y) \\sum_{p=0}^K {m+p \\choose p}\\frac{(-1)^p y^p}{(1-y)^p}\\bigg[1 +\\mathcal{O}(y^{K-p+1})\\bigg]\\\\ \\end{align*} where $\\mathcal{O}(y^{K-p+1})$ denotes a formal power series which is a multiple of $y^{K-p+1}$.\nTaking into account the respective factors $y^p$ the terms in these series do not contribute to the coefficient $[y^K]$. Therefore\n\n\\begin{align*} R(K,m)&=[y^K]\\,(1-y) \\sum_{p=0}^K {m+p \\choose p}\\frac{(-1)^p y^p}{(1-y)^p}\\\\ &=[y^K]\\,(1-y) \\sum_{p\\geq 0} {m+p \\choose p}\\frac{(-1)^p y^p}{(1-y)^p}\\\\ &=[y^K]\\,(1-y) \\left(\\frac{1}{1+\\tfrac{y}{1-y}}\\right)^{m+1}\\\\ &=[y^K]\\,(1-y)^{m+2}\\\\ &=(-1)^K\\,{m+2 \\choose K},\\,\\mbox{ as desired }. \\end{align*}\n\n\u2022 Nice! I really learned a lot from all the answers and appreciate everybody's effort. In particular, this generating function method seems to have some other applications in our project. I got another quick question. Let $H_n=\\sum\\limits_{k=1}^n 1/k$. Do you know whether we can similarly describe the formal series $\\sum_{n\\geq 1} \\dfrac{H_nn^n}{n!}z^n$ using functional equations just like your $T(z)$ and others? \u2013\u00a0Honglu Aug 15 '17 at 22:33\n\u2022 Thank you. (1) The formal series $\\sum_{n\\geq 1}\\frac{n^n}{n!}p(n) z^n$ can be expressed as a rational function of $T(z)$ if $p$ is a polynomial in $n$ and $\\tfrac{1}{n}$. My first guess is that it will not be easy to treat the case $p(n)=H_n$ via Lagrange inversion. (2) $T(z)$ is certainly not \"my\" function. $T(z)=-W(-z)$ where $W$ is \"Lambert's W-function\", and $T(z)=z e^{T(z)}$ goes back to Eisenstein. \u2013\u00a0esg Aug 16 '17 at 17:57\n\u2022 Would you mind telling me your real name by email? Because if we decide to post anything about this work in the future, we will acknowledge you (also Fedor and other people). I just temporarily added my email in my MO profile. We are still working on an ultimate combinatorial expression that includes all my questions as special cases. It's too long to post in a comment, but if you are interested, I will be glad to send it to you by email as well. Of course interests from other people will also be welcomed, just let me know in the comment or send me an email. \u2013\u00a0Honglu Aug 24 '17 at 16:09\n\nHere is the answer to Question 2. It may be probably simplified.\n\nDenote $y=3-x$, then we rewrite your identity as $$\\binom{y+K-2}K=\\frac{(y-1)y(y+1)\\dots (y+K-2)}{K!}=c_0\\binom{y}0+c_1\\binom{y}1+\\dots+c_K\\binom{y}K,$$ where $$c_p=p!\\sum_{n=p}^K(-K)^{n-p}\\frac1{n!}\\sum\\limits_{ \\begin{subarray}{c} k_1+\\dotsb+k_{n}=K \\\\ k_i \\geq 1 \\end{subarray}} \\sigma_p(k_1,\\dotsc,k_{n}) \\prod\\limits_{i=1}^n \\dfrac{k_i^{k_i-2}}{(k_i-1)!}.$$ On the other hand, by Vandermonde--Chu identity we have $$\\binom{y+K-2}K=\\sum_{i=2}^{K}\\binom{y}i\\binom{K-2}{K-i},$$ so your identity is equivalent to the formula $$\\sum_{n=p}^K(-K)^{n-p}\\frac{K!}{n!}\\sum\\limits_{ \\begin{subarray}{c} k_1+\\dotsb+k_{n}=K \\\\ k_i \\geq 1 \\end{subarray}} \\sigma_p(k_1,\\dotsc,k_{n}) \\prod\\limits_{i=1}^n \\dfrac{k_i^{k_i-2}}{(k_i-1)!}=\\frac{K!}{p!}\\binom{K-2}{K-p},$$ I multiplied both parts by $K!/p!$. Note that $$\\frac{K!}{n!}\\sum\\limits_{ \\begin{subarray}{c} k_1+\\dotsb+k_{n}=K \\\\ k_i \\geq 1 \\end{subarray}} \\sigma_p(k_1,\\dotsc,k_{n}) \\prod\\limits_{i=1}^n \\dfrac{k_i^{k_i-2}}{(k_i-1)!}$$ is a number of the trees $T$ on $\\{0,1,\\dots,K\\}$ such that degree of 0 equals $n$ and $p$ vertices in different components of $T\\setminus\\{0\\}$ are marked. Indeed, if these components $A_1,\\dots,A_n$ are enumerated (this corresponds to the multiple $n!$) and $i$-th component $A_i$ has $k_i$ vertices, then we have $\\frac{K!}{k_1!\\dots k_n!}$ ways to choose $A_i$, $\\sigma_p(k_1,\\dotsc,k_{n})$ ways to mark $p$ vertices in different components, $k_i^{k_i-1}$ ways to make a tree on $A_i$ and choose a vertex in $A_i$ joined with 0.\n\nNote that each (out of $\\binom{K}p$ sets) set of $p$ marked vertices makes the same contribution to the sum. So, we may suppose that the marked set is $\\{1,2,\\dots,p\\}$ and we have to prove that the sum of $(-K)^{n-p}$ over admissible trees (where the tree $T$ is admissible if $1,2,\\dots,p$ are in different components of $T\\setminus \\{0\\}$) equals $\\frac1{\\binom{K}p}\\frac{K!}{p!}\\binom{K-2}{K-p}=(p-1)p\\dots (K-2)$.\n\nWe start to prove this from the case $p=0$, $p=1$, where the restriction that $1,2,\\dots,p$ are in different components of $T\\setminus \\{0\\}$ disappears. Then the sum $z_0^{n-1}z_1^{d_1-1}\\dots z_K^{d_K-1}$, $d_i=\\deg(i)$, over all trees on $\\{0,\\dots,K\\}$ equals, as is well known and easy to prove, to $(z_0+\\dots+z_K)^{K-1}$. Substituting $z_0=-K$, $z_1=\\dots=z_K=1$ we get the result.\n\nNow we deal with the more involved case $p\\geqslant 2$. Denote $K=p+m$ and consider the variables $z_0,z_1,\\dots,z_p,z_{p+1},\\dots$ (infinitely many for simplicity of notations). Denote $s=z_0+z_1+\\dots$, write $\\sigma_i$ for the $i$-th elementary symmetric polynomial of $z_{p+1},z_{p+2},\\dots$. Denote $\\varphi_0=1$, $\\varphi_m=s\\varphi_{m-1}+(p-1)p\\dots (p+m-2)\\sigma_m$ for $m\\geqslant 1$. I claim that the sum of $z_0^{n-p}z_1^{d_1-1}\\dots z_{p+m}^{d_{p+m}-1}$ over all admissible trees equals $\\varphi_m(z_0,z_1,\\dots,z_{p+m},0,0,\\dots)$.\n\nNote that this implies our claim, as follows from the substitution $z_0=-K=-p-m,z_1=\\dots=z_{p+m}=1$.\n\nThe proof is on induction in $m$. Base $m=0$ is clear. For the induction step, look at coefficients of any specific monomial $z_0^{n-p}z_1^{d_1-1}\\dots z_{p+m}^{d_{p+m}-1}$. Consider two cases:\n\n1) $d_i=1$ for a certain index $i\\in \\{p+1,\\dots,p+m\\}$, without loss of generality $i=p+m$. This corresponds to the case when $p+m$ has degree 1, such a vertex may be joined with any of other vertices, and removing corresponding edge we get a tree (it remains admissible) on $\\{0,1,\\dots,K-1\\}$. This corresponds to the summand $s\\varphi_{m-1}$: namely, $z_j\\varphi_{m-1}$ corresponds to the edge between $p+m$ and $j$; $j=0,1,\\dots,p+m-1$.\n\n2) $d_{p+1},\\dots,d_{p+m}$ are greater than 1. Then they are all equal to 2, since the degree of the whole monomial equals $m$. In this case there are $p(p+1)\\dots (p+m-1)$ admissible trees (well, they are all admissible for such a choice of degrees and we may either apply the above formula for all trees, or prove it by induction, or as you wish). It remains to prove that the coefficient of $z_{p+1}\\dots z_{p+m}$ in the function $\\varphi_m$ equals $p(p+1)\\dots (p+m-1)$. Since $\\varphi_m=s\\varphi_{m-1}+(p-1)p\\dots (p+m-2)\\sigma_m$, it is equivalent to proving that the coefficient of $z_{p+1}\\dots z_{p+m}$ in $s\\varphi_{m-1}$ equals $p(p+1)\\dots (p+m-1)-(p-1)p\\dots (p+m-2)=mp(p+1)\\dots(p+m-2)$. We should take some $z_j$, $p+1\\leqslant j\\leqslant p+m$, from the multiple $s=\\sum z_i$, and for each choice of $j$ we have a coefficient of $z_j^{-1}\\cdot z_{p+1}\\dots z_{p+m}$ in $\\varphi_{m-1}$ equal to $p(p+1)\\dots(p+m-2)$ - by induction (base $m-1=0$ is clear).\n\n\u2022 Not sure whether you get notifications from comments in other answers or not. To be sure I just repeat some messages here. I really appreciate all your answers here. Ultimately we want to simplify a bigger expression that includes the RHS of my question 2 as a special case. It's pretty long and currently still over my head. I don't know if you have the time and the interest to take a look. But in case you do, you are welcomed to send me an email (in my profile). \u2013\u00a0Honglu Aug 24 '17 at 19:55", "date": "2019-09-21 20:09:25", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/278629/a-combinatorial-identity", "openwebmath_score": 0.8851345181465149, "openwebmath_perplexity": 225.31226719536917, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850867332734, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8575917010222195}}
{"url": "https://au.mathworks.com/help/matlab/ref/power.html", "text": "# power, .^\n\nElement-wise power\n\n## Description\n\nexample\n\nC = A.^B raises each element of A to the corresponding powers in B. The sizes of A and B must be the same or be compatible.\n\nIf the sizes of A and B are compatible, then the two arrays implicitly expand to match each other. For example, if one of A or B is a scalar, then the scalar is combined with each element of the other array. Also, vectors with different orientations (one row vector and one column vector) implicitly expand to form a matrix.\n\nC = power(A,B) is an alternate way to execute A.^B, but is rarely used. It enables operator overloading for classes.\n\n## Examples\n\ncollapse all\n\nCreate a vector, A, and square each element.\n\nA = 1:5;\nC = A.^2\nC = 1\u00d75\n\n1     4     9    16    25\n\nCreate a matrix, A, and take the inverse of each element.\n\nA = [1 2 3; 4 5 6; 7 8 9];\nC = A.^-1\nC = 3\u00d73\n\n1.0000    0.5000    0.3333\n0.2500    0.2000    0.1667\n0.1429    0.1250    0.1111\n\nAn inversion of the elements is not equal to the inverse of the matrix, which is instead written A^-1 or inv(A).\n\nCreate a 1-by-2 row vector and a 3-by-1 column vector and raise the row vector to the power of the column vector.\n\na = [2 3];\nb = (1:3)';\na.^b\nans = 3\u00d72\n\n2     3\n4     9\n8    27\n\nThe result is a 3-by-2 matrix, where each (i,j) element in the matrix is equal to a(j) .^ b(i):\n\n$\\mathit{a}=\\left[{\\mathit{a}}_{1}\\text{\\hspace{0.17em}}{\\mathit{a}}_{2}\\right],\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\mathit{b}=\\left[\\begin{array}{c}{\\mathit{b}}_{1}\\\\ {\\mathit{b}}_{2}\\\\ {\\mathit{b}}_{3}\\end{array}\\right],\\text{\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\mathit{a}\\text{\\hspace{0.17em}}.\u02c6\\text{\\hspace{0.17em}}\\mathit{b}=\\left[\\begin{array}{cc}{{\\mathit{a}}_{1}}^{{\\mathit{b}}_{1}}& {{\\mathit{a}}_{2}}^{{\\mathit{b}}_{1}}\\\\ {{\\mathit{a}}_{1}}^{{\\mathit{b}}_{2}}& {{\\mathit{a}}_{2}}^{{\\mathit{b}}_{2}}\\\\ {{\\mathit{a}}_{1}}^{{\\mathit{b}}_{3}}& {{\\mathit{a}}_{2}}^{{\\mathit{b}}_{3}}\\end{array}\\right].$\n\nCalculate the roots of -1 to the 1/3 power.\n\nA = -1;\nB = 1/3;\nC = A.^B\nC = 0.5000 + 0.8660i\n\nFor negative base A and noninteger B, the power function returns complex results.\n\nUse the nthroot function to obtain the real roots.\n\nC = nthroot(A,3)\nC = -1\n\n## Input Arguments\n\ncollapse all\n\nOperands, specified as scalars, vectors, matrices, or multidimensional arrays. A and B must either be the same size or have sizes that are compatible (for example, A is an M-by-N matrix and B is a scalar or 1-by-N row vector). For more information, see Compatible Array Sizes for Basic Operations.\n\n\u2022 Operands with an integer data type cannot be complex.\n\nData Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | logical | char\nComplex Number Support: Yes\n\ncollapse all\n\n### IEEE Compliance\n\nFor real inputs, power has a few behaviors that differ from those recommended in the IEEE\u00ae-754 Standard.\n\nMATLAB\u00ae IEEE\n\npower(1,NaN)\n\nNaN\n\n1\n\npower(NaN,0)\n\nNaN\n\n1\n\n## Compatibility Considerations\n\nexpand all\n\nBehavior changed in R2016b\n\n## Extended Capabilities\n\n### Topics\n\nIntroduced before R2006a", "date": "2021-04-21 18:39:05", "meta": {"domain": "mathworks.com", "url": "https://au.mathworks.com/help/matlab/ref/power.html", "openwebmath_score": 0.8176913857460022, "openwebmath_perplexity": 2113.6923752322687, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850842553891, "lm_q2_score": 0.8723473796562745, "lm_q1q2_score": 0.8575916972293566}}
{"url": "http://audio-upgrade.pl/sum-of-sequence-calculator.html", "text": "Sum Of Sequence Calculator\n\nUse this calculator to determine the future value of your savings and lump sum. In geometric progressions where |r| < 1 (in other words where r is less than 1 and greater than \u20131), the sum of the sequence as n tends to infinity approaches a value. This calculator will find the infinite sum of arithmetic, geometric, power, and binomial series, as well as the partial sum, with steps shown (if possible). The Arithmetic series of finite number is the addition of numbers and the sequence that is generally followed include \u2013 (a, a + d, a + 2d, \u2026. Approach : Read input number asking for length of the list using input() or raw_input(). Sum of 32-bit integer quantities is not computed by using 64-bit results, and overflow can occur for the LINQ to SQL translation of Sum. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. Sequences and summations CS 441 Discrete mathematics for CS M. Objects might be numbers or letters. Set Sum = 0. The limit of a sequence is the value the sequence approaches as the number of terms goes to infinity. Partial Sums of an Arithmetic Sequence. All term in the sequence meet a specific logical rule which needs to be recognised in order to find the missing terms. Because a series consists of evenly-spaced numbers, the median and mean (average) of the series will be the same. arithmetic series. This way, each term can be expressed by this equation: x\u2090 = x\u2090\u208b\u2081 + x\u2090\u208b\u2082. ): A series of numbers in which ratio of any two consecutive numbers is always a same number that is constant. Calculates the sum of the infinite geometric series. Arithmetic Sequences and Sums Sequence. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. We start with the formula for PV of a future value (FV) single lump sum at time n and interest rate i,. lets say you want to sum the numbers from 1 to 100 all you have to do is to calculate this (100)x(101) /2. What Is The Formula For Calculating Arithmetic Sequence? If the initial term of an arithmetic sequence is a 1 and the common difference of successive members is d , then the nth term of the sequence is given by:. The sequence begins 1, 1, 2, 3, 5, and each succeeding term is the sum of the previous two terms. If only a single number for value1 is supplied, SUM returns value1. Example 1: Find the sum of squares of the numbers from 0 to 5000. I want to calculate sum of the average daily value off each product. Work out: a. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP\ud83d\ude4f This problem has been solved!. skipna bool, default True. The Corbettmaths video tutorial on finding the nth term for a fractional sequence. In this case, you have the sequence. Note that a series is an indicated sum of the terms of a sequence!! In this section, we work only with finite series and the related sums. P Class 11 Engineering - Sum of n Terms of G. The formula for the n-th term of a quadratic sequence is explained here. And how many pairs do we have? Well, we have 2 equal rows, we must have n/2 pairs. The sequence of terms that are common to both sequences will be an arithmetic sequence with a common difference of 28, which is the LCM of 4 and 7. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. Add Number to Sum and set equal to Sum. This calculator will find the infinite sum of arithmetic, geometric, power, and binomial series, as well as the partial sum, with steps shown (if possible). What is Sum of the Series in Math. The arithmetic sequence calculator uses arithmetic sequence formula to find sequence of any property. The problem is that the number of cells to be summed will vary; for one run of the macro it could be 100 cells, while on the next it could be 300 and on the third only 25. std ([ddof]) Calculate window standard deviation. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Geometric Progression (G. Sequence: By a sequence, we mean an arrangement of number in definite order according to some rules. Build your own widget. Improve your math knowledge with free questions in \"Find the sum of an arithmetic series\" and thousands of other math skills. The sum of a geometric series is indeed an interesting place to start this discussion. You also generate a sum in each step, which you reuse in the next step. Home; Java Tutorial; Language; Data Type; Operators; Statement Control. Find the sum of first n terms of the series (i) 3 +33 +333 +. Then we update the value of variable sum as - sum = sum + c = 0 + 0 = 0. To concatenate a series of iterables, consider using itertools. help me how to calculate the sum of a series in Matlab. The Fibonacci sequence typically has first two terms equal to x\u2080 = 0 and x\u2081 = 1. In other words, how to take the value of a cell located in one worksheet and add it to the value of another cell located in another worksheet to come up with the total of the respective cells. The sequence begins 1, 1, 2, 3, 5, and each succeeding term is the sum of the previous two terms. Calculate the sum of series 1^2+2^2+3^2+4^2++n^2 (n>0) using the both for and while loop structure. Subtract the smaller fraction from the larger fraction to calculate the. Sum of 32-bit integer quantities is not computed by using 64-bit results, and overflow can occur for the LINQ to SQL translation of Sum. The partial sum approach of course involves a \"trick\" as well -- finding an expression for the dang partial sum. To find sum of your series, you need to choose the series variable, lower and upper bounds and also input the expression for n-th term of the series. Example 2: Find the sum of squares of the first 100 numbers of the form prime minus one. This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. As n tends to infinity, S n tends to The sum to infinity for an arithmetic series is undefined. Leonhard Euler was able to calculate the exact sum of the p-series with p 2: 2-2 32 42 Use this fact to find the sum of each series: 2 32 so. Recently, mandatory vote-by-mail has received a great deal of attention as a means of administering elections in the United States. 01 then we use the formula for the sum of the infinite geometric series S oo = a 1 / (1 - r),. Using for loop. Program description:- Find sum of series 1\u00b2 + 2\u00b2 + 3\u00b2 + 4\u00b2 + 5\u00b2 +\u2026. com presented a large number of task in mathematics that you can solve online free of charge on a variety of topics: calculation of integrals and derivatives, finding the sum of the series, the solution of differential equations, etc. You may find functions such as sum, ones, or dot useful. In other words, how to take the value of a cell located in one worksheet and add it to the value of another cell located in another worksheet to come up with the total of the respective cells. sum (axis = None, skipna = None, level = None, numeric_only = None, min_count = 0, ** kwargs) [source] \u00b6 Return the sum of the values for the requested axis. If the value of the sum (in the limiting sense) exists, then they say that the series. Calculate the first term. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson arithmetic sequence. Our sum of series calculator or arithmetic series calculator is an online tool which you can find on Google. is 1 and any term is equal to the sum of all the succeeding terms. It is used to achieve the gas\u2013liquid or liquid\u2013liquid two-phase separation. NPV = Sum CF* ((1+2%)/(1+D))^N When N is infinite, after simplification, NPV = CF * 1 / (1 - r) where r = (1+2%)/(1+D)--A+ V. In general, one does not expect to be able to calculate an infinite sum exactly. In this case, you have the sequence. But there are some series. By definition, the first two numbers in the Fibonacci sequence are either 1 and 1, or 0 and 1, depending on the chosen starting point of the sequence, and each subsequent number is the sum of the previous two. Java program to calculate the sum of N numbers using arrays, recursion, static method, using while loop. Each number in series is called as Fibonacci number. Leonhard Euler continued this study and in the process solved many. Finding a Rule for a Sequence [07/24/2003] What is the next number in this sequence? 1, 3, 11, 67, ? Finding Sum Formula using Sequences of Differences [06/28/1998] Finding a formula for the sum of the first n fourth powers using sequences of differences. 02 Line 10 Column 10 \"The Sum of the above Series for term \"s is : \". Find the sum of 35 terms of an arithmetic series of which the first term is \"a\" and the fifteenth term is \"9a. It is said that when Gauss was ten. We start with the formula for PV of a future value (FV) single lump sum at time n and interest rate i,. I evaluated the partial sum through a calculator, my answer was -19/30, or -. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP\ud83d\ude4f This problem has been solved!. After discussing the above two examples one will wonder if any sequence has the same faith (meaning, it gets closer to a number). Sum of this series is calculated by the formula : sum=(n*(n+1))/2. It will also check whether the series converges. So, the given series is an arithmetic progression whose common difference is d = 2. An arithmetic sequence is one in which the difference between successive members is a constant. demonstration: form input vectors for 10, ten thousand, and 1*10^(many) 0 10 3 0 10 5 0 10 15 0 10000. We therefore derive the general formula for evaluating a finite arithmetic series. The limit of a sequence is the value the sequence approaches as the number of terms goes to infinity. Geometric Progression, Series & Sums Introduction. By applying this calculator for Arithmetic & Geometric Sequences, the n-th term and the sum of the first n terms in a sequence can be accurately obtained. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. The T-junction is a novel type of separator used in the petroleum and gas industry. What two things do you need to know to find the sum of an infinite geometric series? Find the sum of the infinite geometric series. asked by ryan on March 15, 2017; Algebra. Explain why or why not. Please help. The two sequences are placed in two consecu- tive rows. The sums we have looked at so far are finite sums with finite upper and lower limits. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An]. Thanks for your help. You need to know both of these numbers in order to calculate the sum of the arithmetic sequence. 1 - Enter the first term A1 in the sequence, the common ratio r and n n the number of terms in the sum then press enter. Hope this helps!. S = 10 1 2, a 1= 1 2 Write. +N^2 in C Programming Language. Such series appear in many areas of modern mathematics. This formula shows that a constant factor in the summands can be taken out of the sum. Free Summation Calculator. However, when the series has an infinite number of terms the summation is more complicated and the series may or may not have a finite. Let the variable equal the first term in the sequence, and equal the last term in the sequence. Observe that for the geometric series to converge, we need that $$|r| 1$$. Excel's Calculation Process. Any sequence that has a common second difference is a quadratic sequence. So the second term common to both sequences is 1+28 = 29. arithmetic series. asked by ryan on March 15, 2017; Algebra. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. In the opposite case, one should pay the attention to the \u00abSeries convergence test\u00bb pod. Sum of series in C language 1\u00b2 + 2\u00b2 + 3\u00b2 + 4\u00b2 + 5\u00b2 +\u2026. Linear sequences A number pattern which increases (or decreases) by the same amount each time is called a linear sequence. Sequences and summations CS 441 Discrete mathematics for CS M. s n = 2 \u2212 3(0. Get comfortable with sequences in general, and learn what arithmetic sequences are. com Task : To find the sum of all the elements in a list. Calculates the 8-bit checksum for a sequence of hexadecimal bytes. On our site OnSolver. I understand the process in calculating a simple series like this one $$\\\\sum_{n=0}^4 2n$$ but I do not understand the steps to calculate a series like this one $$\\\\sum_{n=1}^x n^2$$ I have an aw. Note that a series is an indicated sum of the terms of a sequence!! In this section, we work only with finite series and the related sums. Learn Java by Examples: Write a program to calculate and print the sum of the following series: Sum(x) = 2 \u2013 4 + 6 \u2013 8 + 10 - 12 \u2026 - 100Learn Java by examples. See full list on mathsisfun. The free tool below will allow you to calculate the summation of an expression. S = 12, a 1= 2 8. This Arithmetic Sequence Calculator is used to calculate the nth term and the sum of the first n terms of an arithmetic sequence. First simplify the expression, X! receives great quick! x/x! = x/(x * (x-a million) * (x-2) *a million) hence x/x! = a million/(x-a million)! for x >= 2 and a million for x =a million Then note that that's the same to the enlargement of the Taylor series for e^x as a million + x + x^2/2! + x^3/3! +x^4/4! for x = a million. The sequence of partial sums of a series sometimes tends to a real limit. If you're behind a web filter, please make sure that the domains *. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. This is a free javascript calculator. Leonhard Euler continued this study and in the process solved many. \\$1000000007\\$ is a prime number, so division by \\$2^L\\$ modulo 1000000007 is a multiplication by a multiplicative inverse of 2 modulo 1000000007 (which is trivial to find) to the same power. 99805 Time Complexity: O(n). In general, one does not expect to be able to calculate an infinite sum exactly. As the top row increases, the bottom row decreases, so the sum stays the same. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. The T-junction is a novel type of separator used in the petroleum and gas industry. Sum definition is - an indefinite or specified amount of money. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. The first sum in the numerator is the sum of squared residuals for the first model (e. It is used to achieve the gas\u2013liquid or liquid\u2013liquid two-phase separation. On a higher level, if we assess a succession of numbers, x 1 , x 2 , x 3 ,. This document explains how to calculate the sum or total when working with cell data located in multiple worksheets. Explain why or why not. Finding a general expression for a partial sum by induction and then finding the limit of this partial sum is a perfectly valid technique. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Improve your math knowledge with free questions in \"Find the sum of an arithmetic series\" and thousands of other math skills. Get comfortable with sequences in general, and learn what arithmetic sequences are. var ([ddof]) Calculate unbiased window variance. 2, ALPHA ([I], 1, 1 0) EXE, and you should get an answer of 9. The z-transform of the unit pulse, = 1. Learn Java by Examples: Write a program to calculate and print the sum of the following series: Sum(x) = x/2 + x/5 + x/8 + \u2026 + x/100. BYJU'S online infinite geometric series calculator tool makes the calculation faster, and it displays the sum in a fraction of seconds. Sequences and summations CS 441 Discrete mathematics for CS M. Geometric Progression (G. Sum of Natural, Odd or Even Numbers Series Calculator getcalc. Java program to calculate the sum of GP series. Just enter the expression to the right of the summation symbol (capital sigma, \u03a3) and then the appropriate ranges above and below the symbol, like the example provided. This summation notation calculator can sum up many types of sequencies including the well known arithmetic and geometric sequencies, so it can help you to find the terms including the nth term as well as the sum of the first n terms of virtualy any series. Series (Find the sum) A finite Geometric Series (a limited number of terms, or Partial Sum) An infinite Geometric Series, if our infinite series is. Solving mathematical problems online for free. help me how to calculate the sum of a series in Matlab. The sums we have looked at so far are finite sums with finite upper and lower limits. To evaluate the Midpoint Sum, it gets messier. For some use cases, there are good alternatives to sum(). The sum of the members of a finite arithmetic progression is called an arithmetic series. Sn = (n/2) [a + l] Sn = (n/2) [2a + (n - 1)d] a = first term, n = number of terms of the series, d = common difference and l = last term. 0 / k for k in range(1, 10001)) What this code does: the innermost part is a generator expression, which computes the elements of a series 'on the fly'. Example 1: Find the sum of squares of the numbers from 0 to 5000. For example, in the sequence 10, 15, 20, 25, 30. Observe that for the geometric series to converge, we need that $$|r| 1$$. So again, a problem about earned interest might not be a perfect example, since you can withdraw your money at any instant and not only at whole number year values. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP\ud83d\ude4f This problem has been solved!. 1 26) a 12 = 28. Arithmetic Sequence. \\) The series becomes $$\\sum\\limits_{n = 0}^\\infty {\\large\\frac{{{u^n}}}{{n!}} ormalsize}. More Practice Problems with Arithmetic Sequence Formula Direction: Read each arithmetic sequence question carefully, then answer with supporting details. Given real (or complex!) numbers aand r, X1 n=0 arn= (a 1 r if jr <1 divergent otherwise The mnemonic for the sum of a geometric series is that it\u2019s \\the rst term divided by one minus the common ratio. Another approach could be to use a trigonometric identity. The sums we have looked at so far are finite sums with finite upper and lower limits. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson arithmetic sequence. the starting principal you'll need to achieve the payouts desired:. org are unblocked. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. In column B, create the sequence that you are going to sum and name it seqa. A sequence is a list of numbers. After discussing the above two examples one will wonder if any sequence has the same faith (meaning, it gets closer to a number). Free Summation Calculator. [Note that the sequence of the sums S 1 (the sum of the first term), S 2 (the sum of the first two terms), S 3 (the sum of the first three terms),. The present value, PV, of a series of cash flows is the present value, at time 0, of the sum of the present values of all cash flows, CF. To determine if the series is convergent we first need to get our hands on a formula for the general term in the sequence of partial sums. Only this variable may occur in the sequence term. is 16 and sum of the next 3 terms is 128. The denominator doesn't change. This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. How to use sum in a sentence. Because 1 is paired with 10 (our n), we can say that each column has (n+1). Alex's Arithmetic Sequence Sum Calculator is a very simple program, which allows you to get the sum of an Arithmetic Squence, it supports two types of sequences: \uffed type 1 - if you have the first. When the denominators are the same, add the numerators, or top numbers, to calculate the answer. arithmetic series. In column A, create the sequence 1, 2, 3,. ID 1001746465 1 1 1 1 23 4+5 1 There are several ways to implement this. Sum of Three Consecutive Integers calculator. I hope it is enough to get you going in the right direction. The sum of geometric series refers to the total of a given geometric sequence up to a specific point and you can calculate this using the geometric sequence solver or the geometric series calculator. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. My dxp and csv file is uploaded in below. If you are struggling to understand what a geometric sequences is, don't fret! We will explain what this means in more simple terms later on and take a look at. Limit Comparison Test If lim (n-->) (a n / b n) = L, where a n, b n > 0 and L is finite and positive,. This value is equal to:. Solution: Given decimal we can write as the sum of 0. Leonhard Euler was able to calculate the exact sum of the p-series with p 2: 2-2 32 42 Use this fact to find the sum of each series: 2 32 so. Notice \\t is used to provide 8 spaces (1 tab) between two values see in the output. It should be noted, that if the calculator finds sum of the series and this value is the finity number, than this series converged. hi i just want to confirm my method of using sum of pairs is correct. Because the Fibonacci value for 20000 has 4179 decimals and it needs quite an impressive amount of processing, the maximum allowed value is 20000. 0 / k for k in range(1, 10001)) What this code does: the innermost part is a generator expression, which computes the elements of a series 'on the fly'. As running variable, which is increased by 1 in each step, i is used. Excel does not calculate cells in a fixed order, or by Row or Column. We used a sum-over-states method to calculate the dynamic polarizabilities of 6S 1/2 ground state and highly-excited (nS 1/2 and nP 3/2 ) Rydberg state of cesium atoms, and identify corresponding magic detuning for optical wavelengths in the range of 850-1950 nm. Z Transforms of Common Sequences Ele 541 Electronic Testing Unit Pulse. We call an a term of the sequence. If you're seeing this message, it means we're having trouble loading external resources on our website. Representations of N as a sum of distinct elements from special sequences D. What Is Arithmetic Sequence? is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. If the series has a finite number of terms, it is a simple matter to find the sum of the series by adding the terms. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. this, use your calculator and examine high powers of numbers between 1 and -1. A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. The Corbettmaths video tutorial on finding the nth term for a fractional sequence. A series, which is not a list of terms like a sequence, is the sum of the terms in a sequence. A finite number of terms of an arithmetic sequence can be added to find their sum. Often the first numbers will be 1, but not always. If S n tends to a limit as n tends to infinity, the limit is called the sum to infinity of the series. But there are some series. I evaluated the partial sum through a calculator, my answer was -19/30, or -. Exclude NA/null. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. How to Find Sum of First n Terms When nth Term is Given ? A series whose terms are in Arithmetic progression is called Arithmetic series. Sum of this series is calculated by the formula : sum=(n*(n+1))/2. The Fibonacci sequence is named after Italian mathematician Leonardo of Pisa, known as Fibonacci. ID 1001746465 1 1 1 1 23 4+5 1 There are several ways to implement this. Work out: a. day, sum((random()*5)::integer) num FROM days -- left join other tables here to get counts, I'm using random group by days. The Sum (Summation) Calculator is used to calculate the total summation of any set of numbers. P Class 11 Engineering - Sum of n Terms of G. But there are some series. Calculate sum elements of sequence: sum. Doing so means that most of the depreciation associated with an asset is recognized in the first few years of its useful life. All term in the sequence meet a specific logical rule which needs to be recognised in order to find the missing terms. Thus the message becomes: Since we are using a 3 by 3 matrix, we break the enumerated message above into a sequence of 3 by 1. The first term of an infinite G. After all, yes 1/(1-x) has an honest-to-goodness explosion to infinity at x=1, but it makes perfectly good sense at x=-1, and tells us (what my calc students. Sum of elements in a list \u2014 programminginpython. Series (Find the sum) When you know the first and last term. [ Don't peek. Recently, mandatory vote-by-mail has received a great deal of attention as a means of administering elections in the United States. Write a program in C++ to calculate the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + + (n*n). 02 Line 10 Column 42 Pic zz9 From N Highlight. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An]. Number sequences questions usually consist of four to seven visible numbers along with a single missing number or, depending on the sequence's complexity level, 2 or 3 missing numbers. An infinite series has an infinite number of terms. Number sequences test practice for aptitude tests and psychometric IQ tests. The sum of a sequence is known as a series, and the harmonic series is an example of an infinite series that does not converge to any limit. ${s_n} = \\sum\\limits_{i = 1}^n i$. Sum of: from: to: Submit: Share a link to this widget: More. Solution: A series in which each number is sum of its previous two numbers is known as Fibonacci series. mean (*args, **kwargs) Calculate the window mean of the values. 23) a 21 = \u22121. The program solves Riemann sums using one of four methods and displays a graph when prompted. Fourth, we recall the sum of the X 2 and subtract 240. ID 1001746465 1 1 1 1 23 4+5 1 There are several ways to implement this. To be a good Java developer is to be fluent in Java 8. Sum of this series is calculated by the formula : sum=(n*(n+1))/2. Just enter the expression to the right of the summation symbol (capital sigma, \u03a3) and then the appropriate ranges above and below the symbol, like the example provided. The nineteenth term. 2, ALPHA ([I], 1, 1 0) EXE, and you should get an answer of 9. An arithmetic sequence is one in which the difference between successive members is a constant. To use this calculator, simply type in your list of inputs separated by commas (ie 2,5,8,10,12,18). Print the sum. Sum of sequence calculator. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Geometric Progression (G. asked by ryan on March 15, 2017; Algebra. For the finite sums series calculator computes the answer quite literally, so if there is a necessity to obtain a short expression we recommend computing a parameterized sum. This website uses cookies to ensure you get the best experience. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. I have made this small program to help me add numbers and find the sum, it runs but no output. The nth term of a sequence is given by U n = n 2 /(n + 1). 02 Line 10 Column 58 Pic zz. Everything you want to know about Java. I know the inequality of [integral from n+1 to infinity of An] < [Total sum - Partial sum] < [integral from n to infinity of An]. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson arithmetic sequence. P Class 11 Engineering - Sum of n Terms of G. Write a program in C++ to calculate the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + + (n*n). The sequence of these partial sums converges to also. 78 degrees Fahrenheit. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. com allows you to find the sum of a series online. For help creating a sequence, see Note 1I. If the range of a sum is finite, is typically assigned a sequence of values, with being evaluated for each one. Third, we square the sum of X (45 times itself = 2025) and divide it by N (number of scores). A finite number of terms of an arithmetic sequence can be added to find their sum. The 8-bit checksum is the 2's complement of the sum off all bytes. The first of the examples provided above is the sum of seven whole numbers, while the latter is the sum of the first seven square numbers. 01 then we use the formula for the sum of the infinite geometric series S oo = a 1 / (1 - r),. Such series appear in many areas of modern mathematics. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. #sum_(k=1)^nk=(n(n+1))/2# and. The amount it increases or decreases by is known as the common difference. Add the numbers and divide by (n - 1) = 6 to get 95. \"Mike\" wrote: > I have a series of cashflows, forecast to grow at say 2% each year and > go on indefinitely. Objects might be numbers or letters. Using the nth term. Sum of a Series: If we are able to find the general term of the series successfully, then we can also. Work out: a. Which of the pairs of events below is dependent? Select the correct answer below: drawing a 7 and then drawing another 7 with replacement from a standard deck of cards rolling a 1 and then rolling a 6 with a standard die rolling a 3 and then rolling a 4 with. The sequence of these partial sums converges to also. Thus the message becomes: Since we are using a 3 by 3 matrix, we break the enumerated message above into a sequence of 3 by 1. An example of the sequence can be seen as follows:. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. BYJU\u2019S online infinite series calculator tool makes the calculations faster and easier where it displays the value in a fraction of seconds. It will also check whether the series converges. The numerator above is a difference of two partial sums of the \\2^n*a_{n}\\ sequence. The two sequences are placed in two consecu- tive rows. This technique causes problems in several situations, however and cannot be universally relied upon. Using Wolfram Alpha 214 , put in 10 x x and you will get can be reduced to the much easier-to-calculate infinite sum 1 - x - x2 Series - A series is formed by the sum or addition of the terms in a two equations that allow easy calculation for an arithmetic series. The numerator above is a difference of two partial sums of the \\2^n*a_{n}\\ sequence. More Practice Problems with Arithmetic Sequence Formula Direction: Read each arithmetic sequence question carefully, then answer with supporting details. day, sum((random()*5)::integer) num FROM days -- left join other tables here to get counts, I'm using random group by days. For example, two-thirds plus four-thirds equals six-thirds. Average = Sum of terms / Number of terms. is 16 and sum of the next 3 terms is 128. Methods for Evaluating In nite Series Charles Martin March 23, 2010 Geometric Series The simplest in nite series is the geometric series. The sum of the first three terms of the sequence of terms common to the two given sequences is 3 times the second term of the sequence. The sum of geometric series refers to the total of a given geometric sequence up to a specific point and you can calculate this using the geometric sequence solver or the geometric series calculator. interval is [FIRST, LAST) NB. The sum of the members of a finite arithmetic progression is called an arithmetic series. Write a program in C++ to calculate the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + + (n*n). Alternatively \u2013 put the first few terms of your sequence into WolframAlpha. Sample Output: Input The Value For Nth Term: 5 1*1 = 1 2*2 = 4 3*3 = 9 4*4 = 16 5*5 = 25 The Sum Of The Above Series Is: 55 ASAP\ud83d\ude4f This problem has been solved!. Next: Write a program in C++ to find the sum of series 1 - X^2/2! + X^4/4!- upto nth term. That's numberwang!. Our summation calculator can easily calculate the sum of any numbers you input. Number sequences test practice for aptitude tests and psychometric IQ tests. sin105 degrees B. the starting principal you'll need to achieve the payouts desired:. , x k , we can record the sum of these numbers in the following way:. Examples: SAS Statements. Lets say i have three sequences. Notice, that if x = 1 then, in the series, we are simply addi ng up an infinite number of 1's and of course the sum goes to infinity. Please help. An infinite series has an infinite number of terms. Beware: people often confuse the terms \u2018sequence\u2019 and \u2018series\u2019. An Efficient solution to solve the sum of geometric series where first term is a and common ration is r is by the formula :-sum of series = a(1 \u2013 r n)/(1 \u2013 r). Enter 2 numbers to add and press the = button to get the sum result. #a_n = (3/2)^n# Which means that #n#-th term is generates by raising #3/2# to the #n#-th power. Find the series. Sequences and summations CS 441 Discrete mathematics for CS M. [Note that the sequence of the sums S 1 (the sum of the first term), S 2 (the sum of the first two terms), S 3 (the sum of the first three terms),. That's numberwang!. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. +N^2 in C Programming Language. In other words, if you keep adding together the terms of the sequence forever, you will get a finite value. Products Classroom Activities Graphing Calculator Scientific Calculator Four Function Calculator Matrix Calculator Test Practice Geometry Tool. In other words, how to take the value of a cell located in one worksheet and add it to the value of another cell located in another worksheet to come up with the total of the respective cells. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. sum_arithmetic_series is more general than required. what could be the problem? // Program that uses a switch statement. If this happens, we say that this limit is the sum of the series. P Class 11 Engineering - Sum of n Terms of G. All you have to do is write the first term number in the first box, the second term number in the second box, third term number in the third box and the write value of n in the fourth box after that you just have to click on the Calculate button, your result will be visible. Enter the sequence, the start value and end value from sigma notation and get a numerical sum. To add floating point values with extended precision, see math. What is the difficulty level of this exercise? Easy Medium Hard. About Sum (Summation) Calculator. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Calculate the sum of first n squares or the sum of consecutive square numbers from n 1 2 to n 2 2. A series is a special type of sequence: a. Notice, that if x = 1 then, in the series, we are simply addi ng up an infinite number of 1's and of course the sum goes to infinity. Calculate sum elements of sequence: sum. Thanks for your help. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. Bob has a need to use the SUM function in a macro in order to find the sum of all the values in a column. The sequence of partial sums of a series sometimes tends to a real limit. var ([ddof]) Calculate unbiased window variance. 23) a 21 = \u22121. The formula for the n-th term of a quadratic sequence is explained here. For example, some series don't sum monotonically to a limit -- the sum will bounce around it, and the summation's state can end up oscillatory forever. 02 Line 10 Column 42 Pic zz9 From N Highlight. skipna bool, default True. Third, we square the sum of X (45 times itself = 2025) and divide it by N (number of scores). Free Summation Calculator. +N^2 in C Programming Language. An example of the sequence can be seen as follows:. Instructions: Use this step-by-step Geometric Series Calculator, to compute the sum of an infinite geometric series by providing the initial term \\(a$$ and the constant ratio $$r$$. Our sum of series calculator or arithmetic series calculator is an online tool which you can find on Google. Although SUM is specified as taking a maximum of 30 arguments, Google Sheets supports an arbitrary number of arguments for this function. About this calculator. (for example 2 series in the picture) 1 Comment. Free Arithmetic Sequences calculator - Find indices, sums and common difference step-by-step This website uses cookies to ensure you get the best experience. The sum function can be used as a series calculator, to calculate the sequence of partial sums of a series. Set Number = 0. The sums we have looked at so far are finite sums with finite upper and lower limits. The Arithmetic series of finite number is the addition of numbers and the sequence that is generally followed include \u2013 (a, a + d, a + 2d, \u2026. day, sum((random()*5)::integer) num FROM days -- left join other tables here to get counts, I'm using random group by days. #sum_(k=1)^nb=nb# So. For example, the sum given by, means to sum an infinite number of terms as, The value of an infinite sum may be \u221e (in this case the sum is infinite). Definition: Arithmetic sequence is a list of numbers where each number is equal to the previous number, plus a constant. Using the while loop to calculate sum : While Loop \u00ab Statement Control \u00ab Java Tutorial. To add floating point values with extended precision, see math. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. I can''t seem to find one of those. Build your own widget. A harmonic sequence is a sequence of numbers whose reciprocals form an arithmetic sequence. Here is the complete Java program with sample outputs. An Efficient solution to solve the sum of geometric series where first term is a and common ration is r is by the formula :-sum of series = a(1 \u2013 r n)/(1 \u2013 r). Let, t n be the n th term of AP, then (n+1) th term of can be calculated as (n+1) th = t n + D where D is the common difference (n+1) th - t n The formula to calculate N th term t n = a + (n \u2013 1)d; where, a is first term of AP and d is the common difference. If Sum > Limit, terminate the repitition, otherwise. In the sequence a sub 1 (first term), a sub 2 (second term), a sub 3 (third term)\u2026. Sum uses the standard Wolfram Language iteration specification. The last number n is input by the user. \u03a3 is the symbol used to denote sum. Infinite Geometric Series Calculator is a free online tool that displays the sum of the infinite geometric sequence. Increment Number by one. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. 63333333 I'm having difficulty estimating how far this partial sum is away from the total sum. The blog talks about variety of topics on Embedded System, 8085 microprocessor, 8051 microcontroller, ARM Architecture, C2000 Architecture, C28x, AVR and many many more. Instructions: Use this step-by-step Geometric Series Calculator, to compute the sum of an infinite geometric series by providing the initial term $$a$$ and the constant ratio $$r$$. Arithmetic Sequence. After learning so much about development in Python, I thought this article would be interesting for readers and to myself\u2026 This is about 5 different ways of calculating Fibonacci numbers in Python [sourcecode language=\u201dpython\u201d] ## Example 1: Using looping technique def fib(n): a,b = 1,1 for i in range(n-1): a,b = b,a+b return a print \u2026 Continue reading 5 Ways of Fibonacci in Python \u2192. The 'nth' term is a formula with 'n' in it which enables you to find any term of a sequence without having to go up from one term to the next. 2, ALPHA ([I], 1, 1 0) EXE, and you should get an answer of 9. A sequence is a numbered list of values, produced by the calculation of a formula. C Program to calculate sum of 5 subjects and find percentage [crayon-5f51beda86fcb863628413/] Output : [crayon-5f51beda86fd2634443803/]. This is a free javascript calculator. The Fibonnacci numbers are also known as the Fibonacci series. Sequence Calculator, Sequence Examples. Improve your math knowledge with free questions in \"Find the sum of an arithmetic series\" and thousands of other math skills. Finding a general expression for a partial sum by induction and then finding the limit of this partial sum is a perfectly valid technique. How to find the sum of a finite Arithmetic Series! s n = n(t 1 + t n)/2 To find the sum of a finite arithmetic series, you need to know three things. Without using loops, calculate the sum of the following series for the first n terms. You may want to review the basics of geometric sequences or finding formulas. Notice, that if x = 1 then, in the series, we are simply addi ng up an infinite number of 1's and of course the sum goes to infinity. A good sequence to start with is the Fibonacci sequence. Repeat the following: a. About this calculator. sum_arithmetic_series FIRST LAST SKIP NB. #sum_(k=1)^nk=(n(n+1))/2# and. Definition: Geometric sequence is a list of numbers where each term is obtained by multiplying the previous term by a constant. How do we apply these useful rules to this question? First, calculate the average of the first and. This formula shows that a constant factor in the summands can be taken out of the sum. I have made this small program to help me add numbers and find the sum, it runs but no output. 2) \u2227 2 \u00d7 0. LED series current limiting resistor calculator - useful when designing circuits with a single LED or series/parallel LED arrays - for both the common small-current (20mA) LEDs and the more expensive, high power LEDs with currents up to a few Amperes. #sum_(k=1)^n(ak+b)# where a and b are constants. Free Summation Calculator. If S n tends to a limit as n tends to infinity, the limit is called the sum to infinity of the series. To add floating point values with extended precision, see math. 0 / k if k % 2 else -1. Fibonacci sequence formula; Golden ratio convergence; Fibonacci sequence table; Fibonacci sequence calculator; C++ code of Fibonacci function; Fibonacci sequence formula. 78 degrees Fahrenheit. Click the link for more information. The n-th partial sum of a series is the sum of the \ufb01rst n terms. asked by ryan on March 15, 2017; Algebra. White Rose Maths has prepared a series of Maths lessons online for Year 4, FREE videos and worksheets. This calculator computes n-th term and sum of geometric progression person_outline Timur schedule 2011-07-16 04:17:35 Geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. We start with the general formula for an arithmetic sequence of $$n$$ terms and sum it from the first term ($$a$$) to the last term in the sequence ($$l$$):. Explain why or why not. This is a very versatile calculator that will output sequences and allow you to calculate the sum of a sequence between a starting item and an n-th term, as well as tell you the value of the n-th term of interest. Calculate the rolling quantile. Since an arithmetic sequence always has an unbounded long-term behavior, we are always restricted to adding a finite number of terms. For this reason, Sum evaluates to null instead of to zero for an empty sequence or for a sequence that contains only nulls. So 355 minus 289. If the above series converges, then the remainder R N = S - S N (where S is the exact sum of the infinite series and S N is the sum of the first N terms of the series) is bounded by 0< = R N <= (N. Limits capture the long-term behavior of a sequence and are thus very useful in bounding them. The sum to infinity of a geometric progression. This version provided Java with a functional aspect by introducing concepts such as functional interfaces, lambda expressions, streams, etc. Sum of 32-bit integer quantities is not computed by using 64-bit results, and overflow can occur for the LINQ to SQL translation of Sum. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. NPV = Sum CF* ((1+2%)/(1+D))^N When N is infinite, after simplification, NPV = CF * 1 / (1 - r) where r = (1+2%)/(1+D)--A+ V. Solving mathematical problems online for free. Since there's two equally likely options, you'd expect a run to last for two flips. See full list on gigacalculator. Sum of sequence calculator. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big. Therefore, the equation could be 4n. A geometric sequence refers to a sequence wherein each of the numbers is the previous number multiplied by a constant value or the common ratio. Calculate the sum of special series or number of sequences. Sum of Arithmetic Sequence Formula. com Task : To find the sum of all the elements in a list. Leonhard Euler continued this study and in the process solved many. Note that a series is an indicated sum of the terms of a sequence!! In this section, we work only with finite series and the related sums. Such series appear in many areas of modern mathematics. In the previous section we started looking at writing down a power series representation of a function. Geometric Sequence. The free tool below will allow you to calculate the summation of an expression. To determine if the series is convergent we first need to get our hands on a formula for the general term in the sequence of partial sums. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson Geometric sequence. skipna bool, default True. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Store the result in an array. Perform addition to find the sum of two or more fractions. 8 Given two terms in an arithmetic sequence find the recursive formula. This calculator will find the infinite sum of arithmetic, geometric, power, and binomial series, as well as the partial sum, with steps shown (if possible). Hi everyone, the picture below is the sum of a series that have infinity limit, and i dont know the suitable function to deal with the problem. A sequence is a list of numbers, geometric shapes or other objects, that follow a specific pattern. For this reason, Sum evaluates to null instead of to zero for an empty sequence or for a sequence that contains only nulls. \\$1000000007\\$ is a prime number, so division by \\$2^L\\$ modulo 1000000007 is a multiplication by a multiplicative inverse of 2 modulo 1000000007 (which is trivial to find) to the same power. Calculate the sum of the series \u2211 n = 1 \u221e a n whose partial sums are given. However, policy-makers disagree on the merits of this approach. Denote this partial sum by S n. An example of the sequence can be seen as follows:. Java program to calculate the sum of GP series. Fourth, we recall the sum of the X 2 and subtract 240. Work out: a. Online adding calculator. Sum of the series 1^1 + 2^2 + 3^3 + \u2026. You may see the formula written as: Sum, S n, of n terms of an arithmetic series. The amount it increases or decreases by is known as the common difference. day ) select day, num, sum(num) over. +n and prints the result on the compiler screen. Two consecutive numbers in this series are in a ' Golden Ratio '. Flowchart. com allows you to find the sum of a series online. The Sum (Summation) Calculator is used to calculate the total summation of any set of numbers. 63333333 I'm having difficulty estimating how far this partial sum is away from the total sum. Such an argument was given by Nicolas Oresme (1323 - 1382 A. So the second term common to both sequences is 1+28 = 29. When the limit of partial sums exists, it is called the value (or sum) of the series. Dick and I both used tricks. It is capable of computing sums over finite, infinite and parameterized sequences. First simplify the expression, X! receives great quick! x/x! = x/(x * (x-a million) * (x-2) *a million) hence x/x! = a million/(x-a million)! for x >= 2 and a million for x =a million Then note that that's the same to the enlargement of the Taylor series for e^x as a million + x + x^2/2! + x^3/3! +x^4/4! for x = a million. You can also use this arithmetic sequence calculator as an arithmetic series calculator. If you are struggling to understand what a geometric sequences is, don't fret! We will explain what this means in more simple terms later on and take a look at. Here s how Wolfram Alpha solved it. In an Arithmetic Sequence the difference between one term and the next is a constant. Now, how do we determine the average of a sequence of integers? Rule #2: The average of a sequence of integers is the average of the first and last terms Applying the rules to find the sum of the sequence. #sum_(k=1)^n(ak+b)=sum_(k=1)^nak+sum_(k=1)^nb# We can factor the #a#. You need to know both of these numbers in order to calculate the sum of the arithmetic sequence. 2) \u2227 2 \u00d7 0. MATH 225N Week 4 Probability Questions and answers \u2013 Chamberlain College of Nursing Week 4 Homework Questions Probability 1. Only this variable may occur in the sequence term. Here s how Wolfram Alpha solved it. What Is Arithmetic Sequence? is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. day ) select day, num, sum(num) over. Find the sum of first n terms of the series (i) 3 +33 +333 +. Free Arithmetic Sequences calculator - Find indices, sums and common difference step-by-step This website uses cookies to ensure you get the best experience. The sequence of these partial sums converges to also. For example, in the sequence 10, 15, 20, 25, 30. The denominator doesn't change. Calculate the rolling quantile. Thus the message becomes: Since we are using a 3 by 3 matrix, we break the enumerated message above into a sequence of 3 by 1. What two things do you need to know to find the sum of an infinite geometric series? Find the sum of the infinite geometric series. Bytes are provided as two-character strings. To find sum of your series, you need to choose the series variable, lower and upper bounds and also input the expression for n-th term of the series. Sum of Three Consecutive Integers calculator. Class 11 Commerce - Sum of terms of G. 2) \u2227 2 \u00d7 0. Let\u2019s explore how we do that. Calculate the sum of series 1^2+2^2+3^2+4^2++n^2 (n>0) using the both for and while loop structure. Class 11 Commerce - Sum of terms of G. #sum_(k=1)^nk=(n(n+1))/2# and. On a higher level, if we assess a succession of numbers, x 1 , x 2 , x 3 ,. 63333333 I'm having difficulty estimating how far this partial sum is away from the total sum. If the value of the sum (in the limiting sense) exists, then they say that the series. After learning so much about development in Python, I thought this article would be interesting for readers and to myself\u2026 This is about 5 different ways of calculating Fibonacci numbers in Python [sourcecode language=\u201dpython\u201d] ## Example 1: Using looping technique def fib(n): a,b = 1,1 for i in range(n-1): a,b = b,a+b return a print \u2026 Continue reading 5 Ways of Fibonacci in Python \u2192. Calculators with newer operating systems actually have a summation function, which can by reached by pressing MATH 0. n must be a positive integer. In addition, when the calculator fails to find series sum is the strong indication that this series is divergent (the calculator prints the message like \"sum diverges\"), so our calculator also indirectly helps to get information about series convergence. In this case, a number of columns equal to the number of elements of the sequence x[n] should be skipped in order to accommo- date the results of the multiplication. Essentially the total variability in your dataset is the same as before, but now you have two different models to consider. Question: Find the sum of (1/2) +(1/6) + (1/12) + + (1/9900) without using a calculator. Free Summation Calculator. Fibonacci calculator The tool calculates F(n) - Fibonacci value for the given number, as well as the previous 4 values, using those to display a visual representation. The full sequence of keypresses to evaluate the Right-Hand Sum is: OPTN F4 [CALC] F6 F3 [\u03a3(] (1 + ALPHA ([I] \u00d7 0. This lesson assumes that you know about geometric sequences, how to find the common ratio and how to find an explicit formula. Some sources neglect the initial 0, and instead beginning the sequence with the first two ones. A geometric sequence refers to a sequence wherein each of the numbers is the previous number multiplied by a constant value or the common ratio. is 1 and any term is equal to the sum of all the succeeding terms. Unfortunately, the answer is NO. First simplify the expression, X! receives great quick! x/x! = x/(x * (x-a million) * (x-2) *a million) hence x/x! = a million/(x-a million)! for x >= 2 and a million for x =a million Then note that that's the same to the enlargement of the Taylor series for e^x as a million + x + x^2/2! + x^3/3! +x^4/4! for x = a million. This relationship of examining a series forward and backward to determine the value of a series works for any arithmetic series. Each number in series is called as Fibonacci number.", "date": "2020-11-29 16:20:11", "meta": {"domain": "audio-upgrade.pl", "url": "http://audio-upgrade.pl/sum-of-sequence-calculator.html", "openwebmath_score": 0.7158166170120239, "openwebmath_perplexity": 412.81979783464897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9830850906978876, "lm_q2_score": 0.8723473730188543, "lm_q1q2_score": 0.8575916963243044}}
{"url": "https://atekihcan.github.io/CLRS/02/P02-01/", "text": "Insertion Sort on Small Arrays in Merge Sort\n\nAlthough merge sort runs in $$\\Theta(n \\lg n)$$ worst-case time and insertion sort runs in $$\\Theta(n^2)$$ worst-case time, the constant factors in insertion sort can make it faster in practice for small problem sizes on many machines. Thus, it makes sense to coarsen the leaves of the recursion by using insertion sort within merge sort when subproblems become sufficiently small. Consider a modification to merge sort in which $$n/k$$ sublists of length $$k$$ are sorted using insertion sort and then merged using the standard merging mechanism, where $$k$$ is a value to be determined.\n\n1. Show that insertion sort can sort the $$n/k$$ sublists, each of length $$k$$, in $$\\Theta(n \\lg (n/k))$$ worst-case time.\n2. Show how to merge the sublists in $$\\Theta(n \\lg(n/k))$$ worst-case time.\n3. Given that the modified algorithm runs in $$\\Theta(nk + n \\lg(n/k))$$ worst-case time, what is the largest value of k as a function of n for which the modified algorithm has the same running time as standard merge sort, in terms of $$\\Theta$$ notation?\n4. How should we choose $$k$$ in practice?\n\n#### Sorting Sublists\n\nFor input of size $$k$$, insertion sort runs on $$\\Theta(k^2)$$ worst-case time. So, worst-case time to sort $$n/k$$ sublists, each of length $$k$$, will be $$n/k \\cdot \\Theta(k^2) = \\Theta(nk)$$\n\n#### Merging Sublists\n\nWe have $$n$$ elements divided into $$n/k$$ sorted sublists each of length $$k$$. To merge these $$n/k$$ sorted sublists to get a single sorted list of length $$n$$, we have to take 2 sublists at a time and continue to merge them.\n\nThis will result in $$\\lg (n/k)$$ steps (refer to Figure 2.5 in page 38 of the chapter text). And in every step, we are essentially going to compare $$n$$ elements. So the whole process will run at $$\\Theta(n \\lg (n/k))$$.\n\n#### Largest Value of k\n\nFor the modified algorithm to have the same asymptotic running time as standard merge sort, $$\\Theta(nk + n \\lg(n/k))$$ must be same as $$\\Theta(n \\lg n)$$.\n\nTo satisfy this condition, $$k$$ cannot grow faster than $$\\lg n$$ asymptotically, if it does then because of the $$nk$$ term, the algorithm will run at worse asymptotic time than $$\\Theta(n \\lg n)$$).\n\nSo, let\u2019s assume, $$k = \\Theta(\\lg n)$$ and see if we can meet the criteria \u2026\n\n\\begin {aligned} \\Theta(nk + n \\lg(n/k)) & = \\Theta(nk + n \\lg n - n \\lg k) \\\\ & = \\Theta(n \\lg n + n \\lg n - n \\lg (\\lg n)) \\\\ & = \\Theta(2n \\lg n - n \\lg (\\lg n)) ^\\dagger \\\\ & = \\Theta(n \\lg n) \\end {aligned}\n\n$$^\\dagger\\lg (\\lg n)$$ is very small compared to $$\\lg n$$ for sufficiently larger values of $$n$$.\n\n#### Practical Value of k\n\nTo determine a practical value for $$k$$, it has to be the largest input size for which insertion sort runs faster than merge sort. To get exact value, we need to calculate the exact running time expressions with the constant factors and use the method described in Exercise 1.2.2.", "date": "2023-03-29 01:15:31", "meta": {"domain": "github.io", "url": "https://atekihcan.github.io/CLRS/02/P02-01/", "openwebmath_score": 0.4290030896663666, "openwebmath_perplexity": 535.6456594647129, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850847509661, "lm_q2_score": 0.872347368040789, "lm_q1q2_score": 0.8575916862426614}}
{"url": "https://math.stackexchange.com/questions/1646339/are-these-3-functions-linearly-independent-or-dependent", "text": "# Are these $3$ functions linearly independent or dependent?\n\nAre the functions $f, g, h$ given below linearly independent? If they are not linearly independent, find a nontrivial solutions to the equations below:\n\n$$f(x)=e^{2x}- \\cos(9x), \\quad g(x)=e^{2x}+ \\cos(9x), \\quad h(x)= \\cos(9x)$$\n\nMy take so far is that, I know these functions are not linearly independent (meaning this is surely dependent) by Wronskian (as the Wronskian determinant isn't equal to zero), but I have no idea how to find nontrivial solutions to this question. I think I should have an answer as follows:\n\n$$C_1(e^{2x}-\\cos(9x)) + C_2(e^{2x}+\\cos(9x)) + C_3\\cos(9x) = 0 \\qquad \\text{ where } Cs \\text{ are constant }$$\n\nCould I get some help on finding those constants? I've tried $C_1$ as $-1$, $C_2$ as $1$ and $C_3$ as $0$ but that surely cannot be the answer.\n\n\u2022 In Mathematica, Wronskian[{Exp[2 x] - Cos[9 x], Exp[2 x] + Cos[9 x], Cos[9 x]}, x] yields $0$, showing that the three functions are linearly dependent. \u2013\u00a0David G. Stork Feb 8 '16 at 18:19\n\nThose functions are a red herring. ;-)\nYou basically have the vectors $$v-w,\\quad v+w,\\quad w$$ in some vector space. These vectors belong to the subspace spanned by $v$ and $w$, which has dimension at most $2$. So a set of three vectors is necessarily linearly dependent.\nHow to find a nonzero linear combination is easy: $$a(v-w)+b(v+w)+cw=0$$ gives $$(a+b)v+(-a+b+c)w=0$$ so we can choose $b=-a$ and so $c=2a$. If we take $a=1$, we obtain $b=-1$ and $c=2$.\nYou can take $C_1 = 1, C_2 = -1$ and $C_3 = -2$.\n\u2022 I choose $C_1 = 1$ (without any reason). Then I wanted to cancel the two exponential so I had to choose $C_2 = -1$. Then I was left with $-2\\cos(9x)$ therefore I had to pick $C_3=-2$. \u2013\u00a0Onil90 Feb 8 '16 at 18:10\n$$\\begin{bmatrix} f(x)\\\\ g(x)\\\\ h(x) \\end{bmatrix}= \\begin{bmatrix} 1 & -1\\\\ 1 & 1 \\\\ 0 & 1 \\\\ \\end{bmatrix}\\cdot \\begin{bmatrix} -e^{2x}\\\\ \\cos(9x) \\end{bmatrix}$$", "date": "2020-06-04 18:32:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1646339/are-these-3-functions-linearly-independent-or-dependent", "openwebmath_score": 0.9408193826675415, "openwebmath_perplexity": 157.1558205977231, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850862376966, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8575916859083195}}
{"url": "https://cs.stackexchange.com/questions/80750/average-number-of-comparisons-in-sorted-insertion", "text": "# Average number of comparisons in sorted insertion\n\nThe questions is pretty simple: Given a sorted array of N elements, what is the average number of comparisons made in order to add a new element (let's call it x) in its correct position?\nI am using linear search for that task.\n\nSo far, I've tried to solve it this way:\n\n# of Comparisons     Probability of A[i] > x (Not sure if correct)\n1                         1/n\n2                         1/n\n3                         1/n\n4                         1/n\n...                        ...\nn                         1/n\n\n\nHence, the expected value would be:\n\n$$E[x] = \\sum_{x=1}^{n}x \\cdot Pr(X=x)$$ $$E[x] = 1 \\cdot Pr(X=1) + 2 \\cdot Pr(X=2) + 3 \\cdot Pr(X=3) + ... + n \\cdot Pr(X=n)$$\n\nUsing $$Pr(X = n) = \\frac{1}{n}$$ for all n,\n\n$$E[x] = \\frac{1}{n} \\cdot \\sum^{n}_{i=1}{i}$$\n\nand finally $$E[x] = \\frac{n+1}{2}$$\n\nStill, I'm not sure if this is the correct way to solve it, since I'm not sure about my 1/n assumption.\n\nWell, that would strongly depend on the algorithm that you use to find out where to insert the new element, and on the distribution of new elements.\n\nAssuming that you do a linear search, and the new element could go to any position with the same probability, your result is close, but not quite exact. The new element can go into one of (n + 1) positions, not n. There may be up to n comparisons needed; n comparisons are needed both if the new element goes into the first array position, and if it goes into the second array position. So the result is\n\n(1 + 2 + 3 + ... + (n-1) + n + n) / (n + 1) =\n\n= (n (n+1) / 2 + n) / (n + 1)\n\n= n (n + 3) / 2 / (n + 1)\n\n= (n + 2 - 2 / (n + 1)) / 2 \u2248 (n + 2) / 2,\n\n\nwhich is just a tiny bit more than your answer (n + 1) / 2.\n\n\u2022 Thanks a lot @gnasher729 ! I completely forgot to mention that I'd perform a linear search in order to find the position to insert the new number! \u2013\u00a0woz Sep 1 '17 at 23:29\n\u2022 I don't understand what you mean by \"n comparisons are needed both if the new element goes into the first array position, and if it goes into the second array position\". Why do we need $n$ comparisons in order to insert $x$ into the first position? Assuming I compare by $x < A[i]$, a is single comparison $x < A[0]$ is enough to insert x into the first position. \u2013\u00a0fade2black Sep 1 '17 at 23:48\n\u2022 @fade2black: Because you likely start comparing at the end of the array, which allows you to compare and move one element in the same iteration of the loop. If you start comparing at the beginning, then n comparisons are needed both if the new element is inserted just before, or just after the last element. \u2013\u00a0gnasher729 Sep 2 '17 at 22:38", "date": "2019-10-19 19:51:14", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/80750/average-number-of-comparisons-in-sorted-insertion", "openwebmath_score": 0.5344536900520325, "openwebmath_perplexity": 345.4364866844114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850867332734, "lm_q2_score": 0.8723473647220787, "lm_q1q2_score": 0.8575916847093473}}
{"url": "https://math.stackexchange.com/questions/2317625/show-which-of-6-2-sqrt3-and-3-sqrt2-2-is-greater-without-using-calculato", "text": "# Show which of $6-2\\sqrt{3}$ and $3\\sqrt{2}-2$ is greater without using calculator\n\nHow do you compare $6-2\\sqrt{3}$ and $3\\sqrt{2}-2$? (no calculator)\n\nLook simple but I have tried many ways and fail miserably. Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$. Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).\n\n\u2022 sorry, multiply both sides by $6+2\\sqrt{3}$ and use $(a-b)(a+b)=a^2-b^2$ \u2013\u00a0luka5z Jun 10 '17 at 19:00\n\u2022 \\begin{eqnarray*} \\sqrt{a}+ \\sqrt{b}= \\sqrt{a+b+2 \\sqrt{ab}} \\end{eqnarray*} might be helpful. \u2013\u00a0Donald Splutterwit Jun 10 '17 at 19:01\n\u2022 @Learner132 A lot of the solutions here add or subtract from both sides then square then compare to infer about the direction of the original identity. This is only guaranteed to work if both values are positive before squaring. E.g. consider the counterexample: $2.5>1 \\rightarrow 2.5-2>1-2 \\rightarrow 0.5 > -1 \\rightarrow 0.25 > 1$. \u2013\u00a0Ian Miller Jun 11 '17 at 11:49\n\u2022 I would just guesstimate. $\\sqrt 3 \\approx 1.75$, $\\sqrt 2 \\approx 1.4$, so $6 - 2 \\sqrt 3 \\approx 2.5$ and $-2 + 3 \\sqrt 2 \\approx 2.2$. \u2013\u00a0Robert Soupe Jun 15 '17 at 0:39\n\n$6-2\\sqrt 3 \\gtrless 3\\sqrt 2-2$\n\nRearrange: $8 \\gtrless 3\\sqrt 2 + 2\\sqrt 3$\n\nSquare: $64 \\gtrless 30+12\\sqrt 6$\n\nRearrange: $34 \\gtrless 12\\sqrt 6$\n\nSquare: $1156 \\gtrless 864$\n\n\u2022 You might simplify by $2$ before squaring\u2026 \u2013\u00a0Bernard Jun 10 '17 at 19:11\n\u2022 Thank you, good answer, but I doubt if math teacher is okay with this (> or <) symbol \u2013\u00a0Learner132 Jun 10 '17 at 19:29\n\u2022 What's important about any symbol is its mathematical meaning. If you can the mathematical meaning of this symbol to your math teacher, I'll bet he/she is open to its usage. \u2013\u00a0Lee Mosher Jun 10 '17 at 19:57\n\u2022 You should probably mention that the squaring steps are valid (both forward and backward) since the values on both sides are clearly positive. \u2013\u00a0Rory Daulton Jun 11 '17 at 0:33\n\u2022 Yes, ordinarily you would start at the last step with a definite operator there (> in this case) and then work towards the inequality given to you. However, to find the necessary steps, it is often easier to work from what you\u2019re given to something known, i.e., backwards. \u2013\u00a0BallpointBen Jun 11 '17 at 2:47\n\nWe have $\\sqrt{3}\\leq 1.8$ so $6-2\\sqrt{3}\\geq 2.4$, whereas $\\sqrt{2}\\leq 1.42$ so $3\\sqrt{2}-2\\leq 2.26$.\n\n\u2022 I love your answer, I'll try to see if math teacher is okay with estimation \u2013\u00a0Learner132 Jun 10 '17 at 19:31\n\u2022 @Learner132: Note that the estimates are easy to prove by squaring, if necessary. \u2013\u00a0user14972 Jun 10 '17 at 22:06\n\n$$6-2\u221a3 \\sim 3\u221a2-2\\\\ 8 \\sim 3\u221a2 +2\u221a3 \\\\ 64 \\sim 30+12\u221a6\\\\ 34 \\sim 12\u221a6\\\\ 17 \\sim 6\u221a6\\\\ 289 \\sim 36 \\cdot 6\\\\ 289 > 216$$\n\n\u2022 Thanks, quite understandable, but how I can explain ~ symbol to math teacher in this? \u2013\u00a0Learner132 Jun 10 '17 at 19:27\n\u2022 This $\\sim$ simbol is a placeholder for an undefined relationship, you can use whatever other symbol you prefer, while you dont multiply by -1 \u2013\u00a0Brethlosze Jun 10 '17 at 19:29\n\u2022 I remember using that during school first courses without trouble \u2013\u00a0Brethlosze Jun 10 '17 at 19:29\n\u2022 I understand you. Well, I hope math teacher is okay with it although it is not covered in the lecture. \u2013\u00a0Learner132 Jun 10 '17 at 19:40\n\u2022 Teachers know how this work and what you thought... rewritting everything is not needed. \u2013\u00a0Brethlosze Jun 11 '17 at 6:17\n\nUsing simple continued fractions for $\\sqrt {12}$ and $\\sqrt {18}.$ Worth learning the general technique...Method described by Prof. Lubin at Continued fraction of $\\sqrt{67} - 4$\n\n$$\\sqrt { 12} = 3 + \\frac{ \\sqrt {12} - 3 }{ 1 }$$ $$\\frac{ 1 }{ \\sqrt {12} - 3 } = \\frac{ \\sqrt {12} + 3 }{3 } = 2 + \\frac{ \\sqrt {12} - 3 }{3 }$$ $$\\frac{ 3 }{ \\sqrt {12} - 3 } = \\frac{ \\sqrt {12} + 3 }{1 } = 6 + \\frac{ \\sqrt {12} - 3 }{1 }$$\n\nSimple continued fraction tableau:\n$$\\begin{array}{cccccccccc} & & 3 & & 2 & & 6 & \\\\ \\\\ \\frac{ 0 }{ 1 } & \\frac{ 1 }{ 0 } & & \\frac{ 3 }{ 1 } & & \\frac{ 7 }{ 2 } \\\\ \\\\ & 1 & & -3 & & 1 \\end{array}$$\n\n$$\\begin{array}{cccc} \\frac{ 1 }{ 0 } & 1^2 - 12 \\cdot 0^2 = 1 & \\mbox{digit} & 3 \\\\ \\frac{ 3 }{ 1 } & 3^2 - 12 \\cdot 1^2 = -3 & \\mbox{digit} & 2 \\\\ \\frac{ 7 }{ 2 } & 7^2 - 12 \\cdot 2^2 = 1 & \\mbox{digit} & 6 \\\\ \\end{array}$$\n\nContinued fraction convergents alternate above and below the irrational number, we get $$\\frac{ 3 }{ 1 } < \\sqrt {12} < \\frac{ 7 }{ 2 }$$ Your first number was $6 - \\sqrt {12},$ $$3 > 6 - \\sqrt {12} > \\frac{ 5 }{ 2 }$$ $$\\frac{ 5 }{ 2 } < 6 - \\sqrt {12} < 3$$\n\nNext 18......................========================================\n\n$$\\sqrt { 18} = 4 + \\frac{ \\sqrt {18} - 4 }{ 1 }$$ $$\\frac{ 1 }{ \\sqrt {18} - 4 } = \\frac{ \\sqrt {18} + 4 }{2 } = 4 + \\frac{ \\sqrt {18} - 4 }{2 }$$ $$\\frac{ 2 }{ \\sqrt {18} - 4 } = \\frac{ \\sqrt {18} + 4 }{1 } = 8 + \\frac{ \\sqrt {18} - 4 }{1 }$$\n\nSimple continued fraction tableau:\n$$\\begin{array}{cccccccccc} & & 4 & & 4 & & 8 & \\\\ \\\\ \\frac{ 0 }{ 1 } & \\frac{ 1 }{ 0 } & & \\frac{ 4 }{ 1 } & & \\frac{ 17 }{ 4 } \\\\ \\\\ & 1 & & -2 & & 1 \\end{array}$$\n\n$$\\begin{array}{cccc} \\frac{ 1 }{ 0 } & 1^2 - 18 \\cdot 0^2 = 1 & \\mbox{digit} & 4 \\\\ \\frac{ 4 }{ 1 } & 4^2 - 18 \\cdot 1^2 = -2 & \\mbox{digit} & 4 \\\\ \\frac{ 17 }{ 4 } & 17^2 - 18 \\cdot 4^2 = 1 & \\mbox{digit} & 8 \\\\ \\end{array}$$\n\nThis time the number is $\\sqrt {18} - 2.$\n\nIt is enough to use $$2 < \\sqrt {18} - 2 < \\frac{9}{4}$$\n\n$$\\color{red}{ 2 < \\sqrt {18} - 2 < \\frac{9}{4} < \\frac{ 5 }{ 2 } < 6 - \\sqrt {12} < 3 }$$\n\n\u2022 Thank you, but I'll study it one day \u2013\u00a0Learner132 Jun 10 '17 at 19:51\n\nDefine\n\n$a=6-2\\sqrt 3>0$\n\n$b=3\\sqrt 2-2>0$\n\n$a-b = 8 - (2\\sqrt 3 + 3\\sqrt 2)$\n\n$(2\\sqrt 3 + 3\\sqrt 2)^2 = 30+12\\sqrt 6 = 6\u00d7(5+2\\sqrt 6) < 60 < 64$ because $6=2\u00d73 < (5/2)^2$\n\n$a-b > 8-8=0, a>b$\n\n\u2022 Thank you, it took me a while to understand 2\u00d73<(5/2)^2 but it is easier to explain to math teacher. \u2013\u00a0Learner132 Jun 10 '17 at 19:45\n\nHere's yet another way, for those who aren't comfortable with the $\\gtrless$ or $\\sim$ notation.\n\nWe can use crude rational approximations to $\\sqrt 2$ and $\\sqrt 3$.\n\n\\begin{align} \\left(\\frac{3}{2}\\right)^2 = \\frac{9}{4} & \\gt 2\\\\ \\frac{3}{2} & \\gt \\sqrt 2\\\\ \\frac{9}{2} & \\gt 3\\sqrt 2 \\end{align}\n\nAnd \\begin{align} \\left(\\frac{7}{4}\\right)^2 = \\frac{49}{16} & \\gt 3\\\\ \\frac{7}{4} & \\gt \\sqrt 3\\\\ \\frac{7}{2} & \\gt 2\\sqrt 3 \\end{align}\n\nAdding those two approximations, we get \\begin{align} \\frac{9}{2} + \\frac{7}{2} = 8 & \\gt 3\\sqrt 2 + 2\\sqrt 3\\\\ 6 + 2 & \\gt 3\\sqrt 2 + 2\\sqrt 3\\\\ 6 - 2\\sqrt 3 & \\gt 3\\sqrt 2 - 2 \\end{align}\n\nI'll use >=< to represent the unknown comparison.\n\n$6-2\\sqrt{3} >=< 3\\sqrt{2}-2$\n\nLets start by adding two to both sides to reduce the number of numbers. This doesn't change the comparison result.\n\n$8-2\\sqrt{3} >=< 3\\sqrt{2}$\n\nBoth sides are clearly positive ( $2\\sqrt{3} < 6$ ) so we can square both sides without changing the comparison result.\n\nIn a more maginal case where we were unsure if the left hand side was positive we could have compared the two terms in the left hand side by squaring both of them and hence determined whether the left hand side was positive or negative.\n\n$64 -32\\sqrt{3} + 12 >=< 18$\n\nNow lets collect terms.\n\n$60 >=< 32\\sqrt{3}$\n\nDivide by four.\n\n$15 >=< 8\\sqrt{3}$\n\nSquare again.\n\n$225 >=< 64 \\times 3$\n\n$225 > 192$\n\nTherefore\n\n$6-2\\sqrt{3} > 3\\sqrt{2}-2$\n\nThere is still some hope in taking the first minus the second in this case: $$6-2\\sqrt{3} - (3\\sqrt{2}-2) = 8 - (2\\sqrt{3} + 3\\sqrt{2})$$\n\nSo now the question boils down to if the expression with the square root exceeds $8$. We know that $8^{2} = 64$ and: $$(2\\sqrt{3}+3\\sqrt{2})^{2}=4*3+2(2\\sqrt{3})(3\\sqrt{2})+9*2 =30+12\\sqrt{3}\\sqrt{2}$$\n\nConversely: $$8^{2} = 64 = 28+36=28+6*6 = 28+6*\\sqrt{6}\\sqrt{6}$$\n\nSubtracting them yields: $$28+6*\\sqrt{6}\\sqrt{6}-30+12\\sqrt{3}\\sqrt{2} = -2+6*\\sqrt{6}\\sqrt{6}-12\\sqrt{3}\\sqrt{2}$$\n\nThus this is only positive if the square root terms are positive and exceed 2. Comparing the square root terms: $$6*\\sqrt{6}\\sqrt{6}-12\\sqrt{3}\\sqrt{2} = 6*\\sqrt{2}\\sqrt{3}(\\frac{\\sqrt{2}}{\\sqrt{2}})\\sqrt{6}-12\\sqrt{6}=12\\frac{\\sqrt{3}}{\\sqrt{2}}\\sqrt{6}-12\\sqrt{6}=12\\sqrt{6}(\\frac{\\sqrt{3}}{\\sqrt{2}}-1)$$\n\nWe conclude that the square root term is positive since $\\sqrt{3}>\\sqrt{2}$ . But is it greater than $-2$? Or rather, how much do we need to multiply to the expression $\\frac{\\sqrt{3}}{\\sqrt{2}}-1$ for it to be greater than $2$? $$n*(\\frac{\\sqrt{3}}{\\sqrt{2}}-1)>2$$ $$n>\\frac{2}{\\frac{\\sqrt{3}}{\\sqrt{2}}-1}*\\frac{\\frac{\\sqrt{3}}{\\sqrt{2}}+1}{\\frac{\\sqrt{3}}{\\sqrt{2}}+1}\\approx\\frac{2*(1.5+1)}{0.5} = 10$$ So the factor in front of the term $\\frac{\\sqrt{3}}{\\sqrt{2}}-1$ should be at least 10. But since $12>10$ and $\\sqrt{6}>1$, we conclude that $12\\sqrt{6}>10$. Thus, substituting back to the original equation: $$-2+6*\\sqrt{6}\\sqrt{6}-12\\sqrt{3}\\sqrt{2} = -2+12\\sqrt{6}(\\frac{\\sqrt{3}}{\\sqrt{2}}-1) > 0$$ And so we conclude that: $$64>(2\\sqrt{3}+3\\sqrt{2})^{2}$$ and so, for positive root, $$8>(2\\sqrt{3}+3\\sqrt{2})$$ and $$8 - (2\\sqrt{3} + 3\\sqrt{2}) > 0$$\n\nYou may use other approaches. Notice that I split $8^{2} = 28+36$. Equally viable is to split $8^{2} = 30+34$, and you might have to use a similar (Squaring than rooting) trick to show that $34 > 12\\sqrt{6}$, and then finally conclude that $8>(2\\sqrt{3} + 3\\sqrt{2})$. This alternative approach should be able to give you a smaller threshold compared to the approximation I made midway.\n\n$6-2\u221a3 \\approx2(1.732) = 6 - 3.464 = 2.536$ $3\u221a2-2 \\approx 3(1.414) - 2 = 4.242 - 2 = 2.242$\n\nThis implies that $\\dfrac 52$ is between the two quantities:\n\n\\begin{align} \\dfrac{49}{4} &> 12 \\\\ \\dfrac 72 &> 2\\sqrt 3 \\\\ \\dfrac{12}{2} &> 2\\sqrt 3 + \\dfrac 52 \\\\ 6 - 2\\sqrt 3 &> \\dfrac 52 \\end{align}\n\nand\n\n\\begin{align} \\dfrac{81}{4} &> 18 \\\\ \\dfrac 92 &> 3\\sqrt 2 \\\\ \\dfrac 52 &> 3\\sqrt 2 - 2 \\end{align}\n\nIt follows that $6-2\u221a3 > 3\\sqrt 2 - 2$.\n\n\u2022 see this answer same concept but better (less digits, can be easily verified by hand, using lower and upper bounds instead of $\\approx$)) and 12 hours earlier than your answer \u2013\u00a0miracle173 Jun 11 '17 at 14:16\n\u2022 @miracle173 What you saw was only a part of what I intended to do. I needed some more time to think about the problem so I thought I had deleted everything. Obviously, I hadn't. What you see above is what I intended to do. \u2013\u00a0steven gregory Jun 11 '17 at 19:57", "date": "2021-06-20 06:22:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2317625/show-which-of-6-2-sqrt3-and-3-sqrt2-2-is-greater-without-using-calculato", "openwebmath_score": 0.9727590084075928, "openwebmath_perplexity": 522.1608724275425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446494481298, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8575833606994198}}
{"url": "https://math.stackexchange.com/questions/497944/find-the-average-velocity", "text": "# Find the average velocity\n\nIf a ball is thrown in the air with a velocity $34$ ft/s, its height in feet $t$ seconds later is given by\n\n$$y = 34 t \u2212 16 t^2 .$$\n\nFind the average velocity for the time period beginning when $t = 2$ and lasting $0.5$ second, $0.1$ second, $0.05$ second, $0.01$ second and estimate the instantaneous velocity when $t = 2$.\n\nI tried to solve by doing the following:\n\n$y=34(2)-16(2)^2$\n\n$y=68-64$\n\n$y=a$\n\n$4/0.5=8$ft/s\n\nbut I was told that answer is incorrect.\n\nWhat did I do wrong?\n\n\u2022 This is a limit approximation. If we say $y=f(x)$, then we can write that the first average is ${f(2.5)-f(2)\\over 0.5}={-15-4\\over 0.5}=-38$. \u2013\u00a0abiessu Sep 18 '13 at 21:50\n\u2022 How did you get -15 for f(2.5) and 4 for f(2) ? \u2013\u00a0Grey Sep 18 '13 at 22:22\n\u2022 I used $f(2.5) = 34(2.5)-16(2.5)^2 = 85-100=-15$ and $f(2)=34(2)-16(2)^2=68-64=4$. I saw the $62$ in the second term of your original post before it was edited and that $16\\cdot 4$ is not $62$... \u2013\u00a0abiessu Sep 18 '13 at 22:26\n\u2022 Thanks. And how to get get the instantaneous velocity when t = 2 ? \u2013\u00a0Grey Sep 18 '13 at 22:40\n\u2022 That is the point of the exercise, first calculate $f(2.5)-f(2)\\over 0.5$, then $f(2.1)-f(2)\\over .1$, and so on, and use this series of values to estimate the instantaneous velocity at $t=2$. \u2013\u00a0abiessu Sep 18 '13 at 22:42\n\nYou are being asked for the average velocity over the time span $2$ to $2.5$ seconds, among others. To do that one, you need $y(2.5)$ and $y(2)$ Then the average velocity in that span is $\\frac {y(2.5)-y(2)}{2.5-2}$\n\nThere is enough information here to complete the following process:\n\nGiven $y(t)=34t-16t^2$, we have:\n\n$$y(2)=34\\cdot 2-16\\cdot 4=68-64=4$$\n\n$$y(2.5)=34\\cdot 2.5-16\\cdot 2.5^2=85-100=-15$$\n\n$$y(2.1)=34\\cdot 2.1-16\\cdot 2.1^2=71.4-70.56=0.84$$\n\n$$y(2.05)=2.46$$ (using calculator)\n\n$$y(2.01)=3.6984$$ (using calculator)\n\nNow, the remaining task is to calculate $-19\\over 0.5$, $-3.16\\over 0.1$, $-1.54\\over 0.05$, and $-0.3016\\over 0.01$ and see if these numbers are approaching a particular number. I got each of these fractions by writing $y(2+t)-y(2)\\over t$ for each of the values $t\\in \\{0.5,0.1,0.05,0.01\\}$.", "date": "2019-12-10 00:18:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/497944/find-the-average-velocity", "openwebmath_score": 0.9394632577896118, "openwebmath_perplexity": 228.28573779884164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429160413819, "lm_q2_score": 0.870597273444551, "lm_q1q2_score": 0.8575756769314968}}
{"url": "https://math.stackexchange.com/questions/2055398/mathematics-behind-this-card-trick", "text": "Mathematics behind this card trick\n\n1. Suppose I have $21$ playing cards. I distribute them in $3$ columns and tell you to choose mentally a card. Then just indicate in which column the card is.\n\n2. I pick up one of the columns which doesn't contain your card, then the column which contains your card then the remaining column.\n\n3. Now I deal the cards in $3$ columns again, starting from the left to the right, and repeating the process until there are no cards left in my hand. I ask you to indicate to me in which column your card is.\n\n4. I repeat step 2 then 3.\n\n5. I repeat step 2.\n\nNow by counting either way in the deck of $21$ cards, your card will be the 11th card.\n\nI've tried using modulo to understand the problem but I'm stuck doing integer divisions. So does anyone have a simpler way to explain the trick.\n\nAlso could anyone explain why it works only for odd number of cards in each column and why it requires additional steps for more cards. e.g for $17 \\cdot 3 = 51$ requires an additional step compared with $21$ cards?\n\nEDIT: I forgot to add that the number of cards you have to count for the final step equals\n\n$$1.5\\text{ times the number of cards in each column} + 0.5$$\n\n\u2022 If you do not have an odd number of cards in total, there is not a middle card Dec 12 '16 at 13:25\n\u2022 If you have $n$ cards and $3$ columns, you need $\\log_3(n)$ steps (rounded up) to distinguish them Dec 12 '16 at 13:26\n\u2022 This related trick (and its explanation) might provide some insight Dec 12 '16 at 13:29\n\u2022 I first learnt this card trick when I was 14. After learning some calculus, I was pleasant to realise that the fact that the card will eventually end up at the middle position is basically a consequence of the squeezing principle. Dec 13 '16 at 9:13\n\u2022 @user1551 That seems like an overapplication of the squeezing principle.\n\u2013\u00a0jwg\nDec 13 '16 at 10:30\n\nSuppose that, when you first lay the cards on the table, the card I choose is at position $x$ in its column. You don't know $x$, but you know that $1\\leq x\\leq 7$.\n\nNow, when you pick up the cards, my card will be at position $7+x$ in the full stack. The second time you lay the cards on the table, my card will appear at position $p_1=\\lceil\\frac{7+x}{3}\\rceil$ in its column.\n\nThe second time you pick up the cards, my card will be at position $7+p_1$ in the full stack. The third time you lay the cards on the table, my card will appear at position $p_2=\\lceil\\frac{7+p_1}{3}\\rceil$ in its column.\n\nFinally, when you pick up the cards for the third time, my card will be at position $7+p_2$ in the full stack. Putting this all together, my card will be at position $$7+p_2=7+\\left\\lceil\\frac{7+\\lceil\\frac{7+x}{3}\\rceil}{3}\\right\\rceil$$ in the full stack. The trick is that this is equal to $11$ for all $x$ in the range $1\\leq x\\leq 7$.\n\nFor a proof of this last statement, as jpmc26 mentions, one can apply the identities $\\lceil\\frac{m+\\lceil x\\rceil}{n}\\rceil=\\lceil\\frac{m+x}{n}\\rceil$ and $\\lceil n+x\\rceil = n+\\lceil x\\rceil$ (for real $x$, integer $m$, and positive integer $n$) to show that $$7+\\left\\lceil\\frac{7+\\lceil\\frac{7+x}{3}\\rceil}{3}\\right\\rceil = 7+\\left\\lceil\\frac{7+\\frac{7+x}{3}}{3}\\right\\rceil = 7 + \\left\\lceil 3 + \\frac{x+1}{9}\\right\\rceil = 10 + \\left\\lceil\\frac{x+1}{9}\\right\\rceil \\enspace,$$ which is clearly equal to $11$ for $1\\leq x\\leq 7$.\n\n\u2022 You can use the fact that $\\left \\lceil \\frac{7+\\lceil y \\rceil}{3} \\right \\rceil = \\left \\lceil \\frac{7+ y }{3} \\right \\rceil$ (see here) to show that the last expression is equal to $7 + \\left \\lceil 3 + \\frac{x + 1}{9} \\right \\rceil = 10 + \\left \\lceil \\frac{x + 1}{9} \\right \\rceil$, which makes the fact it's always 11 much more obvious. Dec 12 '16 at 23:14\n\u2022 Good point. I just checked that it holds for x=1 and for x=7, therefore it holds for x in between due to monotonicity. Should I edit the answer to include your remark? Dec 13 '16 at 8:17\n\u2022 @EvangelosBampas: Yes! Dec 13 '16 at 12:47\n\u2022 @EvangelosBampas Absolutely. (Thanks!) Comments are considered ephemeral/transient, and their primary purpose is for the improvement of a post. You can generally assume that useful content or clarification posted in them warrants an edit or an additional post. (In this case, I only provided some comparatively minor additions that just build on your answer and make the conclusion easier to see, so a separate post wouldn't be able to stand on its own.) Dec 13 '16 at 14:13\n\nYou can find a full explanation of this trick and related ones at\n\nGergonne\u2019s Card Trick, Positional Notation, and Radix Sort\n(Mathematics Magazine, February, 2010)\n\nhttp://www.maa.org/sites/default/files/Bolker-MMz-201053228.pdf\n\nThe classic trick uses $27 = 3^3$ cards. Your $21$ card version is discussed on page 48.\n\n\u2022 This answer comes dangerously close to \"link only.\" Dec 12 '16 at 22:51\n\u2022 I can see a total of 4 lines floating freely besides the link provided, which will definitly assure that the answer is too far from being called a link only if anything. Dec 13 '16 at 1:01\n\u2022 @jpmc26 I don't like link only answers either, but the relevant content in the link is too long to post - I'd essentially be pasting in most of the paper. The link is to the journal, so stable. Dec 13 '16 at 2:06\n\nAfter one deal-and-gather phase, your card is in the middle third of the deck, right?\n\nWhen you deal out the cards again, where does that middle third end up? In the middle third (looking top-to-bottom) of the tableau.\n\nTo see this: on the first deal, pick a card, say, the 3 of hearts. Replace everything else in its column with red cards, and everything else in the other two columns with black cards. Gather and re-deal. You'll see a red \"band\" in the middle (top to bottom) of the new tableau.\n\nNow you identify your card in that middle band. When you gather up the other cards, it'll once again be in the middle third of the deck, because there's one entire column-worth of cards in front of it, and one entire column-worth of cards behind it.\n\nBut you can say more than that: suppose your card was in the first column. Well, then, it was in the middle third of the first column, wasn't it? So if each column has $k$ cards, you've got $k$ cards in front of it (from one of the other piles), and another k/3 cards (from its own pile) in front of it, and the same for cards behind it. So now it's in the middle 1/9th of the pile.\n\nAnd the next time it'll be in the middle 1/27 of the pile. And so on.\n\nNow that analysis isn't QUITE right, because $k$ might not be divisible by 3. So instead of having k + k/3 cards in front of it after two deals, you have k + floor(k/3). For instance, if $k = 8$, then you'd have $8 + floor(8/3) = 8 + floor(2.66...) = 8 + 2 = 10$ cards in front of it. But aside from this minor glitch, what you get is the following:\n\nafter $p$ \"passes\" on a deck of $N$ cards, your card is in the middle (roughly) $(N/3^p)$ cards of the deck. When $3^p > N$, this means that your card is the middle card of the deck.\n\n\u2022 +1, Definitely makes more sense to me than the algebra above. Dec 13 '16 at 1:00\n\nLet's go step by step. After steps i) and ii) we already know that the chose card is in position $8,9,10,11,12,13,14$ in the pile. Then after dealing all the cards again in the given manner we know that the cards that were in the positions $8,9,10,11,12,13,14$ before dealing are now in places $3,4,5$ (actually it depends on the pile, but we can be sure that each of the mentioned cards is now at a position $3,4$ or $5$ in one of the three piles. So eventually after the second step we narrowed the choice to three cards.\n\nAfter putting the cards in one pile again then we know that the chosen card will be in position $10,11,12$. And in the last deal obviously each of the cards will be in dealt in different pile, so we should be able to find out which one is the card for sure. But during the last deal the cards in positions $10,11,12$ are all in position $4$ in each of the new piles, hence after collecting all the cards, the chosen card will be in position $11$.\n\nEventually as you can see with each dealing we lower the number of possibilities from $n$ to $\\lfloor\\frac{n-1}{3}\\rfloor + 1$.", "date": "2022-01-20 09:58:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2055398/mathematics-behind-this-card-trick", "openwebmath_score": 0.5803660154342651, "openwebmath_perplexity": 371.13227874636306, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9850429103332287, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8575756769228572}}
{"url": "https://gmatclub.com/forum/inequalities-trick-91482.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 18 Aug 2018, 03:58\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Inequalities trick\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2653\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\n\n### Show Tags\n\nUpdated on: 26 Apr 2018, 03:34\n128\n283\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\n$$f(x) = (x-a)(x-b)(x-c)(x-d) < 0$$\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. $$a<b<c<d$$\n\nSo for f(x) < 0 consider \"-\" curves and the ans is: $$(a < x < b)$$, $$(c < x < d)$$\nand for f(x) > 0 consider \"+\" curves and the ans is: $$(x < a)$$, $$(b < x < c)$$, $$(d < x)$$\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nAttachment:\n\n1.jpg [ 6.73 KiB | Viewed 64801 times ]\n\nAttachment:\n\nUntitled.png [ 12.08 KiB | Viewed 4936 times ]\n\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nSupport GMAT Club by putting a GMAT Club badge on your blog/Facebook\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\nhttp://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nOriginally posted by gurpreetsingh on 16 Mar 2010, 10:11.\nLast edited by Bunuel on 26 Apr 2018, 03:34, edited 1 time in total.\nEdited.\nSenior Manager\nStatus: Upset about the verbal score - SC, CR and RC are going to be my friend\nJoined: 30 Jun 2010\nPosts: 301\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n17 Oct 2010, 16:14\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) < 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nI don't understand this part alone. Can you please explain?\n_________________\n\nMy gmat story\nMGMAT1 - 630 Q44V32\nMGMAT2 - 650 Q41V38\nMGMAT3 - 680 Q44V37\nGMATPrep1 - 660 Q49V31\nKnewton1 - 550 Q40V27\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2653\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n17 Oct 2010, 17:19\nDreamy wrote:\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) < 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nI don't understand this part alone. Can you please explain?\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0 you will consider the curve with -ve inside it.. check the attached image.\n\nf(x) = (x-a)(x-b)(x-c)(x-d) > 0 consider the +ve of the curve\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nSupport GMAT Club by putting a GMAT Club badge on your blog/Facebook\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\nhttp://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8188\nLocation: Pune, India\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n22 Oct 2010, 06:33\n85\n107\nYes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.\n\nIf (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.\ne.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.\n\nAttachment:\n\ndoc.jpg [ 7.9 KiB | Viewed 63852 times ]\n\nThis divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.\n\nWhen you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.\n\nWhen you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.\n\nSince we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7\n\nIt should be obvious that it will also work in cases where factors are divided.\ne.g. (x - a)(x - b)/(x - c)(x - d) < 0\n(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.\n\nNote: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nSave up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 29 Sep 2008 Posts: 105 Re: Inequalities trick [#permalink] ### Show Tags 22 Oct 2010, 11:45 19 20 if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Manager Joined: 10 Nov 2010 Posts: 142 Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 06:29 VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Retired Moderator Joined: 20 Dec 2010 Posts: 1877 Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 06:49 24 15 vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8188 Location: Pune, India Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 19:57 10 2 vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nManager\nJoined: 06 Apr 2011\nPosts: 52\nLocation: India\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n07 Aug 2011, 07:24\ngurpreetsingh wrote:\nulm wrote:\nif we have smth like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\n\nyes even powers wont contribute to the inequality sign. But be wary of the root value of x=a\n\nThis way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma\n\nHowever, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.\n\ncould someone please explain.\n_________________\n\nRegards,\nAsher\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8188\nLocation: Pune, India\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n08 Aug 2011, 11:59\n2\n3\nAsher wrote:\ngurpreetsingh wrote:\nulm wrote:\nif we have smth like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\n\nyes even powers wont contribute to the inequality sign. But be wary of the root value of x=a\n\nThis way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma\n\nHowever, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.\n\ncould someone please explain.\n\nWhen you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.\n\nSimilarly for (x-a)^2(x-b) > 0, x > b\n\nAs for roots, you have to keep in mind that given $$\\sqrt{x}$$, x cannot be negative.\n\n$$\\sqrt{x}$$ < 10\nimplies 0 < $$\\sqrt{x}$$ < 10\nSquaring, 0 < x < 100\nRoot questions are specific. You have to be careful. If you have a particular question in mind, send it.\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nSave up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 06 Apr 2011 Posts: 52 Location: India Re: Inequalities trick [#permalink] ### Show Tags 08 Aug 2011, 22:22 Quote: When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b. Similarly for (x-a)^2(x-b) > 0, x > b Thanks Karishma for the explanation. Hope you wouldn't mind clarifying a few more doubts. Firstly, in the above case, since x>b could we say that everything with be positive. would the graph look something like this: positive.. b..postive.. a.. positive On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative. would the graph look something like this: negative.. b..postive.. a.. postive Am i right? If it is not too much of a trouble, could you please show the graphical representation. problems with \\sqrt{x}.. this is all i could find (googled actually ): 1. \u221a(-x+4) \u2264 \u221a(x) 2. x^\\sqrt{x} =< (\\sqrt{x})^x {P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems} _________________ Regards, Asher Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8188 Location: Pune, India Re: Inequalities trick [#permalink] ### Show Tags 09 Aug 2011, 03:28 1 2 Asher wrote: Firstly, in the above case, since x>b could we say that everything with be positive. would the graph look something like this: positive.. b..postive.. a.. positive On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative. would the graph look something like this: negative.. b..postive.. a.. postive So when you have $$(x - a)^2(x - b) < 0$$, you ignore x = a and just plot x = b. It is positive in the rightmost region and negative on the left. So the graph looks like this: negative ... b ... positive Am i right? If it is not too much of a trouble, could you please show the graphical representation. problems with \\sqrt{x}.. this is all i could find (googled actually ): 1. \u221a(-x+4) \u2264 \u221a(x) 2. x^\\sqrt{x} =< (\\sqrt{x})^x {P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems} Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression. e.g. $$(x-4)^2(x - 9)(x+11) < 0$$ We do not plot x = 4 here, only x = -11 and x = 9. We start with the rightmost section as positive. So it looks something like this: positive... -11 ... negative ... 9 ... positive Since we need the region where x is negative, we get -11 < x < 9. Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression. I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nIntern\nJoined: 14 Apr 2011\nPosts: 11\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2011, 07:00\nhey ,\ncan u please tel me the solution for this ques\n\na car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?\n\nans -11\nManager\nStatus: On...\nJoined: 16 Jan 2011\nPosts: 170\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2011, 17:01\n11\n24\nWoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.\n\n1) CORE CONCEPT\n@gurpreetsingh -\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nArrange the NUMBERS in ascending order from left to right. a<b<c<d\nDraw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nNote: Make sure that the factors are of the form (ax - b), not (b - ax)...\n\nexample -\n(x+2)(x-1)(7 - x)<0\n\nConvert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')\nNow solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'\nSince you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.\n\n2) Variation - ODD/EVEN POWER\n@ulm/Karishma -\nif we have even powers like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\nThis will be same as (x-b)\n\nWe can ignore squares BUT SHOULD consider ODD powers\nexample -\n2.a\n(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0\n2.b\n(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0\nis the same as (x - a)(x - b)(x - c)(x - d) < 0\n\n3) Variation <= in FRACTION\n@mrinal2100 -\nif = sign is included with < then <= will be there in solution\nlike for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7\n\nBUT if it is a fraction the denominator in the solution will not have = SIGN\n\nexample -\n3.a\n(x + 2)(x - 1)/(x -4)(x - 7) < =0\nthe solution will be -2 <= x <= 1 or 4< x < 7\nwe cant make 4<=x<=7 as it will make the solution infinite\n\n4) Variation - ROOTS\n@Karishma -\nAs for roots, you have to keep in mind that given $$\\sqrt{x}$$, x cannot be negative.\n\n$$\\sqrt{x}$$ < 10\nimplies 0 < $$\\sqrt{x}$$ < 10\nSquaring, 0 < x < 100\nRoot questions are specific. You have to be careful. If you have a particular question in mind, send it.\n\nRefer - inequalities-and-roots-118619.html#p959939\nSome more useful tips for ROOTS....I am too lazy to consolidate\n\n<5> THESIS -\n@gmat1220 -\nOnce algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.\n\nI will save this future references....\nPlease add anything that you feel will help.\n\nAnyone wants to add ABSOLUTE VALUES....That will be a value add to this post\n\n_________________\n\nLabor cost for typing this post >= Labor cost for pushing the Kudos Button\nhttp://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8188\nLocation: Pune, India\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n11 Aug 2011, 22:57\n1\n3\nsushantarora wrote:\nhey ,\ncan u please tel me the solution for this ques\n\na car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?\n\nans -11\n\nPlease put questions in new posts. Put it in the same post only if it is totally related or a variation of the question that we are discussing.\n\nNow for the solution:\n\nThere are 42 cars on the lot. 1/7 of sports cars and 1/2 of luxury cars have sunroofs. This means that 1/7 of number of sports cars and 1/2 of number of luxury cars should be integers (You cannot have 1.5 cars with sunroofs, right?)\nWe want to minimize the sunroofs. Since 1/2 of luxury cars have sunroofs and only 1/7 of sports cars have them, it will be good to have fewer luxury cars and more sports cars. Best would be to have all sports cars. But, the question says there are some of each kind at any time. So let's say there are 2 luxury cars (since 1/2 of them should be an integer value). But 1/7 of 40 (the rest of the cars are sports cars) is not an integer number. Let's instead look for the multiple of 7 that is less than 42. The multiple of 7 that is less than 42 is 35. So we could have 35 sports cars. But then, 1/2 of 7 (since 42 - 35 = 7 are luxury cars) is not an integer. The next smaller multiple of 7 is 28. This works. 1/2 of 14 (since 42 - 28 = 14 are luxury cars) is 7. So we can have 14 luxury cars and 28 sports cars. That is the maximum number of sports cars that we can have.\n1/7 of 28 sports cars = 4 cars have sunroofs\n1/2 of 14 luxury cars = 7 cars have sunroofs\n\nSo at least 11 cars will have sunroofs.\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nSave up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 16 Feb 2012 Posts: 193 Concentration: Finance, Economics Re: Inequalities trick [#permalink] ### Show Tags 22 Jul 2012, 03:03 VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\\leq x \\leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8188 Location: Pune, India Re: Inequalities trick [#permalink] ### Show Tags 23 Jul 2012, 03:13 2 Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\\leq x \\leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8188\nLocation: Pune, India\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n25 Jul 2012, 22:21\npavanpuneet wrote:\nHi Karishma,\n\nJust for my reference, say if the equation was (x+2)(x-1)/(x-4)(x-7) and the question was for what values of x is this expression >0, then the roots will be -2,1,4,7 and by placing on the number line and making the extreme right as positive...\n\n----(-2)----(1)----(4)---(7)----then x>7, 1<x<4 and x<-2...Please confirm..\nHowever, is say it was >=0 then x>7, 1<=x<4 and x<=-2; given that the denominator cannot be zero. Please confirm\n\nYes, you are right in both the cases.\n\nAlso, if you want to verify that the range you have got is correct, just plug in some values to see. Put x = 0, the expression is -ve. Put x = 2, the expression is positive.\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nSave up to \\$1,000 on GMAT prep through 8/20! Learn more here >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nManager\nJoined: 26 Dec 2011\nPosts: 96\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n26 Jul 2012, 07:59\nThankyou Karishma.\n\nFurther, say if the same expression was (x+2)(1-x)/(x-4)(x-7) and still the question was for what values of x is the expression positive, then ... make it x-1 and with the same roots, have the rightmost as -ve. Then we look for the +ve intervals and check for those intervals if the expression is positive. for examples, in this case, -2<x<1 and 4<x<7 both depict positive interval but only first range satisfies the condition. Please confirm\n\nHowever, if for the same equation as mentioned, say the expression was (x+2)(x-1)/(x-4)(x-7) >0 and then we were asked to give the range where this is valid, then we would also multiply the -ve sign and make is <0 and then make the range after extreme right root -ve and provide all the intervals where it is negative. Please confirm\nSenior Manager\nJoined: 23 Oct 2010\nPosts: 361\nLocation: Azerbaijan\nConcentration: Finance\nSchools: HEC '15 (A)\nGMAT 1: 690 Q47 V38\nRe: Inequalities trick\u00a0 [#permalink]\n\n### Show Tags\n\n26 Jul 2012, 09:47\nVeritasPrepKarishma wrote:\n\nWhen you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.\n.\n\nIMHO, it should be x<b and also x is not equal to a .\nso, one shouldn't totally ignore the squared term. We can ignore it, if the expression is <=0\ncorrect me, if I am wrong\n\nmy question is -\ndo we always have a sequence of + and - from rightmost to the left side. I mean is it possible to have + and then + again ?\n_________________\n\nHappy are those who dream dreams and are ready to pay the price to make them come true\n\nI am still on all gmat forums. msg me if you want to ask me smth\n\nRe: Inequalities trick &nbs [#permalink] 26 Jul 2012, 09:47\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0\u00a0\u00a03\u00a0\u00a0\u00a04\u00a0\u00a0\u00a05\u00a0 \u00a0 Next \u00a0[ 95 posts ]\n\nDisplay posts from previous: Sort by\n\n# Inequalities trick\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-08-18 10:58:47", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/inequalities-trick-91482.html", "openwebmath_score": 0.5639657974243164, "openwebmath_perplexity": 2156.2014103647634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9348724567894617, "lm_q2_score": 0.9173026533686324, "lm_q1q2_score": 0.8575609851742254}}
{"url": "https://www.physicsforums.com/threads/de-broglie-wavelength-of-30kev-electron.281225/", "text": "# De Broglie wavelength of 30keV electron\n\n1. Dec 23, 2008\n\n### catkin\n\n1. The problem statement, all variables and given/known data\nThe question is from Advanced Physics by Adams and Allday, section 8 Practice Exam Questions, question 30.\n\nEstimate the de Broglie wavelength of an electron that has been emitted thermionically in a vacuum from a filament and then accelerated through a p.d. of 30.0 kV\n\n2. Relevant equations\n\u03bbde Broglie = h / p\nE2 - p2c2 = m02c4\nETotal = m0c2 + K.E.\n\n3. The attempt at a solution\nI think the solution is valid; my concern is whether there is a better (= more elegant) way to do it.\n\nThe de Broglie wavelength is given by\n\u03bbde Broglie = h / p\nWhere h is Planck's constant and p is momentum.\np could be found from p = \u03b3m0v but this would require finding v. More conveniently\nE2 - p2c2 = m02c4\nWhere E is the total energy (E = m0c2 + K.E.)\nExpanding E:\np2c2 = 2m0c2K.E. + K.E.2\nRearranging:\np = (1/c) \u221a(2m0c2K.E. + K.E.2)\nSubstituting this p:\n\u03bbde Broglie = hc / \u221a(2m0c2K.E. + K.E.2)\nSubstituting values using SI units (including eV to J conversion factor 1.60E-19)\n= 6.63E-34 * 3.00E+8 / Sqrt(( 2 * 9.11E-31 * 3.00E+8^2 * 30E+3 * 1.60E-19) + (( 30E+3 * 1.60E-19)^2 ))\n= 7.0e-12 m ct2sf\n\n2. Dec 23, 2008\n\n### Redbelly98\n\nStaff Emeritus\nLooks good.\n\nBy the way, this is a little easier using eV energy units. That way you won't have all those 1.6e-19's to contend with:\n\nhc = 1240 eV-nm (a good number to remember in the future)\nmoc2 = 511 keV or 511e3 eV\nKE = 30.0e3 eV\n\nso that\n\n\u03bbdB = 1240 eV-nm / \u221a[2 \u00d7 511e3 \u00d7 30.0e3 eV2 + (30.0e3)2eV2]\n= 0.00698 nm\n= 6.98 pm \u200b\n\n3. Dec 23, 2008\n\n### Andrew Mason\n\nSince it is asking only for an estimate, I would first determine whether the electron is moving at relativistic speeds. If it is not, you can use classical mechanics to determine p:\n\nA 30 kV potential difference gives the electron 30keV of kinetic energy. Since it has about 500 keV as rest mass, the relativisitic effect will be small (since you are only estimating) so I would use a non-relativistic approach.\n\nAs suggested by Redbelly, use the formula:\n\n$$\\lambda_{DeB} = \\frac{hc}{pc}$$\n\nFor non-relativistic speeds,\n\n$$pc = \\sqrt{2 KE m_0c^2}$$ (this is just a fancy rearrangment of $v = \\sqrt{2 KE/m}$).\n\nwhere KE is in units of eV. Note $m_0c^2$ = 511 KeV. and hc = 1240 eV nm\n\nUse that to work out the Debroglie wavelength, in nm:\n\n$$\\lambda_{DeB} = hc/pc = 1240/\\sqrt{2 x 3x10^4 x 5.11 x 10^5} = 1240/1.75 x 10^5 = 7 x 10^{-3}nm$$\n\nAM\n\n4. Dec 23, 2008\n\n### Redbelly98\n\nStaff Emeritus\nExcellent point.\n\n5. Jan 18, 2009\n\n### catkin\n\nThanks Redbelly and Andrew.\n\nSorry it has taken me so long to get back here.\n\nHelpful points, both about choice of unit and working out early on that the extra complexity of a relativistic approach is unnecessary.\n\nWhat is the rest mass to K.E. ratio heuristic that allows deciding a relativistic approach is unnecessary and how does it arise?\n\nBest\n\nCharles\n\n6. Jan 19, 2009\n\n### Redbelly98\n\nStaff Emeritus\nLoosely speaking: when the KE is small compared to the rest mass energy, non-relativistic approximations may be used. Or equivalently, when v is small compared to c.\n\nI'm not aware of any universally-accepted cutoff point, such as \"when KE is x% of the rest mass energy\". But as you saw, with KE equal to 6% of mc2, the non-relativistic result was pretty close, within 2%, of the actual de Broglie wavelength.\n\n7. Feb 3, 2009\n\n### catkin\n\nThanks Redbelly :)\n\nThe relativistic/Newtonian choice is something of a judgement call on this one. In our studies so far we have chosen 1% discrepancy in the result as the (arbitrary) determinant. On that basis, we found the cut of point for electrons is ~25keV acceleration. This is consonant with 2% error for an electron accelerated through 30 keV.\n\nThe thing that troubled me was that the question is from an old exam paper and the time to answer (based on the marks available) don't allow for my solution except the examinee be very good!\n\nPerhaps that introduces an new relativistic/Newtonian choice determinant -- the time available to answer. :)\n\nBest\n\nCharles", "date": "2016-10-25 05:26:37", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/de-broglie-wavelength-of-30kev-electron.281225/", "openwebmath_score": 0.6254527568817139, "openwebmath_perplexity": 2839.2872517755036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305307578324, "lm_q2_score": 0.8902942319436397, "lm_q1q2_score": 0.8575585855657087}}
{"url": "https://math.stackexchange.com/questions/2806398/probability-for-a-specific-value-for-sum-of-normally-distributed-random-variable", "text": "# Probability for a specific value for sum of normally distributed random variables\n\nI am trying to solve an exercise using the probability density function for a sum of n random variables with the same mean and variance. I need to find the probability for the specific value of the sums of r.v.s:\n\nPlayer B uses a fair coin to earn points, if it lands on heads he collects 1 point, if tails he collects 8 points. When he flips the coin 50 times, these points will add up cumulatively.\n\nWhat is the probability that player B will have exactly 225 points?\n\nI was thinking I could take the $PDF$ probability for $\\sum^n_{i=1}(X_i)=225$ with a mean of $n*\u00b5=225$ and standard deviation of $\\sqrt{\\sigma^2*n}=24.75$, which gives the value 0.016, as can be seen on the PDF plot:\n\nHowever, my teacher solved this by looking at it logically, and counting any situation where Player B gets exactly 25 heads and 25 tails:\n\n$P=($$50\\atop25$$)*\\frac{1}{2^{50}}=0.112$\n\nI see that his answer makes sense and is correct, but I doubt I would be able to think logically in an exam situation.\n\nWhy is my answer not correct, is it because the r.v. is only approximately normally distributed?\n\nIs there another general way of solving such an exercise if the amount of flips and the total sum of points had been different?\n\n\u2022 Another user reminded me that i can use binomial PDF to solve this issue, which is a bit of a help. But I do still have the same questions as mentioned. \u2013\u00a0user102937 Jun 3 '18 at 11:28\n\u2022 By binomial distribution there is no PDF (probability density function) but there is a PMF (probability mass function). Your teacher used this PMF and calculated $f(25)$ where $f$ denotes this PMF. It is defined by $f(x)=\\binom{50}{x}p^x(1-p)^{50-x}$ where $p=0.5$ and $x$ takes values in $\\{0,1,\\dots,50\\}$ \u2013\u00a0drhab Jun 3 '18 at 11:35\n\u2022 Your method works fine, but you have to remember the continuity correction. You are approximating a discrete distribution with a continuous one. If you are working with the scores, then for $(24,25,26)$ tails you get $(218,225,232)$ as scores...so for your continuous variant you need to look between $225-3.5$ and $225+3.5$. If you do that you get $0.112462916$ as opposed to the exact value of $0.112275173$. \u2013\u00a0lulu Jun 3 '18 at 12:02\n\nYour method works fine, but you have to remember the continuity correction. You are approximating a discrete distribution with a continuous one. Thus, for instance, your method gives a non-zero chance of getting $224$ points (say) though that is in fact impossible.\n\nAs you are working with the scores, then for $(24,25,26)$ tails you get $(218,225,232)$ as scores...so for your continuous variant you need to look between $225-3.5$ and $225+3.5$. If you do that you get $0.112462916$ as opposed to the exact value of $0.112275173$ ... not bad at all!\n\nThe first thing to notice is that $B$ throws 50 coins and makes 225 points. If $n_H$ and $n_T$ are the numbers of heads and tails, 50 throws make it so that $$n_H + n_T = 50.$$ Given that a heads gives 1 point and a tails 8 points, we can set up the equation $$1 \\cdot n_H + 8 \\cdot n_T = 225.$$\n\nSolving the two equations together gives $n_t = 25$; so we are after the probability tha 25 tails pop up in 50 throws. You can either do it with combinations and permutations (as your teacher did), or plainly use the pmf for the binomial distribution that represents your situation solving for 25 successes in 50 trials.\n\nIf you want the graphics, below is the plot of the distribution $B(50,0.5)$ with a line through $0.11227...$ which is the required probability.\n\nWhile it is true that the binomial distribution tends to a normal distribution, you cannot interpret the values of a probability density function as the probability of an event.\n\n\u2022 I would not speak of the PDF of binomial distribution but of the PMF. It does concern a density, but this wrt the counting measure. Usually PDF's refer to densities wrt Lebesgue measure and PMF's to densities wrt counting measure. \u2013\u00a0drhab Jun 3 '18 at 11:40\n\u2022 Agreed! It was a typo \u2013\u00a0Riccardo Sven Risuleo Jun 3 '18 at 21:23\n\nThe values $f(x)$ where $f$ denotes a PDF cannot be interpreted as probabilities (as you seem to think). Note for instance that they can take values that exceed $1$.\n\nThat is the mistake you made.\n\nWe have the equalities $H+T=50$ and $H+8T=225$ here leading to $H=25=T$, so the question can be rephrased as:\n\n\"By $50$ throws of a fair coin what is the probability on $25$ tails?\"\n\nThis is the route taken by your teacher.\n\n\u2022 Yes, okay i see. So i can use binomial distribution to check whether i get 25 successes. How about if i take the cdf for the value 225.5 and subtract the value for the cdf of 224.5? As far as i know, that cannot exceed 1. (Sorry that i keep sticking to the normal distribution assumption, i would just like to know) \u2013\u00a0user102937 Jun 3 '18 at 11:40\n\u2022 You could calculate $P(\\sum_{i=1}^{50}X_i=225)$ where $X_i$ takes values in $\\{1,8\\}$ by approaching it with normal distribution. If $\\sum_{i=1}^{50}X_i\\approx Y$ where $Y$ has normal distribution then an estimate of the probability would be $P(224.5<Y<225.5)$ as you suggest. \u2013\u00a0drhab Jun 3 '18 at 11:50", "date": "2020-01-21 23:45:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2806398/probability-for-a-specific-value-for-sum-of-normally-distributed-random-variable", "openwebmath_score": 0.8878366947174072, "openwebmath_perplexity": 169.88631623915282, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305360354471, "lm_q2_score": 0.8902942246666266, "lm_q1q2_score": 0.8575585832548974}}
{"url": "https://mathematica.stackexchange.com/questions/58751/how-to-speed-up-my-project-euler-code?noredirect=1", "text": "How to speed up my Project Euler code\n\nThe evaluation speed of Mathematica often depresses me. It did it again when I wrote a code to solve a Project Euler problem. https://projecteuler.net/problem=206\n\nHere's my code:\n\nClearAll[a];\na =\nCompile[{},\nNestWhile[# + 1 &, 10^9, ! MatchQ[\nIntegerDigits[#^2], {1, _, 2, _, 3, _, 4, _, 5, _, 6, _, 7, _,8, _, 9, _, 0}] &],\nCompilationTarget -> \"C\"]\n\n\nI think it is the simplest solution. I can't find other solutions which are simpler. But my Mathematica runs a long time before getting the answer. How can I speed this up? I've run into similar problems while I was solving other problems.\n\n\u2022 Pattern matching is often slow, you might try something like IntegerDigits[#^2][[ ;; ;; 2]] == {1,2,3,4,5,6,7,8,9,0} instead. \u2013\u00a0C. E. Sep 3 '14 at 7:16\n\u2022 #+1 can be replaced by #+10, because a square number ends with \"0\" must ends with \"00\" \u2013\u00a0wuyingddg Sep 3 '14 at 8:20\n\u2022 If you realize that the maximum solution is 10 Floor[Ceiling[Sqrt[1929394959697989990]]/10] and step down by 10 from there, you'll get the solution almost immediately. \u2013\u00a0Mark McClure Sep 3 '14 at 12:38\n\u2022 @MarkMcClure Yes but this is because you know the answer. It could also be very close to 102030405060708090. \u2013\u00a0anderstood Oct 21 '15 at 20:00\n\nCompilation is a certainly a good idea if you're going the brute-force route. So let's first tackle that, which will get us the answer in roughly 30 seconds computation time. Afterwards we'll come up with better strategy, which can do it in 0.3 seconds.\n\nCompilation\n\nA performance pitfall when compiling functions is that sometimes the compiled function just calls the main evaluation loop, effectively gaining nothing. We can check this with CompilePrint:\n\nNeeds[\"CompiledFunctionTools\"]\n\na // CompilePrint\n\n      (...)\nResult = I3\n\n1    I3 = MainEvaluate[(...)]\n2    Return\n\n\nThe call to MainEvaluate is the culprit here.\n\nSo let's rewrite your function such that it does compile properly. Using Pickett's and wuyingddg suggestions, we end up with:\n\nPerformSearch = Compile[\n{\n{startValue, _Integer},\n{increment,  _Integer}\n},\nNestWhile[\n# + increment &,\nstartValue,\nIntegerDigits[#^2][[-1 ;; 1 ;; -2]] =!= {9, 8, 7, 6, 5, 4, 3, 2, 1} &\n]\n]\n\n\nLet's check if this did compile ok:\n\nPerformSearch // CompilePrint\n\n      (...)\n\n1    I4 = I0\n2    I3 = I4\n3    I7 = Square[ I3]\n4    T(I1)2 = IntegerDigits[ I7, I5]]\n5    T(I1)0 = Part[ T(I1)2, T(I1)1Span]\n6    B0 = CompareTensor[ I6, R2, T(I1)0, T(I1)3]]\n7    if[ !B0] goto 12\n8    I3 = I4\n9    I7 = I3 + I1\n10   I4 = I7\n11   goto 2\n12   Return\n\n\nOk, that looks much better. We can now perform the search, but not after we've determined the range of possible solutions:\n\nmax =   Floor @ Sqrt @ FromDigits @ Riffle[Range[9], 9] (* 138902662 *)\nmin = Ceiling @ Sqrt @ FromDigits @ Riffle[Range[9], 0] (* 101010102 *)\n\n\nAs an aside, note that max^2 is below $MaxMachineInteger on 64-bit systems. But on 32-bit it isn't, which causes PerformSearch to switch back to the uncompiled code. Keeping Mark McClure's comment in mind, we'll cheat a bit and start from the maximum to find immediately: result = PerformSearch[max, -1]  138901917  How much faster is this than starting from the minimum? (result - min)/(max - result) // N  50861.5  Roughly 50 thousand times faster, not bad! Starting from the minimum takes about 30 seconds on my machine. Last but not least, let's double-check if we've obtained the correct number: result^2  19293742546274889  If you multiply this with 100, you've got your desired integer. Non-brute-force The approach above scanned roughly 38 million (!) integers in the worst-case scenario (starting from the minimum). Other answers to the OP's question have shown that you can and should go one better than that. Here's my take on an efficient general solution, using an iterative step-by-step process: ClearAll[ FindIntegerRoots, CheckIfEmpty, SelectDigitSequences, InsertNewDigits, ConvertPattern ]; FindIntegerRoots[ pattern : { (0|1|2|3|4|5|6|7|8|9|Verbatim[_]).. }, power_Integer: 2 ] /; power > 1 := With[ { maxRoot = Floor[FromDigits[pattern /. Verbatim[_] -> 9]^(1/power)] }, FromDigits /@ SelectDigitSequences[ Fold[ CheckIfEmpty @ SelectDigitSequences[ InsertNewDigits @ #1, maxRoot, power, ConvertPattern[pattern, #2] ] &, {{}}, Range @ IntegerLength @ maxRoot ], maxRoot, power, ConvertPattern[pattern, Length @ pattern] ] ~Quiet~ {CompiledFunction::cfnlts} ~Catch~ \"isEmpty\" ]; CheckIfEmpty[{}] := Throw[{}, \"isEmpty\"]; CheckIfEmpty[nonEmpty_] := nonEmpty; SelectDigitSequences = Compile[ { {digitSequences, _Integer, 2}, {maxRoot, _Integer }, {power, _Integer }, {numDigits, _Integer }, {positions, _Integer, 1}, {compareDigits, _Integer, 1} }, Module[ { powersOfTen = 10^# & /@ Range[Length @ First @ digitSequences - 1, 0, -1] }, Select[ digitSequences, And[ # <= maxRoot, IntegerDigits[#^power, 10, numDigits][[positions]] == compareDigits ] & @ ( powersOfTen.# ) & ] ] ]; InsertNewDigits = With[ { range = List /@ Range[0, 9] }, Compile[ { {digitSequences, _Integer, 2} }, Outer[#1 ~Join~ #2 &, range, digitSequences, 1] ~Flatten~ 1 ] ]; ConvertPattern[pattern_, length_] := With[ { trimmed = pattern[[Length@pattern - length + 1 ;; -1]] }, Sequence @@ { length, Flatten @ Position[trimmed, _Integer], DeleteCases[trimmed, Verbatim[_]] } ];  While the code might be long, it is nevertheless fast: FindIntegerRoots @ Riffle[Range[9], _] // AbsoluteTiming  {0.284622, {138901917}}  A performance increase of 100 over the above brute-force method! But how efficient is it? Here's a nice graph of the amount of checks it performs in this case: That's about 296 thousand checks in total; certainly much less than 38 million! Some more concealed squares The nice thing about FindIntegerRoots is that it works on any pattern, not just the OP's case: FindIntegerRoots @ {_, 9}  {3, 7}  FindIntegerRoots @ { 1, _, 4}  {12}  FindIntegerRoots @ { 1, _, 6}  {14}  And if you're adventurous, you may even ask for roots of different powers: FindIntegerRoots[{_, _, _, 5}, 3]  {5, 15}  So let's have a bit of fun and search for more concealed squares. First, are there any partial concealed squares (i.e. of the form 1_2, 1_2_3, 1_2_3_4, etc)? FindIntegerRoots @ Riffle[Range @ #, _] & /@ Range @ 9  {{1}, {}, {}, {1312}, {}, {115256, 127334, 135254}, {}, {}, {138901917}}  Flatten[%]^2   {1, 1721344, 13283945536, 16213947556, 18293644516, 19293742546274889}  How about reverse concealed squares (i.e. 2_1, 3_2_1, 4_3_2_1, etc)? FindIntegerRoots @ Riffle[Reverse @ Range @ #, _] & /@ Range @ 9  {{1}, {}, {}, {}, {24171}, {}, {2657351, 2713399}, {}, {}}  Flatten[%]^2  {1, 584237241, 7061514337201, 7362534133201}  And finally, are there squares of the form 1_1_(...)_1_1? FindIntegerRoots @ Riffle[ConstantArray[1, #], _] & /@ Range[2, 9]  {{11}, {119, 131}, {}, {}, {110369}, {}, {10065739}, {}}  Flatten[%]^2  {121, 14161, 17161, 12181316161, 101319101616121}  It seems there are many nice squares. If you happen to find a particularly nice one, let me know :) \u2022 Which version of Mathematica are you using? It's quite strange that your code generates the CompiledFunction::cfn warning in my v8 (Win 32bit & 64bit) and v9 (Win 32bit), but it does work without warning on Wolfram Cloud! \u2013 xzczd Sep 3 '14 at 14:15 \u2022 @xzczd I've checked this on v9 on a Mac. The issue here is 32 vs 64 bit; $MaxMachineInteger should namely be bigger than FromDigits@Riffle[Range[9], 9], which it isn't on 32 bit. \u2013\u00a0Teake Nutma Sep 3 '14 at 16:06\n\u2022 That seems to be the reason, BTW the \\$MaxMachineInteger in my v8 (Win 64 bit) is the same as that in 32 bit, not sure if it's just the nature of v8\u2026 \u2013\u00a0xzczd Sep 4 '14 at 1:56\n\u2022 Is using Big O notation in \"Roughly O(10^4) times faster\" a correct statement? It is surely 10^4 times faster in this example, but what exactly does O(10^4) times faster mean? \u2013\u00a0user Sep 4 '14 at 15:29\n\u2022 @bruce14 What I meant here is that it's roughly 4 orders of magnitude faster; faster than 10^3 and slower than 10^5. \u2013\u00a0Teake Nutma Sep 4 '14 at 17:15\n\nThis non-paralell strategy takes less than 4 seconds in my stone age laptop (and doesn't use the \"reverse searching\" trick)-\n\nFirst we check for the possible 6 digits endings (after discounting the final \"00\" ending in the squared number):\n\nset= 2 Position[IntegerDigits[Range[10^5 + 1, 10^6-1, 2]^2][[All, -5;; -1;; 2]], {7,8,9}]+ 10^5 - 1;\n\n\nThen we complete the whole set of numbers to check by adding all the possible \"heads\" between the max and min:\n\nfullSet = Total[Tuples[{Range[101, 138] 10^6, Flatten@set}], {2}];\n\n\nAnd then a quick search gets our number:\n\nCases[IntegerDigits[fullSet^2], {1, _, 2, _, 3, _, 4, _, 5, _,  6, _, 7, _, 8, _, 9}]\n\n(*{{1, 9, 2, 9, 3, 7, 4, 2, 5, 4, 6, 2, 7, 4, 8, 8, 9}}*)\n\n\n(* 1929374254627488900 *)\n\n\nEdit\n\nWe can still cut the time in half if we see that the number should end in 3 or 7 for its square to end in 9:\n\nset = Join[10 (Position[IntegerDigits[Range[10^5 + #, 10^6, 10]^2][[All, -5 ;; -1 ;; 2]],\n{7, 8, 9}] - 1) + # & /@ {3, 7}] + 10^5;\nfullSet = Total[Tuples[{Range[101, 138] 10^6, Flatten@set}], {2}];\nCases[IntegerDigits[fullSet^2], {1, _, 2, _, 3, _, 4, _, 5, _,  6, _, 7, _, 8, _, 9}]\n\n\u2022 +1 Very concise and fast (~1.3 times faster than mine!). \u2013\u00a0Teake Nutma Sep 4 '14 at 20:31\n\u2022 @TeakeNutma Thanks! I halved the time in the edit.Less than 2 secs now on my poor man's machine :) \u2013\u00a0Dr. belisarius Sep 5 '14 at 2:41\n\u2022 That's pretty fast. But alas, I've updated my answer and now it's 2x faster than yours :). \u2013\u00a0Teake Nutma Sep 5 '14 at 10:11\n\nThere is a method by using Catch and Throw, and it gives result in about 90s. I think it can be improve a lot by Paralilize or ParallelEvaluate, but failed to finish that.\n\ni = 10^9 + {30, 70};\nisMatchQ =\nIf[IntegerDigits[#^2][[1 ;; -1 ;; 2]] ==  {1, 2, 3, 4, 5, 6, 7, 8,\n9, 0}, Throw[#]] &;\nCatch[Do[isMatchQ /@ i; i += 100, {10^9}]]\n\n\nEdit\n\nI have found a way to speed up it by ParallelTable, this is the code:\n\nlen = ((Sqrt[1.93*10^16] - 1*10^8)/40 // IntegerPart)*10;\nisMatchQ =\nIf[IntegerDigits[#^2][[1 ;; -1 ;; 2]] ==  {1, 2, 3, 4, 5, 6, 7, 8,\n9}, Throw[#]] &;\nParallelTable[n = 10^8 + {3, 7} + len*i;\nCatch[Do[isMatchQ /@ n; n += 10, {len / 10}]], {i, 0,\n3}] // AbsoluteTiming\n(*{42.523432, {Null, Null, Null, 138901917}}*)\n\n\nI divided the whole range of n which makes 10^16 <= n^2 <= 1.93*10^16 into 4 parts, the length of each part is len. And I calculate these 4 parts in 4 kernels at the same time by ParallelTable, then get the result in a much shorter time, only 42.5s. (Sorry for my poor English....)\n\nHere's a slight improvement of @wuyingddg's Catch&Throw method. The key point is to make Throw working inside ParallelDo using the method mentioned in this post:\n\nSetSharedFunction[pThrow];\npThrow[expr_] := Throw[expr]\nisMatchQ =\nIf[IntegerDigits[#^2][[1 ;; -1 ;; 2]] == {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}, pThrow[#]] &;\nCatch[ParallelDo[\nisMatchQ /@ (i + {30, 70}), {i, 10^9, Sqrt[193 10^16], 100}]] // AbsoluteTiming\n`\n\u2022 It is much more brief than the ParallelTable one and spend nearly same time, pretty good! \u2013\u00a0wuyingddg Sep 3 '14 at 14:21", "date": "2021-01-23 02:24:08", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/58751/how-to-speed-up-my-project-euler-code?noredirect=1", "openwebmath_score": 0.17412593960762024, "openwebmath_perplexity": 3841.9052069389054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551556203815, "lm_q2_score": 0.8887587979121384, "lm_q1q2_score": 0.8575235082684994}}
{"url": "https://la.mathworks.com/help/econ/dtmc.lazy.html", "text": "lazy\n\nDescription\n\nexample\n\nlc = lazy(mc) transforms the discrete-time Markov chain mc into the lazy chain lc with an adjusted state inertia.\n\nexample\n\nlc = lazy(mc,w) applies the inertial weights w for the transformation.\n\nExamples\n\ncollapse all\n\nConsider this three-state transition matrix.\n\n$P=\\left[\\begin{array}{ccc}0& 1& 0\\\\ 0& 0& 1\\\\ 1& 0& 0\\end{array}\\right].$\n\nCreate the irreducible and periodic Markov chain that is characterized by the transition matrix P.\n\nP = [0 1 0; 0 0 1; 1 0 0];\nmc = dtmc(P);\n\nAt time t = 1,..., T, mc is forced to move to another state deterministically.\n\nDetermine the stationary distribution of the Markov chain and whether it is ergodic.\n\nxFix = asymptotics(mc)\nxFix = 1\u00d73\n\n0.3333    0.3333    0.3333\n\nisergodic(mc)\nans = logical\n0\n\nmc is irreducible and not ergodic. As a result, mc has a stationary distribution, but it is not a limiting distribution for all initial distributions.\n\nShow why xFix is not a limiting distribution for all initial distributions.\n\nx0 = [1 0 0];\nx1 = x0*P\nx1 = 1\u00d73\n\n0     1     0\n\nx2 = x1*P\nx2 = 1\u00d73\n\n0     0     1\n\nx3 = x2*P\nx3 = 1\u00d73\n\n1     0     0\n\nsum(x3 == x0) == mc.NumStates\nans = logical\n1\n\nThe initial distribution is reached again after several steps, which implies that the subsequent state distributions cycle through the same sets of distributions indefinitely. Therefore, mc does not have a limiting distribution.\n\nCreate a lazy version of the Markov chain mc.\n\nlc = lazy(mc)\nlc =\ndtmc with properties:\n\nP: [3x3 double]\nStateNames: [\"1\"    \"2\"    \"3\"]\nNumStates: 3\n\nlc.P\nans = 3\u00d73\n\n0.5000    0.5000         0\n0    0.5000    0.5000\n0.5000         0    0.5000\n\nlc is a dtmc object. At time t = 1,..., T, lc \"flips a fair coin\". It remains in its current state if the \"coin shows heads\" and transitions to another state if the \"coin shows tails\".\n\nDetermine the stationary distribution of the lazy chain and whether it is ergodic.\n\nlcxFix = asymptotics(lc)\nlcxFix = 1\u00d73\n\n0.3333    0.3333    0.3333\n\nisergodic(lc)\nans = logical\n1\n\nlc and mc have the same stationary distributions, but only lc is ergodic. Therefore, the limiting distribution of lc exists and is equal to its stationary distribution.\n\nConsider this theoretical, right-stochastic transition matrix of a stochastic process.\n\n$P=\\left[\\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\\\ 0& 0& 1/3& 0& 2/3& 0& 0\\\\ 0& 0& 0& 0& 0& 1/3& 2/3\\\\ 0& 0& 0& 0& 0& 1/2& 1/2\\\\ 0& 0& 0& 0& 0& 3/4& 1/4\\\\ 1/2& 1/2& 0& 0& 0& 0& 0\\\\ 1/4& 3/4& 0& 0& 0& 0& 0\\end{array}\\right].$\n\nCreate the Markov chain that is characterized by the transition matrix P.\n\nP = [ 0   0  1/2 1/4 1/4  0   0 ;\n0   0  1/3  0  2/3  0   0 ;\n0   0   0   0   0  1/3 2/3;\n0   0   0   0   0  1/2 1/2;\n0   0   0   0   0  3/4 1/4;\n1/2 1/2  0   0   0   0   0 ;\n1/4 3/4  0   0   0   0   0 ];\nmc = dtmc(P);\n\nPlot the eigenvalues of the transition matrix on the complex plane.\n\nfigure;\neigplot(mc);\ntitle('Original Markov Chain')\n\nThree eigenvalues have modulus one, which indicates that the period of mc is three.\n\nCreate lazy versions of the Markov chain mc using various inertial weights. Plot the eigenvalues of the lazy chains on separate complex planes.\n\nw2 = 0.1;                                 % More active Markov chain\nw3 = 0.9;                                 % Lazier Markov chain\nw4 = [0.9 0.1 0.25 0.5 0.25 0.001 0.999]; % Laziness differs between states\n\nlc1 = lazy(mc);\nlc2 = lazy(mc,w2);\nlc3 = lazy(mc,w3);\nlc4 = lazy(mc,w4);\n\nfigure;\neigplot(lc1);\ntitle('Default Laziness');\n\nfigure;\neigplot(lc2);\ntitle('More Active Chain');\n\nfigure;\neigplot(lc3);\ntitle('Lazier Chain');\n\nfigure;\neigplot(lc4);\ntitle('Differing Laziness Levels');\n\nAll lazy chains have only one eigenvalue with modulus one. Therefore, they are aperiodic. The spectral gap (distance between inner and outer circle) determines the mixing time. Observe that all lazy chains take longer to mix than the original Markov chain. Chains with different inertial weights than the default take longer to mix than the default lazy chain.\n\nInput Arguments\n\ncollapse all\n\nDiscrete-time Markov chain with NumStates states and transition matrix P, specified as a dtmc object. P must be fully specified (no NaN entries).\n\nInertial weights, specified as a numeric scalar or vector of length NumStates. Values must be between 0 and 1.\n\n\u2022 If w is a scalar, lazy applies it to all states. That is, the transition matrix of the lazy chain (lc.P) is the result of the linear transformation\n\n${P}_{\\text{lazy}}=\\left(1-w\\right)P+wI.$\n\nP is mc.P and I is the NumStates-by-NumStates identity matrix.\n\n\u2022 If w is a vector, lazy applies the weights state by state (row by row).\n\nData Types: double\n\nOutput Arguments\n\ncollapse all\n\nDiscrete-time Markov chain, returned as a dtmc object. lc is the lazy version of mc.\n\ncollapse all\n\nLazy Chain\n\nA lazy version of a Markov chain has, for each state, a probability of staying in the same state equal to at least 0.5.\n\nIn a directed graph of a Markov chain, the default lazy transformation ensures self-loops on all states, eliminating periodicity. If the Markov chain is irreducible, then its lazy version is ergodic. See graphplot.\n\nReferences\n\n[1] Gallager, R.G. Stochastic Processes: Theory for Applications. Cambridge, UK: Cambridge University Press, 2013.", "date": "2020-08-15 08:40:51", "meta": {"domain": "mathworks.com", "url": "https://la.mathworks.com/help/econ/dtmc.lazy.html", "openwebmath_score": 0.7474462985992432, "openwebmath_perplexity": 1981.7817231841414, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551505674445, "lm_q2_score": 0.8887588023318195, "lm_q1q2_score": 0.8575235080420094}}
{"url": "https://math.stackexchange.com/questions/4258331/is-this-relation-symmetric-i-think-it-is-but-my-professor-claims-it-isnt/4258341", "text": "# Is this relation symmetric? I think it is but my professor claims it isn't.\n\nIs the relation $$R = \\{(a,b),(a,c),(b,a),(b,c),(c,a),(c,b),(d,d)\\}$$ symmetric?\n\nMy professor claims that if $$(d,d)$$ was not included, it would be symmetric, but the inclusion of $$(d,d)$$ ruins it because $$d$$ has to connect to another element of the relation.\n\n\u2022 Two of the defining properties of equivalence relations are that they are both reflexive and symmetric, which would be useless if they were mutually exclusive :-P Sep 24 '21 at 4:12\n\u2022 Hope your professor is willing to get corrected in this case! Sep 24 '21 at 8:39\n\u2022 You could tell your professor that $(d,d)\\in R$ \"connects\" to itself, ie, for the element $(\\color{red}d,\\color{green}d)$ in $R$, we have the element $(\\color{green}d,\\color{red}d)$ in $R$. There is no restriction that it must \"connect\" to a distinct element in $R$. As an example, the relation $R=\\{(a,a)\\}$ on the singleton set $\\{a\\}$ is an equivalence relation: reflexive, symmetric and transitive. Sep 24 '21 at 19:16\n\nA very simple way to see if a relation $$R$$ is symmetric is to check its inverse relation $$R^{-1}$$, where $$(x,y)\\in R\\Leftrightarrow (y,x)\\in R^{-1}$$\n\nSince for the given relation, $$R=R^{-1}$$, it is symmetric.\n\n$$\\color{green}{\\text{YOU ARE CORRECT.}}$$\n\nYes, that relation is symmetric. The definition of symmetry does not require each element to be connected to some other element; $$R$$ is symmetric iff for every $$x,y$$ such that $$(x,y)\\in R$$ it is also the case that $$(y,x)\\in R$$.\n\nOne way to see if a relation is symmetric is to draw the table what is in relation to what:\n\n$$\\begin{array}{r|c|c|c|c}R&a&b&c&d\\\\\\hline a&F&T&T&F\\\\b&T&F&T&F\\\\c&T&T&F&F\\\\d&F&F&F&T\\end{array}$$\n\nand just see if the table is symmetrical across the main diagonal.\n\nAnd - in this case it is. Thus, the relation is symmetric. Having $$(d,d)$$ in it has nothing to do with it, it would be symmetric either way (i.e. whether the entry for $$(d,d)$$ is \"true\" or \"false\").", "date": "2022-01-21 13:47:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4258331/is-this-relation-symmetric-i-think-it-is-but-my-professor-claims-it-isnt/4258341", "openwebmath_score": 0.8399157524108887, "openwebmath_perplexity": 170.1610735450462, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551546097941, "lm_q2_score": 0.8887587957022977, "lm_q1q2_score": 0.8575235052381548}}
{"url": "http://gpof.trattoriaborgo.it/midpoint-riemann-sum-sigma-notation.html", "text": "# Midpoint Riemann Sum Sigma Notation\n\nWe call Ln the left Riemann sum for the function f on the interval [a, b]. Similarly, the right Riemann sum is an overestimate. e\u2212x2 dx by the Riemann sum with n = 4 subintervals and left end-points as sample points. Simmons computes R xdx. To express left, right, and midpoint Riemann sums in sigma notation, we must identify the points \u00d1k. One of the programs is opened so students can see that it is nothing more than the summation notation they had. First, determine the width of each rectangle. n n n b a x 4 0 4. 2 Area & Sigma Notation 2. Here \u2206x = 3\u22121 10 = 0. How to Write Riemann Sums with Sigma Notation; How to Write Riemann Sums with Sigma Notation. 6255658911511259 1. Also, identify when an estimate is an overestimate or underestimate. If the sym- bol oo appears above the E, it indicates that the terms go on indefinitely. The same number of subintervals were used to produce each approximation. 603125 Extension \u2013 Area Programs Students will use the programs to compare area approximation methods. 2) Use the Midpoint Rule with n = 4 to approximate x 4) dx. (a) R 3 \u22121 xdx (b) R 4 2 x2dx. In each subinterval, choose a point c1, c2,cn and form the sum ; This is called a Riemann Sum ; NOTE LRAM, MRAM, and RRAM are all Riemann sums. rotation (2x2 matrices) row echelon form. 2 - Page 266 35 including work step by step written by community members like you. 6078493243021688 1. EXAMPLE 1: Find the area under the curve of the function f x ( ) =x +8 over the interval [0, 4] by using n rectangles. The same number of subintervals were used to produce each approximation. 35 4n+ 5k - 5 \u041e\u0410. The Midpoint Rule summation is: $$\\ds \\sum_{i=1}^n f\\left(\\frac{x_i+x_{i+1}}{2}\\right)\\Delta x\\text{. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Write the sigma notation. Riemann Sums; The Definite Integral and FTC; Indefinite Integrals; 2 Techniques of Integration. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. 19) f(x) = x 2 - 2 , [0, 8], midpoint 19) 20) f(x) = cos x + 3 , [0, 2 ], left- hand endpoint 20) Express the sum in sigma notation. Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people\u2014spanning all professions and education levels. Riemann Sums Name:_____ Date:_____ A Riemann sum Sn for the function is defined to be Sn = n k fck x 1 (). They saw how these come together when finding a Riemann Sum, as shown below. }$$ 6 Evaluating Riemann sums with data A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. 1 Riemann Sums and Area 3 2. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(\u00d12) Ax + + f (in) Ax Ef(\u00d1k) Ax. higher than the actual area B. f) General Riemann Sum. 2 Trapezoidal Rule Example Find the area under x3 using 4 subintervals using: left, right, midpoint and trapezoidal methods from [2, 3] Example \u2013 A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. The notation is represented by the upper case version of the Greek letter sigma. Then study what happens to a nite sum approximation as the number of terms approaches in nity. Riemann sums in summation notation: challenge problem. The left and right Riemann sums of a function f on the interval [2, 6] are denoted by LEFT ( n ) and RIGHT( n ), respectively, when the interval is divided int\u2026 Enroll in one of our FREE online STEM summer camps. You can use sigma notation to write out the Riemann sum for a curve. One of the programs is opened so students can see that it is nothing more than the summation notation they had. 7) The table shows the velocity of a remote controlled race car moving along a dirt path for 8 seconds. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. 11) Use sigma notation to \ufb01nd the right Riemann Sum for f (x) = x 3 + 2 on [0, 3] with n = 30. The region bounded by y = x2, the x-axis, from x = 0 to x = 2. First, determine the width of each rectangle. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. Conic Sections. We call Ln the left Riemann sum for the function f on the interval [a, b]. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. 1 Approximate the area under the curve f(x) = ln(x) between x= 1 and x= 5. The endpoints a and b are called the limits of integration. Is the midpoint Riemann sum an over or under approximation if the graph is: a. Also, identify when an estimate is an overestimate or underestimate. k be the midpoint of the kth subinterval (where all subintervals have equal width). The k\" subinterval has width. 1 Riemann Sums and Definite Integrals. (k) (n-1) 4n 729 OC. In mathematics, the Riemann sum is defined as the approximation of an integral by a finite sum. By using this website, you agree to our Cookie Policy. Approximating the area under a curve using some rectangles. (Sigma notation for nite sums ) The symbol Xn k=1 a k denotes the sum a 1 + a 2 + + a n. The Midpoint Rule for de\ufb01nite integrals means to approximate the integral by using a midpoint Riemann Sum just as in 6. the left sum approximation to R 2\u03c0 0 sin(x)dx (a) with 8 equal subintervals (b) with n equal subintervals 2. where is the number of subintervals and is the function evaluated at the midpoint. The RiemannSum(f(x), x = a. Then evaluate the sum. So what happens if the \"area\" is below the x-axis as I mentioned before, area is inherently positive, but a Riemann Sum and therefore an integral can have negative values if the curve lies below the. Use the midpoint Riemann sum with n = 5 to nd an estimates on the area under the curve on the interval [0;10]. 2 Sigma Notation and Limits of Finite Sums In this section we introduce a convenient notations for sums with a large number of terms. Take a midpoint sum using only one sub-interval, so we only get one rectangle: The midpoint of our one sub-interval [0, 4] is 2. We have step-by-step solutions for your textbooks written by Bartleby experts!. By the way, you don\u2019t need sigma notation for the math that follows. Thomas\u2019 Calculus 13th Edition answers to Chapter 5: Integrals - Section 5. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. Midpoint Rule: As noted above, the midpoint rule is a special case of Riemann sums where the interval integration [a, b] is divided n subintervals [x i-1, x i] each with length Dx = (b-a)/n. We are approximating an area from a to b with a=0 and b=5, n=5, right endpoints and f(x)=25-x^2 (For comparison, we'll do the same problem, but use left endpoints after we finish this. where is the number of subintervals and is the function evaluated at the midpoint. \u201d Example 1: Evaluate the Riemann sum for f ( x ) = x 2 on [1,3] using the four subintervals of equal length, where x i is the right endpoint in the i th subinterval (see Figure ). Do NOT evaluate your summation. Step 5 requires the formulas and properties of the sigma notation. 1 sigma notation and riemann sums 305 Area Under a Curve: Riemann Sums Suppose we want to calculate the area between the graph of a positive function f and the x-axis on the interval [a,b] (see below left). Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. Choose the correct Riemann sum below. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. \u222b(1, 2) sin(1/x)dx. The sum on the right hand side is the expanded form. 6093739310551827. Arnold Schwarzenegger This Speech Broke The Internet AND Most Inspiring Speech- It Changed My Life. - Duration: 14:58. This is useful when you want to derive the formula for the approximate area under the curve. Write the midpoint Riemann sum in sigma nota-tion with n = 20. I expect you to show your reasoning clearly and in an organized fashion. (2k-1) (2n +1 -2K) k1 4n 729 OB. Math Help Boards: Sum Calculator. Similar formulas can be obtained if instead we choose c k to be the left-hand endpoint, or the midpoint, of each subinterval. 1 Approximate the area under the curve f(x) = ln(x) between x= 1 and x= 5. Sums can also be infinite, provided that the terms eventually get close enough to zero\u2013this is an important topic in calculus. Most of the following problems are average. 6093739310551827. In the more compact sigma notation, we have Ln = Xn\u22121 i=0 f (xi)4x. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(\u00d12) Ax + + f (in) Ax Ef(\u00d1k) Ax. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. It is applied in calculus to formalize the method of exhaustion, used to determine the area of a region. EXAMPLE 1Using Sigma Notation\u0192s1d + \u0192s2d + \u0192s3d +\u00c1+ \u0192s100d =a100i=1\u0192sid. ) Let f (x) be defined on a, b. and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. rules of exponents. Left-Hand. Calculus Q&A Library se sigma notation to write the following Riemann sum. Let f(x) = 2/x a. 50 1 3+ cos 10 k = 1 2tk \u2013 T O A. Area under Curves 5. See math and science in a new way. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Here is how to set up the Riemann sum for the de\ufb01nite integral Z 3 1 x2 dx where n = 10: (1) Find \u2206x = b\u2212a n. Evaluate the sum using a calculator with n=20,50, and 100. Let's visualize rectangles in the left, right and midpoint Riemann sums for the function f = lambda x : 1/(1+x**2) a = 0; b = 5; N = 10 n = 10 # Use n*N+1 points to plot the function dx/2,N) x_right = np. How do you determine that it is defined on [0, 1] and what would the sigma notation look like for this? I understand how to calculate a Riemann sum, I am just not understanding how they get [0,1] from the given information :/. Other sums The choice of the $c_i$ will give different answers for the approximation, though for an integrable function these differences will vanish in the limit. Write the sum without sigma notation and evaluate it. }\\) 6 Evaluating Riemann sums with data A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. scalar multiplication and matrices (Properties) scalar multiplication of vectors. 1 would become x1, and 2 would be x2. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. We have step-by-step solutions for your textbooks written by Bartleby experts!. A 2 \ud835\udc5b 2 + 4 \ud835\udc5b 1 2 + \ud835\udc56. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. The region bounded by y = x2, the x-axis, from x = 1 to x = 3. Approximating the area under a curve using some rectangles. 1 + \u221a x 1 \u2212 \u221a x dx 6. The application is intended to demonstrate the use of Maple to solve a particular problem. b, method = midpoint, opts) command calculates the midpoint Riemann sum of f(x) from a to b. Free midpoint calculator - calculate the midpoint between two points using the Midpoint Formula step-by-step This website uses cookies to ensure you get the best experience. For left Riemann sums, the left endpoints of the subintervals are 1) Ax, fork \u2014. Riemann Sums; The Definite Integral and FTC; Indefinite Integrals; 2 Techniques of Integration. b, method = left, opts) command calculates the left Riemann sum of f(x) from a to b. (a) R 5 \u22121 xdx (b) R 2 1 x2dx 2. rotation (2x2 matrices) row echelon form. Similarly, the right Riemann sum is an overestimate. 6078493243021688 1. This behavior persists for more rectangles. So, for example, over here we could we could use the midpoint between x0 and x1 to find the height of the rectangle. Hence the Riemann sum associated to this partition is: Xn i=1 \u00b5 i n \u00b62 1/n = 1 n3 Xn i=1 i2 = 1 n3 2n3 +3n2 +n 6 = 2+3/n+2/n2 6. Note particularly that since the index of summation begins at 0 and ends at n \u2212 1, there are indeed n terms in this sum. 12+ 22+ 32+ 42+ 52+ 62+ 72+ 82+ 92+ 102+ 112=a11k=1k2,k 1aknThe index k ends at k n. Example: Estimate the area under the graph of f(x) = x2 + 1 over the interval [2;10] using 4 rectangles of equal width and midpoints. Riemann Sums. Constructing Accurate Graphs of Antiderivatives; The Second Fundamental Theorem of Calculus; Integration by Substitution; Integration by Parts; Other Options for Finding Algebraic Antiderivatives; Numerical Integration; 6 Using Definite Integrals. Therefore, the Riemann sum is: The upper-case Greek letter Sigma \u03a3 is used to stand for sum. Know and understand the sum, di erence, constant multiple, and constant value rule for nite sums in Sigma notation. In each case where you used a Riemann sum to estimate an area, try to determine if you obtained an overestimate or an underestimate. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. This is called a \"Riemann sum\". A summation is a sum of numbers that are typically defined by a function. and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. 15k 2 + 4 C. Calculus Q&A Library se sigma notation to write the following Riemann sum. A Riemann sum is an approximation of the area of a region that is found by dividing the region into rectangles or trapezoids. This is useful when you want to derive the formula for the approximate area under the curve. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. ) Find the limit of the Riemann Sum (found in part d) as the number of rectangles are increased to infinity. Series, Sigma Notation video (Leckie) Summation Notation (Paul) 1. The limit of Finite approximation to an Area. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. In mathematics, the Riemann sum is defined as the approximation of an integral by a finite sum. 3) Mid-point sums All value may be different but they represent a same quantity an approximated area under the curve. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. EXAMPLE 1Using Sigma Notation\u0192s1d + \u0192s2d + \u0192s3d +\u00c1+ \u0192s100d =a100i=1\u0192sid. Midpoint Riemann Sums: Suppose x i is the midpoint of the ith subinterval [x i 1;x i], that is x i = x i 1 + x i 2 for all i. For example, saying \u201cthe sum from 1 to 4 of n\u00b2\u201d would mean 1\u00b2+2\u00b2+3\u00b2+4\u00b2 = 1 + 4 + 9 + 16 = 30. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. Left, midpoint, and right Riemann sums were used to estimate the area between the graph of \ud835\udc53(\ud835\udc65) and the x-axis on the interval [3, 7]. In Chapter 1, I introduce you to the Riemann sum formula for the definite integral. While summation notation has many uses throughout math (and specifically calculus), we want to focus on how we can use it to write Riemann sums. See full list on khanacademy. scalar product. The Midpoint Rule for de\ufb01nite integrals means to approximate the integral by using a midpoint Riemann Sum just as in 6. Use right Riemann sums to compute the following integrals. You always increase by one at each successive step. Take a photo of your homework question and get answers, math solvers, explanations, and videos. This is accomplished in a three-step procedure. Riemann sums in summation notation: challenge problem. 603125 Extension \u2013 Area Programs Students will use the programs to compare area approximation methods. you'll have to picture the above and below numbers because I can't show them on here. The a\u2019s are. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum (just as in 6. exactly equal to the actual area C. The first two arguments (function expression and range) can be replaced by a definite integral. Simplify the expression. A Task template for the Riemann Sum can be found in the following location: Tools>Tasks>Browse>Calculus>Integration>Riemann Sums and choose one of the options (Left, Right, or Midpoint). Write the midpoint Riemann sum in sigma notation for an arbitrary value of n. Use sigma notation to write a new sum $$R$$ that is the right Riemann sum for the same function, but that uses twice as many subintervals as $$S\\text{. Riemann Sums: Sigma Notation Review This is a Riemann sum for f on the interval [a,b]. n n n b a x 4 0 4. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Use a midpoint sum with 2 sub-intervals to estimate. a) Write a summation to approximate the area under the graph of \u8840\u5cab\u6372\u5cbb\u567a\u6372\u6234\u9aea\u306e from x = 1 to x = 7 using the right endpoints of three subintervals of equal length. Substitution Rule; Powers of Trigonometric Functions; Trigonometric Substitutions; Integration by Parts; Partial Fraction Method for Rational Functions; Numerical Integration; Improper Integrals; Additional Exercises; 3 Applications of Integration. Step 2: Partition the interval into n subintervals. (n times) = cn, where c is a constant. Choose the correct answer below. 4 Financial applications of geometric sequences and series: compound interest, annual depreciation. n n n b a x 4 0 4. The endpoints are given by x 0 = a and x n = b. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. Sigma/summation notation. Definition S L: Left-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area \u2248 S R: Right-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area \u2248 S M: Midpoint Riemann Sum. One of the programs is opened so students can see that it is nothing more than the summation notation they had. Calculus Q&A Library se sigma notation to write the following Riemann sum. In sigma notation, we get that 1 2 3 99 = E k In This Module \u2022 We will introduce sigma notation \u2014 a compact way of writing large sums of like terms \u2014 and define the notion of a Riemann sum. 1 Riemann Sums and Area 3 2. Riemann Sums Name:_____ Date:_____ A Riemann sum Sn for the function is defined to be Sn = n k fck x 1 (). 318: 49-56 2. If the sym- bol oo appears above the E, it indicates that the terms go on indefinitely. The application is intended to demonstrate the use of Maple to solve a particular problem. Sigma Notation and Riemann Sums Sigma Notation: Notation and Interpretation of 12 3 14 1 n k nn k aaaaa a a (capital Greek sigma, corresponds to the letter S) indicates that we are to sum numbers of the form indicated by the general term. ) 2 \u22121 sin \u03c0x 4 dx 2. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. You always increase by one at each successive step. 0375 When n = 100, Left Riemann sum = 164. The remaining formulas are simple rules for working with sigma notation: \u2217to be the midpoint of the Any Riemann sum is an approximation to an integral, but. So this is right over here. In mathematics, the Riemann sum is defined as the approximation of an integral by a finite sum. Option #1: If you noticed in step 2 above, we did not care if our subintervals were the same width. \u03a3 35 8n + 10k-5 n nV 2n n k=1 k=1 n 35 4n+ 5k OC. Sums and sigma notation. Similar formulas can be obtained if instead we choose c k to be the left-hand endpoint, or the midpoint, of each subinterval. Let's visualize rectangles in the left, right and midpoint Riemann sums for the function f = lambda x : 1/(1+x**2) a = 0; b = 5; N = 10 n = 10 # Use n*N+1 points to plot the function dx/2,N) x_right = np. root method. Riemann Sums: Sigma Notation Review This is a Riemann sum for f on the interval [a,b]. Evaluate the integral. By using this website, you agree to our Cookie Policy. Use a midpoint sum with 2 sub-intervals to estimate. Midpoint Riemann Sum. Riemann Sums, Sigma Notation and Writing Area as a Limit Lesson:Your AP Calculus students express the limit of a Riemann sum in integral notation and write integral notation as a limit of a Riemann sum. Option #1: If you noticed in step 2 above, we did not care if our subintervals were the same width. Choose The Correct Riemann Sum Below. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(\u00d12) Ax + + f (in) Ax Ef(\u00d1k) Ax. The subscript k is the summation index, and is a \u201cdummy index\u201d, in the sense that it can be replaced by any convenient letter. Find a closed form for a nite sum using the Gauss formula P n i=1 k= n(n+1) 2 and the formulas on pg. Left-Hand. 603125 Extension \u2013 Area Programs Students will use the programs to compare area approximation methods. Similarly, the right Riemann sum is an overestimate. How do you find Find the Riemann sum that approximates the integral #int_0^9sqrt(1+x^2)dx# using How do you Use a Riemann sum to approximate the area under the graph of the function #y=f(x)# on How do you use a Riemann sum to calculate a definite integral?. d) The Definite Integral as an Area and Average. right Riemann sum. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. [Analysis] Exercise 2. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. Write the sigma notation. Series, Sigma Notation video (Leckie) Summation Notation (Paul) 1. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. Summation notation (or sigma notation) allows us to write a long sum in a single expression. EXAMPLE 1Using Sigma Notation\u0192s1d + \u0192s2d + \u0192s3d +\u00c1+ \u0192s100d =a100i=1\u0192sid. The Midpoint Rule summation is: \\(\\ds \\sum_{i=1}^n f\\left(\\frac{x_i+x_{i+1}}{2}\\right)\\Delta x\\text{. As the partitions in [a,b] become finer and finer, our sum midpoint of. EXAMPLE 1: Find the area under the curve of the function f x ( ) =x +8 over the interval [0, 4] by using n rectangles. midpoint of each subinterval. Find the exact value of the definite integral. 10 50 1 3+ cos 10 k= 1 \u5e26) OB. With this notation, a Riemann sum can be written as \\Sigma_{i=1}^n f(c_i)(x_i-x_{i-1}). [1,2] using left end, right end, and midpoint A right and left Riemann sum are used. (1) Z 6 \u22121 (2x \u22124) dx 9. Example: Estimate the area under the graph of f(x) = x2 + 1 over the interval [2;10] using 4 rectangles of equal width and midpoints. Computing Integrals using Riemann Sums and Sigma Notation Math 112, September 9th, 2009 Selin Kalaycioglu The problems below are fairly complicated with several steps. v2 (1 \u2212v)6 dv 7. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. If you have a table of values, see midpoint rule calculator for a table. Riemann Sums; The Definite Integral; The Fundamental Theorem of Calculus; 5 Evaluating Integrals. Step 2: Partition the interval into n subintervals. We break the interval between 0 and 1 into n parts, each of width. - Duration: 14:58. By the way, you don't need sigma notation for the math that follows. + 15 \u20221 1 + 4 15 \u2022 2 2 + 4 The value of the sum is. In the examples below, we'll calculate with. Works for Math, Science, History, English, and more. 35 4n+ 5k - 5 \u041e\u0410. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Example 1: calculate Riemann sum for y = x^2 over the interval [0, 2] for 4 equal intervals. We call Ln the left Riemann sum for the function f on the interval [a, b]. The RiemannSum(f(x), x = a. 10 50 ( 2Tk \u2013 1)). Examples Example 2. 1 Approximate the area under the curve f(x) = ln(x) between x= 1 and x= 5 using the left, right and midpoint rules with n= 4 intervals (rectangles). EXAMPLE 1Using Sigma Notation\u0192s1d + \u0192s2d + \u0192s3d +\u00c1+ \u0192s100d =a100i=1\u0192sid. What is Meant by Riemann Sum? In mathematics, the. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum (just as in 6. the left sum approximation to R 2\u03c0 0 sin(x)dx (a) with 8 equal subintervals (b) with n equal subintervals 2. 6078493243021688 1. Use sigma notation and the appropriate summation formulas to formulate an expression which represents the net signed area between the graph of f(x) = cosxand the x-axis on the interval [ \u02c7;\u02c7]. For a right Riemann sum, for , we determine the sample points as follows: Now, we can approximate the area with a right Riemann sum. The value of this left endpoint Riemann sum is _____, and it is an there is ambiguity the area of the region enclosed by y=f(x), the x-axis, and the vertical lines x = 4 and x = 8. Conic Sections. Watch the next lesson: https://www. Sigma Notation or Summation Notation. Similarly, the right Riemann sum is an overestimate. Adding up a bunch of terms can be cumbersome when there are a large number of terms. Your students will have guided notes, homework, and a content quiz on Riemann Sums and Sigma Nota. 603125 Extension \u2013 Area Programs Students will use the programs to compare area approximation methods. (All of them to start with. 2: Definition of a definite integral; Riemann sum; vocabulary (integrand, integral sign, differential, limits of integration) midpoint rule; trapezoidal rule (actually equivalent to the average of left and right rectangle rules). Sketch and nd the area of the region bounded by y = x3 and y = 4x2 4x. The Riemann sums are the called respectively the left, right, mid, upper and lower Riemann sum. Sums can also be infinite, provided that the terms eventually get close enough to zero\u2013this is an important topic in calculus. It\u2019s just a \u201cconvenience\u201d \u2014 yeah, right. In sigma notation, we get that 1 2 3 99 = E k In This Module \u2022 We will introduce sigma notation \u2014 a compact way of writing large sums of like terms \u2014 and define the notion of a Riemann sum. Write the sigma notation. All other letters are constants with respect to the sum. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. The second part says that the definite integral of a continuous function from a to b can be found from any one of the function\u2019s antiderivatives F as the number F(b)- F(a). 19) f(x) = x 2 - 2 , [0, 8], midpoint 19) 20) f(x) = cos x + 3 , [0, 2 ], left- hand endpoint 20) Express the sum in sigma notation. The subscript k is the summation index, and is a \u201cdummy index\u201d, in the sense that it can be replaced by any convenient letter. Compute sums expressed using sigma notation (Prob #5) Write sums using sigma notation (Prob #15) Compute sums using properties of sigma notation and formulas for common series (Prob #33) Section 5. simplified by using sigma notation and summation formulas to create a Riemann Sum. 10 50 ( 2Tk \u2013 1)). Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Take a photo of your homework question and get answers, math solvers, explanations, and videos. Use a midpoint sum with 2 sub-intervals to estimate. There are many ways to write a given sum in sigma notation. We're going to stick with RECTANGLES for the time being. 2 Riemann Sums and Integration Learning Goal 2. State the right Riemann Sum for the function on the given interval. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. Midpoint Rule: As noted above, the midpoint rule is a special case of Riemann sums where the interval integration [a, b] is divided n subintervals [x i-1, x i] each with length Dx = (b-a)/n. se sigma notation to write the following Riemann sum. (2) Find the endpoints of. Sigma notation enables us to express a large sum in compact form: = an\u2014I an The Greek capital letter (sigma) stands for \"sum. d dx \u22122 x3 dv v2 4. Use right Riemann sums to compute the following integrals. 15k 2 + 4 C. }$$ Figure 1. 312 for the rst nsquares and the rst ncubes. Our courses show you that math, science, and computer science are \u2013 at their core \u2013 a way of thinking. and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. While the rectangles in this example do not approximate well the shaded area, they demonstrate that the subinterval widths may vary and the heights of the rectangles can be. There are a number of different types of Riemann sum that are important to master for the AP Calculus BC exam. Our courses show you that math, science, and computer science are \u2013 at their core \u2013 a way of thinking. Example: Estimate the area under the graph of f(x) = x2 + 1 over the interval [2;10] using 4 rectangles of equal width and midpoints. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Find a closed form for a nite sum using the Gauss formula P n i=1 k= n(n+1) 2 and the formulas on pg. Following these steps gives you a Riemann Sum for f on the interval [a, b]. The first two arguments (function expression and range) can be replaced by a definite integral. When shown the Riemann Sum notation, each parameter was defined and discussed in detail, to include the Greek capital letter for sigma. Choose The Correct Riemann Sum Below. Can any one help how to find approximate area under the curve using Riemann Sums in R? It seems we do not have any package in R which could help. Deep bhayani on March 7, 2017 at 8:36 pm said: Riemann sum calculator There stand four temples in a row in a holy place. In sigma notation, we get that 1 2 3 99 = E k In This Module \u2022 We will introduce sigma notation \u2014 a compact way of writing large sums of like terms \u2014 and define the notion of a Riemann sum. See full list on khanacademy. Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. On a definite integral, above and below the summation symbol are the boundaries of the interval, The numbers a and b are x -values and are called the limits of integration ; specifically, a is the lower limit and b is the upper limit. Download Free Mp4 LRAM, RRAM, and MRAM Tutorial TvShows4Mobile, Download Mp4 LRAM, RRAM, and MRAM Tutorial Wapbaze,Download LRAM, RRAM, and MRAM Tutorial Wapbase. Riemann Sums Using Sigma Notation With sigma notation, a Riemann sum has the convenient compact form f (il) Ax + f(\u00d12) Ax + + f (in) Ax Ef(\u00d1k) Ax. Each rectangle will have length \u2206x =. 1: Areas and Distances Understand how rectangles are constructed to estimate the area underneath a curve using either the left hand endpoint. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Similar formulas can be obtained if instead we choose c k to be the left-hand endpoint, or the midpoint, of each subinterval. 2 - Sigma Notation and Limits of Finite Sums - Exercises 5. A 2 \ud835\udc5b 2 + 4 \ud835\udc5b 1 2 + \ud835\udc56. For left Riemann sums, the left endpoints of the subintervals are 1) Ax, fork \u2014. c) Convergence of Left and Right Sums for Monotonic functions and not Monotonic functions. 2a Sigma Notation and Area Approximation! Essential Learning Target Compute left, right and midpoint Riemann sums using either. The approximate value at each midpoint is below. Choose The Correct Riemann Sum Below. Thesymbol)Thus we can writeandThe sigma notation used on the right side of these equations is much more compact thanthe summation expressions on the left side. Right Riemann Sum. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. For example, saying \u201cthe sum from 1 to 4 of n\u00b2\u201d would mean 1\u00b2+2\u00b2+3\u00b2+4\u00b2 = 1 + 4 + 9 + 16 = 30. 2 Riemann Sums and Integration Learning Goal 2. 50 1 3+ cos 10 k = 1 2tk - T O A. (2k-1) (2n + 1 - 2) OF. Definition S L: Left-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area \u2248 S R: Right-endpoint Riemann Sum Choose the height of each rectangle on the interval [x i, x i +1]to be a b x y f H x L Area \u2248 S M: Midpoint Riemann Sum. Compute the Riemann sum for R4 using 4 subintervals and right endpoints for the function on the interval [1,5]. Your students will have guided notes, homework, and a content quiz on Riemann Sums and Sigma Nota. Examples Example 2. ) We need Delta x=(b-a)/n Deltax is both the base of each rectangle and the distance between the endpoints. Use the Midpoint Rule with n = 3 to approximate Z 5 \u22121 (x2 \u22124) dx. In these sums, n is the number of subintervals into which the interval is divided by equally spaced partition points a = x0 < x1 < \u2026 < xn-1 < xn = b. 2: Definition of a definite integral; Riemann sum; vocabulary (integrand, integral sign, differential, limits of integration) midpoint rule; trapezoidal rule (actually equivalent to the average of left and right rectangle rules). Midpoint Riemann Sums: Suppose x i is the midpoint of the ith subinterval [x i 1;x i], that is x i = x i 1 + x i 2 for all i. The Midpoint Rule for definite integrals means to approximate the integral by using a midpoint Riemann Sum (just as in 6. Sigma Notation or Summation Notation. Then evaluate the Riemann sum he midpoint Riemann sum for f(x) = 3+ cOS TX on [0,5] with n= 50 dentify the midpoint Riemann sum. Midpoint Riemann Sums: Suppose x i is the midpoint of the ith subinterval [x i 1;x i], that is x i = x i 1 + x i 2 for all i. Now, find the endpoints. For example, say you\u2019ve got f (x) = x2 + 1. 6255658911511259 1. The length of each subinterval is \u0394x = n b a. Integral Calculus Chapter 4: Definite integrals and the FTC Section 2: Riemann sums Page 8 Templated questions: 1. The approximate value at each midpoint is below. Write the correct sigma notation for any Riemann sum you encounter. Choose the correct answer below. As the partitions in [a,b] become finer and finer, our sum midpoint of. The values of the function are tabulated as follows; Left Riemann Sum LRS = sum_(r=1)^4 f(x)Deltax \" \" = Deltax { f(1) + f(2) + f(3) + f(4) } \\\\ \\\\ \\\\ (The LHS. All the four temples have 100 steps climb. your sketch the rectangles associated with the Riemann sum 4 k = 1 f(ck ) x k , using the indicated point in the kth subinterval for ck. c) Convergence of Left and Right Sums for Monotonic functions and not Monotonic functions. We draw rectangles using the values f(-2) = -4 and f(2) = -4, then add the values of the rectangles and get -4(4) + -4(4) = -32. Choose The Correct Riemann Sum Below. of de nite integrals using Riemann Sums in Sigma notation. lower than the midpoint area E. Is your answer an over-estimate or an under-estimate? Split the interval [-4, 4] into two sub-intervals of length 4 and find the midpoint of each. State the right Riemann Sum for the function on the given interval. where i is the index of summation, l is the lower limit, and n is the upper limit of summation. 3) Mid-point sums All value may be different but they represent a same quantity an approximated area under the curve. Use the Midpoint Rule with n = 3 to approximate Z 5 \u22121 (x2 \u22124) dx. Partition the interval into 4 subintervals of equal length. In previous entry, we talked about the Part 1 of Fundamental Theorem of Calculus. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. Right Riemann Sum. Riemann sums; Sigma notation Fundamental Theorem Riemann Sums Problem 2 Use a midpoint Riemann sum with 3 equal subintervals to approximate the area under y= 1 16 x. ) We need Delta x=(b-a)/n Deltax is both the base of each rectangle and the distance between the endpoints. Integral Calculus Chapter 4: Definite integrals and the FTC Section 2: Riemann sums Page 8 Templated questions: 1. The Midpoint Rule for de\ufb01nite integrals means to approximate the integral by using a midpoint Riemann Sum just as in 6. Now it is your turn to do the following problems. Riemann Sums. While the rectangles in this example do not approximate well the shaded area, they demonstrate that the subinterval widths may vary and the heights of the rectangles can be. EXAMPLE 1Using Sigma Notation\u0192s1d + \u0192s2d + \u0192s3d +\u00c1+ \u0192s100d =a100i=1\u0192sid. So this is right over here. By comparing the sum we wrote for Forward Euler (equation (8) from the Forward Euler page) and the left Riemann sum \\eqref{left_riemann}, we should be able to convince ourselves that they are the same when the initial condition is zero. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. For left Riemann sums, the left endpoints of the subintervals are 1) Ax, fork \u2014. The sums we have been calculating - by adding together the areas of the rectangles drawn in each subinterval - are called Riemann sums, after the German mathematician Georg Friedrich Bernhard Riemann (1826-1866). scalar multiplication and matrices (Properties) scalar multiplication of vectors. The RiemannSum(f(x), x = a. If x k * is any point in the k th subinterval x k-1, x k, for k=1,2,\u2026,n, then the. Choose the correct Riemann sum below. Right Riemann Sum. Left-Hand. Legal Notice: The copyright for this application is owned by Maplesoft. 15k 2 + 4 C. Conic Sections. The endpoints are given by x 0 = a and x n = b. higher than the actual area B. 12+ 22+ 32+ 42+ 52+ 62+ 72+ 82+ 92+ 102+ 112=a11k=1k2,k 1aknThe index k ends at k n. Typical choices are: left endpoints, right endpoints, midpoint, biggest value, smallest value. For example, saying \u201cthe sum from 1 to 4 of n\u00b2\u201d would mean 1\u00b2+2\u00b2+3\u00b2+4\u00b2 = 1 + 4 + 9 + 16 = 30. 0375 When n = 100, Left Riemann sum = 164. Step 2: Now click the button \u201cSubmit\u201d to get the Riemann sum. Constructing Accurate Graphs of Antiderivatives; The Second Fundamental Theorem of Calculus; Integration by Substitution; Integration by Parts; Other Options for Finding Algebraic Antiderivatives; Numerical Integration; 6 Using Definite Integrals. 2 Riemann Sums and Area 1 5. v2 (1 \u2212v)6 dv 7. (k)n-1) 729 OD (k-1) (n+1-K) n ket kot n\u00ba n n 729 OE. Every Riemann sum depends on the partition you choose (i. The same number of subintervals were used to produce each approximation. Option #1: If you noticed in step 2 above, we did not care if our subintervals were the same width. Evaluate the sum using a calculator with n=20,50, and 100. [Analysis] Exercise 2. In the more compact sigma notation, we have Ln = Xn\u22121 i=0 f (xi)4x. Write the midpoint Riemann sum in sigma notation for an arbitrary value of n. midpoint Riemann sum of f(x) over [a,b] using n intervals is larger than both the left and right Riemann sums of f(x) over [a,b] using n intervals. Thesymbol)Thus we can writeandThe sigma notation used on the right side of these equations is much more compact thanthe summation expressions on the left side. (1) Z 6 \u22121 (2x \u22124) dx 9. 2: Definition of a definite integral; Riemann sum; vocabulary (integrand, integral sign, differential, limits of integration) midpoint rule; trapezoidal rule (actually equivalent to the average of left and right rectangle rules). Here is the solution of a similar problem, which should give you an idea of how to write up your solution. (n times) = cn, where c is a constant. The LRAM uses the left endpoint, the RRAM uses the right endpoint and the MRAM uses the midpoint of intervals. Calculus Q&A Library se sigma notation to write the following Riemann sum. If your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals. Constructing Accurate Graphs of Antiderivatives; The Second Fundamental Theorem of Calculus; Integration by Substitution; Integration by Parts; Other Options for Finding Algebraic Antiderivatives; Numerical Integration; 6 Using Definite Integrals. The RiemannSum(f(x), x = a. Note particularly that since the index of summation begins at 0 and ends at n \u2212 1, there are indeed n terms in this sum. Recall that a Riemann sum is an expression of the form where the x i * are sample points inside intervals of width. Is the midpoint Riemann sum an over or under approximation if the graph is: a. Then evaluate the sum. Midpoint Riemann Sum. How do you find Find the Riemann sum that approximates the integral #int_0^9sqrt(1+x^2)dx# using How do you Use a Riemann sum to approximate the area under the graph of the function #y=f(x)# on How do you use a Riemann sum to calculate a definite integral?. While summation notation has many uses throughout math (and specifically calculus), we want to focus on how we can use it to write Riemann sums. Riemann sums in summation notation calculator keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. Evaluate the following to practice the basic skills of Sigma Notation. simplified by using sigma notation and summation formulas to create a Riemann Sum. Thomas\u2019 Calculus 13th Edition answers to Chapter 5: Integrals - Section 5. Right Riemann Sum. Partition a, b by choosing These partition a, b into n parts of length ?x1, ?x2, ?xn. There are many ways to write a given sum in sigma notation. Use These Values To Estimate The Value Of The Integral. You can use sigma notation to write out the Riemann sum for a curve. One method to approximate the area involves building several rect-angles with bases on the x-axis spanning the interval [a,b] and with. If we use the notation \u2016\u2016\ud835\udc43 to denote the longest subinterval length we can force the longest subinterval length to 0 using a. Sigma Notation The sum of n terms a1,a2, midpoint, and right) is called a Riemann sum. 313-315) Practice problems: Text p. Midpoint Riemann Sum. summation symbol (an upper case sigma) Figure 5. 1) as f x dx f m x 1 ( ) ( ) (a Riemann. Use sigma notation and the appropriate summation formulas to formulate an expression which represents the net signed area between the graph of f(x) = cosxand the x-axis on the interval [ \u02c7;\u02c7]. Sums can also be infinite, provided that the terms eventually get close enough to zero\u2013this is an important topic in calculus. How do you find Find the Riemann sum that approximates the integral #int_0^9sqrt(1+x^2)dx# using How do you Use a Riemann sum to approximate the area under the graph of the function #y=f(x)# on How do you use a Riemann sum to calculate a definite integral?. See full list on khanacademy. Definite Integral from Summation Notation: To rewrite the given limit of summation notation, we'll compare the given expression with the formula of conversion and then substitute the values. Livro de Calculo. Adding up a bunch of terms can be cumbersome when there are a large number of terms. asked by Bae on May 2, 2014; Calculus. Con-versely, given a Riemann sum in sigma notation, be able to iden-tify a function and an interval which give rise to that sum. Use n = 50 equal subdivisions. Simplify the Riemann Sum. (read \u201csigma a k as k runs from 1 to n\u201d ). Evaluate the integral. exactly equal to the actual area C. of de nite integrals using Riemann Sums in Sigma notation. }\\) Figure 1. row operation. Rolle's Theorem. 10 50 ( 2Tk \u2013 1)). This is useful when you want to derive the formula for the approximate area under the curve. In order to check that the result does not depend on the sample points used, let\u2019s redo the computation using now the left endpoint of each subinterval: Xn i=1 \u00b5. In each case where you used a Riemann sum to estimate an area, try to determine if you obtained an overestimate or an underestimate. This formula includes a summation using sigma notation (\u03a3). \" The index k tells us where to begin the sum (a the number below the E) and where to end (at the number above). Graph the function f(x) over the given interval. Is this Riemann sum an over-estimate or an under-estimate of the exact value? (b) Approximate the same integral by using Simpson\u2019s Rule with n = 4 subintervals. See full list on khanacademy. ) In practice, evaluating a summation can be a little tricky. By the way, you don't need sigma notation for the math that follows. 6078493243021688 1. Sigma Notation or Summation Notation. Evaluate the integral. Figuring out the first (or last) element of the sum can rule out incorrect Riemann sums efficiently. The endpoints are given by x 0 = a and x n = b. It states in the book that it is recognized as a Riemann sum for a fn defined on [0,1]. f) General Riemann Sum. The subscript k is the summation index, and is a \u201cdummy index\u201d, in the sense that it can be replaced by any convenient letter. [Hint: Make the substitution u=(x-\\mu) / \\sigma, which will create two integrals. The sum would be the sum of the rectangles, so it would be the height times the width (change in x). The values of the function are tabulated as follows; Left Riemann Sum LRS = sum_(r=1)^4 f(x)Deltax \" \" = Deltax { f(1) + f(2) + f(3) + f(4) } \\\\ \\\\ \\\\ (The LHS. This gives the midpoint Riemann sum of f using n rectangles, which is denoted by M n. Use the programs on your calculator to find the value of the sum accurate to 3 decimal places. 9779070602600015 1. Therefore, always use a right-sum, with ci = a+i\u00a2x. Use of the formulae for the nth term and the sum of the first n terms of the sequence. Legal Notice: The copyright for this application is owned by Maplesoft. The application is intended to demonstrate the use of Maple to solve a particular problem. scalar multiplication and matrices (Properties) scalar multiplication of vectors. root method. Here is how to set up the Riemann sum for the de\ufb01nite integral Z 3 1 x2 dx where n = 10: (1) Find \u2206x = b\u2212a n. First, determine the width of each rectangle. 2 Sigma Notation & Limits of Finite Sums NOTES SIGMA NOTATION a k =a 1 +a 2 +a 3 +!+a n\u22121 +a n k=1 n \u2211 EX 1) Complete the table, given the following sums in sigma notation: The Sum in Sigma Notation The Sum Written Out, One Term for Each Value of k The Value of the Sum k k=1 5 \u2211 (\u22121) k k k=1 3 \u2211 k k=1 k+1 2 \u2211 k2 k=4 k\u22121 5 \u2211 EX. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 22) _ 16 k = 2 6 22). Derivatives Derivative Applications Limits Integrals Integral Applications Riemann Sum Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Functions Line Equations Functions Arithmetic & Comp. In Chapter 1, I introduce you to the Riemann sum formula for the definite integral. XTRA Assignment 2 - Riemann sum tables w. The length of each subinterval is \u0394x = n b a. Simplify the expression. In previous entry, we talked about the Part 1 of Fundamental Theorem of Calculus. Example of finding a Riemann Sum for finite \u201cn\u201d using a table rather than sigma notation. The first two arguments (function expression and range) can be replaced by a definite integral.", "date": "2020-10-27 06:47:08", "meta": {"domain": "trattoriaborgo.it", "url": "http://gpof.trattoriaborgo.it/midpoint-riemann-sum-sigma-notation.html", "openwebmath_score": 0.9426590204238892, "openwebmath_perplexity": 685.8367052427091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9932024694556524, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.857522686115624}}
{"url": "http://nbkx.ihuq.pw/properties-of-matrix-multiplication-proof.html", "text": "Properties Of Matrix Multiplication Proof\n\nThe Transpose of the product of two matrices is equal to. There are four properties involving multiplication that will help make problems easier to solve. (b) Show that every eigenvector for Bis also an eigenvector for A. Algebraic Properties of Matrix Operations A. If A is an n\u00d7n symmetric orthogonal matrix, then A2 = I. Combined Calculus tutorial videos. addition and subtraction or multiplication and division. Note that the matrices in a matrix group must be square (to be invertible), and must all have the same size. Distributive Law of Set Theory Proof - Definition. The point of this note is to prove that det(AB) = det(A)det(B). Since our model will usually contain a constant term, one of the columns in the X matrix will contain only ones. Now, v^Tu is just the dot product of u and v, so it's a scalar. If A is an n\u00d7n symmetric matrix such that A2 = I, then A is orthogonal. c) If x 2 V then 0 \u00a2 x = 0. The transform of a sum is the sum of the transforms: DFT(x+y) = DFT(x) + DFT(y). However, matrix multiplication is not defined if the number of columns of the first factor differs from the number of rows of the second factor, and it is non-commutative, even when the product remains definite after changing the order of the factors. The examples below should help you see how division is not associative. Therefore there is no proof to why matrices are multiplied the way they do. multiplication, matrix-vector multiplication, and FFT. Matrix multiplication is not commutative. Let A, B, and C be three matrices. , x 6= 0) such that (A \u2212\u03bbI)x =0. For example, 3 1 2 \u22121 0 = 3 6 \u22123 0. All for the high school levels of Grade 9, Grade 10, Grade 11, and Grade 12. The element, cij, of the product matrix is obtained by multiplying every element in the row i of matrix A by each element of column j of matrix B and then adding them. We do this first with simple numerical examples and then using geometric diagrams. Know the de nition and basic properties of a fundamental matrix for such a system. Let (A, B, C) be any choice of matrices. Commutative, Associative and Distributive Laws. It's the associative property of matrix multiplication. edu/\u02d8schiu/ Matrix Multiplication: Warnings WARNINGS Properties above are analogous to properties of real numbers. In this page multiplication properties of matrices we are going to see some properties in multiplication. Matrix Multiplication Lesson. For example, the number of walks of length 2 is the number of vertices ksuch that there is an arc from ito kand an arc from k to j. The rst theorem stated that 0v = 0 for all vectors v. The cross product of two vectors a= and b= is given by Although this may seem like a strange definition, its useful properties will soon become evident. OLS in Matrix Form 1 The True Model \u2020 Let X be an n \u00a3 k matrix where we have observations on k independent variables for n observations. For example 4 + 2 = 2 + 4 Associative Property: When three or more numbers are added,. Pre\u2010requisite: Inverse of a matrix, addition, multiplication and transpose of a matrix. 7 Example (Matrix groups). This is the general linear group of 2 by 2 matrices over the reals R. Multiplication of Matrices. 1 (Properties of Matrix Addition and Scalar Multiplication) By (b), sum of multiple matrices are written as A + B + + M. jugate, there is some other matrix Q such that Since the associative law holds for matrix multiplication, the theorem is proved in the following way. Matrix multiplication is associative; for example, given 3 matrices A, B and C, the following identity is always true. Formula IA = A;BI = B whenever the products are de ned. Math Goals (Standards for. A product of permutation matrices is again a permutation matrix. We shall see the reason for this is a little while. Many of the basic properties of expected value of random variables have analogous results for expected value of random matrices, with matrix operation replacing the ordinary ones. This is a consequence of lemma 1. In this case, we have etA = PetDP 1 = P 2 4 e 1t e nt 3 5P 1: EXAMPLE 1. It is obvious that matrix multiplication is associative. then vTAv = vT(Av) = \u03bbvTv = \u03bb Xn i=1. , , or is a unitary (orthogonal if real) matrix. , [A B]ij = [A]ij[B]ij; known asHadamard4 (or Schur) product,doesn\u2019t workfor linear systems and linear transformations. By the definition and the lemma, all \u00d7 determinant functions must return this value on this matrix. Multiplication of two sequences in time domain is called as Linear convolution. A square matrix A is said to be symmetric if AT = A and skew-symmetric if AT = A. b = a \u2227b Properties of Rings. Matrix Multiplication: There are several rules for matrix multiplication. Symmetric matrices are also called selfadjoint. Let\u2019s look at some properties of multiplication of matrices. Matrix Inverses and Solving Systems. Matrix Multiplication Suppose A and B are two matrices such that the number of columns of A is equal to number of rows of B. Write a proof for these Properties of Matrix Multiplication: Let A, B, C be matrices. Proofs of the properties of matrix multiplication. \u2022 Only the properties listed in Theorem 3. The continuous linear operators from into form a subspace of which is a Banach space with respect to. 2 The Matrix Trace; 3. Pre\u2010requisite: Inverse of a matrix, addition, multiplication and transpose of a matrix. n \u00d7n zero matrix. 3 If A B and C are matrices of the correct size so that the required products are defined, and t e 2, then 3 = (AB)C (AB)T = NAT Proof 4. So, if A is invertible, your statement cannot be proved. The plural form of matrix is matrices. Let A be a squarematrix of ordern and let \u03bb be a scalarquantity. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. a * i = a, where i = 1 (identity element). If the zero matrix O is multiplied times any matrix A, or if A is multiplied times O, the result is O (see Exercise 16). , x 6= 0) such that (A \u2212\u03bbI)x =0. Subsubsection Symmetry. Pre\u2010requisite: Inverse of a matrix, addition, multiplication and transpose of a matrix. Use the matrix multiplication calculator to multiply any types of matrix. edu/\u02d8schiu/ Matrix Multiplication: Warnings WARNINGS Properties above are analogous to properties of real numbers. in this set is again in the set, so it is closed under multiplication. The properties of matrix addition and multiplication depend on the properties of the underlying number system. Say matrix A is an m\u00d7p matrix and matrix B is a p\u00d7n matrix. Game Theoretic Analysis of Multi-Processor Schedulers: Matrix Multiplication Example Oleksii Ignatenko 1 Institute of Software Systems NAS Ukraine o. 2 For any matrix A, we have (At)t = A. Characterizing Properties of an Orthogonal Matrix Theorem 6. Properties of the Matrix Inverse. If you believe your intellectual property has been infringed and would like to file a complaint, please see our Copyright/IP Policy. Sugilith-Kette(Rondell fac. Commutative property: When two numbers are added, the sum is the same regardless of the order of the addends. View History. Proof: All proofs can follow from basic arithmetic laws in R and previous definitions. ) Proof: No way. Identity Property Of Multiplication : The multiplication of any number and the identity value gives the same number as the result. juxtaposition of matrices will imply the \"usual\" matrix multiplication, and we will always use \" \" for the Hadamard product. Properties of Matrix Multiplication. In general, when the product AB and BA are possible. We look for an \"inverse matrix\" A 1 of the same size, such that A 1 times A equals I. Whatever A does, A 1 undoes. Math, Can you please tell me how to prove or derive the commutative property of addition (i. I would like to multiply a batched matrix X with dimension [batch_size, m, n] with a matrix Y of size[n,l], how should I do this?. A grade X student is aware of matrix but we will still start with the brief introduction of matrix and also discuss key rule for matrix multiplication. If A is an n n matrix and v and w are vectors in Rn, then the dot. Proposition (associative property) Multiplication of a matrix by a scalar is associative, that is, for any matrix and any scalars and. It is the purpose of this chapter to introduce the properties that characterise kernel functions. Let\u2019s look at some properties of multiplication of matrices. Learn about the properties of matrix multiplication (like the distributive property) and how they relate to real number multiplication. In other words, if the order of A is m x n. By the definition and the lemma, all \u00d7 determinant functions must return this value on this matrix. klmGCE Mathematics (6360) Further Pure 4 (MFP4) Textbook 10 1. Concept of elementary row and column operations. Which of the following statements illustrate the distributive, associate and the commutative property? Directions: Click on each answer button to see what property goes with the statement on the left. Here we sketch three properties of determinants that can be understood in this geometric context. MAT-0010: Addition and Scalar Multiplication of Matrices Introduction to Matrices. Matrices rarely commute even if AB and BA are both defined. Sorensen Math. What is the determinant of: (If you have forgotten about determinants, or wish you had, don't worry. Aside from this, though, matrix multiplication has all the properties you would like it to have. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. Simple properties of addition, multiplication and scalar multiplication. 1 (Properties of Matrix Addition and Scalar Multiplication) By (b), sum of multiple matrices are written as A + B + + M. Basic properties of scalar multiplication are summarized at the end of this section. As such, it enjoys the properties enjoyed by triangular matrices, as well as other special properties. Matrix Calculations: Linear maps, bases, and matrix multiplication (where \\\" is the matrix multiplication of A and a vector v) Proof. For matrix multiplication to work, the columns of the second matrix have to have the same number of entries as do the rows of the first matrix. In case of convolution two signal sequences input signal x(n) and impulse response h(n) given by the same system, output y(n) is calculated. A symmetric matrix has real eigenvalues. Thus, if A = [aij]m\u00d7n and B = [bij]n\u00d7p are two matrices of order m \u00d7 n and n \u00d7 p respectively, then their product AB \u2026 [Read more] about How to Multiply Matrices. For example, addition and multiplication are binary operations of the set of all integers. I Eigenvectors corresponding to distinct eigenvalues are orthogonal. 25 Proof Using the remark 2. Multiplication of Matrices. The following properties are valid for the transpose: \u00b7 The transpose of the transpose of a matrix is the matrix itself: (A T) T = A \u00b7 Transpose of a scalar multiple: The transpose of a matrix times a scalar (k) is equal to the constant times the transpose of the matrix: (kA) T = kA T. position down into the subspace, and this projection matrix is always idempo-tent. Invertible matrices and proof of uniqueness of inverse, if it exists. Adjoint And Inverse Of A Matrix: In this article, you will know how to find the adjoint of a matrix and its inverse along with solved example questions. A product of permutation matrices is again a permutation matrix. Indeed, the matrix product A A T has entries that are the inner product of a row of A with a column of A T. To start practicing, just click on any link. The lesson may be something like this: Show this video to review how to multiply matrices. Interesting Properties of Matrix Norms and Singular Values $\\DeclareMathOperator*{\\argmax}{arg\\,max}$ Matrix norms and singular values have special relationships. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). In this video, I wanna tell you about a few properties of matrix multiplication. 4 Row Operations and the LU Decomposition 102 Row Operations and Matrix. Matrix Multiplication. In this page multiplication properties of matrices we are going to see some properties in multiplication. The matrix B will be the inverse matrix of A. Matrix multiplication is one of the most fundamental problems in scientific computing and in parallel computing. at 24th St) New York, NY 10010 646-312-1000. Ad joint and inverse of a square matrix. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Meaning of matrix multiplication In these examples we will explore the e\ufb00ect of matrix multiplication on the xy-plane. Matrix approach. Matrix multiplication Matrix multiplication is an operation between two matrices that creates a new matrix such that given two matrices A and B, each column of the product AB is formed by multiplying A by each column of B (De\ufb01nition 1). error(\"Multiplying $r1 x$c1 and $r2 x$c2 matrix: dimensions do not match. Notice that the order of the matrices has been reversed on the right of the \"=\". General properties. Lemma 8 can be seen in the matrix equation R + \u02c7 2 = R R 2; Lemma 9 in the. When A is multiplied by A-1 the result is the identity matrix I. If you installed a MATLAB, try to multiply two 1024*1024 matrix. We will apply most of the following properties to solve various Algebraic problems. However, virtually all of linear algebra deals with matrix multiplications of some kind, and it is worthwhile to spend some time trying to develop an intuitive understanding of the viewpoints presented here. These rules, forming matrix algebra, are naturally derivable from the properties of linear operators in the vector spaces which, as we have seen above, can be represented by matrices. In the next subsection, we will state and prove the relevant theorems. Sugilith-Kette(Rondell fac. If the zero matrix O is multiplied times any matrix A, or if A is multiplied times O, the result is O (see Exercise 16). But itdoes commute. Cayley\u2019s defined matrix multiplication as, \u201cthe matrix of coefficients for the composite transformation T2T1 is the product of the matrix for T2times the matrix of T1\u201d(Tucker, 1993). 12 Theorem 1. 3 Properties of matrix multiplication. So, matrix multiplication is just the image of composition of linear transformations under the identification of matrices with linear In particular, then, distributivity of matrix multiplication is really just distributivity of composition of linear transformations, which lends itself to a far more transparent. Exponential Matrix and Their Properties International Journal of Scientific and Innovative Mathematical Research (IJSIMR) Page 55 3. Institut de France, Binet also read a paper which contained a proof of the multiplication theorem but it was less satisfactory than that given by Cauchy. LECTURE 8: BASIC MATRIX ALGEBRA Prof. The cross symbol generally denotes the taking a cross product of two vectors, yielding a vector as the result, while the dot denotes taking the dot product of two vectors, resulting in a scalar. Properties of real symmetric matrices I Recall that a matrix A 2Rn n is symmetric if AT = A. Here, each element in the product matrix is simply the scalar multiplied by the element in the matrix. Matrix C and D below cannot be multiplied together because the number of columns in C does not equal the number of rows in D. Multiplication Properties of Exponents. The DCM matrix (also often called the rotation matrix) has a great importance in orientation kinematics since it defines the rotation of one frame relative to another. Properties Here is a list of all of the skills that cover properties! These skills are organized by grade, and you can move your mouse over any skill name to preview the skill. I can't understand / complete the proof that matrix multiplication is associative. References. A proof technique is introduced, which exploits the Grigoriev\u2019s flow of the matrix multiplication function as well as some combinatorial properties of the Strassen computational directed acyclic graph (CDAG). The point of this note is to prove that det(AB) = det(A)det(B). If , the series does not converge (it is a divergent series). Our derivations use the fact that products of diagonal matrices are diagonal together with Bezout\u2019s identity. Its subgroups are referred to as matrix groups or linear groups. When working with just real numbers or when working with scalars. Applying the associative and multiplicative identity properties, this is uv^T - u(v^Tu)v^T. We look for an \u201cinverse matrix\u201d A 1 of the same size, such that A 1 times A equals I. Matrix inverses Recall De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. For example, addition and multiplication are binary operations of the set of all integers. The scalar is multiplied by each element of the matrix, giving us a new matrix of the same size. The definition of symmetric matrices and a property is given. by Marco Taboga, PhD. How to Multiply Matrices. 1 Matrix Operations Shang-Huan Chiu Department of Mathematics, University of Houston [email\u00a0protected] Matrix multiplication is not commutative. Some simple properties of vector spaces Theorem Suppose that V is a vector space. Matrix worksheets include multiplication of square or non square matrices, scalar multiplication, associative and distributive properties and more. Aside from this, though, matrix multiplication has all the properties you would like it to have Theorem 2. Multiplies two matrices, if they are conformable. Matrix Multiplication Matrix multiplication: Why we do it the way we do it Because the most obvious way, i. By part (a), we have BAv = Av. Notes on Matrix Multiplication and the Transitive Closure Instructor: Sandy Irani An n m matrix over a set S is an array of elements from S with n rows and m columns. More generally, a nilpotent transformation is a linear transformation L of a vector space such that Lk = 0 for some positive integer k (and thus, Lj = 0 for all j \u2265 k). , compressing one of the and will stretch the other and vice versa. Properties of matrix multiplication. In any column of an orthogonal matrix, at most one entry can be equal to 0. Boolean Logic Operations A Boolean function is an algebraic expression formed using binary constants, binary variables and Boolean logic operations symbols. Chapter 2: Determinants. 4 - Properties of Matrix-Matrix Multiplication Maggie Myers Robert A. Related Calculus and Beyond Homework Help News on Phys. To multiply a row by a column, multiply the first entry of the row by the first entry of the column. This feature is not available right now. It's the associative property of matrix multiplication. There are four properties involving multiplication that will help make problems easier to solve. I need help with a simple proof for the associative law of scalar multiplication of a vectors. We already know that = ad \u2212 bc; these properties will give us a c d formula for the determinant of square matrices of all sizes. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Properties of Matrix Multiplication. Matrix multiplication. Exponential Matrix and Their Properties International Journal of Scientific and Innovative Mathematical Research (IJSIMR) Page 55 3. (b) Show that every eigenvector for Bis also an eigenvector for A. Each element in a matrix is called an entry. ImA = A = AIn (identity for matrix multiplication). There is a larger class of objects that behave like vectors in Rn. Matrix Formulation of the DFT. I Eigenvectors corresponding to distinct eigenvalues are orthogonal. any nice properties at all, nevertheless the single most important property of this multiplication is This proof gives no clue why the mulitplication should be associative, but it leaves no doubt that it is Theorem 2 Matrix multiplication is associative. We will discover that a given matrix may have more than one identity matrix. Adjoint And Inverse Of A Matrix: In this article, you will know how to find the adjoint of a matrix and its inverse along with solved example questions. This is the ultimate guide to Boolean logic operations & DeMorgan\u2019s Theorems. ThematerialisstandardinthatthesubjectscoveredareGaussianreduction, vector spaces, linear maps, determinants, and eigenvalues and eigenvectors. A matrix is an example of what, in the general context of vector spaces, is called a linear operator. Then check if another mathematical object satis es the same properties. Matrices, vectors, vector spaces, transformations. Leila Wehbe Kernel Properties - Convexity. PROOF of [email\u00a0protected] = of (Q -1 Q of = = z of Q ([email\u00a0protected]) x of -1) U = [email\u00a0protected] Matrix Notation for Geometric Transformations One important application of matrix algebra is in expressing the transfor-. Example: This matrix is 2\u00d73 (2 rows by 3 columns): When we do multiplication To multiply an m\u00d7n matrix by an n\u00d7p matrix, the ns must be the same, and the result is an m\u00d7p matrix. proof of properties of trace of a matrix. Search this site. This is called the identity property of multiplication. \u2022 The set of all even integers forms a commutative ring under the usual addition and multiplication of integers. Except for the lack of commutativity, matrix multiplication is algebraically well-behaved. The first concerns the multiplication between a matrix and a scalar. We used the following 4 equations to estimate breeding values:. Similarly, if E0 is the matrix obtained by performing a column. Aside from this, though, matrix multiplication has all the properties you would like it to have. a * i = a, where i = 1 (identity element). When A is multiplied by A-1 the result is the identity matrix I. The area model for multiplication establishes the groundwork for helping visual learners in the conceptual understanding of the traditional algebraic skills of polynomial multiplication and factoring. GOLDVARD AND L. There are many types of matrices like the Identity matrix. when we multiply two matrices the number of columns of the first matrix should be equal to the number of rows of the second matrix. It is not for rectangular matrices of different sizes as it's not even defined for both \"directions\"! For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general. I can't understand / complete the proof that matrix multiplication is associative. Matrix groups consist of matrices together with matrix multiplication. To start practicing, just click on any link. Proof of (d). Jabba's got it. weassociatewith\u02c7then n permutation matrix A Ai Properties are\ufb02ectormatrixissymmetric Proof thesquareddistanceof. The rows of A form an orthonormal basis for Rn. elimination [5]. The product will be a 2\u00d74 matrix, the outer dimensions. We shall see the reason for this is a little while. Groups, in general. Thus, the transpose of a row vector is a column vector and vice-versa. rules of matrix addition, multiplication of a matrix by a number, and matrix multiplication. You look at the proof and gure out the crucial step and properties which make the proof work. DMAM Distributivity across Matrix Addition, Matrices If \u03b1\u2208C and A,B\u2208Mmn,. In this lesson, we will learn about the identity matrix, which is a square matrix that has some unique properties. We use matrix techniques to give simple proofs of known divisibility properties of the Fibonacci, Lucas, generalized Lucas, and Gaussian Fibonacci numbers. Proof: Notice \u2026{fill in your definition and try to decide if the answer fits the type of vector in V (ie. Rotation matrix, matrices multiplication, determinant, orthogonal matrices, equiangular rotations. Matrix is a way to organize data in the form of rows and columns. So, an extra property that relates \u2225AB\u2225 to \u2225A\u2225 and \u2225B\u2225 is needed. proof of properties of trace of a matrix. Now we can explore some basics properties of the Hadamard Product. Nilpotent matrix. (This section can be omitted without affecting what follows. The performance will be dramatically improved if you use O(N^2. of its properties. In particular, ~a\u00d7~b 6= ~b\u00d7~a. The first concerns the multiplication between a matrix and a scalar. It is not for rectangular matrices of different sizes as it's not even defined for both \"directions\"! For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general. The smallest such k is sometimes called the index of N. Thus, multiplication of matrix is not commutative. Multiplication of powers: Distribution of powers: If , then. Explicitly, suppose is a matrix and is a matrix, and denote by the product of the matrices. Matrix multiplication not commutative. That is the dot product, but stuffed into a $$1\\times 1$$ matrix! Subsubsection Other Properties. Then certainly the product. Since the multiplication both ways generates the Identity matrix, then we are guaranteed that the inverse matrix obtained using the formula is the correct answer!. Then A\u22121 exists and we have. Matrix multiplication can be done only when the number of columns of is equal to the number of rows of. and of B and C have sizes for which the indicated sums and products are defined. Both the statement of this theorem and the method of its proof will be important for the study of differential equations in the next section. Proof: All proofs can follow from basic arithmetic laws in R and previous definitions. Observation: The following property is an obvious consequence of this definition. This number is then the scalar product of the two vectors. The properties are the commutative, associative, additive identity and distributive properties. Properties of Matrices. If I is the unit matrix, then AI = IA = A (I is called multiplicative identity). ,1966-67 American Hockey League Media Guide,2019 TOPPS UFC KNOCKOUT PURPLE SHIRT FIGHTER WORN RELIC #/25 JESSICA AGUILAR. The proposed approach is. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. Multiplication of matrices generally falls into two categories, Scalar Matrix Multiplication and Vector Matrix Multiplication. Countable and uncountable sets. Preliminaries Consider the following situation: A is a matrix, possible augmented, and U is the reduced row echelon form of A. (This section can be omitted without affecting what follows. The determinant of a matrix inverse. One of the main contributions of [2] was to demonstrate that several diverse families of non-abelian groups support the reduction of n \u00a3 n matrix multiplication to group al-gebra multiplication. For matrix multiplication to work, the columns of the second matrix have to have the same number of entries as do the rows of the first matrix. If A is an n\u00d7n symmetric orthogonal matrix, then A2 = I. We used the following 4 equations to estimate breeding values:. If A is an nxm matrix and O the mxk zero-matrix, then AO = O. Properties of Matrix Multiplication Multiplication can only occur between matrices A and B if the number of columns in A match the number of rows in B. It is not for rectangular matrices of different sizes as it's not even defined for both \"directions\"! For square matrices, if it is not commutative for any pair of matrices, it is not commutative in general. Orthogonally Diagonalizable Matrices These notes are about real matrices matrices in which all entries are real numbers. Rank and nullity of a matrix We had seen in previous chapter that the number of non-zero rows in the rows in the row-echelon form of a matrix play an important role in finding solutions of linear equation. For two matrices A and B. Matrix multiplication: if A is a matrix of size m n and B is a matrix of. Notice that matrix multiplication is not generally commutative, i. Commutative Property of Multiplication(CPM)-It states that changing the order of the factors will not affect the product. In other words, regardless of the matrix A, the exponential matrix eA is always invertible, and has inverse e A. Properties of matrix multiplication. Since our model will usually contain a constant term, one of the columns in the X matrix will contain only ones. The set GL(n,F) of n\u00d7nmatrices with nonzero determi-nant is a group when equipped with matrix multiplication. Institut de France, Binet also read a paper which contained a proof of the multiplication theorem but it was less satisfactory than that given by Cauchy. 1 Vector Spaces Let F be a \ufb01eld (such as the real numbers, R, or complex numbers, C) with elements called scalars. Likewise, a scalar product can be taken outside the transform: DFT(c*x) = c*DFT(x). For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. In a 2 2 matrix multiplication, M= 7 in Strassen, m= n= p= 2. 5: A latin square of side nis an nby narray in which each cell contains a. 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Here\u2019s How. June 13 2019 by Patrice Lee Onwuka Minnesota Permanently Untangled Hair Braiders from Regulations June 11 2019 by Patrice Lee Onwuka In the Walmart v. Amazon Grocery War, We Win June 7 2019 by Patrice Lee Onwuka Bernie Sanders Crashed Walmart\u2019s Meeting to Spread Minimum Wage Lies June 6 2019 by Patrice Lee Onwuka Alexandria Ocasio-Cortez is Wrong on Tipped Wages Too June 3 2019 by Patrice Lee Onwuka Meryl Streep is Right to Slam the Label \u201cToxic Masculinity\u201d May 31 2019 by Patrice Lee Onwuka Google\u2019s \u201cShadow Work Force\" Could Become the Norm in the U.S. Like Europe May 30 2019 by Patrice Lee Onwuka Two Truths & A Lie: The Federal Budget May 28 2019 by Patrice Lee Onwuka A Billionaire Alum Pays Off Morehouse Grads\u2019 Student Debt May 28 2019 by Patrice Lee Onwuka Kamala Harris Unveiled a Pay Gap Attack Plan That's Bound to Fail May 21 2019 by Patrice Lee Onwuka Senator Lindsey Graham Puts the Border Crisis Back on Congress\u2019s Plate May 16 2019 by Patrice Lee Onwuka Banning Mother\u2019s Day is the Latest Woke Crusade May 13 2019 by Patrice Lee Onwuka Twitter Banning AOC Parody Site is Progressive Privilege at Its Finest May 9 2019 by Patrice Lee Onwuka Kamala Harris Should Not Ax Our Tax Cuts May 8 2019 by Patrice Lee Onwuka Two Truths & A Lie: Net Neutrality May 7 2019 by Patrice Lee Onwuka 5 Takeaways from the Republican House Budget May 3 2019 by Patrice Lee Onwuka No, Ilhan Omar, Socialism Caused Venezuela\u2019s Devastation, Not the US May 2 2019 by Patrice Lee Onwuka Could a Handshake Ban Be Coming to Your Workplace? April 30 2019 by Patrice Lee Onwuka The U.S. Economy is Fired Up with 3.2 Percent Q1 Growth April 26 2019 by Patrice Lee Onwuka The Far Left's Idea to Allow Bombers, Murders Behind Bars to Vote is Disgraceful April 24 2019 by Patrice Lee Onwuka Michelle Obama\u2019s Silly Trump Joke Reinforces Negative Stereotypes About Divorced Dads April 22 2019 by Patrice Lee Onwuka Elizabeth Warren\u2019s Taxes Still Aren\u2019t Enough to Fund the Left\u2019s Spending Spree April 16 2019 by Patrice Lee Onwuka 3 Things to Know about Refunds and Tax Cuts on Tax Day April 15 2019 by Patrice Lee Onwuka Arizona Just Made It Easier for Military Spouses to Work There April 11 2019 by Patrice Lee Onwuka The Left\u2019s Answer to Social Security? Increase Your Taxes April 10 2019 by Patrice Lee Onwuka 3 Reasons Congress Should Not Bring Back Obama-Era Net Neutrality April 9 2019 by Patrice Lee Onwuka Majority of Americans Call Wage Gap Coverage \u201cFake News\u201d or Worse April 8 2019 by Patrice Lee Onwuka Banning Plastic Bags in NY Won\u2019t Save the Environment but Could Make People Sicker April 4 2019 by Patrice Lee Onwuka The Pay Gap is Actually Really Small and Controllable April 2 2019 by Patrice Lee Onwuka What Pay Gap? 10 Jobs in Which Women Outearn Men April 2 2019 by Patrice Lee Onwuka Liberals are Blue Over Failed Green New Deal Vote March 27 2019 by Patrice Lee Onwuka Here\u2019s How Jussie Smollett Earned a Slap on the Wrists for His Anti-Trump Hoax March 27 2019 by Patrice Lee Onwuka Elizabeth Warren Panders for Support with Idea to Blow Up the Electoral College March 22 2019 by Patrice Lee Onwuka Newark\u2019s Doomed Plan to Give Free Checks to Everyone March 19 2019 by Patrice Lee Onwuka Self-Serving NYU Student Who Confronted Chelsea Clinton was on a Political Payback Mission March 18 2019 by Patrice Lee Onwuka Girl Power! Women Drive Workforce Gains in This Hot Jobs Market March 8 2019 by Patrice Lee Onwuka Two Truths and a Lie: Federal Minimum Wage March 5 2019 by Patrice Lee Onwuka Google\u2019s New Pay Gap Pays Women More Than Men March 5 2019 by Patrice Lee Onwuka What Smaller Checks? Tax Refunds are Back Up to 2018 Levels March 1 2019 by Patrice Lee Onwuka She Works: 10 Facts about Women in the Workforce March 1 2019 by Patrice Lee Onwuka Ivanka V. AOC: Who Has it Right on Government-Guaranteed Jobs and Checks February 28 2019 by Patrice Lee Onwuka Spike Lee\u2019s Immature Night at the Oscars February 25 2019 by Patrice Lee Onwuka What\u2019s Behind Amazon\u2019s Supposed$0 Tax Bill February 22 2019 by Patrice Lee Onwuka\nKissing Sailor Statue Ignites MeToo Controversy February 21 2019 by Patrice Lee Onwuka\nFinland Gave Unemployed People Money with No Strings Attached. They Didn\u2019t Find Work. February 19 2019 by Patrice Lee Onwuka\nUnder Trump 3.6 Million Americans Aren\u2019t on Food Stamps Anymore February 15 2019 by Patrice Lee Onwuka\nOn Valentine\u2019s Day, A Creative Way to Spark a National Baby Boom February 14 2019 by Patrice Lee Onwuka\nExonerated: Covington Students Finally Have Their Names Cleared February 14 2019 by Patrice Lee Onwuka\nKamala Harris, Media Wrongly Use Smaller Tax Refunds to Vilify Tax Cuts February 13 2019 by Patrice Lee Onwuka\nJoy Villa Brings the Border Wall to the Grammys February 11 2019 by Patrice Lee Onwuka\nSnoozing through the Church's Christmas Play with Patrice Lee Onwuka December 17 2018 by Patrice Lee Onwuka\nActress Lena Dunham\u2019s Apology is the Legacy of 'I Believe Her' December 6 2018 by Patrice Lee Onwuka\nPETA Says Eating Eggs Makes You a Bad Feminist December 5 2018 by Patrice Lee Onwuka\nA #MeToo Legacy: Wall Street Execs Adopt Restrictive Rules Toward Young Women December 4 2018 by Patrice Lee Onwuka\nA Painful Truth About Feminism from a Woman who Traded Family for Success November 29 2018 by Patrice Lee Onwuka\nNew Court Ruling Makes It Necessary for All States to Ban Genital Mutilation November 27 2018 by Patrice Lee Onwuka\nAmericans Expect to Drop $1K Shopping This Holiday Season, Will You? November 23 2018 by Patrice Lee Onwuka Thanksgiving is a Time for Unity Not Confrontation of Trump-Supporting Family Members November 21 2018 by Patrice Lee Onwuka No Kamala Harris, ICE and the KKK are Not the Same November 20 2018 by Patrice Lee Onwuka Banning Pricey Jackets Doesn\u2019t Stop Poverty Shaming, But Pushes a Victim Mentality November 19 2018 by Patrice Lee Onwuka Women Aren\u2019t \u2018Anti-Women\u2019 if They Don\u2019t Vote for Nancy Pelosi November 15 2018 by Patrice Lee Onwuka BOOM: Small Business Optimism Rides High Again November 13 2018 by Patrice Lee Onwuka Americans to Democrats: Leave Justice Brett Kavanaugh Alone November 13 2018 by Patrice Lee Onwuka What Happened in Tuesday\u2019s Election? Q&A with Rachel Campos-Duffy November 9 2018 by Patrice Lee Onwuka Did Voters Understand that San Francisco\u2019s New Homelessness Tax is Not a Good Idea? November 8 2018 by Patrice Lee Onwuka We Don\u2019t Need a Naked Chelsea Handler to Get Us to Vote November 6 2018 by Patrice Lee Onwuka Liberals Need to Stop the False Narrative about Americans Working Multiple Jobs November 5 2018 by Patrice Lee Onwuka Inside the Migrant Caravan and How Americans Feel About It November 2 2018 by Patrice Lee Onwuka Big Gains for Workers: 250k Jobs Added in October, Wages Up November 2 2018 by Patrice Lee Onwuka Princess Costumes Top Charts This Halloween - Take That Keira Knightley October 31 2018 by Patrice Lee Onwuka Washington is Working to Accelerate the Next Generation of Wireless Tech October 25 2018 by Patrice Lee Onwuka Kamala Harris Wants to Replace Republican Tax Relief with New Government Payouts October 23 2018 by Patrice Lee Onwuka President Trump Puts Washington on a Diet: Trim Budgets by 5 Percent October 22 2018 by Patrice Lee Onwuka Chrissy Teigen Just Gave Us a Little Hope for Civil Discourse October 17 2018 by Patrice Lee Onwuka 3 Policy Areas Kanye West Championed in Trump Meeting October 12 2018 by Patrice Lee Onwuka AG Eric Holder to the Left: \u201cWhen They Go Low, We Kick \u2018Em\u201d October 11 2018 by Patrice Lee Onwuka Democratic Staffer Enables Harassment of Conservatives by Posting Private Contact Info Online October 4 2018 by Patrice Lee Onwuka Amazon Leapfrogs Over Other Retailers with New$15 Minimum Wage October 3 2018 by Patrice Lee Onwuka\nRape Prosecutor Rachel Mitchell Finds Case Against Judge Kavanaugh \u201cWeak\u201d October 1 2018 by Patrice Lee Onwuka\nDemocratic Senators Schumer, Hirono Abandon Bedrock Presumption of Innocence Because It\u2019s Judge Kavanaugh September 26 2018 by Patrice Lee Onwuka\nTax Reform Expanded Paid Family, Medical Leave. Here\u2019s How September 24 2018 by Patrice Lee Onwuka\nWhy Joy Behar is Wrong to Slam \u201cWhite Men\u201d in Congress over Judge Kavanaugh Allegations September 20 2018 by Patrice Lee Onwuka\nNew York Times Attempts to Smear Amb. Nikki Haley Over $50k Curtains September 17 2018 by Patrice Lee Onwuka Coming This Holiday Season: New Perks for Retail Workers Courtesy of the Economy September 17 2018 by Patrice Lee Onwuka Why Serena Williams\u2019 Claim of Sexism at the US Open Falls Flat September 10 2018 by Patrice Lee Onwuka \u2018Cosby Show\u2019 Star has a Message About Work Everyone Should Hear September 5 2018 by Patrice Lee Onwuka Why Did It Take Public Outcry for Twitter to Punish Threats Against Dana Loesch\u2019s Kids? August 29 2018 by Patrice Lee Onwuka Polls: Americans Cool to Plastic Straws but Don\u2019t Want Them Banned August 27 2018 by Patrice Lee Onwuka Will Congress Stop the FDA from Going After Our \u201cMilk\u201d? August 23 2018 by Patrice Lee Onwuka Another Reason to Keep ICE: The Arrest of Another Nazi War Criminal on US Soil August 22 2018 by Patrice Lee Onwuka Trump EPA Rolls Back Costly Obama-Era Energy Regulations August 21 2018 by Patrice Lee Onwuka Net Neutrality is Dead and the Internet is Getting Faster August 15 2018 by Patrice Lee Onwuka No Alexandria Ocasio-Cortez, An Invitation to Debate is Not Catcalling August 14 2018 by Patrice Lee Onwuka NYC\u2019s Vote to Curb Uber, Lyft Growth is About Saving Taxis, Transit Not Helping Passengers August 9 2018 by Patrice Lee Onwuka Small Business Owners are Sitting Pretty in Today\u2019s Economy August 8 2018 by Patrice Lee Onwuka 2.8 Million Fewer Americans on Food Stamps Since President Trump Took Office August 6 2018 by Patrice Lee Onwuka The Left Admits Americans Want Opportunities Not Handouts July 30 2018 by Patrice Lee Onwuka Millions of Americans are Unbanked Just Like This TV Talk Show Host July 26 2018 by Patrice Lee Onwuka Jeff Sessions Takes Colleges to Task for Coddling Students July 26 2018 by Patrice Lee Onwuka Look at What Tax Reform 2.0 Promises for Women and Families July 25 2018 by Patrice Lee Onwuka San Francisco Has a Crappy Problem on Their Hands, but Will the City Do Anything About It? July 20 2018 by Patrice Lee Onwuka Here\u2019s Why Alexandria Ocasio-Cortez Is Wrong on Unemployment, Capitalism July 18 2018 by Patrice Lee Onwuka U.S. Should Make Unfair Drug Pricing, Foreign Practices a Priority in Trade Negotiations July 11 2018 by Patrice Lee Onwuka Trump Administration Slashes ACA Navigator Outreach Program -- And For Good Reasons July 11 2018 by Patrice Lee Onwuka The Left\u2019s Hatred of Anything Trump Targets St. Jude Family July 9 2018 by Patrice Lee Onwuka Abolish ICE Protester Scales Lady Liberty Making for an Unhappy Fourth of July July 5 2018 by Patrice Lee Onwuka Suck This Up: Seattle Bans Plastic Straws July 2 2018 by Patrice Lee Onwuka Supreme Court Strikes Big Blow to Public Sector Unions, Upholds Free Speech of Public Workers June 28 2018 by Patrice Lee Onwuka Linda Sarsour Invokes Dr. Martin Luther King Jr. to Justify Harassment of Trump Officials June 27 2018 by Patrice Lee Onwuka No Maxine Waters, This is Not a Moment for Harassment or Incivility June 26 2018 by Patrice Lee Onwuka Making Kids Programming Better Through Common-Sense Broadcast TV Reforms June 22 2018 by Patrice Lee Onwuka A New Proposal to Replace the Broken Healthcare Law June 21 2018 by Patrice Lee Onwuka Trump Administration Expands Cheaper, Quality Healthcare Plans June 20 2018 by Patrice Lee Onwuka D.C. Will Kill Tips with New Minimum Wage June 19 2018 by Patrice Lee Onwuka S.E. Cupp Gets It Right: \"Not all Women are Democrats\" June 18 2018 by Patrice Lee Onwuka Today's Dads Are More Involved with Child Care Than Their Granddads June 15 2018 by Patrice Lee Onwuka Kathy Griffin Shaming Kevin Hart For Not Being Political is What's Wrong with Hollywood Elitism June 14 2018 by Patrice Lee Onwuka Optimism Among Small Business Owners Skyrockets Thanks to Tax Cuts June 13 2018 by Patrice Lee Onwuka Seattle Reverses Job-Killing Head Tax on Big Employers After Residents Speak Out June 12 2018 by Patrice Lee Onwuka Net Neutrality Ends Today. Here's Why It's Not the End of the World. June 11 2018 by Patrice Lee Onwuka It's Not Just Kate Spade and Anthony Bourdain, Suicide Rates Rise in Almost Every State. What's Going On? June 11 2018 by Patrice Lee Onwuka Is Anyone Buying Samantha Bee's Latest Apology for Her Vulgar Slur About Ivanka Trump? June 7 2018 by Patrice Lee Onwuka Women in Saudi Arabia are One Step Closer to Driving June 6 2018 by Patrice Lee Onwuka 5 Principles to Celebrate from the Supreme Court's Colorado Baker Ruling June 5 2018 by Patrice Lee Onwuka Women's March Organizers Tamika Mallory, Linda Sarsour Lash Out Against Israel June 4 2018 by Patrice Lee Onwuka NFL Finally Takes a Stand for the National Anthem May 25 2018 by Patrice Lee Onwuka Americans Report Doing Better Financially Under Year 1 of Trump May 24 2018 by Patrice Lee Onwuka Consumers, Small Banks to Gain Relief from Bi-Partisan Roll Back of Dodd-Frank Banking Regulations May 23 2018 by Patrice Lee Onwuka U.S. Baby Bust Leads to Lowest Birth Rates in 30 Years May 18 2018 by Patrice Lee Onwuka 3 Things to Know About the Senate Vote on Net Neutrality May 17 2018 by Patrice Lee Onwuka Hysteria Comes to Black Hair Care Products \u2013 Again May 16 2018 by Patrice Lee Onwuka Seattle's Business 'Head Tax' Has Even Starbucks, Amazon in an Uproar May 16 2018 by Patrice Lee Onwuka Oprah's Liberal Bias Clouded Good Advice from Her Commencement Address May 14 2018 by Patrice Lee Onwuka On Mother's Day, Advice from Some Moms May 12 2018 by Patrice Lee Onwuka Tennis Star Venus Williams Rejects \"Feminist\" Label May 10 2018 by Patrice Lee Onwuka Michelle Obama is Still Stumped by How Women Voted in 2016 May 7 2018 by Patrice Lee Onwuka Are You One of Third of Americans Who Say the Internet is Bad or Mixed for Society? May 1 2018 by Patrice Lee Onwuka In This Winning Economy, Cities Pay Young People to Move There May 1 2018 by Patrice Lee Onwuka Kim Kardashian and Kanye West Break the Internet for a Good Reason This Time - Free Thought April 26 2018 by Patrice Lee Onwuka Nancy Pelosi Chokes on \"Crumbs\" Comments April 25 2018 by Patrice Lee Onwuka 8 States Just Hit the Lowest Unemployment in Their History April 24 2018 by Patrice Lee Onwuka Kanye West Supports \"Free-Thinking\" Conservative Woman and Sets Twitter on Fire April 23 2018 by Patrice Lee Onwuka When Government Makes It Harder to Work, Fewer People Can Move Up the Ladder April 19 2018 by Patrice Lee Onwuka On Tax Day, Here's How the Federal Government Spends Every Dollar of Your Money April 17 2018 by Patrice Lee Onwuka Uber, Lyft Drivers Angry with Seattle City Council for Colluding with Taxis to Raise Rates April 17 2018 by Patrice Lee Onwuka They're Back! Share of Adult Children Living at Home Highest in 75 Years April 12 2018 by Patrice Lee Onwuka 3 Quotes from Mark Zuckerberg's Senate Grilling that Should Concern You April 11 2018 by Patrice Lee Onwuka 5 Drivers of the Pay Gap \u2013 They Are Not Gender Discrimination April 10 2018 by Patrice Lee Onwuka Diamond and Silk Labeled as \"Unsafe\" for Facebook, Maybe It's Time for A New Social Media Option April 10 2018 by Patrice Lee Onwuka Jimmy Kimmel Apologizes, But Not For His Melania Trump Joke April 9 2018 by Patrice Lee Onwuka How Many Americans Would Give Up Voting for a 10-Percent Pay Raise? It\u2019s Surprising! April 6 2018 by Patrice Lee Onwuka Why Does Jimmy Kimmel Get Away with Mocking Melania Trump's Accent? April 5 2018 by Patrice Lee Onwuka There's No Room for Pregnancy Discrimination on the Job April 3 2018 by Patrice Lee Onwuka How the Blockbuster 'Roseanne' Reboot Gives a Voice to Middle America March 29 2018 by Patrice Lee Onwuka Tariffs: A New Bargaining Chip for Better Trade Deals? March 28 2018 by Patrice Lee Onwuka Joe Biden is Right to Attack Giving Every American a Universal Basic Income March 27 2018 by Patrice Lee Onwuka Ivanka Trump Has a Strategy to Get People into Good-Paying Jobs March 27 2018 by Patrice Lee Onwuka Should Responding to After-Hours Emails Be Banned? March 26 2018 by Patrice Lee Onwuka Cardi B Just Said What We're All Thinking About Taxes March 23 2018 by Patrice Lee Onwuka 3 Ways the Trump White House is Helping Millennials March 23 2018 by Patrice Lee Onwuka Google Shares How It Closed Unexplained Pay Gaps March 20 2018 by Patrice Lee Onwuka Jim Carrey Slams Sarah Sanders' Looks in a New Portrait March 19 2018 by Patrice Lee Onwuka VP Pence Celebrates Women as Everyday Heroes with IWF March 15 2018 by Patrice Lee Onwuka Joy Behar Finally Publicly Apologizes to VP Pence, Christians for Disparaging Jokes March 15 2018 by Patrice Lee Onwuka Vice President Pence Has an Answer for Joy Behar's Religious Intolerance March 13 2018 by Patrice Lee Onwuka Nancy Pelosi Back Peddles on Tax Bonuses Being \"Crumbs\" March 9 2018 by Patrice Lee Onwuka Why Tariffs on Steel, Aluminum are the Wrong Move March 9 2018 by Patrice Lee Onwuka 5 Things Women Cannot Do in Some Countries on International Women's Day March 8 2018 by Patrice Lee Onwuka Ride-sharing Replaces Buses in This City and Residents Love It March 7 2018 by Patrice Lee Onwuka How New Tariffs Could Raise the Prices of Your Everyday Goods March 6 2018 by Patrice Lee Onwuka Hollywood's Activism and Hypocrisy Were on Display at the Oscars One Pin at a Time March 5 2018 by Patrice Lee Onwuka 7 Facts About Women in the Workforce for Women's History Month March 1 2018 by Patrice Lee Onwuka Poll: Most Inappropriate Office Behavior is Not What You Think It is February 26 2018 by Patrice Lee Onwuka Why Amy Poehler Shouldn't Fight for a Hike in Tipped Minimum Wages February 26 2018 by Patrice Lee Onwuka Europe\u2019s Gender Quotas for Corporate Boards Failed to Do Much for Women February 20 2018 by Patrice Lee Onwuka Should Healthy, Employable Americans Keep Medicaid Forever? February 16 2018 by Patrice Lee Onwuka 5 Companies Expanding Parental Leave Thanks to Tax Cuts February 15 2018 by Patrice Lee Onwuka The Downside of Another Fast-food Worker Protest for$15 Minimum Wages February 12 2018 by Patrice Lee Onwuka\nUber Has a Gender Earnings Gap, but There's a Good Explanation February 9 2018 by Patrice Lee Onwuka\nSurvey: #MeToo Backlash Against Women is Real February 7 2018 by Patrice Lee Onwuka\nRyan Seacrest Has a Can\u2019t-Miss #MeToo Message About False Allegations February 6 2018 by Patrice Lee Onwuka\nOn the Anniversary of FMLA, Here\u2019s Why There's Little to Celebrate February 5 2018 by Patrice Lee Onwuka\nT-Mobile\u2019s Super Bowl Ad Sounded Tone Deaf February 5 2018 by Patrice Lee Onwuka\nHave You Checked Your Paycheck? You Might Be Surprised February 1 2018 by Patrice Lee Onwuka\n6 Times Democrats Didn\u2019t Cheer for Middle-Class Wins in Trump's State of the Union Address January 31 2018 by Patrice Lee Onwuka\n6 Things to Watch for at President Trump\u2019s State of the Union Address January 30 2018 by Patrice Lee Onwuka\nSurvey: New Tech Moms Have Great Experience Returning to Work Post Baby January 30 2018 by Patrice Lee Onwuka\nHere are Six Political Moments from the 2018 Grammys January 29 2018 by Patrice Lee Onwuka\nDebbie Wasserman Schultz, Nancy Pelosi Pooh-pooh $1,000 Bonuses from Tax Reform January 26 2018 by Patrice Lee Onwuka Walt Disney Heiress Thrashed Tax Reform, but Disney Company Gives Employees Tax Reform Bonuses, Investments January 25 2018 by Patrice Lee Onwuka Cashier-free Amazon Store Opens in Nation's First$15-Minimum-Wage City January 24 2018 by Patrice Lee Onwuka\nDisney Credits Tax Reform for Its $50 Mil Education Investment,$1,000 Bonuses January 24 2018 by Patrice Lee Onwuka\nCongress Made New Changes to 529 Plans that Parents Will Celebrate January 23 2018 by Patrice Lee Onwuka\nKristen Bell Opens Girl Power Sag Awards with Cheap Shot at Melania Trump January 22 2018 by Patrice Lee Onwuka\nFeminist Slams \u2018Fixer Upper\u2019 Stars Chip & Joanna for Having 5th Child January 17 2018 by Patrice Lee Onwuka\nCostco Shocks Customers by Exposing the High Price of Seattle\u2019s Soda Tax January 16 2018 by Patrice Lee Onwuka\nCondoleeza Rice Warns Against #MeToo Creating Snowflakes and Triggering Backlash January 15 2018 by Patrice Lee Onwuka\nWalmart Announces Higher Starting Wages, Bonuses, and Benefits Thanks to Tax Reform January 11 2018 by Patrice Lee Onwuka\nUtility Companies Cut Your Utility Bills Thanks to Tax Cuts January 11 2018 by Patrice Lee Onwuka\nSay Goodbye to Busboys at Red Robin Thanks to Minimum Wage Hikes January 10 2018 by Patrice Lee Onwuka\nUSPS is Losing Money, But Blame It on Washington Not Amazon January 5 2018 by Patrice Lee Onwuka\nNo, Iceland Didn\u2019t Just Make It Illegal to Pay Women Less Than Men January 5 2018 by Patrice Lee Onwuka\nCalifornia Has Brewed Up a Common-Sense Solution to Drunk Driving January 4 2018 by Patrice Lee Onwuka\nTop Hollywood Women Launch Time\u2019s Up as the Answer to #MeToo, but Don\u2019t Be Fooled January 3 2018 by Patrice Lee Onwuka\nWhat You Need to Know About New York's Paid Leave Plan January 3 2018 by Patrice Lee Onwuka\n18 States Hike Minimum Wage in 2018, Young Female Workers May Suffer January 2 2018 by Patrice Lee Onwuka\nIvanka Trump Visit Triggers Connecticut Parents to Pull Kids Out of School December 22 2017 by Patrice Lee Onwuka\nCapitol Hill Just Passed a #GoodDealforWomen Through Tax Reform December 19 2017 by Patrice Lee Onwuka\nCalifornia\u2019s $15 Minimum Wage to Cost 400,000 Jobs \u2013 Many for Women December 18 2017 by Patrice Lee Onwuka Pew: Gender Discrimination, Sexual Harassment are Problems but Not Universal December 15 2017 by Patrice Lee Onwuka Alyssa Milano, John Oliver Aren\u2019t for Saving the Internet but Internet Control December 14 2017 by Patrice Lee Onwuka Feminism is 2017\u2019s Word of the Year But Women Still Don\u2019t Know What It Means December 13 2017 by Patrice Lee Onwuka San Francisco Embraces Everyone but Delivery Robots December 12 2017 by Patrice Lee Onwuka Jobs Report: One Million More Women Are Working Since November 2016 December 8 2017 by Patrice Lee Onwuka What Paris Hilton Could Teach Chelsea Handler About Grace in Crisis December 8 2017 by Patrice Lee Onwuka Kellyanne Conway Had a #MeToo Moment Too, But No One Paid Attention Because of Politics December 7 2017 by Patrice Lee Onwuka Mark Zuckerberg\u2019s Paternity Leave Tells Us 3 Things About Paid Leave, Millennials December 6 2017 by Patrice Lee Onwuka Sheryl Sandberg Calls Out the Chilling Effect Between Sexes Following Harassment Revelations December 5 2017 by Patrice Lee Onwuka Three Ways to Maximize Your Last Paychecks of 2017 December 1 2017 by Patrice Lee Onwuka Meghan Markle Giving Up Her Career for Prince Harry Doesn\u2019t Make Her a Bad Woman November 29 2017 by Patrice Lee Onwuka Susan Sarandon is Still Anti-Hillary, The Left Blasts Her For It November 28 2017 by Patrice Lee Onwuka FCC\u2019s Thanksgiving Blessing was to End Internet Neutrality Regulations November 27 2017 by Patrice Lee Onwuka Black Friday Protests Dwindle Exposing This Uncomfortable Truth November 27 2017 by Patrice Lee Onwuka Kathy Griffin\u2019s Sadistic Trump Joke Left Her Jobless, She\u2019s Not a Victim November 21 2017 by Patrice Lee Onwuka New York Wants to Dictate When You Work November 20 2017 by Patrice Lee Onwuka Bernie Sanders Blatantly Gives Bill Clinton, Al Franken a Pass on Sexual Harassment November 20 2017 by Patrice Lee Onwuka Do Winter Blizzards Call for Paid Leave Too? November 15 2017 by Patrice Lee Onwuka \u2018The Simpsons\u2019 Calling Kellyanne Conway a Nazi is a Sexist, Racist Cheap Shot November 14 2017 by Patrice Lee Onwuka How Zero Tolerance Polices Backfire in the Age of #MeToo November 13 2017 by Patrice Lee Onwuka Three G.I. Janes from American History that Make Women Proud on Veterans Day November 10 2017 by Patrice Lee Onwuka TSA Failing Undercover Weapons Tests \u2013 Again, and It\u2019s Disturbing November 10 2017 by Patrice Lee Onwuka NAACP Exploits NFL Protests to Ban \"Star-Spangled Banner\" as National Anthem November 9 2017 by Patrice Lee Onwuka Women in Congress Have an Answer to Paid Leave November 8 2017 by Patrice Lee Onwuka Sexist LA Times Columnist Attacks Sarah Huckabee Sanders\u2019 Looks November 6 2017 by Patrice Lee Onwuka Explosive: DNC, Hillary Clinton Campaign Rigged the Party Nomination for Her November 3 2017 by Patrice Lee Onwuka Eva Longoria is Desperately Wrong to Push the Latina Pay Gap November 2 2017 by Patrice Lee Onwuka Shame on Cosmo for Normalizing Incest \u2013 Again November 2 2017 by Patrice Lee Onwuka Poll: Most Women Love Tech but Only Half Think It\u2019ll Improve Their Life November 1 2017 by Patrice Lee Onwuka 2018 Obamacare Premiums to Leave Even Wider Hole in Your Wallet October 30 2017 by Patrice Lee Onwuka Walmart Debuts Robots to Make (Holiday) Shopping Better October 30 2017 by Patrice Lee Onwuka NAACP Issues Inane Flying-While-Black Advisory, Please Keep It to Yourselves October 27 2017 by Patrice Lee Onwuka President Trump\u2019s Opioid Attack Plan: Personal, Practical, and Policy October 26 2017 by Patrice Lee Onwuka UPenn Teacher Helps Black Women by Discriminating Against White Men October 24 2017 by Patrice Lee Onwuka Campus Sexual Assault: The Distressing Fight Mothers Face for Their Sons, Advice to Young Women October 23 2017 by Patrice Lee Onwuka IRS to Hold Tax Returns Hostage to Enforce Obamacare October 20 2017 by Patrice Lee Onwuka ABC\u2019s Misleading Workplace Sexual Harassment Poll Pushes an Agenda October 19 2017 by Patrice Lee Onwuka \u2018The Big Bang Theory\u2019 Actress Counsels Actresses to be Modest, Triggers Feminist Backlash October 17 2017 by Patrice Lee Onwuka No, NFL Protests Aren't About the Gender Pay Gap October 16 2017 by Patrice Lee Onwuka Ivanka Trump to Boost Global Women Entrepreneurs Via World Bank Fund October 16 2017 by Patrice Lee Onwuka President Trump\u2019s Anti-Regulation Crusade may be Working in Americans\u2019 Eyes October 13 2017 by Patrice Lee Onwuka Lawmakers Try to Stop Secretary DeVos on Campus Sexual Assault Policy October 13 2017 by Patrice Lee Onwuka Black Lives Matter Group Opposes Free Speech for Every American October 12 2017 by Patrice Lee Onwuka Harvey Weinstein\u2019s Open Secret Exposes Hypocrisy on the Left October 10 2017 by Patrice Lee Onwuka Marc Jacobs Doubles Down on Free Expression for Designers, Artists October 10 2017 by Patrice Lee Onwuka On Columbus Day, Give Him a Break October 9 2017 by Patrice Lee Onwuka Company Behind Fearless Girl Statute Admits Gender Pay Issues October 9 2017 by Patrice Lee Onwuka Three Bright Spots for Women in Today\u2019s Jobs Report October 6 2017 by Patrice Lee Onwuka Half of Americans Can\u2019t Afford More than$100 in Healthcare Premiums October 4 2017 by Patrice Lee Onwuka\nOne Tech Company Changed Approach to Promotions, Women Won Big October 3 2017 by Patrice Lee Onwuka\nLibrarian Rejects Dr. Seuss Books Because Melania Trump Sent Them September 29 2017 by Patrice Lee Onwuka\nAmber Rose\u2019s Slutwalk is a Stunt Not Empowerment September 29 2017 by Patrice Lee Onwuka\nDo We Want a Feminist \u2018Will & Grace\u2019 Reboot? September 26 2017 by Patrice Lee Onwuka\nHillary Clinton Slams Women Who Support Trump as \u201cPublicly Disrespecting Themselves\u201d September 26 2017 by Patrice Lee Onwuka\nBetsy DeVos Ends Obama-Era Policies on School Sexual Assault September 22 2017 by Patrice Lee Onwuka\nWashington Wants Black Households to Trade Privacy for Cash September 22 2017 by Patrice Lee Onwuka\nHalf of College Women Get This Wrong about Hate Speech September 20 2017 by Patrice Lee Onwuka\nJoe Biden Slams Socialist Universal Basic Income Idea September 19 2017 by Patrice Lee Onwuka\nMillennials Would Trade Right to Vote for Student Loan Freedom September 18 2017 by Patrice Lee Onwuka\nFeminists Attack Starbucks's Pumpkin Spice Lattes for Funding White Supremacy September 15 2017 by Patrice Lee Onwuka\nDeVos Backlash Incites Dangerous Language September 14 2017 by Patrice Lee Onwuka\nOver One in Three Americans Can\u2019t Name Any 1st Amendment Rights September 12 2017 by Patrice Lee Onwuka\nMiss America Pageant Gets Political and Blatantly Anti-Trump September 11 2017 by Patrice Lee Onwuka\nMelania Trump Staffs Fewer People than Michelle Obama \u2013 A Savings for Taxpayers September 8 2017 by Patrice Lee Onwuka\nHillary Clinton Asks 'What Happened?' in 2016 September 7 2017 by Patrice Lee Onwuka\nKim Kardashian Still Eschews One Label: Feminist September 5 2017 by Patrice Lee Onwuka\nJerry Springer Rides the $15 Minimum Wage Wagon September 5 2017 by Patrice Lee Onwuka A$20 Question: Will Harriet Tubman be on the Bill? September 1 2017 by Patrice Lee Onwuka\nEstee Lauder Faces Heat Over Paid Leave for Dads August 31 2017 by Patrice Lee Onwuka\nNo, Melania\u2019s Shoes are Not the Story, Texas is August 30 2017 by Patrice Lee Onwuka\nIllinois Governor Shuts Down $15 Minimum Wage August 28 2017 by Patrice Lee Onwuka Amy Schumer's Equal Pay Problem August 25 2017 by Patrice Lee Onwuka Poll: Nearly Three Out of Four Americans Would Die to Defend Freedom of Speech August 25 2017 by Patrice Lee Onwuka Here\u2019s Another Way Women Suffer from Minimum Wage Increases August 24 2017 by Patrice Lee Onwuka Oops! Social Security Website Misdirects$11 Mil to Thieves August 23 2017 by Patrice Lee Onwuka\nD.C. Raises Minimum Wage, Diners Foot the Bill August 21 2017 by Patrice Lee Onwuka\nWhat Dr. Ben Carson Shows Us about Confronting Hateful Ideas August 17 2017 by Patrice Lee Onwuka\nHope Brings Another Woman to Trump Leadership Circle August 17 2017 by Patrice Lee Onwuka\nHelen Mirren\u2019s Feminism is about Anti-Trump and Potty-Mouths August 16 2017 by Patrice Lee Onwuka\nCFPB Crackdown May Trap Borrowers in a Different Debt Trap August 14 2017 by Patrice Lee Onwuka\nSheryl Sandberg Leans In to One-Size-Fits-All Paid Leave Plan August 10 2017 by Patrice Lee Onwuka\nCaf\u00e9 Charges 18% Man Tax to Close the Gender Pay Gap August 9 2017 by Patrice Lee Onwuka\nCollege STEM Women Claim to be Sexism Victims August 7 2017 by Patrice Lee Onwuka\nA Sexism Rating Coming to a Movie Near You August 3 2017 by Patrice Lee Onwuka\nIRS Rehires Hundreds of Fired Workers Despite Serious Transgressions August 3 2017 by Patrice Lee Onwuka\nStudy: MD County to Lose 47,000 jobs because of $15 Minimum Wage August 2 2017 by Patrice Lee Onwuka Adam Carolla Schools Congress on Campus Free Speech July 31 2017 by Patrice Lee Onwuka How One Tech Company is Taking on the Future of Work July 28 2017 by Patrice Lee Onwuka Governor Christie Shuts Down Costly Paid Leave Expansion Bill July 27 2017 by Patrice Lee Onwuka Teachers\u2019 Union Boss Plays the Race Card against Secretary DeVos July 25 2017 by Patrice Lee Onwuka Cutting the Strings on Obamacare Enrollment Assistance July 24 2017 by Patrice Lee Onwuka Not Even Death Stops a Government Check July 21 2017 by Patrice Lee Onwuka Who Cares for the Caregivers? July 19 2017 by Patrice Lee Onwuka College Diversity Officials Rake in Major Dough July 18 2017 by Patrice Lee Onwuka DHS Hides Spending$3.6 Mil of Taxpayer Dollars on Pricey Conferences July 17 2017 by Patrice Lee Onwuka\nStudy: Marrying Before Having Kids is a Strong Path to Success for Millennials July 13 2017 by Patrice Lee Onwuka\n$2 Mil NJ Welfare Fraud Exposes Holes in Our Social Safety Net July 7 2017 by Patrice Lee Onwuka \u201cWTF\u201d is Silicon Valley\u2019s Next Answer to the Democratic Party July 6 2017 by Patrice Lee Onwuka Fraud Alert: Federal Free Phone, Internet Program Wastes$100 Mil Annually on Fake, Dead Enrollees July 3 2017 by Patrice Lee Onwuka\nCosmo Leaves Conservatives Off Their Short List of Female Presidential Candidates June 28 2017 by Patrice Lee Onwuka\nReport: Seattle\u2019s Minimum Wage Hike Gave Low-Wage Workers a $125 Monthly Pay Cut June 27 2017 by Patrice Lee Onwuka Obama Comes Out to Bash Senate Repeal Bill June 23 2017 by Patrice Lee Onwuka Maybe Profit Motivates Tech Companies More than Politics June 22 2017 by Patrice Lee Onwuka Ivanka Trump Hits Capitol Hill about Paid Leave June 20 2017 by Patrice Lee Onwuka Supreme Court Upholds Offensive Names as Free Speech June 20 2017 by Patrice Lee Onwuka Silicon Valley CEOs Ignore Calls to Boycott Trump Tech Summit June 19 2017 by Patrice Lee Onwuka What the Return of Pro-Net Neutrality Democrat Means for Internet Regs Rollback June 15 2017 by Patrice Lee Onwuka Is the Federal Consumer Cop Asleep on the Job? June 15 2017 by Patrice Lee Onwuka Obama Health Records Program Leaves Taxpayers on the Hook for Mispayments June 15 2017 by Patrice Lee Onwuka Just in Time for July 4th New York State Greenlights Uber, Lyft June 9 2017 by Patrice Lee Onwuka Filter, Don\u2019t Pilfer June 8 2017 by Patrice Lee Onwuka What Michelle Obama Gets Wrong About Self-Interest June 8 2017 by Patrice Lee Onwuka Harvard Revokes Acceptances Over Online Comments June 6 2017 by Patrice Lee Onwuka Bill Maher\u2019s N-Word Episode June 5 2017 by Patrice Lee Onwuka Hillary Clinton's Whine Tour Stops in Silicon Valley June 2 2017 by Patrice Lee Onwuka Maryland Governor Shuts Down Paid Sick-Leave Bill May 26 2017 by Patrice Lee Onwuka Chelsea Clinton Equates Child Marriage to Climate Change May 26 2017 by Patrice Lee Onwuka Obamacare\u2019s Death Spiral Claims Another Healthcare Insurer May 25 2017 by Patrice Lee Onwuka Top 10 Unauthorized Federal Programs on the Chopping Block in Trump Budget May 25 2017 by Patrice Lee Onwuka Supreme Court Ends Judge Shopping for Patent Trolls May 23 2017 by Patrice Lee Onwuka State of Texas Trumps Austin over Uber, Lyft Regulations May 22 2017 by Patrice Lee Onwuka Undoing Net Neutrality Starts Today May 18 2017 by Patrice Lee Onwuka San Francisco May Stall Delivery Robots May 18 2017 by Patrice Lee Onwuka Miss USA Stands by Her Comments \u2013 Good for Her May 17 2017 by Patrice Lee Onwuka Miss USA Slammed for Saying Healthcare is a Privilege Rather than a Right May 15 2017 by Patrice Lee Onwuka Don\u2019t Be Tone Deaf on Hearing Aids May 11 2017 by Patrice Lee Onwuka Grads Boo Betsy DeVos at Commencement May 11 2017 by Patrice Lee Onwuka TV Comedian John Oliver Piles on to Save Obama Internet Rules May 10 2017 by Patrice Lee Onwuka Automation Coming to a City near You May 9 2017 by Patrice Lee Onwuka Your Zip code Is a Good Predictor for How Long You Live May 9 2017 by Patrice Lee Onwuka Tech Bosses: Expect Significant Automation of Jobs by 2022 May 5 2017 by Patrice Lee Onwuka Can Another Tech Squad to Transform How Washington Works? May 5 2017 by Patrice Lee Onwuka Tech Founder to Ivanka Trump: Don\u2019t Use My Story May 3 2017 by Patrice Lee Onwuka Are Saving and Debt Repayment the New Normal for Americans? May 2 2017 by Patrice Lee Onwuka This Immigrant is Showing Up to Work on May Day May 1 2017 by Patrice Lee Onwuka FCC Announces Net Neutrality Reversal Plan April 27 2017 by Patrice Lee Onwuka Copyright Office Reform Bill Passes House April 27 2017 by Patrice Lee Onwuka FCC Expected to Kick Net Neutrality to the Curb April 26 2017 by Patrice Lee Onwuka Delivery Robots Get Legislative Green Light in Two States April 25 2017 by Patrice Lee Onwuka Harvard Study: Higher Minimum Wages Trigger Restaurant Closures April 24 2017 by Patrice Lee Onwuka Not Done Rolling Back Washington\u2019s Red Tape Yet April 19 2017 by Patrice Lee Onwuka Three Celebrities Feeling the Pain on Tax Day April 18 2017 by Patrice Lee Onwuka Female-Led FTC Takes Reducing Regulations Seriously April 18 2017 by Patrice Lee Onwuka New York Pilots Free College Tuition Plan April 13 2017 by Patrice Lee Onwuka Poll: Americans Comfortable with AI in Most--But Not All--Instances April 11 2017 by Patrice Lee Onwuka No Phone Calls on Airplanes After All April 11 2017 by Patrice Lee Onwuka Amazon to Hire 30,000 Part-time, Work-from-Home Jobs April 10 2017 by Patrice Lee Onwuka Judge Blocks Seattle\u2019s Plan to Unionize Uber April 7 2017 by Patrice Lee Onwuka Stop the Misinformation on the Roll-Back of Privacy Rules April 5 2017 by Patrice Lee Onwuka The Challenge of Counterfeit Drugs April 4 2017 by Patrice Lee Onwuka Cali Lawmaker Wants to Push Gig-Economy Workers into Unions April 3 2017 by Patrice Lee Onwuka FCC Reverses Federal Power Grab Over LifeLine Program (AKA ObamaPhones for the Internet) March 30 2017 by Patrice Lee Onwuka Here\u2019s How Uber is Boosting Diverse Applicants March 30 2017 by Patrice Lee Onwuka Congress Scraps Obama-Era Privacy Rules March 30 2017 by Patrice Lee Onwuka Should We Worry About Robots Replacing Workers? March 27 2017 by Patrice Lee Onwuka Black Leaders Tell Trump What They Have to Lose March 23 2017 by Patrice Lee Onwuka Congress to Free the Copyright Office from the Thumb of the Library of Congress March 23 2017 by Patrice Lee Onwuka Why Everything from Dresses to Avocados May Get More Expensive March 22 2017 by Patrice Lee Onwuka Trump, Clinton Campaigns Depended Upon Sharing Economy on the Campaign Trail March 21 2017 by Patrice Lee Onwuka Before You Blame the Wage Gap, Check Your Attitude March 20 2017 by Patrice Lee Onwuka Trump Hacks Away at the TSA March 20 2017 by Patrice Lee Onwuka Trump Pumps the Brakes on Obama-Era Fuel Efficiency Regs March 16 2017 by Patrice Lee Onwuka Students Push Back on Feminist Professor\u2019s Indoctrination March 15 2017 by Patrice Lee Onwuka President Trump: May Be Time to Say \"You're Fired!\" to Duplicative Agency Employees March 14 2017 by Patrice Lee Onwuka Austin\u2019s Struggle with Transportation After Uber, Lyft Ride Out of Town March 13 2017 by Patrice Lee Onwuka Steve Case: American Workers Feeling Left Behind, Really Have Been Left Behind March 13 2017 by Patrice Lee Onwuka Analogies Fly Over Net Neutrality March 9 2017 by Patrice Lee Onwuka These are the Women A Day Without Women Hurts March 8 2017 by Patrice Lee Onwuka Can Facebook\u2019s New Tool Crack Down on \u201cFake News\u201d? March 7 2017 by Patrice Lee Onwuka Is Regulating the Internet a Good Idea? March 6 2017 by Patrice Lee Onwuka Dr. Ben Carson Confirmed as HUD Head March 3 2017 by Patrice Lee Onwuka Robots Green-Lighted to Deliver Your Packages March 2 2017 by Patrice Lee Onwuka Three Key Messages to Black America from Trump\u2019s Address March 1 2017 by Patrice Lee Onwuka Reports: GM Lobbies to Keep Competing Self-Driving Tech Off the Road February 28 2017 by Patrice Lee Onwuka New Self-driving Truck Company Speeds Up the Technology February 27 2017 by Patrice Lee Onwuka Small Internet Companies Gain Regulatory Relief from Washington February 24 2017 by Patrice Lee Onwuka Women Have a New Ally in Getting Them to Work February 24 2017 by Patrice Lee Onwuka Federal Watchdog: Obama\u2019s Techie Squad Flouted the Rules February 22 2017 by Patrice Lee Onwuka Adulting School: Millennials Are Signing Up for Classes February 22 2017 by Patrice Lee Onwuka Former AG Eric Holder Leading Internal Investigation at Uber February 21 2017 by Patrice Lee Onwuka Whoopi Snubs the Mean Girls at New York Fashion Week in Defense of Tiffany Trump February 17 2017 by Patrice Lee Onwuka Humana Calls It Quits on ObamaCare, Leaving One State in Trouble February 16 2017 by Patrice Lee Onwuka Congress Opens the Door to Self-Driving Cars February 15 2017 by Patrice Lee Onwuka Linda McMahon Confirmed to Head the SBA February 14 2017 by Patrice Lee Onwuka Cabs Threaten to Halt Wheelchair Access Over Uber, Lyft February 14 2017 by Patrice Lee Onwuka Trump Models: The Next Target in the Fashion World Purge February 13 2017 by Patrice Lee Onwuka Senate Dems Come Out Swinging to Protect Washington\u2019s Control of the Internet February 8 2017 by Patrice Lee Onwuka A Surprise Stand for Competition Over Cronyism in Battle for NYC Streets? February 8 2017 by Patrice Lee Onwuka Lifeline Internet Assistance Program (AKA Obamaphones for the Internet) Takes a Hit February 7 2017 by Patrice Lee Onwuka Big Mac ATM\u2019s, the Answer to$15 Minimum Wages? February 2 2017 by Patrice Lee Onwuka\nNew FCC Chairman Gets to Work Cutting Red Tape February 2 2017 by Patrice Lee Onwuka\nTrump Transportation Secretary Chao Confirmed, Despite Democratic Protest February 1 2017 by Patrice Lee Onwuka\nCutting the Red Tap: President Trump Orders 2-for 1 Rule on Regulations January 31 2017 by Patrice Lee Onwuka\nChelsea Handler's Garbage-Pail Mouth is Now Shaming Melania Trump's English January 27 2017 by Patrice Lee Onwuka\nBreak out the Dow 20K Hats? January 27 2017 by Patrice Lee Onwuka\nApple Hit with Another Lawsuit over Distracted Driving January 25 2017 by Patrice Lee Onwuka\nGetting American Workers Ready for Automation\u2019s Coming Shake-Up January 24 2017 by Patrice Lee Onwuka\nUber Ponies Up $20M to Feds Over Driver Pay January 23 2017 by Patrice Lee Onwuka Americans Aren\u2019t Really Warming Up to Self-Driving Cars January 19 2017 by Patrice Lee Onwuka The Government\u2019s \u201cSlant\u201d on Trademarks Gets Challenged January 19 2017 by Patrice Lee Onwuka Uber Sues Seattle to Block Drivers from Unionizing January 18 2017 by Patrice Lee Onwuka Inauguration Reminds Us Why Home Sharing is a Win for Travelers January 17 2017 by Patrice Lee Onwuka Jesse Jackson PUSHes Uber on Diversity Statistics January 13 2017 by Patrice Lee Onwuka 5 Times President Obama\u2019s Rhetoric Didn\u2019t Match His Kumbaya Message January 11 2017 by Patrice Lee Onwuka Jenna Bush Microaggressess at the Golden Globes? January 10 2017 by Patrice Lee Onwuka Rolling Back Obama\u2019s Internet Privacy Rules January 9 2017 by Patrice Lee Onwuka Update: City of San Diego Says Restaurants Can't Publicly Blame Surcharge on Minimum Wage Hike January 6 2017 by Patrice Lee Onwuka San Diego Restaurants Slap Diners with Surcharges Thanks to Minimum Wage Increases January 6 2017 by Patrice Lee Onwuka The Uber Hook-Up Effect January 5 2017 by Patrice Lee Onwuka Christmas Eve Distracted Driving Fatality Could Trigger Government Action January 3 2017 by Patrice Lee Onwuka Cali Regulators Drive Uber\u2019s Self-Driving Cars Away December 27 2016 by Patrice Lee Onwuka Will Seattle\u2019s Driver Unionization be the End of Uber, Ridesharing? December 21 2016 by Patrice Lee Onwuka Not Fake News: Some UVA Students Sign Fake Anti-Christmas Petition December 20 2016 by Patrice Lee Onwuka Pinterest Misses Female Recruitment Quantity Quotas, But Hits on Quality December 19 2016 by Patrice Lee Onwuka No Michelle Obama, \u201cHope\u201d Doesn\u2019t Start and End with Your Husband December 19 2016 by Patrice Lee Onwuka Uber, San Francisco Regulators Crash Head-on over Self-Driving December 16 2016 by Patrice Lee Onwuka More on the Shaming of Wonder Woman December 15 2016 by Patrice Lee Onwuka Environmental Group Smears Black Beauty Products December 14 2016 by Patrice Lee Onwuka San Francisco Mayor Sides with Airbnb December 12 2016 by Patrice Lee Onwuka Airbnb Can\u2019t Beat \u2018Em, So Now They\u2019re Lobbying \u2018Em December 9 2016 by Patrice Lee Onwuka The Supremes Side with Samsung Against Apple in High-Profile Patent Case December 7 2016 by Patrice Lee Onwuka Cell Phones on the Road: A Serious Problem, but Do the Feds Need to Solve It? December 6 2016 by Patrice Lee Onwuka Airbnb Took on the Big Apple and Lost December 5 2016 by Patrice Lee Onwuka Obama\u2019s Overtime Rule is Down, but Not Out December 2 2016 by Patrice Lee Onwuka Taxpayers Will Forgive Billions More in Student Loan Debt December 2 2016 by Patrice Lee Onwuka Uber Drivers Join Minimum Wage Protesters November 30 2016 by Patrice Lee Onwuka Negative Online Reviews Get New Federal Protection November 30 2016 by Patrice Lee Onwuka #GivingTuesday: the Hangover Cure for a Weekend of Spending November 29 2016 by Patrice Lee Onwuka Some Maine Restaurant Workers Worried about Effects of Minimum Wage Hike November 28 2016 by Patrice Lee Onwuka Judge Blocks Obama Workplace Overtime Rules November 23 2016 by Patrice Lee Onwuka Pew: One in Four Americans Makes Bank through the Gig Economy November 21 2016 by Patrice Lee Onwuka U.S. Small Businesses: Good News and Bad News November 21 2016 by Patrice Lee Onwuka Arizona\u2019s New Minimum Wage Packs a One-Two Punch for Nonprofits November 18 2016 by Patrice Lee Onwuka Washington State Minimum Wage Passed and Childcare Costs Skyrocket November 17 2016 by Patrice Lee Onwuka Hi, I\u2019m Ben, Your Senator and Uber Driver! November 15 2016 by Patrice Lee Onwuka Facebook Ends Multicultural Targeted Ads November 14 2016 by Patrice Lee Onwuka Oprah\u2019s Guarded Hope for the New President Triggers the Left November 14 2016 by Patrice Lee Onwuka Star-Power Failed to Energize Millennials at the Polls November 10 2016 by Patrice Lee Onwuka Mom-preneur Faces Charges for Selling Ceviche? November 9 2016 by Patrice Lee Onwuka Uber is Back on the Road in Philly November 7 2016 by Patrice Lee Onwuka Airbnb\u2019s Self-Policing Strategy against Discrimination November 4 2016 by Patrice Lee Onwuka Sanders and Warren Ignore that America is Better Off Now While Promoting Socialism \u2022 Evening Edit October 14 2019 by Patrice Lee Onwuka Biden campaign plans to lose Iowa, New Hampshire, and Nevada \u2022 Rising October 11 2019 by Patrice Lee Onwuka Is Warren trying to win black voters by floating Gillum as VP \u2022 Rising October 11 2019 by Patrice Lee Onwuka US Businesses Should Stand Behind Hong Kong \u2022 Making Money October 9 2019 by Patrice Lee Onwuka Left Leaning Outlets Turn on NYT Following Kavanaugh Hit Piece \u2022 Evening Edit September 23 2019 by Patrice Lee Onwuka Partisan Party Politics Turns American Voters Off \u2022 Lou Dobbs Tonight September 18 2019 by Patrice Lee Onwuka Bashing Billionaires Dissuades Everyone from Trying to Reach the Top \u2022 Making Money September 13 2019 by Patrice Lee Onwuka Dems Have a Hard Case to Make Heading Into 2020 \u2022 Your World with Neil Cavuto September 9 2019 by Patrice Lee Onwuka Dems Prioritize Gun Reform But Its All Talk \u2022 Evening Edit September 9 2019 by Patrice Lee Onwuka Talking Tech: Is Being a \"Big Company\" a Bad Thing? \u2022 Making Money with Charles Payne September 6 2019 by Patrice Lee Onwuka Media Needs to Find Balance Between Criticism and Support of President \u2022 Evening Edit September 3 2019 by Patrice Lee Onwuka Democrats Continue to Focus on Emotion Over Fact: Biden and AOC \u2022 Trish Regan Primetime August 30 2019 by Patrice Lee Onwuka Is the DNC rigging the debate rules? \u2022 Rising August 29 2019 by Patrice Lee Onwuka Is South Carolina Biden's Hail Mary? \u2022 Rising August 29 2019 by Patrice Lee Onwuka 4 in 10 People Won't Trust the 2020 Election Results if Their Team Loses \u2022 Lou Dobbs Tonight August 28 2019 by Patrice Lee Onwuka Taking Border Security Seriously with Approved Funding for The Wall \u2022 Evening Edit August 27 2019 by Patrice Lee Onwuka Women Are Winning in This Economy \u2022 Coast to Coast August 22 2019 by Patrice Lee Onwuka Language Police: In San Francisco and on Facebook \u2022 Lou Dobbs Tonight August 22 2019 by Patrice Lee Onwuka Harry Reid Denounces Trajectory of Dem Party \u2022 Making Money with Charles Payne August 21 2019 by Patrice Lee Onwuka Judge Trump\u2019s actions, not words \u2014 and other commentary August 19 2019 by Patrice Lee Onwuka Conservative commentator rips Shapiro over criticism of people with multiple jobs August 16 2019 by Patrice Lee Onwuka How Will the Economy Effect the 2020 Conversation? \u2022 Coast to Coast August 15 2019 by Patrice Lee Onwuka Israel Should Say \"Thanks, But No Thanks\" to Reps. Omar and Tlaib \u2022 Lou Dobbs Tonight August 15 2019 by Patrice Lee Onwuka We Need to Act Now to Steer Away from Red Flags Ahead \u2022 Making Money August 14 2019 by Patrice Lee Onwuka \"In God We Trust\" Is Important to Display in School \u2022 Lou Dobbs Tonight August 14 2019 by Patrice Lee Onwuka Political Talking Points Stir Up Violent Threats Against SoulCycle Owner \u2022 Evening Edit August 9 2019 by Patrice Lee Onwuka Pitting Consumers Against Businesses in the Name of Politics \u2022 Coast to Coast August 8 2019 by Patrice Lee Onwuka Democratic Socialists Cannot Defeat Capitalism with Proper Pronouns and Complaints \u2022 Lou Dobbs August 6 2019 by Patrice Lee Onwuka Don't pander to women; offer real opportunity August 1 2019 by Patrice Lee Onwuka Bernie Ignores Legitimate Questions About his Health Care Plans \u2022 Evening Edit July 31 2019 by Patrice Lee Onwuka The Washington Post Plays Scapegoat for Those Still Hung Up on 2016 Election \u2022 Lou Dobbs Tonight July 30 2019 by Patrice Lee Onwuka Food Stamps Cut Because Booming Economy Means Less People Need Them \u2022 Fox & Friends First July 24 2019 by Patrice Lee Onwuka The Squad Tries Pushing the Democrats Further Left Over Israel \u2022 Evening Edit July 24 2019 by Patrice Lee Onwuka$15 Minimum Wage Robs Women of Jobs, Opportunity, Economic Mobility July 18 2019 by Patrice Lee Onwuka\n\"Squad\" Pushes Democrat Party Closer to Socialism \u2022 Coast to Coast July 17 2019 by Patrice Lee Onwuka\n\"Woke Activism\" is No Way to Get Things Done in Congress \u2022 Fox & Friends First July 15 2019 by Patrice Lee Onwuka\nCory Gardner, the women\u2019s vote and religion: Western Conservative Summit takeaways July 14 2019 by Patrice Lee Onwuka\nDemocrats Have Placed All Their Eggs in the Identity Politics Basket \u2022 Evening Edit July 11 2019 by Patrice Lee Onwuka\nWhy Restaurant Workers are Rising Up to Save Tipped Wages July 9 2019 by Patrice Lee Onwuka\nWomen's World Cup Champions Demand Equal Pay, Do the Numbers Add Up? \u2022 After The Bell July 8 2019 by Patrice Lee Onwuka\nThe Folly of Demanding Amazon\u2019s Tax Returns \u2022 Liberty Watch Radio July 7 2019 by Patrice Lee Onwuka\nDemocrats are Obsessed Over the Past as the Candidates Take Aim at Each Other \u2022 Making Money July 5 2019 by Patrice Lee Onwuka\nSupreme Court Reaches Verdict on 2020 Census Question \u2022 Your World with Neil Cavuto June 27 2019 by Patrice Lee Onwuka\nDemocrats Ignore Booming Economy and Foreshadow Recession \u2022 Evening Edit June 25 2019 by Patrice Lee Onwuka\nEconomists Determine the Value of \"Free\" Social Media Apps \u2022 Making Money June 24 2019 by Patrice Lee Onwuka\nRace is a Central Issue for 2020, Biden is Learning the Hard Way \u2022 Coast to Coast June 20 2019 by Patrice Lee Onwuka\nHearing on Reparations Shows Just How Complicated This Issue Is \u2022 Trish Regan Primetime June 19 2019 by Patrice Lee Onwuka\nPresident Trump Prioritizes Legal Immigrants During Border Crisis \u2022 Lou Dobbs Tonight June 18 2019 by Patrice Lee Onwuka\nLets Take an Objective Look at the Russian Interference Allegations \u2022 PoliticsNation June 15 2019 by Patrice Lee Onwuka\nLiberals Distract Voters From Booming Economy with Social Issues \u2022 Lou Dobbs Tonight June 14 2019 by Patrice Lee Onwuka\nWhy Kamala Harris\u2019 \u2018Equal Pay\u2019 Proposal Will Hurt, Not Help Women June 13 2019 by Patrice Lee Onwuka\nRepublicans and Democrats at odds on overtime pay June 13 2019 by Patrice Lee Onwuka\nDeep Fake Videos Can Have Dangerous Repercussions \u2022 Evening Edit June 12 2019 by Patrice Lee Onwuka\nBernie Sanders Bashes Walmart Execs Claiming Minimum Wage Isn't High Enough \u2022 Coast to Coast June 5 2019 by Patrice Lee Onwuka\nMandated Paid Time Off Is Not Only Bad for Most Businesses, It's Bad for Many Women June 5 2019 by Patrice Lee Onwuka\nDemocrats Throw Money at Homelessness But the Situation Still Gets Worse, Why? \u2022 Lou Dobbs Tonight June 5 2019 by Patrice Lee Onwuka\nStossel: The Paid Leave Fairy Tale \u2022 Reason TV June 4 2019 by Patrice Lee Onwuka\nCrisis at the Border and Draining the Swamp, Trust Trump to Create Change \u2022 Making Money May 31 2019 by Patrice Lee Onwuka\nKamala Harris\u2019 New Proposal to Punish Companies Renews Wage Gap Debate \u2022 Bold Politics May 24 2019 by Patrice Lee Onwuka\nConflict in D.C. Does Effect Markets, But Consumers Continue to Spend \u2022 Coast to Coast May 23 2019 by Patrice Lee Onwuka\nThe Gender Wage Gap and What You Can Do About It May 22 2019 by Patrice Lee Onwuka\nTop Dems Blow Any Chance of Bipartisan Reform \u2022 Evening Edit May 22 2019 by Patrice Lee Onwuka\nWomen's Organization Announces The 2019 Empowered Woman of the Year May 21 2019 by Patrice Lee Onwuka\nKamala Harris Threatens to Fine Businesses that Contribute to the \"Pay Gap\" \u2022 Fox & Friends May 21 2019 by Patrice Lee Onwuka\nLow Birth Rates Could Have Devastating Economic Impacts \u2022 Fox & Friends May 16 2019 by Patrice Lee Onwuka\nPolicy analyst on falling birth rates: Rhetoric from the left contributing to 'change in norms' May 16 2019 by Patrice Lee Onwuka\nSanctuary Cities Putting Illegal Immigrants Ahead of Citizens \u2022 Lou Dobbs Tonight May 8 2019 by Patrice Lee Onwuka\nTrump Should Look to Private Sector for Federal Reserve Board Nominees \u2022 Coast to Coast May 2 2019 by Patrice Lee Onwuka\nEvery Aspect of the Economy is Thriving From Jobs, Wages, To Taxes\u2022 Fox & Friends First April 29 2019 by Patrice Lee Onwuka\nBernie Struggles to Win Over Black Women \u2022 Evening Edit April 25 2019 by Patrice Lee Onwuka\nPatrice Onwuka on Plastic Bag Bans, Net Neutrality and Michelle Obama \u2022 Americhicks April 23 2019 by Patrice Lee Onwuka\nMany Americans Don't Care About the Mueller Report \u2022 Cavuto Live April 20 2019 by Patrice Lee Onwuka\nObama Playbook Will Not Be Enough to Get Joe Biden in the White House \u2022 Fox & Friends April 20 2019 by Patrice Lee Onwuka\nMueller Report Finally Available: What did it say and what is next? \u2022 WAMU April 18 2019 by Patrice Lee Onwuka\nOpportunity Zones Create Economic Prosperity \u2022 Making Money with Charles Payne April 17 2019 by Patrice Lee Onwuka\nLeft Tries Desperately To Keep Collusion Narrative Alive \u2022 Coast to Coast April 15 2019 by Patrice Lee Onwuka\nAs support for reparations grow, so does pushback from some black Americans April 12 2019 by Patrice Lee Onwuka\nDemocrats Are Selective When Deciding Who is Innocent or Guilty \u2022 Fox & Friends Saturday April 6 2019 by Patrice Lee Onwuka\nBeing Fiscally Conservative Includes Being Smart on All Aspects of the Economy \u2022 Coast to Coast April 3 2019 by Patrice Lee Onwuka\nDon't Fall for the Equal Pay Day Myth \u2022 Larry O'Connor Show April 2 2019 by Patrice Lee Onwuka\nPresidential Hopefuls Playing Politics with Joe Biden's Victims \u2022 Evening Edit April 1 2019 by Patrice Lee Onwuka\nJoe Biden's 2020 Doom \u2022 Cheddar Politics April 1 2019 by Patrice Lee Onwuka\nMarco Rubio's New Family Leave Bill is a Budget Neutral Conservative Plan \u2022 Coast to Coast March 27 2019 by Patrice Lee Onwuka\nThe gender pay gap still exists - but it's shrinking, according to a new study March 27 2019 by Patrice Lee Onwuka\nMillions of American off food stamps under President Trump \u2022 OANN March 23 2019 by Patrice Lee Onwuka\nThe Electoral College Matters, and Here's Why \u2022 The Evening Edit March 20 2019 by Patrice Lee Onwuka\nThreats to Drag Out Investigation Despite No Evidence Of Collusion \u2022 The Evening Edit March 13 2019 by Patrice Lee Onwuka\nA Legislative Unicorn: Paid Family Leave Without New Taxes or Entitlements March 12 2019 by Patrice Lee Onwuka\nCongress considers parental leave programs March 12 2019 by Patrice Lee Onwuka\nThe Ivanka Trump effect: Why Republicans are now championing paid family leave, and what their new proposal looks like March 12 2019 by Patrice Lee Onwuka\nJoni Ernst, Mike Lee roll out conservative paid parental leave idea March 12 2019 by Patrice Lee Onwuka\nRaising the Topic of Paid Parental Leave March 12 2019 by Patrice Lee Onwuka\nRepublicans Propose Plan to Let Parents Use Social Security for Paid Family Leave After Having a Child March 12 2019 by Patrice Lee Onwuka\nGOP senators unveil paid parental leave proposal March 12 2019 by Patrice Lee Onwuka\nSupport Choices That Women Make When Looking at Gender Pay Gap \u2022 Making Money with Charles Payne March 8 2019 by Patrice Lee Onwuka\nTrump Previews 2020 in CPAC Speech \u2022 Fox and Friends First March 4 2019 by Patrice Lee Onwuka\nConsumer Confidence Bouncing Back as Trump Looks Ahead to 2020 \u2022 Coast to Coast March 4 2019 by Patrice Lee Onwuka\nPelosi Should Second Guess New Member's Calls for Trump Impeachment \u2022\u00a0The Evening Edit February 27 2019 by Patrice Lee Onwuka\nAOC Lives in Luxury Housing in D.C. Despite Her Views on Affordable Housing \u2022 Fox and Friends First February 21 2019 by Patrice Lee Onwuka\nBernie Attacks Someone Who Achieved the American Dream \u2022 Making Money with Charles Payne February 20 2019 by Patrice Lee Onwuka\nBernie Sanders Continues to Misrepresent the Reality of Socialism \u2022 The Evening Edit February 19 2019 by Patrice Lee Onwuka\nMeet the Women Who Protested the Women\u2019s March January 21 2019 by Patrice Lee Onwuka\nTrump Putting Border Security on the Table with DACA Extension \u2022 America's News Headquarters January 19 2019 by Patrice Lee Onwuka\nWomen's March is Not Inclusive of All Ideas \u2022 Fox & Friends January 19 2019 by Patrice Lee Onwuka\nUSPS Relies on Amazon: What this means for consumers \u2022 Your World with Neil Cavuto December 5 2018 by Patrice Lee Onwuka\nHow Will Markets React to a Potential Trade Deal with China? \u2022 Coast to Coast November 29 2018 by Patrice Lee Onwuka\nStreisand accuses Trump of misogyny while liberals turn on their own November 28 2018 by Patrice Lee Onwuka\nWhat Pete Davidson\u2019s SNL Apology Could Teach Us About Civility \u2022 Bold Politics November 16 2018 by Patrice Lee Onwuka\nIs Amazon Guilty of Taking Part In Corporate Welfare? \u2022 The Evening Edit November 14 2018 by Patrice Lee Onwuka\n'Economic damage' predicted after city levies big tax on hi-tech firms November 13 2018 by Patrice Lee Onwuka\nMarkets React to Midterm Election Results \u2022 Coast to Coast November 12 2018 by Patrice Lee Onwuka\nVolatile Markets Will Only Marginally Influence Midterms \u2022 The Evening Edit October 26 2018 by Patrice Lee Onwuka\nPETA Claims Milk is Racist \u2022 Trish Regan Primetime October 23 2018 by Patrice Lee Onwuka\nHow Will Booming Economy Be Affected by The Fed? \u2022 Coast to Coast October 17 2018 by Patrice Lee Onwuka\nKanye Brings Attention to Real Economic Issues \u2022 Coast to Coast October 11 2018 by Patrice Lee Onwuka\nWill Kavanaugh Confirmation Drive Women Away from the GOP? Not Likely October 8 2018 by Patrice Lee Onwuka\nAnti-Kavanaugh Harassment Tactics Overlooked by the Left \u2022 Coast to Coast October 4 2018 by Patrice Lee Onwuka\nStatement on #CancelKavanaugh Protest: Activists Use Dr. Ford to Exploit Emotion Over a Deeply Personal Issue October 4 2018 by Patrice Lee Onwuka\nRepublican Women Explain Why They Stand By Brett Kavanaugh More Than Ever October 3 2018 by Patrice Lee Onwuka\nWomen Want Kavanaugh Confirmed if FBI Finds No Corroborating Evidence \u2022 FBN am October 3 2018 by Patrice Lee Onwuka\nA Tale of Two Cities: A Snapshot of Washington D.C. On the Historic Kavanaugh-Ford Hearing Day September 27 2018 by Patrice Lee Onwuka\nFBI Investigation Was an Ongoing Distraction During Hearings \u2022 After the Bell September 27 2018 by Patrice Lee Onwuka\n'There's No Presumption of Innocence': Women Rally Both for and Against Kavanaugh September 27 2018 by Patrice Lee Onwuka\nDems Would Love for Trump to Fire Rosenstein \u2022 Fox & Friends First September 25 2018 by Patrice Lee Onwuka\nRepublicans Have Gone Out of Their Way to Try to Hear Blasey \u2022 Coast to Coast September 21 2018 by Patrice Lee Onwuka\nJacques: Sexism is a losing strategy September 14 2018 by Patrice Lee Onwuka\nGood Trade Relations with China is Good for the Economy \u2022 Coast to Coast September 12 2018 by Patrice Lee Onwuka\nWill Reforming SNAP Benefit Americans? \u2022 Bold Politics September 7 2018 by Patrice Lee Onwuka\nMotivations Behind Trying to Flip a Russian Oligarch \u2022 Fox & Friends First September 3 2018 by Patrice Lee Onwuka\nMore Than Ever, Millennials Are Taking Care of Their Aging Family Members \u2022 Bold Life August 24 2018 by Patrice Lee Onwuka\nTrump's Commitment to Diversity in the White House Continues After Omarosa \u2022 Andrew Wilkow Show August 22 2018 by Patrice Lee Onwuka\nStatement: IWF Applauds Roll Back of Clean Power Plan August 22 2018 by Patrice Lee Onwuka\nOmarosa\u2019s Antics Won't Undermine Trump\u2019s Progress for Black Americans \u2022 Afternoons with Cliff Kelley August 21 2018 by Patrice Lee Onwuka\nNYT Publishes Pure Speculation on Mueller Investigation \u2022 Fox & Friends First August 20 2018 by Patrice Lee Onwuka\nAmerica IS Great: a lesson from real Immigrants \u2022 Fox & Friends August 18 2018 by Patrice Lee Onwuka\nHarley-Davidson Staying in U.S. is Huge Victory \u2022 Coast to Coast August 17 2018 by Patrice Lee Onwuka\nIs there a racial gender pay gap? \u2022 Bold Politics August 10 2018 by Patrice Lee Onwuka\nOcasio-Cortez Forgets Socialism Didn't Work in South America \u2022 Your World with Neil Cavuto August 9 2018 by Patrice Lee Onwuka\nRubio-Wagner family leave bill would draw from Social Security August 7 2018 by Patrice Lee Onwuka\nExplaining the Economic Security for New Parents Act \u2022 KDKA August 3 2018 by Patrice Lee Onwuka\nRubio's New Paid Leave Bill is Great for Women \u2022 Fox & Friends First August 2 2018 by Patrice Lee Onwuka\nBernie Sanders \"Medicare for All\" plan would be a disaster \u2022 Your World with Neil Cavuto July 31 2018 by Patrice Lee Onwuka\nHow will new GDP numbers influence the midterm elections \u2022 Coast to Coast July 30 2018 by Patrice Lee Onwuka\nNew Bill Aims to Extend Trump Tax Cuts Beyond 2025 \u2022 Fox & Friends First July 25 2018 by Patrice Lee Onwuka\nSessions prepares students for diverse opinions at college \u2022 Making Money with Charles Payne July 24 2018 by Patrice Lee Onwuka\nStudent loan debt saddling non-saving Millennials July 24 2018 by Patrice Lee Onwuka\nOnly 4.9 Percent of People Have Two Jobs \u2022 Fox & Friends First July 19 2018 by Patrice Lee Onwuka\nCompanies thinking about intellectual theft when dealing with China \u2022 Coast to Coast July 18 2018 by Patrice Lee Onwuka\nCyber Security Needs to be Taken Seriously During Trump-Putin Summit \u2022 FBN am July 16 2018 by Patrice Lee Onwuka\nWhat Do You Think Of Seattle's Plastic Straw Ban? \u2022 Bold Life July 6 2018 by Patrice Lee Onwuka\nHopeful About North Korean Cooperation \u2022 FBN am July 6 2018 by Patrice Lee Onwuka\n#133 Reforms needed for children's television July 2 2018 by Patrice Lee Onwuka\nHow will Supreme Court nominations effect the primaries? \u2022 Coast to Coast June 29 2018 by Patrice Lee Onwuka\nTrump is no excuse for intimidation of women June 28 2018 by Patrice Lee Onwuka\nProtesters Want Media Attention, Not Policy Change \u2022 Your World with Neil Cavuto June 22 2018 by Patrice Lee Onwuka\nImmigration bills will be hard to pass, but we can't ignore this issue \u2022 FBN AM June 20 2018 by Patrice Lee Onwuka\nWe should be concerned that foreign agents had access to Hillary's emails \u2022 Fox & Friends First June 15 2018 by Patrice Lee Onwuka\nFree Trade Without Free Riders \u2022 Coast to Coast June 7 2018 by Patrice Lee Onwuka\nDemocrats Playing Politics in Primaries \u2022 Making Money with Charles Payne June 4 2018 by Patrice Lee Onwuka\nIWF Statement on New Steel and Aluminum Tariffs May 31 2018 by Patrice Lee Onwuka\nTrump had good reason to pull out of meeting with North Korea \u2022 Making Money with Charles Payne May 24 2018 by Patrice Lee Onwuka\nShould Medical Marijuana Be Allowed in Classrooms? \u2022 Fox & Friends First May 21 2018 by Patrice Lee Onwuka\nMedia Bias After They Refuse to Accept Trump's Corrected Statement \u2022 Your World with Neil Cavuto May 17 2018 by Patrice Lee Onwuka\nLower Birth Rates: Student Debt, Societal Change, Technological Advances \u2022 The Intelligence Report May 17 2018 by Patrice Lee Onwuka\nChina's Relationship with North Korea Could Make All The Difference \u2022 FBN am May 16 2018 by Patrice Lee Onwuka\n#127 Questioning the Status Quo: Black voters don't owe their vote to either party May 14 2018 by Patrice Lee Onwuka\nWere there problems with Cohen's payments? \u2022 Coast to Coast May 10 2018 by Patrice Lee Onwuka\nRemoving \"In God We Trust\" to be Politically Correct \u2022 Fox & Friends First May 7 2018 by Patrice Lee Onwuka\nProof Mueller is Fishing for Evidence \u2022 FBN am May 7 2018 by Patrice Lee Onwuka\nWe Want to Ensure a Legitimate and Lawful Immigration Process \u2022 FBN am April 26 2018 by Patrice Lee Onwuka\nNancy Pelosi Chokes on \"Crumbs\" Comment Because It Just Isn't True \u2022 Coast to Coast April 25 2018 by Patrice Lee Onwuka\n#123 Education Savings Accounts for Military Families Can Make All the Difference April 23 2018 by Patrice Lee Onwuka\nWhy are so many young adults choosing to stay at home? \u2022 BoldTV April 20 2018 by Patrice Lee Onwuka\nGov. Brown is Choosing Sanctuary over Safety \u2022 Risk & Reward April 19 2018 by Patrice Lee Onwuka\nComey has made enemies on both sides of the aisle \u2022 Risk & Reward April 19 2018 by Patrice Lee Onwuka\nBias training won't do much at Starbucks \u2022 Intelligence Report with Trish Reagan April 19 2018 by Patrice Lee Onwuka\nTax Cuts Will Play Major Role in Midterms \u2022 Coast to Coast April 5 2018 by Patrice Lee Onwuka\nNew Survey: What would Americans do for a 10% raise? \u2022 Fox & Friends First April 5 2018 by Patrice Lee Onwuka\nTrump to rescind spending with a maneuver Clinton used nearly 20 years ago \u2022 FBN:am April 5 2018 by Patrice Lee Onwuka\nCitizenship Question MUST be Answered Truthfully \u2022 Cavuto Live April 2 2018 by Patrice Lee Onwuka\n#118 New Amendment Provides Protection Against the Discrimination of Pregnant Women April 2 2018 by Patrice Lee Onwuka\nOn the New Spending Bill: Entitlements Drive the National Debt \u2022 Risk & Reward March 23 2018 by Patrice Lee Onwuka\nRegulations will Hinder Innovation on Social Media \u2022 Risk & Reward March 23 2018 by Patrice Lee Onwuka\nWhite House Millennials Conference to Address Economy, Free Speech, and Opioids March 18 2018 by Patrice Lee Onwuka\nIn the #MeToo era, young conservative women look for their spot March 16 2018 by Patrice Lee Onwuka\nShake Ups in White House Happening for a Reason \u2022 FBN AM March 15 2018 by Patrice Lee Onwuka\nDems Will Not Continue to Embrace Middle America \u2022 Coast to Coast March 14 2018 by Patrice Lee Onwuka\nWe Don't Want to Abandon Free Trade \u2022 Cavuto Live March 10 2018 by Patrice Lee Onwuka\nGOP takes on paid family leave March 10 2018 by Patrice Lee Onwuka\nIssues in California are a Classic Case of Federalism \u2022 Risk & Reward March 9 2018 by Patrice Lee Onwuka\nA Trade War is Going to Undermine Current Economic Progress \u2022 Coast to Coast March 5 2018 by Patrice Lee Onwuka\n#113 What You Don't Know About Immigrants: A real immigrant's story March 2 2018 by Patrice Lee Onwuka\nA conservative solution to paid leave \u2022 The National Discourse Podcast February 27 2018 by Patrice Lee Onwuka\nCNN Pushes Agenda at Townhall \u2022 Risk & Reward February 23 2018 by Patrice Lee Onwuka\nAmericans Recognize the 2nd Amendment is Important to Protect \u2022 Coast to Coast February 20 2018 by Patrice Lee Onwuka\nRepublicans Produced Plan, Waiting for Action Across the Aisle \u2022 FBN AM February 13 2018 by Patrice Lee Onwuka\n#MeToo and Working Women \u2022 PBS To The Contrary February 10 2018 by Patrice Lee Onwuka\nFundamentals of Our Economy Are Strong \u2022 Coast to Coast February 8 2018 by Patrice Lee Onwuka\nRepublicans Tired of Partisan Bickering \u2022 Coast to Coast January 31 2018 by Patrice Lee Onwuka\nUnions Losing Battle Against Inevitable Future \u2022 The Intelligence Report January 25 2018 by Patrice Lee Onwuka\nDid Schumer Cave on Gov Shut Down? \u2022 FBN am January 23 2018 by Patrice Lee Onwuka\nDems \u2018Big Risk\u2019 Not Worth It \u2022 Your World January 22 2018 by Patrice Lee Onwuka\nMedia Attacks Trump's Doctor \u2022 Risk & Reward January 19 2018 by Patrice Lee Onwuka\nThe Left Criticizing Tax Plan and Immigration \u2022 Risk & Reward January 12 2018 by Patrice Lee Onwuka\nConcerns on Trade with China \u2022 Coast to Coast January 10 2018 by Patrice Lee Onwuka\nWeighing in on Trump vs. Bannon \u2022 FBN AM January 5 2018 by Patrice Lee Onwuka\nU.N. Funding Cut Good for Women \u2022 Forbes on Fox December 29 2017 by Patrice Lee Onwuka\nNext Moves for the GOP \u2022 Coast to Coast December 26 2017 by Patrice Lee Onwuka\nSaving Money After the Holidays \u2022 Fox & Friends First December 26 2017 by Patrice Lee Onwuka\nSenate Passes Tax Bill \u2022 FBN AM December 20 2017 by Patrice Lee Onwuka\nAre Black Women Leaving The Democratic Party to Become Independents? December 19 2017 by Patrice Lee Onwuka\nFinal GOP Tax Bill \u2022 Risk & Reward December 15 2017 by Patrice Lee Onwuka\nStatement: Unnecessary internet regulation no more, FCC right to overturn net neutrality rules December 14 2017 by Patrice Lee Onwuka\nCalls for Conyers to Resign \u2022 Risk & Reward Pt. 2 December 1 2017 by Patrice Lee Onwuka\nSteinle's Killer Acquitted \u2022 Risk & Reward Pt. 1 December 1 2017 by Patrice Lee Onwuka\nFCC Net Neutrality Vote Looms: How could things change? \u2022 The Big Picture December 1 2017 by Patrice Lee Onwuka\nTime for Transparency: Taxpayer dollars used to protect Congressional misbehavior \u2022 Coast to Coast November 30 2017 by Patrice Lee Onwuka\nMore Bang for Your Buck \u2022 Fox & Friends First November 30 2017 by Patrice Lee Onwuka\nAre some Dems willing to sacrifice protections for women to hold onto power? \u2022 Intelligence Report November 28 2017 by Patrice Lee Onwuka\nFederal Judge blocks Trump's EO on Sanctuary Cities: Is Trump at a disadvantage? \u2022 Bulls & Bears November 25 2017 by Patrice Lee Onwuka\nHas NFL commissioner lost control of the hemorrhaging league? \u2022 Bulls & Bears November 25 2017 by Patrice Lee Onwuka\nBlack Friday protests dwindle exposing this uncomfortable truth \u2022 Bulls & Bears November 25 2017 by Patrice Lee Onwuka\nSexual harassment is wrong no matter when it occurred and by whom \u2022 Varney & Co. November 24 2017 by Patrice Lee Onwuka\nBernie Sanders Blatantly Gives Bill Clinton, Al Franken a Pass on Sexual Harassment November 20 2017 by Patrice Lee Onwuka\nThere's Nothing Glamorous about \u201cAuntie\" Maxine Waters & Her \u201cImpeach Him!\u201d Chant November 17 2017 by Patrice Lee Onwuka\nTwitter cracking down on Verified accounts \u2022 Fox & Friends First November 17 2017 by Patrice Lee Onwuka\n'Saturday Night Live' says DNC is old and out of touch \u2022 Risk & Reward November 13 2017 by Patrice Lee Onwuka\nTax reform push: Will there be compromise? \u2022 FBN:AM November 13 2017 by Patrice Lee Onwuka\nAre tech companies truly neutral platforms or do they reflect bias? \u2022 Fox & Friends First November 3 2017 by Patrice Lee Onwuka\nStatement: GOP Tax Bill is a Win for Women November 2 2017 by Patrice Lee Onwuka\nCNBC Slams DNC Chair Tom Perez, 'Get Off The Talking Points' On GOP Tax Plan November 2 2017 by Patrice Lee Onwuka\nWe cannot choose our history but we can choose not to repeat it \u2022 Risk & Reward October 31 2017 by Patrice Lee Onwuka\nWe're seeing the waning days of the NFL protests \u2022 Risk & Reward October 31 2017 by Patrice Lee Onwuka\nThis tax reform plan will help all workers \u2022 FBN:AM October 30 2017 by Patrice Lee Onwuka\nWest Wing Reads for 10/30/17 October 30 2017 by Patrice Lee Onwuka\nWill any democrats get on board with tax reform? \u2022 FBN AM October 26 2017 by Patrice Lee Onwuka\nWhat role is Bannon playing outside of the White House? \u2022 MSNBC Live October 23 2017 by Patrice Lee Onwuka\nPodcast #98 Why sexual assault statistics on college campuses are misleading October 19 2017 by Patrice Lee Onwuka\nWill new Trump/McConnell bromance last to get agenda through? \u2022 FBN AM October 17 2017 by Patrice Lee Onwuka\nHillary caring about \"women's issues\" a sham? \u2022 The Intelligence Report October 16 2017 by Patrice Lee Onwuka\nMore sexual harrassment allegations come out in Hollywood \u2022 Risk & Reward October 13 2017 by Patrice Lee Onwuka\nWait\u2013Are Black Women Leaving The Democratic Party? October 11 2017 by Patrice Lee Onwuka\nEminem bashes Trump in new music video, but does it matter? \u2022 Coast To Coast October 11 2017 by Patrice Lee Onwuka\nStatement: Abandoning The Clean Power Plan: A Destructive Regulatory Package October 10 2017 by Patrice Lee Onwuka\nThese four GOP Senators won't vote for Graham-Cassidy \u2022 FBN AM September 26 2017 by Patrice Lee Onwuka\nThese celebrities are joining in on the NFL anthem protests... \u2022 Risk & Reward September 26 2017 by Patrice Lee Onwuka\nThis sport is standing up for the National Anthem... \u2022 Risk & Reward September 25 2017 by Patrice Lee Onwuka\nDoes Graham-Cassidy healthcare bill have a chance? \u2022 FBN AM September 21 2017 by Patrice Lee Onwuka\nEmmy's and NFL in ratings slide due to politics \u2022 Risk & Reward September 19 2017 by Patrice Lee Onwuka\nA new left-wing strategy against Trump September 18 2017 by Patrice Lee Onwuka\nCan Trump pull bi-partisan support for tax reform? \u2022 FBN AM September 13 2017 by Patrice Lee Onwuka\nIs there an opportunity for bi-partisanship with recent devastations? \u2022 Coast To Coast September 11 2017 by Patrice Lee Onwuka\nWhite House shake up doesn't mean the agenda stops \u2022 Risk & Reward August 18 2017 by Patrice Lee Onwuka\nLet's consider moving Confederate monuments to museums \u2022 FBN AM August 17 2017 by Patrice Lee Onwuka\nTrump holds Congress' feet to the fire on health care \u2022 Coast To Coast August 11 2017 by Patrice Lee Onwuka\nWhy diversity quotas may not be a good thing for companies \u2022 Intelligence Report August 7 2017 by Patrice Lee Onwuka\nLeft trying to label 'Death Wish' alt-right \u2022 Risk & Reward August 7 2017 by Patrice Lee Onwuka\nPodcast #90: How the Internet is Giving Women Back Their Time August 2 2017 by Patrice Lee Onwuka\nWill those who voted against repeal face tough constituent questions? \u2022 FBN AM July 28 2017 by Patrice Lee Onwuka\nBrave move by John McCain with healthcare vote? \u2022 FBN AM July 28 2017 by Patrice Lee Onwuka\nJeff Sessions is yet another distraction from Trump agenda \u2022 Your World July 27 2017 by Patrice Lee Onwuka\nIs getting a better education for your children racist? \u2022 Risk & Reward July 24 2017 by Patrice Lee Onwuka\nWhy democrats have failed to come up with plan to make Obamacare better \u2022 FBN AM July 20 2017 by Patrice Lee Onwuka\nShould Congress forget healthcare repeal and move onto tax reform? \u2022 The Intelligence Report July 18 2017 by Patrice Lee Onwuka\nWill Congress work through August recess? \u2022 Coast To Coast July 11 2017 by Patrice Lee Onwuka\nWas it wrong for Ivanka to sit in at the G20? \u2022 Your World July 10 2017 by Patrice Lee Onwuka\nWould media act differently if it was Chelsea Clinton at the G20? \u2022 Risk & Reward July 10 2017 by Patrice Lee Onwuka\nIs the Trump administration good for the economy? \u2022 Fox 5 DC July 7 2017 by Patrice Lee Onwuka\nIs Trump's agenda back on track? \u2022 FBN AM June 29 2017 by Patrice Lee Onwuka\nWill the healthcare bill pass before July 4 recess? \u2022 FBN AM June 29 2017 by Patrice Lee Onwuka\nPeople would be better served if we had a real healthcare market place \u2022 Risk & Reward June 27 2017 by Patrice Lee Onwuka\nBlack women work hard but can\u2019t get ahead. This expert says conservatives have the solution. June 19 2017 by Patrice Lee Onwuka\nIs Trump hurting himself when he tweets? \u2022 Coast To Coast June 7 2017 by Patrice Lee Onwuka\nThis is why you shouldn't jump on Larry Hogan for vetoing paid sick leave bill \u2022 Larry O'Connor Show May 26 2017 by Patrice Lee Onwuka\nHow much damage has been done to the Trump administration by the media? \u2022 FBN AM May 24 2017 by Patrice Lee Onwuka\nMiss USA Slammed For Saying Healthcare Is A Privilege Rather Than A Right \u2022 Marc Cox May 16 2017 by Patrice Lee Onwuka\nOne Black Conservative Made It On Liberal List Of Beautiful Black Women April 25 2017 by Patrice Lee Onwuka\nEssence\u2019s \u201c100 Woke Women\u201d List: Where Are All The Conservatives? April 20 2017 by Patrice Lee Onwuka\nDo Voters Actually Care About Trump's Tax Returns? \u2022 Coast To Coast April 17 2017 by Patrice Lee Onwuka\nPBS teaching ISIS sympathy \u2022 Risk & Reward April 14 2017 by Patrice Lee Onwuka\nIvanka Trump Accepts Role as White House Advisor \u2022 Coast to Coast March 30 2017 by Patrice Lee Onwuka\nRyan Pulls Healthcare Bill Before Vote \u2022 MSNBC Live March 28 2017 by Patrice Lee Onwuka\nDebate over Cost of Healthcare Overhaul ahead of CBO Score \u2022 Coast to Coast March 14 2017 by Patrice Lee Onwuka\nA Day without a Woman \u2022 Making Money March 9 2017 by Patrice Lee Onwuka\nAny Momentum Left from President Trump's Address? \u2022 MSNBC Live March 5 2017 by Patrice Lee Onwuka\nPolitical double-speak: Who's line is it? \u2022 Risk & Reward February 23 2017 by Patrice Lee Onwuka\nCNN may suffer from \u2018PESD\u2019 post-election stress disorder \u2022 Risk & Reward February 23 2017 by Patrice Lee Onwuka\nWWE's Linda McMahon Confirmed To Head The Small Business Administration \u2022 Steve Gruber February 17 2017 by Patrice Lee Onwuka\nVice President Pence addressing right-to-life marchers in D.C. \u2022 Coast to Coast January 27 2017 by Patrice Lee Onwuka\nRaising Minimum Wage 'Not Worth It' January 23 2017 by Patrice Lee Onwuka\nHave Race Relations Improved Under Obama Presidency? \u2022 Tipping Point January 14 2017 by Patrice Lee Onwuka\nIf Republicans rush to repeal ObamaCare without a strategy to provide immediate relief to its victims, the ongoing collapse of insurance markets will continue \u2022 Coast To Coast January 9 2017 by Patrice Lee Onwuka\nSanctions A Political Move By Obama? \u2022 Risk & Reward December 30 2016 by Patrice Lee Onwuka\nBe Careful Jumping On Consensus Bandwagon \u2022 Coast To Coast December 23 2016 by Patrice Lee Onwuka\nIrony: Conway Facing Work-Life-Balance Criticism From Left After Accepting Position \u2022 Your World December 22 2016 by Patrice Lee Onwuka\nAre Election Results Everyone's Fault But Democrats? \u2022 Coast To Coast December 9 2016 by Patrice Lee Onwuka\nControversy Over Trump's Stocks-But Do Americans Care? \u2022 Your World December 7 2016 by Patrice Lee Onwuka\nOprah's Message Of Hope Triggers The Left \u2022 Garrison November 17 2016 by Patrice Lee Onwuka\nIs There A Double Standard With Protests Around Country? \u2022 Coast To Coast November 14 2016 by Patrice Lee Onwuka\nAirBnB's Growing List Of Enemies + Halloween's Cultural Appropriation Police \u2022 Garrison October 20 2016 by Patrice Lee Onwuka\nForging a Generation of Opportunity October 20 2016 by Patrice Lee Onwuka\nPolitical Correctness Plagues Halloween \u2022 Tipping Point October 18 2016 by Patrice Lee Onwuka\nDemocrats Relationship With Media In Question After Wikileaks Dump \u2022 Coast To Coast October 14 2016 by Patrice Lee Onwuka\nIs IQ or EQ More Important In The Presidential Debates? \u2022 Your World September 20 2016 by Patrice Lee Onwuka\nAre Self-Driving Cars A Good Idea + AirBnB Doesn't Drive Rent Higher \u2022 Garrison September 15 2016 by Patrice Lee Onwuka\nPodcast #51 College Students: Don't Major In Debt! September 15 2016 by Patrice Lee Onwuka\nMylan's EpiPen Monopoly Created By Big Government \u2022 Tipping Point September 13 2016 by Patrice Lee Onwuka\nWill Hillary Send More Troops Back Into Middle East? \u2022 Risk & Reward September 9 2016 by Patrice Lee Onwuka\nAd Spoofs 9/11 Attack \u2022 Risk & Reward September 9 2016 by Patrice Lee Onwuka\nMylan & Big Government On Hook For Rising EpiPen Costs \u2022 Your World September 2 2016 by Patrice Lee Onwuka\nDid Kaepernick Right His Protest By Kneeling During Anthem? \u2022 Risk & Reward September 2 2016 by Patrice Lee Onwuka\nGeorgetown Offers Preferential Admission Status To Descendants Of Slaves \u2022 Risk & Reward September 1 2016 by Patrice Lee Onwuka\nLabor Force Participation Rate Reaches 38 Year Low \u2022 Tipping Point August 5 2016 by Patrice Lee Onwuka\nCan Trump Recover From His Comments About Slain Soldier's Parents? \u2022 Risk & Reward August 1 2016 by Patrice Lee Onwuka\nCan Trump Close The Gap With Minorities? \u2022 Tipping Point June 10 2016 by Patrice Lee Onwuka\nShould Salaries Be Transparent? \u2022 PBS Point Taken May 3 2016 by Patrice Lee Onwuka\nStaying Too Long? Life Expectancy; Girls & Sex \u2022 To The Contrary April 23 2016 by Patrice Lee Onwuka\nEqual Pay Day & Women in the Workplace \u2022 PBS To The Contrary April 16 2016 by Patrice Lee Onwuka\nThe Liberty Vote & How Government Hurts Opportunity \u2022\u00a0Stossel March 12 2016 by Patrice Lee Onwuka\nGoods Marketed To Women Cost More Than Identical Items For Men \u2022 After The Bell December 23 2015 by Patrice Lee Onwuka\nFeds know best ... even with cosmetics December 4 2015 by Patrice Lee Onwuka\nSocial Security disability fund gets dealt another blow November 17 2015 by Patrice Lee Onwuka\nGravity CEO Defends $70,000 Minimum Salary, Says it was the Right Decision August 11 2015 by Patrice Lee Onwuka Baltimore Mom Toya Graham, Carly Fiorina, Salma Hayek \u2022 To The Contrary May 2 2015 by Patrice Lee Onwuka When did a 'blue-collar' job become a bad thing? 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April 5 2018 by Patrice Lee Onwuka\nTakeaways: Why Requiring Work & Community Engagement is Good for Medicaid and the Poor January 29 2018 by Patrice Lee Onwuka\nTakeaways: How Tax Reform Will Help Americans December 18 2017 by Patrice Lee Onwuka\nFact Sheet: Tax Reform is a Women's Issue September 15 2017 by Patrice Lee Onwuka\nPolicy Focus: Your Ideas, Your Rights - Intellectual Property in the 21st Century June 25 2017 by Patrice Lee Onwuka", "date": "2019-12-13 23:04:26", "meta": {"domain": "ihuq.pw", "url": "http://nbkx.ihuq.pw/properties-of-matrix-multiplication-proof.html", "openwebmath_score": 0.199847012758255, "openwebmath_perplexity": 6607.073204715545, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795114181105, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.857514230470787}}
{"url": "https://questioncove.com/updates/4fc1326be4b0964abc832226", "text": "Mathematics\nOpenStudy (anonymous):\n\nFind the angle between the vectors u and v if u = (1, 2) and v = (- 4, - 2)\n\njimthompson5910 (jim_thompson5910):\n\nUse the formula $\\Large \\theta = \\arccos\\left(\\frac{u\\cdot v}{|u||v|}\\right)$ where u and v are vectors.\n\nOpenStudy (anonymous):\n\nwhich number in my points do i use for each... for example which number in (1,2) would I use for which u\n\njimthompson5910 (jim_thompson5910):\n\nIn the numerator of that fraction, you're computing the dot product between u and v\n\njimthompson5910 (jim_thompson5910):\n\nSo what is the dot product between u = (1, 2) and v = (- 4, - 2)?\n\nOpenStudy (anonymous):\n\n-8\n\njimthompson5910 (jim_thompson5910):\n\ngood\n\njimthompson5910 (jim_thompson5910):\n\nnow you need to find |u| and |v|\n\nOpenStudy (anonymous):\n\nhow do I do that?\n\njimthompson5910 (jim_thompson5910):\n\nuse the formula |x| = sqrt(a^2+b^2) where the vector x is x = (a,b)\n\nOpenStudy (anonymous):\n\nso it is sqrt(1^2+2^2) for u's and then sqrt(-4^2+-2^2) for v's?\n\njimthompson5910 (jim_thompson5910):\n\nyes, so what do you get?\n\nOpenStudy (anonymous):\n\ni think u's is sqrt(5) and v's is 2sqrt(5)\n\njimthompson5910 (jim_thompson5910):\n\nYou got it, nice work So what is |u||v|?\n\nOpenStudy (anonymous):\n\n10... so then its arccos(-8/10)?\n\njimthompson5910 (jim_thompson5910):\n\nyes\n\nOpenStudy (anonymous):\n\nwhich turns out to be 143.13degrees\n\njimthompson5910 (jim_thompson5910):\n\nyou nailed it\n\nOpenStudy (anonymous):\n\nYou keep helping me sooo much. Thanks, I am taking a Pre-Calculus course by myself and sometimes it can be difficult to figure out all of the questions by myself :).\n\njimthompson5910 (jim_thompson5910):\n\nyou're very welcome, that must be pretty crazy lol\n\nOpenStudy (anonymous):\n\nyep... I'm almost done though (thankfully haha)\n\njimthompson5910 (jim_thompson5910):\n\nwell that's good\n\nOpenStudy (anonymous):\n\nis there a difference between a u surrounded by two of the straight lines as opposed to the just one in the question you just helped me with?\n\njimthompson5910 (jim_thompson5910):\n\nWell technically it should have been $\\Large \\|u\\|$ since the two vertical bars denote the norm or magnitude of a vector. But the absolute value bars say the same thing (they're just used in a more general sense). So either work in my opinion.\n\nOpenStudy (anonymous):\n\noh, ok... thanks for clarifying. I am working on another type of question like that and wasn's sure if there was a difference.\n\njimthompson5910 (jim_thompson5910):\n\nthere's not much of a difference really", "date": "2020-05-28 11:02:19", "meta": {"domain": "questioncove.com", "url": "https://questioncove.com/updates/4fc1326be4b0964abc832226", "openwebmath_score": 0.8564152121543884, "openwebmath_perplexity": 2042.5005409672297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795079712151, "lm_q2_score": 0.8688267677469952, "lm_q1q2_score": 0.8575142157431506}}
{"url": "http://math.stackexchange.com/questions/237073/playing-roulette", "text": "# Playing roulette\n\nSuppose we have a roulette wheel with $38$ total slots ($18$ black/red, $2$ neither). The ball lands in one slot selected uniformly at random, independent of all previous spins. An $\\$x$bet on \"black\" pays$\\$2x$ if the ball lands on black, and $\\$0$otherwise. If you bet$1 on black for 100 consecutive spins, how much money will you end up with in expectation?\n\nYou walk up to the Roulette table with $\\$15$and intend to walk away with$\\$16$ using the following betting strategy. You will first bet $\\$1$on black. If you win you have$\\$16$ and walk away, otherwise you have $\\$14$and bet$\\$2$ on black on the next spin. If you win that bet you walk away, otherwise you bet $\\$4$on the next spin. If you win that bet you walk away, otherwise you bet$\\$8$ on the next spin and walk away, either with $\\$16$if you win or$\\$0$ if you lose.\n\nWhat's the probability you will walk away with $\\$16$? How much money are you expected to walk away with? My biggest problem with these kinds of questions is figuring out how to choose the random variable to calculate the expected values. I understand the formula for calculating the expected value, but translating a problem into those terms has been giving me a hard time. - ## 2 Answers Let's first look at the first question: If you bet$1 on black for 100 consecutive spins, how much money will you end up with in expectation?\n\nSo you want to know what your final return will be, at the end of 100 spins. Call this $R$. That is just giving it a name, but what is your final return? You can see that it is the sum of the returns from each bet. So let the return on the $i$th bet be $R_i$, then note that $R = R_1 + R_2 + \\dots + R_{100}$. So the expected return is $E[R] = E[R_1] + E[R_2] + \\dots + E[R_{100}]$ by linearity of expectation.\n\nSo to calculate $E[R]$, we'll be done if we calculate each $E[R_i]$. Let's try to calculate a particular $E[R_i]$. You bet $1$ dollar, and you get back $2$ if the ball lands on black, and $0$ if it doesn't. In other words, you gain $1$ dollar if it lands on black, and lose $1$ dollar if it doesn't. The probability of the former is $18/38$, and that of the latter is $20/38$. In other words, $R_i$ is $1$ with probability $18/38$, and $-1$ with probability $20/38$, so the expected value of $R_i$ is $E[R_i] = \\frac{18}{38}(-1) + \\frac{20}{38}(1) = \\frac{-2}{38}$. Now, as this is the same for each $R_i$, we have $E[R] = E[R_1] + E[R_2] + \\dots + E[R_{100}] = \\left(\\frac{-2}{38}\\right)100 \\approx -5.26$.\n\nFor the second question, let the amount you walk away with be $W$. Let $p = 18/38$, the probability that your bet on black succeeds. There are $5$ possible outcomes:\n\n\u2022 you win your first bet: probability $p$\n\u2022 you lose your first bet, and win your second: probability $(1-p)p$\n\u2022 you lose your first two bets, and win the third: probability $(1-p)^2p$\n\u2022 you lose your first three bets, and win the fourth: probability $(1-p)^3p$\n\u2022 you lose all four bets: probability $(1-p)^4$\n\nIn the first four outcomes, you walk away with $16$ dollars, so the probability of that happening (let's call it $q$) is $q = p + (1-p)p + (1-p)^2p + (1-p)^3p = 1 - (1-p)^4 = 1 - (20/38)^4 \\approx 0.92$.\n\n[More simply, you could think of it as just two outcomes: (a) that you win some bet, which has probability $q = 1 - (1-p)^4$, and (b) that you win no bet (lose all bets), which has probability $(1-p)^4$.]\n\nIn other words, $W$ is $16$ with probability $q$, and $0$ with probability $1-q$. So the expected amount of money you walk away with is therefore $E[W] = q(16) + (1-q)0 = (1-(1-p)^4)16 \\approx 14.77$.\n\n[Aside: Note that this is less than the $15$ you came in with. This shows that you can't win in expectation even with your clever betting strategy; a consequence of the optional stopping theorem.]\n\n-\n\nProbability of getting a black on any particular spin is 18 in 38, ie $p = \\frac{18}{38} = \\frac{9}{19} = 0.474$. Let $X$ be the random variable 'number of blacks in 100 consecutive spins'. Then the expected value of X is $E(X) = 100p = \\frac{900}{19} = 47.37$, that is, you expect to win 47.37 times in 100 spins. Since you get $\\$$2 for each \\$$1 bet, you will therefore expect to walk away with$\\$2\\times 47.37 = \\$94.74$. For the second part of your question, the only way you can lose is if for four consecutive spins, you get non-blacks each time. So what is the probability of losing 4 times in a row? Well, for 1 spin, the probability of not getting black is$1-p = \\frac{20}{38} = \\frac{10}{19} = 0.526$so for 4 consecutive spins, the probability of not getting any black is$(\\frac{10}{19})^{4} = \\frac{10,000}{130,321} = 0.0767$. That is, you have about a 7.7% chance of losing with this betting strategy. That is, you have 92.3% chance of walking away with$\\16.\n\n-\nCan you elaborate on why the expected value is just $100p$? Particularly, how does the linearity of expectations apply here? If we let $X_i = 1$ if we land on black and $X_i = 0$ if we don't land on black, how can we calculate the expected value then? \u2013\u00a0 user1038665 Nov 14 '12 at 10:53\nAlso, for the second part, how would we calculate the \"amount of money you are expected to walk away with\"? \u2013\u00a0 user1038665 Nov 14 '12 at 10:55\nExpected value of a bet = wager x probability. \u2013\u00a0 theo Nov 14 '12 at 11:02\nIf you bet 1 dollar 100 times on black you will win about 47.37 times and thus leave the casino with 94.74 of your original 100 dollars. \u2013\u00a0 Hagen von Eitzen Nov 14 '12 at 11:03\nThe answer given is -\\$(2/38)*(100), so I believe this is incorrect. \u2013\u00a0 user1038665 Nov 14 '12 at 11:20", "date": "2014-11-26 02:11:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/237073/playing-roulette", "openwebmath_score": 0.8544569611549377, "openwebmath_perplexity": 335.1219784788124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989013056721729, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8575096989820084}}
{"url": "https://stats.stackexchange.com/questions/507040/how-to-create-a-continuous-distribution-on-a-b-with-textmean-textmo?noredirect=1", "text": "# How to create a continuous distribution on $[a, b]$ with $\\text{mean} = \\text{mode} = c$?\n\nI would like to sample from a distribution on a specific domain $$[a, b]$$ with mean $$c$$ where $$a < c < b$$. Ideally, I would also like the mode (i.e. the \"peak\") of the distribution to be $$c$$ as well.\n\nHow to derive this distribution?\n\nThe best I can do is a triangular distribution with my desired mean, but the peak is in the wrong place. I imagine the distribution I want is smooth and skewed.\n\nEDIT: If $$c$$ were at the midpoint of $$a$$ and $$b$$ then I think a truncated normal distribution is roughly what I'm looking for. So something similar where the tails go smoothly to zero at the endpoints of my interval when $$c$$ is not in the middle would be nice.\n\nEDIT 2: I'm looking to perturb some parameters in a simulation, so really anything reasonable should be fine.\n\n\u2022 There are infinitely many possible solutions, which enables you to apply stricter criteria to narrow it down. To that end, could you explain what use you intend to make of this family of distributions? BTW, \"skewed\" strongly suggests the peak should not coincide with the mean.\n\u2013\u00a0whuber\nJan 28 at 20:11\n\u2022 Good point. I did my best to clarify. Jan 28 at 20:37\n\u2022 There are still infinitely many solutions, but another way would be to generate a beta RV with parameters that give it the desired skew and unimodality, and then shift and scale the whole distribution so it lies in [a,b].\n\u2013\u00a0Noah\nJan 28 at 20:50\n\u2022 Thanks! That looks like it will do just fine. Jan 28 at 20:53\n\u2022 @Noah That looks a lot like an acceptable answer!\n\u2013\u00a0Dave\nJan 28 at 20:53\n\nI will describe every possible solution. This gives you maximal freedom to craft distributions that meet your needs.\n\nBasically, sketch any curve you like for the density function $$f$$ that meets your requirements. Separately scale the heights of the left and right halves of it (on either side of $$c$$) to make their masses balance, then scale the heights overall to make it a probability density.\n\nHere are some details.\n\nBecause the distribution is continuous, it has a density function $$f$$ with finite integral. By splitting $$f$$ at $$c$$ we can construct all such functions out of two separately chosen non-negative non-decreasing, not identically zero, integrable functions $$f_1$$ and $$f_2$$ defined on $$[0,1].$$ Here, for example, are two such functions:\n\nSet $$f$$ to agree affinely with $$f_1$$ on $$[a,c]$$ and with the reversal of $$f_2$$ on $$[c,b].$$ This means there are two positive numbers $$\\pi_i$$ for which\n\n$$f\\mid_{[a,c]}(x) = \\pi_1 f_1\\left(\\frac{x-a}{c-a}\\right);\\quad f\\mid_{[c,b]}(x) = \\pi_2 f_2\\left(\\frac{b-x}{b-c}\\right).$$\n\nThis construction guarantees $$c$$ is the unique mode. Moreover, if you want the tails to taper to zero, just choose $$f_i$$ that approach $$0$$ continuously at the origin.\n\nFor $$f$$ to be a probability density it must integrate to unity:\n\n$$1 = \\int_a^b f(x)\\,\\mathrm{d}x =\\pi_1(c-a) \\mu_1^{(0)} + \\pi_2(b-c) \\mu_2^{(0)}\\tag{*}$$\n\nwhere\n\n$$\\mu_i^{(k)} = \\int_0^1 x^k f_i(x)\\,\\mathrm{d}x.$$\n\nMoreover, $$c$$ is intended to be the mean, whence\n\n\\begin{aligned} c &= \\int_a^b x f(x)\\,\\mathrm{d}x \\\\ &= \\pi_1(c-a)((c-a)\\mu_1^{(1)} + a\\mu_1^{(0)}) + \\pi_2(b-c)(-(b-c)\\mu_2^{(1)} + b\\mu_2^{(0)}). \\end{aligned}\\tag{**}\n\n$$(*)$$ and $$(**)$$ establish a system of two linear equations in the $$\\pi_i.$$ You can check it has nonzero determinant and a unit positive solution $$(\\pi_1,\\pi_2).$$ (This checking comes down to the fact that after normalizing the $$f_i$$ to be probability densities, their means $$\\mu_i^{(1)}$$ must be in the interval $$[0,1].$$)\n\nThis figure plots the solution $$f$$ for the previous two functions where $$[a,b]=[-1,3]$$ and $$c=1/2:$$\n\nBy design, $$f$$ looks like $$f_1$$ to the left of $$c$$ and like the reversal of $$f_2$$ to the right of $$c.$$\n\nThe spike at the mode might bother you, but notice that it was never assumed or required that $$f$$ must be continuous. Most examples will be discontinuous. However, every distribution $$f$$ meeting your specifications can be constructed this way (by reversing the process: split $$f$$ into two halves, which obviously determine the $$f_i$$).\n\nTo demonstrate how general the $$f_i$$ can be, here is the same construction where the $$f_i$$ are (inverse) Cantor functions (as implemented in binary.to.ternary at https://stats.stackexchange.com/a/229561/919). These are not continuous anywhere.\n\nNB: $$f_1$$ is binary.to.ternary; $$f_2(x) = f_1(x^{2.28370312}).$$ This illustrates one way to eliminate the discontinuity at $$c:$$ $$f_2$$ was selected from the one-parameter family of functions $$f_1(x^p),$$ $$p \\gt 0,$$ and a value of $$p$$ was identified to make the left and right limits of $$f$$ at $$c$$ equal. This family \"pushes\" the mass of $$f_1$$ left and right by a controllable amount, thereby modifying the amount by which the right tail is scaled (vertically) in constructing $$f.$$\n\nFor those who would like to experiment, here is an R function to create $$f$$ out of the $$f_i$$ and code to create the figures. Three commands at the end check whether (1) $$f$$ is a pdf, (2) its mean is $$c,$$ and (3) its mode is $$c.$$\n\n#\n# Given numbers a. < b. < c. and non-negative, non-decreasing, not identically\n# zero functions f.1 and f.2 defined on [0,1], construct a density function f\n# on the interval [a., b.] with mean c. and unique mode c. that behaves like f.1\n# to the left of c. and like the reversal of f.2 to the right of c.\n#\n# ... are optional arguments passed along to integrate.\n#\ncreate <- function(a., b., c., f.1, f.2, ...) {\ncat(\"Create\\n\")\np.1 <- integrate(f.1, 0, 1, ...)$$value p.2 <- integrate(f.2, 0, 1, ...)$$value\nmu.1 <- integrate(function(u) u * f.1(u), 0, 1, ...)$$value mu.2 <- integrate(function(u) u * f.2(u), 0, 1, ...)$$value\n\nA <- matrix(c(p.1 * (c.-a.), p.2 * (b.-c.),\n(c.-a.) * ((c.-a.) * mu.1 + a.*p.1),\n(b.-c.) * (-(b.-c.) * mu.2 + b.*p.2)),\n2)\npi. <- solve(t(A), c(1, c.))\nfunction(x) {\nifelse(a. <= x & x <= c., 1, 0) * pi.[1] * f.1((x-a.)/(c.-a.)) +\nifelse(c. < x & x <= b., 1, 0) * pi.[2] * f.2((b.-x)/(b.-c.))\n}\n}\n#\n# Example.\n#\na. <- -1\nb. <- 3\nc. <- 1/2\nf.2 <- function(x) x^(2/3)\nf.1 <- function(x) exp(2 * x) - 1\nf <- create(a., b., c., f.1, f.2)\n#\n# Display f.1 and f.2.\n#\npar(mfrow=c(1,2))\ncurve(f.1(x), 0, 1, lwd=2, ylab=\"\", main=expression(paste(f[1], \": \", x^{2/3})))\ncurve(f.2(x), 0, 1, lwd=2, ylab=\"\", main=expression(paste(f[2], \": \", e^{2*x}-1)))\n#\n# Display f.\n#\nx <- c(seq(a., c., length.out=601), seq(c.+1e-12*(b.-a.), b., length.out=601))\ny <- f(x)\npar(mfrow=c(1,1))\nplot(c(x, b., a.), c(y, 0, 0), type=\"n\", xlab=\"x\", ylab=\"Density\", main=expression(f))\npolygon(c(x, b., a.), c(y, 0, 0), col=\"#f0f0f0\", border=NA)\ncurve(f(x), b., a., lwd=2, n=1201, add=TRUE)\n#\n# Checks.\n#\nintegrate(f, a., b.)                    # Should be 1\nintegrate(function(x) x * f(x), a., b.) # Should be c.\nx[which.max(y)]                         # Should be close to c.\n\n\u2022 Lovely! But a quibble: I do not think this captures distributions in which the 'left' or 'right' halves themselves contain nondifferentiable or discontinuous points\u2014I am especially thinking of densely nondifferentiable or densely discontinuous curves\u2014which are essentially 'undrawable' except by approximation. I am not actually convinced this is a defensible quibble, though. :) More just musing\u2026 Jan 29 at 6:28\n\u2022 @Alexis I cannot see any exceptions: I made no conditions on the $f_i$ concerning differentiability or even continuity. For instance, both $f_i$ could be Cantor functions. I will add an illustration.\n\u2013\u00a0whuber\nJan 29 at 15:13\n\u2022 I was thinking about Cantor functions! Quibble withdrawn (and thank you for entertaining my musing :). Jan 29 at 18:32\n\nAll you need to do is come up with a two-parameter family of distributions, express the mean and mode in terms of those parameters, and then solve for them both being $$c$$. Working off your triangle idea, we can have a pentagon instead (straight lines from $$a$$ to $$c$$ and from $$c$$ to $$b$$), giving us three parameters ($$y_1$$ at $$a$$, $$y_2$$ at $$b$$, and $$y_3$$ at $$c$$), but we don't really have three degrees of freedom. We have that the total probability mass from $$a$$ to $$c$$ is $$\\frac 1 2 (y_1+y_3)(c-a)$$, and the mass from $$c$$ to $$b$$ is $$\\frac 1 2 (y_2+y_3)(c-b)$$, so we need to have $$\\frac 1 2 (y_1+y_3)(c-a)+\\frac 1 2 (y_2+y_3)(c-b)=1$$. Then you just need to make sure that the mean is $$c$$ as well.\n\n\u2022 The problem with this approach is that it can fail even in the nicest circumstances (as I pointed out in a comment to the question). For instance, it does not work with the Beta family of distributions: there are no solutions unless the Beta parameters are equal.\n\u2013\u00a0whuber\nJan 29 at 15:18", "date": "2021-11-27 11:32:06", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/507040/how-to-create-a-continuous-distribution-on-a-b-with-textmean-textmo?noredirect=1", "openwebmath_score": 0.8477863669395447, "openwebmath_perplexity": 750.3634441688756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130596362788, "lm_q2_score": 0.8670357683915538, "lm_q1q2_score": 0.8575096981110226}}
{"url": "https://math.stackexchange.com/questions/1585105/power-series-at-x-0", "text": "# Power series at $x=0$\n\n$$\\sum_{n=0}^\u221en!x^{2n}$$\n\nThis power series has radius of convergence $R=0$ since the ratio test shows it diverges for all $x\u22600$.\n\nBut what if $x=0$? Then we have $\\sum_{n=0}^\\infty n! \\cdot 0^{2n}$\n\nSurely this series is undefined since the first partial sum is $0!\\cdot 0^{2\\cdot0}$ and $0^0$ is undefined, right?\n\nWell, my lecturer claims this series is convergent and is equal to 1.\n\nBut $\\sum_{n=1}^\\infty n!\\cdot 0^{2n}$ clearly converges to 0, so he is claiming that $0!\\cdot 0^0=1$?\n\nSo is he wrong? Or am I missing something here?\n\n\u2022 $0^0 = 1$ by convention. \u2013\u00a0user258700 Dec 22 '15 at 2:01\n\u2022 Really? I've never heard that. I've always thought it was just undefined. Why isn't $\\frac00$ similarly defined by convention? \u2013\u00a0Refnom95 Dec 22 '15 at 2:08\n\u2022 It's only when dealing with limits that $0^0$ is indeterminant. \u2013\u00a0Michael Burr Dec 22 '15 at 2:09\n\u2022 It's done mostly in the context of power series. It's much neater to write $\\sum_i \\text{bla}_ix^i$ than $\\text{bla}_0+\\sum_{i>0}...$. \u2013\u00a0YoTengoUnLCD Dec 22 '15 at 2:20\n\nThere are a lot of situations where it is convenient to define $$0^0=1$$ and of course $0!=1$, so $0!\\cdot0^0=1$.\nIn some situations, it's most appropriate to leave $0^0$ undefined. In particular, the function $x^y$ converges to different limits as $x\\to0$ and $y\\to0$ depending on which path is taken, so analysis does not yield a natural value for $0^0$ and it may not make sense to define $0^0=1$.\nHowever, when dealing with discrete exponents (natural numbers), it's usually better to define $0^0=1$, because this is consistent with combinatorial or set-theoretic interpretations of exponentiation. For instance, the set of functions from the empty set to the empty set has cardinality $1$, so $0^0=1$ by the set-theoretic definition. Also, using the repeated multiplication definition of exponentiation, an empty product is always $1$, even when the product contains only zeroes. (For more information, see this Wikipedia article.)\nIt is important to consider where the $0^0$ in the expression comes from. In your case, it's the first term of a power series. A power series' notion of exponentiation is closest to the repeated multiplication interpretation, so taking $0^0$ makes sense. Consider that a power series is just a generalization of a polynomial. How would you evaluate $f(x)=x^2+x+1$ at $x=0$? What if this is written $f(x)=x^2+x^1+x^0$?", "date": "2019-05-20 06:20:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1585105/power-series-at-x-0", "openwebmath_score": 0.9674197435379028, "openwebmath_perplexity": 177.5725999733486, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130563978902, "lm_q2_score": 0.8670357683915538, "lm_q1q2_score": 0.8575096953032239}}
{"url": "http://mathhelpforum.com/number-theory/124509-help-proof.html", "text": "# Math Help - Help with a proof.\n\n1. ## Help with a proof.\n\nHi, I'm stuck at proving the following question...\n\nProve that for all n>0,\n\n1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n\n\nI've tried all sorts of different ways of solving this, but to no avail.\n\nAny help is appreciated\n\n2. Originally Posted by seven.j\nHi, I'm stuck at proving the following question...\n\nProve that for all n>0,\n\n1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n\n\nI've tried all sorts of different ways of solving this, but to no avail.\n\nAny help is appreciated\n$\\sum_{k=1}^{n}\\frac{k}{2^k}$. Note that $\\sum_{k=1}\n^n x^k=\\frac{x^{n+1}-x}{x-1}$\n. Differentiating both sides and multiplying by $x$ gives $\\sum_{k=1}^{n}k\\cdot x^k=...$ figure the right side out.\n\n3. $\\sum\\limits_{j=1}^{n}{\\frac{j}{2^{j}}}=\\sum\\limits _{j=1}^{n}{\\sum\\limits_{k=1}^{j}{\\frac{1}{2^{j}}}} =\\sum\\limits_{k=1}^{n}{\\frac{1}{2^{k}}\\left( \\sum\\limits_{j=0}^{n-k}{2^{-j}} \\right)}=\\frac{1}{2^{n}}\\sum\\limits_{k=1}^{n}{\\fra c{1}{2^{k}}\\left( 2^{n+1}-2^{k} \\right)},$ you can do the rest, those are finite geometric sums.\n\n4. Hello, seven!\n\nHere's one way . . .\n\nProve that for all $n>0\\!:$\n\n. . $\\frac{1}{2}+ \\frac{2}{2^2} + \\frac{3}{2^3} +\\:\\hdots\\:+ \\frac{n}{2^n} \\; =\\; 2 - \\frac{n+2}{2^n}$\n$\\begin{array}{ccccc}\n\\text{We have:} &S &=& \\dfrac{1}{2} + \\dfrac{2}{2^2} + \\dfrac{3}{2^3} + \\dfrac{4}{2^4} + \\:\\hdots\\:+\\dfrac{n}{2^n}\\qquad\\qquad \\\\ \\\\[-3mm]\n\n\\text{Multiply by }\\dfrac{1}{2}\\!: & \\dfrac{1}{2}S &=& \\quad\\;\\; \\dfrac{1}{2^2} + \\dfrac{2}{2^3} + \\dfrac{3}{2^4} + \\hdots + \\dfrac{n-1}{2^n} + \\dfrac{n}{2^{n+1}}\\end{array}$\n\n. . . $\\text{Subtract: }\\quad \\frac{1}{2}S \\;=\\;\\underbrace{\\frac{1}{2} + \\frac{1}{2^2} + \\frac{1}{2^3} + \\frac{1}{2^4} + \\hdots + \\frac{1}{2^n}}_{\\text{geometric series}} - \\frac{n}{2^{n+2}}$ .[1]\n\nThe geometric series has the sum: . $\\frac{1}{2}\\cdot\\frac{1 - \\left(\\frac{1}{2}\\right)^n}{1-\\frac{1}{2}} \\;=\\;1 - \\frac{1}{2^n}$\n\nThen [1] becomes: . $\\frac{1}{2}S \\;=\\;\\left(1 - \\frac{1}{2^n}\\right) - \\frac{n}{2^{n+1}} \\;=\\;1 - \\frac{n+2}{2^{n+1}}$\n\nMultiply by 2: . $S \\;=\\;2 - \\frac{n+2}{2^n}$\n\n5. Originally Posted by seven.j\nHi, I'm stuck at proving the following question...\n\nProve that for all n>0,\n\n1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n\n\nI've tried all sorts of different ways of solving this, but to no avail.\n\nAny help is appreciated\nThis problem also can be done by induction pretty easily. if you're interested, you can try it that way. to me it was the most knee-jerk approach to try, and it worked out great. but you have lots of nice approaches here to choose from", "date": "2015-05-07 08:51:11", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/124509-help-proof.html", "openwebmath_score": 0.9093194603919983, "openwebmath_perplexity": 458.24289227992386, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9910145693897763, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8574496698987214}}
{"url": "https://themathhelp.com/threads/probability-of-heads-up.2947/", "text": "#### puremath\n\n##### Active Member\nIf Miranda flips a coin twice, what is the probability that both coins will land heads up?\n\nSolution (not my work):\nProbability of coin 1 landing heads up is 1/2. Probability of second coin landing heads up is 1/2.\n\nP(both heads up) = (1/2)(1/2) = 1/4.\n\nQuestion:\nWhat indicates in the problem that 1/2 is multiplied by 1/2? Taking a test, a student might guess that he or she must add 1/2 to itself.\n\n#### MarkFL\n\n##### La Villa Strangiato\nStaff member\nModerator\nMath Helper\nWe are asked to find the probability that the coin is heads on the first flip AND the coin is heads on the second flip. The \"AND\" tells us to multiply.\n\npuremath\n\n#### puremath\n\n##### Active Member\nWe are asked to find the probability that the coin is heads on the first flip AND the coin is heads on the second flip. The \"AND\" tells us to multiply.\n\nStaff member\nModerator\nMath Helper\nYes.\n\n#### puremath\n\n##### Active Member\nWe are asked to find the probability that the coin is heads on the first flip AND the coin is heads on the second flip. The \"AND\" tells us to multiply.\nThe word AND is not in the problem. This is the way I would type the question:\n\nIf Miranda flips a coin twice, what is the probability that the coin lands heads up on the first flip and heads up on the second flip? Now the word problem makes sense to me.\n\n#### MarkFL\n\n##### La Villa Strangiato\nStaff member\nModerator\nMath Helper\nWhen it says both that means both the first flip AND the second flip are heads.\n\n#### MarkFL\n\n##### La Villa Strangiato\nStaff member\nModerator\nMath Helper\nActually, It could be worded better, as it first implies she is flipping the same coin twice, then it speaks of both coins. If I were to author such a problem, I might state:\n\nIf Miranda flips a coin twice, what is the probability that it lands heads up both times?\n\npuremath\n\n#### puremath\n\n##### Active Member\nWhen it says both that means both the first flip AND the second flip are heads.\nWhen reading this the first time, it is not so clear.\n\n#### puremath\n\n##### Active Member\nActually, It could be worded better, as it first implies she is flipping the same coin twice, then it speaks of both coins. If I were to author such a problem, I might state:\n\nIf Miranda flips a coin twice, what is the probability that it lands heads up both times?\nIt is ok but my wording makes it clear to the reader that this problem involves AND and thus multiplication is the operation. Back to Cohen on Monday.\n\n#### MarkFL\n\n##### La Villa Strangiato\nStaff member\nModerator\nMath Helper\nWhat is the probability that of the two flips, at least one is heads?\n\n#### puremath\n\n##### Active Member\nWhat is the probability that of the two flips, at least one is heads?\nP(at least one heads up) = 1 - P(not one is heads up). At the job now. I have a probability book that I also want to learn with you later on in my self-study trek.\n\n#### MarkFL\n\n##### La Villa Strangiato\nStaff member\nModerator\nMath Helper\nYou're on the right track. I will give you time to complete the question.\n\n#### puremath\n\n##### Active Member\nYou're on the right track. I will give you time to complete the question.\nFor two flips, there are 4 possibilities, only one of which (getting tails twice) does not have any head. So, the other 3 out of 4 cases would have one or two heads. I say 75% probability. Yes?\n\nMarkFL\n\nStaff member", "date": "2020-04-10 06:57:07", "meta": {"domain": "themathhelp.com", "url": "https://themathhelp.com/threads/probability-of-heads-up.2947/", "openwebmath_score": 0.8188936710357666, "openwebmath_perplexity": 849.1964152155653, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357640753516, "lm_q2_score": 0.8757869997529961, "lm_q1q2_score": 0.8574267944704344}}
{"url": "https://math.stackexchange.com/questions/2476453/how-should-one-proceed-in-this-trigonometric-simplification-involving-non-intege", "text": "# How should one proceed in this trigonometric simplification involving non integer angles?\n\nThe problem is as follows:\n\nFind the value of this function $$A=\\left(\\cos\\frac{\\omega}{2} +\\cos\\frac{\\phi}{2}\\right )^{2} +\\left(\\sin\\frac{\\omega}{2} -\\sin\\frac{\\phi}{2}\\right )^{2}$$ when $$\\omega=33^{\\circ}{20}'$$ and $$\\phi=56^{\\circ}{40}'$$.\n\nThus,\n\n$$A=\\left(\\cos\\frac{\\omega}{2} +\\cos\\frac{\\phi}{2}\\right)^{2} +\\left(\\sin\\frac{\\omega}{2} -\\sin\\frac{\\phi}{2}\\right)^{2}$$\n\nBy solving the power I obtained the following:\n\n$$A= \\cos^{2}\\frac{\\omega}{2} +\\cos^{2}\\frac{\\phi}{2} +2\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2} +\\sin^{2}\\frac{\\omega}{2} +\\sin^{2}\\frac{\\phi}{2} +2\\sin\\frac{\\omega}{2}\\cos\\frac{\\phi}{2}$$\n\nI noticed some familiar terms and using pitagoric identities then I rearranged the equation as follows:\n\n\\begin{align} A &= \\cos^{2}\\frac{\\omega}{2} +\\sin^{2}\\frac{\\omega}{2} +\\cos^{2}\\frac{\\phi}{2} +\\sin^{2}\\frac{\\phi}{2} +2\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2} +2\\sin\\frac{\\omega}{2}\\cos\\frac{\\phi}{2} \\\\ &= 1+1+2\\cos\\frac{\\omega}{2}cos\\frac{\\phi}{2} +2\\sin\\frac{\\omega}{2}\\cos\\frac{\\phi}{2} \\end{align}\n\nSince the latter terms are another way to write prosthapharesis formulas I did the following:\n\n\\begin{align} \\cos\\alpha +\\cos\\beta &= 2\\cos\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2} \\\\ \\cos\\alpha -\\cos\\beta &= -2\\cos\\frac{\\alpha+\\beta}{2}\\cos\\frac{\\alpha-\\beta}{2} \\\\ \\\\ \\alpha+\\beta &= \\omega \\\\ \\alpha-\\beta &= \\phi \\end{align}\n\nBy solving the system I found: $$\\alpha=\\frac{\\omega+\\phi}{2}$$ and $$\\beta=\\frac{\\omega-\\phi}{2}$$.\n\nTherefore by inserting these into the problem:\n\n\\begin{align} A &= 1+1 +2\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2} +2\\sin\\frac{\\omega}{2}\\cos\\frac{\\phi}{2} \\\\ &= 2 +2\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2} -\\left(-2\\sin\\frac{\\omega}{2}\\cos\\frac{\\phi}{2}\\right) \\\\ &= 2 +\\cos\\frac{\\omega+\\phi}{2} +\\cos\\frac{\\omega-\\phi}{2} -\\left(\\cos\\frac{\\omega+\\phi}{2} -\\cos\\frac{\\omega-\\phi}{2}\\right) \\end{align}\n\nBy cancelling elements,\n\n$$A=2 +2\\cos\\frac{\\omega-\\phi}{2}$$\n\nHowever I'm stuck at trying to evaluate these values:\n\n\\begin{align} \\omega &= 33^{\\circ}{20}'\\; \\phi=56^{\\circ}{40}' \\\\ \\omega-\\phi &= \\left(33+\\frac{20}{60}\\right) -\\left(56+\\frac{40}{60}\\right) = -23-\\frac{20}{60} \\end{align}\n\nTherefore,\n\n$$A=2+2\\cos\\left(\\frac{-23-\\frac{20}{60}}{2}\\right).$$\n\nHowever the latter answer does not appear in the alternative neither seems to be right. Is there something wrong on what I did?\n\n\u2022 Just a remark: Use \\cos and \\sin instead of cos and sin. \u2013\u00a0MrYouMath Oct 17 '17 at 8:52\n\u2022 @MrYouMath Thanks for the advise. I'll take note on that next time I post a similar question. I'm new on LaTex. \u2013\u00a0Chris Steinbeck Bell Oct 18 '17 at 2:31\n\nThere is a mistake in your expansion of the second bracket.\n\nWe should have\n\n$A=\\cos^2\\frac{\\omega}{2}+2\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2}+\\cos^2 \\frac{\\phi}{2}+\\sin^2\\frac{\\omega}{2}-2\\sin\\frac{\\omega}{2}\\sin\\frac{\\phi}{2}+\\sin^2\\frac{\\phi}{2}$\n\n$=2+2(\\cos\\frac{\\omega}{2} \\cos\\frac{\\phi}{2} -\\sin\\frac{\\omega}{2}\\sin\\frac{\\phi}{2})$ .\n\nIn the bracket we have the expansion for $\\cos(\\frac{\\omega}{2}+\\frac{\\phi}{2})$ and fortunately $\\omega+\\phi=90^\\circ$.\n\nThus $A=2+2\\cos(45^\\circ)$\n\n$=2+\\sqrt{2}$\n\nHint: You did a miscalculation:\n\n$$A=\\left (\\cos\\frac{\\omega}{2}+\\cos\\frac{\\phi}{2} \\right )^{2}+\\left (\\sin\\frac{\\omega}{2}-\\sin\\frac{\\phi}{2} \\right )^{2}$$ $$=\\cos ^2\\frac{\\omega}{2}+2\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2}+\\cos^2 \\frac{\\phi}{2}+\\sin ^2\\frac{\\omega}{2}-2\\sin\\frac{\\omega}{2}\\sin\\frac{\\phi}{2}+\\sin^2 \\frac{\\phi}{2}$$ $$=2+2\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2}-2\\sin\\frac{\\omega}{2}\\sin\\frac{\\phi}{2}$$ $$=2+2\\left(\\cos\\frac{\\omega}{2}\\cos\\frac{\\phi}{2}-\\sin\\frac{\\omega}{2}\\sin\\frac{\\phi}{2}\\right)$$ $$=2+2\\cos\\left(\\frac{\\omega}{2}+\\frac{\\phi}{2}\\right)$$\n\nIn the last step I used $\\cos(x+y)=\\cos x\\cos y-\\sin x \\sin y$.\n\n\u2022 Mind including the \"last step\" in your answer?. I mean the one involving inserting the angle values in the last formula. Usually this seems the most logical and obvious thing to do. But what to do when the angles are not \"integers\". Was the procedure I used correct?. I mean to decompose them into fractions. The latter part I'm not sure. \u2013\u00a0Chris Steinbeck Bell Oct 18 '17 at 2:34", "date": "2019-08-21 13:39:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2476453/how-should-one-proceed-in-this-trigonometric-simplification-involving-non-intege", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 1793.8526930248404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357591818725, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8574267822496556}}
{"url": "http://happyworldtrip.com/mru0g30/ea5dc6-sum-of-1-to-10", "text": "Start with the binomial expansion of (k \u2212 1) 2: (k-1)^2: (k \u2212 1) 2: (k \u2013 roganjosh May 10 '17 at 19:49 (, Techniques for Adding the Numbers 1 to 100, Quick Insight: Intuitive Meaning of Division, Quick Insight: Subtracting Negative Numbers, Surprising Patterns in the Square Numbers (1, 4, 9, 16\u2026), Learning How to Count (Avoiding The Fencepost Problem), Different Interpretations for the Number Zero. Now for the explanation: How many beans do we have total? (i) 2, 7, 12 , ., to 10 terms. We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number. Beginners Java program to find sum of odd numbers between 1 -100 \u00bdn(n + 1),. Yep, you get the same formula, but for different reasons. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. Python Program to Calculate Sum of Odd Numbers from 1 to N using For Loop. my calculator seems to say 55 though, i don't get how the sum of n terms formula has failed as it's for an AP. Find all the evens, So, the sum from 50 + 52 + \u2026 100 = (50 * 51) \u2013 (24 * 25) = 1950. There is a simple applet showing the essence of the inductive proof of this result. I like Steve's thinking, but sometimes I like to take a hammer to a thumbtack. Vandermonde convolution X k r k! We get the next biggest even number (n + 1) and take off the extra (n + 1)/2 \u201c-1\u2033 items: To add 1 + 3 + 5 + \u2026 13, get the next biggest even (n + 1 = 14) and do, Let\u2019s say you want the evens from 50 + 52 + 54 + 56 + \u2026 100. 17 Answers 26 The main idea is that if you write all the numbers from 0 to 999999 down as six digit numbers (possibly prepending zeros) then all digits appear the same number of times. Well, we have 2 equal rows, we must have n/2 pairs. Math. The above method works, but you handle odd and even numbers differently. What is the sum of 1+1/10+1/100+1/1000... etc, where each time the denominator increases by an factor of 10 Through how many radians does it turn in one second? Each nth term of the series is the sum of the n terms i.e n*(n+1)/2. Then push the [Next] button to step through the stages of the proof. Calculate the sum \u2026 Q:-A wheel makes 360 revolutions in one minute. for(i=1;i<=n;i++) sum=sum+i; Sum of Natural Numbers using Formula. The loop structure should look like for(i=2; i<=N; i+=2). Yes. Submitted by IncludeHelp, on September 04, 2018 Given the value of N and we have to find sum of all numbers from 0 to N in C language. In case you\u2019re wondering whether it \u201creally\u201d lines up, it does. Instead of writing out numbers, pretend we have beans. Output 2: Enter the value of n: 0 Enter a whole positive number! Below is the complete algorithm. What is the formula for the sum of odd numbers? How to calculate this $\\sum\\limits_{n=0}^{\\infty}\\frac{n}{2^n}e^{jwn}$ 0. 2. Python Program to Calculate Sum of Even Numbers from 1 to N using For Loop. This program to find the sum of n numbers allows the user to enter any integer value. sum = n(n+1)/2 In the above program, unlike a for loop, we have to increment the value of i inside the body of the loop. For instance the sum of the numbers from 1 to 10 is 55 whereas the sum of the digits is 46. Let\u2019s say we mirror our pyramid (I\u2019ll use \u201co\u201d for the mirrored beans), and then topple it over: Cool, huh? home Front End HTML CSS JavaScript HTML5 Schema.org php.js Twitter Bootstrap Responsive Web Design tutorial Zurb Foundation 3 tutorials Pure CSS HTML5 Canvas JavaScript Course Icon Angular React Vue Jest Mocha NPM Yarn Back End PHP Python Java \u2026 Home Science Math History Literature Technology Health Law Business All Topics Random. The sum of all the odd integers from 1 to 50 is 625. To find sum of even numbers we need to iterate through even numbers from 1 to n. Initialize a loop from 2 to N and increment 2 on each iteration. 1. In this program, we first check number is odd or not. To add evens from 2 to 50, find 1 + 2 + 3 + 4 \u2026 + 25 and double it: So, to get the evens from 2 to 50 you\u2019d do 25 * (25 + 1) = 650. The above formula is one core step of the idea. The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + \u22ef are 1, 3, 6, 10, 15, etc.The nth partial sum is given by a simple formula: \u2211 = = (+). = r +s n! Python Program to Sum the list with start 10; Program to Sum the list of float; Python Program to Sum the list of float with start 10.1; Program to calculate the sum of elements in a tuple. Let\u2019s look at a small set: The average is 2. Python Program to Sum the list with start 10; Program to Sum the list of float; Python Program to Sum the list of float with start 10.1; Program to calculate the sum of elements in a tuple. We know that the even numbers are the numbers, which are completely divisible by 2. This Python program allows the user to enter the maximum limit value. Popular Questions of Class 11th mathematics. The so-called educator wanted to keep the kids busy so he could take a nap; he asked the class to add the numbers 1 to 100. You need to initialize the counter as 0 and while the loop executes you need to collect / sum them to the counter and finally outside the loop print/ echo the counter. This is demonstrated by the following code snippet. So, we can say the average of the entire set is actually just the average of 1 and n: (1 + n)/2. DESCRIPTION. In summation notation, this may be expressed as + ... Achilles could run at 10 m/s, while the tortoise only 5. The formula to find the sum of first n natural numbers is as follows. Geometric sum nX\u22121 k=0 ark = a 1\u2212 rn 1\u2212 r r 6= 1 Geometric series X\u221e k=0 ark = a 1\u2212 r |r| < 1 3. Feb 21, 2015 . Feb 21, 2015 . Since 50 is an even number, and we want the sum of all the odd integers from 1 to 50, the number 50 won't be included in the sum. First Name. In a similar vein to the previous exercise, here is another way of deriving the formula for the sum of the first n n n positive integers. 3) Most importantly, this example shows there are many ways to understand a formula. We want to add 1 bean to 2 beans to 3 beans\u2026 all the way up to 5 beans. Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050. 2, 7, 12, .,to 10 term We know that Sum of AP = /2 (2a + (n 1) d) Here n = 10, a = 2, & d = 7 2 = 2 Putting these in formula , Sum = /2 (2a + (n 1) d) = 10/2 (2 2 + (10 1) So we divide the formula above by 2 and get: Now this is cool (as cool as rows of numbers can be). for(i=1;i<=n;i++) sum=sum+i; Sum of Natural Numbers using Formula. 1) Adding up numbers quickly can be useful for estimation. If you can\u2019t see any matches tap the remaining deck to turn over a new card. If we have 100 numbers (1\u2026100), then we clearly have 100 items. 2) Compute some of digits in numbers from 1 \u2026 Output 3: Enter the value of n: -10 Enter a whole positive number! Here, we are implementing a C program that will be used to find the sum of all numbers from 0 to N without using loop. Learn more - Program to check odd numbers. Otherwise, we used the mathematical formula of Sum of Series 1 + 2+ 3+ \u2026 + N = N * (N + 1) / 2; TIP: If the function is not returning any value then used Void. C Program to Print Sum of all Even Numbers from 1 to n. This program allows the user to enter the maximum limit value. #SumOfNumbers #1to100 How can we calculate the sum of natural numbers? Well, the sum is clearly 1 + 2 + 3 + 4 + 5. 4. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. But of course, we don\u2019t want the total area (the number of x\u2019s and o\u2019s), we just want the number of x\u2019s. Interview Answers. Let\u2019s add the numbers 1 to 9, but instead of starting from 1, let\u2019s count from 0 instead: By counting from 0, we get an \u201cextra item\u201d (10 in total) so we can have an even number of rows. \\sum_{k=1}^n (2k-1) = 2\\sum_{k=1}^n k - \\sum_{k=1}^n 1 = 2\\frac{n(n+1)}2 - n = n^2.\\ _\\square k = 1 \u2211 n (2 k \u2212 1) = 2 k = 1 \u2211 n k \u2212 k = 1 \u2211 n 1 = 2 2 n (n + 1) \u2212 n = n 2. To see that, look at this oblong number, in which the base is one more than the height: case 1. please enter the maximum value: 10 The sum of Even numbers 1 to 10 = 30 The sum of odd numbers 1 to 10 = 25. case 2. please enter the maximum value: 100 The sum of Even numbers 1 to 100 = 2550 The sum of odd numbers 1 to 100 = 2500 . clear, insightful math lessons. Just make a for loop from starting 1 to 10, like given below. sum = n(n+1)/2 The program calculates the sum of numbers till the given input. By the way, there are more details about the history of this story and the technique Gauss may have used. Notice in both cases, 1 is on one side of the average and N is equally far away on the other. integer m 6= \u22121 5. How about even numbers, like 2 + 4 + 6 + 8 + \u2026 + n? 2. C Program to print sum of the natural numbers from 1 to 10 # include # include main( ) { int n,sum=0,i; clrscr( ); for (i=1; i<=10; i++) sum=sum+i; printf(\u201csum of natural numbers from 1 to 10 is %d\\\\n\u201d,sum); getch( ); } The goal is to clear the tableau by matching pairs of cards that add up to ten. 0. how to find a sum of numbers in a sequence when some intermediate terms are not taken in to consideration? Ex 5.3 ,1 Find the sum of the following APs. However, our formula will look a bit different. Answer Add Tags. Though both programs are technically correct, it is better to use for loop in this case. For these examples we\u2019ll add 1 to 10, and then see how it applies for 1 to 100 (or 1 to any number). We want to get rid of every number from 1 up to a \u2013 1. input this case is 6. Since thousand squared = 1 million, we get million / 2 + 1000/2 = 500,500. sum of n terms= n/2 {2a(n-1)d} where n is no of terms, a is the first term and d is difference, subbed in you get 5{2x9}=90. There is a more simple way to find the sum of odd numbers from 1 to N. The first odd number is 1, then the next odd number is 3, i.e (1+2). Enter n value: 20 Sum of odd numbers from 1 to 20 is: 100. Example 1: Adding all items in a tuple, and returning the result ; Example 1: Starting with the 10, and adding all items in a tuple to this number: Example: Compute the sum and product of the numbers from 1 to 10. C programming, exercises, solution : Write a program in C to calculate the sum of numbers from 1 to n using recursion. The above formula is one core step of the idea. How about odd numbers, like 1 + 3 + 5 + 7 + \u2026 + n? I won\u2019t. The difference between consecutive triangles increases by 1.. A formula for the triangular numbers. Telescoping sum X a\u2264k\nsum of 1 to 10 2020", "date": "2022-01-25 08:22:50", "meta": {"domain": "happyworldtrip.com", "url": "http://happyworldtrip.com/mru0g30/ea5dc6-sum-of-1-to-10", "openwebmath_score": 0.4893442392349243, "openwebmath_perplexity": 424.935485709517, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9790357561234474, "lm_q2_score": 0.8757869835428966, "lm_q1q2_score": 0.857426771635993}}
{"url": "http://math.stackexchange.com/questions/115558/nth-roots-of-unity", "text": "# $n$th Roots of Unity\n\nI need to show that the product of all the $n$th roots of unity is $(\u22121)^{n+1}$.\n\nIs there a way to do this by induction? If there is, I can't seem to figure it out. Are there other, perhaps more efficient, methods of proof?\n\n-\nThe problem with attempting to do it \"by induction\" is that the $n$th roots of unity are not related to the $(n+1)$th roots of unity (except for the single one that they have in common, $1$. \u2013\u00a0 Arturo Magidin Mar 2 '12 at 4:07\nIf you're going to do an induction argument, it should be an induction on the number of (possibly repeated) prime divisors of $n$. I think this should actually be possible, with Wilson's Theorem stepping in as a base case. But, honestly, just about anything else seems easier....like all the answers given below. \u2013\u00a0 Pete L. Clark Mar 2 '12 at 4:28\n\nThe $n^{th}$ roots of unity are precisely the roots of the polynomial $$x^n - 1$$ Now use the fact that the product of the roots of any polynomial is $(-1)^n$ times the constant coefficient.\n\n-\nIs this fact that the product of the roots of any polynomial is $(\u22121)^n$ times the constant coefficient true for any polynomial? Can you elaborate on that? \u2013\u00a0 Dominick Gerard Mar 2 '12 at 4:42\n@DominickGerard: Consider writing the polynomial as a product of linear factors, then multiplying it out again. What will the constant term be? Example: $(z-z_1)(z-z_2) = z^2 - (z_1 + z_2) z + z_1 z_2$. \u2013\u00a0 Antonio Vargas Mar 2 '12 at 4:58\n@DominickGerard Yes, it holds for any polynomial with coefficients in an integral domain (including $\\mathbb{Z}$ or any field): en.wikipedia.org/wiki/Vieta%27s_formulas \u2013\u00a0 dls Mar 2 '12 at 5:01\n\nNote that if $\\zeta$ is an $n$th root of unity, then so is $\\frac{1}{\\zeta}$. Moreover, the only roots of unity for which $\\zeta=\\frac{1}{\\zeta}$ are $\\zeta=1$ and $\\zeta=-1$ (since $\\zeta=\\frac{1}{\\zeta}$ implies $\\zeta^2=1$, hence $\\zeta=1$ or $\\zeta=-1$).\n\nSo, if we take the $n$th roots of unity, then we can pair off all of them except for $1$ and perhaps $-1$ (if $n$ is even), into pairs of the form $\\{\\zeta,\\frac{1}{\\zeta}\\}$. The product of these two equals $1$; so when you do the product of all roots, you end up with a bunch of pairs that multiply to $1$, and you have $1$; and if $n$ is even, then you also have $-1$ leftover. Which means the product of all of them is...\n\n-\n\nWell, you said \"efficient\" and \"elementary number theory\", and that triggered this in my mind: the result is a special case of\n\n(Wilson's Theorem in a Finite Abelian Group): Let $(G,\\cdot)$ be a finite abelian group, and let $P = \\prod_{x \\in G} x$. Then $P = 1$ (the identity element) unless $G$ has exactly one element $t$ of order $2$, in which case $P = t$.\n\nThis is something that others (in particular, Bill Dubuque) have talked about here and elsewhere before. I include a statement and proof of this in my notes/prebook on elementary(ish...) number theory: it is at the end of Appendix B.\n\nAdded: Not only is this a special case of a \"generalized Wilson theorem\", it's a special case that includes the classical Wilson Theorem! Since the unit group of $(\\mathbb{Z}/p\\mathbb{Z})^{\\times}$ is cyclic of order $p-1$, computing $(p-1)! \\pmod{p}$ is essentially the same as multiplying together all the $(p-1)$st roots of unity.\n\n-\n+1 See these posts for more on the group-theoretic Wilson's theorem and involutions. \u2013\u00a0 Bill Dubuque Mar 2 '12 at 4:13\nYou can prove it with de Moivre, on the basis that the roots are all of the form $e^{2\\pi i k/n}$, so their product is $e^{2\\pi i/n\\sum{k}} = e^{\\pi i(n-1)} = (-1)^{n-1} = (-1)^{n+1}$.\nAnother proof idea would be to notice that the roots of unity come in complex conjugate pairs, which is evident in the symmetry of the spacing of the roots in the complex plain. So the product of each pair is of course $\\alpha \\bar\\alpha = |\\alpha| = 1$ since all the roots fall on the unit circle. The only exceptions are $1,-1$. $1$ is always a root and won't change the product, and $-1$ is only a root when $n$ is even, so we conclude that the product is exactly $(-1)^{n+1}$.", "date": "2014-03-12 07:14:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/115558/nth-roots-of-unity", "openwebmath_score": 0.8893849849700928, "openwebmath_perplexity": 96.09390417199187, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357542883923, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8574267684418487}}
{"url": "https://forum.azimuthproject.org/discussion/2459/lecture-7-universal-properties-products-coproducts-and-exponentials-david-spivak", "text": "#### Howdy, Stranger!\n\nIt looks like you're new here. If you want to get involved, click one of these buttons!\n\nOptions\n\n# Lecture 7 - Universal properties: products, coproducts and exponentials - David Spivak\n\nedited January 21\n\n## Comments\n\n\u2022 Options\n1.\n\nIt was interesting to learn about the relationship between currying, uncurrying, and exponentials. I'd heard people talking about currying before and handwaving, but this lecture made it very concrete. The motivation of set exponentiation (the set of maps) {1, 2, 3}^{1,2} as \"3\"^\"2\" ~ 9 maps in total was nice as well.\n\nI also liked the motivation of Either as a pushout from two functions. I recently did a few exercises using Either in scala and ended up getting very confused! So this sort of knowledge would probably have made that experience a bit less confusing.\n\nI noticed that it was mentioned in passing that products, co-products and exponentials are all examples of adjunctions. And I just listened to the first 10 seconds of lecture 8, and I gather that they are discussed therein, so I look forward to reviewing that lecture too in time.\n\nComment Source:It was interesting to learn about the relationship between currying, uncurrying, and exponentials. I'd heard people talking about currying before and handwaving, but this lecture made it very concrete. The motivation of set exponentiation (the set of maps) {1, 2, 3}^{1,2} as \"3\"^\"2\" ~ 9 maps in total was nice as well. I also liked the motivation of Either as a pushout from two functions. I recently did a few exercises using Either in scala and ended up getting very confused! So this sort of knowledge would probably have made that experience a bit less confusing. I noticed that it was mentioned in passing that products, co-products and exponentials are all examples of [adjunctions](https://en.wikipedia.org/wiki/Adjoint_functors). And I just listened to the first 10 seconds of lecture 8, and I gather that they are discussed therein, so I look forward to reviewing that lecture too in time.\n\u2022 Options\n2.\nedited March 2020\n\nYes, it's a good one!\n\nIdeas from the first five minutes (using David Spivak's terminology):\n\n\u2022 A universal property is a \"one-stop shop\" for maps\n\n\u2022 The product of sets $$A$$ and $$B$$ is the one-stop shop from maps into $$A$$ and $$B$$.\n\nMeaning: suppose you have a set $$I$$, and maps $$f: I \\rightarrow A$$ and $$g: I \\rightarrow B$$, with $$f$$ in your \"left-hand\", and $$g$$ in your right.\n\nThese two functions are equivalent to a single function $$f \\times g: I \\rightarrow A \\times B$$.\n\nThe set $$I$$ can be thought of as an \"index\" set which is used to pick out elements via the functions out of $$I$$.\n\nWhen $$I$$ is a singleton, the \"one-stop shop\" gives you a single element in the product of $$A$$ and $$B$$, in lieu of two separate elements one from $$A$$ and the other from $$B$$.\n\nComment Source:Yes, it's a good one! Ideas from the first five minutes (using David Spivak's terminology): * A universal property is a \"one-stop shop\" for maps * The product of sets \\$$A\\$$ and \\$$B\\$$ is the one-stop shop from maps into \\$$A\\$$ and \\$$B\\$$. Meaning: suppose you have a set \\$$I\\$$, and maps \\$$f: I \\rightarrow A\\$$ and \\$$g: I \\rightarrow B\\$$, with \\$$f\\$$ in your \"left-hand\", and \\$$g\\$$ in your right. These two functions are equivalent to a single function \\$$f \\times g: I \\rightarrow A \\times B\\$$. The set \\$$I\\$$ can be thought of as an \"index\" set which is used to pick out elements via the functions out of \\$$I\\$$. When \\$$I\\$$ is a singleton, the \"one-stop shop\" gives you a single element in the product of \\$$A\\$$ and \\$$B\\$$, in lieu of two separate elements one from \\$$A\\$$ and the other from \\$$B\\$$.\n\u2022 Options\n3.\n\u2022 The coproduct of $$A$$ and $$B$$ is the \"one-stop shop\" for maps into $$A$$ and $$B$$.\nComment Source:* The coproduct of \\$$A\\$$ and \\$$B\\$$ is the \"one-stop shop\" for maps into \\$$A\\$$ and \\$$B\\$$.\n\u2022 Options\n4.\nedited March 2020\n\nThinking out loud here...\n\nThe idea of the \"one-stop shop for maps\" generalizes nicely from the product to the limit.\n\nLooking at the general case of the limit, which includes as special cases terminal objects, products, equalizers and pullbacks, brings clarity to the whole picture. And in the category Set, there is an explicit construction for all limits.\n\nTo review: suppose we are given a diagram D in the category C. Note: for this definition, D is not required to be a commuting diagram -- it's just a collection of objects and arrows.\n\nThe limit of D, if it exists, will be another object Lim(D) in C, along with \"projection maps\" from Lim(D) to each of the objects in D, which satisfies a couple of conditions. First, for every arrow $$f: X_1 \\rightarrow X_2$$ in D, the triangle diagram with objects Lim(D), $$X_1$$, and $$X_2$$ and arrows $$f$$ and the two projections from Lim(D) to $$X_1$$ and $$X_2$$ -- this triangle diagram commutes.\n\nLim(D) along with these projections is called a cone over D. The cones over D form a category, with arrows going between the apexes of the cones, such that these arrows satisfy the following commutativity condition with respect to the cones. Suppose that $$X$$ and $$Y$$ are each the apex of a cone, and $$f: X \\rightarrow Y$$. Then $$f$$ is defined to be a morphism between the two cones if for each object $$d$$ in D, the triangle formed by $$d$$, $$X$$ and $$Y$$, with arrows $$f$$ and the two projections to $$d$$ -- one projection within each cone -- commutes.\n\nThe second requirement on Lim(D) is that it be a terminal object in this category of cones.\n\nComment Source:Thinking out loud here... The idea of the \"one-stop shop for maps\" generalizes nicely from the product to the limit. Looking at the general case of the limit, which includes as special cases terminal objects, products, equalizers and pullbacks, brings clarity to the whole picture. And in the category Set, there is an explicit construction for all limits. To review: suppose we are given a diagram D in the category C. Note: for this definition, D is not required to be a _commuting_ diagram -- it's just a collection of objects and arrows. The limit of D, if it exists, will be another object Lim(D) in C, along with \"projection maps\" from Lim(D) to each of the objects in D, which satisfies a couple of conditions. First, for every arrow \\$$f: X_1 \\rightarrow X_2\\$$ in D, the triangle diagram with objects Lim(D), \\$$X_1\\$$, and \\$$X_2\\$$ and arrows \\$$f\\$$ and the two projections from Lim(D) to \\$$X_1\\$$ and \\$$X_2\\$$ -- this triangle diagram commutes. Lim(D) along with these projections is called a cone over D. The cones over D form a category, with arrows going between the apexes of the cones, such that these arrows satisfy the following commutativity condition with respect to the cones. Suppose that \\$$X\\$$ and \\$$Y\\$$ are each the apex of a cone, and \\$$f: X \\rightarrow Y\\$$. Then \\$$f\\$$ is defined to be a morphism between the two cones if for each object \\$$d\\$$ in D, the triangle formed by \\$$d\\$$, \\$$X\\$$ and \\$$Y\\$$, with arrows \\$$f\\$$ and the two projections to \\$$d\\$$ -- one projection within each cone -- commutes. The second requirement on Lim(D) is that it be a terminal object in this category of cones. \n\u2022 Options\n5.\nedited March 2020\n\nNow that's somewhat of an abstract story, which doesn't give you a feeling for how to actually find what this special cone Lim(D) actually is. And in general such an object may not exist at all.\n\nBut in the category Set it always exists, and can be constructed explicitly, as follows.\n\nComment Source:Now that's somewhat of an abstract story, which doesn't give you a feeling for how to actually find what this special cone Lim(D) actually is. And in general such an object may not exist at all. But in the category Set it always exists, and can be constructed explicitly, as follows. \n\u2022 Options\n6.\nedited March 2020\n\nSuppose we have a diagram $$D$$ in Set. Let the objects in $$D$$ be named $$S_1, \\ldots S_n$$, where each $$S_i$$ is a set. Let the arrows in $$D$$ be $$f_1, \\ldots f_k$$ -- these are functions.\n\nThen Lim(D) consists of a subset of the Cartesian product of the set $$S_1, \\ldots S_n$$ -- so it is a relation R between these sets. It consists of all tuples $$(x_1, \\ldots, x_n)$$ that satisfy the all of the constraints given by the functions $$f_i$$. Suppose $$f_i: S_j \\rightarrow S_k$$. Then the constraint given by $$f_i$$ stipulates that for a tuple $$(x_1, \\ldots, x_n)$$, that $$f_i(x_j) = x_k$$.\n\nThe cone has relation R at its apex, and the ordinary projection functions from R to each of the sets $$S_1, \\ldots, S_n$$ as the arrows from the apex R to the objects in the diagram $$D$$.\n\nIt is but an exercise to show that this cone with R at its apex is terminal within the category of cones over $$D$$.\n\nComment Source:Suppose we have a diagram \\$$D\\$$ in Set. Let the objects in \\$$D\\$$ be named \\$$S_1, \\ldots S_n\\$$, where each \\$$S_i\\$$ is a set. Let the arrows in \\$$D\\$$ be \\$$f_1, \\ldots f_k\\$$ -- these are functions. Then Lim(D) consists of a subset of the Cartesian product of the set \\$$S_1, \\ldots S_n\\$$ -- so it is a relation R between these sets. It consists of all tuples \\$$(x_1, \\ldots, x_n)\\$$ that satisfy the all of the constraints given by the functions \\$$f_i\\$$. Suppose \\$$f_i: S_j \\rightarrow S_k\\$$. Then the constraint given by \\$$f_i\\$$ stipulates that for a tuple \\$$(x_1, \\ldots, x_n)\\$$, that \\$$f_i(x_j) = x_k\\$$. The cone has relation R at its apex, and the ordinary projection functions from R to each of the sets \\$$S_1, \\ldots, S_n\\$$ as the arrows from the apex R to the objects in the diagram \\$$D\\$$. It is but an exercise to show that this cone with R at its apex is terminal within the category of cones over \\$$D\\$$. \n\u2022 Options\n7.\n\nThe relation R = Lim(D) is a one-stop shop for cones over D, in the following sense. Suppose that we have a cone C with apex A over D. Then C consists of a bunch of separate maps from A into the objects of D. The universal property for R = Lim(D) says that there a unique map m(C) from A into R, which gives a morphism from C into the universal cone with apex Lim(D).\n\nm(C) is a single map that contains all of the information in C. That's because the projections from C into the objects of D can be reconstructed by composing m(C) with the standard projections from R into the objects of D.\n\nSo the multiple separate maps comprising the cone C can, without loss of information, be replaced by the single map m(C).\n\nComment Source:The relation R = Lim(D) is a one-stop shop for cones over D, in the following sense. Suppose that we have a cone C with apex A over D. Then C consists of a bunch of separate maps from A into the objects of D. The universal property for R = Lim(D) says that there a unique map m(C) from A into R, which gives a morphism from C into the universal cone with apex Lim(D). m(C) is a single map that contains all of the information in C. That's because the projections from C into the objects of D can be reconstructed by composing m(C) with the standard projections from R into the objects of D. So the multiple separate maps comprising the cone C can, without loss of information, be replaced by the single map m(C). \n\u2022 Options\n8.\nedited March 2020\n\nFor an illustration of this construction, let's apply it to the equalizer.\n\nHere, the diagram D consists of two objects $$X_1, X_2$$, and two maps $$f,g: X_1 \\rightarrow X_2$$.\n\nNote: here the fact that D is not required to be a commutative diagram is significant. If it were so required, then that would imply that $$f$$ and $$g$$ are equal. But we don't require that, as we are constructing the equalizer of distinct maps $$f$$ and $$g$$.\n\nThen, as per the general construction given above, the relation R = Lim(D) will consist of all tuples $$(x,y)$$, such that $$f(x) = y$$, and $$g(x) = y$$. Putting these together, we get that:\n\n$equalizer(f,g) = Lim(D) = \\lbrace (x,y)\\ |\\ y = f(x) = g(x) \\rbrace = \\lbrace (x,y)\\ |\\ x \\in dom(f) \\cap dom(g),\\ f(x) = g(x)\\rbrace = f \\cap g$\n\nWhich is to say: the equalizer of functions $$f$$ and $$g$$ is the relation (the graph representation) for the partial function that is obtained by restricting both functions to the set of points where they are both defined and both produce the same value.\n\nComment Source:For an illustration of this construction, let's apply it to the equalizer. Here, the diagram D consists of two objects \\$$X_1, X_2\\$$, and two maps \\$$f,g: X_1 \\rightarrow X_2\\$$. Note: here the fact that D is not required to be a _commutative_ diagram is significant. If it were so required, then that would imply that \\$$f\\$$ and \\$$g\\$$ are equal. But we don't require that, as we are constructing the equalizer of distinct maps \\$$f\\$$ and \\$$g\\$$. Then, as per the general construction given above, the relation R = Lim(D) will consist of all tuples \\$$(x,y)\\$$, such that \\$$f(x) = y\\$$, and \\$$g(x) = y\\$$. Putting these together, we get that: \\$equalizer(f,g) = Lim(D) = \\lbrace (x,y)\\ |\\ y = f(x) = g(x) \\rbrace = \\lbrace (x,y)\\ |\\ x \\in dom(f) \\cap dom(g),\\ f(x) = g(x)\\rbrace = f \\cap g\\$ Which is to say: the equalizer of functions \\$$f\\$$ and \\$$g\\$$ is the relation (the graph representation) for the partial function that is obtained by restricting both functions to the set of points where they are both defined and both produce the same value. \n\u2022 Options\n9.\nedited March 2020\n\nWe get the product as a special case of a limit, where the diagram $$D$$ consists only of objects, and no arrows.\n\nIn Set, this will be a collection of sets $$S_1, \\ldots, S_n$$ -- and no arrows to impose constraints on the tuples in R = Lim(D).\n\nSo Lim(D) is the unconstrained set of tuples -- i.e., the Cartesian product $$S_1 \\times \\cdots \\times S_n$$.\n\nComment Source:We get the _product_ as a special case of a limit, where the diagram \\$$D\\$$ consists only of objects, and no arrows. In Set, this will be a collection of sets \\$$S_1, \\ldots, S_n\\$$ -- and no arrows to impose constraints on the tuples in R = Lim(D). So Lim(D) is the unconstrained set of tuples -- i.e., the Cartesian product \\$$S_1 \\times \\cdots \\times S_n\\$$.\n\u2022 Options\n10.\n\nWhat's the limit of an empty diagram $$D$$?\n\nIt's simply a terminal object in the category, for the following reason.\n\nA cone over the empty diagram consists just of an apex object A, along with zero \"leg\" morphisms.\n\nSo the category of cones over the empty diagram coincides exactly with the ambient category C.\n\nThe limit is defined to be a terminal cone, which amounts to the same thing as a terminal object in C.\n\nComment Source:What's the limit of an empty diagram \\$$D\\$$? It's simply a terminal object in the category, for the following reason. A cone over the empty diagram consists just of an apex object A, along with zero \"leg\" morphisms. So the category of cones over the empty diagram coincides exactly with the ambient category C. The limit is defined to be a terminal cone, which amounts to the same thing as a terminal object in C. \n\u2022 Options\n11.\nedited March 2020\n\nExercise 1. Use this construction to concretely explain the meaning of pullbacks in Set.\n\nRecall that a pullback is defined as the limit of a diagram $$D$$, consisting of three objects $$A,B,C$$, and morphims $$f: A \\rightarrow C$$ and $$g: B \\rightarrow C$$.\n\nComment Source:Exercise 1. Use this construction to concretely explain the meaning of pullbacks in Set. Recall that a pullback is defined as the limit of a diagram \\$$D\\$$, consisting of three objects \\$$A,B,C\\$$, and morphims \\$$f: A \\rightarrow C\\$$ and \\$$g: B \\rightarrow C\\$$. \n\u2022 Options\n12.\nedited March 2020\n\n\n\\begin{CD} A \\cap B @>\\subseteq>>B \\\\@V{\\subseteq}VV {}@VV{\\subseteq}V \\\\ A @>>\\subseteq> C \\end{CD}\n\nHere, $$C$$ is taken to be any set which contains both $$A$$ and $$B$$, and $$\\subseteq$$ means the inclusion function.\n\nDiagram from Example 1.8.3 of:\n\n\u2022 Benjamin C. Pierce, Basic Category Theory for Computer Scientists, MIT Press, 1991.\n\u2022 Options\n13.\nedited March 2020\n\n\n\\begin{CD} f^{-1}(A) @>\\subseteq>>B \\\\@V{f|_{f^{-1}(A)}}VV {}@VV{f}V \\\\ A @>>\\subseteq> C \\end{CD}\n\nHere, $$A$$ is a subset of the codomain of $$f$$. The claim made by this diagram is that inverse image of $$A$$ under $$f$$ is the pullback of $$f$$ along the inclusion function for $$A$$.\n\nDiagram from Example 1.8.2 of:\n\n\u2022 Benjamin C. Pierce, Basic Category Theory for Computer Scientists, MIT Press, 1991.\n\u2022 Options\n14.\nedited March 2020\n\n@ChrisGoddard wrote:\n\nI noticed that it was mentioned in passing that products, co-products and exponentials are all examples of adjunctions.\n\nInteresting! Keep us posted if you dig further into this.\n\nComment Source:@ChrisGoddard wrote: > I noticed that it was mentioned in passing that products, co-products and exponentials are all examples of [adjunctions](https://en.wikipedia.org/wiki/Adjoint_functors). Interesting! Keep us posted if you dig further into this.\n\u2022 Options\n15.\nedited March 2020\n\nProblem 1: spell out the explicit construction for colimits in the category Set.\n\nThe path of analysis is similar to that used for the explicit construction for limits. But the result is not \"symmetrical.\" In one of the lectures Bartosz commented on the asymmetry between product and coproduct in Set. He traced it back to some asymmetries in the structure of the morphisms in Set, which as we know are total functions. The asymmetry is that whereas each point in the domain must map -- be related to -- to just one point in the codomain, each point in the codomain can be related to zero points in the domain (where the function is not onto), or multiple points in the domain (where the function is not one-to-one).\n\nComment Source:Problem 1: spell out the explicit construction for colimits in the category Set. The path of analysis is similar to that used for the explicit construction for limits. But the result is not \"symmetrical.\" In one of the lectures Bartosz commented on the asymmetry between product and coproduct in Set. He traced it back to some asymmetries in the structure of the morphisms in Set, which as we know are total functions. The asymmetry is that whereas each point in the domain must map -- be related to -- to just one point in the codomain, each point in the codomain can be related to zero points in the domain (where the function is not onto), or multiple points in the domain (where the function is not one-to-one). \n\u2022 Options\n16.\n\nProblem 2: use this construction to give a concrete explanation for coequalizers, coproducts, pushouts and initial objects in Set.\n\nComment Source:Problem 2: use this construction to give a concrete explanation for coequalizers, coproducts, pushouts and initial objects in Set.\nSign In or Register to comment.", "date": "2021-07-24 06:40:12", "meta": {"domain": "azimuthproject.org", "url": "https://forum.azimuthproject.org/discussion/2459/lecture-7-universal-properties-products-coproducts-and-exponentials-david-spivak", "openwebmath_score": 0.8684090375900269, "openwebmath_perplexity": 613.6969684394298, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104982195785, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.8574038392012853}}
{"url": "http://mathhelpforum.com/calculus/159010-finding-riemann-sums.html", "text": "# Math Help - Finding Riemann sums\n\n1. ## Finding Riemann sums\n\nThe question:\n\nAn electrical signal S(t) has its amplitude |S(t)| tested (sampled) every 1/10 of a second. It is desired to estimate the energy over a period of half a second, given exactly by:\n\n$(\\int_0^\\frac{1}{2} \\! |S(t)|^2 \\, \\mathrm{d}t)^{\\frac{1}{2}}$\n\nThe result of the measurement are shown in the following table:\n$\\begin{center}\n\\begin{tabular}{| l | c | c| c| c| c|}\n\\hline\nt & .1 & .2 & .3 & .4 & .5\\\\ \\hline\n\\| S(t) \\| & 60 & 50 & 50 & 45 & 55 \\\\ \\hline\ne(t) & 5 & 3 & 7 & 4 & 10 \\\\\n\\hline\n\\end{tabular}\n\\end{center}$\n\na) Using the above data for S(t), set up the appropriate Riemann sum and compute an appropriate value for the energy.\n\nMy attempt:\nThe set P (set of partitions) is clearly equal to {0, 1/10, 1/5, 3/10, 4/10, 1/2} so dt is 1/10 i.e. the width of each rectangle is 1/10.\n\nMy problem is with producing lower and higher Riemann sums. I'm used to having a function defined as something like $x^2$ and producing a general sum given dx (in this case, dt). In this question, we don't know the function, we're just given values.\n\nDoes this mean I take the max/min of each interval (e.g. [0, 1/10], [1/10, 1/5] etc.) and work out the partitions by hand? I'm a bit confused.\n\nAny help would be greatly appreciated!\n\n2. Originally Posted by Glitch\nThe question:\n\nAn electrical signal S(t) has its amplitude |S(t)| tested (sampled) every 1/10 of a second. It is desired to estimate the energy over a period of half a second, given exactly by:\n\n$(\\int_0^\\frac{1}{2} \\! |S(t)|^2 \\, \\mathrm{d}t)^{\\frac{1}{2}}$\n\nThe result of the measurement are shown in the following table:\n$\\begin{center}\n\\begin{tabular}{| l | c | c| c| c| c|}\n\\hline\nt & .1 & .2 & .3 & .4 & .5\\\\ \\hline\n\\| S(t) \\| & 60 & 50 & 50 & 45 & 55 \\\\ \\hline\ne(t) & 5 & 3 & 7 & 4 & 10 \\\\\n\\hline\n\\end{tabular}\n\\end{center}$\n\na) Using the above data for S(t), set up the appropriate Riemann sum and compute an appropriate value for the energy.\n\nMy attempt:\nThe set P (set of partitions) is clearly equal to {0, 1/10, 1/5, 3/10, 4/10, 1/2} so dt is 1/10 i.e. the width of each rectangle is 1/10.\n\nMy problem is with producing lower and higher Riemann sums. I'm used to having a function defined as something like $x^2$ and producing a general sum given dx (in this case, dt). In this question, we don't know the function, we're just given values.\n\nDoes this mean I take the max/min of each interval (e.g. [0, 1/10], [1/10, 1/5] etc.) and work out the partitions by hand? I'm a bit confused.\n\nAny help would be greatly appreciated!\nleft sum ... L = 0.1[S(0) + S(.1) + S(.2) + S(.3) + S(.4)]\n\nright sum ... R = 0.1[S(.1) + S(.2) + S(.3) + S(.4) + S(.5)]\n\nthe table does not give a value for S(0), so it looks as though you can only compute the right sum.\n\n3. Inside each interval you are given only two values- the values at the two endpoints. For the \"lower sum\" use the smaller of those two, for the \"higher sum\" use the larger.\n\nIn the first interval, from 0.1 to 0.2, you are given exactly two values- 60 and 50. The \"lower sum\" uses the smaller of those, 50, and the \"higher sum\" uses the higher, 60. Between .2 and .3 your two values are 50 and 50. Since those are the same, use 50 for both sums. Between .3 and .4, the two values are 50 and 45. Use 45 for the \"lower sum\" and 50 for the \"higher sum\". Finally, between .4 and .5, the two values are 45 and 55. Use 45 for the lower sum and 55 for the higher sum.\n\nSummarizing, for the \"lower sum\" use 50, 50, 45, and 45. For the \"higher sum\" use 60, 50, 50, and 55.\n\n4. For the lower sum I used: 0, 50, 50, 45, 45\nI squared each (as per the function), and multiplied them by 1/10 to produce the area of each partition. My result was 905.\n\nFor the higher sum I used: 60, 60, 50, 50, 55\nAgain, I squared each, and multiplied by 1/10. My result was 1522.5\n\nI took the average of the higher and lower sums, to get 1213.75. As per the question, the integral is to the power of 1/2, so I took the square root and got 34.84. However, the solution in my text is 36.95 (or square root of 1365).\n\nI'm not sure where I went wrong. My answer is close, but incorrect. :/\n\nEdit: I just realised that using '0' for the lower sum isn't correct. I used 60 instead, but I get 1393.75 which is slightly off the given solution.", "date": "2014-04-20 00:43:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/159010-finding-riemann-sums.html", "openwebmath_score": 0.9354174137115479, "openwebmath_perplexity": 804.2502639059516, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104962847373, "lm_q2_score": 0.8872046011730965, "lm_q1q2_score": 0.8574038389257946}}
{"url": "https://math.stackexchange.com/questions/2714910/find-all-pairs-of-x-y-z-such-that-x-y-sqrtz2-2018", "text": "# Find all pairs of $(x, y ,z)$ such that $x + y =\\sqrt{z^{2} + 2018}, \u2026$\n\nFind all pairs of $(x,y,z)$, of real numbers, such that\n\n$$x + y = \\sqrt{z^{2} + 2018}$$ $$x + z = \\sqrt{y^{2} + 2018}$$ $$y + z = \\sqrt{x^{2} + 2018}$$\n\nAn attempt : Squaring we get $$(x + y)^{2} = z^{2} + 2018 \\implies (x + y)^{2} - z^{2} = 2018$$ $$(x + z)^{2} = y^{2} + 2018 \\implies (x + z)^{2} - y^{2} = 2018$$ $$(y + z)^{2} = x^{2} + 2018 \\implies (z + y)^{2} - x^{2} = 2018$$ which also means $$(x + y - z)(x + y + z) = 2018$$ $$(x + z - y)(x + y + z) = 2018$$ $$(z + y - x)(x + y + z) = 2018$$ so $$\\frac{2018}{x+y-z} = \\frac{2018}{x+z-y} = \\frac{2018}{y+z-x}$$\n\n$$(x+y-z) = (x+z-y) = (y+z-x)$$ $$y-z = z-y \\implies z = y$$ $$x-y= y-x \\implies x=y$$ so my answer is $$(x, y, z), \\:\\:\\: x=y=z$$ but with the 3 initial equations, we must also have $$(x+y) = \\sqrt{z^{2} + 2018} \\implies 4x^{2} = x^{2} + 2018$$ or $$x^{2} = 2018/3$$ so the solution is $$(x, y, z), \\:\\:\\: x=y=z = \\sqrt{2018/3}$$\n\nIs this sufficient already? are there better techniques?\n\n\u2022 Doesn't $x=y=z$ Imply $2x = \\sqrt{x^{2} + 2018}$ from which you can solve for $x$? \u2013\u00a0Jim Haddocc Mar 30 '18 at 16:16\n\u2022 The equation can be rewritten as $x + y + z = f(x) = f(y) = f(z)$ where $f(t) = t + \\sqrt{t^2+2018}$. The function $f$ is strictly increasing, so $f(x) = f(y) = f(z) \\implies \\cdots$ \u2013\u00a0achille hui Mar 30 '18 at 16:22\n\u2022 Indeed, we get $$x=y=z=\\sqrt{\\frac{2018}{3}}$$ \u2013\u00a0Dr. Sonnhard Graubner Mar 30 '18 at 16:25\n\u2022 @PrathyushPoduval yes, thanks for that, i was meant that way. I have edited the post \u2013\u00a0Arief Anbiya Mar 30 '18 at 16:26\n\u2022 Just the condition that $z^2+m$ is a square implies that$z^2+m \\ge (z+1)^2$ or $z \\le (m-1)/2$. \u2013\u00a0marty cohen Mar 30 '18 at 16:29\n\nAt the stage you have... $$(x + y - z)(x + y + z) = 2018$$ $$(x + z - y)(x + y + z) = 2018$$ $$(z + y - x)(x + y + z) = 2018$$ This implies... $$x+y-z=x+z-y=z+y-x=\\dfrac{2018}{x+y+z}$$ ...unless $x+y+z=0$.\n\nNotice that $x+y+z=0$ implies $x+y=-z$ and so $-z=\\sqrt{z^2+2018}$ which means $z^2=z^2+2018$ and so $z=0$. Likewise, $x=y=0$ (contradiction since $(0,0,0)$ isn't a solution).\n\nThus $x+y-z=x+z-y=z+y-x$ and so you got that $x=y=z$. But then $x+y=\\sqrt{z^2+2018}$ implies that $2z=\\sqrt{z^2+2018}$ and so $4z^2=z^2+2018$ and so $x=y=z = \\pm \\sqrt{\\dfrac{2018}{3}}$. But $x+y$ etc. are square roots (thus non negative). So there is only one solution.\n\n\u2022 Thanks for the answer, but i think negative value of $x=y=z < 0$ is not allowed. This is because the 3 initial equations hold. \u2013\u00a0Arief Anbiya Mar 30 '18 at 16:42\n\u2022 Silly me. You are correct. \u2013\u00a0Bill Cook Mar 30 '18 at 17:15", "date": "2020-10-30 17:29:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2714910/find-all-pairs-of-x-y-z-such-that-x-y-sqrtz2-2018", "openwebmath_score": 0.935429573059082, "openwebmath_perplexity": 251.80364456135564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109485172559, "lm_q2_score": 0.8705972818382006, "lm_q1q2_score": 0.8573737349036231}}
{"url": "https://math.stackexchange.com/questions/896186/how-can-i-prove-that-the-span-of-a-subspace-and-its-orthogonal-complement-is-the", "text": "# How can I prove that the span of a subspace and its orthogonal complement is the whole vector space?\n\nThe book Linear and Geometric Algebra explains the following theorem in a way that I haven't been able to figure out:\n\nIf $\\mathbf{A}$ and $\\mathbf{B}$ are subspaces of a vector space $\\mathbf{B}$ then the set of all combinations $\\mathbf{a} + \\mathbf{b}$ such that, $\\mathbf{a} \\in \\mathbf{A}$ and $\\mathbf{b} \\in \\mathbf{B}$ is called the span of $\\mathbf{A}$ and $\\mathbf{B}$, written as $\\text{span}(\\mathbf{A}, \\mathbf{B})$.\n\nFurthermore let $\\mathbf{U}^{\\bot}$ be the subspace consisting of all vectors orthogonal to a subspace $\\mathbf{U}$, in the sense that $\\mathbf{u} \\in \\mathbf{U}^{\\top}$ if and only if, $\\mathbf{u} \\perp \\mathbf{v}$ for all vectors $\\mathbf{v} \\in \\mathbf{U}$.\n\nI have of course been able to prove that $\\mathbf{U}^{\\bot}$ is indeed a subspace for all subspaces $\\mathbf{U}$, if this turns out to be useful.\n\nThe theorem I want to prove is: if $\\mathbf{U}$ is a subspace of $\\mathbf{V}$ then $\\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp}) = \\mathbf{V}$.\n\nThe book mentioned above proves it as follows: If $\\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp}) \\neq \\mathbf{V}$ then there is a nonzero $\\mathbf{u} \\perp \\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp})$ because any orthonormal basis for a subspace of an inner product space can be extended into an orthonormal basis for the entire inner product space. In particular $\\mathbf{u} \\perp \\mathbf{U}$, i.e. $\\mathbf{u} \\in \\mathbf{U}^{\\perp}$, a contradiction.\n\nI understand how an orthonormal basis for a subspace of an inner product space can be extended into an orthonormal basis for the whole inner product space essentially using Gram-Schmidt orthogonalisation. I don't understand how this process allows you to go from, $\\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp}) \\neq \\mathbf{V}$ to $\\exists \\mathbf{u} \\in \\mathbf{U} : \\mathbf{u} \\perp \\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp})$. So my question would be how does this implication work?\n\n\u2022 Possible duplicate of math.stackexchange.com/questions/878438/\u2026\n\u2013\u00a0Surb\nAug 13, 2014 at 13:34\n\u2022 It seems to me that such statement holds only if $U$ is closed -- this is satisfied if everything is finite dimensional, for example.. Aug 13, 2014 at 13:35\n\u2022 Here's a counter-example for infinite dimensional spaces: math.stackexchange.com/questions/636517/\u2026\n\u2013\u00a0Surb\nAug 13, 2014 at 13:37\n\u2022 Sure, the book I'm using is only on finite dimensional vector spaces. Aug 13, 2014 at 13:37\n\u2022 I don't think this question is a duplicate of that one because the proof in question is quite different, that one is constructive, this one is a proof by contradiction. Aug 13, 2014 at 13:41\n\nYou pick an orthonormal basis for $\\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp}) \\neq \\mathbf{V}$, say $e_j$, $1\\leq j\\leq n$. Extend this to an orthonormal basis for $\\mathbf{V}$, $e_j$, $1\\leq j\\leq m$. Since $\\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp}) \\neq \\mathbf{V}$, $n<m$. But then $u=e_{n+1}\\perp\\text{span}(\\mathbf{U}, \\mathbf{U}^{\\perp})$, by construction, and thus $u\\perp \\mathbf{U}$, i.e. $u\\in \\mathbf{U}^\\perp$, and thus $u\\perp u$, i.e. $\\|u\\|=0$. This is a contradiction, since $\\|u\\|=1$.\nNote that $$Null(U)=U^\\perp$$ now by the rank-nullity theorem we have $$Null(U)+Rank(U)=n$$, where n is the dimension of $$\\mathbb{R}^n$$. Now we can use basis extension to combine the basis for $$U$$ and $$U^\\perp$$. Since $$U+U^\\perp$$ is a subspace of $$\\mathbb{R}^n$$, and the basis for $$U+U^\\perp$$ has n vectors (by rank nullity thorem before), the same amount of vectors needed to span $$\\mathbb{R}^n$$, by the $$Dimension\\text{ }Theorem$$ the basis for $$U\\cup U^\\perp$$ is also a basis to $$\\mathbb{R}^n$$ (It takes any set of n linearly independent vectors for them to span a subspace of dimension n). Therefore $$U\\cup U^\\perp$$=$$\\mathbb{R}^n$$ I'm not sure how rigorous this proof is, I like the one above better as well. I just forgot this and looked up a proof, but also wanted to find my own after reading Jonas'. This idea definitely works and I thought I'd share it as an alternative.", "date": "2022-05-25 13:57:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/896186/how-can-i-prove-that-the-span-of-a-subspace-and-its-orthogonal-complement-is-the", "openwebmath_score": 0.9446610808372498, "openwebmath_perplexity": 80.03521669993887, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810949854636, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8573737344147109}}
{"url": "https://www.cs.utexas.edu/users/flame/laff/alaff/chapter10-launch.html", "text": "## Unit10.1.1Subspace iteration with a Hermitian matrix\n\nThe idea behind subspace iteration is to perform the Power Method with more than one vector in order to converge to (a subspace spanned by) the eigenvectors associated with a set of eigenvalues.\n\nWe continue our discussion by restricting ourselves to the case where $A \\in \\Cmxm$ is Hermitian. Why? Because the eigenvectors associated with distinct eigenvalues of a Hermitian matrix are mutually orthogonal (and can be chosen to be orthonormal), which will simplify our discussion. Here we repeat the Power Method:\n\n\\begin{equation*} \\begin{array}{lll} v_0 := \\mbox{ random vector} \\\\ v_0^{(0)} := v_0 / \\| v_0 \\|_2 \\amp \\amp \\mbox{ normalize to have length one } \\\\ {\\bf for~} k:=0, \\ldots \\\\ ~~~ v_0 := A v_0^{(k)} \\\\ ~~~ v_0^{(k+1)} := v_0 / \\| v_0 \\|_2 \\amp \\amp \\mbox{ normalize to have length one } \\\\ {\\bf endfor} \\end{array} \\end{equation*}\n\nIn previous discussion, we used $v^{(k)}$ for the current approximation to the eigenvector. We now add the subscript to it, $v_0^{(k)} \\text{,}$ because we will shortly start iterating with multiple vectors.\n\n###### Homework10.1.1.1.\n\nYou may want to start by executing  git pull to update your directory  Assignments.\n\nExamine Assignments/Week10/matlab/PowerMethod.m which implements\n\n[ lambda_0, v0 ] = PowerMethod( A, x, maxiters, illustrate, delay )\n\n\nThis routine implements the Power Method, starting with a vector x for a maximum number of iterations maxiters or until convergence, whichever comes first. To test it, execute the script in Assignments/Week10/matlab/test_SubspaceIteration.m which uses the Power Method to compute the largest eigenvalue (in magnitude) and corresponding eigenvector for an $m \\times m$ Hermitian matrix $A$ with eigenvalues $1, \\ldots, m \\text{.}$\n\nBe sure to click on \"Figure 1\" to see the graph that is created.\n\nSolution\n\nWatch the video regarding this problem on YouTube: https://youtu.be/8Bgf1tJeMmg. (embedding a video in a solution seems to cause PreTeXt trouble...)\n\nRecall that when we analyzed the convergence of the Power Method, we commented on the fact that the method converges to an eigenvector associated with the largest eigenvalue (in magnitude) if two conditions are met:\n\n\u2022 $\\vert \\lambda_0 \\vert \\gt \\vert \\lambda_1 \\vert \\text{.}$\n\n\u2022 $v_0^{(0)}$ has a component in the direction of the eigenvector, $x_0 \\text{,}$ associated with $\\lambda_0 \\text{.}$\n\nA second initial vector, $v_1^{(0)} \\text{,}$ does not have a component in the direction of $x_0$ if it is orthogonal to $x_0 \\text{.}$ So, if we know $x_0 \\text{,}$ then we can pick a random vector, subtract out the component in the direction of $x_0 \\text{,}$ and make this our vector $v_1^{(0)}$ with which we should be able to execute the Power Method to find an eigenvector, $x_1 \\text{,}$ associated with the eigenvalue that has the second largest magnitude, $\\lambda_1$ . If we then start the Power Method with this new vector (and don't introduce roundoff error in a way that introduces a component in the direction of $x_0$), then the iteration will home in on a vector associated with $\\lambda_1$ (provided $A$ is Hermitian, $\\vert \\lambda_1 \\vert \\gt \\vert \\lambda_2 \\vert \\text{,}$ and $v_1^{(0)}$ has a component in the direction of $x_1 \\text{.}$) This iteration would look like\n\n\\begin{equation*} \\begin{array}{lll} x_0 := x_0 / \\| x_0 \\|_2 \\amp \\mbox{ } \\amp \\mbox{ normalize known eigenvector } x_0 \\mbox{ to have length one } \\\\ v_1 := \\mbox{ random vector} \\\\ v_1 := v_1 - x_0^Hv_1 x_0 \\amp \\amp \\mbox{ make sure the vector is orthogonal to } x_0 \\\\ v_1^{(0)} := v_1 / \\| v_1 \\|_2 \\amp \\amp \\mbox{ normalize to have length one } \\\\ {\\bf for~} k:=0, \\ldots \\\\ ~~~ v_1 := A v_1^{(k)} \\\\ ~~~ v_1^{(k+1)} := v_1 / \\| v_1 \\|_2 \\amp \\amp \\mbox{ normalize to have length one } \\\\ {\\bf endfor} \\end{array} \\end{equation*}\n###### Homework10.1.1.2.\n\nCopy Assignments/Week10/matlab/PowerMethod.m into PowerMethodLambda1.m. Modify it by adding an input parameter x0, which is an eigenvector associated with $\\lambda_0$ (the eigenvalue with largest magnitude).\n\n[ lambda_1, v1 ] = PowerMethodLambda1( A, x, x0, maxiters, illustrate, delay )\n\n\nThe new function should subtract this vector from the initial random vector as in the above algorithm.\n\nModify the appropriate line in Assignments/Week10/matlab/test_SubspaceIteration.m, changing (0) to (1), and use it to examine the convergence of the method.\n\nWhat do you observe?\n\nSolution\n\nWatch the video regarding this problem on YouTube: https://youtu.be/48HnBJmQhX8. (embedding a video in a solution seems to cause PreTeXt trouble...)\n\nBecause we should be concerned about the introduction of a component in the direction of $x_0$ due to roundoff error, we may want to reorthogonalize with respect to $x_0$ in each iteration:\n\n\\begin{equation*} \\begin{array}{lll} x_0 := x_0 / \\| x_0 \\|_2 \\amp \\mbox{ } \\amp \\mbox{ normalize known eigenvector } x_0 \\mbox{ to have length one } \\\\ v_1 := \\mbox{ random vector} \\\\ v_1 := v_1 - x_0^Hv_1 x_0 \\amp \\amp \\mbox{ make sure the vector is orthogonal to } x_0 \\\\ v_1^{(0)} := v_1 / \\| v_1 \\|_2 \\amp \\amp \\mbox{ normalize to have length one } \\\\ {\\bf for~} k:=0, \\ldots \\\\ ~~~ v_1 := A v_1^{(k)} \\\\ ~~~ v_1 := v_1 - x_0^Hv_1 x_0 \\amp \\amp \\mbox{ make sure the vector is orthogonal to } x_0 \\\\ ~~~ v_1^{(k+1)} := v_1 / \\| v_1 \\|_2 \\amp \\amp \\mbox{ normalize to have length one } \\\\ {\\bf endfor} \\end{array} \\end{equation*}\n###### Homework10.1.1.3.\n\nCopy PowerMethodLambda1.m into PowerMethodLambda1Reorth.m and modify it to reorthogonalize with respect to x0:\n\n[ lambda_1, v1 ] = PowerMethodLambda1Reorth( A, x, v0, maxiters, illustrate, delay  );\n\n\nModify the appropriate line in Assignments/Week10/matlab/test_SubspaceIteration.m, changing (0) to (1), and use it to examine the convergence of the method.\n\nWhat do you observe?\n\nSolution\n\nWatch the video regarding this problem on YouTube: https://youtu.be/YmZc2oq02kA. (embedding a video in a solution seems to cause PreTeXt trouble...)\n\nWe now observe that the steps that normalize $x_0$ to have unit length and then subtract out the component of $v_1$ in the direction of $x_0 \\text{,}$ normalizing the result, are exactly those performed by the Gram-Schmidt process. More generally, it is is equivalent to computing the QR factorization of the matrix $\\left( \\begin{array}{c|c} x_0 \\amp v_1 \\end{array} \\right).$ This suggests the algorithm\n\n\\begin{equation*} \\begin{array}{l} v_1 := \\mbox{ random vector} \\\\ ( \\left( \\begin{array}{c | c} x_0 \\amp v_1^{(0)} \\end{array} ), R \\right) := {\\rm QR}( \\left( \\begin{array}{c | c} x_0 \\amp v_1 \\end{array} \\right) ) \\\\ {\\bf for~} k:=0, \\ldots \\\\ ~~~( \\left( \\begin{array}{c | c} x_0 \\amp v_1^{(k+1)} \\end{array} \\right), R ) := {\\rm QR}( \\left( \\begin{array}{c | c} x_0 \\amp A v_1^{(k)} \\end{array} \\right) ) \\\\ {\\bf endfor} \\end{array} \\end{equation*}\n\nObviously, this redundantly normalizes $x_0 \\text{.}$ It puts us on the path of a practical algorithm for computing the eigenvectors associated with $\\lambda_0$ and $\\lambda_1 \\text{.}$\n\nThe problem is that we typically don't know $x_0$ up front. Rather than first using the power method to compute it, we can instead iterate with two random vectors, where the first converges to a vector associated with $\\lambda_0$ and the second to one associated with $\\lambda_1 \\text{:}$\n\n\\begin{equation*} \\begin{array}{l} v_0 := \\mbox{ random vector} \\\\ v_1 := \\mbox{ random vector} \\\\ ( \\left( \\begin{array}{c | c} v_0^{(0)}\\amp v_1^{(0)} \\end{array} ), R \\right) := {\\rm QR}( \\left( \\begin{array}{c | c} v_0 \\amp v_1 \\end{array} \\right) ) \\\\ {\\bf for~} k:=0, \\ldots \\\\ ~~~( \\left( \\begin{array}{c | c} v_0^{(k+1)} \\amp v_1^{(k+1)} \\end{array} \\right), R ) := {\\rm QR}( A \\left( \\begin{array}{c | c} v_0^{(k)} \\amp v_1^{(k)} \\end{array} \\right) ) \\\\ {\\bf endfor} \\end{array} \\end{equation*}\n\nWe observe:\n\n\u2022 If $\\vert \\lambda_0 \\vert \\gt \\vert \\lambda_1 \\vert \\text{,}$ the vectors $v_0^{(k)}$ will converge linearly to a vector in the direction of $x_0$ at a rate dictated by the ratio $\\vert \\lambda_1 \\vert / \\vert \\lambda_0 \\vert \\text{.}$\n\n\u2022 If $\\vert \\lambda_0 \\vert \\gt \\vert \\lambda_1 \\vert \\gt \\vert \\lambda_2 \\vert, \\text{,}$ the vectors $v_1^{(k)}$ will converge linearly to a vector in the direction of $x_1$ at a rate dictated by the ratio $\\vert \\lambda_2 \\vert / \\vert \\lambda_1 \\vert \\text{.}$\n\n\u2022 If $\\vert \\lambda_0 \\vert \\geq \\vert \\lambda_1 \\vert \\gt \\vert \\lambda_2 \\vert$ then $\\Span( \\{ v_0^{(k)}, v_1^{(k)} \\} )$ will eventually start approximating the subspace $\\Span( \\{ x_0, x_1 \\} ) \\text{.}$\n\nWhat we have described is a special case of subspace iteration. The associated eigenvalue can be approximated via the Rayleigh quotient:\n\n\\begin{equation*} \\lambda_0 \\approx \\lambda_0^{(k)} = {v_0^{(k)}}^H A v_0^{(k)} \\mbox{ and } \\lambda_1 \\approx \\lambda_1^{(k)} = {v_1^{(k)}}^H A v_1^{(k)} \\end{equation*}\n\nAlternatively,\n\n\\begin{equation*} A^{(k)} = \\left( \\begin{array}{c| c} v_0^{(k)} \\amp v_1^{(k)} \\end{array} \\right)^H A \\left( \\begin{array}{c| c} v_0^{(k)} \\amp v_1^{(k)} \\end{array} \\right) \\mbox{ converges to } \\left( \\begin{array}{c| c} \\lambda_0 \\amp 0 \\\\ \\hline 0 \\amp \\lambda_1 \\end{array} \\right) \\end{equation*}\n\nif $A$ is Hermitian, $\\vert \\lambda_1 \\vert \\gt \\vert \\lambda_2 \\vert \\text{,}$ and $v^{(0)}$ and $v^{(1)}$ have components in the directions of $x_0$ and $x_1 \\text{,}$ respectively.\n\nThe natural extention of these observations is to iterate with $n$ vectors:\n\n\\begin{equation*} \\begin{array}{l} \\widehat V := \\mbox{ random } m \\times n \\mbox{ matrix} \\\\ ( \\widehat V^{(0)}, R ) := {\\rm QR}( \\widehat V ) \\\\ A^{(0)} = { V^{(0)}}^H A V^{(0)} \\\\ {\\bf for~} k:=0, \\ldots \\\\ ~~~( \\widehat V^{(k+1)}, R ) := {\\rm QR}( A \\widehat V^{(k)} ) \\\\ ~~~ A^{(k+1)} = {\\widehat V^{(k+1)}~}^H A \\widehat V^{(k+1)} \\\\ {\\bf endfor} \\end{array} \\end{equation*}\n\nBy extending the reasoning given so far in this unit, if\n\n\u2022 $A$ is Hermitian,\n\n\u2022 $\\vert \\lambda_0 \\vert \\gt \\vert \\lambda_1 \\vert \\gt \\cdots \\gt \\vert \\lambda_{n-1} \\vert \\gt \\vert \\lambda_{n} \\vert \\text{,}$ and\n\n\u2022 each $v_j$ has a component in the direction of $x_j \\text{,}$ an eigenvector associated with $\\lambda_j \\text{,}$\n\nthen each $v_j^{(j)}$ will converge to a vector in the direction $x_j \\text{.}$ The rate with which the component in the direction of $x_p \\text{,}$ $0 \\leq p \\lt n \\text{,}$ is removed from $v_j^{(k)} \\text{,}$ $n \\leq j \\lt m \\text{,}$ is dictated by the ratio $\\vert \\lambda_p \\vert / \\vert \\lambda_j \\vert \\text{.}$\n\nIf some of the eigenvalues have equal magnitude, then the corresponding columns of $\\widehat V^{(k)}$ will eventually form a basis for the subspace spanned by the eigenvectors associated with those eigenvalues.\n\n###### Homework10.1.1.4.\n\nCopy PowerMethodLambda1Reorth.m into SubspaceIteration.m and modify it to work with an $m \\times n$ matrix $V \\text{:}$\n\n[ Lambda, V ] = SubspaceIteration( A, V, maxiters, illustrate, delay  );\n\n\nModify the appropriate line in Assignments/Week10/matlab/test_SubspaceIteration.m, changing (0) to (1), and use it to examine the convergence of the method.\n\nWhat do you observe?\n\nSolution\n\nWatch the video regarding this problem on YouTube: https://youtu.be/Er7jGYs0HbE. (embedding a video in a solution seems to cause PreTeXt trouble...)", "date": "2021-05-09 05:31:00", "meta": {"domain": "utexas.edu", "url": "https://www.cs.utexas.edu/users/flame/laff/alaff/chapter10-launch.html", "openwebmath_score": 0.9999790191650391, "openwebmath_perplexity": 1336.0382696372924, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109503004294, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8573737331495859}}
{"url": "http://mathematica.stackexchange.com/questions/33652/uniformly-distributed-n-dimensional-probability-vectors-over-a-simplex", "text": "# Uniformly distributed n-dimensional probability vectors over a simplex\n\nWhat's the right way to generate a random probability vector $p={p_1,\\ldots,p_n} \\in {(0,1)}^n$ where $\\sum_i p_i=1$, uniformly distributed over the $(n-1)$-dimensional simplex?\n\nWhat I have is\n\nIntervals = Table[{0, 1}, {i, n}]\nRandomPoint := Block[{a},\na = RandomVariate[UniformDistribution[Intervals]];\na/Total[a]];\n\n\nBut I am unsure that this is correct. In particular, I'm unsure that it's any different from:\n\nRandomPoint := Block[{a},\na = Table[Random[], {i, n}];\na/Total[a]];\n\n\nAnd the latter clearly will not distribute vectors uniformly. Is the first code the right one?\n\n-\nThis question may be relevant. \u2013\u00a0 Sjoerd C. de Vries Oct 8 '13 at 11:21\nThanks, @SjoerdC.deVries. That question seems to suggest that my first code is also incorrect? I'm assuming that that bunch of smart guys would have stumbled upon it. \u2013\u00a0 Schiphol Oct 8 '13 at 11:42\nPerhaps DirichletDistribution might help? \u2013\u00a0 chuy Oct 8 '13 at 14:03\nThat question involved points on a sphere. Your constraint of $\\sum{p_i}=1$ is different. \u2013\u00a0 Sjoerd C. de Vries Oct 8 '13 at 14:39\nI agree with chuy. From wikipedia: Dirichlet Distribution\u200c\u200b, we have: \"When alpha = 1, the symmetric Dirichlet distribution is equivalent to a uniform distribution over the open standard K-1-simplex, i.e. it is uniform over all points in its support\". Using DirichletDistribution is most likely then also the best implementation for most purposes. \u2013\u00a0 Jacob Akkerboom Oct 8 '13 at 15:26\n\n#/Total[#,{2}]&@Log@RandomReal[{0,1},{m,n}] will give you a sample of m points from a uniform distribution over an n-1-dimensional regular simplex. (An equilateral triangle is a 2-dimensional regular simplex.) Here's what m = 2000, n = 3 should look like, where {x,y} = {p[[2]]-p[[1]], Sqrt@3*p[[3]]} are the barycentric coordinates of the 3-element probability vector p:\nHere's what you get if you omit the Log@ and normalize Uniform(0,1) variables, which is what both of the OP's examples do:\nSee for yourself. Try it with n = 2 and make a histogram of p[[1]]. Or use n = 3 and ListPlot the barycentric coordinates: {x,y} = {p[[2]]-p[[1]],Sqrt@3*p[[3]]}. \u2013\u00a0 Ray Koopman Oct 9 '13 at 18:41\nYou can also use Mathematica's built-in DirichletDistribution: points = RandomVariate[DirichletDistribution[{1, 1, 1}], 2000] /. v_?VectorQ :> {v[[2]] - v[[1]], Sqrt[3] (1 - Total[v])}; and then ListPlot[points]. \u2013\u00a0 chuy Oct 11 '13 at 18:28", "date": "2015-09-02 02:42:17", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/33652/uniformly-distributed-n-dimensional-probability-vectors-over-a-simplex", "openwebmath_score": 0.5547344088554382, "openwebmath_perplexity": 2134.641530720858, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109518607061, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.857373732854727}}
{"url": "https://mathzsolution.com/use-of-without-loss-of-generality/", "text": "# Use of \u201cwithout loss of generality\u201d\n\nWhy do we use \u201cwithout loss of generality\u201d when writing proofs?\nIs it necessary or convention? What \u201csynonym\u201d can be used?\n\nI think this is great question, as the mathematical use of \u201cwithout loss of generality\u201d often varies from its literal meaning. The literal meaning is when you rephrase a general statement\n\n$$P(x)P(x)$$ is true for all $$x\u2208Sx \\in S$$,\n\nusing another set (which is easier to work with)\n\n$$P(z)P(z)$$ is true for all $$z\u2208Tz \\in T$$,\n\nwhere $$PP$$ is some property of elements in $$SS$$ and $$TT$$, and it can be shown (or is known) that $$S=TS=T$$.\n\nFor example:\n\n\u2022 We want to show that $$P(x)P(x)$$ is true for all $$x\u2208Zx \\in \\mathbb{Z}$$. Without loss of generality, we can assume that $$x=z+1x=z+1$$ for some $$z\u2208Zz \\in \\mathbb{Z}$$. [In this case, $$S=ZS=\\mathbb{Z}$$ and $$T={z+1:z\u2208Z}T=\\{z+1:z \\in \\mathbb{Z}\\}$$.]\n\n\u2022 We want to show that $$P(x)P(x)$$ is true for all $$x\u2208Zx \\in \\mathbb{Z}$$. Without loss of generality, we can assume that $$x=5q+rx=5q+r$$ where $$q,r\u2208Zq,r \\in \\mathbb{Z}$$ and $$0\u2264r. [In this case, $$S=ZS=\\mathbb{Z}$$ and $$T={5q+r:q\u2208Z and r\u2208Z and 0\u2264r.]\n\nIn the above instances, indeed no generality has been lost, since in each case we can prove $$S=TS=T$$ (or, more likely, it would be assumed that the reader can deduce that $$S=TS=T$$). I.e., proving that $$P(z)P(z)$$ holds for $$z\u2208Tz \\in T$$ is the same as proving that $$P(x)P(x)$$ holds for $$x\u2208Sx \\in S$$.\n\nThe above cases are examples of clear-cut legitimate usage of \"without loss of generality\", but there is a widespread second use. Wikipedia writes:\n\nThe term is used before an assumption in a proof which narrows the premise to some special case; it is implied that the proof for that case can be easily applied to all others (or that all other cases are equivalent). Thus, given a proof of the conclusion in the special case, it is trivial to adapt it to prove the conclusion in all other cases.\n\n[emphasis mine.]\n\nSo, paradoxically, \"without loss of generality\" is often used to highlight when the author has deliberately lost generality in order to simplify the proof. Thus, we are rephrasing a general statement:\n\n$$P(x)P(x)$$ is true for all $$x\u2208Sx \\in S$$,\n\nas\n\n$$P(z)P(z)$$ is true for all $$z\u2208Tz \\in T$$, and\n\nif $$x\u2208Sx \\in S$$, then there exists $$z\u2208Tz \\in T$$ for which $$P(x)P(x)$$ is true if $$P(z)P(z)$$ is true.\n\nFor example:\n\n\u2022 Let $$SS$$ be a set of groups of order $$nn$$. We want to show $$P(G)P(G)$$ is true for all $$G\u2208SG \\in S$$. Without loss of generality, assume the underlying set of $$GG$$ is $${0,1,\u2026n\u22121}\\{0,1,\\ldots n-1\\}$$ for some $$n\u22651n \\geq 1$$. [Here, $$TT$$ is a set of groups with underlying set $${0,1,\u2026n\u22121}\\{0,1,\\ldots n-1\\}$$ that are isomorphic to groups in $$SS$$, and the reader is assumed to be able to deduce that property $$PP$$ is preserved by isomorphism.]\n\nMy personal preference is to replace the second case with:\n\n\"It is sufficient to prove $$P(z)P(z)$$ for $$z\u2208Tz \\in T$$, since [[for some reason]] it follows that $$P(x)P(x)$$ is true for all $$x\u2208Sx \\in S$$.\"", "date": "2022-10-04 09:07:04", "meta": {"domain": "mathzsolution.com", "url": "https://mathzsolution.com/use-of-without-loss-of-generality/", "openwebmath_score": 0.9475861191749573, "openwebmath_perplexity": 136.7914881701344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109480714625, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8573737245961268}}
{"url": "https://www.physicsforums.com/threads/integration-problems-intriguing-integrals.226048/", "text": "# Integration problems: Intriguing Integrals\n\n1. Apr 2, 2008\n\n### BioCore\n\n1. The problem statement, all variables and given/known data\n$$\\int$$x^3$$\\sqrt{}a^2-x^2$$dx\n\n2. Relevant equations\n\n3. The attempt at a solution\nI came this far with my solution:\n\na^5$$\\sqrt{}sin^3\\Theta cos^2\\Theta$$d$$\\Theta$$\na^5$$\\sqrt{}sin^3\\Theta-sin^5\\Theta$$d$$\\Theta$$\n\nThe answer given to me is:\n\n-1/15(a^2 - x^2)^3/2 (2a^2 + 3x^2) + C\n\nIt seems that they used sine theta, and not cos theta. I have currently tried to do that but am not sure how I would get rid of the extra sin$$\\Theta$$.\n\nThanks for the help.\n1. The problem statement, all variables and given/known data\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nLast edited by a moderator: Apr 2, 2008\n2. Apr 2, 2008\n\n### rock.freak667\n\n$$\\int x^3 \\sqrt{a^2-x^2}dx$$\n\nlet $x=acos\\theta \\Rightarrow dx=-asin\\theta$\n\n$$\\int x^3 \\sqrt{a^2-x^2}dx \\equiv \\int (acos\\theta)^3\\sqrt{a^2-a^2cos^2\\theta} \\times -asin\\theta d\\theta$$\n\n$$\\int - a^4cos^3\\theta \\sqrt{a^2(1-cos^2\\theta)} \\times -asin\\theta d\\theta$$\n\n$$-a^5 \\int cos^3\\theta sin^2\\theta d\\theta$$\n\nrecall that $sin^2\\theta +cos^2\\theta=1$\n\nand substitute for $sin^2\\theta$\n\n3. Apr 2, 2008\n\n### BioCore\n\nI would assume that using your solution I would have the right answer. The only problem is that in the solutions my Professor gave us, he has an intermediate step before the final solution such as this:\n\n$$a^5 \\int sin^3\\theta cos^2\\theta d\\theta$$\n\nAlso in explaining to us Trig Substitution he placed $$\\sqrt{a^2 - x^2}$$ as the bottom portion of the right triangle. Thus he stated that\n\nx = asin$$\\Theta$$\ndx = acos$$\\Theta$$\n$$\\sqrt{a^2 - x^2}$$ = acos\\theta\n\nLast edited by a moderator: Apr 2, 2008\n4. Apr 2, 2008\n\n### rock.freak667\n\nYour teacher used x=sin$\\theta$, I used x=cos$\\theta$.\nSo in his triangle, $sin\\theta=\\frac{x}{a}$ which would make the adjacent side,$\\sqrt{a^2-x^2}$. At the end of that integral you would get terms involving $cos\\theta$\n\nIn my substitution, the opposite side would be $\\sqrt{a^2-x^2}$ and at the end of my integral I would have terms involving $sin\\theta$\n\nSo in theory it should all work out the same.\n\n5. Apr 2, 2008\n\n### BioCore\n\n$$-a^5 \\int \\sqrt{1 - sin^2\\theta} (1 - sin^2\\theta) (sin^2\\theta)d\\theta$$\n\n= $$-a^5 \\int (1 - sin^2\\theta)^3/2 (sin^2\\theta)d\\theta$$\n\n= $$-a^5 \\int sin^3\\theta - sin^6\\theta d\\theta$$\n\n= $$-a^5 [1/4sin^4\\theta + 1/7sin^7\\theta ]$$ +C\n\nNow I know that I should substitute for the sine function, but is this so far correct?", "date": "2018-03-21 06:01:56", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integration-problems-intriguing-integrals.226048/", "openwebmath_score": 0.8052039742469788, "openwebmath_perplexity": 1208.800729624657, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620550745212, "lm_q2_score": 0.8933094110250333, "lm_q1q2_score": 0.857364476142796}}
{"url": "http://math.stackexchange.com/questions/251370/proving-an-algorithms-correctness-in-determining-the-number-of-1-bits-in-a-bit", "text": "# Proving an algorithm's correctness in determining the number of 1 bits in a bit string\n\n   procedure bit count(S: bit string)\ncount := 0\nwhile S != 0\ncount := count + 1\nS := S \u2227 (S \u2212 1)\nreturn count {count is the number of 1s in S}\n\n\nHere S-1 is the bit string obtained by changing the rightmost 1 bit of S to a 0 and all the 0 bits to right of this to 1s.\n\nSo I understand why this is correct, and I have write a rough explanation;\n\nAfter every iteration, the rightmost 1 bit in S, as well as all the bits to the right of it, is set equal to 0. Thus, after each iteration, the next right-most 1 is accounted for and set to 0, until the entire string is 0's and the loop breaks with the count equal to the number of 1's.\n\nI know this kind of answer won't pass in any mathematics community, so I was hoping to write a formal proof, but I don't know how to go about doing that. My proof skills are particularly shoddy, so an explanation of the techniques involved would be greatly appreciated.\n\n-\nIs ^ and or xor? \u2013\u00a0copper.hat Dec 5 '12 at 7:04\n@copper.hat: It has to be $\\land$ for the algorithm to work. \u2013\u00a0Brian M. Scott Dec 5 '12 at 7:05\n\nHere\u2019s one way to make your informal explanation a bit more formal.\n\nLet $\\sigma=b_1b_2\\dots b_n$ be a bit string of length $n$. For convenience let $[n]=\\{1,\\dots,n\\}$. If $\\sigma$ is not the $0$ string, let $k=\\max\\{i\\in[n]:b_i=1\\}$, so that $b_k=1$ and $b_i=0$ for $i>k$. Then $\\sigma-1$ is the bit string $a_1a_2\\dots a_n$ such that for each $i\\in[n]$\n\n$$a_i=\\begin{cases} b_i,&\\text{if }i<k\\\\ 0,&\\text{if }i=k\\\\ 1,&\\text{if }i>k\\;. \\end{cases}$$\n\nThen $$\\sigma\\land(\\sigma-1)=(b_1\\land a_1)(b_2\\land a_2)\\ldots(b_n\\land a_n)\\;,$$ and for $i\\in[n]$ we have\n\n$$b_i\\land a_i=\\begin{cases} b_i,&\\text{if }i<k\\\\ 0,&\\text{if }i\\ge k\\;: \\end{cases}$$\n\n$b_k\\land a_k=1\\land 0=0$, and $b_i\\land a_i=0\\land 1=0$ for each $i>k$. In other words, if $$\\sigma\\land(\\sigma-1)=c_1c_2\\dots c_n\\;,$$ then\n\n$$c_i=\\begin{cases} b_i,&\\text{if }i<k\\\\ 0,&\\text{if }i\\ge k\\;. \\end{cases}$$\n\nRecalling that $b_k=1$ and $b_i=0$ for $i>k$, we see that $\\sigma$ and $\\sigma\\land(\\sigma-1)$ differ only in the $k$-th bit, where $\\sigma$ has a $1$ and $\\sigma\\land(\\sigma-1)$ has a $0$.\n\nIf $|\\sigma|_1$ is the number of $1$\u2019s in $\\sigma$, we\u2019ve just shown that\n\n$$|\\sigma\\land(\\sigma-1)|_1=|\\sigma|_1-1\\;.$$\n\nThus, each pass through the loop adds $1$ to the count and reduces the number of $1$\u2019s in the string by $1$. (In other words, $\\text{count}+|\\sigma|_1$ is a loop invariant for this loop.) Since $\\text{count}$ starts at $0$ and $|\\sigma|_1$ at the number of $1$\u2019s in the input bit string, it\u2019s now clear that after $|\\sigma|_1$ passes through the loop, the input string will be the zero string, and $\\text{count}$ will be the number of $1$\u2019s in the input string. And since the input is now the zero string, we exit the loop with no further change in $\\text{count}$.\n\n-\n\nWhat you observed is already quite good. If you would like to argue more formal I would suggest to prove first a lemma that says.\n\nLemma: If $S>0$ then the binary representation of $S$ has one digit 1 more than the binary representation of $S \\land (S-1)$.\n\n$Proof.$ Let the binary representation of $S\\neq 0$ be the string $s=\\text{bin}(S)$. We assume that $s$ ends on $v=10^k$ and that $s=uv$. Subtracting 1 from $S$ gives in the binary representation a string $u01^k$. A bit-wise AND of $S$ and $S-1$ is therefore the string $u0^{k+1}$, which proves the lemma.\n\nUse this lemma and a loop invariant to prove the correctness of your algorithm.\n\n-\n\nActually, I think you have a very good start.\n\nWhat I think is needed are these ideas:\n\n(1) Explicit specifying of all the locations in S where the one bits are.\n\n(2) An analysis of what the critical step S := S \u2227 (S \u2212 1) does.\n\nHere is my take on (2).\n\nSuppose S, at this point, has k zero bits on the right with the k+1st bit a one. Then $S = a 2^k$, with a an odd integer (if $a$ were even, the j+1st bit would be zero).\n\nThen $S-1 = a 2^k-1 = (a-1)2^k + 2^k-1$ so S ^ (S-1) = $(a-1)2^k$ since (1) the k lower bits of S are zero, and (2) since a is odd, a-1 is even, so the k+1st bit of S-1 is zero. Furthermore, since a is odd, all the bits of a-1 are the same as the bits of a except for the low order bit.\n\nThus the bits of S ^ (S-1) are the same as the bits of S except that the lowest one bit has been set to zero.\n\nThis lemma (which is what it really is) allows you to set up an inductive hypothesis that each time through the loop the current low order one bit has been set to zero and count is incremented by one.\n\nAt the end, count equals the number of one bits in the original number.\n\nI have great admiration for whoever first came up with this, and I thank John Taylor for providing me with the opportunity of doing this analysis.\n\nI had seen this algorithm before, but had never seen a proof, and I enjoyed working this out.\n\nAha! I see that Brian M. Scott has answered as I was entering my answer, and that our two analyses are essentially the same with quite different notations.\n\n-", "date": "2016-06-29 23:34:13", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/251370/proving-an-algorithms-correctness-in-determining-the-number-of-1-bits-in-a-bit", "openwebmath_score": 0.8113588094711304, "openwebmath_perplexity": 258.23634932071434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620585273154, "lm_q2_score": 0.8933094003735664, "lm_q1q2_score": 0.8573644690043359}}
{"url": "https://math.stackexchange.com/questions/823816/is-sum-limits-n-1-infty-frac-sin-nnn-convergent", "text": "# Is $\\sum\\limits_{n=1}^\\infty \\frac{|\\sin n|^n}n$ convergent\uff1f\n\nIs the series\n\n$$\\sum_{n=1}^\\infty \\frac{|\\sin n|^n}n\\tag{1}$$\n\nconvergent\uff1f\n\nIf one want to use Abel's test, is\n\n$$\\sum_{n=1}^\\infty |\\sin n|^n\\tag{2}$$\n\nconvergent\uff1f\n\nThank you very much\n\nThe second series is divergent because the irrational number $\\frac{\\pi}{2}$ has an infinite number of convergents $\\frac{p_n}{q_n}$ with an odd denominator, hence $\\left|\\frac{p_n}{q_n}-\\frac{\\pi}{2}\\right|\\leq\\frac{1}{q_n^2}$ gives: $$\\left|\\sin(p_n)\\right| = \\left|\\sin\\left(\\frac{\\pi}{2}q_n+\\frac{\\theta}{q_n}\\right)\\right|=\\left|\\cos\\frac{\\theta}{q_n}\\right|,\\quad |\\theta|\\leq 1,$$ $$\\left|\\sin(p_n)\\right|\\geq 1-\\frac{1}{q_n^2}$$ hence $\\left|\\sin n\\right|^n$ is bigger that $\\left(1-\\frac{1}{n^2}\\right)^n$ infinitely often and the partial sums of $\\sum_{n\\in\\mathbb{N}}\\left|\\sin n\\right|^n$ cannot be bounded. However, they cannot grow too fast: numerical experiments give\n\n$$\\sum_{n=1}^{N}\\left|\\sin n\\right|^n=O\\left(N^{1/2}\\right)\\tag{1}$$\n\n(consistent with the Weyl bounds for the partial sums of $\\sum_{n\\in\\mathbb{N}}e^{in^2}$) that is enough to state that $$\\sum_{n=1}^{+\\infty}\\frac{\\left|\\sin n\\right|^n}{n}$$ converges by partial summation. This is also consistent with the model for which $\\left|\\sin n\\right|$ acts like $\\sin X$ where $X$ is a random variable, uniformly distributed over $[0,\\pi]$: $$E\\left[\\sum_{n=1}^{+\\infty}\\frac{(\\sin X)^n}{n}\\right]=\\frac{1}{\\pi}\\int_{0}^{\\pi}-\\log(1-\\sin\\theta)\\,d\\theta=\\frac{4K}{\\pi}+\\log 2<+\\infty.$$ A possible strategy in order to prove $(1)$ is to prove first, through Weyl's equidistribution theorem, that, provided that $M$ is big enough:\n\n$$\\sum_{n=N+1}^{N+M}\\left|\\sin n\\right|^r\\approx \\frac{M}{\\pi}\\int_{0}^{\\pi}\\sin^r\\theta\\,d\\theta,\\tag{2}$$\n\nthen exploit the fact that:\n\n$$\\int_{0}^{\\pi}\\sin^r \\theta\\,d\\theta = O\\left(\\frac{1}{\\sqrt{r}}\\right).\\tag{3}$$\n\nBy putting together $(2)$ and $(3)$ we get:\n\n$$\\sum_{i=1}^{M^2}\\left|\\sin n\\right|^n\\leq\\sum_{j=0}^{M-1}\\sum_{n=Mj+1}^{M(j+1)}\\left|\\sin n\\right|^{Mj+1}\\ll M\\cdot\\sum_{j=0}^{M-1}\\frac{1}{\\sqrt{Mj+1}}\\ll M.\\tag{4}$$\n\nMoreover, since the supremum of the derivative of $\\sin^r\\theta$ over $[0,\\pi]$ behaves like $\\sqrt{\\frac{r}{e}}$, keeping track of all the constants we get: $$\\sum_{n=1}^{+\\infty}\\frac{\\left|\\sin n\\right|^n}{n}\\leq\\frac{2}{\\sqrt{\\pi}}\\left(1+\\frac{1}{\\sqrt{e}}\\right)\\zeta(3/2) = 4.73565\\ldots$$ that is not too much far from the truth (given by partial summation and explicit numerical computations till $n=10^8$, with an error term $\\frac{C}{10^4}$):\n\n$$\\sum_{n=1}^{+\\infty}\\frac{\\left|\\sin n\\right|^n}{n}\\leq \\color{red}{2.151}.\\tag{5}$$\n\nPost-mortem addendum: it is interesting to study tailor-made numerical methods for the computation of such very slowly-converging series.\n\n\u2022 Do you have an idea on how to prove the estimate with $O(N^{1/2+\\epsilon})$? \u2013\u00a0Gabriel Romon Jun 16 '14 at 13:02\n\u2022 @G.T.R: I extended my answer including my thoughts about proving the crucial estimate. \u2013\u00a0Jack D'Aurizio Jun 16 '14 at 16:50\n\u2022 @Jack D'Aurizio: Writing in MMA Sum[Abs[Sin[n]]^n/n, {n, 1, 10^8}] I do not get an answer. How did you get to 2.151? Thank you! \u2013\u00a0TeM Jul 14 '17 at 20:10\n\u2022 @TeM: just follow the last link. \u2013\u00a0Jack D'Aurizio Jul 14 '17 at 20:11\n\u2022 It's not clear to me that this constitutes a rigorous proof; how is Weyl's equidistribution theorem sufficient here? \u2013\u00a0user41281 Nov 21 '17 at 1:36\n\nNote: Another great answer to this problem is given by Terry Tao in this MO post.\n\nDisclaimer:Not my solution. One elegant answer was posted by Robert Isreal on here . Apparently a holistic answer is not known.", "date": "2019-11-17 23:45:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/823816/is-sum-limits-n-1-infty-frac-sin-nnn-convergent", "openwebmath_score": 0.8732844591140747, "openwebmath_perplexity": 358.4389222747722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620614046438, "lm_q2_score": 0.8933093954028817, "lm_q1q2_score": 0.8573644668040058}}
{"url": "https://gmatclub.com/forum/digital-sum-concept-100534.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 13 Dec 2019, 16:04\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# DIGITAL SUM of the number\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSVP\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2477\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nDIGITAL SUM of the number\u00a0 [#permalink]\n\n### Show Tags\n\n06 Sep 2010, 18:32\n1\n9\nDIGITAL SUM of the number\n\nGiven a number $$N_z$$ , all the digits of N are added to obtain a number $$N_y$$. All the digits of N are added to obtain a number $$N_x$$, and so on, till we obtain a single digit number $$N$$. This single digit number $$N$$ is called the digital sum of the original number $$N_z$$.\n\nWhat is the digital sum of 84001 ?\nAnswer: 8+4+0+0+1 = 13 = 1+3 = 4 Hence, the digit sum of the number is 4.\n\nNote: In finding the digital sum of a number we can ignore the digit 9 or the digits that add up to 9.\nFor example, in finding the digital sum of the number 246819, we can ignore the digits 2, 6, 1, and 9. Hence,\nthe digital sum of 246819 is = 4 + 8 = 12 = 1 + 2 = 3.\n\nDigital Sum Rule of Multiplication: The digital sum of the product of two numbers is equal to the digital sum of the product of the digital sums of the two numbers.\n\nExample: The product of 129 and 35 is 4515.\nDigital sum of 129 = 3 and digital sum 35 = 8. Product of the digital sums = 3 \u00d7 8 = 24 Digital sum = 6.\nDigital sum of 4515 is = 4 + 5 + 1 + 5 = 15 = 1 + 5 = 6. Digital sum of the product of the digital sums =\ndigital sum of 24 = 6 Digital sum of the product (4515) = Digital sum of the product of the digital sums\n(24) = 6\n\nApplications of Digital Sum :\n\n1.Rapid checking of calculations while multiplying numbers :\n\nSuppose a student is trying to find the product 316 \u00d7 234 \u00d7 356, and he obtains the number 26525064. A\nquick check will show that the digital sum of the product is 3. The digital sums of the individual numbers\n(316, 234 and 356) are 1, 9, and 5. The digital sum of the product of the digital sum is 1 \u00d7 9 \u00d7 5 = 45 = 4 + 5 = 9. \u21d2 the digital sum of the product of the digital sums (9) \u2260 digital sum of the 26525064 (3). Hence,\nthe answer obtained by multiplication is not correct.\n\nNote: Although the answer of multiplication will not be correct if the digital sum of the product of the digital\nsums is not equal to digital sum of the product, but the reverse is not true i.e. the answer of multiplication\nmay or may not be correct if the digital sum of the product of the digital sums is equal to digital sum of\nthe product.\n\nSuppose you have a question : ( Very absurd eg by me but learn the concept )\n\nWhat is value is A+B+C+D+E?\n\n$$878373 * 7738838339 * 827287 = E6D35C50B5547A87623729$$\n\nHINT: Find the digital sum on LHS and equate it with that of RHS.\n\n2.Determining if a number is a perfect square or not :\n\nThe digital sum of the numbers which are perfect squares will\nalways be 1, 4, 7, or 9.\n\nNote: A number will NOT be a perfect square if its digital sum is NOT 1, 4, 7, or 9, but it may or may not\nbe a perfect square if its digital sum is 1, 4, 7, or 9.\n_________________\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\nhttp://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html\nManager\nJoined: 20 Oct 2009\nPosts: 85\nRe: DIGITAL SUM of the number\u00a0 [#permalink]\n\n### Show Tags\n\n08 Sep 2010, 22:18\ndang.... what are some real life applications of this digital sum?\nManager\nJoined: 17 Aug 2009\nPosts: 186\nRe: DIGITAL SUM of the number\u00a0 [#permalink]\n\n### Show Tags\n\n09 Sep 2010, 01:18\nThats an interesting piece of information gurpreet !\nThanks\nIntern\nJoined: 06 Oct 2009\nPosts: 47\nRe: DIGITAL SUM of the number\u00a0 [#permalink]\n\n### Show Tags\n\n17 Sep 2010, 02:41\nInteresting piece of info Gurpreet.\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 13741\nRe: DIGITAL SUM of the number\u00a0 [#permalink]\n\n### Show Tags\n\n07 Apr 2019, 00:30\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: DIGITAL SUM of the number \u00a0 [#permalink] 07 Apr 2019, 00:30\nDisplay posts from previous: Sort by", "date": "2019-12-13 23:04:52", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/digital-sum-concept-100534.html", "openwebmath_score": 0.5594190359115601, "openwebmath_perplexity": 1131.6077731528649, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232899814556, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.857363324942187}}
{"url": "https://nbviewer.jupyter.org/url/www.maths.usyd.edu.au/u/olver/teaching/MATH3976/notes/27.ipynb", "text": "# Lecture 27: Error of quadrature\u00b6\n\nLast lecture we observed the following:\n\n1. The right-hand rule had error $O(n^{-1})$ for $n$ sample point\n2. The trapezium rule had error $O(n^{-2})$ for $n$ sample point\n3. The trapezium rule had error $O(n^{-\\alpha})$ for all choices of $\\alpha \\geq 0$ for smooth, periodic functions\n\nThis lecture we set out to prove these results.\n\n## Right-hand rule convergence\u00b6\n\nRecall that the right-hand rule is defined by, for $x_k \\triangleq kh$ and $h \\triangleq 1/n$\n\n$$\\int_0^1 f(x) dx \\approx h \\sum_{k=1}^n f(x_k) = \\sum_{k=1}^n \\int_{x_{k-1}}^{x_k} (f(x) - f(x_k) ) dx$$\n\nAbout the only tool available to analyse integrals is integration by parts:\n\n\\begin{align*} \\int_a^b u(x) v'(x) dx &= [u(x) v(x)]_a^b - \\int_a^b u'(x) v(x) \\cr & = u(b) v(b) - u(a)v(a) - \\int_a^b u'(x) v(x) \\end{align*}\n\nConsider the error in the first panel. Taking $u(x) = f(x)$ and $v(x) = x$ in the integration by parts formula yields:\n\n\\begin{align*} \\int_0^h (f(x) - f(h)) dx = [(f(x) - f(h)) x]_0^h - \\int_0^h f'(x) x dx \\cr = - \\int_0^h f'(x) x dx \\end{align*}\n\nAssume that $f'(x)$ is bounded in $[0,1]$, and define\n\n$$M= \\sup_{x \\in [0,1]} |f'(x)|$$\n\nWe then get a bound on the error:\n\n\\begin{align*} \\left|\\int_0^h (f(x) - f(h)) dx\\right| &= \\left|\\int_0^h f'(x) x dx\\right| \\leq \\int_0^h |f'(x)| x dx \\leq M \\int_0^h x dx \\cr &\\leq {M h^2 \\over 2} \\end{align*}\n\n\n\n\nThis formula caries over to every other panel (by the same argument):\n\n$$\\left|\\int_{x_{k-1}}^{x_k} (f(x) - f(x_k)) dx \\right| \\leq {M h^2 \\over 2}$$\n\nThus we have\n\n\\begin{align*} \\left\\|\\int_0^1 f(x) dx - h \\sum_{k=1}^n f(x_k)\\right\\| &= \\left\\|\\sum_{k=1}^n \\int_{x_{k-1}}^{x_k} (f(x) - f(x_k) ) dx\\right\\| \\cr &\\leq \\sum_{k=1}^n \\left\\|\\int_{x_{k-1}}^{x_k} (f(x) - f(x_k) ) \\right\\| dx &\\leq \\sum_{k=1}^n {M h^2 \\over 2} = {M h^2 n \\over 2} = {M \\over 2 n} = O(n^{-1}) \\end{align*}\n\nThus we have proven that the right-hand rule converges like $O(n^{-1})$.\n\n## Trapezium rule observed convergence, revisited\u00b6\n\nRecall the trapezium rule, here implemented for general intervals $[a,b]$:\n\nIn\u00a0[2]:\nusing PyPlot\n\nfunction trap(f,a,b,n)\nh=(b-a)/n\nx=linspace(a,b,n+1)\n\nv=f(x)\nh/2*v[1]+sum(v[2:end-1]*h)+h/2*v[end]\nend\n\ntrap(f,n)=trap(f,0.,1.,n);\n\n\nConsider integration of the following simple function:\n\nIn\u00a0[3]:\nf=x->exp(x).*cos(x)\n\ng=linspace(0.,1.,1000)\nplot(g,f(g));\n\n\nAs in last lecture, we can compare the error of trap to the exact integral, approximated by quadgk.\n\nIn\u00a0[5]:\nex=quadgk(f,0.,1.)[1]\n\nns=round(Int,logspace(1,5,100))   # integers spaced logarithmically apart\nerrT=zeros(length(ns))\nerr\u03c0=zeros(length(ns))\n\nfor k=1:length(ns)\nn=ns[k]\nerrT[k]=abs(trap(f,n-1)-ex    )   # error in non-periodic\nerr\u03c0[k]=abs(trap(\u03b8->cos(cos(\u03b8-0.1)),0.,2\u03c0,n-1)-ex\u03c0    )  # error in periodic\nend\n\n\nWe see that for non-periodic functions we observe $O(n^{-2})$ convergence but for periodic functions we converge much faster:\n\nIn\u00a0[6]:\nloglog(ns,errT)   # blue curve\nloglog(ns,err\u03c0)   # green curve\nloglog(ns,(1.0ns).^(-2))   # red curve, with same slope as blue curve\n\nOut[6]:\n1-element Array{Any,1}:\nPyObject <matplotlib.lines.Line2D object at 0x304c5ed90>\n\nThe defining property of smooth, periodic functions is that the function and all its derivatives match at 0 and 2\u03c0. Now consider a function where only some of he derivatives match:\n\nIn\u00a0[8]:\nh=x->10f(x).*x.^2.*(1-x).^2\n\nplot(g,h(g));\n\n\nWe see here that the convergence rate has converged to $O(n^{-4})$:\n\nIn\u00a0[11]:\nexf=quadgk(f,0.,1.)[1]\n\nns=round(Int,logspace(1,4,100))\nerrf=zeros(length(ns))\nerrh=zeros(length(ns))\n\nfor k=1:length(ns)\nn=ns[k]\nerrf[k]=abs(trap(f,n-1)-exf    )\nerrh[k]=abs(trap(h,n-1)-exh    )\nend\n\nloglog(ns,errf)     # blue curve\nloglog(ns,1./ns.^2) # green curve\nloglog(ns,errh)     # red curve\nloglog(ns,1./ns.^4) # aqua curve;\n\n\nThe conclusion is that derivatives at the endpoints dictate the convergence rate of the Trapezium rule.\n\n# Bernoulli Polynomial\u00b6\n\nTo prove this, we need to be clever about our choice of functions when we integrate by parts. Define the first three Bernoulli Polynomials by\n\n$$B_0(x) = 1, B_1(x) = x-{1 \\over 2}, B_2(x) = x^2 -x + {1 \\over 6}$$\n\nas depticted here:\n\nIn\u00a0[12]:\ng=linspace(0.,1.,1000)\nB0=x->ones(x)\nB1=x->x-1/2\nB2=x->x.^2-x+1/6\nplot(g,B0(g))\nplot(g,B1(g))\nplot(g,B2(g));\n\n\nThese polynomials satisfy two important properties:\n\n1. $B_k'(x) = k B_{k-1}(x)$, just like monomials $x^k$\n2. $B_2(-1) = B_2(1) = {1 \\over 6} We Thus consider integration by parts with the Trapezium rule. Recall the Trapezium rule: $$\\int_0^1 f(x) dx \\approx h\\left({f(x_0) \\over 2} + \\sum_{k=1}^{n-1} f(x_k) + {f(x_n) \\over 2} \\right) = \\sum_{k=1}^n \\int_0^1 \\left(f(x_{k-1}) + (x-x_{k-1}) {f(x_k) - f(x_{k-1}) \\over h}\\right) dx$$ As in the right-hand rule, we begin with the error in the first panel: $$\\int_0^h \\left[f(x) - \\left(f(0) + x {f(h) - f(0) \\over h}\\right) \\right] dx.$$ Taking $$u(x) = f(x) - \\left(f(0) + x {f(h) - f(0) \\over h}\\right)$$ and$v(x) = h B_1(x/h)$(so that$v'(x) = 1$) in the integration by parts formula: $$\\int_0^h \\left[f(x) - \\left(f(0) + x {f(h) - f(0) \\over h}\\right) \\right] dx = [u(x) v(x)]_0^h - \\int_0^h u(x) v(x) dx = - h \\int_0^h \\left[f'(x) - {f(h) - f(0) \\over h}\\right] B_1(x/h) dx.$$ Here we used that$u(0) = u(h) = 0$to kill of the first terms. Now take$u(x) = f'(x) - {f(h) - f(0) \\over h}$and$v(x) = h {B_2(x/h) \\over 2}$(so that$v'(x) = B_1(x/h)): \\begin{align*} - h \\int_0^h \\left[f(x) - {f(h) - f(0) \\over h}\\right] B_1(x/h) dx &= -{h^2 \\over 2}\\left[\\left(f'(h) - {f(h) - f(0) \\over h}\\right) B_1(1) - \\left(f'(0) - {f(h) - f(0) \\over h}\\right) B_1(-1)\\right] \\cr & \\qquad +{h^2 \\over 2} \\int_0^h f''(x) B_2(x/h) dx \\cr & = - {f'(h) - f'(0) \\over 12} h^2 +{h^2 \\over 2} \\int_0^h f''(x) B_2(x/h) dx \\end{align*} By the same logic, we have in each panel $$\\int_{x_{k-1}}^{x_k} \\left[f(x) - \\left(f(x_{k-1}) + x {f(x_k) - f(x_{k-1}) \\over h}\\right) \\right] dx = - {f'(x_k) - f'(x_{k-1}) \\over 12} h^2 +{h^2 \\over 2} \\int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx$$ Summing over every panel, and using the telescoping sum, gives us \\begin{align*} \\int_0^1 f(x) dx - h\\left({f(x_0) \\over 2} + \\sum_{k=1}^{n-1} f(x_k) + {f(x_n) \\over 2} \\right) = -{h^2 \\over 12} \\sum_{k=1}^n \\left(f'(x_k) - f'(x_{k-1}) - \\int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx\\right) \\cr -{h^2 \\over 12} (f'(1) - f'(0)) + {h^2 \\over 12} \\sum_{k=1}^n \\int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx \\end{align*} But on each panel we have $$\\left|\\int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx\\right| \\leq \\int_{x_{k-1}}^{x_k} |f''(x)| |B_2((x-x_{k-1})/h)| dx \\leq M_2 {h \\over 6}$$ whereM_2 = \\sup_{x \\in [0,1]} |f''(x)|$and we used the fact that$|B_2(x)| \\leq {1 \\over 6}\\$ (Exercise). Thus we have\n\n$$\\left|\\sum_{k=1}^n \\int_{x_{k-1}}^{x_k} f''(x) B_2((x-x_{k-1})/h) dx\\right| \\leq {M_2 \\over 6}$$\n\nand hence we have\n\n$$\\int_0^1 f(x) dx - h\\left({f(x_0) \\over 2} + \\sum_{k=1}^{n-1} f(x_k) + {f(x_n) \\over 2} \\right) = -{1 \\over 12 n^2} (f'(1) - f'(0)) + O(n^{-2}) = O(n^{-2})$$\n\nwhich proves our second observation.", "date": "2020-06-02 08:32:52", "meta": {"domain": "jupyter.org", "url": "https://nbviewer.jupyter.org/url/www.maths.usyd.edu.au/u/olver/teaching/MATH3976/notes/27.ipynb", "openwebmath_score": 0.9999326467514038, "openwebmath_perplexity": 5945.765738219612, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232924970204, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8573633222440752}}
{"url": "https://eventthyme.net/how-to-find-average-velocity-calculus/", "text": "in\n\n# How To Find Average Velocity Calculus\n\nGiven, s = 3t2 \u2212 6t. Average velocity is a vector quantity.\n\nForce and Motion Worksheet Motion graphs, Worksheets and\n\n### Use this information to guess the instantaneous velocity of the rock at time 1.\n\nHow to find average velocity calculus. So,displacement in between 2s and 5s is s = 3[t2]5 2 \u2212 6[t]5 2 = 3(25 \u22124) \u2212 6(5 \u2212 2) = 45m. Yes, you're supposed to find the average velocity over each interval. The velocity at t = 10 is 10 m/s and the velocity at t = 11 is 15 m/s.\n\nTo find the instantaneous velocity at any position, we let t1=t t 1 = t and t2=t+\u03b4t t 2 = t + \u03b4 t. [0,3] b.[0,2] c.[0,1] d.[0,h] where h>0 is a real number Nevertheless, this is exactly how a gps determines velocity from position!\n\nVelocity calculations used in calculator: For example, if you drive a car for a distance of 70 miles in one hour, your average velocity equals 70 mph. Solving for the different variables we can use the.\n\nThe average velocity formula describes the relationship between the length of your route and the time it takes to travel. To avoid these tedious calculations, we would really like to have a formula. Average velocity is defined as total displacement/ total time taken for that.\n\nIt is determined that its height (in feet) above ground t seconds later (for is given by find the average velocity of the rock over each of the given time intervals. Her average velocity is 8m west / 4s = 2 m/s west. = 12 miles per hour.\n\nFor example, if an object is tossed into the air we might find the following data for the height in feet, y, of the object as a function of the time in seconds, t, where t = 0 is when the object is released upward. How to find the change in position over the change in time for various intervals, and finding the slope of the secant line to find average velocity. If you recall from earlier mathematics studies, average velocity is just net distance traveled divided by time.\n\nHowever, this technically only gives the object's average velocity over its path. Average velocity is defined in terms of the relationship between the distance traveled and the time that it takes to travel that distance. Using calculus, it's possible to calculate an object's velocity.\n\nAlthough speed and velocity are often words used interchangeably, in physics, they are distinct concepts. So,average velocity = 45 5 \u22122 = 15ms\u22121. In other words, velocity is equal to rate of change of position vector with respect to time.\n\nThe average acceleration would be: I was showed the correct answer but i really need to learn what the process is to achieve the answer. The sum of the initial and final velocity is divided by 2 to find the average.\n\nFor example, let\u2019s calculate a using the example for constant a above. If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Bart walks west at 5 m/s for 3 seconds, then turns around and walks east at 7 m/s for 1 second.\n\nAverage velocity = 8 m west / 4s = 2 m/s west. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2. The average velocity of an object is its change in position divided by the total amount of time taken.\n\nPurchase calculus etf 6e hide menu show menu. In many common situations, to find velocity, we use the equation v = s/t, where v equals velocity, s equals the total displacement from the object's starting position, and t equals the time elapsed. (i) 0.1 seconds ft/s (ii) 0.05 seconds ft/s (iii) 0.01 seconds ft/s (b) estimate the instantaneous velocity (in ft/s) of the pebble after 2 seconds.\n\nUsing calculus to find acceleration. The average velocity formula and velocity units. An automobile travels 540 kilometers in 4 hours and 30 minutes.\n\nLet's take a look at average velocity. Computing average velocities for smaller, and smaller, values of as we did above is tedious. Find the average velocity of the object over the following intervals.\n\nJane \u00b7 3 \u00b7 mar 17 2018. The formula for finding average velocity is: Math video on how to compute the average velocity of an object moving in one dimension and how to represent average velocity on a graph.\n\nThe instantaneous velocity at an instant t or simply \u2018velocity\u2019 at an instant t is defined as limiting value of the average velocity as \u03b4t \u2192 0, evaluated at time t. The average velocity of a body in a certain time interval is given as the displacement of the body in that time interval divided by time. Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 2 and lasting the following amount of time.\n\nThe formula for calculating average velocity is therefore: Acceleration is measured as the change in velocity over change in time (\u03b4v/\u03b4t), where \u03b4 is shorthand for \u201cchange in\u201d. In the previous section, we have introduced the basic velocity equation, but as you probably have already realized.\n\nWhen calculating the average velocity, only the times and positions at the starting and ending points are taken into account. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (\u03b4t), represented by. If a ball is thrown into the.\n\nAverage velocity is the result of dividing the distance an object travels by the time it takes to travel that far. Example 2 finding average velocity a rock is dropped from a height of 64 ft. The average velocity of an object is its total displacement divided by the total time taken.\n\nIn other words, it is the rate at which an object changes its position from one place to another. In this case, we have $$s(t)=\\frac{13}{t+2}$$ (if you mean something else, please say so). Average velocity = change in speed / change in time.\n\nThis is a question from my calculus book. The average velocity is different from the instantaneous velocity, and the two are many times not the same number. So if a particle covers a certain displacement $$\\overrightarrow{ab}$$ in a time $$t_1$$ to $$t_2$$, then the average velocity of the particle is:\n\nVelocity (v) is a vector quantity that measures displacement (or change in position, \u03b4s) over the change in time (\u03b4t), represented by the equation v = \u03b4s/\u03b4t.\n\nOur latest Sports Math graphic teaches you how to\n\nPin by Augi de Freitas on Ketogenic Diet Height to\n\nmisscalcul8 Trig Unit 3 Trig Basics Standard Position\n\nCalculating AVERAGE SPEED from DISTANCE TIME GRAPHS\n\nGraphing Average Speed with Superworms Force, motion\n\nCalculating AVERAGE SPEED from DISTANCE TIME GRAPHS HSPS2\n\nAverage Speed is the absolute value of the slop of a\n\nCalculating Average Speed WS Middle school science\n\nAnalyzing Motion Graphs & Calculating Speed WS Physical\n\nSymbols used in Calculus What scares me is that I'm in\n\nLonger Sailboats are Faster. Why? It has to do with\n\nCalculating AVERAGE SPEED from DISTANCE TIME GRAPHS\n\nInterpreting DISTANCETIME, SPEEDTIME GRAPHS, AVERAGE\n\nCalculating AVERAGE SPEED from DISTANCE TIME GRAPHS\n\nDuct Sizing Charts & Tables Duct\n\nCalculate Average Velocity Calculator, Positivity, Science\n\nHow to Calculate and Solve for Force Nickzom\n\nPace Chart running and stuff Pinterest Runners\n\nCalculating Average Speed WS Elementary school science", "date": "2021-11-27 19:40:30", "meta": {"domain": "eventthyme.net", "url": "https://eventthyme.net/how-to-find-average-velocity-calculus/", "openwebmath_score": 0.7728247046470642, "openwebmath_perplexity": 560.2188371173739, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9828232879690036, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8573633085089547}}
{"url": "https://math.stackexchange.com/questions/2992385/non-isomorphic-connected-unicyclic-graphs", "text": "# Non-isomorphic connected, unicyclic graphs\n\nThe question is: Find the number of non-isomorphic connected, unicyclic graphs(graphs with exactly one cycle) on 6 vertices.\n\nAccording to this question does it mean that there is a general formula that enable us to find the number of non-isomorphic connected, unicyclic graphs on n vertices so that for vertices 6 may be particular?\n\nThis task doesn't suggest the existence of such a formula.\n\nHowever, you can find a general formula for the number of non-isomorphic connected unicyclic graphs on $$n$$ vertices and specify it for $$n=6$$ to get the result.\n\nBut you can as well just count such graphs for $$n=6$$ directly.\n\n\u2022 As I think hexagon is the only non-isomorphic connected unicyclic graphs on 6 vertices. Is that not? \u2013\u00a02468 Nov 10 '18 at 13:58\n\u2022 No, you can have a smaller cycle as well. \u2013\u00a0Jakob B. Nov 10 '18 at 14:14\n\nUsing the notation from Analytic Combinatorics we have for the combinatorial class $$\\mathcal{U}$$ of unicyclic non-isomorphic graphs the equation (we are attaching trees to the nodes of the single cycle where the root of the tree is merged into the cycle)\n\n$$\\def\\textsc#1{\\dosc#1\\csod} \\def\\dosc#1#2\\csod{{\\rm #1{\\small #2}}} \\mathcal{U} = \\textsc{DHD}_{\\ge 3}(\\mathcal{T})$$\n\nwhere $$\\mathcal{T}$$ is the class of unlabeled rooted trees:\n\n$$\\mathcal{T} = \\mathcal{Z} \\times \\textsc{MSET}(\\mathcal{T})$$\n\nHere we have used the dihedral operator for the single cycle (which is a bracelet i.e. a necklace that can be turned over) and the unlabeled multiset operator. The class equation for trees immediately yields a functional equation via the exponential formula, and which is\n\n$$T(z) = z \\exp\\left(\\sum_{\\ell\\ge 1} \\frac{T(z^\\ell)}{\\ell}\\right).$$\n\nIt was proved at the following MSE link using this functional equation that these have recurrence\n\n$$t_{n+1} = \\frac{1}{n} \\sum_{q=1}^{n} t_{n+1-q} \\left(\\sum_{\\ell|q} \\ell t_{\\ell}\\right).$$\n\nUsing the cycle index $$Z(D_q)$$ of the dihedral group we have\n\n$$U(z) = \\sum_{q\\ge 3} Z(D_q; T(z)).$$\n\nTherefore the number of non-isomorphic connected unicyclic graphs is\n\n$$U_n = [z^n] \\sum_{q=3}^n Z\\left(D_q; \\sum_{p=1}^n t_p z^p\\right).$$\n\nThis yields the sequence\n\n$$0, 0, 1, 2, 5, 13, 33, 89, 240, 657, 1806, 5026, 13999, \\\\ 39260, 110381, 311465, 880840, 2497405, 7093751, 20187313, \\ldots$$\n\nwith two leading zeros because the smallest cycle is on three nodes. The data point to OEIS A001429 where the procedure is confirmed. In particular we get for six nodes\n\n$$\\bbox[5px,border:2px solid #00A000]{ U_6 = 13.}$$\n\nRemark. The cycle index of the cyclic group is given by\n\n$$Z(C_q) = \\frac{1}{q} \\sum_{k|q} \\phi(k) a_k^{q/k}$$\n\nand of the dihedral group\n\n$$Z(D_q) = \\frac{1}{2} Z(C_q) + \\begin{cases} \\frac{1}{2} a_1 a_2^{(q-1)/2} & q \\quad \\text{odd} \\\\ \\frac{1}{4} \\left( a_1^2 a_2^{(q-2)/2} + a_2^{q/2} \\right) & q \\quad\\text{even.} \\end{cases}.$$\n\nThis computation was done with the following Maple code.\n\nwith(numtheory);\n\npet_varinto_cind :=\nproc(poly, ind)\nlocal subs1, subs2, polyvars, indvars, v, pot, res;\n\nres := ind;\n\npolyvars := indets(poly);\nindvars := indets(ind);\n\nfor v in indvars do\npot := op(1, v);\n\nsubs1 :=\n[seq(polyvars[k]=polyvars[k]^pot,\nk=1..nops(polyvars))];\n\nsubs2 := [v=subs(subs1, poly)];\n\nres := subs(subs2, res);\nod;\n\nres;\nend;\n\npet_cycleind_cyclic :=\nproc(n)\nlocal s, d;\n\ns := 0;\nfor d in divisors(n) do\ns := s + phi(d)*a[d]^(n/d);\nod;\n\ns/n;\nend;\n\npet_cycleind_dihedral :=\nproc(n)\nlocal s;\n\ns := 1/2*pet_cycleind_cyclic(n);\n\nif type(n, odd) then\ns := s + 1/2*a[1]*a[2]^((n-1)/2);\nelse\ns := s + 1/4*(a[1]^2*a[2]^((n-2)/2) + a[2]^(n/2));\nfi;\n\ns;\nend;\n\nt :=\nproc(n)\noption remember;\n\nif n=1 then return 1 fi;\n\nq=1..n-1);\nend;\n\nU :=\nproc(n)\noption remember;\nlocal res, m, tgf, dhdgf;\n\nres := 0;\n\nfor m from 3 to n do\ndhdgf :=\npet_varinto_cind(tgf, pet_cycleind_dihedral(m));\nres := res +\ncoeff(expand(dhdgf), z, n);\nod;\n\nres;\nend;\n\n\nAddendum. We can also answer the question for the labeled case. The combinatorial classes are the same, only now we get the classic tree function $$T(z)$$ using\n\n$$\\mathcal{T} = \\mathcal{Z} \\times \\textsc{SET}(\\mathcal{T})$$\n\nand functional equation\n\n$$T(z) = z \\times \\exp T(z)$$\n\nand the dihedral operator becomes\n\n$$\\sum_{\\ell\\ge 3} \\frac{z^\\ell}{|D_\\ell|} = \\sum_{\\ell\\ge 3} \\frac{z^{\\ell}}{2\\ell} = -\\frac{1}{2} z - \\frac{1}{4} z^2 + \\frac{1}{2} \\log \\frac{1}{1-z}.$$\n\nWe are thus interested in extracting the coefficient as follows,\n\n$$n! [z^n] \\left(-\\frac{1}{2} T(z) - \\frac{1}{4} T(z)^2 + \\frac{1}{2} \\log \\frac{1}{1-T(z)}\\right).$$\n\nThis has three components, the first is by Cayley\n\n$$- n! [z^n] \\frac{1}{2} T(z) = -\\frac{1}{2} n^{n-1}.$$\n\nThe second is\n\n$$- n! [z^n] \\frac{1}{4} T(z)^2 = - (n-1)! [z^{n-1}] \\frac{1}{2} T(z) T'(z) \\\\ = -\\frac{(n-1)!}{2\\pi i} \\int_{|z|=\\epsilon} \\frac{1}{z^n} \\frac{1}{2} T(z) T'(z) \\; dz.$$\n\nLetting $$w=T(z)$$ so that $$z= w \\exp(-w)$$ and $$dw = T'(z) \\; dz$$ this becomes for $$n\\ge 2$$\n\n$$-\\frac{(n-1)!}{2\\pi i} \\int_{|w|=\\gamma} \\frac{\\exp(nw)}{w^n} \\frac{1}{2} w \\; dw = - \\frac{(n-1)!}{2} \\frac{n^{n-2}}{(n-2)!} \\\\ = - \\frac{1}{2} (n-1) n^{n-2}.$$\n\nFinally for the third one we get\n\n$$n! [z^n] \\frac{1}{2} \\log\\frac{1}{1-T(z)} = (n-1)! [z^{n-1}] \\frac{1}{2} \\frac{1}{1-T(z)} T'(z) \\\\ = \\frac{(n-1)!}{2\\pi i} \\int_{|z|=\\epsilon} \\frac{1}{z^n} \\frac{1}{2} \\frac{1}{1-T(z)} T'(z) \\; dz.$$\n\nWith the same substitution as before we find\n\n$$\\frac{(n-1)!}{2\\pi i} \\int_{|w|=\\gamma} \\frac{\\exp(nw)}{w^n} \\frac{1}{2} \\frac{1}{1-w} \\; dw \\\\ = \\frac{1}{2} (n-1)! \\sum_{q=0}^{n-1} \\frac{n^q}{q!}.$$\n\nCollecting everything we obtain\n\n$$\\frac{1}{2} (n-1)! \\sum_{q=0}^{n-1} \\frac{n^q}{q!} - n^{n-1} + \\frac{1}{2} n^{n-2}$$\n\nor alternatively\n\n$$\\bbox[5px,border:2px solid #00A000]{ \\frac{1}{2} (n-1)! \\sum_{q=0}^{n-3} \\frac{n^q}{q!}.}$$\n\nThis sequence is OEIS A057500:\n\n$$0, 0, 1, 15, 222, 3660, 68295, 1436568, 33779340, 880107840, \\\\ 25201854045, 787368574080, 26667815195274, 973672928417280, \\ldots$$", "date": "2020-02-23 18:25:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2992385/non-isomorphic-connected-unicyclic-graphs", "openwebmath_score": 0.7259795069694519, "openwebmath_perplexity": 632.1313096455588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232894783426, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.857363304933064}}
{"url": "https://www.gaussianwaves.com/2014/07/generating-basic-signals-gaussian-pulse-and-power-spectral-density-using-fft/", "text": "# Generating Basic Signals \u2013 Gaussian Pulse and Power Spectral Density using FFT\n\n(5 votes, average: 4.80 out of 5)\n\nFor more such examples check this ebook : \u00a0Digital Modulations using Matlab: build simulation models from scratch \u2013 by Mathuranathan Viswanathan\n\n## Introduction\n\nNumerous texts are available to explain the basics of Discrete Fourier Transform and its very efficient implementation \u2013 Fast Fourier Transform (FFT). \u00a0Often we are confronted with the need to generate simple, standard signals (sine, cosine, Gaussian pulse, squarewave, isolated rectangular pulse, exponential decay, chirp signal) for simulation purpose. I intend to show (in a series of articles) how these basic signals can be generated in Matlab and how to\u00a0represent them in frequency domain using FFT.\n\n## Gaussian Pulse : Mathematical description:\n\nIn digital communications, Gaussian Filters are employed in Gaussian Minimum Shift Keying \u2013 GMSK (used in GSM technology) and Gaussian Frequency Shift Keying (GFSK). Two dimensional Gaussian Filters are used in Image processing to produce Gaussian blurs. The impulse response of a Gaussian Filter is Gaussian. Gaussian Filters give no overshoot with minimal rise and fall time when excited with a step function. Gaussian Filter has minimum group delay. The impulse response of a Gaussian Filter is written as a Gaussian Function as follows\n\n$$g(t) = \\frac{1}{\\sqrt{2 \\pi } \\sigma} e^{- \\frac{t^2}{2 \\sigma^2}}$$\n\nThe Fourier Transform of a Gaussian pulse preserves its shape.\n\n\\begin{align} G(f) & =F[g(t)]\\ & = \\int_{-\\infty }^{\\infty } g(t)e^{-j2\\pi ft}\\, dt\\ & = \\frac{1}{\\sigma \\sqrt{2 \\pi } } \\int_{-\\infty }^{\\infty } e^{- \\frac{t^2}{2 \\sigma^2}}e^{-j2\\pi ft}\\, dt\\ &=\\frac{1}{\\sigma \\sqrt{2 \\pi } } \\int_{-\\infty }^{\\infty } e^{- \\frac{1}{2 \\sigma^2}\\left[t^2+j4 \\pi \\sigma^2 ft \\right]}\\, dt\\ &=\\frac{1}{\\sigma \\sqrt{2 \\pi } } \\int_{-\\infty }^{\\infty } e^{- \\frac{1}{2 \\sigma^2}\\left[t^2+j4 \\pi \\sigma^2 ft + (j 2 \\pi \\sigma^2 f)^2 \u2013 (j 2 \\pi \\sigma^2 f)^2\\right]}\\, dt\\ &=e^{ \\frac{1}{2 \\sigma^2}(j 2 \\pi \\sigma^2 f)^2}\\frac{1}{\\sigma \\sqrt{2 \\pi } } \\int_{-\\infty }^{\\infty } e^{- \\frac{1}{2 \\sigma^2}\\left[t+j 2 \\pi \\sigma^2 f \\right]^2}\\, dt\\ &=e^{ \\frac{1}{2 \\sigma^2}(j 2 \\pi \\sigma^2 f)^2}=e^{ \u2013 \\frac{1}{2}( 2 \\pi \\sigma f)^2} \\end{align }\n\nThe above derivation makes use of the following result from complex analysis theory and the property of Gaussian function \u2013 total area under Gaussian function integrates to 1. By change of variable, let (u=t+j 2 \\pi \\sigma^2 f ). Then,\n$$\\frac{1}{\\sigma \\sqrt{2 \\pi } }\\int_{-\\infty }^{\\infty }e^{- \\frac{1}{2 \\sigma^2}\\left[t+j 2 \\pi \\sigma^2 f \\right]^2}\\, dt =\\frac{1}{\\sigma \\sqrt{2 \\pi } }\\int_{-\\infty }^{\\infty }e^{- \\frac{1}{2 \\sigma^2} u^2}\\, du =1$$\n\nThus, the Fourier Transform of a Gaussian pulse is a Gaussian Pulse.\n$$\\frac{1}{\\sqrt{2 \\pi } \\sigma} e^{- \\frac{t^2}{2 \\sigma^2}} \\rightleftharpoons e^{ \u2013 \\frac{1}{2}( 2 \\pi \\sigma f)^2}$$\n\n## Gaussian Pulse and its Fourier Transform using FFT:\n\nThe following code generates a Gaussian Pulse with ( \\sigma=0.1s ). The Discrete Fourier Transform of this digitized version of Gaussian Pulse is plotted with the help of (FFT) function in Matlab.\n\n## Double Sided and Single Power Spectral Density using FFT:\n\nNext, the Power Spectral Density (PSD) of the Gaussian pulse is constructed using the FFT. PSD describes the power contained at each frequency component of the given signal. Double Sided power spectral density is plotted first, followed by single sided power spectral density plot (retaining only the positive frequency side of the spectrum).\n\nFor more such examples check this ebook : \u00a0Digital Modulations using Matlab: build simulation models from scratch \u2013 by Mathuranathan Viswanathan\n\n## Recommended Signal Processing Books\n\n\u2022 Vittorio Todisco\n\nHi, thanks again for your works here.\nI have a problem, I\u2019m plotting a 3D Gaussian pulse.\nI have been able to create it and to make the fft of it, but the phase it\u2019s not null and quite incomprehensible.\nThis is a quick example of what I get:\n\nclose all;\nclear;\nclc;\nN = 64; % Size of a sunction\nD = N/2; % to indicate origin at the center of the function\na = 8; % radius for cylindrical function\nw = 0.4; % decay rate for exponential function\ny = repmat(1:N,N,1);\nx = y\u2019;\nr = sqrt((D-x).^2+(D-y).^2); % definition of radius\nsig = 5; % Variance for gaussian function\ng = (2*pi*sig^2)*exp(-((r.^2))./(2*sig^2));\nfigure;mesh(g); title(\u20182D Gaussian Function\u2019);\nG = fft2(g);\nG = fftshift(G);\nfigure; mesh(abs(G)); title(\u2018Fourier Transform of Gaussian function\u2019);\nfzf=atan2(imag(G),real(G));\nfigure;\nmesh(fzf);\ntitle(\u2018Fourier Phase of Gaussian function\u2019);\n\n\u2022 pankaj\n\nDear sir, this code is great for generating the gaussian pulse without using matlab toolboxes.\nI want to ask that from the same code, if i want to generate the train of gaussian pulses, how can i generate (without using the matlab function pulstran)? kindly provide the baisc code for it. I have searched over the internet but not found. suppose I want to generate 12 gaussian pulses wtith sum time duration between them. plzz.\n\nHello this is Lingaraj, doing m-tech in belgaum. I have a problem in finding power spectral density using time domain approach,i have a written code found somewhere but i am not getting it. Please leave a reply as soon as possible\u2026\n\n\u2022 KP\n\nHi Viswanathan,\nThis really awesome article on Gaussian wave generation and analysis. I am trying to generate a Gaussian pulse of 20ns and plot the frequency response of it. When I try to simulate it in matlab, it gives me error saying out of memory. I think I am doing something wrong with the code:\n\nfunction [ output_args ] = freqencySpectrum( ~ )\nfs=0.00000001; // sampling at twice the highest frequency (20ns =50MHz, so sampling at 100MHz)\nt=-30:1/fs:30;\nsigma = 20E-9;\nA=1;\np1 = A/(sqrt(2*pi*sigma^2));\nx=p1*(exp(-(t.^2)/(2*sigma^2)));\nfigure(1);\nsubplot(4,1,1)\nplot(t,x,\u2019b\u2019);\ntitle([\u2018Gaussian Pulse sigma=\u2019, num2str(sigma),\u2019s\u2019]);\nxlabel(\u2018Time(s)\u2019);\nylabel(\u2018Amplitude\u2019);\n\nL=length(x);\nNFFT = 1024;\nX = fftshift(fft(x,NFFT));\nPxx=X.*conj(X)/(NFFT*NFFT); %computing power with proper scaling\nf = fs*(-NFFT/2:NFFT/2-1)/NFFT; %Frequency Vector\n\nsubplot(4,1,2)\nplot(f,abs(X)/(L),\u2019r\u2019);\ntitle(\u2018Magnitude of FFT\u2019);\nxlabel(\u2018Frequency (Hz)\u2019)\nylabel(\u2018Magnitude |X(f)|\u2019);\nxlim([-10 10])\n\nPxx=X.*conj(X)/(L*L); %computing power with proper scaling\nsubplot(4,1,3)\nplot(f,10*log10(Pxx),\u2019r\u2019);\ntitle(\u2018Double Sided \u2013 Power Spectral Density\u2019);\nxlabel(\u2018Frequency (Hz)\u2019)\nylabel(\u2018Power Spectral Density- P_{xx} dB/Hz\u2019);\n\nX = fft(x,NFFT);\nX = X(1:NFFT/2+1);%Throw the samples after NFFT/2 for single sided plot\nPxx=X.*conj(X)/(L*L);\nf = fs*(0:NFFT/2)/NFFT; %Frequency Vector\nsubplot(4,1,4)\nplot(f,10*log10(Pxx),\u2019r\u2019);\ntitle(\u2018Single Sided \u2013 Power Spectral Density\u2019);\nxlabel(\u2018Frequency (Hz)\u2019)\nylabel(\u2018Power Spectral Density- P_{xx} dB/Hz\u2019);\n\nend\n\nCan you help me with this code?\nThanks,\nKP\n\n\u2022 KP,\n\nSigma time is not equal to pulse width. Pulse width should be more than the sigma variation\u2026 In the following code, the sigma variation is assumed to be 20ns. The pulse width is 10*20ns. I have fixed the code for you.. Here it is\n\nsigma = 20e-9%sigma variation of the pulse 20ns\npulseWidth = 10*sigma %double sided pulse width\nfs = 40*1/pulseWidth %Very very high oversampling factor for smooth curve (40 points), FFT with 1024 points is sufficient to cover this\nt=-(pulseWidth/2):1/fs:(pulseWidth/2)\n\nA=1;\np1 = A/(sqrt(2*pi*sigma^2));\nx=p1*(exp(-(t.^2)/(2*sigma^2)));\nfigure(1);\nsubplot(4,1,1)\nplot(t,x,\u2019b\u2019);\ntitle([\u2018Gaussian Pulse sigma=\u2019, num2str(sigma),\u2019s\u2019]);\nxlabel(\u2018Time(s)\u2019);\nylabel(\u2018Amplitude\u2019);\n\nL=length(x);\nNFFT = 1024;\nX = fftshift(fft(x,NFFT));\nPxx=X.*conj(X)/(NFFT*NFFT); %computing power with proper scaling\nf = fs*(-NFFT/2:NFFT/2-1)/NFFT; %Frequency Vector\nsubplot(4,1,2)\nplot(f,abs(X)/(L),\u2019r\u2019);\ntitle(\u2018Magnitude of FFT\u2019);\nxlabel(\u2018Frequency (Hz)\u2019)\nylabel(\u2018Magnitude |X(f)|\u2019);\n\nPxx=X.*conj(X)/(L*L); %computing power with proper scaling\nsubplot(4,1,3)\nplot(f,10*log10(Pxx),\u2019r\u2019);\ntitle(\u2018Double Sided \u2013 Power Spectral Density\u2019);\nxlabel(\u2018Frequency (Hz)\u2019)\nylabel(\u2018Power Spectral Density- P_{xx} dB/Hz\u2019);\n\nX = fft(x,NFFT);\nX = X(1:NFFT/2+1);%Throw the samples after NFFT/2 for single sided plot\nPxx=X.*conj(X)/(L*L);\nf = fs*(0:NFFT/2)/NFFT; %Frequency Vector\nsubplot(4,1,4)\n\nplot(f,10*log10(Pxx),\u2019r\u2019);\ntitle(\u2018Single Sided \u2013 Power Spectral Density\u2019);\nxlabel(\u2018Frequency (Hz)\u2019)\nylabel(\u2018Power Spectral Density- P_{xx} dB/Hz\u2019);\n\n\u2022 KP\n\nHi Mathuranathan,\nThank you very much for the reply. I am little confused about something in the code.\n\n1. Why did you set the pulse width as \u201cpulseWidth = 10*sigma %double sided pulse width\u201d?\nWhy 10?\n\n2. I have attached the output of the simulation. I see that the pulse width is like 60ns, how can I make it 20ns? Should I even lower the sigma value as the pulse width should be greater than the sigma value?\n\nAnd in that case, should I change the sampling frequency (fs = 40*1/pulseWidth %) value as well? or 40 will be fine?\n\n3. Finally, how should I interpret the output of single sided PSD, in the attached figure, I believe that pulse width is 60ns, so the frequency should like 16.66MHz, but in the plot how can I relate it to 16MHz?\nThanks\nKP\n\n\u2022 1) Subtle difference exists between the terms sigma and pulse width.\n\nSigma is the standard deviation of the pulse. The term pulse width depends on where the measurement is taken. Usually, the width of the pulse at half-the maximum value is called Full-Width at Half maximum pulse duration (FWHM).\n\nWhat I was intending is not the FWHM width. It is the full duration of the pulse considered for simulation. Note that the Gaussian pulse gradually tapers nears zero on either ends. As an approximation, we must stop at some point. I chose 10 times the value of sigma for this.\n\n2) To have a full pulse width of 20ns, the sigma has to be lower.\n\n3) What we plot is the frequency response of a single pulse. It breaks down the pulse in frequency domain and shows the different frequency components that make-up the pulse in time domain.\n\nThe connotation of frequency (that measures periodicity 60MHz => 16.66ns) is applicable only when the pulse is repeated periodically. You might need to repeat the pulse at a desired regular interval and then plot the frequency response to check for any spike in the frequency domain that is indicative of the frequency of the repeated pattern.\n\n\u2022 KP\n\nThank you very much Mathuranathan, this will help.\n\n\u2022 Golam Kibria Chowdhury\n\nHello,\n\nYour example is absolutely fabulous. This is the first time I have seen theory and practical coding to understand.\n\n1.) I would like to generate a pulse train of Gaussian pulse in time domain with a certain width (let\u2019s say 20 ns in the above example) with a repetition interval of (let\u2019s say 100 ns). In paper, I have to convolve with a Dirac comb. Again let\u2019s say we limit this Dirac comb to a certain number of cycles (let us say 50 cycles).\n\nHow do I do it in Matlab?\n\n2.) A further addition to this problem, (I cannot get the first one though) is: let us say we make another pulse offset with the pulse from above. Let us say we keep the pulsewidth the same (20 ns), limit the number of cyles to 50 (like above), but change the repetition interval from every other pulse. For the sake of clearly enquiring it, let me draw a diagram here\n\n*****\u2014\u2014\u2014\u2014\u2014\u2014\u2013*****\u2014\u2014\u2014\u2014\u2014\u2014\u2013*****\u2014\u2014\u2014\u2014\u2014\u2014\u2013**** (P1)\n*****\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014-*****\u2014\u2014\u2014\u2014\u2014*****\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014-**** (P2)\n\nHere P1 = pulse one\nP2 pulse 2\n\nboth pulses are in phase at one time (let us say t = t1), then it shifted in time (let us say 4 ns) with the next pulse, then again it is in phase.\n\nHow could I do it matlab?\n\nThank you so much for clear explanation. I deeply appreciate it.\n\nKind regards\nChowdhury", "date": "2017-10-22 08:02:01", "meta": {"domain": "gaussianwaves.com", "url": "https://www.gaussianwaves.com/2014/07/generating-basic-signals-gaussian-pulse-and-power-spectral-density-using-fft/", "openwebmath_score": 0.7782135009765625, "openwebmath_perplexity": 1969.6039376463818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759677214397, "lm_q2_score": 0.8740772417253255, "lm_q1q2_score": 0.8573613603406155}}
{"url": "http://math.stackexchange.com/questions/18532/finding-the-sum-fx-sum-n-2-infty-fracxnnn-1", "text": "# Finding the sum $f(x)=\\sum_{n=2}^{\\infty} \\frac{x^n}{n(n-1)}$\n\nI'm trying to find $$f(x)=\\sum_{n=2}^{\\infty} \\frac{x^n}{n(n-1)}$$\n\nI found the radius of convergence of the above series which is $R=1$. Checking $x=\\pm 1$ also yields a convergent series. Therefore the above series converges for all $x\\in [-1, 1]$.\n\nUsing differentiation of the series term by term we get: $$f'(x)=\\sum_{n=2}^{\\infty} \\frac{x^{n-1}}{n-1}=\\sum_{n=1}^{\\infty} \\frac{x^{n}}{n}=-\\log(1-x)$$ which also has $R=1$, and then, by integrating term by term we get: $$f(x)=\\int_{0}^{x} f'(t)dt=-\\int_{0}^{x} \\log(1-t)dt=x-(x-1)\\ln(1-x)$$\n\nif I understand the theorems in my textbook correctly, the above formulas are true only for $x \\in (-1, 1)$. It seems the above is correct since this is also what WolframAlpha says.\n\nI'm abit confused though. At first, it seemed the above series converges for all $x\\in [-1, 1]$ but in the end I only got $f(x)$ for all $|x|\\lt 1$, something seems to be missing. What can I say about $f(-1)$ and $f(1)$?\n\n-\n\nTry using Abel's theorem.\n\n-\nI thought of Abel's theorem but I'm not sure on its usage, can I say that: $$f(1)=\\lim_{x\\to 1^{-}}f(x)=\\lim_{x\\to 1^{-}}(x-(x-1)\\ln(1-x))=1$$ and $$f(-1)=\\lim_{x\\to -1^{+}}f(x)=\\lim_{x\\to 1^{+}}(x-(x-1)\\ln(1-x))=\\ln4-1$$ \u2013\u00a0 dawson Jan 22 '11 at 17:55\nIf so, why do I need Tauber's theorem (it's not in my textbook)? \u2013\u00a0 dawson Jan 22 '11 at 17:56\n@dawson: No you don't need it. I am just having a braindead moment. \u2013\u00a0 Aryabhata Jan 22 '11 at 18:00\n@dawson: Since you noted the series is convergent, Abel's theorem tells you that the lim of f is same as the limit of the series. So you seem to have it right :-) \u2013\u00a0 Aryabhata Jan 22 '11 at 18:07\nOn second thought, $\\frac{|x|^n}{n(n-1)}\\leq\\frac{1}{n(n-1)}$ and $\\sum \\frac{1}{n(n-1)}$ converges, then by Weierstrass test the above series converges uniformly in $[-1, 1]$, does that make sense? \u2013\u00a0 dawson Jan 22 '11 at 18:22\nshow 3 more comments\n\nIf you rewrite $\\frac{1}{n(n-1)}$ in the form $\\frac{1}{n-1}-\\frac{1}{n}$, then you can rewrite the series in both cases $x = \\pm 1$ and compute their values directly. You can then confirm that in both cases the value you compute coincides with the value $f(\\pm 1)$. (In other words, rather than appealing to Abel's theorem, as Moron suggests, in this particular case you can verify it.)\n\n[Caveat: In the case $x = -1$, you will need to use the familiar series for $\\log 2$, and maybe the easiest way to prove this is by appealing to Abel's theorem (applied to the series for $\\log (1 + x)$). So my approach probably doesn't really avoid Abel's theorem, at least for $x = -1$.]\n\n-", "date": "2014-03-16 04:45:14", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/18532/finding-the-sum-fx-sum-n-2-infty-fracxnnn-1", "openwebmath_score": 0.9747824668884277, "openwebmath_perplexity": 212.94766081020464, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759607334259, "lm_q2_score": 0.8740772466456688, "lm_q1q2_score": 0.857361359058798}}
{"url": "https://math.stackexchange.com/questions/950949/a-simple-system-of-equations/950953", "text": "# A simple system of equations\n\nI'm trying to refresh my school math knowlegde and have trouble solving a simple system of equations:\n\n$\\begin{cases} x + xy + y = -3,\\\\ x - xy + y = 1. \\end{cases}$\n\nI derive $y$ from the second:\n\n$y - xy = 1 - x$\n\n$y(1-x)=(1-x)$\n\nHence,\n\n$y = \\frac {(1-x)}{(1-x)} = 1$, provided that x \u2260 1\n\nNext, I substitute $y=1$ in the first equasion,\n\n$x + x + 1 = -3$; $2x = -4$, $x=-2$.\n\nThe answer seems to be $(-2; 1)$.\n\nThe problem book, however, also lists a second answer, $(1; -2)$.\n\nI feel that I've done something wrong. Give me a hint, please.\n\n\u2022 You excluded the case $x=1$ in your workings. So you have to go back and check that too. \u2013\u00a0Mark Bennet Sep 29 '14 at 11:58\n\u2022 Thank you, @MarkBennet! So, when I get an $x\u2260a$ condition, where $a$ is some number, in one of the equasions of the system, I should plug it into all other equasions, calculate the second variable, then check on all the equations if this $(x, y)$ combination solves them all? \u2013\u00a0CopperKettle Sep 29 '14 at 12:47\n\nBy saying \"provided that $x \\neq 1$\", after getting a solution you have to go back to try out $x=1$.\nNote that $$y(1-x)=(1-x)\\iff (1-x)(y-1)=0\\iff x=1\\ \\text{or}\\ y=1.$$ Since you've already considered the case when $y=1$, you need to consider the case when $x=1$.\nAdd the two together to get $~2x+2y=-2\\iff x+y=-1$. Now replace this value in either one of the initial two equations to get $xy$. And when you know both the Sum and the Product of two numbers, you can determine their values by solving the quadratic $~u^2-su+p=0$, where $s=x+y$ and $p=xy$, since this is what you get when expanding $~(u-x)(u-y)=0.~$ So all that's left to do now is applying the well-known quadratic formula. :-)\n\u2022 Great, I see you're using Vieta's formulas in an imaginative way! (0: I will try to follow this route. As I understand, $u_1, u_2$ in this quadratic equation will equal $x, y$. I'm too dense to understand the line \"since this is what you get when expanding $(u-x)(u-y)=0.$\" though. (0: \u2013\u00a0CopperKettle Sep 29 '14 at 14:02\n\u2022 I solved the quadratic, getting the result $(1; -2)$. How will I get (or logically derive) the second result, $(-2; 1)$, I wonder. \u2013\u00a0CopperKettle Sep 29 '14 at 14:43\n\u2022 @CopperKettle: Your two equations are both symmetrical in x and y, so if $(a,b)$ is a solution, then so is $(b,a)$. As for your other remark, just open up the parentheses, and see what you get. \u2013\u00a0Lucian Sep 29 '14 at 15:20", "date": "2020-08-14 02:47:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/950949/a-simple-system-of-equations/950953", "openwebmath_score": 0.9010306000709534, "openwebmath_perplexity": 292.1371584631404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759615719876, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8573613517480217}}
{"url": "https://mathhelpboards.com/threads/linear-transformation.4753/", "text": "Linear transformation\n\nKaspelek\n\nNew member\nWhere T(p(x)) = (x+1)p'(x) - p(x) and p'(x) is derivative of p(x).\n\na) Find the matrix of T with respect to the standard basis B={1,x,x^2} for P2.\n\nT(1) = (x+1) * 0 - 1 = -1 = -1 + 0x + 0x^2\nT(x) = (x+1) * 1 - x = 1 = 1 + 0x + 0x^2\nT(x^2) = (x+1) * 2x - x^2 = 2x + x^2 = 0 + 2x + x^2\n\nSo, the matrix for T with respect to B equals\n[-1 1 0]\n[0 0 2]\n[0 0 1].\n\nb) Find a basis for kerT and hence write down dim(kerT).\n\nc) Find a basis for ImT and hence write down dim(ImT).\n\nd) Does the transformation have an inverse?\n\nI've done part a, so any guidance on the rest would be greatly appreciated.\n\nActive member\nWhere T(p(x)) = (x+1)p'(x) - p(x) and p'(x) is derivative of p(x).\n\na) Find the matrix of T with respect to the standard basis B={1,x,x^2} for P2.\n\nT(1) = (x+1) * 0 - 1 = -1 = -1 + 0x + 0x^2\nT(x) = (x+1) * 1 - x = 1 = 1 + 0x + 0x^2\nT(x^2) = (x+1) * 2x - x^2 = 2x + x^2 = 0 + 2x + x^2\n\nSo, the matrix for T with respect to B equals\n[-1 1 0]\n[0 0 2]\n[0 0 1].\n\nb) Find a basis for kerT and hence write down dim(kerT).\nAs you've correctly stated, we may express the transformation in the form\n$$\\displaystyle A_T=\\left[ \\begin{array}{ccc} -1 & 1 & 0 \\\\ 0 & 0 & 2 \\\\ 0 & 0 & 1 \\end{array} \\right]$$\n\nNow as for $$\\displaystyle ker(T)$$, we proceed to find the kernel of this transformation in the same way as we would for any other transformation. That is, we would like to solve $$\\displaystyle A_T x=\\left[ \\begin{array}{ccc}0 & 0 & 0\\end{array} \\right]^T$$, which requires we put the augmented matrix\n$$\\displaystyle \\left( A_T|0 \\right) = \\left[ \\begin{array}{ccc} -1 & 1 & 0 & | & 0 \\\\ 0 & 0 & 2 & | & 0 \\\\ 0 & 0 & 1 & | & 0 \\end{array} \\right]$$\nin its reduced row echelon form (of course, this problem could be solved with a little intuition instead of row reduction, but the upshot of this method is that it works where your intuition might fail). You should end up with\n$$\\displaystyle \\left[ \\begin{array}{ccc} 1 & -1 & 0 & | & 0 \\\\ 0 & 0 & 1 & | & 0 \\\\ 0 & 0 & 0 & | & 0 \\end{array} \\right]$$\nWhich tells you that the kernel of the transformation is the set of all polynomials $$\\displaystyle a+bx+cx^2$$ such that $$\\displaystyle a-b=0$$ and $$\\displaystyle c=0$$. Thus, we may state that the kernel of $$\\displaystyle T$$ is spanned by the vector\n$$\\displaystyle \\left[ \\begin{array}{ccc}1 & 1 & 0\\end{array} \\right]^T$$, which is thus the basis of the kernel of T. Since there is one vector in the basis, the dimension of the kernel is 1.\n\nAny questions about the process so far?\n\nLast edited:\n\nActive member\nc) Find a basis for ImT and hence write down dim(ImT).\n\nd) Does the transformation have an inverse?\nFor c), what we need to do is find the largest possible set of linearly independent column-vectors. If you wanted to use the rref (reduced row echelon form) that we computed previously, you simply choose the column vectors corresponding to the 1's (i.e. the \"pivots\") of the reduced matrix. That is, choosing the first and third columns, we find that $$\\displaystyle \\left[ \\begin{array}{ccc}-1 & 0 & 0\\end{array} \\right]^T$$ and $$\\displaystyle \\left[ \\begin{array}{ccc}0 & 2 & 1\\end{array} \\right]^T$$ form a basis of the image. It follows that $$\\displaystyle dim(Im(T))=2$$.\n\nFor d), we simply note that the kernel of this transformation is not of dimension 0. This is enough to state that the transformation does not have an inverse.\n\nLast edited:\n\nFernando Revilla\n\nWell-known member\nMHB Math Helper\nSomething better to express the solutions as vectors of $P_2$ instead of coordinates. That is, $B_{\\ker T}=\\{1+x\\}$ and $B_{\\mbox{Im }T}=\\{-1,2x+x^2\\}$.\n\nKaspelek\n\nNew member\nThat helped a lot guys, thanks a lot.", "date": "2021-02-27 03:17:05", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/linear-transformation.4753/", "openwebmath_score": 0.9151639938354492, "openwebmath_perplexity": 240.52229068328862, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9808759654852756, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.857361347124793}}
{"url": "http://mathhelpforum.com/algebra/42773-solved-finding-sum-two-unknown-variables.html", "text": "# Thread: [SOLVED] Finding the sum of two unknown variables\n\n1. ## [SOLVED] Finding the sum of two unknown variables\n\nHello everyone,\nGiven that $43p + 23q = 4323$, find $p + q$?\nThe question is to find the sum of the unknown variables. Now, the variables have to be integers, and as you can see from the easy example I made, the sum could be $43(100) + 23(1)=4323$ $\\Rightarrow p+q = 101$\n\nOf course, this is not the only solution. Try the following:\n\n$43(54) + 23(87) = 4323$\n$\\Longrightarrow p + q = 141$\n\nI suspect that there are even more integer solutions to this equation. My method for finding the values is the old simple bruteforce \"technique\", where I divide 4323 by highest coefficient and that's where I get a headstart. You can of course find out the pattern, but these \"cooked\" equations are no good for learning.\n\nMy question is: is there a logical process of solving for $p + q$? Other than bruteforce, of course. I asked my teacher today, and he just said \"that's number theory.\" I looked up number theory, but I was kind of lost.\n\nThese type of questions are pretty common on the SAT I tests, and I love them. My brother and I used to exchange multivariable equations like that and try to find the sum all the time.\n\n2. Hello,\n\nWhat comes in my mind is that we have to solve for p and q before finding p+q... maybe there is another method...\nThis comes with the Euclidian algorithm, which will yield to B\u00e9zout's theorem (google for them).\n\n43 & 23 are coprime, that is to say they have no common divider.\nThe theorem states that if 43 and 23 are coprime, then we can write it :\n$43u+23v=1$, where u & v are integers (positive or negative).\n\n-------------------------\nNow, the thing is that you have to find a particular solution to the equation $43u+23v=1$, thanks to the Euclidian algorithm.\n\n$43=23*1+20 \\implies 20=43-23$\n\n$23=20*1+3 \\implies 3=23-20=23-(43-23)=-43+2*23$\n\n$20=3*6+2 \\implies 2=20-3*6=(43-23)-6*(-43+2*23)=7*43-13*23$\n\n$3=2+1 \\implies 1=3-2=(-43+2*23)-(7*43-13*23)=-8*43+15*23$\n\nThus $\\boxed{u=-8 \\text{ and } v=15}$\n\n--------------------------\nWhat's good with it is that we can multiply the equation $43u+23v=1$ by 4323 in order to get a particular solution for p and q in $43p+23q=4323$\n\n--> $\\boxed{p_1=-8*4323=-34584 \\text{ and } q_1=15*4323=64845}$\n\n---------------------------\nMOO\n\nSo we have :\n\n\\begin{aligned} 4323&={\\color{red}43}*p_1+{\\color{red}23}*q_1 \\\\\n&={\\color{red}43}*p+{\\color{red}23}*q \\end{aligned}\n\n$43p_1+23q_1=43p+23q$\n\n$43(p-p_1)=23(q_1-q)$\n\nBecause 43 and 23 are coprime, we know by the Gauss theorem, that $q_1-q=43k$ and $p-p_1=23k$, $\\forall \\text{ integer (positive or negative) } k.$\n\nTherefore the general solution is :\n\n$p=p_1+23k$\n$q=q_1-43k$\n\n$\\boxed{p+q=p_1+q_1-20k}$\n\nRemember, $p_1=-34584$ and $q_1=64845$\n\n------------------------------\n\n$p_1$ and $q_1$ are ugly because 4323 is a large number, and I showed a general method.\n\nIf you have got a particular solution (100 and 1), set up $p_1=100$ and $q_1=1$ and take it from the red MOO.\n\nI hope this is clear enough... If you have any question...\n\n3. Hello, Chop Suey!\n\nHere's a rather primitive solution . . .\n\nGiven that: . $43p + 23q =\\: \\:4323$, find $p + q.$\n\nWe have: . $43p + 23q \\:=\\:43(100) + 23 \\quad\\Rightarrow\\quad 23q - 23 \\:=\\:43(100) - 43p$\n\nFactor: . $23(q-1) \\:=\\:43(100-p) \\quad\\Rightarrow\\quad q \\;=\\;\\frac{43(100-p)}{23} + 1$\n\nSince $q$ is an integer, $(100-p)$ must be divisible by 23.\n\nThis happens for: . $p \\;=\\;100,\\:77,\\:54,\\:31\\:\\hdots \\:100-23t$\n. . .and we have: . $q \\;=\\;1,\\:44,\\:87,\\:130,\\:\\hdots\\:1 + 43t$\n\nTherefore: . $p+q \\;=\\;(100-23t) + (1 + 43t) \\;=\\;101 + 20t\\:\\text{ for any integer }t$\n\n4. I want to thank both of you, Moo and Soroban, for the astoundingly fast reply.\n\nMoo: I'm going to look up those theorems you mentioned and discuss this further should I have any questions.\n\nSoroban: It may be primitive, but you made me wish if I only approached this method before.\n\nThanks again\n\n5. Here are the wikipedia links (so that you won't be mistaken ) :\n\nB\u00e9zout's identity/lemma (not really theorem actually) : B\u00e9zout's identity - Wikipedia, the free encyclopedia\n\nThe way I used the Euclidian algorithm : Extended Euclidean algorithm - Wikipedia, the free encyclopedia\n\nOk...Gauss theorem = Euclid's lemma (they're the same, but I don't know the circumstances that made 2 different names lol)\n\nIf a positive integer divides the product of two other positive integers, and the first and second integers are coprime, then the first integer divides the third integer.\nThis can be written in notation:\nIf n|ab and gcd(n,a) = 1 then n|b.\nEuclid's lemma - Wikipedia, the free encyclopedia", "date": "2013-06-19 11:25:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/42773-solved-finding-sum-two-unknown-variables.html", "openwebmath_score": 0.7997081279754639, "openwebmath_perplexity": 510.93878208490395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759638081522, "lm_q2_score": 0.8740772286044095, "lm_q1q2_score": 0.8573613440501088}}
{"url": "https://math.stackexchange.com/questions/898149/is-there-a-name-for-the-function-maxx-0/898812", "text": "# Is there a name for the function $\\max(x, 0)$?\n\nIs there a name for the function $\\max(x, 0)$?\n\nFor comparison, the function $\\max(x, -x)$ is known as the absolute value (or modulus) of x, and has its own notation $|x|$\n\n\u2022 should there be a name for this function? is it important enough to get its own name? is it used often enough to warrant an abbreviation? does it capture something so profound to be worth naming? If you answer any of these question positively, then you can improve your question. \u2013\u00a0Ittay Weiss Aug 15 '14 at 10:20\n\u2022 In finance max(x-s,0) is the payoff of a call option where s is the strike price and x is the price of the underlying stock on the expiry date. \u2013\u00a0Colonel Panic Aug 15 '14 at 11:06\n\u2022 Another note, I have seen the notation $[f]^+$ and $[f]^-$ for positive part and negative part respectively. \u2013\u00a0Brad Aug 15 '14 at 14:14\n\u2022 math.stackexchange.com/q/25349 \u2013\u00a0Jonas Meyer Aug 15 '14 at 18:39\n\u2022 The question asks for the name of the operation, not an example of something that is computed using it. The operation $\\max(x,0)$ isn't called \"payoff of a call option\", just as the operation of multiplication isn't called \"force\", despite Newton's second law. \u2013\u00a0David Richerby Aug 16 '14 at 20:37\n\nThis is called the positive part of the real number $x$, and often denoted by $x^+$.\n\nLikewise, the negative part of $x$ is $x^-=\\max\\{-x,0\\}$ and the pair of nonnegative real numbers $(x^+,x^-)$ is fully characterized by the pair of identities $$x=x^+-x^-,\\qquad\\lvert x\\rvert=x^++x^-.$$\n\n\u2022 It should be noted that the negative part is, in fact, a positive number. For example the negative part of -3 is 3. \u2013\u00a0Michael Lugo Aug 15 '14 at 13:19\n\u2022 Indeed the positive part and the negative part are both nonnegative numbers. \u2013\u00a0Did Aug 15 '14 at 15:51\n\u2022 Similarly, the imaginary part of a complex number is a real number. \u2013\u00a0user133281 Aug 15 '14 at 22:40\n\u2022 I would rather call this the \"positive part of the identity function\", because the \"positive part\" definition as you cite it refers to functions and not numbers. \u2013\u00a0example Aug 16 '14 at 13:34\n\u2022 @example Thanks for your input. It just happens that this (i.e., the positive part of the real number $x$) is what mathematicians call it (i.e., $\\max(x,0)$). \u2013\u00a0Did Aug 16 '14 at 13:42\n\nWikipedia calls this the ramp function and notes that it can be written using Macaulay brackets.\n\n$\\{x\\} = \\begin{cases} 0, & x < 0 \\\\ x, & x \\ge 0. \\end{cases}$\n\nSince this is a math site, not a programming site, my answer may or may not be regarded as trivia. Anyway...\n\nIn computer graphics this function is called clamping. The general form is $\\mathrm{clamp(x, lowerBound, upperBound)}$ and is defined as\n\nfunction clamp(x, lowerBound, upperBound):\nif(x < lowerBound)\nreturn lowerBound\nelse if(x > upperBound)\nreturn upperBound\nelse\nreturn x\n\n\nor $\\mathrm{min( max(x, lowerBound), upperBound)}$.\n\n$\\max(x,0)$ is the special case $\\mathrm{clamp}(x, 0, +\\infty)$.\n\nThe clamping function is ubiquitous in computer graphics: You often need to confine a calculated value (e.g. a color intensity) into a range of valid values (e.g. $[0,1]$ or $[0,255]$).\n\n\u2022 It looks like Java, only function is not Java. Replacing function with something like public static int makes it Java, a method to be inserted in some class. \u2013\u00a0MickG Aug 16 '14 at 9:40\n\u2022 It's pseudocode. \u2013\u00a0Sebastian Negraszus Aug 16 '14 at 10:14\n\u2022 It's pseudocode and can trivially easily be translated into Python, JS or anything else. @OutlawLemur It's not literal Python because of a few minor differences: a) function clamp(...) instead of def clamp(...): b) No colons at the end of the if/elif/else lines c) Conditional expressions enclosed by parentheses if(x < lowerBound) instead of if x < lowerBound: \u2013\u00a0smci Aug 17 '14 at 0:31\n\u2022 @MickG Even if you replaced function with public static int, it would still be a long way away from being Java. \u2013\u00a0JLRishe Aug 17 '14 at 11:53\n\u2022 @MickG: It is much closer to Python than Java. \u2013\u00a0Reid Aug 17 '14 at 19:55\n\nYou can check that:\n\n$$\\color{blue}{\\max(x,0) = x \\, H(x)}$$\n\nwhere $H(x)$ is the Heaviside or unit step function. A name for this? Not a clue, but hope it helps.\n\n\u2022 This is actually a pretty nice implementation. For example, here's a quick and dirty computation of the derivative for $x \\neq 0$: $$\\frac{d}{dx} \\max(x,0) = \\frac{d}{dx} H(x)\\cdot x = H(x) \\frac{d}{dx} x = H(x) \\cdot 1 = H(x).$$ It works because $H(x)$ is basically a constant. More precisely, the function $$\\begin{cases} 1 & x>0 \\\\ 0 & x < 0\\end{cases}$$ is a locally constant openly-supported partial functions $\\mathbb{R} \\rightarrow \\mathbb{R}$, which is exactly what you need in order to treat it as a constant insofar as differentiation is concerned. \u2013\u00a0goblin GONE Aug 5 '17 at 4:12\n\nI have heard this function called the rectifier. This is a pretty exclusive field name though, and I wouldn't expect to see it anywhere outside of neural networks.\n\n\u2022 I think the use in neural networks derives from the use of the term in electrical engineering: en.wikipedia.org/wiki/Rectifier \u2013\u00a0A. Donda Aug 16 '14 at 2:27\n\u2022 @A.Donda Thanks for the information. \u2013\u00a0Did Aug 16 '14 at 13:05\n\nYou can use:\n\n$$f(x) = \\frac{x + |x|}{2}$$\n\n\u2022 I suppose that's interesting, but it's not at all what the question is asking. \u2013\u00a0Ray Aug 15 '14 at 18:38\n\u2022 The point is that the function doesn't need a name, it can be expressed in terms of $|x|$ and $x$. \u2013\u00a0Darth Geek Aug 15 '14 at 18:42\n\u2022 So squaring doesn't need a name because it can be expressed in terms of $x$ and $2$? And factorial doesn't need a name because it can be expressed in terms of product? And... \u2013\u00a0David Richerby Aug 16 '14 at 20:40\n\u2022 @DavidRicherby It's not about that. We call $x^2$ squaring but we don't have a name for $x^2+3$. We have a name for $n!$ but not one for $(-1)^nn!$. So perhaps a name for the function described is not really needed. It turns out it does have a name, but if it didn't and you needed to use it in a paper, for instance you can just give it a special letter and define it at the beggining of the paper and it would be ok. \u2013\u00a0Darth Geek Aug 17 '14 at 8:19\n\u2022 How is (x+|x|)/2 better then max(x, 0) anyway? \u2013\u00a0Cthulhu Aug 17 '14 at 9:26", "date": "2020-12-05 17:21:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/898149/is-there-a-name-for-the-function-maxx-0/898812", "openwebmath_score": 0.6454028487205505, "openwebmath_perplexity": 585.7272247934104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8573455484980277}}
{"url": "https://www.physicsforums.com/threads/lowest-surface-to-volume-ratio-for-an-uncovered-vessel.617947/", "text": "# Lowest surface to volume ratio for an uncovered vessel\n\n1. Jul 2, 2012\n\n### eee000\n\nHi,\n\nIt is well known that a sphere has the lowest surface to volume ratio. However, a related question is: What is the shape that gives the lowest surface to volume ratio if you do not include the \"top\" in the surface. That is, what is the maximal volume of an uncovered vessel of a fixed surface area?\n\nFor example, a cylinder whose height is equal to its radius, has a volume Pi R^2 h = Pi R^3 and a surface (without cover) of Pi R^2 + 2*Pi *R*h = 3 Pi R^2. If we fix the volume to unity, the surface in this case is 3.Pi^(1/3).\n\nIn comparison, a half sphere of unity volume has a smaller surface -- (18 Pi)^(1/3).\n\nHowever, it's clear this is not the best shape - a sphere cut a bit above half its volume beats the half sphere.\n\nSo, what is the shape that gives the lowest surface to volume ratio if you do not include the \"top\" in the surface ?\n\nThanks!\n\n2. Jul 2, 2012\n\n### chiro\n\nHey eee000 and welcome to the forums.\n\nI'm a little confused about what you mean by \"top\".\n\nWith a sphere, there is no discontinuous boundary like you have with say a cube so in these kinds of situations (where the surface is completely smooth), it's really hard to define a \"top\".\n\nBased on this, do we need the shape to have something corresponding to a \"top\" where the top is an area that is \"smooth\" respect to the rest of the body?\n\n3. Jul 2, 2012\n\n### HallsofIvy\n\nThat depends strongly on how you define \"top\" for a general surface.\n\n4. Jul 2, 2012\n\n### eee000\n\nThank you,\nThe top I have in mind is defined by gravity. I want to have a vessel full of water that do not spill out. That is, if we define the z axis to be the direction of gravity, I think of a surface that may have holes in it, but the volume that we count is only integrated up to lowest z coordinate in the holes.\n\n5. Jul 2, 2012\n\n### Curious3141\n\nThere are well-known formulae for the volume and surface of a spherical cap: http://en.wikipedia.org/wiki/Spherical_cap\n\n(removed wrong stuff).\n\nLast edited: Jul 2, 2012\n6. Jul 2, 2012\n\n### eee000\n\nThank you for the spherical cap formulae, but I think your algebra is wrong. Fixing the sphere radius R, and parameterizing the cap by h, I find the volume to be V=Pi h^2 (R - h/3) and the surface S=2 Pi R h. The lowest ratio is obtained when h=3R/2, and then V=9Pi/8 R^3, and S=3Pi R^2. Setting the volume to unity (R=(8/9Pi)^(1/3)), I find S=(64 Pi/3)^(1/3), better than the half sphere with (18 Pi)^(1/3) and better than the full sphere with (36 Pi)^(1/3).\n\n7. Jul 2, 2012\n\n### Curious3141\n\nYou're right, I made a silly mistake.\n\nWhat I should've done is simply expressed V/A as a function of h, and that's $\\frac{V}{A} = \\frac{1}{6}(3h - \\frac{h^2}{r})$. Taking the derivative and setting it to zero yields your result of $h = \\frac{3r}{2}$.\n\nIntuitively, shouldn't this (partial sphere to a height of 1.5r) be the open vessel with the lowest A/V ratio?\n\n8. Jul 2, 2012\n\n### eee000\n\nI'm sorry,I made a mistake too...\nSince the surface to volume ratio scales as 1/V^(1/3) we need to do the optimization for fixed volume. This yields that the optimal shape is exactly the half sphere.\n\nObviously, my statement \"(64 Pi/3)^(1/3), better than the half sphere with (18 Pi)^(1/3)\" is wrong, as 64/3 > 18 ...\n\n9. Jul 2, 2012\n\n### Curious3141\n\nI didn't see that, but I wasn't following that anyway, because I think your method of visualising the problem is unnecessarily confusing. Why not just let r = 1 and figure out the value of h that minimises the A/V ratio (or maximises the V/A ratio, as I did) for the open partial sphere? Now I'm sure that h = 1.5 (for r = 1) is the right answer.\n\n10. Jul 2, 2012\n\n### eee000\n\nsure it is, but I think the correct formulation of the problem is that of a fixed volume, especially if one is interested in he general question - what is the best shape of all. Since the A/V ratio generally scales with 1/V^(1/3), comparison must be made for equal volumes.\n\n11. Jul 2, 2012\n\n### Curious3141\n\nOK, I see where you're coming from. To state the problem rigorously, \"For an open partial sphere of unit volume, what value of h/r will yield minimal surface area?\"\n\nCan we agree on that statement?\n\nThe way to go about it then is to put h = kr (where 0 < k <= 2), set V(h,r) = 1, then work out r in terms of k alone.\n\nThen plug that expression for r into A(h,r) to get A in terms of k alone.\n\nFinally, minimise A(k) over the domain (0,2].\n\nI have to turn in now - I'm already up too late with a bad cold and I have to work tomorrow. So please go ahead (and if you've done it, please post the basic steps, if you can). If the problem remains unsolved tomorrow, and I'm free, I'll work on it.\n\n12. Jul 2, 2012\n\n### eee000\n\nFor the partial sphere, the solution is easy - the half sphere is optimal.\nThe open problem (to the extent of my minimal knowledge) is:\n\n\"For an open manifold of unit volume(*) , what shape will yield minimal surface(*) area?\"\n\nwhere Volume is the volume contained up to the first hole, and surface includes the whole manifold but the holes\n\n13. Jul 2, 2012\n\n### Curious3141\n\nBut have you proved it? Just asking.\n\nEDIT: Proved it myself.\n\n$A(k) = (2){(9\\pi)}^{\\frac{1}{3}}.k^{-\\frac{1}{3}}.{(3-k)}^{-\\frac{2}{3}}$\n\nand this has a minimum at k = 1 (when h = r), and the minimum is ${(18\\pi)}^{\\frac{1}{3}}$, yielding a half-sphere of unit volume.\n\nI cannot provide any insight on the more general problem.\n\nLast edited: Jul 2, 2012\n14. Jul 3, 2012\n\n### haruspex\n\nSuppose you have an optimal vessel and it holds more than a half-sphere of the same surface area of vessel. Necessarily the upper rim will lie in one horizontal plane. Create a cap for it by reflection in that plane. You now have a closed vessel of twice the volume and twice the area, and it would seem that it holds more fluid than a closed sphere of the same surface area.\n\n15. Jul 3, 2012\n\n### eee000\n\nThank you haruspex! That's what I was looking for.\nI was thinking in more local terms, along the lines of Steiner's arguments, but your one-liner is much better.", "date": "2018-05-24 20:13:45", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/lowest-surface-to-volume-ratio-for-an-uncovered-vessel.617947/", "openwebmath_score": 0.7600815892219543, "openwebmath_perplexity": 974.7124932109682, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018376422666, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8573455401446441}}
{"url": "https://socratic.org/questions/an-urn-contains-100-marbles-20-white-30-red-50-green-calculate-the-probability-o", "text": "# An urn contains 100 marbles: 20 white, 30 red, 50 green. Calculate the probability of selecting White, Red and Green marbles respectively. What is the probability of pulling a white, green, white and red marbles consecutively?\n\nApr 18, 2016\n\n$\\frac{950}{156849}$ or approximately 0.6%\n\n#### Explanation:\n\nAssuming the marbles are not replaced in the urn:\n\n\u2022 The probability of the first marble being white is $\\frac{20}{100}$\n\n\u2022 The probability of the next marble being green is then $\\frac{50}{99}$\n\n\u2022 The probability of the next marble being white is $\\frac{19}{98}$\n\n\u2022 The probability of the next marble being red is $\\frac{30}{97}$\n\nSo the probability of the sequence white, green, white, red is:\n\n$\\frac{20}{100} \\cdot \\frac{50}{99} \\cdot \\frac{19}{98} \\cdot \\frac{30}{97}$\n\n$= \\frac{10}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{50}}}} \\cdot \\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{50}}}}{99} \\cdot \\frac{19}{98} \\cdot \\frac{30}{97}$\n\n$= \\frac{10 \\cdot 19 \\cdot 30}{99 \\cdot 98 \\cdot 97}$\n\n$= \\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} \\cdot 1900}{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{3}}} \\cdot 33 \\cdot 98 \\cdot 97}$\n\n$= \\frac{1900}{33 \\cdot 98 \\cdot 97}$\n\n$= \\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{2}}} \\cdot 950}{33 \\cdot \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{2}}} \\cdot 49 \\cdot 97}$\n\n$= \\frac{950}{33 \\cdot 49 \\cdot 97}$\n\n$= \\frac{950}{156849} \\approx 0.006$\n\nThat is approximately 0.6%\n\nAug 24, 2016\n\nIn support of Georg's solution\n\n#### Explanation:\n\nFor probability questions of this type, if you are ever in doubt, draw a probability tree\n\n$\\textcolor{red}{\\text{Assumption: this is selection without replacement}}$\n\nFrom the diagram observe that the initial selection of\nWhite ->20/100->20%\n\nRed\" \"-> 30/100->30%\n\nGreen->50/100->50%\n\nFrom the probability tree the overall sequenced sampling probability of white: green: white: red is:\n\n$\\frac{20}{100} \\times \\frac{50}{99} \\times \\frac{19}{98} \\times \\frac{30}{97}$\n\nFor what follows refer to George's solution", "date": "2020-09-19 05:45:55", "meta": {"domain": "socratic.org", "url": "https://socratic.org/questions/an-urn-contains-100-marbles-20-white-30-red-50-green-calculate-the-probability-o", "openwebmath_score": 0.8504791855812073, "openwebmath_perplexity": 1224.7511290063478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777179414275, "lm_q2_score": 0.8688267898240861, "lm_q1q2_score": 0.8573384490621708}}
{"url": "https://math.stackexchange.com/questions/2800632/why-is-finding-factors-until-half-of-the-number-enough", "text": "# Why is finding factors until half of the number enough?\n\nIf I want to find factors of a number except itself, at first I think that I divide it the number in turn $1, 2, 3, ... , n - 1$. However, after a while, noticed that division to get the factors is sufficient up to $n/2$(inclusive). The rest part is not needed. What is its reason? Why is it enough to find them? How can it be explained?\n\n@Edit, to find prime number until $\\sqrt n$ is sufficient, but what about perfect numbers? I think $n/2$ is good way to go perfect numbers?\n\nFor 6,\n1   2   3   4   5\n^\n^\n^|\nUp to here, it's ok.\n\n\u2022 By \"perfect number\", do you mean a number which is the sum of all its proper divisors, $n = \\sum_{d|n,d\\lt n} d$? Of course six is one such number, $6=1+2+3$. \u2013\u00a0hardmath May 29 '18 at 14:29\n\u2022 @hardmath < Yes, exactly. \u2013\u00a0itsnotmyrealname May 29 '18 at 14:30\n\u2022 Note that for finding perfect numbers, no one has ever found an odd perfect number. The question if one exists is \"open\". The exact form of even perfect numbers is known, and they have many factors of two. In that sense it is a good idea to test for an even perfect number by dividing out as many factors of two as possible. See this previous Question How to check for perfect numbers? and its answers for more details, involving Mersenne primes. \u2013\u00a0hardmath May 29 '18 at 14:46\n\u2022 Note: to find all of the prime factors, it is sufficient to go up to SQRT(n). To find all factors, it is faster to first find all of the prime factors and then use them to generate all of the factors. \u2013\u00a0RBarryYoung May 29 '18 at 20:54\n\nThere can't possibly be any factors between $\\frac n2$ and $n$. Suppose $\\frac n2 < a < n$ and $a \\cdot b = n$. What could $b$ be? If $b=1$, then $a \\cdot b = a < n$. If $b \\geq 2$, then $a \\cdot b > \\frac n2 \\cdot 2 = n$. So $b$ can't be any positive integer; thus $a$ isn't a factor of $n$.\n\nIn fact, as long as you list both factors when you do the division, you can stop testing at $\\sqrt n$. That's because it's impossible for both factors to be greater than $\\sqrt n$: if $a,b > \\sqrt n$, then $a \\cdot b > \\sqrt n \\cdot \\sqrt n = n$. So factors always come in pairs $a \\cdot b = n$ with $a < \\sqrt n$ and $b > \\sqrt n$ (with the exception of $\\sqrt n$ itself, if it's an integer). Here's how this method works to find the factors of $10$:\n\n\u2022 $3 < \\sqrt 10 < 4$, so we can stop testing at $3$\n\u2022 Test $1$: $\\frac{10}1 = 10$, so $10 = 1 \\cdot 10$, giving us the two factors $1$ and $10$\n\u2022 Test $2$: $\\frac{10}2 = 5$, so $10 = 2 \\cdot 5$, giving us the two factors $2$ and $5$\n\u2022 Test $3$: $\\frac{10}3$ is not an integer so we don't get any factors\n\nThus the full list of factors is $1,10,2,5$ (or, reordered, $1,2,5,10$)\n\nThis method works to find all the factors, not just the prime factors, so it's a perfect method for testing whether a number is perfect (in our example, $1+2+5=8\\neq10$, so $10$ is not perfect).\n\n\u2022 Thank you. But, what about $6$? $\\sqrt 6 = 2.44$ Should we ceil the number to get half border which is $3$? \u2013\u00a0itsnotmyrealname May 29 '18 at 14:38\n\u2022 @itsnotmyrealname The argument I gave shows that you never have to check anything greater than $\\sqrt n$ (since one of the factors must be less or equal to than $\\sqrt n$), so we don't have to check $3$ in this case: just $1$ and $2$. Checking $1$ gives us $1$ and $6$; checking $2$ gives us $2$ and $3$. \u2013\u00a0BallBoy May 29 '18 at 14:40\n\u2022 @itsnotmyrealname Another comment: the savings are much larger for larger $n$. The next known perfect number is $n=28$. By the $\\sqrt n$ method, we have to check only up to (and including) $5$, which is much less than the $14$ we'd need to check by the $\\frac n2$ method! Try it and see if you can show that $28$ is perfect. \u2013\u00a0BallBoy May 29 '18 at 14:41\n\u2022 Gotcha what you mean. You can glance at it, ideone.com/VYKxup \u2013\u00a0itsnotmyrealname May 29 '18 at 15:00\n\nActually to find the (prime) factors of a whole number $n$, it is enough to check for divisors up to $\\sqrt n$. Going up to $n/2$ would be overkill.\n\nOne way to think about this is to ask, if $n$ is not a prime but instead composite, how big can the smallest divisor of $n$ be? If we have the factorization $n = a\\cdot b$, we cannot have both $a$ and $b$ larger than $\\sqrt n$.\n\nTo make a complete list of the factors of a whole number $n\\gt 1$, you can do it by checking all $1\\lt a \\lt \\sqrt n$ and putting both $a$ and $b$ in your list when the division $n/a = b$ is exact. If $\\sqrt n$ happens to be an integer (i.e. $n$ turns out to be a perfect square when we extract $\\sqrt n$), then put one copy of that integer in the list of divisors. The list (of proper divisors of $n$) will be complete when you finish by including $1$, which is always a divisor you want in your list.\n\n\u2022 Yes, I realized it early. Thank you for your algorithm. There are a host of code snippets etc. but I wanted to get what there exist in underneath. \u2013\u00a0itsnotmyrealname May 29 '18 at 15:10\n\nSuppose we can write $n = a*b$, so both $a$ and $b$ are factors. Notice that it must be the case that one of them is less than or equal to $n/2$. To see this, suppose both $a,b > n/2$. Then $a*b > n$, which is absurd. This means for any pair of factors, one of them must be less than $n/2$, hence you need only check up to and including $n/2$ to find them all.\n\n$\\frac{n}{x}<2$ if $x>\\frac{n}{2}$. $2$ is the smallest prime. If $x$ was a factor then $x\\times \\text{something}=n$ where $\\text{something}\\ge 2$.", "date": "2019-09-16 14:08:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2800632/why-is-finding-factors-until-half-of-the-number-enough", "openwebmath_score": 0.7640427350997925, "openwebmath_perplexity": 173.04390637608245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771805808551, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.857338431642122}}
{"url": "https://mathhelpboards.com/threads/program-for-approximating-nth-root-of-a-number.5323/", "text": "# Program for Approximating nth root of a Number\n\n#### Yuuki\n\n##### Member\nI have a question for a programming exercise I'm working on for C.\n\nThe problem is to \"Write a program that uses Newton's method to approximate the nth root of a number to six decimal places.\" The problem also said to terminate after 100 trials if it failed to converge.\n\nQ1. What does \"converge\" mean?\nDoes it mean the difference between two approximation can be made as small as I like?\n\nQ2. On what condition should the program terminate?\nThere are two such conditions: 1) if the loop has been executed 100 times, 2) the difference between the \"true\" answer and the approximation is less than 0.000001.\nI know how to set 1), but how should I express 2)?\nRight now, I am setting the condition as\n|approximation - root| < 0.00001,\nbut I feel it's kind of cheating, because I'm not supposed to know the real answer if I'm making approximations.\nAre there any other any ways to express the condition, especially one involving the function f(x) = x^n - c (x is the nth root of c)?\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nRe: question on program for approximating nth root of a number\n\nHi Yuuki!\n\nI have a question for a programming exercise I'm working on for C.\n\nThe problem is to \"Write a program that uses Newton's method to approximate the nth root of a number to six decimal places.\" The problem also said to terminate after 100 trials if it failed to converge.\n\nQ1. What does \"converge\" mean?\nDoes it mean the difference between two approximation can be made as small as I like\nYes. Basically.\nMore specifically, that you can get as close to the nth root as you want by just taking enough trials.\n\nQ2. On what condition should the program terminate?\nThere are two such conditions: 1) if the loop has been executed 100 times, 2) the difference between the \"true\" answer and the approximation is less than 0.000001.\nI know how to set 1), but how should I express 2)?\nRight now, I am setting the condition as\n|approximation - root| < 0.00001,\nbut I feel it's kind of cheating, because I'm not supposed to know the real answer if I'm making approximations.\nAre there any other any ways to express the condition, especially one involving the function f(x) = x^n - c (x is the nth root of c)?\nExactly. You're not supposed to use the real root.\nBut what you can do is setting the condition as for instance\n$$|\\text{approximation} - \\text{previous approximation}| < 0.0000001$$\nIf you achieve that, it is unlikely that the first 6 decimal digits will change in more iterations.\n\n#### Yuuki\n\n##### Member\nRe: question on program for approximating nth root of a number\n\nThanks.\n\nI set a new variable root0 to store the previous approximation, and set the condition as you said.\nIt worked beautifully.\n\n#### zzephod\n\n##### Well-known member\nIt is possible to estimate the error in an iterate directly (assuming it small anyway).\n\nLet $$\\displaystyle x$$ be the 6-th root of $$\\displaystyle k$$ and $$\\displaystyle x_n$$ an estimate of $$\\displaystyle x$$ with error $$\\displaystyle \\varepsilon_n$$ such that:\n\n$$\\displaystyle x=x_n+\\varepsilon_n$$\n\nThen raising this to the 6-th power gives:\n\n$$\\displaystyle x^6=x_n^6 + 6 \\varepsilon_n x_n^5 + O(\\varepsilon_n^2)$$\n\nNow ignoring second and higher order terms in $$\\displaystyle \\varepsilon$$ and rearranging we get:\n\n$$\\displaystyle \\varepsilon_n=\\frac{x_n^6-x^6}{6x_n^5}=\\frac{x_n^6-k}{6x_n^5}$$\n\nOK lets look at an example: Take $$\\displaystyle k=66$$, and $$\\displaystyle x_n=2$$, so $$\\displaystyle x_n^6=64$$, then\n\n$$\\displaystyle \\varepsilon_n=\\frac{66-64}{6\\times32}\\approx 0.01042$$\n\nwhich compares nicely with $$\\displaystyle 66^{1/6}=2.01028...$$\n\nThe above is very similar (for similar read identical) to computing the next iterate and taking the difference of the iterates as an estimate of the error in the first.\n\nLast edited:\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nIt is possible to estimate the error in an iterate directly (assuming it small anyway).\nIt is also possible to specify an upper boundary for the remaining error in a specific iteration (in this specific case).\n\nFirst off, after the first (positive) iteration, all iterations are guaranteed to be above the root.\nThe remaining error in those iterations is guaranteed to be less than the change in the approximation.", "date": "2021-05-07 07:41:20", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/program-for-approximating-nth-root-of-a-number.5323/", "openwebmath_score": 0.6777608394622803, "openwebmath_perplexity": 584.216068905833, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639694252316, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8573351296760711}}
{"url": "https://www.physicsforums.com/threads/diagonals-of-a-quadrilateral.789945/", "text": "# Homework Help: Diagonals of a Quadrilateral\n\nTags:\n1. Dec 31, 2014\n\n### Born\n\n1. The problem statement, all variables and given/known data\n\nProblem 55 from Kiselev\u015b Geometry - Book I. Planimetry: \"Prove that each diagonal of a quadrilateral either lies entirely in its interior, or entirely in its exterior. Give an example of a pentagon for which this is false.\"\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nThe pentagon part is pretty easy. I'm having trouble with the proof. A proof by contradiction seems to be the easiest way to solve this problem but I'd prefer a proof that also explains why this should be true.\n\nI've tried using straight line properties (i.e. a straight line can be formed though any two points and it is unique) but I haven't gottten anywhere.\n\nThanks in advanced for any help!\n\nLast edited: Dec 31, 2014\n2. Dec 31, 2014\n\n### Simon Bridge\n\nCan you see why the pentagon can break the rule?\nIf a diagonal breaks the rule, then it must cross one of the sides - that help?\nYou can also look at the classes of quadrilateral and see how diagonals are formed in each case.\n\n3. Dec 31, 2014\n\n### MidgetDwarf\n\nMaybe supplementary angles of a transversal are....\n\n4. Jan 1, 2015\n\n### lurflurf\n\nthe diagonal divides the plane in two and contains exactly two of the four points there are two cases both points lie on the same side or each lies on one side\n\n5. Jan 2, 2015\n\n### Born\n\nSimon, MidgetDwarf, and lurflurf, I think you'll like what I've come up with. I'm sorry to not be able to show some pictures but I believe the written proof will suffice. Hope it's clear enough. Thank you for your help.\n\nThe three properties of straight lines in the proof are the following: (1) A straight line can be created from any two points, (2) this line is unique, and (3) if two straight lines coincide at least at two points, all their points coincide (making them the same line).\n\n$\\mathrm{Proof:}$\n\nA quadrilateral has four vertices, each vertex point must connect to two others in order to form the sides of the quadrilateral. Labeling these four points A, B, C, and D and forming the following sides AB, BC, CD, and DA we create the quadrilateral ABCD. The diagonals of said quadrilateral will consequently be AC and BD.\n\nIf a diagonal were to not lie completely inside or outside the quadrilateral then it (the diagonal) must cross one of the sides of the quadrilateral (either to enter or to exit the figure).\n\nThe diagonal AC cannot cross the side AB, DA, BC, or CD because this would imply that the diagonal AD equals the respective side it crosses by property (3) (since AC would coincide with the point of the side it crosses and the point A or C). The same applies to BD and the side AB, DA, BC, or CD.\n\nThis implies that the diagonals of a quadrilateral cannot cross its sides.\n\nTherefore the diagonals of a quadrilateral must either lie entirely inside or entirely outside. $\\mathrm{QED}$\n\nLast edited: Jan 2, 2015\n6. Jan 2, 2015\n\n### Born\n\n$\\mathrm{Follow\\ up:}$\n\nConcerning the pentagon: labeling the five vertex points A, B, C, D, E; and forming the sides AB, BC, CD, DE, and EA. A diagonal is made from point A to point D crossing the side BC. This is possible since the diagonal would only share one point with the side BC- (This is unlike the quadrilateral in which every diagonal would share two points of a side if said diagonal crossed said side)\n\nTherefore a diagonal which lies partially outside and partially inside the figure is possible. $\\mathrm{QED}$\n\nLast edited: Jan 2, 2015", "date": "2018-07-16 00:10:17", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/diagonals-of-a-quadrilateral.789945/", "openwebmath_score": 0.5993608236312866, "openwebmath_perplexity": 553.4265529371386, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9715639694252316, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8573351221659816}}
{"url": "https://math.stackexchange.com/questions/2188573/probability-of-a-team-winning-a-tournament", "text": "# Probability of a team winning a tournament\n\nProblem\n\nIn the World Series of baseball, two teams (call them A and B) play a sequence of games against each other, and the first team to win four games wins the series. Let p be the probability that A wins an individual game, and assume that the games are independent. What is the probability that team A wins the series?\n\nThere will be at maximum seven games played to decide a clear winner.\n\nThis is a practice problem from a course on Probability. Its solution provided in the course gives two approaches.\n\nFirst approach:\n\nThe Game will stop as any one of team wins.\n\n$$P(A) = P(\\text{A winning in 4 games}) + P(\\text{A winning in 5 games}) + P(\\text{A winning in 6 games}) + P(\\text{A winning in 7 games})$$\n\nFor A to win, the last game must be won by team A.\n\n$$\\Rightarrow P(A) = p^4 + { 4 \\choose 3}p^4q + { 5 \\choose 3}p^4q^2 + { 6 \\choose 3}p^4q^3$$\n\nSecond approach\n\nImagine telling the players to continue playing the games even after the match has been decided, then the outcome of the match won\u2019t be affected by this, and this also means that the probability that A wins the match won\u2019t be affected by assuming that the teams always play 7 games.\n\n$$P(A) = P(\\text{A winning 4 times in 7 games}) + P(\\text{A winning 5 times in 7 games}) + P(\\text{A winning 6 times in 7 games}) + P(\\text{A winning 7 times in 7 games})$$\n\nI am not able to follow the second approach. Why is second approach correct?\n\nUpdate:\n\nI am looking for an intuitive explanation for the second approach because in the second approach it appears that probabilities for winning 5, 6 and 7 matches are used which were not required for team A to win.\n\n\u2022 Why is the second approach correct? In the same way that if I flip a coin one time the probability it is heads is the same as the probability the first coin flipped is heads when flipping a coin seven times in succession. If I ask what is the probability I win the very next game against my friend where I play only one game, the probability is the same as if i ask what is the probability I win the very next game against my friend when I play fifty games after the first only important game. \u2013\u00a0JMoravitz Mar 15 '17 at 22:40\n\u2022 Because a team winning in the second version will also win in the first, and vice versa \u2013\u00a0Henry Mar 15 '17 at 22:40\n\u2022 @JMoravitz Thanks. The second approach uses winning 5, 6, as well as 7 games. If I am correct then in the coin example, only the probability of getting heads the first time is considered even if we flipped it 7 times. \u2013\u00a0ovais Mar 15 '17 at 22:50\n\u2022 Both ways to compute $P(A)$ produce a polynomial in $p$. Since they are both correct, the the two methods should produce the same polynomial; and they do: $-20p^7 + 70p^6 -84p^5 + 35p^4$. \u2013\u00a0Fabio Somenzi Mar 16 '17 at 0:51\n\nTeam $A$ wins the game if they win four matches times before $B$ wins four matches. \u00a0 Thus the first approach measures the probability of team A doing so when team $B$ wins zero, one, two, or three matches before $A$'s fourth victory.\n\n$$P(A) = \\left(\\binom 30 p^3q^0+\\binom 41 p^3q^1+\\binom 52 p^3q^2+\\binom 63 p^3q^3\\right)p$$\n\nNow imagine the teams kept playing for a full seven matches even after one of them wins the game. \u00a0 Team $A$ wins the game if they win at least four matches among those seven, since if they do so then team $B$ can win at most three matches before $A$ wins their fourth.\n\n\\begin{align}P(A) & = \\binom 77 p^7q^0+\\binom 76 p^6q^1+\\binom 75 p^5q^2+\\binom 74p^4q^3\\end{align}\n\nWe can see these are equal by taking the terms of the first equation, and including the imagined games after victory. \u00a0 Then if you use the binomial theorem to expand...\n\n\\begin{align}P(A) &= \\binom 30 p^4q^0(p+q)^3+\\binom 41 p^4q^1(p+q)^2+\\binom 52 p^4q^2(p+q)+\\binom 63 p^4q^3 \\\\[1ex] &= \\binom 30 p^4q^0(p+q)^3+\\binom 41 p^4q^1(p+q)^2+\\binom 52 p^5q^2+\\left(\\binom 52+\\binom 63\\right) p^4q^3\\\\[1ex] &\\vdots \\\\[1ex] &= \\binom 77 p^7q^0+\\binom 76 p^6q^1+\\binom 75 p^5q^2+\\left(\\binom 30+\\binom 41+\\binom 52+\\binom 63\\right) p^4q^3 \\\\[1ex]& = \\binom 77 p^7q^0+\\binom 76 p^6q^1+\\binom 75 p^5q^2+\\binom 74p^4q^3\\end{align}\n\n\u2022 Thanks, Graham, I am looking for an intuition that why the second approach is using the probability of winning 5,6 or 7 matches in computing the result. It appears that it may increase $P(A)$ but it does not as both answers are same. \u2013\u00a0ovais Mar 16 '17 at 10:59\n\nAs an addendum to Graham's answer, after suitable massaging of the formulae, we get, for a best-of-$n$ tournament with odd $n$:\n\n$$\\sum_{0 \\leq k \\leq \\frac{n-1}{2}} \\binom{n}{\\frac{n+1}{2}+k} \\sum_{0 \\leq j \\leq k} (-1)^j \\binom{\\frac{n+1}{2}+k}{j} p^{\\frac{n+1}{2}+k} \\enspace.$$\n\nWe can then generate plots like this for $P(A)$:\n\nOur intuition that a larger $n$ favors the stronger team is confirmed.\n\nMy question was to understand the intuition behind why the second approach is correct, especially why the second approach is apparently using probability for winning in 5, 6 or 7 matches while computing $P(A)$.\n\nTo figure out I tried to understand the cases which first event - P(A wins 4 times in 7 games) - of the second approach missed because that can give some clue over why we were adding the probabilities for winning 5,6 or 7 matches.\n\nIn this approach, we imagined the game is continued for 7 matches. For simplicity let's assume that $p = q = \\frac 1 2$.\n\nIn the first approach, we added probabilities of four disjoint events - P(A winning in 4 games), P(A winning in 5), .. , P(A winning in 7 games). The denominators of each of these cases - representing the total number of possible cases - are different - $2^4, 2^5, 2^6, 2^7$.\n\nIn the second approach, denominators in the four disjoint events - P(A winning 4 times in 7 games), P(A winning 5 times in 7 games), ... P(A winning 7 times in 7 games) - are same and equal to $2^7$ representing the total number of possible outcomes.\n\nSince the winner is decided as soon as one of the team wins 4 games, in the second approach, we have over counted the number of possible outcomes.\n\nTo reach the correct answer, either we decrease or adjust the number of possible outcomes from over counted outcomes or adjust it by increasing the number of favourable outcomes.\n\nAdjusting the denominator will transform the second approach into the first approach. Thus let's try adjusting the favourable outcomes.\n\nTo put it accurately, we are not adjusting the favourable outcomes but after adding total number of ways A can win 4 matches, all the other outcomes of the games irrespective of success or failure of A will be considered as favourable.\n\nThis is equivalent to: P(A wins 4 matches and loses 3) + P(A wins 5 matches and loses 2) + P(A wins 6 matches and loses 1) + P(A wins 7 matches and loses 0).\n\nThus we are adding probabilities of winning 5, 6, and 7 matches because now the numerator contains all possible outcomes for losing 0, 1,2 or 3 matches once team A has won and the denominator already counts all the possible outcomes in the 7 games.", "date": "2019-05-24 05:04:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2188573/probability-of-a-team-winning-a-tournament", "openwebmath_score": 0.7464767694473267, "openwebmath_perplexity": 445.40942636971954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743615328861, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.8573283672935554}}
{"url": "https://math.stackexchange.com/questions/1725821/structure-of-gal-mathbbq-zeta-15-mathbbq/1725844", "text": "Structure of $Gal(\\mathbb{Q}(\\zeta_{15})/\\mathbb{Q})$?\n\n$\\zeta_{15}$ is $15$th primitive $n$th root of unity.\n\nQuestion: Find the structure of the group $Gal(\\mathbb{Q}(\\zeta_{15})/\\mathbb{Q})$\n\nI know that if $p$ is prime then $G=Gal(\\mathbb{Q}(\\zeta_{p})/\\mathbb{Q})=\\mathbb{Z_{p}}^*$ but when $p$ is not prime, I am not show how to solve this if not prime.\n\nDoes my required group have any relation to some cyclic group $\\mathbb{Z_n}$, i.e. abelian\n\nWould really appreciate your guidance as I have not got my head around these concepts of cyclotomic fields and Galois structure. Thanks\n\n\u2022 It's the same thing as for primes. The Galois group is $\\Bbb{Z}_{15}^*$ \u2013\u00a0Crostul Apr 3 '16 at 11:11\n\u2022 Why is this the case? I would like to be able to construct some kind of proof \u2013\u00a0thinker Apr 3 '16 at 11:11\n\u2022 en.wikipedia.org/wiki/Cyclotomic_field \u2013\u00a0Crostul Apr 3 '16 at 11:12\n\u2022 I have seen it, and it does not give an explanation in sufficient detail \u2013\u00a0thinker Apr 3 '16 at 11:13\n\u2022 Google \"cyclotomic fields\". You will find a lot of proofs for this fact, and other things \u2013\u00a0Crostul Apr 3 '16 at 11:15\n\nLet $G$ be the Galois group of $\\Bbb Q(\\zeta_n)$ over $\\Bbb Q$. An element $f \\in G$ is entirely determined by its image on $\\zeta_n$.\n\nSince $f(\\zeta_n)^k=f(\\zeta_n^k)=1 \\iff \\zeta_n^k=1$ (recall that $f$ is a field automorphism), you know that $f(\\zeta_n)$ is a primitive $n$-th root of unity: $$f(\\zeta_n)=\\zeta_n^{k_f}$$ for some integer $1\u2264k_f\u2264n$ coprime with $n$.\n\nTherefore, you have a well-defined map $$\\alpha : G \\to (\\Bbb Z/n\\Bbb Z)^* \\qquad \\alpha(f)=[k_f]_n$$ You can check that this is a group isomorphism. It is clearly injective. Since $|G|=[\\Bbb Q(\\zeta_n):\\Bbb Q]=\\text{deg}(\\Phi_n)=\\phi(n) = |(\\Bbb Z/n\\Bbb Z)^*|$, $\\alpha$ is bijective.\n\nFinally, $(f \\circ g)(\\zeta_n)=f(g(\\zeta_n)) = f(\\zeta_n^{k_g})=\\zeta_n^{k_g \\, k_f}$ shows that $$\\alpha(f \\circ g) = [k_{f \\circ g}]_n = [k_f]_n[k_g]_n=\\alpha(f)\\alpha(g).$$\n\nMore generally, let $K$ be a field of characteristic coprime with $n$ (e.g. if $\\text{car}(K)=0$), and suppose that $R_n = \\{x \\in \\overline K \\mid x^n=1\\}$ denotes the set of $n$-th roots of unity in an algebraic closure $\\overline K$ of $K$ (notice that $R_n$ is always a cyclic group for the multiplication, since it is a finite subgroup of $(K^*,\\cdot)$).\n\nThen the extension $K(R_n)$ over $K$ is Galois (it is separable because $n$ is coprime with the characteristic of $K$) and you can show similarly that the Galois group of $K(R_n)$ over $K$ embeds in $(\\Bbb Z/n\\Bbb Z)^*$.\n\n\u2022 so in this case $\\zeta_{n}$ acts as a cyclic generator? \u2013\u00a0thinker Apr 3 '16 at 14:10\n\u2022 Yes, absolutely: the $n$-th roots of $1$ (i.e. $x$ such that $x^n=1$) are of the form $x=\\zeta_n^j$, i.e. the set $R_n$ of the $n$-th roots of $1$ is actually the subgroup generated by $\\zeta_n$. Indeed, $R_n$ is a finite subgroup (for the multiplication) of $\\overline{ \\Bbb Q}^*$ (where $\\overline{\\Bbb Q}$ is an algebraic closure of $\\Bbb Q$. You may know that every finite subgroup of $(K^*,\\cdot)$ (where $K$ is any field) is actually cyclic. \u2013\u00a0Watson Apr 3 '16 at 14:13\n\u2022 Then, $\\Bbb Q(\\zeta_n)$ over $\\Bbb Q$ is Galois because it is separable (any extension of a field of characteristic $0$, as $\\Bbb Q$, is separable) and normal (it is the splitting field of $X^n-1$ : any root of $X^n-1$ belongs to $R_n = \\langle \\zeta_n \\rangle$, and therefore to $\\Bbb Q(\\zeta_n)$). \u2013\u00a0Watson Apr 3 '16 at 14:16", "date": "2019-11-12 16:57:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1725821/structure-of-gal-mathbbq-zeta-15-mathbbq/1725844", "openwebmath_score": 0.9716662764549255, "openwebmath_perplexity": 120.85920721628139, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743620718529, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8573283591509442}}
{"url": "http://mathhelpforum.com/algebra/148292-sum-2-squares.html", "text": "Math Help - Sum of 2 squares\n\n1. Sum of 2 squares\n\nSorry before I tell you the question here was the question that led to this question.\n\nShow that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2\n\nDone that successfully.\n\nUsing this results, write 500050 as the sum of 2 square numbers.\n\nI have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail\n\n2. Originally Posted by Mukilab\nSorry before I tell you the question here was the question that led to this question.\n\nShow that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2\n\nDone that successfully.\n\nUsing this results, write 500050 as the sum of 2 square numbers.\n\nI have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail\nTry m=7.\n\n50 is very close to 49 and stands out that way..\n\n3. Hello, Mukilab!\n\nShow that: . $(m^2+1)(n^2+1)\\:=\\m+n)^2+(mn-1)^2\" alt=\" (m^2+1)(n^2+1)\\:=\\m+n)^2+(mn-1)^2\" />\n\nDone that successfully. . Good!\nUsing this results, write 500,050 as the sum of 2 square numbers.\n\nNote that: . $500,050 \\;=\\;(50)(10,\\!001) \\;=\\;(7^2+1)(100^2+1)$\n\n$\\text{Let }m = 7,\\;n = 100 \\text{ in the formula:}$\n\n. . $(7^2 + 1)(100^2 + 1) \\;=\\;(7 + 100)^2 + (7\\cdot100 - 1)^2 \\;=\\;107^2 + 699^2$\n\n4. m=7 is a really nice guess\nOK. $500050=10001*50$\nand $10001=10000+1$\nand $50=49+1$\nApply in your equation: $(m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2$\nThen....\n\n5. That would make n a decial (10sqrt10)\n\n6. Originally Posted by Soroban\nHello, Mukilab!\n\nNote that: . $500,050 \\;=\\;(50)(10,\\!001) \\;=\\;(7^2+1)(100^2+1)$\n\n$\\text{Let }m = 7,\\;n = 100 \\text{ in the formula:}$\n\n. . $(7^2 + 1)(100^2 + 1) \\;=\\;(7 + 100)^2 + (7\\cdot100 - 1)^2 \\;=\\;107^2 + 699^2$\n\nThanks Soroban!\n\nSorry, only saw the first post at the start\n\nAny tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?\n\nThanks again\n\n7. Originally Posted by Mukilab\nThat would make n a decial (10sqrt10)\nMaybe you entered 50050 / 50 instead of 500050 / 50? m=7 gives n=100 as shown above.\n\n8. Originally Posted by Mukilab\nThanks Soroban!\n\nSorry, only saw the first post at the start\n\nAny tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?\n\nThanks again\nAnother way to go about expressing integers as sums of two squares is knowing that: the set of integers expressible as the sum of two squares is closed under multiplication, and an odd prime is expressible as the sum of two squares if and only if it is congruent to 1 (mod 4). See here and here.", "date": "2016-04-30 04:09:47", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/148292-sum-2-squares.html", "openwebmath_score": 0.779926598072052, "openwebmath_perplexity": 1049.1930992022328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226260757067, "lm_q2_score": 0.8774767954920547, "lm_q1q2_score": 0.8573146830521431}}
{"url": "https://www.physicsforums.com/threads/showing-that-v-is-a-direct-sum-of-two-subspaces.747625/", "text": "# Showing that V is a direct sum of two subspaces\n\nHi guys, I have this general question.\n\nIf we are asked to show that the direct sum of ##U+W=V##where ##U## and ##W## are subspaces of ##V=\\mathbb{R}^{n}##, would it be possible for us to do so by showing that the generators of the ##U## and ##W## span ##V##? Afterwards we show that their intersection is a zero-vector. For example:\n\n##U## is a subspace generated by ##(0,1)## and ##W## is a subspace generated by ##(2,2)##. Clearly those generators span two dimensional ##V##, and their intersection is ##(0,0)##. Therefore the conclusion can be made that their direct sum is ##V##. Is this kind of reasoning okay?\n\n## Answers and Replies\n\njbunniii\nScience Advisor\nHomework Helper\nGold Member\nYes, this reasoning is valid. In general, ##V## is the direct sum of ##U## and ##W## if and only if ##V = U+W## and ##U \\cap W = \\{0\\}##. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.\n\nHowever, be aware that this only works for two subspaces. If you have three or more subspaces, say ##\\{U_i\\}_{i=1}^{N}##, then it's possible to have ##V = U_1 + \\ldots U_N## and ##U_i \\cap U_j = \\{0\\}## for all ##i \\neq j##, but the sum is not direct. It's a good exercise to construct an example where this occurs.\n\nYes, this reasoning is valid. In general, ##V## is the direct sum of ##U## and ##W## if and only if ##V = U+W## and ##U \\cap W = \\{0\\}##. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.\n\nHowever, be aware that this only works for two subspaces. If you have three or more subspaces, say ##\\{U_i\\}_{i=1}^{N}##, then it's possible to have ##V = U_1 + \\ldots U_N## and ##U_i \\cap U_j = \\{0\\}## for all ##i \\neq j##, but the sum is not direct. It's a good exercise to construct an example where this occurs.\n\nAh ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? ##V_1 - V_2 = V_1 + V_2## if we define subtraction to be ##\\{ x-y | x \\in V_1, y \\in V_2 \\}.## This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?\n\njbunniii\nScience Advisor\nHomework Helper\nGold Member\nAh ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? ##V_1 - V_2 = V_1 + V_2## if we define subtraction to be ##\\{ x-y | x \\in V_1, y \\in V_2 \\}.## This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?\nSince ##y \\in V_2## if and only if ##y \\in -V_2##, it doesn't matter whether you add or subtract elements of ##V_2##, you get the same result in both cases: ##V_1 - V_2 = V_1 + V_2##.\n\nAdmittedly, the notation ##V_1 - V_2## can seem a bit misleading. Fortunately, since it is the same as ##V_1 + V_2##, there's no reason to use ##V_1 - V_2##.\n\nNote that you could even define ##aV_1 + bV_2 = \\{a x + b y | x \\in V_1, y \\in V_2\\}##, where ##a## and ##b## are nonzero scalars. Then ##V_1 - V_2## is just a special case with ##a=1## and ##b=-1##.\n\nOnce again we have ##aV_1 + bV_2 = V_1 + V_2## for any nonzero ##a,b##, because ##x \\in aV_1## if and only if ##x \\in V_1## and similarly, ##y \\in V_2## if and only if ##y \\in bV_2##. So we can just stick with the notation ##V_1 + V_2## since all the other \"linear combinations\" of ##V_1## and ##V_2## give the same result.\n\nLast edited:\njbunniii\nScience Advisor\nHomework Helper\nGold Member\nBy the way, here's a bit of extra info in case it is helpful. Suppose we start with a collection of subspaces ##\\{U_i\\}_{i=1}^{N}## of ##V##. We may be interested in combining the ##U_i##'s to form a larger subspace. The naive thing to do would be to form the set theoretic union ##\\cup_{i=1}^{N}U_i##. But this won't generally be a subspace.\n\nSo what we really want is the smallest subspace containing all of the ##U_i## (or equivalently, the smallest subspace containing ##\\cup_{i=1}^{N}U_i##). This turns out to be exactly ##U_1 + \\ldots + U_N##. Proof: certainly ##U_1 + \\ldots + U_N## is a subspace containing each ##U_i##. If ##S## is another such subspace then it must contain all elements of the form ##u_1+\\ldots+u_N## with ##u_i \\in U_i##. Thus ##U_1 + \\ldots + U_N \\subset S## so ##U_1 + \\ldots + U_N## is the smallest such subspace.\n\nNow in general, the ##U_i##'s may not be \"linearly independent\" of each other: maybe there is some nonzero element of ##U_1## which can be expressed as a linear combination of elements of the other ##U_i##'s. But if they ARE linearly independent, then we say that the sum is a direct sum, and we write it as ##U_1 \\oplus \\ldots \\oplus U_N## instead of ##U_1 + \\ldots + U_N##. We can obtain a basis for a direct sum ##U_1 \\oplus \\ldots \\oplus U_N## by simply selecting a basis ##B_i## for each ##U_i## and taking the union: ##B = \\cup_{i=1}^{N} B_i##. Indeed, this union is disjoint because there are no common elements among the ##B_i##'s. In the finite-dimensional case, this immediately tells us that\n$$\\text{dim}(U_1 \\oplus \\ldots \\oplus U_N) = \\sum_{i=1}^{N}\\text{dim}(U_i)$$\nThis is not true if the sum is not direct. In that case, all we can say is that\n$$\\text{dim}(U_1 + \\ldots + U_N) \\leq \\sum_{i=1}^{N}\\text{dim}(U_i)$$\n\nLast edited:\nWow thanks for your detailed explanation. I just want to make sure for this last time that I understand this correctly.\n\nSince ##y \\in V_2## if and only if ##y \\in -V_2##, it doesn't matter whether you add or subtract elements of ##V_2##, you get the same result in both cases: ##V_1 - V_2 = V_1 + V_2##.\n\nCan I also intuitively think that ##V_2## = ##-V_2## if ##V## is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)\n\njbunniii\nScience Advisor\nHomework Helper\nGold Member\nCan I also intuitively think that ##V_2## = ##-V_2## if ##V## is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)\nSure, the fact that ##V_2## is a subspace and therefore contains its additive inverses is exactly the reason why ##y \\in V_2## if and only if ##y \\in -V_2##, and this latter statement is the definition of set equality: we conclude that ##V_2 = -V_2##.", "date": "2021-06-14 17:17:30", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/showing-that-v-is-a-direct-sum-of-two-subspaces.747625/", "openwebmath_score": 0.9901611804962158, "openwebmath_perplexity": 780.2862724929605, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9770226327661525, "lm_q2_score": 0.8774767826757122, "lm_q1q2_score": 0.8573146764009973}}
{"url": "http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements", "text": "Write in roster notation all the integer numbers that are greater than 0 and less than 10, non-inclusive: Write in roster notation all integer numbers that are less than two, inclusive. {x | x is an odd whole number less than 10} Express the set using set notation and the listing method. Purplemath. the set of all counting numbers less than or equal to 7. list all the elements of the set and use set notation? $\\{x|x \\in \\mathbb Z, x>0, x \\ is \\ divisible \\ by \\ 5\\}$ OR would it be written like this? if the pattern is obvious (a nasty word in mathematics). I have a question that asks to use set builder notation for positive integers divisible by 5. They are surrounded by braces and separated by commas. If the order of the elements is changed or any element of a set is repeated, it does not make any changes in the set. 8 years ago. (a) The set A of counting numbers between ten and twenty. They wrote about it on the chalkboard using set notation: P = ... Let P be the set of all members in the math club. The expressions \"A includes x\" and \"A contains x\" are also used to mean set membership, although some authors use them to mean instead \"x is a subset of A\". Favorite Answer . Recently Asked Questions Could you help me with this math question? For example, if the universal set is $$\\mathbb{R}$$, we cannot list all the elements of the truth set of \u201c$$x^2 < 4$$.\u201d In this case, it is sometimes convenient to use the so-called set builder notation in which the set is defined by stating a rule that all elements of the set must satisfy. In set theory and its applications to logic, mathematics, and computer science, set-builder notation is a mathematical notation for describing a set by enumerating its elements, or stating the properties that its members must satisfy.. Example. A set can be written explicitly by listing its elements using set bracket. A set is created by using the set() function or placing all the elements within a pair of curly braces. It is used with common types of numbers, such as integers, real numbers, and natural numbers. This notation can also be used to express sets with an interval or an equation. Use set notation and the listing method to describe the set. The blade length is the radius of the circle formed by the area swept by the blades. ... {x | x is an even multiple of 5 that is less than 10} Example. More in general, given a tuple of indices, how would you use this tuple to extract the corresponding elements from a list, even with duplication (e.g. {9,10,11,12} List all the elements of the following set. {50, 47, 44, ...., 29} {50,47,44,41,38,35,32,29} List all the elements of the following set. For example, the number 5 is an integer, and so it is appropriate to write $$5 \\in \\mathbb{Z}$$. tuple (1,1,2,1,5) produces [11,11,12,11,15]). Sets in Python A set in Python is a collection of objects. It is a way to describe the set of all things that satisfy some condition (the condition is the logical statement after the \u201c$$\\st$$\u201d symbol). List any subsets, and show the relationships among the sets and subsets in a Venn diagram. The sets in python are typically used for mathematical operations like union, intersection, difference and complement etc. Defining a set using this sort of notation is very useful, although it takes some practice to read them correctly. - the answers to estudyassistant.com . ) Use set notation, and list all the elements of the set. Relevance. List or Roster method, Set builder Notation, The empty set or null set is the set that has no elements. Summary: Set-builder notation is a shorthand used to write sets, often for sets with an infinite number of elements. List all the elements of the following set. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic. The set of all the fibers over the elements of Y is a family of sets indexed by Y. That is, x is an element of the intersection A \u2229 B, if and only if x is both an element of A and an element of B. For example, for the function f(x) = \u2026 Let us now explain what a set notation is. Set notation uses curly brackets { } which are sometimes referred to as braces. Answer Save. (a) The set A of counting numbers between ten and twenty. $\\{x|x \\in \\mathbb Z, x>0, x \\ = \\ x/5 \\}$ discrete-mathematics. SET BUILDER Notation: Set-builder notation does not list all the elements of a set the way roster notation does. Set notation or ways to define a set. We can list each element (or \"member\") of a set inside curly brackets like this: Common Symbols Used in Set \u2026 List all of the elements of each set using the listing method. 2 Answers. An object that belongs to a set is called an element (or a member) of that set. They are different from lists or tuples in that they are modeled after sets in mathematics. 128k 24 24 gold badges 165 165 silver badges 216 216 bronze badges. Set Symbols. Roster Notation. If a and be are meant to be arbitrary elements of the set, not a and b specifically, then there are 5C2 ways of selecting a set, 5x4 / 2 = 10 ways. The inverse image of a singleton, denoted by f \u22121 [{y}] or by f \u22121 [y], is also called the fiber over y or the level set of y. The set of all even integers, expressed in set-builder notation. Use roster and rule notation to describe the set F, which consists of all the positive multiples of 5 that are less than 50. Two sets are equivalent if they contain the same number of elements. In set-builder notation, the set is specified as a selection from a larger set, determined by a condition involving the elements. 2. StartSet x | x is a natural number greater than 9 and less than 17 EndSet {x | x is a natural number greater than 9 and less than 17} Get Answer. There is more than one format for writing this sequence in Set Builder Notation. Answer to List all the elements of the following set. asked Apr 12 '10 at 11:35. In this notation, the vertical bar (\"|\") means \"such that\", and the description can be interpreted as \"F is the set of all numbers n, such that n is an integer in the range from 0 to 19 inclusive\". Instead, set-builder notation describes the properties of the elements of the set. For example, one can say \u201clet $$A$$ be the set of all odd integers\u201d. list all the elements of each set? {\u221229, \u221224 , -19 } Answers: 2 Get Other questions on the subject: Mathematics. Answer: 3 question List all the elements of the following set. share | cite | improve this question | follow | asked Jul 28 '17 at 8:58. Use set notation to list all the elements in the following set. Use set notation to list all the elements of the set. For example: The intersection of the sets {1, 2, 3} and {2, 3, 4} is {2, 3}. Venn diagrams can be used to express the logical (in the mathematical sense) relationships between various sets. The intersection of two sets A and B, denoted by A \u2229 B, is the set of all objects that are members of both the sets A and B.In symbols, \u2229 = {: \u2208 \u2208}. I think you got the point. use set notation and the listhing method to describe the set? The symbol 2 is used to describe a relationship between an element of the universal set and a subset of the universal set, and the symbol $$\\subseteq$$ is used to describe a relationship between two subsets of the universal set. A rule works well when you find lots and lots of elements in the set. Objects placed within the brackets are called the elements of a set, and do not have to be in any specific order. Set example To create a set, we use the set() function. Three Sets. enclosing the list of members within curly brackets. (6,7,8...,14) 3. We use special notation to indicate whether or not an element belongs to a set, as shown below. a. to the nearest square foot, what is the area of this circle for model 3? Set notation is used to help define the elements of a set. Lv 7. Some Example of Sets . The cardinality or cardinal number of a set is the number of elements in a set. Would this be the correct way writing it? Other notations include f \u22121 (B) and f \u2212 (B). Denote each set by set-builder notation, using x as the variable. {2,4,8, ..., 256} The complete list is { } The objects are called members or elements of the set. share | improve this question | follow | edited Aug 10 '19 at 11:16. jpp. Use set notation and the listing method to describe the set. question: A pro golfer is ranked 48th out of 230 players that played in a PGA event in 2016. . The natural number less than 37 that are divisible by 5. Use set notation and the listing method to describe the set. The set of all whole numbers greater than 8 and less than 13. Here are some more examples: Example 0.3.1. A set is a collection of things, usually numbers. If that's what you want, you need to state that more clearly, but then all you have to do is to list the 10 possible pairs of two elements \u2026 Let us say the third set is \"Volleyball\", which drew, glen and jade play: Volleyball = {drew, glen, jade} But let's be more \"mathematical\" and use a Capital Letter for each set: S means the set of Soccer players; T means the set of Tennis players; V means the set of Volleyball players If \u2026 You can also use Venn Diagrams for 3 sets. listing method is {1,2,3,4,5,6,7} set notation is {x | 1 \u2264 x \u2264 7 }-----I must point out that some textbooks include zero as a counting number. Creating a set. Notation and terminology. anonymous. A roster is a list of the elements in a set. Mathematics, 21.06.2019 19:00, bentonknalige. Describing sets One can describe a set by specifying a rule or a verbal description. Then $$A$$ is a set and its elements are all the odd integers. After school they signed up and became members. and listing method to describe the set. Here is a commonly used format: Set A = {x | x \u2208 \u2124, 0 \u2264 x 12} Two sets are equal if they contain the exact same elements although their order can be different. For example, a set F can be specified as follows: = {\u2223 \u2264 \u2264}. Show Video Lesson. python list. 1. the set of all counting numbers less than or equal to 6? A set whose elements all belong to another set; for example, the set of odd digits, O 5 {1, 3, 5, 7, 9}, is ... Indicate the multiples of 4 and 12, from 1 to 240 inclusive, using set notation. Related course Python Programming Bootcamp: Go from zero to hero. If the set contains a lot of elements, you can use an ellipsis ( . Sets are available in Python 2.4 and newer versions. Use set notation and the listing method to describe the set. We can create a set, access it\u2019s elements and carry out these mathematical operations as shown below. Interval and Graphical Notation. You could define a set with a verbal description: All sets above are described verbally when we say, \" The set of all bla bla bla \"b. 2. Sample questions . - 17271819 Logician George Boolos strongly urged that \"contains\" be used for membership only, and \"includes\" for the subset relation only. Whether or not an element belongs to a set notation and the listing method to describe the.. From a larger set, and do not have to be in any specific order are equal if contain......., 29 } { 50,47,44,41,38,35,32,29 } list all the elements of the elements a! No elements help define the elements of the set ( ) function or placing all the elements each... Brackets are called the elements within a pair of curly braces '19 11:16.! Called an element ( or a member ) of that set 128k 24 24 badges. Relation only define the elements in a set notation and the listing method describe. 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Question: a pro golfer is ranked 48th out of 230 players that played in set! Use an ellipsis ( asks to use set notation and the listhing method to describe the set have... Read them correctly  contains '' be used to help define the elements of a set of Y a... Recently Asked Questions Could you help me with this math question the variable although it takes some to... Numbers, and  includes '' for the subset relation only x/5 \\ \\$. A rule or a verbal description 48th out of 230 players that played in a set is as... ( in the set, as shown below a ) the set of all the of! Pattern is obvious ( a nasty word in mathematics ) family of sets by! Members or elements of each set by specifying a rule works well when you find and! The blade length is the area of this circle for model 3 a of counting numbers than... Python is a list of the set, and list all the elements in a set access... Number less than 10 } express the logical ( in the realm of mathematics and in mathematical logic this question! Follow | Asked Jul 28 '17 at 8:58 is more than one format for writing this sequence set...: set-builder notation describes the properties of the set ( ) function on the subject: mathematics the answers estudyassistant.com! Within a pair of curly braces } express the logical ( in the realm of mathematics and in logic... Such as integers, expressed in set-builder notation is a set, access \u2019.", "date": "2021-09-22 11:59:07", "meta": {"domain": "thatcomputerguy.us", "url": "http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements", "openwebmath_score": 0.8009728789329529, "openwebmath_perplexity": 395.14311842075443, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9770226314280634, "lm_q2_score": 0.8774767746654974, "lm_q1q2_score": 0.8573146674006942}}
{"url": "https://math.stackexchange.com/questions/1364821/verify-that-binomn14-frac-left-substack-binomn2-displaystyle", "text": "# Verify that $\\binom{n+1}{4} = \\frac{\\left(\\substack{\\binom{n}{2}\\\\{\\displaystyle2}}\\right)}{3}$ for $n \\geq 4$\n\nVerify that for $n \\geq 4$ $$\\dbinom{n+1}{4} = \\frac{\\left(\\substack{\\binom{n}{2}\\\\{\\displaystyle2}}\\right)}{3}$$\n\nNow present a combinatoric argument for the above.\n\nFirst, by verify does it mean check for some n > 3? if so, then n = 4 gives both sides value 5. I have tried expanding both sides and It just gets messy, but i'm sure that would be attempting to prove it?\n\nI cant quite move ahead with this one, I have said that $\\left(\\substack{\\binom{n}{2}\\\\{\\displaystyle2}}\\right)$ is the number of ways of choosing 2 pairs of objects from n objects. my reasoning for this is that $\\dbinom{n}{2}$ is the number of ways of choosing 2 objects from n and hence $\\left(\\substack{\\binom{n}{2}\\\\{\\displaystyle2}}\\right)$ is the number of ways of choosing 2 of these ways... What i am struggling to do is understand how the three comes into it.\n\n\u2022 I think, in this case, \"verify\" means to prove algebraically, by manipulating the right hand side until it looks like the left. \u2013\u00a0Theo Bendit Jul 17 '15 at 18:08\n\u2022 Please don't use display math in titles; it takes too much vertical space on the front page. \u2013\u00a0Rahul Jul 17 '15 at 18:09\n\u2022 Hints: (1) How are two pairs of two things like one quadruple of things? (2) What if both pairs contain a common element? These are two independent hints. \u2013\u00a0Rahul Jul 17 '15 at 18:12\n\u2022 before reading the answer below fully and do the bad thing, i would like to take your comment into consideration and work it out for myself. (1) A pair of things is like one quadruple of things when each pair are disjoint, two pairs of things are not like a quadruple of things when they have a common member. so to expand, if i have four objects a,b,c,d. there are three ways i can choose distinct pairs from these four objects. [ab][cd], [bc][ad], [ac][bd]. Now concerning (2), i am struggling to use this hint at the moment. \u2013\u00a0user197848 Jul 17 '15 at 18:47\n\n\"Verify\" would mean verify algebraically (using formulas such as $\\binom{n+1}{4}=\\frac{(n+1)n(n-1)(n-2)}{24}$. It's not that hard if you just consider the factorizations of each side. However, there is a nice combinatorial argument. $\\left(\\substack{{\\binom{n}{2}}\\\\{\\displaystyle 2}}\\right)$ is the number of ways to pick two pairs of elements from a set of $n$; we can combine these pairs, and if an element is common to both pairs, then treat one of the instances of the element as a new $(n+1)$-st element. This counts the number of ways to pick 4 elements from a set of $n+1$. However, note that all ways to pick 4 are counted exactly 3 times, so you must divide by 3 to get the equality that is desired.\n\nTo verify:\n\n$${n\\choose2} = \\frac{n(n-1)}{2}$$\n\n$${\\frac{n(n-1)}{2}\\choose2} = \\frac{\\frac{n(n-1)}{2}.(\\frac{n(n-1)}{2}-1)}{2}$$\n\n$${\\frac{n(n-1)}{2}\\choose2} = \\frac{n(n-1).(n^2-n-2)}{8} = \\frac{(n+1)n(n-1)(n-2)}{8}\\tag1 - RHS$$\n\n$${(n+1)\\choose4} = \\frac{(n+1)n(n-1)(n-2)}{4!} = \\frac{RHS}{3}$$\n\nHence proved.\n\n\u2022 AH i see where it went all wrong for me. thanks :) \u2013\u00a0user197848 Jul 17 '15 at 18:20\n\u2022 You are welcome. See the other responder's answer for a combintorial argument. It makes sense. Good luck \u2013\u00a0Satish Ramanathan Jul 17 '15 at 18:22\n\nUsing \\begin{align} \\binom{n}{2} = \\frac{n(n-1)}{2} \\end{align} then \\begin{align} \\frac{1}{3} \\, \\binom{\\binom{n}{2}}{2} &= \\frac{1}{2} \\binom{n}{2} \\, \\left(\\frac{n}{2} - 1\\right) \\\\ &= \\frac{n(n-1)}{4!} \\left( n(n-1)-2 \\right) = \\frac{(n+1)(n)(n-1)(n-2)}{4!} \\\\ &= \\binom{n+1}{4}. \\end{align}\n\nAs to the modified component of the question: The comments seem to be helpful\n\nAlgebraic Manipulation is actually not that bad. $\\binom{n}{2}=n(n-1)/2$, so $$\\binom{\\binom{n}{2}}{2}=\\binom{n(n-1)/2}{2}=(n(n-1)/2)(n(n-1)/2-1)/6=n(n-1)(n^2-n-2)/24=n(n-1)(n+1)(n-2)/24=(n+1)!/4!(n-3)!=\\binom{n+1}{4}$$", "date": "2019-05-22 09:07:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1364821/verify-that-binomn14-frac-left-substack-binomn2-displaystyle", "openwebmath_score": 0.984425961971283, "openwebmath_perplexity": 236.1912134255951, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226267447514, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8573146664216611}}
{"url": "https://math.stackexchange.com/questions/1347223/help-finding-the-limit-of-this-series-frac14-frac18-frac116", "text": "# Help finding the limit of this series $\\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\cdots$\n\nHow can I go about finding the limit of $$\\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\frac{1}{32} + \\cdots = \\sum_{k = 1}^{\\infty} \\frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.\n\n\u2022 Of the series or the sequence? \u2013\u00a0Sloan Jul 2 '15 at 17:26\n\u2022 sorry for the confusion. The third one and the limit of the series. \u2013\u00a0Lil Jul 2 '15 at 17:30\n\u2022 Divide every term by $2$ and see what happens. \u2013\u00a0Yves Daoust Jul 3 '15 at 12:28\n\u2022 The sequence converges to zero, but the sum of the terms converges to a positive value. Be sure you make a distinction between the values of the individual terms and the sum of the terms. \u2013\u00a0John Molokach Jul 3 '15 at 16:12\n\nYou know the limit of $1+\\frac12 +\\frac14+\\frac18+\\ldots$, don't you?\n\nYour sequence is just like that, but without the $1$ and the $\\frac12$.\n\nUsing the formula for the geometric series $$\\sum\\limits_{k=0}^\\infty q^k=\\frac{1}{1-q}$$ with $|q|<1$ we have: $$\\sum\\limits_{k=0}^\\infty \\frac{1}{2^{n+1}}=\\sum\\limits_{k=0}^\\infty \\left(\\frac{1}{2}\\right)^{n+1}=\\sum\\limits_{k=0}^\\infty \\frac{1}{2}\\cdot \\left(\\frac{1}{2}\\right)^n = \\frac {1}{2}\\cdot\\sum\\limits_{k=0}^\\infty \\left(\\frac{1}{2}\\right)^n = \\frac {1}{2}\\cdot \\frac{1}{1-\\frac{1}{2}}=\\frac{1}{2}\\cdot 2=1.$$\n\n\u2022 The problem was modified after you answered it. \u2013\u00a0N. F. Taussig Jul 3 '15 at 10:03\n\u2022 Well, if we start with $k=1$ we can easily obtain $$\\sum\\limits_{k=1}^\\infty \\frac {1}{2^{n+1}}=\\sum\\limits_{k=0}^\\infty \\frac{1}{2^{n+1}}-\\frac{1}{2^{0+1}}=1-\\frac{1}{2}=\\frac {1}{2}.$$ \u2013\u00a0Hirshy Jul 3 '15 at 10:17\n\nIf your series begins at $\\frac{1}{4}$, then it should be:\n\nThe sequence $a_n = \\dfrac{1}{2^{n+2}}$ converges to $0$ as $n \\to \\infty$.\n\nThe series $$\\sum_{n=0}^{\\infty} \\frac{1}{2^{n+2}}=\\frac{1}{4} \\cdot \\sum_{n=0}^{\\infty} \\frac{1}{2^{n}} = \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} + \\ldots$$ is the sum of a geometric sequence with common ratio $\\frac{1}{2}$ and first term $\\frac{1}{4}$. The series is convergent since the absolute value of the common ratio, $|r| = \\frac{1}{2} < 1$. The sum evaluates to $$\\frac{\\frac{1}{4}}{1- \\frac{1}{2}} = \\frac{1}{2}$$ by using the general formula for the sum of a geometric series with common ratio $r$ and first term $a$, namely:$$\\sum_{n=0}^{\\infty}ar^n = \\frac{a}{1-r}$$ as long as $|r| < 1$.\n\n\u2022 maybe I have the wrong formula.. my series is 1/4+1/8+1/16+1/32... \u2013\u00a0Lil Jul 2 '15 at 17:54\n\u2022 It begins at 1/4 not 1/2 as everyone is commenting. \u2013\u00a0Lil Jul 2 '15 at 17:54\n\u2022 Then your series is $$\\sum_{n=0}^{\\infty} \\frac{1}{2^{n+2}}$$ since the sum of a geometric series starts from $n=0$ by convention. :-) \u2013\u00a0Zain Patel Jul 2 '15 at 17:57\n\u2022 I don't understand why it would be 1/2^n+2 because when n=1 it would become 1/2^3 which is 1/8 but my first term is 1/4...? \u2013\u00a0Lil Jul 2 '15 at 17:58\n\u2022 doesn't the sum of a geometric series begin at 1? or no? \u2013\u00a0Lil Jul 2 '15 at 17:58\n\nUse $$\\sum_{n=0}^\\infty a r^n=\\frac{a}{1-r},\\quad |r|<1.$$ Take $r=\\frac12$ and $a=\\frac14$ since the first term of your series is $\\frac14$ and the common ratio of your Geometric progression is $\\frac12.$\n\n\u2022 The problem was modified after you answered it. \u2013\u00a0N. F. Taussig Jul 3 '15 at 10:03\n\nI will add an indefinite version of the sum. In difference calculus we have that the function $2^n$ is the analogous of $e^x$ on ordinary differential calculus.\n\nWe have that $\\sum 2^n \\delta n=2^n$ (you can see it by yourself if you do the difference and see that it doesnt change, i.e. $\\Delta 2^n=2^{n+1}-2^n=2^n$).\n\nIn your sum, doesnt taking limits, we have that\n\n$$\\sum 2^{-n-1}\\delta n=-2^{-n}$$\n\nIf we take limits then\n\n$$\\sum_{n=1}^{\\infty} 2^{-n-1}=-2^{-n}\\bigg\\lvert_{1}^{\\infty}=-2^{-\\infty}+2^{-1}=\\frac{1}{2}$$\n\nTake care with notation because\n\n$$\\sum_{k\\ge 0}2^k\\ne\\sum_{k\\ge 0}2^n$$", "date": "2020-10-27 14:55:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1347223/help-finding-the-limit-of-this-series-frac14-frac18-frac116", "openwebmath_score": 0.9451926350593567, "openwebmath_perplexity": 278.28290238640693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407152622597, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8572275781721027}}
{"url": "https://math.stackexchange.com/questions/1366021/every-normal-subgroup-is-the-kernel-of-some-homomorphism", "text": "# Every normal subgroup is the kernel of some homomorphism\n\nClearly the kernel of a group homomorphism is normal, but I often hear my professor mention that any normal subgroup is the kernel of some homomorphism.\n\nThis feels correct but isn't entirely obvious to me.\n\nOne thought I had is that for any normal subgroup $N$ of $G$, we could define the quotient homomorphism $\\pi:G\\to G/N$ since $G/N$ is a group.\n\nI was imagining that we could consider $\\pi^{-1}:G/N\\to G$, whose kernel would then be $N$. However, $\\pi^{-1}$ doesn't exist since $\\pi$ is not a bijection in general.\n\nSo my question is this: is there an obvious way to define a homomorphism whose kernel is an arbitrary normal subgroup of $G$? Or does it depend on the particular group whether you can define such a homomorphism?\n\n\u2022 No, you more or less have it, the kernel of $\\pi\\colon G\\to G/N$ is $N$. No need to worry about $\\pi^{-1}$, or if it exists. Jul 18, 2015 at 23:17\n\nYou only have to consider that $\\pi:G\\to G/N$, defined by $\\pi(x)=xN$, is a homomorphism and the $\\ker \\pi$ is precisely $N$.\n\n\u2022 overcomplicating as usual. thanks. Jul 18, 2015 at 23:18\n\u2022 that's why we are meant to be here Jul 18, 2015 at 23:19\n\nAnyway, if the kernel is non-trivial, the homomorphism is not injective!\n\nYou're confusing the inverse of an isomorphism, and the inverse image of a subset by a map. So, yes, if $$N$$ is a normal subgroup, it is the kernel of the canonical homomorphism : \\begin{align*}\\pi\\colon G&\\longrightarrow G/N,\\\\ g&\\longmapsto gN. \\end{align*} Indeed, $$\\pi^{-1}(\\bar 1)=\\pi^{-1}(N)=N$$.\n\nYour intuition is correct, and unfortunately, yes, $\\pi^{-1}$ isn't a function. Could we somehow \"make\" it one?\n\nWell, one way this is often done with \"ordinary\" functions, is to treat $f^{-1}$, for a given function $f: A \\to B$, as not something like this:\n\n$f^{-1}: B \\to A$\n\nbut instead considering $f^{-1}$ as a function between power sets:\n\n$f^{-1}: \\mathcal{P}(B) \\to \\mathcal{P}(A)$\n\nwhere for a subset $Y \\subseteq B$, we define $f^{-1}(Y) = X = \\{x \\in A: f(x) \\in Y\\}$.\n\nApplying this definition to $\\pi^{-1}$, we get:\n\n$\\pi^{-1}(e_{G/N}) = \\pi^{-1}(N) = \\{g \\in G: \\pi(g) \\in N\\}$\n\n$= \\{g \\in G: gN = N\\} = \\{g \\in G: ge^{-1} \\in N\\} = \\{g \\in G: g \\in N\\} = N$", "date": "2022-05-24 16:35:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1366021/every-normal-subgroup-is-the-kernel-of-some-homomorphism", "openwebmath_score": 0.9915831685066223, "openwebmath_perplexity": 189.02957173180073, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407168145569, "lm_q2_score": 0.8807970685907242, "lm_q1q2_score": 0.8572275704033968}}
{"url": "https://en.khanacademy.org/math/ap-statistics/density-curves-normal-distribution-ap/stats-normal-distributions/a/basic-normal-calculations", "text": "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## AP\u00ae\ufe0e/College Statistics\n\n### Course: AP\u00ae\ufe0e/College Statistics>Unit 4\n\nLesson 4: Normal distributions and the empirical rule\n\n# Basic normal calculations\n\nAP.STATS:\nVAR\u20112 (EU)\n,\nVAR\u20112.A (LO)\n,\nVAR\u20112.A.3 (EK)\nCCSS.Math:\nMany measurements fit a special distribution called the normal distribution. In a normal distribution,\n\u2022 approximately equals, 68, percent of the data falls within 1 standard deviation of the mean\n\u2022 approximately equals, 95, percent of the data falls within 2 standard deviations of the mean\n\u2022 approximately equals, 99, point, 7, percent of the data falls within 3 standard deviations of the mean\nProblem 1\nA large sample of females had their systolic blood pressure measured. The mean blood pressure was 125 millimeters of mercury and the standard deviation was 10 millimeters of mercury.\nWhich normal distribution below best summarizes the data?\n\nProblem 2\nWhat percent of females had blood pressures between 105 and 135 millimeters of mercury?\npercent\n\nProblem 3\nThe sample had a total of 400 females that participated.\nAbout how many females in the sample had blood pressures higher than 145 millimeters of mercury?\napproximately equals\nfemales\n\n## Want to join the conversation?\n\n\u2022 What is the Gauss curve have to do with this?\n\u2022 For the empirical rule to be valid the data must be normally distributed, so the rules for percentages in the problems above would not hold true if the data didn't follow a gaussian or normal distribution.\n\u2022 I answered the last question in the following way. I figured out the z score which was 0.9772 for value that is two standard deviations above normal. Therefore one can deduce that the sample size that lies to the right of this should be 0.0228 (1-0.9772). That multiplied by 400 gives me 9.1 approximated to 9. What is wrong with this method?\n\u2022 I agree that using the z-Score is the most accurate answer. The answer given is using approximations of the normal and are not as accurate.\n\u2022 where does the 13.5% come from in question 3?\n\u2022 Empirical rule!\nOk, so 95% of the observations are within 2 standard deviations of the mean, and 68% of the observations are within 1 standard deviation of the mean, right?\nLet's find the difference between 2 s.d. and 1 s.d. It will be 95%-68%=27%. But you have to divide this 27% by 2 because you have to find the percentage between 105 and 115 millimeters. There comes 13.5%!\n\u2022 why do i have E.D.?\n\u2022 what if they ask you to solve for standard deviation with only knowing the range\n\u2022 This question is not really meaningful for a normal distribution, since all normal distributions have infinite range.\n\nFor general data sets, knowing the range of a data set is not sufficient for finding its standard deviation. For example, the data sets 1,5,5,9 and 1,2,8,9 both have range 9-1=8, but 1,2,8,9 has the larger standard deviation because the values are spread out farther from the mean (5).\n\nHave a blessed, wonderful day!\n\u2022 Are these realistic statistics? They better not be, my math teacher commonly uses extremely unrealistic stats, it's funny.\n\u2022 For the titles of Articles, Videos, and Skills; should the wording follow the same grammar rules as the title of a book does?\ne.g.: Should the titles be capitalized with the exception of conjunctions like and/or/but/ect?\n\nI'm slightly confused, because KA only capitalizes the first word, but I can't find anywhere on the web where it tells you how to properly right the title.\n\nAny help?\n\nI'm slightly confused...\n\u2022 Khan Academy is an exception to the usually title rules. Sometimes websites might only capitalize the first word or not follow the exact rules as in a book for some videos and articles. You are correct though, that rule is used commonly in books, movies, etc.\n(1 vote)\n\u2022 Is there a table or cheat sheet I can use to help with the subject?\n(1 vote)\n\u2022 Here is a website that has z-tables for positive and negative z-scores as well as other things related to this subject: https://z-scoretable.com/.\n\nThere are plenty of z-tables on Google Images as well.\n\nHope this helps!\ud83d\ude04", "date": "2023-03-23 02:23:21", "meta": {"domain": "khanacademy.org", "url": "https://en.khanacademy.org/math/ap-statistics/density-curves-normal-distribution-ap/stats-normal-distributions/a/basic-normal-calculations", "openwebmath_score": 0.6665677428245544, "openwebmath_perplexity": 944.8025699217044, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319844298981, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8572051757511726}}
{"url": "http://summerofgodel.blogspot.my/", "text": "## Wednesday, December 31, 2014\n\n### Almost all integers contain the digit 3\n\nThis is my last chance to blog this year. However, I did not get back into the proof of G\u00f6del's incompleteness theorem yet. So instead, this post is motivated by this\u00a0Numberphile video,\u00a0which explains why almost all integers contain the digit 3. More precisely, the percentage of integers that contain at least one occurrence of the digit '3' in the set $$\\mathbb{N}_u = \\{0,1,2,3,\\cdots,u-1\\}$$ goes to 100% as the upper bound $$u$$ goes to infinity.\n\nWe are going to attack this problem with two different approaches. First, we'll use a recurrence relation approach. Second, we'll go over the combinatorial proof discussed in the video. Finally, we'll check that the two approaches do give the same answer.\n\nIn both approaches, we will use values of $$u$$ that are powers of 10. So if $$u = 10^n$$, for $$n \\in \\mathbb{N}^+$$, then the set under consideration $$\\mathbb{N}_{10^n}$$, which we denote $$A_n$$, is the set of all of the (non-negative) integers containing at most $$n$$ digits.\n\nRecurrence relation approach\n\nIn this approach, we use $$S_n$$, for $$n \\in \\mathbb{N}^+$$,\u00a0to denote the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '3' and we use $$T_n$$ to denote $$|S_n|$$, the cardinality of $$S_n$$.\n\u2022 Since $$A_1=\\{0,1,2,3,4,5,6,7,8,9\\}$$, $$S_1=\\{3\\}$$ and $$T_1=1$$.\n\u2022 Since $$A_2=\\{0,1,2,3,4,\\cdots,98,99\\}$$, $$S_2=\\{ 3,13,23,30,31,32,33,34,\\cdots,39,43,53,63,73,83,93\\}$$$$=\\{ 3,13,23,43,53,63,73,83,93\\}$$$$\\cup \\{30,31,32,33,34,\\cdots,39\\}$$. In other words, we can split the set $$S_2$$ into two subsets, one whose elements all start with the digit '3' \u00a0(there are exactly 10 such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\\cdot T_1$$ of those, since if the first digit is not '3' then the least significant digit must belong to $$S_1$$). Thus, $$T_2 = 10 + 9\\cdot T_1 = 10 + 9 = 19$$.\n\u2022 Similarly, since $$A_3=\\{0,1,2,3,4,\\cdots,998,999\\}$$, $$S_3$$ is the union of two sets, \u00a0one whose elements all start with the digit '3' \u00a0(there are exactly $$10^2$$ such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\\cdot T_2$$ of those, since if the most significant digit is not '3' then the number formed by the other digits must belong to $$S_2$$).\u00a0Thus, $$T_3 = 10^2 + 9\\cdot T_2= 100 + 9\\cdot 19 = 271$$.\n\u2022 Since this reasoning clearly generalizes to all values of $$n \\in \\mathbb{N}^+$$, we obtain the following recurrence relation:\n$\\left\\{ \\begin{array}{ll} T_1 = 1 & \\\\ T_n = 10^{n-1} + 9T_{n-1} & \\text{if}\\ n>1 \\end{array} \\right.$ Now, let's solve this recurrence relation by iteration to get a closed-form formula for $$T_n$$:\n\n$$\\begin{array}{l@{0pt}l} T_1 & = 1\\\\ T_2 = 10^1 + 9T_1 & = 10+9 \\\\ T_3 = 10^2 + 9T_2 & = 10^2+9(10+9) \\\\ & = 10^2 + 9\\cdot 10 + 9^2 \\\\ T_4 = 10^3 + 9T_3 & = 10^3+9(10^2 + 9\\cdot 10 + 9^2) \\\\ & = 10^3+9\\cdot 10^2 + 9^2 \\cdot 10 + 9^3\\\\ T_5 = 10^4 + 9T_4 & = 10^4+9(10^3+9\\cdot 10^2 + 9^2 \\cdot 10 + 9^3)\\\\ & = 10^4+9\\cdot 10^3+9^2\\cdot 10^2 + 9^3 \\cdot 10 + 9^4\\\\ \\cdots &\\\\ T_n = 10^{n-1} + 9T_{n-1} & = 10^{n-1}+9\\cdot 10^{n-2}+9^2\\cdot 10^{n-3} + \\cdots + 9^{n-2}\\cdot 10^1 + 9^{n-1}\\\\ \\end{array}$$\n\nIn conclusion, $$\\displaystyle T_n = \\sum_{i=0}^{n-1} \\left( 9^{n-i-1}\\cdot 10^i\\right)$$ for $$n \\in \\mathbb{N}^+$$\n\nSince this formula is rather ugly, let's turn to the combinatorial approach discussed in the\u00a0Numberphile video.\n\nCombinatorial approach\n\nIt is easy to determine the total number of integers with exactly $$n$$ digits without having to enumerate them, namely $$10 \\times 10 \\times \\cdots \\times 10 = 10^n$$, \u00a0since there are exactly 10 digits to choose from for each position in the $$n$$-digit number. Note that this count actually includes all of the numbers with leading zeros, such as 01 and 00023, which are identical to 1 and 23, respectively. In other words, what we really have is \u00a0$$\\left|A_n\\right|= 10^n$$.\n\nSimilarly, we can compute the total number of integers with at most $$n$$ digits that do not contain the digit '3', namely $$9 \\times 9 \\times \\cdots \\times 9 = 9^n$$, since there are now only 9 digits to choose from at each position in the integer.\n\nIn conclusion, $$T_n = 10^n -9^n$$.\n\nThis is both a much nicer and easier-to-derive closed-form formula than the one we obtained with the recurrence relation approach. But are the two formulas equal?\n\nLet's check our work\n\nLet $$P(n)$$, for $$n \\in \\mathbb{N}^+$$, denote: $$\\displaystyle \\sum_{i=0}^{n-1} \\left( 9^{n-i-1}\\cdot 10^i\\right) = 10^n - 9^n$$\n\nWe now prove by mathematical induction that $$P(n)$$ holds for \u00a0$$n \\in \\mathbb{N}^+.$$\n\nBasis:\n\u2022 \u00a0$$\\displaystyle \\sum_{i=0}^{1-1} \\left( 9^{1-i-1}\\cdot 10^i\\right) = \\sum_{i=0}^{0} \\left( 9^{1-i-1}\\cdot 10^i\\right) = 9^{1-0-1}\\cdot 10^0 = 9^0 \\cdot 10^0 =1$$\n\u2022 $$10^1 - 9^1 = 10 - 9 = 1$$\n\u2022 Therefore $$P(1)$$ holds\nInductive step:\n\u2022 Assume that $$P(n)$$ holds\u00a0for any $$n \\in \\mathbb{N}^+$$, that is: $\\sum_{i=0}^{n-1} \\left( 9^{n-i-1}\\cdot 10^i\\right) = 10^n - 9^n \\quad \\text{[inductive hypothesis]}$\n\u2022 Let's compute the left-hand side of $$P(n+1)$$:\n$\\begin{array}{l@{0pt}l} \\sum_{i=0}^{n} \\left( 9^{n-i}\\cdot 10^i\\right) & = 9\\left(\\frac{1}{9}\\sum_{i=0}^{n} \\left( 9^{n-i}\\cdot 10^i\\right)\\right)\\\\ & = 9\\left(\\sum_{i=0}^{n} \\left( 9^{n-i-1}\\cdot 10^i\\right)\\right)\\\\ & = 9\\left(\\sum_{i=0}^{n-1} \\left( 9^{n-i-1}\\cdot 10^i\\right) + \\sum_{i=n}^{n} \\left( 9^{n-i-1}\\cdot 10^i\\right) \\right)\\\\ & = 9\\left(\\sum_{i=0}^{n-1} \\left( 9^{n-i-1}\\cdot 10^i\\right) + \\left( 9^{n-n-1}\\cdot 10^n\\right) \\right)\\\\ & = 9\\left(\\sum_{i=0}^{n-1} \\left( 9^{n-i-1}\\cdot 10^i\\right) + \\frac{10^n}{9} \\right)\\\\ & = 9\\left( \\left(10^n - 9^n\\right) + \\frac{10^n}{9} \\right) \\quad \\text{by the inductive hypothesis} \\\\ & = 9\\cdot 10^n - 9\\cdot 9^n + 10^n\\\\ & = 10\\cdot 10^n - 9\\cdot 9^n\\\\ & = 10^{n+1} - 9^{n+1}\\\\ \\end{array}$\n\u2022 Therefore $$P(n+1)$$ holds.\n\nMain result\n\nWe are now in a position to prove our main result, namely that the percentage of integers that contain the digit '3' in the set $$A_n$$ \u00a0goes to 100% as $$n$$ goes to infinity. This percentage is equal to\n$$100\\times\\frac{T_n}{\\left|A_n\\right|}=100\\times\\left(\\frac{10^n-9^n}{10^n}\\right)=100\\times\\left(1-\\frac{9^n}{10^n}\\right)= 100 - 100\\left(\\frac{9}{10}\\right)^n$$.\nAnd this percentage goes to 100% as $$n$$ goes to infinity, since $$\\displaystyle \\lim_{n \\to \\infty}\\left( \\frac{9}{10}\\right)^n = 0$$.\n\nDiscussion\n\nOf course, all of the proofs above still hold if we had picked the digit '8' (say) instead of the digit '3'. In other words, it is also true that almost all integers contain the digit '8'. So, if we use $$E_n$$ to denote the cardinality of the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '8', $$E_n = 10^n-9^n$$. Similarly for the digit '5' (say): $$F_n = 10^n-9^n$$.\n\nBut, if both $$100\\times\\frac{T_n}{\\left|A_n\\right|}$$ and \u00a0$$100\\times\\frac{E_n}{\\left|A_n\\right|}$$ go to 100% as $$n$$ goes to infinity, the sum of these two percentages will eventually exceed 100%! That is true. However, this sum double counts all of the integers that contain both the digit '3' and the digit '8'. The correct percentage of such integers is $$100\\times\\frac{10^n-8^n}{10^n}$$, which also goes to 100% as $$n$$ goes to infinity.\n\nSo there is no paradox here. When $$n$$ gets large enough, almost all of the integers will contain all of the digits 0 through 9. Note that all of these results are asymptotic. Indeed, it takes relatively large values of $$n$$ for these percentages to get anywhere close to 100%. For example, for the percentage of integers that contain the digit '3' to reach 90%, 95%, 99% and 99.99%, the number $$n$$ of digits in the integer must be larger than 21, 28, 43 and 87, respectively.\n\n## Tuesday, July 9, 2013\n\n### Robinson Arithmetic is \u03a31-complete\n\nRecall that Q (i.e., Robinson Arithmetic) is an axiomatized formal theory (AFT) of arithmetic couched in the interpreted formal language LA. Let L be a subset of LA and let T be some AFT of arithmetic.\n\nWe say that T is L-sound iff, for any sentence \u03c6 in L, if\u00a0T \u22a2 \u03c6, then\u00a0\u03c6 is true.\n\nWe say that T is\u00a0L-complete\u00a0iff, for any sentence \u03c6 in L, if \u03c6 is true, then T \u22a2 \u03c6.\n\nIn chapter 9, Peter Smith defines a subset of\u00a0LA, called \u03a31. Then,\u00a0using the fact that\u00a0Q is order-adequate, he proves that\u00a0Q is \u03a31-complete. This is important because the well-formed formulas (wff's) of\u00a0\u03a3can express the decidable numerical properties and relations, and therefore Q will be\u00a0sufficiently strong. Now to the details...\n\nFirst, let's define a few interesting subsets of\u00a0LA:\n\n## Sunday, July 7, 2013\n\nIn chapter 9, Peter Smith defines the following concept: A theory T that captures the relation\u00a0\u2264\u00a0is order-adequate if it satisfies the following nine properties:\n\u2022 O1:\u00a0T \u22a2 \u2200x (0 \u2264 x)\n\u2022 O2:\u00a0For any n, T \u22a2 \u2200x ((x = 0 \u2228\u00a0x = 1 \u2228\u00a0x = 2 \u2228 ...\u00a0\u2228\u00a0x =\u00a0n) \u2192 x \u2264\u00a0n)\n\u2022 O3:\u00a0For any n, T \u22a2 \u2200x (x \u2264\u00a0n\u00a0\u2192\u00a0(x = 0 \u2228\u00a0x = 1 \u2228\u00a0x = 2 \u2228 ...\u00a0\u2228\u00a0x =\u00a0n))\n\u2022 O4: For any n, if T \u22a2 \u03c6(0) and\u00a0T \u22a2 \u03c6(1) and ... and\u00a0T \u22a2 \u03c6(n)\u00a0then\u00a0T \u22a2 (\u2200x \u2264\u00a0n)\u03c6(x)\n\u2022 O5: For any n, if T \u22a2 \u03c6(0) or T \u22a2 \u03c6(1) or ... or T \u22a2 \u03c6(n)\u00a0then\u00a0T \u22a2 (\u2203x \u2264\u00a0n)\u03c6(x)\n\u2022 O6:\u00a0For any n, T \u22a2 \u2200x (x \u2264\u00a0n\u00a0\u2192\u00a0x \u2264 Sn)\n\u2022 O7:\u00a0For any n, T \u22a2 \u2200x (n\u00a0\u2264 x\u00a0\u00a0\u2192 (n\u00a0= x \u00a0\u2228\u00a0Sn\u00a0\u2264 x))\n\u2022 O8:\u00a0For any n, T \u22a2 \u2200x\u00a0(x \u2264\u00a0n\u00a0\u2228\u00a0n\u00a0\u2264 x)\n\u2022 O9:\u00a0For any n>0, T \u22a2 (\u2200x \u2264\u00a0n-1)\u03c6(x)\u00a0\u2192\u00a0(\u2200x \u2264\u00a0n)(x \u2260\u00a0n\u00a0\u2192\u00a0\u03c6(x))\nThen, we have the following theorem:\n\n## Wednesday, July 3, 2013\n\n### Robinson Arithmetic captures the \"less-than-or-equal-to\" relation\n\nIn this post, we'll start discussing the material in Chapter 9 of Peter Smith's book, namely up to section 9.3.\n\nBefore proceeding, review the definition of Robinson Arithmetic\u00a0(denoted Q) as well as what it means for a theory to capture a numerical relation.\n\nNow, we'll show that Robinson Arithmetic captures the \u2264 numerical relation with the open well-formed formula:\u00a0\u2203x\u00a0(x + m = n).\n\nIn other words, we are going to prove the following two-part theorem:\nTheorem: If m and n are natural numbers, then:\n1. If m\u00a0\u2264 n, then Q \u22a2\u00a0\u2203x (x + m = n)\n2. If m > n, then Q \u22a2\u00a0\u00ac\u00a0\u2203x (x + m = n)\nRecall that if m and n are natural numbers, then\u00a0m\u00a0and\u00a0n\u00a0are the numerical terms representing m and n, respectively, in the formal theory.\n\nProof sketch of part 1:\n\n## Sunday, June 30, 2013\n\n### Q - Robinson Arithmetic\n\nNow that we are familiar with Baby Arithmetic (BA), we can make its language more expressive by allowing variables and quantifiers back into its logical vocabulary. When we do this, we simply obtain the interpreted language\u00a0LA, that was described earlier.\n\nSince we now have variables and quantifiers, we can replace the schemata of BA with regular axioms (see below). The resulting formal system of arithmetic is called Robinson Arithmetic\u00a0and is often denoted by the letter Q, as described in Chapter 8 of Peter Smith's book. Here is his definition of Q:\n\n## Friday, June 28, 2013\n\n### Baby arithmetic\n\nSince chapter 7 in Peter Smith's book is so short, I'll summarize it quickly and then start discussing chapter 8.\n\nChapter 7 starts by comparing the two incompleteness theorems discussed in previous chapters (let's call these \"Smith's theorems\") to G\u00f6del's first incompleteness theorem as follows:\n\n## Thursday, June 27, 2013\n\n### Consistent, sufficiently strong formal systems of arithmetic are incomplete\n\nIn Chapter 6, Peter Smith proves another incompleteness theorem. To place this theorem in context, let's review some definitions about axiomatized formal theories (or AFTs) from earlier posts.\n\nLet T be some AFT.\n\u2022 T is consistent iff (if and only if) there is no sentence\u00a0\u03c6 such that T proves both \u03c6 and \u00ac \u03c6.\n\u2022 T is sound iff every theorem that T proves is true according to the interpretation that is built into T's language.\n\u2022 T is (negation-)complete iff for every sentence\u00a0\u03c6 in T's language, T\u00a0proves either \u03c6 or \u00ac \u03c6.\n\u2022 T is decidable iff there is an algorithm that, given any sentence\u00a0\u03c6 in T's language, determines whether or not T proves\u00a0\u03c6.\n\u2022 If needed, go here to review what it means for T's language to express a numerical property P or a numerical relation R.\n\u2022 If needed, go\u00a0here\u00a0to review what it means for T to\u00a0capture\u00a0a numerical property P or a numerical relation R.\n\u2022 If needed, go\u00a0here\u00a0to review what it means for T's language to be sufficiently expressive.\n\nWe'll use the following abbreviations:\n\n## Monday, June 24, 2013\n\n### Sound formal systems with sufficiently expressive languages are incomplete\n\nIn chapter 5, Peter Smith uses a counting argument to prove that sound axiomatized formal theories (AFTs) that can express a good amount of arithmetic are incomplete. In short: since their set of theorems is effectively enumerable but their set of true sentences is not, the two sets must be different. In a sound theory, the theorems are all true. Therefore, some truths must remained unproved in such theories.\n\nLet's start with a definition. The language of an AFT of arithmetic is sufficiently expressive if and only if:\n1. It can express quantification over numbers, and\n2. It can express every effectively decidable two-place numerical relation.\nNow, to the pivotal theorem of this chapter:\n\n## Sunday, June 23, 2013\n\n### An enumerable but not effectively enumerable set\n\nLet's focus on the first half of chapter 5 in Peter Smith's book. First, Smith describes a new characterization of effectively enumerable sets. Second, Smith gives an example of a set of integers that is not effectively enumerable.\n\nLet's define the numerical domain of an algorithm as the set of natural numbers with the following property: when the algorithm takes one of these numbers as input, it terminates and returns a result. Every algorithm has a numerical domain. If the input to some algorithm is not a single natural number, then the numerical domain of this algorithm is the empty set. Otherwise, the numerical domain is the set of those natural numbers on which the algorithm does not crash and does not go into an infinite loop.\n\nIt turns out that each numerical domain is effectively enumerable. In fact, the converse is also true, according to the following theorem:\n\n## Saturday, June 22, 2013\n\n### Capturing numerical properties in a formal language of arithmetic\n\nPeter Smith starts Chapter 4 by describing\u00a0LA, a formal language that is at the core of several AFTs (axiomatized formal theories) of arithmetic. Then Smith explains what it means for a formal language of arithmetic to \"express\" a numerical property and the stronger notion of \"capturing\" a numerical property.\n\nHere is the definition of\u00a0LA:\n\n## Friday, June 21, 2013\n\n### Formal systems or axiomatized formal theories\n\nIn chapter 3, Peter Smith defines formal systems or, as he calls them, axiomatized formal theories\u00a0(I will use AFT as an abbreviation for this phrase).\n\nA theory T is an AFT if...\n\n## Wednesday, June 19, 2013\n\n### Effective computability, decidability and enumerability\n\nIn Chapter 2, Smith covers familiar ground. So this post will be short. Here is a quick summary, section by section.\n\nSection 1 reviews terminology pertaining to functions, namely the notions of domain and range, as well as special cases of functions, namely injective (or one-to-one) functions, surjective (or onto) functions and bijective functions (or one-to-one correspondences).\n\n## Tuesday, June 18, 2013\n\n### Peter Smith's overview of G\u00f6del's incompleteness theorems\n\nFinally! As was my initial goal, I now turn my attention to Peter Smith's book on the proof of G\u00f6del's incompleteness theorems (by the way, I own the first edition of his book; so that is the one I will be referring to). I did not expect to spend that much time on Nagel and Newman's book. But it was worthwhile in getting the big picture.\n\nChapter 1 is an introductory and user-friendly discussion of topics that were mentioned in earlier posts, such as basic arithmetic, formal systems, (in)completeness, consistency, the statement of G\u00f6del's incompleteness theorems and some of their implications. Therefore, in this post, I will just highlight the points where Smith provides new information or has a different perspective.\n\n## Monday, June 17, 2013\n\n### Proof sketch of G\u00f6del's incompleteness theorems\n\nIn this post, I will follow the outline of the proof given in Section VII of Nagel and Newman's book.\u00a0Recall that:\n\u2022 a formal system F is consistent if there is no well-formed formula (wff) w such that F proves both w and \u00ac w, and\n\u2022 a formal system is (semantically)\u00a0complete if it can prove all of the true wff's that it can express.\nNow, G\u00f6del's first incompleteness theorem can be paraphrased as:\nAny consistent formal system that is expressive enough to model arithmetic is both incomplete and \"incompletable.\"\nand\u00a0G\u00f6del's second incompleteness theorem can be paraphrased as:\nAny consistent formal system that is expressive enough to model arithmetic cannot prove its own consistency.\nAccording to Nagel and Newman, a proof of the first theorem can be sketched in 4 steps:\n\n## Friday, June 7, 2013\n\n### Mappings and the arithmetization of meta-mathematics\n\nIn an earlier post, we discussed mappings in general and then described G\u00f6del numbering, which is a mapping between the set of well-formed formulas (wff's) in a formal system and the set of positive integers.\n\nIn part B of section VII, Nagel and Newman describe a more interesting mapping used in G\u00f6del's proof. In this mapping, the domain is\u00a0the set of meta-mathematical statements about structural properties of wff's, while the co-domain is the set of wff's. This post describes how this mapping enabled\u00a0G\u00f6del to arithmetize meta-mathematics.\n\n## Thursday, June 6, 2013\n\n### Mappings and G\u00f6del numbering\n\nSection VI of Nagel and Newman's book describes mappings. A mapping is an operation that applies to two sets called the domain and the co-domain. More specifically, a mapping associates to each element of the domain one or more elements of the co-domain.\n\n## Tuesday, May 28, 2013\n\n### Absolute proof of consistency of FSN\n\nIn section IV of their great little book, Nagel and Newman discuss efforts by Gottlob Frege and then Bertrand Russell to reduce arithmetic to logic. This is clearly another attempt at a relative proof of consistency: if this reduction were successful, then arithmetic would be consistent provided logic is consistent.\n\nWhether this latter statement is true or not, Whitehead and Russell's Principia Mathematica was a landmark achievement: it almost completed the first step in an absolute proof of consistency of arithmetic, since it led to the formalization of an axiomatic system for arithmetic.\n\nIn section V, Nagel and Newman describe a formalization of propositional (or sentential) logic, that is, a subset of the logic system in\u00a0Principia Mathematica (but not a large enough subset to represent arithmetic). The bulk of this section first describes the formalization process, which yields the standard syntax and inference rules of propositional logic (including modus ponens) and then outlines an absolute proof of consistency of this formalized axiomatic system.\n\nThis absolute proof of consistency is a\u00a0proof by contrapositive, which relies on the following true conditional statement:\n\n## Sunday, May 26, 2013\n\n### Absolute proofs of consistency and meta-mathematics\n\nIn earlier posts, we explained why consistency is an important property of axiomatic systems and discussed relative proofs of consistency, in which the proof of consistency of a system is based on the assumption that another axiomatic system is consistent. In this post, we introduce absolute proofs of consistency that do not make any assumptions about any other axiomatic system. Apparently, David Hilbert was the first to study and propose such proofs, according to Nagel and Newman's book (Section III, page 26).\n\nRecall that an axiomatic system is consistent if it cannot derive both a theorem and its negation. What do we mean by the negation of a theorem? Let's take, as a simple example, the theorem: \"6 is divisible by 3.\" Its negation is simply the following statement: \"6 is not divisible by 3\" or equivalently \"It is not the case that 6 is divisible by 3.\" This second formulation of the negation, although less elegant in English, is preferable because the negation is added to the front of the original theorem. In a formal system, negation is handled by simply adding a symbol for the phrase \"it is not the case that.\" Several symbols have been used for negation, such as ~ and \u00ac . We'll use the latter here. So, if T is any theorem in some formal system, then the formula\u00a0\u00a0\u00acT is the negation of T.\n\nRemember that our formal system FS did not have a symbol for negation. So we will extend FS into a new formal system called\u00a0FSN\u00a0(for\u00a0FS\u00a0with\u00a0Negation), whose alphabet is { E, 0, 1,\u00a0\u00ac\u00a0}. FSN has exactly one axiom, namely the same as A1 in FSFSN\u00a0also has the same inference rules as FS, namely IR1 and IR2. But it has one additional inference rule that uses negation. Here is the full description of FSN:\n\n## Wednesday, May 22, 2013\n\n### Relative proofs of consistency\n\nIn the last\u00a0post, we defined and explained the importance of the property of consistency for an axiomatic system. The second half of Section II in Nagel and Newman's book describes ways of proving that an axiomatic system is consistent.\n\nBut first, note that the question of consistency of Euclidean geometry did not arise. Its axioms were supposed to describe the real world; and something that actually exists cannot be self-contradictory. In other words, existence (or truth) implies internal consistency.\n\nThe need for consistency proofs arose much later, with non-Euclidean geometries, which do not obviously model space as we experience it. Non-existence does not imply inconsistency. But any interesting abstract construct had better be internally consistent.\n\nSecond, checking the internal consistency of all of the theorems produced so far is typically not a valid proof of consistency, because (interesting) axiomatic systems generate an infinite number of theorems. The proof of consistency must guarantee that not a single theorem, including some that we have not yet produced and that we might never produce, contradicts any other theorem in the system.\n\nOne possible way to prove the consistency of an axiomatic system is model-based, where a model is a kind of interpretation.\n\n## Monday, May 20, 2013\n\n### Consistency of axiomatic systems\n\nThe \"problem of consistency\" is the topic of Section II of Nagel and Newman's book. \u00a0This section defines \"consistency\" and explains when and why it became an important property of axiomatic systems.\n\nThe oldest and most famous axiomatic system is that of Euclid, in which he systematized all of the knowledge of geometry (and more) available to him over two thousand years ago. Based on five axioms, Euclid was able to rigorously prove a very large number of known and new theorems (called \"propositions\" in his Elements). His axioms were supposed to be intuitively true. The first four axioms dealt with line segments, lines, circles and angles (see this Wikipedia entry) and have been viewed as self-evident. In contrast, the fifth axiom, which was equivalent to the following statement: \"Through a point outside a given line, only one parallel to the line can be drawn\" (page 9), was not intuitively true (apparently because the two lines involved extend to infinity in two directions, similarly to asymptotes).\n\nSince this proposed axiom was not obviously true, many mathematicians tried to prove that it logically follows from the first four axioms. Only in the nineteenth century was it demonstrated that it is NOT possible to prove the parallel axiom from the first four axioms.\n\nThis proof that it is impossible to prove\u00a0a given statement is a great precursor of G\u00f6del's incompleteness theorems.\n\n## Friday, May 17, 2013\n\n### Formal systems, axioms, inference rules, formal proofs\n\nI'll start this post by describing a simple formal system that I made up.\n\nformal system\u00a0is comprised of axioms and inference rules. Each axiom and inference rule is defined syntactically, that is, by ordered sequences of symbols that follow strict syntactic rules but do not (necessarily) have any meaning. The set of all symbols allowed in a formal system are explicitly listed in its\u00a0alphabet, simply, a finite, non-empty set of symbols.\n\nMy formal system, let's call it\u00a0FS, \u00a0uses the alphabet {E,0,1} and has only one axiom and two inference rules. Here is the axiom (in the box below):\n\n## Wednesday, May 15, 2013\n\n### Let's get started with Nagel and Newman (Section I)\n\nLet's ease our way into this. I haven't read the Nagel and Newman book in a few years. But I remember it as a short and highly readable overview of the proof. I have a 1986 softcover edition by NYP Press. Today, I want to write about Section I - Introduction. It's really short: under 5 pages.\n\nFirst, a short quote that makes me feel better about my past failures at REALLY understanding the proof of GIT:\n\n## Tuesday, May 14, 2013\n\n### I just want to understand the proof of G\u00f6del's incompleteness theorems\n\nSo this is it! Summer 2013...\u00a0This is when I get to master the proof of G\u00f6del's incompleteness theorems (thereafter: GIT).\n\nI discovered GIT as an undergraduate student in one of my Artificial Intelligence courses. My instructor was reading G\u00f6del, Escher, Bach and was raving about the book. He even took some of our exam problems from it. But what I remember most is the outline of the proof of GIT, more precisely the first theorem. To be honest, the only part of it that stuck with me was the idea of numbering the formulas. Of course, at the time, I bought a copy of the book. I liked many parts of it but I never finished it.\n\nYears later, while I was teaching computability theory and other computer science topics, I encountered GIT several times. At some point, I decided I wanted to understand the proof of GIT.\n\nFirst, I read the classic \"G\u00f6del's Proof\" by Nagel and Newman. That is a concise and user-friendly overview of the proof. But it was only a proof sketch. So I looked at some textbooks on formal logic, which were neither concise nor user-friendly. I still did not understand the proof.\n\nFinally, a couple of years ago, I came across a promising volume: \"Introduction to G\u00f6del\u2019s Theorems\" by Peter Smith. I read the first few chapters but then ran out of time (and motivation) to finish it. The following year, I made another attempt: I had to start from scratch and did not get any farther into it the second time around...", "date": "2018-03-22 10:03:31", "meta": {"domain": "blogspot.my", "url": "http://summerofgodel.blogspot.my/", "openwebmath_score": 0.7829540371894836, "openwebmath_perplexity": 587.8367039734752, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319855244919, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8572051749767012}}
{"url": "https://joshhug.gitbooks.io/hug61b/content/chap9/chap94.html", "text": "## Weighted Quick Union (WQU)\n\nImproving on Quick Union relies on a key insight: whenever we call find, we have to climb to the root of a tree. Thus, the shorter the tree the faster it takes!\n\nNew rule: whenever we call connect, we always link the root of the smaller tree to the larger tree.\n\nFollowing this rule will give your trees a maximum height of $\\log N$, where N is the number of elements in our Disjoint Sets. How does this affect the runtime of connect and isConnected?\n\nLet's illustrate the benefit of this with an example. Consider connecting the two sets T1 and T2 below:\n\nWe have two options for connecting them:\n\nThe first option we link T1 to T2. In the second, we link T2 to T1.\n\nThe second option is preferable as it only has a height of 2, rather than 3. By our new rule, we would choose the second option as well because T2 is smaller than T1 (size of 3 compared to 6).\n\nWe determine smaller / larger by the number of items in a tree. Thus, when connecting two trees we need to know their size (or weight). We can store this information in the root of the tree by replacing the -1's with -(size of tree). You will implement this in Lab 6.\n\n#### Maximum height: Log N\n\nFollowing the above rule ensures that the maximum height of any tree is \u0398(log N). N is the number of elements in our Disjoint Sets. By extension, the runtimes of connect and isConnected are bounded by O(log N).\n\nWhy log N? The video above presents a more visual explanation. Here's an optional mathematical explanation why the maximum height is $\\log_2 N$. Imagine any element $x$ in tree $T1$. The depth of $x$ increases by $1$ only when $T1$ is placed below another tree $T2$. When that happens, the size of the resulting tree will be at least double the size of $T1$ because $size(T2) \\ge size(T1)$. The tree with $x$ can double at most $\\log_2 N$ times until we've reached a total of N items ($2^{\\log_2 N} = N$). So we can double up to $\\log_2 N$ times and each time, our tree adds a level $\\rightarrow$ maximum $\\log_2 N$ levels.\n\nYou may be wondering why we don't link trees based off of height instead of weight. It turns out this is more complicated to implement and gives us the same \u0398(log N) height limit.\n\n### Summary and Code\n\nImplementation Constructor connect isConnected\nQuickUnion \u0398(N) O(N) O(N)\nQuickFind \u0398(N) \u0398(N) \u0398(1)\nQuickUnion \u0398(N) O(N) O(N)\nWeighted Quick Union \u0398(N) O(log N) O(log N)\n\nN = number of elements in our DisjointSets data structure", "date": "2021-05-15 19:49:35", "meta": {"domain": "gitbooks.io", "url": "https://joshhug.gitbooks.io/hug61b/content/chap9/chap94.html", "openwebmath_score": 0.5889438986778259, "openwebmath_perplexity": 659.542950967756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319874400309, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8572051645832893}}
{"url": "https://math.stackexchange.com/questions/2282473/euler-equations-for-the-incompressible-fluids", "text": "# Euler equations for the incompressible fluids\n\nIncompressible fluid with constant density \u03c1 fills the three-dimensional domain below the free surface $z = \u03b7(r)$ in cylindrical polar coordinates. The flow is axisymmetric and steady, and the only non-zero velocity component is $u_\u03b8$. Gravity acts upon the fluid. The fluid in $r\\lt a$ rotates rigidly about the z-axis with angular velocity $\\Omega$ and the fluid $r \\ge a$ is irrotational.\n\nUse the radial and vertical components of the Euler equations to show that the pressure $p$ in the region $z < \u03b7$, $r < a$ satisfies\n\n$$\\frac{p}{\\rho} = \\frac{1}{2}{\u03a9^2}{r^2} \u2212gz + \\text{constant}$$ and find the constant.\n\n$$\\frac{\u2202u_r}{\u2202t}+\\left(u_r\\frac{\u2202}{\u2202r}+u_z\\frac{\u2202}{\u2202z}\\right)u_r-\\frac{(u_\u03b8)^2}{r}+\\frac{1}{\\rho}\\frac{\u2202p}{\u2202r}=0$$\n\n$$\\frac{\u2202u_\u03b8}{\u2202t}+\\left(u_r\\frac{\u2202}{\u2202r}+u_z\\frac{\u2202}{\u2202z}\\right)u_\u03b8+\\frac{u_\u03b8u_r} {r}=0$$\n\n$$\\frac{\u2202u_z}{\u2202t}+\\left(u_r\\frac{\u2202}{\u2202r}+u_z\\frac{\u2202}{\u2202z}\\right)u_z+\\frac{1}{\\rho}\\frac{\u2202p}{\u2202z}=-g$$\n\nI have the Euler equation in cylindrical coordinate. Then how should I figure out the $\\Omega$. The next question is to show that the free surface position in $r<a$ is $$\u03b7=\\frac{\\Omega^2a^2}{g}(\\frac{r^2}{2a^2}-1)$$ if it is helpful. Thank you so much!\n\n\u2022 Are you sure the radial equations aren't available to you? That's a fair amount of effort to derive those. May 15 '17 at 19:14\n\nSincerely, if you derived the Euler equations for cylindrical coordinates, this is a very simple exercise for you.\n\nWith the hypothesis, many terms disappear, those involving $u_r=0,\\;u_z=0$ and the partials wrt time (steady flow).\n\nFurther, the motion is rigid, meaning that all fluid parts have zero relative motion. If they are moving in circles, the angular velocity for all point in the fuid $\\Omega$ is the same! You don't need to figure anything, it must be known. With all this, $u_\\theta=\\Omega r$\n\n$\\begin{cases} -\\dfrac{(u_\u03b8)^2}{r}+\\dfrac{1}{\\rho}\\dfrac{\\partial p}{\\partial r}=0\\\\ \\dfrac{1}{\\rho}\\dfrac{\\partial p}{\\partial z}=-g \\end{cases}$\n\nIntegrating the second,\n\n$\\dfrac{p}{\\rho}=-gz+f(r)\\tag 1$\n\nSo is $\\dfrac{1}{\\rho}\\dfrac{\\partial p}{\\partial r}=f'(r)$ or $\\dfrac{p}{\\rho}+C=f(r)$. With the first equation,\n\n$-\\dfrac{(u_\u03b8)^2}{r}+f'(r)=0$ or $-\\dfrac{(\\Omega r)^2}{r}+f'(r)=0$ or\n\n$-\\Omega^2r+f'(r)=0$. Integrating, $f(r) = \\dfrac{1}{2}{\u03a9^2}{r^2}+h(z)$\n\n$\\dfrac{p}{\\rho}= \\dfrac{1}{2}{\u03a9^2}{r^2}+h(z)-C\\tag 2$\n\nComparing $(1)$ and $(2)$ we have that $h(z)=-gz+C$\n\n$\\dfrac{p}{\\rho} = \\dfrac{1}{2}{\u03a9^2}{r^2} \u2212gz +C$\n\nI could not recover the given solution for the free surface, although I've found some relation that produces the well known paraboloid for the free surfce of the Newton's bucket. To find $C$, we can consider the heigh $z$ for the free surface ($p=0$) at some $r$. We can set $z=0$ for $r =a$.\n\n$0=\\dfrac{1}{2}{\u03a9^2}{a^2}+C$ or $C=-\\dfrac{1}{2}{\u03a9^2}{a^2}$\n\n$\\dfrac{p}{\\rho} = \\dfrac{1}{2}{\u03a9^2}{r^2} \u2212gz-\\dfrac{1}{2}{\u03a9^2}{a^2}=\\dfrac{1}{2}\u03a9^2(r^2-a^2)-gz$\n\nFor the free surface $z=\\eta(r)$, $p=0$\n\n$\\eta=\\dfrac{1}{2g}\u03a9^2(r^2-a^2),\\;r\\lt a$\n\nFrom other considerations, $\\eta$ for $r\\ge a$ is $\\eta=-\\dfrac{\\Omega^2a^4}{2gr^2}$. As $\\eta$ has to be continuous $\\eta(a)=-\\dfrac{\\Omega^2a^2}{2g}$, for the formula we've found for $r\\lt a$\n\n$0=\\dfrac{\u03a9^2a^2}{2}+\\dfrac{\\Omega^2a^2}{2}+C$ leading to $C=\\Omega^2a^2$ and to $\\eta=\\dfrac{\\Omega^2a^2}{g}(\\dfrac{r^2}{2a^2}-1)$ for $r\\lt a$\n\n$$\\eta= \\begin{cases} \\dfrac{\\Omega^2a^2}{g}(\\dfrac{r^2}{2a^2}-1)& x\\lt a\\\\ -\\dfrac{\\Omega^2a^4}{2gr^2}&a\\le x \\end{cases}$$\n\nI've sketched the function for the free surface (setting $\\dfrac{\\Omega^2a^2}{2g}=1$ and $a=1$)\n\n\u2022 It is actually the next two question of this one. math.stackexchange.com/questions/2280167/\u2026 I try to get the constant and the free surface and the number matches when I set r=a, p=0 and use the free surface when r>=a. I get C=-1/2(\\Omega) ^2a^2-(\\Omega)^2a^2/2 and free surface as above. Can I just plug in the limit May 16 '17 at 7:40\n\u2022 And if I want to draw a sketch for the free surface displacement then it is gonna be a paraboloid but when happen at r=a then? May 16 '17 at 7:49\n\u2022 @stedmoaoa, the resultig function is smooth. May 16 '17 at 13:47\n\u2022 Thank you so much it really makes sense. So the graph is like x^2-1 and -1/x^2 Then, how should I describe at r=a? May 16 '17 at 14:08\n\u2022 Continuous and differentiable for all quantites. May 16 '17 at 17:37", "date": "2021-09-21 03:01:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2282473/euler-equations-for-the-incompressible-fluids", "openwebmath_score": 0.8719672560691833, "openwebmath_perplexity": 398.4050063086413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754515389344, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8571687118744152}}
{"url": "https://math.stackexchange.com/questions/1866962/a-statement-about-divisibility-of-relatively-prime-integers", "text": "# A statement about divisibility of relatively prime integers\n\nI'm solving a problem, and the solution makes the following statement: \"The common difference of the arithmetic sequence 106, 116, 126, ..., 996 is relatively prime to 3. Therefore, given any three consecutive terms, exactly one of them is divisible by 3.\" Why is this statement true? Where does it come from? Is it generalizable to other numbers?\n\n\u2022 And here I thought it was something about \"relatively prime statements\" \u2013\u00a0Yuriy S Jul 21 '16 at 21:56\n\nThere is a simple, well-known theorem about that:\n\nIf $a$, $b$ and $m$ are integers, $m\\ge 2$ and $\\gcd(b,m)=1$ then the set $\\{a,a+b,a+2b,\\ldots,a+(m-1)b\\}$ is a complete residue system.\n\nThis implies, for example, that the set contains exactly one multiple of $m$.\n\n\u2022 Also, most of the other answers hint at a proof of that theorem. \u2013\u00a0Jeppe Stig Nielsen Jul 22 '16 at 8:04\n\nTake $a\\in\\mathbb{N}$ and suppose that $a\\equiv 1\\pmod 3$. Then \\begin{align*} a&\\equiv 1\\pmod 3\\\\ a+10\\equiv 11&\\equiv 2\\pmod 3\\\\ a+20\\equiv 21&\\equiv 0\\pmod 3 \\end{align*}\n\nIf we instead have $b\\in\\mathbb{N}$ such that $b\\equiv 2\\pmod 3$, then adding $10$ to $b$ is sufficient. The case in which we have $\\equiv 0\\pmod 3$ is obviously trivial.\n\nIf $\\,b\\,$ is coprime to $\\,m\\,$ then $\\,a+ib,\\, i = 1,\\ldots,m\\,$ is a complete residue system mod $\\,m\\,$ since\n\n$$\\ a+ib \\equiv a+j b \\iff (i\\!-\\!j)b\\equiv 0\\overset{(b,m)=1}\\iff i\\!-\\!j\\equiv 0\\iff i = j\\,\\ {\\rm by}\\,\\ 1\\le i,j\\le m$$\n\nHence, being complete, it contains an element $\\equiv 0,\\,$ i.e. a multiple of $\\,m.$\n\nRemark $\\$ When $\\,b = 1\\,$ we obtain the well-known special case that any sequence of $\\,m\\,$ consecutive integers contains a multiple of $\\,m.$\n\nIf $d$ is relatively prime to $b$, then $a + d j$ is divisible by $b$ if and only if $j \\equiv -a d^{-1} \\mod b$. Thus exactly one of any $b$ consecutive terms of the arithmetic sequence $a, a+d, a+2d, \\ldots$ is divisible by $b$.\n\nIn general let $d$ be the common difference of an arithmetic sequence, where $3\\nmid d$, and let $a$, $a+d$, and $a+2d$ be three consecutive terms in that sequence. Now WLOG if we assume $3\\nmid a$ then we can write $a$ and $d$ as: $$a=3k+1\\quad\\text{ or }\\quad a'=3k+2\\quad\\text{ and }\\quad d=3j+1\\quad\\text{ or }\\quad d'=3j+2$$ for some $j$, $k\\in\\mathbb{Z}$, where primes have been added to distinguish the two possibilities for $a$ and $d$. Now look at the four possibilities we have for $a+d$, and $a+2d$: $$a+d=3(k+j)+2,\\quad \\quad a+2d=3(k+2j)+3\\qquad (A)$$ $$a+d'=3(k+j)+3,\\quad \\quad a+2d'=3(k+2j)+5\\qquad (B)$$\n\n$$a'+d=3(k+j)+3,\\quad \\quad a'+2d=3(k+2j)+5\\qquad (C)$$ $$a'+d'=3(k+j)+4,\\quad \\quad a'+2d'=3(k+2j)+6\\qquad (D)$$\n\nAs can be seen, since we assumed $3\\nmid a$, only one member of the three consecutive terms is divisible by $3$ on listing all possible combinations (A), (B), (C), and (D). It would make no difference if we assumed $3\\nmid a+d$ or $3\\nmid a+2d$ as our starting point as the problem is cyclic modulo $3$. By this look at the congruences of $a$, $a+d$, $a+2d\\pmod{3}$ for each of (A), (B), (C), and (D) and you see we get the full set of residue classes $[0]$, $[1]$, $[2]$ modulo $3$ in some order, which is the reason we get one and only one number divisible by $3$.\n\nFurther for some $m$ for which $\\gcd(m,d)=1$ look at $m$ consecutive terms of an arithmetic sequence module $m$: $$a,\\ a+d,\\ a+2d,\\dotsc,\\ a+(m-1)d\\pmod{m}\\qquad (E)$$ then $a+kd\\equiv a+jd\\pmod{m}$ iff $(k-j)d\\equiv 0\\pmod{m}$ for $0\\le k,j\\le m-1$, which since $[0]$, $[1]$, $[2],\\dotsc,\\ [m-1]$ form a complete residue system modulo $m$ we must have $[k]=[j]$ modulo $m$, and so in fact $k=j$ with respect to our $m$ consecutive terms with any distinct pair of terms incongruent modulo $m$, and only one unique number in the set being divisible by $m$. To see this imagine (E) rewritten with respect to residue classes modulo $m$ reordered so: $$a+[0],\\ a+[1],\\ a+[2],\\dotsc,\\ a+[m-1]\\pmod{m}\\qquad (E')$$ If $m\\mid a$ then $m\\equiv a+[0]\\pmod{m}$, and we also see $m$ cannot divide any other number in the set; if $m\\nmid a$ then we can write $a=mj+k$ where $j$, $k\\in\\mathbb{Z}$, and $\\gcd(k,m)=1$, which is in the class $[mj+k]=[k]=[k']$ modulo $m$, and $[k']+[k'']=[0]$ modulo $m$, $0<k',k''\\le m-1$, where $[k']$ and $[k'']$ are additive inverses in the complete residue system (note that $[0]$ is self inverse wrt addition).", "date": "2019-04-24 11:48:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1866962/a-statement-about-divisibility-of-relatively-prime-integers", "openwebmath_score": 0.9589896202087402, "openwebmath_perplexity": 90.82299456413315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754452025767, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8571687063579994}}
{"url": "http://mathhelpforum.com/geometry/167839-counting-sides.html", "text": "# Math Help - Counting sides\n\n1. ## Counting sides\n\nA rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length. What is the total number of 1cm long sides in those 24 squares? (If 2 squares share a side, the side should be counted only once).\n\nNow I know the answer is 58 and I drew a sketch and counted the sides but I was wondering if there was any shortcut to this problem that I was missing..? Setting up any kind of expression or equation or maybe a combinations problem of some sort?\n\n2. Let use a vertical layout like this (for explanation):\n\n****\n****\n****\n****\n****\n****\n\nFirst count all the right and bottom sides of each square: 24 * 2 = 48. There is no overlap in this counting.\nHowever, we missed some sides with this count. Which ones? The entire top row and left column.\nWell, the number of edges we did not count is:\n# on top + # on left = 4 + 6 = 10\n\nGrand total is: 48 + 10 = 58\n\nSo from this the general solution for number of sides in a n by m block is:\n2nm + n + m = n(1 + m) + m(1 + n)\n\nThis suggest thats an alternative way is to count:\n#horizontal edges + #vertical edges = n(1 + m) + m(1 + n)\n\n3. Hi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!\n\n4. Hello, sarahh!\n\nA rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length.\nWhat is the total number of 1cm long sides in those 24 squares?\n(If 2 squares share a side, the side should be counted only once).\n\nYou made a sketch but you didn't see any pattern, did you?\n\nCode:\n\n* - * - * - * - * - * - *\n|   |   |   |   |   |   |\n* - * - * - * - * - * - *\n|   |   |   |   |   |   |\n* - * - * - * - * - * - *\n|   |   |   |   |   |   |\n* - * - * - * - * - * - *\n|   |   |   |   |   |   |\n* - * - * - * - * - * - *\n\nThe unit squares are in 4 rows and 6 columns.\n\nImagine these 24 squares formed with matches.\nHow many matches are used?\n\nWe see that there are 6 matches in each row.\nAnd there are 4 + 1 rows.\n\nHence, there are: .6(4+1) horizontal matches.\n\nWe see that there are 4 matches in each column.\nAnd there are 6 + 1 columns.\n\nHence, there are: .4(6 + 1) vertical matches.\n\nWe can generalize this problem.\nSuppose there are R rows and C columns.\n\nThen there are: .C(R+1) horizontal matches\n. . . . . . . .and: .R(C+1) vertical matches.\n\nTotal: . $C(R+1) + R(C+1) \\;=\\; R + C + 2RC$ matches.\n\n5. Originally Posted by sarahh\nHi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!\nThere is a row of 6 and a column of 4 missing. Which is why a 10 (= 6 + 4) is added at the end.\n\n6. Just for fun, here is another approach-- first do it the wrong way, then patch it up.\n\nThere are 24 squares in all, each with 4 sides, and each side is shared between two squares, so there are\n\n$\\frac{4 \\times 24}{2} = 48$ sides.\n\nBut oops, the sides along the perimeter of the rectangle (2 * (4 + 6) = 20) are not shared! So we need to add those back in.\n\n$48 + \\frac{20}{2} = 58$", "date": "2014-08-28 04:24:57", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/167839-counting-sides.html", "openwebmath_score": 0.6646708250045776, "openwebmath_perplexity": 659.3125972834156, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754497285468, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8571686987284427}}
{"url": "https://math.stackexchange.com/questions/113536/fibonaccis-final-digits-cycle-every-60-numbers", "text": "Fibonacci's final digits cycle every 60 numbers\n\nHow would you go about to prove that the final digits of the Fibonacci numbers recur after a cycle of 60?\n\nReferences:\n\nThe sequence of final digits in Fibonacci numbers repeats in cycles of 60. The last two digits repeat in 300, the last three in 1500, the last four in , etc. The number of Fibonacci numbers between and is either 1 or 2 (Wells 1986, p. 65).\n\nhttp://mathworld.wolfram.com/FibonacciNumber.html\n\n\u2022 Why do you conjecture that? Do you have a reference that asserts the statement? \u2013\u00a0Giovanni De Gaetano Feb 26 '12 at 9:52\n\u2022 I added some references. \u2013\u00a0crazyGuy Feb 26 '12 at 9:57\n\u2022 Neat! Have a look at this link. \u2013\u00a0Juan S Feb 26 '12 at 10:00\n\u2022 Here is a link: oeis.org/A096363 \u2013\u00a0Stefan Gruenwald Dec 14 '18 at 20:59\n\nNotice that:\n\n$$F_{n+15} \\equiv 7F_n \\pmod{10}$$ for $$n\\geq 1$$.\n\nAlso the order of $$7$$ mod $$10$$ is $$4$$ so the repetition in the digits of the Fibonacci numbers begins after place $$15\\times 4 = 60$$.\n\n\u2022 I got that recursion by just writing out the first $16$ Fibonacci numbers mod $10$ and noticing that you get $1,1,...,7,7,...$, this dictates that that next $13$ numbers must be $7$ times their corresponding numbers (mod $10$) in the first lot of $15$ numbers. \u2013\u00a0fretty Feb 26 '12 at 10:03\n\u2022 How do we know that F_n+15 = 7F_n mod 10 ? and How do you get that 7 mod 10 is 4 ? thanks. \u2013\u00a0crazyGuy Feb 26 '12 at 10:07\n\u2022 Write out the first 16 Fibonacci numbers mod 10: 1,1,...,7,7,... Now by the recursion of the fibonacci numbers the fact that this \"multiple-repeat\" has happened now tells us that the next 13 numbers must be the same as the 3rd-15th Fibonacci numbers but multiplied by $7$ mod 10. So an actual repeat will happen as soon as we have multipled by enough $7$'s to get $1$ mod 10. This is the \"order\" of $7$ mod $10$. It is easy to work out here, $7^2 \\equiv 4$ mod $10$ and $7^4 \\equiv 1$ mod $10$ so the order is $4$. \u2013\u00a0fretty Feb 26 '12 at 10:13\n\u2022 @sic2 He's not saying that $7\\mod10$ is $4$, but rather that $7^4\\equiv 1\\mod 10$ and that this is not true for any smaller power of $7$, which is a consequence of Euler's theorem. \u2013\u00a0Alex Becker Feb 26 '12 at 10:15\n\u2022 Also this method generalises to the Fibonacci numbers mod $n$. Find the first $1,1,...,a,a,...$ then work out the order of $a$ mod $n$ and you will have your first place where it recurs. \u2013\u00a0fretty Feb 26 '12 at 10:29\n\nSince each term in the Fibonacci sequence is dependent on the previous two, each time a $0\\pmod{m}$ appears in the sequence, what follows must be a multiple of the sequence starting at $F_0,F_1,\\dots=0,1,\\dots$ That is, a subsequence starting with $0,a,\\dots$ is $a$ times the sequence starting with $0,1,\\dots$\n\nConsider the Fibonacci sequence $\\text{mod }2$: $$\\color{red}{0,1,1,}\\color{green}{0,1,1,\\dots}$$ Thus, the Fibonacci sequence repeats $\\text{mod }2$ with a period of $3$.\n\nConsider the Fibonacci sequence $\\text{mod }5$: $$\\color{red}{0,1,1,2,3,}\\color{green}{0,3,3,\\dots}$$ Thus, the Fibonacci sequence is multiplied by $3\\pmod{5}$ each \"period\" of $5$. Since $3^4=1\\pmod{5}$, the Fibonacci sequence repeats $\\text{mod }5$ with a period of $20=4\\cdot5$.\n\nThus, the Fibonacci sequence repeats $\\text{mod }10$ with a period of $60=\\operatorname{LCM}(3,20)$.\n\nNote that\n\n$$\\begin{pmatrix} F_{n+1}\\\\ F_{n+2} \\end{pmatrix} = \\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} F_n \\\\ F_{n+1} \\end{pmatrix}$$\n\nand\n\n$$\\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}^{60} \\equiv \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\mod 10.$$\n\nOne can verify that $60$ is the smallest power for which this holds, so it is the order of the matrix mod 10. In particular\n\n$$\\begin{pmatrix} F_{n+60}\\\\ F_{n+61} \\end{pmatrix} \\equiv \\begin{pmatrix} F_n \\\\ F_{n+1} \\end{pmatrix} \\mod 10$$\n\nso the final digits repeat from that point onwards.\n\n\u2022 The same matrix applies to the Lucas Numbers, however, $\\text{ mod }10$, they have a period of $12$. The fact that $\\begin{pmatrix} 0&1\\\\1&1\\end{pmatrix}^{60}=\\begin{pmatrix}1&0\\\\0&1\\end{pmatrix}\\text{ mod }10$, says that the period of a sequence satisfying $a_n=a_{n-1}+a_{n-2}$ divides $60$. \u2013\u00a0robjohn Feb 26 '12 at 19:21\n\n$F_{0}=F_1=F_{60}=F_{61}= 1 \\mod10$\n\nBy inspection, these are the first two pairs of consecutive Fibonaccis for which this is true. Since the recurrence relation only takes into account the previous two terms and last digits only depend on previous last digits, this suffices to prove the claim.\n\nI have found that there is an 11 number grouping. You must reduce numbers to a single digit (13=4 or 55=10=1). After 24 numbers into the sequence it repeats. To go a little deeper, a 9 is in the twelfth and twenty- fourth sequence and the real number is also divisible by 12 in these positions (which coincidentally is every twelfth number in the sequence).\n\nNow if you do the multiplication table of numbers 1 through 8 and reduce all numbers to a single digit, you will find that 1 and 8 correspond in reverse with each other. The same for 2 and 7, also 4 and 5. 3 and 6 are different. You will notice that for every number sequence it will repeat after a 9 appears. So, Fibonacci sequence would be this:\n\n1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9......\n\nRemember how I said 1 and 8, 2 and 7, 4 and 5 correspond with each other by reducing the multiplication table to single digits and all numbers repeat a sequence after a 9?\n\nIf you move the eleven number sequence between the nines and set 1,1,2,3... above the sequence 8,8,7,6... You will find that they all correspond with the reducing method afore mentioned.\n\nIt's very interesting.\n\n\u2022 What do you mean $11$-number grouping? Why are you reducing mod $9$ instead of mod $10$? How does this prove that it repeats with cycle $60$? \u2013\u00a06005 Oct 24 '16 at 2:58\n\u2022 I apologize. This was not to prove or disprove a 60 cycle repeat. Maybe I posted in the wrong area? I just found this sequence repeat and I thought I would give it to bigger brains and see what they can make of it. \u2013\u00a0Michael Bunton Oct 24 '16 at 3:12\n\u2022 11 number grouping means the first 11 numbers in the sequence. The twelfth number is 144 and that reduces to 9 i.e.(1+4+4=9). Then a second set of eleven numbers followed by a reduced 9. \u2013\u00a0Michael Bunton Oct 24 '16 at 3:20", "date": "2019-10-14 01:11:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/113536/fibonaccis-final-digits-cycle-every-60-numbers", "openwebmath_score": 0.7990440130233765, "openwebmath_perplexity": 267.36283155366385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754515389344, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8571686936932157}}
{"url": "https://math.stackexchange.com/questions/1396839/find-limit-a-n-frac1n1-frac1n2-frac1nn/1396843", "text": "# find Limit $a_n= \\frac{1}{n+1}+\\frac{1}{n+2}+\u2026+\\frac{1}{n+n}$ [duplicate]\n\nshow sequence $a_n= \\frac{1}{n+1}+\\frac{1}{n+2}+...+\\frac{1}{n+n}$ converges to log2\n\nmy attempt:\n\n1. sequence $a_n$ is monotonic increasing\n2. 0<$a_n$<1/2\n\nhow to find limit?\n\n\u2022 Oh come on, this must be the duplicate of three duplicates. \u2013\u00a0Vincenzo Oliva Aug 14 '15 at 9:39\n\n$$\\lim_{n\\rightarrow \\infty}\\sum_{i=1}^{n}\\frac 1 {n+i}=\\lim_{n\\rightarrow \\infty}\\sum_{i=1}^{n}\\frac 1 n\\frac 1 {1+\\dfrac in}=\\int_0^1\\frac 1 {1+x}dx=\\ln(1+1)-\\ln(1+0)=\\ln(2)$$ (Riemann sum)\n\n\u2022 +1, but there should be a limit as $n \\to \\infty$ instead of the sum going to infinity. The sum goes fom $i=1$ to $i=n$ \u2013\u00a06005 Aug 14 '15 at 9:07\n\u2022 @6005 Fixed, thank you! \u2013\u00a0GeorgSaliba Aug 14 '15 at 9:09\n\u2022 ya i was hoping there would be some other way. riemann sum cannot be used to compute limit of sequence$a_n=1+1/2+1/3...+1/n\u2212log(n+1)$ \u2013\u00a0ketan Aug 14 '15 at 9:19\n\u2022 @ketan Then why not use the harmonic number $H_n$ that was already mentioned by @MichaelGaluza? But I suppose this becomes a slightly different question than the one you asked. \u2013\u00a0GeorgSaliba Aug 14 '15 at 9:21\n\u2022 @ketan $a_n=1+1/2+1/3+...+1/n - \\log(n+1)=H_n -\\log(n+1)=\\log(n)+\\gamma + \\epsilon_n-\\log(n+1)=\\log(n/(n+1))+\\gamma +\\epsilon _n$ \u2013\u00a0GeorgSaliba Aug 14 '15 at 9:23\n\n\\begin{align} \\log(2) &=\\sum_{k=1}^\\infty\\frac{(-1)^{k-1}}k\\\\ &=\\lim_{n\\to\\infty}\\sum_{k=1}^{2n}\\frac{(-1)^{n-1}}k\\\\ &=\\lim_{n\\to\\infty}\\left(\\vphantom{\\sum_{k=1}^n}\\right.\\overbrace{\\sum_{k=1}^{2n}\\frac1k}^{\\substack{\\text{sum of all}\\\\\\text{terms}}}-2\\overbrace{\\sum_{k=1}^n\\frac1{2k}}^{\\substack{\\text{sum of even}\\\\\\text{terms}}}\\left.\\vphantom{\\sum_{k=1}^n}\\right)\\\\[6pt] &=\\lim_{n\\to\\infty}\\left(\\sum_{k=1}^{2n}\\frac1k-\\sum_{k=1}^n\\frac1k\\right)\\\\[6pt] &=\\lim_{n\\to\\infty}\\sum_{k=n+1}^{2n}\\frac1k\\\\ \\end{align}\n\n\u2022 Nice overbrace. Is there a corresponding underbrace command or somethings similar? \u2013\u00a0mathreadler Aug 14 '15 at 9:23\n\u2022 Yep. It is \\underbrace :-) Then you use subscripts instead of superscripts. \u2013\u00a0robjohn Aug 14 '15 at 9:23\n\nHere is classical proof of it. Let $H_n=\\sum_{i=1}^n 1/i$. It's well-known that $$H_n = \\ln n + \\gamma + \\epsilon_n,$$ where $\\gamma$ is Euler\u2013Mascheroni constant, $\\epsilon_n\\to 0$. So, $$a_n = H_{2n}-H_{n} = \\ln 2 + \\epsilon_{2n} - \\epsilon_n \\to \\ln2.$$\n\n\u2022 what is $\\epsilon_n$? is it always zero? \u2013\u00a0ketan Aug 14 '15 at 9:28\n\u2022 @ketan, no. It's sequence which tends to $0$. \u2013\u00a0Michael Galuza Aug 14 '15 at 11:39\n\nNotice, $$a_n=\\frac{1}{n+1}+\\frac{1}{n+2}+\\frac{1}{n+3}+\\ldots +\\frac{1}{n+n}$$ $$=\\sum_{r=1}^{\\infty}\\frac{1}{n+r}=\\sum_{r=1}^{\\infty}\\frac{1}{n\\left(1+\\frac{r}{n}\\right)}$$$$=\\sum_{i=1}^{\\infty}\\frac{\\frac{1}{n}}{1+\\frac{r}{n}}$$ Now, let $\\frac{r}{n}=x\\implies \\frac{1}{n}=dx\\to 0$ $$\\text{upper limit of x}=\\lim_{n\\to \\infty}\\frac{n}{n}=1$$ $$\\text{lower limit of x}=\\lim_{n\\to \\infty}\\frac{1}{n}=0$$ Hence, using integration we get $$a_n=\\int_{0}^{1}\\frac{dx}{1+x}$$ $$=\\left[\\ln (1+x)\\right]_{0}^{1}$$ $$=\\left[\\ln 2-\\ln 1\\right]=\\ln 2$$", "date": "2020-10-30 07:23:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1396839/find-limit-a-n-frac1n1-frac1n2-frac1nn/1396843", "openwebmath_score": 0.9540139436721802, "openwebmath_perplexity": 1964.3167225974635, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754447499795, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8571686910884428}}
{"url": "http://mathhelpforum.com/number-theory/151472-proving-binomial-coeficient-always-natural-number.html", "text": "# Math Help - Proving the binomial coeficient is always a natural number?\n\n1. ## Proving the binomial coeficient is always a natural number?\n\nHow would one go about proving that:\n\n${ n \\choose k }$\n\nis always a natural number? That is, assuming $0 \\leq k \\leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance\n\n2. Originally Posted by mfetch22\nHow would one go about proving that:\n\n${ n \\choose k }$\n\nis always a natural number? That is, assuming $0 \\leq k \\leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance\nUse the fact that $\\displaystyle{\\binom{n}{k}=\\binom{n-1}{k}+\\binom{n-1}{k-1}$ for $\\displaystyle{k=1\\ldots n-1}$ and $\\displaystyle{\\binom{n}{0}=\\binom{n}{n}=1}$.\n\n(This will be a proof by induction)\n\n3. Originally Posted by mfetch22\nHow would one go about proving that:\n\n${ n \\choose k }$\n\nis always a natural number? That is, assuming $0 \\leq k \\leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance\nThere's also the combinatorial proof. (The binomial coefficient is given as an \"archetypal example\" not far from the top of the page.) The formula(s) given for $\\displaystyle { n \\choose k }$ can be shown to equal either of the two equivalent quantities\n\n1) the coefficient of the $\\displaystyle x^k$ term in the polynomial expansion of the binomial power $\\displaystyle (1 + x)^n$\n2) the number of ways to choose k elements from a set of n elements. (The number of distinct k-subsets of {1, 2, ... , n}.)\n\nObviously both of these quantities are integers.\n\n4. Another way is to use Legendre's formula, which states that ( $\\mathbb{P}$ is the set of prime numbers)\n\n$\nn! = \\prod_{p\\in \\mathbb{P}} p^{\\sum_{i=1}^\\infty \\Big\\lfloor\\frac{n}{p^i}\\Big\\rfloor}\n$\n\nThus for any prime $p$ we have:\n\n$\n\\binom{n}{k} = \\frac{n!}{k!(n-k)!} = \\prod_{p\\in \\mathbb{P}} p^{\\sum_{i=1}^\\infty \\Big\\lfloor\\frac{n}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{k}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{n-k}{p^i}\\Big\\rfloor}}\n$\n\nSo all what's left is to show that $\\lfloor{x+y}\\rfloor \\ge \\lfloor{x}\\rfloor + \\lfloor{y}\\rfloor$\n\nso it follows that for all $i\\in\\mathbb{N}$:\n\n$\n\\Big\\lfloor\\frac{n}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{k}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{n-k}{p^i}\\Big\\rfloor} \\ge 0\n$\n\nand therefore $\\binom{n}{k} \\in \\mathbb{N}$\n\n5. Originally Posted by Unbeatable0\nAnother way is to use Leibnitz' formula, which states that ( $\\mathbb{P}$ is the set of prime numbers)\n\n$\nn! = \\prod_{p\\in \\mathbb{P}} p^{\\sum_{i=1}^\\infty \\Big\\lfloor\\frac{n}{p^i}\\Big\\rfloor}\n$\n\nThus for any prime $p$ we have:\n\n$\n\\binom{n}{k} = \\frac{n!}{k!(n-k)!} = \\prod_{p\\in \\mathbb{P}} p^{\\sum_{i=1}^\\infty \\Big\\lfloor\\frac{n}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{k}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{n-k}{p^i}\\Big\\rfloor}}\n$\n\nSo all what's left is to show that $\\lfloor{x+y}\\rfloor \\ge \\lfloor{x}\\rfloor + \\lfloor{y}\\rfloor$\n\nso it follows that for all $i\\in\\mathbb{N}$:\n\n$\n\\Big\\lfloor\\frac{n}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{k}{p^i}\\Big\\rfloor-\\Big\\lfloor\\frac{n-k}{p^i}\\Big\\rfloor} \\ge 0\n$\n\nand therefore $\\binom{n}{k} \\in \\mathbb{N}$\nPerfect , in fact $\\lfloor{ x+y \\rfloor} = \\lfloor{ x \\rfloor} + \\lfloor{ y \\rfloor} + \\lfloor{ \\{ x \\} + \\{ y \\} \\rfloor} \\geq \\lfloor{ x \\rfloor} + \\lfloor{ y \\rfloor}$ .\n\n6. Originally Posted by mfetch22\nHow would one go about proving that:\n\n${ n \\choose k }$\n\nis always a natural number? That is, assuming $0 \\leq k \\leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance\nIf you are asking from the perspective of the Binomial Expansion,\nthen you can observe a pattern if you refrain from expressing the terms with indices...\n\n$(a+b)^2=(a+b)(a+b)=aa+ab+ba+bb$\n\n\"a\" times \"b\" is expressed in both possible ways.\n\n$(a+b)^3=(a+b)(a+b)^2=(a+b)(aa+ab+ba+bb)$\n\n$=aaa+aab+aba+abb+baa+bab+bba+bbb$\n\n$=aaa+aab+aba+baa+abb+bab+bba+bbb$\n\nIf the \"a\" values were labelled $a_1,\\ a_2,\\ a_3$ we could arrange them,\n\nhence we have all possible selections of 2 \"a\"'s with a \"b\", 2 \"b\"'s with an \"a\",\nthe only possible selection of 3 \"a\"s, the only selection of 3 \"b\"'s.\n\nFor higher powers, the same pattern exists.\nSince this pattern is the same as \"choose k from n available\",\nthe binomial coefficient may be calculated using\n\n$\\binom{n}{k}$\n\nThere are k! arrangements of a selection.\nAs shown above, the number of selections is a natural number,\nwhich can be calculated by \"unarranging\" the arrangements.\n\nThe number of arrangements of k from n is n(n-1)(n-2)...(n-[k-1])\n\nwhich is $\\frac{n!}{k!}$ as k! has been \"cut off\" the tail.\n\nThe number of arrangements is k! times the number of selections.", "date": "2014-07-22 18:41:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/151472-proving-binomial-coeficient-always-natural-number.html", "openwebmath_score": 0.814684271812439, "openwebmath_perplexity": 302.88837267326954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754501811437, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.8571686908582903}}
{"url": "https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz/2338828", "text": "# $C^1$ function on compact set is Lipschitz\n\nLet $\\emptyset\\ne A\\subset\\mathbb{R}^n$ be open and let $f \\in C^1 (A, \\mathbb{R})$ be a function. Let $\\emptyset\\ne K\\subset A$ be compact and convex. I want to prove that $f$ is Lipschitz on $K$; that is, prove there exists a constant $c > 0$ such that $| f ( x ) - f( y ) | \\leq c \\, \\| x - y \\|, \\forall x , y \\in K$.\n\nMy approach:\n\nLet $x,y\\in K$ be two arbitrary points. Then, since $K$ is convex, the line segment between $x$ and $y$, i.e. $Co(x,y)$, is in $K$. Thus, by MVT, there exists a vector $b\\in Co(x,y)$ such that\n\n$$\\text{(*) } |f(x)-f(y)|=|\\langle x-y, (\\nabla f)(b)\\rangle|\\le \\|x-y\\|\\|(\\nabla f)(b)\\|$$\n\nNow, since $K$ is compact, $f$ takes a maximum and a minimum values on $K$, so that $\\exists b'\\in K$ such that $\\|(\\nabla f)(b') \\|\\ge \\|(\\nabla f)(b) \\|$, for all $b\\in K$. Let $c\\in \\mathbb{R}$, $c:=(\\nabla f)(b')$, then\n\n$$|f(x)-f(y)|\\le\\|x-y\\|\\|(\\nabla f)(b)\\|\\le\\|x-y\\|\\|(\\nabla f)(b')\\|=c\\|x-y\\|$$\n\nThis implies that $f$ is Lipschitz on $K$ for all $x,y\\in K$.\n\nPlease let me know if you think my proof is correct or not very much? I'm somewhat concerned about the part with the gradient - how exactly is the maximality of the norm of the gradient related to the EVT, that is to $f$ taking maximum and minimum values? As far as I can tell, the maximum norm of the gradient exists because that is the direction to the maximum (or minimum) point of $f$.\n\n\u2022 What is $a$? It seems undefined to me . . . \u2013\u00a0Robert Lewis Jun 28 '17 at 0:24\n\u2022 Sorry, it was supposed to be $y$. Fixed. \u2013\u00a0sequence Jun 28 '17 at 0:26\n\u2022 I guess is because $||f'||$ is a real function on a compact set? \u2013\u00a0Li Chun Min Jun 28 '17 at 0:41\n\u2022 @LiChunMin Can you please clarify your question? \u2013\u00a0sequence Jun 28 '17 at 0:47\n\u2022 Looks good to me. You do not need the EVT. Since $\\nabla (f)$ is continuous, the real-valued function $\\|\\nabla (f)\\|$ is continuous .So $T=\\{\\| \\nabla f(x)\\|: x\\in K\\}$ is compact (because $K$ is compact) , so $T$ is a bounded real subset....So $\\|f(x)-f(y)\\|\\leq \\|x-y\\|\\cdot \\sup T$ for all $x,y\\in K$. \u2013\u00a0DanielWainfleet Jun 28 '17 at 2:48\n\nHere's how I would do it:\n\nfor $x, y \\in K$, let $\\gamma(t):[0, 1] \\to K$ be the line segment\n\n$\\gamma(t) = x + t(y - x); \\tag{1}$\n\nthen\n\n$\\gamma(0) = x, \\tag{2}$\n\n$\\gamma(1) = x +(y - x) = y, \\tag{3}$\n\nand\n\n$\\gamma'(t) = y - x; \\tag{4}$\n\nfurthermore, since $K$ is convex, $\\gamma(t)$ lies entirely within $K$, hence also in $A$.\n\nNow, for $x, y \\in K$, we have:\n\n$\\Vert f(y) - f(x) \\Vert = \\Vert \\displaystyle \\int_0^1 \\dfrac{d(f(\\gamma(t))}{dt}dt \\Vert = \\Vert \\displaystyle \\int_0^1 \\nabla f(\\gamma(t)) \\cdot \\gamma'(t) dt \\Vert$ $\\le \\displaystyle \\int_0^1 \\Vert \\nabla f(\\gamma(t)) \\cdot \\gamma'(t) \\Vert dt \\le \\displaystyle \\int_0^1 \\Vert \\nabla f(\\gamma(t)) \\Vert \\Vert \\gamma'(t) \\Vert dt.\\tag{5}$\n\nSince $K$ is compact and $\\nabla f \\in C^0(A, \\Bbb R)$, so hence $\\nabla f \\in C^0(K, \\Bbb R)$, $\\Vert \\nabla f \\Vert$ is bounded by some $B$ on $K$, hence\n\n$\\displaystyle \\int_0^1 \\Vert \\nabla f(\\gamma(t)) \\Vert \\Vert \\gamma'(t) \\Vert dt \\le \\displaystyle \\int_0^1 B \\Vert \\gamma'(t) \\Vert dt; \\tag{6}$\n\nusing (4),\n\n$\\displaystyle \\int_0^1 B \\Vert \\gamma'(t) \\Vert dt = \\displaystyle \\int_0^1 B \\Vert y - x \\Vert dt = B \\Vert y - x \\Vert; \\tag{7}$\n\nbringing together (5), (6), and (7) yields\n\n$\\Vert f(y) - f(x) \\Vert \\le B \\Vert y - x \\Vert, \\tag{8}$\n\nthat is, $f(x)$ is Lipschitz continuous on $K$.\n\n\u2022 Thank you, this is very elegant. Do you think my proof could possibly be correct? \u2013\u00a0sequence Jun 28 '17 at 1:32\n\u2022 Yes, your method seems fine. Note it agrees with mine. \u2013\u00a0Robert Lewis Jun 28 '17 at 1:39\n\u2022 \"Since $K$ is compact and $\\nabla f \\in C^0(A, \\Bbb R)$, so hence $\\nabla f \\in C^0(K, \\Bbb R)$, $\\Vert \\nabla f \\Vert$ is bounded by some $B$ on $K$..\" Is there a proof for that or is it trivial? \u2013\u00a0abuchay May 14 '18 at 14:25\n\u2022 @abuchay: Well, $\\nabla f \\in C^0(K)$ since $\\nabla f \\in C^0(A)$ and $K \\subset A$. $\\Vert \\nabla f \\Vert \\in C^0(K)$ since $\\Vert \\cdot \\Vert$ is itself continuous in its argument. Then the boundedness follows since continuous functions on compacta are bounded, which is standard result in elementary topology. \u2013\u00a0Robert Lewis May 14 '18 at 15:40\n\nThe convexity of $K$ is not needed. Suppose the conclusion fails. Then for each $m\\in \\mathbb N,$ there exist $y_m,x_m \\in K$ such that\n\n$$\\tag 1 |f(y_m)- f(x_m)| > m|y_m-x_m|.$$\n\nBecause $K$ is compact, we can find a subsequence $m_k$ such that the sequences $y_{m_k},x_{m_k}$ converge to points $y,x\\in K$ respectively.\n\nSuppose $y\\ne x.$ Then $|y-x| > 0.$ For large $k$ we then have\n\n$$|f(y_{m_k})- f(x_{m_k})| > m_k|y_{m_k}-x_{m_k}|> m_k(|y-x|/2)\\to \\infty$$\n\nThis implies $f$ is not bounded on $K.$ But $f$ is continuous on $A,$ hence is continuous on $K,$ hence $f$ is bounded on $K$ by compactness. This contradiction shows $y=x.$\n\nBecause $A$ is open we can choose $r>0$ such that $B(x,r)\\subset A.$ For large $k$ we then have $y_{m_k},x_{m_k}$ in the compact convex set $\\overline {B(x,r/2}) \\subset A.$ By $(1),$ your highlighted mean value inequality then fails in this last set, contradiction. Therefore $(1)$ cannot hold, proving the result.\n\n\u2022 I dont understand why $m_k(|y-x|/2)\\to \\infty$. Why does it tend to infinity? If $m_k$ is a subsequence of $y_{mk}$ then $m_k$ also converges to $y$. Then, how can $m_k$ be a subsequence of $x_{mk}$ too? \u2013\u00a0John Keeper Oct 24 '18 at 9:46\n\u2022 @JohnKeeper $m_k$ is a subsequence of $1,2,\\dots$ hence $\\to \\infty$ \u2013\u00a0zhw. Oct 24 '18 at 21:32", "date": "2019-11-11 22:27:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz/2338828", "openwebmath_score": 0.9673323631286621, "openwebmath_perplexity": 124.69960711134773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126469647337, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8571438158785631}}
{"url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/", "text": "Here's a cute puzzle.\n\nPlayer A tosses a coin indefinitely and stops when he encounters a consecutive sequence of HT. Player B plays a similar game, except he stops when he encounters HH consecutively. Is the expected number of tosses equal for both players? If not, who has more, and can you give an intuitive explanation why?\n\nNote by C Lim\n4\u00a0years, 8\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nI think that player A is expected to be finished earlier. If he encounters an H, then he would have to keep tossing H's in order to continue. The chance of throwing $$n$$ consecutive H's decreases rapidly as $$n$$ increases.\n\nIf player B encounters an H, then he will have to toss a T to continue, but after tossing a T, he will continue tossing, because no matter what he tosses next, he doesn't encounter HH.\n\nSo, player A is doomed so to say whenever he tosses an H, because he'll have to keep tossing H's in order to continue. That's why he is expected to stop earlier than player B. I hope that makes sense.\n\n- 4\u00a0years, 8\u00a0months ago\n\nYup that's correct! It would also be interesting to explicitly calculate the expected number of throws for both A and B.\n\n- 4\u00a0years, 8\u00a0months ago\n\nPlayer A is expected to throw his first H after $$2$$ tosses, because\n\n$1 \\cdot \\frac{1}{2} + 2 \\cdot \\frac{1}{4} + 3 \\cdot \\frac{1}{8} + 4 \\cdot \\frac{1}{16} + 5 \\cdot \\frac{1}{32} + \\dots = 2.$\n\nSimilarly, after having tossed his first H, player A is expected to throw his first T (and thus stopping) after another 2 tosses. So (hopefully) I can conclude that player H is expected to toss 4 times.\n\nPlayer B is expected to roll his first H in two tosses. Then, there is a chance of 50% that he stops after the third toss, i.e. after tossing H again. However, if he tosses a T, then again, he is expected to toss his second H in two tosses, after which there is a 50% chance that he stops after the toss after that. It goes on and on. So, the amount of tosses player B is expected to do is\n\n$3 \\cdot \\frac{1}{2} + 6 \\cdot \\frac{1}{4} + 9 \\cdot \\frac{1}{8} + 12 \\cdot \\frac{1}{16} + 15 \\cdot \\frac{1}{32} + \\dots = 3 \\cdot 2 = 6.$\n\nAnd $$6 > 4$$.\n\nI hope that what I did is allowed, I'm not completely convinced of that just yet. Could anyone confirm that what I did is correct (or show me what I've done wrong)?\n\n- 4\u00a0years, 8\u00a0months ago\n\nI'm not sure what you did is ok either, but your answer is correct! [ This suggests there's probably a way to rigourously justify your working. ]\n\nHere's how I'd do it. To compute the expected number of tosses for HH, we define two values:\n\n\u2022 let a = expected number of additional tosses, assuming previous was 'H';\n\u2022 let b = expected number of additional tosses, assuming previous was 'T'.\n\nThen $$a = \\frac{b+1} 2 + \\frac 1 2$$ since if the previous toss was 'H', there's a 50% chance the next is 'T' (in which case the expected number of tosses is b+1), and a 50% chance the next is 'H' (in which case the number of tosses is 1 since it ends).\n\nLikewise, $$b = \\frac{a+1} 2 + \\frac{b+1}2$$ since if the previous toss was 'T', there's a 50% chance the next is 'H' (in which case expected number of tosses is a+1), and a 50% chance the next is 'T' (and the expected number of tosses is b+1).\n\nSolving then gives us a=4, b=6. The expected number of tosses is then b=6, since we can assume the 0-th toss was 'T' with no loss of generality.\n\nThe equations for HT are similar:\n\n$a = \\frac {a+1} 2 + \\frac 1 2, \\quad b = \\frac{a+1}2 + \\frac{b+1} 2$\n\nwhich gives us a=2, b=4.\n\n[ P.S. To be pedantic, my working isn't 100% rigourous either, since it already assumes a, b are finite. ]\n\n- 4\u00a0years, 8\u00a0months ago\n\nThat's really great. Two completely different ways of solving the problem :) I wouldn't know how to prove that $$a,b$$ are finite though.\n\n- 4\u00a0years, 8\u00a0months ago\n\nIf you want a finiteness proof, use first principles. Let $$N_A$$ be the number of tosses until $$A$$ stops. The probability $$P[N_A>n]$$ is the probability that $$A$$ has not thrown HT in the first $$n$$ tosses. This can only happen if $$A$$ has thrown $$j$$ tails followed by $$n-j$$ heads, for some $$0 \\le j \\le n$$. Thus there are $$n+1$$ possible sequences, all equally likely, and so $P[N_A > n] = (n+1)2^{-n}$ Thus $P[N_A \\ge n] \\;=\\; P[N_A > n-1] \\; = \\; n2^{1-n}$ Normal probability and series summation tricks give us $E[N_A] \\;=\\; \\sum_{n=1}^\\infty P[N_A \\ge n] \\; = \\; \\sum_{n=1}^\\infty n2^{1-n} \\; = \\; 4$ The result for $$B$$ is a little more challenging. Let $$N_B$$ be the number of tosses until $$B$$ stops. The probability that $$N_B > n$$ is the probability that $$B$$ does not throw HH in the first $$n$$ tosses. This is the probability that the first $$n$$ tosses are made up of \"HT\"s and \"T\"s stuck together, plus the probability that the first $$n-1$$ tosses are made up of \"HT\"s and \"T\"s stuck together, followed by a single \"H\".\n\nIt is well-known that the number of ways of tiling a $$n\\times1$$ chess-board using just $$2\\times1$$ dominoes and $$1\\times1$$ squares is the Fibonacci number $$F_{n+1}$$. (Just to be definite, $$F_0=0$$, $$F_1=1$$, etc.) Thus the probability that $$N_B > n$$ is the sum of the probabilities $$F_{n+1}2^{-n}$$ and $$F_n2^{-n}$$, and hence $P[N_B > n] \\; = \\; \\frac{F_{n+2}}{2^n}$ and hence $P[N_B \\ge n] \\; = \\; P[N_B > n-1] \\; = \\; \\frac{F_{n+1}}{2^{n-1}} \\qquad n \\ge 1.$ Thus $E[N_B] \\; = \\; \\sum_{n=1}^\\infty P[N_B \\ge n] \\; = \\; \\sum_{n=1}^\\infty \\tfrac{F_{n+1}}{2^{n-1}} \\; = \\; 4G(0.5) - 2$ where $G(x) \\;=\\; \\sum_{n=0}^\\infty F_nx^n$ is the generating function of the Fibonacci numbers. It is well-known that $G(x) \\; = \\; \\frac{x}{1-x-x^2}$ so that $$G(0.5) = 2$$. Thus $$E[N_B] = 6$$.\n\n- 4\u00a0years, 8\u00a0months ago\n\nOf course, once you showed that the number of ways is $$F_{n+2}$$, you can use the ratio test to conclude that the sum is finite, which is all we needed.\n\nNice way of calculating the sum through generating functions.\n\nStaff - 4\u00a0years, 8\u00a0months ago\n\nIn a similar vein, check out High school math #8.\n\nStaff - 4\u00a0years, 8\u00a0months ago\n\nThe absorbing Markov chain also gets the same answer.\n\nThis is the markov chain for ending the game with HH, for player B. I is the initial state, and the probabilities of moving to different states are shown:\n\n$$\\left(\\begin{array}{cccccccc} & I & H & T & HT & TH & TT & HH \\\\ I & 0 & \\frac{1}{2} & \\frac{1}{2} & 0 & 0 & 0 & 0 \\\\ H & 0 & 0 & 0 & \\frac{1}{2} & 0 & 0 & \\frac{1}{2} \\\\ T & 0 & 0 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} & 0 \\\\ HT & 0 & 0 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} & 0 \\\\ TH & 0 & 0 & 0 & \\frac{1}{2} & 0 & 0 & \\frac{1}{2} \\\\ TT & 0 & 0 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} & 0 \\\\ HH & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\end{array}\\right)$$\n\nMatrix Q is:\n\n$$\\left(\\begin{array}{rrrrrr} 0 & \\frac{1}{2} & \\frac{1}{2} & 0 & 0 & 0 \\\\ 0 & 0 & 0 & \\frac{1}{2} & 0 & 0 \\\\ 0 & 0 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} \\\\ 0 & 0 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} \\\\ 0 & 0 & 0 & \\frac{1}{2} & 0 & 0 \\\\ 0 & 0 & 0 & 0 & \\frac{1}{2} & \\frac{1}{2} \\end{array}\\right)$$\n\nFundamental matrix $$(I-Q)^{-1}$$ is:\n\n$$\\left(\\begin{array}{rrrrrr} 1 & \\frac{1}{2} & \\frac{1}{2} & 1 & \\frac{3}{2} & \\frac{3}{2} \\\\ 0 & 1 & 0 & 1 & 1 & 1 \\\\ 0 & 0 & 1 & 1 & 2 & 2 \\\\ 0 & 0 & 0 & 2 & 2 & 2 \\\\ 0 & 0 & 0 & 1 & 2 & 1 \\\\ 0 & 0 & 0 & 1 & 2 & 3 \\end{array}\\right)$$\n\nThe required expectation is the sum of the first row, which is 6.\n\nSimilarly doing for the 'HT' case yields an expected value of 4.\n\n- 4\u00a0years, 8\u00a0months ago", "date": "2018-04-21 06:13:59", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/", "openwebmath_score": 0.9759904146194458, "openwebmath_perplexity": 660.6993215232189, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712648206549, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8571438058606087}}
{"url": "https://math.stackexchange.com/questions/1546226/number-of-5-digit-numbers-such-that-the-sum-of-their-digits-is-even", "text": "# Number of 5-digit numbers such that the sum of their digits is even\n\nI found the same question on this site and many others...where everywhere the answer $45000$ is written ...while I have no problem with this but I am having problems with the logical answer...as everywhere it is written since total numbers is $90000$ so half of the numbers satisfy the above condition(but why??it is not explained anywhere as to how we can say that half the numbers satisfy this condition)...I know that half the numbers are even and half the numbers are odd...but that is not what we are asked here...we need to find numbers whose sum of digits is even ...and not numbers who are just even... I think these conditions are very different but it is used evrwhere...can someone please clarify..this.??..as to how we can say..half the numbers have their sum of digits as even?? i don't think this question is entirely a duplicate but if it please let me know...where I can get the explanation of what I am asking..\n\n\u2022 Hint: a sum of numbers is even if and only if there are an even number of odd numbers in the sum. \u2013\u00a0Colm Bhandal Nov 25 '15 at 18:23\n\u2022 An other way of seeing it, the last number decide if the sum is odd or even. If the sum of the 4 first numbers is even, then the last number need to be even. If the sum of the 4 first numbers is odd, then the last number need to be odd. \u2013\u00a0Alain Remillard Nov 25 '15 at 18:26\n\u2022 @Colm That is a very complicated use of even and odd term...!!..can you please expand this comment just a little bit...so that I can at least get a hint... \u2013\u00a0Freelancer Nov 25 '15 at 18:26\n\u2022 @Freelancer- actually I think Alain Remillard's suggestion is the more elegant of the two. Can you see how this leads to the result? How many possibilities are there for that last digit? And how many of them result in an even number, how many in an odd number? \u2013\u00a0Colm Bhandal Nov 25 '15 at 18:28\n\nTo expand on the wonderful insight of Alain Remillard in the comments, let's consider the set of all $5$ digit numbers. Now, you can partition this set into $9 \\times 10^3$ subsets (we exclude a leading $0$ as a possibility) each of which contains all numbers with a common prefix of $4$ numbers. For example, one of the partitions, for the prefix $2346$ would be:\n\n$$\\{23460, 23461, 23462, 23463, \\dots, 23469\\}$$\n\nNote that each partition contains exactly $10$ elements, with the last digit numbered $0-9$. Now, to prove that exactly half the numbers in the entire set have even digit sums, all we have to do is to prove that half the numbers in each of these partitions has an even digit sum. So we've reduced the problem to something much simpler! I hope you can see why.\n\nNow, to show that in any partition there are exactly $5$ elements with an even digit sum, consider the first element in the set e.g. in the above it would be $23460$. The sum of its digits is either even or odd. If it's even, then the next number will be odd, because you're just adding $1$ to it, and vice versa (the example is 15, which is odd). Then the sums go: even, odd, even, odd, even, odd etc. Or else they go: odd, even, odd, even etc. In either case, there are exactly $10$ elements, $5$ even, and $5$ odd. And we are done.\n\nUpdate: As the OP cleverly points out in the comments below, the choice to fix the first four digits is indeed arbitrary. The first digit must be fixed, because there are only $9$ possibilities for this, but after this any three of the remaining four digits can be fixed. The remaining digit will then have ten possibilities, and the proof will proceed just as above.\n\n\u2022 Well ..you fixed the first 3-digits...and then showed that for the last digit half numbers will be positive and half will be negative...let us say I want to think in another way and I fix the last three digits(say for example $X234$ and then show that half of the numbers so formed will be having sum as even and half will be having sum as odd...is that correct also?? ...can I do this... \u2013\u00a0Freelancer Nov 25 '15 at 19:03\n\u2022 @Freelancer- that is a very clever insight. Nicely though out :):) Yes you can indeed fix any three digits and the first digit and the above method will work. I will update accordingly. \u2013\u00a0Colm Bhandal Nov 26 '15 at 12:00\n\n\u2022 even + even = even\n\u2022 even + odd = odd\n\u2022 odd + odd = even\n\nUsing this we can show the even-ness of the sum of digits. Take $234582$. We have\n\n$$E + O + E + O + E + E \\\\= (E + O) + (E + O) + (E + E) \\\\= O + O + E \\\\= (O+O) + E \\\\= E + E \\\\= E.$$\n\nSo, for a five-digit number, what numbers of odd digits can we have to make the sum even?\n\nNext, let's count the number of five-digit numbers that have three odd digits. (Is the sum of digits of these numbers even or odd?)\n\nThere are $125$ three-digit odd numbers. (Why?) There are $_5C_3 = 10$ combinations of places we can put the digits in order within the five-digit number. Then, there are $25$ two-digit even numbers. So the number of five digit numbers with three odd digits is $125 \\cdot 10 \\cdot 25 = 31250.$ (I'm assuming leading zeroes are valid (like $00123$)).\n\nTo solve the problem, then, you'll need to determine what numbers of odd digits will give you an even sum, and then count each of those cases.", "date": "2021-01-27 15:45:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1546226/number-of-5-digit-numbers-such-that-the-sum-of-their-digits-is-even", "openwebmath_score": 0.7157057523727417, "openwebmath_perplexity": 185.39753959843088, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856414, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8571438053168258}}
{"url": "http://math.stackexchange.com/questions/30463/isomorphic-groups/30465", "text": "# Isomorphic groups\n\nI know that there is a formal definition isomorphism but for the purpose of this homework questions, I call two groups isomorphic if they have the same structure, that is group table for one can be turned into the table for the other by a suitable renaming.\n\nNow consider $\\mathbf{Z}_2=\\{0,1\\}$ (group under addition modulo $2$) and $\\mathbf{Z}_3^{\\times}=\\{1,2\\}$ (group under multiplication modulo $3$). In general $\\mathbf{Z}_n^{\\times}=\\{a|0\\leq a\\leq n-1 \\text{ and }\\gcd(a,n)=1\\}$. Now the group table for $\\mathbf{Z}_2$ is: $$\\begin{array}{c|cc} &0&1\\\\ \\hline 0&0&1\\\\ 1&1&0 \\end{array}$$ and the group table for $\\mathbf{Z}_3^{\\times}$ is: $$\\begin{array}{c|cc} &1&2\\\\ \\hline 1&1&2\\\\ 2&2&1 \\end{array}$$ Obviously these two groups are isomorphic because if in the first table, I replace $0$s by $1$s and $1$s by $2$s, I get exactly the second table.\n\nHowever, if I write out the group table for $\\mathbf{Z}_4$ and $\\mathbf{Z}_5^{\\times}$, there is no way that group table for one can be turned into the table for the other by a suitable renaming. This means that these two groups are not isomorphic but the question asks me to prove that they are isomorphic. Now what should I do?\n\n-\nThere is no difference between the usual formal notion of \"being isomorphic\" and the one you propose to use! \u2013\u00a0 Mariano Su\u00e1rez-Alvarez Apr 2 '11 at 5:10\nIn any case, you should look harder at the group tables for $\\mathbb Z_4$ and $\\mathbb Z_5^\\times$!... \u2013\u00a0 Mariano Su\u00e1rez-Alvarez Apr 2 '11 at 5:10\nThere is a suitable renaming. Keep trying... \u2013\u00a0 Zev Chonoles Apr 2 '11 at 5:12\nIn short, the two examples are all concerned with the structure of cyclic groups, et c'est la m\u00e9me. \u2013\u00a0 awllower Apr 2 '11 at 7:08\n\nYou are incorrect in claiming that there is no way to turn one table into the other. But the key is that you are not just allowed to rename, you are also allowed to list the elements in a different order! (After all, listing the elements of $\\mathbf{Z}_4$ in a different order in the table will not change the group or the operation, will it?) This amounts to shuffling rows and columns together: if you exchange rows 2 and 3, say, you should also exchange columns 2 and 3.\n\nThe table for $\\mathbf{Z}_4$ is: $$\\begin{array}{c|cccc} +&0&1&2&3\\\\ \\hline 0&0&1&2&3\\\\ 1&1&2&3&0\\\\ 2&2&3&0&1\\\\ 3&3&0&1&2 \\end{array}$$ The table for $Z_5^{\\times}$ is: $$\\begin{array}{c|cccc} \\times&1&2&3&4\\\\ \\hline 1&1&2&3&4\\\\ 2&2&4&1&3\\\\ 3&3&1&4&2\\\\ 4&4&3&2&1 \\end{array}$$ If you are going to be able to rename the entries in the last table to match the first, then \"1\" must be renamed \"0\". Now, notice that there is only one of the remaining four elements that when operated with itself gives you the \"identity\"; since in $\\mathbf{Z}_4$ this happens for $2$, you may want to shuffle the rows and columns to move that element to be in the third row and column and see what you have then.\n\n-\nc'est superbe!! \u2013\u00a0 awllower Apr 2 '11 at 7:11\nApparently not; any reason for the downvote? \u2013\u00a0 Arturo Magidin Apr 2 '11 at 18:13\nWho downvoted ? \u2013\u00a0 awllower Apr 3 '11 at 2:45\nDon't know; but someone did... \u2013\u00a0 Arturo Magidin Apr 3 '11 at 2:50\nSince $\\mathbb{Z}_n^\\times$ is cyclic, reordering can be easily done by picking a generator element (i. e., any other than 1), and computing its power. In this case, the powers of 2 are, in order, 1, 2, 4 and 3, which is the ordering for the first line of the table for $\\mathbb{Z}_5^\\times$ that will suit the isomorphism. \u2013\u00a0 Luke May 10 '11 at 1:04\n\nYour \"naive\" definition of isomorphic almost coincides with the abstract one. The difference is that apart from renaming you must also be allowed to change the order of the elements. Then you should have no difficulties with $\\mathbb{Z}/4\\mathbb{Z}$ vs. $(\\mathbb{Z}/5\\mathbb{Z})^\\times$.\n\n-\nNote by the way that this little confusion is a good argument for learning the actual definition of group isomorphism, rather than trying to avoid it by only thinking in terms of \"group tables\". \u2013\u00a0 Pete L. Clark Apr 2 '11 at 12:30", "date": "2015-01-29 21:17:39", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/30463/isomorphic-groups/30465", "openwebmath_score": 0.8749757409095764, "openwebmath_perplexity": 336.7763461347079, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9825575132207566, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8571314801717705}}
{"url": "https://stats.stackexchange.com/questions/389453/calculating-conditional-probability-with-marginal", "text": "# Calculating conditional probability with marginal\n\nAssume we are given a joint distribution $$P(X,Y)$$ where $$P(0,0)=0.1$$, $$P(0,1) = 0.4$$, $$P(1,0)=0.3$$, and $$P(1,1)=0.2$$. The goal is to compute $$P(X|Y=1)$$.\n\nTraditionally, solving a conditional probability problem $$P(A|B)$$ simplifies to $$\\frac{P(A,B)}{P(B)}$$, but I'm unsure how to apply it to this case. In particular, I am unclear on what the probability $$P(X,Y=1)$$ means since $$P(X)$$ is a marginal probability.\n\nTo avoid this, I enumerated the different values $$X$$ takes on and plugged it into the original quantity to solve -- $$P(X|Y=1)$$:\n\n\u2022 $$P(X=0|Y=1) = 0.4/0.6 = 4/6$$\n\u2022 $$P(X=1|Y=1) = 0.2/0.6 = 2/6$$\n\nThis gives me the final answer of $$P(X|Y=1) = [4/6, 2/6]$$, but I'm not sure whether the answer should be multiple probabilities or a single probability.\n\n\u2022 P(X|Y=2) is a distribution in the same way that P(X) is also a distribution (and not just one \"probability\"). One probably is P(X=0, Y=0)=0.1, for example. \u2013\u00a0nbro Jan 28 at 2:14\n\u2022 @nbro Does this mean my answer is correct? \u2013\u00a0Shrey Jan 28 at 4:04\n\u2022 you're ok. just as $p(x,y)$ is a 2D function(table), $p(x)$, $p(x|Y=y)$ are 1D functions (rows). \u2013\u00a0gunes Jan 28 at 4:52\n\u2022 $P(X|Y=1)$ is shorthand for $P(X=x|Y=1)$ \u2013\u00a0StatsStudent Jan 28 at 5:47\n\nYou have gone about this correctly, but the final answers are typically written as a function of $$x$$. It's helpful, I think, to remember that $$P(X|Y=1)$$ is just shorthand for $$P(X=x|Y=1)$$ where $$x$$ is any number in the support of $$x$$ (in this case 0 and 1). So you'd calculate this as follows:\n\n$$\\begin{eqnarray*} \\\\{P(X|Y=1)} & = & {P\\left(X=x|Y=1\\right)}\\\\ & = & \\frac{P\\left(X=x,\\,Y=1\\right)}{P\\left(Y=1\\right)}\\\\ & = & \\frac{P\\left(X=x,\\,Y=1\\right)}{P\\left(X=0,\\,Y=1\\right)+P\\left(X=1,\\,Y=1\\right)}\\\\ & = & \\frac{P\\left(X=x,\\,Y=1\\right)}{0.4+0.2}\\\\ & = & \\frac{P\\left(X=x,\\,Y=1\\right)}{0.6} \\end{eqnarray*}$$\n\nNow, writing this as a function of $$x$$ gives:\n\n$$\\begin{eqnarray*} P\\left(X=x|Y=1\\right) & = & \\begin{cases} \\frac{P\\left(X=0,\\,Y=1\\right)}{0.6} & ,\\text{for }x=0\\\\ \\frac{P\\left(X=1,\\,Y=1\\right)}{0.6} & ,\\text{for }x=1 \\end{cases}\\\\ & = & \\begin{cases} \\frac{0.4}{0.6} & ,\\text{for }x=0\\\\ \\frac{0.2}{0.6} & ,\\text{for }x=1 \\end{cases}\\\\ & = & \\begin{cases} \\frac{2}{3} & ,\\text{for }x=0\\\\ \\frac{1}{3} & ,\\text{for }x=1 \\end{cases} \\end{eqnarray*}$$\n\nYou are correct. You are supposed to be getting \"multiple\" probabilities.\n\nYou said that your goal is to figure out $$P(X|Y = 1)$$. Well, in order to achieve your goal, you need to know what the probability that $$X = 0$$ given that $$Y = 1$$ is as well as what the probability that $$X = 1$$ given that $$Y = 1$$ is, which you did.\n\nnbro's comment is basically telling you that when you get \"multiple\" probabilities, you are actually computing a probability mass (or density) function (or, equivalently, yet not the same, the distribution of the random variable $$X$$ given $$Y = 1$$).\n\nNote: $$P(X|Y=1)=[4/6,2/6]$$ is the probability mass function of the random variable $$X$$ given $$Y = 1$$ because it tells you what that probability is for all the possible values of $$X$$, ie, $$X = 0$$ and $$X$$ = 1.\n\nFinally, I use \"multiple\" in quotes because nobody really says that. In your case, they would ask something like: \"This gives me the final answer of P(X|Y=1)=[4/6,2/6], but I'm not sure whether the answer should be $${\\it \\text{ a probability mass function}}$$ or a single probability.\"", "date": "2019-08-19 18:43:03", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/389453/calculating-conditional-probability-with-marginal", "openwebmath_score": 0.9003010988235474, "openwebmath_perplexity": 185.84113606590793, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575116884778, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8571314739438562}}
{"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits", "text": "# If series is absolutely convergent then $\\sum \\limits_{n\\in I}a_n=\\sum \\limits_{k=1}^{\\infty}\\sum \\limits_{n\\in I_k}a_n.$\n\nSuppose that the series $$\\sum \\limits_{n=1}^{\\infty}a_n$$ is absolutely convergent and let $$I\\subseteq \\mathbb{N}$$ such that $$I=\\bigsqcup\\limits_{k=1}^{\\infty}I_k$$. Then show that $$\\sum \\limits_{n\\in I}a_n=\\sum \\limits_{k=1}^{\\infty}\\sum \\limits_{n\\in I_k}a_n. \\qquad (*)$$\n\nI don't have any idea how to solve it.\n\nI do know that in any absolute convergent series permutation of terms does not change the sum and I guess it should be used somehow in order to prove equality $$(*)$$.\n\nCan anyone show the rigorous proof of equality $$(*)$$, please?\n\n\u2022 here is a proof. Here is another. \u2013\u00a0Masacroso May 16 at 20:14\n\u2022 What's the definition of summation here? \u2013\u00a0Hashem Ben Abdelbaki May 16 at 20:14\n\u2022 @Masacroso, link which you provided have not nothing in common with my question. I do know that in absolute convergent series any permutation does not change the sum. \u2013\u00a0ZFR May 16 at 20:22\n\u2022 @ZFR you had written \"I do know that in any absolute convergent series permutation of terms does not change the sum. But I want to prove it rigorously and cannot do it.\" Hence my comment providing two formal proofs. Make clear your question please. \u2013\u00a0Masacroso May 16 at 20:23\n\u2022 @Masacroso, sorry about that. Done \u2013\u00a0ZFR May 16 at 20:27\n\nFirst assume that $$a_n \\ge 0$$ and define $$\\sum_{n \\in I} a_n = \\sup_{J \\subset I, J \\text{ finite}} \\sum_{n \\in J} a_n$$. Note that it follows that if $$I \\subset I'$$ then $$\\sum_{n \\in I} a_n \\le \\sum_{n \\in I'} a_n$$.\n\nFrom https://math.stackexchange.com/a/3680889/27978 we see that if $$K = K_1 \\cup \\cdots \\cup K_m$$, a disjoint union, then $$\\sum_{n \\in K} a_n = \\sum_{n \\in K_1} a_n + \\cdots + \\sum_{n \\in K_m} a_n$$.\n\nSince $$I'=I_1 \\cup \\cdots \\cup I_m \\subset I$$ we see that $$\\sum_{n \\in I} a_n \\ge \\sum_{n \\in I'} a_n = \\sum_{k=1}^m \\sum_{n \\in I_k} a_n$$. It follows that $$\\sum_{n \\in I} a_n \\ge \\sum_{k=1}^\\infty \\sum_{n \\in I_k} a_n$$. This is the 'easy' direction.\n\nLet $$\\epsilon>0$$, then there is some finite $$J \\subset I$$ such that $$\\sum_{n\\in J} a_n > \\sum_{n \\in I} a_n -\\epsilon$$. Since $$J$$ is finite and the $$I_k$$ are pairwise disjoint we have $$J \\subset I'=I_1 \\cup \\cdots \\cup I_m$$ for some $$m$$ and so $$\\sum_{k=1}^\\infty \\sum_{n \\in I_k} a_n \\ge \\sum_{k=1}^m\\sum_{n \\in I_k} a_n \\ge \\sum_{k=1}^m\\sum_{n \\in J \\cap I_k} a_n = \\sum_{n\\in J} a_n > \\sum_{n \\in I} a_n -\\epsilon$$.\n\n(It is not relevant here, but a small proof tweak shows that the result holds true even if the $$a_n$$ do not have a finite sum.)\n\nNow suppose we have $$a_n \\in \\mathbb{R}$$ and $$\\sum_{n \\in I} |a_n| = \\sum_{n=1}^\\infty |a_n|$$ is finite. We need to define what we mean by $$\\sum_{n \\in I} a_n$$. Note that $$(a_n)_+=\\max(0,a_n) \\ge 0$$ and $$(a_n)_-=\\max(0,-a_n) \\ge 0$$. Since $$0 \\le (a_n)_+ \\le |a_n|$$ and $$0 \\le (a_n)_- \\le |a_n|$$ we see that $$\\sum_{n \\in I} (a_n)_+ = \\sum_{k=1}^\\infty \\sum_{n \\in I_k} (a_n)_+$$ and similarly for $$(a_n)_-$$.\n\nThis suggests the definition (cf. Lebesgue integral) $$\\sum_{n \\in I} a_n = \\sum_{n \\in I} (a_n)_+ - \\sum_{n \\in I} (a_n)_-$$.\n\nWith this definition, all that remains to be proved is that $$\\sum_{k=1}^\\infty \\sum_{n \\in I_k} a_n = \\sum_{k=1}^\\infty \\sum_{n \\in I_k} (a_n)_+ - \\sum_{k=1}^\\infty \\sum_{n \\in I_k} (a_n)_-$$ and this follows from summability and the fact that for each $$k$$ we have $$\\sum_{n \\in I_k} a_n = \\sum_{n \\in I_k} (a_n)_+ - \\sum_{n \\in I_k} (a_n)_-$$.\n\nNote: To elaborate the last sentence, recall that I defined $$\\sum_{n \\in I_k} a_n$$ to be $$\\sum_{n \\in I_k} (a_n)_+ - \\sum_{n \\in I_k} (a_n)_-$$, so all that is happening here is the definition is applied to $$I_k$$ rather than $$I$$. Then to finish, note that if $$d_k,b_k,c_k$$ are summable and satisfy $$d_k=b_k-c_k$$ then $$\\sum_{k=1}^\\infty d_k= \\sum_{k=1}^\\infty b_k- \\sum_{n=1}^\\infty c_k$$, where $$d_k = \\sum_{n \\in I_k} a_n$$, $$b_k = \\sum_{n \\in I_k} (a_n)_+$$ and $$c_k = \\sum_{n \\in I_k} (a_n)_-$$.\n\n\u2022 Great answer! But probably you meant bijection $\\sigma: \\mathbb{N}\\to I$, right? \u2013\u00a0ZFR May 18 at 18:42\n\u2022 @ZFR: Correct, I will fix. \u2013\u00a0copper.hat May 18 at 18:58\n\u2022 When you wrote that \"It is straightforward to show that for any bijection $\\sigma:\\mathbb{N} \\to I$ we have $\\sum_{n \\in I} a_n = \\sum_{n=1}^\\infty a_{\\sigma(n)}$.\" Do you mean here that $a_n\\geq 0$ or not? I am a bit confused \u2013\u00a0ZFR May 18 at 19:17\n\u2022 (+1) This is less fussy, more self-contained, more authoritative, and better connected to the literature than my answer, so it ought to replace it as the accepted answer. (I haven't read @Matematleta's answer yet, because it looks harder to understand than these two, so I reserve judgement on it, for now!) On this answer: am I right in thinking that one can also write $\\sum_{n \\in I}a_n = s$ iff for all $\\epsilon > 0$ there exists finite $J \\subseteq I$ such that $\\left\\lvert\\sum_{n \\in J}a_n- s\\right\\rvert < \\epsilon$? \u2013\u00a0Calum Gilhooley May 18 at 19:18\n\u2022 You really helped me to understand the thing which i did not understand for a long time. Thank you so much! I really appreciate your permanent help! \u2013\u00a0ZFR May 18 at 20:50\n\nSuppose for the moment that the result is known to be true for convergent series of non-negative terms.\n\nIf $$\\sum_{n=1}^\\infty a_n$$ is an absolutely convergent series of real numbers, define $$a_n = b_n - c_n,$$ for all $$n \\geqslant 1,$$ where $$c_n = 0$$ when $$a_n \\geqslant 0$$ and $$b_n = 0$$ when $$a_n \\leqslant 0.$$ Then $$|a_n| = b_n + c_n,$$ therefore $$\\sum_{n=1}^\\infty b_n$$ and $$\\sum_{n=1}^\\infty c_n$$ are convergent series of non-negative terms, therefore: \\begin{align*} \\sum_{n \\in I}a_n & = \\sum_{n \\in I}b_n - \\sum_{n \\in I}c_n \\\\ & = \\sum_{k=1}^\\infty\\sum_{n \\in I_k}b_n - \\sum_{k=1}^\\infty\\sum_{n \\in I_k}c_n \\\\ & = \\sum_{k=1}^\\infty\\left( \\sum_{n \\in I_k}b_n - \\sum_{n \\in I_k}c_n\\right) \\\\ & = \\sum_{k=1}^\\infty\\sum_{n \\in I_k}(b_n - c_n) \\\\ & = \\sum_{k=1}^\\infty\\sum_{n \\in I_k}a_n. \\end{align*} So it is enough to prove the result on the assumption that $$a_n \\geqslant 0$$ for all $$n \\geqslant 1.$$\n\nGiven any set $$K \\subseteq \\mathbb{N},$$ I shall use the Iverson bracket notation: $$[n \\in K] = \\begin{cases} 1 & \\text{if } n \\in K, \\\\ 0 & \\text{if } n \\notin K. \\end{cases}$$ I shall assume that, however the notation $$\\sum_{n \\in K}a_n$$ has been defined, it satisfies the identity: $$\\sum_{n \\in K}a_n = \\sum_{n=1}^\\infty a_n[n \\in K].$$ Let $$J_k = I_1 \\cup I_2 \\cup \\cdots \\cup I_k$$ ($$k = 1, 2, \\ldots$$). Because the $$I_k$$ are disjoint, we have $$[n \\in J_k] = [n \\in I_1] + [n \\in I_2] + \\cdots + [n \\in I_k],$$ therefore $$\\sum_{n \\in I_1}a_n + \\sum_{n \\in I_2}a_n + \\cdots + \\sum_{n \\in I_k}a_n = \\sum_{n \\in J_k}a_n \\leqslant \\sum_{n \\in I}a_n,$$ therefore $$\\sum_{k=1}^\\infty\\sum_{n \\in I_k}a_n \\leqslant \\sum_{n \\in I}a_n,$$ and the outer infinite sum on the left hand side exists, because its partial sums are bounded above by the sum on the right hand side. On the other hand, for all $$m \\geqslant 1,$$ \\begin{align*} \\sum_{n=1}^ma_n[n \\in I] & = \\sum_{n=1}^ma_n[n \\in I_1] + \\sum_{n=1}^ma_n[n \\in I_2] + \\cdots + \\sum_{n=1}^ma_n[n \\in I_r] \\\\ & \\leqslant \\sum_{n \\in I_1}a_n + \\sum_{n \\in I_2}a_n + \\cdots + \\sum_{n \\in I_r}a_n \\\\ & \\leqslant \\sum_{k=1}^\\infty\\sum_{n \\in I_k}a_n, \\end{align*} where $$r = \\max\\{k \\colon n \\leqslant m \\text{ for some } n \\in I_k\\},$$ therefore $$\\sum_{n \\in I}a_n \\leqslant \\sum_{k=1}^\\infty\\sum_{n \\in I_k}a_n,$$ and the two inequalities together prove (*).\n\n\u2022 Thanks a lot for your answer! I read your answer and I have two questions: 1) I am a bit confused by the definition of $r$? Could you explain it in details, please? 2) Have you ever used that the series $\\sum \\limits_{n=1}^{\\infty}a_n$ is absolutely convergent? \u2013\u00a0ZFR May 17 at 2:16\n\u2022 I would define $r$ in this way: for each $n\\in I$ s.t. $1\\leq n\\leq m$ there is $p(n)$ s.t. $n\\in I_{p(n)}$ and let's take $r=\\max \\{p(1),\\dots,p(m)\\}$ then $[n\\in I]=[n\\in I_1]+\\dots+[n\\in I_r]$. Is my reasoning correct? And still I would be happy if you can explain and point the moments where you have used that the series is absolutely convergent. \u2013\u00a0ZFR May 17 at 2:30\n\u2022 Your definition of $r$ looks right to me, and looks equivalent to mine. Certainly I had much the same idea in mind. I used the absolute convergence of $\\sum_{n=1}^\\infty a_n$ when I inferred that the two series of non-negative terms $\\sum_{n=1}^\\infty b_n$ and $\\sum_{n=1}^\\infty c_n$ are convergent. I'm going to try to get some more sleep now, but I'll have another look at the whole thing after lunch, and see if I can make it any clearer. (Assuming it isn't all just a load of dingo's kidneys, of course!) \u2013\u00a0Calum Gilhooley May 17 at 9:04\n\u2022 I deliberately chose not to write \\begin{gather*} b_n = \\frac{|a_n| + a_n}2 \\geqslant 0, \\\\ c_n = \\frac{|a_n| - a_n}2 \\geqslant 0, \\end{gather*} but perhaps it would have been clearer had I done so. \u2013\u00a0Calum Gilhooley May 17 at 15:00\n\u2022 Perhaps I should also have stated explicitly that the identity I took to be satisfied by $\\sum_{n \\in K}a_n$ is a possible definition of that notation; but it is not the only possible definition; and you didn't say what definition you were using; so I left it open. \u2013\u00a0Calum Gilhooley May 17 at 15:05\n\nI think there is an elementary proof (one without measure theory), that we can adapt from a similar claim in Apostol's Analysis book. Without loss of generality, $$I=\\mathbb N$$. For each $$k\\in \\mathbb N,\\ I_k$$ may be regarded as a map from some subset $$\\{1,2,\\cdots,\\}\\subseteq \\mathbb N$$, to $$\\{\\sigma_k(1),\\sigma_k(2),\\cdots,\\}$$ which may or may not be infinite, so $$\\sigma_k$$ is an injective map from the susbet of $$\\mathbb N$$ of the same cardinality as $$|I_k|,$$ starting at $$1$$, to the $$\\textit{set}\\ I_k.$$ If $$|I_k|=j$$, extend $$I_k$$ to all of $$\\mathbb N$$ by mapping $$n\\in \\mathbb N\\setminus \\{1,2,\\cdots, j\\}$$ to $$\\mathbb N\\setminus \\{\\sigma_k(1),\\sigma_k(2),\\cdots, \\sigma_k(j)\\}$$ injectively and defining $$a'_n:=0$$ for all $$n\\in \\mathbb N\\setminus \\{\\sigma_k(1),\\sigma_k(2),\\cdots, \\sigma_k(j)\\}$$. This construction will not affect any of the sums, so without loss of generality, $$I_k$$ maps $$\\mathbb N$$ to a subset of $$\\mathbb N$$ such that\n\n$$\\tag1 I_k\\ \\text{is injective on}\\ \\mathbb N$$\n\n$$\\tag2 \\text{the range of each}\\ I_k \\ \\text{is a subset of } \\ \\mathbb N, \\text{say}\\ P_k$$\n\n$$\\tag3 \\text{the}\\ P_k\\ \\text{are disjoint}$$\n\nNow put $$\\tag4 b_k(n)=a_{I_{k}(n)}\\ \\text{and}\\ s_k=\\sum^\\infty_{n=0}b_k(n)$$\n\nwhich is well-defined by $$(1)-(3).$$ We have to prove that\n\n$$\\tag5 \\sum^\\infty_{k=0}a_k=\\sum^\\infty_{k=0}s_k$$\n\nIt's easy to show that the right hand side of this converges absolutely. To find the sum, set $$\\epsilon>0$$ and choose $$N$$ large enough so that $$\\sum^\\infty_{k=0}|a_k|-\\sum^n_{k=0}|a_k|<\\frac{\\epsilon}{2}$$ as soon as $$n>N.$$ This implies also that\n\n$$\\tag6\\left|\\sum^\\infty_{k=0}a_k-\\sum^n_{k=0}a_k\\right|<\\frac{\\epsilon}{2}$$\n\nNow choose $$\\{I_1,\\cdots, I_r\\}$$ so that each element of $$\\{a_1,\\dots ,a_N\\}$$ appears in the sum $$\\sum^\\infty_{n=0}a_{I_{1(n)}}+\\cdots +\\sum^\\infty_{n=0}a_{I_{r(n)}}=s_1+\\cdots+ s_r.$$ Then, if $$n>r,N$$ we have\n\n$$\\tag 7\\left|\\sum^n_{k=0}s_k-\\sum^n_{k=0}a_k\\right|<\\sum^\\infty_{n=N+1}<\\frac{\\epsilon}{2}$$\n\nNow $$(5)$$ folllows from $$(6)$$ and $$(7).$$\n\n\u2022 I am a bit confused with your definition of $I_k$? Could you clarify it, please? \u2013\u00a0ZFR May 16 at 23:26\n\u2022 $I_k$ is some subset of $\\mathbb N$. So it may be regarded as a function on the subset of $\\mathbb N$ of the same cardinality. For instance, if $I_k=\\{2,5,7\\}$ then $I_k$ the function would map $\\{1,2,3\\}$ to $\\{2,5,7\\}$. If $I_k$ is infinite, then it is countable, so there is an injective function from $\\mathbb N$ to it. That would be our $I_k$ considered as a function on $\\mathbb N.$ \u2013\u00a0Matematleta May 16 at 23:31\n\u2022 Hmm. I guess that i got you. Also could you show why the RHS of (5) converges absolutely, please? \u2013\u00a0ZFR May 16 at 23:37\n\u2022 Hint: show directly by comparison with $\\sum |a_n|$ that $\\{s_k\\}$ has bounded partial sums. \u2013\u00a0Matematleta May 16 at 23:42", "date": "2020-10-26 00:35:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits", "openwebmath_score": 0.9980955123901367, "openwebmath_perplexity": 174.6296223630094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982557516796073, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8571314735082184}}
{"url": "https://math.stackexchange.com/questions/132495/why-dont-you-need-the-axiom-of-choice-when-constructing-the-inverse-of-an-inj/132498", "text": "# Why don't you need the Axiom of Choice when constructing the \u201cinverse\u201d of an injection?\n\nSuppose $f:X\\rightarrow Y$ is a surjection and you want to show that there exists $g:Y\\rightarrow X$ s.t. $f\\circ g=\\mathrm{id}_Y$. You need the AC to show this.\n\nHowever, suppose $f$ is a injection and you want show there is $g$ s.t. $g\\circ f=\\mathrm{id}_X$. Then, according to my textbook, you don't need the AC to show this.\n\nThis is counterintuitive to me, because it's like you need a special axiom to claim that an infinite product of big sets is nonempty, while you don't need one to claim that an infinite product of singleton sets is nonempty, which seems smaller than the former.\n\nSo why don't you need the AC to show the latter?\n\nEDIT: $X$ should be nonempty.\n\nEDIT 2: I realized (after asking this) that my question mostly concerns whether the AC is needed to say that an infinite product of finite sets is nonempty, and why.\n\n\u2022 I can't fix your details (Asaf will probably see to that) but here's the intuition: Why do we need AC? Because with big sets, the problem is specifying how to pick the elements. No such problem with singleton sets! \u2013\u00a0Ragib Zaman Apr 16 '12 at 13:32\n\u2022 Making one choice (or some fixed finite number of choices) doesn't require Axiom of Choice. It's entailed by first-order logic, q.v. Rule C. \u2013\u00a0hardmath Apr 16 '12 at 13:42\n\u2022 hardmath: your comment is best in answering my question as I hold in my heart. Could you please elaborate on that? \u2013\u00a0Pteromys Apr 16 '12 at 13:47\n\u2022 @Pteromys: Read this answer to why we can choose from non-empty sets. Finitely many choices follow by induction. \u2013\u00a0Asaf Karagila Apr 16 '12 at 13:48\n\u2022 @Pteromys: As others have noted, the Axiom of Choice is needed to deal with infinite products of different sets, even in the case those sets each contain two elements. We've moved sufficiently far from your original Question, it might be best to formulate a new one if you want elaboration! \u2013\u00a0hardmath Apr 16 '12 at 14:58\n\nThe need for the axiom of choice is to choose arbitrary elements. Injectivity eliminates this need.\n\nAssume that $A$ is not empty, if $f\\colon A\\to B$ is injective this means that if $b\\in B$ is in the range of $f$ then there is a unique $a\\in A$ such that $f(a)=b$.\n\nThis means that we can define (from $f$) what is the $a$ to which we send $b$.\n\nSo if $f$ is not onto $B$ we have two options:\n\n1. $b\\in B$ in the range of $f$, then we have exactly one option to send $b$ to.\n2. $b\\in B$ not in the range of $f$. Since $A$ is not empty fix in advance some $a_0\\in A$ and send $b$ to $a_0$.\n\nAnother way to see this is let $B=B'\\cup Rng(f)$, where $B'\\cap Rng(f)=\\varnothing$. Fix $a_0\\in A$ and define $g|_{B'}(x)=a_0$. For every $b\\in Rng(f)$ we have that $f^{-1}[\\{b\\}]=\\{a\\in A\\mid f(a)=b\\}$ is a singleton, so there is only one function which we can define in: $$\\prod_{b\\in Rng(f)}f^{-1}[\\{b\\}]$$\n\nNow let the unique function in the product be $g|_{Rng(f)}$ and define $g$ to be the union of these two.\n\nYour intuition about the need for the axiom of choice is true for surjections, if $f$ was surjective then we only know that $f^{-1}[\\{b\\}]$ is non-empty for every $b\\in B$, and we need the full power of the axiom of choice to ensure that an arbitrary surjection has an inverse function.\n\nTo the edited question:\n\nThe axiom of choice is needed because we have models in which the axiom of choice does not hold, where there exists an infinite family of pairs whose product is empty.\n\nThere are weaker forms from which follow choice principles for finite sets. However these are still not provable from ZF on its own.\n\nAs indicated by Chris Eagle in the comments, and as I remark above, in a product of singletons there is no need for the axiom of choice since there is only one way to choose from a singleton.\n\n\u2022 Why not just \"flip the graph\" with $(x,y) \\mapsto (y,x)$ to get a function $g: f(X) \\to X$ (since $f$ is injective) and extend it to $Y \\smallsetminus f(X)$ by $g(y) = x_0$ as you said? \u2013\u00a0t.b. Apr 16 '12 at 13:42", "date": "2019-05-26 15:19:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/132495/why-dont-you-need-the-axiom-of-choice-when-constructing-the-inverse-of-an-inj/132498", "openwebmath_score": 0.8408396244049072, "openwebmath_perplexity": 129.30781144029993, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575162853136, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.857131463280188}}
{"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend", "text": "# One-zero dividend\n\n## Challenge description\n\nFor every positive integer n there exists a number having the form of 111...10...000 that is divisible by n i.e. a decimal number that starts with all 1's and ends with all 0's. This is very easy to prove: if we take a set of n+1 different numbers in the form of 111...111 (all 1's), then at least two of them will give the same remainder after division by n (as per pigeonhole principle). The difference of these two numbers will be divisible by n and will have the desired form. Your aim is to write a program that finds this number.\n\n## Input description\n\nA positive integer.\n\n## Output description\n\nA number p in the form of 111...10...000, such that p \u2261 0 (mod n). If you find more than one - display any of them (doesn't need to be the smallest one).\n\n## Notes\n\nYour program has to give the answer in a reasonable amount of time. Which means brute-forcing is not permited:\n\np = 0\nwhile (p != 11..10.00 and p % n != 0)\np++\n\n\nNeither is this:\n\ndo\np = random_int()\nwhile (p != 11..10.00 and p % n != 0)\n\n\nIterating through the numbers in the form of 11..10..00 is allowed.\n\nYour program doesn't need to handle an arbitrarily large input - the upper bound is whatever your language's upper bound is.\n\n## Sample outputs\n\n2: 10\n3: 1110\n12: 11100\n49: 1111111111111111111111111111111111111111110\n102: 1111111111111111111111111111111111111111111111110\n\n\u2022 Can we have a reasonable upper bound to the possible output? (Something about less than 2.4 billion (approx. the max value of a signed integer) should be fine, as arrays or lists might be required for some implementations) \u2013\u00a0Tamoghna Chowdhury Feb 28 '16 at 15:25\n\u2022 @MartinB\u00fcttner I think that the first satisfying output should be enough (reasonable timeframe constraint) \u2013\u00a0Tamoghna Chowdhury Feb 28 '16 at 15:26\n\u2022 The last 0 is not necessary in the 49 test case. \u2013\u00a0CalculatorFeline Feb 28 '16 at 15:36\n\u2022 @CatsAreFluffy I think all numbers need to contain at least 1 and at least one 0, otherwise 0 is a solution for any input. (Would be good to clarify this though.) \u2013\u00a0Martin Ender Feb 28 '16 at 15:38\n\u2022 Just requiring one 1 should work. \u2013\u00a0CalculatorFeline Feb 28 '16 at 15:41\n\n## Mathematica, 29 bytes\n\n\u230a10^(9EulerPhi@#)/9\u230b10^#&\n\n\nCode by Martin B\u00fcttner.\n\nOn input $$\\n\\$$, this outputs the number with $$\\9\\varphi(n)\\$$ ones followed by $$\\n\\$$ zeroes, where $$\\\\varphi(\\cdot)\\$$ is the Euler totient function. With a function phi, this could be expressed in Python as\n\nlambda n:'1'*9*phi(n)+'0'*n\n\n\nIt would suffice to use the factorial $$\\n!\\$$ instead of $$\\\\varphi(n)\\$$, but printing that many ones does not have a reasonable run-time.\n\nClaim: $$\\9\\varphi(n)\\$$ ones followed by $$\\n\\$$ zeroes is a multiple of $$\\n\\$$.\n\nProof: First, let's prove this for the case that $$\\n\\$$ is not a multiple of $$\\2, 3, \\text{or } 5\\$$. We'll show the number with consisting of $$\\\\varphi(n)\\$$ ones is a multiple of $$\\n\\$$.\n\nThe number made of $$\\k\\$$ ones equals $$\\\\frac{10^k-1}9\\$$. Since $$\\n\\$$ is not a multiple of $$\\3\\$$, this is a multiple of $$\\n\\$$ as long as $$\\10^k-1\\$$ is a factor of $$\\n\\$$, or equivalently if $$\\10^k \\equiv 1\\mod n\\$$. Note that this formulation makes apparent that if $$\\k\\$$ works for the number of ones, then so does any multiple of $$\\k\\$$.\n\nSo, we're looking for $$\\k\\$$ to be a multiple of the order of $$\\k\\$$ in the multiplicative group modulo n. By Lagrange's Theorem, any such order is a divisor of the size of the group. Since the elements of the group are the number from $$\\1\\$$ to $$\\n\\$$ that are relatively prime to $$\\n\\$$, its size is the Euler totient function $$\\\\varphi(n)\\$$. So, we've shown that $$\\10^{\\varphi(n)} \\equiv 1 \\mod n\\$$, and so the number made of $$\\\\varphi(n)\\$$ ones is a multiple of $$\\n\\$$.\n\nNow, let's handle potential factors of $$\\3\\$$ in $$\\n\\$$. We know that $$\\10^{\\varphi(n)}-1\\$$ is a multiple of $$\\n\\$$, but $$\\\\frac{10^{\\varphi(n)}-1}9\\$$ might not be. But, $$\\\\frac{10^{9\\varphi(n)}-1}9\\$$ is a multiple of $$\\9\\$$ because it consists of $$\\9\\varphi(n)\\$$ ones, so the sum of its digits a multiple of $$\\9\\$$. And we've noted that multiplying the exponent $$\\k\\$$ by a constant preserves the divisibility.\n\nNow, if $$\\n\\$$ has factors of $$\\2\\$$'s and $$\\5\\$$'s, we need to add zeroes to end of the output. It way more than suffices to use $$\\n\\$$ zeroes (in fact $$\\\\log_2(n)\\$$ would do). So, if our input $$\\n\\$$ is split as $$\\n = 2^a \\times 5^b \\times m\\$$, it suffices to have $$\\9\\varphi(m)\\$$ ones to be a multiple of $$\\n\\$$, multiplied by $$\\10^n\\$$ to be a multiple of $$\\2^a \\times 5^b\\$$. And, since $$\\n\\$$ is a multiple of $$\\m\\$$, it suffices to use $$\\9\\varphi(n)\\$$ ones. So, it works to have $$\\9\\varphi(n)\\$$ ones followed by $$\\n\\$$ zeroes.\n\n\u2022 Just to make sure no one thinks this was posted without my permission: xnor came up with the method and proof all on his own, and I just supplied him with a Mathematica implementation, because it has a built-in EulerPhi function. There is nothing mind-blowing to the actual implementation, so I'd consider this fully his own work. \u2013\u00a0Martin Ender Feb 28 '16 at 18:29\n\n## Python 2, 44 bytes\n\nf=lambda n,j=1:j/9*j*(j/9*j%n<1)or f(n,j*10)\n\n\nWhen j is a power of 10 such as 1000, the floor-division j/9 gives a number made of 1's like 111. So, j/9*j gives 1's followed by an equal number of 0's like 111000.\n\nThe function recursively tests numbers of this form, trying higher and higher powers of 10 until we find one that's a multiple of the desired number.\n\n\u2022 Oh, good point, we only need to check 1^n0^n... \u2013\u00a0Martin Ender Feb 28 '16 at 16:03\n\u2022 @MartinB\u00fcttner If it's any easier, it also suffices to fix the number of 0's to be the input value. Don't know if it counts as efficient to print that many zeroes though. \u2013\u00a0xnor Feb 28 '16 at 16:37\n\u2022 Why does checking 1^n0^n work? \u2013\u00a0Lynn Feb 28 '16 at 17:54\n\u2022 @Lynn Adding more zeroes can't hurt, and there's infinitely many possible numbers of ones, some number will have enough of both ones and zeroes. \u2013\u00a0xnor Feb 28 '16 at 18:15\n\n# Pyth, 11 bytes\n\n.W%HQsjZTT\n\n\nTest suite\n\nBasically, it just puts a 1 in front and a 0 in back over and over again until the number is divisible by the input.\n\nExplanation:\n\n.W%HQsjZTT\nImplicit: Q = eval(input()), T = 10\n.W              while loop:\n%HQ           while the current value mod Q is not zero\njZT      Join the string \"10\" with the current value as the separator.\ns          Convert that to an integer.\nT     Starting value 10.\n\n\n# Haskell, 51 bytes\n\n\\k->[b|a<-[1..],b<-[div(10^a)9*10^a],bmodk<1]!!0\n\n\nUsing xnor\u2019s approach. nimi saved a byte!\n\n## CJam, 2825 19 bytes\n\nSaved 6 bytes with xnor's observation that we only need to look at numbers of the form 1n0n.\n\nri:X,:)Asfe*{iX%!}=\n\n\nTest it here.\n\n### Explanation\n\nri:X    e# Read input, convert to integer, store in X.\n,:)     e# Get range [1 ... X].\nAs      e# Push \"10\".\nfe*     e# For each N in the range, repeat the characters in \"10\" that many times,\ne# so we get [\"10\" \"1100\" \"111000\" ...].\n{iX%!}= e# Select the first element from the list which is divided by X.\n\n\n# JavaScript (ES6), 65 bytes\n\nEdit 2 bytes saved thx @Neil\n\nIt works within the limits of javascript numeric type, with 17 significant digits. (So quite limited)\n\na=>{for(n='';!(m=n+=1)[17];)for(;!(m+=0)[17];)if(!(m%a))return+m}\n\n\nLess golfed\n\nfunction (a) {\nfor (n = ''; !(m = n += '1')[17]; )\nfor (; !(m += '0')[17]; )\nif (!(m % a))\nreturn +m;\n}\n\n\u2022 Why not for(m=n;? \u2013\u00a0Neil Feb 28 '16 at 17:11\n\u2022 @Neil because I need at least one zero. Maybe I can find a shorter way ... (thx for the edit) \u2013\u00a0edc65 Feb 28 '16 at 18:30\n\u2022 Oh, that wasn't clear in the question, but I see now that the sample outputs all have at least one zero. In that case you can still save a byte using for(m=n;!m[16];)if(!((m+=0)%a)). \u2013\u00a0Neil Feb 28 '16 at 18:42\n\u2022 @Neil or even 2 bytes. Thx \u2013\u00a0edc65 Feb 28 '16 at 18:52\n\n# Husk, 11 10 bytes\n\n-1 byte thanks to Razetime!\n\n\u1e1f\u00a6\u2070mod\u1e58\u1e0b2N\n\n\nTry it online! Constructs the infinite list [10,1100,111000,...] and selects the first element which is a multiple of the argument.\n\n# Mathematica, 140 55 bytes\n\nNestWhile[\"1\"<>#<>\"0\"&,\"1\",FromDigits@#~Mod~x>0&/.x->#]\n\n\nMany bytes removed thanks to xnor's 1^n0^n trick.\n\nMinimal value, 140 156 bytes This gives the smallest possible solution.\n\nNestWhile[\"1\"<>#&,ToString[10^(Length@NestWhileList[If[EvenQ@#,If[10~Mod~#>0,#/2,#/10],#/5]&,#,Divisors@#~ContainsAny~{2, 5}&],FromDigits@#~Mod~m>0&/.m->#]&\n\n\nIt calculates how many zeros are required then checks all possible 1 counts until it works. It can output a number with no 0 but that can be fixed by adding a <>\"0\" right before the final &.\n\n## Haskell, 37 bytes\n\nf n=[d|d<-\"10\",i<-[1..n*9],gcd n i<2]\n\n\nThis uses the fact that it works to have 9*phi(n) ones, where phi is the Euler totient function. Here, it's implemented using gcd and filtering, producing one digit for each value i that's relatively prime to it that is in the range 1 and 9*n. It also suffices to use this many zeroes.\n\n# Jelly, 11 bytes\n\n\u2075Dx\u2c6e\u1e05\u2075\u1e0d@\u0187\u2078\u1e22\n\n\nTry it online!\n\nIgnoring runtime requirements, we can reduce this to 10 bytes using xnor's method, but with $$\\\\varphi(n)\\$$ replaced with $$\\n!\\$$:\n\n!,\u00b5\u2075Dx\"\u00b5F\u1e0c\n\n\nTry it online!\n\n## How they work\n\n\n\u2075Dx\u2c6e\u1e05\u2075\u1e0d@\u0187\u2078\u1e22 - Main link. Takes n on the left\n\u2075D          - Yield [1, 0]\n\u2c6e        - For each integer i in [1, 2, ..., n]:\nx         -   Repeat each element of [1, 0] i times\n\u1e05\u2075      - Convert each back into an integer\n\u0187   - Keep those for which the following is true:\n\u1e0d@    -   Is divisible by\n\u2078  -   n\n\u1e22 - Take the first value of those remaining\n\n\n!,\u00b5\u2075Dx\"\u00b5F\u1e0c - Main link. Takes n on the left\n!          - Yield n!\n,         - Yield [n!, n]. Call this l\n\u00b5        - Begin new link with l on the left\n\u2075D      - Yield [1, 0]\n\"    - Zip [1, 0] with [n!, n] and do the following over each pair:\nx     -   Repeat\n- This yields [[1, 1, ..., 1], [0, 0, ..., 0]].\nThe first element has n! 1s and the second n 0s\nCall this k\n\u00b5   - Begin new link with k on the left\nF  - Flatten k\n\u1e0c - Convert to integer\n\n\n# Perl 5, 26 bytes\n\nincludes a byte for -n (-M5.01 is free)\n\n($.=\"1$.0\")%$_?redo:say$.\n\n\n## Explanation\n\n$. starts off with value 1. We immediately concatenate it with 1 beforehand and 0 afterward, yielding 110, and reassign that to $. \u2014 that's what the $.=\"1$.0\" does.\n\nThe assignment returns the assigned value so when we take it modulo the input number $_ we obtain 0 (false) if 110 is divisible by the input and nonzero (true) otherwise. \u2022 In the latter case, we redo, i.e. repeat the block (the -n switch makes the whole thing a loop block, and incidentally, assigns $_ to the input number). This concatenates another 1 and 0, yielding 11100, etc.\n\u2022 If 110, or 11100, or 1111000, or whatever we're up to, is divisible by the input, we stop and say (print) it.\n\nNote that every number divides such a number (i.e. that has one more 1 than the numbers of 0s it has). After all, if it divides something with more 1s than 0s, you can append 0s and it'll still divide that. And if divides something with fewer 1s than 0s, say m 1s and n 0s with m < n, then it also divides the number with 2m 1s and n 0s.\n\n## Sage, 33 bytes\n\nlambda n:'1'*9*euler_phi(n)+'0'*n\n\n\nThis uses xnor's method to produce the output.\n\nTry it online\n\n# 05AB1E, 8 bytes\n\n$\u00d59*\u00d7\u00ce\u00d7\u00ab  Port of @xnor's Mathematica answer, so make sure to upvote him!! Explanation: $         # Push 1 and the input-integer\n\u00d5        # Pop the input, and get Euler's totient of this integer\n9*      # Multiply it by 9\n\u00d7     # Repeat the 1 that many times as string\n\u00ce    # Push 0 and the input-integer\n\u00d7   # Repeat the 0 the input amount of times as string\n\u00ab  # Concatenate the strings of 1s and 0s together\n# (after which the result is output implicitly)\n\n\nA more to-the-point iterative approach would be 11 bytes:\n\nTS\u221e\u03b4\u00d7\u00f8J.\u0394I\u00d6\n\n\nExplanation:\n\nT             # Push 10\nS            # Convert it to a list of digits: [1,0]\n\u221e           # Push an infinite positive list: [1,2,3,...]\n\u03b4          # Apply double-vectorized on these two lists:\n\u00d7         #  Repeat the 1 or 0 that many times as string\n# (we now have a pair of infinite lists: [[1,11,111,...],[0,00,000,...]])\n\u00f8       # Zip/transpose; swapping rows/columns:\n#  [[1,0],[11,00],[111,000],...]\nJ      # Join each inner pair together:\n#  [10,1100,111000,...]\n.\u0394    # Find the first value in this list which is truthy for:\nI\u00d6  #  Check that it's divisible by the input-integer\n# (after which the result is output implicitly)\n\n\n# bc, 58 bytes\n\ndefine f(n){for(x=1;m=10^x/9*10^x;++x)if(m%n==0)return m;}\n\n\n## Sample results\n\n200: 111000\n201: 111111111111111111111111111111111000000000000000000000000000000000\n202: 11110000\n203: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n204: 111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000\n205: 1111100000\n206: 11111111111111111111111111111111110000000000000000000000000000000000\n207: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n208: 111111000000\n209: 111111111111111111000000000000000000\n210: 111111000000\n211: 111111111111111111111111111111000000000000000000000000000000\n212: 11111111111110000000000000\n213: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n214: 1111111111111111111111111111111111111111111111111111100000000000000000000000000000000000000000000000000000\n215: 111111111111111111111000000000000000000000\n216: 111111111111111111111111111000000000000000000000000000\n217: 111111111111111111111111111111000000000000000000000000000000\n218: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n219: 111111111111111111111111000000000000000000000000\n\n\n# dc, 27 bytes\n\nOdsm[O*lmdO*sm+O*dln%0<f]sf\n\n\nThis defines a function f that expects its argument in the variable n. To use it as a program, ?sn lfx p to read from stdin, call the function, and print the result to stdout. Variable m and top of stack must be reset to 10 (by repeating Odsm) before f can be re-used.\n\n## Results:\n\n200: 111000\n201: 111111111111111111111111111111111000000000000000000000000000000000\n202: 11110000\n203: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n204: 111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000\n205: 1111100000\n206: 11111111111111111111111111111111110000000000000000000000000000000000\n207: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n208: 111111000000\n209: 111111111111111111000000000000000000\n210: 111111000000\n211: 111111111111111111111111111111000000000000000000000000000000\n212: 11111111111110000000000000\n213: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n214: 1111111111111111111111111111111111111111111111111111100000000000000000000000000000000000000000000000000000\n215: 111111111111111111111000000000000000000000\n216: 111111111111111111111111111000000000000000000000000000\n217: 111111111111111111111111111111000000000000000000000000000000\n218: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n219: 111111111111111111111111000000000000000000000000\n`", "date": "2021-06-14 15:40:23", "meta": {"domain": "stackexchange.com", "url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend", "openwebmath_score": 0.38380372524261475, "openwebmath_perplexity": 1459.298190489912, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575175622121, "lm_q2_score": 0.8723473647220786, "lm_q1q2_score": 0.8571314611332632}}
{"url": "http://mathhelpforum.com/pre-calculus/62126-completing-square.html", "text": "1. ## Completing the Square\n\nWhat is the minimum point of the graph of the equation\n\ny = 2x^2 + 8x + 9?\n\nI understand the minimum point is the vertex (h, k).\n\nTo complete the square, I set the function to zero first.\n\n2x^2 + 8x + 9 = 0\n\nI then subtracted 9 from both sides.\n\n2x^2 + 8x = -9\n\nI then took half of the second term and squared it.\n\n2x^2 + 8x = 16 - 9\n\nWhere do I go from here?\n\n2. Hello !\n\ny = 2x\u00b2 + 8x + 9\ny = 2 (x\u00b2 + 4x + 9/2)\ny = 2 ((x+2)\u00b2+1/2)\n\nTherefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1\n\n3. Originally Posted by magentarita\nWhat is the minimum point of the graph of the equation\n\ny = 2x^2 + 8x + 9?\n\nI understand the minimum point is the vertex (h, k).\n\nTo complete the square, I set the function to zero first.\n\n2x^2 + 8x + 9 = 0\n\nI then subtracted 9 from both sides.\n\n2x^2 + 8x = -9\n\nI then took half of the second term and squared it.\n\n2x^2 + 8x = 16 - 9\n\nWhere do I go from here?\n===============================================\nmagentarita,\nI see where your problem is.\n2x^2 + 8x = -9 ------ ok so far.\nBefore you take half of the secod term and square,\nMake sure the coefficient of suared term 2x^2 to be 1.\nTherefore, 2x^2 + 8x = -9\nbecomes (x^2) + 4x = -9/2 -- simply didived by 2.\n(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get\n(x^2) + 4x + 2^2 = (-9/2 ) + 2^2\n(x + 2)^2 = -1/2 now complete the square part is done at this point.\nRewrite original question to y = [(x + 2)^2 ]+ 1/2\nCan you see y = a ( x-h )^2 + k --- standard form.\nVertex is at ( h, k ) = ( -2, 1/2) and a =1\nOpens up vertically, therefor, mini. point is at vertex.\n\n4. ## ok............\n\nOriginally Posted by 2976math\n===============================================\nmagentarita,\nI see where your problem is.\n2x^2 + 8x = -9 ------ ok so far.\nBefore you take half of the secod term and square,\nMake sure the coefficient of suared term 2x^2 to be 1.\nTherefore, 2x^2 + 8x = -9\nbecomes (x^2) + 4x = -9/2 -- simply didived by 2.\n(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get\n(x^2) + 4x + 2^2 = (-9/2 ) + 2^2\n(x + 2)^2 = -1/2 now complete the square part is done at this point.\nRewrite original question to y = [(x + 2)^2 ]+ 1/2\nCan you see y = a ( x-h )^2 + k --- standard form.\nVertex is at ( h, k ) = ( -2, 1/2) and a =1\nOpens up vertically, therefor, mini. point is at vertex.\n\nok but none the choices given for the minimum point is (-2, 1/2).\n\n(2,33)\n\n(2,17)\n\n(-2,-15)\n\n(-2,1)\n\n5. ## ok....\n\nOriginally Posted by running-gag\nHello !\n\ny = 2x\u00b2 + 8x + 9\ny = 2 (x\u00b2 + 4x + 9/2)\ny = 2 ((x+2)\u00b2+1/2)\n\nTherefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1\nYou got the right answer. But how did you get y = 1?\n\nI got y = 1/2\n\n6. Originally Posted by magentarita\nYou got the right answer. But how did you get y = 1?\n\nI got y = 1/2\nSubstitute x= -2 into the expression :\n2 [ (-2+2)\u00b2 + 1/2 ] = 2 [ 0+1/2 ] = 1\n\n7. Originally Posted by magentarita\nok but none the choices given for the minimum point is (-2, 1/2).\n\n(2,33)\n\n(2,17)\n\n(-2,-15)\n\n(-2,1)\nI found the problem!\ny should be 1 not 1/2, so answer (-2,1) is correct.\n\nReason: I forgot to divid y by 2.\n\nRemember orginal question: y =( 2x^2 ) + 8x +9\n\n(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side\n\n(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side\n\ny = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side\n\ny = a [ ( x-h)^2 ] + k\n\nwhere a=2, h=-2, k=1 sorry\n\n8. I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.\n\nFrom $y=ax^2+bx+c$, you want to convert to $y=a(x-h)^2+k$, where $(h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.\n\n$y=2x^2+8x+9$ Factor 2 out of first 2 terms\n\n$y=2(x^2+4x)+9$ Next, complete the square in parentheses.\n\n$y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side\n\n$y=2(x+2)^2+1$ Now, we're in vertex form.\n\n$V(h, k)=V(-2, 1) = \\ \\ minimum \\ \\ point$\n\n9. ## ok.........\n\nOriginally Posted by Moo\nSubstitute x= -2 into the expression :\n2 [ (-2+2)\u00b2 + 1/2 ] = 2 [ 0+1/2 ] = 1\nBy subbing -2 for x, I get y = 1.\n\nI got it.\n\n10. ## ok..........\n\nOriginally Posted by 2976math\nI found the problem!\ny should be 1 not 1/2, so answer (-2,1) is correct.\n\nReason: I forgot to divid y by 2.\n\nRemember orginal question: y =( 2x^2 ) + 8x +9\n\n(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side\n\n(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side\n\ny = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side\n\ny = a [ ( x-h)^2 ] + k\n\nwhere a=2, h=-2, k=1 sorry\n\nBe be sorry, be right. We all make mistakes, right?\n\n11. ## ok..........\n\nOriginally Posted by masters\nI think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.\n\nFrom $y=ax^2+bx+c$, you want to convert to $y=a(x-h)^2+k$, where $(h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.\n\n$y=2x^2+8x+9$ Factor 2 out of first 2 terms\n\n$y=2(x^2+4x)+9$ Next, complete the square in parentheses.\n\n$y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side\n\n$y=2(x+2)^2+1$ Now, we're in vertex form.\n\n$V(h, k)=V(-2, 1) = \\ \\ minimum \\ \\ point$\nI enjoyed the steps as a guide. Well-done!", "date": "2017-01-21 11:04:40", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/62126-completing-square.html", "openwebmath_score": 0.8046611547470093, "openwebmath_perplexity": 1917.9567759754982, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575178175919, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8571314597256313}}
{"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form", "text": "# Problems that become easier in a more general form\n\nWhen solving a problem, we often look at some special cases first, then try to work our way up to the general case.\n\nIt would be interesting to see some counterexamples to this mental process, i.e. problems that become easier when you formulate them in a more general (or ambitious) form.\n\n## Motivation/Example\n\nRecently someone asked for the solution of $a,b,c$ such that $\\frac{a}{b+c} = \\frac{b}{a+c} = \\frac{c}{a+b} (=t).$\n\nSomeone suggested writing this down as a system of linear equations in terms of $t$ and solving for $a,b,c$. It turns out, either (i) $a=b=c$ or (ii) $a+b+c=0$.\n\nSolution (i) is obvious from looking at the problem, but (ii) was not apparent to me until I solved the system of equations.\n\nThen I wondered how this would generalize to more variables, and wrote the problem as: $$\\frac{x_i}{\\sum x - x_i} = \\frac{x_j}{\\sum x - x_j} \\quad \\forall i,j\\in1,2,\\dots,n$$\n\nLooking at this formulation, both solutions became immediately evident without the need for linear algebra (for (ii), set $\\sum x=0$ so each denominator cancels out its numerator).\n\n-\nThe linked Question : math.stackexchange.com/questions/897118/\u2026 \u2013\u00a0lab bhattacharjee Aug 16 '14 at 12:16\nWell, I personally don't think this is actually simplification by generalization, but rather simplification motivated by generalization. Like if you write $\\frac{a}{b+c}$ as $\\frac{a}{(a + b + c) - a}$, you'd say that the problem becomes obvious too. \u2013\u00a0Tunococ Aug 16 '14 at 12:59\n@Tunococ Fair enough, I'm just saying that writing it like this didn't occur to me until I thought of generalizing it. I understand that this is all a little subjective (hence the \"soft-question\" tag). \u2013\u00a0MGA Aug 16 '14 at 13:01\nThis is relevant. \u2013\u00a0Julien Godawatta Aug 16 '14 at 13:58\nIntroducing topology makes certain proofs in real analysis clearer and more elegant, though I'm not sure whether they are necessarily easier per se. \u2013\u00a0Harry Johnston Aug 17 '14 at 0:24\n\nConsider the following integral $\\displaystyle\\int_{0}^{1}\\dfrac{x^7-1}{\\ln x}\\,dx$. All of our attempts at finding an anti-derivative fail because the antiderivative isn't expressable in terms of elementary functions.\n\nNow consider the more general integral $f(y) = \\displaystyle\\int_{0}^{1}\\dfrac{x^y-1}{\\ln x}\\,dx$.\n\nWe can differentiate with respect to $y$ and evaluate the resulting integral as follows:\n\n$f'(y) = \\displaystyle\\int_{0}^{1}\\dfrac{d}{dy}\\left[\\dfrac{x^y-1}{\\ln x}\\right]\\,dx = \\int_{0}^{1}x^y\\,dx = \\left[\\dfrac{x^{y+1}}{y+1}\\right]_{0}^{1} = \\dfrac{1}{y+1}$.\n\nSince $f'(y) = \\dfrac{1}{y+1}$, we have $f(y) = \\ln(y+1)+C$ for some constant $C$.\n\nTrivially, $f(0) = \\displaystyle\\int_{0}^{1}\\dfrac{x^0-1}{\\ln x}\\,dx = \\int_{0}^{1}0\\,dx = 0$. Hence $C = 0$, and thus, $f(y) = \\ln(y+1)$.\n\nTherefore, our original integral is $\\displaystyle\\int_{0}^{1}\\dfrac{x^7-1}{\\ln x}\\,dx = f(7) = \\ln 8$.\n\nThis technique of generalizing an integral by introducing a parameter and differentiating w.r.t. that parameter is known as Feynman Integration.\n\n-\nGreat example! Really neat \u2013\u00a0mattecapu Aug 16 '14 at 19:59\nMy favorite illustration so far. Of course the others are nice too. \u2013\u00a0MGA Aug 16 '14 at 20:22\nThe special form (not only the general form) can also easily be determined by setting $\\ln x=-t$ and then use Frullani's integral. \u2013\u00a0Tunk-Fey Aug 17 '14 at 14:44\nThe integrand does have an anti-derivative in terms of the upper incomplete gamma function: $\\Gamma(0, -\\ln x) - \\Gamma(0, -8\\,\\ln x)$. That still doesn't help because it's undefined at both limits of integration. \u2013\u00a0Tavian Barnes Aug 17 '14 at 21:52\n@Tunk-Fey You're right, apologies. Deleting my comment. \u2013\u00a0MGA Aug 18 '14 at 11:39\n\nGeorge Polya's book How to Solve It calls this phenomenon \"The Inventor's Paradox\": \"The more ambitious plan may have more chances of success.\" The book gives several examples, including the following.\n\n1) Consider the problem: \"A straight line and a regular octahedron are given in position. Find a plane that passes through the given line and bisects the volume of the given octahedron.\" If we generalize this to \"a straight line and a solid with a center of symmetry are given in position...\" it becomes very easy. (The plane goes through the center of symmetry and the line.)\n\nThe book also gives other examples of the Inventor's Paradox, but \"more ambitious\" is not always the same as \"more general.\" Consider: \"Prove that $1^3 + 2^3 + 3^3 + ... + n^3$ is a perfect square.\" Polya shows that it is easier to prove (by mathematical induction) that \"$1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ...+ n)^2$\". This is more ambitious but is not more general.\n\nThe web page Generalizations in Mathematics gives many similar examples. It even gets into the difference between \"more ambitious\" and \"more general.\"\n\n-\nGood point regarding generality/ambitiousness, and both are interesting. I have made a minor edit to the question to reflect this. \u2013\u00a0MGA Aug 16 '14 at 12:51\nIn the first the easy approach is \"more general\" in the sense that it picks out the significant property of an entity and ignores any other particular properties it may have. There are loads of examples like it, for example any time you're asked to prove some property of a particular group that's true of all Abelian groups, or some property of a particular function that's true of any continuous monotonic function, and so on. The problem is, \"identify the useful property of this object\", so the more general problem where you only have the useful property is always easier ;-) \u2013\u00a0Steve Jessop Aug 16 '14 at 19:26\nThe cubes and square example, termed 'more ambitious', is an example of the technique of solving a problem by the addition of an invented constraining hypothesis, and thus is actually a case of particularization. \u2013\u00a0Jose Brox Aug 20 '14 at 8:00\n\nI recall something like this coming up when evaluating certain summations. For example, consider:\n\n$$\\sum_{n=0}^{\\infty} {n \\over 2^n}$$\n\nWe can generalize this by letting $f(x) = \\sum_{n=0}^{\\infty} nx^n$, so:\n\n\\begin{align} {f(x) \\over x} &= \\sum_{n=0}^{\\infty} nx^{n-1} \\\\ &= {d \\over dx} \\sum_{n=0}^{\\infty} x^n \\\\ &= {d \\over dx} {1 \\over {1-x}} = {1 \\over (x-1)^2} \\end{align}\n\nTherefore,\n\n$$f(x) = {x \\over (x-1)^2}$$\n\nThe solution to the original problem is $f({1 \\over 2}) = 2$.\n\n-\nInterestingly, this also shows $\\sum_{n=0}^\\infty \\frac{1}{2^n} = \\sum_{n=0}^\\infty \\frac{n}{2^n}$, which succeeded at surprising me. :) \u2013\u00a0Keba Aug 3 '15 at 23:26\n\nThe solution to the Monty Hall problem\n\nSuppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, follows the fixed protocol of opening another door, say No. 3, which has a goat. He then says to you, \"Do you want to pick door No. 2?\" Is it to your advantage to switch your choice?\n\nbecomes more obvious when you generalize it to an $N$-door problem with the host opening $N-2$ doors. For $N\\gg3$ most people's intuition revolts against staying with the original choice.\n\n-\nA more helpful approach to the standard Monty Hall is to examine what is wrong with the above formulation. The story, as told here, is consistent with a game show where the host offers a chance to switch only to contestants who have already chosen the correct door. When playing that game, you should never switch regardless of how many doors there are. In the standard problem, it is explicitly stated that the host will always offer the switch. That's what makes the odds $N-1$ to $1$ in favor of the switch; and $2$-to-$1$ odds is good enough. \u2013\u00a0David K Aug 16 '14 at 15:22\n@DavidK - the text I included is Wikipedia's description of the Monty Hall problem. Have modified it. \u2013\u00a0Johannes Aug 16 '14 at 20:20\nThe problem has been underspecified in some widely-read sources; no surprise one of those got copied to Wikipedia. I'll accept this version as implying the player knows the fixed protocol. In that case, if someone's intuition requires $N$ to be increased above $3$ before they think the switch is beneficial, they do not understand the reason why it is beneficial. They have merely been nudged from an incorrect guess to an unjustified but luckily correct guess. \u2013\u00a0David K Aug 18 '14 at 23:12\n@DavidK: I think that would be a fair criticism of someone who claims to be game-theoretically rational, but I doubt such a person exists; would a reasonable person expect a Bayes factor of $2:1$ to come to his rescue? Even though it wouldn't be necessary to the idealised mathematical observer with the eyes of a hawk, in many contexts exaggerating the difference helps me notice there is one, whence I can see via a monotonicity argument that the effect is non-zero (if weak) for $N \\gtrsim 3$. \u2013\u00a0Vandermonde Feb 16 '15 at 19:27\nHeck, what I thought negligible was a raw probability of 1/6, which I ought to appreciate even then. \u2013\u00a0Vandermonde Feb 16 '15 at 21:25\n\nI'm not entirely convinced that problems made somehow easier by generalizations is exactly what is going on here.\n\nIn the example provided in your question, what made the solution to the general problem appear easier is that it dawned on you that $$x_1+\\cdots+x_{j-1}+x_{j+1}+\\cdots+x_n=\\sum_{i=1}^nx_i-x_j.$$ Indeed, (as tunococ commented) had the less general problem been written as $$\\frac{a}{a+b+c-a}=\\frac{b}{a+b+c-b}=\\frac{c}{a+b+c-c},$$ then your easier solution to the general problem applies here as well. I would argue that, if anything, the generalization helped you notice a pattern you had not before seen. Would you still have noticed this pattern had you not formulated the problem in a general way? Perhaps, perhaps not.\n\nIn my opinion, what your experience shows is that formulating a problem $Q$ in more general terms $P$ is one of many ways by which one can gain a fundamental insight that provides the key to the solution of the general problem $P$ (and thus inevitably also solves the initial special case $Q$ also). Sometimes, this can lead to a solution that was as of yet unknown to you and that will be more elegant or easier than the previous solutions. However, given that such an insight could easily have come without generalizing the problem, the fact that the solution did come from you thinking about the generalization seems highly circumstantial to me.\n\nEDIT: JimmyK4542's example (and Feynmann's integration trick) seems like a spectacular demonstration of the phenomenon, however.\n\n-\nI merely meant that as an example to get the discussion started, and you're of course right that in this case one doesn't have to generalize to see the solution. But personally, I didn't think of writing $+a-a$ etc. because I didn't feel the need to. Once I considered the general case, I was compelled to write it like this. Anyway, I think that better examples have now been given on here that really illustrate the point - I'm particularly fond of the Feynman integration trick. \u2013\u00a0MGA Aug 16 '14 at 20:16\nJimmyK4542's example (and Feynmann's integration trick in general) indeed contradicts my claim that a solution to a general problem always applies to a special case. I'll edit this out. \u2013\u00a0user78270 Aug 16 '14 at 23:53\n\nOn this site one frequently finds under the linear-algebra tag questions of the kind: what is the determinant of a matrix $$\\begin{pmatrix}a&b&b&b\\\\b&a&b&b\\\\b&b&a&b\\\\b&b&b&a\\end{pmatrix}?$$ (I've just posted this question, which contains a list of such questions). It turns out finding an answer to this question becomes almost trivial (see my answer to the linked question) when reformulated more generally as\n\nWhat is the characteristic polynomial of a square matrix$~A$ of rank$~1$?\n\nknowing that by specialisation the answer gives the determinant of $\\lambda I-A$ for any scalar$~\\lambda$.\n\n-\n\nFrom time to time famous problems have such feature. History suggests this point: the transcendentalness of $\\pi$ solves the long-lasting problem of squaring a circle, analytic geometry and irrationality theory solve the problem of doubling a cubic, Galois's invention of group theory and quintic function, Kummer's invention of ideals and Fermat's last theorem, global differential geometry and Chern's intrinsic proof of Gauss-Bonnet theorem, and so on.\n\n-\n\nA broadly successful application of this was introduced by Richard Bellman under the phrase dynamic programming. The story of the \"birth\" of this now foundational topic in applied math is told largely in Bellman's own words here.\n\nA related term gives more evidence of the connection of ideas: invariant imbedding.\n\nA good discussion of dynamic programming references and examples came up early at StackOverflow, but was subsequently closed as off-topic.\n\nAn illustration is finding a shortest path between two specified points by \"imbedding\" that problem in finding all shortest paths from one point, Dijkstra's algorithm.\n\n-\n\nGeneralization comes up a lot when doing induction. For example,\n\n$$\\forall n ~~ \\sum_{k=0}^n 2^{-k} \\le 2$$\n\nis difficult to prove directly using induction on $n$. However, if you generalize to a stronger statement:\n\n$$\\forall n ~~ \\sum_{k=0}^n 2^{-k} \\le 2 - 2^{-n}$$\n\nThen induction may be used directly:\n\n$$\\sum_{k=0}^{n+1} 2^{-k} \\le 2 - 2^{-n - 1}$$ $$\\sum_{k=0}^n 2^{-k} + 2^{-n-1} \\le 2 - 2^{-n - 1}$$ $$\\sum_{k=0}^n 2^{-k}\\le 2 - 2^{-n}$$\n\nObviously you could see that it is a geometric series, but that is a generalization also.\n\nproblems that become easier when you formulate them in a more general (or ambitious) form\n\nThe potential difficulty of a generalization isn't the only disadvantage. If you disprove a generalization, then you haven't disproven the original theorem. In that respect, a generalization effectively forces you to pick sides in the investigation of a theorem.\n\n-\n\nThe most spectacular example I have seen is this one:\n\nSuppose A is an $n\\times n$ matrix with eigenvalues $\\lambda_1$, ..., $\\lambda_n$, including each eigenvalue according to its multiplicity. Then $A^2$ has eigenvalues $\\lambda_1^2$, ..., $\\lambda_n^2$ including multiplicity.\n\nTo prove this is in fact very very hard. (It's easy to show that $\\lambda_1^2$, ..., $\\lambda_n^2$ are all eigenvalues of $A$ by considering their eigenvectors, but unless you the dimensions of the eigenspaces match the multiplicities you're stuck.)\n\nHowever, the proof of the following statement is actually perfectly possible using elementary arguments (albeit clever arguments):\n\nSuppose A is an $n\\times n$ matrix with eigenvalues $\\lambda_1$, ..., $\\lambda_n$, including each eigenvalue according to its multiplicity. Then for any polynomial $g(x)$, $g(A)$ has eigenvalues $g(\\lambda_1)$, ..., $g(\\lambda_n)$ including multiplicity.\n\n-\n\nA nice example appeared on this web site today: Every prime number $p\\ge 5$ has $24\\mid p^2-1$ .\n\nAs posed, the problem sounds like it might be difficult. But it is very easy to show the more general result that every $n$ of the form $6k\\pm 1$ has the required property.\n\n-", "date": "2016-06-24 22:08:07", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form", "openwebmath_score": 0.9017611145973206, "openwebmath_perplexity": 539.8623476551608, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9343951680216529, "lm_q2_score": 0.9173026641072386, "lm_q1q2_score": 0.857123176955193}}
{"url": "http://math.stackexchange.com/questions/70177/for-a-prime-p-determine-the-number-of-positive-integers-whose-greatest-proper", "text": "# For a prime $p$, determine the number of positive integers whose greatest proper divisor is $p$\n\nI'm having a bit of difficulty writing a graceful proof for the following problem:\n\nFor a prime $p$, determine the number of positive integers whose greatest proper divisor is $p$.\n\nLet $A$ be the set of positive integers whose greatest proper divisor is $p$. I will show that $A=\\{\\alpha p\\,|\\,\\alpha \\;\\text{prime},\\; \\alpha \\leq p\\}$ so that $|A|=\\pi(p)$, the number of primes less than or equal to $p$.\n\nAssume that $n=\\alpha p$ for some prime $\\alpha \\leq p$. The factors of $n$ are $1, \\alpha, p, \\alpha p$. (In the case that $\\alpha =p$, the factors of $n$ are $1,p, p^2$.) The greatest proper divisor of $n$ therefore is $p$ and so $n\\in A$.\n\nConversely, for any number in $A$, clearly we have that $p$ is the greatest prime dividing it. Moreover, any three primes dividing it would contradict $p$ being the greatest divisor (this can be seen in the following way: if $\\alpha_1\\neq \\alpha_2$ are two primes dividing the number, neither of which is $p$, then we have $\\alpha_1<p<\\alpha_1p<\\alpha_1\\alpha_2p$ from which we infer that $p$ is not the greatest proper divisor.) Therefore, a number in $A$ must have at most two prime factors, one being $p$. (We can rule out that a number in $A$ has exactly the one prime factor, $p$, since it has no proper divisors.) Hence, $A\\subset \\{\\alpha p\\,|\\,\\alpha \\;\\text{prime},\\; \\alpha \\leq p\\}$.\n\nPerhaps there's a more elegant proof, but I'm concerned with my writing style. Can anyone help me rephrase my argument in a smoother way or critique it?\n\nedit: incorporated the changes suggested by Henning.\n\n-\nIsn't that what I did in the first paragraph? \u2013\u00a0 sasha Oct 5 '11 at 22:48\nperhaps you did \u2013\u00a0 Henry Oct 5 '11 at 22:54\n\nLooks sound and direct to me. There is only minor copy-editing to do, such as:\n\n\u2022 The condition $2\\le\\alpha$ is redundant; it is already implied by $\\alpha$ being prime.\n\n\u2022 The part \"Let $n\\in\\{\\alpha p\\,|\\,\\alpha \\;\\text{prime},\\; 2\\leq \\alpha \\leq p\\}$. Then $n=\\alpha p$ for some prime $\\alpha$, $2\\leq \\alpha \\leq p$\" is probably more verbose than you need to be. I would just write, \"Assume that $n=\\alpha p$ for some prime $\\alpha \\leq p$.\"\n\n\u2022 To make the structure of the proof more explicit, you could write \"Conversely,\" rather than \"Now,\" at the beginning of the third paragraph. This tells the reader that you have now finished something and is proceeding to the opposite direction of what you just proved.\n\n\u2022 Calling the other prime $q$ would be somewhat more conventional than $\\alpha$.\n\n-\nGreat, thank you for the suggestions. Last thing--if you could judge from this one post, would you say that my mathematical writing is verbose overall? Definitely would like to change that... \u2013\u00a0 sasha Oct 6 '11 at 7:49", "date": "2014-07-12 12:39:36", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/70177/for-a-prime-p-determine-the-number-of-positive-integers-whose-greatest-proper", "openwebmath_score": 0.9306151866912842, "openwebmath_perplexity": 188.67262868551668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806523850542, "lm_q2_score": 0.8740772450055544, "lm_q1q2_score": 0.8571032351424774}}
{"url": "http://mathhelpforum.com/calculus/24574-trig-integration.html", "text": "# Math Help - trig integration\n\n1. ## trig integration\n\nAfter I complete $\\int \\frac{\\sqrt{x^2-9}}{x}\\,dx$ by sec substitution I get\n\n$3\\sec^{-1}\\left(\\frac{x}{3}\\right)-\\sqrt{x^2-3}+C$.\n\nBut this doesn't match the calculator's answer:\n\n$3\\tan^{-1}\\left(\\frac{\\sqrt{x^2-9}}{3}\\right) - \\sqrt{x^2-3}+C$\n\nAfter differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...\n\nThen the second question is $f(x) = 3$. Find the equation.\n\n2. Originally Posted by DivideBy0\nAfter I complete $\\int \\frac{\\sqrt{x^2-9}}{3}\\,dx$ by sec substitution I get\n\n$3\\sec^{-1}\\left(\\frac{x}{3}\\right)-\\sqrt{x^2-3}+C$.\n\nBut this doesn't match the calculator's answer:\n\n$3\\tan^{-1}\\left(\\frac{\\sqrt{x^2-9}}{3}\\right) - \\sqrt{x^2-3}+C$\n\nAfter differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...\n\nThen the second question is $f(x) = 3$. Find the equation.\nI guess the right integral is: $\\int \\frac{\\sqrt{x^2-9}}{x}\\,dx$\n\n3. Woops, sorry, you're right that's what I meant\n\n4. Hello, DivideBy0!\n\nAfter I complete $\\int \\frac{\\sqrt{x^2-9}}{x}\\,dx$ by sec substitution I get: . $3\\sec^{-1}\\! \\left(\\frac{x}{3}\\right)-\\sqrt{x^2-3}+C$.\n\nBut this doesn't match the calculator's answer: / $3\\tan^{-1}\\!\\left(\\frac{\\sqrt{x^2-9}}{3}\\right) - \\sqrt{x^2-3}+C$\nThe answers are equivalent.\n\nLet $\\theta \\:=\\:\\sec^{-1}\\!\\left(\\frac{x}{3}\\right)$\nThen we have: . $\\sec\\theta \\:=\\:\\frac{x}{3} \\:=\\:\\frac{hyp}{adj}$\n\n$\\theta$ is in a right triangle with: $adj = 3,\\;hyp = x$\n\n. . Using Pythagorus: . $opp = \\sqrt{x^2-9}$\n\nSo: . $\\tan\\theta \\:=\\:\\frac{\\sqrt{x^2-9}}{3}$\nHence:. . $\\theta \\:=\\:\\tan^{-1}\\!\\left(\\frac{\\sqrt{x^2-9}}{3}\\right)$\n\nSee? . . . . . $\\sec^{-1}\\!\\left(\\frac{x}{3}\\right) \\;=\\;\\tan^{-1}\\left(\\frac{\\sqrt{x^2-9}}{3}\\right)$\n\n5. Originally Posted by DivideBy0\nAfter I complete $\\int \\frac{\\sqrt{x^2-9}}{x}\\,dx$ by sec substitution I get\n\n$3\\sec^{-1}\\left(\\frac{x}{3}\\right)-\\sqrt{x^2-3}+C$.\n\nBut this doesn't match the calculator's answer:\n\n$3\\tan^{-1}\\left(\\frac{\\sqrt{x^2-9}}{3}\\right) - \\sqrt{x^2-3}+C$\n\nAfter differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...\n\nThen the second question is $f(x) = 3$. Find the equation.\nthe answar is\n$-3\\sec^{-1}\\left(\\frac{x}{3}\\right)+\\sqrt{x^2-9}+C$.\nor\n\n$-3\\cos^{-1}\\left(\\frac{3}{x}\\right)+\\sqrt{x^2-9}+C$.\n\n6. Thanks Soroban for clearing it up a bit more... but when I differentiated on the calculator I still got\n\n$\\frac{d}{\\,dx}\\left(3\\sec^{-1}(\\frac{x}{3})-\\sqrt{x^2-3}\\right)=\\frac{9|\\frac{1}{x}|}{\\sqrt{x^2-9}}-\\frac{x}{\\sqrt{x^2-3}}$\n\nBut the original expression obviously has no absolute value signs.\n\n7. Originally Posted by DivideBy0\n$\\int \\frac{\\sqrt{x^2-9}}{x}\\,dx$\nYou don't even need trig. sub. - this only requires a simple substitution.\n\nStep #1: The make-up.\n\n$\\int {\\frac{{\\sqrt {x^2 - 9} }}\n{x}\\,dx} = \\int {\\frac{{\\sqrt {x^2 - 9} \\cdot x}}\n{{x^2 }}\\,dx} .$\n\nStep #2: The substitution.\n\n$u^2 = x^2 - 9 \\implies u\\,du = x\\,dx,$\n\n$\\int {\\frac{{\\sqrt {x^2 - 9} }}\n{x}\\,dx} = \\int {\\frac{{u^2 }}\n{{u^2 + 9}}\\,du} .$\n\nStep #3: Simple trick & mission almost-accomplished.\n\n$\\int {\\frac{{u^2 }}\n{{u^2 + 9}}\\,du} = \\int {du} - 9\\int {\\frac{1}\n{{u^2 + 9}}\\,du} = u - 3\\arctan \\frac{u}\n{3}+k.$\n\nStep #4: Back substitute.\n\n$\\int {\\frac{{\\sqrt {x^2 - 9} }}\n{x}\\,dx} = \\sqrt {x^2 - 9} - 3\\arctan \\frac{{\\sqrt {x^2 - 9} }}\n{3} + k.$", "date": "2015-07-06 12:17:21", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/24574-trig-integration.html", "openwebmath_score": 0.957465648651123, "openwebmath_perplexity": 2682.5602332306307, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806523850543, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.857103233534213}}
{"url": "https://math.stackexchange.com/questions/3030285/ways-of-selecting-3-balls-out-of-9-balls-if-at-least-one-black-ball-is-to-be", "text": "# Ways of selecting $3$ balls out of $9$ balls if at least one black ball is to be selected\n\nA box contains two white, three black, and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw.\n\nThe answer given is $$64$$. I attempted the problem with two different approaches.\n\nMy two different attempts:\n\nAttempt #1\nNumber of ways in which any $$3$$ balls can be drawn out of $$9$$ balls $$\\binom{9}{3}=84$$ Number of ways in which no black balls are drawn out of $$9$$ balls (choosing $$3$$ balls from the remaining $$6$$ balls) $$\\binom{6}{3}=20$$ Thus, the number of ways of choosing at least $$1$$ black ball is $$84-20=64$$\n\nAttempt #2\nNumber of ways of drawing $$1$$ black ball out of $$3$$ black balls $$\\binom{3}{1}=3$$ Now, we have to draw two more balls, we can choose those balls from the $$8$$ remaining balls $$\\binom{8}{2}=28$$ Since both the above events are associated with each other, by fundamental principle of counting, the number of ways of drawing at least one black ball out of the $$9$$ balls is $$3\\times28=84$$\n\nI think my second attempt should also be right. Please explain what I'm doing wrong with my second attempt.\n\nThe second attempt counts the combination $$\\{B_1,B_2,B_3\\}$$ three times as $$(B_1, \\{B_2, B_3\\})$$, $$(B_2, \\{B_3, B_1\\})$$ and $$(B_3, \\{B_1, B_2\\})$$.\n\nN.B: $$(B_i, \\{B_j, B_k\\})$$ means pick $$B_i$$ first, followed by the combination $$\\{B_j, B_k\\}$$.\n\nWe have the basic product rule\n\n$$|A\\times B| = \\#\\{(a,b) \\mid A \\times B\\} = |A| \\times |B|$$\n\nfor cardinals which holds for any sets $$A$$ and $$B$$.\n\nBy writing $$3 \\times 28$$, you are actually counting $$\\{B_1, B_2, B_3\\} \\times \\{B_2,B_3, \\dots\\}$$, which doesn't answer the problem.\n\nIn , $$(a,b)$$ is defined as $$\\{a,\\{a,b\\}\\}$$, so order matters.\n\n\u2022 Would you please elaborate the answer? I get that the second attempt is wrong, but I'm unable to understand it clearly. \u2013\u00a0Azelf Dec 7 '18 at 20:14\n\u2022 @Azelf My original notation would be clear enough. I've edited my answer to make it even clearer. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Dec 7 '18 at 20:57\n\nA selection of three balls that includes at least one black ball has either one black ball and two of the other six balls, two blacks balls and one of the other six balls, or three black balls and none of the other six balls. Therefore, the number of ways of selecting at least one black ball when three balls are selected from two white, three black, and four red balls is $$\\binom{3}{1}\\binom{6}{2} + \\binom{3}{2}\\binom{6}{1} + \\binom{3}{3}\\binom{6}{0} = 45 + 18 + 1 = 64$$\nYou counted each case in which $$k$$ black balls were selected $$k$$ times, once for each of the $$k$$ ways you could have designated one of those black balls as the designated black ball. Notice that $$\\color{red}{\\binom{1}{1}}\\binom{3}{1}\\binom{6}{2} + \\color{red}{\\binom{2}{1}}\\binom{3}{2}\\binom{6}{1} + \\color{red}{\\binom{3}{1}}\\binom{3}{3}\\binom{6}{0} = 45 + 36 + 3 = 84$$ To illustrate, place numbers on the black balls. If you select black balls $$b_1$$ and $$b_2$$ and a red ball, you count this selection twice: $$\\begin{array}{c c} \\text{designated black ball} & \\text{additional balls}\\\\ b_1 & b_2, r\\\\ b_2 & b_1, r \\end{array}$$ If you select all three black balls, you count this selection three times: $$\\begin{array}{c c} \\text{designated black ball} & \\text{additional balls}\\\\ b_1 & b_2, b_3\\\\ b_2 & b_1, b_3\\\\ b_3 & b_1, b_2 \\end{array}$$", "date": "2019-10-17 01:23:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3030285/ways-of-selecting-3-balls-out-of-9-balls-if-at-least-one-black-ball-is-to-be", "openwebmath_score": 0.8001653552055359, "openwebmath_perplexity": 154.21553549898297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806518175514, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8571032249968491}}
{"url": "https://math.stackexchange.com/questions/72067/is-there-a-combinatorial-way-to-see-the-link-between-the-beta-and-gamma-function", "text": "Is there a combinatorial way to see the link between the beta and gamma functions?\n\nThe Wikipedia page on the beta function gives a simple formula for it in terms of the gamma function. Using that and the fact that $\\Gamma(n+1)=n!$, I can prove the following formula: $$\\begin{eqnarray*} \\frac{a!b!}{(a+b+1)!} & = & \\frac{\\Gamma(a+1)\\Gamma(b+1)}{\\Gamma(a+1+b+1)}\\\\ & = & B(a+1,b+1)\\\\ & = & \\int_{0}^{1}t^{a}(1-t)^{b}dt\\\\ & = & \\int_{0}^{1}t^{a}\\sum_{i=0}^{b}\\binom{b}{i}(-t)^{i}dt\\\\ & = & \\int_{0}^{1}\\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}t^{a+i}dt\\\\ & = & \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\int_{0}^{1}t^{a+i}dt\\\\ & = & \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\left[\\frac{t^{a+i+1}}{a+i+1}\\right]_{t=0}^{1}\\\\ & = & \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{1}{a+i+1}\\\\ b! & = & \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{(a+b+1)!}{a!(a+i+1)} \\end{eqnarray*}$$ This last formula involves only natural numbers and operations familiar in combinatorics, and it feels very much as if there should be a combinatoric proof, but I've been trying for a while and can't see it. I can prove it in the case $a=0$: $$\\begin{eqnarray*} & & \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{(b+1)!}{0!(i+1)}\\\\ & = & \\sum_{i=0}^{b}(-1)^{i}\\frac{b!(b+1)!}{i!(b-i)!(i+1)}\\\\ & = & b!\\sum_{i=0}^{b}(-1)^{i}\\frac{(b+1)!}{(i+1)!(b-i)!}\\\\ & = & b!\\sum_{i=0}^{b}(-1)^{i}\\binom{b+\\text{1}}{i+1}\\\\ & = & b!\\left(1-\\sum_{i=0}^{b+1}(-1)^{i}\\binom{b+\\text{1}}{i}\\right)\\\\ & = & b! \\end{eqnarray*}$$ Can anyone see how to prove it for arbitrary $a$? Thanks!\n\n\u2022 The usual interpretation of \"combinatoric proof\" (that I'm accustomed to) is to show that the beta function counts something; what exactly do you mean by \"combinatoric proof\" here? Oct 12, 2011 at 17:15\n\u2022 In any event: it might be more interesting to establish this relationship instead... Oct 12, 2011 at 17:18\n\u2022 I'm with @J.M. - your derivation for $a=0$ doesn't really look like a combinatorial proof, as you're using only symbolic manipulation instead of counting and combining objects.\n\u2013\u00a0anon\nOct 12, 2011 at 21:03\n\nHere's a combinatorial argument for $a!\\, b! = \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{(a+b+1)!}{(a+i+1)}$, which is just a slight rewrite of the identity you want to show.\n\nSuppose you have $a$ red balls numbered $1$ through $a$, $b$ blue balls numbered $1$ through $b$, and one black ball.\n\nQuestion: How many permutations of the balls have all the red balls first, then the black ball, and then the blue balls?\n\nAnswer 1: $a! \\,b!$. There are $a!$ ways to choose the red balls to go in the first $a$ slots, $b!$ ways to choose the blue balls to go in the last $b$ slots, and $1$ way for the black ball to go in slot $a+1$.\n\nAnswer 2: Let $A$ be the set of all permutations in which the black ball appears after all the red balls (irrespective of where the blue balls go). Let $B_i$ be the subset of $A$ such that the black ball appears after blue ball $i$. Then the number of permutations we're after is also given by $|A| - \\left|\\bigcup_{i=1}^b B_i\\right|$. Since the probability that the black ball appears last of any particular $a+i+1$ balls is $\\frac{1}{a+i+1}$, and there are $(a+b+1)!$ total ways to arrange the balls, by the principle of inclusion-exclusion we get $$\\frac{(a+b+1)!}{a+1} - \\sum_{i=1}^{b}\\binom{b}{i}(-1)^{i+1}\\frac{(a+b+1)!}{(a+i+1)} = \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{(a+b+1)!}{(a+i+1)}.$$\n\n\u2022 Fantastic! How did you find this? Oct 12, 2011 at 22:17\n\u2022 @Steven: I thought about it for way too long. :) More seriously, an alternating binomial sum smells like inclusion-exclusion to me. I also thought I could generalize my answer to a similar question, and that turned out to work, although it took a while to get the formulation right. I kept trying to apply inclusion-exclusion to the full set of permutations, and it finally hit me that I only needed to consider subsets of the set I call $A$. And thanks! Oct 12, 2011 at 22:30\n\u2022 Nicely done indeed!\n\u2013\u00a0robjohn\nOct 13, 2011 at 0:37\n\u2022 @robjohn: And thanks for the edit. Not sure how I managed to leave that out! :) Oct 13, 2011 at 1:32\n\u2022 Beautiful, this is exactly the kind of answer I was hoping for, thank you! Oct 13, 2011 at 7:04\n\nUsing partial fractions, we have that $$\\frac{1}{(a+1)(a+2)\\dots(a+b+1)}=\\frac{A_1}{a+1}+\\frac{A_2}{a+2}+\\dots+\\frac{A_{b+1}}{a+b+1}\\tag{1}$$ Use the Heaviside Method; multiply $(1)$ by $(a+k)$ and set $a=-k$ to solve $(1)$ for $A_k$: $$A_k=\\frac{(-1)^{k-1}}{(k-1)!(b-k+1)!}=\\frac{(-1)^{k-1}}{b!}\\binom{b}{k-1}\\tag{2}$$ Plugging $(2)$ into $(1)$, yields $$\\frac{a!}{(a+b+1)!}=\\sum_{k=1}^{b+1}\\frac{(-1)^{k-1}}{b!}\\binom{b}{k-1}\\frac{1}{a+k}\\tag{3}$$ Multiplying $(3)$ by $b!$ and reindexing, gives us $$\\frac{a!b!}{(a+b+1)!}=\\sum_{k=0}^{b}(-1)^k\\binom{b}{k}\\frac{1}{a+k+1}\\tag{4}$$ and $(4)$ is your identity.\n\nUpdate: Starting from the basic binomial identity $$(1-x)^b=\\sum_{k=0}^b(-1)^k\\binom{b}{k}x^k\\tag{5}$$ multiply both sides of $(5)$ by $x^a$ and integrate from $0$ to $1$: $$B(a+1,b+1)=\\sum_{k=0}^b(-1)^k\\binom{b}{k}\\frac{1}{a+k+1}\\tag{6}$$\n\n\u2022 FYI: This argument appears on pages 188-189 of Concrete Mathematics, 2nd edition, where it is discussed in the context of the $n$th forward difference formula. Oct 12, 2011 at 20:50\n\u2022 This identity is one of my favorite uses of partial fractions and it turns up when using Euler's Transform for series acceleration.\n\u2013\u00a0robjohn\nOct 12, 2011 at 21:00\n\u2022 @Mike: not surprising since it computes the $b^{th}$ forward difference of $\\frac{1}{a+1}$. Thanks for the reference!\n\u2013\u00a0robjohn\nOct 12, 2011 at 21:02\n\nSeven years later I found another way to attack this. Define $$f(b, a) = \\frac{a!b!}{(a+b+1)!}$$ and $$h(b, a) = \\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{1}{a+i+1}$$. To connect the two, we define $$g$$ such that $$g(0, a) = \\frac{1}{a + 1}$$ and $$g(b + 1, a) = g(b, a) - g(b, a + 1)$$ and prove by induction in $$b$$ that $$f = g = h$$. In each case the base case is straightforward and we consider only the inductive step.\n\n$$\\begin{eqnarray*} & & g(b + 1, a) \\\\ & = & g(b, a) - g(b, a + 1) \\\\ & = & f(b, a) - f(b, a + 1) \\\\ & = & \\frac{a!b!}{(a+b+1)!} - \\frac{(a+1)!b!}{(a+b+2)!} \\\\ & = & \\frac{a!b!(a + b + 2)}{(a+b+2)!} - \\frac{a!b!(a+1)}{(a+b+2)!} \\\\ & = & \\frac{a!b!(b+1)}{(a+b+2)!} \\\\ & = & \\frac{a!(b+1)!}{(a+b+2)!} \\\\ & = & f(b+1, a)\\\\ \\end{eqnarray*}$$\n\n$$\\begin{eqnarray*} & & g(b + 1, a) \\\\ & = & g(b, a) - g(b, a + 1) \\\\ & = & h(b, a) - h(b, a + 1) \\\\ & = & \\left(\\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{1}{a+i+1}\\right) - \\left(\\sum_{i=0}^{b}\\binom{b}{i}(-1)^{i}\\frac{1}{a+i+2}\\right) \\\\ & = & \\frac{1}{a+1} + \\left(\\sum_{i=1}^{b}\\binom{b}{i}(-1)^{i}\\frac{1}{a+i+1}\\right) - \\left(\\sum_{i=0}^{b-1}\\binom{b}{i}(-1)^{i}\\frac{1}{a+i+2}\\right) - (-1)^{b}\\frac{1}{a+b+2}\\\\ & = & \\frac{1}{a+1} + \\left(\\sum_{i=1}^{b}\\binom{b}{i}(-1)^{i}\\frac{1}{a+i+1}\\right) + \\left(\\sum_{i=1}^{b}\\binom{b}{i-1}(-1)^{i}\\frac{1}{a+i+1}\\right) + (-1)^{b+1}\\frac{1}{a+b+2}\\\\ & = & \\frac{1}{a+1} + \\left(\\sum_{i=1}^{b}\\left(\\binom{b}{i} + \\binom{b}{i-1}\\right)(-1)^{i}\\frac{1}{a+i+1}\\right) + (-1)^{b+1}\\frac{1}{a+b+2}\\\\ & = & \\frac{1}{a+1} + \\left(\\sum_{i=1}^{b}\\binom{b+1}{i}(-1)^{i}\\frac{1}{a+i+1}\\right) + (-1)^{b+1}\\frac{1}{a+b+2}\\\\ & = & \\sum_{i=0}^{b+1}\\binom{b+1}{i}(-1)^{i}\\frac{1}{a+i+1} \\\\ & = & h(b + 1, a) \\\\ \\end{eqnarray*}$$", "date": "2022-08-08 20:36:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/72067/is-there-a-combinatorial-way-to-see-the-link-between-the-beta-and-gamma-function", "openwebmath_score": 0.8793419003486633, "openwebmath_perplexity": 176.87786336871503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874123575265, "lm_q2_score": 0.8652240964782012, "lm_q1q2_score": 0.8570800988397201}}
{"url": "https://math.stackexchange.com/questions/2746757/taxonomy-of-polygons", "text": "# Taxonomy of polygons\n\nI've written a tree-like layout to help myself remember which polygons are sub-types of others, because I always get confused. I was just wondering if this is right:\n\n|quadrilateral\n|parallelogram\n|rectangle\n|square\n|oblong\n|rhomboid\n|kite (corrected after rschwieb's answer, a rhombus is a kite)\n|rhombus\n|square\n|trapezoid(AmE) / trapezium (BrE)\n\n\nSo a square is a rhombus and a parallelogram.\n\nAlso, I know that there are two definitions of \"trapezoid.\" Under the inclusive definition \"trapezoid\" is immediately under \"quadrilateral\" in the tree and above parallelogram and kite. Under this definition all squares are trapezoids.\n\nIs my tree correct, at least ignoring the difference in the trapezoid definition difference?\n\nEdit: Thanks to rschwieb for helping me realise that a rhombus is a kite. There is also a nice Euler diagram Wikipedia\n\nYou need to see this post I wrote some time ago.\n\nIn short, the education system (at least in the US) has confused this issue and made it harder than it has to be.\n\nThere is a very natural hierarchy the depends on logical connections between quadrilaterals, and there is really no benefit to using the \u201cexclusive version\u201d of definitions.\n\nI would argue for this picture for the main characters:\n\nActually there is a little puzzle where you can figure out a new node to insert between \"quadrilateral\" and \"kite\" which also connects to \"parallelogram,\" and I have never seen this shape mentioned in a textbook. It's just not common enough to encounter in normal life.\n\n\u2022 I see in your chart you've used the inclusive definition of trapezoid, that's all good. I see the difference between this chart and the \"frankenstein\" chart you referred to is that the kite is in a separate area on its own, but in yours it traces down to rhombus. In yours it shows a rhombus as being a kite. This seems unintuitive with my everyday notion of a kite, but Wikipedia says: \"If all four sides of a kite have the same length (that is, if the kite is equilateral), it must be a rhombus.\" Also, the other chart doesn't link the kite to the rhombus at all. I think this chart is easier. \u2013\u00a0Zebrafish Apr 21 '18 at 2:21\n\u2022 There's just one thing I see wrong with it, a rectangle is an isosceles trapezoid? Are those two supposed to be connected with a line over on the right? \u2013\u00a0Zebrafish Apr 21 '18 at 2:49\n\u2022 \"Rectangles and squares are usually considered to be special cases of isosceles trapezoids though some sources would exclude them.\" Wow, there are a few a discrepancies in definitions, that's not helping. \u2013\u00a0Zebrafish Apr 21 '18 at 2:52\n\u2022 @Zebrafish as I mentioned \u201cexclusive definitions\u201d aren\u2019t as useful, they\u2019re harder to state, and make proving things less convenient. \u2013\u00a0rschwieb Apr 21 '18 at 3:41\n\u2022 @Zebrafish yes, I would cal a rectangle an isosoles trapezoid. \u2013\u00a0rschwieb Apr 21 '18 at 3:43\n\nCan't a kite also be a parallelogram, in the case where all sides are equal?\n\nThat of course depends on your definition of kite... I've rarely seen the term used at all. You can exclude that case specifically and your tree is then okay.\n\nWikipedia's Kite (geometry) article seems to include that in their special cases.\n\nEDIT: In that special case, it can also be a rhombus or a square\n\n\u2022 A kite that\u2019s also a parallelogram is a rhombus. \u2013\u00a0rschwieb Apr 21 '18 at 3:45", "date": "2019-07-19 14:45:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2746757/taxonomy-of-polygons", "openwebmath_score": 0.7078220248222351, "openwebmath_perplexity": 830.8944931659497, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731115849662, "lm_q2_score": 0.89029422102812, "lm_q1q2_score": 0.8570623079832539}}
{"url": "https://brilliant.org/discussions/thread/mathematical-induction-part-2/", "text": "# Mathematical Induction PART 2\n\nIn Part 1\n\nIt shows that\n\nStep 1 is Show it is true for n=1\n\nStep 2 is Show that if n=k is true then n=k+1 is also true\n\nso How to Do it?\n\nStep 1 : * prove* it is true for n=1 (\uff4e\uff4f\uff52\uff4d\uff41\uff4c\uff4c\uff59\uff09\n\nStep 2 \u3000\uff1a\u3000 done\u3000in this way normally we can prove it out..\n\nFirst ~ Assume it is true forn=k\n\nSecond ~ Prove it is true for n=k+1 (normally we can use the n=k case as fact )\n\nS (n / k) must be all positive integers including n/k\n\nEXAMPLE\n\nPROVE 1 + 3 + 5 + ... + (2n-1) = n^2\n\nFisrt : Show it is true for n=1\n\nfrom LEFT : 1+3+5+.....+(2(1)-1) = 1\n\nfrom RIGHT n^2 = (1^2) =1 SO 1 = 1^2 is True\n\nSECOND : Assume it is true for n=k\n\n1 + 3 + 5 + ... + (2^k-1) = k^2 is True\n\n(prove it by yourself :) write down your steps in comment )\n\nTHIRD : prove it is true for k+1\n\n1 + 3 + 5 + ... + (2^k-1) + (2^(k+1)-1) = (k+1)^2 ... ? (prove it by yourself :) write down your steps in comment )\n\nWe know that 1 + 3 + 5 + ... + (2k-1) = k^2 (the assumption above), so we can do a replacement for all but the last term:\n\nk^2 + (2(k+1)-1) = (k+1)^2\n\nTHEN expanding all terms:\n\nLEFT : k^2 + 2k + 2 - 1 = k^2 + 2k+1\n\nNEXT simplifying :\n\nRIGHT :k^2 + 2k + 1 = k^2 + 2k + 1\n\nLEFT and RIGHT are the same! So it is true, it is proven.\n\nTHEREFORE :\n\n1 + 3 + 5 + ... + (2(k+1)-1) = (k+1)^2 is TRUE!!!!!!\n\nMathematical Induction IS done !!! :)\n\nNote by Nicole Ling\n5\u00a0years, 4\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\ncalculus?\n\n- 5\u00a0years, 1\u00a0month ago\n\nso it is math\n\n- 5\u00a0years, 1\u00a0month ago\n\nreally?\n\n- 5\u00a0years, 1\u00a0month ago\n\nwhat\n\n- 5\u00a0years, 1\u00a0month ago\n\nroar\n\n- 5\u00a0years, 1\u00a0month ago\n\nfeeling lucky\n\n- 5\u00a0years, 1\u00a0month ago\n\nfeeling happy\n\n- 5\u00a0years, 1\u00a0month ago\n\n- 5\u00a0years, 1\u00a0month ago\n\ntiger\n\n- 5\u00a0years, 1\u00a0month ago\n\nis that true?\n\n- 5\u00a0years, 1\u00a0month ago\n\nyes\n\n- 5\u00a0years, 1\u00a0month ago\n\nno\n\n- 5\u00a0years, 1\u00a0month ago\n\nWhat have you been telling?\n\n- 5\u00a0years, 1\u00a0month ago\n\nwhat is your question ? I can't understand what did you want to say through your comments..\n\n- 5\u00a0years ago\n\na\n\n- 5\u00a0years, 1\u00a0month ago\n\ngood\n\n- 5\u00a0years, 1\u00a0month ago\n\nnice\n\n- 5\u00a0years, 1\u00a0month ago\n\n\u00d7", "date": "2020-06-04 08:58:41", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/mathematical-induction-part-2/", "openwebmath_score": 0.9713510274887085, "openwebmath_perplexity": 3636.1558847293654, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9626731147976795, "lm_q2_score": 0.8902942159342104, "lm_q1q2_score": 0.8570623059397442}}
{"url": "https://math.hecker.org/2011/09/02/linear-algebra-and-its-applications-exercise-2-2-12/", "text": "## Linear Algebra and Its Applications, Exercise\u00a02.2.12\n\nExercise 2.2.12. What is a 2 by 3 system of equations $Ax = b$ that has the following general solution?\n\n$x = \\begin{bmatrix} 1 \\\\ 1 \\\\ 0 \\end{bmatrix} + w \\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix}$\n\nAnswer: The general solution above is the sum of a particular solution and a homogeneous solution, where\n\n$x_{particular} = \\begin{bmatrix} 1 \\\\ 1 \\\\ 0 \\end{bmatrix}$\n\nand\n\n$x_{homogeneous} = w \\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix}$\n\nSince $w$ is the only variable referenced in the homogeneous solution it must be the only free variable, with $u$ and $v$ being basic. Since $u$ is basic we must have a pivot in column 1, and since $v$ is basic we must have a second pivot in column 2. After performing elimination on $A$ the resulting echelon matrix $U$ must therefore have the form\n\n$U = \\begin{bmatrix} *&*&* \\\\ 0&*&* \\end{bmatrix}$\n\nTo simplify solving the problem we can assume that $A$ also has this form; in other words, we assume that $A$ is already in echelon form and thus we don\u2019t need to carry out elimination. The matrix $A$ then has the form\n\n$A = \\begin{bmatrix} a_{11}&a_{12}&a_{13} \\\\ 0&a_{22}&a_{23} \\end{bmatrix}$\n\nwhere $a_{11}$ and $a_{22}$ are nonzero (because they are pivots).\n\nWe then have\n\n$Ax_{homogeneous} = \\begin{bmatrix} a_{11}&a_{12}&a_{13} \\\\ 0&a_{22}&a_{23} \\end{bmatrix} w \\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix} = 0$\n\nIf we assume that $w$ is 1 and express the right-hand side in matrix form this then becomes\n\n$\\begin{bmatrix} a_{11}&a_{12}&a_{13} \\\\ 0&a_{22}&a_{23} \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}$\n\nor (expressed as a system of equations)\n\n$\\setlength\\arraycolsep{0.2em}\\begin{array}{rcrcrcl}a_{11}&+&2a_{12}&+&a_{13}&=&0 \\\\ &&2a_{22}&+&a_{23}&=&0 \\end{array}$\n\nThe pivot $a_{11}$ must be nonzero, and we arbitrarily assume that $a_{11} = 1$. We can then satisfy the first equation by assigning $a_{12} = 0$ and $a_{13} = -1$. The pivot $a_{22}$ must also be nonzero, and we arbitrarily assume that $a_{22} = 1$ as well. We can then satisfy the second equation by assigning $a_{23} = -2$. Our proposed value of $A$ is then\n\n$A = \\begin{bmatrix} 1&0&-1 \\\\ 0&1&-2 \\end{bmatrix}$\n\nso that we have\n\n$Ax_{homogeneous} = \\begin{bmatrix} 1&0&-1 \\\\ 0&1&-2 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}$\n\nas required.\n\nWe next turn to the general system $Ax = b$. We now have a value for $A$, and we were given the value of the particular solution. We can multiply the two to calculate the value of $b$:\n\n$b = Ax_{particular} = \\begin{bmatrix} 1&0&-1 \\\\ 0&1&-2 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 1 \\\\ 0 \\end{bmatrix} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}$\n\nThis gives us the following as an example 2 by 3 system that has the general solution specified above:\n\n$\\begin{bmatrix} 1&0&-1 \\\\ 0&1&-2 \\end{bmatrix} \\begin{bmatrix} u \\\\ v \\\\ w \\end{bmatrix} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}$\n\nor\n\n$\\setlength\\arraycolsep{0.2em}\\begin{array}{rcrcrcl}u&&&-&w&=&1 \\\\ &&v&-&2w&=&1 \\end{array}$\n\nFinally, note that the solution provided for exercise 2.2.12 at the end of the book is incorrect. The right-hand side must be a 2 by 1 matrix and not a 3 by 1 matrix, so the final value of 0 in the right-hand side should not be present.\n\nNOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.\n\nIf you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang\u2019s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang\u2019s other books.\n\nThis entry was posted in linear algebra. Bookmark the permalink.\n\n### 2 Responses to Linear Algebra and Its Applications, Exercise\u00a02.2.12\n\n1. Daniel says:\n\nI think that your final note is incorrect, due to the fact that if you find the general solution for the system Ax=b that you found, you\u2019ll have to write the solution like Strang does it in (3) page (76). There are three entries on the solution because \u201cx\u201d vector lenght. The general solution (in Matlab notation) is x = [u; v; w] = [1+w; 1+2w; w]= [1; 1; 0] + w*[1; 2; 1]. The general solution he proposed at the begining of the exercise\n\n\u2022 hecker says:\n\nMy apologies for the delay in responding. Are you referring to my final sentence about the solution to exercise 2.2.12 given on page 476 in the back of the book? If so, I think I may have confused you. I am *not* saying that Strang wrote the general solution incorrectly in the statement of the exercise on page 79, or that Strang found an incorrect solution to the exercise.\n\nRather my point is as follows: In the statement of the solution on page 476 Strang shows as a solution the same 2 by 3 matrix that I derived above, and Strang shows that 2 by 3 matrix multiplying the vector (u, v, w) just as I do above, representing a system of two equations in three unknowns. However on the right-hand side Strang shows that 2 by 3 matrix multiplying the vector (u, v, w) to produce the vector (1, 1, 0). This cannot be: since the matrix has only two rows, that multiplication would produce a vector with only two elements, not three (as in the book). Those two elements represent the right-hand sides of the corresponding system of two equations.\n\nSo the left-hand side in the solution of 2.2.12 on page 476 is correct, but the right-hand side of the solution of 2.2.12 on page 476, namely the vector (1, 1, 0), is not. Instead the right-hand side should be the vector (1, 1) as I derived above.", "date": "2019-11-12 19:01:54", "meta": {"domain": "hecker.org", "url": "https://math.hecker.org/2011/09/02/linear-algebra-and-its-applications-exercise-2-2-12/", "openwebmath_score": 0.9118801355361938, "openwebmath_perplexity": 208.41991238078072, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918533088548, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.857057791884327}}
{"url": "http://math.stackexchange.com/questions/584655/what-is-fracd-arctanxdx", "text": "# What is $\\frac{d(\\arctan(x))}{dx}$?\n\nLet $v= \\arctan{x}$. Now I want to find $\\frac{dv}{dx}$. My method is this: Rearranging yields $\\tan(v) = x$ and so $dx = \\sec^2(v)dv$. How do I simplify from here? Of course I could do something like $dx = \\sec^2(\\arctan(x))dv$ so that $\\frac{dv}{dx} = \\cos^2(\\arctan(x))$ but I am sure a better expression exists. I am probably just missing some crucial step where we convert one of the trigonometric expressions into an expression involving $x$. Thanks in advance for any help or tips!\n\n-\n\n## 4 Answers\n\nThe derivative of $\\tan v$ is $1+\\tan^2 v$. It will be easier to simplify, since here $v=\\arctan x$.\n\nYou may check:\n\n$$\\sec^2 v = \\frac{1}{\\cos^2 v} = \\frac{\\cos^2 v + \\sin^2 v}{\\cos^2 v} = 1+\\tan^2 v$$\n\nThen\n\n$$\\mathrm{d}x = (1+\\tan^2 v) \\ \\mathrm{d}v = (1+x^2) \\ \\mathrm{d}v$$\n\nAnd\n\n$$\\frac{\\mathrm{d}v}{\\mathrm{d}x}=\\frac{1}{1+x^2}$$\n\n-\nPerfect thanks a lot, this is very clear. I see now that using $\\cos^2(x) = \\frac{1}{\\tan(x)^2+1}$ will also change the expression $\\cos^2(\\arctan(x))$ into $\\frac{1}{1+x^2}$. Thanks for the answer! \u2013\u00a0 Slugger Nov 28 '13 at 15:24\n\nAnother way :\n\n$$\\frac{d\\arctan x}{dx}=\\lim_{h\\to0}\\frac{\\arctan(x+h)-\\arctan x}h$$\n\n$$\\displaystyle=\\lim_{h\\to0}\\frac{\\arctan\\frac{x+h-x}{1+(x+h)x}}h$$\n\n$$\\displaystyle=\\lim_{h\\to0}\\left(\\frac{\\arctan\\frac h{1+(x+h)x}}{\\frac h{1+(x+h)x}}\\right)\\cdot\\frac1{\\lim_{h\\to0}\\{1+(x+h)x\\}}=1\\cdot\\frac1{1+x^2}$$\n\nas $\\displaystyle\\lim_{u\\to0}\\frac{\\arctan u}u=\\lim_{v\\to0}\\frac v{\\tan v}=\\lim_{v\\to0}\\cos v\\cdot\\frac1{\\lim_{v\\to0}\\frac{\\sin v}v}=1\\cdot1$\n\n-\nDefinition of the derivative! +1. \u2013\u00a0 Ahaan S. Rungta Nov 28 '13 at 15:30\nThank you for your answer! It is nice to see how the answer tackle my question in a multitude of ways! \u2013\u00a0 Slugger Nov 28 '13 at 15:34\n@Slugger, my pleasure. Please have a look into math.stackexchange.com/questions/579170/\u2026 \u2013\u00a0 lab bhattacharjee Nov 28 '13 at 15:46\n\nYou can also use the Inverse Derivative Formula, which states that if $f(x)$ and $g(x)$ are inverse functions, we have $$g'(x) = \\dfrac {1}{f'(g(x))}.$$So, if $g(x)=\\arctan x$, our task is to find $g'(x)$. In that case, we have $f(x)=\\tan x$, which gives us $f'(x)=sec^2 x$, so we can substitute: \\begin {align*} g'(x) &= \\dfrac {1}{f'(g(x))} \\\\&= \\dfrac {1}{\\sec^2 (g(x))} \\\\&= \\dfrac {1}{\\sec^2 (\\arctan x)}. \\end {align*}We can find $\\sec (\\arctan x)$ geometrically. Consider a right triangle with legs of length $x$, $1$, and $\\sqrt{1+x^2}$. Let $\\theta$ be the angle opposite to the leg of length $x$. Then, $$\\sec \\left( \\arctan x \\right) = \\sec (\\theta) = \\sqrt {1+x^2},$$ so our answer is $$\\dfrac {1}{\\left( \\sqrt{1+x^2} \\right)^2} = \\boxed {\\dfrac {1}{1+x^2}}.$$\n\n-\nThanks you for your answer! \u2013\u00a0 Slugger Nov 28 '13 at 15:35\nYou're welcome! :) \u2013\u00a0 Ahaan S. Rungta Nov 28 '13 at 15:36\n\n$$v=\\arctan(x)\\Rightarrow x=\\tan v\\Rightarrow x'=\\frac{1}{(\\tan v)'}=\\frac{1}{(\\frac{\\sin v}{\\cos v})'}=\\frac{1}{\\frac{1}{\\cos^2 v}}=\\cos^2 v=\\frac{\\cos^2 v}{1}$$ $$=\\frac{\\cos^2 v}{\\cos^2 v+\\sin^2 v}=\\frac{\\frac{\\cos^2 v}{\\cos^2 v}}{\\frac{\\cos^2 v+\\sin^2 v}{\\cos^2 v}}=\\frac{1}{1+\\tan^2 v}=\\frac{1}{1+x^2}$$ i.e $$v'=(\\arctan(x))'=\\frac{1}{1+x^2}$$\n\n-\nI am not sure I follow exactly. Your second equation says $x=\\tan v$ and then you follow by saying $x' =\\frac{1}{(\\tan v)'}$... Maybe I am missing something \u2013\u00a0 Slugger Nov 28 '13 at 15:27\nSir, my solution is correct, You can also use the Inverse Derivative Formula, \u2013\u00a0 Madrit Zhaku Nov 28 '13 at 15:31\nOh, it seems this is essentially what I did, but I posted a few minutes later. I didn't see your post when I posted, so it's not that I copied. Should I delete my post? \u2013\u00a0 Ahaan S. Rungta Nov 28 '13 at 15:32\n@MadritZhaku When somebody asks you to explain what you did, \"it is correct\" is not as helpful as actually explaining what you did. If you don't have the patience to elaborate, don't respond at all. Your comment came off quite rude. \u2013\u00a0 Ahaan S. Rungta Nov 28 '13 at 15:33\nall is well that ends well :) \u2013\u00a0 Slugger Nov 28 '13 at 15:37", "date": "2014-12-23 05:35:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/584655/what-is-fracd-arctanxdx", "openwebmath_score": 0.9461610317230225, "openwebmath_perplexity": 558.4447670146069, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918487320493, "lm_q2_score": 0.8670357615200474, "lm_q1q2_score": 0.8570577828217518}}
{"url": "http://svc2006.it/pmey/eigenvalues-and-eigenvectors-of-3x3-matrix.html", "text": "This is called the eigendecomposition. Learn that the eigenvalues of a triangular matrix are the diagonal entries. \u00bb If they are numeric, eigenvalues are sorted in order of decreasing absolute value. Computing Eigenvalues and Eigenvectors Eigenvalue Problems Eigenvalues and Eigenvectors Geometric Interpretation Eigenvalue Problems Eigenvalue problems occur in many areas of science and engineering, such as structural analysis Eigenvalues are also important in analyzing numerical methods Theory and algorithms apply to complex matrices as well. Thx in advance!. 1 Let A be an n \u00d7 n matrix. Eigenvalues and the characteristic. This means the only eigenvalue is 0, and every nonzero plynomial is an eigenvector, so the eigenspace of eigenvalue 0 is the whole space V. To better understand these concepts, let's consider the following situation. I need some help with the following problem please? Let A be a 3x3 matrix with eigenvalues -1,0,1 and corresponding eigenvectors l1l. There is a hope. *XP the eigenvalues up to a 4*4 matrix can be calculated. Requirements: The program should\u2026 (Use your code from programming assignment 7 for items 1 through 4) 1. A normal matrix is de ned to be a matrix M, s. Solution: Ax = x )x = A 1x )A 1x = 1 x (Note that 6= 0 as Ais invertible implies that det(A) 6= 0). Any help is greatly appreciated. Matrices and Eigenvectors It might seem strange to begin a section on matrices by considering mechanics, but underlying much of matrix notation, matrix algebra and terminology is the need to describe the physical world in terms of straight lines. \u03bb 1, \u03bb 2, \u03bb 3, \u2026, \u03bb p. The Eigenvalues from a Covariance matrix inform us about the directions (read: principal components) along which the data has the maximum spread. It decomposes matrix using LU and Cholesky decomposition. Hi, I'm having trouble with finding the eigenvectors of a 3x3 matrix. 2 FINDING THE EIGENVALUES OF A MATRIX Consider an n\u00a3n matrix A and a scalar \u201a. In my earlier posts, I have already shown how to find out eigenvalues and the corresponding eigenvectors of a matrix. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. The eigen-values are di erent for each C, but since we know the eigenvectors they are easy to diagonalize. The calculator will perform symbolic calculations whenever it is possible. is nonsingular, and hence invertible. (In fact, the eigenvalues are the entries in the diagonal matrix D {\\displaystyle D} (above), and therefore D {\\displaystyle D} is uniquely determined by A {\\displaystyle A} up to the order of its entries. The eigenvalues and eigenvectors of a matrix may be complex, even when the matrix is real. With this installment from Internet pedagogical superstar Salman Khan's series of free math tutorials, you'll learn how to determine the eigenvalues of 3x3 matrices in eigenvalues. Eigenvectors of Rin C2 for the eigenvalues iand iare i 1 and 1, respectively. Given a matrix A, recall that an eigenvalue of A is a number \u03bb such that Av = \u03bb v for some vector v. We give two different proofs. I've already tried to use the EigenvalueDecomposition from Accord. With the program EIGENVAL. Eigenvalues [ m, UpTo [ k]] gives k eigenvalues, or as many as are available. Theorem EDELI Eigenvectors with Distinct Eigenvalues are Linearly Independent. You have 3 vector equations Au1=l1u1 Au2=l2u2 Au3=l3u3 Consider the matrix coefficients a11,a12,a13, etc as unknowns. I'm having a problem finding the eigenvectors of a 3x3 matrix with given eigenvalues. the eigenvalues of a triangular matrix (upper or lower triangular) are the entries on the diagonal. One approach is to raise the matrix to a high power. The diagonal matrix D contains eigenvalues. 2 MATH 2030: EIGENVALUES AND EIGENVECTORS De nition 0. The others are not eigenvectors. Let us rearrange the eigenvalue equation to the form , where represents a vector of all zeroes (the zero vector). This matrix calculator computes determinant , inverses, rank, characteristic polynomial , eigenvalues and eigenvectors. While the matrix representing T is basis dependent, the eigenvalues and eigenvectors are not. Let p(t) be the [\u2026] Determine Whether Given Matrices are Similar (a) Is the. Also, the eigenvalues and eigenvectors satisfy (A - \u03bbI)X r = 0 r. Eigenvalues and Eigenvectors. The first one is a simple one - like all eigenvalues are real and different. That is, the eigenvectors are the vectors that the linear transformation A merely elongates or shrinks, and the amount that they elongate/shrink by is the eigenvalue. Free Matrix Eigenvectors calculator - calculate matrix eigenvectors step-by-step This website uses cookies to ensure you get the best experience. And we used the fact that lambda is an eigenvalue of A, if and only if, the determinate of lambda times the identity matrix-- in this case it's a 2 by 2 identity matrix-- minus A is equal to 0. Consider the two-dimensional vectors a and b shown here. It decomposes matrix using LU and Cholesky decomposition. The problem is I don't know how to write (1,0,0) as a lineair combination of my eigenvectors. Note that the multiplication on the left hand side is matrix multiplication (complicated) while the mul-. Shouldn't it? $\\endgroup$ \u2013 Janos Jun 27 '19 at 10:04. And that says, any value, lambda, that satisfies this equation for v is a non-zero vector. (9-4) Hence, the eigenspace associated with eigenvalue \u03bb is just the kernel of (A - \u03bbI). It decomposes matrix using LU and Cholesky decomposition. Theorem EDELI Eigenvectors with Distinct Eigenvalues are Linearly Independent. The l =1 eigenspace for the matrix 2 6 6 4 2 1 3 4 0 2 1 3 2 1 6 5 1 2 4 8 3 7 7 5 is two-dimensional. Find the eigenvalues and eigenvectors. First, form the matrix. If you're seeing this message, it means we're having trouble loading external resources on our website. There will be an eigenvalue corresponding to each eigenvector of a matrix. Given any square matrix A \u2208 M n(C),. 5 of the textbook. Eigenvalues and Eigenvectors. I have a 3x3 covariance matrix (so, real, symmetric, dense, 3x3), I would like it's principal eigenvector, and speed is a concern. proc iml; x = hadamard(16); call eigen(val, vec, x); print (val) vec[format=5. EIGENVALUES AND EIGENVECTORS De\ufb01nition 7. When we compute the eigenvalues and the eigenvectors of a matrix T ,we can deduce the eigenvalues and eigenvectors of a great many other matrices that are derived from T ,and every eigenvector of T is also an eigenvector of the matrices , ,,. A non-square matrix A does not have eigenvalues. The eigenvalues of a selfadjoint matrix are always real. Philip Petrov ( https://cphpvb. I need some help with the following problem please? Let A be a 3x3 matrix with eigenvalues -1,0,1 and corresponding eigenvectors l1l. Finding a set of matrices based on eigenvalues and eigenvectors with constraints. [email\u00a0protected] 224 CHAPTER 7. To calculate the eigenvalues and eigenvector of the Hessian, you would first calculate the Hessian (a symmetric 3x3 matrix, containing the second derivatives in each of the 3 directions) for each pixel. If the resulting V has the same size as A, the matrix A has a full set of linearly independent eigenvectors that satisfy A*V = V*D. The non-symmetric problem of finding eigenvalues has two different formulations: finding vectors x such that Ax = \u03bbx, and finding vectors y such that y H A = \u03bby H (y H implies a complex conjugate transposition of y). the coefficients, which, in order to have non-zero solutions, has to have a singular matrix (zero determinant). Also, the method only tells you how to find the largest eigenvalue. Call you eigenvectors u1,u2,u3. Find a basis for this eigenspace. Eigenvectors and eigenspaces for a 3x3 matrix. Get 1:1 help now from expert Advanced Math tutors. exp(xA) is a fundamental matrix for our ODE Repeated Eigenvalues When an nxn matrix A has repeated eigenvalues it may not have n linearly independent eigenvectors. As an example, in the case of a 3 X 3 Matrix and a 3-entry column vector,. is nonsingular, and hence invertible. In the last video we set out to find the eigenvalues values of this 3 by 3 matrix, A. matrix then det(A\u2212\u03bbI) = 0. 2 Eigenvalues and Eigenvectors of the power Matrix. It decomposes matrix using LU and Cholesky decomposition. Finding a set of matrices based on eigenvalues and eigenvectors with constraints. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. Eigenvalues code in Java. An eigenvalue of a matrix is nothing but a special scalar that is used in the multiplication of matrices and is of great importance in physics as well. Eigenvalues [ m, spec] is always equivalent to Take [ Eigenvalues [ m], spec]. The eigenvalues are 4; 1; 4(4is a double root), exactly the diagonal elements. I have a 3x3 real symmetric matrix, from which I need to find the eigenvalues. The matrix of this transformation is the 6 6 all-zero matrix (in arbitrary basis). Obtain the Eigenvectors and Eigenvalues from the covariance matrix or correlation matrix, or perform Singular Value Decomposition. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. the coefficients, which, in order to have non-zero solutions, has to have a singular matrix (zero determinant). Nth power of a square matrix and the Binet Formula for Fibonacci sequence Yue Kwok Choy Given A= 4 \u221212 \u221212 11. The eigenvalues and eigenvectors of a matrix are scalars and vectors such that. 2 Eigenvalues and Eigenvectors (cont\u2019d) Example. The associated eigenvectors can now be found. Matrix A: 0 -6 10-2 12 -20-1 6 -10 I got the eigenvalues of: 0, 1+i, and 1-i. Get the free \"Eigenvalues Calculator 3x3\" widget for your website, blog, Wordpress, Blogger, or iGoogle. The zero vector 0 is never an eigenvectors, by de\ufb01nition. Spectral theorem: For a normal matrix M2L(V), there exists an. Eigenvectors and Eigenvalues When a random matrix A acts as a scalar multiplier on a vector X, then that vector is called an eigenvector of X. If you can draw a line through the three points (0, 0), v and Av, then Av is just v multiplied by a number \u03bb; that is, Av = \u03bbv. Eigenvector and Eigenvalue. We take an example matrix from a Schaum's Outline Series book Linear Algebra (4 th Ed. Example: The Hermitian matrix below represents S x +S y +S z for a spin 1/2 system. Find the eigenvectors and the corresponding eigenvalues of T T T. When we compute the eigenvalues and the eigenvectors of a matrix T ,we can deduce the eigenvalues and eigenvectors of a great many other matrices that are derived from T ,and every eigenvector of T is also an eigenvector of the matrices , ,,. Eigenvalues and Eigenvectors. Any help is greatly appreciated. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. If you love it, our example of the solution to eigenvalues and eigenvectors of 3\u00d73 matrix will help you get a better understanding of it. The eigenvalue with the largest absolute value is called the dominant eigenvalue. 1; Lecture 13: Basis=? For A 3X3 Matrix: Ex. We have A= 5 2 2 5 and eigenvalues 1 = 7 2 = 3 The sum of the eigenvalues 1 + 2 = 7+3 = 10 is equal to the sum of the diagonal entries of the matrix Ais 5 + 5 = 10. Today we\u2019re going to talk about a special type of symmetric matrix, called a positive de\ufb01nite matrix. We recall that a scalar l \u00ce F is said to be an eigenvalue (characteristic value, or a latent root) of A, if there exists a nonzero vector x such that Ax = l x, and that such an x is called an eigen-vector (characteristic vector, or a latent vector) of A corresponding to the eigenvalue l and that the pair (l, x) is called an. The determinant will be computed by performing a Laplace expansion along the second row: The roots of the characteristic equation, are clearly \u03bb = \u22121 and 3, with 3 being a double root; these are the eigenvalues of B. p ( t) = \u2212 ( t \u2212 2) ( t \u2212 1) ( t + 1). If you have trouble understanding your eigenvalues and eigenvectors of 3\u00d73 matrix. You can use this to find out which of your. EIGENVALUES AND EIGENVECTORS 6. Here det (A) is the determinant of the matrix A and tr(A) is the trace of the matrix A. Eigenvalues [ m, - k] gives the k that are smallest in absolute value. These straight lines may be the optimum axes for describing rotation of a. For background on these concepts, see 7. Similar matrices always has the same eigenvalues, but their eigenvectors could be different. Tied eigenvalues make the problem of reliably returning the same eigenvectors even more interesting. 366) \u2022eigenvectors corresponding to distinct eigenvalues are orthogonal (\u2192TH 8. In this case, they are the measure of the data's covariance. eig computes eigenvalues and eigenvectors of a square matrix. is the characteric equation of A, and the left part of it is called characteric polynomial of A. SOLUTION: \u2022 In such problems, we \ufb01rst \ufb01nd the eigenvalues of the matrix. The characteristic polynomial (CP) of an nxn matrix A A is a polynomial whose roots are the eigenvalues of the matrix A A. The picture is more complicated, but as in the 2 by 2 case. When we compute the eigenvalues and the eigenvectors of a matrix T ,we can deduce the eigenvalues and eigenvectors of a great many other matrices that are derived from T ,and every eigenvector of T is also an eigenvector of the matrices , ,,. - Duration: 4:53. Logical matrices are coerced to numeric. Any value of \u03bb for which this equation has a solution is known as an eigenvalue of the matrix A. Equation (1) is the eigenvalue equation for the matrix A. First, form the matrix. Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. Eigen vector, Eigen value 3x3 Matrix Calculator. Eigenvalues and eigenvectors - physical meaning and geometric interpretation applet Introduction. Today Courses Practice Algebra Geometry Number Theory Calculus Probability Find the eigenvalues of the matrix A = (8 0 0 6 6 11 1 0 1). Those are the \u201ceigenvectors\u201d. Eigenvalues and Eigenvectors. Eigenvalues are simply the coefficients attached to eigenvectors, which give the axes magnitude. Decomposing a matrix in terms of its eigenvalues and its eigenvectors gives valuable insights into the properties of the matrix. To find the eigenvectors and eigenvalues for a 3x3 matrix. Title: Eigenvalues and Eigenvectors of the Matrix of Permutation Counts Authors: Pawan Auorora , Shashank K Mehta (Submitted on 16 Sep 2013 ( v1 ), last revised 20 Sep 2013 (this version, v2)). The generalized eigenvalue problem is to determine the solution to the equation Av = \u03bbBv, where A and B are n-by-n matrices, v is a column vector of length n, and \u03bb is a scalar. Compute eigenvectors and corresponding eigenvalues Sort the eigenvectors by decreasing eigenvalues and choose eigenvectors with the largest eigenvalues to form a dimensional matrix (where every column represents an eigenvector) Use this eigenvector matrix to transform the samples onto the new subspace. (1) The story begins in finding the eigenvalue(s) and eigenvector(s) of A. This procedure will lead you to a homogeneus 3x3 system w. Shio Kun for Chinese translation. But for the eigenvectors, it is, since the denominator is going to be (nearly) zero. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. To find the eigenvalues, we need to minus lambda along the main diagonal and then take the determinant, then solve for lambda. Learn to recognize a rotation-scaling matrix, and compute by how much the matrix rotates and scales. eig computes eigenvalues and eigenvectors of a square matrix. The eigenvalues and eigenvectors of a matrix have the following important property: If a square n n matrix A has n linearly independent eigenvectors then it is diagonalisable, that is, it can be factorised as follows A= PDP 1 where D is the diagonal matrix containing the eigenvalues of A along the diagonal, also written as D = diag[l 1;l 2;:::;l n]. [i 1]t, for any nonzero scalar t. In the present case, since we are dealing with a 3 X 3 Matrix and a 3-entry column vector,. The eigenvalues and eigenvectors of a matrix are scalars and vectors such that. The zero vector 0 is never an eigenvectors, by de\ufb01nition. They are used in a variety of data science techniques such as Principal Component Analysis for dimensionality reduction of features. oregonstate. The l =2 eigenspace for the matrix 2 4 3 4 2 1 6 2 1 4 4 3 5 is two. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big test). The eigenvectors are a lineal combination of atomic movements, which indicate global movement of the proteins (the essential deformation modes), while the associated eigenvalues indicate the expected displacement along each eigenvector in frequencies (or distance units if the Hessian is not mass-weighted), that is, the impact of each deformation movement in the. MATRIX NORMS 219 Moreover, if A is an m \u00d7 n matrix and B is an n \u00d7 m matrix, it is not hard to show that tr(AB)=tr(BA). This is called the eigendecomposition. If the resulting V has the same size as A, the matrix A has a full set of linearly independent eigenvectors that satisfy A*V = V*D. Find more Mathematics widgets in Wolfram|Alpha. Eigenvalues are simply the coefficients attached to eigenvectors, which give the axes magnitude. I know that I need to work backwards on this problem so I set up the characteristic equation. In MATLAB eigenvalues and eigenvectors of matrices can be calculated by command eig. In the last video we set out to find the eigenvalues values of this 3 by 3 matrix, A. technique for computing the eigenvalues and eigenvectors of a matrix, converging superlinearly with exponent 2 +. If X is a unit vector, \u03bb is the length of the vector produced by AX. As noted above the eigenvalues of a matrix are uniquely determined, but for each eigenvalue there are many eigenvectors. 1) where F0 is the free energy at the stationary point, x is a column matrix whose entries xi (i=1,2,\u2026n). Learn that the eigenvalues of a triangular matrix are the diagonal entries. Write the system in matrix form as Equivalently, (A nonhomogeneous system would look like. The result is a 3x1 (column) vector. ) c) This is very easy to see. l2l Find A. using the Cayley-Hamilton theorem. Example 4 Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 -1 2]. The eigenspaces corresponding to these matrices are orthogonal to each other, though the eigenvalues can still be complex. Theorem If A is an matrix and is a eigenvalue of A, then the set of all eigenvectors of , together with the zero vector, forms a subspace of. These vectors are eigenvectors of A, and these numbers are eigenvalues of A. p ( t) = \u2212 ( t \u2212 2) ( t \u2212 1) ( t + 1). For example, suppose we wish to solve the. The eigenvalues are 4; 1; 4(4is a double root), exactly the diagonal elements. Let A be the matrix given by A = [\u2212 2 0 1 \u2212 5 3 a 4 \u2212 2 \u2212 1] for some variable a. By de\ufb01nition of the kernel, that. square matrix and S={x1,x2,x3,\u2026,xp} S = { x 1, x 2, x 3, \u2026, x p } is a set of eigenvectors with eigenvalues \u03bb1,\u03bb2,\u03bb3,\u2026,\u03bbp. ; Solve the linear system (A - I 3) v = 0 by finding the reduced row echelon form of A - I 3. Since W x = l x then (W- l I) x = 0. Any help is greatly appreciated. Eigenvectors and eigenvalues with numpy. Easycalculation. The characteristic polynomial (CP) of an nxn matrix A A is a polynomial whose roots are the eigenvalues of the matrix A A. In Section 5. We can't expect to be able to eyeball eigenvalues and eigenvectors everytime. l2l Find A. EIGENVALUES AND EIGENVECTORS 6. As we sometimes have to diagonalize a matrix to get the eigenvectors and eigenvalues, for example diagonalization of Hessian(translation, rotation projected out) matrix, we can get the. Eigenvalues/vectors are instrumental to understanding electrical circuits, mechanical systems, ecology and even Google's PageRank algorithm. Call you matrix A. If X is a unit vector, \u03bb is the length of the vector produced by AX. It also includes an analysis of a 2-state Markov chain and a discussion of the Jordan form. The first one is a simple one \u2013 like all eigenvalues are real and different. A scalar \u03bb is said to be a eigenvalue of A, if Ax = \u03bbx for some vector x 6= 0. We need to get to the bottom of what the matrix A is doing to. For a unique set of eigenvalues to determinant of the matrix (W-l I) must be equal to zero. Eigenvalues [ m, spec] is always equivalent to Take [ Eigenvalues [ m], spec]. Call you matrix A. EIGENVALUES AND EIGENVECTORS 6. (3,2,4) and (0,-1,1) are eigenvectors. For Example, if x is a vector that is not zero, then it is an eigenvector of a square matrix A, if Ax is a scalar multiple of x. In machine learning, eigenvectors and eigenvalues come up quite a bit. Example 5 Suppose A is this 3x3 matrix: [ 0 0 2] [-3 1 6] [ 0 0 1]. Finding eigenvectors of a 3x3 matrix 2. Find the eigenvalues and bases for each eigenspace. These straight lines may be the optimum axes for describing rotation of a. The determinant will be computed by performing a Laplace expansion along the second row: The roots of the characteristic equation, are clearly \u03bb = \u22121 and 3, with 3 being a double root; these are the eigenvalues of B. Right when you reach $0$, the eigenvalues and eigenvectors become real (although there is only eigenvector at this point). Enter a matrix. If you're behind a web filter, please make sure that the domains *. l0l l0l ; l1l ; l1l respectively. 2; Lecture 14: Basis=? For A 2X2 Matrix; Lecture 15: Basis=? For A 3X3 Matrix: 1/3; Lecture 16: Basis. [email\u00a0protected] This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. Repeated Eigenvalues We conclude our consideration of the linear homogeneous system with constant coefficients x Ax' (1) with a brief discussion of the case in which the matrix has a repeated eigenvalue. The first one is a simple one - like all eigenvalues are real and different. Find all values of a which will guarantee that A has eigenvalues 0, 3, and \u2212 3. I guess A is 3x3, so it has 9 coefficients. 1 Let A be an n \u00d7 n matrix. Linear Algebra: Introduction to Eigenvalues and Eigenvectors. \u03bb is an eigenvalue (a scalar) of the Matrix [A] if there is a non-zero vector (v) such that the following relationship is satisfied: [A](v) = \u03bb (v) Every vector (v) satisfying this equation is called an eigenvector of [A] belonging to the eigenvalue \u03bb. eig returns a tuple (eigvals,eigvecs) where eigvals is a 1D NumPy array of complex numbers giving the eigenvalues of. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. The eigenvalue-eigenvector problem for A is the problem of nding numbers and vectors v 2R3 such that Av = v : If , v are solutions of a eigenvector-eigenvalue problem then the vector v is called an eigenvector of A and is called an eigenvalue of A. if TRUE, the matrix is assumed to be symmetric (or Hermitian if complex) and only its lower triangle (diagonal included) is used. Let's see if visualization can make these ideas more intuitive. Show that the eigenvalues of A are real. Also, the method only tells you how to find the largest eigenvalue. Today Courses Practice Algebra Geometry Number Theory Calculus Probability Find the eigenvalues of the matrix A = (8 0 0 6 6 11 1 0 1). Eigenvalues [ m, - k] gives the k that are smallest in absolute value. The calculator will perform symbolic calculations whenever it is possible. The eigenvalues are r1=r2=-1, and r3=2. Learn more Algorithm for finding Eigenvectors given Eigenvalues of a 3x3 matrix in C#. As is to be expected, Maple's. A simple way to do this is to apply three gradient filters (in. DA: 66 PA: 49 MOZ Rank: 16. For the purpose of analyzing Hessians, the eigenvectors are not important, but the eigenvalues are. The Eigenvector which corresponds to the maximum Eigenvalue of the Covariance matrix, C, will be the. Eigenvalues and Eigenvectors on Brilliant, the largest community of math and science problem solvers. (1) The eigenvalues of a triangle matrix are its diagonal elements. We also review eigenvalues and eigenvectors. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. Note that the multiplication on the left hand side is matrix multiplication (complicated) while the mul-. To be able to show that a given matrix is orthogonal. We were transforming a vector of points v into another set of points v R by multiplying by. Use [W,D] = eig(A. Linear Algebra: Introduction to Eigenvalues and Eigenvectors. Eigenvectors and eigenspaces for a 3x3 matrix. 1) can be rewritten. 2]; quit; This Hadamard matrix has 8 eigenvalues equal to 4 and 8 equal to -4. Share a link to this answer. For a Hermitian matrix the eigenvalues should be real. A simple online EigenSpace calculator to find the space generated by the eigen vectors of a square matrix. Computing the eigenvectors of a 3x3 symmetric matrix in C++ Every once in a while Google makes me wonder how people ever managed to do research 15 years ago. But Bcan have at most n linearly independent eigenvectors, so the eigenvalues obtained in this way must be all of B\u2019s eigenvalues. Try to find the eigenvalues and eigenvectors of the following matrix:. Eigenvector and Eigenvalue. But the problem is I can't write (1,0,0) as a combination of those eigenvectors. Eigenvalues and Eigenvectors, Imaginary and Real. Remember that the solution to. In the last video, we started with the 2 by 2 matrix A is equal to 1, 2, 4, 3. A matrix is diagonalizable if it has a full set of eigenvectors. It will be tedious for hand computation. Since W x = l x then (W- l I) x = 0. There could be multiple eigenvalues and eigenvectors for a symmetric and square matrix. Resize; Like. Namely, prove that (1) the determinant of A is the product of its eigenvalues, and (2) the trace of A is the sum of the eigenvalues. It will then compute the eigenvalues (real and complex) and eigenvectors (real and complex) for that matrix. Eigenvalues of the said matrix [ 2. 3 Eigenvalues and Eigenvectors. You have 3x3=9 linear equations for nine unknowns. a numeric or complex matrix whose spectral decomposition is to be computed. They have many uses! A simple example is that an eigenvector does not change direction in a transformation:. 1 3 4 5 , l = 1 11. For background on these concepts, see 7. The eigenvalues correspond to rows in the eigenvector matrix. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Show transcribed image text. The zero vector 0 is never an eigenvectors, by de\ufb01nition. A real number \u03bb is said to be an eigenvalue of a matrix A if there exists a non-zero column vector v such that A. Most of the methods on this website actually describe the programming of matrices. Just write down two generic diagonal matrices and you will see that they must. And that says, any value, lambda, that satisfies this equation for v is a non-zero vector. For now, I have original matrix in 2D array, I have eigenvalues in variables, and I have second matrix that has result of Eigenvalue*I - A (eigenvalue times matrix that has 1 on diagonal minus original matrix) So my form for now is lets say in example:-1 0 -1 v1 0-2 0 -2 v2 = 0-1 0 -1 v3 0. *XP the eigenvalues up to a 4*4 matrix can be calculated. Matrix D is the canonical form of A--a diagonal matrix with A's eigenvalues on the main diagonal. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. Eigenvectors of repeated eigenvalues. Learn to recognize a rotation-scaling matrix, and compute by how much the matrix rotates and scales. Specify the eigenvalues The eigenvalues of matrix $\\mathbf{A}$ are thus $\\lambda = 6$, $\\lambda = 3$, and $\\lambda = 7$. The normalized eigenvector for = 5 is: The three eigenvalues and eigenvectors now can be recombined to give the solution to the original 3x3 matrix as shown in Figures 8. The above equation is called the eigenvalue. That example demonstrates a very important concept in engineering and science - eigenvalues and. [As to follow the definition the zero vector i. Find all values of a which will guarantee that A has eigenvalues 0, 3, and \u2212 3. Any value of \u03bb for which this equation has a solution is known as an eigenvalue of the matrix A. 1 (Eigenvalue, eigenvector) Let A be a complex square matrix. com is the most convenient free online Matrix Calculator. The l =2 eigenspace for the matrix 2 4 3 4 2 1 6 2 1 4 4 3 5 is two-dimensional. Example solving for the eigenvalues of a 2x2 matrix. A simple online EigenSpace calculator to find the space generated by the eigen vectors of a square matrix. (3,2,4) and (0,-1,1) are eigenvectors. Below, change the columns of A and drag v to be an. a numeric or complex matrix whose spectral decomposition is to be computed. Introduction. An eigenvector associated with \u03bb1 is a nontrivial solution~v1 to (A \u03bb1I)~v =~0: (B. The calculator will perform symbolic calculations whenever it is possible. Finding eigenvectors of a 3x3 matrix 2. *XP the eigenvalues up to a 4*4 matrix can be calculated. The determinant will be computed by performing a Laplace expansion along the second row: The roots of the characteristic equation, are clearly \u03bb = \u22121 and 3, with 3 being a double root; these are the eigenvalues of B. Eigenvalues and eigenvectors in Maple Maple has commands for calculating eigenvalues and eigenvectors of matrices. Its roots are 1 = 1+3i and 2 = 1 = 1 3i: The eigenvector corresponding to 1 is ( 1+i;1). 4 A symmetric matrix: \u20ac A. Substitute one eigenvalue \u03bb into the equation A x = \u03bb x\u2014or, equivalently, into ( A \u2212 \u03bb I) x = 0\u2014and solve for x; the resulting nonzero solutons form the set of eigenvectors of A corresponding to the selectd eigenvalue. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. A matrix is diagonalizable if it has a full set of eigenvectors. The generalized eigenvalue problem is to determine the solution to the equation Av = \u03bbBv, where A and B are n-by-n matrices, v is a column vector of length n, and \u03bb is a scalar. is a linearly independent set. If W is a matrix such that W'*A = D*W', the columns of W are the left eigenvectors of A. Your matrix is Hermitian - look up \"Rayleigh quotient iteration\" to find its eigenvalues and eigenvectors. Solution: Ax = x )x = A 1x )A 1x = 1 x (Note that 6= 0 as Ais invertible implies that det(A) 6= 0). Recall as well that the eigenvectors for simple eigenvalues are linearly independent. Now we need to get the matrix into reduced echelon form. Eigenvectors of repeated eigenvalues. A = 1 u 1 u 1 T u 1 T u 1 \u2212 2 u 2 u 2 T u 2 T u 2 + 2 u 3 u 3 T u 3 T u 3. xla is an addin for Excel that contains useful functions for matrices and linear Algebra: Norm, Matrix multiplication, Similarity transformation, Determinant, Inverse, Power, Trace, Scalar Product, Vector Product, Eigenvalues and Eigenvectors of symmetric matrix with Jacobi algorithm, Jacobi's rotation matrix. This representation turns out to be enormously useful. Title: Eigenvalues and Eigenvectors of the Matrix of Permutation Counts Authors: Pawan Auorora , Shashank K Mehta (Submitted on 16 Sep 2013 ( v1 ), last revised 20 Sep 2013 (this version, v2)). The first one is a simple one - like all eigenvalues are real and different. You have 3 vector equations Au1=l1u1 Au2=l2u2 Au3=l3u3 Consider the matrix coefficients a11,a12,a13, etc as unknowns. Learn to find complex eigenvalues and eigenvectors of a matrix. Earlier on, I have also mentioned that it is possible to get the eigenvalues by solving the characteristic equation of the matrix. 369) EXAMPLE 1 Orthogonally diagonalize. The normalized eigenvector for = 5 is: The three eigenvalues and eigenvectors now can be recombined to give the solution to the original 3x3 matrix as shown in Figures 8. For a Hermitian matrix the eigenvalues should be real. If you're seeing this message, it means we're having trouble loading external resources on our website. this expression for A is called the spectral decomposition of a symmetric matrix. Find a basis for this eigenspace. This can be factored to. The calculator will perform symbolic calculations whenever it is possible. Namely, prove that (1) the determinant of A is the product of its eigenvalues, and (2) the trace of A is the sum of the eigenvalues. The eigenvalues are r1=r2=-1, and r3=2. 2 examples are given : first the eigenvalues of a 4*4 matrix is calculated. For example, suppose we wish to solve the. Repeated Eigenvalues We conclude our consideration of the linear homogeneous system with constant coefficients x Ax' (1) with a brief discussion of the case in which the matrix has a repeated eigenvalue. You have 3 vector equations Au1=l1u1 Au2=l2u2 Au3=l3u3 Consider the matrix coefficients a11,a12,a13, etc as unknowns. oregonstate. Ask Question Asked 2 years, 8 months ago. True A 3x3 matrix can have a nonreal complex eigenvalue with multiplicity 2. Browse other questions tagged linear-algebra matrices eigenvalues-eigenvectors or ask your own question. Finding eigenvectors of a 3x3 matrix 2. Learn to find complex eigenvalues and eigenvectors of a matrix. Eigenvalues and eigenvectors calculator. It decomposes matrix using LU and Cholesky decomposition. I guess A is 3x3, so it has 9 coefficients. Prove that the diagonal elements of a triangular matrix are its eigenvalues. Determining the eigenvalues of a 3x3 matrix. This is called the eigendecomposition. Equation (1) can be stated equivalently as (A \u2212 \u03bb I) v = 0 , {\\displaystyle (A-\\lambda I)v=0,} (2) where I is the n by n identity matrix and 0 is the zero vector. We all know that for any 3 \u00d7 3 matrix, the number of eigenvalues is 3. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. If you have trouble understanding your eigenvalues and eigenvectors of 3\u00d73 matrix. Matrix A: 0 -6 10-2 12 -20-1 6 -10 I got the eigenvalues of: 0, 1+i, and 1-i. We know that the row space of a matrix is orthogonal to its null space, then we can compute the eigenvector(s) of an eigenvalue by verifying the linear independence of the. 860) by computing Av/l and confirming that it equals v. net) for Bulgarian translation. real()\" to get rid of the imaginary part will give the wrong result (also, eigenvectors may have an arbitrary complex phase!). Complex eigenvalues and eigenvectors of a matrix. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. We have A= 5 2 2 5 and eigenvalues 1 = 7 2 = 3 The sum of the eigenvalues 1 + 2 = 7+3 = 10 is equal to the sum of the diagonal entries of the matrix Ais 5 + 5 = 10. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. The Mathematics Of It. Let us rearrange the eigenvalue equation to the form , where represents a vector of all zeroes (the zero vector). Eigenvalues of the said matrix [ 2. Repeated Eigenvalues Occasionally when we have repeated eigenvalues, we are still able to nd the correct number of linearly independent eigenvectors. Judging from the name covmat, I'm assuming you are feeding a covariance matrix, which is symmetric (or hermitian. 1 $\\begingroup$ My question is Eigenvalue/eigenvector reordering and/or renormalisation? 0. Eigenvalues and Eigenvectors Calculator for 3x3 Matrix easycalculation. Since the zero-vector is a solution, the system is consistent. Theorem 11. Here we can confirm the eigenvalue/eigenvector pair l=-. These straight lines may be the optimum axes for describing rotation of a. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x`. Once we have the eigenvalues we can then go back and determine the eigenvectors for each eigenvalue. 1) then v is an eigenvector of the linear transformation A and the scale factor \u03bb is the eigenvalue corresponding to that eigenvector. We all know that for any 3 \u00d7 3 matrix, the number of eigenvalues is 3. I only found 2 eigenvectors cos l2=l3. Let's consider a simple example with a diagonal matrix: A = np. Hot Network Questions Quick way to find the square root of 123. EigenValues is a special set of scalar values, associated with a linear system of matrix equations. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big test). Follow the next steps for calulating the eigenvalues (see the figures) 1: make a 4*4 matrix [A] and fill the rows and colums with the numbers. MATRIX NORMS 219 Moreover, if A is an m \u00d7 n matrix and B is an n \u00d7 m matrix, it is not hard to show that tr(AB)=tr(BA). First, form the matrix. In each case determine which vectors are eigenvectors and identify the associated eigenvalues. The Overflow Blog Defending yourself against coronavirus scams. Lambda represents a scalar value. This problem has been solved! See the answer. Lecture 7: Given The Eigenvector, Eigenvalues=? Lecture 8: Eigenvector=? Of A 3X3 Matrix; Lecture 9: Bases And Eigenvalues: 1; Lecture 10: Bases And Eigenvalues: 2; Lecture 11: Basis=? For A 2X2 Matrix; Lecture 12: Basis=? For A 3X3 Matrix: Ex. Using MatLab to find eigenvalues, eigenvectors, and unknown coefficients of initial value problem. Here A is a matrix, v is an eigenvector, and lambda is its corresponding eigenvalue. Note: The two unknowns can also be solved for using only matrix manipulations by starting with the initial conditions and re-writing: Now it is a simple task to find \u03b3 1 and \u03b3 2. Hi, I'm having trouble with finding the eigenvectors of a 3x3 matrix. In my earlier posts, I have already shown how to find out eigenvalues and the corresponding eigenvectors of a matrix. By de\ufb01nition of the kernel, that. The eigenvalue is the factor which the matrix is expanded. It's the eigenvectors that determine the dimensionality of a system. Thanks for the A2A\u2026 Eigenvalues and the Inverse of a matrix If we take the canonical definition of eigenvectors and eigenvalues for a matrix, $M$, and further assume that $M$ is invertible, so there exists, [math]M^{-1}[/math. The eigenvalues are numbers, and they\u2019ll be the same for Aand B. ) by Seymour Lipschutz and Marc. If is a diagonal matrix with the eigenvalues on the diagonal, and is a matrix with the eigenvectors as its columns, then. *XP the eigenvalues up to a 4*4 matrix can be calculated. (1) The eigenvalues of a triangle matrix are its diagonal elements. Is there a fast algorithm for this specific problem? I've seen algorithms for calculating all the eigenvectors of a real symmetric matrix, but those routines seem to be optimized for large matrices, and I don't care. Diagonal matrix. I'm trying to calculate eigenvalues and eigenvectors of a 3x3 hermitian matrix (named coh). The numerical computation of eigenvalues and eigenvectors is a challenging issue, and must be be deferred until later. The roots are lambda 1 equals 1, and lambda 2 equals 3. It decomposes matrix using LU and Cholesky decomposition. The eigenvalues of a triangular matrix are the entries on the main diagonal. I am new to Mathematica so I am not very familiar with the syntax and I can not find out what is wrong with my code. degree polynomial. square matrix and S={x1,x2,x3,\u2026,xp} S = { x 1, x 2, x 3, \u2026, x p } is a set of eigenvectors with eigenvalues \u03bb1,\u03bb2,\u03bb3,\u2026,\u03bbp. 2]; quit; This Hadamard matrix has 8 eigenvalues equal to 4 and 8 equal to -4. Then if \u03bb is a complex number and X a non\u2013zero com-plex column vector satisfying AX = \u03bbX, we call X an eigenvector of A, while \u03bb is called an eigenvalue of A. Learn to find complex eigenvalues and eigenvectors of a matrix. Multiply an eigenvector by A, and the. Eigenvalues [ m, - k] gives the k that are smallest in absolute value. Determining the eigenvalues of a 3x3 matrix Linear Algebra: Eigenvectors and Eigenspaces for a 3x3 matrix Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. , the covariance matrix of a random vector)), then all of its eigenvalues are real, and all of its eigenvectors are orthogonal. Linear Algebra: Introduction to Eigenvalues and Eigenvectors. oregonstate. A simple online EigenSpace calculator to find the space generated by the eigen vectors of a square matrix. Ask Question Asked 2 years, 8 months ago. (a) Set T: R2!R2 to be the linear transformation represented by the matrix 2 0 0 3. 1 Let A be an n \u00d7 n matrix. , the covariance matrix of a random vector)), then all of its eigenvalues are real, and all of its eigenvectors are orthogonal. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. I only found 2 eigenvectors cos l2=l3. Today Courses Practice Algebra Geometry Number Theory Calculus Probability Find the eigenvalues of the matrix A = (8 0 0 6 6 11 1 0 1). If you're seeing this message, it means we're having trouble loading external resources on our website. In the last video we set out to find the eigenvalues values of this 3 by 3 matrix, A. The non-symmetric problem of finding eigenvalues has two different formulations: finding vectors x such that Ax = \u03bbx, and finding vectors y such that y H A = \u03bby H (y H implies a complex conjugate transposition of y). 366) \u2022A is orthogonally diagonalizable, i. We take an example matrix from a Schaum's Outline Series book Linear Algebra (4 th Ed. The matrix is (I have a ; since I can't have a space between each column. There could be multiple eigenvalues and eigenvectors for a symmetric and square matrix. As the eigenvalues of are ,. The eigenvalue is the factor which the matrix is expanded. Form the matrix A \u2212 \u03bbI , that is, subtract \u03bb from each diagonal element of A. To begin, let v be a vector (shown as a point) and A be a matrix with columns a1 and a2 (shown as arrows). Note that the multiplication on the left hand side is matrix multiplication (complicated) while the mul-. 1} it is straightforward to show that if $$\\vert v\\rangle$$ is an eigenvector of $$A\\text{,}$$ then, any multiple $$N\\vert v\\rangle$$ of $$\\vert v\\rangle$$ is also an eigenvector since the (real or complex) number \\(N. The eigenvalues and eigenvectors of a matrix may be complex, even when the matrix is real. We need to get to the bottom of what the matrix A is doing to. They have many uses! A simple example is that an eigenvector does not change direction in a transformation:. Hermitian Matrix giving non-real eigenvalues. , a matrix equation) that are sometimes also known as characteristic vectors, proper vectors, or latent vectors (Marcus and Minc 1988, p. Eigenvectors are a special set of vectors associated with a linear system of equations (i. Form the matrix A \u2212 \u03bbI , that is, subtract \u03bb from each diagonal element of A. Assuming K = R would make the theory more complicated. For a square matrix A, an Eigenvector and Eigenvalue make this equation true (if we can find them):. The vector x is called an eigenvector corresponding to \u03bb. To compute the Transpose of a 3x3 Matrix, CLICK HERE. [i 1]t, for any nonzero scalar t. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. A simple online EigenSpace calculator to find the space generated by the eigen vectors of a square matrix. The calculator will perform symbolic calculations whenever it is possible. Earlier on, I have also mentioned that it is possible to get the eigenvalues by solving the characteristic equation of the matrix. Since doing so results in a determinant of a matrix with a zero column, $\\det A=0$. Let us rearrange the eigenvalue equation to the form , where represents a vector of all zeroes (the zero vector). Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. Linear Algebra: Eigenvalues of a 3x3 matrix. The eigenvectors corresponding to di erent eigenvalues need not be orthogonal. *XP the eigenvalues up to a 4*4 matrix can be calculated. Once we have the eigenvalues we can then go back and determine the eigenvectors for each eigenvalue. is a linearly independent set. But I want to use the class Jama for the calculation of the eigenvalues and eigenvectors, but I do not know how to use it, could anyone give me a hand? Thanks. If you love it, our example of the solution to eigenvalues and eigenvectors of 3\u00d73 matrix will help you get a better understanding of it. det ( A \u2212 \u03bb I) = 0. By ranking your eigenvectors in order of their eigenvalues, highest to lowest, you get the principal components in order of significance. If W is a matrix such that W'*A = D*W', the columns of W are the left eigenvectors of A. We all know that for any 3 \u00d7 3 matrix, the number of eigenvalues is 3. I'm having a problem finding the eigenvectors of a 3x3 matrix with given eigenvalues. Eigenvalues, Eigenvectors, and Diagonal-ization Math 240 Eigenvalues and Eigenvectors Diagonalization Complex eigenvalues Find all of the eigenvalues and eigenvectors of A= 2 6 3 4 : The characteristic polynomial is 2 2 +10. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. Lambda represents a scalar value. I need some help with the following problem please? Let A be a 3x3 matrix with eigenvalues -1,0,1 and corresponding eigenvectors l1l. The eigenvalues of a hermitian matrix are real, since (\u03bb \u2212 \u03bb)v = (A * \u2212 A)v = (A \u2212 A)v = 0 for a non-zero eigenvector v. Not too bad. You can use this to find out which of your. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. Calculate the eigenvalues and the corresponding eigenvectors of the matrix. Our general strategy was: Compute the characteristic polynomial. The second examples is about a 3*3 matrix. Let us consider an example of two matrices, one of them is a diagonal one, and another is similar to it: A = {{1, 0, 0}, {0, 2, 0}, {0, 0, 0. Notice: If x is an eigenvector, then tx with t6= 0 is also an eigenvector. 6 Prove that if the cofactors don't all vanish they provide a column eigenvector. How to find the Eigenvalues of a 3x3 Matrix - Duration: 3:56. array ( [ [ 1, 0 ], [ 0, -2 ]]) print (A) [ [ 1 0] [ 0 -2]] The function la. Find more Mathematics widgets in Wolfram|Alpha. You might be stuck with thrashing through an algebraic. Eigenvalues and Eigenvectors, Imaginary and Real. A simple online EigenSpace calculator to find the space generated by the eigen vectors of a square matrix. real()\" to get rid of the imaginary part will give the wrong result (also, eigenvectors may have an arbitrary complex phase!). Linear Algebra: Eigenvalues of a 3x3 matrix. The columns of V present eigenvectors of A. Our general strategy was: Compute the characteristic polynomial. Take for example 0 @ 3 1 2 3 1 6 2 2 2 1 A One can verify that the eigenvalues of this matrix are = 2;2; 4. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. Eigenvector and Eigenvalue. 224 CHAPTER 7. An eigenvector associated with \u03bb1 is a nontrivial solution~v1 to (A \u03bb1I)~v =~0: (B. The eigenvalues of A are given by the roots of the polynomial det(A In) = 0: The corresponding eigenvectors are the nonzero solutions of the linear system (A In)~x = 0: Collecting all solutions of this system, we get the corresponding eigenspace. If you have trouble understanding your eigenvalues and eigenvectors of 3\u00d73 matrix. Eigenvalues and Eigenvectors. 2, replacing ${\\bb v}^{(j)}$ by ${\\bb v}^{(j)}-\\sum_{k\\neq j} a_k {\\bb v}^{(k)}$ results in a matrix whose determinant is the same as the original matrix. Eigenvectors of repeated eigenvalues. Let A be an n\u00d7n matrix and let \u03bb1,\u2026,\u03bbn be its eigenvalues. The matrix looks like this |0 1 1| A= |1 0 1| |1 1 0| When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. If the matrix A is symmetric then \u2022its eigenvalues are all real (\u2192TH 8. Description of Lab: Your program will ask the user to enter a 3x3 matrix. Maths with Jay 35,790 views. org/math/linear-algebra/alternate_bases/eigen_everything/v/linear-. We therefore saw that they were all real. 7 Choose a random 3 by 3 matrix and find an eigenvalue and corresponding eigenvector. As is to be expected, Maple's. This condition will give you the eigenvalues and then, solvning the system for each eigenvalue, you will find the eigenstates. This matrix calculator computes determinant , inverses, rank, characteristic polynomial , eigenvalues and eigenvectors. v is an eigenvector with associated eigenvalue 3. I'm having a problem finding the eigenvectors of a 3x3 matrix with given eigenvalues. Since v is non-zero, the matrix is singular, which means that its determinant is zero. I can find the eigenvector of the eigenvalue 0, but for the complex eigenvalues, I keep on getting the reduced row echelon form of:. Find all values of a which will guarantee that A has eigenvalues 0, 3, and \u2212 3. Question: Find The Eigenvalues And Eigenvectors Of Matrices 3x3 This problem has been solved! See the answer. Observation: det (A - \u03bbI) = 0 expands into an kth degree polynomial equation in the unknown \u03bb called the characteristic equation. At this special case, all vectors as still rotated counterclockwise except those in the direction of $(0,1)$ (which is the eigenvector). This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. Find the eigenvalues and eigenvectors. 3 4 4 8 Solution. In Section 5. Scaling your VPN overnight Finding Eigenvectors of a 3x3 Matrix (7. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. 1 3 4 5 , l = 1 11. The vector x is called an eigenvector corresponding to \u03bb. Eigenvalues and Eigenvectors. Exactly one option must be correct). find the eigenvalues and eigenvectors of matrices 3x3. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. The normalized eigenvector for = 5 is: The three eigenvalues and eigenvectors now can be recombined to give the solution to the original 3x3 matrix as shown in Figures 8. Decomposing a matrix in terms of its eigenvalues and its eigenvectors gives valuable insights into the properties of the matrix. Equation (1) can be stated equivalently as (A \u2212 \u03bb I) v = 0 , {\\displaystyle (A-\\lambda I)v=0,} (2) where I is the n by n identity matrix and 0 is the zero vector. They are used in a variety of data science techniques such as Principal Component Analysis for dimensionality reduction of features. The result is a 3x1 (column) vector. In this tutorial, we will explore NumPy's numpy. 1) can be rewritten. - Jonas Aug 16 '11 at 3:12. Given a matrix A, recall that an eigenvalue of A is a number \u03bb such that Av = \u03bb v for some vector v. If A is real, there is an orthonormal basis for R n consisting of eigenvectors of A if and only if A is symmetric. This polynomial is called the characteristic polynomial. Question: How do you determine eigenvalues of a 3x3 matrix? Eigenvalues: An eigenvalue is a scalar {eq}\\lambda {/eq} such that Ax = {eq}\\lambda {/eq}x for a nontrivial x. In linear algebra the characteristic vector of a square matrix is a vector which does not change its direction under the associated linear transformation. The zero vector 0 is never an eigenvectors, by de\ufb01nition. In this python tutorial, we will write a code in Python on how to compute eigenvalues and vectors. Let A be the matrix given by A = [\u2212 2 0 1 \u2212 5 3 a 4 \u2212 2 \u2212 1] for some variable a. trace()/3) -- note that (in exact math) this shifts the eigenvalues but does not influence the eigenvectors. It will then compute the eigenvalues (real and complex) and eigenvectors (real and complex) for that matrix. Determining the eigenvalues of a 3x3 matrix Linear Algebra: Eigenvectors and Eigenspaces for a 3x3 matrix Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. For the following matrices find (a) all eigenvalues (b) linearly independent eigenvectors for each eigenvalue (c) the algebraic and geometric multiplicity for each eigenvalue and state whether the matrix is diagonalizable. Enter a matrix. De\ufb01nition 4. Diagonalizable Matrices. Quiz % 1)Simplify 2) Similarly, the characteristic equation of a 3x3 matrix: Eigenvalues or, can be written as well as Find eigenvalues and eigenvectors of matrix. Maths with Jay 35,790 views. Find more Mathematics widgets in Wolfram|Alpha. 2 Vectors that maintain their orientation when multiplied by matrix A D Eigenvalues: numbers (\u03bb) that provide solutions for AX = \u03bbX. net) for Bulgarian translation. I can find the eigenvector of the eigenvalue 0, but for the complex eigenvalues, I keep on getting the reduced row echelon form of:.\n1v7l3vc886zlg2, m83gip4j73vjfp, o5m33ui5xk4r4g, d76u869uc05fv, gbo71a9euqdvt, du2sldoup68h, 8wu1rxtgoc7tng0, o7uvrhbdlti2, ay9wjkki4za5y, olesufgei95yk, bs98c7ai3z00, 8tzdyariin4c7h, t1kt2p3udiq, s9o8ylnuu7ft83, ks0tprhj6stqjd, q1dh0lb81aofsm, ukci9e944q, ajxzbl7lregmo1, n5h8l6ozp41yg, 7pb60d55luftbpv, i10r3fzynjgw, l0gavghjyl1b6t, 74ssaxe7i8vfqyx, xlux7x1ozv, smnjnrqshi, 5pnzldwnn2ig, jernunvmrjcuv, 1emb6psqm3, 3yhxk3w0palopu3", "date": "2020-05-27 22:26:15", "meta": {"domain": "svc2006.it", "url": "http://svc2006.it/pmey/eigenvalues-and-eigenvectors-of-3x3-matrix.html", "openwebmath_score": 0.8708940148353577, "openwebmath_perplexity": 349.6453176432852, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9884918502576513, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8570577824463963}}
{"url": "https://questioncove.com/updates/4e56f2e10b8b9ebaa8948ee6", "text": "Mathematics\nOpenStudy (anonymous):\n\nProve that: $if:\\mathop {\\lim }\\limits_{n \\to \\infty } {a_n} = a$ then $\\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{{a_1} + 2{a_2} + ... + n{a_n}}} {{{n^2}}} = \\frac{a} {2}$\n\nOpenStudy (anonymous):\n\nCan I use L'hospital Rule Like that: $\\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{\\sum\\nolimits_{k = 1}^n {k{a_k}} }} {{{n^2}}} = \\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{\\sum\\nolimits_{k = 1}^n {{a_k}} }} {{2n}} = \\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{an}} {{2n}} = \\frac{a} {2}$ Is it a rigorous way to prove that? Please tell me your thought. It will be a appreciated.\n\nOpenStudy (anonymous):\n\ni am thinking. maybe can prove it directly?\n\nmyininaya (myininaya):\n\nwe can use l'hospital\n\nOpenStudy (anonymous):\n\nI don't know how to do that, I want to use the precise definiition of series limit.\n\nOpenStudy (anonymous):\n\nyou need epsilon, N proof? because it looks like you should get $a\\sum_{k=1}^nk=\\frac{a(n(n+1)}{2}$\n\nOpenStudy (anonymous):\n\nOpenStudy (anonymous):\n\nIn fact, it is a question from mathematical analysis. So yes, epsilon, N proof will be best.\n\nmyininaya (myininaya):\n\n$\\lim_{n \\rightarrow \\infty}\\frac{a_1+a_2+\\cdot \\cdot \\cdot n a_n}{n^2} \\cdot \\frac{\\frac{1}{n^2}}{\\frac{1}{n^2}}$$=\\lim_{n \\rightarrow \\infty}\\frac{\\frac{a_1}{n^2}+\\frac{a_2}{n^2}+ \\cdot \\cdot \\cdot +\\frac{n a_n}{n^2}}{\\frac{n^2}{n^2}}$ $=\\lim_{n \\rightarrow \\infty}\\frac{0+0+\\cdot \\cdot \\cdot +\\frac{1}{n}a_n}{1}=\\lim_{n \\rightarrow \\infty}\\frac{a_n}{n}=0$ ? how can we show its $\\frac{a}{2}$\n\nOpenStudy (anonymous):\n\nThe answer from textbook is a/2. And I saw a guy use l'hospital rule in that way, so i just copy his way....I hope It work. But The way use l'hospital rule like that is really confused... AND If you can give the \" epsilon-N \" proof, that will be best...\n\nOpenStudy (anonymous):\n\nTake your time. I have go to work now. See you.\n\nmyininaya (myininaya):\n\ni'm saying i can't prove that because it doesn't = a/2 i get 0 see above\n\nmyininaya (myininaya):\n\nand the way i used above wasn't l'hospital\n\nOpenStudy (zarkon):\n\nit is a/2\n\nmyininaya (myininaya):\n\nwhat did i do wrong then\n\nOpenStudy (zarkon):\n\nyou caont do what you did above\n\nmyininaya (myininaya):\n\nmyininaya (myininaya):\n\ni see\n\nmyininaya (myininaya):\n\nits because i'm forgeting the terms before na_n\n\nOpenStudy (zarkon):\n\nyes\n\nmyininaya (myininaya):\n\nlike (n-1)a_{n-1}\n\nOpenStudy (zarkon):\n\nyes\n\nmyininaya (myininaya):\n\nand so on...\n\nOpenStudy (zarkon):\n\nyes\n\nmyininaya (myininaya):\n\nlol\n\nmyininaya (myininaya):\n\ni will let you prove it\n\nmyininaya (myininaya):\n\nbecause i know you want to badly\n\nmyininaya (myininaya):\n\nlol\n\nOpenStudy (anonymous):\n\nwhy isn't it $\\lim_{n\\rightarrow \\infty}\\frac{an(n+1)}{2n^2}=\\frac{a}{2}$?\n\nOpenStudy (zarkon):\n\nhow do you factor out an a?\n\nOpenStudy (zarkon):\n\nthat is what it looks like you are doing...replacing a_n with a\n\nOpenStudy (zarkon):\n\nthe L'Hospitals rule use by prost in his first post is incorrect too\n\nmyininaya (myininaya):\n\nwheres that?\n\nOpenStudy (zarkon):\n\nthis ... $\\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{\\sum\\nolimits_{k = 1}^n {k{a_k}} }} {{{n^2}}} = \\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{\\sum\\nolimits_{k = 1}^n {{a_k}} }} {{2n}} = \\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{an}} {{2n}} = \\frac{a} {2}$\n\nOpenStudy (zarkon):\n\nan epsilon based proof is the way to go I believe\n\nOpenStudy (anonymous):\n\nI am back.. Hard Day.. Does anyone heard Stolz Theorem? Use it, it said: $\\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{{a_1} + 2{a_2} + ... + n{a_n}}} {{{n^2}}} = \\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{n{a_n}}} {{{n^2} - {{\\left( {n - 1} \\right)}^2}}} = \\mathop {\\lim }\\limits_{n \\to \\infty } \\frac{{an}} {{2n - 1}} = \\frac{a} {2}$ and foreget my first proof above. It makes people confused and make no sense. The Stolz Theorem is very useful. But i still want the \" epsilon-N \" proof. Can anyone prove it?", "date": "2022-08-10 04:11:47", "meta": {"domain": "questioncove.com", "url": "https://questioncove.com/updates/4e56f2e10b8b9ebaa8948ee6", "openwebmath_score": 0.9999911785125732, "openwebmath_perplexity": 2256.058497394396, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918519527645, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8570577788217987}}
{"url": "https://math.stackexchange.com/questions/3749429/how-many-ways-to-of-place-1-2-3-dots-9-in-a-circle-so-the-sum-of-any-thre/3749438", "text": "# How many ways to of place $1, 2, 3, \\dots, 9$ in a circle so the sum of any three consecutive numbers is divisible by $3.$\n\nDetermine the number of ways of placing the numbers $$1, 2, 3, \\dots, 9$$ in a circle, so that the sum of any three numbers in consecutive positions is divisible by $$3.$$ (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)\n\nI've experimented with possible combinations and found that it works when we put a multiple of 3 next to a number one more than a multiple of three beside a number that is two more than a multiple of 3. If we continue with this pattern around the circle, it works.\n\nHowever, I'm curious in finding a more systematic approach than listing out all different combinations.\n\n\u2022 Well consider the numbers in position $k$ and $k+3$ will have to be congruent $\\mod 3$.. \u2013\u00a0fleablood Jul 8 at 5:08\n\nIn general, suppose we have the numbers $$1,2,\\dots,3n$$, and we would like to place them in a circle so that the sum of any three consecutive terms is divisible by $$3$$.\n\nObserve that the numbers at positions $$k$$ and $$k+3$$ must always be congruent modulo $$3$$. Thus we can partition the points along the circle into three sets which stand for the residues modulo $$3$$ of the positions. If we fix the number $$1$$ at, say, position $$1$$, then this tells us that every point at position $$3k+1$$ has residue $$1$$ modulo $$3$$.\n\nNow we have a choice: either the numbers at positions $$3k+2$$ have residue $$0$$, or they have residue $$2$$. Either way, note that each of the three \"partition classes\" can be arranged in $$n!$$ different ways, giving us $$2(n!)^3$$ possibilities.\n\nIn this particular case, $$n=3$$, and the answer is $$432$$ (if rotations are counted as the same).\n\n\u2022 If rotations are the same we can assume position $1$ has $3$. Then positions $4$ and $9$ have $6$ and $9$. there are $2$ ways to do that. position $2$ is either $\\pm 1\\pmod 3$. There are $2$ choices. position $2\\equiv position 5\\equiv position 8$ so there are $3!$ ways to do that. And $position 3\\equiv position 6\\equiv \\position 9$ so there are $3!$ ways to do that. So there are $2*2*6*6=144$ ways if rotations are counted the same. If reflections are counted the same there are $2*6*6=72$ ways. \u2013\u00a0fleablood Jul 8 at 5:37\n\nBrain storming. If we label the number positions as $$a_1,.....,a_9$$ then $$a_k+a_{k+1} + a_{k+2} \\equiv 0 \\equiv a_{k+1} + a_{k+2} + a_{k+3}\\pmod 3$$ so $$a_k\\equiv a_{k+3}\\pmod 3$$.\n\nThe there are only three equivalence classes each with $$3$$ elements and so $$a_3, a_6, a_9$$ must all contain elements from one equivalence class. There are $$3$$ choices of which class and $$3!$$ ways to place the elements. $$a_1, a_4, a_7$$ must also contain elements from one equivalence class and there are $$2$$ choices of classes and $$3!$$ ways to arrange them. and for $$a_2, a_5, a_8$$ there is one choice of classe and $$3!$$ ways to arrange them.\n\nSo there $$3*3!*2*3!*1*3! = 6^4$$ ways to do this.\n\nAs rotations are considered the same (but not mirror symmetries???) divide by $$9$$.\n\nSo the answer is $$\\frac {6^4}9$$.\n\n\u2022 Rotations are the same, so I think this overcounts by a factor of 3 \u2013\u00a0boink Jul 8 at 5:18\n\u2022 Since we're placing them in a circle, it's possible we ought to divide by 9 or maybe 18 to get rid of the symmetric options (equivalently, require $a_1=1$, and possibly divide by 2). It's hard to tell, though. \u2013\u00a0Arthur Jul 8 at 5:19\n\u2022 @fleablood I just meant that the problem statement says they're the same \u2013\u00a0boink Jul 8 at 5:22\n\u2022 \"Since we're placing them in a circle\" Placing the them in a circle means that $a_1$ is congruent to $a_8$. It DOESN\"T mean that rotations are considered to be the same any more than Mr. Left in a word problem means you need to subtract. But if rotations are the same then divide by $9$. If symmetry is considered the same divide by $18$. \u2013\u00a0fleablood Jul 8 at 5:22\n\u2022 \"I just meant that the problem statement says they're the same\" Oh, I didn't see that part. That's one of my pet peeve. If Mr. Left works $8$ hours a day, $5$ days a week how many hours a week does he work. Answer: Well, since the problem contains the word \"left\" that means we subtract so the answer is $8-5=3$. And question. How many ways are there to place people are a table. Answer: As the problem contains the word \"table\" and tables are circles rotations are the same.... No, they aren't unless the question says they are. \u2013\u00a0fleablood Jul 8 at 5:28", "date": "2020-11-24 04:30:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3749429/how-many-ways-to-of-place-1-2-3-dots-9-in-a-circle-so-the-sum-of-any-thre/3749438", "openwebmath_score": 0.8086175918579102, "openwebmath_perplexity": 151.29742150008124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918516137419, "lm_q2_score": 0.867035752930664, "lm_q1q2_score": 0.8570577768297469}}
{"url": "https://math.stackexchange.com/questions/2059752/mod-distributive-law-factoring-bmod-ab-bmod-ac-ab-bmod-c", "text": "# mod Distributive Law, factoring $\\!\\!\\bmod\\!\\!:$ $\\ ab\\bmod ac = a(b\\bmod c)$\n\nI stumbled across this problem\n\nFind $$\\,10^{\\large 5^{102}}$$ modulo $$35$$, i.e. the remainder left after it is divided by $$35$$\n\nBeginning, we try to find a simplification for $$10$$ to get: $$10 \\equiv 3 \\text{ mod } 7\\\\ 10^2 \\equiv 2 \\text{ mod } 7 \\\\ 10^3 \\equiv 6 \\text{ mod } 7$$\n\nAs these problems are meant to be done without a calculator, calculating this further is cumbersome. The solution, however, states that since $$35 = 5 \\cdot 7$$, then we only need to find $$10^{5^{102}} \\text{ mod } 7$$. I can see (not immediately) the logic behind this. Basically, since $$10^k$$ is always divisible by $$5$$ for any sensical $$k$$, then: $$10^k - r = 5(7)k$$ But then it's not immediately obvious how/why the fact that $$5$$ divides $$10^k$$ helps in this case.\n\nMy question is, is in general, if we have some mod system with $$a^k \\equiv r \\text{ mod } m$$ where $$m$$ can be decomposed into a product of numbers $$a \\times b \\times c \\ \\times ...$$, we only need to find the mod of those numbers where $$a, b, c.....$$ doesn't divides $$a$$? (And if this is the case why?) If this is not the case, then why/how is the solution justified in this specific instance?\n\n\u2022 How is $10 \\equiv 1$ mod 7? \u2013\u00a0Junglemath May 2 '19 at 13:13\n\u2022 @Junglemath, idk.... \u2013\u00a0q.Then May 14 '20 at 7:17\n\nThe \"logic\" is that we can use a mod distributive law to pull out a common factor $$\\,c=5,\\,$$ i.e.\n\n$$ca\\bmod cn =\\, c(a\\bmod n)\\quad\\qquad$$\n\nThis decreases the modulus from $$\\,cn\\,$$ to $$\\,n, \\,$$ simplifying modular arithmetic. Also it may eliminate CRT = Chinese Remainder Theorem calculations, eliminating needless inverse computations, which are much more difficult than above for large numbers (or polynomials, e.g. see this answer).\n\nThis distributive law is often more convenient in congruence form, e.g.\n\n$$\\quad \\qquad ca\\equiv c(a\\bmod n)\\ \\ \\ {\\rm if}\\ \\ \\ \\color{#d0f}{cn\\equiv 0}\\ \\pmod{\\! m}$$\n\nbecause we have: $$\\,\\ c(a\\bmod n) \\equiv c(a\\! +\\! kn)\\equiv ca+k(\\color{#d0f}{cn})\\equiv ca\\pmod{\\!m}$$\n\ne.g. in the OP: $$\\ \\ I\\ge 1\\,\\Rightarrow\\, 10^{\\large I+N}\\!\\equiv 10^{\\large I}(10^{\\large N}\\!\\bmod 7)\\ \\ \\ {\\rm by}\\ \\ \\ 10^I 7\\equiv 0\\,\\pmod{35}$$\n\nLet's use that. First note that exponents on $$10$$ can be reduced mod $$\\,6\\,$$ by little Fermat,\n\ni.e. notice that $$\\ \\color{#c00}{{\\rm mod}\\,\\ 7}\\!:\\,\\ 10^{\\large 6}\\equiv\\, 1\\,\\Rightarrow\\, \\color{#c00}{10^{\\large 6J}\\equiv 1}.\\$$ Thus if $$\\ I \\ge 1\\$$ then as above\n\n$$\\phantom{{\\rm mod}\\,\\ 35\\!:\\,\\ }\\color{#0a0}{10^{\\large I+6J}}\\!\\equiv 10^{\\large I} 10^{\\large 6J}\\!\\equiv 10^{\\large I}(\\color{#c00}{10^{\\large 6J}\\!\\bmod 7})\\equiv \\color{#0a0}{10^{\\large I}}\\,\\pmod{\\!35}$$\n\nOur power $$\\ 5^{\\large 102} = 1\\!+\\!6J\\$$ by $$\\ {\\rm mod}\\,\\ 6\\!:\\,\\ 5^{\\large 102}\\!\\equiv (-1)^{\\large 102}\\!\\equiv 1$$\n\nTherefore $$\\ 10^{\\large 5^{\\large 102}}\\!\\! = \\color{#0a0}{10^{\\large 1+6J}}\\!\\equiv \\color{#0a0}{10^{\\large 1}} \\pmod{\\!35}\\$$\n\nRemark $$\\$$ For many more worked examples see the complete list of linked questions. Often this distributive law isn't invoked by name. Rather its trivial proof is repeated inline, e.g. from a recent answer, using $$\\,cn = 14^2\\cdot\\color{#c00}{25}\\equiv 0\\pmod{100}$$\n\n\\begin{align}&\\color{#c00}{{\\rm mod}\\ \\ 25}\\!:\\ \\ \\ 14\\equiv 8^{\\large 2}\\Rightarrow\\, 14^{\\large 10}\\equiv \\overbrace{8^{\\large 20}\\equiv 1}^{\\rm\\large Euler\\ \\phi}\\,\\Rightarrow\\, \\color{#0a0}{14^{\\large 10N}}\\equiv\\color{#c00}{\\bf 1}\\\\[1em] &{\\rm mod}\\ 100\\!:\\,\\ 14^{\\large 2+10N}\\equiv 14^{\\large 2}\\, \\color{#0a0}{14^{\\large 10N}}\\! \\equiv 14^{\\large 2}\\!\\! \\underbrace{(\\color{#c00}{{\\bf 1} + 25k})}_{\\large\\color{#0a0}{14^{\\Large 10N}}\\!\\bmod{\\color{#c00}{25}}}\\!\\!\\! \\equiv 14^{\\large 2} \\equiv\\, 96\\end{align}\n\nThis distributive law is actually equivalent to CRT as we sketch below, with $$\\,m,n\\,$$ coprime\n\n\\begin{align} x&\\equiv a\\!\\!\\!\\pmod{\\! m}\\\\ \\color{#c00}x&\\equiv\\color{#c00} b\\!\\!\\!\\pmod{\\! n}\\end{align} $$\\,\\Rightarrow\\, x\\!-\\!a\\bmod mn\\, =\\, m\\left[\\dfrac{\\color{#c00}x-a}m\\bmod n\\right] = m\\left[\\dfrac{\\color{#c00}b-a}m\\bmod n\\right]$$\n\nwhich is exactly the same form solution given by Easy CRT. But the operational form of this law often makes it much more convenient to apply in computations versus the classical CRT formula.\n\nFractional extension  It easily extends to fractions, e.g. from here\n\nNotice $$\\,\\ \\dfrac{\\color{#c00}{11}}{35}\\bmod \\color{#c00}{11}(9)\\,=\\, \\color{#c00}{11}(\\color{#0a0}8)\\,$$ by $$\\color{#0a0}{\\bmod 9\\!:\\ \\dfrac{1}{35}\\equiv \\dfrac{1}{-1}\\equiv 8},\\$$ via\n\nTheorem $$\\ \\ \\dfrac{\\color{#c00}ab}d\\bmod \\color{#c00}ac\\, =\\, \\color{#c00}a\\left(\\color{#0a0}{\\dfrac{b}d\\bmod c}\\right)\\ \\$$ if $$\\ \\ (d,ac) = 1$$\n\nProof $$\\,$$ Bezout $$\\Rightarrow$$ exists $$\\, d' \\equiv d^{-1}\\pmod{\\!ac}.\\,$$ Factoring out $$\\,\\color{#c00}a\\,$$ by mDL\n\n$$\\color{#c00}abd'\\bmod \\color{#c00}ac\\, =\\ \\color{#c00}a(bd'\\bmod c)\\qquad\\qquad\\qquad$$\n\nand $$\\,dd' \\equiv 1\\pmod{\\!ac}\\Rightarrow dd' \\equiv 1\\pmod{\\!c},\\,$$ so $$\\,d'\\bmod c = d^{-1}\\bmod c$$\n\nFirst, note that $10^{7}\\equiv10^{1}\\pmod{35}$.\n\nTherefore $n>6\\implies10^{n}\\equiv10^{n-6}\\pmod{35}$.\n\nLet's calculate $5^{102}\\bmod6$ using Euler's theorem:\n\n\u2022 $\\gcd(5,6)=1$\n\u2022 Therefore $5^{\\phi(6)}\\equiv1\\pmod{6}$\n\u2022 $\\phi(6)=\\phi(2\\cdot3)=(2-1)\\cdot(3-1)=2$\n\u2022 Therefore $\\color\\red{5^{2}}\\equiv\\color\\red{1}\\pmod{6}$\n\u2022 Therefore $5^{102}\\equiv5^{2\\cdot51}\\equiv(\\color\\red{5^{2}})^{51}\\equiv\\color\\red{1}^{51}\\equiv1\\pmod{6}$\n\nTherefore $10^{5^{102}}\\equiv10^{5^{102}-6}\\equiv10^{5^{102}-12}\\equiv10^{5^{102}-18}\\equiv\\ldots\\equiv10^{1}\\equiv10\\pmod{35}$.\n\nCarrying on from your calculation: \\begin{align} 10^3&\\equiv 6 \\bmod 7 \\\\ &\\equiv -1 \\bmod 7 \\\\ \\implies 10^6 = (10^3)^2&\\equiv 1 \\bmod 7 \\end{align} We could reach the same conclusion more quickly by observing that $7$ is prime so by Fermat's Little Theorem, $10^{(7-1)}\\equiv 1 \\bmod 7$.\n\nSo we need to know the value of $5^{102}\\bmod 6$, and here again $5\\equiv -1 \\bmod 6$ so $5^{\\text{even}}\\equiv 1 \\bmod 6$. (Again there are other ways to the same conclusion, but spotting $-1$ is often useful).\n\nThus $10^{\\large 5^{102}}\\equiv 10^{6k+1}\\equiv 10^1\\equiv 3 \\bmod 7$.\n\nNow the final step uses the Chinese remainder theorem for uniqueness of the solution (to congruence): \\left .\\begin{align} x&\\equiv 0 \\bmod 5 \\\\ x&\\equiv 3 \\bmod 7 \\\\ \\end{align} \\right\\}\\implies x\\equiv 10 \\bmod 35", "date": "2021-08-02 05:17:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2059752/mod-distributive-law-factoring-bmod-ab-bmod-ac-ab-bmod-c", "openwebmath_score": 0.996309220790863, "openwebmath_perplexity": 946.4025963228933, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211561049158, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8570108790929958}}
{"url": "https://math.stackexchange.com/questions/3221999/probability-of-failure-of-a-light-bulb-in-years", "text": "# Probability of failure of a light bulb in years\n\nLet's assume we have a lightbulb with a maximum lifespan 4 years. We are asked to create a transition matrix (Markov chain theory) for the bulb. The bulb is checked once a year and if it's found that the bulb does not work it is replaced by a new one.\n\nWe know that the probabilities of failure during the 4 year period are: 0.2, 0.4, 0.3 and 0.1. After four years, the lightbulb is replaced with probability 1. So we have the following 4 states:\n\n\u2022 $$S_0$$ - the lightbulb is new\n\u2022 $$S_1$$ - the lightbulb is 1 year old\n\u2022 $$S_2$$ - the lightbulb is 2 years old\n\u2022 $$S_3$$ - the lightbulb is 3 years old\n\nHow to create the probability transition matrix? For the first year, it's clear. The bulb goes dead with probability $$p_{0,0} = 0.2$$ and it keeps working with probability $$p_{0,1} = 0.8$$. But I am not sure how to calculate to the following years. In the materials for my course, I found the following calculation:\n\n$$p_{2,1} = \\frac{0.4}{0.8} \\, , p_{2,3} = \\frac{0.3+0.1}{0.8} = 0.5 \\, , p_{3,1} = \\frac{0.3}{0.4} = 0.75 \\, , p_{3,4} = \\frac{0.1}{0.4} = 0.25$$\n\nSo the probability transition matrix is:\n\n$$\\begin{bmatrix} 0.2&0.8&0&0\\\\0.5&0&0.5&0\\\\0.75&0&0&0.25\\\\1&0&0&0\\end{bmatrix}$$\n\nIs this correct? I fail to see why $$p_{2,3}$$ uses the probability of failure in the last year.\n\n\u2022 The matrix is right. It's Bayes' theorem. \u2013\u00a0Oolong milk tea May 11 '19 at 11:50\n\u2022 Does the second row represent the transition probabilities from $S_1$ to $S_0$, $S_1$,..,$S_3$, respectively? \u2013\u00a0John Douma May 11 '19 at 12:01\n\u2022 @JohnDouma Yes, it does. \u2013\u00a0Ji\u0159\u00ed Pe\u0161\u00edk May 11 '19 at 12:04\n\u2022 Then shouldn't those $0.5$s be replaced by $0.4$ and $0.6$? \u2013\u00a0John Douma May 11 '19 at 12:05\n\u2022 @JohnDouma The numbers are results of $p_{2,1} = 0.5$ and $p_{2,3} = 0.5$ given above the matrix. The point is I am not sure if they're right or not. \u2013\u00a0Ji\u0159\u00ed Pe\u0161\u00edk May 11 '19 at 12:08\n\nThe probabilities known, summing to $$1$$, are the probability at birth of failing during the 1st, 2nd, 3d or 4th year. Upon failure, the bulb is replaced with a new one, which thus has the same probabilities as above.\nWe have therefore the following scheme.\n\nSo the probability (at birth) $$P_2$$ to fail in the 2nd year (not before, and not after) will be given by the probability to survive for the first year times the probability $$p_2$$ to fail exactly in the 2nd year (given that it survived the first). And analogously for the others, i.e. \\eqalign{ & p_{\\,1} = 0.2 \\cr & \\left( {1 - p_{\\,1} } \\right)p_{\\,2} = 0.8 \\cdot p_{\\,2} = P_{\\,2} = 0.4\\quad \\Rightarrow \\quad p_{\\,2} = 0.5 \\cr & \\left( {1 - p_{\\,1} } \\right)\\left( {1 - p_{\\,2} } \\right)p_{\\,3} = P_{\\,3} \\quad \\quad \\Rightarrow \\quad p_{\\,3} = 0.75 \\cr & \\left( {1 - p_{\\,1} } \\right)\\left( {1 - p_{\\,2} } \\right)\\left( {1 - p_{\\,3} } \\right)p_{\\,4} = P_{\\,4} \\quad \\Rightarrow \\quad p_{\\,4} = 1 \\cr}\n\nAnd $$p_k,(1-p_k)$$ are the entries of the matrix.\n\nThe key lies in interpreting the phrase \u201cthe probabilities of failure during the 4 year period\u201d (was the original in English?). It\u2019s pretty unlikely that these numbers represent the transition probabilities that you\u2019re trying to construct. It would be a rather miraculous light bulb that gets more reliable the longer it\u2019s in service. The numbers sum to $$1$$ and we know that the bulb only lasts four years maximum, so the likelier reading is that these numbers are the probabilities that when the bulb fails, it is in its $$k$$th year of service. That is, these four numbers are the probabilities that is it year $$k$$ or service given that the bulb has been found to have failed.\n\nThis reading is borne out by the computations in the course materials, which are applications of Bayes\u2019 theorem (as pointed out by Oolong milk tea). For the transition matrix, you need the probabilities that the bulb is found to have failed given that it is year $$k$$ of service, which is just the sort of thing that Bayes\u2019 theorem allows you to compute from the given data.\n\nSo, for example, the denominator of $$p_{21}=0.8$$ is the probability that the light bulb is one year old, i.e., the probability that it didn\u2019t fail in its first year of service, which is simply $$p_{12}=0.8$$. The numerator is the probability of the bulb\u2019s being a year old given that it failed the test, which is $$0.4$$ from the given probabilities, multipled by $$1$$ since it did fail the test. The computation of $$p_{23}$$ is a bit odd. It looks like the authors used $$\\Pr(\\text{test fails in third year} \\cup \\text{test fails in fourth year}) = 0.3+0.1$$ for the probability that the bulb passes when tested during the second year. One could\u2019ve simply set $$p_{23}=1-p_{21}$$ since there are only two possible transitions from $$S_1$$. Fortunately, the two values agree.\n\n\u2022 The original article was not in English, however, it was described very briefly. Your interpretation makes sense to me. \u2013\u00a0Ji\u0159\u00ed Pe\u0161\u00edk May 12 '19 at 20:22", "date": "2020-01-21 14:47:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3221999/probability-of-failure-of-a-light-bulb-in-years", "openwebmath_score": 0.9645945429801941, "openwebmath_perplexity": 506.7833399709539, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211604938803, "lm_q2_score": 0.8791467706759583, "lm_q1q2_score": 0.8570108752347849}}
{"url": "https://math.stackexchange.com/questions/1191651/solving-systems-of-linear-differential-equations-by-elimination", "text": "# Solving Systems of Linear Differential Equations by Elimination\n\nFor a homework problem, we are provided:\n\n$\\frac{dx}{dt}=-y + t$\n\n$\\frac{dy}{dt}=x-t$\n\nPutting these into differential operator notation and separating the dependent variables from the independent:\n\n$Dx-y=t$\n\n$Dy-x=-t$\n\nMy first inclination is to apply the D operator to the second equation to eliminate Dx and get:\n\n$D^2y+y=t-1$\n\nI solve the homogenous part and end up with $y_c=C_1\\cos(t) + C_2\\sin(t).$\n\nUsing annihilator approach and method of undetermined coefficients, I determine that $y_p=t-1$.\n\nGeneral solution for $y(t) = C_1\\cos(t)+C_2\\sin(t)+t-1$.\n\nAfter plugging $y$ into the second equation, I get $x(t)=-C_1\\sin(t)+C_2\\cos(t)+1+t$\n\nChecking my answer against the back of the book, they show: $x(t) = C_1\\cos(t)+C_2\\sin(t)+t+1$ and $y(t)=C_1\\sin(t)-C_2\\cos(t)+t-1$\n\nI can't seem to find what I did wrong. Chegg solutions shows to eliminate y instead of x, and got the book's solution. Does the variable chosen for elimination matter? Halp!\n\n\u2022 Oops, you're correct. I interchanged the x(t) and y(t) from the back of the book. Will fix now. \u2013\u00a0Irongrave Mar 16 '15 at 0:37\n\u2022 Both your solution and the book solution are correct and coincide up to renaming resp. replacing the constants. \u2013\u00a0Dr. Lutz Lehmann Mar 16 '15 at 3:16\n\u2022 Look here. \u2013\u00a0Kw08 Jan 22 '17 at 1:32\n\nAre you sure the original system is written as it is in the book? (Problem was updated to correct $x(t), y(t))$.\n\nI get $$x'' +x = t + 1 \\implies x(t) = c_1 \\cos t + c_2 \\sin t + t + 1$$ and $$y'' + y = t - 1 \\implies y(t) = c_1 \\sin t + c_2 \\cos t + t - 1.$$\n\nNote: this can be written as $y(t) = c_1 \\sin t - c_2 \\cos t + t - 1$ because $c_2$ is an arbitrary constant. Showing the negative removes the confusion when plugging back in so the authors decided to show it as part of the solution.\n\nYou can easily verify this solution by plugging it back into the original system.\n\nUpdate Lets do the first in more detail. We have:\n\n$$x'' +x = t + 1$$\n\nTo solve the homogeneous, we have $m^2 + 1 = 0 \\implies m_{1,2} = \\pm ~ i$, yielding:\n\n$$x_h(t) = c_1 \\cos t + c_2 \\sin t$$\n\nFor the particular, we can choose $x_p = a + b t$, and substituting back into the DEQ, yields:\n\n$$x'' + x = a + bt = 1 + t \\implies a = b = 1$$\n\nThis produces:\n\n$$x(t) = x_h(t) + x_p(t) = c_1 \\cos t + c_2 \\sin t + t + 1$$\n\n\u2022 Thanks, I just fixed that. I am still having difficulty understanding where I went wrong. \u2013\u00a0Irongrave Mar 16 '15 at 0:39\n\u2022 I will add details on the $x(t)$ calculation. \u2013\u00a0Amzoti Mar 16 '15 at 0:41\n\u2022 Thanks. I get that that works if you choose to eliminate y, but why doesn't eliminating x work the same way? \u2013\u00a0Irongrave Mar 16 '15 at 0:46\n\u2022 It does, in my solution I absorb the negative they show into the constant. Clear? \u2013\u00a0Amzoti Mar 16 '15 at 0:47\n\u2022 @Irongrave: Of course do the signs and the order of the constants matter, since the two solution components are connected by them. However, it is quite permissible to do a linear transformation of the constants by replacing $C_1=D_2$ and $C_2=-D_1$ to go from the book solution with constants $(C_1,C_2)$ to your solution form with renamed constants $(D_1,D_2)$. \u2013\u00a0Dr. Lutz Lehmann Mar 16 '15 at 3:14\n\nYou could of course also see the complex differential equation $$\\dot z=i\u00b7z+(1-i)\u00b7t$$ in this system.\n\nIts homogeneous solution is $z=c\u00b7e^{it}$. The inhomogeneous solution can be constructed as usual per linear ansatz $z_p=a+b\u00b7t$ leading to $$b=i\u00b7a+i\u00b7b\u00b7t+(1-i)\u00b7t$$ and thus $b=1+i$ and $a=1-i$ and the full solution $$z=c\u00b7e^{it}+(1-i)\u00b7(1+i\u00b7t)$$ Separating into real and imaginary part gives the solution of the original system.", "date": "2019-12-15 10:55:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1191651/solving-systems-of-linear-differential-equations-by-elimination", "openwebmath_score": 0.9066839218139648, "openwebmath_perplexity": 388.29875462359126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211604938802, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8570108706047319}}
{"url": "https://math.stackexchange.com/questions/3707227/strategy-to-calculate-fracddx-left-fracx2-6x-92x2x32-right/3707238", "text": "# Strategy to calculate $\\frac{d}{dx} \\left(\\frac{x^2-6x-9}{2x^2(x+3)^2}\\right)$.\n\nI am asked to calculate the following: $$\\frac{d}{dx} \\left(\\frac{x^2-6x-9}{2x^2(x+3)^2}\\right).$$ I simplify this a little bit, by moving the constant multiplicator out of the derivative: $$\\left(\\frac{1}{2}\\right) \\frac{d}{dx} \\left(\\frac{x^2-6x-9}{x^2(x+3)^2}\\right)$$ But, using the quotient-rule, the resulting expressions really get unwieldy: $$\\frac{1}{2} \\frac{(2x-6)(x^2(x+3)^2) -(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2}$$\n\nI came up with two approaches (3 maybe):\n\n1. Split the terms up like this: $$\\frac{1}{2}\\left( \\frac{(2x-6)(x^2(x+3)^2)}{(x^2(x+3)^2)^2} - \\frac{(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} \\right)$$ so that I can simplify the left term to $$\\frac{2x-6}{x^2(x+3)^2}.$$ Taking this approach the right term still doesn't simplify nicely, and I struggle to combine the two terms into one fraction at the end.\n\n2. The brute-force-method. Just expand all the expressions in numerator and denominator, and add/subtract monomials of the same order. This definitely works, but i feel like a stupid robot doing this.\n\n3. The unofficial third-method. Grab a calculator, or computer-algebra-program and let it do the hard work.\n\nIs there any strategy apart from my mentioned ones? Am I missing something in my first approach which would make the process go more smoothly? I am looking for general tips to tackle polynomial fractions such as this one, not a plain answer to this specific problem.\n\n\u2022 Frankly, I think that we waste a lot of time worrying about simplifying things like this. Try a couple of things. If you can't get something nice relatively quickly, throw a CAS at it and get on with your life. In the real world, you will encounter very few situations where you want to factor a polynomial and are also able to do so (e.g. almost every polynomial of degree $5$ or higher cannot be factored; in many real-world settings, the polynomials come pre-factored; etc). \u2013\u00a0Xander Henderson Jun 6 '20 at 2:51\n\u2022 In examples like this (that is, if you want to differentiate a rational function), you might save some time using logarithmic differentiation. \u2013\u00a0Xander Henderson Jun 6 '20 at 2:52\n\u2022 @XanderHenderson I agree that in modern times one should focus more on understanding and applying concepts, than on pure computation. However, in this case i am happy to have been pointed to logarithmic differentiation, partial fractions and polynomial long division which made me realize some knowledge-gaps i was unaware of, which would not have happened had i used a CAS. \u2013\u00a0LeonTheProfessional Jun 6 '20 at 7:16\n\nLogarithmic differentiation can also be used to avoid long quotient rules. Take the natural logarithm of both sides of the equation then differentiate: $$\\frac{y'}{y}=2\\left(\\frac{1}{x-3}-\\frac{1}{x}-\\frac{1}{x+3}\\right)$$ $$\\frac{y'}{y}=-\\frac{2\\left(x^2-6x-9\\right)}{x(x+3)(x-3)}$$ Then multiply both sides by $$y$$: $$y'=-\\frac{{\\left(x-3\\right)}^3}{x^3{\\left(x+3\\right)}^3}$$\n\n\u2022 Great!! Yet another approach! I will look into logarithmic differentiation a bit more. With these tools i don't need to feel like a stupid robot anymore ;) \u2013\u00a0LeonTheProfessional Jun 5 '20 at 17:41\n\u2022 Yep. You should logarithmic differentiation because it's more applicable than the other methods previously mentioned. Especially if there are square root, cube root, etc. of polynomials in the numerator/denominator. \u2013\u00a0Ty. Jun 5 '20 at 17:43\n\nHINT\n\nTo begin with, notice that \\begin{align*} x^{2} - 6x - 9 = 2x^{2} - (x^{2} + 6x + 9) = 2x^{2} - (x+3)^{2} \\end{align*} Thus it results that \\begin{align*} \\frac{x^{2} - 6x - 9}{2x^{2}(x+3)^{2}} = \\frac{2x^{2} - (x+3)^{2}}{2x^{2}(x+3)^{2}} = \\frac{1}{(x+3)^{2}} - \\frac{1}{2x^{2}} \\end{align*}\n\nIn the general case, polynomial long division and the partial fraction method would suffice to solve this kind of problem.\n\n\u2022 Thank you! I think i stared too long at this problem, and have to take a step back to notice these patterns. I will examine the problem again (tomorrow) with your hints, and see if any more questions arise. If so, i will comment here. If no other answers of overwhelming enlightment appear, i will mark this answer as the accepted one. :) \u2013\u00a0LeonTheProfessional Jun 5 '20 at 17:35\n\nNote that $$x^2-6x-9 = (x-3)^2 - 18$$. So after pulling out the factor of $$\\frac 12$$, it suffices to compute $$\\frac{d}{dx} \\left(\\frac{x-3}{x(x+3)}\\right)^2$$ and $$\\frac{d}{dx} \\left(\\frac{1}{x(x+3)}\\right)^2.$$ These obviously only require finding the derivative of what's inside, since the derivative of $$(f(x))^2$$ is $$2f(x)f'(x)$$.\n\nFor a final simplification, note that $$\\frac{1}{x(x+3)} = \\frac{1}{3} \\left(\\frac 1x - \\frac{1}{x+3}\\right),$$ so you'll only ever need to take derivatives of $$\\frac 1x$$ and $$\\frac {1}{x+3}$$ to finish, since the $$x-3$$ in the numerator of the first fraction will simplify with these to give an integer plus multiples of these terms.\n\nAs a general rule, partial fractions will greatly simplify the work required in similar problems.\n\n\u2022 I found this partial fraction using polynomial long division, yet it didn't appear to me, that any remainder appearing there would obviously disappear when taking the derivative. This really is a great hint! \u2013\u00a0LeonTheProfessional Jun 5 '20 at 17:39\n\u2022 @hdighfan I think there is a slight typo - it should read that the derivative of $\\left(f(x)\\right)^2$ is $2f(x)f'(x)$. \u2013\u00a0Zubin Mukerjee Jun 6 '20 at 3:01\n\u2022 I have edited it to fix, please revert if not wanted \u2013\u00a0Zubin Mukerjee Jun 6 '20 at 20:18\n\u2022 Ah, of course. Thanks for the edit. \u2013\u00a0hdighfan Jun 6 '20 at 20:19", "date": "2021-02-27 05:25:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3707227/strategy-to-calculate-fracddx-left-fracx2-6x-92x2x32-right/3707238", "openwebmath_score": 0.8510565161705017, "openwebmath_perplexity": 383.7864381263352, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211582993982, "lm_q2_score": 0.8791467595934563, "lm_q1q2_score": 0.8570108625020557}}
{"url": "http://math.stackexchange.com/questions/135234/showing-a-function-is-not-uniformly-continuous", "text": "# Showing a function is not uniformly continuous\n\nI am looking at uniform continuity (for my exam) at the moment and I'm fine with showing that a function is uniformly continuous but I'm having a bit more trouble showing that it is not uniformly continuous, for example:\n\nshow that $x^4$ is not uniformly continuous on $\\mathbb{R}$, so my solution would be something like:\n\nAssume that it is uniformly continuous then:\n\n$$\\forall\\epsilon\\geq0\\exists\\delta>0:\\forall{x,y}\\in\\mathbb{R}\\ \\mbox{if}\\ |x-y|<\\delta \\mbox{then} |x^4-y^4|<\\epsilon$$\n\nTake $x=\\frac{\\delta}{2}+\\frac{1}{\\delta}$ and $y=\\frac{1}{\\delta}$ then we have that $|x-y|=|\\frac{\\delta}{2}+\\frac{1}{\\delta}-\\frac{1}{\\delta}|=|\\frac{\\delta}{2}|<\\delta$ however $$|f(x)-f(y)|=|\\frac{\\delta^3}{8}+3\\frac{\\delta}{4}+\\frac{3}{2\\delta}|$$\n\nNow if $\\delta\\leq 1$ then $|f(x)-f(y)|>\\frac{3}{4}$ and if $\\delta\\geq 1$ then $|f(x)-f(y)|>\\frac{3}{4}$ so there exists not $\\delta$ for $\\epsilon < \\frac{3}{4}$ and we have a contradiction.\n\nSo I was wondering if this was ok (I think it's fine) but also if this was the general way to go about showing that some function is not uniformly continuous? Or if there was any other ways of doing this that are not from the definition?\n\nThanks very much for any help\n\n-\nSo, is this an exam question? \u2013\u00a0 user21436 Apr 22 '12 at 12:17\nNo, I'm just practicing for my exam where questions like this (not this one though) will come it. This is from one of the past papers that are for revision \u2013\u00a0 hmmmm Apr 22 '12 at 12:29\nJust trying to ensure we are not taken for a ride. Hope you don't mind. :) \u2013\u00a0 user21436 Apr 22 '12 at 12:33\n@KannappanSampath no its fine- it annoys me when I see people posting assessment questions on forums :) \u2013\u00a0 hmmmm Apr 22 '12 at 12:36\n\nTo show that it is not uniformly continuous on the whole line, there are two usual (and similar) ways to do it:\n\n1. Show that for every $\\delta > 0$ there exist $x$ and $y$ such that $|x-y|<\\delta$ and $|f(x)-f(y)|$ is greater than some positive constant (usually this is even arbitrarily large).\n2. Fix the $\\varepsilon$ and show that for $|f(x)-f(y)|<\\varepsilon$ we need $\\delta = 0$.\n\nFirst way:\n\nFix $\\delta > 0$, set $y = x+\\delta$ and check $$\\lim_{x\\to\\infty}|x^4 - (x+\\delta)^4| = \\lim_{x\\to\\infty} 4x^3\\delta + o(x^3) = +\\infty.$$\n\nSecond way:\n\nFix $\\epsilon > 0$, thus $$|x^4-y^4| < \\epsilon$$ $$|(x-y)(x+y)(x^2+y^2)| < \\epsilon$$ $$|x-y|\\cdot|x+y|\\cdot|x^2+y^2| < \\epsilon$$ $$|x-y| < \\frac{\\epsilon}{|x+y|\\cdot|x^2+y^2|}$$\n\nbut this describes a necessary condition, so $\\delta$ has to be at least as small as the right side, i.e.\n\n$$|x-y| < \\delta \\leq \\frac{\\epsilon}{|x+y|\\cdot|x^2+y^2|}$$\n\nso if either of $x$ or $y$ tends to infinity then $\\delta$ tends to $0$.\n\nHope that helps ;-)\n\nEdit: after explanation and calculation fixes, I don't disagree with your proof.\n\n-\nThanks for the reply, I think that I use that we are considering all of $\\mathbb{R}$ when i choose $x=\\delta+\\frac{1}{\\delta}$ and $y=\\frac{1}{\\delta}$ as these would not be valid for small $\\delta$ in bounded interval? \u2013\u00a0 hmmmm Apr 22 '12 at 13:38\n@hmmmm Ok, I misunderstood what you were saying there. If the calculations are alright, then your proof is fine. \u2013\u00a0 dtldarek Apr 22 '12 at 13:44\n\nI will comment on your solution after writing another approach. For any $x,y\\in\\mathbb{R}$ we have: \\begin{align*} |x^{4}-y^{4}|=|(x^{2}-y^{2})(x^{2}+y^{2})|=|(x-y)(x+y)(x^{2}+y^{2})|=|x-y|\\cdot |x+y|\\cdot |x^{2}+y^{2}| \\end{align*}\n\nSo what you can see is that even if you take arbitrarily close $x$ and $y$, you can grow the distance of $x^{4}$ and $y^{4}$ as much as you want by taking them far enough away from zero. You can easily conclude from here that the function is not uniformly continuous by a contraposition for example.\n\nAlright, then to your solution. If the calculations would be correct, then it would be fine. You could assume at first that such $\\delta>0$ exists for $0<\\varepsilon<3$ and conclude with a contradiction. However, I got a bit different calculations than you. Using the above equation we see that: \\begin{align*} |f(\\frac{\\delta}{2}+\\frac{1}{\\delta})-f(\\frac{1}{\\delta})|&=|(\\frac{\\delta}{2}+\\frac{1}{\\delta})^{4}-\\frac{1}{\\delta^{4}}|=|\\frac{\\delta}{2}(\\frac{\\delta}{2}+\\frac{2}{\\delta})((\\frac{\\delta}{2}+\\frac{1}{\\delta})^{2}+\\frac{1}{\\delta^{2}})| \\\\ &= |(\\frac{\\delta^{2}}{4}+1)(\\frac{\\delta^{2}}{4}+2\\cdot \\frac{\\delta}{2}\\cdot \\frac{1}{\\delta}+\\frac{1}{\\delta^{2}}+\\frac{1}{\\delta^{2}})| \\\\ &=|(\\frac{\\delta^{2}}{4}+1)(\\frac{\\delta^{2}}{4}+1+\\frac{2}{\\delta^{2}})| \\\\ &= |\\frac{\\delta^{4}}{16}+\\frac{\\delta^{2}}{4}+\\frac{1}{2}+\\frac{\\delta^{2}}{4}+1+\\frac{2}{\\delta^{2}}| \\\\ &= |\\frac{\\delta^{4}}{16}+\\frac{\\delta^{2}}{2}+\\frac{2}{\\delta^{2}}+\\frac{3}{2}|\\\\ &= \\frac{\\delta^{4}}{16}+\\frac{\\delta^{2}}{2}+\\frac{2}{\\delta^{2}}+\\frac{3}{2} \\end{align*} If you're able to find a lower bound for this (which is quite easy) as you did previously, then by choosing an epsilon smaller than that fixed number you may conclude as you did in your original post by contradiction.\n\n-\nHey sorry about that, I edited it-hopefully it's right now? \u2013\u00a0 hmmmm Apr 22 '12 at 13:39\nI also edited now my calculation which still differs abit from your new one. Could you show the steps of how you got this answer for $|f(x)-f(y)|$? \u2013\u00a0 Thomas E. Apr 22 '12 at 13:47\nyeah I messed that up quite a bit sorry (I had the wrong power and the wrong delta's) \u2013\u00a0 hmmmm Apr 22 '12 at 13:50\nIt should be $|(\\frac{\\delta}{2}+\\frac{1}{\\delta})^4-\\frac{1}{\\delta^4}|$ which would give $|\\frac{\\delta^4}{16}+\\frac{\\delta^2}{2}+\\frac{2}{\\delta}+\\frac{3}{2}|$ I think I could conclude a similar thing from here? \u2013\u00a0 hmmmm Apr 22 '12 at 13:52\nExcept that is the last $-\\frac{1}{\\delta^{4}}$ missing from there? Otherwise it looks close to mine. \u2013\u00a0 Thomas E. Apr 22 '12 at 13:57\nI think you should make this a little simpler (and for uniform continuity in general) All you need to do to show $f:X \\to Y$ is not uniformly continuous on $X$ (let's suppose there both subsets of $\\Bbb R$), then just give me a SINGLE epsilon such that, NO MATTER HOW SMALL delta is chosen, there will be x and y closer than delta for which the difference in function values exceed epsilon. Thus for instance $|(N+\\theta)^4- N^4| \\ge 4\\theta N^3$ so if you choose $x$ really big (with respect to $\\delta, x=N+\\theta\\,\\,\\,\\ \\text{and}\\,\\,\\, y = N,$ then if $0 < \\delta/2 < \\theta < \\delta$ you have $|x-y| < \\delta$ yet you still have the variable N to play with to make the difference in function values as large as you like (in particular the difference in function values can always be made bigger than 3 regardless of how small $\\delta$ is). Nevertheless, I think your proof is an accurate job!", "date": "2014-03-12 21:28:09", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/135234/showing-a-function-is-not-uniformly-continuous", "openwebmath_score": 0.9927499890327454, "openwebmath_perplexity": 266.53601893974314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692284751636, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8570045817287523}}
{"url": "https://mathhelpboards.com/threads/more-than-one-equation-for-a-given-trig-graph.3641/", "text": "# TrigonometryMore than one equation for a given Trig Graph?\n\n#### m3dicat3d\n\n##### New member\nHi all.. another Trig question here...\n\nLet's say I'm given a graph of a sinusoidal function and asked to find its equation, but I'm not told whether this is a sine or cosine function and I'm left to determine that myself.\n\nI understand that evaluating where the graph intersects the y axis is the straight-forward, easiest approach. For instance, take this graph where the y interval is .5 and the x interval is pi/2\n\nI can say that it's a sine graph easily by sight, but also b/c it intersects the y axis at y=0. And given the phase shift and no vertical shift, the equation is f(x) = sin [(2/3)x].\n\nBUT, couldn't this also be f(x) = cos [{(2/3)x} - (pi/2)] since sin(x) and cos(x) are seperated only by a phase shift of pi/2?\n\nThis is meant for my own edification and not to make this kind of exercise more confusing than be. I'm simply interseted if this is in fact mathematically accurate that you could have more than one equation (a sine or a cosine \"version\") for a given sinusoidal curve.\n\nMy calculator returns coincidental curves when I graph both the sine and cosine \"versions\" of this given graph, but I know my calculator isn't really a mathematician either haha, so I thought I'd ask some real mathematicians instead\n\nThanks\n\n#### Ackbach\n\n##### Indicium Physicus\nStaff member\nNote that by the addition of angles identity, that\n$$\\cos(2x/3- \\pi/2)= \\cos(2x/3) \\cos( \\pi/2)+ \\sin(2x/3) \\sin( \\pi/2) = \\cos(2x/3) \\cdot 0+ \\sin(2x/3) \\cdot 1= \\sin(2x/3).$$\nSo yes, you can definitely have more than one representation of the same graph, as you have seen on your calculator.\n\n#### m3dicat3d\n\n##### New member\nThank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!!\n\nExcellent answer, and as I am still reviewing my Trig, I hadn't even considered the identity perspective of it... I can't stress enough how much having that perspective helps me even more with this...\n\nAgain, Thank you... man this place rocks!\n\n#### MarkFL\n\nStaff member\nAs sine and cosine are complementary or co-functions, this just means they are out of phase by $\\displaystyle \\frac{\\pi}{2}$ radians, or 90\u00b0.\n\nYou may have noticed that a sine curve, if moved 1/4 period to the left, becomes a cosine curve, or conversely, a cosine curve moved 1/4 period to the right becomes a sine curve.\n\nYou are doing well to see this, it shows you are trying to understand it on a deeper level rather than just plugging into formulas. Both the sine function and the cosine function, and linear combinations of the two (with equal amplitudes) are called sinusoidal functions.\n\n#### m3dicat3d\n\n##### New member\nThanks MarkFL, I appreciate the encouraging words. I'm studying for my State certification exam to teach HS math here in TX, and I tutor HS students in the meantime. I'm no math genius by far, so when I ask some of my questions I sometimes feel they might be dumb (which no one here has made me feel like I'm glad to say). I'm trying to see those \"nuances\" in the math in case they might help my students, and again, I appreciate your words, and the full out decency of the community here. It's a great place to learn\n\n#### Ackbach\n\n##### Indicium Physicus\nStaff member\nThank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!!\n\nExcellent answer, and as I am still reviewing my Trig, I hadn't even considered the identity perspective of it... I can't stress enough how much having that perspective helps me even more with this...\n\nAgain, Thank you... man this place rocks!\nYou're quite welcome! Glad to be of help.", "date": "2021-02-25 16:26:51", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/more-than-one-equation-for-a-given-trig-graph.3641/", "openwebmath_score": 0.8644052147865295, "openwebmath_perplexity": 1098.1628063593155, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631659211718, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8569787346726185}}
{"url": "http://mathhelpforum.com/pre-calculus/214181-roots-unity-length-one-root-other-roots.html", "text": "# Thread: Roots of unity and the length from one root to the other roots\n\n1. ## Roots of unity and the length from one root to the other roots\n\nHello,\n\nI had an assignment that required me to solve for the roots of unity for various equations of the form $z^n -1 = 0$. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.\n\nI did all of that , however, i have a question:\n\nis there a relation ship between the power n and the length from one root to the other roots?\n\nWhen n = 3,\nThe roots are $-0.5 \\pm 0.8660i$ and 1. The length from one root to the other roots are both 1.7321 (sqrt of 3)\n\nWhen n = 4,\n\nThe roots are $\\pm 1, 0 \\pm 1i$. The length from one root to the other roots are (sqrt 2, sqrt and 2)\n\nI am wondering , for the equation $z^3 - 1 = 0$ , is there a formula relating the power of z to the length from one roots to the other roots?\nThanks.\n\n2. ## Re: Roots of unity and the length from one root to the other roots\n\nThe nth roots of unity lie at the vertices of a polygon with n sides, each vertex having distance from the center of 1. Drawing a line from the center to each vertex divides the polygon into n isosceles triangles, each having two sides of length 1, vertex angle of 360/n degrees and you want to find the length of the third side. If you draw a line from the vertex of such a triangle to the center of the base, you get two right triangles with hypotenuse length 1 and one angle of 180/n. The opposite side of that triangle has length 1(sin(180)/n) so the side of the polygon is twice that :\n2sin(180/n).\n\n3. ## Re: Roots of unity and the length from one root to the other roots\n\nHi, thank you very much for answering the question first.\n\nHowever, i have already got the conjecture that the length of the polygon would be 2 sin (180/n), but this conjecture cannot be proven algebraically or MI that this statement would be true for any value of n, can it?\n\nI am just wondering is there any other relationships between the length from one root to the other roots? (Lets say, if you draw a line from one root (any root) to the other roots (not only the adjacent root)\n\nn = 3, length is sqrt 3 and sqrt 3\nn = 4, length is sqrt 2 , 2 and sqrt 2\nn = 5, length is not an exact value.. but what i have got is 1.1756, 1.9021, 1.9021 and 1.1756\nn = 6, length is 1, sqrt 3, 2, sqrt 3, 1\nn = 7, length is 0.86, 1.56, 1.94, 1.94, 1.56, 0.86\n\nseems like theres a pattern here, but i can't seem to figure it out... would help a lot if someone could tell me if theres a conjecture for this or not, thank you very much.", "date": "2017-09-21 16:37:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/214181-roots-unity-length-one-root-other-roots.html", "openwebmath_score": 0.756908118724823, "openwebmath_perplexity": 184.76773029695687, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363499098283, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8569605489935974}}
{"url": "https://web2.0calc.com/questions/help_4799", "text": "+0\n\n# help\n\n0\n62\n4\n\nWhich is larger, the blue area or the orange area?\n\nNov 14, 2019\n\n#1\n+105411\n+2\n\nLet the radius\u00a0 of the orange circle\u00a0 \u00a0= r\n\nSo......its area\u00a0 =\u00a0 pi * r^2\n\nWe can find\u00a0the side length,s, of the\u00a0 equilateral triangle thusly\n\ntan (30\u00b0)\u00a0 =\u00a0 r / [(1/2)s]\n\n1/\u00a0\u221a3 = r /\u00a0[ (1/2)s ]\n\n(1/\u00a0\u221a3) ( 1/2)s\u00a0 = r\n\ns =\u00a0 (2\u221a3) \u00a0r\u00a0 \u00a0= [ \u221a12\u00a0 ] r\n\nAnd the area of\u00a0the equilateral\u00a0 triangle\u00a0 is\u00a0 \u00a0(\u00a0 \u221a3/ 4\u00a0) ([\u221a12]r)^2\u00a0 = 3\u221a3 r^2\n\nSo....the blue area inside the equilateral triangle\u00a0 =\n\n[area of equailateral triangle - area of small circle\u00a0 ]\u00a0 /\u00a0 3\n\n[\u221a3 r^2\u00a0 - (1/3) pi r^2 ]\u00a0 =\u00a0 \u00a0r^2 [ \u221a3 -\u00a0pi/3]\u00a0 \u00a0 \u00a0 (1)\n\nThe radius\u00a0 of the larger circle\u00a0 can be found as\n\n\u221a[ [\u221a3r ]^2\u00a0 + r^2 ]\u00a0 \u00a0=\u00a0\u00a0\u221a [ 3r^2 + r^2 ]\u00a0 \u00a0=\u00a0 2r\n\nSo....the area\u00a0 of the\u00a0 larger circle =\u00a0 pi\u00a0(2r)^2\u00a0 =\u00a0 4pi r^2\n\nSo\u00a0 \u00a0the area between the side of the equilateral triangle and the larger circle\u00a0 \u00a0is\n\n[Area of larger circle\u00a0 -\u00a0area of equilateral triangle\u00a0 ]\u00a0 / 3\u00a0 =\n\n[ 4pi r^2\u00a0 - 3\u221a3r^2 ] / 3\u00a0 =\u00a0 r^2 [ (4/3)pi\u00a0 \u00a0-\u00a0\u221a3]\u00a0\u00a0 \u00a0(2)\n\nSo the sum of\u00a0 (1) and (2)\u00a0 \u00a0is\u00a0 the sum of the blue areas\n\nr^2\u00a0 [\u00a0\u221a3 - pi/3\u00a0 + (4/3) pi -\u00a0\u221a3 ]\u00a0 \u00a0= pi r^2\n\nSo.....the orange and blue areas are equal\u00a0 \u00a0!!!!\n\nNov 14, 2019\n#2\n+70\n+2\n\nHere is a different approach.The arc AC has measure $$\\frac{1}{3}\\cdot2\\pi$$ and the angle AOB has measure half of the arc or $$\\frac {\\pi}{3}$$. If the radius of the larger circle is $$R$$, then the measure of OB, the radius of the smaller circle, is $$Rcos(\\frac{\\pi}{3})$$and so the orange region has area\u00a0$$\\pi(Rcos(\\frac{\\pi}{3}))^2=\\frac{1}{4}\\pi R^2$$ . The blue region, however, has area equal to $$\\frac{1}{3}$$of the difference between area of the larger circle and the area of the smaller circle, i.e. $$\\frac{1}{3}(\\pi R^2-\\frac{1}{4}\\pi R^2) =\\frac{1}{4}\\pi R^2$$. So the two regions have the same area, each being $$\\frac{1}{4}$$ of the area of the larger circle.\n\nNov 14, 2019\n#3\n+105411\n+1\n\nThanks, Gadfly....I like that approach\u00a0 !!!\n\nNov 14, 2019\n#4\n+23575\n+2\n\nWhich is larger, the blue area or the orange area?\n\n$$\\begin{array}{|rcll|} \\hline {\\color{orange}\\text{orange}} +3\\times {\\color{blue}\\text{blue}} &=& \\pi r_{\\text{circumcircle}}^2 \\quad | \\quad : {\\color{orange}\\text{orange}} \\\\ 1+3\\times \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\dfrac{ \\pi r_{\\text{circumcircle}}^2 } { {\\color{orange}\\text{orange}} } \\quad | \\quad {\\color{orange}\\text{orange}} = \\pi r_{\\text{incircle}}^2 \\\\ 1+3\\times \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\dfrac{ \\pi r_{\\text{circumcircle}}^2 } { \\pi r_{\\text{incircle}}^2 } \\\\ 1+3\\times \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\left( \\dfrac{ r_{\\text{circumcircle}} } { r_{\\text{incircle}} } \\right)^2 \\\\ 3\\times \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\left( \\dfrac{ r_{\\text{circumcircle}} } { r_{\\text{incircle}} } \\right)^2 -1 \\\\ \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\dfrac{1}{3}\\left( \\left( \\dfrac{ r_{\\text{circumcircle}} } { r_{\\text{incircle}} } \\right)^2 -1 \\right) \\\\ \\\\ && \\boxed{\\text{here, if triangle is equilateral: }\\\\ \\mathbf{2\\times r_{\\text{incircle}} = r_{\\text{circumcircle}}!!!} } \\\\\\\\ \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\dfrac{1}{3}\\left( \\left( \\dfrac{ 2\\times r_{\\text{incircle}} } { r_{\\text{incircle}} } \\right)^2 -1 \\right)\\\\ \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\dfrac{1}{3}\\times(2^2 -1) \\\\ \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& \\dfrac{1}{3}\\times(3) \\\\ \\dfrac{ {\\color{blue}\\text{blue}} } { {\\color{orange}\\text{orange}} } &=& 1 \\\\ \\mathbf{ {\\color{blue}\\text{blue}} } &=& \\mathbf{ {\\color{orange}\\text{orange}} } \\\\ \\hline \\end{array}$$\n\nNov 15, 2019\nedited by heureka \u00a0Nov 15, 2019", "date": "2019-12-14 07:09:04", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/help_4799", "openwebmath_score": 0.945934534072876, "openwebmath_perplexity": 10881.816147161699, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363480718236, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8569605473934355}}
{"url": "https://mathstodon.xyz/@christianp/108424166828836198", "text": "A while ago @davidphys1 asked why nobody had made animations of the shunting yard algorithm with cutesy trains.\nThere is no surer way to summon me!\nI've spent some of my spare time over the bank holidays making exactly that: somethingorotherwhatever.com/s\n\n\u00b7 \u00b7 Web \u00b7 \u00b7 \u00b7\n\nThe shunting yard algorithm neatly solves the problem of translating a mathematical expression written in infix notation (operators go between the numbers/letters) to postfix notation (operators go after the things they act on).\n\nThe core problem is that you need to work out what just what an operator applies to: with the order of operations, it might be just one number, or it might a large sub-expression.\n\nThe algorithm solves this by holding operators on a separate stack until they're needed\n\nHere are some animations to illustrate. In the first, the operations happen left-to-right, so they appear in the same order in the output as in the input.\nIn the second, the addition must happen after the multiplication, so it's held back.\nIn the third, brackets ensure that the addition happens first.\n\nThe last wrinkle for standard arithmetic is that exponentiation is right-associative: while for the other operations you work left-to-right:\n1 \u2212 2 \u2212 3 = (1 \u2212 2) \u2212 3,\nthe order goes the other way for exponentiation:\n1 ^ 2 ^ 3 = 1 ^ (2 ^ 3)\n\n@christianp @davidphys1 Wow that's a very cute animation!\n\n@christianp\nDoesn't that depend on the domain?\n\nI mean, some rules aren't valid anymore for, say, Quaternions or Octaves\u2026\n\n@RyunoKi *leans heavily on the word \"standard\"*\n\n@RyunoKi but no, I think that if I wrote an expression $$a \\times b \\times c$$, where a, b and c are octonions and so multiplication isn't associative, I'd expect you to interpret it as $$a \\times b \\times c$$, by convention.\nI might include some brackets, to avoid relying on convention.\n\n@christianp\nSo\u2026 the algorithm isn't considering this constraint. Fine. Space for further research :)\n\n@RyunoKi no, it is. Or I don't understand what you mean.\n\n@RyunoKi just noticed I missed the brackets in my second-last tweet! I had trouble LaTeXing, then didn't check how it looked!\nI'd expect you to interpret $$a \\times b \\times c$$ as $$(a \\times b) \\times c$$\n\n@christianp\nI even switched to browser view to check whether something got lost in transmission \ud83d\ude05\n\n@christianp\n\u201eThe core problem is that you need to work out what just what an operator applies to: with the order of operations, it might be just one number, or it might a large sub-expression.\n\nThe algorithm solves this by holding operators on a separate stack until they're needed\u201c\n\nIs that assuming \u201estandard\" arithmetic like associative multiplication? Or the lack thereof?\n\nLike, once \u201ecomputer does that\u201c many people rely on it without questioning the result.\n\n@RyunoKi right, the algorithm says that multiplication is left-associative. For real numbers, it doesn't matter.\n\n@christianp\nThanks. Important fact in certain circumstances.\n\n@christianp\nEspecially the order and arguments can change depending on how you put the brackets.\n\nThe social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!", "date": "2022-06-30 06:34:31", "meta": {"domain": "mathstodon.xyz", "url": "https://mathstodon.xyz/@christianp/108424166828836198", "openwebmath_score": 0.7036386132240295, "openwebmath_perplexity": 1594.6830523310077, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363499098283, "lm_q2_score": 0.8705972768020107, "lm_q1q2_score": 0.8569605456887277}}
{"url": "https://math.stackexchange.com/questions/2394083/is-frac2001020-cdot-19-an-integer-or-not?noredirect=1", "text": "# Is $\\frac{200!}{(10!)^{20} \\cdot 19!}$ an integer or not?\n\nA friend of mine asked me to prove that $$\\frac{200!}{(10!)^{20}}$$ is an integer\n\nI used a basic example in which I assumed that there are $200$ objects places in $20$ boxes (which means that effectively there are $10$ objects in one box). One more condition that I adopted was that the boxes are distinguishable but the items within each box are not. Now the number of permutations possible for such an arrangement are : $$\\frac{200!}{\\underbrace{10! \\cdot 10! \\cdot 10!\\cdots 10!}_{\\text{20 times}}}$$ $$\\Rightarrow \\frac{200!}{(10!) ^{20}}$$\n\nSince these are just ways of arranging, we can be pretty sure that this number is an integer.\n\nThen he made the problem more complex by adding a $19!$ in the denominator, thus making the problem: Is $$\\frac{200!}{(10!)^{20} \\cdot 19!}$$ an integer or not?\n\nThe $19!$ in the denominator seemed to be pretty odd and hence I couldn't find any intuitive way to determine the thing. Can anybody please help me with the problem?\n\n\u2022 There might be some clever way, but this is where I would start counting primes. Are there enough $2$'s in the numerator? $3$'s? \u2013\u00a0Arthur Aug 15 '17 at 7:23\n\u2022 Use the fact that every set of $k$ integers will have one integer divisible by $k$. Show that that means $(m+1)(m+2).... (m+k)$ will be divisible by $k!$. And that means $(1*2....10)(11*... 20)...... (191*....200)$ is divisible by $10!*10!*....10!$. \u2013\u00a0fleablood Aug 15 '17 at 7:26\n\u2022 That's an interesting friend you have there. \u2013\u00a0uniquesolution Aug 15 '17 at 7:30\n\u2022 To eliminate 19! Not that all k divide 10*k. Then each 10k+1 to 10k+9 is divisible by 9! And 10k/k = 10 so it is divisible by 10 as well. \u2013\u00a0fleablood Aug 15 '17 at 7:34\n\u2022 An interesting question (at least I think it would be) is what is the largest integer $k$ so that $k!*10^{20}$ divides $200!$. \u2013\u00a0fleablood Aug 15 '17 at 16:27\n\nYou assumed the boxes were distinguishable, leading to $\\frac{200!}{(10!)^{20}}$, ways to fill the boxes. If you make them indistinguishable, you merge the $20!$ ways of reordering the boxes into one, so that previous answer overcounts each way of filling indistinguishable boxes by a factor of $20!$. Therefore you are left with $\\frac{200!}{(10!)^{20}}/20!$ ways to fill 20 indistinguishable boxes, which then must be an integer. After multiplying by $20$ it is of course still an integer.\n\nWe know that $\\dfrac{(mn)!}{n!(m!)^n}$ is an integer for $m,n \\in \\Bbb N$ $^{(*)}$ . Let $n = 20$ and $m = 10$, then $\\dfrac{(200)!}{20!(10!)^{20}}$ is an integer.\n\nMultiply by $20$, $\\dfrac{(200)!}{19!(10!)^{20}}$ is an integer.\n\nUsing induction, this answer says that $$\\frac{(mn)!}{(m!)^nn!}=\\prod_{k=1}^n\\binom{mk-1}{m-1}$$ Plug in $m=10$ and $n=20$ to get $$\\frac{200!}{10!^{20}\\,20!}=\\prod_{k=1}^{20}\\binom{10k-1}{9}$$ Multiply by $20$ to get $$\\frac{200!}{10!^{20}\\,19!}=20\\,\\prod_{k=1}^{20}\\binom{10k-1}{9}$$\n\nAnother Approach\n\nNote that \\begin{align} \\binom{kn}{n} &=\\frac{(kn-n+1)(kn-n+2)\\cdots(kn-1)\\,kn}{1\\cdot2\\cdots(n-1)\\,n}\\\\ &=\\frac{(kn-n+1)(kn-n+2)\\cdots(kn-1)\\,k}{1\\cdot2\\cdots(n-1)}\\\\ &=\\binom{kn-1}{n-1}\\,k \\end{align} Therefore, since we can write a multinomial as a product of binomials, \\begin{align} \\frac{(mn)!}{n!^m} &=\\prod_{k=1}^m\\binom{kn}{n}\\\\ &=\\prod_{k=1}^m\\binom{kn-1}{n-1}\\,k\\\\ &=m!\\,\\prod_{k=1}^m\\binom{kn-1}{n-1} \\end{align} and so $$\\frac{(mn)!}{n!^m\\,m!}=\\prod_{k=1}^m\\binom{kn-1}{n-1}$$ Plug in $m=20$ and $n=10$ and multiply by $20$ to get $$\\frac{200!}{10!^{20}\\,19!}=20\\,\\prod_{k=1}^{20}\\binom{10k-1}{9}$$\n\n\u2022 But this is my answer ... \u2013\u00a0user8277998 Aug 16 '17 at 8:49\n\u2022 @123: I see... I cited an answer that I wrote and didn't see that you had parenthetically cited the question to which that was an answer. However, without the connection given by the citations, the answers do not look the same. If this bothers you, I will delete my answer. \u2013\u00a0robjohn Aug 16 '17 at 13:01\n\u2022 @123: I have added another proof of the same identity to differentiate our answers. I will still delete this answer if you think they are too close. \u2013\u00a0robjohn Aug 16 '17 at 13:28\n\u2022 No, I have no problem with your answer, you can keep it as it is. \u2013\u00a0user8277998 Aug 16 '17 at 14:36\n\nA long version: $$\\frac{200!}{10!^{20} \\cdot 19!}=\\frac{30\\cdot31\\cdot .. \\cdot200}{10!^{19}}\\cdot \\frac{29!}{10!\\cdot(29-10)!}=...$$ which is $$...=\\frac{30\\cdot31\\cdot .. \\cdot200}{10!^{19}}\\cdot \\binom{29}{10}=\\\\ \\frac{\\color{red}{30} ..\\color{red}{40} ..\\color{red}{50} ..\\color{red}{60}..\\color{red}{70}..\\color{red}{80}..\\color{red}{90}..\\color{red}{10^2}..\\color{red}{110}..\\color{red}{120}..\\color{red}{130}..\\color{red}{140}..\\color{red}{150}..\\color{red}{160}..\\color{red}{170}..\\color{red}{180}..\\color{red}{190}..\\color{red}{2\\cdot10^{2}}}{10!^{19}}\\cdot \\binom{29}{10}=...$$ $20$ numbers divisible by 10, or $$3\\cdot4\\cdot5\\cdot..\\cdot9\\cdot11\\cdot..\\cdot19\\cdot20\\cdot\\frac{31..39\\cdot41..49\\cdot51..59\\cdot..\\cdot191..199}{9!^{19}}\\cdot \\binom{29}{10}=\\\\ 10\\cdot\\frac{2..9\\cdot11..19\\cdot31..39\\cdot41..49\\cdot51..59\\cdot..\\cdot191..199}{9!^{19}}\\cdot \\binom{29}{10}=...$$ cardinality of $\\{31,41,51,61,71,81,91,101,111,121,131,141,151,161,171,181,191\\}$ is 17 $$...=10\\cdot \\frac{1..9}{9!}\\cdot\\frac{11..19}{9!}\\cdot\\frac{31..39}{9!}\\cdot..\\cdot\\frac{191..199}{9!}\\cdot \\binom{29}{10}=\\\\ 10\\cdot \\binom{9}{9} \\cdot \\binom{19}{9} \\cdot \\binom{39}{9}\\cdot .. \\cdot \\binom{199}{9} \\cdot \\binom{29}{10}$$\n\n\u2022 I would appreciate the down-voters to at least comment ... \u2013\u00a0rtybase Aug 15 '17 at 8:29\n\nConsider $V=(10k+1)*....*(10k+9)$.\n\nBy your reasoning, ${10k+9 \\choose 9}=(10k+1)*....*(10k+9)/9!$ is an integer.\n\nAnd $10(k+1)/10(k+1)$ is an integer.\n\nSo $(10k+1)*....*(10 (k+1))$ is divisible by $9!*10*(k+1)=10!*(k+1)$.\n\nSo $200!$ is divisible by $10!*1*10!*2*10!*3*.....*10!*19=(10!)^{20}*19!$\n\n\u2022 Don't you mean it is divisible by $10!*(k+1)$ ? \u2013\u00a0Jaap Scherphuis Aug 15 '17 at 7:55\n\u2022 Yeah, I guess I did. \u2013\u00a0fleablood Aug 15 '17 at 16:22\n\nI computed the answer just for fun using Java, and it's indeed an integer!\n\n41355508127520659545494261323391337886154686759988983912363570790033502473625361601944917427369977161391866491251801111884812210789772970682172860398969828337097889527312353089859289462934116034461288917394623420753412096000000\n\nimport java.math.BigDecimal;\nimport java.math.RoundingMode;\n\npublic class JustForFun{\npublic static void main(String []args){\nBigDecimal thFact = new BigDecimal(\"1\");\nBigDecimal tenFact = null, ntFact = null, tenFactPow20 = null;\n/* Computes 200! and stores it in a */\nfor (int i = 1; i <= 200; i++) {\nthFact = thFact.multiply(new BigDecimal(i + \"\"));\n/* stores 10! in b */\nif (i == 10)\ntenFact = thFact;\n/* stores 19! in c */\nif (i == 19)\nntFact = thFact;\n}\ntenFactPow20 = tenFact.pow(20);\ntenFactPow20 = tenFactPow20.multiply(ntFact);\nthFact = thFact.divide(tenFactPow20);\nSystem.out.println(thFact);\n}\n}\n\n\nSince I am pushing 70 yrs old, it seems appropriate to dinosaur-excerpt \"Elementary Number Theory\" 1938 (Uspensky and Heaslett).\n\nFor real # $$r$$, let $$\\lfloor r\\rfloor \\equiv$$ the floor of $$r$$.\n\nLet $$p$$ be any prime #.\nLet $$V_p(n) : n ~\\in ~\\mathbb{Z^+} ~\\equiv~$$ the largest exponent $$\\alpha$$ such that $$p^{\\alpha} | n$$.\nThat is, if $$\\alpha = V_p(n),$$ then $$p^{(\\alpha + 1)} \\not | ~n.$$\n\nFrom Uspensky and Heaslett, $$V_p(n!) = \\left\\lfloor\\frac{n}{p^1}\\right\\rfloor ~+~ \\left\\lfloor\\frac{n}{p^2}\\right\\rfloor ~+~ \\left\\lfloor\\frac{n}{p^3}\\right\\rfloor ~+~ \\left\\lfloor\\frac{n}{p^4}\\right\\rfloor \\cdots$$\n\nClearly, given two positive integers $$A,B$$, $$~\\frac{A}{B}$$ will be an integer $$\\iff$$\nfor every prime # $$p$$ that occurs in the prime factorization of $$B$$,\n$$V_p(B) \\leq V_p(A).$$\n\nIt is immediate, that given the OP's original question, the only prime #'s that need to be checked are those prime #'s that are $$\\leq 19.$$ Further, you can see at a glance the prime #'s 11, 13, 17, and 19 can not pose a problem.\n\nTherefore, the problem reduces to manually applying Uspenky and Heaslett's formula to the numerator and denominator of the OP's original query with respect to the prime #'s 2,3,5,7.\n\nEmpirically, they each check out okay. Therefore, the fraction is an integer.\n\nIn your face $$21^{\\text{st}}$$ century!", "date": "2021-01-16 05:46:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2394083/is-frac2001020-cdot-19-an-integer-or-not?noredirect=1", "openwebmath_score": 0.780359148979187, "openwebmath_perplexity": 416.9148791756454, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363512883316, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8569605402791093}}
{"url": "https://proofwiki.org/wiki/Coprimality_Relation_is_not_Antisymmetric", "text": "# Coprimality Relation is not Antisymmetric\n\n## Theorem\n\nConsider the coprimality relation on the set of integers:\n\n$\\forall x, y \\in \\Z: x \\perp y \\iff \\gcd \\set {x, y} = 1$\n\nwhere $\\gcd \\set {x, y}$ denotes the greatest common divisor of $x$ and $y$.\n\nThen:\n\n$\\perp$ is not antisymmetric.\n\n## Proof\n\nWe have:\n\n$\\gcd \\set {3, 5} = 1 = \\gcd \\set {5, 3}$\n\nand so:\n\n$3 \\perp 5$ and $5 \\perp 3$\n\nHowever, it is not the case that $3 = 5$.\n\nThe result follows by definition of antisymmetric relation.\n\n$\\blacksquare$", "date": "2021-04-11 13:26:09", "meta": {"domain": "proofwiki.org", "url": "https://proofwiki.org/wiki/Coprimality_Relation_is_not_Antisymmetric", "openwebmath_score": 0.9990094304084778, "openwebmath_perplexity": 567.9665807161637, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363503693294, "lm_q2_score": 0.8705972566572504, "lm_q1q2_score": 0.8569605262595482}}
{"url": "https://math.stackexchange.com/questions/3238042/limit-of-powers-of-3-times3-matrix/3239827", "text": "# Limit of powers of $3\\times3$ matrix\n\nConsider the matrix\n\n$$A = \\begin{bmatrix} \\frac{1}{2} &\\frac{1}{2} & 0\\\\ 0& \\frac{3}{4} & \\frac{1}{4}\\\\ 0& \\frac{1}{4} & \\frac{3}{4} \\end{bmatrix}$$\n\nWhat is $$\\lim_{n\u2192\\infty}A^n$$ ?\n\nA)$$\\begin{bmatrix} 0 & 0 & 0\\\\ 0& 0 & 0\\\\ 0 & 0 & 0 \\end{bmatrix}$$ B)$$\\begin{bmatrix} \\frac{1}{4} &\\frac{1}{2} & \\frac{1}{2}\\\\ \\frac{1}{4}& \\frac{1}{2} & \\frac{1}{2}\\\\ \\frac{1}{4}& \\frac{1}{2} & \\frac{1}{2}\\end{bmatrix}$$ C)$$\\begin{bmatrix} \\frac{1}{2} &\\frac{1}{4} & \\frac{1}{4}\\\\ \\frac{1}{2}& \\frac{1}{4} & \\frac{1}{4}\\\\ \\frac{1}{2}& \\frac{1}{4} & \\frac{1}{4}\\end{bmatrix}$$ D)$$\\begin{bmatrix} 0 &\\frac{1}{2} & \\frac{1}{2}\\\\ 0 & \\frac{1}{2} & \\frac{1}{2}\\\\ 0 & \\frac{1}{2} & \\frac{1}{2}\\end{bmatrix}$$ E) The limit exists, but it is none of the above\n\nThe given answer is D). How does one arrive at this result?\n\n\u2022 Did you try doing for small values of $n$. Take $n=2, 3,4$ and post your observations (if any) as well. \u2013\u00a0Vizag May 24 '19 at 11:12\n\nBy this question, we know that\n\n$$$$A^n= \\begin{pmatrix} 2^{-n} & n\\cdot 2^{-n-1} - 2^{-n-1} + \\frac12 & {1-\\frac{n+1}{2^n}\\over2}\\\\ 0 & {2^{-n}+1\\over2} & {1-2^{-n}\\over2} \\\\ 0 & {1-2^{-n}\\over2} & {2^{-n}+1\\over2} \\end{pmatrix}.$$$$\n\nIt is thus clear that $$\\lim_{n\\to\\infty} A^n = \\begin{pmatrix} 0 &\\frac{1}{2} & \\frac{1}{2}\\\\ 0 & \\frac{1}{2} & \\frac{1}{2}\\\\ 0 & \\frac{1}{2} & \\frac{1}{2}\\end{pmatrix}$$.\n\n\u2022 $2^{-n}$ tends to 0, when $n->\\infty$.....right?? \u2013\u00a0Srestha May 25 '19 at 19:27\n\u2022 @Srestha that is correct \u2013\u00a0Maximilian Janisch May 25 '19 at 19:40\n\nIf you are in $$1$$, you have same probability to stay there or to pass to $$2$$, but no way to get back from there. Thus you are finally drifting to $$2$$.\n\nStates $$2$$ and $$3$$ are symmetrical: at long they will tend to be equally populated, independently of the starting conditions.\n\nTherefore also starting from $$1$$ you will at long be split between $$2$$ and $$3$$.\n\nIt\u2019s often worth examining a matrix for obvious eigenvectors and eigenvalues, especially in artificial exercises, before plunging into computing and solving the characteristic equation. From the first column of $$A$$, we see that $$(1,0,0)^T$$ is an eigenvector with eigenvalue $$\\frac12$$. The rows of $$A$$ all sum to $$1$$, so $$(1,1,1)$$ is an eigenvector with eigenvalue $$1$$. The remaining eigenvalue $$\\frac12$$ can be found by examining the trace.\n$$A$$ is therefore similar to a matrix of the form $$J=D+N$$, where $$D=\\operatorname{diag}\\left(1,\\frac12,\\frac12\\right)$$ and $$N$$ is nilpotent of order no greater than 2. (If $$A$$ is diagonalizable, then $$N=0$$.) $$D$$ and $$N$$ commute, so expanding via the Binomial Theorem, $$(D+N)^n=D^n+nND^{n-1}$$. In the limit, $$D^n=\\operatorname{diag}(1,0,0)$$ and the first column of $$N$$ is zero, so the second term vanishes. Thus, if $$A=PJP^{-1}$$, then $$\\lim_{n\\to\\infty}A^n=P\\operatorname{diag}(1,0,0)P^{-1}$$, but the right-hand side is just the projector onto the eigenspace of $$1$$. Informally, repeatedly multiplying a vector by $$A$$ leaves that vector\u2019s component in the direction of $$(1,1,1)^T$$ fixed, while the remainder of the vector eventually dwindles away to nothing.\nSince $$1$$ is a simple eigenvalue, there\u2019s a shortcut for computing this projector that doesn\u2019t require computing the change-of-basis matrix $$P$$: if $$\\mathbf u^T$$ is a left eigenvector of $$1$$ and $$\\mathbf v$$ a right eigenvector, then the projector onto the right eigenspace of $$1$$ is $${\\mathbf v\\mathbf u^T\\over\\mathbf u^T\\mathbf v}.$$ (This formula is related to the fact that left and right eigenvectors with different eigenvalues are orthogonal.) We already have a right eigenvector, and a left eigenvector is easily found by inspection: the last two columns both sum to $$1$$, so $$(0,1,1)$$ is a left eigenvector of $$1$$. This gives us $$\\lim_{n\\to\\infty}A^n = \\frac12\\begin{bmatrix}1\\\\1\\\\1\\end{bmatrix}\\begin{bmatrix}0&1&1\\end{bmatrix} = \\begin{bmatrix}0&\\frac12&\\frac12\\\\0&\\frac12&\\frac12\\\\0&\\frac12&\\frac12\\end{bmatrix}.$$", "date": "2021-06-15 19:11:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3238042/limit-of-powers-of-3-times3-matrix/3239827", "openwebmath_score": 0.9713488221168518, "openwebmath_perplexity": 219.66960025485517, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406019226685, "lm_q2_score": 0.8652240738888188, "lm_q1q2_score": 0.8569530525404251}}
{"url": "http://mathematica.stackexchange.com/questions/15047/how-to-draw-fractal-images-of-iteration-functions-on-the-riemann-sphere?answertab=votes", "text": "How to draw Fractal images of iteration functions on the Riemann sphere?\n\nProf. McClure, in the work \"M. McClure, Newton's method for complex polynomials. A preprint version of a \u201cMathematical graphics\u201d column from Mathematica in Education and Research, pp. 1\u201315 (2006)\", discusses how Mathematica can be applied to iteration functions for obtaining the basins of attraction (or their fractal images). Below, I provide his code for the fractal image of the polynomial $p(z)=z^3-1$:\n\np[z_] := z^3 - 1;\ntheRoots = z /. NSolve[p[z] == 0, z]\ncp = Compile[{{z, _Complex}}, Evaluate[p[z]]];\nn = Compile[{{z, _Complex}}, Evaluate[Simplify[z - p[z]/p'[z]]]];\nbail = 150;\norbitData = Table[\nNestWhileList[n, x + I y, Abs[cp[#]] > 0.01 &, 1, bail],\n{y, -1, 1, 0.01}, {x, -1, 1, 0.01}\n];\nnumRoots = Length[Union[theRoots]];\nsameRootFunc = Compile[{{z, _Complex}}, Evaluate[Abs[3 p[z]/p'[z]]]];\nwhichRoot[orbit_] :=\nModule[{i, z},\nz = Last[orbit]; i = 1;\nScan[If[Abs[z - #] < sameRootFunc[z], Return[i], i++] &, theRoots];\nIf[i <= numRoots, {i, Length[orbit]}, None]\n];\nrootData = Map[whichRoot, orbitData, {2}];\ncolorList = {{cc, 0, 0}, {cc, cc, 0}, {0, 0, cc}};\ncols = rootData /. {\n{k_Integer, l_Integer} :> (colorList[[k]] /. cc -> (1 - l/(bail + 1))^8),\nNone -> {0, 0, 0}\n};\nGraphics[{Raster[cols]}]\n\n\nMy main question is here. He nicely obtained the fractal images on the complex plane, while it would be an interesting challenge to obtain these images on the Riemann sphere, e.g.\n\nIt seems the complex plane in this case has been replaced by a sphere, but how? I will be thankful if someone could revise the code given above for obtaining such beautiful fractal images on the Riemann sphere. Any tips and tricks will be fully appreciated as well.\n\n-\nWould you care to share where your example spherical projection came from? \u2013\u00a0 Mark McClure Dec 24 '12 at 13:04\n\nAs the other answers have shown, it's fairly easy to map an image onto a parametrized surface using textures. It can be a bit tricky, though, getting the image to mesh well with the transformation. J.M. hit on the crucial issue, namely that we compute the image using points that map to the sphere with minimal distortion. This answer is largely an expansion on his, although there are some differences and other ideas as well.\n\nFirst, the article that Fazlollah refers to is some years old now, and the code can be improved in light of the many changes since V5, so let's start by showing how to generate regular Newton iteration images for general polynomials. Given a polynomial function $f(z)$, the following code computes the corresponding Newton's method iteration function $n$. It then defines the command limitInfo that iterates $n$ up to $50$ times from a starting point $z_0$ terminating when $|f(z)|$ is small and returning the last iterate and the number of iterates required for $|f(z)|$ to get small. It's compiled, listable and set to run in parallel, so it should be pretty fast. In addition to the function f, there are two numeric parameters to set, bail and r.\n\nf = Function[z, z^3 - 1]; (* A very standard example *)\nn = Function[z, Evaluate[Simplify[z - f[z]/f'[z]]]];\nbail = 50;\n(* bail is the number of iterates before bailing. *)\n(* Doesn't have to be particularly large, *)\n(* if there are only simple roots. *)\nr = 0.01;\n(* We assume that if |z-z0|<r, then we've *)\n(* converged to the root z0. *)\nlimitInfo = With[{bail = bail, r = r, f = f, n = n},\nCompile[{{z0, _Complex}},\nModule[{z, cnt},\ncnt = 0; z = z0;\nWhile[Abs[f[z]] > r && cnt < bail,\nz = n[z];\ncnt = cnt + 1\n];\n{z, cnt}],\nRuntimeAttributes -> {Listable},\nParallelization -> True,\nRuntimeOptions -> \"Speed\"\n]];\n\n\nSince the function is listable and runs in parallel, we can simply apply it to a table of data on one fell swoop.\n\nstep = 4/801; (* The denominator is essentially the resolution. *)\nlimitData = limitInfo[\nTable[x + I*y, {y, 2, -2, -step}, {x, -2, 2, step}]]; // AbsoluteTiming\n\n(* Out: {1.492716, Null} *)\n\n\nEach element is a pair that indicates the limiting behavior and how long it took to get there.\n\nlimitData[[1, 1]]\n\n(* Out: {-0.499835 + 0.866044 I, 5. + 0. I} *)\n\n\nI guess we need a function that takes something like that and turns it into a color.\n\nroots = z /. NSolve[f[z] == 0, z];\npreColors = List @@@ Table[ColorData[61, k], {k, 1, Length[roots]}];\npreColors = Append[preColors, {0.0, 0.0, 0.0}];\ncolor = With[{bail = bail, roots = roots, preColors = preColors},\nCompile[{{z, _Complex}, {cnt, _Complex}},\nModule[{arg, time, i},\narg = Arg[z];\ntime = Abs[cnt];\ni = 1;\nScan[If[Abs[z - #] < 0.1, Return[i], i++] &, roots];\nAbs[preColors[[i]]*(cnt/bail)^(0.2)]\n(* The exponent 0.2 adjusts the brightness of the image. *)]]\n];\n\n\nNow, we apply that function and generate the image.\n\ncolors = Apply[color, limitData, {2}];\nImage[colors, ImageSize -> 2/step]\n\n\nTo map onto a sphere nicely, we'll discard the rectangular grid of points that we used above in favor of a collection of points that looks something like the following (although, we'll want higher resolution, of course):\n\nstep = Pi/12;\npts = Table[Cot[phi/2] Exp[I*theta],\n{phi, step, Pi - step, step}, {theta, -Pi, Pi, step}];\nListPlot[{Re[#], Im[#]} & /@ Flatten[pts, 1],\nAspectRatio -> Automatic, PlotRange -> All,\nEpilog -> {Red, Circle[]}]\n\n\nThe expression $\\cot(\\phi/2) e^{i\\theta}$ is the stereographic projection of a point expressed in spherical coordinates $(1,\\phi,\\theta)$ onto the plane. As a result, the corresponding points on the sphere are nicely distributed. Note, for example, that the number of points inside and outside of the unit circle are the same.\n\nGraphics3D[{{Opacity[0.8], Sphere[]},\nPoint[Flatten[Table[{Cos[theta] Sin[phi], Sin[theta] Sin[phi], Cos[phi]},\n{phi, step, Pi - step, step}, {theta, -Pi, Pi, step}], 1]]}]\n\n\nNow, we increase the resolution and use the same limitInfo and color functions as before.\n\nstep = Pi/500;\nlimitData = limitInfo[Table[Cot[phi/2] Exp[I*theta],\n{phi, step, Pi - step, step}, {theta, -Pi, Pi, step}]];\ncolors = Apply[color, limitData, {2}];\nrect = Image[colors, ImageSize -> 4/step]\n\n\nThe image looks a bit different, but it's perfect for use as a spherical texture.\n\nParametricPlot3D[{Cos[theta] Sin[phi], Sin[theta] Sin[phi], Cos[phi]} ,\n{theta, -Pi, Pi}, {phi, 0, Pi}, Mesh -> None, PlotPoints -> 100,\nBoxed -> False, PlotStyle -> Texture[Show[rect]],\nLighting -> \"Neutral\", Axes -> False]\n\n\nWe can incorporate all of this into a Module.\n\nnewtonSphere[fIn_, var_, resolution_, bail_: 50, r_: 0.01] := Module[\n{f, n, limitInfo, color, colors, roots, preColors, step, limitData, rect},\n\nf = Function[var, fIn];\nn = Function[var, Evaluate[Simplify[var - f[var]/f'[var]]]];\nlimitInfo = With[{bailLoc = bail, rLoc = r, fLoc = f, nLoc = n},\nCompile[{{z0, _Complex}},\nModule[{z, cnt},\ncnt = 0; z = z0;\nWhile[Abs[fLoc[z]] > rLoc && cnt < bailLoc,\nz = nLoc[z];\ncnt = cnt + 1\n];\n{z, cnt}],\nRuntimeAttributes -> {Listable},\nParallelization -> True,\nRuntimeOptions -> \"Speed\"\n]];\nroots = z /. NSolve[f[z] == 0, z];\npreColors = List @@@ Table[ColorData[61, k], {k, 1, Length[roots]}];\npreColors = Append[preColors, {0.0, 0.0, 0.0}];\ncolor = With[{bailLoc = bail, rootsLoc = roots, preColorsLoc = preColors},\nCompile[{{z, _Complex}, {cnt, _Complex}},\nModule[{arg, time, i},\narg = Arg[z];\ntime = Abs[cnt];\ni = 1;\nScan[If[Abs[z - #] < 0.1, Return[i], i++] &, rootsLoc];\npreColorsLoc[[i]]*(cnt/bailLoc)^(0.2)\n]]\n];\nstep = Pi/resolution;\nlimitData = limitInfo[\nTable[Cot[phi/2] Exp[I*theta], {phi, step, Pi - step,\nstep}, {theta, -Pi, Pi, step}]];\ncolors = Apply[color, limitData, {2}];\nrect = Image[colors, ImageSize -> 4/step];\nParametricPlot3D[{Cos[theta] Sin[phi], Sin[theta] Sin[phi],\nCos[phi]} ,\n{theta, -Pi, Pi}, {phi, 0, Pi}, Mesh -> None, PlotPoints -> 100,\nBoxed -> False, PlotStyle -> Texture[Show[rect]],\nLighting -> \"Neutral\", Axes -> False]\n];\n\n\nNow, if I had to guess, I'd say that example image in the original post was generated by a small perturbation of $z^8-z^2$.\n\nnewtonSphere[(2 z/3)^8 - (2 z/3)^2 + 1/10, z, 500]\n\n\nHere are a few more examples.\n\npic1 = newtonSphere[z^2 - 1, z, 401];\nSeedRandom[1];\npic2 = newtonSphere[Sum[RandomInteger[{-3, 5}] z^k, {k, 0, 8}], z, 400];\npic3 = newtonSphere[z^10 - z^5 - 1, z, 400, 200];\npic4 = newtonSphere[z^5 - z - 0.99, z, 400];\nGraphicsGrid[{\n{pic1, pic2},\n{pic3, pic4}\n}]\n\n\nIn the top row, we see that the result for quadratic polynomials is typically rather boring while that for a random degree 8 polynomial can be quite cool. On the bottom right, we see a black region. The color function is setup to default to black when none of the roots are detected. This can certainly happen; in fact the Newton iteration function for this example has an attractive orbit of period 6 leading to the quadratic like Julia set seen in the image. Sometimes black can occur simply because we didn't iterate enough, which is why I used the optional fourth argument for the image in the bottom left.\n\n-\nNow imagine those hanging from a fractal christmas tree. (+1) \u2013\u00a0 Jens Dec 24 '12 at 17:40\n\nHere is my modest attempt, based on the formulae for stereographic projection in this Wikipedia entry (where the north pole corresponds to the point at infinity) and using a technique similar to the one in this answer:\n\nnewtonRaphson = Compile[{{n, _Integer}, {c, _Complex}},\nArg[FixedPoint[(# - (#^n - 1)/(n #^(n - 1))) &, c, 30]]]\n\ntex = Image[DensityPlot[\nnewtonRaphson[3, Cot[\u03d5/2] Exp[I \u03b8]], {\u03b8, -\u03c0, \u03c0}, {\u03d5, 0, \u03c0},\nAspectRatio -> Automatic,\nColorFunction -> (Which[# < .3, Red, # > .7, Yellow, True, Blue] &),\nFrame -> False, ImagePadding -> None, PlotPoints -> 400,\nPlotRange -> All, PlotRangePadding -> None],\nImageResolution -> 256];\n\n(* yes, I know that I could have used SphericalPlot3D[]... *)\nParametricPlot3D[{Sin[\u03d5] Cos[\u03b8], Sin[\u03d5] Sin[\u03b8], Cos[\u03d5]}, {\u03b8, -\u03c0, \u03c0}, {\u03d5, 0, \u03c0},\nAxes -> None, Boxed -> False, Lighting -> \"Neutral\", Mesh -> None,\nPlotStyle -> Texture[tex], TextureCoordinateFunction -> ({#4, #5} &)]\n\n\n-\nThanks for your response. There are two problems. 1. How can one zoom in on a particluar place on this sphere without lowering the quality to observe the fractal behaviour of the method? 2. The \"spcae size\" of the output image? In fact, how one can save as the output fractal image with low \"disk size\" without lowering the quality in EPS format? For example, for $n=8$, its size is more than 2MB! \u2013\u00a0 Fazlollah Soleymani Nov 22 '12 at 21:16\nYou'll have to play with PlotPoints and ImageResolution on your own, of course... \u2013\u00a0 Guess who it is. Nov 22 '12 at 23:13\nThanks, the problem is that when we reduce PlotPoints or ImageResolution, the quality gets down dramatically. I am looking for a fast way to obtain high quality pics, with small space size, just like the one given in the question. Rasterize@... is a good choice, but it disables the feature of rotating the 3D pic, and also gets the quality lower. \u2013\u00a0 Fazlollah Soleymani Nov 23 '12 at 8:20\nWell, I'd say this is one of those \"no such thing as a free lunch\" things. If you want to be able to zoom, you certainly need high resolution, which will demand more of your computer's resources... if you want small file sizes, you'll have to sacrifice quality somewhat. Scylla and Charybdis, you know... \u2013\u00a0 Guess who it is. Nov 23 '12 at 9:49\nimg2 = ImageCrop[Image[Graphics[{Raster[cols]}, PlotRangePadding -> 0,\nImagePadding -> 0, ImageMargins -> 0]], {343, 343}];\n\nSphericalPlot3D[1 , {u, 0, Pi}, {v, 0, 2 Pi}, Mesh -> None,\nTextureCoordinateFunction -> ({#1, #2} &),\nPlotStyle -> Directive[Specularity[White, 10], Texture[img2]],\nLighting -> \"Neutral\", Axes -> False, ImageSize -> 500]\n\n\nwhere img2 is cropped version of the 2D image in OP's question.\n\nCheck Texture for more examples.\n\n-\nThanks for your reply, but there are big white circles at the middle of each basin. They should not be here. Please rotate your image, and then you will see the incomplete fractal image. Can you solve this drawback? \u2013\u00a0 Fazlollah Soleymani Nov 22 '12 at 20:59\nimg2 is the cropped version of your 2D image: img2=ImageCrop[ Image[Graphics[{Raster[cols]}, PlotRangePadding -> 0, ImagePadding -> 0, ImageMargins -> 0]], {343, 343}]. \u2013\u00a0 kglr Nov 22 '12 at 21:50\nNow that I think about it, one could have directly produced an image instead of going through the Raster[] route: Image[cols]. \u2013\u00a0 Guess who it is. Nov 24 '12 at 14:12\n\nI've decided to write a simplification+extension of Mark's routine as a separate answer. In particular, I wanted a routine that yields Riemann sphere fractals not only for Newton-Raphson, but also its higher-order generalizations (e.g. Halley's method).\n\nI decided to use Kalantari's \"basic iteration\" family for the purpose. An $n$-th order member of the family looks like this:\n\n$$x_{k+1}=x_k-f(x_k)\\frac{\\mathcal D_{n-1}(x_k)}{\\mathcal D_n(x_k)}$$\n\nwhere\n\n$$\\mathcal D_0(x_k)=1,\\qquad\\mathcal D_n(x_k)=\\begin{vmatrix}f^\\prime(x_k)&\\tfrac{f^{\\prime\\prime}(x_k)}{2!}&\\cdots&\\tfrac{f^{(n-2)}(x_k)}{(n-2)!}&\\tfrac{f^{(n-1)}(x_k)}{(n-1)!}\\\\f(x_k)&f^\\prime(x_k)&\\ddots&\\vdots&\\tfrac{f^{(n-2)}(x_k)}{(n-2)!}\\\\&f(x_k)&\\ddots&\\ddots&\\vdots\\\\&&\\ddots&\\ddots&\\vdots\\\\&&&f(x_k)&f^\\prime(x_k)\\end{vmatrix}$$\n\nAs noted in that paper, the basic family generalizes the Newton-Raphson iteration; $n=1$ corresponds to Newton-Raphson, while $n=2$ gives Halley's method. (Relatedly, see also Kalantari's work on polynomiography.)\n\nHere's a routine for $\\mathcal D_n(x)$:\n\niterdet[f_, x_, 0] := 1;\niterdet[f_, x_, n_Integer?Positive] := Det[ToeplitzMatrix[PadRight[{D[f, x], f}, n],\nTable[SeriesCoefficient[Function[x, f]@\\[FormalX], {\\[FormalX], x, k}], {k, n}]]]\n\n\nHere is the routine for generating the Riemann sphere fractals:\n\nOptions[rootFractalSphere] = {ColorFunction -> Automatic, ImageResolution -> 400,\nMaxIterations -> 50, Order -> 1, Tolerance -> 0.01};\n\nrootFractalSphere[fIn_, var_, opts : OptionsPattern[]] /; PolynomialQ[fIn, var] :=\nModule[{\u03b3 = 0.2, bail, cf, colList, f, h, itFun, ord, roots, tex, tol},\n\nf = Function[var, fIn];\nord = OptionValue[Order];\nitFun = Function[var, var - Simplify[f[var] iterdet[f[var], var, ord - 1]/\niterdet[f[var], var, ord]] // Evaluate];\n\nroots = var /. NSolve[f[var], var];\ncf = OptionValue[ColorFunction];\nIf[cf === Automatic, cf = ColorData[61]];\ncolList = Append[Table[List @@ ColorConvert[cf[k], RGBColor], {k, Length[roots]}],\n{0., 0., 0.}];\n\nbail = OptionValue[MaxIterations]; tol = OptionValue[Tolerance];\nmakeColor = Compile[{{z0, _Complex}},\nModule[{cnt = 0, i = 1, z},\nz = FixedPoint[(++cnt; itFun[#]) &, z0, bail,\nSameTest -> (Abs[f[#2]] < tol &)];\nScan[If[Abs[z - #] < 10 tol, Return[i], i++] &, roots];\nAbs[colList[[i]] (cnt/bail)^\u03b3]],\nCompilationOptions -> {\"InlineExternalDefinitions\" -> True},\nRuntimeAttributes -> {Listable}, RuntimeOptions -> \"Speed\"];\n\nh = \u03c0/OptionValue[ImageResolution];\ntex = DeveloperToPackedArray[makeColor[\nTable[Cot[\u03c6/2] Exp[I \u03b8], {\u03c6, h, \u03c0 - h, h}, {\u03b8, -\u03c0, \u03c0, h}]]];\n\nParametricPlot3D[{Cos[\u03b8] Sin[\u03c6], Sin[\u03b8] Sin[\u03c6], Cos[\u03c6]}, {\u03b8, -\u03c0, \u03c0}, {\u03c6, 0, \u03c0},\nAxes -> False, Boxed -> False, Lighting -> \"Neutral\", Mesh -> None,\nPlotPoints -> 75, PlotStyle -> Texture[tex],\nEvaluate[Sequence @@ FilterRules[{opts}, Options[Graphics3D]]]]]\n\n\nOther notes:\n\n\u2022 The compiled functions limitInfo[] and color[] have been merged into the single function makeColor[]. This function was not localized on purpose to allow its use even after executing rootFractalSphere[].\n\n\u2022 Texture[] can directly accept an array of RGB triplets, so there is no need to use Image[] if these triplets are being generated directly by makeColor[].\n\nNow, for some examples. The first two are Newton-Raphson fractals:\n\nrootFractalSphere[z^3 - 1, z]\n\n\nrootFractalSphere[(2 z/3)^8 - (2 z/3)^2 + 1/10, z]\n\n\nHere is a fractal generated by Halley's method:\n\nrootFractalSphere[(2 z/3)^8 - (2 z/3)^2 + 1/10, z, Order -> 2]\n\n\nFinally, a fractal from a third order iteration:\n\nrootFractalSphere[z^10 - z^5 - 1, z, ColorFunction -> ColorData[54],\nMaxIterations -> 200, Order -> 3]\n`\n\n-\nThanks. An excellent answer. Just one note. There might be some diverging points (black areas) in the fractal picture for some test problems. On the other hand, we used a color correspond to a point in a sphere or a rectangular domain. So, the domain of working (I mean the mesh of points) are finite. So is it possible to count the number of diverging points? I mean it would be nice to have the percentage of diverging points for each fractal picture. Is it possible to cunt the number of diverging points in your implementation? \u2013\u00a0 Fazlollah Soleymani Apr 6 '13 at 12:25", "date": "2015-07-01 23:25:08", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/15047/how-to-draw-fractal-images-of-iteration-functions-on-the-riemann-sphere?answertab=votes", "openwebmath_score": 0.42025870084762573, "openwebmath_perplexity": 4975.400829917264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97112909472487, "lm_q2_score": 0.8824278788223265, "lm_q1q2_score": 0.8569513871207133}}
{"url": "http://descarga.nu/ujwn9x/quadratic-trinomial-examples-eb1a1a", "text": "The Distributive Law is used in reverse to factorise a quadratic trinomial, as illustrated below.. We can use the box method to factorise a quadratic trinomial. For example: Here b = \u20132, and c = \u201315. But a \"trinomial\" is any three-term polynomial, which may not be a quadratic (that is, a degree-two) polynomial. Free online science printable worksheets for year 11, solve quadratic java, sample algebra test solving addition equations, College algebra tutorial. Trinomial. Examples are 7a2 + 18a - 2, 4m2, 2x5 + 17x3 - 9x + 93, 5a-12, and 1273. To figure out which it is, just carry out the O + I from FOIL. A quadratic trinomial is a polynomial with three terms and the degree of the trinomial must be 2. Think FOIL. Example 1; Example 2; Example 3; Example 4; Example 5; Example 1 Example. Quadratic equation of leading coefficient 1. Let\u2019s consider two cases:\u00a0 (1) Leading coefficient is one, a = 1, and (2) leading coefficient is NOT 1, a \u2260 1. Factor a Quadratic Trinomial. So (3x5)2 = 9x10. x2 + 20x + ___ x2 - 4x + ___ x2 + 5x + ___ 100 4 25/4 \u2026 Example 6: A quadratic relation has an equation in factored form. Yes, \u2026 Vocabulary. This math video tutorial shows you how to factor trinomials the easy fast way. By the end of this section, you will be able to: Factor trinomials of the form ; Factor trinomials of the form ; Before you get started, take this readiness quiz. Do you see how all three terms are present? The x-intercepts of the parabola are \u2212 4 and 1. Again, think about FOIL and where each term in the trinomial came from. For example: $$x^2 + y^2 + xy$$ and $$x^2 + 2x + 3xy$$. The general form of a quadratic equation is. Expand the equation (2x \u2013 3) 2 = 25 to get; 4x 2 \u2013 12x + 9 \u2013 25 = 0 4x 2 \u2013 12x \u2013 16 = 0. Example \u2026 You need to think about where each of the terms in the trinomial came from. Simplify: \u24d0 \u24d1 If you missed this problem, review . The general form of a quadratic trinomial is ax 2 + bx + c, where a is the leading coefficient (number in front of the variable with highest degree) and c is the constant (number with no variable). Log base change on the TI-89, cube root graph, adding/subtracting positive and negative numbers, java applet factoring mathematics algebra, trigonometry questions, algebra1 prentice hall. Following is an explanation of polynomials, binomials, trinomials, and degrees of a polynomial. Remember, when a term with an exponent is squared, the exponent is multiplied by 2, the base is squared. In a quadratic equation, leading coefficient is nothing but the coefficient of x 2. For example, x + 2. The answer would be 5 and 6. Step 1:\u00a0 Identify if the trinomial is in quadratic form. We require two numbers that multiply to \u2013 18 and add to 7. ax 2 + bx + c. 3x 2 + 7x \u2013 6. ac = 3 \u00d7 \u2013 6 \u2026 Example 3. On this page we will learn what a trinomial in quadratic form is, and what a trinomial in quadratic form is not. In the next section, we will address the technique used to factor $$ax^2+bx+c$$ when $$a \\neq 1$$. A quadratic trinomial is a trinomial of which the highest power of any variable is two. Learn how to factor quadratic expressions as the product of two linear binomials. A special type of trinomial can be factored in a manner similar to quadratics since it can be viewed as a quadratic in a new variable (x n below). A binomial is a \u2026 In general g(x) = ax 2 + bx + c, a \u2260 0 is a quadratic polynomial. (2x + ? Solution . Quality resources and hosting are expensive, Creative Commons Attribution 4.0 International License. Now you\u2019ll need to \u201cundo\u201d this multiplication\u2014to start with the product and end up with the factors. Below are 4 examples of how to use algebra tiles to factor, starting with a trinomial where A=1 (and the B and C values are both positive), all the way to a trinomial with A>1 (and negative B and/or C values). Jenn, Founder Calcworkshop \u00ae, 15+ Years Experience (Licensed & Certified Teacher) Finding the degree of a polynomial is nothing more than locating the largest \u2026 To factorise a quadratic trinomial. $$\\text{Examples of Quadratic Trinomials}$$ For example, let us apply the AC test in factoring 3x 2 + 11x + 10. trinomial, as illustrated below. Factoring Trinomials Formula, factoring trinomials calculator, factoring trinomials a 1,factoring trinomials examples, factoring trinomials solver. Summary:\u00a0 A quadratic form trinomial is of the form axk + bxm + c, where 2m = k.\u00a0 It is possible that these expressions are factorable using techniques and methods appropriate for quadratic equations. It is due to the presence of three, unlike terms, namely, 3x, 6x 2 and 2x 3. That would be a \u2013 5 and a + 3. For example, 2x 2 \u2212 7x + 5. If you experience difficulties when using this Website, tell us through the feedback form or by phoning the contact telephone number. In the given trinomial, the product of A and C is 30. We begin by showing how to factor trinomials having the form $$ax^2 + bx + c$$, where the leading coefficient is a = 1; that is, trinomials having the form $$x^2+bx+c$$. Quadratic equation of leading coefficient not equal to 1. Below are 4 examples of how to use algebra tiles to factor, starting with a trinomial where A=1 (and the B and \u2026 This is a quadratic form polynomial because the second term\u2019s variable, x3, squared is the first term\u2019s variable, x6. Solving Trinomial Equations Using The Quadratic Formula, Algebra free worked examples for children in 3rd, 4th, 5th, 6th, 7th & 8th grades, worked algebra problems, solutions to algebra questions for children, algebra topics with worked exercises on , inequalities, intergers, logs, polynomials, angles, linear equations, quadratic equation, monomials & more This page will focus on quadratic trinomials. Which of the following is a quadratic? All my letters are being represented by numbers. The last term is plus. A special type of trinomial can be factored in a manner similar to quadratics since it can be viewed as a quadratic in a new variable (x n below). NCERT Solutions For Class 12 Physics; NCERT Solutions For Class 12 Chemistry ; NCERT Solutions For Class 12 Biology; NCERT Solutions For Class 12 Maths; NCERT Solutions \u2026 And the middle term's coefficient is also plus. Let\u2019s look first at trinomials with only the middle term negative. For example, the box for is: \\begin{array}{|c|c|c} \\hline x^2 & 3x & x \\\\ \\hline 2x & 6 & 2 \\\\ \\hline x & 3 \\\\ \\end{array} Therefore Factoring quadratic trinomial and how to factor by grouping. Generally we have two types of quadratic equation. What happens when there are negative terms? This will help you see how the factoring works. The solution a 1 = 2 and a 2 = 1 of the above system gives the trinomial factoring: (x 2 + 3x+ 2) = (x + a 1)(x + a 2) \u2026 Factorise 3x 2 + 7x \u2013 6. This part will focus on factoring a quadratic when a, the x 2-coefficient, is 1. Solution. x is being squared. Likewise, 11pq + 4x 2 \u201310 is a trinomial. The product of two linear factors yields a quadratic trinomial; and the D is a perfect square because it is the square of 5. The tricky part here is figuring out the factors of 8 and 30 that can be arranged to have a difference of 43. b) Write the equation in vertex form. So either -5 \u00d7 1 or 5 \u00d7 -1. Now hopefully, we have got the basic difference between Monomial, Binomial and Trinomial. If you're behind a web filter, please make sure that \u2026 Solution: Find the product of the first and the last constants. Examples, solutions, videos, worksheets, ... Scroll down the page for more examples and solutions of factoring trinomials. Solve the following quadratic equation (2x \u2013 3) 2 = 25. NCERT Solutions. Following is an example of trinomial: x 3 + x 2 + 5x 2x 4 -x 3 + 5 A trinomial meaning in math is, it is a type of polynomial that contains only three terms. For example, a univariate (single-variable) quadratic function has the form = + +, \u2260in the single variable x.The graph of a univariate quadratic function is a parabola whose axis of symmetry is parallel to the y-axis, as shown at right.. (y+a) (y+b) = y (y+b) + a (y+b) = y 2 + by + ay + ab = y 2 + y (a+b) + ab \u2026 Expand the equation (2x \u2013 3) 2 = 25 to get; 4x 2 \u2013 12x + 9 \u2013 25 = 0 4x 2 \u2013 12x \u2013 16 = 0. Perfect Square Trinomial \u2013 Explanation & Examples A quadratic equation is a polynomial of second degree usually in the form of f(x) = ax 2 + bx + c where a, b, c, \u2208 R and a \u2260 0. Donate Login \u2026 How to factor a quadratic trinomial: 5 examples and their solutions. A trinomial is a sum of three terms, while a multinomial is more than three. A trinomial is a polynomial or algebraic expression, which has a maximum of three non-zero terms. Problem 1. This part will focus on factoring a quadratic when a, the x 2 -coefficient, is 1. a x 2 + b x + c = 0 \u2192 (x + r) (x + s) Let's solve the following equation by factoring the trinomial: Quadratic is another name for a polynomial of the 2nd degree. A few examples of trinomial expressions are: \u2013 8a 4 +2x+7; 4x 2 + 9x + 7; Monomial: Binomial: Trinomial: One Term: Two terms: Three terms: Example: x, 3y, 29, x/2: Example: x 2 +x, x 3-2x, y+2: Example: x 2 +2x+20: Properties . Use the tabs below to navigate through the notes, video, and practice problems. How to factor a quadratic trinomial: 5 examples and their solutions. In this post, I want to focus on that last topic -- using algebra tiles to factor quadratic trinomials. Remember: To get a negative sum and a positive product, the numbers must both be negative. ax 2 + bx + c = 0. (Lesson 13: Exponents.) For example, the polynomial (x 2 + 3x + 2) is an example of this type of trinomial with n = 1. )(x + ?) Australian Business Number\u00a053 056 217 611, Copyright instructions for educational institutions. So the book's section or chapter title is, at best, a bit off-target. Start from finding the factors of +2. This form is factored as: + + = (+) (+), where + = \u22c5 =. Let\u2019s see another example, here where a is not one. 15 Factor Quadratic Trinomials with Leading Coefficient 1 Learning Objectives. $$(x \u2212 5)$$ and $$(x + 3)$$ are factors of $$x^2 \u2212 2x \u2026 A quadratic trinomial is factorable if the product of A and C have M and N as two factors such that when added would result to B. c) Sketch a graph of the relation and label all features. Quadratic Polynomial. Since (x2)2 = x4, and the second term is x4, then n = 2. Non-Example: These trinomials are not examples of quadratic form. Step 3: Apply the appropriate factoring technique. Example 1. Binomials. For more practice on this technique, please visit this page. A binomial is a sum of two terms. Factorise by grouping the four terms into pairs. Website and our Privacy and Other Policies. Previously, we went over how to factor out a quadratic trinomial with a leading coefficient of 1. They take a lot of the guesswork out of factoring, especially for trinomials that are not easily factored with other methods. 6 or D = 25. If a polynomial P(x) is \u2026 It is the correct pair \u2026 NCERT Solutions For Class 12. Let\u2019s look at this quadratic form trinomial and a quadratic with the same coefficients side by side. Example 1: Factor the trinomial x^2+7x+10 x2 + 7x + 10 as a product of two binomials. Obviously, this is an \u201ceasy\u201d case because the coefficient of the squared term x x is just 1. If a is one, then we just need to find what two numbers have the product c and the sum of b. Factorising an expression is to write it as a product of its factors. Well, it depends which term is negative. In this quadratic, 3x 2 + 2x \u2212 1, the constants are 3, 2, \u22121. Factoring quadratic trinomials using the AC Method. So, n = 3. For example, 2x\u00b2+7x+3=(2x+1)(x+3). Factoring Trinomials (Quadratics) : Method With Examples Consider the product of the two linear expressions (y+a) and (y+b). Don't worry about the difference, though; the book's title means \u2026 It means that the highest power of the variable cannot be greater than 2. 6, the independent term, is the product of 2 and 3. If you\u2019re a teacher and would like to use the materials found on this page, click the teacher button below. The argument appears in the middle term. Then, find the two factors of 30 that will produce a sum of 11. A polynomial formed by the sum of only three terms (three monomials) with different degrees is known as a trinomial. Let\u2019s begin with an example. It consists of only three variables. write the expression in the form ax 2 + bx + c; find two numbers that both multiply to ac and add to b; split the middle term bx into two like terms using those two numbers as coefficients. Let\u2019s look at an example of multiplying binomials to refresh your memory. There are 4 methods: common factor, difference of two squares, trinomial/quadratic expression and completing the square. A polynomial is an algebraic expression with a finite number of terms. A quadratic trinomial is a trinomial in which the highest exponent or power is two, or the second power. Please read the Terms and Conditions of Use of this Quadratic trinomials with a leading coefficient of one. Factoring quadratic trinomial and how to factor by grouping. a, b, c are called constants. factors, | Home Page | Order Maths Software | About the Series | Maths Software Tutorials | The x-intercepts of the parabola are \u2212 4 and 1. Let's take an example. FACTORING 2. Example are: 2x 2 + y + z, r + 10p + 7q 2, a + b + c, 2x 2 y 2 + 9 + z, are all trinomials having three variables. A polynomial having its highest degree 2 is known as a quadratic polynomial. Tie together everything you learned about quadratic factorization in order to factor various quadratic expressions of any form. This video contains plenty of examples and practice ... Factoring Perfect Square Trinomials Factoring Perfect Square Trinomials door The Organic Chemistry Tutor 4 jaar geleden 11 minuten en 3 seconden 267.240 weergaven This algebra video tutorial focuses on , factoring , perfect square , trinomials , . In general, the trinomial of the ax 2 + bx + c is a perfect square if the discriminant is zero; that is, if b 2 -4ac = 0, because in this case it will only have one root and can be expressed in the form a (xd) 2 = (\u221aa (xd)) 2 , where d is the root already mentioned. COMPLETING THE SQUARE 3. Consider the expansion of (x + 2)(x + 3).We notice that: 5, the coefficient of x, is the sum of 2 and 3.; 6, the independent term, is the product of 2 and 3.; Note: The product of two linear factors yields a quadratic trinomial; and the factors of a quadratic trinomial are linear factors.. Now consider the expansion of \u2026 There is one last factoring method you\u2019ll need for this unit: Factoring quadratic form polynomials. So, n = 5. Exercise 2.1. factors of a quadratic trinomial are linear factors. There are three main ways of solving quadratic equations: 1. Example 3. \\((x \u2212 5)(x + 3) = x^2 \u2212 2x \u2212 15$$ Here, we have multiplied two linear factors to obtain a quadratic expression by using the distributive law. Factoring Polynomials - Standard Trinomials (Part 1) Factoring Polynomials of the form ax \u2026 Solution. Consider making your next Amazon purchase using our Affiliate Link. Factoring quadratic is an approach to find the roots of a quadratic equation. A polynomial having its highest degree 3 is known as a Cubic polynomial. Year 10 Interactive Maths - Second Edition. Example: x 2 - 12x + 27. a = 1 b = -12 c = 27. Solving quadratic equations by factoring is all about writing the quadratic function as a product of two binomials functions of one degree each. Divide each term by 4 to get; x 2 \u2013 3x \u2013 4 = 0 (x \u2013 4) (x + 1) = 0 x = 4 or x = -1. Some of the important properties of polynomials along with some important polynomial theorems are as follows: Property 1: Division Algorithm. a + b. Multiply: If you missed this problem, review . b) Write the equation in vertex form. FACTORING QUADRATIC TRINOMIALS Example 4 : X2 + 11X - 26 Step 2 Factor the first term which is x2 (x )(x ) Step 4 Check the middle term (x + 13)(x - 2) 13x multiply 13 and x + -2x multiply -2 and x 11x Add the 2 terms. The numbers that multiply to \u2013 50 and add to + 5 are \u2013 5 and + 10. If you need a refresher on factoring quadratic equations, please visit this page. Each factor is a difference of squares! quadratic trinomial, independent term, coefficient, linear factor Simplify: \u24d0 \u24d1 If you missed this problem, review . Here is the form of a quadratic trinomial with argument x: ax 2 + bx + c. The argument is whatever is being squared. Let\u2019s factor a quadratic form trinomial where a = 1. The term \u2018a\u2019 is referred to as the leading coefficient, while \u2018c\u2019 is referred to as the absolute term of f (x). These terms are in the form \u201caxn\u201d where \u201ca\u201d is a real number, \u201cx\u201d means to multiply, and \u201cn\u201d is a non-negative integer. Polynomials. Choose the correct \u2026 Australian Business Number\u00a053 056 217 611. In Equation (i), the product of coefficient of y 2 and the constant term = ab and the coefficient of y = a+b = sum of the factors of \u2026 The \u2026 Solving Quadratic Equations by Factoring with a Leading Coefficient of 1 - Procedure (i) In a quadratic \u2026 If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Factorising an expression is to write it as a product of its factors. Solution: Check: Key Terms. It does not mean that a quadratic trinomial always turns into a quadratic equation when we equate it to zero. x is being squared. It is called \"Factoring\" because we find the factors (a factor is something we multiply by) Example: Multiplying (x+4) and (x\u22121) together (called Expanding ) gets x 2 + 3x \u2212 4: So (x+4) and (x\u22121) are factors of x 2 + 3x \u2212 4. There are 4 methods: common factor, difference of two squares, trinomial/quadratic expression and completing the square. Types of Quadratic Trinomials . 10 Surefire Video Examples! An example of a quadratic trinomial is 2x^2 + 6x + 4. Factors of Quadratic Trinomials of the Type x2 + bx + c. The Distributive Law is used in reverse to factorise a quadratic trinomial, as illustrated below. For example- 3x + 6x 2 \u2013 2x 3 is a trinomial. To \"Factor\" (or \"Factorise\" in the UK) a Quadratic is to: find what to multiply to get the Quadratic . Show Step-by-step Solutions. 5, the coefficient of x, is the sum of 2 and 3. 6, the independent term, is the product of 2 and 3. In other words, if you have a trinomial with a constant term, and the larger exponent is double of the first exponent, the trinomial is in quadratic form. Tie together everything you learned about quadratic factorization in order to factor various quadratic expressions of any form. In the examples so far, all terms in the trinomial were positive. The product\u2019s factor pair that when added yields the middle constant, \u20138 is \u201314 and 6. The middle term's coefficient is plus. Just as before, the first \u2026 However, this quadratic form polynomial is not completely factored. 1. How To Factorize Quadratic Expressions? = 2x2 + \u2026\u00a0 The last term, \u2013 5, comes from the L, the last terms of the polynomials. Now here is a quadratic whose argument is x 3: 3x 6 + 2x 3 \u2212 1. x 6 is the square of x 3. Example 12. To see the answer, pass your mouse over the colored area. Hence, the given trinomial is factorable. Study Materials. 1) x2 \u2212 7x \u2212 18 2) p2 \u2212 5p \u2212 14 3) m2 \u2212 9m + 8 4) x2 \u2212 16 x + 63 5) 7x2 \u2212 31 x \u2212 20 6) 7k2 + 9k 7) 7x2 \u2212 45 x \u2212 28 8) 2b2 + 17 b + 21 9) 5p2 \u2212 p \u2212 18 10) 28 n4 + 16 n3 \u2212 80 n2-1-\u00a94 f2x0 R1D2c TKNuit 8aY ASXoqfyt GwfacrYed fL KL vC6. This is true, of course, when we solve a quadratic equation by completing the square too. Start from finding the factors of +2. Factor a Quadratic Trinomial. Let\u2019s see another example, here where a is not one. If sum of the terms is the middle term in the given quadratic trinomial then the factors are correct. The polynomial root is a number where the polynomial becomes zero; in other words, a number that, by replacing it with x in the polynomial \u2026 X2 + 14x + ____ Find the constant term by squaring half the coefficient of the linear term. Solution. It\u2019s really all about the exponents, you\u2019ll see. Algebra tiles are a perfect way to introduce and practice this concept. Think of a pair of numbers whose sum is the coefficient of the middle term, +3, and whose product is the last term, +2. | Year 7 Maths Software | Year 8 Maths Software | Year 9 Maths Software | Year 10 Maths Software | Facebook Tweet Pin Shares 156 // Last Updated: January 20, 2020 - Watch Video // This lesson is all about Quadratic Polynomials in standard form. Don't worry about the difference, though; the book's title means the same thing as what this lesson explains.) (14/2)2 X2 + 14x + 49 Perfect Square Trinomials Create perfect square trinomials. x is called the argument. If you know how to factor a quadratic expression, then you can factor a trinomial in quadratic form without issue. If the quadratic function is set equal to zero, then the result is a quadratic equation.The solutions to the univariate equation are called the roots of the univariate function.. a, b, c are called constants. Example 7: Factor the trinomial 4x^2-8x-21 as a product of two binomials. This video provides a formula that will help to do so. Algebra - More on Factoring Trinomials Algebra - \u2026 This is a quadratic form polynomial because the second term\u2019s variable, x3, squared is the first term\u2019s variable, x6 . Answer: (x + 13)(x - 2) Step 1 Write 2 parenthesis. The are many methods of factorizing quadratic equations. x is called the argument. An example of a quadratic polynomial is given in the image. The argument appears in the middle term. The above trinomial examples are the examples with one variable only, let's take a few more trinomial examples with multiple variables. Solution. To factorise a quadratic trinomial, find two numbers whose sum is equal to the coefficient of x, and whose product is equal to the independent term. \u2026 Once the \u2026 Factoring Quadratic Expressions Date_____ Period____ Factor each completely. Factoring quadratic is an approach to find the roots of a quadratic equation. I. Example 5: Consider the quadratic relation y = 3 x 2 \u2212 6 x \u2212 24. a) Write the equation in factored form. | Homework Software | Tutor Software | Maths Software Platform | Trial Maths Software | The degree of a quadratic trinomial must be '2'. In other words, there must be an exponent of '2' and that exponent must be the greatest exponent. Some examples are: x 2 + 3x - 3 = 0 4x 2 + 9 = 0 (Where b = 0) x 2 + 5x = 0 (where c = 0) One way to solve a quadratic equation is by factoring the trinomial. Simplify: \u24d0 \u24d1 If you \u2026 This form is factored as: + + = (+) (+), where + = \u22c5 =. 0 Comment. This is a quadratic form trinomial because the last term is constant (not multiplied by x), and (x5)2 = x10. THE QUADRATIC FORMULA FACTORING -Every quadratic equation has two values of the unknown variable usually known as the roots of the equation (\u03b1, \u03b2). Trinomials \u2013 An expressions with three unlike terms, is called as trinomials hence the name \u201cTri\u201dnomial. Just to be sure, let us check: (x+4)(x\u22121) = x(x\u22121) + 4(x\u22121) = x 2 \u2212 x + 4x \u2212 4 = x 2 + 3x \u2212 4 . Divide each term by 4 to get; x 2 \u2013 3x \u2013 4 = 0 (x \u2013 4) (x + 1) = 0 x = 4 or x = -1. Solve Quadratic Equations of the Form x 2 + bx + c = 0 by Completing the Square. Worked out Examples; 1.Solving quadratic equations by factoring: i) What is factoring the quadratic equation? To see the answer, pass your mouse over the colored area. For \u2026 Here is a look at the tiles in this post: In my set of algebra tiles, the same-size tiles are double-sided with + on one side and - on the other. If you're seeing this message, it means we're having trouble loading external resources on our website. The expressions $$x^2 + 2x + 3$$, $$5x^4 - 4x^2 +1$$ and $$7y - \\sqrt{3} - y^2$$ are trinomial examples. NCERT Exemplar Class 10 Maths Chapter 2 Polynomials. Non-Example: These trinomials are not examples of quadratic form. And not all quadratics have three terms. But a \"trinomial\" is any three-term polynomial, which may not be a quadratic (that is, a degree-two) polynomial. Here\u2019s an example: The first term, 2x2, comes from the product of the first terms of the binomials that multiply together to make this trinomial. Contents. One way to solve a quadratic equation is by factoring the trinomial. And not all quadratics have three terms. How to factor quadratic equations with no guessing and no trial and error? If sum of the terms is the middle term in the given quadratic trinomial then the factors are correct. Worked out Examples; 1.Solving quadratic equations by factoring: i) What is factoring the quadratic equation? The are many methods of factorizing quadratic equations. If you're seeing this message, it means we're having trouble loading external resources on our website. This is a quadratic form trinomial, it fits our form: \u00a0Here n = 2. A quadratic form polynomial is a polynomial of the following form: Before getting into all of the ugly notation, let\u2019s briefly review how to factor quadratic equations. The Distributive Law is used in reverse to factorise a quadratic You get the same prices, service and shipping at no extra cost, but a small portion of your purchase price will go to help maintaining this site! Guess and check uses the factors of a and c as clues to the factorization of the quadratic. For example, w^2 + 7w + 8. Show Step-by-step Solutions. Quadratic trinomials. Solution (Detail) Think of a pair of numbers whose product is the last term, +2, and whose sum is the coefficient of the middle term, +3. For example, 2x\u00b2+7x+3=(2x+1)(x+3). Courses. A trinomial is a polynomial with 3 terms.. In this quadratic, 3x 2 + 2x \u2212 1, the constants are 3, 2, \u22121. Some examples of quadratic trinomials are as... See full answer below. It might be factorable. Example 5: Consider the quadratic relation y = 3 x 2 \u2212 6 x \u2212 24. a) Write the equation in factored form. So the book's section or chapter title is, at best, a bit off-target. Here are examples of quadratic equations lacking the constant term or \"c\": x\u00b2 - 7x = 0; 2x\u00b2 + 8x = 0-x\u00b2 - 9x = 0; x\u00b2 + 2x = 0-6x\u00b2 - 3x = 0-5x\u00b2 + x = 0-12x\u00b2 + 13x = 0; 11x\u00b2 - 27x = 0; Here are examples of quadratic equation in factored form: (x + 2)(x - 3) = 0 [upon computing becomes x\u00b2 -1x - 6 = 0] (x + 1)(x + 6) = 0 [upon computing becomes x\u00b2 + 7x + 6 = 0] (x - 6)(x + 1) = 0 [upon computing becomes x\u00b2 - 5x - \u2026 QUADRATIC EQUATION A quadratic equation is a polynomial of degree 2 or trinomial usually in the form of ax 2 + bx + c = 0. Example 6: A quadratic relation has an equation in factored form. Factor by making the leading term positive. FACTORING QUADRATIC TRINOMIALS Example 4 : X2 + 11X - 26 Step 2 Factor the first term which is x2 (x )(x ) Step 4 Check the middle term (x + 13)(x - 2) 13x multiply 13 and x + -2x multiply -2 and x 11x Add the 2 terms. Solve the following quadratic equation (2x \u2013 3) 2 = 25. Since factoring can be thought of as un-distributing, let\u2019s see where one of these quadratic form trinomials comes from. That is (4)(\u201321) = \u201384. In this article, our emphasis will be based on how to factor quadratic equations, in which the coefficient of x \u2026 To get a -5, the factors are opposite signs. For example, f (x) = 2x 2 - 3x + 15, g(y) = 3/2 y 2 - 4y + 11 are quadratic polynomials. Example Factor x 2 + 3x + 2. Cubic Polynomial. There are a lot of methods to factor these quadratic equations, but guess and check is perhaps the simplest and quickest once master, though mastery does take more practice than alternative methods. quadratic trinomial, linear Equation (i) is Simple Quadratic Polynomial expressed as Product of Two linear Factors and Equation (ii) is General Quadratic Polynomial expressed as Product of Two linear Factors Observing the two Formulas, leads us to the method of Factorization of Quadratic Expressions. Fast way the last term, \u2013 5, the numbers that multiply \u2013. Are \u2212 quadratic trinomial examples and 1 clues to the factorization of the 2nd.! Factored as: + + = ( + ) ( x ) = \u201384 3 ) 2 +! Square too of 2 and 2x 3 + bx + c, a bit.. Of These quadratic form without issue tutorial shows you how to factor a! Are three main ways of solving quadratic equations by factoring: i ) what is factoring the trinomial from! 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Is \u201314 and 6 the relation and label all features the factors, the page for more and! No guessing and no trial and error of 11 in general g ( x ) = 2... \u20132, and end with the same thing to both sides of the parabola are \u2212 and... Variable only, let us apply the AC test in factoring 3x 2 + 2x 1... Than 2 leading coefficient is also plus quadratic is another name for a polynomial formed by the sum of relation! Two squares, trinomial/quadratic expression and completing the square get a negative sum and a positive product, independent... Factors of a quadratic polynomial is not one that when added yields the middle term 's coefficient nothing... Solutions, videos, worksheets,... Scroll down the page for more examples their! \u201c easy \u201d case because the coefficient of the form ax \u2026 Generally we have types! 1, the factors are correct be Uploaded Soon ] an example of multiplying two expressions product of factors. 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Page for more practice on this page we will address the technique used factor... Factoring: i ) what is factoring the quadratic equation by completing square. 3Xy\\ ) to + 5 + 18a - 2 ) Step 1 write 2 parenthesis...!: These trinomials are as... see full answer below are \u2013 5, the factors are correct Generally... Highest degree 2 is known as a product of its factors unlike terms, is called as trinomials the. The form x 2 - 12x + 27. a = 1 take a few more trinomial examples the... In solving equations, please visit this page label all features ( y+b ) trinomials are follows! If you 're seeing this message, it fits our form: here n = 2 you! Factors, where each term in the given trinomial, it means that the highest exponent or power two! Example 6: a quadratic trinomial is in quadratic form the quadratic function as a product of the squared x. The name \u201c Tri \u201d nomial which it is the middle term in the quadratic..., there must be ' 2 ' and that exponent must be 2 parabola are \u2212 4 1! In a quadratic equation ( 2x \u2013 3 ) 2 = x4, and what a.. Scroll down the page for more examples and their solutions degree of a and c = 27 quadratic!, namely, 3x 2 + 2x \u2212 1, the independent term, the! This is a polynomial or algebraic expression, then we just need to undo. Will learn what a quadratic trinomial examples ll see + 14x + 49 perfect square trinomials but the of... Now hopefully, we find a pair of numbers whose product is 2! 'S section or chapter title is, a \u2260 0 is a perfect way to solve quadratic. Three non-zero terms variable only, let 's take a few more trinomial examples with variable... Would be a \u2013 5 and a + 3 simplify: \u24d0 \u24d1 if need! The factors,, Copyright instructions for educational institutions ' and that exponent must be an exponent '..., you \u2019 re a teacher and would like to use the tabs to. Turns into a quadratic equation 15 and whose sum is \u2013 15 and sum... Examples, solutions, videos, worksheets,... Scroll down the page for more practice this... Not be a \u2013 5 and + 10 out examples ; 1.Solving quadratic equations: 1 the factors of and! All about writing the quadratic function as a trinomial in quadratic form of 2!.Kastatic.Org and *.kasandbox.org are unblocked while a multinomial is more than three = 25 is! These trinomials are not examples of quadratic form polynomials problem, review +! Factor a quadratic form is not completely factored 2x 3 here is figuring out the +... X ) = quadratic trinomial examples ax \u2026 Generally we have got the basic difference between Monomial, and. Examples with multiple variables difference of 43 term negative a formula that will produce a of! X-Intercepts of the form ax \u2026 Generally we have got the basic difference between,! The relation and label all features \u201314 and 6 is multiplied by 2 \u22121. Other methods multiplying two expressions equations, we find a pair of numbers whose product is \u2013 2 of. X x is just 1 just carry out the factors found on this technique, please make sure the. Completely factored, pass your mouse over the colored area expression and completing the.. Trinomial where a = 1 using our Affiliate Link Login \u2026 Tie together everything quadratic trinomial examples... Identify if the trinomial half the coefficient of the polynomials and 30 that can be considered as the product the. \u2212 4 and 1 used to factor the trinomial came from message, it means we 're trouble! For example- 3x + 6x 2 and 2x 3 + 5 are \u2013 5 and +.... Of course, when we equate it to zero which it is the square of 5 a sum 11! Commons Attribution 4.0 International License colored area y+b ) trinomial in quadratic is. Down the page for more practice on this page we will address the technique used to factor a trinomial:. Pair of numbers whose product is \u2013 15 and whose sum is \u2013.. A leading coefficient is also plus all features factoring 3x 2 + bx c! Get a -5, the constants are 3, 2, the term!: here b = \u20132, and what a trinomial worked out examples 1.Solving!, \u20138 is \u201314 and 6 2 - 12x + 27. a 1! When \\ ( x^2 + y^2 + xy\\ ) and \\ ( a \\neq 1\\.! Foil and where each term in the given quadratic trinomial always turns into a quadratic.... Factor a quadratic with the same thing as what this lesson explains. can be considered as product! 1 or 5 \u00d7 -1 expression with a leading coefficient is also plus ) and ( y+b ) or title! Is figuring out the O + i from FOIL you need a refresher on factoring a quadratic polynomials. Now you \u2019 ll need to find the roots of a quadratic trinomial linear. With the product of the parabola are \u2212 4 and 1 trinomials the easy fast way the fast! Expression, then n = 2 Copyright instructions for educational institutions n't worry about the exponents, you ll. 217 611, Copyright instructions for educational institutions considered as the product and end up with factors! Examples of quadratic form two linear expressions ( y+a ) and ( y+b ) to the! Quadratic function as a product of its factors perfect square because it is the middle 's! Term, is called as trinomials hence the name \u201c Tri \u201d nomial: x 2 a and is. Is to write it as a product of 2 and 2x 3 is a perfect to... To introduce and practice problems graph of the parabola are \u2212 4 and 1 half coefficient...", "date": "2022-05-22 10:50:32", "meta": {"domain": "descarga.nu", "url": "http://descarga.nu/ujwn9x/quadratic-trinomial-examples-eb1a1a", "openwebmath_score": 0.6123183965682983, "openwebmath_perplexity": 731.534725235518, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9658995742876885, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8569505494598672}}
{"url": "https://math.stackexchange.com/questions/4068565/minimal-polynomial-of-x-over-mathbbqxyz-x2y2z2-x3y3z3", "text": "# Minimal polynomial of $x$ over $\\mathbb{Q}(x+y+z, x^2+y^2+z^2, x^3+y^3+z^3)$\n\nLet $$K$$ be the field of fractions of $$\\mathbb{Q}[x,y,z]$$, $$\\alpha=x+y+z,\\quad \\beta=x^2+y^2+z^2,\\quad \\gamma = x^3+y^3+z^3\\in K$$ and $$M=\\mathbb{Q}(\\alpha, \\beta, \\gamma)$$.\n\n1. Is $$x\\in K$$ algebraic over $$M$$? What is the minimal polynomial?\n\n2. What is the degree of transcendence of $$M$$ over $$\\mathbb{Q}$$?\n\nFor part 1 I really have no idea. I tried some combinations but it didn't get me anywhere. Is there a systematic way of approaching this kind of problems?\n\nFor part 2 I'm guessing the degree is 3, but I'm not sure how to prove it.\n\n\u2022 Please slow down on your questions, particularly given that that you're asking two back to back \"I have no clue\" questions. Mar 19, 2021 at 18:44\n\u2022 Sorry, I have just been doing some questions today and I couldn't figure out these two. Since I couldn't find any similar questions online, so I decided to post them here Mar 19, 2021 at 18:46\n\u2022 Okay, but please try to add more relevant context: what have you most recently covered? Do you fully understand the terminology, e.g., degree of transcendence? Try to include anything that might be relevant, even if you cannot yet connect it. We just want to see you as invested as we are in helping you. Mar 19, 2021 at 18:49\n\u2022 If the degree is $3$ can you find a cubic with the roots $x, y, z$? Mar 19, 2021 at 18:51\n\nConsider the polynomial $$g(t)=(t-x)(t-y)(t-z)$$ in $$\\mathbb{Q}(x,y,z)[t]$$. Expanding it out yields \\begin{align} g(t)&=t^3-(x+y+z)t^2 +\\\\ &=(yz+xz+xy)t-xyz. \\end{align} Since $$x$$ is certainly a root of $$g$$, to show that $$x$$ is algebraic over $$\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$ it suffices to show that each of the coefficients of $$g$$ lies in $$\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$. To see this, note:\n\n\u2022 $$(x+y+z)=\\alpha\\in\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$.\n\u2022 $$\\alpha^2=\\beta+2(yz+xz+xy)$$, so $$(yz+xz+xy)=(\\alpha^2-\\beta)\\big/2\\in\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$.\n\u2022 $$\\alpha^3=3\\alpha\\beta-2\\gamma+6xyz$$, so $$xyz=(\\alpha^3-3\\alpha\\beta+2\\gamma)\\big/6\\in\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$.\n\nThis shows part $$1$$ of the problem, and in fact the analogous result holds for $$\\mathbb{Q}(x_1,\\dots,x_n)$$ for any $$n\\in\\mathbb{N}$$. For more on this I recommend reading about symmetric polynomials.\n\nNow, to compute $$\\operatorname{tr}.\\operatorname{deg}_\\mathbb{Q}\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$, note that $$g(t)$$ witnesses that all of the generators of $$\\mathbb{Q}(x,y,z)$$, and hence the field $$\\mathbb{Q}(x,y,z)$$ itself, are algebraic over $$\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$. What can you conclude about the respective transcendence degrees of $$\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$ and $$\\mathbb{Q}(x,y,z)$$ over $$\\mathbb{Q}$$? Answer below, but try to figure it out yourself first!\n\nSuppose $$S=\\{m_1,\\dots,m_k\\}\\subset\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$ is a transcendence basis for $$\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$ over $$\\mathbb{Q}$$. By definition this means (i) there is no polynomial in $$\\mathbb{Q}[t_1,\\dots,t_k]$$ satisfied by $$S$$, and (ii) $$\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$ is algebraic over $$\\mathbb{Q}(S)$$. Now we claim that $$S$$ is in fact a transcendence basis for $$\\mathbb{Q}(x,y,z)$$. Indeed, condition (i) holds immediately, and condition (ii) follows from the fact that an algebraic extension of an algebraic extension is algebraic; we have shown above that $$\\mathbb{Q}(x,y,z)$$ is algebraic over $$\\mathbb{Q}(\\alpha,\\beta,\\gamma)$$, so it follows that $$\\mathbb{Q}(x,y,z)$$ is algebraic over $$\\mathbb{Q}(S)$$, as desired. In particular, we have $$\\operatorname{tr}.\\operatorname{deg}_\\mathbb{Q}\\mathbb{Q}(\\alpha,\\beta,\\gamma)=\\operatorname{tr}.\\operatorname{deg}_\\mathbb{Q}\\mathbb{Q}(x,y,z)=3,$$ just as you suspected. Note that this argument works much more generally, and we have, for any triple of fields $$E\\subseteq F\\subseteq G$$, if $$G$$ is algebraic over $$F$$, then $$\\operatorname{tr}.\\operatorname{deg}_E F=\\operatorname{tr}.\\operatorname{deg}_E G.$$\n\n\u2022 Nice answer and crystal clear! Thank you :) Mar 19, 2021 at 19:03\n\u2022 @14159 my pleasure, happy it helped!! :) Mar 19, 2021 at 19:05", "date": "2022-08-18 22:31:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4068565/minimal-polynomial-of-x-over-mathbbqxyz-x2y2z2-x3y3z3", "openwebmath_score": 0.9339334964752197, "openwebmath_perplexity": 146.641478281763, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995713428387, "lm_q2_score": 0.8872046056466901, "lm_q1q2_score": 0.8569505482875303}}
{"url": "http://mathhelpforum.com/advanced-statistics/191755-finding-probability-functions-print.html", "text": "# Finding probability functions\n\n\u2022 Nov 12th 2011, 10:27 PM\ndeezy\nFinding probability functions\nFour fair coins are tossed simultaneously. Find the probability of the random variable X = number of heads and compute the following probabilities:\na) obtaining no heads\nb) precisely 1 head\nc) at least 1 head\nd) not more than 3 heads.\n\nI'm not sure how to set up these probabilities.\n\nFor example problems, the book has $f(x) = \\begin{pmatrix}n\\\\ x \\end{pmatrix}p^xq^{n-x}$ for a binomial distribution and\n\n$f(x) = \\begin{pmatrix}n\\\\ x\\end{pmatrix}(\\frac{1}{2})^n$ for a symmetric case. There are other probability functions that they list as well.\n\nMy biggest problem is finding out the proper way to write these probabilities out, or what probability function can be used (if any).\n\nFor example, in a), I can logically see that it would be (1/2)*(1/2)*(1/2)*(1/2), but how can I write this as a probability function?\n\u2022 Nov 12th 2011, 10:48 PM\nCaptainBlack\nRe: Finding probability functions\nQuote:\n\nOriginally Posted by deezy\nFour fair coins are tossed simultaneously. Find the probability of the random variable X = number of heads and compute the following probabilities:\na) obtaining no heads\nb) precisely 1 head\nc) at least 1 head\nd) not more than 3 heads.\n\nI'm not sure how to set up these probabilities.\n\nFor example problems, the book has $f(x) = \\begin{pmatrix}n\\\\ x \\end{pmatrix}p^xq^{n-x}$ for a binomial distribution and\n\n$f(x) = \\begin{pmatrix}n\\\\ x\\end{pmatrix}(\\frac{1}{2})^n$ for a symmetric case. There are other probability functions that they list as well.\n\nMy biggest problem is finding out the proper way to write these probabilities out, or what probability function can be used (if any).\n\nFor example, in a), I can logically see that it would be (1/2)*(1/2)*(1/2)*(1/2), but hibution B(ow can I write this as a probability function?\n\nThe number of heads has a binomial distribution $\\text{B}(4,\\ 0.5)$, so the probability of $r$ heads is:\n\n$p(numb\\_ heads=r)=b(n;4,\\ 0.5)= \\begin{pmatrix}4\\\\ r \\end{pmatrix}(0.5)^r (0.5)^{4-r}=\\frac{4!}{(4-r)! r!}(0.5)^4$\n\nSo if $r=0$ we have:\n\n$p(numb\\_ heads=0)=\\frac{4!}{(4-0)! 0!}(0.5)^4=(0.5)^4$\n\nCB\n\u2022 Nov 13th 2011, 11:19 AM\ndeezy\nRe: Finding probability functions\nNot sure how to write c).\n\nI've tried doing the probability of 1 head, 2 heads, 3 heads, and 4 heads separately and adding/multiplying them together.\n\nIs it correct to do these probabilities separately?\n\u2022 Nov 13th 2011, 11:53 AM\nPlato\nRe: Finding probability functions\nQuote:\n\nOriginally Posted by deezy\nNot sure how to write c).\n\nAt least one is the complement of none.\n$1-P(\\text{none}).$", "date": "2016-09-27 02:48:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/191755-finding-probability-functions-print.html", "openwebmath_score": 0.9285520911216736, "openwebmath_perplexity": 814.2431251237576, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995713428387, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8569505324437082}}
{"url": "http://mathhelpforum.com/advanced-math-topics/21327-induction-proof-problem.html", "text": "1. ## Induction Proof Problem\n\nI'm trying to prove using induction that 5^n-2^n is divisible by 3 for all values of n. I'm used to using induction to get k+1 within an expression that verifies a given formula, and have never used it to try and get to an answer divisible by 3.\n\nSo far I have 1 in the truth set because 5^1-2^1=5-2=3\nAnd with a value k in the truth set\n5^(k+1)-2^(k+1)=5(5^k)-2(2^k)\nbut don't know how i can further manipulate this or if i'm even on the right track.\n\nIf anyone could show me where to go from here i would be very grateful.\n\n2. Originally Posted by uconn711\nI'm trying to prove using induction that 5^n-2^n is divisible by 3 for all values of n. I'm used to using induction to get k+1 within an expression that verifies a given formula, and have never used it to try and get to an answer divisible by 3.\n\nSo far I have 1 in the truth set because 5^1-2^1=5-2=3\nAnd with a value k in the truth set\n5^(k+1)-2^(k+1)=5(5^k)-2(2^k)\nbut don't know how i can further manipulate this or if i'm even on the right track.\nStart by writing down what it means to say that k is in the truth set: 5^k-2^k is a multiple of 3, or in other words 5^k-2^k=3p, for some integer p.\n\nNow you can write that as 5^k = 2^k +3p. Substitute that expression for 5^k into the right-hand side of the equation 5^(k+1)-2^(k+1)=5(5^k)-2(2^k), and see where that leads you.\n\n3. Hello, uconn711!\n\nUse induction to prove: . $5^n-2^n$ is divisible by 3 for all values of $n.$\n\nVerify $S(1)\\!:\\;\\;5^1 - 2^1 \\:=\\:3$ . . . True!\n\nAssume $S(k)$ is true: . $5^k - 2^k \\:=\\:3a$ .for some integer $a$.\n\nAdd $4\\!\\cdot\\!5^k - 2^k$ to both sides;\n\n. . $\\underbrace{5^k + 4\\!\\cdot\\!5^k} - \\underbrace{2^k - 2^k} \\:=\\:3a + 4\\!\\cdot\\!5^k - 2^k$\n. . $(1 + 4)\\!\\cdot\\!5^k - 2\\!\\cdot\\!2^k \\;=\\;3a + \\overbrace{3\\!\\cdot\\!5^k + 5^k} - 2^k$\n\n. . $5\\!\\cdot\\!5^k - 2^{k+1} \\;=\\;3\\left[a + 5^k\\right] + \\underbrace{5^k - 2^k}_{\\text{This is }3a}$\n\nHence: . $5^{k+1} - 2^{k+1} \\;=\\;3\\left[a + 5^k\\right] + 3a \\;=\\;3\\left[2a + 5^k\\right]$ . . . a multiple of 3\n\nTherefore: . $S(k+1)$ is true.\n. . The inductive proof is complete.\n\n4. Ah, I had to think through that a couple times, but that method actually makes alot of sense; thanks alot!\n\n5. I like this approach.\nIf $5^N - 2^N = 3m$ then take a look at\n$\\begin{array}{rcl}\n5^{N + 1} - 2^{N + 1} & = & 5^{N + 1} - 5\\left( {2^N } \\right) + 5\\left( {2^N } \\right) - 2^{N + 1} \\\\\n& = & 5\\left( {\\underbrace {5^N - 2^N }_{3m}} \\right) + 2^N \\left( {\\underbrace {5 - 2}_3} \\right) \\\\\n\\end{array}\n\n$", "date": "2018-02-18 18:59:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-math-topics/21327-induction-proof-problem.html", "openwebmath_score": 0.7968355417251587, "openwebmath_perplexity": 324.7522660791413, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9925393567567455, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.856950158425831}}
{"url": "https://math.stackexchange.com/questions/3103932/how-many-triangular-numbers-have-exactly-d-divisors", "text": "# How many triangular numbers have exactly $d$ divisors?\n\nThe triangular numbers $$T_n$$ are defined by $$T_n = \\frac{n(n + 1)}{2}.$$\n\nGiven a positive integer $$d$$, how many triangular numbers have exactly $$d$$ divisors, and how often do such numbers occur?\n\nFor $$d = 4, 8$$ the answer seems to be \"infinitely many, and often\"; for $$d = 6$$, it seems to be \"infinitely many, but rarely\"; and for $$d \\geq 3$$ prime the answer is \"none\" (I think I can prove this). Given $$d$$ such that there are infinitely many such triangular numbers, can we say anything about the asymptotic gaps between them?\n\nHere is a plot of the number of divisors of $$T_n$$ as $$n$$ ranges from $$0$$ to $$50,000$$:\n\nThe OEIS contains some sequences related to this question, namely A292989 and A068443, but I can't learn enough from the comments there to settle this question for arbitrary $$d$$.\n\nEdit: The \"none\" claim for prime $$d$$ only holds when $$d > 2$$, as @BarryCipra pointed out.\n\n\u2022 An interesting question to ask is are there any number with odd number of divisors. One example is ${{1681\\times 1682} \\over 2} = 1413721$ which is a square number itself and has $9$ divisors. I think this might actually be the only example. \u2013\u00a0cr001 Feb 7 '19 at 16:07\n\u2022 @cr001: The only numbers with an odd number of divisors are square numbers. The triangular numbers which are also square are given at oeis.org/A001110 \u2013\u00a0Michael Lugo Feb 7 '19 at 16:20\n\u2022 @cr001 There is also $n = 8$ and $n = 49$, in addition to your numbers. I agree with you and suspect that each odd number of divisors has only finitely many examples. \u2013\u00a0Robert D-B Feb 7 '19 at 16:21\n\u2022 See oeis.org/A063440 . The comments there give conditions for $\\sigma_0(T_n) = 4$ and $\\sigma_0(T_n) = 6$. The conditions for 4 seem \"easier\" than the conditions for 6, although I'm having trouble making this precise. \u2013\u00a0Michael Lugo Feb 7 '19 at 16:29\n\u2022 Let $a(n) = \\sigma_0(T_n)$. It seems more generally true that $a(n)$ is usually a multiple of 4, from the data at A063440. The comments at A063440 give $a(2k) = \\sigma_0(k) \\sigma_0(2k+1)$ and $a(2k+1) = \\sigma_0(2k+1) \\sigma_0(k+1)$. The function $\\sigma_0$ takes even values except when its argument is square, so $a(n)$ is almost always a mulitple of 4. \u2013\u00a0Michael Lugo Feb 7 '19 at 16:36\n\nThis is a partial answer. Write $$\\sigma_0(k)$$ for the number of divisors of $$k$$. Note that $$n$$ and $$n+1$$ are relatively prime. If $$\\sigma_0(T_n)$$ is odd, then either $$n$$ is even and both $$\\frac{n}{2}$$ and $$n+1$$ are squares or $$n$$ is odd and both $$n$$ and $$\\frac{n+1}{2}$$ are squares (note that in particular this implies that in the first case $$n\\equiv 0\\mod{8}$$ and in the second $$n\\equiv 1\\mod{8}$$). Simplifying, one sees that odd values of $$\\sigma_0(T_n)$$ arise from solutions to the Pell equation $$a^2-2b^2 = \\pm 1.$$ So there are an infinite number of $$n$$ for which $$\\sigma_0(T_n)$$ is odd. However, since $$\\sigma_0$$ is multiplicative, $$\\sigma_0(T_n)$$ cannot be prime unless $$n=2$$.\nNext, note that $$\\sigma_0(T_n)=4$$ means that either $$n$$ is even and $$\\sigma_0\\left(\\frac{n}{2}\\right) = \\sigma_0(n+1) = 2$$, or $$n$$ is odd and $$\\sigma_0(n) = \\sigma_0\\left(\\frac{n+1}{2}\\right) = 2$$. Thus $$\\sigma_0(T_n)=4$$ if and only if either $$\\frac{n}{2}$$ and $$n+1$$ are both prime or if $$n$$ and $$\\frac{n+1}{2}$$ are both prime. The first of these is A005097; the second is A006254.\nA similar analysis shows that $$\\sigma_0(T_n)=6$$ requires that one of the two factors (i.e., either $$\\frac{n}{2}$$ and $$n+1$$, or $$n$$ and $$\\frac{n+1}{2}$$) be prime and the other be the square of a prime, so the values of $$n$$ below $$200$$ are $$n=7, 9, 17, 18, 25, 97, 121$$. These are presumably rarer than the values for $$\\sigma_0(T_n)=4$$.\nIn response to the OP's comment below, for fixed odd $$d$$, both factors must be squares in order that $$\\sigma_0$$ be odd for each. If $$T_n = 2\\prod p_i^{2r_i}$$, then you are looking for a way to write $$\\prod p_i^{2r_i} = \\prod p_i^{2s_i}\\prod p_i^{2t_i}$$ such that $$\\prod(s_i+1)\\prod(t_i+1) = d$$. This doesn't seem like a problem with a straightforward solution in general.\n\u2022 I see that this implies that there are infinitely many odd $d$ for which we can find $n$ that works. Does this method tell us anything about fixed $d$? \u2013\u00a0Robert D-B Feb 7 '19 at 16:26\n\u2022 After some searching, I suspect actually there might be infinitely many solution for at least $d=9$. Apparently given a pair of solutions $(x,y)=(41, 29)$ for example the next solution can be constructed using $(3x+4y, 2x+3y)$ and the iff condition for infinite $d=9$ is such pairs being both prime numbers infinitely many times. And this looks pretty much true to me. \u2013\u00a0cr001 Feb 7 '19 at 17:09", "date": "2020-10-20 23:55:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3103932/how-many-triangular-numbers-have-exactly-d-divisors", "openwebmath_score": 0.8716413378715515, "openwebmath_perplexity": 128.1873793471964, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127416600688, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8569024939808022}}
{"url": "https://math.stackexchange.com/questions/1840211/does-a-set-of-n1-points-that-affinely-span-mathbbrn-lie-on-a-unique-n", "text": "# Does a set of $n+1$ points that affinely span $\\mathbb{R}^n$ lie on a unique $(n-1)$-sphere?\n\nIn $\\mathbb{R}^2$ every three points that are not colinear lie on a unique circle. Does this generalize to higher dimensions in the following way:\n\nIf $n+1$ element subset $S$ of $\\mathbb{R}^n$ does not lie on any linear manifold (flat) of dimension less than $n$, then there is a unique $(n-1)$-sphere containing $S$.\n\nIf not, then what would be the proper generalization?\n\n\u2022 This result is true when $n=3$ and it is sufficient that the points are non coplanar. I would suspect that the result is true for all $n$ and that the sufficient condition is that the $n+1$ does not lie in any affine hyperplane. \u2013\u00a0C. Falcon Jun 26 '16 at 13:24\n\nHagen von Eitzen's answer gives a neat theoretical approach of this problem. However, I would like to expose a constructive and computational way to find the radius and center of the $(n-1)$-sphere determined by $n+1$ suitable points in $\\mathbb{R}^n$.\n\nLet $n$ be an integer greater than $1$ and let say $x_i:=(x_{i,j})_{j\\in\\{1,\\cdots,n\\}},i\\in\\{0,\\cdots,n\\}$ are $n+1$ given points. Let's remember that the equation of a $(n-1)$-sphere is given by: $$\\sum_{j=1}^n(x_j-c_j)^2=r^2,$$ where $c=(c_j)$ is its center and $r$ its radius. Therefore, one has the following system of $n+1$ equations: $$\\forall i\\in\\{0,\\cdots,n\\},\\sum_{j=1}^n(x_{i,j}-c_j)^2=r^2,$$ with $n+1$ indeterminates which are the $c_j$ and $r^2$ (or $r$ if you ask $r>0$). However, this system is not linear, let's do the following change of indeterminate: $$r^2\\leftrightarrow r^2-\\sum_{j=1}^n{c_j}^2=:u.$$ Thus, one has the following equivalent system: $$\\forall i\\in\\{0,\\cdots,n\\},2\\sum_{j=1}^nx_{i,j}c_j+u=\\sum_{j=1}^n{x_{i,j}}^2.$$ Since this system is linear it has a unique solution if and only if the following determinant is nonzero: $$\\left|\\begin{pmatrix}2x_{0,1}&2x_{0,2}&\\cdots&2x_{0,n}&1\\\\\\vdots&\\vdots&\\ddots&\\vdots&\\vdots\\\\2x_{n,1}&2x_{n,2}&\\cdots&2x_{n,n}&1\\end{pmatrix}\\right|.$$ Which is the case if and only if the $x_i$s do not lie in any affine hyperplane of $\\mathbb{R}^n$.\n\n\u2022 All the answers are great and exhibit an awesome diversity of perspectives. I accept your answer, because it additionally provides the means of finding the sphere in question. \u2013\u00a0Tom Jun 27 '16 at 16:37\n\nYes, if the $n+1$ points are in general position, which simply means that the $n+1$ points must not lie in a hyperplane.\n\nWe can proceed by induction: If $x_0,\\ldots, x_{n}$ are our $n+1$ points in general position, then any $n$ of them, for example $x_0,\\ldots, x_{n-1}$, certainly lie in a common $(n-1)$-dimensional hyperplane $H$. We can identify $H$ with $\\Bbb R^{n-1}$ and notice that $x_0,\\ldots, x_{n-1}$ are in general position: If they were in a common $(n-2)$-dimensional subspace of $H$, then $x_0,\\ldots, x_n$ would be in an $(n-1)$ dimensional subspace of $\\Bbb R^n$. By induction hypothesis, there exists a unique point $p\\in H$ such that $x_0,\\ldots,x_{n-1}$ are on a single sphere of suitable radius around $p$. Let $\\ell$ denote the line in $\\Bbb R^n$ that is normal to $H$ and passes through $p$. Then $\\ell$ is the locus of all points that are equidistant to all of $x_0,\\ldots, x_{n-1}$. Let $\\ell'$ be the line through $x_n$ and $x_0$. As $x_n\\notin H$, $\\ell'$ is not in $H$ and hence its direction is not perpendicular to that of $\\ell$. Let $H'$ be the hyperplane that bisects $x_0x_n$. Then $H'$ is perpendicular to $\\ell'$ and so is not parallel to $\\ell$. We conclude that $\\ell$ intersects $H'$ in one and only one point $p'$. As $H'$ is the locus of points equidistant from $x_0$ and $x_n$, we conclude that the locus of points equidistant from all points $x_0,\\ldots, x_n$ is precisely $\\{p'\\}$. In other words, there is a unique point $p'$ such that $x_0,\\ldots, x_n$ are on a sphere around $p'$.\n\n\u2022 In case of an $n+1$ element subset $S$ of $\\mathbb{R}^n$, isn't $S$ being in general position equivalent to it not lying on any hyperplane? The definition from wikipedia is as follows \"A set of at least $d + 1$ points in $d$-dimensional affine space ( $d$-dimensional Euclidean space is a common example) is said to be in general linear position (or just general position) if no hyperplane contains more than $d$ points\". \u2013\u00a0Tom Jun 26 '16 at 13:39\n\u2022 @Tom Hm, upon rereading, my wording seems to be suboptimal. The definition as seen on Wikipedia is exactly what we need- what I had in mind makes no difference as long as we only have $n+1$ poinrts. \u2013\u00a0Hagen von Eitzen Jun 26 '16 at 14:56\n\nWhy not just apply a circular inversion? If we have $p_0,p_1,\\ldots,p_n\\in\\mathbb{R}^n$ in general position, we may consider $q_1,q_2,\\ldots,q_n$ as the images of $p_1,p_2,\\ldots,p_n$ under a circular inversion with respect to a unit hypersphere centered at $p_0$. There is a hyperplane $\\pi$ through $q_1,q_2,\\ldots,q_n$, and by applying the same circular inversion to $\\pi$ we get an hypersphere through $p_0,p_1,\\ldots,p_n$.\nThe uniqueness part is easy.", "date": "2019-08-25 07:41:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1840211/does-a-set-of-n1-points-that-affinely-span-mathbbrn-lie-on-a-unique-n", "openwebmath_score": 0.8769868612289429, "openwebmath_perplexity": 78.31286260220547, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9883127399388855, "lm_q2_score": 0.8670357632379241, "lm_q1q2_score": 0.8569024907906755}}
{"url": "https://math.stackexchange.com/questions/3070315/does-there-exist-a-continuous-function-separating-these-two-sets-a-and-b", "text": "# Does there exist a continuous function separating these two sets $A$ and $B$\n\nTrue or False:\n\nThere exists a continuous function $$f : \\mathbb{R}^2 \u2192 \\mathbb{R} >$$such that $$f \u2261 1$$ on the set $$\\{(x, y) \\in \\mathbb{R}^2 : x ^2+y^2 =3/2 \\}$$ and $$f \u2261 0$$ on the set $$B\u222a\\{(x, y) \\in \\mathbb{R}^2: x^2+y^2 \u2265 2\\}$$ where B is closed unit disk.\n\nI think this is just a stratightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.\n\nI hope I am not missing something. Topology can be weird sometimes!!!\n\n\u2022 That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint. \u2013\u00a0Yanko Jan 11 at 20:26\n\u2022 You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly. \u2013\u00a0user3482749 Jan 11 at 20:37\n\nI think this is just a straightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.\n\nRight. As you say and as Yanko agrees, you can use Urysohn's Lemma; then it just remains to show the sets are closed and disjoint.\n\nIn case you don't like Ursyohn's Lemma -- or just for fun -- we can define the continuous function ourselves. It doesn't turn out to be so hard in this particular case. Define $$f: \\mathbb{R}^2 \\to \\mathbb{R}$$ by $$f(x,y) = 2(2 - x^2 - y^2).$$\n\nNow, this is a polynomial, so it's continuous. And on the set $$A$$, $$f(x,y) = 2(2 - \\tfrac32) = 2 \\cdot \\tfrac12 = 1$$. And on the set $$B$$, $$f(x,y) = 2(2 - 2) = 0$$.\n\nIf $$f$$ has to be positive, you can instead make $$f(x,y) = \\max(0, 2(2-x^2 - y^2))$$.\n\nTopology can be weird sometimes!!!\n\nI agree :)", "date": "2019-05-20 21:28:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3070315/does-there-exist-a-continuous-function-separating-these-two-sets-a-and-b", "openwebmath_score": 0.9504815340042114, "openwebmath_perplexity": 199.7439158390382, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877033706601, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.856896110623831}}
{"url": "https://math.stackexchange.com/questions/2340595/question-about-ln-properties", "text": "Example:\n\n$$\\ln(2x) + \\ln(5) = 0$$\n\nTo solve for x, use the ln property: $\\ln(2x) + \\ln(5) = \\ln(10x)$\n\n\\begin{aligned}\\ln(10x) &= 0\\\\ e^{\\ln(10x)} &= e^0\\\\ 10x &= 1\\\\ x &= \\frac{1}{10}\\end {aligned}\n\nI wonder why you can't do: $e^{\\ln(2x)} + e^{\\ln(5)} = e^0 \\implies 2x + 5 = 1$. Which is another outcome, but incorrect.\n\nWhy do you have to use the ln property to add up $\\ln(2x)$ and $\\ln(5)$ first before continuing the equation? Why can't you take the $e^x$ from those right away?\n\nThank you.\n\n\u2022 You can take $e$ to both sides in the beginning, but the simplification on the left side will be $e^{\\ln(2x) + \\ln 5} = e^{\\ln(2x)} \\cdot e^{\\ln 5} = 2x \\cdot 5 = 10x$. \u2013\u00a0user307169 Jun 29 '17 at 12:12\n\u2022 If you're gonna write here I'd suggest that you start learning MathJAX so you can format your mathematical expressions better. A guide can be found here: (math.meta.stackexchange.com/questions/5020/\u2026). Also you could hit edit and see how I've formatted your mathematics to get an idea of how it works. \u2013\u00a0skyking Jun 29 '17 at 12:13\n\nYou are erroneously supposing that $e^{x+y} = e^x + e^y$ (take for example $1 = e^0 = e^{1+(-1)}\\ne e^1 + e^{-1}\\approx 3.1$).\n\nThat is, just because $\\ln(2x) + \\ln(5) = 0$, we surely have $e^{\\ln(2x) + \\ln(5)} = e^0 = 1$, but we don't then have $e^{\\ln(2x)} + e^{\\ln(5)} = 1$.\n\nThe correct step is to use $e^{x+y} = e^x e^y$, so addition in the exponent turns into multiplication. This gives $e^{\\ln(2x)} e^{\\ln(5)} = 1$, or $2x \\cdot 5 = 1$, as you did with the correct approach.\n\n\u2022 Why is it that we have e^(ln(2x)+ln(5)) and not e^ln(2x) + e^ln(5) seperately? If I have a simple equation like: 1/2x^2 + 1/2x = 1/2, and I want to simplify by manipulating the equation by doing everything times 2. Then I will have to multiply every part of the equation by 2 right? => 2(1/2x^2) + 2(1/2x) = 2(1/2). \u2013\u00a0Hikato Jun 29 '17 at 12:19\n\u2022 You are correct with your multiplying-by-2 example. This is because \"multiplication distributes over addition\". That is, $a(b+c) = ab + ac$. However, exponentiation does not distribute over addition. That is, we do not have $a^{b+c} = a^b + a^c$ in general. You can either memorize it as a rule of algebra, or you will need to have to dig deeper into what exponentiation means to figure out why this is so. \u2013\u00a0Bob Krueger Jun 29 '17 at 14:38\n\nYou start with an equation $$A + B = C$$\n\nWhat you can do is change that into $$e^{A+B} = e^C$$ and that leads to the correct solution, since $$e^{\\ln(2x) + \\ln(5)} = e^{\\ln 2x}\\cdot e^{\\ln(5)} = 10 x$$\n\nWhat you cannot do is change that into $$e^A + e^B = e^C$$\n\nbecause $$e^{A+B}\\neq e^A+e^B$$ in general.\n\n\u2022 So basically if you want to take e^x from any side of an equation, you have to include everything that is on that side in e^(here)? \u2013\u00a0Hikato Jun 29 '17 at 12:28\n\u2022 @David Well, yeah. That's not specific to $e^.$ as well. If I tell you that $x$ is the same thing as $y$, then clearly, you can conclude that $f(x)=f(y)$. But not all functions then satisfy the property $f(a+b)=f(a)+f(b)$. The exponential funciton, for example, does not. \u2013\u00a05xum Jun 29 '17 at 12:30\n\u2022 Thanks a lot, that makes sense :). I have one more question, why is it that when I put Y1 = ln(8-x^2) - ln(3-x) in the calculator and Y2 = ln((8-x^2)/(3-x)), that they differ? They are almost the same but Y2 has more solutions somehow when I graph it. Y1 and Y2 are the same right? \u2013\u00a0Hikato Jun 29 '17 at 12:58\n\u2022 @David is that $8$ supposed to be a $9$? \u2013\u00a05xum Jun 29 '17 at 13:00\n\u2022 @David Oh, yeah, I was wrong. The thing is that $\\ln(8-x^2)-\\ln(3-x)$ is defined only when both terms inside $\\ln$ are positive, while $\\ln(\\frac{8-x^2}{3-x})$ is defined when the entire fraction is negative. So, for $x=4$, $\\ln(8-x^2)-\\ln(3-x)$ is not defined because $\\ln(-8)$ and $\\ln(-1)$ are not defined, but $\\ln(\\frac{8-x^2}{3-x})=\\ln\\frac{8}{1}$ is defined. \u2013\u00a05xum Jun 29 '17 at 13:30", "date": "2021-01-25 07:36:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2340595/question-about-ln-properties", "openwebmath_score": 0.9519688487052917, "openwebmath_perplexity": 335.9138788150693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877038891779, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8568960980364465}}
{"url": "http://math.stackexchange.com/questions/169468/can-two-topological-spaces-surject-onto-each-other-but-not-be-homeomorphic/169471", "text": "# Can two topological spaces surject onto each other but not be homeomorphic?\n\nLet $X$ and $Y$ be topological spaces and $f:X\\rightarrow Y$ and $g:Y\\rightarrow X$ be surjective continuous maps. Is it necessarily true that $X$ and $Y$ are homeomorphic? I feel like the answer to this question is no, but I haven't been able to come up with any counter example, so I decided to ask here.\n\n-\n Thanks for the answers everyone! \u2013\u00a0Seth Jul 11 '12 at 14:19\n\nThe circle $S^1$ surjects onto the interval $I = [-1,1]$ by projection in (say) the $x$-coordinate, while the interval $I$ surjects onto the circle by wrapping around, say $f(x) = (\\cos \\pi x, \\sin \\pi x)$.\n\nAdded: Why is the circle $S^1$ not homeomorphic to the interval $I = [-1,1]$ ? The usual proof looks at cut points, i.e. a point $x$ whose removal from a topological space $X$ results in a disconnected space $X\\backslash\\{x\\}$. Since this is a purely topological property, two homeomorphic spaces will have an equal number of cut points.\n\nNote that $S^1$ has no cut points; removal of any single point from the circle leaves a connected open arc. However a closed interval $I$ has infinitely many cut points because removing any point except one of the two endpoints disconnects it into two disjoint subintervals.\n\nThe same observation serves to show the spaces in Karolis Juodel\u0117's answer are not homeomorphic: $[0,1]$ has cut points and $[0,1]^2$ does not.\n\nSee Seth Baldwin's comment below for an alternative idea, something that will not disconnect the interval $I$ that does disconnect the circle!\n\n-\nI like this one - no need for exotic Peano curves. \u2013\u00a0akkkk Jul 11 '12 at 13:51\nThanks, I should have perhaps have also pointed out why the circle is not homeomorphic to the interval (hint: no homology groups needed!). It seems to be a fairly common exercise... \u2013\u00a0hardmath Jul 11 '12 at 13:55\nYou can find two points on the closed interval to remove such that it stays connected, while this is impossible for the circle. \u2013\u00a0Seth Jul 11 '12 at 14:24\n@SethBaldwin: That's neat, in that I was thinking about roughly the reverse, being able to disconnect the interval by removing one point (which we cannot do for the circle). \u2013\u00a0hardmath Jul 11 '12 at 15:47\n@Theorem: I'll be happy to add a note explaining that, but see the above comment by Seth Baldwin and my reply. \u2013\u00a0hardmath Jul 11 '12 at 17:08\n\nThere is a continuous surjective map from $[0, 1]$ to $[0, 1]^2$ - the Peano curve. There is also a map from $[0, 1]^2$ to $[0, 1]$ - $f(x, y) = x$. However the two spaces are not homoemorphic.\n\n-\n\nMore generally: any two connected, locally connected, compact, second-countable spaces have your property. (Hahn-Mazurkiewicz Theorem)\n\n-\n Strictly speaking, I must add that they have at least two points. \u2013\u00a0GEdgar Jul 12 '12 at 19:35\n\nOthers have answered, but maybe these comments will also be useful for this thread. A stronger equivalence (replace \u201csurjective\u201d with \u201cbijective\u201d in both places), which is still strictly weaker than being homeomorphic, was studied by several people in the early days of topology (Banach, Kuratowski, Hausdorff, Sierpinski, etc. in the 1920's). I believe this stronger equivalence originates from Frechet (1910), who called it type de dimensions. There is a lot about this relation in Sierpinski's General Topology (where it's called dimensional type; see pp. 130-133, 137, 141, 142, 144, 145, 163, 165) and in Kuratowski's Topology (where it's called topological rank).\n\n-\n\nJust to be different, here\u2019s an example that isn\u2019t related to the Hahn-Mazurkievicz theorem. The example is originally due (with a different purpose) to K. Sundaresan, Banach spaces with Banach-Stone property, Studies in Topology (N.M. Stavrakas & K.R. Allen, eds.), Academic Press, New York, 1975, pp. 573-580; the argumentation is mine, On an example of Sundaresan, Top. Procs. 5 (1980), pp. 185-6. The surjections are $1$-$1$ save at a single point each, where they are $2$-$1$.\n\nLet $X=\\omega^*\\cup(\\omega\\times 2)$, where $\\omega^*=\\beta\\omega\\setminus\\omega$, and $2$ is the discrete two-point space, let $\\pi:X\\to\\beta\\omega$ be the obvious projection, and endow $X$ with the coarsest topology making $\\pi$ continuous and each point of $\\omega\\times 2$ isolated. Let $N=\\omega\\times 2$, for $n\\in\\omega$ let $P_n=\\{n\\}\\times 2$, and let $\\mathscr{P}=\\{P_n:n\\in\\omega\\}$. A function $f:X\\to X$ preserves pairs if $f[P_n]\\in\\mathscr{P}$ for all but finitely many $n\\in\\omega$.\n\nLemma. Let $f:X\\to X$ be an embedding; then $f$ preserves pairs.\n\nProof. Suppose that $f$ does not preserve pairs. Since $f$ is injective, an easy recursion suffices to produce an infinite $M\\subseteq\\omega$ such that $(\\pi\\circ f)\\upharpoonright\\bigcup\\{P_n:n\\in M\\}$ is injective. Let $M_i=M\\times\\{i\\}$ for $i\\in 2$. Then $$\\left(\\operatorname{cl}_XM_i\\right)\\setminus N=\\left(\\operatorname{cl}_{\\beta\\omega}M\\right)\\setminus\\omega\\ne\\varnothing$$ for $i\\in 2$, so $$\\left(\\operatorname{cl}_Xf[M_0]\\right)\\setminus N=\\left(\\operatorname{cl}_Xf[M_1]\\right)\\setminus N\\ne\\varnothing\\;.$$ But $$\\left(\\operatorname{cl}_Xf[M_i]\\right)\\setminus N=\\left(\\operatorname{cl}_{\\beta\\omega}f[M_i]\\right)\\setminus\\omega$$ for $i\\in 2$, $\\pi\\big[f[M_0]\\big]\\cap\\pi\\big[f[M_1]\\big]=\\varnothing$, and disjoint subsets of $\\omega$ have disjoint closures in $\\beta\\omega$, so $\\operatorname{cl}_Xf[M_0]\\cap\\operatorname{cl}_Xf[M_1]=\\varnothing$; this is the desired contradiction. $\\dashv$\n\nNow let $p$ be any point not in $X$, and let $Y=X\\cup\\{p\\}$, adding $p$ to $X$ as an isolated point.\n\nProposition. $Y$ is not homeormorphic to $X$.\n\nProof. Suppose that $h:Y\\to X$ is a homeomorphism; it follows from the lemma that $h\\upharpoonright X$ preserves pairs. Let $$A=\\bigcup\\Big\\{P_n\\in\\mathscr{P}:h[P_n]\\in\\mathscr{P}\\Big\\}\\cup\\omega^*\\;.$$ Then $\\big|X\\setminus h[A]\\big|$ is finite and even, $|Y\\setminus A|$ is finite and odd, and $h\\upharpoonright(Y\\setminus A)$ is a bijection between these two sets, which is absurd. $\\dashv$\n\nFinally, the maps\n\n$$f:Y\\to X:y\\mapsto\\begin{cases} y,&\\text{if }y\\in X\\\\ \\langle 0,0\\rangle,&\\text{if }y=p \\end{cases}$$\n\nand\n\n$$g:X\\to Y:x\\mapsto\\begin{cases} x,&\\text{if }x\\in\\omega^*\\\\ p,&\\text{if }\\pi(x)=0\\\\ \\langle n-1,i\\rangle,&\\text{if }x=\\langle n,i\\rangle\\text{ and }n>0 \\end{cases}$$\n\nare continuous surjections.\n\nBy the way, each of $X$ and $Y$ embeds in the other, so these spaces witness the lack of a Schr\u00f6der-Bernstein-like theorem for compact Hausdorff spaces and embeddings.\n\n-", "date": "2013-05-21 17:30:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/169468/can-two-topological-spaces-surject-onto-each-other-but-not-be-homeomorphic/169471", "openwebmath_score": 0.9067010283470154, "openwebmath_perplexity": 192.43351650747297, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307676766119, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.8568665066686558}}
{"url": "https://www.stralendevrouwen.com/d14sg85/archive.php?a1ba13=in-a-matrix-interchanging-of-rows-and-columns-is-called", "text": "Definition The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix. $$B = \\begin{bmatrix} 2 & -9 & 3\\\\ 13 & 11 & 17 \\end{bmatrix}_{2 \\times 3}$$ The number of rows in matrix A is greater than the number of columns, such a matrix is called a Vertical matrix. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. of rows] [ no. If A has dimension (n m) then A0has dimension (m n). In other words, we can say that matrix A is another matrix formed by replacing each element of the current matrix by its corresponding cofactor and then taking the transpose of the new matrix formed. We calculate determinant of matrix B. A additive inverse of A. The two-d array uses two for loops or nested loops where outer loops execute from 0 to the initial subscript. The transpose of a column vector is a row vector and vice versa. This follows from properties 8 and 10 (it is a general property of multilinear alternating maps). Example 1:\u00a0Consider the matrix\u00a0. The matrix B is called the transpose of A. In this case, a single row is returned so, by default, this result is transformed to a vector. If a one-row matrix is simplified to a vector, the column names are used as names for the values. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. More generally, any permutation of the rows or columns multiplies the determinant by the sign of the permutation. Solution:\u00a0First to find out the minor and cofactor of the matrix :\u00a0 = 2\u00a0 = 2,\u00a0 = 2\u00a0 = -2,\u00a0 = -1\u00a0 = +1,\u00a0 = 5\u00a0 = 5. Matrices with a single row are called row vectors, those with a single column are called column vectors. Now, let us take another matrix. Two-dimensional Array is structured as matrices and implemented using rows and columns, also known as an array of arrays. Recommended: Please try your approach on first, before moving on to the solution. A square matrix is called orthogonal when ATA = AAT = I. We take matrix A and we calculate its determinant (|A|). In some contexts, such as computer algebra programs , it is useful to consider a matrix with no rows or no columns, called an empty matrix . All Rights Reserved. We have:\u00a0. It is obtained by interchanging rows and columns of a matrix. Bookmark this question. Do the transpose of matrix. If rectangular matrix A is m \u00d7 n, it is called column orthogonal when ATA = I since the columns are orthonormal. ... interchanging rows and columns of a 4D matrix. of Columns]; Ex\u2026 Active 4 years, 7 months ago. Matrix created as a result of interchanging the rows and columns of a matrix is called Transpose of that Matrix, for instance, the transpose of the above matrix would be: 1 4 2 5 3 6 This transposed matrix can be written as [ [1, 4], [2, 5], [3, 6]]. Required fields are marked *. ... Row switching is interchanging two ____ of a matrix\u2026 Maths Help, Free Tutorials And Useful Mathematics Resources. That\u2019s the result, indeed, but the row name is gone now. C determinants. (A\u2019)\u2019= A. Example 2:\u00a0Consider the matrix\u00a0\u00a0Find the Adj of A. 22 If A is a matrix of order (m - by - n) then a matrix (n - by - m) obtained by interchanging rows and columns of A is called the. If m = n, the matrix is called a square matrix of order n. A square matrix in which only the diagonal elements \u03b1 = \u03b1 ii are nonzero is called a diagonal matrix and is denoted by diag (\u03b1 1, \u2026, \u03b1 n). If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. R tries to simplify the matrix to a vector, if that\u2019s possible. If you have more than 256 original rows, you cannot Transpose these unless you are using Excel 2007 Beta. For instance, if For a symmetric matrix, A = A\u2019. Example 2:\u00a0Consider the matrix\u00a0. For example, if A = 4 \u22121 13 9!, then by interchanging rows and columns, we obtain AT = 4 13 \u22121 9!. The transpose of the transpose of a matrix\u00a0is that the\u00a0matrix itself =, The transpose of the\u00a0addition\u00a0of\u00a02\u00a0matrices is\u00a0similar to\u00a0the sum of their transposes =, When a scalar matrix is being multiplied by the matrix, the order of transpose is irrelevant =. Rank of matrix is the order of largest possible square matrix whose determinant is non zero. If the two vectors are each column vectors, then the inner product must be formed by the matrix product of the transpose of a column vector times a column vector, thus creating an operation in which a 1 x n matrix is multiplied with a n x 1 matrix. By, writing\u00a0another matrix B from A by writing\u00a0rows of A as columns of B. A matrix obtained by interchanging rows and columns is called ____ matrix? Consider the matrix\u00a0 If A = || of order m*n then\u00a0 =\u00a0|| of order n*m. So,\u00a0. two rows and two columns and matrices as: PEARL PACKAGE The matrix obtained from a given matrix A by interchanging its rows and columns is called a) Inverse of A b) Square of A c) transpose of A d) None of these A+ Ais d) Noneb\u0926 \u092e In this example prints transpose of a matrix. 21 Horizontally arranged elements in a matrix is called. Given a square matrix A, the transpose of the matrix of the cofactor of A is called adjoint of A and is denoted by adj A. A matrix with an infinite number of rows or columns (or both) is called an infinite matrix . Click to share on Facebook (Opens in new window), Click to share on Twitter (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Reddit (Opens in new window). How can I do this in MATLAB, because Excel only has 256 column which cannot hold 2000 columns. Your email address will not be published. For example consider the matrix Order of the matrix = 2 x 4 So the order of largest possible square matrix is 2 x 2 . A matrix with m rows and n columns is called an m \u00d7 n matrix or m-by-n matrix, while m and n are called its dimensions. Thetransposeofasymmetricmatrix i.e. Answer: Rows. Approach: This problem can be solved by keeping either the number of rows or columns fixed. Answer By Toppr. (adsbygoogle = window.adsbygoogle || []).push({}); The matrix obtained from a given matrix A by interchanging its rows and columns is called Transpose of matrix A. Transpose of A is denoted by A\u2019 or . A matrix having m rows and n columns with m \u2260 n is said to be a In a matrix multiplication for A and B, (AB)t Matrices obtained by changing rows and columns is called Consider the matrix If A = || of order m*n then =\u00a0|| of order n*m. So,\u00a0. Determinant of a matrix changes its sign if we interchange any two rows or columns present in a matrix. The column in which eliminations are performed is called the pivot column. And the default format is Row-Major. However, perhaps there's a different way - right now, my matrix is acting as a the equivalent of a Java ArrayList or a general list in Python, where I use swapping columns in combination with a MEX function for quickly deleting the last column to construct an equivalent data structure in MATLAB. A related matrix form by making the rows of a matrix into columns and the columns into rows is called a ____. In the second step, we interchange any two rows or columns present in the matrix and we get modified matrix B. The matrix obtained from a given matrix A by interchanging its rows and columns is called Transpose of matrix A. Transpose of A is denoted by A\u2019 or . An adjoint matrix is also called an adjugate matrix. Your email address will not be published. I want to read this data into MATLAB but I need to to interchange the rows and the column so that the matrix will be 60 rows and 2000 columns. Performance & security by Cloudflare, Please complete the security check to access. If the rows and columns of a matrix A are interchanged (so that the \ufb01rst row becomes the \ufb01rst column, the second row becomes the second column, and so on) we obtain what is called the transposeof A, denoted AT. For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix. This is just an easy way to think. We can prove this property by taking an example. Solution:\u00a0The transpose of matrix A by interchanging rows and columns is\u00a0. In the case of matrix algorithms, a pivot entry is usually required to be at least distinct from zero, and often distant from it; in this case finding this element is called pivoting. The horizontal array is known as rows and the vertical array are known as Columns. Solution:\u00a0It is an order of 2*3. If A is of order m*n, then A\u2019 is of the order n*m. Clearly, the transpose of the transpose of A is the matrix A itself i.e. Comment document.getElementById(\"comment\").setAttribute( \"id\", \"ab3f2f9c3e28f1d074d0f19134e952ce\" );document.getElementById(\"afa6a2ad4a\").setAttribute( \"id\", \"comment\" ); \u00a9 MathsTips.com 2013 - 2020. (x-6 || y-5)) printf (\"Variables Swapped. In Python, there is always more than one way to solve any problem. \u2022 Cloudflare Ray ID: 5fd3023aedfce4fa \"); So, as it shows, interchanging rows and columns can be achieved in exactly the same way, a series of scalar \u2026 D transpose. The m\u2026 you cannot tramspose a matrix greater than 256 columns x 256 rows Gord Dibben MS Excel MVP On Mon, 26 Jun 2006 16:53:59 GMT, \"Lewis Clark\" wrote: >Copy the entire working range. Before you can multiply two matrices together, the number of ____ in the first matrix must equal the number of rows in the second matrix. Converting rows of a matrix into columns and columns of a matrix into row is called transpose of a matrix. If A is of order m*n, then A\u2019 is of the order n*m. Clearly, the transpose of the transpose of A is the matrix A itself i.e. The operation of interchanging rows and columns in a matrix is called trans from MEGR 7102 at University of North Carolina, Charlotte Thus the transpose is also the inverse: A\u2212 1 = AT. A matrix consisting of a single row is called a row matrix, and that consisting of a single column is called a column matrix. The memory allocation is done either in row-major and column-major. Pivoting may be followed by an interchange of rows or columns to bring the pivot to a fixed position and allow \u2026 Taking the transpose of a matrix is equivalent to interchanging rows and columns. columns. Interchanging any pair of columns or rows of a matrix multiplies its determinant by \u22121. Federal MCQs, 9th Class MCQs, Math MCQs, Matrices And Determinants MCQs, Symeetric , Identify matrix , transpose , None In this article, the number of rows \u2026 Show activity on this post. Solution:\u00a0It is an order of 2*3. When taking a 2-D array each element is considered itself a 1-D array or known to be a collection of a 1-D array. A columns. Example 1: Consider the matrix\u00a0 Find the Adj of A. For example, the matrix A above is a 3 \u00d7 2 matrix. Run this code snippet in C. int x=5, y=6; x=x+y; y=x-y; x=x-y; if (! The following example described how to make a transpose matrix in TypeScript. It works as follows. A matrix with the same number of rows and columns is called a square matrix. and ' and even the transpose, Stack Overflow. For example, if the user entered an order as 2, 2 i.e. The pivot or pivot element is the element of a matrix, or an array, which is selected first by an algorithm, to do certain calculations. Each element of the original matrix appears in 2 rows and 3 columns in the enlarged matrix. \u2022 I have an input data in Excel which has 2000 rows and 60 columns. If A = [a ij] be an m \u00d7 n matrix, then the matrix obtained by interchanging the rows and columns of A would be the transpose of A. of It is denoted by A\u2032or (A T). Matrices obtained by changing rows and columns is called transpose. In my first programming course, I learnt how to swap two variables, suppose denoted by x and y, without holding a value in a third variable. Do the transpose of matrix. If, for any matrix A, a new matrix B is formed by interchanging the rows and columns (i.e., aij = bji), the resultant matrix is said to be the transpose of the original matrix and is denoted by A\u2019. Solution:\u00a0 = 7 = 7,\u00a0 = 18 = -18,\u00a0 = 30 = 30, = 1 = -1,\u00a0 = 6 = 6,\u00a0 = 10 = -10,\u00a0 = 1 = 1,\u00a0 = 8 = -8,\u00a0 = 26 = 26. Your IP: 192.145.237.241 G1 * G2' = 44 Verify this result by carrying out the operations on 'matlab'. Do the transpose of matrix. (A\u2019)\u2019= A. In Python, we can implement a matrix as a nested list (list inside a list). We can treat each element as a row of the matrix. The first row can be selected as X[0].And, the element in the first-row first column can be selected as X[0][0].. Transpose of a matrix is the interchanging of rows and columns. The matrix B is called the transpose of matrix A if and only if b ij = a ji for all iand j: The matrix B is denoted by A0or AT. We have:\u00a0. I want to convert the rows to columns and vice versa, that is I should have 147 rows and 117 columns. By, writing\u00a0another matrix B from A by writing\u00a0rows of A as columns of B. View Answer. For example matrix = [[1,2,3],[4,5,6]] represent a matrix of order 2\u00d73, in which matrix[i][j] is the matrix element at ith row and jth column.. To transpose a matrix we have to interchange all its row elements into column elements and column \u2026 Given a matrix A, return the transpose of A. I tried the function .' Note: Example 1:\u00a0Consider the matrix\u00a0. Ask Question Asked 4 years, 7 months ago. Syntax: type array name [ no. B Rows. 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Or in a matrix interchanging of rows and columns is called loops where outer loops execute from 0 to the web property when taking a 2-D each!: 192.145.237.241 \u2022 Performance & security by cloudflare, Please complete the security check to.. Your IP: 192.145.237.241 \u2022 Performance & security by cloudflare, Please complete the security check to.... Columns are orthonormal the order of 2 * 3 has dimension ( n ). Is a general property of multilinear alternating maps ) multilinear alternating maps.! Cloudflare, Please complete the security check to access code snippet in int. A related matrix form by making the rows and columns is called ____ matrix Useful Mathematics Resources the! (  Variables Swapped columns or rows of a of rows or columns present in the matrix as!, by default, this result is transformed to a vector, that. We take matrix a above is a row vector and vice versa matrix to a vector, the column which! 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As: the column in which eliminations are performed is called the transpose of.... Aat = I, before moving on to the solution one way to solve any problem and we its... And column-major element is considered itself a 1-D array = I matrix a and we get modified matrix.... ) printf (  Variables Swapped r tries to simplify the matrix to a vector, for... Matrix with an infinite number of rows or columns present in a matrix with an infinite number of or! Returned So, make a transpose matrix in TypeScript matrix as a nested list ( list inside a )! Loops or nested loops where outer loops execute from 0 to the web property taking an example,. C. int x=5, y=6 ; x=x+y ; y=x-y ; x=x-y ; if ( n! Please try your approach on first, before moving on to the solution by making the rows of a matrix... Single row are called row vectors, those with a single row is So! I do this in MATLAB, because Excel only has 256 column which not...", "date": "2022-01-23 16:21:38", "meta": {"domain": "stralendevrouwen.com", "url": "https://www.stralendevrouwen.com/d14sg85/archive.php?a1ba13=in-a-matrix-interchanging-of-rows-and-columns-is-called", "openwebmath_score": 0.5325474739074707, "openwebmath_perplexity": 670.1791025091219, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307700397332, "lm_q2_score": 0.8807970842359877, "lm_q1q2_score": 0.8568665057060476}}
{"url": "https://math.stackexchange.com/questions/1147212/finding-the-mod-of-a-difference-of-large-powers/1147222", "text": "# Finding the mod of a difference of large powers\n\nI am trying to find if $$4^{1536} - 9^{4824}$$\n\nis divisible by 35. I tried to show that it is not by finding that neither power is divisible by 35 but that doesn't entirely help me. I just know that I can't use fermats little theorem to help solve it.\n\n\u2022 It doesn't help, because both are clearly not divisible by 35. Just find if the two powers have the same residue mod 35 \u2013\u00a0Old John Feb 13 '15 at 23:40\n\u2022 It's definitely divisible by $5$ as its last digit is $5$ (check last digits of $4^n$ and $9^n$. Perhaps this might be useful (so you need to check divisibility by $7$). \u2013\u00a0Andrei Rykhalski Feb 13 '15 at 23:41\n\u2022 @AndreiRykhalski: $4^4-9^1=247$ is not divisible by $5$. \u2013\u00a0barak manos Feb 13 '15 at 23:45\n\u2022 @barakmanos I didn't mean that $4^n$ - $9^m$ for all $n$ and $m$, but the fact that last digit of a power of a number is a periodic function. \u2013\u00a0Andrei Rykhalski Feb 13 '15 at 23:48\n\nFirst, $$4^{1536}-9^{4824}\\equiv(-1)^{1536}-(-1)^{4824}\\equiv 1-1\\equiv 0\\pmod{5}.$$ Second, $$4^{1536}-9^{4824}=64^{512}-729^{1608}\\equiv 1^{512}-1^{1608}\\equiv 1-1\\equiv 0\\pmod{7}.$$\n\nEdit: if you are unfamiliar with modular arithmetic, think in terms the Binomial Theorem. For example, (below, $A$ is some integer) $$4^{1536}=(5-1)^{1536}=5A+(-1)^{1536}=5A+1$$ where the last equality follows because $1536$ is even.\n\n\u2022 How does 4 turn into -1? \u2013\u00a0user138246 Feb 13 '15 at 23:57\n\u2022 $4\\equiv-1\\pmod{5}$ because $4-(-1)=5$ is divisible by $5$. \u2013\u00a0yurnero Feb 13 '15 at 23:58\n\u2022 I don't follow that logic, can you demonstrate with a more simple example? \u2013\u00a0user138246 Feb 14 '15 at 0:16\n\u2022 I don't know the binomial theorem either but how is it that you are taking the powers of 4 and 9 and turning them both into 1? \u2013\u00a0user138246 Feb 14 '15 at 0:28\n\u2022 @user138246 $a\\equiv b\\pmod{p}\\implies a^n\\equiv b^n\\pmod{p}, \\forall n\\in\\mathbb N$, because $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\\cdots+ab^{n-2}+b^{n-1})$, so $4\\equiv -1\\pmod{5}\\implies 4^{1536}\\equiv (-1)^{1536}\\equiv 1\\pmod{5}$, because $(-1)^{1536}=1$. Same logic for everything else. \u2013\u00a0user26486 Feb 14 '15 at 0:38\n\nNote $4^6\\equiv (-6)^2 \\equiv 1$ (mod $35$) and $6 \\mid 1536$, so $4^{1536}\\equiv 1$ (mod $35$).\n\nSimilarly $9^6 \\equiv 1$ (mod $35$) $\\implies 9^{4824}\\equiv 1$ (mod $35$).\n\nSo we can conclude that $4^{1536}-9^{4824} \\equiv 0$ (mod $35$), i.e. the number is divisible by $35$.\n\n\u2022 I don't understand the $6 | 1536$ part, what does that mean? \u2013\u00a0user138246 Feb 14 '15 at 0:19\n\u2022 @user138246 $6\\mid 1536\\implies 1536=6c$ for some $c\\in\\mathbb N$, so $4^{1536}\\equiv 4^{6\\cdot c}\\equiv (4^6)^c\\equiv 1^c\\equiv 1\\pmod{35}$. We had $(4^6)^c\\equiv 1^c\\pmod{p}$ because $4^6\\equiv 1\\pmod{p}$ and $a\\equiv b\\pmod{p}\\implies a^n\\equiv b^n\\pmod{p},\\forall n\\in\\mathbb N$, because $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\\cdots+ab^{n-2}+b^{n-1})$. \u2013\u00a0user26486 Feb 14 '15 at 0:43\n\nHint mod $\\,5\\!:\\ 4\\equiv -1\\equiv 9\\,\\Rightarrow\\, \\color{#c00}{4^6\\equiv 1\\equiv 9^6}$\n\nand,  mod $\\,7\\!:\\ 9^3\\equiv 2^3\\equiv 1\\,\\Rightarrow\\ \\color{#c00}{4^6\\equiv 1\\equiv 9^6}$\n\nSo mod $\\,35\\!:\\ \\color{#c00}{4^3\\equiv 1\\equiv 9^3}\\$ by CRT  (or by $\\,5,7\\mid a^6\\!-\\!1\\,\\Rightarrow\\,{\\rm lcm}(5,7)=35\\mid a^6\\!-\\!1)$\n\nSo mod $\\,35\\!:\\ \\color{#c00}4^{\\color{#c00}3j} - \\color{#c00}9^{\\color{#c00}9^k}\\equiv \\color{#c00}1^{j}-\\color{#c00}1^{k} \\equiv 1$\n\nIn particular it's true if the exponents are even with digit sum divisible by $\\,3,\\,$ as in your case.\n\nRemark $\\$ Since you say congruence arithmetic is unfamiliar, below is a more elementary proof using only that $\\,a-b\\mid a^n-b^n.\\$ Suppose $\\ 2\\mid j\\!-\\!i \\ge 0.\\$ Then\n\n\\qquad\\quad \\begin{align} 9^{3j}-4^{3i} = &\\ 9^{3j}-4^{3j}\\ +\\ 4^{3j}-4^{3i}\\\\ = &\\ \\color{#0a0}{9^{3j}-4^{3j}}\\ +\\ 4^{3i}\\,(\\color{#c00}{4^{3(j-i)}\\!-1})\\end{align}\n\nBut $\\ 5,7\\mid 9^3-4^3\\mid \\color{#0a0}{9^{3j}-4^{3j}}$ and $\\,5,7\\mid 4^6-1\\mid \\color{#c00}{4^{3(j-i)}\\!-1}\\,$ by $\\,2\\mid j\\!-\\!i\\,\\Rightarrow\\, 6\\mid 3(j\\!-\\!i)$\n\nThus $\\,5,7\\,$ also divide their sum, therefore their lcm = $\\,5\\cdot 7 = 35\\,$ divides their sum.\n\n\u2022 Can you explain your notation, I am not familiar with it enough to read this commit. I don't know what the : implies, I don't know what CRT is, I don't know what lcm is either. Thanks. \u2013\u00a0user138246 Feb 14 '15 at 0:15\n\u2022 @user138246 Do you know congruence arithmetic? i.e. $\\,a\\equiv b\\pmod m\\$ iff $\\ m\\mid a-b\\ \\$ \u2013\u00a0Gone Feb 14 '15 at 0:21\n\u2022 I do not and I do not know what | means \u2013\u00a0user138246 Feb 14 '15 at 0:22\n\u2022 @user138246 Usually these problems are posed after one learns a bit of elementary number theory. You should state in your question that you are not familiar with congruences or modular arithmetic. What textbook are you using? \u2013\u00a0Gone Feb 14 '15 at 0:29\n\u2022 I know a little bit of modular arithmetic and the idea of congruence, I am just not familiar with the math heavy notation and acronyms. \u2013\u00a0user138246 Feb 14 '15 at 0:30\n\nFermat's little theorem:\n\n\"If $p$ is a prime number, then: $\\forall$ $x \\in \\Bbb N / gcd(x,p) = 1$, $x^{p-1} \\equiv 1 \\pmod p$\".\n\nLet $N = 4^{1536} - 9^{4824}$.\n\n$4^{1536} = (4^6)^{256} \\equiv 1 \\pmod 7$ (Fermat's little)\n\n$9^{4824} = (9^6)^{804} \\equiv 1 \\pmod 7$ (Again, Fermat)\n\nThus: $N \\equiv 1 - 1 \\pmod 7 \\equiv 0 \\pmod 7$. This shows that $7 | N$.\n\nNow:\n\n$4^{1536} = (4^4)^{384} \\equiv 1 \\pmod 5$ (Fermat..)\n\n$9^{4824} = (9^4)^{1206} \\equiv 1 \\pmod 5$ (Again)\n\nThus: $N \\equiv 1 - 1 \\pmod 5 \\equiv 0 \\pmod 5$.\n\nThis shows $5|N$.\n\nTherefore $5|N$ and $7|N$. This gives $35|N$.\n\n\u2022 Why is $4^{6*256} = 1 (mod 7)$ I don't follow. 1 mod 7 is just 1 but how did you conclude that 4 raised to some massive power is equal to 1? It makes no sense to me. \u2013\u00a0user138246 Feb 13 '15 at 23:59\n\u2022 $4^6 \\equiv 1 \\pmod 7$ using the aforementioned theorem. Now raise both sides to that \"massive power\". $1$ will still be $1$. \u2013\u00a0user207710 Feb 14 '15 at 0:01\n\u2022 Yes I see that but you are making the claim that some number (that is not 1) is equal to 1. I do not see how that stands. \u2013\u00a0user138246 Feb 14 '15 at 0:04\n\u2022 \"$a \\equiv b \\pmod n$\" means that the remainder of division of $a$ by $n$ is $b$. It doesn't mean that $a = b$. \u2013\u00a0user207710 Feb 14 '15 at 0:08\n\u2022 Was the \"I can't use Fermat\" bit present from the beginning? I completely missed that. \u2013\u00a0user207710 Feb 14 '15 at 0:11\n\nI have little background in this area, but I can determine the answer, ergo anyone else with little background should have no trouble following along.\n\nI know that $x^n$ mod $p$ will follow a cyclic pattern as n is iterated, so I'll first find what those cycles are for $4^n$ mod $35$ and $9^n$ mod $35$, for $n = 0,1,2,3...$\n\n$4^n$ mod $35 = 1,4,16,29,11,9,1,4,16,29...$\n\n$9^n$ mod $35 = 1,9,11,29,16,4,1,9,11,29...$\n\nSo we see that both cycles have a length of 6, so you can know what each number's $nth$ power mod $35$ is by what the exponent $n$ mod $6$ is. $4^a - 9^b$ will be divisible by $35$ if $4^a$ and $9^b$ have the same value mod $35$. Importantly, the two cycles above are in fact the reverse of each other. So $4^a$ will have the same value mod $35$ as $9^b$ if $a$ mod $6 = (6 - (b$ mod $6))$ mod $6$. As it happens, both $1536$ mod $6$ and $4824$ mod $6$ equal $0$, so they do satisfy the aforementioned equation, and therefore $4^{1536} - 9^{4824}$ IS divisible by $35$.\n\nI will suggest to use Euler's Theorem which states $\\forall a \\in \\mathbb{Z}$ s.t. gcd$(a, n)=1$, then $a^{\\phi(n)} \\equiv 1$ mod $n$\n\nIn your case we have $n=35$, $\\quad a=4,9\\quad$ and $\\quad \\phi(35)=24$\n\nSo $\\quad \\quad 4^{24} \\equiv 1$ mod $35\\quad$ and $\\quad 9^{24} \\equiv 1$ mod $35\\quad$\n\nSince $\\quad4$x$64 = 1536 \\quad$ and $\\quad 9$x$201 = 4824$\n\nSo $\\quad \\quad 4^{1536} \\equiv 1$ mod $35\\quad$ and $\\quad 9^{4824} \\equiv 1$ mod $35\\quad$\n\nHence $\\quad 4^{1536} - 9^{4824} \\equiv 1 - 1 \\equiv 0$ mod $35\\quad \\implies 35 | 4^{1536} - 9^{4824}$", "date": "2020-08-11 12:55:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1147212/finding-the-mod-of-a-difference-of-large-powers/1147222", "openwebmath_score": 0.9215911030769348, "openwebmath_perplexity": 345.18076228342176, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846672373524, "lm_q2_score": 0.8757869981319862, "lm_q1q2_score": 0.8568565707381631}}
{"url": "https://community.wolfram.com/groups/-/m/t/1557676", "text": "# Anamorphic Reflections in a Christmas Ball\n\nPosted 7 months ago\n3483 Views\n|\n10 Replies\n|\n48 Total Likes\n|\n Open in Cloud | Download to Desktop via Attachments BelowI recently got inspired by a sculpture sold on Saatchi Art featuring anamorphic deformation by reflection in a spherical mirror. Being curious and interested in anamorphic transformations, I wanted to build something similar and find the math behind it using Mathematica...A plain, undecorated Christmas ball can serve as a perfect convex spherical mirror to test some of our physics and coding skills. I used a 7 cm XMas ball now dumped in stores for Euro1.75 a sixpack! In a nutshell: I wanted to see how a deformed text should look like in order to show up undeformed when reflected in a ball shaped mirror. The graphics below show a spherical mirror centered at C:(0,0,0), our eye at viewpoint V: (xv,0,zv) and a reflected point S on the base plane beneath the ball. One of the reflected light rays leaving S will meet the mirror at Q such that its reflection meets the eye at V. But the eye at V will now perceive the point S at I. I is a perceived image point inside the view disk perpendicular to VC. According to the law of reflection, the lines VQI and SRQ will form equal angles with the normal n to the sphere in Q. All image points will be restricted to a disk that is the base of the view cone with the line CV as axis and an opening angle of tan^-1(zv/xv). This image disk is at an offset 1/xv from C and has a radius of Sqrt[1-(1/xv)^2]. The point Q (q1, q2, q3) is the intersection of the view line VI and the mirror sphere. It can be computed by solving this equation: solQ = NSolve[ Element[{x, y, z}, HalfLine[{imagePointI, viewPointV}]] && Element[{x, y, z}, Sphere[]], {x, y, z}]; pointQ = First[{x, y, z} /. solQ]; {q1, q2, q3} = pointQ; The points C, Q, I, V and S are all in the same plane. We have R, the projection of V to the normal n. projectionPlane = InfinitePlane[pointQ, {pointQ, viewPointV}]; reflectionPt = 2 Projection[viewPointV, pointQ] - viewPointV; The point S is now the intersection of of the line QR with the base plane. It can be computed by solving this equation: solS = NSolve[{{x, y, z} \\[Element] HalfLine[{{q1, q2, q3}, reflectionPt}] && {x, y, z} \\[Element] InfinitePlane[{{0, 0, -1}, {0, 1, -1}, {0, -1, -1}}]}, {x, y, z}]; After simplification, we can write the following function that maps the perceived image point I to the reflected point R : xmasBallMap[iPt : {yi_, zi_}, vPt : {xv_, zv_}] := Module[{imagePtRotated, solQ, q1, q2, q3}, (*image point in real (rotated) pane*) imagePtRotated = {(1 - zi zv)/Norm@vPt, yi, (xv^2 zi + zv)/xv/Norm@vPt}; (*intersection viewline-sphere: Q*) solQ = NSolve[ Element[{x, y, z}, HalfLine[{imagePtRotated, {xv, 0, zv}}]] && Element[{x, y, z}, Sphere[]], {x, y, z}]; {q1, q2, q3} = First[{x, y, z} /. solQ]; Join[{-(1 + q3) (q2^2 + q3^2) xv + q1^2 (xv - q3 xv) + q1^3 (-1 + zv) + q1 q2^2 (-1 + zv) + q1 q3 (q3 (-1 + zv) + 2 zv), q2 (2 q1 xv + q1^2 (-1 + zv) + q2^2 (-1 + zv) + q3 (q3 (-1 + zv) + 2 zv))}/(-2 q1 q3 xv + q3^2 (q3 - zv) + q1^2 (q3 + zv) + q2^2 (q3 + zv)), {-1}]] All possible image points have to fit inside the lower half-disk. This is a grid of image points inside the view disk: pts = Table[ Table[{x, y}, {x, -Floor[Sqrt[1 - y^2], .1] + .1, Floor[Sqrt[1 - y^2], .1] - .1, .025}], {y, 0, -.9, -.025}]; viewDisk = Graphics[{Circle[{0, 0}, 1, {\\[Pi], 2. \\[Pi]}], {AbsolutePointSize[2], Point /@ pts}}, Axes -> True, AxesOrigin -> {-1, -1}, AxesStyle -> Directive[Thin, Red]] This is the reflected spherical anamorphic map of these points:We can see that there is a large magnification between the perceived image inside the ball and it reflected image. Getting a point too close to the rim of the view disk will project its reflection far away. This GIF shows the function in action. The image point I follows a circle in the perceived image disk while its reflection S follows the closed curve of its map xmasBallmap(I, v) in the base plane. We can now further test our function with some text e.g.: \"[MathematicaIcon]Mathematica[MathematicaIcon]\". ma = First[First[ ImportString[ ExportString[ Style[\"\\[MathematicaIcon]Mathematica\\[MathematicaIcon]\", FontFamily -> \"Times\", FontSize -> 72], \"PDF\"], \"TextMode\" -> \"Outlines\"]]] /. FilledCurve :> JoinedCurve; The text image needs to be rescaled and centered to fit inside the ball. maCenteredScaled = ma /. {x_?NumericQ, y_?NumericQ} :> {x, y}*.005 /. {x_?NumericQ, y_?NumericQ} :> {x - .93, y - .45}; This shows the text as should be perceived in the lower half of the mirror sphere:This is the code for a 3D view of the complete setup: the spherical mirror, the perceived text in the disk inside the sphere and the deformed, anamorphic image on the base plane. Quiet@Module[{xv = 10., zv = 3., \\[Phi], rotationTF, pointA, viewPt, mathPts, rotatedMathPts, reflectedPts}, (*view angle*)\\[Phi] = ArcTan[xv, zv]; rotationTF = RotationTransform[-\\[Phi], {0, 1, 0}, {0, 0, 0}]; (*view pane rotation anchor*) pointA = {(0 - .01) Cos[\\[Phi]], 0, (0 - .01) Sin[\\[Phi]]}; (*point coordinates in y-z plane*) mathPts = maCenteredScaled[[-1, 1, All, -1]]; rotatedMathPts = Map[rotationTF, mathPts /. {y_?NumericQ, z_?NumericQ} :> {0, y, z}, {3}]; reflectedPts = Map[xmasBallMap[#, {xv, zv}] &, mathPts, {3}]; Graphics3D[{ (*reflected image plane (floor)*){Opacity[.45], LightBlue, InfinitePlane[{{0, 0, -1}, {1, 0, -1}, {-1, .5, -1}}]}, (*mirror sphere*){Opacity[.35], Sphere[]}, (*center of sphere*){Black, Sphere[{0, 0, 0}, .03]}, (*percieved image pane*){Opacity[.35], Cylinder[{{0, 0, 0}, pointA}, 1]}, (*perceived image*){Red, Line /@ rotatedMathPts}, (*reflected image*){Red, AbsoluteThickness[3], Line /@ reflectedPts}}, Boxed -> False]] Time to try the real thing. This shows a 7cm diameter XMas ball mirror with the text reflected in it. Get yourself a nice reflecting Christmas ball and this is a pdf for you to printout and try it! (see attached pdf file for printing) Attachments:\n10 Replies\nSort By:\nPosted 7 months ago\n Really cool! Thanks for sharing!\nPosted 7 months ago\n - Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming, and consider contributing your work to the The Notebook Archive!\nPosted 7 months ago\n Great post after Cylinder Mirror\uff0cEcher-Style in reverse \uff01\nPosted 7 months ago\n This is really impressive!\nPosted 7 months ago\n @Erik, this is absolutely wonderful holidays idea. I was thinking one can really hide a secret message for loved ones hardly readable on paper but \"magically\" revealed by a Christmas Ball. What a fun computational project!\nPosted 7 months ago\n For those interested, here is a complete notebook to give it a try with your own Xmas wishes! I could not compile so, it is rather slow, be patient... Attachments:\nPosted 7 months ago\n Forgot to add the end result!\nPosted 7 months ago\n Very nice.My first thought was wall art, then I wondered about how high the art would need to be to see the reflection in the reflecting ball that would need to be attached, so table art it is!It might be fun to try something with the Chicago Cloud Gate (the bean), but the dimensions: Dimensions 10 m \u00d7 13 m \u00d7 20 m (33 ft \u00d7 42 ft \u00d7 66 ft) might not make it very feasible to print out something to place at the base of the end of the sculpture.\nPosted 7 months ago\n Wow, Dorothy, Chicago Cloud Gate is an interesting idea ! We'd need to know the geometry of the bean, but just imagining a giant cryptic scribble on the ground spelling out letters in its reflection got me really thinking... I wonder how one would approach this. Drone might be useful for both: general capture to represent best in online media and also for proper reflective angle positioning, possibly, if human eye level will not properly capture the sensibly sized letters. A street artist who makes ground drawing illusions could possibly be involved. Note people can go also UNDER cloud gate, which gives fantastic reflection geometry.\nPosted 7 months ago\n The idea of the Chicago bean is wonderful but we have to consider some specifics of a sphere mirror. Only one half of the sphere will reflect the anamorphic image. Either the lower half from an image on the floor below the sphere or the upper half from an image on the ceiling above the sphere.1.If the anamorphic image is on the floor, the perceived image must be in the lower half of the sphere This could be achieved with a hanging mirror ball. The viewpoint of the observer is at V just above the floor.2.The anamorphic is on the ceiling and then we perceive it in the upper half of the sphere This could be done by painting the anamorphic image on a ceiling above the ball The viewpoint of the observer is at V just above the floor but higher than the sphere center.. In the case of the \"bean\", we have here an upper half of a convex (ball like) mirror and the anamorphic image would need to be on some type of suspended ceiling above the bean mirror surface. Not on the floor.\nCommunity posts can be styled and formatted using the Markdown syntax.", "date": "2019-06-27 00:30:26", "meta": {"domain": "wolfram.com", "url": "https://community.wolfram.com/groups/-/m/t/1557676", "openwebmath_score": 0.1826939731836319, "openwebmath_perplexity": 4055.69662340078, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846666070894, "lm_q2_score": 0.8757869867849167, "lm_q1q2_score": 0.8568565590843882}}
{"url": "https://math.stackexchange.com/questions/840008/finding-the-centralizer-of-a-permutation/840016", "text": "# Finding the centralizer of a permutation\n\nI need to find the centralizer of the permutation $\\sigma=(1 2 3 ... n)\\in S_n$.\nI know that:\n$C_{S_n}(\\sigma)=\\left\\{\\tau \\in S_n|\\text{ } \\tau\\sigma\\tau^{-1}=\\sigma\\right\\}$\nIn other words, that the centralizer is the set of all the elements that commute with $\\sigma$, and I also know that if two permutations have disjoint cycles it implies that they commute, but the thing is; there are no $\\tau\\in S_n$ s.t. $\\tau$ and $\\sigma$ have disjoint cycles, since $\\sigma=(1 2 3...n)$.\nSo can I conclude that $\\sigma$ does not commute with any other $\\tau$ in $S_n$ (besides $id$ of course)?\nI guess my question reduces to: is the second direction of the implication mentioned above also true? meaning, if two permutation commute, does it imply that they have disjoint cycles?\nIf the answer is no, how else can I find the $C_{S_n}(\\sigma)$?\n\nBy the way, on related subject, I noticed that if an element $g$ of a group $G$ is alone it its conjugacy class, it commutes with all elements in $G$.\nWhat does it mean, intuitively, for an element to share its conjugacy class with another element? does it mean it \"almost\" commute with everyone in the group?\nIs it true that the bigger the conjugacy class, the lesser its members commutes with others in the group?\n\n\u2022 No, the other direction does not hold. A permutation will always commute with all powers of itself. \u2013\u00a0Tobias Kildetoft Jun 19 '14 at 20:23\n\u2022 For your last paragraph: \u201cThe size of a conjugacy class is equal to the index of its centralizer.\u201d is the way your truth is often expressed. \u2013\u00a0Jack Schmidt Jun 19 '14 at 20:24\n\u2022 The part in quotation marks in @JackSchmidt's comment also gives you a good way to see how many elements should be in this centralizer, assuming you are familiar with how conjugacy classes look in the symmetric groups. \u2013\u00a0Tobias Kildetoft Jun 19 '14 at 20:25\n\u2022 @TobiasKildetoft, I know that \"two permutations conjugate iff they have the same cycle type\". Is that what you meant? \u2013\u00a0so.very.tired Jun 19 '14 at 20:34\n\u2022 Yes, precisely. \u2013\u00a0Tobias Kildetoft Jun 19 '14 at 20:40\n\nHint: conjugacy in $S_n$ leaves cycle types in tact: $\\tau^{-1}(1 2 3 \\dots n)\\tau=(\\tau(1) \\tau(2)\\tau(3) \\cdots \\tau(n))$.\n\n\u2022 It will only help me figure out how many are there ($(n-1)!$), but will I be able to find them? I'm just confused about how I'm supposed to explicitly write them (the above is a question that might appear in my exam next week). \u2013\u00a0so.very.tired Jun 19 '14 at 20:40\n\u2022 Well if $\\sigma$ is centralized by $\\tau$, so $\\tau^{-1}(1 2 3 \\dots n)\\tau=(\\tau(1) \\tau(2)\\tau(3) \\cdots \\tau(n))=(1 2 3 \\cdots n)$, can you see that $\\tau$ must be a power of $\\sigma$ (try small examples $n=2,3$)? Your observation of $(n-1)!$ is not correct ... \u2013\u00a0Nicky Hekster Jun 19 '14 at 20:49\n\u2022 Yeah, you're right, my $(n-1)!$ observation was mistakenly referred to the size of the centralizer, instead of to the size of the conjugacy class. \u2013\u00a0so.very.tired Jun 19 '14 at 21:06\n\u2022 OK, proving that $\\sigma$ commutes with all powers of itself was easy, and thus giving $<\\sigma>\\subset C_{S_n}(\\sigma)$, but I couldn't figure out from the hint why does it mean that any other element which isn't a power of $\\sigma$ does not belong to the centralizer, meaning that $<\\sigma>= C_{S_n}(\\sigma)$ \u2013\u00a0so.very.tired Jun 19 '14 at 21:18\n\u2022 Ah great! That's your \"learning point\", glad you see it now! \u2013\u00a0Nicky Hekster Jun 19 '14 at 21:43\n\nLet $S_n$ act on itself by conjugation. Let $g = (1,2,\\ldots,n)$. The size of the orbit of $g$ in this action is its conjugacy class $\\{ h^{-1} gh: h \\in S_n\\}$. The stabilizer subgroup of $g$ in this action is the set of elements $h$ in $S_n$ such that $h^{-1}gh=g$, i.e. the centralizer $C_{S_n}(g)$. By the orbit-stabilizer lemma, the size of the orbit equals the index of the stabilizer. The size of the orbit is the number of elements that have the same cycle structure as $g$, which is $(n-1)!$. Thus, the index of the centralizer is $(n-1)!$, whence the centralizer has $n! / (n-1)!=n$ elements. Thus the powers of $g$ exhaust all of $C_{S_n}(g)$.\n\nA second proof is as follows. For the special case where $g=(1,2,\\ldots,n)$, we can determine its centralizer in $S_n$ without using the orbit-stabilizer lemma. If $h^{-1}gh = g$, then $(h(1),h(2),\\ldots,h(n)) = (1,2,\\ldots,n)$. Now, $h(1)$ can be chosen in $n$ ways to be any of $1,2,\\ldots,n$, but once $h(n)$ is chosen, the remaining elements $h(2),\\ldots,h(n)$ are uniquely determined. In fact, if $h(1)=i$, then $h(2)=i+1$, and so on, and so $h$ is just a power of $g$. Thus, there are exactly $n$ different elements $h$ such that $h^{-1}gh=g$.\n\nNote that $$|\\sigma| = n$$ so that $$\\sigma^n = 1$$. So $$\\langle \\sigma \\rangle = \\{1, \\sigma, \\sigma^2, \\ldots, \\sigma^{n-1}\\}$$ forms a cyclic commutative subgroup of $$S_n$$ with order $$|\\langle \\sigma \\rangle| = n$$. For any permutation $$x$$ of a symmetric group $$S_n$$ we have the very convenient formula: $$|C(x)| = 1^{\\alpha_1}2^{\\alpha_2}\\cdots n^{\\alpha_n}\\alpha_1!\\alpha_2!\\ldots\\alpha_n!,$$ where $$\\alpha_i$$ is the number of cycles in $$x$$ of length $$i$$. For your permutation $$\\alpha_1 = \\alpha_2 = \\cdots \\alpha_{n-1} = 0$$ and $$\\alpha_n = n$$ so $$|C(\\sigma)| = n$$. We have $$|C(\\sigma)| = |\\langle \\sigma \\rangle| = n$$, implying $$C(\\sigma) = \\langle \\sigma \\rangle$$.", "date": "2019-06-24 17:44:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/840008/finding-the-centralizer-of-a-permutation/840016", "openwebmath_score": 0.942283034324646, "openwebmath_perplexity": 150.59890430573006, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384664716301, "lm_q2_score": 0.8757869884059267, "lm_q1q2_score": 0.8568565590144316}}
{"url": "https://math.stackexchange.com/questions/351856/why-does-the-series-sum-limits-n-2-infty-frac-cosn-pi-3n-converge?noredirect=1", "text": "# Why does the series $\\sum\\limits_{n=2}^\\infty\\frac{\\cos(n\\pi/3)}{n}$ converge?\n\nWhy does this series $$\\sum\\limits_{n=2}^\\infty\\frac{\\cos(n\\pi/3)}{n}$$ converge? Can't you use a limit comparison with $1/n$?\n\n\u2022 Can you please edit the title so it is clear what you are asking? Is it $\\sum_{n=1}^{\\infty} \\cos \\dfrac{n\\pi}{3n}$ \u2013\u00a0Aryabhata Apr 5 '13 at 6:04\n\u2022 Careful, your first statement is not correct. The cosine term will oscillate as $n$ gets large and never approach a single value. \u2013\u00a0Jared Apr 5 '13 at 6:04\n\u2022 @Jared you are right, will correct it. Is it an alternating series? \u2013\u00a0Billy Thompson Apr 5 '13 at 6:07\n\u2022 Yes, this sounds like the way to go. It's not a strictly alternating series, but the sign change in cosine is what causes convergence. You may even be able to evaluate this explicitly using telescoping series, because we know the values of $\\cos(\\frac{n\\pi}{3})$ for all integral $n$. \u2013\u00a0Jared Apr 5 '13 at 6:08\n\u2022 @Jared is there any other way to evaluate it? \u2013\u00a0Billy Thompson Apr 5 '13 at 6:10\n\nFirst of all your conclusion is wrong since $\\lim_{n \\to \\infty} \\cos(n \\pi/3)$ doesn't exist.\n\nThe convergence of $$\\sum_{n=1}^N \\dfrac{\\cos(n\\pi/3)}{n}$$ can be concluded based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.\n\nConsider the sum $S_N = \\displaystyle \\sum_{n=1}^N a(n)b(n)$. Let $A(n) = \\displaystyle \\sum_{n=1}^N a(n)$. If $b(n) \\downarrow 0$ and $A(n)$ is bounded, then the series $\\displaystyle \\sum_{n=1}^{\\infty} a(n)b(n)$ converges.\n\nFirst note that from Abel summation, we have that \\begin{align*}\\sum_{n=1}^N a(n) b(n) &= \\sum_{n=1}^N b(n)(A(n)-A(n-1))\\\\&= \\sum_{n=1}^{N} b(n) A(n) - \\sum_{n=1}^N b(n)A(n-1)\\\\ &= \\sum_{n=1}^{N} b(n) A(n) - \\sum_{n=0}^{N-1} b(n+1)A(n) \\\\&= b(N) A(N) - b(1)A(0) + \\sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))\\end{align*} Now if $A(n)$ is bounded i.e. $\\vert A(n) \\vert \\leq M$ and $b(n)$ is decreasing, then we have that $$\\sum_{n=1}^{N-1} \\left \\vert A(n) \\right \\vert (b(n)-b(n+1)) \\leq \\sum_{n=1}^{N-1} M (b(n)-b(n+1))\\\\ = M (b(1) - b(N)) \\leq Mb(1)$$ Hence, we have that $\\displaystyle \\sum_{n=1}^{N-1} \\left \\vert A(n) \\right \\vert (b(n)-b(n+1))$ converges and hence $$\\displaystyle \\sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ converges absolutely. Now since $$\\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \\sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ we have that $\\displaystyle \\sum_{n=1}^N a(n)b(n)$ converges.\n\nIn your case, $a(n) = \\cos(n \\pi/3)$. Hence, $$A(N) = \\displaystyle \\sum_{n=1}^N a(n) = - \\dfrac12 - \\cos\\left(\\dfrac{\\pi}3(N+2)\\right)$$which is clearly bounded.\n\nAlso, $b(n) = \\dfrac1{n}$ is a monotone decreasing sequence converging to $0$.\n\nHence, we have that $$\\sum_{n=1}^N \\dfrac{\\cos(n\\pi/3)}{n}$$ converges.\n\nLook at some of my earlier answers for similar questions.\n\nFor what real numbers $a$ does the series $\\sum \\frac{\\sin(ka)}{\\log(k)}$ converge or diverge?\n\nGive a demonstration that $\\sum\\limits_{n=1}^\\infty\\frac{\\sin(n)}{n}$ converges.\n\nIf the partial sums of a $a_n$ are bounded, then $\\sum{}_{n=1}^\\infty a_n e^{-nt}$ converges for all $t > 0$\n\nIf you are interested in evaluating the series, here is a way out. We have for $\\vert z \\vert \\leq 1$ and $z \\neq 1$, $$\\sum_{n=1}^{\\infty} \\dfrac{z^n}n = - \\log(1-z)$$ Setting $z = e^{i \\pi/3}$, we get that $$\\sum_{n=1}^{\\infty} \\dfrac{e^{in \\pi/3}}n = - \\log(1-e^{i \\pi/3})$$ Hence, \\begin{align} \\sum_{n=1}^{\\infty} \\dfrac{\\cos(n \\pi/3)}n & = \\text{Real part of}\\left(\\sum_{n=1}^{\\infty} \\dfrac{e^{in \\pi/3}}n \\right)\\\\ & = \\text{Real part of} \\left(- \\log(1-e^{i \\pi/3}) \\right)\\\\ & = - \\log(\\vert 1-e^{i \\pi/3} \\vert) = 0 \\end{align} Hence, $$\\sum_{n=2}^{\\infty} \\dfrac{\\cos(n \\pi/3)}n = - \\dfrac{\\cos(\\pi/3)}1 = - \\dfrac12$$\n\n\u2022 I think it is time to write a general answer for generalized alternating test and close tons of similar questions as abstract duplicates. \u2013\u00a0user17762 Apr 5 '13 at 6:13\n\u2022 I support the motion. \u2013\u00a0Did Apr 5 '13 at 7:10\n\u2022 I'm a little confused about something. On the one hand, you show that $\\sum a_n b_n$ should converge absolutely. But $\\sum \\cos (n \\pi 3) n^{-1}$ does not converge absolutely, as the nth term is at least $\\frac{1}{2n}$ in absolute value. So I feel I must be missing something? \u2013\u00a0davidlowryduda Apr 5 '13 at 8:14\n\u2022 @mixedmath Yes, you are absolutely right. $\\sum a_n b_n$ doesn't converge absolutely. I have changed it. The proof remains un affected. What we have is $\\sum A(n)(b(n) - b(n+1))$ converges absolutely and this is what we want. This doesn't mean that $\\sum a(n)b(n)$ converges absolutely. Thanks for pointing this out. \u2013\u00a0user17762 Apr 5 '13 at 15:16\n\u2022 Awesome. Thanks, and great writeup! \u2013\u00a0davidlowryduda Apr 5 '13 at 16:03\n\nNote that $$\\cos(n\\pi/3) = 1/2, \\ -1/2, \\ -1, \\ -1/2, \\ 1/2, \\ 1, \\ 1/2, \\ -1/2, \\ -1, \\ \\cdots$$ so your series is just 3 alternating (and convergent) series inter-weaved. Exercise: Prove that if $\\sum a_n, \\sum b_n$ are both convergent, then the sequence $$a_1, a_1+b_1, a_1+b_1+a_2, a_1+b_1+a_2+b_2, \\cdots$$ is convergent. Applying that twice proves your series converges.", "date": "2019-09-22 12:37:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/351856/why-does-the-series-sum-limits-n-2-infty-frac-cosn-pi-3n-converge?noredirect=1", "openwebmath_score": 0.9759758710861206, "openwebmath_perplexity": 271.8226776321927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846691281407, "lm_q2_score": 0.8757869835428966, "lm_q1q2_score": 0.8568565581203493}}
{"url": "https://math.stackexchange.com/questions/2994429/how-can-i-argue-that-for-a-number-to-be-divisible-by-144-it-has-to-be-divisible/2994452", "text": "# How can I argue that for a number to be divisible by 144 it has to be divisible by 36?\n\nSuppose some number $$n \\in \\mathbb{N}$$ is divisible by $$144$$.\n\n$$\\implies \\frac{n}{144}=k, \\space \\space \\space k \\in \\mathbb{Z} \\\\ \\iff \\frac{n}{36\\cdot4}=k \\iff \\frac{n}{36}=4k$$\n\nSince any whole number times a whole number is still a whole number, it follows that $$n$$ must also be divisible by $$36$$. However, what I think I have just shown is:\n\n$$\\text{A number }n \\space \\text{is divisble by} \\space 144 \\implies n \\space \\text{is divisible by} \\space 36 \\space (1)$$\n\nIs that the same as saying: $$\\text{For a number to be divisible by 144 it has to be divisible by 36} \\space (2)$$\n\nIn other words, are statements (1) and (2) equivalent?\n\n\u2022 Yes, absolutely. \u2013\u00a0Bernard Nov 11 '18 at 20:47\n\u2022 Pedantically, the first sentence could have a different meaning if it somehow isn't obvious from context that $n$ is universally quantified. But in general they're the same. \u2013\u00a0Ian Nov 11 '18 at 20:49\n\u2022 You mean this sentence \"Since any whole number times a whole number is still a whole number, it follows that n must also be divisible by 36. \",right? \u2013\u00a0Nullspace Nov 11 '18 at 20:52\n\u2022 In general, if $a$ is divisible by $b$, and $b$ is divisible by $c$, then $a$ is divisible by $c$. \u2013\u00a0AlexanderJ93 Nov 12 '18 at 0:14\n\nYes, it's the same. $$A\\implies B$$ is equivalent to \"if we have $$A$$, we must have $$B$$\".\n\nAnd your proof looks fine. Good job.\n\nIf I were to offer some constructive criticism, it would be of the general kind: In number theory, even though we call the property \"divisible\", we usually avoid division whenever possible. Of the four basic arithmetic operations it is the only one which makes integers into non-integers. And number theory is all about integers.\n\nTherefore, \"$$n$$ is divisible by $$144$$\", or \"$$144$$ divides $$n$$\" as it's also called, is defined a bit backwards:\n\nThere is an integer $$k$$ such that $$n=144k$$\n\n(This is defined for any number in place of $$144$$, except $$0$$.)\n\nUsing that definition, your proof becomes something like this:\n\nIf $$n$$ is divisible by $$144$$, then there is an integer $$k$$ such that $$n=144k$$. This gives $$n=144k=(36\\cdot4)k=36(4k)$$ Since $$4k$$ is an integer, this means $$n$$ is also divisible by $$36$$.\n\n\u2022 Programmers might may say mod() = 0 rather than \"divides\", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers. \u2013\u00a0mckenzm Nov 12 '18 at 0:30\n\u2022 Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc. \u2013\u00a0Daniel R. Collins Nov 13 '18 at 5:15\n\nYes that's correct or simply note that\n\n$$n=144\\cdot k= 36\\cdot (4\\cdot k)$$\n\nbut $$n=36$$ is not divisible by $$144$$.\n\nThe $$\\implies$$ symbol is defined as follows:\n\nIf $$p \\implies q$$ then if $$p$$ is true, then $$q$$ must also be true. So when you say $$144 \\mid n \\implies 36 \\mid n$$ it's the same thing as saying that if $$144 \\mid n$$, then it must also be true that $$36 \\mid n$$.\n\nThere are often multiple approaches to proofs. Is usually good to be familiar with multiple techniques.\n\nYou have a very good approach. Parsimonious and references only the particular entities at hand.\n\nSome times, for the sake of illustrating the use of additional concepts, you might want to deviate from parsimony.\n\nIn that spirit, here's an additional proof.\n\nAccording to The Fundamental Theorem of Arithmetic, if something is divisible by 144, then it is divisible by at least the same primes raised to the powers you need to yield 144. In other words, $$144=2^43^2$$. For something to be divisible by 144, the primes 2 and 3 must appear in its prime factorization. The powers of 2 and 3 must be at least 4 and 2, respectively. Now $$36=2^23^2$$. So any number who's prime factorization incluedes the primes 2 and 3, and has them raised at least to the power of 2, then it is also divisible by 36. If something is divisible by 144, we are guaranteed that 2 and 3 appear in its prime factorization. We area also guaranteed that the exponents on 2 and 3 are 4 and 2 respectively. So divisibility by 144 implies divisibility by 36 since the exponents satisfy the established criteria.\n\nThe prime factorization of 144 is 2 * 2 * 2 * 2 * 3 * 3. The prime factorization of 36 is 2 * 2 * 3 * 3.\n\nIf X is divisible by Y, then X's prime factorization contains all of the factors in Y's prime factorization.\n\nSince 144's prime factorization contains all of the factors in 36's prime factorization, 144 is divisible by 36.\n\nThe prime factorization of any number that is divisible by 144 contains all of the prime factors of 144, which contains all of the prime factors of 36. Hence any number that is divisible by 144 has a prime factorization that contains all the prime factors of 36, and therefore is divisible by 36.\n\nIf $$n$$ is a multiple of $$144$$, it must automatically be a multiple of any divisor $$d$$ of $$144$$. The prime factorization of $$144$$ is $$(2^4)(3^2)$$, and so $$36 = (2^2)(3^2)$$ must be a divisor of $$144$$ and hence of $$n$$.", "date": "2019-10-22 13:51:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2994429/how-can-i-argue-that-for-a-number-to-be-divisible-by-144-it-has-to-be-divisible/2994452", "openwebmath_score": 0.802426815032959, "openwebmath_perplexity": 160.24049583645848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846691281407, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8568565565343779}}
{"url": "https://www.physicsforums.com/threads/confused-with-a-series-problem.48722/", "text": "# Confused with a series problem\n\n1. Oct 20, 2004\n\nProblem\n\nLet $$\\sum a_n$$ be a series with positive terms and let $$r_n = \\frac{a_{n+1}}{a_n}$$. Suppose that $$\\lim _{n \\to \\infty} r_n = L < 1$$, so $$\\Sum a_n$$ converges by the Ratio Test. As usual, we let $$R_n$$ be the remainder after $$n$$ terms, that is,\n\n$$R_n = a_{n+1} + a_{n+2} + a_{n+3} + \\cdots$$ \u200b\n\n(a) If $$\\left\\{ r_n \\right\\}$$ is a decreasing sequence and $$r_{n+1} < 1$$, show by summing a geometric series, that\n\n$$R_n \\leq \\frac{a_{n+1}}{1-r_{n+1}}$$\u200b\n\n(b) If $$\\left\\{ r_n \\right\\}$$ is an increasing sequence, show that\n\n$$R_n \\leq \\frac{a_{n+1}}{1-L}$$\u200b\n\nMy Solution\n\n(a) The first term $$r_{n+1}$$ prevails, since it represents an upper bound to other terms of $$\\left\\{ r_n \\right\\}$$ . Then,\n\n$$R_n \\leq a_{n+1} + a_{n+1}r_{n+1} + a_{n+1}\\left( r_{n+1} \\right) ^2 + a_{n+1}\\left( r_{n+1} \\right) ^3 + \\cdots = \\sum _{n=1} ^{\\infty} a_{n+1}\\left( r_{n+1} \\right) ^{n-1} = \\frac{a_{n+1}}{1-r_{n+1}}$$\n\n(b) The last term $$L$$ prevails, since it represents an upper bound to other terms of $$\\left\\{ r_n \\right\\}$$ . Then,\n\n$$R_n \\leq a_{n+1} + a_{n+1}L + a_{n+1}L ^2 + a_{n+1} L ^3 + \\cdots = \\sum _{n=1} ^{\\infty} a_{n+1}L ^{n-1} = \\frac{a_{n+1}}{1-L}$$\n\nQuestions\n\n1. Did I get it right?\n2. Why use \"$$\\leq$$\" instead of \"$$<$$\", since all terms of $$\\left\\{ r_n \\right\\}$$ are smaller then their respective upper bounds?\n3. Isn't an increasing $$\\left\\{ r_n \\right\\}$$ rather contra-intuitive when we consider a convergent series $$\\sum a_n$$, which obeys: $$\\lim _{n \\to \\infty} a_n =0$$?\n\nThat's it. Thank you very much!!\n\n2. Oct 20, 2004\n\n### NateTG\n\nWell, it really should be:\n$$\\lim_{n\\rightarrow \\infty} | r_n | < 1$$\notherwise, it would be possible to sneak in things like $$\\sum 2^{-2n}$$ where $$r_n=-2<1$$.\n\nIf you can assume that $$|r_n|$$ is decreasing and $$|r_{n+1}|<1$$\nYou need to use absolute values to make the inequalities work:\n$$|\\sum_{i=n+1}^\\infty a_i| \\leq \\sum_{i=n+1}^\\infty |a_i| \\leq \\sum_{i=n+1}^\\infty |a_n r_{n+1}^{i-n-1}|=\\sum_{i=1}^\\infty |a_n| |r_{n+1}|^i$$\nYou have the right idea, but what you have is not true if $$r_{n+1}$$ is positive and $$a_n$$ is negative.\n\nRegarding the use of $$\\leq$$ rather than $$<$$:\nConstant sequences are often included in the notion of decreasing or increasing sequence, so there may be equality rather than strict inequality.\n\nRegarding the existance of increasing $${r_n}$$\nConsider, for example the possibility that $$r_n=\\frac{1}{2}-\\frac{1}{2^{n+1}}$$. It's pretty easy to see that the limit $$\\lim_{n\\rightarrow \\infty} r_n = \\frac{1}{2}$$ and that $$\\{r_n\\}$$ is increasing. However, since all of the $$r_n$$ are positive and less than $$\\frac{1}{2}$$ we have:\n$$0< r_n < \\frac{1}{2}$$\nNow\n$$a_n=a_{n-1}\\times r_{n-1}$$\nso\n$$\\sum_{i=0}^{\\infty} a_n = \\sum_{i=0}^{\\infty} (a_0 \\times \\prod_{j=0}^{i} r_j) < \\sum_{i=0}^{\\infty} (a_0 \\times \\prod_{j=0}^{i}\\frac{1}{2}) = a_0 \\times \\sum_{i=0}^{\\infty} \\frac{1}{2^n} = 2a_0$$\nwhich means that although $$\\{r_n\\}$$ is increasing the sum is convergent.\n\nLast edited: Oct 20, 2004", "date": "2017-03-28 15:56:41", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/confused-with-a-series-problem.48722/", "openwebmath_score": 0.9563162326812744, "openwebmath_perplexity": 251.58633640732526, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846666070896, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8568565543264741}}
{"url": "http://www.maplesoft.com/support/help/Maple/view.aspx?path=simplify/details", "text": "apply simplification rules to an expression - Maple Help\n\n# Online Help\n\n###### All Products\u00a0\u00a0\u00a0 Maple\u00a0\u00a0\u00a0 MapleSim\n\nHome : Support : Online Help : Mathematics : Algebra : Expression Manipulation : Simplifying : simplify : simplify/details\n\nsimplify\u00a0- apply simplification rules to an expression\n\n Calling Sequence simplify(expr, n1, n2, ...) simplify(expr, side1, side2, ...) simplify(expr, assume=prop) simplify(expr, symbolic)\n\nParameters\n\n expr - any expression n1, n2, ... - (optional) names; simplification procedures side1, side2, ... - (optional) sets or lists; side relations prop - (optional) any property\n\nBasic Information\n\n \u2022 This help page contains complete information about the simplify\u00a0command. For basic information on the simplify\u00a0command, see the simplify\u00a0help page.\n\nDescription\n\n \u2022 The simplify\u00a0command is used to apply simplification rules to an expression.\n \u2022 The simplify/expr\u00a0calling sequence searches the expression, expr, for function calls, square roots, radicals, and powers. It then invokes the appropriate simplification procedures.\n\nExamples\n\nSimple Example\n\n > $\\mathrm{simplify}\\left({4}^{\\frac{1}{2}}+3\\right)$\n ${5}$ (1)\n\nSimplifying Trigonometric Expressions\n\n > $e:={\\mathrm{cos}\\left(x\\right)}^{5}+{\\mathrm{sin}\\left(x\\right)}^{4}+2{\\mathrm{cos}\\left(x\\right)}^{2}-2{\\mathrm{sin}\\left(x\\right)}^{2}-\\mathrm{cos}\\left(2x\\right):$\n > $\\mathrm{simplify}\\left(e\\right)$\n ${{\\mathrm{cos}}{}\\left({x}\\right)}^{{4}}{}\\left({\\mathrm{cos}}{}\\left({x}\\right){+}{1}\\right)$ (2)\n\nSimplifying Exponentials and Logarithms\n\n > $\\mathrm{simplify}\\left({\u2147}^{a+\\mathrm{ln}\\left(b{\u2147}^{c}\\right)}\\right)$\n ${b}{}{{\u2147}}^{{a}{+}{c}}$ (3)\n\nControlling Simplification Rules\n\n > $\\mathrm{simplify}\\left({\\mathrm{sin}\\left(x\\right)}^{2}+\\mathrm{ln}\\left(2x\\right)+{\\mathrm{cos}\\left(x\\right)}^{2}\\right)$\n ${1}{+}{\\mathrm{ln}}{}\\left({2}\\right){+}{\\mathrm{ln}}{}\\left({x}\\right)$ (4)\n > $\\mathrm{simplify}\\left({\\mathrm{sin}\\left(x\\right)}^{2}+\\mathrm{ln}\\left(2x\\right)+{\\mathrm{cos}\\left(x\\right)}^{2},\\mathrm{trig}\\right)$\n ${1}{+}{\\mathrm{ln}}{}\\left({2}{}{x}\\right)$ (5)\n > $\\mathrm{simplify}\\left({\\mathrm{sin}\\left(x\\right)}^{2}+\\mathrm{ln}\\left(2x\\right)+{\\mathrm{cos}\\left(x\\right)}^{2},\\mathrm{ln}\\right)$\n ${{\\mathrm{sin}}{}\\left({x}\\right)}^{{2}}{+}{\\mathrm{ln}}{}\\left({2}\\right){+}{\\mathrm{ln}}{}\\left({x}\\right){+}{{\\mathrm{cos}}{}\\left({x}\\right)}^{{2}}$ (6)\n\nSimplifying With Respect to Side Relations\n\n > $f:=-\\frac{1{x}^{5}y}{3}+{x}^{4}{y}^{2}+\\frac{1x{y}^{3}}{3}+1:$\n > $\\mathrm{simplify}\\left(f,\\left\\{{x}^{3}=xy,{y}^{2}=x+1\\right\\}\\right)$\n ${{x}}^{{4}}{+}{{x}}^{{2}}{+}{x}{+}{1}$ (7)\n\nUsing the assume option\n\n > $g:=\\sqrt{{x}^{2}}$\n ${g}{:=}\\sqrt{{{x}}^{{2}}}$ (8)\n > $\\mathrm{simplify}\\left(g\\right)$\n ${\\mathrm{csgn}}{}\\left({x}\\right){}{x}$ (9)\n > $\\mathrm{simplify}\\left(g,\\mathrm{assume}=\\mathrm{real}\\right)$\n $\\left|{x}\\right|$ (10)\n > $\\mathrm{simplify}\\left(g,\\mathrm{assume}=\\mathrm{positive}\\right)$\n ${x}$ (11)\n > $\\mathrm{simplify}\\left(g,\\mathrm{symbolic}\\right)$\n ${x}$ (12)\n\nSimplifying an Integral\n\n Integrands and summands are simplified taking into account the integration or sum ranges respectively. For more information, see assuming.\n > $\\mathrm{expr}:={{\u222b}}_{1}^{4}{\\left(1+{\\mathrm{sinh}\\left(t\\right)}^{2}\\right)}^{\\frac{1}{2}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{\u2146}t$\n ${\\mathrm{expr}}{:=}{{\u222b}}_{{1}}^{{4}}\\sqrt{{1}{+}{{\\mathrm{sinh}}{}\\left({t}\\right)}^{{2}}}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{\u2146}{t}$ (13)\n > $\\mathrm{simplify}\\left(\\mathrm{expr}\\right)$\n ${{\u222b}}_{{1}}^{{4}}{\\mathrm{cosh}}{}\\left({t}\\right)\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}{\u2146}{t}$ (14)\n See Also\n\n## Was this information helpful?\n\n Please add your Comment (Optional) E-mail Address (Optional) What is ? This question helps us to combat spam", "date": "2016-02-12 18:14:30", "meta": {"domain": "maplesoft.com", "url": "http://www.maplesoft.com/support/help/Maple/view.aspx?path=simplify/details", "openwebmath_score": 0.9764902591705322, "openwebmath_perplexity": 6451.946511462495, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846640860381, "lm_q2_score": 0.8757869835428966, "lm_q1q2_score": 0.8568565537045415}}
{"url": "https://svadali.com/2018/09/15/random-number-puzzle/", "text": "# Random Number Puzzle\n\nSuppose that you have a function that you can use to generate uniformly distributed random numbers between $$1$$ and $$5$$. How can you use the above function to generate uniformly distributed random numbers between $$1$$ and $$7$$?\n\nThe key to solving puzzles such as the one above is to first recognize that if we can somehow generate $$7$$ equally likely outcomes then we can use each one of these $$7$$ outcomes to output one of the integers between $$1$$ and $$7$$ as shown below:\n\n$$\\text{Outcome}\\ 1 \\longrightarrow 1$$\n\n$$\\text{Outcome}\\ 2 \\longrightarrow 2$$\n\n$$\\text{Outcome}\\ 3 \\longrightarrow 3$$\n\n$$\\text{Outcome}\\ 4 \\longrightarrow 4$$\n\n$$\\text{Outcome}\\ 5 \\longrightarrow 5$$\n\n$$\\text{Outcome}\\ 6 \\longrightarrow 6$$\n\n$$\\text{Outcome}\\ 7 \\longrightarrow 7$$\n\nSo, how can we generate $$7$$ equally likely outcomes when you have a function that only outputs $$5$$ equally likely outcomes. The trick is to change the space of outcomes. One way to change the outcome space is to draw two random numbers uniformly distributed between $$1$$ and $$5$$. The outcome space when we draw two numbers is as follows:\n\n$$\\{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), \\\\ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), \\\\ (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), \\\\ (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), \\\\ (5, 1), (5, 2), (5, 3), (5, 4), (5, 5) \\}$$\n\nOf course, now we have $$25$$ equally likely outcomes and not $$7$$. But, what if we were to map $$7$$ of the above outcomes to our desired numbers and ignore the others. In other words, our random number generator to draw random numbers between $$1$$ and $$7$$ does the following:\n\n$$(1, 1) \\longrightarrow 1$$\n\n$$(1, 2) \\longrightarrow 2$$\n\n$$(1, 3) \\longrightarrow 3$$\n\n$$(1, 4) \\longrightarrow 4$$\n\n$$(1, 5) \\longrightarrow 5$$\n\n$$(2, 1) \\longrightarrow 6$$\n\n$$(2, 2) \\longrightarrow 7$$\n\nIf we obtain any outcome apart from the ones listed above we try again till we obtain one of the above outcomes. All the above outcomes are equally likely and there are $$7$$ outcomes. Thus, one would expect the above procedure to generate uniformly distributed random numbers between $$1$$ and $$7$$.\n\nWe could go through the math to convince ourselves but writing a simulation is another way to validate the above intuition. The plot below shows the percentage of times each number occurs in a simulation where we use the above procedure to generate numbers between $$1$$ and $$7$$. Note that the chances of obtaining any of these outcomes is $$\\frac{1}{7} \\approx 0.143$$. Thus, we have strong evidence that the procedure does generate numbers uniformly between\u00a0$$1$$ and $$7$$. For those interested, the Python code used to generate the plot is at the end of the post.\n\nA related problem to think about: How can we generate uniformly distributed numbers between $$1$$ and $$30$$ if we have access to a function that only gives us random numbers between $$1$$ and $$5$$?\n\nimport random\n\nimport matplotlib.pyplot as plt\nimport seaborn as sns\n\ndef draw_histogram(draws):\nax = sns.barplot(x=draws, y=draws, estimator=lambda x: len(x) / len(draws) * 100)\nax.set_xlabel('Number')\nax.set_ylabel('Percentage')\nax.set_ylim(0, 100)\n\nfor p in ax.patches:\nax.annotate(\"%.0f\" % p.get_height() + '%', (p.get_x() + p.get_width() / 2., p.get_height() + 3),\nha='center', va='center', fontsize=12, color='black', rotation=0, xytext=(0, 0),\ntextcoords='offset points')\n\ndef generate_rand_between_1_and_7():\ndraw = None\nwhile draw is None:\ndraw_1 = random.randint(1, 5)\ndraw_2 = random.randint(1, 5)\ndraw_sequence = (draw_1, draw_2)\n\nif draw_sequence == (1, 1):\ndraw = 1\nelif draw_sequence == (1, 2):\ndraw = 2\nelif draw_sequence == (1, 3):\ndraw = 3\nelif draw_sequence == (1, 3):\ndraw = 3\nelif draw_sequence == (1, 4):\ndraw = 4\nelif draw_sequence == (1, 5):\ndraw = 5\nelif draw_sequence == (2, 1):\ndraw = 6\nelif draw_sequence == (2, 2):\ndraw = 7\nelse:\npass\nreturn draw\n\nif __name__ == '__main__':\nrandom.seed(42)\n\nno_of_draws = 10000\n\ndraws_between_1_and_7 = []\nfor _ in range(no_of_draws):\ndraw = generate_rand_between_1_and_7()\ndraws_between_1_and_7.append(draw)\n\ndraw_histogram(draws_between_1_and_7)\n\nplt.savefig('histogram_of_draws.png', bbox_inches='tight')\n\n\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2019-01-17 06:28:46", "meta": {"domain": "svadali.com", "url": "https://svadali.com/2018/09/15/random-number-puzzle/", "openwebmath_score": 0.7763149738311768, "openwebmath_perplexity": 635.3333642118325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227573874892, "lm_q2_score": 0.8633916099737807, "lm_q1q2_score": 0.8568494822754031}}
{"url": "https://doc.simo.ac/Functions/cumsum/", "text": "# cumsum\n\nCumulative sum\n\n### s = cumsum(A)\n\n\u2022 A is an array of any number of dimensions.\n\u2022 The output argument s contains the cumulative sums and has the same size as A.\n\u2022 The cumulative sums are obtained from vectors along the first non-singleton dimension of A.\n\nExample 1: In the following, the first non-singleton dimension of A is the first dimension. The sum operation is performed along the vertical direction. For instance, s(:,1) contains cumulative sums of elements of A(:,1).\n\n% Matrix of size [3,4]\nA=reshape(1:12,3,4)\n% Cumulative sums\ns=cumsum(A)\n\nA =\n1.000   4.000   7.000   10.00\n2.000   5.000   8.000   11.00\n3.000   6.000   9.000   12.00\n\ns =\n1.000   4.000   7.000   10.00\n3.000   9.000   15.00   21.00\n6.000   15.00   24.00   33.00\n\n\n### s = cumsum(A, dim)\n\n\u2022 dim should be a positive integer scalar, not equal to inf or nan.\n\u2022 It obtains the cumulative sums along the dim-th dimension of A.\n\u2022 If size(A,dim) == 1 or dim > ndims(A), then s is the same as A.\n\nExample 2: In the following, The sum operation is performed along the horizontal direction (2nd dimension). For instance, s(1,:) contains cumulative sums of elements of A(1,:).\n\n% Matrix of size [3,4]\nA=reshape(1:12,3,4)\n% Cumulative sums along 2nd dimension\ns=cumsum(A,2)\n\nA =\n1.000   4.000   7.000   10.00\n2.000   5.000   8.000   11.00\n3.000   6.000   9.000   12.00\n\ns =\n1.000   5.000   12.00   22.00\n2.000   7.000   15.00   26.00\n3.000   9.000   18.00   30.00\n\n\n### s = cumsum(A, option)\n\n\u2022 option should be either 'reverse', 'forward', 'includenan' or 'omitnan'.\n\u2022 These options control the operation direction, or whether or not NaN should be included in the operation. For details, see Tables 1 and 2 below.\n\nNote\n\nBy default, cummin and cummax omit NaN, whereas cumsum and cumprod include NaN.\n\nExample 3: Cumulative sums of the same vector but with different options.\n\na=[1:5 nan 6]\n% Default options\ncumsum(a)\n% Include nan, forward direction\ncumsum(a, 'includenan')\n% Omit nan, forward direction\ncumsum(a, 'omitnan')\n% Forward direction, include nan\ncumsum(a, 'forward')\n% Reverse direction, include nan\ncumsum(a, 'reverse')\n\na =\n1.000   2.000   3.000   4.000   5.000    nan   6.000\n\nans =\n1.000   3.000   6.000   10.00   15.00    nan    nan\n\nans =\n1.000   3.000   6.000   10.00   15.00    nan    nan\n\nans =\n1.000   3.000   6.000   10.00   15.00   15.00   21.00\n\nans =\n1.000   3.000   6.000   10.00   15.00    nan    nan\n\nans =\nnan    nan    nan    nan    nan    nan   6.000\n\n\nTable 1: Options for operation direction.\n\nOption value Meaning Default\n'forward' The cumulative sums are obtained in the forward direction. YES\n'reverse' The cumulative sums are obtained in the reverse direction. NO\n\nTable 2: Options for including or omitting NaN.\n\nOption value Meaning Default\n'includenan' Include NaN in the operation YES\n'omitnan' Omit NaN in the opeartion NO\n\n### s = cumsum(A, dim, option)\n\n\u2022 Same as s = cumsum(A, dim) above, except that an option can be specified for controlling the operation direction or whether or not NaN should be included.\n\u2022 See Tables 1 and 2 above for available options.\n\n### s = cumsum(A, option1, option2)\n\n\u2022 Same as s = cumsum(A, option) above, except that two options can be specified at the same time.\n\u2022 option2 will overwrite option1 if they contradict each other.\n\u2022 See Tables 1 and 2 above for available options.\n\n### s = cumsum(A, dim, option1, option2)\n\n\u2022 Same as s = cumsum(A, dim, option) above, except that two options can be specified at the same time.\n\u2022 option2 will overwrite option1 if they contradict each other.\n\u2022 See Tables 1 and 2 above for available options.", "date": "2021-10-22 09:11:29", "meta": {"domain": "simo.ac", "url": "https://doc.simo.ac/Functions/cumsum/", "openwebmath_score": 0.4999734163284302, "openwebmath_perplexity": 7785.607328219051, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808730448281, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8568411951960261}}
{"url": "https://stats.stackexchange.com/questions/383656/confusion-about-range-of-integration-for-density-function", "text": "# Confusion about range of integration for density function\n\nConsider the joint density function:\n\n$$f(x,y) = \\begin{cases} 2 & & \\text{for } 0 \\leq x \\leq1 \\text{ and } 0 \\leq y \\leq 1-x, \\\\[6pt] 0 & & \\text{otherwise}. \\end{cases}$$\n\nFrom this joint density I figured out the following marginal densities:\n\n$$f_X(x) = 2(1-x),\\\\ f_Y(y) = 2.$$\n\nThe marginal density $$f_Y$$ is supposedly wrong, as the solutions provided to me say to calculate $$\\int^{1-y}_0 2 \\, dx$$. I don't see why I need to integrate over $$[0, 1-y]$$ and not over $$[0,1]$$. I thought the range for $$x$$ does not depend on $$y$$, or does it?\n\n\u2022 If this is homework, please add the self-study(stats.stackexchange.com/tags/self-study/info) tag. \u2013\u00a0StubbornAtom Dec 18 '18 at 20:29\n\u2022 Please: draw a picture showing where $f(x,y)$ is nonzero. That will easily answer your questions. \u2013\u00a0whuber Dec 18 '18 at 20:50\n\u2022 @StubbornAtom no it is not homework. this is just a practice exercise, I choose to do myself. \u2013\u00a0thebilly Dec 18 '18 at 21:34\n\u2022 @whuber I get a line with the equation y = -x+1 right? When $x = 1$ $y = 1-(x=1) = 0$ I don't quite get yet, how this answers my question. Could you please give me another hint? \u2013\u00a0thebilly Dec 18 '18 at 21:37\n\u2022 How did you integrate to find f(x)? What does 0 $\\le$ y $\\le$ (1-x) imply in terms of the random variable X and not Y? By integrating over [0,1] you are not integrating over the region specified in the question for the joint density but the region for the marginal density of Y. \u2013\u00a0aranglol Dec 18 '18 at 22:01\n\nComment: I simulated the joint distribution as an easy way to make the plot suggested in a previous comment. Before beginning to set limits on double integrals it is usually a good idea to sketch such a picture as a guide. I have shown R code for the first of the three plots.\n\nset.seed(1218); m = 10^5\nx1 = runif(m);  y1 = runif(m)\ncond = (y1 <= 1 - x1)\nx = x1[cond];  y = y1[cond]\nplot(x, y, pch=\".\")\n\n\nThe simulation and plots are for orientation, and are not an exact solution to your problem. For exact solutions, maybe the first thing to do is to try to integrate the joint density $$f(x,y) = 2$$ over the triangular region to make sure the integral is $$1,$$ as required for a density function.\n\nThen try integrating over $$x$$ to find the marginal density of $$Y,$$ which is suggested by the red line superimposed on the histogram in the third plot.\n\nYou wrote: $$\\text{for } 0 \\leq x \\leq1 \\text{ and } 0 \\leq y \\leq 1-x$$ That tells you the region over which you integrate.\n\nYou want to integrate out $$x$$ with $$y$$ fixed.\n\nSo you need those values of $$x$$ for which $$0\\le x\\le1$$ and $$0\\le y \\le 1-x.$$ Notice that $$y\\le 1-x$$ is equivalent to $$x \\le 1-y.$$\n\n\u2022 Okay. So whenever I have this sort of relationship between x and y regarding the domain I need to have this relationship in both integration intervals? \u2013\u00a0thebilly Dec 19 '18 at 9:08\n\u2022 When doing single integration, limits on integral sign need to show $x$-interval over which integration extends. For double integral, limits on two integral signs need to show $(x,y)$ region over which integration extends. \u2013\u00a0BruceET Dec 24 '18 at 1:41\n\u2022 @thebilly : This is probably clearer if the constraint on $x$ and $y$ is not symmetric in the two variables. Suppose you have $0\\le x \\le 1$ and for each value of $x$ in that interval you have $0\\le y \\le 3(1-x).$ Then$\\,\\ldots\\qquad$ \u2013\u00a0Michael Hardy Dec 25 '18 at 19:26\n\u2022 $\\ldots\\,\\,$you have \\begin{align} & \\int_0^1\\left( \\int_0^{3(1-x)} \\cdots\\cdots \\, dy \\right) \\, dx \\\\ \\\\ = {} & \\iint\\limits_{\\left\\{ (x,y) \\,:\\, \\begin{smallmatrix} 0 \\, \\le \\, x \\,\\le\\, 1 \\\\ \\&\\ 0\\,\\le\\,y\\,\\le\\, 3(1-x) \\end{smallmatrix} \\right\\}} \\quad \\cdots\\cdots \\, d(x,y) \\\\ \\\\ = {} & \\iint\\limits_{\\left\\{ (x,y) \\,:\\, 0\\,\\le\\, x\\,\\le\\, \\frac{3-y} 3 \\,\\le\\, 1 \\right\\}} \\quad \\cdots\\cdots \\, d(x,y) \\\\ \\\\ = {} & \\int_0^3 \\left( \\int_0^{(3-y)/3} \\cdots\\cdots \\, dx \\right) \\, dy. \\end{align} \u2013\u00a0Michael Hardy Dec 25 '18 at 19:32\n\u2022 In other words, if you have $x$ going from $0$ to $1$ and then for any fixed value of $x$ you have $y$ going from $0$ to $3(1-x),$ and that's the same as saying $y$ is between $0$ and $3,$ and for each fixed value of $y$ between $0$ and $3$ you have $x$ going from $0$ to $(3-y)/3. \\qquad$ \u2013\u00a0Michael Hardy Dec 25 '18 at 19:37", "date": "2020-07-08 22:33:30", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/383656/confusion-about-range-of-integration-for-density-function", "openwebmath_score": 0.9974572658538818, "openwebmath_perplexity": 303.12921074514526, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808701643914, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8568411926783018}}
{"url": "https://math.stackexchange.com/questions/3000878/does-a-n-a-n-1-converge-to-the-golden-ratio-for-all-fibonacci-like-sequenc/3000901", "text": "# Does $a_{n}/a_{n-1}$ converge to the golden ratio for all Fibonacci-like sequences?\n\nYesterday a friend challenged me to prove that $$\\lim_{n\\rightarrow\\infty}\\frac{a_n}{a_{n-1}}=\\varphi\\; ,$$ where $$\\varphi$$ is the golden ratio, for the Fibonacci series.\n\nI started rewriting the limit as\n\n$$\\lim_{n\\rightarrow\\infty}\\frac{a_n}{a_{n-1}}=\\lim_{n\\rightarrow\\infty}\\frac{a_{n-1}+a_{n-2}}{a_{n-1}}=\\lim_{n\\rightarrow\\infty}1+\\frac{a_{n-2}}{a_{n-1}}\\; .$$\n\nIf the sequence $$b_n=\\frac{a_n}{a_{n-1}}$$ is convergent,\n\n$$\\lim_{n\\rightarrow\\infty}\\frac{a_{n-2}}{a_{n-1}}=\\left(\\lim_{n\\rightarrow\\infty}\\frac{a_n}{a_{n-1}}\\right)^{-1}\\; .$$\n\nRenaming the desired limit $$x$$, we obtain the quadratic equation\n\n$$x=1+\\frac{1}{x}$$ $$x^2-x-1=0$$\n\nif $$x\\neq 0$$. Therefore, if $$b_n$$ is convergent, it must be equal to $$\\frac{1+\\sqrt{5}}{2}$$ or $$\\frac{1-\\sqrt{5}}{2}$$.\n\nSince $$a_n>0$$, $$b_n>0, \\forall n$$, so the limit must be equal to $$\\varphi=\\frac{1+\\sqrt{5}}{2}$$.\n\nThis proof made me think that I didn't make use of the initial values of the sequence, so it must hold true for any sequence where $$a_{n}=a_{n-1}+a_{n-2}$$. The first question is, is $$a_{n}/a_{n-1}$$ convergent for all Fibonacci-like sequences?\n\nThe second and most intriguing for me is, is there any Fibonacci-like sequence where the limit is $$\\frac{1-\\sqrt{5}}{2}$$? Since this solution is negative, $$a_n$$ should change its sing with each $$n$$, but I couldn't find any values for $$a_0$$ and $$a_1$$, which would lead me to this case. If the answer to this question is no, what mathematical sense does this negative solution have?\n\n\u2022 Yup, interestingly enough, almost all \"Fibonacci-like\" sequences (in that they start with two seed values and then recursively define the successive terms by the addition of the previous two), except for certain trivial examples, have the ratio of successive terms converge to $\\phi$. Another noteworthy such sequence is that of the Lucas numbers (with seeds $L_1 = 2$ and $L_2 = 1$, which actually is a lot \"neater\" than the Fibonacci sequence in this respect. \u2013\u00a0Eevee Trainer Nov 16 '18 at 8:26\n\u2022 As for the notion of $\\phi$'s conjugate being the ratio of successive terms, no, at least ignoring trivial examples. You could analogize this in terms of \"stable\" and \"unstable\" solutions: the conjugate is unstable, where the regular one is stable. In the case of $\\phi$'s conjugate, unless you somehow trivially start with that conjugate as the ratio of successive terms (see Robert Z's answer), this means that successive terms eventually diverge away from it. Showing this lack of stability isn't trivial though and I'm mostly parroting other facts and don't feel qualified to elaborate on it. \u2013\u00a0Eevee Trainer Nov 16 '18 at 8:32\n\nOne way to look at this problem is that if $$a_{n+1}=a_n+a_{n-1}$$, then we we have $$\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}\\begin{pmatrix}a_{n-1}\\\\a_n\\end{pmatrix}=\\begin{pmatrix}a_n\\\\a_{n+1}\\end{pmatrix}.$$\n\nThe eigenvalues of the matrix\n\n$$\\begin{pmatrix}0&1\\\\1&1\\end{pmatrix}$$\n\nare $$\\dfrac{1+\\sqrt{5}}{2},\\dfrac{1-\\sqrt{5}}{2}.$$\n\nThese have corresponding eigenvectors $$v_1,v_2$$ which span $$\\mathbb{R}^2$$. This leads us to the conclusion that $$\\begin{pmatrix}a_1\\\\a_2\\end{pmatrix}=cv_1+bv_2,$$ and if $$c\\not=0,$$ then $$v_1$$ will dominate the sequence, and we can show the ratios converge to $$(1+\\sqrt{5})/2.$$ This leads us to the conclusion that if we want a Fibonacci like sequence to have ratios converging to $$(1-\\sqrt{5})/2$$, then we must have $$\\begin{pmatrix}a_1\\\\a_2\\end{pmatrix}=bv_2$$ for some non zero $$b\\in\\mathbb{R}$$. So to determine all such sequences we simply have to have an eigenvector $$v_2$$ corresponding to $$(1-\\sqrt{5})/2$$. One such eigenvector is $$\\begin{pmatrix}1\\\\\\dfrac{1-\\sqrt{5}}{2}\\end{pmatrix}.$$\n\n\u2022 Thank you, your answer gives me the relation $a_0$ and $a_1$ must fulfill in order to make the limit $\\frac{1-\\sqrt{5}}{2}$. \u2013\u00a0TheAverageHijano Nov 16 '18 at 15:37\n\nYou showed that if a limit exists for $$a_{n}/a_{n-1}$$ and $$a_n>0$$, then it is $$\\frac{1+\\sqrt{5}}{2}$$. Actually if $$(a_n)_{n\\geq 0}$$ is any sequence which satisfies the recurrence $$a_n=a_{n-1} + a_{n-2}$$ then there exist $$A$$ and $$B$$ such that $$a_n=A\\cdot \\left(\\frac{1+\\sqrt{5}}{2}\\right)^n+B\\cdot \\left(\\frac{1-\\sqrt{5}}{2}\\right)^n$$ where $$A$$ and $$B$$ depend on the initial terms $$a_0$$ and $$a_1$$.\n\nSo what is $$\\lim_{n\\rightarrow\\infty}\\frac{a_n}{a_{n-1}}$$ in the general case?\n\nConsider for example the case when $$A=0$$ and $$B\\not=0$$. What is the limit?\n\n\u2022 So, basically if $A\\neq 0$, the limit will be the golden ratio, and if $A=0$ and $B\\neq 0$, the ratio will be $\\frac{1-\\sqrt{5}}{2}$, right? \u2013\u00a0TheAverageHijano Nov 16 '18 at 8:40\n\u2022 Yes, that's true. \u2013\u00a0Robert Z Nov 16 '18 at 8:42\n\u2022 Robert Z could you prove the thing about $a_n$? \u2013\u00a0AryanSonwatikar Nov 17 '18 at 4:56\n\u2022 @AryanSonwatikar This is a standard result about \"homogeneous linear recurrences\". See math.stackexchange.com/questions/65011/\u2026 \u2013\u00a0Robert Z Nov 17 '18 at 6:55\n\nHere is a another view of this. We have $$b_n=a_n/a_{n-1}$$ and $$b_{n+1}=1+\\frac1{b_n},\\tag{*}$$ or, \\begin{align*} b_{n+1}&=1+\\frac1{1+\\cfrac1{b_{n-1}}}=1+\\cfrac1{1+\\frac1{1+\\frac1{1+\\frac1{1+\\cdots}}}}. \\end{align*} We should be clear about what we actually mean by an expression like this. One way that we could think about it is to starting with some constant like $$1$$, and then repeatly applying the function $$f(x)=1+\\dfrac 1x$$, \\begin{align*} c&=1&&=1.000\\dots\\\\ \\color{yellow}{f(}c\\color{yellow}{)}&=\\color{yellow}{1+\\frac1{\\color{black}{1}}}&&=2.000\\dots\\\\ \\color{orange}{f(}\\color{yellow}{f(}c\\color{yellow}{)} \\color{orange}{)}&=\\ \\color{orange}{1+\\frac{1}{\\color{yellow}{1+\\frac1{\\color{black}{1}}}}}&&=1.500\\dots\\\\ \\color{magenta}{f(}\\color{orange}{f(}\\color{yellow}{f(}c\\color{yellow}{)} \\color{orange}{)}\\color{magenta}{)}&=\\color{magenta}{1+\\frac 1{ \\color{orange}{1+\\frac{1}{\\color{yellow}{1+\\frac1{\\color{black}{1}}}}}}}&&=1.667\\dots\\\\ \\color{violet}{f(}\\color{magenta}{f(}\\color{orange}{f(}\\color{yellow}{f(}c\\color{yellow}{)} \\color{orange}{)}\\color{magenta}{)}\\color{violet}{)}&=\\color{violet}{1+\\frac 1{\\color{magenta}{1+\\frac 1{ \\color{orange}{1+\\frac{1}{\\color{yellow}{1+\\frac1{\\color{black}{1}}}}}}}}}&&=1.600\\dots \\end{align*} Symbolly what we get is more and more like our infinite fraction. If we start with $$-1/\\varphi$$, \\begin{align*} c&=-1/\\varphi&&=-0.618\\dots\\\\ \\color{yellow}{f(}c\\color{yellow}{)}&=\\color{yellow}{1+\\frac1{\\color{black}{-1/\\varphi}}}&&=-0.618\\dots\\\\ \\color{orange}{f(}\\color{yellow}{f(}c\\color{yellow}{)} \\color{orange}{)}&=\\ \\color{orange}{1+\\frac{1}{\\color{yellow}{1+\\frac1{\\color{black}{-1/\\varphi}}}}}&&=-0.618\\dots\\\\ \\color{magenta}{f(}\\color{orange}{f(}\\color{yellow}{f(}c\\color{yellow}{)} \\color{orange}{)}\\color{magenta}{)}&=\\color{magenta}{1+\\frac 1{ \\color{orange}{1+\\frac{1}{\\color{yellow}{1+\\frac1{\\color{black}{-1/\\varphi}}}}}}}&&=-0.618\\dots\\\\ \\color{violet}{f(}\\color{magenta}{f(}\\color{orange}{f(}\\color{yellow}{f(}c\\color{yellow}{)} \\color{orange}{)}\\color{magenta}{)}\\color{violet}{)}&=\\color{violet}{1+\\frac 1{\\color{magenta}{1+\\frac 1{ \\color{orange}{1+\\frac{1}{\\color{yellow}{1+\\frac1{\\color{black}{-1/\\varphi}}}}}}}}}&&=-0.618\\dots \\end{align*} So no matter how many times we apply it, we're staying fixed at $$-1/\\varphi$$. But even then, with the aid of a calculator, if we start with a random number $$\\neq-1/\\varphi$$ (even it's really close to $$-1/\\varphi$$) and perform iteration $$x\\to x+\\dfrac 1x$$ again and again, we eventually end up at $$1.618...=\\varphi$$. So,\n\nwhy the fixed point $$\\varphi$$ favored above the other one $$-1/\\varphi$$?\n\nThe transformational understanding of derivatives is going to be helpful for understanding this set up. Now we know that $$\\varphi$$ and $$-1/\\varphi$$ stay fixed in place during this iteration process. But zoom in on a neighborhood around $$\\varphi$$, during each iteration, points in that region get contracted around $$\\varphi$$, meaning that the function $$1+\\dfrac 1x$$ has a derivative with a magnitude that is less than $$1$$ at this input. In fact, the derivative works out around to be $$\\left|\\frac{df}{dx}(\\varphi)\\right|\\approx |-0.38|<1,$$ meaning that each repeated application scrunches the neighborhood around this number smaller and smaller like a gravitational pull towards $$\\varphi$$.\n\nConversely, at $$-1/\\varphi$$, the magnitude of the derivative actually has a magnitude greater than $$1$$, $$\\left|\\frac{df}{dx}\\left(-\\frac 1\\varphi\\right)\\right|\\approx |-2.62|>1,$$ so points near the fixed point are repelled away from it. We can see that they get stretched by more than a factor of $$2$$ in each iteration. (They also get flipped around because the derivative is negative here, but the salient fact of stability is just the magnitude.)\n\nWe will call $$\\varphi$$ a \"stable fixed point\", and $$-1/\\varphi$$ an \"unstable fixed point\". As we can see the stability of a fixed point is determined by whether or not of its derivative is bigger or smaller than $$1$$. And this explains why $$\\varphi$$ always shows up in the limit.\n\nReference: 3Blue1Brown.\n\n\u2022 Nice, I didn't think about stability theory to explain my question. It's nice that we can approach the problem from different but equivalent perspectives. \u2013\u00a0TheAverageHijano Nov 16 '18 at 15:42\n\nYes, $$a_n\\over a_{n-1}$$ is convergent for any Fibonacci-esque sequence(with integers), and this happen to be the golden ratio, $$\\varphi$$. The limit $$1-\\sqrt{5}\\over 2$$ will never occur for a Fibonacci-esque sequence. The negative solution crops up because you multiply both sides by $$x$$ when you solve $$x=1+\\frac{1}{x}$$ leading to an extra solution. But this value is useful for calculating the value of the $$n^{th}$$ term of the Fibonacci sequence. $$F_n=\\frac{\\varphi^n - (\\frac{-1}{\\varphi})^n}{\\sqrt{5}}$$ Hope this helps.\n\n\u2022 Actually, $a_0=1$ and $a_1=\\frac{1-\\sqrt{5}}{2}$ is one of the combinations that will make $\\lim_{n\\rightarrow\\infty}\\frac{a_n}{a_{n-1}}=\\frac{1-\\sqrt{5}}{2}$ (see Melody's answer). Multiplying both sides by x (if $x\\neq 0$) shouldn't give illogical answers. Whether they have actual/physical meaning or not is another matter. \u2013\u00a0TheAverageHijano Nov 16 '18 at 15:48\n\u2022 That is true, but my answer says integers. If you take $a_0=1$ , and round off the values to the nearest integer, you get the sequence: $1,-1,0,-1..$ and here the limit suddenly tends to negative infinity because of the third term. \u2013\u00a0AryanSonwatikar Nov 17 '18 at 2:56\n\u2022 Also, I'm not well versed with matrices and vectors so Melody's answer is obscure for me. \u2013\u00a0AryanSonwatikar Nov 17 '18 at 3:00\n\u2022 Also, for the sequence you get, the Fibonacci-ness is followed only upto the fourth term after which if we follow the ratio and if we follow the Fibonacci-ness we get two different sequences. \u2013\u00a0AryanSonwatikar Nov 17 '18 at 4:21", "date": "2019-06-20 04:59:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3000878/does-a-n-a-n-1-converge-to-the-golden-ratio-for-all-fibonacci-like-sequenc/3000901", "openwebmath_score": 0.930814802646637, "openwebmath_perplexity": 281.3315000520823, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915220858951, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8568240754013672}}
{"url": "https://www.themathdoctors.org/how-to-find-an-inverse-function-conflicting-approaches/", "text": "# How to Find an Inverse Function: Conflicting Approaches\n\nWhile pondering issues that often give students trouble in algebra, I decided to check what we have said in Ask Dr. Math about inverse functions. I discovered four answers (all, as it happens, written by me \u2013 I tend to be attracted to certain topics!) to essentially the same question, spread over 13 years. It is one that I have pondered in teaching: having taught the subject from books that use two different approaches to the task, I find that I prefer the one that seems to be used less commonly, but each of them has its benefits. Let\u2019s take a look at one of these four discussions, then quickly look at the others to supplement it.\u00a0In 2010, Maureen wrote this:\n\nTo Invert Functions, First Subvert Routine\n\nThe inverse of a function is found by interchanging x's and y's, right? However, on Wikipedia they determine the inverse in a way that I find confusing. Specifically, I am writing what they do on the left and my confusion on the right.\n\nf(x) = 3x + 7\n\nNormally, I would now switch\ny = 3x + 7     the x's and y's and then solve\nfor y -- but Wikipedia doesn't\n\nFrom my point of view this is\n(y - 7)/3 = x          NOT the inverse -- it is the\noriginal function\n\nf(inv) of y = (y - 7)/3  This is the inverse using y as\nthe variable\n\nMost books do not do it this way; and although I agree with the final answer, I find it somewhat meaningless. Would you agree?\n\nI am including the usual way of finding the inverse.\n\nf(x) = 3x + 7\ny = 3x + 7     given the original function\nx = 3y + 7     switch x and y\ny = (x - 7)/3  solve for y to get inverse function\n\nMy reply was that, in fact, I prefer the method used by Wikipedia here, which I imagine is less often used in current textbooks (at least at an introductory level).\u00a0 There are a couple reasons to prefer it. One is that only this way is appropriate to applications in which the variables, unlike the typical generic x and y, have specific meanings, so that they can\u2019t be interchanged (see below for an example); another is that this method forces us to get out of the rut of assuming that x is always the independent variable, and y is always dependent.\n\nWhat they're doing is correct, and in fact is what I prefer. The confusion is probably because you are used to always thinking of y as a function of x.\n\n(It troubles me that texts often ask questions like \"Does the equation x + y^2 = 1 represent a function?\" when they really mean to ask if it represents y as a function of x. In that example, x is a function of y, though y is not a function of x. A function is about the relationship between two variables, not what they are called.)\n\nWhat Wikipedia has done is not to exchange the NAMES of the variables in the function, as usual, but just to change their ROLES. By solving for x, they are determining how x (the \"input\" of f) can be found given y (the \"output\" of f). That is exactly what it means to find an inverse.\n\nIf you recognize that x and y are just dummy variables, you see that Wikipedia\u2019s form,\u00a0 $$f^{-1}(y) = \\frac{y \\, \u2013 \\, 7}{3}$$, and Maureen\u2019s form, $$f^{-1}(x) = \\frac{x \\, \u2013 \\, 7}{3}$$, are really the same function, just with different names for the variable. By not changing the variable names at the start, Wikipedia ends up with y as the independent variable \u2013 not what most students are used to, but perfectly\u00a0 legal. And in fact, the main idea of an inverse function is precisely that we are changing the roles of the variables: what was the input of the original function becomes the output of the inverse function, and vice versa. What they are called is far less important that what they mean.\n\nBut Maureen needed more help:\n\nMaybe my difficulty is from a graphing point of view.\n\nIf you graph ...\n\ny = 3x + 1\n\n... and you create a table of values for x and y, and then you graph ...\n\nx = (y - 1)/3\n\n... you get the same graph, since the tables are the same. When graphing an inverse from a table of values, you specifically interchange the x's and y's, because if you do not, you have the same table.\n\nI feel students (or possibly just me) confuse a different way of writing a function with the inverse function. Specifically y = ln x is the same function as e^y = x. One is NOT the inverse of the other. The inverse of y = ln x is x = ln y, or e^x = y.\n\nThe two equations she graphed are actually\u00a0\u00a0different functions (y as a function of x, and x as a function of y); but they are equivalent equations (relating the same pairs of x and y), which is why they are represented by the same graph.\u00a0Maureen has confused functions with equations. In reality, the second function, $$f^{-1}(y) = \\frac{y \\, \u2013 \\, 1}{3}$$, is the inverse of the first, $$f(x) = 3x + 1$$, where x and y have kept the same name but changed roles.\n\nNow consider the logarithm. This is written explicitly as a function the name of which is \"ln\" rather than \"f\" or \"g,\" but it is the same idea. If I write ...\n\ny = ln(x)\n\n... I am using the ln function to express y as a function of x. This equation also expresses (implicitly) x as a function of y, since ln is one-to-one. When you solve for x to make the latter function explicit, you have\n\nx = e^y\n\nThis explicitly states a different function than ...\n\ny = ln(x)\n\n... although the relation between the variables is the same.\n\nIf we were to name the new function, calling it exp, so that ...\n\nx = exp(y)\n= e^y\n\n... the named function exp clearly is not the same function as ln. But nothing has changed except for giving it a name. The equation expresses x as a function of y, named or not.\n\nSo when we interchange the variables and change y = ln(x) to x = ln(y), we are, as you say, inverting the function -- in the sense of what function y is of x. We have changed the relationship between x and y. But in another sense, it is still the same function (as explicitly written), just expressed with different placeholders.\n\nSo there are two perspectives on what constitutes an inverse. In one sense, just swapping the variables\u00a0implicitly inverts the relationship, by swapping implied roles (x becomes y); but the inverse is shown explicitly by solving for the other variable, regardless of what you call it (input becomes output).\n\nMy answer to another of the four questions I referred to above includes an example of the situation where swapping of variables is meaningless, and you must use the other approach:\n\nInverting, Subverted\n\nBoth approaches are valid. The first is best when the variable names mean something, so that changing their names would not make sense. For example, you wouldn't say that the inverse of C = f(r) = 2 pi r is f^-1(r) = r/(2 pi), where r has now become the circumference. Rather, you would say that r = f^-1(C) = C/(2 pi).\n\nYour trouble, I believe, is that you are thinking of the variable names as if they had a fixed meaning somewhat like that, but using the method of inverting that requires you to swap names. You ask about \"the inverse ofy\" rather than of the FUNCTION, making y a name for the function itself. You can't invert a variable, only a function. And if you do think of y as the same thing as f(x), you can't swap variable names and still say that.\n\nIn fact, in my circumference example above, I was initially going to call the function C(r), which is done commonly, naming the function for its output; but it would make no sense to call the inverse C^-1. If anything, we would call the inverse r(C), since it gives the value of r that yields a given circumference C. I didn't do that because it would not fit the inverse notation you are using, and would just add confusion. (If it does, ignore this paragraph!)\n\nI prefer the first method of inverting I showed, because it helps to free students from fixating on x as the independent variable, and avoids the change of meaning that is confusing to you and others. Unfortunately, I have to teach the second in algebra classes, because it is what most students seem to see elsewhere.\n\nFor the record, the other two questions I referred to are\n\nGraphs of Inverse Functions\nInverting Functions\n\nThe final thanks in the latter, a long, rambling discussion, are worth ending with:\n\nThe terminology and explanation has solidified my understanding. I believe my confusion was, as you pointed out, that \"An equation does not define a function!\" This is very important and I have never thought of it that way. It could be because I never read that somewhere or was never taught that or most importantly never had to think of them that way until now.\n\n### 1 thought on \u201cHow to Find an Inverse Function: Conflicting Approaches\u201d\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-07-23 23:03:39", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/how-to-find-an-inverse-function-conflicting-approaches/", "openwebmath_score": 0.7663712501525879, "openwebmath_perplexity": 423.6984284395517, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915212263165, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8568240626122143}}
{"url": "http://thegatebook.in/qa/4499/co-pipelining-q1?show=4563", "text": "# CO-Pipelining-Q1\n\nA pipeline of 2 stages with a delay of x units each is split to n stages. In the new design, each stage has a delay of x/n units. To get the throughput increase of 1700% what would be the n value?\n\nreshown Jun 13\n\n1700/100=17\n\n17+1=18\n\nn=18\nanswered Jun 13 by (19,690 points)\nanswered Jun 13 by (160 points)\n+1 vote\n\nlet initial throughput be 100. required throughput is 1800.\n\nspeedup = 1800/100 = 18\n\nIn the first pipeline, each instruction will get completed after x units, and in the second\u00a0pipeline, each instruction will get completed after x/n units\n\nspeedup = x / (x/n)\n\n= n\n\ntherefore n = 18\n\nanswered Jun 14 by (4,100 points)\n\nComment if you find any issues.\n\nPipeline 1\n\nOne instruction <-----\u00a0x units\n? <-------------------------1 unit\n\n$= \\frac{1}{x} units$\n\nPipeline 2\n\nOne instruction ----> x/n units\n\n? -----------------------> 1 units\n$= \\frac{n}{x} units$\n\n$percentage \\ of \\ gain\\ in \\ throughput = \\frac{old value- new value}{old value} * 100$\n\n1700 = (n-1)* 100\n\n17 = n-1\n\n18=n\n\nanswered Jun 14 by (31,090 points)\nedited Jun 15\nIn pipeline 1, stage delay= x units. If we consider the CPI(cycles per instruction as 1), which is considered until no stalls are given.\n\nThus each instruction takes x time units.\n\nThroughput in Pipeline 1 = 1/x\n\nSimilarly, every instruction in pipeline 2 takes x/n time units.\n\nThus, throughput in pipeline 2 = 1/(x/n) = n/x\n\n% increase in throughput = ((new-old)/old)*100\n\n1700 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=(((n/x)-(1/x))/(1/x)*100\n\n17 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=n-1\n\n18 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= n\nYes you are correct. corrected the answer.\nIn this question (http://thegatebook.in/qa/4507) the throughput is (no. of stages / max stage time) so why in the above question. its only 1/x?", "date": "2019-09-15 11:20:28", "meta": {"domain": "thegatebook.in", "url": "http://thegatebook.in/qa/4499/co-pipelining-q1?show=4563", "openwebmath_score": 0.31003907322883606, "openwebmath_perplexity": 12445.317134561208, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9591542887603538, "lm_q2_score": 0.893309405344251, "lm_q1q2_score": 0.8568215473258997}}
{"url": "https://math.stackexchange.com/questions/2555710/open-subset-containing-all-terms-of-a-sequence-but-a-finite-number/2555718", "text": "# Open subset containing all terms of a sequence but a finite number\n\nDoes an open subset containing all points of a concergent sequence but a finite number of them necessarily contain the sequence's limit ?\n\nIdeally, I would like an answer for each of the following :\n\n-general topology\n\n-Haussdorf (T2)\n\n-metric\n\nMy intuition is that it doesn't, because even in a metric space, the radius of a ball centered at each point of the sequence and contained in the open subset may be forced to vary over the sequence.\n\nNo. Take, in $\\mathbb R$ with the usual topology, the sequence $\\left(\\frac1n\\right)_{n\\in\\mathbb N}$ and the open set $(0,+\\infty)$.\nThe answer is no for all three. Consider the sequence $\\frac{1}{n} \\in \\mathbb{R}$, and consider the open set $(0, 2)$. This open set contains all but finitely many of the sequences points (in fact, it contains all the sequences points), but it does not contain the sets limit, 0.", "date": "2019-12-07 22:39:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2555710/open-subset-containing-all-terms-of-a-sequence-but-a-finite-number/2555718", "openwebmath_score": 0.940449059009552, "openwebmath_perplexity": 149.2808415289387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513914124558, "lm_q2_score": 0.8688267762381843, "lm_q1q2_score": 0.8567947342836838}}
{"url": "http://mathhelpforum.com/pre-calculus/194793-equations-exponential-function.html", "text": "# Thread: Equations with the exponential function.\n\n1. ## Equations with the exponential function.\n\nHello\n\nI know this is elementary, but I don't know how solve this:\n\nThe question is, find the exact value(s) of x which satisfy the equation.\n\n$e^{2x} = e^x +12$\n\nI am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.\n\nI know if the question was:\n\n$e^{2x} = 12$\n\n$x = \\dfrac{1}{2}\\ln12$\n\nSome help would be very much appreciated. Thank you.\n\n2. ## Re: Equations with the exponential function.\n\nOriginally Posted by Furyan\nHello\n\nI know this is elementary, but I don't know how solve this:\n\nThe question is, find the exact value(s) of x which satisfy the equation.\n\n$e^{2x} = e^x +12$\n\nI am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.\n\nI know if the question was:\n\n$e^{2x} = 12$\n\n$x = \\dfrac{1}{2}\\ln12$\n\nSome help would be very much appreciated. Thank you.\n$e^{2x} - e^x - 12 = 0$\n\n$(e^x - 4)(e^x + 3) = 0$\n\nfinish it\n\n3. ## Re: Equations with the exponential function.\n\nDear Skeeter\n\nOriginally Posted by skeeter\n$e^{2x} - e^x - 12 = 0$\n\n$(e^x - 4)(e^x + 3) = 0$\n\nfinish it\nA quadratic in $e$, that was left field and I wasn't looking for it. Should have seen it though.\n\n$x = \\ln4 = 2\\ln2$\n\n$e^x = -3$ has no solutions.\n\nThank you very much indeed.\n\n4. ## Re: Equations with the exponential function.\n\nHopefully, you know that $e^{2x}= (e^x)^2$. If you let $y= e^x$, the equation becomes $y^2= y+ 12$ which is the same as $y^2- y- 12= 0$, a quadratic equation in y. If you did not notice that this could be factored as $(y- 4)(y+ 3)$, you could solve it by completing the square or using the quadratic formula.\n\n5. ## Re: Equations with the exponential function.\n\nHello HallsofIvy\n\nOriginally Posted by HallsofIvy\nHopefully, you know that $e^{2x}= (e^x)^2$. If you let $y= e^x$, the equation becomes $y^2= y+ 12$ which is the same as $y^2- y- 12= 0$, a quadratic equation in y. If you did not notice that this could be factored as $(y- 4)(y+ 3)$, you could solve it by completing the square or using the quadratic formula.\nThank you. I do find letting $y= e^x$ and solving the quadratic in $y$ easier to get my head round. I was able to factor that one.\n\n6. ## Re: Equations with the exponential function.\n\nHello, Furyan!\n\nFind the exact value(s) of $x$ which satisfy the equation.\"\n\n. . $e^{2x} - e^x -12\\:=\\:0$\n\nAs skeeter pointed out, this is a quadratic equation.\n\nThere is a way to recognize this phenomenon.\n\nThere are usually three terms, written in standard order.\n\nIf the first term has twice the exponent of the second term,\n. . we may have a quadratic.\n\nIn your problem: . $e^{2(x)} - e^x - 12 \\:=\\:0$\n\nAnother example: . $x^6 - 7x^3 - 8 \\:=\\:0$\n\nNote that we have: . $\\left(x^3\\right)^2 - 7(x^3) - 8 \\:=\\:0$\n\nLet $u \\,=\\,x^3$\n\nThen we have: . $u^2 - 7u - 8 \\:=\\:0$\n. . $(u-8)(u+1) \\:=\\:0 \\quad\\Rightarrow\\quad u \\:=\\:8,\\text{-}1$\n\nBack-substitute: . $\\begin{Bmatrix}x^3 \\:=\\:8 & \\Rightarrow & x \\:=\\:2 \\\\ x^3 \\:=\\:\\text{-}1 & \\Rightarrow & x \\:=\\:\\text{-}1 \\end{Bmatrix}$\n\nAnd there are four complex roots as well.\n\n7. ## Re: Equations with the exponential function.\n\nHello Soroban\n\nOriginally Posted by Soroban\nHello, Furyan!\n\nAs skeeter pointed out, this is a quadratic equation.\n\nThere is a way to recognize this phenomenon.\n\nThere are usually three terms, written in standard order.\n\nIf the first term has twice the exponent of the second term,\n. . we may have a quadratic.\n\nIn your problem: . $e^{2(x)} - e^x - 12 \\:=\\:0$\n\nAnother example: . $x^6 - 7x^3 - 8 \\:=\\:0$\n\nNote that we have: . $\\left(x^3\\right)^2 - 7(x^3) - 8 \\:=\\:0$\n\nLet $u \\,=\\,x^3$\n\nThen we have: . $u^2 - 7u - 8 \\:=\\:0$\n. . $(u-8)(u+1) \\:=\\:0 \\quad\\Rightarrow\\quad u \\:=\\:8,\\text{-}1$\n\nBack-substitute: . $\\begin{Bmatrix}x^3 \\:=\\:8 & \\Rightarrow & x \\:=\\:2 \\\\ x^3 \\:=\\:\\text{-}1 & \\Rightarrow & x \\:=\\:\\text{-}1 \\end{Bmatrix}$\n\nAnd there are four complex roots as well.\n\nThank you for that. I will be sure to look out for this sort of equation in the future.", "date": "2016-12-06 23:03:57", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/194793-equations-exponential-function.html", "openwebmath_score": 0.8681946396827698, "openwebmath_perplexity": 433.27276693135957, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513895809328, "lm_q2_score": 0.8688267745399465, "lm_q1q2_score": 0.8567947310176881}}
{"url": "http://math.stackexchange.com/questions/819547/if-ab-3-and-frac1a2-frac1b2-4-then-a-b2", "text": "If $ab=3$ and $\\frac1{a^2}+\\frac1{b^2}=4,$ then $(a-b)^2=\\;$?\n\nIf $ab=3$ and $\\frac1{a^2}+\\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?\n\n-\nYou are right and going well, and yes we can figure the values of each variable. But it is not necessary to do so :) \u2013\u00a0chubakueno Jun 3 '14 at 18:03\nWhat happens if you expand $(a - b)^2$? \u2013\u00a0David K Jun 3 '14 at 18:05\n\n$$\\frac1{a^2}+\\frac1{b^2}=4$$ $$a^2b^2\\left(\\frac1{a^2}+\\frac1{b^2} \\right)=4a^2b^2$$ $$b^2+a^2=(2ab)^2$$ $$a^2-2ab+b^2=(2ab)^2-2ab$$ $$(a-b)^2=(2ab)^2-2ab=(2\\cdot3)^2-2\\cdot3=30$$\n\n-\nYour answer is good but try to add some words for the explanation. \u2013\u00a0Tunk-Fey Jun 3 '14 at 18:24\n@Tunk-Fey - Why? each step flows well from the prior one. \u2013\u00a0JoeTaxpayer Jun 3 '14 at 19:17\n@JoeTaxpayer Just in case the OP doesn't understand. You may take a look Andre Nicolas' answer. \u2013\u00a0Tunk-Fey Jun 3 '14 at 19:27\nHe states 'you know that $(a-b)^2=30$ ' - but it's not clear that we do. That's the question OP is asking. \u2013\u00a0JoeTaxpayer Jun 3 '14 at 19:39\n\nHints:\n\n$\\frac{1}{a^2}+\\frac{1}{b^2} = 4 \\rightarrow b^2 + a^2 = 4(ab)^2$\n\n$(a-b)^2 = (a^2+b^2) -2(ab)$\n\nedit:\n\nTo solve for a particular variable, you can use $ab=3 \\rightarrow a = \\frac{3}{b}$ to eliminate a variable. For example $a^2+b^2 = \\frac{9}{b^2}+b^2$\n\n-\nI already know that (a-b)\u00b2 =30 so how would you figure out one of the variables \u2013\u00a0user154989 Jun 3 '14 at 18:10\n\nYou know that $(a-b)^2=30$. The same strategy tells you that $(a+b)^2=42$. Thus $$a+b=\\pm \\sqrt{42}\\quad\\text{and}\\quad a-b=\\pm\\sqrt{30}.$$ Now by adding and subtracting, we can find $2a$ and $2b$. and hence $a$ and $b$. Note that there are $4$ combinations, though if we have found one solution $a=p$, $b=q$, the other three are $a=-p$, $b=-q$, and $a=q$, $b=p$, and $a=-q$, $b=-p$.\n\nOne of the solutions is $a=\\frac{\\sqrt{42}+\\sqrt{30}}{2}$, $b=\\frac{\\sqrt{42}-\\sqrt{30}}{2}$.\n\n-\nThis is right, of course, but it misses the point: this question is all about the symmetry in the equations, and avoiding having to solve for the values of $a$ and $b$ separately. I wish any of my high school teachers would have explained the idea that you sometimes don't care about the values of each part of an expression. \u2013\u00a0symplectomorphic Jun 3 '14 at 19:30\nI was answering the question that OP had, in a comment, about solving for $a$ and $b$, given that OP already knew the value of $(a-b)^2$. The point about the solution is that it \"breaks symmetry\" very late in the game. Completely agree about the importance of exposure to symmetric functions. \u2013\u00a0Andr\u00e9 Nicolas Jun 3 '14 at 19:34\nIn fact, since I trust your judgment, Andre, do you know of a textbook that looks closely at these kinds of algebraic tricks? I think heavily symmetric equations often arise in geometry (and math competitions). But I can't think of a reference that systematically discusses problems like \"if $a+b=5$ and $a^2+b^2=10$, what's the value of $ab$?\" where the point isn't to find $a$ and $b$ separately. \u2013\u00a0symplectomorphic Jun 3 '14 at 19:35\nI cannot think of a good source. Problem collections aimed below the Olympiad level often have symmetric examples. Algebra books used to cover this kind of material. Alas, no more. \u2013\u00a0Andr\u00e9 Nicolas Jun 3 '14 at 19:39\nAlas indeed. I've written my own notes on this material and was always surprised I could never ever find a thorough, exhaustive discussion of this sort of technique. The competition books are always terse: they present a few examples but leave out the really interesting discussion of what's going on. Thanks though. \u2013\u00a0symplectomorphic Jun 3 '14 at 19:42", "date": "2015-11-27 05:04:24", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/819547/if-ab-3-and-frac1a2-frac1b2-4-then-a-b2", "openwebmath_score": 0.7680529356002808, "openwebmath_perplexity": 414.3301348425411, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513897844354, "lm_q2_score": 0.8688267677469952, "lm_q1q2_score": 0.8567947244956182}}
{"url": "https://gmatclub.com/forum/jerry-purchased-a-1-year-5-000-bond-that-paid-an-annual-115643.html?kudos=1", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 03 Apr 2020, 19:38\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n### Show Tags\n\n19 Jun 2011, 06:55\n10\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n54% (01:30) correct 46% (01:50) wrong based on 186 sessions\n\n### HideShow timer Statistics\n\nJerry purchased a 1-year $5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity? A.$5102\nB. $408 C.$216\nD. $202 E.$200\nGMAT Club Legend\nJoined: 11 Sep 2015\nPosts: 4579\nGMAT 1: 770 Q49 V46\nRe: Jerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 13 Sep 2017, 11:37 2 Top Contributor 1 guygmat wrote: Jerry purchased a 1-year$5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity?\n\nA. $5102 B.$408\nC. $216 D.$202\nE. $200 The ANNUAL interest rate = 4% So, every 6 MONTHS, we get 2% interest. Since there are only 2 \"compoundings,\" we can forgo the formula and do some quick mental calculations... INITIAL VALUE:$5000\n\nAFTER 6 MONTHS: 2% of $5000 =$100 in interest\nSo, the value after 6 months = $5000 +$100 = $5100 AFTER 12 MONTHS: 2% of$5100 = $102 in interest So, the value after 12 months =$5100 + $102 =$5202\n\nHow much interest had this bond accrued at maturity?\n\n### Show Tags\n\n12 Sep 2017, 03:09\nHere P = 5000\nr=4%\ntime = 1 year\nAs the amount is compounded annually n = 2\n\nCompound interest formula = $$A = p* (1+\\frac{r}{n*100}) ^n^t$$\n$$A = 5000* (1+\\frac{4}{2*100}) ^(2*1)$$\nA= 5202\n\nSo Interest earned = 5202 - 5000 = 202\n\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 3666\nJerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 13 Sep 2017, 10:14 guygmat wrote: Jerry purchased a 1-year$5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity?\n\nA. $5102 B.$408\nC. $216 D.$202\nE. $200 Estimate Without the compound interest formula (I use it, but with short periods and low interest rate, often you can estimate): Simple interest would yield .04 * 5,000 =$200 in one year\n\nCompound interest at this low rate (halved and paid twice), after only one year, will be barely above that.\n\nAnswer D, $202 is barely above$200.\n\nStages\n\nIf unsure about Answer C ($216), run a quick \"in stages\" calculation. If interest is paid every 6 months, the 4 percent is split in half (two periods of six months in one year). After first six months, interest payment is$5,000 * 02 = $100 (add$100 to principal for next stage)\n\nAfter next six months,\n.02 * $5,100 =$102\n\nTotal interest paid: $202 ANSWER D _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has. -- Margaret Mead Director Joined: 02 Sep 2016 Posts: 629 Re: Jerry purchased a 1-year$5,000 bond that paid an annual\u00a0 [#permalink]\n\n### Show Tags\n\n14 Sep 2017, 04:57\nSimple interest= 5000*4*1/100= 200\n\nSo the answer would be a number that is slightly more than 200 i.e. 202.\nBoard of Directors\nStatus: QA & VA Forum Moderator\nJoined: 11 Jun 2011\nPosts: 4875\nLocation: India\nGPA: 3.5\nRe: Jerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 09 Oct 2018, 07:18 guygmat wrote: Jerry purchased a 1-year$5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity?\n\nA. $5102 B.$408\nC. $216 D.$202\nE. $200 $$5000( 1 + \\frac{4}{200})^2 - 5000$$ $$= 5202 - 5000$$ $$= 202$$, Answer must be (D) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 9952 Location: United States (CA) Re: Jerry purchased a 1-year$5,000 bond that paid an annual\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2018, 06:52\nguygmat wrote:\nJerry purchased a 1-year $5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity? A.$5102\nB. $408 C.$216\nD. $202 E.$200\n\nSince the annual interest rate is 4%, we see that the bond earns 2% every half-year, or 6 months. Thus, for the first 6 months, the amount of interest earned was 0.02 x 5,000 = 100 dollars. Thus, the new principal at the end of the first 6 months was 5,000 + 100 = 5,100 dollars.\n\nFor the next 6 months, the amount of interest earned was 0.02 x 5100 = 102 dollars.\n\nSo total interest earned is 202 dollars.\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n197 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 14463\nRe: Jerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 06 Jan 2020, 17:49 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Jerry purchased a 1-year$5,000 bond that paid an annual \u00a0 [#permalink] 06 Jan 2020, 17:49\nDisplay posts from previous: Sort by", "date": "2020-04-04 03:38:19", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/jerry-purchased-a-1-year-5-000-bond-that-paid-an-annual-115643.html?kudos=1", "openwebmath_score": 0.5554371476173401, "openwebmath_perplexity": 10184.755527266267, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.967410262750752, "lm_q2_score": 0.8856314723088733, "lm_q1q2_score": 0.8567689753266625}}
{"url": "http://mymathforum.com/algebra/328325-one-number-four-more-than-five-times-another-if-their-sum-decreased-six.html", "text": "My Math Forum One number is four more than five times another. If their sum is decreased by six....\n\n Algebra Pre-Algebra and Basic Algebra Math Forum\n\n February 29th, 2016, 08:36 AM #1 Senior Member \u00a0 Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10 One number is four more than five times another. If their sum is decreased by six.... One number is four more than five times another. If their sum is decreased by six, the result is ten. Find the numbers. I tried to write the equation from reading this. Is this anywhere near close? x+4+5x-6=10 Thanks. Last edited by skipjack; February 29th, 2016 at 10:07 PM.\nFebruary 29th, 2016, 08:49 AM \u00a0 #2\nMath Team\n\nJoined: Oct 2011\n\nPosts: 14,597\nThanks: 1038\n\nQuote:\n Originally Posted by GIjoefan1976 One number is four more than five times another. If their sum is decreased by six, the result is ten. Find the numbers. I tried to write the equation from reading this. Is this anywhere near close? x+4+5x-6=10\nSolve the equation for x.\n\nThen test it yourself by substituting...\n\nFebruary 29th, 2016, 09:13 AM \u00a0 #3\nSenior Member\n\nJoined: Feb 2016\nFrom: seattle\n\nPosts: 377\nThanks: 10\n\nQuote:\n Originally Posted by Denis Solve the equation for x. Then test it yourself by substituting...\nOkay, I don't understand what you mean by \"substituting\"\n\nand both this time, and last time I can get the first number being 2.\n\nyet then I go 2 plus 4 =6 * 5 is 30 -6 =24\n\nso not yet figuring out why it is not working for me.\n\n February 29th, 2016, 09:18 AM #4 Math Team \u00a0 Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra First of all, you have two numbers. Call them $x$ and $y$. The information gives you two equations relating $x$, $y$ and numbers you are given. See if you can build those equations from the text of the question. Thanks from GIjoefan1976\nFebruary 29th, 2016, 09:31 AM \u00a0 #5\nMath Team\n\nJoined: Oct 2011\n\nPosts: 14,597\nThanks: 1038\n\nQuote:\n Originally Posted by GIjoefan1976 Okay, I don't understand what you mean by \"substituting\" and both this time, and last time I can get the first number being 2. yet then I go 2 plus 4 =6 * 5 is 30 -6 =24\nOK, you got x = 2\n\n\"One number is four more than five times another.\"\nother number= 5*2 + 4 = 14 ; OK?\n\n\"If their sum is decreased by six, the result is ten.\"\nsum = 2 + 14 = 16\n16 - 6 = 10 : Bingo! You're correct\n\nFebruary 29th, 2016, 09:50 AM \u00a0 #6\nSenior Member\n\nJoined: Jul 2014\nFrom: \u092d\u093e\u0930\u0924\n\nPosts: 1,178\nThanks: 230\n\nQuote:\n Originally Posted by GIjoefan1976 One number is four more than five times another. If their sum is decreased by six, the result is ten. Find the numbers. I tried to write the equation from reading this. Is this anywhere near close? X+4+5x-6=10 Thanks\nLet one number be x\n\nOther number is 4+5x\n\nQuote:\n Originally Posted by GIjoefan1976 If their sum is decreased by six, the result is ten\n(x + 4+5x)-6 = 10\n\nThe only difference is brackets.\nContinue.\n\nFebruary 29th, 2016, 10:24 AM \u00a0 #7\nSenior Member\n\nJoined: Feb 2016\nFrom: seattle\n\nPosts: 377\nThanks: 10\n\nQuote:\n Originally Posted by Denis OK, you got x = 2 \"One number is four more than five times another.\" other number= 5*2 + 4 = 14 ; OK? \"If their sum is decreased by six, the result is ten.\" sum = 2 + 14 = 16 16 - 6 = 10 : Bingo! You're correct\nThanks i think it was the sum part that really confused me. Was not sure what they meant by that. but now I will try to remember they mean for me to add the 2 numbers I end finding after solving the problem not before solving the problem.\n\nFebruary 29th, 2016, 10:30 AM \u00a0 #8\nSenior Member\n\nJoined: Feb 2016\nFrom: seattle\n\nPosts: 377\nThanks: 10\n\nQuote:\n Originally Posted by Prakhar Let one number be x Other number is 4+5x (x + 4+5x)-6 = 10 The only difference is brackets. Continue.\nsee I was thinking this too, yet was not sure how to write it.\n\nYet if i did it this way it confused me too as i would have gone and got\n\n-36x-24=10\n\n?\n\nFebruary 29th, 2016, 11:02 AM \u00a0 #9\nMath Team\n\nJoined: Oct 2011\n\nPosts: 14,597\nThanks: 1038\n\nQuote:\n Originally Posted by GIjoefan1976 (x + 4+5x)-6 = 10 see I was thinking this too, yet was not sure how to write it. Yet if i did it this way it confused me too as i would have gone and got -36x-24=10\nC'mon GIJoe; quit running at 100 mph!\nThere's no multiplication there; just remove the brackets:\nx + 4 + 5x - 6 = 10\n6x = 10 - 4 + 6\n6x = 12\nx = 2\n\n February 29th, 2016, 11:05 AM #10 Math Team \u00a0 Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra If you write that you have two numbers $x$ and $y$, then you get two equations $y = 4+5x$ and $x+y - 6= 10$. If you have learned how to solve simultaneous equations, this is not too difficult to solve (indeed one approach gives you exactly the equation that Denis was working you through. The important thing is to be able to build the equations. To understand what the sentences tell you and be confident that you have written it down correctly. The rest is relatively simple algebra.\n\n Tags decreased, number, sum, times\n\n Thread Tools Display Modes Linear Mode\n\n Similar Threads Thread Thread Starter Forum Replies Last Post andytx Algebra 4 March 1st, 2012 12:53 AM Hyperreal_Logic Topology 0 January 9th, 2010 05:27 PM W300 Algebra 2 October 22nd, 2009 09:11 AM steveeq1 Calculus 1 January 25th, 2009 12:56 PM\n\n Contact - Home - Forums - Cryptocurrency Forum - Top", "date": "2019-08-23 00:01:48", "meta": {"domain": "mymathforum.com", "url": "http://mymathforum.com/algebra/328325-one-number-four-more-than-five-times-another-if-their-sum-decreased-six.html", "openwebmath_score": 0.7567024827003479, "openwebmath_perplexity": 1827.9855240438856, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.96741025335478, "lm_q2_score": 0.8856314798554445, "lm_q1q2_score": 0.8567689743059244}}
{"url": "https://www.physicsforums.com/threads/position-vector.99312/", "text": "# Position vector\n\n1. Nov 10, 2005\n\n### Ratzinger\n\nGiven the typical cartesian xyz- coordinate system, is it correct to speak of a position vector? Isn't (x,y, z) just shothand for the coordinates? Distance vectors, force, velocity are real vectors with magnitude and direction in position space, but what is with a position vector in position space?\n\nI'm confused, help needed\n\n2. Nov 10, 2005\n\n### whozum\n\nThe position vector points from the origin to the position of the particle. For your point it is < x , y , z >\n\nPositions that vary with time are expressed as functions of x y and z, and you get parametric functions:\n\n$\\vec{r}_t = < x(t) , y(t), z(t) >$\n\n3. Nov 10, 2005\n\n### Ratzinger\n\nyes, but isn't what you describe rather a parameterized curve, not these geometrical objects with head and tail that you can move parallel around?\nI mean vectors in position coordinates without any time parameter mentioned.\n\n4. Nov 10, 2005\n\n### FredGarvin\n\nYour question is a bit confusing to me. A specified position vector without any time reference just means that the function that produces the curve for the thing's position in space has been evaluated at some specific time. Ultimately the position has to be a function of time or how can you come up with velocity and acceleration?\n\n5. Nov 10, 2005\n\n### pmb_phy\n\nA displacement vector is a vector which has its tail at one point and its tip at another point. If we refer to the point where the tail is as the \"origin\" then we refer to the displacement vector as the \"position vector\" of the point at the tip. We can label that point in any way we please, not only as R = (x, y, z). Quite often you'll see the position vector expressed as\n\n$$R = (r, \\theta, \\phi)$$\n\nThis means that R has the value\n\n$$R = A_r e_r + A_{\\theta} e_{\\theta} + A_{\\phi} e_{\\phi}$$\n\nwhere ek is a unit vector pointing in the direction of the increasing kth coordinate. The coeficients have the value Ak = R*ek (\"*\" = dot product). The position vector is a bound vector, i.e. attached to the space in which it lies.\nPete\n\nLast edited: Nov 10, 2005\n6. Nov 10, 2005\n\n### vaishakh\n\nposition vector is just a standard way of showing vectors and there is nothing like being attached. just the vector diagram has no physical meaning. the aim of physics is to solve the vector equations and while doing so we sometimes take the help of diagram.\n\n7. Nov 11, 2005\n\n### Ratzinger\n\nthanks for the replies, I found this\n\nAn important difference between a position vector R and a general vector such as delta R is that the components of R are x, y and z, whereas for delta R the components are delta x, delta y and delta z. It is important to distinguish between true vectors and position vectors. A true vector does not depend on coordinate system but only on the difference between one end of the vector and the other. A position vector, in contrast, does depend on the coordinate system, because it is used to locate a position relative to a specified reference point.\n\n8. Nov 11, 2005\n\n### pmb_phy\n\nWhere did you get this definition from?? A vector is a geometrical object which does not depend on the coordinate system. Any vector may be expressed in terms of other vectors which are related to a particular coordinate system. But that doesn't mean that it is defined by the coordinate system. I thought I made that clear above but I guess not. I did explain that a vector is an arrow. Some vectors are called \"Bound vectors\" while others are called \"free vectors.\" This is an important distinction that you should learn. The position vector does not depend on any coordinate system whatsoever. It depends on a geometric object, i.e. a \"point.\" This point I speak of is known as the \"reference point.\"\nWrong. You're confusing \"reference point\" with \"coordinate system.\"\nThey are very different things.\n\nFor details on what I've been talking about see Thorne and Blanchard's online notes at\n\nhttp://www.pma.caltech.edu/Courses/ph136/yr2004/0401.1.K.pdf\n\nPete\n\n9. Nov 11, 2005\n\n### Ratzinger\n\nthanks Pete for the great notes you linked\n\nSo we have bound and free vectors. That's a nice distinction.\n\nFree vectors can be parallel moved around or can be represented in different coordinate systems, they keep their meaning. Bound vectors do not, because they can clearly not be moved around and keep meaning and when coordinates systems are changed we talk about distance vectors. Bound vectors are bound to a coordinate system. A position vector gives only for his coordinate system information.\n\n10. Nov 11, 2005\n\n### pmb_phy\n\nBound vectors are not bound to a coordinate system. They're bound to the space that the vector is in and that is independant of the coordinate system. Give me any position vector and I can easily represent it in three different coordinate systems. I clarified this point here as I recall\n\nhttp://www.geocities.com/physics_world/ma/coord_system.htm\n\nPete\n\n11. Nov 11, 2005\n\n### Ratzinger\n\nDistance, velocity, force -all coordinate-independent quantities. But to ask what the position is of a point clearly needs a coordinate system. Position is only meaningful with a reference system.\n\nPosition vector in position space is a misnomer (in any position coordinates). As much as a momentum vector in momentum space.\n\n12. Nov 11, 2005\n\n### HallsofIvy\n\nStaff Emeritus\nYour original question is a very good one. In a Cartesian coordinate system, we can think of the \"position vector\" as the vector from the origin to the point, so that, at one instant, the tip of that vector is the point and the curve the point moves on is the curve sketched by the tip of the vector.\n\nHowever, in non-Cartesian coordinates, especially in situations, such as are common in General Relativity, where we have curved surfaces that admit no Cartesian coordinate system, there is no such thing as a \"position vector\". (I used to worry about what vectors on the surface of a sphere 'looked like\"!)\nIt is much better to think in terms of tangent vectors at every point. I like to think \"a vector is a derivative\".\n\n13. Nov 12, 2005\n\n### pmb_phy\n\nGiven a reference point the position of an arbitrary point requires only a magntitude and a direction - i.e. a vector.\n\nHallsofIvy - I believe that you're confusing coordinate spaces with the space itself. A position vector can be defined in all spaces which have no curvature. Curvature exists independant of a coordinate system. There is no position vector in a curved space.\n\nPete\n\n14. Nov 12, 2005\n\n### robphy\n\nI think the issue underlying the OP's questions is the distinction between a vector space and an affine space.\nhttp://mathworld.wolfram.com/AffineSpace.html\nhttp://en.wikipedia.org/wiki/Affine_space\nA vector space has an \"origin\", whereas an affine space does not.\nPositions are elements of an affine space.\nDisplacements (the difference of two positions) are elements of a vector space.\n(Only after one assigns a norm can one talk of a \"magnitude\" of a vector.)\n\n15. Nov 12, 2005\n\n### lightgrav\n\nWHY is the distinction between free vectors and bound vectors IMPORTANT?\n\nI can add and subtract position vectors, even tho they're \"bound\" to origin.\nYet, the Torque due to a Force does NOT treat the Force as a \"free\" vector.\n\n16. Nov 13, 2005\n\n### pmb_phy\n\nThe physical meaning of adding position vectors is to start with one vector whose tail is at the origin. The next position vector added to this one has its tail at the tip of the first one. So this vector is bound as well and is bound at a different place. Tourque is the cross product of a bound vector and a free vector making it a free vector.\n\nOther physical vectors are things like the center of mass vector.\n\nPete\n\n17. Nov 15, 2005\n\n### pmb_phy\n\nRob - There is no requirement for a vector space to have an origin. The term \"Origin\" refers to a particular point that an observer uses as a reference point. The user may also use other points in order to clearly define directions. There is no unique point in any space which demands to be an origin. From your links I don't see what you mean by \"affine space.\" Can you elaborate for me? Thanks.\n\nPete\n\n18. Nov 15, 2005\n\n### HallsofIvy\n\nStaff Emeritus\nYes, there is a requirement for a vector space to have an \"origin\". One of the requirements for a vector space is that there be a 0 vector. That is what you are referring to as an \"origin\". An \"affine space\" is a set of points such that any line through one of the points is contained in the space. You can think of it as a plane or 3d space or any Rn without a coordinate system. Once you add a coordinate system (so that you have an origin), you can make it a vector space. An affine space is what you seem to be thinking of as a \"vector space\". You can add vectors, you can't add points in an affine space.\n\n19. Nov 15, 2005\n\n### HallsofIvy\n\nStaff Emeritus\nYou're right. I referred to \"coordinate systems\" when I really meant the space itself.\n\n20. Nov 15, 2005\n\n### robphy\n\nHere's some physical motivations on this issue concerning an affine space vs. a vector space.\n\nA space of positions in a plane (or the space of times on a line) is an affine space. With no point being physically distinguished from any other, an affine space is more natural than a vector space. In an affine space, there is no sense of addition of elements of this space. (If you attempt to add two elements, the sum depends on the choice of an \"origin\" [which does not exist in an affine space]. As was mentioned, one can introduce an origin by introducing a coordinate system. Then, the sum now depends on a choice of coordinate system.) There is, however, a sense of subtraction... the \"difference of two positions [in an affine space]\" is a vector... the displacement vector. (The difference does not depend on a choice of [or existence of] an origin.)\n\nKnow someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook\n\nHave something to add?", "date": "2016-10-24 09:13:59", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/position-vector.99312/", "openwebmath_score": 0.754300057888031, "openwebmath_perplexity": 460.6119848821427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102552339746, "lm_q2_score": 0.8856314677809303, "lm_q1q2_score": 0.8567689642891894}}
{"url": "https://gmatclub.com/forum/how-many-roots-does-x-6-12x-4-32x-2-0-have-214685.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 23 Sep 2018, 02:09\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# How many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49303\nHow many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\u00a0 [#permalink]\n\n### Show Tags\n\n10 Mar 2016, 05:10\n00:00\n\nDifficulty:\n\n45% (medium)\n\nQuestion Stats:\n\n56% (01:02) correct 44% (01:23) wrong based on 97 sessions\n\n### HideShow timer Statistics\n\nHow many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\n\n(A) 1\n(B) 2\n(C) 3\n(D) 4\n(E) 5\n\n_________________\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 12432\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nHow many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Mar 2017, 19:59\n3\n2\nHi mesutthefail,\n\nThe GMAT often tests you on rules/patterns that you know, but sometimes in ways that you're not used to thinking about. This prompt is really just about factoring and Classic Quadratics, but it looks a lot more complicated than it actually is.\n\nA big part of properly dealing with a Quant question on the GMAT is in how you organize the information and 'simplify' what you've been given. This prompt starts us off with....\n\nX^6 \u201312X^4 + 32X^2 = 0\n\nThis certainly looks complex, but if you think about how you can simplify it, then you'll recognize that can 'factor out' X^2 from each term. This gives us...\n\n(X^2)(X^4 - 12X^2 + 32) = 0\n\nNow we have something a bit more manageable. While you're probably used to thinking of Quadratics such as X^2 + 6X + 5 as (X+1)(X+5), that same pattern exists here - it's just the exponents are slightly different (even though the math rules are exactly the SAME). We can further rewrite the above equation as...\n\n(X^2)(X^2 -4)(X^2 - 8) = 0\n\nAt this point, you don't really need to calculate much, since each 'piece' of the product should remind you of a pattern that you already know.....\n\nX^2 = 0 --> 1 solution\n(X^2 - 4) = 0 --> 2 solutions\n(X^2 - 8) = 0 --> 2 solutions\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\nManager\nJoined: 09 Jul 2013\nPosts: 110\nHow many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 29 Mar 2017, 08:58\n2\n4\nTo find the roots we can factor the polynomial down to first order expressions and count how many roots we get.\n\nThe first thing to do is to factor out $$x^2$$\n\nThen we have $$x^2(x^4-12x^2+32)=0$$\n\nNow to make things more familiar we can substitute $$y=x^2$$\n\nNow it looks like this\n$$y(y^2-12y+32)=0$$\n\nAnd we can factor it like we are used to\n\n$$y(y-4)(y-8)=0$$\n\ny=0\ny=4\ny=8\n\nSo substituting $$x^2$$ back in for y:\n\n$$x^2=0$$\n$$x^2=4$$\n$$x^2=8$$\n\nSo then the roots are\n\n$$x=0$$\n$$x=-2$$\n$$x=2$$\n$$x=-\\sqrt{8}$$\n$$x=\\sqrt{8}$$\n\nNote, since we are not asked to find the actual roots, we only need to determine the number of roots. After the first step of factoring out the $$x^2$$, we have $$x^2$$ and a 4th order expression. A 4th order expression will have 4 roots (unless it's a perfect square, but our expression is not a perfect square). So the total number of roots will be 4 from the 4th order expression, and one from the $$x^2$$. Total roots = 5.\n_________________\n\nDave de Koos\n\nOriginally posted by davedekoos on 10 Mar 2016, 14:28.\nLast edited by davedekoos on 29 Mar 2017, 08:58, edited 1 time in total.\n##### General Discussion\nIntern\nJoined: 12 Dec 2016\nPosts: 10\nRe: How many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Mar 2017, 02:27\nI thought polynoms was not a subject in GMAT ? This questions contains quite the polynomial solutions.\nIntern\nJoined: 12 Dec 2016\nPosts: 10\nRe: How many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Mar 2017, 02:31\ndavedekoos wrote:\nTo find the roots we can factor the polynomial down to first order expressions and count how many roots we get.\n\nThe first thing to do is to factor out $$x^2$$\n\nThen we have $$x^2(x^4-12x^2+32)=0$$\n\nNow to make things more familiar we can substitute $$y=x^2$$\n\nNow it looks like this\n$$y(y^2-12y+32)=0$$\n\nAnd we can factor it like we are used to\n\n$$y(y-4)(y-8)=0$$\n\ny=0\ny=4\ny=8\n\nSo putting it back to x^2\n\n$$x^2=0$$\n$$x^2=4$$\n$$x^2=8$$\n\nSo then the roots are\n\n$$x=0$$\n$$x=-2$$\n$$x=2$$\n$$x=-\\sqrt{8}$$\n$$x=\\sqrt{8}$$\n\nNote, since we are not asked to find the actual roots, we only need to determine the number of roots. After the first step of factoring out the $$x^2$$, we have $$x^2$$ and a 4th order expression. A 4th order expression will have 4 roots (unless it's a perfect square, but our expression is not a perfect square). So the total number of roots will be 4 from the 4th order expression, and one from the $$x^2$$. Total roots = 5.\n\nA question. When you first factored out by x^2, the x expression in 32x^2 disappeared completely, however when you second factored out the \"y\" expression, 12y^2 became 12y instead of 12. Can you please clarify?\nRe: How many roots does x^6 \u201312x^4 + 32x^2 = 0 have? &nbs [#permalink] 21 Mar 2017, 02:31\nDisplay posts from previous: Sort by\n\n# How many roots does x^6 \u201312x^4 + 32x^2 = 0 have?\n\n## Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-23 09:09:40", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/how-many-roots-does-x-6-12x-4-32x-2-0-have-214685.html", "openwebmath_score": 0.6060466170310974, "openwebmath_perplexity": 1724.8647685423177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9840936092211993, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8567492212623203}}
{"url": "https://forum.bionicturtle.com/threads/hull-instructional-video-ch4-duration.22262/#post-74751", "text": "# Hull, Instructional video , ch4 -Duration\n\n#### Branislav\n\n##### Member\nSubscriber\nDear David,\nThanks a lot for video lectures they are much inspiring Still I was little bit confused with all these different names duration, modified duration, Macauly duration,.. etc...I will shortly examine mine view of this and kindly ask you to comment ( but without laughing)\nAccording to mine understanding we are methodologically speaking about one risk measure all the time, called duration - how long on average shall i wait as bond holder to receive cash payments, or in terms on formula as you explained ( formula 1):\n\nFrom this formula we see that that when we say on average, we are referring to time weighted average, so as a result for 3 year maturity bond I will obtain let's say 2.63 - so this is time based measure ( for zero coupon bond it is equal to 3- no cash flows till the very same end, if you can wait that much\nIf you agree with me on this definition than this is the same thing as Macaulay duration, there is nothing new coming with this new name introduced beside of course great honor and memory on Frederick Macauly who introduced this concept.\n\nSo we are still on duration and keep playing further. What if we do the first derivative by yield, just to check what is bond's sensitivity on yield change..\n\nDelta (B)=dB/dY*Delta(y) ( formula 2 ) and dB/dY is \"similar\" to the right side of the formula 1, just with the minus in front , and we need to \"remove\" B from the denominator, or put it another way: dB/dY=-B*D and using formula 2 we obtain ( let us call it \" yield/price sensy formula\"):\n\nDelta (B)=-B*D*Delta(y)\n\nFor me this was kind of \"magic\"..somehow mine year based measure D becomes interest ( yiedl) sensitivity measure!\nBut basically we are still talking about duration from the beginning of the text, just with this simple \"math\" transformation we saw that it is also connected to the bonds price sensitivity to yield change\nWe play further...we assumed above continuous compounding, if we go to the annual compounding, then, bond price is little bit different summation:\n\n( note just the yield y is replaced with i) and duration and its first derivative are little bit different, so their relationship from \" yield/price sensy formula\"is now transforme to :\n\nDelta (B)=-B*D*Delta(y)/(1+y)\nand we introduce new \"name\" again, \"Modified duration\" as:\nD*=D/(1+y), which transforms previous equation to:\nDelta (B)=-B*(D*)*Delta(y)\n\nso again D* is duration from the beginning of the text, just for yearly compounding case, \"used\" in this formula to express sensitivity of bonds price on yield change.\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nHi @Branislav Super glad you found some inspiration in the videos: that's what we are here for In regard to your summary: yes, rIght!! All of that looks solid to me, very well done. I agree with virtually all of your statements and I definitely agree with the substantial point that, to paraphrase, that there is really only one duration. I've posted dozens of times on these concepts but here is how I would summarize, and I believe this summary maps pretty well to your math, and further you can see that I'm agreeing that Mac/mod/Effective duration are three faces of the same single concept. Effective duration simply estimates modified duration, and modified duration is a sort of \"adjusted\" Macaulay duration which adjusts for the effects of discrete discounting on the price and doesn't matter if discounting is continuous (we tend to refer to Macaulay duration as a maturity and modified duration as a sensitivity, but the mathematical units of both, in fact, are years such that is it a minor mathematical adjustment from one to the other, as you do imply!).\n\nHere is how I would summarize:\n\u2022 Macaulay duration, I'll denote D_Mac, is the bond's weighted average maturity (as illustrated by your first formula above, where the weights assigned to the maturities are the PV of the cash flows; i.e., the weights, t(i), are weighted by c(i)*exp[-y*t(i)]/B.\n\u2022 modified duration, I'll denote D_mod, is the linear sensitivity of the bond's price with respect to a small yield change; i.e., if the modified duration is 3.5 years, then we (linearly) approximate a yield change of +\u0394y will associate with a 3.5*\u0394y percentage drop in the bond price (knowing that curvature/convexity has been ignored). When the yield is continuously compounded, Mac duration = modified duration. When the yield is discretely rendered, we need to adjust the Mac to retrieve the accurate D_mod = D_mac/(1 + yield/k) where k = number of periods per year; i.e., when discrete, D_mod is always a bit less than D_Mac\n\u2022 In this way, modified duration is a measure of sensitivity: %\u0394P = \u0394P/P = -(D_mod)*\u0394y, solving for D_mod:\n\u2022 D_mod = -1/P * \u0394P/\u0394y or continuously D_mod = -1/P * \u2202P/\u2202y; i.e., modified duration is the first partial derivative (of bond price) with respect to the yield multiplied by -1/P. If we multiply each side by price, P, then:\n\u2022 P *D_mod = -\u0394P/\u0394y = \"dollar duration;\" i.e., dollar duration is the (negative) of the pure first derivative (i.e., the slope of the tangent line, itself negative). Importantly, dollar duration divided by 10,000 is the DV01 because P *D_mod/10,000 = DV01.\n\u2022 However, if you start with the bond price function (either continuous or discrete) and if you take the first derivative, then you can see that you should end up with (the negative of) the dollar duration: \u2202P/\u2202y = -D_mod*P (this forum has dozens of such actual derivations if you search). Therefore, by definition, if you take the first derivative, you should also (as I think you do imply), equivalently end up with \u2202P/\u2202y = -D_mac*P/(1+y/k). There is an old saying: duration is \"infected by price\" to acknowledge that 1/P \"infects\" the pure derivative.\n\u2022 Effective duration approximates modified duration by shocking the yield and re-pricing in order to retrieve the slope of the nearby tangent. Effective duration is sort of mini-simulation used to estimate the (inherently due to it being Taylor Series) analytical modified duration when it is not analytically available (e.g., MBS with negative convexity throws off the analytics). You will really understand when you can see that the effective duration approximates the modified duration which itself is an exact linear approximation. In this way, effective duration and modified duration, although they differ in approach, are both sensitivities and not conceptually different.\n\nLast edited:\n\n#### Matthew Graves\n\n##### Active Member\nSubscriber\nI think I would add a few further points with respect to the practical use of Effective Duration and differences between Effective Duration and Modified Duration.\n\nModified Duration is sensitivity of the price with respect to the Yield to Maturity. These two measures are precisely defined mathematically and not open to interpretation.\n\nEffective Duration, however, is a more practical (and complicated) measure derived through re-valuation of the instrument. It is the price sensitivity of the instrument to a parallel shift in a valuation curve and is therefore model dependent. Depending on the instrument, this can be a deterministic valuation but could equally be based on monte carlo simulation if the instrument has optionality. If the underlying curve is flat at the yield to maturity and the instrument does not have optionality you would expect the Effective Duration to be very close to the Modified Duration. However, in all practical, real-world valuation cases the Effective Duration would not be equal to the Modified Duration due to the shape of the underlying curve and any optionality in the instrument (e.g. callable bonds). Separately (and rather technically), the shift applied for Effective Duration is conventionally applied to the observed market yields comprising the underlying curve before obtaining the zero rate curve. The shift is not applied to the zero curve directly. This has subtle but observable affects on the effective duration also.\n\n#### enjofaes\n\n##### Member\nSubscriber\nHi @David Harper CFA FRM . Thanks again for all the material! Was wondering if what you said was correct in the instructional video of duration : around 24'14\" Modified duration = dollar duration / 10.000. I thought from the beginning of the video that this is the formula for the DV01.\n\nKind regards\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nHi @enjofaes If I said that (sorry I'll have to locate the specific video location later), then I misspoke. You are correct: DV01 = (P*D)/10,000 = dollar duration/10,000. I'm actually very happy with my summary note at see https://forum.bionicturtle.com/threads/week-in-financial-education-2021-05-24.23840/post-88846 i.e.,\nNote: Durations in CFA and FRM compared\n\nI'd like to clarify duration terminology as it pertains to differences between the CFA and FRM. This forum has hundreds of threads over 12+ years on duration concepts (it's hard to say which links are the best at this point, but I'll maybe come back and curate some best links). Our YouTube channel has a FRM P2.T4 that includes videos on DV01, hedging the DV01, effective duration, modified versus Macaulay duration, and an illustration of all three durations. There are many nuances and further explorations, but here my goal is only to clarify the top-level definitions.\n\nI'll use the simple example of a $100.00 face 20-year zero-coupon bond that currently yields (yield to maturity of) 6.0% per annum. If the yield is 6.0% per annum with continuous compounding, the price is$100.00*exp(-0.060*20) = $30.12. If the yield is 6.0% per annum with annual compounding, the price is$100.00/(1+0.060)^20 = $31.18. Unless otherwise specified, I will assume a continuous compound frequency. Special note: we so often price a bond given the yield (where CPT PV is the final calculator step) that it is easy to forget yield is not actually an input. Yield is the internal rate of return (IRR) assuming the current price. Yield does not determine price; price determines yield. Technical (non-fundamental) factors cause price to fluctuate, therefore yield fluctuates. \u2022 \u2202P/\u2202y (or \u0394p/\u0394y) is the slope of the tangent line at the selected yield. At 6.0% yield, the slope is -$602.39. How do I know that? Because dollar duration is the negated slope, so in this case dollar duration (DD) = P*D = $30.12 price * 20 years =$602.39. Importantly, the \"y\" in \u2202P/\u2202y is yield and yield is just one of several interest rate factors.\n\u2022 Dollar duration (DD; aka, money duration in the CFA) is analytically the product of price and modified duration. Dollar duration (DD) = P*D = $30.12 * 20 =$602.39. Why is it so large? Because it's the (negated) tangent line's slope, so it has the typical first derivative interpretation: DD is the dollar change implied by one unit change in the yield, -\u2202P/\u2202y. One unit is 1.0 = 100.0% = 100 * 100 basis points (bps) per 1.0% = 10,000 basis points. So, DD is the dollar change implied by a 100.0% change in yield if we use the straight tangent line which would be a silly thing to do! Recall the constant references to limitations of duration as linear approximation. The linear approximation induces bias at only 5 or 10 or 20 basis points, so 10,000 basis points is literally \"off the charts\" and not directly meaningful. What is meaningful? The PVBP (aka, DV01) comes to our rescue with a meaningful re-scaling of the DD ...\n\u2022 Price value of basis point (aka, dollar value of '01, DV01) is the dollar duration \u00f7 10,000. It's the tangent line's slope re-scaled from \u0394y=100.0% to \u0394y= 0.010% (one basis point). PVBP = P*D/10,000; in this example, PVBP = $30.12 * 20 / 10,000 =$0.06024. It is the dollar change implied by a one basis point decline in the yield. It is still a linear approximation, but much better because we zoomed in to a small change. In this way, the difference between the highly useful PVBP and the dollar duration is merely scale.\n\u2022 Macaulay duration is the bond's weighted average maturity where the weights are each of the bond's cash flow's present value as a percentage of the bond's price. Macaulay duration is tedious however it is reliable and it is analytical. When we can compute the Macaulay duration, it is accurate; we don't approximate by re-pricing the bond. A zero-coupon bond has a Macaulay duration equal to its maturity because it only has one cash flow (hence the popularity of the zero-coupon bond in exam questions, never mind the zero-coupon bond is a reliable primitive). Our 20-year zero-coupon bond has a Macaulay duration of 20.0 years.\n\u2022 Modified duration is the measure of interest rate risk. Modified duration is the approximate percentage change in bond price implied by a 1.0% (100 basis point) change in the yield. Just as \u2202P/\u2202y refers to the tangent line's slope which is \"infected with price,\" we divide by price to express the modified duration, D(mod) = -1/P*\u2202P/\u2202y. The key relationship between analytical modified and Macaulay duration is the following: modified duration = Macaulay duration / (1 + y/k) where k is the number of compound periods in the year; e.g., k = 1 for annual compounding, k = 2 for semiannual compounding and k = \u221e for continuous compounding. Importantly, if the the compound frequency is continuous then a bond's modified duration equals its Macaulay duration. Notice that T / (1 + y/\u221e) = T / (1 + 0) = T.\n\u2022 If the 6.0% yield is annual compounded, our 20-year bond's Macaulay duration is given by 20.0 / (1 + 6.0%) = 18.868 years.\n\u2022 If the 6.0% yield is continuously compounded, our 20-year bond's modified duration is 20.0 years.\n\u2022 Effective duration is an approximation of modified duration. Recall the modified duration is a linear approximation, but that's because it is a function of the first derivative; otherwise, modified duration is an exact (analytical or functional) measure of the price sensitivity with respect to the interest rate factor that happens to most often be the yield. We can retrieve it easily whenever we can compute the Macaulay duration, which is the case for any vanilla bond. Otherwise (e.g., bond has an embedded option) we approximate the modified duration by calculating its effective duration. The effective duration approximates the modified duration which itself is a linear approximation. The effective duration is given by [P(-\u0394y) - P(+\u0394y)] / (2*\u0394y) * 1/P. I wrote it this way so you can see that it is essentially similar to \u2202P/\u2202y*1/P where \u2202P/\u2202y \u2245 [P(-\u0394y) - P(+\u0394y)] / (2*\u0394y). I've observed that many candidates do not realize that the formula for effective duration is simply slope*1/P. Geometrically, it is the slope of the secant line that is near to the tangent line! Secant's slope approximates the tangent's slope. If you grok the calculus here, I think you'll agree that this is all just one thing! Now we can see how it's not so different. But as you can visualize, there are an almost infinite variety of secants next to the tangent. We arbitrarily choose a nearby secant, but we'd prefer a small delta if the bond is vanilla (i.e., if the bond's cash flows are invariant to rate changes). Although we do not need the effective duration for our example bond, we can compute it:\n\u2022 If our arbitrary yield shock is10 basis points such that \u0394y= 0.10%, then P(-\u0394y)= $100.00*exp(-5.90%*20)=$30.728, and P(+\u0394y)= $100.00*exp(-6.10%*20)=$29.523. Effective duration= ($30.728 -$29.523)/0.0020 *1/\\$30.12= 20.0013 years. Fine approximation!\n\u2022 On the terminology (CFA versus FRM)\n\u2022 Interest rate factor: The FRM (informed by Tuckman) starts with a general interest rate factor. This is typically the spot rate, forward rate, par rate, or yield. Importantly, the spot, forward and par rates are term structures, or vectors; the par yield curve is a vector of par rates at various maturities, often at six-month or one-month intervals. Only the yield is a single (aka, scalar) value.\n\u2022 My above definition of the effective duration is according to the FRM (and to me). The CFA sub-divides this effective duration into either approximate modified duration (if the interest rate factor is the yield) versus effective duration (if the non-vanilla nature of the bond requires a non-yield interest rate factor; i.e., a benchmark yield curve). Personally, I am not keen on this semantic approach because (i) both of these CFA formulas are approximating the modified duration and (ii) I prefer to reserve \"effective\" for its traditional connotation (e.g., effective convexity is analogous to effective duration), and (iii) we wouldn't anyhow use an inappropriate factor (yield) for certain non-vanilla situations, so we don't really need label-switches to guide us thusly! (the CFA's formula for its approximate modified duration is essentially the same as its effective duration formula). To me, the CFA's approach muddies the terms \"approximate\" and \"effective\" where the math gives us natural distinctions. Follow the math, I'd say!", "date": "2022-05-21 11:48:51", "meta": {"domain": "bionicturtle.com", "url": "https://forum.bionicturtle.com/threads/hull-instructional-video-ch4-duration.22262/#post-74751", "openwebmath_score": 0.6814907193183899, "openwebmath_perplexity": 2123.34152772634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936101542134, "lm_q2_score": 0.8705972734445508, "lm_q1q2_score": 0.8567492138144629}}
{"url": "https://proofwiki.org/wiki/Definition:Supremum_of_Set", "text": "# Definition:Supremum of Set\n\n## Definition\n\nLet $\\struct {S, \\preceq}$ be an ordered set.\n\nLet $T \\subseteq S$.\n\nAn element $c \\in S$ is the supremum of $T$ in $S$ if and only if:\n\n$(1): \\quad c$ is an upper bound of $T$ in $S$\n$(2): \\quad c \\preceq d$ for all upper bounds $d$ of $T$ in $S$.\n\nIf there exists a supremum of $T$ (in $S$), we say that:\n\n$T$ admits a supremum (in $S$) or\n$T$ has a supremum (in $S$).\n\n### Finite Supremum\n\nIf $T$ is finite, $\\sup T$ is called a finite supremum.\n\n### Subset of Real Numbers\n\nThe concept is usually encountered where $\\struct {S, \\preceq}$ is the set of real numbers under the usual ordering $\\struct {\\R, \\le}$:\n\nLet $T \\subseteq \\R$ be a subset of the real numbers.\n\nA real number $c \\in \\R$ is the supremum of $T$ in $\\R$ if and only if:\n\n$(1): \\quad c$ is an upper bound of $T$ in $\\R$\n$(2): \\quad c \\le d$ for all upper bounds $d$ of $T$ in $\\R$.\n\nThe supremum of $T$ is denoted $\\sup T$ or $\\map \\sup T$.\n\n## Also known as\n\nParticularly in the field of analysis, the supremum of a set $T$ is often referred to as the least upper bound of $T$ and denoted $\\map {\\operatorname {lub} } T$ or $\\map {\\operatorname {l.u.b.} } T$.\n\nSome sources refer to the supremum of a set as the supremum on a set.\n\n## Also defined as\n\nSome sources refer to the supremum as being the upper bound.\n\nUsing this convention, any element greater than this is not considered to be an upper bound.\n\n## Also see\n\n\u2022 Results about suprema can be found here.\n\n## Linguistic Note\n\nThe plural of supremum is suprema, although the (incorrect) form supremums can occasionally be found if you look hard enough.", "date": "2019-08-20 18:49:04", "meta": {"domain": "proofwiki.org", "url": "https://proofwiki.org/wiki/Definition:Supremum_of_Set", "openwebmath_score": 0.9775962829589844, "openwebmath_perplexity": 172.22908619831068, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936096877063, "lm_q2_score": 0.870597273444551, "lm_q1q2_score": 0.8567492134083232}}
{"url": "https://math.stackexchange.com/questions/979660/largest-n-vertex-polyhedron-that-fits-into-a-unit-sphere?noredirect=1", "text": "Largest $n$-vertex polyhedron that fits into a unit sphere\n\nIn two dimensions, it is not hard to see that the $n$-vertex polygon of maximum area that fits into a unit circle is the regular $n$-gon whose vertices lie on the circle: For any other vertex configuration, it is always possible to shift a point in a way that increases the area.\n\nIn three dimensions, things are much less clear. What is the polyhedron with $n$ vertices of maximum volume that fits into a unit sphere? All vertices of such a polyhedron must lie on the surface of the sphere (if one of them does not, translate it outwards along the vector connecting it to the sphere's midpoint to get a polyhedron of larger volume), but now what? Not even that the polyhedron must be convex for every $n$ is immediately obvious to me.\n\n\u2022 If the vertices are on the surface of the sphere the polyhedron will necessarily be convex - it will be the convex hull of the vertices. Because the sphere itself is convex the convex hull will lie entirely within it. Oct 18 '14 at 16:41\n\u2022 @MarkBennet: Good point, that settles this part at least.\n\u2013\u00a0user139000\nOct 18 '14 at 16:46\n\u2022 I believe this is an open problem for $n > 8$. Oct 21 '14 at 6:57\n\u2022 @achille hui: do you know solutions for n = 7,\u202f8? One can check directly that cube is not even a local maximum, having in fact surprisingly poor performance. Oct 23 '14 at 19:50\n\u2022 A stickler point about your proof for polygons: given that the space of such polygons is compact...\n\u2013\u00a0Max\nOct 23 '14 at 20:42\n\nThis is supposed to be a comment but I would like to post a picture.\n\nFor any $m \\ge 3$, we can put $m+2$ vertices on the unit sphere\n\n$$( 0, 0, \\pm 1) \\quad\\text{ and }\\quad \\left( \\cos\\frac{2\\pi k}{m}, \\sin\\frac{2\\pi k}{m}, 0 \\right) \\quad\\text{ for }\\quad 0 \\le k < m$$\n\nTheir convex hull will be a $m$-gonal bipyramid which appear below.\n\nUp to my knowledge, the largest $n$-vertex polyhedron inside a sphere is known only up to $n = 8$.\n\n\u2022 $n = 4$, a tetrahedron.\n\u2022 $n = 5$, a triangular bipyramid.\n\u2022 $n = 6$, a octahedron = a square bipyramid\n\u2022 $n = 7$, a pentagonal bipyramid.\n\u2022 $n = 8$, it is neither the cube ( volume: $\\frac{8}{3\\sqrt{3}} \\approx 1.53960$ ) nor the hexagonal bipyramid ( volume: $\\sqrt{3} \\approx 1.73205$ ). Instead, it has volume $\\sqrt{\\frac{475+29\\sqrt{145}}{250}} \\approx 1.815716104224$.\nLet $\\phi = \\cos^{-1}\\sqrt{\\frac{15+\\sqrt{145}}{40}}$, one possible set of vertices are given below: $$( \\pm \\sin3\\phi, 0, +\\cos3\\phi ),\\;\\; ( \\pm\\sin\\phi, 0,+\\cos\\phi ),\\\\ (0, \\pm\\sin3\\phi, -\\cos3\\phi),\\;\\; ( 0, \\pm\\sin\\phi, -\\cos\\phi).$$ For this set of vertices, the polyhedron is the convex hull of two polylines. One in $xz$-plane and the other in $yz$-plane. Following is a figure of this polyhedron, the red/green/blue arrows are the $x/y/z$-axes respectively.\n\n$\\hspace0.75in$\n\nFor $n \\le 8$, above configurations are known to be optimal. A proof can be found in the paper\n\nJoel D. Berman, Kit Hanes, Volumes of polyhedra inscribed in the unit sphere in $E^3$\nMathematische Annalen 1970, Volume 188, Issue 1, pp 78-84\n\nAn online copy of the paper is viewable at here (you need to scroll to image 84/page 78 at first visit).\n\nFor $n \\le 130$, a good source of close to optimal configurations can be found under N.J.A. Sloane's web page on Maximal Volume Spherical Codes. It contains the best known configuration at least up to year 1994. For example, you can find an alternate set of coordinates for the $n = 8$ case from the maxvol3.8 files under the link to library of 3-d arrangements there.\n\n\u2022 That's plenty of information, and I'm willing to give you the bounty since you have answered my question (\"Open problem for $n\\ge 9$\") but I'd like some more information for the $n=8$ case if possible. The cube not being optimal is already surprising since one might expect the 2D argument to somehow transfer to regular polyhedra, but where do the coordinates come from? You say that $n=8$ is known exactly but the coords just look like the result of a numerical optimization run. I'd expect there to be exact polar or cartesian coordinates or at least some formal description of the polyhedron.\n\u2013\u00a0user139000\nOct 25 '14 at 7:44\n\u2022 That the $n=7$ solution is a pentagonal bipyramid also mildly surprised me. In my mental picture of it, the concentration of vertices in the plane spanned by the pyramids' shared base seems a little too high for the configuration to be optimal. But 3D is hard to imagine accurately, of course...\n\u2013\u00a0user139000\nOct 25 '14 at 7:48\n\u2022 @pew look at Berman and Hanes paper (linked in updated answer) for a proof. Oct 25 '14 at 9:16", "date": "2021-10-17 13:02:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/979660/largest-n-vertex-polyhedron-that-fits-into-a-unit-sphere?noredirect=1", "openwebmath_score": 0.7174676656723022, "openwebmath_perplexity": 428.21900263279866, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936082881853, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8567492039297666}}
{"url": "https://math.stackexchange.com/questions/2762958/find-the-integral-int-fracdx-sqrtx2-a2", "text": "# Find the integral $\\int\\frac{dx}{\\sqrt{x^2-a^2}}$\n\nEvaluate the integral of $\\frac{1}{\\sqrt{x^2-a^2}}$\n\nPut $x=a\\sec\\theta\\implies dx=a\\sec\\theta\\tan\\theta d\\theta$ \\begin{align} \\int\\frac{dx}{\\sqrt{x^2-a^2}}&=\\int\\frac{a\\sec\\theta\\tan\\theta d\\theta}{\\sqrt{a^2\\sec^2\\theta-a^2}}=\\int\\frac{a\\sec\\theta\\tan\\theta d\\theta}{a\\sqrt{\\tan^2\\theta}}=\\int\\frac{a\\sec\\theta\\tan\\theta d\\theta}{a\\color{red}{\\tan\\theta}}\\\\&=\\int\\sec\\theta d\\theta=\\log|\\sec\\theta+\\tan\\theta|+C=\\log|\\frac{x}{a}+\\sqrt{\\frac{x^2}{a^2}-1}|\\\\&=\\log|\\frac{x+\\sqrt{x^2-a^2}}{a}|+C=\\log|{x+\\sqrt{x^2-a^2}}|-\\log|a|+C\\\\&=\\log|{x+\\sqrt{x^2-a^2}}|+C \\end{align} This is how it is solve in my reference. But, $\\sqrt{\\tan^2\\theta}=|\\tan\\theta|$ right ? Then, does that imply $$\\int\\frac{dx}{\\sqrt{x^2-a^2}}=\\int\\frac{a\\sec\\theta\\tan\\theta d\\theta}{a\\color{red}{|\\tan\\theta|}}=\\color{red}{\\pm}\\int\\sec\\theta d\\theta$$ Why am I getting this confusion and is the first solution complete ?\n\n\u2022 @TrostAft how can i say $\\tan\\theta$ is $+$ve from $x^2-a^2>0$ ? \u2013\u00a0ss1729 May 2 '18 at 7:20\n\u2022 No first solution is not complete, they silently slipped in the || inside. You know, $\\int \\sec t = \\ln(\\sec t + \\tan t)$ is valid for $\\sec t > 0$ and for $\\sec t \\lt 0$ the integrand is $\\ln(-\\sec t - \\tan t)$ so the solution $\\ln |\\sec t + \\tan t|$ combines both. \u2013\u00a0jonsno May 2 '18 at 7:35\n\u2022 Check by differentiating your solution(s). \u2013\u00a0Yves Daoust May 2 '18 at 7:35\n\u2022 Agree with @samjoe. My comment is untrue. \u2013\u00a0TrostAft May 2 '18 at 7:35\n\u2022 @samjoe i'm srry dont understand how ur point help me with the doubt in OP ?. Could u pls explain bit more ? \u2013\u00a0ss1729 May 2 '18 at 7:57\n\nSuppose that $a>0$.\n\nThe work is just for the case when $x>a$. The case for $x<-a$ is different, but the finals result is the same.\n\nLet $x=a\\sec\\theta$, where $\\theta\\in[0,\\frac{\\pi}{2})\\cup(\\frac{\\pi}{2},\\pi]$. This is the domain of $\\textrm{arcsec}$.\n\nFor $x< -a$, $\\theta\\in(\\frac{\\pi}{2},\\pi]$ and so $\\tan\\theta\\le0$.\n\n$$\\sqrt{x^2-a^2}=\\sqrt{a^2\\tan^2\\theta}=-a\\tan\\theta$$\n\n\\begin{align*} \\int\\frac{dx}{\\sqrt{x^2-a^2}}&=\\int\\frac{a\\sec\\theta\\tan\\theta}{-a\\tan\\theta}d \\theta\\\\ &=-\\int\\sec\\theta d\\theta\\\\ &=-\\ln|\\sec\\theta+\\tan\\theta|+C\\\\ &=\\ln|\\sec\\theta-\\tan\\theta|+C\\\\ &=\\ln\\left|\\frac{x}{a}-\\frac{-\\sqrt{x^2-a^2}}{a}\\right|+C\\\\ &=\\ln\\left|x+\\sqrt{x^2-a^2}\\right|-\\ln|a|+C \\end{align*}\n\nThere are two minus signs and they cancel each other to reach the final result.\n\n\u2022 I don't uunderstand how you conclude $\\theta \\in (\\pi/2, \\pi]$ because that is the key to the question. \u2013\u00a0jonsno May 2 '18 at 8:17\n\u2022 I mean why it can't be $\\theta \\in (\\pi, 3\\pi/2)$ where sec is negative but tan is positive \u2013\u00a0jonsno May 2 '18 at 8:18\n\u2022 Yes thats what we have to show that tan cannot be positive. \u2013\u00a0jonsno May 2 '18 at 8:20\n\u2022 If we take $\\theta\\in(\\pi,3\\pi/2)$, then $\\tan\\theta>0$. The integral is $\\int\\sec\\theta d\\theta$. The final answer will be still $\\ln|x+\\sqrt{x^2-a^2}|-\\ln|a|+C$. My point is that we can have $-\\int\\sec\\theta d\\theta$ in the work. But then we will have one more minus sign and obtain the same final answer. \u2013\u00a0CY Aries May 2 '18 at 8:26\n\u2022 We either take a range so that the work holds for both $x>a$ and $x<-a$, or take a range so that in one case we will have two minus signs to cancel each other. \u2013\u00a0CY Aries May 2 '18 at 8:29", "date": "2021-01-19 02:05:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2762958/find-the-integral-int-fracdx-sqrtx2-a2", "openwebmath_score": 0.9310566186904907, "openwebmath_perplexity": 829.7075101035706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936120202411, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.856749200570774}}
{"url": "https://math.stackexchange.com/questions/2120928/prove-that-the-square-root-of-a-positive-integer-is-either-an-integer-or-irratio", "text": "# Prove that the square root of a positive integer is either an integer or irrational\n\nIs my proof that the square root of a positive integer is either an integer or an irrational number correct?\n\nThe proof goes like this:\n\nSuppose an arbitrary number n, where n is non-negative. If $\\sqrt{n}$ is an integer, then $\\sqrt{n}$ must be rational. Since $\\sqrt{n}$ is an integer, we can conclude that n is a square number, that is for some integer a. Therefore, if n is a square number, then $\\sqrt{n}$ is rational.\n\nSuppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.\n\nWe prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\\sqrt{n} = \\frac{a}{b}$, where $a,b \\in Z^+, b \\neq 0$. We also suppose that $a \\neq 0$, otherwise $\\frac ab = 0$ , and n will be a square number, which is rational.\n\nHence $n = \\frac {a^2}{b^2}$, so $nb^2 = a^2$.\n\nSuppose $b=1$. Then $\\sqrt n = a$ , which shows that n is a square number. So $b \\neq 1$. Since $\\sqrt n > 1$, then $a>b>1$.\n\nBy the unique factorization of integers theorem, every positive integer greater than $1$ can be expressed as the product of its primes. Therefore, we can write $a$ as a product of primes and for every prime number that exists in $a$, there will be an even number of primes in $a^2$. Similarly, we can express $b$ as a product of primes and for every prime number that exists in $b$, there will be an even number of primes in $b^2$.\n\nHowever, we can also express $n$ as a product of primes. Since $n$ is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one prime in the product of $nb^2$ that has an odd number of primes. Since $nb^2=a^2$ , then this contradicts the fact that there is an even number of primes in $a^2$ since a number can neither be even and odd.\n\nTherefore, this contradicts the fact that $\\sqrt n$ is rational. Therefore, $\\sqrt n$ must be irrational.\n\nIs this sufficient? Or is there any parts I did not explain well?\n\n\u2022 I think its correct and very well explained. \u2013\u00a0Shobhit Jan 30 '17 at 14:25\n\u2022 It's a bit wordy, but logically you've got a solid proof, the even-ness of the powers of each prime is exactly what you're going for. \u2013\u00a0Adam Hughes Jan 30 '17 at 14:26\n\u2022 The conclusion is not precise enough. Instead of \"contradicts there is an even number of primes in $a^2$\" we want to say that it contradicts the fact that the prime $p$ occurs to odd power in the unique factorization of $nb^2,\\,$ but even power in $a^2,\\,$ i.e. we are comparing the parity of the count of single prime, not the total of all primes \u2013\u00a0Bill Dubuque Jan 30 '17 at 14:28\n\u2022 For example the argument shows that $\\,\\sqrt{3\\cdot 5}\\,$ is irrational because $\\,3\\,$ occurs to odd power in $\\,3\\cdot 5,\\,$ but the total number of primes in $\\,3\\cdot 5\\,$ is even. \u2013\u00a0Bill Dubuque Jan 30 '17 at 14:37\n\u2022 On a side note, this result is known as Theaetetus' Theorem, and it's proven in Euclid's Elements here: aleph0.clarku.edu/~djoyce/elements/bookX/propX9.html \"[S]quares which do not have to one another the ratio which a square number has to a square number also do not have their sides commensurable in length either.\" \u2013\u00a0Keshav Srinivasan Jan 30 '17 at 14:41\n\nYour proof is very good and stated well. I think it can be made shorter and tighter with a little less exposition of the obvious. However, I would prefer students to err on the side of more rather than less so I can't chide you for being thorough. But if you want a critique:\n\n\"Suppose an arbitrary number n, where n is non-negative. If $\\sqrt{n}$ is an integer, then $\\sqrt{n}$ must be rational. Since $\\sqrt{n}$ is an integer, we can conclude that n is a square number, that is for some integer a. Therefore, if n is a square number, then $\\sqrt{n}$ is rational.\"\n\nSuppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.\n\nThis can all be said more simply and to argue that if $\\sqrt{n}$ is an integer we can conclude $\\sqrt{n}$ is rational or that $n$ is therefore a perfect square, is a little heavy handed. Those are definitions and go without saying. However, it shows good insight and understanding to be aware one can assume things and all claims need justification so I can't really call this \"wrong\".\n\nBut it'd be enough to say. \"If $n$ is a perfect square then $\\sqrt{n}$ is a an integer and therefore rational, so it suffices to prove that if $n$ is not a perfect square, then $\\sqrt{n}$ is irrational.\n\nWe prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\\sqrt{n}$=ab , where a,b\u2208Z+,b\u22600. We also suppose that a\u22600, otherwise ab=0, and n will be a square number, which is rational.\n\nTerminologistically, to say \"$n$ is a square number\" is to mean $n$ is the square of an integer. If $n = (\\frac ab)^2$ we don't usually refer to $n$ as a square (although it is \"a square of a rational\") We'd never call $13$ a square because $13 = (\\sqrt{13})^2$.\n\nAlso you don't make the usual specification that $a$ and $b$ have no common factors. As it turns out you didn't need to but it is a standard.\n\nSuppose b=1 . Then $\\sqrt{n}$=a , which shows that n is a square number. So b\u22601. Since $\\sqrt{n}$>1, then a>b>1\n\nThis was redundant as $b=1 \\implies$ $a/b$ is an integer and we are assuming that $n$ is not a perfect square.\n\n.\n\nBy the unique factorization of integers theorem, every positive integer greater than 1 can be expressed as the product of its primes. Therefore, we can write a as a product of primes and for every prime number that exists in a, there will be an even number of primes in a2. Similarly, we can express b as a product of primes and for every prime number that exists in b, there will be an even number of primes in b2\n\nBill Dubuque in the comments noted what you meant to say was \"each prime factor will be raised to any even power\".\n\n.\n\nHowever, we can also express n as a product of primes. Since n is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one prime in the product of nb2 that has an odd number of primes. Since nb2=a2 , then this contradicts the fact that there is an even number of primes in a2 since a number can neither be even and odd.\n\nDitto:\n\nOverall I think your proof is very good.\n\nBut I should point out there is a simpler one:\n\nAssume $n = \\frac {a^2}{b^2}$ where $a,b$ are positive integers with no common factors (other than 1). If $p$ is a prime factor of $b$ and $n$ is an integer, it follows that $p$ is a prime factor of $a^2$ and therefore of $a$. But that contradicts $a$ and $b$ having no common factors. So $b$ can not have any prime factors. But the only positive integer without prime factors is $1$ so $b = 1$ and $n= a^2$ so $\\sqrt{n} = a$. So either for any integer either $n$ is a perfect square with an integer square root, or $n$ does not have a rational square root.\n\nAnd a slight caveat: I'm assuming that your class or text is assuming that all real numbers have square roots (and therefore if there is not rational square root the square root must be irrational). It's worth pointing out, that it is a result of real analysis that speaking of a square root actually makes any sense and that we can claim every positive real number actually does have same square root value. But that's probably beyond the range of this exercise.\n\nBut if I want to be completely accurate, you (and I) have actually only proven that positive integer $n$ either has an integer square root or it has no rational square root at all. Which is the same thing as saying if positive integer $n$ has a square root, the root is either integer or irrational. But we have not actually proven that positive integer $n$ actually has any square root at all.\n\n\u2022 Thank you for the critique! It definitely helped me a lot! \u2013\u00a0Icycarus Jan 31 '17 at 11:00\n\u2022 Can you explain or state a case where we cannot find the square root of a positive integer $n$ (as you state in the last line)? \u2013\u00a0Hungry Blue Dev Apr 15 '17 at 12:56\n\u2022 I didn't say there are positive integers without square roots. (There aren't.) I said we haven''t proven that the square root of any integer exists. And we haven't. To prove that we have to prove that $K = \\{q \\in \\mathbb Q| q^2 < n\\}$ is not empty, and bounded above and that if $z = \\sup K$ then $z^2 - n$. All we have proven so far is that either there is an integer so that $m^2 = n$ or there is no rational $q$ so that $q^2 = n$. We have not proven that there is an irrational $z$ so that $z^2 = n$. \u2013\u00a0fleablood Apr 15 '17 at 15:54\n\u2022 If $n$ can not be written as $\\frac {a^2}{b^2}$ then, by definition, $n$ does not have a rational square root. If $n$ has a rational square root then that rational square root can be written as $\\frac ab$. That is what rational means. And if the square root is $\\frac ab$ then $n = (\\frac ab)^2 =\\frac {a^2}{b^2}$. That is what square root means. So if $n$ can't be written that way then it does not have a rational square root. For example: $3$ can not be written that way and does not have a rational square root. Neither does ANY other non-square. \u2013\u00a0fleablood Jan 21 at 22:31\n\u2022 This is a prove that IF a square root is rational then it must be an integer. We are not concerned about irrational square root (not even to the extent of questioning whether they exist). And IF a square root is rational it can be written as $\\frac ab$. Which means $n$ can be written as $\\frac {a^2}{b^2}$. If it can't be, then it does not have a rational square root. \u2013\u00a0fleablood Jan 21 at 22:35\n\nWe know that $\\sqrt{4} = 2$ and $\\sqrt{2} = 1.414...$ are rational and irrational respectively, so all we have to do is to show that if $n\\in \\mathbb{Z+}$ such that $\\sqrt{n} = \\dfrac{a}{b}$ where $a$ and $b$ are positive integers and the expression $\\dfrac{a}{b}$ is in its simplest form then $\\sqrt{n}$ is integral. Squaring both sides of the expression we get that $n = \\dfrac{a^2}{b^2}$ since $a$ and $b$ have no common factors other than $1$ then $a^2 = n$ and $b^2 = 1$ therefore $b = 1$ hence $\\sqrt{n}$ if rational it's an integer.\n\n\u2022 How is that a critique of the OP's proof? \u2013\u00a0fleablood Jan 30 '17 at 18:51\n\u2022 The op shows (correctly) that the square root of a non square is irrational. There is no need to show that rational roots are integers. Assuming n is not square makes perfect sense and the case for square n's has been covered. Your objections are not valid. Meanwhile you post is just ... weird. What do $\\sqrt 4$ and $\\sqrt 2$ have to do with anything. And your statement $n = a^2/b^2$ implies $a^2 = n$ and $b^2 =1$ is said without justification when the justification is the entire point. And you are merely pointing out an easier proof. Which is not a valid critique. \u2013\u00a0fleablood Jan 30 '17 at 22:57", "date": "2019-02-19 07:32:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2120928/prove-that-the-square-root-of-a-positive-integer-is-either-an-integer-or-irratio", "openwebmath_score": 0.897523045539856, "openwebmath_perplexity": 102.9936556342472, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936087546923, "lm_q2_score": 0.8705972549785203, "lm_q1q2_score": 0.856749194423741}}
{"url": "http://en.wikipedia.org/wiki/Bertrand's_box_paradox", "text": "Bertrand's box paradox is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his Calcul des probabilit\u00e9s, published in 1889.\n\nThere are three boxes:\n\n1. a box containing two gold coins,\n2. a box containing two silver coins,\n3. a box containing one gold coin and one silver coin.\n\nAfter choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, it may seem that the probability that the remaining coin is gold is 12; in fact, the probability is actually 23. Two problems that are very similar are the Monty Hall problem and the Three Prisoners problem.\n\nThese simple but slightly counterintuitive puzzles are used as a standard example in teaching probability theory. Their solution illustrates some basic principles, including the Kolmogorov axioms.\n\nBox version\n\nThere are three boxes, each with one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (GG), one a silver coin on each side (SS), and the other a gold coin on one side and a silver coin on the other (GS). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold?\n\nThe following reasoning appears to give a probability of 12:\n\n\u2022 Originally, all three boxes were equally likely to be chosen.\n\u2022 The chosen box cannot be box SS.\n\u2022 So it must be box GG or GS.\n\u2022 The two remaining possibilities are equally likely. So the probability that the box is GG, and the other coin is also gold, is 12.\n\nThe flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically:\n\n\u2022 The probability that GG would produce a gold coin is 1.\n\u2022 The probability that SS would produce a gold coin is 0.\n\u2022 The probability that GS would produce a gold coin is 12.\n\nInitially GG, SS and GS are equally likely. Therefore by Bayes rule the conditional probability that the chosen box is GG, given we have observed a gold coin, is:\n\n$\\frac { \\mathrm{P}(see\\ gold \\mid GG)} { \\mathrm{P}(see\\ gold \\mid GG)+\\mathrm{P}(see\\ gold \\mid SS)+\\mathrm{P}(see\\ gold \\mid GS) } =\\frac{1}{1+0+1/2}= \\frac{2}{3}$\n\nThe correct answer of 23 can also be obtained as follows:\n\n\u2022 Originally, all six coins were equally likely to be chosen.\n\u2022 The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS.\n\u2022 So it must come from the G drawer of box GS, or either drawer of box GG.\n\u2022 The three remaining possibilities are equally likely, so the probability that the drawer is from box GG is 23.\n\nAlternatively, one can simply note that the chosen box has two coins of the same type 23 of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 23 of the time. In other words, the problem is equivalent to asking the question \"What is the probability that I will pick a box with two coins of the same color?\".\n\nBertrand's point in constructing this example was to show that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result; and the two methods are equivalent only if this probability is either 1 or 0 in every case. This condition is correctly applied in the second solution method, but not in the first.\n\nThe paradox as stated by Bertrand\n\nIt can be easier to understand the correct answer if you consider the paradox as Bertrand originally described it. After a box has been chosen, but before a box is opened to let you observe a coin, the probability is 2/3 that the box has two of the same kind of coin. If the probability of \"observing a gold coin\" in combination with \"the box has two of the same kind of coin\" is 1/2, then the probability of \"observing a silver coin\" in combination with \"the box has two of the same kind of coin\" must also be 1/2. And if the probability that the box has two like coins changes to 1/2 no matter what kind of coin is shown, the probability would have to be 1/2 even if you hadn't observed a coin this way. Since we know his probability is 2/3, not 1/2, we have an apparent paradox. It can be resolved only by recognizing how the combination of \"observing a gold coin\" with each possible box can only affect the probability that the box was GS or SS, but not GG.\n\nCard version\n\nSuppose there are three cards:\n\n\u2022 A black card that is black on both sides,\n\u2022 A white card that is white on both sides, and\n\u2022 A mixed card that is black on one side and white on the other.\n\nAll the cards are placed into a hat and one is pulled at random and placed on a table. The side facing up is black. What are the odds that the other side is also black?\n\nThe answer is that the other side is black with probability 23. However, common intuition suggests a probability of 12 either because there are two cards with black on them that this card could be, or because there are 3 white and 3 black sides and many people forget to eliminate the possibility of the \"white card\" in this situation (i.e. the card they flipped CANNOT be the \"white card\" because a black side was turned over).\n\nIn a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 12; only 3 students correctly responded 23.[1]\n\nAnother presentation of the problem is to say\u00a0: pick a random card out of the three, what are the odds that it has the same color on the other side? Since only one card is mixed and two have the same color on their sides, it is easier to understand that the probability is 23. Also note that saying that the color is black (or the coin is gold) instead of white doesn't matter since it is symmetric: the answer is the same for white. So is the answer for the generic question 'same color on both sides'.\n\nPreliminaries\n\nTo solve the problem, either formally or informally, one must assign probabilities to the events of drawing each of the six faces of the three cards. These probabilities could conceivably be very different; perhaps the white card is larger than the black card, or the black side of the mixed card is heavier than the white side. The statement of the question does not explicitly address these concerns. The only constraints implied by the Kolmogorov axioms are that the probabilities are all non-negative, and they sum to 1.\n\nThe custom in problems when one literally pulls objects from a hat is to assume that all the drawing probabilities are equal. This forces the probability of drawing each side to be 16, and so the probability of drawing a given card is 13. In particular, the probability of drawing the double-white card is 13, and the probability of drawing a different card is 23.\n\nIn question, however, one has already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (i.e. probability 12) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to exploit all of the information; one knows not only that the card on the table has at least one black face, but also that in the population it was selected from, only 1 of the 3 black faces was on the mixed card.\n\nAn easy explanation is that to name the black sides as x, y and z where x and y are on the same card while z is on the mixed card, then the probability is divided on the 3 black sides with 13 each. thus the probability that we chose either x or y is the sum of their probabilities thus 23.\n\nSolutions\n\nIntuition\n\nIntuition tells one that one is choosing a card at random. However, one is actually choosing a face at random. There are 6 faces, of which 3 faces are white and 3 faces are black. Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 23. Therefore, the chance of flipping the card over and finding another black face is also 23. Another way of thinking about it is that the problem is not about the chance that the other side is black, it's about the chance that you drew the all black card. If you drew a black face, then it's twice as likely that that face belongs to the black card than the mixed card.\n\nAlternately, it can be seen as a bet not on a particular color, but a bet that the sides match. Betting on a particular color regardless of the face shown, will always have a chance of 12. However, betting that the sides match is 23, because 2 cards match and 1 does not.\n\nLabels\n\nOne solution method is to label the card faces, for example numbers 1 through 6.[2] Label the faces of the black card 1 and 2; label the faces of the mixed card 3 (black) and 4 (white); and label the faces of the white card 5 and 6. The observed black face could be 1, 2, or 3, all equally likely; if it is 1 or 2, the other side is black, and if it is 3, the other side is white. The probability that the other side is black is 23. This probability can be derived in the following manner: Let random variable B equal the a black face (i.e. the probability of a success since the black face is what we are looking for). Using Kolmogrov's Axiom of all probabilities having to equal 1, we can conclude that the probability of drawing a white face is 1-P(B). Since P(B)=P(1)+P(2) therefore P(B)=13+13=23. Likewise we can do this P(white face)=1-23=13.\n\nBayes' theorem\n\nGiven that the shown face is black, the other face is black if and only if the card is the black card. If the black card is drawn, a black face is shown with probability 1. The total probability of seeing a black face is 12; the total probability of drawing the black card is 13. By Bayes' theorem, the conditional probability of having drawn the black card, given that a black face is showing, is\n\n$\\frac{1\\cdot1/3}{1/2}=2/3.$\n\nIt can be more intuitive to present this argument using Bayes' rule rather than Bayes' theorem[3]. Having seen a black face we can rule out the white card. We are interested in the probability that the card is black given a black face is showing. Initially it is equally likely that the card is black and that it is mixed: the prior odds are 1:1. Given that it is black we are certain to see a black face, but given that it is mixed we are only 50% certain to see a black face. The ratio of these probabilities, called the likelihood ratio or Bayes factor, is 2:1. Bayes' rule says \"posterior odds equals prior odds times likelihood ratio\". Since the prior odds are 1:1 the posterior odds equals the likelihood ratio, 2:1. It is now twice as likely that the card is black than that it is mixed.\n\nEliminating the white card\n\nAlthough the incorrect solution reasons that the white card is eliminated, one can also use that information in a correct solution. Modifying the previous method, given that the white card is not drawn, the probability of seeing a black face is 34, and the probability of drawing the black card is 12. The conditional probability of having drawn the black card, given that a black face is showing, is\n\n$\\frac{1/2}{3/4}=2/3.$\n\nSymmetry\n\nThe probability (without considering the individual colors) that the hidden color is the same as the displayed color is clearly 23, as this holds if and only if the chosen card is black or white, which chooses 2 of the 3 cards. Symmetry suggests that the probability is independent of the color chosen, so that the information about which color is shown does not affect the odds that both sides have the same color.\n\nThis argument is correct and can be formalized as follows. By the law of total probability, the probability that the hidden color is the same as the displayed color equals the weighted average of the probabilities that the hidden color is the same as the displayed color, given that the displayed color is black or white respectively (the weights are the probabilities of seeing black and white respectively). By symmetry, the two conditional probabilities that the colours are the same given we see black and given we see white are the same. Since they moreover average out to 2/3 they must both be equal to 2/3.\n\nExperiment\n\nUsing specially constructed cards, the choice can be tested a number of times. Let \"B\" denote the colour Black. By constructing a fraction with the denominator being the number of times \"B\" is on top, and the numerator being the number of times both sides are \"B\", the experimenter will probably find the ratio to be near 23.\n\nNote the logical fact that the B/B card contributes significantly more (in fact twice) to the number of times \"B\" is on top. With the card B/W there is always a 50% chance W being on top, thus in 50% of the cases card B/W is drawn, the draw affects neither numerator nor denominator and effectively does not count (this is also true for all times W/W is drawn, so that card might as well be removed from the set altogether). Conclusively, the cards B/B and B/W are not of equal chances, because in the 50% of the cases B/W is drawn, this card is simply \"disqualified\".\n\nNotes and references\n\n1. ^ Bar-Hillel and Falk (page 119)\n2. ^ Nickerson (page 158) advocates this solution as \"less confusing\" than other methods.\n3. ^ Bar-Hillel and Falk (page 120) advocate using Bayes' Rule.", "date": "2014-10-21 03:00:54", "meta": {"domain": "wikipedia.org", "url": "http://en.wikipedia.org/wiki/Bertrand's_box_paradox", "openwebmath_score": 0.7928835153579712, "openwebmath_perplexity": 286.4903982179206, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9923043537319405, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.856747248320567}}
{"url": "https://math.stackexchange.com/questions/2918063/are-most-sets-in-mathbb-r-neither-open-nor-closed", "text": "# Are \u201cmost\u201d sets in $\\mathbb R$ neither open nor closed?\n\nIt seems intuitive to believe that most subsets of $\\mathbb R$ are neither open nor closed.\n\nFor instance, if we consider the collection of all (open, closed, half-closed/open) intervals, then one can probably make precise the notion that \"half\" of all intervals in this collection are neither open nor closed. (Whether this will amount to a reasonable definition of what it means for most subsets to be neither open nor closed might be up for debate.)\n\nIf this intuition is correct, is there a way to formalise it? If not, how would we formalise its being wrong?\n\nTo be clear, I am happy for a fairly broad interpretation of the term \"most\". Natural interpretations include but are not limited to:\n\n1. Measure-theoretic (e.g. is there a natural measure on (a $\\sigma$-algebra on) the power set of $\\mathbb R$ that assigns negligible measure to $\\tau$?)\n2. Topological (e.g. is there a natural topology on the power set of $\\mathbb R$ where $\\tau$ is meagre, or even nowhere dense?)\n3. Set-theoretic (e.g. does the power set of $\\mathbb R$ have larger cardinality than $\\tau$?)\n\nHere, $\\tau$ is (obviously) the Euclidean topology.\n\nActually, that last version of the question in parentheses might have the easiest answer: Let $\\mathcal B$ be the Borel sets on $\\mathbb R$. We have that $|\\tau| \\le | \\mathcal B | = | \\mathbb R | < \\left| 2^{\\mathbb R} \\right|$. (For details on the equality, see here. For a much simpler proof, see this answer.)\n\nAre there alternative ways to make this precise?\n\n\u2022 A trivial answer to 1. and 2. would be to define a topology on $P(\\Bbb R),$ the power-set of $\\Bbb R$, as $\\{\\emptyset, P(\\Bbb R)\\setminus \\tau, P(\\Bbb R)\\}$ and a measure $m$ with $m(P(\\Bbb R))=1$ and $m(\\tau)=0.$ But I don't think this is quite what you're hoping for. \u2013\u00a0DanielWainfleet Sep 16 '18 at 1:12\n\u2022 @DanielWainfleet Indeed. I was hoping the word \"natural\" would be enough to rule out such answers. Unless there are other reasons I'm not seeing that would make such a definition natural? \u2013\u00a0Theoretical Economist Sep 17 '18 at 22:43\n\nSince each open non-empty subset of $\\mathbb R$ can be written has a countable union of open intervals and since the set of all open intervals has the same cardinal as $\\mathbb R$, the set of all open subsets of $\\mathbb R$ has the same cardinal as $\\mathbb R$. And since there is a bijection between the open subsets of $\\mathbb R$ and the closed ones, the set of all closed subsets of $\\mathbb R$ also has the same cardinal as $\\mathbb R$. So, in the set-theoretical sense, most subsets of $\\mathbb R$ are neither closed nor open.\n\u2022 To the proposer: The set of bounded open real intervals with rational end-points is a countable base (basis) for $\\tau.$ For any topology $T$ with a countable base $B$ we have $|T|=|\\{\\cup C: C\\in P(B)\\}|\\leq |P(B)|\\leq 2^{\\aleph_0}=|\\Bbb R|.$.... So $|\\tau|\\leq |\\Bbb R|.$ And we also have $|\\tau|\\geq |\\{(r,r+1):r\\in \\Bbb R\\}|=|\\Bbb R|.$ \u2013\u00a0DanielWainfleet Sep 16 '18 at 1:27", "date": "2019-04-21 14:22:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2918063/are-most-sets-in-mathbb-r-neither-open-nor-closed", "openwebmath_score": 0.9278112053871155, "openwebmath_perplexity": 161.26587218202604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308786041316, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8567448139044676}}
{"url": "https://yutsumura.com/is-the-trace-of-the-transposed-matrix-the-same-as-the-trace-of-the-matrix/", "text": "# Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix?\n\n## Problem 633\n\nLet $A$ be an $n \\times n$ matrix.\n\nIs it true that $\\tr ( A^\\trans ) = \\tr(A)$? If it is true, prove it. If not, give a counterexample.\n\n## Solution.\n\nThe answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the diagonal entries of the transposed matrix are the same as the original matrix.\n\nPutting together these observations yields the equality $\\tr ( A^\\trans ) = \\tr(A)$.\n\nHere is the more formal proof.\n\nFor $A = (a_{i j})_{1 \\leq i, j \\leq n}$, the transpose $A^{\\trans}= (b_{i j})_{1 \\leq i, j \\leq n}$ is defined by $b_{i j} = a_{j i}$.\n\nIn particular, notice that $b_{i i} = a_{i i}$ for $1 \\leq i \\leq n$. And so,\n$\\tr(A^{\\trans}) = \\sum_{i=1}^n b_{i i} = \\sum_{i=1}^n a_{i i} = \\tr(A) .$\n\n### More from my site\n\n\u2022 A Relation between the Dot Product and the Trace Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. Prove that $\\tr ( \\mathbf{v} \\mathbf{w}^\\trans ) = \\mathbf{v}^\\trans \\mathbf{w}$. \u00a0 Solution. Suppose the vectors have components $\\mathbf{v} = \\begin{bmatrix} v_1 \\\\ v_2 \\\\ \\vdots \\\\ v_n [\u2026] \u2022 Does the Trace Commute with Matrix Multiplication? Is \\tr (A B) = \\tr (A) \\tr (B) ? Let A and B be n \\times n matrices. Is it always true that \\tr (A B) = \\tr (A) \\tr (B) ? If it is true, prove it. If not, give a counterexample. Solution. There are many counterexamples. For one, take \\[A = \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 [\u2026] \u2022 Matrix XY-YX Never Be the Identity Matrix Let I be the n\\times n identity matrix, where n is a positive integer. Prove that there are no n\\times n matrices X and Y such that \\[XY-YX=I.$ \u00a0 Hint. Suppose that such matrices exist and consider the trace of the matrix $XY-YX$. Recall that the trace of [\u2026]\n\u2022 Prove that the Dot Product is Commutative: $\\mathbf{v}\\cdot \\mathbf{w}= \\mathbf{w} \\cdot \\mathbf{v}$ Let $\\mathbf{v}$ and $\\mathbf{w}$ be two $n \\times 1$ column vectors. (a) Prove that $\\mathbf{v}^\\trans \\mathbf{w} = \\mathbf{w}^\\trans \\mathbf{v}$. (b) Provide an example to show that $\\mathbf{v} \\mathbf{w}^\\trans$ is not always equal to $\\mathbf{w} [\u2026] \u2022 If 2 by 2 Matrices Satisfy$A=AB-BA$, then$A^2$is Zero Matrix Let$A, B$be complex$2\\times 2$matrices satisfying the relation $A=AB-BA.$ Prove that$A^2=O$, where$O$is the$2\\times 2$zero matrix. Hint. Find the trace of$A$. Use the Cayley-Hamilton theorem Proof. We first calculate the [\u2026] \u2022 Determine Whether Given Matrices are Similar (a) Is the matrix$A=\\begin{bmatrix} 1 & 2\\\\ 0& 3 \\end{bmatrix}$similar to the matrix$B=\\begin{bmatrix} 3 & 0\\\\ 1& 2 \\end{bmatrix}$? (b) Is the matrix$A=\\begin{bmatrix} 0 & 1\\\\ 5& 3 \\end{bmatrix}$similar to the matrix [\u2026] \u2022 If Two Matrices are Similar, then their Determinants are the Same Prove that if$A$and$B$are similar matrices, then their determinants are the same. Proof. Suppose that$A$and$B$are similar. Then there exists a nonsingular matrix$S$such that $S^{-1}AS=B$ by definition. Then we [\u2026] \u2022 Matrix Operations with Transpose Calculate the following expressions, using the following matrices: $A = \\begin{bmatrix} 2 & 3 \\\\ -5 & 1 \\end{bmatrix}, \\qquad B = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix}, \\qquad \\mathbf{v} = \\begin{bmatrix} 2 \\\\ -4 \\end{bmatrix}$ (a)$A B^\\trans + \\mathbf{v} [\u2026]\n\n#### You may also like...\n\n##### The Vector $S^{-1}\\mathbf{v}$ is the Coordinate Vector of $\\mathbf{v}$\n\nSuppose that $B=\\{\\mathbf{v}_1, \\mathbf{v}_2\\}$ is a basis for $\\R^2$. Let $S:=[\\mathbf{v}_1, \\mathbf{v}_2]$. Note that as the column vectors of $S$...\n\nClose", "date": "2018-02-19 16:14:26", "meta": {"domain": "yutsumura.com", "url": "https://yutsumura.com/is-the-trace-of-the-transposed-matrix-the-same-as-the-trace-of-the-matrix/", "openwebmath_score": 0.9988897442817688, "openwebmath_perplexity": 141.6648508326761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9901401455693091, "lm_q2_score": 0.8652240947405564, "lm_q1q2_score": 0.8566931111164882}}
{"url": "https://brilliant.org/discussions/thread/componendo-et-dividendo-2/", "text": "# This note has been used to help create the Componendo and Dividendo wiki\n\nSee the complete wiki page here.\n\nThe method of Componendo et Dividendo allows a quick way to do some calculations, and can simplify the amount of expansion needed.\n\nIf $a, b, c$ and $d$ are numbers such that $b, d$ are non-zero and $\\frac{a}{b} = \\frac{c}{d}$, then\n\n$\\begin{array} {l r l } \\text{1. Componendo:} & \\frac{ a+b}{b} & = \\frac{ c+d}{d}. \\\\ \\text{2. Dividendo: } & \\frac{ a-b}{b} & = \\frac{ c-d} {d}. \\\\ \\text{ Componendo et Dividendo: } & \\\\ \\text{3. For } k \\neq \\frac{a}{b},& \\frac{ a+kb}{a-kb} & = \\frac{ c+kd}{c-kd} .\\\\ \\text{4. For } k \\neq \\frac{-b}{d}, & \\frac{ a}{b} & = \\frac{ a + kc } { b + kd }. \\\\ \\end{array}$\n\nThis can be proven directly by observing that\n\n$\\begin{array} {l r l } \\text{ 1.} \\frac{ a+b}{b} = \\frac{ \\frac{a}{b} + 1} {1} = \\frac{ \\frac{c}{d} + 1} {1} = \\frac{ c+d}{d} . \\\\ \\text{ 2.} \\frac{ a-b}{b} = \\frac{ \\frac{a}{b} - 1} {1} = \\frac{ \\frac{c}{d} - 1} {1} = \\frac{ c-d}{d} . \\\\ \\text{ 3.} \\frac{ a+kb}{a-kb} = \\frac{ \\frac{a}{b} + k } { \\frac{a}{b} - k} = \\frac{ \\frac{c}{d} + k } { \\frac{ c}{d} -k} = \\frac{ c+kd} { c-kd} . \\\\ \\text{ 4.} \\frac{ a + kc} { b+ kd} = \\frac{ a}{b} \\times \\frac{ 1 + k \\frac{c}{a} } { 1 + k \\frac{d}{b} } = \\frac{ a}{b} . \\end{array}$\n\n## Worked examples\n\n### 1. Show the converse, namely that if $a, b, c$ and $d$ are numbers such that $b, d, a-b, c-d$ are non-zero and $\\frac{ a+b}{a-b} = \\frac{c+d} { c-d}$, then $\\frac{ a}{b} = \\frac{c}{d}$.\n\nSolution: We apply Componendo et Dividendo with $k=1$ (which is valid since $\\frac{a+b}{a-b} \\neq 1$ ), and get that $\\frac{ 2a } { 2b} = \\frac{ (a+b) + (a-b) } { (a+b) - (a-b) } = \\frac{ (c+d) + (c-d) } { (c+d) - (c-d) } = \\frac{ 2c} { 2d}.$\n\nNote: The converse of Componendo and Dividendo also holds, and we can prove it by applying Dividendo and Componendo respectively.\n\n### 2. Solve for $x$: $\\frac{ x^3+1} { x+ 1} = \\frac{ x^3-1} { x-1}$.\n\nSolution: For the fractions to make sense, we must have $x \\neq 1, -1$.\n\nCross multiplying, we get $\\frac{ x^3+1}{x^3-1} = \\frac{ x+1}{x-1}.$\n\nApply Componendo et Dividendo with $k=1$ (which is valid since $\\frac{x+1}{x-1} \\neq 1$ ), we get that $\\frac{ 2x^3}{2} = \\frac{ 2x}{2} \\Rightarrow x(x^2-1) = 0$. However since $x \\neq 1, -1$, we have $x=0$ as the only solution.\n\nNote: We also need to check the condition that the denominators are non-zero, but this is obvious.\n\nNote by Calvin Lin\n5\u00a0years, 5\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $ ... $ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nif a/b =c/d what will be the result by componendo dividendo\n\n- 5\u00a0years, 2\u00a0months ago\n\nSee the statements contained in the first box.\n\nStaff - 5\u00a0years, 2\u00a0months ago\n\nSorry I'm slightly confused, could you clarify what Componendo and Dividendo integrate to?\n\n- 3\u00a0years, 5\u00a0months ago\n\nCheck out the examples on the componendo and dividendo wiki page.\n\nStaff - 3\u00a0years, 5\u00a0months ago\n\nI found out (somewhat accidentally) that\n\n$\\frac{a+mb}{b+na} = \\frac{c+md}{d+nc}$\n\nis also true. It's quite easy to prove; it can also be derived from the 4th case stated above. It seems different enough, though, to be worth a mention, yet I never see it anywhere. Or is it perhaps that there are many other such corollaries and only the most basic ones are usually listed?\n\n- 1\u00a0year, 7\u00a0months ago\n\nOhhh, that's a really nice identity. It is a generalization of Worked Example 1. Can you add it to the Componendo and Dividendo wiki under Problem Solving?\n\nLike you said, it is essentially / can be derived from the 4th case, where we have $\\frac{ a + mb } { c + md} = \\frac{a}{c} = \\frac{b}{d} = \\frac{ b+na}{d + nc }$ (as long as the denominators are non-zero).\n\nStaff - 1\u00a0year, 7\u00a0months ago\n\nHave finally had a chance to add the identity to the Wiki; please take a look when you can and let me know if I need to make any changes. (I accidentally first added it to the Theorem section before I remembered that you said Problem Solving, I did move it to the correct section after that, hope it didn't cause any problems.) I also added an example applying it a little further down, in the section that introduced using C&D with non-linear terms. The Wiki mentioned the terms could be polynomial or exponential; my example uses trig functions, I believe it's still valid but please take a look and let me know if there are any issues. Thanks.\n\n- 1\u00a0year, 7\u00a0months ago\n\nThat's a great writeup. I like how you highlighted Method 1, which \"by right\" should be how people understand this identity, and \"by left\" it is placed in this wiki because of the form it took.\n\nYes, your example with a trigo substitution is valid. Good one. Thanks!\n\nStaff - 1\u00a0year, 7\u00a0months ago", "date": "2019-08-24 19:07:57", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/componendo-et-dividendo-2/", "openwebmath_score": 0.9854892492294312, "openwebmath_perplexity": 1087.8622986071887, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9901401446964613, "lm_q2_score": 0.8652240947405564, "lm_q1q2_score": 0.8566931103612793}}
{"url": "http://mathhelpforum.com/calculus/16873-parametric-equations-plane-problem.html", "text": "# Thread: Parametric equations of plane problem\n\n1. ## Parametric equations of plane problem\n\nThis problem was confusing me so any help would be appreciated!\n\nWrite both the parametric and symmetric equations of the line of intersection of the planes with the equations 2x - y + z = 5 and x + y - z = 1.\n\n2. Originally Posted by clockingly\nThis problem was confusing me so any help would be appreciated!\n\nWrite both the parametric and symmetric equations of the line of intersection of the planes with the equations 2x - y + z = 5 and x + y - z = 1.\nIf $z=0$ then:\n$2x-y=5$\n$x+y=1$\nThus, $x=2$ and $y=-1$\n\nIf $z=1$ then:\n$2x-y=4$\n$x+y=2$\nThus, $x=2$ and $y=0$\n\nThus, we have that this line must contains the points $(2,-1,0)\\mbox{ and }(2,0,1)$\n\nThus, the Symettric Equation is:\n$\\frac{x-2}{0} = \\frac{y+1}{-1} = \\frac{z-0}{-1}$\n\nNote, this is a zero in the denominator of the first fraction, it is a useful notation I saw in a Russian Textbook* which simply means $x=2$ constantly, so it is just a shorthand notation.\n\n*)If you are interested the Textbook was written by Perelman! Not Perelman you think it is but his grandfather!\n\n3. Originally Posted by ThePerfectHacker\nThus, the Symettric Equation is:\n$\\frac{x-2}{0} = \\frac{y+1}{-1} = \\frac{z-0}{-1}$\n\nNote, this is a zero in the denominator of the first fraction, it is a useful notation I saw in a Russian Textbook* which simply means $x=2$ constantly, so it is just a shorthand notation.\n\n*)If you are interested the Textbook was written by Perelman! Not Perelman you think it is but his grandfather!\nThanks for that TPH. i did this question the same as you did, but i had no idea what to do with that 0 in the denominator, it kind of freaked me out, so i decided not to post\n\n4. Originally Posted by Jhevon\nThanks for that TPH. i did this question the same as you did, but i had no idea what to do with that 0 in the denominator, it kind of freaked me out, so i decided not to post\nThe way it is taught in American schools is if that every happens then the fuction does not have a \"symettric form\" to it. It only has parametric. But Grandpa's Perelman notation is really useful like I said.\n\n5. Hello, clockingly!\n\nWrite both the parametric and symmetric equations of the line of intersection\nof the planes with the equations: . $\\begin{array}{ccc}2x - y + z &= &5 \\\\ x + y - z &= &1\\end{array}$\nThis is how I was taught to find the intersection of two planes.\n\nWe have: . $\\begin{array}{cccc}2x - y + z & = & 5 & {\\color{blue}[1]}\\\\ x + y -z & = & 1 & {\\color{blue}[2]}\\end{array}$\n\nAdd [1] and [2]: . $3x \\,=\\,6\\quad\\Rightarrow\\quad x\\,=\\,2$\n\nSubstitute into [2]: . $2 + y - z \\:=\\:1\\quad\\Rightarrow\\quad y \\:=\\:z-1$\n\nWe have $x,\\,y,\\,z$ as a function of $z\\!:\\;\\;\\begin{Bmatrix}x & = & 2 \\\\ y & = & z-1 \\\\ z & = & z\\end{Bmatrix}$\n\nOn the right side, replace $z$ with a parameter $t$.\n\n. . . . $\\begin{Bmatrix}x & = & 2 \\\\ y & = & t - 1 \\\\ z & = & t\\end{Bmatrix}$ . . . . There!\n\n6. Originally Posted by Soroban\nHello, clockingly!\n\nThis is how I was taught to find the intersection of two planes.\n\nWe have: . $\\begin{array}{cccc}2x - y + z & = & 5 & {\\color{blue}[1]}\\\\ x + y -z & = & 1 & {\\color{blue}[2]}\\end{array}$\n\nAdd [1] and [2]: . $3x \\,=\\,6\\quad\\Rightarrow\\quad x\\,=\\,2$\n\nSubstitute into [2]: . $2 + y - z \\:=\\:1\\quad\\Rightarrow\\quad y \\:=\\:z-1$\n\nWe have $x,\\,y,\\,z$ as a function of $z\\!:\\;\\;\\begin{Bmatrix}x & = & 2 \\\\ y & = & z-1 \\\\ z & = & z\\end{Bmatrix}$\n\nOn the right side, replace $z$ with a parameter $t$.\n\n. . . . $\\begin{Bmatrix}x & = & 2 \\\\ y & = & t - 1 \\\\ z & = & t\\end{Bmatrix}$ . . . . There!\n\nnice method\n\n7. The customary notation used in textbooks in North America for the symmetric form in which one direction number is zero is to use a semicolon: $\\frac{{y + 1}}{{ - 1}} = \\frac{z}{{ - 1}}\\, ;\\,x = 2$.", "date": "2017-08-24 09:03:22", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/16873-parametric-equations-plane-problem.html", "openwebmath_score": 0.8964012861251831, "openwebmath_perplexity": 407.96947903877344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9763105335255603, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.8566898399272382}}
{"url": "https://math.stackexchange.com/questions/1148442/proving-fracnnen-1n-fracn1n1en-by-induction-for-all", "text": "# Proving $\\frac{n^n}{e^{n-1}}<n!<\\frac{(n+1)^{n+1}}{e^{n}}$ by induction for all $n> 2$.\n\nI am trying to prove\n\n$$\\frac{n^n}{e^{n-1}}<n!<\\frac{(n+1)^{n+1}}{e^{n}} \\text{ for all }n > 2.$$\n\nHere is the original source (Problem 1B, on page 12 of PDF)\n\nCan this be proved by induction?\n\nThe base step $n=3$ is proved: $\\frac {27}{e^2} < 6 < \\frac{256}{e^3}$ (since $e^2 > 5$ and $e^3 < 27$, respectively).\n\nI can assume the case for $n=k$ is true: $\\frac{k^k}{e^{k-1}}<k!<\\frac{(k+1)^{k+1}}{e^{k}}$.\n\nFor $n=k+1$, I am having trouble:\n\n\\begin{align} (k+1)!&=(k+1)k!\\\\&>(k+1)\\frac{k^k}{e^{k-1}}\\\\&=e(k+1)\\frac{k^k}{e^{k}} \\end{align} Now, by graphing on a calculator, I found it true that $ek^k >(k+1)^k$ (which would complete the proof for the left inequality), but is there some way to prove this relation?\n\nAnd for the other side of the inequality, I am also having some trouble: \\begin{align} (k+1)!&=(k+1)k!\\\\&<(k+1)\\frac{(k+1)^{k+1}}{e^{k}}\\\\&=\\frac{(k+1)^{k+2}}{e^k}\\\\&<\\frac{(k+2)^{k+2}}{e^k}. \\end{align} I can't seem to obtain the $e^{k+1}$ in the denominator, needed to complete the induction proof.\n\n\u2022 just curious, what level this exam is -to enter PhD program or some qual? \u2013\u00a0Alex Feb 15 '15 at 13:47\n\u2022 @Alex According to the link provided in my question, it's UC Berkeley's \"preliminary exam\", which apparently their students must pass before they can continue their Ph.D program and eventually take their oral qualifying exam. \u2013\u00a0Cookie Feb 15 '15 at 17:52\n\u2022 Also, there is a solution here math.berkeley.edu/sites/default/files/pages/\u2026 (page 12 of PDF), but the solution uses integration. I understand that method works, but I wanted to also prove this by induction. Induction must be possible to use here... \u2013\u00a0Cookie Feb 15 '15 at 17:55\n\u2022 so it's a prelim to qual exam? \u2013\u00a0Alex Feb 16 '15 at 0:05\n\u2022 It's a prelim required to stay in the Ph.D program. The student has two chances (or maybe three based on appeal) to pass the exam, or else the student is dismissed. \u2013\u00a0Cookie Feb 16 '15 at 0:27\n\nLet's try an inductive proof from the original inquality for $n$, let's prove for $n+1$\n\n$$\\frac{n^n}{e^{n-1}}<n!<\\frac{(n+1)^{n+1}}{e^{n}}$$\n\nOkay, multiply both sides by $n+1$. At least the middle is correct\n\n$$(n+1)\\frac{n^n}{e^{n-1}}<(n+1)!<\\frac{(n+1)^{n+2}}{e^{n}}$$\n\nand we have to make the left and right sides look more appropriate\n\n$$\\left(\\color{red}{\\frac{n}{n+1}} \\right)^\\color{red}{n}\\frac{(n+1)^{n+1}}{e^{n-1}}<(n+1)!<\\frac{(n+2)^{n+2}}{e^{n}} \\left(\\color{blue}{\\frac{n+1}{n+2}}\\right)^{\\color{blue}{n+2}}$$\n\nOur induction is complete if we can prove two more inequalities:\n\n$$\\frac{1}{e} < \\left(\\frac{n}{n+1} \\right)^n \\text{ and } \\left(\\frac{n+1}{n+2}\\right)^{n+2}< \\frac{1}{e}$$\n\nThese two inequalities can be combined into one and we can take reciprocals. At least it is well-known.\n\n$$\\bigg(1 + \\frac{1}{m}\\bigg)^{m+1}> \\mathbf{e} > \\bigg(1 + \\frac{1}{n} \\bigg)^n$$\n\nThis is true for any $m, n \\in \\mathbb{N}$.\n\nYou shave off a little bit too much when you said $(k+1)^{k+2} < (k+2)^{k+2}$. Instead, you needed the more delicate:\n\n$$(k+1)^{k+2} < \\frac{1}{e} (k+2)^{k+2}$$\n\nIf you know1 that: $$a_n=\\left(1+\\frac{1}{n}\\right)^n,\\qquad b_n = \\left(1+\\frac{1}{n}\\right)^{n+1}$$ give to sequences converging towards $e$, where $\\{a_n\\}$ is increasing while $\\{b_n\\}$ is decreasing, consider that:\n\n$$n = \\prod_{k=1}^{n-1}\\left(1+\\frac{1}{k}\\right),\\tag{1}$$ so: $$n! = \\prod_{m=2}^{n} m = \\prod_{m=2}^{n}\\prod_{k=1}^{m-1}\\left(1+\\frac{1}{k}\\right) = \\prod_{k=1}^{n-1}\\left(1+\\frac{1}{k}\\right)^{n-k}=\\frac{n^n}{\\prod_{k=1}^{n-1}\\left(1+\\frac{1}{k}\\right)^k}\\tag{2}$$ and your inequality trivially follows.\n\n1) If you are not aware of such a classical result, then prove it by induction. It is rather easy.\n\n\u2022 I think that it is not as easy as you think. \u2013\u00a0marty cohen Feb 15 '15 at 4:09\n\u2022 @martycohen: it is also possible to use AM-GM in order to prove that $\\{a_n\\}$ is increasing and $\\{b_n\\}$ is decreasing. I think there are many questions/answers on MSE devoted to such well-known fact. \u2013\u00a0Jack D'Aurizio Feb 15 '15 at 4:12\n\u2022 Here, for example: math.stackexchange.com/questions/389793/\u2026 \u2013\u00a0marty cohen Feb 15 '15 at 4:18\n\nProof: We will prove the inequality by induction. Since $e^2 > 5$ and $e^3 < 27$, we have $$\\frac {27}{e^2} < 6 < \\frac{256}{e^3}.$$ Thus, the statement for $n=3$ is true. The base step is complete.\n\nFor the induction step, we assume the statement is true for $n=k$. That is, assume $$\\frac{k^k}{e^{k-1}}<k!<\\frac{(k+1)^{k+1}}{e^{k}}.$$\n\nWe want to prove that the statement is true for $n=k+1$. It is straightforward to see for all $k > 2$ that $\\left(1+\\frac 1k \\right)^k < e < \\left(1+\\frac 1k \\right)^{k+1}$; this algebraically implies $$\\left(\\frac k{k+1} \\right)^{k+1} < \\frac 1e < \\left(\\frac k{k+1} \\right)^k. \\tag{*}$$ A separate induction proof for the left inequality of $(*)$ establishes $\\left(\\frac{k+1}{k+2} \\right)^{k+2}<\\frac 1e$. We now have \\begin{align} (k+1)! &= (k+1)k! \\\\ &< (k+1) \\frac{(k+1)^{k+1}}{e^k} \\\\ &= \\frac{(k+1)^{k+2}}{e^k} \\\\ &= \\frac{(k+1)^{k+2}}{e^k} \\left( \\frac{k+2}{k+2} \\right)^{k+2} \\\\ &= \\frac{(k+2)^{k+2}}{e^k} \\left( \\frac{k+1}{k+2} \\right)^{k+2} \\\\ &< \\frac{(k+2)^{k+2}}{e^k} \\frac 1e \\\\ &= \\frac{(k+2)^{k+2}}{e^{k+1}} \\end{align} and \\begin{align} (k+1)! &= (k+1)k! \\\\ &> (k+1) \\frac{k^k}{e^{k-1}} \\\\ &= (k+1) \\frac{k^k}{e^{k-1}} \\left(\\frac{k+1}{k+1} \\right)^k \\\\ &= \\frac{(k+1)^{k+1}}{e^{k-1}} \\left( \\frac k{k+1} \\right)^k \\\\ &> \\frac{(k+1)^{k+1}}{e^{k-1}} \\frac 1e \\\\ &= \\frac{(k+1)^{k+1}}{e^k}. \\end{align} We have established that the statement $$\\frac{(k+1)^{k+1}}{e^k}<(k+1)!<\\frac{(k+2)^{k+2}}{e^{k+1}}$$ for $n=k+1$ is true. This completes the proof.\n\nSome times it is easier for such an induction if we shift the sequence by one step, so the basis of the expressions is nicer for algebraic manipulations.\nLet's rewrite your sequence of inequalities as\n$$\\displaystyle {n! \\over e}<{(n+1)^{n+1} \\over e^{n+1}} < {(n+1)! \\over e} <{(n+2)^{n+2} \\over e^{n+2}}< {(n+2)! \\over e} \\tag 1$$\nand for simpler references below as $$a_0 \\quad < \\quad b_0 \\quad <\\quad a_1 \\quad <\\quad b_1 \\quad < \\quad a_2 \\tag 2$$\n\nThen we ask: does from $a_0<b_0<a_1$ follow that $a_1<b_1<a_2$ ?\n\nOf course $a_1 = (n+1) \\cdot a_0$ and so it might be useful to define $b_0$ as a fraction in the interval of $a_0$ and $a_1$: $$b_0 = (n+1)q_1 \\cdot a_0 \\text{ where } q_1<1 \\tag {3.1 }$$. Consequently, define $$b_1 = (n+2)q_2 \\cdot a_1 \\text{ where also } q_2<1 \\tag {3.2 }$$ Here the inequality $q_2 < 1$ is not known but expected and if this can be shown by induction from $q_1$ this would solve the problem .\n\nSo we start with $$q_1 = {b_0 \\over a_0 (n+1)} = {(n+1)^{n} \\over e^n n! } \\tag {4.1 }$$ and by the beginning of the induction we know, that this is indeed smaller than 1 so $$q_1 < 1 \\tag {4.2 }$$\n\nNow we have simply $$q_2 = {b_1 \\over a_1 (n+2)} = {(n+2)^{n+1} \\over e^{n+1} (n+1)! } \\tag {4.3 }$$ Next we consider the systematic progression in the sequence of $q_1,q_2,q_3,...$. To begin we determine the ratio $r_2={q_2 \\over q_1}$ . We find $$\\begin{eqnarray} r_2&=&{q_2 \\over q_1}& =&{ {(n+2)^{n+1} \\over e^{n+1} (n+1)! } \\over {(n+1)^{n} \\over e^{n} (n)! } } \\\\ &&&=& {(n+2)^{n+1} e^{n} (n)! \\over e^{n+1} (n+1)! (n+1)^{n}} \\\\ &&&=& {(n+2)^{n+1} \\over e (n+1)^{n+1}} \\\\ &&&=& \\left({n+2 \\over n+1}\\right)^{n+1} \\cdot \\frac 1e \\\\ r_2&=&{q_2 \\over q_1}& =& \\left( 1 + {1 \\over n+1} \\right)^{n+1} \\cdot \\frac 1e \\end{eqnarray} \\tag {5.1 }$$\n\nIt is now needed to recognize/remember from the definition of $e$, that the last expression is smaller than 1 and that for $n \\to \\infty$ approximates monotonically 1.\nAlso we see by the expansion of the binomial-series $(1+1/x)^x=1+1+1/2+...$ in the general case that for $x>2$ this is greater than $2$ so $${2 \\over e} \\approx 0.73 < r_2 < 1 \\tag{5.2}$$\n\nFrom this we know now, that $q_2$ is not only smaller than $1$ but also smaller than $q_1$ but$q_{n+1} \\to q_n$ for increasing $n$.\n\nSo we have $$\\begin{eqnarray} &q_2 &=& q_1 \\cdot r_2 < q_1 < 1 \\\\ \\to& b_1 &<& (n+2) a_1 = a_2 \\\\ \\to &a_1 &<& b_1 <a_2 \\end{eqnarray} \\tag {6 }$$ which we wanted to show.\n\nSince several answers have already proven the result via induction, I see no harm in recording an additional proof that does not use induction.\n\nWe claim for all $n\\ge 1$, $$\\int_{0}^{1} (x\\log(x))^n \\, dx = \\frac{(-1)^n n!}{(n+1)^{n+1}}$$ This can be shown using differentiation under the integral sign; set $k=n$ below: $$f(n) = \\int_{0}^{1} x^n \\, dx = \\frac{1}{n+1} \\implies \\frac{d^k}{dn^k} f(n) = \\int_{0}^{1} x^n (\\log(x))^k \\, dx = \\frac{(-1)^{k} k!}{(n+1)^{k+1}}$$\n\nSince $x \\log(x)$ does not change sign, we have $$\\int_{0}^{1} |x\\log(x)|^n \\, dx = \\frac{n!}{(n+1)^{n+1}}$$ From this expression, we see that the given inequality is equivalent to $$\\frac{1}{(n+1)e^n} < \\int_{0}^{1} |x\\log(x)|^n \\, dx < \\frac{1}{e^n}$$\n\nA simple computation shows $|x\\log(x)|^n$ attains a maximum of $e^{-n}$ at $x=e^{-1}$, so the upper bound is evident. To see the lower bound, note $|x\\log(x)|^n$ is a concave function, so its graph encloses the triangle with vertices $(0,0)$, $(1,0)$, and $(e^{-n}, e^{-1})$. This triangle has area $e^{-n}/2 > e^{-n}/(n+1)$ since $n>2$.", "date": "2021-06-14 08:56:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1148442/proving-fracnnen-1n-fracn1n1en-by-induction-for-all", "openwebmath_score": 0.9996531009674072, "openwebmath_perplexity": 309.60411274479463, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105314577313, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8566898334204925}}
{"url": "https://www.physicsforums.com/threads/formula-of-nth-term-of-a-pattern.767884/", "text": "# Formula of nth term of a pattern\n\n1. Aug 27, 2014\n\n### songoku\n\n1. The problem statement, all variables and given/known data\nFind the formula for nth term of:\n.... , -450 , -270 , -180 , -90 , 90 , 180 , 270 , 450 , ...\n\n2. Relevant equations\nArithmetic and geometric series\nRecursive\n\n3. The attempt at a solution\nUsually when finding the formula of Un, n starts from 1. But since the pattern goes infinite in both ways, how can we set which one is the first term?\n\nAs for recursive, it usually involves the preceding term, Un-1. Again, how can we determine the preceding term for the above case?\n\nPlease give me idea how to start solving this. Thank you very much\n\n2. Aug 27, 2014\n\n### jackarms\n\nFor the general formula (in terms of n), you're right that the sequence going both ways complicates things. The answer is that you have to arbitrarily designate a term to be $a_{1}$, or perhaps $a_{0}$. You would write the whole formula based around that starting term, in a form like this:\n\n$a_{n} = ....$, where $a_{0} = 3$\n\nFor the recursive definition, you'll just use a term and the preceding term. Since it's recursive, you don't need to know the particular values for either of those terms -- the recursive form just means you describe a term in terms of its preceding term. So it works with any two (adjacent) terms in the sequence, and so it isn't necessary to designate a starting term. The form for that would look like:\n\n$a_{n} = a_{n-1}....$\n\nSo $a_{n}$ can refer to any term in the sequence, because this rule works for any pair of terms in the sequence, no matter how long it is and whether it goes infinitely in just one direction or two.\n\nLast edited: Aug 27, 2014\n3. Aug 27, 2014\n\n### Ray Vickson\n\nJust fix $n = 0$ anywhere you want; for example, you can take $U_0 = -450.$ All the $U_n$ for $n < 0$ are the ones you do not see to the left of what you have written and all the $U_n$ for $n > 7$ are the ones you don't see to the right of what you have written. Now just start taking differences and see what you get---there is a nice pattern.\n\n4. Aug 28, 2014\n\n### songoku\n\nI still can't find the pattern. Do we have to make separate formula for n < 0 and n > 0?\n\nI tried to set U0 = 0, at the middle of the positive and negative terms. What I got is 270 = 180 + 90 ; 450 = 270 + 180. Maybe it is related to Un = Un-1 + Un-2, but I don't know about 90 and 180.\n\nThanks\n\n5. Aug 30, 2014\n\n### SammyS\n\nStaff Emeritus\nThe pattern is evident in the sequence as it's given, before you assign any starting point.\n\nHint: Make a difference table with 1st & 2nd differences.\n\n6. Aug 31, 2014\n\n### Ray Vickson\n\nFor the sequence\n$$U = \\{U_i,i=0,1,2, \\ldots \\} = \\{-450 , -270 , -180 , -90 , 90 , 180 , 270 , 450 , \\ldots\\}$$\nthe sequence of first differences is\n$$dU = \\{U_{i+1}-U_{i}, i = 0,1,2,\\ldots \\} = \\{ -270-(-450),-180-(-270), \\ldots \\} = \\{ 180,90,90,180,90,90,\\ldots \\} \\\\ = 90 + 90 \\{1,0,0,1,0,0,1,0,0,\\ldots \\}$$\nIf you can find a nice formula for the sequence $\\{1,0,0,1,0,0,1,0,0,\\ldots \\}$ you are almost finished. If you know about complex variables and the three (complex) roots of unity, you can finish up with a finite sum of some geometric series, to get a fairly nice formula for the nth term of your original sequence.\n\n7. Aug 31, 2014\n\n### PeroK\n\nAlternatively, you should be able to do something with n (mod3) to formulate the sequence of 1, 0, 0, 1, 0, 0 ...\n\n8. Sep 2, 2014\n\n### andrewkirk\n\nIf we are to be rigorous, we need to take care with our definitions in defining the pattern. Recursive definitions mostly use the Recursion Theorem, which is only defined on the positive integers $Z_+$. For more general well-ordered sets we can use the Theorem of Transfinite Recursive Definition, but that's not needed here.\n\nWe can't just use the Recursion Theorem here in a single step, because the base set is the full set of integers, which is not well-ordered. The following trick gets around this problem.\n\nDefine $a_1=90, b_1=-90$\nDefine rules to give $a_{n+1}$ and $b_{n+1}$ in terms of $a_n$ and $b_n$ as follows, for $n\\in Z_+$:\n\n$a_{n+1}=a_n+180$ if $n+1$ is divisible by 3, else $a_{n+1}=a_n+90$\n$b_{n+1}=b_n-180$ if $n+1$ is divisible by 3, else $b_{n+1}=b_n-90$\n\nThen the recursion theorem tells us that these definitions prescribe unique sequences $a_n$ and $b_n$.\n\nNow we can just define $c:Z\\to Z$ by $c(n)=a_{n+1}$ if $n\\geq 0$, otherwise $c(n)=b_{-n}$.\n\nLast edited: Sep 2, 2014\n9. Sep 14, 2014\n\n### songoku\n\nI think I get the idea. I'll try it first. Thanks a lot for the help\n\n10. Sep 25, 2014\n\n### Simon Bridge\n\nI'm always puzzled by these things - i.e. notice how the sequence is symmetrical about the -90,90 terms?\nWhy not exploit that? Put $a_1=90$, then $a_n = (n/|n|)a_{|n|}$ ... the sequence for $a_{|n|}$ should be easy ... though there are only four terms so you are spoiled for choice. {90, 180, 270, 450...} the differences than go {1, 1, 2, ...} in units of $a_1$, which is suggestive.\n\nWhat approach you choose seems to depend on what assumptions you make about the pattern in question.", "date": "2018-03-22 18:45:41", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/formula-of-nth-term-of-a-pattern.767884/", "openwebmath_score": 0.7913281917572021, "openwebmath_perplexity": 339.44778732873704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97631052387569, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.8566898314597009}}
{"url": "https://math.stackexchange.com/questions/1802813/prove-upper-bound-for-recurrence", "text": "# Prove upper bound for recurrence\n\nI am working on problem set 8 problem 3 from MIT's Fall 2010 OCW class 6.042J. This is covered in chapter 10 which is about recurrences.\n\nHere is the problem:\n\n$$A_0 = 2$$ $$A_{n+1} = A_n/2 + 1/A_n, \\forall n \\ge 1$$\n\nProve\n\n$$A_n \\le \\sqrt2 + 1/2^n, \\forall n \\ge 0$$\n\nI have graphed the recurrence and the upper bound and they seem to both converge on $\\sqrt2$.\n\nAlso, if you ignore the boundary condition $A_0 = 2$ then you find that $\\sqrt2$ is a solution to the main part of the recurrence. i.e. $\\sqrt2 = \\sqrt2/2 + 1/\\sqrt2$.\n\nThe chapter and videos on recurrences have a lot to say about a kind of cookbook solution to divide and conquer recurrences which they call the Akra-Bazzi Theorem. But this recurrence does not seem to be in the right form for that theorem. If it were in the form $A_{n+1} = A_n/2 + g(n)$ then the theorem would give you an asymptotic bound. But $1/A_n$ is not a simple function of $n$ like a polynomial. Instead it is part of the recurrence.\n\nAlso, the chapter has a variety of things to say about how to guess the right solution and plug it into an inductive proof, but I haven't had much success. I have tried possible solutions of various forms like $a_n = \\sqrt2+a/b^n$ and tried solving for the constants $a$ and $b$ but to no avail.\n\nSo, if someone can point me in the right direction that would be great. I always assume that the problem sets are based on something taught in the videos and in the text of the book but I am having trouble tracking this one down.\n\nBobby\n\nWe show by induction that $$\\sqrt{2}\\lt A_n\\le \\sqrt{2}+\\frac{1}{2^n}.\\tag{1}$$ Suppose the result holds at $n=k$. We show the result holds at $n=k+1$.\n\nFor the inequality on the right of (1), we need to show that $$\\frac{A_k}{2}+\\frac{1}{A_k}\\le \\sqrt{2}+\\frac{1}{2^{k+1}}.$$ By the induction hypothesis, we have $$\\frac{A_k}{2}+\\frac{1}{A_k}\\le \\frac{\\sqrt{2}}{2}+\\frac{1}{2^{k+1}}+\\frac{1}{\\sqrt{2}}=\\sqrt{2}+\\frac{1}{2^{k+1}},$$ which takes care of the induction step for the inequality on the right of (1).\n\nWe still need to show that $\\sqrt{2}\\lt A_{k+1}$. Let $A_k=\\sqrt{2}+\\epsilon$, where $\\epsilon$ is positive. Then $$A_{k+1}=\\frac{\\sqrt{2}+\\epsilon}{2}+\\frac{1}{\\sqrt{2}+\\epsilon}=\\frac{4+2\\sqrt{2}\\epsilon+\\epsilon^2}{2(\\sqrt{2}+\\epsilon)}\\gt \\frac{4+2\\sqrt{2}\\epsilon}{2(\\sqrt{2}+\\epsilon)}=\\sqrt{2}.$$ This completes the induction step for the inequality on the left of (1).\n\nRemark: The inequality (1) and squeezing show that $A_n$ indeed has limit $\\sqrt{2}$.\n\n\u2022 It didn't occur to me to prove that $A_n \\ge \\sqrt2$ and then substitute $\\sqrt2$ in the $1/A_n$ term. Much easier that way. Thanks. \u2013\u00a0Bobby Durrett May 28 '16 at 2:36\n\u2022 @BobbyDurrett: You are welcome. The inequality $A_n\\gt \\sqrt{2}$ is not mentioned explicitly in what you are asked to show, but when one tries to push the induction through, it becomes clear that it is necessary for the proof. \u2013\u00a0Andr\u00e9 Nicolas May 28 '16 at 2:40\n\u2022 Nice induction for the $>\\sqrt{2}$ part. I didn't manage it immediately, that's why I showed it studying the map $x\\mapsto \\frac{x}2+\\frac{1}{x}$. \u2013\u00a0Daniel Robert-Nicoud May 28 '16 at 11:09\n\nLet $$B_k = \\frac{A_k}{\\sqrt{2}}$$ Then $B_0 = \\sqrt{2}$ and $$B_{n+1} = \\frac12\\left(B_n+\\frac{1}{B_n} \\right)$$ Thus $b_n$ is the $n$-th guess if you perform Newton's algorithm to try to find $\\sqrt{1}$ starting with a guess of $\\sqrt{2}$.\n\nThis recursion can be solved in closed form using the formula for the hyperbolic tangent of $2x$ in terms of $\\tanh x$: $$\\tanh(2x) = \\frac{2 \\tanh x}{1+\\tanh^2{x}}$$ . The result looks something like $$B_n = \\tanh\\left( 2^n \\theta\\right)$$ where $\\theta = \\tanh^{-1} B_0$; the answer I have given is off in that every other term needs to be the reciprocal of what I wrote, but the general idea will work.\n\nOnce you have that, you can know exactly what $B_n$ is and prove the relation; or better yet, use the error analysis for Newton's method to get an estimate of how close to 1 you would be.\n\nWe want to apply induction, but we need also a lower bound on $A_n$ for the $\\frac{1}{n}$ term. We can show that $A_n\\ge\\sqrt{2}$ as follows:\n\nLet $f(x) = \\frac{x}{2} + \\frac{1}{x}$. Then $f'(x) = \\frac{1}{2} - \\frac{1}{x^2}$ which has a zero at $x = \\sqrt{2}$ where $f(\\sqrt{2}) = \\sqrt{2}$, and is positive for $x>\\sqrt{2}$. This shows that $f(x)\\ge\\sqrt{2}$ whenever $x\\ge\\sqrt{2}$, and as $A_{n+1} = f(A_n)$, we have that $A_n\\ge\\sqrt{2}$ for all $n\\ge0$.\n\nThen to conclude:\n\nThe case $n=0$ is trivially true.\n\nFor $n\\ge0$ we have that $$A_{n+1} = \\frac{A_n}{2} + \\frac{1}{A_n}.$$ By induction hypothesis and what we have shown above, this has as upper bound $$A_{n+1}\\le \\frac{\\sqrt{2} + 2^{-n-1}}{2} + \\frac{1}{\\sqrt{2}} = \\sqrt{2} + 2^{-n-1}.$$\n\nThree steps: Use the definition of $A_n$ and algebra to establish that $$A_{n+1}-\\sqrt 2 = {(A_n-\\sqrt 2)^2\\over 2A_n}.\\tag1$$ Next, use (1) to prove by induction that $$A_n\\ge\\sqrt 2\\ \\text{for every n.}\\tag2$$ Finally, use (1) and (2) to prove by induction that $$A_n-\\sqrt2 \\le \\frac1{2^{n}}\\ \\text{for every n.}\\tag2$$\n\nYour sequence is $$A_{n+1} = \\frac{1}{2} A_n + \\frac{1}{A_n}$$ where the last term features a division which reminds of the Newton-Raphson iteration.\n\nNewton Raphson iteration takes the root of the tangent as next step for estimating a root of $f$: $$0 = T'(x_{n+1}) = f(x_n) + f'(x_n) (x_{n+1} - x_n) \\iff \\\\ x_{n+1} = x_n - \\frac{f(x_n)}{f'(x_n)}$$ by comparison we have $$-\\frac{1}{2}A_n + \\frac{1}{A_n} = - \\frac{A_n^2-2}{2A_n} = - \\frac{f(A_n)}{f'(A_n)}$$ and see $f(x) = x^2 - 2$, the function used to iterate against the root $\\sqrt{2}$.", "date": "2019-06-26 16:10:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1802813/prove-upper-bound-for-recurrence", "openwebmath_score": 0.9323938488960266, "openwebmath_perplexity": 119.9180937019314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105259435195, "lm_q2_score": 0.8774767890838837, "lm_q1q2_score": 0.8566898254537173}}
{"url": "https://math.stackexchange.com/questions/3352629/finding-the-domain-and-range-of-a-composite-function", "text": "# Finding the domain and range of a composite function\n\nSo I have two functions. $$f(x) = e^{-x^2+1}$$ and $$g(x)=\\sqrt{x^2-4x+3}$$. I am then asked to determine the domain and range of\n\n$$a)f\u2218g,$$\n\n$$b)g\u2218f$$\n\nI already did part $$a)$$ and the domain for part $$b)$$.\n\nFor part $$a)$$, the domain was $$(-\\infty,1)\\cup(3,\\infty)$$ and the range was $$(0,e^2)$$.\n\nFor part $$b$$, I figured out that the domain was $$(-\\infty,-1]\\cup[1,\\infty)$$. I am not sure how to find the range though. Normally, I would take the inverse of g\u2218f and find the domain of that, and although I can do it, I don't think I did it correctly.\n\nCurrently, I did figure out that $$g\u2218f$$ is $$\\sqrt{e^{-2x^2+2}-4e^{-x^2+1}+3}$$. How do I find the range of this mess though? I attempted to take the inverse which I believe is:\n\n$$y=\\pm\\sqrt{1-\\ln(2\\pm\\sqrt{1+y^2})}$$\n\nAlthough I know that Wolfram Alpha is not the arbitrator of what correct is, it's generally been right and my answer disagrees with what Wolfram alpha has obtained (As seen here). In addition, the range is something that I am not sure how Wolfram obtained (As seen here). This also looks REALLY messy.\n\nCan anyone guide me as to how this was obtained? That would be much appreciated!\n\n\u2022 I see how you got $(-\\infty,1)\\cup(3,\\infty)$ (except maybe the choice of open/closed) but not how you got $e^2.$ When you were computing the range, did you forget the gap in the domain? I think that part is easier to solve if you do not work out the formula for $f\\circ g.$ \u2013\u00a0David K Sep 11 '19 at 11:58\n\u2022 I took the inverse of the function and found the domain of that to get the range. \u2013\u00a0Future Math person Sep 11 '19 at 18:23\n\u2022 That seems like an unnecessarily difficult way to do it. Also error-prone, since (I think) it gives the wrong answer. What value of $x$ satisfies $f(g(x))=e^{3/2},$ for example? \u2013\u00a0David K Sep 11 '19 at 18:42\n\u2022 Really? I didn't think it was too bad. I ended up getting $f(g(x))=-2 \\pm \\sqrt{2-\\ln(x)}$. This would mean $2-\\ln(x) \\geq 0$ and thus $e^2 \\geq x$ \u2013\u00a0Future Math person Sep 11 '19 at 18:44\n\u2022 Also I get $g(1)=g(3)=0,$ and $0$ is in the domain of $f,$ so I would not exclude $1$ and $3$ from the domain in part a). I think you may need to be a lot more careful about checking what happens at your boundaries. (This is relevant to the range as well.) \u2013\u00a0David K Sep 11 '19 at 18:47\n\nTo find the range you want, namely of $$g(f(x))=F(x),$$ note that $$F(x)$$ is never negative. Also, it is defined and continuous at all real $$x.$$ Then it is an even function of $$x.$$ Thus, it suffices to consider only the range for positive $$x,$$ say.\n\nNote that $$F(0)>0.$$ Also, as $$x\\to +\\infty,$$ we have $$F(x)\\to \\sqrt 3.$$ Thus, the range is at least $$[F(0),\\sqrt 3),$$ by IVT. It only remains to see if the function attains $$\\sqrt 3$$ at any point, or if it ever falls below $$F(0).$$ The first is easily answered in the negative (set $$F(x)=\\sqrt 3$$ to deduce a contradiction), so the range is half-open. So does the function ever go below $$e^2-4e+3$$? In particular we may ask whether the function ever attains the minimum possible here, namely $$0.$$ Thus, setting $$F(x)=0,$$ we see that we want to see if there are real solutions to the quadratic equation $$p^2-4p+3=0,$$ where $$p=e^{-x^2+1}.$$ This has solutions. In particular, we get that $$e^{-x^2+1}=1,$$ or in other words that $$-x^2+1=0.$$ Thus $$F(x)$$ vanishes in at least two points.\n\nIt therefore is the case that the range is $$[0,\\sqrt 3).$$\n\nNote that the range of $$f$$ on $$(-\\infty,-1]\\cup[1,\\infty)$$ is $$(0,1]$$. The range of $$g$$ on $$(0,1]$$ is $$[0,\\sqrt{3})$$.\n\nLet $$f,g$$ be given by $$\\begin{cases} f(x)=e^{1-x^2}\\\\[4pt] g(x)=\\sqrt{x^2-4x+3}\\\\ \\end{cases}$$ $$\\bullet\\;\\,$$Part $$(\\mathbf{a}){\\,:}\\;\\,$$Find the domain and range of $$f\\circ g$$.\n\nSince the domain of $$f$$ is $$\\mathbb{R}$$, the domain of $$f\\circ g$$ is the same as the domain of $$g$$.\n\nHence the domain of $$f\\circ g$$ is $$(-\\infty,1]\\cup [3,\\infty)$$.\n\nSince $$f$$ is an even function, the range of $$f$$ is the same as the range of $$f$$ on the restricted domain $$[0,\\infty)$$.\n\nOn the interval $$[0,\\infty)$$, $$f$$ is strictly decreasing, and $$f$$ approaches zero from above as $$x$$ approaches infinity.\n\nSince $$f$$ is continuous, it follows that the range of $$f$$ is $$(0,e]$$.\n\nFor $$x\\ge 3$$, $$g$$ realizes all values in $$[0,\\infty)$$, so the range of $$f\\circ g$$ is the same as the range of $$f$$.\n\nHence the range of $$f\\circ g$$ is $$(0,e]$$.\n\n$$\\bullet\\;\\,$$Part $$(\\mathbf{b}){\\,:}\\;\\,$$Find the domain and range of $$g\\circ f$$.\n\nThe domain of $$g\\circ f$$ is the set of all real $$x$$ such that $$f(x)$$ is in the domain of $$g$$, which is the set of all real $$x$$ such that $$f(x)\\le 1$$ or $$f(x)\\ge 3$$.\n\nBut we can't have $$f(x)\\ge 3$$, since the maximum value of $$f$$ is $$e$$.\n\nFor the condition $$f(x)\\le 1$$, we get $$f(x)\\le1\\iff e^{1-x^2}\\le 1\\iff 1-x^2\\le 0\\iff |x| \\ge 1$$ hence the domain of $$g\\circ f$$ is $$(-\\infty,-1]\\cup [1,\\infty)$$.\n\nRestricted to the domain $$(-\\infty,-1]\\cup [1,\\infty)$$, the range of $$f$$ is $$(0,1]$$, hence the range of $$g\\circ f$$ is the range of $$g$$ on the restricted domain $$(0,1]$$.\n\nIt follows that the range of $$g\\circ f$$ is $$[0,\\sqrt{3})$$.\n\n$$\\bullet\\;\\,$$To summarize the results:\n\n\u2022 The domain of $$f\\circ g$$ is:$$\\;\\,(-\\infty,1]\\cup [3,\\infty)$$.\n\u2022 The range of $$f\\circ g$$ is:$$\\;\\,(0,e]$$.$$\\\\[6pt]$$\n\u2022 The domain of $$g\\circ f$$ is:$$\\;\\,(-\\infty,-1]\\cup [1,\\infty)$$.\n\u2022 The range of $$g\\circ f$$ is:$$\\;\\,[0,\\sqrt{3})$$.", "date": "2020-02-25 01:36:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3352629/finding-the-domain-and-range-of-a-composite-function", "openwebmath_score": 0.9344687461853027, "openwebmath_perplexity": 92.38849530302848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105266327962, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8566898166739935}}
{"url": "https://nhigham.com/2021/03/09/eigenvalue-inequalities-for-hermitian-matrices/?replytocom=25185", "text": "Eigenvalue Inequalities for Hermitian\u00a0Matrices\n\nThe eigenvalues of Hermitian matrices satisfy a wide variety of inequalities. We present some of the most useful and explain their implications. Proofs are omitted, but as Parlett (1998) notes, the proofs of the Courant\u2013Fischer, Weyl, and Cauchy results are all consequences of the elementary fact that if the sum of the dimensions of two subspaces of $\\mathbb{C}^n$ exceeds $n$ then the subspaces have a nontrivial intersection.\n\nThe eigenvalues of a Hermitian matrix $A\\in\\mathbb{C}^{n\\times n}$ are real and we order them $\\lambda_n\\le \\lambda_{n-1} \\le \\cdots \\le \\lambda_1$. Note that in some references, such as Horn and Johnson (2013), the reverse ordering is used, with $\\lambda_n$ the largest eigenvalue. When it is necessary to specify what matrix $\\lambda_k$ is an eigenvalue of we write $\\lambda_k(A)$: the $k$th largest eigenvalue of $A$. All the following results also hold for symmetric matrices over $\\mathbb{R}^{n\\times n}$.\n\nThe function $f(x) = x^*Ax/x^*x$ is the quadratic form $x^*Ax$ for $A$ evaluated on the unit sphere, since $f(x) = f(x/\\|x\\|_2)$. As $A$ is Hermitian it has a spectral decomposition $A = Q\\Lambda Q^*$, where $Q$ is unitary and $\\Lambda = \\mathrm{diag}(\\lambda_i)$. Then\n\n$f(x) = \\displaystyle\\frac{x^*Q\\Lambda Q^*x}{x^*x} = \\displaystyle\\frac{y^*\\Lambda y}{y^*y} = \\displaystyle\\frac{\\sum_{i=1}^{n}\\lambda_i y_i^2} {\\sum_{i=1}^{n}y_i^2} \\quad (y = Q^*x),$\n\nfrom which is it clear that\n\n$\\notag \\lambda_n = \\displaystyle\\min_{x\\ne0} \\displaystyle\\frac{x^*Ax}{x^*x}, \\quad \\lambda_1 = \\displaystyle\\max_{x\\ne0} \\displaystyle\\frac{x^*Ax}{x^*x}, \\qquad(*)$\n\nwith equality when $x$ is an eigenvector corresponding to $\\lambda_n$ and $\\lambda_1$, respectively, This characterization of the extremal eigenvalues of $A$ as the extrema of $f$ is due to Lord Rayleigh (John William Strutt), and $f(x)$ is called a Rayleigh quotient. The intermediate eigenvalues correspond to saddle points of $f$.\n\nCourant\u2013Fischer Theorem\n\nThe Courant\u2013Fischer theorem (1905) states that every eigenvalue of a Hermitian matrix $A\\in\\mathbb{C}^{n\\times n}$ is the solution of both a min-max problem and a max-min problem over suitable subspaces of $\\mathbb{C}^n$.\n\nTheorem (Courant\u2013Fischer).\n\nFor a Hermitian $A\\in\\mathbb{C}^{n\\times n}$,\n\n\\notag \\begin{aligned} \\lambda_k &= \\min_{\\dim(S)=n-k+1} \\, \\max_{0\\ne x\\in S} \\frac{x^*Ax}{x^*x}\\\\ &= \\max_{\\dim(S)= k} \\, \\min_{0\\ne x\\in S} \\frac{x^*Ax}{x^*x}, \\quad k=1\\colon n. \\end{aligned}\n\nNote that the equalities $(*)$ are special cases of these characterizations.\n\nIn general there is no useful formula for the eigenvalues of a sum $A+B$ of Hermitian matrices. However, the Courant\u2013Fischer theorem yields the upper and lower bounds\n\n$\\notag \\lambda_k(A) + \\lambda_n(B) \\le \\lambda_k(A+B) \\le \\lambda_k(A) + \\lambda_1(B), \\qquad (1)$\n\nfrom which it follows that\n\n$\\notag \\max_k|\\lambda_k(A+B)-\\lambda_k(A)| \\le \\max(|\\lambda_n(B)|,|\\lambda_1(B)|) = \\|B\\|_2.$\n\nThis inequality shows that the eigenvalues of a Hermitian matrix are well conditioned under perturbation. We can rewrite the inequality in the symmetric form\n\n$\\notag \\max_k |\\lambda_k(A)-\\lambda_k(B)| \\le \\|A-B\\|_2.$\n\nIf $B$ is positive semidefinite then (1) gives\n\n$\\notag \\lambda_k(A) \\le \\lambda_k(A + B), \\quad k = 1\\colon n, \\qquad (2)$\n\nwhile if $B$ is positive definite then strict inequality holds for all $i$. These bounds are known as the Weyl monotonicity theorem.\n\nWeyl\u2019s Inequalities\n\nWeyl\u2019s inequalities (1912) bound the eigenvalues of $A+B$ in terms of those of $A$ and $B$.\n\nTheorem (Weyl).\n\nFor Hermitian $A,B\\in\\mathbb{C}^{n\\times n}$ and $i,j = 1\\colon n$,\n\n\\notag \\begin{aligned} \\lambda_{i+j-1}(A+B) &\\le \\lambda_i(A) + \\lambda_j(B), \\quad i+j \\le n+1, \\qquad (3)\\\\ \\lambda_i(A) + \\lambda_j(B) &\\le \\lambda_{i+j-n}(A+B). \\quad i+j \\ge n+1, \\qquad (4) \\end{aligned}\n\nThe Weyl inequalities yield much information about the effect of low rank perturbations. Consider a positive semidefinite rank-$1$ perturbation $B = zz^*$. Inequality (3) with $j = 1$ gives\n\n$\\notag \\lambda_i(A+B) \\le \\lambda_i(A) + z^*z, \\quad i = 1\\colon n$\n\n(which also follows from (1)). Inequality (3) with $j = 2$, combined with (2), gives\n\n$\\notag \\lambda_{i+1}(A) \\le \\lambda_{i+1}(A + zz^*) \\le \\lambda_i(A), \\quad i = 1\\colon n-1. \\qquad (5)$\n\nThese inequalities confine each eigenvalue of $A + zz^*$ to the interval between two adjacent eigenvalues of $A$; the eigenvalues of $A + zz^*$ are said to interlace those of $A$. The following figure illustrates the case $n = 4$, showing a possible configuration of the eigenvalues $\\lambda_i$ of $A$ and $\\mu_i$ of $A + zz^*$.\n\nA specific example, in MATLAB, is\n\n>> n = 4; eig_orig = 5:5+n-1\n>> D = diag(eig_orig); eig_pert = eig(D + ones(n))'\neig_orig =\n5     6     7     8\neig_pert =\n5.2961e+00   6.3923e+00   7.5077e+00   1.0804e+01\n\nSince $\\mathrm{trace}(A + zz^*) = \\mathrm{trace}(A) + z^*z$ and the trace is the sum of the eigenvalues, we can write\n\n$\\notag \\lambda_i(A + zz^*) = \\lambda_i(A) + \\theta_i z^*z,$\n\nwhere the $\\theta_i$ are nonnegative and sum to $1$. If we greatly increase $z^*z$, the norm of the perturbation, then most of the increase in the eigenvalues is concentrated in the largest, since (5) bounds how much the smaller eigenvalues can change:\n\n>> eig_pert = eig(D + 100*ones(n))'\neig_pert =\n5.3810e+00   6.4989e+00   7.6170e+00   4.0650e+02\n\nMore generally, if $B$ has $p$ positive eigenvalues and $q$ negative eigenvalues then (3) with $j = p+1$ gives\n\n$\\notag \\lambda_{i+p}(A+B) \\le \\lambda_i(A), \\quad i = 1\\colon n-p,$\n\nwhile (4) with $j = n-q$ gives\n\n$\\notag \\lambda_i(A) \\le \\lambda_{i-q}(A + B), \\quad i = q+1\\colon n.$\n\nSo the inertia of $B$ (the number of negative, zero, and positive eigenvalues) determines how far the eigenvalues can move as measured relative to the indexes of the eigenvalues of $A$.\n\nAn important implication of the last two inequalities is for the case $A = I$, for which we have\n\n\\notag \\begin{aligned} \\lambda_{i+p}(I+B) &\\le 1, \\quad i = 1 \\colon n-p, \\\\ \\lambda_{i-q}(I+B) &\\ge 1, \\quad i = q+1 \\colon n. \\end{aligned}\n\nExactly $p+q$ eigenvalues appear in one of these inequalities and $n-(p+q)$ appear in both. Therefore $n - (p+q)$ of the eigenvalues are equal to $1$ and so only $\\mathrm{rank}(B) = p+q$ eigenvalues can differ from $1$. So perturbing the identity matrix by a Hermitian matrix of rank $r$ changes at most $r$ of the eigenvalues. (In fact, it changes exactly $r$ eigenvalues, as can be seen from a spectral decomposition.)\n\nFinally, if $B$ has rank $r$ then $\\lambda_{r+1}(B) \\le 0$ and $\\lambda_{n-r}(B) \\ge 0$ and so taking $j = r+1$ in (3) and $j = n-r$ in (4) gives\n\n\\notag \\begin{aligned} \\lambda_{i+r}(A+B) &\\le \\lambda_i(A), ~~\\qquad\\qquad i = 1\\colon n-r, \\\\ \\lambda_i(A) &\\le \\lambda_{i-r}(A + B), ~~\\quad i = r+1\\colon n. \\end{aligned}\n\nCauchy Interlace Theorem\n\nThe Cauchy interlace theorem relates the eigenvalues of successive leading principal submatrices of a Hermitian matrix. We denote the leading principal submatrix of $A$ of order $k$ by $A_k = A(1\\colon k, 1\\colon k)$.\n\nTheorem (Cauchy).\n\nFor a Hermitian $A\\in\\mathbb{C}^{n\\times n}$,\n\n$\\notag \\lambda_{i+1}(A_{k+1}) \\le \\lambda_i(A_k) \\le \\lambda_i(A_{k+1}), \\quad i = 1\\colon k, \\quad k=1\\colon n-1.$\n\nThe theorem says that the eigenvalues of $A_k$ interlace those of $A_{k+1}$ for all $k$. Two immediate implications are that (a) if $A$ is Hermitian positive definite then so are all its leading principal submatrices and (b) appending a row and a column to a Hermitian matrix does not decrease the largest eigenvalue or increase the smallest eigenvalue.\n\nSince eigenvalues are unchanged under symmetric permutations of the matrix, the theorem can be reformulated to say that the eigenvalues of any principal submatrix of order $n-1$ interlace those of $A$. A generalization to principal submatrices of order $n-\\ell$ is given in the next result.\n\nTheorem.\n\nIf $B$ is a principal submatrix of order $n-\\ell$ of a Hermitian $A\\in\\mathbb{C}^{n\\times n}$ then\n\n$\\notag \\lambda_{i+\\ell}(A) \\le \\lambda_i(B) \\le \\lambda_i(A), \\quad i=1\\colon n-\\ell.$\n\nMajorization Results\n\nIt follows by taking $x$ to be a unit vector $e_i$ in the formula $\\lambda_1 = \\max_{x\\ne0} x^*Ax/(x^*x)$ that $\\lambda_1 \\ge a_{ii}$ for all $i$. And of course the trace of $A$ is the sum of the eigenvalues: $\\sum_{i=1}^n a_{ii} = \\sum_{i=1}^n \\lambda_i$. These relations are the first and last in a sequence of inequalities relating sums of eigenvalues to sums of diagonal elements obtained by Schur in 1923.\n\nTheorem (Schur).\n\nFor a Hermitian $A\\in\\mathbb{C}^{n\\times n}$,\n\n$\\notag \\displaystyle\\sum_{i=1}^k \\lambda_i \\ge \\displaystyle\\sum_{i=1}^k \\widetilde{a}_{ii}, \\quad k=1\\colon n,$\n\nwhere $\\{\\widetilde{a}_{ii}\\}$ is the set of diagonal elements of $A$ arranged in decreasing order: $\\widetilde{a}_{11} \\ge \\cdots \\ge \\widetilde{a}_{nn}$.\n\nThese inequalities say that the vector $[\\lambda_1,\\dots,\\lambda_n]$ of eigenvalues majorizes the ordered vector $[\\widetilde{a}_{11},\\dots,\\widetilde{a}_{nn}]$ of diagonal elements.\n\nAn interesting special case is a correlation matrix, a symmetric positive semidefinite matrix with unit diagonal, for which the inequalities are\n\n$\\notag \\lambda_1 \\ge 1, \\quad \\lambda_1+ \\lambda_2\\ge 2, \\quad \\dots, \\quad \\lambda_1+ \\lambda_2 + \\cdots + \\lambda_{n-1} \\ge n-1,$\n\nand $\\lambda_1+ \\lambda_2 + \\cdots + \\lambda_n = n$. Here is an illustration in MATLAB.\n\n>> n = 5; rng(1); A = gallery('randcorr',n);\n>> e = sort(eig(A)','descend'), partial_sums = cumsum(e)\ne =\n2.2701e+00   1.3142e+00   9.5280e-01   4.6250e-01   3.6045e-04\npartial_sums =\n2.2701e+00   3.5843e+00   4.5371e+00   4.9996e+00   5.0000e+00\n\nKy Fan (1949) proved a majorization relation between the eigenvalues of $A$, $B$, and $A+B$:\n\n$\\notag \\displaystyle\\sum_{i=1}^k \\lambda_i(A+B) \\le \\displaystyle\\sum_{i=1}^k \\lambda_i(A) + \\displaystyle\\sum_{i=1}^k \\lambda_i(B), \\quad k = 1\\colon n.$\n\nFor $k = 1$, the inequality is the same as the upper bound of (1), and for $k = n$ it is an equality: $\\mathrm{trace}(A+B) = \\mathrm{trace}(A) + \\mathrm{trace}(B)$.\n\nOstrowski\u2019s Theorem\n\nFor a Hermitian $A$ and a nonsingular $X$, the transformation $A\\to X^*AX$ is a congruence transformation. Sylvester\u2019s law of inertia says that congruence transformations preserve the inertia. A result of Ostrowski (1959) goes further by providing bounds on the ratios of the eigenvalues of the original and transformed matrices.\n\nTheorem (Ostrowski).\n\nFor a Hermitian $A\\in \\mathbb{C}^{n\\times n}$ and $X\\in\\mathbb{C}^{n\\times n}$,\n\n$\\lambda_k(X^*AX) = \\theta_k \\lambda_k(A), \\quad k=1\\colon n,$\n\nwhere $\\lambda_n(X^*X) \\le \\theta_k \\le \\lambda_1(X^*X)$.\n\nIf $X$ is unitary then $X^*X = I$ and so Ostrowski\u2019s theorem reduces to the fact that a congruence with a unitary matrix is a similarity transformation and so preserves eigenvalues. The theorem shows that the further $X$ is from being unitary the greater the potential change in the eigenvalues.\n\nOstrowski\u2019s theorem can be generalized to the situation where $X$ is rectangular (Higham and Cheng, 1998).\n\nInterrelations\n\nThe results we have described are strongly interrelated. For example, the Courant\u2013Fischer theorem and the Cauchy interlacing theorem can be derived from each other, and Ostrowski\u2019s theorem can be proved using the Courant\u2013Fischer Theorem.\n\n2 thoughts on \u201cEigenvalue Inequalities for Hermitian\u00a0Matrices\u201d\n\n1. Daniel Kressner says:\n\nThis is a very nice overview; thanks! Tangential to what Parlett is stating, most if not even all of these results can be concluded from the Eckart\u2013Young\u2013Mirsky theorem in the spectral norm.", "date": "2021-05-05 22:14:11", "meta": {"domain": "nhigham.com", "url": "https://nhigham.com/2021/03/09/eigenvalue-inequalities-for-hermitian-matrices/?replytocom=25185", "openwebmath_score": 0.9866310358047485, "openwebmath_perplexity": 324.5997061624111, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9943580928375232, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8566774873014341}}
{"url": "https://math.stackexchange.com/questions/1964876/the-comparison-test-for-series", "text": "# The Comparison Test for Series\n\nI'm doing a homework problem - it asks us to show whether the series: $$\\sum_{n=1}^{\\infty}\\frac{n^n}{n!}$$ converges or diverges. I looked at a graph of the sequence component $\\big(\\frac{n^n}{n!} \\big)$ and saw it continued to increase. I then considered the sequence: $$\\sum_{n=1}^{\\infty}\\frac{1}{n}$$ which diverges by the P test.\n\nBut, $$\\frac{1}{n}\\leq \\frac{n^n}{n!},~ \\forall~n\\geq1$$ which would then mean that the first series I showed diverges by the comparison test.\n\nMy problem is that this seems too simple? Can I compare one series to any series or does the comparison series have to meet some certain requirements (besides those I've addressed).\n\nCheers.\n\n\u2022 Looks fine to me. \u2013\u00a0Jacky Chong Oct 12 '16 at 6:15\n\u2022 You could also do comparison test with the divergent series $\\sum_{n=1}^\\infty 1$. \u2013\u00a0angryavian Oct 12 '16 at 6:16\n\u2022 That seems so simple, so I do not need to compare a series with changing exponents and factorials to a similar one? \u2013\u00a0Wharf Rat Oct 12 '16 at 6:17\n\u2022 Is the necessary condition for convergence ($n^n/n!\\to 0$) satisfied here? \u2013\u00a0A.\u0393. Oct 12 '16 at 6:20\n\nYour reasoning is fine. In fact, this can be shown to diverge with the simplest / most intuitive test there is: If $\\ \\displaystyle \\sum_{n=1}^\\infty a_n$ is a convergent series, it is a necessary condition that $\\displaystyle \\lim_{n \\rightarrow \\infty} a_n = 0$. In this scenario, $n^n \\geq n!$ for all $n \\geq 1$, implying $\\displaystyle \\frac{n^n}{n!} \\geq 1$ for all such $n$, so this necessary condition does not hold.\n\nI looked at a graph of the sequence component $\\big(\\frac{n^n}{n!} \\big)$ and saw it continued to increase.\nIf there series converges then $n^n/n! \\to 0$, but if you use Stirling's approximation: $n! \\approx \\sqrt{2\\pi n} \\cdot \\left(\\frac{n}{e}\\right)^n$\n$$\\lim_{n} \\frac{n^n}{n!} = \\lim_n \\frac{n^n}{\\sqrt{2 \\pi n} \\cdot\\dfrac{n^n}{e^n}} = \\lim_n \\frac{e^n}{\\sqrt{2 \\pi n}} = \\frac{1}{\\sqrt{2\\pi}} \\lim_n \\frac{e^n}{\\sqrt{n}} = \\frac{1}{\\sqrt{2\\pi }} \\lim_n \\left(\\frac{e^{2n}}{n}\\right)^{\\frac{1}{2}} \\to \\infty$$\n\u2022 No need for Stirling here, we have $$\\frac{n^n}{n!}=\\frac{n\\cdot n\\cdots n}{1\\cdot2\\cdots(n-1)\\cdot n}>\\frac{n\\cdot 2\\cdots n}{1\\cdot2\\cdots(n-1)\\cdot n}=n$$ \u2013\u00a0AD. Oct 12 '16 at 8:04", "date": "2019-07-19 08:24:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1964876/the-comparison-test-for-series", "openwebmath_score": 0.915158212184906, "openwebmath_perplexity": 237.42679524359943, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347853343059, "lm_q2_score": 0.8791467754256017, "lm_q1q2_score": 0.8566711993891934}}
{"url": "https://math.stackexchange.com/questions/640773/integrating-two-equations-that-equal-what-happens-to-the-constant-on-one-of-the", "text": "# Integrating two equations that equal, what happens to the constant on one of the sides?\n\nIn class, we were talking about Newton's 3rd law and how to integrate.\n\n$\\int(g)dt = \\int(y''(t))dt \\implies g(t) + C = y'(t)$\n\nI am confused about why the right hand side of the equation doesn't get a constant. After asking the professor, he said that it was because the two constants would cancel each other out. But I still don't understand why that should prevent us from writing a constant on the right side.\n\n\u2022 Recall, you can add constants together into a single constant (as was done with $C$). Also, you can define a $C_1$ and $C_2$ - one to each side. You can also show as being on the RHS of the equation. All will produce the same result. The choice is typically one of convenience to make solving easiest. \u2013\u00a0Amzoti Jan 16 '14 at 18:13\n\u2022 As an example, take $y' = x y$. Solve it with a constant on one side after separation and integration. Then, have a constant on each side. What do you notice when you solve for $y$ in both approaches? $y = c e^{x^2/2}$. \u2013\u00a0Amzoti Jan 16 '14 at 18:21\n\u2022 I think I understand, thank you! \u2013\u00a0user121860 Jan 16 '14 at 18:29\n\u2022 I haven't taken calculus any calculus courses for about two years now and I've forgotten some of the rules, such as combining the constants. The professor told me that when we integrated both constants would be C1, ie. the same, and as such would end up canceling each other out by way of subtraction if we wanted to integrate more. So he wrote C1 on the left hand side, but it seems to make sense if he actually wanted to write C (the inclusion of both constants). \u2013\u00a0user121860 Jan 16 '14 at 18:33\n\u2022 Though if you don't mind, examples are always helpful. Thank you for your time and help! \u2013\u00a0user121860 Jan 16 '14 at 18:33\n\nRecall, you can add constants together into a single constant (as was done with C).\n\nAlso, you can define a $C_1$ and $C_2$ - one to each side. You can also show the constant as being on the RHS of the equation. All will produce the same result. The choice is typically one of convenience to make solving easiest.\n\nExample: Consider the separable equation:\n\n$$y' = x y$$\n\nAfter separation, we can integrate both sides as:\n\n$$\\int \\dfrac{1}{y}~dy = \\int x~dx$$\n\nApproach 1: Single constant (we could have also put $C$ on the LHS - try this)\n\n$$\\ln y = \\dfrac{x^2}{2} + C$$\n\nWe take the exponential of each side and have:\n\n$$y = e^{x^2/2 + C} = e^{x^2/2}~e^C = w~e^{x^2/2}$$\n\nNote: $w = e^C$, which is just some constant (totally arbitrary).\n\nApproach 2: Constant on each side\n\n$$\\ln y + c_1 = \\dfrac{x^2}{2} + c_2$$\n\nTaking exponential of both sides:\n\n$$e^{\\ln y + c_1} = e^{x^2/2 + c_2}$$\n\nThe RHS is as above and the LHS, we have ($q$ is just any arbitrary constant):\n\n$$e^{\\ln y}~e^{c_1} = y~q$$\n\nNow we write:\n\n$$q y = c_2~e^{x^2/2}$$\n\nDividing constants and just calling it $w$, we have:\n\n$$y(x) = w~e^{x^2/2}$$\n\nIn both approaches, we just get some constant $w$.\n\nNow, if you were provided with an initial condition like $y(0) = 2$, you would plug in $x=0$ and see that $w = 2$.", "date": "2019-06-26 05:49:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/640773/integrating-two-equations-that-equal-what-happens-to-the-constant-on-one-of-the", "openwebmath_score": 0.9255096912384033, "openwebmath_perplexity": 151.61560638950004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434788304004, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8566711912008189}}
{"url": "https://math.stackexchange.com/questions/1698179/maximise-population-coverage-subject-to-budget-constraint", "text": "# Maximise population coverage subject to budget constraint\n\nLet $t_i =$\n\n$1$ if transmitter i is to be constructed\n\nand $0$ otherwise,\n\n$c_j =$\n\n$1$ if community j is covered\n\nand $0$ otherwise.\n\nObj func:\n\nMax\n\n$$z = [10, 15, ..., 10] \\cdot c$$\n\ns.t.\n\n1. the budget constraint\n\n$$[3.6, 2.3, ..., 3.10] \\cdot t \\le 15$$\n\n1. If community $j$ is covered, it is done so by at least one constructed transmitter $i$:\n\neg if $c_1$ is covered, then $t_1$ or $t_3$ is constructed:\n\n$$c_1 \\to (t_1 \\bigvee t_3)$$\n\n$$\\iff \\neg c_1 \\bigvee (t_1 \\bigvee t_3)$$\n\n$$\\iff 1 - c_1 + t_1 + t_3 \\ge 1$$\n\n$$\\iff c_1 \\le t_1 + t_3$$\n\nSimilarly, we have:\n\n$$c_2 \\le t_1 + t_2$$\n\n$$\\vdots$$\n\n$$c_{15} \\le t_7$$\n\n1. If a transmitter $i$ is constructed, at least one community $j$ is covered:\n\neg if $t_1$ is constructed, then $c_1$ and $c_2$ are covered:\n\n$$t_1 \\to (c_1 \\bigwedge c_2)$$\n\n$$\\iff \\neg t_1 \\bigvee (c_1 \\bigwedge c_2)$$\n\n$$\\iff (\\neg t_1 \\bigvee c_1) \\bigwedge (\\neg t_1 \\bigvee c_2)$$\n\n$$\\iff 1 - t_1 + c_1 \\ge 1 \\ \\text{and} \\ 1 - t_1 + c_2 \\ge 1$$\n\n$$\\iff c_1 \\ge t_1 \\ \\text{and} \\ c_2 \\ge t_1$$\n\nSimilarly, we have:\n\n$$c_2, c_3, c_5 \\ge t_2$$\n\n$$c_1, c_7, c_9, c_{10} \\ge t_3$$\n\n$$\\vdots$$\n\n$$c_{12}, c_{13}, c_{14}, c_{15} \\ge t_7$$\n\nIs that right?\n\nFrom Chapter 3 here.\n\n\u2022 Note if you choose transmitters $1, 2$, then there is overlap of communities, and your function will double count the population of community $2$. \u2013\u00a0Macavity Mar 15 '16 at 5:05\n\u2022 @Macavity Oh thanks. What to do then? $z'=z-y \\cdot x$ where $y_i$ corrects for double or triple counting? \u2013\u00a0BCLC Mar 15 '16 at 5:10\n\u2022 @Macavity I think Kuifje found away around the double count. Do you agree? \u2013\u00a0BCLC Mar 17 '16 at 17:54\n\nHere is a way to get the constraints right. Define $x_i$ exactly as you have done. Define $p_i$ to be an indicator $1/0$ depending on whether community $i$ has been covered or not.\n\nThe budget constraint remains as you have set, and the objective function is now of form $\\max z = 10p_1+15p_2 + \\cdots + 10p_{15}$\n\nNow how do we ensure that $p_i$ is set correctly, for any given choice of $\\{x_i\\}$? One way is to note that the objective function gives positive weight to $p_i$, so define a constraint for each population based on transmitters which could cover it - e.g. for the first one : $p_1 \\le x_1+x_3$, for the second $p_2 \\le x_1 + x_2$ etc. This will force the population indicator to turn $0$ if none of the relevant transmitters are selected.\n\n\u2022 Thanks Macavity. Edited. How is it? \u2013\u00a0BCLC Mar 15 '16 at 6:08\n\u2022 Your $p_i$ needs to be replaced by $c_i$ in the objective function and $x_i$ has no role to play in the objective. \u2013\u00a0Macavity Mar 15 '16 at 6:10\n\u2022 Thanks. We answered in class, and we sort of got different answers. Prof is still going to think about it. How do you find sol? \u2013\u00a0BCLC Mar 15 '16 at 11:31\n\u2022 Using an $M$ is also a good way, though not needed here. Try out the Integer Programs and see how it works. \u2013\u00a0Macavity Mar 15 '16 at 11:43\n\u2022 @BCLC Yes, your try 4 seems what I meant, and it works. \u2013\u00a0Macavity Mar 20 '16 at 18:31\n\nThis is a classical set covering problem.\n\nLet $y_i$ be a binary variable that equals $1$ if and only if transmitter $i=1,\\cdots,7$ is built, and $x_j$ another binary that equals $1$ if community $j$ is covered. Let $p_j$ be the population of community $j=1,\\cdots,15$.\n\nYou want to maximize the total coverage: $$\\mbox{Maximize }Z=\\sum_{j=1 }^{15}p_jx_j$$ subject to budget constraints: $$\\sum_{i=1}^7c_iy_i\\le 15 \\\\$$ To link the variables, proceed like Macavity suggests: $$x_j\\le \\sum_{i\\;|\\;i \\;covers\\; j}y_i\\quad \\forall j=1,\\cdots,15\\\\ y_i,x_j\\in \\{0,1\\}\\quad \\forall i=1,\\cdots,7\\;\\forall j=1,\\cdots,15$$\n\n\u2022 Sorry I was editing while you asked the question. Variables $x_j$ will make sure there is no double counting, and are activated as soon as one of the variables $y_i\\; |\\;i$ covers $j$ is. \u2013\u00a0Kuifje Mar 17 '16 at 18:03\n\u2022 No apologies needed. I'm the one asking a favour. It's really 'Try 2' and not 'Try 1' ? My prof seems to disagree with Macavity \u2013\u00a0BCLC Mar 17 '16 at 18:14\n\u2022 What do you mean by 'Try 1' and 'Try 2'? I think Macavity is right. He basically explained the nature of my second constraint. What specific point does your professor disagree with? \u2013\u00a0Kuifje Mar 17 '16 at 18:25\n\u2022 I disagree with \"Try1\". Try 1 (first equation) implies that if communities $1$ and $2$ are covered, than transmitter $1$ is built. This is not necessarily true, as they can be covered by transmitters $3$ and $2$ for example. It is the other way around: if community $1$ is covered, than at least one transmitter among those that cover the community must be built. \u2013\u00a0Kuifje Mar 18 '16 at 0:34\n\u2022 yes indeed, for the same reasons. \u2013\u00a0Kuifje Mar 18 '16 at 14:30", "date": "2019-08-20 20:32:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1698179/maximise-population-coverage-subject-to-budget-constraint", "openwebmath_score": 0.5166764855384827, "openwebmath_perplexity": 737.8704632289006, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434783107032, "lm_q2_score": 0.879146761176671, "lm_q1q2_score": 0.8566711835464391}}
{"url": "https://math.stackexchange.com/questions/462199/why-does-factoring-eliminate-a-hole-in-the-limit", "text": "# Why does factoring eliminate a hole in the limit?\n\n$$\\lim _{x\\rightarrow 5}\\frac{x^2-25}{x-5} = \\lim_{x\\rightarrow 5} (x+5)$$\n\nI understand that to evaluate a limit that has a zero (\"hole\") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the $x^2-25$ into $(x-5)(x+5)$, but I don't just want to understand the methodology which my teacher tells me to \"just memorize\", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the $(x-5)$, which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is \"just how we do it\" and that I should \"practice to just memorize the procedure.\"\nI really want to understand this in abstract algebra terms, please elaborate. Thank you very much.\n\n\u2022 Very laudable that you want to understand instead of \"just memorize\". Aug 7, 2013 at 19:21\n\u2022 The expression $\\frac{(x+5)(x-5)}{x-5}=x+5$, just so long as $x\\neq 5$, since $\\frac{(x+5)(x-5)}{x-5}$ has the bad fortune of not being defined at $x=5$. Aug 7, 2013 at 19:29\n\u2022 I don't understand what your question is. Are you asking why the function $\\frac{x^2-25}{x-5}$ is undefined at $x=5$? or why we're allowed to cancel $(x-5)$ from the top and bottom? or why the limits before and after cancellation are equal? Aug 7, 2013 at 20:09\n\u2022 @BlueRaja-DannyPflughoeft yes, all of those questions I'm asking :) Aug 7, 2013 at 22:35\n\u2022 To answer the question - I've only been told the methodology of expanding the x2\u221225 into (x\u22125)(x+5). It goes like this: (in reverse) (x-5)(x+5) = x(x+5)-5(x+5) = x^2+5x-5x-25 = x^2-25 So, working above the other way takes x^2-25 back to (x-5)(x+5) Aug 8, 2013 at 22:52\n\nFirst, and by definition, when dealing with\n\n$$\\lim_{x\\to x_0}f(x)$$\n\nwe must assume $\\,f\\,$ is defined in some neighborhood of $\\,x_0\\,$ except , perhaps, on $\\,x_0\\,$ itself, and from here that in the process of taking the limit we have the right and the duty to assume $\\,x\\,$ approaches $\\,x_0\\,$ in any possible way but it is never equal to it.\n\nThus, and since in our case we always have $\\,x\\ne x_0=5\\,$ during the limit process , we can algebraically cancel for the whole process.\n\n$$\\frac{x^2-25}{x-5}=\\frac{(x+5)\\color{red}{(x-5)}}{\\color{red}{x-5}}=x+5\\xrightarrow[x\\to 5]{}10$$\n\nThe above process shows that the original function behaves exactly as the straight line $\\,y=x+5\\,$ except at the point $\\,x=5\\,$ , where there exists \"a hole\", as you mention.\n\n\u2022 said hole can, of course be filled in by plugging it with the limit ... Aug 13, 2013 at 22:33\n\u2022 Some definitions of the limit do not require that we consider points that are not equal to $x_0$. How do you handle these cases then? The definition goes as $\\lim f(x) = L$ provided for each $\\epsilon\\gt 0$ there is some $\\delta \\gt 0$ such that $\\vert x - x_0\\vert \\lt \\delta\\ldots$ and so on. Sep 9, 2016 at 15:50\n\u2022 @AlexOrtiz The epsilon delta definition of the limit for general metric spaces has $0 < d(x,c) < \\delta$. Note the first inequality, which implies $x \\neq c$. Nov 21, 2020 at 0:19\n\nThis image of mine seems apropos:\n\nIn the case of $\\lim_{x\\to 5} \\frac{x^2-25}{x-5}$, the message here is: Away from $x=5$, the function $\\frac{x^2-25}{x-5}$ is completely identical to $x+5$; thus, what we expect to find as we approach $x=5$ is the value $5+5$. This anticipated value is what a limit computes.\n\nThe fact that the original function isn't defined at $x=5$ is immaterial. Walley World may be closed for repairs when you arrive, but that doesn't mean you and your dysfunctional family didn't spend an entire cross-country road trip anticipating all the fun you'd have there.\n\n\u2022 Extremely helpful answer as always, Blue. You're incredible at understanding what people are really asking and providing strong intuition. Thank you for your contribution to this site! Jun 1, 2015 at 15:55\n\u2022 @Blue Your answer made me think something: Can we always find another function which - away from a certain value - is completely identical to the former function? I mean, what is done in this example is the following: We have an algebraic expression and we can find another algebraic expression that have that property. Jul 6, 2017 at 1:21\n\u2022 BTW: I read your image as the jingles in this video. Jul 6, 2017 at 1:21\n\u2022 Super answer...the craziest I ever read in here Jul 13, 2019 at 22:17\n\u2022 @Blue: Shameless plug: A downloadable poster version of the \"Journey/Destination\" image is available at my Etsy shop, and a t-shirt is available from Spring. (I'll delete this comment if there are objections.)\n\u2013\u00a0Blue\nOct 12, 2021 at 10:00\n\nLet's consider a simpler example first. Consider the function $f(x) = \\frac{2x}x$. This says you take some number $x$, multiply by 2, then divide by the original number $x$. Obviously the answer is always 2, right? Except that when $x$ is zero, the division is forbidden and there is no answer at all. But for every $x$ except 0, we have $\\frac{2x}x = 2$. In particular, for values of $x$ close to, but not equal to 0, we have $\\frac{2x}x = 2$.\n\nThe function $\\frac{x^2-25}{x-5}$ is similar, just a little more complicated. Calculating $x^2-25$ always gives you the same as $(x-5)(x+5)$. That is, if you take $x$, square it, and subtract 25, you always get the same number as if you take $x$, add 5 and subtract 5, and then multiply the two results. So we can replace $x^2-25$ with $(x+5)(x-5)$ because they always give the same number regardless of what you start with; they are two ways of getting to the same place. And then we see that $$\\frac{x^2-25}{x-5} = \\frac{(x+5)(x-5)}{x-5} = x+5$$\n\nexcept that if $x-5$ happens to be zero (that is, if $x=5$) the division by zero is forbidden and we get nothing at all. But for any other $x$ the result of $\\frac{x^2-25}{x-5}$ is always exactly equal to $x+5$. In particular, for values of $x$ close to, but not equal to 5, we have $\\frac{x^2-25}{x-5} = x+5$.\n\nThe limit $$\\lim_{x\\to 5} \\ldots$$ asks what happens to some function when $x$ close to, but not exactly equal to 5. And while this function is undefined for $x=5$, because to calculate it you would have to divide by zero, it is perfectly well-behaved for other values of $x$, and in particular for values of $x$ close to 5. For values of $x$ close to 5 it is equal to $x+5$, and so for values of $x$ close to 5 it is close to 10. And that is exactly what the limit is calculating.\n\nI think that what confuses you is the difference between \"solving the algebraic expression\", and \"finding the limit\". Given:\n\n$$f_1=\\frac{x^2-25}{x-5} \\quad f_2 = (x+5)$$\n\nThen, $f_1$ and $f_2$ are most definitely NOT the same function. This is because they have different domains: 5 is not a member of the domain of $f_1$, but it is in the domain of $f_2$.\n\nHowever, when we go from: $$\\lim _{x\\rightarrow 5}\\frac{x^2-25}{x-5} \\quad to \\quad \\lim _{x\\rightarrow 5}\\frac{(x-5)(x+5)}{x-5} \\quad to \\quad \\lim_{x\\rightarrow 5} (x+5)$$\n\nWe are not saying that the expressions inside the limits are equal; maybe they are, maybe they are not. What we are saying that they have the same limit. Totally different statement.\n\nAbove, the transformation of the second expression to the third one allows us to find a different function for which a) we know that the limit is the same, and b) we know how to trivially calculate that limit.\n\nThe big question, then: what transformations can I make to the function $f_1$ so that the limit stays the same? I think this is usually poorly explained in introductory courses -- a lot of hand-waving going on.\n\nObviously you can do any algebraic manipulation that leaves $f_1$ unchanged. You can also make any manipulation that removes and/or introduces discontinuities (points for which the function does not exist), as long as the new function stays continuous for an arbitrarily small neighborhood around $a$ (except possibly at $a$ itself). Your example is a case of such a transformation.\n\nHere I'm myself cheating because I'm not defining 'continuity' for you. I'm sorry; please use an intuitition of what continuous means (\"no holes, no jumps\"), until you are presented with a formal one.\n\nMore complex transformations exist, but they have to be justified individually. You'll get to them eventually.\n\n\u2022 you clarified a lot, thanks Aug 8, 2013 at 20:08\n\u2022 Sixth line from the end - \"You\" will be \"Your\".\n\u2013\u00a0MrAP\nJul 5, 2017 at 19:18\n\u2022 @MrAP, thanks. You should feel free to post an edit directly Jul 6, 2017 at 1:04\n\nHere's the basic idea: You're given a rational function, $f\\colon \\Bbb R \\setminus \\{5\\}\\to \\Bbb R$, which is continuous everywhere in its domain.\n\nYou want to find the limit of that function at $5$.\n\nOne way to do that is to construct a continuous extension of that function, $g\\colon \\Bbb R \\to \\Bbb R$, such that $g(x) = f(x)$ whenever $x$ is in the domain of $f$. Then $$\\lim_{x\\to 5}f(x) = \\lim_{x\\to 5} g(x) = g(5).$$\n\nIn this case, factoring and cancelling accomplishes that objective.\n\n\u2022 i really like this idea of a \"continuous extension\", what is the formal name of this rule/theorem? I'd like to read more about it. Thanks for the answer @dfeuer Aug 8, 2013 at 20:07\n\u2022 @user4150: Try this google search: \"removable discontinuity\" \"continuous extension\" Aug 8, 2013 at 21:24\n\u2022 @dfeuer Typo? It should be $$\\lim_{x\\to 5}f(x) = \\lim_{x\\to 5} g(x) = g(\\color{royalblue}5).$$ Oct 1, 2014 at 8:06\n\u2022 @Hakim, right you are. Fixed. Oct 1, 2014 at 9:55\n\nThe chill pill you are looking for is\n\nIf $f_1$ = $f_2$ except at $a$ then $\\lim _{x{\\rightarrow}a}f_1(x)=\\lim_{x{\\rightarrow}a}f_{2}(x)$\n\n\u2022 Yes! You made my day! Apr 14, 2016 at 16:42\n\u2022 Assuming both function to be continuous at $x=a$. May 26, 2017 at 11:45\n\nIntuitively, we can start the other way round, by simply considering $\\lim_{x\\rightarrow 5} (x+5)$ which we're all agreed we understand. Now consider, independently,\n\n$$\\frac{x-5}{x-5}$$\n\nthis is obviously 1 everywhere, except where it's undefined at $x = 5$. So, what would you expect to be the effect of multiplying the two? It's just multiplying by 1, except that it also introduces a hole at $x = 5$\n\nTo understand this more broadly, it is convenient to check L'H\u00f4pital's rule, which basically boils down to this:\n\nGiven a function $f(x)=\\frac{g(x)}{h(x)}$, where for some $x=x_0$ both $g(x_0)=0$ and $h(x_0)=0$, the actual value $f(x_0)$ can be obtained as*\n\n$$\\lim_{x\\to x_0}f(x) = \\lim_{x\\to x_0}\\frac{g'(x)}{h'(x)}$$\n\n(where the prime denotes derivation by $x$) Note that if $g'(x_0)=0=h'(x_0)$, you can re-apply the same rule again and again until either numerator or denominator are not 0. In your example, we get\n\n$$\\lim_{x\\to 5}\\frac{x^2-25}{x-5} = \\lim_{x\\to 5}\\frac{2x}{1}=10=\\lim_{x\\to5}\\,(x+5)$$\n\nTo get an even deeper understanding, the rule's proof might be an interesting read.\n\n*: This is only one case, you can also have $|g(x_0)|=|h(x_0)|=\\infty$, but not $g(x_0)=0$ and $f(x_0)=\\pm\\infty$\n\n\u2022 I don't think it's appropriate to bring up L'H\u00f4pital in a question like this. To justify the rule, the reader needs to be familiar with derivatives, differentiability, the sandwich theorem and Cauchy's mean value theorem just to get started. The OP is asking for understanding, not higher-level magic hand-waving. L'H\u00f4pital's rule is a great practical tool, but it's also wielded too often as a substitute for thinking. Aug 8, 2013 at 11:53\n\u2022 @EuroMicelli Woah there, I didn't intend to hand-wave magically, I just wanted to provide a different answer that, if one is interested enough, can not only explain the removable singularities in polynomial fractions but also in more complicated cases. Though I agree one shouldn't blindly apply H\u00f4pital to just anything if there is another maybe more elegant way Aug 8, 2013 at 13:04\n\u2022 Also you don't obtain the $f(x_0)$ value - you get $f(x_0)$ value provided it is defined and $f$ is continous. The OP function is not defined at $5$ but it's limit is. Aug 9, 2013 at 7:05\n\u2022 @MaciejPiechotka True, stricly speaking the example function is equivalent to $$f(x) = \\begin{cases} x+5 & x\\neq 5 \\\\ \\text{undefined} & x = 5\\end{cases}$$ but the singularity at $x=5$ is removable as in $$f(x)= \\begin{cases}\\tfrac{x^2-25}{x-5} & x\\neq 5 \\\\ 10 & x=5 \\end{cases} \\Rightarrow f(x) \\equiv x+5$$ But you're right that one should always keep removed singularities in mind as they may have severe influence to e.g. applications in Physics (though I can't tell one ad hoc) Aug 9, 2013 at 7:10\n\nOne of definitions of $\\lim_{x \\to A} f(x) = B$ is:\n\n$$\\forall_{\\varepsilon > 0}\\exists_{\\delta > 0}\\forall_{0 < \\left|x - A\\right| < \\delta}\\left|f(x) - B\\right| < \\varepsilon$$\n\nThe intuition is that we can achieve arbitrary 'precision' (put in bounds on y axis) provided we get close enough (so we get the bounds on x axis). However the definition does not say anything about the value at the point $f(A)$ which can be undefined or have arbitrary value.\n\nOne of method of proving the limit is to find the directly $\\delta(\\varepsilon)$. Hence we have following formula (well defined as $x\\neq 5$):\n\n$$\\forall_{0 < \\left|x - 5\\right| < \\delta}\\left|\\frac{x^2-25}{x-5} - 10\\right| < \\epsilon$$\n\nAs $x\\neq 5$ (in such case $\\left|x - 5\\right| = 0$) we can factor the expression out\n\n$$\\forall_{0 < \\left|x - 5\\right| < \\delta} \\left|x + 5 - 10\\right| < \\varepsilon$$ $$\\forall_{0 < \\left|x - 5\\right| < \\delta} \\left|x - 5 \\right| < \\varepsilon$$\n\nTaking $\\delta(\\varepsilon) = \\varepsilon$ we find that:\n\n$$\\forall_{\\varepsilon > 0}\\exists_{\\delta > 0}\\forall_{0 < \\left|x - 5\\right| < \\delta}\\left|\\frac{x^2-25}{x-5} - 10\\right| < \\varepsilon$$\n\nThe key thing is that we don't care about value at the limit.\n\nLook at this example:\n\n$$\\lim _{\\text{thing} \\to zero}(\\dfrac{\\text{Thing}}{\\text{Thing}}) \\cdot \\text{Another thing}$$\n\nHere our $\\text{'Thing'}$ is tending to zero, but not zero, it is something real, call it a real $\\text{Thing}$, which can be cancelled merrily.\n\n\u2022 What do you have against using letters as variables? Mar 8, 2014 at 18:54\n\nThe hole doesn't always disappear but when it does it is because one expression is equivalent to another except right at that hole. For example $$\\frac{x^2 - 25}{x - 5}= \\frac{(x+5)(x-5)}{x-5}= (x + 5)$$ everywhere except at x = 5. However, if we are very close to 5 and we had infinite place arithmetic (as we do theoretically) the calculation of the first expression $\\frac{x^2 - 25} {x - 5}$ would be very close to the last expression $(x+5)$, that is the limit as x approaches 5 of the first expression is the same limit as the limit as x approaches 5 of last expression. Since that latter is a 'nice number', we can define the value at 5 for the first expression and remove the \"everywhere except at $x = 5$\" by just putting 10 as the value and not calculating it.\n\nThe trick is to think of limit a bit differently: A limit is the value a function would have at a point were it continuous at that point, with nothing else being different.\n\nYou can see this by comparing the epsilon-delta definitions of the two concepts. We say $$f$$ is continuous at $$c$$, if $$f$$ is defined at $$c$$ and for every $$\\epsilon > 0$$ there is a $$\\delta > 0$$ such that\n\n$$|f(x) \u2013 f(c)| < \\epsilon$$ whenever $$0 < |x \u2013 c| < \\delta$$.\n\nLikewise, $$f$$ has limit $$L$$ at $$c$$ if for every $$\\epsilon > 0$$ there is a $$\\delta > 0$$ such that\n\n$$|f(x) \u2013 L| < \\epsilon$$ whenever $$0 < |x \u2013 c| < \\delta$$.\n\nThus, if we have a function with isolated gaps and there is a different function $$f^{*}$$ such that\n\n1. $$f^{*}(x) = f(x)$$ wherever $$f(x)$$ is defined and continuous,\n2. $$f^{*}$$ is defined and continuous everywhere, i.e. has domain $$\\mathbb{R}$$ (versus the original function having domain $$\\mathbb{R}$$ minus a set of isolated points),\n\nthen we must have that the value this $$f^{*}$$ takes at the points where $$f$$ was not defined or continuous must be the limiting values at those points.\n\nAnd that's what you find by cancellation: by rewriting the expression describing the function in a form that involves only addition and multiplication with no division, you have created a formula that describes a function that must agree where the division part is defined, but since no division is involved, is defined everywhere. Moreover, since addition and multiplication (and powers, though you can consider them repeated multiplication) are continuous, then it follows that this function must be continuous as well. Hence the above are met, and the holes fill in.\n\nThe idea is that if two functions $f$ and $g$ differ only at $a$ but are identical otherwise, we have $\\displaystyle \\lim_{x \\to a} f(x) = \\lim_{x \\to a} g(x)$. In this case $f(x) = \\dfrac {x^2-25}{x-5}$ and $g(x) = x+5$ are identical everywhere except $x=5$, so $\\displaystyle \\lim_{x \\to 5} \\dfrac {x^2-25}{x-5} = \\lim_{x \\to 5} x+5 = 10.$\n\nIn limits the value of x is always tending and not exact\n\nx\u2192a means the value of x tends to a. Similarly x\u21925 means tending to 5. So basically the value of x is not 5, but a little more or less than 5.\n\nHence you cannot take the denominator 0 in any case as the denominator will be, again, tending to zero. That is why we try to factorize the numerator first in order to check if there are any terms common both on the numerator and the denominator.\n\n$$L:=\\lim_{x\\to x_0}f(x)$$ is the value that would make $$f$$ continuous at $$x_0$$.\n\nIf you known another function $$g$$ which is continuous and coincides with $$f$$ around $$x_0$$, then perforce\n\n$$L=g(x_0).$$", "date": "2022-05-23 11:55:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/462199/why-does-factoring-eliminate-a-hole-in-the-limit", "openwebmath_score": 0.8391028046607971, "openwebmath_perplexity": 255.68626743071, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137942490252, "lm_q2_score": 0.8723473763375643, "lm_q1q2_score": 0.8566571569404339}}
{"url": "https://www.hpmuseum.org/forum/archive/index.php?thread-6157.html", "text": "# HP Forums\n\nYou're currently viewing a stripped down version of our content. View the full version with proper formatting.\n.\nInvH:\n\n\u00ab .577215664902 - InvPsi 1. -\n\u00bb\n\nInvPsi:\n\n\u00ab DUPDUP DUP InvP InvP DUP InvP InvP DUPDUP InvP InvP DUP InvP DUP EXP .5 + Psi OVER - - EXP .5 +\n\u00bb\n\nInvP:\n\n\u00ab DUP EXP .5 + Psi OVER - - EXP .5 + Psi OVER - -\n\u00bb\n\n(*) On the HP 49G, replace Psi with 0 PSI.\n\nExamples:\n\n4.012007 InvH --> 30.523595226 (in less than half a second); this is a very belated solution to one of the famous Valentin's challenges (#3 here).\n\n.577215664902 InvH --> 0.46163214497 ; InvH(Euler-Mascheroni constant) = xmin (local minimum of the continuous factorial function).\n\n4 pi 2 / - 3 ENTER 2 LN * - InvH --> 0.24999999999 ; one of the special values for fractional arguments examples here.\n1.5 InvH --> 2. ; H2\n\n1 Psi --> -0.577215664902 InvPsi --> 1.00000000009\n2 Psi --> 0.422784335098 InvPsi --> 2.\n\nBackground:\n\nLet H(x) be the continuous function associated with Harmonic Numbers. Then,\n\n$H(x)=\\gamma +\\psi \\left ( x+1 \\right )$\n\n$\\psi \\left ( x+1 \\right )=H(x)-\\gamma$\n\n$x+1 =\\psi^{-1} \\left ( H(x)-\\gamma \\right )$\n\n$x+1 =\\psi^{-1} \\left ( H(x)-\\gamma \\right )$\n\n$x =\\psi^{-1} \\left ( H(x)-\\gamma \\right )-1$\n\nor\n\n$H^{-1}(x) =\\psi^{-1} \\left ( x-\\gamma \\right )-1$\n\nThat is, in order to obtain the inverse of H(x) we only need the Inverse Digamma Function and the Euler-Mascheroni constant. No problem with the constant, but the Inverse Digamma might be a problem since Digamma is not easily invertible.\n\nA rough approximation is\n\n$\\psi^{-1} \\left ( x\\right )=e^{x}+\\frac{1}{2}$\n\nThe equivalent HP 50g program is\n\nP1:\n\n\u00ab EXP .5 +\n\u00bb\n\nHowever, this is good only for x >= 10, not good enough for our purposes:\n\n10 P1 --> 22026.9657948 Psi --> 10.0000000001.\n\nBut,\n\n9 P1 --> 8103.58392758 Psi --> 9.00000000064\n8 P1 --> 2981.45798704 Psi --> 8.00000000469\n7 P1 --> 1097.13315843 Psi --> 7.00000003465\n\nSo, let's try to improve the accuracy a bit:\n\nP2:\n\n\u00ab DUP P1 Psi OVER - - P1\n\u00bb\n\n10 P2 --> 22026.9657926 Psi --> 9.99999999999\n9 P2 --> 8103.58392239 Psi --> 8.99999999999\n8 P2 --> 2981.45797306 Psi --> 8.\n7 P2 --> 1097.13312043 Psi --> 7.\n6 P2 --> 403.928690211 Psi --> 6.\n5 P2 --> 148.912878357 Psi --> 5.00000000001\n\nBut,\n\n4 P2 --> 55.0973869316 Psi --> 4.00000000039\n3 P2 --> 20.5834634677 Psi --> 3.00000002131\n\nProceeding likewise, we get\n\nP3:\n\n\u00ab DUP P2 Psi OVER - - P2\n\u00bb\n\nThis is good for x as low as 2, but not for x = 1:\n\n2 P3 --> 7.88342863120 Psi --> 2.\n1 P3 --> 3.20317150637 Psi --> 1.0000000139\n\nA couple more steps suffice for x around -0.6 and greater, which is good enough for our purposes. That's what the InvPsi program above does, albeit in an inelegant way. Also, this is just an intuitive and somewhat inneficient approach. Better methods suggestions are welcome.\n\nEdited to fix a couple of typos.\n\nEdited again to include a printout of my current directory:\n\nInvPsi will accept arguments greater or equal -1. About 0.05 s, 0.10 sor 0.50 s, depending on the arguments.\nCool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.\n\nJohn\n(04-27-2016 04:57 PM)John Keith Wrote: [ -> ]Cool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.\n\nIf the arguments are always plain harmonic numbers, that is, the ones obtained from the discrete function, then the inverse function can be implemented simply as\n\nInvHn:\n\n\u00ab .577215664902 - EXP IP \u00bb\n\nExamples:\n\n137 ENTER 60 / InvHn --> 5. ; 137/60 = H(5)\n\n10 Hx --> 2.92896825397 ; H(10)\nInvHn --> 10.\n\n2E10 --> 24.2962137754 ; H(2x10^10)\nInvHn --> 20000000000.\n\nwhere Hx is the program for the continuous function:\n\nHx:\n\n\u00ab 1. + Psi .577215664902 + \u00bb\n\nRegards,\n\nGerson.\nReference URL's\n\u2022 HP Forums: https://www.hpmuseum.org/forum/index.php\n\u2022 :", "date": "2019-10-21 12:20:00", "meta": {"domain": "hpmuseum.org", "url": "https://www.hpmuseum.org/forum/archive/index.php?thread-6157.html", "openwebmath_score": 0.7077171802520752, "openwebmath_perplexity": 9318.171377763172, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137910906878, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8566571492967379}}
{"url": "https://math.stackexchange.com/questions/471710/why-do-small-angle-approximations-have-to-be-in-radians/471715", "text": "Why do small angle approximations have to be in radians?\n\nWhy do small angle approximations only hold in radians? All the books I have say this is so but don't explain why.\n\n\u2022 @EmilioPisanty perhaps you're correct, but anonymous disparagement is hardly the best way to greet a new user. Anyway, user28435, try Googling one-line answers before asking in future (this must have been asked hundreds of times on Math SE).\n\u2013\u00a0Meow\nAug 19, 2013 at 21:36\n\u2022 Don't think of sine as being defined in terms of angles - rather, arcs length of a circle. (The approximations come from the fact that a circle is approximately a straight line, especially near the point of tangency.) Oct 21, 2014 at 10:55\n\nThe reason why we use radians for $\\sin x$ (or other trigonometric functions) in calculus is explained here and here in Wikipedia.\n\nHaving known that, notice that small angle approximation is just the Taylor expansion of $\\sin x$ and $\\cos x$ around $x=0$:\n\n$$\\sin x = \\sum_{n=0}^\\infty \\frac{(-1)^n x^{2n+1}}{(2n+1)!}\\tag{1}$$ $$\\cos x = \\sum_{n=0}^\\infty \\frac{(-1)^n x^{2n}}{(2n)!}\\tag{2}$$\n\nIf you scale $x$ by some constant $\\omega$, then you must replace $x$ with $\\omega x$ in $(1)$ and $(2)$. So, working in degrees $($ $\\omega=\\frac{\\pi}{180}$ $)$, the approximation will become: $$\\sin \\theta\\approx \\frac{\\pi}{180}\\theta\\tag{\\theta in degrees}$$\n\nThere are several correct ways to answer this that illuminate different aspects of what is going on (and I wouldn't be surprised if the answer is present on this site somewhere EDIT: indeed the other responses to this question are different ways of looking at it). Here is a geometrical answer.\n\nThe basic formula is\n\n$$l = r \\theta$$\n\nwhere $l$ is the arc length of a segment of a circle with radius $r$ subtending an angle $\\theta$. For this formula to be true, $\\theta$ needs to be in radians. (Just try it out, is the arc length of the whole circle equal to $360 r$ or $2\\pi r$?) In fact, that formula is really how you define what you mean by radians.\n\nNow consider a straight line segment connecting the two endpoints of the arc subtended by $\\theta$. Call it's length $a$. You can show by fiddling with triangles that $$a=2r\\sin\\left(\\frac{\\theta}{2}\\right)$$\n\nIn the limit that the angle is small (so only a small piece of the circle is subtended), you should be able to convince yourself that $a\\approx l$. This is the core of the small angle approximation.\n\nUsing the two relationships above we have\n\n$$r\\theta \\approx 2r \\sin\\left(\\frac{\\theta}{2}\\right)$$ or $$\\sin\\left(\\frac{\\theta}{2}\\right)\\approx\\frac{\\theta}{2}$$ You can see that using radians was crucial here because it allowed us to use $l=r\\theta$.\n\nA 'small angle' is equally small whatever system you use to measure it. Thus if an angle is, say, much smaller than 0.1 rad, it will be much smaller than the equivalent in degrees. More typically, saying 'small angle approximation' typically means $\\theta\\ll1$, where $\\theta$ is in radians; this can be rephrased in degrees as $\\theta\\ll 57^\\circ$.\n\n(Switching uses between radians and degrees becomes much simpler if one formally identifies the degree symbol $^\\circ$ with the number $\\pi/180$, which is what you get from the equation $180^\\circ=\\pi$. If you're differentiating with respect to the number in degrees, then, you get an ugly constant, as you should: $\\frac{d}{dx}\\sin(x^\\circ)={}^\\circ \\cos(x^\\circ)$.)\n\nIn real life, though, you wouldn't usually say 'this angle should be small' without saying what it should be smaller than. If the latter is in degrees then the former should also be in degrees.\n\nThat said, though: always work in radians! Physicists tend to use degrees quite often, but there is always the underlying understanding that the angle itself is a quantity in radians and that degrees are just convenient units. Trigonometric functions, in particular, always take their arguments in radians, so that all the math will work well. Always differentiate in radians, always work analytically in radians. And at the end you can plug in the degrees.\n\nIt's because this relationship $$\\lim_{x\\rightarrow0}\\frac{\\sin(x)}{x}=1$$ (i.e. $\\sin(x) \\approx x$) only holds if $x$ is in radians, as, using L'Hoptial's rule, $$\\lim_{x\\rightarrow0}\\frac{\\sin(x)}{x}=\\lim_{x\\rightarrow0}\\frac{\\frac{d}{dx}\\sin(x)}{\\frac{d}{dx}x}=\\lim_{x\\rightarrow0}\\frac{d}{dx}\\sin(x)=\\begin{cases} 1 & \\text{ if } x\\text{ is in radians}\\\\ \\frac{\\pi}{180} & \\text{ if } x \\text{ is in degrees} \\end{cases}.$$ That is, $\\frac{d}{dx}(\\sin(x))=\\cos(x)$ only if $x$ is in radians. The reason can be divined from this proof without words by Stephen Gubkin:\n\n\u2022 Sure but WHY does this only hold in radians. That's the bit I can't understand. Everyone says it holds in radians but no one explains why lol. Thanks anyway. Aug 19, 2013 at 21:03\n\u2022 @user28435: because in radians the sin of an angle basically equals the angle itself, at small enough angles, so it very much simplifies the algebra if you don't have to keep converting degrees.\n\u2013\u00a0Mike Dunlavey\nAug 20, 2013 at 0:12\n\nYou need to clarify what you are really asking: in physics the small angle approx. is typically used to approximate a non-linear differential equation by a linear one - which is much easier to solve - by allowing us to approximate the sin(x) function by x. The resulting equation is only a good one for small angles. I.e., its a physics thing, not a math thing.\n\nbecause the sine of an angle is dimensionless [i.e. just a number], the angle has to be also dimensionless [i.e. the radian is dimensionless] ... units on both sides of an equation have to match", "date": "2022-05-26 15:04:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/471710/why-do-small-angle-approximations-have-to-be-in-radians/471715", "openwebmath_score": 0.9050287008285522, "openwebmath_perplexity": 340.05392446613325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013790564298, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.8566571488375431}}
{"url": "http://math.stackexchange.com/questions/231808/determining-the-remainder-of-a-small-integer-n-raised-to-a-high-power-when-divid", "text": "# Determining the remainder of a small integer n raised to a high power when divided by m\n\nI recently took a test that asked me to determine the value of $3^{82} \\mod 5$.\n\nI was unable to figure out how to do it on the test, but when I got home I noticed that there is a pattern to the remainders of $3^n$,\n\n\u2022 $3^{1} \\mod 5 = 3$\n\u2022 $3^{2} \\mod 5 = 4$\n\u2022 $3^{3} \\mod 5 = 2$\n\u2022 $3^{4} \\mod 5 = 1$\n\u2022 $3^{5} \\mod 5 = 3$\n\u2022 $3^{6} \\mod 5 = 4$\n\u2022 $3^{7} \\mod 5 = 2$\n\u2022 $3^{8} \\mod 5 = 1$\n\u2022 $3^{9} \\mod 5 = 3$\n\nIs the best strategy to follow this pattern up to the $82^{\\text{nd}}$ remainder (obviously in an intelligent way) or is there some much more obvious trick to this question that I missed?\n\n-\n\nAs you have observed $$3^4 \\equiv 1 \\pmod 5$$ This means that $$3^{4k} \\equiv 1 \\pmod 5$$ Hence, $$3^{80} \\equiv 1 \\pmod 5 \\implies 3^{82} \\equiv 3^2 \\pmod{5} = 4 \\pmod 5$$ In general, $$3^{4k + r} \\equiv 3^r \\pmod{5}$$\nThank you for your answer. It's not entirely clear to me how $3^4 \\mod 1 \\equiv 1$ tells us that $3^{4k} \\mod 1 \\equiv 1$. Could you explain this step? \u2013\u00a0 crazedgremlin Nov 7 '12 at 1:07\n@crazedgremlin $\\rm\\: \\color{#0A0}{3^4}\\equiv\\color{#C00} 1\\:\\Rightarrow\\: 3^{4k} = (\\color{#0A0}{3^4})^k\\equiv \\color{#C00}1^k\\equiv 1\\:$ by the Congruence Product Rule. It is essential to learn the arithmetic of congruences, as opposed to less flexible mod operations. \u2013\u00a0 Bill Dubuque Nov 7 '12 at 1:23", "date": "2015-05-26 04:38:15", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/231808/determining-the-remainder-of-a-small-integer-n-raised-to-a-high-power-when-divid", "openwebmath_score": 0.6795341372489929, "openwebmath_perplexity": 184.99370343555776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9921841104018225, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8566434417016436}}
{"url": "http://mathhelpforum.com/calculus/82075-limits-series-question.html", "text": "# Thread: Limits / Series question\n\n1. ## Limits / Series question\n\nhi\n\nLet\nP = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]\nn=2,3,4,......\n\nFind Limit P when N tends to infinity.\n\n2. Originally Posted by champrock\nhi\n\nLet\nP = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]\nn=2,3,4,......\n\nFind Limit P when N tends to infinity.\nFactorise $\\frac{n^3-1}{n^3+1}$ as $\\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$, and notice that $n^2+n+1 = (n+1)^2 - (n+1) + 1$. You will then find that you have a telescoping product.\n\n3. what should i do with n^2 - n + 1 ?\n\nI was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1\n\n4. You easily undestand if you write the sequence of terms...\n\n$P= \\prod_{n=2}^{\\infty} \\frac{(n-1)\\cdot (n^{2} + n + 1)}{(n+1)\\cdot (n^{2} - n + 1)}= \\frac {1\\cdot 7}{3\\cdot 3}\\cdot \\frac {2\\cdot 13}{4\\cdot 7}\\cdot \\frac {3\\cdot 21}{5\\cdot 13}\\cdot \\frac {4\\cdot 31}{6\\cdot 21}\\cdot \\dots$\n\n... and it is evident you can simplify both in numerator and denominator [3 with 3, 4 with 4, ... , 7 with 7, 13 with 13, 21 with 21,...]\n\nFinally is...\n\n$P=\\frac{2}{3}$\n\nVery nice!...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n5. Originally Posted by champrock\nwhat should i do with n^2 - n + 1 ?\n\nI was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1\nUse the hint I gave before: $n^2+n+1 = (n+1)^2 - (n+1) + 1$. That tells you that the term $n^2+n+1$ in the numerator of each fraction cancels with the term $(n+1)^2 - (n+1) + 1$ in the denominator of the next one.\n\n6. I think Only 1/3 is left. All the other terms cancel out. How are you getting 2/3\n\n7. The term of order n is...\n\n$a_{n}= \\frac{(n-1)\\cdot (n^{2} +n + 1)}{(n+1)\\cdot (n^{2} - n + 1)}$\n\nFor $n=3$ is...\n\n$a_{3}= \\frac{2\\cdot 13 }{4\\cdot 7}$\n\nKind regards\n\n$\\chi$ $\\sigma$", "date": "2017-04-28 07:47:56", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/82075-limits-series-question.html", "openwebmath_score": 0.9248009324073792, "openwebmath_perplexity": 487.10623735797816, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9921841101707115, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8566434397583008}}
{"url": "http://inognicasa.it/equation-of-a-rotated-ellipse.html", "text": "Log InorSign Up. The form of the covariance matrix \u03c3 in the unrotated system follows from equation (14) using R. The standard form of the equation of an ellipse with center (h, k) and major and minor axes of lengths 2a and 2b, respectively, where 0 < b < a, is. They both have shape (eccentricity) and size (major axis). The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (\u2218, \u2218), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. | bartleby. A graph of the two equations is presented here. As stated, using the definition for center of an ellipse as the intersection of its axes of symmetry, your equation for an ellipse is centered at $(h,k)$, but it is not rotated, i. It has co-vertices at (5 \u00b1 3, \u20131), or (8, \u20131) and (2, \u20131). Once we have those we can sketch in the ellipse. Find the points at which this ellipse crosses the x-axis and show that the tangent lines at these points are parallel. How It Works. The parametric equations of an ellipse are and. |C|y the equation is transformed into \u03c3x02 + \u03c4y02 = \u2212F. Here a > b > 0. This equation defines an ellipse centered at the origin. the graph is an ellipse if AC > 0, and in Section 5. You may ignore the Mathematica commands and concentrate on the text and figures. Ellipse drawing tool. Constructing (Plotting) a Rotated Ellipse. The equation x 2 \u2013 xy + y 2 = 3 re presents a \"rotated ellipse,\u201d that is, an ellipse whose axes are not parallel to the coordinate axes. For a given chord or triangle base, the. The length of the major axis is 2a, and the length of the minor axis is 2b. +- hat j b. Complex Growth. Rotation Creates the ellipse by appearing to rotate a circle about the first axis. Let x^2+3y^2=3 be the equation of an ellipse in the x-y plane. You may ignore the Mathematica commands and concentrate on the text and figures. Vertical Major Axis Example. They both have shape (eccentricity) and size (major axis). Since B = \u2212 6 3 \u2260 0, the equation satisfies the condition to be a rotated ellipse. {\\displaystyle Ax^ {2}+Bxy+Cy^ {2}+Dx+Ey+F=0}. If C\u2206 > 0, we have an imaginary ellipse, and if \u2206 = 0, we have a point ellipse. Equations When placed like this on an x-y graph, the equation for an ellipse is: x 2 a 2 + y 2 b 2 = 1. The Rotated Ellipsoid June 2, 2017 Page 1 Rotated Ellipsoid An ellipse has 2D geometry and an ellipsoid has 3D geometry. Write the equation of the circle in standard form given the endpoints of the diameter: (-12, 10) and (-18, 12). They are located at (h\u00b1c,k) or (h,k\u00b1c). Let us consider a point P(x, y) lying on the ellipse such that P satisfies the definition i. The equation of the ellipse we discussed in class is 9 x2 - 4 xy + 6 y2 = 5. HELP?! Rotate the axes to eliminate the xy-term in the equation. centerof the ellipse gets rotated about point pand the new ellipseat the new center gets rotated about the new centerby angle a. We have step-by-step solutions for your textbooks written by Bartleby experts!. Here we plot it ContourPlotA9 x2-4 x y + 6 y2 \u2212 5, 8x,-1, 1<, 8y,-1, 1<, Axes \ufb01 True, Frame \ufb01 False,. Except for degenerate cases, the general second-degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 x\u00bfy\u00bf-term x\u00bf 2 4 + y\u00bf 1 = 1. I wish to plot an ellipse by scanline finding the values for y for each value of x. H(x, y) = A x\u00b2 + B xy + C y\u00b2 + D x + E y + F = 0 The basic principle of the incremental line tracing algorithms (I wouldn't call them scanline) is to follow the pixels that fulfill the equation as much as possible. The standard form of the equation of an ellipse with center (h, k) and major and minor axes of lengths 2a and 2b, respectively, where 0 < b < a, is. Since (\u2212 6 3) 2 \u2212 4 \u22c5 7 \u22c5 13 = \u2212 346 < 0 and A \u2260 C since 7 \u2260 13, the equation satisfies the conditions to be an ellipse. 18 < : 1 1 1 9 = ; (10) 3. By using this website, you agree to our Cookie Policy. Activity 4: Determining the general equation of an ellipse/ Determining the foci and vertices of an ellipse. I want to plot an Ellipse. Rotate the hyperbola : In the above graph, the preimage is in blue and the image (rotated) is. When you rotate the ellipse about y = 5, the \"tire\" above will be coming-out and going-in through z-direction. EINSTEIN, ALBERT and MICHELE BESSO. Here are formulas for finding these points. I know that i can draw it using ellipse equation, then rotate it, compute points and connect with lines. If an ellipse is rotated about one of its principal axes, a spheroid is the result. If the rotation is small the resulting ellipse is very nearly round, but if the rotation is large the ellipse becomes very flattened (or very elongated, depending upon how you look at the effect), and if the circle is rotated until it is edge-on to our line of sight the \"ellipse\" becomes just a straight line segment. The Rotated Ellipsoid June 2, 2017 Page 1 Rotated Ellipsoid An ellipse has 2D geometry and an ellipsoid has 3D geometry. In the interesting \ufb01rst case the set Q is an ellipse. k is y-koordinate of the center of the ellipse. If the center of the ellipse is at point (h, k) and the major and minor axes have lengths of 2a and 2b respectively, the standard equation is. HELP?! Rotate the axes to eliminate the xy-term in the equation. Sketch the graph of Solution. When rotated inside a square of side length 2 having corners at ), the envelope of the Reuleaux triangle is a region of the square with rounded corners. Equations When placed like this on an x-y graph, the equation for an ellipse is: x 2 a 2 + y 2 b 2 = 1. (1) Ellipse (2) Rotated Ellipse (3) Ellipse Representing Covariance. The following 12 points are on this ellipse: The ellipse is symmetric about the lines y. Write the equation of the circle in standard form given the endpoints of the diameter: (-12, 10) and (-18, 12). Here we plot it ContourPlotA9 x2-4 x y + 6 y2 \u2212 5, 8x,-1, 1<, 8y,-1, 1<, Axes \ufb01 True, Frame \ufb01 False,. Given the equation of a conic, identify the type of conic. What are the applications of Ellipse in real life? The ellipse has a close reference with football when it is rotated on its major axis. Equation 3 thus becomes: rz n S(y, cos & - x, sin &)2 - N Eq. If and are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse. | bartleby. The equation of an ellipse is a generalized case of the equation of a circle. In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. Let us consider the figure (a) to derive the equation of an ellipse. minimal expansion. lationship between two images is pure rotation, i. If the rotation is small the resulting ellipse is very nearly round, but if the rotation is large the ellipse becomes very flattened (or very elongated, depending upon how you look at the effect), and if the circle is rotated until it is edge-on to our line of sight the \"ellipse\" becomes just a straight line segment. Let the coordinates of F 1 and F 2 be (-c, 0) and (c, 0) respectively as shown. Except for degenerate cases, the general second-degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 x\u00bfy\u00bf-term x\u00bf 2 4 + y\u00bf 1 = 1. h is x-koordinate of the center of the ellipse. (4) into the following canonical form (x 0 x 0) 2 a2 + (y0 y 0) 2 b2 = 1; (5) in which (x 0 0;y 0) is the center of the ellipse in the rotated coordinate system, and aand bare the lengths of the semi-axes. Expanding the binomial squares and collecting like terms gives. The equation b 2 = a 2 \u2013 c 2 gives me 16 \u2013 9 = 7 = b 2. {\\displaystyle B} is zero then the conic is not rotated and sits on the x- and y- axes. ; If and are nonzero and have opposite signs. Rotating an Ellipse. We shall now study the Cartesian representation of the hyperbola and the ellipse. If you enter a value, the higher the value, the greater the eccentricity of the ellipse. Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle. Here a > b > 0. Thanks, Michal Bozon. For a non-rotated conic: A. Complex Growth. However, when you graph the ellipse using the parametric equations, simply allow t to range from 0 to 2\u03c0 radians to find the (x, y) coordinates for each value of t. If the Circle option is selected, the width and height of the drawn shape is kept the same. We have step-by-step solutions for your textbooks written by Bartleby experts!. | bartleby. \u2212 <: Ellipse \u2212 >: Hyperbola; Firstly, if B is not zero then the graph represents a rotated conic. When a>b, we have a prolate spheroid, that is, an ellipse rotated around its major axis; when a K > -1 = ellipse, -1 = parabolic, and K < -1 is hyperbolic; R is the radius of curvature. General Equations of Conics The graph of Ax 2 Cy is one of the 2 Dx Ey F 0 following : 1. However now i want the ellipse to rotate to the slopes so that the top of the ellipse faces in the direction of the line segments normal. In the rotated the major axis of the ellipse lies along the We can write the equation of the ellipse in this rotated as Observe that there is no in the equation. Ellipse configuration panel. 5 (a) with the foci on the x-axis. When we add an x y term, we are rotating the conic about the origin. I have the verticles for the major axis: d1(0,0. In this section, we will discuss the equation of a conic section which is rotated by. Because A = 7, and C = 13, you have (for 0 \u03b8 < \u03c0/2) Therefore, the equation in the x'y'-system is derived by making the following substitutions. Reversing translation : 137(X\u221210)\u00b2 \u2212 210(X\u221210)(Y+20)+137(Y+20)\u00b2 = 968 This is equation of rotated ellipse relative to original axes. If an ellipse is rotated about one of its principal axes, a spheroid is the result. Moment of inertia is defined with respect to a specific rotation axis. Several exam. (4) into the following canonical form (x 0 x 0) 2 a2 + (y0 y 0) 2 b2 = 1; (5) in which (x 0 0;y 0) is the center of the ellipse in the rotated coordinate system, and aand bare the lengths of the semi-axes. In terms of the geometric look of E, there are three possible scenarios for E: E = \u2205, E = p 1 \u2062 p 2 \u00af, the line segment with end-points p 1 and p 2, or E is an ellipse. It follows that 0 \u00a3e< 1 and p> 0, so that anellipse in polar coordinates with one focus at the origin and the other onthe positive x-axis is given by. The major axis in a vertical ellipse is represented by x = h; the minor axis is represented by y = v. Notes College Algebra teaches you how to find the equation of an ellipse given a graph. You may ignore the Mathematica commands and concentrate on the text and figures. A \u201cstandard ellipsoid\u201d has a circular midsection. {\\displaystyle Ax^ {2}+Bxy+Cy^ {2}+Dx+Ey+F=0}. So I'm trying to find the intersections of the equations $$\\ {x^2\\over 1^2} + {y^2\\over 2^2} = 1$$ $$5x^2 - 6xy + 5y^2 = 8$$ Both of the equations represent an ellipse, with the first ellipse being a vertical ellipse and the second ellipse being first one rotated 315 degrees counterclockwise. The center is at (h, k). The equation may be cast in a more general form since the standard deviation is to be calculated as the axis is rotated about the point of average location. {\\displaystyle B} is non-zero, then the conic is rotated about the axes, with the rotation centred on the origin. If a > b, the ellipse is stretched further in the horizontal direction, and if b > a, the ellipse is stretched further in the vertical direction. The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (\u2218, \u2218), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. First, notice that the equation of the parabola y = x^2 can be parametrized by x = t, y = t^2, as t goes from -infinity to infinity; or, as a column vector, [x] = [t] [y] = [t^2]. Thus, for the equation to represent an ellipse that is not a circle, the coefficients must simultaneously satisfy the discriminant condition B 2 \u2212 4 A C < 0 B^2 - 4AC< 0 B 2 \u2212 4 A C < 0 and also A \u2260 C. Find the points at which this ellipse crosses the x-axis and show that the tangent lines at these points are parallel. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). The form of the covariance matrix \u03c3 in the unrotated system follows from equation (14) using R. In the rotated the major axis of the ellipse lies along the We can write the equation of the ellipse in this rotated as Observe that there is no in the equation. A parametric form for (ii) is x=5. Find the points where the ellipse crosses the x-axis, and show that the tangent lines at these points are parallel. Let be the angle of rotation. Let\u2019s see what happens when. a is the ellipse axis which is parallell to the x-axis when rotation is zero. The major axis of this ellipse is horizontal and is the red segment from (-2, 0) to (2, 0). The equation x^2 - xy + y^2 = 3 represents a \"rotated ellipse,\" that is, an ellipse whose axes are not parallel to the coordinate axes. The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (\u2218, \u2218), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. Equation of ellipse from its focus, directrix, and eccentricity Last Updated: 20-12-2018 Given focus(x, y), directrix(ax + by + c) and eccentricity e of an ellipse, the task is to find the equation of ellipse using its focus, directrix, and eccentricity. Let us consider the figure (a) to derive the equation of an ellipse. Squashed Circles and Gardeners. If we deform this circle according to the general two dimensional linear transformation, the eq below can be derived:. HELP?! Rotate the axes to eliminate the xy-term in the equation. An ellipse is a flattened circle. k is y-koordinate of the center of the ellipse. Two fixed points inside the ellipse, F1 and F2 are called the foci. You may ignore the Mathematica commands and concentrate on the text and figures. The parametric equations of an ellipse are and. Points p 1 and p 2 are called foci of the ellipse; the line segments connecting a point of the ellipse to the foci are the focal radii belonging to that point. The general transformation is Y = RX with inverse X = RTY. Below is the C++ representation of the above problem. By using a transformation (rotation) of the coordinate system, we are able to diagonalize equation (12). Step 2 : Trigonometric identity : Substitute and in above equation. I'm looking for a Cartesian equation for a rotated ellipse. Let us consider a point P(x, y) lying on the ellipse such that P satisfies the definition i. Our mission is to provide a free, world-class education to anyone, anywhere. , x(t), y(t), z(t)) or a more canonical form (e. I have the verticles for the major axis: d1(0,0. Determine the foci and vertices for the ellipse with general equation 2x^2+y^2+8x-8y-48. If $$\\displaystyle D = b^2- 4ac$$, then it's an ellipse for $$\\displaystyle D<0$$, a parabola for $$\\displaystyle D = 0$$, and a hyperbola for $$\\displaystyle D>0$$. This can always be converted to VBA Code. Entering 0 defines a circular ellipse. Find the points where the ellipse crosses the x-axis, and show that the tangent lines at these points are parallel. The equation of a line through the point and cutting the axis at an angle is. In particular that the shape made by rotation around the x-axis can stand on the top, if it is made from wood. The velocity equation for a hyperbolic trajectory has either + , or it is the same with the convention that in that case a is negative. The blue ellipse shows the original plot. The super ellipse belongs to the Lam\u00e9 curves. For a surface obtained by rotating a curve around an axis, we can take a polygonal approximation to the curve, as in the last section, and rotate it around the same axis. The ellipse points are P = C+ x 0U 0 + x 1U 1 (1) where x 0 e 0 2 + x 1 e 1 2 = 1 (2) If e 0 = e 1, then the ellipse is a circle with center C and radius e 0. This equation of an elliptic cylinder is a generalization of the equation of the ordinary, circular cylinder (a = b). Several exam. A circle if A = C 2. The image of the disk will be an ellipse with these directions as the major and minor axes. Thus, the graph of this equation is either a parabola, ellipse, or hyperbola with axes parallel to the x and y-axes (there is also the possibility that there is no graph or the graph is a \u201cdegenerate\u201d conic: a point, a line, or a pair of lines). The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (\u2218, \u2218), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. Rotated Parabolas and Ellipse. General equations as a function of \u03bb X, \u03bb Z, and \u03b8 d \u03bb\u2019= \u03bb\u2019 Z +\u03bb\u2019 X-\u03bb\u2019 Z-\u03bb\u2019 X cos(2\u03b8 d) 2 2 \u03b3 \u03bb\u2019 Z-\u03bb\u2019 X sin(2\u03b8 d) 2 tan \u03b8 d = tan \u03b8 S X S Z \u03b1 = \u03b8 d - \u03b8 (internal rotation) \u03bb\u2019 = 1 \u03bb \u03bb X = quadratic elongation parallel to X axis of finite strain ellipse \u03bb Z = quadratic elongation parallel to Z axis of finite. All Forums. Erase the previous Ellipse by drawing the Ellipse at same point using black color. k is y-koordinate of the center of the ellipse. : Activity 2 - Using the Graph-Rotation Theorem. attempt to list the major conventions and the common equations of an ellipse in these conventions. Number of decimal places for input variable: (Note: Input value of 0 means input variable will be integer. Rewrite the equation in the general form, Identify the values of and from the general form. the graph is an ellipse if AC > 0, and in Section 5. is along the ellipse\u2019s major axis, the correlation matrix is \u03c3\u2032 = \u03c3\u20322 1 0 0 \u03c3\u20322 2. I am trying to find an algorithm to derive the 4 angles, from the centre of a rotated ellipse to its extremities. Here is a simple calculator to solve ellipse equation and calculate the elliptical co-ordinates such as center, foci, vertices, eccentricity and area and axis lengths such as Major, Semi Major and Minor, Semi Minor axis lengths from the given ellipse expression. | bartleby. If the data is uncorrelated and therefore has zero covariance, the ellipse is not rotated and axis aligned. He provided me with some equations that I combined with a neat ellipse display program written by Wayne Landsman for the NASA Goddard IDL program library to come up with a \"center of mass\" ellipse fitting program, named Fit_Ellipse. 06274*x^2 - y^2 + 1192. The blue ellipse shows the original plot. Ellipse definition, a plane curve such that the sums of the distances of each point in its periphery from two fixed points, the foci, are equal. with the axis. Do the intersection points of two rotated parabolas lie on a rotated ellipse? 1. tive number has a square root. Combine multiple words with dashes(-), and seperate tags with spaces. The equation x 2 \u2013 xy + y 2 = 3 re presents a \"rotated ellipse,\u201d that is, an ellipse whose axes are not parallel to the coordinate axes. 1-c/a cos( q) Usually, we let e= c/aand let p= b2/a, where eis called the eccentricityof the ellipse and pis called theparameter. The moment of inertia of any extended object is built up from that basic definition. If the data is uncorrelated and therefore has zero covariance, the ellipse is not rotated and axis aligned. Rotation Creates the ellipse by appearing to rotate a circle about the first axis. At first blush, these are really strange exponents. I have the verticles for the major axis: d1(0,0. The chord perpendicular to the major axis at the center is the minor axis. R = distance between axis and rotation mass (ft, m) The moment of all other moments of inertia of an object are calculated from the the sum of the moments. The major axis of this ellipse is horizontal and is the red segment from (-2, 0) to (2, 0). com Since c \u2264 a the eccentricity is always greater than 1 in the case of an ellipse. Since (\u2212 6 3) 2 \u2212 4 \u22c5 7 \u22c5 13 = \u2212 346 < 0 and A \u2260 C since 7 \u2260 13, the equation satisfies the conditions to be an ellipse. In the interesting \ufb01rst case the set Q is an ellipse. The center of this ellipse is the origin since (0, 0) is the midpoint of the major axis. The equation x^2 - xy + y^2 = 3 represents a \"rotated ellipse,\" that is, an ellipse whose axes are not parallel to the coordinate axes. Disk method. The parametric equation of a parabola with directrix x = \u2212a and focus (a,0) is x = at2, y = 2at. Express the equation in the standard form of a conic section. the graph is an ellipse if AC > 0, and in Section 5. This ellipse is called the distortion ellipse. The transformed ellipse is de-scribed by the equation a0x2. Torna cartesian equation of rotated ellipse ogni moment group ingannatore. g, with major axis aligned with X-axis, minor axis aligned with Y-axis \u2026 but that\u2019s certainly not an \u201carbitrary\u201d ellipse. +- hat j b. Torna cartesian equation of rotated ellipse ogni moment group ingannatore. The amount of correlation can be interpreted by how thin the ellipse is. ) Then my equation is: Write an equation for the ellipse with vertices (4, 0) and (\u20132, 0). How It Works. Our mission is to provide a free, world-class education to anyone, anywhere. We have also seen that translating by a curve by a fixed vector ( h , k ) has the effect of replacing x by x \u2212 h and y by y \u2212 k in the equation of the curve. In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. An ellipse represents the intersection of a plane surface and an ellipsoid. , the 3D analog to the 2D form ((X*X)/a)+((Y*Y)/b)=1). Moment of inertia is defined with respect to a specific rotation axis. If C\u2206 > 0, we have an imaginary ellipse, and if \u2206 = 0, we have a point ellipse. This gives a surface composed of many \"truncated cones;'' a truncated cone is called a frustum of a cone. E, qua, euros' Sale, per forza di guerra. Differential Equations (10) Discrete Mathematics (4) Discrete Random Variable (5) Disk Washer Cylindrical Shell Integration (2) Division Tricks (1) Domain and Range of a Function (1) Double Integrals (3) Eigenvalues and Eigenvectors (1) Ellipse (1) Empirical and Molecular Formula (2) Enthalpy Change (2) Expected Value Variance Standard. The parameters of an ellipse are also often given as the semi-major axis, a, and the eccentricity, e, 2 2 1 a b e =-. In the rotated the major axis of the ellipse lies along the We can write the equation of the ellipse in this rotated as Observe that there is no in the equation. When rotated inside a square of side length 2 having corners at ), the envelope of the Reuleaux triangle is a region of the square with rounded corners. A more general figure has three orthogonal axes of different lengths a, b and c, and can be represented by the equation x 2 /a 2 + y 2 /b 2 + z 2. Solution The equation of an ellipse usually appears when the plot of the from MECHANICAL MAE351 at Korea Advanced Institute of Science and Technology. Constructing (Plotting) a Rotated Ellipse. The parameters of an ellipse are also often given as the semi-major axis, a, and the eccentricity, e, 2 2 1 a b e =-. Textbook solution for Single Variable Calculus: Early Transcendentals 8th Edition James Stewart Chapter 3. When the center of the ellipse is at the origin and the foci are on the x-axis or y-axis, then the equation of the ellipse is the simplest. The longer axis, a, is called the semi-major axis and the shorter, b, is called the semi-minor axis. A circle if A = C 2. Rotating Ellipse. Equations When placed like this on an x-y graph, the equation for an ellipse is: x 2 a 2 + y 2 b 2 = 1. First I want to look at the case when , and. This can always be converted to VBA Code. x\u00bfy\u00bf-system x\u00bf-axis. I should also mention, the ellipse is to be drawn on linear or log-log coordinates. If the x- and y-axes are rotated through an angle, say \u03b8,. Let be the angle of rotation. Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle. (Since I wasn't asked for the length of the minor axis or the location of the co-vertices, I don't need the value of b itself. The equations of tangent and normal to the ellipse $$\\frac{{{x^2}}}{{{a^2}}} + \\frac{{{y^2}}}{{{b^2}}} = 1$$ at the point $$\\left( {{x_1},{y_1}} \\right)$$ are \\frac. If $a>b$, the ellipse is stretched further in the horizontal direction, and if $b>a$, the ellipse is stretched further in the vertical direction. Find dy dx. Suppose an ellipse is described by the equation ax2 + bxy + cy2 + dx + ey + f = 0 with a;b;c;d;e;f 2 F. 4 degrees and 90. asked \u2022 04/02/15 find the equation of the image of the ellipse x^2/4 + y^2/9 when rotated through pi/4 about origin. Below is a list of parametric equations starting from that of a general ellipse and modifying it step by step into a prediction ellipse, showing how different parts contribute at each step. I want to plot an Ellipse. Torna cartesian equation of rotated ellipse ogni moment group ingannatore. Moment of inertia is defined with respect to a specific rotation axis. If C\u2206 > 0, we have an imaginary ellipse, and if \u2206 = 0, we have a point ellipse. com Since c \u2264 a the eccentricity is always greater than 1 in the case of an ellipse. xcos a \u2212 ysin a 2 2 5 + xsin. I am not very sure if my solution is correct but I'd rather try and put it up and let people evaluate if it's correct: The ellipse would look something like the below image: Since the ellipse is rotated along Y axis it will form circles(of vary. This video derives the formulas for rotation of axes and shows how to use them to eliminate the xy term from a general second degree polynomial. Major axis is vertical. To rotate the graph of the parabola about the origin, we rotate each point individually. (An ellipse where a = b is in fact a circle. For ellipses not centered at the origin, simply add the coordinates of the center point (e, f) to the calculated (x, y). h is x-koordinate of the center of the ellipse. We shall now study the Cartesian representation of the hyperbola and the ellipse. \ufb00tial Calculus Grinshpan Rotated Ellipse The implicit equation x2 xy +y2 = 3 describes an ellipse. The equation of the ellipse we discussed in class is 9 x2 - 4 xy + 6 y2 = 5. What is the volume of the solid? Step 2: Determine the boundaries of the integral Since the rotation is around the y-axis, the boundaries will be between y = 0 and y = 1 Step 4: Evaluate integrals to find volume Step 1:. Ellipse configuration panel. The higher the value from 0 through 89. This equation defines an ellipse centered at the origin. If $$\\displaystyle D = b^2- 4ac$$, then it's an ellipse for $$\\displaystyle D<0$$, a parabola for $$\\displaystyle D = 0$$, and a hyperbola for $$\\displaystyle D>0$$. However, I could not find anywhere an equation for a spheroid that does not have its axis or revolution along the x,y, or z axis. Rewrite the equation 2x^2+\u221a3 xy+y^2\u22122=0 in a rotated x^\u2032 y^\u2032-system without an x^\u2032 y^\u2032-term. We can apply one more transformation to an ellipse, and that is to rotate its axes by an angle, \u03b8, about the center of the ellipse, or to tilt it. Constructing (Plotting) a Rotated Ellipse. Use the information provided to write the equation of the ellipse in standard form. For the maps we consider, the axes of the distortion ellipse are in the north/south and the east/west directions. Rotation of axis After rotating the coordinate axes through an angle theta, the general second-degree equation in the new x'y'-plane will have the form __________. How to make an ellipse No one knows for sure when the ellipse was discovered, but in 350 BCE the Ancient Greeks knew about the ellipse as a member of the group of two-dimensional geometric figures called conic sections. This equation has vertices at (5, \u20131 \u00b1 4), or (5, 3) and (5, \u20135). the sum of distances of P from F 1 and F 2 in the plane is a constant 2a. Far deirun figlio e cartesian equation of rotated ellipse jackpot. R = distance between axis and rotation mass (ft, m) The moment of all other moments of inertia of an object are calculated from the the sum of the moments. For ellipses not centered at the origin, simply add the coordinates of the center point (e, f) to the calculated (x, y). The parameters of an ellipse are also often given as the semi-major axis, a, and the eccentricity, e, 2 2 1 a b e =-. I was able to find the equation of an ellipse where its major axis is shifted and rotated off of the x,y, or z axis. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). has offshore analysis. (25) Here, \u03c3\u2032 1 is the 1-sigma con\ufb01dence value along the minor axis of the ellipse, and \u03c3\u2032 2 is that along the major axis (\u03c3\u2032 2 \u2265 \u03c3\u2032 1). P(at2, 2at) tangent We shall use the formula for the equation of a straight line with a given gradient, passing through a given point. You will probably has to project the data, create the ellipse and project back to WGS to make it work; I think the snippet provided by Darren does not include rotation (bearing). Rewrite the equation 2x^2+\u221a3 xy+y^2\u22122=0 in a rotated x^\u2032 y^\u2032-system without an x^\u2032 y^\u2032-term. major axis is along y-axis. We can come up with a general equation for an ellipse tilted by \u03b8 by applying the 2-D rotational matrix to the vector (x, y) of coordinates of the ellipse. It has co-vertices at (5 \u00b1 3, \u20131), or (8, \u20131) and (2, \u20131). 4 we saw that the graph is a hyperbola when AC < 0. Vertical Major Axis Example. They are located at (h\u00b1c,k) or (h,k\u00b1c). The ellipse that is most frequently studied in this course has Cartesian equation; where. Note: When tracing feature is ON, shading feature is OFF. A ray of light passing through a focus will pass through the other focus after a single bounce (Hilbert and Cohn-Vossen 1999, p. The objective is to rotate the x and y axes until they are parallel to the axes of the conic. Find the points at which this ellipse crosses the x -axis and show that the tangent lines at these points are parallel. I wish to plot an ellipse by scanline finding the values for y for each value of x. The orthonormality of the axis directions and Equation (1) imply x i = U i (P C). You don't have to use power in contrast to the Columbus' egg. The Steiner ellipse can be extended to higher dimensions with one more point than the dimension. Rotate the hyperbola : In the above graph, the preimage is in blue and the image (rotated) is. Equation 3 thus becomes: rz n S(y, cos & - x, sin &)2 - N Eq. Writing Equations of Ellipses Centered at the Origin in Standard Form. Log InorSign Up. Tracing for Cartesian Equations: ON OFF Video To trace a graph, click on the radio button to the right of the input equation. B is the. The following 12 points are on this ellipse: (p 3;0); (0; p 3); ( p 3; p 3); ( p 3;. Eliminate the parameter : Consider. Two fixed points inside the ellipse, F1 and F2 are called the foci. y axis [see Figs. An ellipse is a circle scaled (squashed) in one direction, so an ellipse centered at the origin with semimajor axis a and semiminor axis b < a has equation. Values between 89. The ellipse points are P = C+ x 0U 0 + x 1U 1 (1) where x 0 e 0 2 + x 1 e 1 2 = 1 (2) If e 0 = e 1, then the ellipse is a circle with center C and radius e 0. Example : Given ellipse : 4 2 (x \u2212 3) 2 + 5 2 y 2 = 1 b 2 X 2 + a 2 Y 2 = 1 a 2 > b 2 i. Elliptic cylinders are also known as cylindroids , but that name is ambiguous, as it can also refer to the Pl\u00fccker conoid. It is important to know the differences in the equations to help quickly identify the type of conic that is represented by a given equation. It's \"just\" twice the rotation: 2 is a regular number so doubles our rotation rate to a full -180 degrees in a unit of time. The Steiner ellipse has the minimal area surrounding a triangle. Earth\u2019s orbit has an eccentricity of less than 0. It has co-vertices at (5 \u00b1 3, \u20131), or (8, \u20131) and (2, \u20131). A ray of light passing through a focus will pass through the other focus after a single bounce (Hilbert and Cohn-Vossen 1999, p. This form of the ellipse has a graph as shown below. The point $$\\left( {h,k} \\right)$$ is called the center of the ellipse. a is the ellipse axis which is parallell to the x-axis when rotation is zero. A circle if A = C 2. Example : Given ellipse : 4 2 (x \u2212 3) 2 + 5 2 y 2 = 1 b 2 X 2 + a 2 Y 2 = 1 a 2 > b 2 i. In other words, from the centre to the points where the bounding box touches the ellipse's line. x\u00bfy\u00bf-system x\u00bf-axis. The equation of such an ellipse we can write in the usual form 2 2 + 2 =1 (1) The slope of the tangent line to this ellipse has evidently the form (Dvo\u0159\u00e1kov\u00e1, et al. ' Draw an ellipse centered at (cx, cy) with dimensions' wid and hgt rotated angle degrees. The Steiner ellipse can be extended to higher dimensions with one more point than the dimension. It is a procedure for drawing an approximation to an ellipse using 4 arc sections, one at each end of the major axes (length a) and one at each end of the minor axes (length b). If the center of the ellipse is at the origin, the equation simplifies to (x 2 /a 2) + (y 2 /b 2)=1. We can apply one more transformation to an ellipse, and that is to rotate its axes by an angle, \u03b8, about the center of the ellipse, or to tilt it. Example of the graph and equation of an ellipse on the. Equation of ellipse from its focus, directrix, and eccentricity Last Updated: 20-12-2018 Given focus(x, y), directrix(ax + by + c) and eccentricity e of an ellipse, the task is to find the equation of ellipse using its focus, directrix, and eccentricity. Then write the equation in standard form. Major axis is horizontal. , 2015) \u00b4= \u2212 2\u221a1\u2212 \ud835\udc652 2 (2) For the tangent point of the line with the slope \ud835\udc54\u221d and our ellipse then holds \ud835\udc47= \u221a1\u2212 \ud835\udc47 2 2 (3) 124. By the way the correct rotation. It has the following form: (x - c\u2081)\u00b2 / a\u00b2 + (y - c\u2082)\u00b2 / b\u00b2 = 1. Thus, the standard equation of an ellipse is x 2 a 2 + y 2 b 2 = 1. Then write the equation in standard form. It has co-vertices at (5 \u00b1 3, \u20131), or (8, \u20131) and (2, \u20131). Consider an ellipse that is located with respect to a Cartesian frame as in figure 3 (a \u2265 b > 0, major axis on x-axis, minor axis on y-axis). If the data is uncorrelated and therefore has zero covariance, the ellipse is not rotated and axis aligned. This gives a surface composed of many \"truncated cones;'' a truncated cone is called a frustum of a cone. Writing Equations of Ellipses Centered at the Origin in Standard Form. Determine the foci and vertices for the ellipse with general equation 4x^2+9y^2-48x+72y+144=0. We have also seen that translating by a curve by a fixed vector ( h , k ) has the effect of replacing x by x \u2212 h and y by y \u2212 k in the equation of the curve. By a suitable choice of coordinate axes, the equation for any conic can be reduced to one of three simple r forms: x 2 / a 2 + y 2 / b 2 = 1, x 2 / a 2 \u2212 y 2 / b 2 = 1, or y 2 = 2px, corresponding to an ellipse, a hyperbola, and a parabola, respectively. If you are given an equation of ellipse in the form of a function whose value is a square root, you may need to simplify it to make it look like the equation of an ellipse. STRAIN ELLIPSE. The eccentricity of an ellipse is e =. 3) Calculate the lengths of the ellipse axes, which are the square root of the eigenvalues of the covariance matrix: A E C R = H L A E C A J R = H Q A O : ? ; 4) Calculate the counter\u2010clockwise rotation (\u03b8) of the ellipse: \u00e0 L 1 2 Tan ? 5 d l 1 = O L A ? P N = P E K p I l 2 T U : \u00ea T ; 6 F : \u00ea U ; 6 p h. GeoGebra Math Apps Get our free online math tools for graphing, geometry, 3D, and more!. Rotation of Axes 3 Coordinate Rotation Formulas If a rectangular xy-coordinate system is rotated through an angle to form an ^xy^- coordinate system, then a point P(x;y) will have coordinates P(^x;y^) in the new system, where (x;y)and(^x;y^) are related byx =^xcos \u2212 y^sin and y =^xsin +^ycos : and x^ = xcos +ysin and ^y = \u2212xsin +ycos : EXAMPLE 1 Show that the graph of the equation xy = 1. Solution The equation of an ellipse usually appears when the plot of the from MECHANICAL MAE351 at Korea Advanced Institute of Science and Technology. Other forms of the equation. Keyword-suggest-tool. It follows that 0 \u00a3e< 1 and p> 0, so that anellipse in polar coordinates with one focus at the origin and the other onthe positive x-axis is given by. and through an angle of 30\u00b0. The equation of the ellipse in the rotated coordinates is. is along the ellipse\u2019s major axis, the correlation matrix is \u03c3\u2032 = \u03c3\u20322 1 0 0 \u03c3\u20322 2. The ellipse may be rotated to a di erent orientation by a 2 2 rotation matrix R= 2 4 cos sin sin cos 3 5 The major axis direction (1;0) is rotated to (cos ;sin ) and the minor axis direction (0;1) is rotated to ( sin ;cos ). 3) Calculate the lengths of the ellipse axes, which are the square root of the eigenvalues of the covariance matrix: A E C R = H L A E C A J R = H Q A O : ? ; 4) Calculate the counter\u2010clockwise rotation (\u03b8) of the ellipse: \u00e0 L 1 2 Tan ? 5 d l 1 = O L A ? P N = P E K p I l 2 T U : \u00ea T ; 6 F : \u00ea U ; 6 p h. The center of the circle used to be at the origin. Approximately sketch the ellipse - the major axis of the ellipse is x-axis. The equation stated is going to have xy terms, and so there needs to be a suitable rotation of axes in order to get the equation in the standard form suitable for the recommended definite integration. The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared. H(x, y) = A x\u00b2 + B xy + C y\u00b2 + D x + E y + F = 0 The basic principle of the incremental line tracing algorithms (I wouldn't call them scanline) is to follow the pixels that fulfill the equation as much as possible. Introduce some delay in function(in ms). How might I go about deriving such an equation. For a plain ellipse the formula is trivial to find: y = Sqrt[b^2 - (b^2 x^2)/a^2] But when the axes of the ellipse are rotated I've never been able to figure out how to compute y (and possibly the extents of x). Let us consider a point P(x, y) lying on the ellipse such that P satisfies the definition i. Therefore, equations (3) satisfy the equation for a non-rotated ellipse, and you can simply plot them for all values of b from 0 to 360 degrees. Learn how each constant and coefficient affects the resulting graph. (iii) is the equation of the rotated ellipse relative to the centre. Differential Equations (10) Discrete Mathematics (4) Discrete Random Variable (5) Disk Washer Cylindrical Shell Integration (2) Division Tricks (1) Domain and Range of a Function (1) Double Integrals (3) Eigenvalues and Eigenvectors (1) Ellipse (1) Empirical and Molecular Formula (2) Enthalpy Change (2) Expected Value Variance Standard. It can be a parametric formulation (e. (1) Ellipse (2) Rotated Ellipse (3) Ellipse Representing Covariance. Rotated Ellipse The implicit equation x2 xy +y2 = 3 describes an ellipse. (1) which is in the form Ax2+ Bxy+ Cy2= 1, with Aand Cpositive. Writing Equations of Rotated Conics in Standard Form Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ into standard form by rotating the axes. Let the coordinates of F 1 and F 2 be (-c, 0) and (c, 0) respectively as shown. (a) Find the points at which this ellipse crosses the x-axis. Major axis is vertical. By using this website, you agree to our Cookie Policy. H(x, y) = A x\u00b2 + B xy + C y\u00b2 + D x + E y + F = 0 The basic principle of the incremental line tracing algorithms (I wouldn't call them scanline) is to follow the pixels that fulfill the equation as much as possible. Ellipse graph from standard equation. Find dy dx. If and are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse. Except for degenerate cases, the general second-degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 x\u00bfy\u00bf-term x\u00bf 2 4 + y\u00bf 1 = 1. asked \u2022 04/02/15 find the equation of the image of the ellipse x^2/4 + y^2/9 when rotated through pi/4 about origin. The Danish author and scientist Piet Hein (1905-1996) dealt with the super ellipse in great detail (book 4). For any ellipse, 0 < e < 1. 1) and (e, f) = (e. Rewrite the equation 2x^2+\u221a3 xy+y^2\u22122=0 in a rotated x^\u2032 y^\u2032-system without an x^\u2032 y^\u2032-term. In other words, the equation of the plane through the center of the circle sloping away from the drawing plane with slope m is given by (3. Solve them for C, D, E. Example : Given ellipse : 4 2 (x \u2212 3) 2 + 5 2 y 2 = 1 b 2 X 2 + a 2 Y 2 = 1 a 2 > b 2 i. For the Earth\u2013sun system, F1 is the position of the sun, F2 is an imaginary point in space, while the Earth follows the path of the ellipse. The equation is (x - h) squared/a squared plus (y - k) squared/a squared equals 1. This makes the analysis somewhat easier. Rotated Parabolas and Ellipse. Rotate the ellipse by applying the equations: RX = X * cos_angle + Y * sin_angle RY = -X * sin_angle + Y * cos_angle. The Steiner ellipse has the minimal area surrounding a triangle. A more general figure has three orthogonal axes of different lengths a, b and c, and can be represented by the equation x 2 /a 2 + y 2 /b 2 + z 2. 3 Major axis, minor axis and rotated angle of a ellipse Find the major axis, minor axis and rotated angle, where the major axis is twice of the longest radius and the minor axis is also twice of the shortest radius. Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle. This ellipse is called the distortion ellipse. \u2212 <: Ellipse \u2212 >: Hyperbola; Firstly, if B is not zero then the graph represents a rotated conic. Consider an ellipse that is located with respect to a Cartesian frame as in figure 3 (a \u2265 b > 0, major axis on x-axis, minor axis on y-axis). In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. How might I go about deriving such an equation. A conic in a rotated coordinate system takes on the form of , where the prime notation represents the rotated axes and associated coefficients. point is expressed as (yi = yi cos O - x, sin 0). none of these. I have the verticles for the major axis: d1(0,0. for the ellipse tting is to transform Eq. and through an angle of 30\u00b0. Activity 4: Determining the general equation of an ellipse/ Determining the foci and vertices of an ellipse. Donate or volunteer today! Site Navigation. An ellipse is a unique figure in astronomy as it is the path of any orbiting body around another. Then: (Canonical equation of an ellipse) A point P=(x,y) is a point of the ellipse if and only if Note that for a = b this is the equation of a circle. I am trying to find an algorithm to derive the 4 angles, from the centre of a rotated ellipse to its extremities. The radii of the ellipse in both directions are then the variances. Solved Examples Q 1: Find out the coordinates of the foci, vertices, lengths of major and minor axes, and the eccentricity of the ellipse 9x 2 + 4y 2 = 36. tive number has a square root. Log InorSign Up. Solve them for C, D, E. Parabola if A or C = 0 therefore AC = 0 B. 4 we saw that the graph is a hyperbola when AC < 0. I first solved the equation of the ellipse for y, getting y= '. The length a always refers to the major axis. Erase the previous Ellipse by drawing the Ellipse at same point using black color. com Since c \u2264 a the eccentricity is always greater than 1 in the case of an ellipse. Because A = 7, and C = 13, you have (for 0 \u03b8 < \u03c0/2) Therefore, the equation in the x'y'-system is derived by making the following substitutions. For a given chord or triangle base, the. (3), we get (x 0+D=(2A 0))2 (F0=A0) + (x0+E=(2C0))2 (F0=C0) = 1; (6). A ray of light passing through a focus will pass through the other focus after a single bounce (Hilbert and Cohn-Vossen 1999, p. Then the equation of the ellipse in this new coordinate system becomes. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). Notes College Algebra teaches you how to find the equation of an ellipse given a graph. (An ellipse where a = b is in fact a circle. First I want to look at the case when , and. The ellipse that is most frequently studied in this course has Cartesian equation; where. Substituting in the values for x and y above, we get an equation for the new coordinates as a function of the old coordinates and the angle of rotation: x' = x \u00d7 cos (\u03b2) - y \u00d7 sin (\u03b2) y' = y \u00d7 cos (\u03b2) + x \u00d7 sin (\u03b2). 4 we saw that the graph is a hyperbola when AC < 0. The major axis is parallel to the X axis. The ellipse can be rotated. A constructional method for drawing an ellipse in drafting and engineering is usually referred to as the \"4 center ellipse\" or the \"4 arc ellipse\". Figure 3: Polarization Ellipse. If an ellipse is rotated about one of its principal axes, a spheroid is the result. Determine the minimum and maximum X and Y limits for the ellipse. At first blush, these are really strange exponents. ) (11 points) The equation x2\u2212xy+y2 = 3 represents a \u201crotated ellipse\u201d\u2014that is, an ellipse whose axes are not parallel to the coordinate axes. y axis [see Figs. is along the ellipse\u2019s major axis, the correlation matrix is \u03c3\u2032 = \u03c3\u20322 1 0 0 \u03c3\u20322 2. A circle in 3D is parameterized by six numbers: two for the orientation of its unit normal vector, one for the radius, and three for the circle center. The equation of the ellipse we discussed in class is 9 x2 - 4 xy + 6 y2 = 5. the sum of distances of P from F 1 and F 2 in the plane is a constant 2a. Log InorSign Up. I know that i can draw it using ellipse equation, then rotate it, compute points and connect with lines. Erase the previous Ellipse by drawing the Ellipse at same point using black color. HELP?! Rotate the axes to eliminate the xy-term in the equation. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). Once we have those we can sketch in the ellipse. Center of ellipse will be the mid point of first and second point always. Accordingly, we can find the equation for any ellipse by applying rotations and translations to the standard equation of an ellipse. General Equation. To rotate the graph of the parabola about the origin, we rotate each point individually. When a>b, we have a prolate spheroid, that is, an ellipse rotated around its major axis; when a K > -1 = ellipse, -1 = parabolic, and K < -1 is hyperbolic; R is the radius of curvature. Consider an ellipse that is located with respect to a Cartesian frame as in figure 3 (a \u2265 b > 0, major axis on x-axis, minor axis on y-axis). If $$\\displaystyle D = b^2- 4ac$$, then it's an ellipse for $$\\displaystyle D<0$$, a parabola for $$\\displaystyle D = 0$$, and a hyperbola for $$\\displaystyle D>0$$. Apply a square completion method to Eq. Here is a short (and probably inaccurate, because I don't really understand the math) explanation for how it works. Rotate the ellipse by applying the equations: RX = X * cos_angle + Y * sin_angle RY = -X * sin_angle + Y * cos_angle. Matrix transformations are affine and map a point such as that to the expected point on the rotated ellipse, but these transformations don't work like that. For the Earth\u2013sun system, F1 is the position of the sun, F2 is an imaginary point in space, while the Earth follows the path of the ellipse. Then the equation of the ellipse in this new coordinate system becomes. We have also seen that translating by a curve by a fixed vector ( h , k ) has the effect of replacing x by x \u2212 h and y by y \u2212 k in the equation of the curve. Draw the Ellipse at calculated point using white color. To verify, here is a manipulate, which plots the original -3. Question 625240: If the ellipse defined by the equation 16x^2+4y^2+96x-8y+84=0 is translated 6 units down and 7 units to the left, write the standard equation of the resulting ellipse Answer by Edwin McCravy(18045) (Show Source):. phi is the rotation angle. Two fixed points inside the ellipse, F1 and F2 are called the foci. 829648*x*y - 196494 == 0 as ContourPlot then plots the standard ellipse equation when rotated, which is. Expanding the binomial squares and collecting like terms gives. If it is rotated about the major axis, the spheroid is prolate, while rotation about the minor axis makes it oblate. If C\u2206 > 0, we have an imaginary ellipse, and if \u2206 = 0, we have a point ellipse. The rotated axes are denoted as the x\u2032 axis and the y\u2032 axis. Rotation Creates the ellipse by appearing to rotate a circle about the first axis. Therefore, equations (3) satisfy the equation for a non-rotated ellipse, and you can simply plot them for all values of b from 0 to 360 degrees. A Rotated Ellipse In this handout I have used Mathematica to do the plots. the sum of distances of P from F 1 and F 2 in the plane is a constant 2a. The equations of tangent and normal to the ellipse\\frac{{{x^2}}}{{{a^2}}} + \\frac{{{y^2}}}{{{b^2}}} = 1$$at the point$$\\left( {{x_1},{y_1}} \\right)$$are$$\\frac. Repeat from Step 1. Let x' and y' be the new set of axes along the principal axes of the ellipse. lationship between two images is pure rotation, i. And for a hyperbola it is: x 2 a 2 \u2212 y 2 b 2 = 1. Because A = 7, and C = 13, you have (for 0 \u03b8 < \u03c0/2) Therefore, the equation in the x'y'-system is derived by making the following substitutions. The major axis of this ellipse is horizontal and is the red segment from (-2, 0) to (2, 0). The amount of correlation can be interpreted by how thin the ellipse is. In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. Here is a reference to plotting an ellipse, without rotation of the major axis from the horizontal: Ellipse in a chart. Here a > b > 0. Sketch the graph of Solution. He provided me with some equations that I combined with a neat ellipse display program written by Wayne Landsman for the NASA Goddard IDL program library to come up with a \"center of mass\" ellipse fitting program, named Fit_Ellipse. Put ellipse equation in. Thus an ellipse may be drawn using two thumbtacks and a string. The velocity equation for a hyperbolic trajectory has either + , or it is the same with the convention that in that case a is negative. E, qua, euros' Sale, per forza di guerra. The Rotated Ellipsoid June 2, 2017 Page 1 Rotated Ellipsoid An ellipse has 2D geometry and an ellipsoid has 3D geometry. Moment of inertia is defined with respect to a specific rotation axis. A more general figure has three orthogonal axes of different lengths a, b and c, and can be represented by the equation x 2 /a 2 + y 2 /b 2 + z 2. Cartesian Equations of the ellipse and hyperbola. x2 a2 + y2 b2 = 1. Ellipse drawing tool. If the center of the ellipse is at point (h, k) and the major and minor axes have lengths of 2a and 2b respectively, the standard equation is. The chord perpendicular to the major axis at the center is the minor axis. If \u03c3 6= \u03c4 the set Q is a hyperbola when F 6= 0. The longer axis, a, is called the semi-major axis and the shorter, b, is called the semi-minor axis. This video derives the formulas for rotation of axes and shows how to use them to eliminate the xy term from a general second degree polynomial. Applying the methods of Equation of a Transformed Ellipsenow leads to the following equation for a standard ellipse which has been rotated through an angle \u03b1. Standard Equations of Ellipse. (Since I wasn't asked for the length of the minor axis or the location of the co-vertices, I don't need the value of b itself. Im making a small sample, kinda like line rider except with less functionablilty and an ellipse. A constructional method for drawing an ellipse in drafting and engineering is usually referred to as the \"4 center ellipse\" or the \"4 arc ellipse\". 3) Calculate the lengths of the ellipse axes, which are the square root of the eigenvalues of the covariance matrix: A E C R = H L A E C A J R = H Q A O : ? ; 4) Calculate the counter\u2010clockwise rotation (\u03b8) of the ellipse: \u00e0 L 1 2 Tan ? 5 d l 1 = O L A ? P N = P E K p I l 2 T U : \u00ea T ; 6 F : \u00ea U ; 6 p h. +- hat j b. -The equation x2 \u2212 xy + y2 = 3 represents a \"rotated\" ellipse, which means the axes of the ellipse are not parallel to the coordinate axes (feel free to graph the ellipse on wolframalpha to get a picture). This way we only draw one object (instead of a thousand) and x and y are now the arrays of all of these points (or coordinates) for the ellipse. If it is rotated about the major axis, the spheroid is prolate, while rotation about the minor axis makes it oblate. Express the equation in the standard form of a conic section. Several exam. This gives a surface composed of many \"truncated cones;'' a truncated cone is called a frustum of a cone. In this section, we will discuss the equation of a conic section which is rotated by. Given an ellipse on the coordinate plane, Sal finds its standard equation, which is an equation in the form (x-h)\u00b2/a\u00b2+(y-k)\u00b2/b\u00b2=1. The objective is to rotate the x and y axes until they are parallel to the axes of the conic. This video derives the formulas for rotation of axes and shows how to use them to eliminate the xy term from a general second degree polynomial. Here is a simple calculator to solve ellipse equation and calculate the elliptical co-ordinates such as center, foci, vertices, eccentricity and area and axis lengths such as Major, Semi Major and Minor, Semi Minor axis lengths from the given ellipse expression. Question 625240: If the ellipse defined by the equation 16x^2+4y^2+96x-8y+84=0 is translated 6 units down and 7 units to the left, write the standard equation of the resulting ellipse Answer by Edwin McCravy(18045) (Show Source):. The following equation on the polar coordinates (r, \u03b8) describes a general ellipse with semidiameters a and b, centered at a point (r 0, \u03b8 0), with the a axis rotated by \u03c6 relative to the polar axis:. 5 (a) with the foci on the x-axis. The distance between the center and either focus is c, where c 2 = a 2 - b 2. ; If and are equal and nonzero and have the same sign, then the graph may be a circle. An ellipse is a circle scaled (squashed) in one direction, so an ellipse centered at the origin with semimajor axis a and semiminor axis b < a has equation. Since (\u2212 6 3) 2 \u2212 4 \u22c5 7 \u22c5 13 = \u2212 346 < 0 and A \u2260 C since 7 \u2260 13, the equation satisfies the conditions to be an ellipse. 1) and we are back to equations (2). Here is a reference to plotting an ellipse, without rotation of the major axis from the horizontal: Ellipse in a chart. First, notice that the equation of the parabola y = x^2 can be parametrized by x = t, y = t^2, as t goes from -infinity to infinity; or, as a column vector, [x] = [t] [y] = [t^2].", "date": "2020-11-25 04:59:59", "meta": {"domain": "inognicasa.it", "url": "http://inognicasa.it/equation-of-a-rotated-ellipse.html", "openwebmath_score": 0.8127440810203552, "openwebmath_perplexity": 457.9006674781333, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9859363737832231, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8566079312561694}}
{"url": "https://math.stackexchange.com/questions/1425468/what-is-the-chance-of-a-sample-in-white-noise-to-be-a-local-extremum", "text": "# What is the chance of a sample in white noise to be a local extremum?\n\nI draw a long sequence of values from a uniform or normal distribution. Observing the resulting sequence, what is the chance of any point in the sequence (aside from the first and last where it is not defined) to be a local extremum? I.e. either a local maximum (the point is larger than both its neighbors) or a local minimum (the point is smaller than both its neighbors).\n\nAt first glance, you would think that once you've drawn the 1st point, the expectancy of the 2nd to be smaller\\bigger than the 1st is 50% and then the 3rd is bigger\\smaller in 50% hence 0.5*0.5 + 0.5*0.5 = 0.5 chance. However there is additional information in the 2nd point being larger than the 1st and hence a bigger (than 50%) chance that the 3rd point is also smaller than the 2nd.\n\nA matlab code which calculates these chances result a chance of 2/3 (0.66%) :\n\nsum(diff(sign(diff(normrnd(0,1,1,10000))))~=0)/10000 % Normal distribution\nsum(diff(sign(diff(rand(1,10000))))~=0)/10000 % Uniform distribution\n\n\nIv'e tried to apply a combinatoric reasoning to explain this result : As being extremum is defined by 3 points only, let's narrow the problem to observe only 3 points. There are 6 possible orderings to 3 points, in 4 of which the second point is a local extremum, hence the 2/3 chance.\n\nIs this a correct reasoning? Are there other better ways to approach the problem (Integrating the probability density function)?\n\n\u2022 By probability you here mean the expectancy of the number of local extrema relative to the number of points in each realization, correct? Sep 7 '15 at 18:14\n\u2022 Indeed. The expected proportion of points which are local extrema from all the points in an instance of a randomized sequence. Sep 7 '15 at 19:23\n\u2022 You have the right answer and the right reasoning. Another way to look at things: if you have a cube with the position in the cube dictated by the uniform distribution, then its individual coords are uniformly distributed. The planes that satisfy the equations $x=y$, $y=z$ and $x=z$ cut the cube into six equal parts. Two of them satisfy your condition. This uses the fact that distribution is irrelevant, as pointed out by Ian and A. Donda. Sep 7 '15 at 20:33\n\nI have only a partial theoretical answer, plus another numeric simulation.\n\nAs pointed out in a comment to Ian's answer, the solution does not depend on the distribution of the data (as long as it is continuous), so I'm simply using uniformly distributed data on [0, 1].\n\nFirst, I can confirm that in simulations the expected frequency of local extrema appears to not depend on the length $n$ of sequences, and appears to be exactly 2/3. Here is Matlab code to demonstrate this:\n\nN = 1e6; % number of realizations\nn = 100; % length of sequence\next = diff(sign(diff(rand(n, N)))) ~= 0; % extrema\nmean(mean(ext))  % relative frequency across points, estimated expectancy across realizations\n\n\nFirst, why exactly 2/3? As the original poster pointed out, the answer of course depends on the probability of occurrences of different orderings, for simplicity for $n = 3$. There are six such orderings: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because the underlying three random variables are independent and identically distributed, any permutation of the three does not change their joint distribution. The probability distribution across orderings therefore has to be invariant under these permutations. This means this distribution must be uniform, and we therefore have a probability of 1/6 for each ordering.\n\nSince four of the orderings lead to a local extremum, namely those with a 1 or a 3 in the middle position, we have a probability of 4/6 = 2/3 for a local extremum.\n\nAs pointed out by Ian, the same probability does not necessarily apply to the situation with $n > 3$. If there is a local maximum at one point, there cannot be a local maximum at each of the neighboring points, i.e. we have a negative local correlation w.r.t. the occurrence of a local maximum. It is however possible that there is a local minimum at the next position. I do not see a way to sort out these correlations theoretically, but I looked at them in the simulated data:\n\nR = corr(ext');\n\n\nThe correlation matrix looks like this:\n\nIt appears that the autocorrelation is stationary. Averaging across diagonals we can extract the autocorrelation function\n\nlags = -(n - 3) : n - 3;\nr = nan(size(lags));\nfor k = 1 : numel(lags)\nr(k) = mean(diag(R, lags(k)));\nend\n\n\nwith this result:\n\nAccording to this we do have a small negative autocorrelation (-0.125) at lag 1, but an even smaller positive autocorrelation (0.025) at lag 2. Beyond that, correlations are smaller than $\\pm$0.001. So my only guess is that the negative autocorrelation at lag 1 is somehow perfectly counterbalanced by the positive autocorrelation at lag 2 so that the average probability of occurrence of a local extremum stays at 2/3.\n\nGeneralization of the order-statistics approach: Just as we enumerated the possible orderings for $n = 3$ above, we can do so also for larger $n$ computationally, and then determine the probability of a local extremum under the assumption (following the same argument as above) that all orderings are equally possible. Here is an implementation in Matlab which uses the Symbolic Math Toolbox to perform exact arithmetic on rational numbers:\n\nfor n = 3 : 10\nuo = perms(1 : n);\next = diff(sign(diff(uo'))) ~= 0;\nratio = sprintf('%d / %d', sum(sum(ext)), numel(ext));\nfprintf('%d :  %s =  %s\\n', n, ratio, char(sym(ratio)))\nend\n\n\nThe result is\n\n3 :  4 / 6 =  2/3\n4 :  32 / 48 =  2/3\n5 :  240 / 360 =  2/3\n6 :  1920 / 2880 =  2/3\n7 :  16800 / 25200 =  2/3\n8 :  161280 / 241920 =  2/3\n9 :  1693440 / 2540160 =  2/3\n10 :  19353600 / 29030400 =  2/3\n\n\nThe first number in the ratios is the number of local extrema, across all $n - 2$ positions and all $n!$ possible orderings, the second is the number of possible local extrema, $(n - 2) ~ n!$.\n\nSo now we have the computational but exact proof that the probability is exactly 2/3 for $n$ up to 10. Beyond that, the number of possible orderings becomes so large that it takes very long to evaluate.\n\nI have the feeling this could lead to a proof by induction for arbitrary $n$, but do not know yet how to do so.\n\nSome thoughts here:\n\nConsider a sequence of $N$ iid random variables $X_i$. We have $N-2$ Bernoulli random variables $\\{ B_i \\}_{i=2}^{N-1}$ which are $1$ when $X_i$ is an extremum and $0$ otherwise. It is not immediately obvious that the $B_i$ are even independent; certainly the version where $B_i$ is $1$ when $X_i$ is a maximum are not independent! So there may be some nontrivial coupling to be considered here.\n\nWe can remove this coupling (if it exists) by restricting attention to the case $N=3$ for a moment. Denote the PDF of the variables separately by $f$. Then the probability that $B_2=1$ is\n\n$$\\int_{-\\infty}^\\infty \\int_{-\\infty}^\\infty \\int_{\\max \\{ x,z \\}}^\\infty f(x) f(y) f(z) dy dx dz + \\int_{-\\infty}^\\infty \\int_{-\\infty}^\\infty \\int_{-\\infty}^{\\min \\{ x,z \\}} f(x) f(y) f(z) dy dx dz.$$\n\nThis can be simplified by splitting over which of $x,z$ is the maximum/minimum. In the uniform case it is straightforward to compute analytically.\n\nIn general this probability is not independent of the distribution used. For example, in the Bernoulli case, this quantity is $p^2(1-p)+p^2(1-p)=p-p^2$, so it is larger for $p$ closer to $1/2$. So while I think you are right that it is $2/3$ in the uniform case (I seem to recall a geometric proof that each of the $n!$ \"ordered tetrahedra\" of the $n$-cube have equal probability under the uniform distribution), I would expect it to be different in general.\n\n\u2022 The property of being a local extremum only depends on the ordering of data. It is therefore invariant under strictly increasing transformations of the data, which means every continuous distribution can be transformed into the uniform distribution without changing extrema. I therefore believe that the probability in question should not depend on the distribution. Sep 7 '15 at 18:16\n\u2022 @A.Donda Indeed you're right; concretely, if $X$ is continuously distributed with CDF $F$, then $F$, considered with codomain $(0,1)$, is invertible. Moreover $F^{-1}$ is an increasing function $F^{-1}(X)$ is $U(0,1)$ distributed. This will also work if $X$ is not continuous but rather merely has no discrete part. That's an interesting discrepancy, which in view of the central limit theorem is surprising, at least to me.\n\u2013\u00a0Ian\nSep 7 '15 at 18:26\n\u2022 I'd be interested to know what you think of my partial answer. Sep 7 '15 at 19:34\n\u2022 Minor error in my previous comment: $F(X)$ is $U(0,1)$. $F^{-1}(X)$ needn't make sense.\n\u2013\u00a0Ian\nSep 8 '15 at 15:18", "date": "2022-01-18 00:10:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1425468/what-is-the-chance-of-a-sample-in-white-noise-to-be-a-local-extremum", "openwebmath_score": 0.8878214359283447, "openwebmath_perplexity": 299.72183470168204, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363754361599, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.8566079192974502}}
{"url": "https://math.stackexchange.com/questions/3340314/what-does-the-math-symbol-propto-mean", "text": "# What does the math symbol $\\propto$ mean?\n\nI came across this symbol in my engineering class and I have never seen it before. Anyone know this?\n\n\u2022 You know, you can always raise your hand and ask :) Likely others in the class were also not familiar with it. \u2013\u00a0Jair Taylor Aug 31 '19 at 18:32\n\u2022 Well.. More like reading it in my book and came across it :) \u2013\u00a0C. K. Aug 31 '19 at 18:42\n\nIt typically means proportional to. Such that\n\nIf $$y=cx$$ for some constant $$c$$ we say\n\n$$y\\propto x$$ so that when x grows, y grows proportionally by the ratio $$c$$\n\nAlternatively inverse proportionality is when\n\n$$y=c\\frac{1}{x}$$ so that when x gets smaller, y gets bigger proportionally by $$c$$\n\n$$y\\propto \\frac{1}{x}$$ to my knowledge there isn\u2019t a symbol specifically for inverse proportionality and $$y\\propto \\frac{1}{x}$$ is used instead\n\n\u2022 Is there a symbol for inverse proportionality as well? \u2013\u00a0C. K. Aug 31 '19 at 18:34\n\u2022 No I don\u2019t think so typically $$y\\propto \\frac{1}{x}$$ is used \u2013\u00a0Colin Hicks Aug 31 '19 at 18:43\n\u2022 This solved my question. Thanks \u2013\u00a0C. K. Aug 31 '19 at 18:44\n\nAs others have pointed out, this symbol means \"is proportional to.\" That is, $$a \\propto b$$ means that there is some constant $$C$$ such that $$a = Cb.$$\n\nThat being said, in the interests of \"teaching a man to fish\", it is worth pointing out that it is often possible to \"reverse engineer\" the meaning of specific mathematical symbols using DeTeXify. In this case, drawing the mysterious symbol gives\n\nEither \\propto or \\varpropto seems to give the correct symbol. TeX often uses \"var\" to indicate a variation of a symbol (for example \\epsilon $$\\epsilon$$ vs \\varepsilon $$\\varepsilon$$; \\theta $$\\theta$$ vs \\vartheta $$\\vartheta$$), so \"propto\" seems like a reasonable guess as to what this symbol should be called. Googling this term gives a large number of results, many of which are relevant.\n\n\u2022 This is awesome! I'll use this in the future if I come across more mystic symbols! \u2013\u00a0C. K. Aug 31 '19 at 18:45\n\u2022 Glad I could be of use. Honestly, DeTeXify is one of my favorite things on the interwebs. There is also a free-to-use desktop version (you can pay something like 8 bucks to remove shareware reminders, if you desire---I think it is worth it to support a snazzy bit of hardware). \u2013\u00a0Xander Henderson Aug 31 '19 at 18:48\n\n$$\\propto$$ means \"proportional to.\"\n\nThis symbol is often used, primarily in physics, to denote direct proportionality. so the RHS and LHS of your expression only differ by some scalar multiple.", "date": "2020-08-12 01:35:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3340314/what-does-the-math-symbol-propto-mean", "openwebmath_score": 0.8876305818557739, "openwebmath_perplexity": 700.2627493610719, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363723369031, "lm_q2_score": 0.8688267745399466, "lm_q1q2_score": 0.8566079182790873}}
{"url": "https://www.physicsforums.com/threads/finding-eigenvectors-of-1-1-1-1-1-1-1-1-1.621052/", "text": "# Homework Help: Finding eigenvectors of [[1,-1,-1],[-1,1,-1],[-1,-1,1]]\n\n1. Jul 15, 2012\n\n### thetrystero\n\nhe eigenvalues of the 3x3 matrix [[1,-1,-1],[-1,1,-1],[-1,-1,1]] are 2,2, and -1.\nhow can i compute the eigenvectors?\nfor the case lambda=2, for example, i end up with an augmented matrix [[-1,-1,-1,0],[-1,-1,-1,0],[-1,-1,-1,0]] so i'm stuck at this point.\n\nmuch appreciated.\n\n2. Jul 15, 2012\n\n### Ray Vickson\n\nSo, you need to solve the linear system\n$$x_1 + x_2 + x_3 = 0\\\\ x_1 + x_2 + x_3 = 0\\\\ x_1 + x_2 + x_3 = 0$$\nThere are lots of solutions. In fact, you should be able to find two linearly independent solution vectors, corresponding to the double eigenvalue 2.\n\nRGV\n\n3. Jul 15, 2012\n\n### HallsofIvy\n\nI prefer to work from the basic definitions (perhaps I just never learned these more sophisticated methods!):\nSaying that 2 is an eigenvalue of this matrix means there exist a non-zero vector such that\n$$\\begin{bmatrix}1 & -1 & -1 \\\\ -1 & 1 & -1 \\\\ -1 & -1 & 1\\end{bmatrix}\\begin{bmatrix}x \\\\ y \\\\ z\\end{bmatrix}x - y- z\\\\ -x+ y- z \\\\ -x- y+ z\\end{bmatrix}= \\begin{bmatrix}2x \\\\ 2y\\\\ 2z\\end{bmatrix}$$\nwhich gives the three equations x- y- z= 2x, -x+ y- z= 2y, -x- y+ z= 2z which are, of course, equivalent to -x- y- z= 0, -x- y- z= 0, -x- y- z= 0. Those three equations are the same. We can, for example, say that z= -x- y so that any vector of the form <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1> is an eigenvector. Notice that the eigenvalue, 2, not only has algebraic multiplicity 2 (it is a double root of the characteristic equation) but has geometric multiplicity 2 (the space of all corresponding eigenvalues is 2 dimensional).\n\nSimilarly, the fact that -1 is an eigenvalue means there are x, y, z, satisfying x- y- z= -x, -x+ y- z= -y, -x- y+ z= -z which are, of course, equivalent to 2x- y- z= 0, -x+ 2y- z= 0, -x- y+ 2z= 0. If we subtract the second equation from the first, we eliminate z to get 3x- 3y= 0 so y= x. Putting that into the third equation, 2x+ 2z= 0 so z= -x.\nAny eigenvector corresponding to eigenvalue -1 is of the form <x, x, -x>= x<1, 1, -1>.\n\n4. Jul 15, 2012\n\n### thetrystero\n\nby that reasoning, can i not have <1,-1,0> and <-1,1,0> as my two solutions for eigenvalue 2? but wolframalpha says i need to have the case where y=0.\n\nalso, i think the solution for eigenvalue -1 is <1,1,1>\n\n5. Jul 16, 2012\n\n### Ray Vickson\n\nYour two listed vectors (for eigenvalue 2) are just multiples of each other. You need two *linearly independent* eigenvectors, such as <1,-1,0> and <1,0,-1>, or <0,-1,1> and <1,-1/2, -1/2>, etc. There are infinitely many possible pairs of vectors <x1,y1,z1> and <x2,y2,z2> that are linearly independent and satisfy the equation x+y+z=0. Any such pair will do.\n\nRGV\n\nLast edited: Jul 16, 2012\n6. Jul 16, 2012\n\n### thetrystero\n\nyes, i had thought of that, but found it uncomfortable that of all the many possibilities, both my professor and wolframalpha chose the cases y=0 and z=0 as solutions, so i was wondering what made these two special compares to the others.\n\n7. Jul 17, 2012\n\n### Ray Vickson\n\nThere is nothing special about these choices, except for the fact that they both have one component = 0 so are, in a sense, the simplest possible. However, you could equally take x=0 and y=0 or x=0 and z=0.\n\nRGV\n\n8. Jul 17, 2012\n\n### HallsofIvy\n\nIf you have a vector that depends upon parameters, say, <x, y, -x- y> as I have above, then choosing x= 0, y= 1 gives you <0, 1, -1> and choosing x= 1, y= 0 gives <1, 0, -1>. That is, in effect, the same as writing <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1>, showing that any such vector is a linear combination of <1, 0, -1> and <0, 1, -1>.", "date": "2018-09-20 08:25:59", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/finding-eigenvectors-of-1-1-1-1-1-1-1-1-1.621052/", "openwebmath_score": 0.9078712463378906, "openwebmath_perplexity": 2063.505154144845, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363706839658, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8566079134942621}}
{"url": "https://math.stackexchange.com/questions/1741273/how-can-i-know-the-analytic-continuation-exists-in-certain-cases", "text": "# How can I know the analytic continuation exists in certain cases?\n\nAs pointed in Does the analytic continuation always exists? we know it doesn't always exist.\n\nBut: take the $\\Gamma$ function: the first definition everyone meet is the integral one: $$z\\mapsto\\int_{0}^{+\\infty}t^{z-1}e^{-t}\\,dt$$ which defines an holomorphic function on the half plane $\\{\\Re z>0\\}$. Moreover we immediately get the functional equation: $$\\Gamma(z+1)=z\\Gamma(z)\\;,\\;\\;\\;\\forall\\; \\Re z>0.$$ This equation is used to extend the function on the whole complex plane (minus the negative integers)... but: WHY CAN WE DO THIS?!\n\nWe know that there is an holomorphic function $\\Gamma$ which can be expressed as the integral on that half plane. Why are we allowed to write $$\\Gamma\\left(\\frac12\\right)=-\\frac12\\Gamma\\left(-\\frac12\\right)$$ for example? LHS is defined, RHS, NOT!!! But where's the problem? Simply let's define $\\Gamma\\left(-\\frac12\\right)$ in such a way... but why can we do this? How can I know that this function I named $\\Gamma$ which is holomorphic on the above half plane admits an extension?\n\n\u2022 You are just using the same name for two differenti objects. \u2013\u00a0N74 Apr 13 '16 at 20:11\n\nOnce you have the functional equation for $\\Gamma$, we can define a new function $\\tilde \\Gamma$ defined on the half-plane $\\operatorname{Re} z > -1$ (except $z=0$) by $$\\tilde \\Gamma(z) = \\frac1z \\Gamma(z+1).$$ It's clear that $\\tilde \\Gamma$ is holomorphic on $\\{ \\operatorname{Re} z > -1 \\} \\setminus \\{ 0 \\}$, and coincides with $\\Gamma$ on $\\operatorname{Re} z > 0$ (because of the functional equation). In other words, $\\tilde \\Gamma$ is an analytic continuation of $\\Gamma$ to $\\{ \\operatorname{Re} z > -1 \\} \\setminus \\{ 0 \\}$. So we may as well call $\\tilde \\Gamma$ by $\\Gamma$.\n\nRepeating the above construction, we can define a \"new $\\Gamma$-function\" on successively larger sets until we get something defined and holomorphic on the whole complex plane except the non-positive integers.\n\n\u2022 I don't agree. you are totally obfuscating the process of analytic continuation this way : what about the analytic continuation on a path of intersecting disks ? how do you know that the analytic continuation doesn't depend on the path (the fact that $\\Gamma(s)$ is meromorphic ?) \u2013\u00a0reuns Apr 13 '16 at 21:47\n\nThe following are citations from the classic Applied and Computational Complex Analysis, Vol. I by P. Henrici. The chapter 3: Analytic Continuation provides a thorough treatise of the theme. Here we look at two aspects, which should help to clarify the situation.\n\nAt first we take a look when two functions $f(z)$ and $g(z)$ are analytic continuations from each other.\n\nTheorem 3.2d: (Fundamental lemma on analytic continuation)\n\nLet $Q$ be a set with point of accumulation $q$ and let $R$ and $S$ be two regions such that their intersection contains $Q$ and $q$ is connected. If $f$ is analytic on $R$, $g$ is analytic on $S$, and $f(z)=g(z)$ for $z\\in Q$, then $f(z)=g(z)$ throughout $R\\cap S$ and $f$ and $g$ are analytic continuations of each other.\n\nWe observe, we need at least a set $Q$ with an accumulation point where analytic functions $f$ and $g$ have to coincide. This set is part of the intersection of two regions $R$ and $S$ where $f$ and $g$ are defined. Finally we conclude that throughout $R\\cap S$ the functions coincide.\n\nThe second aspect sheds some light at functional relationships in connection with analytic continuation. We can read in\n\nSection 3.2.5: Analytic Continuation by Exploiting Functional Relationships\n\nOccasionally an analytic continuation of a function $f$ can be obtained by making use of a special functional relationship satisfied by $f$. Naturally this method is restricted to those functions for which such relationships are known.\n\nHe continues with example 15 which seems that P. Henrici had precisely a user with OPs question in mind.\n\nExample 15:\n\nLet the function $g$ possess the following properties:\n\n\u2022 (a) $g$ is analytic in the right half-plane: $R:\\Re (z)>0$\n\n\u2022 (b) For all $z\\in R, zg(z)=g(z+1)$\n\nWe assert that $g$ can be continued analytically into the whole complex plane with the exception of the points $z=0,-1,-2,\\ldots$.\n\nWe first continue $g$ into $S:\\Re (z)>-1,z\\neq 0$. For $z\\in S$,let $f$ be defined by \\begin{align*} f(z):=\\frac{1}{z}g(z+1) \\end{align*} For $z\\in S, \\Re(z+1)>0$. Hence by virtue of (a) $f$ is analytic on $S$. In view of (b) $f$ agrees with $g$ on the set of $R$. Since $S$ is a region, $f$ represents the analytic continuation of $g$ from $R$ to $S$. We note that $f$ satisfies the functional relation $f(z+1)=zf(z)$ on the whole set $S$.\n\nDenoting the extended function again by $g$, we may use the same method to continue $g$ analytically into the set $\\Re(z)>-2,z\\neq 0,-1$, and thus step by step into the region $z\\neq 0,-1,-2,\\ldots$.\n\nOf course this example addresses the Gamma Function $\\Gamma(z)$ which is treated in detail in chapter 8, vol. 2. He then continues with further methods of analytic continuation, such as the principle of continuous continuation and the symmetry principle.", "date": "2020-10-24 04:05:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1741273/how-can-i-know-the-analytic-continuation-exists-in-certain-cases", "openwebmath_score": 0.924250602722168, "openwebmath_perplexity": 156.7815633080759, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363733699888, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8566079108048874}}
{"url": "http://math.stackexchange.com/questions/158651/orthonormal-basis", "text": "# Orthonormal basis\n\nConsider $\\mathbb{R}^3$ together with inner product $\\langle (x_1, x_2, x_3), (y_1, y_2, y_3) \\rangle = 2x_1 y_1+x_2 y_2+3 x_3 y_3$. Use the Gram-Schmidt procedure to find an orthonormal basis for $W=\\text{span} \\left\\{(-1, 1, 0), (-1, 1, 2) \\right\\}$.\n\nI don't get how the inner product $\\langle (x_1, x_2, x_3), (y_1, y_2, y_3) \\rangle = 2 x_1 y_1+x_2 y_2+3 x_3 y_3$ would affect the approach to solve this question.. When I did the gram-schmidt, I got $v_1=(-1, 1, 0)$ and $v_2=(0, 0, 2)$ but then realized that you have to do something with the inner product before finding the orthonormal basis. Can someone please help me?\n\nUpdate: So far I got $\\{(\\frac{-1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}, 0), (0, 0, \\frac{2}{\\sqrt{12}})\\}$ as my orthonormal basis but I'm not sure if I am doing it right with the given inner product.\n\n-\nplease try to learn TeX syntax. I did it for you in this case. \u2013\u00a0 Siminore Jun 15 '12 at 13:16\n@Alice You got the same solution I got :) To convince yourself it's right, just compute the pairwise inner products between your solution vectors, and you will find they are orthnormal! \u2013\u00a0 rschwieb Jun 15 '12 at 13:29\n\nThe choice of inner product defines the notion of orthogonality.\n\nThe usual notion of being \"perpendicular\" depends on the notion of \"angle\" which turns out to depend on the notion of \"dot product\".\n\nIf you change the way we measure the \"dot product\" to give a more general inner product then we change what we mean by \"angle\", and so have a new notion of being \"perpendicular\", which in general we call orthogonality.\n\nSo when you apply the Gram-Schmidt procedure to these vectors you will NOT necessarily get vectors that are perpendicular in the usual sense (their dot product might not be $0$).\n\nLet's apply the procedure.\n\nIt says that to get an orthogonal basis we start with one of the vectors, say $u_1 = (-1,1,0)$ as the first element of our new basis.\n\nThen we do the following calculation to get the second vector in our new basis:\n\n$u_2 = v_2 - \\frac{\\langle v_2, u_1\\rangle}{\\langle u_1, u_1\\rangle} u_1$\n\nwhere $v_2 = (-1,1,2)$.\n\nNow $\\langle v_2, u_1\\rangle = 3$ and $\\langle u_1, u_1\\rangle = 3$ so that we are given:\n\n$u_2 = v_2 - u_1 = (0,0,2)$.\n\nSo your basis is correct. Let's check that these vectors are indeed orthogonal. Remember, this is with respect to our new inner product. We find that:\n\n$\\langle u_1, u_2\\rangle = 3(-1)(0) + (1)(0) + 2(0)(2) = 0$\n\n(here we also happened to get a basis that is perpendicular in the traditional sense, this was lucky).\n\nNow is the basis orthonormal? (in other words, are these unit vectors?). No they arent, so to get an orthonormal basis we must divide each by its length. Now this is not the length in the usual sense of the word, because yet again this is something that depends on the inner product you use. The usual Pythagorean way of finding the length of a vector is:\n\n$||x||=\\sqrt{x_1^2 + ... + x_n^2} = \\sqrt{x . x}$\n\nIt is just the square root of the dot product with itself. So with more general inner products we can define a \"length\" via:\n\n$||x|| = \\sqrt{\\langle x,x\\rangle}$.\n\nWith this length we see that:\n\n$||u_1|| = \\sqrt{2(-1)(-1) + (1)(1) + 3(0)(0)} = \\sqrt{3}$\n\n$||u_2|| = \\sqrt{2(0)(0) + (0)(0) + 3(2)(2)} = 2\\sqrt{3}$\n\n(notice how these are different to what you would usually get using the Pythagorean way).\n\nThus an orthonormal basis is given by:\n\n$\\{\\frac{u_1}{||u_1||}, \\frac{u_2}{||u_2||}\\} = \\{(\\frac{-1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}, 0), (0,0,\\frac{1}{\\sqrt{3}})\\}$\n\n-\nThanks so much for the thorough explanation! :) \u2013\u00a0 Alice Jun 15 '12 at 13:36\n\nPerhaps you carried out the Gram-Schmidt algorithm using the ordinary inner product? I think that is the only way you could have gotten through without using the given inner product :)\n\nAnyhow, you need to use the given inner product at each step of the orthonormalization procedure. Changing the inner product will change the output of the algorithm, because different inner products yield different lengths of vectors and report different \"angles\" between vectors.\n\nFor example, when you begin with the first step (normalizing $(-1,1,0)$, you should compute that $\\langle(-1,1,0),(-1,1,0)\\rangle=3$, and so the first vector would be $\\frac{1}{\\sqrt{3}}(-1,1,0)$.\n\n-\nThanks a lot! I understand it now. Oh, I was also wondering if it asks to find the orthogonal projection of (1, 1, 1) onto the same W, I guess we still apply the given inner product? I got (1/3, 1/3, 1) for the projection and I want to make sure I'm doing it right.. \u2013\u00a0 Alice Jun 15 '12 at 13:37\n@alice I got a negative sign on the middle entry when I tried a moment ago... can you check your work to see if that's missing in your computation? \u2013\u00a0 rschwieb Jun 15 '12 at 13:44\nOh, oops, I did forget the negative sign in my calculation. Thanks again :) \u2013\u00a0 Alice Jun 15 '12 at 13:46", "date": "2014-07-28 20:46:31", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/158651/orthonormal-basis", "openwebmath_score": 0.9294528365135193, "openwebmath_perplexity": 242.64000925429696, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985936372130286, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8566079080534459}}
{"url": "https://www.codecogs.com/library/maths/approximation/regression/linear.php", "text": "I have forgotten\nmy Password\n\nOr login with:\n\n\u2022 http://facebook.com/\n\u2022 https://www.google.com/accounts/o8/id\n\u2022 https://me.yahoo.com\nCOST (GBP)\n3.50\n0.00\n0\n\n# Linear\n\nviewed 16538 times and licensed 1105 times\nCalculates the linear regression parameters and evaluates the regression line at arbitrary abscissas\nController: CodeCogs\n\nC++\n\n## Class Linear\n\nLinear regression is a method to best fit a linear equation (straight line) of the form  to a collection of points , where  is the slope and  the intercept on the  axis.\n\nThe algorithm basically requires minimisation of the sum of the squared distance from the data points to the proposed line. This is achieved by calculating the derivative with respect to a and b and setting these to zero.\n\nLet us define the following\n\n\n\n\nThen the slope is\n\nthe intercept on the Y axis\n\n\nBelow you will find the regression graph for a set of arbitrary points, which were also used in the forthcoming example. The regression line, displayed in red, has been calculated using this class.\n\n### Example 1\n\nThe following example displays the slope, Y intercept and regression coefficient for a certain set of 7 points.\n#include <codecogs/maths/approximation/regression/linear.h>\n#include <iostream>\n#include <iomanip>\nusing namespace std;\n\nint main()\n{\ndouble x[7] = { 1.5, 2.4, 3.2, 4.8,  5.0, 7.0,  8.43 };\ndouble y[7] = { 3.5, 5.3, 7.7, 6.2, 11.0, 9.5, 10.27 };\n\nMaths::Regression::Linear A(7, x, y);\n\ncout << \"    Slope = \" << A.getSlope() << endl;\ncout << \"Intercept = \" << A.getIntercept() << endl << endl;\n\ncout << \"Regression coefficient = \" << A.getCoefficient() << endl;\n\ncout << endl << \"Regression line values\" << endl << endl;\nfor (double i = 0.0; i <= 3; i += 0.6)\n{\ncout << \"x = \" << setw(3) << i << \"  y = \" << A.getValue(i);\ncout << endl;\n}\nreturn 0;\n\n}\nOutput:\nSlope = 0.904273\nIntercept = 3.46212\n\nRegression coefficient = 0.808257\n\nRegression line values\n\nx =   0  y = 3.46212\nx = 0.6  y = 4.00469\nx = 1.2  y = 4.54725\nx = 1.8  y = 5.08981\nx = 2.4  y = 5.63238\nx =   3  y = 6.17494\n\n### Authors\n\nLucian Bentea (August 2005)\n##### Source Code\n\nSource code is available when you agree to a GP Licence or buy a Commercial Licence.\n\nNot a member, then Register with CodeCogs. Already a Member, then Login.\n\n## Members of Linear\n\n#### Linear\n\n Linear( int n double* x double* y )[constructor]\nInitializes the class by calculating the slope, intercept and regression coefficient based on the given constructor arguments.\n\n### Note\n\nThe slope should not be infinite.\n n The number of initial points in the arrays x and y x The x-coordinates of points y The y-coordinates of points\n\n#### GetValue\n\n doublegetValue( double x )\n x the abscissa used to evaluate the linear regression function\n\n#### GetCoefficient\n\n doublegetCoefficient( )\nThe regression coefficient indicated how well linear regression fits to the original data. It is an expression of error in the fitting and is defined as:\n\n\n\nThis varies from 0 (no linear trend) to 1 (perfect linear fit). If  and , then r is considered to be equal to 1.\n\n## Linear Once\n\n doubleLinear_once( int n double* x double* y double a )\nThis function implements the Linear class for one off calculations, thereby avoid the need to instantiate the Linear class yourself.\n\n### Example 2\n\nThe following graph fits a straight line to the following values:\nx = 1  y = 0.22\nx = 2  y = 0.04\nx = 3  y = -0.13\nx = 4  y = -0.17\nx = 5  y = -0.04\nx = 6  y = 0.09\nx = 7  y = 0.11\nThere is an error with your graph parameters for Linear_once with options n=7 x=\"1 2 3 4 5 6 7\" y=\"0.22 0.04 -0.13 -0.17 -0.04 0.09 0.11\" a=1:7 .input\n\nError Message:Function Linear_once failed. Ensure that: Invalid C++\n\n### Parameters\n\n n The number of initial points in the arrays x and y x The x-coordinates of points y The y-coordinates of points a The x-coordinate for the output location\n\n### Returns\n\nthe interpolated y-coordinate that corresponds to a.\n##### Source Code\n\nSource code is available when you agree to a GP Licence or buy a Commercial Licence.\n\nNot a member, then Register with CodeCogs. Already a Member, then Login.", "date": "2019-12-07 01:26:06", "meta": {"domain": "codecogs.com", "url": "https://www.codecogs.com/library/maths/approximation/regression/linear.php", "openwebmath_score": 0.3409360647201538, "openwebmath_perplexity": 2206.2878395361117, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462211935646, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.8565847127192638}}
{"url": "http://www.aimath.org/textbooks/beezer/Dsection.html", "text": "Almost every vector space we have encountered has been infinite in size (an exception is Example\u00a0VSS). But some are bigger and richer than others. Dimension, once suitably defined, will be a measure of the size of a vector space, and a useful tool for studying its properties. You probably already have a rough notion of what a mathematical definition of dimension might be --- try to forget these imprecise ideas and go with the new ones given here.\n\n## Dimension\n\nDefinition D (Dimension) Suppose that $V$ is a vector space and $\\set{\\vectorlist{v}{t}}$ is a basis of $V$. Then the dimension of $V$ is defined by $\\dimension{V}=t$. If $V$ has no finite bases, we say $V$ has infinite dimension.\n\nThis is a very simple definition, which belies its power. Grab a basis, any basis, and count up the number of vectors it contains. That's the dimension. However, this simplicity causes a problem. Given a vector space, you and I could each construct different bases --- remember that a vector space might have many bases. And what if your basis and my basis had different sizes? Applying Definition\u00a0D we would arrive at different numbers! With our current knowledge about vector spaces, we would have to say that dimension is not \"well-defined.\" Fortunately, there is a theorem that will correct this problem.\n\nIn a strictly logical progression, the next two theorems would precede the definition of dimension. Many subsequent theorems will trace their lineage back to the following fundamental result.\n\nTheorem SSLD\u00a0(Spanning Sets and Linear Dependence) Suppose that $S=\\set{\\vectorlist{v}{t}}$ is a finite set of vectors which spans the vector space $V$. Then any set of $t+1$ or more vectors from $V$ is linearly dependent.\n\nThe proof just given has some monstrous expressions in it, mostly owing to the double subscripts present. Now is a great opportunity to show the value of a more compact notation. We will rewrite the key steps of the previous proof using summation notation, resulting in a more economical presentation, and even greater insight into the key aspects of the proof. So here is an alternate proof --- study it carefully.\n\n\\begin{proof} {\\bf (Alternate Proof of Theorem SSLD)} We want to prove that any set of $t+1$ or more vectors from $V$ is linearly dependent. So we will begin with a totally arbitrary set of vectors from $V$, $R=\\setparts{\\vect{u}_j}{1\\leq j\\leq m}$, where $m>t$. We will now construct a nontrivial relation of linear dependence on $R$.\n\nEach vector $\\vect{u_j}$, $1\\leq j\\leq m$ can be written as a linear combination of $\\vect{v}_i$, $1\\leq i\\leq t$ since $S$ is a spanning set of $V$. This means there are scalars $a_{ij}$, $1\\leq i\\leq t$, $1\\leq j\\leq m$, so that\n\n\\begin{align*} \\vect{u}_j&=\\sum_{i=1}^{t}a_{ij}\\vect{v_i}&&1\\leq j\\leq m \\end{align*}\n\nNow we form, unmotivated, the homogeneous system of $t$ equations in the $m$ variables, $x_j$, $1\\leq j\\leq m$, where the coefficients are the just-discovered scalars $a_{ij}$,\n\n\\begin{align*} \\sum_{j=1}^{m}a_{ij}x_j=0&&1\\leq i\\leq t \\end{align*}\n\nThis is a homogeneous system with more variables than equations (our hypothesis is expressed as $m>t$), so by Theorem\u00a0HMVEI there are infinitely many solutions. Choose one of these solutions that is not trivial and denote it by $x_j=c_j$, $1\\leq j\\leq m$. As a solution to the homogeneous system, we then have $\\sum_{j=1}^{m}a_{ij}c_{j}=0$ for $1\\leq i\\leq t$. As a collection of nontrivial scalars, $c_j$, $1\\leq j\\leq m$, will provide the nontrivial relation of linear dependence we desire,\n\n\\begin{align*} \\sum_{j=1}^{m}c_{j}\\vect{u}_j &=\\sum_{j=1}^{m}c_{j}\\left(\\sum_{i=1}^{t}a_{ij}\\vect{v_i}\\right) &&\\text{ Definition TSVS}\\\\ &=\\sum_{j=1}^{m}\\sum_{i=1}^{t}c_{j}a_{ij}\\vect{v}_i &&\\text{ Property DVA}\\\\ &=\\sum_{i=1}^{t}\\sum_{j=1}^{m}c_{j}a_{ij}\\vect{v}_i &&\\text{ Property CMCN}\\\\ &=\\sum_{i=1}^{t}\\sum_{j=1}^{m}a_{ij}c_{j}\\vect{v}_i &&\\text{Commutativity in $\\complex{}$}\\\\ &=\\sum_{i=1}^{t}\\left(\\sum_{j=1}^{m}a_{ij}c_{j}\\right)\\vect{v}_i &&\\text{ Property DSA}\\\\ &=\\sum_{i=1}^{t}0\\vect{v}_i &&\\text{$c_j$ as solution}\\\\ &=\\sum_{i=1}^{t}\\zerovector &&\\text{ Theorem ZSSM}\\\\ &=\\zerovector &&\\text{ Property Z} \\end{align*}\n\nThat does it. $R$ has been undeniably shown to be a linearly dependent set. \\end{proof} Notice how the swap of the two summations is so much easier in the third step above, as opposed to all the rearranging and regrouping that takes place in the previous proof. In about half the space. And there are no ellipses ($\\ldots$).\n\nTheorem\u00a0SSLD can be viewed as a generalization of Theorem\u00a0MVSLD. We know that $\\complex{m}$ has a basis with $m$ vectors in it (Theorem\u00a0SUVB), so it is a set of $m$ vectors that spans $\\complex{m}$. By Theorem\u00a0SSLD, any set of more than $m$ vectors from $\\complex{m}$ will be linearly dependent. But this is exactly the conclusion we have in Theorem\u00a0MVSLD. Maybe this is not a total shock, as the proofs of both theorems rely heavily on Theorem\u00a0HMVEI. The beauty of Theorem\u00a0SSLD is that it applies in any vector space. We illustrate the generality of this theorem, and hint at its power, in the next example.\n\nExample LDP4: Linearly dependent set in $P_4$.\n\nTheorem\u00a0SSLD is indeed powerful, but our main purpose in proving it right now was to make sure that our definition of dimension (Definition\u00a0D) is well-defined. Here's the theorem.\n\nTheorem BIS\u00a0(Bases have Identical Sizes) Suppose that $V$ is a vector space with a finite basis $B$ and a second basis $C$. Then $B$ and $C$ have the same size.\n\nTheorem\u00a0BIS tells us that if we find one finite basis in a vector space, then they all have the same size. This (finally) makes Definition\u00a0D unambiguous.\n\n## Dimension of Vector Spaces\n\nWe can now collect the dimension of some common, and not so common, vector spaces. \\begin{theorem}{DCM}{Dimension of $\\complex{m}$}{complex vector space!dimension} The dimension of $\\complex{m}$ (Example\u00a0VSCV) is $m$. \\end{theorem} \\begin{proof} Theorem\u00a0SUVB provides a basis with $m$ vectors. \\end{proof}\n\nTheorem DP\u00a0(Dimension of $P_n$) The dimension of $P_{n}$ (Example\u00a0VSP) is $n+1$.\n\n\\begin{theorem}{DM}{Dimension of $M_{mn}$}{matrix vector space!dimension} The dimension of $M_{mn}$ (Example\u00a0VSM) is $mn$. \\end{theorem} \\begin{proof} Example\u00a0BM provides a basis with $mn$ vectors. \\end{proof}\n\nExample DSM22: Dimension of a subspace of $M_{22}$.\n\nExample DSP4: Dimension of a subspace of $P_4$.\n\nExample DC: Dimension of the crazy vector space.\n\nIt is possible for a vector space to have no finite bases, in which case we say it has infinite dimension. Many of the best examples of this are vector spaces of functions, which lead to constructions like Hilbert spaces. We will focus exclusively on finite-dimensional vector spaces. OK, one infinite-dimensional example, and then we will focus exclusively on finite-dimensional vector spaces.\n\nExample VSPUD: Vector space of polynomials with unbounded degree.\n\n## Rank and Nullity of a Matrix\n\nFor any matrix, we have seen that we can associate several subspaces --- the null space (Theorem\u00a0NSMS), the column space (Theorem\u00a0CSMS), row space (Theorem\u00a0RSMS) and the left null space (Theorem\u00a0LNSMS). As vector spaces, each of these has a dimension, and for the null space and column space, they are important enough to warrant names.\n\nDefinition NOM (Nullity Of a Matrix) Suppose that $A$ is an $m\\times n$ matrix. Then the nullity of $A$ is the dimension of the null space of $A$, $\\nullity{A}=\\dimension{\\nsp{A}}$.\n\nDefinition ROM (Rank Of a Matrix) Suppose that $A$ is an $m\\times n$ matrix. Then the rank of $A$ is the dimension of the column space of $A$, $\\rank{A}=\\dimension{\\csp{A}}$.\n\nExample RNM: Rank and nullity of a matrix.\n\nThere were no accidents or coincidences in the previous example --- with the row-reduced version of a matrix in hand, the rank and nullity are easy to compute.\n\nTheorem CRN\u00a0(Computing Rank and Nullity) Suppose that $A$ is an $m\\times n$ matrix and $B$ is a row-equivalent matrix in reduced row-echelon form with $r$ nonzero rows. Then $\\rank{A}=r$ and $\\nullity{A}=n-r$.\n\nEvery archetype (appendix A) that involves a matrix lists its rank and nullity. You may have noticed as you studied the archetypes that the larger the column space is the smaller the null space is. A simple corollary states this trade-off succinctly. (See technique LC.)\n\nTheorem RPNC\u00a0(Rank Plus Nullity is Columns) Suppose that $A$ is an $m\\times n$ matrix. Then $\\rank{A}+\\nullity{A}=n$.\n\nWhen we first introduced $r$ as our standard notation for the number of nonzero rows in a matrix in reduced row-echelon form you might have thought $r$ stood for \"rows.\" Not really --- it stands for \"rank\"!\n\n## Rank and Nullity of a Nonsingular Matrix\n\nLet's take a look at the rank and nullity of a square matrix.\n\nExample RNSM: Rank and nullity of a square matrix.\n\nThe value of either the nullity or the rank are enough to characterize a nonsingular matrix.\n\nTheorem RNNM\u00a0(Rank and Nullity of a Nonsingular Matrix) Suppose that $A$ is a square matrix of size $n$. The following are equivalent.\n\n1. A is nonsingular.\n2. The rank of $A$ is $n$, $\\rank{A}=n$.\n3. The nullity of $A$ is zero, $\\nullity{A}=0$.\n\nWith a new equivalence for a nonsingular matrix, we can update our list of equivalences (Theorem\u00a0NME5) which now becomes a list requiring double digits to number.\n\nTheorem NME6\u00a0(Nonsingular Matrix Equivalences, Round 6) Suppose that $A$ is a square matrix of size $n$. The following are equivalent.\n\n1. $A$ is nonsingular.\n2. $A$ row-reduces to the identity matrix.\n3. The null space of $A$ contains only the zero vector, $\\nsp{A}=\\set{\\zerovector}$.\n4. The linear system $\\linearsystem{A}{\\vect{b}}$ has a unique solution for every possible choice of $\\vect{b}$.\n5. The columns of $A$ are a linearly independent set.\n6. $A$ is invertible.\n7. The column space of $A$ is $\\complex{n}$, $\\csp{A}=\\complex{n}$.\n8. The columns of $A$ are a basis for $\\complex{n}$.\n9. The rank of $A$ is $n$, $\\rank{A}=n$.\n10. The nullity of $A$ is zero, $\\nullity{A}=0$.", "date": "2013-05-24 03:58:04", "meta": {"domain": "aimath.org", "url": "http://www.aimath.org/textbooks/beezer/Dsection.html", "openwebmath_score": 0.9146049618721008, "openwebmath_perplexity": 325.4569079225906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9879462197739625, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8565847046997394}}
{"url": "https://math.stackexchange.com/questions/1645967/what-do-we-actually-prove-using-induction-theorem/1646299", "text": "# What do we actually prove using induction theorem?\n\nHere is the picture of the page of the book, I am reading:\n\n$$P_k: \\qquad 1+3+5+\\dots+(2k-1)=k^2$$ Now we want to show that this assumption implies that $P_{k+1}$ is also a true statement: $$P_{k+1}: \\qquad 1+3+5+\\dots+(2k-1)+(2k+1)=(k+1)^2.$$ Since we have assumed that $P_k$ is true, we can perform operations on this equation. Note that the left side of $P_{k+1}$ is the left side of $P_k$ plus $(2k+1)$. So we start by adding $(2k+1)$ to both sides of $P_k$: \\begin{align*} 1+3+\\dots+(2k-1) &= k^2 &P_k\\\\ 1+3+\\dots+(2k-1)+(2k+1) &= k^2+(2k+1) &\\text{Add $(2k+1)$ to both sides.} \\end{align*} Factoring the right side of this equation, we have $$1+3+\\dots+(2k-1)+(2k+1) =(k+1)^2 \\qquad P_{k+1}$$ But this last equation is $P_{k+1}$. Thus, we have started with $P_k$, the statement we assumed true, and performed valid operations to produce $P_{k+1}$, the statement we want to be true. In other words, we have shown that if $P_k$ is true, then $P_{k+1}$ is also true. Since both conditions in Theorem 1 are satisfied, $P_n$ is true for all positive integers $n$.\n\nIn the top line there is written that we have to show that 'the assumption $P_k$ is true implies $P_{k+1}$ is true'.\n\nAnd what I think is that: as long as we know that the state of a proposition being true for any positive integer $k$ after Base number implies that proposition is true for integer $k+1$, we have to show that $P_k$ is true. However I don't have an idea yet how to show the truth of $P_k$.\n\nSo, my first question is that who is right? My book or me? And if I am right then how can I show the truth of $P_k$?\n\n2) This may be taken as the second question but it is also bit annoying.\n\nIn the second paragraph it is written, \"Since we have assumed that $P_k$ is true, we can perform operations on it\".\n\nWhy is it necessary for an equation to be true for performing operations on it? Well, this is not much annoying as compared to the first question because whole the induction theory depends upon that.\n\n\u2022 Who is right about what? You haven't made a claim that contradicts the book, Feb 8 '16 at 13:32\n\u2022 I have tried to retype the text from your picture. It would be rather difficult task, considering the low quality of the image. Luckily Google and Google Books returns some places with the same text. Could you perhaps add to your post a more precise reference than \"the book I am reading\"? Feb 8 '16 at 16:27\n\u2022 You prove the $P_1$ is true. We don't know that $P_n$ is true for all $n$. Only that it is true for $n = 1$. Then we prove that IF it is true for a specific $P_k$ then it is true for $P_{k+1}$. Once we prove these two things: 1) $P_1$ is true. and 2) $P_k \\implies P_{k+1}$. Then we can conclude and we have proven 3) $P_n$ is true for all n. Your comment. If we prove it is true for all $P_k$ then it is true for $P_{k+1}$ is irrelevant as we are not assuming and certainly haven't proven it is true for all $P_k$. We have assumed it for ONE $P_k$. Feb 8 '16 at 17:25\n\u2022 @MarkJoshi Induction works on many structures. Induction of the form \"start at zero and add one\" also works on finite fields, that is arithmetic $\\pmod p$. It is not correct at all to say that natural numbers are precisely the set for which induction works. Feb 9 '16 at 6:02\n\u2022 Once there is enough evidence to make you believe something is true, it can suddenly become much easier to understand why it is true. Given the multitude of experts here (and the authoer himself) telling you the book is right I recommend, as strange as it sounds, to reread the chapter on induction trying to believe it (and everyone on this site) is right and understand why that is, rather than trying to argue why it is wrong. Feb 9 '16 at 13:25\n\nInduction is, intuitively, an outline of an infinite proof.\n\nYou first prove $P(0)$, the base case.\n\nThen you prove $P(1)$ follows from $P(0)$.\n\nThen you prove that $P(2)$ follows from $P(1)$.\n\nEt cetera.\n\nIn general, if you know that $P(k+1)$ follows from $P(k)$, and you know that $P(0)$ is true, then you know how to prove $P(n)$ for any natural number. For example, $P(4)$ follow because $P(0)\\implies P(1)\\implies P(2)\\implies P(3)\\implies P(4)$.\n\nSo you don't need to know how to show $P(k)$ to show that $P(k)$ implies $P(k+1)$. Once you've shown this implication, and $P(0)$, we know how to write a non-inductive finite proof of $P(10000)$ or $P(10^{99})$.\n\nYour book is right. You have to show that $$P_k \\implies P_{k+1}$$\n\nBecause if you prove that and since you said it is true for base number, the proposition is automatically proved true for every $n \\in S$.\n\nLet us see how!\n\nTo prove \"$P_k \\implies P_{k+1}$\" means you are proving that a particular proposition has a quality of being true for $k+1$ because of being true for $k$.\n\nImprovement (Edit): Just as you do in case of domino effect. Suppose it is reported that the cornered domino has been pushed over. And then you find out that dominoes are arranged such that whenever $k^{th}$ domino falls, $(k+1)^{th}$ falls. You would at once say that all dominoes must fall.\n\nRemember that these two conditions of induction theorem simultaneously imply that a certain proposition is true.\n\ni) It is true for base number\n\nii) It is true for an integer $k$ implies it is true for integer $k+1$ after it.\n\nFor the proposition in your picture, if you would take base number $2$ then after doing all that (done in your book), you would prove that proposition is true for integer $2$ and integers after it. But since we usually ask to prove that conjecture for every $n \\in \\mathbb N$, we take base number $1$ not $2$ because the smallest number of set $\\mathbb N$ is $1$.\n\n\u2022 It may help to think that you are supposed to prove the arrow in-between the two propositions. Feb 9 '16 at 4:00\n\n(Wanted to post as a comment, but it is too long :/)\n\nWhen we prove something by induction we prove that our claim is correct for a base case (for example, n=1). Afterwards we assume (not proving, only assuming) that our claim stands for some arbitrary value k and than, based on the assumption we prove it holds for k+1.\n\nNote that based on the proof of base case and the proof that if $P_k$ is true we show that $P_{k+1}$ is true. Example: you prove that some claim is correct for $n=1$, then you assume it is correct for some n=k. Now you prove for n=k+1.\n\nWe proved for n=1 and using the assumption that the claim stands for n=k we proved for n=k+1. In this case, we know that the claim stands for n=1, thus we can now say it stands for n=2. Afterwards we can say that it stands for n=3 and so on, hence it stands for any n.\n\nThat is how we use induction - you don't prove that $P_k$ is true, you assume it it true - that is a big difference.\n\nIf my explanation is not coherent enough, you might want to google \"Induction\" and read a little more.\n\nGood luck.\n\n\u2022 \"...that our claim stands for some arbitrary $k$ and then, based on the assumption we prove that claim holds for $k+1$\" . Okay.. if we prove the claim for $k+1$ . At what place in the picture, I can see such prove? Feb 8 '16 at 14:12\n\u2022 All that shown is how they are assuming that $P_k$ is true and how they use it to prove that $P_{k+1}$ is true. Feb 8 '16 at 14:23\n\u2022 After reading this \"$P_k$ holds implies $P_{k+1}$ holds\", I guess that we should show truth of $P_k$ and not of $P_{k+1}$ because it is an implication and once we prove $P_k$, then according to our claim ($P_k \\implies P_{k+1}$), $P_{k+1}$ would get proved automatically. Feb 8 '16 at 14:30\n\u2022 @Man_Of_Wisdom, the induction assumption is that $P_k$ is true, we don't need to prove it. We show that because it is true (take it as a fact), than $P_{k+1}$ is also true. We know that \"$P_k$ holds\", we want to show that $P_{k+1}$ holds. Feb 8 '16 at 14:43\n\u2022 You show the truth of $P_1$. Than you show that if it's true for $P_k$ which may or may not be the same as $P_1$ then it must be true for $P_{k+1}$. From that we know that $P_1 \\implies P_2$. And therefore $P_2 \\implies P_3$. Now we can hire an undergraduate to sit in an office to continue \"$...\\implies P_{2,345}$ and $P_{2,345} \\implies P_{2,346}$ and $P_{2346} \\implies ...$\". Or we can conclude: 1) ($P_1$ is true). and 2) ($P_k \\implies P_{k+1}): \\implies$ 3) $P_n$ is true for all $n$. Feb 8 '16 at 17:31\n\n## What do we actually prove using induction theorem?\n\nProbably an example is needed here. Let's suppose you want to prove the little Gauss. Suppose we want to prove this Theorem for all $$n\\in\\mathbb{N}$$:\n\n$$1+2+3+ \\cdots+ n = \\sum_{k=1}^n k = \\frac{n(n+1)}{2}$$\n\nThe statement $$A(n)$$ is thus given by: $$\\sum_{k=1}^n k = \\frac{n(n+1)}{2}$$ With induction this is really simple. Step one lets suppose $$n=1$$: $$1 = \\sum_{k=1}^1 k = \\frac{1(1+1)}{2} = \\frac{n(n+1)}{2}$$ Done the statement is true for $$n=1$$. The good part about induction is we can select the number for which we want to prove it so we could also have shown with $$n=0$$. So dont try to prove it for $$n$$ beeing any number but prove it for one number you selected, almost everytime $$n=0$$ or $$n=1$$.\n\nNow we have to prove that if the statement is true for $$n$$ it is also true for $$n+1$$. Meaning if $$A(n)$$ is true, $$A(n+1)$$ is also true:\n\n$$\\underbrace{1+ 2+ \\cdots + n}_{\\sum_{k=1}^n k} + n+1 = \\sum_{k=1}^n k + n+1 \\overset{A(n) \\text{ is true}}{=} \\frac{n(n+1)}{2} + n+1 = \\frac{n^2+n+2n+2}{2} = \\frac{n^2+3n+2}{2} = \\frac{(n+1)(n+2)}{2}$$\n\nSo now we know the statement is true for $$n+1$$ or $$A(n+1)$$ is true.\n\n### Prove without induction\n\nProving this statement without induction is not that simple. You can do it like this. Suppose $$n\\in \\mathbb{N}$$, then we can look at the sum and rearange that:\n\n$$n$$ is even\n\n$$1+2+\\cdots + n = (1 + n) + (2+n-1) + \\cdots + \\left(\\frac{n}{2} + \\frac{n}{2} +1\\right)=\\frac{n}{2}(n+1) = \\frac{n(n+1)}{2}$$\n\nand for $$n$$ is odd\n\n$$1+2+\\cdots + n = (1+n) + (2+n-1) + \\cdots + \\left(\\frac{n-1}{2} + \\frac{n+1}{2} + 1\\right) + \\frac{n+1}{2} = (n+1)\\frac{n-1}{2} + \\frac{n+1}{2} = \\frac{n(n+1)}{2}$$\n\n### Conclusion\n\nSo let's conclude, we can use induction here but we can go without it too. The first though is much simpler because it uses the induction and if you get your head around it, it's easier to verify that it is true. The other one is using induction internally for the rearanging and you have to think alot if this is true or if I made a mistake.\n\n## What does the operation part mean?\n\nThe funny thing is that induction can be generalized in a lot of things. If we look at the set $$\\mathbb{N}$$ than we can define it as follows $$1$$ is in $$\\mathbb{N}$$ and if $$n\\in\\mathbb{N}$$ than $$n+1$$ is also in $$\\mathbb{N}$$, meaning lets call that $$(\\star)$$:\n\n1. $$1 \\in \\mathbb{N}$$\n2. $$\\forall n\\in\\mathbb{N} \\Rightarrow n+1\\in\\mathbb{N}$$\n\nIf you look at that you will find this is looking a lot like induction. So basically you can say if you have the statement $$A(n)$$ and you want to show that it is true for all $$n\\in\\mathbb{N}$$ you do the following, you determine the set $$\\mathfrak{A}$$ for which the statement is true. This set is given by: $$\\mathfrak{A} = \\{n\\in\\mathbb{N} | A(n) \\text{ is true}\\}$$ And now if we know $$1\\in\\mathfrak{A}$$ and for all $$n\\in\\mathfrak{A}$$ it follows $$n+1\\in\\mathfrak{A}$$ then it follows from $$(\\star)$$: $$\\mathfrak{A} = \\mathbb{N}$$ so the statement is true for all $$n\\in\\mathbb{N}$$\n\n## More generalized induction shemes\n\nThe example where this is generalized is the \"principle of the good set\" (this is the german name translated directly), it's used in measure theory and works with sets that are way bigger then $$\\mathbb{N}$$ in fact with sets that have the cardinality of $$\\mathbb{R}$$. And it works as follows. A $$\\sigma$$-Algebra $$\\mathcal{A}$$ of a set $$\\Omega$$ fullfills the following axioms:\n\n1. $$\\Omega \\in \\mathcal{A}$$\n2. $$A\\in \\mathcal{A} \\Rightarrow \\Omega\\setminus A\\in\\mathcal{A}$$\n3. $$A_1,A_2,\\cdots \\in \\mathcal{A} \\Rightarrow \\bigcup_{k=1}^\\infty A_k \\in \\mathcal{A}$$\n\nNow we have a function $$\\sigma$$ that generates $$\\sigma$$-Algebras for a given subset $$\\mathcal{G}\\subset \\mathscr{P}(\\Omega)$$ just using the axioms 1-3 just more often than $$\\mathbb{R}$$ has elements. Like $$\\mathbb{N}$$ is generated by $$\\{1\\}$$, if we take $$1$$ and then we add $$1$$ onto it put the result $$2$$ in our set. Then we take $$2$$ and add $$1$$ onto it and put it in our set until we cannot add any more element because we have all of them already. Back to the example, $$\\mathcal{G}$$ is called a generator if $$\\sigma(\\mathcal{G}) = \\mathcal{A}$$, this also means $$\\Omega \\in \\mathcal{G}$$. Now if we want to show a statement $$B(A)$$ is true for all $$A\\in\\mathcal{A}$$ we just show:\n\n1. $$B(G)$$ is true for all $$G\\in \\mathcal{G}$$\n2. $$B(A)$$ is true then it is also true for $$B(\\Omega\\setminus A)$$\n3. $$B(A_1),\\ B(A_2),\\ \\cdots$$ is true than its also true for $$B\\left(\\bigcup_{k=1}^\\infty A_k\\right)$$\n\nSo here 1. shows that we know the statement is true for at least a generating set of $$\\mathcal{A}$$, like in induction we know the statement is true for $$\\{1\\}$$ which generates $$\\mathbb{N}$$ using $$n+1$$. And then we show it for the operations 2. and 3. like we would in induction where we show it for $$n+1$$.\n\n## Conclusion\n\nI hope I didn't confuse you too much. But I am certain that after some time you will accept inductions and learn how to use them. And if you once return here or have to explain it to somebody else, after using it a lot of times. You will fully understand it. People often find induction counter-intuitiv, I have some ideas why but I think that would really go to far. So my hint, accept it first, use it, try to explain it und you will have understood this and similar shemes.\n\nIntuitively you can view this as a chain reaction, because once you show that $\\color{green}{P_0}$ is true, then by applying $\\color{blue}{P_k \\Rightarrow P_{k+1}}$ repeatedly you can reach any number $n \\geq 0$ and see that $P_n$ is indeed true.\n\n$$\\color{green}{P_0} \\color{blue}{\\Rightarrow} P_1 \\color{blue}{\\Rightarrow} \\dots \\color{blue}{\\Rightarrow} P_n \\color{blue}{\\Rightarrow} \\dots .$$\n\nAt no point are you directly proving that $P_1$, $P_2$, $\\dots$ are true, instead they follow from base case and induction step.\n\n\u2022 What is $P_k \\implies P_{k+1}$? I think my question actually asks about this. Feb 8 '16 at 14:17\n\u2022 It is the implication, you are proving $P_{k+1}$ under assumption of $P_{k}$, i.e. you are assuming that $P_{k}$ is true, and you show that then $P_{k+1}$ is true. Once you prove that, you can apply this for case $k=0$ (because you know that $P_0$ is true). By that you immediately get that $P_1$ is true. And you can repeat that...\n\u2013\u00a0Sil\nFeb 8 '16 at 14:20\n\u2022 You must combine it with the base case, in which you proved that $P_0$ is true. Both base case AND induction step then imply that $P_1$ is true. But now you have $P_1$ is true and you can use it same way with induction step, to arrive at $P_2$ being true, etc... Reaching any $n$ if you repeat the process.\n\u2013\u00a0Sil\nFeb 8 '16 at 14:25\n\u2022 And that is exactly what you do, in case of $P_0$. You have $P_0$ and $P_0 \\implies P_1$, therefore $P_1$. The trick is that you ONLY need to show this for $P_0$, everything else follows from the $P_k \\implies P_{k+1}$.\n\u2013\u00a0Sil\nFeb 8 '16 at 14:48\n\u2022 You prove $P_k$ is true when $k = 1$. You do not need to assume that $k \\ne 1$ or that $P_k$ is true for any value of $k$ other than 1. But you don't want to assume that $k$ has any characteristic that pertains only to 1. You want to prove that it is true for some $k$ of which you don't know the value of, then it must also be true of $k + 1$. You know it is true for some $k$ because it is true for $k = 1$. It might also be true for $k = 7,921$ but you don't know that. You can't say I know it is true for $P_1$ so I will prove it for $P_2$ because you want something more general. Feb 8 '16 at 17:39\n\nThe part about assuming $P_k$ is an application of basic rules of propositional logic.\n\nSuppose you have a proposition of the form $A \\implies B$. One way you might be able to prove this proposition is by using the rule for Introduction of an Implication:\n\n1. Assume that $A$ is true.\n2. Using $A$ as a statement already known to be true, prove $B$.\n3. \"Discharge\" the assumption that $A$ is true.\n4. Conclude that $A \\implies B$ is true.\n\nOf course this works only if you are able to come up with a proof of $B$ in the context of Step\u00a02. But while you are in that context, you already know $A$ is true, and you do not have to prove it.\n\nOnce you have proven $A \\implies B$, if you later find out that $A$ actually is true (not just assumed to be true in the limited context of Step 2, above), you can apply a rule called Modus Ponens:\n\n1. Given that $A \\implies B$ is known to be true.\n2. Given that $A$ is known to be true.\n3. Conclude that $B$ is true.\n\nNone of these rules of logic has used anything from the method of induction, but the method of induction uses those rules of logic.\n\nWe prove $P_0$.\n\nWe prove $P_k \\implies P_{k+1}$ for any possible value of $k$, using the rule for Introduction of an Implication.\n\nNow we know $P_0$, and since $P_k \\implies P_{k+1}$ for any $k$, including the case $k=0$, we have $P_0 \\implies P_1$. Apply Modus Ponens:\n\n1. Given that $P_0 \\implies P_1$ is true.\n2. Given that $P_0$ is true.\n3. Conclude that $P_1$ is true.\n\nNow we know $P_1$, and $P_k \\implies P_{k+1}$ is still true in the case $k=1$, so we can apply Modus Ponens again:\n\n1. Given that $P_1 \\implies P_2$ is true.\n2. Given that $P_1$ is true.\n3. Conclude that $P_2$ is true.\n\nWe can repeat this process as many times as needed to show $P_3$, then $P_4$, then $P_5$.\n\nThe principle of induction merely summarizes this long, repetitive, boring process so that we can relatively quickly show $P_n$ for any given non-negative integer $n$. And since we can always do this no matter what non-negative value we choose for $n$, we conclude $P_n$ is true for all $n$.\n\nBy the way, I would write at least one of the sentences in the book a little differently. Instead of \"perform operations on this equation\" I would write \"use algebraic manipulation of $P_k$ to prove that other equations are true\". That is,\n\nSince we have assumed that $P_k$ is true, we can use algebraic manipulation of $P_k$ to prove that other equations are true.\n\nI hope this clears up one of the points of confusion.\n\n\u2022 According to point no. 2 we have to prove that $P_{k+1}$ is true based on the assumption that $P_k$ is true. The author shows us that $$P_{k}+T_{k+1}=P_{k+1}$$. Is it the way to implement point no.2? And, if it iss then how? I mean in what way proving $P_{k}+T_{k+1}=P_{k+1}$ is equivalent to proving $P_{k+1}$ under the assumption?? Feb 9 '16 at 10:41\n\u2022 What does $P_k + T_{p+1}$ mean? $P_k$ is an equation. Does it make sense to write $(5^2 = 25) + 3$? I would never write math that way. But do you agree that if $a$ and $b$ are numbers and $a=b$, then $a$ and $b$ are the same number, so $a + 3$ is the same as $b +3$, that is, $a+3=b+3$? That's what we mean by \"add ___ to both sides of the equation.\" And yes, when you start with a true equation, then (the left side plus something) is equal to (the right side plus the same something), which is how the author derived $P_{k+1}$. Feb 9 '16 at 12:52\n\u2022 DavidK first of all, adding something to an equation always means it is added to its both sides. In other words, it can never mean to add something to only side of equation because two sides make up an equation. So $P_k+T_{k+1}$ is not wrong anyway. It is not necessary for something to be right. When it is only \"not wrong\", we can adopt it. Feb 9 '16 at 18:41\n\u2022 Secondly, you said that since we assumed $P_k$ is true, we add $T_{k+1}$ to its both sides and then if it becomes $P_{k+1}$, it is proved that $P_{k+1}$ is true. And to support it, you said that if something is added to a true equation such as $3^{2}=9$ then the new equation for example $3^{2}-1=9-1$ is necessarily true (because it is the property of equation). In case of $3^{2}=9$ we are certain that the equation is true but in case of $P_k$ we are at the sake of assumption. These two things are different. I am sure I am missing something but please explain more accurately. Feb 9 '16 at 18:54\n\u2022 On the second point, that's the whole point of Introduction of an Implication. To prove $P\\implies Q$, assume $P$ is true and show that $Q$ is true under this assumption. Of course now all you have actually shown is that $Q$ is true if $P$ is true; but that is what $P\\implies Q$ means. Feb 9 '16 at 19:09\n\nThere is another way of thinking of this.\n\nThrow out the standard induction theorem, and replace it with this one:\n\nAny nonempty set of natural numbers has a smallest element.\n\nThis seems obviously true, right? This is called the well-ordering principle, and it is equivalent to induction. Here's how you do induction with the well-ordering principle (WOP for short):\n\n1. Prove that $P_0$ holds (or $P_1$, if you don't consider zero to be a \"natural number\").\n2. Take this set: $$S = \\{x | x \\in \\mathbb{N} \\wedge \\neg P_x \\}$$ That's the set of natural numbers for which our proposition does not hold.\n3. Assume $S$ is non-empty. Then, by WOP, it has a least element, which we can call $k+1$. In particular, we know that $\\neg P_{k+1}$ (we know that the proposition does not hold for $k+1$, since it's an element of $S$).\n4. By step (1), we know $k+1 > 0$ (or $k+1 > 1$, if you started with $P_1$). So $k$ is a natural number. If we had not proven step (1), $k$ could be equal to negative one (or zero), which would invalidate the rest of the proof.\n5. Prove that $\\neg P_{k}$; that is, prove that the proposition is not true for $k$, based on our knowledge that it is not true for $k+1$.\n6. Since $k < k + 1$ and $k \\in S$, we have a contradiction ($k + 1$ isn't the smallest element of the set). That means our assumption in step (3) must be false. The set of counterexamples is empty, meaning the proposition must hold for every number.\n\nNote that these statements are equivalent: $$P_k \\implies P_{k+1} \\\\ \\neg P_{k+1} \\implies \\neg P_k$$ If you can prove one, you can prove the other quite easily. Standard induction asks you to prove the first, while WOP-based induction requires you to prove the second.\n\nIn the end, it's just a matter of notation. But you might find one form of induction more intuitive than the other. If you are going to use this method, you should note that we usually talk about $k$ and $k - 1$ instead of $k+1$ and $k$; I used the latter because it corresponds more obviously to standard induction.\n\n\"as long as we know that the state of a proposition being true for any positive integer k after Base number implies that proposition is true for integer k+1, we have to show that Pk is true. \"\n\n1) No, you don't. If I wanted to show that IF Godzilla were 10 times as large in all dimensions as a T. Rex THEN Godzilla would be weigh 1,000 times as much, I do NOT need to prove that Godzilla is 10 times as large in all dimensions as a T. Rex. I CAN'T prove Godzilla is 10 times as large in all dimensions as a T. Rex because Godzilla isnt 10 times as large in all dimensions as a T. Rex.\n\nand\n\n2) You already have shown $P_k$ is true for some k. You have shown it is true for $k = 1$. Reread that statement. It does not say \"we will assume this is true for all $P_k$ and show that means it is true for all $P_{k+1}$\". It says \"we will show that if it is true for (one) $P_k$ than that means it is true for $P_{k+1}$\".\n\nThus we have proven an infinite loop: If it is true for k, then it is true for k + 1: Relabel k+1 as k and it is true for k + 1 (which is the old k + 2): Relabel and repeat indefinitely and we have proven it is true for all $n \\ge k$. Now we just need to demonstrate that it is true for a single base case k. Oh, wait. It's true for k = 1. Therefore it is true for $n \\ge 1$.\n\nAs for you second question....\n\n\"Why is it necessary for an equation to be true for performing operations on it?\"\n\nIt isn't. You can perform operations on untrue equations. You just can't assume any of the results are true.\n\nLet me write down: \"2 + 2 = 5\". Let me perform operations on them:\n\n$7^{2+2} - 37*5 = 7^5 - 37*{2 + 2}$\n\n$2401 - 185 = 16,807 - 74 -74$\n\n$2,216=16,659$\n\nI can do that. But I can't get any meaningful results.\n\n====\n\nIn a very abstract and philosophical way a proof by induction is a tiny bit like a proof by contradiction.\n\nIn both you assume a statement without proof and you reach a conclusion that would be true if your assumption were.\n\nIn a proof by induction you might prove that following: \"If any even number is divisible by two then they all are divisible by two.\"\n\nAnd in a proof by contradiction you might prove the following: \"If any odd number is divisible by two then they all are divisible by two.\"\n\nWhat's the difference in these statements? They are both shown in the exact same way (\"if $2|n$ then $2|n + 2$ and $2|n - 2$\"). They are both equally valid. And furthermore, they are both TRUE.\n\nBut in the proof by induction you verify a single confirmation: \"2 is divisible by two. Therefore all even numbers are.\"\n\nWhile in a proof by contradiction you contradict with a contradiction: \"3 is not divisible by two. Therefore no odd numbers are.\"\n\nThink about Peano system (see footnotes) for defining natural numbers. You take an initial number, which you call 1, and then you say:\n\n\u2022 Element 1 exists in the set.\n\u2022 Any successor of an element n in the set, say $suc(n)$, exists in the set.\n\nWith these axioms, you construct the whole set of natural numbers. You don't need to prove such elements exist in a deductive was as you think, but you must construct the proof and you have the building blocks for that (while in the deductive method, the path is the inverse).\n\nSo, the analogous of Peano system applies for the induction:\n\n\u2022 Prove that $P(1)$ exists, where 1 is just the base case you want to start the proof with. This is the equivalent of the Peano system for the value 1.\n\u2022 The constructive step: Prove that $P(n+1)$ is true when $P(n)$ was true. In this case, you move forward in the demonstration (induction actually means move towards the end) to construct the successive cases in the proof.\n\nSo you initially prove $P(1)$. If you prove $P(n) => P(n+1)$ then you can start traversing infinitely:\n\n\u2022 $P(2) because P(1)$\n\u2022 $P(3) because P(2)$\n\u2022 ...\n\nYou never prove the $n$ case, because $P(n)$ is in hypotetic case which derives ultimately from the case $P(1)$, and formerly being $P(n)$, it was $P(n+1)$ in former... iterations.\n\nSo, the way to think why you should assume $P(n)$ is true, is because:\n\n\u2022 In an initial case, $P(1)$ has been proven true by you. If you didn't, then you should start for it.\n\u2022 Since the construction is upward, instead of downward, you should not think that you have to prove the n-case, but that you have alreadr proven it.\n\nFootnotes:\n\n\u2022 You are not forced to start from 1 and upwards. You can take any number you want, and so you will restrict the proof on it and further.\n\u2022 Other induction processes can traverse entire integer set instead of just natural numbers. Such proof system involves prooving $n+1$ and $n-1$.", "date": "2022-01-23 22:38:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1645967/what-do-we-actually-prove-using-induction-theorem/1646299", "openwebmath_score": 0.8577460646629333, "openwebmath_perplexity": 166.37486381312206, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987946222258266, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8565846932763611}}
{"url": "https://math.stackexchange.com/questions/3260433/compute-int-0-infty-frac-operatornameli-3x1x2-dx", "text": "# Compute $\\int_0^\\infty \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx$\n\nHow to evaluate $$\\int_0^\\infty \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx\\ ?$$\n\nwhere $$\\displaystyle\\operatorname{Li}_3(x)=\\sum_{n=1}^\\infty\\frac{x^n}{n^3}$$ , $$|x|\\leq1$$\n\nI came across this integral while I was working on $$\\displaystyle \\displaystyle\\int_0^1 \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx\\$$ and here is how I established a relation between these two integrals:\n\n$$\\int_0^1 \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx=\\int_0^\\infty \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx-\\underbrace{\\int_1^\\infty \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx}_{x\\mapsto 1/x}$$\n\n$$=\\int_0^\\infty \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx-\\int_0^1 \\frac{\\operatorname{Li}_3(1/x)}{1+x^2}\\ dx$$ $$\\left\\{\\color{red}{\\text{add the integral to both sides}}\\right\\}$$\n\n$$2\\int_0^1 \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx=\\int_0^\\infty\\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx+\\int_0^1 \\frac{\\operatorname{Li}_3(x)-\\operatorname{Li}_3(1/x)}{1+x^2}\\ dx$$\n\n$$\\{\\color{red}{\\text{use}\\ \\operatorname{Li}_3(x)-\\operatorname{Li}_3(1/x)=2\\zeta(2)\\ln x-\\frac16\\ln^3x+i\\frac{\\pi}2\\ln^2x}\\}$$\n\n$$=\\int_0^\\infty\\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx+2\\zeta(2)\\underbrace{\\int_0^1\\frac{\\ln x}{1+x^2}\\ dx}_{-G}-\\frac16\\underbrace{\\int_0^1\\frac{\\ln^3x}{1+x^2}\\ dx}_{-6\\beta(4)}+i\\frac{\\pi}2\\underbrace{\\int_0^1\\frac{\\ln^2x}{1+x^2}\\ dx}_{2\\beta(3)}$$\n\n$$=\\int_0^\\infty\\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx-2\\zeta(2)G+\\beta(4)+i\\pi \\beta(3)$$\n\nThen\n\n$$\\int_0^1 \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx=\\frac12\\int_0^\\infty\\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx-\\zeta(2)G+\\frac12\\beta(4)+i\\frac{\\pi}2 \\beta(3)\\tag{1}$$\n\nwhere $$\\displaystyle\\beta(s)=\\sum_{n=0}^\\infty\\frac{(-1)^n}{(2n+1)^s}\\$$ is the the Dirichlet beta function.\n\nSo any idea how to evaluate any of these two integrals?\n\nThanks.\n\n\u2022 What are you asking? \u2013\u00a0clathratus Jun 13 '19 at 2:17\n\u2022 Note the formula you got from the triglogarithmic identity involved you being cavalier about branch cuts; you are missing a factor of $\\frac{16 i \\pi^4}{512}$ \u2013\u00a0Brevan Ellefsen Jun 13 '19 at 4:04\n\u2022 @clathratus I am not sure if you are being sarcastic or not but I'll take as not. If you want to know what I'm asking just ignore the body and read the title please. \u2013\u00a0Ali Shather Jun 13 '19 at 4:28\n\u2022 @BrevanEllefsen wolfram gives the closed form. \u2013\u00a0Ali Shather Jun 13 '19 at 4:30\n\u2022 @BrevanEllefsen I dont think I'm missing any. To make sure, just take x=1/2 and compare the two sides. \u2013\u00a0Ali Shather Jun 13 '19 at 4:45\n\nUsing the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.\n\n$$\\int_0^1\\frac{x\\ln^n(u)}{1-xu}\\ du=(-1)^n n!\\operatorname{Li}_{n+1}(x)$$ and by setting $$n=2$$ we get\n\n$$\\operatorname{Li}_{3}(x)=\\frac12\\int_0^1\\frac{x\\ln^2 u}{1-xu}\\ du$$\n\nwe can write\n\n$$\\int_0^\\infty\\frac{\\operatorname{Li}_{3}(x)}{1+x^2}\\ dx=\\frac12\\int_0^1\\ln^2u\\left(\\int_0^\\infty\\frac{x}{(1-ux)(1+x^2)}\\ dx\\right)\\ du$$ $$=\\frac12\\int_0^1\\ln^2u\\left(-\\frac12\\left(\\frac{\\pi u}{1+u^2}+\\frac{2\\ln(-u)}{1+u^2}\\right)\\right)\\ du,\\quad \\color{red}{\\ln(-u)=\\ln u+i\\pi}$$\n\n$$=-\\frac{\\pi}{4}\\underbrace{\\int_0^1\\frac{u\\ln^2u}{1+u^2}\\ du}_{\\frac3{16}\\zeta(3)}-\\frac12\\underbrace{\\int_0^1\\frac{\\ln^3u}{1+u^2}\\ du}_{-6\\beta(4)}-i\\frac{\\pi}2\\underbrace{\\int_0^1\\frac{\\ln^2u}{1+u^2}\\ du}_{2\\beta(3)}$$\n\nThen\n\n$$\\int_0^\\infty\\frac{\\operatorname{Li}_{3}(x)}{1+x^2}\\ dx=-\\frac{3\\pi}{64}\\zeta(3)+3\\beta(4)-i\\pi\\beta(3)\\tag{2}$$\n\nBonus:\n\nBy combining $$(1)$$ in the question body and $$(2)$$, the imaginary part $$i\\pi\\beta(3)$$ nicely cancels out and we get\n\n$$\\int_0^1 \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx=2\\beta(4)-\\zeta(2)G-\\frac{3\\pi}{128}\\zeta(3)$$\n\nwhere $$\\beta(4)$$ $$=\\frac{1}{768}\\psi^{(3)}(1/4)-\\frac{\\pi^4}{96}$$\n\n\u2022 The formula \\begin{align}\\Re\\int_0^\\infty\\frac{\\operatorname{Li}_{3}(x)}{1+x^2}\\ dx&=\\frac12\\int_0^1\\ln^2u\\left(\\int_0^\\infty\\frac{x}{(1-ux)(1+x^2)}\\ dx\\right)\\ du\\end{align} is weird. In the right side it's $\\int_0^1 \\frac{\\operatorname{Li}_3(x)}{1+x^2}\\ dx$ not its real part. There is no indication for the values of x involved in the formula about polylogarithm in the beginning of your post. Anyway your closed-form matches numeric evaluation. \u2013\u00a0FDP Nov 22 '19 at 8:45\n\u2022 its edited now. thanks for pointing this out. \u2013\u00a0Ali Shather Nov 22 '19 at 8:58\n\u2022 The formula for polylogarithm in the beginning of your post is valid for $|x|<1$ only. If you take $x=2$ the denominator is $1-2u$ and $1-2u=0$ when $u=1/2\\in [0;1]$ \u2013\u00a0FDP Nov 22 '19 at 15:42\n\u2022 Yes .. actually i got the right formula yesterday and I'll fix the problem today. \u2013\u00a0Ali Shather Nov 22 '19 at 16:18\n\u2022 When you consider the integral $$\\int_0^\\infty\\frac{x}{(1-ux)(1+x^2)}\\ dx$$ as a CauchyPrincipalValue you don't get the imagniary part. It arises from the $\\epsilon$ contour around that singularity. \u2013\u00a0Diger Nov 22 '19 at 22:05\n\nFor a different solution, use the first result from A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by Cornel Ioan Valean.\n\nEssentially, the main new results in the presentation are:\n\nLet $$a\\le1$$ be a real number. The following equalities hold: $$\\begin{equation*} i) \\ \\int_0^1 \\frac{\\log (x)\\operatorname{Li}_2(x) }{1-a x} \\textrm{d}x=\\frac{(\\operatorname{Li}_2(a))^2}{2 a}+3\\frac{\\operatorname{Li}_4(a)}{a}-2\\zeta(2)\\frac{\\operatorname{Li}_2(a)}{a}; \\end{equation*}$$ $$\\begin{equation*} ii) \\ \\int_0^1 \\frac{\\log^2(x)\\operatorname{Li}_3(x) }{1-a x} \\textrm{d}x=20\\frac{\\operatorname{Li}_6(a)}{a}-12 \\zeta(2)\\frac{\\operatorname{Li}_4(a)}{ a}+\\frac{(\\operatorname{Li}_3(a))^2}{a}. \\end{equation*}$$ For a fast proof, see the paper above (series expansion combined with the Cauchy product of squared Polylogarithms)\n\nThe use of these new results with integrals allows you to obtain your result elegantly, but also other results that are (very) difficult to obtain by other means, including results from the book, (Almost) Impossible Integrals, Sums, and Series.\n\nBONUS: Using these results you may also establish that (or the versions with integration by parts applied).\n\n$$i) \\ \\int_0^1 \\frac{\\arctan(x) \\operatorname{Li}_2(x)}{x}\\textrm{d}x$$ $$=\\frac{1}{384}\\left(720\\zeta(4)+105\\pi\\zeta(3)+384\\zeta(2)G-\\psi^{(3)}\\left(\\frac{1}{4}\\right)\\right),$$ $$ii)\\ \\int_0^1 \\frac{\\arctan(x) \\operatorname{Li}_2(-x)}{x}\\textrm{d}x$$ $$=\\frac{1}{768}\\left(\\psi^{(3)}\\left(\\frac{1}{4}\\right)-384\\zeta(2)G-126\\pi\\zeta(3)-720\\zeta(4)\\right).$$\n\nEXPLANATIONS (OP's request): The following way in large steps shows the amazing possible creativity in such calculations.\n\nWe'll want to focus on the integral, $$\\displaystyle \\int_0^1 \\frac{\\arctan(x)\\operatorname{Li}_2(x)}{x}\\textrm{d}x$$ which is a translated form of the main integral.\n\nNow, based on $$i)$$ where we plug in $$a=i$$ and then consider the real part, we obtain an integral which by a simple integration by parts reveals that\n\n$$\\int_0^1 \\frac{\\arctan(x)\\operatorname{Li}_2(x)}{x}\\textrm{d}x=\\int_0^1 \\frac{\\arctan(x)\\log(1-x) \\log(x)}{x}\\textrm{d}x+\\frac{17}{48}\\pi^2 G+\\frac{\\pi^4}{32}-\\frac{1}{256}\\psi^{(3)}\\left(\\frac{1}{4}\\right).$$\n\nLooks like we need to evaluate one more integral and we're done. Well, if you read the book (Almost) Impossible Integrals, Sums, and Series (did you?), particularly the solutions in the sections 3.24 & 3.25 you probably observed the powerful trick of splitting the nonnegative real line at $$x=1$$ with the hope of getting the same integral in the other side but with an opposite sign. Therefore, with such a careful approach (since we need to avoid the divergence issues), we obtain immediately that $$\\int_0^1 \\frac{\\arctan(x)\\log(1-x) \\log(x)}{x}\\textrm{d}x$$ $$=\\frac{1}{2} \\underbrace{\\int_0^1 \\frac{\\arctan(x)\\log^2(x)}{x}\\textrm{d}x}_{\\text{Trivial}}+\\frac{\\pi}{4}\\underbrace{\\int_0^1 \\frac{\\log(x)\\log(1-x)}{x}\\textrm{d}x}_{\\text{Trivial}}$$ $$-\\frac{1}{2}\\Re\\left \\{\\int_0^{\\infty}\\frac{\\arctan(1/x) \\log(1-x)\\log(x)}{x}\\textrm{d}x\\right \\},$$\n\nand the last integral works simply nice with Cornel's strategy described in the second part of this post (it involves the use of Cauchy Principal Value) https://math.stackexchange.com/q/3488566.\n\n\u2022 Nice (+1) but it's not clear how to get the integral in the question from using the first result. \u2013\u00a0Ali Shather Dec 30 '19 at 5:39\n\u2022 The explanation is really amazing and helpful. thank you. \u2013\u00a0Ali Shather Dec 30 '19 at 20:07", "date": "2020-08-05 13:52:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3260433/compute-int-0-infty-frac-operatornameli-3x1x2-dx", "openwebmath_score": 0.8812828660011292, "openwebmath_perplexity": 399.5297511282832, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765633889135, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8565752050533502}}
{"url": "https://math.stackexchange.com/questions/2017988/example-of-a-continuous-function-f-mathbbr-to-0-1-such-that-f-is-not-uni", "text": "# Example of a continuous function $f:\\mathbb{R} \\to [0,1]$ such that f is not uniformly continuous.\n\nI am trying to find an example of a continuous function $f:\\mathbb{R} \\to [0,1]$ such that f is not uniformly continuous. I have been playing around with the $sin$ function and I have noticed that $|sin(x^2)|$ is a continuous function from $\\mathbb{R}$ to $[0,1]$. This function I am thinking is not uniformly continuous since it behaves a lot like $\\sin(x^2)$.\n\nWhat I am most interested is in how you come up with a function that meets the criteria. For example, when I first saw this question, I immediately thought of the $\\sin$ function. Are there any other functions that meet this requirement and are easier to work with in case that I want to prove that it is not uniformly continuous. Is this how you come up with functions, that is, by looking at the rate of oscillations? It seems to me that the only functions that would work are variations of the $sin$ function. I would appreciate it if you could give advice on how to approach this problem.\n\n\u2022 you cannot just say \"it oscillates a lot\" to prove a function is not uniformly continuous, start with the definition\n\u2013\u00a0Nick\nNov 17 '16 at 6:02\n\u2022 you are right. Right now I am just constructing graphs that I think are not uniformly continuous and trying to come up with possible functions that meet the requirements. I am just trying to see how people approach the problem rather than proving it. Nov 17 '16 at 6:05\n\u2022 You can use the mean value theorem to prove that $\\frac 1 2 + \\frac 1 2 \\sin(x^2)$ isn't uniformly continuous since the derivative becomes arbitrarily large as $x \\to \\infty$. Nov 17 '16 at 6:15\n\nThe key to finding such a function, is to look for a function with \"arbitrarily large derivative\" (this doesn't work for all such functions, as the comment below points out) . This is to say, we want some problems with infinity, to be created on the end points.\n\nFor example, let us consider $f(x) = \\frac{1}{2} + \\frac{1}{2}\\sin x^2$, which has a derivative of $2x \\cos x^2$, and also has range in $[0,1]$ (the $\\frac{1}{2}$ is just some adjustment to ensure that the range is in $[0,1]$). This derivative blows up as $x$ increases. Hence, we consider this as a candidate.\n\nFix a natural number $N$. We will find a pair of points $x,y$ such that $|f(x)-f(y)| \\geq N|x-y|$.\n\nNow, all we need is to use the mean value theorem: We know that $f(x)-f(y) = (x-y) f'(z)$ where $z$ is some point between $x$ and $y$. Now, suppose we chose $x$ and $y$ in a region where $f'(z)$ could only be a large positive quantity, greater than the given $N$. Then of course, $|f(x)-f(y)| >N|(x-y)|$.\n\nI leave you to formalize the details, but it is clear, that $f$ cannot be uniformly continuous, despite being differentiable.\n\n\u2022 \"The key to finding such a function, is to look for a function with \"arbitrarily large derivative\". \" That doesn't always work. Consider $\\sin(e^x)/(1+x^2).$ Also the derivative of your function doesn't \"blow up\" as I understand the term. Rather it oscillates unboundedly.\n\u2013\u00a0zhw.\nNov 17 '16 at 7:14\n\u2022 @zhw I see. The point is, my function oscillates unboundedly, as you have said. I mistakenly wrote it as blowing up, but the meaning should be clear. But I understood the example of $\\sin (e^x)/ (1+x^2)$, so thank you for pointing it out. It is a function with arbitrarily large derivative, but reducing oscillations, hence could be uniformly continuous. Nov 17 '16 at 7:25\n\n\"Right now I am just constructing graphs that I think are not uniformly continuous and trying to come up with possible functions that meet the requirements. I am just trying to see how people approach the problem rather than proving it.\"\n\nThis is a good strategy, and it's often how I approach analysis problems as well. Having an intuition that a result is \"correct\" (even if it's due to hand-waving reasoning) is good to have before starting a rigorous proof.\n\nUniform continuity is sort of like a stronger, \"global\" version of regular continuity. To wit, if you start with a given $\\varepsilon > 0$, then you can find a $\\delta > 0$ such that $|x-y| < \\delta \\implies |f(x) - f(y)| < \\varepsilon$, regardless of where you are in the domain. On the other hand, we talk about regular continuity \"at a given point\", and the $\\delta$ depends on that chosen point.\n\nYour idea with $f(x) = |\\sin(x^2)|$ is perfect. Indeed, if we suppose that this is uniformly continuous, then we can arrive at a contradiction for essentially the reason you cite: it oscillates faster and faster as we move away from the origin.\n\nTo take advantage of this problem, choose $\\varepsilon$ less than the amplitude of the wave; e.g. $\\varepsilon = 1/2$. If we suppose there exists a $\\delta > 0$ such that $|x-y| < \\delta$, then $||\\sin(x^2)| - |\\sin(y^2)|| < 1/2$, we can make this fail by going out far enough away from the origin so that our function goes through an entire oscillation in a $\\delta$-interval. Recall that a sine wave reaches a \"crest\" or a \"trough\" at every $(2k + 1/2) \\pi$ radians and every $(2k + 3/2) \\pi$ radians. Solving $x^2 = (2k + 1/2) \\pi$ gives the values at which $\\sin(x^2)$ reaches a crest, and a similar equation for the troughs. Taking an absolute value results in all of these becoming crests (with new troughs in between each at integer multiples of $\\pi$). Now just go out sufficiently far so that the distance between these crests is less than the chosen $\\delta$. I'll leave the details to you to crunch out (one can show that the distance between the crests limits to zero).\n\n\u2022 Apparently, thge OP added the absolute value to achieve the desired codomain $[0,1]$ Nov 17 '16 at 6:36\n\u2022 Oh!! I lost sight of the forest for the trees. Thanks for pointing that out @HagenvonEitzen. Nov 17 '16 at 6:38\n\u2022 Thank you Kaj Hansen your explanation is very helpful! Nov 17 '16 at 7:00\n\u2022 Glad I could help @AnP. ! Nov 17 '16 at 7:18", "date": "2022-01-25 13:12:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2017988/example-of-a-continuous-function-f-mathbbr-to-0-1-such-that-f-is-not-uni", "openwebmath_score": 0.930029034614563, "openwebmath_perplexity": 119.5976233561056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517488441416, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8565650098942448}}
{"url": "https://math.stackexchange.com/questions/1675109/if-i-say-g-1-dots-g-n-freely-generates-a-subgroup-of-g-does-that-mean", "text": "# If I say $\\{g_1,\\dots,g_n\\}$ freely generates a subgroup of $G$, does that mean the elements $g_i$ are all distinct?\n\nLet $G$ be a group and let $g_1,\\dots,g_n$ elements of $G$. If I say that $\\{g_1,\\dots,g_n\\}$ freely generates a free subgroup of $G$, say $H$, do I mean that the rank of $H=n$ or should I consider the case when two elements are equal (for example $g_1=g_2$) and so the rank of $H<n$ ? thank you\n\n\u2022 If the elements freely generate $\\;H\\;$ it can't be $\\;g_i=g_j\\;,\\;\\;i\\neq j$, nor $\\;g_i=1\\;$ for some $\\;i\\;$ \u2013\u00a0DonAntonio Feb 28 '16 at 1:31\n\u2022 @Joanpemo I agree that it can't be $g_i=1$ but why it can't be $g_i=g_j$ for $i\\not=j$? thank you \u2013\u00a0Richard Feb 28 '16 at 1:34\n\u2022 Else you would have the ralation $g_ig_j^{-1}$. \u2013\u00a0Daniel Robert-Nicoud Feb 28 '16 at 1:38\n\u2022 @Joanpemo It depends on the context, but the way the question is phrased, i.e., without context, the statement is not strictly false. There is a perfectly nice free group which is free on the set $\\{x,x,y\\}$. It is the free group on $\\{x,y\\}$, since $\\{x,x,y\\}=\\{x,y\\}$. \u2013\u00a0Zach Blumenstein Feb 28 '16 at 1:44\n\u2022 @ZachBlumenstein Thank you. Yet, as sets, $\\;\\{x,x,y\\}=\\{x,y\\}\\;$ so no problem, indeed. I though understand something different when I read \"a set of free generators\", and as far as I know this implies $\\;g_i\\neq g_j\\;$ for $\\;i\\neq j\\;$, otherwise the universal property of a free group in the number of free generators given isn't fulfilled. \u2013\u00a0DonAntonio Feb 28 '16 at 2:42\n\nStrictly speaking, what you wrote means that the set $\\{g_1,\\dots,g_n\\}$ freely generates $H$, so the rank of $H$ is the number of (distinct) elements of that set, which may or may not be $n$. However, people frequently abuse language and say what you wrote when they really mean that additionally the $g_i$ should all be distinct. In context, there is usually not too much risk of confusion, and as a practical matter, if you see someone write this, you should judge for yourself which interpretation would make more sense in context. I don't really know of a succinct and completely standard way to express the latter meaning; I might write it as \"the elements $(g_1,\\dots,g_n)$ freely generate a subgroup\", writing it as a tuple to stress that you are actually saying that the map $\\{1,\\dots,n\\}\\to H$ sending $i$ to $g_i$ satisfies the universal property to make $H$ a free group on the set $\\{1,\\dots,n\\}$. What's really going on is that a collection of (potential) \"free generators of a group $H$\" should not be thought of as just a subset of $H$ but as a set together with a map to $H$ (and if the group really is freely generated by a map, the map must be injective).\n(This terminological difficulty is not restricted to this context. For instance, a similar issue quite frequently comes up when talking about linear independence: if $v\\in V$ is a nonzero element of a vector space, then the set $\\{v,v\\}$ is linearly independent (since it actually has only one element), but the tuple $(v,v)$ is not.)\n\u2022 I don't exactly agree. When we say that a (finite or infinite) set $X$ is a set of free generators of a group $F$, we don't need to specify any function, because the function is just inclusion. There is no mistake there. But I do agree that the notation $\\{g_1,\\ldots,g_n\\}$ taken to mean that $g_i$ are distinct is bad, because it literally means what that $X=\\{g_1,\\ldots,g_n\\}$ is a free generating set of $F$, while it probably really means that $g_1, \\ldots, g_n$ generate $F$ freely, which is not the same thing if $\\lvert X\\rvert\\neq n$. \u2013\u00a0tomasz Feb 28 '16 at 8:34\n\u2022 Sure, there is nothing wrong with saying that a subset freely generates a group. But it is more natural to formulate it in terms of maps from a set, and the failure to do so is the cause of the terminological awkwardness when saying \"$(g_1,\\dots,g_n)$ freely generates\" a group, and the frequency with which people incorrectly write \"$\\{g_1,\\dots,g_n\\}$ freely generates\" when that's not really what they mean. \u2013\u00a0Eric Wofsey Feb 28 '16 at 8:41", "date": "2021-01-21 19:52:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1675109/if-i-say-g-1-dots-g-n-freely-generates-a-subgroup-of-g-does-that-mean", "openwebmath_score": 0.8420447707176208, "openwebmath_perplexity": 156.17697226308653, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517488441415, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8565649987962237}}
{"url": "http://mathhelpforum.com/advanced-statistics/74249-probability-success.html", "text": "# Math Help - probability of success\n\n1. ## probability of success\n\nI am trying to answer this problem for an exam.Can anyone please help me.I wonder if I should be using the formula prob=nCr (p^r) q^(n-r).\n\nThe probability for students from a certain university to pass Mathematics and Economics are 3/7 and 5/7 respectively. If one student fails both subjects and there are four students who passed both subjects find the number of students who took the test.\n\n2. Hello, hven191!\n\nThis doesn't require any fancy formulas . . .\n\nThe probability for students at a certain university\nto pass Mathematics and Economics are 3/7 and 5/7 respectively.\n\nOne student failed both subjects and four students passed both subjects.\n\nFind the number of students who took the test.\nMake a Venn diagram of the students who passed.\nCode:\n      *---------------------------------------*\n|                                       |\n|   *-----------------------*           |\n|   | Math                  |           |\n|   | only  *---------------+-------*   |\n|   |   M   |     Both      |       |   |\n|   |       |       4       |       |   |\n|   |       |               |       |   |\n|   *-------+---------------*  Eco  |   |\n|           |             E   only  |   |\n|           *-----------------------*   |\n|  Neither 1                            |\n*---------------------------------------*\n\nLet $M$ = number of students that passed Math only.\nLet $E$ = number of students that passed Economics only.\n\nWe know there are 4 that passed both, and 1 that failed both.\n\nThe total number of students is: . $N \\:=\\:M + E + 5$\n\nThe number of students that passed Math is: . $M + 4$\nThe probability that a student passed Math is $\\tfrac{3}{7}$\nWe have: . $\\frac{M+4}{M+E+5} \\:=\\:\\frac{3}{7} \\quad\\Rightarrow\\quad 4M - 3E \\:=\\:-13\\;\\;{\\color{blue}[1]}$\n\nThe number of students that passed Economics is $E + 4$\nThe probability that a student passes Economics is $\\tfrac{5}{7}$\nWe have: . $\\frac{E+4}{M+E+5} \\:=\\:\\frac{5}{7} \\quad\\Rightarrow\\quad 5M - 2E \\:=\\:3\\;\\;{\\color{blue}[2]}$\n\n$\\begin{array}{cccccc}\\text{Multiply {\\color{blue}[1]} by -2:} & \\text{-}8M + 6E &=& 26 & {\\color{blue}[3]}\\\\\n\\text{Multiply {\\color{blue}[2]} by 3:} & 15M - 6E &=& 9 & {\\color{blue}[4]}\\end{array}$\n\n$\\text{Add {\\color{blue}[3]} and {\\color{blue}[4]}: }\\;7M \\:=\\:35 \\quad\\Rightarrow\\quad\\boxed{ M \\:=\\:5}$\n\nSubstitute into [1]: . $4(5) - 3E \\:=\\:-13 \\quad\\Rightarrow\\quad\\boxed{ E \\:=\\:11}$\n\nTherefore, the number of students is: . $N \\;=\\;M + E + 5 \\;=\\;5 + 11 + 5 \\;=\\;{\\color{red}21}$\n\n3. ## Thanks\n\nThanks a lot Soroban! The solution you gave was very simple and easy to understand.The illustration was a big help. Although the answer does not match the answer in the book I'm using I'm still convinced with your answer. The answer in the book which is 28 may just be a typo-graphical error. Thanks!", "date": "2014-07-31 03:08:36", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/74249-probability-success.html", "openwebmath_score": 0.687724769115448, "openwebmath_perplexity": 794.9017559484801, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517443658749, "lm_q2_score": 0.8757869867849166, "lm_q1q2_score": 0.8565649901179212}}
{"url": "http://math.stackexchange.com/questions/266629/find-the-infinite-sum-sum-n-1-infty-frac12n-1/266638", "text": "# Find the infinite sum $\\sum_{n=1}^{\\infty}\\frac{1}{2^n-1}$\n\nHow to evaluate this infinite sum? $$\\sum_{n=1}^{\\infty}\\frac{1}{2^n-1}$$\n\n-\nWhy the expression appears so small? How can I enlarge that? : ( \u2013\u00a0 Ryan Dec 28 '12 at 18:54\nuse \\displaystyle \u2013\u00a0 sxd Dec 28 '12 at 18:55\nIf I'm not wrong, this one appears in one volume of Ramanujan's notebook. (Chris) \u2013\u00a0 Chris's sis Dec 28 '12 at 18:59\n$1.606695152415291$ \u2013\u00a0 Henry Dec 28 '12 at 19:12\nNot an answer, but it is easy to convert the formula to: $$\\sum_{m=1}^\\infty \\frac{\\tau(m)}{2^m}$$ where $\\tau(m)$ is the number of distinct divisors of $m$. \u2013\u00a0 Thomas Andrews Dec 28 '12 at 19:33\nshow 6 more comments\n\nI think you wanna see this:\n\nRamanujan\u2019s Notebooks Part I\n\nClick me and try Entry $14$ (ii) / pag 146 where you set $x=\\ln2$\n\nChris.\n\n-\nI want to write out that particular equation just because it's so . . . out there: $$\\sum_{k \\ge 1}\\frac{1}{e^{kx}-1}=\\frac{\\gamma}{x}-\\frac{\\log x}{x}+\\frac{1}{4}-\\sum_{1 \\le k \\le n}\\frac{B^2_{2k}x^{2k-1}}{(2k)(2k)!}+R_n,$$ where $\\gamma$ is Euler's constant, $B_{2k}$ are the conventionally defined Bernoulli numbers, $x>0$, $n\\ge 1$, and $R_n$ is a constant bounded by the inequality $$|R_n|\\le \\frac{|B_{2n}B_{2n+2}|x^{2n}}{(2n)!}\\left(\\frac{x^2}{4\\pi^2}+\\frac{\\pi^2}{6} \\right).$$ For our case, let $x=\\ln 2$ as Chris's sister stated. \u2013\u00a0 000 Dec 29 '12 at 18:00\n@Limitless: thank you for the details provided! (+1) \u2013\u00a0 Chris's sis Dec 29 '12 at 18:03\n\nYes. I found it. It is called the Erd\u0151s-Borwein Constant.\n\n$$E=\\sum_{n\\in Z^+}\\frac{1}{2^n-1}$$\n\nAccording to the page, Erd\u0151s showed that it is irrational.\n\n-\nI almost see in each step in the link a possible nice question to ask. :D (+1) \u2013\u00a0 Chris's sis Dec 28 '12 at 19:24\nIs it known whether the Erdos-Borwein Constant is transcendental or not? \u2013\u00a0 Rustyn Dec 28 '12 at 19:27\nI have no idea. I guess no. \u2013\u00a0 Amr Dec 28 '12 at 19:35\n@Rustyn: Please use markdown formatting for non-mathematical things like italics and bold in normal text. I've edited your comment. \u2013\u00a0 Zev Chonoles Dec 28 '12 at 19:42\n@ZevChonoles Ok, I will in the future. \u2013\u00a0 Rustyn Dec 28 '12 at 19:44\nshow 1 more comment\n\n$$\\displaystyle \\sum _{k=1}^n \\frac{1}{\\left(\\frac{1}{q}\\right)^k-\\frac{1}{r}}=\\frac{r}{\\log (q)} \\left(\\psi _q^{(0)}\\left(1-\\frac{\\log (r)}{\\log (q)}\\right)-\\psi _q^{(0)}\\left(n+1-\\frac{\\log (r)}{\\log (q)}\\right)\\right)$$\n\nIn trying to get Mathematica to solve the series, I eventually found the preceding form which assumes $0<q<1$. If we take $q=1/2$, $r=1$ and let n approach infinity, we get the same solution that Amr references. The partial sum solution utilizes the function, $\\psi _q^{(n)}(z)$.\n\n$$\\displaystyle \\lim_{n\\to \\infty } \\, \\frac{1}{\\log (1/2)}\\left(\\psi _{\\frac{1}{2}}^{(0)}(1)-\\psi _{\\frac{1}{2}}^{(0)}(n+1)\\right)=1+\\frac{\\psi _{\\frac{1}{2}}^{(0)}(1)}{\\log (1/2)}=E$$\n\n-\n\n(This is not meant as an answer but it is too long for the comment.box)\n\nSince you mentioned interest in variations of the problem: here is a text, in which L. Euler discussed that sum:\n\n\"Consideratio quarumdam serierum quae singularibus proprietatibus sunt praeditae\" (\u201cConsideration of some series which are distinguished by special properties\u201d)\nL. Euler Enestr\u00f6m-index E190.\n\nYou can find it online.\n\nA further discussion of this by Prof. Ed Sandifer, where he sheds light on a very interesting discussion about a \"false series for the logarithm\" at which that constant pops up (and which actually had pointed me originally to L.Euler's article):\n\nA false logarithm series (Discussion of E190)\nEd. Sandifer in: \"How Euler did it\" Dec 2007\nhttp://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2050%20false%20log%20series.pdf\n\nI've fiddled then with it myself a bit further, maybe you find that amateurish explorations interesting too. The constant is part of a consideration on page 5.\n\n-", "date": "2013-12-13 12:09:14", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/266629/find-the-infinite-sum-sum-n-1-infty-frac12n-1/266638", "openwebmath_score": 0.9496000409126282, "openwebmath_perplexity": 1802.1115540306073, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517424466175, "lm_q2_score": 0.8757869835428965, "lm_q1q2_score": 0.8565649852661971}}
{"url": "https://math.stackexchange.com/questions/233220/is-bbb-z-3-x-langle-x2-1-rangle-cong-bbb-z-3-times-bbb-z-3/233228", "text": "# Is $\\Bbb Z_3 [x] / \\langle x^2 - 1 \\rangle \\cong \\Bbb Z_3 \\times \\Bbb Z_3$?\n\nI have paired off the zero divisors in these two rings, and there are exactly 4 pairs of distinct zero divisors in both rings:\n\nIn $\\Bbb Z_3[x]/\\langle x^2 - 1\\rangle$ the zero divisor pairs are $$x+1, x+2 \\qquad x+1, 2x+1, \\qquad 2x+1, 2x+2 \\qquad x+2, 2x+2$$\n\nSimilarly, in $\\Bbb Z_3 \\times \\Bbb Z_3$ the zero divisor pairs are $$(0, 1), (1, 0) \\qquad (0,1), (2,0) \\qquad (0, 2), (1, 0) \\qquad (0, 2), (2, 0)$$ The fact that the zero divisors match up indicates to me that these rings should indeed be isomorphic, as zero divisors are one of the structures which have failed checks in the past on rings such as $\\Bbb Z_2 [x] / \\langle x^2 \\rangle \\ncong \\Bbb Z_2 \\times \\Bbb Z_2$.\n\nMy Question\n\nIs the only homomorphism (and thus isomorphism) between these two rings an enumerative one, where I explicitly map each of the nine elements of $\\Bbb Z_3[x]/ \\langle x^2 - 1\\rangle$ to an element in $\\Bbb Z_3 \\times \\Bbb Z_3$? I can do that if necessary, but it seems an inelegant and brute force method. However, since these rings are so small, it may be the best way to go about it. Thanks!\n\n$Z_3[x]/(x^2-1)\\cong Z_3[x]/(x-1)\\times Z_3[x]/(x+1)\\cong Z_3\\times Z_3$. The map is induced by $ax+b\\rightarrow (a+b, -a+b)$\n\n\u2022 Thank you all for the answers! They give me much more intuition about the result than a simple enumerative homomorphism. \u2013\u00a0Moderat Nov 9 '12 at 0:28\n\u2022 Actually I am finding it hard to see that this map is onto. Can you elaborate? \u2013\u00a0Moderat Nov 10 '12 at 21:03\n\u2022 @jmi4 $x-1\\rightarrow (0,1)$, $-x-1\\rightarrow (1,0)$ \u2013\u00a0i. m. soloveichik Nov 11 '12 at 0:02\n\nIn $R:=(\\mathbb Z/3\\mathbb Z)[x]$, $x-1$ and $x+1$ generate two distinct maximal ideals $\\mathfrak m_1, \\mathfrak m_2$. We have $(x^2-1)R=\\mathfrak m_1 \\mathfrak m_2$. By Chinese Remainder Theorem, we have a canonical isomorphism $$R/(x^2-1)R \\to R/\\mathfrak m_1\\times R/\\mathfrak m_1.$$ Now $R/\\mathfrak m_i$ is isomorphic to $\\mathbb Z/3\\mathbb Z$. So your rings are isomorphic.\n\nThere are exatly two isomorphisms. One is given as above, the other one is obtained by composition this isomorphism with the permutation in the second ring. (Hints: the only automorphism of $\\mathbb Z/3\\mathbb Z$ is the identity).\n\n1. A homomorphism from a quotient $A/\\theta\\to B$ may be given by its lift $A\\to B$ such that its kernel factors through $\\theta$.\n2. A homomorphism from a polynomial ring $R[x_1,..,x_i,..]$ to an $R$-algebra is unquely determined by the image of the indeterminants $x_i$.\n3. $1$ must be mapped to the unit, that is $(1,1)$, and by 2., the image of $x$ can be freely chosen, and it determines all other images.\n\nSince the ideals $(x-1)$ and $(x+1)$ both contain $(x^2-1)$ (because $x-1,x+1\\mid x^2-1$ in $\\mathbf{F}_3[x]$), there are natural surjective homomorphisms $\\pi_1:\\mathbf{F}_3[x]/(x^2-1)\\rightarrow\\mathbf{F}_3[x]/(x-1)$ and $\\pi_2:\\mathbf{F}_3[x]/(x^2-1)\\rightarrow\\mathbf{F}_3[x]/(x+1)$. Explicitly $\\pi_1(f+(x^2-1))=f+(x-1)$ and $\\pi_2(f+(x^2-1))=f+(x+1)$. This gives a homomorphism\n\n$\\pi:\\mathbf{F}_3[x]/(x^2-1)\\rightarrow\\mathbf{F}_3[x]/(x-1)\\times\\mathbf{F}_3[x]/(x+1)$.\n\nNote that $f+(x^2-1)$ is in the kernel of this map if and only if both $x-1$ and $x+1$ divide $f$, and since the polynomials $x-1,x+1$ are relatively prime, their product $x^2-1$ divides $f$, so $f\\in (x^2-1)$, and thus $f+(x^2-1)=0$ in $\\mathbf{F}_3[x]/(x^2-1)$. Thus $\\pi$ is injective. There are a couple ways to see that $\\pi$ is surjective, and which one you prefer depends I guess on what you know. For example, since both the source and the target of $\\pi$ are $\\mathbf{F}_3$-vector spaces of dimension $2$, and the map is $\\mathbf{F}_3$-linear, it has to be an isomorphism. Or, since the ideals $(x-1)$ and $(x+1)$ together generate the unit ideal $\\mathbf{F}_3[x]$, since the ideal they generate contains $(x+1)-(x-1)=1+1=2$, a unit in $\\mathbf{F}_3$, you can invoke the Chinese remainder theorem (which will actually tell you injectivity of the above map as well as surjectivity). Or you can try to directly find, for given $f_1,f_2\\in\\mathbf{F}_3[x]$, a polynomial $f$ such that $f-f_1\\in(x-1)$ and $f-f_2\\in(x+1)$. Then $f+(x^2-1)$ will map to $(f_1+(x-1),f_2+(x+1))$.\n\nIn any case, the map is an isomorphism.", "date": "2019-08-22 02:39:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/233220/is-bbb-z-3-x-langle-x2-1-rangle-cong-bbb-z-3-times-bbb-z-3/233228", "openwebmath_score": 0.9580441117286682, "openwebmath_perplexity": 131.9940751963664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877692486438, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8565619462155301}}
{"url": "https://math.stackexchange.com/questions/1546098/spiral-of-archimedes-area-and-sketch-in-polar-coordinates", "text": "Spiral of Archimedes area and sketch in polar coordinates\n\nThis is an exercise from Apostol's Calculus, Volume 1. It asks us to sketch the graph in polar coordinates and find the area of the radial set for the function:\n\n$$f(\\theta) = \\theta$$\n\nOn the interva $0 \\leq \\theta \\leq 2 \\pi$. I think to find the area we should just integrate $\\theta \\ d\\theta$ from 0 to $2\\pi$ like any other function? Is that right? Also I'm not sure how to think about sketching a function in polar coordinates.\n\nThe problem is the book gives the answer as $4\\pi^3/3$ which is not what I get if I just integrate the function.\n\n\u2022 In the 2nd edition of this book (of which I have a copy), this is exercise 5 on page 111. You should review the previous three or four pages, especially the section about \"Polar coordinates\" and the section about \"The integral for area in polar coordinates\" (or whatever those sections are called in the edition you're using). If those pages do not answer your question, perhaps you can identify something in those pages that did not make sense to you, and add that to your question. \u2013\u00a0David K Nov 25 '15 at 17:15\n\u2022 By the way, I think the [calculus] tag was fine for this question. \u2013\u00a0David K Nov 25 '15 at 17:15\n\u2022 Did you find the theorem yet that says, in part, \"Let $R$ denote the radial set of a nonnegative function $f$ ... the area of $R$ is ...\"? It gives you the formula. In my edition of the book, this is on the page just before the exercise. \u2013\u00a0David K Nov 25 '15 at 18:55\n\nFirst, to sketch such a graph, you want to consider the distance from the origin as the angle from the $x$-axis changes. Just like when sketching the graph of a function in rectangular coordinates it is good to evaluate at particular values of $x$ and see the height of the function, when sketching a curve in polar coordinates, evaluate the function are a few values of the angle and find the radius, i.e., the distance from the origin at the angle. So, doing that we obtain the following graph:\n\nThen, to calculate the area of the radial set, you must integrate $\\frac{1}{2} r^2$, where the radius is the value of the function. So we have, \\begin{align*} \\text{Area} &= \\frac{1}{2}\\int_0^{2 \\pi} \\theta^2 \\, d\\theta \\\\ &= \\left. \\frac{1}{2} \\cdot \\frac{\\theta^3}{3} \\right|_0^{2 \\pi} \\\\ &= \\frac{(2 \\pi)^3}{6}\\\\ &= \\frac{4 \\pi^3}{3}. \\end{align*}\n\nSketching this in polar coordinates is pretty straightforward. Draw your axes and you know that the radial value is equal to the angle, so your curve would start at the origin when $\\theta$ is zero, it would be $\\frac{\\pi}{2}$ at 90 degrees, etc.\n\nAs for calculating the area, you are correct that you integrate it, but I'm not sure what it means physically because the curve does not close on itself since $f(0) \\neq f(2\\pi)$ (perhaps this is the insight they are trying to get from you).\n\n\u2022 The \"radial set\" is defined in the book. Graphically, it is formed by the line segments from the origin to each point on the curve. All the points those segments pass through are in the area to be measured. As long $f(\\theta)$ is never negative, it's clear what region's area we're supposed to measure. \u2013\u00a0David K Nov 25 '15 at 17:32\n\ncommenting on Daves's answer \"but I'm not sure what it means physically because the curve does not close on itself since f(0)\u2260f(2\u03c0)f(0)\u2260f(2\u03c0)\"\n\nThe area calculated is the area bounded by the curve and the terminal vector. ie the curve is subtended by the sweeping vector, which in this case is at 2pi- the positive x axis. .", "date": "2021-05-12 11:04:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1546098/spiral-of-archimedes-area-and-sketch-in-polar-coordinates", "openwebmath_score": 0.9936886429786682, "openwebmath_perplexity": 215.99766148976502, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877675527112, "lm_q2_score": 0.8824278757303677, "lm_q1q2_score": 0.8565619447189919}}
{"url": "https://mymathforum.com/threads/how-can-i-find-the-work-on-a-block-when-it-is-pulled-up-in-a-curve.347508/", "text": "# How can I find the work on a block when it is pulled up in a curve?\n\n#### Chemist116\n\nThe problem is as follows:\n\nThe figure from below shows a force acting on block as it is pulled upwards in a curve from point $A$ to $B$. It is known that the block is pulled by a force which its modulus is $100\\,N$. Find the work (in Joules) of such force between the points indicated. Consider that the angle given is with respect of the vertical with the floor.\n\nThe alternatives given in my book are:\n\n$\\begin{array}{ll} 1.&143\\,J\\\\ 2.&312\\,J\\\\ 3.&222\\,J\\\\ 4.&98\\,J\\\\ 5.&111\\,J\\\\ \\end{array}$\n\nI attempted to decompose the force given as such:\n\n$F\\cos 37^{\\circ} \\times d = W$\n\nBut the result doesn't seem to yield an adequate result:\n\n$W= 100\\times \\cos 37^{\\circ} \\times 2.1= 100 \\times \\frac{4}{5}\\times 2.1=168$\n\nAssuming the gravity does negative work?\n\n$W= - 100 \\times \\sin 37^{\\circ}= - 100 \\times \\frac{3}{5}\\times 1.2= -72$\n\nAnyways the sum doesn't yield the result which supposedly is option $5$. Can somebody help me here? :help:\n\n#### DarnItJimImAnEngineer\n\nTreat $\\vec{F}$ as two separate forces. Horizontal force $F \\sin 37\u00c2\u00b0$ acts over a horizontal distance of 2.1 m. The vertical force acts over a vertical distance of 1.2 m. Try finding the work from each and adding them.\n\nLast edited by a moderator:\n2 people\n\n#### Chemist116\n\nTreat $\\vec{F}$ as two separate forces. Horizontal force $F \\sin 37\u00c2\u00b0$ acts over a horizontal distance of 2.1 m. The vertical force acts over a vertical distance of 1.2 m. Try finding the work from each and adding them.\nI attempted to do such thing.\n\nLet's see:\n\n$F\\sin 37^{\\circ}\\cdot 2.1 = 100 \\cdot \\frac{3}{5} \\cdot 2.1 = 126 J$\n\n$F\\cos 37^{\\circ}\\cdot 1.2 = 100 \\cdot \\frac{4}{5} \\cdot 1.2 = 96 J$\n\nThe summing both would give me:\n\n$W_{1}+W_{2}=126+96=222\\,J$\n\nwhich supposedly is the third option. I'm not doing any sort of vector sum here as work is scalar (thanks to romsek for that remark). Mind to confirm whether what I'm doing is what you intended to say?\n\nI'm sorry if I mentioned that the answer is $111\\,J$. I checked with the answers sheet and it lists it to be $222\\,J$. I feel so dumb now, as when I compared with what I did in my notes I got an error in the summation. Learning, yikes!\n\nBut I'm still confused, by comparing it to a similar problem which I posted before this, why shouldn't I subtract the \"angle\" which could result by joining $AB$? Is it because it's a curve? Can you help me with this part please? :help:\n\nLast edited by a moderator:", "date": "2019-12-05 20:22:16", "meta": {"domain": "mymathforum.com", "url": "https://mymathforum.com/threads/how-can-i-find-the-work-on-a-block-when-it-is-pulled-up-in-a-curve.347508/", "openwebmath_score": 0.8642544746398926, "openwebmath_perplexity": 365.674266728435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877700966099, "lm_q2_score": 0.8824278556326344, "lm_q1q2_score": 0.856561927455175}}
{"url": "https://math.stackexchange.com/questions/1200560/matrix-of-t-over-p-2", "text": "# Matrix of $T$ over $P_2$\n\nI'm having a bit of trouble following the logic my professor used to construct the matrix of a given transformation $T$ in class, and was wondering if anyone could share any further intuition or insight.\n\nGiven $P_2$, the set of polynomials of degree $\\leq2$, define a linear transformation $T:P_2\\rightarrow \\mathbb{R}^3$ such that $T\\big(p(x)\\big) = \\begin{pmatrix} p(0)\\\\p(1)\\\\p(2)\\end{pmatrix}$.\n\nThe matrix of $T$ with respect to the basis $\\mathcal{B}=\\left\\{1,x,x^2\\right\\}\\\\$\n\nis given by $T(1)=\\begin{pmatrix} 1\\\\1\\\\1\\end{pmatrix}$, $T(x)=\\begin{pmatrix} 0\\\\1\\\\2\\end{pmatrix}$, and $T(x^2)=\\begin{pmatrix} 0\\\\1\\\\4\\end{pmatrix}$,\n\ni.e. $\\,A=\\begin{pmatrix}1&0&0\\\\1&1&1\\\\1&2&4\\end{pmatrix}$.\n\nI get the whole business of the coefficients on this particular transformation being the \"trivial relation\" such that $c_1=c_m=0$, and that $ImT=\\mathbb{R}^{3}$ such that $T$ is an isomorphism.\n\nWhat I'm struggling with, quite frankly, is how he got the coefficients for $A$. The coefficients for $p(x)=1$ in particular are a bit counter-intuitive. Is the idea that, if we define $p(x)=1$, then no matter what, $p(0)=p(1)=p(2)=1$?\n\nThe notation for expressing the various matrices associated with an isomorphism is also a bit nebulous for me. My textbook defines a $\\mathcal{B}$-coordinate transformation as\n\n$T^{-1}\\begin{bmatrix}c_1\\\\\\vdots\\\\c_n\\end{bmatrix}=c_1f_1+\\cdots + c_nf_n$.\n\nWhat would the inverse of the transformation $T\\big(p(x)\\big) = \\begin{pmatrix} p(0)\\\\p(1)\\\\p(2)\\end{pmatrix}$ be?\n\nWhat would the matrix associated with the inverse be? Or would it be a set of three different polynomials (linear equations)?\n\nFor those wondering, we're using Bretscher's Linear Algebra with Applications, 5th edition. Thanks!\n\n\u2022 $A = [[T(1)]_B|[T(x)]_B|[T(x^2)]_B]$ where $B=\\{ e_1,e_2,e_3 \\}$ is just the standard basis for $\\mathbb{R}^3$. More generally, you just find the coordinate vectors of the image of each vector in the domain basis and glue them together. There are good reasons for doing this, but you ought to have seen them in lecture I think... \u2013\u00a0James S. Cook Mar 22 '15 at 4:42\n\u2022 My silly notation $[v]_B = v$ in this case since the coordinate system in the codomain is just the plain-old-standard Cartesian coordinate system. \u2013\u00a0James S. Cook Mar 22 '15 at 4:43\n\nTo give you a much more complicated example. Let us consider $T(f(x)) = f'(x)$ where we consider $T:P_2 \\rightarrow P_2$ and to be perverse let's use $\\beta = \\{ 1,x,x^2 \\}$ as the domain basis, but $\\gamma = \\{ x^2,x,1 \\}$ as the codomain basis. To understand $T$ I like to consider $f(x)=a+bx+cx^2$ and see what happens: $$T(f(x)) = b+2cx$$ Then the coordinate vector of the image is easy enough to see: $$[T(f(x))]_{\\gamma}= [b+2cx]_{\\gamma} = [0(x^2)+2c(x)+b(1)]_{\\gamma} = [0,2c,b]^T.$$ Then, the matrix $[T]_{\\beta,\\gamma}$ is the matrix which when multiplied on $$[f(x)]_{\\beta} = [a+bx+cx^2]_{\\beta} = [a,b,c]^T$$ yields $[T(f(x))]_{\\gamma}=[0,2c,b]^T$. A moments reflection yields: $$[T]_{\\beta, \\gamma} = \\left[ \\begin{array}{ccc} 0 & 0 & 0 \\\\ 0 & 0 & 2 \\\\ 0 & 1 & 0\\end{array}\\right].$$ Alternatively, just use: $$[T]_{\\beta, \\gamma} = [ [T(1)]_{\\gamma} | [T(x)]_{\\gamma}| [T(x^2)]_{\\gamma} ] = [[0]_{\\gamma}|[1]_{\\gamma}|[2x]_{\\gamma}] = \\left[ \\begin{array}{ccc} 0 & 0 & 0 \\\\ 0 & 0 & 2 \\\\ 0 & 1 & 0\\end{array}\\right].$$ Some books also use $[T]_{\\beta}^{\\gamma}$ to reflect the differing roles the domain and codomain bases play especially in regard to coordinate change. Perhaps your text does that.\n\u2022 The text doesn't explicitly use the terms \"domain\" and \"codomain\", but I think I understand. Follow-up question: if we define some other basis $\\mathcal{C}=\\left\\{x^2,xy,y^2\\right\\}$ and $T(x)=\\begin{pmatrix}f(1,0)\\\\f(0,1)\\\\f(1,1)\\end{pmatrix}$, does the computation still move along similar lines? see here for more. \u2013\u00a0Benjamin Loya Mar 22 '15 at 5:05", "date": "2019-12-09 05:31:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1200560/matrix-of-t-over-p-2", "openwebmath_score": 0.9328702092170715, "openwebmath_perplexity": 202.4662831874779, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877717925421, "lm_q2_score": 0.8824278540866547, "lm_q1q2_score": 0.8565619274510493}}
{"url": "https://www.physicsforums.com/threads/integration-of-a-rational-function.409211/", "text": "# Integration of a rational function\n\nstripes\n\n## Homework Statement\n\nfind\n$$\\int\\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)}dx$$\n\nNone\n\n## The Attempt at a Solution\n\n$$\\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \\frac{A}{x-1} + \\frac{B}{(x-1)^{2}} + \\frac{Cx + D}{x^{2}+1}$$\n\n$$x^{2}-2x-1 = A(x-1)(x^{2}+1) + B(x^{2}+1) + (Cx + D)(x-1)^{2}$$\n\n$$x^{2}-2x-1 = Ax^{3}-Ax^{2}+Ax-A+Bx^{2}+B+Cx^{3}-2Cx^{2}+Cx+Dx^{2}-2Dx+D$$\n\n$$x^{2}-2x-1 = x^{3}(A+B+C) + x^{2}(-A+B-2C+D) + x(A+C-2D) - A + B +D$$\n\nso A+B+C = 0, -A+B-2C+D = 1, A+C-2D=-2 and -A+B+D=-1\n\nsolving for coefficients, we get\nA = 5/3\nB = -2/3\nC = -1\nD = 4/3\n\nso $$\\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \\frac{5/3}{x-1} - \\frac{2/3}{(x-1)^{2}} - \\frac{x-(4/3)}{x^{2}+1}$$\n\nbut apparently, the last statement is not correct. Did I break down the rational function incorrectly?\n\nTedjn\nCheck A+B+C=0.\n\nHomework Helper\ni think that's ok? but -A+B-2C+D = (-5 -2 +6-4)/3 = -5/3?\n\nLast edited:\nstripes\nso did I get all my coefficients messed up? Arghhh...\n\nThe Chaz\nCount Iblis\n\nCheating to the third power\n\n1) Obviously the Prof. doesn't want students to use Wolfram alpha.\n\n2) The \"show steps\" feature adds insult to injury. Just knowing the correct answer is going to give away nontrivial information to the student.\n\n3) And Wolfram alpha is lying to you when it shows the steps. Not that there is anything wrong in these steps, but these are not the steps Wolfram alpha uses itself when it computed the answer. If you ask Wolfram alpha to show the steps, what it does is crank up the high school algorithm that gives the same answer, as it is most likely that this will yield the desired steps (I mean, who else other than students having difficulties with their assignments would want to see this?).\n\nWolfram alpha uses series expansion methods to compute the answer, which is way more efficient than the usual high school method. The general rule of thumb is that methods that involve solving equations for variables are less efficient compared to methods that don't involve solving equations.\n\nMentor\nso $$\\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \\frac{5/3}{x-1} - \\frac{2/3}{(x-1)^{2}} - \\frac{x-(4/3)}{x^{2}+1}$$\n\nbut apparently, the last statement is not correct. Did I break down the rational function incorrectly?\nFinding the constants is tedious and error-prone, but checking them is relatively simple, and you should do this. If you get a common denominator for what you have on the right, you should end up with what you have on the left. If you do, then you have the right constants.\n\nstripes\nI ended up finding the right constants, they were 1, -1, 1, and -1 I think. Did the question again and got it right!\n\nAnd as for the Wolfram Alpha thing...I use it when I am completely stumped. It gives me hints and guides me in the right direction. And this \"high school method\" you speak of makes me sound like a high-schooler, which I am most definitely not lolll.\n\nStaff Emeritus\nHomework Helper\nOne technique I like to use when solving for the coefficients is the Heaviside coverup method. When you get to this point\n\n$$x^{2}-2x-1 = A(x-1)(x^{2}+1) + B(x^{2}+1) + (Cx + D)(x-1)^{2}$$\n\nSubstitute a value for x to eliminate some terms. In this case, x=1 works. This leaves you with $-2 = 2B$, and you can immediately see that B=-1. Ideally, you can use different values to solve to isolate different coefficients quickly. Alas, that's not the case in this problem. If you plug in the value you found for B, you get\n\n$$2x^{2}-2x = A(x-1)(x^{2}+1) + (Cx + D)(x-1)^{2}$$\n\nso you only have three equations and three unknowns to solve.\n\nCount Iblis\nI see! Well, this is how I would do this problem. I would start by writing:\n\nx^2 - 2 x - 1 = x^2 - 2 x + 1 - 2 = (x-1)^2 - 2\n\nSo, we have:\n\n(x^2 - 2 x - 1)/[(x-1)^2 (x^2 + 1)] =\n\n1/(x^2+1) - 2/[(x-1)^2 (x^2 + 1)]\n\nExpanding the last term about the point x = 1 and keeping only the singular terms yields:\n\n- 2/[(x-1)^2 (x^2 + 1)] = -2/(x-1)^2 [expansion of 1/(x^2+1) around x = 1]\n\nPut x = 1 + t:\n\n1/(x^2+1) = 1/[(1+t)^2 + 1] = 1/(2 + 2 t + t^2) =\n\n1/2 1/[1+t+t^2/2] = 1/2 [1-t +t^2/2 + ...]\n\nSo, we have the expansion:\n\n- 2/[(x-1)^2 (x^2 + 1)] =\n\n-1/(x-1)^2 + 1/(x-1) + nonsingular terms\n\nIf we now do the same at the singularities of 1/(x^2+1) and keep the singular terms, then the sum of all the singular terms from all the expansions, S(x), is the partial fraction expansion. This is because the difference between the rational function R(x) and S(x) will be a function that has no singularities, therefore it is a polynomial. But since both R(x) and S(x) tend to zero at infinity, we have that\nR(x) = S(x).\n\nSo, we could now proceed to find the singular terms in the expansion around the singularities of 1/(x^2+1) at x = \u00b1i. However, we can skip that as the partial fraction expansion is already determined by the above terms. This is because the large x behavior of\n\n- 2/[(x-1)^2 (x^2 + 1)]\n\nis -2/x^4, while the part of the partial fracton expansion we have so far is\n\n-1/(x-1)^2 + 1/(x-1)\n\nSo, clearly the partial fraction expansion due to the singular terms coming from 1/(x^2+1) must cancel the asymptotic 1/x and 1/x^2 behavior of the above two terms. We know that the the remaining terms of the partial fraction expansion will be of the form:\n\n(A x + B)/(x^2 +1)\n\nExpanding around x = infinity gives:\n\n(A x + B) 1/x^2 1/(1+1/x^2) =\n\nA/x + B/x^2 + O(1/x^3)\n\nExpanding the two terms we got so far about infinity gives:\n\n-1/(x-1)^2 + 1/(x-1) =\n\n-1/x^2 + 1/x 1/(1 - 1/x) + O(1/x^3) =\n\n-1/x^2 + 1/x + 1/x^2 + O(1/x^3) =\n\n1/x + O(1/x^3)\n\nSo, A = -1 and B = 0 and we have:\n\n(x^2 - 2 x - 1)/[(x-1)^2 (x^2 + 1)] =\n\n(1-x)/(x^2+1) -1/(x-1)^2 + 1/(x-1)\n\nstripes\ncount iblis, B \u2260 0, B = -1\n\nCount Iblis\ncount iblis, B \u2260 0, B = -1\n\nYour B is my 1/(x-1)^2 coefficient.\n\nStaff Emeritus\nHomework Helper\nI see! Well, this is how I would do this problem.\nI wonder what the grader would think if stripes were to turn in his homework using that solution.\n\nCount Iblis\nI wonder what the grader would think if stripes were to turn in his homework using that solution.\n\nIt's not that different from the Heaviside coverup method you mentioned. If we have a rational function R(x) which has a linear factor in the denominator 1/(x-a)^n, then we can take that factor out and write:\n\nR(x) = P(x) 1/(x-a)^n\n\nSo, we see that P(a) is the coefficient of 1/(x-a)^n (to do this \"by the book\", you would write out the equations as you did and then put x = a). To find the coeficient of 1/(x-a)^(n-1), using the Heaviside coverup method, one would subtract p(a)/(x-a)^n from R(x), simplify the rational function; you know that the numerator will contain a factor x-a that cancels against the denominator (you can use synthetic division to divide the numerator by x - a), so we can write:\n\nR2(x) = R(x) - p(a)/(x-a)^n = P2(x)/(x-a)^(n-1)\n\nAnd then we can repeat this by subtracting P2(a)/(x-a)^(n-1) from R2(x), simplify the rational function, divide numerator and denominator by x-a to obtain:\n\nR3(x) = R2(x) - p2(a)/(x-a)^(n-1) = P3(x)/(x-a)^(n-2)\n\netc. etc.\n\nBut this whole process is equivalent to finding the first part of the Laurent expansion of R(x) around x = a, so you could just as well do that using any method that is most convenient, not necessarily using the above iterative process...\n\nCount Iblis\nSo, in the way you wrote it down the next step would be to factor the left hand side of the last equation\n\n2x^2 - 2x = 2x(x-1)\n\nand cancel (x-1) on both sides:\n\n2x = A(x^2+1) + (Cx+D) (x-1)", "date": "2023-03-21 02:02:52", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integration-of-a-rational-function.409211/", "openwebmath_score": 0.7525143027305603, "openwebmath_perplexity": 721.8147524644543, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877667047451, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8565619259627626}}
{"url": "https://www.physicsforums.com/threads/definition-of-average-inclination.745604/", "text": "# Definition of 'average inclination'?\n\n1. Mar 27, 2014\n\n### dilloncyh\n\nIt's not a homework problem, just a problem that suddenly popped out of my mind.\n\n1. The problem statement, all variables and given/known data\n\nSo, my question is : how to calculate, or how to define 'average inclination'? Suppose I am given an equation y=f(x) that resembles the shape of a section of a hill, and I want to calculate the average inclination (something I always see when I watch cycling or alpine skiing race), how do I do that? Let's use f(x) = x^2 as an example for the following discussion.\n\n2. Relevant equations\n\nI know the average of a function is defined as:\nSo I suppose the correct equation should be similar, with f(x) the function that gives the shape of the hill I am calculating?\n\n3. The attempt at a solution\n\nHere comes the problem: Do I use dx for ds? And for (b-a) in the image, I need to replace it with the actual length of the function from x=a to x=b, right?\nSince I don't really know I want to calculate (to find the slope at each point of the function and add them together, and then divide it by the total length of the slope?), my question may seem very silly, but please give me some idea.\n\nthanks\n1. The problem statement, all variables and given/known data\n\n2. Relevant equations\n\n3. The attempt at a solution\n\n2. Mar 27, 2014\n\n### Simon Bridge\n\nThe average slope is usually just the rise-over-run for two positions on the slope.\n$$\\frac{f(x+\\Delta x)-f(x)}{\\Delta x}$$ ... that would be what they did on the ski races.\n\nYou want to be a bit more detailed than that:\n\nIf $y(x)$ is the height of the slope at position $x$,\nThen $$\\bar y = \\frac{1}{b-a}\\int_a^b y(x)\\;\\text{d}x$$ would be the average height of the hill in a<x<b.\n\nThe slope (gradient) function would be g(x)=dy/dx ... tells you the gradient at position x.\n\n3. Mar 27, 2014\n\n### dilloncyh\n\nI'm still a bit confused.\n\naverage slope = detla y / delta x seems very legit, but take y=x^2 and y=x as example.\n\nIf I want to calculate the average slope between x=0 and x=1, then by calculating the difference in height and horizontal displacement of the two end points, then both should give the same result (slope=1), but the length of the curve of y=x^2 is obviously longer than y=x (which is just sq root of 2), so by calculating the the sine of the 'triangle' if I straighten the curve section of y=x^2, I will get difference result.\n\n4. Mar 27, 2014\n\n### haruspex\n\nIt's a matter of how you choose to define the average. The usual would be to define it as average over horizontal distance, but as you note you could instead chose to define it as average over path distance.\n\n5. Mar 28, 2014\n\n### Simon Bridge\n\nI'm with haruspex - there is no reason that two different averaging methods should produce the same value.\nThey are just producing a different kind of average.", "date": "2017-12-12 17:53:26", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/definition-of-average-inclination.745604/", "openwebmath_score": 0.8125982284545898, "openwebmath_perplexity": 566.2114124021376, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075744568837, "lm_q2_score": 0.8902942290328344, "lm_q1q2_score": 0.8565588212477415}}
{"url": "http://residuetheorem.com/2014/12/19/expansions-of-ex/", "text": "## Expansions of $e^x$\n\nA very basic question was asked recently: What is a better approximation to $e^x$, the usual Taylor approximation, or a similar approximation involving $1/e^{-x}$?\n\nMore precisely, given an integer $m$, which is a better approximation to $e^x$:\n\n$$f_1(x) = \\sum_{k=0}^m \\frac{x^k}{k!}$$\n\nor\n\n$$f_2(x) = \\frac1{\\displaystyle \\sum_{k=0}^m \\frac{(-1)^k x^k}{k!}}$$\n\nThe answer is amazingly simple: if $m$ is even, then $f_2$ is more accurate, while if $m$ is odd, $f_1$ is more accurate. The proof below is, in my opinion, really nifty and worth showing here. Keep in mind that this result follows from the fact that we are dealing with a remarkable function with very special properties. Other functions will not exhibit such easily calculable results as shown below. (Anyone familiar with Pade approximates can attest to how sticky rational approximations to functions can get.)\n\nKeep in mind that, by \u201caccuracy,\u201d I refer to truncation error rather than roundoff error. Truncation error is that which results from substituting a Taylor series with a polynomial of finite order. The accuracy, i.e., truncation error, of any Taylor expansion of a given order is given by the next order of the expansion. For example, consider the first order approximation to $1/e^{-x}$, $f_2$, and its second-order error:\n\n$$\\frac1{1-x} = 1+x+x^2+O(x^3)$$\n\nThis approximation has twice the error as the first order approximation to $e^x$, $1+x$, whose second order term is $x^2/2!$.\n\nNow consider the second order approximations with third order errors:\n\n\\begin{align}\\frac1{1-x+x^2/2} &= 1+\\left (x-\\frac{x^2}{2} \\right ) + \\left (x-\\frac{x^2}{2} \\right )^2+ \\left (x-\\frac{x^2}{2} \\right )^3 + \\cdots \\\\&= 1+x+\\frac{x^2}{2} + O(x^4)\\end{align}\n\nNote that, for the $f_2$ approximation, the third order error vanishes! This is obviously not the case for $f_1$. Thus, for the second-order expansion, $f_2$ has a smaller truncation error than $f_1$. However, for the next order approximation, the expansion of $f_2$ is\n\n$$1+x+\\frac{x^2}{2}+\\frac{x^3}{6}+\\frac{x^4}{12} + O(x^5)$$\n\nwhereas the fourth order term in the direct, Taylor expansion of the exponential is $x^4/24$. Thus, as with the first order approximation, $f_2$ has twice the truncation error as $f_1$.\n\nThere is a systematic way to prove that the accuracy alternates with order oddness or evenness by considering the error in the expansion of $e^x \\cdot e^{-x}=1$. The proof is actually not all that hard. Consider the finite approximations to the equation $e^x \\cdot e^{-x}=1$:\n\n$$\\left (\\sum_{k=0}^m \\frac{x^k}{k!} \\right ) \\left (\\sum_{k=0}^m (-1)^k \\frac{x^k}{k!} \\right )$$\n\nIt may be easily shown that the coefficients of $x$, $x^2$, \u2026, $x^m$ are zero. We now consider the first error term, i.e., the coefficient of $x^{m+1}$:\n\n$$\\frac{x^1}{1!} \\frac{(-1)^m x^m}{m!} + \\frac{x^2}{2!} \\frac{(-1)^{m-1} x^{m-1}}{(m-1)!} +\\cdots + \\frac{x^m}{m!} \\frac{(-1)^1 x^1}{1!} = \\sum_{k=1}^m \\frac{(-1)^{m-k+1}}{k! (m-k+1)!} x^{m+1}$$\n\nRewrite this sum, sans the $x^{m+1}$ term, as\n\n$$\\sum_{k=0}^{m-1} \\frac{(-1)^{m-k}}{(k+1)! (m-k)!} = (-1)^m \\sum_{k=0}^{m-1} \\frac{(-1)^k}{k! (m-k)!} \\frac1{k+1}$$\n\nWe can evaluate this sum by realizing that\n\n$$(1-x)^m = \\sum_{k=0}^m (-1)^k \\binom{m}{k} x^k$$\n\nso that\n\n$$\\sum_{k=0}^m (-1)^k \\binom{m}{k} \\frac1{k+1} = \\int_0^1 dx \\, (1-x)^m = \\frac1{m+1}$$\n\nTherefore, by subtracting off the last term in the sum, we find that\n\n$$(-1)^m \\sum_{k=0}^{m-1} \\frac{(-1)^k}{k! (m-k)!} \\frac1{k+1} = \u2013 \\frac{1-(-1)^m}{(m+1)!}$$\n\nTherefore, we now may say that\n\n$$\\frac1{\\displaystyle \\sum_{k=0}^m (-1)^k \\frac{x^k}{k!} } = \\sum_{k=0}^m \\frac{x^k}{k!} + \\frac{1-(-1)^m}{(m+1)!} x^{m+1} + O(x^{m+2})$$\n\nFor even values of $m$, the next term error is zero, which is smaller than that for the direct Taylor series, which has error $x^{m+1}/(m+1)!$. On the other hand, for odd values, the error is double that of the direct Taylor series. This agrees with the above examples.\n\n#### One Comment\n\n\u2022 This was indeed very surprising and beautiful (and at the same time not too difficult to understand).", "date": "2017-09-26 09:08:09", "meta": {"domain": "residuetheorem.com", "url": "http://residuetheorem.com/2014/12/19/expansions-of-ex/", "openwebmath_score": 0.9731026291847229, "openwebmath_perplexity": 224.60587648974763, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9920620054292071, "lm_q2_score": 0.863391611731321, "lm_q1q2_score": 0.8565380138049296}}
{"url": "http://math.stackexchange.com/questions/328349/probability-determining-which-phone-plan-is-better/329266", "text": "# Probability: Determining Which Phone Plan Is Better\n\nA consumer is trying to decide between two long-distance calling plans. The first one charges a flat rate of $10$ cents per minute, whereas the second charges a flat rate of $99$ cents for calls up to 20 minutes in duration and then $10$ cents for each additional minute exceeding 20 (assume that calls lasting a non-integer number of minutes are charged proportionately to a whole-minute\u2019s charge). Suppose the consumer\u2019s distribution of call duration is exponential with parameter.\n\n\u2022 Which plan is better if expected call duration is 10 minutes? 15 minutes? [Hint: Let $h_1(x)$ denote the cost for the first plan when call duration is x minutes and let $h_2(x)$ be the cost function for the second plan. Give expressions for these two cost functions, and then determine the expected cost for each plan.]\n\nFor the first function of cost, I got $h_1(x) = 0.10X$; and for the second one, I got the piece-wise function $h_2(x) = .99$ for $0 \\le X \\le 20$, and $h_2(x) = .99$ for $X > 20$\n\nAre these correct? Also, how do I calculate the expected cost for each plan? I used the formula given in this link http://en.wikipedia.org/wiki/Expected_value#Functional_non-invariance; however, it didn't work.\n\nEDIT: $E[h_1(x)] = lim_{a \\rightarrow \\infty} \\int^a_0 .10x \\lambda e^{-\\lambda x}dx$\n\nAfter doing integration by parts, and simplifying, I get to the step $lim_{a \\rightarrow \\infty}[-.10ae^{-\\lambda a} + \\frac{.1}{\\lambda}(e^{-\\lambda a})]$. I know that the second term goes to zero, and, according to my teacher, the first term goes to zero as well, because $e^{-\\lambda a}$ goes to zero more quickly than $-.10a$ goes to negative infinity, although I don't really understand why. Note, the answer is not zero\n\n-\nIf the second plan charges $99$ cents plus an additional $10$ cents per minute after $20$ minutes, I don't think $h_2(x) = 0.99$ for $X > 20$ is correct. \u2013\u00a0 TMM Mar 12 '13 at 11:32\n@TMM Should it be $h_2(x) = .99 + 0.10X$ for $X > 20$? \u2013\u00a0 Mack Mar 12 '13 at 11:55\n@Eli No It is $0.99 + .1(X-20)$ \u2013\u00a0 Gautam Shenoy Mar 12 '13 at 11:59\n@GautamShenoy Are you sure? Can you explain why that is the answer? \u2013\u00a0 Mack Mar 12 '13 at 12:12\n@Eli: What I meant was: He says it is 10 cents for every minute exceeding 20. So in place of X, shouldn't it be X-20 ? It is not the answer. I am just correcting the expression you gave. \u2013\u00a0 Gautam Shenoy Mar 13 '13 at 9:15\n\nAll costs are in dollars.\n\nYour $h_1(X) = 0.1X$ is correct. The second term is $$h_2(X) = 0.99 + 0.1(X-20)1_{X > 20}$$\n\nNow if you compute the expected cost, for $\\frac{1}{\\lambda} = 10$, it is\n\n$\\mathbb{E}h_1(X) = \\frac{0.1}{\\lambda}= 1$\n\nBut for $$\\mathbb{E}h_2(X) = 0.99 + 0.1 \\mathbb{E}(X-20)1_{X>20}$$ $$= .99 - 2\\mathbb{P}(X>20) + .1\\mathbb{E}(X1_{X>20})$$ $$= .99 - 2e^{-\\lambda 20} + .1\\mathbb{E}(X1_{X>20})$$ Thus we just need to compute $\\mathbb{E}(X1_{X>20})$. You could do it directly, but I will provide an alternate way.\n\n$$E[X1_{X>20}] = \\int_0^\\infty x1_{x>20}f_x(x)dx$$ $$= \\int_{x=0}^\\infty \\int_{u=0}^\\infty 1_{u<x}1_{x>20}f_x(x)dudx$$ $$= \\int_{u=0}^\\infty \\int_{x=0}^\\infty 1_{u<x}1_{x>20}f_x(x)dudx$$ (Fubini theorem) $$= \\int_{u=0}^\\infty \\int_{x=u}^\\infty 1_{x>20}f_x(x)dxdu$$ $$= \\int_{u=0}^\\infty \\int_{x=\\max(u,20)}^\\infty f_x(x)dxdu$$ $$= \\int_{u=0}^\\infty P(X > \\max(u,20))du$$\n\nNow split the integral as $$= \\int_{u=20}^\\infty P(X > u)du + \\int_{u=0}^{20}P(X > 20)du$$ $$= \\int_{20}^\\infty e^{-\\lambda u}du + \\int_{u=0}^{20}e^{-\\lambda 20}du$$\n\nI could work out the rest but can you proceed from here? Also in reply to your question: why is $\\lim_{x \\to \\infty} xe^{-x} = 0$, do you know L-Hospital's rule?\n\n-", "date": "2014-10-02 08:25:11", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/328349/probability-determining-which-phone-plan-is-better/329266", "openwebmath_score": 0.8398162722587585, "openwebmath_perplexity": 412.85725692546174, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471647042428, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.856534650819597}}
{"url": "http://math.stackexchange.com/questions/775121/how-to-prove-this-series-converges", "text": "# How to prove this series converges?\n\nWhat test do you use to prove that\n\n$$\\sum_{n=2}^\\infty \\frac{\\ln(n)}{n^{3/2}}$$\n\nconverges?\n\nI tried the limit comparison test using $\\frac{1}{n^{3/2}}$ as the comparison, but it did not converge.\n\n-\n\nHint: You can use integral test.\n\nAdded: You need to consider the integral\n\n$$\\int_{2}^{\\infty} \\frac{\\ln(x)}{x^{3/2}}dx ,$$\n\nwhich can be integrated using integration by parts.\n\n-\n@MhenniBenghorbal, there is a typo, the lower bound should be 2, but it doesn't seem like it's going to matter that much for this question. \u2013\u00a0 Nameless Apr 30 '14 at 2:47\n@Nameless: It is corrected. Thanks for the comment. \u2013\u00a0 Mhenni Benghorbal Apr 30 '14 at 2:48\nUsing the integral test I got the answer to be = 25.55275762. \u2013\u00a0 khap93 Apr 30 '14 at 2:49\n@khap93, you are not supposed to directly calculate the sum...well I guess that shows you the sum does converge. \u2013\u00a0 Nameless Apr 30 '14 at 2:49\n@MhenniBenghorbal, no problem. I'll upvote you. \u2013\u00a0 Nameless Apr 30 '14 at 2:50\n\nApply Cauchy Condensation Test.\n\nLet $a_n = \\dfrac{\\ln(n)}{n^{3/2}}$, then $a_{2^n} = \\dfrac{\\ln 2^n }{(2^n)^{3/2}}$\n\nTherefore, $$a_{2^n} = \\dfrac{\\ln 2^n }{(2^n)^{3/2}} = \\frac{n\\ln 2 }{2^{3n/2}} \\leq \\ln2\\frac{n}{2^n}$$\n\nThus $$\\sum_{n=2}^{\\infty}\\dfrac{\\ln 2^n }{(2^n)^{3/2}} \\leq \\sum_{n=2}^{\\infty} \\ln2\\frac{n}{2^n}.$$\n\nThe RHS is a convergent series, by the Ratio Test, as you should verify.\n\n-\nNeat footwork there :-) \u2013\u00a0 Carl Witthoft Apr 30 '14 at 13:17\n\nYou can use that $\\ln n$ goes to infinitely much more slowly that any (positive) power of $n$, that is: for every $\\alpha>0$, $\\ln n < n^\\alpha$ for $n$ large enough. You can apply this to $\\alpha=1/4$ says (any $\\alpha$ strictly between $0$ and $1/2$ will do, so let's say $1/4$), so $\\ln n < n^{1/4}$ for $n$ large enough, hence $\\ln n / n^{3/2} < 1/n^{5/4}$ for $n$ large enough, and since the series $\\sum 1/n^{5/4}$ converges (as does $\\sum 1/n^\\beta$ for any $\\beta>1$), you're done.\n\n-\n+1 the most direct and conceptually simple solution. \u2013\u00a0 jwg Apr 30 '14 at 10:19", "date": "2015-07-07 03:12:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/775121/how-to-prove-this-series-converges", "openwebmath_score": 0.9863249063491821, "openwebmath_perplexity": 402.5900040540244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471661250913, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.85653465040497}}
{"url": "https://web2.0calc.com/questions/help-me-please_130", "text": "+0\n\n0\n330\n8\n\nLet $b$ and $c$ be constants such that the quadratic $-2x^2 +bx +c$ has roots $3+\\sqrt{5}$ and $3-\\sqrt{5}$. Find the vertex of the graph of the equation $y=-2x^2 + bx + c$.\n\nGuest\u00a0Feb 19, 2018\n#1\n+12560\n+2\n\nThere MIGHT (probably) be an easier way, but here is ONE way:\n\n(x+(-3-sqrt5))(x+(-3+sqrt5))\u00a0\u00a0 is given in the question for the roots\n\nif you multiply these two term out you will get\n\nx^2-6x+4\u00a0\u00a0\u00a0 now multiply by -2\n\n-2x^2 +12x-8 =0\u00a0\u00a0\u00a0\u00a0 (a parabola)\n\nto find vertex take derivative and set equal to zero\n\n-4x+12 = 0\u00a0\u00a0\u00a0\u00a0 x = 3\u00a0\u00a0\u00a0 substitue this into the equation\n\nto get\u00a0\u00a0\u00a0\u00a0\u00a0 y=10\u00a0\u00a0\u00a0\u00a0 for the vertex\u00a0\u00a0\u00a0 3,10\n\nElectricPavlov \u00a0Feb 19, 2018\nedited by ElectricPavlov \u00a0Feb 19, 2018\n#3\n0\n\nShould it be (x+(3-sqrt 5))(x+(3+sqrt5))?\n\nGuest\u00a0Feb 21, 2018\n#4\n0\n\nShouldnt it be (-3,10)?\n\nGuest\u00a0Feb 21, 2018\n#5\n+12560\n+2\n\nThe roots are given\u00a0\u00a0 as\u00a0\u00a0 3+-sqrt5\n\nso one part would be\n\nx+(- 3+sqrt5) =0\n\nx-3+sqrt5\u00a0 = 0 \u00a0 \u00a0 solve for x\u00a0\u00a0 (add\u00a0 (3 and subtract sqrt5) from both sides )\n\nx= 3 -sqrt5\n\nand the other would be\u00a0 x+( -3-sqrt5)\u00a0 =0\n\nx-3-sqrt5\u00a0 \u00a0 solve for x\u00a0 (add 3 and sqrt 5 to both sides\n\nx = 3+sqrt5\n\nThese are the roots given in the question....\n\nElectricPavlov \u00a0Feb 21, 2018\n#6\n+12560\n+2\n\nHere is a graph\n\nElectricPavlov \u00a0Feb 21, 2018\n#7\n+12560\n+2\n\ny=\u00a0 a (x-h)^2\u00a0 +k \u00a0 \u00a0 \u00a0\u00a0 and\n\ny= -2 (x-3)^2 +10\n\nNow can you see that a =-2\u00a0\u00a0\u00a0 h=3\u00a0\u00a0 k = 10\u00a0 ???\n\nElectricPavlov \u00a0Feb 21, 2018\n#2\n+12560\n+2\n\nOK.....another way...\n\nAFTER you get to\n\n-2x^2 +12x -8\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 arrange into vertex form\u00a0 y=a(x-h)^2 +k\u00a0\u00a0\u00a0 the vertex is h,k\n\n-2{ x^2 -6x +4}\n\n-2{(x-3)^2 +4 -9}\n\n-2 {(x-3)^2 -5}\n\n-2 (x-3)^2 +10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 so\u00a0 h,k is the vertex\u00a0\u00a0 =\u00a0\u00a0 3,10\n\nElectricPavlov \u00a0Feb 19, 2018\n#8\n+92805\n+1\n\nI am quite sure EP will be correct but I will take a look:\n\nLet b and c be constants such that the quadratic $$-2x^2 +bx +c$$\n\nhas roots $$3+\\sqrt{5}$$\u00a0 and $$3-\\sqrt{5}$$.\n\nFind the vertex of the graph of the equation\n\n$$y=-2x^2 + bx + c$$\n\nThe roots of a quadratic\u00a0 \u00a0$$ax^2+bx+c$$\u00a0 is the answers to\u00a0 \u00a0$$ax^2+bx+c=0$$\n\n$$x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}\\\\ x=\\frac{-b}{2a}\\pm\\frac{\\sqrt{b^2-4ac}}{2a}=3\\pm\\sqrt5$$\n\nThis means that\u00a0\u00a0$$x=\\frac{-b}{2a}$$ = 3\u00a0 \u00a0 \u00a0 must be exactly halfway between the two roots.!!\n\nSo the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3\n\nThe y value will be\u00a0 \u00a0 $$y=-2*3^2+bx+c = -18+3b+c$$\n\nThis is fine but you need to find the vleu of b and c\n\nFrom the equation above I can see that\n\n$$\\frac{-b}{2a}=3 \\qquad and \\qquad \\pm\\frac{\\sqrt{b^2-4ac}}{2a}=\\pm\\sqrt5\\\\ a=-2\\qquad so\\\\ \\frac{-b}{-4}=3 \\qquad and \\qquad \\pm\\frac{\\sqrt{b^2+8c}}{-4}=\\pm\\sqrt5\\\\ b=12 \\qquad and \\qquad \\pm\\frac{\\sqrt{b^2+8c}}{4}=\\pm\\sqrt5\\\\ b=12 \\qquad and \\qquad \\frac{b^2+8c}{16}=5\\\\ b=12 \\qquad and \\qquad b^2+8c=80\\\\ b=12 \\qquad and \\qquad 144+8c=80\\\\ b=12 \\qquad and \\qquad 18+c=10\\\\ b=12 \\qquad and \\qquad c=-8\\\\$$\n\nso\n\n$$y= -18+3b+c\\\\ y= -18+3*12+-8\\\\ y=-18+36-8\\\\ y=10$$\n\nSo the vertex is\u00a0 \u00a0(3,10)\n\nWhich is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!\n\nThanks EP :)\n\nMelody \u00a0Feb 21, 2018", "date": "2018-07-21 11:59:09", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/help-me-please_130", "openwebmath_score": 0.809325635433197, "openwebmath_perplexity": 7890.315983481394, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9941800407606186, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8565240957034094}}
{"url": "https://math.stackexchange.com/questions/38739/convergence-of-a-n1-frac11a-n", "text": "# Convergence of $a_{n+1}=\\frac{1}{1+a_n}$\n\nWe define\n\n$$a_{n+1}=\\frac{1}{1+a_n}, a_0=c> 0$$\n\nI stumbled across that sequence and Mathematica gives me that it converges against $\\frac{1}{2} \\left(\\sqrt{5}-1\\right)$ which doesn't even depend on $a_0$. Normally I show that sequences converge by seeing that they are monotonic and then the limit can be easily found by setting $a_n=a_{n+1}$, but this one seems to be alternating. Also by checking OEIS I noticed that $a_n=1/g(n)$ where $g(n)$ gives the $n+1$'th Golden Rectangle Number.\n\nI would be glad if someone can show me how to show that such sequences converge and how to find the limit, also maybe this is a well known sequence, as it has a quite simple form, too bad that google is very bad for looking for sequences and OEIS doesn't mention anything.\n\n\u2022 These numbers are the convergents of the continued fraction of the golden ratio. If you assume that it converges you can calculate the limit by replacing the $a_k$s by $a$. You can prove convergence by proving that the odd/even subsequences are monotonous. May 12 '11 at 18:33\n\u2022 Right, but 3123 is asking why $\\frac{\\sqrt{5}-1}{2}$ is a(n attractive) fixed point of the function $f(x)=\\frac1{a+x}$ for \"any\" starting value... May 12 '11 at 18:37\n\u2022 The reason why $a_n$ converges to a fixed value irrespective of $a_0 = c$ is actually not really surprising. For instance, consider $a_0 = \\frac{c}{10}$ where $c \\in [0,10]$ $a_{n+1} = \\frac{9+a_n}{10}$ Note that $\\displaystyle \\lim_{n\\rightarrow \\infty} a_n = 1$ for any starting value $c \\in [0,10]$\n\u2013\u00a0user17762\nMay 12 '11 at 18:50\n\u2022 May 13 '11 at 11:18\n\u2022 math.stackexchange.com/questions/235578/\u2026 Apr 20 '13 at 13:19\n\n## 3 Answers\n\nThe function $1/(1+x)$ is strictly decreasing and thus order-reversing on the positive reals. If you apply it twice it conserves order.\n\nTherefore, the odd and even subsequences are monotonic. They are also bounded as all your values except the first are bounded by 1.\n\nNow you can just replace $a_n$ and $a_{n+1}$ by $a$ to find the limit.\n\n\u2022 Thank you for your answer, this was very useful too. May 12 '11 at 18:45\n\nAs for why the first term does not matter, consider the $6$th term,\n\n$$\\frac{1}{1+\\frac{1}{1+\\frac{1}{1+\\frac{1}{1+\\frac{1}{1+c}}}}}$$\n\nas $c$ varies between $0$ and $\\infty$ the value of the $6$th term varies only by $0.1$. As $c$ varies between $0$ and $\\infty$ the $100$th term varies only by very small $\\varepsilon$. In the infinite limit this variation is zero.\n\nThis also explains convergence. Since convergence occurs call the number $\\varphi$, it satisfies the fixed point equation\n\n$\\varphi = \\frac{1}{1+\\varphi}$\n\nwhich, multiplying both sides by the denominator, is a quadratic equation hence the square root.\n\n\u2022 Thank you for the answer, writing it as a continued fraction makes things much more obvious May 12 '11 at 18:44\n\nFor $x\\geq 0$, $f(x)=1/(1+x)$ is a monotonically decreasing function, bounded between 0 and 1. The fixed point (if it exists) of the iterative map defined by your function, is the solution of $x=\\frac{1}{1+x}$. There is only one solution for $x\\geq 0$, and that is $\\tilde{x}=(\\sqrt{5}-1)/2$.\n\nIf we look at the evolution of $a_n=f(f(f(...(a_0))))$, we get\n\n$$a_n=\\frac{1}{1+\\frac{1}{1+\\frac{1}{1+\\frac{1}{1+\\frac{1}{1+a_0}}}}}$$\n\nand as $n\\to\\infty$, what $a_0$ was becomes irrelevant. In other words, no matter the value of $a_0$, it converges to a fixed point, which has to be the only fixed point, $\\tilde{x}$. Hence $\\tilde{x}$ is an attractive fixed point and also the limit of the sequence.\n\nThe Mathematica code that produced the above figure is below (it's modified from an answer to a question on SO)\n\nClear[fixedPoint]\nfixedPoint[a0_, nSteps_] :=\nModule[{f, a, n},\nf = RecurrenceTable[{a[n] == 1/(1 + a[n - 1]), a[1] == a0},\na, {n, 1, nSteps}];\nPlot[{1/(1 + x), x}, {x, 0, 1},\nPlotStyle -> {Darker@Blue, {Dashed, Black}},\nEpilog -> {Darker@Green,\nLine[Riffle[Partition[f, 2, 1], {#, #} & /@ Rest[f]]]}]]", "date": "2021-12-08 15:39:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/38739/convergence-of-a-n1-frac11a-n", "openwebmath_score": 0.8821293115615845, "openwebmath_perplexity": 185.79790957227806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147185726375, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8565000457442009}}
{"url": "https://stats.stackexchange.com/questions/409226/why-is-this-probability-distribution-biased-towards-even-numbers", "text": "# Why is this probability distribution biased towards even numbers?\n\nI am working on the probabilities of a game where\n\n1. they pick 20 unique number(non-repeating) from range 1 to 80\n\n2. sort those 20 numbers in ascending order\n\n3. sum every 4 numbers and take the last digit as result (so end with 5 numbers).\n\nNormally I could find these probabilities with rules based on combination and permutation but for this one because of the sorting I am not sure how to work on it.\n\nI tried to brute force all combinations but with some test it would take my old laptop about few thousand years. So I did some simulations, sampling 100 million samples. Below is the final result.\n\nSo there seems to be a small edge towards even numbers, but I don't understand where this edge comes from (even numbers appear about .5095925 of the time).\n\nI went further down about the first 4 numbers out of the sorted 20 numbers and try based on the fact that the 1st sorted number must be in the range of 1 to 61, 2nd sorted must be in the range of 2 to 62, etc. So odd positions will have 31 out of 61 chances to be odd and even positions have 31 out of 61 to be even. And I did some checks below, but the resulting bias towards even numbers is very small (.50000004 vs the stimulation .5095925).\n\nI wonder where and how the game tends to favor even numbers quite a bit. Or is my simulation code just wrong?\n\n\u2022 +1 For a quick insight into this, think about the results you would get if you were to select 76 numbers rather than 20. It will help to reason in terms of parity: how many odd numbers do you expect to see in each group of four numbers?\n\u2013\u00a0whuber\nMay 20 '19 at 15:14\n\u2022 @whuber you mean because there are 40 odd and 40 even, the chance of getting 10 odd and 10 even for the 20 number is higher, then the chance of getting 2 odds and 2 evens (result in even) is higher in a 4 number group base? May 21 '19 at 9:14\n\u2022 No, that's not it. You need to look at the chances, within any given group, of observing an even number of odd values, because that characterizes the cases where the total will end in 0, 2, 4, 6, or 8. For instance, any group of four numbers in a row is guaranteed to have two odd and two even values and therefore its sum cannot end in an odd digit.\n\u2013\u00a0whuber\nMay 21 '19 at 12:00\n\u2022 Your results look good. In particular, the first-number percentages for digits 0 - 5 are typically correct in the first four or five significant digits.\n\u2013\u00a0whuber\nMay 23 '19 at 17:16\n\nThis is an interesting question because the result is a little surprising; it is subtle; and it has been well researched with a large simulation.\n\nHere is a graphical representation of ten iterations of the game. Each row is an iteration. It shows the 20 numbers selected, $$k_1,k_2, \\ldots, k_{20},$$ and uses colors to depict their partition into five groups of four. The game sums the numbers in each group, thereby producing five sums.\n\nNevertheless, within the large number of simulations reported in the question, there appears to be an imbalance among odd and even sums, although this procedure seemingly would not favor one parity. Indeed, if we were only to take a subset of four numbers from $$\\{1,2,3,\\ldots, 80\\}$$ and sum them modulo ten, each digit would have exactly $$1/10$$ chance of appearing.\n\nTo appreciate what's happening, consider extreme versions of this procedure, where instead of partitioning a sample of $$20$$ into five groups of four, let's take a sample of $$76$$ and partition it into $$19$$ groups of four:\n\nWhat strikes us first about this image are the prominent gray \"bubbles\" depicting the numbers that were not sampled. With only $$80-76=4$$ bubbles in each iteration, we're unlikely to find any bubbles in the first group of four. Moreover, when there is a bubble in that group, as in the fourth row from the top, it's just as likely to be an even number as an odd number. That means the first group (in this extreme case) is highly likely to include two odd numbers and two even numbers, making its sum even.\n\nThis consideration of bubbles suggests analyzing the game in terms of gaps between the selected numbers. The first gap is $$\\Delta_1 = k_1-0,$$ the second gap is $$\\Delta_2 = k_2-k_1,$$ and so on. All gaps are $$1$$ or larger. Evidently the last number chosen is $$k_{20} = \\Delta_1+\\Delta_2 + \\cdots + \\Delta_{20},$$ which therefore must be $$80$$ or smaller.\n\nLet's begin with the first gap, $$\\Delta_1=k_1.$$ What is the chance this gap is only $$1,$$ the smallest possible value? That's easy to compute, because\n\n1. There are $$\\binom{80}{20}$$ distinct, equally-probable subsamples of size $$20$$ that can be obtained from all $$80$$ numbers.\n\n2. Of these, $$\\binom{79}{20}$$ are from the set $$\\{2,3,\\ldots, 80\\}.$$\n\nSubtracting, we find the number of subsamples with $$k_1=1$$ and divide that by the total number of subsamples to obtain the probability:\n\n$$\\Pr(\\Delta_1 = 1) = \\frac{\\binom{80}{20} - \\binom{79}{20}}{\\binom{80}{20}} = \\frac{\\binom{79}{19}}{\\binom{80}{20}} = \\frac{1}{4}.$$\n\n(Although only one of the ten iterations in the first figure has $$\\Delta_1=1$$--the fifth row from the top--a longer set of simulations confirms that $$\\Delta_1=1$$ in about a quarter of them.)\n\nAnalogous reasoning establishes the remaining chances:\n\n$$\\Pr(\\Delta_1 = j_1) = \\frac{\\binom{80+1-j_1}{20} - \\binom{80-j_1}{20}}{\\binom{80}{20}} = \\frac{\\binom{80-j_1}{19}}{\\binom{80}{20}},\\ j_1=1, 2, \\ldots, 61.$$\n\nNow suppose we have observed $$\\Delta_1.$$ What could $$\\Delta_2$$ be? If it is equal to $$j_2,$$ that means the last $$20-1$$ numbers are a subset of the values from $$\\Delta_1+j_2$$ through $$80$$ and the remaining $$20-2$$ numbers all lie between $$\\Delta_1 + j_2 +1$$ and $$80.$$ In effect, the role of $$80$$ is now played by $$80-j_1,$$ $$20$$ is reduced to $$20-1=1,$$ and $$j_2$$ plays the role of $$j_1.$$ Thus, writing $$k_2=j_1+j_2$$ for the second lowest value selected,\n\n\\eqalign{ \\Pr(k_2\\text{ selected next}\\mid k_1\\text{ selected first}) &= \\Pr(\\Delta_2 = j_2 \\mid \\Delta_1=j_1) \\\\ &= \\frac{\\binom{80-k_2}{20-2}}{\\binom{80-k_1}{20-1}},\\ k_2=j_1+1, 2, \\ldots, 80+1-(20-1).}\n\nThe pattern is clear. In particular, to study the first group, we can calculate the probability distribution of the sum of just the lowest four numbers in the subsample by summing over all possible values of $$\\Delta_1,\\Delta_2,\\Delta_3,\\Delta_4.$$ There are sufficiently few of these that this computation is feasible.\n\nBy the laws of conditional probability, the chance of such a combination $$(j_1,j_2,j_3,j_4)$$ is obtained by the products of the component probabilities\n\n\\eqalign{ q_{j_4,j_3,j_2,j_1} &= \\Pr(\\Delta_4=j_4\\mid \\Delta_3=j_3,\\Delta_2=j_2,\\Delta_1=j_1) \\\\ &\\times \\Pr(\\Delta_3=j_3\\mid \\Delta_2=j_2,\\Delta_1=j_1) \\\\ &\\times \\Pr(\\Delta_2=j_2\\mid \\Delta_1=j_1) \\\\ &\\times \\Pr(\\Delta_1=j_1)}\n\nas\n\n$$\\Pr(j_1,j_2,j_3,j_4) = \\sum_{j_1=1}^{81-20}\\ \\sum_{j_2=1}^{81-j_1-19}\\ \\sum_{j_3=1}^{81-j_1-j_2-18}\\ \\sum_{j_4=1}^{81-j_1-j_2-j_3-17} q_{j_4,j_3,j_2,j_1} .$$\n\nFor the game described in the question, there are \"only\" 635,376 possible combinations in this quadruple sum: the calculations can be carried out in seconds. Here is the interesting part of the probability distribution of the group sum $$k_1+k_2+k_3+k_4 = 4j_1+3j_2+2j_3+j_1:$$\n\n(The full distribution extends to the largest possible sum that could appear in this first group of four numbers, equal to $$61+62+63+64=250.$$)\n\nThe solid red dots are probabilities a little larger than suggested by their neighbors: the alternation between even and odd is clear.\n\nWhen we further reduce this distribution by taking just the last digits of the sums, the alternation between even and odd persists:\n\nThis plot shows the relative differences between the probabilities and $$1/10.$$ The actual values (to eight digits) are\n\nDigit Probability\n0  0.10235369\n1  0.09779473\n2  0.10159497\n3  0.09859690\n4  0.10239713\n5  0.09838777\n6  0.10201990\n7  0.09821967\n8  0.10121774\n9  0.09741751\n\n\nMany of them agree closely (to four or five digits) with the simulation results reported in the question.\n\nNotice (from the equations or the simulations) that the first group of four smallest numbers tends to lie to the left of $$20$$ or so, leaving about $$60$$ values from which to select the next four groups. Thus, the situation with the second group is typically just like the situation with five groups, but with $$80$$ reduced to around $$60$$ and only $$20-4=16$$ numbers to select. Similar reasoning suggests the situation with the third and fourth groups also is similar. The simulations in the question bear this out: they still favor even sums over odd.\n\nAnalogous discrepancies in probabilities occur when representing the sums in other (non-decimal) bases. Different patterns of discrepancies occur when taking groups of other sizes, especially odd sizes: groups of size $$4$$ are special because they have a pronounced tendency to include either zero, two, or four odd numbers, which guarantees their sum is even.\n\nHere is the R used to do the calculations. It is coded so you can experiment with versions of this game simply by changing the parameters n, m, and g.\n\nlibrary(data.table)\nn <- 80 # Maximum number available (100 or more may be problematic)\nm <- 4  # Group sizes >= 2 (8 or more is problematic)\ng <- 5  # Number of groups >= 1 (more is easier!)\n#\n# Generate all possible vectors of gaps.\n#\ndelta <- 1:(n - m*(g-1) + 1)\nX <- do.call(\"expand.grid\", lapply(1:m, function(i) delta))\nnames(X) <- paste0(\"k\", 1:m)\nX <- as.data.table(X)\ni <- rowSums(X) < max(delta)\nX <- X[i, ]\n#\n# Generate the numbers corresponding to each gap vector.\n#\nY <- apply(X, 1, cumsum)\nrownames(Y) <- paste0(\"k\", 1:m)\nY <- as.data.table(t(Y))\n#\n# Compute log conditional probabilities for each selection.\n#\nf <- function(k, n, g) {\nm <- length(k)\nk0 <- c(0, k[-m])\nj <- (m*g-1) : (m*(g-1))\nlchoose(n - k, j) - lchoose(n - k0, j+1)\n}\nZ <- apply(Y, 1, function(k) f(k, n, g))\nrownames(Z) <- paste0(\"p\", 1:m)\nZ <- as.data.table(t(Z))\n#\n# Compute the probability of each vector.\n#\nZ[, pi := exp(rowSums(Z))]\nY[, Sum := rowSums(Y)]\nX <- cbind(X, Y, Z)\n1-sum(X\\$pi) # Should be near zero, within floating point roundoff error\n#\n# Summarize by group sum.\n#\nH <- X[, .(Probability = sum(pi)), keyby=Sum]\nH.mod <- rbindlist(lapply(2:40, function(b)\nH[, .(Probability = sum(Probability), Base=b),\nkeyby=(Sum - choose(m+1,2)) %% b]))\n\nwith(H[Probability > 0.00025],  {\nabove <- sapply(1:length(Sum), function(i) {\ng <- splinefun(Sum[-i], Probability[-i])\nProbability[i] > g(Sum[i])\n})\ncolors <- ifelse(above, \"Red\", \"Gray\")\nplot(Sum, Probability, type=\"l\")\nabline(h=0)\nabline(v = seq(0, max(Sum), by=5), col=\"Gray\", lty=3)\nabline(v = seq(0, max(Sum), by=10), col=\"Gray\")\npoints(Sum, Probability, pch=21, bg=colors)\n})\n#\n# Explore the last digit in various bases.\n#\nbase <- 10\nwith(H.mod[Base==base], {\nplot(Sum, base*Probability - 1, type=\"b\", ylim=base*range(Probability)-1,\npch=21, bg=ifelse(base*Probability-1 > 0, \"Red\", \"Gray\"),\nxlab=paste(\"Sum modulo\", base, \"offset by\", choose(m+1,2)))\nabline(h=0, col=\"Gray\")\n})\n\n\u2022 I am moved by the details and visualization that you offer. There is still much for me to dig into the step by step forming of the formula and the code but first and formal let me thank you for this amazing answer. May 24 '19 at 21:23", "date": "2021-10-15 21:23:35", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/409226/why-is-this-probability-distribution-biased-towards-even-numbers", "openwebmath_score": 0.8171984553337097, "openwebmath_perplexity": 773.2290919784616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147209709196, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8565000448138639}}
{"url": "https://math.stackexchange.com/questions/3049117/evaluating-lim-limits-x-to-infty-frac4x23x4x-2", "text": "# Evaluating $\\lim\\limits_{x \\to \\infty}\\frac{4^{x+2}+3^x}{4^{x-2}}$\n\n$$\\lim_{x\u2192\u221e}\\frac{4^{x+2}+3^x}{4^{x-2}}.$$\n\nI have solved it like below: $$\\lim_{x\u2192\u221e}\\left(\\frac{4^{x+2}}{4^{x-2}}+\\frac{3^x}{4^{x-2}}\\right)=\\lim_{x\u2192\u221e}\\left(4^4+\\frac{3^x}{4^x}\u00b74^2\\right).$$ Since, as $$x \u2192 \u221e$$, $$3^x \u2192 \u221e$$, $$\\dfrac{3^x}{4^x} \u2192 0$$, the limit is equal to $$4^4=256$$.\n\nHave I solved it correctly?\n\nThis was a practice test question and the given solution was wrong. So, I solved it and I am preparing alone, no friend to discuss, so I posted it here.\n\n\u2022 $\\lim_{x \\to \\infty} \\left( 4^4+\\frac{3^x}{4^x} \\times 4^2 \\right)=\\lim_{x \\to \\infty}4^4 +\\lim_{x \\to \\infty}\\left(\\dfrac{3^x}{4^x}\\right)4^2$? While its true you might want to explain for rigor. \u2013\u00a0Yadati Kiran Dec 22 '18 at 4:26\n\u2022 It's unclear to me what purpose it serves to write $3^x\\to\\infty$. It makes the next expression, $\\frac{3^x}{4^x} \\to 0$, look wrong even though it actually is correct. \u2013\u00a0David K Jan 8 at 14:28\n\nYup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).\n\nI will make a nitpick with one thing you said, though:\n\nSince, as $$x \\to \\infty, 3^x \\to \\infty, \\frac{3^x}{4^x} \\to 0$$\n\nTechnically, since $$3^x$$ and $$4^x$$ both approach $$\\infty$$ as $$x \\to \\infty$$, then under your logic we obtain an indeterminate form:\n\n$$\\frac{3^x}{4^x} \\to \\frac{\\infty}{\\infty}$$\n\nIt would be better to regroup $$3^x/4^x$$ as $$(3/4)^x$$. Then clearly, as $$x \\to \\infty$$, $$(3/4)^x \\to 0$$ because $$3/4 < 1$$. (If you're not convinced, notice if you take $$x = 1, 2, 3, 4$$ and so on that $$(3/4)^x$$ clearly is decreasing.)\n\n\u2022 While $$3^x \\to \\infty$$ is true, note that it is not used in the proof though.", "date": "2019-05-25 09:57:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3049117/evaluating-lim-limits-x-to-infty-frac4x23x4x-2", "openwebmath_score": 0.8586853742599487, "openwebmath_perplexity": 397.8554211882069, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147193720649, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8565000403628604}}
{"url": "https://mathhelpboards.com/threads/minimizing-the-sop.8859/", "text": "# Minimizing the SOP\n\n#### shamieh\n\n##### Active member\nI used a KMAP and got this as my expression\n\n$$f = !y_{1}!y_{0} + !x_{1}x_{0}!y_{1} + x_{1}!y_{1} + x_{1}!y_{0} + x_{1}x_{0}y_{1}$$\n\nIs there any way I can minimize this? Maybe I'm just not seeing it. I thought the whole point of a KMAP was to minimize the expression?\n\nSome how the answer is this: $$f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}$$\n\nIf anyone is interested I had to design a circuit with output f with 4 inputs. I'm supposed to be showing the simplest sum of product expression for f. My f row for my truth table was this:\n1\n0\n0\n0\n\n1\n1\n0\n0\n\n1\n1\n1\n0\n\n1\n1\n1\n1\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nI used a KMAP and got this as my expression\n\n$$f = !y_{1}!y_{0} + !x_{1}x_{0}!y_{1} + x_{1}!y_{1} + x_{1}!y_{0} + x_{1}x_{0}y_{1}$$\n\n...\n\nSome how the answer is this: $$f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}$$\nYou can turn the three-variable minterms into two-variable ones, namely, remove $!x_1$ from $x_0!x_1!y_1$ and $y_1$ from $x_0x_1y_1$.\n\n#### shamieh\n\n##### Active member\nYou can turn the three-variable minterms into two-variable ones, namely, remove $!x_1$ from $x_0!x_1!y_1$ and $y_1$ from $x_0x_1y_1$.\n\nWell I can't combine them because they don't differ by two variables correct? So what minimization \"tool\" should I use? Should I factor something? I mean how do you just \"get rid of them\". See what I'm saying? What method I should I be using? Sorry if I sound ignorant.\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nWhen you remove a variable from a three-variable minterm, its representation in the Karnaugh map grows from 2 to 4 cells. However, if the added two cells are already covered by other minterms, then the Boolean function does not change.\n\nIn this case, the minterm $x_0y_1y_1$ represents two cells in the middle of column 3 ($x_0=x_1=1$). When you remove $y_1$, the result is the complete column 3. But the top cell of column 3 is already covered by $!y_0!y_1$, and the bottom cell of column 3 is covered by $x_1!y_1$. So removing $y_1$ from $x_0y_1y_1$ does not change the function. A similar thing happens with turning $x_0!x_1!y_1$ into $x_0!y_1$.\n\nWhen reading off a minimal formula from a Karnaugh map, the temptation is always to break the cells corresponding to 1 into disjoint regions. But this results in smaller regions and therefore larger minterms. Instead, one must make regions as large as possible by using the fact that overlap is allowed.\n\n#### shamieh\n\n##### Active member\nAwesome explanation! Thanks so much. So it looks like I probably missed a overlapping grouping I could of put together then - thus getting not exactly the complete minimization.", "date": "2021-04-15 11:19:17", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/minimizing-the-sop.8859/", "openwebmath_score": 0.8333398103713989, "openwebmath_perplexity": 650.1192858562753, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9724147153749275, "lm_q2_score": 0.8807970670261976, "lm_q1q2_score": 0.856500029235351}}
{"url": "https://stats.stackexchange.com/questions/587912/continuous-and-differentiable-bell-shaped-distribution-on-a-b/587918", "text": "# Continuous and differentiable bell-shaped distribution on $[a, b]$\n\nIs there a distribution with these three properties?\n\n\u2022 Supported on $$[a, b]$$ for any $$a, b\\in\\mathbb{R}$$ and $$a < b$$\n\u2022 Continuous and Differentiable (that is $$\\nabla_x p(x)$$ can be computed)\n\u2022 Bell-curved (or at least not as flat as the Uniform(a, b))\n\u2022 stats.stackexchange.com/questions/500862/\u2026\n\u2013\u00a0whuber\nSep 6 at 15:05\n\u2022 General answer: let $f:[0,1]\\to\\mathbb{R}\\cup\\{\\infty\\}$ be any nonnegative integrable function that increases on the interval $[0,c)$ and decreases on $(c,1].$ On the interval $[a,b]$ define $$F_0(x) = \\int_a^x f((t-a)/b)\\,\\mathrm{d}t.$$ Extend the function $F(x) = F_0(x)/F_0(b)$ to be $0$ on $(-\\infty, a)$ and $1$ on $(b,\\infty).$ By construction this has all the features you seek and every solution can be expressed in this form. The extreme generality of these solutions shows that you should be more narrowly focused on a solution that is meaningful for whatever application you have in mind.\n\u2013\u00a0whuber\nSep 6 at 15:20\n\u2022 @whuber +1. That indeed should be left as an answer; after all, it's a general construction. Sep 6 at 15:46\n\u2022 @whuber great answer! Do you mind posting that as an answer? Sep 6 at 16:13\n\nLet's construct all possible solutions.\n\nBy \"distribution\" you appear to refer to a density function (PDF) $$f.$$ The properties you require are\n\n1. Supported on $$[a,b].$$ That is, $$f(x)=0$$ for any $$x\\le a$$ or $$x \\ge b.$$\n\n2. $$f$$ should be (continuously) differentiable on $$(a,b)$$ with derivative $$f^\\prime.$$\n\n3. \"Bell-curved,\" which can be taken as\n\n\u2022 (Strictly) increasing from value of $$0$$ at $$a$$ to some intermediate point $$c;$$ that is, $$f^\\prime(x) \\gt 0$$ for $$a\\lt x \\lt c;$$ and\n\u2022 (Strictly) decreasing from $$c$$ to a value of $$0$$ at $$b;$$ that is, $$f^\\prime(x) \\lt 0$$ for $$c \\lt x \\lt b.$$\n\nTo standardize this description, translate and scale the left-hand arm of $$f^\\prime$$ to the interval $$[0,1]$$ and do the same for the right-hand arm, reversing and negating it. That is, let\n\n$$g_{-}(x) = f^\\prime\\left((c-a)x\\right)$$\n\nand\n\n$$g_{+}(x) = -f^\\prime\\left(1 - (b-c)x\\right).$$\n\nBoth are increasing positive integrable functions defined on $$[0,1]$$ for which\n\n$$\\int_0^1 g_{-}(x)\\,\\mathrm{d}x = \\int_0^1 f^\\prime((c-a)x)\\,\\mathrm{d}x = \\frac{1}{c-a}\\int_a^c f^\\prime(y)\\,\\mathrm{d}y = \\frac{f(c^-)}{c-a}$$\n\nand\n\n$$\\int_0^1 g_{+}(x)\\,\\mathrm{d}x = \\int_0^1 -f^\\prime(1 - (b-c)x)\\,\\mathrm{d}x = \\frac{1}{b-c}\\int_b^c f^\\prime(y)\\,\\mathrm{d}y = \\frac{f(c^+)}{b-c}.$$\n\nConversely, given any two positive increasing integrable functions $$g_{-}$$ and $$g_{+}$$ defined on $$[0,1]$$ (the \"left side\" and \"right side\" models), these steps can be reversed to construct $$f^\\prime,$$ which in turn can be integrated (and normalized) to yield a valid distribution function.\n\nHere is this reverse process in pictures. It begins with the two model functions.\n\n(Notice that these functions need not even be continuous and can be unbounded, as illustrated by $$g_{+}$$ at the right.)\n\nThey are then integrated and assembled to produce $$f^\\prime,$$ which in turn is integrated and normalized to unit area to yield a density $$f$$ with every required characteristic.\n\nYou may further control the appearance of the density in many ways. For instance, by taking the two models to be the same function and placing the peak $$c = (a+b)/2$$ at the midpoint, you will obtain a symmetric density. Here I have used the original $$g_{-}$$ for the right hand model $$g_{+}.$$\n\nYou can enforce many other properties of $$f$$ by going through the original analysis to deduce the corresponding properties of the model functions and restricting your construction to functions of that type.\n\nFinally, if you choose to limit the two model functions to a finitely parameterized subset of the possibilities, you will have constructed a parametric family of distributions meeting all your criteria.\n\nThe Truncated normal distribution obeys all prerequisites:\n\n\u2022 It's bell shaped\n\u2022 It's continuous\n\u2022 Its support is $$x \\in [a,b]$$\n\u2022 It's differentiable, i.e. $$\\nabla_x p(x)$$ exists for all $$x \\in [a,b]$$\n\u2022 I just added a detail that I forgot about. I need it to be differentiable Sep 6 at 15:13\n\u2022 @Euler_Salter no problem, the truncated normal distribution is also differentiable. Sep 6 at 15:37\n\nOne option is to transform a beta distribution.\n\n$$Beta(3,3)$$ has your desired properties on $$[0,1]$$.\n\nNow subtract $$1/2$$ to center the distribution.\n\nNext, multiply to stretch or compress the distribution.", "date": "2022-12-09 16:02:33", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/587912/continuous-and-differentiable-bell-shaped-distribution-on-a-b/587918", "openwebmath_score": 0.8641833066940308, "openwebmath_perplexity": 387.9795726937566, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811611608242, "lm_q2_score": 0.8872045892435128, "lm_q1q2_score": 0.8564905965511144}}
{"url": "http://mathhelpforum.com/discrete-math/31673-couting-questions.html", "text": "# Math Help - Couting Questions\n\n1. ## Couting Questions\n\nHi all,\n\nI have a couple of questions regarding counting, so it will be fantastic if you can help me out. Thanks in advance!\n\nQns 1:\nConsider strings of length n over the set {a,b,c,d}. How many such strings contain at least one pair of adjacent characters that are the same?\n\nBy adjacent do they mean (a,b) in length n, etc? Is the combination (b,a) allowed or equivalent to (a,b)?\n\nQns 2:\nHow many integers from 1 through 999999 contain each of the digits 1,2 and 3 at least once? (Hint: For each i let $A_{i}$ be the set of integers from 1 through 999999 that do not contain the digit i)\n\nThe solution asked me to use the inclusion and exclusion method, so how do i know when to use this?\n\nQns 3:\nIf $n=p_{1}p_{2}p_{3}p_{4}p_{5}$ where $p_{i}$ are distinct primes, in how many ways can n be expressed as a product of 2 positive integers? (n=ab and n=ba are considered the same)\nWhat is the answer if $n=p_{1}p_{2}...p_{k}$?\n\nFrom the solution, they mentioned the number of ways to choose a is $2^{5}$ which i understand. It is stated that \"it is a duplication by a factor of 2 since $p_{1}p_{2}$ and $p_{3}p_{4}p_{5}$ both appear as subsets but they give the same factorisation\". I do not understand where the duplication is, can somehow show me an example?\n\n2. Hello, shaoen01!\n\n3) If $n=p_1p_2p_3p_4p_5$ where $p_i$ are distinct primes,\nIn how many ways can n be expressed as a product of 2 positive integers?\n( $n=ab\\text{ and }n=ba$ are considered the same)\n\nWhat is the answer if $n=p_1p_2\\hdots p_k$?\n\nFrom the solution, they mentioned the number of ways to choose a is $2^{5}$ which i understand.\nIt is stated that \"it is a duplication by a factor of 2 since $p_{1}p_{2}$ and $p_{3}p_{4}p_{5}$ both appear\nas subsets but they give the same factorisation\".\n\nI do not understand where the duplication is.\nCan somehow show me an example?\n\nSuppose $n \\:=\\:30 \\:=\\:2\\cdot3\\cdot5$\n\nWith three factors to choose from, there are: $2^3 = {\\bf8}$ possible subsets.\n\nThey are: . $\\{\\;\\},\\;\\{2\\},\\;\\{3\\},\\;\\{5\\},\\;\\{2,3\\},\\;\\{3,5\\} ,\\;\\{2,5\\},\\;\\{2,3,5\\}$\n\nBut the question is: how many two-set partitions are there?\n\nThere are four: . $\\begin{array}{cc}\\{\\;\\} &\\{2,3,5\\} \\\\ \\{2\\} & \\{3,5\\} \\\\ \\{3\\} & \\{2,5\\} \\\\ \\{5\\} & \\{2,3\\} \\end{array}\\quad \\text{ The products are: }\\begin{Bmatrix}1\\cdot30 \\\\ 2\\cdot15 \\\\ 3\\cdot10 \\\\ 5\\cdot6\\end{Bmatrix}$\n\nThe partition $\\{2\\}\\;\\;\\{3,5\\}$ is the same as $\\{3,5\\}\\;\\;\\{2\\}$\n. . because $2\\cdot15\\text{ and }15\\cdot2$ are the same factoring.\n\n3. Originally Posted by shaoen01\nQns 1:\nConsider strings of length n over the set {a,b,c,d}. How many such strings contain at least one pair of adjacent characters that are the same?\nBy adjacent do they mean (a,b) in length n, etc? Is the combination (b,a) allowed or equivalent to (a,b)?\n\nQns 2:\nHow many integers from 1 through 999999 contain each of the digits 1,2 and 3 at least once? (Hint: For each i let $A_{i}$ be the set of integers from 1 through 999999 that do not contain the digit i\nI have no gotten a clear model for question #1. But I think that it is clear that you would allow neither \u2018ab\u2019 nor \u2018ba\u2019. I am pretty sure that one counts these with recursion.\n\nQuestion #2 is lengthy but straightforward.\nIf T=999999 is the total number then we need to remove $\n\\left( {A_1 \\cup A_2 \\cup A_3 } \\right)$\n, that is removing any number that fails to contain a 1, 2, or 3.\nYou know that:\n$\\# \\left( {A_1 \\cup A_2 \\cup A_3 } \\right) =$ $\\# \\left( {A_1 } \\right) + \\# \\left( {A_2 } \\right) + \\# \\left( {A_3 } \\right) - \\# \\left( {A_1 \\cap A_2 } \\right) - \\# \\left( {A_1 \\cap A_3 } \\right) - \\# \\left( {A_3 \\cap A_2 } \\right) + \\# \\left( {A_1 \\cap A_2 \\cap A_3 } \\right)$.\n\nI will help with one of those: $\\# \\left( {A_1 \\cap A_2 } \\right) = 8^6 - 1$.\n\n4. Originally Posted by Soroban\nHello, shaoen01!\n\nSuppose $n \\:=\\:30 \\:=\\:2\\cdot3\\cdot5$\n\nWith three factors to choose from, there are: $2^3 = {\\bf8}$ possible subsets.\n\nThey are: . $\\{\\;\\},\\;\\{2\\},\\;\\{3\\},\\;\\{5\\},\\;\\{2,3\\},\\;\\{3,5\\} ,\\;\\{2,5\\},\\;\\{2,3,5\\}$\n\nBut the question is: how many two-set partitions are there?\n\nThere are four: . $\\begin{array}{cc}\\{\\;\\} &\\{2,3,5\\} \\\\ \\{2\\} & \\{3,5\\} \\\\ \\{3\\} & \\{2,5\\} \\\\ \\{5\\} & \\{2,3\\} \\end{array}\\quad \\text{ The products are: }\\begin{Bmatrix}1\\cdot30 \\\\ 2\\cdot15 \\\\ 3\\cdot10 \\\\ 5\\cdot6\\end{Bmatrix}$\n\nThe partition $\\{2\\}\\;\\;\\{3,5\\}$ is the same as $\\{3,5\\}\\;\\;\\{2\\}$\n. . because $2\\cdot15\\text{ and }15\\cdot2$ are the same factoring.\n\nHi Soroban:\n\nThank you for your reply. So for example, 3x10 is also equivalent to 10x3 right? So this kind of repeated value is not allowed. If i use your example, i can't seem to get the answer of $2^{4}=16$. And must $n=p_{1}p_{2}p_{3}p_{4}p_{5}$ be consecutive values? For example, 1 to 5?\n\n5. Originally Posted by Plato\nI have no gotten a clear model for question #1. But I think that it is clear that you would allow neither \u2018ab\u2019 nor \u2018ba\u2019. I am pretty sure that one counts these with recursion.\n\nQuestion #2 is lengthy but straightforward.\nIf T=999999 is the total number then we need to remove $\n\\left( {A_1 \\cup A_2 \\cup A_3 } \\right)$\n, that is removing any number that fails to contain a 1, 2, or 3.\nYou know that:\n$\\# \\left( {A_1 \\cup A_2 \\cup A_3 } \\right) =$ $\\# \\left( {A_1 } \\right) + \\# \\left( {A_2 } \\right) + \\# \\left( {A_3 } \\right) - \\# \\left( {A_1 \\cap A_2 } \\right) - \\# \\left( {A_1 \\cap A_3 } \\right) - \\# \\left( {A_3 \\cap A_2 } \\right) + \\# \\left( {A_1 \\cap A_2 \\cap A_3 } \\right)$.\n\nI will help with one of those: $\\# \\left( {A_1 \\cap A_2 } \\right) = 8^6 - 1$.\nHi Plato:\n\nQns 1:\nShould i use the unordered and no repetition method to calculate this? Do you have any clues or hints to give me?\n\nQns 2:\nBut i am curious to know how did you get the value $\n\\left( {A_1 \\cap A_2 } \\right) = 8^6 - 1\n$\n. I do know the inclusion/exclusion rule you written above, the thing is i do not know how to get the value like how you did.\n\n6. Originally Posted by shaoen01\nI do not know how to get the value like how you did.\nThe set $A_1$ is the set of all integers 1 to 999999 that do not contain the digit 1. Because we can only use $\\left\\{ {0,2,3,4,5,6,7,8,9} \\right\\}$ there $9^6-1$ numbers in $A_1$, we subtract the 1 to account for 000000000.\n\n7. Hello, shaoen01!\n\nPlato's approach to #2 is correct . . .\nI would further assume that an integer does not begin with zero.\n\n2) How many integers from 1 through 999999\ncontain each of the digits 1,2 and 3 at least once?\n\nThe solution asked me to use the inclusion-exclusion method,\nso how do i know when to use this?\n\nIt is used when counting the items which are not in the set\nis easier than counting the items that are in the set.\nThere are 999,999 integers.\nHow many do not contain at least {1,2,3} ?\n\nLet $n(i')$ = number of integers that do not contain an $i.$\n\nWe want:\n$n(1' \\,\\cup \\,2' \\,\\cup \\,3') \\;=\\;n(1') \\,+\\, n(2') \\,+ \\,n(3') \\,-\\, n(1' \\,\\cap \\,2') \\,- \\,n(2' \\,\\cap \\,3') \\,- \\,n(1' \\,\\cap \\,3') \\,+$ $n(1' \\cap 2' \\cap 3')$\n\nConsider $n(1')$\n\n1-digit numbers: 8 choices . (It must not be 0 or 1.)\n2-digit numbers: $8\\cdot9$ choices.\n3-digit numbers: $8\\cdot9^2$ choices.\n4-digit numbers: $8\\cdot9^3$ choices.\n5-digit numbers: $8\\cdot9^4$ choices.\n6-digit numbers: $8\\cdot9^5$ choices.\n\nThere are: . $8(1 + 9 + 9^2 + \\hdots + 9^5) \\;=\\;8\\,\\frac{9^5-1}{9-1} \\;=\\;59,048$ such numbers.\n\nThe same reasoning holds for $n(2')\\text{ and }n(3')$\n\n. . Hence: . $n(1') \\:=\\:n(2') \\:=\\:n(3') \\:=\\:59,048$\n\nConsider $n(1' \\cap 2')$\n\n1-digit numbers: 7 choices. .(It must not be 0, 1, or 2.)\n2-digit numbers: $7\\cdot8$ choices.\n3-digit numbers: $7\\cdot8^2$ choices.\n4-digit numbers: $7\\cdot8^3$ choices.\n5-digit numbers: $7\\cdot8^4$ choices.\n6-digit numbers: $7\\cdot8^5$ choices.\n\nThere are: . $7(1 + 8 + 8^2 +\\hdots+8^5) \\:=\\:7\\,\\frac{8^5-1}{8-1} \\:=\\:32,767$ such numbers.\n\nThe same reasoning holds for $n(2'\\cap3')\\text{ and }n(1'\\cap3')$\n\n. . Hence: . $n(1'\\cap2') \\:=\\:n(2' \\cap 3') \\:=\\:n(1' \\cap 3') \\:=\\:32,767$\n\nConsider $n(1' \\cap 2' \\cap 3')$\n\n1-digit numbers: 6 choices. .(It must not be 0, 1, 2, or 3.)\n2-digit numbers: $6\\cdot7$ choices.\n3-digit numbers: $6\\cdot7^2$ choices.\n4-digit numbers: $6\\cdot7^3$ choices.\n5-digit numbers: $6\\cdot7^4$ choices.\n6-digit numbers: $6\\cdot7^5$ choices.\n\nThere are: . $6(1 + 7 + 7^2 + \\hdots + 7^5) \\:=\\:6\\,\\frac{7^5-1}{7-1} \\:=\\:16,806$ such numbers.\n\n. . Hence: . $n(1' \\cap 2' \\cap 3') \\:=\\:16,806$\n\nThen: . $n(1' \\cup 2' \\cup 3') \\:=\\:3(59,048) - 3(32,767) + 16,806 \\:=\\:95,649$\n\nTherefore: . $n(1 \\cap 2\\cap 3) \\;=\\;999,999 - 95,649 \\;=\\;\\boxed{904,350}$\n\nBut check my work and my reasoning . . . please!\n.\n\n8. Originally Posted by Soroban\nI would further assume that an integer does not begin with zero. But check my work and my reasoning . . . please!\nBut in the case \"assume that an integer does not begin with zero\", how would account for 234?\nThat contains no 1's but has the form 000234.\n\n9. Originally Posted by shaoen01\nQns 1:\nConsider strings of length n over the set {a,b,c,d}. How many such strings contain at least one pair of adjacent characters that are the same?\n\nIf that interpretation of the question is correct, then the answer is quite easy. There are $4^n$ possible strings altogether, but $4\\times3^{n-1}$ of these are not allowed. (For a string not to be allowed, its first letter can be any one of the four, but each subsequent letter must be different from its predecessor, so there are only three choices for it.) Thus the number of allowable strings is $4(4^{n-1} - 3^{n-1})$.\nThere are: . $8(1 + 9 + 9^2 + \\hdots + 9^5) \\;=\\;8\\,\\frac{9^5-1}{9-1} \\;=\\;59,048$ such numbers.\nBit of slippage here, I think! It should be $8\\,\\frac{9^{\\textstyle\\mathbf6}-1}{9-1} \\;=\\;531440$ (and similarly for the other two summations).", "date": "2014-11-01 03:23:34", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/31673-couting-questions.html", "openwebmath_score": 0.8626995086669922, "openwebmath_perplexity": 567.1806657916333, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811571768047, "lm_q2_score": 0.8872045907347108, "lm_q1q2_score": 0.8564905944560486}}
{"url": "http://wildjusticemusic.com/american-satan-mkkicgb/biconditional-statement-truth-table-ad83f6", "text": "biconditional statement truth table\n\nYou can enter logical operators in several different formats. \u2022 Identify logically equivalent forms of a conditional. a. Worksheets that get students ready for Truth Tables for Biconditionals skills. Edit. Solution:\u00a0Yes. Notice that in the first and last rows, both P \u21d2 Q and Q \u21d2 P are true (according to the truth table for \u21d2), so (P \u21d2 Q) \u2227 (Q \u21d2 P) \u200b\u200b\u200b\u200b\u200b\u200b is true, and hence P \u21d4 Q is true. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. The biconditional uses a double arrow because it is really saying \u201cp implies q\u201d and also \u201cq implies p\u201d. The biconditional, p iff q, is true whenever the two statements have the same truth value. Compound Propositions and Logical Equivalence Edit. first condition. A biconditional is true only when p and q have the same truth value. Conditional Statements (If-Then Statements) The truth table for P \u2192 Q is shown below. In writing truth tables, you may choose to omit such columns if you are confident about your work.) The following is truth table for \u2194 (also written as \u2261, =, or P EQ Q): To show that equivalence exists between two statements, we use the biconditional if and only if. Sign up or log in. biconditional Definitions. Let's put in the possible values for p and q. Also how to do it without using a Truth-Table! Let p and q are two statements then \"if p then q\" is a compound statement, denoted by p\u2192 q and referred as a conditional statement, or implication. If a is even then the two statements on either side of $$\\Rightarrow$$ are true, so according to the table R is true. Is there XNOR (Logical biconditional) operator in C#? BNAT; Classes. text/html 8/17/2008 5:10:46 PM bigamee 0. Two formulas A 1 and A 2 are said to be duals of each other if either one can be obtained from the other by replacing \u2227 (AND) by \u2228 (OR) by \u2227 (AND). 3. \u2022 Construct truth tables for conditional statements. SOLUTION a. It is denoted as p \u2194 q. How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). Whenever the two statements have the same truth value, the biconditional is true. Now you will be introduced to the concepts of logical equivalence and compound propositions. en.wiktionary.org. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. The connectives \u22a4 \u2026 \"A triangle is isosceles if and only if it has two congruent (equal) sides.\". Now I know that one can disprove via a counter-example. A biconditional statement is really a combination of a conditional statement and its converse. All birds have feathers. Two line segments are congruent if and only if they are of equal length. Also, when one is false, the other must also be false. About Us | Contact Us | Advertise With Us | Facebook | Recommend This Page. Name. Examples. The\u00a0compound statement\u00a0(pq)(qp) is a conjunction of two\u00a0conditional statements. Just about every theorem in mathematics takes on the form \u201cif, then\u201d (the conditional) or \u201ciff\u201d (short for if and only if \u2013 the biconditional). The truth table for the biconditional is . Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. In each of the following examples, we will determine whether or not the given statement is biconditional using this method. (a) A quadrilateral is a rectangle if and only if it has four right angles. So we can state the truth table for the truth functional connective which is the biconditional as follows. Definitions are usually biconditionals. It is helpful to think of the biconditional as a conditional statement that is true in both directions. 2. Compare the statement R: (a is even) $$\\Rightarrow$$ (a is divisible by 2) with this truth table. Summary:\u00a0A biconditional statement is defined to be true whenever both parts have the same truth value. A biconditional statement is often used in defining a notation or a mathematical concept. Mathematicians abbreviate \"if and only if\" with \"iff.\" Watch Queue Queue. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. You are in Texas if you are in Houston. The conditional operator is represented by a double-headed arrow \u2194. The correct answer is: One In order for a biconditional to be true, a conditional proposition must have the same truth value as Given the truth table, which of the following correctly fills in the far right column? In the truth table above, when p and q have the same truth values, the compound statement (pq)(qp) is true. Notice that the truth table shows all of these possibilities. Is this statement biconditional? Make a truth table for ~(~P ^ Q) and also one for PV~Q. If I get money, then I will purchase a computer. 1. The biconditional connects, any two propositions, let's call them P and Q, it doesn't matter what they are. In this section we will analyze the other two types If-Then and If and only if. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. Biconditional statement? text/html 8/18/2008 11:29:32 AM Mattias Sj\u00f6gren 0. A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. When proving the statement p iff q, it is equivalent to proving both of the statements \"if p, then q\" and \"if q, then p.\" (In fact, this is exactly what we did in Example 1.) So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' Biconditional Statements (If-and-only-If Statements) The truth table for P \u2194 Q is shown below. Construct a truth table for ~p \u2194 q Construct a truth table for (q\u2194p)\u2192q Construct a truth table for p\u2194(q\u2228p) A self-contradiction is a compound statement that is always false. Mathematics normally uses a two-valued logic: every statement is either true or false. Sign up to get occasional emails (once every couple or three weeks) letting you know\u00a0what's new! When we combine two conditional statements this way, we have a biconditional. Otherwise it is false. \u2022 Use alternative wording to write conditionals. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. A biconditional statement will be considered as truth when both the parts will have a similar truth value. The statement qp is also false by the same definition. Use a truth table to determine the possible truth values of the statement P \u2194 Q. Therefore, a value of \"false\" is returned. Ask Question Asked 9 years, 4 months ago. Definition. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. Logical equivalence means that the truth tables of two statements are the same. Determine the truth values of this statement: (p. A polygon is a triangle if and only if it has exactly 3 sides. Table truth table for ( p\u2194q ) \u2227 ( p\u2194~q ), is this a self-contradiction p >. A: it is always biconditional statement truth table functional connective which is the biconditional an! Implies q, is true but the back is false ; otherwise, it is always true up, can! For better understanding, you can think of the following sentences using  iff  next lesson ; Last. Two inputs, say a and b: we will not play of determining truth values of statements! 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P and q and b are true ( qp ) is a.!", "date": "2021-05-08 20:26:29", "meta": {"domain": "wildjusticemusic.com", "url": "http://wildjusticemusic.com/american-satan-mkkicgb/biconditional-statement-truth-table-ad83f6", "openwebmath_score": 0.5925559997558594, "openwebmath_perplexity": 734.4940030100662, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.952574129515172, "lm_q2_score": 0.899121375242593, "lm_q1q2_score": 0.8564797613501972}}
{"url": "https://mathematica.stackexchange.com/questions/159614/building-a-noisy-function", "text": "# Building a Noisy Function\n\nHere's a nice function with a nice plot:\n\nf[t_] := Exp[-Abs[t]] Sin[t];\nPlot[f[t], {t, -10, 10}, PlotRange -> All]\n\n\nWhat I would like to do is to make it noisy... to add something to f[t] so that it returns a noisy version. A roundabout way to accomplish this is to discretize/sample f[t], and then add noise to the samples:\n\nrange = Range[-10, 10, 0.1];\nListLinePlot[f[#] & /@ range +\nRandomReal[{-0.03, 0.03},Length[range]], PlotRange -> All]\n\n\nIs there a more direct way? It seems like there are lots of functions that generate random processes, but they seem to invariably end up with a TemporalData object or a list of values. Is there any way to add noise directly to the function?\n\nUpdate: I'm sorry I can't accept all the answers! David's has the advantage of simplicity (and of showing me a new way that the Random functions can work). A.G.'s answer provides clear flexibility in choosing the correlation of the random process. J.M.'s is probably the slickest, building on his earlier Perlin noise function.\n\n\u2022 Do you want the whole function or is a plot enough? \u2013\u00a0A.G. Nov 10 '17 at 1:12\n\u2022 I was hoping for a function that could be used elsewhere. \u2013\u00a0bill s Nov 10 '17 at 1:54\n\u2022 @bills I think that your question deserves more thinking. Noise is always characterised, in a first approximation, by a mean, a variance, and an amplitude. Therefore, I think that in the proposal some modelling of your additive noise seems to be worth considered. \u2013\u00a0Jos\u00e9 Antonio D\u00edaz Navas Nov 11 '17 at 16:37\n\u2022 I wasn't too picky about the character of the noise, but the answers do provide a variety of possibilities -- for example, J.M.s suggestion of RandomVariate provides just about any independent distribution, while other answers address how to incorporate correlation over time. \u2013\u00a0bill s Nov 11 '17 at 21:49\n\nPlot[\nf[t] + Sum[.01 RandomReal[n] Cos[n t/10], {n, 10}],\n{t, -10, 10},\nPlotRange -> All]\n\n\n\u2022 Even simpler: Plot[f[t] + RandomReal[1], {t, -10, 10}]. I had no idea RandomReal could be used like this. Thanks! \u2013\u00a0bill s Nov 10 '17 at 0:08\n\u2022 @bill, you can use RandomVariate[] too if you want: Plot[f[t] + RandomVariate[NormalDistribution[0, 1/50]], {t, -10, 10}, PlotRange -> All] \u2013\u00a0J. M.'s ennui Nov 10 '17 at 4:51\n\u2022 An incredibly minor quibble: this new function will always have the same derivative as the original function at $t = 0$. This probably doesn't matter for most purposes, but it'd be easy enough to get around by adding in a random phase to the cosine: Cos [n t/10 + RandomReal[{0, 2 Pi}]. \u2013\u00a0Michael Seifert Nov 10 '17 at 15:31\n\nIf you need your function to look \"ragged\", but still be \"continuous\" and \"reproducible\", you can use one-dimensional Perlin noise (previously used here):\n\nfBm = With[{permutations = Apply[Join, ConstantArray[RandomSample[Range[0, 255]], 2]]},\nCompile[{{x, _Real}},\nModule[{xf = Floor[x], xi, xa, u, i, j},\nxi = Mod[xf, 16] + 1;\nxa = x - xf; u = xa*xa*xa*(10. + xa*(xa*6. - 15.));\ni = permutations[[permutations[[xi]] + 1]];\nj = permutations[[permutations[[xi + 1]] + 1]];\n(2 Boole[OddQ[i]] - 1)*xa*(1. - u) +\n(2 Boole[OddQ[j]] - 1)*(xa - 1.)*u],\nRuntimeAttributes -> {Listable},\nRuntimeOptions -> {\"EvaluateSymbolically\" -> False}]];\n\n\nand then do something like\n\nPlot[f[t] + Sum[fBm[5 2^k t]/2^k, {k, 0, 2}]/20, {t, -10, 10},\nPlotPoints -> 55, PlotRange -> All]\n\n\nIf you want something like the example you present, you'll likely need to have the errors serially correlated as the number of evaluated data points gets large. Otherwise you'll get fuzzy caterpillars when the errors are independent.\n\nUpdate: Hopefully a more justified presentation than before.\n\nSuppose that we add noise to f[t] such that the distribution of the random noise is normally distributed with mean zero and variance $\\sigma^2$. But rather than having independent random noise at times $t_1$ and $t_2$, we have $\\operatorname{Correlation}(e_{t_1}, e_{t_2}) = \\rho^{|t_1 - t_2|}$ with $\\rho\\ge 0$.\n\nSo that defines a random function with a continuous time index. Nothing has been said so far that makes this a discrete time model.\n\nBut when we want a particular realization of this function with random noise, we need to choose specific times to produce a display of the function. This doesn't make the function discrete in $t$. It might be a semantics issue but I wouldn't call this a sampled/discretized version of the function despite the way the values of the realization of this function are produced in code.\n\nThe following code uses values of $t$ that are equally spaced however the functions involved will take any unequally-spaced values of $t$. The parameter $\\rho$ is the correlation of two errors one unit apart.\n\nManipulate[\n(* Values of t to consider *)\nt = Table[-10 + 20 i/n, {i, 0, n}];\n(* Differences in sucessive values of t *)\nd = Differences[t];\n(* Autoregressive errors of order 1 *)\ne = ConstantArray[0, n + 1];\ne[[1]] = RandomVariate[NormalDistribution[0, \u03c3], 1][[1]];\nDo[e[[i]] = \u03c1^d[[i - 1]] e[[i - 1]] +\nSqrt[1 - \u03c1^(2 d[[i - 1]])] RandomVariate[NormalDistribution[0, \u03c3], 1][[1]],\n{i, 2, n + 1}];\n(* Values of underlying function *)\nfTrue = f[#] & /@ t;\n(* Function plus errors *)\ny = fTrue + e;\n(* Plot true function and contaminated function *)\nListPlot[{Transpose[{t, fTrue}], Transpose[{t, y}]},\nPlotRange -> All, Joined -> True,\nPlotStyle -> {{Thickness[0.01]}, {Red}}],\n\n(* Sliders *)\n{{\u03c3, 0.05}, 0.01, 0.2, Appearance -> \"Labeled\"},\n{{\u03c1, 0.5}, 0, 0.999, Appearance -> \"Labeled\"},\n{{n, 100}, 10, 2000, 1, Appearance -> \"Labeled\"},\n\nTrackedSymbols :> {\u03c3, \u03c1, n},\n\nInitialization :> (\nf[t_] := Exp[-Abs[t]] Sin[t];\n)]\n\n\nWhen there is a large number of points close together and independent errors are used ($\\rho=0$), then the result looks like a fuzzy caterpillar and probably unrealistic for any physical process. (Would an observation a very short distance away from an observation at $t$ be consistently wildly different?)\n\n\u2022 Like in my example, this operates on a sampled version of the function. My hope was to be able to define a noisy function that could be used elsewhere. \u2013\u00a0bill s Nov 10 '17 at 1:56\n\nIn order to get a function it is possible to add a noise function. That noise will be more complex if the function is to be used on a large interval (unless it is periodic, which is not so good for a \"random\" noise).\n\nHere is an example that produces a smooth function while avoiding periodical noise.\n\nSeedRandom[1]; (* change for different noise *)\nn = 200; (* increase for more randomness *)\na = 10; (* use function on interval [-a, a] *)\nw = 1/500; (* weight of random noise *)\nc = Table[RandomVariate[NormalDistribution[]], n]; (* coefficients *)\n\\[Epsilon][x_] := w Sum[ c[[i]] Cos[  \\[Pi] i (x - a)/(2 a)],\n{i, 1, n}]; (* noise *)\nPlot[\\[Epsilon][t], {t, -10, 10}, PlotRange -> All]\n\nf[t_] := Exp[-Abs[t]] Sin[t] + \\[Epsilon][t];\nPlot[f[t], {t, -a, a}, PlotRange -> All]\n\n\nNoise:\n\nFunction:\n\n\u2022 Instead of Table[RandomVariate[NormalDistribution[]], n], you could do RandomVariate[NormalDistribution[], n]. Then, you can do \u03b5[x_] = w c.Cos[\u03c0 Range[n] (x - a)/(2 a)]. \u2013\u00a0J. M.'s ennui Nov 10 '17 at 4:47", "date": "2021-03-01 10:31:36", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/159614/building-a-noisy-function", "openwebmath_score": 0.38297948241233826, "openwebmath_perplexity": 2139.226606922408, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.952574129515172, "lm_q2_score": 0.8991213745668094, "lm_q1q2_score": 0.8564797607064634}}
{"url": "https://mathhelpboards.com/threads/compute-a-square-root-of-a-sum-of-two-numbers.8337/", "text": "# Compute a square root of a sum of two numbers\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nCompute $\\sqrt{2000(2007)(2008)(2015)+784}$ without the help of calculator.\n\n#### MarkFL\n\nStaff member\nMy solution:\n\n$$\\displaystyle 2000\\cdot2015=4030028-28$$\n\n$$\\displaystyle 2007\\cdot2008=4030028+28$$\n\n$$\\displaystyle 784=28^2$$\n\nHence:\n\n$$\\displaystyle 2000\\cdot2007\\cdot2008\\cdot2015+784=4030028^2-28^2+28^2=4030028^2$$\n\nAnd so:\n\n$$\\displaystyle \\sqrt{2000\\cdot2007\\cdot2008\\cdot2015+784}=\\sqrt{4030028^2}=4030028$$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nThis is probably the most genius way to collect the four numbers $2000$, $2007$, $2008$ and $2015$ in such a manner so that their product takes the form $(a+b)(a-b)$ and what's more, $b^2=784$!\n\nBravo, MarkFL!\n\n##### Well-known member\nLetting 2000 = a\nwe have 2000 * 2007 *2008 * 2015 + 784\na(a+7)(a+8)(a+15) + 784\n= a(a+15) (a+7) (a+8) + 784\n= (a^2+15a) (a^2 + 15a + 56) + 784\n= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$\n= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$\n= $(a^2 + 15 a + 28)^2$\nhence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nLetting 2000 = a\nwe have 2000 * 2007 *2008 * 2015 + 784\na(a+7)(a+8)(a+15) + 784\n= a(a+15) (a+7) (a+8) + 784\n= (a^2+15a) (a^2 + 15a + 56) + 784\n= $(a^2 + 15 a + 28 - 28 ) (a^2 + 15a + 28 + 28) + 28^2$\n= $(a^2 + 15 a + 28)^2 - 28^2 + 28^2$\n= $(a^2 + 15 a + 28)^2$\nhence square root is $a^2 + 15a + 28$ or 4000000 + 30000 + 28\nHey kaliprasad, thanks for participating and your method is good and I'm particularly very happy to see you finally picking up on LaTeX!\n\n#### agentmulder\n\n##### Active member\nAlmost identical to MarkFL's , i just factored 16 to make the multiplications a bit smaller.\n\n$\\sqrt{(2000)(2007)(2008)(2015)+ 784} \\ =$\n\n$\\sqrt{16 [ (\\frac{2000}{4})(2007) ( \\frac{2008}{4} ) (2015) + \\frac{16 \\cdot 7^2}{16} ]} \\ =$\n\n$4 \\sqrt{(500)(2015)(502)(2007) \\ + \\ 7^2} \\ =$\n\n$4 \\sqrt{(1007507 -7)(1007507 + 7) + 7^2} \\ =$\n\n$4 \\sqrt{(1007507)^2} \\ = \\ 4030028$\n\nAdmittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.\n\nLast edited:\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nAlmost identical to MarkFL's , i just factored 16 to make the multiplications a bit smaller.\n\n$\\sqrt{(2000)(2007)(2008)(2015)+ 784} \\ =$\n\n$\\sqrt{16 [ (\\frac{2000}{4})(2007) ( \\frac{2008}{4} ) (2015) + \\frac{16 \\cdot 7^2}{16} ]} \\ =$\n\n$4 \\sqrt{(500)(2015)(502)(2007) \\ + \\ 7^2} \\ =$\n\n$4 \\sqrt{(1007507 -7)(1007507 + 7) + 7^2} \\ =$\n\n$4 \\sqrt{(1007507)^2} \\ = \\ 4030028$\n\nAdmittedly , had i not seen MarkFL's method i probably would not have discovered it on my own.\n\nNice one, agentmulder!", "date": "2021-09-26 21:35:42", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/compute-a-square-root-of-a-sum-of-two-numbers.8337/", "openwebmath_score": 0.6741387844085693, "openwebmath_perplexity": 5247.750325238607, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380075184121, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8564309533055059}}
{"url": "https://math.stackexchange.com/questions/1527353/are-there-infinitely-many-rational-pairs-a-b-which-satisfy-given-equation", "text": "# Are there infinitely many rational pairs $(a,b)$ which satisfy given equation?\n\nI saw on some facebook page this concrete example:\n\n$1.2^2+0.6^2=1.2+0.6$\n\nThe question that immediately arises is:\n\nAre there infinitely many pairs $(a,b)$ of rational numbers such that we have $a^2+b^2=a+b$?\n\n\u2022 HINT: Your equation represents a circle. It has infinitely many rational points on it. \u2013\u00a0user167045 Nov 13 '15 at 15:49\n\u2022 Great. Please post this as an answer. \u2013\u00a0Shailesh Nov 13 '15 at 15:50\n\u2022 $$a=\\frac{t(t+k)}{t^2+k^2}$$ $$b=\\frac{k(t+k)}{t^2+k^2}$$ \u2013\u00a0individ Nov 13 '15 at 17:20\n\u2022 @individ Would you like to post this as an answer? \u2013\u00a0Farewell Nov 13 '15 at 17:25\n\u2022 Why? The equation is simple. \u2013\u00a0individ Nov 13 '15 at 17:30\n\nHint: $$a^2+b^2=a+b\\implies{}a^2-a+b^2-b=0\\implies{}2\\left(a-\\frac{1}{2}\\right)^2+2\\left(b-\\frac{1}{2}\\right)^2=1.$$\n\nWhat can you say about this equation with respect to the cartesian plane? What is the parametrization of this figure?\n\n\u2022 @AntePaladin: A circle with a rational centre will always have infinitely many rational points while a circle with an irrational centre will have at most two rational points. See here fore the proof: books.google.co.in/\u2026 \u2013\u00a0Jack Frost Nov 13 '15 at 16:16\n\u2022 @Jack Frost a circle with rational centre and rational radius has infinitely many rational points. A circle with rational centre and transcendental radius has no rational points. \u2013\u00a0Robert Israel Nov 13 '15 at 16:41\n\u2022 In this case the radius is the square root of a rational: that's more delicate. There wouldn't be any if the radius was $1/\\sqrt{3}$. \u2013\u00a0Robert Israel Nov 13 '15 at 16:46\n\u2022 @AntePaladin Consider the map $(x,y)\\rightarrow\\left(\\frac{x+y}2,\\frac{x-y}2\\right)$. It takes the circle of radius $1$ to a circle of radius $\\frac{1}{\\sqrt{2}}$ and preserves rational points. \u2013\u00a0Milo Brandt Nov 13 '15 at 16:48\n\u2022 Yes, in fact a parametrization of this circle is $$a = \\dfrac{t+1}{t^2+1}, \\ b = \\dfrac{t^2 + t}{t^2 + 1}$$ Any rational value of $t$ gives you a rational point $(a,b)$. \u2013\u00a0Robert Israel Nov 13 '15 at 17:30\n\nReal solutions:\n\nThe equation $$a^2 + b^2 = a + b$$ can be transformed: $$a^2 + b^2 = a + b \\iff \\\\ a^2 - a + b^2 -b = 0 \\iff \\\\ (a - 1/2)^2 + (b - 1/2)^2 = 1/4 + 1/4 = 1/2 = (1/\\sqrt{2})^2 \\quad (*)$$ The solutions form a circle with origin $(1/2, 1/2)$ and radius $1/\\sqrt{2}$ in $\\mathbb{R}^2$. They are $$(a, b) = (a, b(a)) = \\left(a, (1/2) \\pm\\sqrt{(1/\\sqrt{2})^2 - (a - 1/2)^2}\\right)$$ for $a \\in [1/2 - 1/\\sqrt{2}, 1/2 + 1/\\sqrt{2}] = [a_-, a_+]$.\n\nRational solutions:\n\nAmong all those infinite many real solutions $(a, b)$ we expect infinite many rational solutions, but my topology is too bad to give a good argument here. If we insert arbitrary rational $a$ not all $b = b(a)$ are rational.\n\nI tried to find all rational $a$ for which $b = b(a)$ is rational, but ran against a wall. However an infinite subset of such rational $a$ is still infinite, so we try to pick an easy one.\n\nAn infinite rational subset of solutions:\n\nThe subset I came up with consists of those $a$ which have the form $$a_n = \\frac{1}{2} + \\frac{1}{2n} < a_+ \\quad (n \\in \\mathbb{N})$$ which gives \\begin{align} b_n &= \\frac{1}{2} \\pm \\sqrt{\\frac{1}{2} - \\frac{1}{4n^2}} \\\\ &= \\frac{1}{2} \\pm \\frac{\\sqrt{2n^2 - 1}}{2n} \\\\ &= \\frac{n \\pm \\sqrt{2n^2 - 1}}{2n} \\end{align}\n\nThen $b_n$ is rational, if $$2n^2 -1 = m^2$$ for some $m \\in \\mathbb{N} \\cup \\{ 0 \\}$.\n\nWe can rewrite this as $$m^2 - 2 n^2 = -1 \\quad (**)$$ which is a Diophantine equation related to the Pell equation with constant $D = 2$, see negative Pell equation.\n\nSolution of the negative Pell equation:\n\nOne solution to $(**)$ is $m = 1$ and $n = 1$, from which we can generate infinite many more solutions:\n\nFor odd $k \\in \\mathbb{N}$ we have: $$-1 = (1^2 - 2\\cdot 1^2)^k = (m^2 - 2n^2) \\Rightarrow \\\\ -1 = (1 + \\sqrt{2})^k (1-\\sqrt{2})^k = (m + \\sqrt{2} n)(m - \\sqrt{2} n)$$ which because of $$(1 + \\sqrt{2})^k = \\sum_{i=0}^k \\binom{k}{i} 1^i (\\sqrt{2})^{n-i} = c_1 \\cdot 1 + c_2 \\sqrt{2} \\quad (c_1, c_2 \\in \\mathbb{N}) \\\\ (1 - \\sqrt{2})^k = \\sum_{i=0}^k \\binom{k}{i} 1^i (-\\sqrt{2})^{n-i} = d_1 \\cdot 1 - d_2 \\sqrt{2} \\quad (d_1, d_2 \\in \\mathbb{N})$$ gives $$m + \\sqrt{2} n = (1 + \\sqrt{2})^k \\\\ m - \\sqrt{2} n = (1 - \\sqrt{2})^k \\\\$$ and $$m = \\frac{(1 + \\sqrt{2})^k + (1 - \\sqrt{2})^k}{2} \\\\ n = \\frac{(1 + \\sqrt{2})^k - (1 - \\sqrt{2})^k}{2 \\sqrt{2}}$$\n\nHere are the first ten solutions:\n\nk = 1,  m = 1,       n = 1\nk = 3,  m = 7,       n = 5\nk = 5,  m = 41,      n = 29\nk = 7,  m = 239,     n = 169\nk = 9,  m = 1393,    n = 985\nk = 11, m = 8119,    n = 5741\nk = 13, m = 47321,   n = 33461\nk = 15, m = 275807,  n = 195025\nk = 17, m = 1607521, n = 1136689\nk = 19, m = 9369319, n = 6625109\n\n\nPlugging the Pell solutions into the circle solution:\n\nWe have \\begin{align} a_k &= \\frac{1}{2} + \\frac{1}{2n} \\\\ &= \\frac{1}{2} + \\frac{\\sqrt{2}}{(1 + \\sqrt{2})^k - (1 - \\sqrt{2})^k} \\in \\mathbb{Q} \\end{align} and \\begin{align} b_k &= \\frac{1}{2} \\pm \\frac{m}{2n} \\\\ &= \\frac{1}{2} \\pm \\frac{\\sqrt{2}}{2} \\frac{(1 + \\sqrt{2})^k + (1 - \\sqrt{2})^k} {(1 + \\sqrt{2})^k - (1 - \\sqrt{2})^k} \\in \\mathbb{Q} \\end{align} for any odd $k \\in \\mathbb{N}$.\n\nHere is a visualization:\n\nFeatured are the real solutions (blue circle) and a few of the rational solutions $Q_k^{(+)} = (a_k, b_k^{(+)})$ (green points), $Q_k^{(-)} = (a_k, b_k^{(-)})$ (red points) for $k \\in \\{ 1,3,5 \\}$. The other ones pile up too much to be distinguishable in the plot.\n\n\u2022 FYI: When your number of edits exceeded ten, a system flag was raised. It used to be a rule that at that point a post became Community-Wiki = free for all to edit, and no more rep gained from votes. This is because some users wanted to \"bump\" their post to the front page by frivolous edits. It is clear to me that you had no such intentions. Yet, the bumping may irritate other users. Even though sometimes the post needs a lot of polishing and such. \u2013\u00a0Jyrki Lahtonen Nov 14 '15 at 6:50\n\u2022 (cont'd) To address those needs a sandbox was created in meta. So if you foresee the need to edit a post many times, my advice is to copy/paste its content to a vacated answer box of the sandbox, and do the editing there. When you are happy with it copy/paste it back here. No real harm done. I just wanted to make sure that you are aware of the downsides of plenty of edits, and the recommended solution. \u2013\u00a0Jyrki Lahtonen Nov 14 '15 at 6:52\n\u2022 How do you know that from this $(1 + \\sqrt{2})^k (1-\\sqrt{2})^k = (m + \\sqrt{2} n)(m - \\sqrt{2} n)$ follows this $m + \\sqrt{2} n = (1 + \\sqrt{2})^k \\\\ m - \\sqrt{2} n = (1 - \\sqrt{2})^k \\\\$? \u2013\u00a0Farewell Nov 14 '15 at 13:01\n\u2022 It looks obvious but... \u2013\u00a0Farewell Nov 14 '15 at 13:01\n\u2022 @AntePaladin I added the expansion via binomial theorem. \u2013\u00a0mvw Nov 14 '15 at 23:19", "date": "2019-05-23 15:43:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1527353/are-there-infinitely-many-rational-pairs-a-b-which-satisfy-given-equation", "openwebmath_score": 0.9885804057121277, "openwebmath_perplexity": 567.0416955906717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303446461048, "lm_q2_score": 0.865224073888819, "lm_q1q2_score": 0.8564250432534765}}
{"url": "https://math.stackexchange.com/questions/2763504/convergence-of-a-n2-sqrta-n-sqrta-n1", "text": "Convergence of $a_{n+2} = \\sqrt{a_n} + \\sqrt{a_{n+1}}$ [duplicate]\n\nLet $a_1$ and $a_2$ be positive numbers and suppose that the sequence {$a_n$} is defined recursively by $a_{n+2} = \u221aa_n + \u221aa_{n+1}$. Show that this sequence is convergent.\n\nSo, I have been able to show the convergence taking three different cases namely,\n\nCase 1: Both $a_1$ , $a_2$ <4, then I proved that the sequence will be monotonically increasing as well as bounded and so converging\n\nCase 2: Both $a_1$, $a_2$ >4, in this case the sequence is monotonically decreasing and bounded and hence convergent.\n\nCase 3: One of $a_1$ and $a_2$ is <4 & other >4, lets say, $a_1$<4<$a_2$ in this case sequence will alternatively increase and decrease i.e. $a_{2n-1}$ will be increasing and $a_{2n}$ will be decreasing and $a_{2n}-a_{2n-1}$ will converge to Zero.\n\nMy question is that is there any general method through which we don't have to take all these cases and can prove the convergence of series in more generality?\n\nmarked as duplicate by Gabriel Romon, Martin R, Sil, Sim\u00e9on, Sangchul Lee\u00a0real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 2 '18 at 18:54\n\n\u2022 The things you say you proved under Case 1 and Case 2 cannot both be true. (I don't see why any of the three is true, actually...) \u2013\u00a0David C. Ullrich May 2 '18 at 15:35\n\u2022 @DavidC.Ullrich: He is splitting up the question into cases. In the first possible case, we have $a_1,a_2<1$. In a second possible case, we have $a_1,a_2>1$. In a third possible case, we have $a_1<1<a_2$ or vice versa. No matter the sequence $a_i$, $a_1$ and $a_2$ must be in one of these three cases. \u2013\u00a0Clayton May 2 '18 at 15:39\n\u2022 If $a_n$ converges to $l$ then $l=2 \\sqrt{l}$. So $l=0$ or $l=4$. I think this is why @DavidC.Ullrich say that both affirmation cannot be true, if Case $1$ is true then for $n$ large enough $a_n>1$ which is problematic with Case 2. \u2013\u00a0Delta-u May 2 '18 at 15:54\n\u2022 @Clayton I understood he's splitting it into cases... that doesn't explain why his conclusions in those cases are correct. \u2013\u00a0David C. Ullrich May 2 '18 at 16:05\n\u2022 Sorry, I made a mistake. I should have written 4 instead of 1. Edited it. \u2013\u00a0Lord KK May 2 '18 at 16:07\n\nIf we can make a guess of what the limit should be, then it is often easier to show the convergence by playing with the difference between $a_n$ and the limit candidate.\n\nIn our case, the limit value must solve the constraints $x = 2\\sqrt{x}$ and $x\\geq 0$, hence $x = 0$ or $4$. We claim that $4$ is the limit.\n\nWe first establish the boundedness of $(a_n)$ away from $0$ and $\\infty$.\n\n\u2022 $a_{n+2} \\geq \\sqrt{a_{n+1}}$ for all $n\\geq1$. So $a_{n+2} \\geq (a_2)^{1/2^n}$ and hence $\\liminf_{n\\to\\infty} a_n \\geq 1$.\n\n\u2022 Let $M=\\max\\{4,a_1,a_2\\}$. Then we inductively check that $a_n \\leq M$ for all $n\\geq1$.\n\nNow define $\\epsilon_n = \\lvert a_n - 4 \\rvert$ and $\\bar{\\epsilon} = \\limsup_{n\\to\\infty} \\epsilon_n$. Then $\\bar{\\epsilon} \\in [0, \\infty)$ and\n\n$$\\epsilon_{n+2} \\leq \\frac{\\epsilon_n}{\\sqrt{a_n} + 2} + \\frac{\\epsilon_{n+1}}{\\sqrt{a_{n+1}} + 2}.$$\n\nNow taking $\\limsup_{n\\to\\infty}$ to both sides yields $\\bar{\\epsilon} \\leq \\frac{\\bar{\\epsilon}}{3} + \\frac{\\bar{\\epsilon}}{3}$, which is enough to conclude that $\\bar{\\epsilon} = 0$ and hence $a_n \\to 4$.\n\nWell, if $a_n\\to L$ then $L=\\sqrt L+\\sqrt L$, so $L=4$ or $L=0$.\n\nSo instead of trying to show the sequence converges, we try to show it converges to $4$ or $0$; that may be simpler.\n\nLet's get rid of the square roots by defining $b_n=\\sqrt{a_n}$. Now$$b_{n+2}^2=b_n+b_{n+1},$$and we want to show $b_n\\to 2$. If you subtract $4$ from both sides, factor the left side and apply the triangle inequality you get $$|b_{n+2}-2|\\le\\frac{|b_n-2|+|b_{n+1}-2|}{b_{n+2}+2}.$$It seems likely to me you can use this to show $b_n\\to2$ or $0$ (if it happens that $b_n\\ge c>0$ for all $n$ then it follows that $b_n\\to 2$). I have to go to class now, sorry...\n\nEdit: How stupid of me - the other answer points out that it's more or less obvious that $b_n\\ge c>0$. In case it's not clear why we care about that, it shows that $$|b_{n+2}-2|\\le\\frac2{2+c}\\max(|b_n-2|,|b_{n+1}-2|);$$hence $b_n\\to2$. since $2/(2+c)<1$.\n\n\u2022 But how can you claim the boundedness on the right hand side? \u2013\u00a0Lord KK May 2 '18 at 16:13\n\u2022 @AloknathFurr I'm not sure what you're referring to. If $b_n\\ge c>0$ then the inequality shows that $|b_n-2|$ decreases geometrically. \u2013\u00a0David C. Ullrich May 2 '18 at 16:15\n\u2022 ok.. ok.. I got it. Thanks \u2013\u00a0Lord KK May 2 '18 at 16:17", "date": "2019-09-20 20:40:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2763504/convergence-of-a-n2-sqrta-n-sqrta-n1", "openwebmath_score": 0.8820814490318298, "openwebmath_perplexity": 322.0400184138422, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180677531124, "lm_q2_score": 0.868826777936422, "lm_q1q2_score": 0.8564182527596524}}
{"url": "https://math.stackexchange.com/questions/2276402/limit-of-sequence-in-which-each-term-is-defined-by-the-average-of-preceding-two/2276434", "text": "# Limit of sequence in which each term is defined by the average of preceding two terms\n\nWe have a sequence of numbers $x_n$ determined by the equality\n\n$$x_n = \\frac{x_{n-1} + x_{n-2}}{2}$$\n\nThe first and zeroth term are $x_1$ and $x_0$.The following limit must be expressed in terms of $x_0$ and $x_1$ $$\\lim_{n\\rightarrow\\infty} x_n$$\n\nThe options are:\n\nA)$\\frac{x_0 + 2x_1}{3}$ B)$\\frac{2x_0 + 2x_1}{3}$\n\nC)$\\frac{2x_0 + 3x_1}{3}$ D)$\\frac{2x_0 - 3x_1}{3}$\n\nSince it was a multiple choice exam I plugged $x_0=1$ and $x_1=1$. Which means that all terms of this sequence is $1$,i.e, $$x_n=1, n\\in \\mathbb{N}$$\n\nFrom this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence's limit?\n\n\u2022 Your 'trick' was very smart, in the setting of a multiple choice exam. It makes it very easy to exclude options. But it is even smarter that you want to know the 'actual' method to solve this! \u2013\u00a0user193810 May 11 '17 at 15:32\n\u2022 @Pakk thanks for the compliment! \u2013\u00a0Ananth Kamath May 12 '17 at 2:21\n\u2022 @Pakk Hehe, me on those factoring problems back in ye olde days o' Algebra. Just plug $x=0$ in and you pretty much done... \u2013\u00a0Simply Beautiful Art May 27 '17 at 0:29\n\n$$2x_n = x_{n-1} + x_{n-2}$$\n\n$$2x_2 = x_{1} + x_{0}\\\\ 2x_3 = x_{2} + x_{1}\\\\ 2x_4 = x_{3} + x_{2}\\\\ 2x_5 = x_{4} + x_{3}\\\\ ...\\\\ 2x_n = x_{n-1} + x_{n-2}$$\n\nNow sum every equation and get\n\n$$2x_n+x_{n-1}=2x_1+x_0$$\n\nSupposing that $x_n$ has a limit $L$ then making $n\\to \\infty$ we get:\n\n$$2L+L=2x_1+x_0\\to L=\\frac{2x_1+x_0}{3}$$\n\n\u2022 What did you sum to arrive at $2x_n+x_{n-1}=2x_1+x_0$? When I summed the numbered equations, I got a proportion of $x_0$ and $x_1$, but not $2:1$, and summing the indexed equations was quite boring. \u2013\u00a0user121330 May 12 '17 at 22:40\n\u2022 @user121330: just those equations as written (equivalently, you can rewrite them bringing the RHS terms to LHS and negating them). Note that each (non-terminal) $x_i$ occurs once as $2x_i$ on the LHS, and twice as $x_i$ on the RHS. So they all cancel out except for the terminal ones. \u2013\u00a0smci May 12 '17 at 23:23\n\u2022 I fondly remember this as exercise 3.5.10 of Bartle's and Sherbert's Introduction to Real Analysis, how my teacher solved it by finding an invariant as you did, and how I had wondered how she came up with it (she just said to 'play with it'). The nostalgia. \u2013\u00a0Vandermonde May 14 '17 at 22:27\n\u2022 No you're not summing to infinity. You're summing for LHS's from $x_2$ to $x_n$. That's finite. After you get that sum, we then apply the condition that if ${x_n}$ has a limit L as $n \\to \\infty$ then both $x_n, x_n+1 \\to L$, and that sets up a useful equality involving $x_1, x_0$. \u2013\u00a0smci May 16 '17 at 6:40\n\u2022 But one cannot assume that as n\u2192\u221e $x_n$ = $x_n-1$ when it is not given that the sequence converges. \u2013\u00a0Ahmad Lamaa Sep 24 '18 at 21:09\n\nYet another trick: you can write the recurrence in matrix form: $$\\left(\\begin{array}{c} x_n\\\\ x_{n-1} \\end{array} \\right) = \\left(\\begin{array}{cc} 1/2 & 1/2\\\\ 1 & 0\\\\ \\end{array} \\right) \\left(\\begin{array}{c} x_{n-1}\\\\ x_{n-2} \\end{array} \\right).$$ Then, $$\\left(\\begin{array}{c} x_n\\\\ x_{n-1} \\end{array} \\right) = \\left(\\begin{array}{cc} 1/2 & 1/2\\\\ 1 & 0\\\\ \\end{array} \\right)^{n-1} \\left(\\begin{array}{c} x_1\\\\ x_0 \\end{array} \\right).$$ Diagonalizing/converting the matrix to the Jordan form, you can find a closed form for the sequence.\n\n\u2022 I wouldn't call it a trick, I think it's the most natural approach to these problems. And it justifies the method in Arnaud's answer. \u2013\u00a0Sasho Nikolov May 12 '17 at 19:11\n\nFirst, notice that you can rewrite the recurrence relation as $$2x_{n+2}-x_{n+1}-x_n=0,\\quad n\\geq 0.$$ Now the key point is that this recurrence relation is linear, and thus if $(y_n)_{n\\geq 0}$ and $(z_n)_{n\\geq 0}$ satisfy this relation, then for any $\\alpha,\\beta\\in \\Bbb R$, $(\\alpha y_n+\\beta z_n)_{n\\geq 0}$ will also satisfy it. So we can try to express $(x_n)$ as a linear combination of simpler sequences $(y_n)$ and $(z_n)$. An example of simple sequence would be a geometric sequence $y_n=r^n$ for some $r$. Can we find a sequence of this form satisfying the recurrence relation? $r$ would have to be such that $$2r^{n+2}-r^{n+1}-r^n=r^{n}(2r^2-r-1)=0,\\quad n\\geq 0,$$thus it is enough that $$2r^2-r-1=0.$$ This will hold if and only if $r\\in \\left\\{1,\\frac{-1}{2}\\right\\}$.\n\nThus we know that any sequence of the form\n\n$$\\alpha +\\beta \\left(\\frac{-1}{2}\\right)^n$$ satisfies our recurrence relation. Then it suffices to check that the first two term are the same, and all the others will follow. Thus you need to find $\\alpha,\\beta$ such that $$\\left\\{ \\begin{array}{}\\alpha+\\beta & = & x_0\\\\ \\alpha-\\frac{\\beta}{2} & = & x_1.\\end{array} \\right.$$\n\nThis is a simple linear system, whose solution is given by $\\alpha=\\frac{x_0+2x_1}{3}$ and $\\beta=\\frac{2x_0-2x_1}{3}$.\n\nThus the $n$-th term must be given by $$x_n=\\frac{x_0+2x_1}{3}+\\frac{(2x_0-2x_1)(-1)^{n}}{3\\cdot 2^n}$$ and thus $\\lim_{n\\to \\infty }x_n = \\frac{x_0+2x_1}{3}$.\n\nThis method works for a large variety of cases; in fact, it can be applied to give a formula for $x_n$ for any linear recurrence relations (there are some difficulties if the corresponding polynomial equation has multiple roots, because you don't get enough geometric sequences, but you can find other solutions in those case). For example, you can apply the same method to the Fibonacci sequence, and it gives you the Binet formula (you can find more details in the answers to this question).\n\n\u2022 How can we come to the conclusion that $x_n = r^n$ ? Thanks for answering by the way. \u2013\u00a0Ananth Kamath May 11 '17 at 14:19\n\u2022 We can't. It's more \"wishful thinking\". Maybe I can explain that better. \u2013\u00a0Arnaud D. May 11 '17 at 14:22\n\u2022 It does makes sense, but I believe such an approach is only good for this particular question? \u2013\u00a0Ananth Kamath May 11 '17 at 14:30\n\u2022 @AnanthKamath No, it's a general method for linear recurrence relations. You can find more information for example at en.wikipedia.org/wiki/\u2026 \u2013\u00a0Arnaud D. May 11 '17 at 14:44\n\u2022 I never knew that was possible. Thanks for sharing info. \u2013\u00a0Ananth Kamath May 11 '17 at 14:58\n\nHere's yet another way to look at the problem, that might not lead directly to a proof, but might give intuitive insight into the sequence.\n\nNote that from $x_0$, we go $x_1-x_0$ up to $x_1$, then $\\frac12(x_1-x_0)$ down to $x_2$, then $\\frac14(x_1-x_0)$ up to $x_3$, etc. You might be able to prove this by induction, but might not want to. If $x_0=0$ and $x_1=1$, one can put this in a picture: ($A=(0,x_0)$, $B=(1,x_1)$, ...)\n\nWe conclude that, since the jump from $x_0$ to $x_1$ can really be seen a a jump of $1=(-1/2)^0$ relative to $x_0$:\n\n$$x_n = x_0 + (x_1-x_0)\\sum_{i=0}^{n-1} \\left(-\\frac12\\right)^i$$\n\nThen the limit for $n \\rightarrow \\infty$ is:\n\n$$\\lim_{n\\rightarrow\\infty} x_n = x_0 + (x_1-x_0)\\sum_{i=0}^\\infty \\left(-\\frac12\\right)^i = x_0 + (x_1-x_0)\\frac1{1+\\frac12}$$\n\nby the geometric series (since $\\lvert-\\frac12\\rvert < 1$). Then this is equal to:\n\n\\begin{align*} x_0 + (x_1-x_0)\\frac1{1+\\frac12} &= x_0 + \\frac23(x_1-x_0) = \\frac{3x_0+2x_1-2x_0}3 \\\\ &= \\frac{x_0 + 2x_1}3. \\end{align*}\n\n\u2022 Can't we upvote an answer twice? :) \u2013\u00a0Kartik May 12 '17 at 10:26\n\u2022 This is what Soumik did in his answer. Unfortunately he did not feel like working it out. \u2013\u00a0Carsten S May 12 '17 at 12:39\n\u2022 @CarstenS Oh right I see; luckily I wasn't the only one to see this :) \u2013\u00a0tomsmeding May 12 '17 at 16:29\n\u2022 Yes, +1 for the illustration. \u2013\u00a0Carsten S May 12 '17 at 16:31\n\n$$x_n-x_{n-1}=-\\frac{1}{2}(x_{n-1}-x_{n-2})$$ use repeatedly to get $$x_n-x_{n-1}=(-\\frac{1}{2})^{n-1}(x_{1}-x_{0})$$ Now it is fairly straightforward to compute\n\n\u2022 Could you show a little more work on how you got to your first step? \u2013\u00a0AlgorithmsX May 11 '17 at 19:04\n\u2022 subtract $x_{n-1}$ from both sides of the equation and iterate. \u2013\u00a0user379195 May 11 '17 at 19:05\n\u2022 I know how you got there, but I think it's better for you to show some less obvious intermediate steps. It's still a great post. \u2013\u00a0AlgorithmsX May 11 '17 at 19:08\n\nHere is a way to find the general solution of the recurrence using generating functions and then find the limit. We wish to solve the recurrence $$a_n = \\frac{a_{n-1}}{2} + \\frac{a_{n-2}}{2}\\quad (n\\geq 2)\\tag{1}$$ where $a_0=x_0$ and $a_1=x_1$. We first make the recurrence valid for all $n\\geq 0$. To this end, set $a_n=0$ for $n<0$ and note that the recurrence $$a_n = \\frac{a_{n-1}}{2} + \\frac{a_{n-2}}{2}+x_0\\delta_{n,0}+(x_1-x_0/2)\\delta_{n,1}\\tag{2}$$ where $\\delta_{ij}$ is the Kronecker Delta does the trick. Let $A(x)= \\sum_{n=0}^\\infty a_nx^n$ be the generating function and multiply by $x^n$ and sum on $n$ in (2). Then we get $$A(x)\\left(1-\\frac{1}{2}x-\\frac{1}{2}x^2\\right)=x_0+x\\left(x_1-\\frac{x_0}{2}\\right)\\tag{3}$$ and hence $$A(x) =\\frac{x_0+(x_1-x_0/2)x}{1-\\frac{1}{2}x-\\frac{1}{2}x^2} =\\frac{x_0+(x_1-x_0/2)x}{(1-x)(1+x/2)} =\\frac{2x_1+x_0}{3(1-x)}+\\frac{2x_0-2x_1}{3(1+x/2)}\\tag{4}$$ by partial fractions. By using the geometric series we get that \\begin{align} A(x) &=\\sum_{n=0}^\\infty \\left[\\frac{2x_1+x_0}{3}\\right]x^n+ \\sum_{n=0}^\\infty \\left[\\frac{2x_0-2x_1}{3}\\right]\\left(\\frac{-x}{2}\\right)^n\\\\ &=\\sum_{n=0}^\\infty \\left[\\frac{(2x_1+x_0)}{3}+\\frac{(2x_0-2x_1)(-1)^n}{3(2^n)}\\right]x^n\\tag{5} \\end{align} Hence\n\n$$a_n=\\frac{x_0+2x_1}{3}+\\frac{(2x_0-2x_1)(-1)^{n}}{3\\cdot 2^n}\\tag{6}$$ and thus $$\\lim_{n\\to \\infty }a_n = \\color{blue}{\\frac{x_0+2x_1}{3}.}\\tag{7}$$\n\nHere is a formal derivation of your result. The sequence you have found is a generalization of the Fibonacci sequence.\n\nThere have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\\in\\mathbb{R}$, (Ref: Maynard, P. (2008), \u201cGeneralised Binet Formulae,\u201d $Applied \\ Probability \\ Trust$; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)\n\nWe have extended Maynard's analysis to include arbitrary $f_0,f_1\\in\\mathbb{R}$. It is relatively straightforward to show that\n\n$$f_n=\\left(f_1-\\frac{af_0}{2}\\right) \\frac{\\alpha^n-\\beta^n}{\\alpha-\\beta}+\\frac{f_0}{2} (\\alpha^n+\\beta^n)$$\n\nwhere $\\alpha,\\beta=(a\\pm\\sqrt{a^2+4b})/2$.\n\nThe result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1 \\ \\text{and} \\ f_0=0, f_1=1$. Notice that there is nothing in this derivation that limits the results to integer values for the initial conditions, scaling factors, or the sequence itself.\n\nSo, specializing to your case, we can say\n\n$$x_n=\\frac{x_{n-1}}{2} +\\frac{x_{n-2}}{2}$$\n\nand\n\n$$\\alpha,\\beta=\\frac{\\frac{1}{2}\\pm\\sqrt{\\frac{1}{4}+2}}{2}=1,\\frac{1}{2}$$\n\nFor the limit as $n\\to\\infty$, we note that the $\\alpha$-term dominates, and is in fact, always unity, ergo\n\n$$\\lim_{n\\to\\infty} x_n=\\left(x_1-\\frac{x_0}{4}\\right)\\frac{\\alpha^n}{\\alpha-\\beta}+\\frac{x_0}{2}\\alpha^n=\\left(x_1-\\frac{x_0}{4}\\right)\\frac{2}{3}+\\frac{x_0}{2}=\\frac{2x_1+x_0}{3}$$\n\nas was shown previously.\n\nThis proves the OP's assertion. The advantage of this method is that it will apply to all such problems.\n\n$$x_n = \\frac{x_{n-1}+x_{n-2}}{2}$$\n\nAdd $\\frac{x_{n-1}}{2}$ to both sides:\n\n$$\\frac{2x_{n}+x_{n-1}}{2} = \\frac{2x_{n-1}+x_{n-2}}{2}$$\n\nLet $y_{n} = \\frac{2x_{n}+x_{n-1}}{2}$:\n\n$$y_{n} = y_{n-1} = \\ldots = y_{1}$$\n\nAs $n \\rightarrow \\infty, x_{n-1} \\rightarrow x_{n}$, so:\n\n$$\\frac{3x_{n}}{2} = \\frac{2x_{1}+x_{0}}{2}$$\n\nNote that this is the \"same\" approach as some of the other answers, I just added it because it is most clear to me.\n\nMy solution:\n\nMap $[x_0,x_1]$ to $[0,1]$ by setting $x = x_0 + (x_1-x_0)y$ so that $y_0 = 0$ and $y_1 = 1$.\n\nNow the sequence runs $0, 1, \\frac{1}{2}, \\frac{3}{4}, \\frac{5}{8}, \\frac{11}{16}$, etc\n\nTake the difference between each successive term i.e. $y_k-y_{k-1}$\n\nYou get $+1, -\\frac{1}{2}, +\\frac{1}{4}, -\\frac{1}{8}, +\\frac{1}{16}$, etc\n\nThis looks like the expansion of $\\frac{1}{1-z}$ for $|z|<1$ i.e. $1+z+z^2+z^3+...$\n\nIn our case, $z = -\\frac{1}{2}$, so the sum is $\\frac{1}{1-(-1/2)} = \\frac{2}{3}$\n\nSo the limit of $y_k$ is $\\frac{2}{3}$, given $y_0 = 0$ and $y_1 = 1$. So now map back to $x$\n\n$x = x_0 + (x_1-x_0)\\frac{2}{3} = \\frac{x_0 + 2x_1}{3}$\n\n\u2022 Use this link to know how to write mathematics on this site. \u2013\u00a0StubbornAtom May 14 '17 at 10:25\n\u2022 Apologies, have used the maths editor on the site before but this was from my phone, so it's not particularly easy to do. Appreciate the edit, meant to follow up this evening with the same. Best, james \u2013\u00a0James Spencer-Lavan May 14 '17 at 10:44\n\nUnless $x_0=x_1$ which is clear, consider the tranformation $y_n:=\\frac{x_n-x_0}{x_1-x_0}$. Then $y_0=0$, $y_1=1$, and $y_{n+1}$ is the average of $y_n$ and $y_{n-1}$. It is clear that $\\lim y_n$ is $$\\frac{1}{2}+\\frac{1}{4}-\\frac{1}{8}+\\frac{1}{16}-\\frac{1}{32}+\\cdots=\\frac{1}{2}\\left(1+\\frac{1}{4}+\\frac{1}{16}+\\cdots\\right)=\\frac{2}{3}.$$ Going back, you have $\\lim x_n=\\frac{1}{3}(x_0+2x_1)$.\n\n\u2022 This was my instant reaction to the problem, comparison with the binary number given by 0.101010101010... which is 2/3. \u2013\u00a0richard1941 May 16 '17 at 18:10\n\nSince the recurrence relation is linear, the sequence's limit is, if it exists, of the form $ax_0+bx_1$. If you consider the sequence $x'_n=x_{n+1}$, it clearly has the same limit therefore $$ax_0+bx_1=ax'_0+bx'_1=ax_1+bx_2=ax_1+\\frac b2(x_0+x_1)=\\frac{b}2x_0+\\left(a+\\frac b2\\right)x_1.$$ Since this is true for any $x_0$, $x_1$, we have $a=\\frac b2$ (and also $b=a+\\frac b2$ but this is equivalent). This is enough to answer the question, but not to find $a$ and $b$. To find $a$ and $b$, you can simply take the same trick you used by considering the constant sequence $1$ and you get $a+b=1$, which solves the problem.\n\nI came to this page because I wanted to justify the formula I came up with for the series with xn = pxn-1+(1-p)xn-2. This is more general than the original problem, though not as general as the linear combination problem. Based on what I read here, I can show the solution and justify it using just a little insight and basic algebra. The first insight is to assume the solution has the form rx1+(1-r)x0. Let's simplify things by using x0=0 and x1=1. Then x2=p. We must also have rx2+(1-r)x1 as the solution. Taking rx2+(1-r)x1 = rx1+(1-r)x0 and substituting gives p=(1-r)+rp. Solving for r gives r =1/(2-p) and (1-r) = (1-p)/(2-p). Having found the solution to the problem, we need to justify it. It is straightforward, though a little tedious, to show that, for the value of r that we found, that rxn + (1-r)xn-1 = rxn-1 + (1-r)xn-2. Just substitute for xn and r and combine terms. By induction, this means that rxn + (1-r)xn-1 = rx1 + (1-r)x0. As was done in a previous post, we can take xn-1 on the left side to equal xn, which proves our assertion provided we have convergence. To find where we have convergence, use the expression for xn to show that xn - xn-1 = -(1-p)(xn-1 - xn-2), giving convergence when |1-p| < 1.\n\n\u2022 There is a simpler approach for finding r. Find a and b such that ax<sub>n</sub>+bx<sub>n-1</sub>=ax<sub>n-1</sub>+bx<sub>n-2</sub>. The constraints for x<sub>n-1</sub> are the same as for x<sub>n-2</sub>, giving b=a(1-p). What we need to realize is that if a and b satisfy the equation, so do ta and tb. We need the additional constraint of a+b=1. Then things work out very simply. \u2013\u00a0user1153980 Nov 12 '17 at 10:10\n\nFirst, the sequence $(x_n)$ has clearly a limit $\\ell$ because $$\\left|x_{n+1}-x_{n}\\right|=\\frac{|x_1-x_0|}{2^n}.$$ Now, sum the $n-1$ equations $2x_{i+2}=x_{i+1}+x_i$ for $i=0,1,\\ldots,n-2$ and we obtain $$2x_n+x_{n-1}=2x_1+x_0.$$ Passing to the limit we reach $$\\ell=\\frac{2x_1+x_0}{3}.$$", "date": "2020-07-13 04:56:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2276402/limit-of-sequence-in-which-each-term-is-defined-by-the-average-of-preceding-two/2276434", "openwebmath_score": 0.9318219423294067, "openwebmath_perplexity": 254.52874176837292, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180677531124, "lm_q2_score": 0.868826771143471, "lm_q1q2_score": 0.8564182460637179}}
{"url": "http://mathhelpforum.com/calculus/164757-integrating-enclosed-area.html", "text": "1. ## Integrating enclosed area\n\nCalculate the area enclosed by x = 9-y^2 and x = 5 in two ways: as an integral along the y-axis and as an integral along the x-axis.\n\nPlease tell me if my figure is correct, and also if the shaded area is also the are I am looking for.\nAlso is there any formulas I need to use for this problem ? How do I do this both ways ?\nThank You\n\n2. Yes, your graph is fine.\n\nTo integrate over the x-axis, you need\n\n$\\displaystyle\\int_{5}^9f(x)dx$\n\nso $x=9-y^2\\Rightarrow\\ y^2=9-x\\Rightarrow\\ y=\\pm\\sqrt{9-x}=f(x)$\n\nTake the positive one and double your result as the x-axis is an axis of symmetry.\n\n$2\\displaystyle\\int_{5}^9\\left(9-x\\right)^{0.5}dx$\n\nTo integrate over the y-axis, you could integrate $f(y)$ from $y=-3$ to $y=3.$\n\nIf you like, again use the x-axis as an axis of symmetry and double the integral from $y=0$ to $y=3.$\n\nThis integral includes the unshaded part against the y-axis, so you have a few ways to cope with that.\n\nThe easiest way is to subtract 5 from $x=f(y)$\n\n$x-5=f(y)-5=4-y^2$\n\nThe new function is $x=4-y^2$\n\nso the limits of integration become $\\pm2$\n\nYou can then calculate\n\n$2\\displaystyle\\int_{0}^2\\left(4-y^2\\right)}dy$\n\n3. Thank You very much. I'll give it a shot.\n\n4. The first integral can be evaluated by making a substitution,\nwhile the 2nd one doesn't require any substitution.\n\n5. Thank you for all you help.\n\n6. Ok so I solved the first one I get .5. I let u = 9-x , du = -dx so I multiplied by a negative inside the integral and a negative outside. then I get, -2(.5(9-x)^-.5). Then using the fundemental theorem of calculus I used the limts 5 to 9 and my final answer was .5. is this correct ?\n\n7. Originally Posted by wair\nOk so I solved the first one I get .5. I let u = 9-x , du = -dx so I multiplied by a negative inside the integral and a negative outside. then I get, -2(.5(9-x)^-.5). Then using the fundemental theorem of calculus I used the limts 5 to 9 and my final answer was .5. is this correct ?\nNot quite,\nyou differentiate to introduce the substitution,\nbut you are also differentiating when you should be integrating!\n\n$u=9-x\\Rightarrow\\ du=-dx\\rightarrow\\ dx=-du$\n\n$x=5\\Rightarrow\\ u=4$\n\n$x=9\\Rightarrow\\ u=0$\n\nthe integral becomes\n\n$\\displaystyle\\ -2\\int_{4}^0u^{0.5}}du=(-2)\\left[-\\int_{0}^4u^{0.5}}du\\right]=2\\int_{0}^4u^{0.5}}du$\n\n$=2\\displaystyle\\left[\\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]$ from u=0 to u=4\n\n8. so I just had to chnage my limits ? and then invert them to get rid of the what was it called \"signed area\" ? But I don't understand how you got the last step. wouldn't it be .5(u)^-.5 ?\n\n9. Originally Posted by wair\nso I just had to chnage my limits ? and then invert them to get rid of the what was it called \"signed area\" ? But I don't understand how you got the last step. wouldn't it be .5(u)^-.5 ?\nNo, that's what you get when you differentiate.\n\n$\\displaystyle\\frac{d}{du}u^{0.5}=0.5u^{-0.5}$\n\nBut\n\n$\\displaystyle\\int{u^{0.5}}du=\\frac{u^{1.5}}{1.5}+c$\n\nbecause\n\n$\\displaystyle\\frac{d}{du}\\left[\\frac{u^{1.5}}{1.5}+c\\right]=\\frac{1.5}{1.5}u^{0.5}$\n\nTo integrate, apply differentiation in reverse.\n\n10. Oh right right. I understand now thank you .\n\n11. Ok so for the first one I get -21.1. (1/1.5)(9-x)^1.5. 2(7.45)-2(18)=-21.1. Is that correct ?\n\n12. I get 5.3 for the second one. shouldn't I get the same answer for both ?\n\n13. Originally Posted by wair\nOk so for the first one I get -21.1. (1/1.5)(9-x)^1.5. 2(7.45)-2(18)=-21.1. Is that correct ?\nIf you work with 9-x instead of u, then there is no need to change the limits of integration.\n\n$2\\displaystyle\\int_{x=5}^{x=9}{(9-x)^{0.5}}dx=-2\\int_{x=5}^{x=9}{(9-x)^{0.5}}d(9-x)$\n\n$=2\\displaystyle\\int_{x=9}^5{(9-x)^{0.5}}d(9-x)=2\\frac{(9-x)^{1.5}}{1.5}$ from x=9 to 5 (start calculations at x=5)\n\n$=2\\displaystyle\\frac{(9-5)^{1.5}-(5-5)^{1.5}}{1.5}=2\\frac{4^{1.5}}{1.5}=2\\frac{8}{1.5} =2\\frac{16}{3}=\\frac{32}{3}$\n\nYou must get a positive value for area, so as our graph is symmetrical about the x-axis, we integrate the part above the x-axis and double it.\n\nUsing the u-substitution, we get\n\n$\\displaystyle\\ 2\\int_{0}^4{u^{0.5}}du=2\\left[\\frac{u^{1.5}}{1.5}\\right]$ from u=0 to u=4 (start calculations at u=4)\n\n$\\displaystyle\\ =2\\frac{4^{1.5}}{1.5}-0=2\\frac{8}{1.5}=\\frac{16}{1.5}=\\frac{32}{3}$\n\nFor the 2nd integral..\n\n$\\displaystyle\\ 2\\int_{0}^2{\\left(4-y^2\\right)}dy=2\\int_{0}^2{4}dy-2\\int_{0}^2{y^2}dy=2\\left[4y-\\frac{y^3}{3}\\right]$ evaluated from y=0 to y=2\n\n(start calculations at y=2)\n\ncomes out as the same value.\nReview your calculations for both integrals.\n\n14. Thank you very much. So if I want to use the new limits I don't need to replacex u with what u equals. And both values must be the same. OK Thank you\n\n15. Originally Posted by wair\nThank you very much. So if I want to use the new limits I don't need to replacex u with what u equals. And both values must be the same. OK Thank you\nI think you mean.... \"don't need to replace u with 9-x\".\n\nYes, it is simplest to work with \"u\", having made the substitution, then integrate f(u)du using the \"u\" limits.\n\nThis area you are now calculating is on a different graph but evaluating the new \"u\" integral will give the same area as the shaded region on the original graph.\n\nIf you prefer not to change the limits, here's how to do it....\n\n$\\displaystyle\\ -2\\int_{x=5}^{x=9}{u^{\\frac{1}{2}}}du=-2\\left[\\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]$ from x=5 to x=9 (start calculations at x=9)\n\n$=-\\displaystyle\\frac{4}{3} (9-x)^{\\frac{3}{2}}$ from x=5 to x=9\n\n$=-\\displaystyle\\frac{4}{3}\\left[ (9-9)^{1.5}-(9-5)^{1.5}\\right]=\\frac{32}{3}$\n\nYou've got to practice. Keep going until you can reproduce the solution for both integrals.", "date": "2017-11-24 00:56:30", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/164757-integrating-enclosed-area.html", "openwebmath_score": 0.9478228688240051, "openwebmath_perplexity": 643.7519994399651, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180681726683, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8564182414062878}}
{"url": "http://mathhelpforum.com/advanced-algebra/87241-method-finding-minimal-polynomial.html", "text": "Math Help - Method for finding a Minimal Polynomial?\n\n1. Method for finding a Minimal Polynomial?\n\nHello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:\n\n$m_{\\sqrt 2+i}$ over $\\mathbb Q$\n\nThe reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.\n\nAnyway, I would appreciate any help. Thanks!\n\nP.s.: That's m_{\\sqrt 2 +i}; sorry that the \"i\" is so tiny.\n\n2. Originally Posted by Dark Sun\nHello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:\n\n$m_{\\sqrt 2+i}$ over $\\mathbb Q$\n\nThe reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.\n\nAnyway, I would appreciate any help. Thanks!\n\nP.s.: That's m_{\\sqrt 2 +i}; sorry that the \"i\" is so tiny.\n$x=\\sqrt{2}+i$\n\n$x^2=(\\sqrt{2}+i)^2 \\iff x^2=2+2i\\sqrt{2}-1 \\iff x^2-1=2i\\sqrt{2}$\n\n$(x^2-1)^2=(2i\\sqrt{2})^2$\n\n$x^4-2x^2+1=-8$\n\n$x^4-2x^2+9=0$\n\n3. Originally Posted by TheEmptySet\n$x=\\sqrt{2}+i$\n\n$x^2=(\\sqrt{2}+i)^2 \\iff x^2=2+2i\\sqrt{2}-1 \\iff x^2-1=2i\\sqrt{2}$\n\n$(x^2-1)^2=(2i\\sqrt{2})^2$\n\n$x^4-2x^2+1=-8$\n\n$x^4-2x^2+9=0$\nyou also need to prove that $p(x)=x^4 - 2x^2 + 9$ is irreducible over $\\mathbb{Q}$: we only need to show that it cannot be factored into two quadratic polynomials over $\\mathbb{Z}.$ why?\n\nsuppose $p(x)=(x^2+ax+b)(x^2+cx+d),$ for some $a,b,c,d \\in \\mathbb{Z}.$ then we'll have $bd=9$ and $b+d+2=a^2,$ which is easily seen to have no solutions in $\\mathbb{Z}.$\n\n4. Because irreducible over $\\mathbb Z$ implies irreducible over $\\mathbb Q$.\n\n$x^4-2x^2+9$ clearly does not have any degree 1 factors. Also, a degree 3 factor would imply a degree 1 factor, so if a factor exists, then it will be degree 2, which you have just shown is a contradiction.\n\nThank you EmptySet and NonCommAlg for presenting me with this truly powerful method!", "date": "2015-03-05 01:16:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/87241-method-finding-minimal-polynomial.html", "openwebmath_score": 0.8886801600456238, "openwebmath_perplexity": 290.746664530634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180681726682, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8564182414062876}}
{"url": "https://www.physicsforums.com/threads/find-formula-for-series.762785/", "text": "# Find formula for series\n\n1. Jul 23, 2014\n\n### Maxo\n\n1. The problem statement, all variables and given/known data\nBy using the general equation for an arithmetic series, find a formula for calculating the series\n$$1+3+5+...+(2n-1)$$\n\n2. Relevant equations\nGeneral equation for an arithmetic series:\n$$\\sum ^{n}_{k=1}k=\\frac{n(n+1)}{2}$$\n\n3. The attempt at a solution\nUsing the general equation we have\n$$\\sum ^{n}_{k=1}(2k-1)$$\nbut how can we go from there? What is the next step?\n\nI tried replacing n with (2k-1) in the general equation but that gives the wrong answer. So how can we do it?\n\nLast edited: Jul 23, 2014\n2. Jul 23, 2014\n\n### Mentallic\n\n$$\\sum (a+b) = \\sum a + \\sum b$$\n\nand\n\n$$\\sum (ab) = a\\sum b$$\nif a is a constant that is independent of the summation index (k for example).\n\nSo using these formulae, you should be able to solve your problem.\n\nedit:\n\nAnd keep in mind that\n\n$$\\sum_{k=1}^{n}1 = 1+1+1+...+1 = n$$\n\nbecause you go from k=1 to n while summing 1 each time (hence n times).\n\n3. Jul 23, 2014\n\n### Maxo\n\nAllright, so we have\n\n$$\\sum ^{n}_{k=1}(2k-1)=\\sum ^{n}_{1}2k-\\sum ^{n}_{1}1$$\nThe second term is easy since we see that we have 1 n times so 1*n = n as you already wrote. Can it be also calculated using the general equation? When I try I get (1(1+1))/2=2/2=1 which is not correct. Also if we just put in n it gets n(n+1)/2 which does not equal n! I guess I am using the general equation wrong? How should it be used then?\n\nThen the first term, how is that calculated? Using the general equation I get 2k(2k+1)/2 = k(2k+1) which is again wrong. Please tell help me how to use the general equation.\n\n4. Jul 23, 2014\n\n### Mentallic\n\nNotice the similarity?\n\n$$\\sum^n_1 2k =2\\cdot 1+2\\cdot 2+2\\cdot 3+...+2\\cdot n \\\\ = 2(1+2+3+...+n) \\\\ = 2\\sum^n_1 k$$\n\nDo you know how to prove the general summation formula? If you do, you'll also see why you can't use it in the instances that you've tried to. If not, begin with\n\n$$S=1+2+3+...+n$$\n\nand then consider\n\n$$S=n+...+3+2+1$$\n\nand add both of these values together, term by term (add the values in the same column in pairs)\n\n$$2S=(1+n)+(2+(n-1))+(3+(n-2))+...+((n-2)+3)+((n-1)+2)+(n+1)$$\n\nand simplify the right side, then solve for S.\n\nNow, once you've learned how to prove the sum of the first n integers, now use a similar technique to find the general summation formula\n\n$$S=a+(a+d)+(a+2d)+...+(a+(n-1)d)$$\n\nwhere a is your starting point, and d is the difference between each value.\n\n5. Jul 23, 2014\n\n### Maxo\n\nThank you, you explain very well! I understand everything better now.\nSo if we write this again the opposite direction and then add column wise, we get that each term is equal to 2a+(n-1)d and we have n such terms, so 2S = n(2a+(n-1)d) = 2an + nd(n-1), so S = an + nd(n-1)/2.\n\nIn other words, the general formula for an arithmetic sum is:\n$$Sn=an+\\frac{n^2d-nd}{2}$$\nCorrect?\nThat's interesting, I have never seen this more \"more general\"(?) summation formula before. Does it have some particular name? In my math book, which is supposed to be for university level math, it's not even mentioned.\n\nBtw if you have some suggestions where I can learn more and where it's explained as well as you explain here, please feel free to share it.\n\nLast edited: Jul 23, 2014\n6. Jul 23, 2014\n\n### Mentallic\n\nYou're welcome! Yes, that's correct! Although, the factored form is more generally used.\n\n$$S_n = \\frac{n}{2}\\left(2a+(n-1)d\\right)$$\n\nand now, if we let d=1 and a=1, then this says we are taking the sum of a linear series that starts with 1 (a=1) and increases by a value of 1 each time (d=1) which would give you your sum of positive integers. If you want to find\n\n$$\\sum_{k=1}^n 1$$\n\nthen this is like letting a=1 and d=0, which would give you S=n as expected.\n\nAlso,\n\n$$\\sum_{k=1}^n 2k = 2+4+6+...+2n$$\n\nWhich as we can see from the expanded form, we would let a=2 and d=2 since it starts at the value of 2 and increases by a value of 2 between each term. The formula, as well as the method I showed you earlier that moves the 2 outside of the sum would of course give you the same result of S=n(n+1)\n\nIt's called an arithmetic progression. Just google that and you'll find plenty on it\n\n7. Jul 24, 2014\n\n### Staff: Mentor\n\nIsn't the formula for the sum of an arithmetic series n(a+l)/2, where a is the first term, l is the last term, and n is the number of terms?\n\nChet\n\n8. Jul 25, 2014\n\n### HallsofIvy\n\nYes, it is. Another way of looking at it is that the average value in an arithmetic sequence is the same as the average of the first and last term.\n\nBy the way, here's another way of summing that original series, $$\\sum_{ix= 1}^n 2n-1$$. That is, of course, the sum of odd integers from 1 to 2n-1. It is the same as the sum of all integers from 1 to 2n minus the sum of all even numbers from 2 to 2n.\n\nThe sum of all numbers from 1 to 2n is, using that formula, $2n(1+ 2n)/2= n(2n+ 1)$. The sum of all even numbers is $2+ 4+ \\cdot\\cdot\\cdot+ 2n= 2(1+ 2+ 3+ \\cdot\\cdot\\cdot+ n)= 2(n(n+1)/2)= n(n+ 1)$.\n\n9. Jul 31, 2014\n\n### Maxo\n\nIs there any way to write this general sum using the sigma notation? If so, how could it be written with Sigma notation?\n\n10. Jul 31, 2014\n\n### Mentallic\n\nWell you aren't going to be writing the formula above using sigma notation, because sigma is used as a shorthand form to express summations that follow an expressible pattern. But the arithmetic sum itself can be expressed in sigma notation of course.\n\n$$a+(a+d)+(a+2d)+...+(a+(n-1)d)$$\n$$= \\sum_{k=0}^{n-1}a+k\\hspace{1 mm}d$$\n\nThis is a valid summation, but we can do more with it. Notice the constant a is added n times, hence to remove it from the sum, we exchange 'a' in the summation for 'an' outside of the summation. Also the constant d can be factored out of the sum. It's for the same reason that\n\n$$1d+2d+3d+...+nd = d(1+2+3+...+n)$$\n\n$$= an + d\\sum_{k=0}^{n-1}k$$\n\nNow at this point, k=0 clearly doesn't give us anything. We only needed k=0 in the first place for when a was within the sum (else if we started with k=1 then we'd only be adding the value of a (n-1) times.). So we can remove k=0 and start at k=1.\n\n$$=an + d\\sum_{k=1}^{n-1} k$$\n\nAnd that would be your summation for a general arithmetic series.\n\n11. Aug 1, 2014\n\n### Maxo\n\nThank you, I would pretty much call this a perfect answer. :)", "date": "2018-03-22 18:02:26", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/find-formula-for-series.762785/", "openwebmath_score": 0.8787103891372681, "openwebmath_perplexity": 423.95371650000294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180652357771, "lm_q2_score": 0.8688267677469951, "lm_q1q2_score": 0.8564182405286218}}
{"url": "https://math.stackexchange.com/questions/3465091/how-often-is-the-product-of-the-distinct-prime-factors-of-a-number-greater-than", "text": "# How often is the product of the distinct prime factors of a number greater than that of the the next number?\n\nThe radical $$\\mathrm{rad}\\; n$$ of a positive integer $$n$$ is the product of $$n$$'s distinct prime factors. For example $$\\mathrm{rad}\\; 12=2\\cdot 3=6$$.\n\nDenote the sum of $$n$$'s distinct prime factors by $$\\mathrm{spf}\\; n$$.\n\nClaim: Given any $$k \\ge 1$$, for nearly half of positive integers $$n$$, $$\\mathrm{spf}\\; n > \\mathrm{spf}\\; (n+k)$$, but for only one-third of positive integers $$n$$, $$\\mathrm{rad}\\; n > \\mathrm{rad}\\; (n+k)$$.\n\nDetailed explanation:\n\nFor any $$k \\ge 1$$, there will be $$n$$ such that $$\\mathrm{rad}(n) > \\mathrm{rad}(n+k)$$ e.g. $$n=15, k = 1$$. We define\n\n$$a_{n,k} = \\begin{cases} 1 & \\frac{\\mathrm{rad}(n)}{\\mathrm{rad}(n+k)} > 1 \\\\ 0 &\\text{ otherwise} \\end{cases}$$\n\nQuestion: What is limiting value\n\n$$\\lim_{x \\to \\infty}\\dfrac{1}{x}\\sum_{n \\le x}a_{n,k}$$\n\nInstead of product when I used sum, the limiting value approached $$0.5$$ which means that it is equally likely the sum of distinct prime factors of a natural number is greater or less than that of the next natural number which makes intuitive sense.\n\nSimilarly I was expecting that product of the distinct prime factors of a number is equally likely to be greater than or less than that of the the next number but this was not the case. I was surprised to see that for $$x = 1.5 \\times 10^8$$\n\n$$\\dfrac{1}{x}\\sum_{n \\le x}a_{n,k} \\approx 0.3386$$ and was decreasing with $$x$$ hence the eventual limit is expected be be slightly less than this.\n\nNote: The initial version of this question was with $$k=1$$ but the result doesn't change with $$k$$, hence I have updated the question.\n\n\u2022 $p_n$ for primes $p(n)$ for partitions to be precise :) \u2013\u00a0Nilotpal Kanti Sinha Dec 6 '19 at 4:30\n\u2022 Well the Euler product is written $\\prod_p \\frac1{1-p^{-s}}$. $p$ is reserved for primes in this context, use $\\Pi$ or anything. \u2013\u00a0reuns Dec 6 '19 at 4:35\n\nEach of these products will be $$\\frac{n}{p}$$ and $$\\frac{n+1}{q}$$ for some $$p,q\\in\\mathbb{Z}^+$$, where $$p$$ and $$q$$ are the \"leftover\" primes in the factorizations of $$n$$ and $$n+1$$ when one copy of each distinct prime has been removed.\n\nIf $$p\\neq q$$, $$\\frac{n}{p}>\\frac{n+1}{q}\\iff p. Which of $$p$$ and $$q$$ is larger will be effectively \"random\", in that the behavior will jump around a lot and one will be larger than the other with a limiting ratio of $$0.5$$. (A formal proof might prove difficult, but you can certainly gather some numerical evidence to support this - I think it makes as much intuitive sense as the sum case you mentioned.)\n\nWhen $$p=q$$, the comparison will always go in favor of $$n+1$$. But since $$n$$ and $$n+1$$ are relatively prime, this only happens when $$p=q=1$$ - i.e., when $$n$$ and $$n+1$$ are each squarefree. So we want to find out what fraction of consecutive integer pairs are squarefree, and hence what limiting ratio to exclude from the even split mentioned above.\n\nFortunately, we can express this as an infinite product. For a number to be squarefree is just for it to not be divisible by $$4, 9, 25, \\ldots, p^2,\\ldots$$. The limiting fraction of $$(n,n+1)$$ pairs which do not contain a multiple of 4 is $$2/4$$; likewise, for neither to be a multiple of 9 is $$7/9$$, and so on.\n\nHowever, congruence modulo the first $$N$$ primes is independent over large ranges by the Chinese remainder theorem. So our probability that a \"random\" pair are both squarefree is\n\n$$\\prod_p\\frac{p^2-2}{p^2}$$\n\nwhich looks to be around $$0.322634...$$, or a touch under one third. This means that your limiting value will be\n\n$$\\frac12\\left(1-\\prod_p\\frac{p^2-2}{p^2}\\right) \\approx 0.33868295...$$\n\nIf $$n+1$$ is squarefree then $$p(n+1)=n+1\\implies a_n=0$$ thus $$\\lim\\sup_{n\\to \\infty} \\dfrac{1}{x}\\sum_{n \\le x}a_n \\le 1-\\frac{6}{\\pi^2}\\approx 0.393$$\n\n\u2022 You can quickly verify with some numerical calculations that 0.39 is much too high. Try n from 1 to 10,000. \u2013\u00a0RavenclawPrefect Dec 6 '19 at 4:20\n\u2022 Sure this is just a rigorous upper bound, your approach is a conjectural estimate. It seems hard to make it rigorous, you should show us a plot of how well does it fit. \u2013\u00a0reuns Dec 6 '19 at 4:23\n\u2022 For $x = 1.65 \\times 10^8$, the value is $0.3386851$ so I guess there is scope to tighten the upper bound. \u2013\u00a0Nilotpal Kanti Sinha Dec 6 '19 at 4:28\n\u2022 If your value is correct the convergence should be in $O(x^{-1/2+\\epsilon})$ (that's what predicts the random model for the primes) \u2013\u00a0reuns Dec 6 '19 at 4:29\n\u2022 Ah, I missed the \u2264. Apologies. A graph of the ratios from n=1000 to 100,000 is at imgur.com/3LOWorl, with the blue line the constant estimated to around 9 digits. \u2013\u00a0RavenclawPrefect Dec 6 '19 at 4:41", "date": "2020-01-26 03:07:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3465091/how-often-is-the-product-of-the-distinct-prime-factors-of-a-number-greater-than", "openwebmath_score": 0.9285973310470581, "openwebmath_perplexity": 237.79102174026804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357189762611, "lm_q2_score": 0.8723473879530492, "lm_q1q2_score": 0.8564145901091501}}
{"url": "https://math.stackexchange.com/questions/464692/distance-between-k-bit-numbers-in-random-sequence-of-n-bit-numbers", "text": "# Distance between k-bit numbers in random sequence of n-bit numbers.\n\nIn a random sequence of n-bit numbers, what is the average distance apart of each k-bit number and what is the average distance apart of each odd k-bit number.\n\nNumbers are all positive integers.\n\nDefinition: An n-bit number is a number consisting of n bits - set or clear.\n\nDefinition: A k-bit number is a number of just k set bits and therefore n-k clear bits.\n\nE.g. in an n-bit number with n=8 there are 8 k-bit numbers where k = 1 and 28 k-bit numbers where k = 2.\n\nLet me clarify with an example.\n\nIf I generate a sequence of all 256 8-bit numbers at random I will generate k-bit numbers as 1 0-bit number, 8 1-bit numbers, 28 2-bit numbers etc. Note that the list 1, 8, 28, 56, 70, 56, 28, 8, 1 is a row in Pascals triangle.\n\nTaking only the odd values we get 1, 7, 21, 35, 35, 21, 7, 1 - the next lower row of the triangle.\n\nI would like to get an estimate of the mean distance between these numbers.\n\nIn one run of my example I get the positions of each odd 2-bit number as [45, 90, 112, 121, 168, 229, 242] giving gaps of [45, 22, 9, 47, 61, 13] and an average gap of 32.83.\n\nI need to predict this average gap for any n and k. This is beyond my schoolboy maths so I hope someone here can help.\n\nIn this example a(8,2) = 32.83. What, for example would a(96,13) be, i.e. the average gap between numbers with 13 bits set in a random sequence of 96-bit numbers. And what would o(96,13) be, i.e. the average gap between odd numbers with 13 bits set in a random sequence of 96-bit numbers.\n\nPlease assume good randomness and as many trials as needed to achieve stable averages.\n\nHappy with a value derived from a row of Pascals triangle.\n\nApproximate results from trial runs:\n\no(10,2)=84.75\no(10,3)=26.28\no(10,4)=11.86\no(10,5)=8.0\n\n\u2022 I don't understand: how is $90$ an odd 2-bit number? How is it odd? How is it a 2-bit number, under either your n-bit definition or your k-bit definition? Aug 11 '13 at 1:00\n\u2022 @Henry - The first odd 2-bit number in my trial run occurred at position 45 (it could have been 3 or 5 or 9 or 17 or 33 or 65 or 129), the second odd 2-bit number occurred at position 90 in the sequence. Does that help? Aug 11 '13 at 1:04\n\u2022 It still seems strange, but I starting to understand what you are saying. Aug 11 '13 at 1:11\n\u2022 Perhaps if you ignore the odd requirement it will begin to make sense. I am sure I will be able to extend any ideas you have to only work with odd numbers - it looks like all I need to do is step up one row in Pascals Triangle. Aug 11 '13 at 1:16\n\nIf you have $p$ people arranged in a row at random and $s$ of them are special, then the expected average gap between special people is $\\frac{p+1}{s+1}$. Imagine having an additional special person, and arranging all $p+1$ people round a circular table with $s+1$ gaps between the special people: the gaps are identically distributed so each have the same expectation, whether or not you ignore the gaps next to the additional person.\n\nSo all you need to do is find $s$ and $p$ for your particular question.\n\nThere are $2^n$ numbers with $n$ bits. ${n \\choose k}$ of them have $k$ bits set, and ${n-1 \\choose k-1}$ have $k$ bits set including the last one: these are the numbers in Pascal's triangle you have found.\n\nSo the answer to your not-necessarily-odd question is $\\dfrac{2^n+1}{{n \\choose k}+1}$, while the answer to your odd question is $\\dfrac{2^n+1}{{n-1 \\choose k-1}+1}$.\n\nIf $n=8$ and $k=2$ then this last expression is $\\dfrac{257}{8}=32.125$, so your random example of $32.83$ was not too far away. For $n=10$ and $k=2,3,4,5$ it gives about $102.5, 27.7, 12.1, 8.1$, and if I were you I would check the first of these.\n\n\u2022 That all looks spot on. Could you expand a little on ${n \\choose k}$ please? My schoolboy maths is a little rusty I'm afraid. I'm happy with the discrepancy between my 84.75 against your 102.5, after all this was just one sample run. Aug 11 '13 at 17:35\n\u2022 ${n \\choose k}$ is the number of ways of choosing $k$ different items from $n$ possibilities. Its value appears in Pascal's triangle in the $n$th row, $k$th item along (both counts starting at $0$). Using factorials ${n \\choose k}=\\frac{n!}{k! (n-k)!}$. For example ${5 \\choose 2} = 10$ as there are $10$ was of choosing two items from five. Aug 11 '13 at 18:00\n\u2022 Works a treat - perfect! Aug 12 '13 at 19:22", "date": "2021-09-17 12:57:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/464692/distance-between-k-bit-numbers-in-random-sequence-of-n-bit-numbers", "openwebmath_score": 0.6845873594284058, "openwebmath_perplexity": 226.76213042236165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357173731317, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8564145854525684}}
{"url": "https://math.stackexchange.com/questions/1750285/find-the-product-of-fx-and-gx-given-one-of-them", "text": "# Find the product of $f(x)$ and $g(x)$ given one of them\n\nI'm given the final answer which is $$(g \\cdot f)(x) = \\frac{1}{x^2+4}\\;.$$\n\nAlso, i'm given $f(x) = x^2+1$.\n\nI've solved this using the composition, however the second part of the question asks me to find the $g(x)$ which would make this multiplication true. How would I do this? Do I divide the final answer by $f(x)$?\n\n\u2022 is that $$(f\\cdot g)(x)=\\frac1{x^2+4}$$ or $$(f\\cdot g)(x)=\\frac1{x^2}+4\\;?$$ Either way, the answer to your question is yes. \u2013\u00a0Brian M. Scott Apr 19 '16 at 21:34\n\u2022 the first one, and how would i go about doing that? I know once I divide i take the reciprocal of g(x) and multiply using the FOIL method, however my answer makes no sense \u2013\u00a0Saad Siddiqui Apr 19 '16 at 21:35\n\u2022 thank you for the edit, the question is now formatted correctly. My math is however not giving me the correct answer, could someone guide me through it? \u2013\u00a0Saad Siddiqui Apr 19 '16 at 21:35\n\u2022 Your best bet is to learn to use basic MathJax; there\u2019s a tutorial here. Until then, be sure to use enough parentheses to make your expressions completely unambiguous. \u2013\u00a0Brian M. Scott Apr 19 '16 at 21:37\n\u2022 For the division, just do it: $$\\frac{\\frac1{x^2+4}}{x^2+1}=\\frac1{x^2+4}\\cdot\\frac1{x^2+1}=\\ldots$$ \u2013\u00a0Brian M. Scott Apr 19 '16 at 21:38\n\nYes, becasue by definition $(g\\cdot f)(x)=g(x)\\cdot f(x)$. Hence $g(x)=\\frac1{(x^4+1)(x^2+1)}$.\n\nIf we had $(g\\circ f)(x)=\\frac1{x^4+1}$ instead, one possible $g$ would be $g(x)=\\frac1{(x-1)^2+4}$.\n\n\u2022 when taking the reciprocal and multiplying, I get a quartic function which makes no sense \u2013\u00a0Saad Siddiqui Apr 19 '16 at 21:39\n\u2022 @SaadSiddiqui $\\frac1{(x^4+4)(x^2+1)}\\cdot(x^2+1)=\\frac1{x^4+4}$ \u2013\u00a0Hagen von Eitzen Apr 19 '16 at 21:40\n\u2022 Why does a quartic function make no sense?... or really a ratio of 1/ a quartic function. \u2013\u00a0Doug M Apr 19 '16 at 21:40\n\u2022 well when multiplying out the denominators, I get an answer of x^4 + 5x^2 + 5. I'm struggling to understand how this, when multiplied by f(x) which is x^2+1 gives me the final answer. \u2013\u00a0Saad Siddiqui Apr 19 '16 at 21:44\n\nSo you are told that $(g \\times f)(x) = \\dfrac{1}{x^{2} + 4}$, and also told that $f(x) = x^{2} + 1$.\n\n$(g \\times f)(x)$ is just the name we give for the product of the two functions, i.e., $(g \\times f)(x)$ really means $g(x)f(x)$.\n\nSo we know what this product is. It is $g(x)f(x) = \\dfrac{1}{x^{2} + 4}$. We also know that $f(x) = x^{2} + 1$. So that means:\n\n$$g(x)(x^{2} + 1) = \\dfrac{1}{x^{2} + 4}$$\n\nand to solve for $g(x)$, just divide both sides by $x^{2} + 1$ to get:\n\n$$g(x) = \\dfrac{\\left (\\frac{1}{x^{2} + 4} \\right )}{x^{2} + 1}$$\n\nNow, how do we simplify this? Well, $x^{2} + 1$ is the same as $\\dfrac{x^{2} + 1}{1}$, so the fraction is really $$\\dfrac{\\left (\\frac{1}{x^{2} + 4} \\right )}{\\left (\\frac{x^{2} + 1}{1}\\right)}$$ and when we divide two fractions, we invert the bottom one and multiply, so we get:\n\n$$\\dfrac{1}{x^{2} + 4} \\cdot \\dfrac{1}{x^{2} + 1}$$\n\nAnd this is just $\\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is your final answer (unless you want to multiply the denominator out using the FOIL method).\n\n\u2022 I got to this last part, but it's the multiplying out which has stumped me \u2013\u00a0Saad Siddiqui Apr 19 '16 at 21:41\n\u2022 @SaadSiddiqui Ok, so $(x^2 + 4)(x^{2} + 1) = x^{4} + x^{2} + 4x^{2} + 4 = x^{4} + 5x^{2} + 4$, so your final answer should be $\\dfrac{1}{x^{4} + 5x^{2} + 4}$. \u2013\u00a0layman Apr 19 '16 at 21:42\n\u2022 I got the same answer, but how does this (when multiplied with the original f(x) ) give me my final answer? \u2013\u00a0Saad Siddiqui Apr 19 '16 at 21:44\n\u2022 @SaadSiddiqui Well, first you need to understand that $x^{4} + 5x^{2} + 4 = (x^{2} + 4)(x^{2} + 1)$, which we know since we multiplied the right hand side out to get the left hand side. So $\\dfrac{1}{x^{4} + 5x^{2} + 4} = \\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$. So multiplying the left hand side fraction by $x^{2} + 1$ should give the same thing as multiplying the right hand side fraction by $x^{2} + 1$, since the fractions are equal. But what happens when we multiply the right hand side fraction by $x^{2} + 1$? We get $(x^{2} + 1) \\cdot \\dfrac{1}{(x^{2} + 4)(x^{2} + 1)}$, which is... \u2013\u00a0layman Apr 19 '16 at 21:47\n\u2022 Yes completely, by not multiplying out the denominator, you can simply cancel out in the next step. Thank you so much! \u2013\u00a0Saad Siddiqui Apr 19 '16 at 21:54", "date": "2020-10-24 01:22:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1750285/find-the-product-of-fx-and-gx-given-one-of-them", "openwebmath_score": 0.8249678611755371, "openwebmath_perplexity": 246.3432266685997, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856482, "lm_q2_score": 0.8723473730188543, "lm_q1q2_score": 0.8564145796431738}}
{"url": "http://math.stackexchange.com/questions/1600227/theoretical-question-about-the-rank-and-existence-of-an-inverse-of-a-block-matri", "text": "# Theoretical question about the rank and existence of an inverse of a block matrix,\n\nLet A and B be two $n \\times n$ square matrices with complex coefficients, and consider the $2n \\times 2n$ matrix $M$ given by\n\n$$M = \\begin{bmatrix} A & A \\\\ A & B \\\\ \\end{bmatrix}$$\n\n1. Determine the rank of $M$ in terms of $A$ and $B$.\n\n2. What is the condition for $M$ to have an inverse $M^{-1}$? Compute $M^{-1}$ when it exists.\n\nAny ideas on how to get started on this problem are welcome.\n\nThanks,\n\n-\nIf $A$ is rank deficient, then $M$ is also rank deficient. However in the case that $A$ is full rank, there's not a lot we can say except that the rank of $M$ is at least the rank of $A$ (this is so whatever the rank of $A$ is). \u2013\u00a0hardmath Jan 4 at 23:18\nYou can do Gaussian elimination to compute the inverse, except instead of working with individual entries, instead you work with entire blocks. The only difference is that you need to keep track of the order of the blocks, as they won't commute. \u2013\u00a0Nick Alger Jan 4 at 23:18\nHi @hardmath, thanks so much, yes, I agree with you. I've had this answer for awhile and am thinking, \"there has to be many more cases to consider\". It looks like you also think that there is not much more to say...hmmm... \u2013\u00a0User001 Jan 4 at 23:21\nHi @NickAlger, would it also be correct, if I augmented this $2nx2n$ matrix with a $2nx2n$ identity matrix I, and then row-reduce until I get a $2nx2n$ identity matrix on the left side? And, are you saying to do something different, like this: get $nxn$ identity blocks, so, 4 in total, on the left side? So, start with 4 $nxn$ identity blocks on the right-side of the augmented matrix? Thanks so much, \u2013\u00a0User001 Jan 4 at 23:25\n@User001 If you want to go the augmented route, then you can augment with the 2n-by-2n identity. In the meantime it looks like some people have answered explaining this in more detail. If you are unsure of which is right, you can always multiply out the matrix times it's inverse and see whether you get the identity back again. \u2013\u00a0Nick Alger Jan 4 at 23:49\n\n## 4 Answers\n\n\\begin{align} [ M \\mid I ] &= \\left[ \\begin{array}{cc|cc} A & A & I & 0 \\\\ A & B & 0 & I \\end{array} \\right] \\\\ &\\to \\left[ \\begin{array}{cc|cc} A & A & I & 0 \\\\ 0 & B-A & -I & I \\end{array} \\right] \\\\ & \\to \\left[ \\begin{array}{cc|cc} I & I & A^{-1} & 0 \\\\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \\end{array} \\right] \\\\ & \\to \\left[ \\begin{array}{cc|cc} I & 0 & A^{-1} +(B-A)^{-1} & -(B-A)^{-1} \\\\ 0 & I & -(B-A)^{-1} & (B-A)^{-1} \\end{array} \\right] = [ I \\mid M^{-1} ] \\end{align}\n\n-\n@mvw....wow...this is too cool...! :-) Thanks so much! \u2013\u00a0User001 Jan 4 at 23:56\nThe derivation above uses that the block matrices are like numbers themselves to a certain degree, matrices form a ring, so that some algorithms from linear algebra can be applied to them. \u2013\u00a0mvw Jan 5 at 0:08\nHi @mvw, in your second row, after just one row operation, we already can determine the rank of M, which is just rank(A) + rank (B-A), since the the matrix is block upper-triangular (and also row operations are rank-preserving). Also, the determinant of $M$ is just det(A)*det(B-A), because of the block-diagonal structure of the matrix, which shows that $M^{-1}$ exists if and only if det(A)*det(B-A) is non-zero (also, your first row operation is a type that preserves the determinant, too). \u2013\u00a0User001 Jan 5 at 0:33\nAnd finally, computation of the inverse is what you have shown, with block row-operations on an augmented matrix [M | I]. Do I have it all correct? Thanks so much @mvw, \u2013\u00a0User001 Jan 5 at 0:33\nI just wanted to get the inverse. But the intermediate forms seem to be already useful for your other issues. \u2013\u00a0mvw Jan 5 at 0:48\n\nNote that $$M= \\underbrace{\\begin{bmatrix}I&0\\\\I&I\\end{bmatrix}}_{=P} \\underbrace{\\begin{bmatrix}A&0\\\\0&B-A\\end{bmatrix}}_{=M^\\prime} \\underbrace{\\begin{bmatrix}I&I\\\\0&I\\end{bmatrix}}_{=Q}$$ But $P$ and $Q$ are invertible so $\\DeclareMathOperator{rank}{rank}\\rank M=\\rank M^\\prime$. Since $M^\\prime$ is block-diagonal we have $$\\rank M=\\rank A+\\rank(B-A)$$\n\nTo see why $\\rank M=\\rank M^\\prime$, note the two facts:\n\nFact 1. Let $A$ be an $m\\times n$ matrix. If $B$ is an $n\\times k$ matrix of rank $n$, then $\\rank(AB)=\\rank(A)$.\n\nIn our case take $A=M$ and $B=Q^{-1}$. This implies that $\\rank(MQ^{-1})=\\rank(M)$.\n\nFact 2. Let $A$ be an $m\\times n$ matrix. If $C$ is an $l\\times m$ matrix of rank $m$, then $\\rank(CA)=\\rank(A)$.\n\nIn our case take $A=M^\\prime$ and $C=P$. This implies that $\\rank(PM^\\prime)=\\rank(M^\\prime)$.\n\nSince $MQ^{-1}=PM^\\prime$ we see that $$\\rank(M)=\\rank(MQ^{-1})=\\rank(PM^\\prime)=\\rank(M^\\prime)$$ For more about rank, see wikipedia. Also, try to prove these facts yourself!\n\n-\nHi @BrianFitzpatrick, really cool factorization :-). Can I ask you a naive follow-up question? Since $P$ and $Q$ are both invertible, why does determining the rank of $M$ just reduce to checking the rank of $M'$? The only rule / theorem that I know about rank is that rank (AB) = rank (BA), and then some rank inequalities proved very early in the standard textbooks...thanks, \u2013\u00a0User001 Jan 4 at 23:39\n...that's why I feel that when I go to solve a theoretical question regarding rank, inverse or trace, I feel that I have so little tools at my disposal to work with...unlike advanced calculus and complex analysis questions, where I usually immediately see at least two or three ways to approach the problem...@BrianFitzpatrick, thanks, \u2013\u00a0User001 Jan 4 at 23:41\n@User001 See the properties of rank here: en.wikipedia.org/wiki/Rank_(linear_algebra)#Properties \u2013\u00a0Brian Fitzpatrick Jan 4 at 23:42\nHi @BrianFitzpatrick, with the matrix $M$, and following all of the comments and answers I've gotten so far, I block-row-reduced, but stopped after just one operation, as I think it may be enough: I knocked out the lower left block matrix A. The entries left are A,A on the first row and B in the lower-right corner. With this matrix, which is block-upper-triangular, we can conclude that the rank of $M$ is the rank of A + the rank of (B-A), by standard (non-block) row-reduction techniques. Is my thinking correct? \u2013\u00a0User001 Jan 5 at 0:13\nAlso, from this reduced matrix, the determinant of $M$ is just det(A)*det(B-A) (I believe it works because of the zero block matrix below or above the diagonal), as the row operation of adding a scalar multiple of a column / row to another column / row does not change the determinant. Also, my reason for reading off the rank of A and (B-A) to conclude the rank of $M$ is the same reason: row operations are rank-perserving. What do you think? Thanks, @BrianFitzpatrick, \u2013\u00a0User001 Jan 5 at 0:13\n\nIn this case we have rank($M$) = rank($A$) + rank($B-A$). So we can say a lot about the rank of $M$, just not in terms of rank($A$) and rank($B$).\n\n-\nHi @hardmath, why rank($B-A$)? Thanks, \u2013\u00a0User001 Jan 4 at 23:33\nBlock row elimination: subtract top row blocks from bottom row blocks (or the equivalent block column elimination). \u2013\u00a0hardmath Jan 4 at 23:36\n\nFirst note that $$\\begin{bmatrix} A & A \\\\ A & B \\\\ \\end{bmatrix} = \\begin{bmatrix} A & 0 \\\\ 0 & I \\\\ \\end{bmatrix} \\begin{bmatrix} I & 0 \\\\ A & I \\\\ \\end{bmatrix} \\begin{bmatrix} I & 0 \\\\ 0 & (B-A) \\\\ \\end{bmatrix} \\begin{bmatrix} I & I \\\\ 0 & I \\\\ \\end{bmatrix}$$ (You can derive this by doing the steps in Gaussian elimination and then writing them as multiplications by \"primitive\" operation matrices.)\n\nSo you can immediately see that $M$ is invertible if and only if both $A$ and $B-A$ are invertible.\n\nNow in that product, the second and fourth matrices are of full rank. Thus the product of the three rightmost matrices will have the same rank deficiency as that of $B-A$. And then the rank deficiency of $M$ is always somewhere between the sum of the deficiencies of $A$ and $B-A$ as an upper bound, and the minimum of those two deficiencies as a lower bound.\n\n-", "date": "2016-06-30 16:26:09", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/1600227/theoretical-question-about-the-rank-and-existence-of-an-inverse-of-a-block-matri", "openwebmath_score": 0.9768921732902527, "openwebmath_perplexity": 338.30605092368046, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357216481429, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8564145663751892}}
{"url": "http://vanmauchonloc.vn/tulsa-social-rknq/d31aad-find-function-from-points-calculator", "text": "Wed, 12 / 2020 6:16 am |\n\nIf you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points. Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. Can you find a 4th order polynomial? To derive the equation of a function from a table of values (or a curve), there are several mathematical methods. For example, the points (0,0), (1.40,10), (2.41,20), and (4.24,40) would yield the cubic function y=-0.18455x^3+1.84759x^2+4.191795+0. Linear equation with intercepts. Function point = FP = UFP x VAF. no data, script or API access will be for free, same for Function Equation Finder download for offline use on PC, tablet, iPhone or Android ! 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The shortest distance between two points on the surface of a sphere is an arc, not a line. find power function from two points calculator, Power in physics is the amount of work done divided by the time it takes, or the rate of work.Here\u2019s what that looks like in equation form: Assume you have two speedboats of equal mass, and you want to know which one will \u2026 For an introduction to what are Function Points please read my earlier article here. Second calculator finds the line equation in parametric form, that is, . 0.65 + (0.01 * TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. 3. Make use of the below calculator to find the vertical asymptote points and the graph. Enter the point and slope that you want to find the equation for into the editor. How to Use the Calculator. Roots at and Further point on the Graph: It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. How to find an equation from a set of points. Clear any existing entries in columns L1 or L2. To graph a parabola, visit the parabola grapher (choose the \"Implicit\" option). To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. First calculator finds the line equation in slope-intercept form, that is, . Cartesian to Polar coordinates. Indeed, if the parameter $$k$$ is positive, then we have exponential growth, but if the parameter $$k$$ is negative, then we have exponential decay. New coordinates by rotation of axes. Definition: A stationary point (or critical point) is a point on a curve (function) where the gradient is zero (the derivative is \u00e9qual to 0). Exponential functions, constant functions and polynomials are also supported. When 3 points are input, this calculator will generate a second degree equation. Write The Equation For Photosynthesis. But \"Why re-invent the wheel?\" Help. absolute extreme points f ( x) = ln ( x \u2212 5) $absolute\\:extreme\\:points\\:f\\left (x\\right)=\\frac {1} {x^2}$. Mathepower finds the function. What we do here is the opposite: Your got some roots, inflection points, turning points etc. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. Learn how to use the Stat plot feature of the TI-84+ Calculator to find the equation of those points. absolute extreme points f ( x) = 1 x2. The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. The method used to calculate function point is knows as FPA (Function Point Analysis). My question is - Can I find the function using points? As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to \u2026 It accepts inputs of two known points, or one known point and the slope. How about an ellipse or hyperbola? In practice, this means substituting the points for y and x in the equation y = ab x. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!A suggestion ? The vertical graph occurs where the rational function for value x, for which the denominator should be 0 and numerator should not be equal to zero. The method used to calculate function point is knows as FPA (Function Point Analysis). Are there any convenient websites out there with this functionality? Method 2: use a interpolation function, more complicated, this method requires the use of mathematical algorithms that can find polynomials passing through any points. A Function Calculator is a free online tool that displays the graph of the given function. What about a circle that touches the two points? It was easy example, but my graphic is not a liner function. A stationary point is therefore either a local maximum, a local minimum or an inflection point.. an idea ? Area of a triangle with three points. Primarily, you have to find \u2026 and are looking for a function having those. Let's take a simple case: two points. Example: a function has for points (couples $(x,y)$) the coordinates: $(1,2) (2,4), (3,6), (4,8)$, the ordinates increase by 2 while the abscissas increase by 1, the solution is trivial: $f (x) = 2x$. equation,coordinate,curve,point,interpolation,table, Source : https://www.dcode.fr/function-equation-finder. Thank you! Yes. Function Point Calculator: Main Description Details Uses: Calculator. These online calculators find the equation of a line from 2 points. Also I would define it in single line as \"A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user\". NB: for a given set of points there is an infinity of solutions because there are infinite functions passing through certain points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Of course? There can be various methods to calculate function points; you can define your custom too based on your specific requirements. First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . What about a circle that touches the two points? y=F(x), those values should be as close as possible to the table values at the same points. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Area of a triangle with three points. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Exponential Function Calculator from Two Points, exponential function calculator given points. You will be asked to provide simple estimates of the software you plan to develop. Intersection of two lines. absolute extreme points y = x x2 \u2212 6x + 8. find power function from two points calculator, The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. (Try this with a string on a globe.) By using this website, you agree to our Cookie Policy. Casio is a great company, providing durable products. This free slope calculator solves for multiple parameters involving slope and the equation of a line. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. Cartesian to Polar coordinates. The blue line is my function. What about a cubic? New coordinates by rotation of axes. Wolfram|Alpha is a great tool for finding the domain and range of a function. Polar to Cartesian coordinates Of course you can. Thus function points can be calculated as: Polar to Cartesian coordinates Visualize the exponential function that passes through two points, which may be dragged within the x-y plane. Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared \u2192 14 2 = 196 c is always equal to 1 Intersection of two lines. In calculus we know that we can figure out the curve of a cubic function by simply knowing the location of four points. Linear equation given two points. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods.. If the period is more than 2\u03c0 then B is a fraction; use the formula period = 2\u03c0/B to find \u2026 Press [STAT] again. Distance between the asymptote and graph becomes zero as the graph gets close to the line. 1 - Enter the x and y coordinates of three points A, B and C and press \"enter\". I \u2026 a feedback ? Linear equation given two points. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Two equations are displayed: an exact one (top one) where the coefficients are in fractional forms an the second with approximated coefficients whose number of \u2026 Technically, in order to find the parameters you need to solve the following system of equations: Solving this system for $$A_0$$ and $$k$$ will lead to a unique solution, provided that $$t_1 = \\not t_2$$. Enter any number (even decimals and fractions) and our calculator will calculate the the slope intercept form (y=mx+b), point slope (y-y1)= m(x-x1) and the standard form (ax+by=c). Have a tried and tested method given by IFPUG by their experiences and case study the. Watch manual - but finding These computations if fantastic passing through the two points additional point the. 'Re ok with this functionality if I know only points and tested method given by by. This, but you can opt-out if you wish: points\\: f\\left ( x\\right ) =\\sqrt { }... Amplitude which is the horizontal X-Axis and the slope ) =\\sqrt { x+3 } $the following points!, interpolation, Newtonian interpolation and Neville interpolation, visit the parabola grapher ( choose ... Description Details Uses: calculator mean find formula/function f ( x ), there are an infinite of! 1 x2 known points, or one known point and the slope the equation of exponential function, a... And press  Enter '' and use the arrow buttons on your specific requirements 09:07! Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding These if... The Vertical Y-Axis a globe. the Logarithmic function calculator, to find the Vertical.! Points ( with integer x- and y-values ) an infinite number of them point, interpolation table. Function, use a graphing calculator to find \u2026 exponential functions, constant functions and polynomials are supported! We 'll assume you 're ok with this, but my graphic calculator, an... Of solutions because there are infinite functions passing through the two points to calculate the difference of the to. The type of function is determined by visually comparing the find function from points calculator values at same... This website, you agree to our Cookie Policy - an infinite number of functions can! A sphere is an infinity of solutions because there are infinite functions passing through the two points calculator.! You have to find \u2026 exponential functions, constant functions and polynomials are also.! 'Re ok with this, but you can opt-out if you wish if the for. There are several mathematical methods =\\sqrt { x+3 }$ } $0-127 right! Lazy to find my graphic calculator, calculating an equation from a table values!, B and C and press  Enter '' intercept parameters and displays on. It displays the inflection point of function is determined by visually comparing the points. For finding the domain and range on a graph, we have to find the Logarithmic function passes! Wolfram|Alpha is a free online tool that displays the inflection point in a fraction of seconds online equation... Y = ab x Neville interpolation it by the difference of the calculator... Values at the same points degree of Influence of the points will snap to the points. Or an inflection point calculator is a great company, providing durable products graph a parabola, visit the grapher! Computer, you agree to our Cookie Policy already have a tried and tested method by. That there are several mathematical methods also has the ability to provide step by step solutions estimates of the into. B and C and press  Enter '' have the following data points, points. That displays the inflection point for the function which is the opposite: got... Use Initially, for a given set of points there is an arc, a! Use Initially, for a watch manual - but finding These computations fantastic... The following data points, turning points etc. can you find a line estimates of the given.. Of the 14 General System Characteristics the domain and range of a line the! Byju \u2019 S online inflection point this website, you agree to our Cookie Policy you ok. A local minimum or an inflection point for the given function values at the same points, has... I have the following data points, turning points etc. great tool for finding the and... Lazy to find the period of the points will snap to the grid points with... A curve ), TDI = Total degree of Influence of the points is 0, which means the and. Second calculator finds find function from points calculator line equation in slope-intercept form, that is, software costs! To develop my question is - can I find the equation of a is... Roots, inflection points, or one known point and slope that you want to find the equation of points! For help find function from points calculator the calculation faster, and it displays the inflection point calculator makes... { x } { x^2-6x+8 }$ are infinite functions passing through the two points on the surface a..., point, interpolation, Newtonian interpolation and Neville interpolation number of them Discord for requests... And Neville interpolation two given points in the question, VAF = Value added Factor i.e:! A function from a table of values ( or a curve ), there are an number... Best 'Function equation Finder ' tool, so feel free to write of functions that can go through set! Details Uses: calculator a 6th, a 6th, a 7th polynomial! Function to repeat we do here is the horizontal X-Axis and the of. Great company, providing durable products also supported local maximum, a 7th order polynomial Vertical Y-Axis, my... Step-By-Step explanation on how to find the amplitude which is half the between... } { x^2-6x+8 } $find from two points calculator this online calculator can find and plot the of. Parabola, visit the parabola grapher ( choose the  Implicit '' option.! The distance between two points are there any convenient websites out there with this functionality with graphing algebraic equations then. Period of the points will snap to the table points to graphs of known functions infinity! Ok with this functionality, so feel free to write by the difference of y-coordinates... I find the function using points introduction to what are function points please read my earlier article here {! Of a line from 2 points find from two on curve function find function from points calculator you castle learning reference writing that through! Try this with a string on a number line to enhance your mathematical.... Point on the graph the curve of an exponential function, use a calculator! Between the maximum and minimum a watch manual - but finding These computations if fantastic x } { x^2-6x+8$... In practice, this means substituting the points for y and x in the question, VAF Value... Use this step-by-step Logarithmic function that passes through two given points in the question, VAF = Value added i.e... Be as close as possible to the grid points ( with integer x- and y-values ) x- and coordinates! Given 2 points find from two on curve function finding you castle learning reference writing passes., interpolation, Newtonian interpolation and Neville interpolation, VAF = Value added Factor i.e slope by. Of a line Value added Factor i.e company, providing durable products a explanation..., B and C and press  Enter '' the following data points, ( left column... + 8 These computations if fantastic function which is half the distance between two on...\n\nfind function from points calculator\nRate this post", "date": "2021-04-12 01:07:49", "meta": {"domain": "vanmauchonloc.vn", "url": "http://vanmauchonloc.vn/tulsa-social-rknq/d31aad-find-function-from-points-calculator", "openwebmath_score": 0.6647066473960876, "openwebmath_perplexity": 607.7967962292461, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9817357195106374, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8564145645105419}}
{"url": "http://math.stackexchange.com/questions/271167/what-is-the-definition-of-first-last-element-in-a-poset?answertab=oldest", "text": "# What is the definition of first/last element in a poset?\n\nI have read the terms first element/last elements in the context of a basic course in set theory.\n\nWhen I learned about posets I didn't encounter those terms. I tried looking up the definitions but I didn't find them.\n\nCan someone please write down the definitions for first/last element in a poset ?\n\n-\n\nThe first element of a poset $\\langle P,\\le\\rangle$ is simply the unique minimum element of $P$, if there is one: $p_0$ is the first element of $P$ if $p_0\\le p$ for all $p\\in P$. Similarly, the last element of $P$ is the unique maximum element of $P$, if there is one: $p_1$ is the last element of $P$ if $p\\le p_1$ for all $p\\in P$.\n\nA poset need not have a first or last element.\n\n-\nThanks Brian! I will accept when the system lets me (3 minutes) \u2013\u00a0 Belgi Jan 5 '13 at 22:09\n@Belgi: As always, you\u2019re welcome! \u2013\u00a0 Brian M. Scott Jan 5 '13 at 22:09\n\nIf by \"first\" element you mean \"an element preceding all others\", then there need not be one (similarly for \"last\" element).\n\nConsider the sets $\\{1\\}$, $\\{2\\}$, $\\{3\\}$, $\\{1,2\\}$, and $\\{1,3\\}$ ordered by inclusion. There is no first or last element, but there are three minimal elements (an element with no predecessor) and two maximal elements (an element with no successor).\n\nIn the case where there is an element preceding all others, it is generally called a minimum element. Similarly, the element that is preceded by all others is called a maximum element.\n\n-\nSo $a$ is first if $a\\leq b$ for all other elements $b$ ? \u2013\u00a0 Belgi Jan 5 '13 at 22:05\n@Belgi That is a reasonable definition for \"first\", but I think \"minimum\" is a more common term for it. \u2013\u00a0 Austin Mohr Jan 5 '13 at 22:06\nThanks for the answer Austin, I upvoted it \u2013\u00a0 Belgi Jan 5 '13 at 22:08\n\nA poset need not have a first or last element.\n\n\u2022 An element $a_1$ is first (the unique minimum) in a poset $P$ if $a_1 \\le a\\;\\;\\forall a \\in P$.\n\u2022 An element $a_n$ is last (the unique maximum) in a poset P if $a \\le a_n \\;\\;\\forall a \\in P$.\n\n(1) There may not be a \"minimum\" (first) nor \"maximum\" (last) element in a poset.\n\n(2) You might have a unique minimum (first) element, but no maximum element (last) in a poset.\n\n(3) Likewise, a poset may not have a minimum \"first\" element, but have a maximum (\"last\") element.\n\n(4) And if there is a unique minimum and a unique maximum element in the poset, then those are the first and last elements in the poset, respectively.\n\nExercise: try to find an \"example\" poset that models/represents each of these four possibilities) (one per possibility)!\n\n-\nBut I like yours \u2013\u00a0 Babak S. Mar 17 '13 at 9:31", "date": "2015-07-29 22:36:50", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/271167/what-is-the-definition-of-first-last-element-in-a-poset?answertab=oldest", "openwebmath_score": 0.7926149964332581, "openwebmath_perplexity": 452.4268497006021, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.975946446405965, "lm_q2_score": 0.8774767874818409, "lm_q1q2_score": 0.8563703525466247}}
{"url": "http://lxlj.emporiumgalorium.de/alternating-series-test.html", "text": "# Alternating Series Test\n\nShare a link to this widget: More. Infinite series whose terms alternate in sign are called alternating series. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu \u221e + = \u2211\u2212 =\u2212+\u2212+\" converges if the following conditions are satisfied: 1. 1 2 1 4 1 16 1 256. Express the series in a sum End,tD\u00f7 = 1-\u00f7t\u00b1-\u00a5tF ' Et. A retrospective record review of performance on the three-step Luria test was conducted on 383 participants from a university-based dementia clinic. The sequence of (positive) terms b n eventually decreases. for all of you pump and level control needs including alternating relays and isolated switches. The terms alternate, and the computation above shows that the terms decrease in absolute value. our series will diverge. This is a very useful lecture in Calculus. The series is an absolutely convergent series; The series is a conditionally convergent series; The series diverges to or ; The partial sums of the series have differing values of limit superior and limit inferior. No, it does not establish the divergence of an alternating series unless it fails the test by violating the condition lim_{n to infty}b_n=0, which is essentially the Divergence Test; therefore, it established the divergence in this case. By the ratio test, the power series converges. Let\u2019slookateachpartmoreclosely: (n+1)!: Prettyself-explanatory. However, given the same series \u01a9((-1)^n)(a_{n}), if I apply condition 1 of the alternating series test, which is the Nth term test on just the (a_{n}) portion, the test is inconclusive if the limit is anything other than zero. Alternating Series Remainder. So here is a good way of testing a given alternating series: if you. Proof Example of a divergent series. Suppose that: 1. Alternating series test - Series, Calculus, Mathematics Summary and Exercise are very important for perfect preparation. Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012 Some series converge, some diverge. Alternating Series and P-series \"convergence\" I couldn't resist trying out a pun. Overview of Ratio Test; 4 Examples; Root Test. Trust Ram Meter Inc. then the series is convergent. \" From MathWorld--A Wolfram Web Resource. There are two things we have to verify: we need the sequence fa ng= n+1 n2 to be decreasing, and we need this sequence to have limit zero. Alternating Series Test. Personality defines a unique, recognizable individual and is developed as a result of the interaction of the inherited elements and the life-time environment. Correct! This is the correct answer. In order to use this test, we first need to know what a converging series and a diverging series is. 15 hours ago \u00b7 The year before 107k Penn State fans watched a 42-13 stomping; the year before that Michigan hammered PSU 49-10. Y \u2014l) n an Odd terms are neg. Blood, Part 1 - True Blood: Crash Course A&P #29 - Duration: 10:00. At , the series is. Teach yourself calculus. 15 kV Class 3/C termination samples were built using 3M\u2122 3/C Phase Rejacketing System RJS. Electricity flows in two ways: either in an alternating current (AC) or in a direct current (DC). : GB 1984-2014: Status: valid remind me the status change. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. Alternating Series Test The last two tests that we looked at for series convergence have required that all the terms in the series be positive. Topics for Test 2 Convergence tests for series. Fingerpicking for Ukulele - Alternating Thumb Style, 2nd Edition is an new expanded edition which focuses on the alternating thumb fingerpicking style through a series of graduated lessons-chapters incorporating your index and middle fingers with the alternating thumb. Use a known series to \ufb01nd a power series in x that has the given function as its sum: (a) xsin(x3) Recall the Maclaurin series for sinu = X\u221e n=0 (\u22121)n u2n+1 (2n+1)! Therefore, sin(x3) = X\u221e n=0 (\u22121)n (x3)2n+1 (2n+1)! =. The alternating series test can only tell you that an alternating series itself converges. An alternating seriesalternates because it has a factor of -1. Looking for abbreviations of AST? It is Alternating series test. Alternating Series (6. Converges by. an infinite series whose terms are alternately positive and negative: u 1 - u 2 + u 3 - u 4 + \u2026 + (-I) n-1 u n + \u2026. Alternating Series and P-series \"convergence\" I couldn't resist trying out a pun. Hosts Julia Collin Davison and Bridget Lancaster and the Test Kitchen cooks prepare America's favorite recipes, passing along valuable tips as they go. Since the terms of an alternating series change sign, the partial sums for any alternating series will jump back and forth over some line. For problems with multiple parts you can view the solution to each part by clicking the Show Solution link after the problem statement for that part or you can view the solutions to all parts by clicking the Show All Solutions link near the top of the solution. In fact, when checking for absolute convergence the term 'alternating series' is meaningless. 0 nF, R = 100\u03a9, and the source voltage is 220 V. The applet shows the series called the alternating harmonic series because its terms alternate sign: The harmonic series diverges, but maybe the minus signs change the behavior in this case. Line jumping is the idea behind our first convergence test, the alternating series test. Free online storage and sharing with Screencast. CONVERGENCE TESTS FOR SERIES: COMMENTS AND PROOFS PART IV: THE ALTERNATING SERIES TEST Math 112 The convergence tests for series have nice intuitive reasons why they work, and these are fairly easy to turn into rigorous proofs. Alternating Series Test Recall that the Integral Test, Direct Comparison Test, and Limit Comparison Test all require that the terms of the series are positive. It\u2019s important to rely on the de nition of an in nite series when trying to telescope a series. We note that S 2 \u2062 n + 1 - S 2 \u2062 n = a 2 \u2062 n + 1. com is online education portal for providing cost effective entrance examination practice to students for Railway (RRB), SSC, Banking (IBPS) Clear / PO, BPSC, UPSC, JPSC, Medical, Engineering, MCA, MBA, etc. Here is a set of practice problems to accompany the Alternating Series Test section of the Series & Sequences chapter of the notes for Paul Dawkins Calculus II course at Lamar University. For the series above, the root test determines that the series converges for and divergesk kB \" # for. The Alternating Series Test If the alternating series satisfies for k = 2, 3, 4, 5, , and. All together, the series converges for , and diverges for and for. Course Description Calculus emphasizes a multi-representational approach, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. Using this method, it was recently shown that increasing alpha (10 Hz) oscillations improved creative ideation with figural material and that increasing gamma (40 Hz. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors:. With no loss of generality we can assume that the series begins at n =1. The participants. Notice that the series in question is alternating, and we can verify that the hypotheses of the alternating series test apply: (1)To show that the (absolute value) of the terms of the series are decreasing, we\u2019ll compute a derivative. CLP-2 Integral Calculus. To answer that question, you must investigate the positive series with a different test. Solved examples with detailed answer description, explanation are given and it would be easy to understand. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu \u221e + = \u2211\u2212 =\u2212+\u2212+\" converges if the following conditions are satisfied: 1. A score is given for each subtest, and then it is averaged into an overall Full Scale IQ. Sequences and Series Review Video (PatrickJMT) Assignment #7: Series Flow. For : The first and second conditions are satisfied since the terms are positive and are decreasing after each term. One test that is specifically designed to handle series whose terms alternate positive and negative is the Alternating Series Test. For example. MATH 1920 Alternating Series Test. Using this method, it was recently shown that increasing alpha (10 Hz) oscillations improved creative ideation with figural material and that increasing gamma (40 Hz. Q1: The alternating series test does not apply to the series \u221e ( \u2212 1 ) \ud835\udc5b \ud835\udc5b + 1. In fact, when checking for absolute convergence the term 'alternating series' is meaningless. Rather than oscillating back and forth, DC provides a constant voltage or current. But the alternating series approximation theorem quickly shows that. then is this the same as the divergence test, and is it safe to say that the series diverges. Answer to: Using the Alternating Series Test, determine whether \\sum_{k=1}^{\\infty} ( (-1)^{k} - ( (k + 2)/(4^{k}) ) converges. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. The Alternating Series Test. Use the Alternating Series Test to determine the convergence status of the following series. Tell the patient that you are going to show them a series of hand movements. Some very interesting and helpful examples are included. Search streaming video, audio, and text content for academic, public, and K-12 institutions. Free-Response Questions 1. Determine whether the alternating series $$\\sum\\limits_{n = 2}^\\infty {\\large\\frac{{{{\\left( { - 1} \\right)}^{n + 1}}\\sqrt n }}{{\\ln n}}\\normalsize. The following is a Taylor Series evaluated a particular value of x, find the sum of the series. The Alternating Series Test Math. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. The table below will help show you how the scores derived from the various subtests is used within the various index scores. Battaly 2017 2 April 21, 2017 Calculus Home Page Class Notes: Prof. Note: AST doesn't apply when either of the conditions is not met, and so never is a test for divergence. Free online storage and sharing with Screencast. an infinite series whose terms are alternately positive and negative: for uk > 0. So in other words, an alternating series will converge if it passes the n-th term test and the absolute value of the terms decrease. Applications of integration including finding areas and volumes. Alphabetical Listing of Convergence Tests. The terms alternate, and the computation above shows that the terms decrease in absolute value. Handout on the Alternating Series Test - Part I (Maynooth University) Handout on the Alternating Series Test - Part II (Maynooth University) Handout on the Alternating Series Test - Part III (Maynooth University) Video on Alternating Series (Patrick JMT). a mathematical series in which consecutive terms are alternatively positive and negative\u2026. Therefore, we will have to look at the alternating series to determine if it converges or not. Alternating series, absolute and conditional convergence You have to know the de nition of what it means for a series to be alternating and con-vergent. A quantity that measures how accurately the nth partial sum of an alternating series estimates the sum of the series. Therefore, the sums converge to the same limit if and only if a n \u2192 0 as n \u2192 \u221e. Alternating Series & AS Test Objectives: Be able to describe the convergence of alternating series. However, it can be remedied by choosing any sequence which goes to zero fast enough, and putting it in the place of the zero terms, as you do. an infinite series whose terms are alternately positive and negative: u 1 - u 2 + u 3 - u 4 + \u2026 + (-I) n-1 u n + \u2026. After defining alternating series, we introduce the alternating series test to determine whether such a series converges. series diverges by Limit Comparison, and the original series does not converge absolutely. A retrospective record review of performance on the three-step Luria test was conducted on 383 participants from a university-based dementia clinic. Infinite series whose terms alternate in sign are called alternating series. (This can usually b e done b y insp ection). Proof: Look at the. So this is a geometric series with common ratio r = \u20132. If you need to review this test, please refer to the supplemental notes 23. Infinite series whose terms alternate in sign are called alternating series. Then P\u221e k=1 (\u22121) ka k converges. A divergent alternating series whose terms go to zero. Course Description Calculus emphasizes a multi-representational approach, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. (f) Prove that the alternating harmonic series X1 n=1 ( 1)n n converges. Here is how one can find the derivative of arctan x: The above is a modern proof, Gregory used the derivative of arctan from the work of others. 5 kV through 765 kV and IEEE Standard 82, IEEE Standard Test Procedure for Impulse Voltage Tests on Insulated Conductors. ) Summarizing the above work, we know that 4 is not included, but 6 is. An alternating series is a series whose terms are al-ternately positive and negative. 88 11 Note: and ; therefore, 89 1 b) lim lim 0 8 Therefore. It is one of the most commonly used tests for determining the convergence or divergence of series. It is not obvious that the sequence b n decreases monotonically to 0. Overview of Alternating Series Test; 3 Examples; Conditional and Absolute Convergence for Alternating Series; 2 Examples; Ratio Test. In mathematical analysis, the alternating series test is a method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Converges by ratio test. This test does not prove absolute convergence. toit toshowthatthe series converges. Chroma's 63800 Series AC & DC Electronic Loads include built-in 16-bits precision measurement circuits to measure the steady-state and transient responses for true RMS voltage, true RMS current, true power(P), apparent power(S), reactive power(Q), crest factor, power factor, THDv and peak repetitive current. Run multiple programs, render videos and more \u2014 the Yoga 920 is designed to multitask with ease. The Alternating Series Test If k 1 ak is an Alternating Series and lim 0 k k a and ak eventually becomes strictly decreasing. 01 Single Variable Calculus, Fall 2005 Prof. However, the third condition is not valid since and instead approaches infinity. Overview of Root Test; 3 Examples; Sequences. Since this is an alternating series whose terms decrease to zero, we know that the series converges. The alternating series. In mathematical analysis, the alternating series test is a method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. Free series convergence calculator - test infinite series for convergence step-by-step. Here is how one can find the derivative of arctan x: The above is a modern proof, Gregory used the derivative of arctan from the work of others. Thus, the even and odd series both converge. In this worksheet, we will practice determining whether an alternating series is convergent or divergent using the alternating series test. Series Convergence Flowchart doesa n! 0? Isa n > 0? Diverges by Divergence Test Is it alternating in sign and ja n decreasing? Are there any easy comparisons? Does it feel likea n `looks like' someb n? Try Ratio Test: lim a n+1 a n = c if 0 c < 1 then P a n converges if c > 1 then P a n diverges if c= 1 then test is inconclusive Try Integral. Converges by ratio test. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors:. One test that is specifically designed to handle series whose terms alternate positive and negative is the Alternating Series Test. Several convergence testing methods, such as the alternating series test, ratio test and root test, will be presented. 0 < a n+1 <= a n), and approaching zero, then the alternating series. EX 4 Show converges absolutely. 1 Examples 2 Alternating series test. Alternating series definition is - a mathematical series in which consecutive terms are alternatively positive and negative. Bhagwan Singh Vishwakarma 129,314 views. (b) Prove the Alternating Series Test using the Nested Interval Property (Theorem 1. This is a test which we'll use to show lots of alternating series converge. Here are a bunch of Single Variable Calculus applets. Overview of Alternating Series Test; 3 Examples; Conditional and Absolute Convergence for Alternating Series; 2 Examples; Ratio Test. (LTspice is also called SwitcherCAD by its manufacturer, since they use it primarily for the design of switch mode power supplies (SMPS). Use a known series to \ufb01nd a power series in x that has the given function as its sum: (a) xsin(x3) Recall the Maclaurin series for sinu = X\u221e n=0 (\u22121)n u2n+1 (2n+1)! Therefore, sin(x3) = X\u221e n=0 (\u22121)n (x3)2n+1 (2n+1)! =. ) The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. 5 \u2013 Notes Page 2 of 3 Ex 3. \uc774 \uac10\uc18c\uc218\uc5f4\uc774\ubbc0\ub85c \ub77c\uace0 \ud558\uba74. (g) State the Alternating Series Estimation Theorem. For those that diverge, say which hypotheses of the alternating series test do not. Sequences and Series: Alternating Series Test (MathsCasts). (i) The series (\u22121)n is an alternating. The alternating series test requires that the a n alternate sign, get smaller and approach zero as n approaches infinity, which is true in this case. If s= P ( 1)n 1b. If a k \u2192 0, then X (\u22121)ka k converges. If, as our intuition tells us should be true, the rearrangement does not change the sum, then we have just seen that. AC versus DC. When a series alternates (plus, minus, plus, minus,) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0. 8 = angle Pod sino g 3! as 3! a 7 \" Functions can often be represented by an infinite series. Lecture 27 :Alternating Series The integral test and the comparison test given in previous lectures, apply only to series with positive terms. \u2022 Be sure that you apply the alternating series test only to alternating series. But the alternating series approximation theorem quickly shows that. Part (a) asked students to use the ratio test to determine the interval of convergence for the given Maclaurin series. The integral test, which is my favorite test in general, tends to be awkward with alternating series. 7 Alternating Series, Absolute Convergence notes by Tim Pilachowski So far, we have pretty much limited our attention to series which are positive. The alternating series test is a simple test we can use to find out whether or not an alternating series converges (settles on a certain number). (You probably figured out that with this naked summation. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. The idea of hopping back and forth to a limit is basically. Rearrangements. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. We look at a couple of examples. 7 Alternating Series, Absolute Convergence notes by Tim Pilachowski So far, we have pretty much limited our attention to series which are positive. We're learning alternating series test and the whole class is confused on why the series needs to be decreasing to pass the test. EXPECTED SKILLS: Determine if an alternating series converges using the Alternating Series Test. The un 's are all positive. If it does, then try applying the Ratio Test i. The alternating series simply tells us that the absolute value of each of the terms decreases monotonically, i. Instead, I'll give a more elementary proof of the Alternating series test which does not use Abel's formula (but we will see Abel's formula later). What does alternating personality mean in law?. Just better. An alternating sequence will have numbers that switch back and forth between positive and negative signs. Electricity or \"current\" is nothing but the movement of electrons through a conductor, like a wire. It should be noted that Theorem 1. CLP-2 Integral Calculus. Here are a bunch of Single Variable Calculus applets. 7: Series, alternating series test - solutions Apply the AST (possibly in combination with other tests) and state your conclusion about convergence. Embed this widget \u00bb. Finally, by L'H\u00f4pital's Rule, By the Alternating Series Test, the series converges. of series with positive and negative terms and whether or not they converge. So, given the series look at the limit of the non-alternating part: So, this series converges. Jason Starr. The well-known Leibniz Criterion or alternating series test of convergence of alternating series is generalized for the case when the absolute value of terms of series are \u201cnot absolutely monotonously\u201d convergent to zero. Alexander Street is an imprint of ProQuest that promotes teaching, research, and learning across music, counseling, history, anthropology, drama, film, and more. With the Alternating Series Test, all we need to know to determine convergence of the series is whether the limit of b[n] is zero as n goes to infinity. Alternating series test for convergence. Find more Mathematics widgets in Wolfram|Alpha. ) Note: Some of this was written using SwitcherCad III, and some was written using LTspice IV. 5 kV through 765 kV and IEEE Standard 82, IEEE Standard Test Procedure for Impulse Voltage Tests on Insulated Conductors. There are two main types of current used in most electronic circuits today. The table below will help show you how the scores derived from the various subtests is used within the various index scores. Alternating Series Test If for all n, a n is positive, non-increasing (i. [email protected] This is the logical reasoning questions and answers section on \"Number Series Type 2\" with explanation for various interview, competitive examination and entrance test. A series in which successive terms have opposite signs is called an alternating series. I Absolute convergence test. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. b n+1 \u2264 b n for all n > N. Theorem 4 : (Comparison test ) Suppose 0 \u2022 an \u2022 bn for n \u201a k for some k: Then. You can see some Alternating Series, Absolute and Conditional Convergence - Notes, Engineering, Semester sample questions with examples at the bottom of this page. I believe we sometimes overemphasize the importance of this test because we want to make clear the distinction between absolute convergence and convergence. When a sum does this, we say it \u2018telescopes\u2019. The Absolute-Convergence Test (ACT) For series that have infinitely many negative terms and infinitely many positive terms and that aren't alternating, the tests discussed in previous sections and the AST can't be applied. Definition of alternating personality in the Legal Dictionary - by Free online English dictionary and encyclopedia. The part of the summation makes up the series alternator. Calculus 141, section 9. Suppose that {a i} is a sequence of positive numbers such that a i > a i+1 for all i. Teach yourself calculus. Course Material Related to This Topic:. The lower beds of the Peuquenes ridge, and of the several great lines to the westward of it, are composed of a vast pile, many thousand feet in thickness, of porphyries which have flowed as submarine lavas, alternating with angular and rounded fragments of the same rocks, thrown out of the submarine craters. iii) if \u03c1 = 1, then the test is inconclusive. Line jumping is the idea behind our first convergence test, the alternating series test. The common ratio of a geometric series may be negative, resulting in an alternating sequence. Geometric series. Continuing with post on sequences and series New Series from Old 1 Rewriting using substitution New Series from Old 2 Finding series by differentiating and integrating New Series from Old 3 Rewriting rational expressions as geometric series Geometric Series \u2013 Far Out A look at doing a question the right way and the \u201cwrong\u201d way?\u2026. Hence, the interval of convergence is: (\u22128,10] and the radius convergence is: R = 10. Therefore, the sums converge to the same limit if and only if a n \u2192 0 as n \u2192 \u221e. Alternating Series Test. Absolute Ratio Test Let be a series of nonzero terms and suppose. In a alternating series every term will have a sign different than the term before it. The alternating series simply tells us that the absolute value of each of the terms decreases monotonically, i. (One is the harmonic series; the other can be proved divergent by comparison with the harmonic series. An alternating series is a series whose terms are al-ternately positive and negative. An alternating sequence will have numbers that switch back and forth between positive and negative signs. The short circuit test is maintained by default. Alternating series test listed as AST Alternating series. Then the series converges if both of the following conditions hold. If property 3 is respected but property 1 and/or property 2 do not hold, then the alternating series test is inconclusive. More Alternating Series Examples - Finding whether a given alternating series converges or diverges. Alternating Series (6. The Alternating Series Test. So, the series converges by the alternating series test. \u22121 3 2 4 \u22123 5 4 6 \u22125 7. The test that we are going to look into in this section will be a test for alternating series. The series above is thus an example of an alternating series, and is called the alternating harmonic series. Alternating Series and Leibniz\u2019s Test Let a 1;a 2;a 3;::: be a sequence of positive numbers. In order to use this test, we first need to know what a converging series and a diverging series is. The integral test, which is my favorite test in general, tends to be awkward with alternating series. Overview of Root Test; 3 Examples; Sequences. jxjn+1: Rememberthatjxjrepresentsthedistancebetweenx and0. \u3053\u306e\u30b3\u30f3\u30c6\u30f3\u30c4\u306e\u8868\u793a\u306b\u306f\u3001Adobe Flash Player\u306e\u6700\u65b0\u30d0\u30fc\u30b8\u30e7\u30f3\u304c\u5fc5\u8981\u3067\u3059\u3002 http://www. Warm Up: Find the sum of the infinite series. Alternating Series Test. Does \u2211 (\u22121)\ud835\udc5b3\ud835\udc5b 4\ud835\udc5b\u22121 \u221e \ud835\udc5b=1 converge or diverge? For an alternating series, how close is \ud835\udc60\ud835\udc5b to the sum of the infinite number of terms?. Mathispower4u 70,297 views. Chroma's 63800 Series AC & DC Electronic Loads include built-in 16-bits precision measurement circuits to measure the steady-state and transient responses for true RMS voltage, true RMS current, true power(P), apparent power(S), reactive power(Q), crest factor, power factor, THDv and peak repetitive current. Alternating Current Circuits 5 Open-Ended Problems 57. Topics for Test 2 Convergence tests for series. SERIES CONVERGENCE/ DIVERGENCE FLOW CHART nVergeS TEST FOR DIVERGENCE GEOMETRIC SERIES ALTERNATING SERIES TELESCOPING SERIES ? May to etc. Ratio and Root Tests. 1997 views. Holmes May 1, 2008 The exam will cover sections 8. In this section we introduce alternating series\u2014those series whose terms alternate in sign. Proof: Look at the. The series above is thus an example of an alternating series, and is called the alternating harmonic series. alternating series test. It is important that you verify the conditions of the Alternating Series Test are met; otherwise some-one might not believe your conclusion is valid. An alternating series is any series, , for which the series terms can be written in one of the following two forms. Alternating series convergence: a visual proof Richard H. The Alternating Series Test. alternating with phrase. Incorrect! Remember the conditions of the Alternating Series Test. For : The first and second conditions are satisfied since the terms are positive and are decreasing after each term. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test , Leibniz's rule , or the Leibniz criterion. The alternating series test can tell us if it's safe to open that box. Fourier series are used in the analysis of periodic functions. Solution: Let I1 be the closed interval [0, s1]. It is important that you verify the conditions of the Alternating Series Test are met; otherwise some-one might not believe your conclusion is valid. 1, which asks for a proof of the Alter-nating Series Test using the Cauchy Criterion for series (Theorem 2. The alternating series test says that if the absolute value of each successive term decreases and \\lim_{n\\to\\infty}a_n=0, then the series converges. I Absolute and conditional convergence. LTspice Tutorial Introduction While LTspice is a Windows program, it runs on Linux under Wine as well. 1 Is there a short name for series which satisfy the hypothesis of the alternating series test?. This is a very useful lecture in Calculus. Printing the sum of an alternating series [Basic Python] I've been stuck on this little problem for the best part of today, and it's driving me insane now. Hence, the interval of convergence is: (\u22128,10] and the radius convergence is: R = 10. Write the three rules that are used to satisfy convergence in an alternating series test. Assignment #4: Alternating Series. Therefore, we will have to look at the alternating series to determine if it converges or not. Correct! This is the correct answer. ( 1) Use Alternating Series Test to show con verges. What is alternating personality? Meaning of alternating personality as a legal term. To see how this works, let \\(S$$ be the sum of a convergent alternating series, so. The alternating series test requires that the a n alternate sign, get smaller and approach zero as n approaches infinity, which is true in this case. Theorem (Alternating series test) If the terms of the series \u2211 n = 1 \u221e (-1) n an have the property thatall ofthe an terms are positive and an+1 < an forall n, thenthe series converges. Alternating Series (6. I have a bachelors in Mathematics Education from Slippery Rock University, and a Masters in Administration and Supervision from The College of Notre Dame. Alternating Series Test 1 Alternating Series Test If the terms of the alternating series ( 1)n 1b n b1 b2 b3 n 1 where bn 0 satisfy (1) bn 1 bn for all n 1 (bn is decreasing) (2) lim n bn 0 then the series is convergent. Q1: The alternating series test does not apply to the series \u221e ( \u2212 1 ) \ud835\udc5b \ud835\udc5b + 1. It's also known as the Leibniz's Theorem for alternating series. (Warning: Do not use a multimeter to measure the wall outlets in your home. In this video, Krista King from integralCALC Academy talks about the Alternating Series Test (Calculus problem example). Transcranial alternating current stimulation (tACS) is a non-invasive brain stimulation method that allows to directly modulate brain oscillations of a given frequency. Drag up for fullscreen M M. 01 Single Variable Calculus, Fall 2005 Prof. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test , Leibniz's rule , or the Leibniz criterion. Looking for abbreviations of AST? It is Alternating series test. Mathematics Assignment Help, Steps for alternating series test, Steps for Alternating Series Test Suppose that we have a series \u2211a n and either a n = (-1) n b n or a n = (-1) n+1 b n where b n > 0 for all n. The common ratio of a geometric series may be negative, resulting in an alternating sequence. We can therefore. n satis es the requirements for the alternating series test. However, it is not enough to have having a limit of zero, you also need decreasing, as the following example shows. Solved examples with detailed answer description, explanation are given and it would be easy to understand. You\u2019ll do that one for homework. Alternating Current Circuits 5 Open-Ended Problems 57. We note that S 2 \u2062 n + 1 - S 2 \u2062 n = a 2 \u2062 n + 1. Many, but not all, of the problems will have. Alternating series test for convergence. Alternating Series Estimation Theorem. The fact that sums, products, integrals, antiderivatives of Taylor series are also Taylor series is in 8. 6 Alternating Series and Conditional Convergence Page 1 Theorem 15 - The Alternating Series Test The series 1 12 3 4 1 (1)n n n uuu uu \u221e + = \u2211\u2212 =\u2212+\u2212+\" converges if the following conditions are satisfied: 1. According to the alternating series test, we know that this series converges to some number. Click on the name of the test to get more information on the test. Many series such as 8\u0153\" 8\u0153\" 8\u0153\" \u221e \u221e \u221e # sin 8 \" 88 8 8 8 8\" \u00df \" \u00df \"and do not have all positive terms and thus cannot be investigated using the above mentioned tests. A t-test is one of the most frequently used procedures in statistics. Alternating series have the simplest of sign patterns. The terms alternate. This is a correct reasoning to show the divergence of the above series.", "date": "2019-12-10 08:01:54", "meta": {"domain": "emporiumgalorium.de", "url": "http://lxlj.emporiumgalorium.de/alternating-series-test.html", "openwebmath_score": 0.7549763321876526, "openwebmath_perplexity": 830.88619511814, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.97594644290792, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8563703479136634}}
{"url": "https://math.stackexchange.com/questions/352163/show-that-theres-a-unique-minimum-spanning-tree-if-all-edges-have-different-cos/352212", "text": "# Show that there's a unique minimum spanning tree if all edges have different costs\n\nShow that there's a unique minimum spanning tree (MST) in case the edges' weights are pairwise different $(w(e)\\neq w(f) \\text{ for } e\\neq f)$.\n\nI thought that the proof can be done for example by contradiction, saying that we have $2$ different MST meaning that somewhere was possible to pick from more edges, so $w(e) = w(f)$ for $e\\neq f$, contradiction. Apparently this is not correct.\n\nHow would you show that a graph has a unique MST if all edges have distinct weights?\n\nIf $T_1$ and $T_2$ are distinct minimum spanning trees, then consider the edge of minimum weight among all the edges that are contained in exactly one of $T_1$ or $T_2$. Without loss of generality, this edge appears only in $T_1$, and we can call it $e_1$.\n\nThen $T_2 \\cup \\{ e_1 \\}$ must contain a cycle, and one of the edges of this cycle, call it $e_2$, is not in $T_1$.\n\nSince $e_2$ is a edge different from $e_1$ and is contained in exactly one of $T_1$ or $T_2$, it must be that $w ( e_1 ) < w ( e_2 )$. Note that $T = T_2 \\cup \\{ e_1 \\} \\setminus \\{ e_2 \\}$ is a spanning tree. The total weight of $T$ is smaller than the total weight of $T_2$, but this is a contradiction, since we have supposed that $T_2$ is a minimum spanning tree.\n\n\u2022 How do we know that $w ( e_1 ) < w ( e_2 )$? Mar 4, 2017 at 23:57\n\u2022 @Alexander Because $e_1$ is the edge of least weight that is contained in exactly one of the MSTs, and $e_2$ is a different edge that is contained in exactly one of the MSTs. Mar 5, 2017 at 3:49\n\u2022 Is $e_1$ the edge of least weight out of the edges of $T_1 \\cup T_2$, which is in either $T_1$ or $T_2$? Or is $e_1$ the edge of least weight out of the edges of $T_1 \\bigtriangleup T_2$? Mar 6, 2017 at 1:43\n\u2022 I suspect it's the latter of two, otherwise I don't understand how we know this edge isn't in both $T_1$ and $T_2$. Mar 6, 2017 at 1:47\n\u2022 @Alexander As I said, $e_1$ is the edge of least weight that is contained in exactly one of the MSTs. If it was in both $T_1$ and $T_2$ it would not be contained in exactly one of the MSTs. Mar 6, 2017 at 3:41\n\nA proof using cycle property:\n\nLet $G=(V,E)$ be the original graph.\n\nSuppose there are two distinct MSTs $T_1=(V,E_1)$ and $T_2=(V,E_2)$. Since $T_1$ and $T_2$ are distinct, the sets $E_1-E_2$ and $E_2-E_1$ are not empty, so $\\exists e\\in E_1-E_2$.\n\nSince $e\\notin E_2$, adding it to $T_2$ creates a cycle. By cycle property the most expensive edge of this cycle (call it $e'$) does not belong to any MST. But\n\nIf $e'=e$ then $e'\\in E_1$ (because $e\\in E_1-E_2$)\nIf $e'\\neq e$ then $e'\\in E_2$\n\nBoth cases are contradicting with the fact that $e'$ is not in any MST.\n\nThe best explanation I found was on wikipedia, Proof:\n\n1. Assume the contrary, that there are two different MSTs $A$ and $B$.\n2. Since $A$ and $B$ differ despite containing the same nodes, there is at least one edge that belongs to one but not the other. Among such edges, let e1 be the one with least weight; this choice is unique because the edge weights are all distinct. Without loss of generality, assume $e1$ is in $A$.\n3. As $B$ is a MST, $\\{e1\\}\\cup B$ must contain a cycle $C$.\n4. As a tree, $A$ contains no cycles, therefore $C$ must have an edge $e2$ that is not in $A$.\n5. Since $e1$ was chosen as the unique lowest-weight edge among those belonging to exactly one of $A$ and $B$, the weight of $e2$ must be greater than the weight of $e1$.\n6. Replacing $e2$ with $e1$ in $B$ therefore yields a spanning tree with a smaller weight.\n7. This contradicts the assumption that $B$ is a MST.\n\nMore generally, if the edge weights are not all distinct then only the (multi-)set of weights in minimum spanning trees is certain to be unique; it is the same for all minimum spanning trees [1].\n\nThis is a slightly different (though more lengthy) proof than the other's in that it is an algorithmic proof. We can use Prim's algorithm and demonstrate the proof by studying the spanning tree it builds in such a graph $$G$$. Here is Prim's algorithm:\n\n\u2022 $$V_T \\leftarrow \\{ v_0 \\}$$\n\u2022 $$T_0 \\leftarrow \\emptyset$$\n\u2022 for $$i \\leftarrow 1$$ to $$|V| - 1$$ do:\n\u2022 find an edge $$e_i=(v_i,u_i)$$ of minimum weight such that $$v_i$$ is in $$V_T$$ and $$u_i$$ is not.\n\u2022 $$V_T \\leftarrow V_T \\cup \\{ u_i \\}$$\n\u2022 $$T_i \\leftarrow T_i \\cup \\{ e_i \\}$$\n\nWe demonstrate the following : At iteration $$i$$, the choice $$e_i$$ that Prim's algorithm makes belongs to every minimum spanning tree. We can show this by using the inductive hypothesis:\n\nSuppose $$T_{i-1}$$ is a subset of every minimum spanning tree. Then, $$T_i = T_{i-1} \\cup \\{ e_i \\}$$ is a subset of every minimum spanning tree.\n\n\u2022 Base case: $$T_1 = e_1$$ where $$e_1$$ is incident to $$v_0$$. Suppose that there is some minimum spanning tree $$T$$ that does not contain $$e_1$$. Then, $$T$$ contains another edge, $$\\bar{e}_1$$ incident to $$v_0$$. By the greedy choice of Prim's and the distinctness of weights, $$w(e_1) < w(\\bar{e}_1)$$ and we can replace $$\\bar{e}_1$$ with $$e_1$$ to decrease the weight of $$T$$, which is a contradiction.\n\u2022 Inductive step: Suppose, to the contrary, that there exists some minimum spanning tree $$T$$ which contains the edges in $$T_{i-1}$$ but does not contain $$e_i$$. Consider the set of edges $$T \\setminus T_{i-1}$$, which must be a forest composed of trees $$\\bar{T}^1_{i-1}, \\bar{T}^2_{i-1}, \\ldots, \\bar{T}^k_{i-1},$$ where each tree $$\\bar{T}^t_{i-1}$$ is connected to $$T_{i-1}$$ in $$T$$ by an edge $$e^t_{i-1}$$.\n\nHowever, note that one of those trees must be incident to the vertex $$u_i$$. Suppose that tree is $$\\bar{T}^x_{i-1}$$. Then, by Prim's greedy choice and since the weights of edges are distinct, we know that $$w(e_i) < w(e^x_{i-1})$$. Which means that we replace $$e^x_{i-1}$$ with $$e_i$$ and decrease the weight of $$T$$. Hence, $$T$$ is not a minimum spanning tree. Then, $$e_i$$ must be in every minimum spanning tree and $$T_i$$ must be a subset of every minimum spanning tree.\n\nThus, the minimum spanning tree produced by Prim's algorithm is the only minimum spanning tree of $$G$$.\n\n\u2013\u00a0fhy\nDec 29, 2019 at 12:30\n\nThis is a (slightly) more detailed version of the currently accepted answer\n\n(For the sake of contradiction) Let $$T_1 = (V, E_1)$$ and $$T_2 = (V, E_2)$$ be two distinct MSTs of the graph $$G = (V, E)$$\nNote that both have the same vertex set $$V$$ since both are spanning trees of $$G$$\n\nConsider the set $$E_{\\Delta} = E_1 \\triangle E_2$$\nLet $$e = (u, v)$$ be the edge in $$E_{\\Delta}$$ having the least cost (or weight)\nNote that since all costs are unique, and $$E_{\\Delta}$$ is non-empty, $$e$$ must be unique.\nWithout loss of generality, assume $$e \\in E_1$$\n\nNow, there must be a path $$P$$, with $$e \\notin P$$, in $$T_2$$ connecting $$u$$ and $$v$$, since trees are by definition, connected.\nNote that at least one edge (say $$e'$$) that occurs in $$P$$ must not be in $$E_1$$, thus, $$e' \\in E_{\\Delta}$$\nThis is because, if $$P \\subset E_1$$, $$T_1$$ will contain a cycle formed by the path $$P$$ and the edge $$e$$, this leads to a contradiction, since trees by definition are acyclic.\nNote that by definition of $$e$$ and the fact that all costs are distinct, $$\\text{cost}(e) < \\text{cost}(e')$$\n\nNow, consider $$T_2' = (V, E_2')$$, where $$E_2' = (E_2 \\backslash\\{e'\\})\\cup\\{e\\}$$, this has a strictly lesser total cost than $$T_2$$\nAs $$T_2$$ was a MST, this leads to a contradiction.", "date": "2023-03-21 07:42:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/352163/show-that-theres-a-unique-minimum-spanning-tree-if-all-edges-have-different-cos/352212", "openwebmath_score": 0.8433116674423218, "openwebmath_perplexity": 125.79959717816376, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464485047916, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8563703434437397}}
{"url": "https://thinairmedia.org/at45qi/zat0b.php?931dcd=diagonal-of-parallelogram", "text": "The diagonals of a parallelogram bisect each other. Diagonal of Parallelogram Formula The formula of parallelogram diagonal in terms of sides and cosine \u03b2 (cosine theorem) if x =d 1 and y = d 2 are the diagonals of a parallelogram and a and b are the two sides. Show that it is a rhombus. The shape has the rotational symmetry of the order two. The pair of opposite sides are equal and they are equal in length. DOWNLOAD PDF / PRINT . The diagonals bisect the angles. If they diagonals do indeed bisect the angles which they meet, could you please, in layman's terms, show your proof? Then, substitute 4.8 for in each labeled segment to get a total of 11.2 for the diagonal \u2026 This is the currently selected item. Consecutive angles are supplementary. Diagonals of rectangles and general parallelograms, however, do not. Type your answer here\u2026 Related Topics. The diagonals of a parallelogram bisect each other. Construction of a parallelogram given the length of two diagonals and intersecting angles between them - example Construct a parallelogram whose diagonals are 4cm and 5cm and the angle between them is \u2026 Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. That is, each diagonal cuts the other into two equal parts. 1 answer. These parallelograms have different areas. Learn more about Diagonal of Parallelogram & Diagonal of Parallelogram Formula at Vedantu.com The diagonals of a parallelogram. Area of the parallelogram using Trignometry: $$\\text{ab}$$$$sin(x)$$ where $$\\text{a}$$ and $$\\text{b}$$ are the length of the parallel sides and $$x$$ is the angle between the given sides of the parallelogram. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other). where is the two-dimensional cross product and is the determinant.. As shown by Euclid, if lines parallel to the sides are drawn through any point on a diagonal of a parallelogram, then the parallelograms not containing segments of that diagonal are equal in area (and conversely), so in the above figure, (Johnson 1929).. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram Calculate certain variables of a parallelogram depending on the inputs provided. You get the equation = . Test the conjecture with the diagonals of a rectangle. Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. In the figure below diagonals AC and BD bisect each other. A parallelogram where all angles are right angles is a rectangle! Area of the parallelogram when the diagonals are known: $$\\frac{1}{2} \\times d_{1} \\times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Apply the formula from the Theorem. MCQ in Plane Geometry. These properties concern its sides, angles, and diagonals. If you just look [\u2026] Diagonals divide the parallelogram into two congruent triangles; Diagonals bisect each other; There are three special types of parallelogram, they are: Rectangle; Rhombus; Square; Let us discuss these special parallelograms one by one. You can use the calculator for each formula. In a parallelogram, the sides are 8 cm and 6 cm long. For instance, please refer to the link, does $\\overline{AC}$ bisect $\\angle BAD$ and $\\angle DCB$? A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. Proof: Diagonals of a parallelogram. The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection. If you make the diagonals almost parallel to one another - you will have a parallelogram with height close to zero, and thus an area close to zero. Notice the behavior of the two diagonals. You can rotate the two diagonals around this joint, and form different parallelogram (by connecting the diagonals's end points). See more. Opposite sides are congruent. If \u2220Boc = 90\u00b0 and \u2220Bdc = 50\u00b0, Then \u2220Oab = - Mathematics If \u2220Boc = 90\u00b0 and \u2220Bdc = 50\u00b0, Then \u2220Oab = - Mathematics Question By \u2026 General Quadrilateral; Kite; Rectangle; Rhombus; Square; Discover Resources. Make a conjecture about the diagonals of a parallelogram. The diagonals are perpendicular bisectors of each other. person_outlineTimurschedule 2011-03-28 14:49:28. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . There are three cases when a parallelogram is also another type of quadrilateral. The adjacent angles of the parallelogram are supplementary. The length of the shorter diagonal of a parallelogram is 10.73 . 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And a rhombus with a right angle variables of a rectangle was wondering if within parallelogram the diagonals of parallelogram!: opposite sides are equal in length rectangle and a rhombus with a right angle the conjecture the. Solving in Plane Geometry things that are true about it parallelograms, do not parallelogram,. Ask Doubt, angles, diagonals, height, perimeter and area of parallelograms draw a.. Was wondering if within parallelogram the diagonals of a parallelogram whose angles are right angles is a rectangle the provided. The length of the parallelogram will divide the shape into two similar congruent triangles 18 2... The help of law of cosines opposite sides parallel to each other area of parallelograms diagonal! Has all the properties of the second diagonal of a parallelogram is a rectangle and a rhombus with a angle! Angles as acute and obtuse, the diagonals bisect the angles which they,...\n\nUncategorized", "date": "2021-05-09 13:31:05", "meta": {"domain": "thinairmedia.org", "url": "https://thinairmedia.org/at45qi/zat0b.php?931dcd=diagonal-of-parallelogram", "openwebmath_score": 0.7166275382041931, "openwebmath_perplexity": 854.741629216889, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9669140235181257, "lm_q2_score": 0.8856314632529871, "lm_q1q2_score": 0.8563294814881909}}
{"url": "https://math.stackexchange.com/questions/2729586/monotonically-and-strictly-increasing-functions", "text": "# Monotonically and strictly increasing functions\n\nThis is a question on terminology.\n\nWhat is the difference between a (i) strictly increasing function, and a (ii) monotonically increasing function? Is it that a monotonically increasing function may also include functions that are constant in some intervals, while strictly increasing function must always have a positive derivative where it is defined?\n\nIf so, is it correct to say, that\n\nStrictly increasing functions $\\implies$ monotonically increasing, while the converse is not true? And a strictly increasing function is equivalent to a 'strictly monotonically increasing' function?\n\nThanks.\n\n\u2022 Yes. Given $x>y$, \"monotonically increasing\" means that $f(x)\u2265f(y)$ while \"strictly increasing\" means $f(x)>f(y)$. This means, for example, that a constant function is both monotonically increasing and montonically decreasing. \u2013\u00a0lulu Apr 9 '18 at 16:41\n\u2022 And a strictly increasing function is equivalent to a 'strictly monotonically increasing' function? How do you define \"strictly monotonically increasing\"? I've never heard that. As lulu said, yes for everything else. \u2013\u00a0anderstood Apr 9 '18 at 16:42\n\u2022 To avoid ambiguity, functions satisfying $x\\le y\\implies f(x)\\le f(y)$ are sometimes called non-decreasing. \u2013\u00a0Juli\u00e1n Aguirre Apr 9 '18 at 16:42\n\u2022 Thanks for your clarifications. @anderstood Perhaps I could change that to 'strictly monotone functions', which would refer to strictly increasing or decreasing functions? \u2013\u00a0T J. Kim Apr 9 '18 at 16:46\n\u2022 I would avoid non-standard usage. We already have the phrase \"strictly increasing\", why introduce new terminology for the same thing? \u2013\u00a0lulu Apr 9 '18 at 16:48\n\nYou almost have it right. The condition is better stated without referring to derivatives. A function $f(x)$ is strictly increasing if for all $(x,y)$ such that $y>x$,\n\n$$f(y) > f(x)$$\n\nand is monotonic increasing if for all $(x,y)$ such that $y>x$, $$f(y) \\geq f(x)$$\n\nYour definition involving derivatives would say that the sawtooth $$g(x) = x - \\lfloor x \\rfloor$$ is strictly monotonic (since the derivative is not defined at integer $x$), but it is not monotonic at all.\n\nYour last sentence is completely correct.\n\n\u2022 I see, thanks for your answer. Would the definition involving derivatives hold if $f$ is defined to be differentiable for all x in the domain, in which case the sawtooth function could not be considered? \u2013\u00a0T J. Kim Apr 9 '18 at 16:58\n\nA strictly increasing function: let $$P=\\{x_1,...,x_n; x_i<x_{i+1}\\}$$ then $f(x_i)<f(x_{i+1})$ for all $x_i,x_{i+1} \\in P$. A monotonic increasing function: let $$P=\\{x_1,...,x_n; x_i<x_{i+1}\\}$$ then $f(x_i)\\leq f(x_{i+1})$ for all $x_i,x_{i+1} \\in P$. So a monotonic function can be constant for some interval $(x_k, x_l)$ or can be increasing on that interval too, a strictly increasing funcion has always a greater image provided x increases.", "date": "2019-10-16 09:17:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2729586/monotonically-and-strictly-increasing-functions", "openwebmath_score": 0.8859436511993408, "openwebmath_perplexity": 489.9478911180118, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426420486938, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8563264509683796}}
{"url": "http://mathhelpforum.com/advanced-statistics/38179-product-two-normal-distributions.html", "text": "# Math Help - Product of Two Normal Distributions\n\n1. ## Product of Two Normal Distributions\n\nHello,\n\nI am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.\n\n2. Originally Posted by vioravis\nHello,\n\nI am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.\nNo.\n\nIf the two random variables X and Y are independent, then the pdf of Z = XY is probably (I haven't done the calculation) a Bessel function. See 3. of properties at Normal distribution - Wikipedia, the free encyclopedia.\n\n3. Originally Posted by vioravis\nHello,\n\nI am trying to find the distribution of the product of two normal densities (different means and standard deviations) . Does the product follow normal distribution also? If so, what is the mean and standard deviation of the resultant distribution? Thanks a lot.\nDo you mean \"What is the distribution of the product of two normally distributed RV with means and variances which may be different?\"?\n\nRonL\n\n4. I have the following two densities:\n\nf1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}\nf2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}\n\nSo f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).\n\nDoes f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.\n\n5. In Page 3 of the following link, it is given that the product of two gaussian densities is also gaussian. I think it is same as the one I requested above. I am looking for the derivation of the formula given for the resultant distribution:\n\nconfirmed here as well:\n\nProduct of Two Gaussian PDFs\n\n6. Originally Posted by vioravis\nI have the following two densities:\n\nf1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}\nf2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}\n\nSo f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).\n\nDoes f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.\nIf I understand you correctly, what you're asking boils down to wanting to show that\n\n$-\\frac{(x - \\mu_1)^2}{2\\sigma^2_1} - \\frac{(x - \\mu_2)^2}{2\\sigma^2_2} = -C \\frac{(x - \\mu)^2}{2\\sigma^2}$\n\nand getting the appropriate expressions for $\\mu, ~ \\sigma$ and C in terms of $\\mu_1, ~ \\mu_2, ~ \\sigma_1$ and $\\sigma_2$. Note that $e^C$ becomes part of the normalising constant.\n\nIt's simple to show and get the expressions but tedious to type out.\n\n7. Fantastic,\n\nThanks a lot. Could you direct me to some references instead that show the calculations? or is it possible for your to scan and post the handwritten one instead of typing it? Thank you.\n\n8. Originally Posted by mr fantastic\nIf I understand you correctly, what you're asking boils down to wanting to show that\n\n$-\\frac{(x - \\mu_1)^2}{2\\sigma^2_1} - \\frac{(x - \\mu_2)^2}{2\\sigma^2_2} = -C \\frac{(x - \\mu)^2}{2\\sigma^2}$\n\nand getting the appropriate expressions for $\\mu, ~ \\sigma$ and C in terms of $\\mu_1, ~ \\mu_2, ~ \\sigma_1$ and $\\sigma_2$. Note that $e^C$ becomes part of the normalising constant.\n\nIt's simple to show and get the expressions but tedious to type out.\n$-\\frac{(x - \\mu_1)^2}{2\\sigma^2_1} - \\frac{(x - \\mu_2)^2}{2\\sigma^2_2}$\n\n$= \\frac{-\\sigma^2_2(x - \\mu_1)^2 - \\sigma^2_1 (x - \\mu_2)^2}{2 \\sigma_1^2 \\sigma^2}$\n\n$= \\frac{ -\\sigma^2_2 x^2 + 2\\sigma_2^2 \\mu_1 x - \\mu_1^2 \\sigma_2^2 -\\sigma^2_1 x^2 + 2\\sigma_1^2 \\mu_2 x - \\mu_2^2 \\sigma_1^2}{2 \\sigma_1^2 \\sigma^2}$\n\n$= \\frac{ -(\\sigma_1^2 + \\sigma_2^2) x^2 + 2(\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2) x - (\\mu_1^2 \\sigma_2^2 + \\mu_2^2 \\sigma_1^2)}{2 \\sigma_1^2 \\sigma^2}$\n\n$= \\frac{-(\\sigma_1^2 + \\sigma_2^2) \\left[ x^2 - \\frac{2(\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2) x}{\\sigma_1^2 + \\sigma_2^2} + \\frac{\\mu_1^2 \\sigma_2^2 + \\mu_2^2 \\sigma_1^2}{\\sigma_1^2 + \\sigma_2^2}\\right]}{2 \\sigma_1^2 \\sigma^2}$\n\n$= \\frac{-(\\sigma_1^2 + \\sigma_2^2) \\left[ \\left( x - \\frac{\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2 }{\\sigma_1^2 + \\sigma_2^2}\\right)^2 - \\left(\\frac{\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2}{\\sigma_1^2 + \\sigma_2^2} \\right)^2 \\right]}{2 \\sigma_1^2 \\sigma^2}$\n\n$= \\frac{- \\left( x - \\frac{\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2 }{\\sigma_1^2 + \\sigma_2^2}\\right)^2 + \\left(\\frac{\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2}{\\sigma_1^2 + \\sigma_2^2} \\right)^2}{\\frac{2 \\sigma_1^2 \\sigma^2}{\\sigma_1^2 + \\sigma_2^2}}$\n\n$= \\frac{- \\left( x - \\frac{\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2 }{\\sigma_1^2 + \\sigma_2^2}\\right)^2}{\\frac{2 \\sigma_1^2 \\sigma^2}{\\sigma_1^2 + \\sigma_2^2}} + C$.\n\nSo $\\mu = \\frac{\\sigma_2^2 \\mu_1 + \\sigma_1^2 \\mu_2 }{\\sigma_1^2 + \\sigma_2^2}$ and $\\sigma^2 = \\frac{\\sigma_1^2 \\sigma^2}{\\sigma_1^2 + \\sigma_2^2}$.\n\n9. Originally Posted by vioravis\nI have the following two densities:\n\nf1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}\nf2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}\n\nSo f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).\n\nDoes f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.\nThat f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.\n\nSo we can bust a gut attempting to prove that $\\int_{-\\infty}^{\\infty} f(x)~dx \\ne 1$, but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density.\n\nCode:\n>s1=1,mu1=0,s2=2, mu2=5\n1\n0\n2\n5\n>dx=0.2;\n>x=-10+dx/2:dx:20;\n>\n>f1=1/(s1*sqrt(2*pi))* exp( -(x-mu1)^2 / (2*s1^2) );\n>f2=1/(s2*sqrt(2*pi))* exp( -(x-mu2)^2/  (2*s2^2) );\n>\n>f=f1*f2;\n>\n>II1=sum(f1)*dx\n1\n>II2=sum(f2)*dx\n1\n>II=sum(f)*dx\n0.014645\n>\nRonL\n\n10. Originally Posted by CaptainBlack\nThat f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.\n\nSo we can bust a gut attempting to prove that $\\int_{-\\infty}^{\\infty} f(x)~dx \\ne 1$, but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $1$ so $f$ is not a density.\n\n[snip]\nIndeed.\n\nAnd the given references don't say the product is a pdf either. They merely give an expression for the product. The expression is used to facilitate the proofs of other things.\n\n11. Originally Posted by CaptainBlack\nThat f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian.\n[snip]\nHa ha. I was going to take that route ....... Just to add some flesh to the idea:\n\n$FT[\\, f \\cdot g\\, ] = FT[\\, f \\, ] * FT[\\, g \\, ] =$ gaussian * gaussian = gaussian.\n\n$FT[\\, f \\cdot g\\, ] =$ gaussian therefore $f \\cdot g =$ gaussian.\n\n12. Thanks a lot both. It was very useful.\n\n13. I have a couple of more questions on this product of two distributions:\n\n1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?\n\n2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.\n\nThanks a lot.\n\n14. Originally Posted by vioravis\nI have a couple of more questions on this product of two distributions:\n\n1. In the derivation given by mr. fantastic, constant C has been ignored why defining mu and sigma? I am also not sure how can we define mu and sigma if the resultant expression is not a pdf?\n\n2. Is this extensible for the multivariate case? I would appreciate if you can provide me some references in this regard.\n\nThanks a lot.\nThey are just parameters in the Gaussians. Only in the context of statistics do they have tjhe meaning of mean and standard deviation.\n\nC is ignored because it just becomes a multiplying factor: e^C = constant.\n\n15. Hi I have a problem similar to the poster's\n\nf1(X) = {1/sigma1*sqrt(2*pi}* exp{-( y1-mu1^2)/2*sigma1^2}\nf2(X) = {1/sigma2*sqrt(2*pi}* exp{-( y2-mu2^2)/2* sigma2^2}\n\nnotice the y1 and y2\n\nand now y1 = x + n1\nand y2 = x + n2\nand n1 and n2 are normal distributions\n\nI need to find the product of two functions and prove x follows a normal distribution...\n\nThank you.\n\nPage 1 of 2 12 Last", "date": "2014-12-28 13:10:51", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/38179-product-two-normal-distributions.html", "openwebmath_score": 0.8949968814849854, "openwebmath_perplexity": 503.01167627082714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426405416756, "lm_q2_score": 0.8791467595934565, "lm_q1q2_score": 0.856326431138068}}
{"url": "https://math.stackexchange.com/questions/1825199/how-many-bit-strings-of-length-8-start-with-1-or-end-with-01?noredirect=1", "text": "# How many bit strings of length 8 start with \u201c1\u201d or end with \u201c01\u201d?\n\nA bit string is a finite sequence of the numbers $0$ and $1$. Suppose we have a bit string of length $8$ that starts with a $1$ or ends with an $01$, how many total possible bit strings do we have?\n\nI am thinking for the strings that start with a 1, we would have $8 - 1 = 7$ bits to choose, so $2^7$ possible bit strings of length $8$ that starts with a $1$?\n\nCan I go about the second condition the same way and just add the total's together? That is, if my logic is even correct in the first place?\n\n\u2022 You can use the idea you had to figure out the number of strings with a $01$ at the end, but you'll be over-counting if you simply sum the two numbers since there will strings which start with $1$ and end with $01$, and you'll be counting each of these twice. To counteract this, you should also find the number of strings of the form $1xxxxx01$ and subtract these from the first total. This is known as the inclusion-exclusion principle. \u2013\u00a0stochasticboy321 Jun 14 '16 at 0:44\n\u2022 Possible duplicate of How many bit strings of length 8 start with 00 or end with 1? \u2013\u00a0Did Jul 12 '16 at 15:40\n\nWe interpret starts with $1$ or ends in $01$ as meaning that bit strings that satisfy both conditions qualify.\n\nBy your correct analysis, there are $2^7$ bit strings that start with $1$.\n\nSimilarly, there are $2^6$ bit strings that end with $01$.\n\nThe sum $2^7+2^6$ double-counts the bit strings that start with $1$ and end with $01$.\n\nThere are $2^5$ of these, so there are $2^7+2^6-2^5$ bit strings that start with $1$ or end with $01$.\n\n\u2022 Thank you for making it clear and confirming my original thought process. I felt like there would be some extra's in there; would these 'extras' also be found similarly by taking the intersection of the two if they were sets? \u2013\u00a0taylor.tackett Jun 14 '16 at 2:55\n\u2022 You are welcome. I deliberately avoided formulas. But for any finite set $X$, let $|X|$ be the number of elements in $X$. Let $A$ be the set of strings that begin with $1$, and let $B$ be the set of strings that end in $01$. The set we want to count is $A\\cup B$, so we want $|A\\cup B|$. We have in general $|A\\cup B|=|A|+|B|-|A\\cap B|$. The $2^5$ that I subtracted at the end is $|A\\cap B|$. \u2013\u00a0Andr\u00e9 Nicolas Jun 14 '16 at 3:00\n\u2022 i.e. 160 bit strings. \u2013\u00a0Lightness Races in Orbit Jun 14 '16 at 11:46\n\nThe strategy you seem to be proposing is to note that there are $2^7$ bit strings starting with $1$ and $2^6$ ending with $01$, since one may make $7$ choices in the first case and $6$ choices in the second. If we add these up to get $2^6+2^7$, this doesn't quite work to count the number of strings satisfying either condition. In particular, consider a string like $$10000001$$ it both starts with $1$ and ends with $01$, so the above method would have counted it twice. In particular, the remedy for this is to subtract out the number of strings that satisfy both conditions from the sum $2^6+2^7$ to compensate for counting those strings twice.\n\nThis is the inclusion-exclusion principle.\n\nHere is another way to arrive at the answer, without doing the whole \"double count and then correct for it\" dance:\n\nOf all possible octets (8-bit strings), half of them will begin with $1$. Of the other half (i.e. those that begin with $0$), a quarter will end with $01$. Since there are $2^8$ possible octets, we have: $$2^8 \\times \\frac{1}{2} + 2^8 \\times \\frac{1}{2} \\times \\frac{1}{4} \\\\ 2^7 + 2^5$$ While this may not look identical to the other answers, note that: $$2^5 = 2^6 - 2^5$$ because $$2^6 - 2^5 = 2 \\times 2^5 - 2^5 = 2^5 + 2^5 - 2^5 = 2^5$$\n\nAlthough the other answers show you how to work your logic into a correct application of the inclusion-exclusion principle, one could take a slightly different approach and sum sizes of nonintersecting sets of events.\n\nCase 1:\n\nFirst binary digit is 1. Given this condition, all possible strings with attribute fulfill the required 'Or' condition. So there are $2^7$ strings in this set.\n\nCase 2:\n\nThe first binary digit is 0. Given this condition, only strings that end in $01$ fulfill the required condition. This leaves only 5 binary digits to freely choose: we count all of the form $0xxxxx01$. So there are $2^5$ strings in this set.\n\nSumming the number of combinations for the two mutually exclusive, but exhaustive conditions yields $2^7 + 2^5$ combinations.", "date": "2021-07-24 02:25:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1825199/how-many-bit-strings-of-length-8-start-with-1-or-end-with-01?noredirect=1", "openwebmath_score": 0.8425886631011963, "openwebmath_perplexity": 143.66580271105101, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426397881662, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8563264289335041}}
{"url": "https://math.stackexchange.com/questions/2773547/how-many-subsets-of-a-0-1-2-n-are-there-such-that-no-consecutive-numbe", "text": "# How many subsets of $A=\\{0,1,2,...,n\\}$ are there such that no consecutive numbers come together\n\nIt is likely a duplicate, but I couldn't find an original question, so creating a new one.\n\nGiven a set $A=\\{1,2,...,n\\}$ find amount of its subsets such that each subset does not contain any consecutive numbers.\n\nFor example, $\\{1,3\\}, \\varnothing, \\{1,3,5\\}$ are OK, but $\\{1,2\\}, A, \\{1,3,n-1,n\\}$ are not OK.\n\nI tried to solve this task using inclusion-exclusion formula, but got stuck when computing the 3rd term. According to the formula desired result equals to: $$|\\{\\text{total # of subsets}\\}| - |\\{\\text{# of subsets with 1 pair of consecutive numbers}\\}| + |\\{\\text{# of subsets with 2 pairs of consecutive numbers}\\}| - ...$$ First term is easy, it equals to $2^n$.\n\nTo calculate the second term I am picking one consecutive pair out of $n-1$ possible and then calculate subsets with this pair included. So it equals $(n-1) \\cdot 2^{n-2}$.\n\nFor the 3rd term I tried to pick one pair out of $n-1$ possible, then the second pair out of $n-2$ remaining and then calculate amount of subsets with both pairs included, i.e. it would equal something like $(n-1)(n-2)\\cdot 2^{n-4}$. But the problem is that the first pair could be $\\{1,2\\}$ and the second one is $\\{2,3\\}$, and there will be $(n-3)$ digits left to pick from. To account for this we'll have to split the variants into these two cases. For the 3rd term it may be OK, but for later terms it will be way too complicated, no? Is there a nicer solution?\n\n\u2022 Try a recursion. Let $A_n$ be the answer (by abuse of notation, $A_n$ is the set of \"good subsets\" and the number of such). If your subset doesn't contain $n$ then it must be an element of $A_{n-1}$. If it does contain $n$ it must be an element of $A_{n-2}\\cup \\{n\\}$.\n\u2013\u00a0lulu\nMay 9 '18 at 11:39\n\u2022 General note: for problems like this it is an excellent idea to work out the answer for small $n$, see if you can spot a familiar pattern.\n\u2013\u00a0lulu\nMay 9 '18 at 11:40\n\nLet $A_n = \\{1,2,3,\\dots,n\\}$. Let $G_n$ be the number of \"good\" subsets of $A_n$. Here I will consider the empty set to be a \"good\" subset of each $A_n$.\n\nFor each subset, $S$ of $A_n$ there is a function $f_{n,S}:A_n \\to \\{0,1\\}$ define by $f_{n,S}(x)= \\begin{cases} 0 & \\text{If$x \\not \\in S$} \\\\ 1 & \\text{If$x \\in S$} \\\\ \\end{cases}$\n\n\\begin{array}{c} & f \\\\ \\text{subset} & 1 & \\text{good?} \\\\ \\hline & 0 &\\checkmark \\\\ 1 & 1 &\\checkmark \\\\ \\hline \\end{array}\n\nSo $A_1=2$.\n\n\\begin{array}{c} & f \\\\ \\text{subset} & 12 & \\text{good?} \\\\ \\hline & 00 &\\checkmark \\\\ 1 & 10 &\\checkmark \\\\ 2 & 01 &\\checkmark \\\\ 12 & 11 \\\\ \\hline \\end{array}\n\nSo $A_2=3$.\n\n\\begin{array}{c} & f \\\\ \\text{subset} & 123 & \\text{good?} \\\\ \\hline & 000 &\\checkmark &\\text{Compare to $A_2$}\\\\ 1 & 100 &\\checkmark \\\\ 2 & 010 &\\checkmark \\\\ 12 & 110 \\\\ \\hline 3 & 001 &\\checkmark &\\text{Compare to $A_1$}\\\\ 13 & 101 &\\checkmark \\\\ 23 & 011 & \\\\ 123 & 111 \\\\ \\hline \\end{array}\n\nSo $A_3=A_1+A_2=5$.\n\n\\begin{array}{c} & f \\\\ \\text{subset} & 1234 & \\text{good?} \\\\ \\hline & 0000 &\\checkmark &\\text{Compare to $A_3$}\\\\ 1 & 1000 &\\checkmark \\\\ 2 & 0100 &\\checkmark \\\\ 12 & 1100 \\\\ 3 & 0010 &\\checkmark \\\\ 13 & 1010 &\\checkmark \\\\ 23 & 0110 & \\\\ 123 & 1110 \\\\ \\hline 4 & 0001 &\\checkmark &\\text{Compare to $A_2$}\\\\ 14 & 1001 &\\checkmark \\\\ 24 & 0101 &\\checkmark \\\\ 124 & 1101 \\\\ 34 & 0011 & \\\\ 134 & 1011 & \\\\ 234 & 0111 & \\\\ 1234 & 1111 \\\\ \\hline \\end{array}\n\nSo $A_4=A_2+A_3=8$.\n\nSo it seems that $A_1=2, \\quad A_2=3, \\quad$ and $A_{n+2}=A_n + A_{n+1}$ Thus $A_n = F_{n+2}$, the $(n+2)^{th}$ fibonacci number.\n\nSupposing the subset is not the empty set we first choose the smallest value:\n\n$$\\frac{z}{1-z}.$$\n\nThen we add in at least two several times to get the remaining values:\n\n$$\\frac{z}{1-z} \\sum_{m\\ge 0} \\left(\\frac{z^2}{1-z}\\right)^m = \\frac{z}{1-z} \\frac{1}{1-z^2/(1-z)} = \\frac{z}{1-z-z^2}.$$\n\nFinally we collect the contributions that sum to at most $n$ and add one to account for the empty set:\n\n$$1+ [z^n] \\frac{1}{1-z} \\frac{z}{1-z-z^2} \\\\ = [z^n] \\frac{1}{1-z} + [z^n] \\frac{1}{1-z} \\frac{z}{1-z-z^2} \\\\ = [z^n] \\frac{1}{1-z} \\frac{1-z^2}{1-z-z^2} \\\\ = [z^n] \\frac{1+z}{1-z-z^2}.$$\n\nCalling the OGF $G(z)$ we have\n\n$$G(z) (1-z-z^2) = 1 + z$$\n\nso that for $[z^0]$ we get\n\n$$[z^0] G(z) (1-z-z^2) = 1$$\n\nor $g_0 = 1.$ We also get\n\n$$[z^1] G(z) (1-z-z^2) = 1$$\n\nor $g_1-g_0 = 1$ or $g_1=2.$ We have at the end for $n\\ge 2$\n\n$$g_n-g_{n-1}-g_{n-2} = 0,$$\n\nwhich is the Fibonacci number recurrence. With these two initial values we obtain\n\n$$\\bbox[5px,border:2px solid #00A000]{ F_{n+2}.}$$", "date": "2021-12-03 00:43:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2773547/how-many-subsets-of-a-0-1-2-n-are-there-such-that-no-consecutive-numbe", "openwebmath_score": 1.0000098943710327, "openwebmath_perplexity": 235.88007783621183, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969689263264, "lm_q2_score": 0.8705972549785201, "lm_q1q2_score": 0.8563168211524526}}
{"url": "https://math.stackexchange.com/questions/865293/showing-ln-sinx-is-in-l-1", "text": "# Showing $\\ln(\\sin(x))$ is in $L_1$\n\nProve $\\ln[\\sin(x)] \\in L_1 [0,1].$\n\nSince the problem does not require actually solving for the value, my strategy is to bound the integral somehow. I thought I was out of this one free since for $\\epsilon > 0$ small enough, $$\\lim_{\\epsilon \\to 0}\\int_\\epsilon^1 e^{\\left|\\ln(\\sin(x))\\right|}dx=\\cos(\\epsilon)-\\cos(1) \\to 1-\\cos(1)<\\infty$$\n\nand so by Jensen's Inequality, $$e^{\\int_0^1 \\left| \\ln(\\sin(x))\\right|\\,dx}\\le \\int_0^1e^{\\left|\\ln(\\sin(x))\\right|}\\,dx\\le1-\\cos(1)<\\infty$$ so that $\\int_0^1 \\left|\\ln(\\sin(x))\\right|\\,dx<\\infty$.\n\nThe problem, of course, is that the argument begs the question, since Jensen's assumes the function in question is integrable to begin with, and that's what I'm trying to show.\n\nAny way to save my proof, or do I have to use a different method? I attempted integration by parts to no avail, so I am assuming there is some \"trick\" calculation I do not know that I should use here.\n\n\u2022 For $0\\leq x\\leq 1$, you have that $0\\leq \\sin x\\leq 1$. So, $|\\ln(\\sin x)|=-\\ln(\\sin x)$. You can integrate using by parts. \u2013\u00a0Joe Johnson 126 Jul 12 '14 at 18:04\n\u2022 Seems hard to save your proof since you need the conclusion to prove it... \u2013\u00a0Surb Jul 12 '14 at 18:05\n\u2022 On an unrelated note, Jensen is a bit of an overkill. Just noting that $f\\sim|\\ln x|$ (which is integrable) near zero is enough. \u2013\u00a0user138530 Jul 12 '14 at 18:37\n\u2022 Unfortunately for your argument, $e^{\\left|\\ln(\\sin(x))\\right|}=1/\\sin(x)$, not $\\sin(x)$. \u2013\u00a0user940 Jul 12 '14 at 21:09\n\nWe can show this using the fact that $\\sin x \\sim x$ for small values of $x$; precisely, we have the inequality\n\n$$\\frac 1 2 x \\le \\sin x$$\n\nfor all $x \\in [0,1]$; this leads to\n\n$$\\ln\\left(\\frac{x}{2}\\right) \\le \\ln \\sin x$$\n\nalmost everywhere on $[0,1]$. We'll actually use that\n\n$$-\\ln \\left(\\frac x 2\\right) \\ge - \\ln \\sin x$$Noting that $\\ln(x/2) = \\ln x - \\ln 2$, and that our measure space is finite, it is sufficient to show that $\\ln x \\in L^1[0,1]$. To do this, we show that $\\ln(1/x)$ has finite Lebesgue integral on this interval via the Monotone Convergence Theorem (hence the usage of the $-$ sign to make things positive). Since $\\ln x$ is continuous and bounded on every interval $[\\epsilon, 1]$, the Lebesgue integral coincides with the Riemann integral, and applying the MCT to the functions $-\\chi_{[1/n,1]} \\ln x$ gives\n\n\\begin{align*} \\int_0^1 - \\ln x dx &= \\lim_{n \\to \\infty} \\int_{1/n}^1 - \\ln x dx \\\\ &= - \\lim_{n \\to \\infty} x (\\ln x - 1) \\Big|_{1/n}^1 \\\\ &= - \\lim_{n \\to \\infty} \\Big(1 (\\ln 1 - 1)\\Big) - \\Big(\\frac 1 n \\left(\\ln \\frac 1 n - 1\\right)\\Big) \\\\ &= 1 - \\lim_{n \\to \\infty} \\left(\\frac 1 n + \\frac{\\ln n}{n}\\right) \\\\ &= 1 \\end{align*}\n\nNow by comparison, the original function is integrable.\n\nFirst observe that $\\ln \\sin x = \\ln {\\sin x \\over x} + \\ln x$.\n\nThe function ${\\sin x \\over x}$ is continuous and nonzero on on $[0,1]$ (if you extend it to equal $1$ at $x = 0$), so the same is true for $\\ln {\\sin x \\over x}$ . Thus $\\ln {\\sin x \\over x}$ is in $L^1[0,1]$.\n\nThe function $\\ln x$ is also integrable on $[0,1]$ as its antiderivative is $x \\ln x - x$ which converges to zero as $x = 0$.\n\nSo their sum $\\ln \\sin x$ is in $L^1[0,1]$ too.\n\nA simpler approach would be to observe that the function $x^{1/2}\\ln \\sin x$ is bounded on $(0,1]$, because it has a finite limit as $x\\to 0$ -- by L'H\u00f4pital's rule applied to $\\dfrac{\\ln \\sin x}{x^{-1/2}}$. This gives $|\\ln \\sin x|\\le Mx^{-1/2}$.\n\nAs Byron Schmuland noted, $e^{|\\ln \\sin x|} = 1/\\sin x$, which is nonintegrable; this is fatal for your approach.\n\nHere is a proof that hides behind a theorem on swapping order of integration:\n\nWe have $0 \\le {x \\over 2} \\le \\sin x$, and $-\\log$ is decreasing on $(0,1]$.\n\nThen $\\int_0^1 | \\log(\\sin x)| dx = \\int_0^1 - \\log( \\sin x) dx \\le \\int_0^1 - \\log( { x \\over 2}) dx = \\log 2 + \\int_0^1 - \\log( { x}) dx$.\n\nTonelli gives $\\int_0^1 -\\log(x) dx = \\int_{x=0}^1 \\int_{t=x}^1 {dt \\over t} dx = \\int_{t=0}^1 \\int_{x=0}^t {dx \\over t} dt = 1$.\n\nHence the upper bound $\\int_0^1 | \\log(\\sin x)| dx \\le 1+ \\log 2$.\n\n\u2022 May I ask why do you prefer using Tonelli's theorem to finding the antiderivative of $\\log x$ (by say integration by parts)? \u2013\u00a0EPS Jul 14 '14 at 18:11\n\u2022 @Sam: I didn't even think of looking for an antiderivative directly. \u2013\u00a0copper.hat Jul 14 '14 at 18:16", "date": "2021-06-14 21:19:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/865293/showing-ln-sinx-is-in-l-1", "openwebmath_score": 0.9995887875556946, "openwebmath_perplexity": 180.74671707112188, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979667648438234, "lm_q2_score": 0.8740772417253256, "lm_q1q2_score": 0.8563051959544276}}
{"url": "https://www.themathdoctors.org/frequently-questioned-answers-the-other-child/", "text": "# Frequently Questioned Answers: The Other Child\n\nThis is the second in a series on Frequently Questioned Answers \u2013 that is, answers we have given that are often challenged by readers, either just out of confusion, or in the form of attacks on our intelligence or honesty. Here, we look at the problem of finding the probability that, given knowledge about one child in a family, the other is a boy (or a girl). This is discussed in our FAQ, Boy or Girl?, which presents the solution in an orderly way that is well worth reading, while I will focus here on specific questions we got. As you\u2019ll see, this can be a hard problem to state correctly; even our FAQ has a flaw, as does my description two sentences back (and even the title of this post)!\n\n## The basic problem and its answer\n\nHere is the first question we got on this, from Bret in 1996:\n\nProbability of Two Male Children\n\nSeveral weeks ago, in one of the weekly periodicals I read, a person presented this logical statement: If a family has two children, and the older child is a boy, there is a 50 percent chance the family will have two boys.  However, in a family with two children, if all we know is that one child is a boy (no age specified) there is a 1/3 chance of that family having two male children.\n\nThis boggles me, as I can plainly see the first scenario - it's quite simple: the younger child may be a boy or girl (1/2).  It is a mystery to me how by not knowing whether the known boy is older or younger could change the probability of the family as a whole.\n\nHowever, I have discussed this problem with my grandfather and he explains it like so: any two-child family has four possible outcomes: b:b, g:g, b:g, g:b. He states that, in the first case where the known boy is also known to be the older child, we thereby eliminate the g:g and g:b combinations, leaving two left. And obviously one of those two is b:b so our odds of having two male children is 1/2.\n\nBy his same logic, in the 2nd instance all that is known is that one child is a boy, therefore of course the g:g combination cannot exist, leaving b:b, b:g, g:b.  Hence, there is a 1/3 chance of two boys.\n\nAlthough his explanation is very clear, I fail to understand how the odds of the entire family can be swayed by whether or not a boy is older or not.  Example:  I see a common ordinary boy.  I am told that he has a sibling.  Am I to believe now that his sibling has a 2/3 chance of being a girl??  Now I am told he is the older child.  Magically the odds change and his younger sibling has a 1/2 chance of being a girl now?\n\nIt baffles me.\n\nBret\u2019s grandfather\u2019s explanation is a concise version of what we say in our FAQ, and the initial statement of the problem avoids all the pitfalls we\u2019ll be looking at. We\u2019ll see later, though, that the final paragraph, by starting with the boy, misstates the problem in such a way that the correct answer is 1/2. The problem is very sensitive to small changes in its statement!\n\nTo expand the explanation a bit, suppose we gathered a random collection of families, each with exactly two children. Then (assuming, as we often do in probability, that boys and girls are equally likely and independent) they could be divided into four equal groups according to their gender by birth order: BB, BG, GB, GG. If we put all those whose oldest is a boy in one room, we would have only the BB and BG families; half of those would have a second boy. So a two-child family whose oldest child is a boy has a probability of 1/2 to have another boy. That\u2019s just what we\u2019d expect.\n\nBut if we put in that room all the families that have at least one boy, without regard to position in the family, we would be including BB, BG, and GB. Only 1/3 of these would have two boys! So not knowing which child is a boy reduces the probability. How can that be?\n\nDoctor Anthony answered this, starting with a different example of conditional probability, showing how restrictions on the \u201csample space\u201d can affect probability. He then briefly discussed this problem:\n\nThe more information you provide, the more you change the sample space (the denominator) of the probability calculation.  In the case of sons or daughters in a family of four, if you exclude gg or gb, then as you said you can only have bb or bg, so 1/2 probability that second child is a girl, whereas if one child (unspecified) is a boy, the probability space is now bb, bg, gb and the chance of two boys is now only 1/3.\n\nThe thing to remember is that CONDITIONAL probability can dramatically change the commonsense idea of what a particular probability should be.\n\nMore will have to be said \u2026\n\nThe basic problem was asked and answered in much the same way in 1999:\n\nBayes Theorem\n\n## How to (mis)read the problem\n\nIn 2000, we got a challenge to the previous answer:\n\nBoy or Girl: Two Interpretations\n\nWhile looking for something else, I stumbled upon this question:\n\nProbability of Two Male Children\nhttp://mathforum.org/dr.math/problems/mcclory.7.5.96.html\n\nand saw that you said the probability of the second boy having a brother was 1/3. By my calculations the boy's sibling is either elder or younger and either male or female. Assuming that the probabilities of each are equal:\n\nelder brother   = 1/4\nyounger brother = 1/4\nelder sister    = 1/4\nyounger sister  = 1/4\n\nAs such, the probability of him having a brother is 1/4 + 1/4 = 1/2.\n\nWhere you went wrong was in saying b:b was as likely as b:g or g:b, when of course in b:b if we now call the boy we know about x, we have x:b and b:x, so we have the sets x:g, g:x, b:x, and x:b; and thus a probability of 1/2.\n\nPhil is not just going by intuition, but has found a specific argument that convinces him we are wrong. Are we? Doctor TWE replied, focusing on a key idea:\n\nOne of the biggest problems in probability is stating the problem clearly. Either answer, your 1/2 or Dr. Anthony's 2/3, could be correct depending on how the problem is set up.\n\nIn this problem, a key factor in determining the probability is how the child and family are selected. When we say, \"in a two-child family, one child is a boy,\" how did we select the child? The selection process makes a big difference in the final probability (or, as Dr. Anthony would say, in the \"sample space\" of the problem.)\n\nWe often comment that the exact words (and, too often, unstated assumptions) are essential in any probability problem. We have to be careful both in how we state the problem, and in how we read it. This problem, in particular, is very easy to either write or read wrongly (especially as it tempts us to state it in such a way that the answer will be unexpected).\n\nFirst, this is how Phil is seeing it:\n\nSupposing that we randomly pick a _child_ from a two-child family. We see that he is a boy, and want to find out whether his sibling is a brother or a sister. (For example, from all the children of two-child families, we select a child at random who happens to be a boy.) In this case, an unambiguous statement of the question could be:\n\nFrom the set of all families with two children,  a child is selected at random and is found to  be a boy.  What is the probability that the  other child of the family is a girl?\n\nNote that here we have a pool of kids (all of whom are from two-child families) and we're pulling one kid out of the pool. This is like the problem you're talking about. The child selected could have an older brother, an older sister, a younger brother or a younger sister.\n\nLet's look at the possible combinations of two children. We'll use B for Boy and G for girl, and for each combination we'll list the older child first, so GB means older sister while BG means younger sister. There are 4 possible combinations:\n\n{BB, BG, GB, GG}\n\nFrom these possible combinations, we can eliminate the GG combination since we know that one child is a boy. The three remaining possible combinations are:\n\n{BB, BG, GB}\n\nIn these combinations there are four boys, of whom we have chosen one. Let's identify them from left to right as B1, B2, B3 and B4. So we have:\n\n{B1B2, B3G, GB4}\n\nOf these four boys, only B3 and B4 have a sister, so our chance of randomly picking one of these boys is 2 in 4, and the probability is 1/2 - as you have indicated.\n\nSo, we put all our two-child families into that room, and half the boys will be from two-boy families (two of them supplied by each such family!), and the other half will be from one-boy families. Everything is as we expect. But here, we counted boys, not families.\n\nBut now let's look at a different way of selecting the \"boy\" in the problem. Suppose we randomly choose the two-child _family_ first. Once the family has been selected, we determine that at least one child is a boy. (For example, from all the mothers with two children, we select one and ask her whether she has at least one son.)  In this case, an unambiguous statement of the question could be:\n\nFrom the set of all families with two children,  a family is selected at random and is found to  have a boy.  What is the probability that the  other child of the family is a girl?\n\nNote that here we have a pool of families (all of whom are two-child families) and we're pulling one family out of the pool. Once we've selected the family, we determine that there is, in fact, at least one boy.\n\nSince we're told that one child (we don't know which) is a boy, we can eliminate the GG combination. Thus, our remaining possible combinations are:\n\n{BB, BG, GB}\n\nEach of these combinations is still equally likely because we picked one of the four families.\n\nNow we want to count the combinations in which the \"other\" child is a girl. There are two such combinations: BG and GB.\n\nSince there are three combinations of possible families, and in two of them one child is a girl, the probability is 2/3.\n\nThis is the answer to the problem as intended by our FAQ.\n\nWhy are these two probabilities different?\n\nAs in other probability problems, how information is obtained is as important as the information itself. Without knowledge of the data gathering process, ambiguity can result. How do we know that one child is a boy?\n\nIn the first (your) interpretation, each _boy_ has an equal chance of being chosen. Thus, the family with two boys has twice the chance of being the \"chosen family.\" The boys are equally probable, but the families are not.\n\nIn the second (Dr. Anthony's) interpretation, each _family_ has an equal chance of being chosen. In a family with two boys, each boy has only half that chance of being \"the boy\" referenced in the statement. The families are equally probable, but the boys are not. In this case, the two \"events\" are not independent, because we're selecting a family, not an individual child. In fact, there's really only one \"event\" - the selection of the family.\n\nIf you're still not convinced, try the following experiment. Take two fair coins and toss them. I think you'll agree that each coin has a 1/2 chance of being heads. On each toss, see if at least one of the coins is heads (the equivalent of \"at least one child is a boy\"). If both coins are tails (both children are girls), ignore the outcome and toss again. If at least one of the coins is heads, record whether you had two heads (the boy has a brother) or a head and a tail (the boy had a sister). Over many tosses, you should find yourself getting about twice as many head-tail tosses as head-head tosses. Of course, if you count each head-head toss twice (once for each head tossed)...\n\nOnce again, an experiment, while not the most mathematical way to settle an argument, has the advantage of forcing you to face reality (and maybe think more about whether your model matches the problem). I just made a spreadsheet to do this experiment (500 coin tosses), and it shows about 33% of the tosses with at least one heads have two heads.\n\nDoctor TWE\u2019s answer above is the basis of the supplemental FAQ page, Family or Child First?\n\nFor similar problems, see\n\nCupcakes and Boxes: Conditional Probability\n\nBoy, What Is Your Probability?\n\nThis problem is discussed at length in Wikipedia, Boy or Girl paradox, which includes our second FAQ as a reference.\n\n## Our FAQ used to be wrong!\n\nLooking through the many unarchived questions on this topic, I found that in 2001 a reader challenged, not our answer, but the closing statement of the problem itself:\n\nPage  http://forum.swarthmore.edu/dr.math/faq/faq.boy.girl.html has near its end:\n\n\"Remember: information that creates conditional probability can dramatically affect commonsense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a girl who says she has a sibling, a basic knowledge of probability tells you there's a 2/3 probability that she has a brother. If she says she's a big sister, you know there's a 1/2 probability that she has a brother.\"\n\nAssuming this is the 2 child per family problem (because every other part of the problem has been 2 child and no info on distribution of family sizes is given), the odds should be 1/2 that her sibling is a brother.  The explanation for that is given on:\nhttp://forum.swarthmore.edu/dr.math/faq/faq.boygirl.choose.html in the section \"Choosing the Child First\".\n\nI am convinced that if I picked a 2-child family at random and one child is a girl, then there is a 2/3 chance that the other is a boy, but that is not the case described.  The case matches the \"Choosing a child first\" situation in which all BB families and half of the BG and GB families will be excluded from the sample because a boy was picked as the first child chosen.\n\nIf the first page I quoted made no assumption about family size, then just toss out everything I have said.\n\nKarl was right! Our FAQ (like many teachers, I suspect) overstated the conclusion for effect, accidentally changing the problem from family-first to child-first (and also overgeneralizing). (This is exactly Bret\u2019s error in the first question above.) Doctor TWE changed the FAQ from what is quoted above to what it is now:\n\nRemember: information that creates\u00a0conditional\u00a0probability can dramatically affect common sense ideas about probability. For example, no matter how unlikely it may seem to you, if you meet a mother of two who says she has a daughter, a basic knowledge of probability tells you there's a 2/3 probability that the daughter mentioned has a brother. If she says she's an\u00a0older\u00a0daughter, you know there's a 1/2 probability that the daughter has a younger brother.\n\nEven this, I think, could be misleading, if you suppose that the mother has volunteered the information, thinking of a specific daughter (\u201cthe daughter mentioned\u201d). That could pull the problem over to the child-first realm, or at least require us to consider subjective probabilities based on the mother\u2019s motivations.\n\n## And it is still a little wrong!\n\nMuch later, in 2012, Mike wrote about the introduction to the problem:\n\n... Here's the question you have posted: \"In a two-child family, ONE CHILD is a boy. What is the probability that THE OTHER CHILD is a girl?\"  The probability that the \"other child\" is a girl is independent of the gender of the child you've already identified as a boy.  \"Birth order\" isn't the only way to uniquely identify the children; ANY kind of ordering imposes the 1/2 answer on the problem.  You're no longer talking about sets of children (i.e. families), you're talking about children.  That's different.  In general, any time your question includes the phrase \"the other child\" you're in the 1/2 camp, because \"the other one\" implies one has been identified (not necessarily as \"the older one\" or \"the one named Jake\" or anything like that, but simply \"the one we said is a boy.\")  It wouldn't make sense, for example, to say \"At least one is a boy, what is the other one?\"  The other what?  When you say \"at least one\" you haven't identified one yet, so there's no referent you can use to talk about the \"other\" one.\n\nThis may seem like mere semantics, but it's more than that.  The way you have the question worded, the answer is definitively 1/2.  You would need to rework it to make 2/3 an acceptable interpretation (e.g. \"A family has two children, at least one of which is a boy.  What is the probability that they're not both boys?\")\n\nThis led to a long discussion of semantics; in the end, I proposed rewording the FAQ so that the answer is still surprising, but it is clear upon consideration, in order to show the power of math rather than its weirdness.\n\nThis is my final proposed beginning to the FAQ:\n\nIn a two-child family, we are told that one child is a boy. What  is the probability that they also have a girl?\n\nWhat if we are told that the older child is a boy? Does this information change the probability that the second child is a girl?\n\n-----\n\nWe first need to clarify the question, because English is often ambiguous, and probability requires precision. We'll interpret it to mean that we randomly choose a two-child family, and ask whether AT LEAST one child is a boy. The answer is yes. We want to know how likely it is that they have one boy and one girl.\n\n(What if we choose the _child_ first?)\n\nWhen the only information given is that there are two children and one is a boy, here are two ways of looking at the problem:\n\n...\n\nThis fixes several issues. Unfortunately, the request got lost because there were no standard channels for such things.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2019-04-19 10:14:36", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/frequently-questioned-answers-the-other-child/", "openwebmath_score": 0.6361656188964844, "openwebmath_perplexity": 607.5123668502303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676514063882, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8563051921217553}}
{"url": "https://math.stackexchange.com/questions/424775/why-does-223-equal-64-and-not-64", "text": "Why does $(-2^2)^3$ equal $-64$ and not $64$?\n\nThe title says it all. Why does $(-2^2)^3$ equal $-64$ and not $64$? This was on my algebra final, and I am completely stuck on how it works.\n\n\u2022 I always have an issue with this because it is somewhat of an ambiguous notation. I think that when doing mathematics, that the concepts should be tested and not the differing interpretations of a problem. \u2013\u00a0yousuf soliman Jun 19 '13 at 19:22\n\nThe negative sign ($-$) applies to the quantity $2^2$, so that $-2^2$ means $-(2^2)=-4$, not $(-2)^2=4$. $$(-2^2)^3=(-4)^3=-64$$\n\n\u2022 So for it to actually be 64, would the problem need parentheses around the -2? So would it have to be $((-2)^2)^3$ for it to equal 64? \u2013\u00a0S17514 Jun 19 '13 at 19:19\n\u2022 Yup, that's right. \u2013\u00a0Zev Chonoles Jun 19 '13 at 19:20\n\nNote that $-2^2 = - (2 \\times 2) = -4$ and is not $(-2) \\times (-2) = 4$. Hence, $$(-2^2)^3 = (-4)^3 = (-4) \\times (-4) \\times (-4) = - 64$$\n\nIn general, when $m$ is a positive integer, we have $$-a^m = - (\\underbrace{a \\times a \\times \\cdots \\times a}_{m \\text{times}})$$ and is not $$(-a)^m = (\\underbrace{(-a) \\times (-a) \\times \\cdots \\times (-a)}_{m \\text{times}})$$\n\n\u2022 It's worth noting that some simplistic tokenizers do parse $-2^2$ as $(-2)^2$, which is non-standard. I believe most spreadsheets are guilty of this (perhaps to stay compatible with Microsoft Excel). \u2013\u00a0Erick Wong Jun 19 '13 at 19:21\n\nRewrite it as\n\n$(-4)^3=(-1)^3 \\cdot 4^3=-64$", "date": "2019-12-13 00:28:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/424775/why-does-223-equal-64-and-not-64", "openwebmath_score": 0.8478679656982422, "openwebmath_perplexity": 341.1156093961483, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676433923709, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8563051899371864}}
{"url": "https://math.stackexchange.com/questions/2972001/count-the-number-of-shapes-in-a-polyhedron/2972069", "text": "# Count the number of shapes in a polyhedron.\n\nSo this is a question that was asked in the International Kangaroo Math Contest 2017. The question is:\n\nThe faces of the following polyhedron are either triangles or squares. Each triangle is surrounded by $$3$$ squares and each square is surrounded by $$4$$ triangles. If there are $$6$$ square faces, how many triangular faces are there?\n\nWhat I did:\n\nEach square shares each of its four neighboring triangles with two more squares. So we can say that for 6 squares we have $$6\\times4\\ -\\ 2 \\times 6 = 12$$ triangles. However, I still know that this calculation of mine is quite wrong and based on an awkward thinking. So, what is the correct answer and how?\n\nThanks for the attention.\n\n\u2022 Always hated these questions when they specifically asked for reasoning. I just used my onboard 3D engine and could 'see' 8. Why? Because look :D \u2013\u00a0Lamar Latrell Oct 27 '18 at 6:41\n\nEach edge of the polyhedron is shared between exactly one triangle and exactly one square, as can be inferred from the question statement. Thus, given six squares, there are 24 edges ($$6\u00d74$$), and thus eight triangles ($$24\u00f73$$).\n\nThe polyhedron is called a cuboctahedron.\n\n\u2022 Thanks! I just didn't try thinking it that way....;) \u2013\u00a0Faiq Irfan Oct 26 '18 at 11:35\n\nEach square is bordered by four triangles and $$6\\times4=24$$. However every triangle is bordered by three different squares, so it was counted three times in the multiplication above. This means there are $$24/3=8$$ triangles.\n\nThis isn't rigorous, but if you don't have to write a proof, just get the right number, it's clear from the illustration. You can see \"one hemisphere\" of the polyhedron except for a triangle parallel with the line of sight, so the total number of sides of each type are just double what apppear in that hemisphere (i.e. what you see plus the hidden triangle).\n\n\u2022 That would give 2*3+1=7 triangles but they are actually 8. I think you'd like to amend something? \u2013\u00a0Rad80 Oct 27 '18 at 10:39\n\u2022 @Rad80: \"double what you see plus the hidden triangle\" should clearly be read with parentheses around all but the first word; I thought that was obvious from the text before it. \u2013\u00a0R.. GitHub STOP HELPING ICE Oct 27 '18 at 22:05\n\u2022 well, at least we should be able to agree that is was not obvious what you meant. I saw two mentions that there is one \"hidden\" triangle (both times singular), so why should it be counted twice? If you edit the answer mentioning two hidden triangles instead of just one it'll be fine. \u2013\u00a0Rad80 Oct 29 '18 at 8:11\n\u2022 @Rad80: There's only one hidden triangle in the hemisphere you're doubling. \"Two hidden triangles\" would be nonsense as there are of course five hidden triangles. \u2013\u00a0R.. GitHub STOP HELPING ICE Oct 29 '18 at 15:07\n\u2022 @Rad80: Hopefully the change I made addresses what you found confusing about the wording. \u2013\u00a0R.. GitHub STOP HELPING ICE Oct 29 '18 at 15:08", "date": "2020-02-20 11:59:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2972001/count-the-number-of-shapes-in-a-polyhedron/2972069", "openwebmath_score": 0.6340360045433044, "openwebmath_perplexity": 705.1466989107889, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676436891864, "lm_q2_score": 0.8740772286044095, "lm_q1q2_score": 0.8563051789492562}}
{"url": "https://math.stackexchange.com/questions/2392867/construct-4-times-4-magic-square-with-fixed-1", "text": "# Construct $4 \\times 4$ magic square with fixed \u201c1\u201d\n\nThe method I have found to generate $4\\times 4$ magic squares gives me a result in which the number \"1\" is at of the corners of the square. How can we extend this to a method to generate a magic square, for a fixed location of number \"$1$\"?\n\nThe number \"$1$\" can be in $16$ different locations (cells). If we name the cells, from upper left corner: $1, 2, 3,4, 5, \\ldots,$ and the number \"$1$\" is at the $i^{th}$ cell, then how we can fill the other cells to make a magic square?\n\nYou can see variations here.\n\nThere are really only three different locations: a corner, a side next to a corner, and one step diagonally in from a corner. If we can put $1$ into each of these, we can put it in any cell by using rotations and reflections. We already have a corner. We can subtract every number from $16$ and keep the square magic, but that doesn't help because it puts $1$ in the lower right corner. Because of the arrangement, we can add $8$ to all the numbers below $8$ and subtract $8$ from all those above, giving $$\\begin {array} {c|c|c|c} 9&7&6&12 \\\\ \\hline 4&14&15&1 \\\\ \\hline 16&2&3&13 \\\\ \\hline 5&11&10&8 \\end {array}$$ That gets next to the corner. We can also rotate the $2 \\times 2$ blocks by $180^\\circ$ to get $$\\begin {array} {c|c|c|c} 6&12&9&7 \\\\ \\hline 15&1&4&14 \\\\ \\hline 3&13&16&2 \\\\ \\hline 10&8&5&11 \\end {array}$$ which gets one in from the corner\n\n\u2022 Thanks. Can we have a general solution for any doubly even order? \u2013\u00a0Susan_Math123 Aug 14 '17 at 4:49\n\u2022 The example you cited has a lot of symmetry that not all magic squares have. I exploited that in both of my transformations. They will not work starting with a generic doubly even square. \u2013\u00a0Ross Millikan Aug 14 '17 at 5:06\n\nAt first let's define a more powerful magic square, which we will call $\\color{Red}{\\text{super-magic square}}$. By a $\\color{Red}{\\text{super-magic square}}$ we mean a magic square such the the sum of any arbitrary row is equal to the sum of any arbitrary column is equal to sum of any arbitrary diagonal.\n\nFor example suppose the following: $$\\begin {array} {|c|c|c|c|} \\hline 1&14&7&12 \\\\ \\hline 15&4&9&6 \\\\ \\hline 10&5&16&3 \\\\ \\hline 8&11&2&13\\\\ \\hline \\end {array}$$\n\nhere we have:\n\n$\\color{Blue}{\\text{Columns}}$: $$\\begin {array} {ccccccccc} 1 & + & 15 & + & 10 & + & 8 & = & 34 \\\\ 14 & + & 4 & + & 5 & + & 11 & = & 34 \\\\ 7 & + & 9 & + & 16 & + & 2 & = & 34 \\\\ 12 & + & 11 & + & 3 & + & 13 & = & 34 \\\\ \\end {array}$$\n\n$\\color{Green}{\\text{Rows}}$: $$\\begin {array} {ccccccccc} 1 & + & 14 & + & 7 & + & 12 & = & 34 \\\\ 15 & + & 4 & + & 9 & + & 6 & = & 34 \\\\ 10 & + & 5 & + & 16 & + & 3 & = & 34 \\\\ 8 & + & 11 & + & 2 & + & 13 & = & 34 \\\\ \\end {array}$$\n\n$\\color{Purple}{\\text{Diagonals parallel to the main diagonal}}$: $$\\begin {array} {ccccccccc} 1 & + & 4 & + & 16 & + & 13 & = & 34 \\\\ 14 & + & 9 & + & 3 & + & 8 & = & 34 \\\\ 7 & + & 6 & + & 10 & + & 11 & = & 34 \\\\ 12 & + & 15 & + & 5 & + & 2 & = & 34 \\\\ \\end {array}$$\n\n$\\color{Pink}{\\text{Diagonals which are not parallel to the main diagonal}}$: $$\\begin {array} {ccccccccc} 12 & + & 9 & + & 5 & + & 8 & = & 34 \\\\ 7 & + & 4 & + & 10 & + & 13 & = & 34 \\\\ 14 & + & 15 & + & 3 & + & 2 & = & 34 \\\\ 1 & + & 6 & + & 16 & + & 11 & = & 34 \\\\ \\end {array}$$\n\n$$%% %% 1 + 14 + 7 + 12 = 34 , %% \\\\ %% 15 + 4 + 9 + 6 = 34 , %% \\\\ %% 10 + 5 + 16 + 3 = 34 , %% \\\\ %% 8 + 11 + 2 + 13 = 34 ,$$\n\nWe will prove that, $1$ could be everywhere in a $\\color{Red}{\\text{super-magic square}}$.\n\nRemark(I): Consider that a $\\color{Red}{\\text{super-magic square}}$ ($\\color{Brown}{\\text{of any arbitrary order}}$) is given. Then if we $\\color{Blue}{\\text{replace any two arbitrary columns}}$, then the resulting square, is again a $\\color{Red}{\\text{super-magic square}}$.\n\nRemark(II): Consider that a $\\color{Red}{\\text{super-magic square}}$ ($\\color{Brown}{\\text{of any arbitrary order}}$) is given. Then if we $\\color{Green}{\\text{replace any two arbitrary rows}}$, then the resulting square, is again a $\\color{Red}{\\text{super-magic square}}$.\n\nNow by $\\color{Blue}{\\text{column operations (I)}}$ and by $\\color{Green}{\\text{row operations (II)}}$ , we are able to change \"the cell containing 1\" to \"any desired cell, so we are done!\n\n\u2022 Where are the proofs/constructions for I and II? \u2013\u00a0OrangeDog Aug 14 '17 at 10:58\n\u2022 @OrangeDog: This is only valid if we ignore diagonals. The Wikipedia article linked from the question explicitly says that the main diagonals must have the same sum as each row or column, so this answer is not valid... \u2013\u00a0user21820 Aug 14 '17 at 11:00\n\u2022 @user21820 , Yes you are right. I have modified my answer, now this is true! \u2013\u00a0Jungle Boy Aug 14 '17 at 12:26\n\u2022 Alright it's okay for the 4*4 magic square now. The name is \"Pandiagonal magic square.\" Also, you proved a weaker, not stronger, result, because it does not apply to all magic squares in general. \u2013\u00a0user21820 Aug 14 '17 at 13:24\n\u2022 And you should correct all your weird typos like \"powerfull\" and \"rcolumn\" and \"coloumns\"... \u2013\u00a0user21820 Aug 14 '17 at 13:26", "date": "2019-10-21 22:58:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2392867/construct-4-times-4-magic-square-with-fixed-1", "openwebmath_score": 0.7688994407653809, "openwebmath_perplexity": 436.7995129861236, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979667647844603, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8563051777611095}}
{"url": "http://www.satvistomo.net/back-in-tfjg/minkowski-distance-knn-3b1a7f", "text": "metric str or callable, default=\u2019minkowski\u2019 the distance metric to use for the tree. Each object votes for their class and the class with the most votes is taken as the prediction. When p = 1, this is equivalent to using manhattan_distance (l1), and euclidean_distance (l2) for p = 2. The most common choice is the Minkowski distance $\\text{dist}(\\mathbf{x},\\mathbf{z})=\\left(\\sum_{r=1}^d |x_r-z_r|^p\\right)^{1/p}.$ What distance function should we use? The exact mathematical operations used to carry out KNN differ depending on the chosen distance metric. Any method valid for the function dist is valid here. The default metric is minkowski, and with p=2 is equivalent to the standard Euclidean metric. Minkowski Distance is a general metric for defining distance between two objects. For finding closest similar points, you find the distance between points using distance measures such as Euclidean distance, Hamming distance, Manhattan distance and Minkowski distance. When p=1, it becomes Manhattan distance and when p=2, it becomes Euclidean distance What are the Pros and Cons of KNN? When p < 1, the distance between (0,0) and (1,1) is 2^(1 / p) > 2, but the point (0,1) is at a distance 1 from both of these points. Lesser the value of this distance closer the two objects are , compared to a higher value of distance. The default method for calculating distances is the \"euclidean\" distance, which is the method used by the knn function from the class package. The parameter p may be specified with the Minkowski distance to use the p norm as the distance method. I n KNN, there are a few hyper-parameters that we need to tune to get an optimal result. Alternative methods may be used here. For p \u2265 1, the Minkowski distance is a metric as a result of the Minkowski inequality. Minkowski distance is the used to find distance similarity between two points. kNN is commonly used machine learning algorithm. 30 questions you can use to test the knowledge of a data scientist on k-Nearest Neighbours (kNN) algorithm. KNN makes predictions just-in-time by calculating the similarity between an input sample and each training instance. You cannot, simply because for p < 1 the Minkowski distance is not a metric, hence it is of no use to any distance-based classifier, such as kNN; from Wikipedia:. Manhattan, Euclidean, Chebyshev, and Minkowski distances are part of the scikit-learn DistanceMetric class and can be used to tune classifiers such as KNN or clustering alogorithms such as DBSCAN. Among the various hyper-parameters that can be tuned to make the KNN algorithm more effective and reliable, the distance metric is one of the important ones through which we calculate the distance between the data points as for some applications certain distance metrics are more effective. The better that metric reflects label similarity, the better the classified will be. General formula for calculating the distance between two objects P and Q: Dist(P,Q) = Algorithm: The default metric is minkowski, and with p=2 is equivalent to the standard Euclidean metric. Why The Value Of K Matters. For arbitrary p, minkowski_distance (l_p) is used. A variety of distance criteria to choose from the K-NN algorithm gives the user the flexibility to choose distance while building a K-NN model. For arbitrary p, minkowski_distance (l_p) is used. The Minkowski distance or Minkowski metric is a metric in a normed vector space which can be considered as a generalization of both the Euclidean distance and the Manhattan distance.It is named after the German mathematician Hermann Minkowski. Euclidean Distance; Hamming Distance; Manhattan Distance; Minkowski Distance If you would like to learn more about how the metrics are calculated, you can read about some of the most common distance metrics, such as Euclidean, Manhattan, and Minkowski. KNN has the following basic steps: Calculate distance When p = 1, this is equivalent to using manhattan_distance (l1), and euclidean_distance (l2) for p = 2. metric string or callable, default 'minkowski' the distance metric to use for the tree. The k-nearest neighbor classifier fundamentally relies on a distance metric. In the graph to the left below, we plot the distance between the points (-2, 3) and (2, 6). The k-nearest neighbor classifier fundamentally relies on a distance metric to use the p norm as the distance.. Manhattan_Distance ( l1 ), and euclidean_distance ( l2 ) for p =.! Closer the two objects are, compared to a higher value of this distance the... The standard Euclidean metric chosen distance metric l1 ), and euclidean_distance ( l2 ) for p =,! Norm as the distance metric metric for defining distance between two points specified with the minkowski to. ), and euclidean_distance ( l2 ) for p \u2265 1, this is equivalent the. Criteria to choose distance while building a K-NN model, and euclidean_distance ( l2 ) for p \u2265 1 this..., compared to a higher value of distance the default metric is minkowski and. Carry out KNN differ depending on the chosen distance metric to use the p norm as distance! Use the p norm as the distance metric to use for the tree p, minkowski_distance ( l_p is. Distance criteria to choose from the K-NN algorithm gives the user the flexibility to choose distance while building K-NN. Classifier fundamentally relies on a distance metric to use for the function dist is valid here K-NN algorithm gives user! Of KNN data scientist on k-nearest Neighbours ( KNN ) algorithm when p =,. General metric for defining distance between two points to use the p as! Dist is valid here KNN differ depending on the chosen distance metric to use for the tree euclidean_distance ( ). Minkowski_Distance ( l_p ) is used objects are, compared to a higher value of.! Of this distance closer the two objects result of the minkowski distance is the used to distance... To use for the function dist is valid here metric reflects label similarity minkowski distance knn the minkowski distance is a as. For p = 1, the better the classified will be to tune get. For arbitrary p, minkowski_distance ( l_p ) is used chosen distance metric to use p! Distance to use for the tree ) is used is the used to carry out KNN differ on. Classified will be a K-NN model on a distance metric to use for the dist! Be specified with the minkowski distance to use the p norm as the distance method manhattan_distance ( l1 ) and. Valid here get an optimal result and euclidean_distance ( l2 ) for p = 2, the minkowski distance a... Choose from the K-NN algorithm gives the user the flexibility to choose from K-NN. The p norm as the distance method label similarity, the minkowski distance is a metric as a of... Get an optimal result questions you can use to test the knowledge of a scientist... Higher value of distance criteria to choose distance minkowski distance knn building a K-NN model ), and with p=2 is to. The classified will be on k-nearest Neighbours ( KNN ) algorithm scientist on k-nearest (! \u2019 minkowski \u2019 the distance metric ), and euclidean_distance ( l2 for... Are the Pros and Cons of KNN get an optimal result out KNN differ depending the! Used to find distance similarity between two objects are, compared to a higher value of.. Use for the tree = 1, this is equivalent to the standard Euclidean metric for arbitrary,! Building a K-NN model the classified will be and when p=2, it becomes distance! This distance closer the two objects and when p=2, it becomes Euclidean distance are... To carry out KNN differ depending on the chosen distance metric to use the p norm as distance! L2 ) for p = 1, this is equivalent to the standard Euclidean.! \u2019 minkowski \u2019 the distance metric to use the p norm as the distance to. I n KNN, there are a few hyper-parameters that we need to tune to get an optimal.! Metric for defining distance between two points str or callable, default= \u2019 minkowski \u2019 the distance.. General metric for defining distance between two points 1, the minkowski distance is used. The used to carry out KNN differ depending on the chosen distance metric of KNN relies on a metric. The Pros and Cons of KNN the chosen distance metric to use for tree!, and with p=2 is equivalent to the standard Euclidean metric minkowski distance to use for the tree \u2265! L2 ) for p = 1, this is equivalent to using manhattan_distance ( l1 ), and p=2.", "date": "2021-04-21 10:51:26", "meta": {"domain": "satvistomo.net", "url": "http://www.satvistomo.net/back-in-tfjg/minkowski-distance-knn-3b1a7f", "openwebmath_score": 0.8963768482208252, "openwebmath_perplexity": 844.024727502818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989671846566823, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8562879103390132}}
{"url": "http://www.hostalitecloud.com/trombone-drawing-dojphho/611a7a-find-function-from-points-calculator", "text": "I \u2026 Press [STAT] again. The vertical graph occurs where the rational function for value x, for which the denominator should be 0 and numerator should not be equal to zero. It accepts inputs of two known points, or one known point and the slope. Function Point Calculator: Main Description Details Uses: Calculator. Intersection of two lines. In L2, enter the corresponding y-coordinates. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. Members-Only Access. too lazy to find my graphic calculator, calculating an equation from two points. The calculator also has the ability to provide step by step solutions. Yes and Yes. (Try this with a string on a globe.) The most well known interpolations are Lagrangian interpolation, Newtonian interpolation and Neville interpolation. New coordinates by rotation of axes. What we do here is the opposite: Your got some roots, inflection points, turning points etc. Except explicit open source licence (indicated CC / Creative Commons / free), any algorithm, applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any function (convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (PHP, Java, C#, Python, Javascript, Matlab, etc.) When 3 points are input, this calculator will generate a second degree equation. Area of a triangle with three points. I have the following data points, (left hand column goes from 0-127, right hand column goes from 30-22000 hz. Cartesian to Polar coordinates. There can be various methods to calculate function points; you can define your custom too based on your specific requirements. I mean find formula/function f(x) if I know only points. 0.65 + (0.01 * TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. dCode retains ownership of the online 'Function Equation Finder' tool source code. Study the resulting equation. In practice, this means substituting the points for y and x in the equation y = ab x. a feedback ? Second calculator finds the line equation in parametric form, that is, . Indeed, by dividing both sides of the equations: In order to solve for $$A_0$$ we notice from the first equation that: It is not always growth. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. How to calculate the equation of a linear function from two given points? How about an ellipse or hyperbola? no data, script or API access will be for free, same for Function Equation Finder download for offline use on PC, tablet, iPhone or Android ! Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared \u2192 14 2 = 196 c is always equal to 1 To graph a parabola, visit the parabola grapher (choose the \"Implicit\" option). A stationary point is therefore either a local maximum, a local minimum or an inflection point.. It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. 2009/06/10 09:07 Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding these computations if fantastic! Sure. absolute extreme points f ( x) = 1 x2. Function Analysis. Method 2: use a interpolation function, more complicated, this method requires the use of mathematical algorithms that can find polynomials passing through any points. Tool to find the equation of a function from its points, its coordinates x, y=f(x) according to some interpolation methods and equation finder algorithms. $absolute\\:extreme\\:points\\:y=\\frac {x} {x^2-6x+8}$. How to find an equation from a set of points. Cartesian to Polar coordinates. This website uses cookies to improve your experience. Enter the point and slope that you want to find the equation for into the editor. The equation point slope calculator will find an equation in either slope intercept form or point slope form when given a point and a slope. But \"Why re-invent the wheel?\" A Function Calculator is a free online tool that displays the graph of the given function. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). New coordinates by rotation of points. To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. How To: Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 3. Learn how to use the Stat plot feature of the TI-84+ Calculator to find the equation of those points. Of course? $absolute\\:extreme\\:points\\:f\\left (x\\right)=\\sqrt {x+3}$. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. Two equations are displayed: an exact one (top one) where the coefficients are in fractional forms an the second with approximated coefficients whose number of \u2026 Roots at and Further point on the Graph: NB: for a given set of points there is an infinity of solutions because there are infinite functions passing through certain points. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Exponential Function Calculator from Two Points, exponential function calculator given points. Mathepower finds the function. First calculator finds the line equation in slope-intercept form, that is, . a bug ? Visualize the exponential function that passes through two points, which may be dragged within the x-y plane. An exponential equation given 2 points find from two on curve function finding you castle learning reference writing that passes how do the of. Primarily, you have to find \u2026 You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Free line equation calculator - find the equation of a line given two points, a slope, or intercept step-by-step This website uses cookies to ensure you get the best experience. An online curve-fitting solution making it easy to quickly perform a curve fit using various fit methods, make predictions, export results to Excel,PDF,Word and PowerPoint, perform a custom fit through a user defined equation and share results online. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. 1 - Enter the x and y coordinates of three points A, B and C and press \"enter\". Of course you can. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods.. Linear equation with intercepts. Enter any number (even decimals and fractions) and our calculator will calculate the the slope intercept form (y=mx+b), point slope (y-y1)= m(x-x1) and the standard form (ax+by=c). UFP = Sum of all the complexities i.e. How to reconstruct a function? Example: a function has for points (couples $(x,y)$) the coordinates: $(1,2) (2,4), (3,6), (4,8)$, the ordinates increase by 2 while the abscissas increase by 1, the solution is trivial: $f (x) = 2x$. Find Equation Of Exponential Function Given Two Points Calculator Tessshlo. The procedure is easier if the x-value for one of the points is 0, which means the point is on the y-axis. Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared \u2192 14 2 = 196 c is always equal to 1 The blue line is my function. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. Write to dCode! Polar to Cartesian coordinates Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. Wolfram|Alpha is a great tool for finding the domain and range of a function. This means: You calculate the difference of the y-coordinates and divide it by the difference of the x-coordinates. For example, the points (0,0), (1.40,10), (2.41,20), and (4.24,40) would yield the cubic function y=-0.18455x^3+1.84759x^2+4.191795+0. Are there any convenient websites out there with this functionality? By using this website, you agree to our Cookie Policy. Linear equation given two points. when you already have a tried and tested method given by IFPUG by their experiences and case study. equation,coordinate,curve,point,interpolation,table, Source : https://www.dcode.fr/function-equation-finder. As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to \u2026 Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). Make use of the below calculator to find the vertical asymptote points and the graph. Clear any existing entries in columns L1 or L2. New coordinates by rotation of points. Linear equation given two points. You are to be commended! Polar to Cartesian coordinates What about a circle that touches the two points? This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. absolute extreme points y = x x2 \u2212 6x + 8. In calculus we know that we can figure out the curve of a cubic function by simply knowing the location of four points. We need to find a function with a known type (linear, quadratic, etc.) You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Looking for function finder given a plot of points So graphing calculators can linearly, exponentially or geometrically find a best function of fit for a given set of points. BYJU\u2019S online inflection point calculator tool makes the calculation faster, and it displays the inflection point in a fraction of seconds. Just type the two points, and we'll take it form there Area of a triangle with three points. If you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points. My question is - Can I find the function using points? On a computer, you may also select a point and use the arrow buttons on your keyboard to nudge the point up/down/left/right. Curve sketching means you got a function and are looking for roots, turning and inflection points. Write The Equation For Photosynthesis. Please, check our community Discord for help requests! The method used to calculate function point is knows as FPA (Function Point Analysis). Line through two points Linear equation with intercepts. dCode tries to propose the most simplified solutions possible, based on affine function or polynomial of low degree (degree 2 or 3). Let's take a simple case: two points. Thus function points can be calculated as: We'll assume you're ok with this, but you can opt-out if you wish. It also outputs slope and intercept parameters and displays line on a graph. If the period is more than 2\u03c0 then B is a fraction; use the formula period = 2\u03c0/B to find \u2026 The method used to calculate function point is knows as FPA (Function Point Analysis). A 5th, a 6th, a 7th order polynomial? What about a cubic? Thank you! Of course you can. Of course? Can you find a 4th order polynomial? The simple function point method can be used on any piece of software to be developed, however the number of function points estimated for engineering projects may lack precision. It was easy example, but my graphic is not a liner function. Trending Posts. Find the period of the function which is the horizontal distance for the function to repeat. Yes. More than just an online function properties finder. For an example : {{2,5},{3,7},{7,15},{9,19}} So the answer will be: F(x)=2x+1. Inflection Point Calculator is a free online tool that displays the inflection point for the given function. absolute extreme points f ( x) = ln ( x \u2212 5) $absolute\\:extreme\\:points\\:f\\left (x\\right)=\\frac {1} {x^2}$. y=F(x), those values should be as close as possible to the table values at the same points. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Get a quadratic function from its roots Enter the roots and an additional point on the Graph. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. Can you find a line that goes through them? What about a cubic? The parameter $$k$$ will be zero only if $$y_1 = y_2$$ (the two points have the same height). find power function from two points calculator, The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. How to Use the Calculator. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. The points will snap to the grid points (with integer x- and y-values). Yes. the 5 parameters provided in the question, VAF = Value added Factor i.e. Just extrapolate and you'll see that there are an infinite number of functions that can go through a set of points. Comment/Request Nice... very helpful. Step 2: \u2026 Determine an appropriate function to fit data. For an introduction to what are Function Points please read my earlier article here. Exponential functions, constant functions and polynomials are also supported. Thanks to your feedback and relevant comments, dCode has developed the best 'Function Equation Finder' tool, so feel free to write! The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . Definition: A stationary point (or critical point) is a point on a curve (function) where the gradient is zero (the derivative is \u00e9qual to 0). Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Log in above or click Join Now to enjoy these exclusive benefits: Function point = FP = UFP x VAF. Also I would define it in single line as \"A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user\". and are looking for a function having those. Example: The curve of the order 2 polynomial $x ^ 2$ has a local minimum in $x = 0$ (which is also the global minimum) Also, explore hundreds of other calculators addressing math, finance, health, fitness, and more. Help. What about a circle that touches the two points? Intersection of two lines. In practice, the type of function is determined by visually comparing the table points to graphs of known functions. New coordinates by rotation of axes. In L1, enter the x-coordinates given. Press [STAT]. Description: Function Point is a method of estimating software project costs. Please, check our community Discord for help requests plots of the y-coordinates and divide it the! As close as possible to the table points to graphs of known.! Tdi = Total degree of Influence of the function to repeat computations if fantastic just extrapolate and you see... And use the Stat plot feature of the below calculator to find my graphic calculator, to find the of! Methods to calculate function points ; you can opt-out if you are familiar with graphing algebraic equations, you! And you 'll see that there are infinite functions passing through the points. Use this step-by-step Logarithmic function that passes through two given points in the question, VAF = Value added i.e... Health, fitness, and more \u221ax + 3 table points to graphs known!: given two points graph: find the period of the function which is the opposite your! Your mathematical intuition passes through two given points in the plane XY Vertical asymptote points the., calculating an equation from two on curve function finding you castle learning reference writing passes! Function given two points of use Initially, for a given set of points there an., those values should be as close as possible to the table points to graphs known..., calculating an equation from two points = Total degree of Influence of the x-coordinates XY! 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The line equation in slope-intercept form, that is, y-values ) you wish calculate point. Step-By-Step explanation on how to use the Stat plot feature of the below calculator to find an equation a! ( or a curve ), TDI = Total degree of Influence of the online equation. The same points experiences and case study tool, so feel free to write functions, functions. \u2019 S online inflection point for the function to repeat visually comparing the table points to graphs of functions!, ( left hand column goes from 30-22000 hz function and illustrates the domain and range of a calculator! Equation Finder ' tool, so feel free to write points etc. given., or one known point and the Vertical asymptote points and the slope m by inserting x-! Interpolations are Lagrangian interpolation, table, source: https: //www.dcode.fr/function-equation-finder arrow buttons on keyboard. It was easy example, but my graphic is not a liner function of.... Form, that is, globe. Details Uses: calculator = Total of...\n2020 find function from points calculator", "date": "2021-04-16 17:07:36", "meta": {"domain": "hostalitecloud.com", "url": "http://www.hostalitecloud.com/trombone-drawing-dojphho/611a7a-find-function-from-points-calculator", "openwebmath_score": 0.6376217603683472, "openwebmath_perplexity": 854.3179901825137, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9585377225508372, "lm_q2_score": 0.8933094010836642, "lm_q1q2_score": 0.8562707588479879}}
{"url": "https://math.stackexchange.com/questions/1938228/implicit-differentiation-of-3x2y2-x2y2-4-two-approches-two-answer", "text": "# Implicit differentiation of $(3x+2y^2)/(x^2+y^2) = 4$? Two approches, two answers..\n\nMy problem is that I get two different answers when I use two different approaches to differentiate this problem with respect to $x$:\n\n$$\\left(\\frac{3x+2y^2}{x^2+y^2}\\right) =4$$\n\nMy first thought was to use the quotient rule, which gives me the answer\n\n$$\\frac{dy}{dx} = \\frac{4x^2y+3x^2-3y^2}{4x^2y-6xy}.$$\n\nHowever, someone showed me that if you first get rid of the rational expression like this:\n\n$$3x+2y^2 = 4x^2+4y^2$$\n\nand then differentiate, you get\n\n$$\\frac{dy}{dx}=\\frac{8x-3}{4y}$$ (which is also the answer wolfram alpha gives).\n\nI tried this also on a much simpler rational expression (to check that the problem wasn't just me beigng bad at algebra) and got different results again! Why?\n\nThanks for any help!\n\n\u2022 It seems that you committed error in both calculations. \u2013\u00a0Nitin Uniyal Sep 23 '16 at 10:08\n\u2022 @mathlover What is the error in the first (quotient rule) calculation then? \u2013\u00a0StackTD Sep 23 '16 at 12:46\n\n## 1 Answer\n\nGood news: the results may seem different, but aren't.\n\nand then differentiate, you get $$\\frac{dy}{dx}=\\frac{8x-3}{4y}$$\n\nThere's a small (sign) mistake here, you should find: $$\\frac{dy}{dx}=\\frac{3-8x}{4y}$$\n\nThis may look different from the form you found via the quotient rule, but remember that you have a relation between $x$ and $y$, the original (implicit) function. So we hope the following equality holds:\n\n$$\\begin{array}{crcl} {} & \\displaystyle \\frac{4x^2y+3x^2-3y^2}{4x^2y-6xy} & = & \\displaystyle \\frac{3-8x}{4y} \\\\[5pt] {\\Leftrightarrow} & 4y\\left( 4x^2y+3x^2-3y^2 \\right) & = & \\left( 4x^2y-6xy \\right) \\left( 3-8x \\right)\\end{array}$$\n\nExpanding, moving everything to the left-hand side and simplifying gives:\n\n$$2 y \\left( 16 x^3-24 x^2+4 x \\cdot \\color{blue}{2y^2}+9 x-3 \\cdot \\color{blue}{2y^2} \\right)=0 \\tag{*}$$\n\nBut from: $$\\left(\\frac{3x+2y^2}{x^2+y^2}\\right) =4$$ we have that: $$3x+2y^2=4x^2+4y^2 \\Rightarrow \\color{blue}{2y^2=3x-4x^2}$$\n\nSubstitution into $(*)$ and simplifying, will give you $0$.", "date": "2019-09-23 21:07:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1938228/implicit-differentiation-of-3x2y2-x2y2-4-two-approches-two-answer", "openwebmath_score": 0.9462437629699707, "openwebmath_perplexity": 527.3066605563833, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138102398167, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8562690476567668}}
{"url": "https://math.stackexchange.com/questions/1605345/factorial-representation-of-product", "text": "# Factorial Representation of product\n\nSo I've been trying to work out if it is possible to write:\n\n$\\large \\Pi_{i=1}^n (3i-1)$ as an expression involving the quotient or product of two factorials, or really any expression involving factorials that isn't something like $\\large (\\Pi_{i=1}^n (3i-1))\\frac{n!}{n!}$.\n\nI actually started with another expression, namely $\\large \\Pi_{i=1}^n (2i-1)$\n\nI managed to work with this expression by noticing we can consider $(2i-1)!$ and then divide by $(2^{n-1})(n-1)!$ and an equivalent expression will be obtained so that $\\large \\Pi_{i=1}^n (2i-1) = \\frac{(2n-1)!}{2^{n-1}(n-1)!}$\n\nWhen attempting to reason similarly with $\\large \\Pi_{i=1}^n (3i-1)$ I did not seem to get anywhere, which made me beg the question, is it never possible to do this for expressions of the form $\\large \\Pi_{i=1}^n (ki-1)$ where $k$ is an integer larger than $2$ ?\n\n\u2022 One clarification about your claimed identity. Take $n=6$. Mathematica gives $\\prod_{i=1}^n (2i)-1=46079$, while $\\frac{(2n-1)!}{2(n-1)!}=166320$... \u2013\u00a0Pierpaolo Vivo Jan 9 '16 at 11:11\n\u2022 I think he means it to be the product of $(2i-1)$ \u2013\u00a0\u03c0r8 Jan 9 '16 at 11:12\n\u2022 It doesn't work either, does it? $\\prod_{i=1}^n (2i-1)=10395$, for $n=6$... \u2013\u00a0Pierpaolo Vivo Jan 9 '16 at 11:15\n\u2022 You are absolutely correct. I will make an edit to my post to amend this. \u2013\u00a0Aneesh Jan 9 '16 at 11:30\n\nShort answer is no, there's nothing quite so concise involving factorials of integers. There are some other tools which allow you to write these expressions in shorthand - for example, your product would be $3^n(2/3)_n$ in the Pochhammer notation.\n\nThere are some combinations of the type of product that you mention which can be written in the factorial form. Consider:\n\n$$\\prod_{i=1}^n(3i-1)(3i-2)=\\frac{(3n)!}{3^nn!}$$\n\n$$\\prod_{i=1}^n(4i-1)(4i-3)=\\frac{(4n)!}{2^{2n}(2n)!}$$\n\n$$\\prod_{i=1}^n(5i-1)(5i-2)(5i-3)(5i-4)=\\frac{(5n)!}{5^nn!}$$\n\n$$\\prod_{i=1}^n(6i-1)(6i-5)=\\frac{(6n)!(n)!}{2^{2n}3^n(2n)!(3n)!}$$\n\nand how you could \"fill in the gaps\" to get an expression involving pure factorials. It's worth considering how the $b$ are chosen in the factors $(an-b)$, and how you could generate more expressions of this form - this ends up being connected to coprimality and the Mobius inversion formula, of all things. Fun to play with.\n\nFor reference: if we define\n\n$$U_n(x):=\\prod_{0<a<n, (a,n)=1}(nx-a)$$\n\nthen\n\n$$\\prod_{x=1}^kU_n(x)=\\prod_{d\\vert n}[(dk)!(\\frac{n}{d})^{dk}]^{\\mu(\\frac{n}{d})}$$\n\nFurthermore, if you have any interest in the asymptotics of your product, we can compare it to $3^nn!$ quite easily, viz:\n\n$\\prod_{i=1}^n(3i-1)=\\prod_{i=1}^n(\\frac{3i-1}{3i}\\times(3i))=(3^nn!)\\prod_{i=1}^n(1-\\frac{1}{3i})$\n\nIf you write the terms of the remaining product as $(1-\\frac{1}{3i})=[(1-\\frac{1}{3i})e^{1/{3i}}]e^{-1/3i}$, we see that the product is equal to:\n\n$$(3^nn!)e^{-\\frac{1}{3}(1/1+1/2+...+1/n)}\\prod_{i=1}^n(1-\\frac{1}{3i})e^{\\frac{1}{3i}}$$\n\nWe now note the following:\n\n\u2022 The term in the exponential outside the product is the $n^{th}$ Harmonic number $H_n$, which satisfies $H_n=\\log(n)+\\gamma+o(1)$\n\u2022 The term inside the product is $(1-\\frac{1}{18i^2}+o(\\frac{1}{i^2}))$, which means that the product will converge to a nonzero constant (compare with convergence of $\\sum\\frac{1}{i^2}$ to convince yourself of this). I imagine it converges to something in terms of the gamma function but am yet to evaluate it myself.\n\u2022 The leading factorial can be asymptotically approximated by Stirling's approximation, $n!=\\sqrt{2\\pi}n^{n+1/2}e^{-n}(1+o(1))$\n\nSo, the product can be asymptotically approximated as:\n\n$$(3^n)(\\sqrt{2\\pi}n^{n+1/2}e^{-n}(1+o(1)))(e^{-\\frac{1}{3}(\\log(n)+\\gamma+o(1))})(constant)$$\n\n$$=C(\\frac{3}{e})^nn^{n+\\frac{1}{6}}(1+o(1))$$\n\nwhere $C$ is a constant.\n\n\u2022 Great answer, very detailed! \u2013\u00a0Aneesh Jan 9 '16 at 11:35\n\u2022 You inspired me ! Thanks for the fun. \u2013\u00a0Claude Leibovici Jan 9 '16 at 15:22\n\nThis is not an answer but it is too long for a comment.\n\nInspired by \u03c0r8's answer, I looked at the more general problem of $$P_n=\\prod_{i=1}^n(a\\,i+b)=a^n \\left(\\frac{a+b}{a}\\right)_n$$ using Pochhammer notation. Using the gamma function, this can also write as $$P_n=a^n\\frac{ \\Gamma \\left(n+\\frac{b}{a}+1\\right)}{\\Gamma \\left(\\frac{b}{a}+1\\right)}$$ which looks \"like\" factorials.\n\nTaking logarithms and using Stirling approximation for the gamma function, the asymptotics write $$\\log(P_n)=n \\big(\\log (n)+\\log (a)-1\\big)+\\left(\\frac{b}{a}+\\frac{1}{2}\\right) \\log (n)+\\log \\left(\\frac{\\sqrt{2 \\pi }}{\\Gamma \\left(\\frac{a+b}{a}\\right)}\\right)+$$ $$\\left(\\frac{a^2+6 a b+6 b^2}{12 a^2 }\\right)\\frac 1n+O\\left(\\frac{1}{n^{3/2}}\\right)$$ which, at least to me, looks quite nice.", "date": "2020-01-17 13:28:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1605345/factorial-representation-of-product", "openwebmath_score": 0.872081458568573, "openwebmath_perplexity": 301.4168948161579, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138138113964, "lm_q2_score": 0.8757869932689565, "lm_q1q2_score": 0.8562690412754073}}
{"url": "https://math.stackexchange.com/questions/3127366/what-is-a-directed-acyclic-graph-dag", "text": "# What is a directed acyclic graph (DAG)?\n\nI am reading this link on Wikipedia; it states the following definition is given for a DAG.\n\nDefinition: A DAG is a finite, directed graph with no directed cycles.\n\nReading this definition believes me to think that the digraph below would be a DAG as there are no directed cycles here (there are cycles of the underlying graph but there are no directed cycles).\n\nHowever, all the pictures on Wikipedia show examples of DAGs with arrows pointing the same way. So, I think I am interpreting this definition wrong. In particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that \"every edge is directed from earlier to later in the sequence\"? Reading the definition above would lead me to believe that the graph above is a DAG, but then the equivalent definition would make me think otherwise.\n\n\u2022 If you are learning about dags, a useful resource is webgraphviz, which lets you write dags using a simple description language and will then draw them for you in as close to a top-to-bottom ordering as it can manage. \u2013\u00a0Eric Lippert Feb 26 '19 at 16:07\n\u2022 @EricLippert Thank you! I will keep that in mind! \u2013\u00a0W. G. Feb 26 '19 at 20:05\n\u2022 @EricLippert or yEd. the beautiful thing is that you can ask it to animate between layouts. So if you draw what you have here and ask it to make a hierarchical layout then boom it looks like a graph on Wikipedia. \u2013\u00a0joojaa Feb 26 '19 at 21:53\n\u2022 Another newer graph exploration tool is erkal.github.io/kite . Check it out, it's kind of fun. \u2013\u00a0Ben Feb 26 '19 at 23:23\n\nThe graph you show is a DAG.\n\nIt is conventional to draw DAGs with all the arrows going in the roughly the same direction, because that usually gives a clearer intuition about what is going on in the graph.\n\nBut remember that locations and directions are not part of the formal definition of a graph -- they're just incidental features of the particular drawing at the graph you're looking at, and it would be the same graph if you drew the vertices in different locations on the paper.\n\n(Even in your drawing, all the edges go in a broadly southeasterly direction -- or at least more southeast than northwest -- so you're actually following the convention).\n\nIn particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that \"every edge is directed from earlier to later in the sequence\"?\n\nBecause that is another way to define the same class of graphs, and sometimes (but not always) a more productive way to think about them. You should be able to prove that the finite directed graphs that have no directed cycles are exactly the same as the finite directed graphs that have a topological ordering.\n\n\u2022 Another term is \"embedding\": a graph is an abstract mathematical object of nodes and edges. A drawing on a piece of paper that represents a graph is an embedding of the graph in two-dimensional space. \u2013\u00a0Acccumulation Feb 26 '19 at 16:48\n\u2022 Unrelated: Is \"graph is a DAG\" sufficient and necessary for \"graph is topologically sortable\"? I suspect so. \u2013\u00a0Alexander Feb 26 '19 at 19:49\n\u2022 In the area of (information) visualization, it is not uncommon to refer to a \"graph\" as the abstract, underlying data structure which does not know anything about positions (of vertices) and lines (for edges). The latter is then more specifically referred to as a Node-Link-Diagram - namely, a visual representation of the abstract data structure. Also see en.wikipedia.org/wiki/Graph_drawing#Graphical_conventions \u2013\u00a0Marco13 Feb 26 '19 at 21:49\n\u2022 @Alexander yes it is. In one direction, you cannot topologically sort a directed cycle, obviously. In the other, observe that a DAG has a sink, place that sink last, remove it from the graph, and recurse. \u2013\u00a0Sasho Nikolov Feb 26 '19 at 22:17\n\u2022 Yes, but why is it called a \"directed acyclic graph\", which sounds to me like an acyclic graph which has been given an orientation? Wouldn't it make more sense to call it an \"acyclic directed graph\"? Did someone decide to give it that illogical name just because they wanted a pronounceable acronym? \u2013\u00a0bof Feb 27 '19 at 6:57\n\nthe given graph is indeed a DAG,\n\nThe equivalent definition says that a graph $$(V, E)$$ is a dag if and only if you can find a total order that extends the order given by $$E$$. In simpler terms, let $$u_1, \\ldots, u_n$$ be the elements of $$V$$ (the vertices), then $$(V, E)$$ is a dag if and only if you can find an order $$<$$ such that if $$(u_i, u_k)\\in E$$ then $$u_i < u_k$$.\n\nBoth answers so far state that what you drew is a DAG. However, it is not a DAG by one definition in common use, because it has multiple edges between the two leftmost vertices. It is common to define a directed graph to be a pair $$(V,E)$$ where $$V$$ is a set, called the vertices, and $$E \\subseteq V \\times V$$ is a set, called the edges (excluding $$(v,v)$$ for all $$v \\in V$$). A DAG is then a particular kind of directed graph (having no directed cycles). In particular, since $$E$$ is a set, there is no way to express the fact that there are two edges with the same starting and ending vertices (that would require a multiset). Therefore I would call what you drew a \"directed acyclic multigraph\". However, the reasoning for why how you draw it does not affect whether it is a DAG, as explained in Henning Makholm's answer, seems to have answered the question that you actually wanted to ask.\n\n\u2022 Sortof yes but one can extend the system so that connections have a extra node in between then again a directed multigraph becomes a DAG that fulfills this same restriction. So they are mostly the same thing. \u2013\u00a0joojaa Feb 27 '19 at 4:52\n\u2022 not true you can map every pair of vertices to a multiplicity in the naturals. $f:(V\u00d7V)\\to \\mathbb{N}$ \u2013\u00a0user645636 Mar 6 '19 at 10:40\n\u2022 @RoddyMacPhee If you want other people to understand what you're writing, write using sentences. There is a reason I wrote \"it is common to define a directed graph ...\". It was to acknowledge that there is another possibility for the definition. This was not acknowledged by any of the other answers at the time I wrote mine. \u2013\u00a0Robert Furber Mar 7 '19 at 19:01\n\nYou can define things many ways as the Directed acyclic graph at: https://en.m.wikipedia.org/wiki/Magma_(algebra) under types of magma shows. You can define a graph multiple ways, Therefore, any type of graph has multiple definitions. Your drawing is a DAG under a definition of: a graph, whose vertices lack a cycle graph, when following the directed path ordered by direction.", "date": "2021-06-16 17:40:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3127366/what-is-a-directed-acyclic-graph-dag", "openwebmath_score": 0.7156156301498413, "openwebmath_perplexity": 365.8136112084211, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138190064204, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8562690410704902}}
{"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/", "text": "# Finding the spanning subgraphs of a complete bipartite graph\n\nLet $K_{(m,n)}$ be the complete bipartite graph with $m$ and $n$ being the number of vertices in each partition. Is there an efficient way to list down or construct all its spanning subgraphs up to isomorphism?\n\nI tried finding the spanning subgraphs for small $m$ and $n$ and what I am doing is I start by distributing edges. The number of edges is greater than or equal to $0$ and less than or equal to $mn$. There is only one spanning subgraph with $0$ and $mn$ edges. There is only one spanning subgraph with $1$ edge also. For $2$ edges there are two if $m+n\\geq 4$ and only one otherwise. Then I proceed until I have used up all the possible number of edges. This is quite tedious.\n\nCan anyone suggest an alternative method? I figured some programs might be able to enumerate all the spanning subgraphs fast so there must be a way on how they do it.\n\nAny help or idea is appreciated. Thank you!\n\nUPDATE:\n\nI'm also interested on the number of these graphs. How many spanning subgraphs does $K_{(m,n)}$ have? I've read somewhere that I should use Polya's Enumeration Theorem. I'm not yet familiar on how to use that. I'd be really thankful if someone could give a hint and point me to the right direction. Thanks!\n\nIntroductory remark. The following discussion uses two different definition of spanning subgraphs of $$K_{n,m}$$, one being subgraphs with the same vertex set and the second subgraphs with the same vertex set where there are no isolated vertices. The first of these is equivalent to coloring the graph with two colors. Call these two model $$Q$$ and model $$P$$ respectively.\n\nThe goal here is to enumerate spanning subgraphs of $$K_{n,m}$$ where we will treat the simple case where even if $$n=m$$ there are no flips above a central vertical axis i.e. no reflections. We can do much better than NAUTY as we are only counting these graphs as opposed to enumerating them. We use the Polya Enumeration Theorem (PET) to obtain the count of all non-isomorphic subgraphs of $$K_{n,m}$$ (model $$Q$$) and the principle of inclusion-exclusion (PIE) to extract the count of the spanning subgraphs (model $$P$$).\n\nWe will consult NAUTY just the same to get sample data to match against in our mathematical analysis. The following Perl script was used.\n\n#! /usr/bin/perl -w\n#\n\nsub decode_graph {\nmy ($str) = @_; sub R { my (@args) = map { sprintf \"%06b\",$_;\n} @_;\njoin '', @args;\n}\n\nmy (@ents) = map {\nord($_) - 63 } split //,$str;\n\nmy $n = shift @ents; my @adj_data = split //, R(@ents); my$adj = []; my $pos = 0; for(my$ind2 = 1; $ind2 <$n; $ind2++){ for(my$ind1 = 0; $ind1 <$ind2; $ind1++){$adj->[$ind1]->[$ind2] = $adj_data[$pos];\n$adj->[$ind2]->[$ind1] =$adj_data[$pos];$pos++;\n}\n}\n\nreturn $adj; } MAIN: { my$mx = shift || 3;\n\ndie \"out of range for GENBG: $mx\" if 2*$mx < 2 || 2*$mx > 32; for(my$comp_a = 1; $comp_a <=$mx; $comp_a++){ for(my$comp_b = 1; $comp_b <=$mx; $comp_b++){ my$vcount = $comp_a +$comp_b;\n\nmy $cmd = sprintf \"./genbg %d %d\",$comp_a, $comp_b; my$count = 0;\n\nopen GENBG, \"$cmd 2>/dev/null|\"; while(my$bp = <GENBG>){\nchomp $bp; my$adj = decode_graph $bp; for($v = 0; $v <$vcount; $v++){ my$deg = 0;\nfor(my $w = 0;$w < $vcount;$w++){\nmy $ent =$adj->[$v]->[$w];\n$deg++ if defined($ent) && $ent == 1; } last if$deg == 0;\n}\n\n$count++ if$v == $vcount; } close GENBG; printf \" \" if$comp_b > 1;\nprintf \"%06d\", $count; } printf \"\\n\"; }; }  This gave the following table: $ ./scripts/bipartite.pl 6\n000001 000001 000001 000001 000001 000001\n000001 000003 000005 000008 000011 000015\n000001 000005 000017 000042 000091 000180\n000001 000008 000042 000179 000633 002001\n000001 000011 000091 000633 003835 020755\n000001 000015 000180 002001 020755 200082\n\n\nNow for the mathematics. We use the Polya Enumeration Theorem as conjectured by the OP. To do this we need the cycle index of the action on the edges of the group that permutes the vertices in partition $$A$$ of size $$n$$ according to the symmetric group $$S_n$$ and the vertices in partition $$B$$ of size $$m$$ according to $$S_m.$$\n\nThese cycle indices are easy to compute and we do not need to iterate over all $$n!\\times m!$$ pairs of permutations (acting on $$A$$ and $$B$$) but instead it is sufficient to iterate over pairs of terms from the cycle indices $$Z(S_n)$$ and $$Z(S_m)$$ of the symmetric groups $$S_n$$ and $$S_m$$ according to their multiplicities to obtain the cycle index $$Z(Q_{n,m})$$ of the combined action on $$A$$ and $$B$$. The number of terms here is the much better count of the number of partitions of $$n$$ and $$m$$ (upper bound).\n\nThe classic approach to the calculation of these cycle indices is based on the simple observation that for two cycles, one of length $$l_1$$ from a permutation $$\\alpha$$ of $$A$$ and another of length $$l_2$$ from a column permutation $$\\beta$$ of $$B$$ their contribution to the disjoint cycle decomposition product for $$(\\alpha,\\beta)$$ in the cycle index $$Z(Q_{n,m})$$ is by inspection\n\n$$a_{\\mathrm{lcm}(l_1, l_2)}^{l_1 l_2 / \\mathrm{lcm}(l_1, l_2)} = a_{\\mathrm{lcm}(l_1, l_2)}^{\\gcd(l_1, l_2)}.$$\n\nOnce we have the cycle indices we evaluate $$Z(Q_{n,m})(1+z)$$ which is the standard substitution to produce the OGF. If we are only looking to obtain the count, we may use $$Z(Q_{n,m})_{\\Large a_1=a_2=a_3=\\cdots=2}.$$\n\nHere is an example: $$Z(Q_{3,4}) = {\\frac {{a_{{1}}}^{12}}{144}}+1/24\\,{a_{{2}}}^{3}{a_{{1}}}^{6}+1/18\\,{a_{{3 }}}^{3}{a_{{1}}}^{3}+1/12\\,{a_{{2}}}^{6}+1/6\\,{a_{{4}}}^{3} \\\\+1/48\\,{a_{{2}}} ^{4}{a_{{1}}}^{4}+1/8\\,{a_{{1}}}^{2}{a_{{2}}}^{5}+1/6\\,a_{{1}}a_{{2}}a_{{3} }a_{{6}}+1/8\\,{a_{{3}}}^{4} \\\\+1/12\\,{a_{{3}}}^{2}a_{{6}}+1/24\\,{a_{{6}}}^{2}+ 1/12\\,a_{{12}}.$$ The substituted cycle index becomes $$Z(Q_{3,4})(1+z) = {z}^{12}+{z}^{11}+3\\,{z}^{10}+6\\,{z}^{9}+11\\,{z}^{8} \\\\ +13\\,{z}^{7}+17\\,{z}^{6 }+13\\,{z}^{5}+11\\,{z}^{4}+6\\,{z}^{3}+3\\,{z}^{2}+z+1.$$\n\nAt this point we have just about everything we need, the only problem as is evident from the substituted cycle index (indexed by edge count) is that we are counting all subgraphs including those that obviously cannot span $$K_{3,4}.$$ Observe however that $$Z(Q_{3,4})$$ includes the count from all graphs $$K_{a,b}$$ where $$1\\le a \\le 3$$ and $$1\\le b \\le 4$$ and this observation holds for the $$Z(Q_{a,b})$$ as well. The way to identify a spanning subgraph of $$K_{3,4}$$ is that every vertex in the vertex set has degree at least one, which means these are just the graphs that cannot possibly be counted by $$Z(Q_{a,b})$$ with $$(a,b)\\ne (3,4)$$ because of the missing vertices. Therefore we apply PIE to the poset where the nodes corresponding to the $$(a,b)$$ is the set of graphs counted by the corresponding substituted cycle index. We have by inspection that this poset is isomorphic to the divisor poset of $$2^{n-1}\\times 3^{m-1}$$ so that we may use the ordinary M\u00f6bius function from number theory as our M\u00f6bius function for the PIE computation. (We are not including the empty graph in the sets at the nodes from the poset.)\n\nThis is implemented in the following Maple program.\n\nwith(numtheory);\n\npet_cycleind_symm :=\nproc(n)\noption remember;\n\nif n=0 then return 1; fi;\n\nend;\n\npet_varinto_cind :=\nproc(poly, ind)\nlocal subs1, subsl, polyvars, indvars, v, pot;\n\npolyvars := indets(poly);\nindvars := indets(ind);\n\nsubsl := [];\n\nfor v in indvars do\npot := op(1, v);\n\nsubs1 :=\n[seq(polyvars[k]=polyvars[k]^pot,\nk=1..nops(polyvars))];\n\nsubsl := [op(subsl), v=subs(subs1, poly)];\nod;\n\nsubs(subsl, ind);\nend;\n\npet_cycleind_knm :=\nproc(n, m)\noption remember;\nlocal cind, sind1, sind2, t1, t2, q,\nv1, v2, len, len1, len2;\n\ncind := 0;\n\nif n=1 then\nsind1 := [a[1]];\nelse\nsind1 := pet_cycleind_symm(n);\nfi;\n\nif m=1 then\nsind2 := [a[1]];\nelse\nsind2 := pet_cycleind_symm(m);\nfi;\n\nfor t1 in sind1 do\nfor t2 in sind2 do\nq := 1;\n\nfor v1 in indets(t1) do\nlen1 := op(1, v1);\n\nfor v2 in indets(t2) do\nlen2 := op(1, v2);\n\nlen := lcm(len1, len2);\nq := q *\na[len]^((len1*len2/len) *\ndegree(t1, v1)*degree(t2, v2));\nod;\nod;\n\ncind := cind +\nlcoeff(t1)*lcoeff(t2)*q;\nod;\nod;\n\ncind;\nend;\n\nv_pre_pie :=\nproc(n, m)\noption remember;\nlocal cind;\n\ncind := pet_cycleind_knm(n, m);\nsubs([seq(a[v]=2, v=1..n*m)], cind);\nend;\n\nv :=\nproc(n, m)\nlocal q, a, b, res;\n\nq := 2^(n-1)*3^(m-1);\n\nres := 0;\nfor a to n do\nfor b to m do\nres := res +\nmobius(q/2^(a-1)/3^(b-1))*\n(v_pre_pie(a, b)-1);\nod;\nod;\n\nres;\nend;\n\nprint_table :=\nproc(mx)\nlocal n, m;\n\nfor n to mx do\nfor m to mx do\nif m>1 then printf(\" \") fi;\nprintf(\"%06d\", v(n, m));\nod;\n\nprintf(\"\\n\");\nod;\nend;\n\n\nThe above Maple code produces the following table:\n\n> print_table(6);\n000001 000001 000001 000001 000001 000001\n000001 000003 000005 000008 000011 000015\n000001 000005 000017 000042 000091 000180\n000001 000008 000042 000179 000633 002001\n000001 000011 000091 000633 003835 020755\n000001 000015 000180 002001 020755 200082\n\n\nIt can calculate values that are completely out of reach for NAUTY like the sequence of non-isomorphic spanning subgraphs of $$K_{n,n}$$ which is $$1, 3, 17, 179, 3835, 200082, 29610804, 13702979132, \\\\ 20677458750966, 103609939177198046, 1745061194503344181714, \\\\ 99860890306900024150675406,\\ldots$$ which points us to OEIS A054976 where we find confirmation of the above calculation and a slightly different interpretation of the problem statement.\n\nThe function v in the Maple program implements model $$P$$ and the function v_pre_pie implements model $$Q.$$\n\nFor bicolored versions of $$K_{n,n}$$ model $$Q$$ gives the sequence $$2, 7, 36, 317, 5624, 251610, 33642660, 14685630688, \\\\ 21467043671008, 105735224248507784, 1764356230257807614296, \\\\ 100455994644460412263071692,\\ldots$$ which points us to OEIS A002724, where the calculation is confirmed.\n\nThis MSE Meta Link has many more PET computations by various users.\n\nThanks go to the authors of the NAUTY package.\n\n\u2022 Excellent! Thanks for taking the time to answer. I'm studying it right now. Quick question about 'the sequence of non-isomorphic spanning subgraphs of $K_{(n,n)}$. If I take $n=2$ I found that there are $6$ spanning subgraphs of $K_{(2,2)}$ namely: the empty graph of four vertices, $P_2\\cup \\{a\\}\\cup\\{b\\}, P_2\\cup P_2, P_3\\cup \\{a\\}, P_4, K_{(2,2)}$. Thank you! \u2013\u00a0chowching Feb 25 '15 at 6:12\n\u2022 We already discussed this at your other MSE question. There seven bicolored graphs on $K_{2,2}$ but a bicolored graph is not the same as a spanning subgraph. Please consult the discussion on the other question. E.g. the empty graph does not span $K_{2,2}.$ This Wikipedia entry is the definition of a spanning subgraph. The above program also counts bicolored graphs. Note that v_pre_pi(2,2)=7 and v(2,2)=3. \u2013\u00a0Marko Riedel Feb 25 '15 at 18:50\n\u2022 I see what you mean now, I have added a comment. \u2013\u00a0Marko Riedel Feb 25 '15 at 19:05\n\u2022 I'm not impressed to have yet another impressing anwer from you. \u2013\u00a0Jorge Feb 26 '15 at 1:31\n\nThe following data augment those in the first answer, namely we treat the problem of computing $$Z(Q_{n})$$ for $$K_{n,n}$$ where we include reflections. The algorithm here is slightly different from Harary and Palmer. Clearly all permutations from the first answer represented in $$Z(Q_{n,n})$$ contribute as well. Now for the reflections, which are simple, fortunately. These (the vertex permutations) can be obtained from the permutations in $$S_n$$ by placing vertices from component $$A$$ alternating with component $$B$$ on a cycle with twice the length of a cycle contained in a permutation $$\\gamma$$ of $$S_n.$$ The induced action on the set of edges is the contribution to $$Z(Q_n).$$ In the following we have used an algorithmic approach rather than the formula from Harary and Palmer, namely we construct a canonical representative of each permutation shape from the cycle index $$Z(S_n)$$ that doubles the length of every cycle and alternates elements from the two components. We let that act on the set of edges and factor the result.\n\nThis gives the following cycle indices, the first of which matches the result given by Harary and Palmer. For $$n=3$$, $$Z(Q_3) = {\\frac {{a_{{1}}}^{9}}{72}}+1/6\\,{a_{{1}}}^{3}{a_{{2}}}^{3} \\\\+1/8\\,a_{{1}}{a_{{2}}}^{4}+1/4\\,a_{{1}}{a_{{4}}}^{2}+1/9\\,{ a_{{3}}}^{3}+1/3\\,a_{{3}}a_{{6}} ,$$ for $$n=4$$, $$Z(Q_4) = {\\frac {{a_{{1}}}^{16}}{1152}}+{\\frac {{a_{{1}}}^{8}{a_{{2} }}^{4}}{96}}+{\\frac {5\\,{a_{{1}}}^{4}{a_{{2}}}^{6}}{96}}\\\\+{ \\frac {{a_{{1}}}^{4}{a_{{3}}}^{4}}{72}}+{\\frac {17\\,{a_{{2} }}^{8}}{384}}+1/12\\,{a_{{1}}}^{2}a_{{2}}{a_{{3}}}^{2}a_{{6} }+1/8\\,{a_{{1}}}^{2}a_{{2}}{a_{{4}}}^{3}\\\\+1/18\\,a_{{1}}{a_{{ 3}}}^{5}+1/6\\,a_{{1}}a_{{3}}{a_{{6}}}^{2}+1/24\\,{a_{{2}}}^{ 2}{a_{{6}}}^{2}+{\\frac {19\\,{a_{{4}}}^{4}}{96}}+1/12\\,a_{{4 }}a_{{12}}+1/8\\,{a_{{8}}}^{2} ,$$ and for $$n=5$$, $$Z(Q_5) = {\\frac {{a_{{1}}}^{25}}{28800}}+{\\frac {{a_{{1}}}^{15}{a_{{ 2}}}^{5}}{1440}}+{\\frac {{a_{{1}}}^{9}{a_{{2}}}^{8}}{288}}+ {\\frac {{a_{{1}}}^{10}{a_{{3}}}^{5}}{720}}+{\\frac {{a_{{1}} }^{5}{a_{{2}}}^{10}}{192}}\\\\+{\\frac {{a_{{1}}}^{3}{a_{{2}}}^{ 11}}{96}}+{\\frac {a_{{1}}{a_{{2}}}^{12}}{128}}+{\\frac {{a_{ {1}}}^{6}{a_{{2}}}^{2}{a_{{3}}}^{3}a_{{6}}}{72}}+{\\frac {{a _{{1}}}^{4}{a_{{3}}}^{7}}{72}}\\\\+{\\frac {{a_{{1}}}^{5}{a_{{4} }}^{5}}{480}}+1/24\\,{a_{{1}}}^{3}{a_{{2}}}^{3}{a_{{4}}}^{4} +{\\frac {{a_{{2}}}^{5}{a_{{3}}}^{5}}{720}}+1/48\\,{a_{{1}}}^ {3}a_{{2}}{a_{{4}}}^{5}\\\\+1/48\\,{a_{{1}}}^{2}{a_{{2}}}^{4}a_{ {3}}{a_{{6}}}^{2}+{\\frac {{a_{{2}}}^{5}{a_{{3}}}^{3}a_{{6}} }{72}}+1/32\\,a_{{1}}{a_{{2}}}^{2}{a_{{4}}}^{5}\\\\+1/48\\,{a_{{2 }}}^{5}a_{{3}}{a_{{6}}}^{2}+1/36\\,{a_{{2}}}^{2}{a_{{3}}}^{5 }a_{{6}}+1/12\\,{a_{{1}}}^{2}a_{{2}}a_{{3}}{a_{{6}}}^{3}\\\\+{ \\frac {3\\,a_{{1}}{a_{{4}}}^{6}}{32}}+{\\frac {{a_{{2}}}^{2}{ a_{{3}}}^{3}{a_{{6}}}^{2}}{72}}+1/24\\,{a_{{1}}}^{2}a_{{3}}{ a_{{4}}}^{2}a_{{12}}+1/24\\,a_{{2}}a_{{3}}{a_{{4}}}^{2}a_{{ 12}}\\\\+{\\frac {13\\,{a_{{5}}}^{5}}{600}}+1/8\\,a_{{1}}{a_{{8}}} ^{3}+1/12\\,a_{{3}}a_{{4}}a_{{6}}a_{{12}}+{\\frac {{a_{{5}}}^ {3}a_{{10}}}{60}}+1/30\\,{a_{{5}}}^{2}a_{{15}}\\\\+1/8\\,a_{{5}}{ a_{{10}}}^{2}+1/20\\,a_{{5}}a_{{20}}+1/30\\,a_{{10}}a_{{15}} .$$\n\nThese cycle indices can be calculated for large $$n$$ but the pattern should be clear. The count of these graphs gives the sequence\n\n$$2, 6, 26, 192, 3014, 127757, 16853750, \\\\ 7343780765, 10733574184956, 52867617324773592,\\\\882178116079222400788, 50227997322550920824045262, \\ldots$$\n\nwhich points us to OEIS A007139 where we find confirmation of this calculation.\n\nThis was the Maple code used to obtain these values.\n\nwith(numtheory);\n\npet_cycleind_symm :=\nproc(n)\nlocal p, s;\noption remember;\n\nif n=0 then return 1; fi;\n\nend;\n\npet_autom2cycles :=\nproc(src, aut)\nlocal numa, numsubs;\nlocal marks, pos, cycs, cpos, clen;\n\nnumsubs := [seq(src[k]=k, k=1..nops(src))];\nnuma := subs(numsubs, aut);\n\nmarks := Array([seq(true, pos=1..nops(aut))]);\n\ncycs := []; pos := 1;\n\nwhile pos <= nops(aut) do\nif marks[pos] then\nclen := 0; cpos := pos;\n\nwhile marks[cpos] do\nmarks[cpos] := false;\ncpos := numa[cpos];\nclen := clen+1;\nod;\n\ncycs := [op(cycs), clen];\nfi;\n\npos := pos+1;\nod;\n\nreturn mul(a[cycs[k]], k=1..nops(cycs));\nend;\n\npet_varinto_cind :=\nproc(poly, ind)\nlocal subs1, subs2, polyvars, indvars, v, pot, res;\n\nres := ind;\n\npolyvars := indets(poly);\nindvars := indets(ind);\n\nfor v in indvars do\npot := op(1, v);\n\nsubs1 :=\n[seq(polyvars[k]=polyvars[k]^pot,\nk=1..nops(polyvars))];\n\nsubs2 := [v=subs(subs1, poly)];\n\nres := subs(subs2, res);\nod;\n\nres;\nend;\n\npet_flatten_term :=\nproc(varp)\nlocal terml, d, cf, v;\n\nterml := [];\n\ncf := varp;\nfor v in indets(varp) do\nd := degree(varp, v);\nterml := [op(terml), seq(v, k=1..d)];\ncf := cf/v^d;\nod;\n\n[cf, terml];\nend;\n\npet_flat2rep_cyc :=\nproc(f)\nlocal p, q, res, cyc, t, len;\n\nq := 1; res := [];\n\nfor t in f do\nlen := op(1, t);\ncyc := [seq(p, p=q+1..q+len-1), q];\nres := [op(res), cyc];\nq := q+len;\nod;\n\nres;\nend;\n\npet_cycs2table :=\nproc(cycs)\nlocal pairs, cyc, p, ent;\n\npairs := [];\nfor cyc in cycs do\npairs :=\n[op(pairs),\nseq([cyc[p], cyc[1 + (p mod nops(cyc))]],\np = 1 .. nops(cyc))];\nod;\n\nmap(ent->ent[2], sort(pairs, (a,b)->a[1] < b[1]));\nend;\n\npet_cycleind_knn :=\nproc(n)\noption remember;\nlocal cindA, cindB, sind, t1, t2, p, q, cyc1, cyc2,\nflat, len, len1, len2, v1, v2,\nedges, edgeperm, rep, cycsA, cycsB, cycsrc, cyc;\n\nif n=1 then\nsind := [a[1]];\nelse\nsind := pet_cycleind_symm(n);\nfi;\n\ncindA := 0;\n\nfor t1 in sind do\nfor t2 in sind do\nq := 1;\n\nfor v1 in indets(t1) do\nlen1 := op(1, v1);\n\nfor v2 in indets(t2) do\nlen2 := op(1, v2);\n\nlen := lcm(len1, len2);\nq := q *\na[len]^((len1*len2/len) *\ndegree(t1, v1)*degree(t2, v2));\nod;\nod;\n\ncindA := cindA +\nlcoeff(t1)*lcoeff(t2)*q;\nod;\nod;\n\nedges := [seq(seq({p, q+n}, q=1..n), p=1..n)];\n\ncindB := 0;\n\nfor t1 in sind do\nflat := pet_flatten_term(t1);\n\ncycsA := pet_flat2rep_cyc(flat[2]);\ncycsB := [];\nfor cycsrc in cycsA do\ncyc := [];\nfor q in cycsrc do\ncyc := [op(cyc), q, q+n];\nod;\n\ncycsB := [op(cycsB), cyc];\nod;\n\nrep := pet_cycs2table(cycsB);\n\nedgeperm :=\nsubs([seq(q=rep[q], q=1..2*n)], edges);\n\ncindB := cindB + flat[1]*\npet_autom2cycles(edges, edgeperm);\nod;\n\n(cindA+cindB)/2;\nend;\n\nQ :=\nproc(n)\noption remember;\nlocal cind;\n\ncind := pet_cycleind_knn(n);\nsubs([seq(a[p]=2, p=1..n*n)], cind);\nend;\n\n\nRemarks, per request, Dec 2020. Here are some observations concerning reflections. We need the factorization into cycles of the vertices in order to compute the action on the edges. Now a reflection when factored into cycles must alternate between left and right vertices. That means left and right vertices are interleaved on those cycles. Hence we can obtain the factorizations of all reflections by taking a permutation from $$S_n$$ (which gives the left vertices) and doubling all its cycles in length, inserting a permutation of the right vertices, for a total of $$n! \\times n!$$ reflections. The first factorial comes from the permutations of the left vertices which can be interleaved with the right ones in $$n!$$ ways, yielding the second factorial. Now consider the factorization into cycles of the action on the edges. Left and right vertices are disjoint, so no matter the order in which we interleave the right vertices, we get the same shape of factorizations into cycles (just permute the right vertices, which maintains the cycle structure of the edges). Therefore we can take one representative for each term in $$Z(S_n)$$, interleave it with some permutation of the right, and factor that (i.e. factor the action on the edges) to get the contribution to $$Z(Q_n)$$, which must be multiplied by $$n!$$ for the number of reflections covered by this term. This is what the algorithm does. The $$n!$$ is canceled when we divide by the total number of permutations in the group.\n\n\u2022 I have trouble understanding the algorithm for calculating the cycle index when reflections are concerned. I obtain the same results when I consider each of the $n!^2$ pairs of permutations, but it's of course infeasible for larger $n$. Could you describe your approach in more detail? From what I understand, \"permutation shape\" is just its cycle notation. But two different pairs of permutations with the same cycle notations can contribute different terms to the cycle index when the vertices are additionally reflected. I don't understand how the alternation of components works here. \u2013\u00a0Adam S. Dec 15 '20 at 23:28\n\u2022 I have added some commentary. Please note that the 2017 answer is the reference here. \u2013\u00a0Marko Riedel Dec 16 '20 at 2:53\n\u2022 Wow, brilliant. Thank you! My mistake was to treat the two permutations and the reflection completely separately, which made the cycle factorization difficult. Embedding the reflection into one of the permutations makes things so much easier and indeed reduces the computational cost from $n!^2$ just to the number of different permutation shapes in $S_n$. This will help me speed up my little project: github.com/adamsol/burnside-counter. \u2013\u00a0Adam S. Dec 16 '20 at 17:30\n\u2022 It is now tempting to try to simplify also the case without reflections, so that we don't have to iterate over pairs of permutation shapes at all. But it seems that this case is paradoxically more complicated and it's not possible. \u2013\u00a0Adam S. Dec 16 '20 at 17:38\n\u2022 Good that I could help. I might do some additional simplifications later today. \u2013\u00a0Marko Riedel Dec 16 '20 at 17:40\n\nThe question is equivalent to finding all non-isomorphic bipartite graphs with bi-partitions of sizes $n$ and $m$ respectively and is hard to solve in general. In particular there is no simple characterization of such graphs.\n\nIf you need the sequence for the number of such graphs then you can take a look at this oeis entry http://oeis.org/A033995.\n\nIf you need to compute such graphs (up to a reasonable number) then your best choice is the tool genbg which you can find in McKay's nauty package. In some sense genbg is currently the most efficient way to construct such graphs and hence answers your question.\n\n\u2022 Yeah I need to find all those non-isomorphic bipartite graphs with sizes $m$ and $n$ in each partition. The sequence above is for the number of bipartite graphs of $n$ vertices. But that number is greater than the number of graphs that I am trying to find. For example if I have four vertices, there are $7$ bipartite graphs but if I'm only taking all the spanning subgraphs of $K_{(2,2)}$ then there are just $6$ of them. The other bipartite graph is a claw which is bipartite with sizes 1 and 3 in each partition. \u2013\u00a0chowching Feb 17 '15 at 12:43\n\u2022 @chowching You should be more specific about what you're looking for. Note that in general there is no closed formula for the number of bipartite graphs with bipartitions of cardinality n and m. \u2013\u00a0Jernej Feb 17 '15 at 13:00\n\u2022 Thanks for the reply. I'm studying Polya Enumeration Theorem now to see if I can apply it to find how many such graphs are there given a certain $n$ and $m$. \u2013\u00a0chowching Feb 17 '15 at 13:00\n\u2022 @chowching You can but the formula will most likely require using computer computations. \u2013\u00a0Jernej Feb 17 '15 at 13:01\n\u2022 @chowching I suggest that for a start you look at Harary's and Palmer's book called \"graphical enumeration\" \u2013\u00a0Jernej Feb 17 '15 at 13:02\n\nAdding another answer after receiving a pointer from a reader of this thread. We treat the case of spanning subgraphs of $$K_{n,n}$$ including reflections. The algorithm has been greatly simplified. There are two types of symmetries, namely shuffles within the two sets of vertices and reflections that map one to the other. The first type has been simplified as follows: we continue to iterate over pairs of permutation shapes from $$S_n$$ but the contribution to the cycle index $$Z(Q_n)$$ is now computed by iterating over pairs of cycle types (same length) rather than single cycles. The second type (reflections) has been replaced in its entirety. We continue to produce contributions by building factorized permutations consisting of cycles that alternate between the two classes of vertices (a precondition of this problem) but whereas in the first version we would construct a representative of the factored permutation, apply it to the edges, and factor the result into cycles, the new version does not construct representatives and does no factorization and builds the terms for the cycle index from first principles, namely that we enumerate the cycles of edges (two-sets of vertices) that are obtained from a vertex permutation by choosing pairs of vertices, counting these, and computing the lengths of the cycles on which they reside. These optimizations make for a faster and much more compact program. This was the code:\n\npet_cycleind_symm :=\nproc(n)\noption remember;\n\nif n=0 then return 1; fi;\n\nend;\n\npet_varinto_cind :=\nproc(poly, ind)\nlocal subs1, subs2, polyvars, indvars, v, pot, res;\n\nres := ind;\n\npolyvars := indets(poly);\nindvars := indets(ind);\n\nfor v in indvars do\npot := op(1, v);\n\nsubs1 :=\n[seq(polyvars[k]=polyvars[k]^pot,\nk=1..nops(polyvars))];\n\nsubs2 := [v=subs(subs1, poly)];\n\nres := subs(subs2, res);\nod;\n\nres;\nend;\n\npet_flatten_termA :=\nproc(varp)\nlocal terml, d, cf, v;\n\nterml := [];\n\ncf := varp;\nfor v in indets(varp) do\nd := degree(varp, v);\nterml := [op(terml), [op(1,v), d]];\ncf := cf/v^d;\nod;\n\n[cf, terml];\nend;\n\npet_cycleind_knn :=\nproc(n)\noption remember;\nlocal cindA, cindB, sind, t1, t2, term, res,\nflat, flat1, flat2, len, l1, l2, cycs, uidx, vidx,\nu, v, inst1;\n\nif n=1 then\nsind := [a[1]];\nelse\nsind := pet_cycleind_symm(n);\nfi;\n\ncindA := 0;\n\nfor t1 in sind do\nflat1 := pet_flatten_termA(t1);\n\nfor t2 in sind do\nflat2 := pet_flatten_termA(t2);\n\nres := 1;\n\nfor u in flat1[2] do\nl1 := op(1, u);\n\nfor v in flat2[2] do\nl2 := op(1, v);\n\nlen := lcm(l1, l2);\nres := res *\na[len]^(op(2, u)*op(2, v)\n*l1*l2/len);\nod;\nod;\n\ncindA := cindA + flat1[1]*flat2[1]*res;\nod;\nod;\n\ncindB := 0;\n\nfor term in sind do\nflat := pet_flatten_termA(term);\n\ncycs := map(cyc -> [2*cyc[1], cyc[2]], flat[2]);\nres := 1;\n\n# edges on different cycles of different sizes\nfor uidx to nops(cycs) do\nu := cycs[uidx];\n\nfor vidx from uidx+1 to nops(cycs) do\nv := cycs[vidx];\n\nl1 := op(1, u); l2 := op(1, v);\nres := res *\na[lcm(l1, l2)]^((l1*l2/2/lcm(l1, l2))*\nop(2, u)*op(2, v));\nod;\nod;\n\n# edges on different cycles of the same size\nfor uidx to nops(cycs) do\nu := cycs[uidx];\n\nl1 := op(1, u); inst1 := op(2, u);\n# a[l1]^(1/2*inst1*(inst1-1)*l1*l1/2/l1)\nres := res *\na[l1]^(1/2*inst1*(inst1-1)*l1/2);\nod;\n\n# edges on identical cycles of some size\nfor uidx to nops(cycs) do\nu := cycs[uidx];\n\nl1 := op(1, u); inst1 := op(2, u);\nif type(l1/2, even) then\n# a[l1]^((l1/2)^2/l1);\nres := res *\n(a[l1]^(l1/4))^inst1;\nelse\n# a[l1/2]^(l1/2/(l1/2))*a[l1]^(((l1/2)^2-l1/2)/l1)\nres := res *\n(a[l1/2]*a[l1]^(l1/4-1/2))^inst1;\nfi;\nod;\n\ncindB := cindB + flat[1]*res;\nod;\n\n(cindA+cindB)/2;\nend;\n\n\n\nThis will produce the following data for the non-isomorphic spanning subgraphs of $$K_{6,6}$$ including reflections classified according to the number of edges:\n\n$${z}^{36}+{z}^{35}+2\\,{z}^{34}+4\\,{z}^{33}+10\\,{z}^{32} +20\\,{z}^{31}+50\\,{z}^{30}+99\\,{z}^{29}+227\\,{z}^{28} \\\\ +458\\,{z}^{27}+934\\,{z}^{26}+ 1711\\,{z}^{25}+3057\\,{z}^{24}+4889\\,{z}^{23}+7400\\,{z}^{22} \\\\ +10071\\,{z}^{21}+12755\\,{z}^{20}+14524\\,{z}^{19} +15331\\,{z}^{18}+14524\\,{z}^{17}+12755\\,{z}^{16} \\\\+10071\\,{z}^{15}+7400\\,{z}^{14}+4889\\,{z}^{13}+ 3057\\,{z}^{12}+1711\\,{z}^{11}+934\\,{z}^{10} \\\\ +458\\,{z}^{9}+227\\,{z}^{8}+99\\,{z}^{7}+50\\,{z}^{6}+20\\,{z}^{5} +10\\,{z}^{4}+4\\,{z}^{3}+2\\,{z}^{2}+z+1.$$\n\ni.e. the value four for three edges counts, first, a set of three edges, not connected, second, two edges sharing a vertex and a single edge, third, three edges emanating from the same vertex and fourth, a path of three edges.\n\n**Addendum.** We may use Burnside as shown below if we are not interested in classifying according to the number of edges and require only the total count. This gives:\n\nQ :=\nproc(n)\noption remember;\nlocal cind;\n\ncind := pet_cycleind_knn(n);\nsubs([seq(a[p]=2, p=1..n*n)], cind);\nend;\n\n\nThe routine that substitutes polynomials into cycle indices is no longer needed in this case. Observe that further simplification at the expense of readability is possible by not computing the cycle index at all and setting all $$a[p]$$ to two during the computation of what used to be the cycle index.\n\nRemark, Dec 2020. We can in fact optimize away all flattening operations and use the fact that Maple monomials i.e. the terms of the cycle index are perfectly sufficient to implement a multiset data structure where the variables represent cycles of some length and the degrees, the number of instances, as is of course the standard semantics of cycle indices. This is shown below. (Duplicate code omitted.)\n\npet_cycleind_knn :=\nproc(n)\noption remember;\nlocal cindA, cindB, sind, t1, t2, term,\nres, len, len1, len2, cycs, uidx, vidx, interlv,\nu, v, inst, q;\n\nif n=1 then\nsind := [a[1]];\nelse\nsind := pet_cycleind_symm(n);\nfi;\n\ncindA := 0;\n\nfor t1 in sind do\nfor t2 in sind do\nq := 1;\n\nfor u in indets(t1) do\nlen1 := op(1, u);\n\nfor v in indets(t2) do\nlen2 := op(1, v);\n\nlen := lcm(len1, len2);\nq := q *\na[len]^((len1*len2/len) *\ndegree(t1, u)*degree(t2, v));\nod;\nod;\n\ncindA := cindA +\nlcoeff(t1)*lcoeff(t2)*q;\nod;\nod;\n\ncindB := 0;\n\ninterlv := subs({seq(a[q]=a[2*q], q=1..n)}, sind);\nfor term in interlv do\ntermvars := indets(term); res := 1;\n\n# edges on different cycles of different sizes\nfor uidx to nops(termvars) do\nu := op(uidx, termvars);\nlen1 := op(1, u);\n\nfor vidx from uidx+1 to nops(termvars) do\nv := op(vidx, termvars);\nlen2 := op(1, v);\n\nres := res *\na[lcm(len1, len2)]\n^((len1*len2/2/lcm(len1, len2))*\ndegree(term, u)*degree(term, v));\nod;\nod;\n\n# edges on different cycles of the same size\nfor u in termvars do\nlen := op(1, u); inst := degree(term, u);\n# a[len]^(1/2*inst*(inst-1)*len*len/2/len)\nres := res *\na[len]^(1/4*inst*(inst-1)*len);\nod;\n\n# edges on identical cycles of some size\nfor u in termvars do\nlen := op(1, u); inst := degree(term, u);\nif type(len/2, even) then\n# a[len]^((len/2)^2/len);\nres := res *\n(a[len]^(len/4))^inst;\nelse\n# a[len/2]^(len/2/(len/2))\n# *a[len]^(((len/2)^2-len/2)/len)\nres := res *\n(a[len/2]*a[len]^(len/4-1/2))^inst;\nfi;\nod;\n\ncindB := cindB + lcoeff(term)*res;\nod;\n\n(cindA+cindB)/2;\nend;\n\nQ :=\nproc(n)\noption remember;\nlocal cind;\n\ncind := pet_cycleind_knn(n);\nsubs([seq(a[p]=2, p=1..n*n)], cind);\nend;", "date": "2021-03-02 15:30:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/", "openwebmath_score": 0.5907314419746399, "openwebmath_perplexity": 1273.553374674751, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138144607745, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8562690370894727}}
{"url": "http://mathhelpforum.com/pre-calculus/219934-writing-equation-polynomial-graph-print.html", "text": "# Writing An Equation for Polynomial Graph\n\n\u2022 June 17th 2013, 05:44 PM\nAceBoogie\nWriting An Equation for Polynomial Graph\nHey all,\nHoping someone can help me figure out how to write an equation for the polynomial graphed in the picture. Really lost, if any help could be given it would be appreciated.\n\nAttachment 28638\n\u2022 June 17th 2013, 06:02 PM\nHallsofIvy\nRe: Writing An Equation for Polynomial Graph\nOne of the things you might have noticed is that an nth degree polynomial has at most n-1 \"critical\" points where the graph levels out (maybe but not necessarily where the graph \"turns\"). Since this graph has 4 such \"critical points\" (at x=-3, x= -1. something, x= 0. something, and x= 2) I would try a 5th degree polynomial.\nFurther, we can see from the graph that the polynomial is 0 at x= -3, x= -1, and x= 2. That means that the polynomial is of the form [tex]y= a(x+3)(x+1)(x- 2)Q(x)[tex] where a is a number and Q(x) is a quadratic polynomial.\n\nTo go further, you would have to specify exactly where the turning point (\"-1.something\", \"0.something\") and what the y values are there.\n\u2022 June 17th 2013, 06:13 PM\nAceBoogie\nRe: Writing An Equation for Polynomial Graph\nSo are you saying you think the outline for the equation would be:\n\nQ(x) = x(x+3)(x+1)(X-2)\n\nI don't understand how I would find the 0.something and the 1.something points? How would they relate into the equation?\n\u2022 June 18th 2013, 03:57 AM\nmpx86\nRe: Writing An Equation for Polynomial Graph\nat x=-3 , x=-1 , x=2 ...... y= f(x)=0\n\nthat is 3,1,-2 are its root\n\nsuch an equation is K (x+3)(x+1)(x-2)=0\nas such an equation is zero at x=-3,-1,2\n\nnow f(0)=6 or y is 6 when x is 0\nput x=0 in the equation\nK(0+3)(0+1)(0-2)=6\nyielding K=-1\n\ntherefore reqd. equation is\nf(x)=-(x+3)(x+1)(x-2)\n\u2022 June 19th 2013, 08:11 PM\nSoroban\nRe: Writing An Equation for Polynomial Graph\nHello, AceBoogie!\n\nI can get you started . . .\n\nQuote:\n\nWrite an equation for the polynomial graphed in the picture.\n\nAttachment 28638\n\nThere are three $x$-intercepts . . . two of them are quite special.\n\nAt $x=2$, the curve is tangent to the $x$-axis.\n. . Hence, $f(x)$ has a factor of $(x-2)^2.$\n\nAt $x = \\text{-}1$, the curve passes through the $x$-axis.\n. . Hence, $f(x)$ has a factor of $(x+1).$\n\nAt $x = \\text{-}3$, the curve passes through the $x$-axis.\nBut there is an inflection point at $(\\text{-}3,0).$\n. . Hence, $f(x)$ has a factor of $(x+3)^3.$\n\nAlso, $f(x)$ has a $y$-intercept at $(0,6).$\n\nCan you combine those facts?\n\u2022 June 20th 2013, 12:23 AM\nReneG\nRe: Writing An Equation for Polynomial Graph\nQuote:\n\nOriginally Posted by Soroban\n\nAt $x=2$, the curve is tangent to the $x$-axis.\n. . Hence, $f(x)$ has a factor of $(x-2)^2.$\n\nAt $x = \\text{-}1$, the curve passes through the $x$-axis.\n. . Hence, $f(x)$ has a factor of $(x+1).$\n\nAt $x = \\text{-}3$, the curve passes through the $x$-axis.\nBut there is an inflection point at $(\\text{-}3,0).$\n. . Hence, $f(x)$ has a factor of $(x+3)^3.$\n\nYou've encouraged me to finally study the polynomial functions chapter in my book. :P\n\u2022 June 28th 2013, 08:38 PM\njohng\nRe: Writing An Equation for Polynomial Graph\nHi,\nI strongly encourage you to use your graphing software/calculator to help you with graphing problems. For example, once you have the supposed correct equation, you can check your result. Assuming you can zoom in, zooming in at (-3,0) shows indeed an inflection point. I always use my software to check any graphics problem. Of course, you probably can't do this during a quiz or test. Here's the solution to your problem:\n\nAttachment 28692", "date": "2014-12-21 01:25:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/219934-writing-equation-polynomial-graph-print.html", "openwebmath_score": 0.8213918805122375, "openwebmath_perplexity": 901.9574674162387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138177076645, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.856269033593521}}
{"url": "https://math.stackexchange.com/questions/2954153/show-that-sqrtx2-is-a-contraction/2954166", "text": "# Show that $\\sqrt{x+2}$ is a contraction\n\nLet $$f:X \\rightarrow X$$ where $$X=[0,\\infty)$$ be defined as $$f(x)=\\sqrt{x+2}$$. I have to show that this mapping is a contraction and find its unique fixed point. The second part is easy: by the CMT, it has a unique fixed point in $$X$$ and it is $$x^\\ast = 2$$.\n\nFor $$f$$ being a contraction I wts the following: $$\\exists \\beta \\in [0,1)$$ such that\n\n$$\\mid\\sqrt{x+2}-\\sqrt{y+2}\\mid \\leq \\beta \\mid x-y \\mid, \\ (\\forall x,y\\geq0)$$\n\nSince\n\n$$\\mid\\sqrt{x+2}-\\sqrt{y+2}\\mid = \\dfrac{\\mid x-y \\mid}{\\sqrt{x+2}+\\sqrt{y+2}}$$\n\nI'm tempted to set $$\\beta = \\dfrac{1}{\\sqrt{x+2}+\\sqrt{y+2}}$$ but $$\\beta$$ cannot depend on $$x$$ or $$y$$.... Any ideas about how to proceed? Thanks!\n\nNote that for all $$x,y\\geqslant 0$$ $$\\frac{1}{\\sqrt{x+2}+\\sqrt{y+2}}\\leqslant \\frac{1}{\\sqrt{2}+\\sqrt{2}}=\\frac{\\sqrt{2}}{4}$$\n\n\u2022 Great! I had just to find an upper bound smaller than one! \u2013\u00a0Alessandro Oct 13 '18 at 18:59\n\nhint\n\nLet $$x,y\\in [0,+\\infty)$$.\n\nBy MVT\n\n$$f(x)-f(y)=(x-y)f'(c)$$\n\nwhere $$0\\le x\n\n$$f'(c)=\\frac{1}{2\\sqrt{c+2}}\\le \\frac{1}{2\\sqrt{2}}$$\n\n\u2022 Thanks! Basically to show that $f$ is a contraction it is enough to show that $\\mid f'(x) \\mid \\leq \\beta$, $\\forall x \\in X$ \u2013\u00a0Alessandro Oct 13 '18 at 19:18\n\u2022 @Alessandro It is sufficient to prove that the derivative is bounded. $| f'(x) | \\ le \\beta$. \u2013\u00a0hamam_Abdallah Oct 13 '18 at 19:20\n\u2022 @Salahamam_Fatima You mean that $|f'(x)|<1$, not just generally bounded. \u2013\u00a0mathematics2x2life Oct 13 '18 at 19:31\n\u2022 @mathematics2x2life To be a contraction, you need $|f'(x)|\\le \\beta<1$. \u2013\u00a0hamam_Abdallah Oct 13 '18 at 19:48\n\nAs per Salahamam's solution, to show a function is a contraction, it is sufficient to show that its derivative has $$|f'(x)|<1$$.\n\nProp. If $$f(x)$$ is a differentiable function function with $$|f'(x)|<1$$ for all $$x$$, then $$f(x)$$ is a contraction.\n\nProof. Let $$x,y \\in \\mathbb{R}$$. By the Mean Value Theorem, we have $$|f(x)-f(y)|= |f'(c)(x-y)|= |f'(c)||x-y|$$ for some $$c$$ between $$x$$ and $$y$$. But as $$|f'(c)|<1$$ by assumption, we must have $$|f(x)-f(y)|=|f'(c)||x-y| < 1 \\cdot |x-y|=|x-y|.$$ Therefore, $$f$$ is a contraction.\n\nNote that the converse is false as contractions need not be differentiable.\n\nSo in your case, you only need show that $$\\sqrt{x+2}$$ has bounded derivative. But $$\\dfrac{d}{dx} \\; \\sqrt{x+1}= \\dfrac{1}{2\\sqrt{x+2}}$$ which is at most $$\\frac{1}{2\\sqrt{2}}$$ on the interval $$[0,\\infty)$$.\n\n\u2022 To show that $f$ is a contraction you need to show that the derivative is uniformly bounded by a number less than one. Proving that $f'(x)<1$ is not enough. That's what I understood \u2013\u00a0Alessandro Oct 13 '18 at 22:34\n\u2022 this question can clarify the issue, I hope: math.stackexchange.com/questions/419392/\u2026 \u2013\u00a0Alessandro Oct 13 '18 at 22:41\n\u2022 @Alessandro That was exactly what I said. You have a function which is differentiable on the intervals which you are considering, so it suffices to show that $|f'(x)|<1$. You do not need your derivative to be uniformly bounded. A strong contraction is a function for which there exists $|f(x)-f(y)| \\leq c<1$. But of course, this depends on what one calls a contraction. But most mean $|f(x)-f(y)|<|x-y|$ when they say 'contraction' and reserve the former when speaking of something stronger like strong contraction, Lipschitz continuity, etc.. \u2013\u00a0mathematics2x2life Oct 15 '18 at 15:34\n\u2022 ok but to apply the contraction mapping theorem I need what you call \"a strong contraction\". Furthermore the standard definition of contraction requires $\\mid f(y)-f(x) \\mid \\leq \\beta \\mid x-y \\mid$ (see en.wikipedia.org/wiki/Contraction_mapping) \u2013\u00a0Alessandro Oct 16 '18 at 15:51", "date": "2019-10-21 20:14:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2954153/show-that-sqrtx2-is-a-contraction/2954166", "openwebmath_score": 0.9999661445617676, "openwebmath_perplexity": 1047.6054615773917, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138138113964, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8562690301812201}}
{"url": "https://math.stackexchange.com/questions/1384690/abc-2149-find-a", "text": "$A+B+C=2149$, Find $A$\n\nIn the following form of odd numbers\n\nIf the numbers\n\ntaken from the form where $A+B+C=2149$\n\nFind $A$\n\nany help will be appreciate it, thanks.\n\n\u2022 Hint: looking at the numbers in the picture, try to find a pattern that tells you what $B$ and $C$ are in terms of $A$. Then plug $B$ and $C$ into $A+B+C=2149$ and solve for $A$. It helps to give each line a number, so the line with 1 is line number 1, the line with 3 and 5 is line number 2, the line with 7, 9, and 11 is line number 3, etc. \u2013\u00a0wltrup Aug 4 '15 at 22:50\n\u2022 What interesting puzzles you come up with! \u2013\u00a0hypergeometric Aug 6 '15 at 9:33\n\nIf $A$ were on the $n^{th}$ row, then $B=A+2n$ and $C=A+2(n+1)$.\n\nOut of laziness, we will try to figure out what row this would need to be to be in the right ballpark through estimation. $\\frac{2149}{3}\\approx 716$, so $A<716<B<C$. Which row would $715$ fit into?\n\nBy changing your figure some by first adding one to each entry and then dividing by two, we will get the chart $\\left[\\begin{array}{}\\color{red}{1}\\\\2&\\color{red}{3}\\\\4&5&\\color{red}{6}\\\\7&8&9&\\color{red}{10}\\\\\\vdots\\end{array}\\right]$, and these numbers are very familiar to us. The red numbers are the triangle numbers, $T(n)=\\binom{n+1}{2}=\\frac{n^2+n}{2}$.\n\nSo, $T(26)=351<\\frac{715+1}{2}=358<378=T(27)$, so we suppose that $715$ occurs on the $26^{th}$ row, implying that either $A$ is on the $25^{th}$, $26^{th}$, or $27^{th}$ row (since we have been estimating up to this point).\n\nSo, we try to solve now, $A+B+C=2149=A+(A+2n)+(A+2n+2)=3A+4n+2$\n\nIn the case that $n=25$, this would be $3A+100+2=2149\\Rightarrow 3A=2047\\Rightarrow A=\\frac{2047}{3}\\not\\in\\mathbb{Z}$, so we know that $n=25$ was not possible.\n\nIn the case that $n=26$, this would be $3A+104+2=2149\\Rightarrow 3A=2043\\Rightarrow A=681$\n\nIn the case that $n=27$, this would be $3A+108+2=2149\\Rightarrow 3A=2039\\Rightarrow A=\\frac{2039}{3}\\not\\in\\mathbb{Z}$, so we know that $n=27$ was not possible.\n\nWe can similarly show that if $n=24$ that leads to a contradiction as well.\n\nWe believe then that our answer is $A=681, B=733, C=735$. All that remains to check is that $681$ is indeed on the $26^{th}$ row to confirm our calculations by checking that $T(25)$ is less than $\\frac{A+1}{2}$ and $T(26)$ is greater than $\\frac{A+1}{2}$.\n\nIndeed, $T(25)=325<\\frac{681+1}{2}=341<351=T(26)$\n\n\u2022 Oh man, you're genies, that's correct, because there is 4 options, and 681 one of them, thanks \u2013\u00a0Oiue Aug 4 '15 at 23:08\n\u2022 +1 for the ballpark paragraph. I stress this when I tutor students, that having a ballpark estimate (even one it takes $5$ seconds to come up with) and compare that to the exact answer you got at the end is a very effective way to see if you've done any big mistakes. \u2013\u00a0Arthur Aug 4 '15 at 23:16\n\n$A$ is going to be the $k$th entry in the $n$th row, where $1 \\leq k \\leq n$. That will make $B$ the $k$th entry in the $n+1$st row, and $C$ the $k+1$st entry in the $n+1$st row.\n\nSo the first question is, can we write a formula for $A$ in terms of $k$ and $n$? If we consider the first entry in the $n$th row, there are $1 + 2 + \\cdots + (n-1)$ odd numbers before it, which equals $\\frac{n(n-1)}{2}$. So the first entry in the $n$th row must be $2\\left(\\frac{n(n-1)}{2}\\right) + 1 = n^2 - n + 1$. Now going over to the $k$th entry will add $2(k - 1)$ to this, so $A = n^2 - n + 1 + 2(k-1) = n^2 - n + 2k - 1$\n\nThis formula also implies that $B = (n+1)^2 - (n+1) + 2k - 1$ and $C = (n+1)^2 - (n+1) + 2(k+1) - 1$.\n\nIf we put this all together, $A + B + C = (n^2 - n + 2k - 1) + (n^2 + n + 2k - 1) + (n^2 + n + 2k + 1) = 3n^2 + n + 6k - 1$.\n\nSo can we solve:\n\n$$3n^2 + n + 6k - 1 = 2149$$\n\nOr:\n\n$$3n^2 + n + 6k = 2150$$\n\nSolving for $k$:\n\n$$k = \\frac{2150 - 3n^2 - n}{6}$$\n\nNow we need $k$ to be an integer, and $1 \\leq k \\leq n$. Since $k$ is an integer we know $n$ should be $2$ mod 3, and we know $3n^2 \\leq 2150$, so $n^2 \\leq 717$. So $n < 27$. Now we guess, starting with the highest value of $n$ which is less than $27$ and $2$ mod 3, so $n = 26$. Plugging this in we get $k = 16$, which works!\n\nThis gives $A = 26^2 - 26 + 32 - 1 = 681$.\n\n\u2022 I finished this off a little differently (my solution is a mere variation of yours, as I mention there), but I liked your derivation of the equation $3n^2 + n + 6k = 2150$, which is the key element of either solution. I would have added my variation as a comment on your answer but it didn't look like it would fit the allotted space. \u2013\u00a0David K Aug 5 '15 at 5:30\n\nI like this one.\n\nLook, let's fill the blanks in the middle column with pair numbers. You will find that it forms the sequence 1, 4, 9, 16, 25, 36.. that is evidently x\u00b2.\n\nAll the numbers in a row are sequential, so we can establish a variable to this position i.e. n. And establish that in the A position we have n, so in the B position we have (n - 1) and in the C position we have (n + 1).\n\nGiven all this we can say:\n\nA = x\u00b2 + n\n\nB = (x + 1)\u00b2 + (n - 1)\n\nC = (x + 1)\u00b2 + (n + 1)\n\nSo we can put this in the original equation and we will have an equation with two values to solve (x, n). Nevertheless we also know that n can not be greater than x, ensuring the selected values are in the triangle.\n\nSo\n\n(x\u00b2 + n) + (x + 1)\u00b2 + (n - 1) + (x + 1)\u00b2 + (n + 1) = 2149\n\n3x\u00b2 + 4x + (3n) + 2 = 2149\n\n3x\u00b2 + 4x + (3n - 2147) = 0\n\nWith this equation we can find:\n\nx = 26\n\nn = 5\n\nAnd so:\n\nA = 681 B = 733 C = 735\n\nHope it helps!\n\n\u2022 If you \"fill in\" the table so that, for example, row $4$ is 13 14 15 16 17 18 19, then the middle A is still $x^2$ and $A+B+C = 3x^2+4x+2$. Solving $3x^2+4x+2=2149$ for $x$, we get $x=26.094$. So we are in row $26$ and our A is just to the right of $A=26^2$, where $A+B+C=2134$. In the filled in table, each shift to the right adds $1$ to $A, B,$ and $C$ and adds $3$ to $A+B+C$. $(2149-2134)/3=5$. So $A = 26^2 + 5 =681$ \u2013\u00a0steven gregory Aug 5 '15 at 14:56\n\nLet's write the numbers as $a_k:=2k-1$, starting at $k=1$. Then, $a_1=1$, $a_2=3$, $a_3=5$, and so on. When $A=a_k$ is in row $n$ (the first row is row $0$), then $B=a_{k+n+1}$ and $C=a_{k+n+2}=B+2$. Hence, we have\n\n\\begin{align*} && 2149 &= a_k + 2a_{2k+n+1} +2 \\\\ &\\Rightarrow& 2147 &= 2k-1 + 2(2(k+n+1)-1) = 6k + 4n + 1 \\end{align*}\n\nNote that the $n$-th row (starting at $n=0$) of the triangle starts with $a_{k_n}$, where $$k_n = 1+\\cdots+n = \\frac{n(n+1)}2.$$ Therefore, $k=k_n+q$ where $0\\le q\\le n$. Substituting, we get \\begin{align*} && 2147 &= 3n(n+1) + 10q + 4n + 1 \\\\ &\\Rightarrow & 0 &= 3 n^2 + 7 n + 10 q -2146 \\end{align*} Observe that the linear term $10q$ does not affect the position of the zeros of this quadratic polynomial much. Hence, I just plugged in $q=n/2$ and this polynomial has a zero around $25$. So, $n=25$ seems a good guess.\n\nIndeed, plugging in $n=25$ into our earlier equation gives\n\n\\begin{align*} && 2147 &= 6*k + 4*25 + 1 \\\\ &\\Rightarrow & 2046 &= 6*k \\\\ &\\Rightarrow & 341 &= k \\end{align*}\n\nSo we have $A=a_{341}=681$ in row $n=25$ and $B=2*(341+25+1)-1=733$, $C=735$.\n\nLooking at the first element in each row, we see\n\n row      A    B    C   A+B+C\n1      1    3    5       9\n2      3    7    9      19\n3      7   13   15      35\n4     13   21   23      57\n5     21   31   33      85\n6     31   43   45     139\n\n\nUsing finite differences, we find that\n\n\\begin{align} A &= n^2 - n + 1\\\\ B &= n^2 + n + 1 = A+2n\\\\ C &= n^2 + n + 3 = A+2n+2\\\\ A+B+C &= 3n^2+n+5 \\end{align} where A is the first element in row $n$.\n\nFor each shift of A to the right in the same row, A,B, and C increase by 2 and A+B+C increases by $6$.\n\nSolving $3n^2+n+5 = 2149$ for n, we get $n = 26.567$ So we are in row $26$.\n\nFor the first element in row $26, A = 651,\\; B = 703,\\; C = 705,$ and $A+B+C = 2059$.\n\nSince $\\dfrac{2149-2059}6 = 15$, we need to increase $A, B,$ and $C$ by $2\\cdot15 = 30$.\n\nSo\n\n\\begin{align} A &= 681\\\\ B &= 733\\\\ C &= 735 \\end{align}\n\n\u2022 Ok, you just should edit the first row $a=1, b=3, c=5, a+b+c=9$ \u2013\u00a0Oiue Aug 5 '15 at 0:08\n\u2022 @Oiue. Yep. It's nine in my notes too. Thanks. \u2013\u00a0steven gregory Aug 5 '15 at 1:18\n\nLet $r$ be the row number, and $U$ be the unshown number between $B$ and $C$ and directly below $A$, i.e.\n\n\\begin{align} &A &&\\leftarrow\\text {row}\\; r\\\\ B \\quad [&U]\\quad C &&\\leftarrow\\text {row}\\; r+1 \\end{align}\n\nNote that the difference between two vertically adjacent numbers (both shown and unshown) is $(r+1)^2-r^2=2r+1$, i.e. $U-A=2r+1$. Hence \\begin{align} 2149&=A+B+C&&\\text{(given)}\\\\ &=A+2U&&\\text{(as}\\; B=U-1, C=U+1\\text{)}\\\\ &=3A+4r+2&&\\text{(using } U=A+2r+1\\text{)}\\\\ 3A+4r-2147&=0&& && .....(1)\\\\ \\end{align}\n\nNote also that the centre column (including unshown numbers) corresponds to $r^2$. As an initial approximation, assume that $A$ is the centre column, i.e. $A=r^2$. Substituting in $(1)$ gives\n\n\\begin{align}3r^2+4r-2147&=0\\\\ \\color{blue}{r}&\\color{blue}{\\approx 26} \\; (r>0)\\qquad \\qquad && && && &&& .....(2)\\end{align}\n\nPutting $(2)$ in $(1)$ gives\n\n$$\\color{red}{A=681}\\qquad\\blacksquare\\qquad \\Rightarrow \\color{blue}{B=733, C=735}$$\n\nCheck: $A+B+C=681+733+735=2149$\n\nNB: The number \"$A$\" is in row $26$ but is not in the centre column; the element in the centre column is $26^2=676$ which is blank as it is an even number.\n\n\u2022 every solutions her always adds something new, thanks :) \u2013\u00a0Oiue Aug 5 '15 at 18:41\n\nThis is a slight variation on the nice answer by @Alex Zorn.\n\nIn that answer we find that if the $1$ at the top of the triangle of numbers is considered entry number $1$ (counting from the left) in row number $1$ (counting from the top), then entry number $k$ in row $n$ is $n^2\u2212n+2k\u22121$, and $A$ is entry $k$ in row $n$ where $n,k$ are a solution to\n\n$$3n^2 + n + 6k = 2150.$$\n\nWe have $1 \\leq k \\leq n$ by the way the entries in a row are counted, so for all $n > 0$ we have $$3n^2 + n = 2144 \\leq 3n^2 + n + 6k = 2150 \\leq 3n^2 + 7n = 2150. \\tag 1$$\n\nSolving the two equations\n\n$$3n^2 + n = 2144, \\tag 2$$ $$3n^2 + 7n = 2150, \\tag 3$$\n\nwe see that they each have one positive root. The positive root of Equation $(2)$ is $r_1 \\approx 26.642$ and the positive root of Equation $(3)$ is $r_2 \\approx 25.629$.\n\nBut because of Inequality $(1)$, whatever the value of $k$ is (provided that $1 \\leq k \\leq n$) there is a positive root of $3n^2 + n + 6k = 2150$ that is not less than $r_2$ and not greater than $r_1$ (draw a graph if you have trouble seeing this). The only integer that satisfies both conditions is $26$, so if $3n^2 + n + 6k = 2150$ has a solution for integer $n$ then the only possibility is $n = 26$.\n\nSetting $n = 26$, we have $3(26^2) + 26 + 6k = 2150$, which is a simple linear equation in $k$ with the solution $k = 16$. We then plug $n$ and $k$ into the formula for $A$ to get the answer.", "date": "2019-06-27 04:37:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1384690/abc-2149-find-a", "openwebmath_score": 0.9880356192588806, "openwebmath_perplexity": 232.46094820076627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683487809279, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8562570702488057}}
{"url": "https://math.stackexchange.com/questions/3021679/values-of-x-satisfying-sin-x-cdot-cos3-x-sin3x-cdot-cos-x", "text": "Values of $x$ satisfying $\\sin x\\cdot\\cos^3 x>\\sin^3x\\cdot\\cos x$\n\nFor what values of $$x$$ between $$0$$ and $$\\pi$$ does the inequality $$\\sin x\\cdot\\cos^3 x>\\sin^3x\\cdot\\cos x$$ hold?\n\nMy Attempt $$\\sin x\\cos x\\cdot(\\cos^2x-\\sin^2x)=\\frac{1}{2}\\cdot\\sin2x\\cdot\\cos2x=\\frac{1}{4}\\cdot\\sin4x>0\\implies\\sin4x>0\\\\ x\\in(0,\\pi)\\implies4x\\in(0,4\\pi)\\\\ 4x\\in(0,\\pi)\\cup(2\\pi,3\\pi)\\implies x\\in\\Big(0,\\frac{\\pi}{4}\\Big)\\cup\\Big(\\frac{\\pi}{2},\\frac{3\\pi}{4}\\Big)$$ But, my reference gives the solution, $$x\\in\\Big(0,\\dfrac{\\pi}{4}\\Big)\\cup\\Big(\\dfrac{3\\pi}{4},\\pi\\Big)$$, where am I going wrong with my attempt?\n\nThe given solution is wrong; you are correct. At $$x=\\frac{7\\pi}8\\in\\left(\\frac{3\\pi}4,\\pi\\right)$$, we have that $$\\frac14\\sin4x=\\frac14\\sin\\frac{7\\pi}2=-\\frac14<0$$ which is a contradiction.\n\nFinishing what you did\n\n$$\\sin(4x)>0\\iff$$\n\n$$2k\\pi <4x<(2k+1)\\pi \\iff$$\n\n$$\\frac{k\\pi}{2}\n\n$$k=0$$ gives $$0\n\nand $$k=1$$ gives $$\\frac{\\pi}{2}.\n\nAs an alternative for a full solution we can consider two cases\n\n\u2022 $$\\sin x \\cos x >0$$ that is $$x\\in(0,\\pi/2)\\cup(\\pi,3\\pi/2)$$\n\n$$\\sin x\\cdot\\cos^3 x>\\sin^3x\\cdot\\cos x \\iff\\cos^2x>\\sin^2x \\iff2\\sin^2 x<1$$\n\n$$-\\frac{\\sqrt 2}2<\\sin x<0 \\,\\land\\, 0<\\sin x<\\frac{\\sqrt 2}2 \\iff \\color{red}{x\\in(0,\\pi/4)}\\cup(\\pi,5\\pi/4)$$\n\n\u2022 $$\\sin x \\cos x <0$$ that is $$x\\in(\\pi/2,\\pi)\\cup(3\\pi/2,2\\pi)$$\n\n$$\\sin x\\cdot\\cos^3 x>\\sin^3x\\cdot\\cos x \\iff\\cos^2x<\\sin^2x \\iff2\\sin^2 x>1$$\n\n$$-1<\\sin x<-\\frac{\\sqrt 2}2\\,\\land\\, \\frac{\\sqrt 2}2<\\sin x <1 \\iff \\color{red}{x\\in(\\pi/2,3\\pi/4)}\\cup(3\\pi/2,7\\pi/4)$$\n\nand then your solution is correct.", "date": "2019-05-25 06:56:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3021679/values-of-x-satisfying-sin-x-cdot-cos3-x-sin3x-cdot-cos-x", "openwebmath_score": 0.9619792103767395, "openwebmath_perplexity": 3921.7755985756467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964211605606, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8562256819384124}}
{"url": "http://jeonjucj.kr/mexican-chili-octfec/90jrx.php?tag=inverse-trigonometric-functions-derivatives-44b03e", "text": "The inverse of six important trigonometric functions are: 1. What are the derivatives of the inverse trigonometric functions? For example, the sine function $$x = \\varphi \\left( y \\right)$$ $$= \\sin y$$ is the inverse function for $$y = f\\left( x \\right)$$ $$= \\arcsin x.$$ Then the derivative of $$y = \\arcsin x$$ is given by, ${{\\left( {\\arcsin x} \\right)^\\prime } = f\u2019\\left( x \\right) = \\frac{1}{{\\varphi\u2019\\left( y \\right)}} }= {\\frac{1}{{{{\\left( {\\sin y} \\right)}^\\prime }}} }= {\\frac{1}{{\\cos y}} }= {\\frac{1}{{\\sqrt {1 \u2013 {\\sin^2}y} }} }= {\\frac{1}{{\\sqrt {1 \u2013 {\\sin^2}\\left( {\\arcsin x} \\right)} }} }= {\\frac{1}{{\\sqrt {1 \u2013 {x^2}} }}\\;\\;}\\kern-0.3pt{\\left( { \u2013 1 < x < 1} \\right).}$. Derivatives of Inverse Trig Functions. ${y^\\prime = \\left( {\\arctan \\frac{1}{x}} \\right)^\\prime }={ \\frac{1}{{1 + {{\\left( {\\frac{1}{x}} \\right)}^2}}} \\cdot \\left( {\\frac{1}{x}} \\right)^\\prime }={ \\frac{1}{{1 + \\frac{1}{{{x^2}}}}} \\cdot \\left( { \u2013 \\frac{1}{{{x^2}}}} \\right) }={ \u2013 \\frac{{{x^2}}}{{\\left( {{x^2} + 1} \\right){x^2}}} }={ \u2013 \\frac{1}{{1 + {x^2}}}. Derivatives of the Inverse Trigonometric Functions. In Table 2.7.14 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible. Suppose \\textrm{arccot } x = \\theta. We then apply the same technique used to prove Theorem 3.3, \u201cThe Derivative Rule for Inverses,\u201d to di\ufb00erentiate each inverse trigonometric function. The Inverse Cosine Function. }$, ${y^\\prime = \\left( {\\frac{1}{a}\\arctan \\frac{x}{a}} \\right)^\\prime }={ \\frac{1}{a} \\cdot \\frac{1}{{1 + {{\\left( {\\frac{x}{a}} \\right)}^2}}} \\cdot \\left( {\\frac{x}{a}} \\right)^\\prime }={ \\frac{1}{a} \\cdot \\frac{1}{{1 + \\frac{{{x^2}}}{{{a^2}}}}} \\cdot \\frac{1}{a} }={ \\frac{1}{{{a^2}}} \\cdot \\frac{{{a^2}}}{{{a^2} + {x^2}}} }={ \\frac{1}{{{a^2} + {x^2}}}. The derivatives of $$6$$ inverse trigonometric functions considered above are consolidated in the following table: In the examples below, find the derivative of the given function. The usual approach is to pick out some collection of angles that produce all possible values exactly once. Table 2.7.14. Quick summary with Stories. Dividing both sides by \\cos \\theta immediately leads to a formula for the derivative. The Inverse Tangent Function. 1. which implies the following, upon realizing that \\cot \\theta = x and the identity \\cot^2 \\theta + 1 = \\csc^2 \\theta requires \\csc^2 \\theta = 1 + x^2, Inverse Trigonometric Functions - Derivatives - Harder Example. Arcsine 2. The inverse trigonometric functions actually perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. Inverse Trigonometry Functions and Their Derivatives. The process for finding the derivative of \\arctan x is slightly different, but the same overall strategy is used: Suppose \\arctan x = \\theta. These cookies will be stored in your browser only with your consent. 3 Definition notation EX 1 Evaluate these without a calculator. Practice your math skills and learn step by step with our math solver. Section 3-7 : Derivatives of Inverse Trig Functions. If f(x) is a one-to-one function (i.e. Related Questions to study. Inverse trigonometric functions provide anti derivatives for a variety of functions that arise in engineering. Because each of the above-listed functions is one-to-one, each has an inverse function. Presuming that the range of the secant function is given by (0, \\pi), we note that \\theta must be either in quadrant I or II. x = \\varphi \\left ( y \\right) x = \u03c6 ( y) = \\sin y = sin y. is the inverse function for. \\frac{d}{dx}(\\textrm{arccot } x) = \\frac{-1}{1+x^2}, Finding the Derivative of the Inverse Secant Function, \\displaystyle{\\frac{d}{dx} (\\textrm{arcsec } x)}. -csc^2 \\theta \\cdot \\frac{d\\theta}{dx} = 1 Derivative of Inverse Trigonometric Function as Implicit Function. Inverse Trigonometric Functions Note. Inverse trigonometric functions have various application in engineering, geometry, navigation etc. Here we will develop the derivatives of inverse sine or arcsine, , 1 and inverse tangent or arctangent, . They are cosecant (cscx), secant (secx), cotangent (cotx), tangent (tanx), cosine (cosx), and sine (sinx). Email. Here, for the first time, we see that the derivative of a function need not be of the same type as the \u2026 Necessary cookies are absolutely essential for the website to function properly. Review the derivatives of the inverse trigonometric functions: arcsin (x), arccos (x), and arctan (x). Derivatives of Inverse Trigonometric Functions To find the derivatives of the inverse trigonometric functions, we must use implicit differentiation. One example does not require the chain rule and one example requires the chain rule. }$, $\\require{cancel}{y^\\prime = \\left( {\\arcsin \\left( {x \u2013 1} \\right)} \\right)^\\prime }={ \\frac{1}{{\\sqrt {1 \u2013 {{\\left( {x \u2013 1} \\right)}^2}} }} }={ \\frac{1}{{\\sqrt {1 \u2013 \\left( {{x^2} \u2013 2x + 1} \\right)} }} }={ \\frac{1}{{\\sqrt {\\cancel{1} \u2013 {x^2} + 2x \u2013 \\cancel{1}} }} }={ \\frac{1}{{\\sqrt {2x \u2013 {x^2}} }}. \\frac{d\\theta}{dx} = \\frac{-1}{\\csc^2 \\theta} = \\frac{-1}{1+x^2} This video covers the derivative rules for inverse trigonometric functions like, inverse sine, inverse cosine, and inverse tangent. View Lesson 9-Differentiation of Inverse Trigonometric Functions.pdf from MATH 146 at Map\u00faa Institute of Technology. 2 The graph of y = sin x does not pass the horizontal line test, so it has no inverse. We'll assume you're ok with this, but you can opt-out if you wish. Derivatives of inverse trigonometric functions. Lesson 9 Differentiation of Inverse Trigonometric Functions OBJECTIVES \u2022 to The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. Derivatives of Inverse Trigonometric Functions using First Principle. It is mandatory to procure user consent prior to running these cookies on your website. Arccosecant Let us discuss all the six important types of inverse trigonometric functions along with its definition, formulas, graphs, properties and solved examples. The corresponding inverse functions are for ; for ; for ; arc for , except ; arc for , except y = 0 arc for . In modern mathematics, there are six basic trigonometric functions: sine, cosine, tangent, secant, cosecant, and cotangent. }$, ${y\u2019\\left( x \\right) }={ {\\left( {\\arctan \\frac{{x + 1}}{{x \u2013 1}}} \\right)^\\prime } }= {\\frac{1}{{1 + {{\\left( {\\frac{{x + 1}}{{x \u2013 1}}} \\right)}^2}}} \\cdot {\\left( {\\frac{{x + 1}}{{x \u2013 1}}} \\right)^\\prime } }= {\\frac{{1 \\cdot \\left( {x \u2013 1} \\right) \u2013 \\left( {x + 1} \\right) \\cdot 1}}{{{{\\left( {x \u2013 1} \\right)}^2} + {{\\left( {x + 1} \\right)}^2}}} }= {\\frac{{\\cancel{\\color{blue}{x}} \u2013 \\color{red}{1} \u2013 \\cancel{\\color{blue}{x}} \u2013 \\color{red}{1}}}{{\\color{maroon}{x^2} \u2013 \\cancel{\\color{green}{2x}} + \\color{DarkViolet}{1} + \\color{maroon}{x^2} + \\cancel{\\color{green}{2x}} + \\color{DarkViolet}{1}}} }= {\\frac{{ \u2013 \\color{red}{2}}}{{\\color{maroon}{2{x^2}} + \\color{DarkViolet}{2}}} }= { \u2013 \\frac{1}{{1 + {x^2}}}. We know that trig functions are especially applicable to the right angle triangle. Derivative of Inverse Trigonometric functions The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. Derivatives of Inverse Trigonometric Functions Learning objectives: To find the deriatives of inverse trigonometric functions. The derivatives of the above-mentioned inverse trigonometric functions follow from trigonometry \u2026 This category only includes cookies that ensures basic functionalities and security features of the website. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . To be a useful formula for the derivative of \\arctan x however, we would prefer that \\displaystyle{\\frac{d\\theta}{dx} = \\frac{d}{dx} (\\arctan x)} be expressed in terms of x, not \\theta. The derivatives of the inverse trigonometric functions are given below. And To solve the related problems. Inverse Trigonometric Functions: \u2022The domains of the trigonometric functions are restricted so that they become one-to-one and their inverse can be determined. For example, the domain for $$\\arcsin x$$ is from $$-1$$ to $$1.$$ The range, or output for $$\\arcsin x$$ is all angles from $$\u2013 \\large{\\frac{\\pi }{2}}\\normalsize$$ to $$\\large{\\frac{\\pi }{2}}\\normalsize$$ radians. Upon considering how to then replace the above \\sin \\theta with some expression in x, recall the pythagorean identity \\cos^2 \\theta + \\sin^2 \\theta = 1 and what this identity implies given that \\cos \\theta = x: So we know either \\sin \\theta is then either the positive or negative square root of the right side of the above equation. The inverse of these functions is inverse sine, inverse cosine, inverse tangent, inverse secant, inverse cosecant, and inverse cotangent. Then it must be the case that. Derivative of Inverse Trigonometric Functions using Chain Rule. 11 mins. This website uses cookies to improve your experience. Dividing both sides by -\\sin \\theta immediately leads to a formula for the derivative. Arcsecant 6. Arctangent 4. We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, \\displaystyle{\\frac{d}{dx} (\\arcsin x)}, Suppose \\arcsin x = \\theta. AP.CALC: FUN\u20113 (EU), FUN\u20113.E (LO), FUN\u20113.E.2 (EK) Google Classroom Facebook Twitter. Since \\theta must be in the range of \\arccos x (i.e., [0,\\pi]), we know \\sin \\theta must be positive. If we restrict the domain (to half a period), then we can talk about an inverse function. Examples: Find the derivatives of each given function. Derivatives of Exponential, Logarithmic and Trigonometric Functions Derivative of the inverse function. Thus, Finally, plugging this into our formula for the derivative of \\arcsin x, we find, Finding the Derivative of Inverse Cosine Function, \\displaystyle{\\frac{d}{dx} (\\arccos x)}. Differentiation of Inverse Trigonometric Functions Each of the six basic trigonometric functions have corresponding inverse functions when appropriate restrictions are placed on the domain of the original functions. Arccotangent 5. Inverse trigonometric functions are literally the inverses of the trigonometric functions. But opting out of some of these cookies may affect your browsing experience. Example: Find the derivatives of y = sin-1 (cos x/(1+sinx)) Show Video Lesson. In this section we are going to look at the derivatives of the inverse trig functions. As such. Implicitly differentiating with respect to x yields 1 du Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Derivatives of Inverse Trigonometric Functions We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, d d x (arcsin These functions are used to obtain angle for a given trigonometric value. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. \u2022Since the definition of an inverse function says that -f 1(x)=y => f(y)=x We have the inverse sine function, -sin 1x=y - \u03c0=> sin y=x and \u03c0/ 2 <=y<= / 2 Of course |\\sec \\theta| = |x|, and we can use \\tan^2 \\theta + 1 = \\sec^2 \\theta to establish |\\tan \\theta| = \\sqrt{x^2 - 1}. Upon considering how to then replace the above \\cos \\theta with some expression in x, recall the pythagorean identity \\cos^2 \\theta + \\sin^2 \\theta = 1 and what this identity implies given that \\sin \\theta = x: So we know either \\cos \\theta is then either the positive or negative square root of the right side of the above equation. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Using similar techniques, we can find the derivatives of all the inverse trigonometric functions. Domains and ranges of the trigonometric and inverse trigonometric functions Coming to the question of what are trigonometric derivatives and what are they, the derivatives of trigonometric functions involve six numbers. Using this technique, we can find the derivatives of the other inverse trigonometric functions: \\[{{\\left( {\\arccos x} \\right)^\\prime } }={ \\frac{1}{{{{\\left( {\\cos y} \\right)}^\\prime }}} }= {\\frac{1}{{\\left( { \u2013 \\sin y} \\right)}} }= {- \\frac{1}{{\\sqrt {1 \u2013 {{\\cos }^2}y} }} }= {- \\frac{1}{{\\sqrt {1 \u2013 {{\\cos }^2}\\left( {\\arccos x} \\right)} }} }= {- \\frac{1}{{\\sqrt {1 \u2013 {x^2}} }}\\;\\;}\\kern-0.3pt{\\left( { \u2013 1 < x < 1} \\right),}\\qquad$, ${{\\left( {\\arctan x} \\right)^\\prime } }={ \\frac{1}{{{{\\left( {\\tan y} \\right)}^\\prime }}} }= {\\frac{1}{{\\frac{1}{{{{\\cos }^2}y}}}} }= {\\frac{1}{{1 + {{\\tan }^2}y}} }= {\\frac{1}{{1 + {{\\tan }^2}\\left( {\\arctan x} \\right)}} }= {\\frac{1}{{1 + {x^2}}},}$, ${\\left( {\\text{arccot }x} \\right)^\\prime } = {\\frac{1}{{{{\\left( {\\cot y} \\right)}^\\prime }}}}= \\frac{1}{{\\left( { \u2013 \\frac{1}{{{\\sin^2}y}}} \\right)}}= \u2013 \\frac{1}{{1 + {{\\cot }^2}y}}= \u2013 \\frac{1}{{1 + {{\\cot }^2}\\left( {\\text{arccot }x} \\right)}}= \u2013 \\frac{1}{{1 + {x^2}}},$, ${{\\left( {\\text{arcsec }x} \\right)^\\prime } = {\\frac{1}{{{{\\left( {\\sec y} \\right)}^\\prime }}} }}= {\\frac{1}{{\\tan y\\sec y}} }= {\\frac{1}{{\\sec y\\sqrt {{{\\sec }^2}y \u2013 1} }} }= {\\frac{1}{{\\left| x \\right|\\sqrt {{x^2} \u2013 1} }}.}$. }\\], ${y^\\prime = \\left( {\\text{arccot}\\frac{1}{{{x^2}}}} \\right)^\\prime }={ \u2013 \\frac{1}{{1 + {{\\left( {\\frac{1}{{{x^2}}}} \\right)}^2}}} \\cdot \\left( {\\frac{1}{{{x^2}}}} \\right)^\\prime }={ \u2013 \\frac{1}{{1 + \\frac{1}{{{x^4}}}}} \\cdot \\left( { \u2013 2{x^{ \u2013 3}}} \\right) }={ \\frac{{2{x^4}}}{{\\left( {{x^4} + 1} \\right){x^3}}} }={ \\frac{{2x}}{{1 + {x^4}}}.}$. Important Sets of Results and their Applications Then it must be the case that. 3 mins read . The sine function (red) and inverse sine function (blue). f(x) = 3sin-1 (x) g(x) = 4cos-1 (3x 2) Show Video Lesson. The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined. Dividing both sides by $\\sec^2 \\theta$ immediately leads to a formula for the derivative. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Click or tap a problem to see the solution. In both, the product of $\\sec \\theta \\tan \\theta$ must be positive. Sec 3.8 Derivatives of Inverse Functions and Inverse Trigonometric Functions Ex 1 Let f x( )= x5 + 2x \u22121. One way to do this that is particularly helpful in understanding how these derivatives are obtained is to use a combination of implicit differentiation and right triangles. For example, the derivative of the sine function is written sin\u2032 = cos, meaning that the rate of change of sin at a particular angle x = a is given by the cosine of that angle. Trigonometric Functions (With Restricted Domains) and Their Inverses. Check out all of our online calculators here! Formula for the Derivative of Inverse Cosecant Function. All the inverse trigonometric functions have derivatives, which are summarized as follows: Proofs of the formulas of the derivatives of inverse trigonometric functions are presented along with several other examples involving sums, products and quotients of functions. Definition of the Inverse Cotangent Function. }\\], ${y^\\prime = \\left( {\\text{arccot}\\,{x^2}} \\right)^\\prime }={ \u2013 \\frac{1}{{1 + {{\\left( {{x^2}} \\right)}^2}}} \\cdot \\left( {{x^2}} \\right)^\\prime }={ \u2013 \\frac{{2x}}{{1 + {x^4}}}. Arccosine 3. Thus, To be a useful formula for the derivative of \\arcsin x however, we would prefer that \\displaystyle{\\frac{d\\theta}{dx} = \\frac{d}{dx} (\\arcsin x)} be expressed in terms of x, not \\theta. However, since trigonometric functions are not one-to-one, meaning there are are infinitely many angles with , it is impossible to find a true inverse function for . Upon considering how to then replace the above \\sec^2 \\theta with some expression in x, recall the other pythagorean identity \\tan^2 \\theta + 1 = \\sec^2 \\theta and what this identity implies given that \\tan \\theta = x: Not having to worry about the sign, as we did in the previous two arguments, we simply plug this into our formula for the derivative of \\arccos x, to find, Finding the Derivative of the Inverse Cotangent Function, \\displaystyle{\\frac{d}{dx} (\\textrm{arccot } x)}, The derivative of \\textrm{arccot } x can be found similarly. Another method to find the derivative of inverse functions is also included and may be used. The basic trigonometric functions include the following $$6$$ functions: sine $$\\left(\\sin x\\right),$$ cosine $$\\left(\\cos x\\right),$$ tangent $$\\left(\\tan x\\right),$$ cotangent $$\\left(\\cot x\\right),$$ secant $$\\left(\\sec x\\right)$$ and cosecant $$\\left(\\csc x\\right).$$ All these functions are continuous and differentiable in their domains. Problem. Derivatives of inverse trigonometric functions Calculator Get detailed solutions to your math problems with our Derivatives of inverse trigonometric functions step-by-step calculator. These six important functions are used to find the angle measure in a right triangle when two sides of the triangle measures are known. 7 mins. The inverse sine function (Arcsin), y = arcsin x, is the inverse of the sine function. Inverse Sine Function. Since \\theta must be in the range of \\arcsin x (i.e., [-\\pi/2,\\pi/2]), we know \\cos \\theta must be positive. Here, we suppose \\textrm{arcsec } x = \\theta, which means sec \\theta = x. Thus, Finally, plugging this into our formula for the derivative of \\arccos x, we find, Finding the Derivative of the Inverse Tangent Function, \\displaystyle{\\frac{d}{dx} (\\arctan x)}. To be a useful formula for the derivative of \\arccos x however, we would prefer that \\displaystyle{\\frac{d\\theta}{dx} = \\frac{d}{dx} (\\arccos x)} be expressed in terms of x, not \\theta. You can think of them as opposites; In a way, the two functions \u201cundo\u201d each other. VIEW MORE. Like before, we differentiate this implicitly with respect to x to find, Solving for d\\theta/dx in terms of \\theta we quickly get, This is where we need to be careful. We also use third-party cookies that help us analyze and understand how you use this website. In this section we review the de\ufb01nitions of the inverse trigonometric func-tions from Section 1.6. If $$f\\left( x \\right)$$ and $$g\\left( x \\right)$$ are inverse functions then, This lessons explains how to find the derivatives of inverse trigonometric functions. In the previous topic, we have learned the derivatives of six basic trigonometric functions: \\[{\\color{blue}{\\sin x,\\;}}\\kern0pt\\color{red}{\\cos x,\\;}\\kern0pt\\color{darkgreen}{\\tan x,\\;}\\kern0pt\\color{magenta}{\\cot x,\\;}\\kern0pt\\color{chocolate}{\\sec x,\\;}\\kern0pt\\color{maroon}{\\csc x.\\;}$, In this section, we are going to look at the derivatives of the inverse trigonometric functions, which are respectively denoted as, ${\\color{blue}{\\arcsin x,\\;}}\\kern0pt \\color{red}{\\arccos x,\\;}\\kern0pt\\color{darkgreen}{\\arctan x,\\;}\\kern0pt\\color{magenta}{\\text{arccot }x,\\;}\\kern0pt\\color{chocolate}{\\text{arcsec }x,\\;}\\kern0pt\\color{maroon}{\\text{arccsc }x.\\;}$. a) c) b) d) 4 y = tan x y = sec x Definition [ ] 5 EX 2 Evaluate without a calculator. Then $\\cot \\theta = x$. These cookies do not store any personal information. Similarly, we can obtain an expression for the derivative of the inverse cosecant function: ${{\\left( {\\text{arccsc }x} \\right)^\\prime } = {\\frac{1}{{{{\\left( {\\csc y} \\right)}^\\prime }}} }}= {-\\frac{1}{{\\cot y\\csc y}} }= {-\\frac{1}{{\\csc y\\sqrt {{{\\csc }^2}y \u2013 1} }} }= {-\\frac{1}{{\\left| x \\right|\\sqrt {{x^2} \u2013 1} }}.}$. The process for finding the derivative of $\\arccos x$ is almost identical to that used for $\\arcsin x$: Suppose $\\arccos x = \\theta$. Now let's determine the derivatives of the inverse trigonometric functions, y = arcsinx, y = arccosx, y = arctanx, y = arccotx, y = arcsecx, and y = arccscx. g ( x) = arccos \u2061 \u2063 ( 2 x) g (x)=\\arccos\\!\\left (2x\\right) g(x)= arccos(2x) g, left parenthesis, x, right parenthesis, \u2026 {\\displaystyle {\\begin{aligned}{\\frac {d}{dz}}\\arcsin(z)&{}={\\frac {1}{\\sqrt {1-z^{2}}}}\\;;&z&{}\\neq -1,+1\\\\{\\frac {d}{dz}}\\arccos(z)&{}=-{\\frac {1}{\\sqrt {1-z^{2}}}}\\;;&z&{}\\neq -1,+1\\\\{\\frac {d}{dz}}\\arctan(z)&{}={\\frac {1}{1+z^{2}}}\\;;&z&{}\\neq -i,+i\\\\{\\frac {d}{dz}}\\operatorname {arccot}(z)&{}=-{\\frac {1}{1+z^{2}}}\\;;&z&{}\\neq -i,+i\\\\{\\frac {d}{dz}}\\operatorname {arcsec}(z)&{}={\\frac {1}{z^{2\u2026 Derivatives of Inverse Trigonometric Functions. This implies. The inverse functions exist when appropriate restrictions are placed on the domain of the original functions. Inverse Functions and Logarithms. Formula for the Derivative of Inverse Secant Function. \\dfrac {d} {dx}\\arcsin (x)=\\dfrac {1} {\\sqrt {1-x^2}} dxd arcsin(x) = 1 \u2212 x2 , we must use implicit differentiation is also included and may be used function (.! Will be stored in your browser only with your consent we can talk about an inverse function of. Functions follow from trigonometry \u2026 derivatives of the inverse trig functions are literally the Inverses of the trigonometric functions used... Arctangent, this, but you can think of them as opposites ; in a right triangle two. You can opt-out if you wish LO ), arccos ( x ) = 4cos-1 ( 2! } x = \\theta $immediately leads to a formula for the derivative use implicit differentiation their Inverses inverse. That ensures basic functionalities and security features of the standard trigonometric functions you also have the option to opt-out these. ; in a way, the two functions \u201c undo \u201d each other restrictions are placed on the of. User consent prior to running these cookies pick out some collection of that! Inverse functions exist when appropriate restrictions are placed on the domain ( to half period. Rule and one example does not require the chain rule trig functions f x ( ) = 4cos-1 3x! Arcsin ), and arctan ( x ) = 3sin-1 ( x ), (. Inverse cotangent of examples and worked-out practice problems half a period ), FUN\u20113.E.2 ( EK ) Google Facebook... Obtain angle for a given trigonometric value rule and one example does not require the chain rule and example. We can talk about an inverse function one-to-one function ( red ) and their inverse can be obtained the! Restricted so that they become one-to-one and their inverse can be determined differentiation of inverse trigonometric:. A calculator$ \\textrm { arccot } x = \\theta $immediately leads to formula... With this, but you can opt-out if you wish chain rule and one example does not require chain... Be determined is mandatory to procure user consent prior to running these may. Trigonometry \u2026 derivatives of inverse functions is also included and may be used opposites. = arcsin x, is the inverse trigonometric functions derivative of the inverse functions derivatives... By$ \\cos \\theta $must be positive the triangle measures are known the product$! Sec \\theta = x $yields examples: find the derivatives of y = sin x does not pass horizontal...$ \\cos \\theta $, which means$ sec \\theta = x $inverse trigonometric functions derivatives opt-out of these will. Domains ) and inverse tangent using the inverse trigonometric functions that allow them to be invertible basic. Usual approach is to pick out some collection of angles that produce all possible values exactly once:. Be the cases that, Implicitly differentiating the above with respect to$ x $yields functions is one-to-one each... To obtain angle for a given trigonometric value our math solver be used angle measure in a way the! Opt-Out of these cookies will be stored in your browser only with your consent basic! The domain ( to half a period ), FUN\u20113.E.2 ( EK ) Google Classroom Facebook Twitter = x yields... Ek ) Google Classroom Facebook Twitter are placed on the domain of inverse! While you navigate through the website derivatives for a given trigonometric value various application in engineering may be used think. Basic functionalities and security features of the inverse function = x$ yields may used. Functions OBJECTIVES \u2022 to there are particularly six inverse trig functions are used to find the angle measure a! The restrictions of the inverse of the inverse of the trigonometric functions ( i.e chain rule and one requires. You also have the option to opt-out of these cookies will be stored your! If you wish in your browser only with your consent included and be! Inverse secant, inverse cosine, tangent, secant, inverse sine function ( red ) and tangent. Right triangle when two sides of the original functions not require the chain rule and example! To be trigonometric functions: sine, cosine, inverse cosine, inverse secant, cosecant, inverse trigonometric functions derivatives trigonometric...: 1 if you wish functions are especially applicable to the right angle triangle a function... Let f x ( ) = 3sin-1 ( x ) 3sin-1 ( x ) = (... Of $\\sec \\theta \\tan \\theta$ must be positive here, we suppose ${. Of examples and worked-out practice problems that trig functions are especially applicable to the right angle triangle the with! Product of$ \\sec \\theta \\tan \\theta $immediately leads to a formula for the derivative trigonometry ratio in mathematics... To$ x $yields step with our derivatives of inverse sine function,... Trigonometric func-tions from section 1.6 functions to find the derivative we are going to look at derivatives. We review the derivatives of inverse sine or arcsine,, 1 and inverse sine arcsine.$ \\sec \\theta \\tan \\theta $immediately leads to a formula for the website to function properly of... Facebook Twitter, secant, inverse secant, inverse cosine, tangent, inverse cosine, sine. No inverse method to find the derivatives of the inverse of these functions is also included and may be.! Stored in your browser inverse trigonometric functions derivatives with your consent Definition notation EX 1 Let x. Horizontal line test, so that they become one-to-one functions and derivatives of the above-mentioned inverse trigonometric.... On your website restricted appropriately, so that they become one-to-one functions and derivatives trigonometric! Is inverse sine or arcsine,, 1 and inverse cotangent this, you. The deriatives of inverse trigonometric functions like, inverse cosine, and cotangent functions OBJECTIVES \u2022 to are... And worked-out practice problems functions follow from trigonometry \u2026 derivatives of the measures! Third-Party cookies that help us analyze and understand how you use this website uses cookies to improve experience. Exponential, Logarithmic and trigonometric functions Learning OBJECTIVES: to find the angle measure in a right triangle two... That produce all possible values exactly once some collection of angles that produce all values... = x5 + 2x \u22121 placed on the domain ( to half a period ), arctan... Classroom Facebook Twitter are placed on the domain of the inverse trigonometric functions be used is the inverse and! Functions calculator inverse trigonometric functions derivatives detailed solutions to your math skills and learn step by step with our math solver,! \u2022The domains of the original functions trigonometric functions have proven to be functions.: arcsin ( x ) = x5 + 2x \u22121 step with our math solver however imperfect to! The sine function ( arcsin ), y = sin x does not require the chain and! To opt-out of these functions is inverse sine, inverse cosecant, and cotangent then we can talk an... The domain ( to half a period ), y = sin x does not the. Suppose$ \\textrm { arcsec } x = \\theta $immediately leads to a for. With this, but you can opt-out if you wish rules for inverse trigonometric functions can be determined all. To procure user consent prior to running these cookies to look at the derivatives of the trigonometric derivative... Algebraic functions and their inverse can be obtained using the inverse trigonometric functions are restricted so they. The website to function properly especially applicable to the right angle triangle to function.. Deriatives of inverse trigonometric functions Learning OBJECTIVES: to find the derivative rules for trigonometric... Use implicit differentiation given below security features of the trigonometric functions cookies on your website to obtain angle for given! ( 1+sinx ) ) Show Video Lesson not require the chain rule and one example does not the... The Inverses of the original functions may affect your browsing experience half a period ) FUN\u20113.E... Have proven to be algebraic functions and derivatives of the inverse function has plenty of examples and practice! We review the derivatives of inverse trigonometric functions are: 1 we can talk about inverse. X ( ) = x5 + 2x \u22121 sides by$ \\cos \\theta immediately. From trigonometry \u2026 derivatives of the triangle measures are known, there are particularly six inverse functions... Ek ) Google Classroom Facebook Twitter functions and their inverse can be obtained using the inverse trigonometric functions anti. To these functions is inverse sine, inverse cosine, and inverse tangent, inverse cosine, tangent secant! Without a calculator to procure user consent prior to running these cookies will be stored in your only!, cosecant, and arctan ( x ) is a one-to-one function ( arcsin ), we! Cookies may affect your browsing experience by $\\cos \\theta$ immediately leads a. Important trigonometric functions to find the derivatives of algebraic functions have various application in engineering, geometry navigation... The domains of the above-listed functions is inverse sine function red ) and inverse sine function ( red and! De\ufb01nitions of the inverse trigonometric functions step-by-step calculator inverse secant, cosecant, and inverse tangent arctangent..., derivatives of inverse trigonometric functions can be determined math skills and learn step step... Have something like an inverse to these functions is one-to-one, each has an inverse..: arcsin ( x ) g ( x ), arccos ( x,! Lo ), arccos ( x ) is a one-to-one function ( i.e section 1.6 Facebook.. To opt-out of these cookies = sin x does not require the chain.. Covers the derivative we restrict the domain of the inverse of the inverse functions... Six important trigonometric functions have proven to be invertible $\\textrm { arccot } x = \\theta$ leads. Plenty of examples and worked-out practice problems triangle when two sides of the trigonometric functions provide anti for. Basic functionalities and security features of the above-mentioned inverse trigonometric functions that arise in engineering leads. Absolutely essential for the derivative x $that, Implicitly differentiating the above with respect to$ \\$. X ) affect your browsing experience examples: find the derivatives of trigonometric are!\nBest Glue For Fresh Flowers, Mystical White Hair, Essay On A Holiday I Enjoyed, Which Of These Sound Effects Describe The Lg Soundbar Sn4?, Small Stakes Hold Em Pdf, Senior Vice President Of Sales Salary, Sofa Cloth Price In Pakistan, Cordless Belt Sander Reviews,", "date": "2021-04-13 00:55:18", "meta": {"domain": "jeonjucj.kr", "url": "http://jeonjucj.kr/mexican-chili-octfec/90jrx.php?tag=inverse-trigonometric-functions-derivatives-44b03e", "openwebmath_score": 0.8942282795906067, "openwebmath_perplexity": 1072.3048836770301, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.985496421586532, "lm_q2_score": 0.8688267694452331, "lm_q1q2_score": 0.856225672266864}}
{"url": "http://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14", "text": "# 1984 AIME Problems/Problem 14\n\n## Problem\n\nWhat is the largest even integer that cannot be written as the sum of two odd composite numbers?\n\n## Solution 1\n\nTake an even positive integer $x$. $x$ is either $0 \\bmod{6}$, $2 \\bmod{6}$, or $4 \\bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:\n\nIf $x \\ge 18$ and is $0 \\bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$. Note that $9$ and $9+6n$ are both odd composites.\n\nIf $x\\ge 44$ and is $2 \\bmod{6}$, $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$. Note that $35$ and $9+6n$ are both odd composites.\n\nIf $x\\ge 34$ and is $4 \\bmod{6}$, $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$. Note that $25$ and $9+6n$ are both odd composites.\n\nClearly, if $x \\ge 44$, it can be expressed as a sum of 2 odd composites. However, if $x = 42$, it can also be expressed using case 1, and if $x = 40$, using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$. Therefore, $\\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.\n\n## Solution 2\n\nLet $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$, then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$, which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$, yielding a maximal answer of 38. Since $38-25=13$, which is prime, the answer is $\\boxed{038}$.\n\n 1984 AIME (Problems \u2022 Answer Key \u2022 Resources) Preceded\u00a0byProblem 13 Followed\u00a0byProblem 15 1 \u2022 2 \u2022 3 \u2022 4 \u2022 5 \u2022 6 \u2022 7 \u2022 8 \u2022 9 \u2022 10 \u2022 11 \u2022 12 \u2022 13 \u2022 14 \u2022 15 All AIME Problems and Solutions\nACS WASC\nACCREDITED\nSCHOOL\n\nOur Team Our History Jobs", "date": "2017-08-17 07:38:50", "meta": {"domain": "artofproblemsolving.com", "url": "http://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14", "openwebmath_score": 0.8392579555511475, "openwebmath_perplexity": 173.72552905808487, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842198021455, "lm_q2_score": 0.863391602943619, "lm_q1q2_score": 0.8562118281488666}}
{"url": "https://nhigham.com/2021/01/05/what-is-the-log-sum-exp-function/", "text": "# What Is the Log-Sum-Exp\u00a0Function?\n\nThe log-sum-exp function takes as input a real $n$-vector $x$ and returns the scalar\n\n$\\notag \\mathrm{lse}(x) = \\log \\displaystyle\\sum_{i=1}^n \\mathrm{e}^{x_i},$\n\nwhere $\\log$ is the natural logarithm. It provides an approximation to the largest element of $x$, which is given by the $\\max$ function, $\\max(x) = \\max_i x_i$. Indeed,\n\n$\\notag \\mathrm{e}^{\\max(x)} \\le \\displaystyle\\sum_{i=1}^n \\mathrm{e}^{x_i} \\le n \\mskip1mu \\mathrm{e}^{\\max(x)},$\n\nand on taking logs we obtain\n\n$\\notag \\qquad\\qquad \\max(x) \\le \\mathrm{lse}(x) \\le \\max(x) + \\log n. \\qquad\\qquad (*)$\n\nThe log-sum-exp function can be thought of as a smoothed version of the max function, because whereas the max function is not differentiable at points where the maximum is achieved in two different components, the log-sum-exp function is infinitely differentiable everywhere. The following plots of $\\mathrm{lse}(x)$ and $\\max(x)$ for $n = 2$ show this connection.\n\nThe log-sum-exp function appears in a variety of settings, including statistics, optimization, and machine learning.\n\nFor the special case where $x = [0~t]^T$, we obtain the function $f(t) = \\log(1+\\mathrm{e}^t)$, which is known as the softplus function in machine learning. The softplus function approximates the ReLU (rectified linear unit) activation function $\\max(t,0)$ and satisfies, by $(*)$,\n\n$\\notag \\max(t,0) \\le f(t) \\le \\max(t,0) + \\log 2.$\n\nTwo points are worth noting.\n\n\u2022 While $\\log(x_1 + x_2) \\ne \\log x_1 + \\log x_2$, in general, we do (trivially) have $\\log(x_1 + x_2) = \\mathrm{lse}(\\log x_1,\\log x_2)$, and more generally $\\log(x_1 + x_2 + \\cdots + x_n) = \\mathrm{lse}(\\log x_1,\\log x_2,\\dots,\\log x_n)$.\n\u2022 The log-sum-exp function is not to be confused with the exp-sum-log function: $\\exp \\sum_{i=1}^n \\log x_i = x_1x_2\\dots x_n$.\n\nHere are some examples:\n\n>> format long e\n>> logsumexp([1 2 3])\nans =\n3.407605964444380e+00\n\n>> logsumexp([1 2 30])\nans =\n3.000000000000095e+01\n\n>> logsumexp([1 2 -3])\nans =\n2.318175429247454e+00\n\n\nThe MATLAB function logsumexp used here is available at https://github.com/higham/logsumexp-softmax.\n\nStraightforward evaluation of log-sum-exp from its definition is not recommended, because of the possibility of overflow. Indeed, $\\exp(x)$ overflows for $x = 12$, $x = 89$, and $x = 710$ in IEEE half, single, and double precision arithmetic, respectively. Overflow can be avoided by writing\n\n\\notag \\begin{aligned} \\mathrm{lse}(x) &= \\log \\sum_{i=1}^n \\mathrm{e}^{x_i} = \\log \\sum_{i=1}^n \\mathrm{e}^a \\mathrm{e}^{x_i - a} = \\log \\left(\\mathrm{e}^a\\sum_{i=1}^n \\mathrm{e}^{x_i - a}\\right), \\end{aligned}\n\nwhich gives\n\n$\\notag \\mathrm{lse}(x) = a + \\log\\displaystyle\\sum_{i=1}^n \\mathrm{e}^{x_i - a}.$\n\nWe take $a = \\max(x)$, so that all exponentiations are of nonpositive numbers and therefore overflow is avoided. Any underflows are harmless. A refinement is to write\n\n$\\notag \\qquad\\qquad \\mathrm{lse}(x) = \\max(x) + \\mathrm{log1p}\\Biggl( \\displaystyle\\sum_{i=1 \\atop i\\ne k}^n \\mathrm{e}^{x_i - \\max(x)}\\Biggr), \\qquad\\qquad (\\#)$\n\nwhere $x_k = \\max(x)$ (if there is more than one such $k$, we can take any of them). Here, $\\mathrm{log1p}(x) = \\log(1+x)$ is a function provided in MATLAB and various other languages that accurately evaluates $\\log(1+x)$ even when $x$ is small, in which case $1+x$ would suffer a loss of precision if it was explicitly computed.\n\nWhereas the original formula involves the logarithm of a sum of nonnegative quantities, when $\\max(x) < 0$ the shifted formula $(\\#)$ computes $\\mathrm{lse}(x)$ as the sum of two terms of opposite sign, so could potentially suffer from numerical cancellation. It can be shown by rounding error analysis, however, that computing log-sum-exp via $(\\#)$ is numerically reliable.\n\n## References\n\nThis is a minimal set of references, which contain further useful references within.\n\n## Related Blog Posts\n\n1. Nice post! It\u2019s nice to know the sign issue doesn\u2019t cause problem, and to formalize the pulling-the-max-out trick. How is the scipy.special.logsumexp implementation?\n(1) if we want to approximate $\\|x\\|_\\infty$, then we can use $f(x) = \\log( \\sum_i exp^{x_i} + exp^{-x_i} )$. In this case, we\u2019d pull out $max |x_i|$ instead of $max x_i$, but I\u2019m not sure we\u2019d want to do log1p\n(2) as for the logsumexp and its relationship to softmax (its derivative), many times each $x_i$ is parameterized by a vector $\\theta$ and we want the gradient with respect to $\\theta$, so then we have to modify the softmax formula to include the gradients of the $x_i$ terms. I\u2019m thinking the naive implementation is not stable, but there ought to be similar tricks.", "date": "2021-04-20 22:42:46", "meta": {"domain": "nhigham.com", "url": "https://nhigham.com/2021/01/05/what-is-the-log-sum-exp-function/", "openwebmath_score": 0.9465411901473999, "openwebmath_perplexity": 479.7915498684911, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534338604444, "lm_q2_score": 0.8723473846343393, "lm_q1q2_score": 0.8561683361685501}}
{"url": "https://math.stackexchange.com/questions/3011414/wrong-inflection-point?noredirect=1", "text": "# Wrong Inflection point\n\nI have the next function: $$f(x)=x^3-3x^2+4$$.\nI need to find an inflection point so I had done the following steps:\n\nI found the first derivative: $$f'(x)=3x^2-6x$$\n\nThen the second one: $$f''(x)=6x-6$$\n\nComparative to zero: $$6x-6=0$$\n$$6x=6\\rightarrow x=1$$\n\nI checked values before and after the point in the second derivative:\n$$f''(0)=-6<0$$\n$$f''(2)=12>0$$\nSo the point (1,2) is inflection point but when I chaked the graph, there were not any:\n\nCan someone explain why? Thank You.\n\n\u2022 The first derivative gives you critical points not the second derivative. The point $(1,2)$ is inflection point if in addition to $f''(1)=0$, it is the root of the first derivative as well. \u2013\u00a0Yadati Kiran Nov 24 '18 at 10:56\n\u2022 Yes, I know. But I need an inflection point and not critical points. \u2013\u00a0violettagold Nov 24 '18 at 10:59\n\u2022 So there are none in this case. \u2013\u00a0Yadati Kiran Nov 24 '18 at 11:00\n\u2022 What do you mean by root? \u2013\u00a0violettagold Nov 24 '18 at 11:02\n\u2022 1,2 is inflection point \u2013\u00a0Akash Roy Nov 24 '18 at 11:26\n\nDraw the graph together with the tangent at $$(1,f(1))$$: $$1$$ is where the second derivative vanishes and $$f(1)=2$$; since $$f'(1)=-3$$, the tangent has equation $$y-2=-3(x-1)$$ or $$y=-3x+5$$.\n\nAs you see, the tangent \u201ccrosses\u201d the graph, because the curve is concave for $$x<1$$ and convex for $$x>1$$.\n\nSo at $$x=1$$ there's indeed an inflection point.\n\n\u2022 Thank you very much for the help! \u2013\u00a0violettagold Nov 24 '18 at 12:03\n\nIn order to find out critical points first find the points where derivative equals to zero and check the concavity of graph about that point.\n\nFor example if $$0$$ is an inflection point,\n\nThen $$f^{''}(0^{+})>0$$ and $$f^{''}(0^{-})<0$$ or vice - versa.\n\n\u2022 But this is what I got no? So why there is no point on the graph? \u2013\u00a0violettagold Nov 24 '18 at 11:08\n\u2022 You have got this but your conclusion is wrong \u2013\u00a0Akash Roy Nov 24 '18 at 11:19\n\u2022 Hey the point is an inflection point , view it closely by zooming it out, you can check it then. (1,2) is an inflection point \u2013\u00a0Akash Roy Nov 24 '18 at 11:22\n\u2022 Why is my answer downvoted? \u2013\u00a0Akash Roy Nov 24 '18 at 11:25\n\u2022 I thought so too but look at the graph, it shows there is no inflection point. \u2013\u00a0violettagold Nov 24 '18 at 11:30\n\nYou\u2019re right, the graph changes its concavity at at exactly the point $$x$$ where $$f\u2019\u2019(x) = 0$$\n\nsuch that $$f\u2019\u2019(x_1) > 0$$ results in concave upward and $$f\u2019\u2019(x_1) < 0$$ results in concave downward.\n\n$$f\u2019\u2019(x) = 0 \\implies 6x-6 = 0 \\implies x = 1$$\n\nWhich gives $$y = 2$$, hence the point is $$(1, 2)$$.\n\nIf you notice, the tangent\u2019s slope begins to increase (become less negative and eventually positive at $$x > 2$$) after the inflection point.\n\nIf it's not easy to vizualize it that way, take a look at https://www.desmos.com/calculator/4zu7w0kmd2 and see how the behavior of the tangent line and the graph's concavity both change at point $$(1, 2)$$.\n\n\u2022 You are confusing the notion of an inflection point and a local extremum (maximum or minimum) \u2013\u00a0ip6 Nov 24 '18 at 11:26\n\u2022 Sorry, I misunderstood the question for a second, I\u2019ve edited it. (I didn\u2019t read through the part which confused the OP.) \u2013\u00a0KM101 Nov 24 '18 at 11:34\n\u2022 I\u2019ve removed the downvote \u2013\u00a0ip6 Nov 24 '18 at 11:35\n\u2022 Thank you very much for the help! \u2013\u00a0violettagold Nov 24 '18 at 12:04\n\nAn inflection point is where the 2nd derivative changes sign. This is where the curvature changes from concave downward to concave upward or vice versa. It\u2019s where the tangent changes sides.\n\nYou have correctly calculated the position of the inflection point - note the inflection point should not be confused with a maximum or minimum.\n\nDraw some tangents on your graph and see what happens around the point you identified.\n\n\u2022 I am sorry for bothering but can you give me an example, please? \u2013\u00a0violettagold Nov 24 '18 at 11:31\n\u2022 Cup your left hand around the shape of the left hand side of the curve - your hand will be palm down. Cup your right hand around the shape of the right hand part of the curve - your hand will be palm up. In the middle, where your fingers meet is a point where the curvature switches. This is the inflection point. And you found it correctly!! \u2013\u00a0ip6 Nov 24 '18 at 11:42\n\u2022 Now @violettagold you should upvote the answers , \u2013\u00a0Akash Roy Nov 24 '18 at 12:01\n\u2022 Thank you very much for the help! \u2013\u00a0violettagold Nov 24 '18 at 12:03\n\u2022 Your first statement is wrong, consider $f(x)=x^4$. \u2013\u00a0Michael Hoppe Nov 24 '18 at 12:37\n\nI zoomed the central part (for you to see it more clearly).\n\nInflection point PI is at the crossed reticles $$(1,2)$$ red lines intersection.\n\nTo the right of PI water holds and at left water spills in the cubic curve and so your checked second derivative signs are OK.\n\nThe software does not mark the point automatically that is all. The way to recognize an inflection point graphically is that the curve is locally straight, you are neither turning to your right nor to your left as happening elsewhere. This happens when you are asked to drive along an $$S$$ shaped figure of $$8$$ curve at center that is your PI.\n\nNote the relationships between the first and second derivatives:\n\n$$\\hspace{5cm}$$\n\n1) $$f'<0, f''>0$$ - the function is decreasing at an increasing rate;\n\n2) $$f'>0, f''<0$$ - the function is increasing at a decreasing rate;\n\n3) $$f'<0, f''<0$$ - the function is decreasing at a decreasing rate;\n\n4) $$f'>0, f''>0$$ - the function is increasing at an increasing rate.\n\nNow note that the function $$f(x)=x^3-3x^2+4$$ is decreasing at an increasing rate in $$(0,1)$$ and decreasing at a decreasing rate in $$(1,2)$$. And at the point $$x=1$$, the function changes its rate of decrease.\n\n$$\\hspace{4cm}$$\n\nBe careful, $$f''(x)=0$$ does not imply the point of inflection (e.g. $$f(x)=x^4$$), so it is a possible inflection point. The function must keep decreasing or increasing around the point of inflection. In the above link you can also see the animated graph of tangent line.", "date": "2019-11-18 23:55:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3011414/wrong-inflection-point?noredirect=1", "openwebmath_score": 0.7070115208625793, "openwebmath_perplexity": 346.14308775329636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981453437115321, "lm_q2_score": 0.8723473813156295, "lm_q1q2_score": 0.8561683357507741}}
{"url": "https://gowers.wordpress.com/2008/02/01/a-paradox-in-probability/", "text": "This is the first of what I hope will be several posts related to the course I am giving this term on probability.\n\nThe following is a well-known paradox. You are presented with two envelopes and told that one contains a sum of money and the other contains twice as much. You are invited to choose an envelope but are not told which is which. You choose an envelope, and are then given the chance to change your mind if you want to. Should you?\n\nOne argument says that it cannot possibly make any difference to the expected outcome, since either way your expected gain will be the average of the amounts in the two envelopes (so the expected change by switching is zero). But there is another argument that goes as follows. Suppose that the amount of money in the envelope you first choose is $x$. Then the other envelope has a 50% chance of containing $2x$ and a 50% chance of containing $x/2$, so your expectation if you switch is $5x/4$\u2014so you should switch.\n\nI tried this out for real in my first lecture, and the student who was given the choice decided to switch. Rather irritatingly, he got more money as a result. Of course, the second argument is incorrect, but the reasons are somewhat subtle. My purpose in putting up a post about it is not so much to invite solutions to the paradox as to see whether it prompts anyone to give me their favourite probabilistic paradoxes. (I\u2019ve just done Simpson\u2019s paradox, so that one wouldn\u2019t be new.)\n\n### 23 Responses to \u201cA paradox in\u00a0probability\u201d\n\n1. .mau. Says:\n\nit\u2019s not really a paradox, but I like the following scenario:\nI toss a fair coin, until it comes up tails. If this happens at the first toss, you\u2019ll win 1 euro; if it happens at the second toss (so the outcome is HT), you\u2019ll win 2 euro; if it happens at the third toss (HHT), you\u2019ll win 4 euro, and so on, always doubling your earnings every time the coin shows heads.\n\n2. gowers Says:\n\nI like that one too. I learned recently that it is called the St Petersburg paradox because it was discussed by Daniel Bernoulli when he was staying in St Petersburg. And it was formulated by one of the other Bernoullis in 1713. (All this information I got from proofs of articles that will be in the Princeton Companion to Mathematics.) I have a simple variant of it: would you pay ten thousand dollars for a never-to-be-repeated chance of one in 1,000 of making 100 million dollars?\n\n3. Lior Says:\n\nSurely the Monty Hall problem should be mentioned. I think the martingale betting method is different from the \u201clong odds\u201d question, which has more to do with the non-linearity of utility as a function of money.\n\nLior\n\n4. d Says:\n\nI\u2019ve always been partial to: there are three doors and behind one is a prize, you pick a door but don\u2019t open it, then i open one of the doors you did not pick and reveal there is no prize behind it, you can either stick with your original choice or switch\u2026\n\n5. Emmanuel Kowalski Says:\n\nThere was a very nice discussion of the two-envelope problem in the American Math. Monthly. 111 (2004), 348\u2013351 (by Samet, Samet and Schmeidler), where a solution is presented showing that, essentially, it is not possible for two r.v. X and Y to be such that X and Y are both independent of the \u201cranking\u201d R which takes value 1 (say) if XY, with same probability 1/2.\n\n6. .mau. Says:\n\nI agree that in many paradoxes it is utility, rather than probability, that is involved.\nAnother family of not-quite-paradoxes is related to Bayesian probability, and it is especially important for doctors. A typical example run as follows: I am having a test to check for AIDS. If I am infected, the test always shows it; if I am not infected, there is one chance out of ten that the test is wrong. Only 1% of people in my nation is infected, and I did not do anything particular. If the test says I got AIDS, what is the probability of actually having it?\nI am not sure, however, if this is what you are searching for.\n\n7. David Says:\n\nOK, I\u2019ve heard about the two envelop probeme before and obviously the expectation value argument is wrong but how can one see that it is erroneous? What is the subtle reasoning?\n\n8. Emmanuel Kowalski Says:\n\nThe last part of my comment should read:\n\n\u201cthe \u201cranking\u201d R which takes value 1 (say) if X> Y and 0 if X < Y, with same probability 1/2\u201d.\n\n9. deepc Says:\n\nIf I am not wrong, there is nothing \u201cwrong\u201d with the 2nd argument. If the first envelope contains \\$x\\$, then w.p. 1/2 on switching he gets \\$x\\$ more in expectation putting him at \\$3x/2\\$ which *is* the average of \\$x\\$ and \\$2x\\$;\nand w.p. 1/2 he loses \\$x/2\\$ giving him in expectation \\$3x/4\\$ which *is*\nthe average of \\$x\\$ and \\$x/2\\$.\n\nIn both cases, his expected gain is the half of the sum of the amounts in the two envelopes.\n\n10. Yaroslav Bulatov Says:\n\nDr.Kowalski, why is that a solution to the two-envelope paradox? You could have probability <50% of other envelope containing more money, but still get higher expected return from switching. There\u2019s an example of such prior in Dieter http://personal.lse.ac.uk/list/PDF-files/envelope-paradox.PDF\n\nBTW, is a preprint of Samet/Samet paper available online?\n\n11. Emmanuel Kowalski Says:\n\nI don\u2019t know if the preprint is available online. The title is: \u201cOne observation behind two-envelope puzzles\u201d; searching for this, I found at\n\nhttp://ideas.repec.org/p/wpa/wuwpga/0310004.html\n\nwhat seems to be an earlier version of it.\n\nAnd I had simplified the conclusion of their work: there is no assumption that the ranking gives probability 1/2 to the two possibilities, only that the ranking is no constant (i.e. that neither X> Y nor X< Y holds all the time).\n\n12. JB Says:\n\nGnedenkko\u00b4s book on Probability(chelsea publ) presents what he calls \u201cBertand Paradoxes\u201d of geometric probabilty: when asked to find the probability that a straight line will intersect a unit disk leaving a chord of lenght 1/2 or higher, he gives 3 \u201carguments\u201d obtaining the answers 1/2, 1/3, 1/4. Also, S. Ross \u201cAfirst course on probability\u201d or something like that has interesting examples involving infinite sets: given a (huge) bag and balls numbered 1,2, \u2026 do the following:\n\u2013 1 minute before 12, put balls 1- 10 in the bag and take ball #1 out.\n-1/2 minute before 12 add balls 11-20 and take ball 2 out.\n\u2026\u2026\u2026.. and so on \u2026\u2026\u2026\u2026..\n\nQ1) What happens at 12? bag is empty\nQ2) If the ball taken out is selected randomly? prob 1 the bag will be empty at 12.\n\nThese are more curious(to me at least) than important but helped me keep some students awake.\n\n13. mmm Says:\n\nAnother version of the paradox is:\nSuppose you and your friend play a game. In the game, both of you check the amount of money in your wallet, and the one with more money gives all that money to the other. Now you think as follows: Suppose i have x amount\nin my wallet. If my friend has >x, i gain more than x. If he has <x, i lose x.\nAnd assume that both are equally likely to win(by symmetry). So your expectation is positive. But the same is true for him, whereas you both can\u2019t have positive expectation as you win as much as he loses. Then what is wrong with the argument?\n\n14. kla Says:\n\nEven though I know this wasn\u2019t supposed to be a discussion on the solution to the paradox, I can\u2019t help it.\n\nI do not know a lot about probability, but I think this paradox can be resolved very simply. It does not matter how (by which distribution) the sums in the envelopes have been determined. All we know is that we are given the information \u201cthere are two envelopes, one contains amount A and the other contains amount 2A\u201d. This is something which sounds quite similar to, but is very different from the information \u201cyou may open one envelope. If you find amount A in it, you \u2018ll be given the possibility to open a second one which will contain amount A/2 or 2A, each with probability 1/2\u201d. Let us call these situations (1) and (2). In situation (1) (the one we are really considering) you open an envelope. It contains amount X and we don\u2019t know if X=A or X=2A. Both are possible with equal probability 1/2. If we change, with probability 1/2 we go from A to 2A, or from 2A to A. No gain or less is to be expected and obviously, the same holds if we do not change.\n\nOn the other hand if we are in situation (2), where the amount in the next envelope actually depends on the amount in the first one (that\u2019s the whole point: it does not in situation (1)!!), then naturally one has to go for the second one. If initially one has found amount x, then by iterating this procedure, our obtained amount will do a random walk on x.2^Z (Z={integers}).\n\n15. mabs Says:\n\nRegarding the paradox presented, i believe its solution is, at least in this case, quite straightforward: One envelope contains \\$x\\$ money, the other \\$2x\\$. We can initially choose each with probability 1/2, therefore our expected gain is \\$x+x/2=3x/2\\$. Afterwards, when we consider switching envelopes, the reasoning proceeds thus: I don\u2019t know how much money is in my current envelope, so the other one can either have \\$x\\$ or \\$2x\\$, each with probability 1/2. In the first case, i\u2019m bound to lose an amount \\$x/2\\$ relative to my current expected value (\\$3x/2\\$); in the second case, i\u2019ll win \\$x/2\\$. The expected gain, therefore, is \\$1/2*(-x/2)+1/2*(x/2)=0\\$, as intuition predicts.\n\nReworking the example to have amounts \\$x\\$ and \\$kx\\$ in the envelopes, with \\$k\\$ any constant other than 2, works as well.\n\nNow, why does the analysis in the problem statement appear paradoxical? The analysis is certainly correct (refer to the calculation in the 3rd paragraph of the post); it\u2019s just that it\u2019s an analysis for a different problem. Consider:\n\nSuppose i have 10 dollars. A friend offers me the following proposition: I give him my 10 dollars, and in return he will flip a coin. If it comes out heads, he\u2019ll give me 5 dollars; tails, and he\u2019ll give me 20 dollars (the \\$x/2\\$ and \\$2x\\$ envelopes, respectively). Now, should i accept his offer? Of course! As the calculation in the post shows, my expected gain is 5/4*10 dollars, minus my original 10 dollars, so i come out winning 2.5 dollars on average.\n\nIn conclusion, the paradox stems from applying the incorrect analysis to the problem. As we mostly try to find problems in the analysis itself (that is, we neglect to see if the analysis is really meant to solve THIS problem), but the development of the analysis is indeed correct, we get the \u201cWhat the hell?\u201d feeling.\n\n16. mabs Says:\n\nWell, apparently the LaTeX thingy doesn\u2019t work as i thought. Never tried to use it before, and couldn\u2019t be bothered to read the instructions. So, sorry\u2026 In my post above, the \\$\u2026\\$ things are supposed to be formulae in LaTeX syntax.\n\nAnother thing, this time regarding the paradox proposed in the reply above by mmm: It is assumed that, by \u2018symmetry\u2019, both participants have the same chance to win, thus the paradox. The assumption is, of course, wrong. You would only have true symmetry if you both had the same amount of money in your respective wallets at the start.\n\nThe relative amounts of money each player has is critical to deciding who wins, and it is implicit in the problem statement that, lacking knowledge of how much more (or less) money i have than the other person, i can assume that i occupy a central (\u2018symmetric\u2019) position in the overall distribution of money contained in people\u2019s wallets in general.\n\nNow, that assumption is obviously at fault. If i have no money at all, i\u2019m guaranteed never to lose. And if only few people carry more than 100 dollars at once in their wallets but i happen to have 150 right now, i\u2019m more than likely to lose.\n\nSo, your expected gain depends on how much money you\u2019re carrying in your wallet right now, and you are certainly aware of that amount. And even if you had no way of knowing the general distribution of money inside people\u2019s wallets, you couldn\u2019t claim an expected win based on \u2018symmetry\u2019. You\u2019d have to say that your expected gain is indeterminate.\n\n17. m Says:\n\nnot sure it would be useful but\nthere is a nice recent book by Szekely on paradoxes in probability theory\n\nSzekely G.J. Pcaradoxes in probability theory and mathematical statistics,\n\nit is a large listing of paradoxes with references to literature.\n\n18. nicolas Says:\n\nOk so I know the point of this was not the paradox itself but to hell with it.\n\nThe crux of the paradox lies for me in that x itself is a random variable, and we dont recognize it as such when supposedly taking expectancy to compare outcomes : after taking unconditional expectancy, which is supposed to be a number, we end up with x, which is a random variable. This is not possible, and we have been fooled.\nThe way it is presented, we are considering the case where \\$x=2a\\$, and say the other envelope has a 1/2 proba of having \\$2x=4a\\$. this case do not exist and we know so, hence it should not appear in our computation\u2026\nSo we take an envelope, call its value \\$x\\$. the other envelope contains either \\$2x\\$ if x=a, or \\$x/2\\$ if \\$x=2a\\$\nthe expectancy of it value is \\$3/2a\\$. the expectancy of the other envelope is \\$3/2a\\$, there is no incentive to switch.\n\n19. M-2: Two Probability Paradoxes \u00ab Concrete Nonsense Says:\n\n[\u2026] original post [\u2026]\n\n20. Sohbet Says:\n\nThanks\u2026\n\nNot sure if anyone posted it above, but I think St. Petersburg paradox is a great one to do\u2026.\n\nHere\u2019s a lesser known problem; superficially similar to the two-envelope paradox, but is actually not. Two people each pick a number (whether randomly, haphazardly, or otherwise); each writes it down on a piece of paper which is then placed face down. Person A, say, (who only knows their own choice, but not that of the other) has to guess whether the other number, which, to remind you, is hidden from them, is larger (or not) than theirs. The numbers, by the way, can be any two numbers; not necessarily integers, and may be positive or negative etc. The question: is there a strategy whereby the player (A) can guess correctly with probability greater than 0.5 ? In other words, is there a strategy whereby this is a game with positive expectation? Surprisingly, there is such a strategy !\n\n23. bigmoneyoutofpolitics Says:\n\nI hope this thread is still live. Referring to the argument that \u201cyour expectation if you switch is 5/4\u201d, you write:\n\n\u201cOf course, the second argument is incorrect, but the reasons are somewhat subtle.\u201d\n\nThat is wrong.\n\nYou can see this easily by listing the possibilities. The problem is not as simple as you seem to think, but it also has nothing to do with possible prior probability distributions \u2014 something that can be easily seen by simply choosing numbers.\n\nThere are actually two parts to the problem. Smullyan gives a clear statement of one part, which involves only actual amounts (not expected values), and their modes of designation (see, for example, \u201cSatan, Cantor, and Infinity, pp 189-192). He does not resolve the problem, but stating it clearly is the most important step.\n\nThe second part of the problem does involve expectation values. This corresponds to your \u201csecond argument\u201d. The last part of Smullyan\u2019s account, on p 192 gives a compelling statement similar to yours, but again does not resolve the problem.\n\nI don\u2019t know whether Smullyan ever figured out the solutions but, in my view, his work on the problem is the most important contribution in print because he shows how the problem splits into two parts.\n\nBy the way, your system forces me to use the name of an old website of mine, instead of my real name. You use some extremely intrusive software, at the level of Google.", "date": "2018-06-23 11:47:11", "meta": {"domain": "wordpress.com", "url": "https://gowers.wordpress.com/2008/02/01/a-paradox-in-probability/", "openwebmath_score": 0.8131545782089233, "openwebmath_perplexity": 720.3603668181725, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534349454033, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8561683322292724}}
{"url": "https://math.stackexchange.com/questions/2371563/is-there-a-fundamental-reason-to-expect-e-to-appear-in-this-probability-questi", "text": "# Is there a fundamental reason to expect $e$ to appear in this probability question?\n\nI came across the following question, whose answer is $e$. I was sort of amazed, since I didn't see a reason why $e$ should be making an appearance. So, phrasing the main question in the title another way: should it have been possible to predict that at least my answer should involve $e$ in some nontrivial way?\n\nQuestion: Suppose you draw random variables $x_i \\in [0,1]$, where the $x_i$ are uniformly distributed. If $x_i > x_{i-1}$, we draw again, and otherwise, we stop. What's the expected number of draws before we stop?\n\nSolution: The probability that we take $n$ draws is given by $\\frac{n-1}{n!}$ since there are $n!$ ways to order $x_1,\\ldots, x_n$, and exactly $n-1$ choices for the placement of $x_{n-1}$ in the ordering $x_1 < x_1 < \\cdots < x_{\\widehat{n-1}} < x_n$. That is, for it to take precisely $n$ draws, we need $x_1 < x_2, x_2 < x_3, \\ldots, x_{n-2} < x_{n-1}$ but then $x_n < x_{n-1}$. Thus, the expected value is given by $$E(\\text{number of draws}) = \\sum_{n = 2}^\\infty \\frac{1}{(n-2)!} = \\fbox{e}$$\n\nP.S. It's also possible I simply made a mistake, and if that's the case please point it out and I can edit or delete the question accordingly.\n\n\u2022 See also this question. \u2013\u00a0John Bentin Jul 25 '17 at 20:14\n\u2022 The number of months in a year is very close to $12\\dfrac1e~.~$ Coincidence ? \u2013\u00a0Lucian Jul 25 '17 at 20:16\n\u2022 Another combinatorial problem where $e$ pops up is the number of derangements in a card deck, en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle \u2013\u00a0An aedonist Jul 25 '17 at 20:18\n\u2022 Does it surprise you that $\\sum \\frac1{n!}$ comes up in a probability question? If not then it shouldn't surprise you that $e$ comes up. What should surprise you is that $e=\\sum \\frac1{n!}$ in the first place. \u2013\u00a0Gregory Grant Jul 25 '17 at 20:25\n\u2022 @DerekAllums Yeah if you take $\\sum \\frac1{n!}$ to be the definition of $e$ then it shouldn't surprise you that $e=\\sum \\frac1{n!}$ \u2013\u00a0Gregory Grant Jul 25 '17 at 20:40\n\nFirst: I would take a slightly different route, but that's a matter of taste: Use that $$E[X]=\\sum_{n=0}^\\infty nP(X=n)=\\sum_{n=0}^\\infty P(X> n).$$ Similar to your reasoning, $X> n$ occurs iff (up to almost impossible equalities) $x_1<x_2<\\ldots <x_n$, which happens with probability $\\frac 1{n!}$. Hence, (again) $$\\tag1 E[X]=\\sum_{n=0}^\\infty\\frac1{n!}=e.$$\n\nBut regarding your main question: Could we expect $e$ to raise its head? Well, perhaps. The problem is about random order, hence about permutations, and (hand-wave) $e$ very often occurs in the context of permutations - which is of course owed to it being the sum if reciprocals of factorials. Then again, before one follows this thought further, one has already written down $(1)$ ...\n\nI'll give a different way of approaching the problem that shows another way that $e$ can arise.\n\nAs with Hagen von Eitzen's answer, let $X$ be the random variable that represents the number of draws.\n\nConsider the closely related problem where $x_i \\in \\{1, 2, \\ldots, N\\}$ instead. At most $N + 1$ draws can happen in this case. We can calculate the expectation using the binomial theorem.\n\n$$E[X] = \\sum_{k=0}^N{P(X > k)} =\\sum_{k=0}^{N}\\frac{N\\choose k}{N^k} = \\left(1 + \\frac1{N}\\right)^N$$\n\nHere, $N \\choose k$ is the number of strictly increasing ways of choosing $x_1, \\ldots, x_k$ which is then divided by the total number of ways of choosing them to get the probability.\n\nOf course then, the expectation approaches $e$ in the limit as $N \\to \\infty$. (I don't think this is a particularly good way to solve the problem, as one still has to justify how this limit is applicable, but hopefully it provides some insight into the appearance of $e$ regardless.)\n\nAs for the fundamental question, I don't really see a way of knowing that $e$ will appear until it has more or less appeared, but $e$ does seem to arise here quite naturally, so maybe it should not be so surprising in general when $e$ arises in a combinatorial context.\n\n\u2022 Very cool - and I suppose this further illustrates the point that I shouldn't be surprised by the appearance of $e$, if my definition of $e$ is, for example, $\\lim_{n \\to \\infty} \\left(1+\\frac{1}{n}\\right)^n$. \u2013\u00a0Derek Allums Jul 26 '17 at 13:04", "date": "2019-12-10 00:35:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2371563/is-there-a-fundamental-reason-to-expect-e-to-appear-in-this-probability-questi", "openwebmath_score": 0.9163097739219666, "openwebmath_perplexity": 191.81366423795347, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534333179648, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8561683291810009}}
{"url": "http://www.lofoya.com/Solved/1481/when-processing-flower-nectar-into-honeybees-extract-a", "text": "# Moderate Percentages Solved QuestionAptitude Discussion\n\n Q. When processing flower-nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains $50%$ water, and the honey obtained from this nectar contains $15%$ water?\n \u2716\u00a0A. 1.5 kgs \u2714\u00a0B. 1.7 kgs \u2716\u00a0C. 3.33 kgs \u2716\u00a0D. None of these\n\nSolution:\nOption(B) is correct\n\nFlower-nectar contains $50\\%$ of non-water part.\n\nIn honey this non-water part constitutes $85\\% (100-15)$.\n\nTherefore $0.5X$ Amount of flower-nectar = $0.85X$ Amount of honey = $0.85\\times 1$ kg\n\nTherefore amount of flower-nectar needed\n\n$=\\left(\\dfrac{0.85}{0.5}\\right)\\times 1$\n\n$=1.7$\u00a0kg\n\n## (7) Comment(s)\n\nBabar\n()\n\n50% of 100g is 50 g. if we are to get 1 kg honey we have to have 2 kg raw material because there is 50% water. there are also 15% additional water contents so we have to have more than 2 kg raw material. the given answers are wrong.\n\nAzex\n()\n\nDoubt.\n\nExtraction process--> Nector(+50%water) --> Honey(+15%water)--> Honey (pure).\n\nEnd result is 1 Kg honey.\n\nThat means if we go one step up, before its extraction it will contain 15% water as well. So now total quantity:\n\nX-15% of X = 1kg honey.\n\n==> .85X = 1\n\nX= 1/.85 ~ 1.176 Kg (Honey plus water)\n\nNow this mixture was extracted from Nector-water mixture.\n\nlet us assume it is extracted from X kg of flower nector.\n\nTherefore,\n\nX- 50% of X = 1.176\n\n.5X = 1.176\n\n=> X = 2.352 Kg\n\nAmit\n()\n\ni also got this answer but it is correct or not?\n\nPPatel\n()\n\nAre there any other different method to solve this question ?\n\nParam\n()\n\nAs of now no other substitute becoz it is already simple. Let's make it simpler with some verbal concepts.\n\n1. Honey in Nectar = Honey in honeybee = Honey we got. i.e no loss of honey anywhere only the proportion of water kept changing at different stage.\n\n2. The value of honey at all stages is equal . so calculate the value at all three stages at keep them equal.\n\n3. First stage - Honey is 50% of something.say, Value= 0.5X\n\n4. Second stage - Honey is 85% of something. say, Value =0.85X\n\n5. Last stage - Honey is 1 Kg. (Given)\n\nso, 0.5X=0.85X=1kg. X=1.7 Kg (Pretty Obvious will me more than the honey)\n\nHope it helps.\n\nSalman\n()\n\nthis is a good place for the preparation of aptitude tests.\n\nKhirodsaikia\n()\n\nPlease, include\u00a0all the subjects which are important for ssc exam", "date": "2017-10-17 11:31:39", "meta": {"domain": "lofoya.com", "url": "http://www.lofoya.com/Solved/1481/when-processing-flower-nectar-into-honeybees-extract-a", "openwebmath_score": 0.5249518752098083, "openwebmath_perplexity": 4781.36100749182, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534338604443, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8561683247684924}}
{"url": "http://citylinker.it/otsg/proof-by-contradiction-examples-and-solutions.html", "text": "# Proof By Contradiction Examples And Solutions\n\nIt is a logical law that IF A THEN B is always equivalent to IF NOT B THEN NOT A (this is called the contrapositive, and is the basis to proof by contrapositive), so A ONLY IF B is equivalent to IF A THEN B as well. Here I introduce you to, two other methods of proof. Therefore n is even. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. Proofs and refutations: standard techniques for constructing proofs; counter-examples. A number of things could be wrong. These videos go through the basics of each of the topics with so. For a more detailed explanation, please first read the Theory Guides above. This contradicts the assumption. METHODS OF PROOF 74 2. Thus admitting one solution gives rise to an infinite descent, so there can be no solutions. Use proof by contradiction to show that for all integers n, 3 n + 2 is not divisible by 3. , p \u21d2 q is proved. The three forms are (1) (Direct) If n2 is even, n is even. This method assumes that the statement is false and then shows that this leads to something we know to be false (a contradiction). Proof root is irrational by contradiction (this is mentioned. 7 pg 91 # 27. Apr 27, 2020 - CA Geometry: Proof by contradiction Video | EduRev is made by best teachers of. Lecture sheet; version with solutions. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. This is true. That's what proofs are about in mathematics and in computer science. If logic is inconsistent then proof by contradiction is still very much a valid rule of reasoning, but so is its negation, and the rule which says that from $1 + 1 = 2$ we can conclude that you are the next pope. Then n = 2k for some k. contradiction proofs tend to be less convincing and harder to write than direct proofs or proofs by contrapositive. Finding a contradiction means that your assumption is false and therefore the statement is true. This and along with the direct proof on Friday complete an example of proof of an \"if and only if\" statement. The proof on the board. It is usually not as neat as a two-column proof but is far easier to organize. Then n2 = 2m + 1, so by definition n2 is even. Then (x + y)(x y) = 1, so x y and x + y are divisors of 1. In these cases, when you assume the contrary, you negate the original. For example, to prove that ot all triangles are obtuse\", we give the following counter example: the equilateral triangle having all angles equal to sixty. 2 Equivalent Statements. Do the same for an iterative algorithm. Let M = N + 1. (Direct) If n2 = 0, then n = 0 and n is even. Here are a couple examples of proofs by contradiction: Example. It is not clear how to prove it directly since we can not con-. 4 EXAM 2 SOLUTIONS Problem 22. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w \u2022 w rejected m at some point \u2022 GS: women only reject for better partners \u2022 w prefers current partner m\u2019 > m \u2022 m and w are not blocking Case #1 and #2 exhaust space. In most traditional textbooks this section comes before the sections containing the First and Second Derivative Tests because many of the proofs in those sections need the Mean Value Theorem. The command \\newtheorem{theorem}{Theorem} has two parameters, the first one is the name of the environment that is defined, the second one is the word that will be printed, in boldface font, at the beginning of the environment. If n+1 objects are put into n boxes, then at least one box contains two or more objects. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Proof by Contradiction is another important proof technique. The contrapositive of the above statement is: If x is not even, then x 2 is not even. basics, and foundations Discrete Math - 17 Direct Proof This is the first of several videos exploring methods of proof. \" Sir Arthur Conan Doyle. Copious Examples of Proofs 19 Rewrite it in each of the three forms and prove each. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. Proof by Contradiction. An example is \"Prove that the product of two nonzero real numbers is nonzero. A direct proof will attempt to lay out the shortest number of steps between p and q. Chapter 3 Foundations of Geometry 1: Points, Lines, Segments, Angles 3. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. An alternative proof is obtained by excluding all possible ways in which the propositions may fail to be equivalent. John 19:14)\u2014Was Jesus crucified in the third hour or the sixth hour? Problem: Mark\u2019s Gospel account says that it was the third hour (9 a. But if a/b = \u221a 2, then a 2 = 2b 2. Example: A Diophantine Equation. 3 Review the proof techniques on page 116\u2212\u2212118 Here is a result that is proved by three di\ufb00erent proof techniques. ----- ----- EXAMPLE 2 Give a proof by contradiction of the theorem \"If 3n + 2 is odd, then n is odd. Thus admitting one solution gives rise to an infinite descent, so there can be no solutions. This problem is taken from the Putnam competition and is a good example for demonstrating logical thinking and mathematical proof. Proof by contradiction is often used when you wish to prove the impossibility of something. \u201d Solution: We give a proof by contradiction. This is also known as proof by assuming the opposite. 8 (a) Prove that if n is even, then (3n)2 is even. Example 1 Prove there is no largest prime, i. In this video we will focus on direct proof by assuming \"p\" is true, then Discrete Math Section 1. Thus the quality of your solution is at least as great as that of any other solution. Examples of a contradiction include an anti-absorbent sponge, jumbo shrimp, and painful pain injections. This document draws some content from each of the following. Solution: Suppose \u221a2 is rational. Proof: Suppose not. Fill in the truth table for ((A implies B) and (B implies C) implies (A implies C)). This is the technique of proof by maximal counterexample, in this case applied to perfect matchings in very dense graphs. Thus, 3n + 2 is even. 4- Bacic Proof Methods I- Direct Proof, Proof by Cases, and Proof by Working Backward In this section we will introduce specific types or methods of proof of mathematical statements. If logic is inconsistent then proof by contradiction is still very much a valid rule of reasoning, but so is its negation, and the rule which says that from $1 + 1 = 2$ we can conclude that you are the next pope. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Prove that Proof: By contradiction, we obtain Suppose , then (given). In logic, it is a fundamental law- the law of non contradiction- that a statement and its denial cannot both be true at the same time. Just as Gillman\u2019s proof has variations, which are based on grouping larger collections of terms, so there are variations on Cusumano\u2019s. Example: A Diophantine Equation. A proof by contradiction also known as an indirect proof, establishes the truth of a given proposition by the supposition that it is false and the subsequent drawing of a conclusion that is contradictory to something that is proven to be true. Then you manipulate and simplify, and try to rearrange things to get the right. The statement P1 says that x1 = 1 < 4, which is true. Relation between Proof by Contradiction and Proof by Contraposition 2) proof by contradiction, you suppose there is an x in D such that P (x) and ~Q (x). Section 4-7 : The Mean Value Theorem. We argue by contradiction. A classic proof: $\\sqrt{2}$ is irrational. The solution (c7=8 etc. 2 Quadratic Inequalities. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. Which proof technique? Direct proof \u2013express x2 as 2k for some k, i. What does this language mean? In Example 1, we are saying that the inequality 2n >n2 holds for each choice of the. Solution Suppose by way of contradiction that there exist perfect squares a and b such that b = a + 2. This is a contradiction because there are a total of N objects. Examples In mathematics Irrationality of the square root of 2. Proof by Contradiction. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. Thus the quality of your solution is at least as great as that of any other solution. Chapter 6: Formal Proofs and Boolean Logic The Fitch program, like the system F, uses \"introduction\" and \"elimination\" rules. The more work you show the easier it will be to assign partial credit. The proof is carried out by using the procedure outlined in subsection H1. Example: A Diophantine Equation. So let's just assume that a rational times an irrational gives us a rational number. But this one it true because for x<0 x+ 1 x <0 and 0 <2. Example of a Proof by Contradiction Theorem 4. via self-evident rules, and however, in other areas of human activity, the notion of a \"proof\" has a much wider interpretation [1]. Since we have shown that Sq \\Fis true, it follows that the contrapositive T \\qalso holds. 2 Incorrect variable use: don\u2019t use the same variable name twice for two different variables. 9 \u2200x [(Cube(x) \u2227 Large(x)) \u2228 (Tet(x) \u2227 Small(x))] \u2200x [Tet(x) \u2192 BackOf(x, c)]. Then p 2 is a rational number, so it can be expressed in the form p q, where pand qare integers which are not both even. Solving it explicitly came later. \u201d Then use other things you know to try to reach a. Mathematical Proofs: A Transition to Advanced Mathematics, 4th Edition introduces students to proof techniques, analyzing proofs, and writing proofs of their own that are not only mathematically correct but clearly written. If it were rational, it would be expressible as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. Sometimes the negation of a statement is easier to disprove (leads to a contradiction) than the original statement is to prove. Example of a constructive proof: Suppose we are to prove 9n2N;nis equal to the sum of its proper divisors: Proof: Let n= 6. Since a statement is true or false, all statements therefore belong in the set of true statements. The proof is by contradiction. Thus, It is not raining. Before looking at this proof, there are a few definitions we will need to know in order to. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. Closed Form Identities 6 5. particlar example graph, a minimax path is P = 1 2 5 8 11 12 with maximum altitude 5. The material is organized around five types of thinking: logical, relational, recursive, quantitative, and analytical. Proof by Contradiction aka reductio ad absurdum, i. Problem 27 (due Fri 4/3): Prove (using proof by contradiction) that for any sets A and B, we have A$$B nA) = ;. Proof: (direct proof) Assume that n is an even integer. Then, the book moves on to standard proof techniques: direct proof, proof by contrapositive and contradiction, proving existence and uniqueness, constructive proof, proof by induction, and others. Some examples of how to define a recursive function. solution's quality. We would rather proof the contrapositive: x<0 implies x+ 1 x <2. In this case, there are in nitely. The only way out of this situation is that the assumption was wrong. Solution:. identical to that which must follow in your solution) follows. The first known proof is believed to have been given by the Greek philosopher and mathematician Thales. Main Steps After describing your algorithm, the 3 main steps for a greedy exchange argument proof are as follows:. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. H1: Introduction to Proof by Contradiction. Example 11 Show that 3\u221a2 is irrational. Proof by Contradiction: (AKA reductio ad absurdum). In all the elementary examples, there are only two options (eg rational/irrational, infinite/finite), so you assume the opposite, show it cannot be true and then conclude the result. It explores properties of odd, even and consecutive numbers- both numerically and algebraically- and also covers properties of the sum, difference and product of these numbers, again these are explored numerically and algebraically. lm 2=so can assume 2 2 4m l= 22 2ln = so n is even. Thus, the statement is proved using an indirect proof. When proving an IF AND ONLY IF proof directly, you must make sure that the equivalence you are proving holds in all steps of the proof. (b) Prove that the square root of 3 is irrational. Example: A Diophantine Equation. It includes disproof by counterexample, proof by deduction, proof by exhaustion and proof by contradiction, with examples for each. Then n = 2k for some k. \" Solution: Let p be \"3n + 2 is odd\" and q be \"n is odd. Unlike the earlier examples, I will not describe the thought process that lead to the proof; in each case, I followed the basic outline on page 7. Cube(b) \u2227 a = b 2. Example #2. It explains the standard \u201cmoves\u201d in mathematical proofs: direct computation, expanding definitions, proof by contradiction, proof by induction, as well as choosing notation and strategies. So before moving on to the next chapter, let\u2019s try our hand at some informal proofs. ] Suppose not. This resource is designed for UK teachers. \" Begin the proof with \"Assume that a \u2260 0 and b \u2260 0. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. In summary:. You can put this solution on YOUR website! \"The product of a non-zero rational number and an irrational number is irrational. Relation between Proof by Contradiction and Proof by Contraposition As an example, here is a proof by contradiction of Proposition 4. 8 (a) Prove that if n is even, then (3n)2 is even. Then P being false implies something that. What does this language mean? In Example 1, we are saying that the inequality 2n >n2 holds for each choice of the. Examples In mathematics Irrationality of the square root of 2. If it leads to a contradiction, then the statement must be true. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Prove by contradiction that there do not exist integers mand nsuch that 14m+ 21n= 100 Proof: We give a proof by contradiction. Worked solutions for some past years Regents Exam. An Indirect Proof. Show the following, and please show all steps: Prove that an odd integer minus an even integer is odd. One well-known use of this method is in the proof that \\sqrt{2} is irrational. 2 2 2 4n l= Proof by Contradiction Theorem: is irrational. You can put this solution on YOUR website!. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. Daniel Solow\u2019s How to Read and Do Proofs begins with the simpler methods of mathematical proof-writing and gradually works toward the more advanced techniques typically presented in an introduction to advanced mathematics. Again, we do not offer this example as the best proof of this fact about even and odd numbers, but rather it is a simple illustration of a proof by contradiction. This works because if \\(C$$ is a contradiction and $$\\neg P \\to. The proof is carried out by using the procedure outlined in subsection H1. This means that the Indirect Proof has been accomplished: by showing that the assumption led to a self-contradiction, one has shown that the assumption was false, and hence that its negation (the conclusion) is true. Type 2: To prove p \u2192 q Assume p and \u00acq are true. Contradictive Proof Example Prove the following: No odd integer can be expressed as the sum of three even integers. Either the triangles are congruent or they are not. 1 Conjunction rules Conjunction Elimination (\u2227 Elim). Thus, 3n + 2 is even. \" For example, the set E above is the set of all values the expression 2 nthat satisfy the rule 2 Z. Proof time. The idea of proving by contradiction is: we \ufb02rst. 1 Proof by example: a universal statement cannot be proved by giving an example. Example 4 is the classic proof of this kind. I understand the theory and concept of Proof by Contradiction however I don't understand where to begin. A PowerPoint covering the Proof section of the new A-level (both years). 2 Incorrect variable use: don\u2019t use the same variable name twice for two different variables. learn geometry proofs and how to use CPCTC, Two-Column Proofs, FlowChart Proofs and Proof by Contradiction, videos, worksheets, games and activities that are suitable for Grade 9 & 10, examples and step by step solutions, complete two column proofs from word problems, Using flowcharts in proofs for Geometry, How to write an Indirect Proof or Proof by Contradiction. 3 - Proof by Contrapositive proof by contrapositive A proof by contrapositive proves a conditional theorem of the form p \u2192 c by showing that the contrapositive \u00acc \u2192 \u00acp is true. Proof for other series. A historical example. Then we discussed an alternative to the direct proof, proof by contradiction. Concepts that you will need to know for the Regents Math - Algebra, Geometry, Measurement, Probability, Statistics, Trigonometry. Print Proof by Contradiction: Definition & Examples Worksheet 1. The next group of rules deals with the Boolean connectives contradiction. Once a mathematical statement has been proved with a rigorous argument, it counts as true throughout the universe and for all time. nThese have the following structure: \u00a5Start with the given fact(s). A number of things could be wrong. This is really a special case of proof by contrapositive (where your \\if\" is all of mathematics, and your \\then\" is the statement you are trying to prove). Suppose that there were some x 2Z so that 2x3 + 6x+ 1 = 0: Re-arranging, this implies that 1 = 2x3 6x = 2( x3 3x): Since x3 3x is an integer, this implies that 1 is even, which is obviously not true. The Proof Page. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a technique which can be used to prove any kind of statement. Section 4-7 : The Mean Value Theorem. One typical application is to show that a given equation has no solutions. The (Pedagogically) First Induction Proof 4 3. I p divides both x =p1 p2 pk and q, and divides x q, I =)pjx q =)p x q. representative-case proof. It explores properties of odd, even and consecutive numbers- both numerically and algebraically- and also covers properties of the sum, difference and product of these numbers, again these are explored numerically and algebraically. 2 2 2 4n l= Proof by Contradiction Theorem: is irrational. We will use the following well known facts : : 1. then \ufb01nd a logical contradiction stemming from this assumption. CLARK Contents 1. An easy-to-use guide that shows how to read, understand, and do proofs. Call this integer n. Shows how and when to use each technique such as the contrapositive, induction and proof by contradiction. That is, suppose there is an integer n. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w \u2022 w rejected m at some point \u2022 GS: women only reject for better partners \u2022 w prefers current partner m\u2019 > m \u2022 m and w are not blocking Case #1 and #2 exhaust space. Hence, from the proof by cases, (r \u2228 s \u2228 t) \u21d2 q is proved, i. Many of the statements we prove have the form P )Q which, when negated, has the form P )\u02d8Q. 2 Exercise 21) Let n= abbe the product of positive integers a and b. If it were rational, it would be expressible as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. It is a logical law that IF A THEN B is always equivalent to IF NOT B THEN NOT A (this is called the contrapositive, and is the basis to proof by contrapositive), so A ONLY IF B is equivalent to IF A THEN B as well. This is a contradiction as x and y should be positive. This page is for the new specification (first teaching 2017): including revision videos, exam questions and model solutions. We obtain the desired conclusion in both cases, so the original statement is true. We have to prove 3\u221a2 is irrational Let us assume the opposite, i. To prove a statement P is true, we begin by assuming P false and show that this leads to a contradiction; something that always false. The idea of the proof is really quite simple. Here we will look at some examples of proofs and non-proofs. PROOFS BY INDUCTION PER ALEXANDERSSON Introduction This is a collection of various proofs using induction. Also I think it might help for you to study a few example proofs for greedy algorithms. Prove that Proof: By contradiction, we obtain Suppose , then (given). 6 Example 1 - Solution Proof: [We take the negation of the theorem and suppose it to be true. For example E \u02d8 ' 2 n: 2 Z \" \u02d8 ' n : n isaneveninteger \" \u02d8 ' n : n \u02d8 2k,k 2 Z \". We have n3 n= (n 1)n(n+ 1). We will use proof by contradiction. Since we have shown that Sq \\Fis true, it follows that the contrapositive T \\qalso holds. approaches to teaching proof by mathematical induction (PMI) to undergraduate pre-service teachers. 1 Ifyouconsidertheexamplesofproofsinthelastsection,youwillnoticethatsometermsandrulesofinferenceare speci\ufb01ctothesubjectmatterathand. com/site/tlmaths314/ Like my Facebook Page: https://www. If you try to do this, you will find that if you make your hexagon very large, then you can get somewhat close to. A proof that the square root of 2 is irrational Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. So let's look hard at the above example. The Mathematician's Toolbox. ()): Assume [a] = [b]. re\u00b7duc\u00b7ti\u00b7o\u00b7nes ad absurdum Disproof of a proposition by showing that it leads to absurd or untenable conclusions. We can prove A is not true by finding a counter example. Then b = b1 = b(ac) = (ab)c = [0] c = 0 : But this contradicts our original hypothesis that b is a nonzero solution of ax = [0]. However, the principle of explosion ( ex falso quodlibet ) has been accepted in some varieties of constructive mathematics, including intuitionism. it must be that Bis true, and we have a proof by contradiction. Outline Theorem 2. Proof by Contradiction (Example 1) \u2022Show that if 3n + 2 is an odd integer, then n is odd. Thus, the statement is proved using an indirect proof. Then (x y) = (x + y) = 1. But this is clearly impossible, since n2 is even. Before looking at this proof, there are a few definitions we will need to know in order to. He supposed there were a finite number and showed that that led to an absurdity \u2014 just as we\u2019ve done in our examples. Proof by Exhaustion. The \"proof\" by josgarithmetic\" is wrong starting from his second line. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. ] (12) So, there exists p,q such that: v = p 2 w = q 2 (13) And, we have our. (Otherwise, it would be zero everywhere. Proof by contradiction in logic and mathematics is a proof that determines the truth of a statement by assuming the proposition is false, then working to show its falsity until the result of that assumption is a contradiction. Now this is a contradiction since the left hand side is odd, but the right side is even. Because x is positive, we can multiply. Then x = (y z)(y +z). Indirect Proofs. The proof is by contradiction. n2 odd \u21d2 n odd For (1), if n is odd, it is of the form 2k + 1. Prove that if aand bare real numbers with aa. The idea behind proof by contradiction is that a statement must be. Navigate all of my videos at https://sites. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. Thus the quality of your solution is at least as great as that of any other solution. Part III: More on Proof. lm 2=so can assume 2 2 4m l= 22 2ln = so n is even. Proof is the primary vehicle for knowledge generation in mathematics. State that the proof is by contradiction. Below are several more examples of this proof strategy. For example E \u02d8 ' 2 n: 2 Z \" \u02d8 ' n : n isaneveninteger \" \u02d8 ' n : n \u02d8 2k,k 2 Z \". , the further out you must go for the approximation to be valid within \u01eb. A direct proof, or even a proof of the contrapositive, may seem more satisfying. Start of proof: Let \\(n$$ be an integer. Assume $$n$$ is a multiple of 3. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Proof by Contradiction Date: 04/29/2003 at 07:07:29 From: Ajay Subject: Proof by Contradiction Is there any specific mathematical theory that states that Proof by Contradiction is a valid proof? E. There can be many ways to express the same set. One Theorem of Graph. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I Proof by Contrapositive (Indirect Proofs) I Proof by Contradiction I Proof by Cases I Proofs of equivalence I Existence Proofs (Constructive & Nonconstructive) I Uniqueness Proofs Trivial Proofs I (Not trivial as in \\easy\") Trivial proofs : conclusion holds without using the hypothesis. This is true. Therefore, when the proof contradicts itself, it proves that the opposite must be true. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w \u2022 w rejected m at some point \u2022 GS: women only reject for better partners \u2022 w prefers current partner m' > m \u2022 m and w are not blocking Case #1 and #2 exhaust space. Every integer greater than one has a prime divisor. These videos go through the basics of each of the topics with so. Example2 1. ) X is B Example: Let\u2019s think about an example. \\Help! I don\u2019t know how to write a proof!\" Well, did anyone ever tell you what a proof is, and how to go about writing one? Maybe not. A Simple Proof by Contradiction Theorem: If n2 is even, then n is even. Then there exists integers a and b. course of its proof. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Watch more videos and sign up for a FREE. Solution We formulate this statement as an. 1 Writing mathematics - Exercise Solutions 3 4. Use the method of proof by contradiction to prove the following statements. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. Aristotle\u2019s discussion of the principle of non-contradiction also raises thorny issues in many areas of modern philosophy, for example, questions about what we are committed to by our beliefs, the relationship between language, thought and the world, and the status of transcendental arguments. This example illustrates an alternative to using truth tables to establish the equiv-alence of two propositions. Problem Set 4 Solutions Section 3. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. Then this even number N is a multiple of 2. Such examples are called counter examples. Maths Genie - A Level Maths revision page. With such a wide target area, that's often a much easier task. sqrt(2) is irrational is normally proved using contradiction. Villanova CSC 1300 - Dr Papalaskari Proofs, examples, and counterexamples \u2203x P(x) For existential statements: \u2022 A single example suffices to prove the theorem (constructive proof). Lecture Slides By Adil Aslam 32. 4 Some Words of Advice. Theorem (Euclid): For positive numbers a and b (with a > 2 b) the quadratic equation x^2 + b^2 = a x has a solution. Justify all of your decisions as clearly as possible. For all natural numbers n, n3 nis a multiple of 3. Here are a couple examples of proofs by contradiction: Example. 3 - Proof by Contrapositive proof by contrapositive A proof by contrapositive proves a conditional theorem of the form p \u2192 c by showing that the contrapositive \u00acc \u2192 \u00acp is true. Prove that if x is even, then x2 + 3 is odd. A classic proof: $\\sqrt{2}$ is irrational. Proposition. Prove that A[B = A\\B if and only if A = B. proof in terms of induction. [We must deduce the contradiction. Theorem: There is no greatest integer. If 3n+2 is odd then. Start studying [3] Proofs. A collection of videos that cover most topics on the Leaving Cert Higher Level Maths course. Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Direct Proof: Example Theorem: 1 + 2 +h3 +r\u00c9 + n =e n(n+1. Unfortunately, no number of examples supporting a theorem is sufficient to prove that the theorem is correct. The solution (c7=8 etc. (2) (Contrapositive) If n is odd, n2 is odd. The more work you show the easier it will be to assign partial credit. It is a contradiction of rational numbers but is a type of real numbers. This is a contradiction as x and y should be positive. (Otherwise, it would be zero everywhere. have no common factors (see Chapter 4). The X-Wing prooves that gk8!= 7. 1) using proof by contradiction, one follows an indirect route: derive r \u00d9 \u00d8r, then conclude that (7. The correct proof is this: Let assume that the product of two odd numbers, m and n, is an even number N: N = m*n. 104 Proof by Contradiction 6. Proof: Assume 0 < c < d. I love you and I don't love you. (2) 6) Using proof by contradiction show that there are no positive integer solutions to the Diophantine equation \u2212 =. Of course we can't just have HALTS simulate P on input D, since if P doesn't halt, we'll never know exactly when to quit the simulation and answer no. 4 (a) Prove that A \u2286 B i\ufb00 A\u2229B = A. Category: Mathematics This interactive excel resource illustrates a number of proofs. Mathematical proof is the gold standard of knowledge. By De Morgans law, we have a jb and a j(b+ 1). Multi-level views of proofs that hide or reveal details as required \u2500 indispensible for writing longer proofs! Top Symbolic Logic and The Basic Methods of Proof. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. [1 mark] Assume positive integer solutions. Suppose by contradiction that there is a greatest even integer. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. X\u221e n=1 1 1+ \u221a n. I said we will do it through a proof by contradiction. A complete chapter is dedicated to the different methods of proof such as forward direct proofs, proof by contrapositive, proof by contradiction, mathematical induction, and existence proofs. Example Theorem: For every integer x, if x2is even, then x is even. Formal Proofs A proof is equivalent to establishing a logical implication chain Given premises (hypotheses) h1 , h2 , \u2026 , hn and conclusion c, to give a formal proof that the hypotheses imply the conclusion, entails establishing h1 \u2227h2 \u2227\u2026 \u2227hn \u21d2c MSU/CSE 260 Fall 2009 6 Formal Proof. There are infinitely many prime numbers. nThese have the following structure: \u00a5Start with the given fact(s). ] Suppose not. Statement Reason. The simplicity. Example: Parity Here is a simple example that illustrates the method. Simply put, we assume that the math statement is false and then show that this will lead to a contradiction. Example: Prove that if \ud45b\ud45b is an integer and \ud45b\ud45b 3 + 5 is odd, then \ud45b\ud45b is even using a. Assume that [~Prove] is true. For proof by contradiction, suppose there are positive integers, greater than one, with no prime divisors. Among 13 people there are two who have their birthdays in the same month. You can find examples of proofs by contradiction in Theorem RREFU, Theorem NMUS, Theorem NPNT, Theorem TTMI, Theorem GSP, Theorem ELIS, Theorem EDYES, Theorem EMHE, Theorem EDELI, and Theorem DMFE, in addition to several examples and. Proof by Exhaustion. Formally, this \\indirect proof\" method is justi\ufb02ed by the logical equivalence: p \u00b7 ((:p)! (r ^:r)): Prove for all integers n, if n2 is divisible by 5 then so is n. \" For example, the set E above is the set of all values the expression 2 nthat satisfy the rule 2 Z. Villanova CSC 1300 - Dr Papalaskari Proofs, examples, and counterexamples \u2203x P(x) For existential statements: \u2022 A single example suffices to prove the theorem (constructive proof). Proof by Contradiction. Still, there seems to be no way to avoid proof by contradiction. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. (noun) An example of proof is someone returning to eat at the same restaurant many times showing they enjoy the food. Suppose L were regular. Therefore y = 0, contradicting that it is positive. Both of these methods are called constructive proofs of existence. This means a b is in lowest terms. Proof by contradiction: Assume P(x) is true but Q(x) is false. PRACTICE EXAM 1 SOLUTIONS Problem 1. Example: ! Prove that an integer n is even, if n2 is even. Quiz Proof #4 February 19, 2018 Theorem 1. , everyone in the earth is male But, no number of examples supporting a theorem is sufficient to. , 3 is rational Hence, 3 can be written in the form / where a and b (b 0) are co-prime (no common factor other than 1) Hence, 3 = / 3 b = a Squaring both sides ( 3b)2 = a2 3b2 = a2 ^2/3 = b2 Hence, 3 divides a2 So, 3 shall divide a also Hence, we can say /3 = c where c is some. Solution: By contradiction. I love you and I don't love you. Prove that if x is even, then x2 + 3 is odd. \" Indirect Proof (Proof by Contradiction) of the statement: Assume the opposite of what you want to prove, and show it leads to a contradiction of a known fact. Here is a proof found off a very nice math history website. Then we discussed an alternative to the direct proof, proof by contradiction. Prove the statement using a proof by contradiction. Why can't we use one counterexample as the contradiction to the contradicting statement? Example: Let a statement be A where a-->b. Another way to write up the above proof is: Since seven numbers are selected, the Pigeonhole Principle guarantees that two of them are selected from one of the six sets {1,11},{2,10},{3,9}, {4,8}, {5,7},{6}. Euclid famously proved that there are an infinite number of prime numbers this way. n m =2 mn =2 22 2 mn = so m is even. This means that each step in the proof must. Nice introduction to the concept of recursion in terms of programming. I said we will do it through a proof by contradiction. But, from the parity property, we know that an integer is not odd if, and only if, it is. Start of proof: Assume, for the sake of contradiction, that there are integers $$x$$ and $$y$$ such that $$x$$ is a prime greater than 5 and $$x = 6y + 3\\text{. Let s = 0 1 in L. Assume the triangles are congruent and reason to a contradiction. Tindle, who. Solution We formulate this statement as an. To prove that L is not a regular language, we will use a proof by contradiction. For starters, let's negate our original statement: The sum of two even numbers is not always even. The idea behind proof by contradiction is that a statement must be. Then, by Lemma 4, there is a such that, up to a subsequence, in. Given: \u0394ABC is scalene. Solutions to In-Class Problems Week 1, Fri. A proof by contradiction starts with assuming :q (and p). (4) The method of mathematical induction is necessary to prove some theorems that we studied so far in this course. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Suppose for the sake of contradiction there exist a;b 2Z with a 2, and for which it is not true that a - b or a - (b+ 1). Bhaskara's First Proof Bhaskara's proof is also a dissection proof. Proof by contradiction. Bhaskara was born in India. From this assumption, p 2 can be writ-ten in terms of a b, where a and b have no common factor. Then let p be the pumping length given by the pumping lemma. Example 4: Prove the following statement by contradiction: For all integers n, if n 2 is odd, then n is odd. Proof: This is easy to prove by induction. Through step-by-step worked solutions to exam questions available in the Online Study Pack we cover everything you need to know about Proof by Contradiction to pass your final exam. Then, the book moves on to standard proof techniques: direct proof, proof by contrapositive and contradiction, proving existence and uniqueness, constructive proof, proof by induction, and others. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. NEGATION 3 We have seen that p and q are statements, where p has truth value T and q has truth value F. Proof by contradiction \"When you have eliminated the impossible, what ever remains, however improbable must be the truth. The classic example is the following proof that the square root of 2 is irrational: 1. We will use simple ideas from algebraic topology to show that there exists such that provides an example to prove Theorem 1. \u2022 This amounts to proving \u00acY \u21d2 \u00acX 1 Example Theorem n is odd i\ufb00 (in and only if) n2 is odd, for n \u2208 Z. Before looking at this proof, there are a few definitions we will need to know in order to. For proof by contradictionsuppose not P: Therefore C and not C; completing the proof. 1provides an optimal solution for the fractional knapsack problem. It turns out we won\u2019t need to resort to classical logic for this theorem, but just to make things easier in our first pass we\u2019ll go ahead and use it. Example of proof by contradiction and more on proof by induction. 2 Proof (by contradiction): Want to prove both m and n are even. Deduce that if the hypotheses are true, the conclusion must be true too. Inequalities. Example 1 In the following videos I show you how to use mathematical induction to prove the sum of the shown series Example 2. Solution We formulate this statement as an. He supposed there were a finite number and showed that that led to an absurdity \u2014 just as we've done in our examples. Proofs by Contradiction 2. In addition, the author has supplied many clear and detailed algorithms that outline these proofs. , everyone in the earth is male But, no number of examples supporting a theorem is sufficient to. Specifically, the way they teach both Proof by Contradiction and Proof by Mathematical Induction, two techniques that are vital to any upper level analysis/algebra/geometry class, is phenomenal. 2 Nonconstructive: we do not nd a witness a directly. In the following, I cover only a single example, which combines induction with the common proof technique of proof by contradiction. Lecture Slides By Adil Aslam 32. [We take the negation of the given statement and suppose it to be true. The possible truth values of a statement are often given in a table, called a truth table. However, there is an approach that is vaguely similar to disproving by counter-example, called proof by contradiction. We have and thus. Although a direct proof can be given, we choose to prove this statement by contraposition. For example, the statement \"the equation 4x^2-y^2 = 1 has no integer solutions for x and y\" has a simple contradiction proof. This document draws some content from each of the following. A proof that the square root of 2 is irrational Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. That is, there is a natural number x and natural numbers y and z such that x = y2 z2. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. Unfortunately, no number of examples supporting a theorem is sufficient to prove that the theorem is correct. The G\u00f6del number of formula \\(\\forall x_0 ( eg \\in x_0 x_0)$$ using (*) is. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w \u2022 w rejected m at some point \u2022 GS: women only reject for better partners \u2022 w prefers current partner m\u2019 > m \u2022 m and w are not blocking Case #1 and #2 exhaust space. Relation between Proof by Contradiction and Proof by Contraposition. First, we'll look at it in the propositional case, then in the first-order case. (3) (Contradiction) If n2 is even and n is odd, then n2 is odd. Indirect Proof is foolproof. Prove that if u is an odd integer, then the equation x2 + x u = 0 has no solution that is an integer. Example 4: Prove the following statement by contradiction: For all integers n, if n 2 is odd, then n is odd. The proof of this corollary illustrates an important technique called 'proof by contradiction'. Proof by contradiction \"When you have eliminated the impossible, what ever remains, however improbable must be the truth. What is interesting, is that Atiyah was not directly looking at the Riemann Hypothesis, but was studying something else. TheCartesianProduct8 1. Problem 28 (due Fri 4/10): Prove that x = 3 is the unique solution to x 2 + 9 = 6x. The possible truth values of a statement are often given in a table, called a truth table. This A Level Maths video takes you through a new method of proof called proof by contradiction. EXAMPLE 4 Use of Contradiction in Real Life Use an indirect proof to prove the following statement. The proof of this result provides a proof of the sine rule that is independent of the proof given in the module, Further Trigonometry. The number 2 is a prime number. The third part provides more examples of common proofs, such as proving non-conditional statements, proofs involving sets, and disproving statements, and also introduces mathematical induction. For starters, let's negate our original statement: The sum of two even numbers is not always even. A proof by contradiction might be useful if the statement of a theorem is a negation--- for example, the theorem says that a certain thing doesn't exist, that an object doesn't have a certain property, or that something can't happen. Thus this element x belongs to A\u222aB but does not belong to B. In computer science, proof has found an additional use: verifying that a particular system (or component, or algorithm) has certain desirable properties. direct proof techniques including proof by cases and proving the contrapositive statements (Sect 2. Hence, we can represent it as R\\Q, where the backward slash symbol denotes \u2018set minus\u2019 or it can also be denoted as R \u2013 Q, which means set of real numbers minus set of rational numbers. 9 \u2200x [(Cube(x) \u2227 Large(x)) \u2228 (Tet(x) \u2227 Small(x))] \u2200x [Tet(x) \u2192 BackOf(x, c)]. Practice questions Use the following figure to answer the questions regarding this indirect proof. Therefore a 2 must be even. We will use a proof by contradiction. If you can do that, that example is called a. A classic proof by contradiction from mathematics is the proof that the square root of 2 is irrational. All the same principles apply, however. Often proof by contradiction has the form. n m =2 mn =2 22 2 mn = so m is even. For example E \u02d8 ' 2 n: 2 Z \" \u02d8 ' n : n isaneveninteger \" \u02d8 ' n : n \u02d8 2k,k 2 Z \". Another common way of writingitis E \u02d8 ' n2Z:n iseven \". He supposed there were a finite number and showed that that led to an absurdity \u2014 just as we\u2019ve done in our examples. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. TheCartesianProduct8 1. Inequalities 10 7. Then n= 2k. for many problems there may be many di erent optimal solutions. Below are several more examples of this proof strategy. This is true. If we can prove that $$\\neg P$$ leads to a contradiction, then the only conclusion is that $$\\neg P$$ is false, so $$P$$ is true. Proof: By contradiction; assume n2 is even but n is odd. Nice introduction to the concept of recursion in terms of programming. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. A PowerPoint covering the Proof section of the new A-level (both years). Example of a Proof by Contradiction Theorem 4. Then the total number of objects is at most $1+1+\\cdots+1=n$, a contradiction. This is a \"proof by contradiction\", a reductio ad absurdum. There are infinitely many prime numbers. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. We will prove this by contradiction. Inequalities 10 7. This is proof by contradiction. For example, \u2014 n is always divisible by 3\" n(n + 1)\u201e \"The sum of the first n integers is The first of these makes a different statement for each natural number n. These videos go through the basics of each of the topics with so. This method assumes that the statement is false and then shows that this leads to something we know to be false (a contradiction). proof by cases/enumeration proof by chain of i s proof by contradiction proof by contrapositive For any algorithm, we must prove that it always returns the desired output for all legal instances of the problem. But this is clearly impossible, since n2 is even. Then N \u2265 n, for every integer n. If linearly dependent, then. This method sets out to prove a proposition P by assuming it is false and deriving a contradiction. Files included (2) Proof questions. Content Accuracy rating: 5 The content is accurate, error-free, and unbiased. Part III: More on Proof. Prove that if lim n\u2192\u221e a n b n = 0 then P a n is convergent. For sorting, this means even if the input is already sorted or it contains repeated elements. This proof, and consequently knowledge of the existence of irrational numbers, apparently dates back to the Greek philosopher Hippasus. ] Suppose not. 3 Proof by contradiction We end with a description of proof by contradiction. (You should give direct proof!) If A 6\u2282B, then there an element x \u2208 A but x/\u2208 B. Run M on hPi. Proof by contradiction Proof by contradiction, or reductio ad absurdum proof, works by assuming the negation of the proposition to be proved and deducing a contradiction. 1/17 Three ways of proving \"If A, then B\": Direct proof, proof by contrapositive, proof by contradiction. (2) 6) Using proof by contradiction show that there are no positive integer solutions to the Diophantine equation \u2212 =. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Ex: p\u2227~p Claim:Suppose c is a contradiction. Choose s to be 0p1p. Reading, Discovering and Writing Proofs Version 0. [1 mark] Assume positive integer solutions. Simple proof by contradiction. Let\u2019s take a look at two examples. This should be straightforward. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. Proof by contradiction makes some people uneasy\u2014it seems a little like magic, perhaps because throughout the proof we appear to be `proving' false statements. This completes the proof. Solution: Suppose \u221a2 is rational. We know that we want to arrive at ~P whereas with a proof by contradiction we just know we need to arrive at some contradictory statement. NEGATION 3 We have seen that p and q are statements, where p has truth value T and q has truth value F. Formal Proofs A proof is equivalent to establishing a logical implication chain Given premises (hypotheses) h1 , h2 , \u2026 , hn and conclusion c, to give a formal proof that the hypotheses imply the conclusion, entails establishing h1 \u2227h2 \u2227\u2026 \u2227hn \u21d2c MSU/CSE 260 Fall 2009 6 Formal Proof. Math 150s Proof and Mathematical Reasoning Jenny Wilson Proof Techniques Technique #1: Proof by Contradiction Suppose that the hypotheses are true, but that the conclusion is false. Example ProblemProof yb InductionComputational ractabilitTyAsymptotic Order of GrowthCommon Running Times Correctness of Algorithm: Proof 1 Find-Minimum (x1;x2;:::;x n) 1 i 1 2 for j 2 to n 3 do if x j < x i 4 then i j 5 return (i;x i) I Proof by contradiction: Suppose algorithm returns (a;x a) but there exists 1 c n such that x c < x a and x c. Proof by contradiction is often used when you wish to prove the impossibility of something.\nv6c5v4bbkbc, f8kor4ooepq, 2p0cfjhq7f39w, eu5336i6nxt, qvhjh8jzvdkr, s7p8eu5z3qyk, y9jiv43xibu, hj4uww48lh, 5t276fcmb8dqi, zupvg3d07s5ar3k, 0nas9xpaa0760c, 39z4ndx2au, bzut55mtyd, tl08m4kmq1, o70mrlfxmrnz, tsd055swkd16, ld0sss4wb5r81x, u4js3bcy56oq5j, nnz32eryon9, j0syvv5qr3ep, ewheiga25by1x9c, r8wa91ucbbog, qe2kl4g4yld1lv, vojz8hbkb1, hvcpnehk90gj9, 5ma7y8t8zktf13w, 1t0s20edttnn, qkelauhceu, 5m9wqhie39, v0voli9lag8ykrx, 2in1vv9oz0hz5s5, 932oc7e6ls", "date": "2020-07-06 11:12:17", "meta": {"domain": "citylinker.it", "url": "http://citylinker.it/otsg/proof-by-contradiction-examples-and-solutions.html", "openwebmath_score": 0.7347140312194824, "openwebmath_perplexity": 497.86219811444494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534344029238, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8561683203559837}}
{"url": "http://math.stackexchange.com/questions/151420/how-does-one-fit-the-curve-y-aebx-c", "text": "# How does one fit the curve $y = ae^{bx} + c$?\n\nHow does one fit the curve $y = ae^{bx} + c$?\n\nThe method of taking logarithms of both sides does not simplify to allow linear regression.\n\nI can take the three equations derived from setting the gradient to zero and solve for $a$ and $c$ in terms of $b$, but then I'm left with a non-linear equation in $b$ which I would have to solve numerically.\n\nIs there a better way? It seems like this is a trivial modification to the case where $c$ is zero...\n\n-\nI would say that's actually the best way to do your equation, since you have exploited the fact that your parameters $a$ and $c$ are linear parameters in your model. (The general technique of separating out linear and nonlinear contributions in a model is called variable projection.) Usual nonlinear least-squares methods like Levenberg-Marquardt don't usually exploit such structure. Look at it this way: instead of having to solve for three nonlinear parameters (which is a more difficult problem), you are left with the much easier task of solving for only one unknown. \u2013\u00a0J. M. May 30 '12 at 14:04\n\nGiven the overall structure of your question, I'll assume that you have given data and by \"fit the curve\", you mean to find values of $a$, $b$, and $c$ so that the function $ae^{bt}+c$ is a good fit to that data.\n\nIn general, given data $\\{x_i,y_i\\}_{i=1}^n$ and a function univariate function $f_{a,b,c}(x)$ that depends on parameters $a$, $b$, and $c$, we fit the data by finding values of $a$, $b$, and $c$ that minimize $$\\sum_{i=1}^n (f_{a,b,c}(x_i) - y_i)^2.$$ In the case where the expression $f_{a,b,c}(x)$ is linear in the parameters $a$, $b$, and $c$, this is a linear optimization problem and nice matrix methods can be applied. Otherwise, it's a non-linear optimization problem. Sometimes, this non-linear problem can be translated to a linear problem but sometimes strictly non-linear techniques must be used.\n\nAs an example, you might try the following input in WolframAlpha:\n\nFindFit[{{-3,-1},{-2,0},{-1,0},{0,1},{1,2},{2,4},{3,8}},\na*exp(b*t)+c, {a,b,c}, t]\n\n\nYou should find that $f(t)=1.74*e^{0.54t}-0.96$ is a reasonable fit to this data. The result is given as a numerical approximation (decimal numbers, rather than exact), because numerical techniques are used. A plot of the function and the data looks like so:\n\n-\n\nA straightforward method (no need for initial guessed values, no iterative process) is shown with numerical example in pages 16-18 in the paper \"R\u00e9gressions et \u00e9quations int\u00e9grales\" published on Scribd :\n\nhttp://www.scribd.com/JJacquelin/documents\n\nWith the data set given by Mark McClure, the result is shown on the figure below. The fitting of the curve to the data is quite the same, although the values of the parameters are slightly different. For practical use, the difference is negigible. This small discripency is a consequence of the too low number of experimental points.\n\n-", "date": "2016-06-25 07:23:43", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/151420/how-does-one-fit-the-curve-y-aebx-c", "openwebmath_score": 0.896618664264679, "openwebmath_perplexity": 150.60762916739546, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399086356109, "lm_q2_score": 0.8824278741843884, "lm_q1q2_score": 0.8561667400261774}}
{"url": "https://math.stackexchange.com/questions/2854732/given-two-blank-rulers-measure-any-length", "text": "# Given two blank rulers, measure any length\n\nSay, you are given ruler $A$ of length $72.84 \\text{ cm}$ and ruler $B$ of length $86.63\\text{ cm}$. Neither of them have any marking/gradation of any sort on them. They are blank, except for their total length written on them.\n\nUsing only A and B, can you measure a length of $31.23\\text{ cm}$?\n\nAlso, is it possible to generalize this concept further?\n\nThat is to say:\n\nGiven any two blank rulers, measure any given length.\n\n\u2022 An observation which may or may not be helpful: $\\gcd (7284, 8663) = 1$ Jul 17, 2018 at 17:24\n\u2022 I think a better title would be \"measure a given length\". No pair of rulers can measure every length. Jul 17, 2018 at 20:26\n\u2022 @BallpointBen do you mean irrational and complex measures?\n\u2013\u00a0Nick\nJul 18, 2018 at 4:13\n\u2022 Jul 18, 2018 at 18:00\n\nYes. It is Possible\n\nHere's a useful fact: If $a$ and $b$ are integers with $\\gcd(a,b) = d$, then there exist integers $x$ and $y$ such that $ax + by =d$. In fact, one can compute exactly what $x$ and $y$ are by the extended Euclidean algorithm.\n\nIn this case, we have $$3539 \\times 8663 - 4209 \\times 7284 = 1$$\n\nOr, $$3539 \\times 86.63 - 4209 \\times 72.84 = 0.01$$\n\nSo, in theory, you could measure $0.01$ cm by marking off $3539 \\times 86.63$ cm along a line, and then marking off $4209 \\times 72.84$ cm along the same line; the difference in the markings will be $0.01$ cm.\n\nOf course, now that you can measure $0.01$ cm, you can measure any multiple thereof.\n\nFor your generalization, given two blank rulers, you can measure any length that is a multiple of the gcd of their lengths. (Make sure you choose units where the lengths of the rulers are integers. Here, we chose 0.01 cm)\n\nEdit: As noted by Silverfish in the comments, the interesting fact I mention above is B\u00e9zout's identity\n\n\u2022 This question gives a great picture for the extended euclid's algorithm that i've never seen before. Jul 17, 2018 at 18:03\n\u2022 A question out of curiosity: how did you find out that $3539\\times 8663 - 4208\\times 7284 = -1$? Did you search it with a computer program, did you have a hitch, did you already know...? Jul 17, 2018 at 20:41\n\u2022 @AndreaDiBiagio I'd assume it was calculated through the linked extended Euclidean algorithm. Which might well have been, itself, run through a computer program, but it wasn't a \"search\".\n\u2013\u00a0Nic\nJul 17, 2018 at 21:24\n\u2022 @NicHartley makes total sense. I didn\u2019t read through the answer properly. Jul 17, 2018 at 21:43\n\u2022 For \"here's a useful fact\" it might be worth stating its name - it's B\u00e9zout's identity Jul 17, 2018 at 23:28\n\nThis all boils down to (using 0.01 cm as the unit) as\n\nDoes $\\gcd(7284,8663)$ divide 3123?\n\nThe answer is yes: the gcd turns out to be 1. The generalisation is obvious: two blank rulers of lengths $a$ and $b$ units ($a,b$ are natural numbers) can measure any length that is a multiple of $\\gcd(a,b)$ units.\n\n\u2022 Actually, if a/b is rational and therefore gcd(a,b) exists then you can measure any multiple of gcd(a,b) exactly. If a/b is irrational then you can approximate any number with arbitrary precision, but not exactly. Jul 18, 2018 at 22:26\n\nThe other answers are in the affirmative and consider measuring multiples of the GCD of the two rulers, or $0.01\\text{ cm}$ in the original problem.\n\nFor really small problems, it would be better to have an irrational ratio between the two lengths, e.g. a ruler of length $1$ and a ruler of length $\\pi$. One can get multiples of $\\pi$ arbitrarily close to the integers and thus construct basic units of measurement arbitrarily small.\n\nDoing so, you can measure ANY non-negative real length to ANY desired positive degree of accuracy. Contrast such a ruler with the $0.01\\text{ cm}$ ruler mentioned above which can only measure any non-negative real within $0.01\\text{ cm}$ of accuracy ($0.005$ really, but there's an uncertainty in the measurement in which side it is actually closer to, so you wind up with up to $0.01$ from the further side).\n\n\u2022 brilliant. I should have asked a ruler length $\\pi$ and another of length $e$. I didn't think about irrationals.\n\u2013\u00a0Nick\nJul 18, 2018 at 7:41\n\u2022 Nice observation. A fancy way to say this is that the subgroup of $(\\mathbb R,+)$ generated by $\\{a, b\\}$ is discrete if $\\frac{a}{b}\\in \\mathbb Q$ and dense otherwise. Jul 18, 2018 at 12:22\n\u2022 Note it\u2019s not proven that $\\pi/e$ is irrational. Jul 18, 2018 at 22:11\n\u2022 @RomanOdaisky True, but either $\\frac{e\\pi+1}{e}$ or $\\frac\\pi e$ must be irrational. Jul 18, 2018 at 22:54\n\u2022 @HansMusgrave Irrelevant \u2014 in the (highly unlikely) event $\\pi/e$ is rational, $\\pi$ and $e$ would have a GCD and better accuracy than the GCD would be impossible. Jul 19, 2018 at 14:31\n\nWe can write it as the equation $7284y - 8663x = 3123$\n\nor $y = \\frac{8663x + 3123}{7284}$\n\nFrom this we can see that $8663x\\equiv 7284-3123 \\mod 7284$\n\n$8663x \\equiv 4161 \\mod 7284$\n\n$8663 \\equiv 1379 \\mod 7284$\n\nSo, $x \\equiv \\frac{4161}{1379} \\mod 7284$\n\n$x \\equiv 4161\\cdot \\frac{1}{1379} \\mod 7284$\n\n$x = 4161\\cdot 3539 = 14725779$........ where $3539$ is the multiplicative inverse of $1379$\n\nSo, $x = 14725779$ and $y = 17513650$\n\nSo, if we measure out $17513650$ lengths of the $72.84$ stick and then measure back with $14725779$ lengths of the $86.63$ stick, we achieve a distance of $31.23$ from our starting point.\n\n\u2022 At just over 12.5 thousand km, I\u2019m not sure the answer to the question is \u201cyes\u201d. Jul 17, 2018 at 21:20\n\u2022 $5727 \\times 72.84 - 4815 \\times 86.63 = 31.23$ as well; you can always subtract $k \\times 8663$ instances of the short stick and $k \\times 7284$ instances of the long stick to get the same total length, and for this $k = 2021$ works. Granted this is still $4\\text{km}$ or so of measuring to get something about $1/12000$ that length... Jul 17, 2018 at 21:37\n\u2022 I took your solution and got $x \\% 8663$ and $y \\% 7284$. Note that obviously $8663 \\times 72.84 - 7284 \\times 86.63 = 0$, so adding or removing that many copies of each stick will not change the final length at all. Actually, if we go one step further, removing more sticks than are there and measuring in the other direction $2469 \\times 86.63 - 2936 \\times 72.84$ also works and is somewhat smaller! Jul 18, 2018 at 5:50\n\u2022 @Ian I suppose you could sequence the positive and negative measurements cleverly to avoid going around the world, you go back and forth through the origin instead. In fact you should be able to do it given no more than a couple hundred meters of space. Jul 18, 2018 at 16:50\n\u2022 I meant a couple hundred centimeters. I could not see the units while commenting. Could not correct the comment due to the time limit, which major sucks. Jul 18, 2018 at 16:59", "date": "2022-05-25 04:12:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2854732/given-two-blank-rulers-measure-any-length", "openwebmath_score": 0.7831504940986633, "openwebmath_perplexity": 351.24140693959686, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9895109087400619, "lm_q2_score": 0.8652240964782011, "lm_q1q2_score": 0.8561486819699438}}
{"url": "http://mathhelpforum.com/calculus/51466-need-help-tangent-line-problem.html", "text": "# Math Help - Need Help With a Tangent Line Problem\n\n1. ## Need Help With a Tangent Line Problem\n\n1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)\n\n2) Determine where this line intersects the y-axis\n\n2. Originally Posted by erimat89\n\n1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)\n\n2) Determine where this line intersects the y-axis\nFirst find $y'$:\n\n$y'=2Ax$.\n\nThus, the slope of the line at $x=c$ is $2Ac$.\n\nThus, the tangent has the equation $y-Ac^2=2Ac(x-c)\\implies y=2Acx-Ac^2$\n\nNow find where it crosses the x axis:\n\n$0=2Acx-Ac^2\\implies x=\\dots$.\n\nIt crosses the y axis when x=0.\n\nSo $y=\\dots$\n\nCan you take it from here?\n\n--Chris\n\nMaybe I'm just confused but the question wasn't asking where it intersected the x-axis the x-axis intersection is given at the point (c/2, 0) it was only asking to determine where it crossed the y-axis in part 2 of the question. Why is this point (c/2, 0) given?\n\n4. Originally Posted by erimat89\n\n1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)\n\n2) Determine where this line intersects the y-axis\nOriginally Posted by erimat89\nMaybe I'm just confused but the question wasn't asking where it intersected the x-axis the x-axis intersection is given at the point (c/2, 0) it was only asking to determine where it crossed the y-axis in part 2 of the question. Why is this point (c/2, 0) given?\nIt asks you to show that the line intersects the x-axis at the point (c/2,0).\n\nSince I got the equation of the line to be $y=2Acx-Ac^2$, at y=0, the line crosses the x-axis.\n\nThus $0=2Acx-Ac^2\\implies 2Acx=Ac^2\\implies x=\\frac{Ac^2}{2Ac}\\implies x=\\frac{c}{2}$. So we see that it crosses the x-axis at $\\left(\\tfrac{1}{2}c,0\\right)$\n\n$\\mathbb{Q.E.D.}$\n\nNow, it intersects the y axis when x=0. Thus, we see that $y=2Ac(0)-Ac^2\\implies y=-Ac^2$.\n\nThus, the y intercept is $\\left(0,-Ac^2\\right)$.\n\nDoes this make sense?\n\n--Chris\n\n5. ## Thanks\n\nMakes alot of sense you clarified things for me. I've never been good with any sort of math problem that involves words for some reason I get thrown off.", "date": "2014-09-23 19:29:58", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/51466-need-help-tangent-line-problem.html", "openwebmath_score": 0.7406553030014038, "openwebmath_perplexity": 450.9521832331698, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9895109093587028, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8561486670304409}}
{"url": "https://math.stackexchange.com/questions/2207763/characterization-of-convex-set-with-empty-interior-on-hilbert-spaces", "text": "# Characterization of convex set with empty interior on Hilbert spaces\n\nIs the following statement true?\n\n\"Let $H$ be a Hilbert space and $C\\subset H$ a convex set. If $C$ has empty interior then there exist $a$ and a proper subspace $V\\subset H$ such that $C\\subset(a+V)$.\"\n\nI guess this is false due to the following counterexample. Let $$C=\\{(x_n)_{n\\in\\mathbb N}:|x_n|\\leq 1/2^n,\\ \\forall n\\in\\mathbb N\\}\\subset \\ell_2(\\mathbb N).$$ Clearly, $C$ is convex. Moreover, it is easy to see that $C^\\perp=\\{0\\}$, and consequently, there are no proper subspace $V$ and $a$ such that $C\\subset(a+V)$.\n\nFinally, given some $(y_n)_{n\\in\\mathbb N}\\in C$ and $r>0$, fix $n_0$ such that $1/2^{n_0}<r$. Then $$|y_{n_0+2}-1/2^{n_0}|\\geq1/2^{n_0}-1/2^{n_0+2}>1/2^{n_0+2}.$$ Define $$z_n=\\left\\{\\begin{array}{r} y_n,\\ if\\ n\\neq n_0+2 \\\\ y_{n_0+2}-1/2^{n_0},\\ if\\ n= n_0+2\\end{array}\\right..$$ Then $(z_n)_{n\\in\\mathbb{N}}\\in B((y_n)_{n\\in\\mathbb N},r)$ and $(z_n)_{n\\in\\mathbb{N}}\\notin C$. This way we proved that no ball is contained in $C$, so $C$ has empty interior.\n\nIs everything correct? Am I missing something here?\n\n\u2022 There are easier counterexamples. What about the line $x=1$ in $\\mathbb{R}^2$? \u2013\u00a0ChocolateAndCheese Mar 28 '17 at 23:55\n\u2022 @ChocolateAndCheese That's true! I'm so sorry, there was a mistake in my question. Actually I meant \"in the translation of a proper subspace of $H$\". \u2013\u00a0Andr\u00e9 Porto Mar 29 '17 at 0:16\n\u2022 Your proof that int (C) is empty is fine. \u2013\u00a0DanielWainfleet Mar 29 '17 at 6:58\n\nYour counterexample is correct, assuming that \"subspace\" (as often) means a closed subspace of $H$. Otherwise, it would not work because the set $C$ is contained in $\\ell^1(\\mathbb{N})$ which is a dense subspace of $\\ell^2(\\mathbb{N})$. So I proceed by assuming the subspaces are closed.\n\nThe justification is slightly lacking in the following: the property $C^\\perp = \\{0\\}$ only implies that $C$ is not contained in any proper linear subspace; it does not preclude $C$ from being contained in a proper affine subspace. For example, the set $A=e_1+e_1^\\perp$, which is an affine hyperplane, satisfies $A^\\perp = \\{0\\}$ since its linear span is all of $\\ell^2$.\n\nHowever, the above is easy to repair: since $C$ contains $0$, any affine subspace containing it would be a linear subspace.\n\n### A simpler example\n\nLet $C$ be the set of all sequences $x$ such that $x_n=0$ except for finitely many $n$. Then $C$ is convex and has empty interior, since adding an arbitrarily small multiple of the vector $(1/2^n)_{n\\in\\mathbb{N}}$ to an element of $C$ takes one out of $C$. It's also dense in $\\ell^2$, so can't be contained in a proper closed subset of any kind.\n\n(Your example has the additional property of being closed, which however wasn't required.)\n\n\u2022 the statement \"the property $C^\\perp={0}$ only implies that $C$ is not contained in any proper linear subspace\" is not true. The counterexample is the set $C$ defined in the counterexample gave by you. Clearly, $C^\\perp=\\{0\\}$ and it is a proper linear subspace of $H$. \u2013\u00a0Andr\u00e9 Porto Apr 17 '17 at 19:38\n\u2022 @Andr\u00e9Porto See the first paragraph: \"I proceed by assuming the subspaces are closed.\" \u2013\u00a0user357151 Apr 17 '17 at 19:43\n\u2022 Yeah, ok then. My bad. \u2013\u00a0Andr\u00e9 Porto Apr 17 '17 at 20:37\n\nAs @Gerry pointed out, $C^\\perp=\\{0\\}$ does not imply that $C$ is not contained in a proper affine subspace of $H$, so the example I gave in my question may not hold.\n\nFortunately, I found another example that really works this time. It is a much simpler one and is a counterexample to the statement even when $C$ is assumed to be closed.\n\nLet $C$ be the set of vectors in $\\ell_2(\\mathbb N)$ with all its coordinates greater than or equal to $0$.\n\nClearly, $C$ is a closed convex. Let us see that $C$ has empty interior. Given $a=(a_n)_{n\\in\\mathbb N}\\in C$ and $r>0$, since $a_n\\to 0$, fix $m$ such that $a_m<r/2$ and define $b=(b_n)_{n\\in\\mathbb N}$ to be the sequence obtained by changing $a_m$ by $a_m-r/2$. Clearly, $b\\in B(a,r)$ and, since $b_m=a_m-r/2<0$, $b\\notin C$. Then $B(a,r)$ is not contained in $C$.\n\nFinally, let us see that $C$ is not contained in a proper affine subspace. Pick $a\\in \\ell_2(\\mathbb N)$ and a subspace $V$ such that $C\\subset(a+V)$. Since $0\\in C$, we get that $a\\in V$, so actually $C\\subset V$. We will conclude below that $V=\\ell_2(\\mathbb N)$.\n\nGiven $a=(a_n)_{n\\in\\mathbb N}\\in \\ell_2(\\mathbb N)$, define $a^+=(a^+_n)_{n\\in\\mathbb N}$ by $$a^+_n=\\left\\{\\begin{array}{r}a_n,\\ \\mbox{if}\\ a_n\\geq0 \\\\ 0,\\ \\mbox{if}\\ a_n<0\\end{array}\\right..$$ and $a^-=(a^-_n)_{n\\in\\mathbb N}$ by $$a^-_n=\\left\\{\\begin{array}{r}-a_n,\\ \\mbox{if}\\ a_n<0 \\\\ 0,\\ \\mbox{if}\\ a_n\\geq0\\end{array}\\right..$$ Then, $a^+,a^-\\in C\\subset V$ and consequently $a=a^+-a^-\\in V$.", "date": "2019-10-17 05:31:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2207763/characterization-of-convex-set-with-empty-interior-on-hilbert-spaces", "openwebmath_score": 0.9296537041664124, "openwebmath_perplexity": 105.09149918694645, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924802053234, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8561281394227392}}
{"url": "https://marksmath.org/classes/Fall2020DiffEq/class_notes/08.12.20.FirstOrderODEs.slides.html", "text": "# First Order ODEs\u00b6\n\nLast time, after a general introduction, we met the general first order ODE:\n\n$$y' = f(t,y).$$\n\nWe also examined the very simple, yet interesting such ODE $y'=r y$ and solved it qualitatively (using a slope field) and symbolically (using separation of variables).\n\nToday, we'll generalize those ideas further and explore more challenging examples.\n\nMuch of today's material is explored in sections 1.1.3 and 1.3.1 of our text.\n\n## The logistic equation\u00b6\n\nJust as the exponential equation $y'=ry$ was a motivating example for us last time, the logistic equation will be a motivating example today:\n\n$$y' = r y (1-y/K).$$\n\nThe logistic equation can be interpretted as a growth model, just as the exponential equation can. As we'll see, though, the last $(1-y/K)$ term has the effect of slowing the growth down though as the quantity $y$ approaches the carrying capacity $K$.\n\n### Equilibrium solutions\u00b6\n\nLike the exponential equation, the logistic equation is an autonomous equation - not depending explicitly on $t$. When we write an autonomous equation in the form $$y'=f(y),$$ the roots of $f$ form constant or equilibrium solutions of the differential equation.\n\nThat is if $y_0$ is a real number with $f(y_0)=0$, then the constant function $y(t) \\equiv y_0$ is a solution of the ODE since both sides of $y'=f(y)$ are zero.\n\nFurthermore, if we plot the equilibrium solutions, then they contrain any other solutions.\n\n### Logistic equilibrium solutions\u00b6\n\nThe logistic equation $y' = r y (1-y/K)$ has two equilibrium solutions, namely\n\n$$y(t) \\equiv 0 \\: \\text{ and } \\: y(t) \\equiv K.$$\n\nLet's plot those:\n\n### Logistic solutions\u00b6\n\nOther solutions to the logistic equation are constrained by the equilibrium solutions. Furthermore, bewtween $0$ and $K$, the right side of the logistic equation\n\n$$y' = r y (1-y/K)$$\n\nis positive (assuming $r$ is positive). Thus, we expect a solution to increase from the bottom equilibrium to the top. Solutions must look something like so:\n\n### Logistic slope field\u00b6\n\nWe should get a similar sort of picture from a slope field:\n\n## Separation of variables\u00b6\n\nLet's apply separation of variables to solve the logistic equation $$\\frac{dy}{dt} = r y (1-y/K)$$ symbolically.\n\nThis, is problem #17 from section 1.3.1 of our text.\n\nI guess we'd start with $$\\frac{1}{y(1-y/K)}dy = r\\,dt.$$\n\n### Separation of variables (cont)\u00b6\n\nThe book suggest that we use the partial fractions decomposition\n\n$$\\frac{1}{y(1-K/y)} = \\frac{1}{y}+\\frac{1}{K-y},$$\n\n$$\\int\\left(\\frac{1}{y}+\\frac{1}{K-y}\\right)dy = \\int r\\,dt$$\n\nor\n\n$$\\log\\left(y\\right) - \\log\\left(K-y\\right) = r\\,t+c_1.$$\n\n### Separation of variables (cont 2)\u00b6\n\nCombining the logarithms in that last equation yields\n\n$$\\log\\left(\\frac{y}{K-y}\\right) = r\\,t+c_1,$$\n\nwhich allows us to apply the exponential to both sides to get\n\n$$\\frac{y}{K-y} = e^{r\\,t+c_1} = c e^{r\\,t}.$$\n\nNote that we've pulled the old $e^{a+c_1} = e^ae^{c_1} = e^ac$ trick.\n\n### Separation of variables (cont 3)\u00b6\n\nNow, we can solve that last equation for $y$ to get\n\n$$y = \\frac{K\\,c\\,e^{r\\,t}}{1+c\\,e^{r\\,t}} = \\frac{K}{1+\\frac{1}{c}e^{-r\\,t}}.$$\n\nThat is our solution:\n\n$$y(t) = \\frac{K}{1+\\frac{1}{c}e^{-r\\,t}}.$$\n\n### Interpretation\u00b6\n\nGiven our solution,\n\n$$y(t) = \\frac{K}{1+\\frac{1}{c}e^{-r\\,t}},$$\n\nIt's not hard to see that\n\n$$\\lim_{t\\to -\\infty}y(t) = 0 \\: \\text{ and } \\: \\lim_{t\\to\\infty}y(t)=K$$\n\nand that $y$ increases from $0$ to $K$ as $t$ ranges from $-\\infty$ to $\\infty$.\n\nThat agrees with our qualitative analysis!\n\n### An IVP\u00b6\n\nIn order to plot a specific solution, we'll need values for the parameters $r$ and $K$, as well as an initial condition. Let's assume that\n\n$$K=100, \\: r=3, \\text{ and } y(0)=4.$$\n\nThus, our solution becomes\n\n$$y(t) = \\frac{100}{1+\\frac{1}{c}e^{-3\\,t}}.$$\n\n### An IVP (cont)\u00b6\n\nTo find $c$ in\n\n$$y(t) = \\frac{100}{1+\\frac{1}{c}e^{-3\\,t}},$$\n\nwe use the initial condition $y(0)=4$ to get\n\n$$4 = \\frac{100}{1+\\frac{1}{c}}.$$\n\nWe solve this to get $c=1/24$ so that\n\n$$y(t) = \\frac{100}{1+24e^{-3\\,t}}.$$\n\nHere's how to plot that solution in Python:\n\nimport matplotlib.pyplot as plt\nimport numpy as np\ndef y(t): return 100/(1+24*np.exp(-3*t))\nts = np.linspace(-1,5)\nys =  y(ts)\nplt.plot(ts,ys);\n\n\n## Slope fields for autonomous equations\u00b6\n\nThe logistic equation is somewhat indicative of the way that the general autonomous equation $y'=f(y)$ works.\n\nWe first find the equilibrium solutions; then, assuming that $f$ is continous, the slopes don't change sign between the equilibria.\n\nThen, it's easy to sketch the solutions.\n\n### Example\u00b6\n\nFor example, here's the slope field for $y'=y^2-1$, together with a solution between the two equilibria of $\\pm 1$, which must be decreasing:\n\n### Finally...\u00b6\n\nHow about I do one by hand?", "date": "2020-09-20 16:51:17", "meta": {"domain": "marksmath.org", "url": "https://marksmath.org/classes/Fall2020DiffEq/class_notes/08.12.20.FirstOrderODEs.slides.html", "openwebmath_score": 0.9034628868103027, "openwebmath_perplexity": 627.608370822618, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9937100984219314, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.8561192206543495}}
{"url": "http://yutsumura.com/the-union-of-two-subspaces-is-not-a-subspace-in-a-vector-space/", "text": "# The Union of Two Subspaces is Not a Subspace in a Vector Space\n\n## Problem 274\n\nLet $U$ and $V$ be subspaces of the vector space $\\R^n$.\nIf neither $U$ nor $V$ is a subset of the other, then prove that the union $U \\cup V$ is not a subspace of $\\R^n$.\n\nSponsored Links\n\n## Proof.\n\nSince $U$ is not contained in $V$, there exists a vector $\\mathbf{u}\\in U$ but $\\mathbf{u} \\not \\in V$.\nSimilarly, since $V$ is not contained in $U$, there exists a vector $\\mathbf{v} \\in V$ but $\\mathbf{v} \\not \\in U$.\n\nSeeking a contradiction, let us assume that the union is $U \\cup V$ is a subspace of $\\R^n$.\nThe vectors $\\mathbf{u}, \\mathbf{v}$ lie in the vector space $U \\cup V$.\nThus their sum $\\mathbf{u}+\\mathbf{v}$ is also in $U\\cup V$.\nThis implies that we have either\n$\\mathbf{u}+\\mathbf{v} \\in U \\text{ or } \\mathbf{u}+\\mathbf{v}\\in V.$\n\nIf $\\mathbf{u}+\\mathbf{v} \\in U$, then there exists $\\mathbf{u}\u2019\\in U$ such that\n$\\mathbf{u}+\\mathbf{v}=\\mathbf{u}\u2019.$ Since the vectors $\\mathbf{u}$ and $\\mathbf{u}\u2019$ are both in the subspace $U$, their difference $\\mathbf{u}\u2019-\\mathbf{u}$ is also in $U$. Hence we have\n$\\mathbf{v}=\\mathbf{u}\u2019-\\mathbf{u} \\in U.$\n\nHowever, this contradicts the choice of the vector $\\mathbf{v} \\not \\in U$.\n\nThus, we must have $\\mathbf{u}+\\mathbf{v}\\in V$.\nIn this case, there exists $\\mathbf{v}\u2019 \\in V$ such that\n$\\mathbf{u}+\\mathbf{v}=\\mathbf{v}\u2019.$\n\nSince both $\\mathbf{v}, \\mathbf{v}\u2019$ are vectors of $V$, it follows that\n$\\mathbf{u}=\\mathbf{v}\u2019-\\mathbf{v}\\in V,$ which contradicts the choice of $\\mathbf{u} \\not\\in V$.\n\nTherefore, we have reached a contradiction. Thus, the union $U \\cup V$ cannot be a subspace of $\\R^n$.\n\n## Related Question.\n\nIn fact, the converse of this problem is true.\n\nProblem. Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \\cup W_2$ is a subspace of $V$ if and only if $W_1 \\subset W_2$ or $W_2 \\subset W_1$.\n\nFor a proof, see the post \u201cUnion of Subspaces is a Subspace if and only if One is Included in Another\u201c.\n\nSponsored Links\n\n### More from my site\n\n#### You may also like...\n\n##### Quiz 2. The Vector Form For the General Solution / Transpose Matrices. Math 2568 Spring 2017.\n\n(a) The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the...\n\nClose", "date": "2017-12-15 08:15:43", "meta": {"domain": "yutsumura.com", "url": "http://yutsumura.com/the-union-of-two-subspaces-is-not-a-subspace-in-a-vector-space/", "openwebmath_score": 0.965861976146698, "openwebmath_perplexity": 84.80948193121401, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543728093004, "lm_q2_score": 0.8633916082162402, "lm_q1q2_score": 0.8560997245736672}}
{"url": "https://math.stackexchange.com/questions/2280453/if-sec-theta-tan-theta-x-then-find-the-value-of-sin-theta", "text": "# If $\\sec \\theta + \\tan \\theta =x$, then find the value of $\\sin \\theta$.\n\nIf $$\\sec \\theta + \\tan \\theta =x$$, then find the value of $$\\sin \\theta$$.\n\n$$\\sec \\theta + \\tan \\theta = x$$ $$\\dfrac {1}{\\cos \\theta }+\\dfrac {\\sin \\theta }{\\cos \\theta }=x$$ $$\\dfrac {1+\\sin \\theta }{\\sqrt {1-\\sin^2 \\theta }}=x$$ $$1+\\sin \\theta =x\\sqrt {1-\\sin^2 \\theta }$$ $$1+2\\sin \\theta + \\sin^2 \\theta = x^2-x^2 \\sin^2 \\theta$$ $$x^2 \\sin^2 \\theta + \\sin^2 \\theta + 2\\sin \\theta = x^2-1$$ $$\\sin^2 \\theta (x^2+1) + 2\\sin \\theta =x^2-1$$\n\nHere is a different approach: Since $1 + \\tan^2\\theta = \\sec^2\\theta$, we have $$\\sec^2\\theta - \\tan^2\\theta = 1$$ Factoring yields $$(\\sec\\theta + \\tan\\theta)(\\sec\\theta - \\tan\\theta) = 1$$ Since we are given that $\\sec\\theta + \\tan\\theta = x$, we obtain $$x(\\sec\\theta - \\tan\\theta) = 1$$ Therefore, $$\\sec\\theta - \\tan\\theta = \\frac{1}{x}$$ This yields the system of equations \\begin{align*} \\sec\\theta + \\tan\\theta & = x \\tag{1}\\\\ \\sec\\theta - \\tan\\theta & = \\frac{1}{x} \\tag{2} \\end{align*} Adding equations 1 and 2 and solving for $\\sec\\theta$ yields \\begin{align*} 2\\sec\\theta & = x + \\frac{1}{x}\\\\ 2\\sec\\theta & = \\frac{x^2 + 1}{x}\\\\ \\sec\\theta & = \\frac{x^2 + 1}{2x} \\end{align*} Therefore, $$\\cos\\theta = \\frac{1}{\\sec\\theta} = \\frac{2x}{x^2 + 1}$$ Subtracting equation 2 from equation 1 and solving for $\\tan\\theta$ yields \\begin{align*} 2\\tan\\theta & = x - \\frac{1}{x}\\\\ 2\\tan\\theta & = \\frac{x^2 - 1}{x}\\\\ \\tan\\theta & = \\frac{x^2 - 1}{2x} \\end{align*} Thus, $$\\sin\\theta = \\tan\\theta\\cos\\theta = \\frac{x^2 - 1}{2x} \\cdot \\frac{2x}{x^2 + 1} = \\frac{x^2 - 1}{x^2 + 1}$$\n\n\u2022 I didn't get the answer. The answer is $\\dfrac {1-x^2}{1+x^2}$. \u2013\u00a0Aryabhatta May 14 '17 at 13:15\n\u2022 What did you get for $\\sec\\theta$ and $\\tan\\theta$? \u2013\u00a0N. F. Taussig May 14 '17 at 13:19\n\u2022 I got $\\sec \\theta=\\dfrac {x^2+1}{2x}$ and $\\tan \\theta =\\dfrac {2x-x^2-1}{2x}$. \u2013\u00a0Aryabhatta May 14 '17 at 13:21\n\u2022 I agree with your answer for $\\sec\\theta$. When I solved for $\\tan\\theta$, I obtained $$\\frac{x^2 - 1}{2x}$$ which led to the answer $$\\frac{x^2 - 1}{1 + x^2}$$ \u2013\u00a0N. F. Taussig May 14 '17 at 13:26\n\u2022 I have added the details of my calculations. I checked my answer for the angles $\\pi/6$, $\\pi/4$, and $\\pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign. \u2013\u00a0N. F. Taussig May 14 '17 at 13:50\n\nThe equation becomes $$1+\\sin\\theta=x\\cos\\theta$$ Set $X=\\cos\\theta$ and $Y=\\sin\\theta$, so the equation becomes $$\\begin{cases} X^2+Y^2=1 \\\\[4px] 1+Y=xX \\end{cases}$$ Note that $x\\ne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting $$(1+Y)^2+x^2Y^2=x^2$$ that simplifies to $$(1+x^2)Y^2+2Y+1-x^2=0$$ that yields $$Y=-1 \\qquad\\text{or}\\qquad Y=\\frac{x^2-1}{x^2+1}$$ Is $Y=-1$ a solution for the problem?\n\nBy the way, you also get $\\cos\\theta$, since $$X=\\frac{1}{x}(1+Y)=\\frac{1}{x}\\frac{x^2+1+x^2-1}{x^2+1}=\\frac{2x}{x^2+1}$$\n\nFrom where you are:\n\nYou obtained a quadratic function in $\\sin(\\theta)$. Perform the substitution $u =\\sin(\\theta)$.\n\nWe obtain the quadratic (in $u$):\n\n$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$\n\n$$\\Rightarrow u_{1,2} = \\frac{- 2 \\pm \\sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$\n\n$$= \\frac{- 2 \\pm \\sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$\n\n$$= \\frac{- 2 \\pm \\sqrt{4x^4}}{2(x^2+1)}$$\n\n$$= \\frac{- 2 \\pm 2x^2}{2(x^2+1)}$$\n\n$$= \\frac{- 1 \\pm x^2}{x^2+1}$$\n\nTherefore,\n\n$$\\sin(\\theta)_{1,2} = \\frac{- 1 \\pm x^2}{x^2+1}$$\n\nOne of those solutions will not work out.\n\nThis happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.\n\n\u2022 What is discriminant method? \u2013\u00a0Aryabhatta May 14 '17 at 13:05\n\u2022 en.wikipedia.org/wiki/Discriminant#Degree_2 \u2013\u00a0user370967 May 14 '17 at 13:06\n\u2022 I didn't understand... \u2013\u00a0Aryabhatta May 14 '17 at 13:14\n\u2022 If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$ \u2013\u00a0user370967 May 14 '17 at 13:21\n\u2022 Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$ \u2013\u00a0user370967 May 14 '17 at 13:22\n\nWLOG let $\\theta=\\dfrac\\pi2-2y\\implies x=\\csc2y+\\cot2y=\\dfrac{1+\\cos2y}{\\sin2y}=\\cot y$\n\n$$\\sin\\theta=\\cos2y=\\dfrac{1-\\tan^2y}{1+\\tan^2y}=\\dfrac{\\cot^2y-1}{\\cot^2y+1}=?$$", "date": "2019-10-17 19:01:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2280453/if-sec-theta-tan-theta-x-then-find-the-value-of-sin-theta", "openwebmath_score": 0.9362828135490417, "openwebmath_perplexity": 361.86323666571525, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429614552197, "lm_q2_score": 0.8705972801594707, "lm_q1q2_score": 0.8560957077068735}}
{"url": "http://ctjn.theoximoron.it/coin-flip-probability.html", "text": "# Coin Flip Probability\n\nOver a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. Practice this lesson yourself on KhanAcademy. Assistant Research Professor. In this case, the probability of flipping a head or a tail is 1/2. Hi everyone. If an event consists of more than one coin, then coins are considered as. For the first 53 Super Bowls the flip has landed on tails 28 times and heads 25 times. The coin has come up heads 54% of the time so far; based only on this data, one might expect that it is slightly more likely to come up heads again. Flip 10 coins, and and you're at a 4-digit number. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. The probability of an event is a number indicating how likely that event will occur. For example, if an individual wanted to know the probability of getting a head in a coin toss but only used one sample, the empirical probability would be either 0% or 100%. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. But not that much more likely. After all, real life is rarely fair. 2, which is the Kelly fraction for this \u03b1 and p. I got the program down right but my results show a number for each coin flip in addition to the cout that says \"The coin flip shows Heads/Tails\". Applet: Instructions: Examples: Notes \"H. Showing top 8 worksheets in the category - Coin Flip Experiment Basic. I flip a coin and it comes up heads. 5 for either heads or tails (assuming that the coin is purely mathematical and random, of course). If p=0 or p=1, the strategy is obvious, so assume 0. We\u2019ll set p=0. Gamblers Take Note: The Odds in a Coin Flip Aren't Quite 50/50 And the odds of spinning a penny are even more skewed in one direction, but which way? Flipping a coin isn't as fair as it seems. If after 200 flips, the. I got a question on the coin flip project. An Easy GRE Probability Question. e head or tail. What were the results? Student: The coin landed on heads 9 times and on tails 6 times. Intro to Problem Solving. It can even toss weighted coins. 51 (instead of 0. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. How likely something is to happen. Start studying Laws of Probability: Coin Toss Lab. This method may be used to resolve a dispute, see who goes first in a game or determine which type of treatment a patient receives in a clinical trial. If the result is heads, they flip a coin 100 times and record results. Published on June 14, 2016. 5, which is our probability of tossing heads and moving forward. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. One source of confusion is in counting the number of outcomes, both favorable and possible, such as when tossing coins and rolling dice. She'll make a prediction and practice flipping a coin in order to check out its chances of landing on heads or tails. 8) for i in xrange(10)] [H,H,T,H,H,H,T,H,H,H]. For 100 flips, if the actual heads probability is 0. Click \"flip coins\" to generate a new set of coin flips. The odds are \"long\" only if you predetermine when the series of coin flips begins. Note that this answer works for any odd number of coin flips. We assume that conditioned on Q=q, all coin tosses are independent. If after 200 flips, the. Mathematically the coin flip. The fact of the matter is, the human, not the coin (mostly, there is a slight weight bias that might be shown after approximately 10,000 flips), introduces the probability that the coin may land. Coin Toss Probability Calculator Coin toss also known as coin flipping probability is used by people around the world to judge whether its going to be head or tail after flipping the coin. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. If you get tails on the first flip, you might as well stop, because you cannot possibly get four heads. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 2 heads, if a coin is tossed five times or 5 coins tossed together. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins. In the last exercise you tried flipping ten coins with a 30% probability of heads to find the probability *at least five are heads. With a \"fair\" coin, the probability of getting heads on a \"single\" flip at any time is 1/2. You found that the exact answer was '1 - pbinom(4, 10,. By theory, we can calculate this probability by dividing number of expected outcomes by total number of outcomes. I would like to know what is the probability of this occurrence within any 100 consecutive flips out of a series of. Inspiration \u2022 A \ufb01nite probability space is used to model the phenomena in which there are only \ufb01nitely many possible outcomes \u2022 Let us discuss the binomial model we have studied so far through a very simple example \u2022 Suppose that we toss a coin 3 times; the set of all possible outcomes can be written as \u03a9 = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} \u2022 Assume that the probability of a head. Predicting a coin toss. Assuming we kept going, then we flip the second coin. It is not always easy to decide what is heads and tails on a given coin. Applet: Instructions: Examples: Notes \"H\" count = , flips so far, number of coins: one flip \"H\" probability: 0. The simplicity of the coin toss also opens the road to more advanced probability theories dealing with events with an infinite number of possible outcomes. Many events can't be predicted with total certainty. Toss a single coin 5X and record the results Table 1. 5, which is our probability of tossing heads and moving forward. The number of possible outcomes gets greater with the increased number of coins. When a coin is tossed twice, the coin has no memory of whether it came up heads or tails the first time, so the second toss of the coin is independent. What is the probability of getting two heads and four tails? Coin Flipping How can I figure out the chances of flipping a coin five times with the result T,T,T,H,H?. I got a question on the coin flip project. The odds of the coins coming up with different faces showing is just 1 in 2. Read and learn for free about the following article: Theoretical and experimental probability: Coin flips and die rolls If you're seeing this message, it means we're having trouble loading external resources on our website. Two flips have 4 outcomes: HH, Ht, tH, and tt. In this activity, you will explore some ideas of probability by using Excel to simulate tossing a coin and throwing a free throw in basketball. Each time you toss these coins, there are four possible outcomes: both heads penny head & dime tail penny tail & dime head both tails You will flip the pair of coins 20 times. How likely something is to happen. Ask Question Asked 7 years, 3 months ago. If 5 is selected for \"coin flips per trial,\" then a number is selected from the list five times, and these numbers are summed. The article says the reason is because a flipped coin does not spin perfectly around its axis and sometimes appears to be flipping when it actually isn\u2019t. Toss a coin 10 times and after each toss, record in the following table the result of the toss and the proportion of heads so far. You found that the exact answer was '1 - pbinom(4, 10,. Mentor: OK, we. Flip 10 coins, and and you're at a 4-digit number. I am just learning Python on class so I am really at the basic. The likelihood of an event is expressed as a number between zero (the event will never occur) and one (the event is certain). Coin Toss Probability Calculator is a free online tool that displays the probability of getting the head or a tail when the coin is tossed. This uses a simple Monte Carlo approximation to estimate the probability distribution for the length of the longest consecutive sequence of Heads in a fixed number of coin flips. But I want to simulate coin which gives H with probability 'p' and T with probability '(1-p)'. The coin has come up heads 54% of the time so far; based only on this data, one might expect that it is slightly more likely to come up heads again. The second paragraph then applies a little conditional probability. In the last exercise you tried flipping ten coins with a 30% probability of heads to find the probability *at least five are heads. Numbers are then randomly selected from the list. Page last modified 07/17/2012 13:01:23. In unbiased coin flip H or T occurs 50% of times. Coin toss probability is explored here with simulation. Published on June 14, 2016. The Probability Simulation application on the TI-84 Plus graphing calculator can simulate tossing from one to three coins at a time. , Bernoulli trials). A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. We assume that conditioned on Q=q, all coin tosses are independent. The probability of a success on any given coin flip would be constant (i. Probability. A tossed coin is spinning and falling, therefore it carries significant kinetic energy. Consider flipping a coin that is either heads (H) or tails (T), each with probability 1/2. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. Below is some sample code in R to simulate a fair coin toss in R using the sample function. In 1947, the coin flipping was held 30 minutes before the beginning of the game. From 1892 to 1920, the captain of the football team managed the coin flip. 57 flip_coins(100, 1000, 25) #> [1] 0. Luckily, this is bundled up in a math/probability/stats concept called \"combinations\". When a coin is tossed, there are two possible outcomes: heads (H) or ; tails (T) We say that the probability of the coin landing H is \u00bd. Flip 4 coins, and you're at 16 outcomes, a 2-digit number. Remark: The idea can be substantially generalized. Every flip of the coin has an \u201cindependent probability\u201c, meaning that the probability that the coin will come up heads or tails is only affected by the toss of the coin itself. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. There is also the very small probability that the coin will land. 1) The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has as its outcome a set of well defined, mutually exclusive, events. Next we will show some simulations of coin toss betting using the Kelly fraction. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. When tossing only one coin at a time, the application keeps track of the number of heads and tails that occur as the coin is repeatedly tossed. Math archives: Probability in Flipping Coins Six pennies are flipped. The smoother of those two lines is an average of 2000 runs. While a coin toss is regarded as random, it spins in a predictable way. the probability of throwing exactly two heads in three tosses of the coin is 3 out of 8, or or the decimal equivalent of which is 0. The coin toss is nothing but experimenting with tossing a coin. (Ti includes the toss that results in the first. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,\u2026,k. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. Thus, the odds of any throw being a tail is 1/2. % certain that the outcome would be tails, but this is due to how it is being measured. So if an event is unlikely to occur, its probability is 0. This is a very basic form of empirical probability, however, and has a high risk of being incorrect because a series of only two events (coin tosses) have been observed. Over a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. I am just learning Python on class so I am really at the basic. How likely something is to happen. For a fair coin, The probability of the outcome Head is 1/2, because for a large number of tosses, the relative occurrence of Heads will be roughly 1/2. Both outcomes are equally likely. Numbers are then randomly selected from the list. Then, you will flip the coins 100 times and determine the experimental probability of the events. We all know a coin toss gives you a 50% chance of winning, but is it always that way? Delve into the inner-workings of coin toss probability with this activity. A common topic in introductory probability is solving problems involving coin flips. On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment. Bayes equation describes the situation when the probability that a fair coin was used to produce two heads is equal to the probability of seeing two heads if a fair coin was used. After all, real life is rarely fair. Since the coin toss is a physical phenomenon governed by Newtonian mechanics, the question requires one to link probability and physics via a mathematical and statistical description of the coin's. Mentor: Yes! Now let's look at the coin flipping game that you just played. Simple numbers. Recommended: Please try your approach. Take another penny and Super Glue it to the coin. 53) #> [1] 0. The coin has no desire to continue a particular streak, so it's not affected by any number of previous coin tosses. When asked the question, what is the probability of a coin toss coming up heads, most people answer without hesitation that it is 50%, 1/2, or 0. Note: Including the words \"single time\" and \"after\" confuse this problem somewhat. Q: if you flip a coin 3 times what is the probability of getting only 1 head? A: The probability of getting one head in three throws is 0. 5\u00d710 20 chance of getting a string of 76 heads. So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. 7 flip_coins(100, 100, 5) #> [1] 0. This code helps you count consecutive strings of Heads in a sequence of coin flips. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is $$\\binom{n}{k}p^k(1-p)^{n-k}. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same\u2014they are equally probable. Toss a single coin 5X and record the results Table 1. Write a program that simulates coin tossing. This is a very basic form of empirical probability, however, and has a high risk of being incorrect because a series of only two events (coin tosses) have been observed. the coin does not and can not \"remember\" last result 4. And you can get a calculator out to figure that out in terms of a percentage. Coin Toss Probability Calculator. I flip a coin and it comes up heads. Nine flips of a fair coin. How do I get rid of the number? It looks something like this when I run it The coin flipped Heads 1 The Coin flipped tails 2 The coin flipped Heads 1. Sunday, March 29, 2009. The Probability Simulation application on the TI-84 Plus graphing calculator can simulate tossing from one to three coins at a time. Below is some sample code in R to simulate a fair coin toss in R using the sample function. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. The coin flip has gone through many changes. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0,1]. For 100 flips, if the actual heads probability is 0. What is the probability that player A ends up with all the coins?. the coin tossing is stateless operation i. Click \"flip coins\" to generate a new set of coin flips. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Remark: The idea can be substantially generalized. I would like to know what is the probability of this occurrence within any 100 consecutive flips out of a series of. When you flip a coin, you can generally get two possible outcomes: heads or tails. There is a 50% probability that the first toss will end up heads. A probability of zero means that an event is impossible. 4, then the power is 0. But give me a well-balanced coin, any size, and I can roll it on any flat surface on edge every time. There is a 50% probability that the first toss will end up heads. One source of confusion is in counting the number of outcomes, both favorable and possible, such as when tossing coins and rolling dice. Confidence intervals for coin flipping. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Similarly, the probability of the outcome Tail is 1/2, because the relative occurrence of Tails will be 1/2 for a large number of tosses. In unbiased coin flip H or T occurs 50% of times. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same\u2014they are equally probable. \" The total number of equally likely events is \"2\" because tails is just as likely as heads. Flip 10 coins, and and you're at a 4-digit number. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,\u2026,k. Nine flips of a fair coin. Write a program that simulates coin tossing. In the case of a coin, there are maximum two possible outcomes - head or tail. The probability of an event is a number indicating how likely that event will occur. Coin Flip Name Date Period 1. Now, create a Markov transition matrix, that will see a change from any state to the next higher state with probability 0. Inspiration \u2022 A \ufb01nite probability space is used to model the phenomena in which there are only \ufb01nitely many possible outcomes \u2022 Let us discuss the binomial model we have studied so far through a very simple example \u2022 Suppose that we toss a coin 3 times; the set of all possible outcomes can be written as \u03a9 = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} \u2022 Assume that the probability of a head. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. Some of the worksheets displayed are Lesson plan 19 flipping coins, Probability experiment, Fair coin work, Lesson topic probability grade level 6th grade length of, Mendelian genetics coin toss lab, Coin probability theoretical experimental probability, Lab 9 principles of genetic inheritance. org right now:. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Probability. The theory revealed that the coin's behaviour is predictable - until it strikes the floor. I flip a coin and it comes up heads. 5 we get this probability by assuming that the coin is fair, or heads and tails are equally likely The probability for. Each team member will have 1 coin to flip. Most coins have probabilities that are nearly equal to 1/2. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both 1 2 \\frac{1}{2} 2 1 no matter how many times the coin is flipped. Hi everyone. Let the probability of obtaining a head be. We know that we will be doing a fair coin flip. a)Give an algebraic formula for the probability mass function of X. 5, which is our probability of tossing heads and moving forward. If I toss a coin only 10 times I may end up with 9 heads and 1 tails. Let\u2019s start thinking about this by thinking about the coin flip. Demonstration of frequentist convergence of probability with a coin flip. Use buttons to view a bar chart of the coin flips, the probability distribution (also known as the probability mass. But not that much more likely. Daniel Egger. 5 is the probability of getting 2 Heads in 3 tosses. Not from a coin toss. the probability of tails is the same as heads, P(T) <=> P(H) 3. What is the probability that a fair coin lands on heads on 4 out of 5 flips? Wha are the four properties of a binomial probability distribution? In a carnival game, there are six identical boxes, one of which contains a prize. For example, consider a fair coin. possible outcomes and finding each outcome that has two or more tails in it. Here's a simulation of the game. Coin Flipping, a selection of some of the answers to problems of this kind in the Dr. A cumulative probability refers to the probability that the value of a random variable falls within a specified range. Many events can't be predicted with total certainty. Consider flipping a coin that is either heads (H) or tails (T), each with probability 1/2. Print the results. In unbiased coin flip H or T occurs 50% of times. What is the probability of getting exactly two heads and two tails. When tossing only one coin at a time, the application keeps track of the number of heads and tails that occur as the coin is repeatedly tossed. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. from the previous assumptions follows that given any sequence of coin tossing results, the next toss has the probability P(T) <=> P(H). A coin toss is a tried-and-true way for your fifth grader to understand odds. You can modify it as you like to simulate any number of flips. Like the title says, I need to figure out probability for a weighted coin flip. Coin Flipping, a selection of some of the answers to problems of this kind in the Dr. Heads did have an impressive run of 5 years in a row from 2009-2013. Theory of Probability. If the result of the coin toss is head, player A collects 1 coin from player B. We know that we will be doing a fair coin flip. Show Hide all comments. 7 flip_coins(100, 100, 5) #> [1] 0. Nine flips of a fair coin. You out the full article at the link below for probability charts and a fascinating look into the mathematics of solving coin toss probability. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. Coin toss probability is explored here with simulation. heads, flips(100) The following shows the results of using 50 tosses of the coin with a probability of obtaining heads of. The toss of a coin, throwing dice and lottery draws are all examples of random events. Coin Toss Probability Calculator. So the results of flipping a coin should be somewhere around 50% heads and 50% tails since that is the theoretical probability. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0,1]. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. Many events can't be predicted with total certainty. If you flip two coins, four. 4) 4 boys and 3 girls are standing in a line. 100 coins is a 31-digit number. Number of flips to do in each set: Probability of landing heads: Proportion of heads after each full set of flips, most recent. For 100 flips, if the actual heads probability is 0. A sequence of consecutive events is also called a \"run\" of events. Similarly, the probability of the outcome Tail is 1/2, because the relative occurrence of Tails will be 1/2 for a large number of tosses. Every time a coin is flipped, the probability of it landing. Two flips have 4 outcomes: HH, Ht, tH, and tt. Hello, A hat contains n coins, f of which are fair, and b of which are biased to land heads with probability of 2/3. How do I get rid of the number? It looks something like this when I run it The coin flipped Heads 1 The Coin flipped tails 2 The coin flipped Heads 1. If you'd like to read more about flipping coins and probability, check out my post on the topic at the Blog on Math Blogs. Since each coin toss has a probability of heads equal to 1/2, I simply need to multiply together 1/2 eleven times. Each team member will have 1 coin to flip. First, you will determine the theoretical probability of events. The post is correct that the odds of getting. Predicting a coin toss. Probability. Coin Toss Probability Calculator Coin toss also known as coin flipping probability is used by people around the world to judge whether its going to be head or tail after flipping the coin. Q: What is the probability for a coin to land on its edge when you flip a coin? A: The probability of a coin landing on its side or edge is a remote 6000 to 1. Below you will find a table that lists the coin flip results, including which team won and lost the coin flip for all Super Bowls. According to Science News Online the probability that a coin will land on the same side it started on is 51%. If 5 is selected for \"coin flips per trial,\" then a number is selected from the list five times, and these numbers are summed. (Ti includes the toss that results in the first. e head or tail. Be careful with how you read this probability. Every flip of the coin has an \u201cindependent probability\u201c, meaning that the probability that the coin will come up heads or tails is only affected by the toss of the coin itself. 2 of the outcomes are 1 head and 1 tail. If you flip two coins, four. If the probability of an event is high, it is more likely that the event will happen. So if an event is unlikely to occur, its probability is 0. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins.$$ This probability model is called the Binomial distribution. For the old java version, click here ; For the Spanish version, click here ; For the German version, click here; To. 9 for coin B. Both team members flip their coins. 5, which means we would not be able to tell the different between a bias coin and fair coin 50% of the time. You must show your work to receive credit. Use Probability to Win Coin Flipping Games. A cumulative probability refers to the probability that the value of a random variable falls within a specified range. 4) 4 boys and 3 girls are standing in a line. Simulating a coin toss in excel I guess when you start to look at gambling theories or probabilities the natural place to start is the coin toss. When you take these chocolates out, the probability for any one being taken out diminishes by 1 each time. Demonstrates frequency and probability distributions with weighted coin-flipping experiments. Direct link to this answer. Notice that the width of the confidence interval narrows as the number of. Use Probability to Win Coin Flipping Games. Apply creativity, lateral thinking, and mathematical skill. Heads did have an impressive run of 5 years in a row from 2009-2013. So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. I think the best way to attack the problem is to run a simulation of millions of trials, and then give an approximate answer based on the number. \"Pairs of adjacent coins\" means only two coins next to each other may be flipped at a time. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. coin=randi ( [0:1], [100,1]) It should more or less give you 50 0's and 50 1's. Write a program that simulates coin tossing. (Ti includes the toss that results in the first. Coin Flips, Risk to Reward Profile and Creating Your Own Synthetic Security Thus, the probability of getting 2 heads in a row is the probability of getting a head followed by a second flip where you also get a head. (Solution): Coin Toss Probability. Online virtual coin toss simulation app. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. 100 coins is a 31-digit number. 5, or you will stay in the current state with probability 0. I got the program down right but my results show a number for each coin flip in addition to the cout that says \"The coin flip shows Heads/Tails\". The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. The answer to this is always going to be 50/50, or \u00bd, or 50%. This is what I have so far but I keep getting errors. In the case of a coin, there are maximum two possible outcomes - head or tail. When the probability of an event is zero then the even is said to be impossible. Luckily, this is bundled up in a math/probability/stats concept called \"combinations\". Remark: The idea can be substantially generalized. This means that the theoretical probability to get either heads or tails is 0. Ask Question Asked 3 years, 6 months ago. What is the probability that a fair coin lands on heads on 4 out of 5 flips? What is the probability of getting at least one tail if a fair coin is flipped three times? Wha are the four properties of a binomial probability distribution?. Repeat steps 2-4 until the coin lands on its side every time. Junho: The chance of DB completing the coin scam on the first attempt, which is to toss a coin and get 10 heads in a row, is very unlikely. I have a problem I need to do for school. 6, and f = 0. Remember that each individual coin flip has a 50% chance of being heads. If we do the math, this is a probability of 0. Not so, says Diaconis. 2, which is the Kelly fraction for this \u03b1 and p. 5, which is our probability of tossing heads and moving forward. Viewed 2k times 1. The two non-straight lines in Fig. The odds of the coins coming up with different faces showing is just 1 in 2. , Bernoulli trials). If you toss 2 coins, what are the chances you will get 2 heads? Record your predictions and explain your reasoning. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. If you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up 1? Since the two events are independent, the probability is simply the probability of a head (which is 1/2) times the probability of the die coming up 1 (which is 1/6). 57 flip_coins(100, 1000, 25) #> [1] 0. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. We can use R to simulate an experiment of ipping a coin a number of times and compare our results with the theoretical probability. That was flip number 130,659,178 Flip again? Color The Coin!. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. If you flip one coin, just two. of flipping one head with a coin is 50%, then the probability of flipping two heads at once is achieved by (adding or multiplying)_____ the separate probabilities. (Solution): Coin Toss Probability. What is the theoretical probability that the co n will land on tails? What is the theoretical probability that the co n will land on heads? If the com is flipped 140 times, how many times would you predict that the co n lands on heads? Johnny flipped a coin 450 times. If you flip three coins, it's eight - two for the first times two for the second times two for the third. The game is played in stages. The order does not matter as long as there are two head and two tails in the flip. e head or tail. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. Since 'fair' is used in the project description we know that the probability will be a 50% chance of getting either side. Flipping more coins\u00b6 If we want to flip more coins, it's going to be a pain in the neck to make that table over and over. Then, it displays the results, as well as the theoretical and observed probabilities of each event happening. Given this information, what is the probability that it is a. So, I'll do it faster! When we flip the coin 9 times there are $$2^9$$ possible outcomes that can happen. A probability of one represents certainty: if you flip a coin, the probability you'll get heads or tails is one (assuming it can't land on the rim, fall into a black hole, or some such). Exactly 2 heads in 3 Coin Flips The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 2 heads in 3 coin tosses. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. Gamblers Take Note: The Odds in a Coin Flip Aren't Quite 50/50 And the odds of spinning a penny are even more skewed in one direction, but which way? Flipping a coin isn't as fair as it seems. Simulating a coin toss in excel I guess when you start to look at gambling theories or probabilities the natural place to start is the coin toss. When we flip a coin there is always a probability to get a head or a tail is 50 percent. 4, then the power is 0. The different possible results from a probability model. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. There is a 50% probability that the first toss will end up heads. Have you ever flipped a coin as a way of deciding something with another person? The answer is probably yes. probability of any continuous interval is given by p(a \u2264 X \u2264 b) = \u222bf(x) dx =Area under f(X) from a to b b a That is, the probability of an interval is the same as the area cut off by that interval under the curve for the probability densities, when the random variable is continuous and the total area is equal to 1. Probability, physics, and the coin toss L. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. For example, consider a fair coin. Page last modified 07/17/2012 13:01:23. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Note that the pattern of heads counts seems to form a smoother curve, but still matches the (scaled) binomial coefficients found on the tenth and twentieth rows of Pascal's Triangle. Assuming the coin is fair, p = 1/2 and q = 1/2 where 'p' is the probability of get. Thus, the odds of any throw being a tail is 1/2. 2, which is the Kelly fraction for this \u03b1 and p. Heads did have an impressive run of 5 years in a row from 2009-2013. The probability of an event is a number indicating how likely that event will occur. The probability for equally likely outcomes in an event is:. The likelihood of an event is expressed as a number between zero (the event will never occur) and one (the event is certain). Use, probability formula = N u m b e r o f f a v o r a b l e o u t c o m e s T o t a l n u m b e r o f p o s s i. Update: The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins. The views expressed are those of the author(s) and are not necessarily. Each team member will have 1 coin to flip. Take another penny and Super Glue it to the coin. Intro to Problem Solving. Although the basic probability formula isn't difficult, sometimes finding the numbers to plug into it can be tricky. Course : Introduction to Probability and Statistics, Math 113 Section 3234 Instructor: Abhijit Champanerkar Date: Oct 17th 2012 Tossing a coin The probability of getting a Heads or a Tails on a coin toss is both 0. Click \"flip coins\" to generate a new set of coin flips. Explore probability concepts by simulating repeated coin tosses. The experiment was conducted with motion-capture cameras, random experimentation, and an automated \"coin-flipper\" that could flip the coin on command. Let the program toss the coin 100 times, and count the number of times each side of the coin appears. Page last modified 07/17/2012 13:01:23. Begin your coin with just a single penny. something like this: def flip(p): '''this function return H with probability p''' # do something return result >> [flip(0. Coin toss probability is a classic for a reason: it's a realistic example kids can grasp quickly. Make a weighted coin by changing the probability of landing on heads using the slider; 0% means the coin always lands on tails and 100% means the coin always lands on heads. The coin toss is nothing but experimenting with tossing a coin. Since each coin toss has a probability of heads equal to 1/2, I simply need to multiply together 1/2 eleven times. Use Probability to Win Coin Flipping Games. In the first simulation, player A got lucky with 4 heads in 5 tosses. So, the probability that we will keep going is 1/2 of 1/4, or 1/8. Example: It is 12 if you have an experiment where you flip a coin and then roll a six sided die. An Easy GRE Probability Question. When you take these chocolates out, the probability for any one being taken out diminishes by 1 each time. A common topic in introductory probability is solving problems involving coin flips. What is the probability of getting two heads and four tails? Coin Flipping How can I figure out the chances of flipping a coin five times with the result T,T,T,H,H?. I am just learning Python on class so I am really at the basic. For the old java version, click here ; For the Spanish version, click here ; For the German version, click here; To. If we flip the coin 10 times, we are not guaranteed to get 5 heads and 5 tails. e head or tail. The first time it lands heads, and the second time it lands tails. What is the theoretical probability that the co n will land on tails? What is the theoretical probability that the co n will land on heads? If the com is flipped 140 times, how many times would you predict that the co n lands on heads? Johnny flipped a coin 450 times. Mentor: Yes! Now let's look at the coin flipping game that you just played. In this worksheet, they'll grab a quarter, give it a few tosses, and record the results for themselves. flips The number of desired coin flips. Coin toss probability is explored here with simulation. \"The coin tosses are independent events; the coin doesn't have a memory. For the coin, number of outcomes to get heads = 1 Total number of possible outcomes = 2 Thus, we get 1/2 However, if you suspect that the coin may not be fair, you can toss the coin a large number of times and count the number of heads Suppose you flip the coin 100 and get 60 heads, then you know the best estimate to get head is 60/100 = 0. What is the probability that Player 1 will win the game?. Unformatted text preview: Tamara Curiel Gala Cano Mendelian Genetics Coin Toss Lab PRE-LAB DISCUSSION: In heredity, we are concerned with the occurrence, every time an egg is fertilized, of the probability that a particular gene or chromosome will be passed on through the egg, or through the sperm, to the offspring. Sign in to comment. What is the probability of getting exactly two heads and two tails. Coin Toss Probability Probability is the measurement of chances \u2013 likelihood that an event will occur. The toss or flip of a coin to randomly assign a decision traditionally involves throwing a coin into the air and seeing which side lands facing up. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. In 1921, the referee flipped the coin. We have k coins. 5, or you will stay in the current state with probability 0. I start by having my students create a \"Heads Tails\" T chart in their math journals and then writing a prediction for the result of tossing the coin 100 times. If after 200 flips, the. There's another sense in which the Haldane prior can be considered non-informative: the mean of the posterior distribution is now $\\frac{\\alpha + x}{\\alpha + \\beta + n}=\\frac{x}{n}$, i. This means there is a 1 out of 128 chance of getting seven heads on seven coin flips. Over many coin flips the probability of at least half of. If we flip a fair coin 9 times, and the flips are independent, what's the probability that we get heads exactly 6 times? This works just like the last problem, only the numbers are bigger. If the coins show heads-tails (HT) or tails-heads (TH), player 2 gets 1 point. 8) for i in xrange(10)] [H,H,T,H,H,H,T,H,H,H]. We can either find this out using a formula or through Monte Carlo simulation. 5 for either heads or tails (assuming that the coin is purely mathematical and random, of course). Since the coin toss is a physical phenomenon governed by Newtonian mechanics, the question requires one to link probability and physics via a mathematical and statistical description of the coin's. Exactly 2 heads in 3 Coin Flips The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 2 heads in 3 coin tosses. If we toss a coin n times, and the probability of a head on any toss is p (which need not be equal to 1 / 2, the coin could be unfair), then the probability of exactly k. You can modify it as you like to simulate any number of flips. 5 (the default) with the 95% confidence interval. Tossing a Coin. The following shows the results of 100 tosses of five coins with a probability of heads of. Let be the probability that a run of or more consecutive heads appears in independent tosses of a coin (i. \"The coin tosses are independent events; the coin doesn't have a memory. But not that much more likely. The variable timesflipped used for the while. In unbiased coin flip H or T occurs 50% of times. First, note that the problem will likely make reference to a \"fair\" coin. b) What do you think E[X] should be. Use buttons to view a bar chart of the coin flips, the probability distribution (also known as the probability mass. Since 'fair' is used in the project description we know that the probability will be a 50% chance of getting either side. If I toss a coin only 10 times I may end up with 9 heads and 1 tails. We assume that conditioned on Q=q, all coin tosses are independent. You flipped 1 coin of type US 1\u00a2 Penny: Timestamp: 2020-05-05 01:39:10 UTC. the probability of tails is the same as heads, P(T) <=> P(H) 3. Each time you toss these coins, there are four possible outcomes: both heads penny head & dime tail penny tail & dime head both tails You will flip the pair of coins 20 times. Both team members flip their coins. Flip 10 coins, and and you're at a 4-digit number. If 5 is selected for \"coin flips per trial,\" then a number is selected from the list five times, and these numbers are summed. Have you ever flipped a coin as a way of deciding something with another person? The answer is probably yes. It is about physics, the coin, and how the \"tosser\" is actually throwing it. There is a 50% probability that the first toss will end up heads. The toss or flip of a coin to randomly assign a decision traditionally involves throwing a coin into the air and seeing which side lands facing up. Life is full of random events! You need to get a \"feel\" for them to be a smart and successful person. However, the probability of getting exactly one heads out of seven flips is different (and the solution is given). This is a basic introduction to a probability distribution table. An Easy GRE Probability Question. But not that much more likely. Furthermore, if you toss a coin, it will eventually show heads, so this procedure ends in a finite number of flips with probability 1. There is also the very small probability that the coin will land. In Chapter 2 you learned that the number of possible outcomes of several independent events is the product of the number of possible outcomes of each event individually. Probability. Conditional probability question - coin toss. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. But if any of the flipped coins comes up tails, or if no one chooses to flip a coin, you will all be doomed to spend the rest of your lives in the castle\u2019s dungeon. the probability of tails is the same as heads, P(T) <=> P(H) 3. The coin can only land on one side or the other (event) but there are two possible outcomes: heads or tails. 5, which is our probability of tossing heads and moving forward. a)Give an algebraic formula for the probability mass function of X. Logic problems, Understanding odds, Understanding probability Common Core Standards: Grade 4 Number & Operations: Fractions , Grade 5 Number & Operations in Base Ten , Grade 5 Operations & Algebraic Thinking. Coin Flips, Risk to Reward Profile and Creating Your Own Synthetic Security Thus, the probability of getting 2 heads in a row is the probability of getting a head followed by a second flip where you also get a head. Applet: Instructions: Examples: Notes \"H\" count = , flips so far, number of coins: one flip \"H\" probability: 0. A probability of one means that the event is certain. Probability measures how certain we are a particular event will happen in a specific instance. That was flip number 130,659,178 Flip again? Color The Coin!. Probability: Independent Events. 5, or you will stay in the current state with probability 0. The toss of a coin, throwing dice and lottery draws are all examples of random events. The instructions are written in the handout for the students to understand how to complete the activi. of flipping one head with a coin is 50%, then the probability of flipping two heads at once is achieved by (adding or multiplying)_____ the separate probabilities. When the coin is flipped and the first three flips are heads, the fourth flip still has the probability of \u00bd However, many people misunderstand that the first three flips somehow influence the fourth flip, but they do not. Don't expect the numbers from trials to exactly match the predicted results--especially if you run only a few trials. Were you to toss the coin 100 times, you would get a clearer view of how probable it is that the coin lands on heads each time. A coin toss is a tried-and-true way for your fifth grader to understand odds. We do not know if we will get heads or tails. I got a question on the coin flip project. If we flip the coin 10 times, we are not guaranteed to get 5 heads and 5 tails. Repeat steps 2-4 until the coin lands on its side every time. Expected Value represents the average outcome of a series of random events with identical odds being repeated over a long period of time. Asked in Statistics , Probability. It begins with the two title characters caught in a most unusual coin game. When flipping a coin, the probability of getting a head does not change no matter how many times you flip the coin. Example: It is 12 if you have an experiment where you flip a coin and then roll a six sided die. If the coins show heads-tails (HT) or tails-heads (TH), player 2 gets 1 point. With three coins, all three landing on the same side is 1 in 8. Ask Question Asked 3 years, 6 months ago. The probability for equally likely outcomes in an event is:. In this video, we' ll explore the probability of getting at least one heads in multiple flips of a fair coin. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Explore probability concepts by simulating repeated coin tosses. If the probability of coin flipping head = P The probability of coin flipping tail = 1 - P Now The probability of flipping heads & then tails = Probability of flipping tails & then heads = P(1 - P) Which means to make a fair coin toss we now need 2 flips Player 1 wins if the sequence is HT Player 2 wins if the sequence is TH Any other sequence. Students will flip a coin a total of ten times per trial and record results by simply shading in the space below the realistic-looking coins. Flip 4 coins, and you're at 16 outcomes, a 2-digit number. The number of possible outcomes gets greater with the increased number of coins. Coin Flipper. For example, given 5 trials per experiment and 20 experiments, the program will flip a coin 5 times and record the results 20 times. Become a member and unlock all. The probability of A and B is 1/100. That means I flipped the coin 15 times. It's pretty much a \"coin flip\". The views expressed are those of the author(s) and are not necessarily. (Solution): Coin Toss Probability. Suppose a coin tossed then we get two possible outcomes either a \u2018head\u2019 ( H) or a \u2018tail\u2019 ( T ), and it is impossible to predict whether the result of a toss will be a \u2018head\u2019 or \u2018tail\u2019. However, the probability of getting exactly one heads out of seven flips is different (and the solution is given). But I want to simulate coin which gives H with probability 'p' and T with probability '(1-p)'. The first person to flip heads wins. Flipping coins and the binomial distribution#2 Consider two coins, one fair and one unfair. from the previous assumptions follows that given any sequence of coin tossing results, the next toss has the probability P(T) <=> P(H). Recommended: Please try your approach. Logical Reasoning. Procedure 1: Statistical Probability Reasoning. 5\u00d710 20 chance of getting a string of 76 heads. As long as the coin was not manipulated the theoretical probabilities of both outcomes are the same\u2014they are equally probable. Let us return to the coin flip experiment. If the probability of an event is high, it is more likely that the event will happen. And you can get a calculator out to figure that out in terms of a percentage. This article shows you the steps for solving the most common types of basic questions on this subject. That means I flipped the coin 15 times. I start by having my students create a \"Heads Tails\" T chart in their math journals and then writing a prediction for the result of tossing the coin 100 times. The coin toss is nothing but experimenting with tossing a coin. , the sample frequency of heads, which is the frequentist MLE estimate of $\\theta$ for the Binomial model of the coin flip problem. Unformatted text preview: Tamara Curiel Gala Cano Mendelian Genetics Coin Toss Lab PRE-LAB DISCUSSION: In heredity, we are concerned with the occurrence, every time an egg is fertilized, of the probability that a particular gene or chromosome will be passed on through the egg, or through the sperm, to the offspring. Lends to discussion and discovery of probability, elementary understanding of mode, and graphic organization of information to compare and contrast. Begin your coin with just a single penny. If you flip three coins, it's eight - two for the first times two for the second times two for the third. A sequence of consecutive events is also called a \"run\" of events. You flipped 1 coin of type US 1\u00a2 Penny: Timestamp: 2020-05-05 01:39:10 UTC. It is measured between 0 and 1, inclusive. The probability of coming up heads on the first flip is 1/2. It is about physics, the coin, and how the \"tosser\" is actually throwing it. the coin does not and can not \"remember\" last result 4. For example, consider a fair coin. In the first simulation, player A got lucky with 4 heads in 5 tosses. Remark: The idea can be substantially generalized. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of total results of four flips, 16. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is \\binom{n}{k}p^k(1-p)^{n-k}. Coin Toss: Simulation of a coin toss allowing the user to input the number of flips. Simple numbers. Ask Question Asked 3 years, 6 months ago. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. Based on your flip results, you will infer which of the coins you were given. Recommended: Please try your approach. 5, or you will stay in the current state with probability 0. The game is played in stages. The first person to flip heads wins. \"The coin tosses are independent events; the coin doesn't have a memory. If the probability of flipping heads is 70%, then the list contains 70 ones and 30 zeros. You flipped 1 coin of type US 1\u00a2 Penny: Timestamp: 2020-05-05 01:39:10 UTC. The majority of times, if a coin is heads-up when it is flipped, it will remain heads-up when it lands. First, note that the problem will likely make reference to a \"fair\" coin. In particular, ( 10 3) = 10! 3! 7!. Wait for your coin to dry; Flip (toss) your coin 40 times and record the number of times it lands on its side. The post is correct that the odds of getting. Your function will have to label each of those sequences with a door, since leaving any sequence unlabeled would contradict the assumption that the method always produces a result after F flips. I need to land on heads 3 times or more out of 6, in 80% of all trials. So, I'll do it faster! When we flip the coin 9 times there are $$2^9$$ possible outcomes that can happen. The coin has no desire to continue a particular streak, so it's not affected by any number of previous coin tosses. Coin Toss Activity is a great way for students to have fun and learn about calculating probability. Probability measures how certain we are a particular event will happen in a specific instance. The two non-straight lines in Fig. A single flip of a coin has an uncertain outcome. We could call a Head a success; and a Tail, a failure. In the first simulation, player A got lucky with 4 heads in 5 tosses. For four coin flips, our intuition was probably right: more likely to get two heads. 60 I tried this: P(2H) = 4C2 * 0. 7 flip_coins(100, 100, 5) #> [1] 0. Ask Question Asked 3 years, 6 months ago. The program should call a separate function flip()that takes no arguments and returns 0 for tails and 1 for heads. We all know a coin toss gives you a 50% chance of winning, but is it always that way? Delve into the inner-workings of coin toss probability with this activity. The sum of all possible outcomes is always 1 (or 100%) because it is certain that one of the possible outcomes will happen. Flipping the coin once is a Bernoulli trial, since there are exactly two complementary outcomes (flipping a head and flipping a tail), and they are both 1 2 \\frac{1}{2} 2 1 no matter how many times the coin is flipped. What is the probability that a fair coin lands on heads on 4 out of 5 flips? Wha are the four properties of a binomial probability distribution? In a carnival game, there are six identical boxes, one of which contains a prize. We\u2019ll set p=0. 5, which is our probability of tossing heads and moving forward.", "date": "2020-07-02 21:18:08", "meta": {"domain": "theoximoron.it", "url": "http://ctjn.theoximoron.it/coin-flip-probability.html", "openwebmath_score": 0.7363460659980774, "openwebmath_perplexity": 368.0354943188166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429629196683, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.856095700727982}}
{"url": "http://mathhelpforum.com/advanced-algebra/95329-need-help-elem-linear-algebra-problem.html", "text": "# Math Help - Need Help with Elem. Linear Algebra Problem\n\n1. ## Need Help with Elem. Linear Algebra Problem\n\nExpress the invertible matrix\n\n$\\left(\\begin{array}{ccc}1&2&1\\\\1&0&1\\\\1&1&2\\end{ar ray}\\right)$\n\nas a product of elementary matrices.\n--------------------------\nThe answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!\n\n2. Originally Posted by paupsers\nExpress the invertible matrix\n\n$\\left(\\begin{array}{ccc}1&2&1\\\\1&0&1\\\\1&1&2\\end{ar ray}\\right)$\n\nas a product of elementary matrices.\n--------------------------\nThe answer in the back of the book has a product of 6 matrices. I think I'm supposed to use the identity matrix and perform the same row operations on it as I would to obtain the above matrix... or something like that. Any help with this would be appreciated!\nYes, that's right. A \"row operation\" is one of three kinds:\n1) swap two rows\n2) multiply every member of a row by the same constant.\n3) replace every member of a row by itself plus a constant time the corresponding member of a second row.\n\nAn \"elementary matrix\" is a matrix constructed from the identity matrix by a single row operation.\n\nWhat you want to do is reduce the given matrix to the identity matrix (which is possible because it is invertible) by row operations, writing down the elementary matrix corresponding to each row operation.\n\nFor example, starting from\n$\\left(\\begin{array}{ccc}1&2&1\\\\1&0&1\\\\1&1&2\\end{ar ray}\\right)$\nMy first steps would be to reduce the first column to the column $\\left(\\begin{array}{c}1 \\\\ 0 \\\\ 0\\end{array}\\right)$.\nThe number in the upper left is already 1 so I don't need to change that. I can get 0 in the next row by subtracting the first row from the second row. The elementary matrix corresponding to that is\n$\\left(\\begin{array}{ccc}1 & 0 & 0 \\\\ -1 & 1 & 0 \\\\ 0 & 0 & 1\\end{array}\\right)$\nTo get 0 in the third row, I need to subtract the first row from the third row. That gives the elementary matrix $\\left(\\begin{array}{ccc}1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ -1 & 0 & 1\\end{array}\\right)$\n\nand they convert the matrix to\n$\\left(\\begin{array}{ccc}1 & 2 & 1 \\\\ 0 & -2 & 0 \\\\ 0 & -1 & 1\\end{array}\\right)$\nNow you need to convert the second column properly and then the third. Can you continue?\n\nAfter you have converted the matrix to the identity matrix and written down all the corresponding elementary matrices (in the proper order) so they multiply to give the original matrix.\n\nNormally it would take 9 row operations to reduce a 3 by 3 matrix to the identity matrix and so the matrix would be represented as the product of 9 elementary matrices. Notice that the upper left number was already 1 so we didn't need a row operation for that. Also that 0 now in the second row, third column will stay there we won't need a row operation to change that. I suspect that sort of thing, a 0 or 1 already in the correct position will happen once more to give 6 rather than 9 elementary matrices.\n\n3. This is exactly what I tried doing, but it's not the answer the book gives (although it is similar). Is the solution to this problem unique?\n\n4. No, it is not because you can follow different \"paths\" to row reduce to the identity matrix. What I showed is the most commonly used \"path\".", "date": "2015-06-02 19:53:32", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/95329-need-help-elem-linear-algebra-problem.html", "openwebmath_score": 0.8484682440757751, "openwebmath_perplexity": 178.66534325732104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9833429638959675, "lm_q2_score": 0.8705972583359806, "lm_q1q2_score": 0.8560956883718065}}
{"url": "https://math.stackexchange.com/questions/1583146/transitive-relation", "text": "# Transitive relation\n\nConsider $A$ is a relation de fined on $R$ (real numbers) where $A = \\{(a,b):|a-b|<4, a, b \\in R\\}$. Prove/disprove $A$ is transitive.\n\nI know if $|a-b|<4$ and $|b-c|<4$, then, $|a-c|<4$ ; A is transitive. Can I directly prove it with any counter example such as for $a=6$, $b=3$ and $c=1$ this relation is not transitive because $|6-1|>4$.\n\nIs this suitable for prove or disprove questions of relations?\n\nYour answer is correct. In general a relation $A$ is transitive if for all $a,b,c$ we have $(a,b)\\in A$ and $(b,c)\\in A$ implies $(a,c)\\in A$. That means that if we can find just one counter example (such as $a=6$, $b=3$ and $c=1$) $A$ cannot be transitive.\nYour answer is quite right. The question is basically \"if $a$ is near to $b$, and $b$ is near to $c$, is $a$ near to $c$?\"", "date": "2021-01-19 15:28:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1583146/transitive-relation", "openwebmath_score": 0.921374499797821, "openwebmath_perplexity": 162.6942033835524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750481179169, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.856089478447}}
{"url": "https://math.stackexchange.com/questions/2434494/expansion-in-terms-of-legendre-polynomial", "text": "# Expansion in terms of legendre polynomial\n\nObtain the first three terms in the expansion of function in terms of legendre polynomial F(x) in a series of the form $$F(x) = \\sum_{k=0}^{\\infty} A_k P_k(x)$$ where $$F(x)=\\{\\cos(x) \\text{ for } 0 \\le x \\le \\pi/2 \\$$$$0 \\text{ for } \\frac{\\pi}{2} \\le x \\le \\pi.\\}$$\n\nWhat I know is I have to use legendre's expansion formula i.e,$F(x) =\\sum A_kP_k(x)$ where $-1\u2264x\u22641$ But obviously I cannot use it directly because the range of $x$ differs. I have tried substituting $x=\\cos(\\theta)$ but no success so far.\n\nExtended hint: As already noted in your almost identical question Legendre polynomials you should apply a linear transformation, because the Legendre polynomial have the orthogonality interval $[-1,1]$. With $x=\\frac{\\pi}{2}(t+1)$ and $t=\\frac{2x}{\\pi}-1,$ define $F(x)=:\\tilde{F}(t).$ Now $\\tilde{F}$ is defined on $[-1,1]$ and you can compute $$F(x)=\\tilde{F}(t) = \\sum A_k P_k(t)=\\sum A_k P_k\\left(\\frac{2x}{\\pi}-1\\right)$$ with $$A_k=\\frac{2k+1}{2}\\int_{-1}^1 \\tilde{F}(t) P_k(t) dt$$\nIf you use this approach and compute $A_0, \\dots, A_3$, you get the following graphics with the original function $F(x)$ in red and the approximation in green:\n\u2022 Yes of course, because your function is not defined for $x=-1$. The transformation associates $x=0 \\leftrightarrow t=-1$ and $x=\\pi \\leftrightarrow t=1$. You should compute the expansion for $\\tilde{F}(t), t\\in[-1,1].$ \u2013\u00a0gammatester Sep 18 '17 at 15:08", "date": "2019-08-17 16:58:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2434494/expansion-in-terms-of-legendre-polynomial", "openwebmath_score": 0.9301050305366516, "openwebmath_perplexity": 103.11606074523804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9873750492324249, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8560894709323755}}
{"url": "https://www.physicsforums.com/threads/product-of-two-consecutive-integers.367041/", "text": "# Homework Help: Product of two consecutive integers\n\n1. Jan 4, 2010\n\n### nobahar\n\n1. The problem statement, all variables and given/known data\n\nProve that $$n^2+n$$ is even. Where n is a positive integer.\n\n2. Relevant equations\n\n$$n^2+n$$\n\n3. The attempt at a solution\n\n$$n^2+n = n(n+1)$$\n\nOne of which must be even, and therefore the product of 2 and an integer k.\n\n$$n = 2k, \\left \\left 2*(k(n+1))$$\n\nor\n\n$$n+1 = 2k, \\left \\left 2*(n*k)$$\n\nIs there a better way of doing this? I read this is not an inductive proof; what would this entail?\n\n2. Jan 4, 2010\n\n### rochfor1\n\nThis looks absolutely fine. No need for induction at all.\n\n3. Jan 4, 2010\n\n### HallsofIvy\n\nAs rochfor1 said, that is a perfectly good proof and simpler than \"proof by induction\".\n\nBut since you ask, here goes:\n\nIf n= 1, then $n^2+ n= 1^2+ 1= 2$ which is even.\n\nNow, suppose that $k^2+ k$ is even and look at $(k+1)^2+ (k+1)$\n$(k+1)^2+ (k+1)= k^2+ 2k+ 1+ k+ 1$$= (k^2+ 2k)+ 2k+ 2$. By the induction hypothesis, $k^2+ k$ is even and so $k^2+ k= 2m$ for some integer m (that is the definition of \"even\") so $(k+1)^2+ (k+1)= (k^2+ 2k)+ 2k+ 2$$= 2m+ 2k+ 2= 2(m+k+1)$. Since that is \"2 times an integer\", it is even.\n\nHaving proved that the statement is true for 1 and that \"if it is true for k, it is true for k+1\", by induction, it is true for all positive integers.\n\nYet a third way: If n is a positive integer, it is either even or odd.\n\ncase 1: n is even. Then n= 2m for some integer m. $n^2= 4m^2$ so $n^2+ n= 4m^2+ 2m= 2(m^2+ m)$. Since that is 2 times an integer, it is even.\n\ncase 2: n is odd. Then n= 2m+ 1 for some integer m. $n^2= (2m+1)^2= 4m^2+ 4m+ 1$ so $n^2+ n= 4m^2+ 4m+ 1+ 2m+ 1$$= 4m^2+ 6m+ 2= 2(2m^2+ 3m+ 1)$. Again that is 2 times an integer and so is even.\n\n4. Jan 5, 2010\n\n### nobahar\n\nMany thanks rochfor1 and HallsofIvy!\n\n5. Jan 5, 2010\n\n### icystrike\n\nSimple reasoning is needed.\nFor two consecutive integer , one will be even and another will be odd.\nThus;\n$$n\\equiv0(mod 2)$$\nand\n$$n+1\\equiv1(mod 2)$$\n\nWe may now conclude that $$n(n+1)\\equiv0(mod 2)$$ and means that n\u00b2+n is even.\n\n6. Jan 5, 2010\n\nFile size:\n24.9 KB\nViews:\n388\n7. Jan 5, 2010\n\n### Dick\n\nIf n is odd, n^2 is odd. If n is even, n^2 is even. odd+odd=even and even+even=even.\n\n8. Jan 6, 2010\n\n### ideasrule\n\nOP: Why do you want a proof by induction? The proof you gave is simple and beautiful.\n\n9. Jan 6, 2010\n\n### Staff: Mentor\n\nWhat you say is true, but you can't ascertain which of them will be even and which will be odd. In what follows, you are assuming that n is even and n+1 is odd. That is one of two possible cases, so you work is not complete.\n\n10. Jan 7, 2010\n\n### icystrike\n\nyes that is true. for the other case , it can be proved analogously .", "date": "2018-09-23 23:49:16", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/product-of-two-consecutive-integers.367041/", "openwebmath_score": 0.6292749047279358, "openwebmath_perplexity": 1312.1021854840985, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9873750529474512, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8560894690648703}}
{"url": "http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 07 Jul 2015, 07:46\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Inequalities trick\n\nAuthor Message\nTAGS:\nCurrent Student\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2800\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 196\n\nKudos [?]: 1162 [56] , given: 235\n\nInequalities trick\u00a0[#permalink] \u00a016 Mar 2010, 09:11\n56\nKUDOS\n117\nThis post was\nBOOKMARKED\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\nAttachments\n\n1.jpg [ 6.73 KiB | Viewed 29844 times ]\n\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 5682\nLocation: Pune, India\nFollowers: 1415\n\nKudos [?]: 7357 [43] , given: 186\n\nRe: Inequalities trick\u00a0[#permalink] \u00a022 Oct 2010, 05:33\n43\nKUDOS\nExpert's post\n39\nThis post was\nBOOKMARKED\nYes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.\n\nIf (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.\ne.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.\n\nAttachment:\n\ndoc.jpg [ 7.9 KiB | Viewed 29149 times ]\n\nThis divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.\n\nWhen you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.\n\nWhen you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.\n\nSince we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7\n\nIt should be obvious that it will also work in cases where factors are divided.\ne.g. (x - a)(x - b)/(x - c)(x - d) < 0\n(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.\n\nNote: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2028 Followers: 138 Kudos [?]: 1165 [15] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 15 This post received KUDOS 6 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 150 Followers: 3 Kudos [?]: 49 [11] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 11 This post received KUDOS 6 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [6] , given: 186 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 6 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nManager\nStatus: On...\nJoined: 16 Jan 2011\nPosts: 189\nFollowers: 3\n\nKudos [?]: 42 [5] , given: 62\n\nRe: Inequalities trick\u00a0[#permalink] \u00a010 Aug 2011, 16:01\n5\nKUDOS\n5\nThis post was\nBOOKMARKED\nWoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.\n\n1) CORE CONCEPT\n@gurpreetsingh -\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nArrange the NUMBERS in ascending order from left to right. a<b<c<d\nDraw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nNote: Make sure that the factors are of the form (ax - b), not (b - ax)...\n\nexample -\n(x+2)(x-1)(7 - x)<0\n\nConvert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')\nNow solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'\nSince you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.\n\n2) Variation - ODD/EVEN POWER\n@ulm/Karishma -\nif we have even powers like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\nThis will be same as (x-b)\n\nWe can ignore squares BUT SHOULD consider ODD powers\nexample -\n2.a\n(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0\n2.b\n(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0\nis the same as (x - a)(x - b)(x - c)(x - d) < 0\n\n3) Variation <= in FRACTION\n@mrinal2100 -\nif = sign is included with < then <= will be there in solution\nlike for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7\n\nBUT if it is a fraction the denominator in the solution will not have = SIGN\n\nexample -\n3.a\n(x + 2)(x - 1)/(x -4)(x - 7) < =0\nthe solution will be -2 <= x <= 1 or 4< x < 7\nwe cant make 4<=x<=7 as it will make the solution infinite\n\n4) Variation - ROOTS\n@Karishma -\nAs for roots, you have to keep in mind that given $$\\sqrt{x}$$, x cannot be negative.\n\n$$\\sqrt{x}$$ < 10\nimplies 0 < $$\\sqrt{x}$$ < 10\nSquaring, 0 < x < 100\nRoot questions are specific. You have to be careful. If you have a particular question in mind, send it.\n\nRefer - inequalities-and-roots-118619.html#p959939\nSome more useful tips for ROOTS....I am too lazy to consolidate\n\n<5> THESIS -\n@gmat1220 -\nOnce algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.\n\nI will save this future references....\n\nAnyone wants to add ABSOLUTE VALUES....That will be a value add to this post\n\n_________________\n\nLabor cost for typing this post >= Labor cost for pushing the Kudos Button\nkudos-what-are-they-and-why-we-have-them-94812.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 5682\nLocation: Pune, India\nFollowers: 1415\n\nKudos [?]: 7357 [3] , given: 186\n\nRe: Inequalities trick\u00a0[#permalink] \u00a009 Sep 2013, 22:35\n3\nKUDOS\nExpert's post\nkarannanda wrote:\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nHi Gurpreet,\nThanks for the wonderful method.\nI am trying to understand it so that i can apply it in tests.\nCan you help me in applying this method to the below expression to find range of x.\nx^3 \u2013 4x^5 < 0?\n\nI am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.\n\nBefore you apply the method, ensure that the factors are of the form (x - a)(x - b) etc\n\n$$x^3 - 4x^5 < 0$$\n\n$$x^3 ( 1 - 4x^2) < 0$$\n\n$$x^3(1 - 2x) (1 + 2x) < 0$$\n\n$$4x^3(x - 1/2)(x + 1/2) > 0$$ (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)\n\nNow the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.\nThe solution will be x > 1/2 or -1/2 < x< 0.\n\nCheck out these posts discussing such complications:\nhttp://www.veritasprep.com/blog/2012/06 ... e-factors/\nhttp://www.veritasprep.com/blog/2012/07 ... ns-part-i/\nhttp://www.veritasprep.com/blog/2012/07 ... s-part-ii/\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [2] , given: 186 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 2 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\\leq x \\leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 336\nFollowers: 52\n\nKudos [?]: 380 [2] , given: 81\n\nInequalities trick\u00a0[#permalink] \u00a006 Jan 2015, 02:35\n2\nKUDOS\nExpert's post\n5\nThis post was\nBOOKMARKED\nJust came across this useful discussion.\n\nVeritasPrepKarishma has given a very lucid explanation of how this \u201cwavy line\u201d method works.\n\nI have noticed that there is still a little scope to take this discussion further. So here are my two cents on it.\n\nI would like to highlight an important special case in the application of the Wavy Line Method\n\nWhen there are multiple instances of the same root:\n\nTry to solve the following inequality using the Wavy Line Method:\n\n$$(x-1)^2(x-2)(x-3)(x-4)^3 < 0$$\n\nTo know how you did, compare your wavy line with the correct one below.\n\nDid you notice how this inequality differs from all the examples above?\n\nNotice that two of the four terms had an integral power greater than 1.\n\nHow to draw the wavy line for such expressions?\n\nLet me directly show you how the wavy line would look and then later on the rule behind drawing it.\n\nAttachment:\nFile comment: Observe how the wave bounces back at x = 1.\n\nbounce.png [ 10.4 KiB | Viewed 2175 times ]\n\nNotice that the curve bounced down at the point x = 1. (At every other root, including x = 4 whose power was 3, it was simply passing through them.)\n\nCan you figure out why the wavy line looks like this for this particular inequality?\n\n(Hint: The wavy line for the inequality\n\n$$(x-1)^{38}(x-2)^{57}(x-3)^{15}(x-4)^{27} < 0$$\n\nIs also the same as above)\n\nCome on! Give it a try.\n\nIf you got it right, you\u2019ll see that there are essentially only two rules while drawing a wavy line. (Remember, we\u2019ll refer the region above the number line as positive region and the region below the number line as negative region.)\n\nHow to draw the wavy line?\n\n1. How to start: Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression.\n\n2. How to alternate: In the given expression, if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to \u2013ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region). However, if the power of a term is even, then the wave bounces back into the same region.\n\nNow look back at the above expression and analyze your wavy line. Were you (intuitively) using the above mentioned rules while drawing your wavy line?\n\nSolution\n\nOnce you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to look for the space where the above expression would be less than zero, look for the areas in the wavy line where the curve is below the number line.\n\nSo the correct solution set would simply be {3 < x < 4} U {{x < 2} \u2013 {1}}\n\nIn words, it is the Union of two regions region1 between x = 3 and x = 4 and region2 which is x < 2, excluding the point x = 1.\n\nFood for Thought\n\nNow, try to answer the following questions:\n\n1. Why did we exclude the point x = 1 from the solution set of the last example? (Easy Question)\n2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains?\n\nFoot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases.\n\n- Krishna\n_________________\n\nLast edited by EgmatQuantExpert on 10 Jan 2015, 20:41, edited 1 time in total.\nCurrent Student\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2800\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 196\n\nKudos [?]: 1162 [1] , given: 235\n\nRe: Inequalities trick\u00a0[#permalink] \u00a019 Mar 2010, 11:59\n1\nKUDOS\nttks10 wrote:\nCan u plz explainn the backgoround of this & then the explanation.\n\nThanks\n\ni m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.\n\nsidhu4u wrote:\nI have applied this trick and it seemed to be quite useful.\n\nNice to hear this....good luck.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nMath Forum Moderator\nJoined: 20 Dec 2010\nPosts: 2028\nFollowers: 138\n\nKudos [?]: 1165 [1] , given: 376\n\nRe: Inequalities trick\u00a0[#permalink] \u00a026 May 2011, 19:55\n1\nKUDOS\nchethanjs wrote:\nmrinal2100 wrote:\nif = sign is included with < then <= will be there in solution\nlike for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7\n\nin case when factors are divided then the numerator will contain = sign\n\nlike for (x + 2)(x - 1)/(x -4)(x - 7) < =0\nthe solution will be -2 <= x <= 1 or 4< x < 7\nwe cant make 4<=x<=7 as it will make the solution infinite\n\ncorrect me if i am wrong\n\nCan you please tell me why the solution gets infinite for 4<=x<=7 ?\n\nThanks.\n\n(x -4)(x - 7) is in denominator.\nMaking x=4 or 7 would make the denominator 0 and the entire function undefined.\n\nThus, the range of x can't be either 4 or 7.\n4<=x<=7 would be wrong.\n4<x<7 is correct because now we removed \"=\" sign.\n_________________\nSenior Manager\nJoined: 16 Feb 2012\nPosts: 257\nConcentration: Finance, Economics\nFollowers: 4\n\nKudos [?]: 123 [1] , given: 121\n\nRe: Inequalities trick\u00a0[#permalink] \u00a022 Jul 2012, 02:03\n1\nKUDOS\nVeritasPrepKarishma wrote:\nmrinal2100: Kudos to you for excellent thinking!\n\nCorrect me if I'm wrong.\nIf the lower part of the equation$$\\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\\leq x \\leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.\n_________________\n\nKudos if you like the post!\n\nFailing to plan is planning to fail.\n\nCurrent Student\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2800\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 196\n\nKudos [?]: 1162 [1] , given: 235\n\nRe: Inequalities trick\u00a0[#permalink] \u00a018 Oct 2012, 04:17\n1\nKUDOS\nGMATBaumgartner wrote:\ngurpreetsingh wrote:\nulm wrote:\nif we have smth like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\n\nyes even powers wont contribute to the inequality sign. But be wary of the root value of x=a\n\nHi Gurpreet,\nCould you elaborate what exactly you meant here in highlighted text ?\n\nEven I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.\n\nIf the powers are even then the inequality won't be affected.\n\neg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0\n\njust use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 5682\nLocation: Pune, India\nFollowers: 1415\n\nKudos [?]: 7357 [1] , given: 186\n\nRe: Inequalities trick\u00a0[#permalink] \u00a018 Oct 2012, 09:27\n1\nKUDOS\nExpert's post\nGMATBaumgartner wrote:\nHi Gurpreet,\nCould you elaborate what exactly you meant here in highlighted text ?\n\nEven I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.\n\nIn addition, you can check out this post:\n\nhttp://www.veritasprep.com/blog/2012/07 ... s-part-ii/\n\nI have discussed how to handle powers in it.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25 1 This post received KUDOS Expert's post GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.php DS questions on inequalities: search.php?search_id=tag&tag_id=184 PS questions on inequalities: search.php?search_id=tag&tag_id=189 Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ Current Student Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2800 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 196 Kudos [?]: 1162 [1] , given: 235 Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04 1 This post received KUDOS VeritasPrepKarishma wrote: The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6 - 2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Math Expert Joined: 02 Sep 2009 Posts: 28348 Followers: 4485 Kudos [?]: 45444 [1] , given: 6761 Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7357 [1] , given: 186 Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED PathFinder007 wrote: I have a query. I have following question x^3 - 4x^5 < 0 I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks The factors must be of the form (x - a)(x - b) .... etc x^3 - 4x^5 < 0 x^3 * (1 - 4x^2) < 0 x^3 * (1 - 2x) * (1 + 2x) < 0 x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1) x^3 * 2(x - 1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and -1/2. ____________ - 1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or -1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ The links will give you the theory behind this method in detail. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nIntern\nJoined: 24 Sep 2013\nPosts: 12\nFollowers: 0\n\nKudos [?]: 2 [1] , given: 56\n\nRe: Inequalities trick\u00a0[#permalink] \u00a011 Nov 2014, 08:52\n1\nKUDOS\nFor statement I\n\n(1) (1-2x)(1+x)<0\n\nOnce you get this into the req form\n\n2 (x-1/2) (x+1) > 0\n\nsol +++++ -1 ------ 1/2 +++++\nAs Karishma pointed out ponce in the req form, the rightmost will always be positive and the alternating will happen from there.\n\nSo sol for this x> 1/2 and x<-1\n\nIntegers greater than 1/2 and less than -1 , thus |x| may be >= 1. Unsure\n\nThus Insufficient.\n\nFor Statement II\n\n(2) (1-x)(1+2x)<0\n\nOnce you get this into the req form\n\n2(x+1/2) (x-1) > 0\n\nsol +++++ -1/2 ------ 1 +++++\n\nSo sol for this x>1 and x< -1/2\n\nIntegers greater than 1 and less than -1/2 , thus |x| may be >= 1. Unsure\n\nThus Insufficient.\n\nCombining Both the statements\n\nx<-1 and x>1\n\nThus Integers for this range will give |x| > 1\n\nThus Sufficient.\n\nHope this helps. I am not very confident of my solution though. Its my first solution gmatclub\n\nmayankpant wrote:\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nHi\nCan you please explain this question to me using the graph. I am missing the point when graph is being used here?\n\nIf x is an integer, is |x|>1?\n(1) (1-2x)(1+x)<0\n(2) (1-x)(1+2x)<0\n\nFor me ,the first equations roots are -1 and 1/2. Now I am struggling to get to the correct sign using the graph method here.\n\nSame for second equation: roots are 1 and -1/2 but struggling for the sign.\n\nTHanks\nIntern\nJoined: 08 Mar 2010\nPosts: 6\nFollowers: 0\n\nKudos [?]: 0 [0], given: 4\n\nRe: Inequalities trick\u00a0[#permalink] \u00a016 Mar 2010, 09:46\nCan u plz explainn the backgoround of this & then the explanation.\n\nThanks\nRe: Inequalities trick \u00a0 [#permalink] 16 Mar 2010, 09:46\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0\u00a0\u00a03\u00a0\u00a0\u00a04\u00a0\u00a0\u00a05\u00a0\u00a0\u00a06\u00a0 \u00a0 Next \u00a0[ 105 posts ]\n\nSimilar topics Replies Last post\nSimilar\nTopics:\n127 Tips and Tricks: Inequalities 27 14 Apr 2013, 08:20\ninequality, 11 09 Sep 2007, 12:18\nInequalities 3 09 Dec 2006, 04:28\ninequalities 1 06 Aug 2006, 07:23\ninequalities 3 05 Dec 2005, 11:18\nDisplay posts from previous: Sort by", "date": "2015-07-07 15:46:01", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1", "openwebmath_score": 0.6233364343643188, "openwebmath_perplexity": 1919.4069545627315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305381464927, "lm_q2_score": 0.8887587920192298, "lm_q1q2_score": 0.8560796095191096}}
{"url": "https://math.stackexchange.com/questions/2243813/why-do-we-care-about-the-fact-that-a-series-converges-but-not-what-it-converges", "text": "# Why do we care about the fact that a series converges, but not what it converges to?\n\nIn the 10 tests of convergence/divergent (that I know), them being,\n\nby defn, integral test, div test, comparison test, limit comparison test, gs test, alternating series test, p-series test, root test, ratio test\n\nthe only ones that can tell me what the series converges to is by defn and the gs test.\n\nThe other ones can tell me if the series converges or diverges.\n\nBut I've always wondered, why do we care whether it converges, if we cannot exactly figure out what it converges to? Why is it helpful to know only the behaviour and not the actual value?\n\n\u2022 Of course we care about both, but in practice it's generally a lot easier to decide convergence than to actually compute the sum. A good convergence result often comes with information about the speed of convergence...so that we can use it to compute the series numerically, which is often the best we can manage. \u2013\u00a0lulu Apr 20 '17 at 16:58\n\u2022 There are ways (in specific cases) to prove that if a sequence converges, then it must converge to $L$. For example, some recurrences, if they converge, converge to a fixed point. Also, for many of the tests, you can get estimates on the remaining error. \u2013\u00a0Michael Burr Apr 20 '17 at 16:59\n\u2022 You can numerically compute the sum and see it goes to a certain approximate value, but what if it diverges slowly? The most dangerous and famous example is the harmonic series, which diverges but really slowly. So, knowing if it converges in the first place is essential. \u2013\u00a0AspiringMathematician Apr 20 '17 at 17:10\n\u2022 Well, a professor of mine once said \"Before you go digging around in the mud, you want to know that there's a potato in there\". In context it depends on what you want to do. If the goal is not to actually solve an equation but just to understand a behavior--- say, you want to know if a function is bounded or not. If it is bounded/unbounded you can say such and such will happen. We just want to know the behavior, we don't care what the bound itself is nescessarily. Except when we do. \u2013\u00a0fleablood Apr 20 '17 at 17:15\n\nHere are some ideas:\n\n\u2022 Tests like the integral test actually include bounds on the error. For example, if the integral test applies to $\\sum a_n$, then you could compute $\\sum_{n=1}^Na_n$ and use the integral test on the remainder to bound the remainder.\n\n\u2022 The alternating series test also comes with a bound on the error, so you know how close your partial sum is to the right answer (if you were working on a computer, for example).\n\n\u2022 The ratio test essentially tells you that the the terms (eventually) act like a geometric series. Therefore, if you can get a handle on how close something is to being a geometric series, you can use the geometric series computation to bound the remainder.\n\n\u2022 If you have a recurrence like $x_n=ax_{n-1}+b$, then you can prove that if the sequence converges, then it must converge to a fixed point, i.e., a point that satisfies $L=aL+b$, $L=\\frac{b}{1-a}$. However, to use this, one must first justify why the sequence converges at all.\n\n\u2022 :o I didn't know about this \"bounds on error\" thing... thanks! \u2013\u00a0K Split X Apr 20 '17 at 17:08\n\nIf the series $\\sum_{n=0}^\\infty a_n$ converges, then I already know what it converges to: the actual value is $\\sum_{n=0}^\\infty a_n$.\n\nThere are lots of ways to represent numeric values, and \"the sum of this series\" is one of them. Often, it is a very useful way to write the actual value. Sometimes, it's even the best way we have to do so.\n\nI don't want to replace any of the answers given so far and rather stress just one point: To sometimes consider the questions of convergence independently from the computation of the limit is an important educational step, in order to build up a more abstract point of view on convergence.\n\n\u2022 I guess, there are many reasons to discuss convergence of series as one of the first topics in an analysis course. Analysis of series is a bit more abstract, than analysis of sequences, but simple enough to present elementary proofs for the important theorems. Students see how exchange of limits can lead to wrong results. (This can be also archived with sequences, but harmonic series as counterexample is simpler to remember.) They can to practice how to estimate and handle an abstract term like a sum with unknown entries.\n\nWhy is it important to consider convergence and computation of limit points separate ?\n\n\u2022 Some spaces are constructed to be the completion of certain smaller spaces. For example the real numbers can be constructed as the set of all limit points of rational Cauchy-convergent sequences (with identifying some limit points as being equal). Or sometimes you build function spaces in this way, for example the set of limits of smooth functions with respect to some Sobolev norm. When working in these spaces, it is often important to handle convergence of sequences without using a closed form for the limit point.\n\n\u2022 Many numerical simulations use some kind of series or sequences to approximate complicated terms, like solutions of differential equations. In these situations the computer will compute sufficiently many terms to estimate the exact limit. It is of great interest to prove theoretically, that these series converge to a unique limit. In many cases the limit of the sequence exist only as an abstract object, being a solution of a certain differential equation. It is common the consider the general convergence to some point (~ Stability) independently from conditions to ensure that an existing unique limit is indeed the correct solution (~ Consistency). Therefore I consider it to be an important lesson to see how convergence and computing the limit point can be treated somehow independently from each. It can happen surprisingly easy, that numerical simulations converge to a false solutions if one of both condition is not satisfied, see for example the plots one the first slides here: Slides 3-5.", "date": "2019-06-19 09:23:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2243813/why-do-we-care-about-the-fact-that-a-series-converges-but-not-what-it-converges", "openwebmath_score": 0.8936986327171326, "openwebmath_perplexity": 236.96948826206506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576912786245, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8560460984989582}}
{"url": "https://s108528.gridserver.com/opposite-of-sgretr/difference-between-rhombus-and-square-e6d981", "text": "A rhombus is a quadrilateral with all sides of equal length. Next: Rectangle\u2192 Chapter 3 Class 8 Understanding Quadrilaterals; Concept wise; Rhombus, Rectangle, Square. Square. The main difference between Rhomboid and Rhombus is that the Rhomboid is a parallelogram in which adjacent sides are of unequal lengths and angles are oblique and Rhombus is a quadrilateral in which all sides have the same length. Another interesting thing is that the diagonals (dashed lines) meet in the middle at a right angle. What is the difference between Rhombus and Rectangle? Yes, because a square is just a rhombus where the angles are all right angles. Main Difference . The difference between a kite and a rhombus is that a kite does not always have four equal sides or two pairs of parallel sides like a rhombus. Is Rhombus a Square? Keep in mind that the question \"Is a rhombus a square?\" 1 Descriptions; 2 Rhombus vs Parallelogram; 3 Comparison Chart; Descriptions A rhombus . (c) All angles are equal to 90 degrees. Venn diagrams are schematic diagrams used in logic and in the branch of mathematics known as \u2026 Rectangle Square Example 7 Example 8 Ex 3.4, 5 Ex 3.4, 6 \u2026 The three special parallelograms \u2014 rhombus, rectangle, and square \u2014 are so-called because they\u2019re special cases of the parallelogram. that's the main difference!! Here are the differences between a square and a rhombus. As nouns the difference between rhombus and square is that rhombus is while square is any simple object with four nearly straight and nearly equal sides meeting at nearly right angles. Square\u2026 ADVERTISEMENT. Rhomboid. Trigonometry. Think about what happens when you \"tilt\" a square into a rhombus, keeping the horizontal lines of the square horizontal, but rotating the vertical lines a little bit clockwise. The other opposite sides need not be parallel. Please find the attached file for the answer. In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. On the other hand, if you bisect the diagonals of a rectangle, they bisect each other at equal length but \u2026 To calculate the perimeter, you have 4 x length of one side or 4A, where A is the length of the side. If there are 32.6 million Americans in this age group, find the tot \u2026 al U.S. \u2026 Among these, many people get confused with rhombus and parallelograms and wonder if they are similar or if the terms are used interchangeably. Therefore, the difference between a square and a rhombus is that {eq}\\displaystyle{ \\rm \\boxed{ \\text{1) A square has all sides equal and all angles equal, but a rhombus \u2026 The table below offers a clear distinction between them. The plural is rhombi or rhombuses, and, rarely, rhombbi or rhombbuses (with a double b). On a square, both of the diagonals equal each other, but on a rhombus \u2026 I seems clear that the orientation of the figure has a lot to do with the name applied to it. \u2022 Considering the diagonals; \u2013 The diagonals of the rhombus bisect each other at right angles, and the triangles formed are equilateral. Also opposite sides are parallel and opposite angles are equal. Trapezoid A quadrangle which has only one of the two opposite sides as parallel is called a Trapezoid. WHAT IS DIFFERENCE BETWEEN SQUARE AND RHOMBUS WHAT IS DIFFERENCE BETWEEN SQUARE AND RHOMBUS JUSTIFY YOUR ANSWER pls ans both u will get 28 pt I(t) = 27, 992 * (1.017) ^ t covert to log Americans who are 65 years of age or older make up 13.1% of the total population. To distinguish a rectangle from rhombus following property should be kept in mind: 2. All sides of the rectangle are at right angle while this is not the case with rhombus shape. In other words \u2026 3. While it may be a little bit confusing since a square is also a rectangle, there is one difference that makes it easy to distinguish one from the other. Thus a rhombus is not a square unless the angles are all right angles. Compare the new area to the original area. Its properties are (a) All sides are equal. The rhombus has a square as a special case, and is a special case of a kite and parallelogram. What\u2019s the difference between a square and a rectangle? rhombus . The most common types of quadrilaterals are a square, rectangle, rhombus, parallelogram, trapezium, and kite. Rhombus Rhombus and Kite You are here. Rhombuses, rectangles, and squares 2. If both pair of opposite sides of a \u2026 All the angles of a square are equal \u2026 As nouns the difference between rhomboid and rhombus is that rhomboid is a parallelogram which is neither a rhombus nor a rectangle while rhombus is . 3. \u2013 The \u2026 English (fish) (wikipedia rhombus) Noun \u2026 The top right one, the original \u201cdiamond\u201d is actually a square. 10 thoughts on \u201c Rhombus vs Diamond \u201d howardat58 February 11, 2016 at 2:57 am. As with all quadrilaterals, the sum \u2026 in the above quadrilateral family tree works just like. In this article, let us discuss the difference between rhombus and parallelogram in detail. Square A square is a parallelogram with right angles and equal sides. Rhombus and square have all sides equal, to distinguish a rhombus from square following property should be kept in Each angle of square has to be 90 degrees unlike rhombus. A square must have 4 right angles. (d) The diagonals are equal. Difference Between Rhombus and Parallelogram. (b) Opposite sides are equal and parallel. Unlike rhombus square needs to have all angles equal to 90 degree. A rhombus, also known as \u201crhom\u201d or \u201cdiamond,\u201d is an equilateral quadrilateral, a term that refers to a figure with four parallel sides (the lines will never intersect even if they \u2026 Solved Problems : Home >> Parallelogram >> Difference & Similarity between Square & Parallelogram >> Difference & Similarity between Square & Parallelogram. The Rhombus. To distinguish a rectangle from rhombus following property should be kept in mind: 2. Unlike rhombus only opposite sides of rectangle are equal. Rhombus vs \u2026 Harlon Moss. Area of rhombus = (1/2) x (d\u2081 x d\u2082) Here d\u2081 and d\u2082 are diagonals. A square is a quadrilateral with all sides equal in length and all interior angles right angles. This is true for all squares, so ALL SQUARES ARE RHOMBI (the plural of rhombus). but square is a quadilateral having all 4 sides equal but having all angles of its 4 sides as 90 degree . (In addition, the square is a special case or type of both the rectangle and the rhombus.) means Is every rhombus also always a square? As a adjective square is shaped like a (the polygon). Trigonometry Ratios. A kite is one type of four sided figure known as a quadrilateral. A kite is a four-sided shape that has two sets of adjacent sides that have equal lengths. It is more common to call this shape a rhombus, but some people call it a rhomb or even a diamond. Each angle of rectangle has to be 90 degrees unlike rhombus. Views: 2. A rhombus, on the other hand, does not have any rules about its angles, so there are many many, examples of a rhombus that are not also squares. In contrast, a rhombus is a shape having four equal sides but in a diamond shape. A rhombus is a four-sided shape where all sides have equal length (marked \"s\"). All rhombuses and squares are also kites. A rhombus is a quadrilateral with all sides equal in length. To distinguish a rectangle from rhombus following property should be kept in mind: 2. Same is the case with a rhombus \u2026 In general, a rhombus has the following special properties \u2022 All four sides are equal in length. Other Names. The rhombus is often called a diamond, after the \u2026 Each angle of square has to be 90 degrees unlike rhombus. Updated: February 27, 2019. As you can see in diagram \u2026 The shaded region represents the result of the operation. The different quadrilaterals are square, rectangle, rhombus, parallelogram, kite, trapezium which have some similar characteristics. As a verb square is to adjust so as to align with or place at a right angle to something else. Rhomboid vs. Rhombus. Rhombus and square have all sides equal, to distinguish a rhombus from square following property should be kept in Each angle of square has to be 90 degrees unlike rhombus. Sum of interior angles will be 360 degree Difference between square and rhombus: A square also fits the definition of a rectangle (all angles are 90\u00b0), and a rhombus (all sides are equal length). Each angle of rectangle has to be 90 degrees unlike rhombus. There are many shapes which give the impression of being associated to 1 one other, nevertheless you probably have a take a look at it, there are only some variations between them. 5 Min Read. As a adjective rhomboid is resembling, or shaped like a rhombus or rhomboid. Similarly, the area is A = b x h. Main Difference Between Parallelogram vs Rhombus . \u2022 Area of any can be calculated using the formula base \u00d7height. The square can be considered as a special case of the rhombus, where the internal angles are right angles. Each angle of rectangle has to be 90 degrees unlike rhombus. Unlike rhombus only opposite sides of rectangle are equal. A square, however, has four equal interior angles. Rhombus and Parallelograms are different although they both have four sides and four vertices and look almost similar. (AB=DC=AD=BC) \u2022 The diagonals of the rhombus bisect each other at right angles; diagonals are perpendicular to each other, in addition to the following properties of a \u2026 Area of Parallelogram \u2026 Note that a square is also a rhombus because it has 4 equal sides as well. Rhombus, Rectangle, Square; Rhombus and Kite. Opposite sides of a rhombus are parallel. Rectangle and rhombus are special cases of the parallelograms. rhombus is a quadilateral having all 4 sides equal but not having all angles 90 degree. Learn difference between square & rhombus topic of maths in details explained by subject experts on vedantu.com. The difference is a rhombus is made up of 2 acute and 2 obtuse angles and a square is made up of 4 right angles. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length. Rhombus Square Three dimensional object Trapezium Triangle Quadrilateral. Data-Handling. Answer: Rhombus is a four-sided quadrilateral with all its four sides equal in lengths.Rhombus is also referred to as a slanting square. You can come back to reference it in the future if you have any \u2026 Published: 14 Jan, 2021. \u2022 Rhombus and rectangle are quadrilaterals. CONTINUE READING BELOW. ADVERTISEMENT. Arithmetic Mean Frequency Distribution Table Graphs Median Mode Range . The major difference between rectangle and rhombus is that, the opposite sides of the rectangle are equal. Contents. The properties of square and rhombus are similar, the only distinguishing property is that square has all the angles equal to 90 0, and rhombus does not.. Rhombus is also called an equilateral quadrilateral. This article will explore the difference between a rhombus and a parallelogram. Square\u2026 Opposite angles are congruent in a rhombus. A square however is a rhombus \u2026 Diagram 2. 3. A Rhombus is a quadrilateral: A Square is also a quadrilateral: Opposite sides of Rhombus are parallel to each other : Opposite sides of Square are also parallel to each other: All sides of Rhombus are of equal length: All side of Square are also of equal length: Adjacent Angles of Rhombus are supplementary: Adjacent angles of Square are also supplementary : Diagonals of Rhombus \u2026 Rhomboid is a related term of rhombus. \u2026 Both a square and a rectangle may have their four angles at 90\u00b0, have parallel opposite sides with the same length, and have their diagonals equal in length, but only a square \u2026 Answer: No, a rhombus is not a square. ! Last updated at Sept. 25, 2018 by Teachoo Subscribe to our Youtube Channel - https://you.tube/teachoo. But these shapes are different from each other because of their few properties. The three-level hierarchy you see with. The name \"rhombus\" comes from the Greek word rhombos: a piece of wood whirled on a string to \u2026 Properties of rhombus: It has four equal sides ; Both diagonals are intersecting each other at right angle. 3 difference between Rhombus & square 2 See answers sufiyaanwar200451 sufiyaanwar200451 Answer: Here is four diffence b/w rhombus and square hope it help u. Step-by-step explanation: Purplehidie Purplehidie Answer: The sides of a square are perpendicular to each other whereas the sides of a rhombus are not perpendicular to each other. Unlike rhombus only opposite sides of rectangle are equal. Main Difference. Parallelograms are different although they both have four sides all have the length... H. Main difference between square and a parallelogram with right angles and equal sides ; both are... Rectangle and rhombus are special cases of the figure has a lot to do with the name to... A shape having four equal sides each angle of square has to be 90 unlike... Or place at a right angle almost similar special properties \u2022 all four sides and vertices! Represents the result of the rectangle and rhombus is a quadrilateral with all four. Rectangle\u2192 Chapter 3 Class 8 Understanding Quadrilaterals ; Concept wise ; rhombus rectangle... Sides that have equal lengths referred to as a special case or type four. Angle to something else its sides are equal in lengths.Rhombus is also referred to as a adjective is. Object Trapezium Triangle quadrilateral equal sides to calculate the perimeter, you have x. Diagonals of the rectangle and the triangles formed are equilateral 1/2 ) x ( d\u2081 d\u2082! It has four equal sides two opposite sides of the side Here d\u2081 and d\u2082 are.. Square\u2026 answer: No, a rhombus. square unless the angles are equal Quadrilaterals ; wise... S '' ) shape a rhombus is a four-sided shape where all sides of equal length and kite however! The difference between rhombus and parallelogram in detail not having all angles 90 degree the table below a! In general, a rhombus and parallelogram in detail ( d\u2081 x d\u2082 ) d\u2081! Be 360 degree difference between a rhombus and a rectangle from rhombus following property be! One of the rectangle are equal a slanting square sets of adjacent that. Angles, and the rhombus is a rhombus ( plural rhombi or rhombuses ) a. Chapter 3 Class 8 Understanding Quadrilaterals ; Concept wise ; rhombus, but some call. The question  is a quadrilateral with all sides are equal called a.. To call this shape a rhombus. the Area is a quadilateral having all 4 sides in... Equal sides 1/2 ) x ( d\u2081 x d\u2082 ) Here d\u2081 and are. Have four sides equal in lengths.Rhombus is also referred to as a slanting square parallelogram ; Comparison... Between rhombus and parallelograms and wonder if they are similar or if the terms are used interchangeably in. Right angles of interior angles diamond shape below offers a clear distinction between.! - https: //you.tube/teachoo and all interior angles right angles rectangle from rhombus following property be. Article will explore the difference between square & rhombus topic of maths in explained. Are parallel and opposite angles are equal \u2026 a rhombus is not a square is a special case type! To have all angles of its sides are equal in length d\u2081 and d\u2082 are diagonals you see. Or if the terms are used interchangeably the question  is a quadrilateral the middle a... Its four sides equal but not having all 4 sides equal but not having all 4 equal! Call this shape a rhombus and a parallelogram the branch of mathematics known as difference! Dimensional object Trapezium Triangle quadrilateral shape having four equal interior angles two sets of adjacent sides that have equal.... Not having all angles of a square? Here d\u2081 and d\u2082 are diagonals all of... To align with or place at a right angle to something else rhombuses and! 1 Descriptions ; 2 rhombus vs parallelogram ; 3 Comparison Chart ; Descriptions a rhombus where the internal are! Verb square is just a rhombus where the angles are equal the angles are right.... Thus a rhombus is a four-sided quadrilateral with all sides have equal lengths square is a special case the! Rhombus. to do with the name applied to it \u201c diamond \u201d is a! Addition, the sum \u2026 each angle of rectangle are equal to degree! A related term of rhombus: it has four equal interior angles right angles Considering the diagonals ; the. Clear distinction between them since equilateral means that all of its 4 sides as 90 degree tree works just.! Case or type of both the rectangle are equal angles, and the triangles formed equilateral. Rhombus a square and rhombus: it has four equal interior angles are similar if... Where a is the length of the two opposite sides of the rhombus is that the diagonals ; the. Or type of four sided figure known as \u2026 difference between square and rhombus: it has four equal angles... \u2026 a rhombus and parallelograms are different from each other at right angle to something else ;. The differences between a square however is a four-sided quadrilateral with all sides of the figure has a lot do... Rhombus has the following special properties \u2022 all four sides and four vertices and look almost similar is... And look almost similar diagrams are schematic diagrams used in logic and in the middle at right! 4 sides as 90 degree square and a parallelogram 4A, where the angles... Rectangle has to be 90 degrees or if the terms are used interchangeably call! The major difference between rectangle and rhombus are special cases of the parallelograms square to! This article will explore the difference between a square and a rectangle from rhombus following property should be kept mind... But in a diamond shape details explained by subject experts on vedantu.com their few properties have the same length discuss! As with all sides have equal lengths two sets of adjacent sides that have equal length or 4A, a! Are parallel and opposite angles are right angles is also referred to as a adjective square is just a is!, where a is the length of one side or 4A, where the angles! Any can be considered as a slanting square has only one of the operation the has! And, rarely, rhombbi or rhombbuses ( with a double b ) opposite sides are \u2026... H. Main difference between a square place at a right angle interesting thing is that the question is... People get confused with rhombus shape interesting thing is that the diagonals ( dashed lines ) meet in the at..., rectangle, square ; rhombus and kite, rarely, rhombbi rhombbuses! Used interchangeably often called a trapezoid they both have four sides all have the same.. And equal sides but in a diamond 1/2 ) x ( d\u2081 x d\u2082 ) Here d\u2081 and d\u2082 diagonals. In lengths.Rhombus is also referred to as a adjective square is a quadilateral having all angles degree! To as a slanting square rhombi ( the polygon ) thus a rhombus a. Four sided figure known as \u2026 difference between a square? \u201c diamond \u201d is actually a is. Four equal sides base \u00d7height distinguish a rectangle of one side or 4A, where the are... Triangles formed are equilateral equal sides are special cases of the figure has a to! Used in logic and in the middle at a right angle to else. ( 1/2 ) x ( d\u2081 x d\u2082 ) Here d\u2081 and d\u2082 are diagonals quadrilateral, since equilateral that... Updated at Sept. 25, 2018 by Teachoo Subscribe to our Youtube Channel - https: //you.tube/teachoo this,. Mind: 2 the original \u201c diamond \u201d is actually a square is to adjust as... A rhombus is not a square and a rhombus is that, the original \u201c diamond \u201d is a. A quadrilateral the middle at a right angle to something else as 90 degree or,... Quadilateral having all angles 90 degree to our Youtube Channel - https: //you.tube/teachoo of maths details... Rhombus topic of maths in details explained by subject experts on vedantu.com 2 rhombus parallelogram. ; Concept wise ; rhombus and parallelograms and wonder if they are or. Quadrilateral, since equilateral means that all of its sides are equal \u201d is actually a square are in! And the triangles formed are equilateral a ) all sides equal in and! All angles are all right angles and equal sides but in a diamond shape a term! Is that the question  is a parallelogram difference between rhombus and square a right angle while this is for. Angles of a square however is a four-sided shape that has two sets of adjacent that... And in the middle at a right angle = ( 1/2 ) x ( d\u2081 x d\u2082 ) Here and! Shaped like a ( the polygon ) us discuss the difference between square and rhombus special... Not a square are equal other at right angles and kite to adjust so as to align or. The orientation of the rectangle and rhombus are special cases of the rectangle are at right angles equal! Shapes are different from each other at right angle to something else square rhombus! Rarely, rhombbi or rhombbuses ( with a double b ): Rectangle\u2192 Chapter 3 8. Explained by subject experts on vedantu.com in plane Euclidean geometry, a rhombus and parallelograms are different each... Where the internal angles are equal but square is just a rhombus is a parallelogram maths in explained. Square\u2026 answer: rhombus is not the case with rhombus and kite plural rhombus... Since equilateral means that all of its 4 sides equal but having all angles 90 degree these, people! Comparison Chart ; Descriptions a rhombus or rhomboid a quadilateral having all 4 equal. Have four sides all have the same length after the \u2026 Learn difference between parallelogram vs rhombus. its are! The following special properties \u2022 all four sides all have the same.! Rectangle are equal and d\u2082 are diagonals the rectangle are equal having all angles are all right angles so. For all squares, so all squares, so all squares are (...\n\ndifference between rhombus and square 2021", "date": "2022-05-21 05:17:58", "meta": {"domain": "gridserver.com", "url": "https://s108528.gridserver.com/opposite-of-sgretr/difference-between-rhombus-and-square-e6d981", "openwebmath_score": 0.592010498046875, "openwebmath_perplexity": 901.8399469256652, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9755769099458929, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8560460866291191}}
{"url": "http://mathhelpforum.com/number-theory/132423-jacobi-symbol.html", "text": "# Math Help - Jacobi Symbol\n\n1. ## Jacobi Symbol\n\nEvaluate the Jacobi symbol ((n\u22121)(n+1)/n) for any odd natural number n.\n\nTrying out some numbers, I THINK it alternates between 1 and -1, but how can we PROVE it formally?\n\nAny help is appreciated!\n\n[also under discussion in math link forum]\n\n2. $\\left(\\frac{(n-1)(n+1)}{n}\\right) = \\left(\\frac{n-1}{n}\\right)\\left(\\frac{n+1}{n}\\right) = \\left(\\frac{-1}{n}\\right)\\left(\\frac{1}{n}\\right) = \\left(\\frac{-1}{n}\\right)$\n\nFor $n$ odd, $\\left(\\frac{-1}{n}\\right) = (-1)^\\frac{n-1}{2} = \\begin{cases} \\;\\;\\,1 & \\text{if }n \\equiv 1 \\pmod 4\\\\ -1 &\\text{if }n \\equiv 3 \\pmod 4\\end{cases}$.\n\nFor $n$ even, take $n=2k$, then $\\left(\\frac{-1}{n}\\right) = \\left(\\frac{-1}{2k}\\right) = \\left(\\frac{-1}{2}\\right)\\left(\\frac{-1}{k}\\right) = \\left(\\frac{-1}{k}\\right) = \\begin{cases} \\;\\;\\,1 & \\text{if }k \\equiv 1 \\pmod 4\\\\ -1 &\\text{if }k \\equiv 3 \\pmod 4\\end{cases}$.\n\n3. Originally Posted by chiph588@\n$\\left(\\frac{(n-1)(n+1)}{n}\\right) = \\left(\\frac{n-1}{n}\\right)\\left(\\frac{n+1}{n}\\right) = \\left(\\frac{-1}{n}\\right)\\left(\\frac{1}{n}\\right) = \\left(\\frac{-1}{n}\\right)$\n\nFor $n$ odd, $\\left(\\frac{-1}{n}\\right) = (-1)^\\frac{n-1}{2} = \\begin{cases} \\;\\;\\,1 & \\text{if }n \\equiv 1 \\pmod 4\\\\ -1 &\\text{if }n \\equiv 3 \\pmod 4\\end{cases}$.\n\nFor $n$ even, take $n=2k$, then $\\left(\\frac{-1}{n}\\right) = \\left(\\frac{-1}{2k}\\right) = \\left(\\frac{-1}{2}\\right)\\left(\\frac{-1}{k}\\right) = \\left(\\frac{-1}{k}\\right) = \\begin{cases} \\;\\;\\,1 & \\text{if }k \\equiv 1 \\pmod 4\\\\ -1 &\\text{if }k \\equiv 3 \\pmod 4\\end{cases}$.\nThank you!\n\nBut I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where \"n\" is even??? How is this possible??\n\n4. Originally Posted by kingwinner\nThank you!\n\nBut I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where \"n\" is even??? How is this possible??\nOops! You're right!\n\n5. But I'm concerned with one special case. For the case n=1, ((n\u22121)(n+1)/n)=(0/1)\n\nIs (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?\n\nThanks!\n\n6. Originally Posted by kingwinner\nBut I'm concerned with one special case. For the case n=1, ((n\u22121)(n+1)/n)=(0/1)\n\nIs (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?\n\nThanks!\n$\\left(\\frac{a}{n}\\right)\n= \\begin{cases}\n\\;\\;\\,0\\mbox{ if } \\gcd(a,n) \\ne 1\n\n\\\\\\pm1\\mbox{ if } \\gcd(a,n) = 1\\end{cases}$\n\nSo sub in $0$ and $1$ and see what you get.\n\n7. Originally Posted by chiph588@\n$\\left(\\frac{a}{n}\\right)\n= \\begin{cases}\n\\;\\;\\,0\\mbox{ if } \\gcd(a,n) \\ne 1\n\n\\\\\\pm1\\mbox{ if } \\gcd(a,n) = 1\\end{cases}$\n\nSo sub in $0$ and $1$ and see what you get.\ngcd(0,1)=1, so that rule says that (0/1)=+1 OR -1, but how do we know whether it is +1 or -1?\n\n8. Jacobi symbol\n\nApparently the answer is $0$. I haven't had too much exposure to the Jacobi symbol so I can't really tell you why. Check out the site for yourself though.\n\n9. $\\left( \\frac{0}{1} \\right)=1$ because the set of prime factors of $1$ is empty. if $n > 1$ is an odd integer, then $\\left( \\frac{0}{n} \\right)=0.$\n\n10. Originally Posted by NonCommAlg\n$\\left( \\frac{0}{1} \\right)=1$ because the set of prime factors of $1$ is empty. if $n > 1$ is an odd integer, then $\\left( \\frac{0}{n} \\right)=0.$\nIs there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?\n1 has no prime factorization, so the product is empty. Is it conventional to define the \"empty\" product to be equal to +1??\n\nAlso, is it true that, by definition, (a/1)=1 for any integer a?\n\nCan someone clarify this? Thank you!\n\n11. Originally Posted by kingwinner\nIs there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?\n1 has no prime factorization, so the product is empty. Is it conventional to define the \"empty\" product to be equal to +1??\n\nAlso, is it true that, by definition, (a/1)=1 for any integer a?\n\nCan someone clarify this? Thank you!\n\n12. Originally Posted by NonCommAlg\nIs there any real reason why we define (a/1)=1 for any integer a?\n\nWhy not define it to be 0? why not -1?\n\n13. Originally Posted by kingwinner\nIs there any real reason why we define (a/1)=1 for any integer a?\n\nWhy not define it to be 0? why not -1?", "date": "2016-02-13 14:03:00", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/132423-jacobi-symbol.html", "openwebmath_score": 0.9109496474266052, "openwebmath_perplexity": 609.76899833776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156283, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8560460831969453}}
{"url": "http://math.stackexchange.com/questions/158602/number-of-elements-vs-cardinality-vs-size/158635", "text": "# Number of elements vs cardinality vs size\n\nI have been wondered the definition of cardinality and number of elements. One mathematician told me that one can't said that the cardinality or size of the set $\\{1\\}$ is one, it should be said that the number of elements in the set is one. I guess that his opinion is that there is no term size in mathematics. Is these true? On the other hand, if we have an infinite set like $\\mathbb Z$, can we say that the number of elements of $\\mathbb Z$ is infinite or is the term \"number of elements\" used only in finite sets?\n\n-\nI have no idea what the mathematician was thinking. The cardinality of $\\{1\\}$ is indeed $1$. Of course it\u2019s also true that $1$ is the number of elements in the set. I would probably not use the expression number of elements when the set is infinite, but I\u2019d not complain if someone said \u2018$\\Bbb R$ has $2^\\omega$ elements\u2019 instead of \u2018the cardinality of $\\Bbb R$ is $2^\\omega$\u2019. \u2013\u00a0 Brian M. Scott Jun 15 '12 at 10:51\nFor finite sets there is absolutely no problem; a set has cardinality or size $n$ if it is in bijection with $\\{ 1, 2, ... n \\}$, and it has cardinality or size $0$ if it is empty. (The empty set is unique!) It is silly to prevent yourself from being able to say this. \u2013\u00a0 Qiaochu Yuan Jun 15 '12 at 12:39\n\nConverting my comment to an answer:\n\nI have no idea what the mathematician was thinking: the cardinality of the set $\\{1\\}$ is indeed $1$, and that\u2019s a perfectly normal way to express the fact. Of course it\u2019s also true that $1$ is the number of elements in the set.\n\nI would probably not myself use the expression number of elements when speaking of an infinite set, but \u2018$\\Bbb R$ has $2^\\omega$ elements\u2019 is a perfectly fine synonym of \u2018the cardinality of $\\Bbb R$ is $2^\u03c9$\u2019.\n\n-\nOkay. I guess the mathematician wanted that I won't be his graduate student as he asked me to modify my bachelor thesis so heavily. That was one point where I made a \"mistake\" and got a bad grade. And is the word \"size\" a synonym of the cardinality? \u2013\u00a0 mathenthusiast Jun 15 '12 at 15:52\n@mathenthusiast: I wouldn\u2019t say so: it\u2019s more general, as it can also refer to length, area, volume, etc. \u2013\u00a0 Brian M. Scott Jun 15 '12 at 21:38\n\nThe question you need to ask yourself is \"what is the meaning of the notion number?\"\n\nNatural numbers can be used to measure size, length, etc. while rational numbers measure ratio between two integers, real numbers measure length... what do complex numbers measure?\n\nIn mathematics we allow ourselves to define new ways of measuring things, and usually we refer to the values of these measures as numbers. Indeed there is no real difference between the length of a $100$ meters running track, and a $100$ meters long hot dog...\n\nWhen measuring the size of a set we came up with a clever way to discuss infinite sets. This way is \"cardinality\", and the cardinal of a set represents in a good sense the \"number of elements in the set\". So it has a perfectly good meaning to say that a set has $\\aleph_0$ many elements, or that a set is of size $\\aleph_1$.\n\n-", "date": "2014-12-22 23:34:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/158602/number-of-elements-vs-cardinality-vs-size/158635", "openwebmath_score": 0.87849360704422, "openwebmath_perplexity": 186.55819648630023, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9755769092358048, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.856046082880201}}
{"url": "https://math.stackexchange.com/questions/3210927/integration-by-partial-fractions-int-sec3xdx", "text": "# Integration by Partial Fractions $\\int \\sec^3xdx$\n\nI have looked at multiple ways to do partial fractions to integrate $$\\sec^3x$$, but there is a part where I keep getting stuck when I see the partial fractions get split up.\n\n$$\\int \\frac{\\cos x}{\\cos^4 x} = \\int \\frac{1}{(1-y^2)^2}$$\n\nThe next step I have seen is shown like this on Wikipedia:\n\n$$\\frac{1}{(1-u^2)^2}=\\frac{1/4}{1-u}+\\frac{1/4}{(1-u)^2}+\\frac{1/4}{1+u}+\\frac{1/4}{(1+u)^2}.$$\n\nWhere does the $$\\frac{1}{4}$$ come from in the numerator?\n\nAnd for the denominator, why are there two additional positive fraction decompositions $$1+u$$ and $$(1+u)^2$$?\n\n\u2022 See this page for the general theorem in use. \u2013\u00a0Bernard May 2 at 12:35\n\u2022 \u2013\u00a0cgiovanardi May 2 at 14:54\n\nAnd for the denominator, why are there two additional positive fraction decompositions\n\nBecause $$1-u^2 = (1-u)(1+u).$$ It's the \"difference of squares\" factorization. And so, squaring both sides, we also have $$(1-u^2)^2 = (1-u)^2(1+u)^2$$.\n\nIn partial fractions decomposition, all irreducible factors must appear as denominators, to the highest power that appears, as well as to all lower powers. So that means we need a denominator for $$(1-u)$$ and $$(1-u)^2,$$ but also for $$(1+u)$$ and $$(1+u)^2.$$\n\nWhere does the $$\\frac{1}{4}$$ come from in the numerator?\n\nFor partial fraction decomposition, the numerators must be chosen so that if you were to give every term common denominator and combine the numerators, you would get the starting fraction back. It takes a bit of algebra to find what numerators do the job, but that is where the $$\\dfrac{1}{4}$$ comes from.\n\nBut let's do the algebra. The normal way to find these numerators is to introduce variables for them and solve. Like so:\n\n$$\\dfrac{1}{(1-u^2)^2} = \\dfrac{1}{(1+u)^2(1-u)^2} = \\dfrac{A}{1+u} + \\dfrac{B}{(1+u)^2} + \\dfrac{C}{1-u} + \\dfrac{D}{(1-u)^2}$$\n\nhere the \"variable\" numerators are called $$A, B, C,$$ and $$D.$$\n\nWe solve for them by clearing denominators:\n\n$$1 = A(1-u)^2(1+u) + B(1-u)^2 + C(1-u)(1+u)^2 + D(1+u)^2$$\n\nThen in general you may have to expand all those binomials and match like terms of the resulting polynomials in $$u$$. But a shortcut can be to sub in the values of the roots in $$u$$. For example, at $$u=1,$$ the above equation becomes\n\n$$1 = A(1-1)^2(1+1) + B(1-1)^2 + C(1-1)(1+1)^2+ D(1+1)^2 = D(1+1)^2.$$\n\nSo we have $$D=1/4.$$\n\nSimilarly, by next subbing $$u=-1,$$\n\n$$1 = B(1-(-1))^2$$\n\nSo we have $$B=1/4.$$\n\nTo figure out that $$A$$ and $$C$$ are also $$1/4$$, we can either sub in some other, non-root values of $$u$$, or bite the bullet and do the algebra. (But if this were a partial fractions problem with no powers of irreducibles, we'd be done).\n\nI'll do the algebra, I guess. Expand, and gather like terms in $$u$$, treating the coefficients $$A, B, C,$$ and $$D$$ as variables.\n\n$$1 = A(1-u-u^2+u^3) + B(1-2u+u^2) + C(1+u-u^2-u^3) + D(1+2u+u^2) \\\\ = (A+B+C+D) + (-A-2B+C+2D)u + (-A+B-C+D)u^2 + (A-C)u^3.$$\n\nNow for two polynomials in $$u$$ to be equal for all values of $$u$$, every coefficient must be equal. On the left-hand side we have the polynomial $$1$$, which has a constant coefficient of $$1$$ and all higher coefficients are $$0$$.\n\nSo this gives us the following equations:\n\n$$1 = A+B+C+D\\\\ 0 = -A-2B+C+2D\\\\ 0 = -A+B-C+D\\\\ 0 = A-C.$$\n\nWe could solve this system of four linear equations in four unknowns, using Gaussian elimination or substitution or matrix methods. But first let's remember that we already know that $$B=D=1/4.$$ So we don't have four unknowns any more, our only unknowns left are $$A$$ and $$C$$. By the fourth equation, $$A=C$$, so our only unknown is really $$A$$. The first equation becomes\n\n$$1 = A+B+C+D = A + \\dfrac{1}{4} + A + \\dfrac{1}{4}.$$\n\nSo $$2A=1-\\dfrac{2}{4} = \\dfrac{1}{2}.$$ And so $$A=B=C=D=\\dfrac{1}{4}.$$\n\n\u2022 Oh, seeing the way you wrote that makes sense (going from $1-u^2$ to $(1-u^2)^2)$, just adding ^2 to the factors \u2013\u00a0Evan Kim May 2 at 12:38\n\nThis is actually a problem of partial fractions.\n\nWhen dividing one fraction to partial fractions, all possible cases of denominator should be considered. In this case, the denominator can be factorised to $$(1-u)^2(1+u)^2$$, so there are actually in total 8 possible fractions, with denominators being $$1-u$$, $$(1-u)^2$$, etc. Setting all the numerators to A, B, C, D, etc. for different denominator, layout the expression and compare coefficients, the result should then be that 1 fraction equal to 4 partial fractions.\n\nNote that $$(1-y^2)^2=(y-1)^2(y+1)^2$$. On the other hand, every rational function of the form$$\\frac{p(x)}{(x-r_1)^{m_1}\\times\\cdots\\times(x-r_k)^{m_k}},$$where $$\\deg p(x), can be written as a linear combination of rational functions of the type $$\\frac1{(x-r_j)^N}$$, with $$N\\in\\{1,2,\\ldots,m_j\\}$$. The rest is a matter of computations.", "date": "2019-09-21 17:38:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3210927/integration-by-partial-fractions-int-sec3xdx", "openwebmath_score": 0.9416100382804871, "openwebmath_perplexity": 242.41495961374093, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156283, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8560460816340292}}
{"url": "https://www.khanacademy.org/math/calculus-1/cs1-derivatives-chain-rule-and-other-advanced-topics/cs1-differentiation-using-multiple-rules/v/applying-the-chain-rule-and-product-rule", "text": "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\nMain content\n\n# Applying the chain rule and product\u00a0rule\n\nAP Calc: FUN\u20113 (EU)\n\n## Video transcript\n\n- [Instructor] What we're going to do in this video is try to find the derivative with respect to X of X squared sin of X. All of that to the third power. And what's going to be interesting is that there's multiple ways to tackle it. And I encourage you to pause the video and see if you could work through it on your own. So there's actually multiple techniques. One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to X of something to the third power. So, if I take the derivative it would be the derivative with respect to that something. So that would be three times that something squared times the derivative with respect to X of that something. Where the something, in this case, is X squared sin of X. X squared sin of X. This is just an application of the chain rule. Now, the second part, what would that be? The second part here do this in another color. In orange. Well here, I would apply the product rule. I have a product of two expressions. So I would take the derivative of, let me write this down. So this is gonna be the product rule. Product rule. I would take the derivative of the first expression. So, X, derivative of X squared is two X. Let me write a little bit to the right. This is gonna be two X times the second expression sin of X. Plus the first expression X squared times the derivative of the second one. Cosin of X. That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front. Which actually, let me just rewrite that. So all of this I could rewrite as let's see, this would be three times if I have the product raised to the second power I could take each of them to the second power and then take their products. So X squared squared is X to the fourth. And then sin of X squared is sin squared of X. And then all of that is being multiplied by that. And, if we want, we can algebraically simplify. We can distribute everything out. In which case, what would we get? Well let's see. Three times two is six. X to the fourth times X is X to the fifth. Sin squared of X times sin of X is sin of X to the third power. And then, let's see, three so plus three, X to the fourth times X squared is X to the sixth power. And then I have sin squared of X sin squared of X cosin of X. So there you have it, that's one strategy chain rule first, and then product rule. What would be another strategy pause the video and try to think of it. Well, we could just algebraically use our exponent properties first. In which case, this is going to be equal to the derivative with respect to X of if I'm taking X squared times sin of X to the third power instead I could say X to the third power which is going to be X to the sixth. And then sin of X to the third power. Sin of X to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have if I'm doing the product things to some exponent, well that's the same thing if each of them raised to the exponent and then the product of the two. Now how would we tackle this? Well, I here, I would do the product rule first. So let's do that. So let's do the product rule. So, we're gonna take the derivative of the first expression. So, derivative of X to the sixth is six X to the fifth. Times the second expression. Sin to the third of X. Or, sin of X to the third power. Plus the first X to the sixth times the derivative of the second and I'm just gonna write that D DX of sin of X to the third power. Now, to evaluate this right over here it does definitely make sense to use the chain rule. Use the chain rule. And so what is this going to be? Well I have the derivative of something to the third power. So that's going to be three times that something squared times the derivative of that something. So in this case, the something is sin of X. And the derivative of sin of X is cosin of X. Then I have all of this business over here. I have six X to the fifth. Sin to the third or sin of X to the third power. Plus X to the sixth. And if I were to just simplify this a little bit in fact, you see it very clearly. These two things are equivalent. This this term is exactly equivalent to this term the way it's written. And then this is exactly, if you multiply three times X to the sixth sin of X squared cosin of X. So, the nice thing about math if we're doing things that make logical sense we should get to the same endpoint. But the point here is that there's multiple strategies. You could use a chain rule first and then the product rule. Or you could use a product rule first, and then the chain rule. In this case, you could debate which one is faster. It looks like the one on the right might be a little bit faster. But sometimes these two are pretty close. But sometimes it'll be more clear than not which one is preferable. You really want to minimize the amount of hairiness, the amount the number of steps the chances for careless mistakes you might have.", "date": "2020-08-08 04:12:21", "meta": {"domain": "khanacademy.org", "url": "https://www.khanacademy.org/math/calculus-1/cs1-derivatives-chain-rule-and-other-advanced-topics/cs1-differentiation-using-multiple-rules/v/applying-the-chain-rule-and-product-rule", "openwebmath_score": 0.8288611769676208, "openwebmath_perplexity": 240.07129107804414, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9936116789011081, "lm_q2_score": 0.8615382094310355, "lm_q1q2_score": 0.8560344267102258}}
{"url": "https://math.stackexchange.com/questions/2819906/number-of-rectangles-with-odd-area", "text": "# Number of rectangles with odd area.\n\nWe have a $10\\times 10$ square.\nHow many rectangles with odd area are on the picture?\n\nI say lets choose a vertex first, there are $11\\cdot11=121$ possibilities.\nNow, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\\dfrac{121\\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.\n\nThe width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\\cdot 30=900$ odd area rectangles.\n\nWhat you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.\n\n\u2022 But where was I wrong? \u2013\u00a0Omer Jun 14 '18 at 19:29\n\u2022 @Omer: In counting the number of possible starting vertices (121). Not all of them are valid for all of the sub-cases you count. So, you go wrong at the very first step. Compare your thinking to how Hagen von Eitzen counted the cases, to correct the error. \u2013\u00a0Nominal Animal Jun 14 '18 at 19:55\n\u2022 \"and than assume that each possible odd width and height could be realized with this vertex.\" - no, s/he just assumed that 5 different odd x pairs and 5 different odd y pairs included the x and y of this vertex. If you start at (0,0) then in the x direction you can go to (1,0), (3,0), (5,0), (7,0), (9,0) but if you start at (5,0) you can go to (0,0), (2,0), (4,0), (6,0), (8,0), (10,0), that's 6 choices! \u2013\u00a0immibis Jun 15 '18 at 3:55\n\nThe reasoning is ok, except for this:\n\nThere are 5 possibilities to each\n\nSadly, no. If the chosen vertex is at $(1,1)$ (grid starting at $(0,0)$) you have 6 possibilities for each direction.\n\nIn general, you have 6 possibilities if the coordinate is odd, 5 if it's even.\n\nFix: Because there $36$ all-even-coordinates vertices, $25$ all-odd, and $121-36-25=60$ mixed vertices, the correct counting is\n\n$$36 \\times 5^2+ 25 \\times 6^2 + 60 \\times5 \\times 6=3600$$\n\nDividing by $4$ you get the $900$ rectangles.\n\nI assume you're looking for rectangles with sides parallel to those of the square and vertices that are lattice points. A rectangle with integer sides has odd area iff the sides are both odd. For each pair of odd integers $x, y$ with $1 \\le x,y \\le 9$, there are $(11-x)(11-y)$ possible positions for a rectangle of size $x \\times y$. Thus the number of rectangles is $$\\sum_{x \\in \\{1,3,5,7,9\\}} \\sum_{y \\in \\{1,3,5,7,9\\}} (11-x)(11-y) = 900$$", "date": "2019-09-16 20:29:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2819906/number-of-rectangles-with-odd-area", "openwebmath_score": 0.7377194762229919, "openwebmath_perplexity": 283.9377867944591, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540698633748, "lm_q2_score": 0.8740772433654401, "lm_q1q2_score": 0.8560311056649034}}
{"url": "http://math.stackexchange.com/questions/286315/fundamental-group-of-this-space", "text": "Fundamental group of this space\n\nBased on this question: What is the homology groups of the torus with a sphere inside?\n\nI'm trying to find the fundamental group of this space using the Seifert\u2013van Kampen theorem. If $U$ is the torus and $V$ is the sphere, then $U\\cap V$ is the circle, thus we have the following fundamental groups:\n\n$\\pi_1(U)=\\mathbb Z\\times\\mathbb Z$\n\n$\\pi_1(V)=0$\n\n$\\pi_1(U\\cap V)=\\mathbb Z$\n\nIf we use the group presentation notation we have:\n\n$\\pi_1(U)=\\langle\\alpha,\\beta\\mid \\alpha\\beta=\\beta\\alpha\\rangle$\n\n$\\pi_1(V)=\\langle\\emptyset\\mid\\emptyset\\rangle$\n\n$\\pi_1(U\\cap V)=\\langle\\gamma\\mid\\emptyset\\rangle$\n\nThus using the Seifert\u2013van Kampen theorem:\n\n$\\pi_1(X)=\\langle\\alpha,\\beta\\mid\\alpha\\beta=\\beta\\alpha,\\beta\\rangle$\n\nNote that I added $\\beta$ above because when we turn around the generator of $S^1$ which is $U\\cap V$, we span one of the generators of the torus which is $U$.\n\nThus the fundamental group of this space is $\\mathbb Z\\times \\{0\\}$ which is $\\mathbb Z$ itself.\n\nMy approach is correct?\n\nThanks a lot!\n\n-\nIt looks like that one generator of fundamental group of torus vanishes since you can collapse a close path around the equator. \u2013\u00a0 Sigur Jan 25 '13 at 2:02\n@Sigur yes, you're right. \u2013\u00a0 user42912 Jan 25 '13 at 2:10\n@Sigur but what can you say about my proof? \u2013\u00a0 user42912 Jan 25 '13 at 2:11\n\nThe approach is correct, but you can use a simpler method. The space you care about is homotopic to the wedge sum of two spheres and one circle. Thus the fundamental group is Z. To see why this space is homotopic to the space I said, you just check first that you can contract the upper half-sphere, thus the lower half-sphere becomes to a sphere, and the torus becomes a sphere with its south and north poles glued together. A sphere with its south and north poles glued together is homotopic to a sphere with a line connect its south and north poles. After these deformations you have a space which is what I said above.\n\n-\nThank you very much you helped not only with this question but with the question I linked about homology groups of this space. \u2013\u00a0 user42912 Jan 25 '13 at 11:16\none question, do you know where can I find a book with these constructions (except Hatcher's book)? these constructions aren't so obvious at the first glance. \u2013\u00a0 user42912 Jan 25 '13 at 11:18\n@user42912 I was about to introduce you the Hatcher's book. You just have to learn the first two chapters you will get the idea. There are a lot of deformations of CW cplx. \u2013\u00a0 lee Jan 25 '13 at 11:40\n\nNitpicking, but I suppose you could be more explicit about the relation $\\beta=\\gamma$, which comes from the inclusion $\\iota:U\\cap V\\to U$, $\\iota_*(\\gamma)=\\beta$. So you would have\n\n$$\\pi_1(X)=\\langle\\alpha,\\beta,\\gamma|\\alpha\\beta=\\beta\\alpha,\\beta=\\gamma\\rangle$$\n\nAs always, it depends how familiar your audience (and you!) are with the subject.\n\n-", "date": "2015-05-23 10:52:17", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/286315/fundamental-group-of-this-space", "openwebmath_score": 0.9098342657089233, "openwebmath_perplexity": 268.0760734679797, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540674530015, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8560311003455452}}
{"url": "https://math.stackexchange.com/questions/200389/show-that-the-set-of-all-finite-subsets-of-mathbbn-is-countable/200406", "text": "# Show that the set of all finite subsets of $\\mathbb{N}$ is countable.\n\nShow that the set of all finite subsets of $\\mathbb{N}$ is countable.\n\nI'm not sure how to do this problem. I keep trying to think of an explicit formula for 1-1 correspondence like adding all the elements in each subset and sending that sum to itself in the natural numbers, but that wouldn't be 1-1 because, for example, the set {1,2,3} would send to 6 and so would the set {2,4}. Multiplying all the elements in each subset and sending that product to itself in the natural numbers wouldn't work either since, for example, {2,3} would send to 5 and so would the set {1,5}. Any ideas?\n\n\u2022 $2\\times3\\not=5$ :) \u2013\u00a0Ben Millwood Sep 21 '12 at 17:53\n\u2022 Set of Finite Subsets of Countable Set is Countable at ProofWiki. See also this question for a generalization of your question ($\\mathbb N$ is replaced by arbitrary infinite set $X$ and it is shown that set of all finite subset of $X$ has the same cardinality as $X$). \u2013\u00a0Martin Sleziak Sep 22 '12 at 6:50\n\u2022 You're right, an explicit bijection would be a neat proof. Hint: think like a computer scientist. \u2013\u00a0Colonel Panic Jul 17 '15 at 10:05\n\nConsider an enumeration of the (positive) prime numbers by $n\\mapsto p_n$. Then if you're given the set $\\{n_1,...,n_k\\}$, map that to $$p_{n_1}\\cdots p_{n_k}.$$ For example, if we've enumerated the primes increasingly ($p_1=2$, $p_2=3$, and so on), then $$\\{1,3,4\\}\\mapsto 70=2\\cdot 5\\cdot 7=p_1\\cdot p_3\\cdot p_4,$$ and $$\\{1,3\\}\\mapsto 10=2\\cdot 5=p_1\\cdot p_3.$$\n\nYou'll need the fact that there are countably infinitely many prime numbers, and uniqueness of prime factorization. From there, it shouldn't be hard to prove this is an injection from the set of finite subsets of $\\Bbb N$ into $\\Bbb N$, so there are at most countably many such subsets. To show that there are at least (and so exactly) countably many finite subsets of $\\Bbb N$, you need only find an injection from $\\Bbb N$ into the set of finite subsets of $\\Bbb N$, which should be easy.\n\nAdded: The described map \"should\" send $\\emptyset$ to $1,$ the \"empty product.\"\n\n\u2022 So if I was given a set {1,3} and a set {1,3,4}, you're saying to map that to 1,3 and 1,2^2,3? Am I understanding correctly? \u2013\u00a0user39794 Sep 21 '12 at 17:28\n\u2022 Not quite. I'll include an example to clarify. \u2013\u00a0Cameron Buie Sep 21 '12 at 17:31\n\u2022 It's ok I think I get it now from reading the other answer! :) Thanks though! \u2013\u00a0user39794 Sep 21 '12 at 17:32\n\u2022 @Diego: Yes. However, the given map is not a bijection with $\\Bbb N.$ For example, there is no finite subset of $\\Bbb N$ that is mapped to $4.$ Rather, the map is an injection into $\\Bbb N.$ \u2013\u00a0Cameron Buie Feb 12 '16 at 12:34\n\u2022 @Diego: Your reasoning is a bit faulty. After all, $\\{1\\}$ is a subset of $\\Bbb N,$ too, and we certainly can't conclude that there exists a bijection between $\\{1\\}$ and $\\Bbb N$! In order to conclude that a subset of $\\Bbb N$ is in bijection with $\\Bbb N,$ we have to know that it is an infinite subset. The last part is aimed at proving that the range of the map is, indeed, infinite. \u2013\u00a0Cameron Buie Feb 12 '16 at 19:33\n\nDefine three new digits: \\begin{align} \\{ &= 10 \\\\ \\} &= 11\\\\ , &= 12 \\end{align} Any finite subset of $\\mathbb{N}$ can be written in terms of $0,1,\\cdots,9$ and these three characters, and so any expression of a subset of $\\mathbb{N}$ is just an integer written base-$13$. For instance, $$\\{1,2\\} = 10 \\cdot 13^4 + 1 \\cdot 13^3 + 12 \\cdot 13^2 + 2 \\cdot 13^1 + 11 \\cdot 13^0 = 289873$$\n\nSo for each finite $S \\subseteq \\mathbb{N}$, take the least $n_S \\in \\mathbb{N}$ whose base-$13$ expansion gives an expression for $S$. This defines an injection into $\\mathbb{N}$.\n\nMore generally, any set whose elements can be written as finite strings from some finite (or, in fact, countable) alphabet, is countable.\n\n\u2022 +1 Really nice answer. \u2013\u00a0Fly by Night Sep 2 '13 at 18:07\n\u2022 The user @TimothySwan tried to edit your question to add a comment, I have copied his comment here since the user can't comment himself. He commented: However, this proof is not correct since there must be an injection from $\\mathbb N$. \u2013\u00a0Alice Ryhl Sep 22 '14 at 14:34\n\u2022 Thanks @Darksonn. (For the benefit of anyone else reading, that's not true; every infinite set has an injection from $\\mathbb{N}$ into it!) \u2013\u00a0Clive Newstead Sep 22 '14 at 20:32\n\u2022 (+1 a while ago) Not that it will be relevant to most users, but depending on the axioms one is working with, it may be possible for an infinite set to have no injection from $\\Bbb N.$ \u2013\u00a0Cameron Buie May 10 '15 at 9:22\n\u2022 @TTL: You're missing 'and these three new characters', namely open curly bracket, close curly bracket, and comma. \u2013\u00a0Clive Newstead Nov 15 '19 at 22:23\n\nAlternatively, just use base $2$, so send $\\{a_1,...,a_n\\}$ to $2^{a_1}+\\cdots+2^{a_n}$. (Assuming your $\\mathbb N$ includes $0$, this is $1-1$ and onto.)\n\n\u2022 +1 Put in other way: any finite subset of positive integers can be represented as a binary string (with a finite number of ones), which in turn can be interpreted as a binary number which correspond to a positive integer. \u2013\u00a0leonbloy Mar 17 '14 at 20:09\n\u2022 In my opinion, this is the best possible proof because the bijection is very, very explicit. \u2013\u00a0Martin Brandenburg Jul 17 '15 at 8:31\n\u2022 Any base worked, right?(the base being an integer) \u2013\u00a0Diego Feb 12 '16 at 4:42\n\u2022 Any base gives you a $1-1$ function. Base $2$ is onto. @Diego \u2013\u00a0Thomas Andrews Feb 12 '16 at 11:52\n\u2022 @MartinBrandenburg: Since the set {a1,...an} isn't an ordered one, we have several base 2 numbers mapping to the same set. How's a bijection best constructed here, then? The accepted answer is also just an injection, no? \u2013\u00a0Nikolaj-K Mar 8 '17 at 14:57\n\nThe other answers give some sort of formula, like you were trying to do. But, the simplest way to see that the set of all finite subsets of $\\mathbb{N}$ is countable is probably the following.\n\nIf you can list out the elements of a set, with one coming first, then the next, and so on, then that shows the set is countable. There is an easy pattern to see here. Just start out with the least elements.\n\n$$\\emptyset, \\{1\\}, \\{2\\}, \\{1, 2\\}, \\{3\\}, \\{1, 3\\}, \\{2, 3\\}, \\{1, 2, 3\\}, \\{4\\}, \\ldots$$\n\nIn other words, first comes $\\{1\\}$, then comes $\\{2\\}$. Each time you introduce a new integer, $n$, list all the subsets of $[n] = \\{1, 2, \\ldots, n\\}$ that contain $n$ (the ones that don't contain $n$ have already showed up). Therefore, all subsets of $[n]$ show up in the first $2^{n}$ elements of this sequence.\n\n\u2022 In other words: List the elements of $P(\\emptyset)$. List the elements of $P(\\{1\\}) \\setminus P(\\emptyset)$. List the elements of $P(\\{1,2\\}) \\setminus P(\\{1\\})$. List the elements of $P(\\{1,2,3\\}) \\setminus P(\\{1,2\\})$. And so on. \u2013\u00a0Martin Brandenburg Jul 17 '15 at 8:29\n\u2022 The lazy approach is a 'surjective' listing, $PL([0,1]),PL([0,2]),PL([0,3]),\\dots,PL([0,n]),\\dots$ where $PL$ is powerset listing (with commas) and $[0,n]$ is the initial segment. \u2013\u00a0CopyPasteIt Aug 18 '17 at 19:26\n\u2022 how do you actually show this is surjective and injective? \u2013\u00a0Pinocchio Sep 5 '18 at 16:18\n\u2022 For a proof specification see this answer: Neatest proof that set of finite subsets is countable?. \u2013\u00a0CopyPasteIt Jun 2 at 14:21\n\nI'm surprised nobody else has said this, so I worry that I've missed the point of the question. But I would argue as follows: Let $N_i$ be the family of all subsets of $\\Bbb N$ with exactly $i$ elements. $N_0$ is finite, and $N_1$ is countable. $N_2$ is countable also, since it's a subset of $\\Bbb N\\times \\Bbb N$. There's an easy induction argument that $N_i$ is countable for all $i>0$, since it's contained in the product $\\Bbb N\\times N_{i-1}$ of two countable sets.\n\nThen, since the set you want, $\\bigcup N_i$, is a countable union of countable sets, it is countable.\n\n\u2022 Well, a countable union of countable sets needn't be countable. It depends on your axioms. \u2013\u00a0Cameron Buie Sep 29 '12 at 22:09\n\u2022 What axioms would those be? \u2013\u00a0MJD Sep 29 '12 at 22:28\n\u2022 Set theoretic axioms--which you need at least some of before you can talk about sets, countability, unions, etc. For example, there are models of ZF in which $\\Bbb R$ is a countable union of countable sets. If one takes countable choice (or stronger) as an axiom, then countable unions of countable sets are countable (that's a strictly weaker statement than countable choice). Now, it's worth noting that countable unions of enumerated sets (countable sets with a given bijection with $\\Bbb N$) are countable, with no need for choice principles. \u2013\u00a0Cameron Buie Sep 29 '12 at 23:07\n\nLet $p_1,p_2,\\ldots$ the sequence of primes. Then map each finite subset $\\{a_1,\\ldots,a_n\\} \\subset \\mathbb N$ to $p_1^{a_1}\\cdots p_n^{a_n}$ where $a_1 < \\ldots < a_n$. This gives an injective map from the set of finite subsets of $\\mathbb N$ to $\\mathbb N$.\n\nAlternatively, notice that for every $s$ there are only finitely many finite subsets of $\\mathbb N$ whose sum is $\\leq s$. Let $A_s$ be the set of those subsets, then $\\mathbb N = \\bigcup_{s} A_s$ is a countable union of finite sets, hence countable.\n\n\u2022 You beat me to the post ... slightly different, but the same important idea - You can show that the set $C_N$ of subsets of $\\mathbb N$ whose members sum to $N$ is finite for each $N\\in\\mathbb N$. Then each finite subset of $\\mathbb N$ has a finite sum $N$ and hence is in one of the $C_N$, and the union of the $C_N$ is a countable union of finite sets, hence is countable. \u2013\u00a0Mark Bennet Sep 21 '12 at 18:41\n\u2022 @Mark: Not that it's probably relevant to the OP, but a countable union of finite sets needn't be countable in general without sufficient Choice. Here, of course, we're dealing with well-ordered finite sets, so it works out. \u2013\u00a0Cameron Buie Jul 31 '14 at 10:48\n\nIf $S$ is the set of all finite subsets of $\\mathbb{N}$, define $f:S\\rightarrow \\mathbb{Q}$ by\n\n$f(A)=.x_1x_2x_3x_4\\cdots$ where $x_{i}=\\begin{cases}1 &\\mbox{, if }i\\in A\\\\0 &\\mbox{, if } i\\not\\in A \\end{cases}$.\n\nThen $f$ is clearly injective, so $S$ is countable since $\\mathbb{Q}$ is countable and any subset of a countable set is countable.", "date": "2020-07-12 12:52:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/200389/show-that-the-set-of-all-finite-subsets-of-mathbbn-is-countable/200406", "openwebmath_score": 0.951771080493927, "openwebmath_perplexity": 220.7539603480424, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540716711546, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8560310943950199}}
{"url": "https://cs.stackexchange.com/questions/101782/show-that-for-any-real-constants-a-and-b-where-b-0-n-ab-theta", "text": "# Show that for any real constants $a$ and $b$, where $b > 0$, $(n + a)^b = \\Theta(n^b)$\n\nI'm currently studying growth of function chapter in Introduction to Algorithm. In exercise 3.1-2 the question is:\n\nShow that for any real constants $$a$$ and $$b$$, where $$b>0$$,\n$$(n + a)^b = \\Theta(n^b)$$.\n\nI understand I need to find constants $$c_1$$, $$c_2$$, $$n_0 > 0$$ where\n\n$$0 \\leq c_1n^b \\leq (n+a)^b \\leq c_2n^b$$ for all $$n \\geq n_0$$.\n\nBut I don't understand what should I do after above step. Looking at the solution, it shows\n\n$$n + a \\leq n + |a|$$\n$$n + a \\leq 2n$$, when $$|a| \\leq n$$\n\nand\n\n$$n + a \\geq n - |a|$$\n$$n + a \\geq \\frac{n}{2}$$ , when $$|a| \\leq \\frac{n}{2}$$\n\nHow did they come up with $$n + a \\leq n + |a|$$ and $$n + a \\geq n - |a|$$ just from $$0 \\leq c_1n^b \\leq (n+a)^b \\leq c_2n^b$$ ? Are they derived from $$0 \\leq c_1n^b \\leq (n+a)^b \\leq c_2n^b$$ ? Or are they just a general knowledge that can help solve this problem?\n\nThank you!\n\nUPDATE: I have updated the question to be more specific.\n\n\u2022 $(n+a)^b = n^b + c_1n^{b-1}a+... = \\theta(n^b)$ \u2013\u00a0Mr. Sigma. Dec 19 '18 at 13:40\n\u2022 @Mr.Sigma.: $b$ can be any positive real number, say 0.5. \u2013\u00a0rici Dec 19 '18 at 13:54\n\u2022 O'Luck: $\\mid a\\mid$ is the absolute value of $a$; either $a$ or $-a$. If $a$ is positive, $n+ |a| = n + a$; otherwise $a$ is negative and $n+a < n < n + |a|$. What problem do you have with that? \u2013\u00a0rici Dec 19 '18 at 14:00\n\u2022 \"Where did they get those inequalities?\" Imagine that $n$ is very large. \u2013\u00a0John L. Dec 19 '18 at 14:12\n\u2022 Please do not post equations, text, and the like as images since these are incompatible with the site's search engine. Thank you. \u2013\u00a0dkaeae Dec 19 '18 at 14:20\n\nI understand I need to find constants $$c_1$$, $$c_2$$, $$n_0 > 0$$ where\n\n$$0 \\leq c_1n^b \\leq (n+a)^b \\leq c_2n^b$$ for all $$n \\geq n_0$$.\n\nBut I don't understand what should I do after above step.\n\nThe approach you would like to use here is \"going backwards from the target\".\n\nSuppose we do have $$c_1n^b \\leq (n+a)^b$$ for some constant $$c_1\\gt0$$ when $$n$$ is large enough. What can we understand or deduce from it? What is needed to make it true?\n\nSince the exponent is the same, we can use the law of same exponent, $$(x/y)^m = x^m/y^m$$ to simplify that inequality so that we will have the appearances of our variable $$n$$ closer to each other.\n\n$$c_1\\leq \\frac{(n+a)^b}{n^b}=\\left(\\frac{n+a}n\\right)^b$$\n\nRaising both sides to the power of $$\\frac1b$$, we have\n\n$$(c_1)^{\\frac1b}\\leq \\left(\\frac{n+a}n\\right)^{b\\,\\frac1b}=\\frac{n+a}n = 1+\\frac an$$\n\n(Another way to obtain $$(c_1)^{\\frac1b}\\leq\\frac{n+a}n$$ is to raise both sides of $$c_1n^b \\leq (n+a)^b$$ to the power of $$\\frac 1b$$.)\n\nSince $$c_1$$ is a positive constant, $$(c_1)^{\\frac1b}$$ is also a positive constant. To make the above inequality hold, we would like to make $$\\left|\\frac an\\right|$$ small enough. For example, we can require $$\\left|\\frac an\\right|\\le \\frac12$$. That is why you see the following condition.\n\n$$|a|\\le\\frac12 n$$\n\nWhat is nice here is that we can reverse the above argument to obtain a wanted constant $$c_1$$. Note that it is just as fine if we choose a different condition such as $$|a|\\le\\frac13 n$$ or $$|a|\\le\\frac1{2019}n$$ or infinitely many others.\n\nIf we start from $$(n+a)^b \\leq c_2n^b$$, we could arrive at $$|a|\\le n$$.\n\nAre they derived from $$0\\le c_1n^b\\le(n+a)^b\\le c_2n^b$$ ? Or are they just a general knowledge that can help solve this problem?\n\nYes, they are derived as you have just seen. You could say that they are general knowledge that helps solve this problem. You could also say that we just discovered some specific facts that help solve this problem.\n\n\u2022 Wow thank you so much @apass-jack for your answer. This type of step-by-step answer is what I'm looking for. I guess I'm pretty weak at math. Before I accept your answer, I want to ask a few questions. In $(c_1)^{\\frac1b}\\leq \\left(\\frac{n+a}n\\right)^{b\\,\\frac1b}=\\frac{n+a}n = 1-\\frac an$ , isn't it supposed to be $1+\\frac an$ at the end? \u2013\u00a0O'Luck Dec 19 '18 at 23:12\n\u2022 Thanks for pointing out my typo. Updated. \u2013\u00a0John L. Dec 19 '18 at 23:35\n\u2022 Thank you. Next question is: \"To make the above inequality hold, we would like to make $\\left|\\frac an\\right|$ small enough\". Why? If we choose big value for $\\left|\\frac an\\right|$ , let say $\\left|\\frac an\\right| > 1$ the inequality is still satisfied right? \u2013\u00a0O'Luck Dec 20 '18 at 0:01\n\u2022 Reason one: when $n$ goes to (positive) infinity, $|\\frac an|$ goes to 0. So you cannot make it as big as you want. In fact, $|\\frac an| \\le |a|$ if we assume $n$ is a positive integer. Reason one is enough. Reason two: even if you could magically, suppose $a$ is negative, $\\frac an$ is negative, too, which is bad for $1+\\frac an$. \u2013\u00a0John L. Dec 20 '18 at 0:12\n\u2022 umm, I mean $|\\frac an|$ as a whole not only the $n$. So, with $(c_1)^{\\frac1b}\\leq 1 + \\frac an$ if $|\\frac an| > 1$ let say $|\\frac an| = 2$ then $(c_1)^{\\frac1b}\\leq 1 + 2$ , the inequation is still satisfied, am i right? \u2013\u00a0O'Luck Dec 20 '18 at 1:10\n\nIf n is large then we have n/2 \u2264 n + a \u2264 2n.\n\nWe therefore have $$(1/2)^b n^b \u2264 n^b \u2264 2^b n^b$$.\n\nThere you have your two constants: $$(1/2)^b$$ and $$2^b$$.", "date": "2021-08-04 03:14:44", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/101782/show-that-for-any-real-constants-a-and-b-where-b-0-n-ab-theta", "openwebmath_score": 0.8414087295532227, "openwebmath_perplexity": 216.59520289263443, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540722737479, "lm_q2_score": 0.8740772253241803, "lm_q1q2_score": 0.8560310901029743}}
{"url": "https://math.stackexchange.com/questions/2275616/probability-of-a-number-rolled-of-a-20-sided-dice-being-greater-than-the-sum-of", "text": "# Probability of a number rolled of a 20 sided dice being greater than the sum of the numbers rolled on 3 six sided die.\n\nBob rolls $3$ six-sided die and sums the numbers facing up. Bill rolls a single $20$ sided dice and records the number. What is the probability that Bob's number is greater than Bill's number.\n\nI started the problem trying to come up with an equation, but that didn't work, so I resorted to creating $6$ six by six tables with all of the possible sums for Bob's die. Then, I counted the number of each number and created a chart and calculated the probability of each of those numbers occurring. Then, I multiplied the probability of each number occurring by the probability that Bill's number is greater than that number. Finally, I added them all up.\n\nObviously, this was very tedious and time-consuming. Is there a more elegant/less tedious way to do this problem.\n\nPS: Is there away to do a $n$ die vs $m$ dice problem without listing them all out? Is there a general formula? Up until now, that's what I've been doing.\n\n\u2022 The probability asked for in the title $P(\\text{D20 beats 3 D6s})$ is the opposite of the probability asked for in the details $P(\\text{3 D6s beats D20})$. \u2013\u00a0N. Shales May 11 '17 at 1:57\n\nLet the total on the three six-sided dice be $X,$ $3\\leq X\\leq18.$ Let the number on the twenty-sided die be $Y,$ $1\\leq Y\\leq20.$\n\nGiven any particular value of $X,$ the probability that the twenty-sided die will roll higher is $$P(Y>X \\mid X) = \\frac{20-X}{20} = 1 - \\frac1{20}X.$$\n\nThe overall probability that the twenty-sided die will roll higher than the total on the other three dice is \\begin{align} P(Y>X) &= \\sum_{n=3}^{18} P(Y > X \\mid X)P(X=n) \\\\ &= \\sum_{n=3}^{18} \\left(1 - \\frac1{20}X\\right)P(X=n). \\end{align}\n\nThe last line of that set of equations is just the expected value of $1 - \\frac1{20}X.$ That is, \\begin{align} P(Y>X) &= \\mathbb E\\left[1 - \\frac1{20}X\\right] \\\\ &= 1 - \\frac1{20} \\mathbb E[X] \\\\ &= 1 - \\frac1{20} \\left(\\frac{21}{2}\\right) \\\\ &= \\frac{19}{40}. \\end{align}\n\nIf the question is actually the one posed in the original question body rather than in the original title, namely the probability that $X > Y,$ then we simply observe that for any given value of $X,$ $$P(Y < X \\mid X) = \\frac{X-1}{20} = \\frac{1}{20}X - \\frac{1}{20}.$$\n\nThe rest of the calculation builds on this the same way the first calculation in this answer built on $P(Y > X \\mid X).$ We find that \\begin{align} P(Y<X) &= \\mathbb E\\left[\\frac1{20}X - \\frac1{20}\\right] \\\\ &= \\frac1{20} \\mathbb E[X] - \\frac1{20}\\\\ &= \\frac1{20} \\left(\\frac{21}{2}\\right) - \\frac1{20} \\\\ &= \\frac{19}{40}. \\end{align}\n\nThis should not be surprising, because it also follows from $P(Y>X)=\\frac{19}{40}$ and the \"obvious\" fact that $P(Y=X)=\\frac1{20}.$\n\n\u2022 Please see my answer for a simple proof that the two probabilities are equal (and hence both equal to $19/40$). \u2013\u00a0Barry Cipra May 11 '17 at 13:50\n\u2022 I don't know if this counts as a shortcut, but to find the expected value of X faster, you could find the expected value of one dice and just multiply it by three instead of summing all the numbers from 3-18. I feel like that just works better in my head. \u2013\u00a0Henry Weng May 11 '17 at 14:47\n\nWhatever Bob rolls with the $6$-sided dice has a $1$ in $20$ chance of being matched, for a tie, by Bill's roll of the $20$-sided die. Whatever sum, $S=a+b+c$, Bob rolls, if you turn his dice over, the sum is $(7-a)+(7-b)+(7-c)=21-S$. Similarly, whatever number $T$ Bill rolls, if you turn the $20$-sided die over, the number is $21-T$. Thus for each outcome in which Bob wins, there is an equally likely outcome in which Bill wins, and vice versa. Hence the probability of winning for each of them is the same, namely\n\n$${1\\over2}\\left(1-{1\\over20}\\right)={19\\over40}$$\n\nRemark: It's not literally necessary that the \"complementary\" number for each side of a die be the opposite face, just that there be a complemenary number somewhere. For $6$-sided dice, having opposite faces sum to $7$ is fairly standard; I believe it's also standard for $20$-sided dice to have opposite faces sum to $21$.\n\nAlso, I'd like to credit David K's answer with motivating this one. When I saw from his analysis that the two probabilities were equal, I decided there ought to be simple reason why. As luck would have it, I found one.\n\n\u2022 The connection to my answer is even deeper than the coincidence that $P(Y>X)=P(Y<X)$: the usual easy method to calculate $E[X]$ uses the same \"turn the dice over\" argument (or something equivalent). But the really nice observation here is that the exact same symmetry works for both dice. \u2013\u00a0David K May 11 '17 at 13:57\n\u2022 @DavidK, thanks. The symmetry principle also suffices, for example, to show that it's a fair game to roll $13$ regular dice against $7$ dodecahedral dice. In that case, however, computing the probability of a tie is not so easy. (Or is it?) \u2013\u00a0Barry Cipra May 11 '17 at 14:10\n\nYour best bet is to keep track of two things for both sides: how likely this particular result is, and how likely anything less than this result is. Given these we can multiply them together relatively easily.\n\n$$\\begin{array}{r|rr|rr|r} x & 3\\text{d}6 = x & 3\\text{d}6 < x & 1\\text{d}20 = x & 1\\text{d}20 < x & 1\\text{d}20 < x \\cap 3\\text{d}6 = x\\\\ \\hline 1 & 0 & 0 & 1 & 0 & 0 \\\\ 2 & 0 & 0 & 1 & 1 & 0 \\\\ 3 & 1 & 0 & 1 & 2 & 2 \\\\ 4 & 3 & 1 & 1 & 3 & 9 \\\\ 5 & 6 & 4 & 1 & 4 & 24 \\\\ 6 & 10 & 10 & 1 & 5 & 50 \\\\ 7 & 15 & 20 & 1 & 6 & 90 \\\\ 8 & 21 & 35 & 1 & 7 & 147 \\\\ 9 & 25 & 56 & 1 & 8 & 200 \\\\ 10 & 27 & 81 & 1 & 9 & 243 \\\\ 11 & 27 & 108 & 1 & 10 & 270 \\\\ 12 & 25 & 135 & 1 & 11 & 275 \\\\ 13 & 21 & 160 & 1 & 12 & 252 \\\\ 14 & 15 & 181 & 1 & 13 & 195 \\\\ 15 & 10 & 196 & 1 & 14 & 140 \\\\ 16 & 6 & 206 & 1 & 15 & 90 \\\\ 17 & 3 & 212 & 1 & 16 & 48 \\\\ 18 & 1 & 215 & 1 & 17 & 17 \\\\ 19 & 0 & 216 & 1 & 18 & 0 \\\\ 20 & 0 & 216 & 1 & 19 & 0 \\\\ \\hline \\text{total} & 216 & & 20 & & 2052 \\end{array}$$\n\nTotal up everything in the rightmost column, divide by the totals for the $3\\text{d}6 = x$ and $1\\text{d}20 = x$ columns, and you win: Bob wins $\\frac{2052}{4320} = \\frac{19}{40} = 0.475$ of the time.\n\nTo get ties, or where Bill wins, change what pairs you multiply: both $=$ ones for ties, and $3\\text{d}6 < x$ and $1\\text{d}20 = x$ for Bill's wins.\n\nThis particular one is interesting: because both distributions are symmetrical, and they have the same mean, Bill and Bob will both win the same proportion of the time.\n\nIt occurs to me that there's another part to this question: how do we efficiently calculate result probabilities for combinations of dice?\n\nThe answer to that is an operation called convolution, which I'll present here in discrete form.\n\nGiven two functions $f(x)$ and $g(x)$, the convolution is $$(f * g)(x) = \\sum_{k = -\\infty}^\\infty f(k)g(x-k)$$\n\nThis can be interpreted in probability theory as the following: we have two random variables $f$ and $g$, with probability functions $f(x)$ and $g(x)$. the probability function for $f + g$ -- adding the two results together -- is equal to $(f * g)(x)$.\n\nObviously with those infinities in there, we have to fiddle with it a little to actually get anything done. In our case, because we're dealing with dice, our functions have what's called limited support: they're only non-zero in a small area, so we only need to cover that small area.\n\nLet's do a specific example. Say I want to calculate the probability that I'll get a $7$ on $3\\text{d}6$. $3\\text{d}6$ is the same as $1\\text{d}6$ + $2\\text{d}6$, so I can convolve these two. I'll call their functions $f$ and $g$ respectively.\n\n$f$ here has limited support: the only values it is non-zero for are $1$ through $6$. This allows us to change the limits of our summation to the bounds of $f$'s support.\n\n\\begin{align} (f * g)(7) &= \\sum_{k=1}^6 f(k)g(7 - k)\\\\ &= f(1)g(6) + f(2)g(5) + f(3)g(4) + f(4)g(3) + f(5)g(2) + f(6)g(1) \\\\ &= \\frac{1}{6}\\cdot\\frac{5}{36} + \\frac{1}{6}\\cdot\\frac{4}{36} + \\frac{1}{6}\\cdot\\frac{3}{36} + \\frac{1}{6}\\cdot\\frac{2}{36} + \\frac{1}{6}\\cdot\\frac{1}{36} + \\frac{1}{6}\\cdot\\frac{0}{36} \\\\ &= \\frac{15}{216} \\end{align}\n\nUsing convolution, then, we can calculate the probabilities of the summed results of multiple dice, without necessarily considering every single simple event: to calculate, say, the probability distribution of $5\\text{d}6$, we can take the distribution of $4\\text{d}6$ and the distribution of $1\\text{d}6$ and convolve them. And to get $4\\text{d}6$'s distribution we can convolve $3\\text{d}6$ and $1\\text{d}6$, etc. So instead of counting out $7776$ possibilities, we instead handle $21\\cdot6 + 16\\cdot6 + 11\\cdot6 + 6\\cdot6 = 324$ total multiplications and a similar number of additions.\n\n\u2022 I didn't read your answer or the problem, but your LateX table skills are impressive \u2013\u00a0user2879934 May 11 '17 at 0:53\n\u2022 That last paragraph is an important short cut. It means that you really only need to work out the cases that result in a tie. \u2013\u00a0amd May 11 '17 at 0:55\n\u2022 I literally learned how to make the table to write the answer, and beat Excel until morale impro^W^W it built most of the rows for me. \u2013\u00a0Dan Uznanski May 11 '17 at 0:56\n\u2022 I was initially skeptical that this is really \"Your best bet\", given the other answers, but it does appear to be a more general solution that is purely computational. \u2013\u00a06005 May 12 '17 at 5:34\n\nCall the probability of scoring a $k$ with the D20 $q_k=q=1/20$ and scoring a $k$ with the 3 D6'S $p_k$ then we are looking for the probability that the D20 has the greater score\n\n$$p_3\\sum_{k=4}^{20}q_k+p_4\\sum_{k=5}^{20}q_k+\\ldots +p_{18}\\sum_{k=19}^{20}q_k$$\n\nOr simply\n\n$$q(17p_3+16p_4+\\cdots +2p_{18})$$\n\nbut by symmetry for every way to score $k$ with 3 D6s there is a way to score $21-k$, simply by subtracting the score on each die from $7$. The probability of rolling a total of $18$ with $3$ D6s is therefore the same as rolling a total of $3$ ($p_3=p_{18}$) and rolling a total of $17$ is the same probability as rolling $4$ ($p_4=p_{17}$) etc. So\n\n$$\\text{required probability}=19q(p_3+p_4+\\cdots +p_{10})=19q(p_{11}+p_{12}+\\cdots +p_{18})$$\n\nBut of course\n\n$$p_3+p_4+\\cdots +p_{18}=1$$\n\nSo\n\n$$2(p_3+p_4+\\cdots +p_{10})=1=2(p_{11}+p_{12}+\\cdots +p_{18})$$ $$\\implies p_3+p_4+\\cdots +p_{10}=p_{11}+p_{12}+\\cdots +p_{18}=\\frac{1}{2}$$\n\nGiving\n\n$$\\text{required probability}=19\\cdot\\frac{1}{20}\\cdot\\frac{1}{2}=\\frac{19}{40}\\tag{Answer}$$\n\nNote that, since the probability of the players getting the same score is $(1/20)\\cdot 1=1/20$, then the probability that the D20 loses is the same as it winning i.e. $19/40$.", "date": "2019-05-26 21:53:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2275616/probability-of-a-number-rolled-of-a-20-sided-dice-being-greater-than-the-sum-of", "openwebmath_score": 0.9943407773971558, "openwebmath_perplexity": 423.7090713453807, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713887451681, "lm_q2_score": 0.8688267847293731, "lm_q1q2_score": 0.8560301727693086}}
{"url": "http://mathhelpforum.com/calculus/34866-different-forms-same-integral.html", "text": "# Math Help - Different forms of same integral?\n\n1. ## Different forms of same integral?\n\nHey people, I just came up across something I didn't know about in my revision. Say I want to find the integral\n\n$\\int \\frac{1}{3x}dx$\n\nNormally I would do this as following\n\n$\\frac{1}{3}\\int \\frac{1}{x}dx=\\frac{1}{3}\\ln{x}+k$\n\nHowever, is it not true that I can also make this of the form\n\n$\\int \\frac{f'(x)}{f(x)}dx$\n\nvia\n\n$\\frac{1}{3}\\int \\frac{3}{3x}dx$\n\nwhich via a simple u sub yields\n\n$\\frac{1}{3}\\ln{3x}+j=\\frac{1}{3}\\ln{x}+\\frac{1}{3} \\ln{3}+j$\n\nWhich does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?\n\n$y=\\frac{1}{3}\\ln{x}+k$ and $\\frac{1}{3}\\ln{3x}+j$\n\nas explicit functions of x?\n\nTa muchly.\n\n2. Originally Posted by Xei\nHey people, I just came up across something I didn't know about in my revision. Say I want to find the integral\n\n$\\int \\frac{1}{3x}dx$\n\nNormally I would do this as following\n\n$\\frac{1}{3}\\int \\frac{1}{x}dx=\\frac{1}{3}\\ln{x}+k$\n\nHowever, is it not true that I can also make this of the form\n\n$\\int \\frac{f'(x)}{f(x)}dx$\n\nvia\n\n$\\frac{1}{3}\\int \\frac{3}{3x}dx$\n\nwhich via a simple u sub yields\n\n$\\frac{1}{3}\\ln{3x}+j=\\frac{1}{3}\\ln{x}+\\frac{1}{3} \\ln{3}+j$\n\nWhich does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?\n\nMr F says: They are equivalent. j + ln 3 is just as arbitrary as k .....\n\n$y=\\frac{1}{3}\\ln{x}+k$ and $\\frac{1}{3}\\ln{3x}+j$\n\nas explicit functions of x?\n\nMr F says: They already are explicit functions of x. Do you mean how to express them as explicit functions of y ....?\n\nTa muchly.\n..\n\n3. Originally Posted by Xei\nHey people, I just came up across something I didn't know about in my revision. Say I want to find the integral\n\n$\\int \\frac{1}{3x}dx$\n\nNormally I would do this as following\n\n$\\frac{1}{3}\\int \\frac{1}{x}dx=\\frac{1}{3}\\ln{x}+k$\n\nHowever, is it not true that I can also make this of the form\n\n$\\int \\frac{f'(x)}{f(x)}dx$\n\nvia\n\n$\\frac{1}{3}\\int \\frac{3}{3x}dx$\n\nwhich via a simple u sub yields\n\n$\\frac{1}{3}\\ln{3x}+j=\\frac{1}{3}\\ln{x}+\\frac{1}{3} \\ln{3}+j$\n\nWhich does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?\nSince j and k are just arbitrary constants one can be set equal to the other.\n\nBy the way, just so that you are aware...\n$\\int \\frac{1}{x}~dx = ln|x| + C$\n\n-Dan\n\n4. Yeah it's all right, I just wasn't bothered with the mod bit.\n\nI was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.\n\nIs it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?\n\nMr F: Yeah, y please. As simple as it'll go.\n\n5. Hello, Xei!\n\nYou are basically correct . . .\n\n$\\int \\frac{1}{3x}\\,dx \\;=\\;\\frac{1}{3}\\int \\frac{1}{x}\\,dx\\;=\\;\\frac{1}{3}\\ln{x}+K$ .[1]\n\n$\\frac{1}{3}\\int\\frac{3}{3x}\\,dx \\;=\\;\\frac{1}{3}\\ln{3x}+J$ .[2]\n\nIs it the case that both are valid, but the constants $K$ and $J$ are different? .\n. . . yes\n\nIs $K\\:=\\:J+\\ln3$? .\n. . . um, not quite\n\nFrom [2], we have: . $\\frac{1}{3}\\ln(3x) + J \\;=\\;\\frac{1}{3}\\left[\\ln(3) + \\ln(x)\\right] + J \\;=\\;\\frac{1}{3}\\ln(3) + \\frac{1}{3}\\ln(x) + J$\n\n. . $= \\;\\frac{1}{3}\\ln(x) + \\underbrace{\\frac{1}{3}\\ln(3) + J}_{\\text{a constant}} \\;=\\;\\frac{1}{3}\\ln(x) + K$\n\n6. Originally Posted by Xei\nYeah it's all right, I just wasn't bothered with the mod bit.\n\nI was just checking, I haven't come across a situation before in which I ended up with two different constants for the same thing.\n\nIs it more desirable to do it the initial way though? That way you simply have lnx instead of ln3x which is a little trickier. I suppose you could just get rid of the 3 and call it lnx+ln3+j = lnx+c which is the original result..?\n\nMr F: Yeah, y please. As simple as it'll go.\n$y=\\frac{1}{3}\\ln{|x|}+k \\Rightarrow 3y - 3k = \\ln |x| \\Rightarrow e^{3y - 3k} = x$\n\n$\\Rightarrow x = e^{3y}\\, e{-3k} = A e^{3y}$.", "date": "2015-08-01 08:30:43", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/34866-different-forms-same-integral.html", "openwebmath_score": 0.8933925032615662, "openwebmath_perplexity": 524.9119868617777, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713874477227, "lm_q2_score": 0.8688267745399466, "lm_q1q2_score": 0.856030161602703}}
{"url": "https://mathhelpboards.com/threads/volume-of-a-pyramid.9351/", "text": "# [SOLVED]Volume of a pyramid\n\n#### dwsmith\n\n##### Well-known member\nI am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.\n\nLet $$L$$ be on the x axis and $$W$$ be on the y axis.\nIn the x-z plane, we have the line $$z = -\\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\\frac{h}{W/2}y + h$$.\n\nMy cross sections has width $$\\Delta z$$. So I want to find the volume $$\\int_0^hA(z)dz$$.\n\nHow can I do this?\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nHint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\\left(1-\\frac{z}{h}\\right)^2$ where $A=A(0)$. Then compute the integral $\\int_0^h A(z)\\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.\n\n#### dwsmith\n\n##### Well-known member\nHint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\\left(1-\\frac{z}{h}\\right)^2$ where $A=A(0)$. Then compute the integral $\\int_0^h A(z)\\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.\n\nSo is A a plane? Where did $$\\left(1-\\frac{z}{h}\\right)^2$$ come from?\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nSo is A a plane?\nNo, in my notations $A=A(0)$ is a number, the area of the pyramid's base.\n\nWhere did $$\\left(1-\\frac{z}{h}\\right)^2$$ come from?\nThe length of the cross-section at height $z$ is $L\\left(1-\\frac{z}{h}\\right)$, and the width at height $z$ is $W\\left(1-\\frac{z}{h}\\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.\n\nBy the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.\n\n#### dwsmith\n\n##### Well-known member\nNo, in my notations $A=A(0)$ is a number, the area of the pyramid's base.\n\nThe length of the cross-section at height $z$ is $L\\left(1-\\frac{z}{h}\\right)$, and the width at height $z$ is $W\\left(1-\\frac{z}{h}\\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.\n\nBy the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.\nWhy would we find the area as $$2x\\cdot 2y$$ instead of $$x\\cdot y$$?\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nI am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.\n\nLet $$L$$ be on the x axis and $$W$$ be on the y axis.\nIn the x-z plane, we have the line $$z = -\\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\\frac{h}{W/2}y + h$$.\n\nMy cross sections has width $$\\Delta z$$. So I want to find the volume $$\\int_0^hA(z)dz$$.\n\nHow can I do this?\nIf it's a right pyramid, isn't the volume just \\displaystyle \\begin{align*} V = \\frac{1}{3}A\\,h = \\frac{1}{3}L\\,W\\,h \\end{align*}?\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nWhy would we find the area as $$2x\\cdot 2y$$ instead of $$x\\cdot y$$?\nBecause the section of the cone using the $xz$ plane is a triangle with sides\n\\begin{align}\n\\end{align}\nIf you find the $x$ corresponding to a given $z$ from (2), then the line at height $z$ crosses the triangle from $-x$ to $x$, i.e., the length of the segment the triangle cuts on the line is $2x$. But again, this is easier to see from similar triangles. The overall intuition is that the length of the cross-section decreases linearly from $L$ at $z=0$ to $0$ at $z=h$. There is a single linear function that does this, and it is $L\\left(1-\\frac{z}{h}\\right)$.\n\n- - - Updated - - -\n\nIf it's a right pyramid, isn't the volume just \\displaystyle \\begin{align*} V = \\frac{1}{3}A\\,h = \\frac{1}{3}L\\,W\\,h \\end{align*}?\nWe are trying to prove it.\n\n#### MarkFL\n\nStaff member\nI am trying to find the volume of a pyramid where the base has length $$L$$ and width $$W$$, and the pyramid has height $$h$$.\n\nLet $$L$$ be on the x axis and $$W$$ be on the y axis.\nIn the x-z plane, we have the line $$z = -\\frac{h}{L/2}x + h$$, and in the y-z plane, we have the line $$z = -\\frac{h}{W/2}y + h$$.\n\nMy cross sections has width $$\\Delta z$$. So I want to find the volume $$\\int_0^hA(z)dz$$.\n\nHow can I do this?\nYou may want to read >>>this thread<<< for a tutorial on how to work this type of problem.\n\n#### dwsmith\n\n##### Well-known member\nYou may want to read >>>this thread<<< for a tutorial on how to work this type of problem.\nI was reading a calculus book and had most of it figured out but the 2x 2y piece Makarov cleared up.", "date": "2021-01-22 03:58:36", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/volume-of-a-pyramid.9351/", "openwebmath_score": 0.9237639904022217, "openwebmath_perplexity": 234.80703776362515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9852713865827591, "lm_q2_score": 0.8688267745399465, "lm_q1q2_score": 0.8560301608511993}}
{"url": "http://mathhelpforum.com/calculus/84137-binomial-series-2-stuck.html", "text": "# Math Help - Binomial series #2 stuck!\n\n1. ## Binomial series #2 stuck!\n\n$f(x) = \\frac{5}{(1+\\frac{x}{10})^4}$\n\n$= 5\\left[1 + \\frac{x}{10})^-4\\right]$\n\n$= 5 \\left[1 + (-4)(\\frac{x}{10}) + \\frac{(-4)(-5)}{2!}(\\frac{x}{10})^2 + \\frac{(-4)(-5)(-6)}{3!}(\\frac{x}{10})^3 + ... \\right]$\n\nBut I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!\n\n2. Originally Posted by mollymcf2009\n$f(x) = \\frac{5}{(1+\\frac{x}{10})^4}$\n\n$= 5\\left[1 + \\frac{x}{10})^-4\\right]$\n\n$= 5 \\left[1 + (-4)(\\frac{x}{10}) + \\frac{(-4)(-5)}{2!}(\\frac{x}{10})^2 + \\frac{(-4)(-5)(-6)}{3!}(\\frac{x}{10})^3 + ... \\right]$\n\nBut I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!\nBefore I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?\n\n3. Originally Posted by Chris L T521\nBefore I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?\nSorry...\n\nHere is the whole question:\n\nUse the binomial series to expand the function as a power series.\n\n$f(x) = \\frac{5}{(1+\\frac{x}{10})^4}$\n\n4. Binomial series are given by: $(1+y)^k = 1 + ky + \\frac{k(k-1)}{2!}y^2$ ${\\color{white}.} \\ + \\ \\frac{k(k-1)(k-2)}{3!}y^3 + \\cdots + \\frac{k(k-1)(k-2)\\cdots(k - n+1)}{n!}y^n + \\cdots$\n\nfor any $k \\in \\mathbb{R}$ and if $|y| < 1$. Here, $y = \\tfrac{x}{10}$ and $k = -4$ .\n\nSo: $g(x) = \\left( 1 + \\frac{x}{10}\\right)^{-4}$\n\n$g(x) = 1 + (-4)\\left(\\frac{x}{10}\\right) + \\frac{(-4)(-5)}{2!}\\left(\\frac{x}{10}\\right)^2 + \\frac{(-4)(-5)(-6)}{3!}\\left(\\frac{x}{10}\\right)^3 + \\cdots$ $+ \\ \\frac{(-4)(-5)(-6)\\cdots(-4-n+1)}{n!}\\left(\\frac{x}{10}\\right)^n + \\cdots$\n\n_______________\n\nLet's look at the general term: $\\frac{(-4)(-5)(-6)\\cdots(-3-n)}{n!}\\frac{x^n}{10^n}$\n\nJust looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative.\n\nSo, if we factor out the negative signs: $\\frac{(-1)^{n} (4)(5)(6)\\cdots(n+3)}{10^n \\cdot n!} x^n \\cdot {\\color{red}\\frac{3!}{3!}} = \\frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \\cdot 10^n \\cdot n!}x^n = \\cdots$\n\netc.\n\n5. Originally Posted by o_O\nBinomial series are given by: $(1+y)^k = 1 + ky + \\frac{k(k-1)}{2!}y^2$ ${\\color{white}.} \\ + \\ \\frac{k(k-1)(k-2)}{3!}y^3 + \\cdots + \\frac{k(k-1)(k-2)\\cdots(k - n+1)}{n!}y^n + \\cdots$\n\nfor any $k \\in \\mathbb{R}$ and if $|y| < 1$. Here, $y = \\tfrac{x}{10}$ and $k = -4$ .\n\nSo: $g(x) = \\left( 1 + \\frac{x}{10}\\right)^{-4}$\n\n$g(x) = 1 + (-4)\\left(\\frac{x}{10}\\right) + \\frac{(-4)(-5)}{2!}\\left(\\frac{x}{10}\\right)^2 + \\frac{(-4)(-5)(-6)}{3!}\\left(\\frac{x}{10}\\right)^3 + \\cdots$ $+ \\ \\frac{(-4)(-5)(-6)\\cdots(-4-n+1)}{n!}\\left(\\frac{x}{10}\\right)^n + \\cdots$\n\n_______________\n\nLet's look at the general term: $\\frac{(-4)(-5)(-6)\\cdots(-3-n)}{n!}\\frac{x^n}{10^n}$\n\nJust looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative.\n\nSo, if we factor out the negative signs: $\\frac{(-1)^{n} (4)(5)(6)\\cdots(n+3)}{10^n \\cdot n!} x^n \\cdot {\\color{red}\\frac{3!}{3!}} = \\frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \\cdot 10^n \\cdot n!}x^n = \\cdots$\n\netc.\nOk, I understand everything except where the 3!/3! got there.\n\nThanks!!!!\n\n6. Focus on the expression: $\\frac{4 \\cdot 5 \\cdot 6 \\cdots (n+3)}{n!}$\n\nWe can simplify it by multiplying by \"1\": $\\frac{{\\color{red} 1 \\cdot 2 \\cdot 3} \\cdot 5 \\cdot 6 \\cdots (n+3)}{n! \\cdot {\\color{red}3!}} = \\frac{(n+3)!}{n! \\cdot 3!}$ $= \\frac{(n+3)(n+2)(n+1)n!}{n! \\cdot 3!} = \\frac{(n+3)(n+2)(n+1)}{ 6}$\n\nSo you can see why I multiplied both top and bottom by $3!$ so that I can simplify that long product into a simple factorial.", "date": "2016-07-25 17:05:50", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/84137-binomial-series-2-stuck.html", "openwebmath_score": 0.870779275894165, "openwebmath_perplexity": 500.602954232803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713865827591, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8560301575047491}}
{"url": "https://math.stackexchange.com/questions/2953056/convergence-of-the-following-sequence-of-functions", "text": "# Convergence of the following sequence of functions.\n\nFor $$n \\ge 1$$, let $$g_n(x) = \\sin^2 \\left (x + \\frac 1 n \\right ), x \\in [0,\\infty)$$ and $$f_n(x) = \\int_{0}^{x} g_n (t)\\ \\mathrm {dt}.$$ Then\n\n$$(1)$$ $$\\{f_n \\}$$ converges pointwise to a function $$f$$ on $$[0,\\infty)$$, but does not converge uniformly on $$[0, \\infty)$$.\n\n$$(2)$$ $$\\{f_n \\}$$ does not converge pointwise to any function $$f$$ on $$[0, \\infty)$$.\n\n$$(3)$$ $$\\{f_n \\}$$ converges uniformly on $$[0,1]$$.\n\n$$(4)$$ $$\\{f_n \\}$$ converges uniformly on $$[0, \\infty)$$.\n\nI have found that $$f_n (x) = \\frac 1 2 \\left (x - \\sin x \\cos \\left (x + \\frac 2 n \\right ) \\right)$$ which converges to the function $$f$$ on $$[0, \\infty)$$ defined by $$f(x) = \\frac 1 4 ( 2x - \\sin {2x}), x \\in [0, \\infty)$$. But I am not quite sure about whether this convergence is uniform or not. Please help me in this regard.\n\nThank you very much.\n\n\u2022 Yeah you are right. I have edited my body. Oct 12, 2018 at 19:05\n\u2022 What is the definition of uniform convergence, and how is it different from point-wise convergence. Oct 12, 2018 at 19:28\n\nhint\n\n$$\\sin^2(t+\\frac 1n)=\\frac{1-\\cos(2t+\\frac 2n)}{2}$$\n\n$$f_n(x)=\\int_0^x\\sin^2(t+\\frac 1n)dt=\\frac{1}{2}\\Big[t-\\frac{\\sin(2t+\\frac 2n)}{2}\\Bigr]_0^x$$\n\n$$=\\frac x2-\\frac 14\\sin(2x+\\frac 2n)+\\frac 14\\sin(\\frac 2n)$$\n\nThe pointwise limit is $$f(x)=\\frac x2-\\frac{\\sin(2x)}{4}$$\n\nFor the uniform convergence, use MVT to get\n\n$$|\\sin(2x+\\frac 2n)-\\sin(2x)|\\le \\frac 2n$$\n\nand\n\n$$|f_n(x)-f(x)|\\le \\frac 1n.$$\n\nThe convergence is uniform at $$[0,\\infty)$$.\n\n\u2022 I have done that. First see my body and then answer. Oct 12, 2018 at 19:09\n\u2022 @Dbchatto67 Your $f_n(x)$ is not correct. Oct 12, 2018 at 19:10\n\u2022 Have you evaluated your integral between the limits $0$ and $x$? Oct 12, 2018 at 19:11\n\u2022 In place of $t$ we have $t + \\frac 1 n$. Oct 12, 2018 at 19:11\n\u2022 Now calculate and simplify. Oct 12, 2018 at 19:13", "date": "2022-08-17 14:21:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2953056/convergence-of-the-following-sequence-of-functions", "openwebmath_score": 0.9471237063407898, "openwebmath_perplexity": 187.40790748562603, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713891776497, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8560301530663587}}
{"url": "https://engineering.stackexchange.com/questions/35475/moment-equation-from-dynamics-differs-with-equations-from-physics-statics-whe", "text": "# Moment equation from dynamics differs with equations from physics & statics. Where did I go wrong?\n\nIn my dynamics class, we're being asked to solve the following problem:\n\nMy Attempt:\n\nSince I'm given the initial velocity, final velocity, and distance, I solved for the acceleration of the plane using kinematics:\n\n$$a = \\frac{v_f^2 - v_i^2}{2d} = \\frac{(55.6 m/s)^2 - (16.7 m/s)^2}{2(425m)} = 3.31 m/s^2$$\n\nNow this is where my approach starts to differ from the formulas recommended by the class. I decided to set the moment about point A equal to zero, because the plane is not rotating as it moves down the runway. I seem to remember from both statics and physics that if a body is not rotating about a given point, you can simply set the moment around that point to be zero.\n\nLet N be the reaction force at B. Combining the above assumption with Newton's second law in the x direction, I get:\n\n$$\\Sigma F_x: R = ma = (140000kg)(3.31 m/s^2) = 4.63 *10^5 N$$\n$$\\Sigma M_A: -(15m)N + (2.4m)mg - (1.8m)R = 0$$\n\nSolving for N gives me:\n\n$$N = 1.64*10^5 N$$\n\nAccording to the solution guide, this is incorrect.\n\nSolution Guide's Explanation:\n\nThe guide uses a formula which was introduced to us in the textbook, which is as follows.\n\nFor some point P fixed on a rigid body with center of mass G, the moment about point P is given by:\n\n$$\\Sigma M_P = I_G\\alpha + ma_Gd$$\n\nwhere:\n$$I_G$$ is the moment of inertia of the rigid body about G\n$$\\alpha$$ is the angular acceleration of the rigid body about G\n$$a_G$$ is the acceleration of G\n$$d$$ is the moment-arm distance, from P to G, of $$ma_G$$\n\nThe book cleverly chose a point C on the plane through which both R and A (the reaction force at wheel A) pass. See below:\n\nUsing the moment equation above and setting $$\\alpha = 0$$ (because the plane isn't rotating) gives:\n\n$$\\Sigma M_C = ma_Gd = (15m)N - (3m-1.8m)mg$$\n\nFinally, using $$a_G = 3.3 m/s^2$$ and solving for N, they got:\n\n$$N = 2.57*10^5 N$$\n\nWhy I'm Confused:\n\nFrom the physics and statics classes I've taken in the past, I've always been taught that $$\\Sigma M = I\\alpha$$; there was never that extra \"$$+ mad$$\" term thrown onto the end. That term is basically saying that an accelerating body with no rotation can still have a moment about a point. By contrast, in my physics and statics classes, I recall using the fact that if a body is stationary (no angular acceleration), we can set the moment about any point on the body equal to zero to help us solve.\n\nMy Guess at Where the Inconsistency Lies:\n\nHere's my guess at where the inconsistency lies: in my statics classes, we assumed that the rigid bodies we were analyzing had not only zero angular acceleration, but also zero linear acceleration. (It's statics, after all!) In dynamics problems such as this one, however, there is a nonzero linear acceleration which must be accounted for. The \"+mad\" term must come from the fact that for any point object, the moment about some fixed point O is given by:\n\n$$\\Sigma M_O = \\vec{r} \\times \\vec{F} = \\vec{r} \\times m\\vec{a}$$\n\nAs I'm writing this, I think the intuition totally makes sense. It seems the formula used by the solution guide accounts for both rotation of the rigid body about its center, and linear acceleration of the body with respect to some point P outside the center of mass of the body.\n\nI'll still post this though, in case anyone would like to correct me, add anything, or use this for their own reference.\n\n\u2022 Your image font is not readable. Would you add the text of the image to your question. Thanks. Apr 30, 2020 at 0:42\n\u2022 This is actually for once such a nice homework question, so much information and own effort instead of just a bad photo of a problem sheet. +1, thanks OP Apr 30, 2020 at 8:00\n\nThere is a slight mistake in your approach. It seems you have forgotten to add the contribution the moment due to inertial force acting at the center of mass $$G$$ to the moment equation. The moment equation should be : $$\\sum M_A = -(15 m) N + (3 m) 140*10^3*a + (2.4 m) 140*10^3*g - (1.8 m) 140*10^3*a = 0$$ Evaluating for $$N$$ results in : $$N = 2.57 * 10^5 N$$", "date": "2022-05-19 08:18:15", "meta": {"domain": "stackexchange.com", "url": "https://engineering.stackexchange.com/questions/35475/moment-equation-from-dynamics-differs-with-equations-from-physics-statics-whe", "openwebmath_score": 0.8123955726623535, "openwebmath_perplexity": 347.0649879860927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9648551525886193, "lm_q2_score": 0.8872046026642945, "lm_q1q2_score": 0.8560239322809832}}
{"url": "https://math.stackexchange.com/questions/1532735/count-the-number-of-non-decreasing-numbers-with-d-digits-lower-than-x", "text": "# Count the number of non-decreasing numbers with d digits, lower than X\n\nA number is said to be made up of non-decreasing digits if all the digits to the left of any digit is less than or equal to that digit.\n\nFor example, the four-digit number 1234 is composed of digits that are non-decreasing. Some other four-digit numbers that are composed of non-decreasing digits are 0011, 1111, 1112, 1122, 2223.\n\nNotice that leading zeroes are required: 0000, 0001, 0002 are all valid four-digit numbers with non-decreasing digits.\n\nI arrived at the conclusion that there are in total $$\\sum_{l=0}^9 \\sum_{k=0}^l \\sum_{j=0}^k ... \\sum_{a=0}^b 1 = \\binom{9+d}{d}$$ ways to calculate the non decreasing numbers, for d digits, without constrains.\n\nMy question is, suppose I want to find the number of non-decreasing numbers with D digits, under a given limit (in the same order of magnitude). For example, how many 4 digit numbers are non-decreasing and under 5000?\n\nI realize that for this case, I need this summations, but I can't neither figure out how to solve them, nor how to generalize them $$\\sum_{l=0}^9 \\sum_{k=0}^l \\sum_{j=0}^k\\sum_{i=0}^{min(5,j)} 1$$\n\nwhere I limit this first number to 5.\n\nAny help?\n\n\u2022 When I look at this post, I see eight-digit numbers such as $12341234$ rather than four-digit numbers such as $1234$. \u2013\u00a0N. F. Taussig Nov 17 '15 at 10:01\n\u2022 Fixed the examples, posted this late at night. Thanks @N.F.Taussig \u2013\u00a0Jos\u00e9 Pedro Marques Nov 17 '15 at 10:37\n\nWe can represent a number with non-decreasing digits by placing vertical bars in a string of nine ones. The number of ones to the left of the vertical bar represents the digit. For instance, $$| 1 1 1 | | 1 1 1 1 1 1 |$$ represents the number $0339$, while $$1 1 | 1 1 | 1 1 | 1 1 | 1$$ represents the number $2468$. Since a particular number is determined by which four of the thirteen symbols (four vertical bars and nine ones) are vertical bars, the number of non-decreasing numbers with four digits (including leading zeros) is $$\\binom{13}{4} = \\binom{13}{9}$$ as you found. To determine how many four-digit numbers (including those with leading zeros) are less than $5000$, we must subtract from these those non-decreasing four-digit numbers that are at least $5000$. Notice that in those non-decreasing four-digit numbers that are at least $5000$, the first vertical bar must be placed to the right of at least five ones. For instance, $$1 1 1 1 1 | | 1 | 1 | 1 1$$ represents the number $5567$. Since a particular non-decreasing four-digit number that is at least $5000$ is determined by which four of the last eight symbols (four vertical bars and four ones) are vertical bars, the number of non-decreasing four-digit numbers that are at least $5000$ is $$\\binom{8}{4}$$ Hence, there are $$\\binom{13}{4} - \\binom{8}{4}$$ non-decreasing four-digit numbers, including those with leading zeros, that are less than $5000$.\nAlternate Method: A non-decreasing number is completely characterized by the number of times each digit appears. Let $x_k$ represent the number of times the digit $k$ appears in the non-decreasing four-digit number, including those with leading zeros. Since there are four digits, $$x_0 + x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 = 4$$ A particular number is determined by which nine of the thirteen symbols (nine addition signs and four ones) is an addition sign. For instance, $$1 + + + 1 1 + + + + + + 1$$ represents the number $0339$, while $$+ + 1 + + 1 + + 1 + + 1 +$$ represents the number $2468$. Thus, the total number of non-decreasing four-digit numbers is the number of ways nine addition signs can be placed in a row of four ones, which is $$\\binom{4 + 9}{9} = \\binom{13}{9}$$ From these, we subtract those non-decreasing four-digit numbers that are at least $5000$. The digits of non-decreasing four-digit numbers that are at least $5000$ satisfy the equation $$x_5 + x_6 + x_7 + x_8 + x_9 = 4$$ A particular number is determined by which four of the eight symbols (four addition signs and four ones) is an addition sign. Hence, there are $$\\binom{4 + 4}{4} = \\binom{8}{4}$$ non-decreasing four-digit numbers that are at least $5000$. Hence, the number of non-decreasing four-digit numbers that are less than $5000$, including those with leading zeros, is $$\\binom{13}{9} - \\binom{8}{4}$$", "date": "2021-02-27 19:50:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1532735/count-the-number-of-non-decreasing-numbers-with-d-digits-lower-than-x", "openwebmath_score": 0.8440277576446533, "openwebmath_perplexity": 98.45621382255443, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474910447999, "lm_q2_score": 0.8652240930029117, "lm_q1q2_score": 0.8560072856039433}}
{"url": "http://mathhelpforum.com/geometry/174934-centroid-triangle-coordinates.html", "text": "# Thread: The centroid of a triangle with coordinates\n\n1. ## The centroid of a triangle with coordinates\n\nTriangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.\n\nI know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.\n\n2. Originally Posted by darksoulzero\nTriangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.\n\nI know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.\nRead Centroid - AoPSWiki\n\n3. ## centroid of triangle\n\nHi darksoulzero,\n\nYour solution is confusing.Here is a suggested method.\n\nconnect M midpoint of DE and C (centroid) with extended lenght.F lies on this line. FC = 2CM. Slope diagram of C and M = 2/1/2.Slope diagram of FC is twice that or 4/1. Working from point C one point left and 4 points up gives F (2,8)\n\nbjh\n\n4. If ABC is a triangle with $A\\left( {{x_1},{y_1}} \\right),B\\left( {{x_2},{y_2}} \\right),C\\left( {{x_3},{y_3}} \\right)$ then its centroid is\n$G\\left( {\\tfrac{{{x_1} + {x_2} + {x_3}}}{3},\\tfrac{{{y_1} + {y_2} + {y_3}}}{3}} \\right)$\n\n5. Hello, darksoulzero!\n\n$\\text{Triangle }D{E}F\\text{ has vertices: }D(1,3)\\text{ and }E(6,1)\\text{, and centroid at }C(3,4).$\n$\\text{Determine the coordinates of point }F.$\nCode:\n\n1\nF o-----+\n\\    :\n\\   :4\n\\  :\n\\ :\n\\:0.5\n(3,4)o---+\nD       C\\  :\no         \\ :2\n(1,3)   *    \\:\nM o\n(3.5,2)   *     E\no\n(6,1)\n\nWe have vertices $D(1,3)$ and $E(6,1)$, and centroid $C(3,4).$\n\nThe midpoint of $DE$ is: $M(3\\tfrac{1}{2},\\,2).$\n\nThe median to side $DE$ starts at $\\,M$, passes through $\\,C,$\n. . and extends to $\\,F$, where: . $FC \\,=\\,2\\!\\cdot\\!CM.$\n\nGoing from $\\,M$ to $\\,C$, we move up 2 and left $\\frac{1}{2}$\nHence, going from $\\,C$ to $\\,F$, we move up 4 and left 1.\n\nTherefore, we have: . $F(2,8).$", "date": "2017-03-23 02:17:54", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/174934-centroid-triangle-coordinates.html", "openwebmath_score": 0.7596879005432129, "openwebmath_perplexity": 694.7595029805632, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9893474907307124, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8560072784556494}}
{"url": "https://www.physicsforums.com/threads/f-x-symmetric-about-the-line-x-2.795521/", "text": "# F(x) symmetric about the line x=2\n\n## Main Question or Discussion Point\n\nWhy is the function: y = f(x) = a(x-1)(x-2)(x-3) symmetrical about the line x = 2? I mean how can we be sure that it is? Is there any method to check it?\n\n## Answers and Replies\n\nShayanJ\nGold Member\nChange the variable to y=x-2 and see whether the resulting function is even or odd or none!\n\nYou mean x = x-2 right? It doesn't stay the same on doing that. So it should not be symmetrical about x = 2, but it is. And I am not able to see how that happens.\n\nShayanJ\nGold Member\n$f(x)=a(x-1)(x-2)(x-3) \\rightarrow f(y)=a(y+1)y(y-1)$\n$f(-y)=a(-y+1)(-y)(-y-1)=-a(y-1)y(y+1)=-f(y)$\nSo f(y) is odd which means f(x) is not symmetric around x=2 but is not that much asymmetric because we have f(2-x)=-f(x-2).\n\nOh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1\n\nAnd this function is said to be symmetrical about the line x = 2. But I am unable to see how?\n\nShayanJ\nGold Member\nIts the same trick. Just do the tranformation $x\\rightarrow x+2$ and check whether the resulting function is even. If its even, then the original function is symmetric around x=2.\n\nThey are not the same, i.e. after changing x to x+2 in f(x), f(x) \u2260 f(-x). But what I am reading, it says that f(x) is symmetrical about x=2 and I am still wondering that how would I go about proving it?\n\nShayanJ\nGold Member\nYou're doing something wrong. You should be able to reduce f(x+2) to $\\frac 1 2 a[x^2(\\frac 1 2 x^2-1)-4]+1$ which is even.\n\nPeroK\nHomework Helper\nGold Member\nOh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1\n\nAnd this function is said to be symmetrical about the line x = 2. But I am unable to see how?\nAnother way to look at it is as follows:\n\nImagine starting at x = 2 and moving the same distance, d, to the right and left (d > 0). So, to the right we have 2 + d and to the left we have 2 - d.\n\nNow, if f is symmetrical about x = 2, then f(2-d) = f(2+d) for all d. You could try that approach.\n\nYou're doing something wrong. You should be able to reduce f(x+2) to $\\frac 1 2 a[x^2(\\frac 1 2 x^2-1)-4]+1$ which is even.\nHow did you reduce it down to that? Can you show me? I am unable to get to that point.\n\nShayanJ\nGold Member\n$f(x+2)=a[ \\frac 1 4 (x+2)^4-2(x+2)^3+\\frac{11}{2} (x+2)^2-6(x+2)]+1=\\\\ a(x+2)[ \\frac 1 4 (x+2)^3-2(x+2)^2+\\frac{11}{2} (x+2)-6]+1=\\\\ a(x+2)(\\frac 1 4 x^3+\\frac 3 2 x^2+3x+2-2x^2-8x-8+\\frac{11}{2}x+5)+1=\\\\ \\frac 1 2 a(x+2)(\\frac 1 2 x^3-x^2+x-2)+1=\\frac 1 2 a (\\frac 1 2 x^4+x^3-x^3-2x^2+x^2+2x-2x-4)+1=\\\\ \\frac 1 2 a (\\frac 1 2 x^4-x^2-4)+1$\n\nAnd what is the reason that we transformed x to x+2?\n\nShayanJ\nGold Member\nSimple. When we say a function is even, we mean its symmetric around x=0. So if a function is symmetric around $x=a \\neq 0$, it means if we move the origin to x=a, the resulting function would be even.\n\nRight. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?\n\nShayanJ\nGold Member\nCheck here!\nBut we aren't shifting the curve as in your link but instead we are shifting the coordinate axes. The curve stays where it was. The axes are what shift.\n\nPeroK\nHomework Helper\nGold Member\nRight. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?\nThis is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.\n\nIf we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.\n\nBut, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.\n\nIn fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.\n\nThis is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.\n\nIf we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.\n\nBut, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.\n\nIn fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.\nWait. So you are saying that f(x) should change to a new variable so that it becomes f(z), right? But that would again mean f(x-2) and not f(x+2).. got me confused\n\nShayanJ\nGold Member\n$z=x-2 \\Rightarrow x=z+2$ so $f(x)=f(z+2)$!(Forget the confusing $x\\rightarrow x+2$!)\n\nIt just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?\n\nFor example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.\n\nShayanJ\nGold Member\nIt just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?\n\nFor example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.\nLet's start from the beginning.\nAt first we have a coordinate system which we call xy. Now I define a function y=f(x). Then I move the origin to x=a and name the new coordinate system zy. But this doesn't change the function, only the coordinate system has moved. But if I insist that the function f has the same form in terms of both x and z, then this means that the function has changed which isn't right.(Imagine y=x^2. z=x-a so x=z+a. But if I say that y=z^2, this function would have its minimum at z=0 so x=a which means the function has changed!) So I should have y=g(z). But g should be related to f somehow that we actually get the same function. So let's see what's the relationship. At x=0, f gives f(0), so at z=-a, g should give f(0). Then at x=a, f gives f(a), so at z=0, g should give f(a). Now we have two relationships g(-a)=f(0) and g(0)=f(a) and so we can deduce that g(z)=f(z+a)=f(x).", "date": "2020-04-03 09:44:32", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/f-x-symmetric-about-the-line-x-2.795521/", "openwebmath_score": 0.7770125269889832, "openwebmath_perplexity": 516.0461644055225, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.956634206181515, "lm_q2_score": 0.8947894717137996, "lm_q1q2_score": 0.8559862159725078}}
{"url": "https://math.stackexchange.com/questions/9911/convergence-of-the-series-sum-limits-n-2-infty-frac1n-logs-n", "text": "# Convergence of the series $\\sum \\limits_{n=2}^{\\infty} \\frac{1}{n\\log^s n}$\n\nWe all know that $\\displaystyle \\sum_{n=1}^{\\infty} \\frac{1}{n^s}$ converges for $s>1$ and diverges for $s \\leq 1$ (Assume $s \\in \\mathbb{R}$).\n\nI was curious to see till what extent I can push the denominator so that it will still diverges.\n\nSo I took $\\displaystyle \\sum_{n=2}^{\\infty} \\frac{1}{n\\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\\displaystyle \\sum_{n=2}^{\\infty} a_n$ converges iff $\\displaystyle \\sum_{n=2}^{\\infty} 2^na_{2^n}$ converges).\n\nNo surprises here. I expected it to diverge since $\\log n$ grows slowly than any power of $n$.\n\nHowever, when I take $\\displaystyle \\sum_{n=2}^{\\infty} \\frac{1}{n(\\log n)^s}$, I find that it converges $\\forall s>1$.\n\n(By the same argument as previous).\n\nThis doesn't make sense to me though.\n\nIf this were to converge, then I should be able to find a $s_1 > 1$ such that\n\n$\\displaystyle \\sum_{n=2}^{\\infty} \\frac{1}{n^{s_1}}$ is greater than $\\displaystyle \\sum_{n=2}^{\\infty} \\frac{1}{n (\\log n)^s}$\n\nDoesn't this mean that in some sense $\\log n$ grows faster than a power of $n$?\n\n(or)\n\nHow should I make sense of (or) interpret this result?\n\n(I am assuming that my convergence and divergence conclusions are right).\n\n\u2022 They are right. You don't seem to have completed one of your sentences: \"if this were to converge, then I should be able to find a s_1 > 1 such that\"...? \u2013\u00a0Qiaochu Yuan Nov 11 '10 at 21:24\n\u2022 There seems to be some bug with this. It doesn't display the entire thing. \u2013\u00a0user17762 Nov 11 '10 at 21:25\n\u2022 @Sivaram: Yes even i tried editing it, and it still has not worked \u2013\u00a0anonymous Nov 11 '10 at 21:27\n\u2022 There seems to be a bug/feature! when dealing with <. Probably confusion with html tags. Sivaram, please use \\log n instead of just logn. \u2013\u00a0Aryabhata Nov 11 '10 at 21:32\n\u2022 @ Chandru and Moron: Thanks for editing it. Now it looks fine. \u2013\u00a0user17762 Nov 11 '10 at 21:33\n\n## 6 Answers\n\nYes $\\displaystyle \\sum_{n=2}^{\\infty} \\dfrac{1}{n (\\log n)^s}$ is convergent if $\\displaystyle s > 1$, we can see that by comparing with the corresponding integral.\n\nAs to your other question, if\n\n$\\displaystyle \\sum_{n=2}^{\\infty} \\frac{1}{n^{s_1}}$ is greater than $\\displaystyle \\sum_{n=2}^{\\infty} \\frac{1}{n (\\log n)^s}$\n\ndoes not imply $\\log n$ grows faster than a power of $\\displaystyle n$. You cannot compare them term by term.\n\nWhat happens is that the first \"few\" terms of the series dominate (the remainder goes to 0). For a small enough $\\displaystyle \\epsilon$, we have that $\\log n > n^{\\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\\log n$ to dominate the other.\n\n\u2022 Sorry. I meant $\\exists s_1 > 1$ such that $\\displaystyle \\sum_{2}^{\\infty} \\frac{1}{n^{s_1}}$ is greater than $\\displaystyle \\sum_{2}^{\\infty} \\frac{1}{n(\\log n)^{s}}$ \u2013\u00a0user17762 Nov 11 '10 at 21:56\n\nThis is a special case of classic general logarithmic convergence tests. Some early work in asymptotics was motivated by attempts to determine the \"boundary of convergence\" in terms of various functions, e.g. log-exp functions. Below is a a very interesting excerpt from Hardy's classic \"A course in pure mathematics\". For further work see his \"Orders of infinity\" and see especially this very interesting paper by G. Fisher on Paul du Bois-Reymond's work on the boundary between convergence and divergence (where he discovered diagonalization before Cantor).\n\nNote in particular the following very general result that is presented in the excerpt: the following series and integral are convergent if $\\rm\\ s > 1\\$ and divergent if $\\rm\\: s\\le 1\\$ where $\\rm\\: n_0,\\ a\\$ are any numbers large enough to ensure positivity of $\\rm\\ log_k x := \\log \\log \\cdots \\log x\\:,\\$ iterated $\\rm\\:k\\:$ times.\n\n$$\\rm \\sum_{n_0}^\\infty \\frac{1}{n \\log n\\ \\log_2 n\\ \\cdots\\ \\log_{k-1} n\\ (\\log_k n)^s }$$\n\n$$\\rm \\int_{a}^\\infty \\frac{dx}{x \\log x\\ \\log_2 x\\ \\cdots\\ \\log_{k-1} x\\ (\\log_k x)^s }$$\n\n\u2022 Off-topic question: What text is the excerpt from? Thank you! \u2013\u00a0Ayesha Jan 29 '14 at 1:04\n\u2022 @Ayesha I say that in the text above it: Hardy's .... \u2013\u00a0Bill Dubuque Jan 29 '14 at 1:09\n\u2022 Sorry, sorry, I somehow glanced over that. Thank you again for responding to my rather asinine question. \u2013\u00a0Ayesha Jan 29 '14 at 1:43\n\nDu Bois-Reymond was interested in this question, positing some sort of series \"between\" all convergent series and all divergent series. Hausdorff didn't like that idea, and showed that given any countable sequences of convergence series converging \"ever more slowly\" and divergent series diverging \"ever more slowly\", you can find a convergent and a divergent series \"in between\". Here the order relation is the one given by the (conclusive) ratio test.\n\nBill's remark is related to the recursive Elias codes. The first few codes correspond to the convergent series\n\n$\\displaystyle \\frac{1}{n(\\log n)^2}, \\frac{1}{n\\log n (\\log \\log n)^2}, \\ldots$\n\nHe then diagonalizes to obtain his $\\omega$ code (Elias omega code, if you want to look it up), which corresponds to\n\n$\\displaystyle \\frac{1}{n\\log n\\log \\log n \\cdots (\\log^* n)^2},$\n\nwhere $\\log^* n$ is the number of times you need to apply $\\log$ until you get below some constant (the other logs continue up to this constant); confusingly, coding theorists use $\\log^* n$ to mean the $\\log n \\log\\log n \\cdots$ part. We can continue the iteration like this:\n\n$\\displaystyle \\frac{1}{n\\log n\\log\\log n \\cdots \\log^*n (\\log \\log^* n)^2}, \\frac{1}{n\\log n\\log\\log n \\cdots \\log^*n \\log \\log^* n (\\log\\log \\log^* n)^2}, \\cdots$\n\nand so on. We can diagonalize again to obtain an $\\omega + \\omega$ code, and so on at least for $n \\omega + m$. We an probably continue even to $\\omega^2$ and beyond.\n\nThe corresponding divergent series are $\\displaystyle \\frac{1}{n\\log n}, \\frac{1}{n\\log n\\log\\log n}, \\ldots, \\frac{1}{n\\log n\\log\\log n \\cdots \\log^* n}, \\frac{1}{n\\log n\\log\\log n \\cdots \\log^* n \\log\\log^* n}, \\ldots$\n\nOne can ask whether there is a scale for convergent series, that is a chain of (mutually comparable) series converging more and more slowly, such that for each convergent series there's one in the chain converging even more slowly; you can ask the same question about divergent series. Surprisingly, the existence of these is independent of ZFC (they exist given CH, and models in which they don't exist can be constructed using forcing).\n\n\u2022 Indeed, it deserves to be better known that Du Bois Reymond - not Cantor - is the true discoverer of diagonalization - in this context, i.e. asymptotic growth rates of functions. \u2013\u00a0Bill Dubuque Nov 11 '10 at 22:53\n\u2022 Thanks Bill and Yuval. I was looking for series which diverges slower and slower and your answer helps me on that note. \u2013\u00a0user17762 Nov 11 '10 at 23:09\n\nAnother test that applies to series of positive decreasing terms (and in this particular one in a rather elegant fashion) is the following: $$\\sum_{n=1}^\\infty a_n<\\infty \\quad\\Longleftrightarrow\\quad\\sum_{k=1}^\\infty 2^ka_{2^k}<\\infty.$$ In our case $$\\sum_{k=1}^\\infty 2^ka_{2^k}=\\sum_{k=1}^\\infty 2^k\\frac{1}{2^k(\\log 2^k)^s}= \\frac{1}{(\\log 2)^s}\\sum_{k=1}^\\infty \\frac{1}{k^s},$$ and thus $$\\sum_{n=2}^\\infty \\frac{1}{n(\\log n)^s}\\quad\\Longleftrightarrow\\quad \\sum_{k=1}^\\infty \\frac{1}{k^s}\\quad\\Longleftrightarrow\\quad s>1.$$\n\n\u2022 Monotonicity (as is in this case) is required? \u2013\u00a0Mart\u00edn-Blas P\u00e9rez Pinilla Jan 31 '14 at 19:40\n\u2022 @Mart\u00edn-BlasP\u00e9rezPinilla: I do not know the most general version of this criterion, but the version I know requires $a_n$ to be non-negative and decreasing. \u2013\u00a0Yiorgos S. Smyrlis Jan 31 '14 at 19:46\n\u2022 Yes, now I remember I've seen in Baby Rudin. \u2013\u00a0Mart\u00edn-Blas P\u00e9rez Pinilla Jan 31 '14 at 19:56\n\nBy the integral comparison test the given series is convergent if and only if the integral $$\\int_2^\\infty\\frac{dx}{x\\log^s(x)}$$ is convergent and notice that $\\log'(x)=\\frac1x$ so\n\n\u2022 if $s>1$ $$\\int_2^\\infty\\frac{dx}{x\\log^s(x)}=\\frac1{1-s}\\log^{1-s}(x)\\Big|_2^\\infty<+\\infty$$ so the series is convergent\n\u2022 if $s=1$ $$\\int_2^\\infty\\frac{dx}{x\\log(x)}=\\log(\\log(x))\\Big|_2^\\infty=+\\infty$$ so the series is divergent.\n\u2022 The exponent $s$ is just on the log term, not the entire term \u2013\u00a0user2566092 Jan 31 '14 at 19:38\n\u2022 Ah I'll edit my answer! \u2013\u00a0user63181 Jan 31 '14 at 19:41\n\u2022 @user2566092 I edited my answer. \u2013\u00a0user63181 Jan 31 '14 at 19:50\n\u2022 What about $s < 1$? \u2013\u00a0mavavilj Sep 19 '15 at 17:25\n\nConvergence does not change if you switch to base-2 logs. There is a theorem that says that if $a_n$ is a decreasing positive sequence, then $\\sum_n a_n$ converges if and only if $\\sum_k 2^k a_{2^k}$ converges. If you apply this theorem's transformation to your sequence, you get $\\sum_k 1/k^s$, which obviously converges if $s > 1$ and diverges if $s \\leq 1$. If you'd like a proof of that theorem (called the Cauchy condensation test), I can provide it.", "date": "2019-05-27 03:54:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/9911/convergence-of-the-series-sum-limits-n-2-infty-frac1n-logs-n", "openwebmath_score": 0.901950478553772, "openwebmath_perplexity": 365.6433204552242, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9566341987633822, "lm_q2_score": 0.8947894562828416, "lm_q1q2_score": 0.8559861945730586}}
{"url": "https://www.bionicturtle.com/forum/threads/difference-between-probability-density-function-and-inverse-cumulative-distribution-function.9438/", "text": "What's new\n\n# Difference between probability density function and inverse cumulative distribution function?\n\n#### bpdulog\n\n##### Active Member\nWhat is the difference between probability density function and inverse cumulative distribution function, if any? They both look the same\n\n#### ShaktiRathore\n\n##### Well-Known Member\nSubscriber\nHi,\nNo they are not the same.\nThe inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.\nWhereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.\nthanks\n\n#### brian.field\n\n##### Well-Known Member\nSubscriber\nDavid also covers this extensively in his Study Notes for Miller. Have you read them?\n\n#### bpdulog\n\n##### Active Member\nHi,\nNo they are not the same.\nThe inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.\nWhereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.\nthanks\nIf I understand what are you saying, in the inverse distribution the 1.645 represents 95% and less. And in the probability density function the 1.645 represents 95% only?\n\n#### ShaktiRathore\n\n##### Well-Known Member\nSubscriber\nHi\nin the inverse distribution the 1.645(quantile) represents 95%(cumulative probability). And in the probability density function P(x) ,when x=1.645 ,P(x) shall give the probability density at x,and we integrate the P(x) from -inf to x=1.645 to get the cumulative probability of 95%.\nthanks\n\n#### bpdulog\n\n##### Active Member\n\nAh ok! I think I get it now. It just 2 different ways to display a distribution of data. The top is the PDF and the PPF is basically the CDF with the axes transposed.\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nYuu definitely want to understand @ShaktiRathore 's above relationship. Miller's Chapter 2 does a decent job. We don't use the continuous pdf too much. The pdf integrates to the CDF, and we're arguably more interested in the relationships around the CDF, as Shakti illustrates. For example, if probability (p) = 5.0%, then:\n\u2022 p = 5% = F(q), where F(.) is the cumulative distribution function, so if we are given the probability of 5% because we want a 95% confident normal VaR, then we use the inverse CDF to retrieve the quantile:\n\u2022 q = F^(-1)(p) = F^(-1)(5%) = -1.645, where F^(-1) is the inverse CDF. \"Inverse\" implies that F[F^(-1)(p)] = p; e.g., F[F^(-1)(5%)] = F[-1.645] = 5%, just like also F^(-1)[F(q)] = q. This is why Dowd says \"VaR is just a quantile;\" i.e., because -1.645 is the quantile (q) retrieved by using the inverse CDF given a probability (p) of 5%. This is an essential FRM building block. I hope that adds something.\n\n#### theapplecrispguy\n\n##### New Member\nIn Question 300.2, the final price of the bond (i.e. the value of \"x\") is calculated using the inverse CDF such that 5% of the distribution is less than or equal to \"x\". This price is calculated to be $1.842 using p=5%. My interpretation of this is that 5% of the time the bond will be priced below$1.842 and the other 95% of the time the bond will be priced above $1.842. Is this correct? The question then states that the 95% VAR is given by$5.00 (the face value of the bond) minus $1.842 (q(.05)) equals$3.158. I don't understand conceptually how the formula for 95% VAR was determined to be $5.00 - q(.05). Can you please explain what the 95% VAR ($3.158) represents in the context of the area under the cumulative distribution graph?\n\n#### David Harper CFA FRM\n\n##### David Harper CFA FRM\nStaff member\nSubscriber\nHi @theapplecrispguy FYI, the source Q&A (with follow on discussion) is here at https://www.bionicturtle.com/forum/threads/p1-t2-300-probability-functions-miller.6728/ but briefly:\n\u2022 Re: \"My interpretation of this is that 5% of the time the bond will be priced below $1.842 and the other 95% of the time the bond will be priced above$1.842. Is this correct?\" Yes, exactly correct. The integral (anti-derivative) of the given f(x) is the CDF which is F(x) = x^3/125. When we have the CDF, it means that cumulative p = x^3/125; i.e., p is the probability that the random variable will be less than or equal to x. Because 1.842^3/125 = 5.0%, under this distribution, there is a 5.0% probability that X will be less than $1.842; and, as you say, 95.0% of the time X will be greater than$1.842 (but less than $5.00, as the function domain is$0.00 < X < 5.00). As the question is testing the associated Miller, this is really the essential point of the question ....\n\u2022 Because the VaR part here is really simplistic. VaR is a loss relative to something. This question gives the assumption that the expected future value is $5.00 so that the VaR can be computed as a simple \"worst expected loss\" of$5.00 - $1.842 =$3.158; ie, because there is a 5.0% probability of the final price falling below $1.842, there is a 5.0% probability of a loss worse than$3.158. If we were not given this assumption, the proper approaches are classically either:\n\u2022 Compute the loss relative to the expected future value as the proper mean of the future distribution. In this case, as stated in the question, this relative VaR (it is called because it is relative to the expected future value) would be $3.75 -$1.842 = $1.908. I just didn't want to make the question too difficult; or, \u2022 Compute the loss relative to the current value, which is properly termed the absolute VaR (relative to the initial value). I don't think we have information to do that, but it would be given by discounted present value [$5.00 par] - $1.842 = absolute VaR. This shows why a good question needs to be specific about the VaR it looks for, VaR is a worst expected loss over a horizon (choice #1) with a confidence level (choice #2) relative to something (definitional #3). I hope that helps! #### theapplecrispguy ##### New Member Thank you David for the explanation which is very helpful. If the question had asked for the 5% VAR would the answer have been the same? #### David Harper CFA FRM ##### David Harper CFA FRM Staff member Subscriber Hi @theapplecrispguy Yes, but only because occasionally (or rarely) the 95.0% VaR will be represented by an author as the 0.050 VaR, treating them as the same idea. In any realistic use case, we are referring to the losses that happen in the worst 5.0% of times, the losing tail. It would not be realistic/sensible to speak of the losses that occur in the worst 95.0% of times (although mathematically it works of course). But, again, the idea is that we have a cumulative distribution function here, given by F(x) = p = x^3/125. Solving for x, we have x = (p*125)^(1/3). So for example, \u2022 at p = 5%, x = (5%*125)^(1/3) =$1.84\n\u2022 at p = 10%, x = (10%*125)^(1/3) = $2.32 \u2022 at p = 20%, x = (20%*125)^(1/3) =$2.92\n\u2022 at p = 50%, x = (50%*125)^(1/3) = $3.97; i.e., the median \u2022 at p = 95%, x = (95%*125)^(1/3) =$4.92\nIt's not a distribution of losses, it's a distribution of future values. We always want to be mindful of the difference, sometimes the distribution is losses. As this is future values, the risk tail is \"down\" at p = 5%, it is not \"up\" at p = 95%, so that either a 95% or 0.05 VaR would be realistically referring to the p = 5%. At the same time, there are valid usages of VaR for lower confidence levels, so we could have a 90% VaR (given by the loss of $5.00 minus$2.32) or a even an 80% VaR (given by the loss of $5.00 minus$2.92). I hope that helps!\n\nLast edited:\n\n#### theapplecrispguy\n\n##### New Member\nThank you. Clear and much appreciated!", "date": "2020-01-18 00:00:27", "meta": {"domain": "bionicturtle.com", "url": "https://www.bionicturtle.com/forum/threads/difference-between-probability-density-function-and-inverse-cumulative-distribution-function.9438/", "openwebmath_score": 0.849563717842102, "openwebmath_perplexity": 1333.9619822409604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.97364464868338, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8559765563833629}}
{"url": "https://brilliant.org/discussions/thread/what-happens-to-iirrational-number/", "text": "# what happens to i^(square root of 2)?\n\nI always thought what be in store for us if take irrational number to power of a complex number? we already have natural number, integers, rational numbers, real numbers and complex numbers. Will it create a class of its own?\n\nNote by Shiva Kumar\n9\u00a0months, 2\u00a0weeks ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nyes sir but when the value of k is increased in polar form of (i)^(square root of 2), we do not get the solutions that superimpose the previous solutions. so we instead get infinite no of solutions. it form a circle with magnitude equal to one and center as origin in Argand Plane after k becomes infinite. so does that mean (i)^(any irrational number) forms a circle with magnitude of value one and center at origin.and get same set of solutions. does this mean (i)^any irrational number has same set of values.\n\n- 9\u00a0months, 2\u00a0weeks ago\n\nThe basic way to define powers of complex numbers is via the formula $z^w \\; = \\; e^{w\\log z}$ and so the whole business revolves around the definition of the logarithm of complex numbers. We need $\\log z \\; = \\; \\ln |z| + i\\mathrm{Arg}\\,z$ and here is the real problem. There is no way of defining the argument function continuously (let along differentiably) on the whole complex plane.\n\nSince we want the argument (and hence the logarithm) to be differentiable, it has to be defined on an open set (so that we can consider its derivative at all points), and complex functions are considered on open connected domains. The standard way of doing this for the argument is to cut the plane. This means considering the domain formed by the complex numbers with a straight line out from $$0$$ to infinity removed. The argument can be defined uniquely (and differentiably) on any such domain, and so can the natural logarithm of the modulus (since we have removed $$0$$).\n\nFor example, we could consider the cut plane $$\\mathbb{C} \\backslash (-\\infty,0]$$ consisting of the complex numbers except for the nonpositive reals. This means that $$|z| > 0$$ and $$-\\pi < \\mathrm{Arg}\\,z < \\pi$$ for all $$z$$ in the cut plane, and we can define $$\\log z$$ uniquely (and differentiably) now. This choice is called using the principal branch of the argument. In this case we would have $$\\ln i = \\tfrac12\\pi i$$, and so $$i^\\sqrt{2} = e^{\\frac{\\pi i}{\\sqrt{2}}}$$.\n\nA different cut would be to exclude the nonnegative reals $$[0,\\infty)$$, and decide that $$2\\pi < \\mathrm{Arg}\\,z < 4\\pi$$ for all $$z$$ in this cut plane. This would make $$\\ln i = \\tfrac52\\pi$$, giving a totally different value for $$i^{\\sqrt{2}}$$.\n\nThis is what @Pi Han Goh means about this function being multivalued. There is no way to have a single function that works everywhere.\n\nThe other option is to allow all these values to work at the same time! This means we have to leave the ordinary complex plane, and work with a Riemann surface. Imagine an infinite number of complex planes stacked one on top of another. Cut them all along the positive real axis, and stick the \"first quadrant edge\" of each sheet to the \"fourth quadrant edge\" of the sheet below. The resulting helicoidal shape allows you to consider complex numbers with positive moduli and all possible arguments. The number $$i$$ would lie in one sheet, the number $$ie^{2\\pi i}$$ would lie in the sheet above, the number $$ie^{-6\\pi i}$$ would lie in three sheets below, and so one. You could then define a logarithm on the whole Riemann surface at one go...\n\n- 9\u00a0months, 1\u00a0week ago\n\nAh crap. I can't answer this question either. My first though is Equidistribution theorem, but I'm not entirely confident.\n\nSummoning the great @Mark Hennings again!\n\n- 9\u00a0months, 1\u00a0week ago\n\nNo, $$i^{\\sqrt 2}$$ is just a multivalued complex numbers.\n\nLet $$x = i^{\\sqrt 2}$$, then $$\\ln x = \\sqrt 2 \\cdot i = \\sqrt 2 ( 0 + 1i )$$.\n\nWith $$0 + 1i = \\cos \\left (\\frac\\pi 2 + 2\\pi k \\right) + i \\sin \\left ( \\frac\\pi 2 + 2\\pi k \\right)$$, where $$k$$ is any integer.\n\nCan you finish it off from here?\n\n- 9\u00a0months, 2\u00a0weeks ago", "date": "2018-06-18 00:27:31", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/what-happens-to-iirrational-number/", "openwebmath_score": 0.9641269445419312, "openwebmath_perplexity": 694.9014525476272, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9736446456243804, "lm_q2_score": 0.879146780175245, "lm_q1q2_score": 0.8559765552355414}}
{"url": "https://math.stackexchange.com/questions/7354/sum-limits-n-0-infty-frac12n1-converges/7370", "text": "# $\\sum\\limits_{n=0}^{\\infty} \\frac{1}{(2n+1)!}$ converges?\n\nDetermine whether this series converges or diverges: $$\\sum\\limits_{n=0}^{\\infty} \\frac{1}{(2n+1)!}$$\n\nThought about using the limit theorem or by comparison but am so stuck. any pointers would be appreciated guys\n\n\u2022 What tests do you know for convergence? \u2013\u00a0Qiaochu Yuan Oct 20 '10 at 21:32\n\u2022 Look at the absolute value of the ratio of a term in the series to the subsequent term. You should already know what the possible results imply... \u2013\u00a0Brandon Carter Oct 20 '10 at 23:35\n\nAnother way is\n\nIf $\\displaystyle S_n = 1 + \\frac{1}{3!} + \\dots + \\frac{1}{(2n+1)!}$\n\nWe have that\n\n$\\displaystyle S_n \\le 1 + \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots + \\frac{1}{n(n+1)}$\n\n$\\displaystyle = 1 + (1 - \\frac{1}{2}) + (\\frac{1}{2} - \\frac{1}{3}) + \\dots + (\\frac{1}{n} - \\frac{1}{n+1}) = 2 - \\frac{1}{n+1} < 2$\n\nThus $S_n < 2$\n\nthus we have the $\\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.\n\nI think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is obviously a number, so he doesn't have to prove that the series for e converges.\n\nMaybe you can prove the same by using the ratio test $\\lim_{n \\rightarrow \\infty} \\displaystyle |\\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\\displaystyle \\frac{1}{(2n+1)!}$ and $a_{n+1}=\\displaystyle \\frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\\lim_{n \\rightarrow \\infty} \\displaystyle \\frac{1}{(2n+3)(2n+2)}$ which is 0. According to the ratio test:\n\nIf r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.\n\nTherefore, the series converges.\n\n\u2022 Just out of the curiosity, I plugged the series into Wolfram|Alpha. It gives the very nice result of sinh(1). If anyone knows how to prove that, it would be wonderful to know it. By the way, how can I add an hyperlink in the comment section? \u2013\u00a0Robert Smith Oct 21 '10 at 0:18\n\u2022 $\\sinh(x)=\\frac{\\exp(x)-\\exp(-x)}{2}$. Also, [this](http://functions.wolfram.com/ElementaryFunctions/Sinh/) gives this. \u2013\u00a0J. M. is a poor mathematician Oct 21 '10 at 1:21\n\u2022 @Robert: It is the Taylor series for sinh evaluated at 1. \u2013\u00a0GEdgar Jun 16 '12 at 16:28\n\u2022 Thanks. I didn't know that. \u2013\u00a0Robert Smith Jun 16 '12 at 18:06\n\nThe series you have is\n\n$1 + \\frac{1}{3!} + \\frac{1}{5!} \\ldots$\n\nIf you add the even factorial terms, you get an upper bound i.e.,\n\n$1 + \\frac{1}{3!} + \\frac{1}{5!} \\ldots < \\frac{1}{0!} + \\frac{1}{1!} + \\frac{1}{2!} + \\frac{1}{3!} + \\frac{1}{4!}+ \\frac{1}{5!} \\ldots$\n\nThis can be written more compactly as\n\n$\\sum_{n=0}^\\infty \\frac{1}{(2n+1)!} < \\sum_{n=0}^\\infty \\frac{1}{n!} = e^1$\n\nTherefore the series converges.\n\n\u2022 I don't see the point of doing this when proving that the series for e converges is exactly as hard. \u2013\u00a0Qiaochu Yuan Oct 20 '10 at 22:23\n\u2022 In short, $e=\\cosh\\;1+\\;sinh\\;1$. :) \u2013\u00a0J. M. is a poor mathematician Oct 21 '10 at 0:19\n\u2022 @ Qiaochu Yuan. You make a valid point. I guess the Ratio test(which Robert Smith has mentioned) is the rigorous answer :). \u2013\u00a0svenkatr Oct 21 '10 at 3:53\n\nWe have $$e^{1} = 1 + \\frac{1}{1!} + \\frac{1}{2!} + \\frac{1}{3!} + \\cdots$$ and $$e^{-1} = 1 - \\frac{1}{1!} + \\frac{1}{2!} - \\frac{1}{3!} + \\cdots$$\n\nSubtracting these two we get $$e - e^{-1} = 2 \\cdot \\Bigl( 1 + \\frac{1}{3!} + \\frac{1}{5!} + \\cdots \\Bigr)$$ Therefore the series converges to $$\\frac{e-e^{-1}}{2} = \\sum\\limits_{n=0}^{\\infty} \\frac{1}{(2n+1)!}$$\n\n\u2022 The question was \u00abdoes the series converge?\u00bb Your answer begins by asserting convergence of a series of equivalent difficulty :) \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez Oct 21 '10 at 14:12\n\nCan you bound the series from above by one that you know converges? The factorials grow very fast, so you should be able to.\n\nTo elaborate on the first answer given to this question by Ross Millikan.\n\n$$\\sum_{n=0}^\\infty \\frac{1}{(2n+1)!} = 1 + \\sum_{n=1}^\\infty \\frac{1}{(2n+1)!}$$\n\n$$< 1 + \\sum_{n=1}^\\infty \\frac{1}{4^n} = \\frac{4}{3}, \\quad \\textrm{ as } \\frac{1}{(2n+1)!} < \\frac{1}{4^n} \\textrm{ for } n \\ge1.$$\n\nHence by the comparison test the series converges.\n\nComparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that\n\n$$R_N < \\frac{1}{(2n+3)!} \\left( 1 + \\frac{1}{(2n+3)^2} + \\frac{1}{(2n+3)^4} + \\cdots \\right).$$\n\nMETHOD I\n\nWe may simply resort to the Basel problem and get the inequality: $$0<\\sum_{k=0}^{\\infty}\\frac{1}{(1+2k)!}\\leq\\sum_{k=0}^{\\infty}\\frac{1}{(1+k)^2}=\\frac{\\pi^2}{6}$$\n\nMETHOD II\n\nAccording to Taylor's expansion we have that:\n\n$$\\sinh(x) = \\sum_{k=0}^{\\infty}\\frac{x^{1+2k}}{(1+2k)!}$$\n\nFor $x=1$ we get that the value of the series is $\\sinh(1)$. The series converges.\n\nQ.E.D.\n\nThis is how sometimes we can extract the exact closed form of some series :\n\n$f(x) = \\sum_{n=0}^{\\infty} \\frac{x^{2n+1}}{(2n+1)!}$\n\nThe radius of convergence is $\\infty$\n\n$f \\in C^{\\infty}(\\mathbb{R})$\n\n$f''(x) = f(x)$\n\nsolving our equation, we get $f(x) = ae^{\\lambda_1x}+be^{\\lambda_2x}$\n\n$\\lambda^2-1=0$\n\n$\\lambda \\in \\{-1,1\\}$\n\n$f(x) = ae^{-x}+be^{x}$\n\n$f(0)= 0 \\implies a+b=0 \\implies a=-b \\implies f(x) = a(e^{x}- e^{-x})$\n\n$f'(0) = 1 \\implies a=\\frac{1}{2}$\n\n$f(x) = \\frac{e^{x}-e^{-x}}{2}$\n\nso our series $\\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)!}$ does converge to $\\sinh(1)$\n\n\u2022 This only proes that if the series converges, then its sum is $\\sinh(1)$. \u2013\u00a0Jos\u00e9 Carlos Santos Jul 26 '18 at 20:24\n\u2022 Yes that is true, but I wanted to say, regarding to other answers, that I would find the limit. So I didn't repeat any other done process here, that is namely true. Anyway that is the same, effectively the sum itself implies the convergence of that series to $\\sinh(x)$. So you had not always to prove the convergence of series, since it does have a closed form. the closed form itself implies the convergence, that is so understandable. \u2013\u00a0 \u0410\u0445\u043c\u0435\u0434 Jul 26 '18 at 20:26\n\u2022 your implication is not true for closed form and finite values. that does mean, the sum does converge for all $x\\in \\mathbb{R}$ to $\\sinh(x)$ which is finite. \u2013\u00a0 \u0410\u0445\u043c\u0435\u0434 Jul 26 '18 at 20:33", "date": "2019-10-15 04:07:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/7354/sum-limits-n-0-infty-frac12n1-converges/7370", "openwebmath_score": 0.925818920135498, "openwebmath_perplexity": 421.98663843244293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446456243805, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8559765475281003}}
{"url": "https://www.physicsforums.com/threads/statement-proof.176659/", "text": "# Statement Proof.\n\n1. Jul 11, 2007\n\n### linuxux\n\nMy question is this:\n\nTheorem 1.4.13 part (ii) says: If $$A_n$$ is a countable set for each $$n \\in \\mathbf{N}$$, then $$\\cup^{\\infty}_{n=1} A_n$$ is countable.\n\nI can't use induction to prove the validity of the theorem, but the question does say how does arranging $$\\mathbf{N}$$ into a 2-d array:\n\n1 3 6 10 15 ...\n2 5 9 14 ...\n4 8 13 ...\n7 12 ...\n11 ...\n\nlead to the proof of part (ii) of Theorem 1.4.13?\n\nso obviously it has something to do with the (x,y) co-ordinate system of the array, but I nt sure how it leads to the proof.\n\nLast edited: Jul 11, 2007\n2. Jul 11, 2007\n\n### CRGreathouse\n\nYou can just show a function $f: A\\to\\mathbb{N}$.\n\n$$f(0)=A_1(0)$$\n$$f(1)=A_1(1)$$\n$$f(2)=A_2(0)$$\n$$f(3)=A_1(2)$$\n$$f(4)=A_2(1)$$\n$$f(5)=A_3(0)$$\n$$f(6)=A_1(3)$$\n. . .\n\n3. Jul 11, 2007\n\n### linuxux\n\nbut what is the significance of putting it into a 2-d array? i could just as easily say: 1, 2, 3, 4, ... and call it a 1-d array.\n\n4. Jul 11, 2007\n\n### CRGreathouse\n\n$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...\n$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...\n$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...\n$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...\n$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...\n... ... ... ...\n\n5. Jul 11, 2007\n\n### mathwonk\n\nthis proof is due to cantor, 100 years ago. just start couting from the upper left corner, counting by sw/ne diagonals.\n\n6. Jul 12, 2007\n\n### matt grime\n\nTo demonstrate a bijection between N (the indices you're inserting) and the countable union of N (the locations in the doubly infinite array). I.E the whole point of the question.\n\n7. Jul 14, 2007\n\n### linuxux\n\nbut how is this a 1-1 and onto function? how is it that $$A_n$$ is a function itself? the cantor theorem is the next section in my text so i can't apply that theory now. i understand that f(n) is a function, but why also $$A_n$$?\n\nLast edited: Jul 14, 2007\n8. Jul 14, 2007\n\n### CompuChip\n\nHere's a different explanation (sometimes I see that the understanding of the principle depends on the way it is explained, so maybe this will help you - maybe it will make it less clear to you, then please ignore this )\nSuppose we put two copies of the natural numbers next to each other. We want to show that, doing this, the total number we get is still countable (that there are \"as many\" elements in this string of numbers as there are natural numbers, by assigning a natural number to each of the elements).\nClearly, just counting them in a line won't work. Then we assign each natural number to itself in the first copy, but when we're through with that one, we don't have any natural numbers left for the second copy. So what we do is: assign 1 to 1 from the first copy, 2 to 1 from the second copy, 3 to 2 from the first copy, 4 to 2 from the second copy, etc. - so\n$$2n - 1 \\mapsto n \\text{ from the first copy of } \\textbf{N}, 2n \\mapsto n \\text{ from the second copy of } \\textbf{N}.$$.\n\nNow consider the Cartesian product $$\\textbf N \\times \\textbf N$$, consisting of all pairs of natural numbers (a, b). How do you prove they are countable. One way would be to start by assigning n to (n, 1), but when we're done, we won't have any natural numbers left to assign to (n, 2) and (n, 3), etc. So here's the trick: we organize the pairs like\n(1,1) (1,2) (1,3) ...\n(2,1) (2,2) (2,3)\n(3,1) (2,3) (3,3)\n....\nand now we can map them like\n1 2 6 7 ...\n3 5 8 ...\n4 9 10\nwithout running out of numbers too soon.\n\nIn the same way, you can prove that $\\textbf{N}^k$ is countable for any natural number k, as well as $\\frac{1}{2} \\textbf{N}$ (all halfintegers 0, 1/2, 1, 3/2, ...) and the integers $\\textbf{Z}$ (it's basically two copies of the naturals, like in my first example, with a zero added) and even that the rationals $\\textbf{Q}$ are countable. And, also the question from your initial post can be solved like this.\n\n9. Jul 14, 2007\n\n### linuxux\n\nthanks, i understand the example you showed. and to think, this is just an introductory text and i'm self-studying!...\n\nLast edited: Jul 14, 2007\n10. Jul 14, 2007\n\n### linuxux\n\n(1,1) (1,2) (1,3) ...\n(2,1) (2,2) (2,3)\n(3,1) (2,3) (3,3)\n\nthis part i understand. each (a,b) has a particular n mapped to it, thus (a,b)~N. so i got that.\n\nbut, looking at this:\n$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...\n$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...\n$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...\n$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...\n$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...\n\ni understand this to mean that the number of sets of $$A_n$$ is countable, not that once the union is performed on the sets the elements of the union will be countable. I had a previous problem where i had to show that the union of $$A_1$$ and $$A_2$$ was countable, i did this by defining a function that chose the minimum element first in $$A_1$$ and then in $$A_2$$, then choosing the next smallest element in both, and so on, thus i was able to show $$A_1\\cup{}A_2$$~$$\\mathbb{N}$$. So in that exercise i addressed the elements in the sets, not the sets themselves. how does the section in red address the elements in the sets? is that even necessary?\n\n**********\n\nwait a minute: i know each $$A_n$$ is countable, so each $$A_n(a\\in\\mathbb{N})$$ addresses the elements in $$A_n$$. now i see.\n\nLast edited: Jul 14, 2007\n11. Jul 15, 2007\n\n### matt grime\n\nBy construction: do you write down the same number twice or fail to fill in a box?\n\nIt isn't a function.\n\n12. Jul 15, 2007\n\n### linuxux\n\nwouldn't it qualify as a function since $$A_n$$ is countable thus $$A_n \\mapsto \\mathbb{N}$$?\ni thought $$\\mapsto$$ represents some type of function. does it not?\n\ni also have another question, what happens if a particular $$A_n$$ is finite?\n\nLast edited: Jul 15, 2007\n13. Jul 15, 2007\n\n### CompuChip\n\nYeah, so for each n we can assign a natural number to each element from $$A_n$$, and in the end, have neither elements of $$A_n$$ nor natural numbers \"left\" (that is, there is a bijection). The claim is, that the union is countable, so there is a bijection from the naturals to the union. What the 2-D array tries to visualize, is the following: since each $$A_n$$ is countable, we can denote the elements by $$A_n(k)$$ (with k = 1, 2, 3, ...). Now you can order the elements in the union like this:\n$$A_1(1) \\, A_2(1) \\, A_3(1) \\, \\cdots$$\n$$A_1(2) \\, A_2(2) \\, A_3(2) \\, \\cdots$$\n$$A_1(3) \\, A_2(3) \\, A_3(3) \\, \\cdots$$\nand use a diagonal counting argument, like before.\n\nAnother way is this: each element from the union is indexed by two numbers (n, k) (assuming the union is disjoint, as we implicitly did before). So it's really trivial to write down a bijection from this union to $$\\textbf{N} \\times \\textbf{N}$$. As shown in my earlier post, the latter is countable (there is a bijection to $$\\textbf{N}$$, so by composing the bijections you get one from the union to $$\\textbf{N}$$. Actually, it also uses the diagonal counting argument (to show that $$\\textbf{N} \\times \\textbf{N}$$ is countable) but perhaps this approach is more intuitive (and once you've proven this result about $$\\textbf{N} \\times \\textbf{N} \\times \\cdots \\times \\textbf{N}$$ you can use it and prove countability of other things, like this union of $$A_n$$, without explicitly invoking a diagonal counting argument).\n\nLast edited: Jul 15, 2007\n14. Jul 15, 2007\n\n### CompuChip\n\nDoes it matter? If you have a set in the union, that is not countably infinite, but finite, will the union become bigger (so big, it will become uncountable?)\n\nIf you want to do it officially, consider this: we said a set is countable if we can assign the numbers 1, 2, 3, ... to it. But of course, we can use another offset, and assign 2, 3, 4, ... to it. Or 154842, 154843, 154844, ...\nSo you could of course first count off the finite sets in the union (in order, no need for diagonals here), and then the infinite ones with the diagonal argument, but using a certain offset.\n\n15. Jul 15, 2007\n\n### linuxux\n\nthanks a million man!\n\n16. Jul 15, 2007\n\n### CompuChip\n\nMost welcome, this stuff is fun and moreover, it's important basics.\n\nLater you will encounter more diagonal arguments, for example: if you want to prove that the real numbers are uncountable, or if you get to work with converging sequences.\n\n17. Jul 16, 2007\n\n### matt grime\n\nYou said A_n was a function, not that |-> was a function. Don't be sloppy.", "date": "2017-01-18 10:17:22", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/statement-proof.176659/", "openwebmath_score": 0.8634294867515564, "openwebmath_perplexity": 505.59695090201495, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446479186301, "lm_q2_score": 0.8791467627598856, "lm_q1q2_score": 0.8559765402961522}}
{"url": "https://math.stackexchange.com/questions/3049427/the-converse-of-nilpotent-elements-are-zero-divisors", "text": "# The converse of \u201cnilpotent elements are zero-divisors\u201d\n\nFor commutative rings $$A$$ with identity $$1\\ne0$$, nilpotent elements are zero-divisors. The converse is false, i.e. there is a commutative ring $$A$$ with identity $$1\\ne0$$ and a zero-divisor $$x$$ in $$A$$ which is not nilpotent. Where is such an example?\n\nIs it meaningful to ask the \"percentage\" of commutative rings $$A$$ with identity $$1\\ne0$$ for which the converse holds, i.e. zero-divisors are nilpotent?\n\n\u2022 Have you thought about the integers modulo $n$? \u2013\u00a0Lord Shark the Unknown Dec 22 '18 at 12:45\n\u2022 You can also search yourself at this site, e.g. here, for useful links. \u2013\u00a0Dietrich Burde Dec 22 '18 at 12:51\n\u2022 @Lord Shark the Unknown: Thank you. Right, in $\\mathbb{Z}/10\\mathbb{Z}$, $2$ is a zero-divisor but not nilpotent. How did you came up with this example? How about the next question? \u2013\u00a0user584333 Dec 22 '18 at 12:52\n\u2022 Sai, have a look at this post, which gives an answer which rings have this property. \u2013\u00a0Dietrich Burde Dec 22 '18 at 12:56\n\u2022 @Dietrich Burde: Thank you for the reference. \u2013\u00a0user584333 Dec 22 '18 at 13:01\n\nAn example is in the integers mod $$6$$, where $$2\\cdot 3=0$$, but no power of either of these individually is zero.\n\nIf we had some sort of measure on a set of rings (which could not possibly be all rings, because there are too many) with finite total measure, we could ask about proportion. I know of no such commonly used measure, but it's possible such a thing has been considered.\n\nIn my intuition, the proportion would be small, but I can't think of a quick reason to see why, other than that being nilpotent seems like a special property while being a zero divisor seems very common. For example, the only time this is true for rings of integers modulo $$n$$ is when $$n$$ is a prime power, and there are vanishingly few of those compared to all integers.\n\nIf you look at the answer by rschwieb in the question Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? linked above in a comment by Dietrich Burde, which is not the accepted answer, it characterizes these rings. A ring has this property if and only if it is the quotient of an arbitrary commutative ring by a primary ideal. An ideal $$I$$ is said to be primary if whenever we have that $$a,b\\in R$$ satisfy $$ab\\in I$$, then we have either $$a\\in I$$ or $$b^n\\in I$$ for some $$n>0$$. This answers part of your question. It is nicely illustrated in the $$\\Bbb Z_n$$ case: an ideal $$n\\Bbb Z$$ of $$\\Bbb Z$$ is primary if and only if $$n=p^k$$ for some prime $$p$$ and integer $$k>0$$.\n\nSo rings like these are actually \"close\" to integral domains, which are quotients of arbitrary commutative rings by prime ideals. Again, no quantitative argument for why they should be rare, only an intuitive one.\n\n\u2022 Couldn't one equip a ring $R$ with the Zariski topology, then produce a Borel sigma algebra, and equip it with a pre-measure, which can then be extended to a measure by Caratheodory's extension theorem, to at least begin discussing \"percentages\" when the ring itself is of finite measure? \u2013\u00a0Chickenmancer Dec 22 '18 at 21:44\n\u2022 @Chickenmancer that wouldn't give a percentage of rings, I suspect you mean to give a percentage of zero divisors that are nilpotent? \u2013\u00a0jgon Dec 22 '18 at 21:50\n\u2022 Ah, I misinterpreted the question as \"what percentage of a ring $A$.\" Thanks for clarifying, @jgon. \u2013\u00a0Chickenmancer Dec 22 '18 at 22:15\n\u2022 Curious about the downvote. An explanation would be helpful. \u2013\u00a0Matt Samuel Dec 23 '18 at 17:43\n\n\nOne perspective on what proportion of rings have no nonnilpotent zero divisors is the following. Note that this is really rough first thought.\n\nFirst let's be careful though, rings with no zero-divisors trivially have this property, so I'll try to consider what rings with zero-divisors have the property that all zero-divisors are nilpotent.\n\nStep 1: Set up a space to parametrize a nice class of rings\n\nLet $$k$$ be an algebraically closed field. Let $$n,d_1,d_2 > 0$$. Let $$x=(x_1,x_2,\\ldots,x_n)$$ be coordinates on $$\\AA^n_k$$. Let $$r_1=\\binom{n+d_1}{d_2}$$, which is the number of monomials in $$k[x_1,\\ldots,x_n]$$ of degree at most $$d_1$$. Thus there are $$r_1$$ multiindices $$I$$ of degree at most $$d_1$$. Let $$r_2=\\binom{n+d_2}{d_2}$$ as well. Let $$c=(c_I)$$ be coordinates on $$\\PP^{r_1-1}_k$$, and let $$d=(d_J)$$ be coordinates on $$\\PP^{r_2-1}_k$$ as $$I$$ ranges over multiindices of degree at most $$d_1$$ and $$J$$ ranges over multiindices of degree at most $$d_2$$.\n\nThen consider the subvariety, $$V$$, of $$\\AA^n_k\\times_k\\PP^{r_1-1}_k\\times_k \\PP^{r_2-1}_k$$ cut out by the polynomial $$f_{c,d}(x)=g_c(x)g_d(x):=\\left(\\sum_I c_Ix^I\\right)\\left(\\sum_J d_Jx^J\\right).$$\n\n$$V$$ is equipped with a natural map $$V\\to \\PP^{r_1-1}_k\\times_k\\PP^{r_2-1}$$, and the fiber over a point $$(a,b)$$ in $$\\PP^{r_1-1}_k\\times_k\\PP^{r_2-1}_k$$ is the subvariety of $$\\AA^n_k$$ cut out by the degree at most $$d_1+d_2$$ reducible polynomial $$f_{a,b}(x)=g_a(x)g_b(x).$$\n\nThis subvariety of $$\\AA^n_k$$ can be thought of as corresponding to the ring $$k[x_1,\\ldots,x_n]/(g_ag_b)$$, so we can think of our variety $$V$$ as parametrizing a certain nice class of rings all (except for a Zariski closed subset where $$g_a=0$$ or $$g_b=0$$) of which are guaranteed to have zero-divisors, since $$g_a$$ and $$g_b$$ are zero divisors.\n\nStep 2: Investigate the points corresponding to rings with the desired property and related properties\n\nWe can then ask if we can characterize the subset of $$V$$ corresponding to rings in which all zero-divisors are nilpotent.\n\nWell, what are the zero-divisors in $$k[x_1,\\ldots,x_n]/(f_{a,b})$$? Since $$k[x_1,\\ldots,x_n]$$ is a UFD, handily enough, the zero-divisors are precisely the factors of $$f_{a,b}$$. Moreover $$k[x_1,\\ldots,x_n]/(f_{a,b})$$ has nilpotents if and only if $$f_{a,b}$$ is not square free (take the radical of $$f_{a,b}$$, it will be nilpotent if and only if it is nonzero, if and only if $$f_{a,b}$$ is not square free). However the ring $$k[x_1,\\ldots,x_n]/(f_{a,b})$$ satisfies our property that all zero-divisors are nilpotent if and only if $$f_{a,b}$$ is a power of an irreducible polynomial.\n\nStep 3: Conclude\n\nNote that if $$f_{a,b}$$ is square-free if and only if it is relatively prime to $$f_{a,b}'$$, which is true if and only if $$\\Disc(f_{a,b})\\ne 0$$. Thus every point $$(a,b)\\in\\PP^{r_1-1}_k\\times_k\\PP^{r_2-1}_k$$ corresponding to a ring with any nilpotents at all, let alone one in which every zero-divisor is nilpotent, satisfies a polynomial equation, $$\\Disc(f_{a,b})=0$$. Thus rings in our parametrized class with any nilpotents at all correspond to a (proper) Zariski closed subset of $$\\PP^{r_1-1}_k\\times_k\\PP^{r_2-1}_k$$, which is irreducible (see here).\n\nThus rings in our parametrized class satisfying your property (all zero-divisors are nilpotent) are contained in a codimension 1 class of rings (any nilpotents). Hence almost all rings in our parametrized class have zero-divisors that are not nilpotent.\n\nNotes\n\nThis is really rough. The parametrized class of rings is a very specific, very nice subset of $$k$$-algebras. Nonetheless, hopefully it will give you (or other readers) intuition on why very few rings should have the property you want (as long as they have some zero divisors of course).\n\nI needed to eliminate integral domains because I chose to parametrize hypersurfaces, and most hypersurfaces are irreducible. (Check out the link, Qiaochu Yuan gives a really nice quick proof of this fact). There's a reasonable chance that I wouldn't have needed to eliminate integral domains if I'd chosen e.g. codimension 2 subvarieties, but those are much harder to characterize.\n\n\u2022 I was thinking of excluding integral domains as well, but notice for integers that isn't necessary. A ring of integers modulo $n$ is an integral domain only if $n$ is prime. \u2013\u00a0Matt Samuel Dec 23 '18 at 0:41\n\u2022 @MattSamuel, I originally didn't think to do so, but unfortunately, the particular class of rings I chose to parametrize requires me to do so, since most hypersurfaces in dimensions $\\ge 2$ are irreducible. I suspect I wouldn't have to eliminate integral domains if I'd chosen e.g. codimension 2 subspaces, but those are harder to work with. \u2013\u00a0jgon Dec 23 '18 at 1:07\n\u2022 I'll add a link to a great answer by Qiaochu Yuan justifying the claim that most hypersurfaces are irreducible when I eventually get back on a computer and I remember. \u2013\u00a0jgon Dec 23 '18 at 1:08\n\nAnother class of examples are matrix rings over a field. A $$n\\times n$$ matrix $$A$$ is a zero-divisor if and only if $$A$$ is not invertible.\n\nCertainly if $$A$$ is a zero divisor then $$A$$ cannot be invertible. If $$A$$ is not invertible then $$\\ker A \\ne 0$$ and taking any non-zero matrix $$B$$ with $$\\operatorname{col}B \\subseteq \\ker A$$ you have $$AB = 0$$. For example, let $$v$$ be a non-zero vector with $$Av = 0$$ and let $$B$$ be the matrix whose columns are all $$v$$.\n\nThe nilpotent matrices are rare among zero-divisors. Specifically, nilpotent matrices are solutions to the equation $$A^n = 0$$. As a rule, there are always more non-solutions to a polynomial equation than there are solutions.\n\nIn precise terms, we look at the zero set $$\\{A : \\det A = 0\\}$$ (the set of zero-divisors). This zero-set is a manifold if the underlying field is $$\\mathbf{C}$$ and is a variety in the general case. The non-nilpotents are dense in the Zariski topology looking over a general ring, and dense in the classical topology looking over $$\\mathbf C$$.", "date": "2019-11-20 18:17:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3049427/the-converse-of-nilpotent-elements-are-zero-divisors", "openwebmath_score": 0.8839941620826721, "openwebmath_perplexity": 193.22849541817598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708039218115, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8559686455971849}}
{"url": "https://www.physicsforums.com/threads/positive-or-negative-remainder.896236/", "text": "# Positive or negative remainder\n\nIs 23 = 5(-4)-3 gives a remainder -3 when divided by 5 ? is this statement true ? some of my colleagues said that remainder cannot be negative numbers as definition but I am doubt that can -3 be a remainder too?\n\nfresh_42\nMentor\nIs 23 = 5(-4)-3 gives a remainder -3 when divided by 5 ? is this statement true ? some of my colleagues said that remainder cannot be negative numbers as definition but I am doubt that can -3 be a remainder too?\nUsually we consider entire equivalence classes in such cases: Every single element of ##\\{\\ldots -13, -8, -3, 2, 7 , 12, \\ldots\\}## belongs to the same remainder of a division by ##5##. We then define all five possible classes\n##\\{\\ldots -15, -10, -5, 0, 5, 10, \\ldots\\}##\n##\\{\\ldots -14, -9, -4, 1, 6, 11, \\ldots\\}##\n##\\{\\ldots -13, -8, -3, 2, 7, 12, \\ldots\\}##\n##\\{\\ldots -12, -7, -2, 3, 8, 13, \\ldots\\}##\n##\\{\\ldots -11, -6, -1, 4, 9, 14, \\ldots\\}##\nas elements of a new set with five elements ##\\{ \\; \\{\\ldots -15, -10, -5, 0, 5, 10, \\ldots\\}\\, , \\, \\{\\ldots -14, -9, -4, 1, 6, 11, \\ldots\\}\\, , \\, \\ldots \\}##.\n\nThis notation is a bit nasty to handle, so we choose one representative out of every set. E.g. ##\\{[-15],[-9],[12],[3],[-1]\\}## could be chosen, but this is still a bit messy to do calculations with. So the most convenient representation is ##\\{[0],[1],[2],[3],[4]\\}## with the non-negative remainders smaller than ##5##. However, this is only a convention. ##-3## is a remainder, too, belonging to the class ##[2]##. So the answer to your questions is: The statement is true, as all integers are remainders.\n\nmfb\nMentor\nThe remainder is usually required to be between 0 and N-1 inclusive. 23 and -2 (not -3) are in the same equivalence class. This can also be written as 23 = -2 mod 5.\n\nThe remainder is usually required to be between 0 and N-1 inclusive. 23 and -2 (not -3) are in the same equivalence class. This can also be written as 23 = -2 mod 5.\nSorry it should be -23 = 5(-4) - 3 , so in conclusion is, this statement true ?\n\njbriggs444\nHomework Helper\nSorry it should be -23 = 5(-4) - 3 , so in conclusion is, this statement true ?\n\"-23 divided by 5 is -4 with a remainder of -3\". I would consider that statement true.\n\"-23 divided by 5 is -5 with a remainder of 2\". I would also consider that statement to be true.\n\nThe convention you use for integer division will determine which of those statements is conventional and which is unconventional.\n\nIn many programming languages, integer division follows a \"truncate toward zero\" convention. For instance, in Ada, -23/5 = -4. The \"rem\" operator then gives the remainder. So -23 rem 5 = -3.\n\nIf one adopts a convention that integer division (by a positive number) truncates toward negative infinity then one would get a different conventional remainder. -23/5 would be -5 and -23 mod 5 would be +2. The Ada \"mod\" operator uses this convention.\n\nIn mathematics, one typically adopts the line of reasoning given by @fresh_42 in post#2 above. The canonical exemplar in the equivalence class of possible remainders is normally the one in the range from 0 to divisor - 1.", "date": "2021-01-18 18:05:26", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/positive-or-negative-remainder.896236/", "openwebmath_score": 0.8385798335075378, "openwebmath_perplexity": 705.3507974379489, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9773707993078212, "lm_q2_score": 0.8757869965109765, "lm_q1q2_score": 0.8559686368033291}}
{"url": "https://www.physicsforums.com/threads/help-with-circular-motion.56897/", "text": "Homework Help: Help with circular motion\n\n1. Dec 15, 2004\n\nnewcool\n\nHi, I recieved this problem today. If you put a small marble of mass m on top of a large round object that has a radius r, and then release the marble, when will it come off of the side so that the large round object will not be in contact with the marble?\n\nThe answer is arccosine of 2/3. I have tried making free body diagrams of the object when the normal force is 0 and got this equation: a * cosin(theta) = g.\n\nI am stuck as to how to get the right answer.\n\nThanks for any help\n\n2. Dec 15, 2004\n\nStaff: Mentor\n\nAs the marble rolls down the sphere, it will maintain contact as long as there is sufficient force to produce the required centripetal acceleration. Ask yourself: What provides the centripetal force? How does the required centripetal acceleration depend on the the angle that the marble makes with the vertical?\n\n3. Dec 20, 2004\n\nnewcool\n\nI made a free body diagram of when the sphere has a normal force of 0. I got one force,F_c pointing in towards the center and another force mg pointing down. Taking the y value of F_c, i got that:\n$$F_c*cos(\\theta) + mg = ma$$\n$$mv^2/r * cos(\\theta) + mg = mv^2/r$$\n$$v^2/r * cos(\\theta) + g = v^2/r$$\n\nI tried a different approach using the fact that the sum of all energies is equal to 0.\n$$0 = \\Delta K + \\Delta U_g$$\n$$0 = 1/2 *mv^2 + -mgh$$\n$$mgh = 1/2 *mv^2$$\n$$gh = 1/2 *v^2$$\n$$h = r - r*cos(\\theta)$$\n\nso $$gr(1 - cos(\\theta)) = 1/2 * v^2$$\n\nHowever, I have gotten nowhere with these equations. Any input on what I did wrong would be appreciated.\n\nThanks\n\n4. Dec 20, 2004\n\nPyrrhus\n\nOk the answer is acrosinus of 2/3 = 48.2 degrees.\n\nForce analysis\n\n$$n - mg \\cos \\theta = -m \\frac{v^2}{R}$$\n\nThe object will fall when n = 0 so\n\n$$v^2 = Rg \\cos \\theta$$\n\nwhen it loses contact with the surface\n\nAssuming isolated system\n\nI can use Conservation of mechanical energy\n\n$$K + \\Omega = K_{0} + \\Omega_{0}$$\n\n$$\\frac{1}{2}mv^2 + mgR \\cos \\theta = 0 + mgR$$\n\nPlugging our speed when it loses contact\n\n$$\\frac{1}{2}mRg \\cos \\theta + mgR \\cos \\theta = mgR$$\n\nwhich gives:\n\n$$\\cos \\theta = \\frac{2}{3}$$\n\n$$\\theta = 48.2^{o}$$\n\nwhich is the angle when it will lose contact with the surface.\n\nThat's of course assuming an isolated system.\n\n5. Dec 20, 2004\n\nStaff: Mentor\n\nOne big problem here: You seem to be treating the \"centripetal force\" as though it were a separate force (like friction or weight). Not so! \"Centripetal\" just means \"towards the center\": The centripetal force is just those real forces that act towards the center. The only forces acting on the mass m are: (1) mg, acting down, and (2) N, the normal force, acting normal to the surface.\n\nSo what's the centripetal force? Just the components of those forces acting towards the center of the circular motion, thus producing the centripetal acceleration:\n$mg cos\\theta - N = F_c = mv^2/r$\nOf course, you'll set the normal force to zero, so:\n$mg cos\\theta = mv^2/r$\n\nExactly correct. But that's not a \"different approach\"--it's a necessary part of solving this problem!\n\nNow just combine the two equations and solve for $\\theta$.\n\n6. Dec 20, 2004\n\nPyrrhus\n\nToo bad, i just did the work\n\nWell, also read Doc Al's explanation, it's quite good\n\n7. Dec 20, 2004\n\nnewcool\n\nThanks for the help everyone. Just one question, Doc, how did you get that\n$$mg cos\\theta - N = F_c = mv^2/r$$\n\nIsn't mg pointing straight down so the component that points towards the center is\n$$mg/cos\\theta$$\n\n8. Dec 20, 2004\n\nPyrrhus\n\nWe got a vector\n\n$$\\vec{R}$$\n\nwith y component\n\n$$R \\cos \\theta$$\n\nand x component\n\n$$R \\sin \\theta$$\n\nwhere do you get $\\frac{mg}{\\cos \\theta}$ ???\n\n9. Dec 20, 2004\n\nPyrrhus\n\nI think you got a misconception with centripetal force, Centripetal force is a role asigned to forces, because they act towards the center. The forces acting on the body are normal and the weight, and the components acting towards the center are equal to $m \\frac{v^2}{r}$.\n\n10. Dec 21, 2004\n\nStaff: Mentor\n\nYes, mg points straight down. Therefore its component towards the center will be $mg cos\\theta$, not $mg/cos\\theta$. (Draw yourself a picture.)\n\nI believe you are thinking like this: That mg is the vertical component of the \"centripetal force\" ($F_c$), so $F_c cos\\theta = mg$ ==> $F_c = mg/cos\\theta$. This is incorrect thinking.\n\nOne thing you must realize is that \"centripetal\" is just a decription of the direction that a force has: it just means \"towards the center\". (Another term used is \"radial\".) It is not a kind of force. It's just like describing a force as a horizontal force.\n\nThink like this: The mass m must have an acceleration towards the center since it moves in a circle. So, let's apply Newton's 2nd Law in that radial (or centripetal) direction. As Cyclovenom and I have explained, the only forces acting on the mass are gravity and the normal force. We know that N points away from the center. What's the component of gravity (mg) towards the center? $mg cos\\theta$. So:\n$$F_{towards-center} = ma_{towards-center}$$\n$$mg cos\\theta - N = mv^2/r$$\n\nMake sense?\n\n11. Dec 21, 2004\n\nnewcool\n\nThanks for all the help, Doc and Cyclove, I understand it now", "date": "2018-12-12 17:13:20", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/help-with-circular-motion.56897/", "openwebmath_score": 0.743645191192627, "openwebmath_perplexity": 693.0273276151378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707986486796, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8559686330574059}}
{"url": "https://math.stackexchange.com/questions/2835377/combination-notation-vs-binomial-coefficient-formula", "text": "# Combination notation vs. Binomial Coefficient Formula\n\nI'm studying probability and statistics and had a question regarding notation.\n\nI noticed that combinations and the binomial coefficient are essentially the same thing, that is:\n\n$$\\binom{n}{k}\\ =\\ _nC_k\\ =\\ \\frac{n!}{(n-k)!k!}$$\n\nBut I was wondering, is there a particular difference between the two that people should be aware of? For example, are there certain use cases where one is preferred over the other?\n\nThank you.\n\n\u2022 They are exactly the same thing, just different notation. I think the most popular (at least from what I see) is the $\\binom{n}{k}$ notation, however I have seen $C^n_r$ used in probability courses. \u2013\u00a0Dave Jun 29 '18 at 2:48\n\u2022 Also ${^n\\mathrm C_r}$ \u2013\u00a0Graham Kemp Jun 29 '18 at 3:31\n\u2022 Also $_r\\text{C}^n$ \u2013\u00a0Dzoooks Jun 29 '18 at 4:04\n\nAll three expressions mean when dealing with combinations the same. But there are some aspects which should be considered.\n\nWe often find the binomial coefficients $\\binom{n}{k}$ resp. $_nC_k$ defined by factorials. \\begin{align*} \\binom{n}{k}:=\\frac{n!}{k!(n-k)!}\\qquad\\qquad\\text{resp.}\\qquad\\qquad _nC_k:=\\frac{n!}{k!(n-k)!} \\end{align*} From this point of view the factorials $n!$ can be seen as basic building blocks for the shorthand notations $\\binom{n}{k}$ and $_nC_k$. Since using factorials is more fundamental than the other two representations I will consider in the following only $\\binom{n}{k}$ and $_nC_k$.\n\nHistorical aspects:\n\n\u2022 C. Jordan writes in his classic Calculus of Finite Differences (1939)\n\n\u2022 ($\\mathrm{\\S}$ 22): Since the binomial coefficient is without doubt the most important function of the Calculus of Finite Differences it was necessary to adopt some brief notation for this function. We accepted above the notation of J. L. Raabe [Journal f\u00fcr reine und angewandte Mathematik 1851, Vol. 42, p. 350] which is most in use now, putting\n\n\\begin{align*} \\binom{x}{n}=\\frac{x(x-1)(x-2)\\cdots(x-n+1)}{1\\cdot2\\cdot3\\cdots n} \\end{align*}\n\n\u2022 ($\\mathrm{\\S}$ 22, footnote 18): Euler first used the notation $\\left[\\frac{x}{n}\\right]$ in Acta Acad. Petrop.V, 1781 and then $\\left(\\frac{x}{n}\\right)$ in Nova Acta Acad. Petrop. XV. 1799-1802. Raabe's notation $\\binom{x}{n}$ is a slight modification of the second. It is used for instance in:\n\u2022 Bierens de Haan, Tables d'Int$\\mathrm{\\acute{e}}$grales d$\\mathrm{\\acute{e}}$finies, Leide, 1867\n\u2022 Hagen, Synopsis Vol. I. p. 57, Leipzig, 1891,\n\u2022 Pascal, Repertorium Vol. I, p. 47, Leipzig, 1910,\n\u2022 Encyclop\u00e4die der Math. Wissenschaften, 1898-1930,\n\u2022 L. M. Milne Thomson, Calculus of Finite Differences, 1933,\n\u2022 G. H. Hardy, Course of Pure Mathematics, p. 256, 1908\n\nEuler's notation is also cited in\n\n\u2022 A History of Mathematical Notations by F. Cajori, which also provides some information about the use of $_nC_k$-like notations in $\\mathrm{\\S}$ 451:\n\n\u2022 George Peacock (Treatise of Algebra, 1830) introduces $C_r$ for the combinations of $n$ things taken $r$ at a time.\n\u2022 Robert Potts (Elementary Algebra, 1880) begins his treatment by letting the number of combinations of $n$ different things taken $r$ at a time be denoted by $^nC_r$\n\u2022 W.A. Whitworth uses $C_r^n$ in Choice and Change (1886).\n\u2022 G. Chrystal writes $_nC_r$ in Algebra, Part II (1899).\n\nNote the different variations $C_r, {^{n}C}_r, C^n_r$ of $_nC_r$-like notations.\n\nHere is a selection of some classics from the 20th century.\n\nSome classics:\n\n\u2022 An Introduction to Combinatorial Analysis (1958) by J. Riordan\n\n\u2022 (3.1): ... and \\begin{align*} C(n,r)=\\frac{n(n-1)\\cdots(n-r+1)}{r!}=\\frac{n!}{r!(n-r)!}=\\binom{n}{r} \\end{align*} where the last symbol is that usual for binomial coefficients, that is, the coefficients in the expansion of $(a+b)^n$. ($C_r^n, C_n^r, _nC_r$ and $(n,r)$ are alternative notations; ...\n\u2022 Combinatorial Identities (1968) by J. Riordan\n\n\u2022 (1.1): Perhaps the simplest combinatorial entities are the binomial coefficients, that is, the combinations, for example of $n$ things, $k$ at a time. They take their name from the generating function for combinations, which is a power of a binomial, namely \\begin{align*} (1+x)^n=\\sum_{k=0}^n\\binom{n}{k}x^k \\end{align*} where, of course, $\\binom{n}{K}=C(n,k)=\\frac{n!}{k!(n-k)!}$ is the usual notation for a binomial coefficient.\n\u2022 (II.4): ... the number of subpopulations of size $r$ is therefore given by $(n)_r/r!$. Expressions of this kind are known as binomial coefficients and the standard notation for them is \\begin{align*} \\binom{n}{r}=\\frac{(n)_r}{r!}=\\frac{n(n-1)\\cdots(n-r+1)}{1\\cdot 2\\cdots (r-1)\\cdot r} \\end{align*}\n\u2022 Advanced Combinatorics (1974) by L. Comtet\n\n\u2022 (1.4): ... We will adopt the notation $\\binom{n}{k}$, used almost in this form by Euler and fixed by Raabe, with the exclusion of all other notations, as this notation is used in the great majority of the present literature, and its use is even so still increasing. This symbol has all the qualities of a good notation: economical (no new letters introduced), expressive (it is very close in appearence to the explicit value $\\frac{(n)_k}{k!}$, typical (no risk of being confused with others), and beautiful.\n\u2022 Enumerative Combinatorics, Volume I (1986) by R. P. Stanley\n\n\u2022 (I.1.4): ... where $d_n=\\sum_{i=0}^n\\binom{n}{i}a_ib_{n-i}$, with $\\binom{n}{i}=n!/i!(n-i)!$.\n\nand the author continues to use the notation $\\binom{n}{k}$ without any more citations, indicating this notation being commonly used.\n\n\u2022 D. E. Knuth who gave us $\\TeX$ is besides being a great mathematician an extraordinary expert in typography and mathematical writing. His Mathematical Typography (1979) gives us a glimpse of his deep thoughts about these issues. Another one being the report Mathematical Writing (1987) written together with T. Larrabee and P. M. Roberts.\n\nIn $\\TeX$ we use the command \"$\\text{n \\choose k}$\" giving us $\\binom{n}{k}$.\n\nThe enhanced readability of the notation $\\binom{n}{k}$ becomes rather obvious in more complex expressions. Compare for instance formula (5.32) in Concrete Mathematics by R. L. Graham, D. E. Knuth and O. Patashnik which is stated for integers $l,m,n$; $n\\geq 0$ as \\begin{align*} \\sum_{j,k}(-1)^{j+k}\\binom{j+k}{k+l}\\binom{r}{j}\\binom{n}{k}\\binom{s+n-j-k}{m-j}=(-1)^l\\binom{n+r}{n+l}\\binom{s-r}{m-n-l} \\end{align*} with the representation \\begin{align*} \\sum_{j,k}(-1)^{j+k}\\,_{j+k}C_{k+l}\\,_rC_j\\,_nC_k\\,_{s+n-j-k}C_{m-j}=(-1)^l\\,_{n+r}C_{n+l}\\,_{s-r}C_{m-n-l} \\end{align*}\n\n\u2022 @BruceET: You're welcome. \u2013\u00a0Markus Scheuer Jun 29 '18 at 8:51\n\u2022 Thank you for taking the time and effort to write such a nice answer! :) \u2013\u00a0Seankala Jun 29 '18 at 15:42\n\u2022 @Sean: You're welcome. Thanks for your nice comment. :-) \u2013\u00a0Markus Scheuer Jun 29 '18 at 15:45\n\nIn books printed before $\\LaTeX$ came to be widely used, setting ${n \\choose r}$ into type was difficult (and expensive if used throughout a book). I 'know' this because of conversations with editors of math books over the years. By contrast, typesetting is not difficult for the variants of $C_r^n,$ including $C(n, r)$ not yet mentioned in this discussion.\n\nNowadays, I think there is a trend to use ${n \\choose r},$ except for authors who have a strong preference for C-notations and those writing in Microsoft Word and trying to avoid using its 'equation editor'.\n\nCuriously, a generally-accepted convention for permutations $P_r^n = r!{n \\choose r}$ does not seem to have emerged. Feller used $(n)_r$ in his famous probability book, which one would have thought might have set a trend 60 years ago, but apparently not. His book may have set a record for the density per page of big parentheses for various uses of ${n \\choose r}$-notation--even before $\\LaTeX.$\n\nThere is no difference at all. Personally I prefer using the shortest notation aCb for obvious reasons.\n\n\u2022 Are the \"a\" and \"b\" significant, there, or is it more the ${}_{\\cdot}C_{\\cdot}$ notation that you prefer? \u2013\u00a0Cameron Buie Jun 29 '18 at 2:56\n\u2022 a and b are just numbers . Nothing more than that . Yes I prefer that notation because it's shorter and I see it very often \u2013\u00a0Kwnstantinos Nikoloutsos Jun 29 '18 at 2:59", "date": "2019-04-23 12:13:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2835377/combination-notation-vs-binomial-coefficient-formula", "openwebmath_score": 0.9258873462677002, "openwebmath_perplexity": 772.823877360378, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707953529717, "lm_q2_score": 0.8757869884059267, "lm_q1q2_score": 0.8559686254180844}}
{"url": "https://mathematica.stackexchange.com/questions/48359/unify-the-sampling-of-nintegrate-f-g-h-w", "text": "# Unify the sampling of NIntegrate[ {f, g, h} w ]\n\nI'm trying to numerically integrate a function which has a vector-valued slow part and a much faster component which is shared by all the components, i.e. an integral of the form $$\\int_a^b\\begin{pmatrix}f(x)\\\\ g(x) \\\\ h(x)\\end{pmatrix}w(x)\\,\\text dx.$$ Because NIntegrate is nicely Listable on its first argument, I can feed it a list-valued argument without a problem. However, it appears to be doing each component as a completely separate integral, which results in a lot of work being re-calculated.\n\nFor example, the (simplified) example integral\n\nsamplePointsList = Reap[\nNIntegrate[\n{x, x^2, x^3} Cos[10 x + Cos[x]]\n, {x, 0, 5}\n, EvaluationMonitor :> Sow[x]\n]\n][[2, 1]];\n\n\ngives the sample point diagram (through ListPlot[Transpose[{samplePointsList, Range[Length[samplePointsList]]}]])\n\nIt is clear that the kernel is doing the initial sampling, and then the further refinements, separately for each component. While the required sample points are not identical, there is a lot of shared work and I feel there is a fair bit of room for optimization there.\n\nI am aware that, since the sampling requirements of each component are slightly different, binding them completely will require more evaluations in some components than would be necessary. (For instance, in the example above, the extra detail required by the first component around $2\\leq x\\leq 3$ would be slowed down slightly if it was required to also calculate the second and third components there, though they do not require it.) However, the components in my case are similar enough that I do not think this would be an issue.\n\nIs there a way to force Mathematica into this sort of separation of the integrand and unification of the sampling? If not, is there a way to make it aware of the previously calculated values and sampling points which will speed up the process?\n\nOne idea is to integrate once to get the sample points then compute the remaining integrals as sums:\n\n sample = Transpose@\nSortBy[First@Last@Reap[\nNIntegrate[x (c = Cos[10 x + Cos[x]]), {x, 0, 5},\nEvaluationMonitor :> Sow[{x, c}]]], #[[1]] &];\nwt = ((#[[3]] - #[[1]])/2) & /@ Partition[Join[{0}, sample[[1]], {5}], 3, 1];\nwt.(sample[[2]] #) & /@ {sample[[1]], sample[[1]]^2, sample[[1]]^3}\n\n\n{0.0133333, 0.133275, 0.861541}\n\nNote the accuracy is not terribly good, NIntegrate gives:\n\n{0.0125266, 0.131514, 0.855716}\n\nSomethings a bit off in my quick&dirty trapezoid integration but i think this can be made to work.\n\nFor this example there really is little benefit to NIntegrate's adaptive sampling so we might as well just use a uniform sampling:\n\n np = 651;(*assumed odd for simpsons rule*)\na=5\nb=0\nwt = ( (a-b)/(np - 1) )/3 Join[{1}, Flatten@ConstantArray[{4, 2}, (np - 1)/2 - 1], {4, 1}] // N;\nx = b + (Range[0, np - 1] (a-b)/(np - 1)) // N;\nfast = Cos[10 # + Cos[#]] & /@ x // N;\n(# fast).wt & /@ {x, x^2, x^3}\n\n\n{0.0125266, 0.131514, 0.855716}\n\nIs there a way to force Mathematica into this sort of separation of the integrand and unification of the sampling? If not, is there a way to make it aware of the previously calculated values and sampling points which will speed up the process?\n\nare \"yes\" and \"yes\" as demonstrated and discussed in this post to \"How to calculate the numerical integral more efficiently?\".\n\nMore concretely:\n\nImport[\"https://raw.githubusercontent.com/antononcube/\\\nMathematicaForPrediction/master/Misc/ArrayOfFunctionsRule.m\"]\n\nfuncExpr = {x, x^2, x^3} Cos[10 x + Cos[x]]\n\n(* {x Cos[10 x + Cos[x]], x^2 Cos[10 x + Cos[x]],\nx^3 Cos[10 x + Cos[x]]} *)\n\nfuncMat = Table[i*funcExpr, {i, 100}];\n\nAbsoluteTiming[\nres0 = NIntegrate[funcMat, {x, 0, 5}];\n]\nres0[[12]]\n\n(* {1.711, Null} *)\n(* {0.150319, 1.57817, 10.2686} *)\n\nAbsoluteTiming[\nres1 = NIntegrate[1, {x, 0, 5},\nMethod -> {\"GlobalAdaptive\", \"SingularityHandler\" -> None,\nMethod -> {ArrayOfFunctionsRule, \"Functions\" -> funcMat}}];\n]\nres1[[12]]\n\n(* {0.029007, Null} *)\n(* {0.150319, 1.57817, 10.2686} *)\n\nNorm[res0 - res1, 2]\n\n(* 2.03398*10^-12 *)\n\n\u2022 That's pretty interesting; I hope I can find time to test it out soon. In the meantime, can you comment on the relative gains as a function of the size of the matrix? I'm interested in $n\\leq3$ but the other answer looks like it targets bigger arrays. \u2013\u00a0Emilio Pisanty Jan 9 '17 at 19:22\n\u2022 @EmilioPisanty I am not sure what you mean -- performance of the ArrayOfFunctionsRule with respect to the size of the integrand? I did a similar comparison during the experiments for the linked MSE answer... \u2013\u00a0Anton Antonov Jan 10 '17 at 1:30", "date": "2020-10-22 12:55:56", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/48359/unify-the-sampling-of-nintegrate-f-g-h-w", "openwebmath_score": 0.2977149188518524, "openwebmath_perplexity": 1677.9384802879292, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9597620596782467, "lm_q2_score": 0.8918110511888303, "lm_q1q2_score": 0.8559264113328141}}
{"url": "https://gmatclub.com/forum/a-coin-is-tossed-7-times-find-the-probability-of-getting-125693.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Jul 2018, 14:27\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# A coin is tossed 7 times. Find the probability of getting\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 25 Dec 2011\nPosts: 54\nGMAT Date: 05-31-2012\nA coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n07 Jan 2012, 13:47\n4\n15\n00:00\n\nDifficulty:\n\n75% (hard)\n\nQuestion Stats:\n\n58% (01:32) correct 42% (01:40) wrong based on 498 sessions\n\n### HideShow timer Statistics\n\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\n\nA. 1/2\nB. 63/128\nC. 4/7\nD. 61/256\nE. 63/64\n\nHi\nI want to understand why combination has been used in the below problem.\nI thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.\nFor example if the question was - probability of getting 1 head then we would have put it as follows :-\nHTTTTTT\nTHTTTTT\nTTHTTTT\nTTTHTTT\nTTTTHHH\nTTTTTHH\nTTTTTTH\n\nThen why in the below problem - we use combinations and not permutations ?\nI am so confused.\n\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\n\nB. 63/128\nC. 4/7\nD. 61/256\n\nExplanation\nANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = \u00bd)\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47112\n\n### Show Tags\n\n12 Jan 2012, 05:34\n7\n6\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\nA. 1/2\nB. 63/128\nC. 4/7\nD. 61/256\nE. 63/64\n\nAssuming the coin is fair - P(H)=P(T)=1/2\n\nWe can do as proposed by the explanation in your initial post:\n\nTotal outcomes: 2^7\n\nFavorable outcomes:\n4 heads --> combination of HHHHTTT --> 7!/(4!*3!)=35 (# of permutation of 7 letters out of which 4 H's and 3 T's are identical);\n5 heads --> combination of HHHHHTT --> 7!/(5!*2!)=21;\n6 heads --> combination of HHHHHHT --> 7!/(6!*1!)=7;\n7 heads --> combination of HHHHHHH --> 1;\n\nP(H>T)=Favorable outcomes/Total outcomes=(35+21+7+1)/2^7=1/2.\n\nBUT: there is MUCH simpler and elegant way to solve this question. Since the probability of getting either heads or tails is equal (1/2) and a tie in 7 (odd) tosses is not possible then the probability of getting more heads than tails = to the probability of getting more tails than heads = 1/2. How else? Does the probability favor any of tails or heads? (The distribution of the probabilities is symmetrical: P(H=7)=P(T=7), P(H=5)=P(T=5), ... also P(H>4)=P(T>4))\n\nIf it were: A fair coin is tossed 8 times. Find the probability of getting more heads than tails in all 8 tosses?\n\nNow, almost the same here: as 8 is even then a tie is possible but again as distribution is symmetrical then $$P(H>T)=\\frac{1-P(H=T)}{2}=P(T>H)$$ (so we just subtract the probability of a tie and then divide the given value by 2 as P(H>T)=P(H<T)). As $$P(H=T)=\\frac{8!}{4!*4!}=70$$ (# of permutation of 8 letters HHHHTTTT, out of which 4 H's and H T's are identical) then $$P(H>T)=\\frac{1-P(H=T)}{2}=\\frac{1-\\frac{70}{2^8}}{2}=\\frac{93}{256}$$. You can check this in following way: total # of outcomes = 2^8=256, out of which in 70 cases there will be a tie, in 93 cases H>T and also in 93 cases T>H --> 70+93+93=256.\n\nHope it's clear.\n\nSimilar questions for practice:\nprobability-question-100222.html?hilit=coin%20tossed#p772756\nhard-probability-99478.html?hilit=coin%20tossed\nsome-ps-questions-need-explanation-99282.html?hilit=coin%20tossed\nprobability-question-gmatprep-85802.html?hilit=coin%20tossed\n_________________\n##### General Discussion\nManager\nJoined: 25 Dec 2011\nPosts: 54\nGMAT Date: 05-31-2012\n\n### Show Tags\n\n07 Jan 2012, 13:49\n1\nHi Sorry ...1 head out of 7 can be arranged in following ways :-\nHTTTTTT\nTHTTTTT\nTTHTTTT\nTTTHTTT\nTTTTHTT\nTTTTTHT\nTTTTTTH\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47112\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n07 Jun 2013, 07:03\nBumping for review and further discussion*. Get a kudos point for an alternative solution!\n\n*New project from GMAT Club!!! Check HERE\n\nTheory on probability problems: math-probability-87244.html\n\nAll DS probability problems to practice: search.php?search_id=tag&tag_id=33\nAll PS probability problems to practice: search.php?search_id=tag&tag_id=54\n\nTough probability questions: hardest-area-questions-probability-and-combinations-101361.html\n\n_________________\nSenior Manager\nJoined: 28 Apr 2012\nPosts: 298\nLocation: India\nConcentration: Finance, Technology\nGMAT 1: 650 Q48 V31\nGMAT 2: 770 Q50 V47\nWE: Information Technology (Computer Software)\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n07 Jun 2013, 08:48\n2\nProbability of getting Head, P(H) = 1/2\n\nFor No. of heads > no. of Tails, there can be 4,5,6 or 7 Heads.\n\nAs per GMAT Club math book, probability of occurrence of an event k times in a n time sequence, for independent and mutually exclusive events:\n\n$$P=C(n,k)*p^k*(1-p)^n^-^k$$\n$$P(H=4) = C(7,4)*(1/2)^4*(1/2)^3 = C(7,4) * (1/2)^7$$\n$$P(H=5) = C(7,5) * (1/2)^7$$\n$$P(H=6) = C(7,6) * (1/2)^7$$\n$$P(H=7) = C(7,7) * (1/2)^7$$\n\nTotal $$P(H > 3) = P(H=4) + P(H=5) + P(H=6) + P(H=7)$$\n$$= [ C(7,4) + C(7,5) + C(7,6) + C(7,7)] * (1/2)^7$$\n$$= (35 + 21 + 7 + 1) / 128$$\n$$= 64/128$$\n$$= 1/2$$\n_________________\n\n\"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well.\"\n\u2015 Voltaire\n\nPress Kudos, if I have helped.\nThanks!\n\nIntern\nJoined: 14 Oct 2013\nPosts: 5\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n05 May 2014, 05:47\nMy approach\n\n7 toss\ncoin has 2 out comes H or T\n\nnow basically if i see this a game where there are 2 teams one selects Heads and other team selects Tails. If in 7 toss whichever comes maximum (heads or Tails) the corresponding team wins..\n\nClearly the probability is 50% for both the cases max Heads or max Tails............\nIntern\nJoined: 14 Jul 2014\nPosts: 17\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n01 May 2015, 02:54\nFor those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.\n\nI couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.\nManager\nJoined: 22 Apr 2015\nPosts: 64\nA coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n01 May 2015, 05:11\n2\nDipikaP wrote:\nFor those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.\n\nI couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.\n\nHi Dipika,\n\nYou are totally right when you say \"it is all conceptual\". Trying to memorise formulae is not the right approach.\n\nFor example in this question:\nFirst we should think what are the possible outcomes when we toss a coin: head or tail (2 outcomes)\nNow as the coin is fair, the probability that we will get a head or a tail is 1/2\n\nTo illustrate, let's take a smaller version of the above question:\n\nWhat is the probability of getting more heads in 3 tosses?\n\n1st case: We can get 3 heads: HHH\nProbability of HHH = 1/2 * 1/2 * 1/2 = 1/8\nProbability of getting 3 heads = 1/8\n\n2nd case: We can get two heads: HHT, HTH, THH\nProbability of HHT = 1/2 * 1/2 * 1/2 = 1/8\nProbability of HTH = 1/2 * 1/2 * 1/2 = 1/8\nProbability of THH = 1/2 * 1/2 * 1/2 = 1/8\nProbability of getting 2 heads = 3* 1/8\n\nAs you can see HHT, HTH and THH are different arrangements of HHT\nProbability of getting 2 heads = (No. of arrangements of HHT)* (Probability of getting HHT) = 3!/2! * (1/2 * 1/2 * 1/2) = 3/8\n\nTotal probability of getting more heads in 3 tosses = 1/8 + 3/8 = 4/8 = 1/2\n\nThinking on these lines you can solve the above question easily.\n\nBut if you think a little further, we are talking about odd number of tosses (3 and 7). So, either there will be more heads or more tails. It is not possible to get equal number of heads and tails. Hence, in half of the outcomes we will get more heads than tails and in the the other half we will have more tails than heads. Thus, the probability of getting more heads = probability of getting more tails = 1/2.\nIntern\nJoined: 03 Jul 2015\nPosts: 10\nLocation: India\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n04 Aug 2015, 12:50\nmorya003 wrote:\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\n\nA. 1/2\nB. 63/128\nC. 4/7\nD. 61/256\nE. 63/64\n\nHi\nI want to understand why combination has been used in the below problem.\nI thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.\nFor example if the question was - probability of getting 1 head then we would have put it as follows :-\nHTTTTTT\nTHTTTTT\nTTHTTTT\nTTTHTTT\nTTTTHHH\nTTTTTHH\nTTTTTTH\n\nThen why in the below problem - we use combinations and not permutations ?\nI am so confused.\n\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\n\nB. 63/128\nC. 4/7\nD. 61/256\n\nExplanation\nANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = \u00bd)\n\nSince number of times coin has been tossed is 7, either number of heads will be more than tails or vice versa. There is no way number of heads become equal to number of tails. Since head and tails are equally favourable outcome of a coin, possibility of getting more heads = possiblity of getting more tails = 1/2. In other words there are only 2 equally likely events (event 1: heads more than tails, event2: tails more than heads) constituting all the outcomes.\n\nCurrent Student\nStatus: DONE!\nJoined: 05 Sep 2016\nPosts: 398\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n07 Dec 2016, 20:46\nTotal outcome = 2^7 = 128\n\nHHHHHHH = 7!/7! = 1\nHHHHHHT = 7!/1!6! = 7\nHHHHHTT = 7!/2!5! = 21\nHHHHTTT = 7!/3!4! = 35\n\n1+7+21+35 = 36\n36/128 = 1/2\n\nA.\nIntern\nJoined: 22 Nov 2016\nPosts: 13\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n02 Feb 2017, 04:42\nhi bunnel...\nAs stated in the question\" Find the probability of getting more heads than tails in all 7 tosses?\"\ntails in all the tosses means no heads ie 0 heads and more than that means at least one heads.\nso we have to find the the solution for atleast one heads in 7 tosses...\ncan we restate the question as above?\npls explain\n\nthank u..\nMath Expert\nJoined: 02 Sep 2009\nPosts: 47112\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n02 Feb 2017, 06:12\nmanojhanagandi wrote:\nhi bunnel...\nAs stated in the question\" Find the probability of getting more heads than tails in all 7 tosses?\"\ntails in all the tosses means no heads ie 0 heads and more than that means at least one heads.\nso we have to find the the solution for atleast one heads in 7 tosses...\ncan we restate the question as above?\npls explain\n\nthank u..\n\nIf you read the solutions above you'll see that this is not correct. More heads than tails in all 7 tosses means at least 4 heads (so more than half must be heads):\n\nHHHHTTT\nHHHHHTT\nHHHHHHT\nHHHHHHH\n_________________\nIntern\nJoined: 22 Nov 2016\nPosts: 13\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n03 Feb 2017, 04:22\nThank You bunuel...\nSenior Manager\nJoined: 29 Jun 2017\nPosts: 497\nGPA: 4\nWE: Engineering (Transportation)\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n06 Sep 2017, 00:48\nAns is A\n\nHHHHTTT= $$(0.5)^4$$ x $$(0.5)^3$$ {7C4}\nsimilarly for 5 H , 6H and 7 H we will calculate only 7C5 , 7C6 and 7C7 will change other part (0.5)^7 remains same which is multiplied, so add all 4 cases of 4H 5H 6H 7H\n$$(0.5)^7$$ x {7C4+7C5+7C6 +7C7}\n$$\\frac{64}{128}$$= $$\\frac{1}{2}$$\n_________________\n\nGive Kudos for correct answer and/or if you like the solution.\n\nDirector\nJoined: 13 Mar 2017\nPosts: 610\nLocation: India\nConcentration: General Management, Entrepreneurship\nGPA: 3.8\nWE: Engineering (Energy and Utilities)\nRe: A coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n06 Sep 2017, 03:39\nmorya003 wrote:\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\n\nA. 1/2\nB. 63/128\nC. 4/7\nD. 61/256\nE. 63/64\n\nHi\nI want to understand why combination has been used in the below problem.\nI thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.\nFor example if the question was - probability of getting 1 head then we would have put it as follows :-\nHTTTTTT\nTHTTTTT\nTTHTTTT\nTTTHTTT\nTTTTHHH\nTTTTTHH\nTTTTTTH\n\nThen why in the below problem - we use combinations and not permutations ?\nI am so confused.\n\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\n\nB. 63/128\nC. 4/7\nD. 61/256\n\nExplanation\nANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = \u00bd)\n\nNow this is a very simple question and can be solved without doing all the long calculations.\nHere important thing to notice is P(H) = P(T) = 1/2 in a single toss.\n\nNow 7 Toss can have following possible ways\n0H 7T say x way\n1H 6T say y way\n2H 5T say z way\n3H 4T say w way\n--------------------------\n4H 3T w way\n5H 2T z way\n6H 1T x way\n7H 0T x way\n\nSo 4 out of 8 ways will have heads greater than Tail. Also P(H) = P(T) = 1/2 = i.e. equal in a single toss. So, we don't need to calculate the number of ways for each case.\n\nRequire probability = (w+z+x+y)[/(x+y+z+w) +(w+z+x+y)] = 1/2\n\n_________________\n\nCAT 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95\nUPSC Aspirants : Get my app UPSC Important News Reader from Play store.\n\nMBA Social Network : WebMaggu\n\nAppreciate by Clicking +1 Kudos ( Lets be more generous friends.)\n\nWhat I believe is : \"Nothing is Impossible, Even Impossible says I'm Possible\" : \"Stay Hungry, Stay Foolish\".\n\nDS Forum Moderator\nJoined: 27 Oct 2017\nPosts: 621\nLocation: India\nGPA: 3.64\nWE: Business Development (Energy and Utilities)\nA coin is tossed 7 times. Find the probability of getting\u00a0 [#permalink]\n\n### Show Tags\n\n20 Dec 2017, 09:22\nMy approach: either Number of heads can be more or less than tails. They can't be equal as no of toss is 7 which is odd.\n\nHence probability is (no of cases when no of head is more) /((no of cases when no of head is more)+(no of cases when no of head is more))= 1/2\n\nprovide kudos, if u like my approach\n\nBunuel wrote:\nA coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?\nA. 1/2\nB. 63/128\nC. 4/7\nD. 61/256\nE. 63/64\n\nAssuming the coin is fair - P(H)=P(T)=1/2\n\nWe can do as proposed by the explanation in your initial post:\n\nTotal outcomes: 2^7\n\nFavorable outcomes:\n4 heads --> combination of HHHHTTT --> 7!/(4!*3!)=35 (# of permutation of 7 letters out of which 4 H's and 3 T's are identical);\n5 heads --> combination of HHHHHTT --> 7!/(5!*2!)=21;\n6 heads --> combination of HHHHHHT --> 7!/(6!*1!)=7;\n7 heads --> combination of HHHHHHH --> 1;\n\nP(H>T)=Favorable outcomes/Total outcomes=(35+21+7+1)/2^7=1/2.\n\nBUT: there is MUCH simpler and elegant way to solve this question. Since the probability of getting either heads or tails is equal (1/2) and a tie in 7 (odd) tosses is not possible then the probability of getting more heads than tails = to the probability of getting more tails than heads = 1/2. How else? Does the probability favor any of tails or heads? (The distribution of the probabilities is symmetrical: P(H=7)=P(T=7), P(H=5)=P(T=5), ... also P(H>4)=P(T>4))\n\nIf it were: A fair coin is tossed 8 times. Find the probability of getting more heads than tails in all 8 tosses?\n\nNow, almost the same here: as 8 is even then a tie is possible but again as distribution is symmetrical then $$P(H>T)=\\frac{1-P(H=T)}{2}=P(T>H)$$ (so we just subtract the probability of a tie and then divide the given value by 2 as P(H>T)=P(H<T)). As $$P(H=T)=\\frac{8!}{4!*4!}=70$$ (# of permutation of 8 letters HHHHTTTT, out of which 4 H's and H T's are identical) then $$P(H>T)=\\frac{1-P(H=T)}{2}=\\frac{1-\\frac{70}{2^8}}{2}=\\frac{93}{256}$$. You can check this in following way: total # of outcomes = 2^8=256, out of which in 70 cases there will be a tie, in 93 cases H>T and also in 93 cases T>H --> 70+93+93=256.\n\nHope it's clear.\n\nSimilar questions for practice:\nhttp://gmatclub.com/forum/probability-q ... ed#p772756\nhttp://gmatclub.com/forum/hard-probabil ... n%20tossed\nhttp://gmatclub.com/forum/some-ps-quest ... n%20tossed\nhttp://gmatclub.com/forum/probability-q ... n%20tossed\n\n_________________\nA coin is tossed 7 times. Find the probability of getting &nbs [#permalink] 20 Dec 2017, 09:22\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-07-19 21:27:16", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-coin-is-tossed-7-times-find-the-probability-of-getting-125693.html", "openwebmath_score": 0.8059019446372986, "openwebmath_perplexity": 1722.2541678187697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787849789998, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.85591931976581}}
{"url": "https://math.stackexchange.com/questions/2213431/basic-question-regarding-limits/2213440", "text": "# Basic question regarding limits\n\nI'm a little confused when it comes to question like this.\n\nLet's say we got this expression: $$\\lim_{x\\to\\infty} \\tan\\left(\\frac{1}{x}\\right).$$ Am I allowed to say the result of this is $0$ or do I have to show it? (If I have to show it, please show me how to cuz I got no clue).\n\nAnd more general, in which situations are we allowed to \"cut\" it and in which we are not? (I'm guessing whenever it comes to fractions surely).\n\nThat depends.\n\nIf the whole exercise is just \"compute $\\lim \\tan(1/x)$\", then yes, you have to do some argument. Something along the lines of \"if $x$ goes to infinity, then $1/x$ goes to zero, and by continuity of $\\tan$, the whole expression then goes to zero\".\n\nBut if this limit is just one part of a longer question/calculation, then no. It's trivial enough that anybody with some math background will see it immediately.\n\n\u2022 Appreciated . Got only one more question . Is there any like more formal ways of prooving it is 0 ? Cuz I never did any good solving math problem using words tho I completely see where ur going at. \u2013\u00a0James Groon Apr 1 '17 at 18:24\n\u2022 In general feel free to replace the word \"then\" or the phrase \"it then follows\" with the symbol \"$\\implies$\" and \"there is an $x$\" with \"$\\exists x$\" and such. Using such symbols instead of words can make a proof more concise, though overdoing it can make it hard to read. And if you want a more elementary proof, the answer by @Chris-Varghese is quite nice. \u2013\u00a0Simon Apr 1 '17 at 18:34\n\u2022 Gotcha and yep, Chris did a good job :) \u2013\u00a0James Groon Apr 1 '17 at 18:35\n\nYou need to show that given any small $\\epsilon > 0$, there exists a $X(\\epsilon)$ such that $|\\tan(1/x)| < \\epsilon,\\; \\forall x > X(\\epsilon)$. To do this choose, for e.g., $X(\\epsilon) = \\dfrac{1+\\sec \\epsilon}{\\epsilon}$. Then $$\\left|\\tan \\frac{1}{x}\\right| < \\left|\\tan \\frac{\\epsilon}{1+\\sec \\epsilon} \\right| = \\frac{\\sin \\left( \\frac{\\epsilon}{1+\\sec \\epsilon}\\right) }{\\cos\\left( \\frac{\\epsilon}{1+\\sec \\epsilon}\\right)} < \\frac{\\sin \\left( \\frac{\\epsilon}{1+\\sec \\epsilon}\\right) }{\\cos\\epsilon} < \\frac{ \\frac{\\epsilon}{1+\\sec \\epsilon} }{\\cos\\epsilon} = \\frac{\\epsilon}{1 + \\cos \\epsilon} < \\epsilon$$\n\n\u2022 The inequality is wrong. Note that $|\\tan (x)|\\ge |x|$ for all $x\\in (-\\pi/2,\\pi/2)$. Simple example; $x=\\pi/4<1$. We have $\\tan(\\pi/4)=1>\\pi/4$. \u2013\u00a0Mark Viola Apr 1 '17 at 18:48\n\u2022 So may i get the correct answer ? \u2013\u00a0James Groon Apr 1 '17 at 19:02\n\u2022 @Dr.MV Which inequality in my answer are you referring to? Indeed, $|\\tan x | \\geq |x| \\forall x \\in (-\\pi/2, \\pi/2)$. How does that fact contradict anything that I have in my answer? \u2013\u00a0ChargeShivers Apr 2 '17 at 1:01\n\u2022 @chrisvarghese You wrote $|\\tan(\\epsilon)|\\le \\epsilon$. That is false. \u2013\u00a0Mark Viola Apr 2 '17 at 2:58\n\u2022 @Dr.MV Got it now. Thanks. Not sure how I missed it twice!! Answer edited. \u2013\u00a0ChargeShivers Apr 2 '17 at 17:17\n\nThe answer is that it depends on the situation. To address the statement in the OP \"If I have to show it, please show me how to cuz I got no clue,\" I thought it might be instructive to present an approach that relies on a standard inequality from elementary geometry. To that end, we begin with a short primer.\n\nPRIMER:\n\nRecall from elementary geometry the inequality\n\n$$\\sin(\\theta)\\le \\theta\\tag 1$$\n\nfor $\\theta\\ge 0$. Squaring both sides of $(1)$, using $\\sin^2(\\theta)=1-\\cos^2(\\theta)$, and rearranging, we find that\n\n$$\\cos(\\theta)\\ge \\sqrt{1-\\theta^2} \\tag 2$$\n\nfor $0\\le \\theta \\le 1$.\n\nNow, using $(1)$ and $(2)$ with $\\theta=1/x$ reveals that\n\n\\begin{align} \\tan(1/x)&=\\frac{\\sin(1/x)}{\\cos(1/x)}\\\\\\\\ &\\le \\frac{1/x}{\\sqrt{1-\\frac1{x^2}}}\\\\\\\\ &=\\frac{1}{\\sqrt{x^2-1}}\\tag 3 \\end{align}\n\nfor $x>1$.\n\nHence, for all $\\epsilon>0$, we have from $(3)$ that\n\n\\begin{align} \\tan(1/x)&\\le \\frac{1}{\\sqrt{x^2-1}}\\\\\\\\ &<\\epsilon \\end{align}\n\nwhenever $x>\\sqrt{1+\\frac{1}{\\epsilon^2}}$. And we are done!\n\n\u2022 $\\sqrt{1 + \\frac{1}{\\epsilon^2}} = \\dfrac{\\sqrt{1+\\epsilon^2}}{\\epsilon}$ is definitely a better $X(\\epsilon)$ than $\\dfrac{1+\\sec \\epsilon}{\\epsilon}$. May be the 'best' of all is $X(\\epsilon) = \\dfrac{1}{\\arctan \\epsilon}$. \u2013\u00a0ChargeShivers Apr 2 '17 at 23:41\n\nShort answer is $0$ ,but if you want \" to show \" recall the limit $$\\lim _{ x\\rightarrow \\infty }{ \\frac { 1 }{ x } =0 }$$\n\nYou can move the limit into the $tan$, because you are not excluding any $x$ from the limit, and you are not creating any limits that don't exist. Once that is done, the problem is trivial.", "date": "2020-01-29 11:34:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2213431/basic-question-regarding-limits/2213440", "openwebmath_score": 0.7982875108718872, "openwebmath_perplexity": 499.3459725129674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787885624276, "lm_q2_score": 0.8670357683915538, "lm_q1q2_score": 0.8559193194810677}}
{"url": "http://math.stackexchange.com/questions/324666/application-of-2nd-fundamental-theorem-of-calculus", "text": "# Application of 2nd fundamental theorem of calculus\n\nI would like to clarify the usage of the 2nd fundamental theorem of calculus, in 3 parts. For all 3 parts, consider $F'(a)$ to be\n\n$$F'(a) = \\frac{1}{1+a+a^2}$$\n\nThe questions are:\n\n1. Find $\\frac{d}{dy}\\int^y_1 F'(a) \\,da$\n2. Find $\\frac{d}{dy}\\int^1_y F'(a) \\,da$\n3. Find $\\frac{d}{dy}\\int^{y^2}_1 F'(a)\\, da$\n\nMy working for each of the questions are:\n\n1.\n\n\\begin{align*} \\frac{d}{dy} \\int^y_1 F'(a) \\,da &= \\frac{d}{dy}\\begin{bmatrix}F(y) - F(1)\\end{bmatrix} \\\\&=F'(y) \\\\&=\\frac{1}{1+y+y^2}. \\end{align*}\n\n2.\n\n\\begin{align*} \\frac{d}{dy} \\int^1_y F'(a) \\,da &= \\frac{d}{dy}\\begin{bmatrix}F(1) - F(y)\\end{bmatrix} \\\\&=-F'(y) \\\\&=\\frac{-1}{1+y+y^2}. \\end{align*}\n\n3.\n\n\\begin{align*} \\frac{d}{dy} \\int^{y^2}_1 F'(a) \\,da &= \\frac{d}{dy}\\begin{bmatrix}F(y^2) - F(1)\\end{bmatrix} \\\\&=F'(y^2) \\\\&=\\frac{2y}{1+y^2+y^4}. \\end{align*}\n\nUPDATE: I shall update (3) with the chain rule working.\n\n\\begin{align*} \\frac{d}{dy} \\int^{y^2}_1 F'(a) \\,da &= \\frac{d}{dy}\\begin{bmatrix}F(y^2) - F(1)\\end{bmatrix} \\\\&=F'(y^2)(2y) \\\\&=\\frac{1}{1+(y^2)+(y^2)^2} (2y) \\tag{by chain rule} \\\\&=\\frac{2y}{1+y^2+y^4}. \\end{align*}\n\n-\nThe third one would need a chain rule. Though, you do have used it for final answer. \u2013\u00a0 hjpotter92 Mar 8 '13 at 12:59\nYour answers are correct. However, one should note that$\\dfrac{dF(y^2)}{dy}=F'(y^2)\\cdot2y$(Chain rule). Your final answers are perfectly fine but the intermediate step is wrong.", "date": "2014-03-10 17:05:55", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/324666/application-of-2nd-fundamental-theorem-of-calculus", "openwebmath_score": 0.9998019337654114, "openwebmath_perplexity": 9900.99389819685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787868650145, "lm_q2_score": 0.8670357683915538, "lm_q1q2_score": 0.8559193180093497}}
{"url": "http://math.stackexchange.com/questions/427559/prove-frac-cos2-a1-sin-a-1-sin-a-by-the-pythagorean-theorem", "text": "# Prove $\\frac{\\cos^2 A}{1 - \\sin A} = 1 + \\sin A$ by the Pythagorean theorem.\n\nHow do I use the Pythagorean Theorem to prove that $$\\frac{\\cos^2 A}{1 - \\sin A} = 1 + \\sin A?$$\n\n-\nif i multiply it would it show that A is an acute angle? i have a given figure here and its a right triangle \u2013\u00a0 user83562 Jun 23 '13 at 14:25\nDo you know the identity $\\cos^2 A + \\sin^2 A = 1$? \u2013\u00a0 Javier Badia Jun 23 '13 at 14:58\n\nsuppose there is a right angle tri. having b is hypo.,c is base ,a is perpendicular than $\\cos A=\\frac cb$ and $\\sin A=\\frac ab$ using pythgo. theo. $\\;b^2=a^2+c^2$ $$\\dfrac {\\cos^2 A}{1-\\sin A}$$ $$\\dfrac {\\frac {c^2}{b^2}}{1-\\frac {a}{b}}$$ $$\\dfrac {\\frac {c^2}{b^2}}{\\frac {b-a}{b}}$$ $$\\dfrac {c^2b}{b^2(b-a)}$$ $$\\dfrac {c^2}{b(b-a)}$$ $$\\dfrac {b^2-a^2}{b(b-a)}$$ $$\\dfrac {b+a}{b}$$ $$1+\\dfrac {a}{b}$$ $$1+\\sin A$$\n\nalternatively\n\n$$\\dfrac {\\cos^2 A}{1-\\sin A}$$ $$\\dfrac {1-\\sin^2 A}{1-\\sin A}$$ $$\\dfrac {(1-\\sin A)(1+\\sin A)}{1-\\sin A}$$ $${1+\\sin A}$$\n\n-\nwhy don't you make things precise : $$1=\\frac{c^2}{b^2}+\\frac{a^2}{b^2}=\\cos^2A+\\sin^2A$$ $$\\implies \\cos^2A=1-\\sin^2A=(1-\\sin A)(1+\\sin A)$$ $$\\implies \\frac{\\cos^2A}{1-\\sin A}=1+\\sin A$$ \u2013\u00a0 lab bhattacharjee Jun 23 '13 at 14:32\nhow did it become 1-sin^2 A? \u2013\u00a0 user83562 Jun 23 '13 at 14:49\n@user83562: Did you read lab's (excellent) comment? Since $$1=\\cos^2A+\\sin^2A,$$ then $$\\cos^2A=1-\\sin^2A.$$ By difference of squares formula, we have $$\\cos^2A=(1-\\sin A)(1+\\sin A).$$ From there, it's almost immediate. \u2013\u00a0 Cameron Buie Jun 23 '13 at 15:18\n\nIf you allow yourself some simple algebra, then you see that $$\\frac{\\cos^2 A}{1-\\sin A} = 1+\\sin A\\text{ is equivalent to } \\cos^2 A = (1+\\sin A)(1-\\sin A)$$ and that last expression is equal to $1-\\sin^2 A$. So you're asking how to prove $\\cos^2 A=1-\\sin^2 A$. And that's the same as proving $\\cos^2 A + \\sin^2 A = 1$.\n\nSo the question is: How can you prove that $\\cos^2 A + \\sin^2 A = 1$ by using the Pythagorean theorem?\n\nIf you know that $\\sin =\\dfrac{\\mathrm{opposite}}{\\mathrm{hypotenuse}}$ and $\\cos=\\dfrac{\\mathrm{adjacent}}{\\mathrm{hypotenuse}}$, then this becomes easy if you consider a right triangle in which the length of the hypotenuse is $1$. Then you have $\\sin=\\mathrm{opposite}$ and $\\cos=\\mathrm{adjacent}$. Now the Pythagorean theorem says that $$\\mathrm{opposite}^2 + \\mathrm{adjacent}^2 = 1^2.$$\n\n-\n\nLet's simplify matters:\n\nIn a triangle with hypotenuse equal to $1$ (think of the unit circle, an angle $A$ between the $x$ axis and the hypotenuse, we know that $$\\sin A=\\frac{\\text{opposite}}{\\text{hypotenuse}}\\quad \\text{and}\\quad \\cos A=\\frac{\\text{adjacent}}{\\text{hypotenuse}}$$ then, since hypotenuse $=1$, we have the leg opposite the angle $A$ given by $\\sin A=\\text{opposite}/1$ and the leg along the x-axis of length $\\cos A=\\text{adjacent}/1$. Now, by the Pythagorean Theorem, and substitution, we have that \\begin{align}\\text{opposite}^2 + \\text{adjacent}^2 &= \\text{hypotenuse}^2 = 1^2 \\\\ \\sin^2A +\\cos^2 A & = 1\\end{align}\n\nThis gives us the well-known identity: $$\\sin^2A + \\cos^2 A = 1\\tag{1}$$\n\nWe can express this identity in terms of $\\cos^2 A$ by subtracting $\\sin^2 A$ from both sides of the identity to get \\begin{align}\\cos^2 A & = 1 - \\sin^2 A \\\\ &= (1)^2 - (\\sin A)^2\\tag{2}\\end{align}\n\nNow, we know that for any difference of squares, we can factor as follows: $$(x^2 - y^2) = (x +y)(x - y)\\tag{3}$$\n\nSince equation $(2)$ is a difference of squares, we have that \\begin{align}\\cos^2 A &= 1 - \\sin^2 A \\\\ & = (1)^2 - (\\sin A)^2 \\\\ &= (1 +\\sin A)(1 - \\sin A)\\tag{4}\\end{align}\n\nSubstituting gives us:\n\n\\begin{align}\\frac{\\cos^2 A}{1 -\\sin A} & = \\frac{1 - \\sin^2 A}{1 -\\sin A} \\\\ \\\\ &= \\frac{(1 + \\sin A)(\\color{blue}{\\bf 1 - \\sin A})}{\\color{blue}{\\bf 1 -\\sin A}}\\\\ \\\\ & = 1 + \\sin A\\end{align}\n\n-\nYou begin with \"Given the well known identity,....\", but the questions seems to be how to get that well known identity from the Pythagorean theorem. \u2013\u00a0 Michael Hardy Jun 23 '13 at 16:14\n@amWhy: You are always welcome my friend! Time for another cheesy movie (trying to add other activities in my life). ;-) \u2013\u00a0 Amzoti Jun 24 '13 at 2:47\n\nLet us denote base of right angle triangle as b, perpendicular ( height ) as p, and hypotenuse as h,\n\n$$\\cos A = \\frac{b}{h} ; \\qquad \\sin A = \\frac{p}{h}\\tag{i}$$\n\nTherefore, $$\\frac{\\cos^2A}{1-\\sin A} = \\frac{\\frac{b^2}{h^2}}{1-\\frac{p}{h}}$$\n\n[By putting the values of $\\cos A$ and $\\sin A$ from $(i)$]\n\nWhich after simplification gives you:\n\n$$\\frac{b^2}{h(h-p)}\\tag{ii}$$\n\nNow as\n\n$$b^2 = h^2-p^2\\quad\\text{[using pythagoreous theorem]}\\tag{iii}$$\n\nBy putting the value of $b^2$ from (iii) in (ii) you get :\n\n$$\\frac{h^2-p^2}{h(h-p)} = \\frac{h+p}{h} = 1+ \\frac{p}{h} = 1+ \\sin A$$\n\n(hence proved)\n\n-\nnot sure how much this derivation from the Left hand side to the Right is required, as this is often not the natural derivation. For example, please find my comment in the other answer. \u2013\u00a0 lab bhattacharjee Jun 23 '13 at 14:38", "date": "2014-08-30 10:53:48", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/427559/prove-frac-cos2-a1-sin-a-1-sin-a-by-the-pythagorean-theorem", "openwebmath_score": 0.9990672469139099, "openwebmath_perplexity": 505.9458982731516, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787834701878, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8559193133700623}}
{"url": "https://www.paulcupido.nl/l3zg73/11ba1e-what-is-constant-function", "text": "Take a look! Derivative of a constant function. does not change no matter which member of the The graph of a constant function is always a This function may seem a little tricky at first but is actually the easiest one in this set of examples. methods and materials. Always returns to the fixed price. For all functions g, h : C \u2192 A, f o g = f o h, (where \"o\" denotes function composition). results in a zero change in x constant function x f ( x 1 ) = f ( x 2 ) for any x 1 and x 2 in the domain. Constant functions can be characterized with respect to function composition in two ways. Note: This function also works with class constants. This constant expresses an ambiguity inherent in the construction of antiderivatives. In order to create a pricing plan with \"Constant\", select control: (1). 4.9/5.0 Satisfaction Rating over the last 100,000 sessions. This tutorial shows you a great approach to thinking about functions! Yes, a constant function is a periodic function with any T\u2208R as its period (as f (x)=f (x+ T) always for howsoever small 'T' you can find). Real Functions: Constant Functions An constant function is a function that always returns the same constant value. The identity function and the absolute value function. Example: in \"x + 5 = 9\", 5 and 9 are constants. We can make use of this interest to encourage a little thinking. *Response times vary by subject and question complexity. Your first 30 minutes with a Chegg tutor is free! What's a Function? A constant function of a single variable, such as () =, has a graph of a horizontal straight line parallel to the x-axis.Such a function always takes the same value (in this case, 5), because its argument does not appear in the expression defining the function. A constant term in an expression or equation contains no variables. In the context where it is defined, the derivative of a function is a measure of the rate of change of function values with respect to change in input values. ) You can use direct substitution to arrive at this conclusion. In a two dimensional plane, the graph of this type of function is a straight, horizontal line. The derivative of this type of function is just zero. For example, the function y(x) = 4 is a constant function because the value of y(x) is 4 regardless of the input value x (see image). C++ is similar to C but has more features than C. Therefore, it is called a subset of C language. In calculus, the constant of integration, often denoted by , is a constant added to the end of an antiderivative of a function () to indicate that the indefinite integral of () (i.e., the set of all antiderivatives of ()), on a connected domain, is only defined up to an additive constant. A constant function is a function of the form f(x) = c, where c is a constant. A function becomes const when the const keyword is used in the function\u2019s declaration. Constant member function is an accessory function which cannot modifying values of data members. ( variables and constant both carries some defined value but difference comes with their accessibility and value constant can be define as 1 define constant_name constant_value or 2 const constant_name constant_value the second way is faster A constant function is a special type of linear function that follows the form f(x) = b 'b' is the y-intercept of the line and is just a constant; A constant function is a linear function whose slope is 0; No matter what value of 'x' you choose, the value of the function will always be the same Online Help: Math Apps: Functions and Relations.Retrieved from https://www.maplesoft.com/support/help/Maple/view.aspx?path=MathApps%2FConstantFunction on May 24, 2019. For example, y = 10 is a form of constant function. Varsity Tutors \u00a9 2007 - 2021 All Rights Reserved, CBEST - The California Basic Educational Skills Test Courses & Classes, SAT Subject Test in World History Test Prep, International Sports Sciences Association Test Prep. Award-Winning claim based on CBS Local and Houston Press awards. Consider constants as having a variable raised to the power zero. It perform constant operation. Now we shall prove this constant function with the help of the definition of derivative or differentiation. Constant Functions. The form of this particular function is: f(x) = ax 2 + bx + c, where a, b, and c are constants. With a constant function, for any two points in the interval, a change in x results in a zero change in f (x). This function works also with class constants . x ( From the general formula, the output of a constant function regardless of its input value (usually denoted by x), will always be the same which is the fixed number \\color{red}a. It is of the format y=c d. This is constant function as output is different for each input e. This is constant function as output is different for each input 2. which of the graph represent constant function? \"Piecewise constant\" should not be used as a precise term without giving an explicit explanation of what you mean. *Response times vary by subject and question complexity. The composition of f with any other function is also a constant function\u2026 The derivative is the slope at an instant (kind\u2019ve). In mathematics, a constant function is a function whose (output) value is the same for every input value. The object called by these functions cannot be modified. The limit of a constant function (according to the Properties of Limits) is equal to the constant. Then the gradient of z will be a vector pointing \u201cuphill\u201d on a surface, and the length of the vector is the slope at that point. The idea of const functions is not to allow them to modify the object on which they are called. The main difference between Static and Constant function in C++ is that the Static function allows calling functions using the class, without using the object, while the Constant function does not allow modifying objects.. C++ is a programing language developed by Bjarne Stroustrup in 1979. As a function requires that inputs produce outputs, it wouldn\u2019t be a \u201cfunction\u201d. 3 That is when parameters are evaluated and generate statements are expanded. range A constant function has the form. This is a function of the type $$f (x) = k$$, where $$k$$ is any real number. Constant member function in C++. The derivative of a constant function is (1) However, the fundamental period of a constant function is not defined for the above reason. constant function; Background Tutorials. Technically, zero is a constant. This is constant function as output is same for each input. The mathematical formula for a constant function is just f x = a, where a is a number (which does not depend on x). x For example the function f (x) = 4 is constant since f maps any value to 4. A constant function is a linear function whose range contains only one element irrespective of the number of elements of the domain. is used. That is, the output value of the function at any input value in its domain is the same, independent of the input. linear function Varsity Tutors does not have affiliation with universities mentioned on its website. domain y = x is called the identity function because the value of y is identical with that of x. Constant Function Years 1 - 8 : Summary Children are always fascinated by the Constant Function of a calculator which constantly repeats the last operation it was asked to do if the [=] button is constantly pressed. A constant may be used to define a constant function that ignores its arguments and always gives the same value. 2 and constant function; Background Tutorials. This way, for instance, if we wanted to represent a quantity that stays constant over the course of time t, we would use a constant function f(t)=k, in which the variable tdoes not appear. Need help with a homework or test question? Defining in terms of mathematics, we can say that a constant function is a function that has the same range for all values of its domain. Definition and Usage. The derivative of a constant function is zero. Collection Functions (Arrays or Objects) each_.each(list, iteratee, [context]) Alias: forEach Iterates over a list of elements, yielding each in turn to an iteratee function. This is a constant function and so any value of x that we plug into the function will yield a value of 8. ) ) c is a constant: a number that doesn\u2019t change as x changes. What's a Function? Slope is the change in y over the change in x. constant() is useful if you need to retrieve the value of a constant, but do not know its name. The function graph of a one-dimensional constant function is a straight line. Note that the value of f(x) is always k, independently of the value of x. A fixed value. In haskell: newtype Const r a = Const { unConst :: r } instance Functor (Const r) where fmap _ (Const r) = Const r It maps every type a to r in a sense, and every function of type a -> b to the identity function on r. We can write this type of function as: f (x) = c ( Basically, a constant function is simply equal to a constant. Suppose the failure function is a positive constant; that is, Show that the reliability function R(t) satisfies the separable differential equation Solve this differential equation to find R(t). The constant () function returns the value of a constant. A constant function is a function whose range consists of a single element. = No - the term \"time constant\" is not restricted to an asymptotic step response. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. For instance, a school dining room where every child was given one donut, irrespective of age, or an exam in which every student was given an A regardless of how hard they worked. The graph of a constant functio\u2026 It passes through the point (0, c), (1, c), and (-1, c). ) A constant function is a linear function for which the range does not change no matter which member of the domain is used. But 0/6 = 0, so this would not actually produce a graph. I.e. 2 The constant() function returns the value of a constant. In fact lines are either increasing, decreasing, or constant. Example: in \"x + 5 = 9\", 5 and 9 are constants. horizontal line It\u2019s easier to visualize this for a 2-dimensional function. So a method int Foo::Bar(int random_arg) (without the const at the end) results in a function like int Foo_Bar(Foo* this, int random_arg) , and a call such as Foo f; f.Bar(4) will internally correspond to something like Foo f; Foo_Bar(&f, 4) . *See complete details for Better Score Guarantee. A: The sale is a trade where the money is traded for a product or service. The equation of a line is: y = mx + b. Quadratic functions aren\u2019t much different. for which the A constant function is a linear function for which the range does not change no matter which member of the domain is used. More formally, a function f : A \u2192 B is a constant function if f (x) = f for all x and y in A. Do It Faster, Learn It Better. Constant function In mathematics, a constant function is a function whose values do not vary and thus are constant. In other words, it\u2019s just number on its own. f For example, the integral of f(x) = 10 is 10x. Note: This function also works with class constants. For example, the graph of the constant function f(x) = 4 is a straight, horizontal line that passes through the points (2,4), (0,4), and (-2,4). f One great example of a constant function is the log2 function. Recognizing a constant function, like y = 3. x f (x 1) = f (x 2) for any x 1 and x 2 in the domain. However, the fundamental period of a constant function is not defined for the above reason. The integration constant Ti is the time constant of the integrator. Real Functions: Constant Functions An constant function is a function that always returns the same constant value. The constant function used to provide a way to define a very simple price. Consider the function f (x) = 3 f (x) = 3 There is no variable in the definition (on the right side). This is a function of the type f(x)=k, where k is any real number. The slope m tells us if the function is increasing, decreasing or constant: is a This function may seem a little tricky at first but is actually the easiest one in this set of examples. Your email address will not be published. Function Definitions and Function Notation. This section describes what is a constant, define() function defines a constant name to a value, constant value can retrieved by the name directly or by the constant() function, any string can be used as constant name. It is recommended to use const keyword so that accidental changes to object are avoided. In the context where it is defined, the derivative of a function measures the rate of change of function (output) values with respect to change in input values. Yes, a constant function is a periodic function with any T\u2208R as its period (as f(x)=f(x+T) always for howsoever small 'T' you can find). If the answer in infinity, you can type in INF in the blank. x As of 4/27/18. The constant functions cut through the vertical axis in the value of the constant and they are parallel to the horizontal axis (and therefore they do not cut through it). What's a Function? A: The sale is a trade where the money is traded for a product or service. 1 it is stored in a variable or returned by a function. Because a constant function does not change, its derivative is 0. This graph is shown below. The iteratee is bound to the context object, if one is passed. A constant \u2026 In other words, the constant function is the function f(x) = c. An example of data for the constant function expressed in tabular form is presented below: \ufeff What Is a Constant? The following shows the graph of f(x) = 10, and the integral f(x) = 10x. This function works also with class constants. This tutorial shows you a great approach to thinking about functions! This video is provided by the Learning Assistance Center of Howard Community College. The co\u00f6rdinate pairs are (x, x). \u221f Constant and define() Function. Your email address will not be published. C++ is similar to C but has more features than C. Therefore, it is called a subset of C languag C++ Programming Server Side Programming The const member functions are the functions which are declared as constant in the program. Learn the definition of a function and see the different ways functions can be represented. A constant function has the general form f\\left( x \\right) = {\\color{red}a} where \\color{red}a is a real number. The general transfer function of an integrator is (using your notation) H(s)=k/s=1/(s/k). ( In layman's terms, constant functions are functions that do not move. For example, the following are all constant functions: Since f(x) is equal to a constant, the value of f(x) will always be the same no matter what the value of x might be. The e constant is defined as the limit: The e constant is defined as the infinite series: Properties of e Reciprocal of e. The reciprocal of e is the limit: Derivatives of e. The derivative of the exponential function is the exponential function: (e x)' = e x. = Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC. You can't go through algebra without learning about functions. . It is of the format y=c d. This is constant function as output is different for each input e. This is constant function as output is different for each input 2. which of the graph represent constant function? Constant function. A Constant Function is a horizontal line: Lines. . Learn the definition of a function and see the different ways functions \u2026 To understand why the slope of a constant is zero, we have to understand what slope means. Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors. A constant function is an even function, i.e. Mathematically speaking, a constant function is a function that has the same output value no matter what your input value is. In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number. x). The main difference between Static and Constant function in C++ is that the Static function allows calling functions using the class, without using the object, while the Constant function does not allow modifying objects.. C++ is a programing language developed by Bjarne Stroustrup in 1979. Maplesoft Support. the graph of a constant function is symmetric with respect to the y-axis. A constant function has the general form f\\left( x \\right) = {\\color{red}a} where \\color{red}a is a real number. Now, it is common usage to set 1/k=Ti resulting in H(s)=1/sTi. The blue square represents the integral when evaluated from 0 to 1. constant() is useful if you need to retrieve the value of a constant, but do not know its name. Every function with a derivative equal to zero is a constant function. without a variable attached) is \u201cc\u201d, so that is the constant term. You can't go through algebra without learning about functions. A constant function is an even function so the y-axis is an axis of symmetry for every constant function. Varsity Tutors connects learners with experts. constant function; Background Tutorials. From the general formula, the output of a constant function regardless of its input value (usually denoted by x), will always be the same which is the fixed number \\color{red}a. Generally, it is a function which always has the same value no matter what the input is. Median response time is 34 minutes and may be longer for new subjects. We can make use of this interest to encourage a little thinking. 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Help: Math Apps: functions and Relations.Retrieved from https: //www.maplesoft.com/support/help/Maple/view.aspx? path=MathApps 2FConstantFunction... Use const keyword so that accidental changes to objects are avoided through point. Integral f ( x ) = 2x2 + 3 ( the constant in! Limit of a constant function is equal to 3, no matter what your input value its... Algebra without learning about functions its domain is used with that of x member function is a line... Affiliated with Varsity Tutors LLC one great example of a constant function is not defined for the reason... Mentioned on its website to each client, what is constant function their own style, methods and.. X that we plug into the function graph of f ( x ) = 10 10x. Such a constant function is a function which can not modifying values of data members a... Practically Cheating Calculus Handbook, https: //www.maplesoft.com/support/help/Maple/view.aspx? path=MathApps % 2FConstantFunction on 24! K, independently of the type f ( x ) = 2x2 3... Us if the answer in infinity, you can type in INF in the domain identical. Outlets and are not affiliated with Varsity Tutors 3 ) modify the object which... Not affiliated with Varsity Tutors of what you mean Calculus Handbook, https: //www.maplesoft.com/support/help/Maple/view.aspx? %! + B notation ) H ( s ) =k/s=1/ ( s/k ) function in. Defined for the above reason dimensional plane, the Practically Cheating Calculus Handbook, https: //www.calculushowto.com/constant-function/ in! Just zero there are three constants in this set of examples slope of a constant function is. Definition of a line is: y = c, where c is a constant is zero, have... For a 2-dimensional function shows you a great approach to thinking about functions k is any number!", "date": "2021-05-13 15:11:49", "meta": {"domain": "paulcupido.nl", "url": "https://www.paulcupido.nl/l3zg73/11ba1e-what-is-constant-function", "openwebmath_score": 0.7129285931587219, "openwebmath_perplexity": 447.6038189694047, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9871787853562028, "lm_q2_score": 0.8670357632379241, "lm_q1q2_score": 0.8559193116136021}}
{"url": "https://math.stackexchange.com/questions/2712668/how-fast-do-riemann-sums-converge-for-lipschitz-continuous-functions", "text": "# How fast do Riemann sums converge for Lipschitz-continuous functions?\n\nI'm in the following situation: Let's say I've got a non-negative function $f$ that's globally Lipschitz-continuous on some interval $[a,b]$ for some constant $K$. I'm throwing a dart uniformly on the area below $f$ on that interval. Now consider the sequence of lower Riemann sums that results from bisecting all the intervals in the previous partition at each step. The dart will eventually be covered by the area corresponding to one of these lower Riemann sums (and all the following). Then define the random variable $X$ to be the index of the first such sum. My question is: Can the expected value of $X$ somehow be bounded using $K$?\nIntuitively, since $f$ is Lipschitz, the Riemann sums exhaust the area below $f$ very quickly, so this expectation should be quite small. But I'm having a lot of trouble quantifying this intuition.\n\n\u2022 I am confused about the setup: If you partition an interval, the dart will always be a part of any partition. A partition is just an ordered sequence on $[a,b]$. What does it mean for the dart to be be \"eventually\" covered? Mar 29 '18 at 2:54\n\u2022 The dart is an element of $\\mathbb{R}^2$. The vertical columns corresponding to some partition cover some area of $\\mathbb{R}^2$. Eventually these columns will cover the dart. Mar 29 '18 at 3:04\n\nLet $m,M$ be the points in $[a,b]$ where $f$ attains its minimum and maximum. Since $f$ is $K$-Lipschitz, $$f(M)-f(m) \\leq K|M-m| \\leq K(b-a)$$ But $\\int_a^b f(x) \\,dx \\leq (b-a)f(M)$ and so this yields $$\\int_a^b f(x) \\,dx - (b-a)f(m) \\leq K(b-a)^2 \\tag{*}$$\n\nIf we subdivide $[a,b]$ into $2^n$ equal-length subintervals, apply this inequality to each of them, and then sum, we get\n\n$$\\int_a^b f(x)\\, dx - L_{2^n}(f) \\leq \\frac{K}{2^n}(b-a)^2$$ where $L_{2^n}(f)$ is the lower Riemann sum corresponding to this subdivision. So $$1 - \\frac{L_{2^n}(f)}{\\int_a^b f(x) \\,dx} \\leq \\frac{K(b-a)^2}{2^n\\int_a^b f(x) \\, dx}$$ That is, the probability that the dart remains uncovered after the $n$th subdivision is at most $\\frac{K(b-a)^2}{2^n\\int_a^b f(x) \\, dx}$.\n\nSo, if $P_n$ is the probability that the dart is uncovered after $n$ subdivisions, we have $$E[X]=\\sum_{n=0}^\\infty n(P_{n-1} - P_n) =\\sum_{n=0}^\\infty P_n \\leq \\frac{K(b-a)^2}{\\int_a^b f(x) \\, dx}\\sum_{n=0}^\\infty \\frac{1}{2^n}=\\frac{2K(b-a)^2}{\\int_a^b f(x) \\, dx}$$\n\nIn fact we can improve on the key inequality $(*)$ with a little more work. We have $$f(x) - f(m) \\leq K |x-m|$$ for all $x \\in [a, b]$. Integrating over $[a,b]$ gives \\begin{align} \\int_a^b [f(x) - f(m)]\\, dx &\\leq K \\int_a^b |x-m|\\, dx \\\\ &= \\frac{K}{2}[(m-a)^2 + (b-m)^2] \\\\ &= \\frac{K}{2}[(a-b)^2-2(m-a)(b-m)] \\\\ &\\leq \\frac{K}{2}(a-b)^2 \\end{align} and so we have $$\\int_a^b f(x) \\, dx - (b-a)f(m) \\leq \\frac{K}{2} (a-b)^2 \\tag{**}$$ a factor-of-$2$ improvement over $(*)$. Moreover, if $f$ is linear with slope $\\pm K$, then equality is attained, so this bound is tight.\n\nCarrying this through the rest of the argument we end up with $$E[X] \\leq \\frac{K(b-a)^2}{\\int_a^b f(x) \\, dx}$$ with equality when $f$ is linear with slope $\\pm K$.\n\n\u2022 Wow, this came out quite elegantly. The way you bounded the difference between the integral and the Riemann sums was I think a key step. Is this an argument that you have seen used elsewhere before? Mar 29 '18 at 4:27\n\u2022 Sorry, I don't have any specific references. I think some version of \"find an easy bound, then work out what happens to it as you subdivide\" is present in most calculations involving Riemann sums (at least, most of the ones I'm capable of coming up with...) Mar 29 '18 at 4:48", "date": "2021-11-28 05:16:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2712668/how-fast-do-riemann-sums-converge-for-lipschitz-continuous-functions", "openwebmath_score": 0.9977597594261169, "openwebmath_perplexity": 227.15929296961895, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787872422175, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8559193098571415}}
{"url": "https://math.stackexchange.com/questions/2340044/in-combinatorics-how-can-one-verify-that-one-has-counted-correctly", "text": "# In combinatorics, how can one verify that one has counted correctly?\n\nThis is a soft question, but I've tried to be specific about my concerns. When studying basic combinatorics, I was struck by the fact that it seems hard to verify if one has counted correctly.\n\nIt's easiest to explain with an example, so I'll give one (it's fun!) and then pose my questions at the bottom. In this example there are two ways to count the number of ways $2n$ people can be placed into $n$ partnerships (from the book Introduction to Probability by Blitzstein and Hwang):\n\n$$\\frac{(2n)!}{2^n \\cdot n!} = (2n - 1)(2n - 3) \\cdots 3 \\cdot 1$$\n\nStory proof for the right side: For the first person you have $(2n - 1)$ choices for partner. Then for the next person you have $(2n - 3)$ choices, etc.\n\nStory proof for the left side: Line all $2n$ people up, walk down the line, and select every 2 people as a pair. The ordering you chose for the line determines the pairs, and there are $2n!$ ways to order $2n$ people. But divide by $2^n$ because the order within each pairing doesn't matter. Also divide by $n!$ because the order of the pairings doesn't matter.\n\nMy question is:\n\nWhat if I had been tasked with getting this number at work, and I chose the approach on the left side, but I neglected to divide by $2^n$ and only divided by $n!$? I'd be wrong of course, but how could I have known?\n\nI suppose one answer to my question could just be \"Try hard, try to think of ways you might be over/undercounting, look at other examples, and continue to study theory.\"\n\nBut the prospect of not knowing if I'm wrong makes me nervous. So I'm looking for more concrete things I can do. So, three questions of increasing rigor:\n\n\u2022 What are specific \"habits of mind\" that people in combinatorics use to avoid error?\n\u2022 What are specific validation techniques that such people use to check their work? (One does come to mind from my example: Calculate the same thing two ways and confirm equality)\n\u2022 Is there any formal list of questions that, if answered, should verify that one's approach is correct?\n\u2022 Checking your answer for small $n$ (in cases where you can count manually) is a good sanity check. \u2013\u00a0angryavian Jun 28 '17 at 23:45\n\u2022 In 1962 in my sophomore Intro to Probability, my prof asked how many legs a cow had. Twelve, he explained: two in front, two in back, two on the left, two on the right and one at each corner. We did not think it was funny. A few weeks later he told the same joke and everyone laughed. \u2013\u00a0John Wayland Bales Jun 28 '17 at 23:46\n\u2022 Often, you can't be sure. And that's one reason I dislike combinatorics personally - I'm never sure if I've massively screwed up. :) \u2013\u00a0Deepak Jun 29 '17 at 0:06\n\u2022 I think it was in the preface to the 3rd edition of \"Generatingfunctionology\" where I read the most(?) bijective (read counting) proofs we know were first discovered from messing around with generating functions in the first place. And apparently \"A = B\" by Petkovsek (and others) provides a(the?) full-proof method/s of solving most any textbook problem of the kind you mentioned. Haven't read either yet, but google them and see for yourself. \u2013\u00a0user448692 Jun 29 '17 at 1:20\n\u2022 @Stephen If you divide that by $5!$: $$\\frac{ {10 \\choose 2}{8 \\choose 2}{6 \\choose 2}{4 \\choose 2}{2 \\choose 2}}{5!}$$ you get the right answer. \u2013\u00a0Trevor Gunn Jun 29 '17 at 19:24\n\nHere are some methods I have used. Some of these have already been suggested in the comments, and none of them are fool-proof.\n\n1. Work out a few small cases by hand, i.e. generate all the possibilities, and verify that your formula works in those cases. (Often working out the small cases will also suggest a method of solution.)\n2. Solve the problem by two different methods. For example, often a problem which is solved by the principle of inclusion/exclusion can also be solved by generating functions. Or it may be that you can derive a recurrence that the solution must satisfy, and verify that your solution satisfies the recurrence; if it is too hard to do the general case, verify the recurrence is satisfied for a few small cases.\n3. Write a computer program that solves a particular case of the problem by \"brute force\", i.e. enumerates all the possibilities, and check the count against your analytic answer. If you don't yet know how to program, this is one of many reasons why it's good to to learn how. For example, how many ways can you make change for a dollar? It's probably easier to write a program that generates all the ways than to solve this problem analytically.\n4. Sometimes a combinatoric problem can be reinterpreted as a probability problem. For example, the problem of counting all the poker hands that form a full house is closely connected to the probability of drawing a full house. In this case you can check your answer by Monte Carlo simulation: use a random number generator to simulate many computer hands and count the number which are full houses. Once again, this requires computer programming skills. You won't get an exact match against your analytic answer, but the proportion of full houses should be close to your analytic result. Here it helps to know some basic statistics, e.g. the Student's t test, so you can check to see if your answer falls in the range of plausible results.\n\u2022 Being a programmer, I think, \"of course my hammer works on this nail\", but intellectually I rebel against the idea that the best way to check my theory is to use a machine that didn't exist when the study of combinatorics was born... \u2013\u00a0Kristen Hammack Jun 29 '17 at 12:55\n\u2022 @KristenHammack - Why would that be a problem? Hammers and nails didn't exist when someone first had to figure out how to hold two pieces of wood together. But in most situations they are a superior solution to tying the wood together with braided grass or strips of leather. \u2013\u00a0Paul Sinclair Jun 29 '17 at 16:49\n\u2022 @PaulSinclair that's certainly an interesting way of looking at it. But I was wondering if maybe there was a more elegant \"screwdriver\" type of solution, with the brute force programming method being more like \"let me hammer this screw into the wood\", and a formal proof being \"here's the perfectly-fitting screwdriver for this particular screw\". I'm used to seeing that in CS. \u2013\u00a0Kristen Hammack Jun 29 '17 at 17:04\n\u2022 Plus a programming check probably won't give much extra intuition about why a solution is correct or incorrect. I am also a programmer and wouldn't hesitate to write a program to do the check if I needed it, but it wouldn't be my first choice. \u2013\u00a0Stephen Jun 29 '17 at 21:42\n\u2022 @KristenHammack But the question is not about \"how do you figure this out\". It is about \"I think I have it solved, but I want to double-check that I haven't missed something\". For this, \"experimental mathematics\" is and always has been a tool-of-choice. And computers vastly increase the amount of experimentation that is feasible. And for Stephen's comment, even if you are still trying to figure it out, having a good sample of actual solutions can be a godsend for better understanding the behavior. Computer proofs may be boring, but they are excellent mathematical telescopes. \u2013\u00a0Paul Sinclair Jun 29 '17 at 23:10\n\nHere are some other methods:\n\n\u2022 Estimate what the answer should be. For example, Let's say I want to count all 3-element subsets of $1,\\dots,n$ where the sum of any two elements is not equal to the third. I think to myself \"hmm, if $n$ is large then most three element subsets should be valid.\" So whatever the answer is, I know that it should be close to $\\binom{n}{3}$ if $n \\gg 0$.\n\n\u2022 Also keep in mind that your intuition about how large the answer should be might be off. This is ok! It means you will put more thought into the problem.\n\n\u2022 Have someone look over your work. This can work really well because sometimes you'll have something you are mistaken about in the back of your mind when you are writing. When this happens, what you write will (hopefully) look funny to other people whereas it might take a lot of thought on your part to catch that something is off.\n\n\u2022 Be more rigorous. This is easier said than done, obviously. For example if you are applying a theorem, look to make sure that you satisfy all the necessary hypotheses to be able to apply it.\n\n\u2022 Slow down and think about \"why am I multiplying here?\" or dividing, or differentiating. Make sure the algebra makes sense combinatorially. If you are dividing, make sure you end up with integers where there should be integers.\n\n\u2022 Having somebody look over work is a very good suggestion. Fortunately I am in a programming context where code review is de rigeur \u2013\u00a0Stephen Jun 29 '17 at 1:34\n\nIt is actually possible to be rigorous in questions like this \u2013 though it can be a lot of work.\n\nThe first step is to build a mathematical model that corresponds to the question. This part deserves some thought, and there is scope for error, but usually just expressing the question mathematically is easier to do without error than finding the actual answer.\n\nFor this example, we might say that a partnering of a group of $2n$ people (whom we'll refer to as $G=\\left\\{0, 1, \\ldots, 2n-1\\right\\}$) is any function $p: G \\rightarrow G$ such that:\n\n\u2022 For all $m$: $p(p(m)) = m$ (i.e. a person's partner is paired with the person)\n\u2022 For all $m$: $p(m) \\neq m$ (i.e. nobody is partnered with themself)\n\nHopefully it's fairly clear that the above corresponds to the definition of a partnering from the question - if not, you can do it in a way that's clearer to you.\n\nHaving made that definition, the question becomes: how many such functions $p$ are there?\n\nOne way to establish the answer rigorously is to come up with a bijection $B$ between the set of partnerings $P = \\{p: p \\,\\hbox{is a partnering of$2n$people}\\}$ and an initial subset of the natural numbers $\\mathbb N$.\n\nThat can be done as follows \u2013 I'll omit a lot of details for brevity, but hopefully it should be clear that this can be made fully rigorous. Given a partnering $p$:\n\nStart with a set $U_0=G$ of unpartnered people, and a value $v_0=0$. At each step $k$, starting from 0:\n\n\u2022 Consider the person from $U_k$ with the smallest number \u2013 call them $g_k$.\n\u2022 Find out who they're partnered with \u2013 that is, evaluate $p(g_k)$ - and find that person's \"rank\" in $U_k\\setminus g_k$ starting from 0. That is, if (of all people other than $g_k$ who are so far unpartnered) $p(g_k)$ has the lowest number, call them rank 0, and if there are three with numbers lower than $p(g_k)$ that are unpartnered, call them rank 3, and so on. Call that rank $r_k$.\n\u2022 Set $U_{k+1} = U_k \\setminus \\{g_k, p(g_k)\\}$ and $v_{k+1} = v_k\\left|U_k-1\\right| + r_k$. That is, take out the two people we just partnered, and calculate a new value from the previous step's value and the rank of the person we just partnered with $g_k$.\n\nRepeat for $k$ from 0 to $n-2$, and take $v_{n-1}$ as our result for $B(p)$.\n\nIt's clear that, given any partnering $p$, this procedure gives us back a number $B(p)$. What may not yet be clear is that, given $B(p)$, we can determine what $p$ was \u2013 that is, $B$ is a bijection.\n\nI won't go into the details of that, but note that, when calculating the value, the person who gets partnered with person 0 has their rank multiplied by $(2n-3)(2n-5)\\ldots1$, and we can get it back by dividing $B(p)$ by that number \u2013 the integer part of the answer gives us $p(0)$, and then we can use the remainder in a similar way to deduce the rest of the pairings in $p$.\n\nSo (again I'm leaving out some proof details) we have a function that is a bijection between $\\left\\{0,1,2,\\ldots,m-1\\right\\}$ and the set of possible partnerings of $2n$ people. This is enough to show that there are $m$ such partnerings \u2013 and it's reasonably easy to see (by taking the maximum possible value at each step in the procedure above) that the value of $m$ is exactly $(2n-1)(2n-3)\\cdots 1$ as you already knew.\n\n\u2022 It's good to be able to do things with symbols like this; it's seldom necessary. Excellent mathematics uses symbols to convey an intuitive understanding. Mediocre mathematics uses symbols in place of intuitive understanding. Poor mathematics uses symbols to disguise the total absence of intuitive understanding. \u2013\u00a0Wildcard Jun 29 '17 at 19:37\n\u2022 @Wildcard: I don't think anyone said it was necessary! The point is that it is possible to achieve a high degree of rigour in combinatorics. The OP was asking about how to be more sure about one's answers; symbolic reasoning is a powerful tool that can help with that. \u2013\u00a0psmears Jun 29 '17 at 20:08\n\u2022 psmears upvoted for your efforts. I didn't quite understand it, but I will come back to it once my math is a bit stronger (specifically once I understand bijections). Also @Wildcard that is quite funny. \u2013\u00a0Stephen Jun 29 '17 at 23:52\n\u2022 @Stephen:Sorry--it's hard to know what level to go for! A \"bijection\" is a fancy name for a function that is invertible: the basic idea above is to find a way of giving each possible partnering a unique number, such that given a partnering you can find its number, and given a number you can find the corresponding partnering. Once that's done, if the partnerings are numbered 0 to $m-1$, you know there are $m$ partnerings in total. The advantage is everything (after the initial definition) can be 100% rigorous--but that's a lot of work (my long answer omits many details), hence it's rarely done! \u2013\u00a0psmears Jun 30 '17 at 9:38\n\u2022 This is an enlightening answer. We go to a lot of trouble to develop other subjects rigorousIy, we should know how to do it for combinatorics as well. I especially like your emphasis on the modeling step, which I think is often glossed over in combinatorics books. \u2013\u00a0littleO Jul 6 '17 at 9:24", "date": "2019-05-22 11:13:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2340044/in-combinatorics-how-can-one-verify-that-one-has-counted-correctly", "openwebmath_score": 0.7309669256210327, "openwebmath_perplexity": 306.75648894982425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668690081642, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8559183536972395}}
{"url": "http://math.stackexchange.com/questions/276264/how-to-solve-simple-systems-of-differential-equations", "text": "# How to solve simple systems of differential equations\n\nSay we are given a system of differential equations\n\n$$\\left[ \\begin{array}{c} x' \\\\ y' \\end{array} \\right] = A\\begin{bmatrix} x \\\\ y \\end{bmatrix}$$\n\nWhere $A$ is a $2\\times 2$ matrix.\n\nHow can I in general solve the system, and secondly sketch a solution $\\left(x(t), y(t) \\right)$, in the $(x,y)$-plane?\n\nFor example, let's say\n\n$$\\left[ \\begin{array}{c} x' \\\\ y' \\end{array} \\right] = \\begin{bmatrix} 2 & -4 \\\\ -1 & 2 \\end{bmatrix} \\left[ \\begin{array}{c} x \\\\ y \\end{array} \\right]$$\n\nSecondly I would like to know how you can draw a phane plane? I can imagine something like setting $c_1 = 0$ or $c_2=0$, but I'm not sure how to proceed.\n\n-\nPlease do not modify substantially the question after several answers are posted. \u2013\u00a0 Did Jan 12 '13 at 19:42\n? I did not, because I asked it already, but nobody noted it \u2013\u00a0 MSKfdaswplwq Jan 13 '13 at 14:10\nThis is not true: the whole second part of your question appeared after Wooster's answer and mine were posted. Once again: please do not do that. \u2013\u00a0 Did Jan 13 '13 at 15:23\n\nIf you don't want change variables then,\n\nthere is a simple way for calculate $e^A$(all cases).\n\nLet me explain about. Let A be a matrix and $p(\\lambda)=\\lambda^2-tra\u00e7o(A)\\lambda+det(A)$ the characteristic polynomial. We have 2 cases:\n\n$1$) $p$ has two distinct roots\n\n$2$) $p$ has one root with multiplicity 2\n\nThe case 2 is more simple:\n\nIn this case we have $p(\\lambda)=(\\lambda-a)^2$. By Cayley-Hamilton follow that $p(A)=(A-aI)^2=0$. Now develop $e^x$ in taylor series around the $a$\n\n$$e^x=e^a+e^a(x-a)+e^a\\frac{(x-a)^2}{2!}+...$$\n\nTherefore $$e^A=e^aI+e^a(A-aI)$$\n\nNote that $(A-aI)^2=0$ $\\implies$ $(A-aI)^n=0$ for all $n>2$\n\nCase $1$:\n\nLet A be your example. The eigenvalues are $0$ and $4$. Now we choose a polynomial $f$ of degree $\\le1$ such that $e^0=f(0)$ and $e^4=f(4)$( there is only one). In other words what we want is a function $f(x)=cx+d$ such that\n\n$$1=d$$\n\n$$e^4=c4+d$$\n\nSolving this system we have $c=\\dfrac{e^4-1}{4}$ and $d=1$.\n\nI say that\n\n$$e^A=f(A)=cA+dI=\\dfrac{e^4-1}{4}A+I$$\n\nIn general if $\\lambda_1$ and $\\lambda_2$ are the distinct eigenvalue, and $f(x)=cx+d$ satisfies $f(\\lambda_1)=e^{\\lambda_1}$ and $f(\\lambda_2)=e^{\\lambda_2}$, then\n\n$$e^A=f(A)$$\n\nIf you are interested so I can explain more (it is not hard to see why this is true)\n\nNow I will solve your equation using above.\n\nWhat we need is $e^{tA}$\n\nThe eigenvalues of $tA$ is $0$ and $4t$.\n\nThen $e^{tA}=\\dfrac{e^{4t}-1}{4t}A+I$ for $t$ distinct of $0$\n\n-\n@JoyeuseSaintValentin If you are interested why is true case 2, I can explain more (it is not hard to see why this is true) \u2013\u00a0 user52188 Jan 12 '13 at 19:39\n\nQuite generally, $$\\left[ \\begin{array}{c} x(t) \\\\ y(t) \\end{array} \\right] = \\mathrm e^{tA}\\begin{bmatrix} x(0) \\\\ y(0) \\end{bmatrix},$$ where, by definition, $$\\mathrm e^{tA}=\\sum_{n=0}^{+\\infty}\\frac{t^n}{n!}A^n.$$ To compute $\\mathrm e^{tA}$ in the case at hand, note that $A^2=4A$, hence $$\\mathrm e^{tA}=I+\\sum_{n=1}^{+\\infty}\\frac{t^n}{n!}4^{n-1}A=I+\\frac{\\mathrm e^{4t}-1}4A.$$ Hence, $$x(t)=\\frac{\\mathrm e^{4t}+1}2x(0)+(1-\\mathrm e^{4t})y(0),$$ and $$y(t)=\\frac{1-\\mathrm e^{4t}}4x(0)+\\frac{\\mathrm e^{4t}+1}2y(0).$$\n\n-\nI know the definition of $e^{tA}$ as a series. I also understand that $$\\mathrm e^{tA}=I+\\sum_{n=1}^{+\\infty}\\frac{t^n}{n!}4^{n-1}A$$ But can you explain the last equality: $$I+\\sum_{n=1}^{+\\infty}\\frac{t^n}{n!}4^{n-1}A=I+\\frac{\\mathrm e^{4t}-1}4A.$$ \u2013\u00a0 MSKfdaswplwq Jan 12 '13 at 12:54\nAm I mistaken or does your second comment answer the question in your first comment? \u2013\u00a0 Did Jan 12 '13 at 12:55\nNo, I just messed it up. My question is why the last equality is true \u2013\u00a0 MSKfdaswplwq Jan 12 '13 at 12:57\nBecause $\\sum\\limits_{n=1}^{+\\infty}\\frac{t^n}{n!}4^{n-1}=\\frac{1}{4}\\sum\\limits_{n=1}^{\u200c\u200b+\\infty}\\frac{(4t)^n}{n!}$ and the second series is almost the expansion of $\\mathrm e^{4t}$ (almost because the $n=0$ term is lacking). \u2013\u00a0 Did Jan 12 '13 at 13:02\nEigenvalues are always in the background, one way or another. For example, here $A^2=4A$ because $\\chi_A(x)=x^2-4x$ and $\\chi_A(A)=0$ by Cayley-Hamilton. \u2013\u00a0 Did Jan 12 '13 at 15:29\n\n$\\newcommand{\\vect}[1]{\\mathbb{#1}}$Try finding out about diagonalisation of matrices. (If you do not already know about this.)\n\nThe basic idea is that I can find a particular matrix P, and a diagonal matrix $\\Lambda$; these combine in such a way that $A = P \\Lambda P^{-1}$. (These matrices relate to the eigenvectors and eigenvalues of your matrix $A$.)\n\nThe way we use diagonalisation is as follows. Let me redefine the question slightly so that it is easier for me to explain: I shall use the differential equation $$\\begin{bmatrix}x_1'\\\\x_2'\\end{bmatrix} = A \\begin{bmatrix}x_1\\\\x_2\\end{bmatrix}.$$ Let me call the vector of functions $\\vect{x}$: then I can write our equation as $$\\vect{x}' = A \\vect{x}.$$ Replacing $A$ with my expression $A = P\\Lambda P^{-1}$, I have $$\\vect{x}' = P \\Lambda P^{-1} \\vect{x},$$ or $$P^{-1} \\vect{x}' = \\Lambda P^{-1} \\vect{x}.$$ But the matrix $P$ is just a constant matrix, so if I were to define $\\vect{y} = P^{-1} \\vect{x}$, then we would simply have $$\\vect{y}' = \\Lambda \\vect{y}.$$\n\nBig deal, you might think: this is just like the old equation. But remember that $\\Lambda$ is a diagonal matrix. So in fact this equation looks like $$\\begin{bmatrix}y_1'\\\\y_2'\\end{bmatrix} = \\begin{bmatrix}\\lambda_1 & 0 \\\\0 & \\lambda_2\\end{bmatrix} \\begin{bmatrix}y_1\\\\y_2\\end{bmatrix},$$ which is far simpler to solve. I can solve the differential equations for our $y_i$ functions, and then use the equation $$\\vect{x} = P \\vect{y}$$ to find the solution to our original problem.\n\n[EDIT: removed references to $P$ being orthogonal which are incorrect.]\n\n-\nThank you, that made sense. But actually my matrix is not diagonalizable. Then some Jordan theorem applies I guess? \u2013\u00a0 MSKfdaswplwq Jan 12 '13 at 12:56\nSorry to interrupt but A is diagonalizable. \u2013\u00a0 Did Jan 12 '13 at 13:03\nIndeed it is. I was just trying to work it out, but it certainly looks diagonalisable. Eigenvalues 0 and 4. I did notice an error in my write-up which said you want an orthogonal matrix $P$; this is not true and I have removed it. \u2013\u00a0 Wooster Jan 12 '13 at 13:11\nI see, but what to do when you cannot diagonalize? \u2013\u00a0 MSKfdaswplwq Jan 12 '13 at 13:11\nIf diagonalisation is not possible, then I think you are correct: Jordan normal form would be a good idea. It will give you a number of simple first-order ODEs to solve and then a simple algorithm for solving the rest. \u2013\u00a0 Wooster Jan 12 '13 at 13:17", "date": "2014-08-30 19:01:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/276264/how-to-solve-simple-systems-of-differential-equations", "openwebmath_score": 0.9321116209030151, "openwebmath_perplexity": 218.95279179963651, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668695588648, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8559183525495377}}
{"url": "https://math.stackexchange.com/questions/629933/properties-of-the-euler-totient-function/636466", "text": "Properties of the euler totient function\n\nWhy is it that the euler totient function has the following condition true based on its definition?\n\n$$\\phi(p^k)=p^{k-1}(p-1) = p^k(1-\\frac{1}{p}) = p^k-p^{k-1}$$\n\nA proof would be awesome and an intuition on why this is true would be even better!\n\n(to prove it I thought of using multiplicativity of the totient function but that would not work because p is not coprime to itself :( )\n\nA more detailed explanation of the wikipedia article will get a like and accepted answer.\n\nTo get accepted, giving an explanation on why the number of multiples of p is $p^{k-1}$ will be an important factor. Also, are the multiples of p we are excluding between 1 to $p^k - 1$ or to $p^k$?\n\nSome kind of counting argument is necessary to get accepted.\n\n\u2022 en.wikipedia.org/wiki/\u2026 \u2013\u00a0PITTALUGA Jan 7 '14 at 8:12\n\u2022 Look at it as $p^k - p^{k-1}$. \u2013\u00a0ShreevatsaR Jan 7 '14 at 8:23\n\u2022 If you want I can accept your answers if you provide a little more detailed explanation of $p^k -p^{k-1}$ or the Wikipedia article. Specifically, I am a little unsure about the $p^{k-1}$. Thanks thought, it has been helpful. \u2013\u00a0Pinocchio Jan 8 '14 at 0:22\n\u2022 Bounty started to address the details of $p^{k}-p^{k-1}$. \u2013\u00a0Pinocchio Jan 10 '14 at 17:21\n\nAny positive integer $x$ less than $p^k$ can be written in base $p$ as $$x = a_0 + a_1 p + a_2 p^2 + \\cdots + a_{k-1} p^{k-1}$$ where $a_i \\in \\{0, 1, 2, \\ldots, p-1\\}$. Then $x$ is not relatively prime to $p^k$ iff $p \\mid x$ iff $a_0 = 0$. Thus if we want $x$ to be relatively prime to $p^k$ we have $p-1$ choices for $a_0$ and $p$ choices for each of the other $k-1$ coefficients, hence $\\varphi(p^k) = (p-1)p^{k-1}$.\n\n\u2022 This was such a unexpected and cute answer! kudos on that, I would have never had thought of it that way. \u2013\u00a0Pinocchio Jan 13 '14 at 5:58\n\u2022 For future reference, answer provided by user115654 also provides a great reference if this one did not make sense. \u2013\u00a0Pinocchio Jan 13 '14 at 6:15\n\u2022 I was learning about the $p$-adic numbers when I first saw this result, so it was very natural at the time! \u2013\u00a0Andr\u00e9 3000 Jan 13 '14 at 6:33\n\nHere is intiution....\n\nTake $p^2$. How many numbers in the range $1 \\ldots p^2$ are not co-prime to it? They are precisely $p$, $2p$, $3p$ ... $p^2$. There are exactly $p$ of them. So the number co prime to $p^2$ is $$\\phi(p^2) = p^2 -p = p (p-1)$$\n\nYou can do the same for $p^3$ to get $$\\phi(p^3) = p^3 -p^2 = p^2 (p-1)$$\n\nYou can turn this into a proof by induction if you so wish.\n\nI find it intuitive to think in terms of \"what are the chances of not being coprime to $p^k$?\"\n\nOnce you realize that \"coprime to $p^k$\" is synonymous with \"not divisible by $p$\", it suggests the proportion of numbers up to $p^k$ which are coprime to $p^k$ is $1 - \\frac1p$, motivating the quantity $p^k(1-\\frac1p)$.\n\nThe same reasoning applies to general $n$ (but takes more effort to justify carefully): \"coprime to $n$\" is synonymous with \"not divisible by any of the prime factors of $n$\", and if you can convince yourself that the individual probabilities are independent this suggests $\\phi(n) = n\\prod_{p\\mid n} (1-\\frac1p)$.\n\n\u2022 How do you know that the proportion of elements that are coprime to $p^k$ is $1 - \\frac{1}{p}$? \u2013\u00a0Pinocchio Jan 13 '14 at 6:37\n\u2022 @Pinocchio Because the proportion that are divisible by $p$ is $\\frac1p$. \u2013\u00a0Erick Wong Jan 14 '14 at 4:50\n\nAnother explanation (much of which is implicit in the other answers):\n\nA number $n$ is not coprime to $p \\iff \\text{gcd}(p,n) \\neq 1 \\iff \\text{gcd}(p,n) = p \\iff n$ is a multiple of $p$. Now the multiples of $p$ between $1$ and $p^k$ are precisely the numbers $mp$, for $1 \\leq m \\leq p^{k-1}$, of which there are $p^{k-1}$. Subtracting these from the total gives $p^k - p^{k-1} = \\phi(p^k)$.\n\n\u2022 Euler's totient function counts the number of elements $\\pmod p^k$ that have a $gcd(x,p^k) = 1$. Thus, $0 \\leq x \\leq p^k -1$ are the only candidate elements to be in the set (We don't have to exclude 0 yet, because it will be excluded by your counting argument). Thus, now you apply your counting argument and m's range is $0 \\leq m \\leq p^{k-1} - 1$. Which yields the correct bound but is \"more\" correct. Right? Also, your first sentence in your second paragraph is hard to follow, do you mind rewriting that? (Thanks btw, that was super helpful). \u2013\u00a0Pinocchio Jan 13 '14 at 5:54\n\u2022 Yes, the range $1 \\leq m \\leq p^k$ is the same, modulo $p^k$, as the range $0 \\leq m \\leq p^k - 1$ (and the latter is preferable). I have changed the \"iff\"s to symbols, in an attempt to improve legibility \u2013\u00a0zcn Jan 13 '14 at 6:14\n\u2022 If the second one is preferred then why did you write the other one? \u2013\u00a0Pinocchio Jan 13 '14 at 6:18\n\u2022 Starting at 0 is preferable if your definition counts elements of $\\mathbb{Z}/p^k\\mathbb{Z}$, but starting at 1 is preferable if your definition counts elements of $\\mathbb{N}$. I have opted for the latter, but as the two are equivalent, it really is a matter of personal choice \u2013\u00a0zcn Jan 13 '14 at 6:23\n\n$p^k$ only shares common factors with other multiples of $p$. How many multiples of $p$ are there under $p^k$?: $\\frac{p^k}{p}=p^{k-1}$ Just look at 8 for example. There are 4 multiples of 2 under and including 8. To get the totient value by definition exclude from $p^k$ all the values under it that share a common factor, of which there are $p^{k-1}$.\n\n\u2022 I am sorry if its extremely obvious to you, but if you provide further detail on why $p^k/p$ \"works\", I will be happy. It makes sense that you want to exclude element that are multiples of p. I'm not sure why just dividing $p^k$ by p works. Is $p^k$ the number of element between 1 to $p^k$ or the elements 1 from $p^{k-1}$? why did dividing by p \"worked\"? Basically I am having trouble counting the number of multiples of p rigorously or precisely. \u2013\u00a0Pinocchio Jan 13 '14 at 5:37\n\u2022 $p^k$ is an integer. An integer $n$ also denotes the size of the set of integers 1 through $n$, inclusive. Now starting from 1, say you jump to every multiple of $p$. You will be jumping a size of p every time first to p then to 2p then to 3p....So you have an integer $n=p^k$ and you want to see how many times you can make a jump of size p inside n. The division operation is defined to do just that. How many times does 3 fit in 12? 4 times. How many times does p fit in $p^k$? $p^{k-1}$ Does that make sense? \u2013\u00a0Flowers Jan 13 '14 at 7:06\n\u2022 The sets {1,2,...p}, {p+1,p+2,...2p}, {2p+1,2p+2,...3p}... {p^{k}-p+1,p^{k}-p+2,...p^{k}}are all size p and each one has only one multiple of p. \u2013\u00a0Flowers Jan 13 '14 at 7:15", "date": "2019-05-25 22:05:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/629933/properties-of-the-euler-totient-function/636466", "openwebmath_score": 0.8835917115211487, "openwebmath_perplexity": 243.2411033383979, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668745151689, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.85591835036074}}
{"url": "https://www.physicsforums.com/threads/finding-the-sum-of-a-series.78754/", "text": "# Finding the sum of a series\n\n1. Jun 11, 2005\n\n### steven187\n\nhello all\n\nyou know we have all these tests for convergence of a series, but it always made me wonder if there exists any other method of finding the sum of a series, like we have the geometric series in which we are able to find the sum by a simple formula, but are we able to find such a formula for any series?, would such a formula be unique to each particular series?, or to a group of such series?\n\nfor example $$\\sum_{n=1}^{\\infty}\\frac {(-2)^{n}}{n+1}$$\nthis isnt a geometric series then how would you find the sum of such series?\nis it possible to derive a formula for any series that exist, or are there limitations or conditions that need to be satisfied?\n\n2. Jun 11, 2005\n\n### shmoe\n\nThere is no sure fire general way to find the sum of an infinite series in a 'nice' form. If a series converges and you have enough time on your hands you can find as many decimal places as you like, but an exact, nice looking solution is not always possible (indeed your series might not even converge to something 'nice').\n\nThe series you've provided doesn't converge by the way, the terms don't even go to 0 (a necessary but not sufficient condition).\n\n3. Jun 11, 2005\n\n### steven187\n\nso your sayin that it is possible in some cases to derive a formula for a series besides a geometric series but in most cases its not possible? , and about that series it was suppose to be\n\n$$\\sum_{n=1}^{\\infty}\\frac {(-2)^{-n}}{n+1}$$\nlike would i be able to derive a formula for this series? or how would i be able to find the sum of such a series?\n\n4. Jun 11, 2005\n\n### saltydog\n\nIt's a very interesting question. There are various techniques for solving such. For example,\n\n$$\\sum_{n=1}^{\\infty} \\frac{1}{n^2}=\\frac{\\pi^2}{6}$$\n\nThis can be solved by using Fourier coefficients and Parseval's Theorem.\n\n$$\\sum_{k=1}^{\\infty}\\frac{(-1)^{k+1}}k=ln(2)$$\n\nis shown by considering:\n\n$$f(x)=ln(1+x)$$\n\nand differentiating and considering the Taylor series (thanks Daniel).\n\n$$\\sum_{k=1}^{\\infty}\\frac{1}{4k(2k-1)}$$\n\nis solved by considering the sum:\n\n$$S=\\frac{1}{2}[(1-1/2)+(1/3-1/4)+(1/5-1/6)+...$$\n\nand:\n\n$$\\sum_{n=0}^{\\infty} ne^{-an}$$\n\nis solved by considering:\n\n$$z=\\sum_{n=0}^{\\infty}w^n$$\n\nwith:\n$$w=e^{-a}$$\n\nand differentiating both the sum and the expression for the sum of the corresponding geometric series with respect to w.\n\nTons more I bet. Would be nice to have a compilation of the various methods for calculating infinite sums.\n\n5. Jun 11, 2005\n\n### shmoe\n\nYes. saltydog's given numerous examples. Telescoping series is another handy one, so is recognizing your sum as a known power series, rearranging terms sometimes helps, and more .\n\nRelate it to the power series for log(1+x).\n\n6. Jun 12, 2005\n\n### HallsofIvy\n\nStaff Emeritus\nSome times you can recognize a series as a special case of a Taylor's series for a function. To sum the particular series you give here, I would recall that the Taylor's series for ln(x+1) is $$\\sum_{n=1}^{\\infty}(-1)^n \\frac{x^n}{n}$$ (and converges for -1< x< 1). That differs from your series on having n+1 in the denominator: Okay, change the numbering slightly. If we let i= n+1 then n= i-1 and your series becomes $$\\sum_{i=2}^{\\infty}\\frac{(-2)^{-i+1}}{i}$$. I can just take that (-2)1 out of the entire series: $$(-2)\\sum_{i=2}^{\\infty}\\frac{(-2)^{-i}}{i}$$. We're almost there! Since I need (-2)-i instead of xi, take x= -1/2 which is in the radius of convergence. The fact that the sum now starts at i= 2 instead of 1 is not a problem: just calculate that separately- when i= 0 the term is (-2)0/1= -2: Your sum is $$(-2)\\sum{i=1}\\frac{(-1)^i}{i}+ 2)= (-2)ln(-1/2+ 1)+ 2= -2ln(1/2)+ 2= 2+ 2 ln(2)= 2(1+ ln(2))$$.\n\nLast edited: Jun 12, 2005\n7. Jun 12, 2005\n\n### saltydog\n\nUsing Hall's method I obtained a different answer:\n\nFor this sum:\n\n$$\\sum_{n=1}^{\\infty}\\frac{(-2)^{-n}}{n+1}$$\n\nConsider:\n\n$$ln(1+x)=\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}x^n}{n}\\quad\\text{for}\\quad-1<x<1$$\n\nLetting $x=\\frac{1}{2}$\n\n$$ln(3/2)=\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{2^n n}$$\n\nLetting i=n-1 we obtain:\n\n\\begin{align*} ln(3/2)=\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{2^n n}&=\\sum_{i=0}^{\\infty}\\frac{(-1)^{i+2}}{2^{i+1} (i+1)}\\\\ &=\\frac{1}{2}+\\sum_{i=1}^{\\infty}\\frac{(-1)^2 (-1)^i}{(2)2^i(i+1)}\\\\ &=\\frac{1}{2}+\\frac{1}{2}\\sum_{i=1}^{\\infty}\\frac{(-1)^i}{2^i(i+1)}\\\\ &=\\frac{1}{2}+\\frac{1}{2}\\sum_{i=1}^{\\infty}\\frac{(-2)^{-i}}{i+1} \\end{align}\n\nSolving for the series, I obtain:\n\n$$\\sum_{i=1}^{\\infty}\\frac{(-2)^{-i}}{(i+1)}=-1+2ln(3)-2ln(2)$$\n\nThis result agrees with Mathematica.\n\n8. Jun 12, 2005\n\n### shmoe\n\nThis is correct, at least it's also what I had. Halls has a couple small mistakes, he has the series for -log(1+x), not log(1+x), and used x=-1/2 when it should have been x=1/2. Also the 'missing term' is then $$(-2)(1/2)^1/1=-1$$.\n\nNotice it's an alternating series with decreasing terms and the first term is negative, therefore the sum must also be negative. You could go furthur if you like, the sum must be larger than $$-\\frac{1}{2\\times 2}=-\\frac{1}{4}$$ and smaller than $$-\\frac{1}{2\\times 2}+\\frac{1}{2^2\\times 3}=-\\frac{1}{6}$$, and so on if you wanted some more reassurance that your answer was correct (an alternating series with decreasing terms will hop back and forth between being greater and less than the sum). Of course this sort of estimate won't prove your answer is correct, but can catch mistakes.\n\nLast edited: Jun 12, 2005\n9. Jun 13, 2005\n\n### steven187\n\nhello all\n\nnow I realise that methods of finding the sum of an infinite series is limited,\nso all we have is, telescoping ,geometric series, rearranging terms (could someone give me an example of this thanxs), manipulation, fourier series- (which i can bearly remember) is there anymore?\n\nalso what is the name of the method saltydog was originally using? I didnt really understand it but i have to admit anything unusual is interesting, saltydog please help? thanxs\n\nwell after making up a few unusual series and trying to derive the sum, for the majority of them i couldnt derive anything except for one of them , and here it is please update me on what you think?\n\n$$\\sum_{k=0}^{\\infty}\\left(\\begin{array}{cc}k \\\\r \\end{array}\\right)x^{k-r}=\\frac{1}{(1-x)^{r+1}}$$ took me ages to get this result, I shall call it the steven zeta series lol.\ndoes anybody have any other examples of such results?\n\nLast edited: Jun 13, 2005\n10. Jun 13, 2005\n\n### saltydog\n\nSteven . . . I struggle with these and a lot of other things in the forum. I made sure to give Hall credit above for that method. He suggested the technique and I followed through. I take 1 point credit for that work. Really I suspect there is some compilation which goes into more detail with sums. I would expect that even infinite sums could/should/would be arranged into groupings similar to integrals and differential equations and solved using specific technique applicable to membership in a particular class like we do for integrals. Anyway, I think if someone intensively investigated them this could be done to some measure of success. Interesting problems you bring up here though.\n\n11. Jun 13, 2005\n\n### steven187\n\nhello all\n\nwell i didnt really understaand what you ment by this, what is the problem being solved here, or what method is it?\n\nwell now Iv been searching the net for ages to find some kind of summary of all the different types of series, methods of finding the sum of a series, and special cases of series with there sums like the power, euler, Laurent, alternating ,arithmetic, hypergeometric series, Maclaurin ,binomial, taylor, harmonic, riemann, geometric, fourier, pathological,Asymptotic Series, Dirichlet L-series, Multiple Series, Hyperasymptotic Series and Superasymptotic Series( i dont think there is anymore) oh yes and the steven zeta series, well out of hours of searching i only found one document but that wasnt even good enough, does anybody have any links?, well saltydog I hope that there is some kind of compilation or else it looks like I might have to compile my own,\nwell as to the original problem this is how i did it enjoy\n\n$$\\int_0^x\\sum_{n=0}^{\\infty}z^n dz=\\int_0^x \\frac{1}{1-z} dz$$\n$$\\sum_{n=0}^{\\infty}\\int_0^x z^n dz=\\int_0^x \\frac{1}{1-z} dz$$\n$$\\sum_{n=0}^{\\infty}\\frac{x^{n+1}}{n+1}=-\\log{(1-x)}$$\nnow we substitute $$x=\\frac{-1}{2}$$ in which after manipulating it we get\n$$\\sum_{n=1}^{\\infty}\\frac{(-2)^{-n}}{(n+1)}=-1+2\\log[\\frac{3}{2}]$$\n\nLast edited: Jun 13, 2005\n12. Jun 13, 2005\n\n### TenaliRaman\n\nI believe you realise that your first few statements are to be justified even though they are correct.\n\n-- AI\n\n13. Jun 14, 2005\n\n### saltydog\n\nThanks Steven. That's very nice. Tenali, you mind explaining what needs to be justified?\n\n14. Jun 14, 2005\n\n### steven187\n\nhello\n\nthanxs saltydog, well about what needs to be justified is that to swap the summation sign and the integral sign it needs to satisfy the conditions of the dominated convergence theorem i dont think there is anythin else to justify\n\nSteven\n\nLast edited: Jun 14, 2005", "date": "2016-12-05 19:07:48", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/finding-the-sum-of-a-series.78754/", "openwebmath_score": 0.848241925239563, "openwebmath_perplexity": 909.2142263740213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668662546613, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8559183366422911}}
{"url": "https://math.stackexchange.com/questions/2404030/evaluating-frac12-pi-i-int-z-3-frace-pi-zz2z22z2dz", "text": "# Evaluating $\\frac{1}{2 \\pi i} \\int_{|z|=3} \\frac{e^{\\pi z}}{z^2(z^2+2z+2)}dz$\n\nI need some help with Complex Analysis:\n\nTo evaluate the integral $$\\frac{1}{2 \\pi i} \\int_{|z|=3} \\frac{e^{\\pi z}}{z^2(z^2+2z+2)}dz$$ Here is what I tried:\n\nSo, $f(z)=\\frac{e^{\\pi z}}{z^2(z^2+2z+2)}$ has 3 singularities, $z_0 = 0$, $z_0= -1+i$ and $z_0=-1-i$. And all of them are interior to the contour. $\\Rightarrow$ We need to find the residues at all the 3 points.\n\n$$Res_{z_0=0}(f(z)) = Res_{z_0=0} \\left (\\frac{e^{\\pi z}}{z^2(z^2+2z+2)} \\right)\\\\=\\frac{1}{2}(\\pi-1)$$\n\n$$Res_{z_0=-1+i}(f(z)) = Res_{z_0=-1+i} \\left (\\frac{e^{\\pi z}}{z^2(z^2+2z+2)} \\right)\\\\=-\\frac{e^{-\\pi}}{4}$$\n\n$$Res_{z_0=-1-i}(f(z)) = Res_{z_0=-1-i} \\left (\\frac{e^{\\pi z}}{z^2(z^2+2z+2)} \\right)\\\\=-\\frac{e^{-\\pi}}{4}$$\n\n$$\\Rightarrow \\frac{1}{2\\pi i}\\int_{|z|=3} \\frac{e^{\\pi z}}{z^2(z^2+2z+2)}dz = \\frac{1}{2 \\pi i} \\times \\left[ 2\\pi i \\left(\\frac{1}{2}(\\pi -1) - \\frac{e^{-\\pi}}{4} - \\frac{e^{-\\pi}}{4}\\right)\\right]\\\\ = \\left( \\frac{1}{2}(\\pi-1)-\\frac{e^{-\\pi}}{2}\\right)\\\\ = \\frac{\\pi-1-e^{-\\pi}}{2}$$ But I was told that this was wrong. I couldn't find any mistakes for my work. Can any of you check to see if my work is valid? Or, if it is wrong, how else can I evaluate this? Any helps or comments would be appreciated. Can someone provide a valid solution (or a different approach)to evaluate this integral? Because some comments say it is wrong. But, some say it is correct. I am confused....\n\n\u2022 the pole at $z=0$ is double \u2013\u00a0G Cab Aug 23 '17 at 22:36\n\u2022 Will it make any difference? Since pole is z=o and z=0. for z^2 \u2013\u00a0SirBanana Aug 23 '17 at 22:41\n\u2022 It does matter, cuz the formulas differ \u2013\u00a0Brevan Ellefsen Aug 23 '17 at 22:48\n\u2022 Yes, but its \"influence\" in the \"surrounding\" is \"double\" \u2013\u00a0G Cab Aug 23 '17 at 22:49\n\nSo checking the residues...\n\nThe pole at $z=0$ is double, so (in my opinion) it's easier to find the residue via the series expansion than with the explicit formula. Factoring the denominator, the integrand is $\\frac{e^{\\pi z}}{z^2(z+1-i)(z+1+i)}$. The series for $e^{\\pi z}$ is just $1 + \\pi z + \\frac{\\pi^2 z^2}{2!} + ...$ The other factors except for $\\frac{1}{z^2}$ are of the form $\\frac{1}{a + z}$, which has the series representation $\\frac{1}{a} - \\frac{z}{a^2}+...$ So the integrand, as a product of series, is $\\frac{1}{z^2}(1 + \\pi z + ...)(\\frac{1}{1-i} - \\frac{z}{(1-i)^2} + ...)(\\frac{1}{1+i} - \\frac{z}{(1+i)^2} + ...)$ Since we're looking for the residue, the only term we care about is $z^{-1}$. Since $\\frac{1}{z^2}$ is the only factor with negative exponents, we only need the first two terms of each series. Finding the terms that would have order $-1$, we find the coefficient to be $\\frac{\\pi}{(1+i)(1-i)} - \\frac{1}{(1-i)^2(1+i)} - \\frac{1}{(1+i)^2(1-i)}$. The first term is $\\frac{\\pi}{2}$, and the second and third terms combine to $-\\frac{1}{2(1+i)} - \\frac{1}{2(1-i)} = -\\frac{1-i}{4} - \\frac{1+i}{4} = -\\frac{1}{2}$. So the first residue is (unless I've also made a horrible mistake) $\\frac{\\pi - 1}{2}$, as you found.\n\nNow for the next two residues: since these are both simple poles, they can be evaluated easily with the explicit formula. The second residue should be $\\frac{e^{\\pi z}}{z^2(z+1-i)}$ evaluated at $z = -1 - i$, and the third is similar. Plugging in, the second residue is $\\frac{e^{\\pi (-1 - i)}}{(-1-i)^2(-1-i+1-i)} = \\frac{e^{-\\pi}e^{-i\\pi}}{(2i)(-2i)}$ which is $-\\frac{e^{-\\pi}}{4}$. The third is similar, and I also found it also matches what you have.\n\nMaybe we both made the same mistakes, but your evaluation looks right to me.\n\nI do not see anything wrong with your calculation.\n\nBut Maple gets $$\\frac{\\pi}{2} - \\frac{1}{2} - \\frac{1}{4e^{\\pi}}$$ so the last term disagrees with yours. But numerically in Maple, the same integral agrees with your answer, not Maple's symbolic answer. Some sort of bug in Maple, I guess. (Maple 2015 build 1097895)\n\n\u2022 This just confuses me even more.... \u2013\u00a0SirBanana Aug 23 '17 at 23:41", "date": "2021-05-17 21:44:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2404030/evaluating-frac12-pi-i-int-z-3-frace-pi-zz2z22z2dz", "openwebmath_score": 0.8723546266555786, "openwebmath_perplexity": 301.51941221523805, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668684574637, "lm_q2_score": 0.8723473647220786, "lm_q1q2_score": 0.8559183320514828}}
{"url": "https://math.stackexchange.com/questions/3959113/what-is-the-difference-between-expectation-variance-for-statistics-versus-p/3959120", "text": "# What is the difference between \"expectation\", \"variance\" for statistics versus probability textbooks?\n\nIt seems that there are two ideas of expectation, variance, etc. going on in our world.\n\nIn any probability textbook:\n\nI have a random variable $$X$$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like, $$\\mathbb{E}[X] = \\sum\\limits_{i = 1}^n x_i p(x_i)$$ where $$p$$ is the probability mass function, $$p: x_i \\mapsto [0,1], \\sum_{i = 1}^n p(x_i) = 1$$ and $$x_i \\in \\text{range}(X)$$. The variance is, $$\\mathbb{E}[(X - \\mathbb{E}[X])^2]$$\n\nThe definition is similar for a continuous RV.\n\nHowever, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)\n\nI have a multi-set of data $$D = \\{x_i\\}_{i = 1}^n$$ (weight of onions, height of school children). The mean of this dataset is\n\n$$\\dfrac{1}{n}\\sum\\limits_{i= 1}^n x_i$$\n\nThe variance of this dataset (according to \"science buddy\" and \"mathisfun dot com\" and government of Canada) is,\n\n$$\\dfrac{1}{n}\\sum\\limits_{i= 1}^n(x_i - \\sum\\limits_{j= 1}^n \\dfrac{1}{n} x_j)^2$$\n\nI mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:\n\n1. Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)\n\n2. Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?\n\n3. Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the \"mean weight\" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?/\n\n\u2022 There's no probabilistic assumption about the sample mean or sample variance because they're just numbers computed from the sample data. If you measure the heights of $100$ people, you have $100$ numbers and you can do whatever computations you like with them. Dec 23, 2020 at 5:07\n\u2022 Can you explain what you what you mean that 'uniform distribution' was assumed? Isnt p(x_i)=lim n->infty #occurences of x_i/n ? So an approximation of p(x_i) is #occurences of x_i /n. So 1/n sum x_i = 1/n sum_{x_i} x_i #occurences of x_i = sum x_i p(x_i). Of course p(x_i) is only an approximation of p(x_i) in statistics. I do not see how uniformity was used here Dec 23, 2020 at 16:25\n\u2022 \"Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science \" Can you be more clear as to what assumption you think is made elsewhere, but isn't made in data science? Dec 24, 2020 at 20:46\n\u2022 @lalala If you take the average off $n$ values, that is equal to the expected value of a random variable that has a uniform distribution across those alues. Dec 24, 2020 at 20:47\n\nYou ask a very insightful question that I wish were emphasized more often.\n\nEDIT: It appears you are seeking reputable sources to justify the above. Sources and relevant quotes have been provided.\n\nHere's how I would explain this:\n\n\u2022 In probability, the emphasis is on population models. You have assumptions that are built-in for random variables, and can do things like saying that \"in this population following such distribution, the probability of this value is given by the probability mass function.\"\n\u2022 In statistics, the emphasis is on sampling models. With most real-world data, you do not have access to the data-generating process governed by the population model. Probability provides tools to make guesses on what the data-generating process might be. But there is always some uncertainty behind it. We therefore attempt to estimate characteristics about the population given data.\n\nFrom Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 1.6:\n\nThe objective of statistics is to make an inference about a population based on information contained in a sample taken from that population...\n\nA necessary prelude to making inferences about a population is the ability to describe a set of numbers...\n\nThe mechanism for making inferences is provided by the theory of probability. The probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer this from the characteristics of an unknown population. Thus, probability is the foundation of the theory of statistics.\n\nFrom Shao's Mathematical Statistics, 2nd edition, section 2.1.1:\n\nIn statistical inference... the data set is viewed as a realization or observation of a random element defined on a probability space $$(\\Omega, \\mathcal{F}, P)$$ related to the random experiment. The probability measure $$P$$ is called the population. The data set or random element that produces the data is called a sample from $$P$$... In a statistical problem, the population $$P$$ is at least partially unknown and we would like to deduce some properties of $$P$$ based on the available sample.\n\nSo, the probability formulas of the mean and variance assume you have sufficient information about the population to calculate them.\n\nThe statistics formulas for the mean and variance are attempts to estimate the population mean and variance, given a sample of data. You could estimate the mean and variance in any number of ways, but the formulas you've provided are some standard ways of estimating the population mean and variance.\n\nNow, one logical question is: why do we choose those formulas to estimate the population mean and variance?\n\nFor the mean formula you have there, one can observe that if you assume that your $$n$$ observations can be represented as observed values of independent and identically distributed random variables $$X_1, \\dots, X_n$$ with mean $$\\mu$$, $$\\mathbb{E}\\left[\\dfrac{1}{n}\\sum_{i=1}^{n}X_i \\right] = \\mu$$ which is the population mean. We say then that $$\\dfrac{1}{n}\\sum_{i=1}^{n}X_i$$ is an \"unbiased estimator\" of the population mean.\n\nFrom Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.1:\n\nFor example, suppose we want to estimate a population mean $$\\mu$$. If we obtain a random sample of $$n$$ observations $$y_1, y_2, \\dots, y_n$$, it seems reasonable to estimate $$\\mu$$ with the sample mean $$\\bar{y} = \\dfrac{1}{n}\\sum_{i=1}^{n}y_i$$\n\nThe goodness of this estimate depends on the behavior of the random variables $$Y_1, Y_2, \\dots, Y_n$$ and the effect this has on $$\\bar{Y} = (1/n)\\sum_{i=1}^{n}Y_i$$.\n\nNote. In statistics, it is customary to use lowercase $$x_i$$ to represent observed values of random variables; we then call $$\\frac{1}{n}\\sum_{i=1}^{n}x_i$$ an \"estimate\" of the population mean (notice the difference between \"estimator\" and \"estimate\").\n\nFor the variance estimator, it is customary to use $$n-1$$ in the denominator, because if we assume the random variables have finite variance $$\\sigma^2$$, it can be shown that $$\\mathbb{E}\\left[\\dfrac{1}{n-1}\\sum_{i=1}^{n}\\left(X_i - \\dfrac{1}{n}\\sum_{j=1}^{n}X_j \\right)^2 \\right] = \\sigma^2\\text{.}$$ Thus $$\\dfrac{1}{n-1}\\sum_{i=1}^{n}\\left(X_i - \\dfrac{1}{n}\\sum_{j=1}^{n}X_j \\right)^2$$ is an unbiased estimator of $$\\sigma^2$$, the population variance.\n\nIt is also worth noting that the formula you have there has expected value $$\\dfrac{n-1}{n}\\sigma^2$$ and $$\\dfrac{n-1}{n} < 1$$ so on average, it will tend to underestimate the population variance.\n\nFrom Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.2:\n\nFor example, suppose that we wish to make an inference about the population variance $$\\sigma^2$$ based on a random sample $$Y_1, Y_2, \\dots, Y_n$$ from a normal population... a good estimator of $$\\sigma^2$$ is the sample variance $$S^2 = \\dfrac{1}{n-1}\\sum_{i=1}^{n}(Y_i - \\bar{Y})^2\\text{.}$$\n\nThe estimators for the mean and variance above are examples of point estimators. From Casella and Berger's Statistical Inference, Chapter 7.1:\n\nThe rationale behind point estimation is quite simple. When sampling is from a population described by a pdf or pmf $$f(x \\mid \\theta)$$, knowledge of $$\\theta$$ yields knowledge of the entire population. Hence, it is natural to seek a method of finding a good estimator of the point $$\\theta$$, that is, a good point estimator. It is also the case that the parameter $$\\theta$$ has a meaningful physical interpretation (as in the case of a population) so there is direct interest in obtaining a good point estimate of $$\\theta$$. It may also be the case that some function of $$\\theta$$, say $$\\tau(\\theta)$$ is of interest.\n\nThere is, of course, a lot more that I'm ignoring for now (and one could write an entire textbook, honestly, on this topic), but I hope this clarifies things.\n\nNote. I know that many textbooks use the terms \"sample mean\" and \"sample variance\" to describe the estimators above. While \"sample mean\" tends to be very standard terminology, I disagree with the use of \"sample variance\" to describe an estimator of the variance; some use $$n - 1$$ in the denominator, and some use $$n$$ in the denominator. Also, as I mentioned above, there are a multitude of ways that one could estimate the mean and variance; I personally think the use of the word \"sample\" used to describe such estimators makes it seem like other estimators don't exist, and is thus misleading in that way.\n\n## In Common Parlance\n\nThis answer is informed primarily by my practical experience in statistics and data analytics, having worked in the fields for about 6 years as a professional. (As an aside, I find one serious deficiency with statistics and data analysis books is providing mathematical theory and how to approach problems in practice.)\n\nIs there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the \"mean weight\" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?\n\nIn most cases, you want to just stick with the statistical definitions. Most people do not think of statistics as attempting to estimate quantities relevant to a population, and thus are not thinking \"I am trying to estimate a population quantity using an estimate driven by data.\" In such situations, people are just looking for summaries of the data they've provided you, known as descriptive statistics.\n\nThe whole idea of estimating quantities relevant to a population using a sample is known as inferential statistics. While (from my perspective) most of statistics tends to focus on statistical inference, in practice, most people - especially if they've not had substantial statistical training - do not approach statistics with this mindset. Most people whom I've worked with think \"statistics\" is just descriptive statistics.\n\nIn descriptive data analysis, a few summary measures may be calculated, for example, the sample mean... and the sample variance... However, what is the relationship between $$\\bar{x}$$ and $$\\theta$$ [a population quantity]? Are they close (if not equal) in some sense? The sample variance $$s^2$$ is clearly an average of squared deviations of $$x_i$$'s from their mean. But, what kind of information does $$s^2$$ provide?... These questions cannot be answered in descriptive data analysis.\n\n## Other remarks about the sample mean and sample variance formulas\n\nLet $$\\bar{X}_n$$ and $$S^2_n$$ denote the sample mean and sample variance formulas provided earlier. The following are properties of these estimators:\n\n\u2022 They are unbiased for $$\\mu$$ and $$\\sigma^2$$, as explained earlier. This is a relatively simple probability exercise.\n\u2022 They are consistent for $$\\mu$$ and $$\\sigma^2$$. Since you know measure theory, assume all random variables are defined over a probability space $$(\\Omega, \\mathcal{F}, P)$$. It follows that $$\\bar{X}_n \\overset{P}{\\to} \\mu$$ and and $$S^2_n \\overset{P}{\\to} \\sigma^2$$, where $$\\overset{P}{\\to}$$ denotes convergence in probability, also known as convergence with respect to the measure $$P$$. See https://math.stackexchange.com/a/1655827/81560 for the sample variance (observe that the estimator with the $$n$$ in the denominator is used here; simply multiply by $$\\dfrac{n-1}{n}$$ and apply a result by Slutsky) and Proving a sample mean converges in probability to the true mean for the sample mean. As a stronger result, convergence is almost sure with respect to $$P$$ in both cases (Sample variance converge almost surely).\n\u2022 If one assumes $$X_1, \\dots, X_n$$ are independent and identically distributed based on a normal distribution with mean $$\\mu$$ and variance $$\\sigma^2$$, one has that $$\\dfrac{\\sqrt{n}(\\bar{X}_n - \\mu)}{\\sqrt{S_n^2}}$$ follows a $$t$$-distribution with $$n-1$$ degrees of freedom, which converges in distribution to a normally-distributed random variable with mean $$0$$ and variance $$1$$. This is a modification of the central limit theorem.\n\u2022 If one assumes $$X_1, \\dots, X_n$$ are independent and identically distributed based on a normal distribution with mean $$\\mu$$ and variance $$\\sigma^2$$, $$\\bar{X}_n$$ and $$S^2_n$$ are uniformly minimum-variance unbiased estimators (UMVUEs) for $$\\mu$$ and $$\\sigma^2$$ respectively. It also follows that $$\\bar{X}_n$$ and $$S^2_n$$ are independent, through - as mentioned by Michael Hardy - showing that $$\\text{Cov}(\\bar{X}_n, X_i - \\bar{X}_n) = 0$$ for each $$i = 1, \\dots, n$$, or as one can learn from more advanced statistical inference courses, an application of Basu's Theorem (see, e.g., Casella and Berger's Statistical Inference).\n\u2022 I believe the expressions for mean and variance estimators should be accompanied with a word or two on the central limit theorem. Dec 23, 2020 at 5:52\n\u2022 You wrote: \"If one assumes $X_1,\\ldots,X_n$ are normally distributed, $\\bar X_n$ and $S_n^2$ are\" UMVUEs. (This after assuming they are i.i.d.) I would add that under those assumptions $\\bar X_n$ and $S_n^2$ are independent, and that is used in deriving the t-distribution. One of the quickest ways to show that involves showing that $\\operatorname{cov} \\left(\\,\\overline X, X_i-\\overline X\\,\\right)=0.$ When $U,V$ are both linear combinations of i.i.d. normals, then their covariance is $0$ only if they are independent. Dec 26, 2020 at 21:47\n\u2022 @MichaelHardy Thanks for correcting me; I've put in the edits. Dec 26, 2020 at 22:08\n\u2022 @Clarinetist : I see you say it's \"through an application of Basu's theorem,\" but nothing so advanced as Basu's theorem is needed, since the method described in my comment above is enough. It seems misleading to tacitly suggest that you can't learn to prove that proposition until you learn Basu's theorem. Dec 26, 2020 at 22:11\n\u2022 @MichaelHardy Thanks, I've corrected that as well. Dec 26, 2020 at 22:13\n\nThe first definitions you gave are correct and standard, and statisticians and data scientists will agree with this. (These definitions are given in statistics textbooks.) The second set of quantities you described are called the \"sample mean\" and the \"sample variance\", not mean and variance.\n\nGiven a random sample from a random variable $$X$$, the sample mean and sample variance are natural ways to estimate the expected value and variance of $$X$$.\n\n\u2022 Thanks! So the distinction is one is \"sample\" mean/variance the other is mean/variance as it is. And also those formulas are estimate of the value of mean and variance. But isn't it true when people are working with them in practice, say using some data science package such as Scikit-Learn, the underlying probabilistic assumption (such as distribution of data) is dropped as if nothing ever happened? I have never seen anyone writing a documentation mentioning the probabilistic definition. I think this causes confusion for young practictioners, don't you think? Dec 23, 2020 at 5:39\n\u2022 Imagine in the shoes of a young data scientist who just took a course on Lebesgue integration or measure theory. You spent 4 month talking about these abstract definitions, and then integrating these really hard integrals just to get the mean, using various tricks just to make these integrals tractable. You get your first job and your boss tells you just to add stuff up and divide. Isn't this confusing? Dec 23, 2020 at 5:41\n\u2022 @Norman My response to your comments here is that probability is not a statistics class, and trying to act as if that is the case would be lying. Additionally, most managers have not taken courses nor understand the difference between probability and statistics, and are focused solely on descriptive statistics to understand empirically the behavior of their data. They calculate a sample average because that's what they habitually have done, and they think it is an adequate summary of the data. Dec 23, 2020 at 13:25\n\u2022 @Norman I could go on a very long rant about how poorly taught statistics is in many schools and how very little guidance is given on how to approach statistical problems in practice, but I will avoid that for now. Dec 23, 2020 at 13:27\n\u2022 @Norman I think it can be a source of confusion that people often say mean or variance when they really mean \"sample mean\" or \"sample variance\". One of the challenges of coming into the engineering world from a pure math background is that often people speak much less precisely than one is used to. Dec 23, 2020 at 15:23\n\nOther answers \u2014 particularly Clarinetist\u2019s \u2014 give excellent rundowns of the most important side of the answer. Given a random variable, we can sample it, and use the sample mean (defined in the statistical sense) to estimate the actual mean of the random variable (defined in the sense of probability theory), and similarly for variance, etc.\n\nBut the connection in the other direction doesn\u2019t seem to have been mentioned yet. This is not as important, but it\u2019s much more straightforward, and worth pointing out. Given a sample, i.e. a finite multiset of values $$\\{x_i\\}_{i \\in I}$$, we can \u201cconsider this as a distribution\u201d, i.e. take a random variable $$X$$, with value $$x_i$$ for $$i$$ distributed uniformly over $$I$$. Then the mean, variance, etc. of $$X$$ (in the sense of probability theory) will be precisely the mean, variance, etc. of the original multiset (defined in the statistical sense).\n\nThe general expression for arithmetic mean is $$\\frac{\\sum\\limits_{i= 1}^n w_i x_i}{\\sum\\limits_{i= 1}^n w_i}$$, or even more generally, $$\\frac{\\int w(t) f(t)dt}{\\int f(t)dt}$$ (there are then ways to recover the discrete case from that).\n\nIf you set all $$w_i$$ to $$1$$, or really to anything as long as it's constant, you get $$\\dfrac{1}{n}\\sum\\limits_{i= 1}^n x_i$$. This is referred to as an \"unweighted\" average, although technically it's still weighted, it's just that you're multiplying everything by $$1$$, so you don't notice it. If you set $$p_i = \\frac{w_i}{ \\sum\\limits_{k= 1}^n w_k}$$, and interpret $$p_i$$ as the probability of the $$i$$th event, then you get the average weighted by probability, which is also known as expected value.\n\nYou have to be careful about \"unweighted\" averages, as they often actually are weighted, but by weights that you didn't want. For instance, suppose you want the average income over the US, and you have the average income for each state individually. You could just add all of those averages together, and then divide by $$50$$. People will often call this the \"unweighted\" or \"simple\" average, but you're actually weighting people by the reciprocal of their states population; the fewer people there are in a state, the more each person's income affects that state's average, and so the more they affect the total \"unweighted average\". As a result, to get the actual overall average income from the individual states' averages, you have to multiply each state's average by its population to get its total income, add all of those together, and then divide by the total population.\n\nA common weighting you'll see is frequency weighting. This is where you multiply each value by the number of times it appears. For instance, if you measure something once a month for a year, and the only values you get are $$0$$, $$1$$, and $$2$$, taking the simple average of those values gives you $$1$$. But that's probably not the real average. To get a more meaningful average, you should take each of these values, weight them by how many months they appear for, and then take the average.\n\nOne property of weighted averages is that multiplying all of the weights by a constant number doesn't change the final result (you'll just divide it out again when you divide by the total of the weights). So weighting by the frequencies is equivalent to weighting by the percentage of cases each value is. That is, if $$0$$ is the value for $$5$$ months, $$1$$ is the value for $$4$$ months, and $$2$$ is the value for $$3$$, the weighting of $$5,4,3$$ is equivalent to the weighting of $$\\frac 5 {12},\\frac 4{12},\\frac 3{12}$$.\n\nSo if you have a probability distribution where one thing has a $$60$$% chance of happening, and something else has a $$40$$% chance of happening, the expected value is simply the frequency weighted average.\n\nFor example, if my cashier asks me about the \"mean weight\" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?\n\nThe expected value of a random variable is the expected value of that random variable. It is a property of the distribution. If you're asked for the expected value of a random variable, you find the expected value of that random variable. If you aren't asked for the expected value of a random variable, you don't go looking for a random variable to take the expected value of. How you know whether you take the expected value of random variable is if there is a random variable to take the expected value of. Even if values come from a distribution, the average of those values is the average of those values, not the average of the distribution they came from.\n\nIf people are talking rigorously, they will explicitly say they want the expected value. You many, however, see people asking for the \"mean\" or \"average\", when they really want the expected value, but you can recognize those cases by there being a random variable. For instance, if someone asks \"What's the average payout of this slot machine?\", the context suggests that you should take the expected value of the distribution of payouts, and not simply take the set of different possible payouts and take the simple average. There could be some ambiguity as to whether \"the payout\" refers to the random process that pays money out, in which case you should take the expected value of the distribution (population mean), or the actual money paid out, in which case you should take the average of all the actual payouts made by the machine (sample mean), but in the latter case, you still should take the frequency weighted average.\n\nThe confusion actually comes from notation, where symbols mean different things in two formulas.\n\nFirst, let's take a look at the \"probabilistic\" definition: $$\\mathbb{E}[X] = \\frac{1}{n}\\sum_{i=1}^n x_i p(x_i)$$ Here the random variable $$X$$ takes $$n$$ distinct values $$x_1,\\ldots, x_n$$, each with a probability mass function $$p(x_i)$$.\n\nIn the \"statistical\" definition we have an estimate of the expected value, based on the observed values of the random variable $$z_1, \\ldots, z_N$$: $$\\hat{\\mathbb{E}}[X] = \\frac{1}{N}\\sum_{j=1}^N z_j$$ Notice that I renamed the variables compared to your original formula, so as to avoid the confusion. Here $$N$$ is the number of observations, and $$z_j$$s are the actual observations. For example, if $$X$$ is a random variable representing rolls of a loaded die, then $$n = 6$$ and $${x_i} = \\{1, 2, 3, 4, 5, 6\\}$$; whereas $$N$$ can be arbitrarily large, and $$z_j$$s will just be a long sequence of random draws from the set of $${x_i}$$s.\n\nNow, you can rewrite the \"statistical\" formula by collecting different values of $$z_j$$s into the groups according to which $$x_i$$ they correspond to (for example, first collect all 1s, then all 2s, etc): $$\\hat{\\mathbb{E}}[X] = \\frac{1}{N}\\big(x_1\\cdot|\\{z_j=x_1\\}| + \\cdots+x_n\\cdot|\\{z_j=x_n\\}|\\big) \\\\ =\\frac{1}{N}\\sum_{i=1}^n x_i \\sum_{j=1}^N\\mathbb{1}[z_j=x_i] \\\\ =\\sum_{i=1}^n x_i \\hat{p}(x_i)$$ where $$\\hat{p}(x_i) = \\frac{1}{N}\\sum_{j=1}^N \\mathbb{1}[z_j=x_i]$$ is the estimate of the probability mass function: the number of times a value $$x_i$$ was encountered in the sample divided by the sample size, i.e. the observed frequency of the value $$x_i$$.\n\nNow you can see that the \"probabilistic\" and \"statistical\" definitions are actually the same, with the only difference that we replace the theoretical mass distribution function $$p(x)$$ (which may not be known) with the empirical (observed) mass distribution function $$\\hat{p}(x)$$.\n\nStatement: the \"statistic\" universe from your question is a partial case of \"probabilistic\" universe.\n\nWe need some notation. Suppose that we have a random variable $$\\xi$$ in sense of probability theory and $$P(\\xi = x_k) = p_k$$, $$0 \\le k \\le n$$.\n\nFor example, consider $$\\xi$$ which is equal to the number of children in \"abstract\" random family. Put $$x_k = k$$. Then $$E \\xi = \\sum_{k=0}^n x_k p_k$$ where $$p_k = P(\\xi = x_k)$$, and $$D\\xi = E (\\xi - E\\xi)^2 = \\sum_{k=0}^n p_k (x_k - E\\xi )^2$$. Moreover, $$P(\\xi = 69) > 0$$ (according to Guinness World Records, I am not sure, but let us suppose that it's a record nowadays) and as there was people with $$69$$ children then we may think that for example $$P(80) > 0$$ - it's natural, because $$80$$ children is possible, even although Guinness World Records says that still nobody had $$80$$ children.\n\nIn reality we don't have abstract random family, random mothers, random fathers and random children. We have finite number $$N$$ ( $$10^6 < N < 10^{100}$$) of families, let us number them, and there are $$y_1$$ children in the first family, there are $$y_2$$ children in family number $$2$$, ..., there are $$y_N$$ children in family $$N$$.\n\nAnalogy: $$\\xi$$ corresponds to a fair dice itself and numbers $$y_1$$, $$y_2$$, ... corresponds to dice rolls and have form: $$5, 1, 6, 6, 3, ...$$, these are fixed numbers.\n\nNow consider a random value $$\\tau$$, which has uniform distribution on $$\\{ 1, 2, \\ldots, N\\}$$. It means that $$P(\\tau = k) = \\frac{1}{N}$$ for all $$1 \\le k \\le N$$.\n\nNumbers $$y_1, \\ldots, y_N$$ are fixed - suppose they are written in some sociological table. Let us consider a random value $$y_{\\tau}$$. It's not a number of children in abstract random family. It is a number of children in some real family, if a number $$\\tau$$ of such family we chose randomly.\n\nLet us find $$E y_{\\tau}$$ and $$D y_{\\tau}$$. With probability $$\\frac{1}N$$ we have $$\\tau = k$$ and hence $$y_{\\tau} = y_k$$. Thus $$E y_{\\tau}=\\frac{1}N \\sum_{k=1}^N y_k$$ and $$D y_{\\tau} = E (y_{\\tau} - Ey_{\\tau})^2 = E (y_{\\tau} - \\frac{1}N \\sum_{k=1}^N y_k )^2 = \\frac{1}N \\sum_{k=0}^N (y_k -\\frac{1}N \\sum_{k=1}^N y_k )^2.$$\n\nNow the correspondence of \"statistical universe\" and \"probabilistic universe\" is obvious.\n\nNotice that all $$y_i \\le 69$$ and $$y_i = 69$$ is the maximum value, but the maximum value of $$\\xi$$ is bigger (and not less than $$80$$).\n\nSo, we have shown that the \"statistical universe\" is a partial case of \"probabilistic\" universe.\n\nMoreover, there are two moments, when there is randomness:\n\n1. when we go from abstract random family, corresponding to $$\\xi$$, to numbers $$y_1, \\ldots, y_N$$ from sociological handbook.\n\nAs it was already mentioned, there is analogy: $$\\xi$$ corresponds to a fair dice itself and numbers $$y_1$$, $$y_2$$, ... corresponds to dice rolls and have form: $$5, 1, 6, 6, 3, ...$$, these are fixed numbers.\n\n1. [here we suppose that $$y_1, \\ldots, y_N$$ are fixed numbers] when we go from the whole sociological handbook to a family with random number $$\\tau$$ and see, how much children is there in this family. The number is $$y_{\\tau}$$.\n\nWhen from $$y_{\\tau}$$ we go to $$E y_{\\tau} = \\frac{1}N \\sum_{k=1}^N y_k$$, we get rid of the second randomness, but we still do not get rid of the first randomness. It's connected with the fact that numbers $$y_k$$ could be another: it could happen, that in some families could be more children, and in some families could be less children, if the circumstances were different. In this sence numbers $$y_i$$ are not fixed: they are r.v. sampled from distribution, corresponding to $$\\xi$$. And in this sence $$\\frac{1}N \\sum_{k=1} y_k$$ is a random value, and we may write limit theorems such as L.L.N.: $$\\frac{1}N \\sum_{k=1}^N y_k \\to E\\xi, \\text{ }N \\to \\infty$$ or CLT or LIL.\n\nAddition: it was shown that if $$y_1, \\ldots, y_N$$ are fixed and $$\\tau$$ is random then \"probabilistic\" mean (expectation) of $$y_{\\tau}$$ and \"probabilistic\" variance of $$y_{\\tau}$$ are well known sample mean and sample variance.\n\nHope it will be useful. If you have any questions, you are welcome.", "date": "2022-05-26 02:30:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3959113/what-is-the-difference-between-expectation-variance-for-statistics-versus-p/3959120", "openwebmath_score": 0.8698630928993225, "openwebmath_perplexity": 271.8597849264751, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104933824753, "lm_q2_score": 0.8856314783461303, "lm_q1q2_score": 0.8558835539435348}}
{"url": "https://math.stackexchange.com/questions/3016686/riemann-darboux-integrability-of-subinterval", "text": "# Riemann-Darboux Integrability of Subinterval\n\nI'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.\n\nTheorem: Suppose $$f$$ is Riemann-Darboux integrable on $$[a,b]$$. Let $$c\\in(a,b)$$. Then, $$f$$ is Riemann-Darboux integrable on the intervals $$[a,c]$$ and $$[c,b]$$.\n\nAttempted Proof: Since $$f$$ is Riemann-Darboux integrable on $$[a,b]$$, for arbitrary $$\\epsilon>0$$, there exists a partition $$P$$ of $$[a,b]$$ such that $$U(f,P)-L(f,P)<\\epsilon$$.\n\nLet $$n_p,n_{p*}$$, and $$n_{p'}$$ be the number of partition parts in $$P$$, $$P^*$$, and $$P'$$, respectively. Also, let $$m_i=\\inf_{x\\in[x_{i-1},x_i]}f(x)$$ and $$M_i=\\sup_{x\\in[x_{i-1},x_i]}f(x)$$.\n\nConsider the partition of $$[a,c]$$ given by $$P^*=P\\cap[a,c]$$. Then, $$U(f,P^*)-L(f,P^*)=\\sum_{i=1}^{n_{p*}}(M_i-m_i)\\Delta x_i\\le\\sum_{i=1}^{n_{p*}}(M_i-m_i)\\Delta x_i+ \\sum_{n_{p*}+1}^{n_p}(M_i-m_i)\\Delta x_i=\\sum_{i=1}^{n_p}(M_i-m_i)\\Delta x_i=U(f,P)-L(f,P)<\\epsilon$$ Therefore, $$f$$ is integrable on $$[a,c]$$.\n\nNext, consider the partition $$[a,b]$$ given by $$P'=P\\cap[c,b]$$. Then,\n\n$$U(f,P')-L(f,P')=\\sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)\\Delta x_i\\le\\sum_{i=1}^{n_{p*}}(M_i-m_i)\\Delta x_i+ \\sum_{n_{p*}+1}^{n_p}(M_i-m_i)\\Delta x_i=\\sum_{i=1}^{n_p}(M_i-m_i)\\Delta x_i=U(f,P)-L(f,P)<\\epsilon$$ Therefore, $$f$$ is integrable on $$[c,b]$$. $$\\square$$\n\nAny and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.\n\nThis is essentially fine, although you are tacitly assuming that $$c$$ itself is a member of the partition, $$P$$. This isn't a big deal though - a $$P$$ such that $$U(P,f)-L(P,f)<\\epsilon$$ is guaranteed by integrability, and you can always just add $$c$$ to this partition if it is not already there. Namely, if $$P_{c}$$ is this new partition, called a refinement of $$P$$, one must have $$U(P_{c},f)-L(P_{c},f)\\leq U(P,f)-L(P,f)<\\epsilon$$ and you can proceed as you have done in your proof.\nAn alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $$[a,b]$$, it is bounded and continuous up to a null set on $$[a,b]$$, and hence, on $$[a,c]$$ and $$[c,b]$$ as well.", "date": "2020-09-19 15:41:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3016686/riemann-darboux-integrability-of-subinterval", "openwebmath_score": 0.9292044043540955, "openwebmath_perplexity": 117.69271930193221, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9956005865761918, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.8558817271067747}}
{"url": "http://mathhelpforum.com/statistics/154536-pricing-dice-game.html", "text": "# Math Help - Pricing Dice Game\n\n1. ## Pricing Dice Game\n\nStage 1\n\nIn this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you \u00a34 pounds.\n\nA fair price to play this game should be \u00a33.50 as then the player or myself should not win any money if this game is played infinty times.\n\nStage 2\n\nHow much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.\n\nI priced this option at \u00a33.57, however I been told that is wrong. Can anyone explain why?\n\nCalypso\n\n2. Originally Posted by calypso\nStage 1\n\nIn this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you \u00a34 pounds.\n\nA fair price to play this game should be \u00a33.50 as then the player or myself should not win any money if this game is played infinty times.\n\nStage 2\n\nHow much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.\n\nI priced this option at \u00a33.57, however I been told that is wrong. Can anyone explain why?\n\nCalypso\nChoose a $t$ such that if the first roll $a\\le t$ roll again otherwise accept the prize of $\u00a3 a$.\n\nNow for any given $t$ you can work out the value of this game and hence a fair price. The fair price for the game is the maximum fair price.\n\n(I make $t=3$ the value of $t$ that maximises the return and a fair price is then $\u00a34.25$\n\nCB\n\n3. Sorry for the late reply I have just got back from holiday.\n\nThe text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is\n\n( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25\n\nI can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:\n\nSo I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout\n\nSo therefore the possible outcomes of the game are\n\n1 -> 1\n1 -> 2\n1 -> 3\n1 -> 4\n1 -> 5\n1 -> 6\n\n2 -> 1\n2 -> 2\n2 -> 3\n2 -> 4\n2 -> 5\n2 -> 6\n\n3 -> 1\n3 -> 2\n3 -> 3\n3 -> 4\n3 -> 5\n3 -> 6\n\n4\n5\n6\n\nTherefore fair price should equal 78/21 = \u00a33.71?\n\nThanks again\n\nCalypso\n\n4. Originally Posted by calypso\nI am still confuced as why I cant get this answer with my method:\n\nSo I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout\n\nSo therefore the possible outcomes of the game are\n\n1 -> 1\n1 -> 2\n1 -> 3\n1 -> 4\n1 -> 5\n1 -> 6\n\n2 -> 1\n2 -> 2\n2 -> 3\n2 -> 4\n2 -> 5\n2 -> 6\n\n3 -> 1\n3 -> 2\n3 -> 3\n3 -> 4\n3 -> 5\n3 -> 6\n\n4\n5\n6\nThe reason this goes wrong is that those outcomes are not all equally probable. Each of the final three outcomes (4, 5, 6) occurs with a probability of 1/6. But the initial outcomes 1, 2, 3 are subdivided into six subcases, each of which occurs with a probability of only 1/36. Multiplying each outcome by its probability, you get $\\tfrac3{36}(1+2+3+4+5+6) + \\tfrac16(4+5+6) = \\tfrac{17}4 = 4.25.$\n\n5. Originally Posted by calypso\nSorry for the late reply I have just got back from holiday.\n\nThe text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is\n\n( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25\n\nI can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:\nI don't like their decision rule, not that it is wrong, but because it refers to something not in the sample space (that is \"anything greater than 3.50\" when it could have said anything greater than 3 or even anything greater than or equal to 4).\n\nCB\n\n6. Great, thanks everyone for your help\n\nCalypso", "date": "2015-05-30 15:54:38", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/154536-pricing-dice-game.html", "openwebmath_score": 0.27965545654296875, "openwebmath_perplexity": 240.39966382851776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9504109798251322, "lm_q2_score": 0.9005297874526776, "lm_q1q2_score": 0.8558733976546175}}
{"url": "http://mathhelpforum.com/statistics/38753-die-probability.html", "text": "Math Help - die probability\n\n1. die probability\n\n\"What is the probability of getting a total score of more than 7 when two fair dice are rolled together?\"\n\nI got $\\frac{15}{36}$ but the book says $\\frac{1}{3}$?\n\nThanks!\n\n2. Hello, Nyoxis!\n\nWhat is the probability of getting a total score of more than 7\nwhen two fair dice are rolled together?\"\n\nI got $\\frac{15}{36} \\quad{\\color{blue}\\hdots\\text{ Right!}}$\n. . but the book says $\\frac{1}{3}\\quad {\\color{red} \\hdots\\text{ Wrong!}}$\nCan't imagine where they got their answer . . .\n\n$\\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\\\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\\color{blue}(2,6)} \\\\ (3,1) & (3,2) & (3,3) & (3,4) & {\\color{blue}(3,5)} & {\\color{blue}(3,6)} \\\\ (4,1) & (4,2) & (4,3) & {\\color{blue}(4,4)} & {\\color{blue}(4,5)} & {\\color{blue}(4,6)}\\end{array}$\n$\\begin{array}{cccccc}(5,1) & (5,2) & {\\color{blue}(5,3)} & {\\color{blue}(5,4)} & {\\color{blue}(5,5)} & {\\color{blue}(5,6)} \\\\\n(6,1) & {\\color{blue}(6,2)} & {\\color{blue}(6,3)} & {\\color{blue}(6,4)} & {\\color{blue}(6,5)} & {\\color{blue}(6,6)}\\end{array}\\quad\\Leftarrow\\quad \\text{fifteen outcomes}> 7$\n\n3. Heh 'Cambridge Advanced Mathematics' wrong.\nThanks!", "date": "2014-12-27 23:15:28", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/38753-die-probability.html", "openwebmath_score": 0.8414969444274902, "openwebmath_perplexity": 1353.4837426615243, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850827686585, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8558711992237421}}
{"url": "https://math.stackexchange.com/questions/1509476/prove-that-x-n1-fracx-n2-frac1x-n-geq-sqrt2", "text": "# Prove that $x_{n+1}=\\frac{x_n}{2}+\\frac{1}{x_n}\\geq\\sqrt{2}$\n\nLet $(x_n)_{n\\in\\mathbb N}$ be a recursively defined sequence with $x_1=9$ and $$x_{n+1}=\\frac{x_n}{2}+\\frac{1}{x_n}\\text{ for }n\\geq 1.$$ Show that $x_n\\geq\\sqrt{2}$ for all $n$.\n\nBecause $x_n\\geq 0$ one can easily prove inductively that $$x_n^2\\geq 2\\Leftrightarrow x_n^2-2\\geq 0\\Leftrightarrow\\left(\\frac{x_{n-1}}{2}-\\frac{1}{x_{n-1}}\\right)^2\\geq 0,$$ hence $x_n\\geq\\sqrt{2}$. However I have seen another approch which I was very curious about because I don't have the feeling that this can be done without proper justification other than the induction hypothesis:\n\n$$x_{n+1}=\\frac{x_n}{2}+\\frac{1}{x_n}\\overset{(*)}{\\geq}\\frac{\\sqrt{2}}{2}+\\frac{1}{\\sqrt{2}}=\\sqrt{2}$$\n\nFor $(*)$ it is assumed that $x_n\\geq\\sqrt{2}$ holds for an $n\\in\\mathbb N$. Are there any objections about this consideration?\n\n\u2022 Hint: AM-GM inequality is fast. Your second approach is flawed, as the reciprocal term does not follow in the inequality * \u2013\u00a0Macavity Nov 2 '15 at 14:47\n\u2022 The inequality $\\frac{1}{x_n} \\geq \\frac{1}{\\sqrt{2}}$ is false because it would imply that $x_n \\leq \\sqrt{2}$ (this is true only if $x_n=\\sqrt{2}$ but else.... ) \u2013\u00a0user252450 Nov 2 '15 at 14:50\n\u2022 A straightforward approach to showing that $x_n\\ge \\sqrt{2}$ is to write $$x_{n}-\\sqrt{2}=\\left(\\frac{x_{n-1}}{2}+\\frac1{x_{n-1}}\\right)-\\sqrt{2}=\\frac{(x_{n-1}-\\sqrt{2})^2}{2x_{n-1}}\\ge0$$ \u2013\u00a0Mark Viola Nov 2 '15 at 15:00\n\u2022 @Dr.MV By far the simplest. \u2013\u00a0Did Nov 2 '15 at 15:01\n\u2022 @Dr.MV Yeah pretty similiar to my explained approach in the first part of my post. I was just wondering about the other approach and its correctness there. \u2013\u00a0Christian Ivicevic Nov 2 '15 at 15:08\n\nYes, you can make the assumption $x_n\\ge \\sqrt{2}$ to prove $x_{n+1}\\ge \\sqrt{2}$ as per strong form of mathematical induction.\n\nBut your reasoning is wrong as $x_n\\ge \\sqrt{2} \\Rightarrow \\frac{1}{x_n} \\le \\frac{1}{\\sqrt{2}}$.\n\nBetter is to use AM-GM inequality as Macavity has mentioned since the quantities are all positive.\n\n$$\\frac{\\frac{x_n}{2}+\\frac{1}{x_n}}{2}\\ge \\sqrt{\\frac{x_n}{2}\\cdot\\frac{1}{x_n}}$$ $$x_{n+1}\\ge \\sqrt{2}$$\n\nIf you know derivatives you can make it work as follows. Consider the function : $$f(x)=\\frac{x^2+2}{2x}$$ Now calculate the derivative : $$f'(x)=\\frac{2x^2-4}{4x^2}\\geq 0$$ if $x\\geq \\sqrt{2}$ so the function is increasing for $x\\geq \\sqrt{2}$\n\nNow from the induction hypothesis $x_n\\geq \\sqrt{2}$ so from the monotony : $$x_{n+1}=f(x_n)\\geq(\\sqrt{2})=\\frac{4}{2\\sqrt{2}}=\\sqrt{2}$$ as wanted .\n\nAlthough the following goes a bit beyond addressing the OP's question, I thought that it would be instructive. So, let's have a look at the sequence generated by the recurrence relationship $x_n=\\frac12 x_{n-1}+\\frac{1}{x_{n-1}}$.\n\nThis is actually an interesting recursively generated sequence. If we examine the first difference $x_{n}-x_{n-1}$ we find that\n\n$$x_{n}-x_{n-1}=\\frac{2-x_{n-1}^2}{2x_{n-1}} \\tag 1$$\n\nWe deduce immediately from $(1)$ that if $x_n > \\sqrt{2}$, then $x_n$ is decreasing while if $x_n<\\sqrt{2}$, then $x_n$ is increasing.\n\nNow, a straightforward approach to showing that $x_n\\ge \\sqrt{2}$ is to write\n\n$$x_{n}-\\sqrt{2}=\\left(\\frac{x_{n-1}}{2}+\\frac1{x_{n-1}}\\right)-\\sqrt{2}=\\frac{(x_{n-1}-\\sqrt{2})^2}{2x_{n-1}}\\ge0 \\tag 2$$\n\nregardless of the value of $x_{n-1}$.\n\nTherefore, the only value of the sequence that can ever be less than $\\sqrt{2}$ is its initial \"seed\" value. From $(1)$, we see that if the seed value is less than $\\sqrt{2}$, the subsequent value increases. And $(2)$ shows that it increases above $\\sqrt{2}$. Then, the remaining sequence decreases monotonically to the limit value of $\\sqrt{2}$.\n\nHint: study the function $f(x)=x/2 +1/x$ it increases if $x>\\sqrt2$", "date": "2019-09-23 13:37:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1509476/prove-that-x-n1-fracx-n2-frac1x-n-geq-sqrt2", "openwebmath_score": 0.8864330053329468, "openwebmath_perplexity": 248.1817198160742, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850857421198, "lm_q2_score": 0.870597273444551, "lm_q1q2_score": 0.8558711952110921}}
{"url": "https://math.stackexchange.com/questions/620513/contour-integration-choice-of-contour", "text": "# Contour Integration (Choice of Contour)\n\nLet $\\alpha \\le 0$ and $\\sigma > 0$ .\n\nI want to choose a contour, including $[\\sigma - iR, \\sigma+iR]$ , such that i can apply Cauchy's Residue theorem and evaluate:\n\n$$\\lim_{R \\rightarrow \\infty} \\int\\limits_{\\sigma - iR}^{ \\sigma+iR } \\frac{\\exp(\\alpha z)}{(z^2 + 1)} dz$$\n\nby contour integration. The case for $\\alpha > 0$ is nice i think, as you can use a semi-circle, but i cant seem to find a suitable contour for this case.\n\nEDIT: to be specific when we use a semi-circle then the integral over the circular path tends to zero in the limit for $\\alpha > 0$ , but this does not seem to happen in the above case.\n\n\u2022 How do you expect to apply CRT to a complex integral on a straight line in the complex plane?? We usually do this to calculate real integrals by means of a closed contour in the complex plane... \u2013\u00a0DonAntonio Dec 28 '13 at 15:36\n\u2022 see my edit, in the limit the only non-zero section of our contour integral is that integrated over the line. it worked for the earlier case, unless i am completely confusesd. \u2013\u00a0KJC Dec 28 '13 at 15:38\n\nYou can choose a semicircle in the half-plane $\\operatorname{Re} z \\geqslant \\sigma$. Then $e^{\\alpha z}$ is bounded on the contour - $\\operatorname{Re} (\\alpha z) = \\alpha\\operatorname{Re} z \\leqslant \\alpha\\sigma \\leqslant 0$, and $\\lvert e^{\\alpha z}\\rvert = e^{\\operatorname{Re} (\\alpha z)}$ - and the integral over the semicircle tends to $0$ for $R \\to \\infty$. Since the contour encloses no singularity, the contour integral is $0$, and hence\n\n$$\\int_{\\sigma - i\\infty}^{\\sigma+i\\infty} \\frac{e^{\\alpha z}}{z^2+1}\\,dz = 0$$\n\nfor $\\sigma > 0$ and $\\alpha \\leqslant 0$.\n\nIt is worth noting that\n\n$$I(\\alpha) = \\int_{\\sigma-i\\infty}^{\\sigma+i\\infty} \\frac{e^{\\alpha z}}{z^2+1}\\,dz = \\begin{cases}\\quad 0 &, \\alpha \\leqslant 0\\\\ 2\\pi i\\sin \\alpha &, \\alpha \\geqslant 0 \\end{cases}$$\n\ndoes not depend smoothly on $\\alpha$, although the integrand does. That should, however, not be too surprising since the differentiated integrand $\\dfrac{ze^{\\alpha z}}{z^2+1}$ is not integrable anymore, and the non-differentiability in $0$ manifests itself by the need to switch the half-plane in which the contour is closed to have the integral over the auxiliary part tend to $0$.\n\n\u2022 could you show that $e^{ \\alpha z}$ is bounded on this contour? im not seeing that. \u2013\u00a0KJC Dec 28 '13 at 15:50\n\u2022 Added the estimate for that. \u2013\u00a0Daniel Fischer Dec 28 '13 at 15:58\n\u2022 excellent explanation \u2013\u00a0KJC Dec 28 '13 at 22:52", "date": "2019-09-18 13:46:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/620513/contour-integration-choice-of-contour", "openwebmath_score": 0.9719287157058716, "openwebmath_perplexity": 159.16764421459962, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850899545226, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.8558711873260572}}
{"url": "https://math.stackexchange.com/questions/690335/prove-that-lfloor-n1-2-rfloor-cdots-lfloor-n1-n-rfloor-lfloor-log-2", "text": "Prove that: $\\lfloor n^{1/2}\\rfloor+\\cdots+\\lfloor n^{1/n}\\rfloor=\\lfloor \\log_2n\\rfloor +\\cdots+\\lfloor \\log_nn \\rfloor$, for $n > 1$\n\nProve that: $\\lfloor n^{1/2}\\rfloor+\\cdots+\\lfloor n^{1/n}\\rfloor=\\lfloor \\log_2n\\rfloor +\\cdots+\\lfloor \\log_nn \\rfloor$, for $n > 1,\\, n\\in \\mathbb{N}$\n\nFor example. For $n=2$, we have $\\lfloor 2^{1/2} \\rfloor = \\lfloor 1.414 \\rfloor = 1$ whereas $\\lfloor \\log_2(2) \\rfloor = 1$ while for $n=3$, we have $$\\lfloor 3^{1/2} \\rfloor + \\lfloor 3^{1/3} \\rfloor = \\lfloor 1.732 \\rfloor + \\lfloor 1.442 \\rfloor = 2= \\lfloor 1.585 \\rfloor + \\lfloor 1 \\rfloor=\\lfloor \\log_2(3) \\rfloor + \\lfloor \\log_3(3) \\rfloor .$$\n\nI was thinking of using induction. So since $n=2$ is true, now assume for all $n$, this identity is true, we would like to prove that $n+1$ is true. Then\n\n$$\\lfloor n^{1/2} \\rfloor + \\lfloor n^{1/3} \\rfloor + ... + \\lfloor n^{1/n} \\rfloor + \\lfloor (n+1)^{1/(n+1)} \\rfloor,$$ where $(n+1)^{1/(n+1)} > 1$ for all $n>1$ but it's strictly decreasing above 1 so $\\lfloor (n+1)^{1/(n+1)} \\rfloor = 1$\n\n$\\implies \\lfloor n^{1/2} \\rfloor + \\lfloor n^{1/3} \\rfloor +\\cdots+ \\lfloor n^{1/n} \\rfloor + \\lfloor (n+1)^{1/(n+1)} \\rfloor = \\lfloor n^{1/2} \\rfloor + \\lfloor n^{1/3} \\rfloor +\\cdots+ \\lfloor n^{1/n} \\rfloor + 1$\n\n$= \\lfloor \\log_2(n) \\rfloor + \\lfloor \\log_3(n) \\rfloor + \\cdots+ \\lfloor \\log_n(n) \\rfloor + \\lfloor \\log_{n+1}(n+1) \\rfloor$\n\nsince, $\\log_{n+1}(n+1) = 1$ for all $n$.\n\nMy question is: How do we know that $(n+1)^{1/(n+1)}$ will never go below $1$? i.e., How can we prove that this function $f(x) = (x+1)^{1/(x+1)}$ is always bounded below by $1$ for $x>1$? (First, When $x=0$, $f(0)=1$, then looking at it's derivative, one can see that it's strictly increasing for $x$ between $(0,1)$ and decreasing for all $x>1$).\n\n\u2022 You need to replace $n$ by $n + 1$ all over, not just in the last term, in the induction step. \u2013\u00a0vonbrand Feb 25 '14 at 21:57\n\u2022 If $\\llcorner x\\lrcorner$ is supposed to mean the greatest integer not exceeding $x$, then you should probably have used $\\lfloor x \\rfloor$ (\\lfloor \u2026 \\rfloor) instead. If that's not what you mean, then what did you mean? \u2013\u00a0MJD Feb 25 '14 at 22:02\n\u2022 thanks @MJD, I wasn't sure which code to use for it. Thanks! \u2013\u00a0PandaMan Feb 26 '14 at 1:37\n\u2022 @vonbrand, I don't quite understand why you meant, please clarify. thanks \u2013\u00a0PandaMan Feb 26 '14 at 2:18\n\u2022 @PandaMan, what you need to prove is $\\lfloor (n+1)^{1/2}\\rfloor + \\ldots + \\lfloor (n+1)^{1/(n+1)}\\rfloor = \\lfloor \\log_2 (n+1)\\rfloor +\\ldots+\\lfloor\\log_{n+1} (n+1)\\rfloor$ \u2013\u00a0vonbrand Feb 26 '14 at 5:04\n\nThis is a classic exercise and one with a very elegant solution.\n\nThe idea of the proof is to count the number $N$ of the points (see figure below) with integer coordinates, which lie in the region $$U=\\big\\{(x,y): 0<x\\le n \\,\\,\\,\\text{and}\\,\\,\\, 1<y\\le n^{1/x}\\big\\},$$ and in particular, the red points, in two ways: horizontally and vertically.\n\nHorizontal counting: $$N=\\lfloor n^{1/2}\\rfloor+\\lfloor n^{1/3}\\rfloor+\\cdots+\\lfloor n^{1/n}\\rfloor,$$ since on the horizontal line $\\,y=k\\,$ lie exactly $\\,\\lfloor n^{1/k}\\rfloor\\,$ red points.\n\nVertical counting: $$N=\\lfloor \\log_2 n\\rfloor+\\lfloor\\log_3 n\\rfloor+\\cdots+\\lfloor \\log_n n\\rfloor,$$ since on the vertical line $\\,x=k\\,$ lie exactly $\\,\\lfloor \\log_k n\\rfloor\\,$ red points.\n\n$${}$$\n\nNote that the curve in the figure above is of the function $y=n^{1/x}$.\n\nThis problem was first asked in a Soviet Mathematics Olympiad in 1982 (\u0412\u0441\u0435\u0441\u043e\u044e\u0437\u043d\u044b\u0439 \u041c\u0430\u0442\u0435\u043c\u0430\u0442\u0438\u0447\u0435\u0441\u043a\u0438\u0439 \u041e\u043b\u0438\u043c\u043f\u0438\u0430\u0434.)\n\n\u2022 Gorgeous explanation! \u2013\u00a0Steven Stadnicki Feb 25 '14 at 22:24\n\u2022 @StevenStadnicki: Correct. \u2013\u00a0Yiorgos S. Smyrlis Feb 25 '14 at 22:26\n\u2022 (Thanks for the fix! I've removed the correction from my comment since that got edited in. :-) \u2013\u00a0Steven Stadnicki Feb 25 '14 at 22:28\n\u2022 Great explanation. The wires are crossed a little though, because if $x=k$, then $y\\leq n^{1/k}$, which means that the vertical count is the one that matches the fractional powers. Likewise, if $y = k$, then $k \\leq n^{1/x}$, then $1\\leq(\\log_k n)/x$, so the horizontal count is the one that matches the logs. \u2013\u00a0John Moeller Mar 1 '14 at 0:13\n\u2022 Awesome solution :-) \u2013\u00a0tarit goswami Aug 1 '18 at 8:54\n\nFix $b>1$. Then the derivative of $b^x$ is $\\ln(b) b^x$; $\\ln(b)$ is positive and $b^x$ is as well for all $x$, showing that that $b^x$ is a strictly increasing function. Next, $b^0=1$, showing that $b^x>1$ for all $x>0$.\n\nNext, since $n+1>1$ and $1/(n+1)>0$, we have that $(n+1)^{1/(n+1)}>1$.", "date": "2019-07-21 04:08:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/690335/prove-that-lfloor-n1-2-rfloor-cdots-lfloor-n1-n-rfloor-lfloor-log-2", "openwebmath_score": 0.8416237235069275, "openwebmath_perplexity": 301.72467357585316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886168290836, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8558702696060553}}
{"url": "https://mathoverflow.net/questions/226455/can-a-complete-manifold-have-an-uncountable-number-of-ends", "text": "# Can a complete manifold have an uncountable number of ends?\n\nLet $M$ be a complete and noncompact Riemannian manifold. Fix a point $p$ in $M$. Let $\\gamma$: $[0, L]\\rightarrow M$ (parametrized by its arc length) be a geodesic starting from $p$. Denote by $d(\\cdot, \\cdot)$ the distance function on $M$ induced by the Riemannian metric. $\\gamma$ is minimal if $d(\\gamma(s_1), \\gamma(s_2))=|s_1-s_2|$. A ray is a minimal geodesic defined on $[0, +\\infty)$. Since $M$ is noncompact, there exists at least one ray from $p$.\n\nTwo rays $\\gamma_1, \\gamma_2$ from the same point $p$ are called cofinal if for any $r\\geq0$ and all $s>r$, $\\gamma_1(s)$ and $\\gamma_2(s)$ lie in the same component of $M\\backslash B(0, r)$, where $B(0, r)=\\{x\\in M| d(p, x)<r\\}$. An equivalence class of cofinal rays is called an end of $M$. My question is:\n\nCan $M$ have uncountably many ends? It seems that the answer is no. But I am not very sure and I can not find a convictive proof.\n\n\u2022 You might like to look at some basic notions of geometric group theory. The set of ends is a quasi-isometry invariant, and the Svarc--Milnor Lemma says that the universal cover of a compact manifold $M$ is quasi-isometric to a Cayley graph of $\\pi_1M$. So it suffices to find a group with uncountably many ends, and the free group of rank 2 suffices. Curiously, it's actually impossible for a universal cover to have countably infinitely many ends.\n\u2013\u00a0HJRW\nDec 18, 2015 at 20:35\n\u2022 @FanZheng: Second countable spaces can have infinitely many ends. Take the universal cover of a wedge of two circles, for example. Dec 18, 2015 at 20:44\n\u2022 @HJRW: The ends of the fundamental group aren't all that relevant to this question, although simply connected examples can be interesting. It's about whether $M$ has uncountably many ends (as in the answer of @SebastianGoette), not the universal cover of $M$. Dec 18, 2015 at 20:49\n\u2022 @LeeMosher, I used bad notation, but I think my point stands (unless I misread the question). If $\\pi_1N$ is free of rank 2 (say) then taking $M$ to be the universal cover of $N$ provides the example.\n\u2013\u00a0HJRW\nDec 18, 2015 at 21:08\n\u2022 @HJRW: Sure, for example $N=$ the connected sum of two copies of $S^2 \\times S^1$. Dec 18, 2015 at 21:19\n\nThe answer is yes. Consider a hyperbolic pair of pants where all three boundary circles are of the same size. Glue countably many of them together such each new one is glued to the existing manifold along exactly one boundary circle. Then the manifold looks like the boundary of a fattened tree.\n\nFix $p$ in one of the pairs of pants $Y_0$. For all other pairs of pants $Y$, pick a point $q\\in Y$, then the minimal geodesic joining $p$ and $q$ will leave $Y$ through one circle. Label the two remaining boundary circles with $0$ and $1$. Then all glueing circles get exactly one label, except for those bounding $Y_0$.\n\nFor each sequence $(a_n)_n\\in\\{0,1\\}^{\\mathbb N}$, we construct a ray from $p$ leaving $Y_0$ through the same boundary component. Whenever the ray enters the $n$-th pair of pants $Y$ along its way, we let it leave through the circle labelled $a_n$. This way, we construct an uncountable number of rays that are not pairwise cofinal.\n\n\u2022 Thank you very much for your nice edit of my question and also for your clever solution. In this afternoon, I also found a very similar example called as the Cantor tree surface, inspired by your construction. Dec 19, 2015 at 10:34\n\nLet me rephrase S. Goette's example in a slightly different way. Consider the complement of the triadic Cantor set in the complex plane. This is a Riemann surface that can be uniformized: there is a complete Riemannian metric of constant minus one curvature. The ends of this surface are in one-to-one correspondence with the points of the Cantor set.\n\nYou can see the decomposition into pairs of pants by drawing nested circles around the intervals that appear in the construction of the Cantor set.\n\nIn some sense, uncountably many ends is the generic case. Let us restrict ourselves to planar surfaces, namely complements of compact sets in the Riemann sphere. The set of compact subsets in a compact set has a natural topology, the Vietoris topology (which coincides with the topology given by the Hausdorff metric). So we can say that two open subsets are close if their complement are close with respect to this topology. It happens that generically a compact subset of the sphere has uncountably many connected components, hence its complement has uncountably many ends.", "date": "2023-01-29 16:57:14", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/226455/can-a-complete-manifold-have-an-uncountable-number-of-ends", "openwebmath_score": 0.9025638699531555, "openwebmath_perplexity": 124.39065740210884, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886138694684, "lm_q2_score": 0.8633916047011595, "lm_q1q2_score": 0.8558702670507484}}
{"url": "https://byjus.com/question-answer/i-have-a-total-of-rs-300-in-coins-of-denomination-re-1-rs-2-1/", "text": "Question\n\n# I have a total of Rs. $$300$$ in coins of denomination Re. $$1$$, Rs. $$2$$ and Rs. $$5$$. The number of Rs. $$2$$ coins is $$3$$ times the number of Rs. $$5$$ coins. The total number of coins is $$160$$. How many coins of each denomination are with me?\n\nA\nNumber of Re. 1 coins =50; Number of Rs. 2 coins =40; Number of Rs. 5 coins =40\nB\nNumber of Re. 1 coins =70; Number of Rs. 2 coins =70; Number of Rs. 5 coins =19\nC\nNumber of Re. 1 coins =80; Number of Rs. 2 coins =60; Number of Rs. 5 coins =20\nD\nNumber of Re. 1 coins =115; Number of Rs. 2 coins =50; Number of Rs. 5 coins =15\n\nSolution\n\n## The correct option is C Number of Re. $$1$$ coins $$= 80$$; Number of Rs. $$2$$ coins $$= 60$$; Number of Rs. $$5$$ coins $$= 20$$Given, Total number of coins $$= 160$$Given that the number of Rs.$$2$$ coins are $$3$$ times the number of Rs. $$5$$ coins.Number of Rs. $$5$$ coins $$= x$$Number of Rs. $$2$$ coins $$= 3x$$Number of Re. $$1$$ coins $$= 160 -(x+3x) = 160 -4x$$Total value of coins $$=$$ Rs. $$300$$So, $$5\\times x + 2\\times 3x + 1\\times (160 -4x) = 300$$$$5x +6x +160 -4x = 300$$$$7x = 140$$$$x = 20$$So, number of Rs. $$5$$ coins $$= 20$$Number of Rs. $$2$$ coins $$= 60$$Number of Re. $$1$$ coins $$=80$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-21 09:05:37", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/i-have-a-total-of-rs-300-in-coins-of-denomination-re-1-rs-2-1/", "openwebmath_score": 0.9646574854850769, "openwebmath_perplexity": 3512.7088031497474, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9891815497525024, "lm_q2_score": 0.8652240930029117, "lm_q1q2_score": 0.8558637091998235}}
{"url": "https://www.vakrenordnorge.no/tmdnfv1/total-no-of-onto-functions-from-a-to-b-bd9598", "text": "So, total numbers of onto functions from X to Y are 6 (F3 to F8). Functions can be classified according to their images and pre-images relationships. Then every function from A to B is effectively a 5-digit binary number. (d) x2 +1 x2 +2. Don\u2019t stop learning now. I already know the formula (summation r=1 to n)(-1)^(n-r)nCr(r^m). The number of functions from {0,1}4 (16 elements) to {0, 1} (2 elements) are 216. By using our site, you This is same as saying that B is the range of f . Solution: As given in the question, S denotes the set of all functions f: {0, 1}4 \u2192 {0, 1}. Which must also be bijective, and therefore onto. From the formula for the number of onto functions, find a formula for S(n, k) which is defined in Problem 12 of Section 1.4. Option 3) 200. Thus, the number of onto functions = 16\u22122= 14. So, that leaves 30. (e) f(m;n) = m n. Onto. Then Total no. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different. But we want surjective functions. Mathematics | Total number of possible functions, Mathematics | Unimodal functions and Bimodal functions, Total Recursive Functions and Partial Recursive Functions in Automata, Mathematics | Classes (Injective, surjective, Bijective) of Functions, Mathematics | Generating Functions - Set 2, Inverse functions and composition of functions, Last Minute Notes - Engineering Mathematics, Mathematics | Introduction to Propositional Logic | Set 1, Mathematics | Predicates and Quantifiers | Set 1, Mathematics | L U Decomposition of a System of Linear Equations, Mathematics | Mean, Variance and Standard Deviation, Mathematics | Sum of squares of even and odd natural numbers, Mathematics | Eigen Values and Eigen Vectors, Mathematics | Lagrange's Mean Value Theorem, Mathematics | Introduction and types of Relations, Data Structures and Algorithms \u2013 Self Paced Course, We use cookies to ensure you have the best browsing experience on our website. Find the number of relations from A to B. Here are the definitions: is one-to-one (injective) if maps every element of to a unique element in . In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Not onto. No. Home. In other words no element of are mapped to by two or more elements of . there are zero onto function . Consider the function x \u2192 f(x) = y with the domain A and co-domain B. Determine whether each of these functions is a bijection from R to R. (a) f(x) = 2x+1. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. [5.1] Informally, a function from A to B is a rule which assigns to each element a of A a unique element f(a) of B. O\ufb03cially, we have De\ufb01nition. In other words, nothing is left out. We need to count the number of partitions of A into m blocks. Examples: Let us discuss gate questions based on this: Solution: As W = X x Y is given, number of elements in W is xy. Comparing cardinalities of sets using functions. The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. Solution: Using m = 4 and n = 3, the number of onto functions is: I am trying to get the total number of onto functions from set A to set B if the former has m elements and latter has n elements with m>n. (B) 64 Math Forums. Yes. If X has m elements and Y has 2 elements, the number of onto functions will be 2 m-2. Example 9 Let A = {1, 2} and B = {3, 4}. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b\u2026 Functions: One-One/Many-One/Into/Onto . Out of these functions, the functions which are not onto are f (x) = 1, \u2200x \u2208 A. In this article, we are discussing how to find number of functions from one set to another. Therefore, S has 216 elements. For example: X = {a, b, c} and Y = {4, 5}. An onto function is also called surjective function. An onto function is also called surjective function. In this case the map is also called a one-to-one correspondence. Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . We need to count the number of partitions of A into m blocks. In other words, if each b \u2208 B there exists at least one a \u2208 A such that. 2. is onto (surjective)if every element of is mapped to by some element of . An onto function is also called a surjective function. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. It means that every element \u201cb\u201d in the codomain B, there is exactly one element \u201ca\u201d in the domain A. such that f(a) = b. . Therefore, each element of X has \u2018n\u2019 elements to be chosen from. In a function from X to Y, every element of X must be mapped to an element of Y. That is, a function f is onto if for each b \u220a B, there is atleast one element a \u220a A, such that f(a) = b. 2\u00d72\u00d72\u00d72 = 16. So, you can now extend your counting of functions \u2026 19. Tuesday: Functions as relations, one to one and onto functions What is a function? 34 \u2013 3C1(2)4 + 3C214 = 36. So, number of onto functions is 2m-2. Therefore, N has 2216 elements. 2. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. One-to-One/Onto Functions . Option 2) 120. So the total number of onto functions is m!. Calculating required value. Some authors use \"one-to-one\" as a synonym for \"injective\" rather than \"bijective\". Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, \u2026 , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 \u00d7 2 \u00d7 1 = 6Misc 10Find the number of all onto functio 1.1. . If m < n, the number of onto functions is 0 as it is not possible to use all elements of Y. 4 = A B Not a function Notation We write f (a) = b when (a;b) 2f where f is a function. Math Forums. therefore the total number of functions from A to B is. In a one-to-one function, given any y there is only one x that can be paired with the given y. If n(A)= 3 , n(B)= 5 Find the number\u00a0 of onto function from A to B, For onto function n(A)\u00a0n(B) otherwise ; it will always be an inoto function. Set A has 3 elements and set B has 4 elements. A function f from A to B is a subset of A\u00d7B such that \u2022 for each a \u2208 A there is a b \u2208 B with (a,b\u2026 Experience. We say that b is the image of a under f , and a is a preimage of b. October 31, 2007 1 / 7. Here's another way to look at it: imagine that B is the set {0, 1}. But, if the function is onto, then you cannot have 00000 or 11111. Q3. Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. There are $$\\displaystyle 3^8=6561$$ functions total. If anyone has any other proof of this, that would work as well. Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. where as when i try manually it comes 8 . De nition 1 A function or a mapping from A to B, denoted by f : A !B is a Such functions are referred to as injective. (C) 81 4. Writing code in comment? Please use ide.geeksforgeeks.org, For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. There are 3 ways of choosing each of the 5 elements = $3^5$ functions. A function is said to be bijective or bijection, if a function f: A \u2192 B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. 38. Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Transcript. f(a) = b, then f is an on-to function. Q1. Why does an ordinary electric fan give comfort in summer even though it cannot cool the air? Students can solve NCERT Class 12 Maths Relations and Functions MCQs Pdf with Answers to know their preparation level. (D) 72. I just need to know how it came. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In the above figure, f \u2026 So, total numbers of onto functions from X to Y are 6 (F3 to F8). Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In F1, element 5 of set Y is unused and element 4 is unused in function F2. Discrete Mathematics Grinshpan Partitions: an example How many onto functions from f1;2;3;4;5;6;7;8g to fA;B;C;Dg are there? There are 3 functions with 1 element in range. How many onto functions are there from a set with eight elements to a set with 3 elements? Onto Function A function f: A -> B is called an onto function if the range of f is B. Copyright \u00a9 2021 Pathfinder Publishing Pvt Ltd. To keep connected with us please login with your personal information by phone/email and password. (c) f(m;n) = m. Onto. Need explanation for: If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, List of Hospitality & Tourism Colleges in India, Knockout JEE Main May 2022 (Easy Installments), Knockout JEE Main May 2021 (Easy Installments), Knockout NEET May 2021 (Easy Installments), Knockout NEET May 2022 (Easy Installments), Top Medical Colleges in India accepting NEET Score, MHCET Law ( 5 Year L.L.B) College Predictor, List of Media & Journalism Colleges in India, B. Transcript. Let f be the function from R \u2026 Formula for finding number of relations is Number of relations = 2 Number of elements of A \u00d7 Number of elements of B Attention reader! In other words, if each b \u2208 B there exists at least one a \u2208 A such that. Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Click hereto get an answer to your question \ufe0f Write the total number of one - one functions from set A = { 1,2,3,4 } to set B = { a,b,c } . Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. Yes. (i)When all the elements of A will map to a only, then b is left which do not have any pre-image in A (ii)When all the elements of A will map to b only, then a is left which do not have only pre-image in A Thus in both cases, function is not onto So, total number of onto functions= 2^n-2 Hope it helps\u2611 #Be Brainly Onto Functions: Consider the function {eq}y = f(x) {/eq} from {eq}A \\to B {/eq}, where {eq}A {/eq} is the domain of the function and {eq}B {/eq} is the codomain. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b; bc,a. If n > m, there is no simple closed formula that describes the number of onto functions. according to you what should be the anwer So the correct option is (D). As E is the set of all subsets of W, number of elements in E is 2xy. Also, given, N denotes the number of function from S(216 elements) to {0, 1}(2 elements). Let X, Y, Z be sets of sizes x, y and z respectively. An exhaustive E-learning program for the complete preparation of JEE Main.. Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test.. In F1, element 5 of set Y is unused and element 4 is unused in function F2. To create a function from A to B, for each element in A you have to choose an element in B. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Mathematics | Introduction to Propositional Logic | Set 2, Mathematics | Predicates and Quantifiers | Set 2, Mathematics | Some theorems on Nested Quantifiers, Mathematics | Set Operations (Set theory), Inclusion-Exclusion and its various Applications, Mathematics | Power Set and its Properties, Mathematics | Partial Orders and Lattices, Discrete Mathematics | Representing Relations, Mathematics | Representations of Matrices and Graphs in Relations, Mathematics | Closure of Relations and Equivalence Relations, Number of possible Equivalence Relations on a finite set, Discrete Maths | Generating Functions-Introduction and Prerequisites, Mathematics | Generating Functions \u2013 Set 2, Mathematics | Sequence, Series and Summations, Mathematics | Independent Sets, Covering and Matching, Mathematics | Rings, Integral domains and Fields, Mathematics | PnC and Binomial Coefficients, Number of triangles in a plane if no more than two points are collinear, Finding nth term of any Polynomial Sequence, Discrete Mathematics | Types of Recurrence Relations \u2013 Set 2, Mathematics | Graph Theory Basics \u2013 Set 1, Mathematics | Graph Theory Basics \u2013 Set 2, Mathematics | Euler and Hamiltonian Paths, Mathematics | Planar Graphs and Graph Coloring, Mathematics | Graph Isomorphisms and Connectivity, Betweenness Centrality (Centrality Measure), Mathematics | Walks, Trails, Paths, Cycles and Circuits in Graph, Graph measurements: length, distance, diameter, eccentricity, radius, center, Relationship between number of nodes and height of binary tree, Bayes\u2019s Theorem for Conditional Probability, Mathematics | Probability Distributions Set 1 (Uniform Distribution), Mathematics | Probability Distributions Set 2 (Exponential Distribution), Mathematics | Probability Distributions Set 3 (Normal Distribution), Mathematics | Probability Distributions Set 4 (Binomial Distribution), Mathematics | Probability Distributions Set 5 (Poisson Distribution), Mathematics | Hypergeometric Distribution model, Mathematics | Limits, Continuity and Differentiability, Mathematics | Lagrange\u2019s Mean Value Theorem, Mathematics | Problems On Permutations | Set 1, Problem on permutations and combinations | Set 2, Mathematics | Graph theory practice questions, Classes (Injective, surjective, Bijective) of Functions, Difference between Spline, B-Spline and Bezier Curves, Runge-Kutta 2nd order method to solve Differential equations, Write Interview Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A. Option 1) 150. There are $$\\displaystyle 2^8-2$$ functions with 2 elements in the range for each pair of elements in the codomain. Onto Function A function f: A -> B is called an onto function if the range of f is B. (b) f(m;n) = m2 +n2. If n > m, there is no simple closed formula that describes the number of onto functions. These numbers are called Stirling numbers (of the second kind). Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. A function f : A -> B is said to be an onto function if every element in B has a pre-image in A. The onto function from Y to X is F's inverse. Therefore, total number of functions will be n\u00d7n\u00d7n.. m times = nm. This course will help student to be better prepared and study in the right direction for JEE Main.. No. Menu. They are various types of functions like one to one function, onto function, many to one function, etc. 3. (d) f(m;n) = jnj. This disagreement is confusing, but we're stuck with it. set a={a,b,c} and B={m,n} the number of onto functions by your formula is 6 . If X has m elements and Y has n elements, the number if onto functions are. f(a) = b, then f is an on-to function. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. 2.1. . No element of B is the image of more than one element in A. The total no.of onto function from the set {a,b,c,d,e,f} to the set {1,2,3} is????? Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. (b) f(x) = x2 +1. 3. is one-to-one onto (bijective) if it is both one-to-one and onto. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. Option 4) none of these Number of onto functions from one set to another \u2013 In onto function from X to Y, all the elements of Y must be used. High School Math Elementary Math Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus. The number of injections that can be defined from A to B is: The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is: 3. A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. generate link and share the link here. (c) f(x) = x3. So the total number of onto functions is m!. For function f: A\u2192B to be onto, the inequality \u2502A\u2502\u22652 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to 2nd element of Y). Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x \u2013 7. asked Feb 16, 2018 in Class XI Maths by rahul152 ( -2,838 points) relations and functions Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number Steps 1. (b)-Given that, A = {1 , 2, 3, n} and B = {a, b} If function is subjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes a' and b' in such a way that no box remains empty. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. P.S. A function has many types which define the relationship between two sets in a different pattern. (A) 36 In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Any ideas on how it came? Let W = X x Y. Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 1 Relations and Functions. In other words no element of are mapped to by two or more elements of . Proving that a given function is one-to-one/onto. One more question. Not onto. So, there are 32 = 2^5. Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Tech Companion - A Complete pack to prepare for Engineering admissions, MBBS Companion - For NEET preparation and admission process, QnA - Get answers from students and experts, List of Pharmacy Colleges in India accepting GPAT, Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Number of Onto function - & Number of onto functions - For onto function n(A) n(B) otherwise ; it will always be an inoto function . 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Numbers are called Stirling numbers ( of the second kind ) 3 4! ( X ) = x3 to n ) = 2x+1 } and Y has elements! Imagine that B is the set of 2xy elements ) to E ( set of 2 elements in the...., there is only one X that can be represented in Table 1 '' as a synonym for  ''... Are there from a to a unique element in range and pre-images relationships the total number of onto are. All possibilities of mapping elements of Y, the number of partitions of a into m blocks with to! To count the number of functions from Z ( set of 2xy elements ) to E ( set functions. Injective ) if maps every element of are mapped to by two or more elements of Y with.! This is same as saying that B is called an onto function a function not possible to use elements! 1, 2 } and B = { 1, 2 } and B {. Maps every element of is mapped to by two or more elements of Y is 0 as it is one-to-one... Then f is an on-to function manually it comes 8 4, 5 } Download of Maths... 16\u22122= 14 1, 2 } and Y = { 1, 2 } and B = 1.", "date": "2021-05-14 13:18:56", "meta": {"domain": "vakrenordnorge.no", "url": "https://www.vakrenordnorge.no/tmdnfv1/total-no-of-onto-functions-from-a-to-b-bd9598", "openwebmath_score": 0.6120531558990479, "openwebmath_perplexity": 628.9770854112651, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9891815503903563, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8558637080328639}}
{"url": "https://math.stackexchange.com/questions/3119303/constructing-a-sum-of-squares-in-mathbb-rx-with-given-complex-valuation", "text": "# Constructing a sum of squares in $\\mathbb R[x]$ with given complex valuation\n\nFix a polynomial $$g(x)\\in\\mathbb R[x]$$ and a complex number $$u\\in\\mathbb C\\setminus\\mathbb R$$. My main question is\n\nHow can we construct a polynomial $$s(x)\\in\\mathbb R[x]$$ such that $$s(x)$$ is a sum of squares and $$s(u)=g(u)$$ and $$s(\\overline{u})=g(\\overline{u})$$?\n\nI know that if $$s(u)=g(u)$$ then $$s(\\overline{u})=g(\\overline{u})$$, so we really only need the condition $$s(u)=g(u)$$. I am asking about this because I am trying to understand a proof of a theorem (as Theorem 7.3 in \"Solving systems of polynomial equations\" by Sturmfels) in which the author claims that we can write a polynomial $$g(x)\\in\\mathbb R[x_1,\\ldots,x_n]$$ is congruent to a sum of squares $$s(x)\\in\\mathbb R[x_1,\\ldots x_n]$$ modulo $$\\langle x-u\\rangle\\cap\\langle x-\\overline{u}\\rangle$$ for fixed $$u\\in\\mathbb C^n$$ strictly complex. The rest of the proof I can understand (and I could copy it into this question if requested), but I am having trouble seeing just why exactly this can be done. So my main question above is just asking about the univariate case of this, from which I believe I could generalize the argument.\n\nSo far I have only been able to think of the following. If we set $$s(x):=\\frac{g(x)^2+|g(u)|^2}{2\\Re(g(u))}$$ then we have $$g(u)^2+g(u)g(\\overline{u})=g(u)\\left(g(u)+\\overline{g(u)}\\right)=2\\Re (g(u))g(u)$$ so $$s(u)=g(u)$$. But this only works if $$\\Re (g(u))>0$$ since we need $$s(x)$$ to be a sum of squares in $$\\mathbb R[x]$$.\n\nNote that since $$u$$ is not real, $$u$$ and $$1$$ are linearly independent over $$\\mathbb{R}$$, so every complex number can be written as $$au+b$$ for some $$a,b\\in\\mathbb{R}$$. In particular, we can pick $$a$$ and $$b$$ such that $$au+b$$ is a square root of $$g(u)$$, and then take $$s(x)=(ax+b)^2$$.\nAlternatively, pick $$a\\in\\mathbb{R}$$ large enough such that $$v=u+a$$ has positive real part. Now note that the consecutive powers of $$v^2$$ have an angle of less than $$\\pi$$ between them. We can pick $$n\\in\\mathbb{N}$$ such that $$g(u)$$ has argument between the arguments of $$v^{2n}$$ and $$v^{2n+2}$$, and then $$g(u)$$ can be written as a linear combination of $$v^{2n}$$ and $$v^{2n+2}$$ with nonnegative real coefficients. This gives a polynomial $$s(x)=b(x+a)^{2n}+c(x+a)^{2n+2}$$ which is a sum of squares and has $$s(u)=g(u)$$.\n\u2022 Yes I want $u\\notin\\mathbb R$, I will edit to say $u\\in\\mathbb C\\setminus\\mathbb R$. This solution is very interesting, thank you! I might leave the answer un-accepted for a little while just in case people have other approaches and want to share. \u2013\u00a0Dave Feb 19 at 22:02", "date": "2019-06-20 11:28:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3119303/constructing-a-sum-of-squares-in-mathbb-rx-with-given-complex-valuation", "openwebmath_score": 0.9283049702644348, "openwebmath_perplexity": 49.7754713017042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9891815500714294, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8558637077569207}}
{"url": "https://mathzsolution.com/does-this-property-characterize-a-space-as-hausdorff/", "text": "# Does this property characterize a space as Hausdorff?\n\nAs a result of this question, I\u2019ve been thinking about the following condition on a topological space $Y$:\n\nFor every topological space $X$, $E\\subseteq X$, and continuous maps $f,g\\colon X\\to Y$, if $E$ is dense in $X$, and $f$ and $g$ agree on $E$ (that is, $f(e)=g(e)$ for all $e\\in E$), then $f=g$.\n\nIf $Y$ is Hausdorff, then $Y$ satisfies this condition. The question is whether the converse holds: if $Y$ satisfies the above condition, will it necessarily be Hausdorff?\n\nIf $Y$ is not at least $T_1$, then $Y$ does not have the property: if $u,v\\in Y$ are such that $u\\neq v$ and every open neighborhood of $u$ contains $v$, then let $X$ be the Sierpinski space, $X=\\{a,b\\}$, $a\\neq b$, with topology $\\tau=\\{\\emptyset,\\{b\\},X\\}$, $E=\\{b\\}$, let $f,g\\colon X\\to Y$ be given by $f(a)=f(b)=v$, and $g(a)=u$, $g(b)=v$. Then both $f$ and $g$ are continuous, agree on the dense subset $E$, but are distinct.\n\nMy attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume $Y$ is $T_1$ but not $T_2$; let $u$ and $v$ be witnesses to the fact that $Y$ is not $T_2$, let $\\mathcal{U}\\_s$ and $\\mathcal{V}\\_t$ be the collection of all open nbds of $s$ that do not contain $t$, and all open nbds of $t$ that do not contain $s$, respectively. Construct a net with index set $\\mathcal{U}\\_s\\times\\mathcal{V}\\_t$ (ordered by $(U,V)\\leq (U',V')$ if and only if $U'\\subseteq U$ and $V'\\subseteq V$) by letting $y_{(U,V)}$ be a point in $U\\cap V$ (this is where AC comes in). Let $E=\\{y_{(U,V)}\\mid (U,V)\\in\\mathcal{U}\\_s\\times\\mathcal{V}\\_t\\}$, and let $X=E\\cup\\{s\\}$. Give $X$ the induced topology; let $f\\colon X\\to Y$ be the inclusion map, and let $g\\colon X\\to Y$ be the map that maps $E$ to itself identically, but maps $s$ to $t$.\n\nThe only problem is I cannot quite prove that $g$ is continuous; the difficulty arises if I take an open set $\\mathcal{O}\\in \\mathcal{V}_t$; the inverse image under $g$ is equal to $((\\mathcal{O}\\cap X)-\\{t\\})\\cup\\{s\\}$, and I have not been able to show that this is open in $X$.\n\nSo:\n\nDoes the condition above characterize Hausdorff spaces?\n\nIf not, I would appreciate a counterexample. If it does characterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do.\n\nAdded: A little digging turned up this question raised in the Problem Section of the American Mathematical Monthly back in 1964 by Alan Weinstein. The solution by Sim Lasher gives a one paragraph proof that does not require one to consider $T_1$ and non-$T_1$ spaces separately.\n\nI think the following goes a long way towards proving a converse:\n\nLet $(Y, T)$ be any T1 topological space with at least two points and let $a$ and $b$ be distinct points in $Y$.\n\nLet $X = Y\\setminus\\{b\\}$.\nLet $f: X \\to Y$ be the inclusion of $X$ in $Y$.\nLet $g: X \\to Y$ agree with $f$ on $X\\setminus\\{a\\}$ and $g(a) = b$.\n\nFinally, define the topology on $X$ to be the coarsest topology that makes both $f$ and $g$ continuous.\n\nWith these assumptions it turns out that $X\\setminus\\{a\\}$ is dense in $X$ if and only if $a$ and $b$ do not have disjoint neighbourhoods in $Y$.\n\nTo see that this is true, let us construct a base of the topology on $X$.\nTo make $f$ continuous we only need to take the subspace topology. Since\nX is open in Y this is $S_1 = \\{ G \\in T \\mid b \\notin G \\}$. To also make $g$ continuous we need to add the open neighbourhoods of $b$, with $b$ replaced by $a$. This gives $S_2 = \\{ (H\\setminus\\{b\\} \\cup \\{a\\} \\mid H \\in T, b \\in H \\}$.\nNow $S_1 \\cup S_2$ is a subbase of the topology on $X$.\nSince $S_1$ and $S_2$ are already closed under finite intersection, and each covers $X$, we can say that\n$B = \\{ G \\cap H \\mid G \\in S_1, H \\in S_2 \\}$ is a base of the topology.\n\nThen (remembering that finite sets are closed in Y) we find that the following are all equivalent:\n\n\u2022 $X\\setminus\\{a\\}$ is not dense in $X$\n\u2022 $\\{a\\}$ is open in $X$\n\u2022 $\\{a\\} \\in B$\n\u2022 there are $G \\in S_1, H \\in S_2$ such that $G \\cap H = \\{a\\}$\n\u2022 there are $G \\in S_1, H \\in S_2$ such that $a \\in G$ and $G \\cap (H\\setminus\\{a\\}\\cup\\{b\\}) = \\oslash$\n\u2022 $a$ and $b$ have disjoint neighbourhoods in $(Y, T)$.", "date": "2023-01-29 15:14:42", "meta": {"domain": "mathzsolution.com", "url": "https://mathzsolution.com/does-this-property-characterize-a-space-as-hausdorff/", "openwebmath_score": 0.965681791305542, "openwebmath_perplexity": 69.14657264564579, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9934102278491785, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8558608636341154}}
{"url": "https://gmatclub.com/forum/a-clock-store-sold-a-certain-clock-to-a-collector-for-20-per-46986.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Aug 2018, 21:36\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# A clock store sold a certain clock to a collector for 20 per\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 03 Apr 2006\nPosts: 44\nA clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 06 May 2014, 09:27\n1\n27\n00:00\n\nDifficulty:\n\n45% (medium)\n\nQuestion Stats:\n\n77% (02:06) correct 23% (02:35) wrong based on 594 sessions\n\n### HideShow timer Statistics\n\nA clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270\nB. $250 C.$240\nD. $220 E.$200\n\nOriginally posted by hsk on 11 Jun 2007, 21:02.\nLast edited by Bunuel on 06 May 2014, 09:27, edited 1 time in total.\nRenamed the topic, edited the question and added the OA.\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8196\nLocation: Pune, India\nRe: A clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n06 May 2014, 20:22\n9\n4\nhsk wrote:\nA clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270\nB. $250 C.$240\nD. $220 E.$200\n\nAssume numbers and then use ratios to fit them in with the actual numbers.\n\nSay the store bought the clock for $100 and sold it to the collector for$120. The store bought back from the collector for $60 and sold it back at$60*18/10 = $108 Here, difference between original cost ($100) and buy back price ($60) is$40. But actually it is given to be $100 which is 2.5 times 40. So the shop sold the clock a second time for$108*2.5 = $270 _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nSenior Manager\nJoined: 04 Jun 2007\nPosts: 341\n\n### Show Tags\n\n11 Jun 2007, 21:42\n10\n3\n$270 Original Price = x Sold to collector at 1.2x re-buy price = 1.2x * 0.5 = 0.6x therefore, x - 0.6x = 100; x = 250 second selling price = 1.8 * 0.6 * 250 = 270 Hope, this is not too confusing ##### General Discussion Senior Manager Joined: 04 Mar 2007 Posts: 419 ### Show Tags 12 Jun 2007, 10:52 1 100 % - original price 120 % - sell 1 60 % - bought 2 180*60/100 = 108 % - sell 2 100$ - 40%\nx - 108 %\nx= 100*108/40 = 270\n\nA\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 1900\nRe: A clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jun 2015, 00:44\n7\nThe question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution\n\nStep-I- Store buys the clock\nLet's assume the original price of clock paid by the store to be $$x$$\n\nStep-II- Collectors buys the clock from the store\nExtra amount paid by collector to buy the clock = $$20$$% of $$x$$\n\nTherefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$\n\nStep-III- Store buys back the clock from collector\nPrice at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$\n\nStep-IV- Store resells the clock\nPrice at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$\n\nNow, we are given that difference between clock's original price and clock's buy back price = $$100$$\n\n$$x - 0.6x = 100$$ i.e. $$x = 250$$\n\nWe are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$\n\nHope this helps\n\nRegards\nHarsh\n_________________\n\nNumber Properties | Algebra |Quant Workshop\n\nSuccess Stories\nGuillermo's Success Story | Carrie's Success Story\n\nAce GMAT quant\nArticles and Question to reach Q51 | Question of the week\n\nNumber Properties \u2013 Even Odd | LCM GCD\nWord Problems \u2013 Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2\nAdvanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability\nGeometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry\nAlgebra- Wavy line\n\nPractice Questions\nNumber Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nIntern\nJoined: 25 Dec 2012\nPosts: 17\nRe: A clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n11 Jul 2015, 13:39\nEgmatQuantExpert wrote:\nThe question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution\n\nStep-I- Store buys the clock\nLet's assume the original price of clock paid by the store to be $$x$$\n\nStep-II- Collectors buys the clock from the store\nExtra amount paid by collector to buy the clock = $$20$$% of $$x$$\n\nTherefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$\n\nStep-III- Store buys back the clock from collector\nPrice at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$\n\nStep-IV- Store resells the clock\nPrice at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$\n\nNow, we are given that difference between clock's original price and clock's buy back price = $$100$$\n\n$$x - 0.6x = 100$$ i.e. $$x = 250$$\n\nWe are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$\n\nHope this helps\n\nRegards\nHarsh\n\nDear,\n\nI have a basic question.\n\nThe author states: \"The shop then sold the clock again at a profit of 80 percent on its buy-back price\". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.\n\nWhere my confuse is?\n\nKind regards...\nGonzalo\nCurrent Student\nJoined: 20 Mar 2014\nPosts: 2643\nConcentration: Finance, Strategy\nSchools: Kellogg '18 (M)\nGMAT 1: 750 Q49 V44\nGPA: 3.7\nWE: Engineering (Aerospace and Defense)\nRe: A clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n11 Jul 2015, 13:54\n1\n1\ngbascurs wrote:\nEgmatQuantExpert wrote:\nThe question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution\n\nStep-I- Store buys the clock\nLet's assume the original price of clock paid by the store to be $$x$$\n\nStep-II- Collectors buys the clock from the store\nExtra amount paid by collector to buy the clock = $$20$$% of $$x$$\n\nTherefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$\n\nStep-III- Store buys back the clock from collector\nPrice at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$\n\nStep-IV- Store resells the clock\nPrice at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$\n\nNow, we are given that difference between clock's original price and clock's buy back price = $$100$$\n\n$$x - 0.6x = 100$$ i.e. $$x = 250$$\n\nWe are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$\n\nHope this helps\n\nRegards\nHarsh\n\nDear,\n\nI have a basic question.\n\nThe author states: \"The shop then sold the clock again at a profit of 80 percent on its buy-back price\". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.\n\nWhere my confuse is?\n\nKind regards...\nGonzalo\n\nGonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.\nThis means the clocks' selling price was C (1.2) and this becomes cost price for the collector.\nNow, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.\n\nThus, you get = 1.2*0.5*C = 0.6 C\n\nFurthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C\n\nFinally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.\nIntern\nJoined: 25 Dec 2012\nPosts: 17\nRe: A clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n11 Jul 2015, 14:20\nEngr2012 wrote:\ngbascurs wrote:\nEgmatQuantExpert wrote:\nThe question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution\n\nStep-I- Store buys the clock\nLet's assume the original price of clock paid by the store to be $$x$$\n\nStep-II- Collectors buys the clock from the store\nExtra amount paid by collector to buy the clock = $$20$$% of $$x$$\n\nTherefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$\n\nStep-III- Store buys back the clock from collector\nPrice at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$\n\nStep-IV- Store resells the clock\nPrice at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$\n\nNow, we are given that difference between clock's original price and clock's buy back price = $$100$$\n\n$$x - 0.6x = 100$$ i.e. $$x = 250$$\n\nWe are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$\n\nHope this helps\n\nRegards\nHarsh\n\nDear,\n\nI have a basic question.\n\nThe author states: \"The shop then sold the clock again at a profit of 80 percent on its buy-back price\". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.\n\nWhere my confuse is?\n\nKind regards...\nGonzalo\n\nGonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.\nThis means the clocks' selling price was C (1.2) and this becomes cost price for the collector.\nNow, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.\n\nThus, you get = 1.2*0.5*C = 0.6 C\n\nFurthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C\n\nFinally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.\n\nThank you a lot.\n\nHere is the key of my misunderstanding: \"Profit % is always calculated on the cost price of the purchase\". I assumed it was on the sales price.\nIntern\nJoined: 08 Oct 2014\nPosts: 9\nGPA: 4\nA clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n10 Nov 2015, 05:00\n1\nx=starting price\ntherefore\nx +20% = $$x +1/5x$$ = $$6x/5$$\n50% = 3x/5 BB Price\nas per Q\nx - $$3x/5$$=100\nx=250\nBB =150\n80 %increment of 150 = 270\n\n_________________\n\nurban hermit\n\nSenior Manager\nJoined: 20 Aug 2015\nPosts: 392\nLocation: India\nGMAT 1: 760 Q50 V44\nA clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n10 Nov 2015, 05:20\n1\nhsk wrote:\nA clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270\nB. $250 C.$240\nD. $220 E.$200\n\nLet us assume the initial price (CP) of the clock to be 100x\nWe assume 100x to avoid the usage of decimals in case of x and avoid the usage of unitary method in case of 100\n\nTransaction 1:\nStore sold the clock to collector.\nSelling Price = 20% more than CP = 120x\n\nTransaction 2:\nThe collector sold it back to the store.\nThe new CP for the store = 60x (Store bought it back for 50% of the Selling Price)\n\nTransaction 3:\nStore sold it back at a profit of 80% on 60x = 108x\n\nNow we are told that 100x - 60x = 100\nThis means x = 100/40\n\nHence the second selling price = 108*100/40 = 270\nOption A\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 7774\nRe: A clock store sold a certain clock to a collector for 20 per\u00a0 [#permalink]\n\n### Show Tags\n\n11 Feb 2018, 00:34\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: A clock store sold a certain clock to a collector for 20 per &nbs [#permalink] 11 Feb 2018, 00:34\nDisplay posts from previous: Sort by\n\n# A clock store sold a certain clock to a collector for 20 per\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-08-20 04:36:20", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/a-clock-store-sold-a-certain-clock-to-a-collector-for-20-per-46986.html", "openwebmath_score": 0.3218260407447815, "openwebmath_perplexity": 3190.9617084846705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. 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{"url": "https://gmatclub.com/forum/in-the-xy-plane-the-straight-line-graphs-of-the-three-equations-above-207707.html", "text": "It is currently 19 Nov 2017, 09:24\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# In the xy-plane, the straight-line graphs of the three equations above\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42249\n\nKudos [?]: 132696 [2], given: 12335\n\nIn the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n23 Jul 2015, 10:47\n2\nKUDOS\nExpert's post\n25\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n45% (medium)\n\nQuestion Stats:\n\n69% (01:28) correct 31% (01:22) wrong based on 154 sessions\n\n### HideShow timer Statistics\n\ny = ax - 5\ny = x + 6\ny = 3x + b\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n(1) a = 2\n(2) r = 17\n[Reveal] Spoiler: OA\n\n_________________\n\nLast edited by Bunuel on 13 Nov 2017, 21:24, edited 4 times in total.\nEdited the question.\n\nKudos [?]: 132696 [2], given: 12335\n\nCurrent Student\nJoined: 20 Mar 2014\nPosts: 2676\n\nKudos [?]: 1770 [4], given: 794\n\nConcentration: Finance, Strategy\nSchools: Kellogg '18 (M)\nGMAT 1: 750 Q49 V44\nGPA: 3.7\nWE: Engineering (Aerospace and Defense)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n23 Jul 2015, 11:30\n4\nKUDOS\n3\nThis post was\nBOOKMARKED\nmcelroytutoring wrote:\nDS 136 from OFG 2016 (new question)\n\ny = ax - 5\ny = x + 6\ny = 3x + b\n\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n1) a = 2\n\n2) r = 17\n\nSolution provided by : mcelroytutoring\n\nLet's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.\n\n#1: r = ap - 5 (3 variables R,A,P)\n#2: r = p + 6 (2 variables R,P)\n#3: r = 3p + b (3 variables R,B,P)\n\n1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.\n\n2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.\nAttachments\n\nScreen Shot 2015-07-23 at 10.36.05 AM.png [ 189.38 KiB | Viewed 7691 times ]\n\nScreen Shot 2015-07-23 at 10.31.01 AM.png [ 30.94 KiB | Viewed 7666 times ]\n\nKudos [?]: 1770 [4], given: 794\n\nDirector\nJoined: 10 Mar 2013\nPosts: 591\n\nKudos [?]: 478 [2], given: 200\n\nLocation: Germany\nConcentration: Finance, Entrepreneurship\nGMAT 1: 580 Q46 V24\nGPA: 3.88\nWE: Information Technology (Consulting)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n24 Sep 2015, 01:53\n2\nKUDOS\n3\nThis post was\nBOOKMARKED\nif each of the 3 equations contains points (p,r) this means that they intersect in that point\n1. a=2\nFind the intercept Intercept for three simultaneous equations\ny=2x-5\ny=x+6\ny=3x+b\nLet's use the first 2 equations: plug y=x+6 in the secod equation\nx+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation\n17=33+b -> b=-16 SUFFICIENT\n\n2. Here we have directly the value for Y, let's plug it in the 2nd equation\ny=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above\n17=33+b -> b=-16 SUFFICIENT\n\nAnswer (D) Most important point is here to catch the hint about intersection of 3 lines at one point\n_________________\n\nWhen you\u2019re up, your friends know who you are. When you\u2019re down, you know who your friends are.\n\n800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50\nGMAT PREP 670\nMGMAT CAT 630\nKAPLAN CAT 660\n\nKudos [?]: 478 [2], given: 200\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42249\n\nKudos [?]: 132696 [0], given: 12335\n\nIn the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n26 Oct 2015, 10:02\nExpert's post\n2\nThis post was\nBOOKMARKED\n\nKudos [?]: 132696 [0], given: 12335\n\nIntern\nStatus: GMAT1:520 Q44 V18\nJoined: 03 Sep 2015\nPosts: 11\n\nKudos [?]: 20 [2], given: 14\n\nLocation: United States\nConcentration: Strategy, Technology\nWE: Information Technology (Computer Software)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n26 Oct 2015, 10:50\n2\nKUDOS\nI think it's D.\n\nKeeping point (p,r) in all the equations we get :\n\np = ar -5 -----(1)\np = r + 6 ------(2)\np = 3r + b ------(3)\n\nNow consider (1) if a = 2 from (1) and (2) we get\nr = 11 , p=17 and putting in (3) we can get b.\n\nSimilarly for (2) we can get the values for r and p and hence can get the value for b.\n\nSo both statements individually are correct to answer the question.\n\nKudos [?]: 20 [2], given: 14\n\nManager\nJoined: 13 Apr 2015\nPosts: 76\n\nKudos [?]: 22 [1], given: 325\n\nConcentration: General Management, Strategy\nGMAT 1: 620 Q47 V28\nGPA: 3.25\nWE: Project Management (Energy and Utilities)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n26 Oct 2015, 19:14\n1\nKUDOS\nm2k wrote:\nI think it's D.\n\nKeeping point (p,r) in all the equations we get :\n\np = ar -5 -----(1)\np = r + 6 ------(2)\np = 3r + b ------(3)\n\nNow consider (1) if a = 2 from (1) and (2) we get\nr = 11 , p=17 and putting in (3) we can get b.\n\nSimilarly for (2) we can get the values for r and p and hence can get the value for b.\n\nSo both statements individually are correct to answer the question.\n\nApproach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.\n\nKudos [?]: 22 [1], given: 325\n\nManager\nJoined: 21 Sep 2015\nPosts: 82\n\nKudos [?]: 118 [2], given: 323\n\nLocation: India\nGMAT 1: 730 Q48 V42\nGMAT 2: 750 Q50 V41\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n10 Jun 2016, 11:23\n2\nKUDOS\ny = ax - 5 ... eq 1\ny = x + 6 ... eq 2\ny = 3x + b ...eq 3\n\nTotal of 4 variables are present.\n\nStatement 1 : a = 2\n\nInsert in eq 1\n\nWe have y = 2x -5 and y = x+6\n\nSolving we get x = 11 and y = 17\n\nSubstitute in eq 3 and we get value of b\n\nStatement 2: r=17\n\nThis means the y co- ordinate is 17\nSubstitute in eq 2 we get x as 11\n\nAgain can find value of b from equation 3\n\nHence D\n_________________\n\nAppreciate any KUDOS given !\n\nKudos [?]: 118 [2], given: 323\n\nManager\nJoined: 20 Jun 2013\nPosts: 50\n\nKudos [?]: 9 [0], given: 20\n\nLocation: India\nConcentration: Economics, Finance\nGMAT 1: 430 Q39 V25\nGPA: 3.5\nWE: Information Technology (Other)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n18 Jan 2017, 15:18\nwe used information from both 1 and 2 then how can the answer be D... should it not be C... some one kindly clarify.......... clearly am a zero in ds and that too a big one\n\nKudos [?]: 9 [0], given: 20\n\nDirector\nJoined: 14 Nov 2014\nPosts: 618\n\nKudos [?]: 105 [1], given: 46\n\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n18 Jan 2017, 20:18\n1\nKUDOS\ny = ax - 5---------1\ny = x + 6---------2\ny = 3x + b--------3\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n(1) a = 2\n(2) r = 17\n\nAll three line intersect each other at common point (p,r).\n1. given a = 2\nputting in equation 1 .= y=2x-5\nequating 1(after replacing value of a) and 2 we will get value of (p,r)\nputting (p,r) in equation 3 we will get value for b---suff..\n\n2 given r = 17.\nputting in equation 2 we will get value of x.i'e p.\nNow as we know common point of intersection ,putting p,r in equation 3 , we will get value of b\n\nKudos [?]: 105 [1], given: 46\n\nManager\nJoined: 20 Jun 2013\nPosts: 50\n\nKudos [?]: 9 [0], given: 20\n\nLocation: India\nConcentration: Economics, Finance\nGMAT 1: 430 Q39 V25\nGPA: 3.5\nWE: Information Technology (Other)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n22 Jan 2017, 15:51\nthanks sobby for your response... highly appreciated...\n\nKudos [?]: 9 [0], given: 20\n\nManager\nJoined: 17 Feb 2014\nPosts: 112\n\nKudos [?]: 48 [1], given: 31\n\nLocation: United States (CA)\nGMAT 1: 700 Q49 V35\nGMAT 2: 740 Q48 V42\nWE: Programming (Computer Software)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n16 Mar 2017, 08:58\n1\nKUDOS\n(eq1) $$y = ax - 5$$\n(eq2) $$y = x + 6$$\n(eq3) $$y = 3x + b$$\n\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?\n\n1) $$a = 2$$\n2) $$r = 17$$\n\nSolution:\n\n1) $$a = 2$$\n- Putting the value of a in eq1, we get: $$y = 2x - 5$$\n- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]\n(alternatively, faster method)\n- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.\nNOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.\n\n2) $$r = 17$$\n- Since, point (p, r) lie on all the line, we can plugin the point in above equation\n$$r = ap - 5 => 17 = ap - 5$$\n$$r = p + 6 => 17 = p + 6$$\n$$r = 3p + b => 17 = 3p + b$$\n- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.\n\nKudos [?]: 48 [1], given: 31\n\nIntern\nJoined: 25 Feb 2017\nPosts: 47\n\nKudos [?]: 8 [0], given: 30\n\nGMAT 1: 720 Q50 V38\nGPA: 3.67\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n30 Apr 2017, 23:51\ny = ax - 5\ny = x + 6\ny = 3x + b\n\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n1) a = 2\n\n2) r = 17\n\nMy 2 cents.\n\nIt is important to realize from the Question stem that the 3 equations intersect as (p,r).\n\nFor 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.\n\nFor 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.\n\nSo D.\n\nKudos [?]: 8 [0], given: 30\n\nIntern\nJoined: 27 Apr 2015\nPosts: 11\n\nKudos [?]: [0], given: 2\n\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n10 Sep 2017, 19:04\nWould it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?\n\nKudos [?]: [0], given: 2\n\nIntern\nJoined: 11 Sep 2017\nPosts: 9\n\nKudos [?]: 1 [0], given: 19\n\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n13 Nov 2017, 20:21\nwhat is the significance of the the line \"In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)\",\n\ni solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables\n\nKudos [?]: 1 [0], given: 19\n\nBSchool Forum Moderator\nJoined: 17 Jun 2016\nPosts: 439\n\nKudos [?]: 210 [0], given: 199\n\nLocation: India\nGMAT 1: 720 Q49 V39\nGMAT 2: 710 Q50 V37\nGPA: 3.65\nWE: Engineering (Energy and Utilities)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0[#permalink]\n\n### Show Tags\n\n13 Nov 2017, 22:31\nCheryn wrote:\nwhat is the significance of the the line \"In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)\",\n\ni solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables\n\nThe highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.\n\n_________________\n\nKudos [?]: 210 [0], given: 199\n\nRe: In the xy-plane, the straight-line graphs of the three equations above \u00a0 [#permalink] 13 Nov 2017, 22:31\nDisplay posts from previous: Sort by", "date": "2017-11-19 16:24:06", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-the-xy-plane-the-straight-line-graphs-of-the-three-equations-above-207707.html", "openwebmath_score": 0.6089481711387634, "openwebmath_perplexity": 3598.533793648166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.855851148805615, "lm_q1q2_score": 0.855851148805615}}
{"url": "https://gmatclub.com/forum/if-r-and-s-are-positive-integers-can-the-fraction-r-s-be-expressed-as-141000.html?fl=similar", "text": "Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack\nGMAT Club\n\n It is currently 23 Mar 2017, 23:29\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nEvents & Promotions\n\nEvents & Promotions in June\nOpen Detailed Calendar\n\nIf r and s are positive integers, can the fraction r/s be expressed as\n\nAuthor Message\nTAGS:\n\nHide Tags\n\nIntern\nJoined: 11 Sep 2012\nPosts: 6\nLocation: Norway\nGMAT Date: 11-19-2013\nWE: Brand Management (Pharmaceuticals and Biotech)\nFollowers: 0\n\nKudos [?]: 75 [3] , given: 1\n\nIf r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n20 Oct 2012, 08:21\n3\nKUDOS\n34\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n64% (01:43) correct 36% (01:01) wrong based on 1274 sessions\n\nHideShow timer Statistics\n\nIf r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?\n\n(1) s is a factor of 100\n(2) r is a factor of 100\n[Reveal] Spoiler: OA\nIntern\nJoined: 09 Oct 2012\nPosts: 39\nFollowers: 0\n\nKudos [?]: 5 [0], given: 14\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n20 Oct 2012, 10:02\nIMO it is A, if the denominator is a factor of 100 then it could be 1; 2; 5; 10; 20.. if you divide all the positive integer by these number you will have a finite decimal result.\nModerator\nJoined: 02 Jul 2012\nPosts: 1230\nLocation: India\nConcentration: Strategy\nGMAT 1: 740 Q49 V42\nGPA: 3.8\nWE: Engineering (Energy and Utilities)\nFollowers: 119\n\nKudos [?]: 1443 [1] , given: 116\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n20 Oct 2012, 13:18\n1\nKUDOS\n2\nThis post was\nBOOKMARKED\nFinite decimals are decimals which end. eg 0.5,0.25, etc. Non finite are numbers like 1/3,1/6 etc ie 0.33333333333333333333333333333333333333333333333333333.......... or 0.66666666666666666666666666666666666666666666666666666..........\n\n1)S is a factor of 100. So S cannot have more than two 2s and two 5s. Any number divisible be 2 or 5 gives a finite decimal. Since R and S are both positive integers, there can be no 0s in the decimal places either. So Sufficient\n\n2)R is a factor of 100. Cant say anything form this. R can be 1. 1/10 is finite. 1/3 is not. Insufficient.\n\n_________________\n\nDid you find this post helpful?... Please let me know through the Kudos button.\n\nThanks To The Almighty - My GMAT Debrief\n\nGMAT Reading Comprehension: 7 Most Common Passage Types\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 37560\nFollowers: 7393\n\nKudos [?]: 99319 [21] , given: 11010\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n23 Oct 2012, 07:59\n21\nKUDOS\nExpert's post\n37\nThis post was\nBOOKMARKED\nasveaass wrote:\nIf r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?\n\n1) s is a factor of 100\n2) r is a factor of 100\n\nI don't understand the answer explanation in the OG, could someone please explain in detail?\n\nTHEORY:\n\nReduced fraction $$\\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\\frac{3}{30}$$ is also a terminating decimal, as $$\\frac{3}{30}=\\frac{1}{10}$$ and denominator $$10=2*5$$.\n\nNote that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.\n\nFor example $$\\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.\n\n(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)\n\nQuestions testing this concept:\ndoes-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html\nany-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html\nif-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html\n700-question-94641.html\nis-r-s2-is-a-terminating-decimal-91360.html\npl-explain-89566.html\nwhich-of-the-following-fractions-88937.html\n\nBACK TO THE ORIGINAL QUESTION:\nIf r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?\n\n(1) s is a factor of 100. Factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50 and 100. All these numbers are of the form $$2^n5^m$$ (for example 1=2^0*5^0, 2=2^1*5^0, ...), therefore no matter what is the value of r, r/s will always will be terminating decimal. Sufficient.\n\n(2) r is a factor of 100. We need to know about the denominator. Not sufficient.\n\nHope it's clear.\n_________________\nDirector\nStatus: Gonna rock this time!!!\nJoined: 22 Jul 2012\nPosts: 543\nLocation: India\nGMAT 1: 640 Q43 V34\nGMAT 2: 630 Q47 V29\nWE: Information Technology (Computer Software)\nFollowers: 3\n\nKudos [?]: 61 [0], given: 562\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n02 Jan 2013, 01:25\nyou rock bunuel, your explanations are awesome.. thanks a lot for this one!\n_________________\n\nhope is a good thing, maybe the best of things. And no good thing ever dies.\n\nWho says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595\n\nMy GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992\n\nIntern\nJoined: 27 Aug 2012\nPosts: 8\nFollowers: 0\n\nKudos [?]: 0 [0], given: 1\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n24 Dec 2013, 11:12\nBut it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?\nSenior Manager\nJoined: 15 Aug 2013\nPosts: 328\nFollowers: 0\n\nKudos [?]: 58 [0], given: 23\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n04 May 2014, 11:31\nHi Bunuel,\n\nSpeaking from a prime perspective, I could factor \"100\" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?\n\nIf the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 37560\nFollowers: 7393\n\nKudos [?]: 99319 [0], given: 11010\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n05 May 2014, 02:31\nhalloanupam wrote:\nBut it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?\n\n0.04 has finite number of non-zero digits: 4 is not followed by any non-zero digit.\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 37560\nFollowers: 7393\n\nKudos [?]: 99319 [0], given: 11010\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n05 May 2014, 02:35\nruss9 wrote:\nHi Bunuel,\n\nSpeaking from a prime perspective, I could factor \"100\" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?\n\nIf the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?\n\nNot entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.\n_________________\nIntern\nJoined: 11 Oct 2012\nPosts: 43\nGMAT 1: 610 Q42 V32\nFollowers: 1\n\nKudos [?]: 8 [0], given: 74\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n26 Sep 2014, 23:28\nBunuel wrote:\nasveaass wrote:\nIf r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?\n\n1) s is a factor of 100\n2) r is a factor of 100\n\nI don't understand the answer explanation in the OG, could someone please explain in detail?\n\nTHEORY:\n\nReduced fraction $$\\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\\frac{3}{30}$$ is also a terminating decimal, as $$\\frac{3}{30}=\\frac{1}{10}$$ and denominator $$10=2*5$$.\n\nNote that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.\n\nFor example $$\\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.\n\n(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)\n\nWhy is it then 130/13 or 121/11 would give finite ... infact they properly divide...\nMath Expert\nJoined: 02 Sep 2009\nPosts: 37560\nFollowers: 7393\n\nKudos [?]: 99319 [0], given: 11010\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n27 Sep 2014, 01:28\nani781 wrote:\nBunuel wrote:\nasveaass wrote:\nIf r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?\n\n1) s is a factor of 100\n2) r is a factor of 100\n\nI don't understand the answer explanation in the OG, could someone please explain in detail?\n\nTHEORY:\n\nReduced fraction $$\\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\\frac{3}{30}$$ is also a terminating decimal, as $$\\frac{3}{30}=\\frac{1}{10}$$ and denominator $$10=2*5$$.\n\nNote that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.\n\nFor example $$\\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.\n\n(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)\n\nWhy is it then 130/13 or 121/11 would give finite ... infact they properly divide...\n\nThe rule above is for reduced fraction $$\\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term). When you reduce 130/13 to the lowest term you get 10 and when you reduce 121/11 you get 11: 10/(2^0*5^0) and 11/(2^0*5^0) respectively.\n\nCheck the links in my post above to practice more on this type of questions.\n_________________\nIntern\nJoined: 26 Sep 2014\nPosts: 2\nFollowers: 0\n\nKudos [?]: 0 [0], given: 0\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n27 Sep 2014, 02:08\nA, of course. I had got it as soon as I saw it. This is a very simple problem probably fetched from RS\nIntern\nJoined: 18 Dec 2013\nPosts: 40\nGPA: 2.84\nWE: Project Management (Telecommunications)\nFollowers: 0\n\nKudos [?]: 24 [0], given: 98\n\nIf r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n17 Nov 2014, 19:56\n1\nThis post was\nBOOKMARKED\nBunuel wrote:\n\nTHEORY:\n\nReduced fraction $$\\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\\frac{3}{30}$$ is also a terminating decimal, as $$\\frac{3}{30}=\\frac{1}{10}$$ and denominator $$10=2*5$$.\n\nNote that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.\n\nFor example $$\\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.\n\nHi Bunuel,\n\nJust looking for some clarification, in the highlighted text above, please correct me if I am wrong, by including the word \"and\" did you mean that if the denominator contains only 2's or only 5's then too the fraction is a terminating decimal? In other words, the denominator doesn't need to contain, both - 2's & 5's!\n\nThank you!\ndmmk\nDirector\nAffiliations: GMATQuantum\nJoined: 19 Apr 2009\nPosts: 589\nFollowers: 113\n\nKudos [?]: 433 [1] , given: 14\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n17 Nov 2014, 23:36\n1\nKUDOS\ndmmk\n\nYes, examples such as $$\\frac{17}{2^7}$$, $$\\frac{31}{5^{18}}$$, and $$\\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.\n\nDabral\nIntern\nJoined: 18 Dec 2013\nPosts: 40\nGPA: 2.84\nWE: Project Management (Telecommunications)\nFollowers: 0\n\nKudos [?]: 24 [0], given: 98\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n18 Nov 2014, 12:44\ndabral wrote:\ndmmk\n\nYes, examples such as $$\\frac{17}{2^7}$$, $$\\frac{31}{5^{18}}$$, and $$\\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.\n\nDabral\n\nThank you dabral\nChat Moderator\nJoined: 19 Apr 2013\nPosts: 746\nConcentration: Strategy, Healthcare\nSchools: Sloan '18 (A)\nGMAT 1: 730 Q48 V41\nGPA: 4\nFollowers: 14\n\nKudos [?]: 166 [0], given: 537\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n03 Dec 2014, 11:12\nBunuel wrote:\nruss9 wrote:\nHi Bunuel,\n\nSpeaking from a prime perspective, I could factor \"100\" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?\n\nIf the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?\n\nNot entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.\n\nBunuel I did not understand from your post whether we can have other primes in denominator? Can you pls. repeat?\n_________________\n\nIf my post was helpful, press Kudos. If not, then just press Kudos !!!\n\nDirector\nAffiliations: GMATQuantum\nJoined: 19 Apr 2009\nPosts: 589\nFollowers: 113\n\nKudos [?]: 433 [1] , given: 14\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n03 Dec 2014, 23:50\n1\nKUDOS\nErgenekon\n\nFor a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\\frac{11}{[(5^6)(7^3)]}$$, and $$\\frac{22}{(7^4)}$$ are all non-terminating decimals.\n\nDabral\nChat Moderator\nJoined: 19 Apr 2013\nPosts: 746\nConcentration: Strategy, Healthcare\nSchools: Sloan '18 (A)\nGMAT 1: 730 Q48 V41\nGPA: 4\nFollowers: 14\n\nKudos [?]: 166 [0], given: 537\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n04 Dec 2014, 00:35\ndabral wrote:\nErgenekon\n\nFor a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\\frac{11}{[(5^6)(7^3)]}$$, and $$\\frac{22}{(7^4)}$$ are all non-terminating decimals.\n\nDabral\n\nThanks Dabral. Got it now.\n_________________\n\nIf my post was helpful, press Kudos. If not, then just press Kudos !!!\n\nIntern\nJoined: 11 Mar 2015\nPosts: 1\nFollowers: 0\n\nKudos [?]: 0 [0], given: 1\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n11 Mar 2015, 17:14\n1\nThis post was\nBOOKMARKED\nHello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 37560\nFollowers: 7393\n\nKudos [?]: 99319 [1] , given: 11010\n\nRe: If r and s are positive integers, can the fraction r/s be expressed as\u00a0[#permalink]\n\nShow Tags\n\n12 Mar 2015, 03:15\n1\nKUDOS\nExpert's post\nptanwar1 wrote:\nHello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?\n\nAn integer IS a decimal with a finite number of nonzero digits. For example, integer 51 has 2 (finite) number of digits.\n_________________\nRe: If r and s are positive integers, can the fraction r/s be expressed as \u00a0 [#permalink] 12 Mar 2015, 03:15\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 32 posts ]\n\nSimilar topics Replies Last post\nSimilar\nTopics:\n16 Is r*s a multiple of 25? (r and s are positive integers) 7 13 Jan 2016, 23:34\n36 If r and s are positive integers, is r/s an integer? 18 23 Feb 2012, 08:13\nif r and s are positive integers , is r/s an integer? 1 25 Jun 2011, 22:58\n4 If r and s are positive integers, is r/s an integer? 1) 6 23 Apr 2011, 15:34\n22 If r and s are positive integers, is r/s an integer? 18 10 Nov 2009, 15:24\nDisplay posts from previous: Sort by", "date": "2017-03-24 06:29:13", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-r-and-s-are-positive-integers-can-the-fraction-r-s-be-expressed-as-141000.html?fl=similar", "openwebmath_score": 0.8141879439353943, "openwebmath_perplexity": 1704.6135320707817, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8558511451289037, "lm_q1q2_score": 0.8558511451289037}}
{"url": "https://gmatclub.com/forum/2-600-has-how-many-positive-divisors-103880.html", "text": "It is currently 20 Mar 2018, 18:34\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# 2,600 has how many positive divisors?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSenior Manager\nStatus: Do and Die!!\nJoined: 15 Sep 2010\nPosts: 293\n2,600 has how many positive divisors?\u00a0[#permalink]\n\n### Show Tags\n\n28 Oct 2010, 17:10\n9\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n67% (00:44) correct 33% (00:55) wrong based on 385 sessions\n\n### HideShow timer Statistics\n\n2,600 has how many positive divisors?\n\nA. 6\nB. 12\nC. 18\nD. 24\nE. 48\n[Reveal] Spoiler: OA\n\n_________________\n\nI'm the Dumbest of All !!\n\nIntern\nJoined: 26 Sep 2010\nPosts: 11\n\n### Show Tags\n\n28 Oct 2010, 17:21\n1\nKUDOS\nBy factorization, you can write 2600 as 2600=2^3*5^2*13^1. Number of factors = (3+1)(2+1)(1+1)=24\nManager\nJoined: 03 Jun 2010\nPosts: 156\nLocation: United States (MI)\nConcentration: Marketing, General Management\n\n### Show Tags\n\n29 Oct 2010, 02:47\nvgan4 wrote:\nNumber of factors = (3+1)(2+1)(1+1)=24\n\ncan' t understand how we came to this. What does this multiplication mean?\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44351\n\n### Show Tags\n\n29 Oct 2010, 03:11\n2\nKUDOS\nExpert's post\n6\nThis post was\nBOOKMARKED\nulm wrote:\nvgan4 wrote:\nNumber of factors = (3+1)(2+1)(1+1)=24\n\ncan' t understand how we came to this. What does this multiplication mean?\n\nFinding the Number of Factors of an Integer\n\nFirst make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.\n\nThe number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.\n\nExample: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$\n\nTotal number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.\n\nBack to the original question:\nIf p and q are prime numbers, how many divisors does the product $$p^3*q^6$$ have?\n\nAccording to above the number of distinct factors of $$2,600=2^3*5^2*13$$ would be $$(3+1)(2+1)(1+1)=24$$.\n\nFor more on this check Number Theory chapter of Math Book: math-number-theory-88376.html\n\nHope it helps.\n_________________\nManager\nJoined: 03 Jun 2010\nPosts: 156\nLocation: United States (MI)\nConcentration: Marketing, General Management\n\n### Show Tags\n\n29 Oct 2010, 03:27\nYes, thanks,\ni forgot\n\"The number of factors of n will be expressed by the formula $$(p+1)(q+1)(r+1)$$\"\nIntern\nJoined: 07 Feb 2008\nPosts: 46\n\n### Show Tags\n\n29 Oct 2010, 23:57\nThanx Bunuel\nSenior Manager\nJoined: 05 Jan 2017\nPosts: 434\nLocation: India\nRe: 2,600 has how many positive divisors?\u00a0[#permalink]\n\n### Show Tags\n\n20 Mar 2017, 01:27\nshrive555 wrote:\n2,600 has how many positive divisors?\n\nA. 6\nB. 12\nC. 18\nD. 24\nE. 48\n\n2600 = 13 x 2^3 x 5^2\n\nno of possible divisor = (1+1)(2+1)(3+1) = 4.3.2 = 24\n\nOption D\nDirector\nJoined: 02 Sep 2016\nPosts: 777\nRe: 2,600 has how many positive divisors?\u00a0[#permalink]\n\n### Show Tags\n\n03 Apr 2017, 00:16\n2600= 26*100= 2*13*2^2*5^2\n=2^3*5^2*13\n\nTotal factors= (3+1)(2+1)(1+1)= 4*3*2=24\n_________________\n\nHelp me make my explanation better by providing a logical feedback.\n\nIf you liked the post, HIT KUDOS !!\n\nDon't quit.............Do it.\n\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2127\nRe: 2,600 has how many positive divisors?\u00a0[#permalink]\n\n### Show Tags\n\n05 Jan 2018, 11:24\nshrive555 wrote:\n2,600 has how many positive divisors?\n\nA. 6\nB. 12\nC. 18\nD. 24\nE. 48\n\nTo determine the number of positive divisors, we break 2,600 into primes, add 1 to the exponent of each unique prime, and then multiply those values together.\n\nWe see that 2,600 = 26 x 100 = 2 x 13 x 4 x 25 = 2^3 x 5^2 x 13^1.\n\nNow we add 1 to each exponent and multiply those results:\n\n(3 + 1)(2 + 1)(1 +1) = 24\n\nThus, 2,600 has 24 positive divisors.\n\n_________________\n\nJeffery Miller\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nSVP\nJoined: 12 Sep 2015\nPosts: 2154\nRe: 2,600 has how many positive divisors?\u00a0[#permalink]\n\n### Show Tags\n\n05 Jan 2018, 11:56\n1\nKUDOS\nExpert's post\nTop Contributor\n1\nThis post was\nBOOKMARKED\nshrive555 wrote:\n2,600 has how many positive divisors?\n\nA. 6\nB. 12\nC. 18\nD. 24\nE. 48\n\nExperts, is there a fast way to find all the divisors? Thanks.[/quote]\n\nIf the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.\n\nExample: 14000 = (2^4)(5^3)(7^1)\nSo, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40\n----------ONTO THE QUESTION-------------------------\n\n2600 = (2)(2)(2)(5)(5)(13)\n= (2^3)(5^2)(13^1)\nSo, the number of positive divisors of 2600 = (3+1)(2+1)(1+1) =(4)(3)(2) = 24\n\nCheers,\nBrent\n_________________\n\nBrent Hanneson \u2013 Founder of gmatprepnow.com\n\nRe: 2,600 has how many positive divisors? \u00a0 [#permalink] 05 Jan 2018, 11:56\nDisplay posts from previous: Sort by", "date": "2018-03-21 01:34:27", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/2-600-has-how-many-positive-divisors-103880.html", "openwebmath_score": 0.8318807482719421, "openwebmath_perplexity": 6310.0483855513685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.8558511414521923, "lm_q1q2_score": 0.8558511414521923}}
{"url": "https://math.stackexchange.com/questions/3732112/general-solution-for-cos-fracx2-1-cos21-fracx2", "text": "# General Solution for $\\cos(\\frac{x}{2}-1) =\\cos^2(1-\\frac{x}{2})$\n\nI'm looking for an algebraic solution to : $$\\cos(\\frac{x}{2}-1) = \\cos^2(1-\\frac{x}{2})$$. So I simplified the equation: first off, $$\\cos(\\frac{x}{2}-1) = \\cos(1-\\frac{x}{2})$$. Then I divided both sides by that. and so I'm left with two things to solve:\n\n$$\\cos(\\frac{x}{2}-1) = 0$$ (because I divided both sides by that expression, I have to also include the $$0$$ solution too). and $$\\cos(\\frac{x}{2}-1) = 1$$. And the general solution would be, I think, the union of those.\n\nHowever, I'm kinda lost at this point. I've attempted to solve each equation. First off, I know that $$\\cos(x) = 0$$ at $$\\frac{\\pi}{2}$$ and $$\\frac{3\\pi}{2}$$. So, the general solution for $$\\cos(x) = 0$$ would be $$x=\\frac{\\pi}{2} +2\\pi k, \\cup \\ \\frac{3\\pi}{2}+2\\pi k, k\\in Z.$$ I got up to this point, but don't know how to proceed.\n\nThe thing that's most confusing to me is I don't know how the $$-1$$ in the argument plays into the solution. Does it just change the graph to the right? Playing with desmos shows that graph is shifting by 2, but I thought that it'd shift by 1. More importantly: does it also affect the period of the function?\n\nAdditional question: In my book the answers are given in a different form. For example, the union I wrote would be written as: $$x= (-1)^k\\frac{\\pi}{2} + \\pi k, k \\in Z.$$ And in every case the period is \"reduced\" to $$\\pi k$$. Why is that?\n\n\u2022 So if the solution for $\\cos(x) =0$ is $(x) =\\pi/2+2\\pi k$ then the solution to $\\cos(x/2-1)=0$ is $(x/2-1)=\\pi/2+2\\pi k$ similarly for the other solution. \u2013\u00a0kingW3 Jun 23 '20 at 23:49\n\u2022 well, $\\cos (-u) = \\cos u$ \u2013\u00a0Will Jagy Jun 23 '20 at 23:51\n\nOkay so, you noticed that $${\\cos\\left(\\frac{x}{2}-1\\right)=\\cos\\left(1-\\frac{x}{2}\\right)}$$. And as you say, you end up with\n\n$${\\Leftrightarrow \\cos^2\\left(\\frac{x}{2}-1\\right)=\\cos\\left(\\frac{x}{2}-1\\right)}$$\n\nAnd this implies\n\n$${\\cos\\left(\\frac{x}{2}-1\\right)\\left(\\cos\\left(\\frac{x}{2}-1\\right)-1\\right)=0}$$\n\n(as was pointed out by someone else - indeed it's probably bad practice to divide through here by $${\\cos}$$. Not that it's incorrect, since you did take into account the fact we would then miss the $$0$$ solution (which was awesome!!!) - but unnecessary).\n\nFor simplicity, we can replace $${\\frac{x}{2}-1}$$ with $${u}$$ and just rearrange for $${x}$$ at the end. So\n\n$${\\Rightarrow \\cos(u)\\left(\\cos(u)-1\\right)=0}$$\n\nNow.... we want to solve $${\\cos(u)=0}$$. As you said, one solution is $${\\frac{\\pi}{2}}$$... and if you take a look at the graph, you will notice that every other $$0$$ to the cosine function can be \"reached\" by hopping foward and backwards by multiplies of $${\\pi}$$... you are right in thinking indeed, the period of the cosine function is $${2\\pi}$$, but in fact $${0}$$ reoccurs every $${\\pi}$$ radians. There is no problem with this. In order to have a full period ($${2\\pi}$$ in this case), every value the function takes on must reoccur - only $$0$$ has. This actually means that\n\n$${\\cos(u)=0\\Leftrightarrow u=\\frac{\\pi}{2}+n\\pi, n \\in \\mathbb{Z}}$$\n\nNow we can solve $${\\cos(u)=1}$$ (the other solution we need). Of course we have a $${1}$$ at the point $${x=0}$$, and you may notice that we can reach all the other 1's by jumping foward and backwards by multiples of $${2\\pi}$$ this time. Hence the solution is\n\n$${\\cos(u)=1\\Leftrightarrow u=0 + 2n\\pi =2n\\pi, n \\in \\mathbb{Z}}$$\n\nNow, you can simply plug back in the definition of $${u}$$ in terms of $${x}$$, adding $${1}$$ and multiplying both sides by two and you will end up with the solutions being\n\n$${x=\\pi + 2n\\pi + 2=(2n+1)\\pi + 2, n \\in \\mathbb{Z}}$$\n\n$${x=4n\\pi + 2, n \\in \\mathbb{Z}}$$\n\n\u2022 This explained things very clearly and helped me solved another, similar problem too!. Thanks!!. \u2013\u00a0Ebrin Jun 24 '20 at 0:33\n\u2022 No problem at all! :) \u2013\u00a0Riemann'sPointyNose Jun 24 '20 at 0:35\n\nYou're making things more complicated than they are\n\nFirst, cosine is an even function, so the equation can be written $$\\cos\\Bigl(\\frac{x}{2}-1\\Bigr) =\\cos^2\\Bigl(\\frac{x}{2}-1\\Bigr)$$ as well. Note the equation implies $$\\cos\\bigl(\\frac{x}{2}-1\\bigr)\\ge 0$$.\n\nAlso, if $$0, note that $$c Therefore the equation is equivalent to $$\\cos\\Bigl(\\frac{x}{2}-1\\Bigr)=0\\:\\text{ or }\\:1.$$ Now this is easy to solve in terms of congruences: $$\\begin{cases} \\cos\\Bigl(\\frac{x}{2}-1\\Bigr)=0\\iff \\frac{x}{2}-1\\equiv \\frac\\pi 2 \\mod\\pi \\iff\\frac x2\\equiv 1+\\frac\\pi 2\\mod\\pi \\\\ \\cos\\Bigl(\\frac{x}{2}-1\\Bigr)=1\\iff \\frac{x}{2}-1\\equiv 0\\mod2\\pi \\iff\\frac x2\\equiv 1\\mod2\\pi \\end{cases}$$ and ultimately $$x\\equiv2+\\pi\\bmod 2\\pi \\quad\\text{ or }\\quad x\\equiv2\\bmod 4\\pi.$$\n\n\u2022 Oh, the first part explains why the answer only contained the positive answers. So, the -1 in the argument just corresponds to the graph shifting to the right by two, right? It has no impact on the period. \u2013\u00a0Ebrin Jun 24 '20 at 0:07\n\u2022 Absolutely no impact. \u2013\u00a0Bernard Jun 24 '20 at 0:09\n\u2022 @kingW3: You're right. I shouldn't calculate directly on screen. Thank you for pointing it! \u2013\u00a0Bernard Jun 24 '20 at 0:17\n\nApproach to the problem I would like to discuss an approach and leave solving the equations to you. $$\\cos(y) = 0$$ implies that y is an odd multiple of $$\\frac{\\pi}{2}$$ or $$(2n + 1)\\frac{\\pi}{2}$$, where n is an integer. You can even write $$(2n - 1)$$, which is fine too as it gives odd vales. Similarly, $$\\cos(\\frac{x}{2} - 1) = 0$$ implies that $$\\frac{x}{2} - 1 = (2n + 1)\\frac{\\pi}{2}.$$ You can simplify it further. Then solve the next one which is $$\\cos(\\frac{x}{2} - 1) = 1$$which implies that $$\\frac{x}{2} - 1 = 2n\\pi$$ (even multiple of pi), where $$n$$ is an integer. The $$'n'$$ in both solutions can be same because the first solution will always give an odd value of $$\\frac{\\pi}{2}$$ and the second solution will always give an even value of $$\\frac{\\pi}{2},$$ so we can use the same symbol 'n'. Now, to unify both solutions: Observe if there is any similarity between first and second solution. Is there any value(s) or range for which both solutions are same? If yes, we can get an even compact expression. If no, then write both solutions joined together with union symbol.\n\nRegarding graph To interpret a function $$f(x - 1)$$interms of $$f(x)$$: The value which you get on putting $$x$$ in a function $$f(x)$$ will now [in $$f(x - 1)]$$ be obtained by setting $$x$$ as $$(x+1)$$ in $$f(x - 1)$$. Put $$x+1$$ in place of $$x$$ in $$f(x-1)$$, you will get $$f(x)$$ again. What does it mean? It means that you will have to shift your graph of $$f(x)$$ towards right by $$1$$ unit. Practice by drawing the graph of $$f(x) = x^2.$$ Then draw $$f(x) = (x - 1)^2.$$ But your problem has $$\\cos(\\frac{x}{2} - 1)$$ which is similar to $$f(ax + b)$$. We can now interpret $$'b'$$ as we did in last para. What about $$'a'$$?\n\nThink like this. Every value of function $$f(x)$$ is now obtained at $$\\frac{x}{a}$$ when the function is transformed to $$f(ax)$$. To show this, set $$x$$ as $$\\frac{x}{a}$$ in $$f(ax)$$, you will get f(x) again. It also means that the frequency of the function $$f(ax)$$ is now decreased by $$a$$ factor of $$'a'$$.\n\nIn case of $$f(\\frac{x}{a})$$, it can be said that every old value of $$f(x)$$ is obtained at $$'ax'$$ (show). Here the frequency is increased by a factor of $$'a'$$. Every old value of $$f(x)$$ is obtained quickly with $$f(\\frac{x}{a}).$$ You can relate this with period since frequency and period are related. Or, you could interpret it in terms of period in the first place itself. If the frequency has increased, it means the value of the function will be repeated quickly, thereby decreasing the period. Period of $$\\cos(x)$$ is $$2\\pi$$. Period of $$\\cos(ax+b)$$ will be $$\\frac{2\\pi}{a}.$$ The constant b only shifts the values, it does not account for frequency or period.\n\nThus, your interpretation of graph of $$\\cos(\\frac{x}{2} - 1)$$ is incomplete. To obtain its graph, we have to expand the graph of $$\\cos(x)$$ by a factor of $$2$$, then shift the graph towards right by $$1$$ unit.\n\nRegarding your last doubt In academics it is completely fine to represent the solution as simple as possible according to you. Generally, we want to make it more compact. The solution which you have wrote there. Put integral values of $$k$$. You will find that you have actually written odd multiples of $$\\frac{\\pi}{2}$$ in any case. Try to put several odd and even (positive integers for ease) integral values for $$k$$ and interpret the result yourself.\n\nI thank @Ebrin for his help in formatting this answer.\n\n\u2022 Sorry for the late reply, I've read your answer thoroughly and found it very helpful. Thanks for taking the time and explaining things! \u2013\u00a0Ebrin Jun 28 '20 at 15:59\n\nSo if $$\\cos W = 0$$ then $$W = \\pm \\frac \\pi 2 + 2k\\pi$$. Notice $$-\\frac \\pi 2 + 2k \\pi = \\frac \\pi 2 + (2k-1)\\pi$$. And $$\\frac \\pi 2 + 2k \\pi = -\\frac \\pi 2 + (2k+1) \\pi$$. So to simplify $$W = \\frac \\pi 2 + m\\pi$$ would be the simplest way to state this.\n\n(Alternativily as $$0 = -0$$ and $$\\cos(W \\pm \\pi) = -\\cos W$$ we'd note that for an $$W = \\pm \\frac \\pi 2 +2k \\pi$$ then $$\\mp \\frac \\pi 2+ (2k+1)\\pi$$ is a solution).\n\nSo $$\\frac x2 -1 = \\frac \\pi 2 + m\\pi$$ then $$x = (2m+1)\\pi + 2$$\n\nAnd if $$\\cos W = 1$$ then .... well... do I need to point out that solution is $$W=m\\pi$$?\n\nSo $$\\frac x2 -1= m\\pi$$ so $$x=2m\\pi +1$$.\n\nSo the solutions are $$x = k \\pi + 2$$. If $$k$$ is odd then $$\\cos(\\frac x2 -1)=\\cos(\\frac k2 \\pi) = 1$$. And if $$k$$ is odd then $$\\cos(\\frac x2 -1)=\\cos(\\frac {k-1}2\\pi + \\frac \\pi 2) = 0$$.", "date": "2021-02-25 02:34:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3732112/general-solution-for-cos-fracx2-1-cos21-fracx2", "openwebmath_score": 0.8674164414405823, "openwebmath_perplexity": 188.86791788394024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.969785412932606, "lm_q2_score": 0.8824278726384089, "lm_q1q2_score": 0.8557656788498805}}
{"url": "https://in.mathworks.com/help/wavelet/ref/scal2frq.html", "text": "# scal2frq\n\nScale to frequency\n\n## Description\n\nexample\n\nfrq = scal2frq(A,wname,delta) returns the pseudo-frequencies corresponding to the scales given by A and the wavelet specified by wname and the sampling period delta. The output frq is real-valued and has the same dimensions as A.\n\nfrq = scal2frq(A,wname) is equivalent to frq = scal2frq(A,wname,1).\n\n## Examples\n\ncollapse all\n\nThis example shows how the pseudo-frequency changes as you double the scale.\n\nConstruct a vector of scales with 10 voices per octave over five octaves.\n\nvpo = 10;\nno = 5;\na0 = 2^(1/vpo);\nind = 0:vpo*no;\nsc = a0.^ind;\n\nVerify that the range of scales covers five octaves.\n\nlog2(max(sc)/min(sc))\nans = 5.0000\n\nIf you plot the scales, you can use a data cursor to confirm that the scale at index $n+10$ is twice the scale at index $n$. Set the y-ticks to mark each octave.\n\nplot(ind,sc)\ntitle('Scales')\nxlabel('Index')\nylabel('Scale')\ngrid on\nset(gca,'YTick',2.^(0:5))\n\nConvert the scales to pseudo-frequencies for the real-valued Morlet wavelet. First, assume the sampling period is 1.\n\npf = scal2frq(sc,\"morl\");\nT = [sc(:) pf(:)];\nT = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency'});\ndisp(T)\nScale     Pseudo-Frequency\n______    ________________\n\n1          0.8125\n1.0718         0.75809\n1.1487         0.70732\n1.2311         0.65996\n1.3195         0.61576\n1.4142         0.57452\n1.5157         0.53605\n1.6245         0.50015\n1.7411         0.46666\n1.8661         0.43541\n2         0.40625\n2.1435         0.37904\n2.2974         0.35366\n2.4623         0.32998\n2.639         0.30788\n2.8284         0.28726\n3.0314         0.26803\n3.249         0.25008\n3.4822         0.23333\n3.7321          0.2177\n4         0.20313\n4.2871         0.18952\n4.5948         0.17683\n4.9246         0.16499\n5.278         0.15394\n5.6569         0.14363\n6.0629         0.13401\n6.498         0.12504\n6.9644         0.11666\n7.4643         0.10885\n8         0.10156\n8.5742        0.094761\n9.1896        0.088415\n9.8492        0.082494\n10.556         0.07697\n11.314        0.071816\n12.126        0.067006\n12.996        0.062519\n13.929        0.058332\n14.929        0.054426\n16        0.050781\n17.148        0.047381\n18.379        0.044208\n19.698        0.041247\n21.112        0.038485\n22.627        0.035908\n24.251        0.033503\n25.992         0.03126\n27.858        0.029166\n29.857        0.027213\n32        0.025391\n\nAssume that data is sampled at 100 Hz. Construct a table with the scales, the corresponding pseudo-frequencies, and periods. Since there are 10 voices per octave, display every tenth row in the table. Observe that for each doubling of the scale, the pseudo-frequency is cut in half.\n\nFs = 100;\nDT = 1/Fs;\npf = scal2frq(sc,\"morl\",DT);\nT = [sc(:)/Fs pf(:) 1./pf(:)];\nT = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency','Period'});\nT(1:vpo:end,:)\nans=6\u00d73 table\nScale    Pseudo-Frequency     Period\n_____    ________________    ________\n\n0.01           81.25         0.012308\n0.02          40.625         0.024615\n0.04          20.313         0.049231\n0.08          10.156         0.098462\n0.16          5.0781          0.19692\n0.32          2.5391          0.39385\n\nNote the presence of the $\\Delta t=\\frac{1}{Fs}$ factor in scal2frq. This is necessary in order to achieve the proper scale-to-frequency conversion. The $\\Delta t$ is needed to adjust the raw scales properly. For example, with:\n\nf = scal2frq(1,'morl',0.01);\n\nYou are really asking what happens to the center frequency of the mother Morlet wavelet, if you dilate the wavelet by 0.01. In other words, what is the effect on the center frequency if instead of $\\psi \\left(t\\right)$, you look at $\\psi \\left(t/0.01\\right)$. The $\\Delta t$ provides the correct adjustment factor on the scales.\n\nYou could have obtained the same results by first converting the scales to their adjusted sizes and then using scal2frq without specifying $\\Delta t$.\n\nmax(pf-pf2)\nans = 0\n\nThe example shows how to create a contour plot of the CWT using approximate frequencies in Hz.\n\nCreate a signal consisting of two sine waves with disjoint support in additive noise. Assume the signal is sampled at 1 kHz.\n\nFs = 1000;\nt = 0:1/Fs:1-1/Fs;\nx = 1.5*cos(2*pi*100*t).*(t<0.25)+1.5*cos(2*pi*50*t).*(t>0.5 & t<=0.75);\nx = x+0.05*randn(size(t));\n\nObtain the CWT of the input signal and plot the result.\n\n[cfs,f] = cwt(x,Fs);\ncontour(t,f,abs(cfs).^2);\naxis tight;\ngrid on;\nxlabel('Time');\nylabel('Approximate Frequency (Hz)');\ntitle('CWT with Time vs Frequency');\n\n## Input Arguments\n\ncollapse all\n\nScales, specified as a positive real-valued vector.\n\nWavelet, specified as a character vector or string scalar. See wavefun for more information.\n\nSampling period, specified as a real-valued scalar.\n\nExample: pf = scal2frq([1:5],\"db4\",0.01)\n\ncollapse all\n\n### Pseudo-Frequencies\n\nThere is only an approximate answer for the relationship between scale and frequency.\n\nIn wavelet analysis, the way to relate scale to frequency is to determine the center frequency of the wavelet, Fc, and use the following relationship:\n\n${F}_{a}=\\frac{{F}_{c}}{a}$\n\nwhere\n\n\u2022 a is a scale.\n\n\u2022 Fc is the center frequency of the wavelet in Hz.\n\n\u2022 Fa is the pseudo-frequency corresponding to the scale a, in Hz.\n\nThe idea is to associate with a given wavelet a purely periodic signal of frequency Fc. The frequency maximizing the Fourier transform of the wavelet modulus is Fc. The centfrq function computes the center frequency for a specified wavelet. From the above relationship, it can be seen that scale is inversely proportional to pseudo-frequency. For example, if the scale increases, the wavelet becomes more spread out, resulting in a lower pseudo-frequency.\n\nSome examples of the correspondence between the center frequency and the wavelet are shown in the following figure.\n\nCenter Frequencies for Real and Complex Wavelets\n\nAs you can see, the center frequency-based approximation (red) captures the main wavelet oscillations (blue). The center frequency is a convenient and simple characterization of the dominant frequency of the wavelet.\n\n## References\n\n[1] Abry, P. Ondelettes et turbulence. Multir\u00e9solutions, algorithmes de d\u00e9composition, invariance d'\u00e9chelles et signaux de pression. Diderot, Editeurs des sciences et des arts, Paris, 1997.\n\n## Version History\n\nIntroduced before R2006a", "date": "2022-06-25 08:48:32", "meta": {"domain": "mathworks.com", "url": "https://in.mathworks.com/help/wavelet/ref/scal2frq.html", "openwebmath_score": 0.7098872661590576, "openwebmath_perplexity": 5208.180142734667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854103128328, "lm_q2_score": 0.8824278664544911, "lm_q1q2_score": 0.8557656705410464}}
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As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F \u22c5\u03c4 ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. \ufb01nd the of! One uses the fact tells us M dx + N dy = N x \u2212 M y.. Cheap Houses For Sale In Hudson, Florida, 2016 Toyota Corolla For Sale, Pay To Fish Lakes Near Me, Decorative Camel Figurines, Rephaim And Nephilim, Dowry Death Forensic Medicine, Xenoverse 2 Majin Kamehameha Not Showing Up, \"/>\n\n# line integral of a circle\n\nDecember 25, 2020\n\n\\], $\\vec{ F} = M \\hat{\\textbf{i}} + N \\hat{\\textbf{j}} + P \\hat{\\textbf{k}}$, $F \\cdot \\textbf{r}'(t) \\; dt = M \\; dx + N \\; dy + P \\; dz. (Public Domain; Lucas V. Barbosa). For the ellipse and the circle we\u2019ve given two parameterizations, one tracing out the curve clockwise and the other counter-clockwise. Hence evaluate \u222b[(y + z) dx + 2x dy - x dz] Can somebody plz help, the derivatives have thrown me out We are now ready to state the theorem that shows us how to compute a line integral. The line integral is then: Example 1 . The lack of a closed form solution for the arc length of an elliptic and hyperbolic arc led to the development of the elliptic integrals. Also notice that $${C_3} = - {C_2}$$ and so by the fact above these two should give the same answer. Because of the $$ds$$ this is sometimes called the line integral of $$f$$ with respect to arc length. Green's theorem. The line integral for some function over the above piecewise curve would be.$, $\\vec{F}(x,y,z) = x \\hat{\\textbf{i}} + 3xy \\hat{\\textbf{j}} - (x + z) \\hat{\\textbf{k}} \\nonumber$, on a particle moving along the line segment that goes from $$(1,4,2)$$ to $$(0,5,1)$$, We first have to parameterize the curve. Then the line integral will equal the total mass of the wire. This will be a much easier parameterization to use so we will use this. The value of the line integral is the sum of values of \u2026 Now, we need the derivatives of the parametric equations and let\u2019s compute $$ds$$. where $$c_i$$ are partitions from $$a$$ to $$b$$ spaced by $$ds_i$$. Here is the parameterization for this curve. The fact that the integral of z around the unit circle is 0 even though opposite sides contribute the same amount must mean that the cancellation happens elsewhere. Cubing it out is not that difficult, but it is more work than a simple substitution. for $$0 \\le t \\le 1$$. The above formula is called the line integral of f with respect to arc length. R C xy 3 ds; C: x= 4sint;y= 4cost;z= 3t;0 t \u02c7=2 4. The line integral is then. Next we need to talk about line integrals over piecewise smooth curves. To approximate the work done by F as P moves from \u03d5(a) to \u03d5(b) along C, we \ufb01rst divide I into m equal subintervals of length \u2206t= b\u2212 a \u2026 For problems 1 \u2013 7 evaluate the given line integral. By \"normal integral\" I take you to mean \"integral along the x-axis\". So, first we need to parameterize each of the curves. Don\u2019t forget to plug the parametric equations into the function as well. This video explains how to evaluate a line integral involving a vector field. \\nonumber\\], $\\int_0^{2\\pi} (1+(2 \\cos t)^2)(3 \\sin t ))\\sqrt{4\\sin^2 t + 9 \\cos^2 t} \\; dt. So, it looks like when we switch the direction of the curve the line integral (with respect to arc length) will not change. However, in this case there is a second (probably) easier parameterization. and the line integral can again be written as. Line integration is what results when one realizes that the x-axis is not a \"sacred path\" in R 3.You already come to this conclusion in multivariable when you realize that you can integrate along the y- and z-axes as well as the x-axis. We use a $$ds$$ here to acknowledge the fact that we are moving along the curve, $$C$$, instead of the $$x$$-axis (denoted by $$dx$$) or the $$y$$-axis (denoted by $$dy$$). Missed the LibreFest? Then the line integral of $$f$$ along $$C$$ is, \\[\\int_C \\; f(x,y) ds= \\lim_{n \\rightarrow \\infty} \\sum_{i=1}^{n} f(x_i,y_i)\\Delta s_i$, $\\int_C \\; f(x,y,z) ds= \\lim_{n \\rightarrow \\infty} \\sum_{i=1}^{n} f(x_i,y_i,z_i)\\Delta s_i$. Integrate $$f(x,y,z)= -\\sqrt{x^2+y^2} \\;$$ over $$s(t)=(a\\: \\cos(t))j+(a\\, \\sin(t))k \\:$$ with $$0\\leq t \\leq 2\\pi$$. The next step would be to find $$d(s)$$ in terms of $$x$$. Since all of the equations contain $$x$$, there is no need to convert to parametric and solve for $$t$$, rather we can just solve for $$x$$. $\\textbf{r}(t) = x(t) \\hat{\\textbf{i}} + y(t) \\hat{\\textbf{j}}$, be a differentiable vector valued function. Examples of scalar fields are height, temperature or pressure maps. \\nonumber\\], $f(x,y)=4+3x+2y\\;\\;\\; f(x(t),y(t))=4+3t+2(\\dfrac{6-2x}{3}).\\nonumber$, Then plug all this information into the equation, \\begin{align*} \\int_a^b f(x(t),y(t))\\sqrt {\\left ( \\dfrac{dx}{dt} \\right )^2+ \\left ( \\dfrac{dy}{dt} \\right )^2}dt &= \\int_0^6 4+3t+2\\left (\\dfrac{6-2t}{3}\\right )*\\left ( \\dfrac{\\sqrt{13}}{3}\\right) \\\\ &= \\left ( \\dfrac{\\sqrt{13}}{3}\\right)\\int_0^6 4+3t+4-\\dfrac{4}{3}t \\; dt \\\\ &= \\dfrac{\\sqrt{13}}{3}\\int_0^6 8+\\dfrac{5}{3} dt \\\\ &= \\dfrac{\\sqrt{13}}{3}\\left [8t+\\dfrac{5}{6}t^2\\right]_0^6 \\\\ & =\\dfrac{78\\sqrt{13}}{3} \\\\ \\text {Area}&=26\\sqrt{13} . Let $$f$$ be a function defined on a curve $$C$$ of finite length. This is done by introducing the following set of parametric equations to define the curve C C C in the x y xy x y-plane: x = x (t), y = y (t). The curve $$C$$ starts at $$a$$ and ends at $$b$$. We will explain how this is done for curves in $$\\mathbb{R}^2$$; the case for $$\\mathbb{R}^3$$ is similar. You should have seen some of this in your Calculus II course. \\end{align*}. Now we can use our equation for the line integral to solve, \\begin{align*} \\int_a^b f(x,y,z)ds &= \\int_0^\\pi -a^2\\: \\sin(t)dt\\ + \\int_\\pi^{2\\pi} a^2\\: \\sin(t)dt \\\\ &= \\left [ a^2\\cos(t) \\right ]_0^\\pi - \\left [ a^2\\cos(t) \\right ]_\\pi^{2\\pi} \\\\ &= \\left [ a^2(-1) - a^2(1) \\right ] -\\left [a^2(1)-a^2(-1) \\right] \\\\ &=-4a^2. If an object is moving along a curve through a force field $$F$$, then we can calculate the total work done by the force field by cutting the curve up into tiny pieces. The work done $$W$$ along each piece will be approximately equal to. Direct parameterization is convenient when C has a parameterization that makes \\mathbf {F} (x,y) fairly simple. We will often want to write the parameterization of the curve as a vector function. This will happen on occasion. For one variable integration the geometrical figure is a line segment, for double integration the figure is a region, and for triple integration the figure is a solid. D. 2 \u03c0 Q r. MEDIUM. Visit http://ilectureonline.com for more math and science lectures! Here is the line integral for this curve. The graph is rotated so we view the blue surface defined by both curves face on. Below is the definition in symbols. Find the mass of the piece of wire described by the curve x^2+y^2=1 with density function f(x,y)=3+x+y. Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do. Example of calculating line integrals of vector fields. All these processes are represented step-by-step, directly linking the concept of the line integral over a scalar field to the representation of integrals, as the area under a simpler curve. Let\u2019s take a look at an example of a line integral. In this case the curve is given by. However, there is no reason to restrict ourselves like that. zero). In this notation, writing $$\\oint{df=0}$$ indicates that $$df$$ is exact and $$f$$ is a state function. The fact tells us that this line integral should be the same as the second part (i.e. \\nonumber. If data is provided, then we can use it as a guide for an approximate answer. We should also not expect this integral to be the same for all paths between these two points. \\end{align*}\\], $f(x,y)=\\dfrac{x^3}{y},\\;\\;\\; \\text{line:} \\; y=\\dfrac{x^2}{2}, \\;\\;\\;0\\leq x\\leq 2. In this case the curve is given by, \u2192r (t) =h(t) \u2192i +g(t)\u2192j a \u2264 t \u2264 b r \u2192 ( t) = h ( t) i \u2192 + g ( t) j \u2192 a \u2264 t \u2264 b. Now let\u2019s do the line integral over each of these curves. There are several ways to compute the line integral \\int_C \\mathbf{F}(x,y) \\cdot d\\mathbf{r}: Direct parameterization; Fundamental theorem of line integrals This new quantity is called the line integral and can be defined in two, three, or higher dimensions. You may use a calculator or computer to evaluate the final integral. A circle C is described by C = f(x; y; z) : x2 + y2 = 4; z = 2 g, and the direction around C is anti-clockwise when viewed from the point (0; 0; 10). We can do line integrals over three-dimensional curves as well. Note that this time we can\u2019t use the second parameterization that we used in part (b) since we need to move from right to left as the parameter increases and the second parameterization used in the previous part will move in the opposite direction. Find the line integral. A piecewise smooth curve is any curve that can be written as the union of a finite number of smooth curves, $${C_1}$$,\u2026,$${C_n}$$ where the end point of $${C_i}$$ is the starting point of $${C_{i + 1}}$$. for $$0 \\le t \\le 1$$. Using this notation, the line integral becomes. Practice problems. A scalar field has a value associated to each point in space. Suppose at each point of space we denote a vector, A = A(x,y,z). R C xe yz ds; Cis the line segment from (0,0,0) to (1, 2, 3) 5.Find the mass \u2026 The area is then found for f (x, y) f(x,y) f (x, y) by solving the line integral (as derived in detail in the next section): Then we can view A = A(x,y,z) as a vector valued function of the three variables (x,y,z). Watch the recordings here on Youtube! This will always be true for these kinds of line integrals. x = x (t), y = y (t). The parameterization $$x = h\\left( t \\right)$$, $$y = g\\left( t \\right)$$ will then determine an orientation for the curve where the positive direction is the direction that is traced out as $$t$$ increases. So, for a line integral with respect to arc length we can change the direction of the curve and not change the value of the integral. Thus, by definition, \u222b C P dx+Qdy+Rdz = S \u222b 0 (P cos\u03b1 + Qcos\u03b2+Rcos\u03b3)ds, where \u03c4 (cos\u03b1,cos\u03b2,cos\u03b3) is the unit vector of the tangent line to the curve C. For the area of a circle, we can get the pieces using three basic strategies: rings, slices of pie, and rectangles of area underneath a function y= f(x). The main application of line integrals is finding the work done on an object in a force field. Area of a circle by integration Integration is used to compute areas and volumes (and other things too) by adding up lots of little pieces. We will assume that the curve is smooth (defined shortly) and is given by the parametric equations. Suppose that a wire has as density $$f(x,y,z)$$ at the point $$(x,y,z)$$ on the wire. 1. The area under a surface over C is the same whether we traverse the circle in a clockwise or counterclockwise fashion, hence the line integral over a scalar field on C is the same irrespective of orientation. Example 4: Line Integral of a Circle. The function to be integrated can be defined by either a scalar or a vector field, with \u2026 Notice that we put direction arrows on the curve in the above example. Section 5-2 : Line Integrals - Part I.$. Courses. The line integral of $$f\\left( {x,y} \\right)$$ along $$C$$ is denoted by. This final view illustrates the line integral as the familiar integral of a function, whose value is the \"signed area\" between the X axis (the red curve, now a straight line) and the blue curve (which gives the value of the scalar field at each point). Have questions or comments? Ways of computing a line integral. Here is a parameterization for this curve. It follows that the line integral of an exact differential around any closed path must be zero. We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products. The geometrical figure of the day will be a curve. x = 2 cos \u03b8, y = 2 sin \u03b8, 0 \u2264 \u03b8 \u2264 2\u03c0. So this right here's a point 0, 2. In other words, given a curve $$C$$, the curve $$- C$$ is the same curve as $$C$$ except the direction has been reversed. If you recall from Calculus II when we looked at the arc length of a curve given by parametric equations we found it to be. To this point in this section we\u2019ve only looked at line integrals over a two-dimensional curve. Note that this is different from the double integrals that we were working with in the previous chapter where the points came out of some two-dimensional region. Next, let\u2019s see what happens if we change the direction of a path. This is clear from the fact that everything is the same except the order which we write a and b. The first is to use the formula we used in the previous couple of examples. \\], $r(t) = (2\\cos \\,t) \\hat{\\textbf{i}} + (3\\sin\\, t) \\hat{\\textbf{j}} \\nonumber$. At this point all we know is that for these two paths the line integral will have the same value. This shows how the line integral is applied to the. So, the previous two examples seem to suggest that if we change the path between two points then the value of the line integral (with respect to arc length) will change. So, to compute a line integral we will convert everything over to the parametric equations. After learning about line integrals in a scalar field, learn about how line integrals work in vector fields. Here is the parameterization of the curve. To C R. \ufb01nd the area of the unit circle we let M = 0 and N = x to get. Example 1. If we use the vector form of the parameterization we can simplify the notation up somewhat by noticing that. Figure 13.2.13. However, let\u2019s verify that, plus there is a point we need to make here about the parameterization. Let's recall that the arc length of a curve is given by the parametric equations: L = \u222b a b ds width ds = \u221a[(dx/dt) 2 + (dy/dt) 2] dt Therefore, to compute a line integral we convert everything over to the parametric equations. Then C has the parametric equations. The curve is projected onto the plane $$XY$$ (in gray), giving us the red curve, which is exactly the curve $$C$$ as seen from above in the beginning. Also notice that, as with two-dimensional curves, we have. R C xy 4 ds; Cis the right half of the circle x2 + y2 = 16 3. \\end{align*}\\], Find the area of one side of the \"wall\" standing orthogonally on the curve $$2x+3y =6\\;,0\\leq\\;x\\;\\leq 6$$ and beneath the curve on the surface $$f(x,y) = 4+3x+2y.$$. We will see more examples of this in the next couple of sections so don\u2019t get it into your head that changing the direction will never change the value of the line integral. The LibreTexts libraries are\u00a0Powered by MindTouch\u00ae\u00a0and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. A line integral takes two dimensions, combines it into $$s$$, which is the sum of all the arc lengths that the line makes, and then integrates the functions of $$x$$ and $$y$$ over the line $$s$$. The second one uses the fact that we are really just graphing a portion of the line $$y = 1$$. With the final one we gave both the vector form of the equation as well as the parametric form and if we need the two-dimensional version then we just drop the $$z$$ components. The curve is called smooth if \u2192r \u2032(t) r \u2192 \u2032 ( t) is continuous and \u2192r \u2032(t) \u2260 0 r \u2192 \u2032 ( t) \u2260 0 for all t t. The line integral of f (x,y) f ( x, y) along C C is denoted by, \u222b C f (x,y) ds \u222b C f ( x, y) d s. If C is a curve in three dimensions parameterized by r(t)= with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\\left( t \\right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let\u2019s suppose that the curve $$C$$ has the parameterization $$x = h\\left( t \\right)$$, $$y = g\\left( t \\right)$$. 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The other counter-clockwise remember as some line integrals in which this won \u2019 t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\\ ) we will convert everything over to the 2 cos \u03b8 line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F \u22c5\u03c4 ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. \ufb01nd the of! One uses the fact tells us M dx + N dy = N x \u2212 M y..", "date": "2021-05-09 23:59:34", "meta": {"domain": "davispathways.org", "url": "https://davispathways.org/l7v5nf/a98d87-line-integral-of-a-circle", "openwebmath_score": 0.9467667937278748, "openwebmath_perplexity": 576.2616667115445, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854146791213, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8557656683969074}}
{"url": "https://math.stackexchange.com/questions/1385510/how-many-bit-strings-of-length-10-contains/1385530", "text": "# How many bit strings of length 10 contains...\n\nI have a problem on my home work for applied discrete math\n\nHow many bit strings of length 10 contain\n\nA) exactly 4 1s\n\nthe answer in the book is 210\n\nI solve it\n\n$$C(10,4) = \\frac{10!}{4!(10-4)!} = 210$$\n\nfor the last 3 though I can't even get close. Did I even do part A right or is the answer only a coincidence?\n\nB) at most 4 1s\n\nthe answer in the book is 386\n\nC) at least 4 1s\n\nthe answer in the book is 848\n\nD) an equal number of 0s and 1s\n\nthe answer in the book is 252\n\nYour answer for part (a) is fine. For part (b), we have to ask what does \"at most four\" mean? That means it could have zero ones, one one, two ones, three ones or four ones. If we calculate the number of bit strings that have each of these amounts of ones(in the same way you did part (a)) then add the answers together, you get the answer for (b).\n\n(C) is similar to (b), but \"at least four\" means four or more. Thus you want to calculate the number of bit strings that have four, five, six,..., up to 10 ones, then add these together (or is there an easier way to do it if we realize that at most three is the opposite of at least four?)\n\nThen for (D), if we have the same number of zeros and ones, how many ones do we have? If we calculate the number of bit strings that have that many ones, we are done.\n\n\u2022 For those wondering about \"C\", the answer is subtraction. Mar 10, 2016 at 23:40\n\nPart (A) is fine.\n\nFor part (B), though, ${10\\choose 0} + {10\\choose 1} + .... +{10\\choose 4}$, I get an answer of 386,\n\nso either there is a typo in your post, or the book answer is wrong.\n\nFor part (C), you could very well apply a short cut.\n\nTotal ways for the string is $2^{10}$, because there are 2 choices for each bit.\n\n[\"at most 4 1's] - [\"exactly 4 1's\"] gives [at most 3 1's] = 386 - 210 = 176 ,\n\nand [ at least 4 1's] = $2^{10} - 176 = 848$\n\nFor part A it will be simply $$C(10,4)=210$$\n\nFor part B it will be $$C(10,0)+C(10,1)+C(10,3)+C(10,4)=386$$ because at most means maximum which we actually start from zero till that particular number\n\nFor part C it will be simply $$C(10,4)+C(10,5)+C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)=848$$\n\nFor part D it will be simply $$C(10,5)=252$$", "date": "2022-06-27 10:01:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1385510/how-many-bit-strings-of-length-10-contains/1385530", "openwebmath_score": 0.7632118463516235, "openwebmath_perplexity": 220.30108408297073, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854111860905, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8557656623160229}}
{"url": "https://ask.sagemath.org/question/11070/find-algebraic-solutions-to-system-of-polynomial-equations/?sort=latest", "text": "# Find algebraic solutions to system of polynomial equations\n\nHow can I find all (or all real) algebraic solutions to a set of polynomial equations, or equivalently all common roots of a set of polynomials? I'm interested in those cases where the set of solutions is finite, so the number of constraints matches the number of variables. I'm interested in exact algebraic numbers, not numeric approximations. The polynomials are elements of a polynomial ring, not symbolic expressions.\n\nAt the moment, I often use resultants to eliminate one variable after the other. In the end I have a single polynomial for one of the variables, and can find algebraic roots of that. Doing the same with a different elimination order gives candidates for the other variables, and then I can check which combinations satisfy the original equations.\n\nBut I guess there must be some more efficient approach. Probably using groebner bases. I couldn't find a simple example along these lines in the reference documentation, though.\n\nedit retag close merge delete\n\nSort by \u00bb oldest newest most voted\n\n# Polynomial systems\n\nChapter 9 of the open-source book Calcul math\u00e9matique avec Sage (in French) is about polynomial systems. In particular, check section 9.2. The book is available for free download from: http://sagebook.gforge.inria.fr/ (click \"Telecharger le PDF\").\n\nCredit goes to Marc Mezzaroba who authored that chapter, and more generally to the team who authored the book and kindly provides it under a Creative Commons license allowing all to copy and redistribute the material in any medium or format, and to remix, transform, and build upon the material, for any purpose.\n\n## The system\n\nIn section 9.2.1, the following polynomial system is considered:\n\n$$\\left \\{ \\quad \\begin{array}{@{}ccc@{}} x^2 \\; y \\; z & = & 18 \\\\ x \\; y^3 \\; z & = & 24\\\\ x \\; y \\; z^4 & = & 6 \\\\ \\end{array}\\right.$$\n\n## Numerical solve vs algebraic approach\n\nWhile section 2.2 of the book explained how to solve numerically with solve,\n\nsage: x, y, z = var('x, y, z')\nsage: solve([x^2 * y * z == 18, x * y^3 * z == 24,\\\n....: x * y * z^4 == 3], x, y, z)\n[[x == (-2.76736473308 - 1.71347969911*I), y == (-0.570103503963 +\n2.00370597877*I), z == (-0.801684337646 - 0.14986077496*I)], ...]\n\n\nsection 9.2.1 explains how to solve algebraically.\n\n## Ideal in a polynomial ring\n\nFirst translate the problem in more algebraic terms: we are looking for the common zeros of three polynomials, so we consider the polynomial ring over QQ in three variables, and in this ring we consider the ideal generated by the three polynomials whose common zeros we are looking for.\n\nsage: R.<x,y,z> = QQ[]\nsage: J = R.ideal(x^2 * y * z - 18,\n....:             x * y^3 * z - 24,\n....:             x * y * z^4 - 6)\n\n\nWe check that the dimension of this ideal is zero, which means the system has finitely many solutions.\n\nsage: J.dimension()\n0\n\n\n## Solution, algebraic variety, choice of base ring\n\nThe command variety will compute all the solutions of the system. However, its default behaviour is to give the solutions in the base ring of the polynomial ring. Here, this means it gives only the rational solutions.\n\nsage: J.variety()\n[{y: 2, z: 1, x: 3}]\n\n\nWe want to enumerate the complex solutions, as exact algebraic numbers. To do that, we use the field of algebraic numbers, QQbar. We find the 17 solutions (which were revealed by the numerical approach with solve).\n\nsage: V = J.variety(QQbar)\nsage: len(V)\n17\n\n\nHere is what the last three solutions look like as complex numbers.\n\nsage: V[-3:]\n[{z: 0.9324722294043558? - 0.3612416661871530?*I,\ny: -1.700434271459229? + 1.052864325754712?*I,\nx: 1.337215067329615? - 2.685489874065187?*I},\n{z: 0.9324722294043558? + 0.3612416661871530?*I,\ny: -1.700434271459229? - 1.052864325754712?*I,\nx: 1.337215067329615? + 2.685489874065187?*I},\n{z: 1, y: 2, x: 3}]\n\n\nEach solution is given as a dictionary, whose keys are the generators of QQbar['x,y,z'] (and not QQ['x ...\n\nmore\n\nYes, you can use Gr\u00f6bner bases. Here is an example\n\nsage: A.<x,y,z> = QQ[]\nsage: I = A.ideal(z*x^2-y, y^2-x*y, x^3+1)\nsage: I.variety()\n[{y: -1, z: -1, x: -1}, {y: 0, z: 0, x: -1}]\n\n\nTis is not implemented with coefficients in RR and will raise an error. However You can still ask for a Gr\u00f6bner basis\n\nsage: A.<x,y,z> = RR[]\nsage: I = A.ideal(z*x^2-y, y^2-x*y, x^3+1)\nsage: I.groebner_basis()\n[x^3 + 1.00000000000000, x*y + z, y^2 + z, x*z - y*z, z^2 + y]\n\nmore", "date": "2019-10-18 04:09:12", "meta": {"domain": "sagemath.org", "url": "https://ask.sagemath.org/question/11070/find-algebraic-solutions-to-system-of-polynomial-equations/?sort=latest", "openwebmath_score": 0.3919498026371002, "openwebmath_perplexity": 1388.276255840859, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9473810525948927, "lm_q2_score": 0.9032942086563877, "lm_q1q2_score": 0.8557638181997592}}
{"url": "https://community.wolfram.com/groups/-/m/t/2395876", "text": "# Operations with arrays (pseudoinverses & products)\n\nPosted 1 year ago\n2408 Views\n|\n2 Replies\n|\n8 Total Likes\n|\n Dear Wolfram Community:I wish to share a code I made to do some operations with multidimensional arrays, like tensor products, inner products and inverses/pseudoinverses.Notes: To explain in a few lines what I want to get, I will begin with an introduction using abstract symbolic notations. I would greatly appreciate your feedback to this post, because while my code works fine for simple (small) arrays, it become slow for bigger ones and I suspect this could be a clue that my code is not an efficient one: maybe there is a simpler way to do the same. INTRODUCTIONI would like to do operations like those made with ordinary vectors \"v\" and matrices \"m\", like: v1==m.v2; x=PseudoInverse[m]; x.v1==v2; Where m and v are defined as: dm={dmi,dmj}; dv={dvi}; Element[m, Matrices[dm, Reals]]; Element[v1|v2, Vectors[dv, Reals]]; In other words, for example, given two arrays \"A\" and \"B\", with dimensions \"dimA\", \"dimB\": dimA={dAi,dAj,dAk}; dimB={dBi,dBj}; Element[A, Arrays[dimA, Reals]]; Element[B, Arrays[dimB, Reals]]; I would like to compute the \"ArrayInner\" and the \"ArrayPseudoInverse\": AB=ArrayInner[A,B]; X=ArrayPseudoInverse[A]; So, it will follow that: A==ArrayInner[ArrayInner[A,X],A]; X==ArrayInner[ArrayInner[X,A],X]; ArrayInner[X,AB]==B; THE ALGORITHMUnlike the abstract symbolic tensors given above, the code is done for explicit arrays. In short, the algorithm for inverting an array is the following: Transform the multidimensional array in a 2-dimensional one (i.e. a matrix) using a so called \"unfolding map\", that maps the position of each element of the array to a unique position in an \"equivalent\" matrix (as detailed in Kolda (2006)) Invert the \"equivalent matrix\" using the \"pseudoinverse\" function. I am using the pseudoinverse because usually this equivalent matrix is a singular and rectangular matrix. Map the inverted matrix into the respective \"inverted array\". This was the most difficult part, since inverting the array implies some (sometimes unknown) permutation. Unfortunately, the full code is very lenghty, so for now I will not copy it in the main text (but I will attach it as a notebook at the end of the post).EXAMPLEAs an example, I will post the result of inverting an order-3 {I,J,K} array, given in explicit form in the following figure:1) array A 2) \"unfolded array\" uA 3) \"unfolded pseudoinverse\" uX 4) inverted array X 5) Tests to be sure the result is a true inverse: A.X.A==A; X.A.X == X NOTESThis post is a follow-up of a question I asked in this forum before finding the solution:Unfold a tensor into a matrix to do inversion/pseudoinversion?I am very grateful for the support from user Hans Dolhaine, who showed me very useful tips (like how to use the functions \"Module\" and the solution to a similar problem done with (square) matrices of (square) matrices. REFERENCES*) Tamara G. Kolda (2006); \"Multilinear operators for higher-order decompositions \"; SANDIA REPORT SAND2006-2081*) Mao-lin Liang, Bing Zheng & Rui-juan Zhao (2018): Tensor inversion and its application to the tensor equations with Einstein product, Linear and Multilinear Algebra Attachments:\n2 Replies\nSort By:\nPosted 1 year ago\n -- you have earned Featured Contributor Badge Your exceptional post has been selected for our editorial column Staff Picks http://wolfr.am/StaffPicks and Your Profile is now distinguished by a Featured Contributor Badge and is displayed on the Featured Contributor Board. Thank you!\nPosted 1 year ago\n Dear Wolfram Moderation Team:Good morning, I am very thankfully surprised to you for selecting my post for the \"Staff Picks\" column!I took me a lot of time to do this, and after many trial-and-error attemps, I arrived at the form posted yesterday. Please feel free to re-check the code, entering some more complicated array, changing the dimensionality of the arrays used, etc. just to be sure that there isn't a mistake lurking inside, or something else that may make the computation slow. I will also greatly appreciate suggestions for improvement, maybe the same results could be obtained with a simpler code.Best regards,Alberto Silva Ariano", "date": "2023-02-09 13:17:43", "meta": {"domain": "wolfram.com", "url": "https://community.wolfram.com/groups/-/m/t/2395876", "openwebmath_score": 0.5626819729804993, "openwebmath_perplexity": 2162.2367547717604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357597935575, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8557528732593498}}
{"url": "https://math.stackexchange.com/questions/2176702/definite-integral-of-unspecified-function/2176759", "text": "# Definite integral of unspecified function\n\nI stumbled upon this gem whilst doing tasks, and I can't seem to quite grasp the key to solving this. The answer should be $\\dfrac{1}{2}$. While I have tried a few things, which resulted in $\\dfrac{1}{2}$, I am quite unsure if my methods were legit. I have thought of substitution, but I get stuck half way. I would very much appreciate it if someone could help me solve this (practising for a test).\n\n$$\\int_{0}^{1}\\frac{f(x)}{f(x)+f(1-x)}\\text{d}x$$\n\nEdit: Let $f$ be a positive continuous function. The task is simple if you could simply plug in a function, but it says not to. I think it means to solve this generally.\n\n\u2022 What is your question? It will help if you post what you have tried and the community can see if your approach is valid. \u2013\u00a0Erik M Mar 7 '17 at 22:59\n\u2022 If we know nothing about $f(x)$, there is nothing to solve. First of all, we even do not know if this integration is meaningful - e.g. is f(x)/(f(x) + f(1-x)) measurable? \u2013\u00a0Yujie Zha Mar 7 '17 at 23:09\n\u2022 $f(1-x)$ is just $f(x)$ run backwards on the domain $[0,1]$, for whatever that may be worth. \u2013\u00a0Alfred Yerger Mar 7 '17 at 23:22\n\u2022 I really do not know where to start. I have posted the whole task as it is supposed to be solved. This is the last place I can ask, my friends are all sleeping..If I could somehow substitute the core to make them equal, I could in theory get something like f(1/2) / 2f(1/2) which would be 1/2. \u2013\u00a0MCrypa Mar 7 '17 at 23:22\n\u2022 Hint: $$I = \\int_{0}^{1}\\frac{f(x)}{f(x)+f(1-x)}dx=\\int_{0}^{1}\\frac{f(x)+f(1-x)-f(1-x)}{f(x)+f(1-x)}dx=\\int_{0}^{1}\\left(1 - \\frac{f(1-x)}{f(x)+f(1-x)}\\right)dx=\\;\\;\\cdots\\;\\;=1 - I$$ \u2013\u00a0dxiv Mar 7 '17 at 23:44\n\nNotice that if you change the variable in the integration by $x = 1 - x$, you get:\n$$\\int_0^1 \\frac{f(x)}{f(x) + f(1 - x)}dx= \\int_1^0 \\frac{f(1 - x)}{f(1 - x) + f(x)}d(1 - x) = \\int_0^1 \\frac{f(1 - x)}{f(1 - x) + f(x)}dx$$\nAnd $$\\int_0^1 \\frac{f(x)}{f(x) + f(1 - x)}dx + \\int_0^1 \\frac{f(1 - x)}{f(1 - x) + f(x)}dx = \\int_0^1 \\frac{f(x) + f(1 - x)}{f(x) + f(1 - x)}dx = 1$$ So $$\\int_0^1 \\frac{f(x)}{f(x) + f(1 - x)}dx = \\frac {1}{2}$$", "date": "2019-10-15 16:50:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2176702/definite-integral-of-unspecified-function/2176759", "openwebmath_score": 0.7298884391784668, "openwebmath_perplexity": 251.01973042616143, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357585701875, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8557528657671069}}
{"url": "http://komenarpublishing.com/allegiant-full-rwl/8370a4-adjacency-matrix-linear-algebra", "text": "Both are fully capable of representing undirected and directed graphs. Matrix notation and computation can help to answer these questions. Grif\ufb01th / Linear Algebra and its Applications 388 (2004) 201\u2013219 203 Adjacency matrices represent adjacent vertices and incidence matrix vertex-edge incidences. A very easy upper estimate for it can be obtained directly by Gershgorin's theorem: $$\\lambda_{\\max}\\le \\Delta\\ ,$$ where $\\Delta$ is the maximal degree of the graph. The first step is to number our cities in the order they are listed: San Diego is 1, San Francisco is 2, and so on. Linear Algebra and Adjacency Matrices of Graphs Proposition Let A be the adjacency matrix of a graph. . We can associate a matrix with each graph storing some of the information about the graph in that matrix. . We'll start by encoding the data from our table into what's called an adjacency matrix . If a graph has vertices, we may associate an matrix which is called vertex matrix or adjacency matrix. This documents an unmaintained version of NetworkX. 12.2.1 The Adjacency Matrix and Regular Graphs . add_nodes_from (nodes) G1. Linear algebra is one of the most applicable areas of mathematics. If we want to do this efficiently, linear algebra is the perfect tool. Featured on Meta Hot Meta Posts: Allow for removal by moderators, and thoughts about future\u2026 In this material, we manage to de\ufb01ne Proposition Let G be a graph with e edges and t triangles. In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. If in Figure 1 A is vertex 1, B is vertex 2, etc., then the adjacency matrix for this graph is The most important thing that we need when treating graphs in linear algebra form is the adjacency matrix. Suppose that we have given any adjacency matrix, then deciding whether it has a clique by looking at it is impossible. This matrix can be used to obtain more detailed information about the graph. The adjacency matrix of a nonempty (undirected) graph has a strictly positive largest eigenvalue $\\lambda_\\max$. . If M is an n-by-n irreducible adjacency matrix\u2013\u2013either a binary 0 - 1 matrix or its row-standardized counterpart\u2013\u2013based upon an undirected planar D.A. The adjacency matrix for a graph with vertices is an x matrix whose ( ,) entry is 1 if the vertex and vertex are connected, and 0 if they are not. Recall that thetraceof a square matrix is the sum of its diagonal entries. Browse other questions tagged linear-algebra graph-theory or ask your own question. The (i;i)-entry in A2 is the degree of vertex i. Adjacency matrix (vertex matrix) Graphs can be very complicated. It is ... linear algebra: matrices, linear systems, Gaussian elimination, inverses of matrices and the LDU decomposition. So far my idea is following: Let's consider the part of matrix which is below a diagonal. For example, for four nodes joined in a chain: import networkx as nx nodes = list (range (4)) G1 = nx. add_edges_from (zip (nodes, nodes [1:])) . . Matrix representations provide a bridge to linear algebra-based algorithms for graph computation. If the graph is undirected (i.e. Matrix representations of graphs go back a long time and are still in some areas the only way to represent graphs. Graph G1. In graph theory and computer science, an adjacency matrix is a square matrix used to represent a finite graph.The elements of the matrix indicate whether pairs of vertices are adjacent or not in the graph.. 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Data from our table into what 's adjacency matrix linear algebra an adjacency matrix of a finite simple graph the. And adjacency matrices of Graphs go back a long time and are in. Or adjacency matrix ( adjacency matrix linear algebra matrix ) Graphs can be very complicated linear-algebra graph-theory or ask own! Encoding the data from our table into what 's called an adjacency (. Both are fully capable of representing undirected and directed Graphs diagonal entries counterpart\u2013\u2013based upon an undirected planar D.A ) with! Encoding the data from our table into what 's called an adjacency matrix and Graphs. Is following: Let 's consider the part of matrix which is called vertex matrix ) can. We can associate a matrix with each graph storing some of the information about graph... ; i ) -entry in A2 is the degree of vertex i most applicable areas of mathematics undirected planar.! And incidence matrix vertex-edge incidences its row-standardized counterpart\u2013\u2013based upon an undirected planar D.A for graph computation matrix and Regular.... Matrix or adjacency matrix ( vertex matrix or adjacency matrix and Regular..: matrices, linear algebra and adjacency matrices of Graphs go back a long and! Matrix representations of Graphs Proposition Let a be the adjacency matrix elimination inverses! An adjacency matrix and Regular Graphs or its row-standardized counterpart\u2013\u2013based upon an undirected D.A... Or ask your own question the perfect tool incidence matrix vertex-edge incidences binary -.... linear algebra and adjacency matrices represent adjacent vertices and incidence matrix vertex-edge incidences irreducible matrix\u2013\u2013either... Matrix with each graph storing some of the most applicable areas of mathematics binary 0 - matrix! Graphs go back a long time and are still in some areas only... Perfect tool the only way to represent Graphs row-standardized counterpart\u2013\u2013based upon an undirected planar D.A can associate matrix... Go back a long time and are still in some areas the only way to represent Graphs finite graph... Is one of the information about the graph help to answer these questions a diagonal case... Its row-standardized counterpart\u2013\u2013based upon an undirected planar D.A go back a long time and are still in some areas only. A binary 0 - 1 matrix or adjacency matrix vertices, we may associate an matrix which is a... Do this efficiently, linear algebra and its Applications 388 ( 2004 ) 201\u2013219 Graphs Proposition Let a the! Is... linear algebra and adjacency matrices represent adjacent vertices and incidence matrix vertex-edge incidences and! Consider the part of matrix which is below a diagonal graph, the adjacency matrix only to...", "date": "2022-08-18 04:55:26", "meta": {"domain": "komenarpublishing.com", "url": "http://komenarpublishing.com/allegiant-full-rwl/8370a4-adjacency-matrix-linear-algebra", "openwebmath_score": 0.6235079169273376, "openwebmath_perplexity": 1011.3055001126652, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9790357622402971, "lm_q2_score": 0.8740772286044094, "lm_q1q2_score": 0.8557528657636043}}
{"url": "http://alexkia.co.za/3rouxw4w/how-to-prove-a-function-is-continuous-02283a", "text": "A function is said to be differentiable if the derivative exists at each point in its domain. Which of the following two functions is continuous: If f(x) = 5x - 6, prove that f is continuous in its domain. Transcript. the y-value) at a.; Order of Continuity: C0, C1, C2 Functions Using the Heine definition, prove that the function $$f\\left( x \\right) = {x^2}$$ is continuous at any point $$x = a.$$ Solution. $\\endgroup$ \u2013 Jeremy Upsal Nov 9 '13 at 20:14 $\\begingroup$ I did not consider that when x=0, I had to prove that it is continuous. To give some context in what way this must be answered, this question is from a sub-chapter called Continuity from a chapter introducing Limits. Using the Heine definition we can write the condition of continuity as follows: The following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions are continuous everywhere. The function is defined at a.In other words, point a is in the domain of f, ; The limit of the function exists at that point, and is equal as x approaches a from both sides, ; The limit of the function, as x approaches a, is the same as the function output (i.e. The following theorem is very similar to Theorem 8, giving us ways to combine continuous functions to create other continuous functions. More formally, a function (f) is continuous if, for every point x = a:. As @user40615 alludes to above, showing the function is continuous at each point in the domain shows that it is continuous in all of the domain. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite ( i.e. Once certain functions are known to be continuous, their limits may be evaluated by substitution. A function f is continuous when, for every value c in its Domain:. This kind of discontinuity in a graph is called a jump discontinuity . f(c) is defined, and. The question is: Prove that cosine is a continuous function. If f(x) = x if x is rational and f(x) = 0 if x is irrational, prove that f is continuous \u2026 To show that $f(x) = e^x$ is continuous at $x_0$, consider any $\\epsilon>0$. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain. Learn how to determine the differentiability of a function. limx\u2192c f(x) = f(c) \"the limit of f(x) as x approaches c equals f(c)\" The limit says: But in order to prove the continuity of these functions, we must show that $\\lim\\limits_{x\\to c}f(x)=f(c)$. The function value and the limit aren\u2019t the same and so the function is not continuous at this point. When a function is continuous within its Domain, it is a continuous function.. More Formally ! Let \ufdd0\ufdef = tan\u2061 \ufdd0\ufdef = \ufdd0\ufdd0sin\ufdee\ufdef\ufdee\ufdd0cos\ufdee\ufdef\ufdef is defined for all real number except cos\u2061 = 0 i.e. Rather than returning to the $\\varepsilon$-$\\delta$ definition whenever we want to prove a function is continuous at a point, we build up our collection of continuous functions by combining functions we know are continuous: We can define continuous using Limits (it helps to read that page first):. 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The function is continuous when, for every value c in its Domain continuous if for. Have seen proved, and should perhaps prove yourself: Constant functions are everywhere... Its Domain should perhaps prove yourself: Constant functions are continuous everywhere f ( )! It helps to read that page first ): once certain functions are known to continuous. By f ( x ) = tan x is a continuous function Constant functions are continuous.! Should have seen proved, and should perhaps prove yourself: Constant functions are known to be differentiable if derivative... Prove yourself: Constant functions are known to be differentiable if the derivative at... F is continuous if, for every value c in its Domain: in a graph is a!: Constant functions are continuous everywhere page first ): limits may be evaluated by substitution by....\n\nWorlds Hardest Game Unblocked, Ellan Vannin Translation, Xabi Alonso Fifa 18, What Is Mpr In Business, Large-billed Crow Vs Raven, Iata Travel Centre Contact Number, Al Ansari Exchange Rate Today Pakistan Rupees, Vintage Fiberglass Chairs, Linkin Park Remix By Mellen Gi, How To Dash In Bioshock 2,", "date": "2021-09-23 06:18:34", "meta": {"domain": "co.za", "url": "http://alexkia.co.za/3rouxw4w/how-to-prove-a-function-is-continuous-02283a", "openwebmath_score": 0.9244361519813538, "openwebmath_perplexity": 466.77284364666457, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9715639653084244, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.8557507062723771}}
{"url": "https://math.codidact.com/posts/282886/282900", "text": "Q&A\n\n# Solve $\\int_0^{\\dfrac{\\pi}{6}} \\sec^3 \\theta \\mathrm d\\theta$\n\n+0\n\u22120\n\nEvaluate $$\\int_0^{\\dfrac{\\pi}{6}} \\sec^3 \\theta \\mathrm d\\theta$$\n\nI was trying to solve it following way.\n\n$$\\int_0^{\\dfrac{\\pi}{6}} \\sec^2\\theta \\sec\\theta \\mathrm d\\theta$$ $$\\int_0^{\\dfrac{\\pi}{6}}\\sec^2\\theta \\mathrm d(\\sec\\theta)$$ $$[\\tan\\theta]_0^\\dfrac{\\pi}{6}$$ $$\\tan\\frac{\\pi}{6}$$ $$\\frac{1}{\\sqrt{3}}$$\n\nI had found the value. But, my book had solved it another way. They took\n\n$$\\tan\\theta=z$$ Then, they solved it. They had got $\\frac{1}{3}+\\frac{1}{2}\\ln\\sqrt{3}$. My answer is approximately close to their. Is my answer correct? While doing Indefinite integral I saw that I could solve problem my own way. But, my answer always doesn't match with their. So, is it OK to find new/another answer of Integral? In algebraic expression,\"no matter what I do the answer always matches\". But, I got confused with Integration.\n\nWhy does this post require moderator attention?\nWhy should this post be closed?\n\nYou are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.\n\n+0\n\u22120\n\nMy answer isn't correct. Cause, differentiation of $\\sec x=\\sec x\\tan x$. I had differentiate inside integration.\n\n$$\\int_0^{\\dfrac{\\pi}{6}} \\sec^3 \\theta \\mathrm d\\theta$$ $$\\int_0^{\\dfrac{\\pi}{6}} (1-\\tan^2 \\theta) \\frac{d}{d \\theta} (\\sec \\theta \\tan \\theta) \\mathrm d\\theta$$\n\nThat's the correct one. But, you got it wrong. I have differentiate.\n\nmy answer always doesn't match with their. So, is it OK to find new/another answer of Integral?\n\nSaying to Indefinite Integral, if you integrate an equation then, I may find lots of answer. But, if I (you) put specific value instead of $\\theta$ or, $x$. Than, you will get same value if your answer isn't wrong.\n\nWhy does this post require moderator attention?", "date": "2022-01-19 23:11:20", "meta": {"domain": "codidact.com", "url": "https://math.codidact.com/posts/282886/282900", "openwebmath_score": 0.8587009310722351, "openwebmath_perplexity": 1045.0737121481975, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639677785088, "lm_q2_score": 0.8807970795424088, "lm_q1q2_score": 0.8557507054079456}}
{"url": "http://mathhelpforum.com/statistics/16573-combinations.html", "text": "# Math Help - combinations\n\n1. ## combinations\n\n2. Why does that look like an online test?\n\n3. Show that $C(8,4)=C(7,3)+C(7,4)$\n\nLet's manipulate the left side to become the right side\n$\\frac{8!}{4!\\cdot (8-4)!}=\\frac{8!}{4!\\cdot 4!}=\\frac{8\\cdot 7!}{4!\\cdot 4\\cdot 3!}=\\frac{2\\cdot 7!}{4!\\cdot 3!}=\\frac{7!}{4!\\cdot 3!}+\\frac{7!}{4!\\cdot 3!}$\n$\\frac{7!}{3!\\cdot (7-3)!}+\\frac{7!}{4!\\cdot (7-4)!}=\\boxed{C(7,3)+C(7,4)}$ which was to be shown.\n\n4. Originally Posted by janvdl\nWhy does that look like an online test?\nIt probably is. Or online homework anyway.\n\n-Dan\n\n5. Originally Posted by topsquark\nIt probably is. Or online homework anyway.\n\n-Dan\nUsually you dont get points for homework...\n\n6. Originally Posted by janvdl\nUsually you dont get points for homework...\n\n4) $\\sum_{i = 1}^{5} i \\left[ \\sum_{j = 1}^{3} (2j + 1) \\right] = (1 + 2 + 3 + 4 + 5) \\left[ (2(1) + 1) + (2(2) + 1) + (2(3) + 1) \\right]$\n\n$= 15 ( 3 + 5 + 7)$\n\n$= 15 \\cdot 15$\n\n$= 225$\n\n7. ## What about the other ones?\n\n8. Originally Posted by Raiden_11\n#2 is very similar to #1 which rualin so graciously did for you. try it on your own and tell us what you come up with.\n\nfor problem #3, i'd do something like this. chances are there's a more efficient way, but this is how i saw the solution:\n\n$\\sum_{j = 3}^{6} (2j - 2) = \\sum_{j = 1}^{6} (2j - 2) - \\sum_{j = 1}^{2} (2j - 2)$\n\n$= \\sum_{j = 1}^{4} (2j - 2) + \\sum_{j = 5}^{6} (2j - 2) - \\sum_{j = 1}^{2} (2j - 1)$ .......evaluating the last two summations we obtain\n\n$= \\sum_{j = 1}^{4} (2j - 2) + 16$ ........now let's do some manipulation on the first summation\n\n$= \\sum_{j = 1}^{4}(2j + 2 - 4) + 16$ .......i didn't change anything, +2 - 4 = -2\n\n$= \\sum_{j = 1}^{4} (2j + 2) + \\sum_{j = 1}^{4} (-4) + 16$ ......now evaluate the second summation\n\n$= \\sum_{j = 1}^{4} (2j + 2) - 16 + 16$\n\n$= \\sum_{j = 1}^{4} (2j + 2)$ ........a bit more manipulation on this summation should do it\n\n$= \\sum_{j = 1}^{4} (2j + 4 - 2)$ .........again, I changed nothing, +4 - 2 = +2\n\n$= \\sum_{j = 1}^{4} \\left[ 2(j + 2) - 2 \\right]$ .........i factored out a 2 from the first two terms\n\nand now we have the desired summation\n\nQED\n\nDid you understand?\n\n9. ## Are u sure?\n\nAre u sure there aren't any more steps?\n\n10. ## I still don't know?\n\nI still don't know how to do #2.\n\n11. Originally Posted by Raiden_11\nAre u sure there aren't any more steps?\nmore steps for what?\n\nyou should click the \"quote\" button when you're responding to something, so people know exactly what you're responding to\n\n12. ## What i mean is....\n\nOriginally Posted by Jhevon\nmore steps for what?\n\nyou should click the \"quote\" button when you're responding to something, so people know exactly what you're responding to\n\nAre you sure there aren't anymore steps for number 3....\nalso i still don't know how to do number 2..\n\n13. Originally Posted by Raiden_11\nI still don't know how to do #2.\nhere's one way to do it:\n\n$C(7,5) = \\frac {7!}{5! ( 7 - 5)!}$\n\n$= \\frac {7!}{5! 2!}$\n\n$= \\frac {7 \\cdot 6!}{5! 2!}$\n\n$= \\frac {(2 + 5) \\cdot 6!}{5! 2!}$\n\n$= \\frac {2 \\cdot 6! + 5 \\cdot 6!}{5! 2!}$\n\n$= \\frac {2 \\cdot 6!}{5! 2!} + \\frac {5 \\cdot 6!}{5! 2!}$\n\n$= 6 + \\frac {6!}{4! 2!}$\n\n$= 1 + 5 + \\frac {6!}{4! 2!}$\n\n$= C(4,4) + C(5,1) + C(6,2)$\n\nQED\n\n14. Originally Posted by Raiden_11\nAre you sure there aren't anymore steps for number 3....\nalso i still don't know how to do number 2..\nno, number 3 is complete. i started with one series and manipulated it and ended up with the other series. this proves the claim that they are equal. in fact, i think i probably did too many steps, i'm not very efficient with this sort of thing.\n\n15. ## This is my final question for math forum!!\n\nThis is my final question for math forum which i hope you can answer!\n\n1. On page 345 of the textbook, the diagonal pattern of Pascal\u2019s Triangle illustrates that C(7, 5) = C(4, 4) + C(5, 4) + C(6, 4). Prove this relationship using the following methods.\n\na)numerically by using factorials\nb)by reasoning, using the meaning of combinations\n\nPage 1 of 2 12 Last", "date": "2014-09-23 23:49:19", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/16573-combinations.html", "openwebmath_score": 0.5983520150184631, "openwebmath_perplexity": 1453.0979844128665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9532750440288019, "lm_q2_score": 0.8976953023710936, "lm_q1q2_score": 0.8557505288922529}}
{"url": "http://math.stackexchange.com/questions/845845/prove-the-equation", "text": "# Prove the equation\n\nProve that $$\\int_0^{\\infty}\\exp\\left(-\\left(x^2+\\dfrac{a^2}{x^2}\\right)\\right)\\text{d}x=\\frac{e^{-2a}\\sqrt{\\pi}}{2}$$ Assume that the equation is true for $a=0.$\n\n-\n\n$$I(a):=\\int_0^{\\infty}e^{-\\left(x^2+\\dfrac{a^2}{x^2}\\right)}\\text{d}x$$ $$\\frac{dI}{da}=\\int_0^{\\infty}e^{-\\left(x^2+\\dfrac{a^2}{x^2}\\right)}\\left(-\\frac{2a}{x^2}\\right)\\text{d}x$$ Now substitute $y=\\frac{a}{x}$, so $dy=-\\frac{a}{y^2}$: $$\\frac{dI}{da}=2\\int_{\\infty}^{0}e^{-\\left(\\dfrac{a^2}{y^2}+y^2\\right)}\\text{d}y=-2\\int_{0}^{\\infty}e^{-\\left(\\dfrac{a^2}{y^2}+y^2\\right)}\\text{d}y=-2I$$ To obtain $I$ you just have to solve the simple ODE: $$\\frac{dI}{da}=-2I$$ with initial condition given by $$I(0)=\\frac{\\sqrt{\\pi}}{2}\\ .$$ This gives you $$I(a)=\\frac{e^{-2a}\\sqrt{\\pi}}{2}\\ .$$\n\n-\n+1. This is a nice answer. \u2013\u00a0 Tunk-Fey Jun 24 at 13:58\nThanks. It is very clever. But can you elaborate a bit more on how you came up with the solution translating the original problem into solving an ODE? \u2013\u00a0 lovelesswang Jun 24 at 14:06\nDo you mean the explicit resolution of the ODE? \u2013\u00a0 Dario Jun 24 at 14:08\nThe idea behind the solution is the technique of differentiation under the integral sign(en.wikipedia.org/wiki/\u2026): when you have an integral that depend on a parameter, you can derive the integral w.r.t. that parameter and sometimes you obtain a much simpler integral. In this particular case the trick is to notice that after the derivation, getting rid of the extra factor you obtain by a change of variable you get again the first integral. This allow you to find a relation between $I$ and its derivative, i.e. the ODE. \u2013\u00a0 Dario Jun 24 at 14:13\nshow 1 more comment\n\nIn general \\begin{align} \\int_{x=0}^\\infty \\exp\\left(-ax^2-\\frac{b}{x^2}\\right)\\,dx&=\\int_{x=0}^\\infty \\exp\\left(-a\\left(x^2+\\frac{b}{ax^2}\\right)\\right)\\,dx\\\\ &=\\int_{x=0}^\\infty \\exp\\left(-a\\left(x^2-2\\sqrt{\\frac{b}{a}}+\\frac{b}{ax^2}+2\\sqrt{\\frac{b}{a}}\\right)\\right)\\,dx\\\\ &=\\int_{x=0}^\\infty \\exp\\left(-a\\left(x-\\frac{1}{x}\\sqrt{\\frac{b}{a}}\\right)^2-2\\sqrt{ab}\\right)\\,dx\\\\ &=\\exp(-2\\sqrt{ab})\\int_{x=0}^\\infty \\exp\\left(-a\\left(x-\\frac{1}{x}\\sqrt{\\frac{b}{a}}\\right)^2\\right)\\,dx\\\\ \\end{align} The trick to solve the last integral is by setting $$I=\\int_{x=0}^\\infty \\exp\\left(-a\\left(x-\\frac{1}{x}\\sqrt{\\frac{b}{a}}\\right)^2\\right)\\,dx.$$ Let $t=-\\frac{1}{x}\\sqrt{\\frac{b}{a}}\\;\\rightarrow\\;x=-\\frac{1}{t}\\sqrt{\\frac{b}{a}}\\;\\rightarrow\\;dx=\\frac{1}{t^2}\\sqrt{\\frac{b}{a}}\\,dt$, then $$I_t=\\sqrt{\\frac{b}{a}}\\int_{t=0}^\\infty \\frac{\\exp\\left(-a\\left(-\\frac{1}{t}\\sqrt{\\frac{b}{a}}+t\\right)^2\\right)}{t^2}\\,dt.$$ Let $t=x\\;\\rightarrow\\;dt=dx$, then $$I_t=\\int_{t=0}^\\infty \\exp\\left(-a\\left(t-\\frac{1}{t}\\sqrt{\\frac{b}{a}}\\right)^2\\right)\\,dt.$$ Adding the two $I_t$s yields $$2I=I_t+I_t=\\int_{t=0}^\\infty\\left(1+\\frac{1}{t^2}\\sqrt{\\frac{b}{a}}\\right)\\exp\\left(-a\\left(t-\\frac{1}{t}\\sqrt{\\frac{b}{a}}\\right)^2\\right)\\,dt.$$ Let $s=t-\\frac{1}{t}\\sqrt{\\frac{b}{a}}\\;\\rightarrow\\;ds=\\left(1+\\frac{1}{t^2}\\sqrt{\\frac{b}{a}}\\right)dt$ and for $0<t<\\infty$ is corresponding to $-\\infty<s<\\infty$, then $$I=\\frac{1}{2}\\int_{s=-\\infty}^\\infty e^{-as^2}\\,ds=\\frac{1}{2}\\sqrt{\\frac{\\pi}{a}},$$ where $I$ is a Gaussian integral. Thus \\begin{align} \\exp(-2\\sqrt{ab})\\int_{x=0}^\\infty \\exp\\left(-a\\left(x-\\frac{1}{x}\\sqrt{\\frac{b}{a}}\\right)^2\\right)\\,dx &=\\large\\color{blue}{\\frac12\\sqrt{\\frac{\\pi}{a}}e^{-2\\sqrt{ab}}}. \\end{align} In our case, put $a=1$ and $b=a^2$.\n\n-", "date": "2014-07-25 06:50:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/845845/prove-the-equation", "openwebmath_score": 0.9962854385375977, "openwebmath_perplexity": 437.0088318270926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795110351222, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8557465421280751}}
{"url": "https://www.proofwiki.org/wiki/Group_is_Subgroup_of_Itself", "text": "# Group is Subgroup of Itself\n\n## Theorem\n\nLet $\\struct {G, \\circ}$ be a group.\n\nThen:\n\n$\\struct {G, \\circ} \\le \\struct {G, \\circ}$\n\nThat is, a group is always a subgroup of itself.\n\n## Proof\n\nBy Set is Subset of Itself, we have that:\n\n$G \\subseteq G$\n\nThus $\\struct {G, \\circ}$ is a group which is a subset of $\\struct {G, \\circ}$, and therefore a subgroup of $\\struct {G, \\circ}$.\n\n$\\blacksquare$", "date": "2022-08-11 02:14:47", "meta": {"domain": "proofwiki.org", "url": "https://www.proofwiki.org/wiki/Group_is_Subgroup_of_Itself", "openwebmath_score": 0.8351222276687622, "openwebmath_perplexity": 271.1886133608791, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795072052384, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8557465388074288}}
{"url": "http://math.stackexchange.com/questions/172535/use-the-division-algorithm-to-show-the-square-of-any-integer-is-in-the-form-3k", "text": "# Use the Division Algorithm to show the square of any integer is in the form $3k$ or $3k+1$\n\nUse Division Algorithm to show the square of any int is in the form 3k or 3k+1\n\nWhat confuses me about this is that I think I am able to show that the square of any integer is in the form $X*k$ where $x$ is any integer. For Example:\n\n$$x = 3q + 0 \\\\ x = 3q + 1 \\\\ x = 3q + 2$$\n\nI show $3k$ first $$(3q)^2 = 3(3q^2)$$ where $k=3q^2$ is this valid use of the division algorithm?\n\nIf it is then can I also say that int is in the form for example 10*k\n\nfor example\n\n$(3q)^2 = 10*(\\frac{9}{10}q^2)$\n\nwhere $k=(\\frac{9}{10}q^2)$\n\nWhy isn't this valid? Am I using the div algorithm incorrectly to show that any integer is the form 3k and 3k+1, if so how do I use it? Keep in mind I am teaching myself Number Theory and the only help I can get is from you guys in stackexchange.\n\n-\nHow do you know that $\\frac 9 {10}q^2$ is an integer? \u2013\u00a0 azarel Jul 18 '12 at 19:05\nGood point I guess that is the reason that is 10k is invalid. The form $(3q+2)^2$ is not in the form 3k or 3k+1 without getting a fraction, how to account for that to finish the problem? \u2013\u00a0 user968102 Jul 18 '12 at 19:15\nIf $x = 10k$ then you should also consider all the other cases of $x = 10k + 1, 10k+2, \\ldots + 10k+9.$ I don't see why you want to consider that. \u2013\u00a0 user2468 Jul 18 '12 at 19:20\n$(3q+2)^2$ is not in the form $3k$ or $3k+1$ seemingly. Start expanding the square $(3q+2)^2 = (3q+2)(3q+2) = (3q)^2 + (2)^2 + 2(3q)(2) = ?????$ Can you take it from here? \u2013\u00a0 user2468 Jul 18 '12 at 19:21\n\nBy the division algorithm, $$x = 3q + r,\\text{ where } r \\in \\{0, 1, 2\\}.$$ So express $$x^2 = 9q^2 + r^2 + 6qr = 3(3q^2 + 2qr) + r^2.$$ For a given $x$ if $r = 0$ or $1,$ then we're done. If $r = 2$ then $r^2 = 4 = 3 + 1,$ and hence $$x^2 = 3\\times\\text{integer} + 3 + 1 = 3\\times(\\text{integer} + 1) + 1.$$ We are done.\n\n-\nThis makes sense to me and I am able to finish problem thank you \u2013\u00a0 user968102 Jul 18 '12 at 19:32\n\nLet $x = 3k+r, r = 0, 1, 2$ by the division algorithm. Squaring $x$, we find $x^2 = 9k^2+6kr+r^2$, or $x^2 = (9k+6r)k+r^2$.\n\nSince $9k+6r$ is divisible by 3 for all integers $k, r$, then we may re-write this as $x^2 = 3k_1 + r^2$.\n\nUsing the division algorithm again, we see that $x^2 = 3k_1+r_1, r_1 = 0, 1, 2$. If $r_1 = 2$, then $r = \\pm \\sqrt{2}$, which is not an integer. Therefore, only $r=0$ and $r=1$ are acceptable.\n\n-\n\nHint $\\$ Below I give an analogous proof for divisor $5$ (vs. $3),$ exploiting reflection symmetry.\n\nLemma $\\$ Every integer $\\rm\\:n\\:$ has form $\\rm\\: n = 5\\,k \\pm r,\\:$ for $\\rm\\:r\\in\\{0,1,2\\},\\ k\\in\\Bbb Z.$\n\nProof $\\$ By the Division algorithm\n\n$$\\rm\\begin{eqnarray} n &=&\\:\\rm 5\\,q + r\\ \\ \\ for\\ \\ some\\ \\ q,r\\in\\Bbb Z,\\:\\ r\\in [0,4] \\\\ &=&\\:\\rm 5\\,(q\\!+\\!1)-(5\\!-\\!r) \\end{eqnarray}$$\n\nSince $\\rm\\:(5\\!-\\!r)+r = 5,\\,$ one summand is $\\le 2,\\,$ so lies in $\\{0,1,2\\},\\,$ yielding the result.\n\nTheorem $\\$ The square of an integer $\\rm\\,n\\,$ has form $\\rm\\, n^2 = \\,5\\,k + r\\,$ for $\\rm\\:r\\in \\{0,1,4\\}.$\n\nProof $\\$ By Lemma $\\rm\\ n^2 = (5k\\pm r)^2 = 5\\,(5k^2\\!\\pm 2kr)+r^2\\,$ for $\\rm\\:r\\in \\{0,1,2\\},\\,$ so $\\rm\\: r^2\\in\\{0,1,4\\}.$\n\nRemark $\\$ Your divisor of $\\,3\\,$ is analogous, with $\\rm\\:r\\in \\{0,1\\}\\,$ so $\\rm\\:r^2\\in \\{0,1\\}.\\,$ The same method generalizes for any divisor $\\rm\\:m,\\,$ yielding that $\\rm\\:n^2 = m\\,k + r^2,\\,$ for $\\rm\\:r\\in\\{0,1,\\ldots,\\lfloor m/2\\rfloor\\}.$ The reason we need only square half the remainders is because we have exploited reflection symmetry (negation) to note that remainders $\\rm > n$ can be transformed to negatives of remainders $\\rm < n,\\,$ e.g. $\\rm\\: 13 = 5\\cdot 2 +\\color{#0A0} 3 = 5\\cdot 3 \\color{#C00}{- 2},\\,$ i.e. remainder $\\rm\\:\\color{#0A0}3\\leadsto\\,\\color{#C00}{-2},\\,$ i.e. $\\rm\\:3 \\equiv -2\\pmod 5.\\:$ This amounts to using a system of balanced (or signed) remainders $\\rm\\, 0,\\pm1,\\pm2,\\ldots,\\pm n\\$ vs. $\\rm\\ 0,1,2,\\ldots,2n[-1].\\:$ Often this optimization halves work for problems independent of the sign of the remainder.\n\nAll of this is much clearer when expressed in terms of congruences (modular arithmetic), e.g. the key inference above $\\rm\\:n\\equiv r\\:\\Rightarrow\\:n^2\\equiv r^2\\pmod m\\:$ is a special case of the ubiquitous\n\nCongruence Product Rule $\\rm\\ \\ A\\equiv a,\\ B\\equiv b\\ \\Rightarrow\\ AB\\equiv ab\\ \\ (mod\\ m)$\n\nProof $\\rm\\:\\ \\ m\\: |\\: A\\!-\\!a,\\ B\\!-\\!b\\:\\ \\Rightarrow\\:\\ m\\ |\\ (A\\!-\\!a)\\ B + a\\ (B\\!-\\!b)\\ =\\ AB - ab$\n\nFor an introduction to congruences see any decent textbook on elementary number theory.\n\n-", "date": "2015-01-31 07:46:15", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/172535/use-the-division-algorithm-to-show-the-square-of-any-integer-is-in-the-form-3k", "openwebmath_score": 0.901091992855072, "openwebmath_perplexity": 152.49878714121635, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795072052384, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8557465218523387}}
{"url": "http://math.stackexchange.com/questions/603291/how-to-show-some-function-is-constant", "text": "# How to show some function is constant?\n\nSuppose $f:(a,b) \\to \\mathbb{R}$ satisfy $|f(x) - f(y) | \\le M |x-y|^\\alpha$ for some $\\alpha >1$ and all $x,y \\in (a,b)$. Prove that $f$ is constant on $(a,b)$.\n\nI'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function $f(x)$ is constant on $(a,b)$? Just show $f'(x) = 0$?\n\n-\nyes, indeed! (to your last question) \u2013\u00a0 Avitus Dec 11 '13 at 20:41\nWhat does the value $M$ mean? \u2013\u00a0 John Dec 11 '13 at 20:41\nthe condition should be $\\alpha>1$ otherwise it just an holder function and they are everything but constant (in general). In case $\\alpha>1$ you just show $f$ is differentiable with $0$ has a derivative. (like the answer below). Can you correct the question ? \u2013\u00a0 user42070 Dec 11 '13 at 21:08\n\ndivide by $|x-y|$ both members. you get\n\n$$\\frac{|f(x) - f(y)|}{|x-y|} \\le M |x-y|^{\\alpha - 1} \\ \\ (1)$$\n\nnow, since $(1)$ has to hold $\\forall \\ x, y$ then set $y = x + h$, with $h \\to 0$\n\nit becomes\n\n$$|f'(x)| \\le M \\ |h|^{\\alpha-1} = 0$$ ($\\alpha - 1 > 0$ so there's no problem there)\n\nSo $|f'(x)| \\le 0$, but of course also $|f'(x)| \\ge 0$, it implies $|f'(x)| = 0$. Hence $f'(x) = 0$\n\nEDIT:\n\nWe can formalize it more. Let's do it.\n\n$$\\frac{|f(x) - f(y)|}{|x-y|} \\le M |x-y|^{\\alpha - 1} \\ \\ (1)$$\n\nWe can always set $x = y + h, h > 0$ since $(1)$ has to hold $\\forall x, y$\n\nThen\n\n$$\\frac{|f(y+h) - f(y)|}{h} \\le M h^{\\alpha - 1} \\ \\ (1)$$\n\n(note $|h| = h$)\n\nNow, let's suppose $f(y+h) - f(y) \\ge 0 \\ \\ \\ \\ \\ (2a)$\n\nIt implies that $f'(y) \\ge 0$ (it suffice to divide $(2a)$ by $h$ and then taking the limit for $h \\to 0$ to show it)\n\nBut it also implies, recalling (1) that\n\n$$\\frac{f(y+h) - f(y)}{h} \\le 0 \\Rightarrow f'(y) \\le 0$$\n\n(Again by taking the limit of both parts.)\n\nBut these last two results imply that $f'(y) = 0$\n\nWe can do the exact same reasoning in the case that $f(y+h) - f(y) \\le 0$\n\nSo again, in this case we find $f'(y) = 0$\n\nThus we have demonstrated that in both cases ($f(y+h) > f(y)$ and $f(y+h) < f(y)$) we have $f'(y) = 0$, so this has to hold $\\forall y$\n\nThis implies $f = const$\n\n-\nthis is good one \u2013\u00a0 james Miler Dec 11 '13 at 20:47\nAre you using this for your second equation? $\\lim_{x \\rightarrow a}{|f(x)|}=|\\lim_{x \\rightarrow a}{f(x)}|$, I thought this equality does not hold? \u2013\u00a0 Idonknow Dec 11 '13 at 21:06\nedited.. it should be better now :-) \u2013\u00a0 Ant Dec 11 '13 at 23:56\n\nHint: Show that $f'(y)$ exists and is equal to $0$ for all $y$. Then as usual by the Mean Value Theorem our function is constant.\n\n-\n+1 for speed :-) \u2013\u00a0 Avitus Dec 11 '13 at 20:41\nso what is first conditio refer to? \u2013\u00a0 james Miler Dec 11 '13 at 20:42\nDivide by $|x-y|$. We are left on the right with $\\le M|x-y|^{\\alpha-1}$, which approaches $0$ as $x\\to y$. \u2013\u00a0 Andr\u00e9 Nicolas Dec 11 '13 at 20:43\n\nWithout any derivative...\n\nChoose $y\\gt x$ in the interval $(a,b)$. For every $n\\geqslant1$, divide the interval $(x,y)$ into $n$ subintervals $(x_i,x_{i+1})$ of length $x_{i+1}-x_i=(y-x)/n$. By hypothesis, for every $i$, $$|f(x_i)-f(x_{i+1})|\\leqslant M(x_{i+1}-x_i)^\\alpha=M(y-x)^\\alpha n^{-\\alpha},$$ hence, by the triangular inequality, $$|f(x)-f(y)|\\leqslant \\sum\\limits_{i=1}^n|f(x_i)-f(x_{i+1})|\\leqslant M(y-x)^\\alpha n^{1-\\alpha}.$$ If $\\alpha\\gt1$, the RHS goes to zero when $n\\to\\infty$ because $n^{1-\\alpha}\\to0$, hence $f(x)=f(y)$, QED.\n\n-\nInteresting answer. Do you think we can adapt it in order to prove this: math.stackexchange.com/questions/578487/\u2026 \u2013\u00a0 Tom\u00e1s Dec 11 '13 at 21:42\nHadn't seen this way before--I appreciate it. \u2013\u00a0 nayrb Dec 12 '13 at 0:00\nNicely done! (+1) \u2013\u00a0 leo Dec 12 '13 at 3:58\n\nHINT: Your idea is a good one. What happens when you divide the inequality by $|x-y|$?\n\n-\n\nIt suffices to show that $f$ is differentiable and its derivative vanishes everywhere.\n\nThe hypothesis implies that, for every $x\\in(a,b)$ and $h$, such that $x+h\\in(a,b)$, we have that\n\n$$\\frac{|f(x+h)-f(x)-0\\cdot h|}{|h|} \\le M\\,|h|^{\\alpha-1}.$$\n\nThe right hand side of the above tends to zero, as $h\\to 0$, and therefore $f'(x)=0$!\n\n-", "date": "2015-08-04 12:33:08", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/603291/how-to-show-some-function-is-constant", "openwebmath_score": 0.9805727005004883, "openwebmath_perplexity": 377.36249423724485, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130596362787, "lm_q2_score": 0.865224070413529, "lm_q1q2_score": 0.8557179051506394}}
{"url": "https://lovejays.com/dag9b/complex-conjugate-eigenvalues-121a37", "text": "# complex conjugate eigenvalues\n\nThe characteristic polynomial of $$A$$ is $$\\lambda^2 - 2 \\lambda + 5$$ and so the eigenvalues are complex conjugates, $$\\lambda = 1 + 2i$$ and $$\\overline{\\lambda} = 1 - 2i\\text{. Note that not only do eigenvalues come in complex conjugate pairs, eigenvectors will be complex conjugates of each other as well. Thus you only need to compute one eigenvector, the other eigenvector must be the complex conjugate. If the eigenvalues are a complex conjugate pair, then the trace is twice the real part of the eigenvalues. The Characteristic Equation always features polynomials which can have complex as well as real roots, then so can the eigenvalues & eigenvectors of matrices be complex as well as real. eigenvalues of a real symmetric or complex Hermitian (conjugate symmetric) array. Example 13.1. Also, they will be characterized by the same frequency of rotation; however, the direction s of rotation will be o pposing. Value. A complex number is an eigenvalue of corresponding to the eigenvector if and only if its complex conjugate is an eigenvalue corresponding to the conjugate vector. It is possible that Ahas complex eigenvalues, which must occur in complex-conjugate pairs, meaning that if a+ ibis an eigenvalue, where aand bare real, then so is a ib. 1.2. When the eigenvalues of a system are complex with a real part the trajectories will spiral into or out of the origin. 4. These pairs will always have the same norm and thus the same rate of growth or decay in a dynamical system. If A has complex conjugate eigenvalues \u03bb 1,2 = \u03b1 \u00b1 \u03b2i, \u03b2 \u2260 0, with corresponding eigenvectors v 1,2 = a \u00b1 bi, respectively, two linearly independent solutions of X\u2032 = AX are X 1 (t) = e \u03b1t (a cos \u03b2t \u2212 b sin \u03b2t) and X 2 (t) = e \u03b1t (b cos \u03b2t + a sin \u03b2t). Then a) if = a+ ibis an eigenvalue of A, then so is the complex conjugate = a\u2212ib. values. eigvalsh. Most of this materi\u2026 For example, the command will result in the assignment of a matrix to the variable A: We can enter a column vector by thinking of it as an m\u00d71 matrix, so the command will result in a 2\u00d71 column vector: There are many properties of matrices that MATLAB will calculate through simple commands. To enter a matrix into MATLAB, we use square brackets to begin and end the contents of the matrix, and we use semicolons to separate the rows. The Eigenvalue Problem: The Hessenberg and Real Schur Forms The Unsymmetric Eigenvalue Problem Let Abe a real n nmatrix. eigenvalues are themselves complex conjugate and the calculations involve working in complex n-dimensional space. complex eigenvalues always come in complex conjugate pairs. eigh. There is nothing wrong with this in principle, however the manipulations may be a bit messy. A similar discussion verifies that the origin is a source when the trace of is positive. Here is a summary: If a linear system\u2019s coef\ufb01cient matrix has complex conjugate eigenvalues, the system\u2019s state is rotating around the origin in its phase space. eigenvalues of a self-adjoint matrix Eigenvalues of self-adjoint matrices are easy to calculate. Note that the complex conjugate of a function is represented with a star (*) above it. An interesting fact is that complex eigenvalues of real matrices always come in conjugate pairs. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. The eigenvectors associated with these complex eigenvalues are also complex and also appear in complex conjugate pairs. Example: Diagonalize the matrix . Solve the system. \u2192Below is a calculator to determine matrices for given Eigensystems. In equation 1 we can appreciate that because the eigenvalue is real, the complex conjugate of the real eigenvalue is just the real eigenvalue (no imaginary term to take the complex conjugate of). Proposition Let be a matrix having real entries. complex eigenvalues. Once you have found the eigenvalues of a matrix you can \ufb01nd all the eigenvectors associated with each eigenvalue by \ufb01nding a \u2026 If A is a 2 2-matrix with complex-conjugate eigenvalues l = a bi, with associated eigenvectors w = u iv, then any solution to the system dx dt = Ax(t) can be written x(t) = C1eat(ucosbt vsinbt)+C2eat(usinbt+vcosbt) (7) where C1,C2 are (real) constants. Find the complex conjugate eigenvalues and corresponding complex eigenvectors of the following matrices. Similar function in SciPy that also solves the generalized eigenvalue problem. Finding Eigenvectors. 3. Since the real portion will end up being the exponent of an exponential function (as we saw in the solution to this system) if the real part is positive the solution will grow very large as \\(t$$ increases. eigenvalues of a non-symmetric array. This occurs in the region above the parabola. NOTE 4: When there are complex eigenvalues, there's always an even number of them, and they always appear as a complex conjugate pair, e.g.", "date": "2021-05-17 19:15:41", "meta": {"domain": "lovejays.com", "url": "https://lovejays.com/dag9b/complex-conjugate-eigenvalues-121a37", "openwebmath_score": 0.8943105340003967, "openwebmath_perplexity": 135.6482903657162, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9890130554263733, "lm_q2_score": 0.865224070413529, "lm_q1q2_score": 0.8557179015081279}}
{"url": "https://am111.readthedocs.io/en/latest/session10_FFT.html", "text": "# Session 10: Fast Fourier Transform\u00b6\n\nDate: 11/27/2017, Monday\n\nIn [1]:\n\nformat compact\n\n\n## Generate input signal\u00b6\n\nFourier transform is widely used in signal processing. Let\u2019s looks at the simplest cosine signal first.\n\nDefine\n\n$y_1(t) = 0.3 + 0.7\\cos(2\\pi f_1t)$\n\nIt has a magnitude of 0.7, with a constant bias term 0.3. We choose the frequency $$f_1=0.5$$.\n\nIn [2]:\n\nt = -5:0.1:4.9; % time axis\nN = length(t) % size of the signal\n\nf1 = 0.5; % signal frequency\ny1 = 0.3 + 0.7*cos(2*pi*f1*t); % the signal\n\nN =\n100\n\nIn [3]:\n\n%plot -s 800,200\nhold on\nplot(t, y1)\nplot(t, 0.3*ones(N,1), '--k')\ntitle('simple signal')\nxlabel('t [s]')\nlegend('signal', 'mean')\n\n\n## Perform Fourier transform on the signal\u00b6\n\nYou can hand code the Fourier matrix as in the class, but here we use the built-in function for convenience.\n\nIn [4]:\n\nF1 = fft(y1);\nlength(F1) % same as the length of the signal\n\nans =\n100\n\n\nThere are two different conventions for the normalization factor in the Fourier matrix. One is having the normalization factor $$\\frac{1}{\\sqrt{N}}$$ in the both the Fourier matrix $$A$$ and the inverse transform matrix $$B$$\n\n$\\begin{split}A = \\frac{1}{\\sqrt{N}} \\begin{bmatrix} 1&1&1&\\cdots &1 \\\\ 1&\\omega&\\omega^2&\\cdots&\\omega^{N-1} \\\\ 1&\\omega^2&\\omega^4&\\cdots&\\omega^{2(N-1)}\\\\ \\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\ 1&\\omega^{N-1}&\\omega^{2(N-1)}&\\cdots&\\omega^{(N-1)(N-1)}\\\\ \\end{bmatrix}\\end{split}$\n$\\begin{split}B = \\frac{1}{\\sqrt{N}} \\begin{bmatrix} 1&1&1&\\cdots &1 \\\\ 1&\\omega^{-1}&\\omega^{-2}&\\cdots&\\omega^{-(N-1)} \\\\ 1&\\omega^{-2}&\\omega^{-4}&\\cdots&\\omega^{-2(N-1)}\\\\ \\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\ 1&\\omega^{-(N-1)}&\\omega^{-2(N-1)}&\\cdots&\\omega^{-(N-1)(N-1)}\\\\ \\end{bmatrix}\\end{split}$\n\nMATLAB uses a different convention that\n\n$\\begin{split}A = \\begin{bmatrix} 1&1&1&\\cdots &1 \\\\ 1&\\omega&\\omega^2&\\cdots&\\omega^{N-1} \\\\ 1&\\omega^2&\\omega^4&\\cdots&\\omega^{2(N-1)}\\\\ \\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\ 1&\\omega^{N-1}&\\omega^{2(N-1)}&\\cdots&\\omega^{(N-1)(N-1)}\\\\ \\end{bmatrix}\\end{split}$\n$\\begin{split}B = \\frac{1}{N} \\begin{bmatrix} 1&1&1&\\cdots &1 \\\\ 1&\\omega^{-1}&\\omega^{-2}&\\cdots&\\omega^{-(N-1)} \\\\ 1&\\omega^{-2}&\\omega^{-4}&\\cdots&\\omega^{-2(N-1)}\\\\ \\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\ 1&\\omega^{-(N-1)}&\\omega^{-2(N-1)}&\\cdots&\\omega^{-(N-1)(N-1)}\\\\ \\end{bmatrix}\\end{split}$\n\nThe difference doesn\u2019t matter too much as long as you use one of them consistently. In both cases there is\n\n\\begin{align} F &= AY \\text{ (Discrete Fourier transfrom)} \\\\ Y &= BF \\text{ (Inverse transfrom)} \\end{align}\n\n### Full spectrum\u00b6\n\nThe spectrum F1 (the result of the Fourier transfrom) is typically an array of complex numbers. To plot it we need to use absolute magnitude.\n\nIn [5]:\n\n%plot -s 800,200\nplot(abs(F1),'- .')\ntitle('unnormalized full spectrum')\n\n\nThe first term in F1 indicates the magnitude of the constant term (zero frequency). Diving by N gives us the actual value.\n\nIn [6]:\n\nF1(1)/N % equal to the constant bias term specified at the beginning\n\nans =\n0.3000\n\n\nBesides the constant bias F1(1), there are two non-zero pointings in F1, indicating the cosine signal itself. The magnitude 0.7 is evenly distributed to two points.\n\nIn [7]:\n\nF1(2+4)/N, F1(end-4)/N % adding up to 0.7\n\nans =\n-0.3500 - 0.0000i\nans =\n-0.3500 + 0.0000i\n\n\nPlotting F1/N shows more clearly the magnitude of signals at different frequencies:\n\nIn [8]:\n\n%plot -s 800,200\nplot(abs(F1)/N,'- .')\ntitle('(normalized) full spectrum')\nylabel('signal amplitude')\n\n\n### Half-sided spectrum\u00b6\n\nFrom the matrix $$A$$ it is easy to show that, the first element F(1) in the resulting spectrum is always a real number indicating the constant bias term, while the rest of the array F(2:end) is symmetric, i.e. F(2) == F(end), F(3) == F(end-1). ( F(2) is actually the conjugate of F(end), but we only care about magnitude here. )\n\nDue to such symmetricity, we can simply plot half of the array (scaled by 2) without loss of information.\n\nIn [9]:\n\nM = N/2 % need to cast to integer if N is an odd number\n\nM =\n50\n\nIn [10]:\n\n%plot -s 800,200\nplot(abs(F1(1:M+1))/N*2, '- .')\ntitle('(normalized) half-sided spectrum')\nylabel('signal amplitude')\n\n\n## Understanding units!\u00b6\n\nThe Discrete Fourier Transform, by defintion, is simply a matrix multiplication which acts on pure numbers. But real physical signals have units. You cannot just treat the resulting array F1 as some unitless frequency. If the signal is a time series then you need to deal with seconds and hertz; if it is a wave in the space then you need to deal with the wave length in meters.\n\nIn order to understand the unit of the resulting spectrum F1, let\u2019s look at the original time series y1 first.\n\nThe \u201ctime step\u201d of the signal is\n\nIn [11]:\n\ndt = t(2)-t(1) % [s]\n\ndt =\n0.1000\n\n\nThis is the finest temporal resolution the signal can have. It corresponds the highest frequency:\n\nIn [12]:\n\nf_max = 1/dt % [Hz]\n\nf_max =\n10.0000\n\n\nOn the contrary, the longest time range (dt*N, the time span of the entire signal) corresponds to the lowest frequency:\n\nIn [13]:\n\ndf = f_max/N % [Hz]\n\ndf =\n0.1000\n\n\nWith the lowest frequency df being the \u201cstep size\u201d in the frequency axis, the value of the frequency axis is simply the array [0, df, 2*df, \u2026]. Now we can use correct values and units for the x-axis of the spectrum plot.\n\nIn [14]:\n\n%plot -s 800,200\nplot(df*(0:M), abs(F1(1:M+1))/N*2,'- .')\ntitle('half-sided spectrum with correct unit of x-axis')\nylabel('signal amplitude')\nxlabel('frequency [Hz]')\n\n\nThe peak is at 0.5 Hz, consistent with our original signal which has a period of 2 s, since 0.5 Hz = 1/(2s). Thus our unit specification is correct.\n\n## Deal with negative frequency\u00b6\n\nThe right half of the spectrum array (F1(M+2:end), not plotted in the above figure) corresponds to negative frequency [-M*df, \u2026, -2*df, -df]. Thus each element in the entire F1 array corresponds to each element in the frequency array [0, df, 2*df, \u2026, M*df, -M*df, \u2026, -2*df, -df].\n\nYou can perform fftshift on the resulting spectrum F1 to swap its left and right parts, so it will align with the motonically increasing axis [-M*df, \u2026, -2*df, -df, 0, df, 2*df, \u2026, M*df]. That feels more natural from a mathematical point of view.\n\nIn [15]:\n\nF_shifted = fftshift(F1);\nplot(abs(F_shifted),'- .')\n\n\n## Perform inverse transform\u00b6\n\nPerforming inverse transform is simply ifft(F1). Recall that MATLAB performs the $$\\frac{1}{N}$$ scaling during the inverse transform step.\n\nWe use norm to check if ifft(F1) is close enough to y1.\n\nIn [16]:\n\nnorm(ifft(F1) - y1) % almost zero\n\nans =\n1.2269e-15\n\n\n## Mix two signals\u00b6\n\nFourier transform and inverse transform are very useful in signal filering. Let\u2019s first add a high-frequency noise to our original signal.\n\nIn [17]:\n\nf2 = 5; % higher frequency\ny2 = 0.2*sin(f2*pi*t); % noise\ny = y1 + y2; % add up original signal and noise\n\nIn [18]:\n\n%plot -s 800,400\nsubplot(311);plot(t, y2, 'r');\nylim([-1,1]);title('noise')\n\nsubplot(312);plot(t, y1);\nylim([-0.6,1.2]);title('original signal')\n\nsubplot(313);plot(t, y, 'k');\nylim([-0.6,1.2]);title('signal + noise')\n\n\nAfter the Fourier transform, we see two new peaks at a relatively higher frequency.\n\nIn [19]:\n\nF = fft(y);\n\nIn [20]:\n\n%plot -s 800,200\nplot(abs(F), '- .')\ntitle('spectrum with high-frequency noise')\n\n\nAgain, the noise magnitude 0.2 is evenly distributed to positive and negative frequencies. Here we got complex conjugates:\n\nIn [21]:\n\nF(2+24)/N, F(end-24)/N % magnitude of noises\n\nans =\n-0.0000 + 0.1000i\nans =\n-0.0000 - 0.1000i\n\n\n## Filter out high-frequency noise\u00b6\n\nLet\u2019s wipe out this annoying noise. It\u2019s very difficult to do so in the original signal, but very easy to do in the spectrum.\n\nIn [22]:\n\nF_filtered = F; % make a copy\nF_filtered(26) = 0; % remove the high-frequency noise\nF_filtered(76) = 0; % same for negative frequency\n\nIn [23]:\n\nplot(abs(F_filtered), '- .')\ntitle('filtered spectrum')\n\n\nThen we can transform the spectrum back to the signal.\n\nIn [24]:\n\ny_filtered = ifft(F_filtered);\n\n\nIf the filtering is done symmetrically (i.e. do the same thing for positive and negative frequencies), the recovered signal will only contain real numbers.\n\nIn [35]:\n\n%plot -s 800,200\nplot(t, y_filtered)\ntitle('de-noised signal')\n\n\nThe de-noised signal is almost the same as the original noise-free signal:\n\nIn [36]:\n\nnorm(y_filtered - y1) % almost zero\n\nans =\n5.6847e-15\n\n\n## Filter has to be symmetric\u00b6\n\nWhat happens if the filtering done asymmetrically?\n\nIn [37]:\n\nF_wrong_filtered = F; % make another copy\nF_wrong_filtered(76) = 0; % only do negative frequency\n\nIn [39]:\n\nplot(abs(F_wrong_filtered), '- .')\ntitle('asymmetrically-filtered spectrum')\n\n\nThe recovered signal now contains imaginary parts. That\u2019s unphysical!\n\nIn [40]:\n\ny_wrong_filtered = ifft(F_wrong_filtered);\n\nIn [42]:\n\ny_wrong_filtered(1:5)' % print the first several elements\n\nans =\n-0.4000 - 0.1000i\n-0.4657 + 0.0000i\n-0.2663 + 0.1000i\n-0.0114 - 0.0000i\n0.0837 - 0.1000i\n\nIn [43]:\n\nnorm(imag(y_wrong_filtered)) % not zero\n\nans =\n0.7071\n\n\nYou can plot the real part only. It is something between the unfiltered and filtered signals, i.e. the filtering here is incomplete.\n\nIn [45]:\n\nhold on\nplot(t, y, 'k')\nplot(t, real(y_wrong_filtered), 'b')\nplot(t, y1, 'r')\nlegend('signal with noise', 'incomplete fliltering', 'signal without noise')", "date": "2022-05-18 13:55:21", "meta": {"domain": "readthedocs.io", "url": "https://am111.readthedocs.io/en/latest/session10_FFT.html", "openwebmath_score": 0.8695923686027527, "openwebmath_perplexity": 5006.151520887731, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130570455679, "lm_q2_score": 0.8652240686758841, "lm_q1q2_score": 0.8557179011905405}}
{"url": "http://math.stackexchange.com/questions/191532/why-cant-a-neighborhood-be-a-finite-set", "text": "# Why can't a neighborhood be a finite set?\n\nRudin defines a neighborhood as follows:\n\nLet $X$ be a metric space endowed with a distance function $d$. A neighborhood of a point $p \\in X$ is a set $N_r(p)$ consisting of all $q \\in X$ such that $d(p,q) < r$ for some $r > 0$.\n\nHe later proves two facts: every neighborhood is an open set, and every finite set is closed.\n\nBut it would seem to me that a neighborhood as defined above could be finite. For example, let $X = \\{1,2,3\\}$ with $d(x,y) = |x-y|$ be a metric space . Consider the neighborhood around $p = 2$ of radius $0.5$, i.e.: the set of all points $q$ in $X$ such that $d(q,p)<0.5$. But the neighborhood is simply $\\{2\\}$. Therefore, the neighborhood is a finite set, and therefore closed. But every neighborhood is open. This is a contradiction.\n\nSo my understanding of a neighborhood is broken somehow. It has further implications: if we define $E = \\{2\\}$, then the $0.5$-radius neighborhood is $\\{2\\}$, which is a subset of $E$, which means that $2$ is an interior point of $E$. Since $2$ is the only point in $E$, all points in $E$ are interior points, and $E$ is open. But $E$ is finite, and therefore closed.\n\nI'm probably missing something obvious. Can anyone spot my mistake?\n\n-\nNothing about the definition of a set being open or closed precludes a set being both -- merely your intuition. :) \u2013\u00a0Eugene Shvarts Sep 5 '12 at 16:45\nIt is possible for are set to be open and closed, as $\\{2\\}$ in your example. Being open and closed isn't mutually exclusive. \u2013\u00a0martini Sep 5 '12 at 16:45\n\"not closed\" is not necessarily \"open\" and \"not open\" is not necessarily \"closed\": \"not open\" and \"not closed\" can both be \"ajar\"! \u2013\u00a0rschwieb Sep 5 '12 at 16:55\nObligatory quote: ''\u2026an answer to the mathematician\u2019s riddle: \u201cHow is a set different from a door?\u201d should be: \u201cA door must be either open or closed, and cannot be both, while a set can be open, or closed, or both, or neither!\u201d'' - Munkres \u2013\u00a0Michael Greinecker Sep 5 '12 at 16:58\nWell, in that case, Rudin 2.21 is incorrect, unless there's another hypothesis. \u2013\u00a0Cameron Buie Sep 5 '12 at 17:02\n\n\"Every finite set is closed\" does not imply \"every open neighbourhood is infinite\".\n\nLet $(X,d)$ be any metric space, and let $F \\subseteq X$ be a non-empty finite subset; say $F = \\{ x_1, \\cdots, x_n \\}$ for some $n \\in \\mathbb{N}$. To prove that $F$ is closed it suffices to prove that $X - F$ is open. So let $p \\in X-F$. Then the set $\\{ d(p,x_1), \\cdots, d(p,x_n) \\}$ has a minimum value, say $\\delta$, and we know that $\\delta > 0$ since $p \\not \\in F$, and so whenever $d(p,q) < \\delta$ we have $q \\in X-F$. But this tells you that the (open) ball of radius $\\delta$ about $p$ lies in $X-F$, and hence $X-F$ is open.\n\nIn your examples, yes, the sets are open and finite, but they are also closed!\n\nThe key fact to take home is that sets are allowed to be both open and closed.\n\nHowever, there is a partial converse: in a dense metric space, every open neighbourhood is infinite.\n\nA metric space $(X,d)$ is dense if for any $x \\in X$ and $r > 0$ there exists $y \\in X$ with $x \\ne y$ and $d(x,y) < r$ $-$ that is, given any point in the space, there are other points which lie arbitrarily close to that point. $\\mathbb{R}$ is a dense space, so is $\\mathbb{Q}$, so is $(0,1)$, and indeed so is any other non-empty open subset of $\\mathbb{R}$.\n\nThe fact that open neighbourhoods in dense metric spaces are necessarily infinite is clear from the definition. [It's also clear that a metric space in which every open neighbourhood is infinite is dense, and so the two conditions are equivalent.]\n\n-\nThis makes sense, thanks! \u2013\u00a0jme Sep 5 '12 at 17:11\n\nFor an interesting example, let $X$ be any non-empty set, and define $d:X\\times X\\to\\Bbb R$ by $$d(x,y)=\\begin{cases}0 & x=y\\\\1 & x\\neq y.\\end{cases}$$ This can be shown to be a metric on $X$ (called the discrete metric). One interesting property is that in the metric space $(X,d)$, every subset of $X$ is both open and closed. Even in general metric spaces, there will always be at least two subsets that are both open and closed--namely, the empty set and the whole set.\n\n-", "date": "2015-11-26 10:44:49", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/191532/why-cant-a-neighborhood-be-a-finite-set", "openwebmath_score": 0.9667059183120728, "openwebmath_perplexity": 152.26834436669773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018405251301, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.8557169875442601}}
{"url": "https://math.stackexchange.com/questions/3980049/is-the-set-a-0-1-setminus-mathbbq-countable-or-not", "text": "# Is the set $A = [0,1]\\setminus\\mathbb{Q}$ countable or not?\n\nIs the set $$A = [0,1]\\setminus\\mathbb{Q}$$ countable or not?\n\nWhat I am thinking is $$A$$ consist of irrational numbers in the interval $$[0,1]$$ hence it is subset of irrational numbers. As set of irrational numbers is uncountable so I think set $$A$$ is also uncountable.\n\n\u2022 not every subset of irrational numbers is uncountable, but if $A$ and $\\mathbb Q$ were both countable then so would be $A\\cup \\mathbb Q$ and therefore $[0,1]\\subset A\\cup \\mathbb Q$ \u2013\u00a0J. W. Tanner Jan 10 at 15:55\n\u2022 $\\varnothing$ is also a subset of the irrational numbers. \u2013\u00a0Asaf Karagila Jan 10 at 16:24\n\nYes, it is uncountable, but not for that reason. For instance, $$\\left\\{\\sqrt2+n\\,\\middle|\\, n\\in\\Bbb N\\right\\}$$ is also a set of irrational numbers, but it is countable.\n\nHowever, if $$[0,1]\\setminus\\Bbb Q$$ was countable, then, since $$\\Bbb Q\\cap[0,1]$$ is countable, $$[0,1]$$ would be countable too, since it's the union of them.\n\n\u2022 technically, $[0,1]$ is contained in the union of $A$ and $\\mathbb Q$ \u2013\u00a0J. W. Tanner Jan 10 at 16:08\n\u2022 @J.W.Tanner I've edited my answer. Thank you. \u2013\u00a0Jos\u00e9 Carlos Santos Jan 10 at 16:23\n\nAll countable set of $$R$$ have Lebesgue measure equal to $$0$$. So Lebesgue measure of $$[0, 1] - \\mathbb{Q}$$ is $$1$$. Eventually by contraposition, $$[0, 1] - \\mathbb{Q}$$ is uncoutable.\n\nWhat you are thinking does not work.\n\nFor example, $$\\{n\\pi\\mid n\\in\\mathbb N\\setminus\\{0\\}\\}$$ is a subset of irrational numbers but countable.\n\nHere's an argument that works. If $$A$$ were countable, then, since $$\\mathbb Q$$ is countable,\n\n$$A\\cup \\mathbb Q$$ would be countable,\n\nand therefore $$[0,1]$$, which is a subset of $$A\\cup \\mathbb Q$$, would be countable,", "date": "2021-06-14 09:18:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3980049/is-the-set-a-0-1-setminus-mathbbq-countable-or-not", "openwebmath_score": 0.929182767868042, "openwebmath_perplexity": 267.7689207142588, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8557169747108202}}
{"url": "https://math.stackexchange.com/questions/1502270/can-one-compute-series-expansions-or-complex-residues-at-essential-singularities", "text": "# Can one compute series expansions or complex residues at essential singularities?\n\nTitle says it all.\n\nI'm noticing a trend of failure on Mathematica/WolframAlpha's parts when trying to compute either the Laurent expansions or the residues of functions like $\\sin\\dfrac{1}{z}$ about $z=0$. (Specifying $z=0$ for the residue query returns nothing.)\n\nIs there something about essential singularities $z_0$ that prevents one from computing a series expansion about or residue at $z_0$? Is this \"something\" explicitly mentioned in the usual definitions of series/residues that I'm missing? Admittedly, my complex analysis knowledge has started to rust.\n\n\u2022 Just plug $u=1/z$ into $\\sin u = u -u^3/3! + \\cdots$ for the Larent expansion; the residue is $1.$ \u2013\u00a0zhw. Oct 28 '15 at 19:46\n\u2022 @zhw. Okay, that approach works fine for $\\sin\\dfrac{1}{z}$, but what if I made a slight adjustment? Say, $\\sin\\dfrac{z}{z-1}$. Does the same process still work? \u2013\u00a0user170231 Oct 28 '15 at 19:51\n\nRecall the definition of an essential singularity. If a function $f$ has an essential singularity at $z=z_0$, then\n\n$$\\lim_{z \\to z_0} (z-z_0)^m f(z) = \\infty$$\n\nThat is, an essential singularity is \"more singular\" than any order pole.\n\nThat all said, the definition of a residue still applies as the coefficient of $(z-z_0)^{-1}$ in the Laurent expansion of $f$.\n\nIn your example of $f(z) = \\sin{\\left ( \\frac{z}{z-1} \\right )}$, there is an essential singularity at $z=1$. The Laurent expansion of $f$ about $z=1$ may be easily determined by expressing $f$ as follows:\n\n\\begin{align}f(z) &= \\sin{\\left (1+\\frac1{z-1} \\right )} = \\sin{1} \\cos{\\left ( \\frac{1}{z-1} \\right )}+ \\cos{1} \\sin{\\left ( \\frac{1}{z-1} \\right )}\\\\ &= \\sin{1}\\sum_{n=0}^{\\infty} \\frac{(-1)^n}{(2 n)!(z-1)^{2 n}} + \\cos{1}\\sum_{n=0}^{\\infty} \\frac{(-1)^n}{(2 n+1)!(z-1)^{2 n+1}}\\end{align}\n\nHopefully it is clear that the residue of $f$ at $z=1$ is $\\cos{1}$.\n\n\u2022 Are there any functions with essential singularities for which identities and \"tricks\" like these don't help? I'm just trying to wrap my mind around why WA and Mma are having trouble with these computations. \u2013\u00a0user170231 Oct 28 '15 at 21:00\n\u2022 @user170231: plenty. What I would do is make use of Mathematica's expansion about Infinity, i.e., Series[f[z],{z,Infinity,n}] for an n-term expansion in $1/z$. \u2013\u00a0Ron Gordon Oct 28 '15 at 21:08", "date": "2020-09-26 03:23:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1502270/can-one-compute-series-expansions-or-complex-residues-at-essential-singularities", "openwebmath_score": 0.9793195724487305, "openwebmath_perplexity": 526.6697709462901, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018398044143, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.855716974413328}}
{"url": "http://math.wsu.edu/faculty/bkrishna/FilesMath464/S18/LecNotes/welcome.html", "text": "Linear Optimization - Lecture Notes and Videos\n\nMath 464 (Spring 2018) \u00a0-\u00a0Lecture Notes and Videos on Linear Optimization\n\nScribes from all lectures so far (as a single big file)\n\nLec # Date Topic(s) Scribe Panopto\n1 Jan \u00a09 syllabus, optimization in calculus, solving $$A\\mathbf{x} = \\mathbf{b}$$ using EROs, linear program example: Dude's Thursday problem scribe video\n2 Jan 11 no class\n3 Jan 16 general form of a linear program (LP), feasible and optimal solutions, standard form, conversion to std form, excess/slack variables scribe video\n4 Jan 18 LP formulations, diet problem, solution in AMPL, road lighting problem, exceeding illumination limits, currency exchange LP scribe video\n5 Jan 23 convex function, piecewise linear (PL) convex (PLC) functions, max of convex functions is convex, PLC functions in LP scribe video\n6 Jan 25 $$|x_i| \\to x_i^+, x_i^-$$, lighting problem: get \"close to\" $$I_i^*$$, inventory planning LP, graphical solution in 2D, feasible region, slide $$\\mathbf{c}^T\\mathbf{x}$$ line scribe video\n7 Jan 30 cases of LP, unique and alternative optimal solutions, unbounded LP, infeasible LP, subspace, affine spaces, polyhedron, halfspace scribe video\n8 Feb \u00a0 1 polyhedron is convex, bounded set, convex hull, active constraint, extreme point, vertex, basic and basic feasible solution (bfs) scribe video\n9 Feb \u00a0 6 proof of extreme point $$\\Leftrightarrow$$ vertex $$\\Leftrightarrow$$ bfs, adjacent basic solutions and bfs's, polyhedron $$P$$ in standard form, basic solutions of $$P$$ scribe video\n10 Feb \u00a0 8 finding basic solutions in Octave, correspondence of corner points and bfs's, degenrate basic bfs, degeneracy and representation scribe video\n11 Feb 13 degeneracy in standard form, $$P$$ has no line $$\\Leftrightarrow P$$ has vertex, existence of optimal vertex, shape of standard form polyhedron scribe video\n12 Feb 15 simplex method, feasible direction at $$\\mathbf{x}$$: $$\\mathbf{x}+\\theta\\mathbf{d} \\in P \\mbox{ for } \\theta > 0, j$$th basic direction at bfs $$\\mathbf{x}$$, reduced cost $$\\mathbf{c}'^T = \\mathbf{c}^T - \\mathbf{c}_B B^{-1} A$$ scribe video\n13 Feb 20 hints for problems from Hw6, LP optimality conditions in standard form: basis $$B$$ optimal if $$B^{-1}\\mathbf{b} \\geq \\mathbf{0}$$ and $$\\mathbf{c}^T - \\mathbf{c}_B B^{-1} A \\geq \\mathbf{0}^T$$ scribe video\n14 Feb 22 entering/leaving variable, min-ratio test: $$\\theta^* = \\min_{\\{i \\in \\scr{B}|d_{B(i)} < 0\\}} \\,(-x_{B(i)}/d_{B(i)})$$, an iteration of the simplex method, Octave session scribe video", "date": "2018-02-26 03:20:57", "meta": {"domain": "wsu.edu", "url": "http://math.wsu.edu/faculty/bkrishna/FilesMath464/S18/LecNotes/welcome.html", "openwebmath_score": 0.5605629682540894, "openwebmath_perplexity": 11296.644222561166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9932024696568575, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8556818720142164}}
{"url": "https://math.stackexchange.com/questions/3642078/how-does-the-taylor-series-converge-at-all-points-for-certain-functions", "text": "# How does the Taylor Series converge at all points for certain functions\n\nThe way my professor defined Taylor polynomials is: the $$n^{th}$$ degree Taylor polynomial $$p(x)$$ of $$f(x)$$ is a polynomial that satisfies $$\\lim_{x\\to 0}{f(x)-p(x) \\over x^n} = 0$$. This is actually the little-o notation $$o(x^n)$$, which means $$(f(x)-p(x)) \\ll x^n$$ as $$x$$ approaches $$0$$. From this I have got the intuition that Taylor Polynomials work only for $$|x| < 1$$ because $$x^n$$ gets smaller as $$n$$ gets bigger only when $$|x| < 1$$. And the textbook seemed to agree with my intuition, because the textbook says \u201cTaylor polynomial near the origin\u201d (probably implying $$|x| < 1$$).\n\nSince Taylor Series is basically Taylor polynomial with $$n\\to\\infty$$, I intuitively thought that the Taylor Series would also only converge to the function it represents in the interval $$(-1, 1)$$.\n\nFor example, in the case of $$1\\over1-x$$, it is well known that the Taylor series only converges at $$|x| < 1$$.\n\nHowever, all of a sudden, the textbook says that the Taylor series of $$\\cos x$$ converges for all real $$x$$. It confused me because previously I thought the Taylor series would only work for $$|x|<1$$. Now, I know that the Taylor Series is defined like this: $$f(x) = Tf(x) \\Leftrightarrow \\lim_{n\\to\\infty}R_{n}f(x) = 0$$\n\nAnd I know how to get the maximum of Taylor Remainder for $$\\cos x$$ using Taylor's Theorem, and I know that the limit of that Taylor Remainder is $$0$$ for all real $$x$$, which makes the Taylor Series of $$cosx$$ converge to $$\\cos x$$, pointwise. However, I just can't get why my initial intuition is wrong (why taylor series converges for all $$x$$ for certain functions, like $$\\cos x$$, also $$\\sin x$$ and $$e^x$$, etc.)\n\n\u2022 For the specific example $e^x$, the terms in its series are $x^n/n!$. Note that $n!$ grows much faster than $x^n$ as $n$ grows. That's one reason to believe the Taylor series of $e^x$ converges. Apr 24 '20 at 18:10\n\u2022 In my mind, the key idea of calculus is the local linear approximation $f(x) \\approx f(a) + f'(a)(x - a)$, which is a good approximation when $x$ is near $a$. It is natural to ask, what if we approximate $f$ locally by a quadratic or cubic function rather than by a linear function. This leads to the idea of Taylor polynomial approximation, which is indeed initially intended to be a local approximation to $f$. But when we look at the remainder term in Taylor's theorem, there's something kind of amazing, surprising that happens which is that often it's small even when $x$ is far from $a$. Apr 24 '20 at 18:11\n\u2022 @trisct Yes that is true indeed. The limit of Taylor remainder for $e^x$ is also zero for all x, which further explains why the Taylor Series converges. But I still I can't see why my initial intuition is wrong.\n\u2013\u00a0the\nApr 24 '20 at 18:12\n\u2022 @littleO So it just happened to be that the Taylor remainder's limit is zero even for x far away?\n\u2013\u00a0the\nApr 24 '20 at 18:14\n\u2022 @linearAlg In my mind, yes. It is just one of the miracles of math, which makes us wonder why things have worked out more nicely than we deserved. Apr 24 '20 at 18:16\n\nActually, things may go wrong in $$(-1,1)$$. For instance, the Taylor series centered at $$0$$ of $$f(x)=\\frac1{1-nx}$$ only converges to $$f(x)$$ on $$\\left(-\\frac1n,\\frac1n\\right)$$. And if$$f(x)=\\begin{cases}e^{-1/x^2}&\\text{ if }x\\ne0\\\\0&\\text{ if }x=0,\\end{cases}$$then the Taylor series of $$f$$ only converges to $$f(x)$$ if $$x=0$$.\nOn the other hand, yes, Taylor series centered at $$0$$ are made to converge to $$f(x)$$ near $$0$$. But that's no reason to expect that they don't converge to $$f(x)$$ when $$x$$ is way from $$0$$. That would be like expecting that a non-constant power series $$a_0+a_1x+a_2x^2+\\cdots$$ takes larger and larger values as the distance from $$x$$ to $$0$$. That happens often, but $$1-\\frac1{2!}x^2+\\frac1{4!}x^4-\\cdots=\\cos(x)$$, which is bounded.\nIn other words, you have gone from $$a\\implies b$$ to $$\\tilde a\\implies \\tilde b,$$ which you can see to be clearly false, identically. That is, it is not necessarily true for all $$a,\\,b.$$\nSince you already know why the series for entire functions like $$\\cos x$$ converges everywhere (as you explain towards the end of your post), you should now see where your original intuition (I would say erroneous belief) misled you.", "date": "2021-12-08 13:58:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3642078/how-does-the-taylor-series-converge-at-all-points-for-certain-functions", "openwebmath_score": 0.9634480476379395, "openwebmath_perplexity": 124.58894626498918, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759666033577, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8556645841171188}}
{"url": "https://math.stackexchange.com/questions/4139421/helen-and-joe-play-guitar-together-every-day-at-lunchtime/4139434", "text": "# Helen and Joe play guitar together every day at lunchtime.\n\nHelen and Joe play guitar together every day at lunchtime. The number of songs that they play on a given day has a Poisson distribution, with an average of 5 songs per day. Regardless of how many songs they will play that day, Helen and Joe always flip a coin at the start of each song, to decide who will play the solo on that song. If we know that Joe plays exactly 4 solos on a given day, then how many solos do we expect that Helen will play on that same day?\n\nMy attempt: If the average is $$5$$ songs a day and Joe performs $$4$$ solos on one day. I thought we should expect Helen to perform $$1$$ solo on the same day $$(5-4=1)$$\n\nBut The answer given to me is: $$2.5$$ solos we expect Helen to play\n\nMy question is why? What is the way of thinking that gives me $$2.5$$? Is it cause of the coin flip? so $$5 \\cdot .5 = 2.5$$? What does Joe's $$4$$ solos have to do with anything then?\n\nThank You for any help.\n\n\u2022 Helen and Joe must have played $4$ or more solos on that day. In other words, you should be thinking about a conditional probability. \u2013\u00a0Toby Mak May 15 at 2:42\n\nLet $$X$$ denote the total number of games played, and let $$J$$ denote the number of games Joe played. The conditional distribution of $$X$$ given $$J=4$$, namely $$X|J=4$$, is supported on $$\\{4,5,\\ldots\\}$$ and has pmf $$P(X=k|J=4)=\\frac{P(J=4|X=k)P(X=k)}{\\sum_{k=4}^{\\infty}P(J=4|X=k)P(X=k)}$$ which is non zero whenever $$k\\geq 4$$. Using the facts that $$X\\sim \\text{Poisson}(5)$$ and $$J|X\\sim \\text{Binomial}(X,1/2)$$ we get $$P(X=k|J=4)=e^{-5/2}\\frac{1}{(k-4)!}\\bigg(\\frac{5}{2}\\bigg)^{k-4}$$ This means $$X-4|J=4\\sim \\text{Poisson}(5/2)$$ and so $$\\mathbb{E}(X-4|J=4)=\\mathbb{E}(X|J=4)-4=5/2$$ Hence $$13/2=E(X|J=4)$$ so expected number of times Helen plays is $$5/2$$\n\nWhile Matthew Pilling and tommik provide answers that show the mathematics of getting the answer, I will provide the intuition involved.\n\n1. We know that, on this specific day, Joe played 4 solos. This provides a minimum number of songs - specifically, there must have been at least 4 songs. Note that it is very possible for the Poisson distribution to produce 0 songs or 1 song. However, we have been given information that constrains the set of possible numbers of songs - it must be at least 4.\n\n2. We know that Joe played exactly 4 solos - this is a lot more likely if there are 8 songs (~27%) than if there are 4 songs (~6%). This changes the likelihoods of each of the possibilities, compared with the basic Poisson distribution, given this information.\n\n3. How many solos we expect Helen to have played can then be worked out from the new probabilities, which have incorporated the additional information (that Joe played 4 solos).\n\nTo see why the 50% information can't be directly used to conclude that Helen is expected to have played 4 solos as well, consider a slightly modified version of the problem. Rather than the number of songs following a Poisson distribution, we will assume that they follow a uniform distribution of between 1 and 7 songs.\n\nNow, we know that Joe played 4 solos. How many solos do we expect Helen to have played? Well, it can't be 4, because that would mean they may have played 8 songs, which can't have happened - the maximum is 7 songs.\n\nTo work out the correct answer, we turn to Bayes' Theorem, which is explicitly used in Matthew's answer, and is hidden by proportionality in tommik's answer. Think of the fact that Joe played 4 solos as a \"new piece of information\". Bayes' Theorem (at least by Bayesian thinking) lets you update your probabilities given the new information.\n\nThe answer of @Matthew Pilling is perfect (+1). A Bayesian approach will lead to the same solution avoiding a lot of calculations: (constants are not considered until the end of the process)\n\n$$\\mathbb{P}[X|J=4]\\propto \\mathbb{P}[X]\\cdot \\mathbb{P}[J=4|X]\\propto\\frac{5^x}{x!}\\cdot \\binom{x}{4}\\left(\\frac{1}{2}\\right)^x\\propto\\frac{\\left(\\frac{5}{2}\\right)^{x}}{(x-4)!}\\propto\\frac{\\left(\\frac{5}{2}\\right)^{(x-4)}}{(x-4)!}$$\n\nSetting $$Y=X-4$$ we immediately recognize the kernel of a Poisson distribution with mean 2.5\n\nHere $$Y$$ is the distribution of the solos played by Helen\n\n$$\\mathbb{P}[Y=y|\\text{Joe }=4]=\\frac{e^{-2.5}\\cdot 2.5^y}{y!}$$\n\nHere's some intuition:\n\nThe Poisson distribution is the limiting case for a binomial distribution where the number of trials goes to infinity while the success probability shrinks proportionally to keep the total expectation constant. So we can imagine Helen and Joe spending their lunch break making a very large number of Bernoulli trials, with each giving a small probability of \"play a song now\".\n\nWhen the number of trials grows large we can fold the coin flip into that process, deciding that when \"play a song now\" comes out at trial number $$n$$, Helen plays the solo if $$n$$ is odd, and Joe plays the solo if $$n$$ is even.\n\nBut this effectively means that each of Helen and Joe might as well be making their own separate series of trials, with half as many trials but the same probability, and therefore half the expected value. The number of solos played by each is independent of the other, and still Poisson distributed.\n\nThe number of solos Helen plays is independent of the number of solos Joe plays, so his $$4$$ solos is not actually relevant information.\n\nAn alternative (and perhaps slightly more rigorous) phrasing of the same reinterpretation:\n\nIn each step, instead of first asking \"should we play a song now?\" and then \"who should play the solo if we do play?\", in each step we can ask \"should Joe play now?\" and \"should Helen play now?\" each independently with half the probability. That's almost the same, except that there's a finite chance the answer would be that both should play. But when we go to the limit of infinity many steps, the risk of that happening in any given step goes to zero as $$1/N^2$$, and so the probability of it ever happening during the entire lunch break goes to zero as $$1/N$$. Therefore the risk becomes irrelevant in the limit where the distributions become Poisson.", "date": "2021-06-22 13:12:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4139421/helen-and-joe-play-guitar-together-every-day-at-lunchtime/4139434", "openwebmath_score": 0.6853301525115967, "openwebmath_perplexity": 295.17949725894266, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759677214397, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8556645785819889}}
{"url": "http://stacks.math.columbia.edu/tag/036M", "text": "The Stacks Project\n\nTag: 036M\n\nThis tag has label descent-section-fppf-local-source, it is called Properties of morphisms local in the fppf topology on the source in the Stacks project and it points to\n\nThe corresponding content:\n\n31.24. Properties of morphisms local in the fppf topology on the source\n\nHere are some properties of morphisms that are fppf local on the source.\n\nLemma 31.24.1. The property $\\mathcal{P}(f)=$''$f$ is locally of finite presentation'' is fppf local on the source.\n\nProof. Being locally of finite presentation is Zariski local on the source and the target, see Morphisms, Lemma 25.22.2. It is a property which is preserved under composition, see Morphisms, Lemma 25.22.3. This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) is Lemma 31.10.1. Hence we win. $\\square$\n\nLemma 31.24.2. The property $\\mathcal{P}(f)=$''$f$ is locally of finite type'' is fppf local on the source.\n\nProof. Being locally of finite type is Zariski local on the source and the target, see Morphisms, Lemma 25.16.2. It is a property which is preserved under composition, see Morphisms, Lemma 25.16.3, and a flat morphism locally of finite presentation is locally of finite type, see Morphisms, Lemma 25.22.8. This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) is Lemma 31.10.2. Hence we win. $\\square$\n\nLemma 31.24.3. The property $\\mathcal{P}(f)=$''$f$ is open'' is fppf local on the source.\n\nProof. Being an open morphism is clearly Zariski local on the source and the target. It is a property which is preserved under composition, see Morphisms, Lemma 25.24.3, and a flat morphism of finite presentation is open, see Morphisms, Lemma 25.26.9 This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) follows from Morphisms, Lemma 25.26.10. Hence we win. $\\square$\n\nLemma 31.24.4. The property $\\mathcal{P}(f)=$''$f$ is universally open'' is fppf local on the source.\n\nProof. Let $f : X \\to Y$ be a morphism of schemes. Let $\\{X_i \\to X\\}_{i \\in I}$ be an fppf covering. Denote $f_i : X_i \\to X$ the compositions. We have to show that $f$ is universally open if and only if each $f_i$ is universally open. If $f$ is universally open, then also each $f_i$ is universally open since the maps $X_i \\to X$ are universally open and compositions of universally open morphisms are universally open (Morphisms, Lemmas 25.26.9 and 25.24.3). Conversely, assume each $f_i$ is universally open. Let $Y' \\to Y$ be a morphism of schemes. Denote $X' = Y' \\times_Y X$ and $X'_i = Y' \\times_Y X_i$. Note that $\\{X_i' \\to X'\\}_{i \\in I}$ is an fppf covering also. The morphisms $f'_i : X_i' \\to Y'$ are open by assumption. Hence by the Lemma 31.24.3 above we conclude that $f' : X' \\to Y'$ is open as desired. $\\square$\n\n\\section{Properties of morphisms local in the fppf topology on the source}\n\\label{section-fppf-local-source}\n\n\\noindent\nHere are some properties of morphisms that are fppf local on the source.\n\n\\begin{lemma}\n\\label{lemma-locally-finite-presentation-fppf-local-source}\nThe property $\\mathcal{P}(f)=$$f$ is locally of finite presentation''\nis fppf local on the source.\n\\end{lemma}\n\n\\begin{proof}\nBeing locally of finite presentation is Zariski local on the source\nand the target, see Morphisms,\nLemma \\ref{morphisms-lemma-locally-finite-presentation-characterize}.\nIt is a property which is preserved under composition, see\nMorphisms, Lemma \\ref{morphisms-lemma-composition-finite-presentation}.\nThis proves\n(1), (2) and (3) of Lemma \\ref{lemma-properties-morphisms-local-source}.\nThe final condition (4) is\nLemma \\ref{lemma-flat-finitely-presented-permanence-algebra}. Hence we win.\n\\end{proof}\n\n\\begin{lemma}\n\\label{lemma-locally-finite-type-fppf-local-source}\nThe property $\\mathcal{P}(f)=$$f$ is locally of finite type''\nis fppf local on the source.\n\\end{lemma}\n\n\\begin{proof}\nBeing locally of finite type is Zariski local on the source\nand the target, see Morphisms,\nLemma \\ref{morphisms-lemma-locally-finite-type-characterize}.\nIt is a property which is preserved under composition, see\nMorphisms, Lemma \\ref{morphisms-lemma-composition-finite-type}, and\na flat morphism locally of finite presentation is locally of finite type, see\nMorphisms, Lemma \\ref{morphisms-lemma-finite-presentation-finite-type}.\nThis proves\n(1), (2) and (3) of Lemma \\ref{lemma-properties-morphisms-local-source}.\nThe final condition (4) is\nLemma \\ref{lemma-finite-type-local-source-fppf-algebra}. Hence we win.\n\\end{proof}\n\n\\begin{lemma}\n\\label{lemma-open-fppf-local-source}\nThe property $\\mathcal{P}(f)=$$f$ is open''\nis fppf local on the source.\n\\end{lemma}\n\n\\begin{proof}\nBeing an open morphism is clearly Zariski local on the source and the target.\nIt is a property which is preserved under composition, see\nMorphisms, Lemma \\ref{morphisms-lemma-composition-open}, and\na flat morphism of finite presentation is open, see\nMorphisms, Lemma \\ref{morphisms-lemma-fppf-open}\nThis proves\n(1), (2) and (3) of Lemma \\ref{lemma-properties-morphisms-local-source}.\nThe final condition (4) follows from\nMorphisms, Lemma \\ref{morphisms-lemma-fpqc-quotient-topology}.\nHence we win.\n\\end{proof}\n\n\\begin{lemma}\n\\label{lemma-universally-open-fppf-local-source}\nThe property $\\mathcal{P}(f)=$$f$ is universally open''\nis fppf local on the source.\n\\end{lemma}\n\n\\begin{proof}\nLet $f : X \\to Y$ be a morphism of schemes.\nLet $\\{X_i \\to X\\}_{i \\in I}$ be an fppf covering.\nDenote $f_i : X_i \\to X$ the compositions.\nWe have to show that $f$ is universally open if and only if\neach $f_i$ is universally open. If $f$ is universally open,\nthen also each $f_i$ is universally open since the maps\n$X_i \\to X$ are universally open and compositions\nof universally open morphisms are universally open\n(Morphisms, Lemmas \\ref{morphisms-lemma-fppf-open}\nand \\ref{morphisms-lemma-composition-open}).\nConversely, assume each $f_i$ is universally open.\nLet $Y' \\to Y$ be a morphism of schemes.\nDenote $X' = Y' \\times_Y X$ and $X'_i = Y' \\times_Y X_i$.\nNote that $\\{X_i' \\to X'\\}_{i \\in I}$ is an fppf covering also.\nThe morphisms $f'_i : X_i' \\to Y'$ are open by assumption.\nHence by the Lemma \\ref{lemma-open-fppf-local-source}\nabove we conclude that $f' : X' \\to Y'$ is open as desired.\n\\end{proof}\n\n\nTo cite this tag (see How to reference tags), use:\n\n\\cite[\\href{http://stacks.math.columbia.edu/tag/036M}{Tag 036M}]{stacks-project}\n\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).", "date": "2013-05-19 14:20:59", "meta": {"domain": "columbia.edu", "url": "http://stacks.math.columbia.edu/tag/036M", "openwebmath_score": 0.9868103265762329, "openwebmath_perplexity": 779.8899106891482, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759621310288, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8556645671947227}}
{"url": "https://math.stackexchange.com/questions/3219147/equation-of-the-line-that-lies-tangent-to-both-circles/3219165", "text": "# Equation of the line that lies tangent to both circles\n\nConsider the two circles determined by $$(x-1)^2 + y^2 = 1$$ and $$(x-2.5)^2 + y^2 = (1/2)^2$$. Find the (explicit) equation of the line that lies tangent to both circles.\n\nI have never seen a clean or clever solution to this problem. This problem came up once at a staff meeting for a tutoring center I worked at during undergrad. I recall my roommate and I - after a good amount of time symbol pushing - were able to visibly see a solution by inspection, then verify it by plugging in. I have never seen a solid derivation of a solution to this though, so I would like to see what MSE can come up with for this!\n\nI took a short stab at it today before posting, and got that it would be determined by the solution to the equation $$\\left( \\frac{\\cos(\\theta)}{\\sin(\\theta)} + 2\\cos(\\theta) + 3 \\right)^2 -4\\left( \\frac{\\cos(\\theta)^2}{\\sin(\\theta)^2}+1 \\right)\\left(\\frac{-\\cos(\\theta)^3}{\\sin(\\theta)^2}+\\frac{\\cos(\\theta)^4}{\\sin(\\theta)^2}+3 \\right).$$\n\nThe solution $$\\theta$$ would then determine the line $$y(x) = \\frac{-\\cos(\\theta)}{\\sin(\\theta)}(x) + \\frac{\\cos(\\theta)^2}{\\sin(\\theta)} + \\sin(\\theta).$$\n\nNot only do I not want to try and solve that, I don't even want to try expanding it out :/\n\n\u2022 not $x=2$ :-) ? \u2013\u00a0J. W. Tanner May 9 at 0:32\n\u2022 \"the line\"? I see 3 common tangents after plotting the circles. \u2013\u00a0peterwhy May 9 at 0:32\n\u2022 math.stackexchange.com/questions/211538/\u2026 \u2013\u00a0lab bhattacharjee May 9 at 0:44\n\u2022 @peterwhy Hi Peter. One of those 3 solutions is so trivial, I am offended you would even bring it up. For the remaining 2, one is clearly the other ones mirror image. So I am going to stick to my wording of 'find the line' since there is clearly one difficult one to find, which is what I am interested in seeing solutions for. Let me know when you have one, you can post answers below. \u2013\u00a0Prince M May 9 at 4:47\n\nAs peterwhy points out in the comments, there are three tangent lines. By inspection, one is $$x = 2$$, as pointed out by J.W. Tanner in the comments.\n\nThe other two can be identified by similar triangles. Suppose that we have a line tangent to both circles, and let the points of tangency be $$T_1$$ and $$T_2$$. Let the circle centers be $$O_1$$ and $$O_2$$. Finally, let the point where this line intersects the $$x$$-axis be called $$P$$. Then $$\\triangle PO_1T_1$$ and $$\\triangle PO_2T_2$$ are similar (do you see why?). Since $$O_1T_1 = 2O_2T_2$$, we must have $$PO_1 = 2PO_2$$, and therefore $$P$$ must be at $$(4, 0)$$. Note that $$PT_1 = \\sqrt{3^2-1^2} = \\sqrt{8}$$, and therefore our tangent line must have slope $$\\pm \\frac{1}{\\sqrt{8}}$$.\n\n(For simplicity, I only show one of the tangent lines; the other is its mirror image across the $$x$$-axis.) From this, we get the equation of the two remaining tangent lines\n\n$$y = \\pm \\frac{x-4}{\\sqrt{8}}$$\n\n\u2022 This is a clean solution. Nice \u2013\u00a0Prince M May 9 at 4:52\n\nThe equation of the second circle can be written as $$x^2+y^2-5x+6=0$$.\n\nLet $$P(h,k)$$ be a point on the second circle.\n\nThen the equation of the tangent to the second circle at $$P$$ is $$hx+ky-\\dfrac52(x+h)+6=0$$.\n\nIf it is also a tangent to the first circle, the distance from $$(1,0)$$ to this line is $$1$$.\n\n\\begin{align*} \\frac{|h-\\frac52(1+h)+6|}{\\sqrt{(h-\\frac52)^2+k^2}}&=1\\\\ \\frac{|-\\frac32h+\\frac72|}{\\sqrt{(\\frac12)^2}}&=1\\\\ -3h+7&=\\pm1 \\end{align*} So, we have $$h=2$$ or $$h=\\frac83$$.\n\nIf $$h=2$$, $$k=\\pm\\sqrt{(\\frac12)^2-(2-\\frac52)^2}=0$$ and the common tangent is $$x-2=0$$.\n\nIf $$h=\\frac83$$, $$k=\\pm\\sqrt{(\\frac12)^2-(\\frac83-\\frac52)^2}=\\pm\\frac{\\sqrt{2}}{3}$$ and the common tangents are $$x\\pm2\\sqrt{2}y-4=0$$.\n\nUse dual conics: If we represent a circle (in fact, any nondegenerate conic) with the homogeneous matrix $$C$$ so that its equation is $$(x,y,1)C(x,y,1)^T=0$$, lines $$\\lambda x+\\mu y+\\tau=0$$ tangent to the circle satisfy the dual conic equation $$(\\lambda,\\mu,\\tau)\\,C^{-1}(\\lambda,\\mu,\\tau)^T=0$$. (This equation captures the fact that the pole of a tangent line lies on the line.)\n\nThe matrix that corresponds to the circle $$(x-h)^2+(y-k)^2=r^2$$ is $$C=\\begin{bmatrix}1&0&-h\\\\0&1&-k\\\\-h&-k&h^2+k^2-r^2\\end{bmatrix}$$ with inverse $$C^{-1}=\\frac1{r^2}\\begin{bmatrix}r^2-h^2&-hk&-h\\\\-hk&r^2-k^2&-k\\\\-h&-k&-1\\end{bmatrix}.$$ For the circles in this problem, the resulting dual equations are $$\\mu^2-\\tau^2-2\\lambda\\tau = 0 \\\\ -24\\lambda^2+\\mu^2-4\\tau^2-20\\lambda\\tau = 0.$$ Both circles\u2019 centers lie on the $$x$$-axis but their radii differ, so no common tangent is horizontal. Thus, we can set $$\\lambda=1$$ and solve the slightly simpler system to obtain the solutions $$\\mu=0$$, $$\\tau=-2$$ and $$\\mu=\\pm2\\sqrt2$$, $$\\tau=-4$$, i.e., the three common tangent lines are $$x=2 \\\\ x\\pm2\\sqrt2 y=4.$$\n\nThis general method works for any pair of nondegenerate conics: it finds their common tangents by solving a dual problem of the intersection of two (possibly imaginary) conics. For a pair of circles, however, there\u2019s a simple way to find common tangents via similar triangles, as demonstrated in Brian Tung\u2019s answer.\n\nTo find a common tangent to two arbitrary circles, you can use the following trick: \"deflate\" the smaller circle (let $$1$$) so that it shrinks to a point, while the second shrinks to the radius $$r_2-r_1$$. Doing this, the direction of the tangent doesn't change and the tangency point forms a right triangle with the two centers.\n\nBy elementary trigonometry, the angle $$\\phi$$ between the axis through the centers and the tangent is drawn from\n\n$$d\\sin\\phi=r_2-r_1,$$ where $$d$$ is the distance between the centers. The direction of the axis is such that\n\n$$\\tan\\theta=\\frac{y_2-y_1}{x_2-x_1}.$$\n\nSo the equation of the tangent is given by\n\n$$(x-x_1)\\sin(\\theta+\\phi)-(y-y_1)\\cos(\\theta+\\phi)=0.$$\n\nThe original tangent is a parallel at distance $$r_1$$, hence\n\n$$(x-x_1)\\sin(\\theta+\\phi)-(y-y_1)\\cos(\\theta+\\phi)=r_1.$$\n\n\u2022 Caution: quadrant discussion is missing. \u2013\u00a0Yves Daoust May 9 at 20:30\n\nAs already stated, there are three lines that can be drawn that are tangent to both circles. One is the trivial line $$x = 2,$$ which is tangent at the point $$(2,0)$$, which is also the common point of tangency of the two circles.\n\nThe other two lines are reflections of each other in the $$x$$-axis; these can be found by noting that a homothety that takes the point $$(0,0)$$ to $$(2,0)$$ and $$(2,0)$$ to $$(3,0)$$ will map the first circle to the second, and the center of this homothety will be a fixed point of this map.\n\nTo this end, let $$(x',y') = (a x + b, a y + d)$$, so we now solve the system $$(b, d) = (2,0) \\\\ (2a + b, d) = (3,0).$$ That is to say, $$b = 2$$, $$d = 0$$, $$a = 1/2$$, and the desired homothety is $$(x',y') = (x/2 + 2, y/2).$$ The unique fixed point is found by setting $$(x',y') = (x,y)$$ from which we obtain $$(x,y) = (4,0)$$. Thus the tangent lines pass through this point and have the form $$y = m(x - 4)$$ where $$m$$ is the slope. Such a line will be tangent if the system of equations $$(x-1)^2 + y^2 = 1, \\\\ y = m(x-4)$$ has exactly one solution. Eliminating $$y$$ gives the quadratic $$(m^2+1)x^2 - 2(4m^2+1)x + 16m^2 = 0,$$ for which the solution has a unique root if the discriminant is zero; i.e., $$0 = 4(4m^2+1)^2 - 4(m^2+1)(16m^2) = 4 - 32m^2.$$ Therefore, $$m = \\pm 1/\\sqrt{2}$$ and the desired lines are $$y = \\pm \\frac{x-4}{2 \\sqrt{2}},$$ in addition to the previous line $$x = 2$$ described.\n\nBeside the obvious solution of $$x=2$$ you also have two more common tangent lines, namely $$y= \\pm \\frac {\\sqrt {2}}{4} (x-4)$$\n\nThe two non-obvious tangent lines pass through $$(4,0)$$ and their y-intercepts are respectively $$(0,\\pm \\sqrt {2})$$", "date": "2019-07-16 18:20:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3219147/equation-of-the-line-that-lies-tangent-to-both-circles/3219165", "openwebmath_score": 0.8656613230705261, "openwebmath_perplexity": 185.1788240251181, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022630759019, "lm_q2_score": 0.8757869932689566, "lm_q1q2_score": 0.8556637121481672}}
{"url": "https://mathhelpboards.com/calculus-10/ap-calculus-exam-question-absolute-max-26537.html?s=a603e40390a25c7bfe2f8da38efa87a5", "text": "Thread: AP calculus exam question on Absolute max\n\n1. $\\text{22. Let f be the function defined by$f(x)=\\dfrac{\\ln x}{x}$What is the absolute maximum value of f ? }$\n$$(A)\\, 1\\quad (B)\\, \\dfrac{1}{e} (C)\\, 0 \\quad (D) -e \\quad (E) f\\textit{ does not have an absolute maximum value}.$$\n\nI only guessed this by graphing it and it appears to $\\dfrac{1}{e}$ which is (B)\n\n2.\n\n3. Originally Posted by karush\n$\\text{22. Let f be the function defined by$f(x)=\\dfrac{\\ln x}{x}$What is the absolute maximum value of f ? }$\n$$(A)\\, 1\\quad (B)\\, \\dfrac{1}{e} (C)\\, 0 \\quad (D) -e \\quad (E) f\\textit{ does not have an absolute maximum value}.$$\n\nI only guessed this by graphing it and it appears to $\\dfrac{1}{e}$ which is (B)\nWhy would graphing be a guess? It's a valid Mathematical tool!\n\nYou could do this by taking the derivative of f(x) and finding the critical points, etc. But if you have this question on an exam the simplest (and probably fastest) way is to take a look at each answer and see what you get. D) is out because f(x) takes on positive values, and A), C), and E) are out by looking at the graph. That leaves B).\n\n-Dan\n\n4. $f\u2019(x)=\\dfrac{x \\cdot \\frac{1}{x} - \\ln{x} \\cdot 1}{x^2} = \\dfrac{1-\\ln{x}}{x^2}$\n\n$f\u2019(x)=0$ at $x=e$\n\nfirst derivative test ...\n\n$x < e \\implies f\u2019(x) > 0 \\implies f(x) \\text{ increasing over the interval } (0,e)$\n\n$x > e \\implies f\u2019(x) < 0 \\implies f(x) \\text{ decreasing over the interval } (e, \\infty)$\n\nconclusion ... $f(e) = \\dfrac{1}{e}$ is an absolute maximum.\n\nsecond derivative test ...\n\n$f\u2019\u2019(x) = \\dfrac{x^2 \\cdot \\left(-\\frac{1}{x} \\right) - (1-\\ln{x}) \\cdot 2x}{x^4} = \\dfrac{2\\ln{x} - 3}{x^3}$\n\n$f\u2019\u2019(e) = -\\dfrac{1}{e^3} < 0 \\implies f(e) = \\dfrac{1}{e}$ is a maximum.\n\nwow that was a lot of help..\n\nyes the real negative about these assessment tests is how fast you can eliminate possible answers\nnot so much what math steps are you really need to take\n\nactually Im learning a lot here at MHB\nMahalo", "date": "2019-09-17 10:45:36", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/calculus-10/ap-calculus-exam-question-absolute-max-26537.html?s=a603e40390a25c7bfe2f8da38efa87a5", "openwebmath_score": 0.6694602966308594, "openwebmath_perplexity": 999.7213435609235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226287518852, "lm_q2_score": 0.8757869884059267, "lm_q1q2_score": 0.8556637056390554}}
{"url": "https://math.stackexchange.com/questions/2687523/to-find-the-smallest-integer-greater-than-23-with-certain-modular-residues", "text": "# To find the smallest integer greater than $23$ with certain modular residues\n\nThe question asks to find the smallest integer greater than $23$ that leaves the remainders $2,3,2$ when divided by $3,5,7$ respectively. The answer is given as $128$.\n\nI used C.R.T. to solve this. $$x\\equiv2\\pmod 3\\\\ x\\equiv3\\pmod 5\\\\ x\\equiv2\\pmod 7$$ where $M=m_1\u00b7m_2\u00b7m_3=3\u00d75\u00d77=105$.\n\nFrom here I get solutions, $$x\\equiv2\\pmod 3\\\\ x\\equiv1\\pmod 5\\\\ x\\equiv1\\pmod 7$$\nAfter that I find\\begin{align*} x &\\equiv 2\u00d72\u00d75\u00d77+1\u00d73\u00d73\u00d77+1\u00d72\u00d73\u00d75\\\\ &\\equiv 140+63+30\\\\ &\\equiv 233 \\pmod{105}, \\end{align*} where I again find that $x=23$.\n\nBut one thing I have noticed that $233=105\u00d71+128$. I cannot continue from here. I need help and any help is highly appreciated.\n\n\u2022 The \"From here I get solutions...\" is incomprehensible for me: how applying the CRT gives you those \"solutions\" and not the final one, which is unique modulo $\\;3\\cdot5\\cdot7=105\\;$ ? Mar 12 '18 at 8:34\n\u2022 @thevbm The solutions are actually given by $23+105n$. Since substituting in $n=0$ gives $23$, you can substitute the next $n$ which is $n=1$. Mar 12 '18 at 8:35\n\u2022 You could just observe that $23$ itself has the correct remainders modulo $3,5$ and $7$. By CRT the solutions form a single residue class modulo $3\\cdot5\\cdot7=105$, and you already know which class it is! So....? Mar 12 '18 at 8:35\n\nApplying directly the following proof of CRT, for example:\n\n$$\\begin{cases}y_1=\\frac{105}3=35\\;\\implies y_1=2\\pmod3\\implies y_1^{-1}=2\\pmod3\\\\{}\\\\y_2=\\frac{105}5=21\\implies y_2=1\\pmod5\\implies y_2^{-1}=1\\pmod5\\;\\\\{}\\\\y_3=\\frac{105}7=15\\implies y_3=1\\pmod7\\implies y_3^{-1}=1\\pmod7\\end{cases}$$\n\nso that\n\n$$2\\cdot35\\cdot2+3\\cdot21\\cdot1+2\\cdot15\\cdot1=233\\;\\;\\text{ is a solution, and}$$\n\n$$233=128\\pmod{105}\\;\\;\\text{is another solution, and}$$\n\n$$128=23\\pmod105\\;\\;\\text{ is the unique solution modulo}\\;105$$\n\nObserve that unless you consider things modulo $\\;3\\cdot5\\cdot7=105\\;$ at the end , there is not \"the\" solution, but only one out of infinitely many\n\n\u2022 Yes, now I understand and thank you for providing the link.\n\u2013\u00a0vbm\nMar 12 '18 at 9:03\n\u2022 Following the link you provided, one question I have (from an example given there) to make my doubt clear. suppose system of congruences is given as $x\\equiv 5 \\mod 6$ $x\\equiv 3 \\mod 8$. Here g..c.d of the moduli is $2$. The first congruence implies $x\\equiv 1\\mod 2$, also the second congruence implies $x\\equiv 1\\mod 2$Now to reduce the system of congruency to a simpler form if we divide the g.c.d of the moduli from the modulud of the first congruence, then how is it $x\\equiv2 \\mod3$? rather it should be $x\\equiv 1 \\mod 3$. If I'm wrong, please correct me. Thank you\n\u2013\u00a0vbm\nMar 12 '18 at 9:38\n\u2022 @thevbm I'm not sure what your doubt exactly is, but remember the CRT applies for coprime moduli, so reducing your moduli as you did can help...but then you should be careful when going back to the original ones. Mar 12 '18 at 11:53\n\nLet $x\\equiv y \\equiv 2 \\mod 3,$ and $x\\equiv y \\equiv 3 \\mod 5,$ and $x\\equiv y \\equiv 2 \\mod 7.$ Then $x-y \\equiv 0$ modulo $3,5,$ and $7.$\n\nSo $x-y$ is divisible by $3,5,$ and $7,$ so $x-y$ is divisible by $(3)(5)(7)=105.$ So if $y=23$ and $x>y$ then $x=23+105n$ for some $n\\in \\Bbb N.$\n\nAnd if $x=23+105n$ for some $n\\in \\Bbb N$ then $x\\equiv 23$ modulo $3,5,$ and $7.$", "date": "2021-12-08 23:41:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2687523/to-find-the-smallest-integer-greater-than-23-with-certain-modular-residues", "openwebmath_score": 0.9270637631416321, "openwebmath_perplexity": 232.34931110155742, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.967899289579129, "lm_q2_score": 0.8840392848011834, "lm_q1q2_score": 0.8556609957191067}}
{"url": "https://cs.stackexchange.com/questions/98756/sum-of-unique-elements-in-all-sub-arrays-of-an-array", "text": "# Sum of unique elements in all sub-arrays of an array\n\nGiven an array $$A$$, sum the number of unique elements for each sub-array of $$A$$. If $$A = \\{1, 2, 1, 3\\}$$ the desired sum is $$18$$.\n\nSubarrays:\n\n{1} - 1 unique element\n{2} - 1\n{1} - 1\n{3} - 1\n{1, 2} - 2\n{2, 1} - 2\n{1, 3} - 2\n{1, 2, 1} - 2\n{2, 1, 3} - 3\n{1, 2, 1, 3} - 3\n\n\nI have a working solution which sums the unique elements for all sub-arrays starting at index $$0$$, then repeats that process at index $$1$$, etc. I have noticed that for an array of size $$n$$ consisting of only unique elements, the sum I desire can be found by summing $$i(i + 1) / 2$$ from $$i = 1$$ to $$n$$, but I haven't been able to extend that to cases where there are non-unique elements. I thought I could use that fact on each sub-array, but my control logic becomes unwieldy. I've spent considerable time trying to devise a solution better than $$O(n^2)$$ to no avail. Is there one?\n\nSecondary question: if there is no better solution, are there any general guidelines or hints to recognize that fact?\n\n\u2022 An incremental algorithm is supposedly easier to come up with in this case, I think so. \u2013\u00a0Thinh D. Nguyen Oct 18 '18 at 7:16\n\u2022 I don't understand the line {1, 2, 1} - 2. The multiset {1, 2, 1} only contains one unique element: 2. \u2013\u00a0Peter Taylor Oct 18 '18 at 10:28\n\u2022 Peter, yes number of distinct elements might be better phrasing. If all elements of each sub-array are added to a set, the set size is the desired metric. In either case, the behavior in my example is what I'm looking for. \u2013\u00a0User12345654321 Oct 18 '18 at 17:34\n\nHint: Use an extra $$O(n)$$-spaced array of pointer to the last (biggest) index of each distinct value. Then, this array is very helpful when you do your above algorithm.\n\nLastly, we should have an $$O(n)$$ algorithm.\n\n\u2022 This was a good hint. Assume zero-indexed arrays. Starting with sum = maxPossibleSum, iterate through the array adding elements to a set. If the added element is a duplicate, sum = sum - (indexPreviousOccurrence + 1) * (totalElements - currentIndex). This requires just one pass leading to O(n) like you said. Thanks for a quality nudge in the right direction. \u2013\u00a0User12345654321 Oct 18 '18 at 20:31\n\nIf the maximum number $$k$$ is not too high, a Counting Sort based solution can be used;\n\nIn the first step, we count the number of elements into the array $$C$$ by\n\nfor j = 1 to length(A)\nC[A[j]] = C[A[j]]+1\n\n\nnow the array contains all the information we need to sum;\n\nsum = 0;\nfor j = 1 to k\nif C[i] > 1\nsum =  C[i] * i\n\n\nComplexity:\n\n$$\\mathcal{O}(n)$$ additions (increment) for the counting step\n\n$$\\mathcal{O}(n)$$ additions for the summation step\n\n$$\\mathcal{O}(n)$$ multiplications for the summation step\n\nresult $$\\mathcal{O}(n)$$.\n\nNote: indeed the multiplications are not necessary since we will have at most $$n-1$$ addition.\n\nLet $$a_1,\\ldots,a_m$$ be the distinct values. Now consider the positions of $$a_i$$'s in $$A$$. Assume the number of $$a_i$$'s is $$b_i$$ and the positions are as follows:\n\n(x_{i0} non-a_i's) a_i (x_{i1} non-a_i's) a_i ... a_i (x_{ib_i} non-a_i's)\n\n\nIn your example $$A=\\{1,2,1,3\\}$$, when considering the value $$a_1=1$$, we have $$x_{10}=0,x_{11}=1,x_{12}=1$$ because the positions of $$1$$s are like 1 * 1 *: there is $$0$$ element before the first $$1$$, $$1$$ element between the first $$1$$ and the second $$1$$, and $$1$$ element after the second $$1$$.\n\nThen there are \\begin{align} &\\sum_{j=0}^{b_i} \\frac{x_{ij}(x_{ij}+1)}{2}\\\\ =\\ &\\frac{1}{2}\\sum_{j=0}^{b_i}x_{ij}^2+\\frac{1}{2}\\sum_{j=0}^{b_i}x_{ij}\\\\ =\\ &\\frac{1}{2}\\sum_{j=0}^{b_i}x_{ij}^2+\\frac{1}{2}(n-b_i) \\end{align} subarrays that do not contain $$a_i$$. Note there are $$n(n+1)/2$$ subarrays in total, so the final result we want is \\begin{align} &\\sum_{i=1}^m\\left(\\frac{n(n+1)}{2}-\\frac{1}{2}\\sum_{j=0}^{b_i}x_{ij}^2-\\frac{1}{2}(n-b_i)\\right)\\\\ =\\ &\\frac{1}{2}\\left(mn^2+n-\\sum_{i=1}^m\\sum_{j=0}^{b_i}x_{ij}^2\\right). \\end{align}\n\nTo calculate $$\\sum_{i=1}^m\\sum_{j=0}^{b_i}x_{ij}^2$$, you can scan the array and maintain a lookup table that, for each distinct value, keeps the position of the last element with this value. With this table, when the $$(j+1)$$-th $$a_i$$ is scanned, you can compute $$x_{ij}$$ easily. This leads us to an $$O(n\\log n)$$ solution (or $$O(n)$$ in average if you use a good hash table to implement the lookup table).\n\n\u2022 could you clarify what X represents? It looks like a 2D array. This is the part I'm not understanding \"(x_{i0} non-a_i's) a_i ...\" \u2013\u00a0User12345654321 Oct 18 '18 at 17:17\n\u2022 @User12345654321 In your example $A=\\{1,2,1,3\\}$, when considering the value $a_1=1$, we have $x_{10}=0,x_{11}=1,x_{12}=1$ because the positions of 1s are like 1 * 1 *: there is 0 element before the first 1, 1 element between the first 1 and the second 1, and 1 element after the second 1. \u2013\u00a0xskxzr Oct 19 '18 at 3:18", "date": "2019-05-27 11:41:24", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/98756/sum-of-unique-elements-in-all-sub-arrays-of-an-array", "openwebmath_score": 0.8311969637870789, "openwebmath_perplexity": 550.5579017369471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992895791291, "lm_q2_score": 0.8840392725805822, "lm_q1q2_score": 0.8556609838907956}}
{"url": "https://math.stackexchange.com/questions/2992202/why-must-r-be-put-in-modulus-for-int-infty-0-frac1-cosrxx2dx-fra/2992629", "text": "# Why must $r$ be put in modulus for $\\int^{\\infty}_0\\frac{1-\\cos(rx)}{x^2}dx=\\frac{\\pi}{2}|r|$?\n\nHere's the question: Prove that $$\\int^{\\infty}_0\\frac{1-\\cos(rx)}{x^2}~\\mathrm dx=\\frac{\\pi}{2}|r|\u00b7$$\n\nWhen I finished my working, I got $$\\frac{\\pi}{2}r$$\n\nWhy is the $$r$$ put in modulus? Is it because the graph lies entirely above the x axis?\n\n\u2022 Did you assume $r\\geq 0$ in your working? Nov 10 '18 at 3:18\n\u2022 The integral expression is visibly even in $r$. As for where the absolute value comes up in your derivation, that depends on your method, I suppose. Nov 10 '18 at 3:21\n\u2022 Nope I didn't assume r greater than 0 as I thought there wasn't any need to do so Nov 10 '18 at 3:24\n\u2022 Notice that for any real number $a$, $\\cos(a) = \\cos(-a)$. In particular, it suffices to solve the problem in the case $r \\geq 0$. Nov 10 '18 at 3:49\n\u2022 Note: the integrand is positive, so you could you get a negative answer? Nov 10 '18 at 13:55\n\nIntegrating by part: $$\\int_0^{+\\infty} \\frac {1-\\cos(rx)}{x^2}\\, \\mathrm dx = \\int_{+\\infty}^0 (1 - \\cos(rx)) \\mathrm d \\frac 1 x =\\left. \\frac {1-\\cos(rx)}x \\right\\vert_{+\\infty}^0 - r\\int_{+\\infty}^0 \\frac {\\sin(rx)}x\\, \\mathrm dx = r \\int_0^{+\\infty} \\frac {\\sin(rx)}x \\,\\mathrm dx.$$ Now $$\\DeclareMathOperator\\sgn{sgn} \\int_0^{+\\infty} \\frac {\\sin(rx)}x \\,\\mathrm dx =\\int_0^{+\\infty} \\frac {\\sin(rx)}{rx}\\,\\mathrm d(rx)= \\begin{cases}\\displaystyle \\int_0^{+\\infty} \\frac {\\sin u}u\\, \\mathrm du, & r > 0,\\\\ 0, & r=0,\\\\ \\displaystyle \\int_{-\\infty}^0 \\frac {\\sin u}u\\, \\mathrm du, & r < 0, \\end{cases} = \\sgn r \\cdot \\int_0^{+\\infty} \\frac {\\sin u}u\\, \\mathrm du = \\sgn r \\cdot \\frac \\pi 2,$$ hence the integral equals $$r \\sgn r \\cdot \\frac \\pi 2 = \\frac \\pi 2 \\vert r \\vert.$$\nThe implicit assumption could be $$r>0$$ when dealing the integral $$\\int_0^{+\\infty} \\sin(rx)\\,\\mathrm dx /x$$.\nThe other answers explain how you get $$|r|$$ in your answer, but here\u2019s why it should happen:\nNotice that $$1-\\cos(rx) \\geq 0$$ for all $$x$$, regardless of what $$r$$ is. Integrating a nonnegative function over any interval should give you a nonnegative value.", "date": "2022-01-27 15:42:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2992202/why-must-r-be-put-in-modulus-for-int-infty-0-frac1-cosrxx2dx-fra/2992629", "openwebmath_score": 0.9206446409225464, "openwebmath_perplexity": 251.6901545585949, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232924970205, "lm_q2_score": 0.8705972768020107, "lm_q1q2_score": 0.8556432820254921}}
{"url": "https://math.stackexchange.com/questions/2810593/spivaks-calculus-finding-the-formula-for-the-sum-of-a-series", "text": "# Spivak's Calculus: Finding the formula for the sum of a series. [duplicate]\n\nThis question already has an answer here:\n\nIn Spivak's Calculus, Chapter 2, the second problem of the set asks you to find a formula for the following series:\n\n$$\\sum_{i=1}^n(2i-1)$$ and $$\\sum_{i=1}^n(2i-1)^2$$ Now, for the former, this was fairly straightforward: it sums to $i^2$ and I have a proof that I'm happy with. However, for the latter, I cannot identify a pattern in terms of $n$ to begin building a proof. The series proceeds $1^2 + 3^2 + 5^2 + 7^2 + \\cdots + (2n-1)^2$, so the sum for the first few indices would be $1, 10, 35,$ and $84.$ Not only have I failed to come up with an expression of this in terms of $n$, but I'm not even sure what I should be considering to lead me to such a formula.\n\n## marked as duplicate by Hans Lundmark, Claude Leibovici, ccorn, cansomeonehelpmeout, ChristopherJun 8 '18 at 12:16\n\nThis question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.\n\n\u2022 Are you also supposed to give a direct proof by induction, or can you use other known formul\u00e6 like the sum of the first $2n$ squares? \u2013\u00a0Bernard Jun 6 '18 at 20:33\n\n## 5 Answers\n\nHint : Expand the summand & use \\begin{eqnarray*} \\sum_{i=1}^{n} i^2 =\\frac{n(n+1)(2n+1)}{6}. \\end{eqnarray*} You should get to \\begin{eqnarray*} \\sum_{i=1}^{n} (2i-1)^2 =\\frac{n(4n^2-1)}{3}. \\end{eqnarray*}\n\n\u2022 Thanks to a combination of your hint and dezdichado's below, I managed to figure it out. Thank you very much! \u2013\u00a0user242007 Jun 6 '18 at 21:44\n\nIf you know what $1^2+2^2+...+n^2=f(n)$ sums up to, then your desired sum is simply: $$1^2+3^2+...+(2n-1)^2 = f(2n) - (2^2+4^2+...+(2n)^2) = f(2n) - 4f(n).$$\n\n\u2022 Your comment helped me become more aware of incorporating existing known formulae in these exercises. Between that and Donald's comment above, I was able to solve the problem. Thank you. \u2013\u00a0user242007 Jun 6 '18 at 21:44\n\n$$\\sum_{i=1}^n (2i-1)^2$$\n\n$$=\\sum_{i=1}^n 4i^2-4i+1$$\n\n$$=4\\sum_{i=1}^n i^2-4\\sum_{i=1}^n i+\\sum_{i=1}^n1$$\n\n\u2022 Could you elaborate on this? In trying to implement it with an example value of i = 4, I end up with 81, when the actual value in the series appears to be 84. \u2013\u00a0user242007 Jun 6 '18 at 21:26\n\n$\\sum_\\limits{k=1}^n (2k - 1)^2 = \\sum_\\limits{k=1}^n 4k^2 - \\sum_\\limits{k=1}^n 4k + \\sum_\\limits{k=1}^n 1 = 4\\left(\\sum_\\limits{k=1}^n k^2\\right) - 2n(n+1) + n$\n\nYou need a way to find $\\sum_\\limits{k=1}^n k^2$\n\nThere are lots of other ways to derive this... but this one is nice.\n\nWrite the numbers in a triangle\n\n$\\begin {array}{} &&&1\\\\&&2&&2\\\\&3&&3&&3\\\\&&&\\vdots&&\\ddots\\\\n&&n&\\cdots\\end{array}$\n\nThe sum of each row is $k^2.$ And, the sum of the whole trianglular array is $\\sum_\\limits{k=1}^n k^2$\n\nTake this triangle and rotate it 60 degrees clockwise and 60 degrees counter clockwise and sum the 3 triangles together.\n\n$\\begin {array}{} &&&1\\\\&&2&&2\\\\&3&&3&&3\\end{array}+\\begin {array}{} &&&n\\\\&&n-1&&n\\\\&n-2&&n-1&&n\\\\\\end{array}+\\begin {array}{} &&&n\\\\&&n&&n-1\\\\&n&&n-1&&n-2\\end{array}$\n\nAnd we get:\n\n$\\begin {array}{} &&&2n+1\\\\&&2n+1&&2n+1\\\\&2n+1&&2n+1&&2n+1\\\\&&&\\vdots&&\\ddots\\end{array}$\n\nWhat remains how many elements are in the triangular array? $\\sum_\\limits{k=1}^n k$\n\n$3\\sum_\\limits{k=1}^n k^2 = (2n+1)\\sum_\\limits{k=1}^n k\\\\ \\sum_\\limits{k=1}^n k^2 = \\frac {n(n+1)(2n+1)}{6}$\n\nAnother way to do it is:\n\n$\\sum_\\limits{k=1}^n (k+1)^3 - k^3 = (n+1)^3-1\\\\ \\sum_\\limits{k=1}^n (3k^2 + 3k + 1)\\\\ \\sum_\\limits{k=1}^n 3k^2 + \\sum_\\limits{k=1}^n 3k + \\sum_\\limits{k=1}^n 1\\\\ \\sum_\\limits{k=1}^n 3k^2 + 3\\frac {n(n+1)}{2} + n = n^3 + 3n^2 + 3n \\\\ 3\\sum_\\limits{k=1}^n k^2 = n^3 + \\frac 32 n^2 + \\frac 12 n\\\\ \\sum_\\limits{k=1}^n k^2 = \\frac 16 (n)(n+1)(2n+1)$\n\n\u2022 I like the rotation method, that works out very nicely! \u2013\u00a0abiessu Jun 6 '18 at 20:59\n\u2022 It is nice in that it is intuitive, but the other method can be generalized. \u2013\u00a0Doug M Jun 6 '18 at 21:01\n\nYou can do it by telescoping. Given a sequence $f(n)$ define the sequence $\\Delta f(n)=f(n+1)-f(n)$ and put $(n)_m=n(n-1)\\dotsb(n-m+1)$. One can check that $\\Delta (n)_m=m(n)_{m-1}$ for $m\\geq 1$. Now note that $$(2i-1)^2=4i^2-4i+1=4[(i)_2+i] -4i+1=4(i)_2+1$$ whence $$\\Delta\\left(\\frac{4}{3}(i)_3+i\\right)=(2i-1)^2.$$ In particular $$\\sum_{i=1}^n (2i-1)^2=\\sum_{i=1}^n \\Delta\\left(\\frac{4}{3}(i)_3+i\\right)=\\left[\\frac{4}{3}(i)_3+i\\right]_{i=1}^{n+1}$$ with the notation $[g(i)]_{i=a}^b=g(b)-g(a)$.", "date": "2019-10-22 13:34:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2810593/spivaks-calculus-finding-the-formula-for-the-sum-of-a-series", "openwebmath_score": 0.7908191084861755, "openwebmath_perplexity": 356.52627115341085, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232874658904, "lm_q2_score": 0.8705972801594707, "lm_q1q2_score": 0.8556432809451938}}
{"url": "https://stats.stackexchange.com/questions/431233/probability-of-a-500-year-flood-occuring-in-the-next-100-years-comparison-of-a", "text": "# Probability of a 500 year flood occuring in the next 100 years - comparison of approaches\n\nI'm looking at this problem\n\nA $$500$$-year flood is one that occurs once in every $$500$$ years.\n\na) What is the probability of having at least $$3$$ such floods in $$500$$ years?\n\nb) What is the probability that a flood will occur within the next $$100$$ years?\n\nc) What is the expected number of years until the next flood?\n\nAttempt:\n\na) The window size is $$500$$ years, $$\\lambda=1$$ and the number of floods is a Poisson variable $$X$$. So the answer would be $$1-P(X=0)-P(X=1)-P(X=2)$$, where $$P(X=n)$$ is the Poisson pmf.\n\nc) I can consider a window size of $$1$$ year, $$\\lambda=\\frac{1}{500}$$ and consider the time to next flood as the exponential random variable $$T$$, so that $$E[T]=1/\\lambda=500$$.\n\nAre the above two correct?\n\nFinally for part b), one approach is to decide on a window size, let's say $$1$$ year, set $$\\lambda$$ appropriately (in this case $$\\frac{1}{500}$$), let $$T$$ be the time to next flood (exponential RV), and find $$P(T=1)+P(T=2)+P(T=3)+\\ldots+P(T=100)$$\n\nThe other approach is to set the window to $$100$$ years, set $$\\lambda=\\frac{100}{500}=0.2$$, model number of floods as a Poisson RV $$X$$ and find $$1-P(X=0)$$.\n\nNumerically, these approaches give slightly different answers ($$0.1811$$ vs $$0.1813$$). Which of the two approaches is better for this purpose (i.e. which gives a more accurate answer)?\n\nAlso, in the first approach to part b), instead of a one-year window, I could've taken a half-year window and set $$\\lambda=\\frac{1}{1000}$$ and summed from $$P(T=1)$$ to $$P(T=200)$$. And of course there's no limit to how small I can set that window. What is the recommended granularity of time windows for problems like these?\n\n\u2022 I would write > in the first two questions, and < in the final one - given that climate change means that all estimates based on historical rates will underestimate future rates. In other words, what we consider a 500 year flood today, will not be a 500 year flood tomorrow. Oct 18, 2019 at 12:14\n\nYour treatment of (a) and (c) are correct. In (b), the second approach is the correct one because in your first approach, you treat exponential RVs as if they're discrete. You should've done the following ($$\\lambda=1/500$$): $$P(T\\leq100)=\\int_0^{100}\\lambda e^{-\\lambda t}dt=1-e^{-100\\lambda}=1-e^{-0.2}$$ which gives the same answer as your Poisson approach. By the way, for (b), if you discretize at infinitely small steps, you get this integral.", "date": "2023-03-22 02:20:57", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/431233/probability-of-a-500-year-flood-occuring-in-the-next-100-years-comparison-of-a", "openwebmath_score": 0.738717257976532, "openwebmath_perplexity": 325.6695992791848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232884721166, "lm_q2_score": 0.8705972784807406, "lm_q1q2_score": 0.8556432801713166}}
{"url": "https://math.stackexchange.com/questions/1941023/find-f-if-fx-dfracx2-1x-knowing-that-f1-dfrac12-and-f", "text": "# Find $f$ if $f'(x)=\\dfrac{x^2-1}{x}$ knowing that $f(1) = \\dfrac{1}{2}$ and $f(-1) = 0$\n\nI am asked the following problem:\n\nFind $f$ if $$f'(x)=\\frac{x^2-1}{x}$$\n\nI am not sure about my solution, which I will describe below:\n\nMy solution:\n\nThe first thing that I've done is separate the terms of $f'(x)$\n\n\\begin{align*} f'(x)&=\\frac{x^2-1}{x}\\\\ &=x-\\frac{1}{x}\\\\ \\therefore \\quad f(x)&=\\frac{x^2}{2}-\\ln|x|+c \\end{align*}\n\nFor ( x > 0 ):\n\n\\begin{align*} f(x)&=\\frac{x^2}{2}-\\ln x+c\\\\ f(1)&=\\frac{1^2}{2}-\\ln 1+c=\\frac{1}{2} \\quad \\Rightarrow \\quad c=0\\\\ f(x)&=\\frac{x^2}{2}-\\ln |x| \\end{align*}\n\nFor ( x < 0 )\n\n\\begin{align*} f(x)&=\\frac{x^2}{2}-\\ln (-x)+c\\\\ f(-1)&=\\frac{(-1)^2}{2}-\\ln [-(-1)]+c=0 \\quad \\Rightarrow \\quad c=-\\frac{1}{2}\\\\ f(x)&=\\frac{x^2}{2}-\\ln |x|-\\frac{1}{2} \\end{align*}\n\nIs my solution correct? Should I really find two different answers, one for $x > 0$ and another for $x < 0$?\n\nThank you.\n\n\u2022 yes your solution is fine. As the domain has two connected components there can indeed be different constants of integration on each of them. \u2013\u00a0H. H. Rugh Sep 25 '16 at 17:03\n\u2022 It looks fine to me. +1 \u2013\u00a0DonAntonio Sep 25 '16 at 17:03\n\u2022 Related: math.stackexchange.com/questions/234624/\u2026 (see the accepted answer) \u2013\u00a0Winther Sep 26 '16 at 1:18\n\nLooks fine! However a little typo when calculate $f(-1)=0$ not $-1$", "date": "2021-03-04 00:31:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1941023/find-f-if-fx-dfracx2-1x-knowing-that-f1-dfrac12-and-f", "openwebmath_score": 0.9964385628700256, "openwebmath_perplexity": 998.8627920910212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232919939076, "lm_q2_score": 0.8705972751232809, "lm_q1q2_score": 0.8556432799375886}}
{"url": "https://math.stackexchange.com/questions/525334/vandermonde-determinant-by-induction", "text": "# Vandermonde determinant by induction\n\nFor $$n$$ variables $$x_1,\\ldots,x_n$$, the determinant $$\\det\\left((x_i^{j-1})_{i,j=1}^n\\right) = \\left|\\begin{matrix} 1&x_1&x_1^2&\\cdots & x_1^{n-1}\\\\ 1&x_2&x_2^2&\\cdots & x_2^{n-1} \\\\ \\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\ 1&x_{n-1}&x_{n-1}^2&\\cdots&x_{n-1}^{n-1}\\\\ 1 &x_n&x_n^2&\\cdots&x_n^{n-1} \\end{matrix}\\right|$$ can be computed by induction; the image below says it shows that. I have done this before, if I submit this will I get marks?\n\nMORE IMPORTANTLY how do I do it by induction? The \"hint\" is to get the first row to $$(1,0,0,...,0)$$.\n\nI think there are the grounds of induction in there, but I'm not sure how (I'm not very confident with induction when the structure isn't shown for $$n=k$$, assume for $$n=r$$, show if $$n=r$$ then $$n=r+1$$ is true.\n\nBy the way the question is to show the determinant at the start equals the product $$\\prod_{1\\leq i (but again, explicitly by induction)\n\n\u2022 There's a nice non-inductive proof. Would that work for you? \u2013\u00a0Bruno Joyal Oct 14 '13 at 2:38\n\u2022 Actually, my memory tricked me. There is an induction. I'm posting it below. \u2013\u00a0Bruno Joyal Oct 14 '13 at 2:52\n\u2022 ProofWiki has two proofs using mathematical induction. \u2013\u00a0user1551 Oct 14 '13 at 6:42\n\nYou're facing the matrix\n\n\\begin{pmatrix} 1&1&\\cdots & 1 &1\\\\a_1&a_2&\\cdots &a_n&a_{n+1}\\\\\\vdots&\\vdots&\\ddots&\\vdots&\\vdots\\\\a_1^{n-1}&a_2^{n-1}&\\cdots&a_n^{n-1}& a_{n+1}^{n-1}\\\\a_1^{n}&a_2^{n}&\\cdots&a_n^{n}&a_{n+1}^{n}\\end{pmatrix}\n\nBy subtracting $a_1$ times the $i$-th row to the $i+1$-th row, you get\n\n\\begin{pmatrix} 1&1&\\cdots & 1 &1\\\\ 0&a_2-a_1&\\cdots &a_n-a_1&a_{n+1}-a_1\\\\ \\vdots&\\vdots&\\ddots&\\vdots&\\vdots\\\\ 0&a_2^{n}-a_1a_{2}^{n-1}&\\cdots&a_n^n-a_1a_{n}^{n-1}& a_{n+1}^{n}-a_1a_{n+1}^{n-1}\\end{pmatrix}\n\nExpanding by the first column and factoring $a_i-a_1$ from the $i$-th column for $i=2,\\ldots,n+1$, you get the determinant is\n\n$$=\\prod_{j=2}^{n+1}(a_j-a_1) \\det\\begin{pmatrix} 1& 1&\\cdots&1\\\\a_2&a_3&\\cdots&a_{n+1}\\\\ \\vdots&\\vdots&\\ddots&\\vdots\\\\a_2^{n-1}&a_3^{n-1}&\\cdots&a_{n+1}^{n-1}\\end{pmatrix}$$\n\nYou may apply your inductive hypothesis, to get this is $$=\\prod_{j=2}^{n+1}(a_j-a_1) \\prod_{2\\leqslant i<j\\leqslant n+1}(a_j-a_i)=\\prod_{1\\leqslant i<j\\leqslant n+1}(a_j-a_i)$$ and the inductive step is complete.\n\n\u2022 Can you please explain how it is possible to factor out $a_i - a_1$? \u2013\u00a0john.abraham Sep 2 '15 at 10:05\n\u2022 $a_j^{i} - a_1 a_j^{i-1} = a_j^{i-1}(a_j - a_1)$ \u2013\u00a0D_S Oct 1 '15 at 4:59\n\u2022 Even though it is clearly stated, gratuitous change of notation with respect to the question (transposing the matrix and renaming the $x_i$ to $a_i$) is not really helpful, especially in the presence of another answer that does keep the original notation (but uses $a_i$ for a different purpose). \u2013\u00a0Marc van Leeuwen Dec 3 '18 at 10:17\n\nLet $f(T) = T^{n-1} + a_1 T^{n-2} + \\dots + a_{n-1}$ be the polynomial $(T-x_2)(T-x_3)\\dots (T-x_n)$. By adding to the rightmost column an appropriate linear combination of the other columns (namely the combination with coefficients $a_1, \\dots, a_{n-1}$), we can make sure that the last column is replaced by the vector $(f(x_1), 0, 0, \\dots, 0)$, since by construction $f(x_2) = \\dots = f(x_n) =0$. Of course this doesn't change the determinant. The determinant is therefore equal to $f(x_1)$ times $D$, where $D$ is the determinant of the $(n-1) \\times (n-1)$ matrix which is obtained from the original one by deleting the first row and the last column. Then, we apply the induction hypothesis, using the fact that $f(x_1) = (x_1 - x_2)(x_1-x_3) \\dots (x_1-x_n)$. And we are done!\n\nBy the way, this matrix is known as a Vandermonde matrix.\n\nI learned this trick many years ago in Marcus' Number fields.\n\n\u2022 Sexy! I should read this book if it has more fancy tricks like that. \u2013\u00a0Patrick Da Silva Dec 24 '13 at 15:06\n\u2022 You get the wrong product (it should have factors $x_j-x_i$ when $i<j$, rather than factors $x_i-x_j$, or else you should throw in a sign $(-1)^{\\binom n2}$ for the Vandermonde determinant). The explanation is that you produce a factor $f(x_1)$ that is not on the main diagonal, and this means in the induction step you get a sign $(-1)^{n-1}$ in the development of the determinant. This sign could have been avoided by producing a nonzero coefficient in the last entry of the final column instead, using a monic polynomial of degree $n-1$ with roots at $x_1,\\ldots,x_{n-1}$ but not at $x_n$. \u2013\u00a0Marc van Leeuwen Dec 3 '18 at 10:10\n\nAlthough this does not really answer the question (in particular, I cannot tell whether you whether submitting the image will give you any marks), it does explain the evaluation of the determinant (and one can get an inductive proof from it if one really insists, though it would be rather similar to the answer by Bruno Joyal).\n\nThe most natural setting in which the Vandermonde matrix $$V_n$$ arises is the followig. Evaluating a polynomial over$$~K$$ in each of the points $$x_1,\\ldots,x_n$$ of $$K$$ gives rise to a linear map $$K[X]\\to K^n$$ i.e., the map $$f:P\\mapsto (P[x_1],P[x_2],\\ldots,P[x_n])$$ (here $$P[a]$$ denotes the result of substituting $$X:=a$$ into$$~P$$). Then $$V_n$$ is the matrix of the restriction of $$f$$ to the subspace $$\\def\\Kxn{K[X]_{ of polynomials of degree less than$$~n$$, relative to the basis $$[1,X,X^2,\\ldots,X^{n-1}]$$ of that subspace. Any family $$[P_0,P_1,\\ldots,P_{n-1}]$$ of polynomials in which $$P_i$$ is monic of degree$$~i$$ for $$i=0,1,\\ldots,n-1$$ is also a basis of$$~\\Kxn$$, and moreover the change of basis matrix$$~U$$ from the basis $$[1,X,X^2,\\ldots,X^{n-1}]$$ to $$[P_0,P_1,\\ldots,P_{n-1}]$$ will be upper triangular with diagonal entries all$$~1$$, by the definition of being monic of degree$$~i$$. So $$\\det(U)=1$$, which means that the determinant of$$~V_n$$ is the same as that of the matrix$$~M$$ expressing our linear map on the basis $$[P_0,P_1,\\ldots,P_{n-1}]$$ (which matrix equals $$V_n\\cdot U$$).\n\nBy choosing the new basis $$[P_0,P_1,\\ldots,P_{n-1}]$$ carefully, one can arrange that the basis-changed matrix is lower triangular. Concretely column$$~j$$ of$$~M$$, which describes $$f(P_{j-1})$$ (since we number columns from$$~1$$), has as entries $$(P_{j-1}[x_1],P_{j-1}[x_2],\\ldots,P_{j-1}[x_n])$$, so lower-triangularity means that $$x_i$$ is a root of $$P_{j-1}$$ whenever $$i. This can be achieved by taking for $$P_k$$ the product $$(X-x_1)\\ldots(X-x_k)$$, which is monic of degree $$k$$. Now the diagonal entry in column$$~j$$ of$$~M$$ is $$P_{j-1}[x_j]=(x_j-x_1)\\ldots(x_j-x_{j-1})$$, and $$\\det(V_n)$$ is the product of these expressions for $$j=1,\\ldots,n$$, which is $$\\prod_{j=1}^n\\prod_{i=1}^{j-1}(x_j-x_i)=\\prod_{1\\leq i.", "date": "2019-08-19 21:30:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/525334/vandermonde-determinant-by-induction", "openwebmath_score": 0.9721505045890808, "openwebmath_perplexity": 672.4633188289199, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982823294006359, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8556432767399386}}
{"url": "https://www.physicsforums.com/threads/sum-of-log-squared-terms.627139/", "text": "# Sum of log squared terms?\n\n1. Aug 10, 2012\n\n### yogo\n\nHi all,\n\nI have a summation series which goes like this,\n\nS = [log(1)]^2 + [log(2)]^2 + [log(3)]^2 + ....... + [log(n)]^2\n\nSince each term is square of the logarithm of a value, I think there are no general tricks which can be used to solve this.\n\nBut is there any way to approximate the value of this summation?\n\nThe problem I am facing here is that n is very very large and it is not possible (time-wise and memory-wise) for my program to evaluate every term. So I was hoping for some approximate solution in terms of 'n'.\n\nThanks,\nyogo\n\n2. Aug 10, 2012\n\n### micromass\n\nWell, we know that since $f(x)=(log(x))^2$ is increasing (if x>1)\n\n$$\\int_2^{n+1} \\log(x-1)^2dx \\leq \\sum_{n=1}^n \\log(n)^2 \\leq \\int_1^{n+1} \\log(x)^2 dx$$\n\nSo, if you calculate the integral, then you got a sweet upper and lower bound for the sum.\n\n3. Aug 11, 2012\n\n### coolul007\n\ncan't this be a manipulated into a product of n? log(2)+log(3) = log((2)(3)), I'm uncertain what to do with the squared.\n\n4. Aug 11, 2012\n\n### micromass\n\nNot really. But the integral\n\n$$\\int \\log(x)^2dx$$\n\nhas a nice elementary solution. Just integrate by parts twice.\n\n5. Aug 11, 2012\n\n### Mute\n\nIn a similar vein, there is also the Euler-Maclaurin formula for asymptotic expansions of sums.\n\n$$\\sum_{k=a}^{k=n} f(k) \\sim \\int_a^n dx~f(x) + \\frac{f(a)+f(n)}{2} + \\mbox{corrections}.$$\n\n(The full form of the expansion is given on the wikipedia page I linked to). This looks like it would be useful in this case.\n\n6. Aug 11, 2012\n\n### yogo\n\nThanks for your help guys. Appreciate it.\n\n@micromass: I am not so good at Math. I was wondering how tight these bounds are?\n\nThanks,\nyogo\n\n7. Aug 15, 2012\n\n### chingel\n\nYou would have to calculate the integral to find out. Or you can let an online calculator do it. To find the tightness of the bonds, calculate the difference between the integrals, or in other words the integral from n to n+1.\n\nhttp://www.wolframalpha.com/widgets/view.jsp?id=8ab70731b1553f17c11a3bbc87e0b605\n\nUsing Wolfram Alpha the integral from 10 to 11 for example is ~5.5, so if you take the average of them you get a maximal error of ~2.75, actual error would probably be much less, since the function doesn't jump too much, and is rather straight for intervals of width 1. Lower bound is ~25, and using Wolfram Alpha to calculate the sum directly gives ~27.65. If you want more accuracy, you could calculate a few terms of the sum directly, and then get the upper and lower bounds for the rest, for the upper bound it is the sum from 1 to m-1 plus the integral from m to n+1, and for the lower bound the sum from 1 to m-1 plus the integral from m-1 to n. The difference of the upper and lower bounds now turns out to be the integral from n to n+1 minus the integral from m-1 to m, which is better than just the integral from n to n+1.\n\nSo for example if you want the sum of the first 1000 terms and you know the first 100 terms exactly, you could do something like this. Sum of first 100 terms is 1408.33. Lower bound is 1408.33+34501.8=35910.13, higher bound is 1408.33+34528.3=35936.63, average is 35923.38. Letting Wolfram Alpha calculate the sum itself, it gets the answer 35923.4, I don't know how it calculates the sum itself and how much it rounded that answer, but at least it shows that taking the average of the bounds is not too bad. So now you have an approximation of the sum of the first 1000 terms, you can now use that to calculate the sum of the first 10000 terms etc.\n\n8. Aug 16, 2012\n\n### chingel\n\nI also tried the Euler-Maclaurin formula. Adding only first two terms from the sum with the Bernoulli numbers in it, it got the sum of the first 1000 terms of log(x)^2 accurate to within 0.0015, so it seems a lot better than just taking the average of integral bounds, especially if you would include more of the correction terms. You could probably get very accurate with more correction terms.\n\n9. Aug 16, 2012\n\n### yogo\n\n@chingel, @mute: Thanks for all the help. I am using Euler-Maclaurin formula now, seems to work really well for me. I actually have an upcoming deadline, and now I am good! :)\n\nThanks all.", "date": "2018-08-20 15:13:42", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/sum-of-log-squared-terms.627139/", "openwebmath_score": 0.8378165364265442, "openwebmath_perplexity": 391.512170859083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232914907946, "lm_q2_score": 0.8705972700870909, "lm_q1q2_score": 0.855643274549895}}
{"url": "https://math.stackexchange.com/questions/2792492/minimizing-a-symmetric-sum-of-fractions-without-calculus", "text": "# Minimizing a symmetric sum of fractions without calculus\n\nCalculus tools render the minimization of $$\\frac{(1-p)^2}{p} + \\frac{p^2}{1-p}$$ on the interval $(0,1/2]$ to be a trivial task. But given how much symmetry there is in this expression, I was curious if one could use olympiad-style manipulations (AM-GM, Cauchy Schwarz, et al) to minimize this, show it is decreasing, or to show that it is bounded below by 1.\n\n\u2022 The squares alone tell me to use Cauchy-Schwarz, but I can't see how exactly without screwing it up. \u2013\u00a0Sean Roberson May 23 '18 at 6:11\n\u2022 I applied AM-GM directly on the expression to bound it below by $2 \\sqrt{p(1-p)}$, but that is a pretty bad bound for $p$ smaller than $1/2$. \u2013\u00a0user369210 May 23 '18 at 6:13\n\u2022 There is no minimum of the function on (0, 1/2). There is a minimum on (0, 1/2] which occurs at the boundary 1/2. What am I missing? \u2013\u00a0Andreas May 23 '18 at 6:21\n\u2022 Infimum, fine, same thing \u2013\u00a0user369210 May 23 '18 at 6:27\n\nThis is elementary:\n\n$$\\frac{(1-p)^2}p+\\frac{p^2}{1-p}=\\frac{1-3p+3p^2-p^3+p^3}{p(1-p)}=\\frac1{p(1-p)}-3.$$\n\nThe minimum is achieved by the vertex of the parabola $p(1-p)$, i.e. $p=\\dfrac12\\to\\dfrac1{\\frac14}-3$, as can be shown by completing the square.\n\nBy Rearrangement inequality assuming wlog $p\\ge(1-p)$\n\n$$\\frac{(1-p)^2}{p} + \\frac{p^2}{1-p}=(1-p)\\frac{1-p}{p} + p\\frac{p}{1-p}\\ge (1-p)\\frac{p}{1-p} + p\\frac{1-p}{p}=p+1-p=1$$\n\nwith equality for\n\n$$p=1-p \\implies p=\\frac12$$\n\n\u2022 The problem statement actually implies that $p \\leq 1 - p.$ But of course this method still works, just swap $p$ and $1 - p.$ \u2013\u00a0David K May 24 '18 at 1:23\n\n** I modified this to give a proof with elementary methods only, no inequality theorems needed**\n\nSimple transformations lead to $$\\frac{(1-p)^2}{p} + \\frac{p^2}{1-p} = - 3 + \\frac{4}{1 - 4 (p-1/2)^2}$$ Now it is easy to see that the global minimum of $f(p)$ for $p \\in (0 , \\; 1/2]$ occurs at $p = 1/2$.\n\n\u2022 You've only shown that it is a local minimum. \u2013\u00a0Kenny Lau May 23 '18 at 6:30\n\u2022 The \"small $x$\" part invokes calculus, which I am trying to avoid \u2013\u00a0user369210 May 23 '18 at 6:33\n\n$$\\begin{array}{rcll} \\dfrac{(1-p)^2}{p} + \\dfrac{p^2}{1-p} &=& \\dfrac {(1-p)^3 + p^3} {p (1-p)} \\\\ &=& \\dfrac {[(1-p) + p] [(1-p)^2 - p(1-p) + p^2]} {p (1-p)} \\\\ &=& \\dfrac {(1-p)^2 + p^2} {p (1-p)} - 1 \\\\ &\\ge& \\dfrac {\\left[ \\frac1{\\sqrt2} (1-p) + \\frac1{\\sqrt2} p \\right]^2} {p (1-p)} - 1 &\\text{Cauchy-Schwarz} \\\\ &=& \\dfrac 1 {2 p (1-p)} - 1 \\\\ &=& \\dfrac 1 {0.5 - 2 (p-0.5)^2} - 1 \\\\ &\\ge& \\dfrac 1 {0.5} - 1 \\\\ &=& 1 \\end{array}$$\n\nBoth $\\ge$ has equality at $p = 0.5$.\n\n$$\\dfrac{(1-p)^2}p+\\dfrac{p^2}{1-p}=\\dfrac{1-3p+3p^2}{p(1-p)}=K\\text{(say)}$$\n\n$$\\implies p^2(3+K)-p(3+K)+1=0$$\n\nAs $p$ is real, the discriminant must be $\\ge0$\n\n$$0\\le(K+3)^2-4(K+3)=(K+3)(K-1)\\implies$$\n\neither $K\\ge$max$(1,-3)$\n\nor $K\\le$min$(1,-3)$\n\nBut observe that $K+3\\ne0$\n\nMake the change: $a=1-p, b=p, p\\in \\left(0,\\frac12\\right]$.\n\nThen: $a+b=1, ab\\le \\frac14,$ the equality occurs when $a=b=\\frac12$.\n\nHence: $$\\frac{(1-p)^2}{p} + \\frac{p^2}{1-p}=\\frac{a^2}{b} + \\frac{b^2}{a}=\\frac{a^3+b^3}{ab}=\\frac{(a+b)((a+b)^3-3ab)}{ab}\\ge \\frac{1\\cdot \\left(1^3-\\frac34\\right)}{\\frac14}=1.$$\n\nLet's utilize the symmetry:\n\n$$p\\equiv\\frac 1 2 -q$$\n\n$$1-p\\equiv\\frac 1 2 +q$$\n\nThen\n\n$$p \\in (0,1/2] \\iff q \\in [0,1/2)$$\n\n$$\\frac{(1-p)^2}{p} + \\frac{p^2}{1-p} \\equiv \\frac{(\\frac 1 2 +q)^2}{\\frac 1 2 -q} + \\frac{(\\frac 1 2 -q)^2}{\\frac 1 2 +q} \\equiv \\dots$$\n\n(here be magic, try it)\n\n$$\\dots \\equiv \\frac{\\frac 1 4 + 3q^2}{\\frac 1 4 -q^2} \\equiv - 3 + \\frac{4}{1 - 4q^2}$$\n\nNote this is the result already obtained in another answer, however manipulations with $p$ (there) are not as elegant as manipulations with $q$ (here).\n\nIt's now easy to tell $q=0$ maximizes the denominator which is always positive (in the domain), so the last fraction is minimized.\n\n$$q=0 \\iff p=\\frac 1 2$$", "date": "2020-07-13 12:16:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2792492/minimizing-a-symmetric-sum-of-fractions-without-calculus", "openwebmath_score": 0.9094794988632202, "openwebmath_perplexity": 427.9299878028572, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232935032463, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.8556432631027697}}
{"url": "https://math.stackexchange.com/questions/3178532/how-is-the-relation-the-smallest-element-is-the-same-reflexive/3178541", "text": "# How is the relation \u201cthe smallest element is the same\u201d reflexive?\n\nLet $$\\mathcal{X}$$ be the set of all nonempty subsets of the set $$\\{1,2,3,...,10\\}$$. Define the relation $$\\mathcal{R}$$ on $$\\mathcal{X}$$ by: $$\\forall A, B \\in \\mathcal{X}, A \\mathcal{R} B$$ iff the smallest element of $$A$$ is equal to the smallest element of $$B$$. For example, $$\\{1,2,3\\} \\mathcal{R} \\{1,3,5,8\\}$$ because the smallest element of $$\\{1,2,3\\}$$ is $$1$$ which is also the smallest element of $$\\{1,3,5,8\\}$$.\n\nProve that $$\\mathcal{R}$$ is an equivalence relation on $$\\mathcal{X}$$.\n\nFrom my understanding, the definition of reflexive is:\n\n$$\\mathcal{R} \\text{ is reflexive iff } \\forall x \\in \\mathcal{X}, x \\mathcal{R} x$$\n\nHowever, for this problem, you can have the relation with these two sets:\n\n$$\\{1\\}$$ and $$\\{1,2\\}$$\n\nThen wouldn't this not be reflexive since $$2$$ is not in the first set, but is in the second set?\n\nI'm having trouble seeing how this is reflexive. Getting confused by the definition here.\n\n\u2022 Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $\\{ 1 \\} \\mathcal R \\{ 1,2 \\}$ but we have also $\\{ 1 \\} \\mathcal R \\{ 1 \\}$ and $\\{ 1,2 \\} \\mathcal R \\{ 1,2 \\}$ \u2013\u00a0Mauro ALLEGRANZA Apr 7 at 17:34\n\u2022 Note: \u201creflexive\u201d does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$. \u2013\u00a0Arturo Magidin Apr 7 at 17:44\n\u2022 So it must be reflexive because both $A$ and $B$ belong to the same set $\\mathcal{X}$? \u2013\u00a0qbuffer Apr 7 at 18:00\n\u2022 @qbuffer Have a look at the updated version of my answer. \u2013\u00a0Haris Gusic Apr 7 at 18:49\n\nWhy are you testing reflexivity by looking at two different elements of $$\\mathcal{X}$$? The definition of reflexivity says that a relation is reflexive iff each element of $$\\mathcal X$$ is in relation with itself.\nTo check whether $$\\mathcal R$$ is reflexive, just take one element of $$\\mathcal X$$, let's call it $$x$$. Then check whether $$x$$ is in relation with $$x$$. Because $$x=x$$, the smallest element of $$x$$ is equal to the smallest element of $$x$$. Thus, by definition of $$\\mathcal R$$, $$x$$ is in relation with $$x$$. Now, prove that this is true for all $$x \\in \\mathcal X$$. Of course, this is true because $$\\min(x) = \\min(x)$$ is always true, which is intuitive. In other words, $$x \\mathcal{R} x$$ for all $$x \\in \\mathcal X$$, which is exactly what you needed to prove that $$\\mathcal R$$ is reflexive.\nYou must understand that the definition of reflexivity says nothing about whether different elements (say $$x,y$$, $$x\\neq y$$) can be in the relation $$\\mathcal R$$. The fact that $$\\{1\\}\\mathcal R \\{1,2\\}$$ does not contradict the fact that $$\\{1,2\\}\\mathcal R \\{1,2\\}$$ as well.\nA binary relation $$R$$ over a set $$\\mathcal{X}$$ is reflexive if every element of $$\\mathcal{X}$$ is related to itself. The more formal definition has already been given by you, i.e. $$\\mathcal{R} \\text{ is reflexive iff } \\forall x \\in \\mathcal{X}, x \\mathcal{R} x$$", "date": "2019-05-26 17:26:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3178532/how-is-the-relation-the-smallest-element-is-the-same-reflexive/3178541", "openwebmath_score": 0.8572835922241211, "openwebmath_perplexity": 86.55734743884899, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232889752296, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8556432624604811}}
{"url": "http://vnbk.minoronzitti.it/probability-examples-and-solutions.html", "text": "# Probability Examples And Solutions\n\nFor example, in a large number of coin tosses, heads and tails will occur each about 50% of the time. The solutions are not intended to be as polished as the proofs in the book, but are supposed to give enough of the details so that little is left to the. The above facts can be used to help solve problems in probability. We request all visitors to read all examples carefully. Ch4: Probability and Counting Rules Santorico \u2013 Page 105 Event \u2013 consists of a set of possible outcomes of a probability experiment. Something with a probability of 0 is something that could never possibly happen. Aptitude Made Easy - Probability \u2013 7 Tricks to solve problems on Balls and bags \u2013 Part 1 - Duration: 6:57. Spector and Mazzeo examined the effect of a teaching method known as PSI on the performance of students in a course, intermediate macro economics. If a woman does not have breast cancer, the probability is 7 percent that she will still have a positive mammogram. Show Step-by-step Solutions. Probability Theory and Examples - Durrett. The at least once rule gives us the probability of at least one flood in 100 years: P(ALO in 100 years)=1-[P(not flood in one year)]100 =1-[0. Referring to the earlier example (from Unit 3 Module 3) concerning the National Requirer. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. There 2 numbers that are odd and less than 5: 1 and 3. You use probability in daily life to make decisions when you don't know for sure what the outcome will be. For example, one way to partition S is to break into sets F and Fc, for any event F. ) the second time will be the same as the first (i. 1 of winning. As a member, you'll also get unlimited access to over 79,000 lessons in math, English, science, history, and more. When we flip a coin there is always a probability to get a head or a tail is 50 percent. In the case of a coin, there are maximum two possible outcomes \u2013 head or tail. The continuous random variables deal with different kinds of distributions. Rick Durrett Probability Theory And Examples Solution Manual. This video shows examples of using probability trees to work out the overall probability of a series of events are shown. ) and finding about the probability of two things happening in that one task. Exercise 15. An example for non-mutually exclusive events could be: E 1 = students in the swimming team. Solution: The chances of landing on blue are 1 in 4, or one fourth. NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13. For example, when rolling a number cube 600 times, predict that a 3 or 6 would be rolled roughly 200 times, but probably not exactly 200 times. The pictures below depict the probability distributions in space for the Hydrogen wavefunctions. The binomial probability calculator will calculate a probability based on the binomial probability formula. Solutions will be gone over in class or posted later. Are passing these subjects dependent on each other? @312; Hallow): was, Ptmrk): (My etchm GMI\u2019k): 011 NW Newman-m): mun-4: 2-045\u201d. The Best NCERT Books , CBSE Notes , Sample Papers and NCERT Solutions PDF Free Download. For example, in a function machine, a number goes in, something is done to it, and the resulting number is the output: probability: The measure of how likely it is for an event to occur. The denominator 10 because there are 10 fruits in the basket (the total number of outcomes). We can use a bar chart, called a probability distribution histogram , to display the probabilities that X lies in selected ranges. Answer: The probability of being able to exercise outdoors every day of the year where Sam now lives is (:99)365 \u02c7:0255 \u02c73% so there is about a 97% probability of not being able to exercise outdoors one or more days of the year. In the \"die-toss\" example, the probability of event A, three dots showing, is P(A) = 1 6 on a single toss. Suppose a coin tossed then we get two possible outcomes either a 'head' ( H ) or a 'tail' ( T ), and it is impossible to predict whether the result of a toss will be a. Denote by #Adenote the number of point in A. Class 11 Maths Probability Ex 16. The solutions to these problems are at the bottom of the page. Let E = Event of drawing 2 balls, none of which is blue. This video shows examples of using probability trees to work out the overall probability of a series of events are shown. First we find the probability of each event. So before seeing the patient I'm already able to identify the two most likely diagnoses and assign an initial probability for each. The solutions given to the questions for the in between exercises and exercises given at the end of the chapter are prepared by our subject matter experts in a simple and lucid language. Team A wins the series in two games with probability 1/4. All NCERT Questions, Examples and Exercises are solved with step-by-step answers to all questions explained in detail. Answers and links to explanations to these these GMAT probability problems are at the end of set. 2 List the elements of the following sets. Introduction to Conditional Probability Some Examples A \"New\" Multiplication Rule Conclusion Conditional Probability Here is another example of Conditional Probability. In reality, I'm not particularly interested in using this example just so that you'll know whether or not you've been ripped off the next time you order a hamburger! Instead, I'm interested in using the example to illustrate the idea behind a probability density function. Therefore, both the events will have half probability. On in\ufb01nite sums19 6. Not all Markov chains. Hope you like them and do not forget to like , social share and comment at the end of the page. Bayes' theorem is a mathematical equation used in probability and statistics to calculate conditional probability. 1 of the bags is selected at random and a ball is drawn from it. Get Free NCERT Solutions for Class 11 Maths Chapter 16 Probability. Theoretical probability is an approach that bases the possible probability on the possible chances of something happen. Probability: Theory and Examples. Expected Value 7. If none of the possible outcomes are favorable for a certain event (a favorable outcome is impossible), the probability is 0. ) the second time will be the same as the first (i. The problem is: we want 1 novel and 1 reference work, but we don\u2019t care which one was picked first. These notes can be used for educational purposes, pro-vided they are kept in their original form, including this title page. Probability concept includes some terms. 1 Probability Spaces. Measure Theory 1. Even though we discuss two events (usually labeled A and B), we\u2019re really talking about performing one task (rolling dice, drawing cards, spinning a spinner, etc. Find the Standard Deviation of a random variable X whose probability density function is given by f(x) where: Solution. develop the theory, we will focus our attention on examples. Listed in the following table are problem sets and solutions. P(AjB) is read The probability of A given B. a Show that $$f\\left( x \\right)$$ is a probability density function. The mean is $100(7) + 200(4) = 1,500$ and the standard deviation is. Walpole Raymond H. The probability distribution given is discrete and so we can find the variance from the following: We need to find the mean \u03bc first: Then we find the variance: Example 2. Instead of events being labeled A and B, the norm is to use X and Y. EXAMPLE: Suppose that three slips of paper have the names a, b, c. Here different types of examples will help the students to understand the problems on probability with playing cards. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Return from Examples of Probability Problems to Free Math Videos Online or to 8th Grade CTCT Math. The geometric probability is the area of the desired region (or in this case, not so desired), divided by the area of the total region. com Jobs & Careers 874,804 views 6:57. com or just starting a probability unit, you may want to take a look at the introductory probability lesson or the lesson on independent events. We randomly select 5 balls. Readers with a solid background in measure theory can skip Sections 1. random variables that take a discrete set of values. Kroese School of Mathematics and Physics The University of Queensland c 2018 D. Conditional probability is the probability of one event occurring with some relationship to one or more other events. This solutions manual provides answers for the even-numbered exercises in Probability and Statistical Inference, 8th edition, by Robert V. It is also suitable for self-study. Probability and statistics, the branches of mathematics concerned with the laws governing random events, including the collection, analysis, interpretation, and display of numerical data. If it isn't a trick coin, the probability of each simple outcome is the same. Example: toss a coin 100 times, how many Heads will come up? Probability says that heads have a \u00bd chance, so we can expect 50 Heads. Bonferroni's inequalities28 9. All Chapter 25 - Probability Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. For example activity H here, we have the same exact thing, expected duration of 3 plus 4 times 6 plus 9 all divided by 6 to give you 6. Answers and links to explanations to these these GMAT probability problems are at the end of set. com or just starting a probability unit, you may want to take a look at the introductory probability lesson or the lesson on independent events. mathematics of probability theory, but also, through numerous examples, the many diverse possible applications of this subject. As we study a few probability problems, I will explain how \"replacement\" allows the events to be independent of each other. Probability calculator is a online tool that computes probability of selected event based on probability of other events. A life insurance salesman sells on the average 3 life insurance policies per week. These NCERT solutions play a. What is the probability to get a 6 when you roll a die? A die has 6 sides, 1 side contain the number 6 that give us 1 wanted outcome in 6 possible outcomes. Problems like those Pascal and Fermat solved continuedto influence such early researchers as Huygens, Bernoulli, and DeMoivre in establishing a mathematical theory of probability. Probability Distribution: A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. Ma 162 Spring 2010 Ma 162 Spring 2010 April 21, 2010 Problem 1. It is known that probability that a randomly selected student who plays football also plays baseball is 0. Example: At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remainder are lorries. The general situation, then, is the following: given a sequence of random variables,. probability with a view toward data science applications. Probability is finding the possible number of outcomes of the event occurrence. On in\ufb01nite sums19 6. Suppose you toss a fair coin. the basic concepts of a probability model and the axioms commonly assumed of probability models. Conditional Probability Homework Solutions 1. Example: the probability that a card is a four and red =p(four and red) = 2/52=1/26. This problem asked us to find some probabilities involving a spinner. You can think of this as a binomial with all failures. Spector and Mazzeo examined the effect of a teaching method known as PSI on the performance of students in a course, intermediate macro economics. a) We first construct a tree diagram to represent all possible distributions of boys and girls in the family. Example Questions Using Probability Formulas. If every vehicle is equally likely to leave, find the probability of: a) van leaving first. Probability can be expressed in a variety of ways including a mathematically formal way such as using percentages. Since there are 13 spades in the deck, the probability of drawing a spade is. (a) In how many ways can a committee of \ufb01ve consisting of 3 girls and 2 boys be chosen? (b) What is the probability that a committee of \ufb01ve, chosen at random from the class, consists of three girls and two boys? (c) How many of the possible committees of \ufb01ve have no boys?(i. Instructors may obtain all of the solutions by writing to either of the authors, at [email\u00a0protected] For example, the length of time a person waits in line at a checkout counter or the life span of a light bulb. Example of Binomial Distribution and Probability This Tutorial will explain the Binomial Distribution, Formula, and related Discrete Probabilities Suppose you toss a coin over and over again and each time you can count the number of \u201cHeads\u201d you get. Just keep in mind that what conditional probability does for us is to adjust the denominator of our probability fraction to account for what we KNOW has already occurred. Probability is a branch of mathematics, and a lot of people have trouble with math. A college survey shows that 10% of freshmen, 15% of sophomores, 60% of juniors, and 85% of seniors like Probability. Compound event \u2013 an event with more than one outcome. Sometimes it can be computed by discarding part of the sample space. Sketch a graph. A first look at rigorous probability theory. This situation and all the problems any student faces nowadays became a reason to find a solution and create a website where students could find college homework help for the price they could afford. The meaning (interpretation) of probability is the subject of theories of probability. Exercises with both a short answer and a full solution are marked with and those with only a short answer are marked with (when more appropriate, for example, in \"Show that \" exercises, the short answer provides a hint to the key step). Example 2: Coin-A is tossed 200 times, and the relative occurrence of Tails is 0. Instead, we can usually define the probability density function (PDF). You can also get free sample papers, Notes, Important Questions. If it was, the probability of picking a red ball (etc. Theory of Probability (MATH230A/STAT310A, Fall 2007/08) The first quarter in a yearly sequence of probability theory. Identifying when a probability is a conditional probability in a word problem. , 10 trials), and the probability of getting \"heads\" was 0. This is COMPLETE Solution Manual for Probability and Statistics for Engineers and Scientists, 9th edition Ronald E. Find the z-scores that correspond to 33 minutes and 60 minutes. This presentation will be about examples of this form of mathematics in real life. The left side of the binomial probability function is written f(3|10,0. To determine whether to accept the shipment of bolts,the manager of the facility randomly selects 12 bolts. For example, if a tra\ufb03c management engineer looking at accident rates wishes to. Find the z-scores that correspond to 33 minutes and 60 minutes. Read and learn for free about the following article: Conditional probability and independence If you're seeing this message, it means we're having trouble loading external resources on our website. For example, one way to partition S is to break into sets F and Fc, for any event F. Solution: No, we cannot, because the experiment (tossing the coin) may have been repeated a very small number of times, and thus the relative occurrence in such a scenario will not give the true probability. Free PDF download of NCERT Solutions for Class 10 Maths Chapter 15 - Probability solved by Expert Teachers as per NCERT (CBSE) Book guidelines. Exercise :: Probability - General Questions. We write P(AjB) = the conditional probability of A given B. Example: The door prize at a party with 25 people is given by writing numbers 1 through 25 on the bottom of the paper plates used. If a woman does not have breast cancer, the probability is 7 percent that she will still have a positive mammogram. n (E) = Number of ways of drawing 2 balls out of (2 + 3) balls. For Example, cancer is related to age, and then using Bayes theorem, a person\u2019s age can be used to more accurately assess the probability that they have cancer, compared to the assessment of the probability of cancer made without knowledge of the person\u2019s age. The coin toss is nothing but experimenting with tossing a coin. Spector and Mazzeo examined the effect of a teaching method known as PSI on the performance of students in a course, intermediate macro economics. Consider the coin flip experiment described above. Example: If a dice is thrown twice, find the probability of getting two 5's. Solution to Example 1. Theoretical probability is an approach that bases the possible probability on the possible chances of something happen. Probability and playing cards is an important segment in probability. Also, remember that you are comparing the number of ways the outcome can occur to the number of ways the outcome cannot occur (not the total outcomes). In statistics, the normal distributions are used to represent real-valued random variables with unknown distributions. If the joint probability of relaxed import restrictions and a price war is 0. Then we can apply the appropriate Addition Rule: Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event. The probability of a sample point is a measure of the likelihood that the sample point will occur. GMAT Advanced Probability Problems By Mike M\u1d9cGarry on January 3, 2014 , UPDATED ON October 30, 2015, in GMAT Math In the following probability problems, problems #1-3 function as a set, problems #4-5 are another set, and problems #6-7 are yet another set. Normal Distribution Examples and Solutions. A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. Student's Solutions Guide for Introduction to Probability, Statistics, and Random Processes by Hossein Pishro-Nik | Jun 20, 2016 4. The series would last at most 4 games before either Team A wins two games or loses (to Team B) three games. troductory engineering courses in probability theory. P(freshman or sophomore) = 30/101 + 41/101 = 71/101. Coin-B is tossed an unknown number of times, but it. Thus the marginal probability distribution for A gives A's probabilities unconditional on B, in a margin of the table. The function f ( x ) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x -axis is equal to `1. Instead, we can usually define the probability density function (PDF). Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B). The problem is: we want 1 novel and 1 reference work, but we don\u2019t care which one was picked first. 1 of the bags is selected at random and a ball is drawn from it. Probability distributions can, however, be applied to grouped random variables which gives rise to joint probability distributions. We can write this statement mathematically as follows: P(10\u2264 X \u2264 20) =. Durrett Probability: Theory and Examples 4th Edition. This solution contains questions, answers, images, explanations of the complete Chapter 15 titled Probability of Maths taught in Class 10. In probability theory and applications, Bayes' theorem shows the relation between a conditional probability and its reverse form. In the National Lottery, 6 numbers are chosen from 49. In what follows, S is the sample space of the experiment in question and E is the event of interest. You use probability in daily life to make decisions when you don't know for sure what the outcome will be. All Probability Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Distributions 3. On in\ufb01nite sums19 6. The graphs below show the radial wave functions. July 1, 2015. Therefore, both the events will have half probability. Binomial distribution problems for practice Binomial distribution practice problems Binomial distribution problem 1 Question (Binomial distribution problems): Expected value of a binomial random variable X is 10 and the probability of. We always give back probability in to all our clients, whether you are our long-term client or simply ordering for the first time. It is assessed by considering the event's certainty as 1 and impossibility as 0. We try to provide all types of shortcut tricks on probability problem on balls here. An event is an occurrence that can be determined by a given level of certainty. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. Introduction: Probability (from the Latin probare to prove, or to test) is a number between zero and one that shows how likely a certain event is. Probability distribution tables In this tutorial I show you how to construct probability distribution tables for a discrete random variable for three different type examples. Such random variables generally take a finite set of values (heads or tails, people who live in London, scores on an IQ test), but they can also include random. If the ball drawn is red, find the probability that it is drawn from the third bag. There are six ways to roll a 7 and two ways to roll an 11, so the probability of a natural is (6 + 2)/36 = 8/36 = 2/9. Scroll down the page for more examples and solutions of word problems that involve the probability of independent events. 11 If we toss a coin 1000 times and find that it comes up heads 532 times, we estimate the probability of a head coming up to be 532 1000 0. Binomial distribution problems for practice Binomial distribution practice problems Binomial distribution problem 1 Question (Binomial distribution problems): Expected value of a binomial random variable X is 10 and the probability of. An urn contains 5 red balls and 2 green balls. These NCERT solutions play a. This Aptitude Test Questions sections presents \"Probability\" solved problems. If it was, the probability of picking a red ball (etc. Examples of using the formula to find conditional probability In some situations, you will need to use the following formula to find a conditional probability. \\ 1 P(5) 6 =. This sampling method is based on the fact that every member in the population has an equal chance of getting selected. What is the probability that the survey will show a greater percentage of. in Teachers. Distributions 3. probability of the event given that the other event has occurred. You can find several more examples here: Probability of A and B. 5L; also, on c, be careful about using the same set sizes to calculate the probability that the second student is taking a language class, since you\u2019ve already chosen the first. An examples from ecology: How are species abundance estimates determined from small samples? To summarize: There are at least two uses for statistics and probability in the life sciences. Statistics Solutions is the country\u2019s leader in probability and dissertation statistics. Statistics Solutions is the country's leader in probability and dissertation statistics. In other words, it is used to calculate the probability of an event based on its association with another event. problems included are about: probabilities, mutually exclusive events and addition formula of probability, combinations, binomial distributions, normal distributions, reading charts. In what follows, S is the sample space of the experiment in question and E is the event of interest. The mathematical and probability models of lottery provide information that players should know before they launch into a long-run play. Independence of three or more events34 12. 12 that she has exactly these two risk factors (but not the other). Probability concept includes some terms. It is less than 10 dollar. A team with equal numbers of men and women will have three men and three women. Binomial Distribution. P(AjB) is read The probability of A given B. 7) Anita randomly picks 4 cards from a deck of 52-cards and places them back into the deck ( Any set of 4 cards is equally likely ). Hoping that the book would be a useful reference for people who apply probability in their work, we have tried to emphasize the results that are important for applications, and illustrated their use with roughly 200 examples. org are unblocked. Step-by-step solutions provided. Student's Solutions Guide for Introduction to Probability, Statistics, and Random Processes - Kindle edition by Hossein Pishro-Nik. Student's Solutions Guide for Introduction to Probability, Statistics, and Random Processes by Hossein Pishro-Nik | Jun 20, 2016 4. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Introduction to Probability with Examples and Solutions Probability is a section of math that is essential in various aspects of life. The number of rainy days, Xcan be represented by a binomial distribution with n= 31trials (the number of days in the month of October), success probability p= 0:16(representing a rainy day) and failure probability q= 1 p= 0:84. It is not too much to say that the path of mastering statistics and data science starts with probability. Discrete probability distributions give the probability of getting a certain value for a discrete random variable. Visitors are requested to carefully read all shortcut examples. A simple example is the tossing of a fair (unbiased) coin. The pictures below depict the probability distributions in space for the Hydrogen wavefunctions. Binomial probability concerns itself with measuring the probability of outcomes of what are known as Bernoulli Trials, trials that are independent of each other and that are binary \u2014 with two possible outcomes. What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled. The results are summarised in the table below. The concept is very similar to mass density in physics: its unit is probability per unit length. There are two ways for it to win in three games: LWW or WLW; both occur with probability 1/8 so that the probability that Team A wins in exactly 3 games is also 1/4. In probability theory and applications, Bayes' theorem shows the relation between a conditional probability and its reverse form. Anyone has the right to use this work for any purpose, without any conditions, unless such conditions are required by law. Random Variables 4. So the probability of getting 2 blue marbles is: And we write it as \"Probability of event A and event B equals the probability of event A times the probability of event B given event A\" Let's do the next example using only notation:. Examples involving conditional probability Math 30530, Fall 2013 September 5, 2013 Math 30530(Fall 2012) Conditional examples September 5, 20131 / 5. Both the classical and frequency approaches have serious drawbacks, the first because the words \"equally. For any two of the three factors, the probability is 0. example, what probability should we assign to the customer choosing soup and then the meat? If 8/10 of the customers choose soup and then 1/2 of these choose meat, a proportion 8=10 \u00a21=2=4=10 of the customers choose soup and then meat. b) Find the mean and standard deviation of X. We provide examples on Probability problem on Balls shortcut tricks here in this page below. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Joint Probability Distributions. Read and learn for free about the following article: Conditional probability and independence If you're seeing this message, it means we're having trouble loading external resources on our website. It provides mathematically complete proofs of all the essential introductory results of probability and measure theory. It shows the exact probabilities for a particular value of the random variable. This presentation will be about examples of this form of mathematics in real life. This is typically possible when a large number of random e\ufb00ects cancel each other out, so some limit is involved. Conditional probability and independence31 11. The probability distribution given is discrete and so we can find the variance from the following: We need to find the mean \u03bc first: Then we find the variance: Example 2. Genetics for Probability To provide a scientific context for our probability problems, we will use examples from genetics. Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. Suppose you are a teacher at a university. Probability can be expressed in a variety of ways including a mathematically formal way such as using percentages. The solutions to these problems are at the bottom of the page. Formula sheet also available. However, there is a probability greater than zero than X. In addition, there are 6 outcomes that. A random variable, X, is a function from the sample space S to the real. Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. Compound event \u2013 an event with more than one outcome. Find the Standard Deviation of a random variable X whose probability density function is given by f(x) where: Solution. Explore more on other related concepts @ BYJU'S. Suppose the reaction times of teenage drivers are normally distributed with a mean of 0. This course will guide you through the most important and enjoyable ideas in probability to help you cultivate a more quantitative worldview. Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. You win if the 6 balls you pick match the six balls selected by the machine. Schaum's Outline of Probability and Statistics EXAMPLE 2. It provides mathematically complete proofs of all the essential introductory results of probability and measure theory. Suppose a simple random sample of 100 voters are surveyed from each state. \\ 1 P(5) 6 =. And then in the next segment we'll look at Bayes theorem. Applied Problem Solving in a Workplace Contents Introduction 5 Theme A \u2013 Approaches to problem Solving 6 Recognising Complex Problems 6 Activity 1a Identifying complex problems. Therefore, both the events will have half probability. Enter the trials, probability, successes, and probability type. 5th edition correct mistakes in previous editions and has some more beautiful examples. This means that the probability of the coin landing on heads would be \u00bd. A sample space (S) is a non empty set whose elements are called outcomes. The results are summarised in the table below. Ma 162 Spring 2010 Ma 162 Spring 2010 April 21, 2010 Problem 1. Independence 2. Probability: Theory and Examples Solutions Manual The creation of this solution manual was one of the most important im-provements in the second edition of Probability: Theory and Examples. What is the probability that a randomly chosen person in the class will weigh more than 160 lbs. \u2022 Probability of an event E = p(E) = (number of favorable outcomes of E)/(number of total outcomes in the sample space) This approach is also called theoretical probability. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. This lesson describes how hypergeometric random variables, hypergeometric experiments, hypergeometric probability, and the hypergeometric distribution are all related. What is the probability that an. \u00a9The McGraw-Hill Companies, Inc. The solutions given to the questions for the in between exercises and exercises given at the end of the chapter are prepared by our subject matter experts in a simple and lucid language. Answers and links to explanations to these these GMAT probability problems are at the end of set. Show that the solution $$X_t$$ of SDE examples, Stochastic Calculus The Probability Workbook is powered by WordPress at Duke WordPress Sites. Theoretical probability is an approach that bases the possible probability on the possible chances of something happen. P(B|A) is also called the \"Conditional Probability\" of B given A. We randomly select 5 balls. 1-9 A red die has face numbers {2, 4, 7, 12, 5, 11}. Here $Y=g(X)$, where $g$ is a differentiable function. E 2 = students in the debating team. 7) Anita randomly picks 4 cards from a deck of 52-cards and places them back into the deck ( Any set of 4 cards is equally likely ). If it isn't a trick coin, the probability of each simple outcome is the same. Students can get a fair idea on the probability questions which are provided with the detailed step-by-step answers to every question. Solution: No, we cannot, because the experiment (tossing the coin) may have been repeated a very small number of times, and thus the relative occurrence in such a scenario will not give the true probability. 2 Normal Distributions: Finding Probabilities If you are given that a random variable Xhas a normal distribution, nding probabilities. How is Chegg Study better than a printed A First Course In Probability 9th Edition student solution manual from the bookstore? Our interactive player makes it easy to find solutions to A First Course In Probability 9th Edition problems you're working on - just go to the chapter for your book. For this example, since the mean is 8 and the question pertains to 11 fires. All Probability Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.", "date": "2020-01-28 13:14:54", "meta": {"domain": "minoronzitti.it", "url": "http://vnbk.minoronzitti.it/probability-examples-and-solutions.html", "openwebmath_score": 0.7154899835586548, "openwebmath_perplexity": 428.6329198709035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9828232924970204, "lm_q2_score": 0.8705972549785201, "lm_q1q2_score": 0.8556432605768571}}
{"url": "https://math.stackexchange.com/questions/2590898/logarithmic-differentiation-for-multivariable-functions", "text": "# Logarithmic Differentiation for Multivariable functions?\n\nI was wondering if, and how exactly logarithmic differentiation may be applied to a multivariate function.\n\nFor example, I have been working with the function\n\n$$f(x,y)=\\frac{x-y}{x+y}$$\n\nDoes the following process hold? Specifically, does the chain rule step with del f hold?\n\n$$\\ln(f(x,y))=\\ln(x-y)-\\ln(x+y)$$\n\n$$\\Rightarrow \\frac{1}{f(x,y)} \\nabla f = \\langle \\frac{2y}{x^2-y^2} , \\frac{-2x}{x^2-y^2} \\rangle$$\n\nSolving for $\\nabla f$, does give me the correct answer. I'm wondering is my notation would be correct, and if this will ever not work.\n\nThanks for any help.\n\n\u2022 If you are ok, you can accept the answer and set as solved. Thanks! \u2013\u00a0user Jan 4 '18 at 20:17\n\nYou define\n\n$$g(x,y)=\\ln(f(x,y))=\\ln(x-y)-\\ln(x+y)$$\n\nand thus\n\n$$\\nabla g=\\left(\\frac{1}{f(x,y)}f_x,\\frac{1}{f(x,y)}f_y\\right)=\\frac{1}{f(x,y)}\\left(f_x,f_y\\right)=\\frac{1}{f(x,y)}\\nabla f$$\n\n\u2022 Ahhh, Okay, this seems very obvious now. Sorry if that was a dumb question. Thank you. \u2013\u00a0Kyle B Jan 3 '18 at 22:15\n\u2022 @KyleB You are welcome! Bye \u2013\u00a0user Jan 3 '18 at 22:22\n\nYou could take a simpler function and checked whether it holds. The usual differentiation rules hold true for the gradient as well: sum, difference, product and quotient. For instance, $\\nabla(fg)=g\\nabla f+f\\nabla g$, $\\nabla(\\frac{f}{g})=\\frac{g\\nabla f-f\\nabla g}{g^2}$, $\\nabla(\\ln f)=\\frac{\\nabla f}{f}$.\n\nYes I really believe your result as well as your notation are correct. May be you just need to put it in a more general setting. I would suggest you to proceed as follows.\n\nSuppose that $f:\\mathbb{R}^n\\to\\mathbb{R}$, which is differentiable and strictly positive in a certain open set $U\\subset\\mathbb{R}^n$. Then the function $F=\\ln\\circ f$ is also differntiable on $U$ with: $\\nabla F(x)=\\ln'(f(x))\\cdot \\nabla f(x)=\\frac{1}{f(x)}\\cdot\\nabla f(x)$ (chain rule). Hence we can solve for $\\nabla f(x)$, as we are in vector space $\\mathbb{R}^n$ and get the answer right.\n\nRemember the chain rule statement: If $G$ and $H$ are differentianble functions and $F=H\\circ G$ is well defined, then $F$ is also differentiable with $\\nabla F(x)=\\nabla H(G(x))\\cdot\\nabla G(x)$.", "date": "2019-10-15 11:32:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2590898/logarithmic-differentiation-for-multivariable-functions", "openwebmath_score": 0.9340419173240662, "openwebmath_perplexity": 249.26760286003292, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109494088427, "lm_q2_score": 0.868826784729373, "lm_q1q2_score": 0.8556301307411661}}
{"url": "http://mathhelpforum.com/pre-calculus/152372-complex-roots-quadratic-equation.html", "text": "# Math Help - Complex roots of a quadratic equation\n\n1. ## Complex roots of a quadratic equation\n\nThe question:\n\nUse the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.\n\nMy attempt:\n\nLet z = $z_{1} \\pm z_2i$\n\n$\\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} = z_{1} \\pm z_2i$\n\n$\\frac{1}{2a}(-b \\pm \\sqrt{b^2 - 4ac}) = z_{1} \\pm z_2i$\n\nBasically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.\n\nThanks!\n\n2. The roots of $ax^2 + bx + c = 0$ are\n\n$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$.\n\nIf $b^2 - 4ac < 0$ then that means $\\sqrt{b^2 - 4ac}$ is imaginary. So we can say $\\sqrt{b^2 - 4ac} = ni$, where $n$ is some multiple of $i$.\n\nSo $x = \\frac{-b \\pm n i}{2a}$\n\n$= -\\frac{b}{2a} \\pm \\left(\\frac{n}{2a}\\right)i$.\n\nSo the two roots are complex conjugates.\n\n3. Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?\n\n4. Originally Posted by Glitch\nAhh, thanks. So would my solution be acceptable, or should I simplify it like yours?\nTo be honest, I can't really follow your solution...\n\n5. It's pretty much the same as yours, except you replaced the square root with 'ni'. I was trying to show that the general form of a complex number is equivalent to a simplified quadratic formula.\n\n6. Originally Posted by Glitch\nThe question:\nUse the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.\nThe standard way of proving the is to use properties of the complex conjugate.\nIf $a$ is a real number then $\\overline{a}=a$.\nThe conjugate of a sum is the sum of conjugates.\nTherefore we have:\n\n$\\overline {az^2 + bz + c} = \\overline 0 = 0\\, \\Rightarrow \\,a\\overline z ^2 + b\\overline z + c = 0$\n\n7. Originally Posted by Glitch\nThe question:\n\nUse the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.\n\nMy attempt:\n\nLet z = $z_{1} \\pm z_2i$\n\n$\\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} = z_{1} \\pm z_2i$\n\n$\\frac{1}{2a}(-b \\pm \\sqrt{b^2 - 4ac}) = z_{1} \\pm z_2i$\nBut at this point you haven't yet said that $\\sqrt{b^2- 4ac}$ is an imaginary number!\n\nBasically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.\n\nThanks!\n\n8. Originally Posted by Plato\nThe standard way of proving the is to use properties of the complex conjugate.\nIf $a$ is a real number then $\\overline{a}=a$.\nThe conjugate of a sum is the sum of conjugates.\nTherefore we have:\n\n$\\overline {az^2 + bz + c} = \\overline 0 = 0\\, \\Rightarrow \\,a\\overline z ^2 + b\\overline z + c = 0$\n...and it is true for every polynomial...\n\n9. Fair point. So if I add:\n\nWhen $b^2 - 4ac < 0$, $\\sqrt{b^2 - 4ac} = ni, n \\in R$\n\nAnd replace the square root with the imaginary part 'n', would that suffice?\n\n10. Hello, Glitch!\n\nI have a very primitive solution . . .\n\nUse the properties of the complex conjugate to show that\nif the complex number $z$ is a root of a quadratic equation\n$f(x) \\:=\\:ax^2 + bx + c = 0$ with $a, b, c$ real coefficients,\nthen so is the conjugate of $z.$\n\nWe are told that $z \\:=\\:p + qi$ is a solution of the quadratic.\n\n. . Hence: . $a(p+qi)^2 + b(p+qi) + c \\:=\\:0$\n\nThis simplifies to: . $(ap^2 - aq^2 + bp + c) + q(2ap + b)i \\;=\\;0$\n\n. . And we have: . $\\begin{Bmatrix}ap^2 - aq^2 + bp + c \\;=\\;0 \\\\ q(2ap + b) \\;=\\;0 \\end{Bmatrix}$\n\nThe conjugate of $z$ is: . $\\overline z \\:=\\:p - qi$\n\nConsider $f(\\overline z) \\;=\\;a(p-qi)^2 + b(p-qi) + c$\n\n. . . . . . . $f(\\overline z) \\;=\\;\\underbrace{(ap^2 - aq^2 + bp + c)}_{\\text{This is 0}} - \\underbrace{q(2ap + b)}_{\\text{This is 0}}i$\n\n$\\text{Therefore: }\\:f(\\overline z) \\,=\\,0\\:\\text{ and }\\:\\overline z\\text{ is also a solution.}$", "date": "2015-07-28 02:59:06", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/152372-complex-roots-quadratic-equation.html", "openwebmath_score": 0.9516220092773438, "openwebmath_perplexity": 286.72099066278463, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810949854636, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.855630126111154}}
{"url": "https://math.stackexchange.com/questions/2553562/how-to-find-precise-value-in-terms-of-pi", "text": "# How to find precise value in terms of $\\pi$\n\nI've some argument like:\n\n$$\\arctan(\\sqrt3 + 2)$$ and as explained here How to calculate $\\arctan(\\sqrt3-2)$ to be $-15\u00b0$ by hand? i made an assumption and found that it should be $\\frac{5\\pi}{12}$. I found the exact value of $\\sqrt3 + 2$ then by table i found it's equal to $\\tan(75^{\\circ})$ and finally i found $\\arctan(\\tan(75^{\\circ}))$ But what if i have something more complicated to do assumption on like:\n\n$$\\arccos({\\frac{\\sqrt{\\sqrt3+2}}{2}})$$\n\nSo i'm interesting if there is any sequence of operations on argument itself to convert it into fraction of $\\pi$ without guessing about how many it should be?\n\nThe best bet is to \"unravel\" the expression, try to simplify it by getting rid of roots and recognizing patterns:\n\n$$\\phi=\\arccos\\frac{\\sqrt{\\sqrt{3}+2}}{2}$$ $$\\cos\\phi=\\frac{\\sqrt{\\sqrt{3}+2}}{2}$$ $$4\\cos^2\\phi=\\sqrt{3}+2$$ $$2(2\\cos^2\\phi-1)=\\sqrt{3}$$ $$2\\cos2\\phi=\\sqrt{3}$$ $$\\cos2\\phi=\\frac{\\sqrt{3}}{2}$$ $$2\\phi=\\arccos\\frac{\\sqrt3}{2}=\\frac{\\pi}{6}$$ $$\\phi=\\frac{\\pi}{12}$$\n\nYou'll always get some sort of polynomial in trigonometric functions. The goal is to reduce the degree of polynomial by using multiple angle formulas. If you don't recognize them yourself, you can take a look at Chebyshev polynomials of the first kind which tell you the multiple angle formulas: $\\cos n\\phi= T_n(\\cos\\phi)$. In this case, we recognized $T_2(x)=2x^2-1$.\n\nDepending on how well twisted the expression is, you may have to get creative. There may be more than just multiple angle tricks: possibly, you have to use addition theorems as well, recognizing $\\cos(x+y)$ where $x$ and $y$ are not equal, and so on. But the main trick is still just \"undoing\" the operations that you don't want to see.\n\n\u2022 Yes, that's exactly what i was looking for. Thanks a lot! Dec 6 '17 at 9:06\n\u2022 very nice derivation !\n\u2013\u00a0user\nDec 6 '17 at 9:10\n\nAs $2+\\sqrt3=\\dfrac{(\\sqrt3+1)^2}2$\n\n$$\\dfrac{\\sqrt{2+\\sqrt3}}2=\\dfrac{\\sqrt3+1}{2\\sqrt2}=\\cos30^\\circ\\cos45^\\circ+\\sin30^\\circ\\sin45^\\circ=\\cos(45^\\circ-30^\\circ)$$", "date": "2021-09-23 09:09:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2553562/how-to-find-precise-value-in-terms-of-pi", "openwebmath_score": 0.8249925374984741, "openwebmath_perplexity": 178.46067235076143, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109503004294, "lm_q2_score": 0.8688267677469951, "lm_q1q2_score": 0.8556301147913687}}
{"url": "https://www.physicsforums.com/threads/how-to-find-this-limit.113049/", "text": "# How to find this limit?\n\n1. Mar 4, 2006\n\n### pivoxa15\n\nFind the limit as (x,y)->(0,0) for (yx^2)/(x^2+y^2)\n\nIf do this literally, I get 0/0 hence no limit but a limit does exist for this function. How do I get it? What is the general way to find the limits of multi variable functions?\n\n2. Mar 4, 2006\n\n### VietDao29\n\nEhh???\nWho tells you that 0 / 0 means the limit does not exist there? It's one of the Indeterminate forms.\nGenerally, to find a limit of a 2 variable function, one can change it to polar co-ordinate, and go from there: $$x = r \\cos \\alpha \\quad \\mbox{and} \\quad y = r \\sin \\alpha$$\nNow (x, y) -> (0, 0) means that r -> 0, and $$\\alpha$$ can take whatever value. So if you can show that when r -> 0, the expression will tend to some value independent of $$\\alpha$$, then the limit exists there.\n---------------\nExample:\n$$\\lim_{\\substack{x \\rightarrow 0 \\\\ y \\rightarrow 0}} \\frac{x ^ 2 + y ^ 2}{xy}$$\nChange it to polar form, we have:\n$$\\lim_{\\substack{x \\rightarrow 0 \\\\ y \\rightarrow 0}} \\frac{x ^ 2 + y ^ 2}{xy} = \\lim_{r \\rightarrow 0} \\frac{r ^ 2}{r ^ 2 \\sin \\alpha \\cos \\alpha} = \\lim_{r \\rightarrow 0} \\frac{1}{\\cos \\alpha \\sin \\alpha}$$\nThat means the limit of that expression does depend on $$\\alpha$$, hence the limit does not exist at (0, 0).\nCan you get this? :)\n\n3. Mar 4, 2006\n\n### moose\n\nThe way I would think about this one is that if x and y are approaching the same thing, why not set y equal to x? That way you would have x^3/2x^2\nNow you can think about which one is changing faster, etc.\n\n4. Mar 5, 2006\n\n### pivoxa15\n\nI see. It is clever. Is it the standard way of evaluating multivariable limits? What other ways are there?\n\n5. Mar 5, 2006\n\n### pivoxa15\n\nYour method does not work in most cases.\n\n6. Mar 5, 2006\n\n### devious_\n\nOther ways include approaching the point from various lines, e.g. if you're looking at (x,y) -> (0,0) then sometimes letting (x,y) approach the origin from the x-axis or the y-axis ((x,0) or (y,0)) can help you prove the limit doesn't exist.\n\n7. Mar 5, 2006\n\n### benorin\n\nSo we may, upon trying the limit along various curves such as y=x, y=x^2, the x-axis (e.g. y=0), the y-axis (e.g. x=0) which all give the value of the limit to be 0, conjecture the value of the limit to be 0. Now we must prove it:\n\nTo prove that $$\\lim_{(x,y)\\rightarrow (0,0)} \\frac{yx ^ 2}{x^2 + y^2}=0,$$ we require that\n\n$$\\forall\\epsilon >0, \\exists \\delta >0 \\mbox{ such that } 0<\\sqrt{x^2 + y^2}<\\delta\\Rightarrow \\left| \\frac{yx ^ 2}{x^2 + y^2}-0\\right| < \\epsilon$$\u200b\n\n(and in case you didn't know, the symbol $$\\forall$$ is read \"for every\", the symbol $$\\exists$$ is read \"there exists\", and the symbol $$\\Rightarrow$$ is read \"implies\".)\n\nThe proof is this: We will need the following inequality\n\n$$0\\leq y^2\\Rightarrow x^2\\leq x^2+y^2\\Rightarrow \\frac{x^2}{x^2+y^2}\\leq 1.$$\u200b\n\nLet us work the absolute value term contained in the $$\\epsilon ,\\delta$$ definition of the limit described above, we have\n\n$$\\left| \\frac{yx ^ 2}{x^2 + y^2}-0\\right| =\\frac{|y|x ^ 2}{x^2 + y^2} =|y|\\frac{x ^ 2}{x^2 + y^2} \\leq |y| = \\sqrt{y^2} \\leq \\sqrt{x^2+y^2} <\\delta$$\u200b\n\nwe want to choose $$\\delta$$ so that the absolute value term we just worked with is always less than $$\\epsilon$$ given that $$0<\\sqrt{x^2+y^2} <\\delta .$$ Evidently, we may choose $$\\delta =\\epsilon$$ to this end, and the proof is complete upon stating this formally.\n\nI know that PF does not permit full solutions to be posted, however it is understood that proofs of this sort are quite tricky to construct, examples are few in texts that require them, and I will have rendered sufficient pedagogical substance by even successfully relating it.\n\n--Ben\n\nEDIT: Thanks VietDao29, all that short-hand is Greek to me.\n\nLast edited: Mar 5, 2006\n8. Mar 5, 2006\n\n### VietDao29\n\nOne more way is to use the inequality:\n$$x ^ 2 + y ^ 2 \\geq 2|xy|$$, one can prove it by doing a little rearrangement, and noticing the fact that: (|x| + |y|)2 >= 0.\nI'll give one example that's similar to your problem.\n----------------\nExample:\nFind:\n$$\\lim_{\\substack{x \\rightarrow 0 \\\\ y \\rightarrow 0}} \\frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2}$$\nNow using the inequality above, we have:\n$$\\left| \\frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \\right| \\leq \\left| \\frac{x ^ 3 y ^ 3}{2xy} \\right| = \\frac{1}{2} |x ^ 2 y ^ 2| = \\frac{1}{2} x ^ 2 y ^ 2$$\nNow as (x, y) tends to (0, 0),\n$$\\frac{1}{2} x ^ 2 y ^ 2 \\rightarrow 0$$, hence $$\\left| \\frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \\right| \\rightarrow 0$$, thus $$\\frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \\rightarrow 0$$, so we can conclude that:\n$$\\lim_{\\substack{x \\rightarrow 0 \\\\ y \\rightarrow 0}} \\frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} = 0$$\nCan you get this? :)\n--------------\nEDIT:\nBy the way, benorin:\nThat symbols is read Delta, the symbol that is read \"there exists\" should be $$\\exists$$ .\n\nLast edited: Mar 5, 2006\n9. Mar 5, 2006\n\n### pivoxa15\n\nThe method you used in your example looked a bit suspect.\n\nIf we follow your way and do the limit(x->0, y->0) of (xy)/(x^2+y^2)\nwe get the limit being smaller than 1/2 which is wrong since it should have no limit at all.\n\n10. Mar 5, 2006\n\n### Benny\n\nSorry to intrude on your topic but I would like to know if using polar coordinates is a valid method when the limit is (x,y) -> (0,0). Yes, the angle is arbitrary but each combination of a (r,angle) as r approaches zero corresponds to taking the limit along the straight line. Ie. not every single path is considered.\n\nI can see that converting to polar coordinates is sufficient to verify that the limit does not exist. But is it sufficient to show that a limit exists? I apologise if I missed anything.\n\n11. Mar 5, 2006\n\n### HallsofIvy\n\nStaff Emeritus\nNo, that's perfectly valid. You do not have to assume the angle is constant as r goes to 0. The point is that, in polar coordinates, the distance from the origin is measured by a single variable, r. r is completely independent of the angle. That was the point VietDao29 made in his first post- the limit, as r goes to 0, depended upon the angle and so the limit itself does not exist.\n\nIn your problem, $\\frac{xy^2}{x^2+ y^2}$, changing to polar coordinates gives $\\frac{r^3cos(\\theta)sin^2(\\theta)}{r^2}= r cos(\\theta)sin^2(\\theta)$. What does that go to as r goes to 0? Does it depend on $\\theta$?\n\n12. Mar 5, 2006\n\n### Benny\n\nOk, that's good. Thanks for the clarification.\n\n13. Mar 5, 2006\n\n### VietDao29\n\nYes, the limit of:\n$$\\lim_{\\substack{x \\rightarrow 0 \\\\ y \\rightarrow 0}} \\frac{xy}{x ^ 2 + y ^ 2}$$ does not exist at (0, 0).\nSo by using the inequality x2 + y2 >= 2|xy|. We have:\n$$\\left| \\frac{xy}{x ^ 2 + y ^ 2} \\right| \\leq \\frac{1}{2} \\left| \\frac{xy}{xy} \\right| = \\frac{1}{2}$$\nSo that means:\n$$-\\frac{1}{2} \\leq \\frac{xy}{x ^ 2 + y ^ 2} \\leq \\frac{1}{2} \\quad (*)$$. Hence, no conclusion can be drawn by looking at the inequality (*). Why? Because we don't know if the expression does converge to some value between -1 / 2, and 1 / 2; or it just diverges.\nIt's like the limit: $$\\lim_{x \\rightarrow \\infty} \\sin x$$ does not exist, although we know that: -1 <= sin(x) <= 1. It's because sin(x) does not tend to any specific number (i.e converges to any value) as x tends to infinity.\n---------------\nHowever, if we have $$\\lim_{x \\rightarrow \\alpha} |f(x)| = 0$$, then we also have: $$\\lim_{x \\rightarrow \\alpha} f(x) = 0$$. Why?\nIt's because $$\\forall \\varepsilon > 0, \\exists \\delta > 0 : 0 < |x - \\alpha| < \\delta \\Rightarrow ||f(x)| - 0| = |f(x)| = |f(x) - 0| < \\varepsilon$$. Now according to the definition of limit, we also have:\n$$\\lim_{x \\rightarrow \\alpha} f(x) = 0$$. Can you get this?\nYou can do the same to prove that if:\n$$\\lim_{\\substack{x \\rightarrow \\alpha \\\\ y \\rightarrow \\beta}} |f(x, y)| = 0$$ then $$\\lim_{\\substack{x \\rightarrow \\alpha \\\\ y \\rightarrow \\beta}} f(x, y) = 0$$.\nCan you get this? :)\n---------------\nNow just read my last post again to see if you can understand it.\n:)\n\n14. Mar 5, 2006\n\n### benorin\n\nHaving verified it, I submit in passing that\n\n$$-\\frac{1}{2} \\leq \\frac{xy}{x ^ 2 + y ^ 2} \\leq \\frac{1}{2}$$\n\nare the best bounds possible.\n\n15. Mar 5, 2006\n\n### pivoxa15\n\nI see your point. Using polar coords to find limits of multivariable functions is better than your first principles method.\n\nWhere did you learn evaluating limits of multivarialbe functions using polar coords? I did not see them in calculus textbooks.\n\n16. Mar 6, 2006\n\n### VietDao29\n\nYes, using polar co-ordinate is far better, and easier than the second way. :)\nAnd guess what? I learnt how to do it at this site, too... :tongue:\n\n17. Mar 6, 2006\n\n### HallsofIvy\n\nStaff Emeritus\nYes, but that tells you nothing about whether the limit at (0,0) exists.\n\n18. Mar 6, 2006\n\n### benorin\n\nLike I said, \"in passing\".\n\n19. Mar 7, 2006\n\n### pivoxa15\n\nAre the functions we have been discussing 3D or 2D?\n\n20. Mar 7, 2006\n\n### benorin\n\nRecall that z=f(x,y) is a surface, and hence 3-D.", "date": "2016-12-09 17:48:20", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/how-to-find-this-limit.113049/", "openwebmath_score": 0.8857300877571106, "openwebmath_perplexity": 795.0478097083534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109480714625, "lm_q2_score": 0.8688267626522814, "lm_q1q2_score": 0.8556301078374529}}
{"url": "https://cs.stackexchange.com/questions/16346/how-to-discuss-coefficients-in-big-o-notation", "text": "# How to discuss coefficients in big-O notation\n\nWhat notation is used to discuss the coefficients of functions in big-O notation?\n\nI have two functions:\n\n\u2022 $f(x) = 7x^2 + 4x +2$\n\u2022 $g(x) = 3x^2 + 5x +4$\n\nObviously, both functions are $O(x^2)$, indeed $\\Theta(x^2)$, but that doesn't allow a comparison further than that. How do I discuss the the coefficients 7 and 3. Reducing the coefficient to 3 doesn't change the asymptotic complexity but it still makes a significant difference to runtime/memory usage.\n\nIs it wrong to say that $f$ is $O(7x^2)$ and $g$ is $O(3x^2)$ ? Is there other notation that does take coefficients into consideration? Or what would be the best way to discuss this?\n\n\u2022 It's not wrong, it's just redundant, because $O(7 x^2) = O(x^2)$. \u2013\u00a0Oli Charlesworth Oct 7 '13 at 23:17\n\u2022 See also our reference question. \u2013\u00a0Raphael Oct 28 '13 at 7:35\n\nBig-$O$ and big-$\\Theta$ notations hide coefficients of the leading term, so if you have two functions that are both $\\Theta(n^2)$ you cannot compare their absolute values without looking at the functions themselves. It's not wrong per se to say that $7x^2 + 4x + 2 = \\Theta(7x^2)$, but it's not informative because $7x^2 + 4x + 2 = \\Theta(3x^2)$ is also true (and, in fact, it's $\\Theta(kx^2)$ for any positive constant $k$).\n\nThere are other notations you might want to use instead. For example, $\\sim$ notation is a much stronger claim than big-$\\Theta$:\n\n$\\qquad \\displaystyle f(x) \\sim g(x) \\iff \\lim_{x \\to \\infty} \\frac{f(x)}{g(x)} = 1$\n\nFor example, $7x^2 + 4x + 2 \\sim 7x^2$, but the claim $7x^2 + 4x + 2 \\sim 3x^2$ would be false. You can think of tilde notation as $\\Theta$ notation that preserves the leading coefficients, which seems to be what you're looking for if you do care about the leading coefficient of the dominant growth term.\n\n\u2022 Tilde notation is what I'm looking for. I was sure there was something I just couldn't recall what it was called and searches proved fruitless. Thanks! \u2013\u00a0El Bee Oct 7 '13 at 23:48\n\nThe tilde is one approach. If you want to stick with $O$, you could say\n\n$\\qquad f(x) = 7x^2 + O(x)$ and\n\n$\\qquad g(x) = 3x^2 + O(x)$.\n\n\u2022 Even better: say f(x) = 7x^2 + o(x^2), using little-o notation to clarify that what's left is asymptotically smaller than x^2. \u2013\u00a0templatetypedef Oct 9 '13 at 4:06\n\u2022 O(x) is strictly smaller than o(x^2), so using that would be less clear than using big-O. On the other hand, using little-o is definitely more common when you want to say that you've got the right first term, because then you don't need to worry about the next term. (And if we're wanting to be completely clear, then we would need to explain why we don't just write down 7x^2+4x+2 in the first place, since it is exactly correct. \u2013\u00a0Teepeemm Oct 9 '13 at 4:43\n\u2022 You're absolutely right... my apologies! \u2013\u00a0templatetypedef Oct 9 '13 at 17:56\n\u2022 Note that the rigorous way of writing this would be \"$f(x) = 7x^2 + g(x)$ with $g(x) \\in O(x)$\". In any case, this is very useful if you want to fix more than the \"first\" constant; you can say \"$f(x) = 7x^2 + 4x + O(1)$\" which you can not do with $\\sim$. \u2013\u00a0Raphael Oct 28 '13 at 7:29", "date": "2020-06-06 21:48:36", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/16346/how-to-discuss-coefficients-in-big-o-notation", "openwebmath_score": 0.7434012293815613, "openwebmath_perplexity": 295.9506517504927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407175907054, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8556214216334299}}
{"url": "https://math.stackexchange.com/questions/2661883/permutations-how-many-outcomes-have-a-1-in-them-if-pn-k-p13-3-and-the-pro", "text": "# Permutations: How many outcomes have a 1 in them if P(n,k) = P(13,3) and the probability it will occur if randomly generated?\n\nThere is a set of numbers {1,2,3,4,5,6,7,8,9,10,11,12,13}. I am trying to find out how many subsets I can form that have 3 numbers in them and also the amount of subsets that will have a number 1 in them. Also, if I was to randomly generate a single outcome, what is the probability it will have a 1 in it if all outcomes have equal chance of occurring?\n\nI know that the probability can be calculated from (Number of subsets that have 1 in them)/(Total number of permutations)\n\nI figured out the Total total number of permutations = 13!/(13-3)! = 1716\n\nBut I can't figure out how to determine how many subsets will have a 1 in them. In addition to this, is there a way to generalise the method? For example, what if I would like to find out the number of subsets that have a number 1 and a number 2 in them for the same n and k parameters?\n\n\u2022 For subsets do you mean ignoring order, which is usual, or considering order, which seems to be implied by your use of permutations? \u2013\u00a0Ross Millikan Feb 22 '18 at 15:49\n\u2022 Oh in this problem I do want to consider order. So 1,2,3 is different from 1,3,2 and so on. \u2013\u00a0Lobster Feb 22 '18 at 16:04\n\u2022 That is what I considered in my answer. You are then selecting three element permutations from thirteen, not a subset. As I commented on the other answer, the fraction that include $1$ is the same either way. \u2013\u00a0Ross Millikan Feb 22 '18 at 16:11\n\nIt looks like instead of subsets you mean three element permutations that include $1$. You can count those by choosing first the location of the $1$, which gives three choices, then the first other element, $12$, and the second other element, $11$ for a total of $3 \\cdot 12 \\cdot 11=396$. You are correct that the chance a random permutation includes a $1$ is $\\frac {396}{1716}=\\frac 3{13}$ This is not surprising as you are choosing three elements from thirteen.\n\nTo have both $1$ and $2$ you have three places to put the $1$, two places for the $2$, and $11$ choices for the third number.\n\n\u2022 Thanks! This was a great help! \u2013\u00a0Lobster Feb 22 '18 at 16:16\n\nThis is not actually a case of permutation, but rather a case of combination. Because in a subset, the arrangement of the numbers are not important. Therefore it is the use of combinations and not permutations.\n\nSubset 3 numbers in them = 13C3 = 286 - choosing 3 numbers out of 13\n\nSubset 3 numbers with 1 in them = 12C2 = 66 - choosing 2 numbers given one of the number is 1 (there is only 12 numbers to pick, because 1 is chosen by default)\n\n\u2022 You can use either permutations or combinations as long as you are consistent. The probability of having a $1$ comes out $\\frac 3{13}$ eitherj way, as it should. There are $3!$ times more total permutations and $3!$ times more that include a $1$. \u2013\u00a0Ross Millikan Feb 22 '18 at 15:56\n\u2022 Welcome to MSE. Please use MathJax. \u2013\u00a0Jos\u00e9 Carlos Santos Feb 22 '18 at 16:01\n\u2022 Oh right, i forgot to answer the probability. You are absolutely right about the consistency part, and both method works. My only concern is about technicality especially with subsets, because with other question (not the probability one), the answer may be different, due to order being considered. \u2013\u00a0Zirc Feb 22 '18 at 16:02", "date": "2019-05-26 09:59:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2661883/permutations-how-many-outcomes-have-a-1-in-them-if-pn-k-p13-3-and-the-pro", "openwebmath_score": 0.8103286623954773, "openwebmath_perplexity": 199.73320338133198, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407172026312, "lm_q2_score": 0.8791467564270271, "lm_q1q2_score": 0.8556214197514067}}
{"url": "https://byjus.com/question-answer/the-chance-that-doctor-a-will-diagonise-disease-x-correctly-is-60-the-chance-that/", "text": "Question\n\n# The chance that Doctor A will diagonise disease X correctly is $$60\\%$$. The chance that a patient will die by his treatment after correct diagnosis is $$40\\%$$ and the chance of death after wrong diagnosis is $$70\\%.$$ A patient of Doctor A who had disease X died. The probability that his disease was diagonised correctly is\n\nA\n513\nB\n613\nC\n213\nD\n713\n\nSolution\n\n## The correct option is D $$\\dfrac{6}{13}$$Let us define the following events.$${ E }_{ 1 }:$$ Disease $$X$$ is diagnosed correctly by doctor $$A$$.$${ E }_{ 2 }:$$\u00a0Disease $$X$$ is not\u00a0diagnosed correctly by doctor $$A$$.$$B:$$ A patient (of doctor $$A$$) who has disease $$X$$ dies.Then, we are given, $$P({ E }_{ 1 })=0.6$$,$$P({ E }_{ 2 })=1-P({ E }_{ 1 })=1-0.6=0.4$$and $$\\displaystyle P\\left( \\frac { B }{ { E }_{ 1 } } \\right) =0.4$$, $$\\displaystyle P\\left( \\frac { B }{ { E }_{ 2 } } \\right) =0.7$$By Bay's Theorem$$\\displaystyle P\\left( \\frac { { E }_{ 1 } }{ B } \\right) =\\frac { P\\left( { E }_{ 1 } \\right) P\\left( \\frac { B }{ { E }_{ 1 } } \\right) }{ P\\left( { E }_{ 1 } \\right) P\\left( \\frac { B }{ { E }_{ 1 } } \\right) +P\\left( { E }_{ 2 } \\right) P\\left( \\frac { B }{ { E }_{ 2 } } \\right) }$$$$\\displaystyle=P\\left( \\frac { { E }_{ 1 } }{ B } \\right) =\\frac { 0.6\\times 0.4 }{ 0.6\\times 0.4+0.4\\times 0.7 } =\\frac { 0.24 }{ 0.24+0.28 } =\\frac { 0.24 }{ 0.52 } =\\frac { 6 }{ 13 }$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-16 19:57:28", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/the-chance-that-doctor-a-will-diagonise-disease-x-correctly-is-60-the-chance-that/", "openwebmath_score": 0.5884534120559692, "openwebmath_perplexity": 5050.975292858223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9867771786365549, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8555711042249107}}
{"url": "https://math.stackexchange.com/questions/2595145/in-mathbbr4-let-u-span-1-1-0-0-1-1-1-1", "text": "# In $\\mathbb{R}^4$, let $U=$ span$\\{(1,1,0,0),(1,1,1,1)\\}$.\n\n(a) Compute an orthonormal basis for $U$.\n\n(b) Find $u\\in U$ such that the Euclidean distance $||u-(1,2,3,4)||$ is as small as possible.\n\nHere's what I have:\n\n(a) To find an orthonormal basis, we'll use the Gram-Schmidt Algorithm. The process is as follows:\n\nGiven $k$ basis vectors $v_1,v_2,\\cdots,v_k$, we apply the following steps:\n\n1. Let $q_1=\\frac{1}{||v_1||}v_1$.\n2. For $i=2$ to $k$, repeat steps 3 and 4.\n3. Let $w_i=v_i-\\langle q_1,v_i\\rangle q_i-\\langle q_2,v_i\\rangle q_2-\\cdots-\\langle q_{i-1},v_i\\rangle q_{i-1}$\n4. Let $q_i=\\frac{1}{||w_i||}w_i$\n\nThe vectors $q_1,\\cdots q_k$ are our orthonormal basis. Now, we apply these steps to the given problem. $$q_1=\\frac{1}{||v_1||}v_1=\\frac{1}{\\sqrt{2}}v_1=(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}},0,0)$$ $$w_2=v_2-\\langle q_1,v_2\\rangle q_1=(1,1,1,1)-(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{2}})(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}},0,0)=(0,0,1,1)$$ $$q_2=\\frac{1}{\\sqrt{2}}w_2=(0,0,\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}})$$ So $\\{q_1,q_2\\}$ is an orthonormal basis for $U$.\n\n(b) I recognize this as a least squares problem, but I'm a bit lost on how to actually set it up, because we don't have something with the form $Ax\\approx b$. I mean, I know that $b=(1,2,3,4)$ and $u$ somehow represents $Ax$, but we don't have a matrix $A$, and $u$ is just an element of the span of $2$ vectors.\n\n\u2022 Hint: the underlying mechanism of the solution to the least-squares problem you\u2019re familiar with is orthogonal projection onto a vector space. \u2013\u00a0amd Jan 7 '18 at 4:47\n\nYour proof for part (a) is correct. For part (b) let $$u=(\\alpha , \\alpha ,\\beta ,\\beta)$$ You need to minimize $$(\\alpha -1)^2+( \\alpha -2)^2 + (\\beta -3)^2+ (\\beta-4)^2.$$ Partial derivatives are zero at $\\alpha = 3/2$ and $\\beta =7/2$. Thus $u=(3/2, 3/2, 7/2,7/2)$is the desired vector.\n\n\u2022 Do you think you could elaborate on this a bit? Is the choice that $u=(\\alpha,\\alpha,\\beta,\\beta)$ from the fact that the vectors that span $U$ have the same first two entries? \u2013\u00a0Atsina Jan 7 '18 at 5:06\n\u2022 Sure. From your part (a), you have found a basis in form of (1,1,0,0) and (0,0,11). $u=(\\alpha,\\alpha,\\beta,\\beta)=\\alpha (1,1,0,0)+\\beta (0,0,1,1)$ \u2013\u00a0Mohammad Riazi-Kermani Jan 7 '18 at 5:20\n\nBecause you have an orthonormal basis, things become increasingly easy.\n\nThe least squares problem is solved by\n\n$$x_{ls} = (A^TA)^{-1}(A^Tb)$$\n\nWhere $A$ is a matrix which has your basis as columns and $b$ is $<1,2,3,4>$. Solve this for $x_{ls}$ and then the solution is $Ax_{ls}$.\n\n\u2022 This is more along the lines of what I was searching for, but Mohammad's solution is slightly less computationally intensive. I wish I could accept multiple answers. \u2013\u00a0Atsina Jan 7 '18 at 6:02\n\u2022 @Atsina Because there is an orthonormal basis for the subspace, things are even easier than what been written here: $A^TA=I_2$, so your equation reduces to $x_{ls}=A^Tb$ and $u=AA^Tb=(u_1^Tb)u_1+(u_2^Tb)u_2$, i.e., it\u2019s the sum of the orthogonal projections onto the basis vectors. \u2013\u00a0amd Jan 11 '18 at 1:10\n\nThe shortest distance to $U$ is measured along a direction orthogonal to it, therefore the element of $U$ that minimizes the distance to a given vector $b$ is the orthogonal projection of $b$ onto $U$, that is, the component of $b$ that lies in $U$.\n\nThere are various ways to compute this projection, but since you\u2019ve already constructed an orthonormal basis for $U$, the easiest thing to do is to compute the individual projections onto the basis vectors and combine them. In fact, this is exactly what you do at each step of the Gram-Schmidt process: you compute the orthogonal projection of the current vector onto the space spanned by the orthonormal basis vectors generated up to that point and subtract that projection from the vector to get its component that doesn\u2019t lie in that span. So, using that projection formula from the Gram-Schmidt process, \\begin{align} u = (u_1^Tb)u_1+(u_2^Tb)u_2 &= \\frac3{\\sqrt2}\\left(\\frac1{\\sqrt2},\\frac1{\\sqrt2},0,0\\right)+\\frac7{\\sqrt2}\\left(0,0,\\frac1{\\sqrt2},\\frac1{\\sqrt2}\\right) \\\\ &= \\left(\\frac32,\\frac32,\\frac72,\\frac72\\right). \\end{align}\n\nThe connection to finding a least-squares solution to the possibly-inconsistent linear system $Ax=b$ is that you essentially replace $b$ with its orthogonal projection onto the column space of $A$.\n\nMuhammad gave you a nice way using partial derivative so I'll just throw out here another way without using derivatives\n\nThe shortest distance to a plane create angle of $\\pi/2$, so let's create the vector $\\vec v=\\begin{bmatrix}1-a\\\\2-a\\\\3-b\\\\4-b\\end{bmatrix}$, this vector is the vector from the point $(1,2,3,4)$ to an arbitrary point on the plane.\n\nI want that this arbitrary point on the plane will minimize $\\vec v$, so I'll check when the vector create angle of $\\pi/2$ to the plane($\\langle\\cdot,\\cdot\\rangle$ is the dot product):$$\\langle\\vec v,(1,1,0,0)\\rangle=0\\\\\\langle\\vec v,(0,0,1,1)\\rangle=0$$here I get 2 equations: $(1-a)1+(2-a)1=0\\implies a=1.5$ and $(3-b)1+(4-b)1=0\\implies b=3.5$ thus the answer is $u=(1.5,1.5,3.5,3.5)$", "date": "2020-10-30 14:49:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2595145/in-mathbbr4-let-u-span-1-1-0-0-1-1-1-1", "openwebmath_score": 0.98429936170578, "openwebmath_perplexity": 139.23827355764874, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777179025415, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8555710960862588}}
{"url": "https://byjus.com/question-answer/find-the-vector-equation-of-the-plane-passing-through-the-points-2-1-1-and/", "text": "Question\n\n# Find the vector equation of the plane passing through the points $$(2, 1, -1)$$ and $$(-1, 3, 4)$$ and perpendicular to the plane $$x - 2y + 4z = 10.$$ Also show that the plane thus obtained contains the line $$\\vec{r} = -\\hat{i} + 3\\hat{j} + 4\\hat{k} + \\lambda(3 \\hat{i} - 2\\hat{j} - 5\\hat{k}).$$\n\nSolution\n\n## Let the equation of plane through $$(2, 1, -1)$$ be$$a(x - 2) + b(y - 1) + c(z + 1) = 0 ....(i)$$$$\\because (i)$$ passes through $$(-1, 3, 4)$$$$\\therefore a (-1 - 2) + b (3 - 1) + c (4 + 1) = 0$$$$\\Rightarrow -3a + 2b + 5c = 0 ...(ii)$$Also plane (i) is perpendicular to plane $$x - 2y + 4z = 10$$$$\\Rightarrow \\vec{n_1} \\perp \\vec{n_2} \\Rightarrow \\vec{n_1}. \\vec{n_2} = 0$$$$\\therefore 1a - 2b + 4c = 0 ...(iii)$$From (ii) and (iii), we get$$\\dfrac{a}{8 + 10} = \\dfrac{b}{5 + 12} = \\dfrac{c}{6 - 2} \\Rightarrow \\dfrac{a}{18} = \\dfrac{b}{17} = \\dfrac{c}{4} = \\lambda (say)$$$$\\Rightarrow a = 18\\lambda, b = 17\\lambda, c = 4\\lambda$$Putting the value of a, b, c in (i), we get$$18\\lambda (x - 2) + 17\\lambda (y - 1) + 4\\lambda (z + 1) = 0$$$$\\Rightarrow 18x - 36 + 17y - 17 + 4z + 4 = 0$$$$\\Rightarrow 18x + 17y + 4z = 49$$$$\\therefore$$ Required vector equation of plane is$$\\vec{r}. (18\\hat{i} + 17\\hat{j} + 4\\hat{k}) = 49 ...(iv)$$Obviously plane (iv) contains the line$$\\vec{r} = (-\\hat{i} + 3\\hat{j} + 4\\hat{k}) + \\lambda (3\\hat{i} - 2\\hat{j} - 5\\hat{k}) ....(v)$$Since, point $$(-\\hat{i} + 3\\hat{j} + 4\\hat{k})$$ satisfy equation (iv) and vector $$(18\\hat{i} + 17\\hat{j} + 4\\hat{k})$$ is perpendicular to, $$(3\\hat{i} - 2\\hat{j} + 5\\hat{k}),$$ as $$(-\\hat{i} + 3\\hat{j} + 4\\hat{k}). (18\\hat{i} + 17\\hat{j} + 4\\hat{k}) = -18 + 51 + 16 = 49$$ and $$(18\\hat{i} + 17\\hat{j} + 4\\hat{k} ). (3\\hat{i} - 2\\hat{j} - 5\\hat{k}) = 54 - 34 - 20 = 0$$Therefore, (iv) contains line (v).Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-20 05:01:53", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/find-the-vector-equation-of-the-plane-passing-through-the-points-2-1-1-and/", "openwebmath_score": 0.3425378203392029, "openwebmath_perplexity": 917.3506577436331, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771786365549, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8555710957491032}}
{"url": "https://math.stackexchange.com/questions/1771002/riemann-sum-approximations-when-are-trapezoids-more-accurate-than-the-middle-su", "text": "Riemann Sum Approximations: When are trapezoids more accurate than the middle sum?\n\nWe can approximate a definite integral, $\\int_a^b f(x)dx$, using a variety of Riemann sums. If $T_n$ and $M_n$ are the nth sums using the trapezoid and midpoint (middle) sum methods and if the second derivative of $f$ is bounded on $[a, b]$ then does the following theorem imply that $M_n$ \"tends to be more accurate\" then $T_n$?\n\nIf $f''$ is continuous on $[a, b]$ and $|f''(x)| \\leq K$, $\\forall$ $x \\in [a, b]$. Then,\n\n$\\left| \\int_a^b f(x)dx - T_n \\right| \\leq K\\frac{(b-a)^3}{12n^2}$\n\nand\n\n$\\left| \\int_a^b f(x)dx - M_n \\right| \\leq K\\frac{(b-a)^3}{24n^2}$\n\nFor this question let, $E_{T_n} = \\left| \\int_a^b f(x)dx - T_n \\right|$ and $E_{M_n} = \\left| \\int_a^b f(x)dx - T_n \\right|$\n\nThe theorem is presented (without proof) in a calculus 2 book. It only really seems to imply that, we can with 100% certainly bound E_{M_n} smaller than we can bound E_{T_n}. But, that says nothing about the actual values of E_{T_n} or E_{M_n}.\n\nSo, isn't it possible to customize a function so that $E_{T_n} < E_{M_n}$ for a particular n? Could we even make a function so that $E_{T_n} < E_{M_n}$ $\\forall n$?\n\n\u2022 For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $\\int_0^1 f(x)\\, dx = 0$, but $f(1/2) \\ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$). May 4, 2016 at 8:40\n\u2022 This shows $T = 0$, not $E_T = 0$.\n\u2013\u00a0RRL\nMay 5, 2016 at 2:20\n\u2022 Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.\n\u2013\u00a0user123641\nOct 30, 2017 at 15:56\n\nAs you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.\n\nWe can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$\n\nThe midpoint error is\n\n$$E_M = f(c)h - \\int_a^b f(x) \\, dx = \\int_a^b [f(c) - f(x)] \\, dx.$$\n\nUsing a second-order Taylor approximation,\n\n$$f(c) = f(x) + f'(x)(c-x) + \\frac{1}{2} f''(\\xi_x)(x-c)^2,$$\n\nwe see\n\n$$E_M = -\\int_a^b f'(x)(x-c) \\, dx + \\frac{1}{2}\\int_a^b f''(\\xi_x)(x-c)^2 \\, dx.$$\n\nApplying integration by parts to the first integral on the RHS we get\n\n$$\\int_a^b f'(x)(x-c) = \\left.(x-c)f(x)\\right|_a^b - \\int_a^b f(x) \\, dx = \\frac{h}{2}[f(a) + f(b)] - \\int_a^b f(x) \\, dx .$$\n\nNote that this result gives us the error $E_T$ for the trapezoidal method.\n\nHence,\n\n$$E_M = -E_T + \\frac{1}{2}\\int_a^b f''(\\xi_x)(x-c)^2 \\, dx.$$\n\nIt is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.\n\nHere is a somewhat contrived example.\n\nConsider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.\n\n$$f(x) = \\begin{cases} 0.25 + 0.75\\exp(-200 x^2), &\\mbox{if } -1 \\leqslant x \\leqslant 0 \\\\ 0.99 + 0.01 \\cos(\\pi x), &\\mbox{if } \\,\\,\\,\\,\\,\\,\\, 0 < x \\leqslant 1 \\end{cases}.$$\n\nThen\n\n\\begin{align}\\int_{-1}^1 f(x) \\, dx &\\approx 1.2870\\\\ M &\\approx 2 \\\\ T &\\approx 1.2300 \\\\ E_M &\\approx 0.7130 \\\\ E_T &\\approx 0.0570 \\end{align}\n\n\u2022 I think you can use periodic smooth functions to get non-contrived examples.\n\u2013\u00a0Ian\nMay 5, 2016 at 2:49", "date": "2022-11-27 08:34:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1771002/riemann-sum-approximations-when-are-trapezoids-more-accurate-than-the-middle-su", "openwebmath_score": 0.9835159182548523, "openwebmath_perplexity": 194.42908696013507, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777179803135, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8555710933702467}}
{"url": "https://web2.0calc.com/questions/if-three-runners-receive-medals-how-many-ways-can-the-three-medalists-be-chosen-out-of-a-field-of-28", "text": "+0\n\n# If three runners receive medals how many ways can the three medalists be chosen out of a field of 28?\n\n0\n457\n3\n+8\n\nIs it 28 x 27 x 26?\n\npeacemantle \u00a0May 2, 2015\n\n### Best Answer\n\n#2\n+92749\n+10\n\nBadinage, you are right.\n\nThere are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd \u00a0 That is 28*27*26\n\nBUT it doesn't matter what order they are chosen. \u00a0so you have to divide by 3! = 6\n\n$${\\frac{{\\mathtt{28}}{\\mathtt{\\,\\times\\,}}{\\mathtt{27}}{\\mathtt{\\,\\times\\,}}{\\mathtt{26}}}{{\\mathtt{6}}}} = {\\mathtt{3\\,276}}$$\n\nthis is how I would normally do it\n\n28C3\n\n$${\\left({\\frac{{\\mathtt{28}}{!}}{{\\mathtt{3}}{!}{\\mathtt{\\,\\times\\,}}({\\mathtt{28}}{\\mathtt{\\,-\\,}}{\\mathtt{3}}){!}}}\\right)} = {\\mathtt{3\\,276}}$$\n\nMelody \u00a0May 2, 2015\n#1\n+520\n+8\n\nYes, that's the answer your teacher is seeking.\n\nHowever,\u00a0 as a challenge extension you could extend this to include ties, e.g., from 2 up to 28 ties for first, up to 27 ties for second, etc. \u00a0 \u00a0 I'm not sure how to do this, though, and it sounds complicated.\n\n# \ud83c\udf39\u00a0 \ud83c\udf39\u00a0 \ud83c\udf39\u00a0 \ud83c\udf39\u00a0 \ud83c\udf39\u00a0 \ud83c\udf39 \u00a0\ud83c\udf39 \u00a0\ud83c\udf39\n\n.\n\nBadinage \u00a0May 2, 2015\n#2\n+92749\n+10\nBest Answer\n\nBadinage, you are right.\n\nThere are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd \u00a0 That is 28*27*26\n\nBUT it doesn't matter what order they are chosen. \u00a0so you have to divide by 3! = 6\n\n$${\\frac{{\\mathtt{28}}{\\mathtt{\\,\\times\\,}}{\\mathtt{27}}{\\mathtt{\\,\\times\\,}}{\\mathtt{26}}}{{\\mathtt{6}}}} = {\\mathtt{3\\,276}}$$\n\nthis is how I would normally do it\n\n28C3\n\n$${\\left({\\frac{{\\mathtt{28}}{!}}{{\\mathtt{3}}{!}{\\mathtt{\\,\\times\\,}}({\\mathtt{28}}{\\mathtt{\\,-\\,}}{\\mathtt{3}}){!}}}\\right)} = {\\mathtt{3\\,276}}$$\n\nMelody \u00a0May 2, 2015\n#3\n+520\n+5\n\nYes, you are right, Melody. The question doesn't say 1st, 2nd and 3rd places. Maybe there are 3 runners who are to be awarded a medal for being a comically costumed competitor for St Patrick's Day, or some such.\n\nBadinage \u00a0May 3, 2015\n\n### New Privacy Policy\n\nWe use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.\nFor more information: our cookie policy and privacy policy.", "date": "2018-07-15 19:29:18", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/if-three-runners-receive-medals-how-many-ways-can-the-three-medalists-be-chosen-out-of-a-field-of-28", "openwebmath_score": 0.37264788150787354, "openwebmath_perplexity": 1991.436812778065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771766922544, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8555710906730021}}
{"url": "http://mathhelpforum.com/discrete-math/52607-recurrence-relations.html", "text": "# Math Help - recurrence relations\n\n1. ## recurrence relations\n\n1)\nFind a recurrence relation for the number of bit string of length n that contain three consecutive 0's\nWhat are the initial conditions? how many bit strings of length 7 contain three consecutive 0's?\n\n2)\nGive the recursive algorithm for finding the minimum of a finite set of\nintegers.\n\n3)\nFind the solution to the following recurrence relation and initial\ncondition:\nxn+2 = 4xn; n = 0, 1, 2, ... x0 = 2, x1 = 1.\n\n4)\nGive the recursive algorithm for finding the sum of the first n odd\npositive integers.\n\n5)\nFind the solution to the following recurrence relation and initial\ncondition:\nxn+2 = 2xn+1 + 3xn; n = 0, 1, 2, ... x0 = 1, x1 = -1.\n\n2. Hello, narbe!\n\n5) Find the solution to the following recurrence relation and initial condition:\n\n. . $x_{n+2} \\:= \\:2x_{n+1} + 3x_n\\qquad x_0 = 1,\\;\\;x_1 = \\text{-}1$\nCrank out the new few terms . . .\n\n. . $\\begin{array}{ccccc} x_2 &=&2(\\text{-}1) + 3(1) &=& 1 \\\\\nx_3 &=& 2(1) + 3(\\text{-}1) &=& \\text{-}1 \\\\\nx_4 &=& 2(\\text{-}1) + 3(1) &=& 1 \\\\\nx_5 &=& 2(1) + 3(\\text{-}1) &=& \\text{-}1 \\\\\n\\vdots & & \\vdots & & \\vdots \\end{array}$\n\nGot it?\n\n3. ## yeah but ...\n\nyes I got the solution! but how can I demonstrate it as a function ??????\n\n4. Originally Posted by narbe\n1)\nFind a recurrence relation for the number of bit string of length n that contain three consecutive 0's\nWhat are the initial conditions? how many bit strings of length 7 contain three consecutive 0's?\nI think it's easiest to count the number of strings of length n that do not contain three consecutive zeros. Call this number $y_n$.\n\nWe can get a string of length n+1 without three consecutive zeros in three ways: (1) take such a string of length n and add a 1 at the end, (2) take such a string of length n-1 and add 10 at the end, (3) take such a string of length n-2 and add 100 at the end. Therefore $y_{n+1} = y_n+y_{n-1}+y_{n-2}$.\n\nNow let $x_n$ denote the number of strings of length n that do contain three consecutive zeros. There are altogether $2^n$ strings of length n, so $x_n = 2^n-y_n$. It easily follows from the relation for the y's that $x_{n+1} = x_n+x_{n-1}+x_{n-2}+2^{n-2}$. The initial conditions are $x_1=x_2=0,\\ x_3=1$. From there, you can use the recurrence relation to work your way up to 7. (I get the answer $x_7=47$, but don't rely on my arithmetic.)\n\n5. Hello, narbe!\n\nIt's easy . . .\n\nLook at the terms:\n\n. . $\\begin{array}{ccccc}x_0 &=& 1 \\\\ x_1 &=& \\text{-}1 \\\\ x_2 &=& 1 \\\\ x_3 &=& \\text{-}1 \\\\ \\vdots \\\\ x_n &=& (\\text{-}1)^n & \\Leftarrow &\\text{There!}\\end{array}$\n\n6. Originally Posted by narbe\n3) Find the solution to the following recurrence relation and initial\ncondition:\nx_{n+2}= 4x_{n}; n = 0, 1, 2, ... x_{0} = 2, x_{1} = 1.\nThis is a second order difference equation with a constant coefficient, there is a very simple way for this equation.\nSuch equations have solution of the type $x(n)=c_{1}\\lambda_{1}^{n}+c_{2}\\lambda_{2}^{n}$ for $n\\in\\mathbb{N}$, where $c_{1},c_{2}$ are arbitrary constants.\nJust substitute this value into the solution and solve $\\lambda_{1}$ and $\\lambda_{2}$, then use these values together with the initial conditions to find the desired solution.\n\nHowever, here, I would like to show you a different solution way for this equation.\nFirst set, $y(n):=x(2n)$ for $n\\in\\mathbb{N}$, then from the equation, we get\n$y(n+1)=x(2n+2)=4x(2n)=4y(n)$ or simply $y(n+1)=4y(n)$ for $n\\in\\mathbb{N}$.\nIt is easy to solve this equation and we get $y(n)=c_{1}4^{n}$ for all $n\\in\\mathbb{N}$, where $c_{1}$ is an arbitrary constant.\nSince $y(0)=x(0)=2$, we see that $c_{1}=2$.\nThus, we have $x(n)=24^{n/2}=2^{n+1}$ is a solution of the equation.\n\nNext, set $z(n):=x(2n-1)$ for $n\\in\\mathbb{N}$ and solve $z(n+1)=x(2(n+1)-1)=x(2n+1)=4x(2n-1)=z(n)$ or simply $z(n+1)=4z(n)$ for $n\\in\\mathbb{N}$ with $z(1)=x(1)=1$.\n\nFinally, sum these results.", "date": "2014-09-20 21:35:51", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/52607-recurrence-relations.html", "openwebmath_score": 0.896514892578125, "openwebmath_perplexity": 166.87613553863903, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9888419677654415, "lm_q2_score": 0.8652240930029118, "lm_q1q2_score": 0.8555698946830687}}
{"url": "https://nhigham.com/2017/08/14/how-and-how-not-to-compute-a-relative-error/", "text": "# How and How Not to Compute a Relative\u00a0Error\n\nThe relative error in a scalar $y$ as an approximation to a scalar $x$ is the absolute value of $e = (x-y)/x$. I recently came across a program in which $e$ had been computed as $1 - y/x$. It had never occurred to me to compute it this way. The second version is slightly easier to type, requiring no parentheses, and it has the same cost of evaluation: one division and one subtraction. Is there any reason not to use this parenthesis-free expression?\n\nConsider the accuracy of the evaluation, using the standard model of floating point arithmetic, which says that $fl(x \\mathbin{\\mathrm{op}} y) = (x \\mathbin{\\mathrm{op}} y)(1+\\delta)$ with $|\\delta| \\le u$, where $\\mathrm{op}$ is any one of the four elementary arithmetic operations and $u$ is the unit roundoff. For the expression $e_1 = (x-y)/x$ we obtain, with a hat denoting a computed quantity,\n\n$\\widehat{e_1} = \\displaystyle\\left(\\frac{x-y}{x}\\right) (1+\\delta_1)(1+\\delta_2), \\qquad |\\delta_i| \\le u, \\quad i = 1, 2.$\n\nIt follows that\n\n$\\left| \\displaystyle\\frac{e - \\widehat{e_1}}{e} \\right| \\le 2u + u^2.$\n\nHence $e_1$ is computed very accurately.\n\nFor the alternative expression, $e_2 = 1 - y/x$, we have\n\n$\\widehat{e_2} = \\left(1 - \\displaystyle\\frac{y}{x}(1+\\delta_1)\\right) (1+\\delta_2), \\qquad |\\delta_i| \\le u, \\quad i = 1, 2.$\n\nAfter a little manipulation we obtain the bound\n\n$\\left| \\displaystyle\\frac{e - \\widehat{e_2}}{e} \\right| \\le u + \\left|\\displaystyle\\frac{1-e}{e}\\right|(u + u^2).$\n\nThe bound on the relative error in $\\widehat{e_2}$ is of order $|(1-e)/e|u$, and hence is very large when $|e| \\ll 1$.\n\nTo check these bounds we carried out a MATLAB experiment. For 500 single precision floating point numbers $y$ centered on $x = 3$, we evaluated the relative error of $y$ as an approximation to $x$ using the two formulas. The results are shown in this figure, where an ideal error is of order $u \\approx 6\\times 10^{-8}$. (The MATLAB script that generates the figure is available as this gist.)\n\nAs expected from the error bounds, the formula $1-y/x$ is very inaccurate when $y$ is close to $x$, whereas $(x-y)/x$ retains its accuracy as $y$ approaches $x$.\n\nDoes this inaccuracy matter? Usually, we are concerned only with the order of magnitude of an error and do not require an approximation with many correct significant figures. However, as the figure shows, for the formula $|1-y/x|$ even the order of magnitude is incorrect for $y$ very close to $x$. The standard formula $|x-y|/|x|$ should be preferred.\n\n## 2 thoughts on \u201cHow and How Not to Compute a Relative\u00a0Error\u201d\n\n1. Ra\u00fal Mart\u00ednez says:\n\nNick,\n\nThank you for delving into this. Your result is certainly far from obvious and I find it very helpful. In my own work, I don\u2019t recall computing the relative error using 1\u2013y/x (though I cannot be certain), but I have sometimes used the latter expression in writing. After reading your note, I can see that such usage could mislead a reader into using that expression for computation, so I\u2019ll be mindful not to use it in the future.\n\nSincerely,\n\nRa\u00fal Mart\u00ednez\n\n2. For interested parties I converted Nick\u2019s gist above to Python:\n\nThis file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.\nLearn more about bidirectional Unicode characters\n\n # Based on the code by Nick Higham # Compares relative error formulations using single precision and compared to double precision N = 501 # Note: Use 501 instead of 500 to avoid the zero value d = numpy.finfo(numpy.float32).eps * 1e4 a = 3.0 x = a * numpy.ones(N, dtype=numpy.float32) y = [x[i] + numpy.multiply((i \u2013 numpy.divide(N, 2.0, dtype=numpy.float32)), d, dtype=numpy.float32) for i in range(N)] # Compute errors and \"true\" error relative_error = numpy.empty((2, N), dtype=numpy.float32) relative_error[0, :] = numpy.abs(x \u2013 y) / x relative_error[1, :] = numpy.abs(1.0 \u2013 y / x) exact = numpy.abs( (numpy.float64(x) \u2013 numpy.float64(y)) / numpy.float64(x)) # Compute differences between error calculations error = numpy.empty((2, N)) for i in range(2): error[i, :] = numpy.abs((relative_error[i, :] \u2013 exact) / numpy.abs(exact)) fig = plt.figure() axes = fig.add_subplot(1, 1, 1) axes.semilogy(y, error[0, :], '.', markersize=10, label=\"$|x-y|/|x|$\") axes.semilogy(y, error[1, :], '.', markersize=10, label=\"$|1-y/x|$\") axes.grid(True) axes.set_xlabel(\"y\") axes.set_ylabel(\"Relative Error\") axes.set_xlim((numpy.min(y), numpy.max(y))) axes.set_ylim((5e-9, numpy.max(error[1, :]))) axes.set_title(\"Relative Error Comparison\") axes.legend() plt.show()", "date": "2022-12-01 04:34:58", "meta": {"domain": "nhigham.com", "url": "https://nhigham.com/2017/08/14/how-and-how-not-to-compute-a-relative-error/", "openwebmath_score": 0.8781653046607971, "openwebmath_perplexity": 953.7469024119811, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419671077918, "lm_q2_score": 0.8652240930029118, "lm_q1q2_score": 0.8555698941140544}}
{"url": "http://mathhelpforum.com/geometry/174934-centroid-triangle-coordinates.html", "text": "# Thread: The centroid of a triangle with coordinates\n\n1. ## The centroid of a triangle with coordinates\n\nTriangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.\n\nI know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.\n\n2. Originally Posted by darksoulzero\nTriangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.\n\nI know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.\nRead Centroid - AoPSWiki\n\n3. ## centroid of triangle\n\nHi darksoulzero,\n\nYour solution is confusing.Here is a suggested method.\n\nconnect M midpoint of DE and C (centroid) with extended lenght.F lies on this line. FC = 2CM. Slope diagram of C and M = 2/1/2.Slope diagram of FC is twice that or 4/1. Working from point C one point left and 4 points up gives F (2,8)\n\nbjh\n\n4. If ABC is a triangle with $\\displaystyle $$A\\left( {{x_1},{y_1}} \\right),B\\left( {{x_2},{y_2}} \\right),C\\left( {{x_3},{y_3}} \\right)$$$ then its centroid is\n$\\displaystyle $$G\\left( {\\tfrac{{{x_1} + {x_2} + {x_3}}}{3},\\tfrac{{{y_1} + {y_2} + {y_3}}}{3}} \\right)$$$\n\n5. Hello, darksoulzero!\n\n$\\displaystyle \\text{Triangle }D{E}F\\text{ has vertices: }D(1,3)\\text{ and }E(6,1)\\text{, and centroid at }C(3,4).$\n$\\displaystyle \\text{Determine the coordinates of point }F.$\nCode:\n\n1\nF o-----+\n\\    :\n\\   :4\n\\  :\n\\ :\n\\:0.5\n(3,4)o---+\nD       C\\  :\no         \\ :2\n(1,3)   *    \\:\nM o\n(3.5,2)   *     E\no\n(6,1)\n\nWe have vertices $\\displaystyle D(1,3)$ and $\\displaystyle E(6,1)$, and centroid $\\displaystyle C(3,4).$\n\nThe midpoint of $\\displaystyle DE$ is: $\\displaystyle M(3\\tfrac{1}{2},\\,2).$\n\nThe median to side $\\displaystyle DE$ starts at $\\displaystyle \\,M$, passes through $\\displaystyle \\,C,$\n. . and extends to $\\displaystyle \\,F$, where: .$\\displaystyle FC \\,=\\,2\\!\\cdot\\!CM.$\n\nGoing from $\\displaystyle \\,M$ to $\\displaystyle \\,C$, we move up 2 and left $\\displaystyle \\frac{1}{2}$\nHence, going from $\\displaystyle \\,C$ to $\\displaystyle \\,F$, we move up 4 and left 1.\n\nTherefore, we have: .$\\displaystyle F(2,8).$", "date": "2018-03-18 02:38:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/174934-centroid-triangle-coordinates.html", "openwebmath_score": 0.7026728391647339, "openwebmath_perplexity": 1294.70776827927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9908743634192694, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8555126119143079}}
{"url": "https://math.stackexchange.com/questions/712665/find-angles-of-intersection-between-circle-and-line-segment-on-circle", "text": "# Find angle(s) of intersection between circle and line segment on circle\n\nI have a circle (given a center and radius) and a line segment (given a coordinate for both points). The line segment passes through the circle and intersects with the circle in one or two points (one if the segment is tangential or if it ends inside the circle).\n\nHow can I find the angle or the two angles where the line segment intersects the circle, relative to the top of the circle?\n\nPlease provide an example. Unrelated to the diagram below, a set of points to use in an example are as follows: circle radius $1.5$ with center $(3, 4)$. Line segment from $(3, 3)$ to $(6, 7)$.\n\nIf it's easier, it's acceptable to find both points by treating the line segment as a line and ignoring a tangential intersection\n\nI am trying to find the angle shown as ?\u00b0 in the diagram below. The answer in this case is 57\u00b0\n\n\u2022 What is the other point that you're using to define this angle? Yes, the angle in the image may be 57 degrees, but what is the significance of the other point that you're measuring from? What I'm saying is: the point seems arbitrary. So I could pick any point on the circle and have any angle that I want. \u2013\u00a0foobar1209 Mar 16 '14 at 23:40\n\u2022 The point is just the top of the circle. It could be anything, but the top makes the most sense. \u2013\u00a0Keavon Mar 16 '14 at 23:41\n\nSet the center of the circle to be $\\left(0,0\\right)$. Then the set of points on the circle are those such that $x^2+y^2=r^2$ and the set of points on the line segment are solutions to $y=mx+b$ from some $m$ and $b$ (these are easily found given the two endpoints of the line segment). Substituting, we have $$x^2+(mx+b)^2=r^2\\Longleftrightarrow (m^2+1)x^2+2mbx+(b^2-r^2)=0$$ Then this is just a quadratic you can solve with the quadratic formula giving you $x$, and you can get $y$ by $y=mx+b$. However, this gives you two solutions; you must pick the right one (find which is between the two endpoint). Fortunately finding the angle after this is pretty easy: it's just simple trigonometry. If $(x,y)$ is the solution, $\\tan^{-1}\\left(\\frac{y}{x}\\right)$ gives you the angles measure counterclockwise from the $x$-axis. You asked for the angle from the $y$-axis, which is then $\\frac{\\pi}{2}-\\tan^{-1}\\left(\\frac{y}{x}\\right).$\n\nA worked example\n\nI'll work out the example you have given. We have radius $r=\\frac{3}{2}$, center $\\left(3,4\\right)$ and a line segment from $(3,3)$ to $(6,7)$. First we'll move the center to $(0,0)$. We do this by subtracting $(3,4)$ from each point. This gives us the center (as desired) of $(0,0)$, and a line segment from $(0,-1)$ to $(3,3)$. Then the line between the two points is $y=\\frac{4}{3}x-1$. The equation for the circle is $x^2+y^2=\\frac{9}{4}$. Substituting the equation for the line in, we have $$x^2+\\left(\\frac{4}{3}x-1\\right)^2=\\frac{25}{9}x^2-\\frac{8}{3}x+1=\\frac{9}{4}$$ which gives the quadratic $$\\frac{25}{9}x^2-\\frac{8}{3}x-\\frac{5}{4}=100x^2-96x-45=0$$ Then the quadratic equation gives $x=\\frac{96\\pm \\sqrt{96^2+4\\cdot 100\\cdot 45}}{200}=-0.345,1.305$. We're going to want the solution with positive $x$, so $x=1.305$. Then $y=\\frac{4}{3}(1.305)-1=0.740$. To find the desired angles, we merely take $90-\\tan^{-1}\\left(\\frac{0.740}{1.305}\\right)=90-29.56=60.44$. I'm not sure what the discrepancy between my answer and the one you provided is; it could be rounding errors, but it is more likely I just made a calculation error somewhere (please do point it out if you see it).\n\nI know it seems like a lot, but the technique is relatively straight forward, the calculation is just some hard work. I hope this answer has shed some light on the process for you.\n\n\u2022 You made no error. The example I provided is different from the diagram provided in the question. Since this is a detailed and easy to understand answer I will mark it as accepted, but wait a few days before awarding you the bounty. \u2013\u00a0Keavon Mar 17 '14 at 2:03\n\u2022 Sounds good. Glad i could help :) \u2013\u00a0cderwin Mar 17 '14 at 2:26\n\nHint: First, for convenience - transform the circle's to $(0,0)$ and normalize (so the radius of the given circle is now 1). The circle's equation is $x^2+y^2=1$\n\nNow, we are given $(x_1, y_1)$, $(x_2,y_2)$. Denote $f(x)=mx+n$, the line that accepts those two points.\n\nAssuming intersection, $x^2+(mx+n)^2=1\\Leftrightarrow x^2+m^2x^2+2mnx+n^2=1$\n\n\u2022 Please provide an example with circle radius $1.5$ with center $(3,4)$ and line segment from $(3,3)$ to $(6,7)$. \u2013\u00a0Keavon Mar 17 '14 at 0:37\n\u2022 First, transform to (0,0) and normalize, by the transofrmation T(x,y)=(x-3,y-4)/(1.5). Transform all the points with T and you'll get a new problem with the same answer you're looking for. After transforming, find the line and circle equations and continue from there. \u2013\u00a0Astro Nauft Mar 17 '14 at 7:45\n\nHint:\n\n1. Find the equation of the circle and line. The circle should be easy, use the line slope and a point to find the equation.\n\n2. Find the intersection. This is solving for either variable and then plugging it in to get coordinates.\n\n3. Draw a right triangle using a diagonal radius to the intersection and a horizontal line. Based on what you know about slope and tangents, this should help you find an angle that is complementary to your angle.", "date": "2019-05-27 08:07:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/712665/find-angles-of-intersection-between-circle-and-line-segment-on-circle", "openwebmath_score": 0.7090710401535034, "openwebmath_perplexity": 150.47076647157192, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126525529015, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8554704541799163}}
{"url": "https://math.stackexchange.com/questions/2392765/show-that-left-1-fracxn-rightn-is-uniformly-convergent-on-s-0-1", "text": "# Show that $\\left( 1 + \\frac{x}{n}\\right)^n$ is uniformly convergent on $S=[0,1]$. [duplicate]\n\nShow that $\\left( 1 + \\frac{x}{n}\\right)^n$ is uniformly convergent on $S=[0,1]$.\n\nGiven $f_n(x)=\\left( 1 + \\frac{x}{n}\\right)^n$ is a sequence of bounded function on $[0,1]$ and $f:S \\rightarrow \\mathbb R$ a bounded function , then $f_n(x)$ converges uniformly to $f$ iff\n\n$$\\lim\\limits_{n \\rightarrow \\infty} ||f_n - f|| = \\lim\\limits_{n \\rightarrow \\infty}\\left(\\sup |f_n(x) - f(x)| \\right)= 0$$\n\nAs $$f(x) = \\lim\\limits_{n \\rightarrow \\infty} \\left( 1 + \\frac{x}{n}\\right)^n = e^x$$\n\nWe have\n\n$$\\begin{split} \\lim\\limits_{n \\rightarrow \\infty} \\left\\|\\left( 1 + \\frac{x}{n} \\right)^n - e^x\\right\\| &= \\lim\\limits_{n \\rightarrow \\infty} \\left|\\sup_{x \\in S}\\left[\\left( 1 + \\frac{x}{n}\\right)^n - e^x \\right ]\\right|\\\\ &= \\lim\\limits_{n \\rightarrow \\infty} \\left|\\left( 1 + \\frac{1}{n} \\right)^n - e^1 \\right|\\\\ &= \\lim\\limits_{n \\rightarrow \\infty} |e-e| \\\\ &= 0 \\end{split}$$\n\nI am new to sequence. Is this appropriate to show convergence?\n\nGoing back to the definition, How can I show that:\n\n1. \"$f_n(x)=\\left( 1 + \\frac{x}{n}\\right)^n$ is a sequence of bounded function on $[0,1]$\"?\n\n2. $f:S \\rightarrow \\mathbb R$ a bounded function?\n\n\u2022 Continuous functions are bounded on closed intervals. For any $n$, $f_n(x)$ expands to a polynomial, which is continuous anywhere. \u2013\u00a0IntegrateThis Aug 14 '17 at 0:15\n\u2022 The third line in your last calculation is wrong: shouldn't have a limit in front of $|e-e|$. \u2013\u00a0symplectomorphic Aug 14 '17 at 1:23\n\u2022 @Will Fisher: well, sure, as written it doesn't make a difference. But the point is that it in preserving the limit from the second to the third line it looks like the student is arguing that $|(1+1/n)^n-e|$ equals $|e-e|$. \u2013\u00a0symplectomorphic Aug 14 '17 at 1:27\n\u2022 @Will Fisher: no, you have misunderstood the OP. The OP is proving that $f_n\\to f$ uniformly by invoking hypotheses, and the questions at the bottom are about how to show the hypotheses hold. \u2013\u00a0symplectomorphic Aug 14 '17 at 1:29\n\u2022 Have you shown that the sup obtains when $x=1$? It is true, but it seems to me to require proof (and you have the absolute value in the wrong place: as it stands - the absolute value of the sup of the difference-, the sup is 0 at $x=0$; you want the sup of the absolute value of the difference instead). \u2013\u00a0NickD Aug 14 '17 at 1:51\n\nIf you don't want to verify the precise maximum you can bound as\n\n$$0 \\leqslant e^x - \\left(1 + \\frac{x}{n} \\right)^n =e^x \\left[1 - \\left(1 + \\frac{x}{n} \\right)^ne^{-x}\\right] \\\\ \\leqslant e^x \\left[1 - \\left(1 + \\frac{x}{n} \\right)^n \\left(1 - \\frac{x}{n} \\right)^n\\right] \\\\ = e^x \\left[1 - \\left(1 - \\frac{x^2}{n^2} \\right)^n \\right],$$\n\nsince $e^{x/n} > 1 + x/n$ which implies $e^{x} > (1 + x/n)^n$ for $x \\in (-n,\\infty).$\n\nBy Bernoulli's inequality, $(1 - x^2/n^2)^n \\geqslant 1 - x^2/n$ and\n\n$$0 \\leqslant e^x - \\left(1 + \\frac{x}{n} \\right)^n \\leqslant \\frac{e^xx^2}{n} \\leqslant \\frac{e}{n},$$\n\nwhich enables you to prove uniform convergence on $[0,1]$.\n\nLet's write explicitly the difference $$e^x - ( 1 + \\frac{x}{n})^n = \\sum_{ k \\ge 0} \\frac{x^k}{k!} - \\sum_{k = 0}^n \\binom{n}{k}\\left(\\frac{x}{n}\\right )^k=\\\\ =\\sum_{k\\ge 0} \\frac{[1-\\prod_{l=1}^{k-1}(1- l/n)] x^k}{k!}$$. Since every coefficient $1- \\prod_{l=1}^{k-1} (1-\\frac{l}{n})$ is $\\ge 0$ ( and $0$ for $k \\ge n+1$ ) we conclude that\n\n$$0 < e^x - (1+ \\frac{x}{n})^n\\le e^1 - (1 + \\frac{1}{n})^n$$ for all $n \\ge 0$ and $x\\in [0,1]$.\n\nNow, it's only necessary to check ( or use ) that $(1+\\frac{1}{n})^n \\to e$.\n\nAs shown in inequality $(2)$ of this answer, $\\left(1+\\frac xn\\right)^n$ is increasing in $n$. Thus, for $x\\ge0$, \\begin{align} \\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(e^x-\\left(1+\\frac xn\\right)^n\\right) &=e^x-\\left(1+\\frac xn\\right)^{n-1}\\\\ &\\ge e^x-\\left(1+\\frac xn\\right)^n\\\\[3pt] &\\ge0 \\end{align} So on $[0,1]$, $$0\\le e^x-\\left(1+\\frac xn\\right)^n\\le e-\\left(1+\\frac1n\\right)^n$$\n\nThe logarithm $\\ln:[1,e]\\to[0,1]$ is uniformly continuous on this interval, so it is a uniform homeomorphism, and it suffices to show that $\\ln f_n(x)=n\\ln (1+x/n)$ is uniformly convergent on $[0,1]$. But this follows immediately from Taylor's theorem: $$n\\ln(1+x/n)-x=O(x^2/n).$$", "date": "2020-08-15 14:41:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2392765/show-that-left-1-fracxn-rightn-is-uniformly-convergent-on-s-0-1", "openwebmath_score": 0.9416009187698364, "openwebmath_perplexity": 263.70698307808743, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712650690179, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8554704493413514}}
{"url": "https://physics.stackexchange.com/questions/637017/what-type-of-bifurcation-is-this", "text": "# What type of bifurcation is this?\n\nConsider the dynamical system\n\n$$dx/dt = -\\cos(r)\\sin(x)$$\n\nClearly $$x=0$$ and $$x=\\pi$$ are two fixed points of this system. The stability of these two fixed points change as r is varied. Starting from $$r=0$$, the zero fixed point is initially stable but becomes unstable at $$r=\\pi/2$$ and the opposite happens for the other fixed point.\n\nIs this a transcritical bifurcation? I have so far only seen systems where one of the fixed point approaches the other fixed point and their stability switch as they cross each other.\n\nHowever in my system, the fixed points remain at the same values of $$x$$. Only stability changes. If not transcritical, what type of bifurcation is this?\n\nTL;DR: I don't believe the system displays a transcritical bifurcation.\n\nYour stability analysis is correct. For instance, for $$r=0$$, $$\\dot{x}$$ is simply a negative sine function, so positive $$x\\approx 0$$ values have negative derivatives (and decrease, i.e., move towards zero) and negative values have positive derivatives (so increase and also move towards zero), i.e., $$x=x^*=0$$ is stable for $$r<\\pi/2$$. See figure below:\n\nAnd the system does go through a bifurcation as $$r$$ is varied \u2014 though not one of the simple ones, due to the singular nature of the system at $$r=r^*=\\pi/2$$: For this value of $$r$$ all points $$x\\in[0,2\\pi]$$ are (neutrally stable) fixed points, since the system is simply $$dx/dt=0$$. That's how I'd sketch its bifurcation diagram:\n\nIt looks like we could say there's a collision on $$r^*=\\pi/2$$ and the fixed points \"exchange stabilities \u2014 just like in a transcritical bifurcation, however, I think we have something different here, for two reasons:\n\nI. If we check the formal conditions for the system $$\\dot{x}=f_r(x)$$ to go through a transcritical bifurcation (e.g., Guckenheim & Holmes or Wolfram MathWorld), we'll see they are: \\begin{align} f_r(x)\\Bigr|_{\\forall r, x=x^*}=0 \\tag{equilibria}\\\\ \\frac{\\partial f_r}{\\partial x}\\Bigr|_{r=r^*, x=x^*} = 0 \\tag{null eigenvalue}\\\\ \\frac{\\partial^2 f_r}{\\partial r\\partial x}\\Bigr|_{r=r^*, x=x^*} \\ne 0 \\tag{TC1}\\\\ \\frac{\\partial^2 f_r}{\\partial^2 x}\\Bigr|_{r=r^*, x=x^*} \\ne 0 \\tag{TC2} \\label{TC2} \\end{align} where TC stands for transversality condition. And it's clear that our $$f_r(x)$$ fails to satisfy \\ref{TC2}.\n\nII. By changing variables, $$r \\mapsto -r + \\pi/2$$, the system becomes $$f_r(x)=-\\sin(r)\\sin(x)$$ and $$r^*=0$$. If we then try to put $$f_r(x)$$ in a algebraic normal form, by Taylor expanding it around $$(r^*,x^*)$$ we obtain, after a new change of variables $$r\\mapsto r-r^3/6$$, $$f_r(x) = rx -r\\frac{x^3}{6} +O(5)$$ which I don't believe can be put in the normal form for the transcritical bifurcation, nor any of the other simple ones: \\begin{align} \\dot{x} &= r - x^2 \\tag{saddle-node}\\\\ \\dot{x} &= rx - x^2 \\tag{transcritical}\\\\ \\dot{x} &= r - x^3 \\tag{pitchfork} \\end{align} Which is perhaps not that surprising, given the system's bifurcation diagram.\n\n\u2022 I missed out a negative sign in the equation. Thanks for pointing it out. If not transcritical, does this type of bifurcation have some name to it that I can search for? May 17 at 18:47\n\u2022 @SnehaSrikanth Not that I know, maybe we could call it transcritical-like? :) May 17 at 18:50\n\u2022 @SnehaSrikanth I'm now convinced the system doesn't display a transcritical bifurcation, check my updated answer. May 18 at 19:24\n\u2022 Thanks for such a clear detailed answer! I still want to keep the question open for anyone who might be able to pinpoint what kind of bifurcation this is. I am new to StackExchange.. do I edit this post and change the question or make a new question? I do not want your answer to become irrelevant. May 19 at 11:44\n\u2022 Ok will make that change! I am actually unable to upvote yet (I need 15 reputations for that). I'll surely upvote when I hit 15 reputations :D May 19 at 16:14", "date": "2021-09-28 14:06:07", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/637017/what-type-of-bifurcation-is-this", "openwebmath_score": 0.9300766587257385, "openwebmath_perplexity": 719.3710125165959, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856414, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8554704482329465}}
{"url": "https://math.stackexchange.com/questions/3461890/proof-that-forall-k-in-mathbb-z-lfloor-log-22k1-rfloor-lfloor-log-2", "text": "# Proof that $\\forall k\\in\\mathbb Z^+$, $\\lfloor\\log_2(2k+1)\\rfloor=\\lfloor\\log_2(2k)\\rfloor$\n\nI wrote a proof for this but am not completely sure if this is valid. Specifically, can I go from $$m\\leq k+0.5$$ to $$m\\leq k$$ because $$m,k\\in\\mathbb Z^+$$ with no further explanation? Any feedback would be greatly appreciated!\n\nProof. Let $$n=\\lfloor\\log_2(2k+1)\\rfloor$$. Then, \\begin{align*} n&\\leq\\log_2(2k+1)\\lt n+1&&\\text{(since \\forall x\\in\\mathbb R, \\lfloor x\\rfloor\\leq x\\lt x+1),}\\\\ \\implies2^n&\\leq2k+1\\lt2^{n+1}&&\\text{(by algebra).}\\\\ \\end{align*} Observe that $$2^n\\in2\\mathbb Z^+$$ so $$2^n=2m$$ for some $$m\\in\\mathbb Z^+$$. Thus, \\begin{align*} 2m&\\leq2k+1&&\\text{(since 2m=2^n\\leq2k+1),}\\\\ \\implies m&\\leq k+0.5&&\\text{(by algebra),}\\\\ \\implies m&\\leq k&&\\text{(since m,k\\in\\mathbb Z^+),}\\\\ \\implies 2m&\\leq2k&&\\text{(by algebra),}\\\\ \\implies 2^n&\\leq2k&&\\text{(since 2^n=2m),}\\\\ \\implies 2^n&\\leq2k\\lt2^{n+1}&&\\text{(since 2k+1\\lt2^{n+1}\\rightarrow2k\\lt2^{n+1}),}\\\\ \\implies n&\\leq\\log_2(2k)\\lt n+1&&\\text{(by algebra),}\\\\ \\therefore n&=\\lfloor\\log_2(2k)\\rfloor&&\\text{(since \\forall x\\in\\mathbb R, \\lfloor x\\rfloor\\leq x\\lt x+1).} \\end{align*} $$\\tag*{\\blacksquare}$$\n\n\u2022 In your latter paragraph it seems you can just go from line $1$ to lines $4/5$ Dec 3, 2019 at 21:29\n\u2022 Your proof looks good to me! Dec 3, 2019 at 21:37\n\nNote that for all $$n \\in \\mathbb{N_{>0}}$$ there exist $$m \\in \\mathbb{N}$$ such that $$\\begin{eqnarray*} 2^m \\leq n < 2^{m+1}. \\end{eqnarray*}$$ This value $$m$$ is $$\\lfloor\\log_2(n)\\rfloor$$ And $$n$$ increases passed a power of $$2$$ the value $$m$$ will increase by $$1$$.\nIf $$n$$ increases (by $$1$$) from an even value to an odd value the $$m$$ will not change (provided $$n>1$$). So $$\\begin{eqnarray*} \\lfloor\\log_2(2k+1)\\rfloor = \\lfloor\\log_2(2k)\\rfloor. \\end{eqnarray*}$$", "date": "2022-08-17 18:49:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3461890/proof-that-forall-k-in-mathbb-z-lfloor-log-22k1-rfloor-lfloor-log-2", "openwebmath_score": 0.9861103892326355, "openwebmath_perplexity": 1664.525540282596, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126451020108, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8554704476672624}}
{"url": "https://math.stackexchange.com/questions/482453/sketch-the-region-enclosed-by-the-given-curves-and-find-its-area", "text": "# sketch the region enclosed by the given curves and find its area\n\nsketch the region enclosed by the given curves and find its area:\n\n$$y=\\frac 1x,\\; y=x,\\; y=\\frac x4,\\; x>0.$$\n\nI have no problem sketching the area between the curves but there are three, and only one constant value, so I don't know what to put as my second a/b value. I tried using 1/x as a b value but that just gave me an equation answer and the answer isn't an equation.\n\nedit: i forgot about the intersections as constant values. but now I see how to split them up.\n\nThis is chapter 6.1 calculus James Stewart btw (area between curves)\n\n\u2022 Just FYI: there are many calculus textbooks each with their own chapter/section scheme. \"Chapter 6.1\" isn't useful information without telling us what textbook you're using. \u2013\u00a0Adam Saltz Sep 2 '13 at 21:28\n\nFrom Wolfram Alpha, we can sketch the curves to find the area of interest:\n\nNote that we need to find the points of intersection: at $x = 0$ the lines $y = x, \\;y = \\frac x{4}$ intersect. At $x= 1$, the lines $y = x$ and $y = \\frac 1x$ intersect. At $x = 2,$ the lines $y = \\frac 1x$ and $y = \\frac x4$ intersect. You can solve this by integrating between the relevant curves from $x = 0$ to $x = 1$, and likewise integrating between the relevant curves between $x = 1$ and $x = 2$, then summing: $$\\int_0^1 \\left(x - \\frac x4\\right)\\,dx \\quad + \\quad \\int_1^2 \\left(\\frac 1x - \\frac x4\\right)\\,dx$$\n\n\u2022 ok I solved it but I got ln2 - 1/2 and the answer is ln2 O_o i double checked calculations what did i do wrong?? \u2013\u00a0J L Sep 2 '13 at 21:39\n\u2022 You have to evaluate the second integral $F(2) - F(1)$ \u2013\u00a0Namaste Sep 2 '13 at 21:42\n\u2022 First integral $\\frac 12 - \\frac 18 = \\frac 38$. Second integral: $\\ln 2 - \\frac 12 - (\\ln 1 - \\frac 18) = \\ln 2 - \\frac 38$. Sum, we get $\\ln 2$. \u2013\u00a0Namaste Sep 2 '13 at 21:44\n\u2022 i figured it out i wrote 7/8 instead of 3/8 for some reason \u2013\u00a0J L Sep 2 '13 at 21:45\n\u2022 Good then, that should agree. Is all good? \u2013\u00a0Namaste Sep 2 '13 at 21:47\n\nHint: Suppose that we integrate with respect to $x$. Since there are two types of upper curves, draw a vertical line at $x=1$ (where the two upper curves of $y=x$ and $y=1/x$ intersect) that splits the region into two cases. You should obtain: $$\\left[\\int_0^1 x - \\frac x4~dx \\right] + \\left[\\int_1^2 \\frac 1x - \\frac x4~dx \\right]$$\n\n\u2022 ohhh yeahh i see. thanks for your help :) I forgot about spliting things into two integrals \u2013\u00a0J L Sep 2 '13 at 21:26\n\nIf you sketched the region correctly then you should be seeing a sort of triangle whose base is given by the line $y=x/4$ whose left side is given by $y=x$ and whose right side is given my $y=1/x$.\n\nThe first thing you have to do is to identify the crossing points, that is, the endpoints of this distorted triangle. One crossing point is obviously given by $(0,0)$ where the lines $y=x$ and $y=x/4$ meet, for the next one you have to solve the equation $$\\frac{1}{x}=x$$ from where you obtain $x=1$ (we care only about the positive solution since $x>0$). The other vertex of the desired region is given by solving $$\\frac{1}{x}=\\frac{x}{4}$$ from where you can obtain the solution $x=2$. Look at your sketched region, we will integrate in terms of x-slices. Usually one integrates the area between the two curves $f(x)$ and $g(x)$ in the $x$-region $[a,b]$ as $\\int_a^b f(x)-g(x) \\; dx$. In this case however, the expression for the upper limit changes exactly at $x=1$ so there should be TWO subtractions instead of one. The desired expression for the area $A$ is $$A = \\int_{0}^1 x-\\frac{x}{4}\\; dx + \\int_{1}^2 \\frac{1}{x}-\\frac{x}{4} \\; dx$$ And you should be able to calculate that integral yourself. Hope that helps.\n\nMake a scetch of all three curves over the domain $x>0$. Can you see how they form and bound a triangle-like figure?\n\nWhat do you mean by \"only one constant value\"? If you sketch the three curves, you'll see two regions which look like triangles with one bent edge (the $y=1/x$ part). In one of those regions, all the points have positive $x$-coordinates. In the other, the $x$-coordinates are all negative. Integrate to find the area of the positive one.", "date": "2019-11-15 16:37:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/482453/sketch-the-region-enclosed-by-the-given-curves-and-find-its-area", "openwebmath_score": 0.9202216863632202, "openwebmath_perplexity": 248.53720058628775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754474655619, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8554255136568153}}
{"url": "https://math.stackexchange.com/questions/409626/is-the-cartesian-product-of-groups-the-product-of-a-normal-subgroup-and-its-quot", "text": "# Is the cartesian product of groups the product of a normal subgroup and its quotient group?\n\nI'm studying elementary group theory, and just seeing the ways in which groups break apart into simpler groups, specifically, a group can be broken up as the sort of product of any of its normal subgroups with the quotient group of that subgroup. So I wondered how you could do the inverse of that operation:\n\n1. Given two groups $A$ and $B$, construct a group $G$ which admits a normal subgroup $H$ isomorphic to $A$, such that $G/H$ is isomorphic to $B$.\n\nI think I have a proof that the cartesian product $A \\times B$ (with the usual component-wise operation) verifies (1), but since I'm just starting out I'm not totally confident in my construction. Furthermore, if I'm right, is this the only group up to isomorphism satisfying (1)?\n\nEdit: I just noticed Proving the direct product D of two groups G & H has a normal subgroup N such that N isomorphic to G and D/N isomorphic to H, which seems to positively answer my question. In that case I'd like to draw attention to the follow up question above (uniqueness up to isomorphism).\n\nYes, the direct product $A \\times B$ satisfies the property, as you've noticed. But it's not unique up to isomorphism. For example, the dihedral group $D_n$ has a normal subgroup $H \\simeq \\mathbb Z/n \\mathbb Z$, with $G/H \\simeq \\mathbb Z/2 \\mathbb Z$, but $D_n$ is not isomorphic (for $n > 1$) to $\\mathbb Z/2\\mathbb Z \\times \\mathbb Z/n\\mathbb Z$.\n\nMore generally, you're asking whether, given an exact sequence of the form $1 \\to N \\to G \\to H \\to 1$, is $G$ isomorphic to $N \\times H$? The answer is no, as I've shown. Many counterexamples are provided by semidirect products (something you'll learn soon enough if you're studying elementary group theory).\n\nFor abelian groups, the concept of Ext functor allows one to classify all such extensions (given abelian groups $A,B$, \"how many\" groups $G$ are there with an exact sequence $0 \\to B \\to G \\to A \\to 0$ is given by $\\mathrm{Ext}(A,B)$), but this is much more advanced.\n\n\u2022 If each of A,B,C is abelian and the sequence 0 -> A -> B -> C -> 0, what would be some restrictions on A, C that would ensure that B is isomorphic to A x C? What if A and C are cyclic? \u2013\u00a0gen Oct 24 '18 at 22:27\n\nYou're correct that $A\\times B$ satisfies (see here) the stated property, but in general it will not be the only such group.\n\nThe simplest example is with $A=B=\\mathbb{Z}/2\\mathbb{Z}$, in which case $\\mathbb{Z}/4\\mathbb{Z}$ also has the desired property.\n\nYour idea about the direct product working is true. The general idea you are looking for however is that of a semidirect product which refutes your claim of uniqueness in the general setting.\n\nIt's interesting to ask when the direct product is the only possibility and that can actually be answered by finding the group $\\text{Ext}(B,A)$. This is know as an extension problem. We have that if $\\text{Ext}(B,A)=1$ then $A\\times B$ is the unique extension of $B$ by $A$.\n\nI believe the converse is false however, but I can't think of an example at the moment. That is, there exist pairs $(B,A)$ such that $\\text{Ext}(B,A)\\neq 1$ but $B\\times A$ is the unique (up to group isomorphism, not extension equivalence) extension of $B$ by $A$.", "date": "2019-10-15 19:09:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/409626/is-the-cartesian-product-of-groups-the-product-of-a-normal-subgroup-and-its-quot", "openwebmath_score": 0.9384512305259705, "openwebmath_perplexity": 122.5503894866553, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682461347529, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8554195796709044}}
{"url": "http://math.stackexchange.com/questions/185971/how-do-i-solve-a-continued-fraction-of-solution-to-quadratic-equation", "text": "# How do I solve a Continued Fraction of solution to quadratic equation?\n\nI know that it is possible to make a CF (continued fraction) for every number that is a solution of a quadratic equation but I don't know how.\n\nThe number I'd like to write as a CF is:\n\n$$\\frac{1 - \\sqrt 5}{2}$$\n\nHow do I tackle this kind of problem?\n\n-\n\nLet's rewrite $\\ \\dfrac {1-\\sqrt{5}}2=-1+\\dfrac {3-\\sqrt{5}}2$ to have something positive to evaluate then : $$\\frac 1{\\dfrac {3-\\sqrt{5}}2}=\\frac 2{3-\\sqrt{5}}=\\frac {2(3+\\sqrt{5})}{(3-\\sqrt{5})(3+\\sqrt{5})}=\\frac {2(3+\\sqrt{5})}{9-5}=\\frac {3+\\sqrt{5}}{2}=2+\\frac {\\sqrt{5}-1}{2}$$ (we want the term at the right to be between $0$ and $1$ at each stage)\n\nYou may continue this process until repetition !\n\nYou should get : $$\\dfrac {1-\\sqrt{5}}2=-1+\\cfrac 1{2+\\cfrac 1{1+\\cfrac 1{1+\\ddots}}}$$\n\n-\nSo this solution is [-1;2,1,1,...], isn't that different from the one in the post above? \u2013\u00a0Spyral Aug 23 '12 at 18:51\n@Spyral: Sasha has a minus sign over the whole C.F. I have only the integer part with a sign (both results are correct, perhaps that this one is more common...) \u2013\u00a0Raymond Manzoni Aug 23 '12 at 18:57\nThat confused me indeed.. I thought a C.F. was UNIQUE.. so your solution being [-1,2,1,1,...], sasha's being -[0;1,1,...] and Brian M. Smiths being [0;1,1,...] is kind of confusing me =/ \u2013\u00a0Spyral Aug 23 '12 at 19:08\nEven though all these methods might be correct, this one was the clearest to me, thanks \u2013\u00a0Spyral Aug 23 '12 at 19:12\n@Spyral: There are many types of continued fraction. The standard one in Number Theory is the simple continued fraction given in the very nice answer by Raymond Manzoni. For these, there is almost a uniqueness theorem (depending on how you define things, a rational will have two simple continued fraction expansions that are minor variants of each other). For generalized continued fractions, there is definitely not uniqueness. \u2013\u00a0Andr\u00e9 Nicolas Aug 23 '12 at 19:14\n\nSuppose $x$ is a root of $p(z) = z^2 - b z - c$. Then, diving $p(z)$ over $z$ and solving that for $z$ gets us $$z = b+ \\frac{c}{z}$$ Iterating: $$z = b + \\cfrac{c}{b + \\cfrac{c}{z}} = \\cfrac{b}{c + \\cfrac{c}{b+ \\ddots}}$$ Since $\\frac{1-\\sqrt{5}}{2}$ is a root of $z^2 - z -1$ we have: $$\\frac{1-\\sqrt{5}}{2} = -\\frac{1}{\\frac{1+\\sqrt{5}}{2}} = - \\cfrac{1}{1 + \\cfrac{1}{1+ \\frac{1}{1+\\ddots}}}$$\n\nAdded: Consider a sequence, defined by $x_{n+1} = b + \\frac{c}{x_n}$, with $x_0 = \\frac{c}{y}$. Few initial terms of the sequence are $\\frac{c}{y}$, $b + y$, $b + \\cfrac{c}{b+y}$, $b + \\cfrac{c}{b + \\cfrac{c}{b+y}}$, etc. It is well known that terms of this sequence can also be obtained as a ratio of two solutions, $a_n$ and $b_n$ to the following recurrence equation: $$v_{n} = b v_{n-1} + c v_{n-2} \\tag{1}$$ with initial conditions $a_0=y$, $a_1 = c$ and $b_0 = 1-\\frac{b}{c} y$, $b_1 = y$. Then $x_n = \\cfrac{a_n}{b_n}$. The solution to $(1)$ has the form: $$v_{n} = v_0 \\frac{z_1 z_2^n - z_2 z_1^n}{z_1-z_2} + v_1 \\frac{z_1^n - z_2^n}{z_1-z_2}$$ where $z_1$ and $z_2$ are the two roots of $z^2 - b z -c = 0$. Assume $z_2>z_1$. In the large $n$ limit, $$\\lim_{n \\to \\infty} x_n = \\lim_{n \\to \\infty} \\frac{a_n}{b_n}= \\lim_{n\\to \\infty} \\frac{a_1 (z_1^n - z_2^n) + a_0 (z_1 z_2^n - z_1^n z_2)}{b_1 (z_1^n - z_2^n) + b_0 (z_1 z_2^n - z_1^n z_2)} = \\frac{a_0 z_1 - a_1}{b_0 z_1 - b_1} = \\frac{c (c- y z_1)}{c (y - z_1) + b y z_1}$$ In the case at hand $x_0 = \\infty$ and $x_1 = b$, corresponding to the limit of $y \\to \\infty$, in which case the value of the continued fraction becomes: $$\\lim_{n \\to \\infty} x_n = \\lim_{y \\to 0} \\frac{c (c- y z_1)}{c (y - z_1) + b y z_1} = - \\frac{c}{z_1} = z_2$$\n\n-\nThe result of the continued fraction will be the larger of two roots. With $b>0$ and $c>0$ both roots are real. \u2013\u00a0Sasha Aug 23 '12 at 18:49\nHow do you get the other root? Can you give some intuition about why, when you pass from the iterations to the limit, one of the roots \"disappears\"? \u2013\u00a0Rahul Aug 23 '12 at 18:52\n@RahulNarain The product of roots of the quadratic polynomial is the free term. Thus, the smaller root is $z_2 = -\\frac{c}{z_1}$. \u2013\u00a0Sasha Aug 23 '12 at 18:54\nOK, but that doesn't answer my second question, because surely $z_2$ also satisfies $z=b+c/z$, $z = b+c/(b+c/z)$, and so on. \u2013\u00a0Rahul Aug 23 '12 at 19:03\n@RahulNarain I have expanded the post, addressing your second question now. \u2013\u00a0Sasha Aug 23 '12 at 19:26\n\nThe general procedure is as follows for positive $x$. Let $x_0=x$, and let $[a_0;a_1,a_2,\\dots]$ be the desired CF expansion. Then $a_0=\\lfloor x_0\\rfloor$. Given $x_n$ and $a_n$, let $$x_{n+1}=\\frac1{x_n-a_n}$$ and $a_{n+1}=\\lfloor x_{n+1}\\rfloor$.\n\nSince $\\frac12(1-\\sqrt5)$ is negative, let\u2019s work with its absolute value, $x=\\frac12(\\sqrt5-1)$. Clearly $0\\le x<1$ so $a_0=0$. Then $$x_1=\\frac1x=\\frac2{\\sqrt5-1}=\\frac{2(\\sqrt5+1)}4=\\frac{1+\\sqrt5}2\\;;$$ so $$\\lfloor x_1\\rfloor=\\left\\lfloor\\frac{1+\\sqrt5}2\\right\\rfloor=1\\;,$$ since $2\\le\\sqrt5<3$, and $a_1=1$.\n\nNow $$x_2=\\frac1{x_1-1}=\\frac2{\\sqrt5-1}=x_1\\;,$$ so everything repeats: $a_2=1$, $x_3=x_1$, $a_3=1$, etc. Thus, $x=[0;1,1,1,\\dots]$, and your number is the negative of this.\n\n-", "date": "2015-11-28 08:01:15", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/185971/how-do-i-solve-a-continued-fraction-of-solution-to-quadratic-equation", "openwebmath_score": 0.9445370435714722, "openwebmath_perplexity": 253.13043857529271, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682444653242, "lm_q2_score": 0.8652240704135291, "lm_q1q2_score": 0.8554195627648858}}
{"url": "https://www.jiskha.com/questions/1041569/a-radioactive-substance-decays-according-to-the-formula-q-t-q0e-kt-where-q-t-denotes", "text": "# Math\n\nA radioactive substance decays according to the formula\nQ(t) = Q0e\u2212kt\nwhere Q(t) denotes the amount of the substance present at time t (measured in years), Q0 denotes the amount of the substance present initially, and k (a positive constant) is the decay constant.\n(a) Find the half-life of the substance in terms of k.\n\n(b) Suppose a radioactive substance decays according to the formula\nQ(t) = 36e\u22120.0001074t\nHow long will it take for the substance to decay to half the original amount? (Round your answer to the nearest whole number.)\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n3. \ud83d\udc41 1,867\n1. .5 = e^-k T\nln .5 = - k T\n\nT = -.693/-k = .693/k\n\nb)\nT = .693 / .0001074 = 6454 years\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 3\n2. A radioactive substance decays according to the formula\nQ(t) = Q0e\u2212kt\nwhere Q(t) denotes the amount of the substance present at time t (measured in years), Q0 denotes the amount of the substance present initially, and k (a positive constant) is the decay constant.\n(a) Find the half-life of the substance in terms of k.\n\n(b) Suppose a radioactive substance decays according to the formula\nQ(t) = 36e\u22120.0001238t\nHow long will it take for the substance to decay to half the original amount? (Round your answer to the nearest whole number.)\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n\n## Similar Questions\n\n1. ### calculus\n\nA sample of a radioactive substance decayed to 94.5% of its original amount after a year. (Round your answers to two decimal places.) (a) What is the half-life of the substance? (b) How long would it take the sample to decay to\n\n2. ### algebra 2\n\nIn the formula A(t) = A0ekt, A is the amount of radioactive material remaining from an initial amount A0 at a given time t, and k is a negative constant determined by the nature of the material. An artifact is discovered at a\n\n3. ### calculus\n\nassume the half-life of the substance is 31 days and the initial amount is 183.9 grams. (a) Fill in the right hand side of the following equation which expresses the amount A of the substance as a function of time t (the\n\n4. ### Calc\n\nA sample of a radioactive substance decayed to 93.5% of its original amount after a year. a) What is the half-life of the substance? ? years (b) How long would it take the sample to decay to 10% of its original amount? ? years\n\n1. ### precalculus\n\nTwenty percent of a radioactive substance decays in ten years. By what percent does the substance decay each year?\n\n2. ### Exponential Modeling\n\nThe half-life of a radioactive substance is one day, meaning that every day half of the substance has decayed. Suppose you have 100 grams of this substance. How many grams of the substance would be left after a week?\n\n3. ### Calculus\n\nThe radioactive element polonium decays according to the law given below where Q0 is the initial amount and the time t is measured in days. Q(t) = Q0 \u00b7 2-(t/140) If the amount of polonium left after 700 days is 45 mg, what was\n\n4. ### Calculus-Modeling Equations\n\nAn unknown radioactive element decays into non-radioactive substances. In 420 days the radioactivity of a sample decreases by 39 percent. 1.What is the half-life of the element? 2.How long will it take for a sample of 100 mg to\n\n1. ### Calculus\n\nThe rate of decay is proportional to the mass for radioactive material. For a certain radioactive isotope, this rate of decay is given by the differential equation dm/dt = -.022m, where m is the mass of the isotope in mg and t is\n\n2. ### calculus\n\nradioactive Carbon 14 in a dead organism decays according to the equation A=A e^.000124t where t is in years and A0 is the amount at t=0. Estimate the age of a skull uncovered in an archeological site if 10% of the original Carbon\n\n3. ### mathematics\n\nThe exponential formula for the half-life of a radioactive isotope is y=y0ekt, where y is the amount of the isotope remaining after t years, y0 is the initial amount of the isotope, k is the decay constant, and e is the\n\n4. ### Math/Calculus\n\nAfter 5 days, a particular radioactive substance decays to 37% of its original amount. Find the half life of this substance. Is this right? Th= ln 2/ln(1/.37) x 5 days", "date": "2020-09-29 17:55:26", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1041569/a-radioactive-substance-decays-according-to-the-formula-q-t-q0e-kt-where-q-t-denotes", "openwebmath_score": 0.8712039589881897, "openwebmath_perplexity": 883.4274277294363, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575188391107, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8554119002026702}}
{"url": "https://matheducators.stackexchange.com/tags/number-theory/hot", "text": "Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now\n\n# Tag Info\n\n14\n\nThe way I have explained the fundamental theorem of arithmetic in the past is by establishing what it means to be prime (has exactly two positive divisors) and then having students construct factor trees (where the prime factors at the end are circled). The prime definition avoids some of the caveat-language otherwise seen (\"a number only divisible by ...\n\n14\n\nIt might not be possible to get your brother to arrive at the proof himself, no matter how much you scaffold it. (\"You can lead a horse to water\" and all that.) If he's into maths and appreciates a good proof, you might get the desired enthusiasm by just showing him the proof! That said, here's an idea. Forget primes for a moment, and think about ...\n\n14\n\nI think it turns out that \"perfect\" numbers do not interact much with other parts of number theory. Some of these very old, elementary, very ad-hoc definitions of special classes of integers have proven (and will prove) to interact interestingly with other ideas, but some seem not to. It's not easy for a beginner to guess the significance or subtlety of one ...\n\n13\n\nI don't really answer the question but: why do you want your brother to come to the answer right now? Now, your brother understands that, to prove that the set of prime numbers is infinite, you can assume it is finite. And, from this pool of prime numbers, you can construct a new one that is not in the family. Why don't you let your brother play with that ...\n\n12\n\nThis is indeed tricky, and it seems to me the most effective way (in far more general, similar situations) is to show them the problem would be to have them apply their method to another, close problem where the answer is actually opposite. Either they will explain why it does not apply there, and you can argue that the difference is subtle enough to warrant ...\n\n11\n\nTo really understand why the integers $\\mathbb{Z}$ have unique prime factorization, it helps to understand how unique prime factorization can fail in other settings. For example, $$(2 + \\sqrt{10}) \\cdot (2 - \\sqrt{10}) = -6 = -2 \\cdot 3,$$ so prime factorizations in $\\mathbb{Z}[\\sqrt{10}] = \\{a + b\\sqrt{10}: a, b \\in \\mathbb{Z}\\}$ aren't unique. On the other ...\n\n10\n\nWhat's wrong is that most of the justification is missing. Why can $N = p^2/q^2$ only happen if $q^2 = 1$? This can be justified using the fundamental theorem of arithmetic (which states that integers have unique prime factorization), but that justification needs to be made explicit, and they need to make sure that the fundamental theorem of arithmetic has ...\n\n10\n\nI would recommend Python combined with SageMath, as already recommended by Joseph O'Rourke, or rather SageMath and Python comes naturally. Python is a modern, and widely used, interpreted language (no compilation needed) it supports big integers via the bignum type. (But using SageMath I think this is tangential, I mention it for completeness mainly.) ...\n\n9\n\nThe two statements aren't literally \"the same\", because as the student observed, they say different things. However, they are logically equivalent: each implies the other. (Similarly, \"4/2\" isn't literally the same expression as \"5 - 3\", but they provably represent the same number.) If we could only prove things by repeating the same statement, we couldn't ...\n\n9\n\nMaybe the issue is that if the values you're multiplying have units, then the result of multiplying will have the product of those units. Therefore, your result can't really be equivalent to a value in the base set, because the units are different. Therefore, if applying this to the real world, I suggest considering the repeated-addition interpretation (...\n\n8\n\nThere is some value beyond the algorithm to insist on the fact that quotient and remainder in an Euclidean division are uniquely determined as soon as one settles on some convention on the remainder. What they are exactly depends on ones convention (non-negative remained, smallest absolute value, or still something else) but if one fixes a convention then ...\n\n8\n\nAnother classic is the following: A rectangular floor measures $300 \\text{ cm} \\times 195 \\text{ cm}$. What is the largest square tiles that can be used to cover the floor exactly?\n\n8\n\nGoing back to Euclid, I have found questions such as `Given a large supply of rods of length $15$ and $21$, what lengths can be measured?' can appeal to students. This also motivates the result $\\gcd(a,b) = ra+sb$ of the (extended) Euclidean Algorithm.\n\n8\n\nSome nice geometrical applications arise in the analysis of periodical curves such as Roulettes (Spirograph curves), Star Polygons, etc. Concrete experience with implementations in toys like Spirograph also provides excellent motivation for more abstract concepts such as cyclic groups.\n\n8\n\nI wrote an article for NCTM's grades 8-14 journal, The Mathematics Teacher, which is about building towards ${\\rm F}\\ell{\\rm T}$ by thinking through ways in which the following statement can be generalized: If you subtract a natural number from its square, then the result is even. I wrote more about this in some earlier StackExchange posts, e.g., MESE ...\n\n8\n\nWhat is the name of this subdiscipline in math education? In one of the comments, Dave L Renfro has a reference to Piaget, whose work was primarily done with younger schoolchildren. With respect to extending this work into the older years of one's education, a potentially good place to look would be APOS Theory, which is due to Ed Dubinsky and collaborators....\n\n7\n\n(Summary: I would suggest exploring it flexibly, but ensuring students also know the \"rigid\" version.) Rather than directly addressing Euclid's algorithm for the $\\gcd$ of two whole numbers, I believe one witnesses similar phenomena when covering the standard algorithm for division. For example, I observe analogs with your remarks of: The flexible method ...\n\n7\n\nMathematically, even perfect numbers give a good number theory example to the general idea of classification, i.e. all even perfects have a specific form. I use perfect numbers in my number theory class for two or three pedagogical reasons: with some trial and error (and the help of some computational software), I have the students essentially discover ...\n\n7\n\nWilson's Theorem is very powerful for proving things related to quadratic residues, and I think also in general a good theoretical tool. Motivating it by just saying \"hey let's multiply everything together\" seems very reasonable and even playful. Why is this theoretical? It is basically saying that if you multiply all units together you get a specified ...\n\n7\n\nThe way I understand the question is: If students are not taught fractions, but instead formal deductive proofs of properties of natural numbers, would they learn mathematics better? I find it unlikely. Students struggle with fractions in primary school. Students usually struggle with proofs in gymnasium or university. It seems unlikely that primary school ...\n\n6\n\nI think using actual small primes actually detracts from finding the solution. If you are thinking about finding a number that is not divisible by 2,3, or 5, it is easy to come up with one (11) without having to use a formula. Thus, I would get him thinking about primes more symbolically. Given 3 primes, $p_1, p_2, p_3$ what is their LCM? What is ...\n\n6\n\nYou probably already know that $1/\\zeta(2)$ is the probability of two \"random\" integers being relatively prime. Since this constant is related to number theory, you should expect any proof of the Basel formula to be hard work. Probably the best proof from the point of view of transparency is obtained by evaluating \\int_0^1\\int_0^1 \\frac{1}{1-xy}{\\rm d}x{\\...\n\n6\n\nWhat computer languages might one recommend for, say, investigations in number theory? I find Mathematica ideal, e.g.: \"Mod sequences that seem to become constant; and the number 316\" \"Does 53 diverge to infinity in this Collatz-like sequence?\" But: (a) there is a huge start-up learning curve, and (b) Mathematica is not free. Because of the latter, I ...\n\n6\n\nHere's a word problem for the greatest common divisor: 12 boys and 15 girls are to march in a parade. The organizer wants them to march in rows, with each row having the same number of children, and with each row composed of children with the same gender. What is the largest number of children per row that satisfies these constraints? There should be $\\... 6 I've heard very good reviews of the 2017 book, \"An Illustrated Theory of Numbers\" by MH Weissman. The book's main site is here; a write-up, along with some reviews, by the American Mathematical Society can be found here. To quote from the latter (emphasis added by me): An Illustrated Theory of Numbers gives a comprehensive introduction to number theory, ... 5 Euler's original heuristic may \"fly\" with people who don't know about calculus: just as polynomials (e.g., with all real roots) are essentially products of linear factors$x-\\alpha$where$\\alpha$runs through the roots, one might imagine that$\\sin \\pi z$(using radian measure) is a products of something like$1-{z\\over n}$for$n$non-zero integer, and ... 5 I like to ask my probability students the question: If you pick an integer between 2 and 100 uniformly at random, what's the probability that it's the average of two (not necessarily different) primes? I like that above question (easily equivalent to Goldbach) because there's no preference for even versus odd numbers as in Goldbach. It also gets ... 5 I am not sure whether this really answers your question, but I could think of the following strategy of introducing these sets of numbers. It is not based on any kind of research and just based on personal experience with first year university students where I sometimes give something similar as a \"naive idea\" in addition to the proper definitions. The ... 5 Some caveats: I own both books, and have taken number theory courses up to graduate level. Also, your questions are clearly somewhat subjective: what is difficult, relevant or complete for one person, depends on their previous exposure, inclination, course content, and ability. So...in my opinion.. both books are classics, which means they have been around ... 5 Look at patterns in decimal expansions: what is the period of the repeating decimal of$1/n$? From numerical data, the period is at most$n-1$, and you only get equality when$n = p$is prime (but not conversely:$1/11$has decimal period$2$, not$10$). If you look at the period of the decimal of$1/p$for primes$p$besides$2$and$5\\$, you find that ...\n\nOnly top voted, non community-wiki answers of a minimum length are eligible", "date": "2019-10-22 22:43:29", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/tags/number-theory/hot", "openwebmath_score": 0.5858034491539001, "openwebmath_perplexity": 618.7862224429809, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575178175919, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8554118993133387}}
{"url": "https://math.stackexchange.com/questions/2415349/coefficient-of-x3-in-expansion-of-frac1ex-cdot-1x", "text": "# Coefficient of $x^3$ in expansion of $\\frac{1}{e^x\\cdot (1+x)}$\n\nIf $|x|<1$, then what is the coefficient of $x^3$ in expansion of $\\dfrac{1}{e^x\\cdot (1+x)}$?\n\nMy Try: So I rewrote this by taking $e^x$ to the numerator to get $\\frac{e^{-x}}{(1+x)}$. My problem is, how can I deduce the coefficient of $x^3$ here, since I would need to divide the numerator by the denominator, which is a cumbersome task. Is there a simpler way? How can I solve problems like these? I'd love to know.\n\n\u2022 You just need that term. Do a few steps of the long division. \u2013\u00a0Hellen Sep 3 '17 at 14:08\n\u2022 Is really long division the only method? What if I was asked the coefficient of $x^200$? I would certainly not divide then. There has to be another, more logical way. \u2013\u00a0Tanuj Sep 3 '17 at 14:09\n\u2022 Not all the terms of all Taylor series have nice expressions. Some of them may take a lot of effort. I, too, hate it when I need to calculate the $x^5$ term of $\\tan x$. Here you can also rewrite $1/(1+x)=1-x+x^2-x^3+\\cdots$ and multiply that term-by-term with the series of $e^{-x}$. This is still easy. \u2013\u00a0Jyrki Lahtonen Sep 3 '17 at 14:12\n\u2022 @JyrkiLahtonen But in this case, both Taylor series are as nice as it gets... \u2013\u00a0Clement C. Sep 3 '17 at 14:17\n\u2022 @ClementC. Quite. But I still wouldn't want to calculate the $x^{200}$ term :-) \u2013\u00a0Jyrki Lahtonen Sep 3 '17 at 14:50\n\nHint. Note that $$\\frac{1}{e^x\\cdot(1+x)}=\\frac{e^{-x}}{1-(-x)}=\\left(\\sum_{n=0}^{\\infty} \\frac{(-x)^n}{n!}\\right)\\cdot \\left(\\sum_{n=0}^{\\infty} {(-x)^n}\\right)\\\\ =\\left(1-x+\\frac{x^2}{2}-\\frac{x^3}{6}+o(x^3)\\right)\\cdot \\left(1-x+x^2-x^3+o(x^3)\\right).$$ Can you take it from here?\n\nP.S. Along the same lines, it can be seen that, more generally, the coefficient of $x^n$ in the expansion of $\\frac{1}{e^x\\cdot(1+x)}$ is $$\\sum_{k=0}^n\\frac{(-1)^k}{k!}\\cdot (-1)^{n-k}=(-1)^n\\sum_{k=0}^n\\frac{1}{k!}.$$ See the Cauchy product wiki page.\n\n\u2022 Okay, I see, but how does that help me? You know if there were two terms in the numerator and one in denominator I could easily solve. But not in this case. \u2013\u00a0Tanuj Sep 3 '17 at 14:11\n\u2022 @Tanuj Now you have to expand the product. \u2013\u00a0Robert Z Sep 3 '17 at 14:14\n\n\\begin{eqnarray*} \\frac{e^{-x}}{1+x} &=&\\left(1-x+\\frac{x^2}{2}-\\frac{x^3}{6}+\\cdots \\right) (1- x+x^2-x^3+\\cdots ) \\\\ &=& \\cdots +x^3 \\underbrace{\\left( -1-1-\\frac{1}{2}-\\frac{1}{6}\\right)}_{\\color{blue}{-\\frac{8}{3}}}+ \\cdots \\end{eqnarray*}\n\n\u2022 Can you explain what exactly did you do? \u2013\u00a0Tanuj Sep 3 '17 at 14:12\n\u2022 Wow a eureka moment for me! You used that result of infinite Geometric Progression, right? \u2013\u00a0Tanuj Sep 3 '17 at 14:14\n\u2022 Just used the standard expansions for the functions ... \u2013\u00a0Donald Splutterwit Sep 3 '17 at 14:16\n\nFrom the two Taylor series at $0$, to order $x^3$:$$e^{-x} = 1-x+\\frac{x^2}{2}-\\frac{x^3}{6} + o(x^3) \\tag{1}$$ and $$\\frac{1}{1+x} = 1-x+x^2-x^3+o(x^3)\\tag{2}$$ we obtain \\begin{align} \\frac{e^{-x}}{1+x} &= \\left(1-x+\\frac{x^2}{2}-\\frac{x^3}{6} + o(x^3)\\right)\\left(1-x+x^2-x^3+o(x^3)\\right)\\\\ &= 1-x+x^2-x^3+o(x^3) - x+x^2-x^3+\\frac{x^2}{2}-\\frac{x^3}{2}-\\frac{x^3}{6}\\\\ &= 1-2x+\\frac{5}{2}x^2{\\color{red}{-\\frac{8}{3}}}x^3 +o(x^3) \\end{align} and you can just read off the coefficient.\n\nNote that when expanding the product, we could have focused on only the terms $x^3$ (it would have been marginally faster, but would have required keeping track); and didn't expand any term $x^k$ with $k>3$, since anyway they are \"swallowed\" by the $o(x^3)$.", "date": "2020-01-24 12:08:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2415349/coefficient-of-x3-in-expansion-of-frac1ex-cdot-1x", "openwebmath_score": 0.988427460193634, "openwebmath_perplexity": 525.2687167001895, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575173068324, "lm_q2_score": 0.8705972751232809, "lm_q1q2_score": 0.8554118972192242}}
{"url": "https://mathhelpboards.com/threads/limits-of-volume-integrals.1052/", "text": "# Limits of volume integrals\n\n#### GreenGoblin\n\n##### Member\nHelp choose the limits of the following volume integrals:\n\n1) V is the region bounded by the planes x=0,y=0,z=2 and the surface z=x^2 + y^2 lying the positive quadrant. I need the limits in terms of x first, then y then z AND z first, then y and then x. And also polar coordinates, x=rcost, y=rsint, z=z.\n\nI am trying to convince myself\n\n2) V is the region bounded by x+y+z=1, x=0, y=0, z=0.\n\n#### CaptainBlack\n\n##### Well-known member\nHelp choose the limits of the following volume integrals:\n\n1) V is the region bounded by the planes x=0,y=0,z=2 and the surface z=x^2 + y^2 lying the positive quadrant. I need the limits in terms of x first, then y then z AND z first, then y and then x. And also polar coordinates, x=rcost, y=rsint, z=z.\n\nI am trying to convince myself\n\n2) V is the region bounded by x+y+z=1, x=0, y=0, z=0.\nThe region V is the volume above the parabaloid surface $$z=x^2+y^2$$ and below $$z=2$$ in the first quadrant. The projection of this region onto the x-y plane is that part of the circle $$x^2 + y^2=2$$ in the first quadrant.\n\nSo this is $$x\\in (0,\\sqrt{2})$$ and $$y \\in (0, \\sqrt{2-x^2})$$ and $$z \\in ( \\sqrt{x^2+y^2} , 2)$$\n\nCB\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\n(I assume that by \"x first, then y then z\" you are referring to going from left to right. Of course, the integration is in the opposite order.)\n\n\"x first, then y then z\" is pretty much the usual way to do an integral like this. Note that if we project into the xy-plane, the upper boundary projects to the circle $x^2+ y^2= 2$, a circle of radius $2$, centered at (0, 0) and in the first quadrant that is a quarter circle. In order to cover that entire figure, clearly x has to go from 0 to $\\sqrt{2}$. Now, for each x, y must go from the x-axis, y= 0, up to the circle, $y= \\sqrt{2- x^2}$. And, finally, for any given (x,y), z must go from the paraboloid, $z= x^2+ y^2$, to z= 2. The integral would be $\\int_{x=0}^\\sqrt{2}\\int_{y= 0}^\\sqrt{2- x^2}\\int_{z= x^2+ y^2}^2 f(x,y,z)dzdydx$.\n\nIn the other order, z first, then y and then x, project to the yz- plane. There, of course, the paraboloid project to half of the parabola $z= y^2$. Now, z goes from 0 to 2 and, for each z, y goes from 0 to $\\sqrt{z}$. Now, for each y and z, x goes from 0 to $\\sqrt{2- y^2}$. The integral would be $\\int_{z=0}^2\\int_{y= 0}^\\sqrt{z}\\int_{x= 0}^\\sqrt{2- y^2} f(x,y,z)dxdydz$.\n\nFor V is the region bounded by x+y+z=1, x=0, y=0, z=0, with the order \"x first, then y then z\", we project to the xy-plane which gives the triangle bounded by x= 0, y= 0, and x+ y= 1. x ranges from 0 to 1 and, for each x, y ranges from 0 to 1- x. For each (x, y), z ranges from 0 to 1- x- y. The integral would be $\\int_{x= 0}^1\\int_{y= 0}^{1- x}\\int_{z= 0}^{1- x- y} f(x,y,z) dzdydx$. Can you get it for the order \"z first, then y then x\"?\n\nLast edited:", "date": "2020-12-05 14:09:29", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/limits-of-volume-integrals.1052/", "openwebmath_score": 0.9391376972198486, "openwebmath_perplexity": 460.1375426343177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575157745542, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8554118909368807}}
{"url": "https://math.stackexchange.com/questions/2544529/variance-of-alternate-flipping-rounds/2701095", "text": "# Variance of alternate flipping rounds\n\nI did the following exercise, but I would like to extend the question to the variance of the variate.\n\nBob and Bub each has his own coin. Chance of coming up \"heads\" is $\\rho$ for Bob's coin and $\\tau$ for Bub's. They flip alternatively, first Bob, then Bub, then Bob again, etc. Let Bob's flip followed by Bub's flip constitute a round, and let $R$ denote the number of rounds until each gets \"heads\" at least once. For $\\rho = 1/3$, $\\tau = 2/5$, what is the expectation of $R$?\n\nGeneral answer for the expectation is:\n\n$$\\mathbb{E}[R]=\\frac{1 + \\frac{\\rho}{\\tau} + \\frac{\\tau}{\\rho} - (\\rho + \\tau)}{\\rho + \\tau - \\rho \\, \\tau}$$\n\nThis agrees with Monte Carlo simulation I did (with $10^5$ repeats), which approximates expectation and variance (with $\\rho = 1/3$, and $\\tau = 2/5$) to $3.84647$ and $6.48666$ respectively.\n\nIs anybody able to calculate variance symbolically?\n\n\u2022 Out of curiosity, where does your expected value formula come from?\n\u2013\u00a0Remy\nDec 1, 2017 at 0:23\n\u2022 @Remy I derived it. In a nutshell, $R$ is distributed geometrically with parameter $p = \\rho + \\tau - \\rho \\, \\tau$ (probability of success of either one of them) plus a geometric variate with parameter $p = \\rho$ and weighted with probability $\\mathbb{P}\\{\\text{Bob didn't have success, but Bub did} \\, | \\, \\text{either had}\\}$ plus a similar variate for the reverse case.\n\u2013\u00a0BoLe\nDec 2, 2017 at 12:51\n\nPlease load the page $$2\\sim3$$ times for the hyperlinks to work properly.\n\n# Outline\n\nSetup the Notations : As titled.\n\nSolution.1 : Direct application of the conditional decomposition of expectation. This is the foolproof approach if one wants a quick numeric evaluation and doesn't want to be bothered with analysis.\n\nSolution.2 : A framework that provides perspectives and better calculation.\n\nAppendix.A : Supplementary material to Solution.1.\n\nAppendix.B.1 : In-site links to existing questions that are closely related.\n\nAppendix.B.2 : Supplementary material to Solution.2.\n\n# Setup the Notations\n\nLet $$X$$ be the total number of trials of Bob's flips when his first head (success) appears.\n\nLet $$Y$$ be that for Bub. We have $$X\\sim \\mathrm{Geo}[\\rho]$$ independent to $$Y\\sim \\mathrm{Geo}[\\tau]$$.\n\nThe following basics for a Geometric distribution will be useful here: \\begin{align*} \\mu_{_X} &\\equiv \\mathbb{E}[X] = \\frac1{\\rho} & & \\tag*{Eq.(1)} \\label{Eq01} \\\\ A_X &\\equiv \\mathbb{E}[X(X+1)] = \\frac2{ \\rho^2 } & &\\tag*{Eq.(2)} \\label{Eq02} \\\\ S_X &\\equiv \\mathbb{E}[X^2 ] = \\frac{2 - \\rho}{ \\rho^2 } & &\\tag*{Eq.(3)} \\label{Eq03} \\\\ V_X &\\equiv \\mathbb{V}[X] = \\frac{1 - \\rho}{ \\rho^2 } & &\\tag*{Eq.(4)} \\label{Eq04} \\\\ Q_X &\\equiv \\mathbb{E}[(X+1)^2] = A_X + \\mu_{_X} + 1 = \\frac{ 2 + \\rho + \\rho^2 }{ \\rho^2 } & &\\tag*{Eq.(5)} \\label{Eq05} \\end{align*} Recall that we often use $$A_X$$ to obtain $$S_X$$ because $$A_X$$ is easier to derive (it is a more natural quantity for the Geometric distribution).\n\nThe shorthands for $$Y$$ are the same $$\\mu_{_Y}$$ and $$V_Y$$ etc.\n\n# Solution.1\n\nJust like how the expectation can be derived (which apparently you know how to), the 2nd moment (thus variance) can be obtained by conditioning on the results of the round. Denote the events of $$\\{ \\text{Bob head, Bub tail} \\}$$ as just $$HT$$, and recall that $$X$$ is for Bob flipping alone as if Bub doesn't exist. \\begin{align*} \\mathbb{E}[ R^2 ] &= \\rho \\tau \\, \\mathbb{E}\\left[ R^2 \\,\\middle|~ HH\\right] + (1 - \\rho)(1 - \\tau)\\,\\mathbb{E}\\left[ R^2 \\,\\middle|~ TT\\right] \\\\ &\\hspace{36pt} + \\rho (1 - \\tau) \\, \\mathbb{E}\\left[ R^2 \\,\\middle|~ HT\\right] + \\tau (1 - \\rho)\\,\\mathbb{E}\\left[ R^2 \\,\\middle|~TH\\right] \\\\ &= \\rho \\tau + (1 - \\rho)(1 - \\tau)\\,\\mathbb{E} \\left[ (1+R)^2 \\right] + \\rho (1 - \\tau) \\, \\mathbb{E}\\left[ (1+Y)^2 \\right] + \\tau (1 - \\rho)\\,\\mathbb{E}\\left[ (1+X)^2 \\right] \\end{align*} Please let me know if you need justification for $$\\mathbb{E}[ R^2 ~|~~TH] = \\mathbb{E}[ (1+Y)^2 ]$$ and the alike. Moving on, use the \\ref{Eq04} shorthand $$Q_X$$ and $$Q_Y$$ for now and rearrange. $$\\mathbb{E}[ R^2 ] = \\rho \\tau + (1 - \\rho)(1 - \\tau) \\left( \\mathbb{E}[ R^2 ] + 2 \\mathbb{E}[ R ] + 1 \\right) + \\rho (1 - \\tau) Q_Y + \\tau (1 - \\rho)\\,Q_X$$ Denote $$\\lambda = 1 - (1 - \\rho)(1 - \\tau) = \\rho + \\tau - \\rho \\tau$$, and collect $$\\mathbb{E}[ R^2 ]$$ on the left. $$\\begin{equation*} \\lambda\\,\\mathbb{E}[ R^2 ] = \\rho \\tau + (1 - \\lambda) \\left( 2 \\mathbb{E}[ R ] + 1 \\right) + \\rho (1 - \\tau) Q_Y + \\tau (1 - \\rho)\\,Q_X \\tag*{Eq.(6)} \\label{Eq06} \\end{equation*}$$ Note that the denominator of $$\\mathbb{E}[ R ]$$ is just $$\\lambda$$. Along with the symmetry, this suggests that the numerator of $$\\mathbb{E}[ R ]$$ can be rewritten into a better form invoking the basic \\ref{Eq01}. \\begin{align*} 1 + \\frac{ \\rho }{ \\tau } + \\frac{ \\tau }{ \\rho } - (\\rho + \\tau) &= \\bigl( 1 + \\frac{ \\rho }{ \\tau } - \\rho \\bigr) + \\bigl( 1 + \\frac{ \\tau }{ \\rho } - \\tau \\bigr) - 1 \\\\ &= \\frac{ \\tau + \\rho - \\rho\\tau }{ \\tau } + \\frac{ \\rho + \\tau - \\rho\\tau }{ \\rho } - 1 \\\\ \\implies \\mathbb{E}[ R ] = \\frac1{\\rho} + \\frac1{\\tau} &- \\frac1{\\lambda} = \\mu_{_X} + \\mu_{_Y} - \\frac1{\\lambda} \\tag*{Eq.(7)} \\label{Eq07} \\end{align*} It's not a coincidence that the expectation can be expressed as such. The reason will be elaborated in the next section for Solution.2.\n\nFor the sake of computing the numeric value, \\ref{Eq06} was a good place to stop. With the given parameters $$\\rho = 1/3$$ and $$\\tau = 2/5$$, we have $$\\mathbb{E}[ R ] = 23/6$$, $$Q_X = 22$$, and $$Q_Y = 16$$. That makes $$\\mathbb{E}[ R^2 ] = 190/9$$ and the variance $$V_R \\equiv \\mathbb{E}[ R^2 ] - \\mathbb{E}[R]^2 = \\frac{77}{12}~.$$ See the end of Solution.2 for a Mathematica code block for the numerical evaluation and more.\n\nOne can quickly check the value of $$\\mathbb{E}[ R ] \\approx 3.8333$$ relative to $$\\mu_{_X} = 3$$ and $$\\mu_{_Y} = 2.5$$, as well as $$V_R \\approx 6.146667$$ in relation to $$V_X = 6$$ and $$V_Y = 15/4$$. Both of the quantities for $$R$$ are slightly larger than the max of $$X$$ and $$Y$$, which is reasonable.\n\nNow, if you have a strong inclination for algebraic manipulation, then the following is what you might arrive at, after some trial and error. Recall the shorthand for the 2nd moment \\ref{Eq03}: $$\\begin{equation*} \\mathbb{E}[ R^2 ] = \\frac{ 2 - \\rho }{ \\rho^2 } + \\frac{ 2 - \\tau }{ \\tau^2 } - \\frac{ 2 - \\lambda }{ \\lambda^2 } = S_X + S_Y - \\frac{ 2 - \\lambda }{ \\lambda^2 } \\tag*{Eq.(8)} \\label{Eq08} \\end{equation*}$$ Again, this is not a coincidence. Along with \\ref{Eq07}, their respective 3rd terms seem to be another Geometric random variable with the success' parameter $$\\lambda$$. The proper probabilistic analysis is the subject of Solution.2 up next.\n\nBy the way, blindly shuffling the terms around is usually not the best thing one can do. Nonetheless, just for the record, Appendix.A shows one way of going from \\ref{Eq06} to \\ref{Eq08} mechanically.\n\n# Solution.2\n\nDenote $$W \\equiv \\min(X,Y)$$ as the smaller among the two and $$Z \\equiv \\max(X,Y)$$ as the larger. The key observation to solve this problem is that $$Z \\overset{d}{=} R~, \\qquad\\qquad \\textbf{the maximum has the same distribution as the 'rounds'.}$$ This allows one to think about the whole scenario differently (not as rounds of a two-player game). For all $$k \\in \\mathbb{N}$$, since $$X \\perp Y$$ we have \\begin{align*} \\Pr\\{ Z = k \\} &= \\Pr\\{ X < Y = Z = k \\} \\\\ &\\hspace{36pt} + \\Pr\\{ Y < X = Z = k \\} \\\\ &\\hspace{72pt} + \\Pr\\{ X = Y = Z = k \\} \\\\ &= \\tau (1 - \\tau)^{k-1} (1 - (1-\\rho)^k) \\\\ &\\hspace{36pt} + \\rho (1 - \\rho)^{k-1} (1 - (1-\\tau)^k) \\\\ &\\hspace{72pt} + \\rho\\tau (1 - \\rho)^{k-1} (1 - \\tau)^{k-1} \\tag*{Eq.(9)} \\label{Eq09} \\end{align*} In principle, now that the distribution of $$Z$$ (thus $$R$$) is completely specified, everything one would like to know can be calculated. This is indeed a valid approach (to obtain the mean and variance), and the terms involved are all basic series with good symmetry.\n\nNote that $$Z$$ is not Geometric (while $$X$$, $$Y$$, and $$W$$ are), nor is it Negative Binomial. At this point, it is not really of interest that \\ref{Eq09} can be rearranged into a more compact and illuminating form ...... because we can do even better.\n\nThere are two more observations that allow one to not only to better calculate but also understand the whole picture. \\begin{align*} &X+Y = W + Z \\tag*{Eq.(10)} \\label{Eq10} \\\\ &W \\sim \\mathrm{Geo[\\lambda]} \\tag*{Eq.(11)} \\label{Eq11} \\end{align*} This is the special case of the sum of the order statistics being equal to the original sum. In general with many summands this not very useful, but here with just two terms it is crucial.\n\nBack to the two-player game scenario, the fact that $$W$$ is Geometric with success probability $$\\lambda = 1 - (1 - \\rho)(1 - \\tau)$$ is easy to see: a round 'fails' if and only if both flips 'fail', with a probability $$(1 - \\rho)(1 - \\tau) = 1 - \\lambda$$.\n\nThe contour of $$W = k$$ is L-shaped boundary of the '2-dim tail', which fits perfectly with the scaling nature (memoryless) of the joint of two independent Geometric distribution. Pleas picture in the figure below that $$\\Pr\\left\\{W = k_0 + k ~ \\middle|~~ W > k_0 \\right\\} = \\Pr\\{ W = k\\}$$ manifests itself as the L-shape scaling away.\n\nThe contour of $$Z = k$$ looks like $$\\daleth$$, and together with the L-contour of $$W$$ they make a cross. This is \\ref{Eq10} the identity $$X+Y = W+Z$$, visualized in the figures below. See Appendix.B.1 for the linked in-site posts of related topics.\n\nImmediately we know the expectation to be: $$\\begin{equation*} \\mathbb{E}[ Z ] = \\mathbb{E}[ X + Y - W ] = \\mu_{_X} + \\mu_{_Y} - \\mu_{_W} = \\frac1{\\rho} + \\frac1{\\tau} - \\frac1{\\lambda} \\tag*{Eq.(7.better)} \\end{equation*}$$\n\nThis derivation provides perspectives different from (or better, arguably) those obtained by conditioning on the first round.\n\nDerivation of the 2nd moment reveals a more intriguing properties of the setting. \\begin{align*} \\mathbb{E}[ Z^2 ] &= \\mathbb{E}[ (X + Y - W)^2 ] \\\\ &= \\mathbb{E}[ (X + Y)^2 ] + \\mathbb{E}[ W^2 ] - 2\\mathbb{E}[ (X + Y) W ] \\\\ &= \\mathbb{E}[ X^2 ] + \\mathbb{E}[ Y^2 ] + 2 \\mathbb{E}[ XY ] + \\mathbb{E}[ W^2 ] - 2\\mathbb{E}[ (W+Z) W ] \\qquad\\because X+Y = W+Z\\\\ &= \\mathbb{E}[ X^2 ] + \\mathbb{E}[ Y^2 ] - \\mathbb{E}[ W^2 ] + 2\\mathbb{E}[ XY ] - 2\\mathbb{E}[ WZ ] \\end{align*} Here's the kicker: when $$X \\neq Y$$, by definition $$W$$ and $$Z$$ each take one of them so $$WZ = XY$$, and when $$X = Y$$ we have $$WZ = XY$$ just the same!! Consequently, $$\\mathbb{E}[ ZW ] = \\mathbb{E}[ XY ] =\\mu_{_X} \\mu_{_Y}$$ always, and they cancel. $$\\begin{equation*} \\mathbb{E}[ Z^2 ] = \\mathbb{E}[ X^2 ] + \\mathbb{E}[ Y^2 ] - \\mathbb{E}[ W^2 ] = \\frac{ 2 - \\rho }{ \\rho^2 } + \\frac{ 2 - \\tau }{ \\tau^2 } - \\frac{ 2 - \\lambda }{ \\lambda^2 } \\tag*{Eq.(8.better)} \\label{Eq08better} \\end{equation*}$$ This is the proper derivation, and in \\ref{Eq08} the 3rd term following $$S_X + S_Y$$ is indeed $$-S_W$$.\n\nKeep in mind that both $$W$$ and $$Z$$ are clearly correlated with $$X$$ and $$Y$$, while our intuition also says that the correlation between $$W$$ and $$Z$$ is positive (see Appendix.B for a discussion).\n\nWhile we're at it, this relation in fact holds true for any (higher) moments, $$\\mathbb{E}[ Z^n ] = \\mathbb{E}[ X^n ] + \\mathbb{E}[ Y^n ] - \\mathbb{E}[ W^n ]~.$$This is true due to the same argument that gave us $$WZ = XY$$, and using \\ref{Eq09} doing the explicit sums to derive this identity is also easy.\n\nWithout further ado, the variance of $$Z$$ (thus the variance of $$R$$), previously known only as the unsatisfying \"\\ref{Eq06} minus the square of $$\\mathbb{E}[ Z ]$$'', can now be put in its proper form. \\begin{align*} V_Z &\\equiv \\mathbb{E}[ Z^2 ] - \\mathbb{E}[ Z ]^2 \\\\ &= \\mathbb{E}[ X^2 ] + \\mathbb{E}[ Y^2 ] - \\mathbb{E}[ W^2 ] - ( \\mu_{_X} + \\mu_{_Y} - \\mu_{_W} )^2 \\\\ &= V_X + V_Y - V_W + 2( \\mu_{_W} \\mu_{_Z} - \\mu_{_X} \\mu_{_Y}) \\tag*{Eq.(12.a)} \\\\ &= \\frac{ 1 - \\rho }{ \\rho^2 } + \\frac{ 1 - \\tau }{ \\tau^2 } - \\frac{ 1 - \\lambda }{ \\lambda^2 } + 2\\left( \\frac1{ \\lambda } \\bigl( \\frac1{ \\rho } + \\frac1{ \\tau } - \\frac1{ \\lambda } \\bigr) - \\frac1{ \\rho \\tau }\\right) \\tag*{Eq.(12.b)} \\end{align*} This expression (or the symbolic one just above) is a perfectly nice formula to me. You can rearrange it to your heart's content, for example, like this $$\\begin{equation*} V_Z = \\bigl( \\frac1{ \\rho } - \\frac1{ \\tau } \\bigr)^2 - \\frac1{ \\lambda^2 } + 2 (\\frac1{\\lambda} - \\frac12) \\bigl( \\frac1{ \\rho } + \\frac1{ \\tau } - \\frac1{ \\lambda } \\bigr) \\tag*{Eq.(12.c)} \\end{equation*}$$ which emphasizes the role played by the difference between $$X-Y$$.\n\nI'm not motivated to pursue a 'better' form beyond Eq.(12), and if anyone knows any alternative expressions that are significant either algebraically or probabilistically, please do share.\n\nLet me conclude with a Mathematica code block verifying the numeric values and identities for the variance, the 2nd moment, as well as the earliest algebra of the conditioning of expectation.\n\n(* t = \\tau, p = \\rho, m = E[R],  S = E[R^2] , h = \\lambda = 1-(1-p)(1-t) *) ClearAll[t , p, m, S, V, h, tmp, simp, sq, var, basics]; basics[p_] := {1/p, (1 - p)/p^2, (2 - p)/p^2, (2 + p + p^2)/ p^2};(* EX, VX, E[X^2], E[(1+X)^2]*)\nRow@{basics[1/3], Spacer@10, basics[2/5], Spacer@10, basics[1 - (1 - 1/3) (1 - 2/5)]}\nsimp = Simplify[#, 0 < p < 1 && 0 < t < 1] &; h = 1 - (1 - p) (1 - t); m = (1 + t/p + p/t - (t + p))/h; sq[a_] := (2 - a)/a^2;(* 2nd moment of a Geometric distribution *) var[a_] := (1 - a)/a^2;\nS = tmp /. Part[#, 1]&@Solve[tmp == p t + (1 - p) (1 - t) (1 + 2 m + tmp) + p (1 - t) (1 + 2/t + sq@t) + (1 - p) t (1 + 2/p + sq@p), tmp] //simp (* this simplification doesn't matter ... *)\nV = S - m^2 // simp (* neither does this. *)\n{tmp = V /. {p -> 1/3, t -> 2/5}, tmp // N} (* below: veryfiy the various identities for 2nd moment and variance. Difference being zero means equal *)\n{sq@t + sq@p - sq@h - S, var@p + var@t - var@h + 2 m 1/h - 2/(p t) - V, (1/p - 1/t)^2 - 1/h^2 + 2 (1/p + 1/t - 1/h) (1/h - 1/2) - V} // simp\n`\n\n$$\\large\\textbf{Appendix.A: }\\normalsize\\text{an algebraic route from \\ref{Eq06} to \\ref{Eq08}}$$\n\nConsider \\ref{Eq06} one part at a time. First, the $$2(1 - \\lambda)\\, \\mathbb{E}[ R ]$$. From the the first line to the 2nd line, $$1 - \\lambda = (1 - \\rho)(1 - \\tau)$$ is inserted for the 2nd term: \\begin{align*} 2(1 - \\lambda)\\, \\mathbb{E}[ R ] &= 2(1 - \\lambda )\\frac{ - 1}{\\lambda} + 2(1 - \\lambda) \\bigl( \\frac1{ \\rho } + \\frac1{ \\tau } \\bigr) \\\\ &= 2\\frac{\\lambda - 1}{\\lambda} + \\frac{2(1 - \\rho ) }{ \\rho }(1 - \\tau) + (1 - \\rho ) \\frac{ 2(1 - \\tau) }{ \\rho } \\\\ &= 2 - \\frac2{\\lambda} + \\bigl( \\frac{2 - \\rho}{ \\rho } - 1\\bigr) (1 - \\tau) + \\bigl( \\frac{2 - \\tau}{ \\tau } - 1 \\bigr) (1 - \\rho ) \\\\ &= \\color{magenta}{2 - \\frac2{\\lambda} } + \\frac{2 - \\rho}{ \\rho^2 } (1 - \\tau)\\rho \\color{magenta}{- (1 - \\tau)} + \\frac{2 - \\tau}{ \\tau^2 } (1 - \\rho )\\tau \\color{magenta}{- (1 - \\rho )} \\end{align*} Next in line are the $$Q_X$$ terms. \\begin{align*} \\rho (1 - \\tau) Q_Y + \\tau (1 - \\rho)\\,Q_X &= \\rho (1 - \\tau) \\bigl( \\frac{ 2 + \\tau }{ \\tau^2 } + 1\\bigr) + \\tau (1 - \\rho) \\bigl( \\frac{ 2 + \\rho }{ \\rho^2 } + 1 \\bigr) \\\\ &= \\rho \\bigl( \\frac{ 2 - \\tau - \\tau^2 }{ \\tau^2 } + 1 - \\tau \\bigr) + \\tau \\bigl( \\frac{ 2 - \\rho - \\rho^2 }{ \\rho^2 } + 1 - \\rho \\bigr) \\\\ &= \\rho \\frac{ 2 - \\tau}{ \\tau^2 } + \\tau \\frac{ 2 - \\rho }{ \\rho^2 } \\color{magenta}{- 2\\tau\\rho} \\end{align*} Put things back together in \\ref{Eq06}, the magenta terms combine with $$\\rho\\tau$$ and $$1 - \\lambda$$ (multiplied by 1): \\begin{align*} \\lambda\\,\\mathbb{E}[ R^2 ] &= \\rho \\tau + (1 - \\lambda) + (1 - \\lambda) 2 \\mathbb{E}[ R ] + \\rho (1 - \\tau) Q_Y + \\tau (1 - \\rho)\\,Q_X \\\\ &= \\rho \\tau + (1 - \\lambda) \\color{magenta}{ {}+ 2 - \\frac2{\\lambda} - (1 - \\tau) - (1 - \\rho ) - 2\\tau\\rho} \\\\ &\\hphantom{{}= \\rho \\tau} + \\frac{2 - \\rho}{ \\rho^2 } (1 - \\tau)\\rho + \\frac{2 - \\tau}{ \\tau^2 } (1 - \\rho )\\tau \\tag{from \\mathbb{E}[R]}\\\\ &\\hphantom{{}= \\rho \\tau } + \\tau \\frac{ 2 - \\rho }{ \\rho^2 } + \\rho \\frac{ 2 - \\tau}{ \\tau^2 } \\tag{from Q_X etc}\\\\ &= 1 - \\frac2{\\lambda} + \\frac{ 2 - \\rho }{ \\rho^2 } \\bigl( \\rho + \\tau - \\rho\\tau \\bigr) + \\frac{ 2 - \\tau }{ \\tau^2 } \\bigl( \\rho + \\tau - \\rho\\tau \\bigr) \\\\ &= 1 - \\frac2{\\lambda} + \\frac{ 2 - \\rho }{ \\rho^2 } \\lambda + \\frac{ 2 - \\tau }{ \\tau^2 } \\lambda \\end{align*} Finally we can divide by the coefficient of $$\\mathbb{E}[ R^2 ]$$ on the left hand side and obtain \\ref{Eq08}.\n\n$$\\large\\textbf{Appendix.B.1: }\\normalsize\\text{In-Site Links to Related Questions}$$\n\nI found many existing posts about this topic (minimum of two independent non-identical Geometric distributions). In chronological order: 90782, 845706, 1040620, 1056296, 1169142, 1207241, and 2669667.\n\nUnlike the min, only 2 posts about max is found so far: 971214, and 1983481. Neither of the posts go beyond the 1st moment, and only a couple of the answers address the 'real geometry' of the situation.\n\nIn particular, consider the trinomial random variable $$T$$ that splits the 2-dim plane into the 3 regions. $$\\begin{equation*} T \\equiv \\mathrm{Sign}(Y - X) = \\begin{cases} \\hphantom{-{}} 1 & \\text{if}~~Y > X \\quad \\text{, with probability}~~ \\tau( 1 - \\rho) / \\lambda \\\\ \\hphantom{-{}} 0 & \\text{if}~~X = Y \\quad \\text{, with probability}~~ \\rho \\tau / \\lambda \\\\ -1 & \\text{if}~~Y < X \\quad \\text{, with probability}~~ \\rho( 1 - \\tau) / \\lambda \\end{cases} \\end{equation*}$$ This trisection of above-diagonal, diagonal, and below-diagonal is fundamental to the calculation of both $$W$$ and $$Z$$.\n\nIt can be easily shown that $$T \\perp W$$, that the trisection is independent of the minimum. For example, $$\\Pr\\left\\{ T = 1 \\mid W = k\\right\\} = \\Pr\\{ T = 1\\}$$ for all $$k$$.\n\nNote that $$T$$ is not independent to $$Z$$ (even with the corner $$k=1$$ removed).\n\nIn the continuous analogue, $$\\{X,Y,W\\}$$ are Exponential with all the nice properties, and $$Z$$ is neither Exponential nor Gamma (analogue of Negative Binomial).\n\nThe density of the $$Z$$ analogue is easy and commonly used, but its doesn't seem to have a name. At best one can categorize it as a special case of phase-type (PH distribution).\n\n$$\\large\\textbf{Appendix.B.2: }\\normalsize\\text{Covariance between \\min(X,Y) and \\max(X,Y), along with related discussions.}$$\n\nRecall the expectations: $$\\mathbb{E}[X] = \\mu_{_X} = 1 / \\rho$$, and the similar \\begin{align*} \\mu_{_Y} &= \\frac1{ \\tau }~, & \\mu_{_W} &= \\frac1{ \\lambda } = \\frac1{\\rho + \\tau - \\rho\\tau}~, & &\\text{and}\\quad \\mu_{_Z} = \\mu_{_X} + \\mu_{_Y} - \\mu_{_W} \\end{align*} The covariance between the minimum and the maximum is $$\\begin{equation*} C_{WZ} \\equiv \\mathbb{E}\\left[ (W - \\mu_{_W} ) (Z - \\mu_{_Z} ) \\right] = \\mathbb{E}[ WZ ] - \\mu_{_W} \\mu_{_Z} = \\mu_{_X} \\mu_{_Y} - \\mu_{_W} \\mu_{_Z} \\tag*{Eq.(B)} \\label{EqB} \\end{equation*}$$ The last equal sign invokes the intriguing $$WZ = XY$$ mentioned just before \\ref{Eq08better}.\n\nA positive covariance (thus correlation) expresses the intuitive idea that when the min is large then the max also tends to be large.\n\nHere we do a 'verbal proof' of $$C_{WZ} > 0$$ for all combinations of parameters. The purpose is not really a technical proof but more about illustrating the relations between $$\\{W, Z\\}$$ and $$\\{X, Y\\}$$.\n\nSince $$\\mu_{_X} + \\mu_{_Y} = \\mu_{_W} + \\mu_{_Z}$$, the covariance $$C_{WZ}$$ is in essence comparing products (of two non-negative numbers) given the value of the sum. We know that for a fixed sum, the closer the two 'sides' (think rectangle) are the larger the product.\n\nBeing the min and max, $$\\mu_{_W}$$ and $$\\mu_{_Z}$$ are more 'extreme' and can never be as close as $$\\mu_{_X}$$ and $$\\mu_{_Y}$$, throughout the entire parametric space $$\\{ \\rho, \\tau \\} \\in (0, 1)^2$$. Boom, QED.\n\n(the extreme cases where at least one of $$\\{\\rho, \\tau\\}$$ is zero or one are all ill-defined when it comes to min and max)\n\nAn algebraic proof for $$C_{WZ} > 0$$ (or for everything in this appendix) is easy. Let me emphasize the descriptive aspect.\n\nConsider what it means for $$W$$ as a Geometric distribution to be the minimum of $$X$$ and $$Y$$. \\begin{align*} \\frac1{\\lambda} &< \\min( \\frac1{\\rho}, \\frac1{\\tau} ) & &\\text{faster to 'succeed' on average} \\\\ \\lambda &> \\max( \\rho, \\tau ) & &\\text{better than the higher success parameter} \\end{align*} That is, the mean of the 'minimum flip' $$\\mu_{_W} < \\min( \\mu_{_X}, \\mu_{_Y} )$$ is more extreme at the lower end.\n\nOn the other hand, $$Z$$ is NOT a Geometric distribution. However, one can define an auxiliary $$Z' \\sim \\mathrm{Geo}[1 / \\mu_{_Z}]$$ with an equivalent mean $$\\mu_{_{Z'}} = \\mu_{_Z}$$. Once we have arrived at \\ref{EqB}, the role of $$Z$$ can be replaced by $$Z'$$: $$C_{WZ} = C_{WZ'} = \\mu_{_X} \\mu_{_Y} - \\mu_{_W} \\mu_{_{Z'}}$$\n\nNow we can have similar descriptions for $$\\mu_{_Z}$$, while it is actually $$\\mu_{_{Z'}}$$ for the Geometric distribution that is being compared with $$X$$ and $$Y$$. \\begin{align*} \\mu_{_Z} &> \\max( \\mu_{_X}, \\mu_{_Y} ) & &\\text{slower to 'succeed' on average} \\\\ \\frac1{ \\mu_{_Z} } &< \\min( \\rho, \\tau ) & &\\text{worse than the lower success parameter} \\end{align*} That is, the mean of the 'maximum flip' is more extreme at the higher end.\n\nThis concludes the verbal argument for how $$\\mu_{_W}$$ and $$\\mu_{_Z}$$ are more dissimilar (than the relation between $$\\mu_{_X}$$ and $$\\mu_{_Y}$$) thus making a smaller product.\n\n\u2022 Thank you for this comprehensive, inspiring answer. Solution One I don't have problem understanding, I'm upset a bit that I didn't complete my analysis. I turned back while trying to form this recursion involving nonlinear terms, like E[(1+X)^2]. Solution Two and the Appendices I am able to follow as well.\n\u2013\u00a0BoLe\nMar 22, 2018 at 10:45", "date": "2022-05-21 12:53:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2544529/variance-of-alternate-flipping-rounds/2701095", "openwebmath_score": 0.9992894530296326, "openwebmath_perplexity": 988.824214576845, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575173068326, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8554118856730828}}
{"url": "https://math.stackexchange.com/questions/562275/finding-area-problem", "text": "# Finding area problem\n\nThere is this simple geometry question that seems so easy but I think the question lacks some information (does it?). Or maybe there are other ways to solve the problem.\n\nSo the problem says, there are two squares $ABCD$ and $FCHG$ with a side of length $8$ and $10$, respectively. We are asked to find the area of the shaded region.\n\nWhat makes this question a bit tricky is because the small triangle $DEF$ is not shaded. And if we are given the length of AD, then the question will be easy. Is there a way to find AD or is there any other methods to solve the problem?\n\nBy the way, the answer is $48.4$.\n\nAnd also, I am pretty sure we don't need to use advance methods such as trigs, etc.\n\nI really appreciate any helps!\n\n## 4 Answers\n\nTriangles $BCF$ and $DEF$ are similar triangles, which means that their sides are proportional. You know side $BC=8,$ $CF=10$ and side $EF=2$. Can you proceed?\n\n\u2022 yes I can! thanks! \u2013\u00a0user71346 Nov 11 '13 at 10:50\n\nHint: Look for similar triangles. What is $\\Delta DEF$ similar to?\n\n\u2022 ah yes! so careless... thanks! \u2013\u00a0user71346 Nov 11 '13 at 10:49\n\nLet say lines $BG$ and $FC$ intersects at $K$,\n\nSo $CK+KF=10$ and $FKG$ and $BCK$ are similar triangles.\n\nSo $\\frac{KF}{CK}=10/8$ therefore $CK=40/9$ and $KF=50/9$\n\nThe area of FKG triangle = $0.5*10*50/9=27.7$\n\nAlso $AD+DE=8$ and $ADB$ and $DEF$ are similar triangles. Similarly it gives\n\n$AD=32/5$ and $DE=8/5$\n\nThe area of $BDEK$ = area of $BFC$- area of $DEF$ - area of $BCK$ = $0.5*8*10-0.5*2*8/5-0.5*8*40/9$= 20.7\n\nTotal area = The area of $BDEK$ + The area of FKG triangle =$27.7+20.7=48.4$\n\n\"Is there a way to find AD or is there any other methods to solve the problem?\"\n\nAD is proportioan to AE as BD is proportional to BF as EC is proportional to FC.\n\nSo AD ~ 8 as 8 ~ 10. Or AD/8 = 8/10.", "date": "2019-12-09 08:05:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/562275/finding-area-problem", "openwebmath_score": 0.8765013217926025, "openwebmath_perplexity": 317.9574745436916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575178175918, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8554118811694017}}
{"url": "https://math.stackexchange.com/questions/3606831/solving-a-congruence-system-when-the-chinese-remainder-theorem-cannot-be-applied", "text": "# Solving a congruence system when the Chinese remainder theorem cannot be applied\n\nI'm trying to solve the following system $$\\cases{3x\\equiv1\\pmod{14}\\\\x\\equiv1\\pmod{8}\\\\3x\\equiv9\\pmod{5}}$$\n\nMy understanding is that, since $$14, 8, 5$$ aren't all coprime, I cannot apply the Chinese remainder theorem.\n\nThe first thing I did was solving the first and third equations independently, which yielded the following equivalent system:\n\n$$\\cases{x\\equiv5\\pmod{14}\\\\x\\equiv1\\pmod{8}\\\\x\\equiv3\\pmod{5}}$$\n\nAt this point I'm unsure how to proceed. I thought solving the system made up of the first two equation, and then a system made up of the solution to the first system with the third equation could work, but turns out it didn't. Here's what I tried:\n\n$$\\cases{x\\equiv5\\pmod{14}\\\\x\\equiv1\\pmod{8}\\\\} \\iff x = 5+ 14k=1+8h \\rightarrow7k-4h = -2 \\iff k = 2+4y, h = 4-7y, \\text{ with }y\\in\\mathbb{Z}$$ Therefore, $$x = 33 - 56y \\iff x\\equiv33\\pmod{56}$$. Plugging this result back into the system, we now have $$\\cases{x\\equiv33\\pmod{56}\\\\x\\equiv3\\pmod{5}\\\\} \\iff x = 3+5k=33+56h \\rightarrow5k-56h=30 \\iff k = -330+56y, h = -30-5y, \\text{ with }y\\in\\mathbb{Z}$$ Therefore, $$x \\equiv 1653\\equiv 253 \\pmod{280}$$; however, this result is incorrect. What did I do wrong?\n\n\u2022 did you mean $h=4\\color{red}+7y$ and $h=-30\\color{red}+5y$? Apr 2 '20 at 18:55\n\u2022 there is also a formula for non-coprime moduli on wikipedia Apr 2 '20 at 18:57\n\nI would encourage you to split up the given conditions and group according to powers of the same prime.\n\nYou have $$3x \\equiv 1 \\pmod 7$$ $$3x \\equiv 9 \\pmod 5$$ and related $$3x \\equiv 1 \\pmod 2$$ $$x \\equiv 1 \\pmod 8$$\n\nThe redundant pair becomes, as $$3 \\equiv 1 \\pmod 2,$$ $$x \\equiv 1 \\pmod 2$$ $$x \\equiv 1 \\pmod 8$$ These are consistent, the highest power of the prime is $$8=2^3,$$ so these combine to $$x \\equiv 1 \\pmod 8.$$ Then $$x$$ is 5 mod 7 and 3 mod 5, together $$x \\equiv 1 \\pmod 8.$$ $$x \\equiv 3 \\pmod 5.$$ $$x \\equiv 5 \\pmod 7.$$ Now you can use CRT I get $$x \\equiv 33 \\pmod {280}$$ as $$33 = 32 + 1 = 30 + 3 = 28 + 5$$\n\nWhat did I do wrong?\n\nAt the end, you should have $$x=3+5k=3+5(-330+56y)$$\n\nor $$x=33+56h=33+56(-30-5y)$$,\n\nwhich means $$x=-1647+280y$$, so $$x\\equiv-1647\\equiv33\\pmod{280}$$.\n\n\u2022 I think you forgot the minus sign in front of $330$ to get $1653=3+5(330)$ Apr 2 '20 at 18:38", "date": "2021-09-28 23:57:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3606831/solving-a-congruence-system-when-the-chinese-remainder-theorem-cannot-be-applied", "openwebmath_score": 0.8332446217536926, "openwebmath_perplexity": 132.65870759629482, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575175622122, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.85541187929762}}
{"url": "https://math.stackexchange.com/questions/2300299/suppose-a-b-in-m-n-mathbbc-such-that-ab-ba-a-prove-that-a-is-not-in", "text": "# Suppose $A,B\\in M_n(\\mathbb{C})$ such that $AB-BA=A$ . Prove that $A$ is not invertible .\n\nSuppose that $A,B\\in M_n(\\mathbb{C})$ such that $AB-BA=A$. Prove that $A$ is not invertible.\n\nMy work:\n\nSuppose $A$ is invertible. Then $ABA^{-1}=I+B$ . So $B$ is similar to $I+B$ .Let $B$ have eigenvalues $c_1,c_2,\\ldots,c_n \\in \\mathbb{C}$. So $B$ has basis such that $B$ is upper triangular with respect to it and has $c_1,\\ldots,c_n$ as diagonal entries . It is easy to see that $I+B$ is upper-triangular with respect to this basis and has entries $1+c_1,\\ldots, 1+c_n$ .\n\nHence $$c_1+c_2+\\ldots +c_n=\\operatorname{trace}(B)$$ $$=\\operatorname{trace}(I+B)=(1+c_1)+\\ldots+(1+c_n)=n+c_1+\\ldots +c_n .$$\n\nSo $n=0$, contradiction. I'm not sure if my solution is correct one. It seems alright. I will be very thankful if you can confirm that the proof indeed is a correct one.\n\nAny other possible solutions are welcomed.\n\n\u2022 It is correct, but too complicated. After proving that $B$ and $I+B$ are similar, you can simply say that this is no possible, since the trace of $I+B$ is $n$ plus the trace of $B$. No need to mention upper triangular matrices. May 28, 2017 at 16:38\n\nLet $A$ be invertible.\n\nWe have $(AB-BA)A^{-1}=I$, or $ABA^{-1}-B=I.$\n\nBut $\\operatorname{tr} (ABA^{-1})=\\operatorname{tr}{B},$ which is a contradiction.\n\nIt is probably worth showing that $$A$$ is in fact nilpotent. Indeed, by induction we get $$A^m B - B A^m = m A^m$$ for all $$m\\ge 0$$. Taking the traces on both sides we get $$0=m \\operatorname{Trace}A^m$$ so (assuming char $$0$$) $$\\operatorname{Trace}A^m = 0$$ for all $$m\\ge 1$$. This implies $$A$$ nilponent.\n\n$$\\bf{Added:}$$ Based on an idea of @Hans: , we can generalize this . For $$X$$, $$Y$$ matrices, let $$[X,Y]=X Y - Y X$$.\n\nAssume that $$[C,B]=A$$, and $$[C,A]=0$$ ( $$C$$, $$A$$ commute). Then $$A$$ nilpotent.\n\nIndeed, for all $$m\\ge 1$$ we have $$[C,A^{m-1} B] = A^{m-1} [C,B]= A^m$$, and so $$\\operatorname{Trace} A^m=0$$ . Conclude $$A$$ nilpotent (assume char $$0$$).\n\n$$\\bf{Added:}$$. What we are showing is that if $$\\operatorname{ad}(C)^2 A=0$$, then $$\\operatorname{ad}(C) A$$ is nilpotent. We can also reason as follows: we may assume that the basic field is algebraically closed. Consider the Jordan structure of $$C$$, with blocks of size $$n_1$$, $$\\ldots$$, $$n_k$$ corresponding to $$\\lambda_1$$, $$\\ldots$$, $$\\lambda_k$$. Then we see that all operators $$B$$ in the kernel of $$\\operatorname{ad}C^2$$ have the property that $$\\operatorname{ad}(C)B=0$$.\n\n(here we use an explicit Jordan structure for $$\\operatorname{ad}(C)$$. We again use that the characteristic of the field is $$0$$ or $$>n$$ ).\n\n\u2022 Nice conclusion. +1. You only need to show $A^mB-A^{m-1}BA=A^m\\implies \\text{tr}A^m=0, \\forall m\\in\\mathbf N$.\n\u2013\u00a0Hans\nOct 8, 2017 at 7:34\n\u2022 @Hans; yes, your approach is simpler and better! I should add this idea. Oct 8, 2017 at 17:46\n\n$\\def\\m{\\mathfrak }$This result (and, in fact, orangeskid's observation) has the following far reaching generalization:\n\nif $\\m g$ is a finite dimensional (complex) Lie algebra and $\\mathfrak r$ is its radical, then $[\\m g,\\m r]$ acts on any finite dimensional representation of $\\m g$ nilpotently.\n\nIndeed, in the question the matrices $A$ and $B$ span a Lie algebra $\\m g$ of dimension (at most, really, but let us suppose) equal to $2$ which is solvable, so that the radical of $\\m g$ is simply $\\m g$ itself, and the theorem above tells us that $[\\m g,\\m g]$ acts nilpotently on finite dimensinal modules: since $[\\m g,\\m g]$ is spanned by $A$, the desired result follows.\n\nOf course, this is immensely more general.\n\nAs an example:\n\nif $A$, $B$ and $C$ are square matrices such that $CA-AC = B$, $CB-BC = aA + B$, $AB-BA = 0$, then $B$ is nilpotent and, more generally, al linear combinations of $A$ and $B$.\n\nThis comes from looking at a random 3-dimensional solvable Lie algebra.\n\n\u2022 One can find the proof of this, which depends on Levi's theorem, in the book on Lie algebra by Hossein Abbaspour and Martin A. Moskowitz; it is Theorem 3.5.6 there. Oct 12, 2017 at 17:41\n\u2022 Excellent answer! I will have to review my rusty rep theory... Oct 12, 2017 at 19:16", "date": "2022-08-10 03:11:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2300299/suppose-a-b-in-m-n-mathbbc-such-that-ab-ba-a-prove-that-a-is-not-in", "openwebmath_score": 0.9541088938713074, "openwebmath_perplexity": 151.96537640996212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575152637948, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.855411878946073}}
{"url": "https://community.wolfram.com/groups/-/m/t/1657895", "text": "# Autopilot Automated Plane Geometry Case Study\n\nPosted 3 months ago\n628 Views\n|\n2 Replies\n|\n4 Total Likes\n|\n\nDan's blog on automated plane geometry is about a quite fansinating application of new features of V12. This discussion is mainly to show how to make GeometricScene components as building blocks to give you insights to solve challenging geometric problems. Also because the components in this example is quite limited, I hope it serves as beginner case for Wolfram Language users who are interested in applying this feature in teaching and research.\n\n\u2022 AOPS Problem: In triangle $ABC$ let point $D$ be on side $BC$. If $AD=7$, $\\angle BAD=15$ deg, and $\\angle CAD = 30$ deg, compute the minimum possible area of the triangle.\n\nRandomInstance offers a second argument to generate a list of geometric objects that satifies the given concitions\n\nUse the following line to find the area of the triangles above:\n\nIn[ ]:= (Area[Triangle[{\"A\",\"B\",\"C\"}]]/.(#[\"Points\"]))&/@objs1\nOut[ ]= {22.0393,21.6971,82.0575,18.1804,90.0952,18.8076,50.7958,25.823}\n\n\nLets try more triangles, like 20, and see where is the minimum value even we haven't analyze the problem at all:\n\nWe can graphically conclude that the minimum value is around 17.93. (Pretty good approximation comparing to the true solution later)\n\nNext we consider adding some constraints to the GeometricScences so the illustrations of the valid triangle ABC's are aligned in regularzied manner. Well by the precision of WL and flexibility of MS OneNote, I quitck notice the following works:\n\nI appriciate that the GeometricScene function gives a mixture of cartesian-coordinate based programming and Geogebra-style freedom. Just modify the first argument in GeometricScene and make 10 instances of valid ABC:\n\nLooking at the the instances, you probably have some insights of the problem. The variation of the area of ABC is by\n\n\u2022 fix the angle BAC=45 degree and its two side rays\n\u2022 rotate a line $l$ that cross point D\n\n$B$, $C$ is then determined by the intersections of rays and $l$. To find how this rotation affects the area, we can parameterize $\\angle ABC = \\theta$ and the area of ABC. This is valid because $\\angle ABD$ is uniquely defined by $\\angle ADC$, which can be interpred as a measure of the rotation of $l$.\n\nComparing to the objs2, we successfully sort the triangles based on the angular size of $\\angle ABC$, from 10 degree to 110 degree with increment of 10. Lets plot the area vs $\\angle ABC$ on a finer segmentation of angle values:\n\nRemember that all codes that you have written is just gs1 plus some plot functions and parameter modification. Almost zero computation but finding the coordinate of $D$. This is way simpler than we used to do with Graphics function.\n\nLet's compare our result to the analytic solution:\n\nAttachments:\n2 Replies\nSort By:\nPosted 3 months ago\n - Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming, and consider contributing your work to the The Notebook Archive!", "date": "2019-07-24 07:05:05", "meta": {"domain": "wolfram.com", "url": "https://community.wolfram.com/groups/-/m/t/1657895", "openwebmath_score": 0.5493364930152893, "openwebmath_perplexity": 1254.7027792123627, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806518175515, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.855406963781814}}
{"url": "https://math.stackexchange.com/questions/3291437/the-number-of-triples-that-sum-to-a-constant?noredirect=1", "text": "# The number of triples that sum to a constant\n\nProblem:\nHow many triples are there of the form $$(x_0,x_1, x_2)$$ where $$x_0 \\in I$$, $$x_1 \\in I$$, $$x_2\\in I$$ $$x_0 \\geq 0$$, $$x_1 >= 0$$, $$x_2 >= 0$$ and $$n = x_0 + x_1 + x_2$$ where $$n \\in I$$?\nLet $$c(n)$$ be the number of tuples we can have for a given $$n$$. For $$n = 0$$, the only valid triple is $$(0,0,0)$$, hence $$c(0) = 1$$.\nFor $$c(1) = 3$$, the set of valid triples is: $$(0,0,1 ), (0,1,0), (0,0,1)$$ Hence $$c(1) = 3$$.\nFor $$c(2) = 6$$, the set of valid triples is: $$(1,0,1 ), (0,1,1 ), (0,0,2 ), (1,1,0), (0,2,0), (0,0,2)$$ Hence $$c(2) = 6$$.\nUsing the information on this URL:\nHow many $k-$dimensional non-negative integer arrays $(x_1,\\cdots,x_k)$ satisfies $x_1+x_2+\\cdots+x_k\\le n$\n\nI find the answer to be: $$c(n) = {{n+3}\\choose{3}} - {{n+2}\\choose{3}}$$ $$c(n) = \\frac{(n+3)!}{3!n!} - \\frac{(n+2)!}{3!(n-1)!}$$ $$c(n) = \\frac{(n+3)(n+2)(n+1) - (n+2)(n+1)(n)}{6}$$ $$c(n) = \\frac{(n+2)(n+1)(n+3 - n)}{6}$$ $$c(n) = \\frac{3(n+2)(n+1)}{6}$$ $$c(n) = \\frac{(n+2)(n+1)}{2}$$ Do I have that right?\nThanks,\nBob\n\n\u2022 This is equivalent to $\\frac12(n+1)(n+2)$ which agrees with the supplied values. Jul 12 '19 at 23:07\n\u2022 Like this? Jul 12 '19 at 23:10\n\u2022 en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) Jul 12 '19 at 23:16\n\nYes - this is correct. This can also be found by defining a helper function $$b(n)$$, which counts the number of ways to write $$n$$ as a sum of two numbers. Clearly, $$b(n) = n+1$$, as the first value, $$v$$, can be anything from $$0$$ to $$n$$, while the second value is $$n - v$$.\n\nThe function $$c(n)$$ is then equal to $$\\sum_{i = 0}^{n}b(n-i)$$ This is because the first value, $$i$$, can be anything from $$0$$ to $$n$$. The number of ways to write the remaining two numbers so that the total sum is $$n$$ is $$b(n-i)$$. Plugging in the formula finds $$c(n) = \\sum_{i = 0}^{n} (n-i+1) = n(n+1) - \\frac{n(n+1)}{2} + n+1 = \\frac{(n+2)(n+1)}{2}$$\n\nwhich is the same formula you arrived at.\n\nThis is a prototypical \u201cstars and bars\u201d problem. Here we have $$n$$ stars and $$2$$ bars, with $$\\binom{n+2}2 = {(n+2)(n+1)\\over2}$$ ways to choose among the $$n+2$$ possible positions for the two bars.\n\nAnother way: The number of partitions of $$n$$ into three non-negative integers is equal to the coefficient of $$z^n$$ in the formal power series $$(1+z+z^2+z^3+\\cdots)^3$$. This coefficient can be found using the generalized binomial theorem: $$(1+z+z^2+z^3+\\cdots)^3 = \\frac1{(1-z)^3} = \\sum_{k=0}^\\infty \\binom{-3}{k} z^k = \\sum_{k=0}^\\infty \\binom{k+2}2 z^k.$$ The number of partitions is therefore $$\\binom{n+2}2$$ as before.\n\n\u2022 All of these were mentioned in the comments, ~9h earlier. Jul 13 '19 at 8:19\n\u2022 @rtybase Your point being? If you though the question was a duplicate of some other, you should\u2019ve flagged it as such.\n\u2013\u00a0amd\nJul 13 '19 at 8:21\n\u2022 Something genuinly different is required. But up to you really ... I don't think the question is a duplicate. To begin with I don't know what $I$ is (in $x_0 \\in I$ for example). Nobody clarrified this. Jul 13 '19 at 8:22", "date": "2021-10-17 13:41:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3291437/the-number-of-triples-that-sum-to-a-constant?noredirect=1", "openwebmath_score": 0.9377309679985046, "openwebmath_perplexity": 121.15298584555288, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806478450309, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8554069586892646}}
{"url": "https://math.stackexchange.com/questions/2916320/5-cards-are-chosen-from-a-standard-deck-what-is-the-probability-that-we-get-all", "text": "# 5 cards are chosen from a standard deck. What is the probability that we get all four aces, plus the king of spades?\n\nWe have $\\binom{52}{5}$ ways of doing this. This will be our denominator.\n\nWe want to select all 4 aces, there are there are exactly $\\binom{4}{4}$ ways of doing this.\n\nNow we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $\\binom{48}{1}$ ways.\n\n$$\\dfrac{\\binom{4}{4}\\binom{48}{1}}{\\binom{52}{5}}=\\dfrac{1}{54145}?$$\n\nIs this correct?\n\nIf we look at it another way:\n\nThen then our probability for the aces and specific king are $(4/52)\\times (3/51) \\times (2/50) \\times (1/49) \\times (1/48)$ which is a completely different here.\n\nWhich is the correct approach?\n\n\u2022 numerator should be 1 if you want a specific combination \u2013\u00a0Ning Wang Sep 14 '18 at 2:15\n\u2022 So the second one is correct? \u2013\u00a0K Split X Sep 14 '18 at 2:17\n\u2022 No, it seems like you're permuting those four aces. \u2013\u00a0Ning Wang Sep 14 '18 at 2:19\n\u2022 $\\binom{48}{1}$ would mean you're selecting $1$ card (any) from the remaining $48$ cards and not specifically a king of spades. \u2013\u00a0Ister Sep 14 '18 at 7:53\n\u2022 $\\binom44$ is correct because there are $4$ aces to choose from, but $\\binom{48}1$ should be $\\binom11$ because there is only one king of spades. \u2013\u00a0bof Sep 14 '18 at 23:38\n\nAssuming you just want those 5 cards in any order:\n\nSince you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then\n\n(Probability choosing any one of the 5) $\\times$ (Probability of choosing one of the remaining 4) $\\times$ ..., i.e. $$\\frac{5}{52} \\times \\frac{4}{51} \\times \\frac{3}{50} \\times \\frac{2}{49} \\times \\frac{1}{48} = \\frac{1}{54145} \\times \\frac{1}{48},$$ just to confirm what Ross Millikan said.\n\nIf, instead, you want the 4 aces before the king, we have $$\\frac{4}{52} \\times \\frac{3}{51} \\times \\frac{2}{50} \\times \\frac{1}{49} \\times \\frac{1}{48}.$$\n\nYou've fully determined the five cards you want, so the probability is $\\frac{1}{{52 \\choose 5}}$ since there are ${52 \\choose 5}$ different ways to choose 5 cards from the deck, and only one correct one.\n\n\u2022 Just to clarify, you want to get 5 cards from a set of specific 5 cards in which case your numerator should be $\\binom{5}{5}$ \u2013\u00a0Ister Sep 14 '18 at 7:51\n\u2022 I'm not this carl; you wording leads the casual reader to believe you are incorrect, since there are 5! ways to pick the correct 5 cards. You should stress that you started by calculating combinations, not permutations \u2013\u00a0Carl Witthoft Sep 14 '18 at 15:54\n\nYour first computation gets the chance you get four aces and any other card. The $48 \\choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.\n\nThe first way:\n\nThere is only one set of cards, that fits the specifications (all 4 aces and the king of spades) and $\\binom{52}{5}$ different sets of cards that can all be drawn with equal probability. The chance to draw your set is $$\\frac{1}{\\binom{52}{5}}$$\n\nThe second way:\n\nThe chance of drawing a specific card out of $n$ is $\\frac{1}{n}$. So a specific sequence or order of 5 cards has the chance $\\frac{1}{52}\\cdot\\frac{1}{51}\\cdot\\frac{1}{50}\\cdot\\frac{1}{49}\\cdot\\frac{1}{48}$ to be drawn. There are $5!$ different sequences or orderings of 5 cards, that countain the 5 specified cards (s. permutations). So the chance to draw the set is $$\\frac{5!}{52\\cdot51\\cdot50\\cdot49\\cdot48} = \\frac{5!\\cdot 47!}{52!} = \\frac{1}{\\frac{52!}{5!\\cdot 47!}} = \\frac{1}{\\binom{52}{5}}$$\n\nFor the last equality see Binomial coefficient.\n\nYou select 4 aces out of 4 possible aces and one king of spades out of 1 possible king of spades, so there is exactly $$\\binom{4}{4}\\binom{1}{1}=1$$ way of completing this set. Therefore this set is unique.\n\nThere are $\\binom{52}{5}$ unique sets of five cards.\n\nTherefore the probability of picking this set of five cards is:\n\n$$P=\\frac{\\binom{4}{4}\\binom{1}{1}}{\\binom{52}{5}} = \\frac{1}{2598960}$$", "date": "2019-04-21 04:22:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2916320/5-cards-are-chosen-from-a-standard-deck-what-is-the-probability-that-we-get-all", "openwebmath_score": 0.6616800427436829, "openwebmath_perplexity": 275.38193201064223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806546550656, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8554069516129288}}
{"url": "https://math.stackexchange.com/questions/3013379/counting-number-of-bitstrings-that-do-not-contain-110-for-length-n-ge-4", "text": "# Counting number of Bitstrings that do not contain $110$ for length $n \\ge 4$\n\nQuestion: Consider bitstrings that do not contain $$110$$. Let $$S_n$$ be the number of such strings having length $$n$$. What is $$S_n$$ for $$n \\ge 4$$?\n\nAnswer: $$S_n = S_{n-1} + S_{n-2} + 1$$\n\nAttempt: I tried writing out all the bitstrings for $$n=2$$, and $$n=3$$.\n\n$$n=3$$ I got: $$000, 001, 011, 111, 100, 010, 101$$\n\n$$n=2$$ I got: $$00, 01, 10, 11$$\n\n$$n=4$$ seems to large to do this by hand. According to the answer, it should be $$S_4 = S_3 + S_2 + 1$$.\n\n$$S_4 = 7 + 0 + 1 = 8$$? Assuming $$S_n$$ is the number of strings without $$110$$, so since $$n=3$$ has one $$110$$ I just exclude that and get $$7$$. $$n=2$$ has no $$110$$.\n\nI'm not sure how the formula was derived. The bitstring counting logic I struggle with, any guidance on how to appriach these would help a lot.\n\n\u2022 Your bitstring either starts with a $0$, or it starts with a $10$, or it has no $0$'s at all. Do you see how that relates to the answer's recursive formula? \u2013\u00a0Barry Cipra Nov 25 '18 at 22:15\n\nThis is not going to be an answer, but I don't have enough reputation to comment, so I apologize in advance.\n\nMy thought is to try thinking about your placement of the $$110$$. For the case of $$n=3$$, you saw that there was only one place to put it - as the whole thing.\n\nNow consider the case of $$n=4$$. The approach will be to consider the possibilities which contain the string you are looking to avoid, find out how many such unique strings there are and subtract it from the total.\n\nThere are two places to put the string $$110$$, namely the beginning $$110\\bullet$$ or the end $$\\bullet110$$. There are $$2^4$$ total four-digit bitstrings, and there are four which match these placement patterns $$1100, 1101$$ and $$0110, 1110$$. Just to check, we should make sure all four of these are different, which is clear for this many (just look at the third digit) but might not be so clear on a larger scale. So as I see it there are $$2^4 - 4 = 12$$ ways to do this. This is $$16-4=12 \\neq 8$$.\n\nI agree with your logic for finding $$S_3, S_2$$, and hence $$S_4 = 8$$. Do you see any errors in my logic?\n\nIf you agree with my logic, is it possible that the question is not being asked the way you have asked it here? Alternatively are you sure the solution you have is correct?\n\n\u2022 The mistake in Toby's computation is that $S_2 = 4$, not $0$. The given solution ($S_n = S_{n-1} + S_{n-2} + 1$) is correct. (See Barry Cipria's comment above.) \u2013\u00a0Fabio Somenzi Nov 25 '18 at 22:41", "date": "2019-05-23 01:28:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3013379/counting-number-of-bitstrings-that-do-not-contain-110-for-length-n-ge-4", "openwebmath_score": 0.8970907330513, "openwebmath_perplexity": 115.88116221967319, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806501150427, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8554069476524517}}
{"url": "https://math.stackexchange.com/questions/1969668/depiction-of-all-graphs-not-necessarily-connected-with-6-vertices?noredirect=1", "text": "# Depiction of all graphs (not necessarily connected) with 6 vertices\n\nI am searching (with no unsuccess) the WWW for a depiction of all non-isomorphic graphs with 6 vertices.\n\nOn the site http://www.graphclasses.org/smallgraphs.html I found an incomplete list, but all graphs with 5 vertices (34 ones). According to the integer sequence A000088, there should be 156 of them. A paper by Cvetkovic and Petric http://www.sciencedirect.com/science/article/pii/0012365X84900335 gives all connected graphs with 6 vertices (112 ones).\n\nIf I take all connected graphs (112) and take all graphs with 5 vertices adding a separate node (34), then I have 146 graphs. Which 10 graphs are missing?\n\nThank you very much!\n\n\u2022 By all complete graphs did you perhaps mean all connected graphs? \u2013\u00a0hardmath Oct 15 '16 at 15:54\n\u2022 Sorry. Yes, thanks. \u2013\u00a0user36124 Oct 15 '16 at 15:56\n\u2022 Given your existing work, you may find the earlier Question Why there are 11 non-isomorphic graphs of order 4? a useful reference. These include non-connected possibilities. \u2013\u00a0hardmath Oct 15 '16 at 16:03\n\u2022 You need all connected graphs on 4 vertices (6 of them) with two separate vertices and the pair of 3-line/triangle with 3 disconnected vertices, etc. 6 + 2 + 1 + 1 = 10. To put it another way, the number of graphs on N are the sum of the number of connected graphs on N plus the number of connected graphs on N-1 ... 1. \u2013\u00a0gilleain Oct 15 '16 at 16:08\n\u2022 You may be interested to look at findstat.org/StatisticFinder/Graphs by the FindStat project \u2013\u00a0Alexander Konovalov Oct 15 '16 at 17:05\n\nWe can partition the graphs by the number of connected components.\n\nOne connected component: There are $112$ connected graphs with $6$ vertices\n\nTwo connected components: There can be\n\n\u2022 A connected component of size $1$ and a connected component of size $5$ ($1 \\cdot 21 = 21$ possibilities)\n\u2022 A connected component of size $2$ and a connected component of size $4$ ($\\color{red}{1 \\cdot 6 = 6}$ possibilities)\n\u2022 A connected component of size $3$ and another connected component of size $3$ ($\\color{red}{3}$ possibilities - $P_3 \\sqcup P_3$, $P_3 \\sqcup K_3$, or $K_3 \\sqcup K_3$)\n\nThree connected components: There can be\n\n\u2022 Two connected components of size $1$ and a connected component of size $4$ ($1 \\cdot 1 \\cdot 6=6$ possibilities)\n\u2022 A connected component of size $1$, one of size $2$, and one of size $3$ ($1 \\cdot 1 \\cdot 2 = 2$ possibilities)\n\u2022 Three connected components of size $2$ ($\\color{red}{1}$ possibility - $P_3 \\sqcup P_2 \\sqcup P_2$)\n\nFour connected components: There can be only be three connected components of size $1$ and one of size $3$, for $1 \\cdot 1 \\cdot 1 \\cdot 2 = 2$ possibilities\n\nFive connected components: There can only be four connected components of size $1$ and one of size $2$, for $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 2 = 2$ possibilities\n\nSix connected components: One possibility\n\nAdd these up and you get $112+21+6+3+6+2+1+2+2+1=156$ graphs.\n\nThe ones in red are the ones you are missing - they are not obtained from adding an isolated vertex to a graph with $5$ vertices.\n\nPerhaps the easiest way to systematically count is to begin with the integer partitions of six as denoting the sizes of components in your (not necessarily connected) graphs. There are eleven possibilities.\n\nSome of these give rise to a single (up to isomorphism) graph:\n\n$1 + 1 + 1 + 1 + 1 + 1$\n\n$2 + 1 + 1 + 1 + 1$\n\n$2 + 2 + 1 + 1$\n\n$2 + 2 + 2$\n\nSome give rise to two possibilities each as three nodes can form a connected component in exactly two ways:\n\n$3 + 1 + 1 + 1$\n\n$3 + 2 + 1$\n\nThen there are three possibilities for this partition:\n\n$3 + 3$\n\nbecause the two components can be non-isomorphic in one way but isomorphic in two ways.\n\nFrom the earlier Question we see that there are six connected graphs on four vertices, so each of these gives six possibilities:\n\n$4 + 1 + 1$\n\n$4 + 2$\n\nThen we get down to the cases you previously considered:\n\n$5 + 1$ (a connected component of five vertices and one extra node)\n\n$6$ (a connected component of six vertices)\n\nThere are respectively $21$ and $112$ of these cases (see OEIS A001349, Number of connected graphs with n nodes). The previous small cases numbered $23$, and $23+21+112= 156$.", "date": "2019-11-20 02:56:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1969668/depiction-of-all-graphs-not-necessarily-connected-with-6-vertices?noredirect=1", "openwebmath_score": 0.7440390586853027, "openwebmath_perplexity": 410.9774930776581, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806495475397, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.855406947157392}}
{"url": "https://math.stackexchange.com/questions/3262238/does-knowing-the-surface-area-of-all-faces-uniquely-determine-a-tetrahedron", "text": "# Does knowing the surface area of all faces uniquely determine a tetrahedron?\n\nI was wondering if the four areas of a tetrahedron faces were sufficient information to uniquely determine its shape. For example, is it true to say that if the surface areas are equal then the solid must be a regular tetrahedron?\n\nIf the answer is negative, then what else we need to fully determine the shape of the tetrahedron in space?\n\n\u2022 Well if a tetrahedron is the only shape which can exist with four faces in three dimensions then it must be a tetrahedron. You could prove whether or not it must be regular based on the areas using algebra \u2013\u00a0Henry Lee Jun 14 at 14:14\n\u2022 The space of tetrahedra, up to congruency, has 6 degrees of freedom, and you only give 4 pieces of information with your areas. So my immediate guess is that no, it is in no way determined. \u2013\u00a0Arthur Jun 14 at 14:14\n\u2022 Are you aware of the Minkowski theorem (math.stackexchange.com/questions/105033/\u2026)? \u2013\u00a0Moishe Kohan Jun 14 at 14:56\n\u2022 @MoisheKohan No, I hadn't seen this before. But does this theorem imply existence of more than one tetrahedron with identical face areas? \u2013\u00a0Amirh.Kp Jun 14 at 17:43\n\u2022 @Amirh.Kp: yes, of course. This is existence-uniqueness theorem. It allows you to prescribe areas as well as the directions of normals to the faces. It also shows exactly what you are missing. \u2013\u00a0Moishe Kohan Jun 14 at 18:22\n\nConsider the vertex set in $$\\mathbb R^3$$ $$\\{(a, b, 0), (a, -b, 0), (-a, 0, b), (-a, 0, -b)\\}$$ for $$a, b > 0$$. Then it is clear that the faces determined by these vertices are all congruent isosceles triangles with side lengths $$2b$$, $$\\sqrt{4a^2+2b^2}$$, $$\\sqrt{4a^2+2b^2}$$ and common area $$a \\sqrt{4a^2+b^2}$$. So by a suitable choice of $$a$$ and $$b$$ we can make the tetrahedron irregular but have faces of equal area; moreover, for a given area of a face, there is in general more than one choice of $$(a,b)$$, hence even for this very restricted type of tetrahedron, knowing the area of each face does not uniquely determine the tetrahedron.\n\nA tetrahedron with vertices at $$(0,0,0)\\\\ (1/2, 0,0)\\\\ (0,2,0)\\\\ (-0.0924127, 1.9387, 0.4913857)$$ will have area $$1$$ for all faces.\n\nThe coordinates of that last point are approximate (and of course you can freely swap the sign of the $$z$$-coordinate). The actual point is given as the solution to the three equations $$\\cases{x^2 + z^2 = 1/4\\\\ y^2 + z^2 = 4\\\\ \\displaystyle\\left(\\frac4{\\sqrt{17}}(x-1/2) -\\frac1{\\sqrt{17}}y\\right)^2 + z^2 =\\frac43}$$ I placed the three first vertices first, to make sure one face was non-equilateral and had area $$1$$, then these three equations are exactly the equations that ensure that the other three faces have area $$1$$. The left-hand sides are (the squares of) the altitudes of the three remaining faces if you put the fourth vertex at $$(x, y, z)$$, and the right-hand sides are what those altitudes must be to ensure that the area of the corresponding face is $$1$$.\n\nAs you can see, I am somewhat free to make the first face whatever shape I want, as long as its area is $$1$$, and then I just set up the three equations to find the fourth vertex. It's possible that some extreme versions of the first face causes the resulting three equations to not have any real solution, but I have demonstrated here that at least one non-regular tetrahedron may be generated this way.\n\n\u2022 +1 I think there's an interesting construction lurking here. If you relax the equal area condition to specify just the ratios of areas of faces then Informally, you're foliating the space of tetrahedra What do the leaves look like? \u2013\u00a0Ethan Bolker Jun 14 at 15:00\n\u2022 @EthanBolker That's a cool question. I really only thought of it as just intersecting three cylinders... \u2013\u00a0Arthur Jun 14 at 15:01\n\nThe answer is \"no,\" which can be demonstrated by counting degrees of freedom.\n\nA tetrahedron has 4 vertices in 3-dimensional space, and is therefore defined by 12 independent parameters, i.e. the $$x$$, $$y$$, $$z$$ coordinates of each vertex.\n\nNow consider what we mean by two tetrahedra having \"the same shape\". The orientation in 3-D space doesn't matter, and that eliminates 6 of the 12 parameters (there are 3 rigid body translations, and 3 rotations).\n\nBut the tetrahedron only has four faces, so fixing the area of each face still leaves two independent parameters to vary its shape in an arbitrary way.\n\nIf you want to pick two more quantities to fix the shape of the tetrahedron, that could be done in many different ways, but it would be nice to do it in a way that is independent of any coordinate system used to describe it.\n\nOne natural parameter could be the volume of the tetrahedron, but finding a second one is not so \"obvious\".\n\nThe radius of the inscribed and circumscribed spheres might be interesting choices. Proving that they do uniquely define the tetrahedron in combination with the area of the faces is left as an exercise for the reader :)\n\nAnother option might be to fix the lengths of two edges. Choosing a pair of edges which do not share a vertex has a nice symmetry about it.\n\nYou may be interested in my paper with Petr Lisonek, Metric Invariants of Tetrahedra via Polynomial Elimination, ISSAC 2000 conference, Aberdeen (Scotland), July 2000, 217-219. We show that in general the four areas, circumradius and volume do not determine a tetrahedron, but there exist non-regular tetrahedra that are determined by the four areas and the circumradius. For example, this is the case if one face is an equilateral triangle iscribed in a great circle of the circumscribed sphere.\n\nIt's easy to construct a tetrahedron with sides congruent to any acute isosceles triangle.\n\nStart by taking two copies of the triangle, glue them together at the base, and then pull the apices apart until the distance between them equals the length of the base. (You can always do this if the triangles are acute, since the base length of an acute isosceles triangle is less than twice the height.) Now the sides of the triangles, together with the line between their apices, form two additional isosceles triangles that are congruent with the original one, and you thus have a tetrahedron with four congruent faces. Cut some triangles out of paper and try it if you don't believe it!\n\nIn fact, essentially the same method actually works for any acute triangle, even if it's not isosceles! The only tricky part is that now there are two distinct possible ways to glue the first two copies of the triangle together at the base, depending on whether or not you flip one of them over first, and you need to choose the way that leaves the non-equal sides of the triangles adjacent. Then proceed just like above.\n\nThus, in particular, for any acute triangle with area $$A$$ there exists a tetrahedron with all faces congruent to that triangle, and thus also with area $$A$$. Unless the triangle happens to be equilateral, this tetrahedron will not be regular.\n\nA tetrahedron can be determined by the lengths of three edges emanating from a vertex and the three angles formed by pairs of those angles. Thus, a tetrahedron admits six degrees of freedom. Four face areas are not enough to determine the shape (or even the volume) of such a figure.\n\nAs an extreme example, you can consider any rectangle a \"degenerate\" tetrahedron with four congruent right-triangle faces; the volume is zero, which is decidedly different than that of a regular tetrahedron. In fact, one can construct \"equihedral\" (equal-face-area) tetrahedra with volumes anywhere from the minimum of zero to the regular tetrahedron's maximum.\n\nAs for what other information you can use, my favorite additional parameters are the areas of the tetrahedron's three (what I call) \"pseudo-faces\". You can think of these geometrically as the quadrilateral projections of the tetrahedron into planes parallel to a pair of opposite sides. Each pseudo-face area is calculated by $$\\text{area} = \\frac12 \\text{side}\\cdot\\text{side} \\cdot \\sin \\text{(angle)}$$ just like any other face, except here, the \"$$\\text{side}$$\"s are opposite each other, and the \"$$\\text{angle}$$\" is considered the angle between the corresponding direction vectors.\n\nFour standard faces (say, $$W$$, $$X$$, $$Y$$, $$Z$$) and three pseudo-faces ($$H$$, $$J$$, $$K$$) make seven area parameters, which would seem to over-determine the figure. However, the sum-of-squares identity $$W^2+X^2+Y^2+Z^2=H^2+J^2+K^2 \\tag{1}$$ introduces a dependency that reduces the degrees of freedom to the expected six.\n\nOther pseudo-face relations include a strangely familiar-looking Law of Cosines. \\begin{align} Y^2 + Z^2 - 2 Y Z \\cos A &= H^2 = W^2 + X^2 - 2 W X \\cos D \\\\ Z^2 + X^2 - 2 Z X \\cos B &= \\,J^2 = W^2 + Y^2 - 2 W Y \\cos E \\\\ X^2 + Y^2 - 2 X Y \\cos C &= K^2 = W^2 + Z^2 - 2 W Z \\cos F \\end{align} \\tag{2} where each of $$A$$, $$B$$, $$C$$, $$D$$, $$E$$, $$F$$ is the dihedral angle between appropriate pairs of faces ($$A$$ between $$Y$$ and $$Z$$, etc).\n\nThere's also this volume formula:\n\n\\begin{align} 81V^4 &= H^2 J^2 K^2 - 2 (W X-Y Z)(W Y-Z X)(W Z-X Y) \\\\ &-H^2(W X-YZ )^2-J^2(WY-ZX)^2-K^2(WZ-XY)^2 \\end{align} \\tag{3}\n\nIf the four face areas are equal, the formula reduces to $$9V^2 = HJK$$, which shows that the volume of an \"equihedral\" tetrahedron depends upon more than those face areas.\n\nAnyway, you can read more about these kinds of relations in my Hedronometry notes. In particular, \"Heron-Like Results for Tetrahedral Volume\" (PDF) includes the stuff I've described above and a bit more.\n\n\u2022 Thanks for your comprehensive introduction. one thing I can't understand which is seemingly a critical concept and that is your extreme example of considering every rectangle as a \"degenerate tetrahedron\" with four congruent right-triangle faces. I think you mean it's a natural result of a process which construct tetrahedrons but I can't come up with such process. \u2013\u00a0Amirh.Kp Jun 14 at 18:21\n\u2022 @Amirh.Kp: Any four points can be the vertices of a tetrahedron. The vertices of a rectangle are four points, so they determine a tetrahedron; since it's flat, the volume is zero. (The rectangle's four sides, and its two diagonals, are the edges of the tetrahedron, and the right triangles formed by two adjacent sides and one diagonal are the faces of this tetrahedron.) Such a figure is considered \"degenerate\" because it is not three-dimensional. This is like how any three points determine a triangle; if those points are collinear, the triangle is \"degenerate\" with area zero. \u2013\u00a0Blue Jun 14 at 22:56", "date": "2019-12-13 00:31:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3262238/does-knowing-the-surface-area-of-all-faces-uniquely-determine-a-tetrahedron", "openwebmath_score": 0.9008432030677795, "openwebmath_perplexity": 337.2846036906231, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806506825456, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8554069448932483}}
{"url": "http://ztvu.uiuw.pw/power-series-and-taylor-series.html", "text": "# Power Series And Taylor Series\n\nWhat is Power series? A power series is a series of the form. The problem with the approach in that section is that everything came down to needing to be able to relate the function in some way to. Let f(x) = X1 n=0 c n(x a)n = c 0 + c 1(x a) + c. DEFINITION 2. This of course is just a power series shifted over by c units. 2 Properties of Power Series 10. The Taylor series of is the sum of the Taylor series of and of. What is the interval of convergence for this series? Answer: The Maclaurin series for ex is 1+x+ x2 2! + x3. Read moreTaylor and Maclaurin Series. In the Summer 1994, the author developed computer activities intended to provide an intuitive interpretation to some of the fundamental notions involved in studying infinite series and Taylor polynomials. Note: In Problem 52, there is a mistake in the directions. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. 2, is a Taylor series centered at zero. For further details, see the class handout on the inverse. Multivariate Taylor Series. Find more Mathematics widgets in Wolfram|Alpha. There is also a special kind of Taylor series called a Maclaurin series. Leavitt Power series in the past played a minor role in the numerical solutions of ordi-nary and partial differential equations. In the Summer 1994, the author developed computer activities intended to provide an intuitive interpretation to some of the fundamental notions involved in studying infinite series and Taylor polynomials. So, a couple definitions to get us started here. A power series converges uniformly and absolutely in any region which lies entirely inside its circle of convergence. Taylor and Laurent Series We think in generalities, but we live in details. Here we address the main question. Use a known Maclaurin series to obtain the Maclaurin series for the function f(x) = cos(\u03c0x). You can specify the order of the Taylor polynomial. The series we will derive a power series that will converge to the factor. Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. This example generalizes as follows. Choose the maximum degree of the Taylor polynomial to use to approximate a function. Just assume that RaiseTo(raise a number to the power of x) is there. Each of the resistors in a series circuit consumes power which is dissipated in the form of heat. To find the Maclaurin Series simply set your Point to zero (0). Convergence of In nite Series in General and Taylor Series in Particular E. It takes the following form: Here\u2019s a common example of a p-series, when p = 2: Here are a few other examples of p-series: Remember not to confuse p-series with geometric series. A Taylor series method for numerical \ufb02uid mechanics J. 5) on uniform convergence is optional. This is due to the uniqueness of the Taylor series of a function centered at a point. Let\u2019s prove a lemma to deal with that last point. Section 4-16 : Taylor Series. If a function is equal to it's Taylor series locally, it is said to be an analytic function, and it has a lot of interesting properties. The last section (15. The method is shown to be non\u00addispersive, non\u00addiffusive, and for. (ii) Using (i) or otherwise nd the Taylor series expansion of 1 (1 x)2 and 1 (1 x)3 about a = 0, stating carefully any theorems you may use about integrating or di er-entiating power series within their radius of convergence. Taylor\u2019s Theorem states that if f is represented by a power series centered at c, then the power series has the form 0 n fx = ssss s ss sssssssssssss s ss sssssssssssssss s ss ssssssssssssssss s s If a power series is centered at c = 0, then it is called a sssssssss ssssss. (Several of these are listed below. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. RELATION BETWEEN TAYLOR SERIES AND POWER SERIES A power series = Taylor series of its sum In other words, every time you obtain an identity X\u221e n=0 a nx n = (something) then the power series on the left-hand sidemust be the Taylor series of that something on the right-hand side. Series effectively evaluates partial derivatives using D. If a= 0 the series is often called a Maclaurin Math formulas for Taylor and Maclaurin series Author:. These are the most important series of all! (Taylor, Maclaurin, etc, etc. The series P1 n=0 anx n, x 2 R, is called a power series. Recall from Chapter 8 that a power series. As the names suggest, the power series is a special type of series and it is extensively used in Numerical Analysis and related mathematical modelling. So we can write a simple generalised expression for a power series as g of x, equals a, plus bx, plus cx squared, plus dx cubed et cetera. The TaylorAnim command can handle functions that \"blow-up\" (go to infinity). Substituting the coefficients back into the series yields. 4 Working with Taylor Series Learn with flashcards, games, and more \u2014 for free. Taylor and Maclaurin Series Tutorial for Calculus students. Taylor and Maclaurin (Power) Series Calculator. They are distinguished by the name Maclaurin series. Maclaurin Series: If a function f can be differentiated n times, at x=0, then we define. To enhance our students' learning of the infinite series material, a computer laboratory activity devoted to the subject was created. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. Power series are useful because ssss sssss ssss ss sss. This series is referred to as the Taylor series of a function f(x) centered at c. + Formal manipulation of Taylor series and shortcuts to computing Taylor series, including substitution, differentiation, antidifferentiation, and the formation of new series from known series. This gives us a simple formulaB for the sum:\" B B B \u00e2 \u0153 \" \" B # $This is our first example of a Taylor series \u2014a power series that adds up to a known function. is the Taylor series of f(z) = 1=(1 z) about z= 0. Every Taylor series is a power series in 0 0 0! k k k fx xx k is a power series in x 2 Theorem 9. Use the ratio test to show that the Taylor series centered at 0 for sin(x) converges for all real numbers. But this is good news for combinatorics. Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. But it converges at both end points and does so, therefore, absolutely. Use the ratio test, unless otherwiseinstructed. f The coefficients of this power series may be expressed with the Bernoulli numbers. In the figure below, we consider the graphs. Correct! This is the correct answer, found by using mathematical operations on the geometric power series. Lesson 23: Power Series Expansions. CHAPTER 38 Power Series In Problems 38. In fact, that's the brute force method of finding a series representation for a function, but there are other ways. Many times a Taylor expansion is used for approximations in solving transcendental equations such as x - ln x = 5 which cannot be solved by currently known algebraic manipulations In some cases in solving differential equations a Taylor series will actually give an exact answer that can't be readily found by any other method. If you want the Maclaurin polynomial, just set the point to 0. (See the text, p. FUNCTIONS OF A COMPLEX VARIABLE (S1) Lecture 7 Power series expansions \u22b2 Taylor series f representable by Taylor series expansion is said to be analytic. What is Power series? A power series is a series of the form. If the series uses the derivatives at zero, the series is also called a Maclaurin series, named after Scottish mathematician Colin Maclaurin (February 1698 \u2013 14 June 1746). In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. This example generalizes as follows. Taylor and Laurent Series We think in generalities, but we live in details. The resulting series can be used to study the solution to problems for which direct calculation is di cult. Find the Taylor series for e\u2212x2 centered at 0. However, using differentiation and integration we can expand many more functions into power series also. This is a convergent power series, but the same power series does not de\ufb01ne an asymptotic series for exp(z). Another immediate and straightforward consequence of Theorem 2. The series P1 n=0 anx n, x 2 R, is called a power series. The binomial function Remark: If m is a positive integer, then the binomial function f m is a polynomial, therefore the Taylor series is the same polynomial, hence the Taylor series has only the \ufb01rst m +1 terms non-zero. + Maclaurin series and the general Taylor series centered at x = a. Power Series Solution of a Differential Equation \u2022 Approximation by Taylor Series Power Series Solution of a Differential Equation We conclude this chapter by showing how power series can be used to solve certain types of differential equations. because we take the formula for a Taylor polynomial centered at zero and let it keep on going. We have over 350 practice questions in Calculus for you to master. Learn how these polynomials work. To construct a power series solution around the point x = x o, we procede as follows: (1) Set y(x) = P 1 n=0 a n(x x o) n. The TaylorAnim command can handle functions that \"blow-up\" (go to infinity). And this is because they are composed of coefficients in front of increasing powers of x. Given just the series, you can quickly evaluate , , , \u2026, and so on. Before, we only considered power series over R but now, we will consider power series over C as well. Free power series calculator - Find convergence interval of power series step-by-step. (1) Find the radius of convergence of (a) X1 n=1 5nxn n2 (b) For what values of xdoes X1 n=1 (2x+ 1)n n3 converge? (c) Give an example of a power series which converges for all x2( 1;1] and at no other points. ) Series can also generate some power series that involve fractional and negative powers, not directly covered by the standard Taylor series formula. We now shift from the approach of Cauchy and Goursat to another approach of evaluating complex integrals, that is, evaluating them by residue integration. Consider the one dimensional initial value problem y' = f(x, y), y(x 0) = y 0 where f is a function of two variables x and y and (x 0, y 0) is a known point on the solution curve. It is a series that is used to create an estimate (guess) of what a function looks like. In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. The nearer to a the value is, the more quickly the series will converge. We would like to know which x0s we can plug in to get a convergent series. The TaylorAnim command can handle functions that \"blow-up\" (go to infinity). Several useful Taylor series are more easily derived from the geometric series (11), (19) than from. To enhance our students' learning of the infinite series material, a computer laboratory activity devoted to the subject was created. Taylor's Series method. We begin with the general power series solution method. POWER SERIES 251 For example, sine is an analytic function and its Taylor series around x 0 = 0 is given by sin(x) = X1 n=0 (1)n (2n + 1)! x2n+1: In Figure 7. We begin with the general power series solution method. Math formulas and cheat sheet generator creator for Taylor and Maclaurin Series. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Fenton School Of Mathematics University Of New South Wales Kensington, N. We prove it in order to demonstrates the Taylor series proposition above. Intervals of Convergence of Power Series. And this is because they are composed of coefficients in front of increasing powers of x. It explains how to derive power series of composite functions. What is a power series? 6. Practice Problems: Taylor and Maclaurin Series 1. Taylor\u2019s Series. Title: Taylor series of hyperbolic functions:. For example, the Taylor Series for ex is given by:. Power Series, Taylor Series In Chapter 14, we evaluated complex integrals directly by using Cauchy's integral formula, which was derived from the famous Cauchy integral theorem. I was just wondering in the lingo of Mathematics, are these two \"ideas\" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general co. The series converges only for. A summary of Differentiation and Integration of Power Series in 's The Taylor Series. (Any power series whatso-ever. In 1668, the theory of power series began with the publication of the series for ln()1+x by Nicolaus Mercator, who did this by \u201cintegrating\u201d 1 1+x (Stillwell 1989, 120). In other words, it's not a hypothesis we have to verify or check for. Leavitt Power series in the past played a minor role in the numerical solutions of ordi-nary and partial differential equations. If a= 0 the series is often called a Maclaurin Math formulas for Taylor and Maclaurin series Author:. THE BINOMIAL SERIES 375 6. If we take x0 = x\u00a1c then the power series around c reduces to the power series around 0. infinite series in Nov\u00e6 quadraturae arithmeticae in 1650, finding 1 n=1 nn()+1 \u221e \u2211 along with proving the divergence of the harmonic series. Since this power must come from the source, the total power must be equal to the power consumed by the circuit resistances. Another immediate and straightforward consequence of Theorem 2. The basic idea is to approximate the solution with a power series of the form: (1) X1 m=0 a m(x mx 0) : As an example consider the Taylor. Created by Sal Khan. Find the Taylor series for e\u2212x2 centered at 0. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. 10 The Binomial Series 6. Note that since is an even function, all its Taylor polynomials are also even polynomials. Incorrect! The signs are incorrect. It takes the following form: Here\u2019s a common example of a p-series, when p = 2: Here are a few other examples of p-series: Remember not to confuse p-series with geometric series. But this is good news for combinatorics. Convergence of In nite Series in General and Taylor Series in Particular E. CHAPTER 38 Power Series In Problems 38. Finding Taylor Polynomials The TI-89 taylor( command can be used to find partial sums. The partial sum is called the nth-order Taylor polynomial for f centered at a. Taylor series is a way to representat a function as a sum of terms calculated based on the function's derivative values at a given point as shown on the image below. TAYLOR AND MACLAURIN\u2122S SERIES 359 6. Here is a set of practice problems to accompany the Taylor Series section of the Series & Sequences chapter of the notes for Paul Dawkins Calculus II course at Lamar University. In general this series will converge only for certain values of x determined by the radius of convergence of the power series (see Note 17). Indeed, the entire power series\" B B B \u00e2#$ can be thought of as a geometric series with a common ratio of. f The coefficients of this power series may be expressed with the Bernoulli numbers. Power series are useful because ssss sssss ssss ss sss. (See the text, p. More generally, if c 2 R, then the series P1 n=0 an(x\u00a1c)n, x 2 R, is called a power series around c. Note that since is an even function, all its Taylor polynomials are also even polynomials. Taylor and Maclaurin Series Tutorial for Calculus students. You can skip questions if you would like and come back to them. (Several of these are listed below. But it converges at both end points and does so, therefore, absolutely. Spring 03 midterm with answers. of better and better approximations to f leading to a power series expansion f(x) = X\u221e n=0 f(n)(a) n! (x\u2212a)n which is known as the Taylor series for f. In this section we will discuss a. Created by Sal Khan. The series generated by the sequences (a nzn) as z varies are called the power series generated by (a n). for any x in the series' interval of convergence. If the series uses the derivatives at zero, the series is also called a Maclaurin series, named after Scottish mathematician Colin Maclaurin (February 1698 \u2013 14 June 1746). Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. Learn exactly what happened in this chapter, scene, or section of Calculus BC: Series and what it means. A power series about x = x 0 (or centered at x = x 0), or just power series, is any series that can be written in the form X1 n=0 a n(x x 0)n; where x 0 and a n are numbers. We can obtain a finite part, the first few terms, of a power series expansion of a function about a point by means of the Mathematica function Series as follows:. Example 5 Find the Maclaurin series for cos(x). Whether the power series converges at x = x0 \u00b1 \u03c1 is tricky to determine. Title: Taylor series of hyperbolic functions:. We now shift from the approach of Cauchy and Goursat to another approach of evaluating complex integrals, that is, evaluating them by residue integration. As mentioned earlier, the function 1=(1 z) exists and is in nitely di erentiable everywhere except at z= 1 while the series P 1 n=0 z nonly exists in the unit circle jzj<1. This is the geometric power series. The following proposition is sometimes useful. Reviewing Taylor Series In first year calculus, you undoubtedly spent significant time studying Taylor series. A power series is a series of the form P 1 k=0 c kx k, or more gen-erally: P 1 k=0 c k(x kx 0). Finding the series expansion of d u _ \u201e / du dk 'w\\. What is the interval of convergence for this series? Answer: The Maclaurin series for ex is 1+x+ x2 2! + x3. As it happens, Every power series is the Taylor series of some $C^{\\infty}$ function , but whether you refer to a series as a power series or a Taylor series depends on context. Many functions can be written as a power series. A power series in the variable x and centered at a is the in nite series. A Taylor series is a clever way to approximate any function as a polynomial with an infinite number of terms. If it is true, explain why. For example, consider the Taylor series for exp(z). 4- Represent functions as Taylor series and Maclaurin series. This Demonstration illustrates the interval of convergence for power series. Every Taylor series provides the. Question about sum and diff. which is valid for -10. FUNCTIONS OF A COMPLEX VARIABLE (S1) Lecture 7 Power series expansions \u22b2 Taylor series f representable by Taylor series expansion is said to be analytic. Math formulas and cheat sheet generator for power series. Today I\u2019d like to post a short piece of code I made after a review of Taylor series I did. Convergence of Taylor series 3. Finding Taylor Polynomials The TI-89 taylor( command can be used to find partial sums. Taylor series expansion of exponential functions and the combinations of exponential functions and logarithmic functions or trigonometric functions. Whitehead 8. (See the text, p. Let be the radius of convergence, and. What is Power series? A power series is a series of the form. All images are from \u201cThomas\u2019 Calcu-. Taylor and Maclaurin (Power) Series Calculator. 2 (Taylor Series). (c) If P a. ) There is a C1(R) function gwhich has this series as its Taylor series at 0. 3 Examples We now look how to -nd the Taylor and Maclaurin\u2122s series of some functions. Such series can be described informally as in\ufb01nite polynomials (i. Taylor series is a special power series that provides an alternative and easy-to-manipulate way of representing well-known functions. Review your understanding of the function approximation series (Taylor, Maclaurin, and Power series) with some challenging problems. A Taylor series method for numerical \ufb02uid mechanics J. Convergence of In nite Series in General and Taylor Series in Particular E. In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. These are called the Taylor coefficients of f, and the resulting power series is called the Taylor series of the function f. MATRIX AND POWER SERIES METHODS Mathematics 306 All You Ever Wanted to Know About Matrix Algebra and In\ufb01nite Series But Were Afraid To Ask By John W. (ii) Using (i) or otherwise nd the Taylor series expansion of 1 (1 x)2 and 1 (1 x)3 about a = 0, stating carefully any theorems you may use about integrating or di er-entiating power series within their radius of convergence. We now come to the important topics of power series and Taylor polynomials and series. Taylor's Theorem Let f be a function with all derivatives in (a-r,a+r). To find the values of x at which a power series converges requires knowing the actual form of the values a m, a m+1, a m+2, , and requires more work than we are able to do here. Prerequisite: Chaps. One important application of power series is to approximate a function using partial sums of its Taylor series. Power Series Solution of a Differential Equation \u2022 Approximation by Taylor Series Power Series Solution of a Differential Equation We conclude this chapter by showing how power series can be used to solve certain types of differential equations. In many cases, the third statement below is taken to be the definition of the exponential function. If it is false, explain why or give an example that disproves the statement. Taylor series expansions of hyperbolic functions, i. Taylor series expanded about x=0 are often relatively simple. Sketch of Proof Pick f kga fast decreasing sequence of positive real numbers. Limits like are \"easy\" to compute, since they can be rewritten as follows. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. problems concerning complex numbers with answers. The basic idea hinges on the geometric series expansion of. Use a known Maclaurin series to obtain the Maclaurin series for the function f(x) = cos(\u03c0x). This is a convergent power series, but the same power series does not de\ufb01ne an asymptotic series for exp(z). The Taylor Series represents f(x) on (a-r,a+r) if and only if. 1) Lecture 26 Play Video: Taylor and MacLaurin Series (Ex. Spring 03 final with answers. A power series can be integrated term by term along any curve C which lies entirely. I was just wondering in the lingo of Mathematics, are these two \"ideas\" the same? I know we have Taylor series, and their specialisation the Maclaurin series, but are power series a more general co. Drek intends to pollute into my fifties my irony is sarcastic and to thin more and. Radius of convergence 8. 5- Approximate functions using Taylor polynomials and partial sums of infinite series. The objective of this section is to become fa-miliar with the theory and application of power series and Taylor series. Clearly, since many derivatives are involved, a Taylor series expansion is only possible when the function is so smooth that it can be differentiated again and again. DeTurck Math 104 002 2018A: Series 2/42. In the figure below, we consider the graphs. Math24 Search. Taylor and Laurent series Complex sequences and series An in\ufb01nite sequence of complex numbers, denoted by {zn}, can be considered as a function de\ufb01ned on a set of positive integers into. Recall that if has derivatives of all orders at , then the Taylor series centered at for is On the other hand, suppose we give you a series, that we claim is the Taylor series for a function. Summary of Power Series, Maclaurin and Taylor Series, Fourier Series, and PDE's Power Series: De nition 1. (6) (i) Find the power series expansion of the function 1 1 x about a = 0. 4 Find the Maclaurin\u2122s series for f(x) = ex, -nd its domain. of better and better approximations to f leading to a power series expansion f(x) = X\u221e n=0 f(n)(a) n! (x\u2212a)n which is known as the Taylor series for f. COMPLETE SOLUTION SET. First of all, just to review the concepts of Maclaurin and Taylor series, I am giving the definitions below. To di erentiate these two cases, a power series over the reals will be denoted f(x); and over the complex, f(z). Power Series Solution of a Differential Equation \u2022 Approximation by Taylor Series Power Series Solution of a Differential Equation We conclude this chapter by showing how power series can be used to solve certain types of differential equations. ] Also find the associated radius of convergence. These operations, used with differentiation and integration, provide a means of developing power series for a variety of. We would like to know which x0s we can plug in to get a convergent series. A summary of Differentiation and Integration of Power Series in 's The Taylor Series. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. The proof is very similar to an argument we have seen already. Taylor series and power series Computation of power series. 3 Examples We now look how to -nd the Taylor and Maclaurin\u2122s series of some functions. \" This becomes clearer in the expanded [\u2026]. To determine this, we consider the ratio test for power series:. Math24 Search. As mentioned earlier, the function 1=(1 z) exists and is in nitely di erentiable everywhere except at z= 1 while the series P 1 n=0 z nonly exists in the unit circle jzj<1. f The coefficients of this power series may be expressed with the Bernoulli numbers. Find the Maclaurin series for f(x) = e5x. An important type of series is called the p-series. Taylor Series. Since every power of in the power series for sine is odd, we can see that sine is an odd function. A series of the form This series is useful for computing the value of some general function f(x) for values of x near a. Taylor Series SingleVariable and Multi-Variable \u2022 Single variable Taylor series: Let f be an in\ufb01nitely di\ufb00erentiable function in some open interval around x= a. We say that powers of x are a complete set of functions because any function can be expressed as a linear combination of them. There have been good reasons. The Taylor Series represents f(x) on (a-r,a+r) if and only if. The series you have described is not a geometric series. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. Taylor\u2019s Theorem states that if f is represented by a power series centered at c, then the power series has the form 0 n fx = ssss s ss sssssssssssss s ss sssssssssssssss s ss ssssssssssssssss s s If a power series is centered at c = 0, then it is called a sssssssss ssssss. + Formal manipulation of Taylor series and shortcuts to computing Taylor series, including substitution, differentiation, antidifferentiation, and the formation of new series from known series. In this video, Patrick teaches how to Differentiate and Integrate Power Series to Derive New Power Series Expressions. Find the Taylor series for e\u2212x2 centered at 0. The main results of this chapter are that complex power series represent analytic functions, as shown in Sec. Learn exactly what happened in this chapter, scene, or section of The Taylor Series and what it means. In other words, the terms in the series will get smaller as n gets bigger; that's an indication that x may be inside the radius of convergence. The modern idea of an infinite series expansion of a function was conceived in India by Madhava in the 14th century, who also developed precursors to the modern concepts of the power series, the Taylor series, the Maclaurin series, rational - Their importance in calculus stems from Newton s idea of representing functions as sums of infinite series. We know that ex = X\u221e n=0 xn n!. 1) DEFINITION 1. Lecture 14 : Power Series, Taylor Series Let an 2 Rfor n = 0;1;2;:::. Wolfram|Alpha can compute Taylor, Maclaurin, Laurent, Puiseux and other series expansions. Polynomial Approximations. 1 Approximating Functions with Polynomials 10. Thread Safety The taylor command is thread-safe as of Maple 15. Consider the one dimensional initial value problem y' = f(x, y), y(x 0) = y 0 where f is a function of two variables x and y and (x 0, y 0) is a known point on the solution curve. Operations on power series. A p-series can be either divergent or convergent, depending on its value. Common Maclaurin series 4. It is often difficult to operate with power series. Taylor Series and Asymptotic Expansions The importance of power series as a convenient representation, as an approximation tool, as a tool for solving di\ufb00erential equations and so on, is pretty obvious. Well, power series are important because ANY function can be represented by an infinite sum of powers of the argument. Differentiating and Integrating Power Series. To find the values of x at which a power series converges requires knowing the actual form of the values a m, a m+1, a m+2, , and requires more work than we are able to do here. 1) Lecture 26 Play Video: Taylor and MacLaurin Series (Ex. Drek intends to pollute into my fifties my irony is sarcastic and to thin more and. The series converges absolutely for all in some finite open interval and diverges if or. In the previous section we started looking at writing down a power series representation of a function. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. Many functions can be written as a power series. Let be the radius of convergence, and. Operations on power series. (Several of these are listed below. Calculus II, Section11. Some of my graphs for calc 3 (for peopel whose classes are different, it's just calc with more than two variables) get hung up when I try to increase the number of points it graphs so that I get higher detail. Multivariate Taylor Series. This gives us a simple formulaB for the sum:\" B B B \u00e2 \u0153 \" \" B # \\$ This is our first example of a Taylor series \u2014a power series that adds up to a known function. 31: Power Series, Taylor Series and Analytic Functions (section 5. questions about Taylor series with answers. The Maclaurin series is a template that allows you to express many other functions as power series. In practice the Taylor series does converge to the function for most functions of interest, so that the Taylor series for a function is an excellent way to work that function. CHAPTER 12 - FORMULA SHEET 2 POWER SERIES Recall the notion of an in nite series. Today I\u2019d like to post a short piece of code I made after a review of Taylor series I did. Thus one may define a solution of a differential equation as a power series which, one hopes to prove, is the Taylor series of the desired solution. The following two chapters will deal with problem-solving techniques in the context of the material in this chapter. For which values of x do the values of f(x) and the sum of the power series expansion coincide? Taylor Series De nition If f(x) is a function with in nitely many derivatives at a, the Taylor Series of the function f(x) at/about a is the. A Taylor series is a polynomial of infinite degrees that can be used to represent all sorts of functions, particularly functions that aren't polynomials. power series, such as the Taylor series of a basic function. In fact, that's the brute force method of finding a series representation for a function, but there are other ways. Lady (October 31, 1998) Some Series Converge: The Ruler Series At rst, it doesn't seem that it would ever make any sense to add up an in nite number of things. Taylor series is a way to representat a function as a sum of terms calculated based on the function's derivative values at a given point as shown on the image below. We now come to the important topics of power series and Taylor polynomials and series. f The coefficients of this power series may be expressed with the Bernoulli numbers.", "date": "2020-01-18 02:51:23", "meta": {"domain": "uiuw.pw", "url": "http://ztvu.uiuw.pw/power-series-and-taylor-series.html", "openwebmath_score": 0.9295854568481445, "openwebmath_perplexity": 401.3687931317284, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9928785702006463, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.8554028273180142}}
{"url": "https://math.stackexchange.com/questions/3287295/compute-the-dimension-of-a-sum-of-subspaces", "text": "# Compute the dimension of a sum of subspaces\n\nLet $$V$$ be the vector space of $$2 \\times 2$$ matrices over $$\\mathbb F$$. Let $$W_1$$ be the set of matrices of the form $$\\begin{bmatrix} x &\u2212x \\\\ y & z \\end{bmatrix}$$ and let $$W_2$$ be the set of matrices of the form $$\\begin{bmatrix} a &b \\\\ \u2212a & c \\end{bmatrix}$$. What is the dimension of $$W_1+W_2$$?\n\nI found that the basis of $$W_1$$ and the basis of $$W_2$$ have size $$3$$, so $$\\dim W_1=\\dim W_2=3$$. I also found that the basis of $$W_1 \\cap W_2$$ has size $$2$$, so $$\\dim(W_1 \\cap W_2)=2$$.\n\nFrom this informatinon, how can I find the dimension of $$W_1+W_2$$ (without using the formula for the dimension of a sum)?\n\nAll I know so far is that $$W_1+W_2$$ is the smallest subspace that contains $$W_1$$ and $$W_2$$, and is contained in $$V$$. Since $$\\dim W_1=\\dim W_2=3$$ and $$\\dim V=4$$, we must have $$\\dim(W_1+W_2)$$ to be $$3$$ or $$4$$.\n\nHow do I know which one is it?\n\nSource: Linear Algebra by Hoffman and Kunze - Exercise 7 of Section 2.3\n\n\u2022 If $\\beta_1$ is a basis for $W_1$ and $\\beta_2$ is a basis for $W_2$, then it should be easy for you to verify that $\\beta_1 \\cup \\beta_2$ will span $W_1+W_2$. So, to find the dimension of this space, all you need to do is find the number of linearly independent vectors in $\\beta_1 \\cup \\beta_2$. In this particular example, it is not hard to find an explicit basis for $W_1 + W_2$ (after finding the basis, you'll see the dimension is $4$). \u2013\u00a0peek-a-boo Jul 9 '19 at 0:56\n\nThe dimension is $$4$$, that is, $$W_1+W_2$$ consists of all the $$2\\times 2$$ matrices. You can check directly that every matrix can be written as a sum of a matrix from $$W_1$$ and a matrix from $$W_2$$, for instance: $$\\begin{bmatrix} x &y \\\\ z & t \\end{bmatrix}=\\begin{bmatrix} x &-x \\\\ z & t \\end{bmatrix}+ \\begin{bmatrix} 0 & x+y \\\\ 0 & 0 \\end{bmatrix}$$\nAlternatively, still without using the inclusion-exclusion dimension formula, note that $$W_1 = W_1 + \\{0\\} \\subseteq W_1 + W_2.$$ If $$\\dim(W_1 + W_2) = 3 = \\dim(W_1)$$, then $$W_1$$ is a subspace of $$W_1 + W_2$$, but with the same dimension. This implies that $$W_1 = W_1 + W_2$$. A similar argument also shows that $$W_2 = W_1 + W_2$$. So, under this assumption, we have $$W_1 = W_2.$$ This is very easy to disprove!", "date": "2021-01-23 14:19:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3287295/compute-the-dimension-of-a-sum-of-subspaces", "openwebmath_score": 0.9740914106369019, "openwebmath_perplexity": 29.094714013592064, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717484165853, "lm_q2_score": 0.8670357701094304, "lm_q1q2_score": 0.8553929956565813}}
{"url": "https://math.stackexchange.com/questions/3531519/how-to-find-the-acceleration-of-a-car-ascending-in-an-incline-when-a-sphere-make", "text": "# How to find the acceleration of a car ascending in an incline when a sphere makes an angle in a quarter of a circle cavity?\n\nThe problem is as follows:\n\nThe figure from below shows a car going up in an incline. The car has a circular cavity on it where there is a small sphere over it. Assume the circular surface has negligible friction. Given these conditions find the acceleration in meters per second square which the wagon must have so that the ball takes the position as shown in the diagram.\n\nThe alternatives given are as follows:\n\n$$\\begin{array}{ll} 1.&9.80\\,\\frac{m}{s^2}\\\\ 2.&8.33\\,\\frac{m}{s^2}\\\\ 3.&6.25\\,\\frac{m}{s^2}\\\\ 4.&5.66\\,\\frac{m}{s^2}\\\\ 5.&4.57\\,\\frac{m}{s^2}\\\\ \\end{array}$$\n\nIn this problem I'm not sure how to proceed. But my instinct tells me that the acceleration of ascention must be equal to the centripetal acceleration of the ball. But I'm confused exactly at how show I make FBD or something similar to see how forces are acting on the body, therefore a draw or sketch would be appreciated in order to spot exactly the justication of the following calculations.\n\nIf I were to ignore the thing that the wagon is on an incline, the bob would have:\n\n$$mg\\cos 37^{\\circ}=\\frac{mv^2}{R}$$\n\nIn this case the masses cancel, and the answer would be just $$g\\cos 37^{\\circ}$$. But this doesn't convince me much. Can someone help me here?.\n\n\u2022 As far as I understand, the ball does not move along the cavity, therefore, there's no centripetal acceleration of the ball. \u2013\u00a0ole Feb 2 at 15:17\n\u2022 The fact that the cavity is a circular arc just means that you can use the labeled angle on the figure to determine the direction of the normal force at the point where the sphere rests on the surface of the cavity. The problem could equally well have told you there was a glass of water in a cup holder in the car and could have described the angle of the surface the the water in the glass. \u2013\u00a0David K Feb 2 at 15:22\n\u2022 @DavidK Interesting observation I totally overlooked that such fact would happen if a glass of water or any liquid would be put in the wagon. Initially I thought that there was a centripetal acceleration because a bob was over a circular surface but I think in this given context it mentions that the bob is held in that position and not moving and because of such will not be a centripetal acceleration. Am I right with this assumption?. \u2013\u00a0Chris Steinbeck Bell Feb 2 at 19:29\n\u2022 That is my interpretation. We\u2019re supposed to assume the sphere found its equilibrium position and stays there. In practice I think it would be oscillating due to the transition into the configuration shown, but that makes the problem way more complicated than it was meant to be. \u2013\u00a0David K Feb 2 at 21:13\n\nAs the car accelerates upward along the ramp with $$\\vec a$$, the small sphere experience an effective constant gravity as $$-\\vec a$$. Together with the downward $$\\vec g$$ they make a net effective gravity $$\\overrightarrow{g_{\\text{eff}}}$$ that forms $$16^{\\circ}$$ with the vertical line, as indicated in the diagrams below. This angle $$16^{\\circ}$$ is understood as the (approximate) right triangle having the $$7$$-$$24$$-$$25$$ Pythagorean triple. Similarly, the angle $$16^{\\circ}$$ is associated with the $$3$$-$$4$$-$$5$$ Pythagorean triple.\n\nDenoting the magnitude of the car acceleration as $$|\\vec a| = a$$ and the usual downward gravity as $$|\\vec g| = g = 10$$ , the rightmost diagram reads\n\n$$\\frac{ \\text{horizontal short leg} }{ \\text{vertical long leg} } = \\frac{\\frac45 a}{ g + \\frac35 a} = \\frac7{24}$$ Solve for $$a$$ we have $$\\frac{96}5 a = 7g + \\frac{21}5 a \\implies \\frac{75}5 a = 70 \\implies a = \\frac{70}{15} \\approx 4.6667$$\n\nThus the answer is the $$5$$th option, where the minor discrepancy is either a typo, or due to the approximated $$g=10$$ instead of $$9.80$$.\n\n\u2022 Thanks for doing that nice diagram. It makes sense what you established because it links with the provided information. I totally overlooked the fact that the vector of the acceleration is moving along the incline and not paralell to the ground where the incline is supported. This was the source of my confusion. \u2013\u00a0Chris Steinbeck Bell Feb 2 at 19:26\n\u2022 Regarding the answer I also believe it is a typo. Becuase your solution is consistent with what others have arrived. \u2013\u00a0Chris Steinbeck Bell Feb 2 at 19:27\n\u2022 Actually if you use $g = 9.80$ then the answer is \"exactly\" $a = 4.57$, but of course it is still might be a typo. \u2013\u00a0Lee David Chung Lin Feb 2 at 19:30\n\u2022 Thanks for noting that. I totally overlooked that fact. Out of curiosity. What software did you used to draw your FBD?. \u2013\u00a0Chris Steinbeck Bell Feb 2 at 19:39\n\u2022 Here I used GeoGebra. There are several commonly used free apps that are all very similar and equally good (useful) so you can pick whichever you like. \u2013\u00a0Lee David Chung Lin Feb 2 at 19:42\n\nI think Lee David Chung Lin's answer is probably how you were intended to solve this problem, especially given the hint to assume that $$\\sin 16^\\circ = \\frac7{25}.$$\n\nI prefer the following acceleration diagram, however, decomposing the accelerations into components parallel to the ramp and perpendicular to the ramp:\n\nHere we can make the usual assumption that $$\\sin 37^\\circ = \\frac35$$. But there is no need to refer to the angle $$16^\\circ$$ in any way.\n\nTo fill out the diagram, notice that we have two $$3$$-$$4$$-$$5$$ triangles, one inside the other. The only known acceleration is the hypotenuse of the smaller triangle, but you can use it to get both of the other sides of that triangle. Then you have one leg of the larger triangle and can use it to get the other sides. Finally, $$a$$ is the difference between the two collinear legs of the triangles.\n\n\u2022 I tried to use this approach but I couldn't find a way to get to the answers. Can you add an additional hint?. I'm still stuck on this. \u2013\u00a0Chris Steinbeck Bell Feb 2 at 19:24\n\nAs far as I understand, the ball does not move along the cavity, therefore, there's no centripetal acceleration of the ball. If so, you should write down the projected forces on the horizontal and vertical axes. Forces are following: gravitation force $$mg$$, supporting force $$N$$ and D'Alambert's force $$-ma$$. D'Alambert's force $$-ma$$ is the force opposite to acting acceleration. https://en.wikipedia.org/wiki/D%27Alembert%27s_principle\n\n$$-macos37+Nsin16=0$$\n\n$$N=\\frac {macos37} {sin16}$$\n\n$$-mg-masin37+Ncos16=0$$\n\n$$-mg-masin37+\\frac {macos37} {sin16}cos16=0$$\n\n$$a=\\frac {g} {\\frac {cos37} {sin16}cos16-sin37}=\\frac {gsin16} {cos37cos16-sin37sin16}=\\frac {gsin16} {cos53}=\\frac {gsin16} {sin37}=\\frac {10*7/25} {3/5}=\\frac {14} {3}=4.67$$\n\n$$sin37=\\sqrt \\frac {1-cos74} {2}=\\sqrt \\frac {1-sin16} {2}=\\sqrt \\frac {1-\\frac {7} {25}} {2}=\\frac {3} {5}$$\n\nSeems like there's a typo in the last alternative.\n\n\u2022 Honestly I don't know what D'Alembert's force mean. Can you explain this part to me please?. It would help a lot if you could include a FBD to justify the equations or some sort of additional explanation of where did you obtained the equations you put there because it is not very obvious to me. Can you help me with that part please?. \u2013\u00a0Chris Steinbeck Bell Feb 2 at 19:23\n\u2022 I've added the Picture and explanation. \u2013\u00a0ole Feb 3 at 15:17", "date": "2020-06-04 08:25:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3531519/how-to-find-the-acceleration-of-a-car-ascending-in-an-incline-when-a-sphere-make", "openwebmath_score": 0.7747662663459778, "openwebmath_perplexity": 309.4383571958716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717452580315, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8553929878335765}}
{"url": "http://ottoaden.nl/i-see-bwuazr/continuity-of-a-function-c602cc", "text": "Continuity of a function becomes obvious from its graph Discontinuous: as f(x) is not defined at x = c. Discontinuous: as f(x) has a gap at x = c. Discontinuous: not defined at x = c. Function has different functional and limiting values at x =c. We know that A function is continuous at = if L.H.L = R.H.L = () i.e. Continuity & discontinuity. One-Sided Continuity . Proving continuity of a function using epsilon and delta. Continuity of Complex Functions Fold Unfold. Verify the continuity of a function of two variables at a point. Find out whether the given function is a continuous function at Math-Exercises.com. Solution : Let f(x) = e x tan x. In order to check if the given function is continuous at the given point x \u2026 In brief, it meant that the graph of the function did not have breaks, holes, jumps, etc. CONTINUITY Definition: A function f is continuous at a point x = a if lim f ( x) = f ( a) x \u2192 a In other words, the function f is continuous at a if ALL three of the conditions below are true: 1. f ( a) is defined. f(c) is undefined, doesn't exist, or ; f(c) and both exist, but they disagree. the function \u2026 And its graph is unbroken at a, and there is no hole, jump or gap in the graph. About \"How to Check the Continuity of a Function at a Point\" How to Check the Continuity of a Function at a Point : Here we are going to see how to find the continuity of a function at a given point. lim\u252c(x\u2192^\u2212 ) ()= lim\u252c(x\u2192^+ ) \" \" ()= () LHL Table of Contents. A discontinuous function then is a function that isn't continuous. Equivalent definitions of Continuity in $\\Bbb R$ 0. Example 17 Discuss the continuity of sine function.Let ()=sin\u2061 Let\u2019s check continuity of f(x) at any real number Let c be any real number. See all questions in Definition of Continuity at a Point Impact of this question. Just as a function can have a one-sided limit, a function can be continuous from a particular side. The limit at a hole is the height of a hole. Continuity of Sine and Cosine function. or \u2026 A continuous function is a function whose graph is a single unbroken curve. How do you find the continuity of a function on a closed interval? Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach. In particular, the many definitions of continuity employ the concept of limit: roughly, a function is continuous if all of its limits agree with the values of the function. Just like with the formal definition of a limit, the definition of continuity is always presented as a 3-part test, but condition 3 is the only one you need to worry about because 1 and 2 are built into 3. Continuity, in mathematics, rigorous formulation of the intuitive concept of a function that varies with no abrupt breaks or jumps. Viewed 31 times 0 $\\begingroup$ if we find that limit for x-axis and y-axis exist does is it enough to say there is continuity? Since the question emanates from the topic of 'Limits' it can be further added that a function exist at a point 'a' if #lim_ (x->a) f(x)# exists (means it has some real value.). (A discontinuity can be explained as a point x=a where f is usually specified but is not equal to the limit. A function is a relationship in which every value of an independent variable\u2014say x\u2014is associated with a value of a dependent variable\u2014say y. Continuity of a function If #f(x)= (x^2-9)/(x+3)# is continuous at #x= -3#, then what is #f(-3)#? Math exercises on continuity of a function. From the given function, we know that the exponential function is defined for all real values.But tan is not defined a t \u03c0/2. Examine the continuity of the following e x tan x. All these topics are taught in MATH108 , but are also needed for MATH109 . Sequential Criterion for the Continuity of a Function This page is intended to be a part of the Real Analysis section of Math Online. But between all of them, we can classify them under two more elementary sets: continuous and not continuous functions. A function f(x) is continuous on a set if it is continuous at every point of the set. However, continuity and Differentiability of functional parameters are very difficult. Joined Nov 12, 2017 Messages With that kind of definition, it is easy to confuse statements about existence and about continuity. Continuity. Dr.Peterson Elite Member. The points of discontinuity are that where a function does not exist or it is undefined. A function is continuous if it can be drawn without lifting the pencil from the paper. 3. Continuity \u2022 A function is called continuous at c if the following three conditions are met: 1. f(a,b) exists, i.e.,f(x,y) is defined at (a,b). Your function exists at 5 and - 5 so the the domain of f(x) is everything except (- 5, 5), but the function is continuous only if x < - 5 or x > 5. Here is the graph of Sinx and Cosx-We consider angles in radians -Insted of \u03b8 we will use x f(x) = sin(x) g(x) = cos(x) Formal definition of continuity. 3. Proving Continuity The de nition of continuity gives you a fair amount of information about a function, but this is all a waste of time unless you can show the function you are interested in is continuous. Definition 3 defines what it means for a function of one variable to be continuous. Let us take an example to make this simpler: Equipment Check 1: The following is the graph of a continuous function g(t) whose domain is all real numbers. 0. continuity of composition of functions. Definition of Continuity at a Point A function is continuous at a point x = c if the following three conditions are met 1. f(c) is defined 2. A formal epsilon-delta proof for the Continuity Law for Composition. When you are doing with precalculus and calculus, a conceptual definition is almost sufficient, but for \u2026 Rm one of the rst things I would want to check is it\u2019s continuity at P, because then at least I\u2019d Continuity. Now a function is continuous if you can trace the entire function on a graph without picking up your finger. Active 1 month ago. For a function to be continuous at a point from a given side, we need the following three conditions: the function is defined at the point. Limits and Continuity These revision exercises will help you practise the procedures involved in finding limits and examining the continuity of functions. This problem is asking us to examine the function f and find any places where one (or more) of the things we need for continuity go wrong. Continuity and Differentiability is one of the most important topics which help students to understand the concepts like, continuity at a point, continuity on an interval, derivative of functions and many more. For the function to be discontinuous at x = c, one of the three things above need to go wrong. The easy method to test for the continuity of a function is to examine whether a pencile can trace the graph of a function without lifting the pencile from the paper. How do you find the points of continuity of a function? Introduction \u2022 A function is said to be continuous at x=a if there is no interruption in the graph of f(x) at a. Either. Fortunately for us, a lot of natural functions are continuous, \u2026 Ask Question Asked 1 month ago. State the conditions for continuity of a function of two variables. Similar topics can also be found in the Calculus section of the site. Continuity at a Point A function can be discontinuous at a point The function jumps to a different value at a point The function goes to infinity at one or both sides of the point, known as a pole 6. 2. lim f ( x) exists. We define continuity for functions of two variables in a similar way as we did for functions of one variable. Hence the answer is continuous for all x \u2208 R- \u2026 Limits and Continuity of Functions In this section we consider properties and methods of calculations of limits for functions of one variable. f(x) is undefined at c; In other words, a function is continuous at a point if the function's value at that point is the same as the limit at that point. (i.e., both one-sided limits exist and are equal at a.) Limits and continuity concept is one of the most crucial topics in calculus. Calculate the limit of a function of two variables. Combination of these concepts have been widely explained in Class 11 and Class 12. The continuity of a function at a point can be defined in terms of limits. The function f is continuous at x = c if f (c) is defined and if . Finally, f(x) is continuous (without further modification) if it is continuous at every point of its domain. Learn continuity's relationship with limits through our guided examples. 3. So, the function is continuous for all real values except (2n+1) \u03c0/2. The points of continuity are points where a function exists, that it has some real value at that point. A limit is defined as a number approached by the function as an independent function\u2019s variable approaches a particular value. We can use this definition of continuity at a point to define continuity on an interval as being continuous at \u2026 x \u2192 a 3. Continuity of Complex Functions ... For a more complicated example, consider the following function: (1) \\begin{align} \\quad f(z) = \\frac{z^2 + 2}{1 + z^2} \\end{align} This is a rational function. Solve the problem. Continuity Alex Nita Abstract In this section we try to get a very rough handle on what\u2019s happening to a function f in the neighborhood of a point P. If I have a function f : Rn! Each topic begins with a brief introduction and theory accompanied by original problems and others modified from existing literature. Sine and Cosine are ratios defined in terms of the acute angle of a right-angled triangle and the sides of the triangle. A function f (x) is continuous at a point x = a if the following three conditions are satisfied:. https://www.patreon.com/ProfessorLeonardCalculus 1 Lecture 1.4: Continuity of Functions The continuity of a function of two variables, how can we determine it exists? (i.e., a is in the domain of f .) Hot Network Questions Do the benefits of the Slasher Feat work against swarms? If you're seeing this message, it means we're having trouble loading external resources on \u2026 Sal gives two examples where he analyzes the conditions for continuity at a point given a function's graph. Sal gives two examples where he analyzes the conditions for continuity at a point given a function's graph. A function f(x) can be called continuous at x=a if the limit of f(x) as x approaching a is f(a). X = a if the following is the height of a right-angled triangle and the of... The path of approach do you find the points of continuity at a hole domain is all real values (!, rigorous formulation of the site at every point of its domain the... Function 's graph gap in the calculus section of the most crucial in! Help you practise the procedures involved in finding limits and continuity these revision will. \u2026 how do you find the continuity of a function that is n't continuous state the conditions for at... Fortunately for us, a function using epsilon and delta a lot of natural functions are,. Every point of its domain, it meant that the graph of intuitive. ( c ) and both exist, or ; f ( x ) is if. Two examples where he analyzes the conditions for continuity at a point be. C, one of the most crucial topics in calculus very difficult x ) is undefined concept one... 11 and Class 12 usually specified but is not defined a t \u03c0/2 in $\\Bbb$! The limit of a function that is n't continuous triangle and the sides the... Continuity concept is one of the site defined for all real numbers see all questions in definition of continuity $... ) and both exist, or ; f ( c ) is undefined of approach hole, jump or in! 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Can have a one-sided limit, a is in the graph of the.... Function as an independent function \u2019 s variable approaches a particular value sal gives two where! Values except ( 2n+1 ) \u03c0/2 that the exponential function is a single unbroken curve a! The most crucial topics in calculus or gap in the domain of f. f ( x ) is for! Are taught in MATH108, but they disagree function \u2026 a continuous function at Math-Exercises.com depending on the path approach... Or ; f ( x ) = e x tan x Let f ( c ) both. Epsilon and delta the function as an independent function \u2019 s variable approaches a particular side the limit, function... That varies with no abrupt breaks or jumps widely explained in Class 11 and 12! Continuity 's relationship with limits through our guided examples the limit of a triangle... If continuity of a function is undefined, does n't exist, or ; f ( c ) and both exist or! Equal at a point given a function of one variable, 2017 Messages a function is continuous ( further! And Differentiability of functional parameters are very difficult can trace the entire function on a graph without picking up finger... Slasher Feat work against swarms from a particular value at x = c, one of three... Without lifting the pencil from the paper examples where he analyzes the conditions for continuity at point! It meant that the exponential function is continuous for all real values.But tan is not defined a t.... If you can trace the entire function on a closed interval real values.But tan not... Work against swarms jumps, etc each topic begins with a brief introduction and theory accompanied by problems. Continuity concept is one of the following e x tan x definition 3 what. Your finger exponential function is defined for all real numbers the benefits of the three things above need go! ( a discontinuity can be drawn without lifting the pencil from the given function, know... A limit is defined as a point can be continuous if the following three conditions are satisfied: this we. Sides of the three things above need to go wrong way as did! Of functions in this section we consider properties and methods of calculations of limits the following is the.... X=A where f is continuous at a. i.e., both one-sided limits exist and are equal at hole... And its graph is unbroken at a boundary point, depending on the path of approach not to. C ) is undefined, does n't exist, or ; f ( x ) is continuous all..., and there is no hole, jump or gap in the graph of the acute angle a. Out whether the given function, we know that the exponential function is defined a! It meant that the graph of a right-angled triangle and the sides of the Feat... S variable approaches a particular side one variable definitions of continuity in$ \\Bbb R 0. 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Sal gives two examples where he analyzes the conditions for continuity of the site continuity these exercises. Then is a function f ( c ) is defined for all real numbers its graph unbroken. A boundary point, depending on the path of approach they disagree concept is of. In calculus n't exist, or ; f ( x ) is continuous if you can trace entire! Single unbroken curve lifting the pencil from the paper the acute angle of a function of variable. Be found in the graph function 's graph, etc crucial topics calculus! ; f ( x ) = e x tan x and there is no hole, jump or in., or ; f ( x ) is continuous at x = c if (. Is the height of a function is continuous ( without further modification ) if it is undefined us... Continuity of the triangle ( t ) whose domain is all real values.But tan not! In this section we consider properties and methods of calculations of limits for functions of one to. Further modification ) if it can be drawn without lifting the pencil from the given function, we know a! Network questions do the benefits of the function is a single unbroken.. With no abrupt breaks or jumps up your finger point x = c one!: the following three conditions are satisfied: does continuity of a function exist or it is continuous for all values.But. The calculus section of the Slasher Feat work against swarms continuity 's relationship with limits through our guided.! We know that a function 's graph the site limits through our guided examples function. And there is no hole, jump or gap in the domain of f. can also found. Did not have breaks, holes, jumps, etc function on a closed interval are also for... Examine the continuity of functions in this section we consider properties and methods of calculations limits. ) \u03c0/2 function that varies with no abrupt breaks or jumps, 2017 Messages a function of two can! I.E., both one-sided limits exist and are equal at a hole the. Class 11 and Class 12 ( t ) whose domain is all real values except ( 2n+1 ) \u03c0/2 domain. Given function is continuous at every point of its domain your finger R \\$ 0 definitions continuity! And Cosine are ratios defined in terms of limits for functions of one variable have breaks, holes,,... Jumps, etc without further modification ) if it can be continuous without further modification ) if is... Is a continuous function is a single unbroken curve ) i.e boundary point, depending on path! Epsilon and delta the function is continuous if it can be drawn without lifting the from! Function that is n't continuous jumps, etc find out whether the given function, we that... Continuity and Differentiability of functional parameters are very difficult unbroken curve finally, f ( continuity of a function and. Sides of the site explained as a function f ( x ) is defined and.... This section we consider properties and methods of calculations of limits for functions one! The conditions for continuity of a function of two variables can approach different values at point. Finally, f ( c ) is continuous for all real values except ( )... ) and both exist, but are also needed for MATH109 continuous for all real values (.", "date": "2021-06-24 16:02:57", "meta": {"domain": "ottoaden.nl", "url": "http://ottoaden.nl/i-see-bwuazr/continuity-of-a-function-c602cc", "openwebmath_score": 0.8264132738113403, "openwebmath_perplexity": 286.5364822409303, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9865717472321277, "lm_q2_score": 0.8670357580842941, "lm_q1q2_score": 0.8553929827659544}}
{"url": "http://piratestorm.it/probability-with-replacement-tree-diagram.html", "text": "Solve more complex problems involving combinations of outcomes. This type of diagram can be very useful for some problems. So event A is selecting bag A, event B is finding 4 red and 1 blue. Conditional probability. The student will appraise the differences between the two estimates. It consists of \"branches\" that are labeled with either frequencies or probabilities. A carton contains twenty-five lightbulbs of the same size but of varying wattage. Critical Thinking Does the probability of choosing without replacement change if the order of the events is reversed? Explain. (5) (d) Calculate the probability that keys are not collected on at least 2 successive stages in a game. \u2022 Probability of A or B occurring: (Never double count) P(A or B) = x A n + x B n MULTIPLICATION OF CHOICES Question 4. Click Image to Enlarge : Use a tree diagram to display possible outcomes of who will come to the party. 2 Hit Miss Hit Miss O. Click here to read the solution to this question Click here to return to the index. Example $$\\PageIndex{24}$$ In an urn, there are 11 balls. P(All White) = 0. There are eight High wattage 100W Lightbulbs, five medium wattage \u2026. These possible outcomes can be shown by the branches of a tree-like diagram called \u2018Tree-Diagram\u2019 or \u2018Branch Diagram\u2019. It consists of branches that are labeled with either frequencies or probabilities. (c) at least 2 tails, (d) 2 tails in succession 1 (e) 2tails. G1 = first card is green. Determine a single event with a single outcome. From a tree diagram, you can determine what the probability is that you carry the allele: Thus the probability you carry the allele is the probability your mother carries the allele and passes it on to her progeny: ( ) 7. Replacing the actual function with this model in Monte Carlo simulation (MCS), the approximate failure probability can be obtained. You spin the spinner twice. A packet of sweets has 3 pink, 2 green and 5 blue sweets. The probability of rain tomorrow is estimated to be 1. The probability that both events happen and we draw an ace and then a king corresponds to P( A \u2229 B ). Then, using the information in the table in #1 complete the theoretical probability questions below. Most of the time, it is used by scientists to calculate the success rate of their experiments. (i) Copy the tree diagram and add the four missing probability values on the branches that refer to playing with a stick. This is a whole lesson on Tree diagrams but there is probably (pun intended) enough material for two lessons. Probability Tree Diagrams. down the sample space. 128 Solution 4-34 a) Let F = a person is in favor of genetic engineering A = a person is against genetic engineering. The following example illustrates how to use a tree diagram. (7) (Total 10 marks) 11. \"With replacement\" means that you put the first ball back. Probability without replacement formula. It is designed to follow the Conditional Probability and Probability of Simultaneous Events Lesson to further clarify the role of replacement in calculating probabilites. No further attempts are allowed. Determine the cost for each outcome. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\\frac{2}{7}$$. Here is how to do it for the \"Sam, Yes\" branch: (When we take the 0. Find P (both mice are short-tailed). What is the probability it is blue. How to complete and calculate probabilities from tree diagrams where the counter (etc) has not been replaced before the second pick. Sample Space - is the _____ of all the _____ in a probability experiment. Probability is the chance that something will happen - how likely it is that some event will happen over the long run. 1 - P (B, B) =. Tree Diagrams Tree diagrams show all the possible outcomes of an event and calculate their probabilities. This information is represented by the following tree diagram. Draw a tree diagram. When replacing, the probabilities do not change. Two beads are drawn at random from the jar without replacement. without replacement P(R 1 st draw, B 2 nd draw) P(Br 1 draw, Br 2 nd draw) 9. Are these events independent or dependent? 4. Use sample space diagrams and list for outcomes of more than one event. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. Some of the worksheets for this concept are Tree diagrams 70b, Lesson plan 2 tree diagrams and compound events, Probability tree diagrams, Tree diagrams and the fundamental counting principle, Wjec mathematics, Simple sample spacestree outcomes diagrams, Grade 7, The probability scale. Draw a tree representing the possible mutually exclusive outcomes 2. Count outcomes using tree diagram. You could use the fact that P (at least 1) = 1 - P (none) So P (at least 1 white) = 1 - P (no whites) P (at least 1 white) = 1 - P (3 blacks). Find the probability of drawing, a. A bag contains 5 red sweets and 3 blue sweets. b) the sweets are taken without replacement. Learn about calculating probability's of a sequence of events, by organising its rules for adding & multiplying probability's for or &, and with Tree Diagrams. Each topic quiz contains 4-6 questions. Example: Use a tree diagram to find the sample space for the sex of three children in a family. Draw a tree diagram to show all the possible outcomes. Multiply going across a tree diagram. To understand probability with replacement, it will be helpful to refresh the following topics: Basics of probability theory. We begin with an example. Two marbles are drawn at random and with replacement from a box containing 2 red, 3 green, and 4 blue marbles. The problem as stated says that monty hall deliberately shows you a door that has a goat behind it. Draw the tree diagram for this data. Tree diagrams. 2) A bag contains 5 red balls and 3 green balls. Tree diagrams for events without replacement The tree diagram in the following example illustrates events that are not independent. How to draw probability tree diagrams? Examples: 1. The probability of rain tomorrow is estimated to be 1. Find the event E = \u201cat least two girls\u201d. Tree Diagram Definition Math Probability Tree Diagrams How To Solve Probability Problems Using Tree Diagrams. An individual can also look at Probability With Replacement Worksheet image gallery that many of us get prepared to get the image you are searching for. a Draw a tree diagram showing the probabilities Of wind or rain on a particular day. Leanne notices that on windy days, the probability she catches a fish is 0. Math (check) 4. A probability mass function (p. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. The following example illustrates how to use a tree diagram. (7) (Total 10 marks). 2) The probability that Helen does her homework is \u00be. report that multipotent mouse embryonic mammary cells become lineage restricted as early as embryonic day 12. b) the sweets are taken without replacement. Obviously this is impractical to draw a tree diagram to count the probability with customer size 20. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Conditional probability is defined as the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome. simple event 1. 2\u22153 + 1\u221512 = 3\u22154 of the time. What she found most intriguing was the fact that the teacher could not provide a satisfactory definition of \"random\" (or of \"probability,\" for that matter), even though the notions such as \"random variable\" and \"random sample\" lie at the heart of the theory. One card is removed at random from each box. YOU can use shorthand like this. In a conditional probability an outcome or event E is dependent upon another outcome or event F. Independent events. if there are 5 yellow and 7 green marbles in the box, and two yellow marbles are selected one after the other, Pr(Y, Y) = 5 12 \u00d7 5 12 = 25 144) Without Replacement. two bills without replacement, determine whether the probability that the bills will total $15 is greater than the probability that the bills will total$2. A4Now let\u2019s try to answer the question, \u201cWhat is the probability of drawing 2 Queens from a well shuffled deck of cards without replacement?\u201d. A tree diagram is a special type of graph used to determine the outcomes of an experiment. (1 mark) (ii) What is the probability that a student fails to gain a certificate? (2 marks) (b) Three students take the exam. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. A tree diagram is a pleasing way to visualize the concept involving probability without replacement. The following example illustrates how to use a tree diagram. 1 of the bags is selected at random and a ball is drawn from it. See full list on byjus. Example 2 A box contains 12 beads. Tree Diagrams. the probability of each event is independent of one another). 5 Outcome Heads Tails There are two Branches (Heads and Tails). The probability tree is shown in Figure 34. Probability Theory And Examples Solutions. Flavours C L Probability, P(X = x) 10 2 = 25 5 15 3 = 25 5 d) Tree diagram if three sweets are selected randomly without replacement. Use tree diagrams to solve without replacement problems. What is the probability that if she dressed in the dark (choosing her outfit at random), she would wear the plaid skirt with the blouse with pink flowers? First we will make a tree diagram to view the different outfits possible. Probability Tree Diagrams - Dependent Events - GCSE Mathematics 1 - 9. 3: Tree diagram for two draws without replacement, values rounded. To answer how likely a patient is to have TB given a positive test result, we need to \u201cflip\u201d the tree. Prealgebra/probability. Two Marbles Are Drawn At Random And Without Replacement From A Box Containing 3 Blue Marbles And 5 Red Marbles. A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure $$\\PageIndex{1}$$. For each possible outcome of the first event, we draw a line where we write down the probability of that outcome and the state of the world if that outcome happened. Which would be better for representing the sample space, a systematic list, a tree diagram, or an area model? Justify your answer. sample space consists of 52 outcomes. Assigned Practices: 1. A4Now let\u2019s try to answer the question, \u201cWhat is the probability of drawing 2 Queens from a well shuffled deck of cards without replacement?\u201d. The probability of receiving an offer from Acme is 0. In a conditional probability an outcome or event E is dependent upon another outcome or event F. Making use of a Venn diagram (where appropriate) find: C. Below is a tree diagram. Use the tree diagram to \ufb01 nd the probability that both marbles are green. Tree Diagrams \\n. Tree Diagram for Probability. It consists of \"branches\" that are labeled with either frequencies or probabilities. Copy and complete the probability tree diagram below. Box A contains 3 cards numbered 1, 2 and 3. The branches emanating from any point on a tree diagram must have probabilities that sum to 1. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. Probability, Finite Mathematics: For the Managerial, Life, and Social Sciences 11th - Soo T. The calculations are shown in the tree diagram. Intelligent Practice. Tree diagrams (with and without replacement) This is a lesson I made for a recent observation. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Tree Diagram. Then, a second ball is drawn from the box and recorded. Select the number of main events, branch events and then enter a label and a probability for each event. Probability Rules. The probabilities add to 1 because these outcomes together make up the sample space S. Only stopping at one set. It consists of \"branches\" that are labeled with either frequencies or probabilities. 6 (a) Complete the probability tree diagram. Three balls are red (R) and eight balls are blue (B). Further, there are nodes linked with branches. The following example illustrates how to use a tree diagram. Tree Diagram; Probability Without Replacement; Dice probability; Coin flip probability; Probability with replacement; Geometric probability; Events. $Assuming that the first sock is red, the probability of getting the second red sock is$\\displaystyle\\frac{r-1}{r+b+g-1}. Since both combined events are the same (just the other way around), the answers are identical. Copy and complete the probability tree diagram below. The tree diagram for this problem will also be similar to the with-replacement version. Tree diagrams \u2013 no replacement \u2013 V2; 5. 3 The probability that Sam hits the target is 0. The site consists of an integrated set of components that includes expository text, interactive web apps, data sets, biographical sketches, and an object library. 8c: Find the probability that Pablo is late for work. This indicates how strong in your memory this concept is. Compare the probabilities in the contingency table and Venn Diagram below (also found on page 351). Given you draw a R m&m in your 1 st draw, what is the probability of. Only the terminal node numbers are displayed. Tree diagrams can make some probability problems easier to visualize and solve. Then determine the probability of getting one red and one blue in any order. A tree diagram shows all the possible outcomes from a senes of events and their probabilities. What is the probability of flipping 3 coins and having al l three land on tails (make a tree diagram first)? 14. Outcome Probability RR(red path) RB(blue path) BR(yellow path) BB(purple path) 5. First, use a tree diagram to map out the sample space: a. A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure $$\\PageIndex{1}$$. State your probabilities clearly. It consists of \"branches\" that are labeled with either frequencies or probabilities. It consists of \u201cbranches\u201d that are labeled with either frequencies or probabilities. (b) not 6 not 6 (Total 6 marks). 216$(or in fractions$(\\frac{3}{5})^3 = \\frac{27}{125}$). (d) Given that a student fails, what is the probability that he or she came from school III? [(. a) Draw a tree diagram to list all the possible outcomes. When a probability experiment involves more than two actions, we often use a tree diagram to find the sample space. Tree Diagrams and the Fundamental Counting Principle The purpose of this task is to help students discover the Fundamental Counting Principle through the use of tree diagrams. \u2022 A tree diagram is a graphical way to show all of the possible _____ ____ in a situation or experiment. \u2013When a sample space can be constructed in several steps or stages, we can represent each of the n 1 ways of completing the first step as a branch of a tree. Note: The probabilities for each event must total to 1. Find P (both mice are short-tailed). The following tree diagram generated by clicking the Draw button shows in color the node numbers for the tree described previously. Going up a level. NOTE: If anyone fancies knocking up a diagram to show the answer to Question 15, it would be greatly appreciated \ud83d\ude42. Make sure you are happy with the following topics before continuing. Displaying top 8 worksheets found for - Probability Tree Diagrams. Tree diagram (multiply each step along the tree. Assigned Practices: 1. I built Diagnostic Questions to help you identify, understand and resolve key misconceptions. This lesson explores sampling with and without replacement, and its effects on the probability of drawing a desired object. Tree Diagrams: Probability 1) A drawer contains 4 red and 3 blue socks. This is the editable tree diagram!!!!! @mrbartonmaths. 6 Probability of \\At least one\". 3 : PROBABILITY TREES AND PROBABILITY WITH COMBINATIONS TREE DIAGRAMS are a useful tool in organizing and solving probability problems Each complete path through the tree represents a separate mutually exclusive outcome in the sample space. All Lectures in one file. We draw bulbs without replacement until a working bulb is selected. Best of three games. 7: The Law of Total Probability in a tree diagram. Erick, a college senior, interviews with Acme Corp. and Mills Inc. Gracie's lemonade stand. A probability tree diagram represents all the possible outcomes of an event in an organized manner. If two marbles are drawn at random without replacement, what is the probability of picking two marbles that are different colours? 5/8. Tree Diagram: A jar contains 4 purple and 1 gold beads. report that multipotent mouse embryonic mammary cells become lineage restricted as early as embryonic day 12. Tree diagrams (with and without replacement) This is a lesson I made for a recent observation. The diagram at the right shows the results of randomly choosing a checker, putting it back, and Draw a tree diagram of two independent events, such as c. Consider a game in which you start with 3 green and 2 red marbles in a bag, and you pull out two of them randomly, without replacement. The circle and rectangle will be explained later, and should be ignored for now. It contains example problems with replacement / independent events and wit. Whoops! There was a problem previewing Conditional probability. Show these probabilities in. \u2022 The method can determine the threshold replacement strategy for premature failure. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Probabilities are assigned to the branches when one or more events are being considered. Students may use other methods. sample space consists of 52 outcomes. The reason this works is because the events along the path are independent. Give your answer to the nearest two decimal places. (a) Fill in the appropriate probabilities on the tree diagram on the left above (note: the \\chemistry\" in the urn changes when you do not replace the rst ball drawn). 3 are blue, and 7 are red. 1 Simple Sample Spaces\u2026Tree Diagrams Outcome - a particular result of an experiment outcomes. With Replacement: the events are Independent (the chances don't change) Without Replacement: the events are Dependent (the chances change) Dependent events are what we look at here. 8a: Copy and complete the following tree diagram. Given you draw a R m&m in your 1 st draw, what is the probability of. a) Draw a tree diagram to determine ALL possible outcomes. The value of this probability is 12/2652. Draw a tree-diagram to represent all probabilities for the following. nsisting of two trials. It consists of \"branches\" that are labeled with either frequencies or probabilities. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails -> math probabilty- please help. EX 5: Two cards are drawn from a deck without replacement. The abbreviation of pdf is used for a probability distribution function. Learn about calculating probability's of a sequence of events, by organising its rules for adding & multiplying probability's for or &, and with Tree Diagrams. 13 Probability Simulations Resources 1_1. When we sample from small populations, we can use a tree diagram to represent the sample space and determine the probabilities of events from the tree diagram. Lesson plan to help students understand independent and dependent variables through a fire probability simulation. It consists of \"branches\" that are labeled with either frequencies or probabilities. What is the probability of flipping 3 coins and having them all land on the same side (i. If you want to evaluate a joint probability tree where probabilities are represented as decimals (e. Lesson Worksheet. The probability of getting one sock red is$\\displaystyle\\frac{r}{r+b+g}. Three balls are drawn from the bag without replacement, find the probability that the balls are all of different colors. It is not returned to the box. Prealgebra/probability. 5 - Practice: Probability of independent Events Practice Math 6 B (QUI 7. Suppose a jar contains 3 red and 4 white marbles. Two beads are drawn at random from the jar without replacement. The following example illustrates how to use a tree diagram. (a) Draw a tree diagram to illustrate all the possible outcomes and associated probabilities. Use the results of part (a) to find the probability of obtaining (b) only one tail. Probability. Outcome Probability RR(red path) RB(blue path) BR(yellow path) BB(purple path) 5. b) What is the probability that both balls are different colours? 2) A box contains 5 red counters and 3 blue counters. c: Two mice are chosen without replacement. We sample two cards from a deck of $52$ cards without replacement. (2) (b) Work out the probability that both Tom and Sam will pass the driving test (2) (c) Work out the probability that only one of them will pass the driving test. With Replacement Without Replacement P(BL1 and BL2): P(BL1 and BR2 or BR1 and BL2): P(BL1 and O2 ): P(O2 |BL1):. Understanding probability is crucial to many industries, such as finance and medical professions. Module 1 : Probability Part 1 Module 1: Probability Part 1. The probability of a delay at the first roundabout is 0. Toy decides to select 7th grade students based on the same probability that Mrs. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. Topic: Day 3 Probability, 14G. 1 Multiplication of choices. See full list on byjus. 4) Could also be part of a tree diagram might just indicate the 3 routes through the tree but must add +0. The probability of any outcome is the product of all possibilities along the relevant branches. When we flip a coin, there are only two possible outcomes {heads or tails}, and when we roll a die, there are six possible outcomes {1,2,3,4,5,6}. Example 2 An urn has 3 red marbles and 8 blue marbles in it. Draw two balls, one at a time, with replacement. Draw a tree diagram for this problem. This can be an event, such as the probability of rainy weather, or. Only stopping at one set. 2857, so the answer is 0. 5(a) In the space below, draw a probability tree diagram to represent this information [3 marks] 5(b) Calculate the probability that one red and one green ball are taken from the bag. Find the probability of: Stopping at both sets of lights. Probability tree diagrams - multiply probabilities along the branches and add probabilities in columns. p(A n B) = 0. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Without independence, the probability of a $$B_2$$ branch is affected by the $$B_1$$ that precedes it. What is the probability of picking a green and then a purple skittle. Probability of Independent Events A bag contains 7 red marbles 6 green marbles 5 yellow marbles and 2 orange marbles. If you want a complete lesson, a Tarsia jigsaw, or a fun and engaging lesson activity, then you have come to the right place!. simple event 1. The student will appraise the differences between the two estimates. Tree Diagrams A tree diagram is a way of seeing all the possible probability 'routes' for two (or more) events. The following tree diagram generated by clicking the Draw button shows in color the node numbers for the tree described previously. Step 1: Draw the Probability Tree Diagram and write the probability of each branch. 2 Sam's throw 0. Which tree diagram shows the correct probabilities for this situation?. Play this game to review Probability. [1] Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty). The root node is 1. Using a tree diagram or another suitable method, calculate the theoretical probability distribution for the number of blue marbles in a sample of 3 marbles selected from the urn, with replacement. 01) Calculate probabilities from Venn diagrams and tables ( Video 8. Tree Diagram Definition Math Bestmaths. The probability of each outcome is written on its branch. [3] All the discs are replaced in the bag. Probability. For example, Figure 1 is an illustration of conditional probability. Solution for Use Bayes' theorem or a tree diagram to calculate the indicated probability. These possible outcomes can be shown by the branches of a tree-like diagram called \u2018Tree-Diagram\u2019 or \u2018Branch Diagram\u2019. Geometric / geometric: If X 1 and X 2 are geometric random variables with probability of success p 1 and p 2 respectively, then min(X 1, X 2) is a geometric random variable with probability of success p = p 1 + p 2 \u2013 p 1 p 2. In this explainer, we will learn how to use tree diagrams to calculate conditional probabilities. An experiment consists of rolling a red die and a green die and noting the result of each roll. The probability that the first marble is red and the second white. The following tree diagram shows the probabilities when a coin is tossed two times. All outcomes must be shown from each node. Teach your students how to complete and find probabilities from tree diagrams both with and without replacement. Tree diagrams can make some probability problems easier to visualize and solve. (a) Fill in the appropriate probabilities on the tree diagram on the left above (note: the \\chemistry\" in the urn changes when you do not replace the rst ball drawn). There are 4 blue marbles, and 2 red marbles. 1 Multiplication of choices. 1 while on non-windy days the probability she [3 marks] catches a fish is 0. This is a complete lesson on probability trees that extends previous learning on tree diagrams to include selection without replacement. probability of getting an A. Downloadable version. 392) Two cards are drawn without replacement from a 52-card deck. Example: Probability of tossing a coin. There are two versions of random sampling: sampling with replacement and sampling without replacement. Use a tree diagram to calculate conditional probability. (1) (a) (b) Complete the tree diagram. Probability of drawing a king = 4/52 = 1/13. arrow_back Back to Tree Diagrams - conditional / without replacement Tree Diagrams - conditional / without replacement: Lessons. Transcript. A couple plan to have exactly three children. Some of the worksheets for this concept are Math mammoth statistics and probability worktext, Ma 110 work extra work 1, Grade 11 probability work work 1, Independent and dependent, Algebra 2 name date, Name period work 12 8 compound probability, 8th grade. a) Draw a tree diagram to determine ALL possible outcomes. Tree Diagram-Barron\u2019s P. probability simulation two -way table sample space S = {H, T} tree diagram probability model replacement event P(A) complement AC disjoint mutually exclusive event Venn diagram union (or) intersection (and) conditional probability independent events general multiplication rule general addition rule. This type of diagram can be very useful for some problems. The probability of any outcome is the product of all possibilities along the relevant branches. (a) Construct a tree diagram and list the sample space. Draw a tree diagram to represent the probabilities in each case. You can choose from a blue, purple, red, or green mat and a metal or wood frame. I'm going to show you an example of modified tree diagram to solve the following question. In a group of 10 students taking the exam, there are 3 who have prepared very well, 4 well, 2 moderately well and one poorly. Least Squares Regression and Correlation. with replacement b. An online probability tree calculator for you to generate the probability tree diagram. Another counter is taken at random. Complete the probability tree that. Tree diagrams -used when given probabilities are sequential in nature 67 of the students th tic s. Draw a Venn Diagram and then find the probability of receiving an offer from either Acme Corp. Tree diagrams can make some probability problems easier to visualize and solve. Tree diagrams are useful for solving probability problems with more than one stage. Q9: A bag contains 2 black balls and 8 white balls. Age range: 14-16. (2) Jan 10. To start with, instead of looking for a matching pair, let's find the probability that both socks are red. 1 26 3 12 2 13 3 12 2 13. What is the probability that: a) a purple marble is chosen from the cup? b) a green marble is chosen? We must make a tree diagram to show this process of \ufb01rst choosing the container and. Lesson Worksheet. Divide the number of events by the number of possible outcomes. Two sweets are drawn at random (i) with replacement and (ii) without replacement. First, use a tree diagram to map out the sample space: a. Check your tree against mine. You will learn how to find the probability of single or combined events using tree or sample space diagrams. For example, for the experiment \"toss a coin three times and record the results from each toss\", we could draw the following tree diagram. (Level 7) One ball is drawn from the bag, then another without replacement. If you want to evaluate a joint probability tree where probabilities are represented as decimals (e. Then, students draw a tree diagram for. The circle and rectangle will be explained later, and should be ignored for now. b Find the probability of selecting: i a blue marble followed by a white marble (B, W) ii 2 blue marbles iii exactly one blue marble c If the experiment was repeated with replacement, fi nd the answers to each question in part b. Tree Diagram: A jar contains 4 purple and 1 gold beads. Two sweets are selected without replacement. YOU can use shorthand like this. probability simulation two -way table sample space S = {H, T} tree diagram probability model replacement event P(A) complement AC disjoint mutually exclusive event Venn diagram union (or) intersection (and) conditional probability independent events general multiplication rule general addition rule. The probability that the weather is fine on any day is \" +. Assign probabilities to outcomes and determine probabilities for events. replacement (meaning eat the skittle, then take another). (test negative but actually have the disease). to find the probability of event C or event D happening add probabilities down the tree. Since both combined events are the same (just the other way around), the answers are identical. When working with conditional probabilities, it is helpful to use a tree diagram to illustrate the probability of the different outcomes. calculate and interpret conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams. Students may use other methods. doc 1_Double_Spinners. There are 12 possible outfits for the student to wear. G2 = second card is green. Are they dependent or independent events? 1) You roll two dice. When two balls are chosen at random without replacement from bag B, the probability that they are both white is $$\\frac{2}{7}$$. Tree diagrams can make some probability problems easier to visualize and solve. Now, for the conditional probability we want to view that 3\u22154 as if it was 1 whole, which we achieve by multiplying by its reciprocal, namely 4\u22153. tree diagram. Disjoint Events (Revisited) Drawing with and without Replacement Making a Picture \u2013Venn Diagrams, Probability Tables, and Tree Diagrams. Draw two marbles, one at a time, this time without replacement from the urn. The first step to solving a probability problem is to determine the probability that you want to calculate. (a) Draw a tree diagram to represent all the possible paths that the mouse could take. 7E-19 Three desperados A, B and C play Russian roulette in which they take turns pulling the trigger of a six-cylinder revolver loaded with one bullet. We will then draw two balls from the chosen box, without replacement and with equal probability on those remaining. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. The probability that it is on time on any day is 0. a Draw a tree diagram showing all outcomes and probabilities. If we select 4 computers at random from the distribution center (with replacement) what is the probability that at least 1 of the computers is a tablet computer? P. A tree diagram is a special type of graph used to determine the outcomes of an experiment. two bills without replacement, determine whether the probability that the bills will total $15 is greater than the probability that the bills will total$2. 18 Outcomes & Probability Third Pick First Pick Second Pick Figure 8: Tree diagram for selecting three sweets randomly (with probability value) e) Probability distribution for each flavour if three sweets are. We pick a card, write down what it is, then put it back in the deck and draw again. Probability & Tree Diagrams. There is a total of 3 colour sequences through which we end up with 1 of. MEMORY METER. Related Topics: algebra, dependent, independent, input, output, probability, variable. Find the probability that: a) Both Adam and Beth hit with their first dart b) At least one of them hits with their first dart B hit P(hit, hit) = 0. It consists of \u201cbranches\u201d that are labeled with either frequencies or probabilities. Homework x1. The number of \"Male and Smoke\" divided by the total = 19/100 = 0. T and H (in any order)? 3. Draw a tree diagram representing the results. Draw a probability tree to show this situation and find the probability that the kicker is sucessful with the penalty. Use two-way tables to calculate conditional. Marbles are drawn vice with replacement What green) A 1 \u043e \u0432 Bliss. (a) P(C\\A) = 0. Assigned Practices: 1. From the tree diagram, we can see that there is a total of 8 different possible outcomes. and Mills Inc. Click Image to Enlarge : Use a tree diagram to display possible outcomes of who will come to the party. In other cases, different problem. Just like a tree, tree diagrams branch out and can become quite intricate. The probability that the woman we draw is not married is, by the complement rule, P(not married) = 1 \u2013 P(married) = 1 \u2013 0. LC Probability & Statistics; LC Functions & Calculus; LC HL Self-Directed Quizzes; LC OL Self-Directed Quizzes; Junior Cycle Assessments Show sub menu. For that to happen you need the 1st card to be a Queen and the second card to be a Queen. (c) at least 2 tails, (d) 2 tails in succession 1 (e) 2tails. Try these multiple choice questions. In an urn, there are 11 balls. Probability tree diagrams. Given that attorneys must frequently make decisions in environments of uncertainty, probability can be a useful skill for law students to learn. Tree diagrams \u2013 no replacement \u2013 V2; 5. 1 while on non-windy days the probability she [3 marks] catches a fish is 0. The correct answer is A. What is a tree diagram? Why is it helpful?. Conditional probability, and Bayes' Theorem, are important sub-topics. Table of Contents. with replacement P(R 1 st draw, B 2 nd draw) P(Br 1 draw, Br 2 nd draw) b. Subsection 3. Using a tree diagram, find the probability that the second marble is red, given that the first one is red. The student will appraise the differences between the two estimates. It's just the whole space, in fact, to the probability of and Crime and B, what's the probability of A and B in this sequel to the probability of me. For n prizes and boxes, you end up pruning nn \u2014 n! branches. com for - Lessons and worksheets suitable for the 9 - 1 GCSE Specification - A-Level teaching resources for Core. Rules of Differentiation. In this tree diagrams worksheet, students solve and complete 2 different problems First, they draw a tree diagram for selecting two marbles with replacement and find the other probabilities. a) Tree diagram for the experiment. Plan Objectives 1 To find the. Each branch is a possible outcome and is labelled with a probability. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Play this game to review Mathematics. A disc is chosen at random from the bag and the colour is noted. \"With replacement\" means that you put the first ball back in the. Draw a tree diagram of the situation. Age range: 14-16. A second sweet is then removed. It consists of \"branches\" that are labeled with either frequencies or probabilities. To help understand this, let\u2019s first recall the formula for conditional probability. OLVER EDUCATION. 21 shows these four outcomes and their probabilities. To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. An experiment consists of rolling a red die and a green die and noting the result of each roll. Finally, find the probability of ending at each gate. A tree diagram is a special type of graph used to determine the outcomes of an experiment. The correct answer is A. a) Tree diagram for the experiment. Forexample,ifyoufliponefaircoin, S ={ H , T }where. b) How many outcomes are in the sample space? Exercise 7. Two marbles are selected without replacement. Tree Diagrams can be used to represent the total possible outcomes when you have 2 or more events. Whoops! There was a problem previewing Conditional probability. JC Number; JC Geometry & Trigonometry; JC Algebra & Functions; JC Probability & Statistics; JC Unifying Strand; TY Ideas; TY Analytics Module; Make-a-Mock; Branches; Newsletters; Events; Teacher. Draw a tree diagram to represent the probabilities in each case. Hint: The 4 digits in this probability add to 18. Unit 11 Day 3 Conditional Probability (3). 7, P (A') =. I introduce you to new notation which I would encourage you to do as it will help with conditional probability at. Draw a tree representing the possible mutually exclusive outcomes 2. Draw a probability tree showing the possible outcomes. Learning Outcomes As a result of studying this topic, students will be able to \u2022 understand and use the following terminology: trial, outcome, set of all possible outcomes, relative frequency, event, theoretical probability, If you require probability tree diagram worksheets with answers, or probability maths questions and answers you can. A girls' choir is choosing a concert uniform. Determine the probability of getting 2 heads in two successive tosses of a balanced coin. \ud83d\udea8 Claim your spot here. So we are calculating 99% of 10% which is 0. Let us take note that two cards, one at a time, are drawn at random from the box. Assign students to choose four of their shirts and four pairs of pants. 5(a) In the space below, draw a probability tree diagram to represent this information [3 marks] 5(b) Calculate the probability that one red and one green ball are taken from the bag. WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. Find the probability of the first marble being green and the second marble being yellow. Tree diagrams and conditional probability. 3: Tree diagram for two draws without replacement, values rounded. Tree diagrams are a tool to organize outcomes and probabilities around the structure of the data. The probability that Pat will arrive late is 0. 1 = 1\\$ To find the probability of any path, multiply the probabilities on the corresponding branches. Example 1. (a) Draw a tree diagram and from it write. 1 Sampling without replacement Example 3 A box contains 4 red marbles and 3 white ones. 1 Randomness, Probability, and Simulation (pp. A tree diagram is a special type of graph used to determine the outcomes of an experiment. Then a second marble is chosen at random. One ball is drawn from the bag, then another without replacement. a) Draw a tree diagram for this experiment b) Find the probability that at least one of the two persons favors genetic engineering. Replacement and Probability. A Tree Diagram and Sample Space A tree diagram is a graphic representation of the step by step competion of an experiment showing all possible results of each step. It is generally drawn from a starting point on the left and then branches to the right with each possilbe outcome shown as the end of the next branch or bridge. So, the probability that the student doesn't know the answer AND answers correctly is. Find and create gamified quizzes, lessons, presentations, and flashcards for students, employees, and everyone else. If it does rain, Mudlark will start favourite in the horse race, with probability of winning. Probability Trees RAG. I missed out tree diagrams without replacement. You spin the spinner once. Two are taken from the box, without replacement. Intuitive conditional probability seemingly not working. tree diagram. Two marbles are chosen without replacement. In this video, we will learn how to use tree diagrams to calculate conditional probabilities. In an urn, there are 11 balls. Let's consider another example: Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement. Question 1: Find the probability that a player selects two red counters. Draw a tree diagram representing the results. Using a tree diagram, find the probability that the second marble is red, given that the first one is red. Ron has a bag containing 3 green pears and 4 red pears. b) How many outcomes have a sum of the 2 numbers greater than or equal. iii: Find the probability that both biscuits are plain. Find the probability that: both. How easy is it? Simply open one of the tree diagram templates included, input your information and let SmartDraw do the rest. A tree diagram is a special type of graph used to determine the outcomes of an experiment. I built Diagnostic Questions to help you identify, understand and resolve key misconceptions. So don\u2019t let your student become confused by probability, our probability activities are probably the best resources available. eBook: Read p. Probability: Venn Diagrams and Two-Way Tables. Draw a tree diagram to represent the probabilities in each case. If the weather is fine, the probability that Carlos is late arriving at school is !!*. b) How many outcomes are in the sample space? Exercise 7. 392) Two cards are drawn without replacement from a 52-card deck. \"With replacement\" means that you put the first ball back. Tree diagrams can make some probability problems easier to visualize and solve. This probability is found by using the tree like a probability distribution table, simply identify the leaves that have this event (F), and then sum their probabilities. The probability that it will be windy on a particular day is 0. Make a tree diagram for an experiment that consists of two trials. To answer how likely a patient is to have TB given a positive test result, we need to \u201cflip\u201d the tree. [2] [3] The higher the probability of an event, the more certain we are that the event will occur. Determine the following geometric. MEMORY METER. It consists of \"branches\" that are labeled with either frequencies or probabilities. \u2022 Tree diagram - a diagram which can be helpful in illustrating possible outcomes of an experiment. Example: Probability of tossing a coin. Draw a probability tree showing the possible outcomes. The following example illustrates how to use a tree diagram. Since it's with replacement the first time i'm drawing, the probability would be 2 9 and the second time would also be 2 9 which would be 4 81. Find the probability that: both. SEE MORE : 9. Visit weteachmaths. I don't know how to write out a tree diagram on here, but I think this one is heads -> heads, tails -> math probabilty- please help. (1 mark) (ii) What is the probability that a student fails to gain a certificate? (2 marks) (b) Three students take the exam. Tree diagrams can make some probability problems easier to visualize and solve. The breakdown of the lot size and the sample size in the numerator and denominator of (3. one green ball and one blue ball. Tree Diagrams A tree diagram is a way of seeing all the possible probability 'routes' for two (or more) events. Resource type: Lesson (complete) 4. Therefore, in a family of three children, the probability of having three girls is 1 out of 8. It consists of \"branches\" that are labeled with either frequencies or probabilities. With Replacement Without Replacement P(BL1 and BL2): P(BL1 and BR2 or BR1 and BL2): P(BL1 and O2 ): P(O2 |BL1):. 34, and the probability of selecting a black marble on the first draw is 0. (This path has been drawn on the tree diagram with arrows. the form of a tree diagram or table \u00c6 express the probability of an event as a fraction, a decimal, and a percent independent events \u2022 results for which the outcome of one event has no eff ect on the outcome of another event \u2022 ruler probability \u2022 the likelihood or chance of an event occurring Determining Probabilities Using Tree Diagrams. 5 (since the probability of getting a heads on the first flip is 0. These explanations and tutorials will help you find the probability of all sorts of events, from rolling a number on a die to winning the lottery. We sample two items from the box without replacement. If A and B are independent (that is, the occurrence of a specific one of these two events does not influence the probability of the other event), then. Since nn grows much, much faster, than n!, Z\u2019s algorithm becomes prohibitively tedious in a hurry. Questions 28 \u2013 29 refer to the following probability tree diagram which shows tossing an unfair coin FOLLOWED BY drawing one bead from a cup containing 3 red (R), 4 yellow (Y) and 5 blue (B. ii: Write down the value of b. 368 #6 A plumbing contractor obtains 60% of her boiler circulators from a company whose defect rate is. (iii) the product of the two numbers is at least 5. (2) Jan 10. 7, and the probability that Jamie will pass. \ud83d\udea8 Claim your spot here. It contains example problems with replacement / independent events and wit. The possibilities are: 4 H, 3 H and 1 T (in various orders), 2 H and 2 T (in various orders), 1 H and 3 T (in various orders), or 4 T. Tree Diagrams \u2022Sample spaces can also be described graphically with tree diagrams. This item is taken from IGCSE Mathematics (0580) Paper 43 of May/June 2013. This site is the homepage of the textbook Introduction to Probability, Statistics, and Random Processes by Hossein Pishro-Nik. If it is fine he only has a 1 in 20 chance of winning. The number of \"Male and Smoke\" divided by the total = 19/100 = 0. Submitted by Hannah Yates on 6 March 2017. Draw a tree diagram showing the possible outcomes. 13 Outcomes & Probability Third Pick Second Pick First Pick BBB (0. Construct two tree diagrams (one for with replacement and the other for without replacement) showing the drawing of two M&Ms, one at a time, from the M&Ms you were given, as recorded in the table above.", "date": "2021-08-04 23:15:59", "meta": {"domain": "piratestorm.it", "url": "http://piratestorm.it/probability-with-replacement-tree-diagram.html", "openwebmath_score": 0.6850816011428833, "openwebmath_perplexity": 564.7005215453598, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717436787542, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.8553929746006267}}
{"url": "http://code.jasonbhill.com/category/algorithms/", "text": "A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.\n\nThen, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).\n\nThe result is Pascal\u2019s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?\n\n## Solution Using the Multinomial Theorem\n\nThe generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula\n$(x_1+x_2+\\cdots+x_m)^n=\\sum_{{k_1+k_2+\\cdots+k_m=n}\\atop{0\\le k_i\\le n}}\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)\\prod_{1\\le t\\le m}x_t^{k_t}$\nwhere\n$\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)=\\frac{n!}{k_1!k_2!\\cdots k_m!}.$\nOf course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by\n$\\left({200000}\\atop{k_1,k_2,k_3}\\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$\nwhere $k_1+k_2+k_3=200000$. It\u2019s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there\u2019s no way that we\u2019re going to calculate these coefficients. We only need to know when they\u2019re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.\n\nFirst, we\u2019ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that\n$p(n,d)=\\left\\lfloor\\frac{n}{d}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^2}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^3}\\right\\rfloor+ \\cdots+\\left\\lfloor\\frac{n}{d^r}\\right\\rfloor,$\nwhere $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we\u2019re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we\u2019ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We\u2019ll store two lists:\n$p2=[p(i,2)\\text{ for }1\\le i\\le 200000]\\quad\\text{and}\\quad p5=[p(i,5)\\text{ for }1\\le i\\le 200000].$\nFor a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when\n$p2[k_1]+p2[k_2]+p2[k_3]<199983\\ \\text{and}\\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$\nThat is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.\n\nNow, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\\le k_2\\le k_3$. Under this assumption, we always have\n$k_1\\le\\left\\lfloor\\frac{200000}{3}\\right\\rfloor=66666.$\nWe know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.\n\nIn case 1, we iterate $0\\le k_1\\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).\n\nIn case 2, we iterate $0\\le k_1\\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.\n\nIn case 3, we iterate $0\\le k_1\\le 66666$, and we iterate over $k_2=k_1+a$ where $1\\le a < \\left\\lfloor\\frac{200000-3k_1}{2}\\right\\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).\n\n## Cython Solution\n\nI\u2019ll provide two implementations, the first written in Cython inside Sage. Then, I\u2019ll write a parallel solution in C.\n\n%cython\n\nimport time\nfrom libc.stdlib cimport malloc, free\n\ncdef unsigned long p(unsigned long k, unsigned long d):\ncdef unsigned long power = d\ncdef unsigned long exp = 0\nwhile power <= k:\nexp += k / power\npower *= d\nreturn exp\n\ncdef unsigned long * p_list(unsigned long n, unsigned long d):\ncdef unsigned long i = 0\ncdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long))\nwhile i <= n:\npowers[i] = p(i,d)\ni += 1\nreturn powers\n\nrun_time = time.time()\n\n# form a list of number of times each n! is divisible by 2.\ncdef unsigned long * p2 = p_list(200000,2)\n\n# form a list of number of times each n! is divisible by 5.\ncdef unsigned long * p5 = p_list(200000,5)\n\ncdef unsigned long k1, k2, k3, a\ncdef unsigned long long result = 0\n\nk1 = 0\nwhile k1 <= 66666:\n# case 1: k1 = k2 < k3\nk2 = k1\nk3 = 200000 - k1 - k2\nif 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):\nresult += 3\n# case 2: k1 < k2 = k3\nif k1 % 2 == 0:\nk2 = (200000 - k1)/2\nk3 = k2\nif 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):\nresult += 3\n# case 3: k1 < k2 < k3\na = 1\nwhile 2*a < (200000 - 3*k1):\nk2 = k1 + a\nk3 = 200000 - k1 - k2\nif 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]):\nresult += 6\na += 1\nk1 += 1\n\nfree(p2)\nfree(p5)\n\nelapsed_run = round(time.time() - run_time, 5)\n\nprint \"Result: %s\" % result\nprint \"Runtime: %s seconds (total time: %s seconds)\" % (elapsed_run, elapsed_head)\n\nWhen executed, we find the correct result relatively quickly.\n\nResult: 479742450\nRuntime: 14.62538 seconds (total time: 14.62543 seconds)\n\n## C with OpenMP Solution\n\n#include <stdio.h>\n#include <stdlib.h>\n#include <malloc.h>\n#include <omp.h>\n\n/*****************************************************************************/\n/* function to determine how many factors of 'd' are in 'k!'                 */\n/*****************************************************************************/\nunsigned long p(unsigned long k, unsigned long d) {\nunsigned long power = d;\nunsigned long exp = 0;\nwhile (power <= k) {\nexp += k/power;\npower *= d;\n}\nreturn exp;\n}\n\n/*****************************************************************************/\n/* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer      */\n/*****************************************************************************/\nunsigned long * p_list(unsigned long n, unsigned long d) {\nunsigned long i;\nunsigned long * powers = malloc((n+1)*sizeof(unsigned long));\nfor (i=0;i<=n;i++) powers[i] = p(i,d);\nreturn powers;\n}\n\n/*****************************************************************************/\n/* main                                                                      */\n/*****************************************************************************/\nint main(int argc, char **argv) {\nunsigned long k1, k2, k3, a;\nunsigned long long result = 0;\n\nunsigned long * p2 = p_list(200000, 2);\nunsigned long * p5 = p_list(200000, 5);\n\n#pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result)\nfor (k1=0;k1<66667;k1++) {\n// case 1: k1 = k2 < k3\nk2 = k1;\nk3 = 200000 - k1 - k2;\nif (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {\nresult += 3;\n}\n// case 2: k1 < k2 = k3\nif (k1 % 2 == 0) {\nk2 = (200000 - k1)/2;\nk3 = k2;\nif (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {\nresult += 3;\n}\n}\n// case 3: k1 < k2 < k3\nfor (a=1;2*a<(200000-3*k1);a++) {\nk2 = k1 + a;\nk3 = 200000 - k1 - k2;\nif (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) {\nresult += 6;\n}\n}\n}\n\nfree(p2);\nfree(p5);\n\nprintf(\"result: %lld\\n\", result);\n\nreturn 0;\n}\n\nThis can be compiled and optimized using GCC as follows.\n\n$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result.$ time ./problem-154-omp\nresult: 479742450\n\nreal    0m1.487s\n\nThis appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds.\n\n### Motivation\n\nMany interesting computational problems, such as those on Project Euler require that one find the sum of proper divisors of a given integer. I had a fairly crude brute-force method for doing this, and was subsequently emailed a comment by Bjarki \u00c1g\u00fast Gu\u00f0mundsson who runs the site www.mathblog.dk. He pointed me in the direction of this page and provided some sample code illustrating how such an approach runs asymptotically faster than the approach I had been taking. Awesome! I\u2019m going to expand on that a bit here, providing some mathematical proofs behind the claims and providing code for those who may want to take advantage of this.\n\n### The Mathematics Behind It All\n\nLet the function $\\sigma(n)$ be the sum of divisors for a positive integer $n$. For example,\n$\\sigma(6)=1+2+3+6=12.$\n\nIt should seem obvious that for any prime $p$ we have $\\sigma(p)=1+p$. What about powers of primes? Let $\\alpha\\in\\mathbb{Z}_+$, and then\n$\\sigma(p^\\alpha)=1+p+p^2+\\cdots+p^\\alpha.$\n\nWe\u2019d like to write that in a closed form, i.e., without the \u201c$+\\cdots+$\u201d. We use a standard series trick to do that.\n\n\\begin{align} \\sigma(p^\\alpha) &= 1+p+p^2+\\cdots+p^\\alpha\\cr p\\sigma(p^\\alpha) &= p+p^2+p^3+\\cdots+p^{\\alpha+1}\\cr p\\sigma(p^\\alpha)-\\sigma(p^\\alpha) &= (p+p^2+\\cdots+p^{\\alpha+1})-(1+p+\\cdots+p^\\alpha)\\cr p\\sigma(p^\\alpha)-\\sigma(p^\\alpha) &= p^{\\alpha+1}-1\\cr (p-1)\\sigma(p^\\alpha) &= p^{\\alpha+1}-1\\cr \\sigma(p^\\alpha) &=\\frac{p^{\\alpha+1}-1}{p-1}.\\end{align}\n\nThat solves the problem of finding the sum of divisors for powers of primes. It would be nice if we could show that $\\sigma$ is multiplicative on powers of primes, i.e., that $\\sigma(p_1^{\\alpha_1}p_2^{\\alpha_2})=\\sigma(p_1^{\\alpha_1})\\sigma(p_2^{\\alpha_2})$. We\u2019ll prove that this is the case, and solve the problem in general along the way.\n\nProposition: The function $\\sigma$ is multiplicative on powers of primes.\n\nProof: Let $n$ be a positive integer written (uniquely, by the fundamental theorem of arithmetic) as\n$n=\\prod_{i=1}^m p_i^{\\alpha_i}$\nfor $m$ distinct primes $p_i$ with $\\alpha_i\\in\\mathbb{Z}_+$. Any divisor $k$ of $n$ then has the form\n$k=\\prod_{i=1}^m p_i^{\\beta_i}$\nwhere each $\\beta_i$ satisfies $0\\le\\beta_i\\le\\alpha_i$. Then\n$\\sigma(n)=\\sigma\\left(\\prod_{i=1}^m p_i^{\\alpha_i}\\right)$\nis the sum of all divisors $k$ of $n$ and can be written by summing over all possible combinations of the exponents $\\beta_i$. There are $\\prod_{i=1}^m \\alpha_i$ combinations, and we can form their sum and simplify it as follows.\n\\begin{align}\\sigma(n) &= \\sum_{1\\le i\\le m,\\ 0\\le\\beta_i\\le\\alpha_i}p_1^{\\beta_i}p_2^{\\beta_2}\\cdots p_m^{\\beta_m}\\cr &= \\sum_{\\beta_1=0}^{\\alpha_1}p_1^{\\beta_1}\\left(\\sum_{2\\le i\\le m,\\ 0\\le\\beta_i\\le\\alpha_i}p_2^{\\beta_2}p_3^{\\beta_3}\\cdots p_m^{\\beta_m}\\right)\\cr &= \\sum_{\\beta_1=0}^{\\alpha_1}p_1^{\\beta_1}\\sum_{\\beta_2=0}^{\\alpha_2}p_2^{\\beta_2}\\left(\\sum_{3\\le i\\le m,\\ 0\\le\\beta_i\\le\\alpha_i}p_3^{\\beta_3}p_4^{\\beta_4}\\cdots p_m^{\\beta_m}\\right) \\cr &= \\vdots\\cr &=\\sum_{\\beta_1=0}^{\\alpha_1}p_1^{\\beta_1}\\sum_{\\beta_2=0}^{\\alpha_2}p_2^{\\beta_2}\\sum_{\\beta_3=0}^{\\alpha_3}p_3^{\\beta_3}\\ \\cdots\\ \\sum_{\\beta_m=0}^{\\alpha_m}p_m^{\\beta_m}\\cr &=\\sigma(p_1^{\\alpha_1})\\sigma(p_2^{\\alpha_2})\\cdots\\sigma(p_m^{\\alpha_m}).\\end{align}\nThis completes the proof. Q.E.D.\n\nThus, we now have a formula for the sum of divisors of an arbitrary positive integer $n$ using the factorization of $n$. Namely,\n$\\sigma(n)=\\sigma\\left(\\prod_{i=1}^m p_i^{\\alpha_i}\\right)=\\prod_{i=1}^m\\left(\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}\\right).$\n\nThis is something I use quite a bit for various problems and programming exercises, so I figured I could post it here. It\u2019s a basic post that isn\u2019t advanced at all, but that doesn\u2019t mean that the implementation given below won\u2019t save work for others. The idea is to create a list of primes in C by malloc\u2019ing a sieve, then malloc\u2019ing a list of specific length based on that sieve. The resulting list contains all the primes below a given limit (defined in the code). The first member of the list is an integer representing the length of the list.\n\n#include <stdio.h>\n#include <stdlib.h>\n#include <malloc.h>\n\n#define                 bool    _Bool\n\nstatic unsigned long    prime_limit = 1000000;\n\nunsigned long sqrtld(unsigned long N) {\nint                 b = 1;\nunsigned long       res,s;\nwhile(1<<b<N) b+= 1;\nres = 1<<(b/2 + 1);\nfor(;;) {\ns = (N/res + res)/2;\nif(s>=res) return res;\nres = s;\n}\n}\n\nunsigned long * make_primes(unsigned long limit) {\nunsigned long      *primes;\nunsigned long       i,j;\nunsigned long       s = sqrtld(prime_limit);\nunsigned long       n = 0;\nbool *sieve = malloc((prime_limit + 1) * sizeof(bool));\nsieve[0] = 0;\nsieve[1] = 0;\nfor(i=2; i<=prime_limit; i++) sieve[i] = 1;\nj = 4;\nwhile(j<=prime_limit) {\nsieve[j] = 0;\nj += 2;\n}\nfor(i=3; i<=s; i+=2) {\nif(sieve[i] == 1) {\nj = i * 3;\nwhile(j<=prime_limit) {\nsieve[j] = 0;\nj += 2 * i;\n}\n}\n}\nfor(i=2;i<=prime_limit;i++) if(sieve[i]==1) n += 1;\nprimes = malloc((n + 1) * sizeof(unsigned long));\nprimes[0] = n;\nj = 1;\nfor(i=2;i<=prime_limit;i++) if(sieve[i]==1) {\nprimes[j] = i;\nj++;\n}\nfree(sieve);\nreturn primes;\n}\n\nint main(void) {\n\nunsigned long * primes = make_primes(prime_limit);\n\nprintf(\"There are %ld primes <= %ld\\n\",primes[0],prime_limit);\n\nfree(primes);\nreturn 0;\n}\n\nSay one wanted to form a list of all primes below 1,000,000. That\u2019s what the above program does by default, since \u201cprime_limit = 1000000.\u201d If one compiles this and executes, you would get something like what follows. The timing is relatively respectable.\n\n$gcc -O3 -o prime-sieve prime-sieve.c$ time ./prime-sieve\nThere are 78498 primes <= 1000000\n\nreal\t0m0.008s\nuser\t0m0.004s\nsys\t0m0.000s\n\nThe code is linked here: prime-sieve.c", "date": "2019-04-24 20:41:17", "meta": {"domain": "jasonbhill.com", "url": "http://code.jasonbhill.com/category/algorithms/", "openwebmath_score": 0.8579844236373901, "openwebmath_perplexity": 1756.1766796464285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319868927341, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8553896904395946}}
{"url": "http://mathhelpforum.com/math-challenge-problems/10222-quickie-13-a.html", "text": "# Math Help - Quickie #13\n\n1. ## Quickie #13\n\nThis is a classic (very old).\n\nAn irrational number raised to an irrational power is always irrational.\n\nProve it or provide a counterexample.\n\n2. Seen it. But I did not know it was a classic.\n\n3. Originally Posted by Soroban\nThis is a classic (very old).\n\nAn irrational number raised to an irrational power is always irrational.\n\nProve it or provide a counterexample.\n\nI'll give several counter examples;\n\nFor the most part,\n\n[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.\n\nEX:\n\n[sqrt(2)^(sqrt(2))]^(sqrt(2))\n\n[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)\n\n[2^(sqrt(2)/2)]^(sqrt(2)) = 2\n\n4. Originally Posted by AfterShock\nI'll give several counter examples;\n\nFor the most part,\n\n[sqrt(x)^(sqrt(x))]^(sqrt(x)), where x is even, with the exception of when x is a square.\n\nEX:\n\n[sqrt(2)^(sqrt(2))]^(sqrt(2))\n\n[sqrt(2)^(sqrt(2))] = 2^(sqrt(2)/2)\n\n[2^(sqrt(2)/2)]^(sqrt(2)) = 2\nInteresting, but you are assuming that $\\sqrt{2}^{\\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.\n\n-Dan\n\n5. Originally Posted by topsquark\nInteresting, but you are assuming that $\\sqrt{2}^{\\sqrt{2}}$ is irrational, else this isn't a counterexample. I have little doubt that it is, but this isn't a good counterexample without proving this fact.\n\n-Dan\nFunny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.\n\n6. Originally Posted by ThePerfectHacker\nFunny thing is you do not need to. The assumption is that irrational raised to irrational is always irrational. Thus, he is using this assumption. And arrives at a contradiction.\nAaaaah! I get it now.\n\n-Dan\n\n7. The \"classic\" solution goes like this:\n\nTheorem: . $(\\text{irrational})^{\\text{irrational}}$ can be rational.\n\nProof\n\n$\\text{Consider }a \\:=\\:\\sqrt{2}^{\\sqrt{2}}$\n\nThere are only two possibilities:\n\n. . (1) $a$ is rational.\n\n. . (2) $a$ is irrational.\n\nIf (1) $a$ is rational, the theroem is verified.\n\nIf (2) $a$ is irrational, consider: . $a^{\\sqrt{2}}$\n. . an irrational number raised to an irrational power.\n\nThen we have: . $a^{\\sqrt{2}} \\:=\\:\\left(\\sqrt{2}^{\\sqrt{2}}\\right)^{\\sqrt{2}} \\:=\\:\\left(\\sqrt{2}\\right)^2\\:=\\:2$ . . . a rational number.\n\nEither way, an irrational raised to an irrational power can be rational.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nI've always found this proof amusing.\n\nNote that we still don't know if $\\sqrt{2}^{\\sqrt{2}}$ is rational or irrational\n. . but it doesn't matter . . .\n\n8. I found a different counterexample.\n\n$e$ is irrational.\n\n$\\ln 2$ is irrational (but do not know how to show it).\n\nThen,\n$e^{\\ln 2}=2$", "date": "2016-05-26 14:47:00", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-challenge-problems/10222-quickie-13-a.html", "openwebmath_score": 0.9496710896492004, "openwebmath_perplexity": 515.957030761138, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990731986892734, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8553896834745887}}
{"url": "https://netco-me.com/g0y3k/every-regular-graph-is-complete-graph-bf463e", "text": "A simple non-planar graph with minimum number of vertices is the complete graph K 5. A complete graph is a graph in which every vertex has an edge to all other vertices is called a complete graph, In other words, each pair of graph vertices is connected by an edge. Two further examples are shown in Figure 1.14. A regular graph with vertices of degree k {\\displaystyle k} is called a k {\\displaystyle k} \u2011regular graph or regular graph of degree k {\\displaystyle k}. 3.A graph is k-regular if every vertex has degree k. How do 1-regular graphs look like? If every vertex of a simple graph has the same degree, then the graph is called a regular graph. A simple graph is called regular if every vertex of this graph has the same degree. Before you go through this article, make sure that you have gone through the previous article on various Types of Graphsin Graph Theory. An important property of graphs that is used frequently in graph theory is the degree of each vertex. C Tree. Let $G$ be a regular graph, that is there is some $r$ such that $|\\delta_G(v)|=r$ for all $v\\in V(G)$. 1.3 Find out whether the complete graph, the path and the cycle of order n 1 are bipartite and/or regular. 1.6.Show that if a k-regular bipartite graph with k>0 has a bipartition (X;Y), then jXj= jYj. I think you wanted to ask about a spanning 1-regular graph, also known as a perfect matching or 1-factor. This means that (assuming this is not a multigraph, no self-edges, etc) if you have n vertices, then each vertex has n-1 edges. Regular Graph c) Simple Graph d) Complete Graph \u2026 View Answer ... B Regular graph. Another plural is vertexes. hence, The edge defined as a connection between the two vertices of a graph. That is, if a graph is k-regular, every vertex has degree k. Exercises: Draw all 0-regular graphs with 1 vertex; 2 vertices; 3 vertices. If every vertex in a regular graph has degree k,then the graph is called k-regular. Conjecture 8 : Let G be a 3-regular cyclically 4-edge-connected graph of order n.Then G contains a cycle of length at least cn where c is a positive num- ber. G is said to be regular of degree r (or r-regular) if deg(v) = r for all vertices v in G. Complete graphs of order n are regular of degree n \u2212 1, and empty graphs are regular of degree 0. Explanation: In a regular graph, degrees of all the vertices are equal. Some sources claim that the letter K in this notation stands for the German word komplett, but the German name for a complete graph, vollst\u00e4ndiger Graph, does not contain the letter K, and other sources state that the notation honors the contributions of Kazimierz Kuratowski to graph theory. The problem seems similar to Hamiltonian Path which is NP complete problem for a general graph. How to create a program and program development cycle? yes No Not enough information to decide If Ris the equivalence relation defined by the panition {{1. A complete graph Km is a graph with m vertices, any two of which are adjacent. D n2. {5}. Explanation of Complete Graph with Diagram and Example, Explanation of Abstract Data Types with Diagram and Example, What is One Dimensional Array in Data Structure with Example, What is Singly Linked List? The first example is an example of a complete graph. What is Polynomials Addition using Linked lists With Example. Vertex Cover (VC): A vertex cover in an undirected graph G = (V;E) is a subset of vertices V0 V such that every edge in G has at least one endpoint in V0. therefore, In a directed graph, an edge goes from one vertex, the source, to another, the target, and hence makes the connection in only one direction. I'm not sure about my anwser. In the first, there is a direct path from every single house to every single other house. Every strongly regular graph is symmetric, but not vice versa. B n*n. C nn. Kn has n(n\u00e2\u0088\u00921)/2 edges and is a regular graph of degree n\u00e2\u0088\u00921. (a) every induced subgraph of a complete graph is complete; (b) every subgraph of a bipartite graph is bipartite. In the given graph the degree of every vertex is 3. The vertex is defined as an item in a graph, sometimes referred to as a node, The plural is vertices. definition. The complete graph on n vertices is denoted by Kn. 1.7.Show that, in any group of two or more people, there are always two with exactly the same number of friends inside the group. Theorem 9 : Let G be a 3-connected 3-regular graph , and let S be a set of nine vertices of G.Then G has a cycle which includes every vertex of S. (Aolton et al., 1982; Kelmans and Lomonosov, 1982) A graph of this kind is sometimes said to be an srg(v, k, \u03bb, \u03bc).Strongly regular graphs were introduced by Raj Chandra Bose in 1963.. A complete graph is a graph in which every vertex has an edge to all other vertices is called a complete graph, In other words, each pair of graph vertices is connected by an edge. 4. DEFINITION : Complete graph: In a graph, if there exist an edge between every pair of vertices,then such a graph is called complete graph. A connected graph may not be (and often is not) complete. In the graph, a vertex should have edges with all other vertices, then it called a complete graph. 4)A star graph of order 7. A simple graph with \u2018n\u2019 mutual vertices is called a complete graph and it is denoted by \u2018K n \u2019. The complete graph on n vertices is denoted by Kn. Regular Graphs A graph G is regular if every vertex has the same degree. As the above graph n=7 Ans - Statement p is true. The complete bipartite graph K m, n is planar if and only if m \u2264 2 or n \u2264 2. A complete graph is a graph that has an edge between every single vertex in the graph; we represent a complete graph \u2026 Q.1. Theorem 2.4 If G is a k-regular bipartite graph with k > 0 and the bipartition of G Privacy therefore, The total number of edges of complete graph = 21 = (7)*(7-1)/2. Solution: A 1-regular graph is just a disjoint union of edges (soon to be called a matching). Complete graphs correspond to cliques. regular graph : a regular graph is a graph in which every node has the same degree \u2022 connected graph : a graph is connected if any two points can be joined by a path (a sequence of edges that are pairwise adjacent) MATH3301 EXTREMAL GRAPH THEORY De\ufb02nition: A near regular complete multipartite graph is a complete multipartite graph with orders of its partite sets di\ufb01ering by at most 1. {6} {7}} which of the graphs betov/represents the quotient graph G^R of the graph G represented below. 45 The complete graph K, has... different spanning trees? Some authors exclude graphs which satisfy the definition trivially, namely those graphs which are the disjoint union of one or more equal-sized complete graphs, and their complements, the complete multipartite graphs with equal-sized independent sets. Advantage and Disadvantages. 1 2 3 4 QUESTION 3 Is this graph regular? therefore, in an undirected graph pair of vertices (A, B) and (B, A) represent the same edge. Statement Q Is True. In graph theory, a regular graph is a graph where each vertex has the same number of neighbors; i.e. 1.8.1. A regular graph of degree r is strongly regular if there exist nonnegative integers e, d such that for all vertices u, v the number of vertices \u2026 The complete graph with n graph vertices is denoted mn. A regular directed graph must also satisfy the stronger condition that the indegree and outdegree of each vertex are equal to each other. Regular, Complete and Complete Bipartite. What is the Classification of Data Structure with Diagram, Explanation array data structure and types with diagram, Abstract Data Type algorithm brief Description with example, What is Algorithm Programming? 4.How many (labelled) graphs exist on a given set of nvertices? Important graphs and graph classes De nition. Could you please help me on Discrete-mathematical-structures. A graph and its complement. Shelly has narrowed it down to two different layouts of how she wants the houses to be connected. The set of vertices V(G) = {1, 2, 3, 4, 5} for n 3, the cycle C A regular graph is called n-regular if every vertex in this graph has degree n. Match the values of n (in the right column) for which the graphs (in the left column) are regular? 2)A bipartite graph of order 6. A 2-regular graph is a disjoint union of cycles. A complete graph is connected. Acomplete graphhas an edge between every pair of vertices. What is Data Structures and Algorithms with Explanation? Question: Let Statements P And Q Be As Follows P = \"Every Complete Graph Is Regular.\" Any graph with 8 or less edges is planar. $\\begingroup$ @Igor: I think there's some terminological confusion here - an induced subgraph of a complete graph is a complete graph... $\\endgroup$ \u2013 ndkrempel Jan 17 '11 at 17:25 $\\begingroup$ @ndkrempel: yes, confusion reigns. A symmetric graph is one in which there is a symmetry (graph automorphism) taking any ordered pair of adjacent vertices to any other ordered pair; the Foster census lists all small symmetric 3-regular graphs. Definition: Regular. The complete graph with n graph vertices is denoted mn. In a complete graph, for every two vertices in a graph, there is an edge that directly connects the two. the complete graph with n vertices has calculated by formulas as edges. The vertex cover problem (VC) is: given an undirected graph G and an integer k, does G have a vertex cover of size k? What are the basic data structure operations and Explanation? Every non-empty graph contains such a graph. We have discussed- 1. A single edge connecting two vertices, or in other words the complete graph K 2 on two vertices, is a 1-regular graph. Then, we have $|\\delta_\\bar{G}(v)|=n-r-1$, where $\\bar{G}$ is the complement of $G$ and $n=|V(G)|$. $\\endgroup$ \u2013 Igor Rivin Jan 17 '11 at 17:40 Any graph with 4 or less vertices is planar. Complete Graph. View Answer Answer: Tree ... Answer: The number of edges in walk W 49 If for some positive integer k, degree of vertex d(v)=k for every vertex v of the graph G, then G is called... ? A graph is called Eulerian if it has an Eulerian Cycle and called Semi-Eulerian if it has an Eulerian Path. Fortunately, we can find whether a given graph has a \u2026 graph when it is clear from the context) to mean an isomorphism class of graphs. Regular Graph - A graph in which all the vertices are of equal degree is called a regular graph. Suppose a contractor, Shelly, is creating a neighborhood of six houses that are arranged in such a way that they enclose a forested area. The line graph H of a graph G is a graph the vertices of which correspond to the edges of G, any two vertices of H being adjacent if and\u2026. 1.4 Give the size: 1)of an r-regular graph of order n; 2)of the complete bipartite graph K r;s. For all natural numbers nwe de ne: the complete graph complete graph, K n K n on nvertices as the (unlabeled) graph isomorphic to [n]; [n] 2. They are called 2-Regular Graphs. ... A k-regular graph G is one such that deg(v) = k for all v \u2208G. If all the vertices in a graph are of degree \u2018k\u2019, then it is called as a \u201c k-regular graph \u201c. Aregular graphis agraphwhereevery vertex has the same degree.Therefore, every compl, Let statements p and q be as follows p = \"Every complete graph is regular.\" A nn-2. In this article, we will discuss about Bipartite Graphs. Which of the following statements for a simple graph is correct? In a weighted graph, every edge has a number, it\u2019s called \u201cweight\u201d. Output Result In both the graphs, all the vertices have degree 2. View desktop site. Properties of Regular Graphs: A complete graph N vertices is (N-1) regular. 1.8. Terms complete. In the second, there is a way to get from each of the houses to each of the other houses, but it's not necessarily \u2026 Complete Graph defined as An undirected graph with an edge between every pair of vertices. A graph G is said to be complete if every vertex in G is connected to every other vertex in G. Thus a complete graph G must be connected. therefore, the complete digraph is a directed graph in which every pair of distinct vertices is connected by a pair of unique edges (one in each direction). (Thomassen et al., 1986, et al.) \u2026the graph is called a complete graph (Figure 13B). A K graph. 3)A complete bipartite graph of order 7. The set of edges E(G) = {(1, 2), (1, 4), (1, 5), (2, 3), (3, 4), (3, 5), (1, 3)} To calculate total number of edges with N vertices used formula such as = ( n * ( n \u00e2\u0080\u0093 1 ) ) / 2. Note: An undirected graph represented as a directed graph with two directed edges, one \u201cto\u201d and one \u201cfrom,\u201d for every undirected edge. A graph in which degree of all the vertices is same is called as a regular graph. Regular Graph: A graph is said to be regular or K-regular if all its vertices have the same degree K. A graph whose all vertices have degree 2 is known as a 2-regular \u2026 A graph is a collection of vertices connected to each other through a set of edges. \u00a9 2003-2021 Chegg Inc. All rights reserved. | The complete graph with n vertices is denoted by K n. The Figure shows the graphs K 1 through K 6. D Not a graph. A complete graph K n is planar if and only if n \u2264 4. 2} {3 4}. Statement P Is True. Defined Another way you can say, A complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. every vertex has the same degree or valency. Definition, Example, Explain the algorithm characteristics in data structure, Divide and Conquer Algorithm | Introduction. In this article, we will show that every bipartite graph is 2 chromatic ( chromatic number is 2 ).. A simple graph G is called a Bipartite Graph if the vertices of graph G can be divided into two disjoint sets \u2013 V1 and V2 such that every edge in G connects a vertex in V1 and a vertex in V2. q = \"Every regular graph Is complete\" Select the option below that BEST applies to these statements. In simple words, no edge connects two vertices belonging to the same set. Hence, the complement of $G$ is also regular. An undirected graph is defined as a graph containing an unordered pair of vertices is Know an undirected graph. 1)A 3-regular graph of order at least 5. The study of graphs is known as Graph Theory. And 2-regular graphs? The graphs in the chapter are always regular of degree r, that is, every vertex in the graph is incident to r edges in the graph. Q = \"Every Regular Graph Is Complete\" Select The Option Below That BEST Applies To These Statements. Let Statements P And Q Be As Follows P = \"Every Complete Graph Is Regular.\" Kn For all n \u2026 Both statments are true Neither statement is true QUESTION 2 Find the degree of vertex 5. Statement p is true. 2. Every graph has certain properties that can be used to describe it. therefore, A graph is said to complete or fully connected if there is a path from every vertex to every other vertex. A simple graph }G ={V,E is said to be regular of degree k, or simply k-regular if for each v\u2208V, \u03b4(v) =k. Statement q is true. the complete graph with n vertices has calculated by formulas as edges. & Is planar is same is called a regular graph has the same degree, then jYj! =  every complete graph K n \u2019 BEST Applies to These Statements each vertex are equal to each.. Between the two vertices, then the graph is defined as a connection between two!, 1986, et al. soon to be called a regular graph is complete '' the... Less vertices is planar if and only if n \u2264 4 plural is vertices has... different trees... Operations and explanation to create a program and program development cycle by Kn statments are true Neither statement true. ) = K for all n \u2026 45 the complete graph n vertices is Know an undirected graph pair vertices., is a 1-regular graph is bipartite satisfy the stronger condition that the indegree and of... ) every subgraph of a graph G is one such that deg ( v ) = K for n... Between every pair of vertices K 2 on two vertices, is graph! Soon to be called a regular directed graph must also satisfy the condition! Vertices is denoted by Kn the path and the cycle of order 1! ) every subgraph of a bipartite graph is symmetric, but not vice versa the shows... Two of which are adjacent vertices belonging to the same degree, then graph., then the graph G is regular if every vertex is defined as perfect. The first, there is a disjoint union of edges pair of vertices the graph! Simple graph with 8 or less edges is planar if and only if n \u2264 or. Many ( labelled ) graphs exist on a given set of edges a single edge connecting two vertices, in! Layouts of how she wants the houses to be connected jXj= jYj on! K > 0 has a number, it \u2019 s called \u201c weight \u201d graph n vertices is denoted Kn... \u2018 K \u2019, then the graph is complete ; ( B ) and B! The vertex is defined as a connection between the two vertices belonging the! P =  every complete graph with n vertices is denoted by Kn... a bipartite! ) = K for all v \u2208G ( n\u00e2\u0088\u00921 ) /2 edges and is a collection of vertices a! 6 } { 7 } } which of the graphs, all the vertices are of equal is... And only if n \u2264 4 every vertex has the same degree v \u2208G this graph regular regular. Has an Eulerian cycle and called Semi-Eulerian if it has an Eulerian path an... N graph vertices is denoted mn G represented below not vice versa to Hamiltonian which... By K n. the Figure shows the graphs K 1 through K 6 other words complete. 4 QUESTION 3 is this graph regular with minimum number of vertices denoted! Algorithm characteristics in data structure operations and explanation directed graph must also satisfy the stronger condition the! With K > 0 has a bipartition ( X ; Y ), then it a... Follows P =  every complete graph ( Figure 13B ) of every vertex in a graph..., then it is denoted by \u2018 K \u2019, then the graph G is regular ''! Has certain properties that can be used to describe it bipartite graphs, referred... | Introduction of graphs is known as a graph is called as a connection the... Less edges is planar 2-regular graph is called a regular directed graph must also satisfy the stronger that... Or 1-factor many ( labelled ) graphs exist on a given set of nvertices the graphs, all the are. \u201c weight \u201d a complete graph with 4 or less vertices is Know an undirected graph is a., is a graph G is regular. a node, the edge defined as a node the... Or fully connected if there is a graph with n graph vertices is denoted by K n. Figure! And ( B, a vertex should have edges with all other vertices or. Select the Option below that BEST Applies to These Statements every regular graph is complete graph regular ) a complete graph m. A single edge connecting two vertices of a bipartite graph of degree \u2018 K n is planar given graph degree... To every other vertex spanning trees explanation: in a regular graph induced subgraph of a complete graph n! Then jXj= jYj is also regular. be called a matching ) we will discuss bipartite! K n. the Figure shows the graphs betov/represents the quotient graph G^R of the graphs betov/represents the quotient G^R. N is planar layouts of how she wants the houses to be called a matching ) graph it. 0 has a number, it \u2019 s called \u201c weight \u201d indegree and outdegree of each.. Shows the graphs, all the vertices is denoted by Kn with \u2018 n \u2019 mutual vertices is the graph. Other vertices, any two of which are adjacent operations and explanation matching or 1-factor the to! Of Graphsin graph Theory that BEST Applies to These Statements regular graphs: a complete bipartite graph called... The stronger condition that the indegree and outdegree of each vertex which of the graph, also as... A graph are of degree \u2018 K n is planar if and only if n \u2264 4 the! By K n. the Figure shows the graphs K 1 through K 6 G represented.... Called a regular graph Follows P =  every regular graph a weighted graph, also known as Theory. G^R of the graph is called k-regular Divide and Conquer algorithm | Introduction both the K., there is a 1-regular graph is symmetric, but not vice versa path and the cycle of 7., it \u2019 s called \u201c weight \u201d B, a ) induced... Has n ( n\u00e2\u0088\u00921 ) /2 edges and is a direct path from every vertex has the same.! V ) = K for all v \u2208G plural is vertices the stronger condition that the and. Pair of vertices a regular graph what are the basic data structure, and. Edge connecting two vertices, any two of which are adjacent every regular graph is complete graph an isomorphism of... K 5 { { 1 \u2019, then jXj= jYj al. out whether complete... Edge defined as a graph defined as an item in a weighted,! N is planar minimum number of vertices the quotient graph G^R of the graph said... Or 1-factor if a k-regular graph G is regular if every vertex is 3 ) complete path. Calculated by formulas as edges a given set of edges ( soon to be connected exist on a given of! Have edges with all other vertices, then the graph G is one such that deg ( v =. P and Q be as Follows P =  every complete graph vertex is defined as item... Used frequently in graph Theory is a direct path from every single other house Graphsin graph.... Direct path from every single other house with all other vertices, is a direct path every! Has calculated by formulas as edges for n 3, the path and cycle... A 1-regular graph has... different spanning trees Ris the equivalence relation defined the. K 5 an important property of graphs that is used frequently in graph Theory equal is... \\$ is also regular. said to complete or fully connected if there a! The first example is an example of a graph, a graph G is one such that deg ( )... Graph, every edge has a bipartition ( X ; Y ) then. ) /2 edges and is a direct path from every vertex to every single other house it called regular... To every single house to every single house to every single house to every other vertex other house graph! K > 0 has a bipartition ( X ; Y ), it! As edges seems similar to Hamiltonian path which is NP complete problem for a general graph vertex has the degree. Has certain properties that can be used to describe it but not vice versa graph! A weighted graph, a ) represent the same set has an Eulerian cycle and called Semi-Eulerian if has! Called as a node, the edge defined as an item in regular., et al. Addition using Linked lists with example a direct path from every single other house whether. Symmetric, but not vice versa the quotient graph G^R of the graph is ;... As a \u201c k-regular graph \u201c all other vertices, is a from... Has degree K, then it is called k-regular cycle C a graph is to! Vertex are equal QUESTION: Let Statements P and Q be as Follows =. The study of graphs has narrowed it down to two different layouts of how she wants the houses be... Q be as Follows P =  every regular graph graphhas an edge between every pair vertices. On n vertices is denoted by Kn a vertex should have edges with other... Complete ; ( B ) every induced subgraph of a bipartite graph of degree.! 1-Regular graph is regular. of vertices connected to each other through a set of edges ( to... Denoted by Kn the graphs K 1 through K 6 N-1 ).. Is every regular graph is complete graph, example, Explain the algorithm characteristics in data structure operations and?. S called \u201c weight \u201d { 6 } { 7 } } which of the graph is bipartite the graph!, it \u2019 s called \u201c weight \u201d graph of order 7 statments! Property of graphs is known as graph Theory a vertex should have edges with all other vertices is...", "date": "2022-09-28 12:57:29", "meta": {"domain": "netco-me.com", "url": "https://netco-me.com/g0y3k/every-regular-graph-is-complete-graph-bf463e", "openwebmath_score": 0.6370208263397217, "openwebmath_perplexity": 405.562480765506, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9591542840900507, "lm_q2_score": 0.8918110540642805, "lm_q1q2_score": 0.8553843931046184}}
{"url": "https://mathematica.stackexchange.com/questions/104546/numerically-integrate-a-plotted-function", "text": "# Numerically integrate a plotted function\n\nI used Plot[NIntegrate[...]...] to plot a function of 5 different variables. It took really long. Right now I need to integrate this function one more time over the 5th variable and plot the result. Can someone tell me if there is an option to Nintegrate the plot to speed things up? Maybe there's a way to convert the plot to a list of $x$ and $y$ and then take a sum over $(y2-y1)\\,(x2-x1)$ and list plot it or something similar?\n\nI guess it's not gonna work for me. I have the following function:\n\nf[\u03c1_, \u03d5_, \u03b6_, \u03be_, r_] := 2.*100.*20.*0.02*0.2*\u03c1*Cos[\u03d5]*Sech[20*(\u03c1 - 1.)]^2.*\u03b6*Exp[-\u03b6]*\nBesselJ[0, 100.*\u03b6*Sqrt[r^2. + \u03c1^2. - 2.*r*\u03c1*Cos[\u03d5]]]*(Sqrt[\u03b6^2. + 0.02^2.]*\nCosh[10.*Sqrt[\u03b6^2. + 0.02^2.]*(\u03be + 1.)] + \u03b6*Sinh[10.*Sqrt[\u03b6^2. + 0.02^2.]\n*(\u03be + 1.)])/((2.*\u03b6^2. + 0.02^2.)*Sinh[10.*Sqrt[\u03b6^2. + 0.02^2.]] +\n2.*\u03b6*Sqrt[\u03b6^2. + 0.02^2.]*Cosh[10.*Sqrt[\u03b6^2. + 0.02^2.]])\n\n\nHere the function $f$ have 5 variables: $\\rho,\\ \\phi,\\ \\zeta,\\ \\xi,\\ r.$. First, I need to integrate over $\\rho,\\ \\phi,\\ \\zeta,\\ \\xi.$ and plot $f=f(r).$ It takes about 1 hour. I use as I said before\n\nPlot[NIntegrate[f, {\u03be, -1., 0.}, {\u03b6, 0., \u221e}, {\u03d5, -3.1415, +3.1415}, {\u03c1, 0., \u221e}], {r, 0, 10}]\n\n\nThe next step is I need to integrate one more time over $r$, {r,0,r1}and plot it over {r1,0,7}. I tried what you've said, but I failed. Is your method with NDsolve going to work with this kind of function?\n\n\u2022 See the accepted answer here. \u2013\u00a0Jason B. Jan 21 '16 at 15:32\n\u2022 It seems I misunderstood your question. Can you clarify why you want to tie together the plotting w.r.t r and the integration w.r.t r? Is your motivation performance? I would expect that integrating over all 5 variables in one step with NIntegrate should be much faster than this plot was. You can also speed up plotting by limiting the number of plot points. Set MaxRecursion -> 0 and manually set the desired number of PlotPoints (these symbols can be looked up in the documentation). \u2013\u00a0Szabolcs Jan 21 '16 at 21:35\n\u2022 But it would be even better to not use Plot but build a Table of the values, which you can store for later, and the ListPlot them. If the resolution of the plot is not good enough, you can compute additional points without throwing away the already computed ones. This is not (easily) possible with Plot which will always recompute everything from the start and it's not possible to interrupt it then continue if necessary (at least not without advanced hacks ...) \u2013\u00a0Szabolcs Jan 21 '16 at 21:37\n\nThe trick is to use NDSolve instead of NIntegrate and thus in effect obtain a numerical antiderivative that can be evaluated fast at different points. NIntegrate will only do definite integration, so it needs to be run each time the integration bounds are changed. This is very slow, as you noticed. NDSolve will only need to be run once.\n\n### Slow way (what you used)\n\nWe are going to integrate f:\n\nf[x_] := Sin[x^2]\n\nF[y_?NumericQ] := NIntegrate[f[x], {x, 0, y}]\n\nPlot[F[x], {x, 0, 5}] // AbsoluteTiming\n\n\nIt took 5 seconds on my machine.\n\nNIntegrate[F[x], {x, 0, 5}] // AbsoluteTiming\n(* {0.686052, 2.63519} *)\n\n\nBoth integration and plotting are slow with this method. For complex functions, they can be prohibitively slow.\n\n### Fast way (NDSolve)\n\niF = NDSolveValue[{F2'[x] == f[x], F2[0] == 0}, F2, {x, 0, 5}]\n\n\niF is an interpolating function.\n\nPlot[iF[x], {x, 0, 5}] // AbsoluteTiming\n\n\nHere's a not so widely known trick: the antiderivative of an interpolating function can be computed exactly using Integrate (not NIntegrate). It is returned by Mathematica as another InterpolatingFunction object. This is possible because interpolating functions are really just piecewise polynomials. Derivatives can be taken too.\n\nSo we can compute the second integral directly (and very quickly) as\n\nIntegrate[iF[x], {x, 0, 5}]\n\n(* 2.63519 *)\n\n\nAnother trick: To get the antiderivative as a function object directly, you can also do\n\nDerivative[-1][iF]\n\n\nThe disadvantage of the fast way is that there's less oversight about precision loss and numerical errors. To make sure that the final integration result is accurate enough (despite error accumulation), you must make sure that NDSolve computes iF with sufficient accuracy.\n\n\u2022 \"Here's a not so widely known trick\": in some sense, this trick should be obvious, but it clearly isn't. +1 for showing the tricks! \u2013\u00a0march Jan 21 '16 at 17:33\n\u2022 I added some details. Could you check it out? \u2013\u00a0LexRomah Jan 21 '16 at 19:10\n\u2022 @LexRomah Right, this method won't work for you. :-( \u2013\u00a0Szabolcs Jan 21 '16 at 21:37\n\u2022 sorry, but doesn't your method work for a function of several variables? \u2013\u00a0illuminates Dec 3 '17 at 16:41", "date": "2020-09-22 12:04:58", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/104546/numerically-integrate-a-plotted-function", "openwebmath_score": 0.47845658659935, "openwebmath_perplexity": 1106.0991318544334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9591542805873231, "lm_q2_score": 0.8918110440002044, "lm_q1q2_score": 0.8553843803278456}}
{"url": "http://mathhelpforum.com/calculus/172638-half-life.html", "text": "# Math Help - Half-life.\n\n1. ## Half-life.\n\nA radioactive substance has a half-life of 20 days.\n\na) How much time is required so that only $\\frac{1}{32}$ of the original amount remains?\nb) Find the rate of decay at this time.\n\nMy attempts:\na) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\\frac{1}{8}$. After 80 days it decays to $\\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\\frac{1}{32}/MATH] of its original mass.\nb) $f(t) = \\frac{1}{32}(\\frac{1}{2})^{\\frac{t}{20}$\n$f'(t) =\\frac{1}{32}(\\frac{1}{2})^{\\frac{t}{20}} \u00d7 (ln\\frac{1}{2}) \u00d7 (\\frac{1}{20})$\n$f'(100) =\\frac{1}{32}(\\frac{1}{2})^{\\frac{100}{20}} \u00d7 (ln\\frac{1}{2}) \u00d7 (\\frac{1}{20})$\n$f'(100) = -3.384 \u00d7 10^{-5}$ is the rate of decay when $t = 100$ days.\n\nI just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.\n\n2. Originally Posted by Pupil\nA radioactive substance has a half-life of 20 days.\n\na) How much time is required so that only $\\frac{1}{32}$ of the original amount remains?\nb) Find the rate of decay at this time.\n\nMy attempts:\na) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\\frac{1}{8}$. After 80 days it decays to $\\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\\frac{1}{32}/MATH] of its original mass.\nb) $f(t) = \\frac{1}{32}(\\frac{1}{2})^{\\frac{t}{20}$\n$f'(t) =\\frac{1}{32}(\\frac{1}{2})^{\\frac{t}{20}} \u00d7 (ln\\frac{1}{2}) \u00d7 (\\frac{1}{20})$\n$f'(100) =\\frac{1}{32}(\\frac{1}{2})^{\\frac{100}{20}} \u00d7 (ln\\frac{1}{2}) \u00d7 (\\frac{1}{20})$\n$f'(100) = -3.384 \u00d7 10^{-5}$ is the rate of decay when $t = 100$ days.\n\nI just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.\nThe first part is correct However you could have solved it this way\n\n$\\displaystyle \\left( \\frac{1}{2}\\right)^\\frac{x}{20}=\\frac{1}{32}=\\left ( \\frac{1}{2}\\right)^5 \\iff \\frac{x}{20}=5 \\implies x=100$\n\nFor the 2nd part all we know is\n\n$\\displaystyle f(t)=A_0\\left( \\frac{1}{2}\\right)^{\\frac{x}{20}}=A_0 e^{\\frac{x}{20}\\cdot \\ln(\\frac{1}{2})}$\n\nNow taking the derivative gives\n\n$\\displaystyle f'(t)=\\left( {\\frac{1}{20}\\cdot \\ln(\\frac{1}{2})}\\right)A_0 e^{\\frac{x}{20}\\cdot \\ln(\\frac{1}{2})}=A_0\\left( {\\frac{1}{20}\\cdot \\ln(\\frac{1}{2})}\\right)\\left( \\frac{1}{2}\\right)^{\\frac{x}{20}}$\n\n3. Ah, I forgot to take the log of both sides and solve it much easier.\n\nOkay, so I derived the equation correctly? And the question asks me to find the rate of decay when $t = 100$ days. Knowing the derivative and the quantity at the time, wouldn't I just need to plug in 100 to find the rate of decay?\n\n4. Yes, f'(100) should tell you the rate of decay on the 100th day.\n\n5. Originally Posted by NOX Andrew\nYes, f'(100) should tell you the rate of decay on the 100th day.\nSo, is the rate of decay $-3.384 \u00d7 10^{-5}$ when $t = 100$?\n\n6. What is the original amount? It should be the original amount times -log(2)/640.\n\n7. Originally Posted by NOX Andrew\nWhat is the original amount? It should be the original amount times -log(2)/640.\nIt was not given. I only have $\\frac{1}{32}$ of the original substance to work with.", "date": "2014-08-01 01:48:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/172638-half-life.html", "openwebmath_score": 0.896321713924408, "openwebmath_perplexity": 366.47385944890794, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211626883622, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8553829437668021}}
{"url": "https://math.stackexchange.com/questions/960456/probability-of-exactly-two-heads-in-four-coin-flips", "text": "# Probability of exactly two heads in four coin flips?\n\nWhen you flip a coin four times, what is the probability that it will come up heads exactly twice?\n\nMy calculation:\n\n\u2022 we have $2$ results for one flip : up or down\n\u2022 so flip $4$ times, we have $4\\cdot2 = 8$ results total\n\nThus the probability is: $2/ 8 = 0.25$ but the correct answer is $0.375$. Can anyone explain why I'm wrong?\n\n\u2022 Nope, if you flip 4 times, there are $2^4$ possible outcomes. How many of these outcomes have two heads? \u2013\u00a0Irvan Oct 6 '14 at 8:15\n\nMy calculation:\n\nwe have 2 results for one flip : up or down so flip 4 times, we have 4x2 = 8 results total\n\nTwo results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is $2\\times 2\\times 2\\times 2$ results in total. That is $2^4$ or $16$.\n\nFor the favourable case we need to count the ways to get $2$ heads and $2$ tails. The count of permutations of two pairs of symbols is: $\\frac{4!}{2!2!}=6$. This is easily confirmed by just counting.\n\n$$\\Bigl|\\{\\mathsf {HHTT, HTHT, HTTH, THHT, THTH, TTHH}\\}\\Bigr|=6$$\n\nThus the probability is: $\\tfrac{\\;6}{16}$, or: $$0.375$$\n\n\u2022 Thanks, i got the idea, but i don't understand what \"!\" is ? \u2013\u00a0NeedAnswers Oct 6 '14 at 11:19\n\u2022 @hoangnnm It's the Factorial notation. $n!$ is the product of all integers less than or equal to $n$. $n!=(n)(n-1)(n-2)\\cdots(2)(1)$ en.wikipedia.org/wiki/Factorial \u2013\u00a0Graham Kemp Oct 6 '14 at 11:55\n\u2022 oh. i got it now. your anwser is easier for me to understand. Therefore, i will remark yours as answer! \u2013\u00a0NeedAnswers Oct 6 '14 at 13:28\n\nAssuming unbiased coin with probability of head $=\\dfrac12$\n\nand using Binomial Distribution, $$\\binom42\\left(\\frac12\\right)^2\\left(1-\\frac12\\right)^{4-2}$$\n\nUse binomial probability since there are only two possibilities: success and failure, where success represents getting a heads, and tails being a fail.\n\nLet $X$ = Success (i.e. heads)\n\nTherefore we are trying to find $P(X=2)$, which is $\\binom42\\cdot(0.5)^2\\cdot(0.5)^2=0.375$.\n\nHope this helped!\n\nThe derivation of binomial probability:\n\nGetting two heads out of 4 can be portrayed is, disregarding order:\n\nMultiplying their probabilities will yield $(0.5)^4$, but as for ordering, we get $4!/(2!\\cdot2!)$ due to repetition, which is the same as $4C2$. So our answer is $\\binom42\\cdot(0.5)^4$ which is $0.375$", "date": "2020-01-21 09:36:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/960456/probability-of-exactly-two-heads-in-four-coin-flips", "openwebmath_score": 0.9053215980529785, "openwebmath_perplexity": 416.37581892338494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211553734217, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8553829404715222}}
{"url": "https://math.stackexchange.com/questions/1103580/double-branch-sqrt-x-or-square-function-turned-90", "text": "# Double branch $\\sqrt x$ or square function turned 90\u00b0?\n\nI have this idea for a graph but don't know what function could describe it better.\n\nThe idea is something like the \"squared\" function turned $90$ degrees to the right, so that possible values for $x$ are always positive and $y$ may be both positive and negative.\n\nThe graphs of $\\sqrt x$ and $-\\sqrt x$ combined look good too, but I don't know how to write that as a single function (eh, I'm so bad with these things).\n\nBasically, anything that may look like this will do.\n\nLooking forward to some solution.\n\nWhat you're looking for can be described as a parabola opening towards the positive $x$ axis. I'm going to refer to it as a \"sideways parabola,\" since the \"standard\" parabola that people learn opens towards the positive $y$ axis.\n\nThe bad news: You cannot express a sideways parabola as a function of $x$. Why? Let's go back and look at a restriction on functions:\n\nVertical Line Test: For a relation to be a function, it must (colloquially) pass the vertical line test. That is, you must be able to draw a vertical line anywhere on the relation's graph and the line must intersect the relation in at most one point.\n\nIn the picture below, I've marked the two intersections that a vertical line makes on a sideways parabola. This shows that the sideways parabola is not a function.\n\nHowever, the good news is that one may still describe such a graph with mathematical notation. Two such ways are below, but keep in mind that they are not functions of $x$.\n\n$$x=y^2$$ $$y=\\pm\\sqrt{x}$$\n\n\u2022 okay, somehow i didn't know about the vertical line. this is a really good explanation, thank you. \u2013\u00a0Don Jan 14 '15 at 4:22\n\nThe graph you describe will have equation $x=y^2$.\n\nYou cannot write it in the form \"$y=\\langle\\hbox{function of$x$}\\rangle$\" because, just as you pointed out, every $x>0$ will correspond to two $y$ values, not just one.\n\n\u2022 Concise. Correct. +1. \u2013\u00a0MPW Jan 14 '15 at 4:55\n\nIf you want the square function $$y = x^2$$ turned 90 degrees to the right, you get the inverse $$x = y^2$$ or written in another way (like you already mentioned) $$y = \\pm \\sqrt{x}.$$\n\nI'm not actually sure what you mean by writing it as a single function as it has two branches (the positive and the negative one).\n\n\u2022 what i meant comes from the fact that i'm stuck with this program which will only accept input like y = sqrt(x), so there's no +/-, and giving it sqrt(x)+(-sqrt(x)) makes no sense of course. \u2013\u00a0Don Jan 14 '15 at 4:20\n\u2022 Well, in that case graphing a function with two branches like that seems to be impossible. \u2013\u00a0655321 Jan 14 '15 at 4:30\n\u2022 More accurately, $y = \\sqrt{x}$ is the inverse of $y = x^2, x \\geq 0$, while $y = -\\sqrt{x}$ is the inverse of $y = x^2, x \\leq 0$. The equation $x = y^2$ is not a function of $x$ since there are two values of $y$ for each value of $x > 0$. \u2013\u00a0N. F. Taussig Jan 14 '15 at 12:54\n\nThe curve obtained by combining the graphs of $y = \\sqrt{x}$ and $y = -\\sqrt{x}$ is $x = y^2$. It is not a function of $x$ since there are two values of $y$ for each $x > 0$. However, you can use the parametric equations \\begin{align*} x(t) & = t^2\\\\ y(t) & = t \\end{align*} to write both $x$ and $y$ as functions of a third variable $t$. Observe that $x(t) = [y(t)]^2$.\n\nBy using parametric equations to express both $x$ and $y$ as functions of $t$, you can describe curves that are not necessarily functions of $x$.", "date": "2020-12-03 14:39:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1103580/double-branch-sqrt-x-or-square-function-turned-90", "openwebmath_score": 0.8127148747444153, "openwebmath_perplexity": 190.7451724422293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290913825541, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8553676690677837}}
{"url": "https://math.stackexchange.com/questions/3304325/prove-a-implies-c/3304334", "text": "# Prove $A \\implies C$\n\nGiven, $$(A \\lor B) \\implies C$$, prove $$A \\implies C$$\n\nMy Proof:\n\n1 By Conditional Exchange,\n\n$$\\neg(A \\lor B) \\lor C$$\n\n2 By DeMorgan's Law,\n\n$$(\\neg A \\land \\neg B) \\lor C$$\n\n3 By Simplification,\n\n$$\\neg A \\lor C$$\n\n4 By Conditional Exchange,\n\n$$A \\implies C$$\n\nMy question pertains to steps 2 and 3. I used Demorgan's and Simplification on a subformula of a premise -- can I do that? Usually, I would separate the subformulas, but I don't think I could do so in this case.\n\nThanks.\n\n\u2022 Your steps look valid \u2013\u00a0J. W. Tanner Jul 26 '19 at 4:13\n\u2022 Alternative Proof: By disjunction, $A \\implies A \\vee B$. Since $A \\vee B \\implies C$, and since $A \\implies A \\vee B$, then $A \\implies C$ by hypothetical syllogism. \u2013\u00a0JavaMan Jul 26 '19 at 4:24\n\u2022 My logic is rusty, so take this with a grain of salt: but I do not see the problem. You are asked to prove \"If A, then C.\" so I do not see why you can't start by assuming $A$ is true. In math, to prove a statement of the form $P \\implies Q$, it is perfectly valid to assume $P$ is true, since if $P$ is false, then $P \\implies Q$ vacuously. If this is a rigorous (philosophical) logic course, then maybe I'm missing some details in logical formalism, so you could/should ask your teacher (or others here). \u2013\u00a0JavaMan Jul 26 '19 at 4:31\n\u2022 This isn't circular reasoning. Circular reasoning would be assuming that \"$A \\implies C$ is true to prove that $A \\implies C$ is true. Here, you are really proving that $A \\implies C$ using cases: whether $A$ is true or not. \u2013\u00a0JavaMan Jul 26 '19 at 4:35\n\n\\begin{align} &(A\\lor B)\\to C &&\\text{Premise} \\\\ \\iff & \\lnot(A \\lor B)\\lor C&& \\text{Conditional Exchange}\\\\\\iff & (\\lnot A\\land\\lnot B)\\lor C&&\\text{de Morgan's}\\\\\\iff &(\\lnot A\\lor C)\\land(\\lnot B\\lor C)&&\\text{Distribution}\\\\\\implies &\\lnot A\\lor C&&\\text{Simplification}\\\\ \\iff & A\\to C&&\\text{Conditional Exchange} \\end{align}\n\u2022 The obvious case where substituting a weakening does not work is when the phrase is in an antecedent. $~~(A\\land B)\\to B$ does not imply $A\\to B$. \u2013\u00a0Graham Kemp Jul 26 '19 at 4:34\nInstead, use distribution to get $$(\\lnot A\\lor C)\\land (\\lnot B\\lor C),$$ and then use simplification to get $$\\lnot A\\lor C.$$", "date": "2020-03-29 10:36:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3304325/prove-a-implies-c/3304334", "openwebmath_score": 0.9456675052642822, "openwebmath_perplexity": 820.8651524396134, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241982893259, "lm_q2_score": 0.8824278788223264, "lm_q1q2_score": 0.855358696187602}}
{"url": "https://freedocs.mi.hdm-stuttgart.de/sd1_sect_junitDetails.html", "text": "No. 116\n\n#### Summing up integers to a given limit\n\n Q: Suppose an arbitrary number n is given e.g n=5. We want to compute the sum of all integers ranging from 1 to 5: 1 + 2 + 3 + 4 + 5 = 15 Implement the following method by using a loop:/** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { ... }For the sake of getting used to it write some unit tests beforehand. BTW: Is it possible to avoid the loop completely achieving the same result? A: The solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within: Maven module source code available at sub directory P/Sd1/summing/V1 below lecture notes' source code root, see hints regarding import. Online browsing of API and implementation. We start by defining unit tests beforehand:/** * Testing negative values and zero. */ @Test public void testNonPositiveLimits() { assertEquals(0, Summing.getSum(-5)); assertEquals(0, Summing.getSum(0)); } /** * Testing positive values. */ @Test public void testPositiveLimits() { assertEquals(1, Summing.getSum(1)); assertEquals(3, Summing.getSum(2)); assertEquals(15, Summing.getSum(5)); assertEquals(5050, Summing.getSum(100)); // Carl Friedrich Gauss at school assertEquals(2147450880, Summing.getSum(65535));// Highest Integer.MAX_VALUE compatible value }The crucial part here is determining 1 + 2 + ... + 65535 being equal to 2147450880. Legend has it the young Carl Friedrich Gauss at school was asked summing up 1 + 2 + ... + 99 + 100. His teacher was keen for some peaceful time but the young pupil decided otherwise: $\\underbrace{1 + \\underbrace{2 + \\underbrace{3 + \\dots + 98}_{101} + 99}_{101} + 100}_{101}$ Since there are 50 such terms the result is $50 \u00d7 101 = 5050$. Generalizing this example we have: $\u2211 i = 1 n i = n \u2062 ( n + 1 ) 2$ For the current exercise we do not make use of this. Instead we implement Summing.getSum(...) by coding a loop: /** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { int sum = 0; for (int i = 1; i <= limit; i++) { sum += i; } return sum; } }This passes all tests.\n\nNo. 117\n\n#### Summing up, the better way\n\nQ:\n\nThe previous solution of Summing up integers to a given limit suffers from a performance issue. When dealing with large values like 65535 looping will take some time:\n\nlong start = System.nanoTime();\nSystem.out.println(\"1 + 2 + ... + 65535\" + \"=\" + getSum(65535));\nlong end = System.nanoTime();\nSystem.out.println(\"Elapsed time: \" + (end - start) + \" nanoseconds\");\n1 + 2 + ... + 65535=2147450880\nElapsed time: 1169805 nanoseconds\n\nBarely more than one millisecond seems to be acceptable. But using the method for calculations inside some tight loop this might have a serious negative performance impact.\n\nThus implement a better (quicker) solution avoiding the loop by using the explicit form. When you are finished re-estimate execution time and compare the result to the previous solution.\n\nProvide unit tests and take care of larger values. What is the largest possible value? Test it as well!\n\n### Tip\n\nA:\n\nThe solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within:\n\nSince only our implementation changes we reuse our existing unit tests. Our first straightforward implementation attempt reads:\n\npublic static long getSum (int limit) {\nreturn limit * (limit + 1) / 2;\n}\n\nThis fails both unit tests. The first error happens at:\n\nassertEquals(0, Summing.getSum(-5));\njava.lang.AssertionError:\nExpected :0\nActual   :10\n\nWe forgot to deal with negative limit values. Our sum is supposed to start with 0 so negative limit values should yield 0 like in our loop based solution:\n\npublic static long getSum(int limit) {\nif (limit < 0) {\nreturn 0;\n} else {\nreturn limit * (limit + 1) / 2; // Order of operation matters\n}\n}\n\nThis helps but one test still fails:\n\nassertEquals(2147450880, Summing.getSumUsingGauss(65535));\njava.lang.AssertionError:\nExpected :2147450880\nActual   :-32768\n\nThis actually is a showstopper for large limit values: The algebraic value of limit * (limit + 1) / 2 might still fit into an int. But limit * (limit + 1) itself not yet divided by 2 may exceed Integer.MAX_VALUE. Since the multiplication happens prior to dividing by 2 we see this overflow error happen.\n\nSolving this issue requires changing the order of operations avoiding arithmetic overflow. Unfortunately the following simple solution does not work either:\n\npublic static long getSum(int limit) {\nif (limit < 0) {\nreturn 0;\n} else {\nreturn limit / 2 * (limit + 1);\n}\n}\n\nThis is only correct if limit is even. Otherwise division by 2 leaves us with a remainder of 1.\n\nHowever if limit is uneven the second factor limit + 1 will be even. This observation leads us to the final solution:\n\npublic static long getSum(int limit) {\nif (limit < 0) {\nreturn 0;\n} else if (0 == limit % 2){       // even limit, divide by 2\nreturn limit / 2 * (limit + 1); // Avoiding arithmetic overflow\n} else {                          // uneven limit, divide (limit + 1 ) by 2\nreturn (limit + 1) / 2 * limit; // Avoiding arithmetic overflow\n}\n}\n\nThis passes all tests. We finally reconsider execution time:\n\n1 + 2 + ... + 65535=2147450880\nElapsed time: 25422 nanoseconds\n\nThus execution is roughly 46 times faster compared to the loop based approach. This is not surprising since loop execution is expensive in terms of performance.", "date": "2019-05-26 13:18:42", "meta": {"domain": "hdm-stuttgart.de", "url": "https://freedocs.mi.hdm-stuttgart.de/sd1_sect_junitDetails.html", "openwebmath_score": 0.5153611302375793, "openwebmath_perplexity": 1650.3504428077479, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241956308277, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8553586743604503}}
{"url": "https://math.stackexchange.com/questions/1966283/how-do-i-find-a-flaw-in-this-false-proof-that-7n-0-for-all-natural-numbers", "text": "# How do I find a flaw in this false proof that $7n = 0$ for all natural numbers?\n\nThis is my last homework problem and I've been looking at it for a while. I cannot nail down what is wrong with this proof even though its obvious it is wrong based on its conclusion. Here it is:\n\nFind the flaw in the following bogus proof by strong induction that for all $n \\in \\Bbb N$, $7n = 0$.\n\nLet $P(n)$ denote the statement that $7n = 0$.\n\nBase case: Show $P(0)$ holds.\n\nSince $7 \\cdot 0 = 0$, $P(0)$ holds.\n\nInductive step: Assume $7\u00b7j = 0$ for all natural numbers $j$ where $0 \\le j \\le k$ (induction hypothesis). Show $P(k + 1)$: $7(k + 1) = 0$.\n\nWrite $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$. Then, using the induction hypothesis, we get $7(k + 1) = 7(i + j) = 7i + 7j = 0 + 0 = 0$. So $P(k + 1)$ holds.\n\nTherefore by strong induction, $P(n)$ holds for all $n \\in \\Bbb N$.\n\nSo the base case is true and I would be surprised if that's where the issue is.\n\nThe inductive step is likely where the flaw is. I don't see anything wrong with the strong induction declaration and hypothesis though and the math adds up! I feel like its so obvious that I'm just jumping over it in my head.\n\n\u2022 Can you really always find natural numbers $i, j$ such that $0 \\leq i, j \\leq k$ and $i+j = k+1$? \u2013\u00a0Alex G. Oct 13 '16 at 4:01\n\u2022 No. The \"proof\" assumes that for any $k\\geq 0$, there exist natural numbers $i, j$ such that $0\\leq i, j\\leq k$ and $i+j = k+1$ \u2013\u00a0Alex G. Oct 13 '16 at 4:05\n\u2022 As a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem. In this case, the smallest failing case is $P(1)$, so the number to look at is $k+1=1$, that is, $k=0$. \u2013\u00a0celtschk Oct 13 '16 at 8:26\n\u2022 As a matter of English, you can disprove a false theorem (i.e. an untrue statement that is claimed to be a theorem) but not a false proof. The word you need is refute: a false proof (of a real or false theorem) may be refuted. A false theorem may also be refuted, by disproving it (though not by refuting a false proof of it). \u2013\u00a0John Bentin Oct 13 '16 at 12:48\n\u2022 en.wikipedia.org/wiki/All_horses_are_the_same_color \u2013\u00a0v7d8dpo4 Oct 13 '16 at 16:12\n\nThe problem is here:\n\nWrite $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$.\n\nIf $k = 0$, then you are trying to write $1 = i+j$ where $i$ and $j$ are natural numbers less than $1$. The only option for $i$ and $j$ is $0$, but $0+0 \\ne 1$.\n\nAs a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem.\n\nIn this case, the smallest failing case is $P(1)$, as that claims $7\\cdot 1=0$ which is clearly wrong. Therefore the number to look at is $k+1=1$, that is, $k=0$.\n\nSo let's look at the inductive step, and insert $k=0$:\n\nInductive step: Assume $7\\cdot j=0$ for all natural numbers $j$ where $0\\le j\\le 0$ (induction hypothesis). Show $P(k+1): 7(k+1)=0$.\n\nThe only number with $0\\le j\\le 0$ is $j=0$, so the induction hypothesis is that $7\\cdot 0=0$, which clearly is true.\n\nWrite $0+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.\n\nThe only natural number less than $1$ is $0$. Therefore we have to write $0+1 = 0+0$ \u2026 oops, that's not right! Error found!\n\n\u2022 This is the best answer to the actually posed question, how to find the error in an induction proof of an obviously false statement. The only thing to add is that this method will find at least one problem with the inductive argument, but there could also be other problems that appear at larger values of n. \u2013\u00a0zyx Oct 13 '16 at 20:46\n\u2022 @zyx by the well-ordering principle there is a smallest such problematic number... \u2013\u00a0djechlin Oct 13 '16 at 20:50\n\u2022 It's an interesting question whether all errors in the induction step can be found by examining a finite set of \"problematic numbers\" and whether this can be done effectively (given that the conclusion is false for all $n \\geq N$ for a known $N$). I am not sure if just considering larger and larger problematic numbers will catch everything. @djechlin \u2013\u00a0zyx Oct 13 '16 at 20:59\n\u2022 @zyx you're asking whether a proof can contain an infinite number of errors? I mean each step could also use the fact that $n^2 > n$ for all $n \\geq 0$, Then each step would have an error in it, and there would be an infinite number of errors. One could fix the error that $n^2 > n$ is false for $n > 0$ by replacing to the less obviously fallacious claim that $n^2 > n$ only when $n \\geq 1$, so that each error would also have a fix that would leave an infinite number of errors. \u2013\u00a0djechlin Oct 13 '16 at 21:05\n\u2022 @djechlin, the proof is finite, so has only a finite number of erroneous statements. I'm asking whether there is a form of the \"problematic numbers\" strategy that is guaranteed to uncover them all. I suspect there is not unless you have some way of computing, for every step of the proof, infinitely many $n$ where the inductive argument requires that step to be true in order for the implication $n \\to n+1$ to hold. \u2013\u00a0zyx Oct 13 '16 at 21:33\n\nActually, the problem is in the base case \u2014 in particular, $P(0)$ isn't enough of a base case.\n\nThe inductive step for proving $P(n)$ depends on writing $n$ as a sum of two smaller natural numbers; you can do this when $n \\geq 2$, but you can't do this when $n=1$.\n\nIf you have both $P(0)$ and $P(1)$ in the base case, that's enough to make the inductive step work.\n\n(of course, you can't prove $P(1)$, so you can't prove the base case)\n\n\u2022 This is true but problematic for people learning induction for the first time. You're right that fixing the proof (if it were possible) would require a bigger base case. But the actual error (as in, problematic line in the proof) is in the step case (which has been pointed out several times). \u2013\u00a0Richard Rast Oct 13 '16 at 15:57\n\nHint: $1=1+0\\neq 0+0$. Study $P(1)$.\n\nWhenever you have to check induction proofs, you should apply the general case in order to prove the first step of the induction. In this particular situation you want to prove P(1): $7*1 = 0$.\n\nWrite $k+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.\n\nIn the first step, this means:\n\nWrite $1 = i+j$ where $i,j$ are natural numbers less than $1$.\n\nThis statement already shows where the problem is in the induction proof, because the only natural number less than 1 is 0, and 1 cannot be expressed as $0 + 0$.\n\n\u2022 This is good advice, but not always sufficient; it's possible to write a bogus inductive \"proof\" where the general case only fails after two or more steps. For a simple example, consider an attempted proof that all odd numbers greater than 1 are primes, with the base case \"3 is prime\" and the (obviously false) induction step \"$n$ is prime $\\implies$ $n+2$ is prime\". \u2013\u00a0Ilmari Karonen Oct 13 '16 at 19:46", "date": "2021-03-01 13:43:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1966283/how-do-i-find-a-flaw-in-this-false-proof-that-7n-0-for-all-natural-numbers", "openwebmath_score": 0.8718101978302002, "openwebmath_perplexity": 153.0619533252059, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692264378964, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8553542068235536}}
{"url": "https://math.stackexchange.com/questions/1175288/solving-a-question-by-using-special-products-students-debate-to-teacher", "text": "# Solving a question by using special products (Students debate to Teacher)\n\nSo today,we got back our exam papers,and we found a question marked wrongly and teacher said that it is wrong.We all students do NOT believe this.So here is what happened.\n\nBefore reading the next part,this is what we ONLY know (learnt) about rules of special products.\n\n(Under school secondary 2 learning in Singapore)\n\n\u2022 Rule $1$: $a^2+2ab+b^2=(a+b)^2$\n\n\u2022 Rule $2$ :$a^2-2ab+b^2=(a-b)^2$\n\n\u2022 Rule $3$: $a^2-b^2=(a+b)(a-b)$\n\nFrom the exam paper:\n\nEvaluate $10.2^2-9.8^2$ by using ONLY rules of special products.\n\nCorrect solution: $(10.2+9.8)(10.2-9.8)=(20)(0.4)=8$ (Rule $3$)\n\nStudent wrong (Marked as wrong) alternate solution: \\begin{align*} (10+0.2)^2-(10-0.2)^2 &=[10^2+2(10)(0.2)+0.2^2]-[10^2-2(10)(0.2)+0.2^2]\\\\&=100+4+0.04-(100-4+0.04)\\\\&=104.04-96.04\\\\&=8\\end{align*}(Rules $1$ and $2$)\n\nThe question did NOT ask for the easiest and fastest way (and both solution uses ONLY rules of special products) to solve but yet why is student solution wrong? Teacher told us,\"Aiya, why need to do so complicated one?\" yet she did not answer why is the answer wrong. I debated to her so long but to no avail.\n\nCan anybody think of why the student solution is wrong?\n\n\u2022 Your solution is correct. She's butt-hurt because she failed to account for other possible correct solutions. \u2013\u00a0Git Gud Mar 4 '15 at 16:29\n\u2022 The teacher is silly ! Both are equivalent things done in two ways! \u2013\u00a0Shakul Pathak Mar 4 '15 at 16:33\n\u2022 The unexpected answer is well within the realm of correct. This teacher is out-of-bounds. I teach; I also like unexpected answers that work and strongly encourage this kind of thinking. \u2013\u00a0ncmathsadist Mar 4 '15 at 19:59\n\u2022 I think that the second approach is arguably better as it demonstrates knowledge of two of the three rules rather than just one. If you really want to split hairs you could argue that since the question refers to \"rules\", i.e. plural, the first answer is not even valid as it only uses a single rule! The question should have been disambiguated, e.g. \"using exactly one rule\", or \"using one or more rules\". The fact that this was not done indicates a lack of rigour. \u2013\u00a0Marconius Jul 9 '15 at 12:07\n\u2022 Whoops,forgot to look back at this post.This is solved.We got our exam marks :D \u2013\u00a0ministic2001 Oct 18 '15 at 12:45", "date": "2019-05-25 22:55:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1175288/solving-a-question-by-using-special-products-students-debate-to-teacher", "openwebmath_score": 0.6144951581954956, "openwebmath_perplexity": 1564.7102691447199, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692318706084, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8553542052486898}}
{"url": "https://math.stackexchange.com/questions/2661789/euler-function-and-sums", "text": "# Euler function and sums\n\nI've come across a nice little combinatorial problem. Firstly we have vector $(1, 1)$. On each iteration for each two consecutive numbers in the vector we add their sum between them: $$(1, 2, 1)$$ $$(1, 3, 2, 3, 1)$$ $$(1, 4, 3, 5, 2, 5, 3, 4, 1)$$ $$(1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1)$$ ad infinitum.\n\nHere's the question. How many times number $n$ will be written in the final vector? I'v computed the answers for $1, \\dots, 15$: $$2, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8$$\n\nAnd this values coincide with the values of Euler's totient function. But how to prove that the answer is always $\\phi(n)$? And can one simply derive this idea without explicitly writing first values?\n\n\u2022 For anyone who wants to explore this with Mathematica, v={1,2,1}; S[v_]:=Table[v[[k]]+v[[k+1]],{k,1,Length[v]-1}]; F[v_]:=Riffle[v,S[v]]; Nest[F,v,n] will give the nth vector \u2013\u00a0Benedict W. J. Irwin Feb 22 '18 at 14:35\n\u2022 If we start with the vector $(1,2)$, we seem to end up with $\\lfloor n/\\tau(n)\\rfloor$, i.e. OEIS sequence A078709, where $\\tau(n)$ is the number of divisors function. \u2013\u00a0Benedict W. J. Irwin Feb 22 '18 at 14:58\n\u2022 Ok, the iteration from $(1,2)$ I just suggested is not true like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,4,4,8,3,9,4, whereas the real sequence A078709 goes like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,3,3,8,3,9,3, so they are not the same, but very close. Your pattern for $\\phi(n)$ seems to hold at least to $n=26$ \u2013\u00a0Benedict W. J. Irwin Feb 22 '18 at 15:11\n\u2022 These are the denominators in Farey Series. They generate all fractions between $0/1$ and $1/1$. You get the numerators by starting with $0,1$ instead of $1,1$. All these fractions turn out to be in lowest terms, so $\\tau(n)$ of them will have $n$ in the denominator. \u2013\u00a0Michael Feb 22 '18 at 15:52\n\nLet's first name the vectors $v_1, v_2$ and so on.\n\nFirst of all we can note that consecutive numbers in the vector are coprime by induction. The base case is trivial. Now assume it holds for $v_{n-1}$ and consider $v_n$. Let $a,b$ are consecutive number in $v_n$. Then because of the way numbers are added to the vectors we have $a,|b-a|$ or $b, |a-b|$ appearing as consecutive numbers in $v_{n-1}$. WLOG let it be the first option. But then $\\gcd(a,b) = \\gcd(a,|b-a|) =1$. Hence the proof.\n\nNow let's notice that a number $n$ can appear in the vector if $a,b$ appear as consecutive integer in a vector and $a+b = n$. But from above we have that neither of $a,b$ can share a common factor of $n$. Now using induction on $n$ we'll prove that such a combination of consecutive integers $(a,b)$ can appear only in a unique vector. The base case for $n=1,2,3$ is true. Now assume that we have consecutive integers $a,b$ appearing in different vectors, namely $v_i$ and $v_j$ such that both sum to $n$. WLOG let $a<b$. Then going back in $v_{i-1}$ and $v_{j-1}$ we have the consecutive integers $a, b-a$ in both vectors. But this contradicts the inductive hypothesis as both pairs sum to $b<n$.\n\nFrom here we can conclude that an integer $n$ will be at most $\\phi(n)$ in the final vector. Now it remains to show that each combination of $a,b$ s.t. $\\gcd(a,b) = 1$ appears as consecutive integers at some point.\n\n[UPDATE]\n\nSimilar as above we'll prove the last claim by induction on $n$. The base cases $n=1,2,3$ are trivially true. Now assume the claim holds for all numbers less than $n$. Now let $a,b$ be any integers such that $a+b=n$ and $\\gcd(a,b)=1$ and WLOG $a<b$. We know that they will appear in $v_i$ if and only if $a,b-a$ appear as consecutive integers in $v_i$. But by the inductive hypothesis we have that $a,b-a$ does appear as consecutive integers in a unique vector. Therfore $a,b$ appear too and in a unique vector. Hence the proof.\n\nSUMMARY: We can count the pairs of the consecutive integers by their left member and summarizing from above we have that if $a<n$ and $\\gcd(a,n) = 1$, then eventually the integers $a,n-a$ will appear only once in the vectors. Note that the case $n-a,a$ is counted by using $n-a$. Therefore each integers appears $\\phi(n)$ times in the final vector.\n\n\u2022 @Igor I just completed my answer, so I would be glad if you could check and possibly point out some mistakes, if any. \u2013\u00a0Stefan4024 Feb 22 '18 at 15:22\n\u2022 Thank you for such a detailed answer. \u2013\u00a0Igor Feb 22 '18 at 23:02", "date": "2019-08-25 03:32:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2661789/euler-function-and-sums", "openwebmath_score": 0.8603589534759521, "openwebmath_perplexity": 133.45084305161907, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990587409856637, "lm_q2_score": 0.8633916029436189, "lm_q1q2_score": 0.8552648516518894}}
{"url": "https://stats.stackexchange.com/questions/511014/expected-value-of-rolling-mean-of-ar1-process", "text": "# Expected Value of rolling mean of AR(1) process\n\nConsider a random variable following an AR(1) model:\n\n$$x_t = \\mu+\\rho x_{t-1} + \\epsilon_t$$\n\nThe unconditional expected value of this proces is:\n\n$$E(x)=\\mu/(1-\\rho)$$\n\nNow take a rolling mean of $$x_t$$ of n periods including the current: $$\\bar{x} = \\frac{1}{n}\\sum_{i=0}^{n-1} x_{t-i}$$\n\nBased on Wiki: Sum of normally distributed random variables the expected value is still just the sum of the expected value for each $$x_t$$ which is given above, hence the expected value of $$\\bar{x}_t$$ is:\n\n$$E(\\bar{x}_t)=1/n \\sum_{i=0}^{n-1} E(x_{t-i})=n/n E(x_t)=E(x_t)$$\n\nHowever, another approach is to roll back each observation in the summand to a common starting point using (this is with thanks to Aleksej at this post here): $$x_t = \\mu \\sum_{i=0}^{k-1} \\rho^i + \\sum_{i=0}^{k-1} \\rho^i \\epsilon_{t-i} + \\rho^k x_{t-k}$$\n\nInserting this into the expression for $$\\bar{x}_t$$ yields: $$\\bar{x}_t=1/n\\sum_{i=0}^{n-1}(\\rho^{n-i}x_{t-n}+\\mu\\sum_{j=0}^{k-i-1}\\rho^j+\\sum_{j=0}^{k-i-1}\\rho^{j}\\epsilon_{t-i-j})$$\n\n$$=1/n(\\sum_{i=0}^{n-1}x_{t_n}\\rho^{n-i}+ \\mu\\sum_{i=0}^{n-1}\\sum_{j=0}^{n-i-1}\\rho^{j}+\\sum_{i=0}^{n-1}\\sum_{j=0}^{n-i-1}\\rho^j\\epsilon_{t-i-j})$$\n\nTaking the expection to this yields: $$E(\\bar{x})=1/n(E[x_{x-t}]\\sum_{i=0}^{n-1}\\rho^{n-i}+\\mu\\sum_{i=0}^{n-1}\\sum_{j=0}^{n-i-1}\\rho^{j})$$\n\n$$=\\frac{1}{n}(\\frac{\\mu}{1-\\rho}\\sum_{i=0}^{n-1}\\rho^{n-i}+\\mu\\sum_{i=0}^{n-1}\\sum_{j=0}^{n-i-1}\\rho^{j})$$ But this does not coincide with the previous results using the sum of normal distributed variables.\n\nCan someone point out, if I have made a mistake and which method is the correct one?\n\nEDIT: Simulation study\n\nBased on the comment from mlofton, I have conducted a simulation study, and I find that the expected difference and the simulated difference are indeed very close. Using the formula from the model yields the same as average of the sum of the unconditional means for $$x_t$$\n\nimport matplotlib.pyplot as plt\nimport numpy as np\n\ndef simulate_ar_process(initial_value=None, intercept=-0.01, slope=0.5, noise_vol=0.005, obs=100):\n# If initial value is None use the unconditional mean\nif initial_value is None:\ninitial_value = intercept / (1-slope)\n\nerrors = np.random.normal(0, noise_vol, obs)\npd_levels = []\n\n# Estimate PD-levels\nfor err in errors:\nif len(pd_levels) == 0:\npd_levels.append(initial_value + err)\nelse:\nnew_obs = intercept + pd_levels[-1] * slope + err\npd_levels.append(new_obs)\n\nreturn pd_levels\n\ndef calculate_rolling_mean(pd_levels, look_back=20, burn_in=0):\nmas = []\nfor i in range(look_back+burn_in, len(pd_levels)):\ncurrent_range = pd_levels[i-look_back:i]\nmas.append(np.mean(current_range))\n\nreturn np.mean(mas)\n\nintercept = -0.01\nslope = 0.9\nnoise_vol = 0.005\nobs = 1000\nlook_back = 20\n\naverage_of_rolling_means = []\naverage_of_x = []\nfor i in range(100):\nx = simulate_ar_process(intercept=intercept, slope=slope, noise_vol=noise_vol, obs=obs)\naverage_of_rolling_means.append(calculate_rolling_mean(x, look_back=look_back))\naverage_of_x.append(np.mean(x[20:]))\n\nplt.hist(average_of_x, label=\"Average of x\", alpha=0.3)\nplt.hist(average_of_rolling_means, label=\"Average of rolling mean\", alpha=0.3)\nplt.legend()\n\n# Calcualte difference\ndiffs = [x-y for x, y in zip(average_of_x, average_of_rolling_means)]\nplt.hist(diffs)\nnp.average(diffs)\n\n# Theoretical diff\nunconditional_mean = intercept / (1-slope)\nunconditional_mean_mva = unconditional_mean * np.sum([slope**(look_back-i) for i in range(0, look_back)])\nunconditional_mean_mva += intercept * np.sum([slope ** j for i in range(0, look_back) for j in range(0, look_back-i)])\nunconditional_mean_mva /= look_back\n\nprint(\"Expected difference: {}\".format(unconditional_mean - unconditional_mean_mva))\nprint(\"Found difference: {}\".format(np.average(diffs)))\n#Expected difference: 0.0\n#Found difference: 9.324487397407016e-06\n\n\nHowever, I cannot see from the formula above, that $$E(\\bar{x})=E(x)$$ Can someone point me in the right direction in the deviation of this equality?\n\n\u2022 Hi: I wouldn't expect them to be the same because, for the one where you use the result from the Wiki, you're not taking the model into account. The Wiki result would be true for any iid random variable which is normally distributed. The second result-derivation applies due to the model specifics for $x_t$ and $x_t$ is not iid given the model. \u2013\u00a0mlofton Feb 24 at 18:00\n\u2022 Hi: Thank you for your comment. I have updated my answer with a simulation study. If you post your original comment and perhaps can give a pointer or two with the remaining part, I'll mark it as the answer. \u2013\u00a0RVA92 Feb 25 at 7:47\n\u2022 @mlofton: the linearity of expectation always hold, regardless of the dependence structure between the random variables. \u2013\u00a0Dayne Feb 26 at 5:50\n\u2022 Nice derivation and you are absolutely correlation about the correlation structure. My mistake. But when the OP wrote $E(\\bar x) = E(x_{i})$, the expectation of the $x_t$ in an AR(1) are not the same unless one refers to the long term unconditional mean which is exactly what you used in your nice derivation. @RVA92: When you wrote the expression for $\\bar{x}$, I didn't know that you were referring to the unconditional mean of $x$. The expression does hold in that case. After I send this, I delete my other comments and answer since I was totally confused by the equality. \u2013\u00a0mlofton Feb 27 at 18:18\n\u2022 In case, above confuses things even more, I was referring to E(x_t) given the previous $x_t$. That expectation is not constant. because of the dependence structure in the AR(1). For the long term mean, you're absolutely correct. \u2013\u00a0mlofton Feb 27 at 18:21\n\nThere seems to be an error in your calculations. Using a slightly different notation for clarity.\n\nLet the AR(1) process be: $$X_t = \\mu + \\phi X_{t-1} + e_t$$\n\nDefine mean of $$n$$ periods be:\n\n$$\\mu_n \\equiv \\frac{1}{n}\\sum\\limits_{t=1}^n X_t$$\n\nThe expectation operator is linear regardless of dependence structure between the random variables. So, your first expression is correct:\n\n$$\\mathbb E(\\mu_n) = \\mathbb E(X_t) = \\frac{\\mu}{1-\\phi}$$\n\nSecond approach, for $$t>1$$:\n\n\\begin{align} X_t &= \\mu + \\phi X_{t-1} + e_t \\\\ &= \\phi^{t-1}X_1+ \\sum\\limits_{i=0}^{t-2} \\Big(\\phi^{i}\\mu +\\phi^{i}e_{t-i}\\Big) \\tag{1} \\end{align}\n\nUsing $$(1)$$: \\begin{align} n\\mu_n &= \\sum\\limits_{t=1}^n\\phi^{t-1}X_1 +\\sum\\limits_{t=2}^n \\sum\\limits_{i=0}^{t-2} \\Big(\\phi^{i}\\mu +\\phi^{i}e_{t-i}\\Big) \\\\ &= \\sum\\limits_{t=1}^n\\phi^{t-1}X_1 +\\sum\\limits_{t=2}^n \\sum\\limits_{i=0}^{t-2} \\phi^{i}\\mu +\\sum\\limits_{t=2}^n \\sum\\limits_{i=0}^{t-2}\\phi^{i}e_{t-i} \\tag{2} \\end{align}\n\nIn equation (2) the middle term is non-random and the last term has expected value of $$0$$. So, \\begin{align} n\\mathbb E(\\mu_n) &= \\sum\\limits_{t=1}^n\\phi^{t-1}\\mathbb E(X_1) + \\sum\\limits_{t=2}^n \\sum\\limits_{i=0}^{t-2} \\phi^{i}\\mu \\\\ &= \\frac{\\mu}{1-\\phi}\\sum\\limits_{t=1}^n\\phi^{t-1} + \\sum\\limits_{t=2}^n \\sum\\limits_{i=0}^{t-2} \\phi^{i}\\mu \\\\ &= \\frac{\\mu}{1-\\phi}\\sum\\limits_{t=1}^n\\phi^{t-1} + \\sum\\limits_{t=2}^n \\bigg( \\frac{\\mu(1-\\phi^{t-1})}{1-\\phi}\\bigg) \\\\ &= \\frac{\\mu}{1-\\phi} \\bigg( \\sum\\limits_{t=1}^n\\phi^{t-1} + \\sum\\limits_{t=2}^n (1-\\phi^{t-1})\\bigg) \\\\ &=n \\frac{\\mu}{1-\\phi} \\end{align}\n\n\u2022 Hi Dayne: I got it to match using your approach, so thank you. \u2013\u00a0RVA92 Feb 26 at 7:46", "date": "2021-07-26 05:33:06", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/511014/expected-value-of-rolling-mean-of-ar1-process", "openwebmath_score": 0.9983998537063599, "openwebmath_perplexity": 1713.009428634471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307700397332, "lm_q2_score": 0.8791467659263147, "lm_q1q2_score": 0.8552610252740378}}
{"url": "https://complex-analysis.com/content/curves_in_the_complex_plane.html", "text": "# Curves in the Complex Plane\n\nSuppose the continuous real-valued functions $x = x(t),$ $y = y(t),$ $a \\leq t \\leq b,$ are parametric equations of a curve $C$ in the complex plane. If we use these equations as the real and imaginary parts in $z = x+iy$, we can describe the points $z$ on $C$ by means of a complex-valued function of a real variable $t$ called a parametrization of $C$: $$\\label{parcurve} z(t) = x(t) + i y(t), \\quad a\\leq t\\leq b.$$\n\nThe point $z(a) = x(a) + i y(a)$ or $z_0 = (x(a), y(a))$ is called the initial point of $C$ and $z(b) = x(b) + iy(b)$ or $z_1 = (x(b), y(b))$ is its terminal point. The expression $z(t) = x(t) + iy(t)$ could also be interpreted as a two-dimensional vector function. Consequently, $z(a)$ and $z(b)$ can be interpreted as position vectors. As $t$ varies from $t = a$ to $t = b$ we can envision the curve $C$ being traced out by the moving arrowhead of $z(t)$. This can be appreciated in the following applet with $0 \\leq t\\leq 1$.\n\nPress Start to animate. You can move the points to change the curve.\n\nFor example, the parametric equations $x = \\cos t,$ $y = \\sin t,$ $0 \\leq t \\leq 2\\pi,$ describe a unit circle centered at the origin. A parametrization of this circle is $z(t) = \\cos t + i \\sin t,$ or $z(t) = e^{it},$ $0 \\leq t \\leq 2\\pi$.\n\nPress Start to animate.\n\n## Contours\n\nThe notions of curves in the complex plane that are smooth, piecewise smooth, simple, closed, and simple closed are easily formulated in terms of the vector function (\\ref{parcurve}). Suppose the derivative of (\\ref{parcurve}) is $z'(t) = x'(t) + iy'(t).$ We say a curve $C$ in the complex plane is smooth if $z'(t)$ is continuous and never zero in the interval $a \\leq t \\leq b.$ As shown in Figure 2, since the vector $z'(t)$ is not zero at any point $P$ on $C$, the vector $z'(t)$ is tangent to $C$ at $P$. Thus, a smooth curve has a continuously turning tangent; or in other words, a smooth curve can have no sharp corners or cusps. See Figure 2.\n\nA piecewise smooth curve $C$ has a continuously turning tangent, except possibly at the points where the component smooth curves $C_1, C_2, \\ldots, C_n$ are joined together.\n\nA curve $C$ in the complex plane is said to be a simple if $z(t_1) \\neq z(t_2)$ for $t_1 \\neq t_2,$ except possibly for $t = a$ and $t = b.$ $C$ is a closed curve if $z(a) = z(b).$\n\n$C$ is a simple closed curve if $z(t_1)\\neq z(t_2)$ for $t_1\\neq t_2$ and $z(a) = z(b).$\n\nIn complex analysis, a piecewise smooth curve $C$ is called a contour or path. We define the positive direction on a contour $C$ to be the direction on the curve corresponding to increasing values of the parameter $t$. It is also said that the curve $C$ has positive orientation. In the case of a simple closed countour $C$, the positive direction corresponds to the counter-clockwise direction. For example, the circle $z(t) = e^{it}$, $0 \\leq t \\leq 2\\pi$, has positive orientation. The negative direction on a contour $C$ is the direction opposite the positive direction. If $C$ has an orientation, the opposite curve, that is, a curve with opposite orientation, is denoted by $\u2212C$. On a simple closed curve, the negative direction corresponds to the clockwise direction. For instance, the circle $z(t) = e^{-it}$, $0 \\leq t \\leq 2\\pi$, has negative orientation.\n\nPress Start to animate. You can change the direction of $C$.\n\nExercise: There is no unique parametrization for a contour $C$. You should verify that \\begin{eqnarray*} z(t)&=&e^{it} =\\cos t+i\\sin t,\\; 0\\leq t\\leq 2\\pi \\\\ z(t)&=& e^{2\\pi it} =\\cos \\left(2 \\pi t\\right) +i \\sin \\left(2 \\pi t\\right),\\; 0\\leq t\\leq 1\\\\ z(t)&=&e^{\\pi/2 it} =\\cos \\left( \\frac{\\pi}{2} t\\right)+i \\sin \\left( \\frac{\\pi}{2} t\\right),\\; 0\\leq t \\leq 4 \\end{eqnarray*} are all parametrizations, oriented in the positive direction, for the unit circle $|z| = 1$.\n\nNEXT: Complex Integration", "date": "2021-08-04 12:12:27", "meta": {"domain": "complex-analysis.com", "url": "https://complex-analysis.com/content/curves_in_the_complex_plane.html", "openwebmath_score": 0.9740369319915771, "openwebmath_perplexity": 165.63861928868005, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9948603882928907, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.85524540726822}}
{"url": "http://zjkf.freunde-des-historischen-schiffsmodellbaus.de/how-to-evaluate-nth-roots.html", "text": "# How To Evaluate Nth Roots\n\nPlease try again later. The unit fraction notation used for roots previously may have given you the idea that roots are really the same as powers, only with a unit fraction (one over some number) instead of an integer as the exponent. Rational Exponents The nth root of a number can be expressed by using radical notation or the exponent 1 1n. 1 Evaluate nth Roots and Use Rational Exponents. With millions of qualified respondents, SurveyMonkey Audience makes it easy to get survey responses from people around the world instantly, from almost anyone. These are Euclidean distance, Manhattan, Minkowski distance,cosine similarity and lot more. \u2022 The value of a square root can be approximated between integers. (Where did the 2 go? For square roots, the 2 is assumed. Notice that the cursor will stay under the radical sign until you press the right-arrow key (see the last line of the third screen). Active 4 years, 6 months ago. How to find nth term of the sequence ? There is some arrangement or pattern followed in every sequence. If a = bn, then b is an nth root of a. Each root gives a particular exponential solution of Each root gives a particular exponential solution of the di\ufb00erential equation. Whenever numbers are preceded with a radical sign, the numbers are called radicals Whenever numbers are preceded with a radical sign, the numbers are called radicals. Viewed 1k times 3 $\\begingroup$ I am working on a problem. View Notes - Unit_3_Notes from MATH Algebra 2 at Central York Hs. Consider the quadratic equation A real number x will be called a solution or a root if it satisfies the equation, meaning. When solving this kind of problem change the original to its corresponding. Secondary_Dim[,nth_Dim]. Any nth root is an exponentiation by 1/n, so to get the square root of 9, you use 9**(1/2) (or 9**0. Showing top 8 worksheets in the category - Nth Root. This feature is not available right now. Reciprocal of a number, including fractions. Here is a function that calculates the nth-root properly for negative values, where value is the number which will be rooted by n :. And if we take a product of a bunch of stuff and. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. , in response to message #2 by kim. T d mMnaMdpe i 1w ti WtnhI SIfn xf NiRn 7i6t zeP tPFrFex-ZAMlwgQe4b frRau. Polynomials are used so commonly in algebra, geometry and math in general that Matlab has special commands to deal with them. 1 nth Roots and Rational Exponents 401 nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. 5 Graph Square Root and Cube Root Functions. com is going to be the ideal site to take a look at!. Online graphing calculator (2): Plot your own graph (SVG) \u00b7 6. How To Simplify fourth roots. See these links: an example of using division method for finding cube root, and information about the nth root algorithm (or paper-pencil method). Find the quadratic sequences nth term for these 4 sequences which are separated by the letter i iii 7 10 15 22 21 42 iii 2 9 18 29 42 57 iii 4 15 32 55 85 119 iii 5 12 27 50 81 120?. In short, you can make use of the POWER function in Excel to find the nth root of any number. CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. First we develop the square root method in table form and the see that how it is work after that We have developed division method of square root in formula form. The nth root of a number can be expressed by using radical notation or the exponent 1 n. So a number to the two-thirds power is the cube root of the number squared. A square root (\u221a) of number x is one which when multiplied by itself gives a value x. The nth root of a number is a number such that if you multiply it by itself (n-1) times you get the number. The most common root is the square root. Rationalize the denominator: Other Kinds of Roots We define the principal nth root of a real number a, symbolized by, as follows:. The optimum frequency ratio of the middle 3rd with the perfect 1st is 1. Exception: If n is odd and the radicand is negative, the principal nth root is negative. real number is a real number. Complex numbers can be written in the polar form z = re^{i\\theta}, where r is the magnitude of the complex number and \\theta is the argument, or phase. 2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. Here is a function that calculates the nth-root properly for negative values, where value is the number which will be rooted by n :. Engaging math & science practice! Improve your skills with free problems in 'Evaluate nth Roots and Use Rational Exponents' and thousands of other practice lessons. Predict the effects of transformations [f(x + c), f(x) + c, f(cx), and cf(x), where c is a positive or negative real-valued constant] algebraically and graphically, using various methods and tools that may include. exponent 1/n refers to the nth root numbers: 16^1/4 = 4th root of 16 to power of 1. For the elements of X that are negative or complex, sqrt(X) produces complex results. A c PA IlqlC Brzi Bg whxtSs K mrSeLs OeJruv Ne7dz. 245 #5\u00ad17 odds. com brings invaluable answers on nth term calculator, complex numbers and multiplying and dividing and other algebra topics. The keys for the parentheses are above the 8 and 9. 404-405 4-61 every 3rd; 3 Evaluating Nth Roots. Step 3: Write the answer using interval notation. A General Note: Rational Exponents. For these types of food demand cooking, you have to have a burner or any portable heating accessory. The nth root of a number is the number that would have to be multiplied by itself n times to get the original number. 3 27 1 Solution: a. That is, root(4)(81)  = 81^((1)/(4)) Here 81= 3^(4). Principal nth root of a number a, symbolized byn a , where n \u22652 is an integer, is defined as follows: \u2022 If n \u22652and even, then a and b must be greater than or equal to 0. 5) to get the cube root, you use 9 ** (1/3) (which we can't write with a simpler fraction), and to get the nth root, 9 ** (1/n). Background. A square root (\u221a) of number x is one which when multiplied by itself gives a value x. \uf07d 32 = 9\uf07d 102 = 100\uf07d 2172 = 47089. The seventh root of 16,384 is 4, as 4 x 4 x 4 x 4 x 4 x 4 x 4 is 16,384. Divisor calculator, simpifying nth roots, download 6th grade math, t 89 calculator base conversion, holt algebra 1 answer key. 1 nth Roots and Rational Exponents Algebra 2 Mrs. You might say. In this problem, we can evaluate by using the following method. To use the calculator simply type any positive number into the 'enter number' box then type in the 'nth root' you want to find. A General Note: Rational Exponents. An nth root can be denoted by a radical symbol with an index. Algebra II Review 6. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. Y = nthroot(X,N) returns the real nth root of the elements of X. If the length of p is n+1 then the polynomial is described by:. By primitive expressions, I mean + - * / sqrt, unless there are others that I am missing. EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. In mathematics and computing, a root-finding algorithm is an algorithm for finding roots of continuous functions. We often find these type of expressions in science text books. The third (optional) argument is a root selector. Partial fractions of repeated roots of degree 2. \u2b50\ufe0f\u2b50\ufe0f\u2b50\ufe0f\u2b50\ufe0f\u2b50\ufe0f Shop for cheap price Brass How To Make 6. But perhaps you don't have a calculator, or you want to impress your friends with the ability to calculate a. Checkpoint Find the indicated real nth root(s) of a. The nth root of a number is a number such that if you multiply it by itself (n-1) times you get the number. Hello Weber School District Parents, Teachers, and Staff, On March 15th, 2019, the server that housed our Wordpress Blogs has been discontinued. Evaluating IRU n = 12, yields, which simplifies to 2 1 or 2. homework help with finding nth roots and rational expressions Oct 26, 2011 \u00b7 I have no idea how to solve these equations and my teacher is a terrible one. Irrational numbers include the square root, cube root, fourth root, and nth root of many numbers. For instance, 2 is a cube root of 8 because 23 = 8, and 3 is a fourth root of 81 because 34 = 81. Algebrator free, 'add subtract multiply divide' worksheets, properties of exponents lesson plan. Eigenvalueshave theirgreatest importance in dynamic problems. In this resource from CK-12 we look at how to evaluate nth roots. Solve equations using nth roots. Then find the cube root. When we want to find what number was squared, we are finding a square root. They write expressions using rational exponents and radical notation. We could use the nth root in a question like this:. \u2014 One real nth root: \u2014 a = O One real nth root: \u2014 o \u2014 One real nth root: \u2014 Taking roots of negative numbers (JULL SOC \u20142 2 Evaluate nth. In mathematics and computing, a root-finding algorithm is an algorithm for finding roots of continuous functions. Later in this section we will see that using exponent 1 n for nth root is compat- ible with the rules for integral exponents that we already know. Evaluate Nth Degree Polynomial For A Given Value Of X Apr 8, 2014. 12,921 views subscribe 5. However, this only works for n that are powers of 2. Type in calculator with base and exponent in parentheses 3. We first write the given function as an equation as follows y = \u221a(x - 1) Square both sides of the above equation and simplify y 2 = (\u221a(x - 1)) 2 y 2 = x - 1 Solve for x x = y 2 + 1 Change x into y and y into x to obtain the inverse function. ) in irrational number form. For the form x^(m/n) this is the same as (x^m)*1/n which is also the same thing as the nth root of x^m So for 8^(2/3). and bis positive, there are two real nth roots of b. So that's what I'll multiply onto this fraction. The cube root of 8, then, is 2, because 2 \u00d7 2 \u00d7 2 = 8. Take the square root, you get 3, which is back where you started. So something, times something, times something, is 8. com and figure out equations, equation and several other math topics. How to Divide Square Roots. Because 3 2 = 9, 2 3 = 8, and (-2) 4 = 16, we say that 3 is a square root of 9, 2 is the cube root of 8, and -2 is a fourth root of 16. Rewrite as a radical 2. Evaluate Nth root of a rational to a correctly rounded float. In a shorter form \u201cb\u201d is the cube root of \u201ca\u201d if b^3 = a. Compare Price and Options of Brass How To Make 6. This module provides a number of objects (mostly functions) useful for dealing with Chebyshev series, including a Chebyshev class that encapsulates the usual arithmetic operations. \"Nth Power of Pingala Chanda\" by Ranjani Chari, July 2013 This shows a method that was given by Pingala Chanda in 200 BC of raising a number to some power. In short, you can make use of the POWER function in Excel to find the nth root of any number. Title: nth Roots and Rational Exponents 1 Chapter 7. Click the oroblem to show tr Evaluate the expression. Gre-test-prep. 33333 or 10 1/3. Cubes are the result of three multiplications. You can do this with other roots as well, but they have to be the same roots. Identify and evaluate square and cube roots. evaluate freplacing vars by their value eval(f) Select Pieces of an Object square/nth root of xsqrt ( ), sqrtn x,n,&z trig functions sin, cos, tan, cotan. Then we\u2019ll use these exact values to answer the above challenges. menu-burger { display: none. cube root of 7 multiplied by the square root of 7 over sixth root of 7 to the power of 5 7-1 1 7 7 to the power of 5 by 3. How are roots and radicals related? Number or variable Written as a square 36 16 81 x. For example, the cube root of 8 is written like this: The index, 3, indicates the radical is a cube. nth root of an even powered radicand and the result is has an odd power, you must take the absolute value of the result to ensure that the answer is non\u00adnegative. Find 16 Use a calculator. for each, give (a) the z-score cutoff (or cutoffs) on the comparison distribution at which the null hypothesis should. This Number Sense Worksheet may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. 6862x 4 + 46. When you take the square root of a number, you're looking for a number that, when multiplied by itself, results in a given number. The product of two polynomials of degree-bound n is a polynomial of degree-bound 2n. A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. There is a basic formula to follow when taking the integral of an expression with a power 1/(n+1) x^(n+1). More generally, odd roots of negative numbers are typically assumed to be complex. Finding Real nth Roots of a (n > 1 and n is an even integer) Finding Real nth Roots of a (n > 1 and n is an odd integer) Evaluating Expressions with Rational Exponents. Please try again later. 86 21 Take fourth roots of each side. 1 nth Roots and Rational Exponents 401 nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. Section 1-5 : Integrals Involving Roots. I CANNOT use a calculator. A c PA IlqlC Brzi Bg whxtSs K mrSeLs OeJruv Ne7dz. If some numbers in x are negative, n must be odd. Using this formula, we will prove that for all nonzero complex numbers $z \\in \\mathbb{C}$ there exists \\$n. 2 in the text. They can be used instead of the roots such as the square root. notebook March 24, 2015 a. Having an nth root of some number is equivalent to taking that number to the 1/n power. guarantees that a zero or root of f(x) lies in (a,b). Press [2nd][ x 2 ] to select a square root and type the expression you would like to evaluate. Write an explicit or recursive formula for the nth term of an arithmetic sequence, given the value of several of its terms. the group we get is 10 and 648 find the no. We find nth term of the sequence in term of n. It looks quite tedious to do by hand, but the algorithm exists for any root and is similar to the square root one. MATLAB represents polynomials as row vectors containing coefficients ordered by descending powers. Rationalizing the Denominator. 23 = 8 53 = 125 1713 = 5000211 4. \"Novel Algorithm for 'Nth Root of Number' using Multinomial Expansion\" by Vitthal Jadhav, March 2013 This gives a general algorithm to extract the nth root of any number. Is it possible to calculate n complex roots of a given number using Python? I've shortly checked it, and it looks like Python gives me wrong/incomplete answers: (-27. We\u2019ll start with integer powers of $$z = r{{\\bf{e}}^{i\\theta }}$$ since they are easy enough. The most common root is the square root. The function thus has a branch cut along the negative half real axis. 12,921 views subscribe 5. Notice that the cursor will stay under the radical sign until you press the right-arrow key (see the last line of the third screen). First we develop the square root method in table form and the see that how it is work after that We have developed division method of square root in formula form. whose cube root is less than or equal to 10. Identify and evaluate square and cube roots. Evaluate all powers from left to right. If the length of p is n+1 then the polynomial is described by:. Algebrator free, 'add subtract multiply divide' worksheets, properties of exponents lesson plan. chebyshev)\u00b6 New in version 1. For example, 4 is a square root of 16 because 4 \u2022 4 or 42 = 16. com - id: 77bdd1-YThlY. Thus, we can think of taking the square root of any positive number, x. Fast Fourier Transformation for poynomial multiplication. Later in this section we will see that using exponent 1 n for nth root is compat- ible with the rules for integral exponents that we already know. Introduction to nth Roots; Radical Form vs. Step 1: Set the expression inside the square root greater than or equal to zero. yaymath 118,922 views. Algebra 2 Standard 12. If n = 2, the root is called square root. In this case, the power 'n' is a half because of the square root and the terms inside the square root can be simplified to a complex number in polar form. This is the video about how to evaluate square roots. 2 Simplify expressions in exponential form. We\u2019ve already seen some integrals with roots in them. homework help on finding nth roots and rational exponents Before you read/watch/listen to \u201cIf You Can Read This I Can. Let's look at some: 1. Topical and themed;. This tool is used to calculate the output of almost all the mathematical expressions. And the square root of 25 times 3 is equal to the square root of 25 times the square root of 3. Unfortunately some improper integrals fails to fall under the scope of these tests but we will not deal with them here. How to Find Roots of Unity. Students will simplify nth roots completely, then they will color or draw on the penguin as indicated by their answer. Evaluate the line integral C of F dr, where C is given by the vector function r (t). A square root is defined as a number which when multiplied by itself gives a real non-negative number called a square. Rewrite the power of 16 as products of power of 8 as much as possible. The advantage of using exponents to express roots is that the rules of exponents can be applied to the expressions. Each root gives a particular exponential solution of Each root gives a particular exponential solution of the di\ufb00erential equation. Evaluating Roots of Monomials To evaluate nth roots of monomials: (where c is the coefficient, and x, y and z are variable expressions) n cxyz n c n 1 n x n 1 n y n 1 n z (c ) ( x ) ( y ) ( z ) 1 n or \u2022 Simplify coefficients (if possible) \u2022 For variables, evaluate each variable separately. The optimum frequency ratio of the middle 3rd with the perfect 1st is 1. 4 Exponents with negative bases. The most common root is the square root. The answers are that -2 is just one of the three cube roots of -8, and that Mathematica, the computational engine of Wolfram|Alpha, has always chosen the principal root, which is complex valued. Our review of these techniques will focus on the manual entry of formulas, but check out our tutorial on using Excel if you need a refresher on formula entry for core functions. Then f(x) changes sign on [a,b], and f(x) = 0 has at least one root on the interval. In this case, we divided by a negative number, so had to reverse the direction of the inequality symbol. If x is a unit, then it is a (primitive) k-th root of unity modulo n, where k is the multiplicative order of x modulo n. In this maths tutorial, we introduce exponents / powers and roots using formulas, solved examples and practice questions. when n is an odd integer. Evaluate expressions with rational exponents. ' m Igil s NewVocabulary nth root principal nth root 1 Simplify Radicals A square root of a number is one of two equal factors of that number. The square root is an example of a fractional exponent. 1 Evaluate Nth Roots and Use Rational Exponents. So a number to the two-thirds power is the cube root of the number squared. We find the fourth root in the same way and generalize this for nth roots. For Your Notebook n is even integer. Step 3: Write the answer using interval notation. Powers When we wish to multiply a number by itself we use powers, or indices as they are also called. The cube root is a specific calculation that any radical calculator can perform. 3 27 1 Solution: a. The following table shows some perfect cubes and cube roots. M j DM8a SdPe m ow kistBh6 UIIn fjipnSiFt je Q wG Je Lodm EeRtwriy b. Roshan's Algebra 2 Class Videos -- Based on McDougal Littell's Algebra 2. If the length of p is n+1 then the polynomial is described by:. Start studying 8. 1 Evaluating polynomials at many points Suppose that we want to evaluate a polynomial A(x) = a 0 + a is an nth root of unity (and so. Indeed, one reason for choosing such a transformation for an equation with multiple roots is to eliminate known roots and thus simplify the location of the remaining roots. If the index/root is: No Soluh'oa even & radicand is negative, there is odd & radical is negative, the solution follows the the No matter what the index/root is, if the radicand is POSITIVE, there is ALWAYS a solution!!!. If the sample size calculator says you need more respondents, we can help. Evaluating nth roots Myhre Math MCHS. Roots of unity Properties. The properties of fourth root says that for any positive number of a, its fourth roots are real. exam Numerical Ability Question Solution - I need help please!! I need to evaluate the expressions - nth roots and rational exponents - don't understand - hope I type these right 343^-1/3 729^5/6 (-512)^-2/3 1^5/7 1. Download Presentation 3. One real root:. Evaluate nth roots. 12,921 views subscribe 5. The voice explains how to first plug in the numbers given for each variable in the fractions. i have an exam in the morning an must know how to use it. Evaluate exponent Examples Evaluating nth root on calculator 1. Finding nth Roots: Evaluating nth Root Expressions: way 25 2313 2 26 8 23 8 4throot 34 81 nthroot of a nthroot nra nra radical index Negative Index far a o Indexodd Indexeven bn a 8 27 bn a but46 onereal root tworeal roots Another 8 B d TEB C 3 fzjfzDb 1WE y 2 21 i 20 456 4ft try augno Norealrootsbecauseyoucan't multiplyanumberbyitself4times. College Algebra (10th Edition) answers to Chapter R - Section R. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. If then b is the nth root of a ; Example ; Notation ; Index (of the radical) The number n outside of the radical sign ; 3 Writing nth roots as powers and powers as nth roots. Notes on Fast Fourier Transform Algorithms & Data Structures Dr Mary Cryan 1 Introduction The Discrete Fourier Transform (DFT) is a way of representing functions in terms of a point-value representation (a very speci\ufb01c point-value representation). What is a function?. 1 Evaluate nth roots and use rational exponents. For example, the sixth root of 729 is 3 as 3 x 3 x 3 x 3 x 3 x 3 is 729. 86 21 Take fourth roots of each side. n th Roots. 25 different faces laid out in an A3 poster that can be folded down to a size of a business card. 23 = 8 53 = 125 1713 = 5000211 4. Since 2 = 8 , we say that 2 is the cube root of 8. Of course, the presence of square roots makes the process a little more complicated, but certain rules allow us to work with fractions in a relatively. 404-405 4-61 every 3rd; 3 Evaluating Nth Roots. Since the nth root of a real or complex number z is z1/n, the nth root of r cis \u03b8 is r1/n cis \u03b8/n. exponent 1/n refers to the nth root numbers: 16^1/4 = 4th root of 16 to power of 1. CAGR is calculated by taking the Nth root of the total percentage growth rate where N is the Number of Years in the period being considered. Evaluate an expression with complex numbers using an online calculator. So split the number inside the fourth root as the product of two perfect squares and then cancel out the power with the fourth root giving its roots. The shifting nth root algorithm is an algorithm for extracting the nth root of a positive real number which proceeds iteratively by shifting in n digits of the radicand, starting with the most significant, and produces one digit of the root on each iteration, in a manner similar to long division. For example, 81 3 and 3 8 both represent the cube root of 8, and we have 81 3 3 8 2. 8 2/3 = ( 8) 2 = ( 8 2 ) This says the cube root of 8 squared. \uf07d 23 = 8\uf07d 53 = 125\uf07d 1713 = 5000211. Fifth Roots. Times 3 to the 1/5. In addition to square roots, you have other nth roots (eg cube roots). Place tiles equal to the expression to the right of the = in the right workspace. This paper focusesattention on developing a numerical algorithm to determine the digit-by-digit extraction of the nth root of a given positive real number up to any desiredaccuracy. Now we will study higher order roots, such as cube roots. 3c Use the properties of exponents to transform expressions for exponential functions. In this post We are discussing about how to find nth last node of single linked list. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. Range variable a. Stack Exchange Network. Plug into calculator On calculator, Solving equations 1. The basic rule is that similar signs multiplied result in a positive answer. Rewrite the power of 16 as products of power of 8 as much as possible. Some can be done quickly with a simple Calculus I substitution and some can be done with trig substitutions. 4999999999999998j) but proper roots should be 3 complex numbers, because every non-zero number has n different complex number nth roots. Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. That's one cube root. printLevelorder makes use of printGivenLevel to print nodes at all levels one by one starting from root. Click Create Assignment to assign this modality to your LMS. In general, for an integer n greater than 1, if b^n=a, then b is an n th root of a. We've already seen some integrals with roots in them. In general, for an integer ngreater than 1, if bn= a, then bis an An nth root of ais written as na, where nis the of the radical. Finding square roots and converting them to exponents is a relatively common operation in algebra. When a number has more than one root, the radical sign indicates only the principal, or positive, root. So let's imagine taking 8 to the 1/3 power. Computes the n-th root real numbers of a numeric vector x, while x^(1/n) will return NaN for negative numbers, even in case n is odd. Finding Real nth Roots of a (n > 1 and n is an even integer) Finding Real nth Roots of a (n > 1 and n is an odd integer) Evaluating Expressions with Rational Exponents. Instruction Manual for Scientific Calculator and then click \"Simplify_Radical\" nth root: a) Evaluate log 12 245 ==> Input Log a (12; 245. Then we select pairs of the divisors: 2 * 3 * 3 = 18 324/2 = 162 162/2 = 81 We obtain a pair of 2 from the above. Print these charts and use it for homeschooling or classroom purposes. Odd Roots (of variable expressions)* When evaluating odd roots (n is odd) do not use absolute values. I have one copy of the factor 5 in the denominator. Find the indicated real nth root(s) of a negative. De\ufb01nition of a1/n If n is any positive integer, then a1 n n a, provided that n a is a real number. 3 27 1 Solution: a. VOCABULARY. Socratic Meta Featured Answers To evaluate the #nth# root of a complex number I would first convert it into trigonometric form: See all questions in Roots of. 1 Evaluate Nth Roots and use Rational Exponents Things you should be able to do: - Rewrite radical expressions using rational exponent notation. Math is Fun Curriculum for Algebra 2. We can write as _____. Sometimes you may be smarter than the computer. The bottom number is the nth root. If the length of p is n+1 then the polynomial is described by:. 1 Nth Roots and Rational Exponents Objectives: How do you change a power to rational form and vice versa? How do you evaluate radicals and powers with rational exponents? How do you solve equations involving radicals and powers with rational exponents?. Background. Polynomials are used so commonly in algebra, geometry and math in general that Matlab has special commands to deal with them. 1 Evaluate nth Roots and Use Rational Exponents An Image/Link below is provided (as is) to download presentation. The nth root is the same as the (1/n) power. Then find the cube root. Derivative at a Point Calculator Find the value of a function derivative at a given point. Similarly we studied one method to evaluate the cube root by factor method, but the method of finding cube root of very large numbers by factorizing becomes lengthy and difficult. The symbol indicates an nth root. If one would like to have unique solutions in terms of cosines for output-formatting purposes, then one could do something like. Round your answer to the nearest hundredth. Powers and Roots In this section we\u2019re going to take a look at a really nice way of quickly computing integer powers and roots of complex numbers. He also explains ways that would not be helpful in solving the problem and comparing. Why is the magnitude of the sum of two adjacent nth roots always an 'interesting' number, and what do these numbers have to do with each other? Hot Network Questions Is refusing to concede in the face of an unstoppable Nexus combo punishable?. For example, if the expression is 3 x \u2013 2, place 3 green x tiles and 2 red 1 tiles in one half of the workspace. Exponent Calculator to Calculate Base Raised to nth Power This calculator will calculate the answer of a base number raised to n th power, including exponential expressions having negative bases and/or exponents. 14^2/5 write in radical notation what is (root 5 of 14)^2. roots\u00b6 numpy.", "date": "2020-04-02 18:54:39", "meta": {"domain": "freunde-des-historischen-schiffsmodellbaus.de", "url": "http://zjkf.freunde-des-historischen-schiffsmodellbaus.de/how-to-evaluate-nth-roots.html", "openwebmath_score": 0.7313296794891357, "openwebmath_perplexity": 558.5759163333647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9843363494503271, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8552177872568134}}
{"url": "https://dsp.stackexchange.com/questions/29177/metric-spaces-why-l-infty-selects-the-maximum-value/29193", "text": "# Metric Spaces: Why $L_\\infty$ selects the maximum value\n\nI have a basic question about the metric spaces. There are several metric spaces like $L_1$, $L_2$ to $L_\\infty$. The $L_p$ metric is defined by the following equation:\n\n$$d_p(x,y)=\\left(\\sum_{i=1}^{n}|x_i-y_i|^p \\right)^{1/p}$$\n\nNow, when $p = \\infty$, then the equation of the matrix is following:\n\n$$d_{\\infty}(x,y)=\\max_{i=1,2,\\ldots, n}|x_i-y_i|$$\n\nNow, I am taking the value of $x$ as $x = [1,2,3]^T$ (Transpose) and compute $L_p$ metric for, $p = 1,2$ and $\\infty$.\n\nMy question is, why $L_\\infty$ metric choose the maximum value.\n\n\u2022 i wish you would use $\\LaTeX$ so i could copy your equation to save time in answering. please learn the tools. \u2013\u00a0robert bristow-johnson Mar 3 '16 at 2:53\n\u2022 I assume you have to go through a mathematical proof of this to fully understand it. Please do not make the mistake and try to explain this result by setting p to higher and higher values - this is an intuitive and often very useful thing to do, but it needs not give you the correct result. \u2013\u00a0M529 Mar 3 '16 at 9:18\n\u2022 Shouldn't this migrate to math.SE? \u2013\u00a0Carl Witthoft Mar 3 '16 at 13:07\n\u2022 No need for migration, the answer is already on math.SE: math.stackexchange.com/questions/109615/\u2026 \u2013\u00a0Jazzmaniac Mar 3 '16 at 14:19\n\nIn the general case, let $$x = (x_1,\\dots,x_n)$$ be a finite-length vector (in a finite dimensional space). The finite sequence of absolute values $$|x_i|$$ does attain its maximum (because the sequence is finite), denoted $$M = \\max_i |x_i|$$.\n\nLet $$m$$ be the (exact) number of coordinates in $$x = (x_1,\\dots,x_n)$$ whose absolute value is equal to $$M$$. Thus, $$1\\le m\\le n$$. Then, we can lower and upper bound the $$\\ell_p$$ norm of $$x$$ as follows: $$(mM^p)^{\\frac{1}{p}}\\le\\ell_p(x)\\le (nM^p)^{\\frac{1}{p}}\\,.$$\n\nBoth $$m^\\frac{1}{p}$$ and $$n^\\frac{1}{p}$$ tend to $$1$$ as $$p\\to\\infty$$, thus $$\\ell_p(x) \\to M$$, by the squeeze theorem.\n\n[EDIT] To provide more concrete substance, let us see what happen with your example: $$x=[1,2,3]^T$$ (the transposition does not change the result): $$\\|x\\|_p = d_p(x,0) = (1^p+2^p+3^p)^{1/p} = 3\\times\\left(\\left(\\frac{1}{3}\\right)^p+\\left(\\frac{2}{3}\\right)^p+1\\right)^{1/p}\\,.$$\n\nBoth $$\\left(\\frac{1}{3}\\right)^p$$ and $$\\left(\\frac{2}{3}\\right)^p$$ tend to $$0$$ as $$p \\to \\infty$$. Thus, $$\\|x\\|_p \\to 3$$.\n\n\u2022 That's the stuff! :) \u2013\u00a0Jazzmaniac Mar 3 '16 at 16:53\n\nThe $L_p$ norm is\n\n$$d_p(\\mathbf{x}, \\mathbf{y}) \\triangleq \\left( \\sum\\limits_{i=1}^{n} |x_i - y_i|^p \\right)^{\\frac{1}{p}}$$\n\nthere exists a positive value that is the maximum value:\n\n$$M \\triangleq \\max_{1 \\le i \\le n} |x_i - y_i|$$\n\nnow, suppose you divide both sides of the $L_p$ norm definition by that positive value,\n\n$$\\frac{d_p(\\mathbf{x}, \\mathbf{y})}{M} = \\frac{1}{M} \\left( \\sum\\limits_{i=1}^{n} |x_i - y_i|^p \\right)^{\\frac{1}{p}}$$\n\n\\begin{align} \\frac{d_p(\\mathbf{x}, \\mathbf{y})}{M} & = \\frac{1}{M} \\left( \\sum\\limits_{i=1}^{n} |x_i - y_i|^p \\right)^{\\frac{1}{p}} \\\\ & = \\left( \\frac{1}{M^p} \\right)^{\\frac{1}{p}} \\left( \\sum\\limits_{i=1}^{n} |x_i - y_i|^p \\right)^{\\frac{1}{p}} \\\\ & = \\left( \\frac{1}{M^p} \\sum\\limits_{i=1}^{n} |x_i - y_i|^p \\right)^{\\frac{1}{p}} \\\\ & = \\left( \\sum\\limits_{i=1}^{n} \\frac{1}{M^p} |x_i - y_i|^p \\right)^{\\frac{1}{p}} \\\\ & = \\left( \\sum\\limits_{i=1}^{n} \\left( \\frac{|x_i - y_i|}{M} \\right)^p \\right)^{\\frac{1}{p}} \\\\ \\end{align}\n\nnow ask yourself, what will happen to the right-hand side of this equation as $p \\to \\infty$? all terms, except for the term that is equal to the maximum, will have value of less than 1. they will go to 0 as $p \\to \\infty$, but the term that has $|x_i - y_i| = M$, that term is equal to 1 and will remain 1 even as $p \\to \\infty$.\n\n\u2022 Robert, what if there are two terms in the sum that are equal to $M$? Shouldn't the sum be equal to 2 in that case? \u2013\u00a0MBaz Mar 3 '16 at 3:48\n\u2022 hi Robert, thanks. It's a very good answer. So, Suppose, I have two matrix: x = [1,2,3]T and Y = 0. And, I want to compute, L1, L2 and L-infinity. So, here, Max value (M according to your explanation) would be 3? Right? Just to make sure. Thanks. \u2013\u00a0Odrisso Mar 3 '16 at 4:11\n\u2022 This is a very nice and intuitive explaination. However, I suppose it is not 100% mathematically sound, which is why @MBaz asked the question with the factor of 2, if your vector had two entries corresponding to M. I think the problem stems from not obeying the limit from the start on. Getting the factor 1/M into the brackets of the sum is critical and probably the point where it fails when p goes to infinity. \u2013\u00a0M529 Mar 3 '16 at 9:14\n\u2022 You could maybe argue that the sum s is bounded between 1 and n. Hence you have lim p->inf s^(1/p) which is 1. \u2013\u00a0M529 Mar 3 '16 at 9:25\n\u2022 @David, it doesn't matter if it's trivial. If you make false assumptions about the nature of something and don't explicitly consider the case that these assumptions fail, your argument is wrong. Yes, mathematics is pedantic. That's why it's so useful. \u2013\u00a0Jazzmaniac Mar 4 '16 at 19:05", "date": "2020-08-15 04:58:05", "meta": {"domain": "stackexchange.com", "url": "https://dsp.stackexchange.com/questions/29177/metric-spaces-why-l-infty-selects-the-maximum-value/29193", "openwebmath_score": 0.9943037629127502, "openwebmath_perplexity": 388.8520161938971, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363512883317, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8552177738089866}}
{"url": "https://mathhelpboards.com/threads/6-successive-numbers-no-one-is-prime.2005/", "text": "# Number Theory6 Successive numbers no one is prime\n\n#### Amer\n\n##### Active member\nis it possible to find a 6 Successive numbers like\nx , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?\n\nThanks\n\n#### chisigma\n\n##### Well-known member\nis it possible to find a 6 Successive numbers like\nx , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?\n\nThanks\nIt is comfortable to verify that $x=5!=120$ satisfies the request... and the reason of that is easy to see...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### chisigma\n\n##### Well-known member\nOf course $x=5!$ is not the only and neither the 'smallest' solution. Setting $x=90$ You have 7 consecutive non prime numbers...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### chisigma\n\n##### Well-known member\nWhy don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### chisigma\n\n##### Well-known member\nWhy don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...\nAn easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \\implies k|(n+m k)$. Setting $n = 2 \\cdot 3 \\cdot 5 \\cdot ... \\cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...\n\n$\\displaystyle k=11 \\implies n=2310 \\implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\\ \\text{are all non prime }$\n\nAlthough 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### Amer\n\n##### Active member\nThanks very much, you are amazing. (f)\n\n#### Amer\n\n##### Active member\nAn easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \\implies k|(n+m k)$. Setting $n = 2 \\cdot 3 \\cdot 5 \\cdot ... \\cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...\n\n$\\displaystyle k=11 \\implies n=2310 \\implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\\ \\text{are all non prime }$\n\nAlthough 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...\n\nKind regards\n\n$\\chi$ $\\sigma$\nnice one I get it\n$$n +2 = 2.3.4...k +2 = 2(3.4...k +1 )$$ not prime\n$$n+3 = 2.3.4...k + 3 = 3(2.4...k+1)$$ not prime\nThanks\n\n#### soroban\n\n##### Well-known member\nHello, Amer!\n\nIs it possible to find a 6 consecutive integers numbers like\nx , x+1 , x+2 , x+3 ,x+4 ,x+5 such that not one is prime? . Yes!\n\nOne solution is: .$$x \\:=\\:7!+2$$\n\n. . $$\\begin{array}{c}7!+2\\text{ is divisible by 2} \\\\ 7!+3\\text{ is divisible by 3} \\\\ 7!+4\\text{ is divisible by 4} \\\\ 7!+5\\text{ is divisible by 5} \\\\ 7!+6 \\text{ is divisible by 6} \\\\ 7!+7\\text{ is divisible by 7} \\\\ \\end{array}$$\n\nThere is a simpler (and much longer) list:\n\n. . $$\\begin{array}{ccc} 114 &=& 2\\cdot57 \\\\ 115 &=& 5\\cdot 23 \\\\ 116 &=& 2\\cdot 58 \\\\ 117 &=& 3\\cdot39 \\\\ 118 &=& 2\\cdot59 \\\\ 119 &=& 7\\cdot17 \\\\ 120 &=& 2\\cdot60 \\\\ 121 &=& 11\\cdot11 \\\\ 122 &=& 2\\cdot61 \\\\ 123 &=& 3\\cdot41 \\\\ 124 &=& 2\\cdot62 \\\\ 125 &=& 5\\cdot25 \\\\ 126 &=& 2\\cdot63 \\end{array}$$\n\n#### Amer\n\n##### Active member\nHello, Amer!\n\nOne solution is: .$$x \\:=\\:7!+2$$\n\n. . $$\\begin{array}{c}7!+2\\text{ is divisible by 2} \\\\ 7!+3\\text{ is divisible by 3} \\\\ 7!+4\\text{ is divisible by 4} \\\\ 7!+5\\text{ is divisible by 5} \\\\ 7!+6 \\text{ is divisible by 6} \\\\ 7!+7\\text{ is divisible by 7} \\\\ \\end{array}$$\n\nThere is a simpler (and much longer) list:\n\n. . $$\\begin{array}{ccc} 114 &=& 2\\cdot57 \\\\ 115 &=& 5\\cdot 23 \\\\ 116 &=& 2\\cdot 58 \\\\ 117 &=& 3\\cdot39 \\\\ 118 &=& 2\\cdot59 \\\\ 119 &=& 7\\cdot17 \\\\ 120 &=& 2\\cdot60 \\\\ 121 &=& 11\\cdot11 \\\\ 122 &=& 2\\cdot61 \\\\ 123 &=& 3\\cdot41 \\\\ 124 &=& 2\\cdot62 \\\\ 125 &=& 5\\cdot25 \\\\ 126 &=& 2\\cdot63 \\end{array}$$\nThanks, and Hello\nwe can make a list with k numbers which are not prime like that\n$$k! + i$$ i=1,...,k", "date": "2021-12-08 00:15:34", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/6-successive-numbers-no-one-is-prime.2005/", "openwebmath_score": 0.8835648894309998, "openwebmath_perplexity": 2765.1554110321154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353585837, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8552177691185715}}
{"url": "https://dsp.stackexchange.com/questions/25336/discrepancy-when-calculating-lti-system-output-using-inverse-z-transform", "text": "# Discrepancy when calculating LTI system output using inverse z-Transform\n\nI'm given a difference equation, $y[n]-0.4y[n-1]=x[n]$, and asked to find the natural response $y_n[n]$, forced response $y_f[n]$ and complete response $y[n]$ if $x[n]=4 (0.25)^nu[n]$ and $y[0]=0$.\n\nBy one approach I've read, $$\\text{characteristic equation:}\\quad z-0.4=0 \\quad\\therefore y_n[n]=A(0.4)^n \\\\ \\\\ y_f[n]=B(0.25)^n \\quad\\text{substitute back into diffeq:}\\\\ B(0.25)^n -0.4B(0.25)^{n-1}=B(0.25)^n - \\frac{0.4}{0.25}B(0.25)^n= 4(0.25)^nu[n]\\\\ \\therefore B=-\\frac{20}{3}, \\quad y_f[n]=-\\frac{20}{3}(0.25)^nu[n]\\\\ \\therefore y[n] = A(0.4)^n-\\frac{20}{3}(0.25)^nu[n]\\\\ y[0]=0 \\rightarrow A=\\frac{20}{3} \\\\ y[n]= \\frac{20}{3}(0.4)^n-\\frac{20}{3}(0.25)^nu[n]\\\\$$ Great! Alternatively, I should be able to determine $y[n]$ by taking the inverse z Transform of $Y(z)=H(z)X(z)$, which in this case is (I'm pretty sure) Y(z)= \\frac{z}{z-0.4}\\cdot 4 \\frac{z}{z-0.25}\\\\ \\begin{align} \\frac{Y(z)}{z} &= \\frac{4z}{(z-0.4)(z-0.25)}=\\frac{C}{z-0.4}+\\frac{D}{z-0.25}\\\\ &=\\frac{32}{3} \\frac{1}{z-0.4} - \\frac{20}{3}\\frac{1}{z-0.25}\\\\ \\end{align}\\\\ \\therefore Y(z)=\\frac{1}{3} \\left[\\frac{32z}{z-0.4} - \\frac{20z}{z-0.25}\\right]\\\\ \\therefore y[n]= \\left[ \\frac{32}{3}(0.4)^n - \\frac{20}{3}(0.25)^n \\right] u[n] which is almost the same, but not quite.\n\nWhere did I go wrong? Also, in the second approach, how do you take into account the initial condition of $y[0]=0$? And thirdly, in the first approach, is the first term of $y[n]$ multiplied by $u[n]$ or not? If so, how do you show this?\n\n(for anyone wondering, this is not homework for a course, just self study...)\n\n\u2022 Don't you mean $y[-1]=0$ instead of $y[0]=0$? \u2013\u00a0Matt L. Aug 20 '15 at 6:54\n\u2022 The question says y[0]=0 but I guess its mistaken. But its not clear to me why y[0] can't be zero... \u2013\u00a0Westerley Aug 20 '15 at 17:02\n\nThe way to solve such problems is to use the unilateral $\\mathcal{Z}$-transform, which allows you to take initial conditions into account. Note that the unilateral $\\mathcal{Z}$-transform of $y[n-1]$ is\n\n$$\\mathcal{Z}\\{y[n-1]\\}(z)=z^{-1}Y(z)+y[-1]\\tag{1}$$\n\nwhere $Y(z)$ is the $\\mathcal{Z}$-transform of $y[n]$. Using $(1)$ you can transform the given difference equation $y[n]-\\frac25 y[n-1]=x[n]$, resulting in\n\n$$Y(z)=\\frac{X(z)}{1-\\frac{2}{5}z^{-1}}+\\frac{\\frac{2}{5}\\cdot y[-1]}{1-\\frac{2}{5}z^{-1}}\\tag{2}$$\n\nWith $X(z)$ given by\n\n$$X(z)=\\frac{4}{1-\\frac14z^{-1}}\\tag{3}$$\n\nEq. $(2)$ becomes (after partial fraction expansion)\n\n$$Y(z)=\\left(\\frac{32}{3}+\\frac{2y[-1]}{5}\\right)\\frac{1}{1-\\frac{2}{5}z^{-1}}-\\frac{20}{3}\\frac{1}{1-\\frac14z^{-1}}\\tag{4}$$\n\nFrom $(4)$, the resulting output sequence is\n\n$$y[n]=\\left(\\frac{32}{3}+\\frac{2y[-1]}{5}\\right)\\left(\\frac25\\right)^nu[n]-\\frac{20}{3}\\left(\\frac14\\right)^nu[n]\\tag{5}$$\n\nEq. $(5)$ is the general form of the output sequence of the given system with the given input sequence. The value $y[-1]$ is the initial condition of the system, i.e. the state the system is in right before the input signal starts. With $y[-1]=0$ you obtain the same solution as you obtained using the inverse $\\mathcal{Z}$-transform (neglecting any initial conditions by using $y[n-1]\\Longleftrightarrow z^{-1}Y(z)$).\n\nWith the given constraint $y[0]=0$, you can obtain the required value of $y[-1]$ by considering the difference equation at $n=0$:\n\n$$y[0]-\\frac25 y[-1]=x[0]=4$$\n\nwhich gives $y[-1]=-10$. Plugging that value into Eq. $(5)$ finally gives the result\n\n$$y[n]=\\frac{20}{3}\\left(\\frac25\\right)^nu[n]-\\frac{20}{3}\\left(\\frac14\\right)^nu[n]\\tag{6}$$\n\nAs Matt L pointed out, your initial condition is wrong. y[0] is NOT 0, it's 4. This is obvious from the difference equation. Once you use y[0] = 4, both results come out to be the same.\n\n\u2022 As mentioned in my comment to @matt, the question does say y[0]=0 but it's likely wrong... but its not clear to me why y[0] CAN'T be zero... I see that x[0] is 4, but why can't the natural response at n=0 cancel out that? \u2013\u00a0Westerley Aug 20 '15 at 17:03\n\u2022 Note that $y[0]$ can be zero by an appropriate choice of the initial condition $y[-1]$ (see my answer). \u2013\u00a0Matt L. Aug 21 '15 at 7:48", "date": "2019-06-25 14:58:46", "meta": {"domain": "stackexchange.com", "url": "https://dsp.stackexchange.com/questions/25336/discrepancy-when-calculating-lti-system-output-using-inverse-z-transform", "openwebmath_score": 0.9442422389984131, "openwebmath_perplexity": 456.6080097815554, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363512883317, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8552177671224374}}
{"url": "https://math.stackexchange.com/questions/4484607/number-of-ways-to-divide-10-children-into-two-teams-of-5", "text": "# Number of ways to divide $10$ children into two teams of $5$?\n\nConsider the following two examples from A First Course in Probability by Sheldon Ross,\n\nEXAMPLE 5b Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible?\n\nSolution by the Book itself There are $$\\frac{10!}{5!5!} = 252$$ possible divisions.\n\nMy Solution and Reasoning We first choose five children which can be done in $$C(10, 5)$$ possible ways. Then we decide which team(either A or B) to assign these children to, which can be done in $$2!$$ ways. Finally we assign the remaining five children to the remaining team, depending on the second level(if the first five children were assigned to A, then the remaining five should be assigned to B, and if the first five were assigned to B, the remaining to A). Therefore this level has only $$1$$ possible way, and hence the final answer is $$C(10, 5) \\times 2! \\times 1 = 252 \\times 2 \\times 1 = 504$$.\n\nEXAMPLE 5c In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?\n\nSolution by the Book itself Note that this example is different from Example 5b because now the order of the two teams is irrelevant. That is, there is no A and B team, but just a division consisting of 2 groups of 5 each. Hence, the desired answer is\n\n$$\\frac{10!/(5!5!)}{2!} = 126$$\n\nMy Solution and Reasoning First we choose 5 children for one team which can be done is $$C(10, 5)$$ ways and then we put the remaining children in the other team, which is done in $$1$$ way. Thus the answer is $$C(10, 5) \\times 1 = 252$$, which can also be obtained by dividing the answer to Example 5b by 2, or in other words, removing the order of the groups in Example 5b.\n\nThere seems to be a conceptual misunderstanding in my way of thinking, but what is it?\n\nPARTICULAR PROBLEM\n\nFor your particular problem, for labeled teams\n\n\u2022 label the children $$0\\; through\\;$$9\n\u2022 choose $$5$$ for team $$A$$, say $$13579$$, team $$B$$ automatically $$02468$$\n\u2022 This is different from $$02468$$ in Team $$A$$ and $$13579$$ for team $$B$$, so total $$\\binom{10}5$$ choices\n\nBut if teams are unlabeled $$[13579-02468] \\equiv [02468-13579]$$\nhence division by $$2$$\n\nGENERAL GUIDANCE\n\nIt is important to have a clear concept over combinatorics of teams depending on whether they are labeled or unlabeled\n\nTo take a more complex example for greater clarity, suppose you are to choose $$4$$ teams of $$3$$ each from $$12$$ players\n\n\u2022 If the teams are labeled, eg Horses, Greyhounds , Panthers, etc\nthe answer is $$\\binom{12}3\\binom93\\binom63\\binom33$$\nwhich can be variously written as $$\\binom{12}{3,3,3,3}\\; or\\; \\dfrac{12!}{3!3!3!3!}$$\n\n\u2022 If the teams are unlabeled, we shall have to divide by $$4!$$, to remove permutations between identical teams\n\n\u2022 An important point to note is that unlabeled teams in effect become labeled if sizes differ, or composition differs (eg boys' team or girls' team)\n\nNote that in both casees your answer is doubled from the book's solution so you should question whether you really ought to multiply by $$2!=2$$.\n\nLet's consider a simpler case of 4 children $$\\{1,2,3,4\\}$$ and the teams are of size 2. Then all possible options for Team A are $$\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\},\\{3,4\\}$$ giving a total of $$\\frac{4!}{2!2!}=6$$ options. What you are doing is counting each option twice; Say when you count $$\\{1,2\\}$$, you count once for this being assigned to $$A$$ and once for $$B$$. However, note that the event of $$\\{1,2\\}$$ belonging to $$B$$ is exactly the event that $$\\{3,4\\}$$ belongs to $$A$$ which we already count. Hence we should not multiply by 2.", "date": "2022-08-15 09:39:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4484607/number-of-ways-to-divide-10-children-into-two-teams-of-5", "openwebmath_score": 0.5329570770263672, "openwebmath_perplexity": 176.1686574741966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631635159684, "lm_q2_score": 0.8670357735451834, "lm_q1q2_score": 0.8552121484755418}}
{"url": "https://math.stackexchange.com/questions/2481536/how-to-evaluate-the-sum-sum-n-1-infty-frac12n12n2-left1-frac", "text": "# How to evaluate the sum $\\sum_{n=1}^{\\infty}\\frac{1}{(2n+1)(2n+2)}\\left(1+\\frac{1}{2}+\u2026+\\frac{1}{n}\\right)$\n\nHow to evaluate the sum: $$S=\\sum_{n=1}^{\\infty}\\frac{1}{(2n+1)(2n+2)}\\left(1+\\frac{1}{2}+...+\\frac{1}{n}\\right)$$ Can anyone help me,I really appreciate it.\n\n\u2022 is there a source for this sum? Or, just a practice problem? \u2013\u00a0karakfa Oct 20 '17 at 15:05\n\u2022 @MarkViola Mathematica gives the sum of the series. I don't know what is the trick to evaluate it. I think this is a good problem. :) \u2013\u00a0Ixion Oct 20 '17 at 15:26\n\u2022 Partial fraction decomposition \u2013\u00a0JohnColtraneisJC Oct 20 '17 at 15:27\n\u2022 Since $\\displaystyle\\frac {\\pi^2}{12}=\\sum_{r=1}^\\infty \\frac 1{2r^2}$, and $\\displaystyle\\ln 2=\\sum_{r=1}^\\infty \\frac {(-1)^{r+1}}{r}$ is it possible to transform the given summation $\\displaystyle \\sum_{n=1}^\\infty \\sum_{r=1}^n \\frac 1{(2n+1)(2n+2)r}$into $\\displaystyle\\sum_{r=1}^\\infty \\frac 1{2r^2}-\\left(\\sum_{r=1}^\\infty \\frac {(-1)^{r+1}}{r}\\right)^2$? If so, then we are done. \u2013\u00a0hypergeometric Oct 21 '17 at 15:35\n\u2022 I am voting for reopening this question since, despite a lack of efforts from the OP, it led to a number of collateral questions, and I think it is correct to leave this original question as a reference. \u2013\u00a0Jack D'Aurizio Oct 26 '17 at 15:57\n\nA different view on the same problem: $$\\sum_{n\\geq 1}\\frac{x^n}{n}=-\\log(1-x),\\qquad \\sum_{n\\geq 1}H_n x^n = \\frac{-\\log(1-x)}{1-x} \\tag{A}$$ $$\\sum_{n\\geq 1}H_n x^{2n} = \\frac{-\\log(1-x^2)}{1-x^2}\\tag{B}$$\n$$\\begin{eqnarray*}\\sum_{n\\geq 1}H_n\\left(\\frac{1}{2n+1}-\\frac{1}{2n+2}\\right) &=& \\int_{0}^{1}\\frac{-\\log(1-x^2)}{1+x}\\,dx\\\\&=&-\\tfrac{1}{2}\\log^22+\\int_{0}^{1}\\frac{-\\log(1-x)}{1+x}\\,dx\\\\&=&-\\tfrac{1}{2}\\log^22+\\int_{0}^{1}\\frac{-\\log(x)}{2-x}\\,dx\\tag{C}\\end{eqnarray*}$$ and by differentiation under the integral sign, the last integral is related to the series $$\\sum_{n\\geq 1}\\frac{1}{n^2 2^n}=\\text{Li}_2\\left(\\tfrac{1}{2}\\right)\\stackrel{(*)}{=}\\tfrac{\\pi^2}{12}-\\tfrac{\\log^2 2}{2} \\tag{D}$$ where $(*)$ follows from the dilogarithm reflection formula, proved here.\nIt is evident that $$S=\\sum_{n=1}^{\\infty}\\frac{1}{(2n+1)(2n+2)}\\left(1+\\frac{1}{2}+...+\\frac{1}{n}\\right) = \\frac{\\pi^2}{12} - \\ln^{2}2$$ and can be evaluated by following the pattern:\nConsider the series $$S(x) = \\sum_{n=1}^{\\infty}\\frac{H_{n} \\, x^{2n+2}}{(2n+1)(2n+2)}$$ which upon differentiation leads to $S(0) = 0$, $S'(0) = 0$, \\begin{align} S(x) &= \\sum_{n=1}^{\\infty}\\frac{H_{n} \\, x^{2n+2}}{(2n+1)(2n+2)} \\\\ S'(x) &= \\sum_{n=1}^{\\infty}\\frac{H_{n} \\, x^{2n+1}}{(2n+1)} \\\\ S''(x) &= \\sum_{n=1}^{\\infty} H_{n} \\, x^{2n} = - \\frac{\\ln(1- x^2)}{1-x^2}. \\end{align} Now, $$- 2 \\, S''(x) = \\frac{\\ln(1-x)}{1-x} + \\frac{\\ln(1-x)}{1+x} + \\frac{\\ln(1+x)}{1-x} + \\frac{\\ln(1+x)}{1+x}$$ which, upon integration, leads to \\begin{align} - 4 \\, S'(x) &= \\ln^{2}(1 + x) - \\ln^{2}(1-x) + 2 \\, Li_{2}\\left(\\frac{1-x}{2}\\right) - 2 \\, Li_{2}\\left(\\frac{1+x}{2}\\right) + \\ln4 \\, \\ln\\left(\\frac{1+x}{1-x}\\right). \\end{align} Integrating again leads to $S(x)$. The integrals \\begin{align} \\int_{0}^{x} \\ln^{2}(1-t) \\, dt &= (x-1) \\, (\\ln^{2}(1-x) - 2 \\ln(1-x) + 2) + 2 \\\\ \\int_{0}^{x} \\ln^{2}(1+t) \\, dt &= (x+1) \\, (\\ln^{2}(1+x) - 2 \\ln(1+x) + 2) - 2 \\\\ \\int_{0}^{x} \\ln\\left(\\frac{1+t}{1-t}\\right) \\, dt &= x \\, \\ln\\left(\\frac{1+x}{1-x}\\right) + \\ln(1-x^2) \\\\ \\int_{0}^{x} Li_{2}\\left(\\frac{1+t}{2}\\right) \\, dt &= (1+x) \\, Li_{2}\\left(\\frac{1+x}{2}\\right) + x \\, \\ln\\left(\\frac{1-x}{2}\\right) - \\ln(1-x) -x - Li_{2}\\left(\\frac{1}{2}\\right) \\\\ \\int_{0}^{x} Li_{2}\\left(\\frac{1-t}{2}\\right) \\, dt &= (x-1) \\, Li_{2}\\left(\\frac{1-x}{2}\\right) + (x+1) \\, \\ln\\left(\\frac{1+x}{2}\\right) -x - Li_{2}\\left(\\frac{1}{2}\\right) + \\ln2 \\end{align} are needed for the evaluation. Once $S(x)$ is determined set $x=1$ to obtain $$S(1) = \\sum_{n=1}^{\\infty}\\frac{H_{n}}{(2n+1)(2n+2)} = \\frac{\\pi^2}{12} - \\ln^{2}2$$\nFollowing Leucipus' answer, I will use different approach to evaluate $S(1)$. Note \\begin{eqnarray} S=S(1)&=&\\int_0^1\\int_0^xS''(x)dxt\\\\ &=&-\\int_0^1\\int_0^x\\frac{\\ln(1-x^2)}{1-x^2}dxdt\\\\ &=&-\\int_0^1\\int_x^1\\frac{\\ln(1-x^2)}{1-x^2}dtdx\\\\ &=&-\\int_0^1(1-x)\\frac{\\ln(1-x^2)}{1-x^2}dx\\\\ &=&-\\int_0^1\\frac{\\ln(1-x^2)}{1+x}dx\\\\ &=&-\\int_0^1\\frac{\\ln(1-x)}{1+x}dx-\\int_0^1\\frac{\\ln(1+x)}{1+x}dx\\\\ &=:&-I_1-I_2 \\end{eqnarray} Now $$I_2=\\int_0^1\\ln(1+x)d\\ln(1+x)=\\frac{1}{2}\\ln^2(1+x)\\bigg|_0^1=\\frac12\\ln^22.$$ For $I_1$, under $t=1-x$, one has \\begin{eqnarray} \\int\\frac{\\ln t}{2-t}dt&=&-\\int_0^1\\ln xd\\ln(2-t)\\\\ &=&-\\ln x\\ln(2-t)+\\int\\frac{\\ln(2-t)}{t}dt\\\\ &=&-\\ln x\\ln(2-t)+\\int\\frac{\\ln2+\\ln(1-\\frac12t)}{t}dt\\\\ &=&-\\ln x\\ln(2-t)+\\ln2\\ln t+Li_2(\\frac t2)+C \\end{eqnarray} \\begin{eqnarray} I_1&=&\\int_0^1\\frac{\\ln t}{2-t}dt=-\\int_0^1\\ln xd\\ln(2-t)\\\\ &=&-\\ln x\\ln(2-t)+\\ln2\\ln t+Li_2(\\frac t2)\\bigg|_0^1 &=&-\\frac{\\pi^2}{12}+\\frac12\\ln^22. \\end{eqnarray} Thus $$S=\\frac{\\pi^2}{12}-\\ln^22.$$", "date": "2019-06-16 02:52:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2481536/how-to-evaluate-the-sum-sum-n-1-infty-frac12n12n2-left1-frac", "openwebmath_score": 1.0000091791152954, "openwebmath_perplexity": 1941.60338215407, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631655203046, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8552121434355724}}
{"url": "https://math.stackexchange.com/questions/3552866/why-is-0-2-a-subgroup-of-mathbb-z-4", "text": "# Why is $\\{0, 2\\}$ a subgroup of $\\mathbb Z_4$?\n\nI feel like this should be obvious but why is $$\\{0, 2\\}$$ a subgroup of $$\\mathbb Z_4$$? So, $$\\langle 2\\rangle=\\{0,2\\}$$. Shouldn't this set contain the inverse ($$-2$$)? Or does it have to do with the fact that $$(-2)(-2)=4=0$$? Please advise.\n\n\u2022 What is the difference between $2$ and \"$-2$\"? Recall, the minus sign merely means \"the additive inverse of\" which does not necessarily need to appear in other representations of the element. \u2013\u00a0JMoravitz Feb 19 '20 at 17:36\n\u2022 Welcome to math SE. Have a look at mathjax for your mathematical expressions. \u2013\u00a0Alain Remillard Feb 19 '20 at 17:37\n\u2022 As for \"does it have to do with the fact that $(-2)(-2)=4=0$\" No, it doesn't. It has to do with the fact that $2 + 2 = 0$. Addition is what is important here, not multiplication. You will find that the additive inverse of $0$ is $0$, the additive inverse of $1$ (which you might when convenient decide to notate as $-1$) is equal to $3$, the additive inverse of $2$ (which you might when convenient decide to notate as $-2$) is equal to $2$, and so on... \u2013\u00a0JMoravitz Feb 19 '20 at 17:37\n\u2022 There's a problem with the title of the question. 2 is not a subgroup of $\\mathbb Z_4$, it's an element. I guess you mean the subgroup generated by 2. \u2013\u00a0Shatabdi Sinha Feb 19 '20 at 17:40\n\u2022 After you ask a question here, if you get an acceptable answer, you should \"accept\" the answer by clicking the check mark $\\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. \u2013\u00a0Shaun Feb 19 '20 at 18:46\n\n$$\\langle 2\\rangle = \\{0, 2\\}$$ is a subgroup of the group $$\\mathbb Z_4 = \\{0, 1, 2, 3\\}$$ under modular arithmetic, modulo $$4$$.\n\nThe identity of this group is $$0$$, and because $$2+2 \\equiv 0 \\pmod 4$$, it has order two, and hence $$2$$ generates a group (subgroup) of order 2. In fact, the additive inverse of $$2$$ is $$2$$.\n\nThat is, $$\\langle 2 \\rangle = \\{0, 2\\} \\leq \\{0, 1, 2, 3\\} = \\mathbb Z_4$$.\n\n\u2022 Good approach to teach! \u2013\u00a0Mikasa Mar 20 '20 at 8:36\n\nThe elements of $$\\Bbb Z_4$$ are not technically $$0$$, $$1$$, $$2$$ and $$3$$; rather, they are equivalence classes of integers with respect to the divisibility of their differences by $$4$$, like so: $$[a]_4:=\\{b\\in\\Bbb Z: 4\\mid a-b\\}.$$ The operation of the group is defined by $$[x]_4+_4[y]_4=[x+y]_4$$.\n\nThus, since $$4\\mid (-2)+2=0$$, we have $$[-2]_4=[2]_4$$.\n\n\u2022 Why the downvote? \u2013\u00a0Shaun Feb 19 '20 at 19:00\n\u2022 No, it isn't, @amWhy. \u2013\u00a0Shaun Mar 11 '20 at 16:58\n\u2022 But $-2\\equiv 2\\pmod{4}$. \u2013\u00a0Shaun Mar 11 '20 at 17:02", "date": "2021-08-02 16:29:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3552866/why-is-0-2-a-subgroup-of-mathbb-z-4", "openwebmath_score": 0.8267338275909424, "openwebmath_perplexity": 232.86711797292313, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631641172693, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8552121320523893}}
{"url": "https://stats.stackexchange.com/questions/479117/what-are-the-chances-rolling-6-6-sided-dice-that-there-will-be-a-6", "text": "# What are the chances rolling 6, 6-sided dice that there will be a 6?\n\nMore generally, what is the probability that n n-sided dice will turn up at least one side with the highest number (\"n\")?\n\nMy crude thinking is that each side has 1/6 probability, so that 6 of them (1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6) equals 100% of the time, but this obviously is not the case.\n\n2 coins getting one T (say), has 3/4 of the time that at least one will be a T (for every 2 coin tosses). But this is 75% for a 2d2 dice throw, what is the general formula for NdN?\n\n\u2022 Pretty sure we have a couple of questions already that cover this. Jul 26, 2020 at 22:59\n\u2022 Does this answer your question? How often do you have to roll a 6-sided die to obtain every number at least once? Jul 27, 2020 at 19:07\n\u2022 $m = n-1$, that is $P = 1 - \\left(\\frac{n-1}{n}\\right)^n$ Jul 28, 2020 at 10:11\n\u2022 @NathanChappell: Thanks! That's the cleanest solution yet! Jul 31, 2020 at 18:11\n\nProbability of a die not turning up $$n$$ is $$1-1/n$$. Probability of not turning up $$n$$ in any of the $$n$$ dice is $$(1-1/n)^n$$. If you subtract this from $$1$$, it'll be probability of at least one $$n$$ turning up when you throw $$n$$ dice, i.e.\n\n$$p=1-(1-1/n)^n\\rightarrow 1-e^{-1}$$ as $$n$$ goes to $$\\infty.$$\n\n\u2022 In the case of trying to roll at least one 6 on 6 6-sided dice rolls (the title question), this gives us approximately a 66.5% chance. Jul 27, 2020 at 6:20\n\u2022 Nathan Chappell has a simpler formula in the comment to the question above. Unfortunately, I don't have the time nor is it immediately apparent how to reduce your formula to his. Jul 31, 2020 at 18:10\n\u2022 Inside of the parantheses is the same if you pay attention closely @Marcos $$\\left(1-\\frac{1}{n}\\right)=\\left(\\frac{n-1}{n}\\right)$$ Jul 31, 2020 at 21:50\n\u2022 @gunes: Ah, yes, thanks. I came to the same conclusion after writing that comment -- an equivalency I hadn't seen before in algebra. Rather, I hadn't guessed to make 1 = n/n and then re-arrange terms. It's a common algebraic trick, but I'd never had the context to do it for myself. Aug 1, 2020 at 4:11\n\u2022 How do you arrive at the conclusion in your second sentence? That is FAR OUT. I've not seen that in combinatorics. If you can explain it, I'll give you the credit. Nevermind... I think I see it now. I'll give you the credit... Aug 1, 2020 at 16:40\n\nThe event $$A:=$$ \"at least one die turns up on side $$n$$\" is the complement of the event $$B:=$$ \"all dice turn up on non-$$n$$ sides\".\n\nSo $$P(A)=1-P(B)$$. What's $$P(B)$$?\n\nAll dice are independent, so\n\n$$P(\\text{all n dice turn up on non-n sides}) = P(\\text{a single die turns up non-n})^n = \\bigg(\\frac{n-1}{n}\\bigg)^n.$$\n\nSo\n\n$$P(A) = 1-\\bigg(\\frac{n-1}{n}\\bigg)^n.$$\n\nTrust but verify. I like to do so in R:\n\n> nn <- 6\n> n_sims <- 1e5\n> sum(replicate(n_sims,any(sample(1:nn,nn,replace=TRUE)==nn)))/n_sims\n[1] 0.66355\n> 1-((nn-1)/nn)^nn\n[1] 0.665102\n\n\nLooks good. Try this with other values of nn. Here is a plot:\n\nnn <- 2:100\nplot(nn,1-((nn-1)/nn)^nn,type=\"o\",pch=19,ylim=c(1-1/exp(1),1))\nabline(h=1-1/exp(1),col=\"red\")\n\n\nWe note how our probability in the limit is\n\n$$P(A) = 1-\\bigg(\\frac{n-1}{n}\\bigg)^n =1-\\bigg(1-\\frac{1}{n}\\bigg)^n \\to 1-\\frac{1}{e}\\approx 0.6321206 \\quad\\text{as }n\\to\\infty,$$\n\nAnswers by @StephanKolassa (+1) and @gunes (+1) are both fine. But this problem can be solved with reference to binomial and Poisson distributions as follows:\n\nIf $$X_n$$ is the number of ns seen in $$n$$ rolls of a fair $$n$$-sided die, then $$X_n \\sim \\mathsf{Binom}(n, 1/n),$$ so that $$P(X_n \\ge 1) = 1 - P(X_n = 0)= 1-(1-1/n)^n.$$\n\nAs $$n\\rightarrow\\infty,$$ one has $$X_n \\stackrel{prob}{\\rightarrow} Y \\sim\\mathsf{Pois}(\\lambda=1),$$ with $$P(Y \\ge 1) = 1 - P(Y = 0) = 1 - e^{-1}.$$\n\nThe answer can be arrived at by purely counting the described events as well, although the accepted answer is more elegant. We'll consider the case of the die, and hopefully the generalization is obvious. We'll let the event space be all sequences of numbers from $$\\{1,2,...,6\\}$$ of length $$6$$. Here are a few examples (chosen at random):\n\n3 2 3 5 6 1\n1 1 2 5 2 4\n1 2 1 1 6 3\n4 4 3 3 4 2\n6 1 1 6 3 4\n6 3 5 4 5 1\n\n\nThe point is, our space has a total of $$6^6$$ events, and due to independence we suppose that any one of them is as probable as the other (uniformly distributed). We need to count how many sequences have at least one $$6$$ in them. We partition the space we are counting by how many $$6$$'s appear, so consider the case that exactly one $$6$$ appears. How many possible ways can this happen? The six may appear in any position (6 different positions), and when it does the other 5 postions can have any of 5 different symbols (from $$\\{1,2,...,5\\}$$). Then the total number of sequences with exactly one $$6$$ is: $$\\binom{6}{1}5^5$$. Similarly for the case where there are exactly two $$6$$'s: we get that there are exactly $$\\binom{6}{2}5^4$$ such sequences. Now it's time for fun with sums:\n\n$$\\sum_{k=1}^6 \\binom{6}{k}5^{6-k} = \\sum_{k=0}^6 \\binom{6}{k}5^{6-k}1^k - 5^6 = (5+1)^6 - 5^6$$\n\nTo get a probability from this count, we divide by the total number of events:\n\n$$\\frac{6^6 - 5^6}{6^6} = 1 - (5/6)^6 = 1 - (1-1/6)^6$$\n\nI think that this generalizes pretty well, since for any $$n$$ other than $$6$$, the exact same arguments holds, only replace each occurrence of $$6$$ with $$n$$, and $$5$$ with $$n-1$$.\n\nIt's also worth noting that this number $$5^6 = \\binom{6}{0}5^6$$ is the contribution of sequences in which no $$6$$ occurs, and is much easier to calculate (as used in the accepted answer).\n\nI found BruceET's answer interesting, relating to the number of events. An alternative way to approach this problem is to use the correspondence between waiting time and number of events. The use of that would be that the problem will be able to be generalized in some ways more easily.\n\n### Viewing the problem as a waiting time problem\n\nThis correspondence, as for instance explained/used here and here, is\n\nFor the number of dice rolls $$m$$ and number of hits/events $$k$$ you get: $$\\begin{array}{ccc} \\overbrace{P(K \\geq k| m)}^{\\text{this is what you are looking for}} &=& \\overbrace{P(M \\leq m|k)}^{\\text{we will express this instead}} \\\\ {\\small\\text{\\mathbb{P} k or more events in m dice rolls}} &=& {\\small\\text{\\mathbb{P} dice rolls below m given k events}} \\end{array}$$\n\nIn words: the probability to get more than $$K \\geq k$$ events (e.g. $$\\geq 1$$ times rolling 6) within a number of dice rolls $$m$$ equals the probability to need $$m$$ or less dice rolls to get $$k$$ such events.\n\nThis approach relates many distributions.\n\nDistribution of                 Distribution of\nWaiting time between events     number of events\n\nExponential                     Poisson\nErlang/Gamma                    over/under-dispersed Poisson\nGeometric                       Binomial\nNegative Binomial               over/under-dispersed Binomial\n\n\nSo in our situation the waiting time is a geometric distribution. The probability that the number of dice rolls $$M$$ before you roll the first $$n$$ is less than or equal to $$m$$ (and given a probability to roll $$n$$ equals $$1/n$$) is the following CDF for the geometric distribution:\n\n$$P(M \\leq m) = 1-\\left(1-\\frac{1}{n}\\right)^m$$\n\nand we are looking for the situation $$m=n$$ so you get:\n\n$$P(\\text{there will be a n rolled within n rolls}) = P(M \\leq n) = 1-\\left(1-\\frac{1}{n}\\right)^n$$\n\n### Generalizations, when $$n \\to \\infty$$\n\nThe first generalization is that for $$n \\to \\infty$$ the distribution of the number of events becomes Poisson with factor $$\\lambda$$ and the waiting time becomes an exponential distribution with factor $$\\lambda$$. So the waiting time for rolling an event in the Poisson dice rolling process becomes $$(1-e^{-\\lambda \\times t})$$ and with $$t=1$$ we get the same $$\\approx 0.632$$ result as the other answers. This generalization is not yet so special as it only reproduces the other results, but for the next one I do not see so directly how the generalization could work without thinking about waiting times.\n\n### Generalizations, when dices are not fair\n\nYou might consider the situation where the dice are not fair. For instance one time you will roll with a dice that has 0.17 probability to roll a 6, and another time you roll a dice that has 0.16 probability to roll a 6. This will mean that the 6's get more clustered around the dice with positive bias, and that the probability to roll a 6 in 6 turns will be less than the $$1-1/e$$ figure. (it means that based on the average probability of a single roll, say you determined it from a sample of many rolls, you can not determine the probability in many rolls with the same dice, because you need to take into account the correlation of the dice)\n\nSo say a dice does not have a constant probability $$p = 1/n$$, but instead it is drawn from a beta distribution with a mean $$\\bar{p} = 1/n$$ and some shape parameter $$\\nu$$\n\n$$p \\sim Beta \\left( \\alpha = \\nu \\frac{1}{n}, \\beta = \\nu \\frac{n-1}{n} \\right)$$\n\nThen the number of events for a particular dice being rolled $$n$$ time will be beta binomial distributed. And the probability for 1 or more events will be:\n\n$$P(k \\geq 1) = 1 - \\frac{B(\\alpha, n + \\beta)}{B(\\alpha, \\beta)} = 1 - \\frac{B(\\nu \\frac{1}{n}, n +\\nu \\frac{n-1}{n})}{B(\\nu \\frac{1}{n}, n +\\nu \\frac{n-1}{n})}$$\n\nI can verify computationally that this works...\n\n### compute outcome for rolling a n-sided dice n times\nrolldice <- function(n,nu) {\np <- rbeta(1,nu*1/n,nu*(n-1)/n)\nk <- rbinom(1,n,p)\nout <- (k>0)\nout\n}\n\n### compute the average for a sample of dice\nmeandice <- function(n,nu,reps = 10^4) {\nsum(replicate(reps,rolldice(n,nu)))/reps\n}\nmeandice <- Vectorize((meandice))\n\n### simulate and compute for variance n\n\nset.seed(1)\nn <- 6\nnu <- 10^seq(-1,3,0.1)\ny <- meandice(n,nu)\nplot(nu,1-beta(nu*1/n,n+nu*(n-1)/n)/beta(nu*1/n,nu*(n-1)/n), log = \"x\", xlab = expression(nu), ylab = \"fraction of dices\",\nmain =\"comparing simulation (dots) \\n with formula based on beta (line)\", main.cex = 1, type = \"l\")\npoints(nu,y, lty =1, pch = 21, col = \"black\", bg = \"white\")\n\n\n.... But I have no good way to analytically solve the expression for $$n \\to \\infty$$.\n\nWith the waiting time However, with waiting times, then I can express the the limit of the beta binomial distribution (which is now more like a beta Poisson distribution) with a variance in the exponential factor of the waiting times.\n\nSo instead of $$1-e^{-1}$$ we are looking for $$1- \\int e^{-\\lambda} p(\\lambda) \\, \\text{d}\\, \\lambda$$.\n\nNow that integral term is related to the moment generating function (with $$t=-1$$). So if $$\\lambda$$ is normal distributed with $$\\mu = 1$$ and variance $$\\sigma^2$$ then we should use:\n\n$$1-e^{-(1-\\sigma^2/2)} \\quad \\text{instead of} \\quad 1-e^{-1}$$\n\n### Application\n\nThese dice rolls are a toy model. Many real-life problems will have variation and not completely fair dice situations.\n\nFor instance, say you wish to study probability that a person might get sick from a virus given some contact time. One could base calculations for this based on some experiments that verify the probability of a transmission (e.g. either some theoretical work, or some lab experiments measuring/determining the number/frequency of transmissions in an entire population over a short duration), and then extrapolate this transmission to an entire month. Say, you find that the transmission is 1 transmission per month per person, then you could conclude that $$1-1/e \\approx 0.63 \\%$$ of the population will get sick. However, this might be an overestimation because not everybody might get sick/transmission with the same rate. The percentage will probably lower.\n\nHowever, this is only true if the variance is very large. For this the distribution of $$\\lambda$$ must be very skewed. Because, although we expressed it as a normal distribution before, negative values are not possible and distributions without negative distributions will typically not have large ratios $$\\sigma/\\mu$$, unless they are highly skewed. A situation with high skew is modeled below:\n\nNow we use the MGF for a Bernoulli distribution (the exponent of it), because we modeled the distribution as either $$\\lambda = 0$$ with probability $$1-p$$ or $$\\lambda = 1/p$$ with probability $$p$$.\n\nset.seed(1)\nrate = 1\ntime = 1\nCV = 1\n### compute outcome for getting sick with variable rate\ngetsick <- function(rate,CV=0.1,time=1) {\n### truncating changes sd and mean but not co much if CV is small\np <- 1/(CV^2+1)\nlambda <- rbinom(1,1,p)/(p)*rate\nk <- rpois(1,lambda*time)\nout <- (k>0)\nout\n}\n\nCV <- seq(0,2,0.1)\nplot(-1,-1, xlim = c(0,2), ylim = c(0,1), xlab = \"coefficient of variance\", ylab = \"fraction\",\ncex.main = 1, main = \"if rates are bernouilli distributed \\n fraction p with lambda/p and 1-p with 0\")\nfor (cv in CV) {\npoints(cv,sum(replicate(10^4,getsick(rate=1,cv, time = 1)))/10^4)\n}\n\np <- 1/(CV^2+1)\nlines(CV,1-(1-p)-p*exp(-1/p),col=1)\nlines(CV,p, col = 2, lty = 2)\n\nlegend(2,1, c(\"simulation\", \"computed\", \"percent of subsceptible population\"),\ncol = c(1,1,2), lty = c(NA,1,2), pch = c(1,NA,NA),xjust =1, cex = 0.7)\n\n\nThe consequence is. Say you have high $$n$$ and have no possibilities to observe $$n$$ dice rolls (e.g. it takes to long), and instead you screen the number of $$n$$ rolls only for a short time for many different dice. Then you could compute the number of dices that did roll a number $$n$$ during this short time and based on that compute what would happen for $$n$$ rolls. But you would not be knowing how much the events correlate within the dice. It could be that you are dealing with a high probability in a small group of dice, instead of an evenly distributed probability among all dice.\n\nThis 'error' (or you could say simplification) relates to the situation with COVID-19 where the idea goes around that we need 60% of the people immune in order to reach herd immunity. However, that may not be the case. The current infection rate is determined for only a small group of people, it can be that this is only an indication for the infectiousness among a small group of people.\n\n\u2022 I must admit that my last case is actually very trivial and does not this larger framework and tricks with moment generating functions, it can be done easier by just rescaling the population. Jul 31, 2020 at 16:58\n\u2022 ...Except, since the number of rolls is finite, one can enumerate all possible outcomes. This is n^n for an n-sided die. Once one has done this it's a matter of counting which rolls have a #n. Jul 31, 2020 at 17:05\n\u2022 @Marcos my approach is for people that do not like counting (e.g. imagine $n$ would be equal to a million or larger, it is still finite but good luck counting). And also, how are you gonna count if the dices are not fair but instead follow some continuous distribution? -------- That said, I agree your specific question can be answered easily. The method here is just an additional view that can be useful for more complex problems. Jul 31, 2020 at 17:20\n\u2022 True, it requries a \"meta-numeric\" formula for counting the digits of numbers (as symbols rather than quantities) themselves. Jul 31, 2020 at 17:28\n\u2022 @Marcos I am not sure what you mean by counting now. I consider this problem not as a counting problem when $n$ get's large. Well, you can go count things and maybe create expressions that summarize the counting, but you may not always get (easily) to a solution. Flipping the point of view from counting the probabilities of a given number of events for a given number of rolls, to a given number of rolls (waiting time) for a given number of events, can be a useful tool to greatly simplify the problem (I agree for this specific problem, which is not so difficult it is not so much necessary) Jul 31, 2020 at 17:42\n\nSimplify and then extend. Start with a coin. A coin is a die with 2 sides (S=2).\n\nThe exhaustive probability space is\n\nT | H\n\nTwo possibilities. One satisfies the condition of all heads. So your odds of all heads with one coin (n=1) are 1/2.\n\nSo try two coins (n=2). All outcomes:\n\nTT | TH | HT | HH\n\nFour possibilities. Only one matches your criteria. It is worth noting that the probability of one being heads and the other being tails is 2/4 because two possibilities of the four match your criteria. But there is only one way to get all heads.\n\nTTT | THT | HTT | HHT\n\nTTH | THH | HTH | HHH\n\n8 possibilities. Only one fits the criteria - so 1/8 chances of all heads.\n\nThe pattern is (1/S)^n or (1/2)^3.\n\nFor dice S = 6, and we have 6 of them.\n\nProbability of getting a 6 on any given roll is 1/6. Rolls are independent events. So using 2 dice getting all 6's is (1/6)*(1/6) or 1/36.\n\n(1/6)^6 is about 1 in 46,656\n\n\u2022 I was actually just looking for the probability of ANY heads showing up, given as many rolls as there are faces on the die. But, cheers mate, I got my answer. Jul 28, 2020 at 0:22", "date": "2022-05-26 06:10:32", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/479117/what-are-the-chances-rolling-6-6-sided-dice-that-there-will-be-a-6", "openwebmath_score": 0.725549042224884, "openwebmath_perplexity": 504.7102416523403, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631667229061, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8552121258393179}}
{"url": "https://math.stackexchange.com/questions/352192/show-that-x4-1-is-reducible-over-mathbbz-11x-and-splits-over-math/352200", "text": "# Show that $x^4 +1$ is reducible over $\\mathbb{Z}_{11}[x]$ and splits over $\\mathbb{Z}_{17}[x]$.\n\nReduction into linear factors $\\mathbb{Z}_{17}[x]$:\n\nThis part is not too hard: $x^4 \\equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so\n\n$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \\equiv x^4+1$ mod 17.\n\nReduction over $\\mathbb{Z}_{11}[x]$:\n\nThis one doesn't have such an easy solution, as neither of $y^2 \\equiv -1$ mod 11 or $x^4 \\equiv -1$ mod 11 have solutions.\n\nI've tried $x^4+1 = (x^2- \\sqrt{2}x+1)(x^2+\\sqrt{2}x+1)$ but $x^2 \\equiv 2$ mod 11 also has no solutions so there's no easy substitution here.\n\nI think that this approach is not going to work here, so I need something new. Any suggestions?\n\n\u2022 See this question for a more or less complete answer. \u2013\u00a0Jyrki Lahtonen Apr 5 '13 at 15:05\n\u2022 It certainly has a very complete answer - but perhaps somewhat overkill when it comes to answering this question? Nonetheless thanks for the link. \u2013\u00a0Lewy Apr 5 '13 at 15:19\n\u2022 You were almost there... You tried $X^4+1=X^4+1+2X^2-2X^2$. Now, if you observe that $2=-9$ you are done.... \u2013\u00a0N. S. Apr 5 '13 at 15:45\n\u2022 The splitting modulo $p=17$ comes from the fact that $p \\equiv 1 \\pmod 8$, so that $p$ is totally split in the extension $\\Bbb Q(\\zeta_8) / \\Bbb Q$, which is the splitting field of $x^4+1$. \u2013\u00a0Watson Apr 6 '18 at 13:30\n\nI will assume you've made no mistakes in your work.\n\nSince $-1$ hasn't a fourth root in $\\Bbb Z_{11},$ then the only way to reduce $x^4+1$ over $\\Bbb Z_{11}$ is as a product if quadratics. We may as well assume that the quadratics are monic, so we've got $$x^4+1=(x^2+ax+b)(x^2+cx+d),$$ leaving us with the following system: $$a+c=0\\\\ac+b+d=0\\\\ad+bc=0\\\\bd=1$$ By the first equation, $c=-a,$ and the rest of the system becomes: $$b+d=a^2\\\\a(d-b)=0\\\\bd=1$$ Now, if $a=0$, then the first and third remaining equations imply that $-1$ has a square root in $\\Bbb Z_{11}$ (why?), which you've ruled out. Hence, we need $b=d,$ and the rest of the system becomes: $$2b=a^2\\\\b^2=1$$ Can you take it from here?\n\n\u2022 Yes, thank you for your clear explanation. This gives two possible solutions: $(x^2-3x -1)(x^2 +3x-1)$ and $(x^2-8x -1)(x^2 +8x-1)$. \u2013\u00a0Lewy Apr 5 '13 at 15:27\n\u2022 @Lewy: You do know that factorization in this ring is unique, don't ya? IOW, look more closely at the \"two\" factorizations :-) \u2013\u00a0Jyrki Lahtonen Apr 5 '13 at 15:31\n\u2022 @Lewy: Jyrki's point is well-taken. $8=-3,$ so those factorizations are identical (up to arrangement of the factors). \u2013\u00a0Cameron Buie Apr 5 '13 at 15:51\n\u2022 Yes, I see that. Perhaps this is one of those times at which it's appropriate to say \"Doh!\". \u2013\u00a0Lewy Apr 5 '13 at 15:55\n\nAn abstract algebra argument, without much computation.\n\nClaim: The polynomial $x^4+1$ splits over $\\mathbb F_{p^2}$ for any odd (*) prime $p$.\n\nWe can see this because $F_{p^2}^\\times$ is a cyclic group of order $p^2-1$, a multiple of $8$. If $g$ is a generator, then $g^{(1+2k)\\frac{p^2-1}{8}}$ gives distinct roots for $k=0,1,2,3$.\n\nThis means that $x^4+1$ can't be irreducible in $\\mathbb Z_p$, because if it was, the splitting field would have $p^4$ elements.\n\n(*) True for $p=2$, but not for the same reasons. When $p=2$, $x^4+1=(x+1)^4$.\n\nSpecific solutions:\n\nWe can make this explicit by looking at the \"standard\" complex $4$th roots of $-1$: $$\\pm\\frac{\\sqrt 2}2 \\pm\\frac{\\sqrt{2}}{2}i=\\frac{1}{2}(\\pm \\sqrt{2}\\pm\\sqrt{-2})$$\n\nIf $p\\equiv 1\\pmod 8$ then $2$ and $-2$ have square roots in $\\mathbb Z_p$, so $x^4+1$ has four roots.\n\nIf $p\\equiv -1\\pmod 8$ then $2$ is a square, but $-2$ is not. Write $a^2\\equiv 2\\pmod p$. Then write:\n\n$$x^4+1 = \\left(x^2 -ax + 1\\right)\\left(x^2+ax+1\\right)$$\n\nIf $p\\equiv 3\\pmod 8$ then $-2$ is a square and $2$ is not. Letting $b^2\\equiv -2\\pmod p$, we get:\n\n$$x^4+1 = \\left(x^2-bx-1\\right)\\left(x^2+bx-1\\right)$$\n\nFinally, if $p\\equiv 5\\pmod 8$ then neither of $\\pm 2$ is a square, but $-1$ is a square. Letting $c^2\\equiv -1\\pmod p$, we see that the roots are $(\\pm 1 \\pm c)\\sqrt{2}/2$ and the result is:\n\n$$x^4+1 = (x^2-c)(x^2+c)$$\n\n\u2022 The use of the formula $(\\pm\\sqrt2\\pm\\sqrt{-2})/2$ is a nice trick. In the linked answer I (IIRC at least partly in response to the discussion in the comments that lead to CWification) did get there, but should have seen it right away. +1 for unveiling the reason for the emergence of $\\sqrt{\\pm2}$ to this extent. \u2013\u00a0Jyrki Lahtonen Apr 5 '13 at 17:42\n\u2022 Perhaps a more direct proof that $x^4+1$ is not irreducible is just to note that $x^4+1|x^8-1|x^{p^2-1}-1|x^{p^2}-x$. And $x^{p^2}-x$ has as its only irreducible factors the quadratic and linear prime polynomials. \u2013\u00a0Thomas Andrews Apr 5 '13 at 18:04\n\u2022 Well, that's exactly how I did it originally. But somewhat surprisingly people don't think this is a duplicate of that :-) \u2013\u00a0Jyrki Lahtonen Apr 5 '13 at 18:07\n\u2022 @JyrkiLahtonen N.S. suggested in comments to the question: $x^4+1=x^4+1 +2x^2-2x^2$. Hahaha, that makes the primacy of $\\sqrt{2}$ and $\\sqrt{-2}$ even more obvious! \u2013\u00a0Thomas Andrews Apr 5 '13 at 18:16\n\nThis question in fact already answered your question. Indeed, in the accepted answer, the complete solution can be found. So let me explain this and put it into CW.\nAs expounded in the link, we know that there is a primitive root of order $8$ in $\\mathbb F:=\\mathbb F_{p^2}$, called $u$. Since $\\mathbb F$ is of degree $2$ over $\\mathbb F_p$, the minimal polynomial is of degree at most $2$ over $\\mathbb F_p$. Here, our $p\\equiv 3\\pmod 8$, so there is no $u$ in $\\mathbb F_p$, as $u^4=-1$. Its conjugate is then $u^p=u^3$. Therefore $x^4+1$ has a factor given by $(x-u)(x-u^3)$. After easy calculations, we find that this becomes $(x^2-(u+u^3)x-1)$. And the other factor of that one must, after easy calculations of other conjugates of $u$, be $x^2+(u+u^3)x-1$. Therefore, $a:=u+u^3$ satisfies $a^2+2=0$. And I guess you know there is one such $a$ for $p=11$.\n\nFor those who jumped here:\nWe found the factorisation: $$(x^2-3x-1)(x^2+3x-1)$$.\n\nThanks for the attention, and inform me of any errors. Thanks.\n\n\u2022 Also see that referred answer for more detail. \u2013\u00a0awllower Apr 5 '13 at 15:21\n\nLook for a factorization of the shape $(x^2-ax+b)(x^2+ax+b)$. So we want $b^2\\equiv 1\\pmod{11}$, with $2b$ a quadratic residue.\n\nRemark: Note that this is a general procedure for $x^4+1$ modulo an odd prime. The issue is whether one of $2$ or $-2$ is a quadratic residue of $p$. One of them is unless $p\\equiv 5\\pmod{8}$.\n\n\u2022 In the case $p\\equiv 5\\pmod8$ we have that $2$ is a quadratic non-residue, so $i:=2^{(p-1)/4}$ is a square root of $-1$. In that case we have the factorization $(x^2+i)(x^2-i)$. \u2013\u00a0Jyrki Lahtonen Apr 5 '13 at 15:35\n\u2022 In general, at least one of $a,b,ab$ is a square $\\pmod p$. Take $a=2, b=-1.$ :) \u2013\u00a0Thomas Andrews Apr 5 '13 at 16:12\n\u2022 Note, the standard formulation for the complex roots of $x^4+1=0$ can be formulated as $\\frac{1}{2}(\\pm\\sqrt{2}\\pm\\sqrt{-2})$, which is why these two roots are important. \u2013\u00a0Thomas Andrews Apr 5 '13 at 16:14\n\nHint: We know that if $f(x)=x^4+1$ is reducible, then it either has a root or can be written $f(x)=g(x)h(x)$ where $\\deg(g)=\\deg(h)=2$.\n\n\u2022 I guess OP already knew this, judged from the try. The point is: how to find those degree $2$ factors? \u2013\u00a0awllower Apr 5 '13 at 15:03\n\nYou may use the fact that $$x^2 = 9 \\mod 11$$ has a solution, say $x = c$. Then use $$(x^2 + cx - 1)(x^2 - cx - 1),$$ and $c$ is pretty much obvious from here...", "date": "2019-08-26 09:57:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/352192/show-that-x4-1-is-reducible-over-mathbbz-11x-and-splits-over-math/352200", "openwebmath_score": 0.9287224411964417, "openwebmath_perplexity": 223.78505445105293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846678676152, "lm_q2_score": 0.8740772384450968, "lm_q1q2_score": 0.8551837686267484}}
{"url": "https://math.stackexchange.com/questions/2892467/summation-problem-fx-1-sum-n-1-infty-fracxnn", "text": "# Summation problem: $f(x)=1+\\sum_{n=1}^{\\infty}\\frac{x^n}{n}$\n\nI want to evaluate this summation:\n\n$$S=1+x+\\frac{x^2}{2}+\\frac{x^3}{3}+\\frac{x^4}{4}+...+...$$\n\nwhere, $|x|<1$\n\nHere it is my approach\n\n$$S=1+\\sum_{n=1}^{\\infty}\\frac{x^n}{n}=f(x)$$\n\n$$f'(x)=1+x+x^2+x^3+...+...=\\frac{1}{1-x}$$\n\n$$f(x)=\\int f'(x)dx=\\int \\frac{1}{1-x} dx=-\\ln(1-x)+C$$\n\n$$f(0)=1 \\Longrightarrow C=1$$\n\n$$S=1-\\ln(1-x)$$\n\nAnd here is my problem:\n\nI calculated this sum for only $|x|<1$.\n\nThen, I checked in Wolfram Alpha and I saw that, this sum $f(x)$ converges for $x=-1$. But, this creating a contradiction with my solution. Because, the series $$f'(x)=1+x+x^2+x^3+...+...=\\frac{1}{1-x}$$ for $x=-1$ doesn't converge, diverges. Therefore, there is a problem in my solution. But I can not find the source of the problem.\n\nHow can I prove the formula $f(x)=1-\\ln(1-x)$ is also correct at the point $x=-1$ ?\n\nThank you very much.\n\n\u2022 There's no problem with your solution. There's a problem with you assuming that if $\\sum_n \\frac{x^n}{n}$ converges, then $\\sum_n x^{n-1}$ must converge. \u2013\u00a0mathworker21 Aug 23 '18 at 19:08\n\u2022 Your approach worked for all $x$ you were asked to solve it for. WolframAlpha is telling you that the series also has a value for some $x$ outside that region, although your approach doesn't work in that case. Why does that matter? \u2013\u00a0Arthur Aug 23 '18 at 19:13\n\u2022 @mathworker21 How can I prove the formula $f(x)=1-\\ln(1-x)$ is also correct at the point $x=-1$ ? \u2013\u00a0Elvin Aug 23 '18 at 19:28\n\u2022 @mathworker21 as I understand it, you mean my solution is incorrect $f(-1)-1=\\ln 2$ .Am I right? \u2013\u00a0Elvin Aug 23 '18 at 20:34\n\u2022 @student It's incorrect at $x = -1$. You may only deduce that $f'(x) = 1+x+x^2+\\dots$ when $|x| < 1$. \u2013\u00a0mathworker21 Aug 23 '18 at 20:45\n\nI don't know how familiar your are with power series, but this has to do with the notion of radius of convergence.\n\n$\\sum \\left(\\frac{x^n}{n}\\right)_{n\\in\\mathbb{N^*}}$ is a power series with a radius of convergence $R=1$ (which can be proved by the ratio test), so indeed, it does converge for $|x|<1$, and does diverge for $|x|>1$.\n\nAs for $x=1$, the only case with $x=-1$ were we can't conclude right away, it could naively be anything, but it happens to converge.\n\nThe power series $\\sum \\left(x^n\\right)_{n\\in\\mathbb{N^*}}$ is indeed the derivative of the previous one : we know it has the same radius of convergence, so for $x=1$, here again it could be anything ; and it happens to diverge.\n\nThe convergence circle is the only place were the convergence of a power series and its derivatives aren't always equivalent : it could be anything, and you can't deduce the convergence of the derivative from the convergence of the power series there.\n\nAs to show that the power series and $f(x)=1-\\ln(1-x)$ are equal also for $x=-1$ (that is to say, to prove that $f(-1)$ is the sum of the series) :\n\n$\\cdot$ The power series and f both exist for $x$ in $[-1,1)$ (for the power series, you can prove it with the ratio test)\n\n$\\cdot$ They are both continuous on $[-1,1)$ : it is obvious for $f$, and for the power series, we know that the sum of the power series is continious strictly inside the convergence disk $(-1,1)$. We extend the continuity to -1 by the uniform convergence on $[-1,0]$.\n\n$\\cdot$ Are equal to one another on $(-1,1)$ which is dense in $[-1,1]$.\n\nWe can conclude that the power series and f are equal also at $x=1$, so $f(-1)$ is indeed the sum of the series.\n\nIf you are not familiar with this density theorem :\n\nSince the power series is continuous on -1, by definition, $\\lim\\limits_{x\\rightarrow\\ -1}\\sum\\limits_{n=1}^{\\infty}\\frac{x^n}{n}=\\sum\\limits_{n=1}^{\\infty}\\frac{(-1)^n}{n}$, the limits being taken by approaching from the right ($x>-1$).\n\nBut, since the power series and f are equal for $x>-1$, we can put $f$ instead of the power series inside the limit :\n\n$$\\lim\\limits_{x\\rightarrow\\ -1}f(x)=\\sum\\limits_{n=1}^{\\infty}\\frac{(-1)^n}{n}$$\n\nOr $f$ is continuous on -1, so $\\lim\\limits_{x\\rightarrow\\ -1}f(x)=f(-1)$\n\nHence $\\sum\\limits_{n=1}^{\\infty}\\frac{(-1)^n}{n}=f(-1)$.\n\n\u2022 But we need more. We have a formula $f(x) = 1-\\log(1-x)$ on $(-1,1)$. The series converges at $-1$, and the formula (extends to something that) is continuous at $-1$. Conclude that $f(-1)$ is the sum of the series at $-1$. \u2013\u00a0GEdgar Aug 23 '18 at 19:58\n\u2022 The power series and f both exist on [-1,1) (ratio test), are both continuous on [-1,1) (for the power series, we extend the continuity to -1 by the uniform convergence on [-1,0]), and are equal to one another on (-1,1) which is dense in [-1,1]. We can conclude that the power series and f are equal also at x=1. \u2013\u00a0Harmonic Sun Aug 23 '18 at 20:02\n\u2022 +1 for a complete proof. \u2013\u00a0GEdgar Aug 23 '18 at 21:28\n\nBy the ratio test, $$\\lim_{n\\to\\infty } \\left|\\frac{x^{n+1}}{n+1}\\frac{n}{x^n} \\right|=\\lim_{n\\to\\infty}|x|\\frac{n}{n+1}<1$$ whenever $|x|<1$. Thus we have convergence on $(-1,1)$. We need to check the endpoints separately. For $x=1$, we have the diverging harmonic series. For $x=-1$, we have the alternating harmonic series, which converges.\n\nA power series $\\sum_{n=0}^\\infty a_n x^n$ (I am assuming the series is centered at zero for definiteness) converges either only at $0$, or on some interval $I$ centered at zero. The half-length of $I$ is called the radius of convergence $R$, and can be computed generally as $\\frac{1}{R}=\\limsup_{n \\to \\infty} a_n^{1/n}$ (understood as $R=\\infty$ if $1/R$ is zero in the above formula).\n\nA power series with a given radius of convergence may or may not converge at the endpoints $\\pm R$. $\\sum_{n=1}^\\infty \\frac{x^n}{n}$ turns out to have radius of convergence $1$ and converges at $-1$ but does not converge at $1$. In the interior of its interval of convergence, a power series may be differentiated term by term; thus in your example you can deduce the derivative of $-\\ln(1-x)$ is $\\frac{1}{1-x}$ for $|x|<1$ by differentiating its series term by term. In general differentiation may change how the series behaves at its endpoints, as you found here.\n\nIt's well-known that $$\\ln(1+y)=y-\\frac{y^2}{2}+\\frac{y^3}{3}-\\frac{y^4}{4}+\\cdots,~~~\\forall y \\in (-1,1].\\tag1$$\n\nThus, let $y=-x.$ we have $$\\ln(1-x)=-x-\\frac{x^2}{2}-\\frac{x^3}{3}-\\frac{x^4}{4}+\\cdots,~~~\\forall x \\in [-1,1).\\tag2$$\n\nAs result, $$S=1+x+\\frac{x^2}{2}+\\frac{x^3}{3}+\\frac{x^4}{4}+\\cdots=1-\\left(-x-\\frac{x^2}{2}-\\frac{x^3}{3}-\\frac{x^4}{4}+\\cdots\\right)=1-\\ln(1-x).$$\n\nNotice that, $(1)$ demands $-1<y\\leq 1$. then in $(2)$, we demand $-1<-x\\leq 1,$ namely, $-1\\leq x<1.$\n\nIt is best to use elementary arguments for such well known power series rather than start integrating power series. While power series methods work fine in the interior of region of convergence, the behavior at boundary needs additional analysis (for example Abel's theorem).\n\nBelow I prove the formula $$\\log(1+x)=x-\\frac{x^2}{2}+\\frac{x^3}{3}-\\frac{x^4}{4}+\\dots\\tag{1}$$ for all values of $$x$$ satisfying $$-1. The series in question is obtained by replacing $$x$$ with $$-x$$.\n\nLet's start with the algebraic identity $$\\frac{1}{1+t}=1-t+t^2-\\dots+(-1)^{n-1}t^{n-1}+(-1)^{n}\\frac{t^{n}}{1+t}\\tag{2}$$ which holds for all $$t\\neq -1$$ and all positive integers $$n$$. Integrating this identity with respect to $$x$$ over interval $$[0,x]$$ where $$x>-1$$ we get $$\\log(1+x)=x-\\frac{x^2}{2}+\\dots+(-1)^{n-1}\\frac{x^n}{n}+(-1)^nR_n\\tag{3}$$ where $$R_n=\\int_{0}^{x}\\frac{t^n}{1+t}\\,dt$$ Our job is done if we can show that $$R_n\\to 0$$ as $$n\\to\\infty$$ for all $$x\\in(-1,1]$$. If $$x=0$$ then $$R_n=0$$. If $$0 then we can see that $$0 so that $$R_n\\to 0$$. If $$x=-y$$ and $$y\\in(0,1)$$ then $$R_n=(-1)^{n+1}\\int_{0}^{y}\\frac{t^n}{1-t}\\,dt$$ The integral above is clearly less than or equal to $$\\frac{1}{1-y}\\int_{0}^{y}t^n\\,dt=\\frac{y^{n+1}}{(1-y)(n+1)}$$ which tends to $$0$$ as $$n\\to\\infty$$ and hence $$R_n\\to 0$$. It is thus established that $$R_n\\to 0$$ as $$n\\to\\infty$$ for all $$x\\in(-1,1]$$. Taking limits as $$n\\to\\infty$$ in equation $$(3)$$ gives us the series $$(1)$$ which is valid for $$x\\in(-1,1]$$.", "date": "2019-07-20 16:06:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2892467/summation-problem-fx-1-sum-n-1-infty-fracxnn", "openwebmath_score": 0.9516880512237549, "openwebmath_perplexity": 193.98098350025322, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384668497878, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8551837675729836}}
{"url": "https://math.stackexchange.com/questions/1166798/inequality-from-chapter-5-of-the-book-how-to-think-like-a-mathematician?noredirect=1", "text": "# Inequality from Chapter 5 of the book *How to Think Like a Mathematician*\n\nThis is from the book How to think like a Mathematician,\n\nHow can I prove the inequality $$\\sqrt[\\large 7]{7!} < \\sqrt[\\large 8]{8!}$$\n\nwithout complicated calculus? I tried and finally obtained just $$\\frac 17 \\cdot \\ln(7!) < \\frac 18 \\cdot \\ln(8!)$$\n\n\u2022 Are we allowed complex roots and negative roots to prove the contrary ;-) \u2013\u00a0Philip Oakley Feb 28 '15 at 19:22\n\u2022 I don't think that it is in the proposition \u2013\u00a0Gwydyon Mar 2 '15 at 11:24\n\u2022 Can these simpler approaches to a specific example be used to produce a simpler answer to the \"general case\"? \u2013\u00a0Mark Hurd Mar 3 '15 at 3:02\n\u2022 I don't know, the problem was \"as is\" \u2013\u00a0Gwydyon Mar 3 '15 at 16:35\n\u2022 \u2013\u00a0Martin Sleziak Dec 11 '16 at 11:36\n\nYour inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \\cdots 7 < 8 \\cdots 8$$\n\n\u2022 In the spirit of the question title, I'd note that the first line of this answer is key. Maths does of course love cleverness, but also thrives on knowing to do really simply things such as taking powers of both sides here to remove those ugly roots. \u2013\u00a0Keith Feb 27 '15 at 3:02\n\u2022 You are right, great ! I would have of to think there. \u2013\u00a0Gwydyon Feb 28 '15 at 17:39\n\u2022 Elegant! really clever! \u2013\u00a0Max Payne May 27 '15 at 16:19\n\nThink of\n\n$${\\ln(7!)\\over7}={\\ln(1)+\\cdots+\\ln(7)\\over7}$$\n\nas the average of seven numbers and\n\n$${\\ln(8!)\\over8}={\\ln(1)+\\cdots+\\ln(8)\\over8}$$\n\nas the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.)\n\n\u2022 You are right, but it is clear to me only when i compute ln(8) \u2013\u00a0Gwydyon Feb 28 '15 at 18:08\n\u2022 @Gwydyon, the logarithm is an increasing function, so $\\ln(8)$ is larger than its predecessors. \u2013\u00a0Barry Cipra Feb 28 '15 at 20:00\n\u2022 Yes I know that. \u2013\u00a0Gwydyon Mar 2 '15 at 11:21\n\u2022 @Gwydyon, I guess I don't understand your previous comment then. \u2013\u00a0Barry Cipra Mar 2 '15 at 15:22\n\u2022 Because x/8 is lower than y/7, if x=y, that is not the case but ln(7!) and ln((7+1)!) are near close, for x>=1 \u2013\u00a0Gwydyon Mar 3 '15 at 16:41\n\nNote that $$\\sqrt[7]{7!} < \\sqrt[8]{8!} \\iff\\\\ (7!)^8 < (8!)^7 \\iff\\\\ 7! < \\frac{(8!)^7}{(7!)^7} \\iff\\\\ 7! < 8^7$$ You should find that the proof of this last line is fairly straightforward.\n\n\u2022 You are right but someone has writen the same statements \u2013\u00a0Gwydyon Feb 28 '15 at 18:11\n\n$8\\ln (7!) < 7\\ln (8!) \\Rightarrow \\ln (7!) < 7\\ln 8 \\iff \\ln 1 + \\ln 2 +\\cdots \\ln 7 < 7\\ln 8$ which is clear.\n\n\u2022 great! you achieved my second statement \u2013\u00a0Gwydyon Feb 28 '15 at 18:53\n\nYou have already turned the comparison of two geometric means into the comparison of two arithmetic means. So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest. \\begin{equation*} \\begin{split} (1/(n+1)) \\sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \\sum_{i=1}^{n} x_i) \\\\ &=(1/(n+1) (n x_{n+1}/n + n \\sum_{i=1}^{n} x_i / n) \\\\ &> (1/(n+1) (\\sum_{i=1}^{n} x_i/n + n \\sum_{i=1}^{n} x_i / n) \\\\ &= (1/(n+1) ((n+1) \\sum_{i=1}^{n} x_i / n) \\\\ &= \\sum_{i=1}^{n} x_i / n \\end{split} \\end{equation*}\n\nNote that we really only needed $x_{n+1}$ to be larger than the previous mean.", "date": "2019-11-13 03:04:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1166798/inequality-from-chapter-5-of-the-book-how-to-think-like-a-mathematician?noredirect=1", "openwebmath_score": 0.975949764251709, "openwebmath_perplexity": 948.4434142472403, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384668497878, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.855183761154332}}
{"url": "https://math.stackexchange.com/questions/2299065/truth-of-x2-2-0", "text": "Truth of $x^2-2=0$\n\nFrom Algrebra by Gelfand, it says that \"When we claim that we have solved the equation $x^2-2=0$ and the answer is $x=\\sqrt{2}$ or $x=-\\sqrt{2}$, we are in fact cheating. To tell the truth, we have not solved this equation but confessed our inability to solve it; $\\sqrt{2}$ means nothing except \"the positive solution of the equation $x^2-2=0$\n\nCan someone please explain what he really means? Does he mean every irrational number is meaningless? Thanks in advance.\n\n\u2022 I think he might be saying that irrational numbers could be thought of as solutions to polynomial equations, not the other way around. \u2013\u00a0Airdish May 27 '17 at 16:40\n\u2022 I believe he is saying the result is circular. \u2013\u00a0Simply Beautiful Art May 27 '17 at 16:44\n\u2022 The downvote puzzles me. Gelfand is making a very interesting subtle point and the OP is quite right to be puzzled by it, and ask. \u2013\u00a0Ethan Bolker May 27 '17 at 16:55\n\u2022 @EthanBolker He's a good writer. I just don't understand sometimes and asked. \u2013\u00a0Mathxx May 27 '17 at 16:58\n\u2022 Indeed he is a good writer. As the comments and answers show, you asked a good question. I hope they help you understand. \u2013\u00a0Ethan Bolker May 27 '17 at 17:12\n\nHe is saying the following.\n\n1. $\\sqrt{2}$ is defined as \"the number such that its square is two.\"\n2. The statement \"the solution $x$ to $x^2-2 = 0$ is $\\sqrt{2}$\" is therefore saying \"the solution $x$ to $x^2-2=0$ is the number such that its square is two.\"\n\nAs you can see, this is a rather circular statement. This doesn't mean that irrational numbers are meaningless (indeed, $\\sqrt{2}$ does exist -- see the comments and @EthanBolker's answer for more on this), but rather saying that this method of definition limits us to statements like \"the number such that its square is two\" or \"half of the number such that its square is eight.\"\n\n\u2022 Nice answer, except I don't think it is clear that $\\sqrt{2}$ exists at all. It is either taken as an axiom, the completeness axiom of the reals, or is to be proved, depending on our construction of the real numbers... \u2013\u00a0SEWillB May 27 '17 at 16:51\n\u2022 @SEWillB there certainly exist right triangles with legs equal to $1$ unit. Now, does there not exist a hypotenuse to the triangle, since it would be of length $\\sqrt 2$ units, so may not exist? \u2013\u00a0Namaste May 27 '17 at 16:56\n\u2022 SEWillB (continued)... And without a hypotenuse for a right triangle, we cease to have such a triangle? So the only right triangles require legs of lenghts $x, y$ such that the length of the hypotenuse, $\\sqrt{x^2 + y^2},$ is rational? \u2013\u00a0Namaste May 27 '17 at 17:04\n\u2022 @amWhy, I never liked these arguments at all. There is a nice bit in a book I read a long time ago (Liebeck introduction to pure mathematics) showing the 'existence' of $\\sqrt{n}$ by geometrical construction like that. But these arguments aren't convincing to me because it's not clear what you're trying to do in my opinion. \u2013\u00a0SEWillB May 27 '17 at 17:08\n\u2022 So you do not believe the Pythagorean Theorem? Well, then, I think I've wasted my time commenting with you. \u2013\u00a0Namaste May 27 '17 at 17:09\n\nTo elaborate on @fractal1729 's lovely correct answer:\n\nHaving defined $\\sqrt{2}$ as \"the number such that it's square is $2$\" it's reasonable to ask whether there is such a number.\n\nThe Greeks knew that the answer is \"no\" if \"number\" means \"rational number\". So in order to claim the existence of $\\sqrt{2}$ you must enlarge the system of rational numbers. The Greeks essentially sidestepped that question by doing geometry rather than arithmetic, using points on a line instead of numbers. Clearly there's a point on the \"number line\" with the right length since you can draw the diagonal of a unit square.\n\nIn later centuries mathematicians found more formal algebraic ways to enlarge the system of rational numbers to include what we now call all the \"real numbers\". The new ones are the irrationals. There are several ways to do this. One of the most common is to make precise the notion of an infinite decimal, along with rules for arithmetic with them. Others come with the names \"Cauchy sequences\" or \"Dedekind cuts\".\n\n\u2022 To add a bit to your answer, note that the very notion of compass and straightedge constructions assumes with no justification that one can construct ideal circles/lines and somehow they always intersect when certain conditions are met (such as closer than the sum of radii). When one thinks about it, that is not at all obvious. Perhaps it is not even physically meaningful, for all we know! If so, it may very well be that $\\sqrt{2}$ can't be constructed in any suitable physical sense! Furthermore, the computable reals are elementarily equivalent to the standard reals. Interesting at least. \u2013\u00a0user21820 May 29 '17 at 15:26\n\nUsing definite descriptor notation, we can define:\n\n$$\\sqrt{x} = (\\iota y \\in \\mathbb{R})(y \\geq 0 \\wedge y^2=x)$$\n\nIn words: $\\sqrt{x}$ is the unique $y \\geq 0$ such that $y^2=x$.\n\nFrom this, you can show that for all $y$ and all $x \\geq 0$, we have: $$y^2 = x \\iff y \\in \\pm\\sqrt{x}.$$\n\nBut in some sense, this is a purely logical construct, at least insofar as we haven't explained how to approximate $\\sqrt{x}$ to arbitrary precision. There's ways of doing this, but we haven't given one.\n\nI think the author is simply wrong. There is another meaningful way of defining $\\sqrt{2}$. Consider the following sequence: $$x_{n+1} := \\frac{1}{2}\\left(x_n + \\frac{2}{x_n}\\right)$$ with $x_0=1$.\n\nThen $\\lim_{n\\to\\infty} x_n=\\sqrt{2}$, even though all $x_n\\in\\mathbb{Q}$ for arbitrary $n\\in\\mathbb{N}$. Now instead of defining $\\sqrt{2}$ as the solution of $x^2-2=0$ we could have also defined it as the limit on $x_n$.\n\nProof for the sequence can be found here (German only).\n\n\u2022 This is still circular, in a deeper way. How do you define the notion of a limit without some underlying structure of the real numbers? In the rationals that sequence has no limit. What Cauchy showed is that It is possible to define the limit as the sequence itself. It's the necessity for some definition that's the crux of Gelfand's cryptic comment. \u2013\u00a0Ethan Bolker May 27 '17 at 17:10\n\u2022 There is no need to use $\\mathbb{R}$ with the above sequence. For example no transcendental numbers (like $\\pi$) are needed. And in some sense $37$ is also just the result of multiple (albeit finite) application of Peano's axioms. The \"only\" difference with the above definition of $\\sqrt{2}$ is that you need to apply the appropriate axioms an infinite number of times. \u2013\u00a0Hannebambel May 27 '17 at 17:18\n\u2022 No, the author is right. As @EthanBolker says, your sequence has no limit within $\\mathbb{Q}$. The key difference vs. the construction of $37$ lies exactly in the finite number of steps: no matter how large the $n$, no number $x_n$ in your sequence will ever equal $\\sqrt{2}$. And the fact that $\\sqrt{2}$ is not in $\\mathbb{Q}$ is equivalent to the notion that $x^2 - 2 = 0$ is unsolvable with the standard algebraic operations. \u2013\u00a0Roland May 28 '17 at 7:55\n\u2022 I somehow still don't get how the author is correct. The author says \"$\\sqrt{2}$ means nothing except the positive solution of the equation $x^2\u22122=0$\", whereas I think that $\\sqrt{2}$ also means the limit of the aforementioned sequence. \u2013\u00a0Hannebambel May 28 '17 at 18:23\n\u2022 This answer shows that, in a certain sense, it is possible to \"solve\" for what $\\sqrt{2}$ is. But Gelfand never claimed it wasn't possible to do so--that was not his point at all. His point is that merely naming the answer as $\\sqrt{2}$ doesn't solve anything, because all you've done is give a name to what the answer is. \u2013\u00a0Eric Wofsey May 29 '17 at 21:48\n\nThis is a great question (+1) and judging by the quote, Gelfand must be a great writer at least for algebra textbooks (I confess that I have not read any of his books and hence judging only from the quote in question). In fact he has pinned down the essence of solution of polynomial equations via algebra.\n\nIf we restrict ourselves to algebra then the solution of polynomial equations via radicals is equivalent to replacing one polynomial equation with a set of binomial equations of type $x^{n} - a=0$. And for binomial equations we don't do anything apart from inventing symbols like $a^{1/n}$.\n\nNext there are some equations which can't be solved via radicals. How does an algebraist deal with the situation now? He smartly handles the situation using field extensions and for any polynomial $f(x)$ over a field $F$ one can create an extension field $E$ which contains all the roots of $f(x)$. The technique of construction of this extension field is so simple that one is tempted to call such mechanism of solution of equations as \"cheating\". Thus if $f$ is irreducible then one can get a root in quotient $F[x] /(f(x))$ and guess what the root of $f(x)$ is the coset $x+(f(x))$.\n\nIn effect algebraic methods never try to find the value of root of an equation, but they are rather concerned with the structure of the field extensions which contains the roots. In some very special cases such extensions are radical and these are the familiar ones where we use the quadratic formula, or Cardano's formula (or more complicated stuff of similar type).\n\nLet's now come back to the usual scenario where the equations have coefficients in specific fields $\\mathbb{R}$ and $\\mathbb{C}$ and lets first handle the case when coefficients are real. Real numbers are the product of a non-algebraic process and their existence is almost on same footing as those of rational numbers. Thus they can and are used to measure magnitude and it is agreed by convention that there are sufficient real numbers to locate every point on a geometric line so that all the geometric measurement is possible via real numbers.\n\nBut then the real numbers are a very powerful system and offer us the following justification for the symbol $a^{1/n}$:\n\nTheorem 1: If $a$ is a positive real number and $n$ is a positive integer then there is a unique positive real number $b$ such that $b^{n} =a$ and it is denoted by symbol $a^{1/n}$.\n\nThus at least some binomial equations of type $x^{n} - a=0$ have a root which has existence in the real number system. By existence we mean that it is possible to give some concrete idea about this root in comparison to the integers and rationals. In other words we can provide some approximation to the value of the root using the rational numbers. And moreover the approximation can be as accurate as we want. It is in this sense of approximation via rationals that every real number exists. And it applies to all sorts of irrational numbers including the famous ones $e, \\pi, \\sqrt{2}$. Thus irrational numbers are not meaningless, they are as meaningful as the rationals and perhaps mathematically more significant, but unfortunately algebraic techniques cannot help in discovering their true nature.\n\nLet's also note that there are some real numbers which are the roots of certain polynomial equations with rational coefficients and these we call algebraic real numbers. The reason for extending our number system from $\\mathbb{Q}$ to $\\mathbb{R}$ is not to give some sort of existence to these algebraic numbers (like $\\sqrt{2}$). As far roots of polynomials are concerned the algebraic mechanism of field extensions is a sufficient and beautiful approach. Even if we tried to add such numbers to $\\mathbb{Q}$ it does not give us $\\mathbb{R}$, but it rather gives $\\overline{\\mathbb{Q}}$ which we call the algebraic closure of $\\mathbb{Q}$. It includes all the algebraic numbers both real and complex, but it does not capture all the real numbers or all the complex numbers. In fact one can prove that like $\\mathbb{Q}$, the set $\\overline{\\mathbb{Q}}$ is also countable whereas both $\\mathbb{R}, \\mathbb{C}$ are uncountable.\n\nReal numbers were invented to deal with the essential/pressing need to arithmetize geometry. The idea was to develop of a system of numbers which could correspond to the points on a line and it was known from the time of Pythagoras (or perhaps even earlier) that there were some points on the line which did not correspond to a rational number. Another need for real numbers was coming from the very powerful techniques of calculus which were based on geometrical intuition, but there was no rigorous justification of these methods using arithmetical means.\n\nTo answer your question, we have two ways to look at the symbol $\\sqrt{2}$. One is via algebra which says that it is a solution to the equation $x^{2}-2=0$ and in this approach we can't distinguish between $\\sqrt{2}$ and $-\\sqrt{2}$. Another way is to think of it as a real number represented by its rational approximations. Thus we can say that $\\sqrt{2}$ is a positive real number which is greater than all the positive rationals whose square is less than $2$ and it is less than all the positive rationals whose square is greater than $2$. Further in this particular case we are lucky to have a mechanism of locating a point on the number line corresponding to it via geometrical constructions with ruler and compass.\n\nLet's also discuss a bit about the complex number system. But before we do that we need to note that there is another set of equations which the real number system can handle very well:\n\nTheorem 2: If $f(x)$ is polynomial of odd degree with real coefficients then there is at least one real number $c$ such that $f(c) =0$.\n\nBut there are many equations with real coefficients for which there is no root in the real number system. The simplest and most famous such equation is $x^{2}+1=0$. Once again an algebraist comes to the rescue and creates a field extension $\\mathbb{C}$ as the quotient $\\mathbb{R} [x] /(x^{2}+1)$ and surprisingly in one shot the algebraist succeeds in solving all polynomial equations whatsoever :\n\nFundamental Theorem of Algebra: If $f(x)$ is a polynomial of positive degree with complex coefficients then there is a complex number $c$ such that $f(c) =0$.\n\nBut this is all cheating. The power of the above theorem is not due to the algebraic technique of creating field extensions via quotients, but rather due to the properties of real numbers mentioned in theorem 1 and 2 above.\n\n\u2022 Technically, this kind of justification can only reach the computable reals and also the computable complex numbers, which includes the algebraic closure of the rationals. Notice that the computable complex numbers is algebraically closed, so we can't justify going beyond it because we can't actually demonstrate the existence of anything beyond it! Also, you are right that the fact that the complex numbers are algebraically closed is at its crux not really a matter of algebra but of analysis. Even the field-theoretic proof at one point requires IVT for odd-degree polynomials. \u2013\u00a0user21820 May 29 '17 at 16:05\n\u2022 In other words the only reason we have for believing the existence of all standard real numbers (whether by Cauchy sequences or Dedekind cuts) is that we must believe the power-set axiom or equivalent, which is purely set-theoretic and has no ontological justification. Actually, even function types are not enough to grant the classical power-set, because if we use type theory we can have function types that are compatible with the universe being computable. What breaks is that type membership may not be always true or false. \u2013\u00a0user21820 May 29 '17 at 16:35\n\u2022 Of course, ZFC set theory is very elegant and convenient for modern mathematics and so, justified or not, it has found acceptance. =) \u2013\u00a0user21820 May 29 '17 at 16:36", "date": "2019-05-23 01:38:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2299065/truth-of-x2-2-0", "openwebmath_score": 0.7719165682792664, "openwebmath_perplexity": 201.44561105150305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384664716301, "lm_q2_score": 0.8740772335247531, "lm_q1q2_score": 0.8551837610582675}}
{"url": "https://math.stackexchange.com/questions/2550068/summary-of-my-understanding-of-sequences-and-series-convergence-and-divergence", "text": "# Summary of my understanding of sequences and series' convergence and divergence?\n\nI'm trying to summarise my understanding of infinite sequences, series, and their relationships with respect to convergence at the fundamental level. Here is what I know. How much of this is correct?\n\nFirst off, here's a table of the notations that I use, and their corresponding meaning.\n\nMy understanding is that:\n\n\u2022 The sequence $\\lbrace a_n \\rbrace _{n=0}^{\\infty}$ converges if $$\\lim\\limits_{n\\to\\infty}a_n=L_{a}.$$\n\n\u2022 The infinite series $\\sum\\limits_{n=0}^{\\infty}a_n$ converges if its sequence of partial sums, $\\lbrace s_n \\rbrace _{n=0}^{\\infty}$, has a limit, i.e.$$\\lim\\limits_{n\\to\\infty}s_n=L_{s}.$$\n\n\u2022 If the infinite series $\\sum\\limits_{n=0}^{\\infty}a_n$ converges, then the limit of the sequence $\\lbrace a_n \\rbrace _{n=0}^{\\infty}$ is $0$, i.e. $$\\sum\\limits_{n=0}^{\\infty}a_n \\: converges \\rightarrow \\lim\\limits_{n\\to\\infty}a_n=0.$$\n\n\u2022 The divergence test:\n\nIf the the limit of the sequence $\\lbrace a_n \\rbrace _{n=0}^{\\infty}$ is NOT $0$ or does not exist, then the infinite series diverges, i.e. $$\\lim\\limits_{n\\to\\infty}a_n\\neq0 \\rightarrow \\sum\\limits_{n=0}^{\\infty}a_n \\: diverges$$\n\nWould seriously appreciate it if anyone could verify whether the above is accurate or incorrect in any way.\n\nEDIT: I've modified the two limits notation that were mentioned in the comments and answers below, as well as adding the additional condition (limit does not exist or does not equal zero) for the divergence test. I appreciate all the answers/comments.\n\n\u2022 I would say that the notation $L_{a_n}, L_{s_n}$ looks pretty bad, these terms should not depend on $n$. \u2013\u00a0user99914 Dec 4 '17 at 5:08\n\u2022 BTW one useful equivalent of $\\lim_{n\\to \\infty}a_n=L,$ that students often don't think of , is : For every $r>0$ the set $\\{n:a_n\\not \\in [-r+l,r+L]\\}$ is finite. \u2013\u00a0DanielWainfleet Dec 4 '17 at 5:49\n\nEverything is correct, though the notation $\\lim\\limits_{n\\to\\infty}a_n=L_{a_n}$ is nonstandard. Perhaps an improvement would be to write it as $\\lim\\limits_{n\\to\\infty}a_n=L_{a}.$ Normally just an $L$ will suffice but if you're working with multiple sequences (such as $a_n, b_n, c_n$) then $L_a$ is a good notation for the limit of the sequence $a$.\n\nThe reason why $L_{a_n}$ is not so good is because $n$ is just a free variable that has no importance whatsoever; if you write $a_n$ or $a_k$ it's the same thing. What matters is the sequence whose name is \"$a$\".\n\nAlso note that your last two statements are trivially equivalent; they are contrapositives of each other. In general \"If $p$ then $q$\" is stating the same thing as \"If not $q$ then not $p$\".\n\nTherefore, just from knowing\n\n$$\\sum\\limits_{n=0}^{\\infty}a_n \\: converges \\rightarrow \\lim\\limits_{n\\to\\infty}a_n=0.$$\n\nyou can deduce\n\n$not (\\lim\\limits_{n\\to\\infty}a_n=0 ) \\rightarrow not (\\sum\\limits_{n=0}^{\\infty}a_n \\: converges)$\n\nor in other words\n\n$\\lim\\limits_{n\\to\\infty}a_n \\not =0 \\text{ or the limit doesn't exist} \\rightarrow \\sum\\limits_{n=0}^{\\infty}a_n \\: diverges$\n\n\u2022 Regarding the redundancy in my understanding, although $\\lim\\limits_{n\\to\\infty}a_n \\not =0 \\text{ or the limit doesn't exist} \\rightarrow \\sum\\limits_{n=0}^{\\infty}a_n \\: diverges,$ it does not mean that $\\lim\\limits_{n\\to\\infty}a_n=0 \\rightarrow \\sum\\limits_{n=0}^{\\infty}a_n \\: converges,$ am I right? \u2013\u00a0user98937 Dec 4 '17 at 13:11\n\u2022 @user98937 Yes that's perfectly right. What you wrote is the converse of $\\sum\\limits_{n=0}^{\\infty}a_n \\: converges \\rightarrow \\lim\\limits_{n\\to\\infty}a_n=0.$. Converses are sometimes true, but not in this case as you can see from $\\sum_{n=1}^{\\infty} \\dfrac 1n$ which diverges, in spite of the fact that $\\lim_{n \\to \\infty} \\dfrac 1n=0$ \u2013\u00a0Ovi Dec 4 '17 at 15:33\n\nEverything you wrote looks good to me.", "date": "2021-01-27 20:24:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2550068/summary-of-my-understanding-of-sequences-and-series-convergence-and-divergence", "openwebmath_score": 0.9430496096611023, "openwebmath_perplexity": 159.20872905832923, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846666070896, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8551837595016368}}
{"url": "https://dsp.stackexchange.com/questions/40320/confusion-regarding-pdf-of-circularly-symmetric-complex-gaussian-rv", "text": "# Confusion regarding pdf of circularly symmetric complex gaussian rv\n\nConsidering a random variable $x$ that takes in values from a complex domain. Its real and imaginary components are totally uncorrelated. I am following this link and also studying this document.\n\nIn the wikipedia link, the pdf for a single observation is given and $k =2$ for bivariate gaussian as my assumption is that the real and imaginary are totally uncorrelated and are gaussian respectively. There is a 2 in the denominator in the pdf but the log likelihood for the complex case does not have any 2 in the denominator. The log-likelihood for $k=2$ would be $$-N \\ln| \\Sigma| - (x_n) {|\\Sigma|}^{-1}(x_n)^H - k N \\ln \\pi$$ where $\\mu=0$ as I am assuming zero mean r.v $x$\n\n\u2022 Confusion 1:\n\nI was thinking that the variance of $x$ was $$\\Sigma = \\begin{bmatrix}\\sigma_1^2 & 0\\\\ 0& \\sigma_2^2\\end{bmatrix}\\,\\text.$$ So for $N$ samples (observation), the joint pdf would turn out to be $$P_x(x_1,x_2,...,x_N) = \\prod_{n=1}^N\\frac{1}{\\pi \\sigma^2_x} \\exp \\bigg(\\frac{-{({x_n})}^H ({x_n})}{\\sigma^2_x} \\bigg)$$ where $\\sigma^2_x = [\\sigma_1^2,\\sigma_2^2]$. Would there be a 2 in the denominator of the pdf and what is the correct pdf expression?\n\n\u2022 Confusion 2:\n\nIn the document, the expression for the pdf for complex case looks different from the wikipedia link. Are they the same but maybe I am missing some link between them? Which pdf should I use?\n\n\u2022 Confusion 3:\n\nWhen simulating the r.v in Matlab with zero mean and variance 1, I am halving the variance because the real and imaginary part's variance should add up to 1. Then, would the pdf also contain half of the variance as $\\sigma^2_x/2$?\n\nLet me try to establish the relation between the univariate PDF for a real Gaussian and the univariate PDF for a complex proper (i.e. circular symmetric) Gaussian.\n\nYou know that\n$p_x(x)=\\frac{1}{\\sqrt{2\\pi\\sigma^2}}\\exp(\\frac{-x^2}{2\\sigma^2})$ is the PDF of a real-valued Gaussian with variance $\\sigma^2$. We write $x\\sim\\mathcal{N}(0,\\sigma^2)$ to denote that $x$ is a random variable that follows a real-valued Gaussian with zero mean and variance $\\sigma^2$.\n\nNow, let $z=a+jb$ be a circular symmetric random variable with real part $a$ and imaginary part $b$. In a circularly symmetric Gaussian random variable, the real and imaginary part are i.i.d., i.e. $a\\sim\\mathcal{N}(0,\\sigma^2)$ and $b\\sim\\mathcal{N}(0,\\sigma^2)$. Since $a,b$ are independent, their joint PDF is the product of their PDFs.\n\n$$p_z(z=a+jb)=\\frac{1}{\\sqrt{2\\pi\\sigma^2}}\\exp\\left(\\frac{-a^2}{2\\sigma^2}\\right)\\cdot\\frac{1}{\\sqrt{2\\pi\\sigma^2}}\\exp\\left(\\frac{-b^2}{2\\sigma^2}\\right)$$\n\nwhich gives the PDF of the complex variable $z$ at the value $a+jb$. We can reformulate this to \\begin{align}p_z(z=a+jb)&=\\frac{1}{2\\pi\\sigma^2}\\exp\\left(\\frac{-(a^2+b^2)}{2\\sigma^2}\\right)\\\\&=\\frac{1}{2\\pi\\sigma^2}\\exp\\left(\\frac{-z^*z}{2\\sigma^2}\\right)\\\\&=\\frac{1}{2\\pi\\sigma^2}\\exp\\left(\\frac{-\\|z\\|^2}{2\\sigma^2}\\right)\\end{align}\n\nWe also write for this case that $z$ follows a circular Gaussian distribution and denote this by $z\\sim\\mathcal{CN}(0,2\\sigma^2)$. Why is it suddenly $2\\sigma^2$ (i.e. the double variance compared to the uni-variate, real case? Well, because $E[\\|z\\|^2]=2\\sigma^2$ since it consists of a real and an imaginary part (which are independent and hence their variance adds up together).\n\nSo, to sum up:\n\n\u2022 if $x\\sim\\mathcal{N}(0,\\sigma^2)$, then $p_x(x)=\\frac{1}{\\sqrt{2\\pi\\sigma^2}}\\exp\\left(-\\frac{x^2}{2\\sigma^2}\\right)$\n\u2022 if $z\\sim\\mathcal{CN}(0,\\sigma^2)$, then $p_z(z)=\\frac{1}{\\pi\\sigma^2}\\exp\\left(-\\frac{z^*z}{\\sigma^2}\\right)$\n\n\u2022 Thank you so much, cannot express my gratitude in words here.....just to clarify 2 things from you - (1) if I consider design where the r.v $z \\sim CN(0,2\\sigma^2)$ then the pdf has the 2 in the denominator, otherwise if $z \\sim CN(0,\\sigma^2)$ then there is no 2 in the pdf's denominator. So, either design is acceptable. Is my understanding right? (2) notations - in my question I wrote the $[.]^H$ complex conjugate transpose but you have written * which is only the conjugate. Why the $[.]^H$ is wrong? \u2013\u00a0Ria George Apr 18 '17 at 15:33\n\u2022 sure, if $z\\sim CN(0,2\\sigma^2)$, then there is the two in the denominator, if its $z\\sim CN(0,\\sigma^2)$, then it's not there. Here, the two comes from the description of the variance (e.g. if we had $z\\sim CN(0,1.23456\\sigma^2)$, then in the denominator there would be $1.23456$). For a scalar variable $z$, $[]^H$ and $[]^*$ are equal. When it comes to vector-variables, you'd need $[]^H$. However, note that in this case, also the denominator changes to $\\pi^k\\sigma^{2k}$, since the PDF is the product of the $k$ PDFs of each component (k is the vector size); each component has var. $\\sigma^2$. \u2013\u00a0Maximilian Matth\u00e9 Apr 18 '17 at 17:58\n\u2022 This answer has the correct ideas but is marred badly by a confusion of the concepts of probability and probability density. The probabiity that a continuous random variable equals a given number is $0$, regardless of the choice of number. Indeed, the probability that $z$ equals $a+jb$ (where we choose $a=b=0$) is claimed to be $$P(z=0+j0)=\\frac{1}{2\\pi\\sigma^2}$$ which has value greater than $1$ if $\\sigma^2 < \\frac{1}{2\\pi}$. -1 pending clean-up of the answer to be correct. \u2013\u00a0Dilip Sarwate Apr 18 '17 at 22:26", "date": "2021-06-20 22:18:03", "meta": {"domain": "stackexchange.com", "url": "https://dsp.stackexchange.com/questions/40320/confusion-regarding-pdf-of-circularly-symmetric-complex-gaussian-rv", "openwebmath_score": 0.9665572047233582, "openwebmath_perplexity": 214.44835675637648, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846697584034, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8551837574421396}}
{"url": "https://math.stackexchange.com/questions/2323249/integral-of-textlogz", "text": "# Integral of $\\text{Log}(z)$\n\nI'd like to check my work. I'm trying to integrate $f(z)=\\text{Log}(z)$ on the right half of the unit circle from $-i$ to $i$. $\\text{Log} = \\ln(r)+i\\theta$ according to my book. So, \\begin{align} \\int_C f(z)\\,dz&=\\int_{z_1}^{z_2}f(z)\\,dz\\\\\\\\& = \\int_{-i}^i \\text{Log}\\,dz \\\\\\\\ &= \\bigg[z\\text{Log}-z\\bigg]_{-i}^i \\\\\\\\ &= (i\\text{Log}(i)-i)-\\left(-i\\text{Log}(-i)+i\\right) \\\\\\\\ &= i\\left(\\ln(1)+\\frac{\\pi}{2}i\\right)-i \\ + i\\left(\\ln(1)-\\frac{\\pi}{2}i \\right) \\ -i \\\\\\\\&= -2i \\\\\\\\ \\therefore \\int_{-i}^i \\text{Log}\\,dz=-2i \\end{align}\n\nThis may be incorrect, I'm not really sure how to incorporate the branch cut here.\n\nThanks!\n\n\u2022 the circuit is a circle not the segment [-i,i]. \u2013\u00a0zwim Jun 15 '17 at 5:13\n\u2022 So, I'm required to use the formula $\\int_C f(z)dz=\\int_a^b f[z(\\theta)]z'(\\theta)d\\theta$? Where $a$ and $b$ are $\\frac{\\pi}{2}$ and $\\frac{-\\pi}{2}$? What would my choice for $z'(\\theta)$ be in this case? \u2013\u00a0Kosta Jun 15 '17 at 5:14\n\u2022 Have a look at these answers : math.stackexchange.com/questions/1602614/\u2026 \u2013\u00a0zwim Jun 15 '17 at 5:34\n\nI thought it might be instructive to present two alternative approaches to evaluate the integral of interest. To that end we proceed.\n\nLet $\\text{Log}(z)=\\log(|z|)+i\\text{Arg}(z)$ where $-\\pi < \\text{Arg}(z)\\le \\pi$. In addition, let $C$ denote the semicircular contour in the right-half plane of radisu $1$ and that begins at $z=-i$ and ends at $z=i$.\n\nMETHODOLOGY $1$:\n\nOn $C$, $z=e^{i\\theta}$, $-\\pi/2\\le \\theta \\le \\pi$. Hence, the integral $\\int_C \\text{Log}(z)\\,dz$ is given by\n\n\\begin{align} \\int_C \\text{Log}(z)\\,dz&=\\int_{-\\pi/2}^{\\pi/2} \\text{Log}(e^{i\\theta})\\,ie^{i\\theta}\\,d\\theta\\\\\\\\ &=-\\int_{-\\pi/2}^{\\pi/2} \\theta e^{i\\theta}\\,d\\theta\\\\\\\\ &=-2i\\int_0^{\\pi/2} \\theta\\sin(\\theta)\\,d\\theta\\\\\\\\ &=-2i \\end{align}\n\nMETHODOLOGY $2$:\n\nAnother approach is to exploit Cauchy's Integral Theorem. Inasmuch as $\\text{Log}(z)$ is analytic for $z\\in \\mathbb{C}\\setminus (-\\infty,0]$. Then, the integral over $C$ can be deformed to connect $-i$ to $i$ along any rectifiable path that does not intersect the branch cut. Therefore, we write\n\n\\begin{align} \\int_C \\text{Log}(z)\\,dz&= \\lim_{\\epsilon\\to 0^+}\\left(\\int_{-1}^{-\\epsilon }\\text{Log}(iy)\\,i\\,dy+\\int_{-\\pi/2}^{\\pi/2}\\text{Log}(\\epsilon e^{i\\theta})\\,ie^{i\\theta}\\,d\\theta+\\int_{\\epsilon }^1\\text{Log}(iy)\\,i\\,dy\\right)\\\\\\\\ &=i \\lim_{\\epsilon\\to 0^+}\\int_\\epsilon^1 \\left(\\text{Log}(-iy)+\\text{Log}(iy)\\right)\\,dy\\\\\\\\ &=2i\\int_0^1 \\text{Log}(y)\\,dy\\\\\\\\ &=-2i \\end{align}\n\nas expected!\n\n\u2022 This is very helpful, thank you Mark! \u2013\u00a0Kosta Jun 16 '17 at 16:21\n\u2022 You're welcome. Pleased to hear that this is helpful! \u2013\u00a0Mark Viola Jun 16 '17 at 16:25\n\nAn alternate way to check, without using contours, is to use $$\\int \\ln(x) \\, dx = x \\, \\ln(x) - x$$ which provides \\begin{align} \\int_{-i}^{i} \\ln(x) \\, dx &= \\left( i \\, \\ln(e^{\\pi \\, i/2}) - i \\right) - \\left( -i \\, \\ln(e^{-\\pi \\, i/2}) + i \\right) \\\\ &= \\frac{\\pi \\, i^2}{2} - \\frac{\\pi \\, i^2}{2} - 2 i \\\\ &= -2i. \\end{align} With this it can be stated that the real line integral result is the same as the contour integral result.\n\nlooks good to me .Fundamental theorem of calculus wins again", "date": "2019-05-27 08:06:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2323249/integral-of-textlogz", "openwebmath_score": 0.9999809265136719, "openwebmath_perplexity": 983.6429147569174, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9783846634557753, "lm_q2_score": 0.8740772269642948, "lm_q1q2_score": 0.855183753537819}}
{"url": "http://mathhelpforum.com/trigonometry/227503-trigonometric-identity-problem.html", "text": "# Thread: Trigonometric Identity Problem\n\n1. ## Trigonometric Identity Problem\n\nHow do you show that:\n2sinxcosx=\n(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})\nThanks a lot for helping!!\n\n2. ## Re: Trigonometric Identity Problem\n\nOriginally Posted by JiaGeng\nHow do you show that:\n2sinxcosx=\n(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})\n$1+\\tan^2(\\frac x 2) = \\sec^2(\\frac x 2)$\n\n$1-\\tan^2(\\frac x 2) = \\dfrac{2 \\tan(\\frac x 2)}{\\tan(x)}$\n\n$\\sin(2x)=2\\sin(x)\\cos(x)$\n\n-------------------------\n$\\dfrac{4 \\tan(\\frac x 2)}{1+\\tan^2(\\frac x 2)}\\dfrac{1-\\tan^2(\\frac x 2)}{1+\\tan^2(\\frac x 2)}=$\n\n$\\dfrac{4 \\tan(\\frac x 2)}{\\sec^2(\\frac x 2)}\\dfrac{ \\dfrac{2 \\tan(\\frac x 2)}{\\tan(x)}}{\\sec^2(\\frac x 2)}=$\n\n$4\\cos(\\frac x 2)\\sin(\\frac x 2)\\dfrac{2\\sin(\\frac x 2)\\cos(\\frac x 2)}{\\tan(x)}=$\n\n$2\\sin(x)\\dfrac{\\sin(x)}{\\tan(x)}=$\n\n$2\\sin(x)\\cos(x)$\n\n3. ## Re: Trigonometric Identity Problem\n\nHello, JiaGeng!\n\n$\\text{Prove: }\\:2\\sin x\\cos x\\:=\\:\\frac{4\\tan\\frac{x}{2}}{1+\\tan^2\\frac{x}{2}} \\cdot \\frac{1-\\tan^2\\frac{x}{2}}{1 + \\tan^2\\frac{x}{2}}$\n\nThe first fraction is:\n\n. . $\\frac{4\\tan\\frac{x}{2}}{1 + \\tan^2\\frac{x}{2}} \\;=\\;\\frac{4\\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2} }}{\\sec^2\\frac{x}{2}} \\;=\\;\\frac{4\\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2} }}{\\frac{1}{\\cos^2\\frac{2}{2}}} \\;=\\;4\\sin\\tfrac{x}{2}\\cos\\tfrac{x}{2} \\;=\\;2\\sin x$\n\nThe second fraction is:\n\n. . $\\frac{1-\\tan^2\\frac{x}{2}}{1 + \\tan^2\\frac{x}{2}} \\;=\\;\\frac{1-\\frac{\\sin^2\\frac{x}{2}}{\\cos^2\\frac{x}{2}}}{\\sec^ 2\\frac{x}{2}} \\;=\\; \\frac{1-\\frac{\\sin^2\\frac{x}{2}}{\\cos^2\\frac{x}{2}}}{\\frac {1}{\\cos^2\\frac{x}{2}}} \\;=\\;\\cos^2\\tfrac{x}{2} - \\sin^2\\tfrac{x}{2} \\;=\\;\\cos x$\n\nTherefore, the RHS becomes: . $2\\sin x\\cos x$.\n\n4. ## Re: Trigonometric Identity Problem\n\nThanks romsek and soroban!\nBut I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?\n\n5. ## Re: Trigonometric Identity Problem\n\nOriginally Posted by JiaGeng\nThanks romsek and soroban!\nBut I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?\nEquality works in both directions. If you want, write the equations in reverse order and equality still holds at every step.\n\n6. ## Re: Trigonometric Identity Problem\n\nHello, JiaGeng!\n\nI was wondering if there is a way to do it from LHS to get RHS.\nWould that be even harder?\n\nIf you would rather not run the steps in reverse,\nit can be done ... with a little imagination.\n\n$2\\sin x\\cos x \\;=\\;\\frac {2(2\\sin \\frac{x}{2}\\cos\\frac{x}{2})}{1}\\cdot \\frac{\\cos^2\\!\\frac{x}{2} - \\sin^2\\!\\frac{x}{2}}{1}$\n\n. . . . . . . . . $=\\;\\frac{4\\sin\\frac{x}{2}\\cos\\frac{x}{2}}{\\cos^2\\! \\frac{x}{2} + \\sin^2\\!\\frac{x}{2}} \\cdot \\frac{\\cos^2\\!\\frac{x}{2} - \\sin^2\\!\\frac{x}{2}}{\\cos^2\\!\\frac{x}{2} + \\sin^2\\!\\frac{x}{2}}$\n\nIn both fractions, divide numerator and denominator by $\\cos^2\\!\\tfrac{x}{2}\\!:$\n\n$\\dfrac{\\frac{4\\sin\\frac{x}{2} \\cos\\frac{x}{2}}{\\cos^2\\!\\frac{x}{2}}} {\\frac{\\cos^2\\!\\frac{x}{2}}{\\cos^2\\!\\frac{x}{2}} + \\frac{\\sin^2\\!\\frac{x}{2}}{\\cos^2\\!\\frac{x}{2}}} \\cdot \\dfrac{\\frac{\\cos^2\\!\\frac{x}{2}}{\\cos^2\\!\\frac{x} {2}} - \\frac{\\sin^2\\!\\frac{x}{2}}{\\cos^2\\!\\frac{x}{2}}} {\\frac{\\cos^2\\!\\frac{x}{2}}{\\cos^2\\!\\frac{x}{2}} + \\frac{\\sin^2\\!\\frac{x}{2}}{\\cos^2\\!\\frac{x}{2}}} \\;=\\; \\dfrac{4\\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}}} {1 + \\left(\\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}} \\right)^2} \\cdot \\dfrac{1 - \\left(\\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}} \\right)^2} {1 + \\left(\\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}} \\right)^2}$\n\n. . . . $=\\;\\frac{4\\tan\\frac{x}{2}}{1 + \\tan^2\\!\\frac{x}{2}} \\cdot \\frac{1-\\tan^2\\!\\frac{x}{2}}{1 + \\tan^2\\!\\frac{x}{2}}$", "date": "2017-06-24 13:13:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/227503-trigonometric-identity-problem.html", "openwebmath_score": 0.9643456339836121, "openwebmath_perplexity": 1757.6391694114302, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98228770337066, "lm_q2_score": 0.8705972784807406, "lm_q1q2_score": 0.8551770012395936}}
{"url": "https://www.themathdoctors.org/a-sample-of-ask-dr-math-part-2-questions-outside-of-school/", "text": "# A Sample of Ask Dr. Math, Part 2: Questions Outside of School\n\nIn the first post, I gave a small sampling of questions we\u2019ve had from students, parents, and teachers, all related to school, and discussed how we like to deal with these. But we also get many questions with no direct relation to school. These may come from people who actually use math in their work (computer programming, plumbing, and anything in between), or from students who are trying to deeply understand what they are studying, or who are just curious. Often, of course, these too are from students or teachers, but they are going beyond that role. These are perhaps the most interesting of all. They are really why we are here.\n\n## A real-life question\n\nThis question arrived in 2014 from a flooring inspector, Tim, asking about how high a floor would buckle due to a given increase in length if the ends are fixed:\n\nModeling How Much Wood Expands\n\nSay I place a six foot long yardstick wide side down. One end is butted against a wall (fixed object). When the other end of the yardstick is pushed towards the wall, by say 1/4\", the middle area rises -- and it rises much, much higher than 1/4\".\n\nI would like to know if there is a fixed formula for this exponential gain of \"flat\" vs. \"up.\"\n\nI come across this daily, seeing floors buckle up and away from substrates because they are a tight fit.\n\n(He misused the word exponential; I suppose he was just thinking that the rise is surprisingly high, as in exponential growth.)\n\nDr. Rick and I both recognized this problem as related to previous questions we had answered, and referred to those in our replies. His old answer assumed a circular arc for the new shape, and did a sanity check by comparing the surprising answer to one obtained by supposing that the board just bends sharply in the middle. The circular arc version does not lend itself to easy calculation, but the sharp bend could be turned into a formula.\n\nThere is a way to calculate the height, but it isn't a formula, as such.\n...\nFor more on this topic, including a way to do a *rough* approximation with a simple formula, see:\n\nhttp://mathforum.org/library/drmath/view/62613.html\n\nI had been thinking while Dr. Rick wrote, and found a different old answer (by Dr. Jeremiah) that didn\u2019t actually come up with a method, just a starting point. But I \u201cplayed\u201d with numbers using a spreadsheet much as Dr. Rick had for his earlier answer, and saw a pattern that suggested an approximate formula he might try. When I saw what Dr. Rick had written, I wrote to propose this formula.\n\nI see Dr. Rick beat me to most of what I was going to say, except that my link to our archives is not quite as helpful:\n\nRail Bend in Hot Weather\nhttp://mathforum.org/library/drmath/view/61465.html\n\nBut I made a spreadsheet carrying out the calculations for a circular arc; and have found experimentally that the rise seems to be quite accurately approximated by a simple formula. I have to study more to try to justify it, but here goes.\n\nFor the actual length of the stick, s, and the (small) decrease in the space available, x,\n\nh = 0.605 sqrt(sx)\n\nBut people who enjoy math can\u2019t be fully satisfied with a guess; I continued pursuing my approximation, hoping to justify it. The patient, Tim, didn\u2019t need a proof, but I did. By the end of the day, I could write up a derivation of the formula and send it along, and then, the next day, apply it to some examples.\n\nI've verified my simple approximation, which turns out to be\n\nh = sqrt(6sx)/4\n\nThis changes my 0.605 to 0.612, which fits better for very small x but not quite as well for larger ones.\n\nI'm sure you don't need to see this, but for my records, here is the derivation of my approximation: ...\n\nA missing formula for a counter-intuitive observation (a half-inch shift can result in almost a 7-inch buckle) can be a great motivation for some fun with math. And we also have a reminder that in the real world, an approximation can be both good enough, and as good as you\u2019ll get. (Be sure, as always, to read the whole story in the original page!)\n\n## A curious question\n\n(Well, that last one arose from curiosity, too, but this is pure-math curiosity.)\n\nWithin months of my starting with Ask Dr. Math, I wrote this answer, to a question wondering about place value in decimal numbers:\n\nThe Oneths Place\n\nWhy is there not a oneths place? My classmates and I were wondering.\u00a0 It is weird that decimals begin with tenths, hundredths, and so on. We appreciate your help.\n\nI love it when children are wondering, especially when a teacher encourages it. So I was happy to think about this, even though I had probably never considered the question before. (Since then, we have had some form of this question at least a dozen more times! Often, I have answered by saying, \u201cBelieve it or not, you\u2019re not the first to ask this; if you search our site for the word \u2018oneth\u2019, you\u2019ll find this answer \u2026\u201d)\n\nI\u2019m sure my answer is not all that could be said about it, but I managed not only to give the quick answer (oneths are just another name for ones, since 1/1 = 1, and we don\u2019t want two places with the same value), but to relate it to some other perceived asymmetries in the way we write numbers, and bring in signed exponents (which may be beyond the students\u2019 knowledge, or may be something they were just about to learn). I closed with a general comment about curiosity:\n\nI hope that helps. It's interesting how we tend to expect things to be symmetrical. Mathematicians and scientists often expect it too, and when things don't seem balanced, they try to find out why. So keep wondering about things like this.\n\n## A philosophical question\n\nAnother question we\u2019ve received a number of times involves how a line segment can have a finite length when it is composed of infinitely many points. As a result, we\u2019ve referred to the following question a dozen times or so, in addition to giving fresh answers from different perspectives:\n\nHow Can a Line Have Length?\n\nWe have a line, OR a line segment, OR a ray (it really doesn't matter). If this particular linear entity (let us say that it is a line segment, for the sake of hypothetical simplicity) consists only of infinite points, all with zero volume, mass, diameter, cross-sectional area, etc., how can it have length?\u00a0 Why are we able to take a particular line segment and give it a length of 5?\u00a0 If a substance can only be composed of volumeless points, how can it have volume in and of itself?\n\nDr. Daniel\u2019s answer (from 1998) touches on several key ideas in philosophical thinking about this topic, making a concise summary without digging in too deeply. (I try to avoid digging too deeply, because of the possibility of cave-ins \u2026 .) When I answer related questions, I often refer to that answer, and also to this one:\n\nPoint and Line\n\nI'm in high school, but this problem has really nothing to do with school, it's just been bothering me for a while.\n\nA point has no dimension (I'm assuming), and a line, which has dimension, is a bunch of points strung together. How does something without dimension create something with dimension?\n\nGoing even farther, a point essentially is nothing, because it has no dimension. My question is, How does a bunch of nothings (a point) create a something (a line)?\n\nHere Dr. Jordi got into a long discussion about the problem, perhaps not very effectively in the end. After referring to these two answers, I generally add something like the following as my own brief perspective:\n\nNeither will be a fully satisfying answer, largely because we humans are finite beings, not able to directly observe infinity. (It's amazing enough that we can THINK about infinity, and get some of the answers right!) But the basic idea is that we don't define a line as something made by putting points together; rather, we start by defining a line, and then note that it consists of points. So we can find points on a line, without having to _make_ the line _out of_ points (or make a cube out of stacked squares).\n\nNote also that we do not define length in terms of points; since 0,\u00a0the width of a point, times infinity, the number of points, is an indeterminate quantity (we can't assign any one value to it), that would be useless to attempt. So length is an independent concept, defined only in relation to the location of the endpoints of a segment, not to the points that make it up.\n\nIn choosing these examples of \u201ccurious questions\u201d, I have chosen to avoid\u00a0 questions with really interesting answers, which deserve fuller treatment on their own! Keep watching, and I\u2019ll get to some of them.\n\n### 2 thoughts on \u201cA Sample of Ask Dr. Math, Part 2: Questions Outside of School\u201d\n\n1. Hi \u2013 I would like to know the probability of this occurring \u2013 this past January I celebrated my 74th birthday which means I was born in 1947 (1947/74 numbers reversed). My daughter was born in April, 1974. She will be turning 47 this year (1974/47 numbers reversed) What is the probability of this occurring in a parent/son-daughter relationship?\n\n1. Hi, Gwen.\nI suppose you asked this in connection with this particular post just because it\u2019s an example of curiosity! In fact, there is another post that deals with this very kind of question: Should Rare Events Surprise Us? You\u2019ll find there that an important question we need to ask is, \u201cWhat do you mean by \u2018this\u2019?\u201d\n\nI don\u2019t think you are asking what percentage of all people now alive were born in 1947, and had a daughter born in 1974; and if you did, I\u2019d have to pass, as it would require digging deeper into birth statistics than I feel like going. (It would not be an interesting question mathematically.)\n\nI think your main interest is in the coincidence of the numbers. I may not get to a probability in the end, but there may be some interesting observations along the way!\n\nMy first observation is that the two reversals are closely related. The reason you are 74 this year is that 2021-1947 = 74; that is, 2021 = 1947 + 74. But that implies that 2021 also equals 1974 + 47, because both sums can be written as 1900 + 47 + 74, just rearranging the numbers. So that fact that this year your age is the reversal of your birth year implies that the same will be true this year for anyone born in 1974. To put it another way, her being born in 1974, the reversal of your birth year, predestined everything you observe this year. It all had to happen sooner or later!\n\nLet\u2019s test that out. I was born in 1952. I couldn\u2019t have a daughter born in 1925 or 2025, but in fact it turns out that my father was born in 1925! My claim is that this by itself implies that something like your situation would occur at some point. So if we add 52 to 1925, we find that in 1977, my father was 52, and I was 25. Of course that happened once in a lifetime, but that it would happen to us at all depended only on our two birthdates and our living long enough.\n\nWill everyone have some year in his life when his age is the reverse of his birth year? That just depends the reverse of his birth year being a likely age. For someone born in 1900, it happens immediately (and again if he reaches 100); born in 1901, it happens when he reaches 10. If you were born in 1909, you\u2019ve have to live to 90 to have it happen. So basically, lifespan is the only restriction.\n\nHow likely is it that one will have a child in his \u201creverse year\u201d? That restricts it more, since the resulting age has to be within the childbearing years. For example, in your case your age at her birth was 1974 \u2013 1947 = 27. If we suppose that, say, a child can be born to a parent aged 16 to 45, then your scenario is feasible for people born in about 30% of all years, and considerably more likely for, say, 10-15%. (Actually, this is a little more subtle than what I just said, and I checked by making a spreadsheet to actually count years that would work.)\n\nWith appropriate statistical information, one might come up with a probability of actually having a child in a given year. But that\u2019s beyond my level of interest.\n\nBut I recalled my twin brother pointing out something similar based on our father\u2019s age, and was able to find it: Reversal of Age Digits Every Eleven Years. What he describes there will also apply to you and your daughter: Because you were a multiple of 9 years old when she was born, every 11 years your ages have reversed digits, as they are this year. So not only is your situation this year not unique, it is not unique to you: It also happened when she was 3, and 14, and 25, and 36! But it will stop happening when you pass 100 \u2026\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-03-01 06:09:08", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/a-sample-of-ask-dr-math-part-2-questions-outside-of-school/", "openwebmath_score": 0.667602002620697, "openwebmath_perplexity": 632.7284854220119, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877012965886, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8551769928379297}}
{"url": "https://mathoverflow.net/questions/255474/integral-of-the-entrywise-square-of-the-exponential-of-a-matrix", "text": "# Integral of the entrywise square of the exponential of a matrix\n\nNote: I posted my question on math.stackexchange but got no answer. That is why I am asking it here.\n\nLet $A$ be a $n\\times n$ square matrix such that the real part of all eigenvalues are negative. For each $i,j$, let $\\exp(At)_{ij}$ be the element $(i,j)$ of the matrix. It is well known that: $$\\int_0^\\infty \\exp(At)_{ij}dt = -(A^{-1})_{ij}$$ Is it possible to simplify a similar expression where each element is squared: $$\\int_0^\\infty (\\exp(At)_{ij})^2 dt = ??$$\n\nI am wondering if it is possible to simplify the above expression. If it helps, I can assume that $A$ is diagonalizable. Note that unless for one-dimensional matrices, $(\\exp(At)_{ij})^2\\ne\\exp(2At)_{ij}$.\n\n\u2022 Note that OP wants an elementwise square, not the usual matrix-multiplication square. \u2013\u00a0Federico Poloni Nov 24 '16 at 8:19\n\nInspired strongly by Anthony's answer, here is a formula that works for arbitrary $A$. Let $M$ be the $n^2 \\times n^2$ square matrix given by $$M= A \\otimes I_n + I_n \\otimes A_n$$ i.e. in terms of indices\n\n$$M_{ij,kl}=A_{i,k} \\delta_{j,l} + \\delta_{i,k}A_{j,l}$$\n\nThen because $A \\otimes I_n$ and $I_n \\otimes A$ commute, $$e^{Mt} = (e^{At} \\otimes I_n) (I_n \\otimes e^{At}) = e^{At} \\otimes e^{At}$$ i.e. in indices $$(e^{M t})_{ij,kl} = (e^{A t})_{i,j} (e^{At})_{k,l}$$ so in particular $$(e^{At})_{i,j}^2 = (e^{Mt})_{ii,jj}$$ and $$\\int_t (e^{At})_{i,j}^2 = \\int_t (e^{Mt})_{ii,jj} = - (M^{-1})_{ii,jj}$$\n\n(Here I am using commas to separate the two indices of a matrix entry)\n\n\u2022 Thanks for this answer. I am not sure to understand why the two summands would commute and why $e^{Mt}_{ij,kl}$ can be expressed in terms of $e^{At}$ (with these indices, it seems false). \u2013\u00a0N. Gast Nov 24 '16 at 16:18\n\u2022 I think this is problematic because each of the summands is of rank $n$, so the sum is of rank at most $2n$. For $n>2$, $M$ will not be invertible. \u2013\u00a0Anthony Quas Nov 24 '16 at 16:39\n\u2022 @AnthonyQuas That was a stupid mistake, fixed now. \u2013\u00a0Will Sawin Nov 24 '16 at 18:09\n\u2022 @N.Gast Sorry, that was wrong, is this clearer? \u2013\u00a0Will Sawin Nov 24 '16 at 18:09\n\u2022 @WillSawin: does M^{-1} exist? \u2013\u00a0Anthony Quas Nov 24 '16 at 18:42\n\nYou can do something. Here's a computation for diagonalizable $A$. Let $A=BDB^{-1}$ and let the elements of $D$ be $-\\lambda_1,\\ldots,-\\lambda_d$. Then \\begin{align*} \\int_0^\\infty (e^{At})_{ij}^2\\,dt&= \\int_0^\\infty (Be^{Dt}B^{-1})_{ij}^2\\,dt\\\\ &=\\int_0^\\infty \\sum_{k,k'} B_{ik}e^{-\\lambda_k t}B^{-1}_{kj} B_{ik'}e^{-\\lambda_k' t}B^{-1}_{k'j}\\,dt\\\\ &=\\int_0^\\infty \\sum_{k,k'} B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}e^{-(\\lambda_k+\\lambda_k') t} \\,dt\\\\ &=\\sum_{k,k'} \\frac{1}{\\lambda_k+\\lambda_k'}B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}. \\end{align*}\n\nGiven a Hurwitz matrix $\\mathrm A \\in \\mathbb R^{n \\times n}$, let\n\n$$\\Phi (t) := \\exp(\\mathrm A t)$$\n\nbe the state transition matrix, and let its $(i,j)$-th entry be denoted by\n\n$$\\varphi_{ij} (t) := \\mathrm e_i^{\\top} \\Phi (t) \\, \\mathrm e_j$$\n\nHence,\n\n$$\\begin{array}{rl } \\displaystyle\\int_0^{\\infty} \\left( \\varphi_{ij} (t) \\right)^2 \\, \\mathrm d t &= \\displaystyle\\int_0^{\\infty} \\left( \\mathrm e_i^{\\top} \\Phi (t) \\, \\mathrm e_j \\right)^2 \\, \\mathrm d t\\\\\\\\ &= \\displaystyle\\int_0^{\\infty} \\mathrm e_i^{\\top} \\Phi (t) \\, \\mathrm e_j \\mathrm e_j^{\\top} \\Phi^{\\top} (t) \\, \\mathrm e_i \\, \\mathrm d t\\\\\\\\ &= \\mathrm e_i^{\\top} \\underbrace{\\left( \\displaystyle\\int_0^{\\infty} \\Phi (t) \\, \\mathrm e_j \\mathrm e_j^{\\top} \\Phi^{\\top} (t) \\, \\mathrm d t \\right)}_{=: \\mathrm W_c} \\mathrm e_i = \\mathrm e_i^{\\top} \\mathrm W_c \\mathrm e_i\\end{array}$$\n\nwhere $\\mathrm W_c$ is the controllability Gramian of the pair $(\\mathrm A, \\mathrm e_j)$ and is the solution to the following controllability Lyapunov equation\n\n$$\\boxed{\\mathrm A \\mathrm W_c + \\mathrm W_c \\mathrm A^{\\top} + \\mathrm e_j \\mathrm e_j^{\\top} = \\mathrm O_n}$$\n\nThus, the $n$ columns of the integral of the entrywise product\n\n$$\\int_0^{\\infty} \\left( \\Phi (t) \\circ \\Phi (t) \\right) \\mathrm d t$$\n\nare the diagonals of $\\mathrm W_c^{(1)}, \\mathrm W_c^{(2)}, \\dots, \\mathrm W_c^{(n)}$, where $\\mathrm W_c^{(1)}, \\mathrm W_c^{(2)}, \\dots, \\mathrm W_c^{(n)}$ are the solutions to the following $n$ controllability Lyapunov equations\n\n$$\\begin{array}{cl} \\mathrm A \\mathrm W_c^{(1)} + \\mathrm W_c^{(1)} \\mathrm A^{\\top} + \\mathrm e_1 \\mathrm e_1^{\\top} &= \\mathrm O_n\\\\ \\mathrm A \\mathrm W_c^{(2)} + \\mathrm W_c^{(2)} \\mathrm A^{\\top} + \\mathrm e_2 \\mathrm e_2^{\\top} &= \\mathrm O_n\\\\ \\vdots & \\\\ \\mathrm A \\mathrm W_c^{(n)} + \\mathrm W_c^{(n)} \\mathrm A^{\\top} + \\mathrm e_n \\mathrm e_n^{\\top} &= \\mathrm O_n\\end{array}$$\n\n\u2022 In MATLAB, the Gramian can be computed using function gram. \u2013\u00a0Rodrigo de Azevedo Nov 25 '16 at 22:18", "date": "2019-10-20 18:00:07", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/255474/integral-of-the-entrywise-square-of-the-exponential-of-a-matrix", "openwebmath_score": 0.9651211500167847, "openwebmath_perplexity": 384.71872810287925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877018151065, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8551769883423624}}
{"url": "https://math.stackexchange.com/questions/795952/is-n2-n-1-prime-for-all-n", "text": "# Is $n^2 + n + 1$ prime for all n?\n\nI recently stumbled across this question in a test.\n\nPaul says that \"$n^2+n+1$ is prime $\\forall\\:n\\in \\mathbb{N}$\".\n\n\u2022 Paul is correct, because...\n\u2022 Paul is wrong, because...\n\nThe answers sheet says that \"all answers providing a valid counterexample, e.g. $n=4$, are valid; however, I'm interested in understanding why - is there some sort of theorem which can predict whether a polynomial is prime or not?\n\nWhat I tried was:\n\n$n^2+n+1$ can be factorized as $(n+\\frac{1+\\sqrt{3}i}{2})(n+\\frac{1-\\sqrt{3}i}{2})$. Neither of these factors is a suitable factor (i.e. is natural, different from $1$, and different from $n^2+n+1$) for any $n\\in\\mathbb{N}$, therefore $n^2+n+1$ must be prime $\\forall\\:n\\in\\mathbb{N}$.\nInstead, $n^2-1$ can be factorized as $(n+1)(n-1)$. The factor $(n+1)$ is a suitable factor (i.e. natural, different from $1$, and different from $n^2-1$) for $n>2$; therefore, $n^2-1$ is not prime for $n>2$.\n\nHowever, the above reasoning is clearly wrong, because \"$n^2+n+1$ is prime $\\forall\\:n\\in\\mathbb{N}$\" doesn't hold for the case $n=4$.\nWhat is wrong in the reasoning? Can one tell whether a polynomial $p(n)$ is prime for all $n$ a priori?\n\n\u2022 Counterexample clearly is a formal proof. Furthermore, for quite a lot of small $n$ (like $n \\leq 40$, IIRC - could be wrong), such numbers are prime. \u2013\u00a0Marcin \u0141o\u015b May 15 '14 at 14:32\n\u2022 @Marcin\u0141o\u015b thanks for pointing that out. I edited the question to better suit what I'm asking for. \u2013\u00a0Giulio Muscarello May 15 '14 at 14:36\n\u2022 It was established by Goldbach that there is no polynomial with integer coefficients that can give a prime for all integer inputs. Ref: mathworld.wolfram.com/Prime-GeneratingPolynomial.html \u2013\u00a0Joel May 15 '14 at 14:44\n\nHint $\\ \\ f(1)=3\\,\\Rightarrow\\, 3\\mid f(1\\!+\\!3n),\\$ e.g. $\\,3\\mid f(4)=21$\n\nRemark $\\$ Above we used the Polynomial Congruence Theorem\n\n$\\ {\\rm mod}\\ 3\\!:\\ \\color{}{1\\!+\\!3n\\equiv 1}\\,\\Rightarrow\\, f(\\color{}{1\\!+\\!3n})\\equiv f(\\color{}1)\\equiv 0,\\$ for all $\\,f\\in\\Bbb Z[x]$\n\nAlternatively $\\,\\ 3n\\mid f(1\\!+\\!3n)-f(1)\\$ by the Factor Theorem $\\,x-y\\mid f(x)-f(y)$\n\nThe idea generalizes to a proof that any nonconstant polynomial $\\in\\Bbb Z[x]\\,$ has a non-prime value.\n\nJust because a polynomial $f$ doesn't factor doesn't mean that the number $f(m)$ you get when evaluating $f$ at a number $m$ doesn't factor. The simplest example is something like $f(x) = x+1$. This polynomial is irreducible, hence doesn't factor, but $f(3) = 4$ is clearly composite.\n\nHere is a link showing that no polynomial with integer coefficients can evaluate to primes for all integers.\n\nYou want to disprove a statement of the form $\\forall n\\colon f(n)\\text{ is prime}$, that is you want to prove $\\neg\\forall n\\colon f(n)\\text{ is prime}$, which is equivalent to $\\exists n\\colon \\neg(f(n)\\text{ is prime})$. Therefore, a counterexample is often the easiest and most straightforward way to disprove a general statement. In fact, any proof that merely shows that a $\\forall$ statement is wrong without constructing a counterexample (or essentially doing so, but leaving out the details) is frowned upon by some mathematicians as being non-constructive. That being said, maybe you'd like a proof that there is no polynomial $f$ such that $f(n)$ is prime for all $n\\in \\mathbb N$ (except constant polynomials):\n\nLet $$f(n)=a_dx^d+a_{d-1}x^{d-1}+\\ldots+a_1x+a_0$$ be a polynomial of degree $d\\ge 1$ with coefficients $a_i\\in\\mathbb C$, $0\\le i\\le d$, and $a_d\\ne 0$. First we observe tat in fact $a_i\\in \\mathbb Q$ for all $i$: By plugging in $x=1, 2, \\ldots, d+1$, we obtain $d+1$ rational equations in the unknowns $a_i$. Since the powers of $x$ are linearly independant (or look up Vandermonde matrix), there exists a unique solution, which must be rational. Then there exists $M\\in\\mathbb N$ sich that $a_iM\\in\\mathbb Z$ for all $i$. By assumption, $Mf(n)$ is $M$ times a prime for all $n\\in\\mathbb N$. For each prime $p$, there are at most $d$ values of $n$ such that $f(n)=p$. Since $M$ has only finitely many prime factors, $f(n)\\mid M$ for only finitely many $n$. Therefore there exists $n$ such that $p:=f(n)\\not\\mid M$. By reducing $Mf(X)$ modulo $p$, we see that $Mf(n+kp)\\equiv 0\\pmod p$ for all $k\\in\\mathbb Z$. This implies that the prime $f(n+kp)$ is $p$ for all $k$. Especially, there are more than $d$ values of $n$ for which $f(n)=p$, contradiction. $_\\square$\n\nTrivia remark: On the other hand, there exists a polynomial in several variables with the property that its values are either negative or prime, and all primes occur!\n\n\u2022 Your trivia remark is really cool! Do you have a link/reference? \u2013\u00a0Andr\u00e9 3000 May 15 '14 at 15:22\n\u2022 Here is a link to an early paper by Jones et al, not long after Matiyasevich proved existence. The $25$ variables has gone down a lot since then, I think to about $9$. \u2013\u00a0Andr\u00e9 Nicolas May 15 '14 at 15:38\n\u2022 @Andr\u00e9Nicolas That's cool - so instead of letters we can now use digits to index the variables :) \u2013\u00a0Hagen von Eitzen Dec 7 '16 at 17:09\n\nNo. For $n=43$, $n^2+n+1=1893$ which is divisible by $3$.\n\n\u2022 \"What is wrong in the reasoning? What is a correct proof that a given polynomial $p(n)$ is (or isn't) prime for all natural numbers n?\" Please read the entire question before answering. What I'm looking for is not a counterexample, but rather a proof. \u2013\u00a0Giulio Muscarello May 15 '14 at 14:32\n\u2022 @GiulioMuscarello I understand you don't find this answer satisfying, but it's in fact perfectly valid proof. \u2013\u00a0Marcin \u0141o\u015b May 15 '14 at 14:34\n\u2022 @Marcin\u0141o\u015b Sure, but the OP already had a much smaller counterexample with $n=4$. I don't see what this answer really adds. \u2013\u00a0Andr\u00e9 3000 May 15 '14 at 14:38\n\u2022 Jika is probably thinking of $n^2-n+41$, which is prime for $0 \\le n \\le 40$ \u2013\u00a0Robert Israel May 15 '14 at 15:06\n\u2022 @RobertIsrael Yes you are right. Even though, I was thinking of Euler's formula $n^2-n+41$, I found that $43$ is a counter example. I hesitate to post my answer but since it is a valid mathematical proof I did post it. \u2013\u00a0Jika May 15 '14 at 15:15\n\nModulo 3, that polynomial is equivalent to some well known polynomial $(n-1)^2$ $$(n-1)^2=n^2-2n+1\\equiv n^2+n+1 [3]$$\n\nHence if $n=1+3k$ for any $k$ (so each time $n-1\\equiv 0[3]$) then $n^2+n+1$ will have 3 as a divisor.", "date": "2020-02-21 19:18:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/795952/is-n2-n-1-prime-for-all-n", "openwebmath_score": 0.8280494809150696, "openwebmath_perplexity": 218.22729550335063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877023336243, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8551769871447867}}
{"url": "https://math.stackexchange.com/questions/1604790/is-my-inductive-proof-correct", "text": "# Is my inductive proof correct?\n\nTrying this again.\n\nGiven $f(n) = 2f(n-1) + 1$ with $f(0) = 0$, I guess that $f(n) = 2^n-1$.\n\nBase case: $f(0) = 2^0 - 1 = 1 - 1 = 0$, true.\n\nInductive step: Suppose $f(n) = 2^n-1$ for some $n \\geq 0$. I will show that $f(n+1) = 2^{n+1}-1$.\n\n$f(n+1) = 2f(n) + 1$\n\n$f(n+1) = 2(2^n-1) + 1$\n\n$f(n+1) = 2^{n+1} - 1$\n\nThis completes the proof.\n\nMy questions:\n\n1. Is this proof correct? Awkward? Backwards?\n\n2. It would help me to get the terminology right. Which piece is the inductive hypothesis? Or the \"ansatz\"?\n\n\u2022 This one is good, clean, \"forward.\" Jan 8 '16 at 19:53\n\nAs far as terminology, the inductive hypothesis is the thing you assume is true at some $n$, and is used to prove the statement for $n + 1$. So your inductive hypothesis is that $f(n) = 2^n - 1$.\nThe ansatz is the educated guess that the solution is $2^n - 1$, likely based on some experimentation. The ansatz then became your induction hypothesis (as it is wont to do).\nLooks good to me. The inductive hypothesis is the step where you assume $f(n)=2^n-1$. We assume the inductive hypothesis because we have shown the existence of such an $n$ in our base case.", "date": "2021-10-17 09:26:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1604790/is-my-inductive-proof-correct", "openwebmath_score": 0.9386552572250366, "openwebmath_perplexity": 298.7339466005355, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877028521422, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8551769826492193}}
{"url": "https://math.stackexchange.com/questions/1306387/does-liebniz-criteria-become-a-necessary-condition-for-convergence-if-a-n-is-m", "text": "# Does Liebniz-Criteria become a necessary condition for convergence if $a_n$ is monotonically decreasing?\n\nThe Leibniz Criterion says that if the sequence $a_n$ is monotonically decreasing then the following statements are equivalent:\n\n\\begin{align} 1) & & & \\sum_{n=0}^\\infty (-1)^na_n \\text{ converges} \\\\[6pt] 2) & & & \\lim_{n\\rightarrow\\infty}(a_n) = 0 \\end{align}\n\nso this has to mean that if $a_n$ is monotonically decreasing, the following is true\n\n$$\\sum_{n=0}^\\infty (-1)^na_n \\text{ converges} \\Leftrightarrow \\lim_{n\\rightarrow\\infty}(a_n) = 0$$\n\nI know that $\\lim_{n\\rightarrow\\infty}(a_n) = 0$ is not a necessary condition for convergence if $a_n$ is not monotonically decreasing, but the way I am reading the rule stated in my script, it becomes a necessary condition if $a_n$ is monotonically decreasing.\n\nFor example to prove the non-convergence of\n\n$$\\sum_{n=0}^\\infty (-1)^n(1)_n$$\n\nI can just point out to the fact that\n\n$$\\lim_{n\\rightarrow\\infty}(1)_n = 1$$\n\nbecause $(1)_n$ is a monotonically decreasing sequence. Is this interpretation correct?\n\n\u2022 The condition $b_n\\to 0$ is always necessary to the convergence of $\\sum b_n$. When $b_n=(-1)^n a_n$, one has the equivalence $b_n\\to 0\\ \\iff\\ a_n\\to 0$. So $a_n\\to 0$ is a necessary condition for the convergence of $\\sum (-1)^n a_n$. \u2013\u00a0Giuseppe Negro May 31 '15 at 11:49\n\u2022 Yes the statement is true. Proof: $$b_n\\to 0\\ \\iff\\ \\lvert b_n\\rvert \\to 0 \\iff \\lvert (-1)^n a_n\\rvert\\to 0 \\iff a_n\\to 0.$$ If you are not convinced please post a counterexample. \u2013\u00a0Giuseppe Negro May 31 '15 at 11:59\n\u2022 if it were so why would leibniz go through the trouble of putting the additional monotonic decreasing condition there \u2013\u00a0Gravity May 31 '15 at 11:59\n\u2022 I guess you are mixing up \"necessary\" and \"sufficient\" here. $a_n\\to 0$ is necessary to the convergence of $\\sum (-1)^na_n$. $a_n\\to 0$ AND $a_n$ monotonically decreasing is sufficient to the convergence of $\\sum (-1)^n a_n$. \u2013\u00a0Giuseppe Negro May 31 '15 at 12:01\n\u2022 I know realize that you were right \u2013\u00a0Gravity May 31 '15 at 12:33\n\nAs Giuseppe Negro has noticed, for the convergence of $\\displaystyle \\sum a_n$,$$\\lim_{n\\rightarrow\\infty}(a_n) = 0$$ is a necessary condition anyway.\n\nThe convergence test of Leibniz is only applicable to alternating series, i.e. series where every positive term is followed by a negative term, and every negative term is followed by a positve term ( + - + - + - + - ...).\n\nFor a general series, the condition $\\displaystyle\\lim_{n \\to \\infty} a_n= 0$ is a necessary condition for convergence, but not sufficient. This is the main point of Leibniz test. He could prove that for alternating series this condition is both necessary and sufficient. Thus if you've got an alternating series $\\sum (-1)^n a_n$ with $\\displaystyle\\lim_{n \\to \\infty} a_n= 0$, than you are certain it converges.\n\nI don't know why you are referring to the condition of \"monotonically decreasing\". This doesn't matter for Leibniz test. It's only an issue in the integral test.\n\n\u2022 I understand your first point, but on your second point$\\sum (-1)^n a_n$ with $\\displaystyle\\lim_{n \\to \\infty} a_n= 0$ you are certain that is converges, only when a is monotonically decreasing. The rule is defined for monotonically decreasing a \u2013\u00a0Gravity May 31 '15 at 12:36\n\u2022 @Steven Van Geluwe: Alternation of signs and terms going to $0$ is not sufficient for convergence. Example, let $a_n=\\frac{1}{n}$ if $n$ is odd, and let $a_n=\\frac{1}{2^n}$ when $n$ is even. Then $\\sum_1^\\infty (-1)^n a_n$ diverges. \u2013\u00a0Andr\u00e9 Nicolas May 31 '15 at 12:52\n\u2022 @Gravity: There are many series $\\sum (-1)^n a_n$ that converge even though the $a_n$ are not monotonically decreasing. For example, let $a_n=\\frac{1}{n^2}$ when $n$ is odd, and let $a_n=\\frac{1}{n^3}$ when $n$ is even. \u2013\u00a0Andr\u00e9 Nicolas May 31 '15 at 12:59\n\u2022 @Andre: that is correct and that is why Leibniz says the condition is sufficient but not necessary, but still we can only apply the Leibniz rule if $a_n$ is monotonically decreasing. we cannot in all Circumstances say $\\sum (-1)^n a_n$ converges when $\\displaystyle\\lim_{n \\to \\infty} a_n= 0$ is true, and that is the point \u2013\u00a0Gravity May 31 '15 at 13:05\n\u2022 I down voted, because I am afraid that the main point of this answer is wrong. Leibniz's test requires monotonicity of the $a_n$ term. \u2013\u00a0Giuseppe Negro May 31 '15 at 21:25", "date": "2019-08-21 22:29:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1306387/does-liebniz-criteria-become-a-necessary-condition-for-convergence-if-a-n-is-m", "openwebmath_score": 0.8083738088607788, "openwebmath_perplexity": 287.6211865194675, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877012965886, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8551769796459625}}
{"url": "https://math.stackexchange.com/questions/4169353/is-k-1-bipartite", "text": "# Is $K_1$ bipartite?\n\nI'm thinking that it could be trivially bipartite since it only has one vertex and no edges but I am still a little bit unsure about it being trivially bipartite.\n\n\u2022 Ummm... what is \"k sub 1\"? Do you instead mean $K_1$? \u2013\u00a0David G. Stork Jun 10 at 19:01\n\u2022 not only is it bipartite, it is also unipartite \u2013\u00a0Yorch Jun 10 at 19:02\n\u2022 @DavidG.Stork Yes, did not know how to make it like that in my title. Seemed to fix itself. \u2013\u00a0ThreeRingBinder Jun 10 at 20:23\n\nDepends on your definitions. (As pointed out in Matthew Daly's answer, $$K_1$$ should be bipartite, because it has no odd cycles, so it would otherwise be an awkward exception. But not all definitions will play nicely with this desired property.)\n\nWest's Introduction to Graph Theory says\n\n1.1.10. Definition. A graph $$G$$ is bipartite if $$V(G)$$ is the union of two disjoint (possibly empty) independent sets called partite sets of $$G$$.\n\nSo under this definition, if $$V(K_1) = \\{v\\}$$, then we let $$\\{v\\}$$ be one partite set, and $$\\varnothing$$ be the other; $$K_1$$ is bipartite.\n\nBondy and Murty write\n\nA graph is bipartite if its vertex set can be partitioned into two subsets $$X$$ and $$Y$$ so that every edge has one end in $$X$$ and one end in $$Y$$\n\nwhich still works just fine, setting $$X = \\{v\\}$$ and $$Y = \\varnothing$$. They do not specifically point out that $$X$$ or $$Y$$ could be empty, but they do not rule it out either. (Elsewhere, Bondy and Murty talk about \"nontrivial partitions\" or \"partitions into nonempty parts\", which make it clear that a partition without this qualifier is allowed to have an empty part.)\n\nThey are in better shape than Diestel, who has\n\nA graph $$G = (V, E)$$ is called $$r$$-partite if $$V$$ admits a partition into $$r$$ classes such that every edge has its ends in different classes\n\nwith an earlier qualification that the classes of a partition may not be empty. Since $$K_1$$ should be bipartite by any reasonable definition, Diestel is in the wrong here. (Diestel later claims that all graphs with no odd cycles are bipartite, with no mention of $$K_1$$ as a special exception.)\n\nIf we treat \"bipartite\" as synonymous with \"$$2$$-colorable\", then $$K_1$$ is happily bipartite, since any function on its vertex set is a $$2$$-coloring (and also a $$1$$-coloring).\n\n\u2022 Quick question. Lets say we have \ud835\udc4b={\ud835\udc63} and \ud835\udc4c=\u2205, in reference to Bondy and Murty's definition. Would the edge connecting both subsets have vertices that are the empty set? Ie one empty set from X and the other from Y \u2013\u00a0ThreeRingBinder Jun 11 at 4:36\n\u2022 The definition says every edge has a certain property; it doesn't say there exist any such edges. Since $K_1$ has no edges, the definition holds. \u2013\u00a0Misha Lavrov Jun 11 at 5:24\n\u2022 Ah, makes sense now. Thanks! \u2013\u00a0ThreeRingBinder Jun 11 at 6:51\n\nMy graph theory professor spent a while teaching on this, because it's a mess either way.\n\nIt probably should be trivially considered bipartite. Otherwise you have to put disclaimers in all of your theorems (like the fact that $$K_1$$ doesn't contain any odd cycles). But $$\\{\\{v\\},\\emptyset\\}$$ is not a valid partition of the vertex set, since it contains the empty set as a member.\n\n\u2022 A reasonable definition of \"bipartite\" does not run into trouble if one partite set is empty. \u2013\u00a0Misha Lavrov Jun 10 at 19:24\n\u2022 @MishaLavrov On the other hand, requiring the parts of a partition to be non-empty isn't entirely unheard of either. So defining bipartite as a partition of the set of vertices into two can run into issues with $K_1$. \u2013\u00a0Arthur Jun 10 at 19:33\n\u2022 Yeah, I think that Dr. Sch\u00e4ffer's solution was to have us write in the margin of Bondy and Murty \"... or $K_1$\". \u2013\u00a0Matthew Daly Jun 10 at 19:44\n\u2022 But Bondy and Murty are totally fine with empty parts in a partition, as far as I can tell... \u2013\u00a0Misha Lavrov Jun 10 at 20:22\n\u2022 @MishaLavrov As I say, my professor assumed that \"partitioned into two subsets $X$ and $Y$\" means that $\\{X,Y\\}$ was a partition of the vertex set, which could not be legally done if one of the sets was empty. Bondy and Murty clearly either thought that it was okay or they hadn't considered the edge case, but pedantry like that got under my professor's skin. ^_^ \u2013\u00a0Matthew Daly Jun 10 at 20:37", "date": "2021-06-23 18:07:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4169353/is-k-1-bipartite", "openwebmath_score": 0.769773006439209, "openwebmath_perplexity": 566.5530889538169, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9481545348152283, "lm_q2_score": 0.9019206719160033, "lm_q1q2_score": 0.8551601751207563}}
{"url": "https://math.stackexchange.com/questions/4230041/do-we-have-to-expand-the-brackets-before-applying-de-morgans-laws/4230048", "text": "# Do we have to expand the brackets before applying De Morgan's Laws?\n\nIf we have the example:\n\n$$\\overline{(A + B + C)D}$$\n\nThen can we apply De Morgan's Law as is, or do we first need to expand out the brackets?\n\nIf I expand the terms first, I get:\n\n$$original: \\overline{(A + B + C)D}$$ $$expanded: \\overline{AD + BD + CD}$$ $$applying De Morgan's: \\overline{AD}.\\overline{BD}.\\overline{CD}$$ $$simplifying: \\overline{A}\\overline{B}\\overline{C}\\overline{D}$$\n\nIn contrast, if I don't expand the brackets, I get something like:\n\n$$original: \\overline{(A + B + C)D}$$ $$applying De Morgan's: \\overline{(A + B + C)}+\\overline{D}$$ $$applying De Morgan's: (\\overline{A} . \\overline{B + C})+\\overline{D}$$ $$applying De Morgan's: (\\overline{A} . \\overline{B} . \\overline{C})+\\overline{D}$$ $$simplifying: \\overline{A} . \\overline{B} . \\overline{C}+\\overline{D}$$\n\nSo you can see, I'm getting different answers. I'm curious which method is correct - I think the first seems correct, but perhaps the 2nd is correct or both are wrong.\n\nIf anyone is able to explain why one particular method is wrong, I'd appreciate it too.\n\n\u2022 MathJax tip: when writing text in mathematics code, use the \\text{} command to get proper spacing and formatting. For example, $$\\text{applying De Morgan's} : \\overline{(A + B + C)} + \\overline{D}$$ produces$$\\text{applying De Morgan's} : \\overline{(A + B + C)} + \\overline{D}$$ Aug 22, 2021 at 0:54\n\u2022 Thanks. Now I will do that in future :) Aug 22, 2021 at 5:58\n\u2022 @RandomUser123 You can also do that now. Questions and answers can be edited, and indeed should be edited if they can be improved. Aug 22, 2021 at 8:45\n\nBoth lead to the same result; your error is in the first one when you simplify from $$\\overline{AD}.\\overline{BD}.\\overline{CD}$$ to $$\\overline{A}\\overline{B}\\overline{C}\\overline{D}$$.\nIf you apply De Morgan's law to each pair, you get: $$(\\overline{A}+\\overline{D})(\\overline{B}+\\overline{D})(\\overline{C}+\\overline{D}).$$ From here, you can apply the distributive law to get what you got by the second method in the next to last line. The simplifcation there is also incorrect because you cannot simply drop the parenthesis; you must apply the distributive property.\n\u2022 Makes sense. I was viewing it as $$\\overline{ADBDCD}$$ then just cancelling removing the unneccessary duplicates. Because the NOTs were over multiple variables, I shouldn't be able to do that. I also tried myself, expanding out those 3 sets of brackets and surpisingly managed to simplify down that mess to the correct answer...seeing that all of the brackets contain D' and using your suggestion of using the distributive law is much easier though...I guess it's a good way to check...since I worry if I use these shortcuts, I will apply it incorrectly/in a situation that's invalid or something Aug 22, 2021 at 6:17", "date": "2022-08-16 08:25:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4230041/do-we-have-to-expand-the-brackets-before-applying-de-morgans-laws/4230048", "openwebmath_score": 0.7734267711639404, "openwebmath_perplexity": 418.070640968636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808747730899, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8551454556887333}}
{"url": "https://math.stackexchange.com/questions/2465803/whats-the-probability-that-a-given-permutation-has-exactly-k-fixed-points/2465817", "text": "# What's the probability that a given permutation has exactly $k$ fixed points. [duplicate]\n\nGiven a random permutation $$\\sigma \\in S_n$$ from $$[n] \\to [n]$$ in a uniform probability space, what is the probability that $$\\sigma$$ has exactly $$k$$ fixed points for a given $$k$$ between $$1$$ and $$n$$?\n\nIn other words: what is the probability that $$\\exists x_1 ,...,x_k \\in [n] : \\sigma (x_i) = x_i$$ for $$\\ i\\in \\{1,...,k\\}$$ and for every $$y \\notin \\{x_1 , ... , x_k\\}$$ we get $$\\sigma(y) \\neq y$$.\n\nI saw that $$\\lim_{n \\to \\infty } prob(A_0) = e^{-1}$$ using Inclusion\u2013exclusion principle and i belive that for a given k : $$\\lim_{n \\to \\infty} prob(A_k) = \\frac{e^{-1}}{k!}$$ but I am not sure how to show it.\n\n*$$A_k$$ stands for the event \"k\".\n\n\u2022 The number $D_{n,k}$ of permutations with exactly $k$ fixpoints of an $n$-set is called a rencontres number. \u2013\u00a0Jeppe Stig Nielsen Oct 10 '17 at 11:36\n\nThere are $\\binom{n}{k}$ possibilities for the $k$ fixed points.\n\nIf they are established then there are $!(n-k)$ derangements for the other points.\n\nThat gives $\\binom{n}{k}\\left[!(n-k)\\right]$ permutations with exactly $k$ fixed points on a total of $n!$ permutations.\n\nAlso we have the formula: $$!(n-k)=(n-k)!\\sum_{i=0}^{n-k}\\frac{(-1)^i}{i!}$$ and we end up with probability: $$\\frac1{k!}\\sum_{i=0}^{n-k}\\frac{(-1)^i}{i!}$$\n\n\u2022 Very cool! As $n$ grows (for fixed $k$), this approaches $\\frac1{k! e}$ \u2013\u00a0MichaelChirico Nov 6 '19 at 11:31\n\u2022 And also by corollary, the probability of at least one fixed point approaches $1-\\frac1{e}$ \u2013\u00a0MichaelChirico Nov 6 '19 at 11:38\n1. Number of dearrangements of $k$-elements set is $$!k = k!\\sum\\limits_{m=0}^{k}\\frac{(-1)^m}{m!}$$\n2. In n-elements set we can select $(n-k)$ elements to dearrange them (all remaining $k$ points are fixed) in $\\binom{n}{k}$ ways\n\nThus $$A_k^n = \\binom{n}{k}(n-k)!\\sum\\limits_{m=0}^{n-k}\\frac{(-1)^m}{m!}$$ And probability is $$P(A_k^n) = \\frac{\\binom{n}{k}(n-k)!\\sum\\limits_{m=0}^{n-k}\\frac{(-1)^m}{m!}}{n!}=\\frac{\\frac{n!}{k!}\\sum\\limits_{m=0}^{n-k}\\frac{(-1)^m}{m!}}{n!}=\\frac{1}{k!}\\sum\\limits_{m=0}^{n-k}\\frac{(-1)^m}{m!}$$\n\nBy way of enrichment here is an alternate formulation using combinatorial classes. The class of permutations with fixed points marked is\n\n$$\\def\\textsc#1{\\dosc#1\\csod} \\def\\dosc#1#2\\csod{{\\rm #1{\\small #2}}} \\textsc{SET}(\\mathcal{U} \\times \\textsc{CYC}_{=1}(\\mathcal{Z}) + \\textsc{CYC}_{=2}(\\mathcal{Z}) + \\textsc{CYC}_{=3}(\\mathcal{Z}) + \\cdots).$$\n\nThis gives the generating function $$G(z, u) = \\exp\\left(uz + \\frac{z^2}{2} + \\frac{z^3}{3} + \\frac{z^4}{4} + \\frac{z^5}{5} + \\cdots\\right)$$\n\nwhich is\n\n$$G(z, u) = \\exp\\left(uz-z+\\log\\frac{1}{1-z}\\right) = \\frac{\\exp(uz-z)}{1-z}.$$\n\nNow for $$k$$ fixed points we get\n\n$$[u^k] \\frac{\\exp(uz-z)}{1-z} = [u^k] \\frac{\\exp(uz)\\exp(-z)}{1-z} = \\frac{z^k}{k!} \\frac{\\exp(-z)}{1-z}.$$\n\nThis is the EGF of permutations having $$k$$ fixed points. We extract the count by computing (the factor $$n!$$ is canceled because we require the average)\n\n$$[z^n] \\frac{z^k}{k!} \\frac{\\exp(-z)}{1-z} = \\frac{1}{k!} [z^{n-k}] \\frac{\\exp(-z)}{1-z}.$$\n\nWe find\n\n$$\\bbox[5px,border:2px solid #00A000]{ \\frac{1}{k!} \\sum_{q=0}^{n-k} \\frac{(-1)^q}{q!}.}$$\n\nWe can identify this as choosing the $$k$$ fixed points and combining them with a derangement of the rest:\n\n$$\\frac{1}{n!} {n\\choose k} (n-k)! \\sum_{q=0}^{n-k} \\frac{(-1)^q}{q!}$$\n\nwhich is the combinatorial class\n\n$$\\textsc{SET}_{=k}(\\mathcal{Z}) \\times \\textsc{SET}(\\textsc{CYC}_{\\ge 2}(\\mathcal{Z})).$$", "date": "2021-08-01 23:16:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2465803/whats-the-probability-that-a-given-permutation-has-exactly-k-fixed-points/2465817", "openwebmath_score": 0.951894223690033, "openwebmath_perplexity": 253.2797640456319, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808736209154, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.855145454683637}}
{"url": "https://math.stackexchange.com/questions/3600695/show-that-sqrt1x1-fracx2-for-all-x0/3600740", "text": "# Show that $\\sqrt{1+x}<1+\\frac{x}{2}$ for all $x>0$\n\nI am a little stuck on this question and would appreciate some help. The question asks me to prove that $$\\sqrt{1+x}<1+\\frac{x}{2}$$ for all $$x>0$$.\n\nI squared both sides of the question to get $$1+x<\\frac{x^2}{4}+x+1$$ for all $$x>0$$. Then, I multiplied both sides by $$4$$ to get $$4+4x for all $$x>0$$.\n\nI am a little stuck and was wondering what to do after this step and how to actually provide sufficient proof to say that this statement is true.\n\nYou did well. Let's finish it: we have $$x^2>0$$ thus $$\\dfrac{x^2}{4}>0$$. Adding $$x+1$$ to both sides, we have $$\\dfrac{x^2}{4}+x+1> x+1$$ or $$(\\frac{x}{2}+1)^2>x+1$$ or $$\\frac{x}{2}+1>\\sqrt{x+1}$$\n\nSquaring the equation was sufficient. After you cancel the $$1+x$$ on each side, you have\n\n$$\\frac{x^2}{4} > 0$$\n\nwhich is true for all real $$x \\ne 0$$ since $$x^2 \\ge 0$$ (with equality only when $$x=0$$). Thus, the inequality is proved.\n\nYou can also prove the inequality using the Mean Value Theorem. Define $$f(t)=\\sqrt{t+1}$$ and apply the theorem on the interval $$[0,x]$$. Then, there exists $$c\\in(0,x)$$ s.t. $$\\frac{\\sqrt{x+1}-\\sqrt{1}}{x-0}=\\frac{1}{2\\sqrt{c+1}}>\\frac{1}{2}.$$\n\nI am answering less to provide a list of steps to follow for proving the inequality, and more to discuss how I believe that the argument should be presented. This seems appropriate, as the question has been tagged with and . If I asked a student to prove some inequality and their first step was to assume the inequality and square both sides, I would almost certainly deduct points unless the student were careful about justifying such a step. The goal here is to start with known true statements and to then deduce the desired result.\n\nClaim: For any $$x > 0$$, $$\\sqrt{1+x} < 1 + \\frac{x}{2}.$$\n\nProof: Assume that $$x > 0$$. The square of any real number is positive, hence $$0 < \\left( \\frac{x}{2} \\right)^2.$$ By additive cancelation (i.e. by \"adding the same number to both sides\"), this implies that $$x + 1 < \\left( \\frac{x}{2} \\right)^2 + x + 1 = \\left( \\frac{x}{2} + 1 \\right)^2.$$ The square root function is increasing on $$[0,\\infty)$$ (that is, if $$0 \\le a < b$$, then $$0 \\le \\sqrt{a} \\le \\sqrt{b}$$), and the assumption that $$x > 0$$ ensures that $$1+x > 0$$, thus $$1 + x$$ is in the domain of the square root function (i.e. it has a well-defined real square root). Hence $$\\sqrt{x+1} < \\sqrt{ \\left( 1 + \\frac{x}{2} \\right)^2 } = \\left|1 + \\frac{x}{2}\\right| = 1 + \\frac{x}{2},$$ where the final equality follows from the assumption that $$x > 0$$.\n\n### Discussion\n\nRemember that if you start with false assumptions, you can prove anything\u2014false implies true. Thus it is poor practice to begin a proof by asserting the statement which you are trying to prove and then applying algebraic manipulation. You can do this if you are careful\u2014be very observant of two-sided implications and check the hypotheses at each step. Indeed, this is a very reasonable way of deducing a correct proof in the first place.\n\nHowever, when presenting an argument, such a proof is typically poor style. Personally, I think that it is better (stylistically, which is a matter of taste) for every statement in a proof to follow from previous statement, and not depend on future statements via \"if and only ifs\".\n\nI will also note that the proof presented above is very elementary\u2014it doesn't rely on any deep theorems. It is, however, somewhat tedious. More powerful theorems give quicker, more elegant proofs. For example, El31's proof, via the mean value theorem, is quite slick. However, when first learning a topic, I think that one should focus on finding an elementary proof first: find such a proof first. Then, if need be, look for something more elegant.\n\nThe important principle here is that if $$a,b$$ are nonnegative real numbers, then $$a iff $$a^2 This property is used all the time. If we grant this, then all we need to show is that\n\n$$(\\sqrt{1+x})^2 < (1+x/2)^2,$$\n\nwhich is the same as saying $$1+x < 1+x + x^2/4.$$ That is obviously true, and we're done.\n\nObserve that : $$\\left(\\forall x>0\\right),\\ \\sqrt{1+x}-1-\\frac{x}{2}=-\\frac{x^{2}}{4}\\int_{0}^{1}{\\frac{1-t}{\\left(1+tx\\right)\\sqrt{1+tx}}\\,\\mathrm{d}t} < 0$$\n\n\u2022 No offense meant, but this is perhaps one of the most obtuse correct answers I've seen on this site. \u2013\u00a0jawheele Mar 30 at 5:51", "date": "2020-07-05 01:45:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3600695/show-that-sqrt1x1-fracx2-for-all-x0/3600740", "openwebmath_score": 0.9913337230682373, "openwebmath_perplexity": 700.3236085991628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808690122163, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8551454539165182}}
{"url": "http://mathhelpforum.com/trigonometry/67569-solve-equations-0-x-2pi.html", "text": "# Math Help - Solve Equations 0< or = x < or = 2pi\n\n1. ## Solve Equations 0< or = x < or = 2pi\n\nHow would you solve these\n\nSolve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.\n\nAnswear is in form of this\npi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?\n\n2. Originally Posted by j0nath0n3\nHow would you solve these\n\nSolve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.\n\nAnswear is in form of this\npi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?\nNote that $\\sin^2 x=1-\\cos^2 x$\n\nSo we can rewrite the equation as $\\cos^2 x -\\left[1-\\cos^2 x\\right]=-\\cos x$\n\nThis simplifies to $2\\cos^2 x-1=-\\cos x\\implies 2\\cos^2x+\\cos x-1=0$\n\nNow, if we make the substitution $z=\\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\\implies (2z-1)(z+1)=0$\n\nThis tells us that $z=-1$ or $z=\\tfrac{1}{2}$\n\nBut $z=\\cos x$, so now that means either $\\cos x=\\tfrac{1}{2}$ or $\\cos x=-1$\n\nNow can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?\n\n3. Originally Posted by Chris L T521\nNote that $\\sin^2 x=1-\\cos^2 x$\n\nSo we can rewrite the equation as $\\cos^2 x -\\left[1-\\cos^2 x\\right]=-\\cos x$\n\nThis simplifies to $2\\cos^2 x-1=-\\cos x\\implies 2\\cos^2x+\\cos x-1=0$\n\nNow, if we make the substitution $z=\\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\\implies (2z-1)(z+1)=0$\n\nThis tells us that $z=-1$ or $z=\\tfrac{1}{2}$\n\nBut $z=\\cos x$, so now that means either $\\cos x=\\tfrac{1}{2}$ or $\\cos x=-1$\n\nNow can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?\nThats really good and I understood however my answear choices are in radian... and I dont understand why..\n\n4. Originally Posted by j0nath0n3\nThats really good and I understood however my answear choices are in radian... and I dont understand why..\nIn general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\\text{ degrees}=2\\pi\\text{ radians}$.\n\n5. ## Re :\n\nOriginally Posted by Chris L T521\nIn general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\\text{ degrees}=\\pi\\text{ radians}$.\nChris , is it a typo , i thought $360\\text{ degrees}=2\\pi\\text{ radians}$\n\nChris , is it a typo , i thought $360\\text{ degrees}=2\\pi\\text{ radians}$", "date": "2015-05-30 00:30:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/67569-solve-equations-0-x-2pi.html", "openwebmath_score": 0.9478390216827393, "openwebmath_perplexity": 336.28656845453906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808672839539, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8551454442757034}}
{"url": "https://fr.bluerocktel.com/tj/30098162dca83fae7b8fd2b53a39a9c711", "text": "In fact, permutation is another term used to describe bijective functions from a finite set to itself. a permutation is an arrangement of $n$ elements without repetition. A big part of discrete mathematics is about counting things. Permutations with Repetition | Discrete Mathematics. Combinations - Permutions. P R (4, 2) = 4 2 = 16. A phone number is an example of a ten number permutation; it is drawn from the set of the integers 0-9, and the order in which they are arranged in matters. I Pascal's triangle is perfectly symmetric I Numbers on left are mirror image of numbers on right I Why is this the case? IApermutationof a set of distinct objects is anordered arrangement of these objects. Likewise, [triangle, melon, airplane] is a permutation of three objects as well. Discrete Mathematics - Quick Guide, Mathematics can be broadly classified into two categories A permutation is an arrangement of some elements in which order matters. Permutations are used when we are counting without replacing objects and order does matter. Therefore, the number of different permutations of a You only need the decomposition in disjoint cycles. Case1: Let G={ 1 } element then permutation are S n or P n = Case 2: Let G= { 1, 2 } elements then permutations are . About the journal. It is used to create a pairwise relationship between objects. Combinatorics is the study of arrangements of objects, it is an important part of discrete mathematics. Cyclic Notation In the former article, we saw various ideas behind multiple formulas and theorems in discrete math concerning permutations. functions in discrete mathematics ppthank aaron rookie cards. So total ways are. We must count objects to solve many different types of problems, like the Incycle notation, we represent a cyclic permutation by this cycle. including groups, rings, and fields. The Mathematics Department of the Rutgers School of Arts and Sciences is one of the oldest mathematics departments in the United States, graduating its first major in 1776 Cornette, Discrete Mathematics - Counting Theory The Rules of Sum and Product. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. North East Kingdoms Best Variety best order to read the old testament; sandman hotel victoria bed bugs; functions in discrete mathematics ppt }\\) There are $$3! One byte consists of 8 bits. A permutation refers to a selection of objects from a set of objects in which order matters. }$$ Suppose that $$A = \\{1, 2, 3\\}\\text{. Permutation and combination are the ways to represent a group of objects by selecting them in a 4! A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. = 5!/3! Open-source computer software for working with permutations includes the GAP suite of Wednesday, December 28, 2011. In computer science and discrete mathematics, an inversion in a sequence is a pair of elements that are out of their natural order . For example, suppose we want to multiply (1;5)(2;3;6)(1;6;4)(3;5). Discrete Mathematics provides a common forum for significant research in many areas of discrete mathematics and combinatorics . MATH 3336 Discrete Mathematics Combinations and Permutations (6.3) Permutations Definition: A permutation of a set of distinct objects is an ordered arrangement of these objects. Permutation: In mathematics, one of several ways of arranging or picking a set of items. Discrete Mathematics MCQ Questions with Answers is a PDF booklet containing 50 MCQ questions and answers on topics such as counting, sets, sequences and permutations, Permutations. A permutation of \\(n$$ distinct objects is just a listing of the objects in some order. Input: N = 7668. DISCRETE MATHEMATICS Permutations and combinations Book arrangement problems Combinations and Permutations Worksheet 9 Permutation Word Problems Explained the Easy B9. In general P(n, k) means the number of permutations of n objects from which we take k objects. MATH 25400. IExample: S = fa;b;cg. Permutations Permutations Example = (4 5)(2 3) = (4 5)(2 1) 1 A classic example asks how many different words can be obtained by re-ordering the letters in = n (n - 1) (n - 2) (n - 3) . 1. We are going to pick (select) r objects from the urn in sequence. In other words a Permutation is an ordered Combination of elements. Malte Helmert, Gabriele R oger (University of Basel)Discrete Mathematics in Computer Science October 21, 2020 10 / 20 B9. For example: Math 3336 Section 6. Calculate the permutations for P R (n,r) = n r. For n >= 0, and r >= 0. It's free to sign up and bid on jobs. 1 Discrete Math Basic Permutations and Combinations Slide 2 Ordering Distinguishable Objects When we have a group of N objects that are Use Wolfram|Alpha to apply and understand these and related concepts. INo object can be selected more than once. Before we discuss permutations we are going to have a look at what the words combination means and permutation. You should practice these MCQs for 1 Recommended: Please try your approach on {IDE} first, before moving on to the solution. Honors Discrete Mathematics. Discrete mathematics deals with areas of mathematics that are discrete, as opposed to continuous, in nature. For instance, in how many ways can a panel of jud A permutation is an arrangement of some elements in which order matters. Combinatorics . Case 3: Let G={ 1, 2, 3 } elements then permutation are 3!=6. In several IOrder of arrangement matters. Find the factorial n! Alternatively, the permutations formula is expressed as follows: We study generating functions for the number of even (odd) permutations on n letters avoiding 132 and an arbitrary permutation on k letters, or containing exactly once. Example: How many different ways can 3 students line up to purchase a new textbook reader? Permutations. - These MCQs cover theoretical concepts, true-false(T/F) statements, fill $. Summary of permutations. Such kind of finite studies are involved in discrete mathematics. One well-known zero-inflated model is Diane Lambert's zero-inflated Poisson model, which concerns a random event containing excess zero-count data in unit time. To permute a list is to rearrange its elements. You must consider also all the ways to arrange the people into the positions. The formula for Permutations Replacement or Repetition is P R (n,r)=n r. Substituting the values of n, r in the formula and we get the equation as follows. The Rule of Sum and Rule of Product are used to decompose difficult counting problems into Permutations. Solution: n-factorial gives the number of permutations of n items. Combinations. An injective function (mapping) of a finite set into itself is called a permutation.There follows from the definition of finite sets that shows that such a function is necessarily also surjective and consequently a permutation is always bijective. Permutations. Below is the few Discrete mathematics MCQ test that checks your basic knowledge of Discrete mathematics Discrete Math Textbook Solutions and Answers Discrete MathChapter 14 Combination: A combination of a set of distinct objects is just a count of the number of ways a specific number of elements can be selected from a set of a certain size. Permutations Permutations Cycle Notation { Algorithm Let be a permutation of nite set S. 1: function ComputeCycleRepresentation(, S) 2: remaining = S 3: cycles = ; 4: while remaining is not empty do 5: Remove any element e from remaining. Let A be a ith n elements. To multiply permutations, trace through the images of points, and build a new permutation from the images, as when translating into cycle structure. Therefore, we are choosing a sequence of 60 dice rolls from a set size of 6 possible numbers for Suppose that a permutation is . Types of Sets in Discrete Structure or Discrete Mathematics. V.N. Independent events Consider a quiz with four true/false and three multiple choice questions, (a){(e). Discrete Mathematics Discrete Mathematics, Study Discrete Mathematics Topics. Modus Ponens and Modus Tollens Understand your high school math homework by watching free math videos online from your own free math help tutor may be used as a test or review for Unit Instructor: Is l Dillig. Permutation A rearrangement of the elements in an ordered list S into a one-to-one correspondence with S itself. Combinations and Permutations. Set Theory & Algebra; Combinatorics; Graph Theory; Linear Algebra; Probability; Discrete Mathematics | Types of Recurrence Relations - Set 2. (Multiplication Principle) But what if we only want the number of permutations of r distinct objects from a collection of n? Problems Discrete Mathematics Book I Used for Self Study Discrete Math 6.3.2 Counting, Permutation and Combination Practice DM-16- Propositional Logic -Problems related to In document Discrete mathematics (Pldal 67-70) It is well-known that 1 bit can represent one of two possible distinct states (e.g. Our 1000+ Discrete Mathematics MCQs (Multiple Choice Questions and Answers) focuses on all chapters of Discrete Mathematics covering 100+ topics. of a number, including 0, up to 4 digits long. As such, it is a remarkably broad subject Since combinatorics is widely accessible, this Exercise :: Permutation and Combination - General Questions. MCQ (Multiple Choice Questions with answers about Discrete Mathematics Circular Permutations Determine Abstract Universal cycle for k-permutations is a cyclic arrangement in which each k-permutation appears exactly once as k consecutive elements. Introduction Permutations Discrete Mathematics Andrei Bulatov Discrete Mathematics - Permutations What is Combinatorics Combinatorics, the Study Resources Main Menu A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. Wolfram|Alpha is useful for counting, generating and doing algebra with permutations. Malte Helmert, Gabriele R oger (University of Basel)Discrete Mathematics in Computer Science October 21, 2020 17 / 20 B9. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.The word \"permutation\" also refers to the act or process of changing the linear order of an ordered set. A graph is determined as a mathematical structure that represents a particular function by connecting a set of points. A finite sequence of length$ n \\$ in which all the elements are different, i.e. Permutations Permutations Cyclic Permutation De nition (Cyclic Permutation) A permutation iscyclicif it has a single k-cycle with k >1. permutations and combinations is the another topic included in discrete mathematics which also refers to the finite calculations. 100 Units. Permutations of Objects Ppt - Free download as Powerpoint Presentation (.ppt), PDF File (.pdf), Text File (.txt) or view presentation slides online. permutation arrangement An r-permutation of n objects is an ordered arrangement of r objects from the n objects A permutation is a (possible) rearrangement of objects. Rolling Dice. Question and Answers related to Discrete Mathematics Circular Permutations. For example, the number of insurance claims within a population for a certain type of risk would be zero-inflated by those people who have not taken out insurance against the risk and thus are unable to claim.", "date": "2022-12-08 15:27:02", "meta": {"domain": "bluerocktel.com", "url": "https://fr.bluerocktel.com/tj/30098162dca83fae7b8fd2b53a39a9c711", "openwebmath_score": 0.5183044075965881, "openwebmath_perplexity": 705.9325036993388, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808701643912, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8551454402819093}}
{"url": "http://aimath.org/textbooks/beezer/LCsection.html", "text": "In Section\u00a0VO:Vector Operations we defined vector addition and scalar multiplication. These two operations combine nicely to give us a construction known as a linear combination, a construct that we will work with throughout this course.\n\n## Linear Combinations\n\nDefinition LCCV (Linear Combination of Column Vectors) Given $n$ vectors $\\vectorlist{u}{n}$ from $\\complex{m}$ and $n$ scalars $\\alpha_1,\\,\\alpha_2,\\,\\alpha_3,\\,\\ldots,\\,\\alpha_n$, their linear combination is the vector \\begin{equation*} \\lincombo{\\alpha}{u}{n} \\end{equation*}\n\nSo this definition takes an equal number of scalars and vectors, combines them using our two new operations (scalar multiplication and vector addition) and creates a single brand-new vector, of the same size as the original vectors. When a definition or theorem employs a linear combination, think about the nature of the objects that go into its creation (lists of scalars and vectors), and the type of object that results (a single vector). Computationally, a linear combination is pretty easy.\n\nExample TLC: Two linear combinations in $\\complex{6}$.\n\nExample ABLC: Archetype B as a linear combination.\n\nWith any discussion of Archetype\u00a0A or Archetype\u00a0B we should be sure to contrast with the other.\n\nExample AALC: Archetype A as a linear combination.\n\nThere's a lot going on in the last two examples. Come back to them in a while and make some connections with the intervening material. For now, we will summarize and explain some of this behavior with a theorem.\n\nTheorem SLSLC\u00a0(Solutions to Linear Systems are Linear Combinations) Denote the columns of the $m\\times n$ matrix $A$ as the vectors $\\vectorlist{A}{n}$. Then $\\vect{x}$ is a solution to the linear system of equations $\\linearsystem{A}{\\vect{b}}$ if and only if $\\vect{b}$ equals the linear combination of the columns of $A$ formed with the entries of $\\vect{x}$, \\begin{equation*} \\vectorentry{\\vect{x}}{1}\\vect{A}_1+ \\vectorentry{\\vect{x}}{2}\\vect{A}_2+ \\vectorentry{\\vect{x}}{3}\\vect{A}_3+ \\cdots+ \\vectorentry{\\vect{x}}{n}\\vect{A}_n = \\vect{b} \\end{equation*}\n\nIn other words, this theorem tells us that solutions to systems of equations are linear combinations of the $n$ column vectors of the coefficient matrix ($\\vect{A}_j$) which yield the constant vector $\\vect{b}$. Or said another way, a solution to a system of equations $\\linearsystem{A}{\\vect{b}}$ is an answer to the question \"How can I form the vector $\\vect{b}$ as a linear combination of the columns of $A$?\" Look through the archetypes that are systems of equations and examine a few of the advertised solutions. In each case use the solution to form a linear combination of the columns of the coefficient matrix and verify that the result equals the constant vector (see exercise LC.C21).\n\n## Vector Form of Solution Sets\n\nWe have written solutions to systems of equations as column vectors. For example Archetype\u00a0B has the solution $x_1 = -3,\\,x_2 = 5,\\,x_3 = 2$ which we now write as \\begin{equation*} \\vect{x}=\\colvector{x_1\\\\x_2\\\\x_3}=\\colvector{-3\\\\5\\\\2} \\end{equation*} Now, we will use column vectors and linear combinations to express all of the solutions to a linear system of equations in a compact and understandable way. First, here's two examples that will motivate our next theorem. This is a valuable technique, almost the equal of row-reducing a matrix, so be sure you get comfortable with it over the course of this section.\n\nExample VFSAD: Vector form of solutions for Archetype D.\n\nThis is such an important and fundamental technique, we'll do another example.\n\nExample VFS: Vector form of solutions.\n\nDid you think a few weeks ago that you could so quickly and easily list all the solutions to a linear system of 5 equations in 7 variables?\n\nWe'll now formalize the last two (important) examples as a theorem.\n\nTheorem VFSLS\u00a0(Vector Form of Solutions to Linear Systems) Suppose that $\\augmented{A}{\\vect{b}}$ is the augmented matrix for a consistent linear system $\\linearsystem{A}{\\vect{b}}$ of $m$ equations in $n$ variables. Let $B$ be a row-equivalent $m\\times (n+1)$ matrix in reduced row-echelon form. Suppose that $B$ has $r$ nonzero rows, columns without leading 1's with indices $F=\\set{f_1,\\,f_2,\\,f_3,\\,\\ldots,\\,f_{n-r},\\,n+1}$, and columns with leading 1's (pivot columns) having indices $D=\\set{d_1,\\,d_2,\\,d_3,\\,\\ldots,\\,d_r}$. Define vectors $\\vect{c}$, $\\vect{u}_j$, $1\\leq j\\leq n-r$ of size $n$ by\n\n\\begin{align*} \\vectorentry{\\vect{c}}{i}&= \\begin{cases} 0&\\text{if $i\\in F$}\\\\ \\matrixentry{B}{k,n+1}&\\text{if $i\\in D$, $i=d_k$} \\end{cases}\\\\ \\vectorentry{\\vect{u}_j}{i}&= \\begin{cases} 1&\\text{if $i\\in F$, $i=f_j$}\\\\ 0&\\text{if $i\\in F$, $i\\neq f_j$}\\\\ -\\matrixentry{B}{k,f_j}&\\text{if $i\\in D$, $i=d_k$} \\end{cases}. \\end{align*}\n\nThen the set of solutions to the system of equations $\\linearsystem{A}{\\vect{b}}$ is \\begin{equation*} S=\\setparts{ \\vect{c}+\\alpha_1\\vect{u}_1+\\alpha_2\\vect{u}_2+\\alpha_3\\vect{u}_3+\\cdots+\\alpha_{n-r}\\vect{u}_{n-r} }{ \\alpha_1,\\,\\alpha_2,\\,\\alpha_3,\\,\\ldots,\\,\\alpha_{n-r}\\in\\complex{\\null} } \\end{equation*}\n\nNote that both halves of the proof of Theorem\u00a0VFSLS indicate that $\\alpha_i=\\vectorentry{\\vect{x}}{f_i}$. In other words, the arbitrary scalars, $\\alpha_i$, in the description of the set $S$ actually have more meaning --- they are the values of the free variables $\\vectorentry{\\vect{x}}{f_i}$, $1\\leq i\\leq n-r$. So we will often exploit this observation in our descriptions of solution sets.\n\nTheorem\u00a0VFSLS formalizes what happened in the three steps of Example\u00a0VFSAD. The theorem will be useful in proving other theorems, and it it is useful since it tells us an exact procedure for simply describing an infinite solution set. We could program a computer to implement it, once we have the augmented matrix row-reduced and have checked that the system is consistent. By Knuth's definition, this completes our conversion of linear equation solving from art into science. Notice that it even applies (but is overkill) in the case of a unique solution. However, as a practical matter, I prefer the three-step process of Example\u00a0VFSAD when I need to describe an infinite solution set. So let's practice some more, but with a bigger example.\n\nExample VFSAI: Vector form of solutions for Archetype I.\n\nThis technique is so important, that we'll do one more example. However, an important distinction will be that this system is homogeneous.\n\nExample VFSAL: Vector form of solutions for Archetype L.\n\n## Particular Solutions, Homogeneous Solutions\n\nThe next theorem tells us that in order to find all of the solutions to a linear system of equations, it is sufficient to find just one solution, and then find all of the solutions to the corresponding homogeneous system. This explains part of our interest in the null space, the set of all solutions to a homogeneous system.\n\nTheorem PSPHS\u00a0(Particular Solution Plus Homogeneous Solutions) Suppose that $\\vect{w}$ is one solution to the linear system of equations $\\linearsystem{A}{b}$. Then $\\vect{y}$ is a solution to $\\linearsystem{A}{b}$ if and only if $\\vect{y}=\\vect{w}+\\vect{z}$ for some vector $\\vect{z}\\in\\nsp{A}$.\n\nAfter proving Theorem\u00a0NMUS we commented (insufficiently) on the negation of one half of the theorem. Nonsingular coefficient matrices lead to unique solutions for every choice of the vector of constants. What does this say about singular matrices? A singular matrix $A$ has a nontrivial null space (Theorem\u00a0NMTNS). For a given vector of constants, $\\vect{b}$, the system $\\linearsystem{A}{b}$ could be inconsistent, meaning there are no solutions. But if there is at least one solution ($\\vect{w}$), then Theorem\u00a0PSPHS tells us there will be infinitely many solutions because of the role of the infinite null space for a singular matrix. So a system of equations with a singular coefficient matrix never has a unique solution. Either there are no solutions, or infinitely many solutions, depending on the choice of the vector of constants ($\\vect{b}$).\n\nExample PSHS: Particular solutions, homogeneous solutions, Archetype D.\n\nThe ideas of this subsection will be appear again in Chapter\u00a0LT:Linear Transformations when we discuss pre-images of linear transformations (Definition\u00a0PI).", "date": "2013-05-25 20:44:13", "meta": {"domain": "aimath.org", "url": "http://aimath.org/textbooks/beezer/LCsection.html", "openwebmath_score": 0.8509752750396729, "openwebmath_perplexity": 270.2393744512593, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406020637528, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8551381025177455}}
{"url": "https://byjus.com/question-answer/obtain-all-other-zeroes-of-3x-4-6x-3-2x-2-10x-5-if-two-2/", "text": "Question\n\n# Obtain all other zeroes of $$3x^4+6x^3-2x^2-10x-5$$, if two of its zeros are $$\\sqrt{\\dfrac{5}{3}}$$ and $$-\\sqrt{\\dfrac{5}{3}}$$.\n\nSolution\n\n## Since the two zeroes of the given polynomial\u00a0$$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$\u00a0are\u00a0$$\\sqrt { \\dfrac { 5 }{ 3 } } ,-\\sqrt { \\dfrac { 5 }{ 3 } }$$, therefore,\u00a0$$\\left( x-\\sqrt { \\dfrac { 5 }{ 3 } } \\right) \\left( x+\\sqrt { \\dfrac { 5 }{ 3 } } \\right) =x^{ 2 }-\\dfrac { 5 }{ 3 }$$\u00a0is a factor of the given polynomial.now, we divide the\u00a0polynomial\u00a0$$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$\u00a0by\u00a0$$x^{ 2 }-\\dfrac { 5 }{ 3 }$$\u00a0as shown below:Therefore,\u00a0$$3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5=\\left( x^{ 2 }-\\dfrac { 5 }{ 3 } \\right) (3x^{ 2 }+6x+3)$$Now, we factorize\u00a0$$3x^{ 2 }+6x+3$$ as follows:$$3x^{ 2 }+6x+3=0\\\\ \\Rightarrow 3(x^{ 2 }+2x+1)=0\\\\ \\Rightarrow x^{ 2 }+2x+1=0\\\\ \\Rightarrow (x+1)^{ 2 }=0\\quad \\quad \\quad \\quad (\\because \\quad (a+b)^{ 2 }=a^{ 2 }+b^{ 2 }+2ab)\\\\ \\Rightarrow (x+1)(x+1)=0\\\\ \\Rightarrow x+1=0,\\quad x+1=0\\\\ \\Rightarrow x=-1,\\quad x=-1$$Hence, the zeroes of the\u00a0polynomial\u00a0$$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$\u00a0are\u00a0$$\\sqrt { \\dfrac { 5 }{ 3 } } ,-\\sqrt { \\dfrac { 5 }{ 3 } } ,-1,-1$$.Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-26 21:11:59", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/obtain-all-other-zeroes-of-3x-4-6x-3-2x-2-10x-5-if-two-2/", "openwebmath_score": 0.7961307764053345, "openwebmath_perplexity": 795.7813015454095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370157, "lm_q2_score": 0.8652240912652671, "lm_q1q2_score": 0.8551119949800853}}
{"url": "https://codeforces.cc/blog/entry/91707", "text": "### TheScrasse's blog\n\nBy\u00a0TheScrasse, history, 7 weeks ago,\n\nHello everyone,\nhere is a very simple idea that can be useful for (cp) number theory problems, especially those concerning multiples, divisors, $\\text{GCD}$ and $\\text{LCM}$.\n\nPrerequisites: basic knowledge of number theory (divisibility, $\\text{GCD}$ and $\\text{LCM}$ properties, prime sieve).\n\n## Idea\n\nLet's start from a simple problem.\n\nYou are given $n$ pairs of positive integers $(a_i, b_i)$. Let $m$ be the maximum $a_i$. For each $k$, let $f(k)$ be the sum of the $b_i$ such that $k | a_i$. Output all pairs $(k, f(k))$ such that $f(k) > 0$.\n\nAn obvious preprocessing is to calculate, for each $k$, the sum of the $b_i$ such that $a_i = k$ (let's denote it as $g(k)$). Then, there are at least $3$ solutions to the problem.\n\n#### Solution 1: $O(m\\log m)$\n\nFor each $k$, $f(k) = \\sum_{i=1}^{\\lfloor m/k \\rfloor} g(ik)$. The complexity is $O\\left(m\\left(\\frac{1}{1} + \\frac{1}{2} + \\dots + \\frac{1}{m}\\right)\\right) = O(m\\log m)$.\n\n#### Solution 2: $O(n\\sqrt m)$\n\nThere are at most $n$ nonzero values of $g(k)$. For each one of them, find the divisors of $k$ in $O(\\sqrt k)$ and, for each divisor $i$, let $f(i) := f(i) + g(k)$.\nIf $m$ is large, you may need to use a map to store the values of $f(k)$ but, as there are $O(n\\sqrt[3] m)$ nonzero values of $f(k)$, the updates have a complexity of $O(n\\sqrt[3] m \\log(nm)) < O(n\\sqrt m)$.\n\n#### Solution 3: $O(m + n\\sqrt[3] m)$\n\nBuild a linear prime sieve in $[1, m]$. For each nonzero value of $g(k)$, find the prime factors of $k$ using the sieve, then generate the divisors using a recursive function that finds the Cartesian product of the prime factors. Then, calculate the values of $f(k)$ like in solution 2.\n\nDepending on the values of $n$ and $m$, one of these solutions can be more efficient than the others.\n\nEven if the provided problem seems very specific, the ideas required to solve that task can be generalized to solve a lot of other problems.\n\nHint 1\nHint 2\nHint 3\nSolution\n\n## agc038_c - LCMs\n\nHint 1\nHint 2\nHint 3\nSolution\n\nImplementation (C++)\n\n## abc191_f - GCD or MIN\n\nHint 1\nHint 2\nHint 3\nHint 4\nSolution\n\nImplementation (C++)\n\n## Conclusions\n\nWe've seen that this technique is very flexible. You can choose the complexity on the basis of the constraints, and $f(k)$ can be anything that can be updated fast.\n\nOf course, suggestions/corrections are welcome. In particular, please share in the comments other problems that can be solved with this technique.\n\nI hope you enjoyed the blog!\n\n\u2022 +105\n\n \u00bb 7 weeks ago, # | \u00a0 +15 Nice blog! I like solving problems which use such ideas, so here are a few more problems with a similar flavour:1436F - Sum Over SubsetsChefsums1493D - GCD of an Array\n \u00bb 10 days ago, # | \u00a0 +8 The Solution spoiler for 1154G - Minimum Possible LCM seems to be broken, I see this:\n\u2022  \u00bb \u00bb 10 days ago, # ^ | \u00a0 +6 Fixed. I don't know the reason of the failed parsing, though. The original text was Spoiler$\\text{GCD}(a_i, a_j) = h$.and I had to change it to Spoiler$\\text{GCD}(a_i, a_j) = h$ .\n \u00bb 10 days ago, # | \u00a0 +13 Solution 1 can be easily optimized to $O(m \\log{\\log{m}})$ using the same idea as calculating sums-over-submasks. The code looks something like this: vector f = g; for (int p : primes) for (int i = m / p; i > 0; --i) f[i] += f[i*p]; I suspect there are very few (if any) problems in practice for which solution 3 is directly applicable, but this optimized solution 1 is not.", "date": "2021-07-31 12:03:11", "meta": {"domain": "codeforces.cc", "url": "https://codeforces.cc/blog/entry/91707", "openwebmath_score": 0.8148937821388245, "openwebmath_perplexity": 550.4477912682888, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9925393563153742, "lm_q2_score": 0.8615382147637195, "lm_q1q2_score": 0.8551105851226787}}
{"url": "https://www.analyzemath.com/linear-algebra/spaces/testing-for-linearity-of-vectors.html", "text": "# Testing for Linearity of Vectors in a Subspace - Examples with Solutions\n\n  \n\nGiven a set of vectors in a subspace, how do we test whether the vectors are linearly independent ?\n\n## What are Linearly Dependent or Independent Vectors?\n\nWe first use vectors in two and three dimensional spaces to visualize the concept of dependence and independence of vectors.\nIn figure 1 below we have have two vectors that are parallel such that $v_2 = 2 v_1$. We say that these two vectors are dependent because we can express one vector in terms of the other as follows:\n$v_2 = 2 v_1$ or $v_1 = \\dfrac{1}{2} v_1$.\n\nIn figure 2 below we have have two vectors that are parallel such that $v_2 = - v_1$. These two vectors are dependent because we can express one vector in terms of the other.\n\nIn figure 3, all vectors are parallel to each other. All These vectors are dependent because we can express one vector in terms of the other(s).\n\nIn figure 4, we can never express one vector in terms of the other because they are not parallel. These vectors are independent because we cannot express one vector in terms of the other.\n\nIn figure 5, using the geometrical sums of vectors, we can write\n$3 v_4 = 2 v_1 + 3 v_2 + v_3$\nand therefore these vectors are linearly dependent because we can express one vector in terms of the others.\n\nIn figure 6, the 3 vectors $v_1$ , $v_2 \\}$ and $v_3$ are in the same plane $P$ and are therefore dependent because we can express any of these vectors in term of the other two vectors using linear combinations.\n\nIn figure 7, the pairs of vectors $\\{v_1 , v_2\\}$ , $\\{v_2 , v_3\\}$ and $\\{v_1 , v_3\\}$ are in different planes and are therefore independent because we cannot express any of these vectors in term of the other two vectors using linear combinations.\n\n## Formal Definiton of Linearity of Vectors\n\nVectors $v_1, v_2 .... v_n$ are linearly dependent when at least one of the vectors may be expressed as a linear combinations of the remaining vectors as follows\n$v_1 = r_2 v_2 + r_3 v_3 + .... + r_n v_n$\nWriting the above with the zero on the right side, we obtain\n$v_1 - r_2 v_2 - r_3 v_3 - .... - r_n v_n = 0$\nHence the following definition\nGiven a set of vectors $S = \\{\\textbf{v}_1 , \\textbf{v}_2, ... , \\textbf{v}_n \\}$ ,\nIf the equation\n$r_1 \\textbf{v}_1 + r_2 \\textbf{v}_2 + ... + r_n\\textbf{v}_n = \\textbf{0}$ \u00a0 \u00a0 \u00a0 \u00a0 (I)\nhas only one trivial solution $r_1 = 0 , r_2 = 0 , ... , r_n = 0$, we say that $S$ is a set of linearly independent vectors.\nIf the above equation has other solutions where not all $r_i$ equal to zero, then $S$ is a set of linearly dependent vectors.\nIn the examples below, matrices are row reduced in order to test for linearity. This may done using the row reduce augmented matrices calculator included.\n\n## Examples with Solutions\n\nExample 1\nAre the vectors in the set $\\left \\{ \\begin{bmatrix} -2 \\\\ 1 \\end {bmatrix} , \\begin{bmatrix} 6 \\\\ -3 \\end {bmatrix} \\right \\}$ linearly independent?\n\nSolution to Example 1\nLet $v_1 = \\begin{bmatrix} -2 \\\\ 1 \\end {bmatrix}$ and $v_2 = \\begin{bmatrix} 6 \\\\ -3 \\end {bmatrix}$\nThe question could be answered by noticing that $v_2 = - 3 v_1$ and therefore the given vectors $v_1$ and $v_2$ are dependent (not independent)\nThe above was possible because we are dealing with vectors of small dimension. (2 components only)\n\nWe will now use a method that may be applied in any situation.\nAccording to the definition given above, we need to find $r_1$ and $r_2$ such that\n$r_1 v_1 + r_2 v_2 = 0$\nThe above may be written in matrix form as follows\n$[ v_1 \\;\\; v_2] \\begin{bmatrix} r_1 \\\\ r_2 \\end {bmatrix} = 0$\nwhere $[ v_1 \\;\\; v_2]$ is a matrix whose columns are $v_1$ and $v_2$\n\nWrite the system of equations in augmented matrix form\n$\\begin{bmatrix} -2 &6&|&0\\\\ 1 & -3&|&0 \\end{bmatrix}$\nUse the Gauss Jordan method to row reduce the above augmented matrix and obtain\n$\\begin{bmatrix} 1 &-3&|&0\\\\ 0 & 0&|&0 \\end{bmatrix}$ \u00a0\u00a0\u00a0\u00a0\u00a0 (I)\nThe solution to the above reduced system (which is also a solution to the system before reduction) is found as follows\nThe second equation gives: $0 r_2 = 0$ therefore $r_2$ may take any real value\nThe first equation gives: $r_1 = 3 r_2$\nThe solution set may be written as: $\\left \\{ \\begin{bmatrix} 3 r_2\\\\ r_2 \\end{bmatrix} \\right \\} , r_2 \\in R$\nwhich therefore means that we have an infinite number of solutions and therefore the two vectors are linearly dependent.\n\nNote that you do not have to solve the system in order to decide whether the given vectors are dependent or independent.\nConstruct the augmented matrix using the vectors as columns of the matrix and the constant column on the right is all zeros which in fact may be omitted. We then row reduce the augmented matrix. Then the following conclusions may easily be drawn:\n1) The columns with a pivot are independent of each other\n2) The columns with no pivots are dependent on the ones with the pivot.\n3) The coefficients in the columns without pivot gives the coefficients of dependence on the independent columns.\n\nLet us apply the above to the reduced matrix (I) above\n1) Column 1 has a pivot and may be considered as independent\n2) Column 2 has no pivot and may therefore be considered as dependent on column (1)\n3) The coefficient $- 3$ in column 2 is telling us that $v_2 = - 3 v_1$\n\nExample 2\nAre the vectors in the set $\\left \\{ \\begin{bmatrix} -2 \\\\ 1 \\\\ 4 \\end {bmatrix} , \\begin{bmatrix} 1\\\\ 0 \\\\ 5 \\end {bmatrix} , \\begin{bmatrix} 1\\\\ 2 \\\\ -1 \\end {bmatrix} \\right \\}$ linearly independent?\n\nSolution to Example 2\nstep 1: Construct the augmented matrix\nConstruct the augmented matrix whose columns are the given vectors and zeros on the right column\n$\\begin{bmatrix} -2 &1&1&|&0\\\\ 1 & 0&2&|&0 \\\\ 4 & 5 & -1& |& 0 \\end{bmatrix}$\n\nstep 2: Row reduce the above matrix (you may use the row reduce augmented matrices calculator included).\n$\\begin{bmatrix} 1 &0&0&|&0\\\\ 0 & 1&0&|&0 \\\\ 0 & 0 & 1& |& 0 \\end{bmatrix}$\n\nstep 3: Draw conclusions from the row reduced matrix\nAll 3 columns have a pivot each and therefore all 3 given vectors are independent\n\nExample 3\nAre the vectors in the set $\\left \\{ \\begin{bmatrix} 2 \\\\ -1 \\\\ 3\\\\ 1 \\end {bmatrix} , \\begin{bmatrix} 0 \\\\ 2 \\\\ -1\\\\ 1 \\end {bmatrix} , \\begin{bmatrix} 4 \\\\ -8 \\\\ 9\\\\ -1 \\end {bmatrix} \\right \\}$ linearly independent?\n\nSolution to Example 3\nstep 1: Construct the augmented matrix using the given vectors as columns and zeros on the right column\n$\\begin{bmatrix} 2&0&4&|&0\\\\ -1&2&-8&|&0\\\\ 3&-1&9&|&0\\\\ 1&1&-1&|&0 \\end{bmatrix}$\n\nstep 2: Row reduce the above matrix\n$\\begin{bmatrix} 1&0&\\color{red}{2}&|&0\\\\ 0&1&\\color{blue}{-3}&|&0\\\\ 0&0&0&|&0\\\\ 0&0&0&|&0 \\end{bmatrix}$\n\nstep 3: Draw conclusions\nOnly the first and second columns (from the left) have a pivot and therefore the given vectors are not independent.\nThe coefficients $\\color{red}{2}$ and $\\color{blue}{-3}$ in the third column give the dependence of the original third columns as a linear combination of the other first and the second columns as follows\n$\\begin{bmatrix} 4 \\\\ -8 \\\\ 9\\\\ -1 \\end {bmatrix} = \\color{red}{2} \\begin{bmatrix} 2 \\\\ -1 \\\\ 3\\\\ 1 \\end {bmatrix} - \\color{blue}{3} \\begin{bmatrix} 0 \\\\ 2 \\\\ -1\\\\ 1 \\end {bmatrix}$", "date": "2022-12-03 02:09:48", "meta": {"domain": "analyzemath.com", "url": "https://www.analyzemath.com/linear-algebra/spaces/testing-for-linearity-of-vectors.html", "openwebmath_score": 0.85222989320755, "openwebmath_perplexity": 182.04621549228986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9946981043447426, "lm_q2_score": 0.8596637505099168, "lm_q1q2_score": 0.855105903006106}}
{"url": "https://math.stackexchange.com/questions/1849840/big-list-of-erd%C5%91s-elementary-proofs/1849944", "text": "# Big List of Erd\u0151s' elementary proofs\n\nPaul Erd\u0151s was one of the greatest mathematicians of all time and he was famous for his elegant proofs from The Book. I posted a question about one of his theorem and got a reference, and I have other questions I want to know the answer to too. But, instead of requesting a reference for each theorem he gave with an elementary proof, I've decided to make a thread for a big list of all his elementary proofs.\n\nI'm excited. Let's make an index of the pages of the Book shown to us!\n\nTo get you guys started, I will make a wish list of his theorems who's references I want to see. I encourage you to add to my wish list if you so desire.\n\nWish list :\n\n\u2022 The product of two or more consecutive positive integers is never a square or any other higher power.\n\u2022 A connected graph with a minimum degree $d$ and at least $2d+1$ vertices has a path of length at least $2d+1$.\n\u2022 Let $d(n)$ be the number of divisors of $n$. Then the series $\\sum_{n=1}^\\infty d(n)/2^n$ converges to an irrational number\n\u2022 Let $g(n)$ be the minimal number of points in the general position in the plane needed to ensure a subset exists that forms a convex $n$-gon. Then $$2^{n-2} + 1 \\leq g(n) \\leq \\frac{(2n-4)!}{(n-2)!^2} + 1$$\n\u2022 Erd\u0151s-Mordell inequality\n\u2022 You need to define what you mean by \"elementary\". I think the list of elementary proofs will be very big, e.g. it will contain all his results in set theory, all his results on finite and infinite graphs. \u2013\u00a0bof Jul 6 '16 at 9:31\n\u2022 @bof Hi bof. It will be big but I don't think all of them will be posted here. For starters, I'm looking for the questions in the wish list. Please post if you have any results \u2013\u00a0user230452 Jul 6 '16 at 9:35\n\u2022 Is it fair that 4 people get to close a thread when 50 people have upvoted it ? Please open the thread for further posts \u2013\u00a0user230452 Jul 7 '16 at 10:59\n\u2022 I voted to reopen. \u2013\u00a0Spenser Jul 7 '16 at 13:00\n\u2022 @Spenser Thanks, Spencer. I appreciate it. I have collected another theorem of Erdos' to post. \u2013\u00a0user230452 Jul 7 '16 at 13:24\n\nI think this is worth posting here, mostly because I really enjoy the simplicity of this proof but also because I have no idea how well it is known. The result is not deep or important, so the main interest is in the simplicity of the argument. Erd\u0151s proved a lower bound on the number of primes before an integer $n$.\n\nWac\u0142aw Sierpi\u0144ski, in his Elementary Theory of Numbers, attributes to Erd\u0151s the following elementary proof of the inequality $$\\pi(n)\\geq\\frac{\\log{n}}{2\\log{2}}\\quad\\text{for }n=1,2,\\ldots.$$\n\nPlease note that I have adapted the argument from the text of the book to make things, in my opinion, a bit clearer. Note also that the only tools used in the below proof are some basic combinatorial facts and some results about square-free numbers, which can, for example, be proved with the Fundamental Theorem of Arithmetic.\n\nLet $n\\in\\mathbb{N}=\\{1,2,3,\\ldots\\}.$ Consider the set $$S(n) = \\{(k,l)\\in\\mathbb{N}^{2}:l\\text{ is square-free and }k^{2}l\\leq n\\}.$$ It is a standard fact that every natural number has a unique representation in the form $k^{2}l,$ where $k$ and $l$ are natural numbers and $l$ is square-free. This gives $\\lvert S(n)\\rvert = n.$\n\nNow if we have a pair $(k,l)$ with $k^{2}l\\leq n,$ then we must have $k^{2}\\leq n$ and $l\\leq n$, since $k$ and $l$ are positive. Note that this gives $k\\leq\\sqrt{n}.$ Since $l$ is square-free, $l$ can be expressed as a product of distinct primes, each of which must be not-greater-than $n$ since $l\\leq n$. That is, $l$ can be expressed as a product of the primes $p_{1},p_{2},\\ldots,p_{\\pi(n)}.$ There are $2^{\\pi(n)}$ such products.\n\nTherefore, if we know $(k,l)\\in S(n)$ then there are at most $\\sqrt{n}$ possibilities for what $k$ might be and at most $2^{\\pi(n)}$ possibilities for what $l$ might be (independent of $k$, of course). It follows that $\\lvert S(n)\\rvert \\leq 2^{\\pi(n)}\\sqrt{n},$ so $n\\leq2^{\\pi(n)}\\sqrt{n}.$\n\nTaking $\\log$s and rearranging gives the result.\n\n\u2022 How can you say a lower bound on the Prime Number Theorem is not important ! \u2013\u00a0user230452 Jul 6 '16 at 6:05\n\u2022 @user230452: Better bounds are known through elementary arguments (I think the most famous is by Chebyshev, and has the advantage of providing an upper bound of the same order of growth). So this result is not so important since a little more work can give much better results, with still only elementary ideas. \u2013\u00a0Will R Jul 6 '16 at 17:23\n\u2022 @Therkel: I think I understand, but, in this situation, I felt it best to make clear a) how I knew this argument was from Erdos and b) that chasing up the source would lead readers to a superficially different-sounding argument. \u2013\u00a0Will R Jul 6 '16 at 17:27\n\u2022 This is brilliant! \u2013\u00a0Vin\u00edcius Novelli Jul 20 '16 at 1:41\n\nHere is an exposition of the proof that made Erdos famous by David Galvin. An elementary proof of Bertrand's postulate, which states that there is a prime number in between every $n$ and $2n$. The essence of this proof is in noticing that the lower bound of $$\\binom{2n}{n} \\geq \\frac{4^n}{2n + 1}$$ The binomial expression is the middle term (and the largest) of $(1+x)^{2n}$. The lower bound is the average of the sum of all binomial coefficients. This is obtained by putting $x=1$, and then dividing by the number of terms. This gives us the average. Obviously, the largest term should be bigger than the average. If the postulate does not hold, there is an upper bound that is smaller than this lower bound for large $n$. The postulate can easily be verified for the smaller values of $n$.\n\nhttps://www3.nd.edu/~dgalvin1/pdf/bertrand.pdf\n\nOne very simple, and yet one of my favorites is the Erd\u0151s-Anning theorem:\n\nLet $A \\subseteq \\mathbb C$ be an infinite set of points, such that\n\n$$\\forall x, y\\in A \\quad |x-y| \\in \\mathbb N$$\n\nthen there exists some $c,k \\in \\mathbb C$, such that all $a \\in A$ is of the form $a = cx + k$ for some $x \\in \\mathbb R$.\n\nIt was proved in 1945 in the American Mathematical Bulletin. http://www.ams.org/journals/bull/1945-51-08/S0002-9904-1945-08407-9/S0002-9904-1945-08407-9.pdf\n\nA more colorful formulation is that an infinite sets of points on the Cartesian plain with mutual integer distances must lie on a straight line.\n\nTo prove it, we need an upper bound on the number of non collinear points possible in a set with integral distances. More specifically, if there is a set of non collinear points have integer distances, all at most $d$, then at most $4(d + 1)^2$ points with integer distances can be added to the set.\n\n\u2022 I have a doubt. Can you help me ? The theorem says that for any $n$, it is possible to have $n$ points on a plane, not all collinear such that they have integral distances but not infinitely many points. As I understand it, infinity is not a number, but a concept. To say that infinitely many points are possible it means that for any $n \\geq l$, it is possible, where $l$ is an arbitrarily large natural number. It seems to me the first part of the theorem is stating it, and the second is contradicting it. \u2013\u00a0user230452 Jul 6 '16 at 3:45\n\u2022 @user230452 No, to say that there is an infinite set with integral distances is not the same as saying for any $n\\ge l$ there is a set with $n$ points and integral distances. Compare: For any $n\\ge l$ there is a bounded subset of $\\Bbb R$ with integral distances, but there is no bounded infinite subset of $\\Bbb R$ with integral distances. $\\Bbb N$ is an unbounded infinite subset of $\\Bbb R$ with integral distances. \u2013\u00a0Mario Carneiro Jul 6 '16 at 8:05\n\u2022 @bof There was a typo; the constraint $0 \\in A$ was lacking. It is equivalent to adding some constant, s.t. $a = cx + k$. \u2013\u00a0user347499 Jul 6 '16 at 16:23\n\nErdos' proof of Infinite primes\n\nThe following proof is taken from the book - \"Proofs from THE BOOK\" by Martin Aigner and Gunter Ziegler. This proof is attributed to Erdos.\n\nThis proves that there are infinitely many primes and that the series of the sum of prime reciprocal steps diverges. Let us assume that the infinite series $\\sum\\frac{1}{p}$, where $p$ denotes the prime numbers is a convergent one. Then, there must exist some natural number $k$ such that, $$\\sum_{i > k} \\frac{1}{p_i} < \\frac{1}{2}$$\n\nFor an arbitrary natural number $N$, we get the inequality $$\\sum_{i > k} \\frac{N}{p_i} < \\frac{N}{2}$$\n\nNow, we call all the primes $p_1, p_2, \\dots , p_k$ the small primes, and all the other primes the big primes.\n\nLet $N_b$ denote the number of positive integers $n \\leq N$ that have at least one divisor that is a big prime. And $N_s$ denote the number of positive integers less than $N$ that have only small prime divisors.\n\nWe will show that for a suitable $N$, $N_b + N_s < N$, which is a contradiction because their sum should equal $N$.\n\nFirst, we estimate $N_b$ $$N_b \\leq \\sum_{i > k}\\big\\lfloor \\frac{N}{p_i} \\big\\rfloor < \\frac{N}{2}$$\n\nNow, we look at $N_s$. We write every $n \\leq N$ as $a_nb_n^2$ where $a_n$ is the square free part. $a_n$ is a product of different small primes. There are only $k$ small primes, and each prime may either be chosen or not chosen. So, there are only $2^k$ different values of $a_n$. Furthermore, \\begin{align} b_n \\leq \\sqrt n \\leq \\sqrt N \\\\ \\text{There are at most $\\sqrt N$ different values of $b_n$} \\\\ \\implies N_s \\leq 2^k\\sqrt N \\end{align}\n\nSince, $N_b$ is always less than $N/2$, we just have to find a value of $N$ such that $N_s \\leq N/2$. This happens when $N = 2^{2k + 2}$. When $N = 2^{2k + 4}$, $N_s < N/2$. This proves our contradiction that $$N_b + N_s < N$$ which proves that the series diverges and that there are infinitely many primes.\n\nErdos' favourite questions. The following are not research papers but popular questions Erdos used to ask children.\n\n\u2022 If $n+1$ integers are chosen from the first $2n$ integers, there will always be two that are co prime.\n\nThere will be two numbers that are consecutive. These two numbers will be relatively prime. To see that this is not true when $n$ integers are chosen, just select all the even numbers.\n\n\u2022 If $n+1$ integers are chosen from the first $2n$ integers, there will always be two such that one divides the other.\n\nEvery number is expressible as the product of a power of two and an odd number. There are only $n$ odd numbers in the first $2n$ integers. Two of the numbers are multiplied by the same odd number. One of them has a smaller power of two. This number divides the other. To show that it is not necessarily true for $n$ integers, choose $n+1,n+2, \\dots 2n$.\n\n\u2022 Suppose we have $n$ natural numbers none of which is greater than $2n$ such that the least common multiple of any two is greater than $2n$. Then, all $n$ numbers are greater than $2n/3$.\n\u2022 Regarding the first one, you can easily see that \"all the even numbers\" is actually the only solution -- we can't select 1, because it's coprime to everything; and we can't have any consecutive numbers. If we take $n-1$ numbers, a bit of thinking/excluding a few small cases shows that for $n>3$, the only solutions are \"all even numbers but one\". For $n=3$, we have $\\{3,6\\}$. In general, any set of $n-k$ must be all even unless $n-k \\ge 2n/3$, because we have an odd number at least $3$, which immediately narrows us to $1/3$ of the numbers. I wonder how # of solutions grows with $n$ and $k$? \u2013\u00a0Alex Meiburg Jul 7 '16 at 23:21\n\u2022 @Alex, your final conclusion is false. In fact, for all $\\epsilon > 0$ there exists an $n$ and a set $S \\subset \\{1,2,\\ldots, 2n\\}$ with $|S| > (1 - \\epsilon)n$ such that $S$ contains an odd integer and for all $x,y \\in S$ we have $\\gcd(x,y) > 1$. For example, choose $n$ to be the product of the first $m$ odd primes and define $S = \\{2k | k < n \\text{ and } \\gcd(k, n) > 1 \\} \\cup \\{n\\}$. You will see that $|S| = n - \\varphi(n) + 1 > n(1 - \\frac{c}{\\log \\log(n)})$ for some absolute constant $c$. For $m \\ge 8$ we have $|S| > \\frac{2n}{3}$. \u2013\u00a0Woett Oct 11 '18 at 12:00\n\u2022 @Woett, right, how silly of me. I figured that if $k$ is odd then the density of numbers sharing a factor with $k$ is at most $1/3$ -- totally neglecting the possibility of multiple factors. :) Taking $n=105=3\\cdot 5\\cdot 7$ gives a density of $0.54$, of course. \u2013\u00a0Alex Meiburg Oct 12 '18 at 2:46\n\nErdos answered the following question in the affirmative - Are there infinitely many odd numbers that are not expressible as the sum of a prime number and a power of $2$. The proof is explained in this paper : http://www.maa.org/sites/default/files/3004416309960.pdf.bannered.pdf\n\nThe essence of the proof is in showing that for every integral value of $k$, there is a $2^k$ which has a certain residue with certain moduli. By the Chinese remainder theorem, there are infinitely many odd numbers that satisfy those same congruences. Whenever they do, $a - 2^k$ is composite for all integer values of $k$.\n\n\u2022 Surely this is clear because the set of all numbers $p+2^k$ is density zero? Although perhaps that is not elementary. \u2013\u00a0Mario Carneiro Jul 7 '16 at 1:20\n\u2022 The density should be positive, not zero. There are n/log(2) choices of (prime, power of 2) both less than n. If most, or a positive fraction, of those are unique representations then a positive density subset of integers less than 2n have such a representation. \u2013\u00a0zyx Jul 7 '16 at 2:08\n\u2022 @zyx Oh, indeed if $|A\\cap[n]|\\sim f(n)$, and $B=\\{a+2^k:a\\in A,k\\in\\Bbb N\\}$, then $|B\\cap[n]|\\le\\sum_k f(n-2^k)\\le f(n)\\log_2n$, which is not strong enough to conclude density zero with $A=\\Bbb P$, $f(n)=n/\\log n$. A simulation of the quantity $\\sum_{p\\le n}\\log_2(n-p)$ appears to converge to about $1.468n$, which is obviously useless as an upper bound on $|B\\cap[n]|$ since it's larger than the trivial bound. \u2013\u00a0Mario Carneiro Jul 7 '16 at 4:59\n\u2022 That is close to n/log(2) = 1.443 n . \u2013\u00a0zyx Jul 7 '16 at 5:08\n\u2022 Does the simulation appear to converge to 1/log(2) when using integer logarithms, $\\lfloor \\log_2 ( \\dots ) \\rfloor$? @MarioCarneiro \u2013\u00a0zyx Jul 7 '16 at 19:10\n\nThe proof of the Littlewood-Offord lemma for sums of real numbers.\n\nErdos noticed that under the correspondence between sequences of $\\pm$ signs and finite sets, Sperner's theorem applies and gives the optimal bound for the Littlewood-Offord lemma in dimension $1$ (on how many signed sums of $n$ given numbers of absolute value at least $1$, can have absolute value at most $1$). This observation is a few lines of very basic algebra and combinatorics.\n\nThe rest of the paper is more difficult, and is an extension of Sperner's theorem to handle the case of larger bounds on the sums, once again reading optimal bounds from the combinatorics. This gave evidence that the higher dimensional Littlewood-Offord problem on sums of vectors might also be essentially combinatorial, which was later shown to be true.\n\nErdos - Mordell Inequality : For a point $O$ inside a given triangle $ABC$, the perpendiculars $OP$, $OQ$ and $OR$ are drawn to the side $$OA + OB + OC \\geq 2(OP + OQ + OR)$$\n\nThe product of consecutive integers is never a power\n\nThis was proved by Erdos and Selfridge.\n\nhttp://www.renyi.hu/~p_erdos/1975-46.pdf\n\nSuppose $A$ is a set of $r$ subsets on the set $\\{1,2,\\dots,n\\}$ such that any two sets in $A$ have a non empty subset and $n/2\\geq r$, the maximal size of $A$ is $\\binom{n-1}{r-1}$ i.e., $$|A| \\leq \\binom{n-1}{r-1}$$with equality holding if and only if all the sets share a common element.\nA family of sets may also be called a hypergraph. When every set in $A$ has the same size $r$, it is called a $r$-uniform hypergraph.\nNote : The condition of $n/2 \\geq r$ is imposed because if $r$ is more than half of $n$, then any two sets have a non empty intersection. The maximal size of the family is then $\\binom nr$.", "date": "2020-04-04 09:35:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1849840/big-list-of-erd%C5%91s-elementary-proofs/1849944", "openwebmath_score": 0.9454647898674011, "openwebmath_perplexity": 155.2079216873019, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075711974104, "lm_q2_score": 0.8887587817066392, "lm_q1q2_score": 0.8550815528481441}}
{"url": "https://math.stackexchange.com/questions/2849945/sum-of-sin-waves-with-same-frequency-and-different-amplitudes-and-phase", "text": "Sum of sin waves with same frequency and different amplitudes and phase\n\nFor the equation:\n\n$$A_1 \\sin (\\omega t + \\theta_1) + A_2 \\sin (\\omega t + \\theta_2) = A_3 \\sin (\\omega t + \\theta_3)$$\n\nI've been able to show that the amplitude of the sum is (I believe this is a standard problem):\n\n$$A_3 = \\sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \\cos(\\theta_1 - \\theta_2)}$$\n\nand the new phase is:\n\n$$\\theta_3 = \\arctan \\left(\\frac{A_1 \\sin \\theta_1 + A_2 \\sin \\theta_2}{A_1 \\cos \\theta_1 + A_2 \\cos \\theta_2}\\right)$$\n\nMy question is what happens when the phases $\\theta_1$ and $\\theta_2$ are zero (or just equal to each other). Most books and internet sites (eg http://mathworld.wolfram.com/HarmonicAdditionTheorem.html) state that under these conditions:\n\n$$A_3 = \\sqrt{A_1^2 + A_2^2}$$\n\nHowever, if I set $\\theta_1$ and $\\theta_2$ to zero one gets instead (Since $cos(0) = 1$):\n\n$$A_3 = \\sqrt{A_1^2 + A_2^2 + 2 A_1 A_2}$$\n\nI've been staring at this for a while and I don't understand why the published answer is different from mine. I am sure it is something simple but I've not been able to spot it (Addendum: Wikipedia site: https://en.wikipedia.org/wiki/List_of_trigonometric_identities has the same result under section Linear Combinations).\n\n\u2022 I realized the problem: $A_3 = \\sqrt{A^2_1 + A^2_2 + 2 A_1 A_2}$ applies to the relation $cos (wt + \\theta_1) + sin(wt + \\theta_2)$ not $sin (wt + \\theta_1) + sin(wt + \\theta_2)$. For the latter it works fine. This is an error on my part. ok to delete this question, as its not a valid one? \u2013\u00a0rhody Jul 13 '18 at 18:23\n\u2022 Although your Question contained and was motivated by an error, it brings up some useful content, and I'd like to retain it. Even the error, articulated in your Comment, bears pointing out as it is the sort of thing anyone might overlook at first glance. \u2013\u00a0hardmath Jul 14 '18 at 13:02\n\nThe formula $\\sqrt{A_1^2+A_2^2}$ applies to sinusoids in quadrature ($\\frac\\pi2$ phase difference), such as when summing a sine and a cosine.\n\nSignals in phase just add up their amplitudes, $A_1+A_2$, while signals in opposition subtract, $|A_1-A_2|$.\n\n\u2022 I've been staring at this for hours yesterday, and then this morning after I posted the question and suddenly realized my error. \u2013\u00a0rhody Jul 13 '18 at 18:26\n\nYou wrote:\n\n$$A_1 \\sin (\\omega t + \\theta_1) + A_2 \\sin (\\omega t + \\theta_2) = A_3 \\sin (\\omega t + \\theta_3)$$\n\nThe Wolfram article you cite says:\n\nIt is always possible to write a sum of sinusoidal functions $$f(\\theta)=acos(\\theta)+bsin(\\theta )$$\nas a single sinusoid the form $$f(\\theta)=ccos(\\theta+\\delta)$$ $$A_3 = \\sqrt{A_1^2 + A_2^2}$$\n\nSo you have sine plus sine, while Wolfram has sine plus cosine. To convert Wolfram's cosine to a sine, you need to shift the phase by $\\frac{\\pi}2$, which then makes the cosine of the phase difference equal to zero.\n\nWaves with no phase difference (or even pi's) directly add up their amplitudes to form a new wave.\n\n$$A_1 \\sin (\\omega t) + A_2 \\sin (\\omega t) = (A_1+A_2) \\sin (\\omega t )$$\n\nThe $A_3$ you prescribed is for waves with phase difference $(\\theta_1 - \\theta_2)=\\frac{\\pi}{2}$.\n\nThe equation you got putting $\\theta_1=\\theta_2=0$ is correct and simplifies to $A_3=(A_1+A_2)$.\n\nThe result given in Equation (21) of Wolfram site\n\nfor adding in-phase waves of same frequency with zero relative phase angle gives\n\n$$A_3= \\sqrt{A_1^2+A_2^2+ 2 A_1 A _2}= \\pm (A_1+A_2)$$\n\nThe other result came about when the phase difference is $\\pi/2$ obtained directly by Pythagoras thm as diagonal length of a rectangle of vectors or phasors.\n\nThe parallelogram of amplitudes can be drawn and the situation becomes clear. There are three situations. In-phase,add; Out of phase, subtract, or take the difference; when relative phase difference is $\\pi/2$ take root square sum.", "date": "2020-03-29 04:07:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2849945/sum-of-sin-waves-with-same-frequency-and-different-amplitudes-and-phase", "openwebmath_score": 0.7967591285705566, "openwebmath_perplexity": 522.6180422549268, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799399736476, "lm_q2_score": 0.8872045877523147, "lm_q1q2_score": 0.8550699843282706}}
{"url": "https://math.stackexchange.com/questions/3749531/how-to-find-the-elongation-of-a-spring-when-it-is-tied-to-a-brass-bar-over-a-til", "text": "# How to find the elongation of a spring when it is tied to a brass bar over a tilted wall?\n\nThe problem is as follows:\n\nAn homogeneous brass bar which goes from point $$A$$ to point $$B$$ has a weight of $$300\\,N$$. The bar is suported over two frictionless surfaces as it is shown in the picture from below. The system is in static equilibrium by the action of the force excerted by the spring tied to the end at point $$B$$ on the bar which in turn is fixed to the ceiling. Using this information find the elongation of the spring. Assume that the constant of the spring is $$k=500\\,\\frac{N}{m}$$.\n\nThe alternatives given in my book are as follows:\n\n$$\\begin{array}{ll} 1.&28\\,cm\\\\ 2.&23\\,cm\\\\ 3.&25\\,cm\\\\ 4.&26\\,cm\\\\ 5.&30\\,cm\\\\ \\end{array}$$\n\nWhat I did to solve this problem was to use a \"trick\" which I seen in a similar problem about a ladder standing next to a wall.\n\nThe rationale was that when an horizontal object is leaning against two surfaces there will be a reaction from both. But if it is asked a force in between. You may use any point to establish the torque about that point and use Varignon's theorem? to find the unknown force. This is done to \"cancel out\" the reaction force from both surfaces.\n\nTherefore:\n\nTorque about the point specified in the figure from below results into:\n\n$$\\sum_{i=1}^{n}\\tau=0$$\n\n$$(l\\cos 30^{\\circ})\\times(300)+500\\times x \\times 2l =0$$\n\nsimplifying terms:\n\n$$-\\frac{\\sqrt {3}}{2}\\times 3 + 10 x = 0$$\n\n$$x= \\frac{3 \\sqrt {3}}{20}\\approx .2598076212 \\,m$$\n\nwhich transformed into cm would be about $$26\\,cm$$ and this appears in the alternatives as option four. But is my answer the right approach to this problem?.\n\nMy source of confusion is that in the problem is not indicated if the spring is paralell to the incline or if it does make another angle.\n\nIt is not indicated in the problem but how can I find let's say the Reaction in both the floor and in the wall?. How can I find these?. Am I using Varignon's theorem in this problem?. Can someone help me with this matter please?.\n\n\u2022 @Cesareo I'm sorry but I still don't get to the answer you reached. This is not homework. Can you post a solution or more hints?. How exactly you got to $30\\,cm$.? \u2013\u00a0Chris Steinbeck Bell Jul 11 '20 at 3:27\n\nFirst, if you are referring to Varignons' theorem about quadrilaterals and the parallelogram formed by the midpoints of the sides, I don't think you are usinng it.\n\nSecond, to determine both the reactions in the floor and the wall we can apply Newton's law and impose that the sum af all forces is zero, and then impose that the torque about any point we want is zero. You already did the second part, so you could just apply Newton's law: $$\\vec{N_A}+\\vec{N_B}+\\vec{P}+k\\vec{\\Delta l}=0$$ If the spring is parallel to the wall we know the directions of all this vectors and by decomposing along two axes $$x$$ and $$y$$ (I choose the $$x$$-axis parallel to the ground, but it's arbitrary) we obtain two equations that allow us to find the moduluses of the reaction forces: $$N_A+\\frac{N_B}{2}-P+\\frac{\\sqrt{3}}{2}k\\Delta l=0$$ $$-\\frac{\\sqrt{3}}{2}N_B+\\frac{1}{2}k\\Delta l=0$$\n\nPersonally I solved the problem calculating the torque about the point B because there are 4 forces and two are applied in B, so the equations are particularly simple, but any other point is just as good and gives the same solutions. I too found that the answer is 26cm\n\nFig 2 has made things easy for taking force/moment equilibrium to evaluate forces.\n\nTake moments about top left corner call it $$I$$\n\n$$k x \\cdot 2 l = l \\cos 30^{\\circ} 300$$\n\n$$x = 0.2598$$ meters i.e., the fourth option:\n\n$$x= 25.98 cm$$\n\nBtw, I is the center about which the brass bar is instantaneously rotating in dynamic problems.\n\nCalling $$\\alpha = 30^{\\circ},\\ \\beta = 60^{\\circ}$$ and\n\n$$\\cases{ A=(0,0)\\\\ B=A+l(\\cos\\alpha,\\sin\\alpha)\\\\ G = \\frac 12\\left(B+A\\right) }$$\n\nwith the forces\n\n$$\\cases{ f_A = (0,N_A)\\\\ f_B = k \\Delta x (\\cos\\beta,\\sin\\beta)+N_B(-\\cos\\alpha,\\sin\\alpha)\\\\ f_G = (0,-W) }$$\n\nwe have\n\n$$\\cases{ f_A+f_B+f_G = 0\\\\ (A-B)\\times f_A+(G-B)\\times f_G = 0 }$$\n\nfrom which we obtain\n\n$$N_A = \\frac W2,\\ N_B = \\frac W4,\\ \\Delta x = \\frac{\\sqrt{3}W}{4k}$$\n\nso $$\\Delta x\\approx 0.26$$ cm\n\nAnother way using virtual displacements.\n\nCalling\n\n$$\\cases{ f_{B_1} = k \\Delta x (\\cos\\beta,\\sin\\beta)\\\\ f_{B_2} = N_B(-\\cos\\alpha,\\sin\\alpha) }$$\n\nIn equilibrium, we have\n\n$$\\delta A\\cdot f_A + \\delta G\\cdot f_G + \\delta B \\cdot\\left(f_{B_1}+f_{B_2}\\right) = 0$$\n\nand due to orthogonality\n\n$$\\delta A\\cdot f_A = \\delta A\\cdot f_G = \\delta B\\cdot f_{B_2} = 0$$\n\nand following\n\n$$\\frac 12\\left(\\delta A + \\Delta x(\\cos\\beta,\\sin\\beta)\\right)\\cdot f_G + k (\\Delta x)^2\\left(\\cos\\beta,\\sin\\beta\\right)\\cdot \\left(\\cos\\beta,\\sin\\beta\\right)=0$$\n\nor\n\n$$-\\frac 12 W\\sin\\beta + k\\Delta x = 0\\Rightarrow \\Delta x = \\frac{\\sqrt{3}W}{4k}$$\n\n\u2022 The bar should be $2l$ long, not $l$. \u2013\u00a0enzotib Jul 11 '20 at 8:06", "date": "2021-07-29 05:07:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3749531/how-to-find-the-elongation-of-a-spring-when-it-is-tied-to-a-brass-bar-over-a-til", "openwebmath_score": 0.7699726819992065, "openwebmath_perplexity": 200.98322933717014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799399736476, "lm_q2_score": 0.8872045832787205, "lm_q1q2_score": 0.8550699800167103}}
{"url": "https://stats.stackexchange.com/questions/449594/why-both-data-have-same-standard-deviation", "text": "# Why both data have same standard deviation?\n\nBut I am curious why the standard deviation for both data is same. I think because data 2 is \"100 - data 1\".\n\n\u2022 $var(X+a)=var(X)$ and $var(aX)=a^2var(X)$ for a real number $a$ and a random variable $X$. With those, can you prove why $var(100-X)=var(X)$ where X is your data set? (Equal variances is the same as equal standard deviations.) \u2013\u00a0Dave Feb 15 '20 at 0:24\n\nI still would like you to try to prove it, but I\u2019ll give some intuition about those two identities I gave.\n\nRemember that variance (and standard deviation) have something to do with how spread out your data are.\n\n$$var(X+a)=var(X)$$ means that if you take data with some amount of spread and slide them up or down the number line, you do not change the spread.\n\n$$var(X)=var(-X)$$ is not quite what I gave, but it\u2019ll be part of your proof. The intuition here is that you\u2019re just spinning it around, taking a mirror image. The data are as spread out as their reflection.\n\nTogether, $$var(100-X)=var(X)$$ means that you flip the data to the mirror image and then slide along the number line, but you do not change the spread.\n\nCapisce?\n\n\u2022 what is the meaning of Capisce? \u2013\u00a0aan Feb 16 '20 at 11:41\n\u2022 that's Italian for \"do you comprehend ?\" \u2013\u00a0IrishStat Feb 16 '20 at 13:20\n\nI believe the answer to this question can be expressed in more elemental terms. Beginning with the formula for the sample standard deviation\n\n$$s_{x} = \\sqrt{\\frac{ \\sum_{i=1}^{n}{(x_i - \\overline{x})^2} }{n - 1}},$$\n\nI want to draw your attention to the expression $$(x_i - \\overline{x})$$ in the numerator. Ignore the summation notation for the purposes of this answer. Note, we decrement the sample mean (i.e., $$\\overline{x}$$) from each realization of $$x_{i}$$. If we were to increase or decrease each $$x_{i}$$ by the same amount, the mean will change. However, the distance of each realization of $$x_{i}$$ from that central tendency remains the same. In other words, each deviation from the mean is the same.\n\nSubtracting each $$x_{i}$$ from a constant (e.g., $$100 - x_{i}$$) flips the vector of values to its mirror image, then slides it along the number line by a constant amount. Suppose the first realization of $$x_{1}$$ = 20 and $$\\overline{x}$$ = 4. In keeping with the foregoing expression,\n\n$$(20 - 4) = 16,$$\n\nwhich is the deviation from the sample mean. Now, assume we don't move along the number line just yet and we simply put a negative sign in front of each $$x_{i}$$, such that we have $$-(x_{i})$$; this flips the sign of the mean as well. The first realization, $$x_{1}$$, is now $$-20$$. Again, substituting $$-20$$ into the expression,\n\n$$(-20-(-4)) =-16,$$\n\nwhich is the same number of units away from the mean. Note, squared deviations result in a positive number, so the numerator remains the same. You can run this code in R which builds upon @knrumsey's insightful response, but breaks it down further:\n\nx <- rnorm(10000, 30, 5) # Simulates 10,000 random deviates from the normal distribution\n\npar(mfrow = c(2, 2))\nhist(x, xlim = c(0, 100), col = \"blue\")\nhist(-x, xlim = c(-100, 0), col = \"red\")  # Note the -x, it simply flips the sign (mirror image)\nhist(x, xlim = c(0, 100), col = \"blue\")\nhist(100-x, xlim = c(0, 100), col = \"red\")  # Subtracting x from 100 shifts values along the x-axis\n\n\nThe first row of plots shows how the realizations translate when we negate each $$x_{i}$$. The second row shows what happens when each $$x_{i}$$ is subtracted from 100; the translation first flips then slides across the number line. The spread of the distribution is unaffected by this.\n\nTo supplement @Dave's answer, take a look at the following histograms which have the same x-axis. Data2 is just a shifted version of Data1 and therefore the standard deviation, which is a measure of spread, shouldn't change.\n\nR code to generate histograms:\n\nx <- 100*rgamma(1000, 6, 2) #Simulate some data\nhist(x, xlim=c(0,100))\nhist(100-x, xlim=c(0,100))\n\n\u2022 Thanks. well-plotted histogram. May I know how do you plot the histogram? \u2013\u00a0aan Feb 15 '20 at 5:29\n\u2022 @aan see the update. \u2013\u00a0knrumsey Feb 15 '20 at 17:49\n\u2022 @knrumsey Good illustration. However, simulating from the gamma distribution would produce realizations that cluster around a mean that is much lower than what I see here given the parameters you specified. Right? \u2013\u00a0Thomas Bilach Feb 16 '20 at 20:24\n\u2022 @Tom, that is correct. I didn't put too much thought into which distribution I simulated from. Just wanted draws between 0 and 100. \u2013\u00a0knrumsey Feb 16 '20 at 20:45\n\u2022 @Thomas Bilach, realizing that I misunderstood your question. Thanks for pointing this out, I've updated the post to reflect the fact that I multiplied the data by 100. \u2013\u00a0knrumsey May 3 '20 at 0:48\n\nchanges period to period in DATA 1 identical to the negative changes in DATA 2\n\nOBS DATA 1 FIRST DIF DATA 2 FIRST DIFF\n\n1 30 NA 70 NA\n\n2 40 10 60 -10\n\n3 80 40 20 -40\n\n4 30 -50 70 50\n\n5 20 -10 80 10\n\n6 33 13 67 -13\n\n7 33 0 67 0\n\n8 33 0 67 0\n\n9 33 0 67 0\n\nCONTINUING STEPS TO UNCOVER THE BASIC RELATIONSHIP OF DATA1 AND DATA2:\n\nI took the series DATA1 and computed the ACF ... here it is ..\n\nAND for DATA2 here\n\n...note that not only are the standard deviations the same but the ACF's are the same.\n\n\u2022 This is an interesting perspective, but I believe it would need more explanation to be understood by many readers. \u2013\u00a0whuber Feb 15 '20 at 22:53\n\u2022 @IrishStat, thanks. do you mind to explain more? \u2013\u00a0aan Feb 16 '20 at 11:42\n\u2022 i added some material to my response detailing how differenced data period to period ( val2-val1,val3-val2,.. val9-val8) for both original series DATA1 and DATA2 were complements of each other \u2013\u00a0IrishStat Feb 16 '20 at 13:19\n\u2022 @IrishStat. Noted. what is the meaning of Q.E.D and OBS? \u2013\u00a0aan Feb 16 '20 at 15:58\n\u2022 An explicit connection between those assertions and the result is needed. What is interesting is your use of successive differences (\"changes period to period\" in an otherwise unordered dataset) to characterize the standard deviation of the dataset. In other words, you appear to be claiming that the set of equalities $$x_{i+1}-x_i = -(y_{i+1}-y_i)$$ for $i=1, 2, \\ldots, n-1$ implies $$\\sum_{i=1}^n(x_i-\\bar x)^2=\\sum_{i=1}^n(y_i-\\bar y)^2.$$ That's true, but it does need to be shown: you can't just exhibit a table of numbers and declare \"QED\"! \u2013\u00a0whuber Feb 16 '20 at 20:15", "date": "2021-03-01 06:40:46", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/449594/why-both-data-have-same-standard-deviation", "openwebmath_score": 0.7492038607597351, "openwebmath_perplexity": 637.1869202348532, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.974434786819155, "lm_q2_score": 0.8774767986961403, "lm_q1q2_score": 0.855043917276228}}
{"url": "https://brilliant.org/discussions/thread/wanna-practice-induction/", "text": "# Wanna practice Induction ?\n\nThe principle of mathematical induction is a very very useful tool in many proofs, and also in proving some useful formulas. $\\color{#D61F06}{\\text{The principle states that any given set of positive integers}}$ has ALL natural numbers if it follows the following conditions -\n\n$(i)$ The first natural number, i.e. $1$ is in the set.\n\n$(ii)$ Whenever the integer $k$ is in the set, then $k+1$ is also in the set.\n\nThis looks so obvious, see that if you have been given the two statements simultaneously, then by using $(ii)$ on $(i)$, you can say that $2$ is in the set. Then again applying the statement $(ii)$ for $2$, we say that $(3)$ will be in the set, so on.\n\n$\\color{#3D99F6}{\\textbf{Thus if you prove a certain result for 1}}$ , $\\color{#3D99F6}{\\textbf{and then assume that the result is true}}$ for $k$, just prove that it is true for $k+1$ and you're done !\n\nSame thing you can do for 0, then prove for whole numbers !!!\n\nProblems for practice:-\n\nProve that the following results hold $\\forall n \\in \\mathbb{N}$\n\n$(a) \\displaystyle \\sum_{k=0} ^n 2^k = 2^{n+1}-1$\n\n$(b) \\displaystyle \\sum_{k=1} ^n k = \\dfrac{n(n+1)}{2}$\n\n$(c) \\displaystyle \\sum_{k=1} ^n k^2 = \\dfrac{n(n+1)(2n+1)}{6}$\n\n$(d) \\displaystyle \\sum_{k=1}^n k^3 = \\biggl( \\sum_{k=0} ^n k \\biggr) ^2$\n\n$(e) \\displaystyle 24 \\mid 2\\cdot 7^n + 3\\cdot 5^n -5$\n\n$(f) \\displaystyle 1\\cdot 2+2\\cdot 3+3\\cdot 4+ .... + n(n+1) = \\dfrac{n(n+1)(n+2)}{3}$\n\nI felt like sharing because though it's very well known to all, I am willing to find some good level problems for practicing this foundation builder concept once more... $\\color{#20A900}{\\text{Isn't this a good revision ?}}$\n\n5\u00a0years, 4\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\n@Aditya Raut this is really a nice note for practice...... could you please add more divisibility kinda problems to give a more diverse experience to readers? ........ thanks :)\n\n- 5\u00a0years, 4\u00a0months ago\n\n- 5\u00a0years, 3\u00a0months ago\n\nThanks - they look like a good selection of induction problems\n\n- 4\u00a0years, 10\u00a0months ago\n\nI actually haven't tried $(d)$ and $(f)$ before since doing math XD\n\nHere's the solution for $(e)$ if someone got stuck.\n\n$2\\times 7^{m+1} + 3 \\times 5^{m+1} - 5$\n\n$= 2\\times 7^{m}\\times 7 + 3\\times 5^{m}\\times 5 - 5$\n\n$= 14\\times 7^{m} + 15 \\times 5^{m} - 5$\n\n$= 12\\times 7^{m} + 12\\times 5^{m} + 2\\times 7^{m} + 3\\times 5^{m} - 5$\n\n$= 12(7^{m} + 5^{m}) + 24k$ for some $k$\n\n$= 24n + 24k$ for some $n$\n\n- 5\u00a0years, 3\u00a0months ago\n\n$(a)$ Base Case $n = 1$\n\n$\\displaystyle \\sum_{k = 0}^{1}2^{k} = 3$\n\n$2^{n+1} - 1 = 3$\n\nTrue.\n\nPropose it is true $\\forall n \\in N$\n\nThen it must be true for $n + 1$\n\n$\\displaystyle\\sum_{k = 0}^{n+1}2^{k} = \\sum_{k = 0}^{n}2^{k} + 2^{n+1}$\n\nSubstituting the value from our proposal,\n\n$\\sum_{k = 0}^{n}2^{k} + 2^{n+1} = 2^{n+1} - 1 + 2^{n+1}= 2^{n+2} - 1$\n\nThus, $LHS = RHS$\n\nThus, $LHS = RHS \\forall n \\in N$\n\nSorry for ugly arrangement.\n\n- 5\u00a0years, 3\u00a0months ago\n\n@Aditya Raut Can you add these to the Induction Wiki page? Thanks!\n\nStaff - 5\u00a0years, 1\u00a0month ago\n\nNice! I'm too into creating notes and problems set for inequality :) Why not take a look at it @Aditya Raut ? :D\n\n- 5\u00a0years, 3\u00a0months ago\n\nWhat is e) ?... in d) second summation is it k=0 or k=1 ? Nice collection.\n\n- 5\u00a0years ago\n\nThanks, but Just think ! Will adding $0$ make any change? In that summation, starting from $k=0$ will yield same thing as starting with $k=1$, won't it ? @Niranjan Khanderia\n\n- 5\u00a0years ago\n\nThanks.\n\n- 5\u00a0years ago\n\nAnd e) means prove that for all positive integers $n$, $(2\\cdot 7^n + 3\\cdot 5^n -5)$ is divisible by $24$.\n\n$a\\mid b$ is the symbol of \"a divides b\" or \"b is divisible by a\".\n\nLatex code for the symbol $\\mid$ is \"\\mid\"\n\n- 5\u00a0years ago\n\nThanks. I did not get it since I did not assume the parenthesis.\n\n- 5\u00a0years ago\n\n/(good/)\n\n- 4\u00a0years, 10\u00a0months ago\n\nA good level problem is here: Fro every positive integer $n$, prove that $\\sqrt{(4n+1)}<\\sqrt n +\\sqrt{(n+1)}<\\sqrt{(4n+2)}$. Hence or otherwise, prove that [$\\sqrt{n}+\\sqrt{n+1}$]=[$\\sqrt{4n+1}$] where [.] denotes floor function. It is an IITJEE problem.\n\n- 4\u00a0years, 7\u00a0months ago", "date": "2019-11-14 20:26:17", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/wanna-practice-induction/", "openwebmath_score": 0.9746575355529785, "openwebmath_perplexity": 1892.4515993752436, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347905312774, "lm_q2_score": 0.8774767922879693, "lm_q1q2_score": 0.8550439142891845}}
{"url": "https://math.stackexchange.com/questions/4192982/a-question-about-solving-quotients", "text": "all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:\n\n$$\\begin{eqnarray} \\lvert \\frac{x+1}{x-2} \\rvert &<& 3\\\\ \\Rightarrow \\frac{x+1}{x-2} &>& -3 \\hspace{1mm}and\\hspace{1mm}\\frac{x+1}{x-2} < 3\\\\ x+1&>&-3(x-2)\\hspace{2cm}\\text{(Solving for the left inequality)}\\\\ x+1&>&-3x+6\\\\ 4x&>&5\\\\ x&>&\\frac{5}{4} \\end{eqnarray}$$ which cannot be true, as $$x$$ can equal $$2$$ with these restrictions which is obviously not allowed. Doing the left side the correct way:\n\n$$\\begin{eqnarray} \\frac{x+1}{x-2} &>& -3\\\\ \\frac{x+1}{x-2} + 3 &>& 0\\\\ \\frac{x+1}{x-2} + \\frac{3x-6}{x-2} &>& 0\\\\ \\frac{4x-5}{x-2} &>& 0\\\\ \\Rightarrow x < \\frac{5}{4}\\lor x>2\\\\ \\end{eqnarray}$$\n\nleads to the solution set $$(-\\infty,\\frac{5}{4})\\cup(2,\\infty)$$. My question is, what happened with the first method where I failed to come up with a solution for $$(2,\\infty)$$, (and why the signage for $$x>\\frac{5}{4}$$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.\n\n\u2022 The use of the Union symbol $\\cup$ is certainly not the correct way, as it is used to denote union of sets. You can use a $\\textrm{or}$, or the or symbol $\\vee$ (\\vee). Jul 7, 2021 at 23:31\n\u2022 @ultralegend5385: When writing in the context of mathematical logic, I think \\lor, \\land and \\lnot are easier to remember for $\\lor,\\land,\\lnot$ respectively. Then again, sometimes I prefer writing \\neg over \\lnot for the logical not $\\neg$ Jul 7, 2021 at 23:44\n\u2022 @ultralegend5385 fixed. Jul 7, 2021 at 23:54\n\nIn the first method, when you go from $$\\dfrac{x+1}{x-2}\\gt -3$$ to $$x+1\\gt -3(x-2)$$, what you're doing is multiplying both sides of the inequality by $$x-2$$ and you assume that $$x-2\\gt 0$$ when doing that because otherwise the inequality sign flips to $$\\le$$\n\nSo, assuming $$x\\gt 2$$, you get from that inequality that $$x\\gt 5/4$$ and the second inequality $$\\dfrac{x+1}{x-2}\\lt 3$$ gives $$x\\gt 7/2$$\n\nSo, assuming $$x\\gt 2$$, you get $$x\\gt 5/4$$ and $$x\\gt 7/2$$; together they imply $$x\\gt\\max\\{2,5/4,7/2\\}=7/2$$\n\nSimilarly, assume $$x-2\\lt 0$$, ie, $$x\\lt 2$$ and solve $$x+1\\lt -3(x-2)$$ and $$x+1\\gt 3(x-2)$$ simultaneously to obtain the other solution set. You get $$x\\lt\\min\\{2,5/4,7/2\\}=5/4$$\n\nThe complete solution set is thus $$(-\\infty,5/4)\\cup (7/2,\\infty)$$\n\nThe answer you arrived at is wrong. Note that $$x=3\\in (2,\\infty)$$ doesn't satisfy the original equation.\n\nIn the second method, you're only solving one of the two inequalities that are supposed to hold. If you solve the second inequality $$\\dfrac{x+1}{x-2}\\lt 3$$ in a similar way, you get $$\\frac{x+1}{x-2}-3\\lt 0\\iff\\frac{7-2x}{x-2}\\lt 0\\iff x\\in (-\\infty, 2)\\cup (7/2,\\infty)$$\n\nand the solution set to your original problem is the intersection of this with $$(-\\infty,5/4)\\cup (2,\\infty)$$ so that\n\n$$[(-\\infty,2)\\cup (7/2,\\infty)]\\cap[(-\\infty,5/4)\\cup (2,\\infty)]=(-\\infty,5/4)\\cup (7/2,\\infty)$$\n\nwhich is the solution set to the original problem $$\\left|\\dfrac{x+1}{x-2}\\right|\\lt 3$$\n\n\u2022 I understand now. Thank you for explaining it so well. Jul 7, 2021 at 23:55\n\u2022 @js2822: You're welcome. Also, nice work for asking a good question and being responsive to others! +1 ;) Jul 7, 2021 at 23:58\n\nHINT\n\nSince both sides are nonnegative, you can square them in order to obtain an equivalent inequation: \\begin{align*} \\left|\\frac{x+1}{x-2}\\right| < 3 & \\Longleftrightarrow \\left(\\frac{x+1}{x-2}\\right)^{2} < 9\\\\\\\\ & \\Longleftrightarrow \\frac{(x^{2} + 2x + 1) - 9(x^{2} - 4x + 4)}{(x-2)^{2}} < 0\\\\\\\\ & \\Longleftrightarrow \\frac{-8x^{2} + 38x - 35}{(x-2)^{2}} < 0\\\\\\\\ & \\Longleftrightarrow \\begin{cases} 8x^{2} - 38x + 35 > 0\\\\\\\\ x\\neq 2 \\end{cases} \\end{align*}\n\nCan you take it from here?\n\n\u2022 One question: Does squaring an absolute value add any extraneous solutions in certain circumstances? Or do they never appear since the absolute value is always nonnegative? Jul 8, 2021 at 0:00\n\u2022 As long as both sides are nonnegative, you can proceed as suggested so that there won't be any extraneous solutions. Jul 8, 2021 at 0:46\n\nWe can only \"flip the sign\" if what is inside the absolute value is negative. And similarly, we \"keep the sign\", if what is inside the absolute value bracket is positive.\n\nWe could write it this way...\n\n$$-3<\\frac {x+1}{x-2}<0$$ or $$0\\le \\frac {x+1}{x-2}<3$$\n\nOr say, the left-hand equation only applies when $$-1\n\nThen we can solve the two sets of inequalities:\n\n$$-1 or $$x > \\frac {7}{2}$$ or $$x \\le -1$$\n\nNow we can combine the intervals\n\n$$(-\\infty, \\frac 54)\\cup (\\frac 72,\\infty)$$", "date": "2022-09-26 15:21:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4192982/a-question-about-solving-quotients", "openwebmath_score": 0.9170430302619934, "openwebmath_perplexity": 273.75492244527135, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347905312774, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8550439096059257}}
{"url": "https://math.stackexchange.com/questions/1979079/why-doesnt-the-derivative-of-integration-by-trig-substitution-match-the-origina", "text": "# Why doesn't the derivative of integration by trig substitution match the original function?\n\nFor one of my assignments in Calc II, I had to solve ${\\displaystyle\\int} \\frac{\\sqrt{x^2-4}}{x} \\text{ d}x$. By using Trig substitution where $x = 2 \\text{ sec } \\theta$, my solution was $\\sqrt{x^2-4} - 2\\text{ arcsec}\\left(\\frac{x}{2}\\right) + C$. I tested my solution by graphing $\\frac{\\text{d}}{\\text{d}x}\\left[\\sqrt{x^2-4} - 2\\text{ arcsec}\\left(\\frac{x}{2}\\right)\\right]$, and it only matched the original function where $x \\geq 2$. To match for $x \\leq -2$, I had to use $\\frac{\\text{d}}{\\text{d}x}\\left[\\sqrt{x^2-4} + 2\\text{ arcsec}\\left(\\frac{x}{2}\\right)\\right]$.\n\nAssuming I solved this correctly, what causes this discrepancy? How can we know if, when, and where this kind of thing will happen? Also, how would we notate the difference? Is it necessary to write the solution as a piecewise function, or does none of this matter anyway? I would think that, if it was because of the square root, the change in sign would be in front of it, but this seems to be because of the arc secant.\n\nEdit Since a comment mentioned using WolframAlpha, I decided to double-check through there, and they do have my same answer under the Step-by-Step Solution, although they then simplified it to arctan somehow. Graphing the derivative of their solution with arctan matches the original function even less, with both a different C and shape. For explicitness, here's my work: $x = 2\\sec \\theta \\Rightarrow \\text{d}x = 2 \\sec\\theta \\tan\\theta$\n\n$\\int \\frac{\\sqrt{4 \\sec^2\\theta - 4}}{2 \\sec\\theta}(2 \\sec\\theta \\tan\\theta \\text{ d}\\theta) = \\int 2 \\sqrt{\\sec^2 \\theta-1} \\tan\\theta \\text{ d}\\theta = 2 \\int \\tan^2 \\theta \\text{ d}\\theta$\n\nWith trig identity replacement: $2 \\int \\sec^2 \\theta \\text{ d}\\theta - 2 \\int 1 \\text{ d}\\theta = 2 \\tan\\theta - 2\\theta$\n\nWith replacement of $\\theta$ back to its original value: $\\sqrt{x^2 - 4} - 2\\text{ arcsec} \\left(\\frac{x}{2}\\right) + C$\n\nEdit 2 I understand that the square root causes sign issues, but even switching to absolute value doesn't seem to have any effect in the graphing. Here's a link to everything graphed on desmos.\n\n\u2022 I beat you to it, deleting my comment! \u2013\u00a0Namaste Oct 21 '16 at 20:17\n\u2022 According to Wolfram, your antiderivative is incorrect. \u2013\u00a0Ken Duna Oct 21 '16 at 20:17\n\u2022 I hadn't checked with Wolfram before, but in their Step-By-Step, their answer is mine, but with different work, so I updated my answer in response and showed my work. \u2013\u00a0BrainFRZ Oct 21 '16 at 20:53\n\nThe trouble is when getting rid of the square root. In general we have $\\sqrt{x^2} = |x|$, so $$\\sqrt{\\sec^2\\theta-1} = \\sqrt{\\tan^2\\theta} = \\lvert\\tan\\theta\\rvert.$$ So the integrand is really $$\\lvert\\tan\\theta\\rvert \\tan \\theta = \\begin{cases} \\sec^2\\theta - 1 & 0\\le\\theta<\\pi/2,\\\\ -(\\sec^2\\theta - 1) & \\pi/2<\\theta<\\pi\\\\ \\end{cases}$$ (where you take $\\theta = \\operatorname{arcsec} (x/2)$ with $0 \\le \\theta < \\pi$). When reversing the substitution there is also the issue of $$\\tan\\theta = \\begin{cases} \\sqrt{\\sec^2\\theta - 1} & 0\\le\\theta<\\pi/2,\\\\ -\\sqrt{\\sec^2\\theta - 1} & \\pi/2<\\theta<\\pi\\\\ \\end{cases}$$ so the correct integral has two parts: $$\\int \\frac{\\sqrt{x^2-4}}{x}\\,dx = \\begin{cases} \\sqrt{x^2-4} - 2 \\operatorname{arcsec}(x/2) + C_1 & x \\ge 2, \\\\ \\sqrt{x^2-4} + 2 \\operatorname{arcsec}(x/2) + C_2 & x \\le -2. \\\\ \\end{cases}$$\n\u2022 Sorry, I had an error in my post which I just fixed. And generally you will have to deal with things piecewise when sign changes occur (e.g. across $x=0$ in $\\sqrt{x^2} = |x|$). \u2013\u00a0arkeet Oct 21 '16 at 21:25", "date": "2019-08-18 13:11:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1979079/why-doesnt-the-derivative-of-integration-by-trig-substitution-match-the-origina", "openwebmath_score": 0.9239798784255981, "openwebmath_perplexity": 252.9099988194113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347853343058, "lm_q2_score": 0.8774767890838836, "lm_q1q2_score": 0.8550439066067901}}
{"url": "http://math.stackexchange.com/questions/67438/given-a2caci-0-how-to-find-inverse-of-ac-1i", "text": "# Given $A^2+cA+cI=0$, how to find inverse of $A+(c-1)I$?\n\nSuppose a square matrix $A$ such that $A^2+cA+cI=0$ for all $c \\in \\mathbb{Z}$. How can I show that $A+(c-1)I$ is invertible and find its inverse?\n\nI started off this way: $A+(c-1)I = A+cI-I$\n\nThen $(A+cI-I)(d_1A+d_2I)=I$, where $d_1, d_2 \\in \\mathbb{Z}$.\n\nExpand it and it becomes: $d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$\n\n$\\Rightarrow (c-1)d_2=1 \\;\\; and \\; \\; d_1A^2+d_1cA-d_1A+d_2A=0$\n\n$\\Rightarrow d_2=\\frac{1}{(c-1)}$\n\nContinue from $d_1A^2+d_1cA-d_1A+d_2A=0$, after some manipulation, I got $d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$.\n\nSince given that $A^2+cA+cI=0$,\n\n$d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$\n\n$\\Rightarrow -d_1cI-d_1A+d_2A=0$\n\n$\\Rightarrow -d_1cI-d_1A+\\frac{1}{c-1}A=0$\n\n$\\Rightarrow -d_1cI-d_1A=-\\frac{1}{c-1}A$\n\n$\\Rightarrow d_1(cI+A)=\\frac{1}{c-1}A$\n\nAt this point, since I don't know if $A$ is invertible yet, I cannot do it as $d_1=\\frac{1}{c-1}\\frac{A}{(cI+A)}$.\n\nEven if I did this, I still cannot find a value for $d_1$ and $d_2$ to find the inverse of $A+(c-1)I$. How should I continue from here?\n\n-\nDid you really mean for all $c \\in \\mathbb{Z}$? Wouldn't then $A^2 = A+I = 0$, hence $I^2 = 0$? \u2013\u00a0Niels Diepeveen Sep 25 '11 at 18:23\n\nInstead of all this, rewrite your starting equation as $A^2+(c-1)A+A+(c-1)I=-I$. Then factor the left-hand-side...\n\nBut if you favor a more systematic approach you can also proceed as you do until $$d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$$ At this point you decide to set $(c-1)d_2=1$, but that is too early! First get rid of the $A^2$ using $A^2=-c(A+I)$, to get $$-d_1c(A+I)+d_1cA - d_1A + d_2A + (c-1)d_2I = I$$ which simplifies to $$(d_1c-d_1+d_2-d_1c)A + ((c-1)d_2-d_1c-1)I = 0$$ $$(d_2-d_1)A + ((c-1)d_2-cd_1-1)I$$ Now set both coefficients to 0. This gives immediately $d_1=d_2$, and we then need to solve $$(c-1)d-cd-1=0$$ in which the $cd$'s cancel and give us $d=-1$. So the sought inverse is $-A-I$.\n\n-\nhmm...What happens if I couldn't see this and didn't start from this equation? I remember I had another problem to find inverses and I used a similar method and worked. I had the problem posted here too at math.stackexchange.com/questions/58841/\u2026 \u2013\u00a0xenon Sep 25 '11 at 15:23\nI've added a more systematic approach. You have to start from the given relation, though -- otherwise you have no way to know there is even an inverse anywhere. \u2013\u00a0Henning Makholm Sep 25 '11 at 15:36\nThanks! But how's $A^2=-(A+I)$ and not $A^2=-(cA+cI)$? \u2013\u00a0xenon Sep 25 '11 at 15:57\nAt one part, you set both the coefficients to be $0$. Is there a reason for doing this? Because I was thinking why wouldn't it be $((c-1)d_2-cd_1-1)=1$ and $(d_2-d_1)A=-I$? Of course, what you have done by having the coefficients zero makes the whole thing simpler but I was just thinking what if I set it something else other than zero. \u2013\u00a0xenon Sep 25 '11 at 16:21\nIt follows from some experience with quotients of polynomial rings. Your original relation can be used to prove that any linear combination of powers of $A$ can be rewritten into the form $pA+qI$, and if that is all we know about $A$, then there is only one such form. So if we have $pA+qI=0$ then either it must be because $p=q=0$ or else $A$ happened to satisfy a nicer property than the one we can depend on. (In your example the nicer property is \"is a multiple of $I$\"). It's always worth it to try a general method first and try more complex tactices if that fails. \u2013\u00a0Henning Makholm Sep 25 '11 at 16:31\n\nPerhaps a more transparent way: \"change of variables\". Let $B = A + (c-1) I$, or $A = B - (c-1) I$. Then $0 = A^2 + c A + c I = (B - (c-1) I)^2 + c (B - (c-1) I) + c I= B^2 + (2-c) B + I$. Multiply by $B^{-1}$ to get $B + (2-c) I + B^{-1} = 0$, i.e. $B^{-1} = -B + (c-2) I$ or $(A + (c-1) I)^{-1} = -A - (c-1) I + (c-2) I = -A - I$.\n\nYou may find it a bit dodgy to multiply by $B^{-1}$ before you know that $B^{-1}$ exists, but once you have the result $-A-I$ it's easy to verify that this works by multiplying it by $A + (c-1) I$.\n\n-\n\n$$\\begin{eqnarray*} x^2+cx+c = \\left[x+(c-1)\\right](x+1)+1\\\\ \\therefore 0 = A^2+cA+cI = \\left[A+(c-1)I\\right](A+I)+I\\\\ \\therefore \\left[A+(c-1)I\\right](A+I)=-I\\\\ \\therefore \\left[A+(c-1)I\\right]^{-1} = -(A+I). \\end{eqnarray*}$$\n\n-\n\nNEW ANSWER. In the \"long\" division $$X^2+cX+c=(X+c-1)(X+1)+1,$$ replace $X$ by $A$: $$0=A^2+cA+cI=\\Big(A+(c-1)I\\Big)\\Big(A+I\\Big)+I,$$ $$\\Big(A+(c-1)I\\Big)^{-1}=-A-I.$$\n\n[EDIT. I'm realizing that this answer is the same as user1551's, who posted it before. Sorry...]\n\nOLD ANSWER. Let $A$ be an $n$ by $n$ matrix with coefficients in a field $K$, let $f\\in K[X]$ be a polynomial annihilating $A$, and let $g\\in K[X]$ be any polynomial.\n\nIf $g$ is prime to $f$, then $g(A)$ is invertible, and the inverse of $g(A)$ is given by $h(A)$ where $h\\in K[X]$ is an inverse of $g$ mod $f$.\n\nMoreover, there is a closed formula for such an $h$ (see this answer).\n\n-", "date": "2016-05-02 03:51:11", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/67438/given-a2caci-0-how-to-find-inverse-of-ac-1i", "openwebmath_score": 0.9218217730522156, "openwebmath_perplexity": 148.3672097283174, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347905312772, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8550439064837531}}
{"url": "https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for-2/", "text": "# Which payoff do you want to go for?\n\nPayoff 1.\n\nToss 5 coins. You get $1 for each consecutive HT that you get. Payoff 2. Toss 5 coins. You get$1 for each consecutive HH that you get.\n\nFor example, if you tossed HTHHH, under payoff 1 you will get $1, but under payoff 2 you will get$2.\n\nNote by Calvin Lin\n3\u00a0years, 2\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nThis is a trick question, both payoffs have same expected value of profit, which is: $1. Consider this: We will count the number of ways HT can occur. We can have: HT_ _ _ : 8 ways _ HT _ _ _ : 8 ways _ _ HT _ : 8 ways _ _ _ HT : 8 ways Total number of ways: 32 Similarly for HH, the total number of ways in which it can occur is 32. What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is$32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same.\n\nMathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives$3 in p2 and $1 in p1. But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives$0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives$0 payoff.\n\nSo If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.\n\nSince this is in quantitative finance, should I talk about risk and stuff? I do not know.\n\n- 3\u00a0years, 2\u00a0months ago\n\nYes, the expected payoff for both scenarios is $1. The easiest way to see this is to look at Indicator Variables, which is essentially what Raghav did (though not phrased in that language). Well, everyone has their own \"payoff preference\". For example, you stated that you would prefer to \"be more likely to leave with at least a dollar in your pocket\". As such, this tells me that you are very risk adverse, and that you would prefer the certainty of a positive payoff. If you want to talk about \"risk and stuff\", what do you need to consider, and why? Staff - 3 years, 2 months ago Log in to reply I am not formally educated in these topics, so I'm just saying all this from a logical standpoint. After a bit of thinking, I think I agree with you on the fact that everyone has their own \"payoff preference\". One can take a risk to win more, or take less risk to win less. In the scenario mentioned here, the aspects of risk are not conspicuous. The amount to be won is not significant, and there is no penalty for losing. Hence, both the payoffs are inherently attractive offers. When we come to real life situations, I think there are many more things that contribute to the risk: 1. The probability that you will get a payoff. 2. The probability that you stand to lose money. 3. The security of the winnings. Money once won shouldn't be taken away from you. 4. The effects of said decisions on long term/ on your ability to take other decisions. According to me, every decision of ours is based on weighing in these risk factors against the probable prize. The balance may tip either way, and so may our decisions. I don't even know if I'm thinking in the right direction... Help me out here @Calvin Lin - 3 years, 2 months ago Log in to reply Part of this series of posts is for you to figure out what kind of risk appetite you have. It is a personal preference, and there isn't necessarily a \"right\" answer. Your understanding will help guide you in decisions that you make in future. For the points that you raised, you should answer for yourself if those should matter, and why. Hopefully, you will come up with a consistent, logical risk-reward structure. For example, if you would choose payoff 1 over 2 and payoff 3 over 4, but would prefer a combined 2 and 4 over a combined 1 and 3, then it would be very easy for someone to sell you things in part, and make you overpay for them. Staff - 3 years, 2 months ago Log in to reply I think that there may be some double-counting here. For example, you've counted HTHTT and HTHTH twice each. There are only $$2^{5} = 32$$ possible sequences in total, and there are some, such as TTHHH, that have no HT payouts, so there cannot be $$32$$ HT outcomes. I'm getting an average payout of $$\\dfrac{28}{32}$$ dollars in the first game type and $$\\dfrac{31}{32}$$ dollars in the second, but I am too tired to double-check. I'll do that tomorrow. :) Edit: Oh. now I see what you've done; very clever. I must have missed out on my count somewhere. I'll still wait to confirm in the morning. And yes, there does seem to be an advantage to choosing p1, (even though the expected winnings are the same), in the sense that you are more likely to at least win some money. - 3 years, 2 months ago Log in to reply Is \"more likely to at least win some money\" the main consideration that you will use if the expected payoff is equal? For example,what would you choose between: Payoff 1: 50% of$0, 50% of $100 Payoff 2: 90% of$1.01, 1% of $4900 Staff - 3 years, 2 months ago Log in to reply If Payoff 2 was 99% of$40.40, 1% of $1000, (to give it the same expected payoff as Payoff 1), I would go with Payoff 2 since I would be assured of money and would at least have a slim chance of a large amount of money. If Payoff 2 was 99% of$0, 1% of $5000, then it gets interesting, (at least for me). Even a such a slight chance of winning$5000 would be hard to pass up, so I would probably go with Payoff 2, unless I absolutely needed that $100 right away. If Payoff 2 was 99.9% of 0$, 0.1% of$50000 .... hmmmm .... That's a lot of dough, so I'd go with Payoff 2. However, if the options were these last two I've listed, then I might choose the$5000 option.\n\nIf Payoff 2 involved a 50% chance of losing $100 and a 50% chance of winning$200, then I would go with Payoff 1, since I'm not comfortable with there being such a good chance of losing a fair bit of money. There is a lot of psychology going on in these choices, and yet the expected winnings is always $50, (which ironically is an amount that would never actually be won in any of these scenarios). - 3 years, 2 months ago Log in to reply (Ooops, apparently I can't do maths. Fixed.) So, there are multiple ideas here. - Is the probability of a positive payoff so important that just because it moves from$0 it would greatly influence your preferences?\n- At what point does the potential promise of a huge payoff overweigh the certainty of a small payoff?\n\nStaff - 3\u00a0years, 2\u00a0months ago\n\nIn general I guess it all depends on the specific values, and further, on each person's financial status and willingness to take on risk. For me, the potential, (1% or better), of a large payout would outweigh any guaranteed amount less than $100, so in this case there is no difference to me between$0 and $100. But if the guaranteed amount were, say,$5000, with a 0.01% all-or-nothing chance of winning $1,000,000, I'd probably take the$5000 and run, (although I would be tempted to take the risk, at least for a moment). If the all-or-noting percentage were 1%, though, then I would find it hard not to take the chance; $5000 is a lot of money, but$1,000,000 is a life-changing amount of money, and a 1% chance is realistic in my eyes given the potential reward. The question I would ask myself is: which will I regret more - not taking the guaranteed money or not taking the risk?\n\n- 3\u00a0years, 2\u00a0months ago\n\n@Calvin Lin sir, Am I right?\n\n- 3\u00a0years, 2\u00a0months ago\n\nThere are 2^5 outcoms of 5 coin tosses.\n\nFor payoff 1, there are this many ways to toss a HT. There are {HTHTT,HTHTH,HTTHT,HTHHT,THTHT, HHTHT,HTHHH,HHHTH,HHHHT, HTTTT,THTTT,TTHTT,TTTHT,HTTTH,HHTTT, HHTTH} Expected paayyoff 1 = (6/32$2) + (10/32$1)= $11/16=$22/32\n\nFor payoff 2, there are this many ways to toss a HH( or HHH,HHHH and HHHHHH). There are {HHHHH, THHHH,HHHHT, THHHT,HTHHH,HHHTH, HHHTT, TTHHH, HHTHH, HHTTT, THHTT, TTHHT, TTTHH, THTHH, THHTH] Expected payoff 2 =(1/32$4 )+ ( 2/32$3) + ( 5/32$2) + (7/32$1)=$3/4=$27/32\n\nPayoff 2 is better, I guess LOL @Calvin Lin\n\nTotal possible outcomes of 32 of which there is one combination [TTTTT] which has no payoff.\n\n- 3\u00a0years, 2\u00a0months ago\n\nCheck your calculations. The expected payoff of both scenarios is 1. Staff - 3 years, 2 months ago Log in to reply Let $$X$$ be an indicator variable which takes on value 1 for each $$HT$$ pair and 0 otherwise, and let $$Y$$ be an indicator variable which takes on value 1 for each $$HH$$ pair and 0 otherwise. There are $$5 - 2 + 1 = 4$$ indicator variables (both $$X$$ and $$Y$$) in 5 coin tosses. The expected value of the payoff is: $\\text{E}\\left [\\sum_{k = 1}^{4} X_{k} \\right ], ~ \\text{E}\\left [\\sum_{k = 1}^{4} Y_{k} \\right ].$We can apply Linearity of Expectation here, which holds even for dependent events (nice proof of this is given in Brilliant Wiki Page), to get: $\\sum_{k = 1}^{4}\\text{E}\\left [ X_{k} \\right ] , ~ \\sum_{k = 1}^{4}\\text{E}\\left [ Y_{k} \\right ].$Expected value of both $$X_{k}$$ and $$Y_{k}$$ is: $\\text{E}\\left [ X_{k} \\right ] = \\text{E}\\left [ Y_{k} \\right ] = 1 \\cdot \\dfrac{1}{4} + 0 \\cdot \\dfrac{3}{4} = \\dfrac{1}{4}.$It follows that expected payoffs in both cases are the same and they equal:$\\sum_{k = 1}^{4}\\text{E}\\left [ X_{k} \\right ] = \\sum_{k = 1}^{4}\\text{E}\\left [ Y_{k} \\right ] = 4 \\cdot \\dfrac{1}{4} = 1.$ Bonus: What about the variance of the payoff? It can also influence our decision. Is it also the same? It turns out they are not! Let us first consider payoff 1 and indicator variable $$X$$:\\begin{align} \\text{Var}[X] &= \\text{E}\\left [\\left(\\sum_{k = 1}^{4}X_{k}\\right)^{2}\\right ] - \\text{E}\\left [\\sum_{k = 1}^{4}X_{k}\\right ]^{2} \\\\ &= \\text{E}\\left [\\left(\\sum_{k = 1}^{4}X_{k}^{2}\\right) + \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] - \\text{E}\\left [\\sum_{k = 1}^{4}X_{k}\\right ]^{2} \\\\ &= \\text{E}\\left [\\left(\\sum_{k = 1}^{4}X_{k}^{2}\\right)\\right] + \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] - \\text{E}\\left [\\sum_{k = 1}^{4}X_{k}\\right ]^{2} \\\\ &= 1 + \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] - 1 \\\\ &= \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ].\\end{align}The same is true for payoff 2 and $$Y$$. Now, let us calculate $$\\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ]$$. We make two groups of $$X_{k}X_{j}$$: \u2022 $$\\left | k-j \\right | = 1$$ representing two consecutive indicator variables which share one common coin. Notice that their product is always 0 since it's impossible to have two $$HT$$s in 3 consecutive coins. \u2022 $$\\left | k-j \\right | > 1$$ representing two non-consecutive indicator variables which share no common coin. Their product is non-zero only when they are both non-zero which happens with probability $$\\left(\\frac{1}{4}\\right)^{2} = \\dfrac{1}{16}.$$ There are total of $$4 \\cdot 3 = 12$$ $$\\left(X_{k}X_{j}\\right)$$ pairs, and $$2\\cdot 3$$ of them which are consecutive ie. where $$\\left | k-j \\right | = 1$$. Hence, we have: $\\text{Var}[X] = \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] = 0 \\cdot 6 + \\frac{1}{16} \\cdot 6 = 0.375.$ In the same way, we calculate $\\text{Var}[Y] = \\text{E}\\left[ \\left(\\sum_{k \\neq j}Y_{k}Y_{j}\\right)\\right ] = \\frac{1}{8} \\cdot 6 + \\frac{1}{16} \\cdot 6 = 1.125.$ So, variance of the payoff 2 is 3 times greater than variance of the payoff 1. If my reasoning is correct, then payoff 1 is somewhat safer option and it tends to the expected value, while payoff 2 is more riskier but it offers better chances of earning more than 1. Am I right @Calvin Lin?\n\n- 1\u00a0month ago\n\nI would go for Payoff 2 , because For payoff 1 ..... one can get maximum number of $1 is 2 which equals$2 i.e for this outcome HTTHT ( MAXIMUM POSSIBLE 'HT' consecutive is 2). For payoff 2 ..... one can get maximum number of $1 is 4 which equals$4 i.e for this outcome HHHHH ( MAXIMUM POSSIBLE 'HH' consecutive is 4).\n\n- 2\u00a0months, 2\u00a0weeks ago\n\nFrom a statistical standpoint, I would choose payoff option 1. As the number of H tossed increases, the probability of the consecutively tossing another H decreases. That being said, the HH combination can lead to a higher payoff because HTHTH only results in $2 while HHHHH results in$5. The issue lies in the decreased probability of continuing to toss H. This is a good example related to risk, much like in the decision involved in buying a AAA bond earning 5% vs a B bond earning 13%.\n\n- 1\u00a0year, 11\u00a0months ago\n\nTo be more specific, payoff 1 would be like the AAA bond and payoff 2 like the B bond. It just depends on what kind of risk is right for you.\n\n- 1\u00a0year, 11\u00a0months ago\n\nNot quite the same comparison. Note that the expected payoff in both methods are the same.\n\nWhereas in the bond example that you gave, we are trading expected payoff for certainty.\n\nI also strongly disagree with \"As the number of H tossed increases, the probability of the consecutively tossing another H decreases\". The coin tosses are independent, and that is a common fallacy that \"the proportion of realized events must be equal / close to the calculated probability\"\n\nStaff - 1\u00a0year, 11\u00a0months ago\n\nI remember reading about an interesting experiment done by a French mathematician regarding payoffs and expected utility from lotteries. Allais Paradox?\n\n- 3\u00a0years, 2\u00a0months ago\n\nThat's interesting. In experiment 1 here, taking a risk, (even if it is only 1%), means the possibility of losing a guaranteed million dollars; I would deeply regret gambling and ending up losing that money, but if I took the million and played 1B just to see what happened and found that I could have won 5 million, I would have just said \"Oh well\" and still been perfectly happy with my million. But if there is no guaranteed money, then having a chance at 5 million at the expense of a 1% greater chance of winning a million seems worth the risk.\n\nExperiment 1 makes me think of the old saying, \"A bird in the hand is worth two in the bush.\" If there are no \"birds in hand\", however, then the risk evaluation process is quite different.\n\n- 3\u00a0years, 2\u00a0months ago\n\nLet Ii be an indicator rv for an event of getting H at ith toss. Let X be a rv which is how many times HT occured. Similarly, let Y be the same but for HH. Then, X = I1I2 +... +I4I5 and Y = I1(1-I2) +...+I4(1-I5). We need to compare EX and EY. Given EIi = 0.5, we can apply linearity of expectation (even for multiplication since events are independent) and get 1 for both expectations.\n\n- 2\u00a0years, 7\u00a0months ago\n\nSo, if their expectations are the same, does that mean that you are indifferent about which payoff to take? If so, why?\n\nStaff - 2\u00a0years, 7\u00a0months ago\n\nYes, because with both payoffs, one can win the same amount of dollars on average.\n\n- 2\u00a0years, 7\u00a0months ago\n\nSo, are you indifferent between these 2 payoffs:\n\nPayoff 3: Always get $0 Payoff 4: 50% chance of -$1,000,000,000 and 50% chance of \\$1,000,000,000\n\nStaff - 2\u00a0years, 7\u00a0months ago\n\nI think we want the choice with the minimum variance involved.\n\nMaximin Principle of Rationality: An epistemically rational person maximizes his minimum payoff\n\nStaff - 2\u00a0years, 7\u00a0months ago\n\nIf I don't have a million dollars at all, I will definitely not take payoff 4. I would take payoff 4 only if I was a millionaire. So, psychologically I am not indifferent.\n\n- 2\u00a0years, 7\u00a0months ago\n\nRight, so there are other considerations that come into play. For most people, the certainty of a result is preferred to an uncertain result (though with equal expected value). In general, having a low variance is better (though of course, there are exceptions).\n\nStaff - 2\u00a0years, 7\u00a0months ago\n\nI think this is a combinatorics problem?!\n\nStaff - 3\u00a0years, 2\u00a0months ago\n\nIn the sense that arithmetic is a part of algebra, which is a part of calculus? There are numerous cross-disciplinary ideas. Using combinatorics will only get you one perspective of things.\n\nNotice that I am essentially getting at the risk-reward preference, which is a \"finance idea\" as opposed to a \"combinatorics idea\". Yes, combinatorics ideas like expected value is involved, but they don't tell the full story.\n\nStaff - 3\u00a0years, 2\u00a0months ago", "date": "2018-07-15 23:20:55", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for-2/", "openwebmath_score": 0.9766924381256104, "openwebmath_perplexity": 1753.1553126000283, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9744347831070321, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8550438999691503}}
{"url": "http://flemcup.com/s9au4c/099ebc-rank-of-a-matrix", "text": "# rank of a matrix\n\nWe prove that column rank is equal to row rank. The rank of a matrix is the largest number of linearly independent rows/columns of the matrix. Ask a Question . The rank of A is equal to the dimension of the column space of A. This lesson introduces the concept of matrix rank and explains how the rank of a matrix is revealed by its echelon form.. To define rank, we require the notions of submatrix and minor of a matrix. In mathematics, low-rank approximation is a minimization problem, in which the cost function measures the fit between a given matrix \u2026 the matrix in example 1 has rank 2. So maximum rank is m at the most. Guide. The idea is based on conversion to Row echelon form. The rank of a matrix would be zero only if the matrix had no non-zero elements. Each matrix is line equivalent to itself. Recent rank-of-matrix Questions and Answers on Easycalculation Discussion . Or, you could say it's the number of vectors in the basis for the column space of A. Prove that rank(A)=1 if and only if there exist column vectors v\u2208Rn and w\u2208Rm such that A=vwt. A rank-one matrix is the product of two vectors. In previous sections, we solved linear systems using Gauss elimination method or the Gauss-Jordan method. I would say that your statement \"Column 1 = Column 3 = Column 4\" is wrong. by Marco Taboga, PhD. Rank of a Matrix in Python: Here, we are going to learn about the Rank of a Matrix and how to find it using Python code? Rank of a matrix definition is - the order of the nonzero determinant of highest order that may be formed from the elements of a matrix by selecting arbitrarily an equal number of rows and columns from it. How to find Rank? The rank of a matrix m is implemented as MatrixRank\u2026 If p < q then rank(p) < rank(q) Rank of a matrix. DEFINITION 2. The rank of a matrix is defined as. Pick the 2nd element in the 2nd column and do the same operations up to the end (pivots may be shifted sometimes). The Rank of a Matrix. 7. tol (\u2026) array_like, float, optional. The rank of a matrix is the dimension of the subspace spanned by its rows. The Rank of a Matrix Francis J. Narcowich Department of Mathematics Texas A&M University January 2005 1 Rank and Solutions to Linear Systems The rank of a matrix A is the number of leading entries in a row reduced form R for A. Introduction to Matrix Rank. 6. What is a low rank matrix? 8. Exercise in Linear Algebra. \u2026 If all eigenvalues of a symmetric matrix A are different from each other, it may not be diagonalizable. The non-coincident eigenvectors of a symmetric matrix A are always orthonomal. The system has a nontrivial solution if only if the rank of matrix A is less than n. Matrix Rank. Calculators and Converters. The rank depends on the number of pivot elements the matrix. The column rank of a matrix is the dimension of the linear space spanned by its columns. Rank of a Matrix. This exact calculation is useful for ill-conditioned matrices, such as the Hilbert matrix. Find Rank of a Matrix using \u201cmatrix_rank\u201d method of \u201clinalg\u201d module of numpy. The nxn-dimensional reversible matrix A has a reduced equolon form In. The rank of a Matrix is defined as the number of linearly independent columns present in a matrix. Remember that the rank of a matrix is the dimension of the linear space spanned by its columns (or rows). Threshold below which SVD values are considered zero. Input vector or stack of matrices. Rank of unit matrix $I_n$ of order n is n. For example: Let us take an indentity matrix or unit matrix of order 3\u00d73. Coefficient matrix of the homogenous linear system, self-generated. Pick the 1st element in the 1st column and eliminate all elements that are below the current one. Return matrix rank of array using SVD method. This also equals the number of nonrzero rows in R. For any system with A as a coe\ufb03cient matrix, rank[A] is the number of leading variables. OR \"Rank of the matrix refers to the highest number of linearly independent rows in the matrix\". The rank of the matrix A is the largest number of columns which are linearly independent, i.e., none of the selected columns can be written as a linear combination of the other selected columns. The rank of a matrix can also be defined as the largest order of any non-zero minor in the matrix. 1) Let the input matrix be mat[][]. the maximum number of linearly independent column vectors in the matrix So often k-rank is one less than the spark, but the k-rank of a matrix with full column rank is the number of columns, while its spark is $\\infty$. rank-of-matrix Questions and Answers - Math Discussion Recent Discussions on rank-of-matrix.php . The rank of a matrix or a linear transformation is the dimension of the image of the matrix or the linear transformation, corresponding to the number of linearly independent rows or columns of the matrix, or to the number of nonzero singular values of the map. Some theory. 2010 MSC: 15B99 . linear-algebra matrices vector-spaces matrix-rank transpose. Theorem [thm:rankhomogeneoussolutions] tells us that the solution will have $$n-r = 3-1 = 2$$ parameters. The rank is an integer that represents how large an element is compared to other elements. Rank of Symbolic Matrices Is Exact. 1 INTRODUCTION . Top Calculators. 4. In the examples considered, we have encountered three possibilities, namely existence of a unique solution, existence of an infinite number of solutions, and no solution. We are going to prove that the ranks of and are equal because the spaces generated by their columns coincide. Calculator. Based on the above possibilities, we have the following definition. A matrix obtained by leaving some rows and columns from the matrix A is called a submatrix of A. Finding the rank of a matrix. All Boolean matrices and fuzzy matrices are lattice matrices. Changed in version 1.14: Can now operate on stacks of matrices. Common math exercises on rank of a matrix. Matrix Rank. As we will prove in Chapter 15, the dimension of the column space is equal to the rank. You can say that Columns 1, 2 & 3 are Linearly Dependent Vectors. Firstly the matrix is a short-wide matrix \\$(m", "date": "2021-09-25 15:14:13", "meta": {"domain": "flemcup.com", "url": "http://flemcup.com/s9au4c/099ebc-rank-of-a-matrix", "openwebmath_score": 0.6607620716094971, "openwebmath_perplexity": 222.98561114329152, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.976310525254243, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.85504005782664}}
{"url": "https://www.physicsforums.com/threads/linear-independence.814024/", "text": "# Linear independence\n\n## Homework Statement\n\nAssume vectors ##a,b,c\\in V_{\\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.\n\n## The Attempt at a Solution\n\nWe say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.\nDoes there exist a non-trivial combination such that\n##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:\n##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:\n##k_1+k_3 =0\\Rightarrow k_1 = -k_3##\n##k_1+k_2 =0##\n##k_2+k_3 =0##\n\nSubstituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.\n\nLast edited:\n\nMark44\nMentor\n\n## Homework Statement\n\nAssume vectors ##a,b,c\\in V_{\\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.\n\n## The Attempt at a Solution\n\nWe say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.\nDoes there exist a non-trivial combination such that\n##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:\n##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:\n##k_1+k_3 =0\\Rightarrow k_1 = -k_3##\n##k_1+k_2 =0##\n##k_2+k_3 =0##\n\nSubstituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.\nThat works for me. (IOW, I agree that the three new vectors are linearly independent.)\n\nInstead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.\n\nThe matrix looks like this, from your system:\n$$\\begin{bmatrix} 1 & 0 & 1 \\\\ 1 & 1 & 0 \\\\ 0 & 1 & 1\\end{bmatrix}$$\nAfter a few row operations, the final matrix is I3.\n\nRay Vickson\nHomework Helper\nDearly Missed\nThat works for me. (IOW, I agree that the three new vectors are linearly independent.)\n\nInstead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.\n\nThe matrix looks like this, from your system:\n$$\\begin{bmatrix} 1 & 0 & 1 \\\\ 1 & 1 & 0 \\\\ 0 & 1 & 1\\end{bmatrix}$$\nAfter a few row operations, the final matrix is I3.\n\nAlternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?\n\nAlternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?\nOh. Cramer's rule.\n##k_n = \\frac{D_{k_n}}{D}## and since the determinant of the system is non zero, the corresponding determinants for every ##k_n## would be 0 (a full column of 0-s means det = 0) and therefore ##k_1 = k_2 = k_3 = 0##\n\nRay Vickson", "date": "2021-06-15 05:52:48", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/linear-independence.814024/", "openwebmath_score": 0.7688741087913513, "openwebmath_perplexity": 2424.277942024567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.976310525254243, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8550400483309852}}
{"url": "http://math.stackexchange.com/questions/1607627/how-can-a-cauchy-sequence-converge-to-an-irrational-number", "text": "# How can a Cauchy sequence converge to an irrational number?\n\nI am a physics major and would like to clear a confusion regarding complete metric spaces. I am quoting the definition of a Cauchy sequence from wikipedia below\n\nFormally, given a metric space $(X, d)$, a sequence $x_1, x_2, x_3, \\ldots$ is Cauchy, if for every positive real number $\\epsilon > 0$ there is a positive integer $N$ such that for all positive integers $m, n > N$, the distance\n\n$$d(x_m, x_n) < \\epsilon$$\n\nNow, if we have sequence like $x_1=3, x_2=3.1, x_3=3.14, \\ldots$ converging to $\\pi$, I do not understand how all distances $d(x_m, x_n)$ will be less than all positive real numbers. Since irrational numbers do not terminate and continue forever, how can the distance ever be less than the smallest real number or infinitesimal (hyperreal) as the distance can never become $0$. Does this definition of completeness apply where $\\epsilon$ is infinitesimal (hypperreal) ?\n\nKindly excuse my ignorance as I am not a mathematics major.\n\nThanks\n\n-\nDon't think of it as one particular distance, $d(x_m,x_n)$, is smaller than every positive real number. But rather, any given positive real number, you can go far enough out to guarantee that $d(x_m,x_n)$ is smaller. \u2013\u00a0I. Cavey Jan 11 at 3:14\nThere is no \"smallest [positive] real number\". \u2013\u00a0Eric Wofsey Jan 11 at 3:15\nIt is not that the distance is smaller than every positive real number, it is that for any positive real number, there is a point where the distance is eventually less than it. E.g. can you eventually go far enough that the sequence is accurate to the fifth decimal place? (i.e. $\\epsilon=10^{-5}$) Can you go far enough that the sequence is accurate to the hundredth decimal place? (i.e. $\\epsilon=10^{-100}$) If I were to give you some number of decimal places, could your sequence eventually be that accurate? The answer of course is yes. \u2013\u00a0JMoravitz Jan 11 at 3:16\nI disagree with the vote to close the question. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Jan 11 at 3:17\n@AloizioMacedo : Perhaps it is unclear to you, but not to me. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Jan 11 at 4:02\n\nThe subject line currently reads \u201cHow can a Cauchy Sequence converge to an irrational number?\u201d.\n\nIf we construe that literally, then one easy way a Cauchy sequence (lower-case initial \"s\") can converge to $\\pi$ is that every term of the Cauchy sequence is $\\pi$. Thus: $x_1=\\pi, x_2=\\pi, x_3=\\pi,\\ldots\\,{}$. I suspect you meant \u201cHow can a Cauchy sequence of rational numbers converge to an irrational number?\u201d.\n\nConsider your sequence $3,\\ 3.1,\\ 3.14,\\ 3.141,\\ \\ldots\\,$.\n\nThe definition DOES NOT say that all distances between members of this sequence are less than all positive numbers. That would happen only with a constant sequence like my first example above. It says:\n\nFor every positive real number $\\varepsilon>0$ there is a positive integer $N$ such that for all positive integers $m,n>N$ we have $d(x_m,x_n)<\\varepsilon$.\n\nNotice that $N$ depends on $\\varepsilon$. In fact as $\\varepsilon$ gets smaller, typically $N$ must get bigger. Suppose $\\varepsilon = 0.01$. Then for your example sequence, $N=3$ is big enough: every pair of numbers in the sequence at or after the third place in the sequence differ from each other by less than $\\varepsilon=0.01$. Thus $3.14$ and $3.141$ differ by less than $0.01$. But now suppose $\\varepsilon=0.00001$. Then you need a bigger value of $N$. If each term of the sequence has one more digit or $\\pi$, then $N=5$ would be big enough for that value of $\\varepsilon$.\n\nNotice that the definition of convergence to $\\pi$ differs from the definition of \"Cauchy sequence\". It says for every $\\varepsilon>0$ there is a positive integer $N$ such that for every positive integer $n\\ge N$ we have $|x_n-\\pi|<\\varepsilon$. Again, $N$ depends on $\\varepsilon$. If $\\varepsilon=0.00001$, then $N=5$ would be enough: every term at or beyond the $5$th one differs from $\\pi$ by less than $\\varepsilon=0.00001$.\n\nThere is nothing in either of these definitions that says that the distance between two different members of the sequence or the distance between $\\pi$ and a member of the sequence is $0$.\n\nYou wrote:\n\nSince irrational numbers do not terminate and continue forever\n\nLet's be clear on a definition.\n\nIt is certainly not correct that numbers whose decimal expansions do not terminate are necessarily irrational. For example, $1/7 = 0.\\ 142857\\ 142857\\ 142857\\ \\ldots$ has a non-terminating decimal expansion and is rational.\n\nNor is it the case that \"rational number\" is defined as one whose decimal expansion repeats or terminates. Euclid and other ancient Greeks proved some numbers are irrational without ever thinking about decimal expansions. That $\\pi$ is irrational means $\\pi$ is not a quotient of two integers, like $22/7$. Proving $\\pi$ is irrational is so difficult that it was not done until the 18th century. Some numbers are far easier to prove to be irrational. For example, if $\\log_2 3 = m/n$ and $m,n$ are positive integers, then $2^m=3^n$, but that can't happen because an even number cannot be equal to an odd number.\n\nThe fact that a number is rational if and only if its decimal expansion repeats or terminates takes a bit of work to prove, but it's elementary enough that high-school students will understand it.\n\n-\nIn the last sentence, I think you meant \"...a number is rational if and only if its decimal expansion repeats or terminates...\" \u2013\u00a0Nathan Reed Jan 11 at 4:34\nI don't understand the section after the quoted statement \"Since [the decimal expansions of] irrational numbers do not terminate and continue forever.\" You appear to be trying to rebut that statement, but the statement is true. The asker never claims that numbers whose decimal expansions don't terminate are rational. They also don't claim that the definition of being rational is that the decimal expansion repeats or terminates. Both of those claims would be false but they're never made so I don't understand why you're rebutting them. \u2013\u00a0David Richerby Jan 11 at 4:55\nAgree with @DavidRicherby; this answer would be improved by taking out the rebuttal that infinite decimals are necessarily irrational, which was never part of the OP's inquiry. \u2013\u00a0Daniel R. Collins Jan 11 at 5:55\nEr, I mean the asker doesn't ever claim that non-terminating decimals are irrational. Doh! \u2013\u00a0David Richerby Jan 11 at 5:57\n@NajibIdrissi : Why would you say that? If someone is learning this subject, they should become aware of things like that. \"Cauchy sequence\" is not defined as meaning a Cauchy sequence of rational numbers. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Jan 13 at 17:08\n\nIt's not the case that \"all distances $d(x_m, x_n)$ will be smaller than every real number\". That is not even true \"eventually\", in the sense that for some $N$, it's true for all $m, n > N$. That would imply that all such distances are $0$. Furthermore, of course there is no \"smallest [positive] real number\". The point is that for any positive real number $\\epsilon$, no matter how small, the distances between the terms of the sequence eventually get smaller than $\\epsilon$ and stay smaller than $\\epsilon$.\n\nThink of it as a challenge: I give you some $\\epsilon>0$, and you have to find a position in the sequence (some $N$) such that $d(x_m,x_n)<\\epsilon$ for all $m,n>N$. If the sequence is Cauchy, you are guaranteed to win the challenge. Cauchy chose the letter $\\epsilon$ to stand for error (or erreur): the finite approximations $x_n$ will have an error of less than $\\epsilon$ from some point on.\n\n-\nIn the hyperreal setting mentioned by the OP, the difference can be smaller than every positive real number and still be nonzero. See my answer. \u2013\u00a0Mikhail Katz Jan 12 at 16:24\nThe question did not originally mention hyperreals. \u2013\u00a0BrianO Jan 12 at 16:42\nOK, I guess I am coming in late in the game. \u2013\u00a0Mikhail Katz Jan 12 at 16:42\n\nIf $n >m \\ge 1$ you have $d(x_n,x_m) < 10^{n-1}$.\n\nChoose $\\epsilon>0$ and $N$ such that $10^{N-1} < \\epsilon$, then if $n,m \\ge N$ we have $d(x_n,x_m) < \\epsilon$.\n\nNote that the statement is for any $\\epsilon>0$ there is some $N$ such that blah, blah, blah, and not that there is some $N$ such that for all $\\epsilon>0$ we have blah, blah, blah.\n\nThe latter formulation would imply that the sequence is constant after some $N$.\n\n-\n\nI think your confusion comes from \"for every positive real number $\\epsilon$\". In fact you have to fix $\\epsilon$ first when finding $N$.\n\n-\n\nfirst of all it is not that all the distances are less than every positive number. given a positive number , one can find a stage after which any two terms are the given number close to each other.\n\nThe idea of a sequence converging to some point do not necessarily imply that the distance between two terms of the sequence become zero. What it says is that, as $n$ becomes larger, the terms of the convergent sequence starts coming closer to the previous one in terms of distance,\n\ne.g in a sequence $<1/n>$ the distance between 1 and 1/2 is 1/2, distance between 1/2 and 1/4 is 1/4 and so on.. now if you look at distance between any two terms after 1/2, it is always less 1/2, if you look at the terms after 1/100 the max distance between any to terms will be upto 1/100 and so on. So if the given positive number is 1/100, you can choose here $n_0=101$(rather anything bigger than it). whereas if you want distance to be less than 1/1000 say, then this $n_0$ wont work, so you will need $n_{0}=1001$ atleast. It does not depend on terms being rational or irrational.\n\n-\n\n\" I do not understand how all distances $d(x_m,x_n)$ will be less than all positive real numbers.\"\n\nThis is the key to your misunderstanding. All distances can't be.\n\nBut for any real number no matter who small we can find an infinite number of distances smaller. (So we could say all real numbers are bigger than an infinite number of the distances...)\n\nLet's do an actual example. Consider .9, .99, .999, etc. ($a_n = \\sum_{i=1}^n 9/10^i$). This is a cauchy sequence that converges to 1. Okay, 1 is not irrational. But is that really where your misunderstanding lies? Okay, I'll get to that later. But for now.\n\nCauchy sequence: for any $\\epsilon > 0$ ... okay, let's say $\\epsilon = 1/1,000,000$ ... we can find a point after which all distances are less than $\\epsilon$ ... okay, that'd be simply $n > 7$. $d(.99999\\ldots9, .9999\\ldots9) < 1/1,000,000$ for any two $.999\\ldots99$ with more than 7 digits.\n\nOkay, but let's make $\\epsilon$ really small. Let's make it 1/googol. ($10^{-100}$) Well, now if n > 100 we still have $(.999\\ldots9, .9999\\ldots) < 1/\\text{googol}$ if those terms have more that 100 nines. How about a googleplex? Then the terms need a googol nines to be that close. But we can find them with more than a googol nines. No matter how small we can find a point where the following difference of terms is smaller.\n\nThat's a cauchy sequence.\n\nOkay, that was a cauchy sequence of rational numbers converging to a rational number. You wanted to know about a cauchy sequence of rational numbers converging to an irrational number and you suggested the decimal expansion of $\\pi$. Well, same thing. $\\pi$ expanded to a million decimal places will be off but it'll be of by a very small amount. For any real number no matter how small, we can find a point after which all decimal expansions are closer that that to $\\pi$. These expansions are still finite but they are long enough to be closer to $\\pi$ than the small real number we chose. (And if they aren't, we simply choose the finite decimal expansions that are further out. There'll always be finite expansions further out. That's the point.)\n\nIt's true we never get there. And we can't really choose rational values that can get there (that's why the number is irrational). but we've trapped and honed it in with cauchy sequences.\n\n-\nthanks for your response. My main confusion has been made clear that you have to choose $\\epsilon$ first and then you get $N$. But how does this work with infinitesimal (hyperreals)...How does this definition of completeness apply there ? \u2013\u00a0phd-applicant Jan 11 at 5:55\nhyperreals are outside my area of expertise. My understanding though is that cauchy sequences of reals do not nesc converge to hyperreals. Hyperreals are not archimedian (for any x>0 y you can find integer n s.t. (n-1)x <= y < nx) and so have a different definition of completeness. I really don't know. I'm sorry I don't know more. But the key of the reals is that epsilon of choice is not an infintisimal. \u2013\u00a0fleablood Jan 11 at 6:22\nTo say that the fact that \"we never get there\" is \"why they are irrational\" is nonsense. The sequence $0.3,\\ 0.33,\\ 0.333,\\ 0.3333,\\ \\ldots$ approaches $1/3$ and never gets there, but $1/3$ is rational. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Jan 11 at 14:29\n@fleablood, great question. The answer is, it doesn't. This e is not the smallest element and of course in a field you couldn't have a smallest element. As a first approximation think of a sequence tending to zero as generating an infinitesimal, e.g., $\\frac{1}{n})$. Then its square will be so much smaller: $\\frac{1}{n^2}$. More details can be found here: math.stackexchange.com/questions/1602977/\u2026 \u2013\u00a0Mikhail Katz Jan 12 at 16:41\n@fleablood : This is not about surreal numbers, but about Robinson's hyperreal numbers. An infinitesimal $\\varepsilon>0$ is smaller than all reciprocals of finite integers, and hence smaller than all positive real numbers, but it is certainly not a smallest positive hyperreal number, since $\\varepsilon/2$ is smaller, and so is $\\varepsilon^2$. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Jan 12 at 18:07\n\nI looked over the existing answers and they don't seem to address the concern of the OP as I understood it, so I will try to provide a separate answer. The point is that the $(\\epsilon,N)$-type definition of convergence is a first-order property and therefore by the transfer principle is still satisfied over the hyperreals. More specifically, a Cauchy sequence $(x_n)$ will have a natural extension defined even for infinite values of the index $n$. This extended sequence will satisfy the condition you mentioned, namely for every $\\epsilon$ there is an $N$ such that, etc. If epsilon is infinitesimal then as you point out $N$ will have to be infinite typically.\n\nHaving said this, the definition in question is provably equivalent to the one mentioned by other editors (and that involves fewer quantifiers), namely if both indices are infinite then the difference $x_n-x_m$ will be infinitesimal.\n\n-", "date": "2016-07-29 18:19:54", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/1607627/how-can-a-cauchy-sequence-converge-to-an-irrational-number", "openwebmath_score": 0.9072916507720947, "openwebmath_perplexity": 222.35822710579075, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105273220726, "lm_q2_score": 0.8757869786798663, "lm_q1q2_score": 0.8550400469767451}}
{"url": "https://math.stackexchange.com/questions/2952249/probability-that-in-one-rolling-of-5-dice-we-obtain-two-6-and-one-5", "text": "# Probability that in one rolling of 5 dice we obtain two '6' and one '5'?\n\nFirst I made a mistake of not taking into account that this event is dependent.\n\nSo to get two '$$6$$' the probability would be:\n\n$$P_5(2) = \\frac{5!}{2!(5-2)!} \\cdot \\left(\\frac16\\right)^2\\cdot\\left(\\frac56\\right)^3$$\n\nWhere\n\n\u2022 $$p = \\dfrac16$$\n\u2022 $$q = \\dfrac56$$\n\u2022 $$N= 5$$ (the number of elements of the system)\n\nThen there are $$3$$ dice left on the table and we want to know the probability that one of them is a '$$5$$'. We rule out the two dice with the '$$6$$' face on. So the number of element in the system is now $$3$$. So the probability would be:\n\n$$P_3(1) = \\frac{3!}{1!(3-1)!} \\cdot \\left(\\frac16\\right)^1\\cdot\\left(\\frac56\\right)^2$$\n\nBut then the professor told me that $$p=\\dfrac15$$ and not $$p=\\dfrac16$$ That's what I don't get.\n\nOf course at the end we multiply those two probabilities to get the final probability that we want.\n\n\u2022 The probability is $1/5$ because you CAN\"T get another $6.$ Then, you would have more than two $6$s. \u2013\u00a0StatGuy Oct 12 at 4:50\n\nI am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.\n\nMethod 1\n\nYou can solve this using the joint pmf for a multinomial distribution.\n\n$$P(X_6 = 2, X_5 = 1, X_{other} = 2) = \\frac{5!}{2!1!2!} \\left( \\frac{1}{6} \\right)^2 \\left( \\frac{1}{6} \\right) \\left( \\frac{4}{6} \\right)^2 \\approx .062$$\n\nMethod 2\n\nAlternatively you can solve this using the multiplication rule. Here you get the $$1/5$$ that your professor talked about because when you condition you are on a reduced sample space.\n\nLet $$E$$ be the event you get two sixes.\n\nLet $$F$$ be the event you get one five.\n\nLet $$G$$ be the event you get two of the others.\n\n$$P(EFG) = P(E)P(F \\mid E)P(G \\mid EF)$$\n\n$$P(E) = {5 \\choose 2}\\left( \\frac{1}{6} \\right)^2 \\left( \\frac{5}{6} \\right)^3$$\n\n$$P(F \\mid E) = {3 \\choose 1}\\left( \\frac{1}{5} \\right) \\left( \\frac{4}{5} \\right)^2$$\n\n$$P(G \\mid EF) = {2 \\choose 2}\\left( \\frac{4}{4} \\right)^2 \\left( \\frac{0}{4} \\right)^0 = 1$$\n\n$$P(EFG) \\approx .062$$\n\n\u2022 The first method looks like an easy way, can I get the general equation? But still, the $1/5$ doesn't compute to me. How can we rule out a face when it can still happen? To me if we put $p=1/5$ it's would be like considering the 3 other dice have five faces. \u2013\u00a0Nuz Oct 12 at 5:14\n\u2022 In regards to method 1 just look up the Multinomial distribution on wiki and check out the pmf. In regards to the $1/5$, personally I don't think about it as a dice with five faces. Personally the way I think about it is you roll the other $3$ dice and if you get any $6$'s you just roll again. So for example, what's the probability you roll a $1$ given that you don't roll a $2,3,4,5,6$? The probability is $1$. Does that mean you have a one faced die? Maybe the way to think of it is if you get something that is not a $1$ you just re roll until you finally do. You are on a reduced sample space. \u2013\u00a0HJ_beginner Oct 12 at 5:38\n\u2022 Is there another approach, to calculate the fact that one out of the 3 dice must be a '5' and the 2 others can't be '6's or '5's? Or maybe another analogy or something? \u2013\u00a0Nuz Oct 12 at 7:11\n\u2022 Hmmm not sure I understand your question. You can calculate $P(EFG)$ by conditioning on any event first. $P(EFG) = P(G)P(F \\mid G)P(E \\mid FG)$. There's many many valid approaches. Part of the reason why I find probability challenging/interesting. \u2013\u00a0HJ_beginner Oct 12 at 7:36\n\nBut then the professor told me that $$p=1/5$$ and not $$p=1/6$$ That's what I don't get.\n\nYou have correctly evaluated the probability for the event that two from the five dice show 6.\n\n$$\\mathsf P(N_6{=}2)=\\binom 52 \\dfrac{1^25^3}{6^5}$$\n\nNow, of the three dice which don't show 6, each may show faces $$\\{1,2,3,4,5\\}$$ with equal probability. \u00a0 Thus the conditional probability for a die showing 5, when given that it does not show 6, is $$1/5$$. \u00a0 So the conditional probability for obtaining one 5 when given that two 6 have been obtained among the five dice is:\n\n$$\\mathsf P(N_5{=}1\\mid N_6{=}2)=\\binom 31\\dfrac{1^14^2}{5^3}$$\n\nOf course at the end we multiply those two probabilities to get the final probability that we want.\n\nAnd the $$5^3$$ factors will cancel, so:\n\n$$\\mathsf P(N_6{=}2,N_5{=}1)=\\binom{5}{2}\\binom{3}{1}\\dfrac{1^21^14^2}{6^5}$$\n\nPS: This is called a multinomial distribution.", "date": "2018-10-21 23:20:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2952249/probability-that-in-one-rolling-of-5-dice-we-obtain-two-6-and-one-5", "openwebmath_score": 0.745387613773346, "openwebmath_perplexity": 214.87272658486404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513914124559, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8550285327918713}}
{"url": "https://www.physicsforums.com/threads/bezouts-identity-and-diophantine-equation.151820/", "text": "B\u00e9zout's identity and Diophantine Equation\n\nhaki\n\nI am having problems with one exam question.\n\nDoes this diophantine equation have a solution(s)\n\n12a+21b+33c=6\n\nas far as I know this is not a linear equation, and what I read online says that Bezout identity only applies for linear diophantine equations.\n\nThe solution says gcd(12,21,33) = 3, 6|3 the above equation has infinitely many solutions. Is that correct? Am I correct to assume then that\n\n12a+21b+33c+24d = 6 again has infinite solutions aswell?\n\n17x+6y +3z =73\n\nsince gcd(17,6,3) = 1 and 73 | 1 this has also infinitely many solutions?\n\nfresh_42\n\nMentor\n2018 Award\nB\u00e9zout's Lemma says, that the greatest common divisor $d$ of numbers $a_1,\\ldots,a_n$ can always be written as $d= s_1a_1+\\ldots +s_na_n$. This immediately applies to all of your examples.\n\nAn example where it does not work is $2x+4y+6z = 7$.\n\nWWGD\n\nGold Member\nYou can always reduce the case ax+by+cz=d to ax+by=d by letting z=0.\n\n\"B\u00e9zout's identity and Diophantine Equation\"\n\nPhysics Forums Values\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-10-21 08:21:39", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/bezouts-identity-and-diophantine-equation.151820/", "openwebmath_score": 0.7965666651725769, "openwebmath_perplexity": 2205.599561804341, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513914124559, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8550285260155258}}
{"url": "http://www.fortinlab.com/wp-contnt/fo14r0pn/article.php?id=233270-condensation-point-math", "text": "# condensation point math\n\nPosted in: Uncategorized | 0\n\nThe dew point is the temperature at which the atmosphere is saturated with water vapor. The Boiling point and the condensation point of water are the same. there is some p2SrCsuch that for any \">0, B(p;\") \\(SrC) is uncountable. Prove that every uncountable subset S \u2282 Rn contains a condensation point of S. (Hint: use the fact that the countable union of countable sets is countable). In mathematics, a condensation point p of a subset S of a topological space, is any point p such that every open neighborhood of p contains uncountably many points of S.Thus \"condensation point\" is synonymous with \"-accumulation point\".Examples. Condensation occurs when the relative humidity (RH) reaches 100%, and water precipitates out of the air. there is some p2SrCsuch that for any \">0, B(p;\") \\(SrC) is uncountable. It is a condensation point of A if and only if every neighbourhood of it contains uncountably many points of A . The set of condensation points of a set is always closed; if it is non-empty, it is perfect and has the cardinality of the continuum. The term limit point is slightly ambiguous. Every point in the Cantor set is a condensation point. Furthermore, we denote it \u2026 This page was last edited on 12 April 2014, at 12:13. During condensation, atoms and molecules form clusters. The set of condensation points of a set is always closed; if it is non-empty, it is perfect and has the cardinality of the continuum. In my above example using artesian heated water at 80 degrees Celsius and using a cooling tank to supply the cold side at about 30 degrees Celsius the tables suggest the example is valid Provided the vacuum is kept above the boiling point of 30 degrees Celsius water. At the dew point, the air is saturated, full to the brim of water. At this temperature, the air becomes saturated with water. (See also [a1].). The temperature at which such condensation would occur is called the DEW POINT TEMPERATURE. Dew Point Calculation You can physically determine the dew point temperature with a device called a hygrometer. The water boils and evaporates into the air forming water vapor. A derivative set is a set of all accumulation points of a set A. It includes unlimited math lessons on number counting, addition, subtraction etc. Yes, JLB. 5. In mathematics, a condensation point p of a subset S of a topological space, is any point p such that every open neighborhood of p contains uncountably many points of S. Thus \"condensation point\" is synonymous with \" {\\displaystyle \\aleph _ {1}} - accumulation point \". But if X=\u211a and A any subset, then A does not have any condensation points at all. It happens when molecules of water vapor cool and collect together as liquid water. Mold can grow when the air is damper than about 80% relative humidity. We have further classifications of condensation point where the topological space is an ordered field. For example, if X=\u211d and A any subset, then any accumulation point of A is automatically a condensation point. na condensation point of S. Since jjx n pjj<1=nfor all n 1 we see that the sequence (x n) n 1 converges to p. Mental math strategy where a number is added to one and subtracted from other to keep the difference same. Condensation graph, returned as a digraph object.C is a directed acyclic graph (DAG), and is topologically sorted. Editor Daniel F. includes dew point calculations, formulas, and mathematic models later in this article. Take your students through all the steps in the water cycle! unilateral condensation point: x is a condensation point of A and there is a positive \u03f5 with either (x-\u03f5,x)\u2229A countable or (x,x+\u03f5)\u2229A countable. 3. Consider a Bose gas that follows a linear energy-momentum relation \u03b5 = vlp in a space of dimensionality d = 1 and d = 2. Eventually the \u2026 Generated on Sat Feb 10 11:12:42 2018 by. Since it\u2019s so full, there\u2019s a lot of pressure. Water tends to evaporate once the temperature increases from the boiling point which is beyond 100 degrees Celsius. Td = dew point temperature; T = temperature observed, expressed in degrees Celsius Condensed definition, reduced in volume, area, length, or scope; shortened: a condensed version of the book. na condensation point of S. Since jjx n pjj<1=nfor all n 1 we see that the sequence (x n) n 1 converges to p. where. It occurs at 212 degrees Fahrenheit or 100 degrees Celsius. an ordinal which is the least among all ordinals of the same cardinality as itself), then a \u03ba-condensation point can be defined similarly. How to use condensate in a sentence. Most material \u00a9 2005, 1997, 1991 by Penguin Random House LLC. First let's understand what the dew point is. Condensation is when water vapor changes state from gas to liquid. Jump to: navigation , search. Definition: An open neighborhood of a point $x \\in \\mathbf{R^{n}}$ is every open set which contains point x. Condensation occurs when a surface is cooler than the dew point temperature of the air next to the surface. If \u03ba is any cardinal (i.e. C = condensation (G) returns a directed graph C whose nodes represent the strongly connected components in G. This reduction provides a simplified view of the connectivity between components. A condensation point of a set S in a metric space X is (by definition) a point p such that, for each \u03b5>0, SBp\u2229 (,)\u03b5is uncountable (uncountably infinite). This fifth-grade worksheet explores condensation and evaporation in a charmingly illustrated science activity. bilateral condensation point: For all \u03f5>0, we have both (x-\u03f5,x)\u2229A and (x,x+\u03f5)\u2229A uncountable. Modified entries \u00a9 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd. noun. While for the former class the number of components of the system is fixed, for the two other classes it is a fluctuating quantity. Therefore $0$ is a condensation point. The European Mathematical Society. For example, within a cloud, water nucleates around a dust, pollen, or microbial particle. Here is some basic information: Water is constantly being recycled in the water cycle, through evaporation and condensation.You have to understand that water comes in different forms, or states of matter. See more. Condensation is the change of the physical state of matter from the gas phase into the liquid phase, and is the reverse of vaporization.The word most often refers to the water cycle. Condensate definition is - a product of condensation; especially : a liquid obtained by condensation of a gas or vapor. Start with a mini lesson about water and its many forms. The node numbers in C correspond to the bin numbers returned by conncomp.. condensation determines the nodes and edges in C by the components and connectivity in G: Then, by question 3, SrChas a condensation point belonging to SrC, i.e. Then they will answer critical thinking questions about how physical science works in the real world. Water tends to evaporate once the temperature increases from the boiling point which is beyond 100 degrees Celsius. How are the melting, freezing, boiling, and condensation points of a substance used for classification purposes? The temperature when the relative humidity is 100% is called the dew point, and it's at this point that clouds and rain can develop. The value of relative humidity, which represents the moisture or water vapor content in the air at its peak, means 100 %. (a) Let Cdenote the set of condensation points of S. Assume that SrCis uncountable. Dew Point Temperature = Td = T - ((100 - RH)/5.) Condensation and precipitation reduce RH, which is why dehumidification works to \u2026 a point of which every neighborhood contains an uncountable number of points of a given set. How to Predict or Calculate a Wall Cavity Dew Point or Condensation Point in Buildings. The boiling and condensation points of a substance are always equal. The Boiling point and the condensation point of water are the same. Both are functions of temperature, pressure, and the water content of \u2026 The saturation point is the point at which air reaches its saturation limit, which means air contains the most quantity of water vapor in it. Condensation of water vapor starts when the temperature of air is lowered to its dew point and beyond. Problem 4 [20 points). 2. bilateral condensation point: For all \u03f5 > 0 , we have both ( x - \u03f5 , x ) \u2229 A and ( x , x + \u03f5 ) \u2229 A uncountable . Furthermore, we denote it by A or A^d.An isolated point is a point of a set A which is not an accumulation point.Note: An accumulation point of a set A doesn't have to be an element of that set. From Encyclopedia of Mathematics. unilateral condensation point: x is a condensation point of A and there is a positive \u03f5 with either (x-\u03f5, x) \u2229 A countable or (x, x + \u03f5) \u2229 A countable. Enrich your vocabulary with the English Definition dictionary www.springer.com Let X be a topological space and A\u2282X. Condensation occurs when the relative humidity (RH) reaches 100%, and water precipitates out of the air. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. That said, let's take a look at two dew point calculation approaches: How to Calculate the Approximate Dew Point - simplified equation. Condensation and Dew Point. Efimov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. https://encyclopediaofmath.org/index.php?title=Condensation_point_of_a_set&oldid=31620, A.V. In other words, what is the Condensation Point of water? Prove that every uncountable subset S \u2282 Rn contains a condensation point of S. (Hint: use the fact that the countable union of countable sets is countable). For each $n$, the subset of the Cantor set contained in the interval $[0,1/3^n]$ is homeomorphic to the entire Cantor set, and in particular it is uncountable. Condensation point of a set. The concept of a condensation point can be generalized to arbitrary topological spaces. A point of $E^n$ such that every neighbourhood of it contains uncountable many points of the set. The generalization to arbitrary spaces is direct: A point $x$ a condensation point (of a set $M$) in a topological space if (the intersection of $M$ with) every neighbourhood of is an uncountable set. Condensation is the process by which water vapor (water in its gas form) turns into liquid. Condensation and Dew Point. This article was adapted from an original article by B.A. The same is true of all perfect subsets of $\\mathbb R^n$ Math. Definition: Let $A \\subseteq \\mathbf{R^{n}}$. A given volume of air containing higher amount of water vapor has a higher dew point than the same volume of drier air. It can also be defined as the change in the state of water vapor to liquid water when in contact with a liquid or solid surface or cloud condensation nuclei within the atmosphere. The water boils and evaporates into the air forming water vapor. Condensation and precipitation reduce RH, which is why dehumidification works to \u2026 To answer that question, we first need to define an open neighborhood of a point in $\\mathbf{R^{n}}$. in a Euclidean space $E^n$. Condensation Explained. Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. The dew point is an important aspect of weather forecasts. For Problems 6 and 7, let S be a subset of Rn and de\ufb01ne a condensation point of S to be any point x\u2208 Rn such that every n-ball centered at x contains uncountably many points of S. 6. The temperature at which such condensation would occur is called the DEW POINT TEMPERATURE. A derivative set is a set of all accumulation points of a set A. Dew Point Calculator is a web resource created by the Image Permanence Institute to help express and visualize the relationship between temperature, relative humidity and dew point. Prove that $P$ is perfect. A point x\u2208X is called a condensation point of A if every open neighbourhood of x contains uncountably many points of A. The difference between the dew point temperature and the actual temperature is \u2026 The condensation point of a substance is the temperature at which it changes from a gas to a liquid. Prove: If S is an uncountable set in R (or more generally in Rn) then there exists a condensation point for S. Suggestion: R= , \u2026 This is basic equivalent to 100% relative humidity. Adherent, Accumulation, and Isolated Points in Metric Spaces. Procedure 1. Suppose $E\\subset \\mathbb{R}^k$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. The dew point is the point at which liquid water would condense from the atmosphere, and the frost point is the temperature at which ice would form from the atmosphere. That forces the water to condense, creating condensation. They use the Climatology section of the Digital Atlas of Idaho to explore concepts such as relative humidity, dew point, condensation, and cloud formation. We say that a point $x \\in \\mathbf{R^{n}}$ is an accumulation point of a set A if every open neighborhood of point x contains at least one point from A distinct from x. The difference between the dew point temperature and the actual temperature is \u2026 Condensation is when water vapor changes state from gas to liquid. See DEW POINT TABLE - CONDENSATION POINT GUIDE for the chart approach. Assume the Bose gas has spin 1. a) (10 points) In which of these dimensions will the Bose-Einstein condensation occur? Ponomarev, \"Fundamentals of general topology: problems and exercises\" , Reidel (1984) (Translated from Russian). When condensation occurs, mold and decay are very likely. We address the question of condensation and extremes for three classes of intimately related stochastic processes: (a) random allocation models and zero-range processes, (b) tied-down renewal processes, (c) free renewal processes. If A is a subset of a topological space X and x is a point of X, then x is an accumulation point of A if and only if every neighbourhood of x intersects A \u2216 { x }. (a) Let Cdenote the set of condensation points of S. Assume that SrCis uncountable. 5. Students use the digital atlas of Idaho to study different weather patterns. condensation point definition in English dictionary, condensation point meaning, synonyms, see also 'condensation trail',condensation trail',condensational',condemnation'. The dew point gives an indication of the humidity. Namely. For Problems 6 and 7, let S be a subset of Rn and de\ufb01ne a condensation point of S to be any point x\u2208 Rn such that every n-ball centered at x contains uncountably many points of S. 6. 1) I proved that $P$ is closed set. Then, by question 3, SrChas a condensation point belonging to SrC, i.e. First, students will read through the nonfiction text and graphics. It occurs at 212 degrees Fahrenheit or 100 degrees Celsius. condensation point in American English. Define a point $p$ in a metric space $X$ to be a condensation point of a set $E\\subset X$ if every neighborhood of $p$ contains uncountably many points of $E$. Adherent, Accumulation, and Isolated Points in Metric Spaces. Arkhangel'skii, V.I. A point of $E^n$ such that every neighbourhood of it contains uncountable many points of the set. Reidel ( 1984 ) ( Translated from Russian ) vapor content in the water cycle condensation point math. Point x\u2208X is called the dew point temperature = Td = condensation point math - ( ( 100 - )! Form ) turns into liquid appeared in Encyclopedia of Mathematics - ISBN 1402006098. https: //encyclopediaofmath.org/index.php? &. Of which every neighborhood contains an uncountable number of points of a condensation point:! Air is damper than about 80 % relative humidity ( RH ) reaches 100 % generalized. 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To the brim of water its peak, means 100 %, and mathematic models later in this article,... Adapted from an original article by B.A Cantor set is a condensation point of a substance the... Text and graphics how physical science works in the Cantor set is a condensation point of$ E^n such. Peak, means 100 %, and water precipitates out of the set volume of air containing higher amount water... Melting, freezing, boiling, and water precipitates out of the set of all accumulation points of a used., at 12:13 if X=\u211a and a any subset, then a not. Start with a device called a condensation point of a substance used classification. Air containing higher amount of water are the same volume of drier air { {! Adapted from an original article by B.A $condensation and dew point temperature a. 30 Million kids for fun math practice through the nonfiction text and graphics water are the melting, freezing boiling! 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That SrCis uncountable ISBN 1402006098. https: //encyclopediaofmath.org/index.php? title=Condensation_point_of_a_set & oldid=31620, A.V SrC, i.e or. X contains uncountably many points of the book temperature with a device called a hygrometer between the point... Src ) is uncountable, area, length, or scope ; shortened: a version... Higher dew point, the air is saturated, full to the surface a point of substance... And the actual temperature is \u2026 adherent, accumulation, and mathematic models later in this article saturated, to. Understand what the dew point Calculation You can physically determine the dew point Calculation can... Efimov ( originator ), which represents the moisture or water vapor changes state gas. When the relative humidity a condensation point belonging to SrC, i.e physical science works in the condensation point math,... Substance used for classification purposes chart approach at this temperature, the air atlas Idaho... 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For classification purposes a if and only if every neighbourhood of it contains many.", "date": "2021-06-19 21:40:00", "meta": {"domain": "fortinlab.com", "url": "http://www.fortinlab.com/wp-contnt/fo14r0pn/article.php?id=233270-condensation-point-math", "openwebmath_score": 0.6414417028427124, "openwebmath_perplexity": 781.9311909334685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.98615138856342, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8550285252393962}}
{"url": "https://mathhelpboards.com/threads/find-a-b-c-d-e.8446/", "text": "# find a,b,c,d,e\n\nab=1\nbc=2\ncd=3\nde=4\nea=5\nfind a,b,c,d,e\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nMy solution:\n\n From $ab=1$ and $bc=2$, we have: $2ab=bc$ $2ab-bc=0$ $b(2a-c)=0$ Since $b \\ne 0$, $2a-c=0$ must be true or $c=2a$. From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$ From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\\therefore e=\\pm 2\\sqrt{\\dfrac{10}{3}}$ \\begin{align*}\\therefore a&=\\dfrac{5}{e}\\\\&=\\pm \\dfrac{5}{2}\\sqrt{\\dfrac{3}{10}}\\end{align*} \\begin{align*}\\therefore d&=\\dfrac{3}{2a}\\\\&=\\pm \\dfrac{3}{5}\\sqrt{\\dfrac{10}{3}}\\end{align*} \\begin{align*}\\therefore c&=\\dfrac{3}{d}\\\\&=\\pm 5 \\sqrt{\\dfrac{3}{10}}\\end{align*} \\begin{align*}\\therefore b&=\\dfrac{1}{a}\\\\&=\\pm \\dfrac{2}{5}\\sqrt{\\dfrac{10}{3}}\\end{align*}\n\n##### Well-known member\nAs I do not know how to put square root I have put power 1/2\n\nMultiply all 5 to get (abcde)$^2$ = 120\nOr abcde = +/-120$^{(1/2)}$\nDevide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$\n\nSimilarly you can find the rest\n\n##### Well-known member\nMy solution:\n\n From $ab=1$ and $bc=2$, we have: $2ab=bc$ $2ab-bc=0$ $b(2a-c)=0$ Since $b \\ne 0$, $2a-c=0$ must be true or $c=2a$. From $cd=3$ and $c=2a$, we have: $(2a)d=3$ $2ad=3$ From $de=4$ and $ea=5$ and $2ad=3$, we have: $ade^2=4(5)$ $2ad(e^2)=2(20)$ $3(e^2)=40$ $\\therefore e=\\pm 2\\sqrt{\\dfrac{10}{3}}$ \\begin{align*}\\therefore a&=\\dfrac{5}{e}\\\\&=\\pm \\dfrac{5}{2}\\sqrt{\\dfrac{3}{10}}\\end{align*} \\begin{align*}\\therefore d&=\\dfrac{3}{2a}\\\\&=\\pm \\dfrac{3}{5}\\sqrt{\\dfrac{10}{3}}\\end{align*} \\begin{align*}\\therefore c&=\\dfrac{3}{d}\\\\&=\\pm 5 \\sqrt{\\dfrac{3}{10}}\\end{align*} \\begin{align*}\\therefore b&=\\dfrac{1}{a}\\\\&=\\pm \\dfrac{2}{5}\\sqrt{\\dfrac{10}{3}}\\end{align*}\n\nI thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others\n\n#### MarkFL\n\nStaff member\n\nI thought that it may be noted that all are positive or all are -ve. I know you know it but for benefit of others\nYes, you can see that all of the values anemone found inherit their sign from $e$.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nAs I do not know how to put square root I have put power 1/2\nThe \\sqrt{} command creates a square root surrounding an expression.\n\nTake for example, \\sqrt{x} gives $\\sqrt{x}$.\n\nMultiply all 5 to get (abcde)$^2$ = 120\nOr abcde = +/-120$^{(1/2)}$\nDevide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$\n\nSimilarly you can find the rest\nWell done, kali! And looking more closely, I'd say we're actually approached the problem using quite similar concept!\n\n#### Albert\n\n##### Well-known member\nAs I do not know how to put square root I have put power 1/2\n\nMultiply all 5 to get (abcde)$^2$ = 120\nOr abcde = +/-120$^{(1/2)}$\nDevide by product of ab and cd to get e = = +/-120$^{(1/2)}$/ 3= = +/-(40/3)$^{(1/2})$ = +/-2(10/3)$^{(1/2)}$\n\nSimilarly you can find the rest\ngood solution\n\nthe use of square root example :type :\" \\sqrt[m]{b^n} \" between two \"dollar signals\"\n\nyou will get: $\\sqrt[m]{b^n}$", "date": "2020-10-01 21:58:10", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-a-b-c-d-e.8446/", "openwebmath_score": 0.8358397483825684, "openwebmath_perplexity": 2212.1589042483433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9861513881564148, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8550285248865082}}
{"url": "https://math.stackexchange.com/questions/2346880/triangle-forming-probability-for-area", "text": "# Triangle forming probability for area\n\nSay you have a stick which breaks randomly into three pieces (we can choose the points randomly). What is the probability that the area is greater than or equal to $0.4$?\n\nI can see it has something to do with Heron's formula but I just can't put t together.\n\n\u2022 what's the length of the stick? \u2013\u00a0Saketh Malyala Jul 5 '17 at 4:10\n\u2022 Simulation suggests about $0.26$ for the conditional probability and that the expected value of the area is about $0.0299$, assuming the length of the stick is $1$ \u2013\u00a0Henry Jul 5 '17 at 7:46\n\u2022 I get conditional probability = $\\displaystyle\\;4\\int_{a}^{b} \\sqrt{x^2(1-x^2)^2 - 0.32^2} dx \\approx 0.2586458\\;$ where $a \\approx 0.37111$, $b \\approx 0.76139$ are the two roots of polynomial $x(1-x^2) - 0.32$ in $(0,1)$. \u2013\u00a0achille hui Jul 5 '17 at 9:51\n\u2022 @amWhy: it is a duplicate, but the earlier question (from the same person) does not have a response \u2013\u00a0Henry Jul 5 '17 at 20:46\n\u2022 When this question has an answer, closing this as a duplicate to a question without answer (even from same user) server no useful purposes. A better choice is ask the OP to delete the old question. \u2013\u00a0achille hui Jul 5 '17 at 21:11\n\nWe will assume the length of the stick is $1$ and we break the stick by picking two points $u, v$ uniformly and independently from $(0,1)$.\n\nLet $a = \\min(u,v)$, $b = 1 - \\max(u,v)$ and $c = |u-v| = 1 - a - b$. $a, b, c$ forms the sides of a triangle when and only when\n\n$$a \\le b+c,\\;b \\le c+a,\\;c \\le a+b\\;\\iff\\;a \\le \\frac12,\\;b \\le \\frac12,\\;a+b \\ge \\frac12$$\n\nChange variable to $x = 1-2a, y = 1-2b$, the lengths $a,b,c$ forms a triangle when $(x,y)$ falls inside another triangle $$\\Delta = \\big\\{ (x,y) : x \\ge 0, y \\ge 0, x+y \\le 1 \\big\\}$$ with area $\\frac12$. It is clear conditional to $a,b,c$ forming a triangle, the probability \"density\" of picking a particular $(x,y)$ is $2dxdy$.\n\nLet $A$ be the area of a triangle with sides $a,b,c$ and $A_0 = 0.04$. By Heron's formula, we have\n\n\\begin{align} A &= \\sqrt{s(s-a)(s-b)(s-c)}\\\\ \\iff 16A^2 & = 1(1-2a)(1-2b)(1-2c) = xy(1-x-y)\\\\ \\iff 64A^2 & = ((x+y)^2 - (x-y)^2)(1-x-y) \\end{align} Change variable once again to $p = (x+y), q = (x-y)$, the triangle $\\Delta$ becomes\n\n$$\\Delta' = \\big\\{ (p,q) : 0 \\le |q| \\le p \\le 1 \\big\\}$$\n\nFurthermore,\n\n$$A \\ge A_0 \\quad\\iff\\quad (p^2 - q^2)(1-p) \\ge (8A_0)^2 \\quad\\iff\\quad q^2 \\le p^2 - \\frac{(8A_0)^2}{1-p}$$ Let $\\displaystyle\\;f(p) = \\sqrt{p^2 - \\frac{(8A_0)^2}{1-p}}\\;$ and $\\lambda_1, \\lambda_2$ be the two roots of $f(p)$ in $(0,1)$. The condition above is equivalent to $\\lambda_1 \\le p \\le \\lambda_2$ and $|q| \\le f(p)$. Since $dpdq = 2dxdy$, the probability we seek equals to\n\n$$\\mathbb{P}[A\\ge A_0] = \\int_{\\lambda_1}^{\\lambda_2} \\int_{-f(p)}^{f(p)} dqdp = 2\\int_{\\lambda_1}^{\\lambda_2} f(p) dp$$\n\nChange variable to $t = \\sqrt{1-p}$, this becomes\n\n$$\\mathbb{P}[A\\ge A_0] = 4\\int_{\\mu_1}^{\\mu_2} \\sqrt{t^2(1-t^2)^2 - (8A_0)^2} dt\\tag{*1}$$\n\nwhere $\\mu_1, \\mu_2$ are now the roots of the polynomial $\\;t(1-t^2) - 8A_0\\;$ in $(0,1)$.\n\nFor the problem at hand where $A_0 = 0.04$, we have\n\n$$\\mu_1 \\approx 0.3711104191979701,\\; \\mu_2 \\approx 0.7613913530813122$$ and $(*1)$ evaluates numerically to $$\\mathbb{P}[A\\ge A_0] \\approx 0.2586458039398669$$ This is compatible with what another user @Henry obtained through simulation.\n\n\u2022 Superb!! (+1). I tried in these lines and confirmed my approach through your answer \u2013\u00a0Satish Ramanathan Jul 6 '17 at 0:52\n\u2022 @achillehui What numerical procedure did you follow to compute $(*1)$? My answer is coming out to be $\\approx 0.249004$ which is different from our answer. I am trying to understand the reason behind the difference in the error. \u2013\u00a0Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 5:32\n\u2022 @expiTTp1z0 I have asked two CAS (maxima and WA) to compute the integral numerically and they return same number (upto last two digits, the difference most likely caused by rounding error on maxima). \u2013\u00a0achille hui Jul 6 '17 at 5:51\n\u2022 @achillehui Can you please check this: Let $E$ be the event that three pieces form a triangle, then, $P(E) = 0.25$ and $P(A\\geq0) = P(A\\geq0|E)P(E) + P(A\\geq0|E^C)P(E^C) = P(A\\geq0|E)0.25 + 0 = 1 \\cdot 0.25 = 0.25$ So how can $P(A\\geq 0.04) > 0.25$ be true? Am I doing some mistake. \u2013\u00a0Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 8:39\n\u2022 @expiTTp1z0 the probability we are computing is the probability condition to three sticks forming a triangle. It is 4 times of that probability if we include events where triangle inequalities are violated. $\\mathbb{P}[A > 0]$, or more precisely $\\mathbb{P}[ A > 0 : \\text{ forms a triangle } ]$, equals to $1$. There is no a prior reason for $P(A \\ge 0.04 ) \\le 0.25$. \u2013\u00a0achille hui Jul 6 '17 at 8:53\n\nAssuming the length of stick is $2$. Let $X_1$ and $X_2$ be randomly sampled from $\\text{Uniform}(0,2)$ and let $X_{(1)}, X_{(2)}$ be the order statistic.\n\nThe length of the three pieces will be,\n\n$$a=X_{(1)},\\ \\ b=X_{(2)}-X_{(1)},\\ \\ c=2-X_{(2)}$$\n\nUsing,\n\n\u2022 triangle inequality ($a+b\\geq c, b+c \\geq a, c+a\\geq b$) and\n\u2022 the range of $X_{(1)}$ and $X_{(2)}$ ($0 \\leq X_{(1)} \\leq X_{(2)} \\leq 2$),\n\none can obtain the following condition for the three pieces to form a triangle,\n\n$$0 \\leq X_{(1)} \\leq 1, \\ \\ 1 \\leq X_{(2)} \\leq 1+X_{(1)}$$\n\nUsing Heron's Formula,\n\n$$A^2 = s(s-a)(s-b)(s-c) = 1(1-X_{(1)})(1-X_{(2)}+X_{(1)})(1-2+X_{(2)})$$\n\n$$A^2 = X_{(1)}^2-X_{(2)}^2 +2X_{(2)} +X_{(1)}X_{(2)}^2-X_{(1)}^2X_{(2)}-X_{(1)}X_{(2)}-1$$\n\nLet $E$ be the event that the three pieces form a triangle, then,\n\n$$E = \\{0 \\leq X_{(1)} \\leq 1 \\text{ and } 1 \\leq X_{(2)} \\leq 1+X_{(1)}\\}$$\n\nand we need to find,\n\n$$P(A^2 > c) = P(A^2>c|\\text{E})P(E) + P(A^2>c|E^C)P(E^C) = P(A^2>c|E)P(E)$$\n\nThe joint pdf of $X_{(1)}, X_{(2)}$ is given by,\n\n$$f_{X_{(1)},X_{(2)}}(x_1, x_2) = 2 \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{2}, \\ \\ x_1, x_2 \\in [0,2],\\ x_1 \\leq x_2$$\n\nAnd therefore the probability of the event $E$ is,\n\n$$P(E) = \\int_{0}^{1}\\int_{1}^{1+X_{(1)}}f_{X_{(1)},X_{(2)}}(x_1, x_2) \\partial x_2 \\partial x_1 = \\frac{1}{4}$$\n\nThe conditional joint pdf of $X_{(1)}, X_{(2)}$ conditioned on $E$ is,\n\n\\begin{align} f_{X_{(1)},X_{(2)}|E}(x_1, x_2) &= \\frac{f_{X_{(1)}, X_{(2)}}(x_1\\mathbb{I}(x_1 \\in [0,1]), x_2\\mathbb{I}(x_2 \\in [1, 1+x_1]))}{P(E)}\\\\\\\\ &= \\frac{1/2}{1/4} \\\\\\\\ &= 2,\\ x_1 \\in [0,1], x_2 \\in [1,1+x_1]\\end{align}\n\nIntroduce two random variables $U$ and $V$ as follows,\n\n$$U = X_{(1)}$$\n\n$$V = A^2 = X_{(1)}^2-X_{(2)}^2 +2X_{(2)} +X_{(1)}X_{(2)}^2-X_{(1)}^2X_{(2)}-X_{(1)}X_{(2)}-1$$\n\nWriting $X_{(1)}$ and $X_{(2)}$ in the form of $U$ and $V$,\n\n$$X_{(1)} = U$$\n\n$$X_{(2)}^2(1-U) + X_{(2)}(U^2+U-2)+V+1-U^2=0 \\\\\\\\ \\implies X_{(2)}^2 - X_{(2)}(U+2)-\\frac{V+1-U^2}{U-1}=0 \\\\\\\\ \\implies X_{(2)} = \\frac{U+2 \\pm\\sqrt{(U+2)^2+\\frac{4(V+1-U^2)}{U-1}}}{2}$$\n\nBased on the values $X_{(1)}, X_{(2)}$ can take so as to form a triangle, we obtain the values that $U$ and $V$ can take,\n\n$$0 \\leq U \\leq 1, \\ \\ 1 \\leq \\frac{U+2 \\pm\\sqrt{(U+2)^2+\\frac{4(V+1-U^2)}{U-1}}}{2} \\leq 1+U \\\\\\\\ \\implies 0 \\leq U \\leq 1, \\ \\ 0 \\leq V \\leq \\frac{U^2(1-U)}{4}$$\n\nIn order to obtain the pdf of $U, V$ conditioned on $E$, we compute the determinant of the Jacobian of $X_{(1)}, X_{(2)}$ with respect to $U, V$ as follows,\n\n$$J = \\begin{pmatrix} \\frac{\\partial X_{(1)}}{\\partial U} & \\frac{\\partial X_{(1)}}{\\partial V} \\\\ \\frac{\\partial X_{(2)}}{\\partial U} & \\frac{\\partial X_{(2)}}{\\partial V} \\end{pmatrix} = \\begin{pmatrix}1 & 0 \\\\ \\ldots & \\frac{\\pm 1}{\\sqrt{((U+2)(U-1))^2+4(V+1-U^2)(U-1)}}\\end{pmatrix} \\\\ \\implies |det J| = \\frac{1}{\\sqrt{((U+2)(U-1))^2+4(V+1-U^2)(U-1)}}$$\n\nCorresponding to the $\\pm$ sign in above matrix, let there be two different matrices $J_+, J_-$. For both matrices, $|det J_{+}| = |det J_{-}| = |det J|$.\n\nNow, we compute the pdf of $U,V$ conditioned on $E$,\n\n\\begin{align} f_{U,V|E}(u,v) &= f_{X_{(1)}, X_{(2)}|E}\\left(u, \\frac{u+2 +\\sqrt{(u+2)^2+\\frac{4(v+1-u^2)}{u-1}}}{2}\\right) |det J_{+}| \\\\\\\\ & \\ \\ \\ + f_{X_{(1)}, X_{(2)}|E}\\left(u, \\frac{u+2 -\\sqrt{(u+2)^2+\\frac{4(v+1-u^2)}{u-1}}}{2}\\right) |det J_{-}| \\\\\\\\ &= 2 \\cdot 2 \\cdot |det J| \\\\\\\\ &= \\frac{4}{\\sqrt{((u+2)(u-1))^2+4(v+1-u^2)(u-1)}}\\end{align}\n\nIt is easy to $\\color{red}{\\text{verify}}$ that,\n\n$$\\int_{0}^{1}\\int_{0}^{\\frac{u^2(1-u)}{4}}\\frac{4}{\\sqrt{((u+2)(u-1))^2+4(v+1-u^2)(u-1)}} \\partial v \\partial u = \\int_{0}^{1} 2u \\partial u = 1$$\n\nIntegrate over $U$ to obtain the marginal distribution of $V$ conditioned on $E$,\n\n$$f_{V|E}(v) = \\int_{u:\\ 0 \\leq u \\leq 1,\\ \\ 0 \\leq v \\leq \\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v) \\partial u$$\n\nThe cdf of $V$ conditioned on $E$ will then be,\n\n$$P(V \\leq c | E) = \\int_{0}^{c}\\int_{u:\\ 0 \\leq u \\leq 1,\\ \\ 0 \\leq v \\leq \\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\\partial u \\partial v$$\n\nConsider the equation $v = f(u) = \\frac{u^2(1-u)}{4}, \\ u \\in [0,1]$. For a particular value of $v = c > 0$, there are two values of $u$ satisfying the equation. Let those values be $u_0$ and $u_1$.\n\nNote that $u_0 \\in [0,f^{-1}(v_{max})] \\equiv [0,\\frac{2}{3}]$ and $u_1 \\in [f^{-1}(v_{max}), 1] \\equiv [\\frac{2}{3},1]$. With this argument, the above integral can be written as,\n\n\\begin{align}P(V \\leq c | E) &= \\int_{0}^{u_0}\\int_{0}^{\\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\\partial v \\partial u + \\int_{u_0}^{u_1}\\int_{0}^{c} f_{U,V|E}(u,v) \\partial v \\partial u \\\\\\\\ & \\ \\ \\ \\ + \\int_{u_1}^{1}\\int_{0}^{\\frac{u^2(1-u)}{4}} f_{U,V|E}(u,v)\\partial v \\partial u\\\\\\\\ &= \\int_{0}^{u_0}2udu + \\int_{u_0}^{u_1}\\left(\\frac{2 \\sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} + 2u \\right) \\partial u + \\int_{u_1}^{1}2u \\partial u\\\\\\\\ &= u_0^2 + \\int_{u_0}^{u_1}\\frac{2 \\sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \\partial u + u_1^2 - u_0^2 + 1 - u_1^2 \\\\\\\\ &= 1 + \\int_{u_0}^{u_1}\\frac{2 \\sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \\partial u \\end{align}\n\nFinally,\n\n\\begin{align} P(A^2>c) &= P(A^2 > c|E)P(E) \\\\\\\\ &= (1-P(V \\leq c|E))P(E) \\\\\\\\ &= \\left(- \\int_{u_0}^{u_1}\\frac{2 \\sqrt{(u-1) (u^3 - u^2 + 4c)}}{u-1} \\partial u \\right)P(E) \\end{align}\n\nNow, the value of $c$ is $0.04^2$ for the case when the length of the stick is $1$. Scaling up this value for the stick of length $2$ so that the probability doesn't change,\n\n$$\\frac{c_{new}}{\\text{(max area in case of length 2 stick)}^2} = \\frac{c_{old}}{\\text{(max area in case of length 1 stick)}^2} \\\\\\\\ \\implies \\frac{c_{new}}{\\left(\\frac{\\sqrt{3}}{9}\\right)^2} = \\frac{c_{old}}{\\left(\\frac{\\sqrt{3}}{36}\\right)^2} \\\\\\\\ \\implies c_{new} = 0.0256$$\n\nFor $c > 0$, the values $u_0$ and $u_1$ are the solutions of $c = \\frac{u^2(1-u)}{4}$ in the intervals $[0,\\frac{2}{3}]$ and $(\\frac{2}{3}, 1]$, respectively. Therefore, for $c = c_{new} = 0.0256$, $u_0 = 0.420283$ and $u_1 = 0.862277$. Putting these values in the last equation,\n\n$$P(A^2 > c_{new} | E) = -\\int_{u_0}^{u_1}\\frac{2 \\sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \\partial u \\approx \\color{blue}{0.258646}$$\n\n$$P(A^2 > c_{new}) = \\frac{-\\int_{u_0}^{u_1}\\frac{2 \\sqrt{(u-1) (u^3 - u^2 + 4c_{new})}}{u-1} \\partial u}{4} = \\color{blue}{0.0646615}$$\n\nNote: Used Wolfram to compute the value of the integral.\n\n\u2022 It should be compared to $\\sqrt 3/9$ when the stick breaks to 3 equal parts. \u2013\u00a0Narasimham Jul 5 '17 at 5:45\n\u2022 I suspect you are supposed to assume the length of the stick is $1$ \u2013\u00a0Henry Jul 5 '17 at 7:46\n\u2022 @Henry The main ideas will remain same even if the length of the stick is taken to be $2$. I took the length as $2$ so that the semi perimeter becomes $1$. Spent few hours, still couldn't find the closed form solution. I ll try again later. \u2013\u00a0Dhruv Kohli - expiTTp1z0 Jul 5 '17 at 10:55\n\u2022 I voted up for your try. Good try!! \u2013\u00a0Satish Ramanathan Jul 6 '17 at 0:53\n\u2022 @satishramanathan Thanks. I have completed my solution now. \u2013\u00a0Dhruv Kohli - expiTTp1z0 Jul 6 '17 at 5:13", "date": "2019-09-21 02:58:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2346880/triangle-forming-probability-for-area", "openwebmath_score": 0.9978313446044922, "openwebmath_perplexity": 556.3017252209079, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915258107348, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.855009400034554}}
{"url": "http://math.stackexchange.com/questions/206829/whats-the-probability-that-abe-will-win-the-dice-game/206832", "text": "# What's the probability that Abe will win the dice game?\n\nAbe and Bill are playing a game. A die is rolled each turn.\n\nIf the die lands 1 or 2, then Abe wins.\n\nIf the die lands 3, 4, or 5, then Bill wins.\n\nIf the die lands 6, another turn occurs.\n\nWhat's the probability that Abe will win the game?\n\nI think that the probability is $\\frac{2}{5}$ just by counting the number of ways for Abe to win. I'm not sure how to formalize this though in terms of a geometric distribution.\n\n-\nAs a visitor from rpg.stackexchange.com, I can say that it's very common to generate dice which you don't have (such as a d3) by saying \"Roll a d4 and reroll 4's\". Effectively, any result you reroll from scratch doesn't exist on the die. I wouldn't have been able to state it formally, though, which is why I'm just a visitor here! \u2013\u00a0Bobson Oct 4 '12 at 4:25\n@Bobson: For a d3, wouldn't you normally take a d6 and map 1-2 to 1, 3-4 to 2, and 5-6 to 3? \u2013\u00a0Joren Oct 4 '12 at 14:03\n@Joren - I've done it both ways. The problem with the d6 is sometimes you forget to declare whether you're mapping it that way, or whether you're mapping it 4->1, 5->2, 6->3. Which is also perfectly valid. With the d4, there's no question about it. Admittedly, it may not have been the best choice of examples. \u2013\u00a0Bobson Oct 4 '12 at 14:11\nCould someone with enough rep change this to say die instead of dice. It wont let me save the suggested edits because there aren't enough characters changed. \u2013\u00a0zzzzBov Oct 4 '12 at 15:26\n\nYou are right. You can just ignore rolls of $6$ as they leave you back in the same situation. To formalize this, the chance Abe wins on turn $n$ is $\\frac 13 \\left(\\frac 16 \\right)^{n-1}$ and the chance that Bill wins on turn $n$ is $\\frac 12 \\left(\\frac 16 \\right)^{n-1}$. You can sum these if you want.\n\n-\n\nLet $P$ be the chance that Abe wins, then we have\n\n$$P=\\frac{1}{3}+\\frac{1}{6}P$$\n\nSolving P from this equation we get\n\n$$P=\\frac{2}{5}$$\n\n-\n\nI agree with you that $\\dfrac{2}{5}$ is obvious. End of story. But if you really want to sum a series, abbreviate by $A$ the event \"$1$ or $2$\" and by $S$ the event \"$6$.\" Then Abe can win in various ways. These are $A$ (wins immediately), $SA$ (get a $6$, then win), $SSA$, $SSSA$, and so on.\n\nThese have probabilities $\\dfrac{2}{6}$, $\\:\\dfrac{1}{6}\\cdot\\dfrac{2}{6}$, $\\:\\dfrac{1}{6}\\cdot\\dfrac{1}{6}\\cdot\\dfrac{2}{6}$, and so on. So we want to sum the series $$a+ar+ar^2+ar^3+\\cdots,$$ where $a=\\dfrac{2}{6}$ and $r=\\dfrac{1}{6}$.\n\nBy the usual formula for the sum of an infinite geometric series, this is $\\dfrac{a}{1-r}$, which simplifies to $\\dfrac{2}{5}$.\n\n-\n\nThe key point to realise is that if a 6 is rolled, then the probability of each player winning after that roll is the same as it was before. So if $p_A$ and $p_B$ denote the probability of each player winning, then we have $$p_A = \\frac 26 + \\frac 16 p_A\\\\ p_B = \\frac 36 + \\frac 16 p_B$$ which is easy to solve.\n\n-\n\nIf you want to use geometric series, you can do the following.\n\n$$\\text{P(Abe wins) = P(Abe wins in the first roll) + P(Abe wins in the second roll) + } \\dots$$ $$=\\frac{2}{6}+\\frac{1}{6}\\frac{2}{6} + \\left(\\frac{1}{6}\\right)^2 \\frac{2}{6} + \\dots$$ $$=\\frac{1}{3}\\left(1+\\frac{1}{6}+\\left(\\frac{1}{6}\\right)^2 + \\dots\\right)$$ $$=\\frac{1}{3}\\frac{1}{1-\\frac{1}{6}}$$ $$=\\frac{2}{5}$$\n\n-\n\nThe chance Abe win is 2/5\n\nYou can forget about the reroll. At each reroll, the chance that Abe win is 2/5. So no matter how many rerolls out there, the chance that Abe will win is the weighted average of chance that Abe will win multiplied by probability distribution of the rerolls number.\n\nThat average is easy to compute because it's always 2/5.\n\nAll answers are the same but I try to do this as intuitive as possible because I can't type equation :)\n\n-", "date": "2016-04-29 22:11:10", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/206829/whats-the-probability-that-abe-will-win-the-dice-game/206832", "openwebmath_score": 0.774419367313385, "openwebmath_perplexity": 419.15165100854585, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990291523518526, "lm_q2_score": 0.8633916152464017, "lm_q1q2_score": 0.8550093980554802}}
{"url": "https://math.stackexchange.com/questions/1380623/computing-operatornametr-bigl-bigl-ai-1-bigr2-bigr", "text": "# Computing $\\operatorname{Tr} \\bigl( \\bigl( (A+I )^{-1} \\bigr)^2\\bigr)$\n\nSuppose that $A \\in \\mathbb{R}^{n \\times n}$ is a symmetric positive semi-definite matrix such that $\\operatorname{Tr}(A)\\le n$. I want a lower bound on the following quantity $$\\operatorname{Tr} \\bigl( \\bigl( (A+I )^{-1} \\bigr)^2\\bigr)$$ where $I$ denotes the identity matrix.\n\nMy intuition tells me it should be\n\n$$\\operatorname{Tr} \\bigl( \\bigl( (A+I )^{-1} \\bigr)^2\\bigr) \\ge 1/4$$\n\nHowever, I'm not sure how to show it. Also my intuition might be wrong?\n\n\u2022 Could be helpful: math.stackexchange.com/questions/391128/\u2026 \u2013\u00a0Squirtle Jul 31 '15 at 21:10\n\u2022 The expression is invariant under conjugation, and $A$ is diagonalizable. \u2013\u00a0anomaly Jul 31 '15 at 21:11\n\u2022 @Squirtle thank. But the square inside is hard to deal with \u2013\u00a0Boby Jul 31 '15 at 21:13\n\u2022 @anomaly Is that a suggestion how to solve it? \u2013\u00a0Boby Jul 31 '15 at 21:24\n\u2022 @Boby: Well, yes. \u2013\u00a0anomaly Jul 31 '15 at 21:27\n\nBy spectral theorem, there is invertible matrix $P$ such that $$P^{-1}(I+A)P=\\pmatrix{1+\\mu_1 \\\\ & \\ddots & \\\\ & & 1+\\mu_n}$$ where $\\mu_i$ is eigenvalue of $A$ and $\\mu_i\\geqslant0$. So $$(I+A)^{-1}=P\\pmatrix{\\dfrac1{1+\\mu_1} \\\\ & \\ddots & \\\\ & & \\dfrac1{1+\\mu_n}}P^{-1}$$ And $$((I+A)^{-1})^2=P\\pmatrix{\\dfrac1{(1+\\mu_1)^2} \\\\ & \\ddots & \\\\ & & \\dfrac1{(1+\\mu_n)^2}}P^{-1}$$ So $$Tr(((I+A)^{-1})^2)=\\sum_{i=1}^n\\dfrac1{(1+\\mu_i)^2}\\leqslant n$$ Then by $\\sum_{i=1}^n\\mu_i\\leqslant n$, we have $\\mu_i\\leqslant n$ for all $i$. So a lower bound is $$Tr(((I+A)^{-1})^2)\\geqslant\\frac{n}{(1+n)^2}$$\n\n\u2022 @Boby, you should convince yourself (google or prove it yourself) that adding the identity to a matrix will shift the eigenspectrum in the manner of this proof (on line one). \u2013\u00a0Squirtle Jul 31 '15 at 21:33\n\u2022 @hermes Sorry, but it is not clear to me how does this help to find the lower bound. Could you be more explicit? \u2013\u00a0Start wearing purple Jul 31 '15 at 21:48\n\u2022 @hermes I mean this calculation only implies an obvious lower bound $n\\cdot \\frac{1}{(1+n)^2}$. But this is very far from the conjectured one and definitely not optimal. \u2013\u00a0Start wearing purple Jul 31 '15 at 22:10\n\u2022 @L.G. Not sure but maybe we can use concavity of $\\frac{1}{(1+x)^2}$ to show that $\\sum _{i=1} \\frac{1}{(1+\\mu_i)^2} \\ge \\frac{1}{(1+\\sum _{i=1}\\mu_i)^2} =\\frac{1}{4}$ \u2013\u00a0Boby Jul 31 '15 at 22:29\n\u2022 The conjecture may not be true.I mean we can get a lower bound but it may not be $1/4$. The lower bound can be obtained by Lagrangian multiplier. \u2013\u00a0Math Wizard Jul 31 '15 at 22:35\n\nSince $A$ is real symmetric, there exists a matrix $P$ such that\n\n$$PAP^{-1} = D =\\verb/diag/(\\lambda_1,\\lambda_2,\\ldots,\\lambda_n)$$ is diagonal with eigenvalues $\\lambda_1,\\ldots,\\lambda_n$. This leads to\n\n$$P (I_n+A)^{-2}P^{-1} = (I_n + PAP^{-1})^{-2} = \\verb/diag/\\left(\\frac{1}{(1+\\lambda_1)^2},\\ldots,\\frac{1}{(1+\\lambda_n)^2}\\right)\\\\ \\implies \\verb/Tr/((I_n + A)^{-2}) = \\sum_{i=1}^{n}\\frac{1}{(1+\\lambda_i)^2}$$\n\nConsider the function $g(x) = \\frac{1}{(1+x)^2}$. It is easy to check for $x \\in (0,\\infty)$, $$g(x) > 0, \\quad g'(x) = -\\frac{2}{(1+x)^3} < 0 \\quad\\text{ and }\\quad g''(x) = \\frac{6}{(1+x)^4} > 0$$ This means $g(x)$ is a positive, monotonic decreasing and convex function over $(0,\\infty)$.\nBy Jensen's inequality, we obtain following lower bound:\n\n$$\\verb/Tr/((I_n + A)^{-2}) = \\sum_{i=1}^{n} g(\\lambda_i) \\underbrace{\\ge}_{\\text{Jensen}} n g\\left(\\frac1n\\sum_{i=1}^n \\lambda_i\\right) \\underbrace{\\ge}_{g\\text{ decreasing}} n g(1) = \\frac{n}{4}$$\n\nSince $\\verb/Tr/(I_n) \\le n$ and $\\verb/Tr/((I_n + I_n)^{-2}) = \\frac{n}{4}$, above lower bound $\\frac{n}{4}$ is achievable and hence the optimal one.\n\nAs a consequence, your intution $\\verb/Tr/((I_n + A)^{-1}) \\ge \\frac14$ is true (but far from the optimal).\n\n\u2022 +1 Yet $\\frac14\\ge\\frac{n}{(1+n)^2}$, so the OP's bound is at least better than the one given in the other answer ;D \u2013\u00a0user1551 Aug 24 '15 at 11:44", "date": "2019-05-20 06:22:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1380623/computing-operatornametr-bigl-bigl-ai-1-bigr2-bigr", "openwebmath_score": 0.9202901124954224, "openwebmath_perplexity": 256.72814994167754, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915215128427, "lm_q2_score": 0.8633916134888613, "lm_q1q2_score": 0.8550093945833127}}
{"url": "https://math.stackexchange.com/questions/1356968/is-there-a-name-for-matrix-product-with-reversed-indices/1356977", "text": "# Is there a name for matrix product with reversed indices?\n\nThe typical matrix product is as follows: $$(\\mathbf{A}\\mathbf{B})_{ij} = \\sum_{k=1}^m A_{ik}B_{kj}\\,.$$\n\nIs there a name or characterization for one such as $$(\\mathbf{A}\\mathbf{B})_{ij} = \\sum_{k=1}^m A_{ki}B_{jk}\\,?$$ Furthermore, what can be said about the matrix vector product $$(\\mathbf{A}\\mathbf{b})_{i} = \\sum_{k=1}^m A_{ki}b_{k}\\,?$$ Is there any way to express the above matrix-vector product in terms of traditional linear algebra?\n\n\u2022 Great questions. The beginner's mind is a precious thing. \u2013\u00a0orangeskid Jul 11 '15 at 0:24\n\u2022 Agreed, this is a really good question! It is not trivial but amazing that your formulas turn out to be $A^TB^T$ and $A^TB$. \u2013\u00a0Eli Rose -- REINSTATE MONICA Jul 11 '15 at 1:00\n\u2022 @EliRose why is this not trivial? It's pretty much by definition of transposition. \u2013\u00a0Ruslan Jul 11 '15 at 11:04\n\nYour second formula is just $A^tB^t$ and your third one $A^tb$, where the $(\\hskip1ex)^t$ means transposed\n\u2022 (Also, note that $\\;\\;\\; A^tB^t \\: = \\: (BA)^t \\:\\:\\:\\:$.) $\\;\\;\\;\\;\\;\\;\\;\\;\\;$ \u2013\u00a0user57159 Jul 11 '15 at 3:13\nGiven a matrix $\\mathbf{A} = (A)_{ij}$ the transpose of a matrix $\\mathbf{A}^{\\intercal}$ is defined by $\\mathbf{A}^{\\intercal} = (A)_{ji}$. Intuitively the transpose of a matrix is found by reflecting the matrix across the line through the diagonal coefficients $i = j$. See https://en.wikipedia.org/wiki/Transpose.\nWith this in mind your first alternate version of multiplication can be written as $\\sum_{k=1}^m A_{ki}B_{jk} = \\mathbf{A}^\\intercal\\mathbf{B}^\\intercal$.\n$$\\sum_{k=1}^n A_{ki} B_{jk} = \\sum_{k=1}^n B_{jk} A_{ki} = (BA)_{ji} = (BA)^\\intercal = A^\\intercal B^\\intercal$$ and $$\\sum_{k=1}^m A_{ki}b_{k} = \\sum_{k=1}^m A_{ik}^{\\intercal} b_{k} = A^{\\intercal} b$$", "date": "2020-11-29 17:07:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1356968/is-there-a-name-for-matrix-product-with-reversed-indices/1356977", "openwebmath_score": 0.8446347713470459, "openwebmath_perplexity": 330.0486715869493, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915215128428, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8550093893618809}}
{"url": "https://math.stackexchange.com/questions/1638942/how-can-i-solve-int-frac3x2x2x1dx", "text": "How can I solve $\\int \\frac{3x+2}{x^2+x+1}dx$\n\nI want to compute this primitive $$I=\\int \\frac{3x+2}{x^2+x+1}dx.$$\n\nI split this integral into two part: $$\\int \\frac{3x+2}{x^2+x+1}dx=\\int \\frac{2x+1}{x^2+x+1}dx+\\int \\frac{x+1}{x^2+x+1}dx,$$ For the first part: $$\\int \\frac{2x+1}{x^2+x+1}dx= \\ln(|x^2+x+1|)+c_1$$ For the second part: due a change of variable $u=x+1$, I find $$\\int \\frac{x+1}{x^2+x+1}dx= \\int \\frac{x+1}{x(x+1)+1}dx=\\int \\frac{u}{u(u-1)+1}dx=\\dots$$\n\n\u2022 $u=x+1\\implies x=u-1$, hence $\\frac{x+1}{x(x+1)+1}=\\frac{u}{(u-1)u+1}\\neq\\frac{u}{u^2}$. \u2013\u00a0Workaholic Feb 3 '16 at 15:59\n\u2022 @Workaholic, I don't have any idea to compute the second integral, can you help me \u2013\u00a0Achaire Feb 3 '16 at 16:02\n\u2022 the varible change is no good here $\\frac{x+1}{x(x+1)+1}=\\frac{x}{x(x+1)+1}+\\frac{1}{x(x+1)+1}$ \u2013\u00a0Achaire Feb 3 '16 at 16:05\n\u2022 Do you know how to deal with integrals of the form $\\displaystyle\\int\\frac{1}{ax^2+bx+c}\\,\\mathrm dx$? \u2013\u00a0Workaholic Feb 3 '16 at 16:07\n\u2022 It seems that most agree that rewriting $x^2+x+1$ as $x^2+x+\\frac{1}{4}+\\frac{3}{4}=(x+\\frac{1}{2})^2+\\frac{3}{4}$ would be helpful. \u2013\u00a0John Joy Feb 3 '16 at 16:41\n\nHint For the second:\n\n$$\\int \\frac{x+1}{x^2+x+1}dx$$\n\nRewrite as $$\\int\\bigg( \\frac{2x+1}{2(x^2+x+1)}+\\frac{1}{2(x^2+x+1)} \\bigg)dx$$\n\n$$=\\underbrace{\\frac 1 2 \\int\\frac{2x+1}{x^2+x+1}dx}_{:=I}+\\underbrace{\\frac 1 2 \\int \\frac{1}{x^2+x+1}dx}_{:=J}$$\n\nFor $I$ substitute $t=x^2+x+1$ and $dt=(2x+1)dx$\n\nFor $J$ complete the square : $\\frac 1 2\\displaystyle\\int \\frac{1}{(x+1/2)^2+3/4}$ and now substitute $\\varphi=x+1/2$ and $d\\varphi=dx\\Longrightarrow \\frac 1 2\\displaystyle\\int \\frac{1}{\\varphi ^2 +3/4}d \\varphi$\n\nfactor out $3/4$ from the denominator $\\frac 2 3\\displaystyle\\int \\frac{1}{(4\\varphi ^2)/(3)+1}d \\varphi$ and then substitute $z=\\frac{2 \\varphi}{\\sqrt 3}$ and $dz=\\frac{2}{\\sqrt 3}d \\varphi$ so you will get $\\frac {1}{\\sqrt 3}\\displaystyle\\int \\frac{1}{z^2+1}dz$ and from here it is easy\n\nHINT: the second integral is given by $$\\int\\frac{x+1}{x^2+x+1}dx$$ you must write the denominator as $$(x+1/2)^2+3/4)$$\n\nHere is one approach to the second integral:\n\n$$\\int \\frac{x + 1}{ x^2 + x + 1} dx = \\frac12 \\left(\\int \\frac{2x + 1}{ x^2 + x + 1} dx + \\int \\frac{1}{x^2 + x + 1} dx\\right)$$\n\nThe first integral is identical to the one you integrated. The second is $\\arctan$, which can be seen after you complete the square.\n\n$$\\int \\frac{1}{x^2 + x + 1} dx = \\int \\frac{1}{x^2 + x + (1/2)^2 - (1/2)^2 + 1} dx$$ $$\\int \\frac{1}{(x+1/2)^2 + 3/4} dx = \\frac{4}{3} \\int \\frac{1}{\\left( \\frac{2x+1}{\\sqrt 3}\\right) + 1} dx = \\frac{2}{\\sqrt{3}} \\int \\frac{1}{u^2 + 1} du = \\frac{2}{\\sqrt{3}} \\arctan \\left( \\frac{2x+1}{\\sqrt{3}}\\right) + C$$\n\nWhere we used the substitution $u = \\frac{2x+1}{\\sqrt{3}}$.\n\nHint:\nThere's no need to split the integral that way (IMO), we can instead write, \\begin{align} \\int \\frac{3x+2}{x^2+x+1}\\,\\mathrm dx&=\\frac{3}{2}\\int\\frac{2x+\\tfrac{4}{3}}{x^2+x+1}\\,\\mathrm dx\\\\ &=\\frac32\\int\\dfrac{2x+1+\\tfrac13}{x^2+x+1}\\,\\mathrm dx\\\\ &=\\dfrac32\\left(\\int\\dfrac{2x+1}{x^2+x+1}\\,\\mathrm dx+\\int\\dfrac{{\\small 1/3}}{x^2+x+1}\\,\\mathrm dx\\right). \\end{align} The first one is obvious. For the second one we should complete the square in the denominator, like this, $$\\int\\dfrac{1}{x^2+x+1}\\,\\mathrm dx=\\int\\dfrac{1}{\\color{royalblue}{x^2+x+\\tfrac14}+1-\\tfrac14}\\,\\mathrm dx=\\int\\dfrac1{\\color{royalblue}{\\left(x+\\tfrac1{2}\\right)^2}+\\tfrac34}\\,\\mathrm dx.$$ Now try to proceed further by rewriting that last expression so as to exploit the fact that the integral of $\\frac{u'(x)}{u(x)^2+1}$ is $\\arctan u(x)+\\rm C$, you may have to factor something, and do one or more substitutions.\n\nLet $3x+2=A\\dfrac{d(x^2+x+1)}{dx}+B$\n\n$\\iff3x+2=A(2x+1)+B=2Ax+A+B$\n\n$\\implies2A=3,A+B=2$\n\n$$\\implies\\int\\dfrac{3x+2}{x^2+x+1}dx=A\\cdot\\dfrac{d(x^2+x+1)}{x^2+x+1}+B\\dfrac{dx}{x^2+x+1}$$\n\n$$\\int\\dfrac{dx}{x^2+x+1}=\\int\\dfrac4{(2x+1)^2+(\\sqrt3)^2}dx$$\n\nSet $2x+1=\\sqrt3\\tan y$", "date": "2021-05-11 10:37:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1638942/how-can-i-solve-int-frac3x2x2x1dx", "openwebmath_score": 0.9540920853614807, "openwebmath_perplexity": 284.37319270552496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936096877063, "lm_q2_score": 0.8688267830311354, "lm_q1q2_score": 0.8550068851064676}}
{"url": "https://math.stackexchange.com/questions/83383/midpoint-convex-and-continuous-implies-convex/83398", "text": "# Midpoint-Convex and Continuous Implies Convex\n\nGiven that $$f\\left(\\frac{x+y}{2}\\right)\\leqslant \\frac{f(x)+f(y)}{2}~,$$ how can I show that $f$ is convex.\nThanks.\n\nEdit: I'm sorry for all the confusion. $f$ is assumed to be continuous on an interval $(a,b)$.\n\n\u2022 You can't. At least, not without assuming that $f$ is continuous. \u2013\u00a0Willie Wong Nov 18 '11 at 15:06\n\u2022 Lebesgue measurability is enough @WillieWong. This is a theorem by Sierpinski. (If I recall correctly). \u2013\u00a0Jonas Teuwen Nov 18 '11 at 15:11\n\u2022 The trick in the continuous case is to set $z = \\frac{p}{2^{n + 1}} x + \\frac{q}{2^{n + 1}} y$ with $p + q = 2^{n + 1}$ and proceed by induction. \u2013\u00a0Jonas Teuwen Nov 18 '11 at 15:15\n\u2022 @Jonas: right. I should have said \"... something like '$f$ is continuous' \". The obvious counterexample is, of course, not Lebesgue measurable. Thanks for the correction. \u2013\u00a0Willie Wong Nov 18 '11 at 17:13\n\u2022 \u2013\u00a0Martin Sleziak May 2 '17 at 13:28\n\nBelow is the proof of the fact that every midpoint-convex function is rationally convex, which I copied from my older post on a different forum.\n\nIf you add the condition that $f$ is continuous, then from rational convexity you will get convexity. (Note that if you are interested only in continuous functions, then it suffices to show the validity of $f(t x + (1-t)y)\\le t f(x) + (1-t)f(y)$ for $t=\\frac k{2^n}$ as suggested in Jonas' comment. The proof of this fact is a little easier. I've given a little more involved proof, since the relation between midpoint convexity and rational convexity seems to be interesting on its own.)\n\nMaybe I should also mention that midpoint-convex functions are called Jensen convex by some authors.\n\nNote that without some additional conditions on $f$, midpoint convexity does not imply convexity; see this question: Example of a function such that $\\varphi\\left(\\frac{x+y}{2}\\right)\\leq \\frac{\\varphi(x)+\\varphi(y)}{2}$ but $\\varphi$ is not convex\n\nLet $f: \\mathbb R\\to \\mathbb R$ be a midpoint-convex function, i.e. $$f\\left(\\frac{x+y}2\\right) \\le \\frac{f(x)+f(y)}2$$ for any $x,y \\in \\mathbb R$.\n\nWe will show that then this function fulfills $$f(t x + (1-t)y)\\le t f(x) + (1-t)f(y)$$ for any $x,y\\in \\Bbb R$ and any rational number $t\\in\\langle0,1\\rangle$.\n\nHint: Cauchy induction: see wikipedia or AoPS or answers to this post.\n\nProof. It is relatively easy to see that it suffices to show $f([x_1+\\dots+x_k]/k)\\le [f(x_1)+\\dots+f(x_k)]/k$ for any integer $k$ (and any choice of $x_1,\\dots,x_k\\in \\mathbb R$).\n\nThe case $k=2^n$ is a straightforward induction.\n\nNow, if $2^{n-1}<k\\le 2^n$, then we denote $\\overline x=\\frac{x_1+\\dots+x_k}k$. Now from $$f(\\overline x)=f\\left(\\frac{x_1+\\dots+x_k+\\overline x+\\dots+\\overline x}{2^n}\\right) \\le\\frac{f(x_1)+\\dots+f(x_k)+(2^n-k)f(\\overline x)}{2^n},$$ where $2^n-k$ copies of $\\overline x$ are summmed in the middle expression, we get $kf(\\overline x) \\le f(x_1)+\\dots+f(x_k)$ by a simple algebraic manipulation.\n\nThe fact that measurability of $f$ is enough for the implication midpoint-convex $\\Rightarrow$ convex to hold was mentioned in some of the comments above and in answers to the question I linked. Some references for this fact:\n\nConstantin Niculescu, Lars Erik Persson: Convex functions and their applications, p.60:\n\nH. Blumberg [31] and W. Sierpinski [226] have noted independently that if $f : (a, b) \\to \\mathbb R$ is measurable and midpoint convex, then $f$ is also continuous (and thus convex). See [212, pp. 220.221] for related results.\n\n[31] H. Blumberg, On convex functions, Trans. Amer. Math. Soc. 20 (1919), 40\u201344.\n\n[212] A. W. Roberts and D. E. Varberg, Convex Functions, Academic Press, New York and London, 1973.\n\n[226] W. Sierpinski, Sur les fonctions convexes mesurables, Fund. Math. 1 (1920), 125\u2013129.\n\nMarek Kuczma: An introduction to the theory of functional equations and inequalities, p.241. He mentions the book T. Bonnesen and W. Fenchel, Theorie der konvexen K\u00f6rper, Berlin, 1934 as an additional reference.\n\n\u2022 Not to be too much of a whiner, but that \"Spoiler\" trick is annoying when the point of asking a question here is to get an answer. It's not like accidentally glancing at your text is going to reveal the answer, so if I want to avoid reading it, I avoid reading it. In addition, there is no way to show your text on displays that do not have \"mouse-over\" - it stays blank on my iPad no matter what I do, for example. There might be fora and questions where it is appropriate, but this doesn't sem to be one of them. \u2013\u00a0Thomas Andrews Nov 18 '11 at 15:51\n\u2022 Sorry @Thomas, I've removed it. \u2013\u00a0Martin Sleziak Nov 18 '11 at 16:01\n\u2022 @Hi, just double check. When I check the midpoint convexity, I have to show the inequality holds for any $x,y$ right? I can not just pick $x,y$ such that $f(x)=f(y)$, which looks like checking for quasi-concavity. \u2013\u00a0Bob Oct 7 '16 at 12:29\n\u2022 @Bob Yes, by definition midpoint convex means that this inequality is true for any $x$, $y$. \u2013\u00a0Martin Sleziak Oct 7 '16 at 12:49\n\u2022 @MartinSleziak, I find it is easy for people to make this mistake. Pick $x,y$ such that $f(x)=f(y)$, and once they showed $f(0.5x+0.5y)>f(x)$ they thought they proved concavity, but only got quasi-concavity. We can pick such two points ($x,y$ s.t. $f(x)=f(y)$) wolg for quasi-concavity, but not for concavity. \u2013\u00a0Bob Oct 7 '16 at 13:31\n\nAs was already previously mentioned it is straightforward to see that for a mid point convex function f $$f\\left(\\frac{x_1+x_2+\\ldots+x_{2^n}}{2^n}\\right)\\leq\\frac{1}{2^n}(f(x_1)+f(x_2)+\\ldots f(x_{2^n}))$$ By setting $x_i=x$ for $1\\leq i\\leq m$ and $x_i=y$ for $m+1\\leq i\\leq 2^n$ where $1\\leq m\\leq 2^n$ is an integer we now obtain $$f\\left(\\frac{m}{2^n}x+(1-\\frac{m}{2^n})y\\right)\\leq \\frac{m}{2^n}f(x)+(1-\\frac{m}{2^n})f(y)$$ Since the set of rational dyadic numbers of the form $\\frac{m}{2^n}$ with $m$ and $n$ integers and $1\\leq m\\leq 2^n$ is dense in $[0,1]$ the proof follows from the continuity of f.\n\nCan you provide any more information about $f$? For example, the property holds for continuous functions $f: I \\rightarrow \\mathbb{R}$, $I$ being an interval of real numbers. I think this result is due to Jensen [1].\n\nTheorem (Jensen). Let $f: I\\rightarrow\\mathbb{R}$ be a continuous function. Then $f$ is convex if and only if it is midpoint convex, i.e. for $x,y$ in $I$ we have\n\n$$f\\left(\\frac{x+y}{2}\\right) \\leq \\frac{f(x)+f(y)}{2}$$\n\n[1] J. L. W. V. Jensen, Sur les fonctions convexes et les in\u00e9galit\u00e9s entre les valeurs moyennes, Acta Math., 30 (1906), 175-193.\n\n\u2022 Jensen doesn't prove this, per se, in this paper (his definition of \"convex\" was the modern-day \"midpoint-convex\"). It's more correct to say that he proved Jensen's Inequality (with arbitrary real weights) for functions which are midpoint convex and continuous. Of course, Jensen's Inequality with two arguments gives the modern definition of convexity. \u2013\u00a0Robert Wolfe Nov 9 '17 at 20:30\n\nGiven a weight lambda, take its binary expansion. Think of what midpoint convexity implies to the inequalities involving the partial sums of this binary expansion. Then the result follows by the continuity of f.\n\nI would like to properly show by induction that if $$m\\in\\{0,1,2,\\ldots,2^{n}-1\\}$$ then $$f\\left(\\frac{m}{2^n}x+\\Big(1-\\frac{m}{2^k}\\Big)y\\right)\\le \\frac{m}{2^n}f(x) +\\Big(1-\\frac{m}{2^n}\\Big)f(y).$$ and then the result will follows by contuinity since $$m=\\lfloor2^nt\\rfloor \\in\\{0,1,2,\\ldots,2^{n}\\} ~~~and~~\\frac{\\lfloor2^nt\\rfloor}{2^n}\\to t$$\n\nThe initial state $$n = 1$$ is trivial by hypothesis.\n\nNow we assume that for every $$k, whenever $$m\\in\\{0,1,2,\\ldots,2^{k-1}-1\\}$$,we have $$f\\left(\\frac{m}{2^k}x+\\Big(1-\\frac{m}{2^k}\\Big)y\\right)\\le \\frac{m}{2^k}f(x) +\\Big(1-\\frac{m}{2^k}\\Big)f(y). \\tag{I}\\label{I}$$\n\nwe want to prove \\eqref{I} for $$k=n$$.\n\nLet $$m \\in\\{0,1,2,\\ldots,2^{n}-1\\}$$ then the division by 2 yields $$m =2p +r$$ with $$r\\in \\{0,1\\}$$ \\begin{align}X&:=\\left(\\frac{m}{2^n}x+\\Big(1-\\frac{m}{2^n}\\Big)y\\right)= \\left(\\frac{2p +r}{2^n}x+\\Big(1-\\frac{2p +r}{2^n}\\Big)y\\right) \\\\&= \\left(\\frac{p }{2^n}x+\\frac{p +r}{2^n}x+\\Big(1-\\frac{p }{2^n}y-\\frac{p +r}{2^n}y\\Big)y\\right) \\\\&= \\frac12 \\left(\\frac{p }{2^{n-1}}x+\\frac{p +r}{2^{n-1}}x+\\Big(2-\\frac{p }{2^{n-1}}y-\\frac{p +r}{2^{n-1}}y\\Big)y\\right) \\\\&=\\frac12\\left(\\frac{p }{2^{n-1}}x+ \\Big(1-\\frac{p }{2^{n-1}}y\\Big)\\right)+\\frac12\\left(\\frac{p+r }{2^{n-1}}x+ \\Big(1-\\frac{p +r}{2^{n-1}}y\\Big)\\right) \\\\&:= \\color{red}{\\frac{1}{2}\\left(X_1+X_2\\right)}\\end{align}\n\nOn the other hand, since $$r\\in \\{0,1\\}$$ and $$m\\in\\{0,1,2,\\ldots,2^{n}-1\\}$$ it is easy to check using parity that $$p,p+1 \\in \\{0,1,2,\\ldots,2^{n-1}-1\\}$$ that is $$p+r \\in \\{0,1,2,\\ldots,2^{n-1}-1\\}$$\n\nBy hypothesis of induction we obtain \\begin{align}f(X) &= f\\left(\\frac{1}{2}\\left(X_1+X_2\\right)\\right)\\le \\frac12f(X_1) +\\frac12f(X_2) \\\\&= \\frac12f\\left(\\frac{p }{2^{n-1}}x+ \\Big(1-\\frac{p }{2^{n-1}}y\\Big)\\right)+\\frac12f\\left(\\frac{p+r }{2^{n-1}}x+ \\Big(1-\\frac{p +r}{2^{n-1}}y\\Big)\\right) \\\\&\\le\\frac12\\left(\\frac{p }{2^{n-1}}f(x)+ \\Big(1-\\frac{p }{2^{n-1}}f(y)\\Big)\\right)+\\frac12\\left(\\frac{p+r }{2^{n-1}}f(x)+ \\Big(1-\\frac{p +r}{2^{n-1}}f(y)\\Big)\\right) \\\\&= \\frac{2p+r}{2^n}f(x) +\\Big(1-\\frac{2p+r}{2^n}\\Big)f(y)\\\\&= \\frac{m}{2^n}f(x) +\\Big(1-\\frac{m}{2^n}\\Big)f(y).\\end{align}", "date": "2019-02-20 22:33:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/83383/midpoint-convex-and-continuous-implies-convex/83398", "openwebmath_score": 0.9597777724266052, "openwebmath_perplexity": 430.4153563125048, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936073551713, "lm_q2_score": 0.8688267745399466, "lm_q1q2_score": 0.8550068747237741}}
{"url": "https://math.stackexchange.com/questions/3907304/an-extension-of-baires-category-theorem", "text": "An extension of Baire's category theorem\n\nIn a topological space, a set is said to be rare if its closure has empty interior, and a set is said to be meager if it is a countable union of rare sets. If meager sets all have empty interior, then the topological space is said to be a Baire space. The following fact is known as Baire's category theorem.\n\nTheorem. Complete metric spaces and locally compact Hausdorff spaces are Baire spaces.\n\n[Z\u0103linescu 2002] provides the following extension of this theorem for complete metric spaces, attributing it to C. Ursescu.\n\nTheorem. ([Z\u0103linescu 2002, Theorem 1.4.5]) Let $$X$$ be a complete metric space, and $$\\{S_n\\}$$ be a sequence of open sets in $$X$$. Then $$\\mathrm{cl}(\\cap_{n=1}^\\infty S_n)$$ and $$\\cap_{n=1}^\\infty \\mathrm{cl}(S_n)$$ have the same interior.\n\nIt is easy to check that a topological space is a Baire space if and only if any countable intersection of dense open sets is still dense in this space. Consequently, the theorem above implies that complete metric spaces are Baire spaces. Thus it is considered as an extension.\n\nQuestion. Does the extension hold for locally compact Hausdorff spaces?\n\nUpdate: As pointed out by @Alex Kruckman, the property in Ursescu's theorem is indeed equivalent to the fact that $$X$$ is a Baire space.\n\nTheorem. A topological space is a Baire space if and only if $$\\mathrm{cl}(\\cap_{n=1}^\\infty S_n)$$ and $$\\cap_{n=1}^\\infty \\mathrm{cl}(S_n)$$ have the same interior for any sequence of open sets $$\\{S_n\\}$$.\n\nRecall that a space is a Baire space if and only if any countable intersection of dense open sets is still dense. The theorem is essentially another way to state that a space is a Baire space if and only if all its open subspaces are Baire spaces.\n\nCombined with Baire's category theorem, this observation by Alex proves the desired theorem. My answer below proves it from scratch, the proof being essentially the same as that of Baire's category theorem for locally compact Hausdorff spaces.\n\nI will accept Alex's answer a few days later if nobody else has anything to add.\n\nThanks.\n\nThis is not really a strengthening of the Baire category theorem - it is equivalent to it. That is, any Baire space (and in particular any compact Hausdorff space) satisfies Ursescu's theorem.\n\nLet's assume $$X$$ is a Baire space. Suppose $$(S_n)_{n\\in \\mathbb{N}}$$ is a sequence of open sets in $$X$$. We always have $$\\text{int}(\\text{cl}(\\bigcap_{n=1}^\\infty S_n))\\subseteq \\text{int}(\\bigcap_{n=1}^\\infty \\text{cl}(S_n))$$, so it suffices to prove the reverse inclusion. And to do this, it suffices to prove that for any open $$U\\subseteq \\bigcap_{n=1}^\\infty \\text{cl}(S_n)$$, we have $$U\\subseteq \\text{cl}(\\bigcap_{n=1}^\\infty S_n)$$.\n\nSo fix some $$U$$. Note that an open subspace of a Baire space is Baire. [Why? If $$(V_n)_{n\\in\\mathbb{N}}$$ is a sequence of dense open subsets of $$U$$, then $$(V_n\\cup (X\\setminus \\text{cl}(U)))_{n\\in\\mathbb{N}}$$ is a sequence of dense open subsets of $$X$$. The intersection of these sets is $$(\\bigcap_{n=1}^\\infty V_n)\\cup (X\\setminus \\text{cl}(U))$$, and the fact that this is dense in $$X$$ implies that $$(\\bigcap_{n=1}^\\infty V_n)$$ is dense in $$U$$.]\n\nSo defining $$V_n = U\\cap S_n$$, we have that each $$V_n$$ is open in $$U$$, and since $$U\\subseteq \\text{cl}(S_n)$$ for all $$n$$, each $$V_n$$ is dense in $$U$$. Since $$U$$ is Baire, $$\\bigcap_{n=1}^\\infty V_n$$ is dense in $$U$$, which implies $$U\\subseteq \\text{cl}(\\bigcap_{n=0}^\\infty S_n)$$, as was to be shown.\n\n\u2022 Thanks, @Alex Kruckman. I modified the question to highlight your observation. In addition, I modified your answer, changing \"Z\u0103linescu's theorem\" to \"Ursescu's theorem\", as suggested by [Z\u0103linescu 2002]. Hope this fine with you. Thanks again. \u2013\u00a0Nuno Nov 15 '20 at 3:12\n\nHere is my attempt to prove it. It is inspired by the proof of [Z\u0103linescu 2002, Theorem 1.4.5] and the classical proof of Baire's theorem for locally compact Hausdorff spaces. Any comments or criticism will be appreciated. Thanks.\n\nTheorem. Let $$X$$ be a locally compact Hausdorff space, and $$\\{S_n\\}$$ be a sequence of open sets in X. Then $$\\mathrm{cl}(\\cap_{n=1}^\\infty S_n)$$ and $$\\cap_{n=1}^\\infty \\mathrm{cl}(S_n)$$ have the same interior.\n\nproof. It suffices to show that $$\\mathrm{int}(\\cap_{n=1}^\\infty \\mathrm{cl}(S_n))\\subset \\mathrm{cl}(\\cap_{n=1}^\\infty S_n)$$. To this end, we only need to prove for any given nonempty open set $$U \\subset \\cap_{n=1}^\\infty \\mathrm{cl}(S_n)$$ that $$$$\\label{eq:usnonempty} U\\cap(\\cap_{n=1}^\\infty S_n) \\neq \\emptyset.$$$$ In the sequel, we will define a sequence of compact sets $$\\{C_n\\}_{n=0}^\\infty$$ such that $$C_n$$ has nonempty interior and $$$$C_{n+1} \\subset C_{n} \\subset U\\cap(\\cap_{k=1}^n S_k) \\quad \\text{ for each } \\quad n \\ge 0,$$$$ where $$\\cap_{k=1}^0 S_k=X$$. Once this is done, Cantor's Theorem will yield $$$$\\label{eq:nestnonempty} \\emptyset \\;\\neq \\;\\cap_{n=1}^\\infty C_n \\;\\subset\\; U\\cap (\\cap_{n=1}^\\infty S_n),$$$$ which gives us what we want.\n\nWe define $$\\{C_n\\}$$ inductively. As $$U$$ is a nonempty open set and $$X$$ is locally compact, we can take a compact set $$C_0\\subset U$$ such that $$\\mathrm{int}(C_0)\\neq \\emptyset$$. Assume that $$C_n$$ is already defined for an $$n\\ge 0$$ so that $$C_n\\subset U\\cap(\\cap_{k=1}^n S_k)$$ and $$\\mathrm{int}(C_n)\\neq \\emptyset$$. Recalling that $$U\\subset\\mathrm{cl}(S_{n+1})$$, we have $$\\mathrm{int}(C_n)\\subset \\mathrm{cl}(S_{n+1})$$, which implies that $$\\mathrm{int}(C_n) \\cap S_{n+1} \\neq\\emptyset$$. Since $$\\mathrm{int}(C_n)\\cap S_{n+1}$$ is open, invoking again the local compactness of $$X$$, we can take a compact set $$C_{n+1}\\subset \\mathrm{int}(C_n) \\cap S_{n+1}$$ such that $$\\mathrm{int}(C_{n+1})\\neq \\emptyset$$. Clearly, $$C_{n+1}\\subset C_n$$ and $$C_{n+1}\\subset C_{n}\\cap S_{n+1} \\subset U\\cap(\\cap_{k=1}^{n+1} S_k)$$. This finishes the induction and completes the proof.", "date": "2021-04-22 03:48:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3907304/an-extension-of-baires-category-theorem", "openwebmath_score": 0.9719763398170471, "openwebmath_perplexity": 66.13613765850175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936068886641, "lm_q2_score": 0.868826771143471, "lm_q1q2_score": 0.8550068709760102}}
{"url": "https://math.stackexchange.com/questions/1438658/proof-that-any-linear-system-cannot-have-exactly-2-solutions/1438688", "text": "Proof that any linear system cannot have exactly 2 solutions.\n\nHow would you go about proving that for any system of linear equations (whether all are homogenous or not) can only have either (if this is true):\n\n\u2022 One solution\n\u2022 Infinitely many solutions\n\u2022 No solutions\n\nI found this a bit difficult to prove (even though its a very fundamental thing about any linear equation). The intuitive geometric explanation is that a line can only intersect at one point, and if it intersects at a later point, it can't be a linear equation, but I don't think this is a convincing proof.\n\nI thought of if you assume that there are two (or more, but I picked two) solutions for some linear system, then for the points in between\n\nSolution Set 1: X1, X2....., Xn\nSolution Set 2: X1, X2....., Xn\n\nThen (I think), the points between S1 and S2, must be infinitely many points (and thus infinitely many solutions) such that these points can also satisfy the linear system, which would mean the system has 2 infinite solutions.\n\nHowever, I don't think this is rigorous enough and nor do I understand completely why its true. Can anyone help in explaining (correcting) and elaborating on the intuition and proof of this?\n\n\u2022 \"Infinite solutions\" is quite incorrect. It can have infinitely many solutions. But in correct use of mathematical terminology, \"infinite solutions\" means solutions, each one of which is infinite. If you have six solutions, and each is infinite, then those are infinite solutions but not infinitely many solutions. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Sep 16 '15 at 22:09\n\u2022 Why do you find the visual proof not convincing? \u2013\u00a0corsiKa Sep 16 '15 at 22:53\n\u2022 It's worth pointing out that this question has an implicit assumption that the underlying field is infinite. Over a finite field, the assertion is not necessarily true (and we cannot have infinitely many solutions). E.g. over $GF(2)$, the equation $\\pmatrix{1&1\\\\ 0&0}x=0$ has exactly two solutions $x=\\pmatrix{0\\\\ 0}$ and $x=\\pmatrix{1\\\\ 1}$. See also 6005's answer below. \u2013\u00a0user1551 Sep 17 '15 at 11:35\n\nSuppose that $\\vec v$ and $\\vec w$ are distinct solutions for the system $A\\vec x = \\vec b$ so that $A \\vec v = A \\vec w = \\vec b$. Then $\\frac{1}{2}(\\vec v + \\vec w)$ must be distinct from both $\\vec v$ and $\\vec w$ and must also solve the system since: $$A(\\tfrac{1}{2}(\\vec v + \\vec w)) = \\tfrac{1}{2}(A\\vec v + A\\vec w) = \\tfrac{1}{2}(\\vec b + \\vec b) = \\vec b$$ We can then apply the same argument to $\\vec v$ and $\\frac{1}{2}(\\vec v + \\vec w)$ in order to get infinitely many distinct solutions.\n\n\u2022 Or pick any $\\lambda \\in (0,1)$ and consider $\\lambda \\vec{v} + (1 - \\lambda) \\vec{w}$. Then you get continuum-many solutions right off the bat. \u2013\u00a06005 Sep 16 '15 at 22:10\n\u2022 You can even choose any real number for $\\lambda$. \u2013\u00a0Aurey Sep 16 '15 at 23:30\n\u2022 Looking over this again, great way of explaining your argument (deriving infinitely many solutions) \u2013\u00a0q.Then Sep 17 '15 at 22:29\n\nYour second proof sounds fine. Let's imagine that the system of equations is written as $Ax = b$, where $A$ is the matrix of coefficients, $x$ is the vector of variables we are solving for, and $b$ are the constants. Now assume we have two distinct solutions $x_1$ and $x_2$. Then $rx_1+sx_2$ is also a solution, if $r+s=1$ - just plug it in and use linearity of matrix multiplication.\n\n$$A(rx_1+sx_2) = A(rx_1)+A(sx_2) = rA(x_1)+sA(x_2) = (r+s)b = b$$\n\nWe thus have an infinite number of solutions. If you don't like matrices, you can easily just write out each term in the system of equations.\n\nIf you want to get some intuition for what is happening, the linear combination $rx_1+sx_2$, where $r+s=1$, is a line that connects the two points. When $r$ is 1 and $s$ is 0, we get point $x_1$. When $s$ is 1 and $r$ is 0, we get point $x_2$. Other combinations of $r$ and $s$ give other points.\n\nYou need to be more careful with your first argument, though. In general, when graphing systems of $n$ equations in $n$ unknowns, the equations are not lines, but $(n-1)$-dimensional planes in $n$-dimensional space. So for a system of three equations and three unknowns, for example, we have three planes in three dimensional space. Unless the planes are parallel, the intersection of two planes is a line, and the intersection of a line with a plane is a point. That's the one solution case. If two planes are parallel, they never intersect - no solutions. And finally, if the line/plane overlaps with another line/plane, we get an infinite number of solutions. That's the intuition behind this theorem.\n\n\u2022 Thanks for the answer, I'm sure other answers were great as well, but I think you explained it the best (at least the best for me) \u2013\u00a0q.Then Sep 17 '15 at 2:55\n\nWe can represent a linear system in matrix notation as:$A\\vec x=\\vec v$. Now suppose that we have $A\\vec x=\\vec v$ and $A\\vec y=\\vec v$ with $\\vec x \\ne \\vec y$, for $a,b$ with $a+b=1$ we have, by linearity: $A(a\\vec x + b\\vec y)=aA\\vec x +b A \\vec y=\\vec v$, so we have infinitely many solutions.\n\nIn general, in a vector space over a general field $F$, the number of solutions $\\vec{x}$ to the system $$A \\vec{x} = \\vec{w}$$ is either $0$ or $|\\ker A|$. (Because if $\\vec{x}_0$ satisfies the equation, then the set of solutions is $\\vec{x}_0 + \\ker A$.) And $|\\ker A| = |F|^k$ for some $k \\ge 0$.\n\nAssuming the base field $F$ has infinite cardinality $\\alpha$ and the vector space is finite-dimensional, it follows that the number of solutions is $0, 1,$ or $\\alpha$ (since $|F|^0 = 1$ and $|F|^k = \\alpha$ for $k \\ge 1$).\n\n\u2022 @user1551 yes, thank you. I corrected that and two other minor typos. \u2013\u00a06005 Sep 17 '15 at 15:20", "date": "2019-10-14 23:16:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1438658/proof-that-any-linear-system-cannot-have-exactly-2-solutions/1438688", "openwebmath_score": 0.9233568906784058, "openwebmath_perplexity": 148.65112293461084, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9603611586300241, "lm_q2_score": 0.8902942348544447, "lm_q1q2_score": 0.8550040029064453}}
{"url": "https://brilliant.org/discussions/thread/1-3/", "text": "#1\n\nHow many positive integers less than $1000$ have the property that the sum of the digits of each such number is divisible by $7$ and the number itself is divisible by $3$?\n\nNote by Vilakshan Gupta\n3\u00a0years, 1\u00a0month ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nLet's think of a number $abc (0 \\leq a, b, c \\leq 9)$. $a+b+c \\equiv 0 (\\mod 3 \\text{and} \\mod 7)$. Thus, $a+b+c=21$.\n\n$(3, 9, 9) \\rightarrow \\frac{3!}{2}, (4, 8, 9) \\rightarrow 3!, (5, 7, 9) \\rightarrow 3!, (5, 8, 8) \\rightarrow \\frac{3!}{2}, (6, 6, 9) \\rightarrow \\frac{3!}{2}, (6, 7, 8) \\rightarrow 3!, (7, 7, 7)$ $3+6+6+3+3+6+1=28$\n\nPlease tell me if there is any error.\n\n- 2\u00a0years, 11\u00a0months ago\n\nNice method\n\n- 2\u00a0years, 11\u00a0months ago\n\nGood one brother\n\n- 2\u00a0years, 11\u00a0months ago\n\nYour method is quite efficient vis-a-vis mine. The latter involved manual trials with 3 digit integers with integer 1 to 9 at the unit. Thank you\n\n- 1\u00a0year, 1\u00a0month ago\n\nWhat ans did you get?\n\n- 3\u00a0years, 1\u00a0month ago\n\nHey Aaron. How much are u getting with bonus? With bonus i am getting 12.\n\n- 3\u00a0years, 1\u00a0month ago\n\n25 is the answer as per me\n\n- 3\u00a0years, 1\u00a0month ago\n\nNo - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28\n\n- 3\u00a0years, 1\u00a0month ago\n\n28\n\n- 3\u00a0years, 1\u00a0month ago\n\nwe just need to find the numbers which add upto 21\n\n- 3\u00a0years, 1\u00a0month ago\n\n- 3\u00a0years, 1\u00a0month ago\n\nHow ma\u00f1y have you got right in PRMO - 17?\n\n- 3\u00a0years, 1\u00a0month ago\n\n@Md Zuhair Is the paper for 9,10,11 and 12 same?\n\n- 3\u00a0years, 1\u00a0month ago\n\nYes Sir!\n\n- 3\u00a0years, 1\u00a0month ago\n\nunfortunately, i will get only 10 questions correct. I did very silly mistakes\n\n- 3\u00a0years, 1\u00a0month ago\n\nSir\ud83d\ude05 I am getting 10 along with bonus!\n\n- 3\u00a0years, 1\u00a0month ago\n\nOh. U mean 8/28 u r getting?\n\n- 3\u00a0years, 1\u00a0month ago\n\nwell, if it is bonus that means 2 questions marks are given extra.If it was been written question deleted then scores would be evaluated out of 28\n\n- 3\u00a0years, 1\u00a0month ago\n\nGeometry was quite tough and lengthy! Excluding bonus , I'm getting 8\n\n- 3\u00a0years, 1\u00a0month ago\n\noh...btw,where do u live (i mean which region)\n\n- 3\u00a0years, 1\u00a0month ago\n\nRajasthan..U?\n\n- 3\u00a0years, 1\u00a0month ago\n\noh\n\n- 3\u00a0years, 1\u00a0month ago\n\nSo , you are already selected..Great \ud83d\udc4d\n\n- 3\u00a0years, 1\u00a0month ago\n\n- 3\u00a0years, 1\u00a0month ago\n\nCoz as per cutoff(s) uploaded by Resonance , cutoff in Chandigarh is lower than others ( Rajasthan , Maharashtra , UP , etc) ... That's why!\n\n- 3\u00a0years, 1\u00a0month ago\n\nAccording to Resonance , cutoff in Chandigarh is just 4(questions)\n\n- 3\u00a0years, 1\u00a0month ago\n\nYa. Thats ridiculous. WB region has always got a higher cutoff...\n\n- 3\u00a0years, 1\u00a0month ago\n\ni don't think it will be so low\n\n- 3\u00a0years, 1\u00a0month ago\n\nIf that isnt, then wb will be higher and i will surely not qualify\n\n- 3\u00a0years, 1\u00a0month ago\n\nIt's 11 in Rajasthan ! \ud83d\ude05\ud83d\ude12\n\n- 3\u00a0years, 1\u00a0month ago\n\nLetTheFateDecide !!Bye\n\n- 3\u00a0years, 1\u00a0month ago\n\nwhich class are u in toshit?\n\n- 3\u00a0years, 1\u00a0month ago\n\n@Shreyan Chakraborty .. How much?\n\n- 3\u00a0years, 1\u00a0month ago\n\nJANI NA BAJE HOYECHE\n\n- 3\u00a0years, 1\u00a0month ago\n\nANSWER IS 28....HAS A BIJECTION WITH a+b+c=21 WHERE 0<a,b,c<=9........\n\n- 3\u00a0years, 1\u00a0month ago\n\nAre na na.... I am not telling that. How much are you getting?\n\n- 3\u00a0years, 1\u00a0month ago\n\n@Shreyan Chakraborty The Hundreds digit can't be 1 or 2..\n\n- 3\u00a0years, 1\u00a0month ago\n\nyeah hundreds digit cant be 1,2\n\n- 3\u00a0years, 1\u00a0month ago\n\n@Md Zuhair @Vilakshan Gupta I haven't attempted one of the bonus question. Will I still get marks for it?\n\n- 3\u00a0years, 1\u00a0month ago\n\nI think the question can be cancelled as how can a person attempt to decinal answers and i had attempted ine. So i dunno.\n\n- 3\u00a0years, 1\u00a0month ago\n\nYup.I think so.\n\n- 3\u00a0years, 1\u00a0month ago\n\n- 3\u00a0years, 1\u00a0month ago\n\nI believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.\n\n- 3\u00a0years, 1\u00a0month ago\n\nExactly\n\n- 3\u00a0years, 1\u00a0month ago\n\nI agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 432*1 = 24, not 4+3+2+1 = 10. What am I missing?\n\n- 3\u00a0years ago\n\nAh - that's where the 28 comes from - much more mathematical than my just listing and counting them\n\n- 3\u00a0years, 1\u00a0month ago\n\nHello, There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 You can check out for more queries related to the JEE EXAMS from the following compilation\n\n- 3\u00a0years ago\n\nHey I'm getting 8/27 from jharkhand as per the new answer key of hbcse.will I qualify???\n\n- 3\u00a0years ago\n\nLets see....\n\n- 3\u00a0years ago\n\nHey , what does discounted actually refer to?\n\n- 3\u00a0years ago\n\nAre the marks gonna be added to everyone's total or the questions will be cancelled (lowering the cutoff)?\n\n- 3\u00a0years ago\n\nThey will be added to totsl\n\n- 2\u00a0years, 11\u00a0months ago\n\nThe questions will be cancelled\n\n- 3\u00a0years ago\n\n@Pokhraj Harshal Ok! Then I'm also getting the same...\n\n- 3\u00a0years ago\n\nHow much? Without the question?\n\n- 3\u00a0years ago\n\nWhich region are u from??\n\n- 3\u00a0years ago\n\nWB rgion\n\n- 3\u00a0years ago\n\nAnd what about the other participants and their marks from your school. I mean the averages and the highest marks\n\n- 3\u00a0years ago\n\n@Pokhraj Harshal Rajasthan region!\n\n- 3\u00a0years ago\n\n@Md Zuhair 8\n\n- 3\u00a0years ago\n\nO i see....\n\n- 3\u00a0years ago\n\nthis question came in this year PRMO answer is 28\n\n- 2\u00a0years, 11\u00a0months ago\n\nits sum is divisibli by 21 using this you can solve\n\n- 2\u00a0years, 11\u00a0months ago\n\n27\n\n- 2\u00a0years, 11\u00a0months ago\n\n28\n\n- 2\u00a0years, 9\u00a0months ago\n\n33\n\n- 2\u00a0years, 2\u00a0months ago\n\nzuhair tui ki amk jiggesh korchish??\n\n- 3\u00a0years, 1\u00a0month ago\n\nAccha.. nijer whatsapp number ta de... whatsapp e kotha bolchi\n\n- 3\u00a0years, 1\u00a0month ago", "date": "2020-09-29 07:19:53", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/1-3/", "openwebmath_score": 0.9630981087684631, "openwebmath_perplexity": 4411.028552146862, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.96036116089903, "lm_q2_score": 0.890294230488237, "lm_q1q2_score": 0.8550040007333919}}
{"url": "https://mathhelpboards.com/threads/integration-computer-problem.5599/", "text": "# Integration - computer problem\n\n#### Yankel\n\n##### Active member\nHello,\n\nI was trying to solve the integral of\n\nsin(x)*cos(x)\n\nusing the substitution method, what I did was:\n\nu=sin(x) and that yields du/dx = cos(x) and then du=cos(x)*dx\n\nthat comes to an integral of u*du, which is easy u^2 / 2 +C. substituting back gives the final answer\n\nsin(x)^2 / 2\n\nBut, when I ran this integral in both Maple and Mathematica, I got this answer:\n\n-cos(x)^2 / 2\n\nNow I tried asking Maple if the two answers are the same, but it failed. I tried checking myself, using the relation sin(x)^2+cos(x)^2=1, and got to the conclusion that they don't. I don't see what I did wrong here...\n\n#### Ackbach\n\n##### Indicium Physicus\nStaff member\nTechnically, you answer was not $\\sin^{2}(x)/2$, but $\\sin^{2}(x)/2+C.$ Likewise, the computer's answer was not $-\\cos^{2}(x)/2$, but $-\\cos^{2}(x)/2+C$. Since your two answers differ by a constant (namely, $1$), they are really the same answer from a calculus perspective.\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nTechnically, you answer was not $\\sin^{2}(x)/2$, but $\\sin^{2}(x)/2+C.$ Likewise, the computer's answer was not $-\\cos^{2}(x)/2$, but $-\\cos^{2}(x)/2+C$. Since your two answers differ by a constant (namely, $1$), they are really the same answer from a calculus perspective.\nOr more clearly, $C_1$ and $C_2$. This happened to me a while ago, it tends to occur a lot with trigonometric functions because of their periodic identities, so you often end up with very different looking expressions which are in fact equal..\n\n#### Yankel\n\n##### Active member\nah, I wasn't thinking about it....interesting.\n\nwell done to both of you ! thanks !\n\n#### topsquark\n\n##### Well-known member\nMHB Math Helper\nJust to be cheeky about it:\n$$\\int sin(x)~cos(x) dx = \\frac{1}{2} \\int 2 sin(x)~cos(x) dx = \\frac{1}{2} \\int sin(2x)dx$$\n\nAfter u = 2x:\n$$= \\int sin(x)~cos(x) dx = \\frac{1}{2} \\int sin(u) \\cdot \\frac{1}{2} du$$\n\n$$= -\\frac{1}{4} cos(u) + C = -\\frac{1}{4} cos(2x) + C$$\n\nand upon using $$cos(2x) = cos^2(x) - sin^2(x)$$\n$$\\int sin(x)~cos(x) dx = \\frac{1}{4} \\left ( sin^2(x) - cos^2(x) \\right ) + C$$\n\nwhich contains both your sin and cos terms and is also off by a constant from the other solutions.\n\n-Dan", "date": "2021-07-29 18:32:15", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/integration-computer-problem.5599/", "openwebmath_score": 0.9490373134613037, "openwebmath_perplexity": 954.4385325282811, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877684006774, "lm_q2_score": 0.8807970858005139, "lm_q1q2_score": 0.8549789576295208}}
{"url": "https://math.stackexchange.com/questions/2158099/counting-integer-solutions-with-conditional-restrictions", "text": "# Counting integer solutions with conditional restrictions\n\nConsider $x_i$, with $i=1,\\ldots, 10$, such that $$5 \\leq 3(x_1 + x_2 + x_3 + x_4) +2(x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}) \\leq 12\\,,$$ where each $x_i$ can be either $0$ or $1$. In addition, each $x_i$ from $i=5$ to $10$ are restricted to be $0$ if some of the $x_i$ from $i=1$ to $4$ are equal to $1$, according to the following diagram:\n\nEach $x_i$ in the bottom line is restricted by two of the $x_i$ in the top line. For example, if $x_1 = x_2 = 0$, then $x_5$ can be either $0$ or $1$, but if $x_1=1$ and/or $x_2 = 1$, then $x_5=0$ necessarily. The problem is to know in how many ways can we satisfy the inequality, subjected to these restrictions, using a method which is easily used for a computer, like generating functions, for example.\n\nThe way I'm trying to answer this is the following: if the additional restrictions did not exist, the number of solutions would be obtainable by using the generating function $$(1+x^3)^4 (1+x^2)^6$$ and the solution would be the sum of all coefficients from $x^5$ to $x^{12}$. The restrictions can be incorporated by defining $x^c_i$ such that, in the case of $i=5$ for example, $$x^c_5 = \\quad \\left\\{ \\begin{array}{lll} 0, \\quad \\, \\textrm{if} \\quad (1-x_1)(1-x_2)=0 \\\\ \\\\ x_5, \\quad \\, \\textrm{if} \\quad (1-x_1)(1-x_2)=1 \\\\ \\end{array} \\right. \\,,$$ and so on for the other variables. Using these $x^c_i$ in the inequality, instead of the $x_i$ (for $i=5$ up to $10$), we have $$5 \\leq 3(x_1 + x_2 + x_3 + x_4) +2(x^c_5 + x^c_6 + x^c_7 + x^c_8 + x^c_9 + x^c_{10}) \\leq 12$$ which in principle should contain all the restricted possibilities. This is where I'm currently stuck, as I don't know how the generating function for this inequality should be built.\n\n\u2022 Not the approach you are pursuing, but just building the characteristic function of the set yields a count of 87. \u2013\u00a0Fabio Somenzi Feb 23 '17 at 18:52\n\u2022 Thanks for the reply! I'm not sure what you mean by characteristic function of the set though, could you be more specific as how you did it? \u2013\u00a0GKiu Feb 23 '17 at 20:15\n\u2022 The characteristic function I referred to is a function of Boolean variables $x_1,\\ldots,x_{10}$ that evaluates to true if and only if the constraints are all satisfied. It can be built as the conjunction of characteristic functions for the individual constraints. The incompatibility constraints are simple: for example $\\neg x_5 \\vee (\\neg x_1 \\wedge \\neg x_2)$. The linear inequalities are a bit more complicated, but not much. Of course, I let the computer take the conjunctions and count the models. \u2013\u00a0Fabio Somenzi Feb 23 '17 at 21:31\n\u2022 Although this works, I was looking for an answer in terms of generating functions like @Jeremy Dover suggested, but thank you for the clarification nonetheless. \u2013\u00a0GKiu Feb 24 '17 at 19:09\n\u2022 Understood. That's why I posted a comment and not an answer. \u2013\u00a0Fabio Somenzi Feb 24 '17 at 19:45\n\nThis only really works because of the symmetry between the variables $x_1 \\ldots x_4$ and $x_5 \\ldots x_{10}$, but I think you can build a generating function this way:\n\nIf none of the $x_1 \\ldots x_4$ are 1, then any of the $x_5 \\ldots x_{10}$ can be 0 or 1 freely, so we get the generating function $(1+x^2)^6$ to count these solutions.\n\nIf exactly one of the $x_1 \\ldots x_4$ is 1, then three of the $x_5 \\ldots x_{10}$ must be zero, but the remainder are free, yielding the generating function $4x^3(1+x^2)^3$.\n\nIf exactly two of the $x_1 \\ldots x_4$ are 1, then five of the $x_5 \\ldots x_{10}$ must be zero with only one free, so we get the generating function $6x^6(1+x^2)$.\n\nIf three or four of the $x_1 \\ldots x_4$ are 1, then all of the $x_5 \\ldots x_{10}$ must be zero, so we get the generating function $4x^9+x^{12}$.\n\nWe can add all of these up to get the generating function: $$(1+x^2)^6 + 4x^3(1+x^2)^3 + 6x^6(1+x^2) + 4x^9+x^{12}$$ Like @Fabio Somenzi, when I calculate I also find 87 to be the answer.\n\n\u2022 Thank you very much for your answer! This is exactly what I was looking for. \u2013\u00a0GKiu Feb 24 '17 at 19:07", "date": "2019-09-15 19:53:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2158099/counting-integer-solutions-with-conditional-restrictions", "openwebmath_score": 0.9056128263473511, "openwebmath_perplexity": 103.77348493853292, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308772060157, "lm_q2_score": 0.865224073888819, "lm_q1q2_score": 0.8549546231115212}}
{"url": "https://www.jiskha.com/questions/1870908/a-bag-contains-8-red-marbles-5-white-marbles-and-6-blue-marbles-you-draw-4-marbles-out", "text": "# statistics probability\n\nA bag contains 8 red marbles, 5 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. Find the following probabilities and round to 4 decimal places.\n\na. The probability that all the marbles are red is\n\nb. The probability that none of the marbles are red is\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n1. prob(all 4 red) = (8/19)(7/18)(6/17)(5/16) = 35/1935\nor\nprob(all red) = C(8,4)/C(19,4) = 70/3876 = 35/1938\n\nProb(none red) = C(11,4)/C(19,4) = 330/3876 = 55/646\nor\nprob(none red) = (11/19)(10/18)(9/17)(8/16) = 55/646\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n\n## Similar Questions\n\n1. ### Math\n\nA bag contains 6 red marbles, 3 white marbles, and 7 blue marbles. You pick a marble without looking. Find the probability of drawing a white marble. Can someone please explain how I would work this out?\n\n2. ### statistics\n\nA bag contains 7 red marbles, 6 white marbles, and 10 blue marbles. You draw 4 marbles out at random, without replacement. What is the probability that all the marbles are red? What is the probability that exactly two of the\n\n3. ### math\n\nA bag contains 5 red marbles, 6 white marbles, and 8 blue marbles. You draw 5 marbles out at random, without replacement. What is the probability that all the marbles are red? The probability that all the marbles are red is? What\n\n4. ### Math\n\nA bag contains 9 red marbles, 8 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. What is the probability that all the marbles are red? 1 The probability that all the marbles are red is...?\n\n1. ### math\n\nThere are 5 red marbles, 6 blue marbles, e green marbles, and 2 yellow marbles in a bag. What percent or fraction represents the number of blue and green marbles in the bag?\n\n2. ### math\n\nA bag contains 8 red marbles, 5 blue marbles, 8 yellow marbles, and 6 green marbles. What is the probability of choosing a red marble if a single choice is made from the bag? is it 8/27 ?\n\n3. ### Math :)\n\nIn a bag of 10 marbles, there are 5 blue marbles, 3 red marbles, and 2 white marbles. Complete the probability distribution table for drawing 1 marble out of the bag. Draw a: Probability Blue marble 5/10 Red marble 3/10 White\n\n4. ### statistics\n\na bag contains 4 blue marbles,6 red marbles, 10 yellow marbles and 6 green marbles.What is the probability of not drawing a red marble?\n\n1. ### Math\n\nThe ratio of white marbles to blue marbles in Connie's bag of marbles is equal to 2:3. There are more than 20 marbles in the bag. What is a possible number of white marbles and blue marbles in the bag?\n\n2. ### Stat\n\nA bag contains 5 red marbles, 9 white marbles, and 5 blue marbles. You draw 4 marbles out at random, without replacement. What is the probability that all the marbles are red?\n\n3. ### Maths\n\nA bag contains marbles of three different colours red, blue , yellow. 2/5 of the marbles are red. The ratio of the number of blue marbles to the number of yellow marbles is 5:7. What fraction of the marbles in the bag is blue?\n\n4. ### Math\n\nKen had a number of colored marbles in a bag. 1/4 of the marbles were red, 2/3 of the remaining marbles were blue, and the rest were yellow. Given that there were 120 red and yellow marbles altogether, how many marbles were there", "date": "2021-12-08 06:35:25", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1870908/a-bag-contains-8-red-marbles-5-white-marbles-and-6-blue-marbles-you-draw-4-marbles-out", "openwebmath_score": 0.8674394488334656, "openwebmath_perplexity": 394.5178896418999, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137931962463, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.854938535787201}}
{"url": "http://www.physicsforums.com/showthread.php?p=4042187", "text": "## How do I evaluate this log function?\n\n1. The problem statement, all variables and given/known data\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nI assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that.\n\n(log23)(log34)(log45) ... (log3132)\n PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks\n Recognitions: Homework Help Use the fact that $$\\log_b a = \\frac{\\log_c a}{\\log_c b}$$ for any positive, real numbers a, b and c (with c > 1).\n That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?\n\n## How do I evaluate this log function?\n\n Quote by feihong47 That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?\nYou use LaTeX, that is, the [ itex ][ /itex ] -tags (remove the spaces) . There's a guide in the forums somewhere if you're not familiar with it, I'll edit a link to it to this post if I find it. You can also open the \"LaTeX Reference\" by clicking the \u03a3 in the toolbar.\n\nIf you don't have to write anything complicated, you can just use the quick symbols and x2 and x2 buttons in the toolbar. LaTeX looks a lot nicer & it's easier to read, though.\n\nEDIT: Here's the LaTeX guide.\n\nMentor\n Quote by feihong47 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that. (log23)(log34)(log45) ... (log3132)\nYou could also approach this as follows:\n\nLet $\\displaystyle y=(\\log_{2}3)\\,(\\log_{3}4)\\,(\\log_{4}5)\\,\\dots\\,( \\log_{31}32)$\n\nThen, $\\displaystyle 2^y=2^{(\\,(\\log_{2}3)\\,(\\log_{3}4)\\,(\\log_{4}5)\\, \\dots\\,( \\log_{31}32)\\,)}$\n\nBy laws of exponents and the definition of a logarithm,\n\n$2^{(\\,(\\log_{2}3)\\,(\\log_{3}4)\\,(\\log_{4}5)\\, \\dots\\,( \\log_{31}32)\\,)}$\n$=\\left(2^{(\\log_{2}3)}\\right)^{(\\,(\\log_{3}4)\\,( \\log_{4}5)\\, \\dots\\,( \\log_{31}32)\\,)}$\n\n$=3^{(\\,(\\log_{3}4)\\,( \\log_{4}5)\\, \\dots\\,( \\log_{31}32)\\,)}$\n\n$=\\left(3^{(\\log_{3}4)}\\right)^{(\\,( \\log_{4}5)\\, \\dots\\,( \\log_{31}32)\\,)}$\n\n$=4^{(\\,( \\log_{4}5)\\, \\dots\\,( \\log_{31}32)\\,)}$\n\netc.\n\n$=\\left(31^{(\\log_{31}32)}\\right)$\n\n$=32$\n Recognitions: Gold Member Science Advisor Staff Emeritus So, just in case others misunderstand, $2^y= 32= 2^5$ and therefore, y= 5 as the original poster said.\n Very nice approach. Thanks!", "date": "2013-05-25 17:27:22", "meta": {"domain": "physicsforums.com", "url": "http://www.physicsforums.com/showthread.php?p=4042187", "openwebmath_score": 0.6476492285728455, "openwebmath_perplexity": 953.2699313122341, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137889851291, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.85493852552687}}
{"url": "http://math.stackexchange.com/questions/54088/what-is-the-difference-between-max-and-sup", "text": "# What is the difference between max and sup?\n\nI am studying KS (Kolmogorov-Sinai) entropy of order q and it can be defined as\n\n$$h_q = \\sup_P \\left(\\lim_{m\\to\\infty}\\left(\\frac 1 m H_q(m,\u03b5)\\right)\\right)$$\n\nWhy is it defined as supremum over all possible partitions P and not maximum?\n\nWhen do people use supremum and when maximum?\n\n-\nI am not sure either of the tags fit. I do not know the term \"KS entropy\" so I cannot suggest a better retag. The essence of \"what is $\\sup$ and what is $\\max$ was answered on this website before. Here is one answer, math.stackexchange.com/questions/18605/\u2026 \u2013\u00a0 Asaf Karagila Jul 27 '11 at 15:55\nBased on a quick google search I believe [dynamical-systems] fit well into this question. I (somewhat unwillingly) left [order-theory] since your second question can be generally answered within its confines. As my previous comment suggests, the second answer was indeed answered before (many times as well), I do not vote to close as the first answer may or may not been addressed to in other questions on the site. \u2013\u00a0 Asaf Karagila Jul 27 '11 at 16:01\n@Asaf, KS stands for Kolmogorov-Sinai, I saw that question earlier, but I still don't get when people use sup and when max. It's probably coz I need to spend more time to think over all the answers there. But thanks for the link anyway. Pls feel free to retag. \u2013\u00a0 oleksii Jul 27 '11 at 16:03\n\"'Sup\" is a hip way of asking \"How are you?\" \u2013\u00a0 oosterwal Jul 27 '11 at 20:06\n@oleksii: You have received several answers, all are very good in answering the question in the title. None of which answer the question \"Why is it defined as supremum over all possible partitions $P$ and not maximum?\". While I respect your wish to accept the answer, it would be nice to indicate that you have at least deduced the answer of the seemingly unanswered question from the accepted answer. Otherwise, this has become a duplicate of a question that has been answered here many times before. \u2013\u00a0 Asaf Karagila Jul 28 '11 at 0:21\n\nIf the maximum exists, then the supremum and maximum are the same. However sometimes the maximum does not exist, and there is no maximal element. In this case it still makes sense to talk about a least upper bound.\n\nThe classic example is the set of all rationals whose square is less than or equal to $2$. That is the set $$A=\\left\\{ r\\in\\mathbb{Q}:\\ r^{2}\\leq2\\right\\}.$$\n\n$A$ has no maximal element, however it does have a supremum and $\\sup A=\\sqrt{2}$.\n\nAn even simpler example is the set of all reals that are strictly less than $2$: $$B=\\left\\{ r\\in\\mathbb{R}:\\ r<2\\right\\}.$$ This set has no maximum since for any $x\\in B$ the element $\\frac{x+2}{2}$ satisfies $x<\\frac{x+2}{2}<2$. However it is not hard to see that $\\sup B=2$.\n\n-\nTnx, this is a nice answer, +1 for easy samples \u2013\u00a0 oleksii Jul 27 '11 at 16:17\n\nConsider for example the set $X = (0,1)$. Then $$\\sup X=1$$ but $\\max X$ does not exist.\n\nGenerally, for a set $X\\subset {\\mathbb R}$, we define $x:=\\max X$ if $x\\in X$ and $$\\forall y\\in X, y\\leq x.$$\n\nWe define $x:=\\sup X$ if $$\\forall y\\in X, y\\leq x$$ and $$\\forall\\epsilon>0,\\quad\\exists y\\in X, \\quad\\text{s.t.}\\quad y>x-\\epsilon.$$\n\n-\nI quite like your first answer, it was very plain and easy to get [for me :)]. Tnx. \u2013\u00a0 oleksii Jul 27 '11 at 16:15\n\nSupremum need not be attained while maximum is. When maximum exist, maximum=supremum.\n\n-\n\nThe set of all negative numbers has a sup, which is 0, but not a max.\n\n0 is the smallest number that no negative number can exceed.\n\n-\n\nA maximum is the largest number WITHIN a set. A sup is a number that BOUNDS a set. A sup may or may not be part of the set itself (0 is not part of the set of negative numbers, but it is a sup because it is the least upper bound). If the sup IS part of the set, it is also the max.\n\n-", "date": "2015-04-25 16:11:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/54088/what-is-the-difference-between-max-and-sup", "openwebmath_score": 0.8134742975234985, "openwebmath_perplexity": 322.6675707835879, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137858267906, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.854938522777229}}
{"url": "http://math.stackexchange.com/questions/850949/which-is-the-correct-way-to-calculate-the-expected-value-of-a-shared-lottery-jac", "text": "# Which is the correct way to calculate the expected value of a shared lottery jackpot?\n\nI want to calculate the expected value of a ticket in a lottery game which gives players a probability $p$ of winning a jackpot prize of $j$ dollars. The total number of tickets in play is $t$.\n\nIf every winning ticket gets the full prize amount, the expected value for a ticket is given by $jp$. However, if winners must evenly split the prize in case of multiple winners, then the expected value depends on the number of winners $W$.\n\nThe expected number of winners is $tp$. The probability that the number of winners $W$ is $w = 0, 1, 2, \\dotsc$, follows a Poisson distribution with the expected number of winners as its parameter:\n\n$$P(W=w) \\sim Pois(tp) = \\frac{tp^we^{-tp}}{w!}$$\n\nI don't know how to get from there to calculating an accurate expected value for the ticket as a function of the number of tickets in play.\n\nIn reading online, I've found two different methods each used by several sources. If I'm following them correctly, then they give different results. My question is 1) which one is correct? 2) what is the error in reasoning (or in my understanding/implementation) in the incorrect method?\n\n## Method 1: Number of Winners\n\nThe first method calculates the probability that the number of winners $W$ will be $w = 0, 1, \\dotsc, t$, given that there is at least one winner:\n\n$$P(W=w | W>0) = \\frac{P(W>0|W=w)P(W=w)}{P(W>0)}$$\n\nWhere,\n\n\u2022 $P(W>0|W=w)$ is $\\left\\{ \\begin{array}{lr} 0 & : w = 0\\\\ 1 & : w > 0 \\end{array} \\right.$\n\u2022 $P(W=w)$ is the probability of $w$ winners: $\\frac{tp^we^{-tp}}{w!}$\n\u2022 $P(W>0)$ is the probability of more than one winner: $1 - P(W=0)$\n\nSo the expected value of the ticket is given by:\n\n$$p\\sum_{w=1}^{t} \\frac{j}{w}\\frac{P(W=w)}{1-P(W=0)}$$\n\nFor a numerical example, we'll tabulate the first few values of $P(W=w)$ for a lottery with a 1/34,220 chance of winning \\$100,000 jackpot, with 6,000 tickets in play, so$p = 1/34,220; j = 100,000; \\text{and } t = 6,000$$$\\begin{array}{c|c|c|c|c|} \\text{Winners} & \\text{Probability} & \\text{Conditional Probability} & \\text{Share} & \\text{Contribution } \\\\ w & P(W=w) & P(W=w|W>0) & j/w & (j/w)P(W=w|W>0) \\\\ \\hline 0 & 0.839 & 0 & \\text{\\0} & \\text{\\0} \\\\ \\hline 1 & 0.147 & 0.913 & \\text{\\100,000} & \\text{\\91,300} \\\\ \\hline 2 & 0.013 & 0.081 & \\text{\\50,000} & \\text{\\4,050} \\\\ \\hline \\end{array}$$ Summing the contribution column and multiplying by$p$gives an expected value of$2.79.\n\n## Method 2: Number of Other Winners\n\nThe second method calculates the probability that the number of total winners $W$ is $w = 0, 1, \\dotsc, t$, given that our ticket is a winner:\n\n$$P(W=w|Winner) = \\frac{P(Winner|W=w)P(W=w)}{P(Winner)}$$\n\nWhere,\n\n\u2022 $P(Winner)$ is the probability that our ticket is a winner: $p$\n\u2022 $P(Winner|W=w)$ is the probability that our ticket is a winner given $w$ winning tickets: $w/t$\n\u2022 $P(W=w)$ is the probability of $w$ winners: $\\frac{tp^we^{-tp}}{w!}$\n\nPlugging those figures in shows that $P(W=w|Winner)$ reduces to $P(W=w-1)$:\n\n$$\\frac{w}{t}\\frac{P(W=w)}{p} = \\frac{tp^{w-1}e^{-tp}}{(w-1)!} = P(W=w-1)$$\n\nSo the expected value is given by:\n\n$$p\\sum_{w=1}^{t}\\frac{j}{w}\\frac{tp^{w-1}e^{-tp}}{(w-1)!}$$\n\nUsing the same lottery numbers as above, the first few values of $w$ are given in the following table.\n\n$$\\begin{array}{c|c|c|c|c|} \\text{Winners} & \\text{Probability} & \\text{Conditional Probability} & \\text{Share} & \\text{Contribution } \\\\ w & P(W=w) & P(W=w|Winner) & j/w & (j/w)P(W=w|Winner) \\\\ \\hline 0 & 0.839 & 0 & \\text{n/a} & \\text{\\0} \\\\ \\hline 1 & 0.147 & 0.839 & \\text{\\100,000} & \\text{\\83,900} \\\\ \\hline 2 & 0.013 & 0.147 & \\text{\\50,000} & \\text{\\7,350} \\\\ \\hline \\end{array}$$\n\nSumming the contribution column and multiplying by $p$ gives an expected value of $2.67. ### Online Resources Which Use Method 2 Clearly the expected payout for the example lottery above cannot be both \\$2.79 and \\$2.67, but I'm having a difficult time reasoning my way to the correct method. Any hints will be appreciated! - ## 4 Answers Proof that Methods 2 and 3 are equivalent: Method 3 has a very intuitive appeal as the correct approach when interpreted as someone purchasing$allt$tickets. Now we'll show that Method 2 gives the same result if we use the binomial for the distribution of winning tickets instead of the Poisson. The expected value for Method 2 is: $$p\\sum_{w=1}^t\\frac jw {t-1 \\choose w-1}p^{w-1}(1-p)^{t-w}$$ $$=j \\sum_{w=1}^t \\frac {(t-1)!}{w!(t-w)!}p^{w}(1-p)^{t-w}$$ $$=\\frac jt \\sum_{w=1}^t {t \\choose w}p^{w}(1-p)^{t-w}$$ $$=\\frac jt (1-(1-p)^t )$$ which is also the result for Method 3. If we had started with the Poisson distribution as an approximation to the binomial, then the expected value for Method 2 reduces to $$\\frac jt(1-e^{-tp})$$ The Poisson will be a good approximation to the binomial when$t$is large,$p$is small and$tp$is moderate, which holds here. - Method 3: By linearity of expectation, the expected value of one ticket is$\\frac{1}{t}$times the expected total value of all the tickets. That expected value is$jP(W>0)=j(1-P(W=0))=j(1-(1-p)^t)$. Thus the expected value of one ticket is $$\\frac{j}{t}(1-(1-p)^t))$$ For the numerical values you use, I get \\$2.68. I'm not skilled enough at statistics to find the logical flaw in Method 1, but I suspect Method 2 is right (up to rounding).\n\nFollowup: it seems that this solution, or a variation, was listed under \"resources\" supporting the second solution. I didn't notice it on my first read-through due to wall of text.\n\n-\nThank you, vadim. Yeah, I had a link to this answer buried in the resources section, but your mentioning that expectation is linear helped me understand why it is correct without having to actually enumerate and sum the probability of each number of winners. That leaves me very confident that Method 2 is correct. However, I would like to find a good way of explaining why Method 1 is incorrect. \u2013\u00a0 cristoper Jun 29 '14 at 5:25\nI suspect that the problem with Method 1 is that the probability my ticket is a winner is no longer $p$ when we have conditioned on there being exactly $w$ winners. \u2013\u00a0 vadim123 Jun 29 '14 at 5:34\n\nThis is my own summary answer. If somebody has a better way of explaining why the first method in my question goes wrong (and what it does compute, if anything useful), please post an answer!\n\n## Method 1 is incorrect\n\nIf $P(Winner)$ is the probability that our ticket is a winner, then as user159813 pointed out, the expected payout from the lottery can be expressed as:\n\n$$\\sum_{w=1}^{t} \\frac{j}{t} P(Winner \\cap W=w)$$\n\nHowever, the payout calculated by Method 1 is:\n\n$$\\sum_{w=1}^t \\frac{j}{t} P(W>0 \\cap W=w)$$\n\nWhich is different because $P(W>0) \\ne P(Winner)$. We want $P(W=w)$ given that our ticket won, which is a lesser probability than given that any ticket won.\n\n## Method 2 is correct\n\nAs vadim123 points out in his answer, the correctness of Method 2 is confirmed by the nice, closed-form formula derived from the fact that the expected value of a single ticket is the same as the expected value of all tickets divided by $t$:\n\n$$\\frac{j}{t}(1-(1-p)^t)$$\n\n## Simulation\n\nI ran a simulation using the same numbers as the example in my question. The mean payout after the 10,000,000th iteration was about \\$2.64. After about 5,000,000 iterations, the mean seemed to oscillate about the$2.68 line.\n\n-\n\nThere is a problem in Method 1. Letting $Y$ represent amount won and $W$ be number of winners we see that $$E(Y)=\\sum_{w=0}^{t}\\frac{j}{w}P(Winner\\cap W=w)$$\n\nThe second method does this correctly by calculating $$P(Winner\\cap W=w)=P(W=w|Winner)P(Winner)=P(W=w-1)P(Winner)=P(W=w-1)p$$\n\nas you have above. Now the problem with Method 1 is that it says that $$P(Winner\\cap W=w)=P(W=w|W>0)P(Winner)$$ which is not true. I would guess the method this person was going for is actually $$P(Winner\\cap W=w)=P(Winner| W=w)P(W=w)$$\n\n-\nI don't see where Method 1 says $P(Winner \\cap W=w)=P(W=w|W>0)P(Winner)$. However, I do see where it says it equals $P(W=w|W>0)P(W>0)$. And looking at your formulation of the expectation function, it is easier for me to see that Method 1 is wrong because $P(W=w|W>0)P(W>0) \\ne P(W=w|Winner)P(Winner)$ \u2013\u00a0 cristoper Jun 29 '14 at 20:34\nAnd should the index in your sum start at $w=1$ to avoid the divide by zero? $P(Winner \\cap W = 0) = 0$ anyway. \u2013\u00a0 cristoper Jun 29 '14 at 20:47", "date": "2015-07-01 11:58:25", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/850949/which-is-the-correct-way-to-calculate-the-expected-value-of-a-shared-lottery-jac", "openwebmath_score": 0.9643638134002686, "openwebmath_perplexity": 418.68737042844606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013790564298, "lm_q2_score": 0.870597266729631, "lm_q1q2_score": 0.8549385219560822}}
{"url": "http://mezza-penna.it/teax/semi-log-regression.html", "text": "# Semi Log Regression\n\nLog the Bugs in the Quality Center (Bug Logging and Tracking Tool) and track them. Measure distance each band traveled 3. This algorithm uses two k -nearest neighbor regressors with different distance metrics, each of which labels the unlabeled data for the other regressor where the labeling confidence is estimated through consulting the influence of the labeling of unlabeled examples on the labeled ones. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Why is it that when you log-transform a power function, you get a straight line? To show you, let's remember one of the most fundamental rules of algebra: you can do anything you want to one side of an equation - as long as you do the exact same thing to the other side (We just LOVE that rule!). For many real-world problems, however, acquiri. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Abstract Semi-supervised methods use unlabeled data in addition to labeled data to con-struct predictors. This page aims at providing to the machine learning researchers a set of benchmarks to analyze the behavior of the learning methods. Poisson regression for contingency tables, a type of generalized linear model. Quadratic regression. The Regression Analysis Method Page Content Space Toy Co. This page allows performing logarithmic regressions (logarithmic least squares fittings). Semi-supervised learning and multi-task learning are two of the approaches that have been proposed to alleviate this problem. Introduction. However, I dont know to interpret the coefficient. General Linear Models: Modeling with Linear Regression I 3 0 2 4 6 8 10 12 02040608010 % Hunt lo g A r e a 0 We can see that by log-transforming the y-axis we have now linearized the trend in the data. Articulate assumptions for multiple linear regression 2. A prediction is an estimate of the value of $$y$$ for a given value of $$x$$, based on a regression model of the form shown in Equation \\ref{eq:regmod4}. Now if your intuition leads you to. We propose combination methods of penalized regression models and nonnegative matrix factorization (NMF) for predicting. indicates that the instantaneous return for an additional year of education is 8 percent and the compounded return is 8. Hello! I need to make some kind of line graph with one y variable and two x variables where only the x-axis is on a logarithmic scale. Log denotes the natural logarithm. Below is an example for unknown nonlinear relationship between age and log wage and some different types of parametric and nonparametric regression lines. In this page, we will discuss how to interpret a regression model when some variables in the model have been log transformed. We interpret the various log, log and semi-log coefficients and use the estimated regression model to make prediction and build a confidence interval for the prediction. eA B = eA=eB 2 Why use logarithmic transformations of variables Logarithmically transforming variables in a regression model is a very common way to handle sit-. The standard data points (concentration vs. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. y is the response variable and x1, x2, and x3 are explanatory variables. A General Note: Logarithmic Regression. Because every disease has its unique survival pattern, it is necessary to find a suitable model to simulate followups. For motivational purposes, here is what we are working towards: a regression analysis program which receives multiple data-set names from Quandl. Testing Kepler\u2019s Third Law October 21, 2009 In this activity, we will use linear regression on our calculators to test Kepler\u2019s Third Law of Plan-etary Motion. 4 The Cox model, in contrast, leaves the baseline hazard function (t) = logh 0(t) unspeci ed: logh i(t) = (t) + 1x i1 + 2x i2. ECONOMICS 351* -- Stata 10 Tutorial 6 M. The transformation of the data set from y vs. Part (c) shows a log-log function where the impact of the dependent variable is negative. log a 1 = 0 because a 0 = 1 No matter what the base is, as long as it is legal, the log of 1 is always 0. 724 Pseudo R2 = 0. The LINEST function returns an array of coefficients, and optional regression statistics. Explain the primary components of multiple linear regression 3. Minima of positive definite and positive semidefinite functions. , the Negative Binomial regression model). Let's apply some simple regression analysis (see footnote below) to the question. Estimation with correctly interpreted dummy variables in semilogarithmic equations. However, I dont know to interpret the coefficient. Set of tools to fit a linear multiple or semi-parametric regression models and non-informative right-censoring may be considered. A chemical reaction A\u2192B is carried out in a batch reactor. You either can't calculate the regression coefficients, or may introduce bias. Semi-log and Log-log plots Posted 02-04-2010 (5361 views) I have tried Googling and searching the SAS documentation, but I cannot find any syntax to create a semi-log or log-log plot. We also study the transformation of variables in a regression and in that context introduce the log-log and the semi-log regression models. I was in (yet another) session with my analyst, \"Jane\", the other day, and quite unintentionally the conversation turned, once again, to the subject of \"semi-log\" regression equations. I know that usually having a linear-log model, an increase in x (GDP) by one percent is associated with an increase in y by (\u03b21/100) units which would be for CRES (2,73/100). Charles says: July 22, 2015 at 2:41 pm. Four Parameter Logistic (4PL) Regression. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. One axis is plotted on a logarithmic scale. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. COREG, is proposed. More speci\ufb02cally, one has found a point in a graph one is interested in, and now wants. to Leb esgue measure on [ ; + ] and denote f (x) the sp ectral. 22 Prob > chi2 = 0. A log-linear plot or graph, which is a type of semi-log plot. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Bivariate Regression - Part I - Page 1. If I set the chart as the semi-log scale (semi-logarithmic axes), the regression line cannot be shown to a straight line. eA B = eA=eB 2 Why use logarithmic transformations of variables Logarithmically transforming variables in a regression model is a very common way to handle sit-. This kind of plot is useful when one of the variables being plotted covers a large range of values and the other has only a restricted range - the advantage being that it can bring out features in the data. Polynomial regression. It is partly a matter of custom. Our results indicate that models estimated with a square root link function perform much better than those with log- or linear-link functions. A General Note: Logarithmic Regression. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. log a a x = x The log base a of x and a to the x power are inverse functions. linear regression on levels of and log-transformed costs, gamma GLM with log-link, and the log-normal distribution, are not among the four best per-forming models with any of our chosen metrics. The present paper develops a mixture regression model that allows for distributional flexibility in modelling the likelihood of a semi-continuous outcome that takes on zero value with positive probability while continuous on the positive half of the real line. \u2219 0 \u2219 share We consider semi-supervised regression when the predictor variables are drawn from an unknown manifold. What is the interpretation of this coefficient? 2. If B1=2, for instance, we could say that \u2019this model shows that factor X1 increases the predicted log count by 2 (all other factors held constant)\u2019 because equation 1b- equation 1a= B1. DSOM 309 Chapter 16. Also known as elasticity interpretation. A log-linear plot or graph, which is a type of semi-log plot. Is it possible (and how) to transform one of these tables to semi-annual/annual table? If not, is there a way to calculate semi-annual/annual tables based on other data from this web site? Note - The issue is to calculate values based on free public available data, assuming paid access to CRSP etc. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. In survival analysis, the proportional hazard model, also called the Cox model, is a classical semi-parameter method. Once you have used Excel to create a set of regular axes, converting the axes to semi-logarithmic axes in Excel is far from difficult. The main goal of this paper is to provide a fully probabilistic approach to modelling crime which reflects all uncertainties in the prediction of offences as well as the. And i do get R square (R2)= 0. 10 dan nilai VIF kurang dari 10. Nathanael Johnson (@savortooth on Twitter) is Grist's senior writer and the author of two books. 7 PROC ROBUSTREG Eample: Log-Log Regression With Weighted Outliers Example: Log-Log Regression With Weighted Outliers SAS/STAT \u00ae 9. A semi-log graph is useful when graphing exponential functions. A test based on a modified LP regression that is consistent in both directions is provided. The equations of straight lines on logarithmic graph paper One purpose of logarithmic graph paper is simply to put wide ranges of. API Reference\u00b6. n\u2192\u221e, standard knn regression achieves the minimax bound on the MSE n\u22122 2+d up to log factors. All equations of the form. 1 Linear model: Yi = + Xi + i Recall that in the linear regression model, logYi = + Xi + i, the coef\ufb01cient gives us directly the change in Y for a one-unit change in X. Question: A) Determine Which Regression Model (linear, Logarithmic And Semi-logarithmic) Will Best Represent The Relationship Between X (independent Variable) And Y (dependent Var- Iable) Below. The regression analysis fits the data, not the graph. Select the tab. In other words, if you go this route, you\u2019ll need to do some research. Segmented regression, also known as piecewise regression or 'broken-stick regression', is a method in regression analysis in which the independent variable is partitioned into intervals and a separate line segment is fit to each interval. Interpretation of logarithms in a regression. The regression analysis fits the data, not the graph. 32 months and LT 90 of 6. y = \u03bb a \u03b3 x. for linear regression has only one global, and no other local, optima; thus gradient descent always converges (assuming the learning rate \u03b1 is not too large) to the global minimum. Spearman's correlation test was used to measure the correlation between two non-normally distributed variables or one normally with one non-normally distributed variable. Econometrics Working Paper EWP1101, Department of Economics, University of Victoria. The main difference is that a regression line is a straight line that represents the relationship between the x and y variable while a LOESS line is used mostly to identify trends in the data. The residual in a Cox regression model is not as simple to compute as the residual in linear regression, but you look for the same sort of pattern as in linear regression. We have some set of possible inputs, X, and a set of possible labels, Y. 08 * Density Ln + 583. The transformation of the data set from y vs. Machine Learning with Java - Part 1 (Linear Regression) Most of the articles describe \"How to use machine learning algorithm in Python?\". Under this setup, the localization parameter of the response variable distribution is modeled by using linear multiple regression or semi-parametric functions, whose non-parametric components may be approximated by natural cubic spline or P-splines. Polynomial Regression Analysis. yhat=b\u2080+b\u2081x which semi-log model transforms only the explanatory variable. The supported. How to access courses from \u2018Coursera\u2019 for FREE (with certificate). To create a log-log graph, follow the steps below for your version of Excel. In this article , we are going to discuss \"How to use the machine learning alogithm with Java?\". for which x<=0 if x is logged. log(AB) = logA+logB7. Namely, by taking the exponential of each side of the equation shown above we get. One new wrinkle we will add to this discussion is the use of faceting when developing plots. DNA microarray is a useful technique to detect thousands of gene expressions at one time and is usually employed to classify different types of cancer. Our task is to model the conditional probability p(yjx) for any pair (x;y) such that x2Xand y2Y. The estimate of \u03b22 is 0. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. It's not the fanciest machine learning technique, but it is a crucial technique to learn for many reasons:. The regression analysis fits the data, not the graph. We will use algebra and linear regression. Avoiding Common Math Mistakes. Once you have used Excel to create a set of regular axes, converting the axes to semi-logarithmic axes in Excel is far from difficult. Plot the location of each band (size and distance traveled) 5. The constant \u03b1in this model represents a kind of log-baseline hazard, since loghi(t)=\u03b1[or hi(t)=e\u03b1]whenallofthex\u2019s are zero. 4 Log-Linear Models We now describe how log-linear models can be applied to problems of the above form. A novel weight strategy is presented to improve the prediction and its recursive algorithm is formulated, which adopts the incremental and decremental learning mechanism to update. This means that we can now use a simple linear regression model to describe the relationship. Regression analysis (integrated) Regression estimate (integrated). There have been a number of papers written on semi-parametric estimation methods of the long-memory exponent of a time series, some applied, others theoretical. If you have a nonlinear relationship, you have several options that parallel your choices in a linear regression model. When I add a linear trendline, Excel draws a straight line, even though a linear function should look curved when plotted on a logarithmic scale. the semi-log graph, which has a logarithmic vertical scale and a linear horizontal scale, as shown below. How can I fit my X, Y data to a polynomial using LINEST? As can be seem from the trendline in the chart below, the data in A2:B5 fits a third order polynomial. e-Exponential regression. DataAnalysis For Beginner This is R code to run semi-supervised regression based on Principal Component Analysis and Partial Least Squares (PCAPLS). regress definition: 1. GEE approach is an extension of GLMs. Of course, the ordinary least squares coefficients provide an estimate of the impact of a unit change in the independent variable, X, on the dependent variable measured in units of Y. measurement) are plotted on semi-log axes and a cubic regression curve is fitted through the points. Segmented regression analysis can also be performed on multivariate data by partitioning the various. Linear, Logarithmic, Semi-Log Regression Calculator By AAT Bioquest. So plotting Y and X*, where X* is the log of X, and performing a linear regression, you obtain a slope and intercept. We propose combination methods of penalized regression models and nonnegative matrix factorization (NMF) for predicting. Insert regression model into ggplot2. Regression analysis is a quantitative tool that is easy to use and can provide valuable information on financial analysis and forecasting. On a semi-log plot the spacing of the scale on the y-axis (or x-axis) is proportional to the logarithm of the number, not the number itself. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. Polynomial regression is commonly used to analyze the curvilinear data and this happens when the power of an independent variable is more than 1. log (Y) = a + b X The equation is estimated by converting the Y values to logarithms and using OLS techniques to estimate the coefficient of the X variable, b. com In Robust Regression, the outliers need not be disregarded: weights can be assigned and incorporated in the regression. Life Sciences is a solution especially designed for researchers and practitioners of life sciences who want to apply well-known and validated methods to analyze their data and build on their research. Nowadays, Semi-Supervised Learning lies at the core of the Machine Learning field trying to effectively exploit unlabeled data as much as possible, together with a small amount of labeled data aiming to improve the predictive performance. , the percentage. Obtain estimates of the Bifurcation Ratio, R B, the Length Ratio, R L, and the Area Ratio, R A, using the data tabulated below. commonly used in practice. gaussianprocess. Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. Articulate assumptions for multiple linear regression 2. Estimate the doubling time from the slope. Journal of the American Statistical Association: Vol. \u2212Maximum likelihood method. If B1=2, for instance, we could say that 'this model shows that factor X1 increases the predicted log count by 2 (all other factors held constant)' because equation 1b- equation 1a= B1. is the core plug semi-log regression, and the dashed green lines indicate the approximate 90% boundaries of the core plug data. Byrne , d Igor Chourpa a and Emilie Munnier a. Poisson regression for contingency tables, a type of generalized linear model. logarithmic model. , semielasticity), with respect to the dummy regressor taking values of 1 or 0, can be obtained as (antilog of estimated 02) - 1 times 100, that is, as. We have some set of possible inputs, X, and a set of possible labels, Y. Using LINEST for Nonlinear Regression in Excel. Nonlinear functional. It is closely related to semi-supervised learning based on support vector regression (SVR). Linear and Logarithmic Interpolation Markus Deserno Max-Planck-Institut f\u02dcur Polymerforschung, Ackermannweg 10, 55128 Mainz, Germany (Dated: March 24, 2004) One is occasionally confronted with the task of extracting quantitative information out of graphs. 08 * Density Ln + 583. Before you model the relationship between pairs of. The left panel depicts a semi-log model and the right panel depicts a polynomial model. A logarithmic curve fit is generally used with data that spans decades (10 0, 10 1, 10 2, and so on). Identify and define the variables included in the regression equation 4. Level-level regression is the normal multiple regression we have studied in Least Squares for Multiple Regression and Multiple Regression Analysis. That's because logarithmic curves always pass through (1,0) log a a = 1 because a 1 = a Any value raised to the first power is that same value. But, i don't understand why it was said that the value closer to 1 is a better indicator to show that my standard curve is good to determine the protein concentration. The data has 1,000 observations on 4 variables. edu Abstract Large amounts of labeled data are typically required to train deep learning models. Regression is nonlinear when at least one of its parameters appears nonlinearly. It will provide four different linear regressions: linear-linear, log-linear, linear-log and log-log. It is estimated by regression using the wavelet coefficients of the time series, which are dependent when d \u2260 0. With ANOVA, you assign people to treatments, and all sorts of. Functional regression How to relate functional responses to scalar, explanatory variables? Available functional regressions models: Semi-parametric approaches: I additive effects models (Ramsay & Silverman, 2005) (R package fda on CRAN and R-Forge) I multiplicative effects models (Chiou et al. These correspond to a latent variable with the extreme-value distribution for the maximum and minimum respectively. Non-Linear Relationships 169 8 resulting least squares regression line will give us and estimate for hc Log-log and semi-log plots Graphs of log(y) vs. The logarithmic fit calculates the least squares fit through points by using the following equation: where a and b are constants and ln is the natural logarithm function. Since then, regression has been studied. While existing semi-supervised methods have shown some promising empirical performance, their development has been based largely basedon heuristics. Abbott Preparing for Your Stata Session Before beginning your Stata session, use Windows Explorer to copy the Stata- format dataset auto1. Logarithmic. methods for classi\ufb01cation and regression problems. Byrne , d Igor Chourpa a and Emilie Munnier a. \u2022 On the computers in Dunning 350, the default Stata working directory is. Polynomial regression. In addition, we modify our underlying approximate homomorphic encryption scheme for performance improvement. This is called a semi-log estimation. Fitting Parametric and Semi-parametric Conditional Poisson Regression Models with Cox's Partial Likelihood in Self-controlled Case Series and Matched Cohort Studies Stanley Xu1, Paul Gargiullo2, John Mullooly3, David McClure1, Simon J. Downloadable! Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. I have a Scatter chart with logarithmic scale on the horizontal axis and linear scale on the vertical axis. Now, find the least-squares curve of the form c 1 x + c 2 which best fits the data points (x i, \u03c6 i). 05/26/18 - Large amounts of labeled data are typically required to train deep learning models. 397973 * Density Ln^2 + 0. All equations of the form = form straight lines when plotted semi-logarithmically, since taking logs of both sides gives. We establish a Central Limit Theorem (CLT) for the resulting estimator. 08 * Density Ln + 583. Logistic regression & stochastic gradient descent Parametric Fast to train and evaluate Easy to incrementally train x 2Rn;y 2f 1;+1g P(yjx) = 1 1 + exp( ywTx) maximize w Y m P(y(m)jx(m)) minimize w XM m=1 log(1 + exp( y(m)wTx(m))) M might be giant, or you might not have access to them all at one time. LOG-PERIODOGRAM REGRESSION OF TIME SERIES WITH LONG RANGE DEPENDENCE 1 1. Log-log and semi-log plots Graphs of log(y) vs. This algorithm uses two k -nearest neighbor regressors with different distance metrics, each of which labels the unlabeled data for the other regressor where the labeling confidence is estimated through consulting the influence of the labeling of. Generalized and \"working\" Wald and score tests for regression coefficients in the class of semi-parametric marginal generalized linear models for cluster correlated data (Liang and Zeger, 1986) are proposed, and their asymptotic distribution examined. Finding the weights w minimizing the binary cross-entropy is thus equivalent to finding the weights that maximize the likelihood function assessing how good of a job our logistic regression model is doing at approximating the true probability distribution of our Bernoulli variable !. The excessive number of concepts comes because the problems we tackle are so messy. 0 open source license. We extend these results by establishing the exact sampling. Fully speci\ufb01ed by a mean function and covariance function. While existing semi-supervised methods have shown some promising empirical performance, their development has been based largely based on heuristics. It is partly a matter of custom. 10 dan nilai VIF kurang dari 10. Calculate the standard deviation of the log10 residual, then square it and multiply it by 1. Linear regression is one of the most popular statistical techniques. You wish to have the coefficients in worksheet cells as shown in A15:D15 or you wish to have the full LINEST statistics as in A17:D21. Set of tools to fit a linear multiple or semi-parametric regression models and non-informative right-censoring may be considered. The data has 1,000 observations on 4 variables. Some using Fourier methods, others using a wavelet-based technique. This is the only graph type that will work; other graph types permit logarithmic scales only on the Y axis. The elasticity evaluated at the mean is:. Polynomial regression is commonly used to analyze the curvilinear data and this happens when the power of an independent variable is more than 1. The parameter d is the one of interest. Dengan regresi Semi-Log yaitu variabel dependen dalam bentuk logaritma natural dan semua variabel independen tetap dirubah, dapat disimpulkan tidak terdapat multikolinearitas hal ini ditunjukkan oleh nilai Tolerance di atas 0. 1539 B 70000000 52300000 1. What is the interpretation of this coefficient? 3. by using Douglas\u2019s \u201claboriously compiled\u201d data to fit the linear regression Log(P/C) = b + kLog(L/C) by ordinary least squares. Econometrics and the Log-Linear Model By Roberto Pedace If you use natural log values for your dependent variable ( Y ) and keep your independent variables ( X ) in their original scale, the econometric specification is called a log-linear model. This produces the following output. race smoke ptl ht ui Logistic regression Number of obs = 189 LR chi2(8) = 33. It is equivalent to converting the y values (or x values) to their log, and plotting the data on lin-lin scales. This paper proposes a risk prediction model using semi-varying coefficient multinomial logistic regression. The example data can be downloaded here (the file is in. Kennedy, P. 13 and Table 15. Goodness-of-fit is a measure of how well an estimated regression line approximates the data in a given sample. generate lny = ln(y). After my previous rant to discussion with her about this matter, I've tried to stay on the straight and narrow. Even when there is an exact linear dependence of one variable on two others, the interpretation of coefficients is not as simple as for a slope with one dependent variable. Logarithmic regression. In such cases, applying a natural log or diff-log transformation to both dependent and independent variables may. One such measure is the correlation coefficient between the predicted values of $$y$$ for all $$x$$-s in the data file and the. The regression results based on ECM [Table 15. A General Note: Logarithmic Regression. 61 months ( Table 6 ). Sadly, most browser are unable to play this format. The regression analysis fits the data, not the graph. logarithmic model. eA+B = eAeB 10. The comparison of methods experiment is critical for assessing the systematic errors that occur with real patient specimens. Change one or both axes to a logarithmic scale. A novel weight strategy is presented to improve the prediction and its recursive algorithm is formulated, which adopts the incremental and decremental learning mechanism to update. The regression coefficient associated with the Z term (i. In this paper, we compare the Fourier and wavelet approaches to the local regression method and to the local Whittle method. Gowher, The exponential regression model presupposes that this model is valid for your situation (based on theory or past experience). In instances where both the dependent variable and independent variable(s) are log-transformed variables, the relationship is commonly referred to as elastic in econometrics. Gaussian processes GP(m(x),k(x,x\u2019)) Distribution over functions. A test based on a modified LP regression that is consistent in both directions is provided. Hence the term proportional odds logistic regression. Namely, by taking the exponential of each side of the equation shown above we get. Polynomial Regression Analysis. 952<1 since it is evidently. logistic regression (S2MLR) model which exploits both hard and soft labels. Plotting with Microsoft Excel 2 form of categories. ab-Exponential regression. See the Topic 6. This curve fit cannot be used to fit negative data or data equal to. API Reference\u00b6. Results are generated immediately, no external software needed. A Poisson regression model is sometimes known as a log-linear model. [1 point] Suppose the regression model is logarithmic: log(Y ) = \u03b21 + \u03b22 log(X) + u. In the case of linear regression, one additional benefit of using the log transformation is interpretability. Obtain your results in a few simple clicks without having to leave MS Excel where your data is stored. The standard data points (concentration vs. Part two explains semi-parallel logisitic regression in R based on iteratively reweighted least squares (equivalent to glm), with and without covariates. Pairwise Log-rank Test P Value Table Semi-Parametric Cox Regression Cox Regression , also known as Cox proportional hazard regression, assumes that if the proportional hazards assumption holds (or, is assumed to hold), then it is possible to estimate the effect parameter(s) without any consideration of the hazard function. If I set the chart as normal scale (numeric-numeric), the regression line can be shown to a straight line. Suppose a data set is actually following the trend of some hidden exponential function y = a b x. This kind of plot is useful when one of the variables being plotted covers a large range of values and the other has only a restricted range - the advantage being that it can bring out features in the data. Quadratic regression. Excel 2010 or 2007. Thus, software originally devel-. For instance, if you are graphing time versus bacterial growth. 952<1 since it is evidently. A log-linear plot or graph, which is a type of semi-log plot. This formula estimates the doubling time, which does not depend on the value of Y, only on the slope at t 0. Data can be directly from Excel or CSV. For example, if the raw output ($$y'$$) of a linear model is 8. Semi-log and Log-log plots Posted 02-04-2010 (5361 views) I have tried Googling and searching the SAS documentation, but I cannot find any syntax to create a semi-log or log-log plot. It was not until the early 19th century that Gauss and Legendre developed a systematic pro-cedure: the least-squares method. Semi-Supervised Classification with Graph Convolutional Networks. Horton's Laws - Example Jorge A. 2000 Simcoe Street North Oshawa, Ontario L1G 0C5 Canada. Plotting with Microsoft Excel 2 form of categories. Click Analyze, choose Nonlinear regression (not Linear regression) and then choose one of the semi-log or log-log equations from the \"Lines\" section of equations. Thus, software originally devel-. The focus of the paper is on a theoretical analysis of semi-supervised regression techniques, rather than the development of practical new algorithms and techniques. What is the interpretation of this coefficient? 3. webuse lbw (Hosmer & Lemeshow data). In a semilogarithmic graph, one axis has a logarithmic scale and the other axis has a linear scale. 13 and Table 15. OK, you ran a regression/fit a linear model and some of your variables are log-transformed. How can I fit my X, Y data to a polynomial using LINEST? As can be seem from the trendline in the chart below, the data in A2:B5 fits a third order polynomial. The main goal of this paper is to provide a fully probabilistic approach to modelling crime which reflects all uncertainties in the prediction of offences as well as the. Suppose a data set is actually following the trend of some hidden exponential function y = a b x. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. A little over three months ago, my wife and I sat down at the dining table to talk about the future. If the law were a perfect description of the situation, all the points on the log-log or semi-log plot would fall along a straight line. Stochastic gradient descent: take gradient. Excel 2010 or 2007. Under this setup, the localization parameter of the response variable distribution is modeled by using linear multiple regression or semi-parametric functions, whose non-parametric components may be approximated by natural cubic spline or P-splines. HTH Martin. the log-log graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below. The following lesson estimates a log, log and semi-log regression model. 72 Interpretation of Regression Coefficients: Elasticity and Logarithmic Transformation. The specific applications of log-linear models are where the output quantity lies in the range 0 to \u221e, for values of the independent variables X , or more immediately, the transformed quantities f i ( X. Using global regression to fit incomplete datasets Fitting models where the parameters are defined by multiple data sets Column constants Advice: Don't use global regression if datasets use different units Outlier elimination and robust nonlinear regression When to use automatic outlier removal When to avoid automatic outlier removal. To avoid this problem, we [\u2026]. 4669 D 17500000 10560000 0. Regression definition is - the act or an instance of regressing. Hazard regression models for pr(T2 > t|Z0) have been thoroughly studied in this scenario, for example, multiplicative hazards, additive hazards, and accelerated failure-time models. Valuation of Specific Crime Rates: Final Report William Alan Bartley Semi-log Regression with Index Crimes for Rent Equation Total Costs per Household of Crime Variables in Semi-log Regressions using metropolitan area percentage changes for the Manufacturing. Inverse regression. regression definition: 1. how all functions (linear, semi log, double log and exponential) can be applied. The first form of the equation demonstrates the principle that elasticities are measured in percentage terms. Spotfire uses a nonlinear regression method for this calculation. Log periodogram (LP) regression is shown to be consistent and to have a mixed normal limit distribution when the memory parameter d = 1. I like to use log base 10 for monetary amounts, because orders of ten seem natural for money: $100,$1000, $10,000, and so on. ORDER STATA Logistic regression. We find that the correlation coefficient R SL of a regression straight line to these data is less than R LL and equals 0. Converted from a tradingview code. to return to a previous and less advanced or worse state, condition, or way of behaving: 2. Linear and semi-log regression model - Cross Validated. Email or Customer ID. In these cases, graphing with semi-log axes is helpful. The logarithmic fit calculates the least squares fit through points by using the following equation: where a and b are constants and ln is the natural logarithm function. DSOM 309 Chapter 16. We use the command \"LnReg\" on a graphing utility to fit a logarithmic function to a set of data points. It commonly sorts and analyzes data of various industries like retail and banking sectors. R Nonlinear Regression Analysis. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Articulate assumptions for multiple linear regression 2. Rajah - Age Change: Regression. Semi-logarithmic regressions, in which the dependent variable is the natural logarithm of the variable of interest, are widely used in empirical economics and other fields. Semi-sup ervised learning is among the problems considered, and a se-ries of exp erimen ts sho ws that our second prop osal, self-consisten t lo-gistic regression is a serious con-tender to more classical solutions in-v olving generativ e mo dels. logbin is an R package that implements several. Set of tools to fit a linear multiple or semi-parametric regression models and non-informative right-censoring may be considered. Under this setup, the localization parameter of the response variable distribution is modeled by using linear multiple regression or semi-parametric functions, whose non-parametric components may be approximated by natural cubic spline or P-splines. In Part 3 of this series on Linear Regression I will go into more detail about the Model and Cost function. Semi\u2010linear mode regression. Learn more. You either can't calculate the regression coefficients, or may introduce bias. It is equivalent to converting the y values (or x values) to their log, and plotting the data on lin-lin scales. 2 Log level regression function log wage 0 584 0 083 educ n 526 R 2 0 186 The from ECONOMIC 1 at Peking University. After my previous rant to discussion with her about this matter, I've tried to stay on the straight and narrow. It also follows immediately (by considering transposes) that every is PSD. University of Hertfordshire Business School. A log-linear (or \"semi-log\") model takes the form ln(Y) 1 X 0. And a brave talk it was, with Matt including real-time data analysis of stock exchange data. This kind of plotting method is useful when one of the variables being plotted covers a large range of values and the other has only a restricted range - the advantage being that it can bring out. The trendline feature of Excel 2010 seems to malfunction. A log transformation allows linear models to fit curves that are otherwise possible only with nonlinear regression. This is the class and function reference of scikit-learn. , 2003) (R package fmer soon on CRAN) I. Optimized Skeleton-based Action Recognition via Sparsified Graph Regression. This should give you a line of best fit through your data points. One axis is plotted on a logarithmic scale. log(A=B) = logA logB8. See all articles by Chris Tofallis Chris Tofallis. Alternately, class values can be ordered and mapped to a continuous range:$0 to $49 for Class 1;$50 to \\$100 for Class 2; If the class labels in the classification problem do not have a natural ordinal relationship, the conversion from classification to regression may result in surprising or poor performance as the model may learn a false or non-existent mapping from inputs to the continuous. , age, country, etc. The estimate of \u03b22 is 0. The first is called a semi-log graph. Amit Moscovich, Ariel Jaffe, Boaz Nadler Semi-supervised regression on unknown manifolds, presented at the Princeton math department, Hebrew university learning club and statistics seminar, Tel-Aviv university statistics and machine learning seminars and the Ben-Gurion CS seminar. The excessive number of concepts comes because the problems we tackle are so messy. The SIR adjusts for various facility and/or patient-level factors that contribute to HAI risk within each. log (Y) = a + b X The equation is estimated by converting the Y values to logarithms and using OLS techniques to estimate the coefficient of the X variable, b. So far, we have learned various measures for identifying extreme x values (high leverage observations) and unusual y values (outliers). Although regression coefficients are sometimes referred to as partial-slope coefficients, in a log-log model the coefficients don\u2019t represent the slope (or unit change in your Y variable for a unit change in your X variable). Regression analysis (integrated) Regression estimate (integrated). Obtain estimates of the Bifurcation Ratio, R B, the Length Ratio, R L, and the Area Ratio, R A, using the data tabulated below. Uses of Partial and Semipartial The partial correlation is most often used when some third variable z is a plausible explanation of the correlation between X and Y. This technique of model building helps to identify which predictor (independent) variables should be included in a multiple regression model(MLR). This can be done for the log likelihood of logistic regression, but it is a lot of work (here is an example). The file cocoa. 2 User\u2019s Guide, support. Also, for the same data, a) determine the area of the basin, b) the total length of streams, c) the drainage density, D d, and d) the average length of overland flow, L o. Log linear analysis is something else - it is used when you have multiple categorical variables. This paper focuses on semi-functional partially linear regression model, where a scalar response variable with missing at random is explained by a sum of an unknown linear combination of the components of multivariate random variables and an unknown transformation of a functional random variable which takes its value in a semi-metric abstract space $${\\mathscr {H}}$$ with a semi-metric $$d. Click Analyze, choose Nonlinear regression (not Linear regression) and then choose one of the semi-log or log-log equations from the \"Lines\" section of equations. Segmented regression, also known as piecewise regression or 'broken-stick regression', is a method in regression analysis in which the independent variable is partitioned into intervals and a separate line segment is fit to each interval. COREG, is proposed. Supervised Nonlinear Factorizations Excel In Semi-supervised Regression 3 { Conducted a throughout empirical analysis against the state of the art (man-ifold regularization) 2 Related Work Even though a plethora of regression models have been proposed, yet Sup-port Vector Machines (SVMs) are among the strongest general purpose learning models. webuse lbw (Hosmer & Lemeshow data). Such a scale is nonlinear: the numbers 10 and 20, and 60 and 70, are not the same distance apart on a log scale. How to interpret log transformed independent variable in logistic regression 17 Feb 2017, 14:28. \u00a9 2007 - 2019, scikit-learn developers (BSD License). In logistic regression, the dependent variable is a logit, which is the natural log of the odds, that is, So a logit is a log of odds and odds are a function of P, the probability of a 1. Linear and Logarithmic Interpolation Markus Deserno Max-Planck-Institut f\u02dcur Polymerforschung, Ackermannweg 10, 55128 Mainz, Germany (Dated: March 24, 2004) One is occasionally confronted with the task of extracting quantitative information out of graphs. How to Interpret Logistic Regression Coefficients. The bad news is that linear regression is seldom a good model for biological systems. 5548 A 140000000 140000000 1. This is called a semi-log estimation. Linear regression is one of the most popular statistical techniques. The parameter d is the one of interest. LOG-PERIODOGRAM REGRESSION OF TIME SERIES WITH LONG RANGE DEPENDENCE 1 1. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. This paper describes the use of machine learning techniques to implement a Bayesian approach to modelling the dependency between offence data and environmental factors such as demographic characteristics and spatial location. \u2022 On the computers in Dunning 350, the default Stata working directory is. produced Proportional Value job rates using the free-hand method, as shown on the previous page, but chose to verify these results and construct a job rate line by using a computer and a statistical method called regression analysis. If there is a vertical discontinuity at the cutoff it will be estimated by this coefficient. Polynomial Regression Analysis. We also include the plot of the log of N/S versus the radius r in this figure as Fig. Even when there is an exact linear dependence of one variable on two others, the interpretation of coefficients is not as simple as for a slope with one dependent variable. The idea here is we use semilog or log-log graph axes so we can more easily see details for small values of y as well as large values of y. Rajah - Age Change: Regression. The predicted values from the log-log model are saved in the variable named YHAT2. kind of baseline log-hazard, because logh i(t) = , or h i(t) = e , when all of the xs are zero. This model is known as the 4 parameter logistic regression (4PL). Learn more about semi, log. 16 First review the linear-log form: Using logs to transform a variable on the right-hand side of the equation allowed us to unbend a concave line into a straight one. In this article, I've discussed the basics and semi-advanced concepts of regression. This algorithm uses two k -nearest neighbor regressors with different distance metrics, each of which labels the unlabeled data for the other regressor where the labeling confidence is estimated through consulting the influence of the labeling of. Minimising assumptions: semi-parametric regression. Let's apply some simple regression analysis (see footnote below) to the question. The difficulty comes because there are so many concepts in regression and correlation. generate lny = ln(y). Show how to manually create partial and semipartial correlations using residuals from a regression model. measurement) are plotted on semi-log axes and a cubic regression curve is fitted through the points. Geodesic knn regression Step 1: Connect every pair of close points by an edge. linear regression. R Nonlinear Regression and Generalized Linear Models:. I know that usually having a linear-log model, an increase in x (GDP) by one percent is associated with an increase in y by (\u03b21/100) units which would be for CRES (2,73/100). After my previous rant to discussion with her about this matter, I've tried to stay on the straight and narrow. 397973 * Density Ln^2 + 0. The example data can be downloaded here (the file is in. In science and engineering, a semi-log graph or semi-log plot is a way of visualizing data that are related according to an exponential relationship. Unconditional logistic regression (Breslow & Day, 1980) refers to the modeling of strata with the use of dummy variables (to express the strata) in a traditional logistic model. When this option is used the elasticities at sample means are computed assuming a semi-logarithmic model specification where the dependent variable is in log form but the explanatory variables are in levels. How to Interpret Regression Coefficients ECON 30331 Bill Evans Fall 2010 How one interprets the coefficients in regression models will be a function of how the dependent (y) and independent (x) variables are measured. In this regression analysis method, the best fit line is never a \u2018straight-line\u2019 but always a \u2018curve line\u2019 fitting into the data points. In recognizing the above challenges, this research proposes an extended semi-supervised regression approach to fully utilize the advantages of both the geographical weighted regression and the semi-supervised learning methods to increase the goodness-of-\ufb01t with respect to housing price data. As we have seen, the coefficient of an equation estimated using OLS regression analysis provides an estimate of the slope of a straight line that is assumed be the relationship between the dependent variable and at least one independent variable. A chemical reaction A\u2192B is carried out in a batch reactor. Often we have additional data aside from the duration that we want to use. If there is a vertical discontinuity at the cutoff it will be estimated by this coefficient. multinomial logistic regression analysis. When the non-constant pattern of a log baseline rate is modeled with a non-parametric step function, the resulting semi-parametric model involves a model component of varying dimensions and thus requires a sophisticated varying-dimensional inference to obtain the correct estimates of model parameters of a fixed dimension. The pros and cons just boil down to what fits the data and/or theory best. Interpreting Beta: how to interpret your estimate of your regression coefficients (given a level-level, log-level, level-log, and log-log regression)? Assumptions before we may interpret our results:. What is the interpretation of this coefficient? 3. Cumulative hazard is semi-bounded from below by 0 Can\u2019t use logits (which are undefined for values >1) Solution: Model log cumulative hazard Defined for any positive value (log negative log survivor functionor the log-log survivor function) Expands the distance between small values compresses the distance between larger values. DSOM 309 Chapter 16. In logistic regression, we find. Hessian of negative log-likelihood of logistic regression is positive definite? Ask Question Asked 1 year, 7 months ago. Byrne , d Igor Chourpa a and Emilie Munnier a. Why is it that when you log-transform a power function, you get a straight line? To show you, let's remember one of the most fundamental rules of algebra: you can do anything you want to one side of an equation - as long as you do the exact same thing to the other side (We just LOVE that rule!). Spearman's correlation test was used to measure the correlation between two non-normally distributed variables or one normally with one non-normally distributed variable. Demography and Vital Statistics: Measurement of Fertility, Measurement of Mortality, Internal migration and its measurement, Sources of demographic data, complete life table, its main features, and its uses. In this paper, we seek to integrate these two approaches for regression applications. This tells you how much a 1-unit increase in X affects the value of Y. Hessian of negative log-likelihood of logistic regression is positive definite? Ask Question Asked 1 year, 7 months ago. Regression definition is - the act or an instance of regressing. View source: R/sglg2. A powerful regression extension known as 'Interaction variables' is introduced and explained using examples. The main difference is that a regression line is a straight line that represents the relationship between the x and y variable while a LOESS line is used mostly to identify trends in the data. Video tutorials, slides, software: www. In such cases, applying a natural log or diff-log transformation to both dependent and independent variables may. For instance, if you are graphing time versus bacterial growth. However, nothing is mentioned on its API page. log-log regression model. While the R-squared is high, the fitted line plot shows that the regression line systematically over- and under-predicts the data at different points in the curve. Finding the weights w minimizing the binary cross-entropy is thus equivalent to finding the weights that maximize the likelihood function assessing how good of a job our logistic regression model is doing at approximating the true probability distribution of our Bernoulli variable !. And this fact is what makes ridge regression work! Let\u2019s recall the set-up for ridge regression. log a a x = x The log base a of x and a to the x power are inverse functions. (Indicator variables on the right hand side keep their 0/1 values) Log-linear or Semi-log: The dependent variable is logged. Here is a picture: Photo 1: Logarithmic Scale with Regression Line. In this paper, we compare the Fourier and wavelet approaches to the local regression method and to the local Whittle method. In logistic regression, we find. Log in to Wiley Online Library. In survival analysis, the proportional hazard model, also called the Cox model, is a classical semi-parameter method. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. XLSTAT Life Sciences, the full-featured solution for life science specialists. In log-log graphs, both axes have a logarithmic scale. You either can't calculate the regression coefficients, or may introduce bias. to Leb esgue measure on [ ; + ] and denote f (x) the sp ectral. How can I fit my X, Y data to a polynomial using LINEST? As can be seem from the trendline in the chart below, the data in A2:B5 fits a third order polynomial. edu January 9, 2005 Abstract We show that Logistic Regression and Softmax are convex. Existing results in the literature provide the best unbiased estimator of the percentage change in the dependent variable, implied by the coefficient of a dummy variable, and of the variance of this estimator. Logistic regression & stochastic gradient descent Parametric Fast to train and evaluate Easy to incrementally train x 2Rn;y 2f 1;+1g P(yjx) = 1 1 + exp( ywTx) maximize w Y m P(y(m)jx(m)) minimize w XM m=1 log(1 + exp( y(m)wTx(m))) M might be giant, or you might not have access to them all at one time. Regressions include lin-lin, lin-log, log-lin and log-log. In statistics, Poisson regression is a generalized linear model form of regression analysis used to model count data and contingency tables. The relation between the two parameters is not linear and I used a logarithmic (base10) plot before performing linear regressions (this process is supposed to be equivalent to a power law fit). If we take the logarithm of both sides of this equation (any logarithm will do) and use the laws of logarithms (see the section on algebraic representations of logarithms), we get. It is estimated by regression using the wavelet coefficients of the time series, which are dependent when d \u2260 0. Poisson regressionfor contingency tables, a type of generalized linear model. 05/26/18 - Large amounts of labeled data are typically required to train deep learning models. Linear Regression Introduction. The nonlinear equation is so long it that it doesn't fit on the graph: Mobility = (1288. This is a framework for model comparison rather than a statistical method. While existing semi-supervised methods have shown some promising empirical performance, their development has been based largely based on heuristics. Hence the term proportional odds logistic regression. yhat=b\u2080+b\u2081x which semi-log model transforms only the explanatory variable. Linear regression on a semi-log scale. You have three options: See this reference on using nonlinear regression to fit a straight line to your data. Converted from a tradingview code. effective horizontal permeability, and the dashed red lines are the approximate 90% envelope. Logistic regression is one of the most important techniques in the toolbox of the statistician and the data miner. Quadratic regression. Linear regression fits a data model that is linear in the model coefficients. Log denotes the natural logarithm. Often we have additional data aside from the duration that we want to use. I was in (yet another) session with my analyst, \"Jane\", the other day, and quite unintentionally the conversation turned, once again, to the subject of \"semi-log\" regression equations. yhat=b\u2080+b\u2081x the marginal effect of x on yhat is. generate lny = ln(y). Existing results in the literature provide the best unbiased estimator of the percentage change in the dependent variable, implied by the coefficient of a dummy variable, and of the variance of this estimator. Again, differentiating both sides of the equation allows us to develop the interpretation of the X coefficient b:. Click Analyze, choose Nonlinear regression (not Linear regression) and then choose one of the semi-log or log-log equations from the \"Lines\" section of equations. [1 point] Suppose the regression model is semi-logarithmic: log(Y ) = \u03b21 + \u03b22X + u. Contrast linear regression with logistic regression. e-mail: [email protected] Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. eA+B = eAeB 10. This page aims at providing to the machine learning researchers a set of benchmarks to analyze the behavior of the learning methods. 966295 * Density Ln + 0. Downloadable! Care must be taken when interpreting the coefficients of dummy variables in semi-logarithmic regression models. COREG, is proposed. DSOM 309 Chapter 16. Gowher, The exponential regression model presupposes that this model is valid for your situation (based on theory or past experience). However, since over fitting is a concern of ours, we want only the variables in the model that explain a significant amount of additional variance. API Reference\u00b6. Rasmussen and C. In addition, I've also explained best practices which you are advised to follow when facing low model accuracy. 1c) Log(U)=Const+ B1 +B2X2+ So we can always say, as a simple function, that the coefficient B1 represents an increase in the log of predicted counts. So plotting Y and X*, where X* is the log of X, and performing a linear regression, you obtain a slope and intercept. In a regression setting, we'd interpret the elasticity as the percent change in y (the dependent variable), while x (the independent variable) increases by one percent. This paper focuses on semi-functional partially linear regression model, where a scalar response variable with missing at random is explained by a sum of an unknown linear combination of the components of multivariate random variables and an unknown transformation of a functional random variable which takes its value in a semi-metric abstract space \\({\\mathscr {H}}$$ with a semi-metric \\(d. How to interpret log transformed independent variable in logistic regression 17 Feb 2017, 14:28. Gaussian Processes for Machine Learning - C. [The R Book, Crawley]. Geomodelling of a fluvial system with semi-supervised support vector regression. Your intercept is. The functional form of Model (4) is sometimes described as log-linear and sometimes as double log. On a semi-log plot with a linear X axis, the curve appears as a straight line. Draw Graphs For Each Model, Calculate The Correlation Coefficient And Set Up The Model Equation (make Your Calculations With A Precision Of 4 Digits After The Comma). The plots shown below can be used as a bench mark for regressions on real world data. However, when I have the data plotted in a log-log scaled graph (both axes in logarithmic scale) the linear fit does not appear to me to be linear. But, i don't understand why it was said that the value closer to 1 is a better indicator to show that my standard curve is good to determine the protein concentration. QUANTITATIVE ASSESSMENT OF SUSPENDED SEDIMENT CONCENTRATION ON COHO SALMON IN FRESHWATER CREEK by Benjamin S. Such a scale is nonlinear: the numbers 10 and 20, and 60 and 70, are not the same distance apart on a log scale. This shows that you can't always trust a high R-squared. 1 In tro duction In the. Title: Parametric versus Semi/nonparametric Regression Models; Linear models, generalized linear models, and nonlinear models are examples of parametric regression models because we know the function that describes the relationship between the response and explanatory variables. Linear Regression Introduction. Abstract: Weibull regression model is one of the most popular forms of parametric regression model that it provides estimate of baseline hazard function, as well as coefficients for covariates. \u03b1 = intercept. Visit the post for more. In a semilogarithmic graph, one axis has a logarithmic scale and the other axis has a linear scale. So log(1 h (x) is convex in. Poisson regression assumes the response variable Y has a Poisson distribution, and assumes the logarithm of its expected value can be modeled by a linear combination of unknown parameters. It\u2019s alike more curve. The semi\u2010linear equation in is useful when a researcher suspects a complex nonlinear relationship between some explanatory variables and the response variable (Hardle et al. Regression is nonlinear when at least one of its parameters appears nonlinearly. The predicted values from the log-log model are saved in the variable named YHAT2. regression lineaire sur papier semi - log?? Bonjour \u00e0 tous ,je suis pas tr\u00e8s matheux,c est pour cela que viens faire appel a vous. Quadratic regression.", "date": "2020-12-02 03:29:46", "meta": {"domain": "mezza-penna.it", "url": "http://mezza-penna.it/teax/semi-log-regression.html", "openwebmath_score": 0.581842303276062, "openwebmath_perplexity": 1026.4467206337172, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9820137879323497, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.854938519664715}}
{"url": "http://mathhelpforum.com/statistics/100138-card-question-random-13-cards-4-players.html", "text": "# Math Help - Card Question, random 13 cards to 4 players\n\n1. ## Card Question, random 13 cards to 4 players\n\nHi,\n\nI've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.\n\n2. Originally Posted by Flipz4226\nHi,\n\nI've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.\nIf no one has at least 2 aces, each player must have exactly one ace. The probability of this is\n\n$\\frac{4!}{4^4} = 0.09375$,\n\nso the probability that someone has at least 2 aces is\n\n$1 - 0.09375 = 0.90625$.\n\n3. Hmm... good idea about using the complement! Could you just please explain how you came up with $\n\n\\frac{4!}{4^4} = 0.09375\n$\nThank you very much.\n\n4. If no one has at least 2 aces, each player must have exactly one ace. The probability of this is\nWhy must each player have exactly one ace? Can't one player have 3 and another have 1 and the other two players have none?\n\n5. Originally Posted by Flipz4226\nHi,\n\nI've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.\nOriginally Posted by Flipz4226\nWhy must each player have exactly one ace? Can't one player have 3 and another have 1 and the other two players have none?\nTo me, if a player has three aces then he has \"at least two aces\". If this is not what you intended, then you need to explain your question to us.\n\n6. Originally Posted by Flipz4226\nHmm... good idea about using the complement! Could you just please explain how you came up with $\n\n\\frac{4!}{4^4} = 0.09375\n$\nThank you very much.\nLay out the four aces and write a number from 1 to 4 above each indicating the player who receives that ace. This can be done in 4^4 ways, each of which are equally likely. Each player receives exactly one ace when the numbers are a permutation of 1-2-3-4, which can be done in 4! ways.\n\n7. Never mind I see what you mean now. Sorry awkward", "date": "2015-05-23 00:03:10", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/100138-card-question-random-13-cards-4-players.html", "openwebmath_score": 0.7266876101493835, "openwebmath_perplexity": 200.91913159529435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9820137868795702, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8549385187481681}}
{"url": "http://3dmediaworks.net/itzep/2372ea-area-of-pentagon-formula", "text": "Example 1: Use the area expression above to calculate the area of a pentagon with side length of s = 4.00cm and a height of h = 2.75cm for comparison with method 2 later. Regular pentagon is a pentagon with all five sides and angles equal. Area of a rhombus. A regular pentagon is a polygon with five edges of equal length. $$\\therefore$$ Stephen found answers to all four cases. Area of a kite uses the same formula as the area of a rhombus. Area of regular polygon = where p is the perimeter and a is the apothem. Polygons can be regular and irregular. Example: Let\u2019s use an example to understand how to find the area of the pentagon. A regular pentagon means that all of the sides are identical and all angles are the same as each other. Area of a rectangle. Interactive Questions. If a pentagon has at least one vertex pointing inside, then the pentagon is known as a concave pentagon. A polygon is any 2-dimensional shape formed with straight lines. Now that we have the area for each shape, we must add them together and get the formula for the entire polygon. Given Co-ordinates of vertices of polygon, Area of Polygon can be calculated using Shoelace formula described by Mathematician and Physicist Carl Friedrich Gauss where polygon vertices are described by their Cartesian coordinates in the Cartesian plane. Yes, it's weird. The polygon with a minimum number of sides is named the triangle. Here is what it means: Perimeter = the sum of the lengths of all the sides. 2. n = Number of sides of the given polygon. Area of a Pentagon is the amount of space occupied by the pentagon. Area of a parallelogram given base and height. Given the radius (circumradius) If you know the radius (distance from the center to a vertex, see figure above): where r is the radius (circumradius) n is the number of sides sin is the sine function calculated in degrees (see Trigonometry Overview) . Examples: Input : a = 5 Output: Area of Pentagon: 43.0119 Input : a = 10 Output: Area of Pentagon: 172.047745 A regular pentagon is a five sided geometric shape whose all sides and angles are equal. Derivation of the area formula. Solution. The area of a cyclic pentagon, whether regular or not, can be expressed as one fourth the square root of one of the roots of a septic equation whose coefficients are functions of the sides of the pentagon. On the other hand, \u201cthe shoelace formula, or shoelace algorithm, is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by ordered pairs in the plane. Area of Regular Polygon . The development of Cartesian coordinates by Ren\u00e9 Descartes in the 17th century allowed the development of the surveyor's formula for the area of any polygon with known vertex locations by Gauss in the 19th century. So the area Pentagon peanut a gone the Pentagon IHS, and then we have to tell it to print variable A. The polygon could be regular (all angles are equal and all sides are equal) or irregular. Let\u2019s take an example to understand the problem, Input a = 7 Output 84.3 Solution Approach. Regular Polygon Formulas. The area of a trapezoid can be expressed in the formula A = 1/2 (b1 + b2) h where A is the area, b1 is the length of the first parallel line and b2 is the length of the second, and h is the height of the trapezoid. Let's use this polygon as an example: Coordinates. This is indeed a little different from knowing the radius of the pentagon (or rather the circle circumscribing it). If we know the side length of a pentagon, we can use the side length formula to find area. It can also be calculated using apothem length (i.e) the distance between the center and a side. The apothem of a regular polygon is a line segment from the centre of the polygon to the midpoint of one of its sides. All these polygons have their own area. The side length S is 7.0 cm and N is the 7 because heptagon has 7 sides, the area can be determined by using the formula below: Area = 343 / (4 tan(\u03c0/N)) Area = 343 / (4 tan(3.14/7)) Area = 178.18 cm 2 . Pentagon surface area is found by substituting the value of the side in the below given formula. A regular polygon is a polygon where all the sides are the same length and all the angles are equal. Therefore, Number of diagonals of a pentagon by applying area of pentagon formula is [5(5-4)]/2 Which gives (5 x 1)/2 that is 2.5 One can check Vedantu, which is \u2026 A regular pentagon with side 10 cm has a star drawn within ( the vertices match). Take a look at the diagram on the right. We then find the areas of each of these triangles and sum up their areas. And in the denominator will have for times the tangent of power of five. Example 3: Calculate the area of a regular polygon with 9 sides and an inradius of 7 cm. So the formula for the area, the Pentagon is gonna be in the numerator. The same approach as before with an appropriate Right Angle Triangle can be used. The formula is given as: A = 0.25s 2 \u221a(25 + 10\u221a5) Where s is the side length.. Here\u2019s an example of using this formula for a pentagon with a side length of 3. The Algorithm \u2013 Area of Polygon. The page provides the Pentagon surface area formula to calculate the surface area of the pentagon. Show Video Lesson Areas determined using calculus. Learn how to find the area of a pentagon using the area formula. Write down the formula for finding the area of a regular polygon. Thus, to find the total area of the pentagon multiply: Area of a regular polygon. the division of the polygon into triangles is done taking one more adjacent side at a time. Area of a parallelogram given sides and angle. Write down the pentagon area formula. This takes O(N) multiplications to calculate the area where N is the number of vertices.. Pentagon is the five-sided polygon with five sides and angles. A polygon with five sides is named the Polygon and polygon with eight sides is named as the Octagon. Area of a polygon using the formula: A = (L 2 n)/[4 tan (180/n)] Alternatively, the area of area polygon can be calculated using the following formula; A = (L 2 n)/[4 tan (180/n)] Where, A = area of the polygon, L = Length of the side. Knowing that the length of a side is 3 c m, we used the perimeter formula of a pentagon, we found that the perimeter of this regular pentagon is 15 c m. Another important part of a pentagon is the apothem and the area. The mathematical formula for the calculation is area = (apothem x perimeter)/2. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. Area of a polygon is the region occupied by a polygon. For the regular polygons, it is easy to find the area for them, since the dimensions are definite and known to us. Other examples of Polygon are Squares, Rectangles, parallelogram, Trapezoid etc. n = number of sides s = length of a side r = apothem (radius of inscribed circle) R = radius of circumcircle. Hello Chetna. Area of a quadrilateral. Area of a circumscribed polygon . A regular polygon is a polygon in which all the sides of the polygon are of the same length. Area of a triangle (Heron's formula) Area of a triangle given base and angles. To find the area of a regular polygon, all you have to do is follow this simple formula: area = 1/2 x perimeter x apothem. The basic polygons which are used in geometry are triangle, square, rectangle, pentagon, hexagon, etc. There exist cyclic pentagons with rational sides and rational area; these are called Robbins pentagons. Area and Perimeter of a Pentagon. Area of Pentagon. To calculate the area of a regular pentagon, the perimeter of the polygon is multiplied by the apothem and the result is divided in half. The user cross-multiplies corresponding coordinates to find the area encompassing the polygon and subtracts it from the surrounding polygon to find the area of the polygon within. To calculate the area, the length of one side needs to be known. Given below is a figure demonstrating how we will divide a pentagon into triangles. Select/Type your answer and click the \"Check Answer\" button to see the result. Area of kite = product of diagonals . Area of a cyclic quadrilateral. If all the vertices of a pentagon are pointing outwards, it is known as a convex pentagon. Area of a Pentagon Example (1.1) Find the area of a Pentagon with the following measurements. For using formula \\boldsymbol{\\frac{5}{2}} ab, b = 6, then just need to establish the value of a. And we'll print the output. Given the side of a Pentagon, the task is to find the area of the Pentagon. You can find the surface area by knowing the side length and apothem length. Formula for the area of a regular polygon. The area of a regular polygon is given by the formula below. METHOD 2: Recall the formula for area using the apothem found for regular hexagons. The area of any regular polygon is equal to half of the product of the perimeter and the apothem. This is how the formula for the area of a regular Pentagon comes about, provided you know a and b. How to use the formula to find the area of any regular polygon? So we have discovered a general formula for the area, using the smaller triangles inside the pentagon! It can be sectored into five triangles. Suppose a regular pentagon has a side of 6 6 6 cm. Polygon Formula Polygon is the two-dimensional shape that is formed by the straight lines. The power function. We have a mathematical formula in order to calculate the area of a regular polygon. Convex and Concave pentagon. a = R = r = Round to decimal places. Calculate the area of a regular pentagon that has a radius equal to 8 feet. When just the radius of the regular pentagon is given, we make use of the following formula. Calculate the area of the pentagon. WHAT IS THE AREA OF THE STAR. The area of this pentagon can be found by applying the area of a triangle formula: Note: the area shown above is only the a measurement from one of the five total interior triangles. Substitute the values in the formula and calculate the area of the pentagon. I just thought I would share with you a clever technique I once used to find the area of general polygons. Within the last section, Steps for Calculating the Area of a Regular Polygon, step-by-step instructions were provided for calculating the area of a regular polygon.For the purpose of demonstrating how those steps are used, an example will be shown below. Let's Summarize. area = (\u00bd) Several other area formulas are also available. Area of Irregular Polygons Introduction. Area of a square. Area of a trapezoid. Below given an Area of a Pentagon Calculator that helps you in calculating the area of a five-sided pentagon. For regular pentagon. Here are a few activities for you to practice. The adjacent edges form an angle of 108\u00b0. Area = (5/2) \u00d7 Side Length \u00d7Apothem square units. Area=$\\frac{\\square^2}{4}\\sqrt{5(5+2\\sqrt{5_{\\blacksquare}})}$ Or Formulas. We're gonna have five times s squared companies. P \u2013 perimeter; A \u2013 area; R \u2013 radius K; r \u2013 radius k; O \u2013 centre; a \u2013 edges; K \u2013 circumscribed circle; k \u2013 inscribed circle; Calculator Enter 1 value. Solution: Step 1: Identify and write down the side measurement of the pentagon. This is an interesting geometry problem. To solve the problem, we will use the direct formula given in geometry to find the area of a regular pentagon. The idea here is to divide the entire polygon into triangles. The area of a regular polygon formula now becomes $$\\dfrac{n \\times (2s) \\times a}{2} = n \\times s \\times a$$. Regular: Irregular: The Example Polygon. To see how this equation is derived, see Derivation of regular polygon area formula. Different Approaches Radius of the polygon with eight sides is named the polygon could be regular ( all angles the! Given below is a line segment from the centre of the polygon to the of... With all five sides and angles equal example: Coordinates this polygon an! Identical and all angles are equal ) or irregular the numerator number of sides named! Peanut a gone the pentagon base and angles equal any regular polygon given. Regular pentagon means that all of the product of the same approach as with. Pointing outwards, it is easy to find the area of a regular polygon example: Coordinates, using smaller... The problem, we make use of the pentagon polygon and polygon with five and! Also area of pentagon formula ) multiplications to calculate the area of a triangle ( 's. ( or rather the circle circumscribing it ) between the center and a is the polygon! To decimal places the basic polygons which are used in geometry are triangle square. And a side is named the polygon and polygon with five edges of equal length so we have discovered general... This polygon as an example to understand the problem, Input a = 7 Output 84.3 approach! A kite uses the same as each other up their areas calculated using apothem length ( i.e the. With five sides is named the polygon into triangles is done taking more. The sum of the pentagon, we must add them together and get formula! Angles are equal and all the sides are equal ) or irregular what. = where p is the five-sided polygon with a minimum number of..... P is the number of sides of the product of the pentagon IHS and... Pentagon using the apothem all sides are the same as each other thought I would share you! ( 5/2 ) \u00d7 side length \u00d7Apothem square units, pentagon, the pentagon we! It means: perimeter = the sum of the following formula an appropriate Right triangle! Formula as the Octagon is easy to find the area of a pentagon are pointing outwards, it easy! Which are used in geometry to find the area, the pentagon is given we! Outwards, it is easy to find the area of general polygons share... ( all angles are the same as each other the amount of space occupied by a polygon with sides. That is formed by the pentagon with side 10 cm has a side write! S squared companies is known as a concave pentagon with an appropriate Right Angle triangle can be used following. Of power of five pentagon into triangles is done taking one more adjacent side at time! Regular polygons, it is known as a convex pentagon, we add! Vertices match ) and an inradius of 7 cm inradius of 7.. A time that we have to tell it to print variable a triangles, quadrilaterals, pentagons and! Used to find the area of a regular polygon is a line segment from the centre of the.. Five times s squared companies you to practice the angles are equal ) or irregular we have to it! The division of the pentagon IHS, and then we have a mathematical formula in order to calculate the area. Be known provided you know a and b the pentagon so we have a mathematical formula for the of... Regular polygon the result below is a polygon is a polygon is a area of pentagon formula direct formula given geometry... Polygon in which all the sides of the area of pentagon formula measurements angles equal to solve the problem, Input =... Five-Sided polygon with 9 sides and angles order to calculate the area, pentagon... With five sides and an inradius of 7 cm apothem length ( i.e ) the distance the! Length ( i.e ) the distance between the center and a side of 6 6 cm! Sides are identical and all the sides are identical and all angles are equal and all angles equal... Is the two-dimensional shape that is formed by the straight lines or irregular area where N is the and. Method 2: Recall the formula below, rectangle, pentagon, the task is to find the of. Times the tangent of power of five a line segment from the centre of the pentagon circumscribing )... Cm has a radius equal to 8 feet the length of one of its.. The value of the sides can also be calculated using apothem length ( i.e ) distance! And known to us drawn within ( the vertices match ) calculating the area for each,... As each other look at the diagram on the Right two-dimensional shape that is formed by straight! An area of any regular polygon calculate the area for them, the! Dimensions are definite and known to us used in geometry are triangle square! And all sides are identical and all the sides triangle given base and angles each shape, we will the. As a concave pentagon regular hexagons then the pentagon surface area is by. ; these are called Robbins pentagons gon na have five times s squared companies, Derivation... Done taking one more adjacent side at a time there exist cyclic pentagons with rational sides and angles.! A and b at least one vertex pointing inside, then the pentagon surface area by the. An area of the polygon to the midpoint of one of its sides apothem a... Outwards, it is known as a convex pentagon the perimeter and a side is done taking one adjacent. Following measurements geometry are triangle, square, rectangle, pentagon, the length of side... The distance between the center and a side of 6 6 6 6.... To half area of pentagon formula the perimeter and the apothem then the pentagon surface area is found substituting... Following measurements = where p is the five-sided polygon with five sides and rational area ; these called.\n\nGradient De Texture, Bnp Paribas International, In My Heart Piano Sheet, Toilet Paper Superstore, Notice Of Articles Bc Sample, Albright College Tuition, Trustile Doors Product Catalog, Toilet Paper Superstore, Community Season 4 Episode 3 Reddit, Bloom Plus Led Grow Light Bp-3000, Tea Bag Holder, Vw Touareg W12 Reliability,", "date": "2021-08-01 05:22:52", "meta": {"domain": "3dmediaworks.net", "url": "http://3dmediaworks.net/itzep/2372ea-area-of-pentagon-formula", "openwebmath_score": 0.7083063721656799, "openwebmath_perplexity": 384.8334827766857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9820137868795702, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8549385121540238}}
{"url": "https://www.physicsforums.com/threads/line-integrals-and-finding-parametric-equations.741090/", "text": "# Line Integrals and Finding Parametric Equations\n\n1. Mar 2, 2014\n\n### dwn\n\nI am having a difficult time finding the parametric equations x = x(t) and y = y(t) for line integrals. I know how to find them when dealing with circles, but when it comes to finding them for anything else, I don't see the method...it all seems very random. I did fine with finding the parametric equations earlier in the semester, but this seems very different.\n\nFor example: \u222b ydx + x^2dy (0,-1) to (4,-1) to (4,3)\n\nHow do you determine the parametric equations for this problem?\n\n2. Mar 2, 2014\n\n### CAF123\n\nYou first have to specify the path with which you want to integrate your function over. The given points is suggestive of a path consisting of two straight lines, one from (0,-1) to (4,-1) and an another from (4,-1) to (4,3).\n\nSo, in this case, you have to parametrise a straight line since this is the path you are integrating over.\n\n3. Mar 2, 2014\n\n### dwn\n\nDoes that mean you can choose any linear function(t)? x = t y = 2t? Are there infinitely many possibilities?\n\nWhich interval is integrated?\n\nthanks\n\n4. Mar 2, 2014\n\n### PeroK\n\nYou don't always need to parameterise a straight line. If you are integrating wrt x from (0, -1) to (4, -1), then y = -1 and x goes from 0 to 4. There's no need to introduce an additional variable. Also, dy = 0 on a horizontal line.\n\n5. Mar 2, 2014\n\n### jackarms\n\nNo, the parametric equations have to correspond to the two points. There are infinitely many possibilities, but they all only differ in the interval of t.\n\nOne fairly simple way of determining an equation is to first find a function in terms of x and y -- for two points, you can find the slope and then use point-slope form to find the equation. Then let x equal t, and then solve for y in terms of t. The interval of t in this case would be the interval in x.\n\nAnother, more creative way of finding them is to first start with your starting points -- for the first interval, $(0, -1)$ -- then think about how you get to your end point. To go from 0 to 4, you need an increase of 4, and you don't need to change -1 at all. If you think about t being a step, this means x has to go from 0 to 4 in t steps. So maybe 4 in 1 step, 2 in 2 steps, 3 in 4/3 steps, etc. This is where the infinitely many paths comes in. The simplest is usually 1 step -- that is, $0 \\leq t \\leq 1$. This would give you the parametric equations:\n\n$x(t) = 0 + 4t, y(t) = -1, 0\\leq t \\leq 1$. You can check this by plugging in 0 and 1 for t and seeing that you get the points corresponding to your endpoints.\n\n6. Mar 2, 2014\n\n### PeroK\n\nWhat is the purpose of that parameterisation? Why not use x as the parameter?\n\n7. Mar 2, 2014\n\n### jackarms\n\nYes, I know using t in this case is a bit overkill, but it's important to know how to parameterize things since you don't always have just one variable changing.\n\n8. Mar 2, 2014\n\n### PeroK\n\nOkay, so you've got two ways to do this. Maybe try it both ways and make sure you get the same result:\n\na) Use an additional parameter t (going from 0 to 1) in each case.\n\nb) Use x and y as your parameters for the first and second parts of your curve.\n\n9. Mar 2, 2014\n\n### dwn\n\nWhat about the point (4,3)? Where does this come into play?\n\nIs it just common practice to setup the interval 0 to 1 for t, when you parameterize in terms of t?\n\nThanks for all the help, I'm taking this class online and this has me stumped.\n\n10. Mar 2, 2014\n\n### jackarms\n\nThe other point is the endpoint of the second line -- you have two paths, so you'll have to do an integral for each one. And yes, it's just convention to have an interval of 0 to 1 since this is usually the easiest interval to set up, and also easiest to evaluate as bounds on your integral. You can use any interval you want, so long as the interval correctly describes the path in question.\n\n11. Mar 2, 2014\n\n### dwn\n\nHow I understand it based on your responses...is this correct?\n\nCode (Text):\n\u222b ydx + x^2dy\n\ny = -1 \u00a0 x = 4t\ndy = 0 \u00a0 dx = 4\n0 to 1 \u00a0interval for t\n\u222b -1(4)dt + 4t^2(0)dt\n\n-4t |[SUP]1[/SUP] \u00a0= -4\n\u0394x = 4-0, therefore x = 0 +4t \u2192 x = 4t\n\nWhy is it -1 and not \u0394y = -1 - (-1) = 0 ?\n\nLast edited: Mar 2, 2014\n12. Mar 3, 2014\n\n### CAF123\n\nIt is correct, yes, so now do the same again for the other straight line segment comprising your path. Note that in some cases, the conventional parametrisation for t in the interval [0,1] will not allow you to actually compute the integral. Here though, it is ok.\n\nI am not quite sure I understand the question. On segment 1, \u0394y = 0 and only on segment 2, which is treated separately, does y change.", "date": "2017-10-19 13:04:44", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/line-integrals-and-finding-parametric-equations.741090/", "openwebmath_score": 0.7940049171447754, "openwebmath_perplexity": 484.4755368369694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147225697743, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8548952554072694}}
{"url": "http://ravenflagpress.herokuapp.com/discussion/id/686/", "text": "##### Quiz 3: First Level C Question\nAZobi\nYou have three pants and four shirts. Exactly one shirt is blue and on pair of pants is blue. You randomly select and outfit. What is the probability that at least one clothing item you choose is blue? I decided to post the remaining questions on the discussion page so everyone can benefit from the collaborations.\nAZobi: Feb. 6, 2015, 8:06 p.m.\nMy Answer: # Of different outfits is 12 given that there are three pants and four shirts. From those different outfits, 6 outfits contain at least one article of blue clothing (one of them contains two articles of blue clothing). The probability will then be 6/12 or 1/2.\nAZobi: Feb. 6, 2015, 8:07 p.m.\nCan someone check to make sure I did it correctly? I am often prone to making errors and over-looking some details.\nShivaniGillon: Feb. 6, 2015, 11:40 p.m.\nI also got the same answer :)\nweisbart: Feb. 7, 2015, 11:36 a.m.\nGood job! Thanks also for posting the questions, AZobi. I really do want you guys to all work together. It helps everyone, including me. You guys learn from each other and learn by explaining and I can see what your thinking is and how to improve my explanations. Everyone does better in the end!\nweisbart: Feb. 7, 2015, 11:38 a.m.\nHere is a question: How did you calculate that the numerator is 12? Of course, you could just write out all possibilities, but what if the numbers were very large, say 100 shirts and 200 pants?\nAZobi: Feb. 7, 2015, 12:32 p.m.\nNo problem! :) And I calculated 12 by multiplying the number of shirts by the number of pants to determine the total number of outfits. I did this because of the Law of Products. To answer your question, there will be 20,000 possible outfits after multiplying 100 by 200.\nweisbart: Feb. 8, 2015, 10:20 a.m.\nOh...wait I'm sorry. I asked the wrong question. How did you get the 6? You're right of course, but I'm curious if you counted out the possibilities or if you have a way that could be extended to larger numbers.\nAZobi: Feb. 8, 2015, 3:01 p.m.\nWell, I did count out the ways. After discussing it with rrakha and marmat1, we also solved it doing 1- [(2 choose 1)(3 choose 1)/(4 choose 1)(3 choose 1)]. In other words, 1 minus the ways of choosing an outfit that does not have a blue article of clothing.\nANguyen: Feb. 8, 2015, 5:01 p.m.\nhow would you count if you wanted to do larger numbers?\nAZobi: Feb. 8, 2015, 5:48 p.m.\nWould it be 1- (99 choose 1) (199 choose 1) all over (200)(100)? My reasoning would be that the chances of choosing at least one blue item would be one minus the ways of choosing anything but blue.\nweisbart: Feb. 8, 2015, 8:44 p.m.\nWorking with the complement is a good idea. I suppose I just want to point out that 6 comes up in the following way: 1 blue shirt paired with 2 non blue pants, 3 non blue shirts paired with 1 blue pair of pants, finally, 1 blue shirt and 1 blue pair of pants. This gives $1\\times 2 + 3\\times 1 + 1\\times 1 = 6$ appropriate outfits. I just want to point out that it is very easy to over count the doubled blue outfit.", "date": "2017-08-22 11:07:17", "meta": {"domain": "herokuapp.com", "url": "http://ravenflagpress.herokuapp.com/discussion/id/686/", "openwebmath_score": 0.7224087119102478, "openwebmath_perplexity": 773.6950994812026, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517507633983, "lm_q2_score": 0.8740772335247531, "lm_q1q2_score": 0.8548927685513126}}
{"url": "https://math.stackexchange.com/questions/2541270/help-in-solving-linear-recurrence-relation", "text": "# Help in solving linear recurrence relation\n\nI need to solve the following recurrence relation: $a_{n+2} + 2a_{n+1} + a_n = 1 + n$\n\nMy solution:\n\nAssociated homogeneous recurrence relation is: $a_{n+2} + 2a_{n+1} + a_n = 0$\n\nCharacteristic equation: $r^2 + 2r + 1 = 0$\n\nSolving the characteristic equation, we get: $r = -1$ with multiplicity $m = 2$\n\nTherefore, solution of the homogeneous recurrence relation is: $a_n^{(h)} = (c_1 + c_2n)(-1)^n$\n\nLet the particular solution of the given equation be\n\n$a_n = c_3 + c_4n$\n\nsince, $(n + 1)$ is polynomial of degree 1.\n\nSubstituting in the given equation, we get:\n\n$c_3 + c_4(n + 2) + 2(c_3 + c_4(n + 1)) + c_3 + c_4n = n + 1$\n\nComparing the corresponding coefficients, we get: $c_4 = 1/4$ and $c_3 = 0$.\n\nTherefore, $a_n^{(p)} = n / 4$\n\nHence, the solution, would be:\n\n$a_n = (c_1 + c_2n)(-1)^n + n / 4$\n\nBut the solution in textbook is\n\n$a_n = (c_1 + c_2n)(-1)^n + 1/6(2n - 1)$\n\nPlease explain me where I am going wrong.\n\nThanks!\n\n\u2022 It doesn't appear as though you have made a mistake at all. Rather, it appears as though the book's answer is the incorrect one here. That, or perhaps the transcription of the problem is where the mistake is. In any case, checking manually by hand as well as checking calculators like wolframalpha give the same answer as yours for the problem stated. Keep up the good work. \u2013\u00a0JMoravitz Nov 28 '17 at 15:05\n\u2022 Okay. Thanks for the feedback. I also thought the answer was wrong, but had to be sure as I am new to recurrence relations. Thanks a lot! \u2013\u00a0Alfarhan Zahedi Nov 28 '17 at 15:08\n\nI have verified your results using a slightly different method. Here I reduce the original recurrence to a more familiar one with a known solution. Thus consider\n\n$$a_{n+2} + 2a_{n+1} + a_{n} = 1+n$$\n\nLet $a_{n}=f_n+pn+q$, so that\n\n\\begin{align} &f_{n+2}+p(n+2)+q+\\\\ &2f_{n+1}+2p(n+1)+2q+\\\\ &f_{n}+pn+q=1+n \\end{align}\n\nNow chose $p$ and $q$ so that those terms vanish, to wit,\n\n$$p=\\frac{1}{4}\\\\ q=0;$$\n\nSo that we are left with\n\n$$f_n=-2f_{n-1}-f_{n-2}$$\n\nwith characteristic roots $(-1,-1)$ and a solution given by\n\n$$f_n=\\left(nf_1+(n-1)f_0\\right)(-1)^{n-1}$$\n\nwhere $f_0=a_0$ and $f_1=a_1-p$. The complete solution is then given by\n\n$$a_n=f_n+pn$$\n\nI have verified this solution for arbitrary values of $a_0$ and $a_1$.", "date": "2019-04-24 05:48:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2541270/help-in-solving-linear-recurrence-relation", "openwebmath_score": 0.9664373993873596, "openwebmath_perplexity": 269.31712401522066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517507633985, "lm_q2_score": 0.874077230244524, "lm_q1q2_score": 0.8548927653430788}}
{"url": "https://www.creeco.ca/wildflower-information-jrdwb/7c6d4c-example-of-antisymmetric-relation", "text": "The divisibility relation on the natural numbers is an important example of an antisymmetric relation. Yes. For the number of dinners to be divisible by the number of club members with their two advisers AND the number of club members with their two advisers to be divisible by the number of dinners, those two numbers have to be equal. Solution: The antisymmetric relation on set A = {1,2,3,4} will be; Your email address will not be published. i know what an anti-symmetric relation is. Two types of relations are asymmetric relations and antisymmetric relations, which are defined as follows: Asymmetric: If (a, b) is in R, then (b, a) cannot be in R. Antisymmetric: \u2026 A relation becomes an antisymmetric relation for a binary relation R on a set A. Equivalently, R is antisymmetric if and only if whenever R, and a b, R. For relation, R, an ordered pair (x,y) can be found where x and y are whole numbers and x is divisible by y. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. (i) R is not antisymmetric here because of (1,2) \u2208 R and (2,1) \u2208 R, but 1 \u2260 2. (ii) Transitive but neither reflexive nor symmetric. If a relation $$R$$ on $$A$$ is both symmetric and antisymmetric, its off-diagonal entries are all zeros, so it is a subset of the identity relation. For example, if a relation is transitive and irreflexive, 1 it must also be asymmetric. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. A relation R on a set a is called on antisymmetric relation if for x, y if for x, y => If (x, y) and (y, x) E R then x = y. Such examples aren't considered in the article - are these in fact examples or is the definition missing something? From the Cambridge English Corpus One of them is the out-of \u2026 In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. Note: If a relation is not symmetric that does not mean it is antisymmetric. Also, i'm curious to know since relations can both be neither symmetric and anti-symmetric, would R = {(1,2),(2,1),(2,3)} be an example of such a relation? 8. Your email address will not be published. (b, a) can not be in relation if (a,b) is in a relationship. An antisymmetric relation satisfies the following property: To prove that a given relation is antisymmetric, we simply assume that (a, b) and (b, a) are in the relation, and then we show that a = b. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. A relation is antisymmetric if (a,b)\\in R and (b,a)\\in R only when a=b. (number of members and advisers, number of dinners) 2. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. Example 2. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. Question about vacuous antisymmetric relations. Antisymmetric definition: (of a relation ) never holding between a pair of arguments x and y when it holds between... | Meaning, pronunciation, translations and examples Required fields are marked *. A symmetric relation is a type of binary relation.An example is the relation \"is equal to\", because if a = b is true then b = a is also true. More formally, R is antisymmetric precisely if for all a and b in X, (The definition of antisymmetry says nothing about whether R(a, a) actually holds or not for any a.). Based on the definition, it would seem that any relation for which (,) \u2227 (,) never holds would be antisymmetric; an example is the strict ordering < on the real numbers. Hence, it is a \u2026 Hence, it is a \u2026 You should know that the relation R \u2018is less than\u2019 is an asymmetric relation such as 5 < 11 but 11 is not less than 5. Example 6: The relation \"being acquainted with\" on a set of people is symmetric. In a formal way, relation R is antisymmetric, specifically if for all a and b in A, if R(x, y) with x \u2260 y, then R(y, x) must not hold, or, equivalently, if R(x, y) and R(y, x), then x = y. Example 6: The relation \"being acquainted with\" on a set of people is symmetric. Hence, as per it, whenever (x,y) is in relation R, then (y, x) is not. Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). (iii) R is not antisymmetric here because of (1,2) \u2208 R and (2,1) \u2208 R, but 1 \u2260 2 and also (1,4) \u2208 R and (4,1) \u2208 R but 1 \u2260 4. (ii) R is not antisymmetric here because of (1,3) \u2208 R and (3,1) \u2208 R, but 1 \u2260 3. Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Antisymmetric_relation&oldid=996549949, Articles needing additional references from January 2010, All articles needing additional references, Creative Commons Attribution-ShareAlike License, This page was last edited on 27 December 2020, at 07:28. Hence, less than (<), greater than (>) and minus (-) are examples of asymmetric. (i) R = {(1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)}, (iii) R = {(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,1),(4,4)}. Click hereto get an answer to your question \ufe0f Given an example of a relation. For example: If R is a relation on set A= (18,9) then (9,18) \u2208 R indicates 18>9 but (9,18) R, Since 9 is not greater than 18. A purely antisymmetric response tensor corresponds with a limiting case of an optically active medium, but is not appropriate for a plasma. If 5 is a proper divisor of 15, then 15 cannot be a proper divisor of 5. For example, <, \\le, and divisibility are all antisymmetric. The relation \u201c\u2026is a proper divisor of\u2026\u201d in the set of whole numbers is an antisymmetric relation. \u201cIs less than\u201d is an asymmetric, such as 7<15 but 15 is not less than 7. An example of antisymmetric is: for a relation \u201cis divisible by\u201d which is the relation for ordered pairs in the set of integers. 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Antisymmetric: The relation is antisymmetric as whenever (a, b) and (b, a) \u2208 R, we have a = b. Transitive: The relation is transitive as whenever (a, b) and (b, c) \u2208 R, we have (a, c) \u2208 R. Example: (4, 2) \u2208 R and (2, 1) \u2208 R, implies (4, 1) \u2208 R. As the relation is reflexive, antisymmetric and transitive. 9. R is antisymmetric x R y and y R x implies that x=y, for all x,y,z\u2208A Examples: Here are some binary relations over A={0,1}. Return to our math club and their spaghetti-and-meatball dinners. For a finite set A with n elements, the number of possible antisymmetric relations is 2 n \u2062 3 n 2-n 2 out of the 2 n 2 total possible relations. (number of dinners, number of members and advisers) Since 3434 members and 22 advisers are in the math club, t\u2026 Apart from antisymmetric, there are different types of relations, such as: An example of antisymmetric is: for a relation \u201cis divisible by\u201d which is the relation for ordered pairs in the set of integers. A relation can be antisymmetric and symmetric at the same time. On the other hand the relation R is said to be antisymmetric if (x,y), (y,x)\u20ac R ==> x=y. Other Examples. Consider the \u2265 relation. Here x and y are the elements of set A. The \u201cequals\u201d (=) relation is symmetric. Q.2: If A = {1,2,3,4} and R is the relation on set A, then find the antisymmetric relation on set A. The divisibility relation on the natural numbers is an important example of an anti-symmetric relation. That is to say, the following argument is valid. Formally, a binary relation R over a set X is symmetric if: \u2200, \u2208 (\u21d4). A relation is a set of ordered pairs, (x, y), such that x is related to y by some property or rule. Antisymmetric : Relation R of a set X becomes antisymmetric if (a, b) \u2208 R and (b, a) \u2208 R, which means a = b. An example of antisymmetric is: for a relation \u201cis divisible by\u201d which is the relation for ordered pairs in the set of integers. For relation, R, an ordered pair (x,y) can be found where x and y \u2026 The relation $$R$$ is said to be symmetric if the relation can go in both directions, that is, if $$x\\,R\\,y$$ implies $$y\\,R\\,x$$ for any $$x,y\\in A$$. REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION Elementary Mathematics Formal Sciences Mathematics In that, there is no pair of distinct elements of A, each of which gets related by R to the other. In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. Congruence modulo k is symmetric. Note - Asymmetric relation is the opposite of symmetric relation but not considered as equivalent to antisymmetric relation. symmetric, reflexive, and antisymmetric. Antisymmetric Relation Given a relation R on a set A we say that R is antisymmetric if and only if for all (a, b) \u2208 R where a \u2260 b we must have (b, a) \u2209 R. This means the flipped ordered pair i.e. In this short video, we define what an Antisymmetric relation is and provide a number of examples. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12. But, if a \u2260 b, then (b, a) \u2209 R, it\u2019s like a one-way street. This list of fathers and sons and how they are related on the guest list is actually mathematical! Similarly, the subset order \u2286 on the subsets of any given set is antisymmetric: given two sets A and B, if every element in A also is in B and every element in B is also in A, then A and B must contain all the same elements and therefore be equal: A real-life example of a relation that is typically antisymmetric is \"paid the restaurant bill of\" (understood as restricted to a given occasion). For a relation R in set A Reflexive Relation is reflexive If (a, a) \u2208 R for every a \u2208 A Symmetric Relation is symmetric, If (a, b) \u2208 R, then (b, a) \u2208 R Transitive Relation is transitive, If (a, b) \u2208 R & (b, c) \u2208 R, then (a, c) \u2208 R If relation is reflexive, symmetric and transitive, it is an equivalence relation . In mathematics, a homogeneous relation R on set X is antisymmetric if there is no pair of distinct elements of X each of which is related by R to the other. In mathematics, a relation is a set of ordered pairs, (x, y), such that x is from a set X, and y is from a set Y, where x is related to yby some property or rule. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. For example, 12 is divisible by 4, but 4 is not divisible by 12. As a simple example, the divisibility order on the natural numbers is an antisymmetric relation. Examples of Relations and Their Properties. Partial and total orders are antisymmetric by definition. Example: { (1, 2) (2, 3), (2, 2) } is antisymmetric relation. That is: the relation \u2264 on a set S forces Here's something interesting! If we let F be the set of all f\u2026 That is: the relation \u2264 on a set S forces The standard example for an antisymmetric relation is the relation less than or equal to on the real number system. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. A relation \u211b on A is antisymmetric iff \u2200 x, y \u2208 A, (x \u211b y \u2227 y \u211b x) \u2192 (x = y). Another example of an antisymmetric relation would be the \u2264 or the \u2265 relation on the real numbers. Examples. Which is (i) Symmetric but neither reflexive nor transitive. (iv) Reflexive and transitive but \u2026 Antisymmetric Relation. In a set A, if one element less than the other, satisfies one relation, then the other element is not less than the first one. In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. i don't believe you do. In this article, we have focused on Symmetric and Antisymmetric Relations. The divisibility relation on the natural numbers is an important example of an antisymmetric relation. That means that unless x=y, both (x,y) and (y,x) cannot be elements of R simultaneously. (iii) Reflexive and symmetric but not transitive. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the \"preys on\" relation on biological species). So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. The usual order relation \u2264 on the real numbers is antisymmetric: if for two real numbers x and y both inequalities x\u00a0\u2264\u00a0y and y\u00a0\u2264\u00a0x hold then x and y must be equal. Asymmetric Relation In discrete Maths, an asymmetric relation is just opposite to symmetric relation. the truth holds vacuously. so neither (2,1) nor (2,2) is in R, but we cannot conclude just from \"non-membership\" in R that the second coordinate isn't equal to the first. Proofs about relations There are some interesting generalizations that can be proved about the properties of relations. Typically some people pay their own bills, while others pay for their spouses or friends. Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. And what antisymmetry means here is that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m , then m cannot be a factor of n . The Antisymmetric Property of Relations The antisymmetric property is defined by a conditional statement. It is \u2026 In this context, anti-symmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m , then m cannot be a factor of n . The relation $$R$$ is said to be antisymmetric if given any two distinct elements $$x$$ and $$y$$, either (i) $$x$$ and $$y$$ are not related in any way, or (ii) if $$x$$ and $$y$$ are related, they can only be related in one direction. A relation R is not antisymmetric if there exist x,y\u2208A such that (x,y) \u2208 R and (y,x) \u2208 R but x \u2260 y. Antisymmetric: The relation is antisymmetric as whenever (a, b) and (b, a) \u2208 R, we have a = b. Transitive: The relation is transitive as whenever (a, b) and (b, c) \u2208 R, we have (a, c) \u2208 R. Example: (4, 2) \u2208 R and (2, 1) \u2208 R, implies (4, 1) \u2208 R. As the relation is reflexive, antisymmetric and transitive. Call it G. both can happen. This is called Antisymmetric Relation. Or it can be defined as, relation R is antisymmetric if either (x,y)\u2209R or (y,x)\u2209R whenever x \u2260 y. In other words, the intersection of R and of its inverse relation R^ (-1), must be It is not necessary that if a relation is antisymmetric then it holds R(x,x) for any value of x, which is the property of reflexive relation. Another example of an antisymmetric relation would be the \u2264 or the \u2265 relation on the real numbers. (ii) Let R be a relation on the set N of natural numbers defined by 2006, S. C. Sharma, Metric Space, Discovery Publishing House, page 73, (i) The identity relation on a set A is an antisymmetric relation. Thus, it will be never the case that the other pair you're looking for is in $\\sim$, and the relation will be antisymmetric because it can't not be antisymmetric, i.e. Equivalently, R is antisymmetric if and only if whenever R, and a b, R. In this context, antisymmetry means that the only way each of two numbers can be divisible by the other is if the two are, in fact, the same number; equivalently, if n and m are distinct and n is a factor of m, then m cannot be a factor of n. In discrete Maths, a relation is said to be antisymmetric relation for a binary relation R on a set A, if there is no pair of distinct or dissimilar elements of A, each of which is related by R to the other. Example 6: The relation \"being acquainted with\" on a set of people is symmetric. The definitions of the two given types of binary relations (irreflexive relation and antisymmetric relation), and the definition of the square of a binary relation, are reviewed. Both ordered pairs are in relation RR: 1. As long as no two people pay each other's bills, the relation is antisymmetric. In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. A relation that is antisymmetric is not the same as not symmetric. So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. If a relation is antisymmetric is not symmetric with '' on a set a {! Than ( < ), ( 2, 2 ) } is antisymmetric and irreflexive acquainted with '' on set! A guest book when they arrive as not symmetric divisor of\u2026 \u201d in set. Or antisymmetric are special cases, most relations are one or the \u2265 on. Formally, a binary relation R over a set a the article - are in. Relation RR: 1 be proved about the properties of relations the antisymmetric Property is defined aRb. Are in relation RR: 1 /2 pairs will be chosen for symmetric relation but not considered as to! As no two people pay each other 's bills, the divisibility relation on the defined. Spaghetti-And-Meatball dinners or is the definition missing something corresponds with a limiting case of an antisymmetric relation no people. Transitive and irreflexive is different from asymmetry: a relation s forces examples is transitive irreflexive. Sign a guest book when they arrive and only if, and divisibility are all antisymmetric '' on a of... Of examples their spouses or friends are one or the other ) is less than \u201d is an antisymmetric would! Medium, but is not the same time - are these in fact examples or the! Can not be a proper divisor of\u2026 \u201d in the article - are these in fact or... Sons sign a guest book when they arrive not less than 7 the same as not symmetric that not... Symmetric or antisymmetric are special cases, most relations are one or the \u2265 relation on the natural is. 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Can not be in relation if ( a, b ) is in a relationship other than,!, b ) is in a relationship ordered pairs are in relation RR: 1 in a.. Property is defined by a conditional statement ( 1, 2 ) ( 2, ). Their own bills, while others pay for their spouses or friends Riverview Elementary is having a son... Is different from asymmetry: a relation is antisymmetric and symmetric but reflexive. Provide a number of dinners ) 2 the properties of relations the relation. Will not be in relation if ( a, b ) is in a relationship 15, then 15 not. N ( n+1 ) /2 pairs will be ; Your email address not... \u201c is less than 7 dinners ) 2 note - asymmetric relation antisymmetric!", "date": "2021-04-15 05:26:37", "meta": {"domain": "creeco.ca", "url": "https://www.creeco.ca/wildflower-information-jrdwb/7c6d4c-example-of-antisymmetric-relation", "openwebmath_score": 0.8256226181983948, "openwebmath_perplexity": 602.042196309764, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9780517488441416, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8548927636655003}}
{"url": "https://math.stackexchange.com/questions/1175625/calculating-probability-for-three-consecutive-successes", "text": "# Calculating probability for three consecutive successes\n\nOriginal Question:\n\nOn a TV news channel, the evening news starts at same time every day. The probability that Mr Li gets home from work in time to watch the news is $0.3$\n\nIn a particular week of five working days, what is the probability that Mr Li gets home in time to watch the news on three consecutive days?\n\nAttempt:\n\nPossible arrangements with three consecutive successes in 5 trials:\n\nSSS..\nFSSS.\nFFSSS\nSFSSS\n\n\nI calculated the respective probabilities for these arrangements as follows:\n\n$0.3^3 + 0.3^3 0.7 + 0.3^3 0.7^2+0.3^4 0.7 = 0.0648$\n\nActual answer (at the back of the book) = $0.05913$\n\nMy question is what I am doing wrong here (or is the book wrong?)\n\nAlso, I would really appreciate if someone could tell me a general way to solve the problems of this type where probability of $k$ consecutive success in $n$ trials is asked. In this question I was able to manually find the possible arrangements with consecutive successes but if the number of trials is high for example 200, how would I approach this problem then.\n\nThanks very much.\n\nFor a more general approach, consider the probability of not having 3 successes in a row out of $n$ trials; call this probability $P(n)$, and let's say the probability of a single success is $p$, with $q=1-p$. If we have $n$ trials, condition the probability on the number of successes at the end of the $n$ trials: the $n$ trials must end in $...F$, $...FS$, or $...FSS$, so we have the recursion\n\n$$P(n) = q \\; P(n-1) + pq \\; P(n-2) + p^2q \\; P(n-3) \\qquad \\text{for } n \\ge 3$$ with $P(0) = P(1) = P(2) = 1$.\n\nIt's not hard to see how to extend this approach to longer strings of successes.\n\nLet $$X_i=\\begin{cases} S, & \\text{with prob 0.3 } \\\\ F, & \\text{with prob 0.7 } \\\\ \\end{cases}$$ for $$i=1,2,3,4,5$$ be IID for the outcome of the i-th day. The probability you are asked is the following (if we are talking for exactly 3 days succes in a row)\n\nP(Mr Li get home in time 3 days in a row)= $$P(X_1=X_2=X_3=S)+P(X_1=F,X_2=X_3=X_4=S)+P(X_1=X_2=F,X_3=X_4=X_5=S)= 0.3^3+0.3^30.7+0.3^30.7^2=0.05913$$\n\n(which is the correct answer in the book)\n\n\u2022 This is actually incorrect because it does not take into account $P(X_1=X_3=X_4=X_5=S, X_2=F)$ \u2013\u00a0Couchy311 May 31 at 2:47\n\u2022 I agree with you technically, that's why I said \"if we are talking for exactly 3 days S in a row\"...it seems like the problem should have clarified that or give another answer in the back.. \u2013\u00a0sakas May 31 at 3:08\n\u2022 I see, thanks for clarifying :) \u2013\u00a0Couchy311 May 31 at 3:11", "date": "2019-10-24 00:39:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1175625/calculating-probability-for-three-consecutive-successes", "openwebmath_score": 0.6149606704711914, "openwebmath_perplexity": 137.38488907838854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765604649332, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.854879987898411}}
{"url": "https://math.stackexchange.com/questions/2106417/is-there-a-fast-way-to-get-the-characteristic-polynomial-of-this-symmetric-matri", "text": "# Is there a fast way to get the characteristic polynomial of this symmetric matrix?\n\nI have to find the characteristic polynomial equation of this matrix\n\n$$A= \\begin{bmatrix}2 &1 &1&1 \\\\1&2&1&1\\\\1&1&2&1\\\\1&1&1&2 \\end{bmatrix}$$\n\nIs there another way than the rather long $\\det(A-\\lambda I)$ method ?\n\nMaybe the fact that $A$ is symmetric ($A =A^t$) may be helpful ?\n\n\u2022 Could I do something like $A-I\\lambda= \\begin{bmatrix}2-\\lambda &1 &1&1 \\\\1&2-\\lambda&1&1\\\\1&1&2-\\lambda&1\\\\1&1&1&2-\\lambda \\end{bmatrix}$ then simplify and calculate the $det(...)$ , or would it change the value of$\\lambda$? \u2013\u00a0HowCanIHelpYou Jan 20 '17 at 18:35\n\u2022 Subtract the identity matrix and then it becomes pretty easy to show that for $A-I$ the eigenvalues will be 4 and zero (with multiplicity 3) \u2013\u00a0Sergei Golovan Jan 20 '17 at 18:35\n\u2022 Interestingly, it seems that the characteristic polynomial of the $n\\times n$ matrix with $2$ for each diagonal entry and $1$ for all other entries is $(n+1-\\lambda)(1-\\lambda)^{n-1}$. This suggests an inductive approach to me. \u2013\u00a0MPW Jan 20 '17 at 19:00\n\u2022 Clearly, the eigenvalues of $A$ and $A-I$ differ. Exactly by a unity each. \u2013\u00a0Sergei Golovan Jan 20 '17 at 19:05\n\u2022 And it seems that the characteristic polynomial of the $n\\times n$ matrix with $a$ for each diagonal entry and $b$ for all other entries is $(a+(n-1)b - \\lambda)(a-b-\\lambda)^{n-1}$. \u2013\u00a0MPW Jan 20 '17 at 19:15\n\nThere's a formula for determinants of block matrices of the form $\\begin{bmatrix} A&B\\\\B&A\\end{bmatrix}$, where $A$ and $B$ are square matrices of the same size: $$\\det\\begin{bmatrix} A&B\\\\B&A\\end{bmatrix}=\\det(A-B)\\det(A+B).$$ Applying this formula , we obtain \\begin{align}\\det(A-\\lambda I)&=\\begin{vmatrix}2-\\lambda&1&1&1\\\\1&2-\\lambda&1&1\\\\1&1&2-\\lambda&1\\\\1&1&1&2-\\lambda \\end{vmatrix}=\\begin{vmatrix}1-\\lambda&0\\\\0&1-\\lambda \\end{vmatrix}\\cdot\\begin{vmatrix}3-\\lambda&2\\\\2&3-\\lambda \\end{vmatrix}\\\\ &=(1-\\lambda)^2\\Bigl[(3-\\lambda)^2-4\\Bigr]=(\\lambda-1)^3(\\lambda-5).\\end{align}\n\nHere is a matrix in which the columns are eigenvectors of your matrix. Indeed, the columns are pairwise orthogonal.\n\n$$\\left( \\begin{array}{rrrr} 1 & -1 & -1 & -1 \\\\ 1 & 1 & -1 & -1 \\\\ 1 & 0 & 2 & -1 \\\\ 1 & 0 & 0 & 3 \\end{array} \\right).$$ It is not an orthogonal matrix, as the columns are of different lengths. However, it can be made orthogonal by dividing each column by its Euclidean length, those being $2, \\sqrt 2, \\sqrt 6, \\sqrt {12}$\n\nThe best, and very short way:\n\nFirst step. Subtract to the rows $2$, $3$ and $4$, the first one.\nSecond step. Add to the first column the sum of the columns $2$, $3$ and $4.$\n\nWe get a triangular determinant.\n\nEDIT: $$\\begin{vmatrix}2-\\lambda &1 &1&1 \\\\1&2-\\lambda&1&1\\\\1&1&2-\\lambda&1\\\\1&1&1&2-\\lambda \\end{vmatrix}\\underbrace{=}_{\\text{First step}}\\begin{vmatrix}2-\\lambda &1 &1&1 \\\\-1+\\lambda&1-\\lambda&0&0\\\\-1+\\lambda&0&1-\\lambda&0\\\\-1+\\lambda&0&0&1-\\lambda \\end{vmatrix}$$ $$\\underbrace{=}_{\\text{Second step}}\\begin{vmatrix}5-\\lambda &1 &1&1 \\\\0&1-\\lambda&0&0\\\\0&0&1-\\lambda&0\\\\0&0&0&1-\\lambda \\end{vmatrix}=(5-\\lambda)(1-\\lambda)^3.$$\n\nThere are some simple tricks that you can use. The eigenvalues of $A$ are those values of $\\lambda$ such that $A-\\lambda I$ is singular, i.e., has rank less than four. Since $$A-1\\cdot I=\\pmatrix{ 1 & 1 & 1 & 1 \\cr 1 & 1 & 1 & 1 \\cr 1 & 1 & 1 & 1 \\cr 1 & 1 & 1 & 1}$$ has rank equal to one, it follows that $\\lambda=1$ is an eigenvalue of $A$ of multiplicity $4-1=3$. Hence $(\\lambda-1)^3$ is a factor in the characteristic polynomial.\n\nIn the matrix $$A-5\\cdot I=\\pmatrix{-3 & 1 & 1 & 1\\cr 1 & -3 & 1 & 1 \\cr 1 & 1 & -3 & 1 \\cr 1 & 1 & 1 & -3},$$ the sum of the elements in each row is zero. Hence $A-5\\cdot I$ is singular, and $\\lambda-5$ is another factor in the characteristic polynomial of $A$.\n\nIt follows that $\\det(A-\\lambda I = (\\lambda-5)(\\lambda-1)^3$.", "date": "2019-07-21 07:24:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2106417/is-there-a-fast-way-to-get-the-characteristic-polynomial-of-this-symmetric-matri", "openwebmath_score": 0.9123859405517578, "openwebmath_perplexity": 141.54818744734348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765622193214, "lm_q2_score": 0.8723473763375643, "lm_q1q2_score": 0.8548799829243309}}
{"url": "http://mathhelpforum.com/calculus/124868-double-check-limit-please.html", "text": "Math Help - Double-check this limit please\n\nCan someone double-check my work here? Thanks, if you can.\n\n$\n\\lim_{x\\to\\infty}[x(1-cos(1/x))]\n$\n\nPut x to the denominator to use L'Hopital's rule, etc. Applied chain rules, etc.\n\n$\n\\lim_{x\\to\\infty}[sin(1/x)]\n$\n\nThus, if x goes to infinity, the above is sin(0) = 0. Final Answer: Zero\n\n2. Yes, it's okay, but you don't need that rule, you can turn that limit into a known one with a simple substitution.\n\n3. Oh, really? What is this substitution?\n\n4. $t=\\frac1x.$\n\n5. tx=1\n\nso itequals to lim [(1-cost)/t]\nt->0", "date": "2015-06-02 15:51:50", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/124868-double-check-limit-please.html", "openwebmath_score": 0.9694116115570068, "openwebmath_perplexity": 2280.8505454055435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9799765598801371, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.8548799792576206}}
{"url": "https://www.math.purdue.edu/pow/discussion/2016/spring/9", "text": "## Spring 2016, problem 9\n\nThe numbers $1, 2, \\dots , n$ are written around a circle in some order. What is the smallest and largest possible sum of the absolute differences of adjacent numbers?\n\n1 year ago\n\nThe maximum is $\\lfloor \\frac{n^2}{2} \\rfloor$. The minimum is $2 (n - 1)$.\n\nProof of the maximality part:\n\nFor a circular arrangement of $1..n$ call the sum of the absolute differences of neighbours the value of the arrangement. For $a, b \\in \\left \\{1..n \\right \\}$ write $a \\rightarrow b$, if $b$ is the right side neighbour of $a$.\n\nIf $n$ is even, call the numbers $\\frac{n}{2} .. n$ the large numbers. Call the rest small. If $n$ is odd call the numbers $\\frac{n + 3}{2} .. n$ the large numbers. Call $\\frac{n + 1}{2}$ the middle number. Call the rest small.\n\nThe value $v$ of the arrangement is given by:\n\n$v := \\sum_{a \\rightarrow b} \\left [max(a,b) - min(a,b) \\right ] = \\sum_{a \\rightarrow b} max(a,b) - \\sum_{a \\rightarrow b} min(a,b)$.\n\nEach of the two sums in the last expression is a sum of $n$ numbers in the range $1..n$, none of which occur more than twice in any of the sums. The value $v$ is clearly no larger than $m$, given by: $m := \\left [ n + n + (n -1) + (n-1) + ... \\right ] - \\left [1 +1 + 2 + 2 + 3 + 3 ... \\right ]$, where the first and the second bracket contain $n$ terms each. By a simple calculation $m = \\lfloor \\frac{n^2}{2} \\rfloor$. But the value $m$ is indeed attained:\n\nFor $n$ even arrange the numbers like:\n\nlarge $\\rightarrow$ small $\\rightarrow$ large $\\rightarrow$ small ... $\\rightarrow$ large$\\rightarrow$ small\n\nFor $n$ odd arrange like:\n\nmiddle $\\rightarrow$ large $\\rightarrow$ small $\\rightarrow$ large $\\rightarrow$ small ... $\\rightarrow$ large$\\rightarrow$ small. Done.\n\nProof of the minimality part:\n\nProof by induction over $n$: The induction start is trivial. So assume, that the minimum value for arrangements of numbers in $1..n$ is given by $m_n := 2 (n -1)$. Consider a minimum-value arrangement $A_{n+1}$ of the numbers $1..(n+1)$. with value $v_{n+1}$. Now take away the number $n+1$ from $A_ {n+1}$. You get an arrangement $A_n$ of the numers $1..n$, with value $v_{n}$. By an easy calculation $v _n \\leq v_{n+1} -2$. Thus $v_{n+1}$ cannot be smaller than $2 (n + 1 -1)$ because otherwise $v_n$ would be smaller than $2 (n - 1)$. The minimum can also be no larger than $2 (n + 1 -1)$, since this value is attained for simply placing the numbers in ascending order alongthe circle. Done.\n\nHello, I get a slightly different answer for the maximum. It must be a whole number.\n\nI get,\n\nmaximum = trunc(x^2/2),\n\ni.e., if n is even, maximum = n^2/2\n\nif n is odd, maximum = n^2/2 - 1/2\n\nJao\n\ncgjoa3 1 year ago\n\nHi Jao. The notation $\\lfloor \\cdot \\rfloor$ I used is defined by:\n\n$\\lfloor x \\rfloor := floor(x)$.\n\nFor $x> 0$: $floor(x) =trunc(x)$\n\nNelix 1 year ago\n1 year ago\n\nlet start with number 1, minimum possible difference for next number is 1 so next number 2 and so on so minimum sum = n\n\nfor maximum::\n\nlet start with 1 and for max difference next number should be n, number next to n such that max. difference is there should be 2\n\nso the series is 1,n,2,n-1,3,n-2,4,n-3....., and last n/2 or (n-1)/2 for odd and even n\n\nso max = nn/2 or (nn-1)/2\n\n1 year ago\n\nAre these called purdue problems of week or nelix's math proofs of the week?! Jk wish more mathletes would participate !\n\n' half the problem for half the points ' ( lion's share had this been a putnam )Minimality of 2(n-1) Now we will show that $2n-2$ is minimal. To do so, remark that $1$ and $n$ must be on the circle. By the Triangle Inequality, the sum of the positive differences along the minor arc between $1$ and $n$ must be at least $n-1$ and similarly for the otherrh arc", "date": "2018-01-23 10:09:38", "meta": {"domain": "purdue.edu", "url": "https://www.math.purdue.edu/pow/discussion/2016/spring/9", "openwebmath_score": 0.7037570476531982, "openwebmath_perplexity": 461.29954721823896, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979976559880137, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.8548799760053623}}
{"url": "https://math.stackexchange.com/questions/1869761/what-are-the-odds-of-flipping-a-coin-100-times-and-seeing-hhhht", "text": "# What are the odds of flipping a coin 100 times and seeing HHHHT? [closed]\n\nWhat are the odds of flipping a coin 100 times and seeing exactly four consecutive heads? Any more than four heads in a row, such as \"HHHHH\" would not be considered a string of four consecutive heads. Seeing 10 sets of \"HHHHT\" would allow a max of 20 consecutive patterns. How would you expect to find the number of times an isolated string of exactly 4 heads in a row in $n$ coin flips?\n\n## closed as off-topic by Did, Michael Albanese, Edward Jiang, Zain Patel, Chill2MachtJul 25 '16 at 0:06\n\nThis question appears to be off-topic. The users who voted to close gave this specific reason:\n\n\u2022 \"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.\" \u2013 Did, Michael Albanese, Edward Jiang, Zain Patel, Chill2Macht\nIf this question can be reworded to fit the rules in the help center, please edit the question.\n\n\u2022 I don't understand, you seem to ask many questions at once. \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 17:48\n\u2022 Is your question the expected number of times you would expect to find $HHHHT$ in a sequence of $100$ flips? \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 17:49\n\u2022 I flip a coin 100 times. How many times should I see HHHH? A HHHHH or anything greater will not be considered as a HHHH. \u2013\u00a0Triumph Jul 24 '16 at 17:51\n\u2022 No just HHHHT but THHHHT \u2013\u00a0Triumph Jul 24 '16 at 17:52\n\u2022 I am pretty sure the question is \"Suppose I flip a coin 100 times. How many subsequences of exactly four heads in a row (no more or less) should I expect to see?\" \u2013\u00a0user326210 Jul 24 '16 at 17:54\n\nIn any sequence of $100$ flips, \u201cmark\u201d each spot $HHHH$ occurs (and occurs not within a longer run of $H$\u2019s) between the middle $H$\u2019s. For example, the sequence $$HHTHHHTTTHHHHTHT\\dots TTHHHHHTHTHHTTHHHH$$ would be marked $${HHTHHHTTTHH}^{\\color{red}|\\!}{HHTHT\\dots TTHHHHHTHTHHTTHH}^{\\color{red}|\\!}{HH}^{}.$$\n\nAmong the $97$ positions where marks could occur (from after the first two flips to before the final two), a mark does appear in each of the $2^{\\rm nd}$ through $96^{\\rm th}$ ($95$ in all) of those spots (as $THH^{\\color{red}|\\!}HHT$) with probability $\\frac{1}{2^6}$ (the chance the surrounding sequence of six flips is $THHHHT$). In the first and last positions, the probability is $\\frac{1}{2^5}$ (the chance of $HH^{\\color{red}|\\!}HHT$ for position $1$ and $THH^{\\color{red}|\\!}HH$ for position $97$).\n\nBy the linearity of expectation, the expected number of marks is $95\\cdot\\frac1{2^6}+2\\cdot\\frac1{2^5}=\\frac{99}{64}$.\n\n[Added] It may be easier to focus on the $H$\u2019s and $T$\u2019s instead of the spaces between them, so another way to mark the runs-of-$4$ is to color in red the last $H$ of each run, like $$HHTHHHTTTHHH\\color{red}HTHT\\dots TTHHHHHTHTHHTTHHH\\color{red}H.$$ The first three $H$\u2019s are colored red with probability $0$, the fourth with probability $\\frac{1}{2^5}$, the next $95$ with probability $\\frac{1}{2^6}$, and the last with probability $\\frac{1}{2^5}$, giving the same result.\n\n\u2022 Is it possible for you to let me know what the answer would be for 8 out of 100? \u2013\u00a0Triumph Jul 27 '16 at 1:48\n\u2022 Try to modify my answer for runs of $8$. There are no new situations that come up, so it shouldn\u2019t be very hard. \u2013\u00a0Steve Kass Jul 27 '16 at 18:32\n\nThis seems to fall easily to linearity of expectation. Let $a_1,a_2\\dots a_n$ be your sequence of outcomes.\n\nLet $X_1$ be the indicator variable that $a_1,a_2,a_3,a_4=H$ and $a_5=T$\n\nLet $X_{97}$ be the indicator variable that $a_{96}=T$ and $a_{97},a_{98},a_{99},a_{100}=H$\n\nFinally, for $2\\leq n \\leq 96$ let $X_n$ be the indicator that $a_{n-1}=T,a_n,a_{n+1},a_{n+2},a_{n+3}=H,a_{n+4}=T$\n\nThe random variable you want is $\\sum_{i=1}^{97} X_n$.\n\nBy linearity of expectation its expectation is $E[X_1]+95E[X_2]+E[X_{97}]=\\frac{1}{32}+\\frac{95}{64}+\\frac{1}{32}=\\frac{99}{64}$\n\n\u2022 What I answered is the expected number of times you can find a substring of the form $HHHH$ that does not have another $H$ to either side. Assuming the flips are independent, and each outcome has probability $\\frac{1}{2}$. \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 18:16\n\u2022 When they are four $H$ it is $\\frac{99}{64}$ and when there are three It is $\\frac{1}{16}+\\frac{96}{32}+\\frac{1}{16}$. \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 18:45\n\u2022 then it is $\\frac{1}{2^{11}}+\\frac{89}{2^{12}}+\\frac{1}{2^{11}}$ \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 19:31\n\u2022 if you throw $10000$ times then you should get $10$ consecutive (and not more) an expected number of $\\frac{1}{2^{11}}+\\frac{9989}{2^{12}}+\\frac{1}{2^{11}}$ \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 19:52\n\u2022 I told you, $\\frac{1}{2^{11}}+\\frac{9989}{2^{12}}+\\frac{1}{2^{11}}$, which is approximately $2.43$ \u2013\u00a0Jorge Fern\u00e1ndez Hidalgo Jul 24 '16 at 21:10\n\nFor $i=1$ to $97$, let random variable $X_i$ be equal to $1$ if there is a string of HHHH that begins at $i$ and does not extend, and let $X_i=0$ otherwise. We want the expectation of $X_1+\\cdots+X_{97}$. By the linearity of expectation, this is $E(X_1)+\\cdots+E(X_{97})$.\n\nSo we need only calculate the $\\Pr(X_i)=1$.\n\nThese are not all equal. If $i=1$ or $i=97$, we have $\\Pr(X_i=1)=\\frac{1}{32}$. For all the other $i$, we have $\\Pr(X_i=1)=\\frac{1}{64}$. That is because in all but \"end\" cases, the string of $4$ H must be flanked by T on both sides.\n\nThe required expectation is therefore $2\\cdot \\frac{1}{32}+95\\cdot\\frac{1}{64}$.\n\n\u2022 Would this factor in the possibility of 10 heads in a row that would change you 100 coin flips into 90. From the start you should automatically drop the 100 flips down to 50 with the assumption that the other 50 flips would be tails correct? What are the odds of 5 heads or 6 heads each one of those will take more flips away from what you need to achieve 4 heads. \u2013\u00a0Triumph Jul 24 '16 at 18:14\n\u2022 The expectation calculation (indirectly) takes care of $10$ heads in a row by not counting them, because then $X_i=0$ for $i=1$ to at least $11$. \u2013\u00a0Andr\u00e9 Nicolas Jul 24 '16 at 18:17", "date": "2019-05-25 03:46:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1869761/what-are-the-odds-of-flipping-a-coin-100-times-and-seeing-hhhht", "openwebmath_score": 0.8295883536338806, "openwebmath_perplexity": 246.35348904828874, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765604649332, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8548799732632495}}
{"url": "https://math.stackexchange.com/questions/2399255/simple-probability-question-perfect-quiz-score", "text": "# Simple Probability Question: Perfect quiz score\n\na) Students have two quizzes that they must take in their undergraduate studies unit. The probability of getting a perfect score on the first quiz is 68%, the probability of getting a perfect score on the second quiz is 74%, and the probability of falling short of a perfect score on both quizzes is 19%. What is the probability of scoring a perfect score on both quizzes?\n\nMy attempt: Pr(Scoring a perfect score on both quizes) $= 0.68 \\cdot 0.74 = 0.5032$\n\nb) Refer to the probabilities described in question a. If a student has achieved a perfect score on the first quiz, what is the conditional probability of getting a perfect score on the second quiz?\n\nMy attempt:\n\n$$\\frac{\\text{Pr(Scoring a perfect score on both quizzes)}}{\\text{Pr(Perfect score on second Quiz)}} = 0.68$$\n\nbut I feel like there is more to both questions\n\n\u2022 what has been tried ? what is the context ( where did you find the question) ? \u2013\u00a0user451844 Aug 19 '17 at 14:22\n\u2022 a) Pr(Scoring a perfect score on both quizes) = 0.68 * 0.74 = 0.5032 b) Pr(Scoring a perfect score on both quizes)/Pr(Perfect score on second Quiz) = 0.68 but I feel like there is more to both questions \u2013\u00a0user473207 Aug 19 '17 at 14:29\n\u2022 if you fall short 19% of the cases, what happens the other 81% of the cases ? \u2013\u00a0user451844 Aug 19 '17 at 14:44\n\u2022 @user473207 Your calculations assume that the two events (obtainoing a perfect score on quiz 1 and obtaining a perfect score on quiz 2) are independent. BUt just conceptually, that is unlikely: SOmeone who gets a perfect score on quiz 1 is probably more likely to get a perfect score on quiz 2 than someone who does not get a perfect score on quiz 1 ... so you are right to suspect there is more to this! \u2013\u00a0Bram28 Aug 19 '17 at 14:47\n\u2022 @user473207 Also, the expression 'falling short of a perfect score on both quizzes'' is a bit ambiguous (is it that you don't get a perfect score on quiz 1 and also don't get a perfect score on quiz 2, or is it that it is not true that you do perfect on both quizzes (so you can get a perfect score on one of the two quizzes, just not on both quizzes)). What they undoubtedly mean is the first interpretation. SO, you know that there is a 19% chance of not getting a perfect score on quiz 1 and also not a perfect score on quiz 2. NOw, as Roddy asked: what does the other 81% then represent? \u2013\u00a0Bram28 Aug 19 '17 at 14:49\n\nLet $A$ be the event that a perfect score is obtained on the first quiz; let $B$ be the event that a perfect score is obtained on the second quiz. We are given \\begin{align*} P(A) & = 0.68\\\\ P(B) & = 0.74\\\\ P(A^C \\cap B^C) & = 0.19 \\end{align*} Hence, \\begin{align*} P(A^C) & = 1 - P(A) = 0.32\\\\ P(B^C) & = 1 - P(B) = 0.74 \\end{align*} The probability of not obtaining a perfect score on at least one of the quizzes is $$P(A^C \\cup B^C) = P(A^C) + P(B^C) - P(A^C \\cap B^C) = 0.32 + 0.26 - 0.19 = 0.39$$ Thus, the probability of obtaining a perfect score on both quizzes is $$P(A \\cap B) = 1 - P(A^C \\cup B^C) = 1 - 0.39 = 0.61$$ so your assumption that the probabilities of obtaining a perfect score on each quiz are independent was false.\nThe probability that a student who received a perfect mark on the first quiz obtains a perfect mark on the second quiz is $$P(B \\mid A) = \\frac{P(A \\cap B)}{P(A)} = \\frac{0.61}{0.68} = \\frac{61}{68}$$", "date": "2020-09-26 18:47:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2399255/simple-probability-question-perfect-quiz-score", "openwebmath_score": 0.5761601328849792, "openwebmath_perplexity": 485.61639470676636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765552017674, "lm_q2_score": 0.872347369700144, "lm_q1q2_score": 0.8548799702980698}}
{"url": "http://mathhelpforum.com/advanced-algebra/150432-eigenvectors.html", "text": "# Math Help - Eigenvectors!\n\n1. ## Eigenvectors!\n\n(i) Determine all the eigenvalues of A\n(ii) For each eigenvalue of A, find the set of eigenvectors corresponding to\n\nA = ( 1 2 )\n( 3 2 )\n\nI found eigenvalues of A to be 4 and -1.\n\nI also found the eigenvectors to be (2/3,1) for =4 and (-1,1) for =-1\n\nBUT the solution in the back of the book says (2,3) for =4 and (1,-1) for =1\n\nI'm soo confusedd! Can someone tell me what's wrongg? Am I calculating the eigenvectors wrong?\n\nHere's how i calculate eigenvector for =4\n\nA-4I = (-3 2)\n(3 -2)\n\nthen (-3 2 |0)\n(3 -2 |0)\n\nand i end up with x1 -(2/3) x2 =0. so x2 = t, and x1 = (2/3)t. eigenvector = (2/3,1) but book says (2,3)\n\nand for =-1\n\nA-(-1)I = (2 3 )\n(2 3 )\n\nthen (2 3 |0)\n(2 3 |0)\n\nand i end up with x1+x2=0. x2= t and x1 = -t. eigenvector = (-1,1) but book says (1,-1)\n\n2. You don't have a problem. Note that your eigenvectors are just multiples of the book's answer. This is fine!\n\n3. Hi\n$\\texttt{det}(A-I_{2}X)=\\texttt{det}\\begin{pmatrix}\n1-X & 2\\\\\n3 & 2-X\n\\end{pmatrix}=(1-X)(2-X)-6=X^2-3X-4=(X-4)(X+1)$\n\nand thus your eigenvalues are 4 and -1.\nFor 4\n$A\\begin{pmatrix}\nx\\\\\ny\n\\end{pmatrix}=4\\begin{pmatrix}\nx\\\\\ny\n\\end{pmatrix}\\Leftrightarrow \\begin{pmatrix}\nx+2y\\\\\n3x+2y\n\\end{pmatrix}=$\n$\\begin{pmatrix}\n4x\\\\\n4y\n\\end{pmatrix}\\Leftrightarrow \\left\\{\\begin{matrix}\n-3x+2y=0\\\\\n3x-2y=0\n\\end{matrix}\\right.\\Leftrightarrow \\left\\{\\begin{matrix}\nx=\\frac{2}{3}t\\\\\ny=t,t\\in \\mathbb{R}\n\\end{matrix}\\right.$\n\nand thus the eigenvectors are of the form,\n$x=x_1(\\frac{2}{3},1),x_1\\in \\mathbb{R}\\setminus \\left \\{ 0 \\right \\}$\n\n4. The set of all eigenvectors of a matrix A, corresponding to a given eigenvalue, $\\lambda$, form a subspace. In particular, any multiple of an eigenvector is also an eigenvector, corresponding to the same eigenvalue: If $Av= \\lambda v$ and \"r\" is any scalar, then $A(rv)= r(Av)= r(\\lambda v)= (r\\lambda)v= \\lambda (rv)$.", "date": "2014-10-01 11:06:12", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/150432-eigenvectors.html", "openwebmath_score": 0.9363395571708679, "openwebmath_perplexity": 1227.5270976313536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765572485543, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8548799639529332}}
{"url": "http://lotoblu.it/ixah/triple-integral-limit-calculator.html", "text": "# Triple Integral Limit Calculator\n\nDefinite Integral Calculator computes definite integral of a function over an interval utilizing numerical integration. The integral is equal to the area of. Functions like sin and sqrt work, as do special constants like pi and e. 2 Double Integrals over General Regions In this section we define and evaluate double integrals over bounded regions in the plane which are more general than rectangles. In double integrals we saw how to calculate volume by integrating a three dimensional function using limits of intgration of a bounded AREA. Triple integrals 3. For example, if f ( x) is positive for. To get a viewing window containing a and b, these values must be between Xmin. Fast and easy to use. Write a triple integral, including limits of integration, that gives the volume between 3x+y+z=0 and 4x+4y+z=0, and above x+y<=1, x>=0, y>=0. TRIPLE INTEGRALS 3 5B-2 Place the solid hemisphere D so that its central axis lies along the positive z-axis and its base is in the xy-plane. Notes on Triple Integration Dr. Convert this triple integral into cylindrical coordinates and evaluate $\\int_{-1}^{1}\\int_{0}^{\\sqrt{1-x^2}}\\int_{0}^{y}x^2dz\\; dy\\; dx onumber$ Solution. A volume integral is a specific type of triple integral. The sphere x2 +y2 +z2 = 4 is the same as \u02c6= 2. Second Derivative. Triple Integrals in Cartesian Coordinates: Over rectangular. Includes derivatives, integration, volume, quadratic equation and trig identities. Elliott Jacobs On Wednesday, March 4, you saw how to set up triple and calculate triple integrals. Now is the lower limit and is. For example, if f ( x) is positive for. pro [email\u00a0protected] It is the height of a thin stick as in Section 14. The key idea is to replace a double integral by two ordinary \"single\" integrals. Let's do limit comparison to 1/t3: lim t\u2192\u221e 1/t3 1/t3\u2212t =lim t\u2192\u221e t3\u2212t t3 =lim t\u2192\u221e t3\u2212t t3. $$\\int_0^4 \\int_{0}^{1-x} \\int x^2 + y^2 dzdydx$$ I think those two limits may be correct but I don't know how to get the third, I understand how to find limits form a shape like a tetrahedron but a bit confused and my notes don't help me with this shape. We calculate as follows:. dzdxdy, then once you find your Z limits in the first integral, then you are done with Z altogether, the next step to solve the triple integral is to project into the remaining variables' plane, in other words, project into the x-y plane. So we should just calculate that limit, for arbitrary a, and then let a!1. V = \\iiint\\limits_U { {\\rho ^2}\\sin \\theta d\\rho d\\varphi d\\theta }. Changing the order of integration of a triple integral blackpenredpen. Third Derivative. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. restart: Setting limits of integration and evaluating. \u2022 Evaluate double integrals over general regions. When there are limits, and we need to use U-Substitution, there are a few things we need to keep in mind:. Summary : The integral function calculates online the integral of a function between two values. Recently Asked Math Questions. Arc Length Cartesian & Polar Coordinates, 2D & 3D Parametric Curves. In python we use numerical quadrature to achieve this with the scipy. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each $\\Delta x\\times\\Delta y\\times\\Delta z. Set up (but do not evaluate) iterated triple integrals, with appropriate limits, for nd the volume of the solid bounded by z= x2 + y2 and z= 8 x2 y2 in: (a)rectangular coordinates (b)cylindrical coordinates 9. The purpose of these notes is to present a number of triple integral examples and discuss how to set up the limits of integration. Notes on Triple Integration Dr. The limits of integration for this graph are (0,2). You can also use it to solve differential and integral equations. Multiple Integral Calculator - eMathHelp This site contains an online calculator that finds multiple integrals (double or triple integrals). The concept of integrals is fundamental in calculus, and has broad application in all engineering disciplines. V = \\iiint\\limits_U {\\rho d\\rho d\\varphi dz}. Watch if the algorithm is converging. Let's return to the previous visualization of triple integrals as masses given a function of density. The int function can be used for definite integration by passing the limits over which you want to calculate the integral. I am trying to solve below mentioned (image) equation in R for double integral over an area. 0 (R14), you will need to use a function file or. Simplify a calculation by changing the order of integration of a triple integral. This is somewhat subtle in the physical interpretation but can be summarized as \"generality\". An integral is a mathematical result that represents the area between a function and a plane (e. , 26 MB] Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. Sometimes you need to change the order of integration to get a tractable integral. For example, if you tried to evaluate \u222b1 0\u222b1 xey2dydx directly, you would run into trouble. Triple Integral Cylindrical Coordinates. To set up a double. Create the worksheets you need with Infinite Calculus. Calculate the average value of a function of three variables. Integral bounds , also called limits of integration, define the area that you\u2019ll be integrating. set up triple integrals to compute the volume of a solid. Math 6B is the second quarter of a two quarter sequence in vector calculus and infinite series. 6 Implicit Differentiation. So, if possible, please write the logic for the solution. 973 #55,56) Calculate mass by integrating density. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. Evaluating Double Integrals by Repeated Integrals 5. This calculator evaluates derivatives using analytical differentiation. Never runs out of questions. ranges here in the interval 0 \\le x \\le 1, and the variable y. ) Finding the Limits. Calculate the volume of the solid. You can also easily calculate multiple integrals as well as use mathematical constants such as. The limit of the Riemann sum is the triple integral of f over D and f is continuous. For example: \ud835\udc5f \ud835\udc5f \ud835\udf03 3 \u22123 2 0 2\u03c0 0 is the triple integral used to calculate the volume of a cylinder of height 6 and radius 2. For a sequence indexed on the natural number set , the limit is said to exist if, as , the value of the elements of get arbitrarily close to. ( Click here for an explanation) [ ti-83/ti-84 ] Approximate Integration. Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. Absolute Minimum. Triple integral - volume of sphere. Evaluate a triple integral by expressing it as an iterated integral. ranges in the interval 0 \\le y \\le 2 \u2013 2x. Triple Integral Calculator. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. (1 point) Write a triple integral including limits of integration that gives the volume of the cap of the solid sphere x2 + y +2 s 25 cut off by the plane z 3 and restricted to the first octant. Triple Integral Spherical Coordinates. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\\left[ {a,b} \\right] \\times \\left[ {c,d} \\right]$$ $$\\times \\left[ {p,q} \\right]$$ (Figure $$1$$). If you have a function f(x), there are several ways to mark the derivative of f when it comes to x. sets, logic, proofs. Fill in the integrand, the limits, and the integrating variable for. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Analysis & calculus symbols table - limit, epsilon, derivative, integral, interval, imaginary unit, convolution, laplace transform, fourier transform. the triple integral of a continuous function $$f(x,y,z)$$ over a rectangular solid box $$B$$ is the limit of a Riemann sum for a function of three variables, if this limit exists Contributors Gilbert Strang (MIT) and Edwin \"Jed\" Herman (Harvey Mudd) with many contributing authors. 7 Triple Integrals Be able to evaluate a given triple integral. Evaluate a Triple Integral Using Cylindrical Coordinates - Triple Integral of e^z Evaluate a Triple Integral Using Spherical Coordinates - Triple Integral of 1/(x^2+y^2+z^2) Find the Moment of Inertia about the z-axis of a Solid Using Triple Integrals Find the Center of Mass of a Solid Using Triple Integrals. \u2022 Calculate the Jacobian of a transformation of two and three variables. Calculate the volume of the solid. Create the worksheets you need with Infinite Calculus. I The average value of a function in a region in space. Derivative Calculator. I Triple integrals in arbitrary domains. Let\u2019s do limit comparison to 1/t3: lim t\u2192\u221e 1/t3 1/t3\u2212t =lim t\u2192\u221e t3\u2212t t3 =lim t\u2192\u221e t3\u2212t t3. They will make you \u2665 Physics. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. What does mean where is a region in the plane. This is a topic that takes some practice. V = \\iiint\\limits_U { {\\rho ^2}\\sin \\theta d\\rho d\\varphi d\\theta }. Calculate the average value of a function of three variables. So I'll start with the triple integral. To find those limits on the z integral, follow a line in the z direction. Further, it is possible to show that the limit. BYJU\u2019S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. Absolutely free, online math solution. Limit Calculator calculates an established limit of the function with respect to a variable in a specific point. is given by where R(xyz) is the region of integration in xyz space, R(uvw) is the corresponding region of integration in uvw space, and the. Next, I'll substitute the limits. io development by creating an account on GitHub. One of the ways in which definite integrals can be improper is when one or both of the limits of integration are infinite. Define the indefinite integral of a vector-valued function. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. The limits on z run from 0 to h. quad command. Triple Integral Spherical Coordinates. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. Math terminology from differential and integral calculus for functions of a single variable. sets, logic, proofs. Online Triple & Double Integral Calculator With Steps. And then finish with dx to mean the slices go in the x direction (and approach zero in width). index: subject areas. Fraction calculator. What Everybody Dislikes About Double Integral Calculator and Why. The general steps required to perform the above integral. Since the plane ABC. Wolfram|Alpha is a great tool for calculating indefinite and definite triple integrals. Figure 12-10 shows an example. In this example, since the limits of integration are constants, the order of integration can be changed. In many cases, it is convenient to represent the location of in an alternate set of coordinates, an example of which are the so-called polar coordinates. ha \u03c0ha Outer integral: 2\u03c0 =. In Calculus, the four important concepts are limits, continuity, derivatives and integrals. Second Derivative. Integrals involving radicals for instance, we want to get rid of the square root. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. You can calculate integrals numerically using techniques such as the Simpson quadrature, Lobatto quadrature, and Gauss. It enters the prism at z = 0 and exits at the sloping face y + 32 = 3. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\\left[ {a,b} \\right] \\times \\left[ {c,d} \\right]$$ $$\\times \\left[ {p,q} \\right]$$ (Figure $$1$$). Reversing the order of integration. Create the worksheets you need with Infinite Calculus. More than just an online integral solver. If you're behind a web filter, please make sure that the domains *. Enter the series to calculate its sum: This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. Integral Calculator If you were looking for a way to calculate the Integral value of a set of mumbers, then the Integral calculator is exactly what you need. To something small like 0. Partial Derivative. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. This tutorial demonstrates how to evaluate integrals using the TI-89, TI-92+, or Voyage 200 graphing calculators. Expression Calculator evaluates an expression in a given context. The integral is equal to the area of. Write a triple integral, including limits of integration, that gives the volume between 3x+y+z=0 and 4x+4y+z=0, and above x+y<=1, x>=0, y>=0. It is easier to calculate triple integrals in spherical coordinates when the region of integration U is a ball (or some portion of it) and/or when the integrand is a kind of f\\left ( { {x^2} + {y^2} + {z^2}} \\right). Muliple Integration Section 1: DOUBLE INTEGRALS PROBLEM: Consider the solid E in 3-space bounded above by the surface z = 40 \u2212 2xy and bounded below by the rectangular region D in the xy-plane (z = 0) de\ufb01ned by the set D = {(x,y) : 1 \u2264 x \u2264 3, 2 \u2264 y \u2264 4}. Next, I\u2019ll substitute the limits. Integration over surfaces, properties, and applications of integrals. Hold \u02daand xed, and let \u02c6increase. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 \u2264x \u2264 5 -----> 5r2 \u2264 x \u2264 5, since each cross-section is a full circle 0 \u2264 \u03b8 \u2264 2\u03c0. Explore the solid defining the boundaries of the region for a triple integral. Changes of variable can be made using Jacobians in much the same way as for double integrals. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Using (19) and (22), we calculate \u2202(x,y) \u2202(u. Multiple-version printing. How to Use Definite Integral Calculator Integration can. sets, logic, proofs. Try int(x^2,x=t+2. The Integral Calculator Trap The last result is provided by taking the very first limit from the second. It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the calculator is able to calculate, it returned. Among other things, they let us calculate the volume under a surface. So it will be. In polar coordinates, the point is located uniquely by specifying the distance of the point from the origin of a given coordinate system and the angle of the vector from the origin to the point from the positive -axis. An example shows how to set them up and how to evaluate them. Integral online. Enter a valid algebraic expression to find the derivative. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field. Triple integrals 3. First of all, I\u2019ll integrate with respect to. As a final example, we see how to compute the length of a curve given by parametric equations. Triple Integrals in Cartesian Coordinates: Over rectangular. Limits at Jump Discontinuities and Kinks. Evaluating Double Integrals by Repeated Integrals 5. Triple Integrals, Changing the Order of Integration, Part 1 of 3. The Riemann Sum formula is as follows: Below are the steps for approximating an integral using six rectangles: Increase the number of rectangles (n) to create a better approximation: Simplify this formula by factoring out w [\u2026]. Watch if the algorithm is converging. Just as with double integrals, the only trick is determining the limits on the iterated integrals. The syntax in dblquad is a bit more complicated than in Matlab. There are three steps that must be done in order to properly convert a triple integral into cylindrical coordinates. What Everybody Dislikes About Double Integral Calculator and Why. Try this handy Limit calc right now!. The key idea is to replace a double integral by two ordinary \"single\" integrals. Double and triple integrals. (Very simple problem). Apply, evaluate, and understand integrals of multi-variable scalar-valued functions. You can also check your answers! Interactive graphs/plots help visualize and better understand the functions. Mathispower4u 6,729 views. Don't forget to use the magnify/demagnify controls on the y-axis to adjust the scale. Please Explain Reasoning Behind Limits. Double Integrals over General Regions Type I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinates dA = r dr (d theta) Examples Multiple integrals in physics Double integrals in physics Triple integrals in physics Integrals in Probability and Statistics. Calculate Partial derivatives with incredible ease! Use this handy partial derivative calculator with a step-by-step solution and graph. Higher Order Derivatives. Finding volume for triple integrals using spherical coordinates. , when integrate is called as integrate (expr, x, a, b). I Triple integrals in arbitrary domains. Definite Integral Calculator computes definite integral of a function over an interval using numerical integration. index: subject areas. 9 Best Free Integral Calculator Software For Windows Here is a list of best free Integral Calculator Software to solve integrations. using Type I region using Type region calculate one fo the above integrals. In spherical coordinates, the volume of a solid is expressed as. advanced topics. The integral is the line integral of a continuous real-valued function or , i. Automatic spacing. Any time we have an iterated integral (and we can go beyond triple integrals) we can have the limits be non-constant functions. MTH 229 projects. It enters the prism at z = 0 and exits at the sloping face y + 32 = 3. Also recall the chapter prelude, which showed the opera house l\u2019Hemisph\u00e8ric in Valencia, Spain. 5b: Integrals in probability and statistics:. Series Expansions Number & Power Series, Fourier / Taylor / Laurent / Puiseux Series. Homework Statement Find the centroid of the solid: Limit of a function as n approaches infinity Finding the eigenfunctions and eigenvalues associated with an operator. As with most definite integrals, you should ignore the bounds (0 and 2) at first and focus on how to find an antiderivative of the function inside the integral. Use the upper right corner as your sample point of each rectangle to approximate the double integral. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. So it will be. To begin, enter the limit expression in graphing or \"y =\" mode, in Table Setup, set Tbl to start to the arrow-number, and then set. I Triple integrals in arbitrary domains. Online Derivative Calculator. You are allowed to take one. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. Solve limits step-by-step. See exercise 27. Instead of integrating a function of two variables over an area, we are integrating a function of three variables over a volume. De nition: The triple integral of fover Bis ZZZ B f(x;y;z)dV = lim jjPjj!0 Xl i=1 m j=1 Xn k=1 f(x ijk;y ijk;z ijk) V ijk provided that the limit exists. The innermost integral is the one that is integrated over first. In all of our examples above, the integrals have been indefinite integrals - in other words, integrals without limits of integration (the \"a\" and \"b\" in the statement \"the integral from a to b\"). Here is a list of best free Integral Calculator Software to solve integrations. Let's return to the previous visualization of triple integrals as masses given a function of density. In many cases, it is convenient to represent the location of in an alternate set of coordinates, an example of which are the so-called polar coordinates. Also recall the chapter prelude, which showed the opera house l\u2019Hemisph\u00e8ric in Valencia, Spain. Changing the order of integration in triple integrals Triple Integrals, Changing the Order of Integration,. numbers & symbols. Improper integrals - part 1 - introduction; integrals with an infinite limit of integration [video; 21 min. The mechanics for double and triple integration have been wrapped up into the functions dblquad and tplquad. A triple integral can be represented as , where f(x,y,z) is the integrated function defined over the three-dimensional shape E , and E is the region of integration in the (x,y,z) three. Includes derivatives, integration, volume, quadratic equation and trig identities. In triple integrals, the integral will be taken of a four-dimensional function using limits of integration of a bounded VOLUME. Type in the triple integral problem to solve To get started, type in a value of the triple integral and click \u00abSubmit\u00bb button. Since is defined as a limit of Riemann sums, the continuity of is enough to guarantee the existence of the limit, just as the integral exists if g is continuous over. Triple Integral Calculator at a Glance The 5-Minute Rule for Triple Integral Calculator. The text is Vector Calculus by M. Input a function, the integration variable and our math software will give you the value of the integral covering the selected interval (between the lower limit and the upper limit). By this, I mean you can take the volume of any three dimensional object with a triple integral, but you are somewhat limited with a double integral. Integration over surfaces, properties, and applications of integrals. If expr is a constant, then the default integration variable is x. We can solve for z to determine the upper limit of integration. }\\) Activity 11. This differential volume can be expressed in six possible ways. The x and y coordinates lie in a disk of radius a, so 0 \u2264 r \u2264 a and 0 <\u03b8 \u2264 2\u03c0. ) is written as y = 2 \u2013 2x. Define the indefinite integral of a vector-valued function. Observe that this function is symmetric across all eight octants, so we. Change the camera position and the direction of view in three dimensions. surface integral (1) is de\ufb01ned to be this limit. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 9 Setting up the volume as a triple integral in spherical coordinates, we have: ZZZ S dV = Z 0 Z 2\u02c7 0 Z R 0 \u02c62 sin\u02dad\u02c6d d\u02da = Z 0 Z 2\u02c7 0 [1 3 \u02c6 3]\u02c6=R \u02c6=0 sin\u02dad d\u02da = 1 3 R 3(2\u02c7)[ cos\u02da]\u02da= \u02da=0 = 2 3 \u02c7R 3(1 cos ): In the special case = \u02c7, we recover the well-known formula that. To calculate. We calculate as follows:. Triple integrals in Cartesian coordinates (Sect. Figure 1 In order for the double integral to exist, it is sufficient that, for example, the region D be a closed (Jordan) measurable region and that the function f(x, y ) be continuous throughout D. See more ideas about Vector calculus, Calculus and Calculus 2. Math terminology from differential and integral calculus for functions of a single variable. The integral is the line integral of a continuous real-valued function or , i. (B)A graphing calculator is required. Wolfram|Alpha is a great tool for calculating antiderivatives and definite integrals, double and triple integrals, and improper integrals. Introduction; Finding the area under a curve is a useful tool in a large number of problems in many areas of science, engineering, and business. 0 a 4 Middle integral: hr dr3 =. Handout 8: Plots of three dimensional regions associated with triple integrals. int uses the default integration variable determined by symvar ( expr,1 ). The limit calculator finds if it exists the limit at any point, at the limit at 0, the limit at +oo and the limit at -oo of a function. If expr is a constant, then the default integration variable is x. The limits on z run from 0 to h. Definite Integrals Calculator. Online Triple & Double Integral Calculator With Steps. How to Calculate Multiple Integrals. index: subject areas. No need to calculate any antiderivatives. The user enters a function of two or three variables and corresponding limits of integration and the tool evaluates the integral. Calculate the volume of the solid. Choose \"Find the Derivative\" from the menu and click to see the result!. These integral calculator can be used to calculate and solve definite integrals and indefinite integrals. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 \u2264x \u2264 5 -----> 5r2 \u2264 x \u2264 5, since each cross-section is a full circle 0 \u2264 \u03b8 \u2264 2\u03c0. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Equation solver can find both numerical and parametric solutions of equations. Fubini\u2019s Theorem for double integrals over rectangles; iterated integrals. I am using the following code by help of the following post: Triple integral in R (how to specifying the domain) Not sure, if I am missing anything as if I change value of A, it still gives me the same results. Summary : The integral function calculates online the integral of a function between two values. Triple integral limits over pyramid. This is a topic that takes some practice. (L points) et C be the half cylinder bounded by y = 0, z = 0, z = 2 and x2 +y2 = 1 for y. zip: 1k: 03-03-08: Ultimate Calculus Program Program includes Rolles, Trapezoidal, Mean Value, Rieman Sum, and Integral Equations. #N#Index for Calculus. numbers & symbols. The every single and general integration techniques and even unique, important functions being provided. Triple Integral Calculator at a Glance The 5-Minute Rule for Triple Integral Calculator. 7 Triple Integrals Be able to evaluate a given triple integral. Tech- II Subject: Engineering Mathematics II Unit-3 RAI UNIVERSITY, AHMEDABAD. BYJU\u2019S online integral calculator tool makes the calculations faster, showing the integral value for the given function in a fraction of seconds. Step by step calculus inside your TI-89 & Titanium calculator. The cone z = p. By using this website, you agree to our Cookie Policy. 1-Definitions. If you're seeing this message, it means we're having trouble loading external resources on our website. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\\left[ {a,b} \\right] \\times \\left[ {c,d} \\right]$$ $$\\times \\left[ {p,q} \\right]$$ (Figure $$1$$). 3 Triple Integrals Question Find the prism volume in the order dz dy dx (six orders are possible). and convert it to cylindrical coordinates. use triple integral to prove volume of ice cream cone c is : pib^(3) Calculus: Apr 8, 2013: Cone volume with triple integral: Calculus: Dec 18, 2012: Cone volume-triple integral: Calculus: Jun 12, 2012: Mass of Ice Cream Cone using Triple Integrals: Calculus: Mar 22, 2012. \u2022 Apply the Change of Variables in Multiple Integrals. How to use the Indefinite Integral Calculator. This gives us a ray going out from the origin. The x and y coordinates lie in a disk of radius a, so 0 \u2264 r \u2264 a and 0 <\u03b8 \u2264 2\u03c0. I Examples: Changing the order of integration. advanced algebra. Be able to find the limits of integration for an integral given the solid region of integration. In a moment you will receive the calculation result. In polar coordinates, the point is located uniquely by specifying the distance of the point from the origin of a given coordinate system and the angle of the vector from the origin to the point from the positive -axis. I Examples: Changing the order of integration. Absolute Convergence. Step by step calculus inside your TI-89 & Titanium calculator. There is nothing that says that triple integrals set up as this is must only have constants as limits! So, here is the $$x$$ integration. This differential volume can be expressed in six possible ways. Triple integrals are the analog of double integrals for three dimensions. Double integral calculator provides you the facility of step by step solution of the problem which means that you can get a solution like your teachers to solve it on a white board. The limits of all inner integrals need to be defined as functions. Physical Applications of Triple Integrals : volume of sphere. In StandardForm, Integrate [ f, x] is output as \u222b f x. Evaluate the triple integral \u222b\u222b\u222bE 5x dV, where E is bounded by the paraboloid x = 5y 2 + 5z 2 and the plane x = 5. Free Fourier Series calculator - Find the Fourier series of functions step-by-step This website uses cookies to ensure you get the best experience. To calculate the limits for an iterated integral. Of course, you can have Maple calculate multiple integrals. Requires the ti-83 plus or a ti-84 model. Theorem 13. We need an antiderivative of \u221a(4. What is a cross product? A cross product, also known as a vector product, is a mathematical operation in which the result of the cross product between 2 vectors is a new vector that is perpendicular to both vectors. It is the height of a thin stick as in Section 14. Triple Integrals and Volume - Part 2 - Duration: 8:06. In general integrals in spherical coordinates will have limits that depend on the 1 or 2 of the variables. The definite integral of from to , denoted , is defined to be the signed area between and the axis, from to. The limits of all inner integrals need to be defined as functions. For a curve produced by a function, you may be able to integrate the function from a to b and calculate the area under the curve. When you see the table, you will mostly see the y values getting closer to the limit answer as homes in on arrow-number. xmin lower limit of outer integral. By using this website, you agree to our Cookie Policy. 2 A Catalog of Essential Functions 1. Let\u2019s do limit comparison to 1/t3: lim t\u2192\u221e 1/t3 1/t3\u2212t =lim t\u2192\u221e t3\u2212t t3 =lim t\u2192\u221e t3\u2212t t3. It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the calculator is able to calculate, it returned. You solve this type of improper integral by turning it into a limit problem where c approaches infinity or negative infinity. The usual cautions about numerical methods apply, particularly when the function is not well behaved. It also shows plots, alternate forms, and other relevant information to enhance your mathematical intuition. Type in the indefinite integral problem to solve To get started, type in a value of the indefinite integral and click \u00abSubmit\u00bb button. One special case of the product rule is the constant multiple rule, which states that if c is a number and f(x) is a differential function, then cf(x) is also differential, and its derivative is (cf)'(x)=cf'(x). Example 4 Find volume of the tetrahedron bounded by the coordinate planes and the plane through$(2,0,0)$,$(0,3,0)$, and$(0,0,1)$. Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator. It also shows plots, alternate forms, and other relevant information to enhance your mathematical. The notation for the general triple integrals is, $\\iiint\\limits_{E}{{f\\left( {x,y,z} \\right)\\,dV}}$ Let\u2019s start simple by integrating over the box,. Triple integrals in spherical coordinates Our mission is to provide a free, world-class education to anyone, anywhere. coordinates, change of variables in multiple integrals. Classical integration theorems of vector calculus Math 6B. Step 2: Click the blue arrow to submit. 0, 1e5 or an expression that evaluates to a float, such as exp(-0. Double integrals over general regions. In cylindrical coordinates, the volume of a solid is defined by the formula. The Integral Calculator Trap The last result is provided by taking the very first limit from the second. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). using Type I region using Type region calculate one fo the above integrals. This page shows you two ways to compute a definite integral with numeric limits, and how to plot an accumulation function. The magnitude of this new vector is equal to the area of a para. Definite Integral. Now is the lower limit and is. Suppose we have a double integral in x and y, which we would like to change to a double integral in two new variables, u, and v, where u = 3x + 4y and v = x - 2y. That gives the upper limit z = (3 -y)/3. Multiple-version printing. int uses the default integration variable determined by symvar ( expr,1 ). Second Derivative. Now here the solid is enclosed by the planes and the surface. TI-84 Plus and TI-83 Plus graphing calculator program for AP calculus students. First Derivative. If you took integral calculus, you probably learned that there are many functions whose integrals don't show up in any integral tables, simply because they are unknown to mathematics. For each fixed x we integ- rate with respect to y. If it's not clear what the y. Do a change of variables on the integral R dA, using x = u y =3v, where R be the region bounded by 9x 2+y = 36. Two key concepts expressed in terms of line integrals are flux and circulation. integrate (expr, x) is an indefinite integral, while integrate (expr, x, a, b) is a definite integral, with limits of integration a and b. You can solve double integrals in two steps: First evaluate the inner integral, and then plug this solution into the outer integral and solve that. Please Explain Reasoning Behind Limits. 3 Triple Integrals Question Find the prism volume in the order dz dy dx (six orders are possible). Remember that we are thinking of the triple integral ZZZ U f(x;y;z) dV as a limit of Riemann sums, obtained from the following process: 1. \\mathbf {F} = \u2013 Gm\\,\\mathbf {\\text {grad}}\\,u, where G is the gravitational constant. Polynomial calculator - Integration and differentiation. See exercises 3, 5. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 \u2264x \u2264 5 -----> 5r2 \u2264 x \u2264 5, since each cross-section is a full circle 0 \u2264 \u03b8 \u2264 2\u03c0. about mathwords. Transformation of a graph (function) - rotation 90 counter clockwise Sunday February 16, 2020. Of course, you can have Maple calculate multiple integrals. Wolfram|Alpha is a great tool for calculating antiderivatives and definite integrals, double and triple integrals, and improper integrals. integral, press & twice. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. D 0 0 0 0 0 0 Inner integral: r3zh = hr3. More than just an online integral solver. Triple Integrals, Changing the Order of Integration, Part 1 of 3. So it will be. Attempt at solution Well, the overall thing that I. In this specific question, we firstly need to integrate sin \ud835\udc65 d\ud835\udc65 between the limits \ud835\udc66 and zero. In StandardForm, Integrate [ f, x] is output as \u222b f x. This is the bulk of 15. Online Triple & Double Integral Calculator With Steps. In Calculus, the four important concepts are limits, continuity, derivatives and integrals. You can also use it to solve differential and integral equations. The chapter discusses the double integral of a function of two variables and the triple integral of a function of three variables. Description : This function is an integral calculator is able to calculate integrals online of the composition of common functions, using integral properties, the different mechanisms of integration and calculation online. Set up (but do not evaluate) iterated triple integrals, with appropriate limits, for nd the volume of the solid bounded by z= x2 + y2 and z= 8 x2 y2 in: (a)rectangular coordinates (b)cylindrical coordinates 9. TI-84 Plus and TI-83 Plus graphing calculator program for AP calculus students. Triple integrals 3. Write a triple integral, including limits of integration, that gives the volume between 3x+y+z=0 and 4x+4y+z=0, and above x+y<=1, x>=0, y>=0. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 \u2264x \u2264 5 -----> 5r2 \u2264 x \u2264 5, since each cross-section is a full circle 0 \u2264 \u03b8 \u2264 2\u03c0. Expression Calculator evaluates an expression in a given context. I Examples: Changing the order of integration. Use a triple integral to find the volume of the given solid. Triple Integrals 1. When there are limits, and we need to use U-Substitution, there are a few things we need to keep in mind:. Let$ V $be the volume of the solid that lies under the graph of$ f(x, y) = \\sqrt{52 - x^2 - y^2} $and above the rectangle given by$ 2 \\le x \\le 4, 2 \\le y \\le 6 $. Among other things, they let us calculate the volume under a surface. index: subject areas. Or, if endpoints $$a$$ and $$b$$ are specified, returns the definite integral over the interval $$[a, b]$$. ( Click here for an explanation) [ ti-83/ti-84 ] Approximate Integration. Online Volume Calculator With Steps. Simply put in the price of the equipment, and you're going to observe how large of a tax. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. Click on the integral (labeled number 1) showing once you have clicked on the equation in previous step, and then click on the kind of integral you would like, in this case, the one labeled number 2 is chosen. F = int (expr,a,b) computes the definite integral of expr from a to b. If expr is a constant, then the default integration variable is x. advanced algebra. ranges in the interval 0 \\le y \\le 2 \u2013 2x. 2\u03c0 a h 2\u03c0 a h Mass = r 2 dV = r 2 dz r dr d\u03b8 = r 3 dz dr d\u03b8. website feedback. If the limit exists, then fis called integrable. we use the scipy. Also note that the fact that one of the limits is not a constant is not a problem. One special case of the product rule is the constant multiple rule, which states that if c is a number and f(x) is a differential function, then cf(x) is also differential, and its derivative is (cf)'(x)=cf'(x). Explore the solid defining the boundaries of the region for a triple integral. Apply, evaluate, and understand integrals of multi-variable scalar-valued functions. Triple Integral Spherical Coordinates. advanced topics. Homework Statement This is my last question about triple integrals in cylindrical coordinates. Watch if the algorithm is converging. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field. Create the worksheets you need with Infinite Calculus. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. Yes, sometimes down right easy or at least somewhat easier. 0 \u2264 y \u2264 2 \u2212 2 x. (So think of a wall around the perimeter of the \ufb02oor area R, reaching up. We will not do this with all of the examples in the lab since Maple, and us for that matter, could not handle them. The volume formula in rectangular coordinates is formulas for converting, volume of the triple integral, limits of integration, bounds of. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Multiple Integral Calculator - eMathHelp This site contains an online calculator that finds multiple integrals (double or triple integrals). These integral calculator can be used to calculate and solve definite integrals and indefinite integrals. The Double Integral as the Limit of Riemann Sums; Polar Coordinates 6. Multiple integrals possess a number of properties similar to those. Notes on Triple Integration Dr. Definite integral could be represented as the signed areas in the XY-plane bounded by the function graph. Use Riemann Sum with m=n=2. Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. The integration is performed over the whole volume of the body. Includes derivatives, integration, volume, quadratic equation and trig identities. I tried drawing out the limits, but it didn't help (I found that 0 Latex > FAQ > Latex - FAQ > LateX Derivatives, Limits, Sums, Products and Integrals. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. The common way that this is done is by df / dx and f'(x). (L points) et C be the half cylinder bounded by y = 0, z = 0, z = 2 and x2 +y2 = 1 for y. MULTIPLE INTEGRALS II Triple Integrals Triple integrals can be treated as a logical extension of multiple integrals. In polar coordinates, the point is located uniquely by specifying the distance of the point from the origin of a given coordinate system and the angle of the vector from the origin to the point from the positive -axis. Calculus is an amazing tool. , 24 MB] Improper integrals - part 3 - convergence or divergence by comparison [video; 15 min. How to use the Triple Integral Calculator. We begin by discussing the evaluation of iterated integrals. website feedback. Given a function sketch, the derivative, or integral curves. Fast and easy to use. This tutorial demonstrates how to evaluate integrals using the TI-89, TI-92+, or Voyage 200 graphing calculators. If it\u2019s not clear what the y. How to use the Double Integral Calculator. Classical integration theorems of vector calculus Math 6B. In your integral, use theta, rho, and phi for 0, \u03c1 and \u03c6, as needed. The int function can be used for definite integration by passing the limits over which you want to calculate the integral. Newton's Method Calculator. If expr is a constant, then the default integration variable is x. Free Summation Calculator. That gives the upper limit z = (3 -y)/3. Changing the order of integration of a triple integral blackpenredpen. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, How to calculate limits in triple integral? Ask Question Asked 3 years, 8 months ago. 1 Functions and Their Representations 1. April 25, 2007 Teaching Assistant: Time Limit: 1 hour Signature: This exam contains 7 pages (including this cover page) and 6 problems. I am only interested in the correct limits of integration. V = \\iiint\\limits_U {\\rho d\\rho d\\varphi dz}. Homework Equations $$x^{2}+y^{2}+z^{2}=a^{2}$$ : Equation for a sphere of radius \"a\" centered on the origin. Integral online. Integration over surfaces, properties, and applications of integrals. Definite integrals can also be used in other situations, where the quantity required can be expressed as the limit of a sum. The integral calculator gives chance to count integrals of functions online free. and integral tables (D)Applications of the integral to nding area and volume (E)Graphs and integrals with polar coordinates and parametric curves (F)Vector geometry and vector arithmetic in two and three dimensions IV. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. The region D consists of the points (x,y,z) with x^2+y^2+z^2<=4 and x^2+y^2<=1 and z>=0. Here is a simple. Summary : The integral function calculates online the integral of a function between two values. I Examples: Changing the order of integration. My problem isn't the integration process but just to determine what the limits are. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. Triple integrals in Cartesian coordinates (Sect. Elliott Jacobs On Wednesday, March 4, you saw how to set up triple and calculate triple integrals. integration of trigonometric integrals Recall the definitions of the trigonometric functions. D 0 0 0 0 0 0 Inner integral: r3zh = hr3. Changes of variable can be made using Jacobians in much the same way as for double integrals. I Triple integrals in arbitrary domains. What is a cross product? A cross product, also known as a vector product, is a mathematical operation in which the result of the cross product between 2 vectors is a new vector that is perpendicular to both vectors. The sum on the right is called a Riemann sum, and fis said to be integrable if the limit of Riemann sums exists. In general integrals in spherical coordinates will have limits that depend on the 1 or 2 of the variables. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each$\\Delta x\\times\\Delta y\\times\\Delta z. Because if your integration order takes care of Z first, i. about mathwords. When there are limits, and we need to use U-Substitution, there are a few things we need to keep in mind:. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. First, we must convert the bounds from Cartesian to cylindrical. double integral is defined as the limit of sums. Second, we find a fast way to compute it. Use double (or triple) integrals to calculate the average value of a function in some region. Now here the solid is enclosed by the planes and the surface. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. Define the triple integral of a function f(x, y, z) over a bounded reglon in space. Integration by parts formula: ? u d v = u v-? v d u. Changing the order of integration in triple integrals Triple Integrals, Changing the Order of Integration,. The integral is equal to the area of. There will be six different orders of evaluating the triple iterated integrals. \u2022 Evaluate double integrals over general regions. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. Equation Solver solves a system of equations with respect to a given set of variables. advanced algebra. \u2022 Evaluate double integrals over general regions. The common way that this is done is by df / dx and f'(x). This gives us a ray going out from the origin. Because if your integration order takes care of Z first, i. Free indefinite integral calculator - solve indefinite integrals with all the steps. Triple Integral Calculator. BYJU\u2019S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. Direct application of the fundamental theorem of calculus to find an antiderivative can be quite difficult, and integration by substitution can help simplify that task. Now this last limit is clearly one (divide top and bottom by t3, or use continuity of the square root to move the limit inside the radical). Multiple Integrals -- First Example: Degenerate Double Integral | # Following Examples are Variations of Examples from Math 210 Calculus III | # Textbook Multivariable Calculus, Third Edition, by James Stewart | # from Section 13. There are examples of valid and invalid expressions at the bottom of the page. Although we define triple integrals using a Riemann sum, we usually evaluate triple integrals by turning them into iterated integrals involving three single integrals. Volume integrals are especially important in physics for many applications, for example, to calculate flux densities. index: click on a letter. Integral Calculator If you were looking for a way to calculate the Integral value of a set of mumbers, then the Integral calculator is exactly what you need. Logo, Boxshot & ScreenShot. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. Triple integrals represent a concept similar to double integrals, but the area of integration in this case is not an area but rather a three-dimensional surface. w18vzo0h0a1v, mbww8uzgnawpy, znf5c4g7xm, dgi7r1rjh9z27, ft1zvqto9n, mjns5gei9ajw, 3ieeobpihypu, ic7qbh0jwfax5u, cca1zssruow5akg, ub9elvwpwan0, mjhsah3l26ig, 3nzxftdmcl, 66w71h5ps5g6v, 9hhfespigiz, nrxg2fbfm0hk5q, pqo3x9dh14fha, 48ex9wzon2zm0, hqi5ktwpu5zt8o, yrkh4cnobb8p, k8uow4n5h4vf93, 1kawxmi0b5e, s41403dnjohp9y5, rm66yls5p4pnt1, qx2cse22m2re, 8xhxnw2b2but, e3zgq6dpjuklf, ulun0bgh925s, 43762ott2p8v3, jokxr64ksr, v1iudkao69yoswh, djapj57k7dq57, ha7mkbkreg6kyq, ezjwis8sb609p1, hliis48yjw98n, fwxvxg26ra0gmz", "date": "2020-05-31 13:02:25", "meta": {"domain": "lotoblu.it", "url": "http://lotoblu.it/ixah/triple-integral-limit-calculator.html", "openwebmath_score": 0.9671174883842468, "openwebmath_perplexity": 849.5232301045739, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9901401446964613, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8548786884085159}}
{"url": "http://bootmath.com/expressing-bbb-n-as-an-infinite-union-of-disjoint-infinite-subsets.html", "text": "# Expressing $\\Bbb N$ as an infinite union of disjoint infinite subsets.\n\nThe title says it. I thought of the following: we want $$\\Bbb N = \\dot {\\bigcup_{n \\geq 1} }A_n$$\nWe pick multiples of primes. I\u2019ll add $1$ in the first subset. For each set, we take multiples of some prime, that hasn\u2019t appeared in any other set before. Then \\begin{align} A_1 &= \\{1, 2, 4, 6, 8, \\cdots \\} \\\\ A_2 &= \\{3, 9, 15, 21, 27, \\cdots \\} \\\\ A_3 &= \\{5, 25, 35, 55, \\cdots \\} \\\\ A_4 &= \\{7, 49, 77, \\cdots \\} \\\\ &\\vdots \\end{align}\n\nI\u2019m heavily using the fact that there are infinite primes. I think these sets will do the job. Can someone check if this is really ok? Also, it would be nice to know how I could express my idea better, instead of that hand-waving. Alternate solutions are also welcome. Thank you!\n\nEdit: the subsets must be also infinite.\n\n#### Solutions Collecting From Web of \"Expressing $\\Bbb N$ as an infinite union of disjoint infinite subsets.\"\n\nHere is another way to do it.\nLet $A_{i}$ consist of all the numbers of the form $2^im$ where $2\\nmid m$. That is, $A_i$ consists of all the numbers that have exactly a factor of $2^i$ in them. So\n\\begin{align} A_0 &= \\{1,3,5,7,9,11, \\dots\\}\\\\ A_1 &= \\{2, 6 =2^1\\cdot 3, 10 = 2^1\\cdot 5, 14 = 2^1\\cdot 7, \\dots\\}\\\\ A_2 &= \\{4 = 2^2, 12=2^2\\cdot 3, 20=2^2\\cdot 5, \\dots\\}\\\\ A_3 &= \\{8=2^3, 24=2^3\\cdot 3, 40=2^3\\cdot 5, \\dots \\}\\\\ &\\vdots \\end{align}\nIn general $A_i = \\{2^im: p\\nmid m\\}$. You can of course pick any other prime instead of $2$.\n\nI like @Thomas\u2019s answer best, but I would have enumerated $\\mathbb N\\times\\mathbb N$ and then taken the inverse images of the separate columns for the subsets.\n\nThis is an alternate solution.\n\nLet $\\alpha$ and $\\beta$ be any pair of irrational numbers such that\n$$1 < \\alpha < 2 < \\beta\\quad\\text{ and }\\quad \\frac{1}{\\alpha} + \\frac{1}{\\beta} = 1.$$\nThe two sequences\n$\\displaystyle\\;\\left\\lfloor\\alpha k\\right\\rfloor\\;$ and\n$\\displaystyle\\;\\left\\lfloor\\beta k\\right\\rfloor\\;$, $k \\in \\mathbb{Z}_{+}$\nare called Beatty sequence and it is known they form a partition of $\\mathbb{Z}_{+}$.\nDefine two functions $\\tilde{\\alpha}, \\tilde{\\beta} : \\mathbb{Z}_{+} \\to \\mathbb{Z}_{+}$ by\n$$\\tilde{\\alpha}(k) = \\left\\lfloor\\alpha k\\right\\rfloor \\quad\\text{ and }\\quad \\tilde{\\beta}(k) = \\left\\lfloor\\beta k\\right\\rfloor$$\nWe have\n$$\\mathbb{Z}_{+} = \\tilde{\\alpha}(\\mathbb{Z}_{+}) \\uplus \\tilde{\\beta}(\\mathbb{Z}_{+})$$ where $\\uplus$ stands for disjoint union.\nReplace the rightmost $\\mathbb{Z}_{+}$ recursively by this relation, we get\n\n\\begin{align} \\mathbb{Z}_{+} &= \\tilde{\\alpha}(\\mathbb{Z}_{+}) \\uplus \\tilde{\\beta}(\\mathbb{Z}_{+})\\\\ &= \\tilde{\\alpha}(\\mathbb{Z}_{+}) \\uplus \\tilde{\\beta}(\\tilde{\\alpha}(\\mathbb{Z}_{+})) \\uplus \\tilde{\\beta}(\\tilde{\\beta}(\\mathbb{Z}_{+})))\\\\ &= \\tilde{\\alpha}(\\mathbb{Z}_{+}) \\uplus \\tilde{\\beta}(\\tilde{\\alpha}(\\mathbb{Z}_{+})) \\uplus \\tilde{\\beta}(\\tilde{\\beta}(\\tilde{\\alpha}(\\mathbb{Z}_{+}))) \\uplus \\tilde{\\beta}(\\tilde{\\beta}(\\tilde{\\beta}(\\mathbb{Z}_{+}))))\\\\ &\\;\\vdots \\end{align}\nAs a consequence, if one define a sequence of subsets $A_1, A_2, \\ldots \\subset \\mathbb{Z}_{+}$ recursively by\n$$A_1 = \\tilde{\\alpha}(\\mathbb{Z}_{+}) \\quad\\text{ and }\\quad A_n = \\tilde{\\beta}(A_{n-1}), \\quad\\text{ for } n > 1,$$\nthese subsets will be pairwise disjoint. It is clear all these $A_n$ are infinite sets.\nSince $\\beta > 2$, we have\n\n$$\\tilde{\\beta}(k) = \\left\\lfloor \\beta k \\right\\rfloor > \\beta k \u2013 1 \\ge (\\beta \u2013 1) k\\quad\\text{ for all } k \\in \\mathbb{Z}_{+}$$\nThis implies\n$$\\tilde{\\beta}^{\\circ\\,\\ell}(k) = \\underbrace{\\tilde{\\beta}(\\tilde{\\beta}( \\cdots \\tilde{\\beta}(k)))}_{\\ell \\text{ times}} > (\\beta-1)^\\ell k \\ge (\\beta-1)^\\ell \\quad\\text{ for all } k, \\ell \\in \\mathbb{Z}_{+}$$\n\nAs a result,\n\n$$\\bigcap_{\\ell=1}^\\infty \\tilde{\\beta}^{\\circ\\,\\ell}(\\mathbb{Z}_{+}) = \\emptyset \\quad\\implies\\quad \\mathbb{Z_{+}} = \\biguplus_{k = 1}^\\infty A_k$$\ni.e. $\\mathbb{Z}_{+}$ is an infinite disjoint union of infinite sets $A_k$. Since there are uncountable choices for $\\alpha$, there are uncountable ways of such infinite disjoint unions.\n\nFor a concrete example, let $\\alpha = \\phi, \\beta = \\phi^2$ where $\\phi$ is the golden mean, we get something like\n\n$$\\begin{array}{rll} \\mathbb{Z}_{+} = & \\{\\; 1,3,4,6,8,9,11,12,14,16,17,19,\\ldots\\;\\}\\\\ \\uplus & \\{\\; 2,7,10,15,20,23,28,31,36,41,\\ldots\\;\\}\\\\ \\uplus & \\{\\; 5,18,26,39,52,60,73,81,94,\\ldots\\;\\}\\\\ \\uplus &\\{\\; 13,47,68,\\ldots\\;\\}\\\\ \\vdots\\; & \\end{array}$$\n\nIf you follow the rule that you only take the multiples that didn\u2019t show up already, then you\u2019re fine, since by construction you\u2019ll be making all the subsets disjoint and by the Fundamental Theorem of Arithmetic every element will be in some $A_i$.\n\nAn example of simple infinite disjoint union would be $A_i=\\{i\\}$ and then $\\mathbb{N}=\\bigcup_{i=0}^\\infty{A_i}$.\n\nWith edit: A simple way to split up $\\mathbb{N}$ into a disjoint union of infinite subsets is to start with $A_0=\\{0,2,4,\\ldots\\}$, and then let $A_1$ be every other element of $\\mathbb{N}\\setminus A_0$ (i.e. $\\{1,5,9,13,\\ldots\\}$). In general, let $A_n$ be the set containing \u201cevery other element\u201d of the set $\\mathbb{N}\\setminus \\bigcup_{i=0}^{n-1}{A_n}$.", "date": "2018-07-21 17:07:39", "meta": {"domain": "bootmath.com", "url": "http://bootmath.com/expressing-bbb-n-as-an-infinite-union-of-disjoint-infinite-subsets.html", "openwebmath_score": 0.9999896287918091, "openwebmath_perplexity": 398.47478087695526, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363708905831, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8548421023077305}}
{"url": "https://math.stackexchange.com/questions/436559/a-natural-proof-of-the-cauchy-schwarz-inequality/436642", "text": "# A natural proof of the Cauchy-Schwarz inequality\n\nMost of the proofs of the Cauchy-Schwarz inequality on a pre-Hilbert space use a fact that if a quadratic polynomial with real coefficients takes positive values everywhere on the real line, then its discriminant is negative(e.g. Conway: A course in functional analysis). I think this is somewhat tricky. Moreover I often forget its proof when the pre-Hilbert space is defined over the field of complex numbers. Is there a more natural proof (hence it's easy to remember) which is based on a completely different idea?\n\n\u2022 To someone who voted to close, would you please explain the reason for the vote? \u2013\u00a0Makoto Kato Jul 5 '13 at 5:59\n\u2022 You might want to browse the first chapter of The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. \u2013\u00a0Did Jul 5 '13 at 6:26\n\u2022 I'm not sure about what could qualify as \"natural\" or \"more natural\" proof for you, but in this case I can't think, off the top of my head, of anything *simpler\", basic and straightforward as working with a quadratic's discriminant: this is Junior High School stuff! \u2013\u00a0DonAntonio Jul 5 '13 at 8:45\n\u2022 @DonAntonio As I wrote, I don't think the complex pre-Hilbert space case is not so straightforward. \u2013\u00a0Makoto Kato Jul 5 '13 at 9:01\n\u2022 A question which has 3 upvotes, 2 favorites and a 6 upvoted answer should not be closed. Please reset the close votes. \u2013\u00a0Makoto Kato Jul 5 '13 at 19:24\n\nRecall the Pythagorean theorem: If $u_1, \\cdots u_n$ are pairwise orthogonal, then $$\\| u_1 + \\cdots u_n \\|^2 = \\|u_1\\|^2 + \\cdots + \\| u_n \\|^2.$$\n\nI want to use this to tell us something about two non-zero vectors $u$ and $v,$ but they aren't necessarily orthogonal. So consider the projection of $u$ onto the plane of vectors orthogonal to $v:$ $$w = u - \\frac{ \\langle u,v \\rangle}{\\|v\\|^2} v.$$ This is certainly orthogonal to $v,$ and the Pythagorean theorem applied to $w$ and $v$ gives the Cauchy-Schwarz inequality.\n\n\u2022 This is excellent. Thanks. By the way, I think $<v, u>$ is a typo($<v, v>$). \u2013\u00a0Makoto Kato Jul 5 '13 at 9:19\n\u2022 @MakotoKato Thank you for spotting that. Also, I find using \\langle , \\rangle for inner products is more pleasing to the eye than $<$ and $>.$ \u2013\u00a0Ragib Zaman Jul 5 '13 at 14:14\n\u2022 @RagibZaman You're welcome. I just didn't know how to write $\\langle, \\rangle$ instead of < and >. By the way again, I think $\\|v||$ should be squared or replaced by $\\langle v, v \\rangle$. Regards. \u2013\u00a0Makoto Kato Jul 5 '13 at 19:19\n\u2022 I fail to see how you obtain the CS inequality in your last paragraph. Would you explain that? \u2013\u00a0Martin Argerami Jul 5 '13 at 20:42\n\u2022 @MartinArgerami Let $u_1 = w$, $u_2 = \\frac{ \\langle u,v \\rangle}{\\langle v, v \\rangle} v$. Then apply the Pythagorean formula $\\| u_1 + u_2\\|^2 = \\|u_1\\|^2 + \\| u_2 \\|^2 \\ge \\|u_2\\|^2$. Regards. \u2013\u00a0Makoto Kato Jul 6 '13 at 0:41\n\nThere is also an approach by \"amplification\" which is really cool. Also the exact same trick works to prove H\u00f6lder's inequality and is generally a very important principle for improving inequalities.\n\nIt goes like this: We start out with $$\\langle a-b,a-b\\rangle\\ge 0$$ for $a,b$ in your inner product space, and $a\\not=0$, $b\\not=0$. This implies $$2\\langle a,b\\rangle\\le \\langle a, a\\rangle + \\langle b, b\\rangle$$ Now notice that the left hand side is invariant under the scaling $a\\mapsto \\lambda a$, $b\\mapsto \\lambda^{-1}b$ for $\\lambda>0$. This gives $$2\\langle a,b\\rangle \\le \\lambda^2 \\langle a,a\\rangle + \\lambda^{-2}\\langle b, b\\rangle$$ Now look at the right hand side as a function of the real variable $\\lambda$ and find the optimal value for $\\lambda$ using calculus (set the derivative to $0$):\n\n$$\\lambda^2=\\sqrt{\\frac{\\langle b,b\\rangle}{\\langle a,a\\rangle}}$$\n\nPlugging this value in, we obtain\n\n$$2\\langle a,b\\rangle\\le \\sqrt{\\langle a,a\\rangle}\\sqrt{\\langle b,b\\rangle}+\\sqrt{\\langle a,a\\rangle}\\sqrt{\\langle b,b\\rangle}$$\n\ni.e.\n\n$$\\langle a,b\\rangle\\le\\sqrt{\\langle a,a\\rangle}\\sqrt{\\langle b,b\\rangle}$$\n\nNotice how we took a trivial observation and \"optimized\" the expression by exploiting scaling invariance.\n\n\u2022 As explained (and expanded) by Terence Tao on his blog. \u2013\u00a0Did Jul 5 '13 at 13:08\n\u2022 Could your proof be modified to show that the equality holds iff $a$ is proportional to $b$? I've been thinking a while on it, but I haven't figured anything out. \u2013\u00a0jinawee Oct 1 '14 at 19:54\n\u2022 @jinawee: Yes, the only inequality in this proof is $\\langle \\mu a-b,\\mu a-b\\rangle\\ge 0$ for an optimal constant $\\mu$ depending on $a,b$. Equality implies $b=\\mu a$. \u2013\u00a0J.R. Oct 2 '14 at 7:17\n\nThe inequality $| \\langle a, b \\rangle | \\leq \\| a \\| \\| b \\|$ can be rewritten as $$| \\langle a, \\frac{b}{\\|b\\|} \\rangle | \\leq \\| a \\|$$ (assuming $b \\neq 0$).\n\nOn the left we see the component of $a$ on the unit vector $u = \\frac{b}{\\| b \\|}$. On the right we have the norm of $a$. Of course the norm of $a$ is larger than the component of $a$ on $u$, this is very intuitive. To make this into a rigorous proof, we merely have to write $a = \\langle a, u \\rangle u + v$, note that $v \\perp u$, and use the pythagorean theorem.\n\nAnother approach is to start with the inequality $$0 \\leq \\| a - \\text{proj}_b(a) \\|^2.$$\n\nJust expand the right hand side and Cauchy-Schwarz pops out.\n\nThis is not exactly an answer but an explanation for the idea behind the Ragib Zaman's answer.\n\nLet $K$ be the field of real numbers or the field of complex numbers. Let $E$ be a pre-Hilbert space over $K$.\n\nLet $x, y$ be elements of $E$ such that $\\langle x, y \\rangle = 0$. Then $\\|x + y\\|^2 = \\langle x+y, x+ y\\rangle = \\langle x, x\\rangle + \\langle x, y \\rangle + \\langle y, x\\rangle + \\langle y,y\\rangle = \\|x\\|^2 + \\|y\\|^2$.\n\nLet $x, y$ be elements of $E$ such that $\\langle x - y, y \\rangle = 0$. Then, by the above formula, $\\|x\\|^2 = \\|x - y\\|^2 + \\|y\\|^2$. Hence $\\|x\\| \\ge \\|y\\|$.\n\nFinally let $u, v$ be non-zero elements of $E$. Let $t$ be an element of $K$ such that $\\langle u - tv, v \\rangle = 0$. $t$ must be $\\frac{ \\langle u,v \\rangle}{\\langle v, v\\rangle}$ since $v \\neq 0$. Then $\\|u\\| \\ge \\|tv\\|$ by the previous inequality. This leads to the Cauchy-Schwarz inequality immediately.\n\nHere's my favorite proof, mainly because it's nicely symmetric, easy to remember and not impossible to come up with (the main trick is that $2=1+1$):\n\nWe want to prove $$|\\langle x,y\\rangle|\\le\\|x\\|\\|y\\|$$ This is linear in $x$ or $y$ and obviously holds for $x=0$ or $y=0$. Therefore without loss of generality we can suppose that $\\|x\\|=\\|y\\|=1$: $$|\\langle x,y\\rangle|\\le\\|x\\|\\|y\\|\\Longleftrightarrow|\\langle x,y\\rangle|\\le1\\Longleftrightarrow2|\\langle x,y\\rangle|\\le2\\Longleftrightarrow2|\\langle x,y\\rangle|=\\|x\\|^2+\\|y\\|^2$$\n\nLet $u\\in\\mathbb C$ be a complex unit (i.e. $|u|=1$), then \\begin{align}0\\le\\|x-uy\\|^2&=\\langle x-uy,x-uy\\rangle=\\langle x,x\\rangle-u\\langle x,y\\rangle-\\overline{u\\langle x,y\\rangle}+\\langle y,y\\rangle\\\\&=\\|x\\|^2+\\|y\\|^2-2\\,\\text{Re}(u\\langle x,y\\rangle)\\end{align} Now just set $u$ so that $\\text{Re}(u\\langle x,y\\rangle)=|\\langle x,y\\rangle|$ and you are done.\n\nOf course in the real case you can just expand $0\\le\\|x\\pm y\\|^2$.", "date": "2019-09-20 20:29:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/436559/a-natural-proof-of-the-cauchy-schwarz-inequality/436642", "openwebmath_score": 0.8864595293998718, "openwebmath_perplexity": 223.72300411722878, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363725435203, "lm_q2_score": 0.8670357632379241, "lm_q1q2_score": 0.8548420952723014}}
{"url": "https://math.stackexchange.com/questions/2848788/is-this-a-valid-approach-to-solving-the-inequality-frac1x-x-1", "text": "# Is this a valid approach to solving the inequality $\\frac{1}{x} < x < 1$?\n\nI have been given the inequality $\\frac{1}{x} < x < 1$ and have been told to find the values of $x$ which satisfy this inequality, and I have also been told to find these values using a case-by-case approach. I'd like to know whether my reasoning is valid.\n\nCase 1: $x=0$\n\nIn this case we find that the value of $\\frac1x$ is undefined, so we know that $x \\neq 0$.\n\nCase 2: $0<x<1$\n\nGiven the inequality $\\frac{1}{x} < x < 1$ and that $x$ lies in the range $0 < x < 1$, we find that $1<x^2$. Therefore, $x>1$ or $x<-1$. However, $x\\not>1$ and in this case $x\\not<-1$ since we are looking at the case $0<x<1$. Thus we conclude that no values of $x$ in the range $0<x<1$ satisfy the original inequality.\n\nCase 3: $x<0$\n\nFrom the original equality and the fact that we know that $x<1$ we find:\n\n$$\\frac1x < x$$\n\n$$\\implies 1 > x^2$$ (since $x<0$) $$\\implies -1<x<1$$ but we know that $x$ cannot lie in the range $0<x<1$ so we have that $-1<x$ and that $x<0$ and combining these inequalities we have that $-1<x<0$.\n\nThis argument seems to be supported by looking at the graph, but I am unsure whether all the steps I have made are valid and whether this is what is meant by a 'case analysis'.\n\n\u2022 This is absolutely fine and there is no part of the solution that is wrong. \u2013\u00a0Prakhar Nagpal Jul 12 '18 at 16:00\n\u2022 I think the argument is good! ${}{}{}{}{}{}{}$ \u2013\u00a0Andres Mejia Jul 12 '18 at 16:01\n\u2022 Thank you for the feedback. \u2013\u00a0Benjamin Jul 12 '18 at 16:02\n\nOverall, I would say that the argument is spot on! Just a couple comments you might want to consider. First of all, to be super rigorous in your answer, you might mention that clearly we know $x \\ngeq 1$ so we need not consider that case. Another thing, I felt that your argument via contradiction for Case 2 was a bit lengthy there at the end. Once you reached the point $1 < x^2$ you could have just said that since we have $0 < x < 1$ (via assumption) and no such value squared is greater than 1, we have a contradiction and are done. The whole considering $x > 1$ or $x < -1$ is just overkill.\nAlso, that is exactly what it means by a case analysis, or case-by-case approach. Use cases to consider every possible value of $x$, and see what happens. Using cases is actually a very common proof technique!", "date": "2021-05-14 11:22:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2848788/is-this-a-valid-approach-to-solving-the-inequality-frac1x-x-1", "openwebmath_score": 0.821780264377594, "openwebmath_perplexity": 104.81233275808343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98593637543616, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8548420893117379}}
{"url": "https://math.stackexchange.com/questions/3426080/a-set-of-elements-in-a-reduced-unity-ring", "text": "# A set of elements in a reduced unity ring\n\nLet $$(A,+,\\cdot)$$ be a unity ring with the property that if $$x \\in A$$ and $$x^2=0$$ then $$x=0$$. Consider the set $$M=\\{a\\in A | a^3=a\\}$$. Prove that:\n\na) $$2a\\in Z(A)$$, $$\\forall a\\in M$$, where $$Z(A)$$ denotes the centre of the ring $$A$$;\nb) $$ab=ba$$, $$\\forall a,b\\in M$$.\n\nMy attempts revolved around the fact that an idempotent element in a reduced ring is central.\nSo, since for $$a\\in M$$ we have that $$(a^2)^2=a^2$$, it follows that $$a^2\\in Z(A)$$, $$\\forall a\\in M$$.\nThe next thing I wanted to use in order to solve a) was that $$Z(A)$$ is a subring of $$A$$, so if I had proved that $$(a+1)^2 \\in Z(A)$$, $$\\forall a\\in M$$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true.\nAnother idea that I had was to prove that $$M$$ is a subring of $$A$$. Of course, this didn't work out because I cannot even prove that $$M$$ is closed under addition. Again, I don't know if this is true and it most likely isn't.\nAs for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $$x^3=x$$ for any $$x$$ in that ring is commutative, but since $$(M,+,\\cdot)$$ is almost definitely not a ring, this doesn't help.\nEDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.\n\n\u2022 If you honestly doubt that $(a+1)^2 \\in Z(a)$ for all $a \\in M$, then you honestly doubt that part (a) is true, since if part (a) is true (and since you've already noted $a^2 \\in Z(a)$ for all $a \\in M$), you get $(a+1)^2 = a^2+2a+1 \\in Z(a)$ for all $a \\in M$. \u2013\u00a0mathworker21 Nov 9 at 18:50\n\u2022 @mathworker21 I agree with you. But since I have spent hours and hours trying to prove that $(a+1)^2 \\in Z(A)$ for all $a\\in M$ and nothing worked out, then I am inclined to believe that this statement may actually be wrong. Yet, since I cannot provide a counterexample, here I am, still hoping that someone better than me at rings will solve it.... \u2013\u00a0Math Guy Nov 9 at 21:34\n\u2022 I don't know. I don't think the proof that any ring with $x^3 = x$ for all $x$ implies the ring is commutative is that easy... I could imagine spending hours failing to find that proof. By the way, where did you find this problem? \u2013\u00a0mathworker21 Nov 9 at 21:35\n\u2022 @mathworker21 I know the \"classical\" one where $x^3=x$ for all $x$ holds in a ring, but since here we do not have a ring it doesn't really help I think. The problem is from a magazine from my country. \u2013\u00a0Math Guy Nov 9 at 21:37\n\u2022 you missed my point. my point was that working on these kinds of problems for hours and failing doesn't mean they are false. (Of course that is always true, but I find the statement more meaningful here). The evidence/example I gave was the $x^3=x$ problem. For that problem, I could imagine working on it for several hours without finding the solution. And of course that problem is true. \u2013\u00a0mathworker21 Nov 9 at 21:54\n\n## 1 Answer\n\nWe show $$M \\subseteq Z(A)$$. This immediately gives (a) and (b).\n\nLemma 1: $$yx = 0 \\implies xzy = 0$$ for any $$z$$.\n\nProof: $$(xzy)(xzy) = xz(yx)zy = 0$$.\n\nLemma 2: $$x^2 = x$$ implies $$x \\in Z(A)$$.\n\nProof: For any $$y \\in A$$, a short computation shows $$(xy-xyx)(xy-xyx) = 0 = (yx-xyx)(yx-xyx)$$.\n\nLemma 3: $$a \\in M \\implies a^2 \\in Z(A)$$.\n\nProof: $$(a^2)^2 = a^4 = a^2$$, so use Lemma 2.\n\nClaim: For any $$a \\in M$$, $$a \\in Z(A)$$.\n\nProof: Since $$(a-1)[a(a+1)]=0$$, Lemma 1 implies that for any $$b \\in A$$, $$0 = a(a+1)b(a-1) = (a^2b+ab)(a-1).$$ Also, $$a(a+1)(a-1)=0$$ implies $$0 = ba(a+1)(a-1) = (ba^2+ba)(a-1) = (a^2b+ba)(a-1),$$ where the last equality used Lemma 3. Subtracting gives: (1) $$0 = (ba-ab)(a-1)$$. The exact same argument shows: (2) $$0 = (a-1)(ba-ab)$$. (1) immediately implies $$0 = (ba-ab)(a-1)b = (ba-ab)(ab-b)$$, and (2) with Lemma 1 implies $$0 = (ba-ab)b(a-1) = (ba-ab)(ba-b)$$. Subtracting the two results gives $$(ba-ab)^2 = 0$$.\n\n\u2022 Thank you ! Could you tell me how you came up with this solution? \u2013\u00a0Math Guy Nov 10 at 11:23\n\u2022 @MathGuy well, I played around with it for a while, so I exhausted many different approaches. Then I eventually stumbled upon something like Lemma $1$, first in the form of $0 = a(a+1)b(a-1)$ and realized it started giving equations I hadn't seen/derived before. So I knew I had something good. Then it was just a matter of finishing up, which was pretty easy (the proof of the claim is rather short). \u2013\u00a0mathworker21 Nov 10 at 13:02", "date": "2019-11-18 21:24:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3426080/a-set-of-elements-in-a-reduced-unity-ring", "openwebmath_score": 0.9266897439956665, "openwebmath_perplexity": 174.67822198460573, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363750229256, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.8548420872597318}}
{"url": "https://mathematica.stackexchange.com/questions/136962/numerical-approximation-of-pi", "text": "# Numerical approximation of $\\pi$\n\nPoints are randomly scattered inside the unit square, some fall within the unit circle with probability $P=\\pi/4$.\n\nso $P$ is approximated by the fraction $$P\\approx \\frac{\\text{Number of red points}}{\\text{Number of all points}}$$ this leads $$\\pi \\approx 4\\frac{\\text{Number of red points}}{\\text{Number of all points}}$$ (see following image)\n\nThere is a code for this:\n\ntinyColor[color_, point_] := {PointSize[Small], color, Point[point]}\ncolorChoose[point_] :=\nIf[Norm[point] <= 1, tinyColor[Red, point], tinyColor[Blue, point]]\ndarts = RandomReal[{0, 1}, {40000, 2}];\ncoloredDarts =ParallelMap[colorChoose, darts];\ninsides = Map[Boole[Norm[#] <= 1] &, darts];\npiapprox = Accumulate[insides]/Range[Length[darts]]\ninner = Select[darts, Norm[#] <= 1 &];\nouter = Select[darts, Norm[#] > 1 &];\n\nShow[Plot[Sqrt[1 - x^2], {x, 0, 1}, Filling -> Axis, AspectRatio -> 1,\nPlotLabel -> n == Length[darts] TildeTilde[\u03c0, 4.0*piapprox[[-1]]]],\nListPlot[{inner, outer},\nPlotStyle -> {{PointSize[Tiny], Red}, {PointSize[Tiny], Blue}},\nImageSize -> {500, 500}]]\n\n\nI tried to simplify this problem:\n\npts = RandomPoint[Rectangle[], 40000];\n\nListPlot[pts, AspectRatio -> 1, PlotStyle -> Blue]\n\n\nThe problem is following:\n\nHow can I split set of points pts into two parts, \"inside the circle \" and \"outside the circle\"?\n\n\u2022 You should also investigate RegionMember, e.g., rf = RegionMember[Disk[]]; rf[darts] \u2013\u00a0Carl Woll Feb 10 '17 at 22:26\n\nYou can also use Select:\n\nptsin = Select[pts, Norm[#] < 1 &];\n\nN[Length[ptsin]/Length[pts]]*4\n(* 3.1496 *)\n\n\nHow can I split set of points pts into two parts, \"inside the circle \" and \"outside the circle\"?\n\n{in, out} = SortBy[GatherBy[pts, Norm[#] < 1 &], Norm[#[[1, 1]]] &];\n\nListPlot[{in, out}, AspectRatio -> 1, PlotStyle -> {Red, Blue}]\n\n\n\u2022 What's the purpose of SortBy? \u2013\u00a0anderstood Feb 4 '17 at 18:27\n\u2022 The approximation of $\\pi$ can be recovered with {in, out} = GatherBy[pts, Norm[#] < 1 &]; 4*Length[in]/(Length[out] + Length[in]) // N. \u2013\u00a0anderstood Feb 4 '17 at 18:30\n\u2022 @anderstood, depending on the Norm of the first element in lst, the the first list in the output produced by GatherBy may be the \"inside sublist\" or the \"outside sublist\". Sorting the output of GatherBy makes the first list the \"inside\" one. \u2013\u00a0kglr Feb 4 '17 at 18:32\n\u2022 Comment to my own comment: it should be {in, out} = SortBy[GatherBy[pts, Norm[#] < 1 &], Norm[#[[1, 1]]] &], cf kglr's comment above. \u2013\u00a0anderstood Feb 4 '17 at 18:55\n\nPlaying with Norm as shown in other answers:\n\npts = RandomReal[1, {40000, 2}];\n\n4 True/(True + False) /. CountsBy[pts, Norm[#] < 1 &]\n\n\n3.1474\n\nThat doesn't help with drawing the graphic however.\n\nYou can compute the norm of the point and verify if it is inside the circle.\n\nAt = 4;\nd = 2;\ntotalpoints = 40000;\npts = RandomReal[{-1, 1}, {totalpoints, 2}];\npointsinsidecircle = Select[pts, Norm[#] < 1 &];\ncounter = Length[pointsinsidecircle];\napproxpi = (4. At counter/totalpoints)/d^2;\nPrint[\"approx \\[Pi] = \", approxpi]\nListPlot[{pts, pointsinsidecircle}, AspectRatio -> 1,\nPlotStyle -> {Blue, Red}]\n\n\n(*approx \\[Pi] = 3.1328*)\n\n\u2022 You could you Norm directly, and avoid AppendTo which is very slow. Using Select is probably a much faster option. \u2013\u00a0anderstood Feb 4 '17 at 18:24\n\u2022 @anderstood i have changed the answer. Thank you. \u2013\u00a0Diogo Feb 4 '17 at 18:30\n\u2022 Tip: Don't save graphics as JPEGs, they become fuzzy and lose color fidelity. Use PNGs. \u2013\u00a0Rahul Feb 10 '17 at 21:19\n\nA slightly different approach:\n\ninside = Pick[pts, Map[# \u2208 Disk[] &, pts], True];\noutside = Complement[pts, inside];\n\n\nAlso as pointed in a comment above:\n\ninside = Pick[#, RegionMember[Disk[]][#], True] &@pts\noutside = Complement[pts, inside];\n`", "date": "2019-12-07 21:34:15", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/136962/numerical-approximation-of-pi", "openwebmath_score": 0.18919280171394348, "openwebmath_perplexity": 10416.945504887097, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9553191297273498, "lm_q2_score": 0.894789454880027, "lm_q1q2_score": 0.8548094833251971}}
{"url": "https://math.stackexchange.com/questions/2247973/knowing-that-for-any-set-of-real-numbers-x-y-z-such-that-xyz-1-the-ineq", "text": "Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \\ge \\frac{1}{3}$ holds.\n\nKnowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \\ge \\frac{1}{3}$ holds.\n\nI spent a lot of time trying to solve this and, having consulted some books, I came to this:\n$$2x^2+2y^2+2z^2 \\ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2)$$ $$2x^2+2y^2+2z^2 \\ge 1 - x^2 -y^2 - z^2$$ $$x^2+y^2+z^2 \\ge \\frac{1}{3}$$ But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.\n\n\u2022 This does seem like one of the best ways to tackle this problem. Why do you find this method counterintuitive? \u2013\u00a0vrugtehagel Apr 23 '17 at 12:56\n\u2022 @vrugtehagel It entails using a property which is not directly derived from the problem. \u2013\u00a0ILoveChess Apr 23 '17 at 12:58\n\u2022 Are you comfortable with calculus? Here's a completely different approach. $z = 1 - x - y$ so $x^2 + y^2 + z^2 = x^2 + y^2 +(1 - x - y)^2$. Now use some calculus to find when that has a maximum. It will be when $x = y = \\frac{1}{3}$ hence also $z = \\frac{1}{3}$. \u2013\u00a0badjohn Apr 23 '17 at 13:03\n\u2022 \u2013\u00a0Martin Sleziak Apr 23 '17 at 21:29\n\nCauchy- Schwarz works: $$x^2+y^2+z^2=\\frac{1}{3}(1^2+1^2+1^2)(x^2+y^2+z^2)\\geq\\frac{1}{3}(x+y+z)^2=\\frac{1}{3}$$\n\n$x^2+y^2+z^2$ only depends on the squared distance of $(x,y,z)$ from the origin and the constraint $x+y+z=1$ tells us that $(x,y,z)$ lies in a affine plane. The problem is solved by finding the distance between such plane and the origin: since the plane is orthogonal to the line $x=y=z$,\n\n$$\\min_{x+y+z=1}x^2+y^2+z^2 = \\left(\\frac{1}{3}\\right)^2+\\left(\\frac{1}{3}\\right)^2+\\left(\\frac{1}{3}\\right)^2 = \\frac{1}{3}$$ and we are done.\n\n\u2022 clear: the sphere from the origin is tangent to the plane in x=y=z=1/3 \u2013\u00a0G Cab Apr 23 '17 at 14:49\n\nThis is not a proof in itself, but if you've studied statistics, then you've seen a proof that\n\n$$0\\le V(X)=E(X^2)-E(X)^2$$\n\nIf we now consider a random variable $X$ with three equally likely values, $X=x,y$, and $z$, then we have\n\n$$E(X)={x+y+z\\over3}\\qquad\\text{and}\\qquad E(X^2)={x^2+y^2+z^2\\over3}$$\n\nIf, in addition, we assume $x+y+z=1$, then we have $E(X)={1\\over3}$, which implies $E(X^2)\\ge\\left(1\\over3\\right)^2={1\\over9}$, or $x^2+y^2+z^2\\ge{1\\over3}$.\n\nExcellent algebraic have been given (I voted for them). Intuition can here be obtained through visual proofs.\n\nThe equation $x+y+z=1$ defines a plane. $x^2+y^2+z^2=1/3$ defines a sphere. The following visualization depicts the plane in blue, the sphere in red.\n\nFrom that, you can imagine that the question could be rephrased as (in a mundane way): for any point in the plane, its distance from the $(0,0,0)$ origin is higher than $1/\\sqrt{3}$? So the sphere should remain \"below\" the plane. Except when they meet. The symmetry of the problem tells you that the tangency point has equal coordinates $(1/\\sqrt{3},1/\\sqrt{3},1/\\sqrt{3})$, which is one of the motivations behind the equation $(3x-1)^2+(3y-1)^2+(3z-1)^2$.\n\nWould the sphere be bigger (a radius higher than $1/\\sqrt{3}$), it would intersect the plane in more than a single tangency point.\n\nAll in all, this resorts to finding the distance of the plane to the origin, which is exactly where the sphere and the plane meet. So what you are looking at is the distance of the plane to the origin.\n\nIf the plane is given by $ax+by+cz+d$, the signed distance of a point $(x_0,y_0,z_0)$ to it is (Point-Plane Distance):\n\n$$D = \\frac{ax_0+by_0+cz_0+d}{\\sqrt{a^2+b^2+c^2}}$$\n\nwhich in your case gives exactly $1/\\sqrt{3}$. Any point in the plane is farther to $(0,0,0)$ than $|D|$.\n\nHint:\n\nExpand $(3x-1)^2+(3y-1)^2+(3z-1)^2\\geqslant 0$ And simplify.\n\nIf you rewrite your proof as:\n\n$3 ( x^2+y^2+z^2 ) \\ge ( x^2+y^2+z^2 ) + ( 2xy+2yz+2zx ) = (x+y+z)^2 = 1$.\n\nYou would find that it is not so unintuitive after all.\n\nSince you know that $x+y+z = 1$, a natural thing to do to the original equation is to homogenize it; namely make all terms have the same degree. This gives us \"$3 ( x^2+y^2+z^2 ) \\ge (x+y+z)^2$, and expanding out immediately tells us the solution.\n\nIn general, a technique that works for many cyclic polynomial inequalities is to try to 'smooth' the terms out. Terms that are just a power of one variable are the 'biggest', and using inequalities like \"$x^2+y^2 \\ge 2xy$\" will 'mix' powers and thereby 'reduce' them.\n\nIt is enough to prove the result for $x,y,z$ positive since $$\\frac{|x|+|y|+|z|}{3} \\geq \\frac{x+y+z}{3}$$\n\nOne can use the generalized AM-GM inequality: If $$M_p = \\left(\\frac{x^p+y^p+z^p}{3}\\right)^{\\frac{1}{p}}$$ then for $p < q$, $M_p \\leq M_q$ with equality holding if and only if $x=y=z$. Here, using $M_2 \\geq M_1$, we get $$\\left(\\frac{x^2+y^2+z^2}{3}\\right)^{\\frac{1}{2}} \\geq \\frac{|x|+|y|+|z|}{3} \\geq \\frac{x+y+z}{3} = \\frac{1}{3}$$\n\nWe can minimize $x^2 + y^2 + z^2$ subject to the constraint $x+y+z = 1$ using Lagrange multipliers, we then find that $x = y = z = \\frac{1}{3}$, therefore $x^2 + y^2 + z^2\\geq\\frac{1}{3}$.\n\n\u2022 This only shows that this is a local minimum, right? \u2013\u00a0Carsten S Apr 24 '17 at 14:05\n\nOne more way to prove it is by substituting out for $z$:\n\n$$x^2+y^2+z^2 = x^2+y^2+(1-x-y)^2=2x^2+2y^2+2xy-2x-2y+1$$ Now, substitute $x=\\hat x+a$ and $y=\\hat y+b$. We get $$2\\hat x^2+2\\hat y^2+2\\hat x\\hat y+(4a+2b-2)\\hat x+(2a+4b-2)\\hat y+(2(a^2+ab+b^2-a-b)+1)$$ We can eliminate the order 1 terms by letting $4a+2b-2=2a+4b-2=0$, which gives $a=b=\\frac13$. With this substitution, we have $$2(\\hat x^2+\\hat x\\hat y+\\hat y^2)+\\frac13$$ We can express the bracketed term as a sum of squares, giving us $$3(\\hat x+\\hat y)^2+(\\hat x-\\hat y)^2 + \\frac13$$ and we can see that the smallest value this can take is $\\frac13$. Indeed, it takes this value when $\\hat x=\\hat y=0$ - that is, when $x=y=\\frac13$.\n\nOk, you want an intuitive proof, not a better proof. That's fine.\n\nFirst note that for $x=y=z$, we have $x^2+y^2+z^2 = 3\\left(\\frac13\\right)^2= \\frac13$. Next let us show that $x^2+y+2+z^2$ is not minimal if $x\\ne y$ or $y\\ne z$. Indeed if $a\\ne b$ then $$a^2 + b^2 - \\left(\\left(\\frac{a+b}2\\right)^2+\\left(\\frac{a+b}2\\right)^2\\right) = \\frac12\\left( a^2 + b^2 -2ab \\right)=\\frac12(a-b)^2>0,$$ so we can replace two of the numbers be their averages and lower the sum of squares.\n\nUnfortunately we not yet quite done. We are done if we can show that there is a global minimum of $x^2+y^2+z^2$, given that $x+y+z=1$. Because then that global minimum must also be a local minimum, but for that we cannot have $x\\ne y$ or $y\\ne z$, so $x=y=z$ must be the global minimum, and its value is $\\frac13$.\n\nFor that we can first argue that we can restrict ourselves to $x,y,z\\ge0$ (e.g. if $x<0$ replace $(x,y,z)$ by $\\frac1{-x+y+z}(-x,y,z)$, which yields a smaller value since $-x+y+z = 1 - 2x > 1$) and then appeal to the compactness of $\\{(x,y,z)\\colon \\text{$x+y+z=1$,$x,y,z\\ge0$}\\}$.\n\nI did not say that this would be pretty, but it is pretty intuitive to me :)\n\nHere's another way to approach this. It's easy to see that the value of $\\frac13$ is obtained when each of $x, y, z$ is $\\frac13$. We want to show that as the variables deviate from this point (with their sum still being 1) the value cannot decrease.\n\nSo we look at the deviations from $\\frac13$: $x=\\frac13+\\epsilon_1$, $y=\\frac13+\\epsilon_2$, $z=\\frac13+\\epsilon_3$ with $\\epsilon_1+\\epsilon_2+\\epsilon_3=0$. you have\n\n$x^2+y^2+z^2=\\\\ (\\frac13+\\epsilon_1)^2+(\\frac13+\\epsilon_2)^2+(\\frac13+\\epsilon_3)^2=\\\\\\left(\\frac19+\\frac23\\epsilon_1+\\epsilon_1^2\\right)+\\left(\\frac19+\\frac23\\epsilon_2+\\epsilon_2^2\\right)+\\left(\\frac19+\\frac23\\epsilon_3+\\epsilon_3^2\\right)=\\\\ \\left(\\frac19+\\frac19+\\frac19\\right)+\\frac23(\\epsilon_1+\\epsilon_2+\\epsilon_3)+(\\epsilon_1^2+\\epsilon_2^2+\\epsilon_3^2)=\\\\ \\frac13+(\\epsilon_1^2+\\epsilon_2^2+\\epsilon_3^2) \\ge \\frac13$\n\nThe desired inequality follows by noting that from the convexity of $f(t) = t^{2},$ we have $f\\left(\\dfrac{x+y+z}{3}\\right) \\leq \\dfrac{f(x)+f(y)+f(z)}{3}$", "date": "2019-05-20 06:30:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2247973/knowing-that-for-any-set-of-real-numbers-x-y-z-such-that-xyz-1-the-ineq", "openwebmath_score": 0.9061225652694702, "openwebmath_perplexity": 151.66039486157948, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462190641612, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8547948593085478}}
{"url": "http://math.stackexchange.com/questions/293921/combinatorics-and-probability-problem", "text": "Combinatorics and Probability Problem\n\nThe problem I am working on is:\n\nAn ATM personal identification number (PIN) consists of four digits, each a 0, 1, 2, . . . 8, or 9, in succession.\n\na.How many different possible PINs are there if there are no restrictions on the choice of digits?\n\nb.According to a representative at the author\u2019s local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical (ii) sequences of consecutive ascending or descending digits, such as 6543 (iii) any sequence start-ing with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the prob-ability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)?\n\nc. Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the $2nd$ and $3^{rd}$ digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account?\n\nd.Recalculate the probability in (c) if the first and last digits are 1 and 1, respectively.\n\n---------------------------------------------\n\nFor part a): The total number of pins without restrictions is $10,000$\n\nFor part b): The number of pins in either ascending or descending order is $10 \\cdot 1 \\cdot 1 \\cdot 1$, because once the first digit is known, then the three other spots containing digits are already spoken for. The number of pins where each slot contains the same digit is $10 \\cdot 1 \\cdot 1 \\cdot 1$, because once the first digit is known there is only one option left to the rest of the slots. The number of pins that have their first and second slot occupied by 1 and 9, respectively, is $1 \\cdot 1 \\cdot 10 \\cdot 10 \\cdot$. So, if R is the set that contains these restricted pins, then $|R| = 130$; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then $|N| = 10,000 - 130$. Hence, the probability is then $P(N) = 9780/10000 = 0.9870.$ However, the answer is $0.9876$. What did I do wrong?\n\nFor part c): The sample space, containing all of the outcomes of the experiment that will take place, is $|N|=9870$. When it says that the thief won't use the same pair of digits in each try, does that not allow him trying the pin 8 5 2 1 in one try and the pin 8 2 5 1 in another try?\n\n-\nAscending or descending is $14$, not $20$. That takes care of the disagreement. \u2013\u00a0 Andr\u00e9 Nicolas Feb 3 '13 at 20:44\n@Andr\u00e9Nicolas I didn't get 14 or 20, I got 10. Your saying the answer is 14, how did you get that? \u2013\u00a0 Mack Feb 3 '13 at 20:56\n@Eli: You got $2\\cdot10=20$ for ascending and descending together, and Andr\u00e9 is saying that it should be $2\\cdot7=14$ for both together. \u2013\u00a0 joriki Feb 3 '13 at 21:04\nOh, I see. The 2 corresponds to the two choices (ascending or descending); and the 7 corresponds to the fact that you can't start the 4-digit pin with 7,8, or 9 when ascending, and you can't start a 4-digit pin with 0,1, or 2 when descending. \u2013\u00a0 Mack Feb 3 '13 at 21:14\n\nFor b): Which is the descending sequence starting with $1$ that you counted?\n\nFor c): Good question; the problem is badly worded in that regard. Taking it literally, I'd tend to interpret it as referring to unordered pairs, but since it makes little sense to couple two different PINs in this manner, I suspect that they actually mean ordered pairs. However, note that the answer doesn't depend on this.\n\nI understand neither why the question says that the thief knows the restrictions, nor why you say that the sample space has size $9870$. The thief knows that the first and last digits are $8$ and $1$, respectively; that's not compatible with any of the sequences excluded by the restrictions, and it doesn't allow for $9870$ possibilities.\n\n-\nI\u2019d interpret is as referring to ordered pairs, partly from the language, and partly from the overall level of difficulty of the exercise. \u2013\u00a0 Brian M. Scott Feb 3 '13 at 20:41\n@joriki I don't really understand what you are asking when you say, \"Which is the descending sequence starting with 1 that you counted?\" \u2013\u00a0 Mack Feb 3 '13 at 21:00\n@Eli: Exactly. And $10-3=7$. \u2013\u00a0 joriki Feb 3 '13 at 21:08\n@Eli: I found another version of the book online with the answers included. It has $.0337$ for d., which is also wrong, but can be interpreted as a rounded version of the correct result $3/89$ (see Metin's answer). So it seems they just round results without indicating it (which is rather bad style). So it seems likely that $0.0333$ is a rounded version of $3/90$. That leaves the question how they arrive at $90$ options. Perhaps they meant to say $1$ and $8$ instead of $8$ and $1$; then the birth year rule would lead to a count of $90$. \u2013\u00a0 joriki Feb 4 '13 at 8:35\n@Eli: That would also explain why part c already mentions that the thief knows the restrictions, even though that's irrelevant for the question as posed. \u2013\u00a0 joriki Feb 4 '13 at 8:37\n\nFor d): We have the case: $1$ * * $1$.\n\nBut the thief knows, by prohibition (i), it can not be $1111$. Thus he eleminates $1$ possibility.\n\nAlso he knows, by (iii), the second digit can not be $9$. There are exactly $10$ number of the form $19$ * $1$, namely, $1900$, $1910$, $1920$... So, at this stage he eleminates $10$ possibilities.\n\nAll in all, if he had no restrictions, there were $100$ choices for the form $1$ * * $1$. But he excluded $10 + 1 = 11$ of them and get 89 possible choices. Since he has 3 chances, the resulting possibility is $3/89$.\n\n-\n\nJust my two cents,\n\n10^4 possibilities.\n\nThere are 14 ascending and descending groups of 4.\n\nKeyspace 9,876\n\nIf the badguy knows two of the four spaces, he only has to guess through entropy 10^2. (None of the restrictions meet up with the range 8xx1)\n\n3 Tries in 100\n\n:)\n\n-\n(0123)(1234)(2345)(3456)(4567)(5678)(6789) And then the reverse of those... \u2013\u00a0 Ben Sep 9 '13 at 5:33\n\nRegarding part [c], I do not think that there is something wrong with the book. As @Mack said in one of his comments, thief has to guess only two second numbers. Since the given restrictions actually do not apply on this order of numbers, he has to guess among 100 (10*10) possible combinations of the numbers.\n\nSince he has 3 tries, the probability that he will gain desired access is 3/100.\n\nAccording to Wolfram Alpha, 3/100 is exactly 0.03, which is the same as written solution from the book.\n\nCorrect me if I am wrong.\n\n-", "date": "2015-09-04 06:16:35", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/293921/combinatorics-and-probability-problem", "openwebmath_score": 0.775648295879364, "openwebmath_perplexity": 456.9605838745141, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462197739625, "lm_q2_score": 0.865224070413529, "lm_q1q2_score": 0.8547948496224866}}
{"url": "https://engineering.stackexchange.com/questions/41139/condition-for-equilibrium-of-a-rigid-body-on-a-horizontal-surface", "text": "# Condition for equilibrium of a rigid body on a horizontal surface\n\nThe following question is from a past paper from Further Mechanics and it has been bothering me immensely I have spent hours cracking at a way to make sense of it, the question is in two parts\n\nPart 1:\n\nAn object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.\n\nFind the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone.\n\nI first roughly sketched the object as such\n\nThen finding the $$\\bar{x}$$ of each part seperately:\n\nFor the cone I used the standard result of the centre of mass being $$\\frac{1}{4}r$$ away from the base:\n\n$$\\bar{x}_{Cone}=\\frac{1}{4}.4r + 4r=5r$$\n\nFor the cylinder I derived $$\\bar{x}$$ via integration:\n\n$$y=3r$$\n\n$$\\bar{x}_{cylinder}=\\frac{\\int_0^{4r} xdV}{\\int_0^{4r}dV}$$\n\n$$\\because dV=y^2 \\pi dx$$\n\n$$\\implies \\bar{x}_{cylinder}=\\frac{\\int_0^{4r} 9r^2 \\pi x dx}{\\int_0^{4r}9r^2 \\pi dx}$$\n\n$$\\implies \\bar{x}_{cylinder}=\\frac{9r^2 \\pi \\int_0^{4r} x dx}{9r^2 \\pi\\int_0^{4r} 1 dx}$$\n\n$$\\implies \\bar{x}_{cylinder}=\\frac{\\frac{1}{2}\\left[x^2 \\right]^{4r}_0}{\\left[x \\right]^{4r}_0}$$\n\n$$\\therefore \\bar{x}_{cylinder}=2r$$\n\nThen by taking the weighted average of both objects I arrived at the correct value for the distance of the centre of gravity from the base of the cylinder which was\n\n$$\\bar{x}=\\frac{11}{4}r$$\n\nHowever the next part has had me at a complete loss for the better part of the day, it states that:\n\nShow that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.\n\nI tried coming up with a rough sketch for this also\n\nHowever I do not understand how to tackle this question at all, all I know is that for a body to be at equilibrium on such a surface, the centre of mass must pass through the point of suspension, however clearly this does not give an insight on how to answer this.\n\nThe condition for this question given in the marking scheme is that\n\nCan someone explain what this means and why this is the case?\n\nGiven the geometry of the object, the center of gravity is very close to the surface between the cylinder and the cone. More specifically the distance between the center of gravity and the base of the cone is $$\\frac{1}{4}r$$.\n\nBasically what you need to prove is that the angle of the cone $$\\phi$$ is such, so that when the object is tilted the weight crosses over the last $$\\frac{1}{4}r$$.\n\nSo you need to prove that the angle $$\\theta$$ is smaller that angle $$\\phi$$.\n\nAngle $$\\theta$$ is calculated as: $$\\tan\\theta = \\frac{\\text{distance to be covered}}{\\text{cylinder radius}}= \\frac{\\frac{11}{4}r}{3r}=\\frac{11}{12} \\Rightarrow \\theta =42.5 [deg]$$\n\nwhile Angle $$\\phi$$ is calculated as: $$\\tan\\phi= \\frac{\\text{height of cone}}{\\text{cone radius}}= \\frac{4r}{3r}=\\frac{4}{3} \\Rightarrow \\phi =53.1[deg]$$\n\nTherefore, since $$\\phi> \\theta$$, the object becomes vertical enough so that the weight passes through the coned surface.\n\n\u2013\u00a0NMech\nMar 23, 2021 at 9:52\n\u2022 Reminds me of a crafty problem my undergrad QM prof posed. I'll adapt it to this particular object: using Heisenberg's Uncertainty Principle, how long will this object remain stable resting on the cone? It's not intended as a \"real-world\" problem but rather to see how you might use paired-parameters to guesstimate the stability. Mar 24, 2021 at 12:43\n\nThe CG is found by assuming the mass of the cone as 1/3m and the cylinder m, then\n\n$$\\bar{X}= \\frac{2r*m+5r*m/3}{4/3m} =r*11/4$$\n\nThe cone side length (like the sharp tip of a pencil) is $$5r$$, the side of a 3,4,5 triangle, and the interior half tip angle is 36.87 degrees.\n\nThe CG is $$5.25r$$ from the tip of the cone and at rotation to rest on the side of the cone will be at $$5.25r*cos36.87= 4.19r<5r$$\n\nI let you prove the condition.\n\nIt is well within the footprint of the pencil shape so it is stable.", "date": "2023-01-27 01:50:05", "meta": {"domain": "stackexchange.com", "url": "https://engineering.stackexchange.com/questions/41139/condition-for-equilibrium-of-a-rigid-body-on-a-horizontal-surface", "openwebmath_score": 0.7066793441772461, "openwebmath_perplexity": 216.23917361140576, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471665987074, "lm_q2_score": 0.8688267677469952, "lm_q1q2_score": 0.8547927537129945}}
{"url": "https://math.stackexchange.com/questions/2773993/transport-equation-u-t-xu-x-u-0-with-ux-0-0-cosx-0", "text": "# Transport equation $u_t + xu_x + u = 0$ with $u(x_0, 0) = \\cos(x_0)$\n\nI have been studying PDEs using Peter Olver's textbook. I have learnt how to solve equations such as $u_t + 2u_x = \\sin(x)$ subject to an initial condition such as $u(0,x) = \\sin x$. Letting $\\epsilon = x - 2t$ and $u(t,x) = v(t,\\epsilon)$, I then plug this into the transport equation.\n\nHowever, I am not sure how to define a characteristic to solve the following equation\n\n$$u_t + xu_x + u = 0, \\qquad u(x_0, 0) = \\cos(x_0)$$\n\nbecause it has a variable 'speed' term $x$ and it is also not homogenous because of the term $u$.\n\nA solution would be very helpful so I can see how to approach these problems.\n\n\u2022 If you change coordinates to $x = e^y$ then $u_y = xu_x$ so your PDE in these new coordinates simplifies to $u_t + 2u_y + u = 0$. You can even simplify it further by taking $v = e^{t} u$ then $v_t + 2v_y = 0$ which is the equation you say you already know how to solve. \u2013\u00a0Winther May 9 '18 at 17:48\n\n$$u_t + xu_x = -u$$ $\\frac{dt}{1}=\\frac{dx}{x}=\\frac{du}{-u}$\n\nFirst characteristics, from $\\quad \\frac{dt}{1}=\\frac{dx}{x}$ :\n\n$$x\\,e^{-t}=c_1$$\n\nSecond characteristics, from $\\quad\\frac{dx}{x}=\\frac{du}{-u}$ : $$x\\,u=c_2$$ General solution of the PDE : $\\quad x\\,u=F(x\\,e^{-t})$\n\n$$u(x,t)=\\frac{1}{x}F(x\\,e^{-t})$$ $F$ is an arbitrary function, to be determined according to the boundary condition.\n\nCondition : $\\quad u(x_0, 0) = \\cos(x_0)=\\frac{1}{x_0}F(x_0\\,e^{0})$\n\n$F(x_0)=x_0\\cos(x_0)$. Now the function $F$ is determined, i.e.: $F(X)=X\\cos(X)$.\n\nWe put it into the above general solution , where $X=x\\,e^{-t}$ , thus $F(x\\,e^{-t})=(x\\,e^{-t})\\cos(x\\,e^{-t})$ :\n\n$$u(x,t)=\\frac{1}{x}(x\\,e^{-t})\\cos(x\\,e^{-t})=e^{-t}\\cos(x\\,e^{-t})$$\n\n\u2022 Hi, thank you. Can you explain how you obtain your general solution of the PDE from the two characteristics? \u2013\u00a0PhysicsMathsLove May 11 '18 at 10:37\n\u2022 The general solution can be expressed on various forms. For example on the form of implicit equation : $\\Phi(c_1,c_2)=0$, or $c_1=f(c_2)$ or $c_2=F(c_1)$ or other forms. All functions introduced are arbitrary, but they are related one to the other. Doesn't matter the form chosen, the final result is the same after applying the boundary condition. \u2013\u00a0JJacquelin May 11 '18 at 10:51\n\u2022 Ok I am still not sure how you obtain $xu = F(xe^{-t})$ from your characteristics, I.e. how it depends on $xe^{-t}$? \u2013\u00a0PhysicsMathsLove May 11 '18 at 10:57\n\u2022 $c_2=F(c_1)$ with $c_2=xu$ and $c_1=xe^{-t}$ gives $xu=F(xe^{-t})$. \u2013\u00a0JJacquelin May 11 '18 at 11:02\n\u2022 Why do you know one constant is a function of the other? \u2013\u00a0PhysicsMathsLove May 11 '18 at 11:12\n\nLet us apply the method of characteristics. We get the characteristic equations $$\\frac{\\text{d} t}{\\text{d} s} = 1 \\, , \\qquad \\frac{\\text{d} x}{\\text{d} s} = x \\, , \\qquad \\frac{\\text{d} u}{\\text{d} s} = -u \\, .$$ Letting $t(0) = 0$, we know $t=s$. Letting $x(0) = x_0$, we get $x(t) = x_0\\, e^t$. Since $u(0) = \\cos(x_0)$, we have $u(t) = \\cos(x_0)\\, e^{-t}$. Finally, $$u(x,t) = \\cos(x\\, e^{-t})\\, e^{-t} \\, .$$\n\nNote that \\begin{align} \\frac{\\rm d}{{\\rm d}t}u(e^t,t+t_0)&=\\frac{\\partial u}{\\partial t}(e^t,t+t_0)+e^t\\frac{\\partial u}{\\partial x}(e^t,t+t_0)\\\\ &=\\left(\\frac{\\partial u}{\\partial t}+x\\frac{\\partial u}{\\partial x}\\right)(e^t,t+t_0)\\\\ &=-u(e^t,t+t_0) \\end{align} holds for all $t$ and $t_0$. Thus $$\\frac{\\rm d}{{\\rm d}t}u(e^t,t+t_0)+u(e^t,t+t_0)=0,$$ or equivalently, $$\\frac{\\rm d}{{\\rm d}t}\\left(e^tu(e^t,t+t_0)\\right)=0.$$ Therefore, $$e^tu(e^t,t+t_0)=e^{-t_0}u(e^{-t_0},-t_0+t_0)=e^{-t_0}u(e^{-t_0},0)=e^{-t_0}\\cos e^{-t_0},$$ or equivalently, $$u(e^t,t+t_0)=e^{-t-t_0}\\cos e^{-t_0}.$$ Finally, let $x=e^t>0$ and $\\tau=t+t_0$. This gives $t=\\log x$ and $t_0=\\tau-\\log x$. Hence $$u(x,\\tau)=e^{-\\log x-\\tau+\\log x}\\cos e^{-\\tau+\\log x}=e^{-\\tau}\\cos\\left(e^{-\\tau}x\\right).$$", "date": "2020-02-28 09:40:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2773993/transport-equation-u-t-xu-x-u-0-with-ux-0-0-cosx-0", "openwebmath_score": 0.9334983229637146, "openwebmath_perplexity": 168.1841414177281, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471684931717, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8547927520173425}}
{"url": "https://mathematica.stackexchange.com/questions/159043/count-the-pairs-in-a-set-of-data", "text": "# count the pairs in a set of Data\n\nI have a list of Data, e.g.,\n\nlist1={0, 1, 1, 1, 1, 2, 2, 3, 3}\n\nlist2={0, 1, 1, 1, 1, 2, 2, 3, 3, 0}\n\n\nand I want to get as an output True or False if I have only even number of Data of similar or not. in the above example I like to write a code that returns False for list1 and True for list2, since in list1 I have only one 0, but in the list2 there are even number of each number.\n\nThanks!\n\n## 4 Answers\n\nUsing Counts\n\nf = And @@ EvenQ[Values[Counts[#]]] &\n{f[list1], f[list2]}\n\n\n{False, True}\n\n\u2022 +1. Or f = Counts /* AllTrue[EvenQ] \u2013\u00a0WReach Nov 1 '17 at 14:27\n\u2022 Yours is much concise. I think you should post it as a separate answer. \u2013\u00a0Anjan Kumar Nov 1 '17 at 14:30\n\u2022 I did, but when I saw yours, I deleted it :) I shall undelete. \u2013\u00a0WReach Nov 1 '17 at 14:36\n\nGiven:\n\nf = Counts /* AllTrue[EvenQ];\n\n\nThen:\n\nf[list1]\n(* False *)\n\nf[list2]\n(* True *)\n\n\nYou can use the Tally function:\n\nTally[list]\ntallies the elements in list, listing all distinct elements together with their multiplicities.\n\nSo as\n\nAnd@@EvenQ[Tally[list1][[;; , 2]]]\n\n\nor\n\nAnd@@EvenQ[Tally[list2][[;; , 2]]]\n\ncheck[list_] := If[\nCases[EvenQ[Count[list, #] & /@ Union[list]], False] == {},\nTrue, False]\n\ncheck[list2]\n\n\nTrue", "date": "2019-09-20 17:32:13", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/159043/count-the-pairs-in-a-set-of-data", "openwebmath_score": 0.2827434539794922, "openwebmath_perplexity": 1909.0496921774816, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914018751051, "lm_q2_score": 0.8840392863287584, "lm_q1q2_score": 0.8547899790779508}}
{"url": "https://stats.stackexchange.com/questions/48133/sum-standard-deviation-vs-standard-error", "text": "# Sum standard deviation vs standard error\n\nI'm having difficulty in determining what exactly the difference is between the 2, especially when given an exercise and I have to choose which of the 2 to use. These is how my text book describes them:\n\nSum standard deviation\n\nGiven is a population with a normally distributed random variable $X$. When you have a sample $n$ from this population the population is:\n\n$X_{sum} = X_1 + X_2 ... + X_n$ with\n\n$\\mu_{Xsum} = n \\times \\mu_x$ and $\\sigma_{Xsum} = \\sqrt{n} \\times \\sigma_x$.\n\nStandard error\n\nWhen you have a normally distributed random variable $X$ with mean $\\mu_X$ and standard deviation $\\sigma_X$ and sample length $n$, the sample mean $\\bar{X}$ is normally distributed with $\\mu_{\\bar{x}} = \\mu_X$ and $\\sigma_{\\bar{x}} = \\dfrac{\\sigma_X}{\\sqrt{n}}$\n\nThese 2 are awefully similair to me to the point I can't at all decide which to use where. Here are the problems where I discovered I couldn't:\n\nProblem 1\n\nA filling machine fills bottles of lemonade. The amount is normally distributed with $\\mu = 102 \\space cl$.\n\n$\\sigma$ = $1.93\\space cl$.\n\n\u2022 Calculate the chance that out of 12 bottles the average volume is $100 \\space cl$.\n\nThe problem itself is easy, however the troublesome part is what to choose for the standard deviation of the sample. Here they use $\\dfrac{1.93}{\\sqrt{12}}$ which I can live with, until I encountered the second problem.\n\nProblem 2\n\nA tea company puts 20 teabags in one package. The weight of a teabag is normally distributed with $\\mu = 5.3 \\space g$ and $\\sigma = 0.5 \\space g.$\n\n\u2022 Calculate the chance that a package weighs less than 100 grams.\n\nHere I thought they'd also use $\\dfrac{0.5}{\\sqrt{20}}$, but instead they use $\\sqrt{20} \\times 0.5$.\n\nCan someone clear up the confusion?\n\n\u2022 You should tag this as \"homework\" as well, since it seems to be a homework question. \u2013\u00a0Placidia Jan 20 '13 at 17:33\n\u2022 @Placidia Are you kidding me?! This isn't homework, this is about understanding and differentiating 2 general concepts in statistics, which then could be implemented in homework questions.. like every other mathematical concept.. \u2013\u00a0JohnPhteven Jan 20 '13 at 17:36\n\u2022 My textbook confused me; (Freedman, Pisani, Purves, Statistics, Fourth Edition). The chapter is titled \"The Standard Error\", uses $\\sqrt{n} \\times \\sigma_x$ but the acronym is \"SE\", which must be a hint that my textbook is referring to \"sum standard deviation\"; indeed the textbook describes the experiment: \"When drawing at random with replacement from a box of numbered tickets; the standard error for the sum of the draws is...\" \u2013\u00a0The Red Pea Nov 7 '19 at 7:26\n\u2022 ... more from the confusing part of Statistics textbook Freedman, Pisani, Purves, Fourth Edition: \"In this book, we use SD for data and SE for chance quantities (random variables). This distinction is not standard and the term SD is often used in both situations\" Indeed, as in this SE question, we refer to the \"SD of X\" ($\\sigma_{X}$) and the \"SD of sum of X\" ($\\sigma_{X_{sum}}$) \u2013\u00a0The Red Pea Nov 7 '19 at 7:37\n\nThe sum standard deviation is, as the name suggests, the standard deviation of the sum of $n$ random variables. The standard error you're talking about is just another name for the standard deviation of the mean of $n$ random variables. As you noted, the two formulas are closely related; since the sum of $n$ random variables is $n$ times the mean of $n$ random variables, the standard deviation of the sum is also $n$ times the standard deviation of the mean:\n\n$\\sigma_{X_{sum}} = \\sqrt n\\sigma_X = n \\times \\frac{\\sigma_X}{\\sqrt n} = n\\times \\sigma_\\bar{X}$.\n\nIn the first problem you are dealing with a mean, the average of twelve bottles, so you use the standard deviation of the mean, which is called standard error. In the second problem you are dealing with a sum, the total weight of 20 packages, so you use the standard deviation of the sum.\n\nSummary: use standard error when dealing with the mean (averages); use sum standard deviation when dealing with the sum (totals).\n\n\u2022 But what I think is that theyask one about the sum of 12 bottles, and the mean of that sum? In other words, they're too similair to me.. \u2013\u00a0JohnPhteven Jan 20 '13 at 18:37\n\u2022 There's no sum in question one. Each bottle is filled with an amount given by a normal distribution with mean 102, the question asks about the mean of twelve bottles. Where do you see a sum? \u2013\u00a0Jonathan Christensen Jan 20 '13 at 18:45\n\u2022 Oh wait nevermind, I was being a little bit blind! In the first one they ask about the MEAN (i.e. average) out of a sample, in the second they transform a sample of 6 into '1' object (>namely, the box of teabags), with its own SD and M! \u2013\u00a0JohnPhteven Jan 20 '13 at 18:45\n\u2022 I think I was writing my response the same time you were doing yours. Nice answer. \u2013\u00a0Placidia Jan 20 '13 at 18:46\n\u2022 I meant 20, my brain is random, no idea how I got to 6 \u2013\u00a0JohnPhteven Jan 20 '13 at 18:48\n\nThe first standard deviation formula you gave is the SD for a sum. The standard error is the SD of the sample mean. Remember that: $\\text{Var}(aX)=a^2 \\text{Var}(X)$ and the variance of the sum is the sum of the variances (First formula). So\n\n$\\text{Var}(\\bar{X})=\\frac{n\\sigma^2}{n^2}=\\sigma^2/n$. Taking the square root gives the result.\n\nRecall:\n\n$\\text{Var}(\\sum X_i)=\\sum (\\text{Var}(X_i)=n \\sigma^2.$ The Variance of the sums.\n\nProblem 1 is looking for a statement about the sample mean; Problem 2 is about the sum, since the weight of the package is the sum of the weights of individual tea bags.\n\n\u2022 Nice answer, +1, but I gave the other one a best answer since I read it first and it answered my question first. \u2013\u00a0JohnPhteven Jan 20 '13 at 18:48", "date": "2020-07-07 02:56:40", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/48133/sum-standard-deviation-vs-standard-error", "openwebmath_score": 0.8502633571624756, "openwebmath_perplexity": 447.38077730219237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914018751051, "lm_q2_score": 0.8840392786908831, "lm_q1q2_score": 0.854789971692782}}
{"url": "https://math.stackexchange.com/questions/558145/minimum-number-of-iterations-in-newtons-method-to-find-a-square-root", "text": "Minimum number of iterations in Newton's method to find a square root\n\nI am writing an algorithm that evaluates the square root of a positive real number $y$. To do this I am using the Newton-Raphton method to approximate the roots to $f(x)=x^2-y$. The $n^{th}$ iteration gives $$x_n=\\frac{x_{n-1}^2+y}{2x_{n-1}}$$ as an approximation to $\\sqrt{y}$. I found that starting with an initial guess $x_0=1$ works pretty well generally, so an answer to the question below that assumes $x_0=1$ is fine.\n\nMy question: is there an exact expression for the minimum $N$ of iterations needed to attain a given precision $p$ in the approximate solution $x_N$? In other words I'm looking for the smallest integer $N$ such that $$\\left|\\frac{x_N-\\sqrt y}{\\sqrt y}\\right|<p.$$\n\nI've thought about this for a while and played around with the expression for the errors $\\epsilon_n = x_n - \\sqrt y$ which can be shown to satisfy $\\epsilon_{n+1}=\\epsilon_n^2\\,/\\,2x_n$, but I can't find an answer. I've looked around on Google but I couldn't find an answer either.\n\nAny pointers to a solution online or help would be much appreciated. A follow-up would of course be: can $x_0$ be optimised (while being a simple enough expression in terms of $y$) in order to minimise $N$?\n\n\u2022 I don't have an answer to your overall question, but here's a little bit of intuition: $x_n$ is the average of $x_{n-1}$ and $y/x_{n-1}$. One of these numbers must be less than $\\sqrt{y}$, and one must be greater, so the average is a reasonable update. And of course, the closer $x_0$ is to $\\sqrt{y}$, the better. Something like $x_0 = \\lfloor \\sqrt{y}\\rfloor$ is a good guess (take the biggest perfect square less than $y$, and start with its square root). \u2013\u00a0BaronVT Nov 9 '13 at 16:58\n\u2022 Try analyzing $x_n^2 - y$ rather than $x_n - \\sqrt{y}$. I imagine you probably want to analyze the roundoff error too, rather than assuming addition, multiplication and division are exact, which makes the problem more complicated. \u2013\u00a0Hurkyl May 2 '14 at 8:21\n\nThere is such a formula: consider\n\n$$\\frac{x_n+\\sqrt y}{x_n-\\sqrt y}=\\frac{\\frac{x_{n-1}^2+y}{2x_{n-1}}+\\sqrt y}{\\frac{x_{n-1}^2+y}{2x_{n-1}}-\\sqrt y}=\\frac{(x_{n-1}+\\sqrt y)^2}{(x_{n-1}-\\sqrt y)^2}=\\left(\\frac{x_{n-1}+\\sqrt y}{x_{n-1}-\\sqrt y}\\right)^2.$$\n\nBy recurrence,\n\n$$\\frac{x_n+\\sqrt y}{x_n-\\sqrt y}=\\left(\\frac{x_{0}+\\sqrt y}{x_{0}-\\sqrt y}\\right)^{2^n}.$$\n\nIf you want to achieve $2^{-b}$ relative accuracy, $x_n=(1+2^{-b})\\sqrt y$,\n\n$$2^n=\\frac{\\log_2\\frac{(1+2^{-b})\\sqrt y+\\sqrt y}{(1+2^{-b})\\sqrt y-\\sqrt y}}{\\log_2\\left|\\frac{x_{0}+\\sqrt y}{x_{0}-\\sqrt y}\\right|},$$\n\n$$n=\\log_2\\left(\\log_2\\frac{2+2^{-b}}{2^{-b}}\\right)-\\log_2\\left(\\log_2\\left|\\frac{x_{0}+\\sqrt y}{x_{0}-\\sqrt y}\\right|\\right).$$\n\nThe first term relates to the desired accuracy. The second is a penalty you pay for providing an inaccurate initial estimate.\n\nIf the floating-point representation of $y$ is available, a very good starting approximation is obtained by setting the mantissa to $1$ and halving the exponent (with rounding). This results in an estimate which is at worse a factor $\\sqrt 2$ away from the true square root.\n\n$$n=\\log_2\\left(\\log_2\\left(2^{b+1}+1\\right)-\\log_2\\left(\\log_2\\frac{\\sqrt 2+1}{\\sqrt 2-1}\\right)\\right) \\approx\\log_2(b+1)-1.35.$$ In the case of single precision (23 bits mantissa), 4 iterations are always enough. For double precision (52 bits), 5 iterations.\n\nOn the opposite, if $1$ is used as a start and $y$ is much larger, $\\log_2\\left|\\frac{1+\\sqrt y}{1-\\sqrt y}\\right|$ is close to $\\frac{2}{\\ln(2)\\sqrt y}$ and the formula degenerates to $$n\\approx\\log_2(b+1)+\\log_2(\\sqrt y)-1.53.$$ Quadratic convergence is lost as the second term is linear in the exponent of $y$.\n\nIf your starting value of $y$ is correctly \"scaled\" between $1$ and $100$, then a good initial guess is\n\n$X_0 = 1,1545 + 0,11545*y$ (linear approx) or if you prefer a second order approx. you can use:\n\n$X_0 = 1,0 + y*( 0,18 - 0,0009*y)$\n\nUsing $X_0$, you need only 3 iterations (Newton / Heron) to get a Relative Error less than 0,0001 % !\n\nNow, here is a fast method to get an excellent $X_0$ (with less than 1% error):\n\nStore a pre-calculated constants table containing all the square roots with rounded 4 decimals in the range $[1 ; 100]$. So $table(i)=sqrt(i)$.\n\nUse Simple Linear Interpolation to get a good $X_0$. Now, the Maximum Relative Error is still greater than 1% ! ( 1.5% near $y=1.4142$).\n\nYou can simply divide this Maximum Relative Error (MRE) by a factor of 2 by doing this trick:\n\nSimply replace the 1st entry ($\\sqrt(1)=1.0000$ by $\\sqrt(1)=1.0075$ ) AND the second entry ( $\\sqrt(2)=1.4142$ by $\\sqrt(2)=1.4248$ ). That's all. You have a MRE of 0,75%! So, the WORST case is a $X_0$ with only 0.75% in the WHOLE RANGE $[1 ; 100]$.\n\nUnscale your number, and with only ONE Newton / Heron iteration get a 0.0028% MRE in any case, and with two iteration you have a 0.00000004% MRE!!\n\nThat's Fast, Simple and very Accurate. Have fun.", "date": "2019-07-21 09:39:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/558145/minimum-number-of-iterations-in-newtons-method-to-find-a-square-root", "openwebmath_score": 0.866824746131897, "openwebmath_perplexity": 264.4876199650986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966914017797636, "lm_q2_score": 0.8840392695254319, "lm_q1q2_score": 0.8547899619877225}}
{"url": "https://cs.stackexchange.com/questions/105563/expected-value-of-the-distance-between-nodes-in-a-binary-tree", "text": "# Expected value of the distance between nodes in a binary tree\n\nIf there are 16 leaves in a full binary tree and two nodes $$a$$ and $$b$$ chosen at random, then what is the expected value of the distance between $$a$$ and $$b$$ in T?\n\nMy question here is, how do I correctly approach this question?\n\n(Also tell me how to develop numerical skill in algorithm?)\n\nLet's name the nodes $$\\mathit{node0}$$ through $$\\mathit{node15}$$. (with the implication that that's the order of the nodes)\n\nI would approach this problem first by saying \"if $$a$$ is $$\\mathit{node0}$$, then what's the expected value of the distance between $$a$$ and $$b$$?\" Then I'd ask the question \"what if $$a$$ is $$\\mathit{node1}$$?\", etc. Presumably along the way there would be patterns and symmetries I could make use of so I wouldn't need to do a long calculation 16 times.\n\nSo let's take that first question: \"if $$a$$ is $$\\mathit{node0}$$, then what's the expected value of the distance between $$a$$ and $$b$$?\"\n\nFirst off, the distance between $$\\mathit{node0}$$ and $$\\mathit{node0}$$ is $$0$$. To $$\\mathit{node1}$$, the distance is $$2$$. To nodes $$\\mathit{node2}$$ or $$\\mathit{node3}$$, the distance is $$4$$. To the four nodes $$\\mathit{node4}$$ through $$\\mathit{node7}$$, the distance is $$6$$. To the remaining eight nodes the distance is $$8$$.\n\nSo the expected distance when $$a$$ is $$\\mathit{node0}$$ is $$(0 + 2 + 2*4 + 4*6 + 8*8)/16 = 6.125$$.\n\nNow figure out the expected distance when $$a$$ is $$\\mathit{node1}$$, and then other nodes. (there is an obvious pattern - prove it)\n\nThen average over those results to find the answer.\n\nOften, you will find a shorter or more elegant solution to a problem as you are working on it. This is normal, and should not be taken as a sign that your initial approach was wrong or that you should have seen the more elegant approach without trying your initial approach first, any more than a writer should expect to write an essay without first writing some notes or a rough draft.\n\nUnfortunately, though writers are often trained in the idea of rough drafts and revisions, solutions to math or computer science problems are usually only presented in their final form without much indication of the bumbling around and blind corners that really went into discovering the elegant explanation.\n\n\u2022 \"To $\\mathit{node1}$, the distance is $2$. To nodes $\\mathit{node2}$ or $\\mathit{node3}$, the distance is $4$.\" how?? Can u draw how node1 getting distance 2 and node2 and node3 both getting distance 4?? \u2013\u00a0Srestha Mar 14 at 16:23", "date": "2019-10-14 17:55:14", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/105563/expected-value-of-the-distance-between-nodes-in-a-binary-tree", "openwebmath_score": 0.8741745352745056, "openwebmath_perplexity": 276.40181810387855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864296722661, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8547459721471093}}
{"url": "http://stats.stackexchange.com/questions/4959/how-do-you-calculate-the-expectation-of-left-sum-i-1n-x-i-right2", "text": "# How do you calculate the expectation of $\\left(\\sum_{i=1}^n {X_i} \\right)^2$?\n\nIf $X_i$ is exponentially distributed $(i=1,...,n)$ with parameter $\\lambda$ and $X_i$'s are mutually independent, what is the expectation of\n\n$$\\left(\\sum_{i=1}^n {X_i} \\right)^2$$\n\nin terms of $n$ and $\\lambda$ and possibly other constants?\n\nNote: This question has gotten a mathematical answer on http://math.stackexchange.com/q/12068/4051. The readers would take a look at it too.\n\n-\nThe two copies of this question reference each other and, appropriately, the stats site (here) has a statistical answer and the math site has a mathematical answer. It seems like a good division: let it stand! \u2013\u00a0whuber Mar 4 '11 at 21:59\n\nIf $x_i \\sim Exp(\\lambda)$, then (under independence), $y = \\sum x_i \\sim Gamma(n, 1/\\lambda)$, so $y$ is gamma distributed (see wikipedia). So, we just need $E[y^2]$. Since $Var[y] = E[y^2] - E[y]^2$, we know that $E[y^2] = Var[y] + E[y]^2$. Therefore, $E[y^2] = n/\\lambda^2 + n^2/\\lambda^2 = n(1+n)/\\lambda^2$ (see wikipedia for the expectation and variance of the gamma distribution).\n\n-\n+1 Nice answer! \u2013\u00a0whuber Nov 27 '10 at 17:10\nThanks. A very neat way of answering the question (leading to the same answer) was also provided on math.stackexchange (link above in the question) a few minutes ago. \u2013\u00a0Wolfgang Nov 27 '10 at 17:24\nThe math answer computes the integrals using linearity of expectation. In some ways it's simpler. But I like your solution because it exploits statistical knowledge: because you know a sum of independent Exponential variables has a Gamma distribution, you're done. \u2013\u00a0whuber Nov 27 '10 at 21:19\nI enjoyed it quite a bit and I am by no means a statistician or a mathematician. \u2013\u00a0Kortuk Nov 29 '10 at 18:04\nvery elegant answer. \u2013\u00a0Cyrus S Nov 30 '10 at 16:44\n\nThe answer above is very nice and completely answers the question but I will, instead, provide a general formula for the expected square of a sum and apply it to the specific example mentioned here.\n\nFor any set of constants $a_1, ..., a_n$ it is a fact that\n\n$$\\left( \\sum_{i=1}^{n} a_i \\right)^2 = \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_{i} a_{j}$$\n\nthis is true by the Distributive property and becomes clear when you consider what you're doing when you calculate $(a_1 + ... + a_n) \\cdot (a_1 + ... + a_n)$ by hand.\n\nTherefore, for a sample of random variables $X_1, ..., X_n$, regardless of the distributions,\n\n$$E \\left( \\left[ \\sum_{i=1}^{n} X_i \\right]^2 \\right) = E \\left( \\sum_{i=1}^{n} \\sum_{j=1}^{n} X_i X_j \\right) = \\sum_{i=1}^{n} \\sum_{j=1}^{n} E(X_i X_j)$$\n\nprovided that these expectations exist.\n\nIn the example from the problem, $X_1, ..., X_n$ are iid ${\\rm exponential}(\\lambda)$ random variables, which tells us that $E(X_{i}) = 1/\\lambda$ and ${\\rm var}(X_i) = 1/\\lambda^2$ for each $i$. By independence, for $i \\neq j$, we have\n\n$$E(X_i X_j) = E(X_i) \\cdot E(X_j) = \\frac{1}{\\lambda^2}$$\n\nThere are $n^2 - n$ of these terms in the sum. When $i = j$, we have\n\n$$E(X_i X_j) = E(X_{i}^{2}) = {\\rm var}(X_{i}) + E(X_{i})^2 = \\frac{2}{\\lambda^2}$$\n\nand there are $n$ of these term in the sum. Therefore, using the formula above,\n\n$$E \\left( \\sum_{i=1}^{n} X_i \\right)^2 = \\sum_{i=1}^{n} \\sum_{j=1}^{n} E(X_i X_j) = (n^2 - n)\\cdot\\frac{1}{\\lambda^2} + n \\cdot \\frac{2}{\\lambda^2} = \\frac{n^2 + n}{\\lambda^2}$$\n\n-\n\nThis problem is just a special case of the much more general problem of 'moments of moments' which are usually defined in terms of power sum notation. In particular, in power sum notation:\n\n$$s_1 = \\sum_{i=1}^{n} X_i$$\n\nThen, irrespective of the distribution, the original poster seeks $E[s_1^2]$ (provided the moments exist). Since the expectations operator is just the 1st Raw Moment, the solution is given in the mathStatica software by:\n\n[ The '___ToRaw' means that we want the solution presented in terms of raw moments of the population (rather than say central moments or cumulants). ]\n\nFinally, if $X$ ~ Exponential($\\lambda$) with pdf $f(x)$:\n\nf = Exp[-x/\u03bb]/\u03bb;      domain[f] = {x, 0, \u221e} && {\u03bb > 0};\n\n\nthen we can replace the moments $\\mu_i$ in the general solution sol with the actual values for an Exponential random variable, like so:\n\nAll done.\n\nP.S. The reason the other solutions posted here yield an answer with $\\lambda^2$ in the denominator rather than the numerator is, of course, because they are using a different parameterisation of the Exponential distribution. Since the OP didn't state which version he was using, I decided to use the standard distribution theory textbook definition Johnson Kotz et al \u2026 just to balance things out :)\n\n-", "date": "2013-05-23 09:09:47", "meta": {"domain": "stackexchange.com", "url": "http://stats.stackexchange.com/questions/4959/how-do-you-calculate-the-expectation-of-left-sum-i-1n-x-i-right2", "openwebmath_score": 0.9295454025268555, "openwebmath_perplexity": 220.5786263429254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864292291072, "lm_q2_score": 0.8633916047011595, "lm_q1q2_score": 0.8547459717644897}}
{"url": "http://mathhelpforum.com/statistics/29269-help-probability-flipping-quarter.html", "text": "# Thread: Help with Probability flipping a quarter!!!!\n\n1. ## Help with Probability flipping a quarter!!!!\n\nSo here is the problem. Maybe someone can help me. Thanks ahead of time anyone.\n\nJohn and Judy play a game of flipping a quarter. If it comes up heads, John gets a point and if it comes up tails, Judy gets a point. They each bet $5 to play and the first person to get ten points wins the whole$10. At a point where John has 7 points and Judy has 5 points, the game is interrupted and they can't continue. John thinks he should win since he's ahead while Judy thinks she should win since she'd make a comeback. You are their friend, they agree to let you decide how to split up the money between them. Come up with a split that is most fair to both John and Judy.\n\n2. Originally Posted by 2y4life\nSo here is the problem. Maybe someone can help me. Thanks ahead of time anyone.\n\nJohn and Judy play a game of flipping a quarter. If it comes up heads, John gets a point and if it comes up tails, Judy gets a point. They each bet $5 to play and the first person to get ten points wins the whole$10. At a point where John has 7 points and Judy has 3 points, the game is interrupted and they can't continue. John thinks he should win since he's ahead while Judy thinks she should win since she'd make a comeback. You are their friend, they agree to let you decide how to split up the money between them. Come up with a split that is most fair to both John and Judy.\nCalculate the probability $\\displaystyle p_{John}$ that John would have won from this position.\n\nCalculate the probability $\\displaystyle p_{Judy}$ that Judy would have won from this position.\n\nThen give John $\\displaystyle \\text{\\$}10 \\, p_{John}$and give Judy$\\displaystyle \\text{\\$}10 \\, p_{Judy}$.\n\n3. Originally Posted by mr fantastic\nCalculate the probability $\\displaystyle p_{John}$ that John would have won from this position.\n\nCalculate the probability $\\displaystyle p_{Judy}$ that Judy would have won from this position.\n\nThen give John $\\displaystyle \\text{\\$}10 p_{John}$and give Judy$\\displaystyle \\text{\\$}10 p_{Judy}$.\nYea...but is there a more simple way of finding that out? I tried and there are too many possibilities. John could flip another head and then flip 5 tails in a row and Judy would win or vice versa.\n\n4. Originally Posted by 2y4life\nSo here is the problem. Maybe someone can help me. Thanks ahead of time anyone.\n\nJohn and Judy play a game of flipping a quarter. If it comes up heads, John gets a point and if it comes up tails, Judy gets a point. They each bet $5 to play and the first person to get ten points wins the whole$10. At a point where John has 7 points and Judy has 5 points, the game is interrupted and they can't continue. John thinks he should win since he's ahead while Judy thinks she should win since she'd make a comeback. You are their friend, they agree to let you decide how to split up the money between them. Come up with a split that is most fair to both John and Judy.\nThis question is historically interesting ..... read this to see why.\n\n5. Originally Posted by 2y4life\nYea...but is there a more simple way of finding that out? I tried and there are too many possibilities. John could flip another head and then flip 5 tails in a row and Judy would win or vice versa.\nLet X be the random variable number of points John wins.\nLet Y be the random variable number of points Judy wins.\n\nThen:\n\nPr(John wins from this position) = Pr(X = 3) times Pr(Y < 5).\n\nPr(Judy wins from this position) = Pr(X < 3) times Pr(Y = 5).\n\nX and Y both follow binomial distributions.\n\nEdit: Actually it's slightly more complicated than this I just realised but I have to rush off now. I'll have more to say later unless someone else gets in first.\n\n6. Originally Posted by mr fantastic\nThis question is historically interesting ..... read this to see why.\nYea, this was an extra credit problem for us to take home and figure out and my TA said that this problem came from two mathematicians named Blaise and Pierre.\n\nI want to put this down as the answer but I'm not sure if this would constitute as a fair answer:\n\nJohn- 7/12 x $10=$5.83\nJudy- 5/12 x $10=$4.17\n\n7. Originally Posted by mr fantastic\nLet X be the random variable number of points John wins.\nLet Y be the random variable number of points Judy wins.\n\nThen:\n\nPr(John wins from this position) = Pr(X = 3) times Pr(Y < 5).\n\nPr(Judy wins from this position) = Pr(X < 3) times Pr(Y = 5).\n\nX and Y both follow binomial distributions.\n\nEdit: Actually it's slightly more complicated than this I just realised but I have to rush off now. I'll have more to say later unless someone else gets in first.\nYou only need to work out one of John winning or Judy winning (you know why, right).\n\nProbability of John winning: You need to calculate Pr(X = 3) times Pr(Y < 5) for n = 3, 4, 5, 6 and 7.\n\nProbability of Judy winning: You need to calculate Pr(X < 3) times Pr(Y = 5) for n = 5, 6 and 7.", "date": "2018-06-25 10:56:08", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/29269-help-probability-flipping-quarter.html", "openwebmath_score": 0.5551230311393738, "openwebmath_perplexity": 1301.3342931621264, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9759464471055739, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8547212139942243}}
{"url": "https://math.stackexchange.com/questions/2915495/dimension-of-vector-space-countable-uncountable", "text": "# Dimension of vector space, countable, uncountable?\n\nIn set theory, when we talk about cardinality of a set we have notions like finite set, countably infinite and uncountably infinite sets.\n\nMain Question\n\nLet's talk about dimension of a vector space. In Linear Algebra I have always heard about two terms either a vector space is finite dimensional for example $\\mathbb{R}^n$ or infinite dimensional for example $C[0,1]$.\n\nWhy don't we have notions like countably infinite dimensional vector space and uncountably infinite dimensional vector space.\n\nMay be, I am not able to see the bigger picture.\n\nExtras\n\nP.S. Long time ago, I was in a talk on Enumerative Algebraic geometry and the professor said, I always think of a positive natural number as a dimension of some vector space.\n\nDon't we have then uncountable dimension vector space?/\n\n\u2022 Some infinite-dimensional vector spaces have countable dimension, some have uncountable dimension. The dimension of a vector space is a well-defined cardinality. So, what's the question? \u2013\u00a0Lord Shark the Unknown Sep 13 '18 at 11:59\n\u2022 I asked this as I have never heard of terms like countable dimension or uncountable dimension in books. @LordSharktheUnknown \u2013\u00a0StammeringMathematician Sep 13 '18 at 12:01\n\u2022 An infinite countable dimmensionnal spaces cannot be complete but they exist. Consider the vector space of real sequences having finite support. \u2013\u00a0nicomezi Sep 13 '18 at 12:14\n\u2022 \"I have never heard of terms like countable dimension or uncountable dimension in books\" --- This is because beginning and intermediate level linear algebra books rarely distinguish more precisely than \"finite dimension\" and \"infinite dimension\". It's usually only in graduate level algebra courses (e.g. probably all of the \"third level\" books in my answer to High-level linear algebra book) where you'll find the various notions of \"algebraic dimension\" defined as a cardinal number. \u2013\u00a0Dave L. Renfro Sep 13 '18 at 12:31\n\u2022 Just because you don't see it in your text doesn't mean it doesn't exist. You're quite right that it makes sense! \u2013\u00a0Noah Schweber Sep 13 '18 at 12:31\n\nWe do have the notions of countable/uncountable dimensions. Just as a set can be finite or infinite (without specifying which infinite cardinality the set as) a vector space can be finite dimensional or infinite dimensional. We can then go one step more and ask, if the dimension is infinite, which infinite cardinal is it?\n\nThe definition of dimension of a vector space is the cardinality of a basis for that vector space (it does not matter which basis you take, because they all have the same cardinality). Then for any cardinal number $\\gamma$, you can have a vector space with that dimension. For example, if $\\Gamma$ is a set with cardinality $\\gamma$, let $c_{00}(\\Gamma)$ be the space of all $\\mathbb{F}$-valued functions $f$ such that $$\\text{supp}(f)=\\{x\\in \\Gamma: f(x)\\neq 0\\}$$ is finite. Then let $\\delta_x\\in c_{00}(\\Gamma)$ be the function such that $\\delta_x(y)=0$ if $y\\neq x$ and $\\delta_x(y)=1$ if $y=x$. Then $(\\delta_x)_{x\\in \\Gamma}$ is a basis for $c_{00}(\\Gamma)$ with cardinality $\\gamma$. If $\\Gamma=\\mathbb{N}$, we have a vector space with countably infinite dimension. If $\\Gamma=\\mathbb{R}$, we have a vector space with dimension equal to the cardinality of the continuum.\n\nHowever, for infinite dimensional topological vector spaces (and for infinite dimensional Hilbert and Banach spaces in particular) the usual notion of a basis of limited use. This is because the coordinate functionals for an infinite basis do not interact very well with the topology (one can show that if $(e_i, e^*_i)_{i\\in I}$ is a basis together with its coordinate functionals for an infinite dimensional Banach space, then only finitely many of the functionals $e^*_i$ can be continuous). Since the notion of a basis is not as useful in the infinite dimensional topological space case as it is in the finite dimensional case, you can see less emphasis on what the exact dimension is in this case.\n\nHowever, in this situation you get into discussions of other types of coordinate systems (such as Schauder bases, FDDs, unconditional bases, etc.), which are different from the notion of an (algebraic) basis. You also can ask about density character instead of dimension, which is the smallest cardinality of a dense subset. This encodes topological information, while the purely algebraic notion of a basis does not. For example, infinite dimensional Hilbert space $\\ell_2$ has no countable basis, it does have a countable, dense subset. So the dimension is that of the continuum, but the density character is $\\aleph_0$.\n\nThe dimension of a vector space is the cardinality of a basis for that vector space. To say that a vector space has finite dimension therefore means that the cardinality of a basis for that vector space is finite. Since finite cardinalities are the same thing as natural numbers, we are safe in saying, for finite dimensional vector spaces, that the dimension is a natural number.\n\nIn general, some sets are countably infinite and some sets are uncountably infinite. So, applying this to those sets which happen to be bases of vector spaces, some vector spaces have countably infinite bases and therefore countably infinite dimension, and other vector space have uncountable infinite bases and therefore uncountably infinite dimension.\n\nAn example of a vector space over $\\mathbb R$ of countably infinite dimension is $\\mathbb R^{\\infty}$ which is the space of infinite sequences of real numbers such that all but finitely terms in the sequence are equal to $0$. A countably infinite basis consists of $(1,0,0,0,...)$, $(0,1,0,0,...)$, $(0,0,1,0,...)$ and so on.\n\nAn example of a vector space over $\\mathbb R$ of uncountably infinite dimension is the one you mention in your question, $C[0,1]$.\n\n\u2022 That's okay. I do often write things wrong the first time. \u2013\u00a0Lee Mosher Sep 13 '18 at 12:18\n\u2022 ... all (but finitely many) terms are zero. \u2013\u00a0Lee Mosher Sep 13 '18 at 12:19\n\u2022 Had a hard time understanding your sentence, sorry for incovenience. \u2013\u00a0nicomezi Sep 13 '18 at 12:22\n\nFor the sake of a real world example: The electron in a hydrogen atom can take on countably many states. Each state is a basis vector for the span of possible electron states of hydrogen. There's are bound states and bound states are typically discrete. The relevant Schrodinger Equation also permits scattering states which have a continuous spectrum of possible energy states implying an uncountable vector space spanning the possible scattering states.\n\nThe following theorem is an example where we need to distinguish between countably/uncountably infinite dimensional vector spaces. Some important theorems, such as Hilbert Nullstallensatz, can be deduced from it.\n\nLet $$A$$ be an associative, not necessarily commutative, $$\\mathbb{C}-$$algebra with unit. For $$a \\in A$$ define $$\\text{Spec } a = \\{\\lambda \\in \\mathbb{C} | a-\\lambda \\text{ is not invertible}\\}$$ Assume that $$A$$ has no more than countable dimension over $$\\mathbb{C}$$. Then\n\n(a) If $$A$$ is a division algebra, then $$A=\\mathbb{C}$$\n\n(b) For all $$a \\in A$$ we have $$\\text{Spec } A \\neq \\emptyset$$; furthermore, $$a \\in A$$ is nilpotent if and only if $$\\text{Spec } A = \\{0\\}$$\n\n(Adapted from Representation Theory and Complex Geometry by Chriss and Ginzburg, theorem 2.1.1.)\n\nThe proof uses the fact that for any $$a \\in A$$, $$\\{(a - \\lambda)^{-1} | \\lambda \\in \\mathbb{C}\\}$$ is an uncountable family of elements of $$A$$. But $$A$$ has only countable dimension over $$\\mathbb{C}$$, so they are not linearly indenpendent over $$\\mathbb{C}$$.\n\nA weak version of Hilbert Nullstellensatz:\n\nLet $$A$$ be a finitely generated commutative algebra over $$\\mathbb{C}$$. Then any maximal ideal of $$A$$ is the kernel of an algebra homomorphism $$A \\to \\mathbb{C}$$\n\nThis follows directly from the above theorem: $$A$$ finitely generated $$\\Longrightarrow$$ $$A$$ has countable dimension over $$\\mathbb{C}$$ $$\\Longrightarrow$$ $$A/\\mathfrak{m}$$ has countable dimension over $$\\mathbb{C}$$ $$\\Longrightarrow$$ $$A/\\mathfrak{m}=\\mathbb{C}$$\n\n(We can also deduce the strong version of Nullstellensatz from it but need more argument.)\n\n\u2022 Welcome the Mathematics Stack Exchange! A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. \u2013\u00a0dantopa Dec 28 '18 at 4:57", "date": "2019-09-18 12:05:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2915495/dimension-of-vector-space-countable-uncountable", "openwebmath_score": 0.9293712973594666, "openwebmath_perplexity": 164.142783292367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464506036182, "lm_q2_score": 0.8757869981319863, "lm_q1q2_score": 0.8547212123117096}}
{"url": "http://openstudy.com/updates/55a2f823e4b05670bbb5452a", "text": "## anonymous one year ago HELP ME PLEASE Which of the following exponential functions goes through the points (1, 6) and (2, 12)? f(x) = 3(2)x f(x) = 2(3)x f(x) = 3(2)\u2212x f(x) = 2(3)\u2212x\n\n1. ybarrap\n\nPlug in x=1 and see if it equals 6 Plug in x=2 and see if it equals 12 Make sense?\n\n2. anonymous\n\nnot really? can you guide me through it? (xD I dont want to be an answer-hogger/wanter.. I dont want the answer, just explanations :) ) @freckles\n\n3. anonymous\n\nDo you understand that coordinates are in the form (x,y)?\n\n4. anonymous\n\nyes. @BMF96\n\n5. anonymous\n\nf(x) = y\n\n6. anonymous\n\nWhat don't you understand?\n\n7. jim_thompson5910\n\nIf $\\Large f(x) = 7(3)^x$ (for example), then what is the value of f(x) when x = 2? In other words, what is f(2) equal to?\n\n8. anonymous\n\nf(2)=441?? #CONFUSED\n\n9. jim_thompson5910\n\nReplace each x with 2 $\\Large f(x) = 7(3)^x$ $\\Large f(2) = 7(3)^2$ $\\Large f(2) = 7(9)$ $\\Large f(2) = 63$ Do you see how I got f(2) to be equal to 63?\n\n10. jim_thompson5910\n\nbtw you square first and then you multiply\n\n11. anonymous\n\nok, I forgot that rule :) (like DUHR, Bella, get a grip)\n\n12. jim_thompson5910\n\nSince $$\\Large f({\\color{red}{2}}) = {\\color{blue}{63}}$$ from my example, this means the point $$\\Large ({\\color{red}{x}},{\\color{blue}{y}})=({\\color{red}{2}},{\\color{blue}{63}})$$ lies on the function curve of f(x)\n\n13. anonymous\n\nokayyy....\n\n14. anonymous\n\n$$f(x)=y= \\begin{cases} 3(2)^x\\\\ 2(3)^x\\\\ 3(2)^{-x}\\\\ 2(3)^{-x} \\end{cases}\\qquad \\qquad \\begin{array}{llll} x&y \\\\\\hline\\\\ {\\color{brown}{ 1}}&3(2)^{\\color{brown}{ 1}}\\\\ &2(3)^{\\color{brown}{ 1}}\\\\ &3(2)^{-{\\color{brown}{ 1}}}\\\\ &2(3)^{-{\\color{brown}{ 1}}}\\\\ {\\color{brown}{ 2}}&3(2)^{\\color{brown}{ 2}}\\\\ &2(3)^{\\color{brown}{ 2}}\\\\ &3(2)^{-{\\color{brown}{ 2}}}\\\\ &2(3)^{-{\\color{brown}{ 2}}} \\end{array}$$\n\n15. jim_thompson5910\n\nso what ybarrap said at the top, you plug in each x coordinate into each function and see if you get the correct corresponding y coordinates Let's say we pick choice B at random $\\Large f(x) = 2(3)^x$ The first point is (1,6). To test if this point lies on the function f(x) curve, we plug in x = 1 and see if y = 6 pops out $\\Large f(x) = 2(3)^x$ $\\Large f(1) = 2(3)^1$ $\\Large f(1) = 2(3)$ $\\Large f(1) = 6$ It does, so (1,6) is definitely on this curve. How about (2,12)? Let's check $\\Large f(x) = 2(3)^x$ $\\Large f(2) = 2(3)^2$ $\\Large f(2) = 2(9)$ $\\Large f(2) = 18$ Nope. The point (2,18) actually lies on this function curve and NOT (2,12). So we can rule out choice B.\n\n16. jim_thompson5910\n\nSo what just happened was that I've proven that the function $$\\Large f(x) = 2(3)^x$$ does NOT go through both points (1,6) and (2,12).\n\n17. anonymous\n\nokay, so its A, C, or D lol.... so lets try to rule out A...\n\n18. anonymous\n\n@jim_thompson5910\n\n19. jim_thompson5910\n\nwhat did you get so far in checking choice A?\n\n20. anonymous\n\nthat A is, in fact NOT the answer!?\n\n21. jim_thompson5910\n\nif x = 1, then what is f(1) ?\n\n22. anonymous\n\nf\n\n23. anonymous\n\nWait, so it IS A!!!!!\n\n24. jim_thompson5910\n\n$\\Large f(x) = 3(2)^x$ $\\Large f(1) = 3(2)^1$ $\\Large f(1) = \\underline{ \\ \\ \\ \\ \\ \\ \\ } \\text{ (fill in the blank)}$\n\n25. anonymous\n\n6\n\n26. jim_thompson5910\n\n27. anonymous\n\n12\n\n28. jim_thompson5910\n\nGood. Choice A is definitely the answer. As practice, why not go through C and D and eliminate them. With choice C, if x = 1, then what is f(x) equal to?", "date": "2016-10-21 22:13:56", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/55a2f823e4b05670bbb5452a", "openwebmath_score": 0.6945479512214661, "openwebmath_perplexity": 1359.5582736869017, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464474553783, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8547212079725032}}
{"url": "https://stats.stackexchange.com/questions/214127/simulating-random-variables-from-a-discrete-distribution", "text": "# Simulating random variables from a discrete distribution\n\nI have the following discrete distribution where $p$ is a known constant:\n\n$p(x,p)= \\frac{(1-p)^3}{p(1+p)}x^2p^x , (0<p<1), x=0, 1, 2, \\ldots$ .\n\nHow can I sample from this distribution?\n\n\u2013\u00a0Tim\nMay 23, 2016 at 15:03\n\u2022 Does this even sum to 1 for some value of $p$? (I doubt so) May 23, 2016 at 15:19\n\u2022 @Elvis It does sum to 1. May 23, 2016 at 15:41\n\u2022 @MatthewGunn My bad. May 24, 2016 at 6:43\n\nThis answer develops a simple procedure to generate values from this distribution. It illustrates the procedure, analyzes its scope of application (that is, for which $p$ it might be considered a practical method), and provides executable code.\n\n### The Idea\n\nBecause\n\n$$x^2 = 2\\binom{x}{2} + \\binom{x}{1},$$\n\nconsider the distributions $f_{p;m}$ given by\n\n$$f_{p;m}(x) \\propto \\binom{x}{m-1}p^x$$\n\nfor $m=3$ and $m=2$.\n\nA recent thread on inverse sampling demonstrates that these distributions count the number of observations of independent Bernoulli$(1-p)$ variables needed before first seeing $m$ successes, with $x+1$ equal to that number. It also shows that the normalizing constant is\n\n$$C(p;m)=\\sum_{x=m-1}^\\infty \\binom{x}{m-1}p^x = \\frac{p^{m-1}}{(1-p)^m}.$$\n\nConsider the probabilities in the question,\n\n$$x^2 p^x = \\left( 2\\binom{x}{2} + \\binom{x}{1} \\right)p^x = 2 \\binom{x}{2}p^x + \\binom{x}{1} p^x =2 C(p;3) f_{p;3}(x) + C(p;2) f_{p;2}(x).$$\n\nConsequently, the given distribution is a mixture of $f_{p;3}$ and $f_{p;2}$. The proportions are as $$2C(p;3):C(p;2) = 2p:(1-p).$$ It is simple to sample from a mixture: generate an independent uniform variate $u$ and draw $x$ from $f_{p;2}$ when $u \\lt (1-p)/(2p+1-p)$; that is, when $u(1+p) \\lt 1-p$, and otherwise draw $x$ from $f_{p;3}$.\n\n(It is evident that this method generalizes: many probability distributions where the chance of $x$ is of the form $P(x)p^x$ for a polynomial $P$, such as $P(x)=x^2$ here, can be expressed as a mixture of these inverse-sampling distributions.)\n\n### The Algorithm\n\nThese considerations lead to the following simple algorithm to generate one realization of the desired distribution:\n\nLet U ~ Uniform(0,1+p)\nIf (U < 1-p) then m = 2 else m = 3\nx = 0\nWhile (m > 0) {\nx = x + 1\nLet Z ~ Bernoulli(1-p)\nm = m - Z\n}\nReturn x-1\n\n\nThese histograms show simulations (based on 100,000 iterations) and the true distribution for a range of values of $p$.\n\n### Analysis\n\nHow efficient is this? The expectation of $x+1$ under the distribution $f_{p;m}$ is readily computed; it equals $m/(1-p)$. Therefore the expected number of trials (that is, values of Z to generate in the algorithm) is\n\n$$\\left((1-p) \\frac{2}{1-p} + (2p) \\frac{3}{1-p}\\right) / (1-p+2p) = 2 \\frac{1+2p}{1-p^2}.$$\n\nAdd one more for generating U. The total is close to $3$ for small values of $p$. As $p$ approaches $1$, this count asymptotically is\n\n$$1 + 2\\frac{1 + 2p}{(1-p)(1+p)} \\approx \\frac{3}{1-p}.$$\n\nThis shows us that the algorithm will, on the average, be reasonably quick for $p \\lt 2/3$ (taking up to ten easy steps) and not too bad for $p \\lt 0.97$ (taking under a hundred steps).\n\n### Code\n\nHere is the R code used to implement the algorithm and produce the figures. A $\\chi^2$ test will show that the simulated results do not differ significantly from the expected frequencies.\n\nsample <- function(p) {\nm <- ifelse(runif(1, max=1+p) < 1-p, 2, 3)\nx <- 0\nwhile (m > 0) {\nx <- x + 1\nm <- m - (runif(1) > p)\n}\nreturn(x-1)\n}\nn <- 1e5\nset.seed(17)\npar(mfcol=c(2,3))\nfor (p in c(1/5, 1/2, 9/10)) {\n\n# Simulate and summarize.\nx <- replicate(n, sample(p))\ny <- table(x)\n\n# Compute the true distribution for comparison.\nk <- as.numeric(names(y))\ntheta <- sapply(k, function(i) i^2 * p^i) * (1-p)^3 / (p^2 + p)\nnames(theta) <- names(y)\n\n# Plot both.\nbarplot(y/n, main=paste(\"Simulation for\", format(p, digits=2)),\nborder=\"#00000010\")\nbarplot(theta, main=paste(\"Distribution for\", format(p, digits=2)),\nborder=\"#00000010\")\n}\n\n\n@dsaxton's approach is known as inverse transform sampling and is probably the way to go for a problem like this. To be a bit more explicit, the approach is:\n\n1. Draw $u$ from uniform distribution on (0,1).\n2. Compute $x = F^{-1}(u)$ where $F^{-1}$ is the inverse of the cumulative distribution function.\n\nComputing $x = F^{-1}(u)$ is equivalent to finding the integer $x$ that is the solution to: $$\\text{minimize} \\quad x \\quad \\text{subject to} \\sum_{j=0}^x \\frac{(1 - p)^3}{p(1+p)} j^2p^j \\geq u$$\n\nQuick pseudo code to do this numerically:\n\n1. Construct a vector $\\boldsymbol{m}$ such that $m_j = \\frac{(1 - p)^3}{p(1+p)} j^2p^j$.\n2. Create a vector $\\boldsymbol{c}$ such that $c_j = \\sum_{k=0}^j m_j$.\n3. Find the minimum index $x$ such that $c_x \\geq u$.\n\nDraw $u$ from a uniform$(0, 1)$ distribution and let $x$ be the smallest value of $k$ for which $\\sum_{j=0}^{k} \\frac{(1 - p)^3}{p (1 + p)} j^2 p^j > u$. Then $x$ will be a realization from the desired distribution.\n\n\u2022 May be clearer to write $x = \\min k \\text{ subject to } \\sum_j^k \\ldots > u$. As written, it's not terribly clear what is being minimized and that $x$ equals the optimal $k$. May 23, 2016 at 16:10", "date": "2022-05-24 03:25:51", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/214127/simulating-random-variables-from-a-discrete-distribution", "openwebmath_score": 0.7729891538619995, "openwebmath_perplexity": 671.7193666075293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464443071381, "lm_q2_score": 0.8757869965109765, "lm_q1q2_score": 0.8547212052153154}}
{"url": "https://math.stackexchange.com/questions/2566206/notation-for-cartesian-product-except-one-set/2566431", "text": "# Notation for cartesian product except one set?\n\nLet's say I have a list of sets $S_i$, for $i=1,\\ldots,n$. We often write the cartesian product of all these sets, with the exception of $S_k$ as:\n\n$$S=S_1\\times\\cdots\\times S_{k-1}\\times S_{k+1}\\times\\cdots\\times S_n$$\n\nIs there a more succinct way to write it?\n\nIn Wikipedia (https://en.wikipedia.org/wiki/Cartesian_product), I found something, which might be what you are looking for: $\\prod_{n=1}^k \\Bbb{R} = \\Bbb{R}\\times \\Bbb{R} \\times\\cdots\\times \\Bbb{R} = \\Bbb{R}^k$. So maybe something like this one is also valid: $$\\prod_{\\scriptstyle i = 1\\atop\\scriptstyle i \\ne k}^nS_i$$\n\nwhere $S_i$ is the $i^\\text{th}$ set of the list you mentioned.\n\n\u2022 Possibly \\prod_{i = 1\\atop i \\ne k}^n would be better? (with \\atop separating lines instead of a comma) \u2013\u00a0CiaPan Dec 14 '17 at 11:15\n\u2022 Yes, you are right. Thank you for the formatting :) \u2013\u00a0ArsenBerk Dec 14 '17 at 11:16\n\u2022 I changed $\\Bbb{R}$ x $\\Bbb{R}$ x ... x $\\Bbb{R}$ to $\\Bbb{R} \\times \\Bbb{R} \\times \\cdots \\times \\Bbb{R}.$ That is proper MathJax usage. $\\qquad$ \u2013\u00a0Michael Hardy Dec 14 '17 at 13:13\n\u2022 In addition to the atop business, another way to notate the indices would be with $\\in$. I could see that being especially convenient if these indices are already in a set or are repeated throughout the work. If $E=\\{1,\\cdots,n\\}\\setminus\\{k\\}$ then you can have $$\\prod_{i\\in E}S_i$$ \u2013\u00a0gen-\u2124 ready to perish Dec 14 '17 at 13:42\n\u2022 Another formatting comment: if you look carefully, your subscripts on the product symbol are in a smaller font than the superscript. IMHO it's better to avoid this by using \\prod_{\\scriptstyle i = 1\\atop\\scriptstyle i \\ne k}^n: compare$$\\prod_{i = 1\\atop i \\ne k}^nS_i\\quad\\hbox{and}\\quad \\prod_{\\scriptstyle i = 1\\atop\\scriptstyle i \\ne k}^nS_i$$ \u2013\u00a0David Dec 15 '17 at 1:36\n\nI have seen a notation for this kind of construction during some of my math lectures (but can't find a reference right now). This was mostly in the context of differential forms (e.g. interior product with vector), but can be applied to your case: $$S_1\\times \\dotsm \\times \\widehat{S_k} \\times\\dotsm \\times S_n := S_1\\times \\dotsm \\times S_{k-1}\\times S_{k+1} \\times \\dotsm S_n$$ The hat denotes the factor to be omitted. Note that this is not a universally standard notation, so even the professors that used it defined it at some point early in the lecture.\n\n\u2022 This is what I would use, along with a parenthetical remark along the lines of \"where the hat denotes the factor to be omitted.\" I very seldom see anything as formal as in the other answers. \u2013\u00a0Matthew Leingang Dec 14 '17 at 15:00\n\u2022 Hatcher uses a notation similar to this in his text Algebraic Topology (see page 105 for an example). \u2013\u00a0Xander Henderson Dec 17 '17 at 2:52\n\nIn general, we can write\n\n$$S_1 \\times \\dots \\times S_n := \\prod_{i=1}^n S_i$$\n\nand then we can apply all conventions we are used to.\n\nAs for your question, this can be written as:\n\n$$\\prod_{i = 1 \\atop i \\neq k}^n S_i$$\n\nAlthough it might not be common in set theory, it is common for game theorists to write $S_{-i}$ for $S_1 \\times \\cdots \\times S_{i-1} \\times S_{i+1} \\times \\cdots \\times S_n$. See page 15 of chapter one of Osborne and Rubinstein's text on game theory, for example.\n\nThat notation is useful in game theory because, if $S_j$ represents the set of strategies available to player $j$, then one often needs to describe how all players except player $i$ have acted. Such a description will be a member of $S_i$. In particular, the notation becomes useful in defining a Nash equilibrium. See https://en.wikipedia.org/wiki/Nash_equilibrium.", "date": "2020-10-01 13:42:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2566206/notation-for-cartesian-product-except-one-set/2566431", "openwebmath_score": 0.8885844945907593, "openwebmath_perplexity": 314.3137492617829, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464492044004, "lm_q2_score": 0.8757869867849167, "lm_q1q2_score": 0.8547212000121607}}
{"url": "http://radiojuventudnerja.es/f3y8d/euclidean-distance-between-two-vectors-32de3b", "text": "2017) and the quantum hierarchical clustering algorithm based on quantum Euclidean estimator (Kong, Lai, and Xiong 2017) has been implemented. Applying the formula given above we get that: \\begin{align} d(\\vec{u}, \\vec{v}) = \\| \\vec{u} - \\vec{v} \\| \\\\ d(\\vec{u}, \\vec{v}) = \\| \\vec{u} - \\vec{w} +\\vec{w} - \\vec{v} \\| \\\\ d(\\vec{u}, \\vec{v}) = \\| (\\vec{u} - \\vec{w}) + (\\vec{w} - \\vec{v}) \\| \\\\ d(\\vec{u}, \\vec{v}) \\leq || (\\vec{u} - \\vec{w}) || + || (\\vec{w} - \\vec{v}) \\| \\\\ d(\\vec{u}, \\vec{v}) \\leq d(\\vec{u}, \\vec{w}) + d(\\vec{w}, \\vec{v}) \\quad \\blacksquare \\end{align}, \\begin{align} d(\\vec{u}, \\vec{v}) = \\| \\vec{u} - \\vec{v} \\| = \\sqrt{(2-1)^2 + (3+2)^2 + (4-1)^2 + (2-3)^2} \\\\ d(\\vec{u}, \\vec{v}) = \\| \\vec{u} - \\vec{v} \\| = \\sqrt{1 + 25 + 9 + 1} \\\\ d(\\vec{u}, \\vec{v}) = \\| \\vec{u} - \\vec{v} \\| = \\sqrt{36} \\\\ d(\\vec{u}, \\vec{v}) = \\| \\vec{u} - \\vec{v} \\| = 6 \\end{align}, Unless otherwise stated, the content of this page is licensed under. . The reason for this is because whatever the values of the variables for each individual, the standardized values are always equal to 0.707106781 ! The Euclidean distance between two random points [ x 1 , x 2 , . Determine the Euclidean distance between. The Pythagorean Theorem can be used to calculate the distance between two points, as shown in the figure below. The distance between two vectors v and w is the length of the difference vector v - w. There are many different distance functions that you will encounter in the world. In mathematics, the Euclidean distance between two points in Euclidean space is the length of a line segment between the two points. The euclidean distance matrix is matrix the contains the euclidean distance between each point across both matrices. A generalized term for the Euclidean norm is the L2 norm or L2 distance. Append content without editing the whole page source. u = < -2 , 3> . Solution. Wikidot.com Terms of Service - what you can, what you should not etc. The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two\u00c2\u00a0 In mathematics, the Euclidean distance or Euclidean metric is the \"ordinary\" straight-line distance between two points in Euclidean space. Otherwise, columns that have large values will dominate the distance measure. pdist2 is an alias for distmat, while pdist(X) is \u2026 You are most likely to use Euclidean distance when calculating the distance between two rows of data that have numerical values, such a floating point or integer values. If you want to discuss contents of this page - this is the easiest way to do it. If columns have values with differing scales, it is common to normalize or standardize the numerical values across all columns prior to calculating the Euclidean distance. To calculate the Euclidean distance between two vectors in Python, we can use the numpy.linalg.norm function: Applying the formula given above we get that: (2) \\begin {align} d (\\vec {u}, \\vec {v}) = \\| \\vec {u} - \\vec {v} \\| = \\sqrt { (2-1)^2 + (3+2)^2 + (4-1)^2 + (2-3)^2} \\\\ d (\\vec {u}, \\vec {v}) = \\| \\vec {u} - \\vec {v} \\| = \\sqrt {1 + 25 + 9 + 1} \\\\ d (\\vec {u}, \\vec {v}) = \\| \\vec {u} - \\vec {v} \\| = \\sqrt {36} \\\\ d (\\vec {u}, \\vec {v}) = \\| \\vec {u} - \\vec {v} \\| = 6 \u2026 Find out what you can do. 1 Suppose that d is very large. ml-distance-euclidean. Check out how this page has evolved in the past. By using this metric, you can get a sense of how similar two documents or words are. Computes the Euclidean distance between a pair of numeric vectors. View wiki source for this page without editing. A generalized term for the Euclidean norm is the L2 norm or L2 distance. Recall that the squared Euclidean distance between any two vectors a and b is simply the sum of the square component-wise differences. View and manage file attachments for this page. With this distance, Euclidean space becomes a metric space. Older literature refers to the metric as the Pythagorean metric. $d(\\vec{u}, \\vec{v}) = \\| \\vec{u} - \\vec{v} \\| = \\sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 ... (u_n - v_n)^2}$, $d(\\vec{u}, \\vec{v}) = d(\\vec{v}, \\vec{u})$, $d(\\vec{u}, \\vec{v}) = || \\vec{u} - \\vec{v} || = \\sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 ... (u_n - v_n)^2}$, $d(\\vec{v}, \\vec{u}) = || \\vec{v} - \\vec{u} || = \\sqrt{(v_1 - u_1)^2 + (v_2 - u_2)^2 ... (v_n - u_n)^2}$, $(u_i - v_i)^2 = u_i^2 - 2u_iv_i + v_i^2 = v_i^2 - 2u_iv_i + 2u_i^2 = (v_i - u_i)^2$, $\\vec{u}, \\vec{v}, \\vec{w} \\in \\mathbb{R}^n$, $d(\\vec{u}, \\vec{v}) \\leq d(\\vec{u}, \\vec{w}) + d(\\vec{w}, \\vec{v})$, Creative Commons Attribution-ShareAlike 3.0 License. The primary takeaways here are that the Euclidean distance is basically the length of the straight line that's connects two vectors. 3.8 Digression on Length and Distance in Vector Spaces. Using our above cluster example, we\u2019re going to calculate the adjusted distance between a \u2026 Euclidean metric is the \u201cordinary\u201d straight-line distance between two points. Euclidean distance between two vectors, or between column vectors of two matrices. , y d ] is radicaltp radicalvertex radicalvertex radicalbt d summationdisplay i =1 ( x i \u2212 y i ) 2 Here, each x i and y i is a random variable chosen uniformly in the range 0 to 1. This victory. The corresponding loss function is the squared error loss (SEL), and places progressively greater weight on larger errors. The shortest path distance is a straight line. And now we can take the norm. Computes the Euclidean distance between a pair of numeric vectors. The answers/resolutions are collected from stackoverflow, are licensed under Creative Commons Attribution-ShareAlike license. We will now look at some properties of the distance between points in $\\mathbb{R}^n$. It corresponds to the L2-norm of the difference between the two vectors. The Euclidean distance between 1-D arrays u and v, is defined as maximum: Maximum distance between two components of x and y (supremum norm) manhattan: Absolute distance between the two vectors (1 \u2026 I have the two image values G= [1x72] and G1 = [1x72]. And these is the square root off 14. A little confusing if you're new to this idea, but it \u2026 D = \u221a [ ( X2-X1)^2 + (Y2-Y1)^2) Where D is the distance. We determine the distance between the two vectors. if p = (p1, p2) and q = (q1, q2) then the distance is given by. u of the two vectors. Notify administrators if there is objectionable content in this page. Solution to example 1: v . is: Deriving the Euclidean distance between two data points involves computing the square root of the sum of the squares of the differences between corresponding values. ... Percentile. u, is v . Let\u2019s assume OA, OB and OC are three vectors as illustrated in the figure 1. It can be computed as: A vector space where Euclidean distances can be measured, such as , , , is called a Euclidean vector space. General Wikidot.com documentation and help section. The squared Euclidean distance is therefore d(x\u00c2\u00a0 SquaredEuclideanDistance is equivalent to the squared Norm of a difference: The square root of SquaredEuclideanDistance is EuclideanDistance : Variance as a SquaredEuclideanDistance from the Mean : Euclidean distance, Euclidean distance. . So there is a bias towards the integer element. The following formula is used to calculate the euclidean distance between points. View/set parent page (used for creating breadcrumbs and structured layout). Two squared, lost three square until as one. $\\vec {u} = (2, 3, 4, 2)$. I need to calculate the two image distance value. The distance between two points is the length of the path connecting them. Source: R/L2_Distance.R Quickly calculates and returns the Euclidean distances between m vectors in one set and n vectors in another. {\\displaystyle \\left\\|\\mathbf {a} \\right\\|= {\\sqrt {a_ {1}^ {2}+a_ {2}^ {2}+a_ {3}^ {2}}}} which is a consequence of the Pythagorean theorem since the basis vectors e1, e2, e3 are orthogonal unit vectors. Most vector spaces in machine learning belong to this category. Accepted Answer: Jan Euclidean distance of two vector. By using this formula as distance, Euclidean space becomes a metric space. $\\vec {v} = (1, -2, 1, 3)$. Each set of vectors is given as the columns of a matrix. Installation $npm install ml-distance-euclidean. Euclidean distance.$\\begingroup$Even in infinitely many dimensions, any two vectors determine a subspace of dimension at most$2$: therefore the (Euclidean) relationships that hold in two dimensions among pairs of vectors hold entirely without any change at all in any number of higher dimensions, too. Squared Euclidean Distance, Let x,y\u00e2\u0088\u0088Rn. This system utilizes Locality sensitive hashing (LSH) [50] for efficient visual feature matching.$\\endgroup$\u2013 whuber \u2666 Oct 2 '13 at 15:23 The formula for this distance between a point X ( X 1 , X 2 , etc.) In simple terms, Euclidean distance is the shortest between the 2 points irrespective of the dimensions. Y = cdist(XA, XB, 'sqeuclidean') The Euclidean distance between two vectors, A and B, is calculated as: Euclidean distance = \u221a \u03a3(A i-B i) 2. You want to find the Euclidean distance between two vectors. We can then use this function to find the Euclidean distance between any two vectors: #define two vectors a <- c(2, 6, 7, 7, 5, 13, 14, 17, 11, 8) b <- c(3, 5, 5, 3, 7, 12, 13, 19, 22, 7) #calculate Euclidean distance between vectors euclidean(a, b) [1] 12.40967 The Euclidean distance between the two vectors turns out to be 12.40967. Computing the Distance Between Two Vectors Problem. Find the Distance Between Two Vectors if the Lengths and the Dot , Let a and b be n-dimensional vectors with length 1 and the inner product of a and b is -1/2. \u2014 Page 135, D\u2026 w 1 = [ 1 + i 1 \u00e2\u0088\u0092 i 0], w 2 = [ \u00e2\u0088\u0092 i 0 2 \u00e2\u0088\u0092 i], w 3 = [ 2 + i 1 \u00e2\u0088\u0092 3 i 2 i]. In mathematics, the Euclidean distance or Euclidean metric is the \"ordinary\" distance between two\u00c2 (geometry) The distance between two points defined as the square root of the sum of the squares of the differences between the corresponding coordinates of the points; for example, in two-dimensional Euclidean geometry, the Euclidean distance between two points a = (a x, a y) and b = (b x, b y) is defined as: What does euclidean distance mean?, In the spatial power covariance structure, unequal spacing is measured by the Euclidean distance d \u00e2\u008c\u00a2 j j \u00e2\u0080\u00b2 , defined as the absolute difference between two\u00c2 In mathematics, the Euclidean distance or Euclidean metric is the \"ordinary\" distance between two points that one would measure with a ruler, and is given by the Pythagorean formula. So the norm of the vector to three minus one is just the square root off. = v1 u1 + v2 u2 NOTE that the result of the dot product is a scalar. It is the most obvious way of representing distance between two points. We here use \"Euclidean Distance\" in which we have the Pythagorean theorem. The result is a positive distance value. (we are skipping the last step, taking the square root, just to make the examples easy) \u2016 a \u2016 = a 1 2 + a 2 2 + a 3 2. Euclidean distance. We will derive some special properties of distance in Euclidean n-space thusly. The associated norm is called the Euclidean norm. The average distance between a pair of points is 1/3. First, determine the coordinates of point 1. Active 1 year, 1 month ago. And that to get the Euclidean distance, you have to calculate the norm of the difference between the vectors that you are comparing. . Definition of normalized Euclidean distance, According to Wolfram Alpha, and the following answer from cross validated, the normalized Eucledean distance is defined by: enter image\u00c2 In mathematics, the Euclidean distance or Euclidean metric is the \"ordinary\" straight-line distance between two points in Euclidean space. This is helpful\u00c2 variables, the normalized Euclidean distance would be 31.627. So this is the distance between these two vectors. Both implementations provide an exponential speedup during the calculation of the distance between two vectors i.e. These names come from the ancient Greek mathematicians Euclid and Pythagoras, although Euclid did not represent distances as numbers, and the connection from the Pythagorean theorem to distance calculation wa Y1 and Y2 are the y-coordinates. Copyright \u00a9document.write(new Date().getFullYear()); All Rights Reserved, How to make a search form with multiple search options in PHP, Google Drive API list files in folder v3 python, React component control another component, How to retrieve data from many-to-many relationship in hibernate, How to make Android app fit all screen sizes. Discussion. Okay, then we need to compute the design off the angle that these two vectors forms. Given some vectors$\\vec{u}, \\vec{v} \\in \\mathbb{R}^n$, we denote the distance between those two points in the following manner. Euclidean Distance Formula. <4 , 6>. sample 20 1 0 0 0 1 0 1 0 1 0 0 1 0 0 The squared Euclidean distance sums the squared differences between these two vectors: if there is an agreement (there are two matches in this example) there is zero sum of squared differences, but if there is a discrepancy there are two differences, +1 and \u20131, which give a sum of squares of 2. X1 and X2 are the x-coordinates. The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two points. With this distance, Euclidean space becomes a metric space. Watch headings for an \"edit\" link when available. As such, it is also known as the Euclidean norm as it is calculated as the Euclidean distance from the origin. Euclidean Distance. Click here to edit contents of this page. For three dimension 1, formula is. Determine the Euclidean distance between$\\vec{u} = (2, 3, 4, 2)$and$\\vec{v} = (1, -2, 1, 3)$. This process is used to normalize the features\u00c2 Now I would like to compute the euclidean distance between x and y. I think the integer element is a problem because all other elements can get very close but the integer element has always spacings of ones. Brief review of Euclidean distance. Euclidean distance, Euclidean distances, which coincide with our most basic physical idea of squared distance between two vectors x = [ x1 x2 ] and y = [ y1 y2 ] is the sum of\u00c2 The Euclidean distance function measures the \u00e2\u0080\u0098as-the-crow-flies\u00e2\u0080\u0099 distance. In this article to find the Euclidean distance, we will use the NumPy library. Change the name (also URL address, possibly the category) of the page. Basic Examples (2) Euclidean distance between two vectors: Euclidean distance between numeric vectors: If not passed, it is automatically computed. Euclidean distance Suppose w 4 is [\u00e2\u0080\u00a6] Construction of a Symmetric Matrix whose Inverse Matrix is Itself Let v be a nonzero vector in R n . The associated norm is called the Euclidean norm. Usage EuclideanDistance(x, y) Arguments x. Numeric vector containing the first time series. Glossary, Freebase(1.00 / 1 vote)Rate this definition: Euclidean distance. Euclidean and Euclidean Squared Distance Metrics, Alternatively the Euclidean distance can be calculated by taking the square root of equation 2. First, here is the component-wise equation for the Euclidean distance (also called the \u201cL2\u201d distance) between two vectors, x and y: Let\u2019s modify this to account for the different variances. The standardized Euclidean distance between two n-vectors u and v is $\\sqrt{\\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i\u2019th components of the points. Let\u2019s discuss a few ways to find Euclidean distance by NumPy library. How to calculate normalized euclidean distance on , Meaning of this formula is the following: Distance between two vectors where there lengths have been scaled to have unit norm. It can be calculated from the Cartesian coordinates of the points using the Pythagorean theorem, therefore occasionally being called the Pythagorean distance. u = < v1 , v2 > . See pages that link to and include this page. and. This library used for manipulating multidimensional array in a very efficient way. Available distance measures are (written for two vectors x and y): euclidean: Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)). and a point Y ( Y 1 , Y 2 , etc.) their Dot Product of Two Vectors The dot product of two vectors v = < v1 , v2 > and u = denoted v . I've been reading that the Euclidean distance between two points, and the dot product of the\u00c2 Dot Product, Lengths, and Distances of Complex Vectors For this problem, use the complex vectors. In a 3 dimensional plane, the distance between points (X 1 , \u2026 Sometimes we will want to calculate the distance between two vectors or points. The points are arranged as m n -dimensional row vectors in the matrix X. Y = cdist (XA, XB, 'minkowski', p) (Zhou et al. gives the Euclidean distance between vectors u and v. Details. Compute the euclidean distance between two vectors. Understand normalized squared euclidean distance?, Try to use z-score normalization on each set (subtract the mean and divide by standard deviation. Ask Question Asked 1 year, 1 month ago. Before using various cluster programs, the proper data treatment is\u00e2\u0080\u008b\u00c2 Squared Euclidean distance is of central importance in estimating parameters of statistical models, where it is used in the method of least squares, a standard approach to regression analysis. Computes Euclidean distance between two vectors A and B as: ||A-B|| = sqrt ( ||A||^2 + ||B||^2 - 2*A.B ) and vectorizes to rows of two matrices (or vectors). With this distance, Euclidean space becomes a metric space. Older literature refers to the metric as the Pythagorean metric. Something does not work as expected? Example 1: Vectors v and u are given by their components as follows v = < -2 , 3> and u = < 4 , 6> Find the dot product v . . The Euclidean distance between two points in either the plane or 3-dimensional space measures the length of a segment connecting the two In mathematics, the Euclidean distance or Euclidean metric is the \"ordinary\" straight-line distance between two points in Euclidean space. The points A, B and C form an equilateral triangle. In this presentation we shall see how to represent the distance between two vectors. , x d ] and [ y 1 , y 2 , . API Euclidean Distance Between Two Matrices. The Euclidean distance d is defined as d(x,y)=\u00e2\u0088\u009an\u00e2\u0088\u0091i=1(xi\u00e2\u0088\u0092yi)2. . Directly comparing the Euclidean distance between two visual feature vectors in the high dimension feature space is not scalable. linear-algebra vectors. How to calculate euclidean distance. The associated norm is called the Euclidean norm. Click here to toggle editing of individual sections of the page (if possible). Euclidean distancecalculates the distance between two real-valued vectors. The length of the vector a can be computed with the Euclidean norm. In \u211d, the Euclidean distance between two vectors and is always defined. ||v||2 = sqrt(a1\u00b2 + a2\u00b2 + a3\u00b2) scipy.spatial.distance.euclidean\u00b6 scipy.spatial.distance.euclidean(u, v) [source] \u00b6 Computes the Euclidean distance between two 1-D arrays. Compute distance between each pair of the two Y = cdist (XA, XB, 'euclidean') Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. The most obvious way of representing distance between two vectors, euclidean distance between two vectors column. ) =\u00e2\u0088\u009an\u00e2\u0088\u0091i=1 ( xi\u00e2\u0088\u0092yi ) 2 need to compute the design off the angle that these two vectors v... Get a sense of how similar two documents or words are are collected from stackoverflow, are licensed under Commons! Library used for creating breadcrumbs and structured layout ) squared Euclidean distance between vectors... Set ( subtract the mean and divide by standard deviation easiest way to do it glossary, (! Calculation of the page ( if possible ) be 31.627 used for creating breadcrumbs and structured layout ) here toggle., -2, 1, -2, 1 month ago n vectors in high. Metrics, Alternatively the Euclidean distance between vectors u and v, is defined as d (,. 135, D\u2026 Euclidean distance matrix is matrix the contains the Euclidean distance Euclidean distancecalculates the distance between two.. Result of the straight line that 's connects two vectors i.e containing the first time..  Euclidean distance between a pair of points is 1/3 and OC are three vectors illustrated. Attribution-Sharealike license 'sqeuclidean ' ) Brief review of Euclidean distance?, Try to use z-score on. Also URL address, possibly the category ) of the page ( used for euclidean distance between two vectors... You want to calculate the distance between a pair of points is 1/3 Locality hashing... Euclidean distance between a \u2026 linear-algebra vectors ( 1, -2, 1, x 2, creating breadcrumbs structured. Usage EuclideanDistance ( x, y ) Arguments x. numeric vector containing the first series! Here are that the result of the page ( if possible ) has evolved the. ) Where d is defined as d ( x, y ) =\u00e2\u0088\u009an\u00e2\u0088\u0091i=1 xi\u00e2\u0088\u0092yi! Most obvious way of representing distance between two points in$ \\mathbb { R } ^n $1 month.... Distance Euclidean distancecalculates the distance between two vectors an  edit '' link when.! Distance would be 31.627 evolved in the figure 1 the columns of a line between! ] for efficient visual feature vectors in one set and n vectors in the past are comparing corresponding function. Therefore occasionally being called the Pythagorean theorem, therefore occasionally being called the Pythagorean distance dimension feature is! Points a, B and C form an equilateral triangle and that to get the Euclidean distance can calculated! Returns the Euclidean distance from the origin older literature refers to the metric the... V. Details by taking the square root of equation 2 array in a very efficient way - this helpful\u00c2. Image values G= [ 1x72 ] and G1 = [ 1x72 ] and G1 = [ 1x72 ] and =. Include this page - this is because whatever the values of the page ( used for multidimensional... Connects two vectors in Python, we can use the numpy.linalg.norm function: Euclidean distance, you can what... V } = ( p1, p2 ) and q = ( p1, p2 ) and q = q1! And divide by standard deviation function is the L2 norm or L2.. Under Creative Commons Attribution-ShareAlike license d = \u221a [ ( X2-X1 ) ^2 (. At some properties of the straight line that 's connects two vectors coordinates the! Loss function is the easiest way to do it to 0.707106781 \\vec { v } = q1... Vector a can be calculated by taking the square root off, B and C form an equilateral.! Euclidean and Euclidean squared distance Metrics, Alternatively the Euclidean distance between points. Watch headings for an  edit '' link when available angle that these two vectors year,,! We can use the numpy.linalg.norm euclidean distance between two vectors: Euclidean distance between two points XA, XB, 'sqeuclidean )., 2 )$ use z-score normalization on each set ( subtract the mean and divide standard. [ y 1, 3, 4, 2 ) $) Rate this definition: Euclidean distance two... X ( x, y 2, etc. norm as it is the \u201c ordinary straight-line. 1 2 + a 3 2 Euclidean metric is the most obvious way of representing distance between pair! 1 year, 1 month ago in another computes the Euclidean distance euclidean distance between two vectors in which have! Subtract the mean and divide by standard deviation that these two vectors, or between column of. Understand normalized squared Euclidean distance is basically the length of the vector to three one!, etc. usage EuclideanDistance ( x, y ) =\u00e2\u0088\u009an\u00e2\u0088\u0091i=1 ( xi\u00e2\u0088\u0092yi ) 2 norm as it calculated... Metric as the Euclidean norm is the \u201c ordinary \u201d straight-line distance between a \u2026 linear-algebra vectors you to! The NumPy library d ] and G1 = [ 1x72 ] category of! 2 )$ for an  edit '' link when available structured layout.! For creating breadcrumbs and structured layout ) the Pythagorean theorem can be with! I have the Pythagorean metric can get a sense of how similar two documents or words.! How similar two documents or words are the design off the angle that these two vectors squared distance! Understand normalized squared Euclidean distance d is defined as ( Zhou et al  distance. Zhou et al an  edit '' link when available between these two vectors forms theorem be... Here are that the squared error loss ( SEL ), and places greater! Normalized squared Euclidean distance in mathematics, the standardized values are always equal to 0.707106781 between points library used creating. Of points is 1/3, it is calculated as the columns of a line segment between the vectors you. Distance can be used to calculate the Euclidean distance between any two vectors or points ) Rate definition! Feature space is the L2 norm or L2 distance is because whatever the values the. 'S connects two vectors the distance between points in $\\mathbb { R } ^n.... = \u221a [ ( X2-X1 ) ^2 ) Where d is defined as ( Zhou et al ordinary! This is because whatever the values of the dimensions ^n$ Where d is defined d... Terms of Service - what you can, what you can, what can! Subtract the mean and divide by standard deviation v } = ( q1, q2 then. Numeric vectors formula is used to calculate the adjusted distance between points the variables for each individual the! As one irrespective of the dot product is a scalar, x 2, we here use  Euclidean is... By using this formula as distance, Euclidean space becomes a metric space Alternatively the Euclidean distance in. Points using the Pythagorean theorem going to calculate the distance Arguments x. numeric vector containing the first time series root. The L2 norm or L2 distance the average distance between two points and returns the Euclidean distance between 1-D u... = cdist ( XA, XB, 'sqeuclidean ' ) Brief review of Euclidean distance between point! Similar two documents or words are R/L2_Distance.R Quickly calculates and returns the norm. + ( Y2-Y1 ) ^2 ) Where d is the L2 norm or L2 distance that you comparing. L2 norm or L2 distance linear-algebra vectors points [ x 1, 3, 4, 2 \\$... Do it > = v1 u1 + v2 u2 NOTE that the squared Euclidean distance between points, can! R/L2_Distance.R Quickly calculates and returns the Euclidean distance between two random points [ 1! Points [ x 1, x d ] and [ y 1 3... Of a line segment between the 2 points irrespective of the difference between the two points the ordinary... In another two visual feature matching here are that the Euclidean distance be...\n\nBridge Mountain Climb, Malvan Village Ghost Story, Graham Greene Rdr2, Bd Script Font Canva, Vintage Toy Tractor Price Guide, Software Developer No Experience, Fire Pit Insert Round, Classroom Leadership Pdf, Police Officer Hard Skills, How To Separate Succulent Arrangement,", "date": "2021-04-20 07:51:11", "meta": {"domain": "radiojuventudnerja.es", "url": "http://radiojuventudnerja.es/f3y8d/euclidean-distance-between-two-vectors-32de3b", "openwebmath_score": 0.9609122276306152, "openwebmath_perplexity": 866.690683796926, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464513032269, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8547211955222094}}
{"url": "https://math.stackexchange.com/questions/801679/help-needed-to-derive-combinatorics-formula", "text": "# Help needed to derive combinatorics formula.\n\nI am having troubles understanding a combinatorics formula. I would appreciate any ideas or hints, leading to an explanation how this formula might be derived. I came across the formula reading a book on statistical physics. Unfortunately the formula was given without derivation.\n\nSuppose a total number of objects $N$ and a set of '$m$' boxes are given. To each of these boxes, fraction $n_i$ of the $N$ objects might be assigned. With the condition that\n\n$$\\sum_{i=1}^m n_i = N$$\n\nAccording to my textbook the total number of ways to generate certain distribution $W\\{n_i\\}$ is given by\n\n$$W\\{n_i\\} = \\frac{N!}{\\prod_i n_i!},$$\n\nwhere $i$ runs over all boxes. I have troubles understanding how the formula for $W\\{n_i\\}$ is derived.\n\nTo make things more clear consider the case with four boxes, where\n\n$$N=5\\\\ n_1=2\\\\ n_2=2\\\\ n_3=1\\\\ n_4=0\\\\$$\n\nIn this particular case one can generate the following distributions over the four boxes\n\n$$|1,2|3,4|5|0|\\\\ |1,3|2,5|4|0|\\\\ |2,5|1,4|3|0|\\\\ ......$$\n\nThe textbook goes further and defines the total probability $W_{tot}\\{n_i\\}$ of finding the distribution $\\{n_i\\}$\n\n$$W\\{n_i\\} = N!\\prod_i\\frac{\\omega_i^{n_i}}{ n_i!}.$$\n\nWhere $w_i$ is the probability of finding one particular object in a certain box. Both of the above formulas are not clear to me, therefore I would need your help to understand them.\n\nThank you for reading my question.\n\nBest Regards,\n\nAlex\n\n\u2022 In short, you're dealing with a multinomial distribution. \u2013\u00a0Graham Kemp May 20 '14 at 1:42\n\nI have troubles understanding how the formula for $W\\{n_i\\}$ is derived.\n\nIt's just the multinomial coefficient. ${n\\choose k_1,k_1,\\ldots k_m} = \\frac{n!}{k_1!k_2!\\cdots k_m!}$\n\nYou have $N$ distinct objects sorted into $m$ distinct groups, where the arrangement inside each group does not matter. The number of objects within any group $i$ is $n_i$\n\nThere are $N!$ ways to rearrange $N$ distinct objects. There are $n_i!$ ways to rearrange all objects in box $i$. But the order of objects in each box does not matter, so all arrangements with the same objects in a box are equivalent.\n\nThus we divide the total permutations by the size of the equivalent cases: $$\\dfrac{N!}{n_1!n_2!n_3!\\cdots n_m!}=\\dfrac{N!}{\\prod\\limits_{i=1}^m n_i!}$$\n\nExample: Sort 4 different balls into 3 boxes so that the last box contains 2 balls. Let us count the ways.\n\nThere are $4!$ ways to arrange 4 different balls. But these arrangements can be divided into pairs where the same balls go into the last box, just in different orders, and the order of the balls in the last box does not matter.\n\nSo there are $\\frac{4!}{1!1!2!}$ ways to sort the balls into these boxes.\n\nThe textbook goes further and defines the total probability $W_\\text{tot}\\{n_i\\}$ of finding the distribution $\\{n_i\\}$.\n\nAnd here we have a multinomial distribution. This is analogous to the binomial distribution::\n\n$$X\\sim\\mathcal{B}(n,p) \\iff \\mathrm{P}(X=x)={n\\choose x}p^x (1-p)^{n-x}$$\n\nWhich is the probability of $x$ successes in $n$ trials, with probability $p$ of an individual success.\n\nThis is analogous to sorting objects into 2 boxes, with probability $p_1$ of going into the first box and probability $p_2=(1-p_1)$ of any ball going into the second. So:\n\n$$P(X_1=n_1, X_2=n_2 : n_1+n_2=N) = \\frac{N!}{n_1!n_2!} p_1^{n_1}p_2^{n_2}$$\n\nNow just extend this to the case of $m$ boxes.\n\nSo, let us take $N$ balls to drop into $m$ boxes, such that the probability of a ball landing in any box $i$ is $\\omega_i$; where $\\sum\\limits_{i=1}^m \\omega_i = 1$.\n\nNow the probability that box $1$ contains $n_1$ balls, and box $2$ contains $n_2$ balls, and so on, and so on is: $$W_\\text{tot}\\{n_i\\} \\\\ = P(X_1=n_1, X_2=n_2, \\ldots, X_m=n_m) \\\\ = {N\\choose n_1,n_2,\\ldots, n_m} \\omega_1^{n_1} \\cdot \\omega_2^{n_2} \\cdots \\omega_m^{n_m} \\\\ = \\frac{N!}{\\prod\\limits_{i=1}^m n_i!} \\prod\\limits_{i=1}^m \\omega_i^{n_i} \\\\ = N!\\prod\\limits_{i=1}^m \\frac{\\omega_i^{n_i}}{n_i!}$$\n\n\u2022 Graham, thank you for answering my question and writing this rigorous and well-structured reply. I was also thinking about the multinomial coefficient. However, I totally failed to see the connection to the binomial distribution. An at the end I have one last question, suppose the ordering of the object in the boxes mattered, in this case the factorial form the denominator of $W\\{n_i\\}$ and $W_{tot}\\{n_i\\}$ should be removed. Is this correct? \u2013\u00a0Alexander Cska May 20 '14 at 7:41\n\u2022 The reason order inside the boxes doesn't matter is because ultimately you are only concerned with the number of items in the boxes, not their identity. If the identity of each item was important, then you wouldn't use multinomials at all. There would be just $1$ way to arrange each unique sequence; and the probability of that specific arrangement would be: $\\prod\\limits_{i=1}^{m} \\omega_i^{n_i}$ \u2013\u00a0Graham Kemp May 20 '14 at 8:13", "date": "2019-07-15 18:02:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/801679/help-needed-to-derive-combinatorics-formula", "openwebmath_score": 0.7588387131690979, "openwebmath_perplexity": 181.93141513568673, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464429079201, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8547211897517281}}
{"url": "http://math.stackexchange.com/questions/117860/how-to-define-sparseness-of-a-vector", "text": "# How to define sparseness of a vector?\n\nI would like to construct a measure to calculate the sparseness of a vector of length $k$.\n\nLet $X = [x_i]$ be a vector of length $k$ such that there exist an $x_i \\neq 0$ . Assume $x_i \\geq 0$ for all $i$.\n\nOne such measure I came across is defined as $$\\frac{\\sqrt{k} - \\frac{\\|X\\|_1}{{\\|X\\|_2}}} {\\sqrt{k} -1}\\;,$$ where $\\|X\\|_1$ is $L_1$ norm and $\\|X\\|_2$ is $L_2$ norm.\n\nHere, $\\operatorname{Sparseness}(X) = 0$ whenever the vector is dense (all components are equal and non-zero) and $\\operatorname{Sparseness}(X) = 1$ whenever the vector is sparse (only one component is non zero).\n\nThis post only explains the when $0$ and $1$ achieved by the above mentioned measure.\n\nIs there any other function defining the sparseness of the vector.\n\n-\nIsn't your sparseness function 1 for a sparse vector and 0 for a dense vector? \u2013\u00a0 Christian Rau Mar 8 '12 at 10:51\nIt doesnt matter.we can achieve that by defining the measure as 1-Sparseness($X$). \u2013\u00a0 Learner Mar 8 '12 at 11:02\nI know, I just wanted to make clear, that your explanation as it stands is wrong, if Sparseness(X) is indeed defined as above. \u2013\u00a0 Christian Rau Mar 8 '12 at 11:20\nYeah. I should have used different name. \u2013\u00a0 Learner Mar 8 '12 at 11:24\nSorry if my comments are a bit confusing. It isn't the name sparseness that bothers me, it's the hard fact, that your above function (the one with the $\\sqrt{k}$) is 1 for a sparse vector and 0 for a dense vector (no matter how you name it). \u2013\u00a0 Christian Rau Mar 8 '12 at 11:28\n\nYou could of course generalize your current measure\n\n\\begin{align} S(X) = \\frac{\\frac{k^{(1/m)}}{k^{(1/n)}} -\\frac{\\|X\\|_m}{\\|X\\|_n} } {\\frac{k^{(1/m)}}{k^{(1/n)}}-1} \\end{align}\n\nwhile preserving your properties you specified.\n\nAn interesting special case could be $m = 1, n \\to \\infty$, in which case the expression simplifies to\n\n$$S(X) = \\frac{k-\\frac{\\|X\\|_1}{\\|X\\|_c}}{k-1}$$\n\nwhere $c = \\infty$, (for some reason, mathjax refused to render when I inserted $\\infty$ directly in the fraction)\n\n-\nGood that you generalized it. I never thought in those lines. However, i didnt get what you meant by \"interesting special case\" as $n$ goes to infinity. Can you please elaborate more on that. \u2013\u00a0 Learner Mar 9 '12 at 4:25\n@Learner: I added a description of what you obtain for that special case, I'll leave it to you to decide what to do with it. \u2013\u00a0 Mikael \u00d6hman Mar 9 '12 at 19:47\nEven if something fails to parse correctly in a mathjax preview it should work when you actually submit it and refresh. \u2013\u00a0 anon Mar 12 '12 at 11:22\n\nThere is a definition of sparsity, which is used (amongst others) in the compressed sensing literature, see e.g. here.\n\nA vector $x\\in \\mathbb{C}^k$ is called $s$-sparse, if $|| x ||_0 = |\\text{supp}(x)| \\leq s$, that is, it has at most $s$ non-zero entries. Denote by $\\Sigma_s$ the set of all such vectors. Then, the $s$-term approximation error of a vector $x\\in \\mathbb{C}^k$ is defined as $$\\sigma_s(x)_p = \\min_{y\\in\\Sigma_s} ||x-y||_p.$$\n\nNow this quantity equals $0$, if your vector $x$ is $s$-sparse, and will be greater than $0$ otherwise. Note that you now have two parameters $s$ and $p$ to tune this \"measure\". Clearly, you get your definition of sparsity if you set $s=1$.\n\n-\n\nDisclaimer: This post considers the case in which you do not mind some computational effort to get nice sparseness value. For something new, please skip to part 2.\n\nPart 1\n\nI agree with Mikael, that kind of generalization is nice, what's more, with Mikael's formula it is intuitive from where it came from: the most basic notion for vector sparseness would be $$\\frac{\\text{number of indices k such that }X_k = 0}{\\text{total number of indices}}.$$\n\nHowever, by this definition $\\langle 0, 0, \\ldots, 0\\rangle$ is sparse, but vector $\\langle c, c, \\ldots, c \\rangle$ is not. Still, it is easy to fix it: $$\\frac{\\text{number of indices on which }X_k - c = 0}{\\text{total number of indices}}\\,,$$ where $c$ is some average of $X$, e.g. $c = \\|X\\|_\\infty$. The problem with this measure is that it is not easy to count the number of indices. To alleviate for that, we could approximate the number of indices by $\\frac{\\|X\\|_1}{\\|X\\|_\\infty}$. Naturally we need some normalization, and by that we arrive at Mikael's special case: $$\\frac{k-\\frac{\\|X\\|_1}{\\|X\\|_\\infty}}{k-1}.$$\n\nBut the average we took as an example $x = \\|X\\|_\\infty$ isn't the only one. Similarly we could approximate the number of indices in different fashions: $\\frac{\\|X\\|_m}{\\|X\\|_n}$ would do for any $m < n$, and the normalization is just $$\\frac{\\frac{\\|C\\|_m}{\\|C\\|_n}-\\frac{\\|X\\|_1}{\\|X\\|_\\infty}}{R_\\max-R_\\min}, R_\\max = \\frac{\\|C\\|_m}{\\|C\\|_n}, R_\\min = \\frac{\\|D\\|_m}{\\|D\\|_n},$$ where $C = \\langle c, c, \\ldots, c \\rangle$ and $D = \\langle c, 0, 0, \\ldots, 0 \\rangle$ for any $c \\neq 0$.\n\nPart 2\n\nStill, this measure is rather weird, because intuitively it more depends on the sizes of the values, than how many different numbers there are. I am not sure that this is a property we would want to have. There is a different measure that can take that into account, i.e. measure based on entropy. Interpreting $X$ as the samples, one can calculate $$-\\sum_i P(X = i) \\log_k P(X = i) .$$\n\nTo soften this a bit just pick any distribution you want (best specific to your application), e.g. $F_\\mu = N(\\mu, \\sigma^{2})$, set $$F = \\frac{1}{k}\\sum_i F_{X_i},$$ and then calculate differential entropy ($f$ is the density function of $F$): $$-\\int_\\mathbb{R} f(x) \\ln f(x) \\ dx$$ or even better relative entropy if you have some reference measure (e.g. the very same $F_\\mu$ adjusted a bit might do the trick). Of course, all this has to be scaled to $[0,1]$ what makes the formulas even nastier, however, in my opinion, it catches the notion of sparseness pretty good. Finally, you can combine the two approaches in infinitely many ways, to get even more sparseness models!\n\n-", "date": "2015-05-23 14:58:49", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/117860/how-to-define-sparseness-of-a-vector", "openwebmath_score": 0.951259434223175, "openwebmath_perplexity": 349.0096419936737, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426405416754, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8546997977300903}}
{"url": "https://math.stackexchange.com/questions/295250/geometric-interpretations-of-matrix-inverses", "text": "# Geometric interpretations of matrix inverses\n\nLet $A$ be an invertible $n \\times n$ matrix. Suppose we interpret each row of $A$ as a point in $\\mathbb{R}^n$; then these $n$ points define a unique hyperplane in $\\mathbb{R}^n$ that passes through each point (this hyperplane does not intersect the origin).\n\nUnder this geometric interpretation, $A^{-1}$ has an interesting property: the normal vector to the hyperplane is given by the row sums of $A^{-1}$ (i.e. $A^{-1} \\cdot 1$, where $1 = \\langle 1, \\dots, 1 \\rangle^T$).\n\nWithin this geometric interpretation of $A$, what other interesting properties does $A^{-1}$ have? Do the individual entries of $A^{-1}$ have geometric meaning? How about the column sums?\n\nHere is a visual answer for the $2\\times 2$ case.\n\n\u2022 Plot the row (or column) vectors, $a_1, a_2$ of $A$ in $\\mathcal{R}^2$ to visualize $A$. The area of the parallelogram they form is of course $\\det(A)$.\n\u2022 In the same space, plot the row (or column) vectors $a^1, a^2$ of $A^{-1}$, and the area of their parallelogram is then $\\det(A^{-1}) = 1/ \\det(A)$.\n\u2022 The relationship between the two illustrates various properties of the matrix inverse.\n\nAn example is shown in the picture below, which comes from the matrix (in R notation)\n\nA <- matrix(c(2, 1,\n1, 2), nrow=2, byrow=TRUE)\n\n\nIn the R package matlib I recently added a vignette illustrating this with the following diagram for this matrix.\n\nThus, we can see:\n\n\u2022 The shape of $A^{-1}$ is a $90^o$ rotation of the shape of $A$.\n\n\u2022 $A^{-1}$ is small in the directions where $A$ is large\n\n\u2022 The vector $a^2$ is at right angles to $a_1$ and $a^1$ is at right angles to $a_2$\n\n\u2022 If we multiplied $A$ by a constant $k$ to make its determinant larger (by a factor of $k^2$), the inverse would have to be divided by the same factor to preserve $A A^{-1} = I$.\n\nI wondered whether these properties depend on symmetry of $A$, so here is another example, for the matrix A <- matrix(c(2, 1, 1, 1), nrow=2), where $\\det(A)=1$.\n\nIt would be interesting to extend this to other properties and to the $3 \\times 3$ case, which I leave to others.\n\n\u2022 Not all your observations are true in the general case. The 90\u00b0 thing for example. You just got lucky with your numbers. For example $A = [[-1, 0], [-3, 2]]$, $A^{-1} = [[-1, -0. ], [-1.5, 0.5]]$ $A_1 \\cdot A^{-1}_2 = -1.5 \\neq 0$. Note: internal brackets are rows. \u2013\u00a0user3578468 May 25 at 2:42\n\u2022 Please provide statements of which of my observations are not true in the general case and why. \u2013\u00a0user101089 May 25 at 22:13\n\u2022 But I did give you an example. It shows that \"The vector $a_2$ is at right angles to $a_1$ and $a_1$ is at right angles to $a_2$\" is not always true. \u2013\u00a0user3578468 May 27 at 10:50\n\nIt turns out that an answer for the $3 \\times 3$ case has similar properties and is also illuminating.\n\n\u2022 Start with a unit cube, representing the identity matrix. Show its transformation by a matrix $A$ as the corresponding transformation of the cube.\n\n\u2022 This also illustrates the determinant, det(A), as the volume of the transformed cube, and the relationship between $A$ and $A^{-1}$.\n\nIn R, using the matlib and rgl package, the unit cube is specified as\n\nlibrary(rgl)\nlibrary(matlib)\n# cube, with each face colored differently\ncolors <- rep(2:7, each=4)\nc3d <- cube3d()\n# make it a unit cube at the origin\nc3d <- scale3d(translate3d(c3d, 1, 1, 1),\n.5, .5, .5)\n\n\nA $3 \\times 3$ matrix $A$ with $\\det(A)=2$ is\n\nA <- matrix(c( 1, 0, 1,\n0, 2, 0,\n1, 0, 2), nrow=3, ncol=3)\n\n\nExtending the 2D idea from the answer above of drawing the images of $A$ and $A^{-1}$ together to 3D, we get the following, best viewed as an animated graphic. The faces of the parallelpiped representing $A^{1}$ are colored identically to those of $A$, so you can see the mapping from one to the other.\n\n\u2022 Great visualization! \u2013\u00a0Vincent Oct 17 '18 at 13:32", "date": "2019-08-23 15:38:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/295250/geometric-interpretations-of-matrix-inverses", "openwebmath_score": 0.8924434781074524, "openwebmath_perplexity": 206.99232533675527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357232512719, "lm_q2_score": 0.8705972768020107, "lm_q1q2_score": 0.8546964472018097}}
{"url": "http://math.stackexchange.com/questions/124300/finding-inverse-of-polynomial-in-a-field/124307", "text": "# Finding inverse of polynomial in a field\n\nI'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take:\n\nIn $\\frac{\\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$.\n\nMy understanding is that one needs to use the (Extended?) Euclidean Algorithm and Bezout's Identity. Here's what I currently have:\n\nProceeding with Euclid's algorithm:\n\n$$x^3 + 2x + 1 =(x^2 + 1)(x) + (x + 1) \\\\\\\\ x^2 + 1 = (x + 1)(2 + x) + 2$$\n\nWe stop here because 2 is invertible in $\\mathbb{Z}_3[x]$. We rewrite it using a congruence:\n\n$$(x+1)(2+x) \\equiv 2 mod(x^2+1)$$\n\nI don't understand the high level concepts sufficiently well and I'm lost from here. Thoughts?\n\nWikipedia has a page on this we a decent explanation, but it's still not clear in my mind.\n\nNote that this question has almost the same title, but it's a level of abstraction higher. It doesn't help me, as I don't understand the basic concepts.\n\nThanks.\n\n-\nIf you can write $2$ as $(x^3+2x+1)f(x)+(x^2+1)g(x)$, then $2g(x)$ is an inverse to $x^2+1$ in $\\mathbb Z_3[x]/(x^3+2x+1)$. Does it help? \u2013\u00a0 Pierre-Yves Gaillard Mar 25 '12 at 16:20\n\nWrite $f := x^3+2x+1$ and $g := x^2+1$. We want to find the inverse of $g$ in the field $\\mathbb F_3[x]/(f)$ (I prefer to write $\\mathbb F_3$ instead of $\\mathbb Z_3$ to avoid confusion with the 3-adic integers), i.e. we are looking for a polynomial $h$ such that $gh \\equiv 1 \\pmod f$, or equivalently $gh+kf=1$ for some $k\\in \\mathbb F_3[x]$. The Euclidean algorithm can be used to find $h$ and $k$: $$f = x\\cdot g+(x+1)\\\\ g = (x+2)\\cdot(x+1) + 2\\\\ (x+1) = (2x)\\cdot2 + 1$$ Working backwards, we find $$1 = (x+1)-(2x)\\cdot 2\\\\ = (x+1)-(2x)(g-(x+2)(x+1))\\\\ = (2x^2+x+1)(x+1)-(2x)g\\\\ = (2x^2+x+1)(f-xg)-(2x)g\\\\ = (2x^2+x+1)f- (x^3+2x^2)g\\\\ = (2x^2+x+1)f - (2x^3+x^2)g\\\\ = (2x^2+x+1)f + (x^3+2x^2)g.$$ So, the inverse of $g$ modulo $f$ is $h = x^3+2x^2 \\pmod f = 2x^2+x+2 \\pmod f$.\n\n-\nGreat, I understand. Thank you. When using Maple, however, I find a different result to the Extended Euclidean Algorithm ($(x^3+2x+1)f + (2x^2+2+x)f$). Therefore, I find $2x^2+2+x$ to be the inverse, which is different than what you find. Is this normal? (integers only have one inverse, is this different for polynomials?) \u2013\u00a0 David Chouinard Mar 25 '12 at 16:55\nYou are right, there is only one inverse. However, since we are working modulo $f$, it is only determined up to multiples of $f$ (technically, the solution is not a polynomial but rather a residue class of polynomials). Note that $(x^3+2x^2) = (2x^2+x+2)+f$, so the solutions are in fact equivalent. (I just edited my answer to include also the reduced solution $2x^2+x+2$.) \u2013\u00a0 marlu Mar 25 '12 at 17:03\nPretty obvious, don't know why I missed that. Thanks. (@anon, marlu updated his response after I replied) \u2013\u00a0 David Chouinard Mar 25 '12 at 17:11\n@David: Sorry, for some reason there is no edit history recorded on the answer (not sure how that's possible...), so I didn't know it was edited. \u2013\u00a0 anon Mar 25 '12 at 17:19\n@anon Yes, that's strange. The answer seems to have been edited after 5 minutes but there is no history. I've flagged for mods, in case there may be a bug. \u2013\u00a0 Bill Dubuque Mar 25 '12 at 17:41\n\nThe goal of the Extended Euclidean algorithm is to compute polynomials $\\rm\\:A,B\\:$ such that $\\rm\\: A\\: (x^2+1)\\: +\\: B\\:(x^3+2x+1) = 1.\\:$ This implies that $\\rm\\:A\\:(x^2+1) \\equiv 1\\pmod{x^3+2x+1},\\:$ hence $\\rm\\:(x^2+1)^{-1}\\equiv A \\pmod{x^3+2x+1}.\\:$\n\nGenerally, the simplest way to compute $\\rm\\:A,B\\:$ is analogous to Gaussian elimination in linear algebra: append an identity matrix to accumulate elementary row operations, e.g. see my post here, which gives a very detailed example (for integers, but the same method works over any domain having a division / Euclidean algorithm). This method is easier to memorize and less error-prone than the alternative \"back-substitution\" method often proposed.\n\nThis is a special-case of Hermite/Smith row/column reduction of matrices to triangular/diagonal normal form, using the division/Euclidean algorithm to reduce entries modulo pivots. Though one can understand this knowing only the analogous linear algebra elimination techniques, it will become clearer when one studies modules - which, informally, generalize vector spaces by allowing coefficients from rings vs. fields. In particular, these results are studied when one studies normal forms for finitely-generated modules over a PID, e.g. when one studies linear systems of equations with coefficients in the non-field! polynomial ring $\\rm F[x],$ for $\\rm F$ a field, as above.\n\n-\nYes, this makes a lot of sense. In this case, I found $A = 2x^2+2+x$ and $B = (x^3+2x+1)$. Therefore, $2x^2+2+x$ is the inverse. However, I don't understand why this isn't in contradiction with Pierre-Yves Gaillard's comment? \u2013\u00a0 David Chouinard Mar 25 '12 at 16:41\n@David Pierre is writing $2$ (not $1$) as a linear combination, then scaling that by $\\rm\\:2^{-1}\\equiv 2\\pmod{3},\\:$ so to get $1$ as a linear combination. \u2013\u00a0 Bill Dubuque Mar 25 '12 at 16:50\nUnderstood, thanks. \u2013\u00a0 David Chouinard Mar 25 '12 at 17:00\n\nThe same algorithm used to solve the linear diophantine equation can be used here. $$\\begin{array}{c} &&x&x-1&(x+1)/2\\\\ \\hline 1&0&1&1-x&(x^2+1)/2\\\\ 0&1&-x&x^2-x+1&-(x^3+2x+1)/2\\\\ x^3+2x+1&x^2+1&x+1&2&0 \\end{array}$$ Thus, $$(1-x)(x^3+2x+1)+(x^2-x+1)(x^2+1)=2$$ Thus, the inverse of $x^2+1$ is $\\tfrac12(x^2-x+1)$ mod $x^3+2x+1$.\n\n-\nThat's the same as the method I mentioned - which is more conceptually viewed from a linear algebra (module) perspective, where it is a special case of Hermite/Smith row/column reduction of matrices to triangular/diagonal normal form, using the division/Euclidean algorithm to reduce entries mod pivots. \u2013\u00a0 Bill Dubuque Mar 25 '12 at 17:20\nThe Euclidean algorithm begins with two polynomials $r^{(0)}(x)$ and $r^{(1)}(x)$ such that $\\deg r^{(0)}(x) > \\deg r^{(1)}(x)$ and then iteratively finds quotient polynomials $q^{(1)}(x), q^{(2)}(x), \\ldots$ and remainder polynomials $r^{(2)}(x), r^{(3)}(x), \\ldots$ of successively smaller degrees via division \\begin{align*} r^{(0)}(x) &= q^{(1)}(x)\\cdot r^{(1)}(x) + r^{(2)}(x)\\\\ r^{(1)}(x) &= q^{(2)}(x)\\cdot r^{(2)}(x) + r^{(3)}(x)\\\\ \\vdots\\qquad &= \\qquad\\qquad\\vdots \\end{align*} One version of the Extended Euclidean Algorithm also finds pairs of polynomials $(s^{(0)}(x),t^{(0)}(x)), (s^{(1)}(x),t^{(1)}(x)), (s^{(2)}(x),t^{(2)}(x)) \\ldots$ where $(s^{(0)}(x),t^{(0)}(x)) = (1,0)$ and $(s^{(1)}(x),t^{(1)}(x)) = (0,1)$ that satisfy the generalized Bezout identity $$s^{(i)}(x)\\cdot r^{(0)}(x) + t^{(i)}(x)\\cdot r^{(1)}(x) = r^{(i)}(x).$$\nThese polynomials satisfy the \"same\" recursion as the remainder polynomials, viz., \\begin{align*} r^{(i+1)}(x) &= r^{(i-1)}(x) - q^{(i)}(x)\\cdot r^{(i)}(x)\\\\ s^{(i+1)}(x) &= s^{(i-1)}(x) - q^{(i)}(x)\\cdot s^{(i)}(x)\\\\ t^{(i+1)}(x) &= t^{(i-1)}(x) - q^{(i)}(x)\\cdot t^{(i)}(x)\\\\ \\end{align*}\nThis form of the extended Euclidean algorithm is useful in practical applications since only two polynomials $r, s,$ and $t$ need to be remembered with each new $(i+1)$-th polynomial replacing the $(i-1)$-th polynomial which is no longer needed.\nIn your instance, you have $r^{(0)}(x) = x^3 + 2x+1$ and $r^{(1)}(x) = x^2 + 1$. You have already computed the quotient and remainder sequence ending with $r^{(3)}(x) = 2$. Now compute $t^{(2)}(x)$ and $t^{(3)}(x)$ iteratively using the sequence of quotient polynomials and write \\begin{align*} s^{(3)}(x)\\cdot (x^3 + 2x + 1) + t^{(3)}(x)\\cdot(x^2 + 1) &= 2\\\\ -s^{(3)}(x)\\cdot (x^3 + 2x + 1) - t^{(3)}(x)\\cdot(x^2 + 1) &= 1\\\\ (-t^{(3)}(x))\\cdot (x^2 + 1) &= 1 ~ \\mod (x^3 + 2x + 1) \\end{align*} and deduce that the multiplicative inverse of $x^2 + 1$ in $\\mathbb Z_3[x]/(x^3 + 2x + 1)$ is $-t^{(3)}(x)$. Note that the $s^{(i)}(x)$ sequence does not need to be computed at all if all that one needs is the inverse.", "date": "2014-03-07 10:13:45", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/124300/finding-inverse-of-polynomial-in-a-field/124307", "openwebmath_score": 0.9932308197021484, "openwebmath_perplexity": 372.49844595970956, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357179075082, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8546964458456819}}
{"url": "https://rayboy.org/3sqln99/unique-left-inverse-5f932b", "text": "Note the subtle difference! As f is a right inverse to g, it is a full inverse to g. So, f is an inverse to f is an inverse to Then t t t has many left inverses but no right inverses (because t t t is injective but not surjective). eralization of the inverse of a matrix. >> save hide report. If the function is one-to-one, there will be a unique inverse. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). If S S S is a set with an associative binary operation \u2217 * \u2217 with an identity element, and an element a \u2208 S a\\in S a \u2208 S has a left inverse b b b and a right inverse c, c, c, then b = c b=c b = c and a a a has a unique left, right, and two-sided inverse. Note that other left inverses (for example, A\u00a1L = [3; \u00a11]) satisfy properties (P1), (P2), and (P4) but not (P3). When working in the real numbers, the equation ax=b could be solved for x by dividing bothsides of the equation by a to get x=b/a, as long as a wasn't zero. Active 2 years, 7 months ago. G is called a left inverse for a matrix if 7\u201a8 E GE\u0153M 8 \u00d0 \u00d1so must be G 8\u201a7 It turns out that the matrix above has E no left inverse (see below). Let e e e be the identity. example. << /S /GoTo /D [9 0 R /Fit ] >> We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. Subtraction was defined in terms of addition and division was defined in terms ofmultiplication. For any elements a, b, c, x \u2208 G we have: 1. 36 0 obj << Let\u2019s recall the definitions real quick, I\u2019ll try to explain each of them and then state how they are all related. %\ufffd\ufffd\ufffd\ufffd Then they satisfy $AB=BA=I \\tag{*}$ and Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. Remark When A is invertible, we denote its inverse \u2026 Theorem. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. In a monoid, if an element has a right inverse\u2026 Show Instructions. \ufffd\ufffd\ufffd In a monoid, if an element has a left inverse, it can have at most one right inverse; moreover, if the right inverse exists, it must be equal to the left inverse, and is thus a two-sided inverse. 87 0 obj <>/Filter/FlateDecode/ID[<60DDF7F936364B419866FBDF5084AEDB><33A0036193072C4B9116D6C95BA3C158>]/Index[53 73]/Info 52 0 R/Length 149/Prev 149168/Root 54 0 R/Size 126/Type/XRef/W[1 3 1]>>stream Yes. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Some easy corollaries: 1. Stack Exchange Network. In matrix algebra, the inverse of a matrix is defined only for square matrices, and if a matrix is singular, it does not have an inverse.. An associative * on a set G with unique right identity and left inverse proof enough for it to be a group ?Also would a right identity with a unique left inverse be a group as well then with the same . given $$n\\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Two-sided inverse is unique if it exists in monoid 2. Then 1 (AB) ij = A i B j, 2 (AB) i = A i B, 3 (AB) j = AB j, 4 (ABC) ij = A i BC j. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. Still another characterization of A+ is given in the following theorem whose proof can be found on p. 19 in Albert, A., Regression and the Moore-Penrose Pseudoinverse, Aca-demic Press, New York, 1972. I know that left inverses are unique if the function is surjective but I don't know if left inverses are always unique for non-surjective functions too. Generalized inverse Michael Friendly 2020-10-29. This thread is archived. Actually, trying to prove uniqueness of left inverses leads to dramatic failure! New comments cannot be posted and votes cannot be cast. Ask Question Asked 4 years, 10 months ago. wqhh\ufffd\ufffdllf\ufffd)eK\ufffdy\ufffdI\ufffd\ufffdbq\ufffd(\ufffd\ufffd\ufffd\ufffd\ufffd\u00c3.4-\ufffd{xe\ufffd\ufffd8\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdb\ufffdc[\ufffd\ufffd\ufffd\u00f6\ufffd\ufffd\ufffd\ufffdTBYb\ufffd\u02834\ufffd\ufffd\ufffd&\ufffd1\ufffd\ufffd\ufffd\ufffdo[{cK\ufffdsAt\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd3\ufffd'vp=\u007f\ufffd$\ufffd\ufffd$\ufffdi.\ufffd\ufffdj8@\ufffdg\ufffdUQ\ufffd\ufffd\ufffd>\ufffd\ufffdg\ufffdlI&\ufffdOuL\ufffd\ufffd*\ufffd\ufffd\ufffdwCu\ufffd0 \ufffd]l\ufffd In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. For any elements a, b, c, x \u2208 G we have: 1. This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. Let (G, \u2295) be a gyrogroup. (We say B is an inverse of A.) One consequence of (1.2) is that AGAG=AG and GAGA=GA. Right inverse If A has full row rank, then r = m. The nullspace of AT contains only the zero vector; the rows of A are independent. /Filter /FlateDecode 5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about \"the\" inverse of f). Proof. Suppose that there are two inverse matrices $B$ and $C$ of the matrix $A$. Hence it is bijective. stream It's an interesting exercise that if $a$ is a left unit that is not a right uni Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego If E has a right inverse, it is not necessarily unique. numpy.unique\u00b6 numpy.unique (ar, return_index = False, return_inverse = False, return_counts = False, axis = None) [source] \u00b6 Find the unique elements of an array. Proposition If the inverse of a matrix exists, then it is unique. Theorem A.63 A generalized inverse always exists although it is not unique in general. So to prove the uniqueness, suppose that you have two inverse matrices $B$ and $C$ and show that in fact $B=C$. 100% Upvoted. Left-cancellative Loop (algebra) , an algebraic structure with identity element where every element has a unique left and right inverse Retraction (category theory) , a left inverse of some morphism However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. Matrix Multiplication Notation. Let $f \\colon X \\longrightarrow Y$ be a function. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y \u2208 U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. Let G G G be a group. 0 A.12 Generalized Inverse De\ufb01nition A.62 Let A be an m \u00d7 n-matrix. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. 125 0 obj <>stream '+o\ufffdf P0\ufffd\ufffd\ufffd'\ufffd,\ufffd\\\ufffd y\ufffd\ufffd\ufffd\ufffdbf\\\ufffd; wx.\ufffd\ufffd\";MY\ufffd}\ufffd\ufffd\ufffd\ufffd\u0625\ufffd The left inverse tells you how to exactly retrace your steps, if you managed to get to a destination \u2013 \u201cSome places might be unreachable, but I can always put you on the return flight\u201d The right inverse tells you where you might have come from, for any possible destination \u2013 \u201cAll places are reachable, but I can't put you on the If A is invertible, then its inverse is unique. share. If BA = I then B is a left inverse of A and A is a right inverse of B. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. 6 comments. From this example we see that even when they exist, one-sided inverses need not be unique. See the lecture notesfor the relevant definitions. The following theorem says that if has aright andE Eboth a left inverse, then must be square. Recall also that this gives a unique inverse. g = finverse(f,var) ... finverse does not issue a warning when the inverse is not unique. In gen-eral, a square matrix P that satis\ufb02es P2 = P is called a projection matrix. best. Then a matrix A\u2212: n \u00d7 m is said to be a generalized inverse of A if AA\u2212A = A holds (see Rao (1973a, p. 24). inverse Proof (\u21d2): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (\u21d0): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). endobj Then a matrix A\u2212: n \u00d7 m is said to be a generalized inverse of A if AA\u2212A = A holds (see Rao (1973a, p. 24). x\ufffd\ufffdXKo#7\ufffd\ufffdW\ufffdhE\ufffd[\u05e2\ufffd\ufffdE\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd:v\ufffd4q\ufffd\ufffd\ufffd/)\ufffdc\ufffd\ufffd\ufffd\ufffd>~\"%\ufffd\ufffdd\ufffd\ufffdN\ufffd\ufffd8\ufffdw(LY\u027d2L:\ufffdAZv\ufffdb\ufffd\ufffd\u065e\u0473G\ufffd\ufffd\ufffd8>\ufffd\ufffd\ufffd\ufffd'\ufffd\ufffdx\ufffd\u067crc\ufffd\ufffd\u007f>?\ufffd\ufffd[\ufffd\ufffd?\ufffd'\ufffd\ufffd\ufffd(%#R\ufffd\ufffd1 .\ufffd-7\ufffd;6\ufffdSg#>Q\ufffd\ufffd7\ufffd##\u03e5 \"\ufffd[\ufffd \ufffd\ufffd\ufffdN)&Q \ufffd\ufffdM\ufffd\ufffd\ufffdYy\ufffd\ufffd?A\ufffd\ufffd\ufffd\ufffd4\ufffd\u03e0H\ufffd%\ufffdf\ufffd\ufffd0a;N\ufffdM\ufffd,\ufffd!{\ufffd\ufffdy\ufffd<8(t1\u0199\ufffdzi\ufffd\ufffd\ufffde\ufffd\ufffdA\ufffd\ufffd(;p*\ufffd\ufffd\ufffd\ufffdV\ufffdJ\u069b,\ufffdt~\ufffdd\ufffd\ufffd\u0318H9\ufffd\ufffd\ufffd\ufffd/\ufffd\ufffd_a\ufffd\ufffd\ufffdv\ufffd68gq\"\ufffd\ufffd\ufffdD\ufffd|a5\ufffd\ufffd\ufffd\ufffdP|Jv\ufffd\ufffdl1j\ufffd\ufffdx\ufffd\ufffd&\u07baN\ufffd\ufffd\ufffd\ufffdV\"\ufffd\ufffd\ufffd\"\ufffd\ufffd\u007f\ufffd\ufffd}! Proof: Assume rank(A)=r. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. The equation Ax = b always has at least one solution; the nullspace of A has dimension n \u2212 m, so there will be 53 0 obj <> endobj Theorem 2.16 First Gyrogroup Properties. Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. Theorem A.63 A generalized inverse always exists although it is not unique in general. Let (G, \u2295) be a gyrogroup. A i denotes the i-th row of A and A j denotes the j-th column of A. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Show Instructions. h\ufffdbbdb\ufffd \ufffd\ufffd \ufffd9D\ufffdH\ufffd_ \ufffd\ufffdDj*\ufffdHE\ufffd8\ufffd,\ufffd&f\ufffd\ufffdL[\ufffdz\ufffdH\ufffdW\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffdHU{\ufffd\ufffdZ \ufffd(\ufffd \ufffd\ufffd \ufffd\ufffdA\ufffd\ufffdO0\ufffd lZ'\ufffd\ufffd\ufffd\ufffd{,\ufffd\ufffd.\ufffdl\ufffd\\\ufffd\ufffd@\ufffd\ufffd\ufffdOL@\ufffd\ufffd\ufffdq\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd %PDF-1.6 %\ufffd\ufffd\ufffd\ufffd Left inverse if and only if right inverse We now want to use the results above about solutions to Ax = b to show that a square matrix A has a left inverse if and only if it has a right inverse. Returns the sorted unique elements of an array. u (b 1 , b 2 , b 3 , \u2026) = (b 2 , b 3 , \u2026). There are three optional outputs in addition to the unique elements: Thus both AG and GA are projection matrices. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Theorem 2.16 First Gyrogroup Properties. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective (4x1\ufffd@\ufffdy\ufffd,(\ufffd\ufffd\ufffd\ufffd.\ufffdBY\ufffd\ufffd\u29c67G\ufffd\u07f1Zb\ufffd?\ufffd\ufffd,\ufffd\ufffdT\ufffd\ufffd9o\ufffd\ufffdH0\ufffd(1q\ufffd\ufffd\ufffd\ufffdD\ufffd \ufffd;:\ufffd\ufffdvK{Y\ufffdwY\ufffd/\ufffd\ufffd\ufffd5\ufffd\ufffd\ufffd\ufffd\ufffdc\ufffdiZl\ufffdB\\\\\ufffd\ufffdL\ufffdbE\ufffd\ufffd\ufffd8;\ufffd!\ufffd#\ufffd*)\ufffdL\ufffd{\ufffdM\ufffd\ufffddU\u04426\ufffd\ufffd\ufffd%\ufffdV^\ufffd\ufffd\ufffd\ufffdZW\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdf\ufffd4R\ufffdp\ufffdp\ufffdb\ufffd\ufffdx\ufffd\ufffd\ufffd.L\ufffd\ufffd1sh\ufffd\ufffdY\ufffd\u007fU\ufffd\ufffd\ufffd\ufffd! Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. One of its left inverses is the reverse shift operator u (b 1, b 2, b 3, \u2026) = (b 2, b 3, \u2026). In mathematics, and in particular, algebra, a generalized inverse of an element x is an element y that has some properties of an inverse element but not necessarily all of them. \ufffdn\ufffd\ufffd\ufffd\ufffd\ufffdr\ufffd\ufffd\ufffd\ufffd6\ufffd\ufffd\ufffdd}\ufffd\ufffd\ufffdwF>\ufffdG\ufffd/\ufffd\ufffdk\ufffd K\ufffdT\ufffdSE\ufffd\ufffd\ufffd\ufffd \ufffd&\u02ac\ufffdRbl\ufffdj\ufffd\ufffd|\ufffdTx\ufffd\ufffd)\ufffd\ufffdRdy\ufffdY ? In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. If the function is one-to-one, there will be a unique inverse. left A rectangular matrix can\u2019t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. Remark Not all square matrices are invertible. This is no accident ! u(b_1,b_2,b_3,\\ldots) = (b_2,b_3,\\ldots). Let A;B;C be matrices of orders m n;n p, and p q respectively. If f contains more than one variable, use the next syntax to specify the independent variable. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. Note that other left h\ufffdb\ufffdy\ufffd\ufffd\ufffd cca\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\u0650\ufffd q\ufffd\ufffd\ufffd#\ufffd!\ufffdA\ufffd\u046cQ\ufffda\ufffd\ufffd\ufffd[\ufffd50\ufffdF\ufffd\ufffd3&9'\ufffd\ufffd0 qp\ufffd(R\ufffd&\ufffda\ufffds4\ufffdp\ufffd[\ufffd\ufffd\ufffdf^'w\ufffdP&\u07b6 7\ufffd\ufffd,\ufffd\ufffd\ufffd[T\ufffd+\ufffdJ\ufffd\ufffd\ufffd\ufffd9\ufffd$\ufffd\ufffd4r\ufffd:4';m$\ufffd\ufffd#\ufffds\ufffdOj\ufffdL\u00cc\ufffdcY{-\ufffdXTA\u06bd\ufffdBEOpr\ufffdl\ufffdT\ufffd\ufffdf1\ufffdM\ufffd1$\ufffd\ufffd\u0421\ufffd\ufffd6I\ufffd\ufffd\u048e)w It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. %%EOF This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. Recall that$B$is the inverse matrix if it satisfies $AB=BA=I,$ where$I$is the identity matrix. Proof: Assume rank(A)=r. Hello! (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) A.12 Generalized Inverse De\ufb01nition A.62 Let A be an m \u00d7 n-matrix. Yes. /Length 1425 endstream endobj startxref Generalized inverses can be defined in any mathematical structure that involves associative multiplication, that is, in a semigroup.This article describes generalized inverses of a matrix. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent \u2026 Viewed 1k times 3. Sort by. Let\u2019s recall the definitions real quick, I\u2019ll try to explain each of them and then state how they are all related. This preview shows page 275 - 279 out of 401 pages.. By Proposition 5.15.5, g has a unique right inverse, which is equal to its unique inverse. 8 0 obj h\ufffd\ufffd[[\ufffd\u06f6\ufffd+|l\\wp\ufffd\ufffd\u07dd\ufffdN\\\ufffd\ufffd&\ufffd\u4052\ufffd]\ufffd\ufffd%\"e\ufffd\ufffd\ufffd{>\ufffd\ufffdHJZi\ufffdk\ufffdm\ufffd \ufffdwnt.I\ufffd%. g = finverse(f) returns the inverse of function f, such that f(g(x)) = x. inverse. By using this website, you agree to our Cookie Policy. If is a left inverse and a right inverse of , for all \u2208, () = ((()) = (). Let $f \\colon X \\longrightarrow Y$ be a function. (An example of a function with no inverse on either side is the zero transformation on .) \ufffd\ufffd\ufffd\ufffdE\ufffdO]{z^\ufffd\ufffd\ufffdh%\ufffdw\ufffd-\ufffdB,E\ufffd\\J\ufffd\u008b\ufffd|\ufffdY\\2z)\ufffd\ufffd\ufffd\ufffd\ufffdME\ufffd\ufffd5\ufffd\ufffd\ufffd@5\ufffd\ufffdq\ufffd\ufffd|7P\ufffd\ufffd\ufffd@\ufffd\ufffd\ufffd\ufffd\ufffd&\ufffd\ufffd5\ufffd9\ufffdq#\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdh\ufffd>R\u04b9\ufffd/\ufffdZ1\ufffd&\ufffdcu6\ufffd\ufffdB\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffde\ufffd^BXx\ufffd\ufffd\ufffdr\ufffd\ufffd=\ufffdE\ufffd_\ufffd \ufffd\ufffd\ufffdTm\ufffd\ufffdz\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd8g\ufffd~t.i}\ufffd\ufffd\ufffd\u07ee:>;\ufffdPG\ufffdpaH\ufffdT. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. %PDF-1.4 Let f : A \u2192 B be a function with a left inverse h : B \u2192 A and a right inverse g : B \u2192 A. U-semigroups The Moore-Penrose pseudoinverse is de\ufb02ned for any matrix and is unique. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by \u2192 \u21a6 \u22c5 \u2192 has the two-sided inverse \u2192 \u21a6 (/) \u22c5 \u2192.In this subsection we will focus on two-sided inverses. 3. See Also. Outside semigroup theory, a unique inverse as defined in this section is sometimes called a quasi-inverse. 11.1. When a is invertible, then \\ ( M\\ ) is called a right inverse ( a inverse...  5x  is equivalent to  5 * x  they,! Of the matrix$ a $, if it exists in monoid 2 matrix or its has! ( Generalized inverses are unique is you impose more conditions on G see... ), then it is unique use the next syntax to specify the variable. Matrices$ b $and$ c $of the matrix$ a unique left inverse n. An m \u00d7 n-matrix matrix $a$ be unique votes can not be posted and votes not! General, you can skip the multiplication sign, so  5x  is equivalent ! Has aright andE Eboth a left inverse and the right inverse ( a two-sided inverse?. More conditions on G ; see Section 3 below. [ math ] f x. In u that minimizes the distance from b to any point in u b to any point in.... And right inverse of a and a j denotes the i-th row of a matrix exists then... I-Th row of a function, you can skip the multiplication sign, so  . \\Longrightarrow Y [ /math ] be a gyrogroup is not necessarily commutative ; i.e 3 \u2026... Any matrix and is unique in u that minimizes the distance from b to any point in that! That satis\ufb02es P2 = p is called a projection matrix ( a two-sided inverse is not unique 3.... It necessarily have a unique left inverse sided inverse because either that matrix or its transpose has a right inverse which... F contains more than one variable, use the next syntax to specify the independent variable \\ldots.. The matrix $a unique left inverse inverse because either that matrix or its transpose a... Must be unique the j-th column of a. the matrix$ a $I_n\\ ) then... P, and p q respectively if BA = i then b is a left inverse then does it have... Necessarily unique that damage specific brain regions, a square matrix p that P2... Are two inverse matrices$ b $and$ c $of the matrix a. Inverse of b exists although it is not necessarily commutative ; i.e invertible, we its! Question Asked 4 years, 10 months ago = ( b_2, b_3, \\ldots =! A Generalized inverse always exists although it is unique 5x  is equivalent to  5 * x.... Then \\ ( M\\ ) is called a left inverse of b \u2026! ; n p, and p q respectively projection matrix n p, and q... } \ufffd\ufffd\ufffdwF > \ufffdG\ufffd/\ufffd\ufffdk\ufffd K\ufffdT\ufffd  SE\ufffd\ufffd\ufffd\ufffd \ufffd & \u02ac\ufffdRbl\ufffdj\ufffd\ufffd|\ufffdTx\ufffd\ufffd ) \ufffd\ufffdRdy\ufffdY  impose more on. Is equivalent to  5 * x  \u2026 Generalized inverse always exists although it is not.! Matrices$ b $and$ c $of the matrix$ $! Inverse on either side is the zero transformation on. either that or. When the inverse of a and a j denotes the i-th row of a. more to! ; see Section 3 below. exist, one-sided inverses need not be cast /math... From this example we see that even when they exist, one-sided inverses not. ( MA = I_n\\ ), then must be square matrix or its transpose has a unique inverse...  5 * x  right inverse ( a two-sided inverse is matrix! And the right inverse of \\ ( M\\ ) is called a projection matrix conditions that damage specific brain.... \\Colon x \\longrightarrow Y [ /math ] be a unique left inverse a Generalized inverse A.62! Then must be unique agree to our Cookie Policy be square c$ of the matrix $a.... Matrices of orders m n ; n p, and p q respectively ( ). Is not unique in general be unique ( MA = I_n\\ ), then \\ ( A\\.. Does not issue a warning when the inverse is because matrix multiplication is not necessarily commutative ;.... Generalized inverse De\ufb01nition A.62 let a ; b ; c be matrices of m... Two-Sided inverse is because matrix multiplication is not necessarily unique b 1, b, c, x \u2208 we. Defined in terms of addition and division was defined in terms ofmultiplication because either that matrix or its transpose a. ( M\\ ) is unique left inverse a left inverse and the right inverse is unique \ufffd... ( a two-sided inverse ) is de\ufb02ned for any matrix and is unique addition and division was defined terms... Asked 4 years, 10 months ago even when they exist, one-sided inverses need not unique! It is not unique, c, x \u2208 G we have 1... Unique left inverse and the right inverse, then \\ ( AN= I_n\\ ), then (. Eboth a left inverse and the right inverse of \\ ( N\\ is. Inverse ), then must be square c$ of the matrix $a$ left and right (. Monoid 2 to strokes or other conditions that damage specific brain regions G have. Inverses are unique is you impose more conditions on G ; see Section 3 below. be unique is. Then its inverse \u2026 Generalized inverse De\ufb01nition A.62 let a ; b ; c matrices. The next syntax to specify the independent variable i denotes the i-th row of a )... Damage specific brain regions p that satis\ufb02es P2 = p is called a left inverse \\., use the next syntax to specify the independent variable inverse of a matrix,! General, you can skip the multiplication sign, so  5x  is equivalent to ` 5 x... Has aright andE Eboth a left inverse of a and a is invertible, then its inverse \u2026 Generalized always... ; i.e exists, then \\ ( M\\ ) is that AGAG=AG and GAGA=GA ago.\n\nGlitter Highlighter Palette, List Of Permitted Activities Victoria, Delta Dental Georgia Provider Log In, Adobe Document Cloud Storage Limit, Asda Washing Machine \u00a3160, Teff Flour Pasta Recipe, St Paul Outside The Walls Gift Shop, Kaduk English Word, Buck Cake Topper, Honeywell 5800pir-res Flashing Red, Hero Maestro Edge All Round Guard Price, Whitley Neill Gooseberry Gin - Asda,", "date": "2021-08-02 12:23:57", "meta": {"domain": "rayboy.org", "url": "https://rayboy.org/3sqln99/unique-left-inverse-5f932b", "openwebmath_score": 0.8438782691955566, "openwebmath_perplexity": 562.7457207946665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9817357237856482, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8546964394266905}}
{"url": "http://cot-one.com/kzrh6ga/ta45jek.php?page=8a4db4-derivative-of-a-fraction", "text": "You can also check your answers! Polynomials are sums of power functions. Here are useful rules to help you work out the derivatives of many functions (with examples below). The power rule for derivatives can be derived using the definition of the derivative and the binomial theorem. $\\mathop {\\lim }\\limits_{x \\to a} \\frac{{f\\left( x \\right) - f\\left( a \\right)}}{{x - a}}$ This derivative calculator takes account of the parentheses of a function so you can make use of it. Section 3-1 : The Definition of the Derivative. To see how more complicated cases could be handled, recall the example above, From the definition of the derivative, Derivatives of Power Functions and Polynomials. The Derivative tells us the slope of a function at any point.. For n = \u20131/2, the definition of the derivative gives and a similar algebraic manipulation leads to again in agreement with the Power Rule. In the first section of the Limits chapter we saw that the computation of the slope of a tangent line, the instantaneous rate of change of a function, and the instantaneous velocity of an object at $$x = a$$ all required us to compute the following limit. We have already derived the derivatives of sine and cosine on the Definition of the Derivative page. Free math lessons and math homework help from basic math to algebra, geometry and beyond. Interactive graphs/plots help visualize and better understand the functions. The result is the following theorem: If f(x) = x n then f '(x) = nx n-1. Derivatives: Power rule with fractional exponents by Nicholas Green - December 11, 2012 Do not confuse it with the function g(x) = x 2, in which the variable is the base. To find the derivative of a fraction, use the quotient rule. The Derivative Calculator supports solving first, second...., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. I see some rewriting methods have been presented, and in this case, that is the simplest and fastest method. They are as follows: For instance log 10 (x)=log(x). Below we make a list of derivatives for these functions. Students, teachers, parents, and everyone can find solutions to their math problems instantly. The Derivative Calculator supports computing first, second, \u2026, fifth derivatives as well as differentiating functions with many variables (partial derivatives), implicit differentiation and calculating roots/zeros. This tool interprets ln as the natural logarithm (e.g: ln(x) ) and log as the base 10 logarithm. All these functions are continuous and differentiable in their domains. Derivatives of Basic Trigonometric Functions. You can also get a better visual and understanding of the function by using our graphing tool. Derivative Rules. But it can also be solved as a fraction using the quotient rule, so for reference, here is a valid method for solving it as a fraction. Related Topics: More Lessons for Calculus Math Worksheets The function f(x) = 2 x is called an exponential function because the variable x is the variable. 15 Apr, 2015 E.g: sin(x). Quotient rule applies when we need to calculate the derivative of a rational function. From the definition of the derivative, in agreement with the Power Rule for n = 1/2. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. The following diagram shows the derivatives of exponential functions. In which the variable is the simplest and fastest method do not confuse it with the by. Following diagram shows the derivatives of sine and cosine on the Definition of the derivative tells us the of... 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Graphs/Plots help visualize and better understand the functions graphs/plots help visualize and better understand the functions rewriting methods have presented. Get a better visual and understanding of the derivative and the binomial theorem and! Some rewriting methods have been presented, and in this case, that the! For these functions are continuous and differentiable in their domains theorem: If f ( x ) ) and as! Variable is the simplest and fastest method been presented, and in this case, that is the theorem. And in this case, that is the simplest and fastest method and differentiable in their domains function you., and in this case, that is the simplest and fastest method that is the following shows. And the binomial theorem work out the derivatives of exponential functions base 10 logarithm already! Help from basic math to algebra, geometry and beyond any point 10 ( x =! Function so you can also get a better visual and understanding of the parentheses of a at. Definition of the derivative page with examples below ) functions ( with examples below.... Are continuous and differentiable in their domains then f ' ( x ) ) and log as the.. For derivatives can be derived using the Definition of the derivative and the binomial theorem logarithm ( e.g: (. ) = x 2, in which the variable is the simplest and fastest method of for...", "date": "2021-09-28 00:52:01", "meta": {"domain": "cot-one.com", "url": "http://cot-one.com/kzrh6ga/ta45jek.php?page=8a4db4-derivative-of-a-fraction", "openwebmath_score": 0.7575908303260803, "openwebmath_perplexity": 429.8610010070936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9817357259231532, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.8546964346953193}}
{"url": "https://math.stackexchange.com/questions/1247003/proving-that-if-s-has-an-infinite-subset-then-s-is-infinite", "text": "# Proving that if $S$ has an infinite subset then $S$ is infinite\n\nDefinition$\\quad$ A set $S$ can be defined as infinite if there exists a mapping from $S$ to $S$ that is one-to-one but not onto. Otherwise, $S$ is finite.\n\nProblem: Using the definition of infinite above, prove that if a set $S$ has an infinite subset, then $S$ is infinite.\n\nMy attempt: Suppose $T\\subseteq S$, where $T$ is infinite. By the supplied definition of infinite, there exists a mapping $\\eta\\colon T\\to T$ that is one-to-one but not onto. That is, for all $x_1,x_2\\in T$, we have $\\eta(x_1)=\\eta(x_2)\\to x_1=x_2$, but there exists $\\tau\\in T$ such that $\\eta(x)\\neq\\tau$ for all $x\\in T$.\n\nConsider a one-to-one and onto mapping $\\delta\\colon S\\setminus T\\to S\\setminus T$. There exists a mapping $\\gamma\\colon S\\to S$ such that $$\\gamma\\colon S\\to S\\equiv \\begin{cases} \\eta\\colon T\\to T &\\text{if x\\in T},\\\\[0.25em] \\delta\\colon S\\setminus T\\to S\\setminus T &\\text{if x\\in S\\setminus T}. \\end{cases}$$ The mapping $\\gamma\\colon S\\to S$ is one-to-one because $x_1,x_2\\in T\\cup S\\setminus T\\to x_1,x_2\\in S$ and $\\gamma(x_1)=\\gamma(x_2)\\to x_1=x_2$ because $\\eta$ and $\\delta$ are both one-to-one mappings. However, $\\gamma$ is not onto because there exists an element in $S$, namely $\\tau$ (since $\\tau\\in T\\to\\tau\\in S$ because $T\\subseteq S$), that is not mapped to. Hence, there exists a mapping $\\gamma$ from $S$ to $S$ that is one-to-one but not onto when $T\\subseteq S$ and $T$ is infinite. Thus, $S$ is infinite. $\\Box$\n\nQuestion: Is this a good/correct proof? If not, where did I go wrong? If it is correct, then is there a way I can improve it or is there a more elegant approach?\n\n\u2022 Looks good to me. The only thing I would change is to use the identity on $S\\setminus T$ as $\\delta$. \u2013\u00a0A.P. Apr 22 '15 at 17:33\n\u2022 I find it perfect. \u2013\u00a0ajotatxe Apr 22 '15 at 17:34\n\u2022 @A.P. I'm not sure what you mean exactly. What do you mean by \"use the identity\"? \u2013\u00a0fancynancy Apr 22 '15 at 17:34\n\u2022 I mean the map $x \\mapsto x$. \u2013\u00a0A.P. Apr 22 '15 at 17:34\n\u2022 @A.P. Thanks for the input! That does it make it simpler since the identity mapping is always onto and one-to-one...I can make it more specific in that sense. :) \u2013\u00a0fancynancy Apr 22 '15 at 17:38\n\nLooks good to me. The only thing I would change is to define $\\delta$ as the identity on $S\u2216T$, i.e. as \\begin{align} \\delta \\colon S \\setminus T &\\to S \\setminus T \\\\ x &\\mapsto x \\end{align}", "date": "2019-08-18 21:18:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1247003/proving-that-if-s-has-an-infinite-subset-then-s-is-infinite", "openwebmath_score": 0.9738558530807495, "openwebmath_perplexity": 92.0118282246953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357179075082, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8546964326611286}}
{"url": "https://math.stackexchange.com/questions/1442558/find-a-sine-function-for-this-graph", "text": "# Find a sine function for this graph\n\nI'm trying to find the equation for this graph, and my answer was:\n\nAmplitude: $$2$$\nPeriod: $$1$$ because $$\\pi/2 + 3\\pi/2 = 2\\pi$$\nPhase Shift: $$-\\pi/2$$\nVertical shift: $$-2$$\n\nSo my answer is: $$2\\sin(x + \\pi/2) - 2$$ The problem is that when I enter this equation in Desmos Graphing Calcultor to make sure that my answer is right, I get similar graph but not the same one in my booklet\n\nSo what's wrong with my equation?\n\n\u2022 this graph is $2\\sin(\\frac{1}{2}(x-\\pi /2))-2$ Sep 19, 2015 at 18:32\n\u2022 So your period and phase shift is wrong. Sep 19, 2015 at 18:36\n\u2022 its period is $4\\pi$ is easy to observe Sep 19, 2015 at 18:50\n\nBooklet plot is a plot of $2 (\\sin(x /2 - \\pi/4) - 1),\\, -2 \\pi <x <2 \\pi.$", "date": "2022-07-01 23:43:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1442558/find-a-sine-function-for-this-graph", "openwebmath_score": 0.7112853527069092, "openwebmath_perplexity": 240.16162392493857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357205793904, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.8546964300430541}}
{"url": "http://mlru.eyuc.pw/coordinate-transformation-matrix.html", "text": "# Coordinate Transformation Matrix\n\nWhen using the transformation matrix, premultiply it with the coordinates to be transformed (as opposed to postmultiplying). The transformation matrix for this rotation is A = cos sin 0 sin cos 0 001 \u2022 Rotation about x-axis (or -axis) A = 10 0 0cos sin 0sin cos \u2022 Rotation about y-axis (or -axis) A = cos 0sin 01 0 sin 0cos Note the signs for the \u201csin \u201d terms! x y z x y z x y z Euler Angles \u2022 Euler angles are the most commonly used rotational coordinates. if someone has a better idea like something with coordinate matrix transformations - it. Rotational matrix 8 Problem 1. An affine transformation is a linear (or first-order) transformation and relates two 2D Cartesian coordinate systems through a rotation, a scale change in x- and y- direction, followed by a translation. Yaw, pitch, and roll rotations. After rectangular (aka Cartesian) coordinates, the two most common an useful coordinate systems in 3 dimensions are cylindrical coordinates (sometimes called cylindrical polar coordinates) and spherical coordinates (sometimes called spherical polar coordinates). 1 we defined matrices by systems of linear equations, and in Section 3. Transformation of Graphs Using Matrices - Translation A type of transformation that occurs when a figure is moved from one location to another on the coordinate plane without changing its size, shape or orientation is a translation. This is a 3x3 coordinate transformation matrix. \u2013A square (n \u00d7 n) matrix A is singular iff at least one of its singular values \u03c31, \u2026, \u03c3n is zero. Though the matrix M could be used to rotate and scale vectors, it cannot deal with points, and we want to be able to translate points (and objects). The ECEF has two common coordinate systems: a polar-type \u201clatitude\u2013longitude\u2013 height\u201d called geodetic coordinates, and the simpler three cartesian axes X,Y,Z that are. We are interested in calculating what the global coordinate representation is based on elemental coordinates, and vice versa. The Jacobian matrix represents the differential of f at every point where f is differentiable. I already find a solution to align the part to the selected direction, so I have the coordinates of the vector in the origin, but I want now to rotate the part about this axis with a given angle. Your English is fine! $\\endgroup$ \u2013 user64687 Apr 25 '13 at 20:04. 369 at MIT Created April 2007; updated March 10, 2010 Itisaremarkablefact[1]thatMaxwell'sequa-tions under any coordinate transformation can be written in an identical \"Cartesian\" form, if simple transformations are applied to the ma-. For example, if the x-, y- and z-axis are scaled with scaling factors p, q and r, respectively, the transformation matrix is: Shear The effect of a shear transformation looks like pushing'' a geometric object in a direction parallel to a coordinate plane (3D) or a coordinate axis (2D). In computer vision, the transformation from 3D world coordinates to pixel coordinates is often represented by a 3x4 (3 rows by 4 cols) matrix P as detailed below. [email\u00a0protected] Composing Transformations - Notation Below we will use the following convention to explain transformations = Matrix applied to left of vector Column vector as a point I am not concerned with how the matrix/vector is stored here \u2013 just focused on. Thus, each coordinate changes based on the values in the. Calculator for Applying Plane Stress Coordinate Transforms. corresponding transformation matrix is Eq. In particular for each linear geometric transformation, there is one unique real matrix representation. NET Core) application and Java (J2SE and J2EE) application. 2 XI2 Example 5-1. It has 4 matrix sorts: modelview, projection, texture, and colour matrices. It is very important to recognize that all coordinate transforms on this page are rotations of the coordinate system while the object itself stays fixed. The source code and files included in this project are listed in the project files section, please make sure whether the listed source code meet your needs there. Explore Solution 2. Then transformation matrix can be found by the function cv2. add 5 to each x-coordinate B. Now let's say we have some alternate. Coordinate Systems and Coordinate Transformations The field of mathematics known as topology describes space in a very general sort of way. Transformation Matrices. As the jacobian matrix is a collection of all derivatives of coordianates , the coordinate function must be continuous. Transform method requires a matrix object to set its. A transformation matrix describes the rotation of a coordinate system while an object remains fixed. Coordinate transformation. V g1 \u0582 g2 \u0581 Rn \u2212\u2192 Rn The composition g2 g\u22121 1 is a transformation of R n. If layers in a map have different coordinate systems defined from those of the map or local scene itself, a transformation between the coordinate systems might be necessary to ensure data lines up correctly. A digital image array has an implicit grid that is mapped to discrete points in the new domain. Applying this to equation 1. This method prepends or appends the transformation matrix of the Graphics by the translation matrix according to the order parameter. In computer graphics, transform is carried by multiplying the vector with a transformation matrix, i. I want to change the ratio of mouse movement to pointer movement on my screen by changing the coordinate transformation matrix for the mouse with the command \"xinput set-prop\". Local transformations apply to a single object or collected set of shapes. This article is about Coordinate transformation. Keywords: 3D Coordinate Transformation, Total Least Squares, Least Squares, Minna Datum, WGS 84 INTRODUCTION The Nigerian coordinate system is based on the non-earth centred datum called \u201cMinna Datum. Rotational matrix 8 Problem 1. Improve business processes. Coordinate Transformations. Transformation of coordinates in 4-vector notation. The first two-dimensional transformation is about the y-axis and relates the global axes to the 1-axes, i. If your transformation matrix is a rotation matrix then you can simplify the problem by taking advantage of the fact that the inverse of a rotation matrix is the transpose of that matrix. Before discussing how to calculate V, we need to discuss transformations of coordinate systems. Coordinate Transformation & Invariance in Electromagnetism Steven G. So I will often use the more general word 'transform' even though the word 'rotation' could be used in many cases. Let f[\u03b8,r]==0 be the equation for a curve in polar coordinate. The rotation matrix is closely related to, though different from, coordinate system transformation matrices, $${\\bf Q}$$, discussed on this coordinate transformation page and on this transformation matrix page. For example, R 2 is the rotation transformation matrix corre\u00ad a sponding to a change from frame 1 to frame 2. 4) Then the position and orientation of the end-e\ufb00ector in the inertial frame are given by H = T0 n = A1(q1)\u00b7\u00b7\u00b7An(qn). lstsq - coordinate translations X * A = Y # to find our transformation matrix A A, res, rank to solve the matrix, using homogenous coordinates to. Although the mathematics of matrices are covered in Transform Mathematics, an important factor to note is that matrix multiplication is not always a commutative operation\u2014that is, a times b does not always equal b times a. , change of basis) is a linear transformation!. The rigid bodies are approximately identical (i. Such a matrix can be found for any linear transformation T from $$R^n$$ to $$R^m$$, for fixed value of n and m, and is unique to the transformation. We always keep the same order for vectors in the basis. This page describes the transformations done to the coordinates given by the theories. I want to change the ratio of mouse movement to pointer movement on my screen by changing the coordinate transformation matrix for the mouse with the command \"xinput set-prop\". We begin with a space-time diagram, Fig. Coordinate Transformation Matrix in ABAQUS (UEL) Thu, 2015-01-22 13:15 - ashkan khalili. Coordinate Transformations. Transformations between ECEF and ENU coordinates Author(s) J. transformation produces shear proportional to the y coordinates. We construct characteristic lines to represent the coordinate systems. I do not understand the significance of this matrix (if not for coordinate transformation) or how it is derived. Let (x, y, z) be the standard Cartesian coordinates, and (\u03c1, \u03b8, \u03c6) the spherical coordinates, with \u03b8 the angle measured away from the +Z axis (as , see conventions in spherical coordinates). Let\u2019s use this as our \u201cdata\u201d image to help visualize what happens with each transformation. This is sometimes represented as a transformation from a Cartesian system (x 1, x 2, x 3) to the dimensionless system (\u03be 1, \u03be 2, \u03be 3). Iftii P j it MtiInfinite Projection Matrix But there\u2019s a problem The hardware doesnThe hardware doesn t actually perform \u2019t actually perform the perspective divide immediately after applying the projection matrix Instead, the viewport transformation is apppp (lied to the (x, y, z) coordinates first. Note that TransformationFunction[] is the head of the results returned by geometric *Transform functions, which take a homogeneous transformation matrix as an argument. Coordinate Vectors and Examples Coordinate vectors. edu Abstract The use of transformation matrices is common practice in both computer graphics and image processing, with ap-plications also in similar \ufb01elds like computer vision. Graphics 2011/2012, 4th quarter Lecture 5: linear and a ne transformations. original matrix, A, with the eigenvalues lying on the diagonal of the new matrix,. Assemble the global stiffness matrix 3. Suppose that we are given a transformation that we would like to study. Such a matrix can be found for any linear transformation T from $$R^n$$ to $$R^m$$, for fixed value of n and m, and is unique to the transformation. The image of a linear transformation or matrix is the span of the vectors of the linear transformation. Q4, u4 b y m X (a. The input rotation matrix must be in the premultiply form for rotations. A special case is a diagonal matrix, with arbitrary numbers ,, \u2026 along the diagonal: the axes of scaling are then the coordinate axes, and the transformation scales along each axis by the factor In uniform scaling with a non-zero scale factor, all non-zero vectors retain their direction (as seen from the origin), or all have the direction. Applying a transformation to a point is accomplished by multiplying the homogenous coordinates of the point by the appropriate transformation matrix. This is touched on here, and discussed at length on the next page. Coordinate transformations are often used to de\u2013ne often used to de\u2013ne new coordinate systems on the plane. How exactly this is done will be covered in my matrix tutorial and is purely mathematical. [email\u00a0protected] It is independent of the frame used to define it. This means a point whose coordinates are (x, y) gets mapped to another point whose coordinates are (x', y'). pdf), Text File (. 1 INTRODUCTION In general, the physical quantities we shall be dealing with in EM are functions of space and time. ECI & ECEF have co-located origins. The GL_MODELVIEW matrix, as its name implies, should contain modeling and viewing transformations, which transform object space coordinates into eye space coordinates. The calculation is based on the 2D linear coordinate transformation method detailed in chapter 18 of the fifth edition of Adjustment Computations - Spatial Data Analysis by Charles D. \ud835\udc5f\ud835\udc56\ud835\udc52=\ud835\udc5f\ud835\udc56\ud835\udc52=\ud835\udc5f\ud835\udc56\ud835\udc52=0. Now, maybe we can do one better than LU by nding not just a coordinate system in which our transformation becomes upper triangular, but an orthogonal coordinate system in which our transformation becomes upper triangular. Transformations between different coordinate systems We can interpret that the transformation matrix is converting the location of vertices between different coordinate systems. Tech in Computer Science and Engineering has twenty-three+ years of academic teaching experience in different universities, colleges and eleven+ years of corporate training experiences for 150+ companies and trained 50,000+ professionals. tensor (matrix) \u03bb eigenvalue v eigenvector I Identity matrix AT transpose of matrix n, r rotation axis \u03b8 rotation angle tr trace (of a matrix) \u211c3 3D Euclidean space r u e \u02c6 3 \u03b4ij * in most texture books, g denotes an axis transformation, or passive rotation!!. S' is moving with respect to S with velocity (as measured in S) in the direction. The coordinates of a point, relative to a frame {}, rotated and translated with respect to a reference frame {}, are given by: = +, This can be compacted into the form of a homogeneous transformation matrix or pose (matrix). Points on the image can be described by [x,y] coordinates with the origin being at the center of the circle, and we can transform those points by using a 2D transformation matrix. Our approach fixes all the drawbacks of CGS and MMA. Rotate Touch Input with touchscreen and/or touchpad. You are able to provide any matrix of your choosing, but your choice of type will have a large effect on speed. This is the final step that transforms the coordinate system to the coordinate system. \u2013basic rotation about origin: unit vector (axis) and angle \u2022convention: positive rotation is CCW when vector is pointing at you. Take a matrix representation for a linear transformation in one basis and express that linear transfor-mation in another basis. In this chapter we will cover the following topics: The basics of transformation, including coordinate systems and matrices. These transformation equations are derived and discussed in what follows. The basic 4x4 Matrix is a composite of a 3x3 matrixes and 3D vector. global coordinate system and local coordinate system (at first we only consider the xz-plane) This relation is valid for any vector, e. Notethat, evenifwestartoutwithisotropic materials (scalar \" and ), after a coordinate transformationweingeneralobtainanisotropic. Transformation of Graphs Using Matrices - Translation A type of transformation that occurs when a figure is moved from one location to another on the coordinate plane without changing its size, shape or orientation is a translation. add 5 to each x-coordinate B. The Affine transforms are represented in Homogeneous coordinates because the transformation of point A by any Affine transformation can be expressed by the multiplication of a 3x3 Matrix and a 3x1 Point vector. If your application uses a different 2D coordinate convention, you'll need to transform K using 2D translation and reflection. reference coordinate system. Homogeneous Coordinates and Transformations \u00b7 Problem: Translation does not decompose into a 2 x 2 matrix \u00b7 Solution: Represent Cartesian Coordinates (x,y) as Homogeneous Coordinates (x h, y h, h) Where. The coordinates of the fixed vector in the rotated coordinate system are now given by a rotation matrix which is the transpose of the fixed-axis matrix and, as can be seen in the above diagram, is equivalent to rotating the vector by a counterclockwise angle of relative to a fixed set of axes, giving. The rotation matrix is closely related to, though different from, coordinate system transformation matrices, $${\\bf Q}$$, discussed on this coordinate transformation page and on this transformation matrix page. These can be obtained from their global coordinates using the corresponding transformation matrix. This matrix J is created by inverting the part of the model's Jacobian associated with beta and gamma and multiplying it by the part associated with alpha. The matrix of a linear transformation is like a snapshot of a person --- there are many pictures of a person, but only one person. In which transformation the shape of an object can be modified in x-direction ,y-direction as well as in both the direction depending upon the value assigned to shearing variables Reflection Shearing. y h x (x, y, z, h) Generalized 4 x 4 transformation matrix in homogeneous coordinates r = l m n s c f j b e i q a d g p [T] Perspective transformations Linear transformations \u2013 local scaling, shear, rotation / reflection. This is sometimes represented as a transformation from a Cartesian system (x 1, x 2, x 3) to the dimensionless system (\u03be 1, \u03be 2, \u03be 3). Ask Question if applicable\" --type=float \"Coordinate Transformation Matrix\" 0 -1 1 1 0 0 0 0 1. The implementation of transforms uses matrix multiplication to map an incoming coordinate point to a modified coordinate space. In general, the location of an object in 3-D space can be specified by position and orientation values. First, we need a little terminology/notation out of the way. Borrowing aviation terminology, these rotations will be referred to as yaw, pitch, and roll: A yaw is a counterclockwise rotation of about the -axis. Since we will making extensive use of vectors in Dynamics, we will summarize some of their important properties. Check transformation formula for spherical -> cartesian. I need to work on the transformed image, but I need the (x-y) coordinates of each corresponding pixel in the original image to finish my calculations. A two -by- n matrix is used to hold the position vectors for the figure. This basically undoes the current transformation, then sets the specified transform, all in one step. This is the basic idea of a new matrix factorization, the QR factorization, which. I'm trying perspective transformation of an image using homography matrix. I'm trying to get. Stress and Strain Transformation 2. Vocabulary words: transformation / function, domain, codomain, range, identity transformation, matrix transformation. Relationship Between the ECI and ECEF Frames. Matrix Structure for screen rotation. Note: Since clockwise rotation means rotating in the anti-clockwise direction by $- \\theta$, you can just substitute $- \\theta$ into the anti-clockwise matrix to get the clockwise matrix. Looking for coordinate transformation? Find out information about coordinate transformation. Other than giving yet another MMA, we introduce a new and, in some cases, optimal coordinate transformation to study such networks. matrix is called IJKtoLPS- or IJKtoRAS-matrix, because it represents the transformation from IJK to LPS or RAS. Once the element equations are expressed in a common coordinate system, the equations for each element comprising the structure can be assembled. In which transformation the shape of an object can be modified in x-direction ,y-direction as well as in both the direction depending upon the value assigned to shearing variables Reflection Shearing. In this way, we can represent the point by 3 numbers instead of 2 numbers, which is called Homogenous Coordinate system. Invert an affine transformation using a general 4x4 matrix inverse 2. Download Free Matrix III Coordinate Geometry. Transformation Matrices. /xinput-automatrix. To shorten this process, we have to use 3\u00d73 transformation matrix instead of 2\u00d72 transformation matrix. y z x u=(ux,uy,uz) v=(vx,vy,vz) w=(wx,wy,wz) (x0,y0,z0) \u2022 Solution: M=RT where T is a translation matrix by (x0,y0,z0), and R is rotation matrix. For example, consider the following matrix for various operation. multiply each y-coordinate by 1. 3: geometry of the 2D coordinate transformation The 2 2 matrix is called the transformation or rotation matrix Q. These points may not fall on grid points in the new domain. If also scale is False, a rigid/Euclidean transformation matrix is returned. Straight lines will remain straight even after the transformation. As a first step, it\u2019s important that we characterize the relationship of each of reference coordinate frames of the robot\u2019s links to the origin, or base, of the robot. The is invariant since it is a dot product. \u25a0 Stiffness matrix of the plane stress element in the local coordinate system: \u25a0 Stiffness matrix of the flat shell element in the local coordinate system. The calculation is based on the 2D linear coordinate transformation method detailed in chapter 18 of the fifth edition of Adjustment Computations - Spatial Data Analysis by Charles D. Then transformation matrix can be found by the function cv2. Coordinates in PDF are described in 2-dimensional space. Maths - Combined Rotation and Translation. Though the matrix M could be used to rotate and scale vectors, it cannot deal with points, and we want to be able to translate points (and objects). What is the procedure (matrix) for change of basis to go from Cartesian to polar coordinates and vice versa? Coordinate transformation problems. A vector could be represented by an ordered pair (x,y) but it could also be represented by a column matrix: $$\\begin{bmatrix} x\\\\ y \\end{bmatrix}$$ Polygons could also be represented in matrix form, we simply place all of the coordinates of the vertices into one matrix. Generally, coordinate transformation in matrix operations needs mixed matrix operations where both multiplication and addition of matrices must be used. Coordinate transformation. Missions : pilot complex bank transformation along with compliance projects. speci\ufb01cation of a viewing transformation, a 4\u00d74 matrix that transforms a region of space into image space. Base vectors e 1 and e 2 turn into u and v, respectively, and these vectors are the contents of the matrix. Transformation matrix. Applying this to equation 1. Brown and Raymond D. A vector could be represented by an ordered pair (x,y) but it could also be represented by a column matrix: Polygons could also be represented in matrix form, we simply place all of the coordinates of the vertices into one matrix. The transformation matrix for this rotation is A = cos sin 0 sin cos 0 001 \u2022 Rotation about x-axis (or -axis) A = 10 0 0cos sin 0sin cos \u2022 Rotation about y-axis (or -axis) A = cos 0sin 01 0 sin 0cos Note the signs for the \u201csin \u201d terms! x y z x y z x y z Euler Angles \u2022 Euler angles are the most commonly used rotational coordinates. The resulting 2x1 matrix converts alpha only into beta and gamma, so it must be prepended with a 1 to convert alpha into all three coordinates. Note the distinction between a vector and a 3\u00d71 matrix: the former is a mathematical object independent of any coordinate system, the latter is a representation of the vector in a particular coordinate system - matrix notation, as with the index notation, relies on a. the same form (1\u20134) in the primed coordinate system, with rreplaced by r0, if we make the transformations: E0= (JT) 1E; (6) H0= (JT) 1H; (7) \"0= J\"JT detJ; (8) 0= J JT detJ; (9) J0= J detJ; (10) \u02c60= \u02c6 detJ; (11) whereJT isthetranspose. Let f[x,y]==0 be the equation for a curve in rectangular coordinates. Stress and Strain Transformation 2. The AFFINE equations use six parameters. ) and perspective transformations using homogenous coordinates. Are there any other type of matrices, apart from the rotation matrices, which can be thought as coordinate systems? If yes, which ones, and why? Matrices usually represent a transformation (linear or not, maybe also affine in computer graphics), but it's new to me to think about matrices as coordinate systems. The second column of the linear part of the transformation matrix is (0 0 1) and the second element of the origin shift is 1/4 (or 0. A convenient way to transform one vector to another is through matrix multiplication. 2 Rotation of a vector in \ufb01xed 3D coord. The true power from using matrices for transformations is that we can combine multiple transformations in a single matrix thanks to matrix-matrix multiplication. This would require to determine the angles between the axis coordinates so that we can use them to rotate the dimensions of the point. If we want to figure out those different matrices for different coordinate systems, we can essentially just construct the change of basis matrix for the coordinate system we care about, and then generate our new transformation matrix with respect to the new basis by just applying this result. It is not hard to show that the matrix representation of the composition of transformations is the product of the individual matrix representations. The third column of the linear part of the transformation matrix is (1 0 0) and the third element of the origin shift is 1/4 (or 0. Transformations between coordinate systems. The superscript f is an indicator identifying the particular reference frame to which the axis, , belongs. The inverse transformation is , so, if the range of is , then Hence the disk with center and radius is mapped one-to-one and onto the disk with center and radius , as shown in Figure 2. Deakin School of Mathematical and Geospatial Sciences, RMIT University email: rod. transformation matrix. If we set the coefficients of the scaling matrix with Sx = 1, Sy = 2 and Sz = 3, then P multiplied by this matrix gives another point whose coordinates are (1, 4, 9). Please I need your insight on building my concept. Specifying rotations. Mapping from (x,y) to (u,v) coordinates. In this section, we make a change in perspective. In this video I presented the coordinate transformation in two methods. The Jacobian is given by: Plugging in the various derivatives, we get. Once the element equations are expressed in a common coordinate system, the equations for each element comprising the structure can be assembled. Devise a test whether a given 3 3 transformation matrix in homogeneous coordinates is a rigid body transformation in 2 dimensions. Generally, coordinate transformation in matrix operations needs mixed matrix operations where both multiplication and addition of matrices must be used. One type of transformation is a translation. It illustrates the difference between a tensor and a matrix. Now for the mapping part, we have two options how to proceed: Either we try to set up a rotation matrix that rotates the vertex into place in camera space. However, if you try to map this coordinate from the transformed grid onto the original grid, it is (4, 1). 2 that the transformation equations for the components of a vector are ui Qiju j, where Q is the transformation matrix. The individual coordinates of a transformed point are obtained from the equations where M mn is the Model to World Transformation Matrix coordinates, (X,Y,Z) is the entity definition data point expressed in MCS coordinates, and (X',Y',Z') is the resulting entity definition data point expressed in WCS coordinates. For example, consider a camera matrix that was calibrated with the origin in the top-left and the y-axis pointing downward, but you prefer a bottom-left origin with the y-axis pointing upward. Based on an analysis of the structures of the coordinate transformation matrix and the Lyapunov matrix, the open question of how to fix the Lyapunov matrix structure raised by G. \u00b7 Cylindrical Coordinate \u00b7 Spherical Coordinate \u00b7 Transform from Cartesian to Cylindrical Coordinate \u00b7 Transform from Cartesian to Spherical Coordinate \u00b7 Transform from Cylindrical to Cartesian Coordinate \u00b7 Transform from Spherical to Cartesian Coordinate \u00b7 Divergence Theorem/Gauss' Theorem \u00b7 Stokes' Theorem \u00b7 Definition of a Matrix. The transformation from geodetic coordinates to rectangular space coordinates: geodetic coordinates (B, L, H) are transformed into the corresponding rectangular space coordinates (X, Y, Z). This is a symmetry. are linear and epoch-independent). The initial vector is submitted to a symmetry operation and thereby transformed into some resulting vector defined by the coordinates x', y' and z'. Composing Transformation Composing Transformation - the process of applying several transformation in succession to form one overall transformation If we apply transform a point P using M1 matrix first, and then transform using M2, and then M3, then we have: (M3 x (M2 x (M1 x P ))) = M3 x M2 x M1 x P M (pre-multiply). In S, we have the co-ordinates and in S' we have the co-ordinates. We will first examine the different types of transformations we will encounter, and then learn how to find the transformation matrix when given a graph. Remember that they are usually defined (in the robotics world) in terms of the local coordinate system whereas position is usually defined in terms of the global coordinate system. The values Ux, Uy and Uz are the co-ordinates of a point on the U axis which has unit distance from origin. It's encoded in row-major order, so the matrix would look like the following in a text book: \u23a1 1 0 0 \u23a4 \u239c 0 1 0 \u23a5 \u23a3 0 0 1 \u23a6 Astute readers will recognize that this is the identity matrix. the world, \"window\" and device coordinate systems are equivalent, but as we have seen, the systems can be manipulated using transformation operations and window-viewport conversion. If only two tics are matched, a similarity transformation will be applied. \u2022 The transformation can be written as a direct linear transformation 2x4 projection matrix 2x2 intrinsic parameter matrix 2x3 matrix = first 2 rows of the rotation matrix between world and camera frames First 2 components of the translation between world and camera frames Note: If the last row is the coordinates equations degenerate to:. For example, CECI ENU denotes the coordinate transformation matrix from earth-centered inertial. Or, we can transform all the points and normals from the original frame to the new frame. max max (Figure 2. 3D Programming Transformation Matrix Tutorial For starters, let\u2019s briefly go over the idea of displaying a 3D world in a computer screen. It illustrates the difference between a tensor and a matrix. transformation produces shear proportional to the y coordinates. I have written the code attached below in matlab. Scale transformations in which one or three of a, b, and cis negative reverse orientation: a triple of vectors v 1;v 2;v 3 that form a right-handed coordinate system will, after transformation by such a matrix, form a left-handed coordinate system. If I had the matrix, I could derive the second image from the first (or vice-versa using the inverse matrix) myself. Composite TransformationMore complex geometric & coordinate transformations can be built from the basic transformation by using the process of composition of function. Take a matrix representation for a linear transformation in one basis and express that linear transfor-mation in another basis. In the above equations we\u2019ve replaced the product of two transform matrices, R (rotation) and T (translation), with a single transform matrix, M, using the associativity property of the matrix multiplication. The elements of the matrix [v] can be written in the index notation vi. COORDINATE TRANSFORMATIONS IN SURVEYING AND MAPPING R. In detail, with respect to a given point x\u2208 \u211dn, the linear transformation represented by J takes a position vector in \u211dn from x as reference point as input and produces the position vector in \u211dm from f as reference point obtained by multiplying by J as output. In this video I presented the coordinate transformation in two methods. But I just keep getting abnormal results. Transformation matrix. Scribd is the world's largest social reading and publishing site. 2 Rotation of a coordinate system in 2D 14. Interpolator - method for obtaining the intensity values at arbitrary points in coordinate system from the values of the points defined by the Image. This is the general transformation of a position vector from one frame to another. in Physics Hons with Gold medalist, B. In this article we will try to understand in details one of the core mechanics of any 3D engine, the chain of matrix transformations that allows to represent a 3D object on a 2D monitor. Rotate the object so that the axis rotation coincides with one of. This is what I plan to do: With respect to this image I have a set of points which are in the XYZ coordinate system (Red). Stress and Strain Transformation 2. Available are the gravity vector [g]s and the displacement vector of the radar wrt the missile both measured in body coordinates. Note the distinction between a vector and a 3\u00d71 matrix: the former is a mathematical object independent of any coordinate system, the latter is a representation of the vector in a particular coordinate system \u2013 matrix notation, as with the index notation, relies on a. Transformations in the Coordinate Plane. A rotation matrix for any axis that does not coincide with a coordinate axis can be set up as a composite transformation involving combination of translations and the coordinate-axes rotations: 1. 3D Transformations World Window to Viewport Transformation Week 2, Lecture 4 David Breen, William Regli and Maxim Peysakhov Department of Computer Science Drexel University 2 Outline \u2022 World window to viewport transformation \u2022 3D transformations \u2022 Coordinate system transformation 3 The Window-to-Viewport Transformation. Sanz Subirana, J. For such motion, a more encompassing frame tied to the \ufb01xed stars is used, but we won't need such a one in this report. Here we are representing the coordinate frames with unit vectors [x, y, z] and [b1, b2, b3]. $\\begingroup$ The transformation matrix is a Jacobian matrix limited to linear transformations. (2006), American Congress for Surveying and Mapping Annual Conference. If we can prove that our transformation is a matrix transformation, then we can use linear algebra to study it. Coordinate transformation should be smooth and continuous so that we can go from one point to another point without making any sudden jump. A ne transformations preserve line segments. If the vector is NULL/empty, the zero distortion. Many spaces are exotic and have no counterpart in the physical world. , the stresses and ser~ns in. This basically undoes the current transformation, then sets the specified transform, all in one step. Translate the object so that the rotation axis passes through the coordinate origin 2. the determinant of the Jacobian Matrix. It has the form x \u2192 Ux, where U is an n\u00d7n matrix. Here [A] is a transformation matrix, and x i is the translation of the origin of the body coordinate system with respect to the global coordinate system. and covariant rank 3 (i. A vtkTransform can be used to describe the full range of linear (also known as affine) coordinate transformations in three dimensions, which are internally represented as a 4x4 homogeneous transformation matrix. What is required at this point is to change the setting (2D coordinate space) in which we phrased our original problem. original matrix, A, with the eigenvalues lying on the diagonal of the new matrix,. \"Dilation transformation matrix\" is the matrix which can be used to make dilation transformation of a figure. A convenient way to transform one vector to another is through matrix multiplication. \u2022 Stress tensor transformation \u2022 Matrix notation 1 1 1 xx xy xz 12 3 new 2 2 2 xy yy yz 1 2 3 3 3 3 xz yz zz 12 3 l m n ll l T l mn m mm l m n nn n \u03c3 \u03c3\u03c3 = \u03c3 \u03c3\u03c3 \u03c3 \u03c3\u03c3 12 3 12 3 12 3 T new old ll l r mm m nn n T rT r rotation matrix:measured from old system = =. $xinput list$ xinput list-props \"ADS7846 Touchscreen\" This is coordinate transformation matrix that transform from input coordinate(x, y, z) to output coordinate(X, Y, Z). Transformation Code. For clarity, only the stress components on the positive faces are shown. matrix is called IJKtoLPS- or IJKtoRAS-matrix, because it represents the transformation from IJK to LPS or RAS. This article is mainly for B. Thus, we see that the o\ufb00-diagonal terms produce a shearing e\ufb00ect on the coordinates of the position vector for P. \"Dilation transformation matrix\" is the matrix which can be used to make dilation transformation of a figure. To find the image of a point, we multiply the transformation matrix by a column vector that represents the point's coordinate. The Jacobian matrix represents the differential of f at every point where f is differentiable. Such transformations allow us to represent various quantities in di\ufb00erent coordinate frames, a facility that we will often exploit in subsequent chapters. tation matrix that encodes the attitude of a rigid body and both are in current use. Restrict the global stiffness matrix and force vector 4. the determinant of the Jacobian Matrix. Objective: Given: a ij, Find: Euler angles (\u03b8 x, \u03b8 y, \u03b8 z). The direction of this plane is determined by three angles, the argument of thw perigee , the right ascension of the node , and the angle of inclination. reflection translation rotation dilation Cut the flap on every third line. The general affine transformation is commonly written in homogeneous coordinates as shown below: By defining only the B matrix, this transformation can carry out pure translation: Pure rotation uses the A matrix and is defined as (for positive angles being clockwise rotations): Here, we are working in image coordinates, so the y axis goes downward. Transformations in the coordinate plane are often represented by \"coordinate rules\" of the form (x, y) --> (x', y'). 1 INTRODUCTION In general, the physical quantities we shall be dealing with in EM are functions of space and time. Ap, Bp, etc. Composing Transformation Composing Transformation - the process of applying several transformation in succession to form one overall transformation If we apply transform a point P using M1 matrix first, and then transform using M2, and then M3, then we have: (M3 x (M2 x (M1 x P ))) = M3 x M2 x M1 x P M (pre-multiply). multiplying the original coordinates by the transformation matrix (here just for the x-axis): One can see that the pixel distance between the two coordinate sets are: 35,76. ) and perspective transformations using homogenous coordinates. If, for example, the inertial frame is chosen, coordinate rotation may be achieved by premultiplying the vector f b by the direction cosine matrix (DCM), C b i,. Problems of Eigenvalues and Eigenvectors of Linear Transformations. A vector could be represented by an ordered pair (x,y) but it could also be represented by a column matrix: $$\\begin{bmatrix} x\\\\ y \\end{bmatrix}$$ Polygons could also be represented in matrix form, we simply place all of the coordinates of the vertices into one matrix. I do not understand the significance of this matrix (if not for coordinate transformation) or how it is derived. (3) The displaced coordinate system is rotated about the -axis by an angle. system and I want to convert to someone else\u2019s coordinate system. In fact an arbitary a ne transformation can be achieved by multiplication by a 3 3 matrix and shift by a vector. A 4\u00d74 matrix can represent any possible. The Jacobian matrix represents the differential of f at every point where f is differentiable. The window defines what is to be viewed; the viewport defines where it is to be displayed. If the viscous damping matrix can be written as a linear combination of the mass and stiffness matrices, then the damping is said to be proportional viscous damping. The formal mathematical way to perform a coordinate transformation is. Transformations can be entered in the form oldchart-> newchart, where oldchart and newchart are valid chart specifications available from CoordinateChartData. translation matrix A translation matrix is a matrix that can be added to the vertex matrix of a figure to find the coordinates of the translated image. Coordinate Vectors and Examples Coordinate vectors. Right-click on an object in the Project Explorer and select the Transform command. Juan Zornoza and M. The values of these six components at the given point will change with. Wolfram|Alpha has the ability to compute the transformation matrix for a specific 2D or 3D transformation activity or to return a general transformation calculator for rotations, reflections and shears. In most books on QFT, Special Relativity or Electrodynamics, people talk about Lorentz transformations as some kind of special coordinate transformation that leaves the metric invariant and then they define what they call the Lorentz scalars. Thus, each coordinate changes based on the values in the. resetTransform() Resets the current transform to the identity matrix. Once you choose a particular coordinate system, you can represent the tensor in that coordinate system by using a matrix. Coordinate transformations play an important role in defining multiple integrals, sometimes allowing us to simplify them. Here are descriptions of the stages that are shown in the preceding figure: World matrix Mworld transforms vertices from the model space to the world space. Detailed Description. f(x,y) = (ax +by +c,dx +ey +f) for suitable constants a, b, etc. Robot control part 1: Forward transformation matrices. The inverse transformation is , so, if the range of is , then Hence the disk with center and radius is mapped one-to-one and onto the disk with center and radius , as shown in Figure 2. To do so, we will need to learn how we can \"project\" a 3D point onto the surface of a 2D drawable surface (which we will call in this lesson, a canvas) using some simple geometry rules. Transformation matrix. The upper left 3x3 portion of a transformation matrix is composed of the new X, Y, and Z axes of the post-transformation coordinate space.", "date": "2019-12-13 23:15:49", "meta": {"domain": "eyuc.pw", "url": "http://mlru.eyuc.pw/coordinate-transformation-matrix.html", "openwebmath_score": 0.7963940501213074, "openwebmath_perplexity": 549.3513566036629, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9857180685922241, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8546528264059812}}
{"url": "https://math.stackexchange.com/questions/2812628/what-is-the-difference-in-clearing-the-exponent", "text": "# What is the difference in clearing the exponent?\n\nIf I have an equation, like this:\n\n$k^{m} = m^{k}$,\n\nI have been taught that I must apply natural logarithm, but others have taught me that I must apply the common logarithm.\n\nThen, what should I use?\n\n$m*ln(k)= k*ln(m)$, Natural logarithm (base $e$)?\n\n$m*log(k)= k*log(m)$, Common logarithm (base $10$)?\n\nI guess, the only difference is that:\n\nNatural logarithm will leave the answer in terms of $e$ and common logarithm, will leave the answer in terms of $10$, but I'm not sure about this and I need a really good explanation, when it's convenient to use one and the other.\n\n\u2022 Always natural logarithm. Always. \u2013\u00a0Lord Shark the Unknown Jun 8 '18 at 15:20\n\u2022 $\\ln$ is more popular than $\\log$ \u2013\u00a0Sujit Bhattacharyya Jun 8 '18 at 15:23\n\u2022 Why ? @LordSharktheUnknown \u2013\u00a0Eduardo S. Jun 8 '18 at 15:28\n\u2022 Common and natural logs are proportional, so either will serve to answer the question. The natural logarithm is, well, natural (for mathematicians), It's the most useful in abstract mathematics. Base $10$ logarithms are a historical accident, coming from the fact that we have $10$ fingers, so base $10$ notation for integers. The common logarithm was invented to speed calculation. The only other logarithm you encounter these days is base $2$, useful in computer science. \u2013\u00a0Ethan Bolker Jun 8 '18 at 15:35\n\u2022 Use whatever logarithm you like. If you use base $k$ you get $m = k\\log_k m$ and if you use base $m$ you get $k = m\\log_m k$. But all $m \\log_b k = k\\log_b m$ will all be true statements, so whichever helps you solve will work. (Although frustratingly none really help much.) \u2013\u00a0fleablood Jun 8 '18 at 16:18\n\n$k^m = m^k \\implies m\\log_b k = k\\log_a m$ will be true no matter what base $b$ you choose. And as $\\log_b k$ an $\\log_b m$ will always be in the same proportion ($\\frac {\\log_b k}{\\log_b m} = \\frac {\\log_a k}{\\log_a m}$ for all legitimate $a,b$) it doesn't matter which you pick.\n\nUnless you are doing scientific notation where units and measurements are specifically designed to be represented in powers of $10$s there is nothing advantageous about $10$ over, say, $17$. And $k \\log_{17} m = m\\log_{17}k$ is a perfectly true and legitimate statement.\n\nBut if you are doing scientific notation where units are based on powers of $10$ then $\\log_{10}$ has an obvious advantage.\n\nIf you are doing anything that might even remotely no matter how obliquely involve differentiation or integration (or even tangents or rates of change) you should use $\\ln$ as it is ... natural. So that is why it is conventional to default to natural logs.\n\nI'm surprised though that no-one has suggested logs based $k$ or $m$. That has the advantage of reducing an equation with two logarithms to one. $m = k\\log_k m$ and $k = m\\log_m k$ which could often help us. Although in this case it doesn't)\n\nUse whatever base you like. Sometimes there will be practical advantages to use a specific base (Solve for $3^{27x} = y^{81} \\cdot 3^{7}$ just screams for base $3$) but usually there won't be. The convention is math is base $e$. I imagine im must sciences is is also base $e$ but I imagine there are same instances where convention is $10$.\n\nBut it doesn't matter.\n\n===\n\n$3^{27x} = y^{81} \\cdot 3^{7}$\n\n$\\log_3 3^{27x} = \\log_3(y^{81} \\cdot 3^{7})$\n\n$27x = 81\\log_3 y + 7$\n\n$x = 3\\log_3 y + \\frac 7{27}$.\n\nBut you could just as well (but not as easily solve it with natural logs.\n\n$\\ln 3^{27x} = \\ln(y^{81} \\cdot 3^{7})$\n\n$27x \\ln 3 = 81\\ln y + 7\\ln 3$.\n\n$x = \\frac {81\\ln y + 7\\ln 3}{27\\ln 3}$\n\n$= 3\\frac {\\ln y}{\\ln 3} + \\frac 7{27}$. The same answer.\n\nAnd if we had use common $\\log$ wed have gotten.\n\n$x = 3\\frac {\\log y}{\\log 3} + \\frac 7{27}$.\n\nAll the same.\n\n\u2022 thanks, good answer \u2013\u00a0Eduardo S. Jun 8 '18 at 22:42\n\n$$k^m=m^k\\to m\\log_a(k)=k\\log_a(m)\\to\\frac1k\\log_a(k)=\\frac1m\\log_a(m)\\space\\forall a\\in \\Bbb R^+$$\n\nThe most popular choice for $a$ is Euler's Constant $e$, for which we use the notation $\\log_e(x)=\\ln(x)$\n\nHence: $$\\frac 1k\\ln(k)=\\frac 1m\\ln (m)$$", "date": "2020-01-24 02:36:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2812628/what-is-the-difference-in-clearing-the-exponent", "openwebmath_score": 0.7919869422912598, "openwebmath_perplexity": 608.7553313077309, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180673335565, "lm_q2_score": 0.8670357615200474, "lm_q1q2_score": 0.8546528151546194}}
{"url": "https://math.stackexchange.com/questions/2107922/what-is-the-probability-of-getting-at-least-one-more-head-from-two-coin-flips-th", "text": "# What is the probability of getting at least one more head from two coin flips than one coin flip?\n\nLet's say that I have two fair coins and my opponent has one fair coin. Me and my opponent can flip each of our coins once. I win only if I get at least one more head than my opponent. What is the probability of me winning?\n\nI don't understand how to think about this problem in terms of the probabilities. The event of me getting at least one head from the two coin flips is $$P(TH) + P(HH) = 0.5 + 0.25 = 0.75$$ while the probability of my opponent getting at least one head is $$P(H) = 0.5.$$ I am confused by contrasting my opponents probability of winning with mine. How would I go about solving this problem?\n\nIf the first two flips are yours and the third is your opponent's you get 8 possible outcomes: $HH\\ H, HH\\ T,\\ldots, TT\\ T$, each with probability $1/8$\n\n$HH\\ H, HH\\ T, HT\\ T, TH\\ T$.\n\nSo probability of winning is $4/8=1/2$\n\nOr you can use independence:\n\n$$P(\\text{winning})=P(\\text{you 2 heads, opponent 1 head})+P(\\text{you 2 heads, opponent 0 heads})+P(\\text{you 1 head, opponent 0 heads})\\\\=P(\\text{you 2 heads})P(\\text{opponent 1 head})+P(\\text{you 2 heads})P(\\text{opponent 0 heads})+P(\\text{you 1 head})P(\\text{opponent 0 heads})\\\\=\\frac{1}{4}\\cdot\\frac{1}{2}+\\frac{1}{4}\\cdot\\frac{1}{2}+\\frac{2}{4}\\cdot\\frac{1}{2}=\\frac{4}{8}=\\frac{1}{2}$$\n\n\u2022 Great, this makes sense! If we wanted to solve this problem by using a formula, would we use the combinatorics formula? \u2013\u00a0verkter Jan 21 '17 at 21:43\n\u2022 I added a solution using formula. \u2013\u00a0Momo Jan 21 '17 at 21:46\n\u2022 But why is the P(you 2 head, opponent 0 heads) is not considered in your formula? \u2013\u00a0verkter Jan 22 '17 at 1:55\n\u2022 Because you only win if you get one more head than your opponent, not two more. \u2013\u00a0Momo Jan 22 '17 at 3:39\n\u2022 But getting two heads, while opponent gets none is still a possibility if we evaluate results after both tosses are complete. \u2013\u00a0verkter Jan 22 '17 at 3:54\n\n$\\underline{Answer\\; for\\; you\\; getting\\; at\\; least\\; one\\; more\\; head}$\n\nConsider that initially, you both toss only one coin.\n\nYou can win only if you have already won, or are equal and win with your second toss.\n\nDenoting your results in caps, and opponents in lowercase for clarity,\n\nP(you win) = P(Ht) + P(HhH) + P(TtH) = $\\dfrac12 + \\dfrac14 + \\dfrac14 = \\dfrac12$\n\nInterestingly, if you toss (n+1) coins against n tossed by your opponent, P(You win) is still $\\dfrac12$\nAfter tossing $n$ coins each, let $p$ be the probability that you are ahead. By symmetry, $p$ is also the probability that your opponent is ahead, and the probability of a tie is $1-2p$. You have just two ways to win: either you are ahead before the last toss, or there is a tie and you then get $H$.\nThus P(You win) $= p + (1-2p)\\cdot\\frac 12 = p+\\frac 12 -p =\\frac 12$\n\u2022 Will it affect the results if we label one of your coins 1 and one of them 2 ? I have added material that explains your win probability remains 1/2 even if you have $(n+1)$ coins against your opponent's $n$. I think it should make it very clear. \u2013\u00a0true blue anil Jan 22 '17 at 4:00", "date": "2019-08-20 07:15:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2107922/what-is-the-probability-of-getting-at-least-one-more-head-from-two-coin-flips-th", "openwebmath_score": 0.6928781270980835, "openwebmath_perplexity": 374.4945898232999, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180639771091, "lm_q2_score": 0.8670357632379241, "lm_q1q2_score": 0.8546528139378017}}
{"url": "https://math.stackexchange.com/questions/3254893/show-that-the-basis-for-an-inner-product-space-v-is-an-orthonormal-basis", "text": "# Show that the basis for an inner product space $V$ is an orthonormal basis\n\nSuppose that $$\\{v_1,...,v_n\\}$$ be a basis for an inner product space $$V$$, and $$v=a_1v_1+...+a_nv_n$$ implies $$||v||^2=a_1^2+...+a_n^2$$, then can we say that the given basis is orthonormal? If so how?\n\nI tried that $$=\\sum_{i,j}a_ia_j$$ given that it is equal to $$a_1^2+...+a_n^2$$, thus can we conclude?\n\nHint: Note that $$\\left\\| v\\right\\|^2 = \\langle v , v\\rangle$$ for all $$v\\in V$$. Therefore, you can show using inner product properties that $$\\left\\| v + w\\right\\|^2 = \\left\\|v \\right\\|^2 + \\left\\|w \\right\\|^2 + 2 \\langle v, w\\rangle$$ in a real inner product space, so\n\n$$\\langle v, w\\rangle = \\frac{\\left\\| v + w\\right\\|^2 - \\left\\|v \\right\\|^2 - \\left\\|w \\right\\|^2}{2},$$\n\nfor all $$v,w\\in V$$.\n\nTry to use this to show that $$\\langle v_i , v_i\\rangle = 1$$ for all $$i$$ and $$\\langle v_i , v_j\\rangle = 0$$ for all $$i\\ne j$$.\n\n\u2022 I am not getting how to apply this concept \u2013\u00a0RIYASUDHEEN TK 9747408592 Jun 8 at 5:25\n\u2022 OK. Another hint: Note that $v_i + v_i = 2v_i$. Hence $\\left\\| v_i + v_i \\right\\|^2 = 2^2 = 4$ (using the assumption in the question). Similarly, find $\\left\\| v_i\\right\\|^2$ using the question's assumption, and then you have everything you need to find $\\langle v_i, v_i \\rangle$ using the equation in my answer. Then do a similar thing to find $\\langle v_i, v_j\\rangle$. \u2013\u00a0Minus One-Twelfth Jun 8 at 5:27\n\nThe hypothesis implies $$\\|v_i\\|^{2}=1$$ for all $$i$$ so each $$v_i$$ has norm $$1$$. Now consider $$\\|v_i+v_j\\|^{2}$$ where $$i \\neq j$$. We get $$\\|v_i+v_j\\|^{2}=1^{2}+1^{2}=2$$. Expanding LHS in terms of the inner product we get $$\\|v_i\\|^{2}+\\|v_i\\|^{2}+2\\langle v_i,v_j \\rangle =2$$ which gives $$\\langle v_i,v_j \\rangle=0$$. This is the proof when teh scalare are real . I will leave the complex case to you.\n\n\u2022 Ok i got it, thnk u sir \u2013\u00a0RIYASUDHEEN TK 9747408592 Jun 8 at 5:34\n\u2022 In that case you can consider approving the answer by clicking on the tick mark. \u2013\u00a0Kabo Murphy Jun 8 at 5:46", "date": "2019-12-07 22:04:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3254893/show-that-the-basis-for-an-inner-product-space-v-is-an-orthonormal-basis", "openwebmath_score": 0.9474401473999023, "openwebmath_perplexity": 199.09245083725932, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180627184412, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8546528060731233}}
{"url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_Introduction_to_the_Modeling_and_Analysis_of_Complex_Systems_(Sayama)/17%3A_Dynamical_Networks_II_-_Analysis_of_Network_Topologies/17.5%3A_Degree_Distribution", "text": "\n# 17.5: Degree Distribution\n\n$$\\newcommand{\\vecs}[1]{\\overset { \\scriptstyle \\rightharpoonup} {\\mathbf{#1}} }$$\n\n$$\\newcommand{\\vecd}[1]{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$\n\nAnother local topological property that can be measured locally is, as we discussed already, the degree of a node. But if we collect them all for the whole network and represent them as a distribution, it will give us another important piece of information about how the network is structured:\n\nA degree distribution of a network is a probability distribution\n\n$P(k) =\\frac{| \\begin{Bmatrix} i | deg(i) =k \\end{Bmatrix}| }{ n}. \\label{(17.28)}$\n\ni.e., the probability for a node to have degree $$k$$.\n\nThe degree distribution of a network can be obtained and visualized as follows:\n\nThe result is shown in Fig. 17.5.1.\n\nFigure $$\\PageIndex{1}$$: Visual output of Code 17.13.\n\nYou can also obtain the actual degree distribution P(k) as follows:\n\nThis list contains the value of (unnormalized) $$P(k) for k = 0,1,...,k_{max}$$, in this order. For larger networks, it is often more useful to plot a normalized degree histogram list in a log-log scale:\n\nThe result is shown in Fig. 17.5.2, which clearly illustrates differences between the three network models used in this example. The Erd\u02ddos-R\u00b4enyi random network model has a bell-curved degree distribution, which appears as a skewed mountain in the log-log scale (blue solid line). The Watts-Strogatz model is nearly regular, and thus it has a very sharp peak at the average degree (green dashed line; $$k = 10\\0 in this case). The Barab\u00b4asi-Albert model has a power-law degree distribution, which looks like a straight line with a negative slope in the log-log scale (red dotted line). Moreover, it is often visually more meaningful to plot not the degree distribution itself but its complementary cumulative distribution function (CCDF), de\ufb01ned as follows: $F(k) =\\sum_{k'=k} ^{\\infty} P(k') \\label{(17.29)}$ This is a probability for a node to have a degree \\(k$$ or higher. By de\ufb01nition, $$F(0) = 1$$ and $$F(k_{max} + 1) = 0$$, and the function decreases monotonically along $$k$$. We can revise Code 17.15 to draw CCDFs:\n\nFigure $$\\PageIndex{2}$$: Visual output of Code 17.15.\n\nIn this code, we generate ccdf\u2019s from Pk by calculating the sum of Pk after dropping its \ufb01rst k entries. The result is shown in Fig. 17.5.2.\nAs you can see in the \ufb01gure, the power law degree distribution remains as a straight line in the CCDF plot too, because $$F(k)$$ will still be a power function of$$k$$, as shown below:\n\nFigure $$\\PageIndex{3}$$: Visual output of Code 17.16.\n\nF(k ) \\begin{align} \\sum_{k'=k} ^{\\infty} {P(k')} = \\sum_{k'=k} ^{infty}{ak' ^{-\\gamma}} \\label{(17.30)} \\\\ \\approx \\int_{k}^{infty} ak^{-\\gamma}dk' = {\\begin{bmatrix} \\frac{ak^{J-\\gamma+1}}{-\\gamma + 1}\\end{bmatrix}}^{\\infty}_{k} = \\frac{0-ak^{-\\gamma+1}}{-\\gamma +1} \\label{(17.31)} \\\\ = \\frac{a}{-\\gamma -1}k ^{-(\\gamma -1)} \\label{(17.32)} \\end{align}\nThis result shows that the scaling exponent of $$F(k)$$ for a power law degree distribution is less than that of the original distribution by 1, which can be visually seen by comparing their slopes between Figs. 17.5.2 and 17.5.3.\n\nExercise $$\\PageIndex{1}$$\n\nImport a large network data set of your choice from Mark Newman\u2019s Network Data website: http://www-personal.umich.edu/~mejn/netdata/\n\nPlot the degree distribution of the network, as well as its CCDF. Determine whether the network is more similar to a random, a small-world, or a scale-free network model.\n\nIf the network\u2019s degree distribution shows a power law behavior, you can estimate its scaling exponent from the distribution by simple linear regression. You should use a CCDF of the degree distribution for this purpose, because CCDFs are less noisy than the original degree distributions. Here is an example of scaling exponent estimation applied to a Barab\u00b4asi-Albert network, where the linregress function in SciPy\u2019s stats module is used for linear regression:\n\nIn the second code block, the domain and ccdf were converted to log scales for linear \ufb01tting. Also, note that the original ccdf contained values for all k\u2019s, even for those for which $$P(k) = 0$$. This would cause unnecessary biases in the linear regression toward the higher k end where actual samples were very sparse. To avoid this, only the data points where the value of $$F$$ changed (i.e., where there were actual nodes with degree k) are collected in the logkdata and logFdata lists.\n\nThe result is shown in Fig. 17.5.4, and also the following output comes out to the terminal, which indicates that this was a pretty good \ufb01t:\n\nFigure $$\\PageIndex{4}$$: Visual output of Code 17.17.\nAccording to this result, the CCDF had a negative exponent of about -1.97. Since this value corresponds to$$\u2212(\u03b3\u2212$$1$$)$$, the actual scaling exponent $$\u03b3$$ is about 2.97, which is pretty close to its theoretical value, 3.\n\nExercise $$\\PageIndex{2}$$\n\nObtain a large network data set whose degree distribution appears to follow a power law, from any source (there are tons available online, including Mark Newman\u2019s that was introduced before). Then estimate its scaling exponent using linear regression.", "date": "2019-08-24 18:56:57", "meta": {"domain": "libretexts.org", "url": "https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_Introduction_to_the_Modeling_and_Analysis_of_Complex_Systems_(Sayama)/17%3A_Dynamical_Networks_II_-_Analysis_of_Network_Topologies/17.5%3A_Degree_Distribution", "openwebmath_score": 0.8542770147323608, "openwebmath_perplexity": 941.3247233118972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9877587218253717, "lm_q2_score": 0.8652240930029118, "lm_q1q2_score": 0.8546326441970727}}
{"url": "http://math.stackexchange.com/questions/62126/proof-that-6n-always-has-a-last-digit-of-6/62133", "text": "# Proof that $6^n$ always has a last digit of $6$\n\nWithout being proficient in math at all, I have figured out, by looking at series of numbers, that $6$ in the $n$-th power always seems to end with the digit $6$.\n\nAnyone here willing to link me to a proof?\n\nI've been searching google, without luck, probably because I used the wrong keywords.\n\n-\n@Stijn: your comment seems rather obnoxious. The OP did not say that he has \"proved\" anything; he made an observation which he thinks \"seems to always\" hold. And he's right... \u2013\u00a0 Pete L. Clark Sep 6 '11 at 0:47\n@Stijn: Ragnar said \"seems\", so s/he knows it's not a proof. :) \u2013\u00a0 Guess who it is. Sep 6 '11 at 1:19\n@J.M. A p.c. \"s/he\" in combination with a person called Ragnar is making me chuckle... \u2013\u00a0 t.b. Sep 6 '11 at 1:27\n@Theo: I've been \"victimized\" by ladies using \"manly\" names on the Internet, so I'm covering myself just in case. :D \u2013\u00a0 Guess who it is. Sep 6 '11 at 1:34\nMy comment wasn't meant to be obnoxious. I'll delete it if it comes across as such. \u2013\u00a0 Stijn Sep 6 '11 at 8:40\n\nWe can prove it using mathematical induction.\n\nClaim: $6^n\\equiv 6\\bmod 10$ for all $n\\in\\mathbb{N}$ (the symbol $\\mathbb{N}$ denotes the natural numbers, and $\\bmod 10$ means we are using modular arithmetic with a modulus of 10).\n\nBase case (i.e., showing it's true for $n=1$): $$6^1\\equiv 6\\bmod 10\\qquad\\checkmark$$\n\nInduction step (i.e., showing that, if it is true for $n=k$, then it is true for $n=k+1$):\n\n$$6^k\\equiv 6\\bmod 10\\implies 6^{k+1}\\equiv 6^k\\cdot 6\\equiv6\\cdot 6\\equiv 36\\equiv 6\\bmod 10\\qquad\\qquad\\checkmark$$\n\n-\n\nIf you multiply any two integers whose last digit is 6, you get an integer whose last digit is 6: $$\\begin{array} {} & {} & {} & \\bullet & \\bullet & \\bullet & \\bullet & \\bullet & 6 \\\\ \\times & {} & {} &\\bullet & \\bullet & \\bullet & \\bullet & \\bullet & 6 \\\\ \\hline {} & \\bullet & \\bullet & \\bullet & \\bullet & \\bullet & \\bullet & \\bullet & 6 \\end{array}$$ (Get 36, and carry the \"3\", etc.)\n\nTo put it another way, if the last digit is 6, then the number is $(10\\times\\text{something}) + 6$. So \\begin{align} & {} \\qquad \\Big((10\\times\\text{something}) + 6\\Big) \\times \\Big((10\\times\\text{something}) + 6\\Big) \\\\ & = \\Big((10\\times\\text{something})\\times (10\\times\\text{something})\\Big) \\\\ & {} \\quad + \\Big((10\\times\\text{something})\\times 6\\Big) + \\Big((10\\times\\text{something})\\times 6\\Big) + 36 \\\\ & = \\Big(10\\times \\text{something}\\Big) +36 \\\\ & = \\Big(10\\times \\text{something} \\Big) + 6. \\end{align}\n\n-\n\nHINT $\\rm\\ \\ 6-1\\ |\\ 6^k-1,\\$ so $\\rm\\:\\ 2,5\\ |\\ 6^n-6\\ \\Rightarrow\\ 10\\ |\\ 6^n - 6\\:,\\$ i.e. $\\rm\\ 6^n\\ =\\ 6 + 10\\ k\\:$ for $\\rm\\:k\\in\\mathbb Z\\:.$\n\nAlternatively: $\\rm\\ mod\\ 10:\\ \\ 6^n\\equiv 6\\$ since it is $\\rm\\ 0^n \\equiv 0\\pmod 2,\\ \\ 1^n \\equiv 1\\pmod 5$\n\nSimilarly odd $\\rm\\:b\\: \\Rightarrow\\: (b+1)^n\\equiv b+1\\pmod{2\\:b}\\:,\\:$ so $\\rm\\:(b+1)^n\\:$ has last digit $\\rm\\:b+1\\:$ in radix $\\rm\\:2\\:b\\:.$\n\nNOTE how modular arithmetic reduces the induction to the trivial inductions $\\rm\\ 0^n = 0,\\ 1^n = 1\\:.$ This is a prototypical example of the sort of simplification afforded by reducing arithmetical problems to their counterparts in the simpler arithmetical rings of integers $\\rm\\:(mod\\ m)\\:.\\:$\n\n-\nIn some contexts, this would be a good way to answer this question. But given the way the question was phrased on this occasion, I wouldn't have considered it probable that this is one of those. \u2013\u00a0 Michael Hardy Sep 6 '11 at 1:34\n@Mic Surely the OP can grok the first proof. The rest requires only basic knowledge of modular arithmetic - which it seems is known to the OP given the accepted answer. Even if was not known to the OP, it is known to many other readers. The site is for all to learn - not just OP's. So I disagree. \u2013\u00a0 Bill Dubuque Sep 6 '11 at 1:58\nThanks for the answer, I understand it. Actually, I stumbled upon seeing the series, trying to prove $5 | 6^k-1$. My professor has since pointed me in the right direction, using mathematical induction. \u2013\u00a0 Ragnar123 Sep 6 '11 at 18:20\n@Ragnar Thanks for the feedback. Should anything I write be unclear, please feel free to ask further questions. I am always happy to elaborate. \u2013\u00a0 Bill Dubuque Sep 6 '11 at 18:35\n\nThis follows from the more general result that the product of two numbers ending with digit 6 also ends with digit 6. This can be proved in an elementary way: $$(10x+6)\\cdot(10y+6) = 100xy + 60x +60y + 36 = 10(10xy+6x +6y +3) + 6 = 10z+6$$\n\nOf course, avoiding all these letters is what congruences are all about.\n\n-\nOn the other hand, for the purposes of this question, I prefer this to the mod solutions. :) \u2013\u00a0 Guess who it is. Sep 6 '11 at 1:21\nBTW, the same holds for numbers ending with 1 or 5, by the same reasoning. \u2013\u00a0 lhf Sep 6 '11 at 1:51\n\n$6 \\times 6 \\equiv 6 \\pmod{10}$.\n\nOr, more elementarily put, think back to the pen-and-paper multiplication algorithm. When you multiply something by 6, the only part of the original number that can affect the last digit of the result is the last digit of the original. If you start with something that ends in 6, you get 36 for the last position, write 6 down and carry the 3. But no matter what happens after the carry, it cannot affect the final 6 that you've already produced.\n\n-", "date": "2015-08-01 22:30:12", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/62126/proof-that-6n-always-has-a-last-digit-of-6/62133", "openwebmath_score": 0.9736328721046448, "openwebmath_perplexity": 723.4563474177974, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587239874877, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8546326374859187}}
{"url": "https://www2.warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/year2/ma249/", "text": "# MA249 Algebra 2: Groups and Rings\n\nLecturer: Dmitriy Rumynin\n\nTerm(s): Term 2\n\nStatus for Mathematics students: Core for Year 2 mathematics students. It could be suitable as a usual or unusual option for non-maths students\n\nCommitment: 30 lectures.\n\nAssessment: Assignments (15%), two-hour examination (85%)\n\nPrerequisites: MA132 Foundations (MA138 Sets and Numbers for non-maths students), MA106 Linear Algebra, and MA251 Algebra I: Advanced Linear Algebra\n\nLeads To: The results of this module are used in several modules including: MA377 Rings and Modules, MA3A6 Algebraic Number Theory, MA453 Lie Algebras, MA3G6 Commutative Algebra, MA3D5 Galois Theory, MA3E1 Group and Representations, and MA3J3 Bifurcations Catastrophes and Symmetry, although unfortunately not all of these modules are offered every year.\n\nContent: This is an introductory abstract algebra module. As the title suggests, the two main objects of study are groups and rings. You already know that a group is a set with one binary operation. Examples include groups of permutations and groups of non-singular matrices. Rings are sets with two binary operations, addition and multiplication. The most notable example is the set of integers with addition and multiplication, but you will also be familiar already with rings of polynomials. We will develop the theories of groups and rings.\n\nSome of the results proved in MA242 Algebra I: Advanced Linear Algebra for abelian groups are true for groups in general. These include Lagrange's Theorem, which says that the order of a subgroup of a finite group divides the order of the group. We defined quotient groups $G/H$for abelian groups in Algebra I, but for general groups these can only be defined for certain special types of subgroups H of G, known as normal subgroups. We can then prove the isomorphism theorems for groups in general. An analogous situation occurs in rings. For certain substructures I of rings R, known as ideals, we can define the quotient ring $R/I$, and again we get corresponding isomorphism theorems.\n\nOther results to be discussed include the Orbit-Stabiliser Theorem for groups acting as permutations of finite sets, the Chinese Remainder Theorem, and Gauss' theorem on unique factorisation in polynomial rings.\n\nAims: To study abstract algebraic structures, their examples and applications.\n\nObjectives: By the end of the module the student should know several fundamental results about groups and rings as well as be able to manipulate with them.\n\nBooks:\n\nComplete lecture notes for the module will be available from the General Office soon after the beginning of the spring term, and will appear on the module resources page towards the end of term.\n\nOne possible book is\n\nNiels Lauritzen, Concrete Abstract Algebra, Cambridge University Press.\n\nRecommended Syllabus\n\nYear 1 regs and modules\nG100 G103 GL11 G1NC\n\nYear 2 regs and modules\nG100 G103 GL11 G1NC\n\nYear 3 regs and modules\nG100 G103\n\nYear 4 regs and modules\nG103\n\nPast Exams\nCore module averages", "date": "2017-11-19 04:57:00", "meta": {"domain": "ac.uk", "url": "https://www2.warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/year2/ma249/", "openwebmath_score": 0.6629374623298645, "openwebmath_perplexity": 1046.8813450362948, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587268703082, "lm_q2_score": 0.8652240808393983, "lm_q1q2_score": 0.8546326365474567}}
{"url": "https://byjus.com/question-answer/mr-dupont-is-a-professional-wine-taster-when-given-a-french-wine-he-will-identify/", "text": "Question\n\n# Mr. Dupont is a professional wine taster. When given a French wine, he will identify it with probability $$0.9$$ correctly as French and will mistake it for a Californian wine with probability $$0.1$$. When given a Californian wine, he will identify it with probability $$0.8$$ correctly as Californian and will mistake it for a French wine with probability \u00a0$$0.2$$. Suppose that Mr. Dupont is given ten unlabelled glasses of wine, three with French and seven with Californian wines. He randomly picks a glass, tries the wine and solemnly says : \"French\". The probability that the wine he tasted was Californian, is nearly equal to\n\nA\n0.14\nB\n0.24\nC\n0.34\nD\n0.44\n\nSolution\n\n## The correct option is C $$0.34$$Given,There are 7 californian wine glasses and 3 french wine glasses.The probability of selecting French wine glass, $$P(FG)=\\frac{3}{10}$$The probability of selecting California wine glass, $$P(CG)=\\frac{7}{10}$$\u00a0When given french wine,The probability of Dupont to say correctly as french wine, $$P(F)=0.9$$ When given french wine,The probability of Dupont to say wrongly as Californian wine, $$P(\\overline F)=0.1$$ When given Californian wine,The probability of Dupont to say correctly as Californian wine, $$P(F)=0.8$$ When given Californian wine,The probability of Dupont to say wrongly as french wine, $$P(\\overline C)=0.2$$$$\\therefore$$ The probability that Dupont says selected glass as French wine, $$P(A)=$$The probability of selecting french wine glass and will say $$correctly$$ as $$french$$ wine $$+$$ Probability of selecting $$californian$$ wine glass and saying $$wrongly$$ it as $$French$$ wine. $$=P(FG)*P(F)+P(CG)*P(\\overline C)$$$$=\\displaystyle\\frac{3}{10}*0.9+\\displaystyle\\frac{7}{10}*0.2=0.041$$$$\\therefore$$ The probability that Dupont says selected glass as French wine Given it as Californian=$$\\displaystyle\\frac{P(CG)*P(\\overline C)}{P(A)}=\\displaystyle\\frac{(\\frac{7}{10}*0.2)}{0.41}=0.341$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-18 10:52:15", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/mr-dupont-is-a-professional-wine-taster-when-given-a-french-wine-he-will-identify/", "openwebmath_score": 0.6559938788414001, "openwebmath_perplexity": 7188.401210617611, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9877587257892505, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8546326356120997}}
{"url": "https://math.stackexchange.com/questions/3191303/show-that-x2k1-mid-x2l-1-when-kl", "text": "# Show that $(x^{2^k}+1)\\mid (x^{2^l}-1)$, when $k<l$ [duplicate]\n\nI found this in some notes from a course in number theory. How do i work to solve this?\n\n\u2022 Write $x^{2^k}=y$. Can you show that $y+1\\mid y^2-1\\mid y^{2^{\\ell-k}}-1$? \u2013\u00a0Jyrki Lahtonen Apr 17 '19 at 17:15\n\u2022 Meaning that the question is reduced to this near duplicate. \u2013\u00a0Jyrki Lahtonen Apr 17 '19 at 17:22\n\u2022 Also, I highly recommend that you take a look at our guide for new askers. The question is a bit lacking in the context department. If you could convince us that you are not just trying to get somebody to do your homework we would feel a lot better about the question. \u2013\u00a0Jyrki Lahtonen Apr 17 '19 at 17:24\n\n$$x^{2^l}-1=(x^{2^{l-1}}+1)(x^{2^{l-1}}-1)$$ Continue to expand the right hand factor until you get $$x^{2^l}-1=(x-1)\\prod_{k=0}^{l-1} \\left(x^{2^k}+1\\right)$$ Which is obviously divisible by $$x^{2^k}+1$$ for all $$0\\le k\\lt l$$.\n$$\\bmod\\, x^{\\large 2^{\\Large K}}\\!\\!+1\\!:\\ \\ \\color{#c00}{x^{\\large 2^{\\Large K}}\\!\\!\\equiv -1}\\,\\Rightarrow\\, x^{\\large 2^{\\Large K+N}}\\!\\!\\equiv (\\color{#c00}{x^{\\large 2^{\\Large K}}})^{\\large 2^{\\Large N}}\\!\\!\\equiv (\\color{#c00}{-1})^{\\large 2^{\\Large N}}\\!\\!\\equiv 1\\$$ $$\\!\\!\\overbrace{{\\rm when} \\ \\ N> 0}^{\\large K\\, <\\, K+N\\, =:\\, L_{\\phantom{I_I}}}$$\nRemark  It's a special case of $$\\ x^{\\large K}\\!+1\\mid x^{\\large 2K}\\!-1\\mid x^{\\large 2KN}\\!-1,\\,$$ also provable by mod\n$$\\bmod\\, x^{\\large K}\\!+1\\!:\\ \\ \\color{#c00}{x^{\\large K} \\!\\!\\equiv -1}\\,\\Rightarrow\\, x^{\\large 2KN} \\!\\equiv (\\color{#c00}{x^{\\large K}})^{\\large 2N}\\!\\equiv (\\color{#c00}{-1})^{\\large 2N}\\!\\equiv 1\\$$\n\u2022 If you learn to reason by $\\!\\bmod$ as above then you don't have to remember motley divisibility formulas - they occur very naturally as special cases of general results (e.g. abobe that $\\,(-1)^{2N} = 1)\\ \\$ \u2013\u00a0Gone Apr 17 '19 at 17:40", "date": "2020-08-10 05:36:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3191303/show-that-x2k1-mid-x2l-1-when-kl", "openwebmath_score": 0.5675240159034729, "openwebmath_perplexity": 331.7757473594485, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303416461329, "lm_q2_score": 0.8633916064587, "lm_q1q2_score": 0.8546112087954185}}
{"url": "https://brilliant.org/discussions/thread/the-mysterious-fractions-f/", "text": "# The mysterious fractions\n\nWhat value do you get when you convert $\\frac {1}{81}$to decimal? You get $0.0123456790123456790...$.\n\nWhat value do you get when you convert $\\frac {1}{9801}$ to decimal? You get $0.000102030405060708091011...9697990001$.\n\nWhat value do you get when you convert $\\frac {1}{998001}$ to decimal? You get $0.000001002003...100101102...996997999000...$.\n\nThese decimals list every $n$ digit numbers (81 is 1, 9801 is 2, 998001 is 3, etc.) apart from the second last number. There is a pattern to find one of these fractions.\n\nCan you see something special about the denominators? $81$ is $9^2$, $9801$ is $99^2$, $998001$ is $999^2$.\n\nThis means that if you did $\\frac {1}{99980001}$ you would get $0.0000000100020003...9996999799990000...$.\n\nCan you find a fraction that, when converted to decimal, lists every $n$ digit number?\n\nNote by Sharky Kesa\n6\u00a0years, 7\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nWatch the numberphile video which explains this very well at :http://www.youtube.com/watch?v=daro6K6mym8\n\n- 6\u00a0years, 7\u00a0months ago\n\nThis is a result of generating functions, and plugging $x=\\frac{1}{10}$ into the function. For example, the generating function for the Fibonacci sequence is $\\frac{x}{1-x-x^2}$. If we plug in $x=\\frac{1}{10}$, we get $\\frac{1/10}{1-(1/10)-(1/10)^2}=\\frac{10}{100-10-1}=\\frac{10}{89}=0.11235\\dots$.\n\nTo answer your question, sure you can. If you want to list every $n$-digit number, you'll want to have the sum $\\sum_{i=1}^\\infty i\\times10^{-in-n}$. Recall that the generating function for $i$ is $\\frac{1}{(1-x)^2}$, so we'll have $\\frac{10^{-n}}{(1-10^{-n})^2}$.\n\n- 6\u00a0years, 7\u00a0months ago\n\nThough, for the Fibonacci sequence, note that with $x = \\frac{1}{10}$, you 'add' the tens digit to the preceding units digit, so you don't get the sequence of $0.112358132134\\ldots$, but instead $\\frac{10}{89} = 0.1123595\\ldots$. Ah, if only patterns were verified by checking the first 5 terms.\n\nHere's a spinoff question.\n\nIs $0.112358132134 \\ldots$ rational or irrational?\n\nStaff - 6\u00a0years, 7\u00a0months ago\n\nYou see the effects of carrying the digit, yes, but it seemed more magical to post the first 5 digits.\n\n- 6\u00a0years, 7\u00a0months ago\n\nIrrational. I would provide a proof,but then I would be guilty of stealing your answer from MSE.\n\n- 6\u00a0years, 7\u00a0months ago", "date": "2020-08-11 07:10:07", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/the-mysterious-fractions-f/", "openwebmath_score": 0.9803304672241211, "openwebmath_perplexity": 1379.2252638521986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9796676514063882, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8546104907565976}}
{"url": "http://ekodev3.com/d2bial/0736a6-kinetic-energy-depends-on", "text": "It\u2019s essentially the energy of an object due to its vibrational motion. Kinetic energy is the energy of motion. The amount of kinetic energy an object has depends on the mass of the object and the speed of the object. An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity. Kinetic energy review. C. only velocity. Sort by: Top Voted. Kinetic energy definition, the energy of a body or a system with respect to the motion of the body or of the particles in the system. The stopping potential depends on the kinetic energy of the electrons, which is affected only by the frequency of incoming light and not by its intensity. The kinetic energy of an object depends on two things: the objects _____ and _____. answer choices . As you increase temp of gas then K.E will also increase. Practice: Using the kinetic energy equation. Calculating kinetic energy. Kinetic Energy: energy of motion; Depends on: mass, velocity; III. Mon to Sat - 10 AM to 7 PM For any content/service related issues please contact on this number . Translational kinetic energy depends on motion through space. kinetic energy A bowling ball has kinetic energy when it is moving. Thus, the higher the mass and velocity of a body, the greater the kinetic energy attained. Ok, so the kinetic energy of an object is supposed to be proportional to the mass of the object, and to the square of the speed of the object, and speed is basically the magnitude of velocity, right? 8788563422. A spring has more potential energy when it is compressed or stretched. Kinetic energy depends on...? As such, it can be concluded that the average kinetic energy of the molecules in a thermalized sample of gas depends only on the temperature. Yes. The amount of kinetic energy stored in a body depends on its mass and its speed. 2. They do not have extent in space, the only difference between a gamma and an infrared photon is in the energy. Which point has the most kinetic energy, assuming friction and air resistance are negligible? In the photoelectric effect, explain why the stopping potential depends on the frequency of the light but not on the intensity. For more such videos please go to http://vidhyasangam.com/Videos.html Height, mass and gravitational strength. This question is off-topic. The kinetic energy is then 1 2 m i v i 2, and summation gives the total kinetic energy of the body: E kin = 1 2 m 1 v 1 2 + 1 2 m 1 v 2 2 + \u22ef = 1 2 \u03a9 2 ( m 1 r 1 2 + m 2 r 2 2 + \u22ef ) . Friction and movement. D. mass and velocity. Using the kinetic energy equation. Kinetic energy depends on the mass and velocity of the body in motion, with the velocity contributing more to the overall kinetic energy of the body. If you count the reduction in kinetic energy of the train together with the increase in kinetic energy of the ball, the sum is the same regardless of what reference frame you choose. Our mission is to provide a free, world-class education to anyone, anywhere. 5,000 J. solve : \u2026 Kinetic energy is one of several types of energy that an object can possess. Using the kinetic energy equation. K.E is directly proportional to temp. C. Height. Does the kinetic energy change depending on the direction of the velocity? Only velocity. B. c. D. Tags: Question 12 . $\\begingroup$ @user248881 In the standard model of particle physics all photons are point particles, i,e, have no size. It is given as K E = a s 2, where a is a constant. kinetic energy synonyms, kinetic energy pronunciation, kinetic energy translation, English dictionary definition of kinetic energy. The kinetic energy of a moving object can be calculated using the equation: Kinetic energy = $$\\frac{1}{2}$$ x mass x (speed) 2. 2 See answers Nonportrit Nonportrit The amount of kinetic energy an object has depends on its \"speed.\" Only mass. You are very important to us. Linear Kinetic energy of a object is dependent upon the linear speed of an object and its mass( we can use the scalar term \u201cspeed\u201d as energy is a scalar quantity and is independent of the direction of motion of body). Kinetic energy depends on A. only mass. When the particles are moving very fast, we feel the substance and say \"That's hot!\". The equation for kinetic energy is *KE = \u00bdmv**2*, so the greater the mass and the greater the velocity, the greater the kinetic energy. are solved by group of students and teacher of Class 11, which is also the largest student community of Class 11. The E=h\u03bd. Kinetic energy is energy a object has while its in motion so the key word is speed. Thus kinetic energy of the emitted photoelectrons depends on wavelength, frequency of the incident photon and work function of the metal but does not depend on the intensity. Relevance. Mass is weight because of how much weight a object has would decrease the speed. A. The Questions and Answers of The kinetic energy 'K' of a particle moving in a straight line depends up on the distance 's' as K=as square, force acting on the particle is ??? While kinetic energy is not an invariant in classical mechanics, the gain or loss in kinetic energy due to internal forces within a system is an invariant. The Planck constant transforms energy to a number that miraculously is the frequency of the light that will be built up by zillion such photons. This is because the temperature of a substance depends on the kinetic energy of the particles. Examples of Kinetic Energy: 1. Active 4 years, 10 months ago. Potential energy, stored energy that depends upon the relative position of various parts of a system. 2 $\\begingroup$ Closed. D. Mass and velocity. It is the energy stored in a moving body. The unit for energy is joules. Definition of potential energy, gravitational potential energy, elastic potential energy and other forms of potential energy. The force acting on the particle is [closed] Ask Question Asked 4 years, 10 months ago. B. height. Kinetic energy is the energy of motion. Unlike forces, energy is a scalar, so direction doesn\u2019t matter.-1*xample%\"& A 1250 kg car moves with at a speed of 25 m/s. \u2026 Hope this helps! It is not currently accepting answers. *kinetic energy depends on an objects mass and speed . Which has more kinetic energy, the regular car or the race car? 2 Answers. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered. Kinetic Energy (K.E) = (\u00bd)mv 2. m= Mass in Kilograms Kinetic Energy. A. For example, if all the particles in a system have the same velocity, the system is undergoing translational motion and has translational kinetic energy. The kinetic energy of an object depends on both its mass and velocity, with its velocity playing a much greater role. Up Next. The kinetic energy of a particle is a single quantity, but the kinetic energy of a system of particles can sometimes be divided into various types, depending on the system and its motion. answer choices . A 90 kg man climbs 9.47 m up a rope. III. See more. Kinetic Energy. Which would hurt more, getting hit by a 15 foot wave or a 2 foot wave? Define kinetic energy. The sum in the parentheses depends on the rigid body concerned (its size, shape and mass distribution) and on \u2026 SURVEY . A golf club striking a golf ball is a great example of this form of energy. An object\u2019s kinetic energy (*KE*) depends on its mass (*m*) and velocity (*v*). The amount of kinetic energy an object has depends on its. Particles which have more kinetic energy will move faster than particles which have less kinetic energy. Kinetic energy depends on an objects density and speed . (T) only. The race car has more KE because it has a higher velocity. I also have to put, once again, why I chose the answer I choose. Next lesson. Q. Gravitational potential energy depends on the _____ and _____ of the object. What is the kinetic energy of the car? The amount of kinetic energy that it possesses depends on how much mass is moving and how fast \u2026 Ans; 2as? So simple explanations are helpful. Kinetic energy is one type of energy store. What is his gravitational potential energy? correct the following statement : The equation used to calculate the kinetic energy is KE = 2 mv2 . In my book given that, Average K.E is independent of pressure, Volume or nature of the gas. If an object is moving, then it possesses kinetic energy. B. Kinetic energy. Viewed 4k times 1. Work-energy theorem. Vibrational Kinetic Energy Vibrational kinetic energy is, unsurprisingly, caused by objects vibrating. Answer Save. * KE = 1 mv 2 ____ 2. solve : mass = 100 kg velocity = 10 m/s. Average kinetic energy of a molecule is = (3/2)kT Avg. 2. It is also important to recognize that the most probable, average, and RMS kinetic energy terms that can be derived from the Kinetic Molecular Theory do not depend on the mass of the molecules (Table 2.4.1). CHRIS Q. Lv \u2026 Solution: Above the threshold frequency, the maximum kinetic energy of the emitted photoelectrons depends on the frequency of the incident light, but is independent of the intensity of the incident light so long as the latter is not too high . You Try%\"& 1. Is compressed or stretched why the stopping potential depends on an kinetic energy depends on and! Any content/service related issues please contact on this number a circle of radius R depends on its mass its... Then it possesses kinetic energy a bowling ball has kinetic energy stored in a moving body pronunciation, energy... Of pressure, Volume or nature of the particles the amount of energy. Has a higher velocity space, kinetic energy depends on regular car or the race car has more KE because has! In motion so the key word is speed. \u2026 potential energy, friction... Of potential energy, the only difference between a gamma and an infrared photon is in the photoelectric,... Again, why I chose the answer I choose energy is, unsurprisingly, caused by objects.... K.E will also increase both its mass and speed. energy that depends upon relative! 2, where a is a constant Question Asked 4 years, 10 months.! Possesses kinetic energy is KE = 1 mv 2 ____ 2. solve mass! In a body, the higher the mass and velocity, with its velocity playing a much greater.... The _____ and _____ of the object, stored energy that an object to! 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Of gas then K.E will also increase contact on this number its vibrational motion a golf ball a...", "date": "2021-05-06 22:44:17", "meta": {"domain": "ekodev3.com", "url": "http://ekodev3.com/d2bial/0736a6-kinetic-energy-depends-on", "openwebmath_score": 0.43632879853248596, "openwebmath_perplexity": 508.16888157696775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9304582516374121, "lm_q2_score": 0.9184802406781396, "lm_q1q2_score": 0.8546075189048913}}
{"url": "http://math.stackexchange.com/tags/exponentiation/hot", "text": "# Tag Info\n\n54\n\nWe just have to show that $a^{a-b} \\ge b^{a-b}$. This is equivalent to $(\\frac{a}{b})^{a-b} \\ge 1$. If $a \\ge b$, then $\\frac{a}{b} \\ge 1$, Also $a-b \\ge 0$. A number greater than $1$ raised to a positive exponent is clearly greater than $1$. If $a \\le b$, then $\\frac{a}{b}\\leq 1$. $a-b\\leq 0$. A positive number less than $1$ raised to a negative ...\n\n43\n\nRaise them both to the power of $6$. Since they are both positive, their order will be preserved and you will get: $$\\left({\\dfrac{1}{2}}\\right)^2=\\frac{1}{4} > \\frac{1}{27}=\\left({\\dfrac{1}{3}}\\right)^3$$\n\n39\n\nNo need to do any calculations at all: since we are talking about numbers between $0$ and $1$, a cube root is larger than a square root: $$\\Bigl(\\frac12\\Bigr)^{1/3}>\\Bigl(\\frac12\\Bigr)^{1/2}>\\Bigl(\\frac13\\Bigr)^{1/2}\\ .$$\n\n25\n\n$$\\log(a^a b^b)=a \\log a + b \\log b$$ $$\\log(a^b b^a)=a \\log b + b \\log a$$ Thus, by the rearrangement inequality, because $\\log$ is strictly increasing, $$\\log(a^a b^b)\\geq \\log(a^b b^a)$$ Similarly, because $\\log$ is strictly increasing, $$a^a b^b \\geq a^b b^a.$$\n\n14\n\nFor any $x \\in \\mathbb{R}$, $2^x > 0$ and $4^x > 0$, therefore $$2^x + 4^x + 12 > 0 + 0 + 12 = 12 > 0$$ Therefore there is no real solution.\n\n12\n\nMy favorite paper about $x^x$ is The $x^x$ Spindle, which appeared in Mathematics Magazine back in 1996. The main idea is to visualize the fact that we can write it as $$x^x = e^{x (\\ln(x)+2k\\pi i)}.$$ Note that for each choice of $k$, we get a different branch of the logarithm. Given any real number $x$, most of these branches will be complex valued. ...\n\n10\n\n$$n^{\\frac{1}{n}} \\leq (n+9)^{\\frac{1}{n}} \\leq (2n)^\\frac{1}{n}$$ For $n$ greater than, say, $9$. Apply \"squeeze theorem\"\n\n8\n\n$$\\left(\\frac ab\\right)^{a-b}-1=\\frac{a^ab^b-a^bb^a}{b^aa^b}$$ If $a=b, \\left(\\dfrac ab\\right)^{a-b}=1$ Else if $a>b;\\dfrac ab>1$ and $a-b>0\\implies \\left(\\dfrac ab\\right)^{a-b}>1$ Similarly if $a<b$\n\n6\n\nThere are a few things going on. Jump to the end for discussion of complex numbers and what exactly WolframAlpha is plotting. When you plot the function, WolframAlpha and most plotting systems don't restrict themselves to special inputs like \"only rational numbers with odd denominator.\" Even if you try using an odd denominator, most computational software ...\n\n5\n\nWhen is $x^y > y^x$ ? When $x^{1/x} > y^{1/y}$. Let's look at the function $x^{1/x}$. Differentiating, we find it has a maximum at $x=e$. Since $1/2$ and $1/3$ are both less than $e$, the one that's nearer wins. So $(1/2)^2 > (1/3)^3$, so $(1/2)^{1/3} > (1/3)^{1/2}$. But more to the point, this shows that $e^\\pi > \\pi^e$, which might be a lot ...\n\n5\n\nGiven that both $a$ and $b$ are positive integers, let us consider the case where $b > a$. $b$ can be expressed as $a+x$, where $x$ is some positive integer. to prove $a^a b^b > a^b b^a$,we need to prove that $a^a b^b - a^b b^a > 0$ Rewrite $a^a b^b - a^b b^a$, by substituting $b = (a+x)$ $= a^a (a+x)^{a+x} - a^{a+x} (a+x)^a$ ...\n\n5\n\nThis inequality is equivalent to $a\\ln a+b\\ln b\\geq a\\ln b+b\\ln a$, which is obvious by Rearrangement or it's $(a-b)(\\ln a-\\ln b)\\geq$, which is a proof of Rearrangement.\n\n5\n\nlet $u = 2^x$, then $4^x = (2^2)^x = 2^{2x} = 2^{2x} = (2^x)^2 = u^2$. Thus, $2^x + 4^x + 12 = 0$ becomes, $$u + u^2 + 12 = 0$$ Using the quadratic equation will solve $u$, which is really $2^x$. To solve for the $x$, just take $\\log$ on both sides of the solution, then after rearranging, you should be able to solve for $x$.\n\n5\n\nYou could try the binomial expansion of $(1+0.05)^{10}$ and stop calculating terms after they become small enough to not affect your required degree of accuracy\n\n4\n\nOf the several different but related definitions for exponentiation, the one that accepts a rational exponent should be phrased to exclude negative bases, in order to avoid exactly this problem. In fact, the only one of the usually encountered definitions that give meaning to a negative integer to a non-integral power is the one for complex numbers, which ...\n\n4\n\nIf you don't want to depend on the \"trick\" of raising to the sixth power, you can compare the logs: $\\frac 13 \\log \\frac 12=\\frac {- \\log 2}3$ and $\\frac 12 \\log \\frac 13=\\frac {-\\log 3}2$ Now $\\frac 12 \\gt \\frac 13$ and $\\log 3 \\gt \\log 2$, so $\\frac {\\log 3}2 \\gt \\frac {\\log 2}3, \\frac {-\\log 3}2 \\lt \\frac {-\\log 2}3,\\left(\\frac{1}{3}\\right)^{\\frac{1}{2}} ... 4 The following is a variation of the visualization of the function$x^xthat I described in this answer. It's not clear to me how to explain it to middle school kids, though. Specifically, if you want to explain the WolframAlpha output to middle schoolers, then they've got to know that $$(-2)^x = e^{x\\log(-2)} = e^{x(\\log(2) + i\\pi)} = 2^x e^{xi\\pi} = ... 3 ((\\frac{1}{2})^{\\frac{1}{3}})^6=(\\frac{1}{2})^2=\\frac{1}{4} ((\\frac{1}{3})^{\\frac{1}{2}})^6=(\\frac{1}{3})^3=\\frac{1}{27} So as it is obvious from the above relations, ((\\frac{1}{2})^{\\frac{1}{3}})^6>((\\frac{1}{3})^{\\frac{1}{2}})^6, so we can say (\\frac{1}{2})^{\\frac{1}{3}}>(\\frac{1}{3})^{\\frac{1}{2}} 3$$\\left(\\frac{1}{2}\\right)^{\\frac{1}{3}}=\\frac{\\sqrt[3]1}{\\sqrt[3]2}=\\frac1{\\sqrt[3]2}\\left(\\frac{1}{3}\\right)^{\\frac{1}{2}}=\\frac{\\sqrt1}{\\sqrt3}=\\frac1{\\sqrt3}$$Now it is obvious that$$\\sqrt[3]2<\\sqrt3$$Thus$$\\frac1{\\sqrt[3]2}>\\frac1{\\sqrt3}$$3 Let's write$$ \\begin{align} &\\log\\log n = L_2(n), \\\\ &\\log\\log\\log n = L_3(n), \\\\ &\\log\\log\\log\\log n = L_4(n). \\end{align} $$Then from x^{x^x} = n we get$$ x\\log x + L_2(x) = L_2(n), \\tag{1} $$and so, since x \\to \\infty as n \\to \\infty, we have$$ x\\log x \\sim L_2(n) \\tag{2} $$as n \\to \\infty. Taking logs of this yields$$ ... 3 $$a^a \\ b^b \\;?\\; a^b \\ b^a \\\\ \\frac{a^a}{b^a} \\;?\\; \\frac{a^b}{b^b} \\\\ \\left(\\frac{a}{b}\\right)^a \\;?\\; \\left(\\frac{a}{b}\\right)^b \\\\ \\left(\\frac{a}{b}\\right)^{a-b} \\;?\\; 1$$ ifa \\ge b$, then$c = \\frac{a}{b} \\ge 1$, and$d = a-b \\ge 0$. Thus$c^d \\ge 1$, so$?$is$\\ge$. if$a < b\\$, then: $$\\left(\\frac{a}{b}\\right)^{a-b} \\\\ = ... 3 Hint: Suppose that \\sqrt{1}+\\sqrt{2} is of the shape c^r, where c is a positive rational and r=\\frac{m}{n} where m and n are integers, with n\\gt 0. Show by taking the n-th power of both sides that this implies that \\sqrt{2} is rational. 3 You have 2 terms of 2^{n+1}, meaning you have$$2\\cdot 2^{n+1} -1 = 2^{(n+1)+1} - 1 = 2^{n+2}-1$$3 By the difference of perfect squares.$$\\large(a+b)(a-b) = (a+b)a - (a+b)b = a^2+ab-ab-b^2=a^2-b^2$$We just have to let a = 2^6 and b=2^3. 2 See OEIS sequence A074981 and references there. 10 does have a solution as 13^3-3^7, but apparently no solutions are known for 6 and 14. 2 I don't have an answer to your question, but I did search through quite a few set theory books this morning and I made notes of what I found in case you or others are interested. The topic seems less covered in books than I expected, and I suspect you'll have to consult journal articles to find much of significance (unless you can read Hessenberg's and ... 2 Note that$$2^{n+1}+2^{n+1}=2\\cdot 2^{n+1}=2^1\\cdot 2^{n+1}=2^{n+1+1}=2^{n+2}.$$2 Take \\log:$$ \\log L = \\lim_{n\\to\\infty}\\log(n+9)^{\\frac{1}{n}} = \\lim_{n\\to\\infty}\\frac{\\log(n+9)}n = \\cdots $$2 Let$$f(t)=\\left(\\frac{b^{t+1}-a^{t+1}}{b-a}\\right)^{1/t} =\\{g(t)\\}^{1/t}$$and if \\lim_{t\\to 0}f(t)=L, then$$\\begin{aligned}\\log L &= \\log\\left\\{\\lim_{t \\to 0}\\left(\\frac{b^{t+1}-a^{t+1}}{b-a}\\right)^{1/t}\\right\\}\\\\ &= \\lim_{t \\to 0}\\,\\log\\left(\\frac{b^{t+1}-a^{t+1}}{b-a}\\right)^{1/t}\\text{ (by continuity of log)}\\\\ &= \\lim_{t \\to ...\n\nOnly top voted, non community-wiki answers of a minimum length are eligible", "date": "2014-12-20 11:32:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/tags/exponentiation/hot", "openwebmath_score": 0.9898709058761597, "openwebmath_perplexity": 342.38327436318394, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138190064204, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8545974029245538}}
{"url": "https://math.stackexchange.com/questions/2220058/solve-sin-x-frac12", "text": "Solve $\\sin x = \\frac{1}{2}$\n\nTo solve this I did $x = \\arcsin(\\frac{1}{2}) = \\frac{\\pi}{6}$\n\nHowever, my book states that the solution is $$x = \\frac{\\pi}{6}+k2\\pi \\lor x = \\frac{5\\pi}{6}+k2\\pi, k \\in \\mathbb{Z}$$\n\nWhat did I do wrong?\n\n\u2022 well, arcsin always gives you only one solution, but of course sin is periodic so there's a lot more solutions than the one the arcsin gives you.. as a matter of fact, the arcsin always gives you the unique solution between $-\\pi /2$ and $\\pi/2$ \u2013\u00a0Sebastian Schulz Apr 6 '17 at 0:18\n\nThe arcsine function $$\\arcsin x: [-1, 1] \\to \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right]$$ is defined by $$\\arcsin x = y \\iff \\sin y = x, y \\in \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right]$$ By finding $$x = \\arcsin\\left(\\frac{1}{2}\\right) = \\frac{\\pi}{6}$$ you found the unique value of $x$ in the interval $[-\\pi/2, \\pi/2]$ such that $\\sin x = 1/2$. However, the solution set of the equation $$\\sin x = \\frac{1}{2}$$ is the set of all angles that have sine $1/2$.\n\nAn angle in standard position (vertex at the origin, initial side on the positive $x$-axis) has sine $1/2$ if the terminal side of the angle intersects the unit circle at a point that has $y$-coordinate $1/2$. You have shown that one such angle is $\\pi/6$. To find the others, consider the following diagram.\n\nTwo angles have the same sine if the $y$-coordinates of the terminal side of the angle are equal. Thus, by symmetry, $\\sin(\\pi - \\theta) = \\sin\\theta$. Also, coterminal angles have the same sine since they intersect the unit circle at the same point. Hence, $$\\sin\\theta = \\sin\\varphi$$ if $$\\varphi = \\theta + 2k\\pi, k \\in \\mathbb{Z}$$ or $$\\varphi = \\pi - \\theta + 2k\\pi, k \\in \\mathbb{Z}$$ We wish to solve the equation $$\\sin x = \\frac{1}{2}$$ Since a particular solution is $$x = \\arcsin\\left(\\frac{1}{2}\\right) = \\frac{\\pi}{6}$$ we need to find all values of $x$ that satisfy the equation $$\\sin x = \\sin\\left(\\frac{\\pi}{6}\\right)$$ Using the formulas above for the solution of the equation $\\sin\\theta = \\sin\\varphi$, we obtain $$x = \\frac{\\pi}{6} + 2k\\pi, k \\in \\mathbb{Z}$$ or \\begin{align*} x & = \\pi - \\frac{\\pi}{6} + 2k\\pi, k \\in \\mathbb{Z}\\\\ & = \\frac{5\\pi}{6} + 2k\\pi, k \\in \\mathbb{Z} \\end{align*}\n\nSo you are right that a solution to $sin(x)=\\frac{1}{2}$ is $x=\\frac{\\pi}{6}$, however, note that this is far from the only solution. If we look at the geometric definition of the $sin(x)$ function, we see that there are many values of $x$ for which the y-value of the unit circle are 1. The first is $x=\\frac{\\pi}{6}$, however, that is only the coordinate in the first quadrant. There is also a coordinate $(\\frac{-\\sqrt3}{2},\\frac{1}{2})$ which will come when $x=\\frac{5\\pi}{6}$. Now also note that a rotation of $2\\pi$ will get you back to the same value, and so the solutions are not only $x=\\frac{\\pi}{6}$ and $x=\\frac{5\\pi}{6}$, but $x=\\frac{\\pi}{6}+2\\pi k$ and $x=\\frac{5\\pi}{6}+2\\pi k, k \\in \\mathbb{Z}$.\n\n\u2022 So, for every equation of this kind there are always two solutions? \u2013\u00a0Mark Read Apr 6 '17 at 0:23\n\u2022 For essentially all equations with $sin(x)=a$ for some value a, yes. The only exceptions are the maximum and minimum values of $f(x)=sin(x)$, 1 and -1. Have a look at this. You can see that within any one period of the sin(x) function there will be 2 points of for a non-zero value between 0 and 1 (as shown with $y=\\frac{1}{2}$). The only exceptions are y=-1, y=1 (as they are extremities) and y=0 (though I'm not completely sure if that counts as it passes through three times if the domain is 0 to 2pi inclusive). \u2013\u00a0Cameron Eggins Apr 6 '17 at 0:38\n\nHint:\n\nUse the properties: $\\;\\sin(\\pi-x)=\\sin x$ and $\\sin x$ has period $2\\pi$. There results the equation $\\sin x=\\sin \\theta$ has as solutions: $$\\begin{cases} x\\equiv\\theta\\\\x\\equiv\\pi-\\theta \\end{cases}\\mod2\\pi$$", "date": "2019-11-19 13:58:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2220058/solve-sin-x-frac12", "openwebmath_score": 0.9675387740135193, "openwebmath_perplexity": 99.11042745362886, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138105645059, "lm_q2_score": 0.874077222043951, "lm_q1q2_score": 0.854597371492229}}
{"url": "https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-64.html", "text": "CSIR JUNE 2011 PART C QUESTION 64 SOLUTION\n\nLet $X$ denotes the two-point set $\\{0,1\\}$ and write $X_j = \\{0,1\\}$ for every $j=0,1,2,\\dots$. Let $Y = \\prod\\limits_{j=1}^{\\infty} X_j$. Which of the following is/are true?\n1) $Y$ is a countable set,\n2) $\\text{Card} \\,Y = \\text{Card} \\,[0,1]$,\n3) $\\cup_{n=1}^{\\infty}(\\prod\\limits_{j=1}^nX_j)$ is uncountable,\n4) $Y$ is uncountable.\nSolution: option 1: (False) option 4: (True) Let $x$ be an element of $Y$, then $x = (x_1,x_2,x_3,\\dots)$ an infinite tuple with entries $0$ and $1$. This can be identified with a $0-1$ sequence $\\{x_n\\}$ naturally. This defines a bijection between the set $Y$ and the set of all $0-1$ sequences (which is denoted by $\\{0,1\\}^{\\Bbb N}$)\nClaim: $\\{0,1\\}^{\\Bbb N}$ is uncountable where $Y^X$ is the notation for the set of all functions from $X$ to $Y$.\nSuppose this set is countable, then we can enumerate its elements and $\\{0,1\\}^{\\Bbb N} = \\{\\overline x_1, \\overline x_2, \\overline x_3 \\dots,\\}$ where $\\overline x_i$s\u00a0are $0-1$ sequences. We will use Cantor's diagonalization argument to get a contradiction. We will construct a $0-1$ sequence $\\overline y$ which is not listed above. This will prove that the above enumeration is not complete and hence the set $\\{0,1\\}^{\\Bbb N}$ is uncountable. Define the ith term of the sequence $\\overline y$ to be $\\begin{cases}0 \\text{ if the$i$th term of$\\overline x_i$is 1}\\\\ 1 \\text{ otherwise }\\end{cases}$. Now, this $0-1$ sequence $\\overline y$ is different from the $\\overline x_i$ in the $i$th position. So $\\overline y \\ne \\overline x_i$ for any $i$ and hence $\\overline y \\notin \\{0,1\\}^{\\Bbb N}$. Contradiction.\nThis shows that $Y$ is uncountable.\noption 2: (True). Consider the binary representation of any element of $[0,1]$, this will be of the form $0.a_1a_2a_3\\dots$ where $a_i$s are either $0$ or $1$ which can be identified with a $0-1$ sequence. This gives the required bijection.\noption 3: (False) Finite product of countable set is countable and a countable union of countable sets is countable. This shows that the set given in option 3 is countable(Note that an infinite product of countable set is uncountable. Even infinite product of two-element set is uncountable which is proved in option 1)\n\nNBHM 2020 PART A Question 4 Solution $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$\nEvaluate : $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx = \\int_{-\\infty}^{\\inft...", "date": "2022-09-25 02:29:41", "meta": {"domain": "theindianmathematician.com", "url": "https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-64.html", "openwebmath_score": 0.9823499917984009, "openwebmath_perplexity": 128.2309636066876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380082333993, "lm_q2_score": 0.8615382165412808, "lm_q1q2_score": 0.8545925025329132}}
{"url": "http://math.stackexchange.com/questions/128405/partial-order-relations-and-total-order-relations", "text": "# Partial Order Relations and Total Order Relations\n\nQuestion:\nA={1,2,3}\n1) How many partial order relations can be induced over A ?\n2) How many total order relations can be induced over A ?\n3) Does A exist a transitive relation?\n\nI guess total order relations over A is P(3,3)=3!=6, but I don't know how to count partial order relations over A.\n\nAny help will be greatly appreciated!\n\n[UPDATE]\nAccording to Scott's reply, I get the following result,\nType 1: 3!=6\nType 2: 3\nType 3: 3\nType 4: 3\nType 5: 3!=6\n\nSo there are 21(6+3+3+3+6) partial orders on A.\nIs it right? I hope someone can check it.thanks.\n[UPDATE2]\nThanks for Henry's help.\n\nType 4:6\nType 5:1\nSo there are 19(6+3+3+6+1) partial orders on A.\nI am not sure I am right but I hope to get corrected.\n\n3)Does A exist a transitive relation?\nHowever, I don't understand why <3,3> is included in the relation.\n\n-\nPlease show what you've tried so far so we can better help you. \u2013\u00a0Austin Mohr Apr 5 '12 at 16:00\nYou have types 1, 2 and 3 correctly counted but there are not three of type 4 (there are more), nor six of type 5 (there are fewer). \u2013\u00a0Henry Apr 6 '12 at 8:37\nHi @Matt. You can upvote and/or accept helpful answers. See here how to do it: meta.math.stackexchange.com/questions/3286/\u2026 I thought the answer below was quite helpful so I upvoted it. \u2013\u00a0Rudy the Reindeer Apr 6 '12 at 15:05\nYour updated counts for types for types (4) and (5) are correct. In type (4) there are $3$ ways to choose the loner, and then $2$ ways to order the other two elements, for a total of $3\\cdot 2=6$. In type (5) it doesn\u2019t matter how you label the three elements, it\u2019s still the same order: the only ordered pairs in it are $\\langle 1,1\\rangle,\\langle 2,2\\rangle$, and $\\langle 3,3\\rangle$. \u2013\u00a0Brian M. Scott Apr 6 '12 at 18:33\n\nCounting the partial order relations is a bit messy. Perhaps the most straightforward way is to organize them by their \u2018shapes\u2019. In the following diagrams the order is from bottom to top.\n\n1. They can be linear:\n\n *\n|\n*\n|\n*\n\n2. They have a minimum element but no maximum:\n\n *   *\n\\ /\n*\n\n3. They can have a maximum but no minimum:\n\n   *\n/ \\\n*   *\n\n4. They can have two related elements and one unrelated element:\n\n *\n|   *\n*\n\n5. They can have three unrelated elements:\n\n *   *   *\n\n\nYou\u2019ve already counted the partial orders that are linear: there are indeed $3!=6$ of them. Now you just have to count the partial orders of types (2)-(5) above. To get you started, in type (2) any one of the three elements of $A$ can be the minimum element. Once you\u2019ve chosen that, however, the partial order is completely determined:\n\n    1   2                  2   1\n\\ /        and         \\ /\n3                      3\n\n\nare the same partial order, just drawn differently. Thus, there are $3$ partial orders of type (2) on $A$.\n\n-", "date": "2015-11-30 03:04:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/128405/partial-order-relations-and-total-order-relations", "openwebmath_score": 0.7986985445022583, "openwebmath_perplexity": 396.9805351110838, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380089483864, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8545924960959794}}
{"url": "http://insd.chiavette-usb-personalizzate.it/contradiction-truth-table.html", "text": "Such statements are called tautologies. \u2022 Statement Variable - a variable that represents any proposition (by convention we use lower-case letters 'p', 'q', 'r', 's', etc. notebook 9 September 30, 2015 KEY CONCEPTS A self\u00adcontradiction is a compound statement that is always FALSE. Working backward, attempt to avoid a contradiction as you derive the truth values of the separate components 3. That is, a statement is something that has a truth To make a truth table, start with columns corresponding to the most basic statements (usually represented by letters). Truth Tables - Tautology and Contradiction. A tautology is a compound statement that is true for all possibilities in a truth table i. Tautology and Contradiction ! A tautology is a compound proposition that is always true. The first way follows from the truth table definition of conjunction and implication: (P and \u00acP) is false. , it is true in all worlds, e. Every proposition is assumed to be either true or false and the truth or falsity of each proposition is said to be its truth-value. \" (2) Either p. Rather than constructing the entire truth table, we can simply check whether it is possible for the proposition to be false, and then check whether it is possible for the proposition to be true. A TT-contingent sentence comes out true on at least one row of its truth-table and false on at least one row. A truth table column which consists entirely of T's indicates a situation where the proposition is true no matter whether the individual propositions of which it is composed are true or false. If you know how to make a truth-table, great: you're almost there! For every statement that you work out on a truth-table, there are three possible outcomes: The statement is True in all rows. Show 39 related questions 18M. In this post, I will briefly discuss tautologies and contradictions in symbolic logic. Hence, it is a TT-contradiction. Compare truth tables for logic forms of two statements: 1. A truth table is a mathematical table used to determine if a compound statement is true or false. The amsthm package provides three predefined theorem styles: plain, definition and remark. First, write the argument as you would if you were going to do a full table. A compound proposition that is always true (no matter what the truth values of the propositions that occur in it), is called a tautology. The first case agrees with all combinations of truth. The Lord of Non-Contradiction: An Argument for God from Logic James N. p q _ TTT TFT FTT FFF In this module we will often use truth tables. In fact, the laws of logic stated in Section 3. Neff, 2018 1 Truth Tables Please do the following exercises individually. Note that if you claim that a proposition is a tautology, then you must argue( by using truth tables or otherwise) that it is true for every assignment of truth values to the propositional variables; if you claim that it is false for every assignment of truth values to the propositional variables; and if. ;) \u2013 user20153 Sep 18 '16 at 23:43. Use truth tables to explain why. This is an interesting option to consider, but then we might need to consider why the method of constructing truth tables tells us that the law of excluded middle holds, if it actually doesn\u2019t. Is q implies p true for this row? Does true imply true? Yeah. Truth Table Description. Logically Equivalent Statements, Tautologies, & Contradictions; Definition 27. No matter what the individual parts are, the result is a true statement; a tautology is always true. But this is too complex even for modern computers for large problems. The rate of growth in a truth table rows as a functions of the number of propositions is shown in the table below. When doing truth tables, a result can occur called a tautology. ! A contingency is neither a tautology nor a contradiction. Exam 1 Answers: Logic and Proof September 17, 2012 Instructions: Please answer each question completely, and show all of your work. When you define a new theorem-like environment with ewtheorem, it is given the style currently in effect. These are answers from the 12th edition of Hurley. That is, a statement and its negation can never have the same truth value. Truth Table. A\u2192 (B \u2192 A) b. Complete the truth table shown elow. A propositional form that is false in all rows of its truth table is a contradiction. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. We start by listing all the possible combinations for p and q: Finally, we use the disjunction rule on. Truth Table Calculator,propositions,conjunction,disjunction,negation,logical equivalence. We start by listing all the possible truth value combinations for A, B, and C. 1) And two De-Morgan rules: (1. That is, the propositions having nothing but 0s i. Question 1. 3 Learn the basic rules of natural deduction, including rules of. The propositional calculus, as it is also known, is a staple of first-year university logic courses. A proposition that is neither a tautology nor a contradiction is From the following truth table may use truth tables or properties of logical equivalences). By Using Truth Table Un 1. The naive, and intuitively correct, axioms of set theory are the Comprehension Schema and Extensionality Principle:. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. If there are. \u2022 The truth table for a compound proposition: table with entries (rows) for all possible combinations of truth values of elementary propositions. The expression \"p or not p\" is true under any circumstance, so it is a tautology. Assume x is a particular real number. \u00ac \u2227\u00ac \u2192 ( \u2227\u00ac )\u2194 ( \u2192 ) b. Now, Number of rows in the truth table will always be equal to the total numbers of distinct combination of truth values of boolean variables (i. Each entry in the 3rd column of the truth table has 2 possible values (T/F). We can use a mathematical function to calculate that n elemental propositions produce L(n) groups of truth values. Expert Answer. Use the truth tables method to determine whether p!(q^:q) and :pare logically equivalent. The rate of growth in a truth table rows as a functions of the number of propositions is shown in the table below. You can put this solution on YOUR website! Construct a truth table for q -> ~p Rule for the conditional -> : If the truth value of what is on the left of -> is T and the truth value of what is on the right of -> is F, then the truth value of -> is F. A tautology in math (and logic) is a compound statement (premise and conclusion) that always produces truth. Moving around an indirect table on the computer screen is very much like moving around a regular table except that a truth value entered from the keyboard may be erased by placing the cursor on it and pressing delete or the spacebar. Truth tables, equivalences, and proof by contradiction We use the word \\statement\" interchangeably with the word \\sentence\", and we agree that a statement can be true or false or neither, but a statement cannot be simultaneously true and false. if you are necessarily led into a contradiction, the argument is valid. Logical Symbols are used to connect to simple statements, to define a compound statement and this process is called as logical operations. Show that q is false. This is an interesting option to consider, but then we might need to consider why the method of constructing truth tables tells us that the law of excluded middle holds, if it actually doesn\u2019t. Argument Form: \u00acP \u2192 F0 where F0 is a contradiction. What is a tautology? Please provide an example of a resulting truth table that yields a tautology. Typically, the writer will skip to this combination (assume P is false and Q is true) and derive his contradiction from those two statements and then stops. Prove the following statement by contradiction: There is no integer solution to the equation x 2 \u2013 5 = 0. A contradiction is a statement that is always false. Truth Tables for Conditional and Biconditional Statements Construct a truth table for a conditional statement and determine its truth value Construct a truth table for a biconditional statement and determine its truth value Self-Contradictions, Tautologies, and Implications Identify self-contradictions, tautologies, and implications. This is an interesting option to consider, but then we might need to consider why the method of constructing truth tables tells us that the law of excluded middle holds, if it actually doesn't. Why, then, O brawling love! O loving hate! O anything, of nothing first create! O heavy lightness! Serious vanity! Mis-shapen chaos of well-seeming forms!. Notice that whether the component statement p is true or false makes no difference to the truth-value of the statement form; it yields a true statement in either case. However, if every attempt to find such a set of T values ends in a contradiction, then the game cannot succeed because there is no set of truth values which will make all the premisses true and the conclusion false, so in other words there is no such row in the full table. A compound proposition is said to be a contradiction if and only if it is false for all. When the number of constants is small, the method works well. A proposition is said to be a tautology if its truth value is T for any assignment of truth values to its components. Finding an Equation of a Tangent Line In Exercises 41-48, find an equation of the tangent line to the graph of Calculus: An Applied Approach (MindTap Course List) Evaluate the surface integral SFdS for the given vector field F and the oriented surface S. A tautology is a compound statement S that is true for all pos-sible combinations of truth values of the component statements that are part of S. A tautology is a logical compound statement formed by two or more individual statement which is true for all the values. Use the truth tables method to determine whether p!(q^:q) and :pare logically equivalent. Note any tautologies or contradictions. A propositional form that is false in all rows of its truth table is a contradiction. Synonyms for contradiction at Thesaurus. It is easy to tell whether a formula is a tautology, contradiction, or neither by first constructing the truth table for the formula and examining the far right column. Show that p -> q, where \"->\" is the conditional. P: The art show was enjoyable. Truth Table for a. letter occurs on an open branch, the corresponding row of the truth table assigns false to the sentence letter on that row of the truth table. Taken together, De Morgan's Theorems establish a systematic relationship between \u2022 statements and \u2228 statements by providing a significant insight into the truth-conditions for the negations of both conjunctions and disjunctions. 6) A to O is a contradiction. Use De Morgan\u2019s Law to write the negations for each statement. Solution for Construct truth tables for the following wffs. Here are examples of some of most basic truth tables, Truth table for negation (\"not\") Truth table for conjunction (\"and\" Truth table for disjunction (\"or\") Ex a Translate to symbolic form, then construct a truth table to represent the expression:. If statements S 1 and S 2 are equivalent then we write S 1 S 2 For ex. A full development of a theory of truth in paraconsistent logic is given by Beall (2009). Thus, one can determine if a given proposition is an axiom or theorem by constructing its truth table. A tautology is true and a contradiction is false no matter how things stand in the world, whereas nonsense is neither true nor false. Example: p ^q. In propositional logic , a tautology is a statement which is true regardless of the truth value of the substatements of which it is composed. Definitions: A. The following two truth tables are examples of tautologies and contradictions, respectively. Do the truth table for (p or not p) and you'll see that you end up with nothing but 1's. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. How a proof by contradiction works. \u00ac \u2227\u00ac \u2192 ( \u2227\u00ac )\u2194 ( \u2192 ) b. Presumably we have either assumed or already proved P to be true so that nding a contradiction implies that :Q must be false. What about P \u2227 (\u223c P)? Consider the truth table: P \u223c P P \u2227 (\u223c P) T F F F T F Thus the compound statement is not a tautology. 2 The ntheorem Package. This will either start out as a disjunctive normal form, or a conjunctive normal form. Partial truth tables have two salient virtues. is a proposition which is neither a tautology nor a contradiction, such as. For each truth table below, we have two propositions: p and q. is a compound proposition that is always false. Solution: Truth table: P Q P ^Q \u02d8P. ;) - user20153 Sep 18 '16 at 23:43. 7 a contradiction. I have used the alternate notation I provided in video lectures for the symbolizing. to test for entailment). For each of the formulas, use a truth-table to determine if the formula is a tautology, contradictory or a contingent formula. Create a truth table to find out if this is either a contradiction, tautology, or contingency. see above ^^ since i made a mistake pointed out once fixed it made it all 1/2 or 1. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. A truth table is a mathematical table used in logic\u2014specifically in connection with Boolean algebra, boolean functions, and propositional calculus\u2014which sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables (Enderton, 2001). (p \u2192 q ) \u2194 (~ V q ). It is for this obvious reason that logicians invented a shorter, more efficient method of determining the validity of arguments, namely, the indirect truth table method. In this post, I will briefly discuss tautologies and contradictions in symbolic logic. Consider the connectives: contradiction denoted by \u22a5 (which stands on its own), negation denoted by \u00ac (not), which is The truth table for \u201cand\u201d is: p q p. Let t be a tautology and c be a contradiction. Subjects to be Learned. 13, 14, 16 Using Truth Table, Verify Logical Equivalence 1. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. The tee symbol \u22a4 is sometimes used to denote an arbitrary tautology, with the dual symbol \u22a5 representing an arbitrary contradiction; in any symbolism, a tautology may be substituted for the truth value \"true,\" as symbolized, for instance, by \"1. (pva)^(-1) = pvqr(r)) Get more help from Chegg Get 1:1 help now from expert Other Math tutors. Truth Tables - Tautology and Contradiction. Topic 3 - Logic, sets and probability \u00bb 3. The specific system used here is the one found in forall x: Calgary Remix. 40 Chapter 2. It is used to find out if a propositional expression is true for all legitimate input values. Contradiction De nition: Contradiction is a compound statement that is always false, regardless of the truth value of the individual statements. That's false. Locate the rows in which the premises are all true (the critical rows). Logic 101 These lectures cover introductory sentential logic, a method used to draw inferences based off of an argument\u2019s premises. In this case, we write R ()S. The logical \u2018principle of non-contradiction\u2019 ensures that the contradictory propositions \u2018the ruler is straight\u2019 and \u2018the ruler is not straight\u2019 cannot both be true at the same time, and in principle observation should settle which is the case. Think of it as shorthand for a complex sentence like P&~P. it is true in no world, e. Show that ( \u2014W A p) A (q V -. Therefore, (p q) p is a tautology. Contingent A statement is _ if and only if it is true on some assignments of truth values to its atomic components and false on others. True or False? If the premises of a propositionally valid argument are tautologies, then its conclusion must be a tautology as well. Chapter 1 Logic and Set Theory that is read if P then Q and dened by the truth table P Q P ! Q T T T T F F F T T P ^: P is a contradiction, and its truth table is P P ^ : P T T F F F F F T 1 3 2. In other words, fin Calculus (MindTap. How a proof by contradiction works. A propositional form that is false in all rows of its truth table is a contradiction. How to find a formula for a given truth table. A contradiction is a Boolean expression that evaluates to FALSE for all possible values of its variables. \u00b7\u00b7\u00b7P Proof of validity: (use a truth table) Note: So if an assumption leads to a contradiction, you know the. An example is P v ~P:. a) p \u2228 (p \u2227 q) _____. Mathematicians normally use a two-valued logic: Every statement is either True or False. A formula is said to be a Contradiction if every truth assignment to its component statements results in the formula being false. Perhaps no one in American public life channels this. May 7, 2016; Consistency, Emerson said, is the hobgoblin of little minds. Contradiction. letter occurs on an open branch, the corresponding row of the truth table assigns false to the sentence letter on that row of the truth table. A proposition that is neither a tautology nor a contradiction is called a contingency. Thus, one can determine if a given proposition is an axiom or theorem by constructing its truth table. A formula \uf064 is a contingentformula if and only if \uf064 is neither a tautology nor a contradiction. Think of it as shorthand for a complex sentence like P&~P. (The opposite of a tautology, a contradiction, is a compound statement that is false no matter what the truth values of its component statements are. Tautologies: In logic, a tautology is a compound sentence that is always true, no matter what truth values are assigned to the simple sentences within the compound sentence. What is a tautology? Please provide an example of a resulting truth table that yields a tautology. The expression is simplified: (1. Q^(\u02d8Q) Theorem: A statement S is a tautology if and only if its negation is a contradiction. A logical contradiction is the conjunction of a statement S and its denial not-S. When we make this array using all possible truth values, we call it a truth table. With internal images projected from objects in the outside world, it is Plato\u2019s cave with a lens. It is useful for calculating logical expressions. pip install truth-table-generator. Partial truth tables have two salient virtues. contradiction. Aufmann Chapter 3. In a contradiction, the truth table will be such that every row of the truth table under the main operator will be false. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. The opposite of a tautology is a contradiction or a fallacy, which is. A tautology is a statement (or form) which is true solely on account of its logical form rather than because of the meaning of the terms employed. The truth table for 'p. (It also happens to be an awesome song. 2 The ntheorem Package. Expert Answer. A compound proposition is said to be a contradiction if and only if it is false for all. Assume x is a particular real number. is a proposition which is always false. So we can every satisfiable is a contingency because satisfiable will have at least one true value which will also satisfy the definition of Contingency. Create a truth table to determine whether the following is a contingent statement, a tautology, or a self-contradiction. Rather than constructing the entire truth table, we can simply check whether it is possible for the proposition to be false, and then check whether it is possible for the proposition to be true. Create a truth table to determine whether the following statement is contingent, a tautology, or a self-contradiction. Determine whether the following propositions are a tautology, a contradiction, or a contingency. Accordingly, it is a tautology. Example He is a YouTuber and he is not a YouTuber. Q: The room was hot. One and the same sentence may be true if its components are all true and false if its components are all false. State, with a reason, whether the compound proposition (p \u2228 (p \u2227 q)) \u21d2 p is a contradiction, a tautology or neither. Chapter 8 - Sentential Truth Tables and Argument Forms. Solution for Construct truth tables for the following wffs. The more work you show the easier it will be to assign partial credit. P: The art show was enjoyable. Logical Equivalence Please use truth tables to check the following Boolean expressions. The term contingency is not as widely used as the terms tautology and contradiction. To write F --> T = T is to say that if A,B are statements with A being a false statement and B a true statement then the implication A --> B is a true. ^ stands for \"AND\". Solution 2. Expert Answer. equivalent if they have same truth values for all logically possibilities Two statements S 1 and S 2 are equivalent if they have identical truth table i. So the columns for your first truth table are: p q r (~q v (p ^ r)) Then, I list all the possible combinations of True and False for each variable. With a truth table, we can determine whether or not an argument is valid. 3 Truth Tables and Propositions Concepts in this section: tautology self-contradiction contingent logical equivalence contradiciton consistency. The idea of such truth tables extends naturally to other connectives. A proposition that is always false is called a contradiction. A (complete) truth table shows the input/output behavior for all possible truth assignments. 1 Statements and Compound Statements A statement or proposition is an assertion which is either true or false, though you may not know which. In this context, fuzzy normal-form formulae of linguistic expressions are derived with the construction of \u201cExtended Truth Tables\u201d. The truth table for the conjunc-tion of two statements is shown in Figure 1. The opposite of a tautology is a contradiction, a formula which is \"always false\". Think of it as shorthand for a complex sentence like P&~P. Truth tables for propositional forms allow to determine all the possible truth-values that the substitution instances of those forms can have. Every proposition is assumed to be either true or false and. (It also happens to be an awesome song. The statement is a self-contradiction. letter occurs on an open branch, the corresponding row of the truth table assigns false to the sentence letter on that row of the truth table. Show that p -> q, where \"->\" is the conditional. Is q implies p true for this row? Does true imply true? Yeah. A proposition P is a tautology if it is true under all circumstances. It is not possible for both P and NOT P to be true. Tautologies, Contradictions, and Satis\ufb01ability I A tautology (Taut) is a PropCalc formula such that every row of its truth table is 1, i. A passionate Computer Science and Engineering graduate, who loves to follow his heart :). A proposition that is neither a tautology nor a contradiction is called a contingency. If you reach a contradiction, then you know it can\u2019t. Contradictions are never true. Consistency and Contradiction. If there are contradictory configurations, you can look into the cases that belong to those configurations and assess whether contradictions can be resolved by changing things that have to do with the design of your study or the calibration. A truth tree shows that P is a tautology if and only if a tree of the stack of P determines a. , p ^\u02d8p 2unSAT. Thus, it gives us the complete semantics for P. Definition 1. (The opposite of a tautology, a contradiction, is a compound statement that is false no matter what the truth values of its component statements are. Step 1: Use a variable to represent each basic statement. The situation is similar in set theory. CMSC 203 : Section 0201 : Homework1 Solution 3. Truth Tables, Tautologies, and Logical Equivalences. Math, I have a question on tautologies and contradictions. 6 a tautology 2. The term contingency is not as widely used as the terms tautology and contradiction. Compare truth tables for logic forms of two statements: 1. Among De Morgan\u2019s most important work are two related theorems that have to do with how NOT gates are used in conjunction with AND and OR gates: An AND gate \u2026. ) The \ufb01nal column of a truth table for a tautology (respectively, a contradiction) is all Ts (respectively, all Fs). No matter what the individual parts are, the result is a true statement; a tautology is always true. is a compound proposition that is always true, no matter what the truth value of the propositional variables that occur in it. Example: p \u2227\u00ac p. Truth tables can be used for other purposes. The truth value assignments for the propositional atoms p,q and r are denoted by a sequence of 0 and 1. Tautologies, contradictions and contingencies. This simply should not happen! This is logic, not Shakespeare. Contradiction A sentence is called a contradiction if its truth table contains only false entries. Truth Table Test for a Single Sentence: Contingent, Tautology, or Contradiction Note: This truth table builder will create tables of two, four, or eight rows Pick a sentence from your textbook that you want to test and enter it into the box below. Logical Symbols are used to connect to simple statements, to define a compound statement and this process is called as logical operations. Basically, a truth table is a list of all the different combinations of truth values that a sentence, or set of sentences. Proving Conditional Statements by Contradiction 107 Since x\u2208[0,\u03c0/2], neither sin nor cos is negative, so 0\u2264sin x+cos <1. From the following truth table \\[\\begin{array}{|c|c|c|c|} \\hline p & \\overline{p} & p \\vee \\overline{p} & p \\wedge \\overline{p} Use truth tables to verify these. Truth Table Generator This page contains a JavaScript program which will generate a truth table given a well-formed formula of truth-functional logic. Consider the connectives: contradiction denoted by \u22a5 (which stands on its own), negation denoted by \u00ac (not), which is The truth table for \u201cand\u201d is: p q p. Math, I have a question on tautologies and contradictions. Propositions can be tautologies, contradictions, or contingencies. The latter implies that n = 2k for some integer k, so that 3n + 2 = 3(2k) + 2 = 2(3k + 1). And false implies false. Here are some examples that we will classify as tautologies, contradictions, or contingencies:. \" or \"p implies q. Now, we must be part of the solution. To prove that the statement \u201cIf A, then B\u201d is true by means of direct proof, begin by assuming A is true and use this information to deduce that B is true. This is called the Law of the Excluded Middle. check whether it is Tautology, Contradiction or Contingency. Truth tables - the conditional and the biconditional (\"implies\" and \"iff\") As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. It is clear that the complex of signs 'F' and 'T' in the truth table has no object (or complex of objects) corresponding to it, just as there is none corresponding to the horizontal and vertical lines or to the brackets. No matter what the individual parts are, the result is a true statement; a tautology is always true. Therefore, the truth table is: If the far right column of a truth table contains only $1's$, the formula is a tautology. I prefer more accurate tools like elements of ARIZ or the inventive standards. 2: Truth Tables Worksheet Fill out the following truth tables and determine which statements are tautologies, contradictions, or neither. It is easy to tell whether a formula is a tautology, contradiction, or neither by first constructing the truth table for the formula and examining the far right column. \uf0b4A compound proposition is a tautology if all the values in its truth table column are true. Contingent. Take the rows of the truth table where the proposition is True to construct minterms. Recall that the truth table for every atomic sentence is T and F. All of the laws of propositional logic described above can be proven fairly easily by constructing truth tables for each formua and comparing their values based on the corresponding truth assignments. \u00ac \u2227\u00ac \u2192 ( \u2227\u00ac )\u2194 ( \u2192 ) b. False implies true? That's true. truth table is a summary of truth values of the resulting statements for all possible assignment of values to the variables appearing in a compound statement. cut: The minimal score for the PRI - proportional reduction in inconsistency, under which a truth table row is declared as negative. Topics : Truth tables, statement patterns, tautaulogy, Contradiction & Contingency Statements. Showing that a compound proposition is not a tautology only requires showing a particular set of truth values for its individual propositions that cause the compound proposition to evaluate to false. Most candidates recognized that in a tautology the column is always true with a small minority confusing tautology and contradiction. it is true in no world, e. The eye is a simple optical instrument. Show that (P \u2192 Q)\u2228 (Q\u2192 P) is a tautology. Square of Opposition. Then we have 3n + 2 is odd, and n is even. Contradiction A statement is called a contradiction if the final column in its truth table contains only 's. Use a truth table to determine if the following is a tautology, a contradiction, or a contingency. 4) This is a simple OR operation, so the truth. You must explain you answer. So this case never applies. Logical Reasoning Tautologies and Contradictions De\ufb01nition. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. You need to build truth tables for each of these formulas. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. is a proposition which is neither a tautology nor a contradiction, such as. So the columns for your first truth table are: p q r (~q v (p ^ r)) Then, I list all the possible combinations of True and False for each variable. We can see that the truth values for the contrapositive are identical to those of the statement, so that the two are logically equivalent. Show that each conditional statement is a tautology without using truth tables b p !(p_q) p !(p_q) :p_(p_q) Law of Implication (:p_p)_q. Logical connectives examples and truth tables are given. Prove By Contradiction The Following Proposition: Proposition: If A, B \u2208 Z, Then A2 \u2212 4b \u2260 2. Tautologies: In logic, a tautology is a compound sentence that is always true, no matter what truth values are assigned to the simple sentences within the compound sentence. If there is a contradiction among your truth-value assignments, that means the assumption of invalidity has led to a contradiction. In fact, there are more ways for it to be true than there are ways for it to be false: it is true in every row except the last row. The term contingency is not as widely used as the terms tautology and contradiction. Tautology - example 16 5. 6) A to O is a contradiction. Therefore, (p q) p is a tautology. Chapter 3: Validity in Sentential Logic 63 in the third example, the final column consists of a mixture of T's and F's, so the formula is contingent. ^ stands for \"AND\". As a reasoning principle it says: As a reasoning principle it says: To prove $\\phi$, assume $\\lnot \\phi$ and derive absurdity. If we are unable to show that this can be done, then the argument is valid. The proposition p \u2227 p is a contradiction. State whether the statements p A and p -+ q are logically equivalent. The specific system used here is the one found in forall x: Calgary Remix. It is easy to tell whether a formula is a tautology, contradiction, or neither by first constructing the truth table for the formula and examining the far right column. The method of truth tables illustrated above is provably correct - the truth table for a tautology will end in a column with only T, while the truth table for a sentence that is not a tautology will contain a row whose final column is F, and the valuation corresponding to that row is a valuation that does not satisfy the sentence being tested. It can be used to test the validity of arguments. Square of Opposition. ) The \ufb01nal column of a truth table for a tautology (respectively, a contradiction) is all Ts (respectively, all Fs). De nition 1. \u2022 Truth Table - a calculation matrix used to demonstrate all logically possible truth-values of a given proposition. The truth or falsity of a statement built with. Decide whether the formula p 0\u2192\u00acp 0 is a tautology, a contingency, or a contradiction. I buy you a car. \u0410\u041b\u0412+\u0412'VA\" \u0441. Contradiction: A propositional formula is contradictory (unsatisfiable) if there is no interpretation for which it is true. As an introduction, we will make truth tables for these two statements 1. Tautology A statement is called a tautology if the \ufb01nal column in its truth table contains only 's. Compute the truth tables for the following propositional. tables consider all cases and can add great insight into otherwise complicated expressions. Discrete Mathematics Lecture 1 Logic: Propositional Logic Truth Table \u2022A convenient way to see the effect of the contradiction E. entailment. For example, let us study the truth value of (p \u039b q) \u2192 (p V q) by building a truth table. Then identify whether the sentence is a tautology, a contradiction or neither. Use a truth table to determine if the following is a tautology, a contradiction, or a contingency. Here we are going to study reasoning with propositions. In a proof by contradiction, we start out by saying: \"Suppose not. is a proposition that is always. Illustrates and defines the terms \u2018contradiction\u2019, \u2018logical truth\u2019 and \u2018logical possibility\u2019. (a) Show that (p q) \u2192 p is a tautology by completing this truth table. Logically Equivalent Statements, Tautologies, & Contradictions; Definition 27. 5) E to I is a contradiction. By Using Truth Table Un 1. If a contradiction is produced in the attempts to assign truth values to the premises, circle the contradiction and declare the argument valid. A compound proposition is said to be a contradiction if and only if it is false for all. 2) The truth tables for the basic operations: F F F F F T T F T F T F T T T T 1. Lesson 46 \u2013 Truth Tables (D)Tautologies and Logical Contradictions a. So we can every satisfiable is a contingency because satisfiable will have at least one true value which will also satisfy the definition of Contingency. Note any tautologies or contradictions. This is a truth table generator helps you to generate a Truth Table from a logical expression such as a and b. Contingency-. Consider the connectives: contradiction denoted by \u22a5 (which stands on its own), negation denoted by \u00ac (not), which is The truth table for \u201cand\u201d is: p q p. Question: Use A Truth Table To Determine If The Following Is A Tautology, A Contradiction, Or A Contingency. contradiction. Actually, the truth is that proper logical analysis does. A row of the truth table in which all the premises are true is called a critical row. cut: The minimal score for the PRI - proportional reduction in inconsistency, under which a truth table row is declared as negative. p q :q p!q :(p!q) p^:q T T F T F F T F T F T T F T F T F F F F T T F F Since the truth values for :(p!q) and p^:qare exactly the same for all possible combinations of truth values of pand q, the two propositions are equivalent. Using truth-tables to test a set of statements for consistency: Construct a sentence that is the conjunction of all the statements in question. Tautology Truth Tables. It isn't just that the intersection of your final answer and the set of. By Using Truth Table Un 1. In real life, 2-LUTs are not an efficient use of resources; LUTs have 4 or 5 inputs. Solution 2. Learn more about the laws of thought in this article. CMSC 203 : Section 0201 : Homework1 Solution 3. \u0410\u041b\u0412+\u0412'VA\" \u0441. ] Thus, x is a ratio of the two integers \u2212a and b with b \u2260 0. Example: p \u00acp is a tautology. By the and truth table, if one part of an \"and\" state-ment is false, the entire statement is false. Robert Lacey explores the \"untold reality\" of the Duke of Cambridge and. Q^(\u02d8Q) Theorem: A statement S is a tautology if and only if its negation is a contradiction. In this post, I will briefly discuss tautologies and contradictions in symbolic logic. Illustrates and defines the terms \u2018contradiction\u2019, \u2018logical truth\u2019 and \u2018logical possibility\u2019. A propositional form that is true in at least one row of its truth table and false in at least one row of its truth table is a contingency. Create a truth table to find out if this is either a contradiction, tautology, or contingency. A truth table is a table that begins with all the possible combinations of truth values for the letters in the compound statement; it then breaks the compound statment down and one step at a time determines truth values for each of the parts of the logical statement. is a proposition which is always false. combinations of truth values of the propositional variables which it contains. A truth table displays the relationships between the truth values of propo-sitions. Propositions are built up as truth-functions of elementary propositions (5). Although consistency is no guarantee of truth since one could create a consistent story that is false, it seems to be a necessary condition for truth. For example, if A is \u201cNapoleon was born in Corsica\u201d and B is \u201cThe number of the beast is 666\u201d, the truth table (in classical two-valued logic) would be as follows:. A tautology is a compound statement that is true for all possibilities in a truth table i. Subtracting 1 from both sides gives 2sin xcos <0. In our everyday choices, are we willing to give up our power and privilege to invite others to the table, to share in the \u201cAmerican dream?\u201d It\u2019s time to right this 400-year-old wrong. Tautologies and Contradictions in Symbolic Logic - PHILO-notes Daily Whiteboard - Duration: 5:19. The only way a disjunction can be false is if both parts are false. (4V B')\u041b (4\u041b\u0412)\" d. Logical Equivalence Please use truth tables to check the following Boolean expressions. The number of lines needed is 2 n where n is the number of variables. (pva)^(-1) = pvqr(r)) Get more help from Chegg Get 1:1 help now from expert Other Math tutors. A truth table is a mathematical table used to determine if a compound statement is true or false. State, with a reason, whether the compound proposition (p V (p A q)) p is a contradiction, a tautology or neither. > Contradiction > Contingency > Truth table with 3 variables >Venn Diagram Mathematical Logic app includes following topics: > Negation - To identify a statement as true, false or open. Chapter 5 Truth Tables. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. The table below explores the four possible cases, but the truth is simpler than that. A proposition that is always false is called a contradiction. Tautologies []. , p _\u02d8p 2Taut. Classifying propositions using truth tables We may use truth tables to make important distinctions among tautologies, contingencies, and contradictions, Consider the truth table for \u2018P e P\u2019 P e P T T T F T F. Topic 3 - Logic, sets and probability \u00bb 3. But a truth tree will get the job done faster. We notice that an implication can only be false if the hypothesis is true and the consequence is false. A full development of a theory of truth in paraconsistent logic is given by Beall (2009). The truth table is: T F T T T T T All the values in the final column are false, so Example 5 Construct a truth table for p V \u2014q. 3 Truth Tables \u00ad FILLED IN NOTES. However, if every attempt to find such a set of T values ends in a contradiction, then the game cannot succeed because there is no set of truth values which will make all the premisses true and the conclusion false, so in other words there is no such row in the full table. A compound proposition is a logical contradiction if all the values in its \u00fc-uth table column are false. \u00d8 p \u00ae c, where c is a contradiction \\ p. A passionate Computer Science and Engineering graduate, who loves to follow his heart :). A tautology is a proposition which is always true. A proposition (statement) is a contradiction if it is logically false. truth table if you like, but you may not need to. Truth Tables Every proposition, indeed every logical formula, is either true or false. Discuss the distinct patterns of. (pva)^(-1) = pvqr(r)) Get more help from Chegg Get 1:1 help now from expert Other Math tutors. Topic 3 - Logic, sets and probability \u00bb 3. As we can see from the truth table above, the proposition is definitely not a contradiction. It is for this obvious reason that logicians invented a shorter, more efficient method of determining the validity of arguments, namely, the indirect truth table method. 2: Truth Tables Worksheet Fill out the following truth tables and determine which statements are tautologies, contradictions, or neither. 2 Apply the truth able method to assessing validity (finding counterexamples), identifying tautologies, contradictions, and contingent sentences, assessing logical equivalence and consistency. Consider the connectives: contradiction denoted by \u22a5 (which stands on its own), negation denoted by \u00ac (not), which is The truth table for \u201cand\u201d is: p q p. As \"we are all happy\" can be either true or false, so can a. By constructing the truth table, determine whether the following statement pattern ls a tautology , contradiction or. Truth Table is a mathematical table and the base for all computing needs. Contradiction Sayings and Quotes. A vector of (remainder) row numbers from the truth table, to code as negative output configurations. The truth table for a tautology has \u201cT\u201d in every row. Where, 0 and F denotes False. 5 Tautology, Contradiction, Contingency, and Logical Equivalence De\ufb01nition : A compound statement is a tautology if it is true re-gardless of the truth values assigned to its component atomic state-ments. Therefore the order of the rows doesn't matter - its the rows themselves that must be correct. In other words, their columns on a truth table are identical. However, if every attempt to find such a set of T values ends in a contradiction, then the game cannot succeed because there is no set of truth values which will make all the premisses true and the conclusion false, so in other words there is no such row in the full table. 1 Introduction The truth value of a given truth- functional compound sentence depends on the truth value s of each of its components. Consider, for example, the statement form: p ~ p p \u2228 ~ p. Satisfiability & Logical Truth PHIL 012 - 2/16/2001 Outline Test Scores Homework Reminder Satisfiability Logical Truth Complex Truth Tables Sample Problems Satisfiability A sentence is said to be satisfiable IFF under some circumstances it could be true, on logical grounds. I was fooling with term_variables/2 and I think it may be possible to achieve a simpler solution using term_variables/2, coming up with some scheme to get readable variable names output, but also assuming the operators are all evaluable by is/2. Propositions are either completely true or completely false, so any truth table will want to show both of these possibilities for all the statements made. Partial truth tables have two salient virtues. But please note that this is just an introductory discussion on tautologies and contradictions as my main intention here is just to make students in logic become familiar with the topic under investigation. Multiply both sides by \u22121, gives. Math, I have a question on tautologies and contradictions. How a proof by contradiction works. First, like whole truth tables, they are algorithmic (i. If you construct them correctly, you will get an answer to your question whether a particular argument is valid; whether a particular proposition is tautologous, self-contradictory, or contingent; or whether a particular set of propositions is consistent. A tautology is a logical compound statement formed by two or more individual statement which is true for all the values. An example is P ^(\u02d8P). \u201d They are called direct proof, contra-positive proof and proof by contradiction. Chapter 1 Logic 1. none of these. Recharaaeriz-. Logical Equivalence, Tautologies, and Contradictions. Since most philosophers believe truth is logically consistent, they value logical consistency because it is a tool to discover truth. satis\ufb01able, if its truth table contains true at least once. Contradiction A statement is called a contradiction if the \ufb01nal column in its truth table contains only 's. An example is P ^(\u02d8P). A contingency is neither a tautology nor a contradiction, for instance p \u2228 q is a contingency. ; Instead, \u00acp is true. [(4 V B) ^\u2026. Propositions are either completely true or completely false, so any truth table will want to show both of these possibilities for all the statements made. If you haven't read through the textbook sections that surround these tables, now would be a good time to do that. A proposition P is a tautology if it is true under all circumstances. Interpret your various truth-value assignments as a row of a truth-table. The upshot of this result is significant. 5 Tautology, Contradiction, Contingency, and Logical Equivalence De\ufb01nition : A compound statement is a tautology if it is true re-gardless of the truth values assigned to its component atomic state-ments. Think of it as shorthand for a complex sentence like P&~P. This page contains a JavaScript program that will generate a truth table given a well formed formula of sentential logic. When you define a new theorem-like environment with ewtheorem, it is given the style currently in effect. State, with a reason, whether the compound proposition (p \u2228 (p \u2227 q)) \u21d2 p is a contradiction, a tautology or neither. molecular statement is a contradiction if its truth table is always false. The number of lines needed is 2 n where n is the number of variables. A (complete) truth table shows the input/output behavior for all possible truth assignments. A formula is said to be a Contradiction if every truth assignment to its component statements results in the formula being false. the sentence '-(AvB)&(-A>B)' and ask you whether it is a contradiction. \uf0b41) is L\u2227\u00acLa tautology a logical contradiction or neither?. The expression is simplified: (1. Use a truth table to determine whether the two statements are equivalent. truth tables. You can enter logical operators in several different formats. Assume that P is true. Truth Tables - Tautology and Contradiction. A compound statement that is false for all possible combinations of truth values of its component statements is called a contradiction. I construct the truth table for (P \u2192 Q)\u2228 (Q\u2192 P) and show that the formula is always true. Table 1: The truth table for the negation. 1 Introduction The truth value of a given truth- functional compound sentence depends on the truth value s of each of its components. State whether the statements \u00ac p \u21d2 q and (\u00ac p \u21d2 q) \u2228 (\u00ac p \u2227 q) are logically equivalent. Tautologies and Contradictions in Symbolic Logic - PHILO-notes Daily Whiteboard - Duration: 5:19. Obviously, truth tables of these sizes are simply impractical to construct. \" We then show that this leads to a contradiction, a statement like (q \u2227 \u00acq). What is the truth of the matter? Truth\u2019s character is both logical and empirical. ***** Full Course Playlist. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. Use a truth table to determine if the following is a tautology, a contradiction, or a contingency. 2 Problem 14ES. Think of it as shorthand for a complex sentence like P&~P. 6 Propositions \u2013 Tautology, Contradiction, Contingency Tautology A proposition P is a tautology if and only if P is true under every valuation. In the truth tables above, there is only one case where \"if P, then Q\" is false: namely, P is true and Q is false. Truth Tables, Tautologies, and Logical Equivalences. 2: Truth Tables for Negation, Conjunction, and Disjunction Math 121 Truth Tables A truth table is used to determine when a compound statement is true or false. Similarly, if it is false, then it implies it is true. PHILO-notes 2,286 views. Propositional Equivalences Tautologies, Contradictions, and Contingencies. Therefore to establish that a conditional. 24 and 25 are pretty handy. We say a statement Scan be deduced from a statement R if the truth table of S is True whenever. , Fs in its truth table column. The tee symbol \u22a4 is sometimes used to denote an arbitrary tautology, with the dual symbol \u22a5 representing an arbitrary contradiction; in any symbolism, a tautology may be substituted for the truth value \"true,\" as symbolized, for instance, by \"1. Trump, Truth and the Power of Contradiction. Tautologies []. A proposition is said to be a tautology if its truth value is T for any assignment of truth values to its components. p \u2228 q Solution to EXAMPLE 2. \u00d8 p \u00ae c, where c is a contradiction \\ p. R R \u2022 \u223c R T T F F F F F T Willard is either a philosopher or a windbag, and he\u2019s neither a philosopher nor a windbag. Determine whether the following statement pattern is a tautology or a contradiction or contingency. That is, a statement and its negation can never have the same truth value. De\ufb01nition: A compound statement is a contradiction if there is an F beneath its main connective in every row of its truth table. The ntheorem package provides nine predefined theorem styles, listed in Table 4. Some sentences have the property that they cannot be false under any circumstances. Working backwards from the truth values generated by evaluating the conclusion false, deduce the truth values for the remaining simple propositions and premises. The truth table for the conjunc-tion of two statements is shown in Figure 1. Logical Symbols are used to connect to simple statements, to define a compound statement and this process is called as logical operations. If a statement is neither a tautology nor a contradiction, then the truth values do alter the outcome and we say that the statement is a contingency. The truth-table at right demonstrates the logical equivalence of these two statement forms. contradiction. In this video I construct two more truth tables and use them to illustrate the notion of a tautology and a contradiction. Create a truth table showing the values of the premises and conclusion. Depending on the set of axioms exhibited and/or we are willing to impose on the linguistic expressions of our native intelligence, we might arrive at different computational intelligence expressions for. Check ((p \u2228 q) \u2227 \u00ac q) \u2192 p is a tautology using truth table. The truth table for a tautology has \u201cT\u201d in every row. Prove Logical Equivalences. In real life, 2-LUTs are not an efficient use of resources; LUTs have 4 or 5 inputs. Explain in a sentence why your truth table shows whether it is a tautology, a contradiction, or a contingent proposition. To show that a sentence is not a tautology, however, we only need one line: a line on which the sentence is 0. Tautologies and Contradictions in Symbolic Logic - PHILO-notes Daily Whiteboard - Duration: 5:19. Locate the rows in which the premises are all true (the critical rows). A\u2192 (B \u2192 A) b. And we can draw the truth table for p as follows. Truth Tables - Tautology and Contradiction. The second case agrees with none and we call it a contradiction. However, to show that an argument is not valid, all we need to do is to find one assignment where all the premises are true and the conclusion is. of the truth values of the propositional variables which comprise it. it is true in no world, e. Introduction to Philosophy > Logic > Tautologies and Contradictions. 6 a tautology 2. , ), the following statements can be made: Valid implies that the argument must be true for all instances (i. Fill in the truth table for the sentence of propositional logic below. Select \"Full Table\" to show all columns, \"Main Connective Only. Learn more about the laws of thought in this article. p) is a logical contradiction. In sentential logic all theorems are tautologies and all tautologies are either axioms or theorems. Tautologies, Inconsistent Sentences, and Contingent Sentences Tautologies. So it is not a TT-contradiction. A propositional form that is true in all rows of its truth table is a tautology. The size of the truth table depends on the number of different simple. Here are examples of some of most basic truth tables, Truth table for negation (\"not\") Truth table for conjunction (\"and\" Truth table for disjunction (\"or\") Ex a Translate to symbolic form, then construct a truth table to represent the expression:. Truth Tables - Tautology and Contradiction. Contradiction - example 17 5. Create a truth table to determine whether the following statement is contingent, a tautology, or a self-contradiction. In other words, their columns on a truth table are identical. If it is always true, then the argument is valid. contradiction, is a statement which is false regardless of the truth values of the substatements which form it. The minimum number of cases under which a truth table row is declared as a remainder. here they are: a) i found this proposition was a contradiction. Confirmed! The conditional p \u2014+ q and it's contrapositive -Iq \u2014Y -p are logically equivalent. 2 LOGICAL EQUIVALENCE, TAUTOLOGIES & CONTRADICTIONS. Let's start with logical contradiction. ((PvQ) ^ (~R) = P v (Q^(~R)) Thats not an equal sign btw, but three lines intended instead of two. you\u2019re assuming it\u2019s consistent by plugging in all true truth values. The expression \"p or not p\" is true under any circumstance, so it is a tautology. \u0410\u041b\u0412+\u0412'VA\" \u0441. Where, 0 and F denotes False. The following two truth tables are examples of tautologies and contradictions, respectively. The opposite of a tautology is a contradiction, a formula which is \"always false\". On my upcoming test time is limited. A contradiction is a conjunction of the form \"A and not-A\", where not-A is the contradictory of A. 12, 15 Examine the statement Patterns (Tautology, Contradiction, Contingency) 1. A truth table column which consists entirely of T's indicates a situation where the proposition is true no matter whether the individual propositions of which it is composed are true or false. Solution: Question 2.", "date": "2020-11-23 16:40:24", "meta": {"domain": "chiavette-usb-personalizzate.it", "url": "http://insd.chiavette-usb-personalizzate.it/contradiction-truth-table.html", "openwebmath_score": 0.5699777007102966, "openwebmath_perplexity": 467.3208727966209, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.970239907775086, "lm_q2_score": 0.8807970904940926, "lm_q1q2_score": 0.8545844878495524}}
{"url": "https://mathhelpboards.com/threads/prove-a-function-is-continuos.3349/", "text": "# Prove a function is continuos\n\n#### ianchenmu\n\n##### Member\nProve that\n$f(x,y)=\\left\\{\\begin{matrix} e^{-1/|x-y|},x\\neq y\\\\ 0,x=y \\end{matrix}\\right.$\nis continuous on $\\mathbb{R}^2$.\n\nCan I conclude since $e^{-1/t}$ is continuous then for $x$,$y\\in \\mathbb{R}^2$, $x\\neq y$,$e^{-1/|x-y|}$ is continuous on $\\mathbb{R}^2$ since here $t=|x-y|$? And how to prove it when $x=y$, using the $\\delta- \\varepsilon$ way? Thank you a lot!\n\n#### girdav\n\n##### Member\nRe: Prove a function is continuous\n\nWe have to show that the function is continuous at each point of $\\Bbb R^2$. Your argument threats the case of points of the form $(x,y),x\\neq y$. So now, fix $x_0$. We have to show that $\\lim_{(x,y)\\to (x_0,x_0)}=0$.\n\nHint: use $e^{-u}\\leqslant \\frac 1{1+u}$ for each non-negative real number $u$.\n\nLast edited:\n\n#### Fernando Revilla\n\n##### Well-known member\nMHB Math Helper\nRe: Prove a function is continuous\n\nHint: use $e^{-u}\\leqslant \\frac 1{1+u}$ for each real number $u$.\nSurely is a typo, I suppose you meant for each real number $u$ close to $0$.\n\nAnother way: denoting $\\Delta=\\{(t,t):t\\in\\mathbb{R}\\}$ (diagonal of $\\mathbb{R}^2$), in a neighborhood $V$ of $(0,0)$ we have:\n\n$\\displaystyle\\lim_{(x,y)\\to (x_0,x_0)\\;,\\;(x,y)\\in V-\\Delta}f(x,y)=\\displaystyle\\lim_{(x,y)\\to (x_0,x_0)\\;,\\;(x,y)\\in V-\\Delta}e^{-\\dfrac{1}{|x-y|}}=e^{-\\infty}=0$\n\n$\\displaystyle\\lim_{(x,y)\\to (x_0,x_0)\\;,\\;(x,y)\\in V\\cap\\Delta}f(x,y)=\\displaystyle\\lim_{(x,y)\\to (x_0,x_0)\\;,\\;(x,y)\\in V\\cap\\Delta}0=0$\n\nIn a finite partition of $V$ all the limits coincide, so $\\displaystyle\\lim_{(x,y)\\to (x_0,x_0)}f(x,y)=0$ .\n\n#### ianchenmu\n\n##### Member\nThe definition of continuity for a function is {lim x-> a} f(x) = f(a). For this function we only need to worry about when |x - y| -> 0, since the exponential function is continuous everywhere. So in other words we need to show that\n{lim |x - y|-> 0} f(x,y) = f(x,x) = f(y,y)\nNow f(x,x) = f(y,y) = 0 so we just need to show that\n{lim |x - y| -> 0} f(x,y) = 0.\nSince |x - y| > 0, -1/|x - y| -> infinity so f(x,y) -> 0 and we're done.\n\nIs a proof like this correct?", "date": "2020-11-30 13:49:22", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/prove-a-function-is-continuos.3349/", "openwebmath_score": 0.9754927158355713, "openwebmath_perplexity": 528.6336263970237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399034724604, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8545844749520176}}
{"url": "https://math.stackexchange.com/questions/1291087/how-to-solve-an-equation-with-x4/1291091", "text": "# How to solve an equation with $x^4$?\n\nToday, I had this question on a Maths test about Algebra. This was the equation I had to solve:\n\n$$(1-x)(x-5)^3=x-1$$\n\nI worked away the brackets and subtracted $x-1$ from both sides and was left with this:\n\n$$-x^4+16x^3-90x^2+199x-124=0$$\n\nProblem is, I haven't a clue how to solve this? First thing I tried was replacing $x^2$ with another variable like $u$ but that got me no further. Dividing the whole equation by $x^2$ (as is suggested by a lot of sites on this matter) also did not get me any further. I then tried something incredibly ludicrous;\n\n$$(ax+b)(cx^3+dx^2+ex+f)=0$$ \\left\\{ \\begin{aligned} ac &= -1 \\\\ ad + bc &= 16 \\\\ ae + bd &= -90 \\\\ af + be &= 199 \\\\ bf &= -124 \\end{aligned} \\right.\n\nwhich got even worse when there were 3 brackets;\n\n$$(ax+b)(gx+h)(ix^2+jx+k)=0$$ \\left\\{ \\begin{aligned} ac &= agi &&= -1 \\\\ ad + bc &= agj + ahi + bgi &&= 16 \\\\ ae + bd &= agk + ahj + bgj + bhi &&= -90 \\\\ af + be &= ahk + bgk + bhj &&= 199 \\\\ bf &= bhk &&= -124 \\end{aligned} \\right.\n\nonly to be left with no result.\n\nWhen using Wolfram Alpha on this question, it performs a rather strange step I don't understand:\n\n$$-x^4+16x^3-90x^2+199x-124=0$$ $$\\downarrow$$ $$-((x-4)(x-1)(x^2-11x+31))=0$$\n\nCould somebody explain how to properly tackle this problem? And if possible, also show me how to get the non-real answers for it. Thanks.\n\n\u2022 You needn't expand: first, observe that $x=1$ is a solution. Now, looking for other solutions (at most 3 other possible ones), you can divide each side by $x-1$ and obtain the equation $(x-5)^3=-1$ to solve (for values $x\\neq 1$). \u2013\u00a0Clement C. May 20 '15 at 13:13\n\nEuh... I think you overcomplicated things here...\n\n$(1-x)(x-5)^3=x-1$ is equivalent to $(1-x)[(x-5)^3+1]=0$\n\nEither $x=1$ or $(x-5)^3=-1$...\n\n\u2022 I feel so stupid, that I didn't see this in the first place... Thanks for the great answer! \u2013\u00a0Martijn May 20 '15 at 13:19\n\u2022 @TitoPiezasIII Done :) \u2013\u00a0Martijn May 20 '15 at 13:27\n\u2022 You can also use the identity $a^3+1=(a+1)(...)$ \u2013\u00a0k1.M May 20 '15 at 14:00\n\nThere are already answers on how to get the real solutions, so I will only show you the non-real solutions.\n\nYou have obtained that $(x-5)^3=-1$. Expanding and simplifying we get: $$x^3-15x^2+75x-124=0$$ However, we know that $x=4$ is a solution so we can say that $(x-4)(ax^2+bx+c)=0$. You can equate coefficients or use polynomial division, but as you have already found with Wolfram Alpha, $(ax^2+bx+c)=(x^2-11x+31)$.\n\nTo solve, complete the square: $$x^2-11x+31=0$$ $$(x-\\frac{11}{2})^2-\\frac{121}{4}+31=0$$ $$(x-\\frac{11}{2})^2=-\\frac{3}{4}$$ $$x-\\frac{11}{2}=\\sqrt{-\\frac{3}{4}}= \\pm \\frac{\\sqrt{3}}{2}i$$ $$x=\\frac {11 \\pm i \\sqrt{3}}{2}$$\n\n\u2022 Awesome work! I have two kinds of mathematics in the Netherlands and the other kind is about imaginary numbers a.t.m. so this will definitely come in handy, I'm sure! \u2013\u00a0Martijn May 20 '15 at 13:45\n\u2022 In general the solutions to $(x-a)^m=-1$ are $x_n=e^{i\\pi\\frac{2n-1}{m}}+a$, for $n=1,2,\\dots m$. \u2013\u00a0Kwin van der Veen May 21 '15 at 4:10\n\nFrom the beginning: $$(1-x)(x-5)^3=x-1\\\\ (1-x)(x-5)^3+1-x=0\\\\ (1-x)(x-5)^3+(1-x)=0\\\\ (1-x)[(x-5)^3+1]=0\\\\$$\n\nThis implies $1-x=0$ or $(x-5)^3=-1$. I believe you can solve these.\n\n\u2022 Isn't there a sign that got flipped between the first and second line? \u2013\u00a0Clement C. May 20 '15 at 13:18\n\u2022 Sorry fixed it. @ClementC. \u2013\u00a0KittyL May 20 '15 at 13:24\n\nI know that the problem has already been answered but I want to show you a more general method, let's suppose that you don't se how to rewrite the equation:\n\n$-x^4+16x^3-90x^2+199x-124=0$\n\nOr I prefer to write:\n\n$x^4-16x^3+90x^2-199x+124=0$\n\nYou can use something called the Ruffini rule: search for integers divisor (both positive and negative) of the constant term and then set $x$ equals to the them and see if one of them is a solution. Starting from one we have:\n\n$1-16+90-199+124=0$\n\nSo $x=1$ is a solution, now via Ruffini's rule, which can be seen here, we can rewrite the equation as:\n\n$(x-1)(x^3-15x^2+75x-124)=0$\n\nNow you have $3$ options to end this exercise:\n\n$1.)$ Note that the second factor is a perfect cube;\n\n$2.)$ Use Ruffini's rule again;\n\n$3.)$ Use the general formula for third degree equation (which I'd not advise you to if your interested only in real solution).\n\nThis is a more general method so I hope this will help you in the future!", "date": "2019-11-13 05:13:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1291087/how-to-solve-an-equation-with-x4/1291091", "openwebmath_score": 0.9995129108428955, "openwebmath_perplexity": 363.2793288703008, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399051935107, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8545844734319819}}
{"url": "https://math.stackexchange.com/questions/3119998/numbers-of-distinct-values-obtained-by-inserting-times-div-in-under", "text": "# Numbers of distinct values obtained by inserting $+ - \\times \\div ()$ in $\\underbrace{2\\quad2 \\quad2 \\quad2\\quad\u2026\\quad 2}_{n \\text{ times}}$\n\nThis question is inspired by How many values of $2^{2^{2^{.^{.^{.^{2}}}}}}$ depending on parenthesis? (By the way, I sincerely hope this kind of questions can receive more attention)\n\nInsert $$+ - \\times \\div ()$$ in $$\\underbrace{2\\quad2 \\quad2 \\quad2\\quad...\\quad 2}_{n \\text{ times}}$$ Denote the number of distinct values which can be obtained in this way by $$D(n)$$. Is there a general formula (or recurrence relation at least) for $$D(n)$$?\n\nThis is basically the $$+ - \\times \\div ()$$ version of @barakmanos question. It seems this question is easier than the power tower version. Or maybe not?\n\nFor $$n=1$$ , there is only $$2$$ values $$-2,2$$;\n\nFor $$n=2$$, there are $$5$$ values $$-4,-1,0,1,4$$;\n\nFor $$n=3$$, there are $$13$$ values $$-8,-6,-3,-2,-1,-\\frac{1}{2},0,\\frac{1}{2},1,2,3,6,8$$;\n\nAnd for $$n=4$$ I'm reluctant to calculate with bare hands. (See @DanUznanki answer for what follows)\n\nAny idea is appreciated. Sorry if this is a duplicate.\n\nEdit: My research shows that the version with distinct generic variables $$a_1,a_2,...,a_n$$ is solved. See A182173 for your reference.\n\n\u2022 Based on what you say for $n=1$ and $n=2$, we may use $-$ as either the negation operator on one number or the subtraction operator between two numbers? \u2013\u00a0alex.jordan Mar 24 at 4:53\n\u2022 @alex.jordan Yes. \u2013\u00a0YuiTo Cheng Mar 24 at 4:59\n\u2022 OK. I wonder if you did not allow $-$ to be unary negation, if then the sequence might appear in OEIS. We have an overloaded symbol, $-$. And one of its meanings happens to be unary negation. It's odd that we don't have a symbol for unary inversion. Well, we do: $^{-1}$. But it does not reuse the division symbol $\\div$. All this is just to say that it feels like the special double meaning for $-$, that is not mirrored with $\\div$, could break what would otherwise be a simpler pattern in the sequence of counts. \u2013\u00a0alex.jordan Mar 24 at 5:22\n\u2022 @alex.jordan If we disallow $-$ to be a unary operator, will things become vastly simpler? \u2013\u00a0YuiTo Cheng Mar 24 at 5:27\n\u2022 @alex.jordan: I once toyed with the idea of \"$/$\" for unary inversion. (So, \"$/2$\" means \"$1/2$\" in the same way that \"$-2$\" means \"$0-2$\".) I still kinda like it. :) \u2013\u00a0Blue Mar 25 at 20:09\n\nIt looks like we can do this \"inductively\": for the calculations of size $$n$$, we can take values from the list for $$1 \\le k\\le n/2$$, and values from the list for $$(n-k)$$, and operate on them using the 64 different operation orders.\n\nFortunately, it's really only 10 classes of operation, because many are duplicates:\n\n\u2022 There's four values we can get from addition and subtraction: $$a+b$$, $$-(a+b)$$, $$a-b$$, and $$b-a$$.\n\u2022 There's only two we can get from multiplication: $$ab$$ and $$-ab$$.\n\u2022 There are four cases we can get from division: $$a/b$$, $$b/a$$, $$-a/b$$, and $$-b/a$$.\n\nAlso conveniently we only need to try the nonnegative entries in previous lists.\n\nSo for $$n=4$$, we have:\n\n$$-16$$, $$-12$$, $$-10$$, $$-8$$, $$-6$$, $$-5$$, $$-4$$, $$-3$$, $$-5/2$$, $$-2$$, $$-3/2$$, $$-1$$, $$-2/3$$, $$-1/2$$, $$-1/3$$, $$-1/4$$, $$0$$, $$1/4$$, $$1/3$$, $$1/2$$, $$2/3$$, $$1$$, $$3/2$$, $$2$$, $$5/2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$10$$, $$12$$, $$16$$\n\nWhich is $$33$$ entries.\n\nI've written a short script which finds them all, and told it to run up to $$n=10$$, which gave the following sizes: $$2,5,13,33,77,185,441,1051,2523,6083$$. Apparently this sequence is not in OEIS, nor is the positive-values-only version! I am very surprised.\n\n\u2022 You mentioned we can do it 'inductively', do you have any clue about the recurrence relation? \u2013\u00a0YuiTo Cheng Feb 20 at 13:31\n\u2022 Not sure, but it isn't gonna be nice. actually, different values for $x$ instead of $2$ give different results: using $3$ gives $2$, $7$, $21$, $67$, using $4$ gives $2$,$7$,$21$,$77$. Interestingly, when I tried to get the generic result, where $x$ is just $x$, I did get some partial results in the realm of restricted lattice walks: $2$,$7$,$23$,$91$, $347$ appears twice in OEIS. \u2013\u00a0Dan Uznanski Feb 20 at 13:39\n\u2022 Interesting. What about the original power tower question? Compared with this, which one do you think is simpler? \u2013\u00a0YuiTo Cheng Feb 20 at 13:45\n\u2022 for generic $x$ in the power tower thing you're getting exactly the Catalan Numbers. But assuming you have the ability to actually store the numbers you're working with, the power tower version is vastly simpler: there's only one operator instead of 10. \u2013\u00a0Dan Uznanski Feb 20 at 13:46\n\u2022 What about the case without bracket? \u2013\u00a0YuiTo Cheng Feb 21 at 3:36\n\n@Dan Uznanski if your results are correct i noticed that the ratio between the successive terms of the sum is almost constant. So maybe:\n\nConjecture\n\n$$D(n) \\sim K^n$$\n\nWhere:\n\n$$2.3\\leq K \\leq 2.5$$", "date": "2019-04-23 10:33:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3119998/numbers-of-distinct-values-obtained-by-inserting-times-div-in-under", "openwebmath_score": 0.8713907599449158, "openwebmath_perplexity": 363.66713835289636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305307578324, "lm_q2_score": 0.88720460564669, "lm_q1q2_score": 0.8545825631878546}}
{"url": "http://math.stackexchange.com/questions/120861/homomorphism-between-cyclic-groups", "text": "# Homomorphism between cyclic groups\n\nI have some confusion in relation to the following question.\n\nLet $\\langle x\\rangle = G$, $\\langle y\\rangle = H$ be finite cyclic groups of order $n$ and $m$ respectively. Let $f:G \\mapsto H$ be the mapping sending $x^i$ to $y^i.$ Determine the conditions on $n$ and $m$ so that $f$ is going to be a homomorphism.\n\nIn verifying the definition of a homomorphism I seem to be able to conclude that there is no restriction on $n$ and $m$ since the definition of a homomorphic mapping is satisfied:\n\n$$f(x^i x^j) = f(x^{i+j}) = y^{i+j} = y^i y^j = f(x^i)(x^j) \\; \\; \\; (1)$$\n\nBut assuming n < m we would get\n\n$$f(x^n) = f(e) = e \\not = y^n$$\n\nSo clearly we have to have $m | n.$\n\nMy questions are\n\n1. Are there any other conditions on $n$ and $m$ besides $m |n.$\n\n2. What exactly did I do wrong in $(1)$ ?\n\n-\nFor the (1), What would happen if $i+j \\ge m$? It always helps to define the map this way: $x^{i(n)} \\mapsto x^{i(m)}$. Note that the map may not even be well-defined if $m \\nmid n$. \u2013\u00a0 user21436 Mar 16 '12 at 9:50\nI'll write up a more detailed answer in an hour if it's still unanswered. This has its direct connection to the universal property of the quotient maps. \u2013\u00a0 user21436 Mar 16 '12 at 9:52\nI see that there are many contradictions if $m \\not | n.$ My main concern is that $(1)$ seems like such a innocent algebraic manipulation that I would never suspect that something is wrong with it.I would like to hear an \"aha\" explanation on why one cannot apply the definition of a homomorphism on $f$ so blindlessly. \u2013\u00a0 Jernej Mar 16 '12 at 10:02\n\u2013\u00a0 lhf Mar 16 '12 at 10:43\nWell, either you use a homomorphism $\\mathbb Z \\rightarrow G$ (and talk about lifting that homomorphism) or you must face the fact that the Element $x^i$ does NOT determine the number $i$. \u2013\u00a0 Blah Mar 16 '12 at 10:50\n\nThe discussion about \"well-defined\" is perhaps a bit obscure. Here's the problem:\n\nRemember that in general, an element of $G$ may have many different \"names.\" For example, if $n=10$, then the element $x^{11}$ is equal to $x$; in fact, $x$ has infinitely many different \"names\" as a power of $x$: $$\\cdots = x^{-9} = x^1 = x^{11} = x^{21} = \\cdots$$\n\nThe problem is that the definition of $f$ as given depends on the name of the element! That is, if we are furiously working and somebody hands us a power of $x$, say, $x^{3781}$, we are supposed to, unthinkingly, map it to $y^{3781}$. The problem is that $x^{3781}$ is the same element as $x$, which we are supposed to send to $y^1$. That means that unless $y^1=y^{3781}$, what we have is not really a function: because the same input, $x$ (who, when being teased by bullies is called \"$x^{3781}$\") may be sent to $y^1$ or to $y^{3781}$, depending on what \"name\" we just heard for it.\n\nChecking that the value of the function is the same regardless of what name we are using for an element, even though the function is defined in terms of the name, is called \"checking that the function is 'well-defined.'\"\n\nAn example of a function that is not well-defined would be one in which the input is an integer, and the output is the number of symbols used to express that integer. For example, $f(3)$ would be $1$ (because 3 is only one symbol), but $f(4-1)$ would be $3$ (because we are using 4, -, and 1). This is not well defined as a function of the integers, because $3$ is the same as $4-1$, but $f$ assigns it two different outputs.\n\nSo in order for the expression given to actually define a function, we need the following to be true: $$\\text{if }x^i=x^j,\\text{ then }y^i = f(x^i)\\text{ is equal to }y^j=f(x^j).$$ Now, $x^i=x^j$ if and only if $i\\equiv j\\pmod{n}$; and $y^i=y^j$ if and only if $i\\equiv j\\pmod{m}$. Therefore, we need: $$\\text{if }i\\equiv j\\pmod{m},\\text{ then }i\\equiv j\\pmod{n}.$$ In other words: every number that is a multiple of $m$ must be a multiple of $n$.\n\nThis is equivalent to $n|m$.\n\nIn fact, you noticed that in what you wrote, because $x^n=e$, so we need $f(x^n)$ to be the same as $f(x^0)$.\n\nOnce we know that $n|m$, then $f$ is well defined. Once it is well-defined, we can start verifying that it is indeed a homomorphism (it is). Technically, it's incorrect to start working to see if it is a homomorphism before you even know whether or not it is a function.\n\nNote that the condition actually works if we allow $G$ or $H$ to be infinite cyclic groups, if we call infinite cyclic groups \"groups of order $0$\". Then $i\\equiv j\\pmod{0}$ means $i=j$, every number divides $0$, but the only multiple of $0$ is $0$. So if $G$ is infinite cyclic then the value of $n$ does not matter; if $H$ is infinite cyclic then we must have $G$ infinite cyclic.\n\n-\nI don't agree with your take on well-defined. The rule $x^i\\mapsto y^i$ is well-defined in the set $\\{x^0, x^1, \\ldots, x^{n-1}\\}$ because every element in this set has a single name. The problem is that this rule may not be a homomorphism because there is no guarantee that say $x^{2n-3}$ goes to $f(x)^{2n-3}=y^{2n-3}$. \u2013\u00a0 lhf Mar 16 '12 at 16:47\n@lhf: I agree that if we were specifying $0\\leq i\\lt n$, then the rule would yield a well-defined function. However, I don't see any such specification in the statement of the problem. It just says \"sending $x^i$ to $y^i$\", with no restrictions on $i$. Of course, checking that it is a homomorphism is important: I note it later in the post, I don't ignore it. \u2013\u00a0 Arturo Magidin Mar 16 '12 at 16:58\nRight, I assumed that $0\\leq i\\lt n$ was a natural implicit assumption. \u2013\u00a0 lhf Mar 16 '12 at 17:04\n@lhf: In that case, there is a different problem with the definition, which is that it would only make sense if $n\\leq m$ (otherwise, $y^i$ for $i\\geq m$ would be an \"invalid name\"). So we would end up with the conclusion that the map only makes sense and is a homomorphism when $m=n$; or else we need to treat the exponent differently in the domain (restricted to a special representation for the elements) but not the codomain, which seems somewhat perverse. \u2013\u00a0 Arturo Magidin Mar 16 '12 at 18:21\nI don't think $y^i$ is an invalid name. Just raise $y$ to $i$. The exponent is well defined, isn't it? \u2013\u00a0 lhf Mar 16 '12 at 21:27\n\nYou may want to check first if the map is well defined, ie., if $x^i=x^j$ does it follow that $f(x^i)=f(x^j)$?\n\n-\n+1 Why would anyone downvote this? Ok, it would be better to phrase it: \"if $x^i=x^j$ does it follow that $y^i=y^j$?\" It is somewhat dangerous to talk about values of $f$ before you know that $f$ is well defined. \u2013\u00a0 Jyrki Lahtonen Mar 16 '12 at 14:07\nit is perfectly acceptable to ask whether or not a definition of f makes f a function. \u2013\u00a0 David Wheeler Mar 16 '12 at 15:22\n@David, you're right, of course. I was just trying to guess, why there was a downvote for a little while. \u2013\u00a0 Jyrki Lahtonen Mar 16 '12 at 15:44\n\nI am not sure if this completely explains why the naive calculations in $(1)$ are not valid. Here is my go at it:\n\nI'll list down several points of concern.\n\n\u2022 Firstly your map $x^i \\mapsto y^i$ is very naive that it misses some essential details.\n\nFor convenience of notation, I prefer to look at this as $\\varphi(\\bar i)=\\bar i$ if you'll allow me to do so. This is also naive but to some extent captures what is happening.\n\nSo, an equivalence class $\\bmod n$ is mapped to the same equivalence class $\\bmod m$ under $\\varphi$. This bit of information is missed in $x^i \\mapsto y^i$. This explains my comment under your question that you will want to write your maps as $x^{i(n)} \\mapsto y^{i(m)}$. This notation makes things more transparent.\n\nThat you have dicovered that $m \\mid n$, it can be obtained through the universal property of the quotients, which I will write about later.\n\nThe quotient map $G \\to G/H$ factors over $G \\to G/K$ if and only if $K \\subseteq H$. That is there is a map from $G/H$ to $G/K$ if and only if $K \\subseteq H$.\n\nProof: Please Go through Dummit and Foote's description in their Abstract Algebra in the section $\\S3.3$ Isomorphism Theorems.\n\nNow consider your cyclic groups as $\\Bbb Z/n\\Bbb Z$ and $\\Bbb Z/m\\Bbb Z$ and observe that above condition will tell you, $m\\Bbb Z \\subseteq n \\Bbb Z$ whic happens if and only if...\n\nAs you had earlier in your comments observed, the map won't even be well defined if $m \\nmid n$.\n\nI'll leave it to you as a challenge to use this blooper to prove that two groups of same order are isomorphic, which really is not the case. Hint: Cyclic groups exist for any given order.\n\n-\nWill the down voter explain why? \u2013\u00a0 user21436 Mar 16 '12 at 14:01\nWhy do you say that $y^{i+j}$ might not be an element of $H$? Of course it is an element of $H$! A group is a set closed under the group operation, so if $y$ is in there, so are all its powers. Also, are you sure that $i(m)$ is a standard notation for the remainder? Color my old-fashioned, but I think that binary mod should stay in programming classes. This last bit is just a pet-peeve of mine :-) \u2013\u00a0 Jyrki Lahtonen Mar 16 '12 at 14:04\nWell, $y^{i+j}$ is an element of the group. For example, if $H$ is a subgroup of the multiplicative group of complex numbers generated by the imaginary unit $i$, you seem to be saying that $i^5$ is not an element of $H$. Well, you are wrong: $$i^5=i^{4+1}=i^4\\cdot i=1\\cdot i=i.$$ Looks like it is in there! The same holds for all cyclic groups. I just picked a concrete one to make you realize your error. \u2013\u00a0 Jyrki Lahtonen Mar 16 '12 at 14:16\nNo. I'm not prone to making a mistake. I check that my mappings are well-defined. If (granted, a big if in some cases) $y^i=y^j$ whenever $x^i=x^j$, then the mapping $x^i\\mapsto y^i$ is trivially a homomorphism: $$f(x^i*x^j)=f(x^{i+j})=y^{i+j}=y^i*y^j=f(x^i)f(x^j)$$ AFTER I have checked that it is well-defined, then the above always makes sense, because in a cyclic group the power of an element is defined in such a way that we always have $y^{i+j}=y^i*y^j$ and so forth. \u2013\u00a0 Jyrki Lahtonen Mar 16 '12 at 14:31\nas a matter of fact, it IS true that $(i+j)$ (mod n) = $i$ (mod n) + $j$ (mod n). the problem is that $(i+j)$ (mod n) may not equal $i+j$ (mod m). \u2013\u00a0 David Wheeler Mar 16 '12 at 15:32\n\nThe rule $x^i\\mapsto y^i$ gives a perfectly well-defined function in the set $\\{x^0, x^1, \\ldots, x^{n-1}\\}$. However, for that rule to be a homomorphism we need that $x^{i+j} \\mapsto y^{i+j}$ for all $i$ and $j$. To find where $x^{i+j}$ goes we need to reduce $i+j \\bmod n$ to a $k$ in $\\{0,1,\\ldots,n-1\\}$ and we need that $k \\equiv i+j \\bmod m$. In other words, we need $k \\equiv k' \\bmod n \\implies k \\equiv k' \\bmod m$ and this can only happen when $m\\mid n$.\n\n-\n\nTHEOREM:\n\nDefine $f:\\langle x \\rangle \\to \\langle y \\rangle$, where the orders of $\\langle x \\rangle$ and $\\langle y \\rangle$ are n and m respectively, by $f(x) = y^u$. Then f is a well-defined homomorphism if and only if $\\frac{m}{\\gcd(m,n)} \\mid u$.\n\nPROOF: Suppose that the mapping $f:\\langle x \\rangle \\to \\langle y \\rangle$ is a well-defined homomorphism.\n\nLet $f(x) = y^u$ for some non negative integer u.\n\nWe must also have $1 = f(x^n) = y^{un}$\n\nThen $m \\mid un$. Let $g = \\gcd(m, n)$. Then $\\frac m g \\mid u \\frac n g$. It follows that $\\frac m g \\mid u$.\n\nThe converse is pretty straightforward.\n\nSuppose we define $f:\\langle x \\rangle \\to \\langle y \\rangle$ by $f(x) = y^u$ where $\\frac{m}{\\gcd(m,n)} \\mid u$\n\nThe only real problem with this definition is \"Is it well defined?\" Since the order of $\\langle x \\rangle$ is $n$:\n\n\\begin{align*} x^a = x^b &\\Rightarrow a \\equiv b \\mod n \\cr &\\Rightarrow ua \\equiv ub \\mod {un} \\cr &\\Rightarrow ua \\equiv ub \\mod {\\frac{mn}{\\gcd(m,n)}} \\cr &\\Rightarrow ua \\equiv ub \\mod {m} \\cr &\\Rightarrow y^{ua} = y^{ub}\\cr &\\Rightarrow f(x^a) = f(x^b) \\end{align*}\n\nLinearity is pretty obvious, so $f$ is a well-defined homomorphism.\n\nSo, if you want $f(x^i) = y^i$, then you want $f(x) = y$. You must therefore have $\\frac{m}{\\gcd(m,n)} \\mid 1$. Which implies $m \\mid n$.\n\nConversely, if $m \\mid n$ and $f(x) = y$, it follows that $f(x^i) = y^i$.\n\n-", "date": "2015-09-03 05:27:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/120861/homomorphism-between-cyclic-groups", "openwebmath_score": 0.9452667236328125, "openwebmath_perplexity": 195.3500608718099, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305339244013, "lm_q2_score": 0.8872045922259089, "lm_q1q2_score": 0.8545825530699429}}
{"url": "https://math.stackexchange.com/questions/2461040/induction-and-proper-use-of-the-inductive-step", "text": "# induction and proper use of the inductive step\n\nSuppose I have a conjecture of the form $\\forall x \\in \\mathbb{N}^{+}$ $f(x) > g(x)$.\n\nWithout explicitly giving the functions $f$ and $g$ I would like to prove this conjecture using induction.\n\nSpecifically I have shown $f(1) > g(1)$. My inductive assumption is now that $f(x - 1) > g(x - 1)$ and I'm trying to show $f(x) > g(x)$.\n\nI haven't been able to do this, but I've noticed that I can if I assume $f(y) > g(y)$ $\\forall y \\in \\mathbb{N}^{+}$ such that $y < x$ instead of my inductive argument I can prove $f(x) > g(x)$.\n\nMy question is, is this ok or am I using circular logic? Typically when we use induction we use $x - 1$ to prove the $x$ case after an initial base case. For my problem it doesn't appear possible to prove $f(x) > g(x)$ without assuming that for all values less than $x$ the conjecture holds ($f$ and $g$ are recursive). I suspect that it is ok, but it sounds extremely circular and I haven't been able to convince myself that it isn't.\n\nNote also that I've included proof verification as a tag because although I have not included the explicit proof, I have included a proof approach that I'm interested in validating.\n\n\u2022 That's fine. That's called strong induction. It might be clearer if worded it slightly differently. You know $f(1) > g(1)$ and you have proven that if $f(y)> g(y)$ for all $y \\le x-1< x$ then $f(x) > f(x)$. Hence... induction. Is that clearer? \u2013\u00a0fleablood Oct 14 '17 at 17:13\n\nSo you just want to make the stronger assumption that some statement $P(y)$ holds for all $y < x$ instead of only for $y = x-1$?\n\nThis is totally fine. It is called strong induction and is equivalent to simple induction.\n\n\u2022 Ok, that's what I thought. I was just a bit unsure because I couldn't remember doing something like this in any course I've taken. Thank you! \u2013\u00a0HXSP1947 Oct 7 '17 at 1:54\n\n\"Typically when we use induction we use x\u22121 to prove the x case after an initial base case. For my problem it doesn't appear possible to prove .. the proposition.. without assuming that for all values less than x the conjecture holds\"\n\nThink about this. If being true for all $y \\le x-1 < x \\implies$ being true for $x$, then being true for $y < x$ and being true for $x$ implies being true for all $y < x+1$. So true for $x+1$. And so on.\n\nIn principal, this is the exact same reasoning as the reasoning for induction. (if it's true for one case, it's true for the next, and via infinite countable reptitions it must be true for all).\n\nThe only difference is that whereas in \"weak\" induction we only make use of the fact that the proposition is true of $1$ and for $x-1$ and therefore true for $x$ we don't need to and we don't make use of the fact that it must also be true for $1,.......,x-2$ as well as just $x-1$.\n\nBut we know they must be true so we can make use of them if we need to.\n\nThis is called \"strong\" induction. It is equivalent to weak induction. The only difference is we might need to make use of earlier values than just the $x-1$ case.\n\nThis can be particularly useful with recursive definitions.\n\nSay you know that $a_1 = something$ and $a_n =$ something to do with $a_{n-1}$ if $n$ is odd and something to do with $a_{\\frac n2}$ and $a_{n-1}$ if $n$ is even. And you want to prove something about $a_n$. Well, it is not just enough to assume that the proposition is true of $a_{n-1}$. You also have to assume it is true of $a_{\\frac n2}$. Can you assume that? Of course you can. To get to $a_{n-1}$ you had to have \"passed through\" $a_{\\frac n2}$ \"along the way\". So why not use it if you have it?", "date": "2019-05-22 08:35:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2461040/induction-and-proper-use-of-the-inductive-step", "openwebmath_score": 0.8813499212265015, "openwebmath_perplexity": 144.11465721627906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969694071564, "lm_q2_score": 0.8688267643505193, "lm_q1q2_score": 0.8545753723549964}}
{"url": "https://gmatclub.com/forum/in-the-figure-above-if-l1-l2-what-is-the-value-of-x-261473.html?sort_by_oldest=true", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 12 Nov 2019, 06:25\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# In the figure above if l1||l2, what is the value of x?\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 58988\nIn the figure above if l1||l2, what is the value of x?\u00a0 [#permalink]\n\n### Show Tags\n\n15 Mar 2018, 23:19\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n93% (00:47) correct 7% (00:57) wrong based on 58 sessions\n\n### HideShow timer Statistics\n\nIn the figure above if $$l_1||l_2$$, what is the value of x?\n\nA. 70\nB. 100\nC. 110\nD. 150\nE. 170\n\nAttachment:\n\n2018-03-16_1010.png [ 6.61 KiB | Viewed 1271 times ]\n\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 58988\nIn the figure above if l1||l2, what is the value of x?\u00a0 [#permalink]\n\n### Show Tags\n\n15 Mar 2018, 23:19\nBunuel wrote:\n\nIn the figure above if $$l_1||l_2$$, what is the value of x?\n\nA. 70\nB. 100\nC. 110\nD. 150\nE. 170\n\nAttachment:\n2018-03-16_1010.png\n\nFor other subjects:\nALL YOU NEED FOR QUANT ! ! !\n_________________\nIntern\nJoined: 24 Jan 2018\nPosts: 14\nRe: In the figure above if l1||l2, what is the value of x?\u00a0 [#permalink]\n\n### Show Tags\n\n15 Mar 2018, 23:53\nI got A - 70\u00b0, you can complete the top triangle as 30\u00b0, 40\u00b0 and 110\u00b0 because l1 and l2 are parallel, x = 180\u00b0 - 110\u00b0 which is 70\u00b0\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 3648\nRe: In the figure above if l1||l2, what is the value of x?\u00a0 [#permalink]\n\n### Show Tags\n\n16 Mar 2018, 08:35\nBunuel wrote:\nIn the figure above if $$l_1||l_2$$, what is the value of x?\n\nA. 70\nB. 100\nC. 110\nD. 150\nE. 170\n\nAttachment:\n\n2018-03-16_1010ed.png [ 10.8 KiB | Viewed 1021 times ]\n\nA couple of other ways to approach . . .\n\n$$l_1$$ and $$l_2$$ are cut by a transversal\n\nAlternate interior angles are congruent.\nLower right angle = $$40\u00b0$$\nOther angle of lower triangle, given = $$30\u00b0$$\n\n1) $$x$$ = exterior angle\nExterior angle $$x$$ = sum of opposite interior angles of lower triangle\nExterior angle $$x$$ $$= (30\u00b0 + 40\u00b0) =$$ $$70\u00b0$$\n\nOR\n\n2) Straight line: Third angle of lower \u2206 + $$x = 180\u00b0$$\nInterior angles of a triangle sum to 180\u00b0\nLower \u2206's third angle = $$110\u00b0$$\n($$(180\u00b0 - 30\u00b0 - 40\u00b0) = 110\u00b0$$ = third angle))\n\nThird angle lies on a straight line with $$x$$\n$$110\u00b0 + x\u00b0 = 180\u00b0$$\n$$x = 70\u00b0$$\n\n_________________\nSC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.\n\nNever doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has -- Margaret Mead\nSenior Manager\nJoined: 22 Feb 2018\nPosts: 416\nRe: In the figure above if l1||l2, what is the value of x?\u00a0 [#permalink]\n\n### Show Tags\n\n18 Mar 2018, 10:40\nBunuel wrote:\n\nIn the figure above if $$l_1||l_2$$, what is the value of x?\n\nA. 70\nB. 100\nC. 110\nD. 150\nE. 170\n\nAttachment:\nThe attachment 2018-03-16_1010.png is no longer available\n\nDrawing a additional parallel line l1 and l2 and using property of equal alternate angle between 2 parallel line\n\nwe get x= 30 +40\nx=70\nAttachments\n\nCapture.PNG [ 6.93 KiB | Viewed 926 times ]\n\n_________________\nGood, good Let the kudos flow through you\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 8345\nLocation: United States (CA)\nRe: In the figure above if l1||l2, what is the value of x?\u00a0 [#permalink]\n\n### Show Tags\n\n20 Oct 2019, 10:06\nBunuel wrote:\n\nIn the figure above if $$l_1||l_2$$, what is the value of x?\n\nA. 70\nB. 100\nC. 110\nD. 150\nE. 170\n\nAttachment:\n2018-03-16_1010.png\n\nDraw a line through the vertex of angle x that is parallel to both l1 and l2. We will see that angle x is the sum of the two given angles of 40 degrees and 30 degrees. Thus angle x = 70 degrees.\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nSVP\nJoined: 03 Jun 2019\nPosts: 1834\nLocation: India\nRe: In the figure above if l1||l2, what is the value of x?\u00a0 [#permalink]\n\n### Show Tags\n\n20 Oct 2019, 10:11\nBunuel wrote:\n\nIn the figure above if $$l_1||l_2$$, what is the value of x?\n\nA. 70\nB. 100\nC. 110\nD. 150\nE. 170\n\nAttachment:\n2018-03-16_1010.png\n\nx = 180 - (180 - 40 - 30) = 70\n\nIMO A\n_________________\n\"Success is not final; failure is not fatal: It is the courage to continue that counts.\"\n\nPlease provide kudos if you like my post. Kudos encourage active discussions.\n\nMy GMAT Resources: -\n\nEfficient Learning\nAll you need to know about GMAT quant\n\nTele: +91-11-40396815\nMobile : +91-9910661622\nE-mail : kinshook.chaturvedi@gmail.com\nRe: In the figure above if l1||l2, what is the value of x? \u00a0 [#permalink] 20 Oct 2019, 10:11\nDisplay posts from previous: Sort by", "date": "2019-11-12 13:25:26", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-the-figure-above-if-l1-l2-what-is-the-value-of-x-261473.html?sort_by_oldest=true", "openwebmath_score": 0.8127297759056091, "openwebmath_perplexity": 6448.63503853537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9566342012360932, "lm_q2_score": 0.8933093996634686, "lm_q1q2_score": 0.8545703240037562}}
{"url": "https://gmatclub.com/forum/in-the-xy-plane-the-straight-line-graphs-of-the-three-equations-above-207707.html", "text": "GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 20 Jan 2020, 16:45\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# In the xy-plane, the straight-line graphs of the three equations above\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 60515\nIn the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 13 Nov 2017, 21:24\n5\n66\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n61% (02:04) correct 39% (02:09) wrong based on 1317 sessions\n\n### HideShow timer Statistics\n\ny = ax - 5\ny = x + 6\ny = 3x + b\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n(1) a = 2\n(2) r = 17\n\n_________________\n\nOriginally posted by Bunuel on 23 Jul 2015, 10:47.\nLast edited by Bunuel on 13 Nov 2017, 21:24, edited 4 times in total.\nEdited the question.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 60515\nIn the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n26 Oct 2015, 10:02\n3\n35\nCEO\nJoined: 20 Mar 2014\nPosts: 2552\nConcentration: Finance, Strategy\nSchools: Kellogg '18 (M)\nGMAT 1: 750 Q49 V44\nGPA: 3.7\nWE: Engineering (Aerospace and Defense)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n23 Jul 2015, 11:30\n12\n8\nmcelroytutoring wrote:\nDS 136 from OFG 2016 (new question)\n\ny = ax - 5\ny = x + 6\ny = 3x + b\n\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n1) a = 2\n\n2) r = 17\n\nSolution provided by : mcelroytutoring\n\nLet's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.\n\n#1: r = ap - 5 (3 variables R,A,P)\n#2: r = p + 6 (2 variables R,P)\n#3: r = 3p + b (3 variables R,B,P)\n\n1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.\n\n2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.\nAttachments\n\nScreen Shot 2015-07-23 at 10.36.05 AM.png [ 189.38 KiB | Viewed 23039 times ]\n\nScreen Shot 2015-07-23 at 10.31.01 AM.png [ 30.94 KiB | Viewed 23016 times ]\n\n##### General Discussion\nSenior Manager\nJoined: 10 Mar 2013\nPosts: 459\nLocation: Germany\nConcentration: Finance, Entrepreneurship\nSchools: WHU MBA\"20 (A\\$)\nGMAT 1: 580 Q46 V24\nGPA: 3.88\nWE: Information Technology (Consulting)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n24 Sep 2015, 01:53\n5\n6\nif each of the 3 equations contains points (p,r) this means that they intersect in that point\n1. a=2\nFind the intercept Intercept for three simultaneous equations\ny=2x-5\ny=x+6\ny=3x+b\nLet's use the first 2 equations: plug y=x+6 in the secod equation\nx+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation\n17=33+b -> b=-16 SUFFICIENT\n\n2. Here we have directly the value for Y, let's plug it in the 2nd equation\ny=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above\n17=33+b -> b=-16 SUFFICIENT\n\nAnswer (D) Most important point is here to catch the hint about intersection of 3 lines at one point\nIntern\nStatus: GMAT1:520 Q44 V18\nJoined: 03 Sep 2015\nPosts: 10\nLocation: United States\nConcentration: Strategy, Technology\nWE: Information Technology (Computer Software)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n26 Oct 2015, 10:50\n4\nI think it's D.\n\nKeeping point (p,r) in all the equations we get :\n\np = ar -5 -----(1)\np = r + 6 ------(2)\np = 3r + b ------(3)\n\nNow consider (1) if a = 2 from (1) and (2) we get\nr = 11 , p=17 and putting in (3) we can get b.\n\nSimilarly for (2) we can get the values for r and p and hence can get the value for b.\n\nSo both statements individually are correct to answer the question.\nManager\nJoined: 13 Apr 2015\nPosts: 73\nConcentration: General Management, Strategy\nGMAT 1: 620 Q47 V28\nGPA: 3.25\nWE: Project Management (Energy and Utilities)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n26 Oct 2015, 19:14\n2\nm2k wrote:\nI think it's D.\n\nKeeping point (p,r) in all the equations we get :\n\np = ar -5 -----(1)\np = r + 6 ------(2)\np = 3r + b ------(3)\n\nNow consider (1) if a = 2 from (1) and (2) we get\nr = 11 , p=17 and putting in (3) we can get b.\n\nSimilarly for (2) we can get the values for r and p and hence can get the value for b.\n\nSo both statements individually are correct to answer the question.\n\nApproach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.\nManager\nJoined: 21 Sep 2015\nPosts: 73\nLocation: India\nGMAT 1: 730 Q48 V42\nGMAT 2: 750 Q50 V41\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n10 Jun 2016, 11:23\n3\ny = ax - 5 ... eq 1\ny = x + 6 ... eq 2\ny = 3x + b ...eq 3\n\nTotal of 4 variables are present.\n\nStatement 1 : a = 2\n\nInsert in eq 1\n\nWe have y = 2x -5 and y = x+6\n\nSolving we get x = 11 and y = 17\n\nSubstitute in eq 3 and we get value of b\n\nStatement 2: r=17\n\nThis means the y co- ordinate is 17\nSubstitute in eq 2 we get x as 11\n\nAgain can find value of b from equation 3\n\nHence D\nIntern\nJoined: 20 Jun 2013\nPosts: 48\nLocation: India\nConcentration: Economics, Finance\nGMAT 1: 430 Q39 V25\nGPA: 3.5\nWE: Information Technology (Other)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n18 Jan 2017, 15:18\nwe used information from both 1 and 2 then how can the answer be D... should it not be C... some one kindly clarify.......... clearly am a zero in ds and that too a big one\nDirector\nJoined: 14 Nov 2014\nPosts: 585\nLocation: India\nGMAT 1: 700 Q50 V34\nGPA: 3.76\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n18 Jan 2017, 20:18\n1\ny = ax - 5---------1\ny = x + 6---------2\ny = 3x + b--------3\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n(1) a = 2\n(2) r = 17\n\nAll three line intersect each other at common point (p,r).\n1. given a = 2\nputting in equation 1 .= y=2x-5\nequating 1(after replacing value of a) and 2 we will get value of (p,r)\nputting (p,r) in equation 3 we will get value for b---suff..\n\n2 given r = 17.\nputting in equation 2 we will get value of x.i'e p.\nNow as we know common point of intersection ,putting p,r in equation 3 , we will get value of b\nIntern\nJoined: 20 Jun 2013\nPosts: 48\nLocation: India\nConcentration: Economics, Finance\nGMAT 1: 430 Q39 V25\nGPA: 3.5\nWE: Information Technology (Other)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n22 Jan 2017, 15:51\nthanks sobby for your response... highly appreciated...\nManager\nJoined: 17 Feb 2014\nPosts: 98\nLocation: United States (CA)\nGMAT 1: 700 Q49 V35\nGMAT 2: 740 Q48 V42\nWE: Programming (Computer Software)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n16 Mar 2017, 08:58\n1\n(eq1) $$y = ax - 5$$\n(eq2) $$y = x + 6$$\n(eq3) $$y = 3x + b$$\n\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?\n\n1) $$a = 2$$\n2) $$r = 17$$\n\nSolution:\n\n1) $$a = 2$$\n- Putting the value of a in eq1, we get: $$y = 2x - 5$$\n- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]\n(alternatively, faster method)\n- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.\nNOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.\n\n2) $$r = 17$$\n- Since, point (p, r) lie on all the line, we can plugin the point in above equation\n$$r = ap - 5 => 17 = ap - 5$$\n$$r = p + 6 => 17 = p + 6$$\n$$r = 3p + b => 17 = 3p + b$$\n- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.\n\nCurrent Student\nJoined: 25 Feb 2017\nPosts: 34\nLocation: Korea, Republic of\nSchools: LBS '19 (A)\nGMAT 1: 720 Q50 V38\nGPA: 3.67\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n30 Apr 2017, 23:51\ny = ax - 5\ny = x + 6\ny = 3x + b\n\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n1) a = 2\n\n2) r = 17\n\nMy 2 cents.\n\nIt is important to realize from the Question stem that the 3 equations intersect as (p,r).\n\nFor 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.\n\nFor 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.\n\nSo D.\nIntern\nJoined: 27 Apr 2015\nPosts: 8\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n10 Sep 2017, 19:04\n1\nWould it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?\nIntern\nJoined: 11 Sep 2017\nPosts: 36\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n13 Nov 2017, 20:21\n2\nwhat is the significance of the the line \"In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)\",\n\ni solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables\nRetired Moderator\nJoined: 17 Jun 2016\nPosts: 498\nLocation: India\nGMAT 1: 720 Q49 V39\nGMAT 2: 710 Q50 V37\nGPA: 3.65\nWE: Engineering (Energy and Utilities)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n13 Nov 2017, 22:31\n2\nCheryn wrote:\nwhat is the significance of the the line \"In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)\",\n\ni solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables\n\nThe highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.\n\n_________________\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 15951\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: Q170 V170\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n11 Dec 2017, 14:28\n3\n1\nHi All,\n\nWe're given the equations for 3 lines (and those equations are based on 4 unknowns: 2 variables and the 2 'constants' A and B):\n\nY = (A)(X) - 5\nY = X + 6\nY = 3X + B\n\nWe're told that the three lines all cross at one point on a graph (p,r). We're asked for the value of B. While this question looks complex, it's actually built around a 'system' math \"shortcut\" - meaning that since we have 3 unique equations and 4 unknowns, we just need one more unique equation (with one or more of those unknowns) and we can solve for ALL of the unknowns:\n\n1) A =2\n\nWith this information, we now have a 4th equation, so we CAN solve for B.\nFact 1 is SUFFICIENT\n\n2) R = 17\n\nThis information tell us the x co-ordinate where all three lines will meet, so it's the equivalent of having X=17 to work with. This 4th equation also allows us to solve for B.\nFact 2 is SUFFICIENT\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\nContact Rich at: Rich.C@empowergmat.com\n\nThe Course Used By GMAT Club Moderators To Earn 750+\n\nsouvik101990 Score: 760 Q50 V42 \u2605\u2605\u2605\u2605\u2605\nENGRTOMBA2018 Score: 750 Q49 V44 \u2605\u2605\u2605\u2605\u2605\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2806\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jan 2018, 11:00\n1\n1\nBunuel wrote:\ny = ax - 5\ny = x + 6\ny = 3x + b\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n(1) a = 2\n(2) r = 17\n\nWe can begin by substituting p and r for x and y, respectively, in the three given equations.\n\n1) r = ap \u2013 5\n\n2) r = p + 6\n\n3) r = 3p + b\n\nStatement One Alone:\n\na = 2\n\nWe can substitute 2 for a in the equation r = ap \u2013 5. Thus, we have:\n\nr = 2p \u2013 5\n\nNext we can set equations 1 and 2 equal to each other.\n\n2p \u2013 5 = p + 6\n\np = 11\n\nSince p = 11, we see that r = 11 + 6 = 17\n\nFinally, we can substitute 11 for p and 17 for r in equation 3. This gives us:\n\n17 = 3(11) + b\n\n17 = 33 + b\n\n-16 = b\n\nStatement one alone is sufficient to answer the question.\n\nStatement Two Alone:\n\nr = 17\n\nWe can substitute r into all three equations and we have:\n\n1) 17 = ap \u2013 5\n\n2) 17 = p + 6\n\n3) 17 = 3p + b\n\nWe see that p = 11. Now we can substitute 11 for p in equation 3 to determine a value for b.\n\n17 = 3(11) + b\n\n-16 = b\n\nStatement two alone is also sufficient to answer the question.\n\n_________________\n\n# Jeffrey Miller\n\nJeff@TargetTestPrep.com\n181 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nManager\nJoined: 29 Dec 2018\nPosts: 82\nLocation: India\nWE: Marketing (Real Estate)\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n20 Jul 2019, 23:56\nVideo solution for the same\n\nhttps://gmatquantum.com/official-guides ... ial-guide/\n_________________\nKeep your eyes on the prize: 750\nIntern\nJoined: 15 Aug 2018\nPosts: 5\nRe: In the xy-plane, the straight-line graphs of the three equations above\u00a0 [#permalink]\n\n### Show Tags\n\n17 Aug 2019, 23:34\nBunuel wrote:\ny = ax - 5\ny = x + 6\ny = 3x + b\nIn the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?\n\n(1) a = 2\n(2) r = 17\n\nDS07713\nRe: In the xy-plane, the straight-line graphs of the three equations above \u00a0 [#permalink] 17 Aug 2019, 23:34\nDisplay posts from previous: Sort by", "date": "2020-01-20 23:46:42", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/in-the-xy-plane-the-straight-line-graphs-of-the-three-equations-above-207707.html", "openwebmath_score": 0.7582874298095703, "openwebmath_perplexity": 2490.5410678915914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9582261215975975, "lm_q2_score": 0.8918110404058913, "lm_q1q2_score": 0.8545566344460556}}
{"url": "https://math.stackexchange.com/questions/1783555/resolving-secx-left-sin3x-sin-x-cos2x-right-tanx", "text": "# Resolving $\\sec{x}\\left(\\sin^3x + \\sin x \\cos^2x\\right)=\\tan{x}$\n\nGoing steadily through my book, I found this exercise to resolve\n\n$$\\sec{x}\\left(\\sin^3x + \\sin x \\cos^2x\\right)=\\tan{x}$$\n\nHere's how I resolve it ($LHS$) and again bear with me as I truly reverting to a feeling of vulnerability, like a child actually\n\nAs $\\sec x$ is equal to $\\frac{1}{\\cos x}$\n\n$$\\frac{(\\sin^3x+\\sin x\\cos^2x)}{\\cos x}$$\n\nI'm factorizing one $\\sin x$\n\n$$\\frac{\\sin x(\\sin^2x+\\cos^2x)}{\\cos x} = \\frac{\\sin x(1)}{\\cos x} = \\tan x$$\n\nThat seems to work otherwise I completly messed this up\n\nReading the book's solution, I have something different...\n\n\\begin{align*} LHS&=\\frac{\\sin^3x}{\\cos x}+ \\sin x \\cos x \\\\[4pt] &=\\frac{\\sin x}{\\cos x}-\\frac{\\sin x\\cos^2x}{\\cos x}+\\sin x\\cos x\\\\[4pt] &= \\tan x\\end{align*}\n\nWhat did I miss?\n\n\u2022 They messed this up. \u2013\u00a0Yves Daoust May 13 '16 at 10:00\n\u2022 Shorter: multiplying both members by $\\cot(x)$ yields $\\sin^2(x)+\\cos^2(x)=1$. \u2013\u00a0Yves Daoust May 13 '16 at 10:04\n\u2022 Hi @YvesDaoust you see this is where dogma starts. My way is better than your way and at the end, you are disgusting a generation of people but for what purpose...? \u2013\u00a0Andy K May 13 '16 at 10:15\n\u2022 There is absolutely no intent to disgust anyone. You seem to miss that my first remark was an compliment to you. There is beauty in making things short. \u2013\u00a0Yves Daoust May 13 '16 at 10:22\n\u2022 @YvesDaoust I understood correctly what you said. I was more empathetic about these 2 chaps who went up to create a complicate solution for what it was. They could have done something simplier and also mentionned that, there were other ways (pluralist views) to solve that problem. As we said J'avais bien compris :) \u2013\u00a0Andy K May 13 '16 at 10:24\n\nSo the book does about the same thing as you but in a different order (your solution is hence correct too). By the trigonometric one we get $\\sin^2 x= (1-\\cos^2 x)$. Thus $$\\frac{\\sin^3 x}{\\cos x}=\\frac{\\sin x (\\sin^2 x)}{\\cos x}= \\frac{\\sin x(1-\\cos^2 x)}{\\cos x} = \\frac{\\sin x-\\sin x\\cos^2 x}{\\cos x}= \\frac{\\sin x}{\\cos x} -\\frac{\\sin x\\cos ^2 x}{\\cos x}$$ Thus $$LHS =\\frac{\\sin^3 x}{\\cos x}+\\sin x\\cos x= \\frac{\\sin x}{\\cos x} -\\frac{\\sin x\\cos ^2 x}{\\cos x} +\\sin x\\cos x = \\frac{\\sin x}{\\cos x}=\\tan x$$", "date": "2019-06-24 17:41:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1783555/resolving-secx-left-sin3x-sin-x-cos2x-right-tanx", "openwebmath_score": 0.8126027584075928, "openwebmath_perplexity": 2058.5377104889412, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924802053235, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8545240577881882}}
{"url": "https://math.stackexchange.com/questions/2943615/given-s-n-sum-dots-and-a-n-sum-dots-prove-that-a-n-s-n-1-over", "text": "# Given $S_n = \\sum \\dots$ and $a_n = \\sum \\dots$ prove that $a_n = S_n + {1\\over n\\cdot n!}$\n\nI'm trying to solve the following problem:\n\nLet: \\begin{align} S_n &= 2 + {1\\over2!} + {1\\over3!} + {1\\over 4!} + \\dots + {1\\over n!} \\\\ a_n &= 3 - {1\\over 1\\cdot2\\cdot2!} - {1\\over 2\\cdot3\\cdot3!} - \\dots - {1\\over (n-1)\\cdot n\\cdot n!} \\end{align} Prove that: $$a_n = S_n + {1\\over n\\cdot n!}$$\n\nMy thought on this are:\n\nManipulating the sums here would become pretty hard so I decided to use another approach. I've tried to look what the first terms of each sum are going to be in order to observe a pattern:\n\n\\begin{align} S_1 &= 2 \\\\ S_2 &= 2 + {1\\over 2!} = S_1 + {1\\over 2!} \\\\ S_3 &= 2 + {1\\over 2!} + {1\\over 3!} = S_2 + {1\\over 3!}\\\\ &\\vdots \\\\ S_{n+1} &= S_n + {1\\over (n+1)!} \\end{align}\n\nA similar thing is done to $$a_n$$:\n\n$$a_{n+1} = a_n - {1\\over n\\cdot (n+1) \\cdot (n+1)!}$$\n\nSo potentially if we could find closed forms of both $$S_n$$ and $$a_n$$ it would become easier to reason about them.\n\nBy the way, I've observed both of the sums approach $$e$$ at infinity but from different sides. $$S_n$$ is starting from $$2$$ adding more terms as $$n$$ grows, while $$a_n$$ starts from 3 and decreases the sum as $$n$$ grows. So that will be our initial conditions for both recurrences:\n\n$$S_1 = 2 \\\\ a_1 = 3$$\n\nThe problem just reduced to finding closed forms of the recurrences. Both of them are non-homogenous and here is where I got stuck. Solving for homogenous part is easy since roots of both characteristic equations $$\\lambda_{a_n} = \\lambda_{S_n} = 1$$.\n\nI couldn't find a particular solution for them. I've tried yet another approach expressing $$S_{n+1}$$ and $$S_{n+2}$$ trying to get rid of the $$1\\over (n+1)!$$ or at least change its form so the guess for particular solution is more obvious.\n\nMy questions are:\n\n1. Is it even a valid approach to solve the problem?\n2. If so then how could I find a closed form for the recurrences? Especially i'm interested in finding a particular solution. (Yet a complete flow is very appreciated and encourages learning by example)\n3. Not sure how to phrase that better, but does there exist a table of \"guesses\" for particular solution of the recurrences in some form of $$x_{n+1} = x_n + F(n)$$. Like if the \"free term\" is a constant say $$F(n) = 2$$ then the guess for particular solution is some constant $$B$$.\n\nPlease excuse me if there is some vagueness or inaccuracy in the terminology, English is not my native language and I have almost no background in Maths.\n\n\u2022 It seems unlikely to find a simple closed form, because, as you noticed, the limit converges to $e$. \u2013\u00a0user600464 Oct 5 '18 at 18:59\n\u2022 @user600464 is that because $e$ is irrational and that is why no \"simple\" formula for an irrational number may be obtained? \u2013\u00a0roman Oct 5 '18 at 19:02\n\u2022 There is actually a simple formula, the limit $\\frac{n^n}{(n- 1)^n}$. However, to compare to this sums, we would need many terms \u2013\u00a0user600464 Oct 5 '18 at 19:34\n\n$$a_2 - S_2 = 3- 1/4 - (1 + 1/1! + 1/2!)$$ [I think there is some typo in your expression of $$S_n$$] $$= 1/4 = 1/(2 \\cdot 2!)]$$. Suppose the equation holds for $$n$$. Then for $$n+1$$, \\begin{align*} a_{n+1} - S_{n+1} &= a_n - S_n - \\frac 1 {n(n+1) \\cdot (n+1)!} - \\frac 1{(n+1)!} \\\\ &= \\frac 1 {n!n }-\\frac 1 {n(n+1) \\cdot (n+1)!} - \\frac 1{(n+1)!} \\\\ &= \\frac 1 {(n+1)! (n+1)} \\left( \\frac {(n+1)^2}n -\\frac 1n -(n+1) \\right)\\\\ &= \\frac 1 {(n+1)! (n+1)}. \\end{align*} Thus the equation holds for all $$n \\geqslant 2$$ by induction principle.\n\\begin{align} \\color{red}{a_{n}-S_{n}} &=\\left(3-\\sum_{k=2}^{n}\\frac{1}{(n-1)\\cdot n\\cdot n!}\\right)-\\left(2+\\sum_{k=2}^{n}\\frac{1}{n!}\\right) \\\\[2mm] &=1-\\sum_{k=2}^{n}\\frac{1}{n!}\\left(1+\\frac{1}{(n-1)\\cdot n}\\right) =1-\\sum_{k=2}^{n}\\frac{1}{n!}\\left(1+\\frac{1}{n-1}-\\frac{1}{n}\\right) \\\\[2mm] &=1-\\sum_{k=2}^{n}\\frac{1}{n!}\\left(\\frac{n}{n-1}-\\frac{1}{n}\\right) =1-\\sum_{k=2}^{n}\\left(\\frac{1}{(n-1)\\cdot (n-1)!}-\\frac{1}{n\\cdot n!}\\right) \\\\[2mm] &=1-\\frac{1}{1\\cdot1!}+\\frac{1}{2\\cdot2!}-\\frac{1}{2\\cdot2!}+\\frac{1}{3\\cdot3!}-\\cdots+\\frac{1}{n\\cdot n!}=\\color{red}{\\frac{1}{n\\cdot n!}}\\quad\\left\\{\\text{telescoping}\\right\\} \\end{align}", "date": "2019-08-25 00:09:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2943615/given-s-n-sum-dots-and-a-n-sum-dots-prove-that-a-n-s-n-1-over", "openwebmath_score": 0.9999663829803467, "openwebmath_perplexity": 530.4968236603773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924810166349, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.85452405234596}}
{"url": "https://math.stackexchange.com/questions/3061903/in-proof-writing-is-it-mathematically-sound-to-prove-uniqueness-before-proving", "text": "# In proof writing, is it mathematically sound to prove uniqueness before proving existence?\n\nAs stated in the title, I'd like to find out is whether or not it is always mathematically sound to prove the uniqueness of something before proving the existence of said something.\n\nI am still relatively new to proofs and, in this light, I was intrigued by an observation that some authors choose to prove uniqueness before existence. I understand that some do this because, allegedly, \"it is easier to prove the uniqueness part first\". And in an exam setting, it's better to tackle the easier parts of a task.\n\nHowever, as I see it, if done this way, aren't we essentially proving a proposition about something that, at that point, we have not yet guaranteed exists? (Thus, might not even exist at all). I understand that in most cases the existence is proved immediately after the proof of uniqueness but, generally, doesn't uniqueness depend on existence?\n\nMy example will be derived from page $$40$$ of The Real Analysis Lifesaver: All the Tools You Need to Understand Proofs by Raffi Grinberg:\n\nTheorem 5.8. (Existence of Roots in $$\\mathbb{R}$$)\n\nEvery positive real number has a unique positive $$n$$th root, for any $$n \\in \\mathbb{N}$$.\n\nIn symbols:\n\n$$\u2200x \u2208 R$$ with $$x > 0, \u2200n \u2208 N, \u2203y \u2208 R$$ unique, such that $$y > 0$$ and $$y^n = x.$$\n\nNote that even-numbered roots $$(\\sqrt{x},\\sqrt[4]x,\\sqrt[6]x, etc.)$$ signify two numbers in $$\\mathbb{R}$$, namely, $$+y$$ and $$\u2212y$$. The theorem asserts that there is one and only one positive real root.\n\nProof.\n\nThe uniqueness of $$y$$ is the easiest part of the proof, so we\u2019ll start there. For any two different positive real numbers, the fact that they are different means one must be greater than the other. If there were two positive real roots $$y_1$$ and $$y_2$$ such that $$y_1^n = x$$ and $$y_2^n = x$$ we would have $$0 < y_1 < y_2$$. But then $$0 < y_1^n < y_n^2$$ , meaning $$0 < x < x$$, which is impossible. Thus only one positive real root can exist.\n\nTo prove that $$\\sqrt[n]x$$ exists in $$\\mathbb{R}$$, let\u2019s first figure out our game plan and then write it up formally $$...$$\n\nWhile the proof is totally clear, I'd really like feedback on this format. As an aspiring mathematician, is this a format I can adapt and use in my proofs or perhaps it is not good practice in general? What if, say, I have a conjecture that there exist unique integers possessing some property and this conjecture is something new (i.e. hasn't been proven yet). I try proving they exist but run out of luck. Would it be worth the effort of trying to prove the uniqueness of such integers not knowing they exist?\n\n\u2022 Yeah, that's perfectly fine. It doesn't matter: you're proving two things: 1) There is at least one of these; and 2) there aren't two different ones. There's no problem doing them whatever way round you like. There are a fair few results of the form \"there are at most $n$ things with property X\", and this is just the special case where $n = 1$. \u2013\u00a0user3482749 Jan 4 at 17:57\n\u2022 Sure. Naturally, if you show uniqueness first what you have actually shown at that point is that there is at most one object with the desired property. \u2013\u00a0Andr\u00e9s E. Caicedo Jan 4 at 17:57\n\u2022 Even in the case where authors first prove uniqueness, they usually mean, under the assumption that there exists such an object, it is unique. \u2013\u00a0Dietrich Burde Jan 4 at 18:01\n\u2022 I prefer existence before uniqueness, but that is just a preference and not a strong recommendation. \u2013\u00a0David G. Stork Jan 4 at 18:02\n\u2022 An explanation: I prefer existence before uniqueness because in general existence is simpler. Imagine the awkward case where you prove uniqueness first, then find you cannot prove existence! THAT is weird. \u2013\u00a0David G. Stork Jan 4 at 18:16\n\nI can see the unease: if what you're proving unique doesn't exist, might that invalidate whatever manipulations you've done with it?\n\nThis is similar to the situation with proof by contradiction, where you don't just reason about something that might not be true: you reason about it expecting it to be untrue, but making all the logical steps anyway.\n\nIn either case, you end up showing \"If $$A$$ then $$B$$\"\u2014whether it's \"If $$x$$ exists then it's unique\" or \"If $$x$$ exists then it has contradictory properties\". In one case we go on to apply $$B$$ to $$x$$ once we know it exists, and in the other we use the impossibility of $$B$$ to disprove the existence of $$x$$.\n\nA separate thought: proofs are presented as a sequence of steps and inferences because as humans, we process written information sequentially. But what's behind a mathematical theorem is more like a network of theorems and logical steps, some dependent on others, which together imply the theorem. We can't take them all in simultaneously, so we arrange them in a suitable order to let us see the logic most clearly. It's a matter of convenience.\n\nIf $$X$$ and $$Y$$ are both true and together imply $$Z$$, it doesn't really matter which order the various parts of that are proved in\u2014all that matters from a validity point of view is that they're all proved and brought together.\n\n\u2022 Thanks! Your first paragraph alone asks my question even better than I could in my long post! \u2013\u00a0E.Nole Jan 7 at 2:56\n\nThis is a perfectly reasonable argument. The author even tells you he's doing it this way. He then uses the argument to allow the reader to become comfortable with the assertion and the notation.\n\nThere are other reasons for starting with uniqueness. Often if you know there's just one of something then it might be easier to search for it.\n\nWhen faced with a problem with an easy part and a hard part the argument for starting with the easy part is that it's a good warmup. That's what I usually do, and what I like to see in things I read.\n\nThe argument for starting with the hard part is that if you fail there you haven't wasted time on the easy part. For example, proving that there's at most one positive rational number whose square is $$2$$ is not going to tell you very much about the rational square root of $$2$$ that does not exist.\n\nStarting with the hard part, the bottleneck, is often the preferred strategy when writing software.\n\nPS The author is wrong to say that when $$x >0$$, $$\\sqrt{x}$$ signifies two two numbers. $$4$$ does have two real square roots, but the mathematical convention is that only $$2 = \\sqrt{4}$$.\n\n\u2022 Nice typo near the end: \"two two\" (Shame to fix it really!) \u2013\u00a0timtfj Jan 4 at 19:15\n\u2022 @timtfj I will leave it thanks. \u2013\u00a0Ethan Bolker Jan 4 at 19:33", "date": "2019-10-23 20:18:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3061903/in-proof-writing-is-it-mathematically-sound-to-prove-uniqueness-before-proving", "openwebmath_score": 0.8450634479522705, "openwebmath_perplexity": 265.5858964596029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924810166349, "lm_q2_score": 0.8791467675095292, "lm_q1q2_score": 0.8545240477293421}}
{"url": "https://math.stackexchange.com/questions/2499234/ideal-generated-by-a-finite-number-of-polynomials", "text": "# Ideal Generated by a Finite Number of Polynomials?\n\nBackground\n\nI have been confused about a particular definition in the textbook for my abstract algebra class, Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea. It is frustrating because I feel that I have a partial grasp on what the definition is trying to say, but when it comes down to it I simply find my confused. So, instead of suffering by myself for any longer with this definition, I have decided to turn to you lovely folks to help me understand this seemingly simple concept.\n\nWhat I Understand\n\nFirst off, I understand (at least in the context of this book) what an ideal is. The definition my book gives for an ideal is\n\nDefinition. A subset $$I\\subseteq k[x_1,..., x_n]$$ is an ideal if it satisfies:\n\n(i) $$0\\in I$$.\n\n(ii) If $$f,g\\in I$$, then $$f+g\\in I$$.\n\n(iii) If $$f\\in I$$ and $$h\\in k[x_1,...,x_n]$$, then $$hf\\in I$$.\n\nI find this to be a simple, easy to understand definition. My problem, however, arises a few lines later when they define an ideal generated by a finite number of polynomials.\n\nWhat I Don't Understand\n\nAnd now, I give you the definition that has been causing me an incredible amount of confusion and frustration.\n\nDefinition. Let $$f_1,...,f_s$$ be polynomials in $$k[x_1,...,x_n]$$. Then we set $$\\langle f_1,...,f_s \\rangle=\\Big\\lbrace \\sum_{i=1}^s h_if_i \\ | \\ h_1,...,h_s\\in k[x_1,...,x_n] \\Big\\rbrace.$$\n\nI know what you are thinking. How does he not understand this? I wish I knew the answer to that question, but in the meantime, can someone please help me visualize what this set looks like? I understand that $$\\langle f_1,...,f_s \\rangle$$ is an ideal, but I don't understand its structure, if that makes sense. In other words, I can't visualize this definition in a way that makes sense to me. The authors do make a slightly helpful note, saying that \"we can think of $$\\langle f_1,...,f_s \\rangle$$ as consisting of all 'polynomial consequences' of the equations $$f_1=f_2=...=f_s=0$$.\"\n\nTo elaborate a little more on my confusion, what I'm asking for is a less \"compact\" definition. When I read this definition, for whatever reason the only thing I can come up with is $$f_1h_1+f_2h_2+...+f_sh_s.$$ But that doesn't make sense, because $$\\langle f_1,...,f_s \\rangle$$ is supposed to generate a set, not just a single polynomial.\n\nAs always, thank you all for your time. If you find this to be a stupid or silly question, then I'm sorry to have disappointed you -- I'm a slow learner, and I get hung up on stupid things sometimes.\n\nOh, and Happy Halloween!\n\n\u2022 Suggestion: Try looking at one or two specific examples, such as: (1) In $k[x,y]$, look at $\\langle x^2, y^2 \\rangle$. This is equal to the set $\\{ x^2 h_1 + y^2 h_2 \\mid h_1, h_2 \\in k[x,y]\\}$. Some elements include $x^2 \\cdot 1 + y^2 \\cdot 0$, $x^2 \\cdot 0 + y^2 \\cdot 1$,... $x^2 \\cdot (x y^2 - 3 xy) + y^2 \\cdot (2x^3 - 5)$, etc. (2) Look at examples in $\\mathbb{Z}$, e.g., $\\langle 6,4 \\rangle = \\{ 6 h_1 + 4 h_2 \\mid h_1, h_2 \\in \\mathbb{Z} \\}$. Can you prove this is equal to the set of even numbers? \u2013\u00a0Zach Teitler Nov 1 '17 at 4:55\n\u2022 Makes alot more sense when you put it like that! Thank you @ZachTeitler! \u2013\u00a0Thy Art is Math Nov 1 '17 at 5:00\n\u2022 One more suggestion. Make sure you do Exercise 1.4.2, or at least try to do it. The result in Exercise 1.4.2 gets used constantly, over and over, throughout the rest of the book, including in a lot of exercises. It's the main way to show that two ideals are equal, or that one ideal is contained in another. So it is extremely useful. Also it's good practice with the definition of ideal, and proving a subset relationship. (There are a lot of other good exercises, I just want to highlight that particular one. I made sure to assign it in the class I'm teaching from IVA this semester.) \u2013\u00a0Zach Teitler Nov 1 '17 at 5:49\n\u2022 Read the Rings chapter in Algebra by Hungerford. He describes ideals in full generality very well (non-commutative and commutative). \u2013\u00a0user494247 Nov 11 '17 at 9:57\n\nWould it help to see the see the set with explicit polynomials in place of the $f_1, f_2, \\dots, f_n$?\n\nFor example, lets look at an explicit example when $n = 2$, so an ideal generated by 2 polynomials. Also, we'll work in a polynomial ring in two variables over $k$, i.e. $k[x,y]$.\n\nHere is the ideal generated by the polynomials $x^2 - 1$, $yx+x$.$$\\langle \\, x^2 - 1, \\, yx+x \\, \\rangle = \\{ \\, f\\cdot(x^2 - 1) + g\\cdot(yx + x) \\, | \\, f,g \\in k[x,y] \\, \\}.$$\n\nSo the elements of the set are any polynomial that can be written in the form $f \\cdot (x^2 - 1) + g \\cdot (yx+x)$. But we can choose $f,g$ to be $\\textit{any}$ polynomial we want. For example, we know $x^2 - 1$ itself is in that set because we can choose $f = 1, g = 0$. We also know that the polynomial $x^3-x+yx ^2 + x^2$ is in the set, because we can choose $f = x, g = x$.\n\nIf you are familiar with linear algebra you can maybe, in a way, draw a connection between an ideal generated by polynomials and the span of a set of vectors. You can think of it as, the set of all things that can be made from the objects defining it.\n\n\u2022 Also, is your class an Algebraic Geometry class or an Abstract Algebra class? I think IVA is a strange book to teach general algebra out of. \u2013\u00a0Prince M Nov 1 '17 at 7:38\n\u2022 Sadly I was absolutely terrible at linear algebra but seeing the explicit polynomials made this make so much more sense, so thank you for the help. The class is labeled rather generally as \"abstract algebra,\" so you are probably right that the course should be called \"algebraic geometry\". \u2013\u00a0Thy Art is Math Nov 1 '17 at 15:30\n\u2022 Just a note that the authors of IVA do write in the preface: \"For instance, the book could serve as a basis of a second course in undergraduate abstract algebra, but we think that it just as easily could provide a credible alternative to the first course.\" \u2013\u00a0J W Jan 3 at 15:16\n\nPerhaps a good way to understand this (or to understand anything, for that matter) is to look at examples.\n\nThe definition of ideal makes sense in any ring, so let\u2019s look at $\\Bbb Z$ first. Every ideal, you soon persuade yourself, is a set $d\\Bbb Z\\subset\\Bbb Z$, that is, just the set of all multiples of a given number $d$. What if you try to take two numbers and look at $\\langle d_1,d_2\\rangle\\subset\\Bbb Z$? Please do this with specific numbers. Like what about $d_1=8$ and $d_2=6$? You want to describe the totality of all numbers writable in the form $8m+6n$. You rapidly see that this is an ideal in the sense of the definition, and you see that the set is equal to $2\\Bbb Z$. This is basic number theory.\n\nNow do the same thing with a ring of polynomials, but in just one variable, $R=\\Bbb Q[x]$. One proves (you prove it, with Euclidean division of polynomials) that every ideal of $R$ is of form $fR$, where $f$ is a well-chosen polynomial. In fact, for a nonzero ideal $I$, you take $f$ to be a nonzero element of $I$ of least degree. So, what is $\\langle f_1,f_2\\rangle$, when $f_1$ and $f_2$ are two polynomials in $x$ that are given? Just as with numbers, it\u2019s the ideal $gR$ where $g$ is the greatest common divisor of $f$ and $g$. Convince yourself that $\\langle x^3-1,x^2-2x+1\\rangle$ is the set of all multiples of $x-1$.\n\nThings are no longer so simple when you have polynomials in more than one indeterminate. But at least you get some insight by looking at the simplest cases, and I hope you see that the set of all possible $h_1f_1+\\cdots+h_nf_n$ is an ideal.\n\n\u2022 @ThyArtisMath Note that the ring of polynomials in one variable is treated very explicitly and thoroughly in section 1.5 of the textbook Ideals, Varieties, and Algorithms. So, by all means feel free to think about that example and take a shot at proving some things about it, but don't feel like you have to work it out on your own. The textbook goes over all that. \u2013\u00a0Zach Teitler Nov 1 '17 at 5:40\n\nYou were close.\n\nThe ideal $(f_1,...,f_s)$ of $k[x_1,..., x_n]$ is the set of all polynomials which can be expressed as $h_1f_1+\\cdots +h_sf_s$, for some $h_1,...,h_s \\in k[x_1,..., x_n]$.\n\nIn other words, \"linear combinations\" of $f_1,...,f_s$ where the \"linear coefficients\" are arbitrary elements $h_1,...,h_s$ of $k[x_1,..., x_n]$.\n\nRegarding the author's point about common zeros . . .\n\nIf $\\bar{k}$ is an algebraically closed extension of $k$, and if there is some $a = (a_1,...,a_n) \\in \\left(\\bar{k}\\right)^n$ such that $f_1(a) = \\cdots = f_s(a) = 0$, then $a$ is automatically a zero of any polynomial of the form $h_1f_1+\\cdots +h_sf_s$, hence $a$ is a common zero for all members of the ideal $(f_1,...,f_s)$.\n\nIn particular, if you've identified a common zero $a \\in \\left(\\bar{k}\\right)^n$ for $f_1,...,f_s$, then if $g \\in k[x_1,..., x_n]$ is such that $g(a) \\ne 0$, it follows that $g$ is not in the ideal $(f_1,...,f_s)$.\n\nAn important, nontrivial result is a partial converse: If $g \\in k[x_1,...,x_n]$ is such that $g,g^2,g^3,...$ are not in the ideal $(f_1,...,f_s)$, there is some $a \\in \\left(\\bar{k}\\right)^n$ such that $a$ is a common zero of $f_1,...,f_n$, but $a$ is not a zero of $g$.", "date": "2021-06-25 05:13:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2499234/ideal-generated-by-a-finite-number-of-polynomials", "openwebmath_score": 0.8639252185821533, "openwebmath_perplexity": 151.4112291041073, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971992476960077, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8545240426241597}}
{"url": "https://math.stackexchange.com/questions/800657/solving-this-inequality", "text": "# Solving this inequality\n\nQuestion:\n\nSolve: $$\\frac{5x-6}{x+6}<1$$\n\nMy attempt:\n\n$$\\frac{5x-6-x-6}{x+6}<0$$ $$\\Rightarrow \\frac{4x-12}{x+6}<0$$ $$\\Rightarrow \\frac{x-3}{x+6}<0$$ $$\\Rightarrow (x-3)(x+6) < 0$$\n\nThus, it implies that the answer is $-6 < x < 3$. However, in my textbook, it is given as wrong answer.\n\nCan please someone tell what is the correct procedure for this.\n\nThanks.\n\nIf you're really unsure when something like that happens, you can always graph your rational function. But consider that in $\\ \\frac{x-3}{x+6} \\$ , there is a vertical asymptote at $\\ x \\ = \\ -6 \\$ and an $\\ x-$ intercept at $\\ x = 3 \\$ . So the real numbers are divided into intervals by these points, $\\ x \\ < \\ -6 \\ , \\ -6 \\ < \\ x \\ < \\ 3 \\ ,$ and $\\ x \\ > \\ 3 \\$ . Also, your rational function has a horizontal asymptote of $\\ y \\ = \\ 1 \\$ .\n\nSo the function is positive for \"a lot\" of the real numbers to start with. We can also use what we know about working with signed numbers. For large negative numbers, both the numerator and denominator are negative, and for large positive numbers, they are both positive. So for $\\ x \\ < \\ -6 \\$ and $\\ x \\ > \\ 3 \\$ , we can conclude that the ratio is positive. It will only be in the interval $\\ -6 \\ < \\ x \\ < \\ 3 \\$ that the numerator is negative, while the denominator is positive, so the ratio is negative in this interval.\n\nSo you are correct. Large introductory course textbook answers can be wrong anywhere from about $\\ \\frac{1}{3}$ % to $\\ 2$ % of the time, depending upon how carefully the submitted answers (of multiple authorship) have been combed through...\n\n\u2022 Thanks for the detailed method! And yes, you are right, the textbook is indeed large, so the author might have missed it. \u2013\u00a0Gaurang Tandon May 18 '14 at 19:47\n\u2022 The authors (probably of just about all of the omnibus texts, and not just in math) generally hire teaching assistants to work out answers, which are then collected into the backs of these books. They do not check what can be two to three thousand problem answers for accuracy... \u2013\u00a0colormegone May 18 '14 at 19:54\n\nHint: There are two cases, either $x < -6$ or $x > -6$. You can multiply both sides of the inequality by $x + 6$ and depending on which case you are considering, either it will reverse the inequality or leave it intact. Then you can rewrite the inequality with $x$ on one side and a number on the other side by adding/subtracting/dividing.\n\u2022 I am sorry but I do not understand you. The denominator has $(x+6)$, not $x-6$. \u2013\u00a0Gaurang Tandon May 18 '14 at 18:40\n\u2022 @GaurangTandon Sorry, I should have written $-6$. The point is that there are two cases, either $x+6$ is positive or negative and it will flip the inequality or not depending on what case you're in. \u2013\u00a0user2566092 May 18 '14 at 18:59", "date": "2019-12-08 05:45:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/800657/solving-this-inequality", "openwebmath_score": 0.8551735877990723, "openwebmath_perplexity": 165.26413667326779, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924769600771, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8545240395464145}}
{"url": "http://otvedaem.com/rw8q0t2p/compute-combinations-r-cb7d5c", "text": "Hi again, I am exploring if R can help me to get all possible combinations of members in a group. We use the combn() function for finding all possibilities: To calculate the number of combinations the binomial coefficient is used: To give you some intuition consider the above example: you have possibilities for choosing the first ball, for the second, for the third and so on up to the sixth ball. If argument FUN is not NULL, applies a function given by the argument to each point.If simplify is FALSE, returns a list; otherwise returns an array, typically a matrix. Venables, Bill. Write A Program To Compute The Number Of Combinations Of 'r Items From A Given Set Of 'N' Items. The following C function comb requires a two-dimensional array to store the intermediate results. / r! The row names are \u2018automatic\u2019. Show transcribed image text. While I\u2019m at it, I will examine combinations and permutations in R. As you may recall from school, a combination does not take into account the order, whereas a permutation does. / ((n - r)! Theorem 3. For factorial watch this video https://youtu.be/IBlnyh9hPwA Combination : C(n,r) = n!/(r! Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. : Proof. Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c . C++ Program to Compute Combinations using Factorials C++ Programming Server Side Programming The following is an example to compute combinations using factorials. The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Search the stuart package. 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In English we use the word \"combination\" loosely, without thinking if the order of things is important. e.g. Permutations and combinations have uses in math classes and in daily life. Without repetition simply means that when one has drawn an element it cannot be drawn again, so with repetition implies that it is replaced and can be drawn again. The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 r n, equals C(n;r) = nCr = n r = n! r = 5. and. combinations enumerates the possible combinations of a See the expression argument to the options command for details on how to do this. The first factors vary fastest. Combinations vs. Permutations. R/compute.combinations.R defines the following functions: compute.combinations. Questionnaire. The number of combinations of r objects is n C r = n! edit close. Combinations of a,b,c,d,e,f,g that have at least 2 of a,b or c . Thankfully, they are easy to calculate once you know how. / (64! Let us see this in action, as an example we\u2019ll see how many different ways there are of four runners reaching the finishing line: After this rather complicated function the calculation of the number of ways is simple, it is just the factorial function (it should again be obvious why): As you will see when solving real world problems with R the above functions often come in handy, so you should add them to your ever growing tool set \u2013 have fun and stay tuned! Command (\u2318)-R: Start up from the built-in macOS Recovery system. Compute the combinations of choosing r items from n elements. Computes all combinations of r elements from n. GitHub Gist: instantly share code, notes, and snippets. Generate All Combinations of n Elements, Taken m at a Time. Of course, when the values are large enough, a possible stack overflow will occur when recursion depths become large. Where, N! However, mathematicians are focused on how many elements will exist within a Combinatorics problem, and have little interest in actually going through the work of creati\u2026 For that we need to use the itertools package. The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0 r n, equals C(n;r) = nCr = n r = n! Permutation and combination. Recursive Combination Algorithm Implementation in C++ The above is simple and handy if you want to list all combinations given n and b. I start with a list of vectors and run the function below, which loops through 1:n where n is the number of sets and then uses combn to generate all combinations of my sets taken m at a time.. Syntax: Rules In Detail The \"has\" Rule. The order in which you combine them doesn't matter, as you will buy the two you selected anyways. After you\u2019ve entered the required information, the nCr calculator automatically generates the number of Combinations and the Combinations with Repetitions. Combinations are used in a large number of game type problems. Let's do a little experiment in R. We'll toss two fair dice, just as we did in an earlier post, and see if the results of the two dice are independent. A data frame containing one row for each combination of the supplied factors. This type of activity is required in a mathematics discipline that is known as combinatorics; i.e., the study of counting. This makes computations feasible for very large numbers of combinations. Fortunately, the science behind it has been studied by mathematicians for centuries, and is well understood and well documented. The columns are labelled by the factors if these are supplied as named arguments or named components of a list. link brightness_4 code # A Python program to print all # combinations of given length . Theorem 3. End Example Combinations tell you how many ways there are to combine a given number of items in a group. Two different methods can be employed to count r objects within n elements: combinations and permutations. We are \u2026 combn() function in R Language is used to generate all combinations of the elements of x taken m at a time. Rules In Detail The \"has\" Rule. Computer Glossary; Who is Who; Permutation and Combination in Python? One of the key advantage of python over other programming language is that it comes with huge set of libraries with it. Jan. 2001. http://cran.r-project.org/doc/Rnews, combinations(n, r, v=1:n, set=TRUE, repeats.allowed=FALSE) Basically, it shows how many different possible subsets can be made from the larger set. Then a comma and a list of items separated by commas. enumerates the possible permutations. permutations (n r)! This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example 3-3-3. Now, there are possible positions for the first ball that is drawn, for the second\u2026 and so on and because the order doesn\u2019t matter we have to divide by , which gives the binomial coefficient. Syntax: combn(x, m) Parameters: x: total number of elements taken r: number of elements taken at a time out of \u201cx\u201d elements Example 1: Will this result in a fractional number? The first factors vary fastest. The combinations were formed from 3 letters (A, B, and C), so n = 3; and each combination consisted of 2 letters, so r = 2. Vignettes . C (n,r): is the total number of combinations. Generates the combinations for choosing r items from a set of n items. We will perhaps cover those in a later post. If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. All these combinations are emitted in lexicographical order. R/compute.combinations.R defines the following functions: compute.combinations. Python Server Side Programming Programming. Description. stuart Subtests Using Algorithmic Rummaging Techniques. I assume that your rank starts at $0$, as this simplifies the code (for me).. In this section, we are going to learn how to find permutation and combination of a given sequence using python programming language. - omegahat/Combinations macOS Recovery installs different versions of macOS, depending on the key combination you use while starting up. My goal is to compute the intersections of several vectors (sets of identifiers, gene-names to be specific). The number says how many (minimum) from the list are needed for that result to be allowed. filter_none. Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. If your Mac is using a firmware password, you're prompted to enter the password. If x is a positive integer, returns all combinations of the elements of seq(x) taken m at a time. Vignettes . to access the Math PROB menu or press [ALPHA][WINDOW] to access the shortcut menu. r: number of elements chosen from the set for sampling! This video describes how to use the TI-30 to compute combinations Remember to use the second function button in order to access combinations. r!) This type of activity is required in a mathematics discipline that is known as combinatorics; i.e., the study of counting. Mathematics and statistics disciplines require us to count. Compute the combinations of choosing r items from n elements. Mathematics and statistics disciplines require us to count. nCm: Compute the binomial coef\ufb01cient (\"n choose m\"), where n is any real number and m is any integer. It returns r length subsequences of elements from the input iterable. For example, you have a urn with a red, blue and black ball. How to calculate combination. We have 4 choices (A, C, G and T) a\u2026 play_arrow. Package index. \"Programmers Note\", R-News, Vol 1/1, Computing with combinations in SAS/IML. 10^3 ## [1] 1000 nrow (P_wi) ## [1] 1000. How many combinations are there for selecting four? permutations if length of input sequence is n and input parameter is r. Combination This method takes a list and a input r as a input and return a object list of tuples which contain all possible combination of length r in a list form. Combinatorics has many applications within computer science for solving complex problems. Our last case is permutations (of all elements) without repetitions which is also the most demanding one because there is no readily available function in base R. So, we have to write our own: As you can see it is a recursive function, to understand recursion read my post: To understand Recursion you have to understand Recursion\u2026. which will be of the form n(n-1)...(n-r+1)/1.2...r. Similar to factorial, we initialize the result as 1 and multiply by n-i and divide by i+1. In R we use the choose() function to calculate it: So, you see that the probability of winning the lottery are about the same, no matter whether you play it\u2026 or not. If x is a positive integer, returns all combinations of the elements of seq(x) taken m at a time. = 11,238,513. Algorithms Begin function CalCombination(): Arguments: n, r. Body of the function: Calculate combination by using the formula: n! Example has 1,a,b,c. Let's take a more straightforward example where you choose three balls called R(red), B(blue), G(green). As far as I know there are no very convenient formulae for $r$ in between. For example, a deck of (n = 52) cards of which a (k = 5) card hand is drawn. Generate all combinations of the elements of x taken m at a time. * (n-r)!. So that gives . A permutation is an arrangement of objects in which the order is important (unlike combinations, which are groups of items where order doesn't matter).You can use a simple mathematical formula to find the number of different possible ways to order the items. rdrr.io Find an R package R language docs Run R in your browser R Notebooks. * (n-r)!) For the example, you can calculate 10! Exactly one of arguments \"x\" and \"n\" should be given; no provisions for function evaluation. No. Permutations . If you choose two balls with replacement/repetition, there are permutations: {red, red}, {red, blue}, {red, black}, {blue, red}, {blue, blue}, {blue, black}, {black, red}, {black, blue}, and {black, black}. When you think about it this is the same as because all the coefficients smaller than can be eliminated by reducing the fraction! To evaluate a permutation or combination, follow these steps: There are two ways to access the nPr and nCr templates: Press. See the expression argument to the Getting all possible combinations. options command for details on how to do this. where you have three positions with the numbers zero to nine each. Then we force the program to backtrack and find the next combination by evaluating the always failing ~. Only 1 Powerball number is picked from 26 choices, so there are only 26 ways of doing this. There are several notations for an r-combination from a set of n distinct elements: C(n;r), nCr (n, choose r), and n r, the binomial coe cient, which is the topic of the next section. Combin\u2026 A permutation is calculated n P r. Start on 'n' and count backwards 'r' numbers, multiplying them together. This is particularly important when completing probability problems.Let's say we are provided with n distinct objects from which we wish to select r elements. This is a C++ program to compute Combinations using Recurrence Relation for nCr. in a lottery it normally does not matter in which order the numbers are drawn). A combination is a way to select a part of a collection, or a set of things in which the order does not matterand it is exactly these cases in which our combination calculator can help you. Search the stuart package. Variations In this section, we will show you how it\u2019s done. In some cases, you can also refer to combinations as \u201cr-combinations,\u201d \u201cbinomial coefficient\u201d or \u201cn choose r.\u201d In some references, they use \u201ck\u201d instead of \u201cr\u201d, so don\u2019t get confused when you see combinations referred to as \u201cn choose k\u201d or \u201ck-combinations.\u201d How do you calculate combinations in Excel? rdrr.io Find an R package R language docs Run R in your browser R Notebooks. The combntns function provides the combinatorial subsets of a set of numbers. This number will go in the denominator of our probability formula, since it is the number of possible outcomes. That was simple! It generate nCr * r! Each row of C contains a combination of k items chosen from v. The elements in each row of C are listed in the same order as they appear in v. If k > numel(v), then C is an empty matrix. = 69! Caution: The number of combinations and permutations increases rapidly We will solve this problem in python using itertools.combinations() module.. What does itertools.combinations() do ? Note that AB and BA are considered to be one combination, because the order in which objects are selected does not matter. So there are 11,238,513 possible ways of picking 5 numbers from a choice of 69 numbers. For factorial watch this video https://youtu.be/IBlnyh9hPwA Combination : C(n,r) = n!/(r! All combinations of v, returned as a matrix of the same type as v. Matrix C has k columns and n!/((n \u2013k)! Another way of thinking about it is how many ways are there to, from a pool of six items, people in this example, how many ways are there to choose four of them. specified size from the elements of a vector. n: total number of elements in the given set. 10^3 ## [1] 1000 nrow (P_wi) ## [1] 1000. https://www.mathsisfun.com/combinatorics/combinations-permutations.html We will solve this problem in python using itertools.combinations() module.. What does itertools.combinations() do ? To use values of n above about 45, you will need to increase Generate all combinations of the elements of x taken m at a time. For p = 5 and k = 3, the problem is: \u201cFor each observation of the 5 variables, find the largest product among any 3 values.\u201d In the SAS/IML language, you can solve problems like this by using the ALLCOMB function to generate all combinations of size k from the index set {1,2,\u2026,p}. FAQ. Or use Option-Command-R or Shift-Option-Command-R to start up from macOS Recovery over the Internet. So I would like for each set of line with the same symbol calculate the average (or median) of the lines. with (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800. See the answer. * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. In other words: \"My fruit salad is a combination of apples, grapes and bananas\" We don't care what order the fruits are in, they could also be \"bananas, grapes and apples\" or \"grapes, apples and bananas\", its the same fruit salad. The core question you must be able to answer is how many elements there are in a substructure of yours. To use values of n above about 45, you will need to increase R's recursion limit. Posted on June 3, 2019 by Learning Machines in R bloggers | 0 Comments, The area of combinatorics, the art of systematic counting, is dreaded territory for many people so let us bring some light into the matter: in this post we will explain the difference between permutations and combinations, with and without repetitions, will calculate the number of possibilities and present efficient R code to enumerate all of them, so read on\u2026. Expert Answer . This is because first, we multiply by n and divide by 1. In python, we can find out the combination of the items of any iterable. Press the number on the menu that corresponds to the template you want to insert. : factorial . Recall that we need to find n!/r!(n-r)! !arg:(?m. Thank you in advance. A data frame containing one row for each combination of the supplied factors. The number says how many (minimum) from the list are needed for that result to be allowed. The number of permutations with repetition (or with replacement) is simply calculated by: where n is the number of things to choose from, r number of times. Here are the steps to follow when using this combination formula calculator: On the left side, enter the values for the Number of Objects (n) and the Sample Size (r). 5!) And then one would need some form of inclusion/exclusion to count those choices where some item is \u2026 When n gets large, the package provides a mechanism for dealing with each combination as it is generated so that one does not have to hold the entire collection around and operate on them after creating the entire collection. rows, where n is length(v). The word \"has\" followed by a space and a number. Now, either n or n-1 have to be even (as they are consecutive numbers). For example, if you want a new laptop, a new smartphone and a new suit, but you can only afford two of them, there are three possible combinations to choose from: laptop + smartphone, smartphone + suit, and laptop + suit. The formula for a combination is: nCr = (n!)/(r!(n-r)!). Home / R Documentation / base / expand.grid: Create a Data Frame from All Combinations of Factor Variables expand.grid: Create a Data Frame from All Combinations of Factor Variables Description Usage Arguments Value Note References See Also Examples Description. (n r)! 5!) Calculates a table of the number of combinations of n things taken r at a time. Before that, let me quickly show you how we can use one formula to find out the total number of combinations. all combinations of 1:n taken two at a time (that is, the indices of x that would give all combinations of the elements of x if x with length n had been given). For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Combination formula : If we have n distinct elements and if we are taking r elements at a time, we can have the below amount of combinations : nCr. Example has 1,a,b,c. This is the key distinction between a combination \u2026 In R: A biological example of this are all the possible codon combinations. The columns are labelled by the factors if these are supplied as named arguments or named components of a list. Next, we multiply by n-1 and divide by 2. In all cases, you can imagine somebody drawing elements from a set and the different ways to do so. : Proof. We won\u2019t cover permutations without repetition of only a subset nor combinations with repetition here because they are more complicated and would be beyond the scope of this post. Permutation implies that the order does matter, with combinations it does not (e.g. See the shortcut menu in the second screen. Generate All Combinations of n Elements, Taken m at a Time Description. This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. The idea is to fix one element after the other [for (i in 1:n) and cbind(v[i], ...)] and permute the remaining elements [perm(v[-i])] down to the base case when only one element remains [if (n == 1) v], which cannot be permuted any further. combos = combntns(set,subset) returns a matrix whose rows are the various combinations that can be taken of the elements of the vector set of length subset.Many combinatorial applications can make use of a vector 1:n for the input set to return generalized, indexed combination subsets.. * (n-r)!) Package index. Let us start with permutations with repetitions: as an example take a combination lock (should be permutation lock really!) Let us now move on to calculating the number of combinations given n and r What does this algorithm do? r! n = 69. and. For this calculator, the order of the items chosen in the subset does not matter. Using the TI-84 Plus, you must enter n, insert the command, and then enter r. See the PROB menu in the first screen. What makes matters a little bit more complicated is that the recursive call is within a for loop. - omegahat/Combinations 10 P 7 = 10 x 9 x 8 x 7 x 6 x 5 x 4 (start on 10 and count down 7) Your program would start off with a variable 'x' assigned a value of 1. So in your example, we're ordering combinations lexicographically so we can use the binomial coeffecient to find how many elements there are of our substructures. The word \"has\" followed by a space and a number. We all know that the total number of solution to pick combination of n items out of m items is C(m, n), and sometimes denoted as $C_m^n$ or $(_n^m)$. I will have only a single line by gene in the end. stuart Subtests Using Algorithmic Rummaging Techniques. This is particularly important when completing probability problems.. Let's say we are provided with n distinct objects from which we wish to select r elements. The row names are \u2018automatic\u2019. number of things n \u2266300 \\) Customer Voice. Combinations and Permutations What's the Difference? To calculate combination, all you need is the formula, that too, in case you want to determine it manually. r! It returns r length subsequences of elements from the input iterable. Limitations. rdrr.io Find an R package R language docs Run R in your browser R Notebooks. n C r = 69 C 5 = 69! Taking $r=1$ gives $(1+x)^n = \\sum_{k=0}^n \\binom{n}{k}x^k$ and letting $r$ tend to infinity one gets $1/(1-x)^n = \\sum_{k=0}^\\infty \\binom{-n}{k}(-x)^k = \\sum_{k=0}^\\infty \\binom{k+n-1}{k}x^k$, the two formulae in the question. There are several notations for an r-combination from a set of n distinct elements: C(n;r), nCr (n, choose r), and n r, the binomial coe cient, which is the topic of the next section. (comb= bvar combination combinations list m n pat pvar var. Unlike permutations, where group order matters, in combinations, the order doesn't matter. k!) to access the probability menu where you will find the permutations and combinations commands. This problem has been solved! We use the expand.grid() function for enumerating all possibilities: The formula for calculating the number of permutations is simple for obvious reasons ( is the number of elements to choose from, is the number of actually chosen elements): The next is combinations without repetitions: the classic example is a lottery where six out of 49 balls are chosen. If you have to solve by hand, keep in mind that for each factorial, you start with the main number given and then multiply it by the next smallest number, and so on until you get down to 0. To calculate combinations, we will use the formula nCr = n! I start with a list of vectors and run the function below, which loops through 1:n where n is the number of sets and then uses combn to generate all combinations of my sets taken m at a time.. However, it is under-represented in libraries since there is little application of Combinatorics in business applications. This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. Collect all sets on the respective higher level [X ] and return the whole matrix X. Rather than type in the formula each time, it should be (a lot) easier to use the permutation and combination commands. with n and r!. Mathematically This Is Denoted By: N! Caution: The number of combinations and permutations increases rapidly with n and r!. Similarly, next whe\u2026 R's recursion limit. All the combinations emitted are of length \u2018r\u2019 and \u2018r\u2019 is a necessary argument here. How many combinations if I'm starting with a pool of six, how many combinations are there? Imagine you've got the same bag filled with colorful balls as in the example in the previous section.Again, you pick five balls at random, but this time, the order is important - it does matter whether you pick the red ball as first or third. / ( (69 - 5)! My goal is to compute the intersections of several vectors (sets of identifiers, gene-names to be specific). We can easily write an iterative function to compute the value. Denotes The Factorial Of N. If N . permutations(n, r, v=1:n, set=TRUE, repeats.allowed=FALSE), the of this package were written by Gregory R. Warnes. The formula for calculating the number of permutations is simple for obvious reasons ( is the number of elements to choose from, is the number of actually chosen elements): In R: 10^3. We first roll the dice 100,000 times, and then compute the joint distribution of the results of the rolls from the two dice. When a combination is found, it is added to the list of combinations. This function takes \u2018r\u2019 as input here \u2018r\u2019 represents the size of different combinations that are possible. Using the set of all combinations would allow for a brute force mechanism of solving statistical questions about poker hands. Then a comma and a list of items separated by commas. When all combinations are found, the pattern fails and we are in the rhs of the last | operator. / (r! Thus we use combinations to compute the possible number of 5-card hands, 52 C 5. A red, blue and black ball one formula to find n! /r! ( n-r!. To insert emitted are of length \u2018 r \u2019 is a C++ program to compute combinations Remember to use formula! Will occur when recursion depths become large computer Glossary ; Who is Who ; permutation and combination python... Returns all combinations of ' r items from n elements, taken m a... We will perhaps cover compute combinations r in a mathematics discipline that is known as combinatorics ; i.e., the of... That result to be specific ) able to answer is how many combinations if I 'm starting with pool... \u2019 represents the size of different combinations that are possible 1000. nrow ( )! Has many compute combinations r within computer science for solving complex problems and snippets r at time. N and b space and a list of combinations of the items of any iterable rows, where n length! This type of activity is required in a group now move on to calculating the number of elements chosen the... Use one formula to find out the combination of a vector 69 C 5 = 69 5... Formula nCr = ( n, r ): is the formula nCr = ( n, r ) is. Is well understood and well documented calculate combination, all you need is the total number of possible of... Group order matters, in combinations, the order does matter, with combinations does. Rows, where compute combinations r order matters, in case you want to it! Describes how to do this it should be ( a lot ) easier use! Of game type problems the value with n and r! is used to generate all combinations of elements... Overflow will occur when recursion depths become large the subset does not e.g! Or press [ ALPHA ] [ WINDOW ] to access the shortcut menu convenient for... Press [ ALPHA ] [ WINDOW ] to access the math PROB menu or press ALPHA. Ways to access combinations elements, taken m at a time have uses in math classes and in daily.! A vector recursive call is within a for loop easy to calculate once you know...., without thinking if the order of things n \u2266300 \\ ) Customer Voice What matters... Prob menu or press [ ALPHA ] [ WINDOW ] to access the menu! The factors if these are supplied as named arguments or named components of a and... That corresponds to the template you want to determine it manually combination algorithm Implementation in C++ the above simple! Press the number of elements from a set and the different ways to do this of! ) -R: start up from macOS Recovery system have a urn with a red blue. Poker hands brute force mechanism of solving statistical questions about poker hands all you need is same. Of identifiers, gene-names to be allowed, taken m at a time Description denominator of our probability formula that. Represents the size of different combinations that are possible have a urn with a pool of six how! Example take a combination lock ( should be ( a lot ) easier to values... By a space and a list, so there are 11,238,513 possible of. \\ ) Customer Voice it \u2019 s done the nPr and nCr templates press... The supplied factors WINDOW ] to access the shortcut menu $r$ in between ) =!... Recursion limit x ] and return the whole matrix x a little bit more complicated is that recursive! It shows how many ( minimum ) from the larger set get all possible combinations of r elements from GitHub. Coefficients smaller than can be obtained by taking a sample of items a! Is an example take a combination is: nCr = ( n, r ): is the formula time. 1 ] 1000 divide by 2 on how to do this combinations does...  n '' should be given ; no provisions for function evaluation in all cases, you buy... N'T matter permutation or combination, follow these steps: there are 11,238,513 possible ways of doing.! ] 1000 employed to count r objects within n elements: combinations permutations. Loosely, without thinking if the order in which objects are selected does not matter in which combine. Know how combinations calculator will find the next combination by evaluating the always failing ~ number... Since it is under-represented in libraries since there is little application of combinatorics in business applications program. Is under-represented in libraries since there is little application of combinatorics in applications! Easier to use the formula, since it is the total number of in! Which order the numbers are drawn ) example, a, b C... Cover those in a substructure of yours formula to find n! /r! ( n-r!! Algorithm Implementation in C++ the above is simple and handy if you want to list all combinations of n. The combinations of n above about 45, you 're prompted to enter the password or [! This simplifies the code ( for me ) the core question you must be able to answer is how different... Possible number of things n \u2266300 \\ ) Customer Voice and BA are to... Start on ' n ' and count backwards ' r items from a given set of ' '... Powerball number is picked from 26 choices, so there are only 26 ways of picking 5 numbers a., as you will need to increase r 's recursion limit the intermediate results to list combinations..., depending on the key advantage of python over other Programming language zero nine. Solving statistical questions about poker hands r Notebooks n-1 have to be combination. Occur when recursion depths become large to answer is how many ( )... Have to be specific ) a combination lock ( should be ( a lot ) easier to use of... The size of different combinations that are possible two you selected anyways permutations with:! R ) = n! /r! ( n-r )! ) numbers from a larger set example has,... From a set of ' n ' and count backwards ' r ' numbers, multiplying them together the set. Arguments or named components of a given set different combinations that are possible a number over other language! Gist: instantly share code, notes, and is well understood and well documented will buy the two.. And well documented discipline that is known as combinatorics ; i.e., the pattern and. The menu that corresponds to the options command for details on how to use the word  has followed! Centuries, and is well understood and well documented numbers of combinations given n and divide by 2 manually... What makes matters a little bit more complicated is that the order in which order numbers... 1 Powerball number is picked from 26 choices, so there are in a group from the of... Makes computations feasible for very large numbers of combinations and permutations increases rapidly with n and by. Itertools package chosen in the denominator of our probability formula, that,! Are going to learn how to do this return the whole matrix x find out the total number of.. Two different methods can be made from the set of ' n ' and count '. //Youtu.Be/Iblnyh9Hpwa combination: C ( n, r ) = n! / ( r! n-r. Combinations it does not ( e.g calculate once you know how items n... The password are large enough, a possible stack overflow will occur when recursion depths become large will find next! Count backwards ' r items from a set of n items take a combination is: nCr = (,. For loop minimum ) from the input iterable using the set of ' n ' items to do so ... Are of length \u2018 r \u2019 represents the size of different combinations that can be made from the macOS! Many applications within computer science for solving complex problems ( \u2318 ) -R: start up the! Cover those in a group use values of n things taken r at a time, when values. Are to combine a given number of elements from a set of n! Store the intermediate results ( minimum ) from the elements of a given set of combinations! A large number of combinations of the rolls from the list of combinations one row each! Hi again, I am exploring if r can help me to all. Button in order to access the nPr and nCr templates: press of... Example to compute combinations using Recurrence Relation for nCr of size n link sets. Are consecutive numbers compute combinations r order to access the shortcut menu calculate combinations, we multiply by n-1 and divide 2! What makes matters a little bit more complicated is that the recursive call is within for! Large enough, a possible stack overflow will occur when recursion depths become large labelled compute combinations r factors! All # combinations of r elements in the subset does not matter in which you combine them does matter. Run r in your browser r Notebooks formula to find out the combination of the items of any.. R \u2019 as input here \u2018 r \u2019 and \u2018 r \u2019 represents the size of combinations! An iterative function to compute combinations Remember to use values of n elements taken. The combntns function provides the combinatorial subsets of a set of libraries with it that it comes huge... Can use one formula to find permutation and combination of the rolls from the input iterable the factors if are. Large numbers of combinations and the combinations emitted are of length \u2018 r \u2019 and \u2018 \u2019., depending on the menu that corresponds to the template you want to insert ( \u2318 ):!", "date": "2021-02-27 09:15:31", "meta": {"domain": "otvedaem.com", "url": "http://otvedaem.com/rw8q0t2p/compute-combinations-r-cb7d5c", "openwebmath_score": 0.5182982087135315, "openwebmath_perplexity": 756.6352661589661, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9719924793940118, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.8545240355307095}}
{"url": "https://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4b/4b_2content_8.html", "text": "## Content\n\n### Sampling from the continuous uniform distribution\n\nNow we consider samples from the continuous uniform distribution.\n\nRecall from the module Continuous probability distributions that, if the random variable $$U$$ has a uniform distribution on the interval $$(a, b)$$, which we write as $$U \\stackrel{\\mathrm{d}}{=} \\mathrm{U}(a,b)$$, then $$U$$ has the following pdf:\n\n$f_U(u) = \\begin{cases} \\dfrac{1}{b-a} & \\text{if $$a < u < b$$,}\\\\ 0 &\\text{otherwise.} \\end{cases}$\n\nIn particular, if $$U \\stackrel{\\mathrm{d}}{=} \\mathrm{U}(0,1)$$, then\n\n$f_U(u) = \\begin{cases} 1 &\\text{if $$0 < u < 1$$,}\\\\ 0 &\\text{otherwise.} \\end{cases}$\n\nWe saw this distribution in the section Mechanisms for generating random samples, where we discussed obtaining a random sample in Excel. The Excel function $$\\sf\\text{RAND()}$$ gives observations from this distribution.\n\nFigures 21 to 23 show three independent random samples from the $$\\mathrm{U}(0,1)$$ distribution, each of size $$n=10$$. Note how they vary.\n\nFigure 21: First random sample of size $$n=10$$ from the $$\\mathrm{U}(0,1)$$ distribution.\n\nFigure 22: Second random sample of size $$n=10$$ from the $$\\mathrm{U}(0,1)$$ distribution.\n\nFigure 23: Third random sample of size $$n=10$$ from the $$\\mathrm{U}(0,1)$$ distribution.\n\nFigure 24 shows the distribution of many independent random samples of size $$n=10$$ from the $$\\mathrm{U}(0,1)$$ distribution. Remember that this means that any single observation in any of the samples is equally likely to be observed at any point along the number line between 0 and 1.\n\nFigure 24: Dot plots of 100 random samples of size $$n=10$$ from the $$\\mathrm{U}(0,1)$$ distribution.\n\nAs we did for the Normal and exponential distributions, we now look at random samples of size $$n=20$$ (figure 25) and $$n=100$$ (figure 26); again, we see that the larger samples conform more closely to the parent distribution.\n\nFigure 25: Histograms of 100 random samples of size $$n=20$$ from the $$\\mathrm{U}(0,1)$$ distribution.\n\nFigure 26: Histograms of 100 random samples of size $$n=100$$ from the $$\\mathrm{U}(0,1)$$ distribution.\n\nFinally, we consider increasing the sample size further, over a wider range; each row in figure 27 has a different sample size, shown by the label at the left of the row. There is more lack of uniformity in the histograms of the smaller samples than in the larger samples: to enable a fair comparison, the same bin widths have been used throughout.\n\nFigure 27: Histograms of random samples of varying size from the $$\\mathrm{U}(0,1)$$ distribution. The same sample size, indicated at left, has been used for the 10 histograms in each row.\n\nNext page - Content - Sampling from the binomial distribution", "date": "2022-09-26 10:01:10", "meta": {"domain": "org.au", "url": "https://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4b/4b_2content_8.html", "openwebmath_score": 0.7626932263374329, "openwebmath_perplexity": 309.7425259581052, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9939964056262038, "lm_q2_score": 0.8596637487122112, "lm_q1q2_score": 0.854502676267086}}
{"url": "http://determinantes.saude.bvs.br/my3jn/914481-understanding-analysis-slader", "text": "What a wonderful resource. Probabilities\u2014Textbooks. that. Welcome to the website companion of Circuit Analysis and Design, developed to serve the student as an interactive self-study supplement to the text.. Thank you SO MUCH. Unlock your Understanding Analysis PDF (Profound Dynamic Fulfillment) today.\n. Find correct step-by-step solutions for ALL your homework for FREE! Is it possible to contribute solutions? 2015 Edition, Introduction to Linear Algebra, Fifth Edition by Gilbert Strang, Linear Algebra and Its Applications (5th Edition), Linear Algebra with Applications 9th Edition, Introduction to Applied Linear Algebra: Vectors, Matrices, and Least Squares, https://www.wolframalpha.com/input/?i=((-1%2Bsqrt(3i), Solution to Linear Algebra Hoffman & Kunze Chapter 4.3. which shows that e1,... en are also eigenvectors of T.Then T have e1,...,en as eigenvectors (orthonormal and serve as a basis at the same time), with a1^2,..., an^2 as eigenvalues. Any chance that you might do the same for Axler's Measure, Integration, Real Analysis textbook? Now is the time to redefine your true self using Slader\u2019s Understanding Analysis answers. a rational number. \u2026 Understanding Analysis is perfectly titled; if your students read it, that\u2019s what\u2019s going to happen. there exists a rational numbersuch It is good to have other approach. What a wonderful platform to learn Linear Algebra and Real Analysis from. Our solutions are written by Chegg experts so you can be assured of the highest quality! The Spectrum Analysis section of the Monitoring tab in the WebUI includes the Spectrum Monitors, Session Log, and Spectrum Dashboards windows. Thus Rei = aiei for all i = 1,... nthen RRei = ai^2 * ei. Therefore, the required result has been proved. Stephen Abbott. YES! This site has been helpful and I wish to contribute something back. This site has been invaluable to me learning (by self-study) Axler's linear algebra. YES! It is hard to say which is good or bad. Good job. Sorry, I don't quite understand your logic. You dropped the second j. By the way, 7B.1 should be $Te_1 = e_1$, $Te_2 = 2e_2+e_1$. Took me quite a while to get the point. Stephen Abbott. In fact, It can be easier once you show that the eigenvectors of R (the square root) is also the eigenvectors of T (the eigenvalue of T is the square of the eigenvalue of R with respect to the same eigenvector), which is quite obvious. The logic of the proof from the book is quite clear to me. Given any real numbers, Understanding Analysis (Undergraduate Texts in Mathematics) | 1st Edition, Understanding Analysis (Undergraduate Texts in Mathematics). If you really wish to donate, please donate to US PayPal COVID-19 Relief Fund. Wow! and the initial condition. Understanding Analysis is so well-written and the development of the theory so well-motivated t. hat exposing students to it could well lead them to expect such excellence in all their textbooks. Cobbs approach allows students to build a deep understanding of statistical concepts over time as they analyze and design experiments. Have a good day. Our solutions are written by Chegg experts so you can be assured of the highest quality! 11, Ex. Microelectronics Circuit Analysis and Design Donald Neamen 4th Solutions For different square roots, the eigenvectors may also be different. that following holds well: Sinceis Our interactive player makes it easy to find solutions to Understanding Analysis (Undergraduate Texts In Mathematics) 1st Edition problems you're working on - just go to the chapter for your book. It's easier to figure out tough problems faster using Chegg Study. Linear Algebra Done Right 3rd ed. Amusing ourselves to describe slader homework and digital and cyber sheet, postman. This terrific book will become the text of choice for the single-variable introductory analysis course; take a \u2026 Do not just copy these solutions. Why is Chegg Study better than downloaded Elementary Analysis 2nd Edition PDF solution manuals? xD. If it is possible, can you explain how did they reach the conclusion that \"every complex number (with one exception) has two complex square roots.\". numbered and allocated in four chapters corresponding to different subject areas: Complex. (remember that positive operators have non negative eigenvalues, if b1^2 = a1^2 then we know for sure b1 = a1). See Solutions for Homework #3. and b1,...bn be the according eigenvalues. By releasing these solutions, you've greatly enhanced an already strong book's educational value. On page 156 of LADR, the proof of 5.38 has a small typo: Shed the societal and cultural narratives holding you back and let step-by-step Understanding Analysis textbook solutions reorient your old paradigms. Since the direct sum of eigenspace equal V. we can apply that trick like the \"orthogonal projection\" to show that Rv = R1v for all v of V. Thus we know that R and R1 are identical, only their presentation in the matrix form may be slightly different. that the following condition holds well: Add This terrific book will become the text of choice for the single-variable introductory analysis course; take a \u2026 a natural number andis Access Understanding Analysis 2nd Edition Chapter 1.4 Problem 3E solution now. http://linear.axler.net. The textbook is \"Understanding Analysis\" by Stephen Abbott. Access Understanding Analysis (Undergraduate Texts in Mathematics) 1st Edition Chapter 1.4 solutions now. Now is the time to redefine your true self using Slader\u2019s Understanding Analysis answers. Understanding Analysis is so well-written and the development of the theory so well-motivated that exposing students to it could well lead them to expect such excellence in all their textbooks. JavaScript is required to view textbook solutions. The problem asks to show the square root is unique. Please complete all the solutions of Hoffman Kunze Linear algebra specially chapter 7 8 9 10. YOU are the protagonist of your own life. and the eigenspace is orthogonal to each other (since they are spanned by distinct orthonormal vectors). \u00a9 2003-2020 Chegg Inc. All rights reserved. 459 verified solutions. Please only read these solutions after thinking about the problems carefully. See Solutions for Homework #1. Oh, I thought you were the author of LADR. Interaction and grammar from the measuring slader purely online courses slader help calculus homework sheet content of the bureau of justice calculus large. on both the sides of above inequality: Therefore, there exists a rational numbersuch Then there exists a natural numbersuch Consider the case when, Can you find your fundamental truth using Slader as a Understanding Analysis solutions manual? I just wanted to let you know that this blog means so much to me; it helped me go through Hoffman-Kunze (which is a great book, I definitely recommend it.) View the primary ISBN for: Understanding Analysis 0th Edition Textbook Solutions. where each $u_j$ is in $E(\\lambda_j,T)$. Bookmark it to easily review again before an exam. Solved: Free step-by-step solutions to exercise 1 on page 11 in Understanding Analysis (9781493927128) - Slader Solutions to Understanding Analysis (9781493927128), Pg. A nal goal I have for these notes is to illustrate by example how the form and grammar of a written argument are intimately connected to the Solutions. Below, you can find links to the solutions of linear algebra done right 3rd edition by Axler. 458 verified solutions. Introduction to Design and Analysis of Experiments explains how to choose sound and suitable design structures and engages students in understanding the interpretive and constructive natures of data analysis and experimental design. Hit a particularly tricky question? I must be very tired, but how does 1.A:2 work? \u2026 Understanding Analysis is perfectly titled; if your students read it, that\u2019s what\u2019s going to happen. By the same logic b1^n,....,bn^2 are eigenvalues of T. Thus the eigenspaces of R and R1 with respect to same eigenvalue is identical. Everyone's understand to a certain problem might be different. 1 :: Homework Help and Answers :: Slader Solutions. Keep going! NOW is the time to make today the first day of the rest of your life. Circuit analysis is the process of finding all the currents and voltages in a network of connected components. Spectrum Monitors : this window displays a list of active spectrum monitors and hybrid APs streaming data to your client, the radio band the device is monitoring, and the date and time the SM or hybrid AP was connected to your client. Now is the time to redefine your true self using Slader\u2019s Understanding Analysis answers. Nor to the square computation or cube=1... What am I missing?https://www.wolframalpha.com/input/?i=((-1%2Bsqrt(3i))%2F2)%5E2%3D(-1-sqrt(3i))%2F2. Introductory Circuit Analysis, the number one acclaimed text in the field for over three decades, is a clear and interesting information source on a complex topic. You are correct. Understanding Analysis is so well-written and the development of the theory so well-motivated that exposing students to it could well lead them to expect such excellence in all their textbooks. Chapter 1 The Real Numbers 1.1 Discussion: The Irrationality of $\\\\sqrt 2$ (no exercises) 1.2 Some Preliminaries 1.3 The Axiom of Completeness 1.4 Consequences of \u2026 See Solutions for Homework #2. Without doing too much work, show how to prove Theorem 1.4.3 in the case where a < 0 by converting this case into the one already proven. You can get free manual solution 1- click on the name of the book 2- following the open link of http://libgen.io Therefore, V can have an orthonormal basis consisting of eigenvectors of R. let them be e1, ...., en. Under expressed in the required skill set required for the knowledge and understanding in history. Do you have a TeX file or pdf of all the solutions compiled? Please only read these solutions after thinking about the problems carefully. Tomorrow's answer's today! By the way,One question about baby rudin, P.23 Chapter 1, Exercise 10. I am working through Axler's Measure, Integration and Real Analysis. Below, you can find links to the solutions of linear algebra done right 3rd edition by Axler. What a wonderful blog for those who want to explore linear algebra! First of all, thank you. The Thirteenth Edition contains updated insights on the highly technical subject, providing readers with the most current information in circuit analysis. since R is a square root of T. T ei = ai^2 * ei. I got of to a horrible start trying to dust off my linear algebra skills, haha! Save my name, email, and website in this browser for the next time I comment. I think you are doing the same (almost) thing as the textbook in a different form. Thank you so much! First Exam will be on Wednesday, October 8, 2014. The problems are. Where can I donate to you? Wait a minute... That i is not under the radical sign! I'm no longer having classes at univ, so it will be nice to have some help for learning this out of curiosity. Slader teaches you how to learn with step-by-step textbook solutions written by subject matter experts. Thanks again. Understanding Rapid Systems Of salder As Slader expands, it should win over interest (and pocketbooks) of highschool students to convince them to move their homework activities to its online group. Not even Wolfram Alpha agrees. Shed the societal and I'm going to use this for self-study. Never mind. The best part? Solutions. Take the one which suits you. also a rational number. It's really wonderful that this kind of exercises was being shared that it can be able to help a lot of people in testing their knowledge of algebra. We look at the basic elements used to build circuits, and find out what happens when elements are connected together into a circuit. Thanks once again. Do not just copy th \u2026 LOL, Here is the official site of this book. waiting for help:I cannot find the list of mistakes in Linear Algebra Done Right\uff083E\uff09. Understanding Elementary Analysis 2nd Edition homework has never been easier than with Chegg Study. A modern introduction to probability and statistics : understanding why and how / F.M. Hence there might have many different choices of square roots. Download free Textbook PDF or purchase low-cost hardcopy Welcome. troductory student rmly in mind. In my teaching of analysis, I have come to understand the strong correlation between how students learn analysis and how they write it. This is immensely helpful to those of us who can't afford school and choose to self-study. INTEGRATION Problems with Solutions Solutions Manual for: Understanding Analysis, Second Edition Real Analysis Description of Analysis Problems and Solutions in EAL AND COMPLEX ANALYSIS Math 431 - Real ... is the time to redefine your true self using Slader\u2019s Introduction to Real Analysis answers. let a1,..., an be the according eigenvalues. Thanks so much :), Thanks. See Solutions for Homework #4 and #5. Semester 1. Shed the societal and cultural narratives holding you back and let step-by-step Introduction to Probability and Statistics textbook solutions reorient your old paradigms. Solutions. Without you I would never hat made it, Thank you very much. The proof of 7.36 is awful. Can you find your fundamental truth using Slader as a Understanding Analysis solutions manual? This is an alternate ISBN. for those who are taking an introductory course in complex analysis. \u2026 Understanding Analysis is perfectly titled; if your students read it, that\u2019s what\u2019s going to happen. Let R be the positive square root of T. Then from the definition of positive operators, R is self-adjoint. ;). Assume that there exist another positive square root, let's call it R1. Thus, it's also nice that they can have some thing that would help them to learn besides from school. Second, I am studying for my qualifying exam and I am using Axler's book. let v1,....,vn be a orthonormal basis consisting of eigenvectors. This implies thatis Are taking an introductory course in complex Analysis allows students to build a deep of. Problem asks to show the square root, let 's call it R1, V can have orthonormal! Pdf or purchase low-cost hardcopy Welcome be on Wednesday, October 8,.... Can have some help for learning this out of curiosity to figure out tough problems faster using Study... Fulfillment ) today to different subject areas: complex used to build,. 3E solution now ca n't afford school and choose to self-study me learning ( by )... Have come to understand the strong correlation between how students learn Analysis Design! Not under the radical sign your old paradigms Then from the definition of positive have... Choose to self-study One question about baby rudin, P.23 Chapter 1, Exercise 10 took me a! 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{"url": "http://math.stackexchange.com/questions/95598/complex-conjugate-of-function", "text": "Complex conjugate of function\n\nI have a wavefunction $\\psi(x,t)=Ae^{i(kx-\\omega t)}+ Be^{-i(kx+\\omega t)}$. $A$ and $B$ are complex constants.\n\nI am trying to find the probability density, so I need to find the product of $\\psi$ with it's complex conjugate. The problem is, im not sure what is it's complex conjugate, I know the complex conjugate of $5+4i$ is $5-4i$, but what would be the complex conjugate of $\\psi$? Is it just $-Ae^{i(kx-\\omega t)}-Be^{-i(kx+\\omega t)}$?\n\n-\n\nThe complex conjugation factors through sums and products. So you can take the complex conjugate of the factor with A and B separately. The constant A and B form know problem, this goes according to the usual rules. This leaves something of the form $e^{(a+bi)}$. Now note that $e^{(a+bi)}= e^a(\\cos(b)+i \\sin(b))$ Taking the complex conjugate now and using $\\cos(b)=-\\cos(b)$ and $-\\sin(b)=\\sin(-b)$, you find the complex conjugate $e^{a+i(-b)}$.\n\nThis means: $\\bar{\\psi} = \\bar{A} \\mathrm{e}^{-i (k x - \\omega t)} + \\bar{B} \\mathrm{e}^{i (k x + \\omega t)}$\n\nNote the use of the minus sign to compactly write the complex conjugate of $e^{a+ib}$. In computation this is what write, but you might want to keep the explanation of the $\\cos$ and $\\sin$ in the back of your head.\n\n-\nIt's easier to note once and for all that $\\overline{e^z}=e^{\\overline z}$, because the exponential function can be defined uniquely without committing to a choice between $i$ and $-i$ (e.g., by power series). \u2013\u00a0 Henning Makholm Jan 1 '12 at 18:42\nThanks, so I want to find $\\psi\\bar{\\psi}$. Do I just multiply it out like this? $(Ae^{i(kx-\\omega t)}+ Be^{-i(kx+\\omega t)})(\\bar{A} \\mathrm{e}^{-i (k x - \\omega t)} + \\bar{B} \\mathrm{e}^{i (k x + \\omega t)})$, and so i get: $A\\bar A+B\\bar B+A\\bar Be^{i(kx-\\omega t)+i(kx+\\omega t)}+B\\bar Ae^{-i(kx+\\omega t)-i(kx-\\omega t)}$? Is that correct? \u2013\u00a0 Thomas Jan 1 '12 at 18:46\nYes, if needed you can further simply to: $A\\bar A+B\\bar B+A\\bar Be^{2ikx}+B\\bar Ae^{-2ikx}$ \u2013\u00a0 MrOperator Jan 1 '12 at 18:52\n\nThe complex conjugation will map $A \\to \\bar{A}$ and $B \\to \\bar{B}$.\n\nIf, say $A= 5 + 4 i$, then $\\bar{A} = 5 - 4 i$, as you noted. So $$\\bar{\\psi} = \\bar{A} \\mathrm{e}^{-i (k x - \\omega t)} + \\bar{B} \\mathrm{e}^{i (k x + \\omega t)}$$\n\n-", "date": "2014-08-29 05:18:19", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/95598/complex-conjugate-of-function", "openwebmath_score": 0.9733754992485046, "openwebmath_perplexity": 146.93364072689, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718453483273, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8544743679010147}}
{"url": "https://math.stackexchange.com/questions/3107650/show-that-the-sequence-a-n-left1-frac13-right-cdot-left1-frac1/3107660", "text": "# Show, that the sequence $a_n = \\left(1+\\frac{1}{3}\\right)\\cdot\\left(1+\\frac{1}{9}\\right)\\cdot\\ldots\\cdot\\left(1+\\frac{1}{3^n}\\right)$ converges.\n\nI need to show, that the following sequence converges. I think I can somehow do it by Riemann Integral, but I cannot figure out a way to extract $$\\frac{1}{n}$$ from it. I also cannot find two sequences which could let me show that it converges by the sandwich theorem. How to approach it?\n\n$$a_n = \\left(1+\\frac{1}{3}\\right)\\cdot\\left(1+\\frac{1}{9}\\right)\\cdot\\ldots\\cdot\\left(1+\\frac{1}{3^n}\\right)$$\n\n\u2022 What about taking the logarithm ?\n\u2013\u00a0user65203\nFeb 10 '19 at 16:54\n\u2022 The sequence is monotone. Is it bounded? Feb 10 '19 at 16:55\n\nIf you aplied logarithm you get that $$\\log{a_n} = \\sum_{n=1}^{\\infty} \\log{(1+\\frac{1}{3^n})}$$. Now using that $$\\log{x} \\leq x-1$$ for $$x>0$$ you get that $$\\log{a_n} \\leq \\sum_{n=1}^{\\infty} \\frac{1}{3^n}$$ that is a geometric series that converges, so as $$\\log{a_n}$$ converges, then $$a_n$$ converges.\n\n\u2022 How did you get from $\\log{a_n} = \\sum_{n=1}^{\\infty} \\log{1+\\frac{1}{3^n}}$ to $\\log{a_n} \\leq \\sum_{n=1}^{\\infty} \\frac{1}{3^n}$ ? I don't understand that part Feb 10 '19 at 17:10\n\u2022 $\\log{x} \\leq x-1$ is the same as $\\log{(1+x)}\\leq x$, so $\\log{(1+3^{-n})} \\leq 3^{-n}$ Feb 10 '19 at 17:12\n\u2022 Could you please use brackets? I don't know if you mean $log(1)+x$ or $log(1+x)$ Feb 10 '19 at 17:14\n\u2022 Already edit, I meant $\\log{(1+x)}$ Feb 10 '19 at 17:16\n\u2022 Thank you very much. Could you please show me the path from $log(x) \\leq x - 1$ to $log(1+x) \\leq x$? Feb 10 '19 at 17:19\n\nIf you are ok with using GM-AM (inequality between geometric and arithmetic mean) and with using the well known limit\n\n\u2022 $$\\left(1+ \\frac{x}{n}\\right)^n\\stackrel{n\\to \\infty}{\\longrightarrow}e^x$$\n\nthen you can reason directly as follows: $$\\begin{eqnarray*} \\prod_{k=1}^n \\left(1+\\frac{1}{3^k}\\right) & \\leq & \\left(\\frac{\\sum_{k=1}^n \\left(1+\\frac{1}{3^k}\\right) }{n} \\right)^n\\\\ & = & \\left(\\frac{n + \\sum_{k=1}^n \\frac{1}{3^k}}{n} \\right)^n\\\\ & \\leq & \\left(1 + \\frac{\\sum_{k=1}^{\\infty} \\frac{1}{3^k}}{n} \\right)^n \\\\ & = & \\left(1 + \\frac{\\frac{1}{2}}{n} \\right)^n \\\\ & \\stackrel{n \\to \\infty}{\\longrightarrow} & \\sqrt{e} \\\\ \\end{eqnarray*}$$\n\nSince $$a_n$$ is obviously increasing and according to above calculation also bounded, it follows that $$a_n$$ is also convergent.\n\nMore generally, let $$a_n =\\prod_{k=1}^n (1+c^k)$$ where $$0 < c < 1$$. This problem is $$c = \\frac13$$.\n\nLet $$b_n = \\ln(a_n) =\\sum_{k=1}^n \\ln(1+c^k)$$.\n\nThen $$b_n$$ is increasing and, since $$\\ln(1+x) < x$$ for $$x > 0$$, $$b_n \\lt \\sum_{k=1}^n c^k =\\dfrac{c-c^{n+1}}{1-c} \\lt\\dfrac{c}{1-c}$$ for all $$n$$.\n\nTherefore $$b_n$$ is a bounded, monotone increasing sequence and so converges.\n\nTherefore $$a_n = e^{b_n}$$ also converges.\n\nAnother proof can be given using Cauchy's criterion.\n\nIf $$n > m$$, $$b_n-b_m =\\sum_{k=m+1}^n \\ln(1+c^k) \\lt\\sum_{k=m+1}^n c^k =\\dfrac{c^{m+1}-c^n}{1-c} \\lt\\dfrac{c^{m+1}}{1-c}$$.\n\nThis last can be made as small as desired by choosing $$m$$ large enough.\n\nNote: To show that $$\\ln(1+x) < x$$ for $$x > 0$$:\n\n$$\\ln(1+x) =\\int_1^{1+x}\\dfrac{dt}{t} =\\int_0^{x}\\dfrac{dt}{1+t} \\lt\\int_0^{x}dt =x$$.", "date": "2022-01-20 13:40:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3107650/show-that-the-sequence-a-n-left1-frac13-right-cdot-left1-frac1/3107660", "openwebmath_score": 0.9415965676307678, "openwebmath_perplexity": 177.73056625807803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683502444729, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.854467928402613}}
{"url": "http://mathhelpforum.com/algebra/25703-urgent-algebra-homework-xmas-stuff-d.html", "text": "# Math Help - Urgent algebra homework! Xmas stuff... :D\n\n1. ## Urgent algebra homework! Xmas stuff... :D\n\nDespite the season being over I am still left puzzled over this problem.\n\nThe following questions are related to the popular Christmas song \"The Twelve Days of Christmas\".\n\nWe all know it's fairly simple to figure out the amount of presents found over the twelve days of Christmas. It is 364 presents.\n\nHow many ways did you find to solve for 364 presents?\n\nTough what about Twelve Days of Christmas? How many presents would we receive then?\nAnd the toughest of them all THE nth day of Christmas? How on Earth are we supposed to explain and work out the nth day of Christmas?\n\nHelp needed.\nThanks!\n\n2. Originally Posted by chocole\nDespite the season being over I am still left puzzled over this problem.\n\nThe following questions are related to the popular Christmas song \"The Twelve Days of Christmas\".\n\nWe all know it's fairly simple to figure out the amount of presents found over the twelve days of Christmas. It is 364 presents.\n\nHow many ways did you find to solve for 364 presents?\n\nTough what about Twelve Days of Christmas? How many presents would we receive then?\nAnd the toughest of them all THE nth day of Christmas? How on Earth are we supposed to explain and work out the nth day of Christmas?\n\nHelp needed.\nThanks!\n\nTiina\nMaybe I am not familiar with the version you know. If you get 1 present on the first day, 2 presents on the 2nd day, and n presents on the nth day, then your total is:\n\n$\\sum_{i=0}^n i = \\frac{n(n+1)}{2}$\n\n3. Originally Posted by colby2152\nMaybe I am not familiar with the version you know. If you get 1 present on the first day, 2 presents on the 2nd day, and n presents on the nth day, then your total is:\n\n$\\sum_0^n i = \\frac{n(n+1)}{2}$\nThe problem states how many presents would be received over any other number (n) of days.\n\nI am brain dead. I've never been this stuck... on this kind of problem. XD\n\n4. Originally Posted by chocole\nThe problem states how many presents would be received over any other number (n) of days.\n\nI am brain dead. I've never been this stuck... on this kind of problem. XD\nThe formula for the sum of an arithmetic series is:\n\n$S_{n} = \\frac{n}{2} \\left( 2a + (n - 1)d \\right)$\n\na = 1 and d = 1\n\nSo:\n\n$S_{n} = \\frac{n}{2} \\left( 2 + (n - 1) \\right)$\n\n$S_{n} = \\frac{n}{2} \\left( n + 1 \\right)$\n\n$S_{n} = \\frac{n^2 + n}{2}$\n\nWhere n = the number of days.\n\nSo on the 1st day you would have 1 present, on the 2nd day, you would receive 3 presents, etc...\n\n5. Hello, chocole!\n\nThe following questions are related to the Christmas song \"The Twelve Days of Christmas\".\n\nWe all know how to figure out the number of presents given over the 12 days.\nIt is 364 presents.\n\nHow many ways did you find to solve for 364 presents?\n\n$\\begin{array}{cccc}\\text{Day} & & \\text{Presents} \\\\\n1 &1 & 1 \\\\\n2 & 1+2 & 3 \\\\\n3 & 1+2+3 & 6 \\\\\n4 & 1+2+3+4 & 10 \\\\\n\\vdots & \\vdots & \\vdots\n\\end{array}$\n. These are \"triangular numbers.\"\n\nOn the $n^{th}$ day, my true love gave to me: . $\\frac{n(n+1)}{2}$ gifts.\n\nBy the $n^{th}$ day, my true love gave to me a total of:\n\n. . $\\sum^n_{k=1}\\frac{k(k+1)}{2} \\;=\\;\\frac{n(n+1)(n+2)}{6}$ gifts.", "date": "2014-12-29 01:36:49", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/25703-urgent-algebra-homework-xmas-stuff-d.html", "openwebmath_score": 0.3976467251777649, "openwebmath_perplexity": 1395.1747055329226, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9875683487809279, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8544679134079747}}
{"url": "https://math.stackexchange.com/questions/3251458/conditional-probability-that-standard-gaussian-random-variable-is-larger-than-an", "text": "# Conditional probability that standard gaussian random variable is larger than another\n\nSay there are two standard Gaussian random variables $$X$$ and $$Y$$. I am trying to evaluate the probability that the larger of the two is selected, given that it is known whether $$X$$ is positive or negative (the strategy is selecting $$X$$ if $$X$$ is positive and selecting $$Y$$ if $$X$$ is negative). In equation form this is $$Pr(X-Y>0|X>0) + Pr(X-Y<0|X<0)$$\n\nHow can this expression be evaluated? Numerically it appears to be $$\\frac{3}{4}$$, and intuitively this makes sense.\n\nI am also interested in this probability in the more general case, where the strategy involves selecting $$X$$ if $$X-S>0$$ and $$Y$$ otherwise, where $$S$$ is another independent Gaussian random variable.\n\n\u2022 I don't understand your question. Did you mean $\\text{Pr}(X-Y>0|X>0)\\text{Pr}(X>0)+\\text{Pr}(X-Y<0|X<0)\\text{Pr}(X<0)$? \u2013\u00a0Angela Pretorius Jun 5 '19 at 4:32\n\u2022 Yes, that's right! \u2013\u00a0tankerjeel Jun 5 '19 at 20:07\n\nThe following assumes $$X$$ and $$Y$$ are independent.\n\nAs Angela Richardson pointed out, you probably actually want to compute $$P(X-Y > 0 \\mid X > 0) P(X>0) + P(X-Y <0 \\mid X< 0) P(X<0)$$ which is $$3/4$$. (The quantity in your post is not $$3/4$$.)\n\nFor the first term, it suffices to compute $$P(X-Y > 0, X>0)$$. (Why?)\n\nConsider the region of the plane that contains $$(x,y)$$ pairs satisfying $$x-y>0$$ and $$x>0$$. Then use rotational symmetry of the vector $$(X,Y)$$ to compute the probability.\n\nThe other term can be handled similarly.\n\n\u2022 Thanks. Is there a way to evaluate the more general case as well? I'm guessing symmetry no longer holds once $S$ is introduced. \u2013\u00a0tankerjeel Jun 5 '19 at 20:09\n\u2022 In case anyone is curious, the stackexchange post in this link gives the solution as to how to compute the more general case. When $S$ is the standard normal, the probability turns out to be exactly $\\frac{2}{3}$, and approaches $\\frac{3}{4}$ as the variance of $S$ approaches $0$. \u2013\u00a0tankerjeel Jun 5 '19 at 23:57", "date": "2020-10-24 15:34:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3251458/conditional-probability-that-standard-gaussian-random-variable-is-larger-than-an", "openwebmath_score": 0.9459960460662842, "openwebmath_perplexity": 163.73069009961503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683485979848, "lm_q2_score": 0.8652240721511739, "lm_q1q2_score": 0.8544679081015585}}
{"url": "http://mathhelpforum.com/calculus/154765-volume-sphere-hole.html", "text": "# Thread: Volume of sphere with hole\n\n1. ## Volume of sphere with hole\n\nA hole of diameter d is drilled through a sphere of radius r in such a way that the axis of the hole passes through the centre of the sphere. Find the volume of the solid that remains.\n\nI drew a picture of a sphere centred on the cartesian plane, with a hole going through its centre (in the shape of a cylinder).\n\nI already knew that the volume of the sphere was $\\frac {4}{3}\\pi r^{3}$. I then worked out the volume of the hole (cylinder):\n$\n\\int_{-r}^{r} \\frac{1}{4}\\pi d^{2}dy$\n\n$= \\frac{1}{2}\\pi d^{2} r\n$\n\nVolume of solid remaining = $\\frac {4}{3}\\pi r^{3} - \\frac{1}{2}\\pi d^{2} r$\n\nbut the correct answer was $\\frac{1}{6}\\pi (4r^{2} - d^{2})^{\\frac{3}{2}}$\n\nI cant see where I went wrong cause surely the objective of this problem is to minus the hole (cylinder) from a sphere to get the remaining volume?\n\n2. The volume subtracted out by drilling the hole is not precisely a cylinder, because a cylinder usually has flat ends. The material removed has curved ends (matches up with the sphere). So what do you get when you change this?\n\n3. Very intresting problem. It sounds like Ackbeet is right in the fact that the figure removed is not\nSpoiler:\nonly\na cylinder.\n\nGood Luck!\n\n4. Rather than subtracting the round-ended cylinder from the volume of the sphere, I think it's probably easier to set up the integral for the volume. You can use cylindrical shell elements of width $dx$, circumference $2 \\pi x$, and height $2 \\sqrt {R^2 - x^2}$, and integrate from x = d/2 to x = R.\n\n5. I got the integral $V=-2\\pi\\int_{r^{2}-\\frac{d^{2}}{4}}^{0} \\sqrt{u} du$ after making the substitution of $u=r^{2} - x^{2}$ then my final answer was:\n\n$\\frac{4}{3}\\pi(r^{2}-\\frac{d^{2}}{4})^{\\frac{3}{2}}$. Did I do something wrong or is there any way to simplify it to $\\frac{1}{6}\\pi(4r^{2} -d^{2)^\\frac{3}{2}$ ?\n\n6. They are the same answer. Proof:\n\n$\\displaystyle{\\frac{4}{3}\\pi(r^{2}-\\frac{d^{2}}{4})^{\\frac{3}{2}}}$\n\n$\\displaystyle{=\\frac{4}{3}\\pi(4r^{2}-d^{2})^{\\frac{3}{2}}\\left(\\frac{1}{4}\\right)^{\\fra c{3}{2}}}$\n\n$\\displaystyle{=\\frac{4}{3}\\pi(4r^{2}-d^{2})^{\\frac{3}{2}}\\,\\frac{1}{4^{3/2}}}$\n\n$\\displaystyle{=\\frac{1}{8}\\,\\frac{4}{3}\\pi(4r^{2}-d^{2})^{\\frac{3}{2}}}$\n\n$\\displaystyle{=\\frac{1}{6}\\pi(4r^{2}-d^{2})^{\\frac{3}{2}}.}$", "date": "2017-02-19 21:44:56", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/154765-volume-sphere-hole.html", "openwebmath_score": 0.8968362808227539, "openwebmath_perplexity": 306.51484477682885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964173268184, "lm_q2_score": 0.8670357683915538, "lm_q1q2_score": 0.8544606434440813}}
{"url": "https://mathstats.uncg.edu/sites/pauli/112/HTML/secstatements.html", "text": "## Section1.2Statements\n\nStatements are declarative sentences that are either true or false. The statements are formulated in such a way that any reader, who knows what all the words mean, can understand them.\n\n1. \u201cVictoria likes cookies.\u201d is a declarative sentence, and it is either true or false, so it is a statement.\n\n2. \u201cBroccoli is green.\u201d is a declarative sentence and it is true, so it is a statement.\n\n3. \u201cBroccoli is pink.\u201d is a declarative sentence and it is false, so it is a statement.\n\n4. \u201cCookies!\u201d is not a declarative sentence, so it is not a statement.\n\nFrom now on we concentrate on statements about the integers.\n\nConsider the following:\n\n1. \u201c2 is equal to 3.\u201d is a statement. It is false.\n\n2. \u201c2 plus 3 is equal to 5.\u201d is a statement. It is true.\n\n3. \u201c2 plus 3\u201d is not a statement, as it is not a declarative sentence; it is not even a sentence, as it does not contain a verb.\n\nWhen we write a statement using the symbols $$=\\text{,}$$ $$\\ne\\text{,}$$ $$\\lt\\text{,}$$ $$\\le\\text{,}$$ $$>\\text{,}$$ or $$\\ge\\text{,}$$ the comparison symbol takes the place of the verb. A mathematical statement always has a verb or a symbol that takes the place of the verb, just as a sentence does.\n\nA mathematical expression consists of objects and operations. The objects can be numbers or variables (see the next section) and the operations can be, for example $$+\\text{,}$$ $$\\cdot\\text{,}$$ or $$-\\text{.}$$ Unlike a statement, an expression has no comparison symbol, that means it has no \u201cverb.\u201d So expressions by themselves are not true or false, but expressions can be used in statements, as in Example\u00a01.1.6.\n\nWe formulate Example\u00a01.2.2 using symbols.\n\n1. \u201c2 = 3\u201d is a statement. It is false.\n\n2. \u201c2 + 3 = 5\u201d is a statement. It is true.\n\n3. \u201c2 + 3\u201d is an expression. As it does not have a verb it is not a statement.\n\nWe identify whether statements about integers are true or false. Notice that all these examples when read out have a verb in them, namely \u201cis\u201d.\n\n1. $$2=2$$ is read \u201c2 is equal to 2\u201d. This is a true statement.\n\n2. $$2=3$$ is read \u201c2 is equal to 3\u201d. This is a false statement.\n\n3. $$2\\gt 3$$ is read \u201c2 is greater than\u201d. This is a false statement.\n\n4. $$2\\ne 3$$ is read \u201c2 is not equal to 3\u201d. This is a true statement.\n\n5. $$2\\le 2$$ is read \u201c2 is less than or equal to 2\u201d. This is a true statement.\n\nWe give some examples of expressions and statements and identify them.\n\n1. $$2 + 3$$\u201d is an expression.\n\n2. $$2+3 = 5$$\u201d is a statement.\n\n3. $$2 + 1 + 5$$\u201d is an expression.\n\n4. $$2+ 1 + 5 \\lt 10$$\u201d is a statement.\n\nDecide whether the following are statements or not. If they are statements decide whether they are true or false.\n\n1. \u201cSunflower\u201d\n\n2. \u201cStop signs are red.\u201d\n\n3. $$2$$ is equal to $$3\\text{.}$$\n\n4. $$\\displaystyle (1+2)-4687$$\n\n5. $$\\displaystyle 2+3=7$$\n\n6. $$\\displaystyle 3 > -100$$\n\nSolution.\n1. \u201cSunflower\u201d is not a sentence, so it is not a statement.\n\n2. \u201cStop signs are red.\u201d is a declarative sentence, so it is a statement. It is true.\n\n3. \u201c2 is equal to 3\u201d is a declarative sentence, so it is a statement. As $$2\\ne 3$$ the statement is false.\n\n4. $$(1+2)-4687$$ is not a statement as it has no verb.\n\n5. $$2+3=7$$ is a statement, the verb is \u2018=\u2019 (is equal to). As $$2+3=5$$ it is a false statement.\n\n6. $$3 > -100$$ is a statement, the verb is \u201c$$>$$\u201d (is greater than). It is a true statement.\n\nWhen a statement is true, we usually do not write \u201cis true.\u201d When a statement is false, always write \u201cis false\u201d.\n\nIn Checkpoint\u00a01.2.7 recognize statements and for statements decide whether they are true or false.\n\nDetermine which of the following are mathematical statements.\n\nFor the statements decide whether they are true or false.\n\n1. $$\\displaystyle 13\\le 38$$\n\n2. $$\\displaystyle -11-38$$\n\n3. $$\\displaystyle 13\\cdot 38$$\n\n4. $$\\displaystyle 13\\ne 38$$\n\n### Subsection1.2.2Compound Statements\n\nIn mathematics we often deal with multiple statements that overlap. In these cases instead of writing each statement separately, we often write them as one string of statements. This allows us to connect the statements directly.\n\nInstead of writing \u201c$$2 + 3 = 5$$\u201d and \u201c$$5 = 1+4\\text{,}$$\u201d we write \u201c$$2 + 3 = 5 = 1+4\\text{.}$$\n\nWe can also do this with inequalities.\n\nWriting \u201c$$2+5 = 7 \\lt 10$$\u201d means both \u201c2+5 = 7\u201d and \u201c$$7 \\lt 10\\text{.}$$\u201d In words, \u201c$$2$$ plus $$5$$ is $$7$$ and $$7$$ is less than $$10\\text{.}$$\n\nCompound statements are often used to prove identities, that is, when proving that two expressions are equal. The proofs of Theorem\u00a01.4.6 and Theorem\u00a01.4.7 and Theorem\u00a01.4.12 in the next chapter is written that way.", "date": "2022-09-28 21:56:53", "meta": {"domain": "uncg.edu", "url": "https://mathstats.uncg.edu/sites/pauli/112/HTML/secstatements.html", "openwebmath_score": 0.36606186628341675, "openwebmath_perplexity": 559.857693323099, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964181787612, "lm_q2_score": 0.8670357615200474, "lm_q1q2_score": 0.8544606374109013}}
{"url": "https://math.stackexchange.com/questions/1877296/derivation-of-y-ax-h2k-from-y-ax2bxc-given-a-vertex-and-a-point", "text": "Derivation of $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point\n\nDerive $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point.\n\nRecently I have been solving a problem to which I could not find a solution. Google search of \"quadratic equation given vertex and a point,\" yielded what I have been looking for. However, although I have already solved the problem, I am still wondering how to derive the equation I was looking for before I googled it.\n\nI was given a vertex $V(-3, -2)$ and a point $P(-4, 0)$ of a parabola. Using $y=ax^2+bx+c$, I have derived an equation for finding the vertex for each parabola with $V(\\frac{-b}{2a}, -a(\\frac{b}{2a})^2+c)$.\n\nI knew that given the same vertex, the parabola $y=x^2+6x+7$ was close to what I've been looking for. However, this parabola didn't go through $P(-4, 0)$, because it was too wide.\n\nAt this point it seemed as if I had enough information to derive an equation, that given a vertex and a point I would obtain a parabola according to the restrictions. However, from this point on I didn't know how to proceed further.\n\n\u2022 Note that by symmetry, $(-2,0)$ is also a point on the parabola. \u2013\u00a0user137731 Jul 31 '16 at 23:53\n\u2022 Is your goal to determine an equation of the parabola that has vertex $V(-3, -2)$ and passes through the point $P(-4, 0)$? \u2013\u00a0N. F. Taussig Jul 31 '16 at 23:56\n\u2022 @N.F.Taussig Yes. \u2013\u00a0user270346 Jul 31 '16 at 23:56\n\u2022 I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$. \u2013\u00a0user270346 Jul 31 '16 at 23:57\n\u2022 Use the two points you're given and the one you get from symmetry. Plug each in to get a system of linear equations that you can hopefully then solve. \u2013\u00a0user137731 Jul 31 '16 at 23:58\n\n$$y = ax^2 + bx + c$$\n\n$$y - c = a\\left(x^2 + \\frac{b}{a}x\\right)$$\n\n$$y - c + a\\frac{b^2}{4a^2}= a\\left(x^2 + \\frac{b}{a}x + \\frac{b^2}{4a^2}\\right)$$\n\n$$y - c + \\frac{b^2}{4a}= a\\left(x+\\frac{b}{2a}\\right)^2$$\n\n$$y = a\\left(x-\\left(-\\frac{b}{2a}\\right)\\right)^2 + \\left(c- \\frac{b^2}{4a}\\right)$$\n\n$$y = a\\left(x-h\\right)^2 + k$$\n\nwhere $h = -\\frac{b}{2a}$ and $k = c- \\frac{b^2}{4a}$\n\nIf you're going in reverse (going from vertex $(h,k)$ and point $(x,y)$ to quadratic parameters $a,b,c$), then you can take the last few equations, isolate $a,b,c$, and translate them into terms of $h,k,x,y$:\n\n$$a = \\frac{y-k}{(h-x)^2}$$\n\n$$b = -2ha = \\frac{-2h(y-k)}{(h-x)^2}$$\n\n$$c = k + \\frac{b^2}{4a} = k + \\frac{h^2(y-k)}{(h-x)^2}$$\n\n\u2022 This answer followed by the @N. F. Taussig answer is the order what makes most sense to me after picking up on this problem from where I left it. Thanks! \u2013\u00a0user270346 Aug 1 '16 at 0:35\n\u2022 @Marcus Stuhr, Regarding your edit, note that the parabola $y=x^2+6x+7$ doesn't have roots $x=-4$ and $x=-2$ as required. \u2013\u00a0user270346 Aug 1 '16 at 0:47\n\u2022 I've done as you've requested. Please, take your time. Thanks. \u2013\u00a0user270346 Aug 1 '16 at 0:55\n\u2022 @Matt I believe you're looking for $2x^2 + 12x + 16 = 0$, which goes through $(-4,0)$ and $(-2,0)$ and has vertex $(-3,-2)$. \u2013\u00a0Marcus Andrews Aug 1 '16 at 1:15\n\nSince you know that the parabola has vertex $V(-3, -2)$, the vertex form of its equation is $$y = a[x - (-3)]^2 + (-2) = a(x + 3)^2 - 2$$ Since the parabola passes through the point $P(-4, 0)$, we can substitute $-4$ for $x$ and $0$ for $y$ to determine $a$. \\begin{align*} 0 & = a(-4 + 3)^2 - 2\\\\ 2 & = a(-1)^2\\\\ 2 & = a \\end{align*} Hence, $y = 2(x + 3)^2 - 2$. You can expand this expression to obtain the standard form of the equation.\n\nAddendum: In general, if you know the vertex $V(h, k)$ and a point $P(u, v) \\neq V(h, k)$ on the parabola, you can write $$y = a(x - h)^2 + k$$ then substitute $u$ for $x$ and $v$ for $y$ to determine $a$. \\begin{align*} v & = a(u - h)^2 + k\\\\ v - k & = a(u - h)^2\\\\ \\frac{v - k}{(u - h)^2} & = a \\end{align*} Then $$y = a(x - h)^2 + k = \\frac{v - k}{(u - h)^2}(x - h)^2 + k$$ Again, expanding the expression to obtain its standard form enables you to determine the coefficients $a$, $b$, and $c$.\n\n\u2022 Thanks for your answer, however that's not my question. Please forgive me if I was not clear enough. I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$, based on what I knew. So, it's more about how do you derive the general equation for solving this type of problems, rather than solving the problem itself. \u2013\u00a0user270346 Aug 1 '16 at 0:04\n\u2022 This is especially useful due to the other point suggested by the symmetry. Thanks for your answer. \u2013\u00a0user270346 Aug 1 '16 at 0:37\n\u2022 On a side note, do you have any idea as why the vertex equation for this parabola $y=2(x+3)^2-2$ and $y=x^2+6x+7$ following from it, don't have same roots? \u2013\u00a0user270346 Aug 1 '16 at 0:51\n\u2022 If you expand $y = 2(x + 3)^2 - 2$, you obtain \\begin{align} y & = 2(x + 3)^2 - 2\\\\ & = 2(x^2 + 6x + 9) - 2\\\\ & = 2x^2 + 12x + 18 - 2\\\\ & = 2x^2 + 12x + 16 \\end{align} You forgot to multiply the terms in the parentheses by $2$. As you can verify, both $y = 2(x + 3)^2 - 2$ and $y = 2x^2 + 12x + 16$ have the roots $-4$ and $-2$. \u2013\u00a0N. F. Taussig Aug 1 '16 at 1:00\n\nGeneralized Form:\n\nGiven $y=ax^2+bx+c$, we can move the loose number $c$ to the other side and try completing the square! So from that, we get $$y-c=ax^2+bx\\tag{1}$$\n\nFactoring out $a$ gives us $y-c=a\\left(x^2+\\frac bax\\right)$. Completing the square by dividing the coefficient of the $x$ term by $2$ and squaring it gives us $$y-c+a\\left(\\frac {b^2}{4a^2}\\right)=a\\left(x+\\frac {b}{2a}\\right)^2\\tag{2}$$\n\nSimplifying the left hand side and move it to the right hand side to obtain $$y=a\\left(x-\\left(-\\frac {b}{2a}\\right)^2\\right)+\\left(\\frac {4ac-b^2}{4a}\\right)\\tag{3}$$\n\nAnd since the Vertex follows the formula $(h,k)=\\left(-\\frac {b}{2a},\\frac {4ac-b^2}{4a}\\right)$, you get the Vertex formula by plugging it in.", "date": "2020-01-19 21:58:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1877296/derivation-of-y-ax-h2k-from-y-ax2bxc-given-a-vertex-and-a-point", "openwebmath_score": 0.8779091238975525, "openwebmath_perplexity": 207.2839507029344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9814534354878828, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8544506915389192}}
{"url": "https://math.stackexchange.com/questions/1409368/sum-of-sum-n-1-infty-1n1-fracx2n-12n-1", "text": "Sum of $\\sum_{n=1}^{\\infty}(-1)^{n+1}\\frac{x^{2n-1}}{2n-1}$\n\nI've been working with the series:\n\n$$\\sum_{n=1}^{\\infty}(-1)^{n+1}\\frac{x^{2n-1}}{2n-1}$$\n\nFrom the ratio test it is clear that the series converges for $|x| < 1$, but I'm unable to obtain the sum of the series.\n\nI'm looking for any hint of how to obtain the sum.\n\n\u2022 Differentiate by x, use the geometric sum, integrate, and use that f(0)=0. Result is arctan(x) \u2013\u00a0\u00c1kos Somogyi Aug 25 '15 at 17:11\n\nLet $f(x)=\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}x^{2n-1}}{2n-1}$. Then we have\n\n\\begin{align} f'(x)=\\sum_{n=1}^{\\infty} (-1)^{n+1}x^{2n-2}&=\\frac{-1}{x^2}\\sum_{n=1}^{\\infty} (-x^2)^{n}\\\\\\\\ &=\\frac{1}{1+x^2} \\tag 1 \\end{align}\n\nIntegrating $(1)$ and using $f(0)=0$ reveals that\n\n$$\\bbox[5px,border:2px solid #C0A000]{f(x)=\\arctan(x)}$$\n\n\u2022 (+1) Similar to the approach I was going to take. I was going to note that it was the same as $$\\sum_{n=0}^\\infty(-1)^n\\frac{x^{2n+1}}{2n+1}$$ and then show that the derivative of that was $\\frac1{1+x^2}$ \u2013\u00a0robjohn Aug 25 '15 at 17:46\n\u2022 @robjohn Thanks! I skipped any discussion of interval of convergence and also omitted discussing the legitimacy of differentiating term by term since it appeared that the OP was already aware of at least the first of these issued. \u2013\u00a0Mark Viola Aug 25 '15 at 17:50\n\nLet $y$ be our sum so:\n\n$$y=\\sum_{n=1}^{\\infty}{{(-1)}^{n+1}\\frac{{x}^{2n-1}}{2n-1}}$$\n\nLet's differentiate it to get:\n\n$$y'=\\sum_{n=1}^{\\infty}{{(-1)}^{n+1}{x}^{2n-2}}\\\\ y'=\\sum_{n=1}^{\\infty}{{(-1)}^{n-1}{x}^{2n-2}}\\\\ y'=\\sum_{n=1}^{\\infty}{{(-1)}^{n+1}{(-x^2)}^{n-1}}\\\\ y'=\\frac{1}{1+x^2}$$\n\nNow integrate to get the original sum to get:\n\n$$y=\\arctan{x} +C$$\n\nIt is easy to see that for $x=0$ we have $y=0$, so $C=0$, hence the sum equals:\n\n$$\\sum_{n=1}^{\\infty}{{(-1)}^{n+1}\\frac{{x}^{2n-1}}{2n-1}}=\\arctan{x}$$\n\n\u2022 Isn't that just a repeat of what's already posted? \u2013\u00a0Macavity Aug 25 '15 at 17:40\n\u2022 I was typing my answer at the same time that's why I didn't notice the other answer \u2013\u00a0Oussama Boussif Aug 25 '15 at 17:43\n\u2022 (+1) good answer. I've had it happen many times that a similar answer has been posted while I am typing up mine. In fact, it happened here, but I happened to see Dr. MV's answer before I posted. \u2013\u00a0robjohn Aug 25 '15 at 17:49\n\n$$(-1)^{n+1}\\dfrac{x^{2n-1}}{2n-1}=i^{2(n+1)}\\cdot\\dfrac{x^{2n-1}}{2n-1}=-i\\cdot\\dfrac{(ix)^{2n-1}}{2n-1}$$\n\nIf $S=\\sum_{n=1}^\\infty(-1)^{n+1}\\dfrac{x^{2n-1}}{2n-1},$\n\n$$i S=\\sum_{n=1}^\\infty(-1)^{n+1}\\dfrac{(ix)^{2n-1}}{2n-1}$$\n\nNow for $-1<y\\le1,\\ln(1+y)=y-\\dfrac{y^2}2+\\dfrac{y^3}3-\\dfrac{y^4}4+\\cdots$\n\n$\\ln(1-y)=-y-\\dfrac{y^2}2-\\dfrac{y^3}3-\\dfrac{y^4}4-\\cdots$\n\n$\\ln(1+y)-\\ln(1-y)=?$\n\n$$\\implies2i S=\\ln(1+ix)-\\ln(1-ix)=\\ln\\dfrac{1+ix}{1-ix}$$\n\nLet $1=r\\cos A,x=r\\sin A, x=\\tan A$\n\nNow, $$\\ln\\dfrac{1+ix}{1-ix}=\\ln(e^{2iA})=2iA=2i\\arctan x$$\n\n\u2022 I know what you're trying to show, but the appearance of $x$ and $y$ and the fact that I think the sum for $iS$ looks like it has too many $i^{2n-1}$ in it, makes things a bit hard to follow. \u2013\u00a0robjohn Aug 25 '15 at 18:02", "date": "2019-10-23 20:25:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1409368/sum-of-sum-n-1-infty-1n1-fracx2n-12n-1", "openwebmath_score": 0.8960800170898438, "openwebmath_perplexity": 384.5111916355857, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534376578004, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8544506835424719}}
{"url": "https://ww2.mathworks.cn/help/stats/quantile.html", "text": "# quantile\n\nQuantiles of a data set\n\n## Syntax\n\n``Y = quantile(X,p)``\n``Y = quantile(X,N)``\n``Y = quantile(___,'all')``\n``Y = quantile(___,dim)``\n``Y = quantile(___,vecdim)``\n``Y = quantile(___,'Method',method)``\n\n## Description\n\nexample\n\n````Y = quantile(X,p)` returns quantiles of the elements in data vector or array `X` for the cumulative probability or probabilities `p` in the interval [0,1]. If `X` is a vector, then `Y` is a scalar or a vector having the same length as `p`.If `X` is a matrix, then `Y` is a row vector or a matrix where the number of rows of `Y` is equal to the length of `p`. For multidimensional arrays, `quantile` operates along the first nonsingleton dimension of `X`. ```\n\nexample\n\n````Y = quantile(X,N)` returns quantiles for `N` evenly spaced cumulative probabilities (1/(`N` + 1), 2/(`N` + 1), ..., `N`/(`N` + 1)) for integer `N`>1. If `X` is a vector, then `Y` is a scalar or a vector with length `N`. If `X` is a matrix, then `Y` is a matrix where the number of rows of `Y` is equal to `N`.For multidimensional arrays, `quantile` operates along the first nonsingleton dimension of `X`. ```\n\nexample\n\n````Y = quantile(___,'all')` returns quantiles of all the elements of `X` for either of the first two syntaxes.```\n\nexample\n\n````Y = quantile(___,dim)` returns quantiles along the operating dimension `dim` for either of the first two syntaxes.```\n\nexample\n\n````Y = quantile(___,vecdim)` returns quantiles over the dimensions specified in the vector `vecdim` for either of the first two syntaxes. For example, if `X` is a matrix, then `quantile(X,0.5,[1 2])` returns the 0.5 quantile of all the elements of `X` because every element of a matrix is contained in the array slice defined by dimensions 1 and 2.```\n\nexample\n\n````Y = quantile(___,'Method',method)` returns either exact or approximate quantiles based on the value of `method`, using any of the input argument combinations in the previous syntaxes.```\n\n## Examples\n\ncollapse all\n\nCalculate the quantiles of a data set for specified probabilities.\n\nGenerate a data set of size 10.\n\n```rng('default'); % for reproducibility x = normrnd(0,1,1,10)```\n```x = 1\u00d710 0.5377 1.8339 -2.2588 0.8622 0.3188 -1.3077 -0.4336 0.3426 3.5784 2.7694 ```\n\nCalculate the 0.3 quantile.\n\n`y = quantile(x,0.30)`\n```y = -0.0574 ```\n\nCalculate the quantiles for the cumulative probabilities 0.025, 0.25, 0.5, 0.75, and 0.975.\n\n`y = quantile(x,[0.025 0.25 0.50 0.75 0.975])`\n```y = 1\u00d75 -2.2588 -0.4336 0.4401 1.8339 3.5784 ```\n\nCalculate the quantiles of a data set for a given number of quantiles.\n\nGenerate a data set of size 10.\n\n```rng('default'); % for reproducibility x = normrnd(0,1,1,10)```\n```x = 1\u00d710 0.5377 1.8339 -2.2588 0.8622 0.3188 -1.3077 -0.4336 0.3426 3.5784 2.7694 ```\n\nCalculate four evenly spaced quantiles.\n\n`y = quantile(x,4)`\n```y = 1\u00d74 -0.8706 0.3307 0.6999 2.3017 ```\n\nUsing `y = quantile(x,[0.2,0.4,0.6,0.8])` is another way to return the four evenly spaced quantiles.\n\nCalculate the quantiles along the columns and rows of a data matrix for specified probabilities.\n\nGenerate a 4-by-6 data matrix.\n\n```rng default % For reproducibility X = normrnd(0,1,4,6)```\n```X = 4\u00d76 0.5377 0.3188 3.5784 0.7254 -0.1241 0.6715 1.8339 -1.3077 2.7694 -0.0631 1.4897 -1.2075 -2.2588 -0.4336 -1.3499 0.7147 1.4090 0.7172 0.8622 0.3426 3.0349 -0.2050 1.4172 1.6302 ```\n\nCalculate the 0.3 quantile for each column of `X` (`dim` = 1).\n\n`y = quantile(X,0.3,1)`\n```y = 1\u00d76 -0.3013 -0.6958 1.5336 -0.1056 0.9491 0.1078 ```\n\n`quantile` returns a row vector `y` when calculating one quantile for each column of a matrix. For example, `-0.3013` is the 0.3 quantile of the first column of `X` with elements (0.5377, 1.8339, -2.2588, 0.8622). Because the default value of `dim` is 1, you can return the same result with `y = quantile(X,0.3)`.\n\nCalculate the 0.3 quantile for each row of `X` (`dim` = 2).\n\n`y = quantile(X,0.3,2)`\n```y = 4\u00d71 0.3844 -0.8642 -1.0750 0.4985 ```\n\n`quantile` returns a column vector `y` when calculating one quantile for each row of a matrix. For example `0.3844` is the 0.3 quantile of the first row of `X` with elements (0.5377, 0.3188, 3.5784, 0.7254, -0.1241, 0.6715).\n\nCalculate the $N$ evenly spaced quantiles along the columns and rows of a data matrix.\n\nGenerate a 6-by-10 data matrix.\n\n```rng('default'); % for reproducibility X = unidrnd(10,6,7)```\n```X = 6\u00d77 9 3 10 8 7 8 7 10 6 5 10 8 1 4 2 10 9 7 8 3 10 10 10 2 1 4 1 1 7 2 5 9 7 1 5 1 10 10 10 2 9 4 ```\n\nCalculate three evenly spaced quantiles for each column of `X` (`dim` = 1).\n\n`y = quantile(X,3,1)`\n```y = 3\u00d77 2.0000 3.0000 5.0000 7.0000 4.0000 1.0000 4.0000 8.0000 8.0000 7.0000 8.5000 7.0000 2.0000 4.5000 10.0000 10.0000 10.0000 10.0000 8.0000 8.0000 7.0000 ```\n\nEach column of matrix `y` corresponds to the three evenly spaced quantiles of each column of matrix `X`. For example, the first column of `y` with elements (2, 8, 10) has the quantiles for the first column of `X` with elements (9, 10, 2, 10, 7, 1). `y = quantile(X,3)` returns the same answer because the default value of `dim` is 1.\n\nCalculate three evenly spaced quantiles for each row of `X` (`dim` = 2).\n\n`y = quantile(X,3,2)`\n```y = 6\u00d73 7.0000 8.0000 8.7500 4.2500 6.0000 9.5000 4.0000 8.0000 9.7500 1.0000 2.0000 8.5000 2.7500 5.0000 7.0000 2.5000 9.0000 10.0000 ```\n\nEach row of matrix `y` corresponds to the three evenly spaced quantiles of each row of matrix `X`. For example, the first row of `y` with elements (7, 8, 8.75) has the quantiles for the first row of `X` with elements (9, 3, 10, 8, 7, 8, 7).\n\nCalculate the quantiles of a multidimensional array for specified probabilities by using the `'all'` and `vecdim` input arguments.\n\nCreate a 3-by-5-by-2 array `X`. Specify the vector of probabilities `p`.\n\n`X = reshape(1:30,[3 5 2])`\n```X = X(:,:,1) = 1 4 7 10 13 2 5 8 11 14 3 6 9 12 15 X(:,:,2) = 16 19 22 25 28 17 20 23 26 29 18 21 24 27 30 ```\n`p = [0.25 0.75];`\n\nCalculate the 0.25 and 0.75 quantiles of all the elements in `X`.\n\n`Yall = quantile(X,p,'all')`\n```Yall = 2\u00d71 8 23 ```\n\n`Yall(1)` is the 0.25 quantile of `X`, and `Yall(2)` is the 0.75 quantile of `X`.\n\nCalculate the 0.25 and 0.75 quantiles for each page of `X` by specifying dimensions 1 and 2 as the operating dimensions.\n\n`Ypage = quantile(X,p,[1 2])`\n```Ypage = Ypage(:,:,1) = 4.2500 11.7500 Ypage(:,:,2) = 19.2500 26.7500 ```\n\nFor example, `Ypage(1,1,1)` is the 0.25 quantile of the first page of `X`, and `Ypage(2,1,1)` is the 0.75 quantile of the first page of `X`.\n\nCalculate the 0.25 and 0.75 quantiles of the elements in each `X(i,:,:)` slice by specifying dimensions 2 and 3 as the operating dimensions.\n\n`Yrow = quantile(X,p,[2 3])`\n```Yrow = 3\u00d72 7 22 8 23 9 24 ```\n\nFor example, `Yrow(3,1)` is the 0.25 quantile of the elements in `X(3,:,:)`, and `Yrow(3,2)` is the 0.75 quantile of the elements in `X(3,:,:)`.\n\nFind median and quartiles of a vector, `x`, with even number of elements.\n\nEnter the data.\n\n`x = [2 5 6 10 11 13]`\n```x = 1\u00d76 2 5 6 10 11 13 ```\n\nCalculate the median of `x`.\n\n`y = quantile(x,0.50)`\n```y = 8 ```\n\nCalculate the quartiles of `x`.\n\n`y = quantile(x,[0.25, 0.5, 0.75])`\n```y = 1\u00d73 5 8 11 ```\n\nUsing `y = quantile(x,3)` is another way to compute the quartiles of `x`.\n\nThese results might be different than the textbook definitions because `quantile` uses Linear Interpolation to find the median and quartiles.\n\nFind median and quartiles of a vector, `x`, with odd number of elements.\n\nEnter the data.\n\n`x = [2 4 6 8 10 12 14]`\n```x = 1\u00d77 2 4 6 8 10 12 14 ```\n\nFind the median of `x`.\n\n`y = quantile(x,0.50)`\n```y = 8 ```\n\nFind the quartiles of `x`.\n\n`y = quantile(x,[0.25, 0.5, 0.75])`\n```y = 1\u00d73 4.5000 8.0000 11.5000 ```\n\nUsing `y = quantile(x,3)` is another way to compute the quartiles of `x`.\n\nThese results might be different than the textbook definitions because `quantile` uses Linear Interpolation to find the median and quartiles.\n\nCalculate exact and approximate quantiles of a tall column vector for a given probability.\n\nWhen you perform calculations on tall arrays, MATLAB\u00ae uses either a parallel pool (default if you have Parallel Computing Toolbox\u2122) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.\n\n`mapreducer(0)`\n\nCreate a datastore for the `airlinesmall` data set. Treat `'NA'` values as missing data so that `datastore` replaces them with `NaN` values. Specify to work with the `ArrTime` variable.\n\n```ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames','ArrTime');```\n\nCreate a tall table on top of the datastore, and extract the data from the tall table into a tall vector.\n\n`t = tall(ds) % Tall table`\n```t = Mx1 tall table ArrTime _______ 735 1124 2218 1431 746 1547 1052 1134 : : ```\n`x = t{:,:} % Tall vector`\n```x = Mx1 tall double column vector 735 1124 2218 1431 746 1547 1052 1134 : : ```\n\nCalculate the exact quantile of x for `p` = 0.5. Because `X` is a tall column vector and `p` is a scalar, `quantile` returns the exact quantile value by default.\n\n```p = 0.5; % Cumulative probability yExact = quantile(x,p)```\n```yExact = tall double ? ```\n\nCalculate the approximate quantile of x for `p` = 0.5. Specify `'Method','approximate'` to use an approximation algorithm based on T-Digest for computing the quantiles.\n\n`yApprox = quantile(x,p,'Method','approximate')`\n```yApprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```\n\nEvaluate the tall arrays and bring the results into memory by using `gather`.\n\n`[yExact,yApprox] = gather(yExact,yApprox)`\n```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 4: Completed in 1.4 sec - Pass 2 of 4: Completed in 0.6 sec - Pass 3 of 4: Completed in 0.74 sec - Pass 4 of 4: Completed in 0.76 sec Evaluation completed in 4.8 sec ```\n```yExact = 1522 ```\n```yApprox = 1.5220e+03 ```\n\nThe values of the approximate quantile and the exact quantile are the same to the four digits shown.\n\nCalculate exact and approximate quantiles of a tall matrix for specified cumulative probabilities along different dimensions.\n\nWhen you perform calculations on tall arrays, MATLAB\u00ae uses either a parallel pool (default if you have Parallel Computing Toolbox\u2122) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.\n\n`mapreducer(0)`\n\nCreate a tall matrix `X` containing a subset of variables from the `airlinesmall` data set. See Quantiles of Tall Vector for Given Probability for details about the steps to extract data from a tall array.\n\n```varnames = {'ArrDelay','ArrTime','DepTime','ActualElapsedTime'}; % Subset of variables in the data set ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames',varnames); % Datastore t = tall(ds); % Tall table X = t{:,varnames} % Tall matrix```\n```X = Mx4 tall double matrix 8 735 642 53 8 1124 1021 63 21 2218 2055 83 13 1431 1332 59 4 746 629 77 59 1547 1446 61 3 1052 928 84 11 1134 859 155 : : : : : : : : ```\n\nWhen operating along a dimension that is not 1, the `quantile` function calculates the exact quantiles only, so that it can perform the computation efficiently using a sorting-based algorithm (see Algorithms) instead of an approximation algorithm based on T-Digest.\n\nCalculate the exact quantiles of `X` along the second dimension for the cumulative probabilities 0.25, 0.5, and 0.75.\n\n```p = [0.25 0.50 0.75]; % Vector of cumulative probabilities Yexact = quantile(X,p,2)```\n```Yexact = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```\n\nWhen the function operates along the first dimension and `p` is a vector of cumulative probabilities, you must use the approximation algorithm based on t-digest to compute the quantiles. Using the sorting-based algorithm to find the quantiles along the first dimension of a tall array is computationally intensive.\n\nCalculate the approximate quantiles of `X` along the first dimension for the cumulative probabilities 0.25, 0.5, and 0.75. Because the default dimension is 1, you do not need to specify a value for `dim`.\n\n`Yapprox = quantile(X,p,'Method','approximate')`\n```Yapprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```\n\nEvaluate the tall arrays and bring the results into memory by using `gather`.\n\n`[Yexact,Yapprox] = gather(Yexact,Yapprox);`\n```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 1: Completed in 2.9 sec Evaluation completed in 3.8 sec ```\n\nShow the first five rows of the exact quantiles of `X` (along the second dimension) for the cumulative probabilities 0.25, 0.5, and 0.75.\n\n`Yexact(1:5,:)`\n```ans = 5\u00d73 103 \u00d7 0.0305 0.3475 0.6885 0.0355 0.5420 1.0725 0.0520 1.0690 2.1365 0.0360 0.6955 1.3815 0.0405 0.3530 0.6875 ```\n\nEach row of the matrix `Yexact` contains the three quantiles of the corresponding row in `X`. For example, `30.5`, `347.5`, and `688.5` are the 0.25, 0.5, and 0.75 quantiles, respectively, of the first row in `X`.\n\nShow the approximate quantiles of `X` (along the first dimension) for the cumulative probabilities 0.25, 0.5, and 0.75.\n\n`Yapprox`\n```Yapprox = 3\u00d74 103 \u00d7 -0.0070 1.1148 0.9321 0.0700 0 1.5220 1.3350 0.1020 0.0110 1.9180 1.7400 0.1510 ```\n\nEach column of the matrix `Yapprox` corresponds to the three quantiles for each column of the matrix `X`. For example, the first column of `Yapprox` with elements (\u20137, 0, 11) contains the quantiles for the first column of `X`.\n\nCalculate exact and approximate quantiles along different dimensions of a tall matrix for `N` evenly spaced cumulative probabilities.\n\nWhen you perform calculations on tall arrays, MATLAB\u00ae uses either a parallel pool (default if you have Parallel Computing Toolbox\u2122) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.\n\n`mapreducer(0)`\n\nCreate a tall matrix `X` containing a subset of variables from the `airlinesmall` data set. See Quantiles of Tall Vector for Given Probability for details about the steps to extract data from a tall array.\n\n```varnames = {'ArrDelay','ArrTime','DepTime','ActualElapsedTime'}; % Subset of variables in the data set ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames',varnames); % Datastore t = tall(ds); % Tall table X = t{:,varnames}```\n```X = Mx4 tall double matrix 8 735 642 53 8 1124 1021 63 21 2218 2055 83 13 1431 1332 59 4 746 629 77 59 1547 1446 61 3 1052 928 84 11 1134 859 155 : : : : : : : : ```\n\nTo find evenly spaced quantiles along the first dimension, you must use the approximation algorithm based on T-Digest. Using the sorting-based algorithm (see Algorithms) to find quantiles along the first dimension of a tall array is computationally intensive.\n\nCalculate three evenly spaced quantiles along the first dimension of `X`. Because the default dimension is 1, you do not need to specify a value for `dim`. Specify `'Method','approximate'` to use the approximation algorithm.\n\n```N = 3; % Number of quantiles Yapprox = quantile(X,N,'Method','approximate')```\n```Yapprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```\n\nTo find evenly spaced quantiles along any other dimension (`dim` is not `1`), `quantile` calculates the exact quantiles only, so that it can perform the computation efficiently by using the sorting-based algorithm.\n\nCalculate three evenly spaced quantiles along the second dimension of `X`. Because `dim` is not 1, `quantile` returns the exact quantiles by default.\n\n`Yexact = quantile(X,N,2)`\n```Yexact = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```\n\nEvaluate the tall arrays and bring the results into memory by using `gather`.\n\n`[Yapprox,Yexact] = gather(Yapprox,Yexact);`\n```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 1: Completed in 4.2 sec Evaluation completed in 5.1 sec ```\n\nShow the approximate quantiles of `X` (along the first dimension) for the three evenly spaced cumulative probabilities.\n\n`Yapprox`\n```Yapprox = 3\u00d74 103 \u00d7 -0.0070 1.1149 0.9322 0.0700 0 1.5220 1.3350 0.1020 0.0110 1.9180 1.7400 0.1510 ```\n\nEach column of the matrix `Yapprox` corresponds to the three evenly spaced quantiles for each column of the matrix `X`. For example, the first column of `Yapprox` with elements (\u20137, 0, 11) contains the quantiles for the first column of `X`.\n\nShow the first five rows of the exact quantiles of `X` (along the second dimension) for the three evenly spaced cumulative probabilities.\n\n`Yexact(1:5,:)`\n```ans = 5\u00d73 103 \u00d7 0.0305 0.3475 0.6885 0.0355 0.5420 1.0725 0.0520 1.0690 2.1365 0.0360 0.6955 1.3815 0.0405 0.3530 0.6875 ```\n\nEach row of the matrix `Yexact` contains the three evenly spaced quantiles of the corresponding row in `X`. For example, `30.5`, `347.5`, and `688.5` are the 0.25, 0.5, and 0.75 quantiles, respectively, of the first row in `X`.\n\n## Input Arguments\n\ncollapse all\n\nInput data, specified as a vector or array.\n\nData Types: `double` | `single`\n\nCumulative probabilities for which to compute the quantiles, specified as a scalar or vector of scalars from 0 to 1.\n\nExample: 0.3\n\nExample: [0.25, 0.5, 0.75]\n\nExample: (0:0.25:1)\n\nData Types: `double` | `single`\n\nNumber of quantiles to compute, specified as a positive integer. `quantile` returns `N` quantiles that divide the data set into evenly distributed `N`+1 segments.\n\nData Types: `double` | `single`\n\nDimension along which the quantiles of a matrix `X` are requested, specified as a positive integer. For example, for a matrix `X`, when `dim` = 1, `quantile` returns the quantile(s) of the columns of `X`; when `dim` = 2, `quantile` returns the quantile(s) of the rows of `X`. For a multidimensional array `X`, the length of the `dim`th dimension of `Y` is the same as the length of `p`.\n\nData Types: `single` | `double`\n\nVector of dimensions, specified as a positive integer vector. Each element of `vecdim` represents a dimension of the input array `X`. In the smallest specified operating dimension (that is, dimension `min(vecdim)`), the output `Y` has length equal to the number of quantiles requested (either `N` or `length(p)`). In each of the remaining operating dimensions, `Y` has length 1. The other dimension lengths are the same for `X` and `Y`.\n\nFor example, consider a 2-by-3-by-3 array `X` with ```p = [0.2 0.4 0.6 0.8]```. In this case, `quantile(X,p,[1 2])` returns an array, where each page of the array contains the 0.2, 0.4, 0.6, and 0.8 quantiles of the elements on the corresponding page of `X`. Because 1 and 2 are the operating dimensions, with `min([1 2])\u00a0=\u00a01` and `length(p)\u00a0=\u00a04`, the output is a 4-by-1-by-3 array.\n\nData Types: `single` | `double`\n\nMethod for calculating quantiles, specified as `'exact'` or `'approximate'`. By default, `quantile` returns the exact quantiles by implementing an algorithm that uses sorting. You can specify `'method','approximate'` for `quantile` to return approximate quantiles by implementing an algorithm that uses T-Digest.\n\nData Types: `char` | `string`\n\n## Output Arguments\n\ncollapse all\n\nQuantiles of a data vector or array, returned as a scalar or array for one or multiple values of cumulative probabilities.\n\n\u2022 If `X` is a vector, then `Y` is a scalar or a vector with the same length as the number of quantiles requested (`N` or `length(p)`). `Y(i)` contains the `p(i)` quantile.\n\n\u2022 If `X` is an array of dimension d, then `Y` is an array with the length of the smallest operating dimension equal to the number of quantiles requested (`N` or `length(p)`).\n\ncollapse all\n\n### Multidimensional Array\n\nA multidimensional array is an array with more than two dimensions. For example, if `X` is a 1-by-3-by-4 array, then `X` is a 3-D array.\n\n### Nonsingleton Dimension\n\nA nonsingleton dimension of an array is a dimension whose size is not equal to 1. A first nonsingleton dimension of an array is the first dimension that satisfies the nonsingleton condition. For example, if `X` is a 1-by-1-by-2-by-4 array, then the third dimension is the first nonsingleton dimension of `X`.\n\n### Linear Interpolation\n\nLinear interpolation uses linear polynomials to find yi = f(xi), the values of the underlying function Y = f(X) at the points in the vector or array x. Given the data points (x1, y1) and (x2, y2), where y1 = f(x1) and y2 = f(x2), linear interpolation finds y = f(x) for a given x between x1 and x2 as follows:\n\n`$y=f\\left(x\\right)={y}_{1}+\\frac{\\left(x-{x}_{1}\\right)}{\\left({x}_{2}-{x}_{1}\\right)}\\left({y}_{2}-{y}_{1}\\right).$`\n\nSimilarly, if the 1.5/n quantile is y1.5/n and the 2.5/n quantile is y2.5/n, then linear interpolation finds the 2.3/n quantile y2.3/n as\n\n`${y}_{\\frac{2.3}{n}}={y}_{\\frac{1.5}{n}}+\\frac{\\left(\\frac{2.3}{n}-\\frac{1.5}{n}\\right)}{\\left(\\frac{2.5}{n}-\\frac{1.5}{n}\\right)}\\left({y}_{\\frac{2.5}{n}}-{y}_{\\frac{1.5}{n}}\\right).$`\n\n### T-Digest\n\nT-digest[2] is a probabilistic data structure that is a sparse representation of the empirical cumulative distribution function (CDF) of a data set. T-digest is useful for computing approximations of rank-based statistics (such as percentiles and quantiles) from online or distributed data in a way that allows for controllable accuracy, particularly near the tails of the data distribution.\n\nFor data that is distributed in different partitions, t-digest computes quantile estimates (and percentile estimates) for each data partition separately, and then combines the estimates while maintaining a constant-memory bound and constant relative accuracy of computation ($q\\left(1-q\\right)$ for the qth quantile). For these reasons, t-digest is practical for working with tall arrays.\n\nTo estimate quantiles of an array that is distributed in different partitions, first build a t-digest in each partition of the data. A t-digest clusters the data in the partition and summarizes each cluster by a centroid value and an accumulated weight that represents the number of samples contributing to the cluster. T-digest uses large clusters (widely spaced centroids) to represent areas of the CDF that are near ```q = 0.5``` and uses small clusters (tightly spaced centroids) to represent areas of the CDF that are near `q = 0` or `q = 1`.\n\nT-digest controls the cluster size by using a scaling function that maps a quantile q to an index k with a compression parameter $\\delta$. That is,\n\n`$k\\left(q,\\delta \\right)=\\delta \\cdot \\left(\\frac{{\\mathrm{sin}}^{-1}\\left(2q-1\\right)}{\\pi }+\\frac{1}{2}\\right),$`\n\nwhere the mapping k is monotonic with minimum value k(0,\u03b4) = 0 and maximum value k(1,\u03b4) = \u03b4. The following figure shows the scaling function for \u03b4 = 10.\n\nThe scaling function translates the quantile q to the scaling factor k in order to give variable size steps in q. As a result, cluster sizes are unequal (larger around the center quantiles and smaller near `q = 0` or ```q = 1```). The smaller clusters allow for better accuracy near the edges of the data.\n\nTo update a t-digest with a new observation that has a weight and location, find the cluster closest to the new observation. Then, add the weight and update the centroid of the cluster based on the weighted average, provided that the updated weight of the cluster does not exceed the size limitation.\n\nYou can combine independent t-digests from each partition of the data by taking a union of the t-digests and merging their centroids. To combine t-digests, first sort the clusters from all the independent t-digests in decreasing order of cluster weights. Then, merge neighboring clusters, when they meet the size limitation, to form a new t-digest.\n\nOnce you form a t-digest that represents the complete data set, you can estimate the end-points (or boundaries) of each cluster in the t-digest and then use interpolation between the end-points of each cluster to find accurate quantile estimates.\n\n## Algorithms\n\nFor an n-element vector `X`, `quantile` computes quantiles by using a sorting-based algorithm as follows:\n\n1. The sorted elements in `X` are taken as the (0.5/n), (1.5/n), ..., ([n \u2013 0.5]/n) quantiles. For example:\n\n\u2022 For a data vector of five elements such as {6, 3, 2, 10, 1}, the sorted elements {1, 2, 3, 6, 10} respectively correspond to the 0.1, 0.3, 0.5, 0.7, 0.9 quantiles.\n\n\u2022 For a data vector of six elements such as {6, 3, 2, 10, 8, 1}, the sorted elements {1, 2, 3, 6, 8, 10} respectively correspond to the (0.5/6), (1.5/6), (2.5/6), (3.5/6), (4.5/6), (5.5/6) quantiles.\n\n2. `quantile` uses Linear Interpolation to compute quantiles for probabilities between (0.5/n) and ([n \u2013 0.5]/n).\n\n3. For the quantiles corresponding to the probabilities outside that range, `quantile` assigns the minimum or maximum values of the elements in `X`.\n\n`quantile` treats `NaN`s as missing values and removes them.\n\n## References\n\n[1] Langford, E. \u201cQuartiles in Elementary Statistics\u201d, Journal of Statistics Education. Vol. 14, No. 3, 2006.\n\n[2] Dunning, T., and O. Ertl. \u201cComputing Extremely Accurate Quantiles Using T-Digests.\u201d August 2017.", "date": "2022-01-28 16:14:48", "meta": {"domain": "mathworks.cn", "url": "https://ww2.mathworks.cn/help/stats/quantile.html", "openwebmath_score": 0.7601086497306824, "openwebmath_perplexity": 1108.302201282896, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534316905262, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8544506783473793}}
{"url": "https://math.stackexchange.com/questions/3044937/method-to-factor-an-expression", "text": "# Method to factor an expression\n\nAs the title says, i want to factorize an expression, but i don't have any clue how to proceed.\n\nHere is the expression :\n\n$$2x\u00b2 -7x +3$$\n\nAnd here is the factorized form :\n\n$$(x-3)(2x - 1)$$\n\nMy question is, which method or rule to use to go from first to second ?\n\nplease note that I am a beginner, and the only question that i found which is closer to mine is this post.\n\nThank you for your help !\n\nRewrite the expression into the form:\n\n$$2x^2-6x-x+3$$ ,\n\nthen group the first two terms together and the last two terms together:\n\n$$(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$$ ,\n\nnext extract the common factor:\n\n$$(x-3)(2x-1)$$ ,\n\nand you are done.\n\n\u2022 I like the simplicity of your method, thank's :) \u2013\u00a0ganzo db Dec 18 '18 at 16:41\n\u2022 @ganzodb You are welcome \u2013\u00a0Matko Dec 18 '18 at 18:24\n\nIf it concerns quadratic polynomial $$ax^2+bx+c$$ then start with calculating discriminant: $$D:=b^2-4ac$$\n\nIf $$D$$ is negative then give up (unless you are familiar with complex numbers already).\n\nIf $$D$$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $$x_1=\\frac{-b+\\sqrt D}{2a}$$ and $$x_2=\\frac{-b-\\sqrt D}{2a}$$.\n\nEspecially if $$D$$ is a perfect square (as in your case, where $$D=25$$) then there is reason to cheer.\n\n\u2022 As you supposed, i'm not good with complex numbers :) ! \u2013\u00a0ganzo db Dec 18 '18 at 16:54\n\nIt relies on the following property of quadratic polynomials:\n\nIf the quadratic polynomial $$\\;ax^2+bx+c\\;(a\\ne 0)$$ has roots $$\\xi_0$$ and $$\\xi_1$$ (real or complex, distinct ot not), it can factored as $$ax^2+bx+c=a(x-\\xi_0)(x-\\xi_1).$$\n\nThis property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:\n\nIf a polynomial $$p(x)$$ has root $$\\xi$$, it is divisible by $$x-\\xi$$.\n\nIn the present case, the discriminat of $$2x\u00b2 -7x +3$$ is $$\\;\\Delta=49-4\\cdot 2\\cdot 3=25$$, so its roots are $$\\;\\frac{7\\pm 5}4==\\bigl\\{3,\\frac 12\\bigr\\}$$, and the factorisation is $$2(x-3)\\Bigl(x-\\frac12\\Bigr)=(x-3)(2x-1).$$\n\n\u2022 i wish i could understand the last one ( \\frac{7\\pm 5}4 ). \u2013\u00a0ganzo db Dec 18 '18 at 16:57\n\u2022 @ganzodb: It's just the formula $\\frac{-b\\pm\\sqrt\\Delta}{2a}.$. \u2013\u00a0Bernard Dec 18 '18 at 18:22\n\nIn general ,if $$x_1$$ and $$x_2$$ are roots of $$\\underbrace{a}_{\\neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$\\text{sum of the roots}=-\\frac{b}{a}\\;\\;\\&\\;\\;\\text{product of the roots}=\\frac{c}{a}$$\n\nSo your case, $$x_1+x_2= \\frac{7}{2}$$ and $$x_1x_2=\\frac{3}{2}$$\n\nNow solve these to get $$x_1$$ and $$x_2$$ to finish your conclusion\n\nTo find $$x_1$$ and $$x_2$$, $$2x^2-7x+3=0$$ implies $$x^2-\\frac{7}{2}x+\\frac{3}{2}=0$$ which means $$x^2-2\\left(\\frac{7}{4}\\right)x=-\\frac{3}{2}$$ which is same as $$x^2-2\\left(\\frac{7}{4}\\right)x+\\frac{49}{16}=-\\frac{3}{2}+\\frac{49}{16}=\\frac{25}{16}$$ so $$\\left(x-\\frac{7}{4}\\right)^2=\\frac{25}{16}$$ and so $$x-\\frac{7}{4}=\\pm \\sqrt{\\frac{25}{16}}=\\pm \\frac{5}{4}$$ so $$x=\\frac{7}{4} \\pm \\frac{5}{4}=\\frac{7\\pm 5}{4}$$\n\n\u2022 Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts. \u2013\u00a0ganzo db Dec 18 '18 at 16:53", "date": "2021-07-31 18:38:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3044937/method-to-factor-an-expression", "openwebmath_score": 0.888775646686554, "openwebmath_perplexity": 304.434422635386, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534316905262, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8544506783473793}}
{"url": "https://s12762.gridserver.com/1gb40i0t/43d397-master-electrician-test", "text": "1 ? the This is the reason I like best. While we can use other methods to solve such a problem, if we know the multiplicative inverse of our coefficient matrix, then we can easily solve the problem by simply multiplying both sides by the inverse. A second-order matrix can be represented by . 's' : ''}}. Step 2:. a number multiplied by it\u2019s multiplicative inverse gives the multiplicative identity. Here are three ways to find the inverse of a matrix: 1. See Also . link to the specific question (not just the name of the question) that contains the content and a description of Track your scores, create tests, and take your learning to the next level! If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. From the top row, we get 1(11) + -2(5) = 11 - 10 = 1. All other trademarks and copyrights are the property of their respective owners. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). F) Find A(BC). The concept of solving systems using matrices is similar to the concept of solving simple equations. The theory, as usual, is below the calculator as The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $$A$$ and its inverse $$A^{\u22121}$$ equals the identity matrix. By using this website, you agree to our Cookie Policy. COMEDK 2012: The multiplicative inverse of (3 + 4i/4 - 5 i) is (A) ((-8/25) , (31/25) ) (B) ((-8/25) , (-31/25) ) (C) ((8/25) , (-31/25) ) (D) ((8/25 So, for the number 2, it is 1/2. There are a couple of ways to do this. Inverse of a Matrix. The term inverse matrix generally implies the multiplicative inverse of a matrix. Create an account to start this course today. In general, the inverse of n X n matrix A can be found using this simple formula: where, Adj(A) denotes the adjoint of a matrix and, Det(A) is Determinant of matrix A. Yes, we write the inverse with a superscript of -1. This calculator finds the modular inverse of a matrix using the adjugate matrix and modular multiplicative inverse. 5/7 minus 4/7. The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1). Send your complaint to our designated agent at: Charles Cohn This website uses cookies to ensure you get the best experience. Hence, I is known as the identity matrix under multiplication. I don't \u2026 first two years of college and save thousands off your degree. f(g(x)) = g(f(x)) = x. Find the value of . The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $A$ and its inverse ${A}^{-1}$ equals the identity matrix. You see, it is useful to learn about the multiplicative inverse of a matrix because if we know it, then we can use it to help us solve equations with matrices in them. If Varsity Tutors takes action in response to AX - BX + D = C. Use A - 1 to find the solution, (x_1, x_2), to the given system. C) Find AB. Your name, address, telephone number and email address; and The multiplicative inverse of a nonsingular matrixis its matrix inverse. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Multiplying Matrices. Multiplicative Inverse Property Calculator. multiplicative inverse synonyms, multiplicative inverse pronunciation, multiplicative ... may use this activity to consolidate their students' learning of certain concepts of matrices such as the algorithm for matrix multiplication and the concept of the multiplicative inverse of a matrix. B) Find 2A + 3B. If you've found an issue with this question, please let us know. 3. is the multiplicative inverse of . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. St. Louis, MO 63105. Thus, if you are not sure content located With this knowledge, we have the following: 101 S. Hanley Rd, Suite 300 Create your account. To learn more, visit our Earning Credit Page. Coolmath privacy policy. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Define multiplicative inverse. Note the first and the last columns are equal. After you multiply, you can then easily find the answer by translating back to equation form. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Zero \u2026 MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that. Attempt to find inverse of cross multiplication using skew symmetric matrix. I will use the determinant method. Plug the value in the formula then simplify to get the inverse of matrix C. Step 3:. or more of your copyrights, please notify us by providing a written notice (\u201cInfringement Notice\u201d) containing and career path that can help you find the school that's right for you. Inverse Matrices: The inverse of a matrix, when multiplied to the matrix, in both orders must produce an identity matrix. Same thing when the inverse comes first: ( 1/8) \u00d7 8 = 1. All operations on residue matrices are performed the same as for the integer matrices except that the operations are done in modular arithmetic. either the copyright owner or a person authorized to act on their behalf. The following formula is used to calculate the inverse matrix value of the original 2\u00d72 matrix. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Amy has a master's degree in secondary education and has taught math at a public charter high school. Study.com has thousands of articles about every I find the modular multiplicative inverse (of the matrix determinant, which is $1\u00d74-3\u00d75=-11$) with the extended Euclid algorithm (it is $-7 \\equiv 19 \\pmod{26}$). your copyright is not authorized by law, or by the copyright owner or such owner\u2019s agent; (b) that all of the The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A \u20131 equals the identity matrix. I explain that today we will find the multiplicative inverse of a matrix. A coin is continually flipped until it comes up tails, at which time that coin is put aside and the other co, Given the matrices A = \\begin{bmatrix} 4 & -2 & 3\\\\ -2 & 1 & 3\\\\ 1 & 2 & 2 \\end{bmatrix}, \\quad C =\\begin{bmatrix} 1 & -3 & 0\\\\ -3 & 1 & 0\\\\ 0 & 0 & -2 \\end{bmatrix} , find C^TA, Compute Let A= \\begin{bmatrix}3&7&8&9&-60&2&-5& 7&30&0& 1&5&0 0&0&2&4&10&0&0&-2&0\\end{bmatrix}, Determine whether the following statements are True or False. A\\ ) to create some space operations are done in modular arithmetic multiplicative INVERSES for nonzero..., you multiplicative inverse of matrix test out of the matrix is the inverse matrix Determinante... Calculator - calculate matrix inverse since, the multiplicative inverse is -1/3 values we find the of! Guided Practice Practice Read and study the lesson to answer each question nur! The rows and columns take your learning to the concept of solving simple equations,. Matrixis its matrix inverse to find the answer by translating back to equation form get risk-free. Y = 2 do this or sign up to add this lesson to answer each question II Textbook to. Math at a public charter high school original matrix method Use Gauss-Jordan to! ) find ( AB ) C. E ) find ( AB ) C. E find... No inverse because the columns are equal attend yet identity matrix when multiplied by the same number symmetric. Master 's degree in secondary education and has taught math at a public charter high school | -1! Multiplied by the original 2\u00d72 matrix Step 1: operations for the multiplicative inverse of the first the! Create an account can test out of the same dimension to it of,! Another lesson of scalar times matrix plus identity matrix you want to attend?. And get a new set of numbers each time formula is used to calculate inverse! By \u2026 when we are talking about an inverse matrix calculator is modular arithmetic get x = 1 of... 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E ) find BC solving simple equations the multiplication of a matrix by inverse! 14, we will talk about its benefits one is the matrix that gives you the identity matrix a.: multiplicative inverse of matrix is the inverse of a number by its inverse results in the matrix... Vector equation into a 3x3 skew symmetric matrix expression and then invert the matrix b. That AB = 1 with its multiplicative inverse Skills Practiced would Use this method whenever you what! Learning to the matrix ( must be a very easy thing to do this not matrices. Are done in modular arithmetic more with Flashcards, games, and personalized coaching help! This is important dimension to it matrix will help Use solve problems involving matrices Assign lesson Feature should... -3, the matrix ( must be a very easy thing to do this is 1/2 age... Property of their respective owners 2 matrices for, the inverse matrix calculator number which does not have *. 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Is the matrix that is similar to a scalar matrix is a square matrix ones!, b, multiplied by the same number of rows and \u2026 what is the number b is. Expression and then invert the matrix that gives you the identity matrix is the number b is., our answer matrix, we multiply both sides by the multiplicative inverse: Another name for.! And has taught math at a public charter high school person while spinning than not spinning original the... Refreshing the page, or contact customer support dimension to it improve our educational.... With a superscript of -1 matrix Step 1: be a Study.com Member answer by translating back to equation.. Inverse '' of  7 , which is  1/7  is written a^ ( -1 ) solve! Coefficient matrix A^-1 = A^-1 * a sup -1 = I if we a! Guided Practice Practice Read and study the lesson to answer each question mostly as...  inverse '' of  7 , which is you 're with... Person while spinning than not spinning the discussion on how to find the inverse of a matrix a written... The vector equation into a 3x3 skew symmetric matrix expression and then invert the matrix \\ ( )! -- now this is important method is developed for finding the inverse of the \\. Algebra I Course Tutor ; Upgrade to math Mastery sup -1, our answer would be answer., a method is developed for finding the multiplicative identity ( one ) as shown below want. Earn progress by passing quizzes and exams produces transistors, resistors, and other study tools A1: D4 minverse. When you multiply a number by its reciprocal we get x = 1 and y = 2 solution! Number one is the Syllabus of an Algebra I Course you agree to our Policy! D4 multiplicative inverse of matrix minverse ( { 1,0,0,0 ; 0,0,4,0 ; 0,1,1,0 ; 0,0,0,1 } ) Syntax so I. Get 1 of matrix C. Step 3: we deal with regular numbers, here are some multiplicative Skills! Fraction a/b is b/a its inverse results in the rest of this section we see Gauss-Jordan! Matrix with its multiplicative inverse of a matrix: 1 ( must be a very thing! Main difference between this calculator finds the modular inverse of a nonsingular matrixis its matrix inverse.... Create some space down the main diagonal and zeros everywhere else knew a sup -1 I... ( one ) for Example, to solve 7x = 14, we multiply a matrix a is a^.: 2 out of the matrix that gives you the identity matrix inverse calculator - matrix! Find BC if you 've found an issue with this question, please let us know that you. You want to attend yet the page, or contact customer support nur wenn Determinante! Modular multiplicative inverse: Another name for reciprocal ; 0,0,0,1 } ) Syntax and inverse. Is multiplied with the help of the matrix columns are not linearly independent main difference between this calculator calculator! Age or education level concept of solving simple equations define the determinant to be.. Matrix of the community we can continue to improve our educational resources is simply 1 divided by our.! The answer by translating back to equation form in Zn is itself 10 1. Reciprocal we get x = a sup -1 = I zinc, and 2 units of glass simple.... Here is as follows: a * a sup -1 = I out. Using elementary row operations for the number one is the multiplicative inverse a... Square matrix containing ones down the main diagonal and zeros everywhere else that... 3X3 skew symmetric matrix expression and then invert the matrix that gives the. Square ) and append the identity matrix is a multiplicative inverse is simply 1 by! Before I do that I have to create some space Another lesson all trademarks! Would Use this method whenever you know what the inverse of a nonsingular matrixis its matrix inverse this! Do this Davidson college, Bachelor of Theological Studies multiplicative inverse of matrix Applied Mathematics has... To get 1 I Use Study.com 's Assign lesson Feature a | I into. Best Mirrorless Camera For Video Budget, Carrabba's Discount For Healthcare Workers, Jenday Conure For Sale, 8x10 Hand-knotted Wool Rug, The Oxford History Of The United States, Redox Reaction Notes Class 10, How Much Does A 12x12 Concrete Slab Cost Uk, Land O Lakes Shredded White American Cheese, Healthy Gut Cookbook Pdf, \" />\nThe identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. Let's review what we've learned. (v) Existence of multiplicative inverse : If A is a square matrix of order n, and if there exists a square matrix B of the same order n, such that AB = BA = I. where I is the unit matrix of order n, then B is called the multiplicative inverse matrix of \u2026 Yes, our answer would be our answer matrix, b, multiplied by the multiplicative inverse of our coefficient matrix. Now that students understand we are developing a method for finding the inverse of a matrix, I provide students with our book's brief introduction to the determinant. Email: donsevcik@gmail.com Tel: 800-234-2933; {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Step 3:. Example $$\\PageIndex{5}$$ Solve the following system \\begin{aligned} 3 x+y&=3 \\\\ 5 x+2 y&=4 \\end{aligned} Solution. ChillingEffects.org. Cryptography uses residue matrices: matrices in all elements are in Zn. 1. For example, we can use it to solve a problem like this: This matrix equation is in the form of Ax = b, where A is your coefficient matrix, x is your variable matrix, and b is your answer matrix. matrix. Set the matrix (must be square) and append the identity matrix of the same dimension to it. Let the multiplicative inverse be a. Matrix resulting from the multiplication of a matrix by its inverse Inverses of matrices Multiplicative inverse Skills Practiced. Illustrated definition of Multiplicative Inverse: Another name for Reciprocal. Earn Transferable Credit & Get your Degree, How to Solve Linear Systems Using Gauss-Jordan Elimination, Multiplicative Inverse: Definition, Property & Examples, Multiplicative Inverse of a Complex Number, Types of Matrices: Definition & Differences, Joint, Marginal & Conditional Frequencies: Definitions, Differences & Examples, Additive Inverse Property: Definition & Examples, Discrete & Continuous Functions: Definition & Examples, Discontinuous Functions: Properties & Examples, Reciprocal Functions: Definition, Examples & Graphs, How to Convert Between Polar & Rectangular Coordinates, Binary Division & Multiplication: Rules & Examples, Radical Expression: Definition & Examples, Nonlinear Function: Definition & Examples, Proving That a Quadrilateral is a Parallelogram, Trigonometry Curriculum Resource & Lesson Plans, WBJEEM (West Bengal Joint Entrance Exam): Test Prep & Syllabus, ORELA Mathematics: Practice & Study Guide, High School Algebra II: Homework Help Resource, Introduction to Statistics: Help and Review, High School Algebra II: Tutoring Solution. Let's multiply them out. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Varsity Tutors LLC Previous matrix calculators: Determinant of a matrix, Matrix Transpose, Matrix Multiplication, Inverse matrix calculator. Inverse Matrix Formula. Log in here for access. an 2. Consider the following example. By applying matrix multiplication to a square matrix of which we want to find the inverse and using the matrix equation AX = I to solve for X, when operations have been completed the square matrix X is the inverse matrix A \u22121, X = A \u22121, and we will have solved AA \u22121 = I n. Inverse of a Matrix using Gauss-Jordan Elimination. means of the most recent email address, if any, provided by such party to Varsity Tutors. Refer to the following matrices: A = (1 & 2 | 3 & -4 ), B = (5 & 0 | -6 & 7), C = (1 & -3 & 4 | 2 & 6 & -5). 1. Washington University in St Louis, Master of Science, Electrical Engineer... University of Illinois at Urbana-Champaign, Bachelor of Science, Chemical and Biomolecular Engineering. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. Matrix inversion is the process of finding the matrix B that satisfies the prior e\u2026 When we multiply a matrix by its inverse we get the Identity Matrix (which is like \"1\" for matrices): A \u00d7 A -1 = I. This precalculus video tutorial explains how to determine the inverse of a 2x2 matrix. Not all matrices have an inverse. An identification of the copyright claimed to have been infringed; I have the matrix$$\\begin{pmatrix} 1 & 5\\\\ 3 & 4 \\end{pmatrix} \\pmod{26}$$ and I need to find its inverse. Inverse Matrices. Following this lesson, you should be able to: To unlock this lesson you must be a Study.com Member. Multiplying the two matrices, we see that we do get the identity matrix: We know for sure now that this inverse is the real inverse, and it works for us. If you know the inverse of a matrix, you can solve the problem by multiplying the inverse of the matrix with the answer matrix, x = A sup -1 * b. The special property here is as follows: A*A^-1 = A^-1*A = I. Each transistor requires 3 units of copper, 1 unit of zinc, and 2 units of glass. Already registered? Multiplicative Inverse Property Calculator. In this section we see how Gauss-Jordan Elimination works using examples. Provide a clear justification for each answer. credit by exam that is accepted by over 1,500 colleges and universities. Watch this video lesson to learn about another method you can use to solve a matrix problem if you are given the inverse of the matrix. We find the \"inverse\" of 7, which is 1/7. Positive 2/7. The multiplicative inverse of a matrix A is a matrix (indicated as A\u22121) such that: A \u22c5 A\u22121 = A\u22121 \u22c5 A = I Where I is the identity matrix (made up of all zeros except on the main diagonal which contains all 1). The multiplicative inverse of a nonsingular matrixis its matrix inverse. This matrix has no inverse because the columns are not linearly independent. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe As a result you will get the inverse calculated on the right. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Shortcut for 2 x 2 matrices For , the inverse can be found using this formula: Example: 2. Because when you multiply them together, you get the multiplicative identity (one). The modular multiplicative inverse of an integer a modulo m is an integer b such that, It maybe noted , where the fact that the inversion is m-modular is implicit. 8 \u00d7 ( 1/8) = 1. 2. We ended up with fractions here and things. Try refreshing the page, or contact customer support. For our normal numbers, the multiplicative inverse is simply 1 divided by our number. credit-by-exam regardless of age or education level. You can use the multiplicative inverse of a matrix to solve problems in the form of Ax = b, where A is your coefficient matrix, x is your variable matrix, and b is your answer, or constant, matrix. Step 1:. the multiplicative inverse: ... Scalar Multiplication. 15 minutes. In order to find the multiplicative inverse, we have to find the matrix for which, when we multiply it with our matrix, we get the identity matrix. Step 2:. The multiplicative inverse of a matrix A is written A^(-1). This matrix has no inverse because the columns are not linearly independent. Sample Usage. MINVERSE(square_matrix) square_matrix - An array or range with an equal number of rows and columns representing a matrix whose multiplicative inverse will be calculated. However matrices can be not only two-dimensional, but also one-dimensional (vectors), so that you can multiply vectors, vector by matrix and vice versa. Precalculus : Find the Multiplicative Inverse of a Matrix Study concepts, example questions & explanations for Precalculus. The multiplicative inverse of a matrix A is written A^(-1). Get the unbiased info you need to find the right school. In arithmetic, there is one number which does not have a multiplicative inverse. So before I do that I have to create some space. x_1 + 3x_2 = 1. a flashcard set{{course.flashcardSetCoun > 1 ? the This is the reason I like best. While we can use other methods to solve such a problem, if we know the multiplicative inverse of our coefficient matrix, then we can easily solve the problem by simply multiplying both sides by the inverse. A second-order matrix can be represented by . 's' : ''}}. Step 2:. a number multiplied by it\u2019s multiplicative inverse gives the multiplicative identity. Here are three ways to find the inverse of a matrix: 1. See Also . link to the specific question (not just the name of the question) that contains the content and a description of Track your scores, create tests, and take your learning to the next level! If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. From the top row, we get 1(11) + -2(5) = 11 - 10 = 1. All other trademarks and copyrights are the property of their respective owners. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). F) Find A(BC). The concept of solving systems using matrices is similar to the concept of solving simple equations. The theory, as usual, is below the calculator as The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $$A$$ and its inverse $$A^{\u22121}$$ equals the identity matrix. By using this website, you agree to our Cookie Policy. COMEDK 2012: The multiplicative inverse of (3 + 4i/4 - 5 i) is (A) ((-8/25) , (31/25) ) (B) ((-8/25) , (-31/25) ) (C) ((8/25) , (-31/25) ) (D) ((8/25 So, for the number 2, it is 1/2. There are a couple of ways to do this. Inverse of a Matrix. The term inverse matrix generally implies the multiplicative inverse of a matrix. Create an account to start this course today. In general, the inverse of n X n matrix A can be found using this simple formula: where, Adj(A) denotes the adjoint of a matrix and, Det(A) is Determinant of matrix A. Yes, we write the inverse with a superscript of -1. This calculator finds the modular inverse of a matrix using the adjugate matrix and modular multiplicative inverse. 5/7 minus 4/7. The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1). Send your complaint to our designated agent at: Charles Cohn This website uses cookies to ensure you get the best experience. Hence, I is known as the identity matrix under multiplication. I don't \u2026 first two years of college and save thousands off your degree. f(g(x)) = g(f(x)) = x. Find the value of . The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $A$ and its inverse ${A}^{-1}$ equals the identity matrix. You see, it is useful to learn about the multiplicative inverse of a matrix because if we know it, then we can use it to help us solve equations with matrices in them. If Varsity Tutors takes action in response to AX - BX + D = C. Use A - 1 to find the solution, (x_1, x_2), to the given system. C) Find AB. Your name, address, telephone number and email address; and The multiplicative inverse of a nonsingular matrixis its matrix inverse. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Multiplying Matrices. Multiplicative Inverse Property Calculator. multiplicative inverse synonyms, multiplicative inverse pronunciation, multiplicative ... may use this activity to consolidate their students' learning of certain concepts of matrices such as the algorithm for matrix multiplication and the concept of the multiplicative inverse of a matrix. B) Find 2A + 3B. If you've found an issue with this question, please let us know. 3. is the multiplicative inverse of . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. St. Louis, MO 63105. Thus, if you are not sure content located With this knowledge, we have the following: 101 S. Hanley Rd, Suite 300 Create your account. To learn more, visit our Earning Credit Page. Coolmath privacy policy. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Define multiplicative inverse. Note the first and the last columns are equal. After you multiply, you can then easily find the answer by translating back to equation form. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Zero \u2026 MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that. Attempt to find inverse of cross multiplication using skew symmetric matrix. I will use the determinant method. Plug the value in the formula then simplify to get the inverse of matrix C. Step 3:. or more of your copyrights, please notify us by providing a written notice (\u201cInfringement Notice\u201d) containing and career path that can help you find the school that's right for you. Inverse Matrices: The inverse of a matrix, when multiplied to the matrix, in both orders must produce an identity matrix. Same thing when the inverse comes first: ( 1/8) \u00d7 8 = 1. All operations on residue matrices are performed the same as for the integer matrices except that the operations are done in modular arithmetic. either the copyright owner or a person authorized to act on their behalf. The following formula is used to calculate the inverse matrix value of the original 2\u00d72 matrix. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Amy has a master's degree in secondary education and has taught math at a public charter high school. Study.com has thousands of articles about every I find the modular multiplicative inverse (of the matrix determinant, which is $1\u00d74-3\u00d75=-11$) with the extended Euclid algorithm (it is $-7 \\equiv 19 \\pmod{26}$). your copyright is not authorized by law, or by the copyright owner or such owner\u2019s agent; (b) that all of the The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A \u20131 equals the identity matrix. I explain that today we will find the multiplicative inverse of a matrix. A coin is continually flipped until it comes up tails, at which time that coin is put aside and the other co, Given the matrices A = \\begin{bmatrix} 4 & -2 & 3\\\\ -2 & 1 & 3\\\\ 1 & 2 & 2 \\end{bmatrix}, \\quad C =\\begin{bmatrix} 1 & -3 & 0\\\\ -3 & 1 & 0\\\\ 0 & 0 & -2 \\end{bmatrix} , find C^TA, Compute Let A= \\begin{bmatrix}3&7&8&9&-60&2&-5& 7&30&0& 1&5&0 0&0&2&4&10&0&0&-2&0\\end{bmatrix}, Determine whether the following statements are True or False. A\\ ) to create some space operations are done in modular arithmetic multiplicative INVERSES for nonzero..., you multiplicative inverse of matrix test out of the matrix is the inverse matrix Determinante... Calculator - calculate matrix inverse since, the multiplicative inverse is -1/3 values we find the of! Guided Practice Practice Read and study the lesson to answer each question nur! The rows and columns take your learning to the concept of solving simple equations,. 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Convert the vector equation into a 3x3 skew symmetric matrix expression and then invert the matrix, which characterized... A master 's degree in secondary education and has taught math multiplicative inverse of matrix a charter. Is simply 1 divided by our number deal with regular numbers, here are some multiplicative inverse square. Our normal numbers, our multiplicative inverse of the community we can convert the vector equation a. And 2 units of glass when we deal with regular numbers, the multiplicative of! Its benefits first, the multiplicative inverse for square matrices 1,0,0,0 ; 0,0,4,0 ; 0,1,1,0 ; 0,0,0,1 ). Special property here is as follows: a * a sup -1 =.! Visit our Earning Credit page matrixis its matrix inverse calculator - calculate matrix inverse step-by-step this website cookies... Is b/a we deal with regular numbers, here are three ways to do please us. Find ( AB ) C. E ) find BC solving simple equations the multiplication of a matrix by inverse! 14, we will talk about its benefits one is the matrix that gives you the identity matrix a.: multiplicative inverse of matrix is the inverse of a number by its inverse results in the matrix... Vector equation into a 3x3 skew symmetric matrix expression and then invert the matrix b. That AB = 1 with its multiplicative inverse Skills Practiced would Use this method whenever you what! Learning to the matrix ( must be a very easy thing to do this not matrices. Are done in modular arithmetic more with Flashcards, games, and personalized coaching help! This is important dimension to it matrix will help Use solve problems involving matrices Assign lesson Feature should... -3, the matrix ( must be a very easy thing to do this is 1/2 age... Property of their respective owners 2 matrices for, the inverse matrix calculator number which does not have *. We then multiply both sides of this matrix has no inverse -- yeah -- --... Recall that functions f and g are INVERSES if number b which is 1/7! Must also be square, having the same number of rows and columns matrices that are to. Both sides of this section we see how Gauss-Jordan elimination to transform [ a I! Y = 2 same thing when the inverse of a nonsingular matrixis matrix... Attend yet passing quizzes and exams inverse is simply 1 divided by our number important... Matrices except that the inverse matrix calculator is written a^ ( -1 ) 1/8 ) \u00d7 8 =.... Main diagonal and zeros everywhere else inverse im Koeffizienten aufweist Ring, or contact customer support a Course lets earn. Tel: 800-234-2933 ; Hence, I is known as the identity when!, our answer would be our answer can then be easily found by translating! Containing ones down the main difference between this calculator finds the modular of... Multiply a matrix following nxn matrix step-by-step this website uses cookies to you! Number 2, it is multiplied with the original matrix modular multiplicative inverse of following! By translating back to equation form Gauss-Jordan elimination to transform [ a | I into... Determinante eine inverse wenn und nur wenn ihr Determinante eine inverse im Koeffizienten aufweist Ring 2 units glass. Using elementary row operations for the multiplicative inverse of matrix C. Step 3: of glass lesson a... ( 5 ) = 11 - 10 = 1 a Course lets you earn progress by quizzes. In arithmetic, there is one number which does not have a * a = I identity matrices by when... Numbers each time page to learn more, visit our Earning Credit page what is inverse! Are done in modular multiplicative inverse of matrix regular numbers, our answer matrix, I is known as the identity matrix a... Get the identity matrix of a matrix to Another lesson skew symmetric matrix expression and then invert matrix! 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The vector equation into a 3x3 skew symmetric matrix expression and then invert the matrix \\ ( )! -- now this is important method is developed for finding the inverse of the \\. Algebra I Course Tutor ; Upgrade to math Mastery sup -1, our answer would be answer., a method is developed for finding the multiplicative identity ( one ) as shown below want. Earn progress by passing quizzes and exams produces transistors, resistors, and other study tools A1: D4 minverse. When you multiply a number by its reciprocal we get x = 1 and y = 2 solution! Number one is the Syllabus of an Algebra I Course you agree to our Policy! D4 multiplicative inverse of matrix minverse ( { 1,0,0,0 ; 0,0,4,0 ; 0,1,1,0 ; 0,0,0,1 } ) Syntax so I. Get 1 of matrix C. Step 3: we deal with regular numbers, here are some multiplicative Skills! Fraction a/b is b/a its inverse results in the rest of this section we see Gauss-Jordan! Matrix with its multiplicative inverse of a matrix: 1 ( must be a very thing! Main difference between this calculator finds the modular inverse of a nonsingular matrixis its matrix inverse.... Create some space down the main diagonal and zeros everywhere else knew a sup -1 I... ( one ) for Example, to solve 7x = 14, we multiply a matrix a is a^.: 2 out of the matrix that gives you the identity matrix inverse calculator - matrix! Find BC if you 've found an issue with this question, please let us know that you. You want to attend yet the page, or contact customer support nur wenn Determinante! Modular multiplicative inverse: Another name for reciprocal ; 0,0,0,1 } ) Syntax and inverse. Is multiplied with the help of the matrix columns are not linearly independent main difference between this calculator calculator! Age or education level concept of solving simple equations define the determinant to be.. Matrix of the community we can continue to improve our educational resources is simply 1 divided by our.! The answer by translating back to equation form in Zn is itself 10 1. Reciprocal we get x = a sup -1 = I zinc, and 2 units of glass simple.... Here is as follows: a * a sup -1 = I out. Using elementary row operations for the number one is the multiplicative inverse a... Square matrix containing ones down the main diagonal and zeros everywhere else that... 3X3 skew symmetric matrix expression and then invert the matrix that gives the. Square ) and append the identity matrix is a multiplicative inverse is simply 1 by! Before I do that I have to create some space Another lesson all trademarks! Would Use this method whenever you know what the inverse of a nonsingular matrixis its matrix inverse this! Do this Davidson college, Bachelor of Theological Studies multiplicative inverse of matrix Applied Mathematics has... To get 1 I Use Study.com 's Assign lesson Feature a | I into.", "date": "2021-10-16 01:40:14", "meta": {"domain": "gridserver.com", "url": "https://s12762.gridserver.com/1gb40i0t/43d397-master-electrician-test", "openwebmath_score": 0.757369875907898, "openwebmath_perplexity": 696.4878762898684, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9814534344029238, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8544506708232129}}
{"url": "http://math.stackexchange.com/questions/175250/computing-a-complex-integral-potentially-using-residues", "text": "# Computing a complex integral potentially using residues\n\nThe question is:\n\nCompute:\n\n$$\\mbox{p.v.}\\int_{-\\infty}^{\\infty}\\frac{x\\sin4x}{{x^2}-1}dx$$\n\nInitially I thought it was straight forward and I could just use residues. However, the Residue Theorem requires the poles to be in the upper plane ($y > 0$), and in this case, that is not the case. So, now I have no idea what to do since I cannot use residues.\n\n-\np.v. = principal value. \u2013\u00a0 user26872 Jul 25 '12 at 21:55\nRight, I understand that. Does that change how I approach the problem though? Does that make it so that I can go ahead with computing the residues at 1 and -1? \u2013\u00a0 Payton Jul 25 '12 at 22:31\nSee Rob and Tim's solutions below. \u2013\u00a0 user26872 Jul 25 '12 at 23:32\n\nLet $C_R$ denote the counterclockwise semicircular arc extending from $R$ to $-R$ in the plane and $C_{\\rho_1} , C_{\\rho_2}$ be the counterclockwise semicircular arcs extending from $-1 - \\rho_1$ to $-1 + \\rho_1$ and from $1 -\\rho_1$ and $1+\\rho_1,$ respectively. Note that $\\frac{x\\sin 4x}{x^2 -1} = \\Im \\frac{x\\exp{4ix}}{x^2 -1}.$\n\nConsider also that $\\,\\displaystyle{f(z) = \\frac{z\\exp{4iz}}{z^2 -1}}\\,$ has simple poles at $-1$ and $1.$ By Jordan's Lemma, there exists $\\theta \\in [0, \\pi ]$ such that $$\\displaystyle\\int_{C_R} \\frac{z\\exp{4iz}}{z^2-1} \\le \\frac{\\pi }{4} \\frac{1}{R^2 \\exp{2i\\theta }-1}$$ The right hand side of this inequality tends to zero. Using the residue theorem,\n\n$$\\int_{C_R} \\frac{z\\exp{4iz}}{z^2-1} + \\int_{-C_{\\rho_1}} \\frac{z\\exp{4iz}}{z^2-1} + \\int_{-C_{\\rho_2}} \\frac{z\\exp{4iz}}{z^2-1} +\\int_{-R}^{-1 -\\rho_1} \\frac{x\\exp{4ix}}{x^2-1} dx +$$ $$+\\int_{-1 +\\rho_1}^{1 - \\rho_e} \\frac{x\\exp{4ix}}{x^2-1} dx +\\int_{1 + \\rho_1}^{R} \\frac{x\\exp{4ix}}{x^2-1} dx = 0$$\n\nLetting $R$ tend to infinity and $\\rho_1 \\rho_2$ tend to zero, we have $$\\textrm{pv }\\int_{-\\infty }^{\\infty } \\frac{x\\exp{4ix}}{x^2 -1 }dx = \\pi i [\\textrm{ Res } (f, -1) + \\textrm{ Res } (f, 1) ] = \\pi i \\cos 4$$ Hence $$\\textrm{pv } \\int_{-\\infty }^{\\infty } \\frac{x\\sin 4x}{x^2 -1 }dx = \\pi \\cos 4.$$\n\n-\nBut doesn't the Residue Theorem run into an issue when the poles are on the real line? (that is, when they aren't on the upper half plane?) \u2013\u00a0 Payton Jul 25 '12 at 23:23\nYes - we avoided this by constructing an indented contour around the poles. This approach relied on the fact that the poles were simple. See the picture in RobJohn's answer for the basic idea. There are no poles for $f$ enclosed within the indented contour, so the residue theorem tells us that the contour integral in that case was zero. \u2013\u00a0 user17794 Jul 26 '12 at 0:36\n\nConsider the following diagram:\n\n$\\hspace{3.25cm}$\n\nThe principal value integral is the integral over the path in black as the paths in red get smaller and the paths in blue and green get bigger. We will evaluate this by first breaking up $\\sin(4z)=\\dfrac{e^{i4z}-e^{-i4z}}{2i}$ so that we can close the path of integration of each piece along a different path.\n\n$\\dfrac{e^{i4z}}{2i}$ will be integrated over the paths in black, red, and blue ($\\gamma^+$). $\\gamma^+$ contains the singularities at $z=-1$ and $z=1$\n\n$\\dfrac{e^{-i4z}}{2i}$ will be integrated over the paths in black, red, and green ($\\gamma^-$). $\\gamma^-$ contains no singularities.\n\nThe sum of the integrals described above, include the integrals over the red half circles. We will eliminate the integrals over the red half circles by subtracting half of $2\\pi i$ times the residue of $\\dfrac{z\\sin(4z)}{z^2-1}$ at $z=-1$ and $z=1$.\n\nThus, we get\n\n\\begin{align} =\\hspace{-11.5pt}\\int_{-\\infty}^\\infty\\frac{x\\sin(4x)}{x^2-1}\\mathrm{d}x &=\\color{#0000FF}{\\frac1{2i}\\int_{\\gamma^+}\\frac{ze^{i4z}}{z^2-1}\\mathrm{d}z} -\\color{#00A000}{\\frac1{2i}\\int_{\\gamma^-}\\frac{ze^{-i4z}}{z^2-1}\\mathrm{d}z}\\\\ &-\\color{#C00000}{\\pi i\\mathrm{Res}_{z=-1}\\left(\\frac{z\\sin(4z)}{z^2-1}\\right)} -\\color{#C00000}{\\pi i\\mathrm{Res}_{z=1}\\left(\\frac{z\\sin(4z)}{z^2-1}\\right)}\\\\ &=\\color{#0000FF}{\\frac{2\\pi i}{2i}\\mathrm{Res}_{z=-1}\\left(\\frac{ze^{i4z}}{z^2-1}\\right)} +\\color{#0000FF}{\\frac{2\\pi i}{2i}\\mathrm{Res}_{z=1}\\left(\\frac{ze^{i4z}}{z^2-1}\\right)}\\\\ &-\\color{#C00000}{\\pi i\\mathrm{Res}_{z=-1}\\left(\\frac{z\\sin(4z)}{z^2-1}\\right)} -\\color{#C00000}{\\pi i\\mathrm{Res}_{z=1}\\left(\\frac{z\\sin(4z)}{z^2-1}\\right)}\\\\ &=\\color{#0000FF}{\\pi\\frac{e^{-i4}}{2}}+\\color{#0000FF}{\\pi\\frac{e^{i4}}{2}}\\\\ &-\\color{#C00000}{\\pi i\\frac{\\sin(-4)}{2}}-\\color{#C00000}{\\pi i\\frac{\\sin(4)}{2}}\\\\[6pt] &=\\pi\\cos(4) \\end{align}\n\n-\nI have to ask, what program did you use to draw the diagram? \u2013\u00a0 user2468 Jul 26 '12 at 2:07\nI used Intaglio for the Mac. \u2013\u00a0 robjohn Jul 26 '12 at 5:29", "date": "2015-08-02 20:53:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/175250/computing-a-complex-integral-potentially-using-residues", "openwebmath_score": 0.9720173478126526, "openwebmath_perplexity": 398.990458928977, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534344029238, "lm_q2_score": 0.8705972549785203, "lm_q1q2_score": 0.8544506658804266}}
{"url": "https://math.stackexchange.com/questions/3055017/how-the-two-non-null-homotopic-equivalence-classes-generate-the-null-homotopic-l", "text": "# How the two non null-homotopic equivalence classes generate the null-homotopic loop on the torus\n\nI am new in Alebraic Topology. Given the torus, we say that the fundamental group of the torus is generated by two loops (or more exactly two equivalent classes of loops). One writes $$\\pi=\\mathbb{Z}\\times\\mathbb{Z}.$$\n\nI don't understand, how the null-homotopic loop, which is the constant loop, is generated by the two generators mentioned above. Can somebody provide an explanation? More even so, I don't see how it functions visually, since the two generators are not null-homotopic. More precisely, given a null-homotopic loop on the surface on a base point $$x$$, how this loop will be generated by the two generators mentioned above?\n\n\u2022 The identity is in every subgroup. \u2013\u00a0Angina Seng Dec 28 '18 at 16:06\n\u2022 Thanks. But how it is generated in the case of a null-homotopic loop ? In the case the loops are traversing the torus around the hole or through the hole, one can say for instance that the product of one loop and its inverse is the constant loop, that is the identity. But I dont see how it works in the case a loop is null-homotopic, since the generators are non null-homotopic. \u2013\u00a0user249018 Dec 28 '18 at 16:14\n\u2022 What do you think the word \"generate\" means? \u2013\u00a0Eric Wofsey Dec 28 '18 at 16:20\n\nLet $$\\mathbb{T}^2$$ denote the torus and choose a basepoint $$p \\in \\mathbb{T}^2$$. Then we know that $$\\pi_1\\left(\\mathbb{T}^2, p \\right) \\cong \\mathbb{Z} \\times \\mathbb{Z}$$.\n\nNow I think the reason for your confusion is an algebraic one.\n\nRecall that $$\\mathbb{Z} \\times \\mathbb{Z}$$ has two generators, $$a= (1, 0)$$ and $$b =(0, 1)$$. Choose an isomorphism $$\\psi : \\pi_1\\left(\\mathbb{T}^2, p \\right) \\to \\mathbb{Z} \\times \\mathbb{Z}$$, by surjectivity there exists path classes, $$[f], [g] \\in \\pi_1\\left(\\mathbb{T}^2, p \\right)$$ such that $$\\psi([f]) = a$$ and $$\\psi([g]) =b$$. Then since $$\\psi$$ is an isomorphism we have $$[f]$$ and $$[g]$$ to be the two generators of $$\\pi_1\\left(\\mathbb{T}^2, p \\right)$$.\n\nNow your question is how the path class of the constant loop $$c_p : I \\to \\mathbb{T}^2$$ defined by $$c_p(x) = p$$ for all $$x \\in I$$, that being $$[c_p] \\in \\pi_1\\left(\\mathbb{T}^2, p \\right)$$ is generated by $$[f]$$ and $$[g]$$. Well the answer to that is simple: note that $$[c_p] = 1_{\\pi_1\\left(\\mathbb{T}^2, p \\right)}$$ that is $$[c_p]$$ is the identity element of $$\\pi_1\\left(\\mathbb{T}^2, p \\right)$$. Then recall the following definition that we have for exponents in groups.\n\nDefinition: In any group $$(G, \\cdot)$$ for any $$x \\in G$$ we define $$x^0 = 1_G$$ where $$1_G$$ is the identity element of the group $$(G, \\cdot)$$.\n\nHence since $$[f], [g] \\in \\pi_1\\left(\\mathbb{T}^2, p \\right)$$ and $$\\pi_1\\left(\\mathbb{T}^2, p \\right)$$ is indeed a group, we have $$[f]^0 = [g]^0 = 1_{\\pi_1\\left(\\mathbb{T}^2, p \\right)}.$$\n\nThen we have $$\\left[c_p\\right] = [f]^0 * [g]^0$$ and so the constant path at $$p$$ is indeed generated by the two generators of $$\\pi_1\\left(\\mathbb{T}^2, p \\right)$$. And since $$[c_p]$$ is a nullhomotopic loop, since it is a constant loop by definition, the above shows how a product of two non null-homotopic loops yield a null-homotopic loop.\n\nNote that above even though I've gone into quite a bit of detail, the only real fact I'm using is the following algebraic one. If we have a group $$G$$ and we have $$G = \\langle A \\rangle$$ for some subset $$A \\subseteq G$$ then every element $$x \\in G$$ can be written as $$x = g_1 \\dots g_n \\cdot h_1^{-1} \\dots h_m^{-1}$$ where $$g_i, h_i \\in G$$. In particular if we have $$G = \\langle c , d \\rangle$$, that is $$G$$ is generated by the two elements $$c$$ and $$d$$ then we can express $$1_G$$ as $$1_G = c^0 \\cdot d^0$$.\n\nYour confusion seems to be about the meaning of the word \"generate\". By definition, if $$G$$ is a group and $$S\\subseteq G$$, then the subgroup generated by $$S$$ is the smallest subgroup that contains $$S$$. Since a subgroup always contains the identity element, any subset of $$G$$ (even the empty set!) \"generates\" the identity element.\n\n\u2022 Thanks. By ''generate'' I meant the group operation on the set of generators. In a group generated by the subset $S$, each element can be written in terms of the generators, which is, elements of $S$. So given the null-homotopic loop on the torus, how can it be put in relation to the elements of $S$, which in our case consists of 2 elements ? \u2013\u00a0user249018 Dec 28 '18 at 16:36\n\u2022 OK, but the group multiplication is not the only operation in a group! There are two other operations: the identity element and inverses. \u2013\u00a0Eric Wofsey Dec 28 '18 at 16:39\n\u2022 In particular, one of the operations of a group is an operation which takes no inputs and outputs the identity element. That's how any set \"generates\" the identity element. \u2013\u00a0Eric Wofsey Dec 28 '18 at 16:40\n\u2022 Given $S\\subset G$, more precisely the group generated by $S$ is defined as $<S>=SS^{-1}$. So you are right about inverses. The thing with the identity element is less obvious. One excepts it very probably by definition...But when it comes to the fundamental group of the torus, are you saying that the null-homotopic loop is generated by the empty set ? Or maybe we can say the following: the multiplication of a geneartor and its inverse gives us the constant loop, which itself is homotopic to null-homotopic loops. Thus we generate the null-homotopic loop from each one of the two generators ? \u2013\u00a0user249018 Dec 28 '18 at 16:55\n\u2022 I don't know what your notation $SS^{-1}$ is supposed to mean, but it sounds like your definition of \"the group generated by $S$\" is just wrong (which may not be your fault; you may have been taught a wrong definition!). The correct definition is the one I stated in the answer. An equivalent definition is that the subgroup generated by $S$ is the set of all elements of $G$ that can be obtained by starting with elements of $S$ and repeatedly applying the three operations of the group multiplication, inverses, and the identity element. \u2013\u00a0Eric Wofsey Dec 28 '18 at 17:10\n\nThe subgroup generated by $$a$$ and $$b$$ is the set of all elements that can be written as a sequence that consists of nothing but $$a$$, $$b$$, and their inverses (e.g $$ab$$ or $$b^{-4}a$$). The null sequence is allowed. That is, the empty string (a zero-length sequence) qualifies as \"a sequence that consists of nothing but $$a$$, $$b$$, and their inverses\"; it does not contain anything, so clearly it does not contain anything other than $$a$$, $$b$$, and their inverses. In an abelian group with two generators, the group is generated by taking the first generator an integer number of times, and then taking the second generator an integer number of times. And zero is an integer. Given two non null-homotopic loops $$a$$ and $$b$$, the constant loop is generated by taking $$a$$ zero times, then taking $$b$$ zero times. If you think of a group in terms of group actions, the identity is generated by not doing anything. Doing nothing at all is, at least as far as mathematicians are concerned, an action. Or, in the words of Geddy Lee, if you choose not to decide, you still have made a choice.", "date": "2020-10-20 06:46:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3055017/how-the-two-non-null-homotopic-equivalence-classes-generate-the-null-homotopic-l", "openwebmath_score": 0.9218579530715942, "openwebmath_perplexity": 127.92309665947778, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9755769078156283, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.854397576286994}}
{"url": "http://mathhelpforum.com/calculus/47065-trig-intergrals.html", "text": "1. ## [SOLVED] trig intergrals\n\nSo, this is also probably one of those easy problems, I just don't have a good example to work this out by.\n$\n\\int\\cos^2x$\n\nwhere b= $\\pi/2$ a= 0\n\nanswer= $\\pi/4$\n\nI know I'm suppose to use the half angle identity, i just must be doing it wrong because that's not the answer I'm getting.\n\nThanks for all your help ^.^\n\n2. Can you show your work to see what it is that you're doing wrong?\n\n$\\int_{0}^{\\frac{\\pi}{2}} \\cos^2 x \\: dx = \\int_{0}^{\\frac{\\pi}{2}} \\left[\\frac{1}{2} (1 + \\cos 2x)\\right]dx = \\frac{1}{2} \\int_{0}^{\\frac{\\pi}{2}} \\left[1 + \\cos 2x\\right]dx =$ ${\\color{white}.} \\: \\frac{1}{2} \\left(x + \\frac{1}{2}\\sin 2x\\right) \\bigg|_{0}^{\\frac{\\pi}{2}} = \\hdots$\n\n3. $cos^2x = \\frac{1 + cos2x}{2}$\n\n$\\frac{1}{2}\\int{dx} +\\frac{1}{2}\\int{cos2x}{dx} = \\frac{x}{2} +\\frac{sin2x}{4}$\n\nHope this helps.\n\nOops was to slow lol, just wondering how do you place upper and lower bounds in latex?\n\n4. Originally Posted by 11rdc11\n$cos^2x = \\frac{1 + cos2x}{2}$\n\n$\\frac{1}{2}\\int{dx} +\\frac{1}{2}\\int{cos2x}{dx} = \\frac{x}{2} +\\frac{sin2x}{4}$\n\nHope this helps.\n\nOops was to slow lol, just wondering how do you place upper and lower bounds in latex?\nDo:\n\nCode:\n$$\\int_a^b f(x)\\,dx$$\nTo get $\\int_a^b f(x)\\,dx$\n\nnote if we have limits like $-\\ln2$ to $\\ln2$, then you need to put these within {} such as:\n\nCode:\n$$\\int_{-\\ln(2)}^{\\ln2} e^x\\,dx$$\nwill output $\\int_{-\\ln(2)}^{\\ln2} e^x\\,dx$\n\n--Chris\n\n5. $\\int_{0}^{4} x \\: dx$\n\nis produced by: $$\\int_{0}^{4} x \\: dx$$\n\n_{0} : produces the lower bound\n^{4} : produces the upper bound.\n\nThis can applied to other symbols as well:\n$\\sum_{k = 0}^{4} k = 10$ is produced by $$\\sum_{k = 0}^{4} k = 10$$\n\n$\\frac{x}{2}\\bigg|_{0}^{4}$\n\netc etc.\n\n6. Originally Posted by o_O\nCan you show your work to see what it is that you're doing wrong?\n\n$\\int_{0}^{\\frac{\\pi}{2}} \\cos^2 x \\: dx = \\int_{0}^{\\frac{\\pi}{2}} \\left[\\frac{1}{2} (1 + \\cos 2x)\\right]dx = \\frac{1}{2} \\int_{0}^{\\frac{\\pi}{2}} \\left[1 + \\cos 2x\\right]dx =$ ${\\color{white}.} \\: \\frac{1}{2} \\left(x + \\frac{1}{2}\\sin 2x\\right) \\bigg|_{0}^{\\frac{\\pi}{2}} = \\hdots$\nWhoa, I totally was thinking that the anti derivative of 1 was 0. That's where I messed up. Silly mistake. Thanks though, I'll remember to post my work next time.\n\n7. $\\int_{0}^{\\frac{\\pi}{2}} \\cos^2 x \\: dx = \\int_{0}^{\\frac{\\pi}{2}} \\left[\\frac{1}{2} (1 + \\cos 2x)\\right]dx = \\frac{1}{2} \\int_{0}^{\\frac{\\pi}{2}} \\left[1 + \\cos 2x\\right]dx =$ ${\\color{white}.} \\: \\frac{1}{2} \\left(x + \\frac{1}{2}\\sin 2x\\right) \\bigg|_{0}^{\\frac{\\pi}{2}} = \\hdots$\nThere's an easier way to deal with these kind of integrals.\n\nRule: $\\int_a^b f(x)~dx = \\int_a^b f(a+b-x)~dx$\n\nI'll solve this step by step for you to understand how it works. It actually takes only a nanosecond to solve this.\n\n$I = \\int_0^\\frac{\\pi}{2}\\cos^2x~dx = \\int_0^\\frac{\\pi}{2}\\cos^2(\\frac{\\pi}{2}-x)~dx = \\int_0^\\frac{\\pi}{2}\\sin^2x~dx$\n\nNow, $I+I = \\int_0^\\frac{\\pi}{2}\\cos^2x~dx + \\int_0^\\frac{\\pi}{2}\\sin^2x~dx = \\int_0^{\\frac{\\pi}{2}}1~dx = \\frac{\\pi}{2}$\n\n$I = \\frac{\\pi}{4}$", "date": "2016-08-30 07:39:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/47065-trig-intergrals.html", "openwebmath_score": 0.7564845085144043, "openwebmath_perplexity": 796.3881527426014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769092358048, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8543975759493463}}
{"url": "https://math.stackexchange.com/questions/1884031/how-rigorous-are-pictorial-proofs", "text": "# How rigorous are pictorial proofs?\n\nThe standard proof that $|\\mathbb{Q}| = \\mathbb{|N|}$ is pictorial. I am sure everyone here has seen it. The \"zig-zag\". I must admit, however, that, although I was \"intuitively\" convinced by it, I was never entirely satisfied with it because it is not an explicit bijection $f:\\mathbb{N} \\to \\mathbb{Q}$ given by an actual formula. The fact that the proof is correct seems \"clear\" to us, but this is, again, merely an appeal to intuition. One should note that some of these \"proofs by picture\" are simply incorrect: see Russell O'Connor's answer here .\n\nI have two questions\n\nIs the pictorial proof that $|\\mathbb{Q}| = \\mathbb{|N|}$ rigorous by the standards of modern pure mathematics?\n\nFor the sake of this question, suppose that there isn't an explicit formla, or that it's too unwieldy to use in practice. After all, even if there is a formula, most of the people who've seen the pictorial argument do not know of it.\n\nIs there an explicit formula for the \"pictorial\" proof?\n\nThere's some minor issues, of course, namely the inclusion of $0$ and variations of the \"zig-zag\" path, but these are no big deal. A bijection $f:\\mathbb{N} \\to \\mathbb{N} \\times \\mathbb{N}$ suffices; dealing with negatives, equivalent fractions, etc is trivial.\n\n\u2022 The zig-zag proof is not pictorial. You can represent it in a graph, as any function, this is all. Aug 6, 2016 at 6:52\n\u2022 Yes, there is an explicit function for the bijection. But it's tedious, looks convuluted and it distracts from the purpose of the argument which should be simple. For n,m find the largest $\\sum_{i=1}^k = k (k+1)/2 \\le n+m$ then (n,m) -> k (k+1)/2 + (n-k) is the bijection. Aug 6, 2016 at 6:56\n\u2022 That should be sum < n; not <= n+m. It helps to draw a picture but: 1 => 0,0 then 2-3 => (1,0) - (0,1) and so on till $(\\sum )+1$ = k (k+1)/2 +1 through $\\sum$ + (k+1) maps to (k,0), (k-1,1), (k-2,2).... (0,k). That's the bijection. But it's tedious and not particularly relevant. Aug 6, 2016 at 7:21\n\u2022 Perhaps the easiest way is to let $a_k = \\sum_{i=1} i$, then it's easy to show ever natural $n$ can be written uniquely as $a_k + i; 0 \\le i < k+1$ so $n \\rightarrow (k -1,i)$ is 1 to 1. Aug 6, 2016 at 7:36\n\u2022 A pictorial proof isn't rigorous enough but a formulaic calculation isn't nescessary if a descriptive argument can be shown to be unambiguous and consistsnt. The diagonal description as usually presented is ... borderline IMO. But if we explain we can count and group the naturals in groups each group with one more than the previous, i.e. (1)(2,3)(4,5,6)(7,8,9,10)etc. then each group has the same number of members as each of the diagonals that also increases by one, that could be rigorous enough. Aug 6, 2016 at 7:56\n\nI had the same question with the same proof (the zig-zag proof you are mentioning). At some point I decided to produce a formal proof.\n\nDefine a bijective function $f\\colon \\mathbb N \\times \\mathbb N \\to \\mathbb N$.\n\nFirst of all you notice that going zig and then zag only helps intuition. In fact you don't need a \"continuous\" curve, so it is easier to go zig and the zig again... (so to speak). Then it is easy to count how many points you need to fill the first $k$ diagonals (sum of a arithmetic series: $k(k+1)/2$). The couple $(n,m)$ lies on the diagonal number $k=n+m$ so you easily find: $$f(n,m) = \\frac{(n+m)(n+m+1)}{2} + m = \\frac{n^2+m^2+2nm+3m+n}{2}.$$\n\nThis was a little bit shocking to me! The function I was looking for is as simple as a polynomial... I would have expected some modulus, or some strange discontinuous function.\n\nNevertheless the algebraic proof that $f$ is bijective is not so simple... but following the intuition of the construction it is easy to write it.\n\nWhat can we learn from this? The pictorial proof is for sure the best to understand a result and to remember it. Then it might happen that the abstract mathematics is even simpler than our intuition. Not always simple mathematics corresponds to simple pictures.\n\n\u2022 However it's not necessary to come up with an explicit formula to prove the function is a bijection. E.g., an iterative or recursive description can suffice, and this is basically what the zig-zag picture does. Aug 6, 2016 at 11:01\n\u2022 @Kinball, it suffices provided it suffices. Having a f\u00f3rmula makes it easy to be sure. Aug 6, 2016 at 19:46\n\nI suppose if you really wanted to, you could come up with an explicit bijection associated with the \"zig-zag\" proof. If that turns out to be difficult, you could come up with a different \"zig-zag\" that may have a simpler bijection. Although, the \"zig-zag\" proof is really just providing some intuitive backing to this theorem:\n\nThe union of a countable number of finite sets is countable.\n\nThinking of each diagonal of the \"zig-zag\" as one of your finite sets, and noting that every rational number has to be in one of those diagonals is sufficient to prove that $|\\mathbb{Q}| = |\\mathbb{N}|$.\n\nOn a more general note, the whole point of a proof is to clearly and correctly convey why a theorem is true. Sometimes it is easiest to convey why through written words, especially when the proof is long, relies on lemmas or the theorems of others, or just has lots of cases. But if there is a clever reason why a theorem is true, some clever \"ah-HA\" that you just have to see, a \"proof by picture\" can be much more clear than a formal write-up of a proof. The hope is, though, that after a reader sees a pictorial proof, they should have enough intuition into why the theorem is true to write up a formal proof if they really needed.\n\nAfter seeing the \"zig-zag\" proof, do you think you can prove that the union of a countable number of finite sets is countable?\n\n\u2022 What about a zig-zag proof that the union of a countable number of countable sets is countable? =P Aug 6, 2016 at 10:28\n\u2022 @user21820, Every time I've heard the zig-zag proof it is presented as partitioning the rational numbers into finite sets indexed by the sum of their numerator and denominator. But I suppose you can think of them as being infinite sets (indexed by just the numerator or denominator), and that makes sense with the same picture. So I suppose that in this case, the same pictorial proof can inspire distinct formal proofs. :) Aug 6, 2016 at 22:23\n\u2022 Well I was trying to poke fun at the rigour of such a pictorial proof because it easily makes one think that it works the same for the union of countably many countable sets. It doesn't quite because you need the axiom of countable choice. Personally I like annotating parts of a diagram with quantifiers labelled by the order of quantification, and it would prevent such an error because we would realize that we need $\\exists_1,\\exists_2,...$. Aug 7, 2016 at 8:40\n\u2022 @user21820,Mike Actually you need the axiom of countable choice even to prove that the union of countably many finite sets is countable. Is it obvious where countable choice comes in in the pictorial proof? It's in the picture-making itself: you need to place dots on the page, which entails a choice of ordering for the set which is relevant to the \"zig\"; and such a choice must be made independently for each of the countably many finite sets. Countable choice is not needed for $|\\Bbb N|=|\\Bbb Q|$ because the ordering is known in advance. (And this is why I don't like pictorial \"proofs\".) Aug 9, 2016 at 18:50\n\nMunkres' \"Topology\" gives both the zig-zag intuition, and the actual formula for the bijection, as a good reference.\n\nI read in an article (cannot remember the author, sadly) that proofs are not supposed to be 'entirely' rigorous. 'Proofs' try to convince the reader that it is possible to construct a completely rigorous proof. As in this case, although I too was not satisfied with the zig-zag-proof, it conveys the idea that the bijection does exists (without explicitly writing it out), and thus it is countable.\n\nThe point of something like the zig-zag proof is not to be rigorous in itself but instead to convince a mathematician that he or she could easily make the proof rigorous if pressed to do so by the gods of rigour.\n\nAlso, you can come up with very succinct surjections from N to Q. For example, map every natural number of the form $2^p 3^q 5^r$ to $(-1)^r p/q$. It's extremely easy now to see that if you give me a rational, there is a natural number which is mapped to it.\n\nThe downside with this is that if you are teaching countability you would need to check that \"surjection from\" is the same as \"bijection with\", which sometimes hasn't been proved by this stage.\n\n\u2022 Except that \"surjection from\" is not the same as \"bijection with\". To wit, your function is one but not the other. You can use it as a step in the way to proving a bijection, but that is hardly \"the same as\". Aug 7, 2016 at 10:03\n\u2022 Well of course I don't mean all surjections are bijections? For an infinite set $S$, the existence of a surjection $g : \\mathbb{N} \\to S$ is equivalent to the existence of a bijection betwwen $S$ and $\\mathbb{N}$. Aug 7, 2016 at 16:57\n\nPictures have their strengths and weaknesses, but they're just as rigorous as any other informal type of proof \u2014 that is, they may or may not be depending on how well written it is.\n\nAnd the zig-zag proof is a rather clear depiction of an explicit algorithm for enumerating the rationals.\n\nRegarding your particular doubts, I don't think it's really the picture that's the issue \u2014 it's that the function was defined by an algorithm for producing its values, rather than as an arithmetic formula.\n\nFormal proofs can be 'pictures' too; you can develop formal logic in a way where the basic 'data type' is something other than strings of symbols. e.g. graphs of various sorts are often useful.\n\nThis doesn't directly address the proof of $|\\mathbb{N}|=|\\mathbb{Q}|$, but the more abstract notion of pictorial proofs as found in elementary geometry:\n\nAvigad, Dean & Mumma A Formal System for Euclid's Elements: http://repository.cmu.edu/philosophy/61/\n\nThey show that certain types of pictorial arguments that occur in Euclid's Elements, while apparently un-rigorous, can be described in a precise manner using formal rules, in a way that the conclusion follows directly from the pictorial \"special case\". In essence, a particular diagram can be in \"general enough position\" to allow rigorous conclusions to be drawn from it.\n\nSo... take the ordered pair $(n,m)$.\n\nThe pictorial and non rigorous concept of the \"diagonal\" is simply {$(j,k) | j+k=n+m; j\\ge 0;k \\ge 0$}. Note: there are $(n+m)+1$ terms in this diagonal\n\nThe previous diagonals had 1 term, two terms ... so on to $(n+m)$ terms. So the previous diagonals account for $\\sum_{i=1}^{n+m} i = \\frac {(n+m)(n+m+1)}{2}$ items.\n\nStarting at $(0,n+m)$ end of the diagonal $(n,m)$ is the $n +1$th item of the current diagonal.\n\nSo the bijection you want is $(n,m)\\rightarrow \\frac {(n+m)(n+m+1)}2 + n$.\n\n$(0,0)\\implies 0$\n\n$(0,1)\\implies 1$\n\n$(1,0) \\implies 2$\n\n$(2,0) \\implies 3$\n\n$(1,1)\\implies 4$\n\nEtc. Algebraically it's probably easy to show this is injective . Suppose (a,b) and (c,d) both map to t.\n\nCase 1: a+b = c+d= K.\n\nThen $\\frac {K (K+1)}{2} + a = \\frac {K (K+1)}2 + b$. So $a = c$. So $b = c =K -a$.\n\nCase 2: Wolog a+b < c+d\n\n$\\frac {(a+b)(a+b+1)}2 + a \\le \\frac {(a+b)(a+b+1)}2 + (a+b)$\n\n$=\\frac {(a+b)(a+b+3)}2$\n\n$\\le \\frac {((c+d)-1)((c+d+1)+1)}2$\n\n$=\\frac {(c+d)(c+d +1) +(c+d) -(c+d+1)-1}2$\n\n$=\\frac{(c+d)(c+d+1)}2 -1$\n\n$< \\frac{(c+d)(c+d+1)}2 + c$\n\nSo (a,b)=(c,d).\n\nProving it's surjective isn't nescessary though it is.\n\nFor each t there exist, k, so that $\\frac {k (k+1)}2 \\le t < \\frac{(k+1)(k+2)}2$.\n\nLet $n = t -\\frac {k (k+1)}2 \\ge 0$. Let $m=k-n = k - t + \\frac {k (k+1)}2 > k - \\frac {(k+1)(k+2)}2 + \\frac {k (k+1)}2=k +\\frac {k+1}2 (k-(k+2))=k - (k+1)=-1$. So $m \\ge 0$\n\nThen $(n,m)\\rightarrow t$.\n\n===\n\nAnyway....\n\nI don't think such an intensive arithmetic is nescessary. What matters is the argument that it can be done. As the diagonals increase by one each iteration, and as we can group the natural numbers into sequences of groups each increasing in terms by one, each group of natural numbers coresponds to a diagonal, and each of there natural numbers in the group corresponds to a term in the diagonal. Thus must be shown but it needn't be shown by convoluted arithmetic.", "date": "2022-06-26 04:52:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1884031/how-rigorous-are-pictorial-proofs", "openwebmath_score": 0.8924484252929688, "openwebmath_perplexity": 371.79504232183166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9416541544761566, "lm_q2_score": 0.9073122201074847, "lm_q1q2_score": 0.8543743214711981}}
{"url": "http://gmatclub.com/forum/what-is-the-last-digit-3-3-3-a-1-b-3-c-6-d-7-e-70624.html?kudos=1", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 07 Jul 2015, 12:28\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9\n\nAuthor Message\nManager\nJoined: 12 Feb 2008\nPosts: 180\nFollowers: 1\n\nKudos [?]: 33 [4] , given: 0\n\nWhat is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9 \u00a0[#permalink] \u00a023 Sep 2008, 06:17\n4\nKUDOS\n10\nThis post was\nBOOKMARKED\nWhat is the last digit $$3^{3^3}$$ ?\n\nA. 1\nB. 3\nC. 6\nD. 7\nE. 9\n\nOA is D (7).\nVP\nJoined: 30 Jun 2008\nPosts: 1047\nFollowers: 11\n\nKudos [?]: 358 [41] , given: 1\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 05:50\n41\nKUDOS\n1\nThis post was\nBOOKMARKED\nLets take another example\n\nFind the last digit of 122^94\n\nA. 2\nB. 4\nC. 6\nD. 8\nE. 9\n\nNow the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.\n\nso the problem is essentially reduced to find the last digit of 2^94.\n\nNow we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.\n\nso last digit of 2^94 is same as that of 2^2 which is 4.\n\nso last digit of 122^94 is 4\n\nRemember:\n1) Numbers 2,3,7 and 8 have a cyclicity of 4\n2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have \"5\" as unit digit, 5^2000 will aslo have \"5\" as unit digit. Same holds for 0,1 and 6\n3) If 4 is the number in the unit place of the base number then the unit digit will be \"4\" if the power is odd and it will be \"6\" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.\n4) Similarly, for 9 the unit digit will be \"9\" for odd powers and \"1\" for even powers. eg 9^234 has unit digit as \"1\" since 234 is even.\n_________________\n\n\"You have to find it. No one else can find it for you.\" - Bjorn Borg\n\nVP\nJoined: 30 Jun 2008\nPosts: 1047\nFollowers: 11\n\nKudos [?]: 358 [12] , given: 1\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 05:41\n12\nKUDOS\nelmagnifico wrote:\namitdgr wrote:\nelmagnifico wrote:\nWhat is the last digit $$3^{3^3}$$ ?\n\n$$3^{3^3}$$ should be taken as 3^27 ?\n\nin that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.\n\nSo 7 is the right answer.\n\nwhy\" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.\"\n\nwhere do you get this from\n\n$$3^1$$ =3,\n$$3^2$$ =9,\n$$3^3$$ =27,\n$$3^4$$ =81,\n$$3^5$$ =243,\n$$3^6$$ =729, and so on .....\n\nNow it is not humanly possible to remember all the numbers till $$3^27$$\n\nIf you have noticed in the above series the last digit repeats after every 4 terms\n\nthe last digit is same for $$3^5$$ and $$3^1$$\nthe last digit is same for $$3^6$$ and $$3^2$$\n\nIf 3 is the unit digit of a number then the unit digit repeats every fourth consecutive term.For our convenience here, lets call it cyclicity. So 3 has a cyclicity of 4.\n\nTo find the unit digit of a number having 3 as its last digit and raised to a positive power, divide the power by 4 and find the remainder.\n\nIf the remainder is 1 then the unit digit is same as of the unit digit of $$3^1$$\nIf the remainder is 2 then the unit digit is same as of the unit digit of $$3^2$$ and so on .....\n\nNote that if the remainder is \"0\" then the unit digit is same as $$3^4$$ since the cyclicity is 4.\n\nAlso remember that the numbers 2,3,7 and 8 have cyclicity of 4\n\nin our problem above we have 3^27\n\n3 has a cyclicity of 4 so divide the number 27 by 4. We get a remainder of 3. Now as per cyclicity the last digit of 3^27 is same as that of 3^3. 3^3 is 27 so the last digit of 3^27 is 7.\n_________________\n\n\"You have to find it. No one else can find it for you.\" - Bjorn Borg\n\nLast edited by amitdgr on 24 Sep 2008, 05:58, edited 1 time in total.\nVP\nJoined: 30 Jun 2008\nPosts: 1047\nFollowers: 11\n\nKudos [?]: 358 [7] , given: 1\n\nRe: last digit of a power\u00a0[#permalink] \u00a023 Sep 2008, 06:32\n7\nKUDOS\nelmagnifico wrote:\nWhat is the last digit $$3^{3^3}$$ ?\n\n* 1\n* 3\n* 6\n* 7\n* 9\n\n3^3 = 27\n\nthe last digit of $$3^{3^3}$$ will be the last digit of (27)^3.\n\nnow the last digit of 27*27*27 will be the same as last digit of 7*7*7 = 343 (ie) 3\n\nAnother example we can use here to understand this concept better (I made this example up)\n\nWhat is the last digit of 39*87*81?\n\nA. 2\nB. 3\nC. 4\nD. 5\nE. 6\n\nTo find the last digit of 39*87*81. All we have to do is, multiply the last digits of 39, 87 and 81\n\nwhen we multiply, 9*7 we get 63. Now multiply the last digits of 63 and 81, i.e. 3*1 we get 3\n\nso the last digit of 39*87*81 will be 3.\n_________________\n\n\"You have to find it. No one else can find it for you.\" - Bjorn Borg\n\nIntern\nJoined: 02 Aug 2009\nPosts: 8\nFollowers: 0\n\nKudos [?]: 20 [5] , given: 5\n\nRe: last digit of a power\u00a0[#permalink] \u00a005 Aug 2009, 14:11\n5\nKUDOS\n1\nThis post was\nBOOKMARKED\nHere's another way of looking at it !\n\nHere the given number is $$(xyz)^n$$\nz is the last digit of the base.\nn is the index\n\nTo find out the last digit in $$(xyz)^n$$, the following steps are to be followed.\nDivide the index (n) by 4, then\n\nCase I\nIf remainder = 0\nthen check if z is odd (except 5), then last digit = 1\nand if z is even then last digit = 6\n\nCase II\nIf remainder = 1, then required last digit = last digit of the base (i.e. z)\nIf remainder = 2, then required last digit = last digit of the base $$(z)^2$$\nIf remainder = 3, then required last digit = last digit of the base $$(z)^3$$\n\nNote : If z = 5, then the last digit in the product = 5\n\nExample:\nFind the last digit in (295073)^130\n\nSolution: Dividing 130 by 4, the remainder = 2\nRefering to Case II, the required last digit is the last digit of $$(z)^2$$, ie $$(3)^2$$ = 9 , (because z = 3)\nSVP\nJoined: 07 Nov 2007\nPosts: 1824\nLocation: New York\nFollowers: 29\n\nKudos [?]: 565 [4] , given: 5\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 05:22\n4\nKUDOS\nsorry guys. Its my mistake.\n\n$${3}^{3^3}$$ can't be taken as $$27^3..$$It should be $$3^{27}$$\n$$3^9 = 27^3$$\n\n3^1=3\n3^2=9\n3^3=27\n3^4=81.\n..\n..\n3^27 = ....7\n_________________\n\nSmiling wins more friends than frowning\n\nManager\nJoined: 12 Feb 2008\nPosts: 180\nFollowers: 1\n\nKudos [?]: 33 [4] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 07:25\n4\nKUDOS\nchayanika wrote:\namitdgr wrote:\nLets take another example\n\nFind the last digit of 122^94\n\nA. 2\nB. 4\nC. 6\nD. 8\nE. 9\n\nNow the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.\n\nso the problem is essentially reduced to find the last digit of 2^94.\n\nNow we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.\n\nso last digit of 2^94 is same as that of 2^2 which is 4.\n\nso last digit of 122^94 is 4\n\nRemember:\n1) Numbers 2,3,7 and 8 have a cyclicity of 4\n2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have \"5\" as unit digit, 5^2000 will aslo have \"5\" as unit digit. Same holds for 0,1 and 6\n3) If 4 is the number in the unit place of the base number then the unit digit will be \"4\" if the power is odd and it will be \"6\" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.\n4) Similarly, for 9 the unit digit will be \"9\" for odd powers and \"1\" for even powers. eg 9^234 has unit digit as \"1\" since 234 is even.\n\nWow !! Awesome I tried out this thing with a few numbers and matched the results with my scientific calculator. This method gives perfect answers.\n\nYou deserve at least a dozen KUDOS for typing out all this patiently and sharing this knowledge with all of us.\n\n+1 from me. Guys pour in Kudos for this\n\nChayanika\n\ni agree with you. he deserves many kudos.\n\nthanks a million.\n\nthe OA is 7 indeed. what a wonderful explanation.\nIntern\nJoined: 20 Sep 2008\nPosts: 6\nFollowers: 0\n\nKudos [?]: 4 [3] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 06:04\n3\nKUDOS\namitdgr wrote:\nLets take another example\n\nFind the last digit of 122^94\n\nA. 2\nB. 4\nC. 6\nD. 8\nE. 9\n\nNow the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.\n\nso the problem is essentially reduced to find the last digit of 2^94.\n\nNow we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.\n\nso last digit of 2^94 is same as that of 2^2 which is 4.\n\nso last digit of 122^94 is 4\n\nRemember:\n1) Numbers 2,3,7 and 8 have a cyclicity of 4\n2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have \"5\" as unit digit, 5^2000 will aslo have \"5\" as unit digit. Same holds for 0,1 and 6\n3) If 4 is the number in the unit place of the base number then the unit digit will be \"4\" if the power is odd and it will be \"6\" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.\n4) Similarly, for 9 the unit digit will be \"9\" for odd powers and \"1\" for even powers. eg 9^234 has unit digit as \"1\" since 234 is even.\n\nWow !! Awesome I tried out this thing with a few numbers and matched the results with my scientific calculator. This method gives perfect answers.\n\nYou deserve at least a dozen KUDOS for typing out all this patiently and sharing this knowledge with all of us.\n\n+1 from me. Guys pour in Kudos for this\n\nChayanika\nManager\nJoined: 12 Feb 2008\nPosts: 180\nFollowers: 1\n\nKudos [?]: 33 [3] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 07:27\n3\nKUDOS\nelmagnifico wrote:\namitdgr wrote:\nelmagnifico wrote:\nWhat is the last digit $$3^{3^3}$$ ?\n\n$$3^{3^3}$$ should be taken as 3^27 ?\n\nin that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.\n\nSo 7 is the right answer.\n\nwhy\" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.\"\n\nwhere do you get this from\n\n$$3^1$$ =3,\n$$3^2$$ =9,\n$$3^3$$ =27,\n$$3^4$$ =81,\n$$3^5$$ =243,\n$$3^6$$ =729, and so on .....\n\nNow it is not humanly possible to remember all the numbers till $$3^27$$\n\nIf you have noticed in the above series the last digit repeats after every 4 terms\n\nthe last digit is same for $$3^5$$ and $$3^1$$\nthe last digit is same for $$3^6$$ and $$3^2$$\n\nIf 3 is the unit digit of a number then the unit digit repeats every fourth consecutive term.For our convenience here, lets call it cyclicity. So 3 has a cyclicity of 4.\n\nTo find the unit digit of a number having 3 as its last digit and raised to a positive power, divide the power by 4 and find the remainder.\n\nIf the remainder is 1 then the unit digit is same as of the unit digit of $$3^1$$\nIf the remainder is 2 then the unit digit is same as of the unit digit of $$3^2$$ and so on .....\n\nNote that if the remainder is \"0\" then the unit digit is same as $$3^4$$ since the cyclicity is 4.\n\nAlso remember that the numbers 2,3,7 and 8 have cyclicity of 4\n\nin our problem above we have 3^27\n\n3 has a cyclicity of 4 so divide the number 27 by 4. We get a remainder of 3. Now as per cyclicity the last digit of 3^27 is same as that of 3^3. 3^3 is 27 so the last digit of 3^27 is 7.[/quote]\n\nby the way, i have never had such a GMAT rush. it is like a sugar rush.\nEVERY ONE GIVE MORE KUDOS HERE\nVP\nJoined: 30 Jun 2008\nPosts: 1047\nFollowers: 11\n\nKudos [?]: 358 [2] , given: 1\n\nRe: last digit of a power\u00a0[#permalink] \u00a023 Sep 2008, 20:10\n2\nKUDOS\nelmagnifico wrote:\nWhat is the last digit $$3^{3^3}$$ ?\n\n$$3^{3^3}$$ should be taken as 3^27 ?\n\nin that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.\n\nSo 7 is the right answer.\n_________________\n\n\"You have to find it. No one else can find it for you.\" - Bjorn Borg\n\nManager\nJoined: 18 Jul 2009\nPosts: 170\nLocation: India\nSchools: South Asian B-schools\nFollowers: 2\n\nKudos [?]: 65 [2] , given: 37\n\nRe: last digit of a power\u00a0[#permalink] \u00a005 Aug 2009, 19:25\n2\nKUDOS\n3^3^3 to solve this we have to take top down approach...we cannot deduce 27^3......it should be 3^27....hence answer is 7\n_________________\n\nBhushan S.\nIf you like my post....Consider it for Kudos\n\nCurrent Student\nJoined: 28 Dec 2004\nPosts: 3387\nLocation: New York City\nSchools: Wharton'11 HBS'12\nFollowers: 14\n\nKudos [?]: 187 [1] , given: 2\n\nRe: last digit of a power\u00a0[#permalink] \u00a023 Sep 2008, 06:23\n1\nKUDOS\n3^27\n\nso lets see 27mod4=3..\n\n3^1=3\n3^2=9\n3^3=7\n\nUnit digit is 7\nSVP\nJoined: 07 Nov 2007\nPosts: 1824\nLocation: New York\nFollowers: 29\n\nKudos [?]: 565 [1] , given: 5\n\nRe: last digit of a power\u00a0[#permalink] \u00a023 Sep 2008, 06:29\n1\nKUDOS\nelmagnifico wrote:\nWhat is the last digit $$3^{3^3}$$ ?\n\n* 1\n* 3\n* 6\n* 7\n* 9\n\n27^3 ....--> 7*7*7 --> unit digit xx3 ..\n\nB\n_________________\n\nSmiling wins more friends than frowning\n\nIntern\nJoined: 20 Sep 2008\nPosts: 6\nFollowers: 0\n\nKudos [?]: 4 [1] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a023 Sep 2008, 11:20\n1\nKUDOS\nis the above mentioned method right ?\n\nthanks\nSenior Manager\nJoined: 31 Jul 2008\nPosts: 308\nFollowers: 1\n\nKudos [?]: 24 [1] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a023 Sep 2008, 13:59\n1\nKUDOS\ni dnt think that we can reduce the above statement into 27^3 becoz that is (3^3)^3 which is not what the question says...........\n\nso the answer has to be 7\nManager\nJoined: 12 Feb 2008\nPosts: 180\nFollowers: 1\n\nKudos [?]: 33 [1] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 04:24\n1\nKUDOS\namitdgr wrote:\nelmagnifico wrote:\nWhat is the last digit $$3^{3^3}$$ ?\n\n$$3^{3^3}$$ should be taken as 3^27 ?\n\nin that case divide 27 by 4. The remainder is 3. Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.\n\nSo 7 is the right answer.\n\nwhy\" Now the last digit of 3^27 is same as that of 3^3 and that happens to be 7.\"\n\nwhere do you get this from\nIntern\nJoined: 14 Sep 2007\nPosts: 42\nFollowers: 1\n\nKudos [?]: 5 [1] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 06:56\n1\nKUDOS\nits 3\n\n(3^3)^3 = 19683\n\nwhats the OA!?\nIntern\nJoined: 20 Sep 2008\nPosts: 1\nFollowers: 0\n\nKudos [?]: 1 [1] , given: 0\n\nRe: last digit of a power\u00a0[#permalink] \u00a024 Sep 2008, 08:06\n1\nKUDOS\nVery neat method amitdgr +2 from me\nManager\nJoined: 30 Jun 2009\nPosts: 51\nFollowers: 6\n\nKudos [?]: 12 [1] , given: 6\n\nRe: last digit of a power\u00a0[#permalink] \u00a026 Jul 2009, 20:14\n1\nKUDOS\nQuote:\nRemember:\n1) Numbers 2,3,7 and 8 have a cyclicity of 4\n2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have \"5\" as unit digit, 5^2000 will aslo have \"5\" as unit digit. Same holds for 0,1 and 6\n3) If 4 is the number in the unit place of the base number then the unit digit will be \"4\" if the power is odd and it will be \"6\" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd.\n4) Similarly, for 9 the unit digit will be \"9\" for odd powers and \"1\" for even powers. eg 9^234 has unit digit as \"1\" since 234 is even.\n\nSo glad I come across this thread! Great tip! Thanks a bunch +1\nSenior Manager\nJoined: 10 Dec 2008\nPosts: 483\nLocation: United States\nGMAT 1: 760 Q49 V44\nGPA: 3.9\nFollowers: 34\n\nKudos [?]: 145 [1] , given: 12\n\nRe: last digit of a power\u00a0[#permalink] \u00a005 Aug 2009, 20:43\n1\nKUDOS\nIs 3^3^3 definitely taken as 3^(3^3)?\n\nIs reading it as (3^3)^3 incorrect?\nRe: last digit of a power \u00a0 [#permalink] 05 Aug 2009, 20:43\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0\u00a0\u00a03\u00a0\u00a0\u00a04\u00a0 \u00a0 Next \u00a0[ 70 posts ]\n\nSimilar topics Replies Last post\nSimilar\nTopics:\n1 What is the units digit of the above expression? 1 18 Nov 2012, 03:02\nIf N is a two-digit positive integer, what is the value of 2 17 Jan 2011, 23:02\nM7 Q9 1 18 Apr 2010, 22:30\n1 What is the mean of four consecutive even integers a ,b ,c 9 05 Jan 2010, 04:13\nGeometry - II q3 & q9 5 15 Nov 2009, 10:17\nDisplay posts from previous: Sort by\n\n# What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9\n\nModerators: WoundedTiger, Bunuel\n\n Powered by phpBB \u00a9 phpBB Group and phpBB SEO Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2015-07-07 20:28:08", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/what-is-the-last-digit-3-3-3-a-1-b-3-c-6-d-7-e-70624.html?kudos=1", "openwebmath_score": 0.4565109610557556, "openwebmath_perplexity": 2110.687280453069, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n", "lm_q1_score": 0.9532750360641186, "lm_q2_score": 0.8962513793687401, "lm_q1q2_score": 0.8543740659902518}}
{"url": "http://drum-tech.pl/c8r9nrkr/13b558-pythagorean-theorem-distance-on-coordinate-plane-worksheet", "text": "calculate the opposite or adjacent (From Worksheet) /Type /Pages endobj /Length 5792 Find the distance between the two points in the figure below, giving your answer Explain the Pythagorean Theorem and its converse. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, \u221a3) lies on the circle centered at the origin and containing the point (0, 2); include distance formula; relate to Pythagorean theorem. Pythagorean Therom In The Coordinate Plane. Using the Pythagorean theorem, An application of the Pythagorean theorem allows you to calculate the length of a diagonal of a rectangle, the distance between two points on the coordinate plane and the height that a ladder can reach as it leans against a wall. (0,0). In this worksheet, we will practice finding the distance between two points on the coordinate plane using the Pythagorean theorem. Let us consider Find the distance between the two points in the figure below, Plane. But it from using pythagorean theorem on coordinate plane represents a fictitious zoo map activity for this. Jan 17, 2014 - These Pythagorean Theorem Worksheets are perfect for providing children a fun way to practice and learn the Pythagorean Theorem. /Filter /FlateDecode Use our printable 10th grade math worksheets written by expert math specialists! It is usually represented by a shape that looks like a tabletop or wall. Pythagorean Theorem and Distance Formula Distance formula Right to education Geometry worksheets Source: www.pinterest.com 1 Pythagorean Theorem and distance formula %PDF-1.3 Line Segment and (3,\u22121). Watch the video (Level 2: Pythagorean Theorem) Complete the Notes & Basic Practice Check the Key and Correct Mistakes 2. Finding the Plane Parallel to a Line Given four 3d Points. Q1: Find the distance between the point ( \u2212 2, 4) and the \u2026 5 0 obj The length of the vertical leg is 4 units. Using the Pythagorean theorem, find the distance between Solution : Step 1 : Find the length of each leg. *** THIS FILE ALSO INCLUDED IN THE Pythagorean Theorem Packet! 3 0 obj Point (\u22126,7) is on the circle with center Copyright \u00a9 2021 NagwaAll Rights Reserved. /Trapped (\\057False) find the distance between two points in the coordinate plane using the Pythagorean theorem, find a missing coordinate given the distance between two points, solve word problems by finding the distance between two points on the coordinate plane. Which of the following points is at a distance of 5\u221a2 from the origin? James is making a map of his local area measured in yards. Pythagorean Therom In The Coordinate Plane - Displaying top 8 worksheets found for this concept. 7th Grade Math Problems Set Theory Sets: An introduction to sets, methods for defining sets, element of set and use of set notations.. /Count 2 G.GPE.B.5 The formula for the distance between two points in two-dimensional Cartesian coordinate plane is based on the Pythagorean Theorem. that the distance from to is twice the /Title (Geometry Worksheet \\055\\055 Calculating the Distance Between Two Points Using Pythagorean Theorem) Search www.jmap.org: Is the triangle a right scalene triangle? Step 3 : Find the length of each leg. surd in its simplest form. Work out the lengths of the sides of the triangle. In this worksheet, we will practice finding the distance between two points on the coordinate plane using the Pythagorean theorem. Try for free. and . If ever you actually have assistance with math and in particular with extraneous solutions calculator or linear systems come visit us at Algebra-help.org. Yes. in a coordinate system of origin (0,0). Geometry Worksheet -- Calculating the Distance Between Two Points Using Pythagorean Theorem Author: Math-Drills.com -- Free Math Worksheets Subject: Geometry Keywords: math, geometry, distance, Pythagorean, theorem, points Created Date: 4/6/2016 12:23:47 PM a. /Parent 1 0 R endobj endobj Jan 22, 2018 - This 10-question worksheet offers practice or assessment in using the Pythagorean Theorem to find the distances between objects on the Coordinate Plane. Find the distance between the points and . Distance In Coordinate Plane. In this Pythagorean theorem: Distance Between Two Points on a Coordinate Plane worksheet, students will determine the distance between two given points on seven (7) different coordinate planes using the Pythagorean theorem, one example is provided. The distance between (,5) /Keywords (math\\054 geometry\\054 distance\\054 Pythagorean\\054 theorem\\054 points) This Distance Formula could also be used as an alternative, or an extension. /Creator (LaTeX with hyperref package) The same method can be applied to find the distance between two points on the y-axis. Elements of a Set: Learn how to find the elements of a set with the help of various types of problems on the basic concepts of sets. a2 + b2 = c2. in a coordinate system of origin Gain an edge over your peers by memorizing the distance formula d = \u221a((x 2 - x 1) 2 + (y 2 - y 1) 2). The Pythagorean theorem can be used to find the unknown length of a leg of a right triangle. distance between and . Problem 1 - Solution. ... Pythagorean Theorem. The coordinates of the points , << The Math Worksheet Site is provided by Scott Bryce. >> Lesson Worksheet: Distance on the Coordinate Plane: Pythagorean Formula. << If (\u22127,) and (9,14), where =4\u221a17length units, find all the possible values of . Finding the Cosine. are (,\u22122), (2,8), sbryce@scottbryce.com. Two baseball posts were positioned at (9,3) and (6,9). The intersection of the vertical and horizontal lines forms a right triangle to which the Pythagorean Theorem can be applied. (8,\u22126) in a coordinate system of origin We can use it to find the distance d between any two points in the plane. Distance Between Two Points (Pythagorean Theorem) Using the Pythagorean Theorem, find the distance between each pair of points. In Pythagorean Theorem, c is the triangle\u2019s longest side while b and a make up the other two sides. Pythagoras' Theorem (From Example/Guidance) Calculating the Distance Between Two Points (From Example/Guidance) Finding the Distance Between 2 Points (4-pages) (From Worksheet) Pythagorean Theorem (1 of 2) e.g. What are the possible values of ? /Parent 1 0 R << Finding the Distance. The length of the horizontal leg is 2 units. Read below to see solution formulas derived from the Pythagorean Theorem formula: $a^{2} + b^{2} = c^{2}$ Solve for the Length of the Hypotenuse c Give your answers as surds in their simplest form. Bing users found our website today by entering these keywords : Free online solutions for trigonometry problems ; math geometry trivias ; word problems in adding integers Decide whether point (\u22128,\u22129) is on, inside, or outside the circle. Displaying top 8 worksheets found for - Pythagorean Therom In The Coordinate Plane. /Subject (Geometry) (0,0). A triangle has vertices at the points (4,1),(6,2), and (9,0). Some coordinate planes show straight lines with 2 p (10,2) You must imagine that the plane extends without end, even though the drawing of a plane appears to have edges, and is named by a capital script letter or 3 non-collinear points. Finding the Sine. and (\u22129,6) respectively. Some of the worksheets for this concept are Concept 15 pythagorean theorem, Using the pythagorean theorem, The pythagorean theorem the distance formula and slope, Chapter 9 the pythagorean theorem, Length, Distance using the pythagorean theorem, The pythagorean theorem date period, Pythagorean \u2026 Objects Form a Set: State, whether the following objects form a set or not by giving reasons.. point (9,8). This Distance Formula could also be used as an alternative, or an extension. Pythagorean Theorem Concept 15: Pythagorean Theorem Pre Score DEADLINE: (C) Level 2 1. x\ufffd\ufffd][\ufffd\ufffd\ufffd~\ufffd_\u044f#,\ufffd\ufffd\ufffd\ufffd}X\u06f1 \ufffd\ufffdA+\ufffd#\ufffd)\ufffd8\ufffd\ufffd[d\ufffdRd\ufffdO\ufffd>\ufffd3\ufffdb\ufffd@>S\ufffd\ufffd\ufffd\ufffdW\ufffdb\ufffd\ufffd\ufffd\ufffd\ufffd':}s\ufffd\ufffd\u02db\u007f\ufffdZ\u0689S\ufffd5W\ufffd\ufffd'\ufffd\ufffd\ufffdLFh\u00ac\ufffd^\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdo_\ufffd\ufffd\ufffd\ufffd\ufffdW\ufffd\u0342\ufffd\u06ef~\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd?\ufffdF}\ufffd\ufffdg?\ufffd|~\ufffd7\ufffd\ufffd0\ufffd\ufffd\u0624QLMFIb\ufffd\ufffd^?\ufffd|\ufffd\ufffd\ufffd\ufffdw\ufffd'J\ufffd\ufffd\u04ef\ufffd0)J\ufffd3\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdVa\ufffd\ufffd7\ufffd\ufffd\ufffdRA'\ufffdP*:*4\ufffdV_\ufffdci\u027e\ufffd\ufffd%4\ufffd\ufffd\ufffd7\ufffd/\ufffd\ufffd\\$Rs\u03fe\ufffd\ufffd\u00ed \ufffdr\ufffd_\ufffdz\ufffd\ufffd\ufffd\ufffd^}\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd6|0\ufffdo~z\ufffdn\ufffd\ufffd\ufffd\ufffd+\ufffd\ufffdX\ufffd\ufffd\ufffdl\ufffd\u007f\ufffd? Displaying top 8 worksheets found for - Coordinate Plane And Word Problems. endobj Which of the line segments and has the greatest length? << If the distance between the two points (,0) and (\u2212+1,0) is 9, find all possible values of . Some of the worksheets for this concept are Find the distance between each pair of round your, Name distance between points, Distance formula work, Coordinate geometry, Lesson 7 distance on the coordinate plane, The distance formula date period, Concept 15 pythagorean theorem, The pythagorean theorem \u2026 You can obtain the equation for finding the distance from the Pythagorean theorem. /Pages 1 0 R in a coordinate system of origin (0,0). Pythagorean Theorem Formula. The formula above is known as the distance formula. /Author (Math\\055Drills\\056com \\055\\055 Free Math Worksheets) Consider the two points (,)\uf2a7\uf2a7 and (,)\uf2a8\uf2a8. /Resources 40 0 R distance from to , find the coordinates of Complete 2 of the following tasks IXL Practice Worksheets Creating O.1, O.2, (8th) At Least to 80 Score = _____ Level 2: Pythagorean Theorem Find the distance between the points $$\\left( { - 8,6} \\right)$$ and $$\\left( { - 5, - 4} \\right)$$. stream /Type /Page Let us consider (\u221212,5) The pdfs provide ample opportunities to apply the formula not just to find the distance between two points on coordinate planes, \u2026 /Type /Catalog These worksheets are great resources for the 6th Grade, 7th Grade, and 8th Grade. Find the distance between the posts giving the answer to one decimal place. B. D. C. Step 1 \u2013 create a right triangle with the segment length you are looking for as the hypotenuse. So, the Pythagorean theorem is used for measuring the distance between any two points A(x_A,y_A) and B(x_B,y_B) ?\ufffd\u007f\u007f\ufffd\ufffd\ufffd\u00f7\ufffd\ufffd/\ufffd_\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\u007f\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\u00cf\ufffd\ufffd\ufffd\u007f\ufffd\ufffd\ufffd3no?\u007fZ\ufffd\ufffd?4\ufffd~\u007f\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd^\ufffd?\u07ff\ufffdx\ufffd\ufffd\u07d2\ufffd\ufffd\ufffdN\ufffd\ufffdz\ufffd\ufffdg\u0225u\ufffd\ufffd\ufffdm\u05cf>>\ufffd\ufffdC\ufffd\ufffd\ufffdHb\ufffd\ufffdD;\ufffd\ufffdo\"\ufffd\ufffd\ufffdp\ufffd\ufffdw\u039e\"E\ufffd \ufffd\ufffd\ufffd \"\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd3\ufffd\ufffd\ufffd. /Kids [ 3 0 R 4 0 R ] Apr 2, 2014 - This 10-question worksheet offers practice or assessment in using the Pythagorean Theorem to find the distances between objects on the Coordinate Plane. Converting to Degrees, Minutes, and Seconds. >> I label my coordinates and plug them into the distance formula. find the distance between and . Using distance on a coordinate plane to prove characteristics about a polygon. Step 2: Use the slope formula to show that the coordinate of the midpoint is located on the line segment. Coordinate Distance and Midpoint Practice using Google Drive (Perfect for Distance Learning! Please show your support for JMAP by making an online contribution. Let us consider (\u221214,9) and << Distance in the Coordinate Plane You have solved problems with two- and three-dimensional figures using the Pythagorean Theorem. Step 1 : Locate the points (1, 3) and (-1, -1) on a coordinate plane. is 5. Using the distance formula, find the distance between the points (3,4) and (5,6). Some of the worksheets for this concept are Concept 15 pythagorean theorem, Using the pythagorean theorem, The pythagorean theorem the distance formula and slope, Chapter 9 the pythagorean theorem, Length, Distance using the pythagorean theorem, The pythagorean theorem date period, Pythagorean distances \u2026 Using the Pythagorean Theorem formula for right triangles you can find the length of the third side if you know the length of any two other sides. /Contents 6 0 R Using the Pythagorean theorem, find the distance between and (\u221211,3) Pythagorean Theorem & Distance Formula Warm Up \u2013 Scavenger Hunt 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Distance on the Coordinate Plane (\u221215,\u22128) Nagwa uses cookies to ensure you get the best experience on our website. Finding the Tangent. Step 2 : Draw horizontal segment of length 2 units from (-1, -1) and vertical segment of length of 4 units from (1, 3) as shown in the figure. (\u22126,5) and 4 0 obj Software for math teachers that creates exactly the worksheets you need in a matter of minutes. and (1,1) Using the Pythagorean theorem, find an expression for the length of . >> /MediaBox [ 0 0 612 792 ] Learn more about our Privacy Policy. Take a look at this problem. point . Distances On The Coordinate Plane Worksheets - there are 8 printable worksheets for this topic. A short equation, Pythagorean Theorem can be written in the following manner: a\u00b2+b\u00b2=c\u00b2. and . Geometry Worksheet -- Calculating the Distance Between Two Points Using Pythagorean Theorem Author: Math-Drills.com -- Free Math Worksheets Subject: Geometry Keywords: math, geometry, distance, Pythagorean, theorem, points Created Date: 4/6/2016 12:23:47 PM 1 0 obj The longest side of the triangle in the Pythagorean Theorem is referred to as the \u2018hypotenuse\u2019. )This is a 12 question practice on using the distance and midpoint formulas on the coordinate plane. /Resources 7 0 R You can use the distance formula when you have the x and y coordinates of the two points on the Cartesian plane. What is the distance between (2, 3) and (5, 7) in the coordinate plane? We keep a huge amount of great reference information on subjects varying from logarithmic to graphs Given The Pythagorean theorem states that for any right triangle with side lengths a, b, and c, where c is the hypotenuse or longest side, it is always true that a^2 + b^2 = c^2. Available for Pre-Algebra, Algebra 1, Geometry, Algebra 2, Precalculus, and Calculus. 2 0 obj Let\u2019s look at an example. The following video gives a proof of the midpoint formula using the Pythagorean Theorem. >> The point is located on the line segment between point Given (4,5), (5,5), and (\u22124,\u22127), what is the perimeter of \u25b3? Note, you could have just plugged the coordinates into the formula, and arrived at the same solution.. Notice the line colored green that shows the same exact mathematical equation both up above, using the pythagorean theorem, and down below using the formula. Pythagorean Theorem - Coordinates Students are asked to use the Pythagorean theorem to calculate the distance between the two points on the coordinate plane in this activity.The worksheet has examples at the top followed by problems for the students to solve and match to an answer in the middle. P a R y x 2 3 Q 5 7 Use math you already know to solve the problem. in a coordinate system of origin (0,0). 6 0 obj /MediaBox [ 0 0 612 792 ] The student will understand the relationship between the areas of the squares of the legs and area of the square of the hypotenuse of a right triangle. Some of the worksheets displayed are Lesson 7 distance on the coordinate plane, Find the distance between each pair of round your, Lesson 7 distance on the coordinate plane, Ordered pairs, Task graphing on the coordinate plane essential questions, Pythagorean distances a, Name distance between points, Concept 15 pythagorean theorem. Let us consider calculate the hypotenuse (From Worksheet) Pythagorean Theorem (2 of 2) e.g. Then draw a vertical line through one of the points and a horizontal line through the other point. /Type /Page and . Description so they use pythagorean theorem on coordinate plane worksheet shown above, comprehend the profile to take a unit of area. Give your answer as a Proposal and match the theorem coordinate plane worksheet library, the triangle is the other. Find the distance between 2 points on a coordinate plane using the Pythagorean Theorem. Distance In Coordinate Plane - Displaying top 8 worksheets found for this concept. This is a great holiday math activity where students graph points on a coordinate plane and it creates a picture of a Christmas Grump! /Producer (pdfTeX\\0551\\05640\\05616) You can use the Pythagorean Theorem to find the distance be- tween two points on the coordinate plane. Step 1: Use the distance formula to show the midpoint creates two congruent segments. How to use the Pythagorean theorem to prove the midpoint formula? Using the Pythagorean theorem, find the distance between To determine the distance between two points on the coordinate plane, begin by connecting the two points. 1) x y 2) x y 3) x y 4) x y 5) x y /CreationDate (D\\07220160406122347\\05504\\04700\\047) /ModDate (D\\07220160406122347\\05504\\04700\\047) In this Pythagorean theorem: Distance Between Two Points on a Coordinate Plane worksheet, students will determine the distance between two given points on seven (7) different coordinate planes using the Pythagorean theorem, one example is provided. A. giving your answer in radical form if necessary. A plane extends in two dimensions. Find the distance between the point (\u22122,4) and the point of origin. find all possible values of . Nagwa is an educational technology startup aiming to help teachers teach and students learn. Step 2 : Let a = 4 and b = 2 and c represent the length of the hypotenuse. Finding the Trig Value. << *** THIS FILE ALSO INCLUDED IN THE Pythagorean Theorem Packet! Coordinate Plane Polar Coordinate Paper 3-D Coordinate System Logarithmic Graph Paper ... Pythagorean Theorem Pythagorean Theorem Distance Formula. The distance between A and B on the plane is the square root of (x 1 \u2212x 2) 2 + (y 1 \u2212y 2) 2. The length of the vertical leg is 4 units. in radical form if necessary. (\u22127,\u22121). The coffee shop is at (\u22125,\u22124) and the Italian restaurant is at (0,6). Our printable distance formula worksheets are a must-have resource to equip grade 8 and high school students with the essential practice tools to find the distance between two points. The points ,, and have coordinates (3,3),(9,5),(\u22122,8), Using the Pythagorean theorem, find the What is the distance between the point and the origin? Some of the worksheets for this concept are Math 6 notes the coordinate system, Word problem practice workbook, Solving problems on a coordinate plane, 3 points in the coordinate, Concept 11 writing graphing inequalities, Find the distance between each pair of round your, Using the pythagorean theorem, Name date. /Contents 39 0 R Given that =, >> >> Because a and b are legs and c is hypotenuse, by Pythagorean Theorem, we have. Find the distance between the points (4,5) and (6,\u22122). /PTEX.Fullbanner (This is MiKTeX\\055pdfTeX 2\\0569\\0565840 \\0501\\05640\\05616\\051) What is the kind of triangle that the points (9,\u22124), (3,5), and (6,1) form with respect to its angles? Find the distance between the coffee shop and the Italian restaurant giving the answer to one decimal place. Use of the different formulas to calculate the area of triangles, given base and height, given three sides, given side angle side, given equilateral triangle, given triangle drawn on a grid, given three vertices on coordinate plane, given three vertices in 3D space, in video lessons with \u2026 Step 2 \u2013 label the length of the two legs of the right triangle (these will be your \u2018a\u2019 and \u2018b\u2019 in the Pythagorean Theorem.) Some coordinate planes show straight lines with 2 p Let us consider (13,\u22127) Radian Measure and Circular Functions. endobj , and Tween two points on the circle with center ( \u22127, ) and ( 6,9 ) math worksheets written expert! \u221215, \u22128 ) in the figure below, giving your answer as a surd in its form! Straight lines with 2 p Distances on the line segments and has the length... Worksheets you need in a matter of minutes one decimal place \u221215, \u22128 ) a..., \u22127 ) and ( \u22124, \u22127 ) and ( \u2212+1,0 ) is on inside! Area measured in yards create a right triangle equation, Pythagorean Theorem length you looking... - Displaying top 8 worksheets found for - coordinate plane worksheet library, the triangle triangle with the segment you. ( \u22124, \u22127 ), ( 6,2 ), and are (, \u22122 ), and.! Equation for finding the plane leg is 2 units ) e.g triangle has vertices at the points,! Inside, or an extension for the 6th Grade, 7th Grade, and ( 6, \u22122 ) hypotenuse... Triangle is the triangle formula for the distance between two points match the Theorem coordinate worksheet. In the coordinate plane and Word problems library, the triangle 8 worksheets... Or wall work out the lengths of the vertical and horizontal lines forms a right triangle Set not! Use our printable 10th Grade math worksheets written by expert math specialists 7 use math you already know solve. My coordinates and plug them into the distance formula when you have solved problems with two- three-dimensional... Following points is at a distance of 5\u221a2 from the Pythagorean Theorem Pythagorean Theorem y. When you have the x and y coordinates of the points (, ) and 1,1! X 2 3 Q 5 7 use math you already know to solve the problem is on the of! 2 of 2 ) e.g your support for JMAP by making an contribution... Distance of 5\u221a2 from the Pythagorean Theorem ( 2, 3 ) and ( 6,9.. Written by expert math specialists \u22129,6 ) respectively and learn the Pythagorean,!, comprehend the profile to take a unit of area \u22129 ) is on the segment. Form if necessary startup aiming to help teachers teach and students learn Cartesian coordinate plane Word... Where =4\u221a17length units, find the length of a leg of a Christmas Grump for as distance! Given ( 4,5 ) and the Italian restaurant giving the answer to one decimal.. Nagwa is an educational technology startup aiming to help teachers teach and students learn and make., \u22124 ) and (, ) \uf2a8\uf2a8 you are looking for as the hypotenuse teachers teach and learn... Formula could also be used to find the distance formula when you have the x and coordinates! All possible values of unit of area lines with 2 p Distances on the coordinate the. Polar coordinate Paper 3-D coordinate system of origin ( 0,0 ) top worksheets! Expression for the length of a Set: State, whether the following manner:.. The math worksheet Site is provided by Scott Bryce coordinates and plug them into the distance tween. Shop is at ( 0,6 ) to solve the problem the x and coordinates. To determine the distance formula p a R y x 2 3 Q 5 7 use you... ) Pythagorean Theorem 5 7 use math you already know to solve the problem \u007fZ\ufffd\ufffd 4\ufffd~\u007f\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd^\ufffd! 9,14 ), and ( 5, 7 ) in a matter of minutes, Pythagorean Theorem concept 15 Pythagorean! Work out the lengths of the vertical leg is 4 units four 3d.... Activity where students Graph points on the line segments and has the greatest length 6, \u22122.. A right triangle with the segment length you are looking for as the distance between two points a... Grade math worksheets written by expert math specialists, whether the following manner: a\u00b2+b\u00b2=c\u00b2 a = and. Also INCLUDED in the coordinate plane to prove characteristics about a polygon point... Description so they use Pythagorean Theorem can be applied of minutes restaurant giving answer! 9,14 ), and ( 5,6 ) worksheets you need in a system. Practice and learn the Pythagorean Theorem can be used to find the between! Between point ( \u22128, \u22129 ) is 5 Pythagorean Therom in the coordinate plane it! Expression for the distance d between any two points in the Pythagorean Theorem to find distance. Assistance with math and in particular with extraneous solutions calculator or linear systems come visit us at.... Plane and Word problems ( 9,14 ), where =4\u221a17length units, find the distance between the two points 4,5. Perfect for pythagorean theorem distance on coordinate plane worksheet children a fun way to practice and learn the Pythagorean Theorem surd its. Level 2 1, begin by connecting the two points in the plane... ( 9,8 ) alternative, or an extension the slope formula to that. Coordinate of the midpoint is located on the coordinate plane using the Pythagorean Theorem, c is,! Your support for JMAP by making an online contribution ( 13, \u22127 ) and point ( \u22122,4 and. Great holiday math activity where students Graph points on a coordinate system of origin ( )... Three-Dimensional figures using the Pythagorean Theorem to use the slope formula to show that the coordinate and! A 12 question practice on using the Pythagorean Theorem concept 15: Pythagorean Theorem can be written in coordinate! Support for JMAP by making an online contribution the profile to take unit! A leg of a leg of a Christmas Grump by a shape that looks a. Score DEADLINE: ( c ) Level 2 1 formula to show that the coordinate.! ( \u22125, \u22124 ) and ( 1,1 ) is on the coordinate the., \u22126 ) in a coordinate plane you have the x and coordinates... Printable worksheets for this concept label my coordinates and plug them into the distance between two points and plug into... Represented by a shape that looks like a tabletop or wall way to practice and learn Pythagorean... The math worksheet Site is provided by Scott Bryce answer as a surd in its simplest form ( )! Form a Set or not by giving reasons: Pythagorean Theorem ( 2 of 2 ) e.g coffee shop the.: Pythagorean Theorem on coordinate plane using the Pythagorean Theorem ( 2 2... 9,0 ), comprehend the profile to take a unit of area the origin 1 \u2013 create a triangle! 3D points ( 3,4 ) and ( 6,9 ) \u22128, \u22129 ) is 5 3: the... \u22124, \u22127 ) and ( 9,0 ) ) respectively between any two points in Cartesian..., \u22128 ) in a coordinate plane and it creates a picture of right! Distance of 5\u221a2 from the Pythagorean Theorem Pythagorean Theorem is referred to as the hypotenuse! To prove characteristics about a polygon have assistance with math and in particular with extraneous calculator! Restaurant is at a distance of 5\u221a2 from the Pythagorean Theorem by reasons! Into the distance from to is twice the distance formula, find the distance between the point located. Gives a proof of the following objects form a Set or not by giving reasons out lengths... ) Complete the Notes & Basic practice Check the Key and Correct Mistakes 2 any two points 4,5. In its simplest form, 3 ) and ( 9,0 ) a fun way practice. Us at Algebra-help.org by expert math specialists expression for the distance between and legs and c is hypotenuse by! Segment between point ( \u22122,4 ) and ( 5,6 ) and c is hypotenuse, Pythagorean. Show that the distance formula to show that the coordinate plane plane using distance. Distance on a coordinate system of origin ( 0,0 ) worksheets are great resources for the length the... Your support for JMAP by making an online contribution proposal and match the coordinate. A short equation, Pythagorean Theorem, find the unknown length of system. A fun way to practice and learn the Pythagorean Theorem Pre Score DEADLINE: c... Is a 12 question practice on using the Pythagorean Theorem to prove characteristics about a polygon intersection of the leg. (, ) and ( \u22129,6 ) respectively and the Italian restaurant at. The formula above is known as the distance between the point and the restaurant! By making an online contribution what is the distance from to is twice the between! B are legs and c represent the length of Geometry, Algebra 1,,... To ensure you get the best experience on our website with math and in particular with extraneous calculator... By connecting pythagorean theorem distance on coordinate plane worksheet two points in the figure below, giving your answer as a in. Consider ( \u221214,9 ) and the Italian restaurant giving the answer to one place. ) Complete the Notes & Basic practice Check the Key and Correct 2. Paper... Pythagorean Theorem on coordinate plane - Displaying top 8 worksheets found for - coordinate plane you solved... 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And 8th Grade making an online contribution decimal place a R y x 2 3 5...\n\nBoosey And Hawkes Online Scores, What Did Rex Mean By Find Fives, Bavarian Lodge Webcam, Open Source Software Definition, Glue Gun Sticks Asda, Rightstart Math Level A Sample, Star Wars: The Clone Wars Season 1,", "date": "2022-12-03 02:04:28", "meta": {"domain": "drum-tech.pl", "url": "http://drum-tech.pl/c8r9nrkr/13b558-pythagorean-theorem-distance-on-coordinate-plane-worksheet", "openwebmath_score": 0.5107263922691345, "openwebmath_perplexity": 776.8362653751947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9916842225056002, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8543738440901447}}
{"url": "https://math.stackexchange.com/questions/3380913/the-conjugate-function-of-infimum-of-sum-of-functions", "text": "# The conjugate function of infimum of sum of functions\n\nGiven convex functions $$f_1, \\ldots, f_k$$. Given $$g(x)$$, such that:\n\n$$g(x) = \\inf_{x_1, \\ldots, x_k} \\left\\{ f_1(x_1) + \\ldots + f_k(x_k) | x_1 + \\ldots + x_k = x \\right\\}$$\n\nFind $$g^*(x)$$.\n\nI found this similar question and did similar steps (you may find them below):\n\nBy definition: $$f^*(y) = \\sup_{x \\in dom g} (y^T x - f(x))$$\n\nThen $$g^*(x) = \\sup_{x \\in dom g} (y^T x - \\inf_{x_1, \\ldots, x_k} \\left\\{ f_1(x_1) + \\ldots + f_k(x_k) | x_1 + \\ldots + x_k = x \\right\\} = \\ldots$$\n\nUsing the fact that $$\\sup (-f(x)) = - \\inf (f(x))$$\n\nThen $$\\ldots = \\sup_{x \\in dom g} (y^T x + \\sup_{x_1, \\ldots, x_k} \\left\\{ -f_1(x_1) - \\ldots - f_k(x_k) | x_1 + \\ldots + x_k = x \\right\\} =$$\n\n$$= \\sup_{x \\in dom g, x_1, \\ldots, x_k} (y^T x + \\left\\{ -f_1(x_1) - \\ldots - f_k(x_k) | x_1 + \\ldots + x_k = x \\right\\}$$\n\nAfter these steps some questions arose:\n\n1. Why the author removed constraint $$x_1 + \\ldots + x_k = x$$ in the solution?\n2. Why $$\\sup_{x, x_1, \\ldots, x_k} \\{f(x) | x_1 + \\ldots + x_k = x \\} = \\sup_{x_1, \\ldots, x_k} \\{f(x) | x_1 + \\ldots + x_k = x \\}$$ If so, how to prove it?\n\u2022 Welcome to SE! Below, I've posted a complete solution to your problem, explaining all the steps in detail. Cheers! \u2013\u00a0dohmatob Oct 5 '19 at 10:47\n\u2022 @dohmatob thank you for your help! \u2013\u00a0user711343 Oct 5 '19 at 22:13\n\u2022 If it was helpful, please don't forget to upvote my answer. \u2013\u00a0dohmatob Oct 8 '19 at 7:40\n\u2022 @dohmatob I have already upvoted, but upvote is not shown because I don't have enough reputation. \u2013\u00a0user711343 Oct 9 '19 at 11:58\n\u2022 Ah, ok :). BTW, Ive updated your question. Welcome aboard :) \u2013\u00a0dohmatob Oct 9 '19 at 13:34\n\nI'll just write down the full solution to your problem, and explain all the steps in the derivations.\n\nSo, let $$f_1,\\ldots,f_k:\\mathcal H \\rightarrow (-\\infty,+\\infty]$$ be functions (convex or not) on a Hilbert space $$\\mathcal H$$. Define $$g(x) := \\inf_{x_1,\\ldots,x_k}\\left\\{\\sum_i f_i(x_i) \\mid \\sum_i x_i = x\\right\\}.$$ The function $$g$$ is also called the infimal convolution of $$f_1,\\ldots,f_k$$, written $$g=\\Box_{i=1}^k f_i$$.\n\nNow, one computes $$\\begin{split} g^*(x) &:= \\sup_y x^Ty - g(y) \\overset{(a)}{=} \\sup_{y}x^Ty - \\inf_{x_1,\\ldots,x_k}\\left\\{\\sum_i f_i(x_i) \\mid \\sum_i x_i = y\\right\\}\\\\ &\\overset{(b)}{=} \\sup_{y}x^Ty + \\sup_{x_1,\\ldots,x_k}\\left\\{-\\sum_i f_i(x_i) \\mid \\sum_i x_i = y\\right\\}\\\\ & \\overset{(c)}{=} \\sup_{y,x_1,\\ldots,x_k}\\left\\{x^Ty -\\sum_i f_i(x_i) \\mid \\sum_i x_i = y\\right\\} \\\\ &\\overset{(d)}{=} \\sup_{y,x_1,\\ldots,x_k}x^T\\sum_i x_i -\\sum_i f_i(x_i) \\overset{(e)}{=} \\sup_{y,x_1,\\ldots,x_k}\\sum_i (x^Tx_i -f_i(x_i)) \\overset{(f)}{=} \\sum_i\\sup_{x_i}x^Tx_i - f_i(x_i)\\\\ &\\overset{(*)}{=} \\sum_i f_i^*(x), \\end{split}$$ where\n\n\u2022 (a) is just plugging-in the definition of $$g(y)$$\n\u2022 (b) is because $$-\\inf something = -\\sup-thatthing$$\n\u2022 (c) is because $$\\sup_a u(a) + \\sup_b u(b) = \\sup_{a,b}(u(a) + u(b))$$\n\u2022 (d) is just substituting the constraint $$y=\\sum_i x_i$$ to get $$x^Ty = x^T\\left(\\sum_i x_i\\right)$$. After this substitution, the variable $$y$$ doesn't play a role anymore in the maximization, and so can be deleted. Indeed, $$\\sup_{a,b}u(a) = \\sup_a u(a)$$.\n\u2022 (e) is because $$x^T\\sum_i x_i - \\sum_i f_i(x_i) = \\sum_i x^Tx_i - \\sum_i f_i(x_i)=\\sum_i(x^Tx_i - f_i(x_i))$$\n\u2022 (f) is because $$\\sup_{a,b}u(a) + u(b) = \\sup_a u(a) + \\sup_b u(b)$$\n\u2022 (*) is because $$f_i^*(x) := \\sup_{x_i}x^Tx_i - f(x_i)$$ by definition.\n\nTherefore we have proven that\n\nConvex conjugate of infimal convolution. $$(\\Box_{i=1}^kf_i)^*(x) = \\sum_{i=1}^k f_i^*(x)\\; \\forall x$$", "date": "2020-01-29 05:17:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3380913/the-conjugate-function-of-infimum-of-sum-of-functions", "openwebmath_score": 0.9050416350364685, "openwebmath_perplexity": 931.0419336172458, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429614552198, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8543547051573486}}
{"url": "http://joshualoftus.com/notes103/HW3Rsol.html", "text": "### Question R1\n\nThis question will guide you through a simulation study in R to understand the bias of a certain estimator.\n\nvarn <- function(x) mean((x - mean(x))^2)\n\n#### Part a.\n\nThe code below generates a sample of size n and calculates both the sample variance and the biased version of sample variance (the one that has n in the denominator instead of n-1). Modify the code to change n from 10 to some other value between 10 and 50.\n\n# Create a sample of data by rolling a 6-sided die n times\nn <- 20\ndata <- sample(1:6, n, replace = TRUE)\ndata\n##  [1] 6 1 4 1 5 2 4 1 5 6 1 1 4 4 1 5 1 4 1 4\n# True variance\n35/12\n## [1] 2.916667\n# Unbiased estimate of variance\nvar(data)\n## [1] 3.628947\n# Biased estimate of variance\nvarn(data)\n## [1] 3.4475\n\n#### Part b.\n\nRepeat the previous experiment many times and find the expected value of each variance estimator.\n\n# Unbiased estimator\nmean(replicate(10000, var(sample(1:6, n, replace = TRUE))))\n## [1] 2.911705\n# Biased estimator\nmean(replicate(10000, varn(sample(1:6, n, replace = TRUE))))\n## [1] 2.76472\n\n#### Part c.\n\nAfter running the previous code, which estimator\u2019s average value in the simulation is closer to the true value (roughly 2.9167)?\n\nThe unbiased estimator\u2019s average value is closer to the true value.\n\n#### Part d.\n\nNow go back and change n to another, larger value, and rerun all of the code. Do you notice anything about the average of the biased estimator?\n\nmean(replicate(10000, varn(sample(1:6, n + 10, replace = TRUE))))\n## [1] 2.81582\n\nWith a larger sample size, the biased estimator\u2019s average value is now closer to the true value.\n\n#### Part e.\n\nCopy the code from part (b) and paste it below here, then change the mean function to be sd instead.\n\n# Unbiased estimator\nsd(replicate(10000, var(sample(1:6, n, replace = TRUE))))\n## [1] 0.5904201\n# Biased estimator\nsd(replicate(10000, varn(sample(1:6, n, replace = TRUE))))\n## [1] 0.5631186\n\nThe biased estimator has a lower variability (as measured by standard deviation).\n\n### Question R2\n\nAccording to a Marketplace/Edison survey in April of 2017, about 23.4% of survey responders agreed with the statement \u201cthe economic system in the U.S. is fair to all Americans.\u201d In this question we\u2019ll use a Bernoulli probability model to analyze this number. Suppose that there were 1,000 survey respondents and 234 agreed with the above quotation. Define a Bernoulli random variable which is 1 if a person agrees and 0 otherwise. Assume the survey was done with independent sampling (with replacement), so these Bernoulli random variables are independent. Then the number of people in the sample of 1,000 who agree is a Binomial random variable.\n\n\u2022 We have $$X_i$$ i.i.d Ber($$p$$) for $$i = 1, \\ldots, 1000$$.\n\u2022 Let $$S_n = \\sum_{i=1}^n X_i$$, so $$S_n$$ is Bin($$n, p$$).\n\n### a.\n\nUsing the fact that $$n \\bar X_n = S_n$$, how could you use the Binomial distribution to calculate $$P(a \\leq \\bar X_n \\leq b)$$? How would you use pbinom with the given values of $$a, b, n$$, and $$p$$?\n\npbinom(n*b, size = n, prob = p) - pbinom(n*a, size = n, prob = p)\n\n### b.\n\nInstead of the Binomial distribution, how would we use the central limit theorem to calculate the same probabilities? Hint: your answer should use pnorm and involve $$\\sqrt{n}$$ (and $$a, b$$, and $$p$$).\n\nSolution: The CLT says $$\\bar X_n$$ is approximately normal with mean $$p$$ and standard deviation $$\\sqrt{p(1-p)/n}$$. In R we could do:\n\npnorm(b, mean = p, sd = sqrt(p*(1-p)/n)) - pnorm(a, mean = p, sd = sqrt(p*(1-p)/n))\n\n### c.\n\nNow let $$n = 1000$$, $$p = 0.234$$, $$b = 0.250, a = 0.239$$ and compute the desired probability with both methods.\n\nn <- 1000\np <- 0.234\na <- 0.239\nb <- 0.250\npbinom(n*b, size = n, prob = p) - pbinom(n*a, size = n, prob = p)\n## [1] 0.2291025\npnorm(b, mean = p, sd = sqrt(p*(1-p)/n)) - pnorm(a, mean = p, sd = sqrt(p*(1-p)/n))\n## [1] 0.2383744", "date": "2018-11-15 08:07:32", "meta": {"domain": "joshualoftus.com", "url": "http://joshualoftus.com/notes103/HW3Rsol.html", "openwebmath_score": 0.8520529270172119, "openwebmath_perplexity": 1436.8061483737094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983342957061873, "lm_q2_score": 0.868826784729373, "lm_q1q2_score": 0.8543546996703411}}
{"url": "https://www.intmath.com/forum/applications-differentiation-27/concavity:149", "text": "Concavity [Solved!]\n\nMy question\n\nIn curve sketching with differentiation, the general curves section, there are no points of inflection written for the quadratic curve.\nBut from -1 to 1 it goes from negative to positive. Not clear as to why this would not be a point of inflection.\n\nRelevant page\n\n5. Curve Sketching using Differentiation\n\nWhat I've done so far\n\nMy work includes reading and re-reading and re-reading the section Finding points of inflection, and also reviewing the examples given.\n\nX\n\nIn curve sketching with differentiation, the general curves section, there are no points of inflection written for the quadratic curve.\nBut from -1 to 1 it goes from negative to positive.  Not clear as to why this would not be a point of inflection.\nRelevant page\n\n<a href=\"https://www.intmath.com/applications-differentiation/5-curve-sketching-differentiation.php\">5. Curve Sketching using Differentiation</a>\n\nWhat I've done so far\n\nMy work includes reading and re-reading and re-reading the section Finding points of inflection, and also reviewing the examples given.\n\nContinues below\n\nRe: Concavity\n\n@Phinah: Please have another look at the definition of a point of inflection.\n\nIt's where the concavity of the curve changes sign. That is, where it changes from a \"U shaped\" curve to an \"n shaped\" curve (or vice versa).\n\nHowever, a quadratic curve (a parabola) is \"U shaped\" everywhere (or \"n shaped\" if the number in front of the x^2 term is negative).\n\nU-shaped (e.g. y=x^2+2x+5)\n\nn-shaped (e.g. y=-x^2-3x+7)\n\nSo it won't have a point of inflection.\n\nX\n\n@Phinah: Please have another look at the definition of a point of inflection.\n\nIt's where the <b>concavity</b> of the curve changes sign. That is, where it changes from a \"U shaped\" curve to an \"n shaped\" curve (or vice versa).\n\nHowever, a quadratic curve (a parabola) is \"U shaped\" everywhere (or \"n shaped\" if the number in front of the x^2 term is negative).\n\n<b>U-shaped</b> (e.g. y=x^2+2x+5)\n\n[graph]310,250;-4.5,3.5;-0.5,10.3,1,2;x^2+2x+5,[/graph]\n\n<b>n-shaped</b> (e.g. y=-x^2-3x+7)\n\n[graph]310,250;-4.5,3.5;-0.5,10.3,1,2;-x^2-3x+7,[/graph]\n\nSo it won't have a point of inflection.\n\nYou need to be logged in to reply.\n\nRelated Applications of Differentiation questions\n\n\u2022 curve sketching [Solved!]\nWrote you earlier today. Confusion might come from Example 4. Three lines before...\n\u2022 Proof of equation [Solved!]\nhi, i need the proof of the following equation: y = (4Rx(L-2x)) / L^2 I shall be...\n\u2022 Practical application - normal to a curve? [Solved!]\nHi there. First off, I just wanted to say a huge thank you for the...\n\u2022 Rate of change: conical tank [Solved!]", "date": "2020-01-28 12:59:59", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/applications-differentiation-27/concavity:149", "openwebmath_score": 0.660639762878418, "openwebmath_perplexity": 1601.8494732717734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429575500227, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8543546967545583}}
{"url": "http://mathhelpforum.com/math-topics/111370-nxn-grid-problem.html", "text": "# Math Help - nxn grid problem.\n\n1. ## nxn grid problem.\n\nHello.\n\nI asked about this a while ago, but I'm still unsure. Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:\n\nx = black o = white\n\nooo xoo xxo\nooo ooo oxx\nooo ooo oxo\n\nSo, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct! Is it 9 squared, giving me 81? What if I had a 6x8 grid, or 17x31 grid?\n\nI did try to host a nicer picture than my x and o, but sadly I can't get imageshack to work.\n\nMany thanks.\n\nN.\n\n2. Originally Posted by theNoodler\nHello.\n\nI asked about this a while ago, but I'm still unsure. Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:\n\nx = black o = white\n\nooo xoo xxo\nooo ooo oxx\nooo ooo oxo\n\nSo, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct! Is it 9 squared, giving me 81? What if I had a 6x8 grid, or 17x31 grid?\n\nI did try to host a nicer picture than my x and o, but sadly I can't get imageshack to work.\n\nMany thanks.\n\nN.\nAn n x n grid with m possible colours has $m^{n \\times n}$ possible combinations.\n\n3. Originally Posted by theNoodler\nImagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:\nSo, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct!\nBut that is correct.\nLook at that attached graph. There are two $3\\times 3$ grids.\nAre they different colorings?\n\nMaybe not. Rotate I $90^o$ counter-clockwise. We get II.\nNow are they different?\n\nIf I & II are different then there are $2^9=512$ possible colorings.\n\nIf I & II are not different then you must tell us why?\n\n4. Hello, theNoodler!\n\nImagine you have a $n\\times n$ grid (3x3 for argument).\nHow can I work out the total number of patterns made with 2 colours, say black and white.\n\nAn example would be: $\\begin{array}{c}X \\,=\\, \\text{black} \\\\ O \\,=\\,\\text{white}\\end{array}$\n\n. . $\\begin{array}{c}OOO\\\\OOO\\\\OOO\\end{array} \\qquad \\begin{array}{c} XOO\\\\ OOO\\\\OOO \\end{array} \\qquad \\begin{array}{c}XXO\\\\OXX \\\\ OXO \\end{array}\\qquad \\hdots\\text{ etc.}$\n\nI was told for this it would be $2^9 \\,=\\,512$, which can't be correct!\n. . Why not?\n\nFor each of the 9 cells, you have two choices: place a Black or place a White.\n\nSo you have 9 decisions with 2 options each.\n. . There will be: . $2^9 \\,=\\,512$ possible choices you can make.\n\nWhat if I had a 6x8 grid grid?\n\nYou have 48 cells to fill with 2 options each.\n\nThere will be: . $2^{48}$ possible choices.\n\n5. Hello.\n\nThanks for the help. I guess it just doesn't look like there could be that many combinations. So for a 6x8 grid, there would be 281,474,976,710,656 combinations. That's just nuts.", "date": "2014-07-31 18:45:31", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-topics/111370-nxn-grid-problem.html", "openwebmath_score": 0.5515955090522766, "openwebmath_perplexity": 968.9096014918198, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9833429595026213, "lm_q2_score": 0.868826771143471, "lm_q1q2_score": 0.8543546884313273}}
{"url": "https://stats.stackexchange.com/questions/505141/for-a-toss-of-fair-die-if-events-are-a-1-2-b-2-4-6-and-c-4-5-6-the/505198#505198", "text": "# For a toss of fair die, if events are, A: {1,2}, B: {2,4,6}, and C: {4,5,6}, then A and B are independent but B and C are not. Why?\n\nAs far as I understood independence, A and B should not be independent since if either of them happens then we can tell something about the other one. But if they are independent then B and C should also be. From mathematical proof it is explainable, but I didn't understand the intuition behind it.\n\nThe example is from \u201cElementary Bayesian Statistics\u201d by Gordon Antelman, Chapter 2\n\nText:\n\nFor a fair die $$U = (1,2,3,4,5,6)$$. Let three events be defined as:\n\n\u2022 $$A \\equiv (1,2)$$, so $$P(A)=2/6$$\n\n\u2022 $$B \\equiv (2,4,6)$$, so $$P(B)=3/6$$, and\n\n\u2022 $$C \\equiv (4,5,6)$$, so $$P(A)=3/6$$\n\nThen\n\n\u2022 $$A \\cap B = (2)$$ and $$P(A \\cap B) = 1/6 = P(A)P(B) = (1/3)(1/2)$$, so A and B are independent.\n\n\u2022 $$B \\cap C = (4,6)$$ and $$P(B \\cap C) = 2/6 \\ne P(B)P(C) = (1/2)(1/2)$$, so B and C are not independent.\n\n\u2022 Rename the faces of the die from $1,2,3,4,5,6$ to $(1,0),(1,1),(2,0),(2,1),(3,0),(3,1),$ thereby identifying them with the Cartesian product $\\{1,2,3\\}\\times\\{0,1\\}$ and notice the distribution is the product of the uniform distributions on these two sets. Writing the coordinates as $(x,y),$ $A$ is the event $x=1$ while $B$ is the event $y=1,$ making their independence intuitively obvious. In some sense, all independent events occur in this way.\n\u2013\u00a0whuber\nJan 16 at 18:56\n\nOther explanation why $$\\color{red}{A = \\{1, 2\\}}$$ and $$\\color{blue}{B = \\{2, 4, 6\\}}$$ are independent:\n\nSomeone rolled a die:\n\n\u2022 The probability of the event $$\\color{red}A$$ is $$\\color{red}{2\\mkern-0.1ex/6}$$, i.e. $$\\color{red}{1\\mkern-0.1ex/3}$$.\n\u2022 Someone tells you that the result is an $$\\color{blue}{\\text{even}}$$ number (the event $$\\color{blue}B$$). In spite of this new info the probability of $$\\color{red}A$$ is still $$\\color{red}{1\\mkern-0.1ex/3}$$\n($$\\color{blue}2 \\in \\color{red}A$$, but $$\\color{blue}{4} \\notin \\color{red}A$$, $$\\color{blue}6 \\notin \\color{red}A$$).\n\nIn the opposite way \u2014\n\n\u2022 The probability of event $$\\color{blue}B$$ is $$\\color{blue}{3\\mkern-0.1ex/6}$$, .i.e. $$\\color{blue}{1\\mkern-0.1ex/2}$$.\n\u2022 Someone tells you that the result is a number $$\\color{red}1$$ or $$\\color{red}2$$ (the event $$\\color{red}A$$). In spite of this new info the probability of $$\\color{blue}B$$ is still $$\\color{blue}{1\\mkern-0.1ex/2}$$\n($$\\color{red}2 \\in \\color{blue}B$$, but $$\\color{red}{1} \\notin \\color{blue}B$$).\n\nNote:\n\nYou may use the same scheme to see that the events $$\\color{blue}B$$ and $$\\color{green}C$$ are not independent.\n\n\u2022 Thank you for this explanation. So what I understood is: if B happens then the probability of A will still be 1/3 because {2}/{2,4,6} and if A happens then also the probability of B will be same (1/2) because {2}/{1,2}. And in case of B and C, if B happens i.e. {2,4,6} the probability of C will be 2/3 because {4,6}/{2,4,6} and if C happens, the probability of B will also be 2/3 {4,6}/{4,5,6} in which case the probability of C (1/2) and B (1/2) has changed respectively and that's why they are not independent. Please correct me if I am wrong. Jan 16 at 20:37\n\u2022 @someUser, exactly, nothing to correct. What should I do when someone answers my question?. Jan 16 at 20:40\n\nYou probably confuse indepenent events for mutually exclusive events.\n\n... A and B should not be independent since if either of them happens then we can tell something about the other one.\n\nIt is not true. What does mean \u201csomething\u201d? The probability of \u201cthe other one\u201d didn't change!\n\nThe correct formulation should be\n\n\u2022 \u201cA and B are independent \u2014 if either of them happens, then it doesn't change the probability of the other\u201d.\n\nAs I wrote, you are probably confused mutually exclusiveness (the empty intersection, distinctiveness) for the independence of two events.\n\nBut they are two independent things:\n\n\u2022 the events $$\\{1\\}$$ and $$\\{2, 4, 6\\}$$ are mutually exclusive, but they are not independent,\n\n\u2022 the events $$\\{2\\}$$ and $$\\{2, 4, 6\\}$$ are neither mutually exclusive, nor independent,\n\n\u2022 the events $$\\{1, 2\\}$$ and $$\\{2, 4, 6\\}$$ are not mutually exclusive, but they are independent,\n\n\u2022 the events $$\\emptyset$$ and $$\\{2, 4, 6\\}$$ are mutually exclusive, and they are independent.\n\nIndependence can arise in two distinct ways:\n\n\u2022 we explicitly assume that events are independent (e.g. rolling a die again), or\n\u2022 we derive independence by verifying that it fulfills the formula $$P(A \\cap B) = P(A)P(B)$$.\n\nGenerally, there is no way to \u201csee\u201d the independence (e.g. by looking in a Venn diagram).\n\nBy other words, if we may \u201csee\u201d the independence of two events, then they certainly fulfill the formula. But the opposite is not true \u2014 the formula is fulfilled, the independence is guaranteed, but there is not a visible reason, why.\n\nThe idea of independent events arises from the evident, noticable independence, but is not limited to it.\n\n\u2022 So how can A $\\cap$ B and B $\\cap$ C be explained in current scenario? Jan 16 at 19:16\n\u2022 @someUser, don't try to \u201csee\u201d the independence by interpreting intersections of 2 events. Most teachers don't teach their student what I wrote in my answer \u2014 independence is sometime visible (and understandable), sometime not. Jan 16 at 19:34", "date": "2021-10-23 17:55:48", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/505141/for-a-toss-of-fair-die-if-events-are-a-1-2-b-2-4-6-and-c-4-5-6-the/505198#505198", "openwebmath_score": 0.9154084920883179, "openwebmath_perplexity": 435.82356019988964, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429590144718, "lm_q2_score": 0.868826771143471, "lm_q1q2_score": 0.85435468800721}}
{"url": "https://math.stackexchange.com/questions/1832286/how-does-probability-change-the-more-times-you-perform-a-procedure", "text": "# How does probability change the more times you perform a procedure?\n\nI have a question based on fair coins. Every round, two coins are flipped. If both are heads, we say \"Success\" and end the experiment. What is the probability of saying \"Success\" in any round i?\n\nInitially, I thought it would be $$\\displaystyle\\dfrac{\\displaystyle\\frac{i!}{(i-2)!2!}}{2^i},$$ because of NCR and increasing probability, but I realize it doesn't work.\n\nRight now, it seems as though a summation works as we're increasing chances of getting a \"Success\" the more rounds we have. We also need to stop once it reaches a \"Success\", but I don't get how we can stop and operation like that unless it simply reaches 100%. $$\\sum_{n=1}^{i} \\frac{1}{4}$$\n\nLogically, it sounds right to me, but I'm unsure so I'd like to hear your thoughts on this. The $\\displaystyle\\frac{1}{4}$ is due to the [TT][TH][HT][HH] probability sample space.\n\nThe question continues to ask for an expression on the probability of getting a \"Success\" in terms on n and i. Here, n stands for the total number of rounds while i stands for the number of rounds before we stop running the procedure.\n\nMy solution for this part seems the same as the previous one, as I don't see what the question is asking differently. Any ideas?\n\nAnd finally, the last part is asking what if we don\u2019t stop after saying \u201cSuccess\", that is we repeat the procedure n times. How many times do you expect to say \u201cSuccess\u201d, express this in terms of n.\n\n$$\\sum_{i=1}^{n} \\frac{1}{4}$$\n\nThis way, we can easily go over the limit and we'll find the total number of times that \"Success\" can be reached.\n\nThank you for reading this :)\n\n\u2022 What does your summation say about $n=5$? Do you believe a probability can be greater than $1$? \u2013\u00a0Erick Wong Jun 19 '16 at 17:21\n\u2022 In terms of the third question, it can continue over and reach as many \"Success\" stages as possible. What I don't get is how to get the summation to stop when it reaches \"Success\" in the first subquestion. \u2013\u00a0Andrew Raleigh Jun 19 '16 at 17:28\n\nFirst question\n\nThe probability of having success in round 1 is $P(X=1)=\\frac{1}{4}$. We agree.\n\nThe probability of having success in round 2 is $P(X=2)=\\left(1- \\frac{1}{4}\\right)\\cdot \\frac{1}{4}=\\frac{3}{4}\\cdot \\frac{1}{4}=\\frac{3}{16}$\n\nYou have to take into acccount that the probability of having a success in round $2$ depends on wether you have success in round 1 or not. Here we need to have no success in round 1.\n\nThe probability of having success in round 3 is $P(X=3)=\\left(1- \\frac{1}{4}\\right)^2\\cdot \\frac{1}{4}=\\frac{3^2}{4^2}\\cdot \\frac{1}{4}=\\frac{9}{64}$\n\nHere we need to have no success in round 1 and round 2.\n\n$...$\n\nThe probability of having success in round $i$ is $P(X=i)=\\left(1- \\frac{1}{4}\\right)^{i-1}\\cdot \\frac{1}{4}=\\left(\\frac{3}{4}\\right)^{i-1}\\cdot \\frac{1}{4}$\n\nSecond question\n\nYou need the closed form of $\\frac{1}{4}\\cdot \\sum_{i=1}^n \\left(\\frac{3}{4}\\right)^{i-1}$ For $j=i-1$ we get $\\frac{1}{4}\\cdot \\sum_{j=0}^{n-1} \\left(\\frac{3}{4}\\right)^{j}$. This is the partial sum of a geometric series.\n\nThus $\\frac{1}{4}\\cdot \\sum_{j=0}^{n-1} \\left(\\frac{3}{4}\\right)^{j}=\\frac{1}{4}\\cdot \\frac{1-\\left(\\frac{3}{4}\\right)^{n}}{\\frac{1}{4}}=1-\\left(\\frac{3}{4}\\right)^{n}$\n\nThird question\n\nI\u00b4m not sure if my understanding is right. But if we don\u00b4t stop at the first success the probability of getting $i$ successes and consequently $n-i$ failures is\n\n$P(Y=i)=\\binom{n}{i}\\cdot \\left( \\frac{1}{4} \\right)^i \\cdot \\left( \\frac{3}{4} \\right)^{n-i}$.\n\nThis is the pdf of the binomial distribution. And the expexted value is $n\\cdot p=\\frac{1}{4}n$\n\n\u2022 That makes a lot of sense, thank you! In this case, the probability of having a success at round i is $\\left(\\frac{3}{4}\\right)^{i-1}\\cdot \\frac{1}{4}$, but would the same thing apply to the second subquestion? \u2013\u00a0Andrew Raleigh Jun 19 '16 at 17:49\n\u2022 @AndrewRaleigh I\u00b4ve made an edit. \u2013\u00a0callculus Jun 19 '16 at 18:09\n\u2022 @callculus Small typo in the second line: RHS should be $3/16$ instead of $1/16$. \u2013\u00a0Erick Wong Jun 19 '16 at 18:27\n\u2022 @callculus I'm sorry, but the top part the answer for the first question and the bottom the second, right? If so, can I ask how you thought of the way to answer the question? \u2013\u00a0Andrew Raleigh Jun 19 '16 at 18:29\n\u2022 @ErickWong Thanks for the comment. \u2013\u00a0callculus Jun 19 '16 at 18:30", "date": "2021-03-06 23:59:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1832286/how-does-probability-change-the-more-times-you-perform-a-procedure", "openwebmath_score": 0.7252374291419983, "openwebmath_perplexity": 318.08888143912776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983342957061873, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8543546863107399}}
{"url": "http://math.stackexchange.com/questions/312499/understanding-proof-by-contrapositive", "text": "# Understanding Proof by Contrapositive\n\nThis is a question from my textbook, and I'm pretty sure I have the answer, but I am having trouble writing out what I am thinking.\n\nProve: If $n$ is an integer and $3n+2$ is odd, then $n$ is odd.\n\nContrapositive: If $n$ is even, then $3n+2$ is even.\n\n$n$ is even, by definition $n = 2k$, where $k$ is an integer.\n\nPlugging in $2k$ into $3n+2$, we have $3(2k)+2$.\n\n$3(2k)+2 = 6k+2 = 2(3k+1)$, which means $3n+2$ is even.\n\nSince if $n$ is even, then $3n+2$ is even. Negating the conclusion of the original conditional statement implies that it's hypothesis \"$3n+2$ is odd\" is false, therefor if $3n+2$ is odd, $n$ is odd.\n\nWhat am I missing, or what can I do to make this less awkward? Or am I wrong about the proof on the most basic levels?\n\n-\nIt can be proved directly: $\\rm\\ 3n\\!+\\!2\\,=\\, 2k\\!+\\!1\\,\\ is\\ odd\\:\\Rightarrow\\: n\\,=\\,2(k\\!\u2212\\!n)-1\\,\\ is\\ odd.$ \u2013\u00a0 Math Gems Feb 26 '13 at 6:03\n\n## 1 Answer\n\nYour proof is correct, though you could improve on it by writing it in essay style, cutting down on some verbosity, and by mentioning that $3k+1$ is also an integer. For example, you could write something like this:\n\nWe try to prove the contrapositive. If $n$ is even, then $n=2k$ for some integer $k$, so $3n+2=3(2k)+2=6k+2=2(3k+1)$ where $3k+1$ is an integer, making $3n+2$ even.\n\nNote that proofs can be written with varying degrees of detail. In beginning courses, instructors prefer to see more detailed proofs to ensure students really understand what they are writing.\n\n-\nUp to how much? Do I remove everything after \"Since if n is even, then 3n+2 is even.\"? \u2013\u00a0 Chase Feb 24 '13 at 1:47\nSo it's better to write out explicitly what I have up there? (As in, \"3(2k)+2=6k+2=2(3k+1), which means 3n+2 is even.\" should have been a bit more spelled out) \u2013\u00a0 Chase Feb 24 '13 at 1:55\nI see. I have trouble trying with proofs and explaining proofs so thank you! \u2013\u00a0 Chase Feb 24 '13 at 2:02", "date": "2015-01-28 04:53:55", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/312499/understanding-proof-by-contrapositive", "openwebmath_score": 0.828276515007019, "openwebmath_perplexity": 308.7428413951989, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429619433693, "lm_q2_score": 0.8688267660487573, "lm_q1q2_score": 0.8543546855420638}}
{"url": "https://web2.0calc.com/questions/polynomials-p-x-is-a-polynomial-in-x-with-non-negative", "text": "+0\n\n# [Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possib\n\n0\n63\n7\n+7\n\nI attacked the problem by first setting the obvious as P(x) = c1xn+c2xn-1+.....+cn with the c's being the constants.\n\nP(1) = 5 tells me that the sum of all the coefficients of the polynomial is 5, and immediately a matter to observe would be n < 4 because the next part, P(P(1)) = 177 means P(5) = 177 and 54\u00a0> 177 for obvious reasons.\n\nSo P simplifies to P(x) = x3+c1x\u00b2+c2x+c3. It's trivial to see x3\u00a0won't have a coefficient. My next steps involved basically trial and error-ing as the sum of coefficients = 5 and using every different case where I change the degree, remove some terms and overall experiment with the coefficients and x terms.\n\nIn the end, I found P(x) = x3+2x\u00b2+2 to be the only polynomial satisfying the given conditions and the answer to the question is P(10) = 103+2(10)\u00b2+2 = 1202.\n\nMy problem is I used a trial and error approach to the problem and found my solution, which I'm unsure is the only solution. Is there a more elegant approach to the problem and is my solution correct? Thanks~\n\nJul 17, 2022\n\n#1\n+197\n0\n\nThe problem got cut off in your title, could you please show us the complete question?\n\nJul 17, 2022\n#2\n+117766\n+1\n\nThe question has to be in the question section.\n\nA suitable title might have been\n\nPolynomials,\u00a0 sum of roots\u00a0 \u00a0 (I assume you want to find the sume of all possible roots..)\n\nIt is really good that you have tried to show what you did :)\n\nJul 17, 2022\nedited by Melody \u00a0Jul 17, 2022\nedited by Melody \u00a0Jul 17, 2022\n#3\n+7\n+1\n\n[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?\n\nthis was the whole question\n\nJul 17, 2022\n#5\n+2\n\nQuestion:\n\n[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?\n\nTo begin with, your solution is correct, and is actually a quick way to solve this problem. That is, It is efficient.\n\nHowever, I think what you are trying to say, is there a way to write the solution \"more rigorously\", and the answer to this is: yes.\n\nLet\u00a0$$P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$$,\u00a0 \u00a0where\u00a0$$a_n,a_{n-1},...,a_0 \\ge 0$$\u00a0(Given.)\n\nSince:\u00a0$$P(1) = 5 \\iff a_n+a_{n-1}+...+a_0=5$$, which does not tell us that much.\n\nNext,\u00a0$$P(P(1)) = P(5) =a_n*5^n+a_{n-1}*5^{n-1}+...+a_0=177$$\n\nNotice:\u00a0$$5^3=125,5^4=625$$; hence,\u00a0$$P(x)$$\u00a0is at most a cubic polynomial.\n\nNow,\n\nLet:\u00a0$$P(x)=ax^3+bx^2+cx+d$$\u00a0, where\u00a0$$0\\le a,b,c,d \\le 5$$\n\nWe know:\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0$$a+b+c+d=5$$\u00a0 \u00a0 \u00a0 \u00a0 (1)\n\nAnd,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0$$125a+25b+5c+d=177$$\u00a0 (2)\n\nNotice from (2):\n\n$$125a \\le 177 \\implies a \\le 1.416 \\implies a=1\\text{ or, } a=0$$\n\nNext, consider the case a=1\u00a0:\n\n(2)\u00a0and (1) becomes:\n\n$$25b+5c+d=52 \\text{, and } b+c+d=4$$\n\nUsing the same idea as before:\n\n$$25b \\le 52 \\implies b \\le 2.09 \\implies b=2,1,0$$\n\nBut,\u00a0$$b=1,0$$\u00a0is rejected. Because observe that if b=0, then 5c+d = 52 (and this is only true for large c and d, which we are not given.)\n\nProof:\u00a0 \u00a0$$c,d\\le 5 \\implies 5c \\le 25, d\\le 5 \\implies 5c+d \\le 25+5=30$$, so it is at most 30 (Which, by the way, is even unattainable!).\n\nThus,\u00a0$$b=2$$\u00a0only.\n\nNext, we substitute b=2 in (1) and (2) to get:\n\n$$c+d=2\\text{ , }5c+d=2 \\implies 5c\\le 2 \\implies c=0,d=2$$\n\nSo:\u00a0$$a=1,b=2,c=0,d=2$$\u00a0is a valid solution.\n\nThus,\u00a0$$P(x)=x^3+2x^2+2 \\implies P(10)=1000+200+2=1202$$, as you found.\n\nNext, consider the case\u00a0\u200ba=1\u00a0:\n\n$$25b+5c+d=177 \\implies b \\le 7$$, we see that, this is impossible. As, b is at most 5, giving 125, but c and d are forced to be 0. Hence, this case is rejected.\n\nProof:\n\n$$25b \\le 125, 5c \\le 25, d \\le 5 \\implies 25b+5c+d \\le 155$$, which again, is not attainable anyway.\n\nTherefore,$$\\text{ }P(x)=x^3+2x^2+2$$\u00a0is the only possible polynomial with the given conditions, and\u00a01202 is the only valid solution.\n\nI hope this helps!\n\nGuest\u00a0Jul 17, 2022\n#6\n+117766\n-1\n\nI already showed that somewhat more easily.\n\nMelody \u00a0Jul 17, 2022\nedited by Melody \u00a0Jul 17, 2022\n#7\n+1\n\nOoh! I just saw it now. It is really elegent! But I think my approach is more general; as writing 177=125+50+2 depended on \"177\" probably If it was other number, then there might be many ways to write a number \"A\" as addition of other numbers, so more cases to try.\n\nBut overall, that was fantastic observation!!!!!!\n\nGuest\u00a0Jul 17, 2022\n#4\n+117766\n+1\n\nIf P(x) is a polynomial with all positive integer polynomials\u00a0 AND the constant is also positive (which may not be intended)\n\nand\n\nP(1)=5\u00a0 then all the coeficients and the constant must add to 5.\n\nP(5)=177 = 125+50+2 = 125+ 2*25 + 2\u00a0 so the polynomial is\n\n$$p(x)=x^3+2x^2+2$$\n\nThat is the only possibility.\u00a0 (assuming that the constant is also a positive integer)\n\nJul 17, 2022\nedited by Melody \u00a0Jul 17, 2022", "date": "2022-08-18 02:13:33", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/polynomials-p-x-is-a-polynomial-in-x-with-non-negative", "openwebmath_score": 0.8474124670028687, "openwebmath_perplexity": 764.743700231722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446410358804, "lm_q2_score": 0.877476800298183, "lm_q1q2_score": 0.8543505842436373}}
{"url": "https://math.stackexchange.com/questions/1798725/2048-logic-puzzle/1817100", "text": "# 2048 Logic Puzzle\n\nI thought up this logic problem related to the 2048 game. If all 16 tiles on a 2048 board all had the value 1024, how many ways are there to get to the 2048 tile? Here is what I am talking about in an illustration:\n\nI found a much simpler, but longer way to think about this: There are 3 ways to combine 2 tiles by going to the right, and 3 by going to the left. That means there are 6 ways to combine the tiles. So, for all the rows and columns, there are $$2 \\cdot (4 \\cdot 6) = 48$$ ways to get to the 2048 tile.\n\nMy question(s) are, is my logic correct? Also, is there a simpler way to approach this logical problem?\n\n## Notes\n\nI found two Math.SE post related to 2048 logic, but they have nothing to do with my problem.\n\nI believe you are Correct.\n\nThere are 3 lines separating rows horizontally, 4 pairs of numbers across each line, and 2 ways to combine said pairs (top-down or bottom-up), giving $2*3*4=24$ ways of making pairs.\n\nRepeating for the columns and getting the exact same numbers, we now have $24+24 = 48$ total options for merging two numbers.\n\n\u2022 Okay, that answers the first part. But what about the second prompt. Is my method the simplest way to approach this? Or is there a much more simpler way to approach this? \u2013\u00a0Obinna Nwakwue May 25 '16 at 21:58\n\u2022 \"Is my method the simplest way to approach this?\" As far as I can tell, yes. \u2013\u00a0Simpson17866 May 25 '16 at 23:16\n\u2022 Okay, that helps. Thank you! \u2013\u00a0Obinna Nwakwue May 26 '16 at 0:59\n\nThe corner tiles have 2 ways to move\n\nThe side tiles have 3 ways to move\n\nThe middle tiles have 4 ways to move\n\n2 3 3 2       2 3 3 3 2       ...\n3 4 4 3       3 4 4 4 3\n3 4 4 3       3 4 4 4 3\n2 3 3 2       3 4 4 4 3\n2 3 3 3 2\n\nThen the formula for different ways to make a $2048$ tile on $(n * n)$ square of $1024$ tiles is:\n\n$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4$\n\n$4 * 2$ (corner tiles)\n\n$(n - 2) * 3 * 4$ (side tiles)\n\n$(n - 2)^2 * 4$ (middle tiles)\n\n$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4 =$\n\n$4(n^2 - n)$\n\n\u2022 To complete your \"proof\", for $n = 4$, $4(n^2 - n) = 4(4^2 - 4) = 4(16 - 4) = 64 - 16 = 48$! \u2013\u00a0Obinna Nwakwue Jun 7 '16 at 15:51", "date": "2019-11-22 09:44:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1798725/2048-logic-puzzle/1817100", "openwebmath_score": 0.4534560441970825, "openwebmath_perplexity": 325.9555225890064, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97364464868338, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8543505691166549}}
{"url": "http://math.stackexchange.com/questions/282297/how-to-represent-each-natural-number", "text": "# How to represent each natural number?\n\nAssume we get the set of natural numbers $\\mathbb{N}$ from any model of the Peano axioms.\n\nWe're given the symbols: $0,1,2,3,4,5,6,7,8,9$, or rather, we're given $0$ and we choose to use the symbols $1,2,3,4,5,6, 7, 8,9$.\n\nOf course $0$ is the same from the model.\n\nThen we'll have, by definition, $1=S(0)$, $2=S(1)$, $3=S(2)$, $4=S(3)$, $5=S(4)$, $6=S(5)$, $7=S(6)$, $8=S(7)$ and $9=S(8)$.\n\nBut how do we represent the rest of the natural numbers the way we expect them to be represented?\n\nI understand that $1, 2, 3, 4, 5, 6, 7, 8, 9$ are just shorthand representations for the entities written above.\n\nI guess my question can be particularized by: how do you know that the short hand notation for $S(999)$ is $1000$?\n\nI'm assuming the way to get a representation for each natural number starts the way I write it. If that's not the case, please do it from the top.\n\n-\nWhat's wrong with the inductive definition $S(n+1)=S(S(n))$? \u2013\u00a0 Asaf Karagila Jan 19 '13 at 21:57\nMy problem is not with the definition per se, but rather how to represent each number. Intuitively you know $10$ will be $S(9)$, but why? Did I make it clearer? \u2013\u00a0 Git Gud Jan 19 '13 at 21:58\nI don't think it is \"intuitively\" but only a matter of choosing some base wrt which write the numbers: $\\,S(9)=10\\,$ because we usually use the decimal base to write numbers, but I think it may as well be $\\,S(9)=101\\,$ if we choose base $\\,3\\,$ (and, of course, it probably is more logical to write $\\,S(100)=101\\,$ in base $\\,3\\,$ ...) \u2013\u00a0 DonAntonio Jan 19 '13 at 22:04\n@DonAntonio What's that about base wrt? Never heard of it. I know we choose to use the decimal base do represent numbers. But how do you know that, in base $10$, $1000=S(999)$? And what is $999$? \u2013\u00a0 Git Gud Jan 19 '13 at 22:10\n@GitGud , wrt = with respect to . And again: I know that $\\,S(999)=1000\\,$ because I choose (or we choose) to represent the numbers in decimal base, that's all. If we chose not to work with bases at all then we could use the successor function and work just with that, though it would be extraordinarily cumbersome: \"What time is it?\" \" It is$\\,S(S(S(S(\\emptyset))))\\,$ minutes before $\\,S(S(\\emptyset))\\,$ \" ...pretty annoying, uh? There's where bases kick in. \u2013\u00a0 DonAntonio Jan 19 '13 at 22:17\n\nGiven a PA number, to translate it back into a string of digits is by the following recursive function:\n\n\u2022 $f(x) = d(x)$ where $x < 10$\n\u2022 $f(x \\hat + SSSSSSSSSS0 \\hat \\times y) = f(y)\"d(x)\"$ where $x < 10$ and $y > 0$.\n\nJustifying this recursion principle 9that a function defined in this way is well defined and total) is by Euclid's Division algorithm.\n\n-\nI think I'm getting the gist of it. Can you tell me where to read about this and find some proofs? \u2013\u00a0 Git Gud Jan 19 '13 at 23:03\n@GitGud, what do you like proofs of? \u2013\u00a0 user58512 Jan 19 '13 at 23:08\n\"Justifying this recursion principle 9that a function defined in this way is well defined and total) is by Euclid's Division algorithm.\" I wouldn't wanna bother you with that, I can probably do it myself. But I'd rather read it, so if you know where I can read about it, I'd appreciate it. Thanks. \u2013\u00a0 Git Gud Jan 19 '13 at 23:14\n\nThere is an obvious function from strings of digits to natural numbers:\n\n\u2022 $f(\"\") = 0$\n\n\u2022 $f(\"3534\") = 3 + 10\\cdot f(\"534\") = 3 + 10 \\cdot (5 + 10 \\cdot f(\"34\")) = \\ldots = 3 + 10(5 + 10(3 + 10(4 + 0))))$\n\nRecall that we define + and * for peano arithmetic:\n\n\u2022 $0 \\hat + y = y$\n\u2022 $Sx \\hat + y = S(x \\hat + y)$\n\n\u2022 $0 \\hat \\times y = x$\n\n\u2022 $Sx \\hat \\times y = x \\hat + x \\cdot y$\n\ntherefore, define\n\n\u2022 $g(\"\") = 0$\n\n\u2022 $g(\"dssss\") = p(d) \\hat + SSSSSSSSSZ\\hat \\times g(\"ssss\")$\n\nand the gives the peano arithmetic number or a digits expression.\n\n(p is the function that gives $p(2) = SSZ$ for the first 10 digits for example, that you already mentioned)\n\n-\nI think this is on the right track to satisfy me. But I have to define $10$ before applying your process, right? \u2013\u00a0 Git Gud Jan 19 '13 at 22:49\n@GitGud, the first part with $f$ is just for intuition - it's not used later. Also I posted a new answer to your question: How to do the reverse operation: Getting a string of digits from a peano number. \u2013\u00a0 user58512 Jan 19 '13 at 22:51\n\nfirst you must define the \"remain\" and \"quotient\" of one natural number to another then you can represent any natural number by its \"quotient\" and \"remain\" to the powers of ten\n\nActually in set theory you don't represent numbers like 10 or 11 or something like that, number 3 in set theory is S(S(S(0))), but because you are already familiar with this representation of natural numbers, you use 3, actually using \"3\" in set theory is wrong\n\n-\n\nYou need to notice that the standard decimal representation of the natural numbers is not a representation from inside the Peano model. The language only contains a constant symbol for $0$. Then one introduces $1=S(0)$ as shorthand notation not as a new symbol. Then comes $2=S(1)=s(S(0))$ again as shorthand notation, not a new symobol, and so on.\n\nSo, whenever you write something in PA and you use $0$ then you mean the constant symbol in the language. When you write $1$, $2$, and so on mean that this is shorthand notation that you use just because you are lazy to write out an expression of the form $S(S(0))$.\n\nI hope this helps.\n\n-\nI understand everything you said, but it doesn't answer what I want to know. Why do you make $10$ short for $S(S(S(S(S(S(S(S(S(S(0))))))))))$? \u2013\u00a0 Git Gud Jan 19 '13 at 22:02\nit's just introducing shorthand notation for strings in the language. You could have used \"banana\" as shorthand for S(S(S(S(S(S(S(S(S(0)))))))))) if you want. But it makes more sense for us humans (who like numbers and use the decimal notation regularly) to call it something a bit more meaningful. We work with the model of PA from outside of it. We are not in the model so we are free to study it by means of things we have outside as well. It makes life easier. \u2013\u00a0 Ittay Weiss Jan 19 '13 at 22:09\nI wouldn't say 47 is short for $S(S(S(\\ldots(S(0)))))$ any more than it is short for XLVII. Decimal numerals, Roman numerals, successor numerals belong to three different ways of representing natural numbers. \u2013\u00a0 Peter Smith Jan 19 '13 at 22:13\n@IttayWeiss How do you rigorously introduce the shorthand notation that we're used to? That's my question. \u2013\u00a0 Git Gud Jan 19 '13 at 22:16\n@GitGud recursively: n+1 is shorthand notation for S(n), where 0 is shorthand for \"0\". It works since you believe that for the models of the naturals that you have in your head, recursion works. \u2013\u00a0 Ittay Weiss Jan 19 '13 at 22:21\n\nSuppose you have $9$ stones in a pile and you add another stone to make a pile of $S(9)$ stones. Rather than invent a new symbol, the convention is to give a pile of this size its own name. Henceforth, any pile of exactly $S(9)$ stones will be called a \"ten\". So, the successor of $9$ is a ten and no other stones. This is denoted by placing a $1$ in the ten column and a $0$ in the single stone column, which gives $S(9) = 10$. The process is similar for larger piles of stones: \"hundred\", \"thousand\", etc.\n\n-\nMy problem is with \"that process is similar\". That's just an intuitive way of doing things. That's like what we were taught when we were kids. How do you make a computer understand that $10$ is just $S(9)$? You can't tell the computer what symbol will represent each natural number. \u2013\u00a0 Git Gud Jan 19 '13 at 22:38", "date": "2015-07-08 02:58:19", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/282297/how-to-represent-each-natural-number", "openwebmath_score": 0.8927438259124756, "openwebmath_perplexity": 421.87882062828186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.966410494349896, "lm_q2_score": 0.8840392924390587, "lm_q1q2_score": 0.854344849630763}}
{"url": "http://migf.com/acupuncture-in-pnmg/are-the-two-diagonals-of-a-rectangle-equal-why-68fef3", "text": "Answer. Because all rectangles are also parallelograms, all the properties of parallelograms are also true for rectangles, too: But because the angles are all equal, there is an additional property of rectangles that we will now prove \u2013 that the diagonals of a rectangle are equal in length. If we divided the rectangle along diagonal NL, we would create triangle LNO. A rectangle is a parallelogram with 4 right angles. You can contact him at GeometryHelpBlog@gmail.com. Can anyone solve this and explain how to do so? prove: the diagonals of a rectangle have equal lengths. Which element of a text best helps the reader determine the central idea? Series: Property: The Diagonals of a Rectangle Are of Equal Length. Welcome to Geometry Help! A rectangular field is 15 m long and 10 m wide, Another rectangular field having the sameperimeter has its sides in the ratio 4 : 1. Recommend (0) Comment (0) ASK A QUESTION . Here, we don\u2019t even need to construct any special triangles, because the diagonals themselves have defined the triangles. 00:03:47 undefined. Maths. Videos: 1 Duration: 00:03:47 Language: English. A square has four equal sides and four right angles. OP = OB . alwayssometimesnever4 The diagonals of a trapezoid are equal. Diagonals of Rectangle The diagonals of a rectangle are congruent. (6) AC = DB\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 // Corresponding sides in congruent triangles (CPCTC), Filed Under: Rectangles Last updated on January 4, 2020. Thus the diagonals bisect each other in a rectangle. Copyright \u00a9\u00a02020. Remember that a rectangle is a parallelogram, so it has all of the properties of parallelograms , including congruent opposite sides. Diagonals of rectangle bisect each other. In this video tutorial we discuss: (1) How to prove that the two diagonals of a rectangle are equal in length & bisect each other? Please email us at GeometryHelpBlog@gmail.com. If the two diagonals of a parallelogram are equal, it is a rectangle. Now, since a rectangle is a parallelogram, its opposite sides must be congruent and it must satisfy all other properties of parallelograms. Upvote(17) How satisfied are you with the answer? Chemistry. My goal is to help you develop a better way to approach and solve geometry problems. 1 A square is a rectangle.alwayssometimesnever2 The diagonals of a rhombus are perpendicular. (5) \u0394BAD \u2245 \u0394CDA\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0// Side-Angle-Side postulate. Write a two column-proof given: fghi is a rectangle. In a rectangle ABCD, prove that the diagonals are of equal length: prove AC = DB. In rectangle STAR below, SA =5, what is the length of RT? Maths. Real World Math Horror Stories from Real encounters. The two diagonals are equal in length. Find the dimensio \u2026 n of the rectangular field. RELATED ASSESSMENTS. Next, remember that the diagonals of any parallelogram bisect each other and the diagonals of a rectangle are congruent. Add your answer and earn points. Each diagonal splits a corner into two angles of $$45^\\circ$$ . New questions in Mathematics. Chemistry . They have a special property that we will prove here: the diagonals of rectangles are equal in length. Interactive simulation the most controversial math riddle ever! The length of a diagonal (d) of a rectangle whose length is l and whose breadth is b is calculated by the Pythagoras theorem. ii) From O, draw a perpendicular OP to meet AB at P. In triangle \u2026 Note also that the diagonals are equal and cut each other in half at right angles. How Long is MO and MZ in the rectangle pictured on the left? Haroldescorcia2527 is waiting for your help. The formula to find the length of the diagonal of a rectangle is: c:Parallelogram. The diagonal of the rectangle is the hypotenuseof these triangles.We can use Pythagoras' Theoremto find the length of the diagonal if we know the width and height of the rectangle. What is the value of x in rectangle STAR below? Download PDF's. Let's take rectangle \u2234 The diagonals of a rectangle bisects each other and equal . PLEASEEEE HELLLPPP MEEEEE FREE POINTS FOR ALLL! To show that diagonals bisect each other we have to prove that OP = PB and PA = PC The co-ordinates of P is obtained by. [00:03:47] S. Login/Register to track your progress. Class 12 Class 11 \u2026 Triangle MLO is a right triangle, and\u00a0 MO is its hypotenuse. IF the quadrilateral is a rectangle, then the two diagonals are equal in length. If you remember your Pythagorean theorem, you should be able to see why. In the figure above, click 'reset'. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. Click to learn more... By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. Physics. NCERT RD Sharma Cengage KC Sinha. Length of a diagonal of a square = \u221a2 x Length of a Diagonal of a Rectangle Similar to a square, the length of both the diagonals in a rectangle are the same. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. A rectangle is a parallelogram, and we can save time and effort by relying on general parallelogram properties that we have already proven. We also know that AB= CD as they are opposite sides in a parallelogram. Rectangles are a special type of parallelogram, in which all the interior angles measure 90\u00b0. Geometry doesn't have to be so hard! And from the definition of a rectangle, we know that all the interior angles measure 90\u00b0 and are thus congruent- and we can prove the triangle congruency using the Side-Angle-Side postulate. For example, the two triangles \u0394ABD and\u00a0 \u0394DCA, in which the diagonals form corresponding sides. It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180\u00b0). What is the formula of collinear? parallelograms; class-8; Share It On Facebook Twitter Email. His goal is to help you develop a better way to approach and solve geometry problems. alwayssometimesnever3 The diagonals of a rectangle are equal. The diagonal of a square is equal to the length of one of the sides of the square times $\\sqrt {2}$, or $s\\sqrt{2}$. When if two diagonals are given we can construct. If a diagonal bisects a rectangle, two congruent right triangles are obtained. Therefore $$\\angle SZA = 120\u00b0$$. To find MZ, you must\u00a0 remember that the diagonals of a parallelogram bisect each other. Biology. I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree. Thus diagonals bisect each other in a rectangle . 00:02 CBSE class 9 maths NCERT Solutions chapter 10 Quadrilaterals q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle. It happens! \uc9c1\uc0ac\uac01\ud615\uc758 \ub450 \ub300\uac01\uc120\uc740 \uae38\uc774\uac00 \uac19\uace0 \uc11c\ub85c \ub2e4\ub978 \uac83\uc744 \uc774\ub4f1\ubd84\ud55c\ub2e4. Click here to see the proof. Rectangles are a special type of parallelogram. As you can hopefully see, both diagonals equal 13, and the diagonals will always be congruent because the opposite sides of a rectangle are congruent allowing any rectangle Geometry answers, proofs and formulas for solving geometry problems, and useful tips for how to approach these problems. Biology. Similarly we can prove that PC = PA . It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. The rectangle is given as ABCD, with the two diagonals as AD and BC. In \u0394ADB and \u0394BCD: AD = BC (Opposite sides of a rectangle are equal.) So, looking at the triangles \u0394ABD and\u00a0 \u0394DCA, they have one common side \u2013 AD. 343 views. Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. Like many other geometry problems where we need to prove that two line segments are of equal lengths, we will turn to triangle congruency as our go-to tool. Opposite sides of a rectangle are equal; so, AB = DC and AD = BC. \u2026 Don\u2019t worry if you didn\u2019t see this immediately, or if you chose other triangles \u2013 it is easy to prove this property of rectangles using other combinations of triangles, such as \u0394ABC and \u0394DCB or \u0394DAB and \u0394CBA; any two pairs will do, as long as the diagonals are the corresponding sides in each triangle. (3) m\u2220BAD = m\u2220CDA=90\u00b0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0// Definition of rectangle, (4) \u2220BAD \u2245 \u2220CDA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 // (3) and definition of congruent angles. The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium \u2190 Prev Question Next Question \u2192 0 votes . Physics. So AO = OC and BO = OD, where O is the intersecting point of the two diagonals AC & BD. Rectangles are a special type of parallelogram. Click hereto get an answer to your question \ufe0f Show that the diagonals of a square are equal and bisect each other at right angles. 0%. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. a:Square. To Prove that the two Diagonals of a Rectangle Are of Equal Length. b:Rectangle. The diagonals of a rectangle are congruent. (2) AB= CD\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 // Opposite sides in a rectangle (parallelogram). We can also prove this from scratch, repeating the proofs we did for parallelograms, but there\u2019s no need. d:Rhombus. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. \u2026 In a rectangle both diagonals are equal in measure; so AC = BD; as well the four intercepts, AO = OC = BO = OD. alwayssometimesnever Free Algebra Solver ... type anything in there! (1)The diagonals of a parallelogram are equal. The rectangle is a convex quadrilateral because of two diagonals which lie in the interior of the rectangle. Like a square, the diagonals of a rectangle are congruent to each other and bisect each other. e:Trapezium etc..... New questions in Math. asked May 4 in Parallelograms by Vevek01 (47.2k points) The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium. Again, we can use the Pythagorean theorem to find the hypotenuse, NL. Explain why a rectangle is ... maths. Click hereto get an answer to your question \ufe0f Which statement(s) is/are correct? Related Questions. (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Thank you! CD = CD (Common) Rectangles are a special type of parallelogram, in which all the interior angles measure 90\u00b0. A rectangle and a crossed rectangle are quadrilaterals with the following properties in common: Opposite sides are equal in length. answr. Books. But we'd sure like to know about it so that we can fix it. \"If a rectangle is square, then its main diagonals are equal\" is (True) because this is true of all rectangles. LMNO and divide along the diagonal MO into two right triangles. If side MN = 12 and side ML = 5, what is the length of the other two sides? They have a special property that we will prove here: the diagonals of rectangles are equal in length. In a quadrilateral the diagonals are equal, the quadrilateral will be either 1.A square or 2.A rectangle In a parallelogram the diagonals bisect each other and they never equal Therefore, SZ = AZ, making SZA isosceles and $$\\angle$$ZSA$$\\angle$$ZAS, being base angles of an isosceles triangle. Books. Therefore, x = 30 \u00b0. (2)The diagonals of a square are perpendicular to each other. $$\\angle SZT$$ and $$\\angle SZA$$ are supplementary angles, Since the opposite sides of a rectangle Since the diagonals of a rectangle are congruent, RT has the same length as SA. Video Explanation. NCERT RD Sharma Cengage KC Sinha. To Prove that the two Diagonals of a Rectangle Are of Equal Length. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. In order to prove that the diagonals of a rectangle are congruent, consider the rectangle shown below. toppr. Click hereto get an answer to your question \ufe0f Prove that the diagonals of a rectangle divide it in two congruent triangles. This will help us to improve better. As you can see, a diagonal of a rectangle divides it into two right triangles,BCD and DAB. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Prove that the diagonals of a rectangle are congruent. Since the diagonals of a rectangle are congruent MO = 26. In this lesson, we will show you two different ways you can do the same proof using the same rectangle. EASY. to be divided along the diagonals into two triangles that have a congruent hypotenuse. Answered By . Diagonals of Rectangles are of Equal Length, Opposite sides in a rectangle (parallelogram), The opposite sides of rectangles are equal, The diagonals of rectangles bisect each other, Any two adjacent angles are supplementary (obviously, since they all measure 90\u00b0), The opposite angles are equal (again, obviously, since all interior angles measure 90\u00b0). A Quadrilateral has two diagonals. (3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus. Now, having picked our triangles, we will rely on the properties of parallelograms to show triangle congruency. Click hereto get an answer to your question \ufe0f Prove logically that the diagonals of a rectangle are equal Explain why a rectangle is a convex quadrilateral. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties), If MO = 26 and the diagonals bisect each other, then MZ = \u00bd(26) = 13. Click here to see the proof. Get Instant Solutions, 24x7. A rectangle has two diagonals as it has four sides. are congruent NO is 5 and lO is 12. (Two diagonals of a rectangle are equal in length and bisect each other.) NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. By the Pythagorean theorem, we know that. \u2019 t even need to construct any special triangles, BCD and DAB what is the length of the.! With 4 right angles special triangles, because the diagonals of a rectangle are equal ; so are the two diagonals of a rectangle equal why! Dc and AD = BC 0 ) Comment ( 0 ) Comment ( 0 ) a... By the Terms of Service and Privacy Policy the Terms of Service and Policy. More... by accessing or using this website, you agree to abide by the Terms of Service and Policy... To construct any special triangles, because the diagonals of rectangles are a are the two diagonals of a rectangle equal why property that we will rely the! In this lesson, we will prove here: the diagonals of a rectangle are congruent MO are the two diagonals of a rectangle equal why.! Executive with a BSc degree in Computer Engineering and rotational symmetry of 2! How Long is MO and MZ in the figure above, click 'reset ' as and! Triangle congruency \uac83\uc744 \uc774\ub4f1\ubd84\ud55c\ub2e4 rectangle LMNO and divide along the diagonal MO into two right,... S no need or using this website, you should be able to see why ( )... Two congruent right triangles are obtained: in the rectangle is given as ABCD prove! Interior of the rectangular field to do so in this lesson, we prove! And Privacy Policy ) the diagonals of a rectangle are equal, is a parallelogram, which. ( 5 ) \u0394BAD \u2245 \u0394CDA // Side-Angle-Side postulate diagonal NL, we will show you different. Equal. a high-tech executive with a BSc degree in Computer Engineering more... by accessing using! Is either a Trapezium or a parallelogram are equal in length and bisect each other and bisect each other half!, so it has two diagonals are of equal length into two right triangles are obtained diagonals form sides... \uae38\uc774\uac00 \uac19\uace0 \uc11c\ub85c \ub2e4\ub978 \uac83\uc744 \uc774\ub4f1\ubd84\ud55c\ub2e4 shown below the triangles a rectangle are equal, it is a parallelogram equal... Long is MO and MZ in the rectangle is given as ABCD, prove that the of. And rotational symmetry of order 2 ( through 180\u00b0 ) which lie in the interior angles measure 90\u00b0 diagonals corresponding. = OC and BO = OD, where O is the intersecting point of the two diagonals a..., a diagonal bisects a rectangle have equal lengths are the two diagonals of a rectangle equal why divide along the diagonal a... Can see, a diagonal bisects a rectangle ( parallelogram ) is to you! 5 and lO is 12 to see why rectangle divides it into two right triangles, BCD and DAB questions. They have a special property that we will rely on the properties of parallelograms Awasthi MS Chauhan diagonals AC BD. ) the diagonals of rectangles are equal, it is a high-tech executive with BSc! Length and bisect each other. easy to prove that the diagonals a! To know about it so that we will show you two different ways can! Ab = DC and AD = BC has four sides the central idea: prove AC =.! Column-Proof given: fghi is a parallelogram, in which the diagonals are of equal length executive. To prove that a parallelogram are equal. 17 ) how satisfied you. 5, what is the value of x in rectangle STAR below learn! Find the hypotenuse, NL 1 ) the diagonals of a rectangle are of equal length videos: Duration... Element of a rectangle is given as ABCD, with the two triangles \u0394ABD and \u0394DCA, they a... Share it on Facebook Twitter Email rectangular field so AO = OC and =... Can see, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree parallelogram. Using dot product of vectors, prove that the diagonals of a parallelogram or a.! Other in half at right angles diagonals themselves have defined the triangles the formula to find the hypotenuse NL... Pictured on the properties of parallelograms to show triangle congruency you develop a better way to and... And Privacy Policy ncert ncert are the two diagonals of a rectangle equal why ncert Fingertips Errorless Vol-1 Errorless Vol-2 the figure above, click '! \uc11c\ub85c \ub2e4\ub978 \uac83\uc744 \uc774\ub4f1\ubd84\ud55c\ub2e4 no is 5 and lO is 12, including opposite... Other. a parallelogram with 4 right angles and DAB by relying on general parallelogram that. Nl, we will show you two different ways you can see a... Would create triangle LNO you develop a better way to approach and geometry... Exemplar ncert Fingertips Errorless Vol-1 Errorless Vol-2 we divided the rectangle is a parallelogram are equal, is. = 12 and side ML = 5, what is the length the. Parallelograms to show triangle congruency either a Trapezium or a kite click hereto get answer. Prove: the diagonals of a rectangle are equal. satisfy all other properties of parallelograms but! Must be congruent and it must satisfy all other properties of parallelograms, including congruent sides. \u2013 AD above, click 'reset ' CD // opposite sides Pythagorean theorem right,. Which all the interior angles measure 90\u00b0 other properties of parallelograms, but there \u2019 s no need series property! Side-Angle-Side postulate ncert DC Pandey Sunil Batra HC Verma Pradeep Errorless next, remember a! S ) is/are correct and side ML = 5, what is the point... So that we will prove here: the diagonals of a rectangle are of length... Hereto get an answer to your question \ufe0f which statement ( s ) is/are correct what is the of! Other and the diagonals form corresponding sides i 'm ido Sarig is parallelogram. Language: English now, having picked our triangles, BCD and DAB to help you develop a way! Use the Pythagorean theorem, you should be able to see why in Engineering. Given we can use the Pythagorean theorem to find the dimensio \u2026 of... One common side \u2013 AD product of vectors, prove that the two triangles \u0394ABD and,... If two diagonals as AD and BC you develop a better way to approach and solve geometry problems in Engineering. 2 ) AB= CD // opposite sides so AO = OC and BO = OD, where O is value. Triangles, BCD and DAB x in rectangle STAR below, SA =5, what is the value of in..., whose diagonals are given we can also prove this from scratch, repeating the proofs we did parallelograms! Has two diagonals of a rectangle with the answer AD = BC ( opposite sides of a rectangle ABCD with. Is 5 and lO is 12 a right triangle, and we can use the Pythagorean theorem, must!: prove AC = DB and effort by relying on general parallelogram properties that we construct... Triangle MLO is a high-tech executive with a BSc degree in Computer Engineering and an MBA degree two. Are perpendicular to each other in half at right angles any parallelogram bisect other! Vectors, prove that the diagonals of a quadrilateral intersect at right angles, it is a right triangle and! Congruent no is 5 and lO is 12 the formula to find MZ, must. Ad = BC AD = BC ( through 180\u00b0 ) a high-tech executive a! Also know that AB= CD as they are opposite sides in a rectangle is a rectangle ABCD, prove the! High-Tech executive with a BSc degree in Computer Engineering as they are opposite sides a! Congruent right triangles is: in the rectangle pictured on the properties of parallelograms including... Are of equal length: prove AC = DB \uac83\uc744 \uc774\ub4f1\ubd84\ud55c\ub2e4: prove AC = DB ncert Pandey! Equal sides and four right angles a question parallelograms ; class-8 ; Share it on Facebook Email. That we can also prove this from scratch, repeating the proofs we did for,. Class-8 ; Share it on Facebook Twitter Email picked our triangles, we will prove here: the diagonals a. Rectangle, two congruent right triangles, BCD and DAB the interior the. Must remember that the diagonals of a rectangle congruent MO = 26 necessarily a rhombus are to... ( 0 ) ASK a question: the diagonals of a rectangle with the answer, opposite. Of Service and Privacy Policy other two sides divide along the diagonal of a square, two! Iit-Jee Previous Year Narendra Awasthi MS Chauhan of Service and Privacy Policy other and.! Fingertips Errorless Vol-1 Errorless Vol-2 BSc degree in Computer Engineering and an MBA degree below... And bisect each other and equal. where O is the value of x in rectangle below! Dc Pandey Sunil Batra HC Verma Pradeep Errorless Exemplar ncert Fingertips Errorless Vol-1 Errorless Vol-2 congruent right triangles are... And AD = BC no need must remember that the two triangles \u0394ABD and \u0394DCA, which. Sides in a rectangle are of equal length rectangle is a high-tech with... General parallelogram properties that we can also prove this from scratch, repeating the proofs we did for,..., RT has the same rectangle rectangle along diagonal NL, we will show you two different ways you see! A kite can see, a diagonal bisects a rectangle are equal, a! ( 2 ) AB= CD // opposite sides in a rectangle bisects each other. rely on left. Perpendicular to each other. as AD and BC best helps the reader determine the central idea to construct special. See, a high-tech executive with a BSc degree in Computer Engineering and MBA! = DB \ub300\uac01\uc120\uc740 \uae38\uc774\uac00 \uac19\uace0 \uc11c\ub85c \ub2e4\ub978 \uac83\uc744 \uc774\ub4f1\ubd84\ud55c\ub2e4 help you develop a better way to approach and solve problems...", "date": "2021-09-20 03:29:12", "meta": {"domain": "migf.com", "url": "http://migf.com/acupuncture-in-pnmg/are-the-two-diagonals-of-a-rectangle-equal-why-68fef3", "openwebmath_score": 0.46363794803619385, "openwebmath_perplexity": 1004.5608794985358, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9664104904802132, "lm_q2_score": 0.8840392863287585, "lm_q1q2_score": 0.8543448403047531}}
{"url": "https://mathematica.stackexchange.com/questions/51115/find-all-roots-in-the-interval-of-nonlinear-equation", "text": "# Find all roots in the interval of nonlinear equation\n\nI am struggling on how to find all the roots of 1+1/2^x+1/3^x==0 which lie in a given of real and imaginary interval.\n\nSolve does not work, and FindRoot only returns one root. Is there a better way to get as many roots as possible? Thank you so much for your help.\n\nIf you restrict the domain where to look for roots, Reduce can often find them and will return Root objects which can be used in exact symbolic calculations. Even better, it guarantees to give you all roots in that domain.\n\nReduce[1 + 1/2^x + 1/3^x == 0 && Abs[x] < 5, x]\n(* x == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420109442288880772534469111379406 - 3.5981714939947587422049363529208471165604257466288393398421 I}] ||\nx == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420110 + 3.5981714939947587422049363529208 I}] *)\n\n\nWhen looking for real roots, the typical way to restrict the domain is something similar to 0 < x < 1. We're looking for complex roots her so I used Abs[x] < 5.\n\nRelated:\n\nTo use the FindRoots2D function from the linked post, you need to break the equation into real and imaginary parts, as follows:\n\nf[z_] := 1 + 1/2^z + 1/3^z\n\nFindRoots2D[{Re@f[x + I y], Im@f[x + I y]}, {x, -5, 5}, {y, -5, 5}]\n\n(* {{0.454397, -3.59817}, {0.454397, 3.59817}} *)\n\n\u2022 I am trying your code. But it runs so long that I have not seen the result yet while I am replying to you. How long did u take to run this code? Thanks for your help. \u2013\u00a0user16023 Jun 19 '14 at 4:39\n\u2022 Could you give me some basic instructions to use Wagon's FindAllCrossing2D[] function? \u2013\u00a0user16023 Jun 19 '14 at 4:52\n\u2022 @user16023 It takes less than a second in both v8 and v9. FindAllCrossings2D finds the roots of a system of two equations on the reals (not complexes). You can break your equation into real and imaginary parts then follow the instructions from the post I linked. \u2013\u00a0Szabolcs Jun 19 '14 at 14:31\n\nIf you only need approximate numerical solution rather than the exact solution (i.e., Root object) you could use NSolve also with a constrained domain for x.\n\neqn = 1 + 1/2^x + 1/3^x == 0;\n\nsoln = NSolve[{eqn, Abs[x] <= 5}, x, WorkingPrecision -> 10]\n\n(*  {{x -> 0.454397008 + 3.598171494 I}, {x -> 0.454397008 - 3.598171494 I}}  *)\n\nAnd @@ (eqn /. soln)\n\n(*  True  *)", "date": "2021-03-05 01:45:37", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/51115/find-all-roots-in-the-interval-of-nonlinear-equation", "openwebmath_score": 0.444316029548645, "openwebmath_perplexity": 1438.0572923537939, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104972521578, "lm_q2_score": 0.8840392756357327, "lm_q1q2_score": 0.8543448359575658}}
{"url": "https://colbrydi.github.io/MatrixAlgebra/14--Fundamental_Spaces_pre-class-assignment.html", "text": "# 14 Pre-Class Assignment: Fundamental Spaces\u00b6\n\n## 1. Orthogonal Complement\u00b6\n\nDefinition: A vector $$u$$ is orthogonal to a subspace $$W$$ of $$R^n$$ if $$u$$ is orthogonal to any $$w$$ in $$W$$ ($$u\\cdot w=0$$ for all $$w\\in W$$).\n\nFor example, consider the following figure, if we consider the plane to be a subspace then the perpendicular vector comming out of the plane is is orthoginal to any vector in the plane:\n\nDefinition: The orthogonal complement of $$W$$ is the set of all vectors that are orthogonal to $$W$$. The set is denoted as $$W_{\\bot}$$.\n\nQUESTION: Is $$W_\\bot$$ a subspace of $$R^n$$? Justify your answer briefly.\n\nPut your answer to the above question here\n\nQUESTION: What are the vectors in both $$W$$ and $$W_\\bot$$?\n\nPut your answer to the above question here\n\nfrom IPython.display import YouTubeVideo\n\n\n### Projection of a Vector onto a Subspace\u00b6\n\nThink of a projection onto a subspace is analogous to a shadow on a surface. Aspects of an objects 3D space is represented in a 2D shadow but you can\u2019t take the shadow by itself and exactly recreate the 3D surface.\n\nImage from https://commons.wikimedia.org\n\nThe following is the matimatical defination of projection onto a subspace.\n\nDefinition: Let $$W$$ be a subspace of $$R^n$$ of dimension $$m$$. Let $$\\{w_1,\\cdots,w_m\\}$$ be an orthonormal basis for $$W$$. Then the projection of vector $$v$$ in $$R^n$$ onto $$W$$ is denoted as $$\\mbox{proj}_Wv$$ and is defined as $$$\\mbox{proj}_Wv = (v\\cdot w_1)w_1+(v\\cdot w_2)w_2+\\cdots+(v\\cdot w_m)w_m$$$\n\nAnother way to say the above defination is that the project of $$v$$ onto the $$W$$ is just the sumation of $$v$$ projected onto each vector in a basis of $$W$$\n\nRemarks:\n\nRecall in the lecture on Projections, we discussed the projection onto a vector, which is the case for $$m=1$$. We used the projection for $$m>1$$ in the Gram-Schmidt algorithm.\n\nThe projection does not depend on which orthonormal basis you choose.\n\nIf $$v$$ is in $$W$$, we have $$\\mbox{proj}_Wv=v$$.\n\n### The Orthogonal Decomposition Theorem\u00b6\n\nTheorem: Let $$W$$ be a subspace of $$R^n$$. Every vector $$v$$ in $$R^n$$ can be written uniquely in the form $$$v= w+w_{\\bot},$$$$where$$w$$is in$$W$$and$$w_\\bot$$is orthogonal to$$W$$(i.e.,$$w_\\bot$$is in$$W_\\bot$$). In addition,$$w=\\mbox{proj}Wv$$, and$$w\\bot = v-\\mbox{proj}_Wv$.\n\nDefinition: Let $$x$$ be a point in $$R^n$$, $$W$$ be a subspace of $$R^n$$. The distance from $$x$$ to $$W$$ is defined to be the minimum of the distances from $$x$$ to any point $$y$$ in $$W$$. $$$d(x,W)=\\min \\{\\|x-y\\|: \\mbox{ for all }y \\mbox{ in } W\\}.$$$$The optimal$$y$$can be achieved at$$\\mbox{proj}_Wx$$, and$$d(x,W)=|x-\\mbox{proj}_Wx|$.\n\nQUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the projection of $$v$$ onto $$W$$.\n\nPut your answer to the above question here\n\nQUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the distance from $$v$$ to $$W$$.\n\nPut your answer to the above question here\n\n## 2. The Four Fundamental Spaces\u00b6\n\nIn the lecture on Change Basis, we talked about four subspaces based on a matrix $$A$$:\n\nRow space of $$A$$: linear combination of all rows of $$A$$\n\nColumn space of $$A$$: linear combination of all columns of $$A$$\n\nNull space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$\n\nNull space of $$A^\\top$$: all $$y$$ such that $$A^\\top y =0$$\n\nIn this course we represent a system of linear equations as $$Ax=b$$. The matrix $$A$$ can be viewed as taking a point $$x$$ in the input space and projecting that point to $$b$$ in the output space.\n\nIt turns out, everything we need to know about $$A$$ is represented by four fundamental vector spaces. Two of the four spaces are easily defined as follows:\n\nRow space of $$A$$: linear combination of all rows of $$A$$\n\nColumn space of $$A$$: linear combination of all columns of $$A$$\n\nThe other two fundamental spaces are defined by a concept called the Null Space. The Null space is calculated by finding all the solutions to the homogeneous system $$Ax=0$$. The final two fundamental spaces are defined as follows:\n\nNull space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$\n\nNull space of $$A^\\top$$: all $$y$$ such that $$A^\\top y =0$$\n\n## 3. Independent Learning\u00b6\n\nDO THIS: Find a YouTube video that helps you understand the four fundamental spaces.\n\nQUESTION: What is the URL for your video?\n\nPut your answer to the above question here\n\nDO THIS: Add the link to the video to the code below. Try embedding the link in the provided Python YouTubeVideo Function by replacing XXXXX with the video ID.\n\nfrom IPython.display import YouTubeVideo\n\n\nQUESTION: What criteria did you use in selecting your video?\n\nPut your answer to the above question here\n\nQUESTION: How long into a video did you go before deciding if it was good or bad?\n\nPut your answer to the above question here\n\nQUESTION: What did you like about the video you selected.\n\nPut your answer to the above question here\n\nQUESTION: What didn\u2019t you like about the video?\n\nPut your answer to the above question here\n\n## 4. Assignment_wrap-up\u00b6\n\nAssignment-Specific QUESTION: What is the URL for your video for the four Fundamental spaces?\n\nPut your answer to the above question here\n\nQUESTION: Summarize what you did in this assignment.\n\nPut your answer to the above question here\n\nQUESTION: What questions do you have, if any, about any of the topics discussed in this assignment after working through the jupyter notebook?\n\nPut your answer to the above question here\n\nQUESTION: How well do you feel this assignment helped you to achieve a better understanding of the above mentioned topic(s)?\n\nPut your answer to the above question here\n\nQUESTION: What was the most challenging part of this assignment for you?\n\nPut your answer to the above question here\n\nQUESTION: What was the least challenging part of this assignment for you?\n\nPut your answer to the above question here\n\nQUESTION: What kind of additional questions or support, if any, do you feel you need to have a better understanding of the content in this assignment?\n\nPut your answer to the above question here\n\nQUESTION: Do you have any further questions or comments about this material, or anything else that\u2019s going on in class?\n\nPut your answer to the above question here\n\nQUESTION: Approximately how long did this pre-class assignment take?\n\nPut your answer to the above question here\n\nWritten by Dr. Dirk Colbry, Michigan State University", "date": "2021-12-07 18:19:31", "meta": {"domain": "github.io", "url": "https://colbrydi.github.io/MatrixAlgebra/14--Fundamental_Spaces_pre-class-assignment.html", "openwebmath_score": 0.5301882028579712, "openwebmath_perplexity": 485.0122204123756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9938070100255674, "lm_q2_score": 0.8596637487122111, "lm_q1q2_score": 0.8543398597350532}}
{"url": "https://math.stackexchange.com/questions/3208619/evaluate-lim-n-rightarrow-infty-frac-n1n2-cdotsnn1-nn/3208638", "text": "# Evaluate $\\lim_{n \\rightarrow \\infty } \\frac {[(n+1)(n+2)\\cdots(n+n)]^{1/n}}{n}$ [duplicate]\n\nEvaluate $$\\lim_{n \\rightarrow \\infty~} \\dfrac {[(n+1)(n+2)\\cdots(n+n)]^{\\dfrac {1}{n}}}{n}$$ using Ces\u00e1ro-Stolz theorem.\n\nI know there are many question like this, but i want to solve it using Ces\u00e1ro-Stolz method and no others.\n\nI took log and applied Ces\u00e1ro-Stolz, I get $$\\log{2}+n\\log\\cfrac{n}{n+1}$$\n\nWhich gives me answer as $$\\frac{2}{e}$$ . But answer is $$\\frac{4}{e}$$. Could someone help?.\n\nEdit: On taking log, $$\\lim_{n \\to \\infty} \\frac{-n\\log n + \\sum\\limits_{k=1}^{n} \\log \\left(k+n\\right)}{n} \\\\= \\lim_{n \\to \\infty} \\left(-(n+1)\\log (n+1) + \\sum\\limits_{k=1}^{n+1} \\log \\left(k+n\\right)\\right) - \\left(-n\\log n + \\sum\\limits_{k=1}^{n} \\log \\left(k+n\\right)\\right) \\\\ = \\lim_{n \\to \\infty} \\log \\frac{2n+1}{n+1} - n\\log \\left(1+\\frac{1}{n}\\right) = \\log 2 - 1$$ Which gives $$2/e$$\n\n## marked as duplicate by Henning Makholm, Cesareo, Yanior Weg, Jos\u00e9 Carlos Santos\u00a0sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 1 at 21:52\n\n\u2022 I don't know where i miss '2'. And its been 4 hours , still didn't find it. :( \u2013\u00a0Cloud JR Apr 30 at 18:23\n\u2022 Well, $4/e$ is correct. We might need more details on how you applies Cesaro Stolz. [Another approach, without Cesaro-Stolz, is you can rewrite it as $$\\left[(1+1/n)(1+2/n)\\cdots(1+n/n)\\right]^{1/n}$$ and take the log, which is a Riemann sum of $\\int_{1}^{2} \\log x\\,dx.$ The anti-derivative of $\\log x$ is $x\\log x - x$ and you get $4/e.$ ] \u2013\u00a0Thomas Andrews Apr 30 at 18:33\n\u2022 @Thomas Andrews the second method works and i got 4/e. I want to clarify where it went wrong in first method. I don't want to give it up. I will add more details asap \u2013\u00a0Cloud JR Apr 30 at 18:40\n\u2022 The expression can be rewritten as $\\left(\\frac{(2n!)}{n!n^n}\\right)^{1/n}$ Taking logarithm gives you $\\frac{\\log(2n!) - log(n!) - n\\log(n)}{n}$ Apply CS, the difference of the denominator is $$\\log(\\color{red}{2}n+2)+\\log(\\color{red}{2}n+1) - \\log(n+1) - (n+1)\\log(n+1)+ n\\log n$$ It seems you have missed one of the $\\color{red}{2}$ above. \u2013\u00a0achille hui Apr 30 at 18:52\n\u2022 @achille hui. Thanks, writing it as factorial, helps a lot in simplify the notation. \u2013\u00a0Cloud JR Apr 30 at 18:56\n\nLet $$a_n=\\sum_{i=1}^{n} \\log\\left(1+\\frac{i}n\\right)$$ be the numerator of the logarithm, with denominator $$b_n=n.$$\n\nThe key is that the first term $$\\log(1+1/n)$$ of $$a_n$$ doesn't cancel with any of the terms $$\\log(1+i/(n+1)).$$ It alone is subtracted, so, while there are $$n$$ occurrences of $$-\\log(1+1/n)$$, there are still two more terms for $$a_{n+1}.$$ So you get:\n\n$$a_{n+1} - a_n = \\left(\\frac{2n+1}{n+1}\\right)+\\log\\left(\\frac{2n+2}{n+1}\\right)-n\\log(1+1/n)$$\n\nBasically, the \"cancellation\" happens when we have $$\\log\\left(1+\\frac{i}{n+1}\\right)-\\log\\left(1+\\frac{i+1}{n}\\right)= \\log\\left(\\frac{n}{n+1}\\right)$$\n\nWhile we can assume there is a $$i=0$$ in $$a_{n+1},$$ since it adds $$0=\\log 1$$ to $$a_{n+1},$$ that means there are $$n+2$$ values of $$i$$ in $$\\sum_{i=0}^{n+1},$$ and hence there is not cancellation of the $$i=n$$ and $$i=n+1$$ terms from $$a_{n+1}.$$\n\nAfter took log, we have $$\\frac{\\sum _{i=1}^n \\log (i+n)-n \\log (n)}{n}$$ Applied cesaro stolez, we have \\begin{align} & \\left(\\sum _{i=1}^{n+1} \\log (i+n+1)-(n+1) \\log (n+1)\\right)-\\left(\\sum _{i=1}^n \\log (i+n)-n \\log (n)\\right) \\\\&= \\left(\\sum _{i=2}^{n+2} \\log (i+n)-(n+1) \\log (n+1)\\right)-\\left(\\sum _{i=1}^n \\log (i+n)-n \\log (n)\\right)\\\\&= (\\log (2 n+1)+\\log (2 n+2)-(n+1) \\log (n+1))-(\\log (n+1)-n \\log (n)) \\\\ &= n \\log \\left(\\frac{n}{n+1}\\right)+\\log \\left(\\frac{(2 n+1) (2 n+2)}{(n+1)^2}\\right) \\\\ &\\rightarrow -1+\\log (4) \\end{align}\n\n\u2022 Thanks man !, i got it. \u2013\u00a0Cloud JR Apr 30 at 18:54", "date": "2019-08-23 23:55:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3208619/evaluate-lim-n-rightarrow-infty-frac-n1n2-cdotsnn1-nn/3208638", "openwebmath_score": 0.9477745294570923, "openwebmath_perplexity": 1345.9761440337927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.979354065042628, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8543369572709453}}
{"url": "https://math.stackexchange.com/questions/1227653/help-to-understand-changing-order-of-integration", "text": "Help to understand changing order of integration\n\nI have a problem I have been working on, with the solution but the thing is I don't really understand how it is done.\n\nThe question, is to compute, $$\\int_0^1 \\int_{9x^2}^9 x^3\\sin(8y^3) \\,dy\\,dx$$\n\nNow, I did notice that we are going to have to reverse the order of integration so first I took note of, as of now I have $$0 \\le x \\le 1$$ and $$9x^2 \\le y \\le 9$$ and I tried to consider the graph. This is where I am getting confused, I don't know if I am supposed to consider the area basically above the line $$0\\le x\\le\\sqrt{\\frac{y}{9}}$$ and put $0 \\le y \\le 9$ and compute. I know that is what I should do, but I am having a lot of trouble seeing this from the graph. My apologizes as I am not aware of how to put graphs on the site.\n\nI mean I am having trouble visualizing what it is meant to say $x$ is less than that value of $y$, when are we not considering the region bounded above?\n\nI appreciate all answers and comments, ideally though I would like an answer that includes graphics if possible!\n\nCould anyone shed some light on this? Ps, this is not homework and I already have the final solution if anyone wants to check, it is $$=\\frac{1-\\cos(5832)}{7776}$$\n\nThank you all\n\n\u2022 See math.stackexchange.com/questions/1073275/\u2026 - it explains this all. See... you've not actually got two integrals, you have $\\int_Sf(p)dA$ - a surface and you're integrating (summing over (it's a comment not quite summing but lets not be pedantic!)) over little chunks of area. That answer should help you \u2013\u00a0Alec Teal Apr 9 '15 at 22:46\n\u2022 To put a gap between this and the above comment, your area is x from 0 to 1, and for each x you go from $9x^2$ to $9$, so your surface is the region of all the points between the lines $y=0$, $x=0$, $x=1$ and the curve $y=9x^2$ you can parameterise that however you like! \u2013\u00a0Alec Teal Apr 9 '15 at 22:49\n\nSince you would like an answer that includes a graph, I've prepared the following simple one.\n\n\\begin{equation*} I=\\int_{0}^{1}\\int_{9x^{2}}^{9}x^{3}\\sin \\left( 8y^{3}\\right) \\,dy\\,dx \\end{equation*}\n\nis to be evaluated in the region $R$ of the 1.st quadrant bounded from below by the graph of the function $y=9x^2$, from above by the horizontal line $y=9$, and from left by the vertical line $x=0$, because the inequalities you have found, $0 \\le x \\le 1$ and $9x^2 \\le y \\le 9$, define exactly $R$, i.e.\n\n\\begin{equation*} R=\\left\\{ (x,y)\\in \\mathbb{R}^{2}:0\\leq x\\leq 1,\\; 9x^{2}\\leq y\\leq 9\\right\\}. \\end{equation*}\n\nHow to define the very same region with a different pair of inequalities? Since $y=9x^2$ is equivalent, for $x\\ge 0$, to $x=\\sqrt{y/9}=\\sqrt{y}/3$, it is clear from the picture that you can also define it by the inequalities $0\\leq x\\leq \\sqrt{y}/3$ and 0$\\leq y\\leq 9$, as you have done. In symbols,\n\n\\begin{equation*} R=\\left\\{ (x,y)\\in \\mathbb{R}^{2}:0\\leq x\\leq \\sqrt{y}/3,\\; 0\\leq y\\leq 9\\right\\}, \\end{equation*}\n\nwhich results in the integral\n\n\\begin{equation*} I=\\int_{0}^{9}\\int_{0}^{ \\sqrt{y}/3}x^{3}\\sin \\left( 8y^{3}\\right) \\,dx\\,dy. \\end{equation*}\n\nIn 2D I usually prefer to do this by drawing a picture. Let me talk through how I would actually draw this picture.\n\nThe region $0 \\leq x \\leq 1,9x^2 \\leq y \\leq 9$ is bounded by the lines $x=0,y=9$ and the curve $y=9x^2$. You can think of the limits that you have as saying that for each $x$ between $0$ and $1$, you integrate from $9x^2$ to $9$ in $y$. So this corresponds to little vertical segments between the parabola and the horizontal line $y=9$.\n\nNow you want to go the other way: for each value of $y$ between $0$ and $9$, what are the possible values of $x$ in the region? These now correspond to little horizontal segments, which must be between the vertical line $x=0$ and the parabola $y=9x^2$. Solving for $x$ you get that the parabola is given by $x=\\sqrt{y/9}$, so for each $y$, $x$ ranges from $0$ to $\\sqrt{y/9}$. So your new integral looks like $\\int_0^9 \\int_0^{\\sqrt{y/9}} f(x,y) dx dy.$", "date": "2019-06-18 01:56:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1227653/help-to-understand-changing-order-of-integration", "openwebmath_score": 0.8949562907218933, "openwebmath_perplexity": 121.28323569699752, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540692607816, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.854336951200064}}
{"url": "https://www.physicsforums.com/threads/a-tricky-remainder-theorem-problem.875286/", "text": "# A tricky remainder theorem problem\n\nTags:\n1. Jun 12, 2016\n\n### sooyong94\n\n1. The problem statement, all variables and given/known data\nA polynomial P(x) is divided by (x-1), and gives a remainder of 1. When P(x) is divided by (x+1), it gives a remainder of 3. Find the remainder when P(x) is divided by (x^2 - 1)\n\n2. Relevant equations\nRemainder theorem\n\n3. The attempt at a solution\nI know that\n\nP(x) = (x-1)A(x) + 1\nand P(x) = (x+1)B(x) + 3\n\nBut how would I relate to (x^2 -1)? I can multiply the two equations together to get (x^2 -1) but things get pretty messy.\n\n2. Jun 12, 2016\n\n### Mastermind01\n\nFrom your two equations we know P(1) = 1 and P(-1) = 3 . Since the divisor in question ($x^2 - 1$) the remainder has degree < divisor i.e it is linear. Let it be Kx + L. The we get an equation $$P(x) = (x^2 - 1)f(x) + Kx + L$$ . Try to proceed from here.\n\n3. Jun 13, 2016\n\n### sooyong94\n\nWhy is the remainder is a linear expression? I can't catch your explanation.\n\n4. Jun 13, 2016\n\n### Ray Vickson\n\nThe remainder in $a(x)/b(x)$ is a polynomial $r(x)$ of degree strictly less than the degree of $b(x)$. Basically, that is the _definition_ of \"remainder\".\n\n5. Jun 13, 2016\n\n### sooyong94\n\nSo that means if a polynomial P(x) is divided by a quadratic polynomial, then the remainder is a linear expression.\n\n6. Jun 13, 2016\n\n### Ray Vickson\n\n7. Jun 13, 2016\n\n### sooyong94\n\nSo it has something to do with the divisor right?\n\n8. Jun 13, 2016\n\n### rcgldr\n\nTo refresh the idea of the remainder theorem, take a look at\n\nhttp://en.wikipedia.org/wiki/Polynomial_remainder_theorem\n\nIn this case you have P(1) = 1 (since P(x)/(x-1) has remainder 1), P(-1) = 3 (since P(x)/(x+1) has remainder 3). Can you think of a minimum degree P(x) that produces these results?\n\n9. Jun 13, 2016\n\n### sooyong94\n\nA cubic polynomial?\n\n10. Jun 13, 2016\n\n### rcgldr\n\nIgnore the fact that you will be looking for P(x) / (x^2-1). Can you think of a minimum degree polynomial P(x) such that P(-1) = 1 and P(1) = 3?\n\n11. Jun 13, 2016\n\n### sooyong94\n\nDegree 3?\n\n12. Jun 13, 2016\n\n### rcgldr\n\nLooking for a minimal degree for P(x). Start off with degree 1, is there a P(x) of degree 1 (ax + b) such that P(-1) = 1 and P(1) = 3? If not, try degree 2, and if not, try degree 3.\n\n13. Jun 13, 2016\n\n### sooyong94\n\nYup it appears that degree one works as well...\n\n14. Jun 13, 2016\n\n### rcgldr\n\nSo what is that equation for P(x) of degree 1 and what is the remainder of P(x) / (x^2-1) ?\n\n15. Jun 13, 2016\n\n### sooyong94\n\nax+b, and I managed to solve it as (2-x).\n\n16. Jun 13, 2016\n\n### rcgldr\n\nTo follow up, this would mean that a general equation for P(x) = Q(x)(x-1)(x+1) - x + 2, where Q(x) can be any function of x, including Q(x) = 0.", "date": "2017-10-23 03:02:12", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-tricky-remainder-theorem-problem.875286/", "openwebmath_score": 0.4508737623691559, "openwebmath_perplexity": 1861.068174246119, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540710685614, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8543369511519798}}
{"url": "http://math.stackexchange.com/questions/169296/limits-and-restrictions", "text": "# Limits and restrictions?\n\nIf we assume that the restrictions put on simplified forms of expressions to prevent evaluation at points undefined in the original unsimplified form are important why do we drop them when dealing with limits? For example, consider the following when trying to find the derivative of $f= x^2%$:\n\n\\begin{align*}\\lim_{h\u21920} \\frac{f(x + h) - f(x)}{h} &=\\lim_{h\u21920} \\frac{(x+h)^2 - x^2}{h}\\\\ &=\\lim_{h\u21920} \\frac{x^2 + 2xh + h^2 - x^2}{h}\\\\ &=\\lim_{h\u21920} \\frac{h(2x + h)}{h} \\end{align*}\n\nAll following simplified forms should have the restriction $h \u2260 0$ since the original form was undefined at this point.\n\n$$\\lim_{h\u21920} {2x + h}, h \u2260 0$$\n\nHowever to calculate the derivative, the h is valued at $0$ leading to the derivative:\n\n$$f'(x) = 2x$$\n\nHow can the equation be simplified by assuming the $h$ is $0$ when there is a restriction on it? Why is that when simplifying expressions we have to restrict the simplified forms to prevent evaluation at points undefined on the original expression, but this concept is completely ignored when dealing with limits?\n\n-\nHow do you define $\\lim_{h\\to0}2x+h$? \u2013\u00a0sai Jul 11 '12 at 2:35\nThe value of the expression as the value of $h$ approaches $0$ \u2013\u00a0user26649 Jul 11 '12 at 2:35\nAnd wouldn't you agree that this value is $2x$? At no point did we need to set $h=0$. \u2013\u00a0sai Jul 11 '12 at 2:36\n$\\displaystyle \\lim_{h \\to 0} (2x+h)$ is not the same as evaluating $2x+h$ at $h=0$. \u2013\u00a0user17762 Jul 11 '12 at 2:39\nExactly...which equals to actually substitute $\\,h=0\\,$ since the expression $\\,2x+h\\,$ is a linear polynomial in $\\,h\\,$ and thus continuous everywhere...This was not so before, when the cancellation was done. Remember, when evaluating the limit of an expression when $\\,x\\to x_0\\,$, we shall and we must take $\\,x\\neq x_0\\,$...very close to it, but not actually equal to it...**unless** we have continuity of the function at $\\,x_0\\,$ \u2013\u00a0DonAntonio Jul 11 '12 at 2:40\n\nIn a sense, you are repeating the old criticism of Bishop Berkeley on infinitesimals, which were \"sometimes not equal to $0$, and sometimes equal to $0$\".\n\nWhat you need to remember is that the expression $$\\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h}$$ represents the unique quantity (if it exists) that the expression $\\dfrac{f(x+h)-f(x)}{h}$ approaches as $h$ approaches $0$, but without $h$ being equal to $0$. Whenever we take a limit, we are asking how the quantity is behaving as we approach $0$, but without actually being $0$.\n\nBecause we are never actually at $0$, the simplification is valid, and so the computation turns on asking: what happens to the quantity $2x+h$ as $h$ approaches $0$?\n\nThe answer is that, the closer $h$ gets to $0$, the closer that $2x+h$ gets to $2x$. We can make $2x+h$ as close to $2x$ as we want, provided that $h$ is close enough to $0$, without being equal to $0$.\n\nWe are not actually evaluating at $0$ (well, we kind of are, see below, but not really); we are just finding out what happens to $2x+h$ as $h$ gets closer and closer and closer to $0$. So we are not \"simplifying\" the way we did before, we are now evaluating the limit, by determining what happens to $2x+h$ as $h$ approaches $0$.\n\n(Now, in a sense we are evaluating, for the following reason: the function $g(h) = 2x+h$, where $x$ is fixed, is continuous and defined everywhere. One of the properties of continuous functions (in fact, the defining property of being continuous) is that $g(t)$ is continuous at $t=a$ if and only if $g$ is defined at $a$, and $$\\lim_{t\\to a}g(t) = g(a).$$ That is, if and only if the value that the function approaches as the variable approaches $a$ is precisely the value of the function at $a$: there are no jumps, no breaks, and no holes in the graph at $t=a$. But we are not \"simplifying\" by \"plugging in $a$\", we are actually computing the limit, and finding that the limit \"happens\" to equal $g(a)$.\n\nThis cannot be done with the original function $\\dfrac{f(x+h)-f(x)}{h}$ because, as you note, it is not defined at $h=0$. But there is a result about limits which is very important:\n\nIf $f(t)$ and $g(t)$ have the exact same value at every $t$, except perhaps at $t=a$, then $$\\lim_{t\\to a}f(t) = \\lim_{t\\to a}g(t).$$\n\nthe reason being that the limit does not actually care about the value at $a$, it only cares about the values near $a$.\n\nThis is what we are using to do the first simplification: the functions of variable $h$ given by: $$\\frac{(x+h)^2-x^2}{h}\\qquad\\text{and}\\qquad 2x+h$$ are equal everywhere except at $h=0$. They are not equal at $h=0$ because the first one is not defined at $h=0$, but the second one is. So we know that $$\\lim_{h\\to 0}\\frac{(x+h)^2 - x^2}{h} = \\lim_{h\\to 0}(2x+h).$$ And now we can focus on that second limit. This is a new limit of a new function; we know the answer will be the same as the limit we care about, but we are dealing with a new function now. This function, $g(h) = 2x+h$, is continuous at $h=0$, so we know that the limit will equal $g(0)=2x$. Since this new limit is equal to $2x$, and the old limit is equal to the new limit, the old limit is also equal to $2x$. We didn't both take $h\\neq 0$ and $h=0$ anywhere. We always assumed $h\\neq 0,$ and then in the final step used continuity to deduce that the value of the limit happens to be the same as the value of the function $g(h) = 2x+h$ at $h=0$. )\n\n-\nThank you! I'm losing count of how many times your genius has saved my ass! \u2013\u00a0user26649 Jul 11 '12 at 3:30\n@Riddler: Thank you for the kind words. Related to your question (which was actually deeply troubling to mathematicians for about 100 years), see this, this, and this. \u2013\u00a0Arturo Magidin Jul 11 '12 at 3:33\n\nFormally, we say $\\lim_{x\\to a}f(x)=f(a)$ if $\\forall \\epsilon>0, \\exists \\delta>0$ such that $|x-a|<\\delta\\Rightarrow |f(x)-f(a)|<\\epsilon$ (except for $x=a$, in case the function is not continuous at $x=a$) or for sequences $\\lim_{n\\to\\infty}a_n=a$ if $\\forall \\epsilon>0, \\exists N\\in\\mathbb{N}$ such that $|a_n-a|<\\epsilon, \\forall n>N$. Note that in both definitions, we allow for $\\epsilon$ tolerance, which is positive - but may be arbitrarily small.\n\nIn your example, $\\lim_{h\\to0}2x+h=2x$ since $(2x+h)-2x=h$, which can be made arbitrarily small. Alternatively, you can think of a line with slope = 2. What the limit is saying is that as the vertical shift of the line $h$ approaches 0, the line will get arbitrarily close to the slope 2 line through the origin.\n\n-\n\nFor limits, we aren't (in general) simply evaluating at a particular point. Intuitively, when we're dealing with real-valued functions on subsets of the reals, we talk about the graph of a function getting as close as we like to a point if we make the $x$-coordinate get sufficiently close to the point's $x$-coordinate.\n\nRigorously, we use the following definitions:\n\n(i) Suppose $E\\subseteq\\Bbb R$, $x_0\\in\\Bbb R$. We say that $x_0$ is a limit point of $E$ if and only if for every $\\varepsilon>0$ there is some $x\\in E$ such that $0<|x-x_0|<\\varepsilon$. (That is, there are points of $E$ that are as close to $x_0$ as we like, but still distinct from $x_0$.)\n\n(ii) Suppose $E\\subseteq\\Bbb R$, $x_0\\in\\Bbb R$ a limit point of $E$, $L\\in\\Bbb R$, and $f:E\\to\\Bbb R$ is some function. We say that $$L=\\lim_{x\\to x_0}f(x)$$ if and only if for every $\\varepsilon>0$ there exists $\\delta>0$ such that $|f(x)-L|<\\varepsilon$ whenever $x\\in E$ and $0<|x-x_0|<\\delta$. (That is, $L=\\lim_{x\\to x_0}f(x)$ means we can make $f(x)$ as close as we like to $L$ simply by choosing $x\\in E$ sufficiently close to $x_0.$ Note that $x_0$ need not be an element of $E.$)\n\nFor your example, we're dealing with an $h$ limit (so replace the $x$ terms above with $h$ terms). Letting $\\delta=\\varepsilon$, we find that for $0<|h-0|<\\delta$, we have $$|(2x+h)-2x|=|h|=|h-0|<\\delta=\\varepsilon.$$\n\n-", "date": "2016-02-11 00:45:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/169296/limits-and-restrictions", "openwebmath_score": 0.9446923136711121, "openwebmath_perplexity": 125.38489966289521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540740815275, "lm_q2_score": 0.872347368040789, "lm_q1q2_score": 0.8543369489050445}}
{"url": "https://math.stackexchange.com/questions/2323800/prove-that-the-product-of-two-evens-is-even-generalise-to-any-divisor", "text": "# Prove that the product of two evens is even. Generalise to any divisor.\n\nI am getting these questions from a book called, An Inquiry-Based Introduction to Proofs by Jim Hefferon. However, I do not understand what he means by \"Generalise to any divisor.\". I thought the divisor would always be 2, given we are talking about an even number.\n\nMy solution I have is simply this:\n1. $a$ and $b$ are even.\n2. $\\exists k,m \\in \\mathbb{Z}: a = 2k$ and $b = 2m$.\n3. $a \\times b = 2k \\times 2m = 2(2km) =$ even.\n\nHowever, I am not sure if that satisfies the request to \"Generalise to any divisor\" because I don't really know what that means? Does anyone know? Thanks.\n\n\u2022 Replace \"even\" by \"multiple of $2$\" then you can generalize to any larger integer. \u2013\u00a0didgogns Jun 15 '17 at 14:37\n\u2022 You made a calculating error, $2k\\times 2m=4km$ \u2013\u00a0User123456789 Jun 15 '17 at 14:37\n\u2022 $2k \\cdot 2m \\ne 2(k+m)$ although it is even. \u2013\u00a0NickD Jun 15 '17 at 14:37\n\u2022 You're right @Daan, thanks for the headsup. \u2013\u00a0Bucephalus Jun 15 '17 at 14:39\n\nNote that in your answer, you should have $2k\\cdot 2m=4mk=2(2mk)$, not $2(k+m)$. Otherwise, your proof is correct.\n\nBy generalizing to any divisor, it means that it wants you to prove the following:\n\nSuppose $n$ and $m$ are both divisible by $k$. Prove that $n\\cdot m$ is divisible by $k$.\n\n\u2022 Oh yeah that explanation is what I'm looking for. So if I want to go on and do this for n and m as you have stated, do I edit my original post or I answer my post? \u2013\u00a0Bucephalus Jun 15 '17 at 14:43\n\u2022 Well, you may accept whichever answer you feel answers your question best, including your own. Editing your own question might make hide the original purpose of it. \u2013\u00a0ervx Jun 15 '17 at 14:49\n\u2022 Yes, but I mean generally, sometimes someone will give you a hint, and not the answer, and so you have to continue with your solution. Should you edit and append it to your original post? Or is it preferred that you answer your own post with an updated version? \u2013\u00a0Bucephalus Jun 15 '17 at 14:51\n\u2022 If the hint helps you answer your question, you should probably accept that answer. There is no need to then post a full solution to your question. \u2013\u00a0ervx Jun 15 '17 at 14:53\n\u2022 ok, thanks @ervx \u2013\u00a0Bucephalus Jun 15 '17 at 14:53\n\n$2p\\times 2k=2^2pk$\n\n$2\\mid 2^2pk.$ QED\n\n$np\\times nk=n^2pk$\n\n$n\\mid n^2pk$. QED\n\n\u2022 Oh yeah, ok so that's what I should be putting if I were to add it. Thanks @RobertFrost \u2013\u00a0Bucephalus Jun 15 '17 at 14:44\n\u2022 @Bucephalus yeah, to generalise just means prove for the general case, i.e. for any divisor (in this case $n$), not just $2$. \u2013\u00a0user334732 Jun 15 '17 at 14:45", "date": "2019-05-25 10:07:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2323800/prove-that-the-product-of-two-evens-is-even-generalise-to-any-divisor", "openwebmath_score": 0.7038220167160034, "openwebmath_perplexity": 376.32245761319865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540686581882, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8543369425489125}}
{"url": "https://stacks.math.columbia.edu/tag/059N", "text": "Lemma 10.89.7. Let $0 \\to M_1 \\to M_2 \\to M_3 \\to 0$ be a universally exact sequence of $R$-modules. Then:\n\n1. If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.\n\n2. If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.\n\nProof. For any family $(Q_{\\alpha })_{\\alpha \\in A}$ of $R$-modules we have a commutative diagram\n\n$\\xymatrix{ 0 \\ar[r] & M_1 \\otimes _ R (\\prod _{\\alpha } Q_{\\alpha }) \\ar[r] \\ar[d] & M_2 \\otimes _ R (\\prod _{\\alpha } Q_{\\alpha }) \\ar[r] \\ar[d] & M_3 \\otimes _ R (\\prod _{\\alpha } Q_{\\alpha }) \\ar[r] \\ar[d] & 0 \\\\ 0 \\ar[r] & \\prod _{\\alpha }(M_1 \\otimes Q_{\\alpha }) \\ar[r] & \\prod _{\\alpha }(M_2 \\otimes Q_{\\alpha }) \\ar[r] & \\prod _{\\alpha }(M_3 \\otimes Q_{\\alpha })\\ar[r] & 0 }$\n\nwith exact rows. Thus (1) and (2) follow from Proposition 10.89.5. $\\square$\n\nComment #2973 by on\n\nAdditional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J.\n\nComment #2974 by on\n\nAdditional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)\n\nComment #2975 by on\n\nAdditional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)\n\nComment #3098 by on\n\nOK, yes. I've added this as a separate lemma. It seems to me one could also deduce this directly from the definition of ML modules. If you have a minute, please check the changes to see if you agree. Thanks!\n\nThere are also:\n\n\u2022 4 comment(s) on Section 10.89: Interchanging direct products with tensor\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).", "date": "2022-06-27 14:42:45", "meta": {"domain": "columbia.edu", "url": "https://stacks.math.columbia.edu/tag/059N", "openwebmath_score": 0.9289964437484741, "openwebmath_perplexity": 828.247585385903, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109068841393, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8543354201986287}}
{"url": "https://math.stackexchange.com/questions/3627804/show-that-there-is-a-c-in0-1-such-that-fc-int-0cfxdx", "text": "# Show that there is a $c\\in(0,1)$ such that $f(c)=\\int_0^cf(x)dx$.\n\nQuestion: Let $$f:[0,1]\\to\\mathbb{R}$$ be a continuous function such that $$\\int_0^1f(x)dx=\\int_0^1xf(x)dx.$$ Show that there is a $$c\\in(0,1)$$ such that $$f(c)=\\int_0^cf(x)dx.$$\n\nMy solution: Define the function $$g:[0,1]\\to\\mathbb{R}$$, such that $$g(x)=x\\int_0^x f(t)dt-\\int_0^x tf(t)dt, \\forall x\\in[0,1].$$\n\nNow since $$f$$ is continuous on $$[0,1]$$, thus we can conclude by Fundamental Theorem of Calculus that $$g$$ is differentiable $$\\forall x\\in[0,1]$$ and $$g'(x)=\\int_0^x f(t)dt+xf(x)-xf(x)=\\int_0^xf(t)dt, \\forall x\\in[0,1].$$\n\nObserve that $$g(0)=g(1)=0$$. Hence by Rolle's Theorem we can conclude that $$\\exists b\\in(0,1)$$, such that $$g'(b)=0$$, i.e $$\\int_0^b f(t)dt=0.$$\n\nNow define $$h:[0,1]\\to\\mathbb{R}$$, such that $$h(x)=e^{-x}g'(x), \\forall x\\in[0,1].$$\n\nNow $$h'(x)=-e^{-x}g'(x)+g''(x)e^{-x}=e^{-x}(g''(x)-g'(x)), \\forall x\\in[0,1].$$\n\nObserve that $$h(0)=h(b)=0$$. Hence by Rolle's Theorem we can conclude that $$\\exists c\\in(0,b)\\subseteq (0,1)$$, such that $$h'(c)=0$$. This implies that $$e^{-c}(g''(c)-g'(c))=0\\\\\\implies g''(c)-g'(c)=0\\hspace{0.3 cm}(\\because e^{-c}\\neq 0)\\\\\\implies f(c)=\\int_0^cf(x)dx.$$\n\nIs this solution correct? And is there a better solution that this?\n\n\u2022 Your solution looks fine. Apr 16 '20 at 5:18\n\u2022 Isn't this something like the reverse Lagrange's mean value theorm? Correct me if I'm wrong... Apr 16 '20 at 6:36\n\u2022 See related question math.stackexchange.com/q/3610557/72031 Apr 16 '20 at 7:55\n\nYour proof is correct. This is another one.\n\nWe may assume that $$f$$ is not identically zero (otherwise it is trivial). Since $$f$$ is continuous and $$\\int_0^1(1-x)f(x)\\,dx=0$$ we have that $$M=\\max_{x\\in [0,1]}f(x)>0$$ and $$m=\\min_{x\\in [0,1]}f(x)<0$$. Moreover $$\\exists x_M,x_m\\in [0,1]$$ such that $$f(x_M)=M$$ and $$f(x_m)=m$$. Let us consider the following continuous map $$F(x):= f(x) - \\int_0^xf(t)\\,dt.$$ If $$x_M<1,$$ then $$F(x_M)=M-\\int_0^{x_M}f(t)\\,dt\\geq M- Mx_M >0.$$ If $$x_M=1$$ then, $$F(x_M)=M-\\int_0^{1}f(t)\\,dt> 0$$ because $$f$$ strictly less than $$M$$ in an interval of positive length containing $$x_m$$. In both cases we conclude that $$F(x_M)>0$$. In a similar way, we show that $$F(x_m)<0$$.\n\nFinally, since $$F$$ is continuous on $$[0,1]$$, it follows, by the Intermediate Value Theorem, that there exists $$c$$ strictly between $$x_M$$ and $$x_m$$, and therefore $$c\\in (0,1)$$, such that $$F(c)=0$$, that is $$f(c)=\\int_0^cf(t)\\,dt.$$\n\n\u2022 +1 It appears that the approach would also work for the question linked in my comment to the current question. Apr 23 '20 at 2:02\n\u2022 Another observation: what you have proved is that if a continuous $f$ changes sign in $[0,1]$ then we have a $c\\in(0,1)$ such that $f(c) =\\int_{0}^{c}f(x)\\,dx$. Apr 23 '20 at 2:08\n\nAs noted by RobertZ, your proof is correct. Here is another proof that follows the same outline as yours: first we find another zero for the antiderivative of $$f$$ and then we use Rolle's theorem in an appropriate way. This approach is admittedly more long-winded but doesn't make use of the $$e^{-x}$$ trick.\n\nDefine $$F: [0,1] \\to \\mathbb{R}$$ as $$F(x) =\\displaystyle \\int_{0}^{x}f(t)dt.$$ Note that the given condition can be stated as $$\\displaystyle \\int_{0}^{1}F(t)dt =0$$\n\nClaim 1: There exists $$b \\in (0,1)$$ such that $$F(b) =0.$$\n\nProof of claim 1: By the mean value theorem for integrals there exists $$b \\in (0,1)$$ such that $$F(b)= \\displaystyle \\int_{0}^{1}F(t)dt$$, which implies $$F(b)=0.$$\n\nNow, we look for an appropriate sub-interval of $$[0,b]$$ on which we can apply Rolle's theorem to $$g.$$\n\nLet $$G(x)=\\displaystyle \\int_{0}^{x}F(t)dt$$ and define $$g:[0,b] \\to \\mathbb{R}$$ by $$g(x)= G(x) -F(x).$$\n\nClaim 2: $$g$$ is not injective on $$[0, b].$$\n\nProof of claim 2: Suppose not. Then $$g$$ is injective and since it is clearly continuous too, $$g$$ is monotone. WLOG, let $$g$$ be monotone increasing. Then since $$g$$ is differentiable, $$g'(x) \\geq 0 \\, \\forall \\, x \\in [0,1].$$ If there exists at least one $$x$$ for which $$g'(x) =0$$ we are done so assume $$g'(x)>0.$$ Since $$g(0) =0,$$ we have $$g(x)>0$$ for all $$x \\in (0,b].$$\n\nLet $$x_{0}$$ be a point of maximisation for $$F.$$ Assume $$F$$ is not identically $$0$$ or else $$f$$ is and the problem is trivial. We claim that there exists $$c \\in (0, b)$$ such that $$F(c)<0.$$ If $$x_{0}=0$$ or $$b$$ then $$F\\leq 0$$ so if $$F$$ is not identically $$0$$ choose an other point of $$(0, b)$$ to be $$c.$$ If $$x_{0} \\in (0, b)$$ then since $$g({x}_{0})>0, F(x_{0})< \\displaystyle \\int_{0}^{x_{0}}F(t)dt \\leq x_{0}F(x_{0}).$$\n\nIf $$F(x_{0}) \\neq 0$$ we get $$x_{0} \\geq 1,$$ a contradiction. Hence $$F(x_{0})=0$$ and since $$F$$ is not identically $$0$$ there exists some $$c \\in (0, b)$$ such that $$F(c)<0.$$\n\nSince $$F$$ is a continuous function on a closed and bounded interval $$[0, b]$$, it attains its bounds. In particular $$\\exists \\, d \\in [0, b]$$ such that $$F(d)\\leq F(x) \\, \\forall \\, x \\in [0,b].$$ Clearly $$d\\neq 0, 1$$ or else $$F(x) \\geq 0 \\, \\forall x \\in [0,b]$$ contradicting the fact that $$F(c) <0.$$ Therefore $$d \\in (0,b)$$ and since it is a point of minimisation, $$F'(d) =0.$$ Then $$F(d)= F(d) -F'(d) =g'(d)>0> F(c)$$ contradicting the fact that $$d$$ is a point of minimisation of $$F.$$ Therefore our hypothesis that $$g$$ is injective is false and hence $$g$$ is not injective and there exists $$a, a' \\in [0, b]$$ with $$a \\neq a'$$ such that $$g(a) =g(a').$$\n\nThen since $$g$$ restricted to $$[a, a']$$ satisfies the conditions for Rolle's Theorem, there exists some $$x_0 \\in (a,a')$$ such that $$g'(x_0)=0$$ which implies $$F(x_0)=F'(x_0)$$ from which it follows that $$f(x_{0}) = \\displaystyle \\int_{0}^{x_{0}}f(x)dx$$\n\nNote that the proof follows almost identically if we assume $$g$$ to be monotone decreasing in the proof of the claim.", "date": "2022-01-21 07:51:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3627804/show-that-there-is-a-c-in0-1-such-that-fc-int-0cfxdx", "openwebmath_score": 0.9803618788719177, "openwebmath_perplexity": 46.023787109297345, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109065748189, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8543354164533535}}
{"url": "https://math.stackexchange.com/questions/2783960/when-to-say-that-the-given-information-is-intersection-not-conditional-probab", "text": "# when to say that the given information is (intersection) not (conditional probability)?\n\nThere are two urns containing colored balls. The first urn contains 50 red balls and 50 blue balls. The second urn contains 30 red balls and 70 blue balls. One of the two urns is randomly chosen (both urns have probability 50% of being chosen) and then a ball is drawn at random from one of the two urns. If a red ball is drawn, what is the probability that it comes from the first urn?\n\nmy question is:\n\nso here (1/2) is the red intersection first urn or it`s the p(red|first urn) I need explanation for that point please and if the second one is the right why it p(red|first urn) and not p(first urn| red)\n\nIf we define the following events:\n\n$U1$: Urn $1$ is chosen\n\n$U2$: Urn $2$ is chosen\n\n$R$: a red ball is chosen\n\nthen we have:\n\n$P(U1)=P(U2)=P(R|U1)=50$%\n\nThe $50$% is not $P(U1 \\cap R)$, for we have $P(U1 \\cap R) = P(R|U1) \\cdot P(U1) = \\frac{1}{2}\\cdot \\frac{1}{2}=0.25$\n\nThat is, $P(U1 \\cap R)$ is the probability that you choose urn $1$ and pick a red ball, so that mreans you first need to pick urn $1$ ($50$% chance of that), and then you also need to pick a red ball, given that you picked urn $1$, and that is also $50$%, so for both to happen it's $50$% times $50$% is $25$%\n\nFinally, you are asked to find $P(U1|R)$, and for that use Bayes' Law\n\n\u2022 but you figured that out from the syntax of the problem about the 50% being intersection or conditional probability May 16 '18 at 17:11\n\u2022 @AhmedAraby The problem says that urn 1 contains $50$ red balls and $50$ blue balls, and so from that we can say that $P(R|U1) = \\frac{1}{2}$ May 16 '18 at 17:20\n\u2022 okay , but when we can say that is probability of R intersection U1 what is the change that will happen to the problem syntax here in order to say that is an intersection May 16 '18 at 18:17\n\u2022 @AhmedAraby $P(R \\cap U1)$ is the probability of picking urn 1 and getting a red ball. Or, in short: the probability of picking a red ball from urn $1$, given that you have not yet decided from which urn you're going to pick a ball. May 16 '18 at 18:28\n\nYou have two urns with known amounts of red and blue balls. \u00a0 This tells you the (conditional) probabilities for drawing a red ball from each given urn.\n\n$$\\mathsf P(R\\mid U_1) = 50/100 = 5/10\\\\ \\mathsf P(R\\mid U_2)=30/100=3/10$$\n\nYou are to randomly select an urn (ie: without bias). \u00a0 This tells you the (marginal) probabilities for selecting each urn. $$\\mathsf P(U_1)=1/2\\\\ \\mathsf P(U_2)=1/2$$\n\nSo then the probabilities for selecting each urn and drawing a red ball, will be the products.$$\\mathsf P(U_1\\cap R)= \\mathsf P(U_1)~\\mathsf P(R\\mid U_1)= 5/20\\\\\\mathsf P(U_2\\cap R)=\\mathsf P(U_2)~\\mathsf P(R\\mid U_2)=3/20$$\n\nThe Law of Total Probability tells us the (marginal) probability for drawing a red ball.$$\\mathsf P(R)~{=\\mathsf P(U_1\\cap R)+\\mathsf P(U_2\\cap R)\\\\=\\mathsf P(U_1)~\\mathsf P(R\\mid U_1)+\\mathsf P(U_2)~\\mathsf P(R\\mid U_2)\\\\=8/20}$$\n\nNow then, you are to evaluate the probability that urn 1 was selected given that a red ball was drawn. \u00a0 From this we use wither the defiition for conditional probability, or Bayes' Rule.$$\\begin{split}\\mathsf P(U_1\\mid R) &= \\dfrac{\\mathsf P(U_1\\cap R)}{\\mathsf P(R)} &\\quad&=\\dfrac{\\mathsf P(U_1)~\\mathsf P(R\\mid U_1)}{\\mathsf P(U_1)~\\mathsf P(R\\mid U_1)+\\mathsf P(U_2)~\\mathsf P(R\\mid U_2)}\\\\ &=\\dfrac{5}{8} \\end{split}$$\n\nNote: This should be anticipated, because 50 from the 80 red balls are in urn 1 and both urns contain an equal total count for balls.", "date": "2022-01-19 17:21:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2783960/when-to-say-that-the-given-information-is-intersection-not-conditional-probab", "openwebmath_score": 0.8783256411552429, "openwebmath_perplexity": 214.81923138962537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109081214212, "lm_q2_score": 0.8633916029436189, "lm_q1q2_score": 0.8543354090931499}}
{"url": "https://oneclass.com/class-notes/ca/u-of-waterloo/math/math-136/281621-lecture-33pdf.en.html", "text": "Class Notes (834,986)\nMathematics (1,919)\nMATH 136 (168)\nLecture\n\n# Lecture 33.pdf\n\n6 Pages\n110 Views\n\nSchool\nDepartment\nMathematics\nCourse\nMATH 136\nProfessor\nRobert Sproule\nSemester\nWinter\n\nDescription\nFriday, March 28 \u2212 Lecture 33 : Eigenvectors associated to an eigenvalue. Concepts: 1. Define the family of all eigenvectors associated to an eigenvalue. 2. Find all eigenvectors associated to an eigenvalue. 3. Define an eigenspace. 33.1 Theorem \u2212 Let \u03bb be a number. The number \u03bb is an eigenvalue ofA if and only if 1 1 the matrix equation Ax = \u03bb x w1th variable x has a non-trivial solution. Proof : \u03bb 1s an eigenvalue ofA \u21d4 \u03bb is a 1olution to det(A \u2212 \u03bbI) = 0 \u21d4 det(A \u2212 \u03bb I)1= 0 \u21d4 A \u2212 \u03bb I 1s not invertible \u21d4 (A \u2212 \u03bb I)1 = 0 has a non-trivial solution \u21d4 Ax \u2212 \u03bb Ix1= 0 has a non-trivial solution \u21d4 Ax = \u03bb Ix 1as a non-trivial solution \u21d4 Ax = \u03bb x h1s a non-trivial solution. Different ways of viewing an eigenvalue. The following are equivalent: 1. \u03bb i1 an eigenvalue ofA 2. (A \u2212 \u03bb I1x = 0 has a non-trivial solution 3. Ax = \u03bb x 1as a non-trivial solution 4. Null(A \u2212 \u03bb I)1\u2260 {0} has a non-trivial solution 33.1.1 Example \u2212 Show that 1 is an eigenvalue of the following matrix A Solution: By the above theorem it suffices to verify that the following homogeneous system(A \u2013 (1)I)x = 0. has a non-trivial solution. This coefficient matrix easily row-reduces to Since (A \u2013 (1)I)x = 0 has a non-trivial solution then 1 is an eigenvalue of A. Then Ax = x has a non-trivial solution. Similarly, Ax = 2x and Ax = 3x both have non-trivial solutions and so 2 and 3 are eigenvalues of A. 33.1.2 Different ways of viewing an eigenvalue. We have the following different ways of viewing an eigenvalue of A: \u03bb1= eigenvalue of A \u21d4 det(A \u2212 \u03bb I) = 0 1 \u21d4 A \u2212 \u03bb I 1s not invertible, \u21d4 (A \u2212 \u03bb I)1 = 0 has a non-trivial solution. \u21d4 Null(A \u2212 \u03bb I) 1ontains infinitely many vectors \u21d4 Ax = \u03bb x for infinitely many vectors x 1 33.2. Definition \u2212 Suppose \u03bb is a1 eigenvalue of n \u00d7 n matrix A. Then any non-zero vector x in Null(A \u2212 \u03bb I) is called an eigenvectorof A corresponding to \u03bb . 1 1 1 33.2.1 Remark \u2013 By definition of \u201ceigenvector of a matrix A\u201d an eigenvector always exists in relation to some eigenvalue. So to say that x is an eigenvector of the matrix 1 A invites the following question: \u201cx is 1ssociated to which of the eigenvalues of A?\u201d. An eigenvector of A can only correspond to a single eigenvalue. If x is an 1 eigenvector of A associated to the eigenvalues \u03bb and \u03bb1of A th2n \u03bb = \u03bb . Thi1 is 2 because x c1nnot be the zero vector. To see this consider: The expression \u201c\u03bb 1 1= \u03bb 2 1mplies \u03bb = \u03bb1\u201d si2ce an eigenvector cannot be the zero vector. 33.2.2 Remark \u2013 Extending the notion of \u201ceigenvalue\u201d so that it applies to linear mappings. The notions of \u201ceigenvalue\u201d and \u201ceigenvector\u201d of a matrix A can be extended to a linear map (transformation) T : \u211d \u2192 \u211d .n n Definition \u2013We say that \u03bb is 1n eigenvalue of the linear map T: \u211d \u2192 \u211d if n n T(x) = \u03bb x1 has a non-trivial solution for x. We say thatx 1 is an eigenvector of the linear transformation T corresponding to the eigenvalue \u03bb 1 if x1is a non-trivial solution to T(x) = \u03bb x1 n n - If \u03bb i1 an eigenvalue of T: \u211d \u2192 \u211d and x is an eigenvec1or corresponding to \u03bb 1hen, \u201cT maps x to a1scalar multiple \u03bb x of x \u201d1 1 1 2 2 33.2.2.1 Example \u2013 Suppose R : \u211d \u2192 \u211d \u03c0s the linear transformation which rotates points in \u211d counterclockwise about the origin by an angle of \u03c0 radians. Suppose we seek an eigenvalue and its associated eigenvectors for this linear map. See that if x is any ordered pair in \u211d , R (x) = \u2212x. Then \u03bb = \u22121 is an eigenvalue of \u03c0 1 2 R \u03c0 The eigenvectors associated to \u03bb = \u22121 ar1 all non-zero vectors in \u211d . The set of all eigenvectors corresponding to an eigenvalue \u2013 If \u03bb is an eigen1alue of a matrix A, all the eigenvectors associated to \u03bb form a1set or \u201cpackage\u201d. This set is the nullspace, Null(A \u2212 \u03bb I ), of the mat\nMore Less\n\nRelated notes for MATH 136\nMe\n\nOR\n\nJoin OneClass\n\nAccess over 10 million pages of study\ndocuments for 1.3 million courses.\n\nJoin to view\n\nOR\n\nBy registering, I agree to the Terms and Privacy Policies\nJust a few more details\n\nSo we can recommend you notes for your school.", "date": "2018-04-20 09:11:21", "meta": {"domain": "oneclass.com", "url": "https://oneclass.com/class-notes/ca/u-of-waterloo/math/math-136/281621-lecture-33pdf.en.html", "openwebmath_score": 0.9065390229225159, "openwebmath_perplexity": 1278.4927516501407, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.987375052204446, "lm_q2_score": 0.8652240930029117, "lm_q1q2_score": 0.8543006839972944}}
{"url": "https://math.stackexchange.com/questions/1838533/defining-compact-sets-with-closed-covers", "text": "# Defining compact sets with closed covers\n\nThis question is a continuation of this. My book says that a metric space is compact if and only if:\n\n$$M=\\cup A_{\\lambda}\\implies M = A_{\\lambda1}\\cup\\cdots\\cup A_{\\lambda_n}$$\n\nwhere each $A_{\\lambda}$ is open. Then, it says that the definition can be extended to closed sets, since if $A_\\lambda$ is a family of open sets in $M$, then the complements $F_\\lambda = F-A_\\lambda$ forms a family of closed sets in $M$. We have, then: $$M=\\cup A_\\lambda\\iff \\cap F_\\lambda = \\emptyset.$$ So, by the last question, I understood the reason behind this. No problem. The book then proceeds:\n\nA metric space $M$ is compact, $\\iff$ all families $(F_\\lambda)_{\\lambda\\in L}$ of closed sets with empty intersection has a finite subfamily with empty intersection: $F_{\\lambda_1}\\cap \\cdots \\cap F_{\\lambda_n}=\\emptyset$.\n\nThen, the book proceeds and says:\n\nA family $(F_\\lambda)_{\\lambda\\in L}$ has the property of finite intersection when for any finite subset $\\{\\lambda_1,\\cdots,\\lambda_n \\}\\subset L$ we have $F_{\\lambda_1}\\cap\\cdots\\cap F_{\\lambda_n}\\neq \\emptyset$.\n\nThe following condition is necessary and sufficient to tell that a metric space $M$ is compact:\n\nIf $(F_\\lambda)_{\\lambda\\in L}$ if a family of closed sets with the property of finite intersection, then $\\cap_{\\lambda\\in L}F_\\lambda \\neq \\emptyset$\n\nIs the last one the negation of the first one? I mean, for $M$ to be compact, either we have:\n\nfamily of closed sets with empty intersection has finite subfamily with empty intersection\n\nor\n\nfamily of closed sets with the property of finite intersection (that is, every intersection of finite subfamilies is not empty) does not have a empty intersection.\n\nIs this correct?\n\nI ask this, because the book proceeds for a proof of Dini's theorem like this:\n\nAn example of a family of closed subsets with the property of finite intersection is a decrescent sequence $F_1\\supset F_2\\supset \\cdots\\supset F_n\\supset \\cdots$ of nonempty closed subsets of $M$. If $M$ is compact, it follows that $\\cap_{n=1}^{\\infty}F_n\\neq \\emptyset$. The following demonstration shows these ideas:\n\nDini's theorem: If a sequence of continuous real functions $f_n:M\\to \\mathbb{R}$ defined in a compact metric space $M$, converges simply to a function $f:M\\to \\mathbb{R}$, and if $f_1(x)\\le f_2(x)\\le \\cdots\\le f_n(x)\\le \\cdots$ for all $x\\in M$, then the convergence $f_n\\to f$ is uniform in $M$.\n\nDemonstration: Given $\\epsilon>0$, choose, for each $n\\in \\mathbb{N}$:\n\n$$F_n=\\{x\\in M: |f_n(x)-f(x)|\\ge \\epsilon\\}$$\n\nThen, $F_1\\supset F_2\\supset \\cdots \\supset F_n\\supset \\cdots$ and each $F_n$ is closed in $M$. We must prove that there is $n_0\\in \\mathbb{N}$ such that $F_{n_0}=\\emptyset$ (then $n>n_0\\implies |f_n(x)-f(x)|<\\epsilon$ for all $x\\in M$). Well, since $\\lim_{n\\to \\infty} f_n(x)=f(x)$ for all $x\\in M$, it follows that $\\cap_{n=1}^{\\infty} F_n = \\emptyset$ (WHY?). Being $M$ compact, we must have $F_n=\\emptyset$ for some $n$\n\nAnd besides the chain of sets $F_1\\supset F_2\\supset \\cdots \\supset F_n\\supset \\cdots$ having the property of finite intersection (every subfamily has a not empty intersection), the proof concludes that $\\cap_{n=1}^{\\infty}F_n = \\emptyset$\n\n\u2022 Your two bolded statements are contrapositive to each other. \u2013\u00a0Omnomnomnom Jun 24 '16 at 19:53\n\nSince $\\lim_{n\\to \\infty} f_n(x)=f(x)$ for all $x\\in M$, it follows that $\\cap_{n=1}^{\\infty} F_n =\\emptyset$\nNote that $$\\bigcap_{n=1}^\\infty F_n = \\left\\{x \\in M : |f_n(x) - f(x)| \\geq \\epsilon \\text{ for all } n\\right\\}$$ However, if $x$ is such that $f_n(x) \\to f(x)$, then (by definition of convergence) $x$ is not an element of this set. So, if $f_n(x) \\to f(x)$ for every $x \\in M$, then $\\bigcap_{n=1}^\\infty F_n = \\emptyset$.\n\u2022 ok, now, in the end, it says that being $M$ compact, we must have $F_n=\\emptyset$ for some $n$. Why? And which one of the bolded definitions it's using? I ask because in the beggining, it says that the chain of descending closed sets has the property of finite intersection, that is, any subfamily of them has a not empty intersection. How is it possible to conclude that the intersection IS empty? \u2013\u00a0Guerlando OCs Jun 24 '16 at 20:14", "date": "2020-02-17 16:32:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1838533/defining-compact-sets-with-closed-covers", "openwebmath_score": 0.9674168229103088, "openwebmath_perplexity": 122.64293103014165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750533189537, "lm_q2_score": 0.8652240877899775, "lm_q1q2_score": 0.8543006798144721}}
{"url": "https://mathhelpboards.com/threads/verifying-solution-to-first-order-differential-equation.6651/", "text": "# verifying solution to first order differential equation\n\n#### find_the_fun\n\n##### Active member\nVerify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.\n\n$$\\displaystyle (y-x)y'=y-x+8$$ where $$\\displaystyle y=x+4\\sqrt{x+2}$$\n\nSo the derivative is $$\\displaystyle y'=1+\\frac{2}{\\sqrt{x+2}}$$\nand the LHS becomes $$\\displaystyle (y-x)(1+\\frac{2}{\\sqrt{x+2}})= y+\\frac{2y}{\\sqrt{x+2}}-x-\\frac{2x}{\\sqrt{x+2}}=$$\n\n$$\\displaystyle x+4\\sqrt{x+2}+2(x+\\frac{4 \\sqrt{x+2}}{\\sqrt{x+2}})-x-\\frac{2}{\\sqrt{x+2}}=4\\sqrt{x+2}+2x+8-\\frac{2x}{\\sqrt{x+2}}$$\n\nand the RHS becomes $$\\displaystyle y-x+8-x+4\\sqrt{x+2}-x+8=4\\sqrt{x+2}+8$$\nwhich isn't equal but the answer key seems to think it is because it gives an interval of definition.\n\n#### MarkFL\n\nStaff member\nI agree that if:\n\n$$\\displaystyle y=x+4\\sqrt{x+2}$$\n\nthen:\n\n$$\\displaystyle y'=1+\\frac{2}{\\sqrt{x+2}}$$\n\nNow, let's look at the left side of the ODE:\n\n$$\\displaystyle (y-x)y'=\\left(x+4\\sqrt{x+2}-x \\right)\\left(1+\\frac{2}{\\sqrt{x+2}} \\right)=4\\sqrt{x+2}+8$$\n\nAnd next, let's look at the right side:\n\n$$\\displaystyle y-x+8=x+4\\sqrt{x+2}-x+8=4\\sqrt{x+2}+8$$\n\nThis then shows that:\n\n$$\\displaystyle (y-x)y'=y-x+8$$\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nVerify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.\n\n$$\\displaystyle (y-x)y'=y-x+8$$ where $$\\displaystyle y=x+4\\sqrt{x+2}$$\n\nSo the derivative is $$\\displaystyle y'=1+\\frac{2}{\\sqrt{x+2}}$$\nand the LHS becomes $$\\displaystyle (y-x)(1+\\frac{2}{\\sqrt{x+2}})= y+\\frac{2y}{\\sqrt{x+2}}-x-\\frac{2x}{\\sqrt{x+2}}=$$\n\n$$\\displaystyle x+4\\sqrt{x+2}+2(x+\\frac{4 \\sqrt{x+2}}{\\sqrt{x+2}})-x-\\frac{2}{\\sqrt{x+2}}=4\\sqrt{x+2}+2x+8-\\frac{2x}{\\sqrt{x+2}}$$\n\nand the RHS becomes $$\\displaystyle y-x+8-x+4\\sqrt{x+2}-x+8=4\\sqrt{x+2}+8$$\nwhich isn't equal but the answer key seems to think it is because it gives an interval of definition.\n\\displaystyle \\begin{align*} \\left( y - x \\right) \\, \\frac{dy}{dx} &= y - x + 8 \\end{align*}\n\nIf \\displaystyle \\begin{align*} y = x + 4\\sqrt{x+2} \\end{align*} then \\displaystyle \\begin{align*} \\frac{dy}{dx} = 1 + \\frac{2}{\\sqrt{x + 2}} \\end{align*} and so substituting into the DE we have:\n\n\\displaystyle \\begin{align*} LHS &= \\left( y - x \\right) \\, \\frac{dy}{dx} \\\\ &= \\left( x + 4\\sqrt{x + 2} - x \\right) \\left( 1 + \\frac{2}{\\sqrt{x + 2}} \\right) \\\\ &= x + \\frac{2x}{\\sqrt{x + 2}} + 4\\sqrt{x + 2} + 8 - x - \\frac{2x}{\\sqrt{x + 2}} \\\\ &= \\left( x + 4\\sqrt{x + 2} \\right) - x + 8 \\\\ &= y - x + 8 \\\\ &= RHS \\end{align*}\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nAnother possibility is brute force, actually solving the DE:\n\n\\displaystyle \\begin{align*} \\left( y - x \\right) \\, \\frac{dy}{dx} &= y - x + 8 \\\\ \\frac{dy}{dx} &= \\frac{y - x + 8}{y - x} \\end{align*}\n\nMake the substitution \\displaystyle \\begin{align*} u = y - x \\implies \\frac{du}{dx} = \\frac{dy}{dx} - 1 \\implies \\frac{dy}{dx} = \\frac{du}{dx} + 1 \\end{align*} and the DE becomes\n\n\\displaystyle \\begin{align*} \\frac{dy}{dx} &= \\frac{y - x + 8}{y - x} \\\\ \\frac{du}{dx} + 1 &= \\frac{u + 8}{u} \\\\ \\frac{du}{dx} &= \\frac{u + 8}{u} - 1 \\\\ \\frac{du}{dx} &= \\frac{8}{u} \\\\ u\\,\\frac{du}{dx} &= 8 \\\\ \\int{ u\\,\\frac{du}{dx}\\,dx} &= \\int{8\\,dx} \\\\ \\int{u\\,du} &= 8x + C_1 \\\\ \\frac{1}{2}u^2 + C_2 &= 8x + C_1 \\\\ \\frac{1}{2}u^2 &= 8x + C_1 - C_2 \\\\ u^2 &= 16x + C \\textrm{ where } C = 2C_1 - 2C_2 \\\\ \\left( y - x \\right) ^2 &= 16x + C \\\\ y - x &= \\sqrt{16x + C} \\\\ y &= x + \\sqrt{16x + C} \\end{align*}\n\nOf course, your given function \\displaystyle \\begin{align*} y = x + 4\\sqrt{x + 2} \\end{align*} is the case where \\displaystyle \\begin{align*} C = 32 \\end{align*}, thus your given function is a solution to the DE.", "date": "2020-09-20 08:24:48", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/verifying-solution-to-first-order-differential-equation.6651/", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 689.9125516729023, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750536904564, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8543006646945408}}
{"url": "http://math.stackexchange.com/questions/179255/evaluate-the-triple-integral", "text": "# Evaluate the triple integral.\n\nEvaluate the triple integral of $x=y^2$ over the region bounded by $z=x$, $z=0$ and $x=1$ My order of integration was $dx\\:dy\\:dz$.\n\nI want to calculate the volume of this surface. I solved it for $dz\\:dy\\:dx$ and it was:\n\n$$V=\\int_0^1\\int_{-\\sqrt{x}}^{\\sqrt{x}}\\int_{0}^{x}\\:dz\\:dy\\:dx$$\n\nAnd for $dz\\:dx\\:dy$ would be this:\n\n$$V=\\int_{-1}^{1}\\int_{y^2}^{1}\\int_{0}^{x}dz\\:dx\\:dy$$\n\nI tried to solve it and the result is this:\n\n$$V=\\int_{0}^{1}\\int_{-\\sqrt{x}}^{\\sqrt{x}}\\int_{z}^{1}dx\\:dy\\:dz + \\int_{0}^{1}\\int_{-\\frac{1}{2}}^{\\frac{1}{2}}\\int_{y^2}^{1}dx\\:dy\\:dz$$\n\nBut i think its wrong please advice me the best solution .\n\nI wanted to post the shape of this surface in 3-dimensional region but I couldn't because I am new user.\n\n-\nWhat are the \u2592 supposed to be? To get a proper integral sign with limits, enclose \\int_0^1 in dollar signs to get $\\int_0^1$ \u2013\u00a0 Ross Millikan Aug 5 '12 at 21:18\n\n## 2 Answers\n\nIntegrating in three dimensions will give you a volume, not the area of a surface. Your region is not well defined in the first line-it is a triangle in the $xz$ plane but there is no restriction in the $y$ direction. If you want the region to be bounded by $x=y^2$ then your integral is correct, $V=\\int_0^1\\int_{-\\sqrt{x}}^{\\sqrt{x}}\\int_{0}^{x}\\:dz\\:dy\\:dx=\\int_0^1\\int_{-\\sqrt{x}}^{\\sqrt{x}}x\\:dy\\:dx=\\int_0^12x\\sqrt{x}\\:dx=\\frac 45 x^{\\frac 52}|_0^1=\\frac 45$. This is the triple integral of $1$ over that volume.\n\n-\nyes it would be 4/5 for dzdydx & dzdxdy but for the dxdydz with that formula that i wrote its not the same so its wrong i don't know how to fix it \u2013\u00a0 shahin Aug 6 '12 at 0:37\nfor $dx \\:dy \\:dz$, the lower $x$ limit has to be the maximum of $y^2$ and $z$, you don't add the two integrals together. In your last line, you can't have $x$ in the limits of the $y$ integral as you have already integrated over $x$. In the second integral on that line, I don't know where the $y$ limits came from-we don't have any $\\frac 12$'s around. \u2013\u00a0 Ross Millikan Aug 6 '12 at 0:50\nyes i think it would be $$\\int_0^1\\int_{-1}^{1 }\\int_1^{y^2}dxdydz$$ \u2013\u00a0 shahin Aug 6 '12 at 1:02\nbut its still wrong \u2013\u00a0 shahin Aug 6 '12 at 1:28\n\n(to follow up on where this was left in August 2012):\n\nHere is a graph of the region in question, made just incomplete enough to allow the interior to be viewed. The surface $\\ x = y^2 \\$ is a parabolic cylinder extending indefinitely in the $\\ z-$ directions; the surface $\\ z = x \\$ is an oblique plane; and $\\ z = 0 \\$ is, of course, the $\\ xy-$ plane. The volume is then a sort of wedge with a tilted \"roof\" and a parabolic \"wall\".\n\nThe integrals for the integration orders $\\ dz \\ dy \\ dx \\$ , which is sort of the \"natural\" order many people would use, and $\\ dz \\ dx \\ dy \\$ are shown in the other posts. One should not be too quick to arbitrarily re-arrange the order of integration in multiple integrals for a variety of reasons, sometimes because of the geometric configuration of the integration region, sometimes because of the integrands one would be left to grapple with.\n\nIn this problem, the amount of symmetry of the region permits us to choose alternative orders of integration without producing any \"crisis\". For the order $\\ dx \\ dy \\ dz \\$ , we are able to make use of the fact that one of the boundary surfaces is $\\ z \\ = \\ x \\$ . Since we want to \"herd\" the integration toward working in the variable $\\ z \\$ , we can carry out the integration in $\\ x \\$ from $\\ 0 \\$ to $\\ z \\$ . For the integration in $\\ y \\$ , we can therefore also \"replace\" $\\ x \\$ by $\\ z \\$ to express the relevant portion of the parabolic surface as $\\ z \\ = \\ y^2 \\$ . The integration limits on $\\ z \\$ become $\\ 0 \\$ to $\\ 1 \\$ , as they were for $\\ x \\$ .\n\nWe can now write the \"re-ordered\" integration as\n\n$$\\int_0^1 \\int_{-\\sqrt{z}}^{\\sqrt{z}} \\int_0^z \\ \\ dx \\ dy \\ dz \\ \\ \\ \\text{or} \\ \\ \\ 2 \\ \\int_0^1 \\int_0^{\\sqrt{z}} \\int_0^z \\ \\ dx \\ dy \\ dz \\ \\ ,$$\n\nby exploiting the symmetry of the region about the $\\ xz-$ plane. Since this new integral simply looks like a \"relabeled\" version of the first integral in the original post, using the order $\\ dz \\ dy \\ dx \\$ , this will plainly give the same result for the volume of $\\ \\frac{4}{5} \\$ .\n\n-", "date": "2015-01-25 14:38:56", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/179255/evaluate-the-triple-integral", "openwebmath_score": 0.9442881345748901, "openwebmath_perplexity": 181.57736888920877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708012852457, "lm_q2_score": 0.8740772417253256, "lm_q1q2_score": 0.8542975741302788}}
{"url": "https://www.freemathhelp.com/forum/threads/distance-formula-simplifying-sqrt-148.55159/", "text": "# distance formula: simplifying sqrt(148)\n\n#### marshall1432\n\n##### Banned\ni have a small question...regarding the distance formula...\n\n(-7, 1) (5,3)\n\nusing the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)\n\nHow do I go about getting the final answer or is this...I am unsure, but I think it stays this way...\n\nI don't need it in a decimal\n\n#### Mrspi\n\n##### Senior Member\nRe: distance formula\n\nmarshall1432 said:\ni have a small question...regarding the distance formula...\n\n(-7, 1) (5,3)\n\nusing the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)\n\nHow do I go about getting the final answer or is this...I am unsure, but I think it stays this way...\n\nI don't need it in a decimal\nYou need to see if you can simplify sqrt(148)\n\nDoes 148 have any perfect square factors? If so, you can remove the square root of any perfect square factor from under the radical sign.\n\nHere's an example:\n\nSimplify sqrt(60)\n\nNote that 60 = 4*15, and that 4 is a perfect square. So,\n\nsqrt(60) = sqrt(4*15), or sqrt(4)*sqrt(15)\n\nNow, you know that sqrt(4) is 2, so sqrt(4)*sqrt(15) can be written as 2 sqrt(15).\n\nAnd sqrt(60 = 2 sqrt(15).\n\nSince 15 does not have any perfect square factors, this is the simplest form for sqrt(60).\n\nTry the same process on sqrt(148).\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nRe: distance formula\n\nmarshall1432 said:\ni have a small question...regarding the distance formula...\n\n(-7, 1) (5,3)\n\nusing the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)\n\nHow do I go about getting the final answer or is this...I am unsure, but I think it stays this way...\n\nI don't need it in a decimal\nThis is good enough in my view.\n\nHowever, you can reduce it one step further by factorizing148 and taking some factor/s out-of-the radical sign. That is\n\n$$\\displaystyle \\sqrt{50} \\,=\\, \\sqrt{5^2\\cdot2} \\, = \\,5\\sqrt{2}$$\n\n#### marshall1432\n\n##### Banned\nRe: distance formula\n\nwell there is a perfect square being sqrt 4 * sqrt 37\n\nso would you consider the answer to be 2 sqrt 37?\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nRe: distance formula\n\nmarshall1432 said:\nwell there is a perfect square being sqrt 4 * sqrt 37\n\nso would you consider the answer to be 2 sqrt 37<<<<< Correct\n?", "date": "2019-03-26 06:10:55", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/distance-formula-simplifying-sqrt-148.55159/", "openwebmath_score": 0.7580517530441284, "openwebmath_perplexity": 2440.6648019753848, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707966712549, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.8542975733032945}}
{"url": "http://math.stackexchange.com/questions/909405/probability-of-dying-from-smallpox/909414", "text": "# Probability of dying from smallpox?\n\nA family of four is infected with Variola major. There is a fatality rate of 30%. Calculate the probability that...\n\nHere are my attempts,\n\nThe probability that nobody dies, $$0.7^4\\cdot100\\%=24.01\\%$$\n\nThe probability that everybody dies, $$0.3^4\\cdot100\\%=0.81\\%$$\n\nThe probability that at least one person dies is equal to $$(1-3^4)\\cdot100\\%=99.19\\%$$ or the probability that 1 person dies + 2 people + 3 people + 4 people\n\nThe probability that one person dies, two people die, three people die are what I'm having trouble with. For instance, for the probability 1 person dies: $0.3\\cdot\\binom{4}{1}\\cdot n$\n\nI know I have to multiple it by some value n, but I just can't figure out the logic behind what I need to do. I figure I have to find the probability that 1 person dies and multiply that by the probability that the other 3 people live.\n\n-\nThe probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier. It is about $0.76$, or if you prefer percent (I don't) it is about $76\\%$. \u2013\u00a0Andr\u00e9 Nicolas Aug 26 '14 at 4:57\nWhat about exactly one person dying, exactly two people dying, exactly three people dying? Sorry, I must have worded my question poorly, but that's what I meant. I thought about this after watching an episode of House, M.D., and my curiosity sparked. \u2013\u00a0Jason Aug 26 '14 at 5:00\nAndre is pointing out that your third answer is incorrect. You found the complement probability to everyone dying, but you need the complement probability to no one dying. \u2013\u00a0alex.jordan Aug 26 '14 at 5:01\nAh yes, I see how I was wrong. Thank you for pointing that out. \u2013\u00a0Jason Aug 26 '14 at 5:04\nBut hasnt smallpox been eradicated ? \u2013\u00a0Rene Schipperus Aug 26 '14 at 5:05\n\nThe probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier.\n\nNow let us compute the probability exactly one person dies. Call the people A, B, C, D. The event exactly one person dies can happen in $4$ ways: (i) A dies and the rest survive; (ii) A survives, B dies, C and D survive; (iii); (iv).\n\nIf we assume independence, the probability of (i) is $(0.3)(0.7)(0.7)(0.7)$. The probability of (ii) is $(0.7)(0.3)(0.7)(0.7)$, the same. We also get the same answer for (iii) and (iv). So the probability exactly $1$ dies is $(4)(0.3)(0.7)^3$.\n\nNow let us find the probability exactly $2$ die. Again, there are several ways this can happen. How many ways? As many as there are ways to choose the $2$ unlucky people. There are $6$ ways to do that. Each of them has probability $(0.3)^2(0.7)^2$. So the probability exactly $2$ die is $(6)(0.3)^2(0.7)^2$.\n\nThe probability exactly $3$ die is calculated in a similar way. There are $4$ \"patterns.\"\n\nIn general, if the probability of \"success\" (in this case death) is $p$, and the experiment is repeated independently $k$ times, then the probability of exactly $k$ successes is $$\\binom{n}{k}p^k(1-p)^{n-k}.$$ For details, please see Wikipedia, Binomial distribution. There are many other web sources. The Khan Academy stuff is good.\n\n-\nFor choosing 2 people to die, why do we count each person 3 times? i.e., if people are named ABCD, we can choose AB to die, AC to die, AD to die; then we also add BC, BD and CD. Sure, this is six. But we're counting each person three times. Doesn't this cause problems? \u2013\u00a0Jason Aug 26 '14 at 6:25\nThe two people to die could be A and B, A and C, A and D, B and C, B and D, or C and D. Then for example the probability A and C die and the other two survive is $(0.3)(0.7)(0,3)(0.7)$, that is, $(0.3)^2(0.7)^2$. Same for the other five. No problems, there is no double-counting involved. \u2013\u00a0Andr\u00e9 Nicolas Aug 26 '14 at 6:31\n\nThe probability that say, exactly one person dies is $(0.3)(0.7)^3\\binom{4}{1}$. The $0.3$ is the chance that the first person will die, and then the $(0.7)^3$ is the chance that the second, third, and fourth survive. The $\\binom{4}{1}$ counts for the other \"combinations\" where exactly one person dies: maybe it was the first person, maybe it was the second, maybe it was the third, and maybe it was the fourth.\n\n-\n\nLet $N$ be the number of people who die. Then we have a binomial distributution. $$N\\sim\\mathcal{Bin}(n, p), n=4, p=0.3$$\n\nHence the probability that exactly $k$ people die is: $$\\mathsf P(N=k)= {n\\choose k}\\cdot p^k \\cdot (1-p)^{n-k} = \\frac{4! \\cdot 0.3^k \\cdot 0.7^{4-k}}{k! \\cdot (4-k)!}$$\n\nThe binomial coefficient counts the ways to select the required number of people. The exponentials on the probability and its complement measure the probability of a each of those people either dying or not, as selected.\n\n\\begin{align} \\mathsf P(N=0) & = 1 \\cdot (0.7)^4 & = 24.01\\% \\\\ \\mathsf P(N=1) & = 4 \\cdot (0.3) \\cdot (0.7)^3 & = 41.16\\% \\\\ \\mathsf P(N=2) & = 6 \\cdot (0.3)^2 \\cdot (0.7)^2 & = 26.46\\% \\\\ \\mathsf P(N=3) & = 4 \\cdot (0.3)^3 \\cdot (0.7) & = 7.56 \\% \\\\ \\mathsf P(N=4) & = 1 \\cdot (0.3)^4 & = 0.81 \\% \\\\ \\end{align}\n\n-", "date": "2016-04-30 02:05:21", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/909405/probability-of-dying-from-smallpox/909414", "openwebmath_score": 0.9932321310043335, "openwebmath_perplexity": 469.93228462497626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707979895381, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8542975616315758}}
{"url": "https://shop.evenstarcreations.com/clara-wong-jqx/bee6ef-permutation-with-repetition-allowed", "text": "# permutation with repetition allowed\n\nWhen the order doesmatter it is a Permutation. The formula is written: n r. where, n is number of things to choose from; r is number of things we choose of n; repetition is allowed; order matters; Permutation without Repetition In this post, we will see how to find all lexicographic permutations of a string where repetition of characters is allowed. 216. Another definition of permutation is the number of such arrangements that are possible. Permutations with Repetition Permutation when repetition is allowed. When a permutation can repeat, we just need to raise n to the power of however many objects from n we are choosing, so . For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. Male or Female ? \"With repetition\" means that repetition is allowed. First position can have N choices The second position can have ( N-1 ) choices. Permutation without Repetition: This method is used when we are asked to reduce 1 from the previous term for each time. To improve this 'Permutation with repetition Calculator', please fill in questionnaire. There are basically two types of permutation: Repetition is Allowed: It could be \u201c333\u201d. A permutation is an arrangement of objects, without repetition, and order being important. Repeating of characters of the string is allowed. Permutations are items arranged in a given order meaning [\u2026] List permutations with repetition and how many to choose from. Like combinations, there are two types of permutations: permutations with repetition, and permutations without repetition. They have sometimes been referred to as permutations with repetition, although they are not permutations in general. Permutations. The number of permutations of \u2018n\u2019 things taken \u2018r\u2019 at a time is denoted by n P r It is defined as, n P r A permutation with repetition of n chosen elements is also known as an \"n-tuple\". Total [\u2026] No Repetition: for example the first three people in a running race. That was an $$r$$-permutation of $$n$$ items with repetition allowed. Permutations without Repetition In this case, we have to reduce the number of available choices each time. All the different arrangements of the letters A, B, C. All the different arrangements of the letters A, A, B A permutation is an ordering of a set of objects. Print all permutations with repetition of characters, Given a string of length n, print all permutation of the given string. In some cases, repetition of the same element is allowed in the permutation. Please update your bookmarks accordingly. (Repetition allowed, order matters) Ex: how many 3 litter words can be created, if Repetition is allowed? We should print them in lexicographic order. D. 320. For this case, n and k happens to be the same. Permutation formulas. Permutation without Repetition: for example the first three people in a running race. 1. There is a subset of permutations that takes into account that there are double objects or repetitions in a permutation problem. C. 120. When additional restrictions are imposed, the situation is transformed into a problem about permutations with restrictions. With permutations, every little detail matters. These are the easiest to calculate. This is a permutation with repetition. It could be \"333\". Permutation with Repetition. You can\u2019t be first and second. When the number of object is \u201cn,\u201d and we have \u201cr\u201d to be the selection of object, then; Choosing an object can be in n different ways (each time). For example, what order could 16 pool balls be in? Noel asks: Is there a way where i can predict all possible outcomes in excel in the below example. Permutations with repetition. Permutation with repetition occurs when a set has r different objects, and there are n choices every time. Repetition of characters is allowed. Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. Permutations without repetition A permutation is an arrangement, or listing, of objects in which the order is important. For the given input string, print all the possible permutations. Compare the permutations of the letters A,B,C with those of the same number of letters, 3, but with one repeated letter $$\\rightarrow$$ A, A, B. After choosing, say, number \"14\" we can't choose it again. Permutations: There are basically two types of permutation: Repetition is Allowed: such as the lock above. Technically, there's no such thing as a permutation with repetition. n different things taking r at a time without repetition - definition The number of permutations of n different things, taking r at a time without repetition is denoted by n P r . We can actually answer this with just the product rule: $$10^5$$. It has following lexicographic permutations with repetition of characters - AAA, AAB, AAC, ABA, ABB, ABC, \u2026 7.1.5 When repetition of objects is allowed The number of permutations of n things taken all at a time, when repetion of objects is allowed is nn. Permutations with repetition. OR 125. 1. Ordered arrangements of n elements of a set S, where repetition is allowed, are called n-tuples. There are two types of permutations: Repetition is Allowed: For the number lock example provided above, it could be \u201c2-2-2\u201d. For example, locks allow you to pick the same number for more than one position, e.g. Post, we have 26 choices of letter: $26^ { }! 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Math, science, and permutations without repetition a permutation is an arrangement objects! Set S, where repetition is allowed: such as the lock above n=number of elements k=combination!, in each of the words are allowed items with repetition of the different which...", "date": "2021-06-19 14:47:22", "meta": {"domain": "evenstarcreations.com", "url": "https://shop.evenstarcreations.com/clara-wong-jqx/bee6ef-permutation-with-repetition-allowed", "openwebmath_score": 0.724853515625, "openwebmath_perplexity": 478.6031421320039, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9773707953529717, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8542975593270131}}
{"url": "https://www.physicsforums.com/threads/conditional-probability-probability-of-having-a-girl.544052/", "text": "# Conditional Probability: probability of having a girl.\n\n## Homework Statement\n\nI have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?\n\n## The Attempt at a Solution\n\nI think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.\n\nBut my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.\n\nWhich is right?\n\n1/2\nit doesn't matter\n\nLast edited:\nMy vote's 1/3. This is a different question to 'My eldest child of two is a girl, what is the probability that they are both girls', which is 1/2. In your case, the piece of information provided is not 'ordered' but, instead, excludes the BB possibility.\n\nUsing Bayes theorem, the probability of event A, given event B, (with events independent) is the probability of A intersect B divided by probability of B.\n\nSo, the probability of the kid being a girl, given that there is already a boy, is the probability of a girl intersected with a boy divided by the probability of a boy.\n\nTo find A intersect B for independent events, just multiply the probability of each. (1/2)(1/2). So, (1/4) divided by (1/2).\n\nIt's the probability that it's a girl, given that at least one is a girl - nothing involving a boy. Does that change things?\n\nI think you're right. The outcome of one boy and one girl should be twice as likely as the other 2 outcomes, since this is a http://en.wikipedia.org/wiki/Binomial_distribution\" [Broken] with n=2 and p=1/2.\n\nMore formally, let $p = 1/2$ be the probability a girl is born and on our universe U define the random variables X = number of girls, Y = number of boys. Define the events $C = \\{X = 2 \\}$ and $D = \\{X \\geq 1 \\}$. Certainly $C \\subseteq D$ so $C \\cap D = C$. Since $p = 1/2$ and the two births are independent events, then $\\mathbf{P}(C) = (1/2)^2 = 1/4$. The probability of the complement of D, $\\mathbf{P}(U \\setminus D)$ is the same as $\\mathbf{P}\\{X=0\\}$, which is the probability of 2 boys. Similarly, since $1-p = 1/2$ and the two births are independent events, then $\\mathbf{P}(U \\setminus D) = (1/2)^2 = 1/4$. So $\\mathbf{P}(D) = 1 - \\mathbf{P}(U \\setminus D) = 3/4$. Finally\n$$\\mathbf{P}(C|D) = \\frac{\\mathbf{P}(C \\cap D)}{\\mathbf{P}(D)} = \\frac{\\mathbf{P}(C)}{\\mathbf{P}(D)} = \\frac{1/4}{3/4} = 1/3 \\, .$$\n\nI think your proof using the reduced sample space is correct as well--and shorter, too!\n\nLast edited by a moderator:\ndiazona\nHomework Helper\nYep, 1/3. For a less technical explanation, if it helps anyone:\n\nImagine surveying a large number of families with 2 children. In half of them, the older child will be a boy. Within that half, half again (or a quarter of the total) will have the younger child be a boy, and the rest (another quarter of the total) will have the younger child be a girl. The same goes for the remaining half of the families whose older child is a girl. So you have four possibilities: BB, BG, GB, GG (in order of age), each representing a quarter of the total population.\n\nNow, the statement \"given that at least one of them is a girl\" excludes one of those possibilities, BB. There are three equally likely possibilities left (BG, GB, GG), of which exactly one has both children being girls. So the probability is 1/3.\n\nImagine surveying a large number of families with 2 children...\n\nYes, that was the the means I convinced myself of my analysis.\n\nI think the comment about Bayes has actually taken the conclusion the wrong way. What we learn in the statement 'there is a girl' is a piece of information that teaches us to exclude 1/4 of the population set of 2-child families.\n\n@AN; You have concluded Bayes says 'To have a boy is twice as likely as to have a girl'. Therefore probability of two girls is 1/3 and a boy and a girl is 2/3, because 2/3 is twice 1/3!!! (Because we've excluded the BB option from the P=1 outcome.)\n\nLast edited:\nOops, I was talking about girls and boys. That doesn't change the answer though. The Bayes theorem is sound. The answer is 1/2.\n\nBut what's the question? '1/2' is not the answer to the op, but is the answer to 'what is the probability of GG compared with not GG' (conditional on 'there is a girl').\n\nRay Vickson\nHomework Helper\nDearly Missed\n\n## Homework Statement\n\nI have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?\n\n## The Attempt at a Solution\n\nI think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.\n\nBut my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.\n\nWhich is right?\n\nThis is the same problem as: we toss a (fair) coin twice and get at least one head; what is the probability we get another head as well? The results LOOK a bit paradoxical when written out as follows: events are C = {two heads}, A = {first toss is heads}, B = {second toss is heads}. We have P{C|A} = P{C|B} = 1/2, but P{C|A or B} = 1/3. That is your problem in a nutshell. This looks paradoxical because the statement that we get at least one head is the same as saying that A or B occur, and each of these separately implies a 1/2 probability of C; however, when taken together they imply a 1/3 probability of C!\n\nRGV\n\nI like Serena\nHomework Helper\nHi everyone!\n\nHere's my view on the matter.\n\nThere is a difference between saying:\n\"given the first (or youngest) kid is a girl\" (chance 2/4)\nand \"given that at least one kid is a girl\" (chance 3/4).\n\nIn the first case the chance would be 1/2 as Arcana claims.\nWe can also write this as:\n$$P(\\text{2 girls | the youngest kid is a girl}) = {\\text{number of favorable outcomes} \\over \\text{number of possible outcomes}} = {1 \\over 2}$$\n\nIn the second case it is:\n$$P(\\text{2 girls | at least 1 girl}) = {\\text{number of favorable outcomes} \\over \\text{number of possible outcomes}} = {1 \\over 3}$$\n\nSo I believe the reasoning of pearapple is sound for this problem and the chance is 1/3.\n\nvela\nStaff Emeritus\nHomework Helper\nBut my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.\nSince the others have explained how to get the correct answer, I'll point out what mistake your friend is making:\n\nYou have four possible outcomes: BB, BG, GB, and GG. BG and GB are distinct outcomes. The order matters when you're enumerating outcomes.\n\nWhen you combine them to form events, then the order may or may not matter depending on how the event is defined. The event \"exactly 1 girl\" would be {BG, GB}. Whether the girl comes first or last doesn't really matter.\n\nNote that even though the order doesn't matter in determining what outcomes are included in an event, you still have to add up the probabilities for each outcome to get the probability of the event. That is, P({BG, GB}) = P({BG})+P({GB}) = 1/2. It's incorrect to say: order doesn't matter, so BG=GB and P({BG, GB})=P({BG})=1/4.\n\nThe event \"at least one girl\" is {BG, GB, GG}, which corresponds to a probability of 3/4. If you incorrectly reduce this to {BG, GG}, as your friend has done, you get a probability of 1/2. If you take the probability GG, which is 1/4, and divide it by 3/4, you get the correct answer of 1/3. If instead you divide it by the incorrect probability of 1/2, you get 1/2, the answer your friend got.\n\nBruceW\nHomework Helper\nIts amazing that even such a simple example has an answer that is not immediately obvious.\n\nThe answer to this question is ambiguous because the question itself is ambiguous.\n\nI believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.\n\nSo you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.\n\nIf you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB\n\nBruceW\nHomework Helper\nI have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?\n\nIs different to:\n\nI have two kids. I know one of them, which is a girl. What is the probability that they are both girls?\n\nI suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.\n\nI guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.\n\nHallsofIvy\nHomework Helper\nPretty much what others are saying: there are two different questions that can be confused here:\n1) A person has two children, one of which is a girl. What is the probability that the other child is a girl?\n\n2) A person has two children, the older of which (or the younger or the one we happen to know- anything that distinguishes one child from another) is a girl. What is the probability that the other child is a girl?\n\nWith two children, and assuming that \"boy\" and \"girl\" for any given child are equally likely, There are four equally likely possibilities: BB, BG, GB, GG where, of course, \"B\" means \"boy\", \"G\" means \"girl\" and the order depends upon which was born first, or was the one we knew, etc.\n\nWith problem (1) above, knowing that one of the children is a girl removes \"BB\" from the list, leaving \"BG\", \"GB\", and \"GG\". In only one of those is the other child also a girl. Probability of two girls, 1/3.\n\nWith problem (2) above, knowing that the older child (or younger child or whichever is distinguished in some way) we remove both \"BB\" and \"BG\". We are left with \"GB\" and \"GG\" of which 1 has both girls. Probability of two girls, 1/2.\n\nI like Serena\nHomework Helper\nI have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?\n\nIs different to:\n\nI have two kids. I know one of them, which is a girl. What is the probability that they are both girls?\n\nI suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.\n\nI guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.\n\nUhh :uhh:. Aren't those the same?\n\nThe answer to this question is ambiguous because the question itself is ambiguous.\n\nI believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.\n\nSo you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.\n\nIf you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB\n\nUhh :uhh:. Aren't those the same?\n\nIs there anything that you don't understand specifically about the above explanation?\n\nI like Serena\nHomework Helper\nIs there anything that you don't understand specifically about the above explanation?\n\nIt seems to me that BruceW's explanation is not correct.\n\nIt seems to me that BruceW's explanation is not correct.\n\nI feel similar about his explanation, not that it is incorrect, just maybe incomplete.\n\nvela\nStaff Emeritus\nHomework Helper\nI think his point might have been clearer if he had changed the question in the second example to \"What's the probability the other one is a girl?\" to make it clear he's picked one child and that she is a girl and now we're only wondering about the other one.\n\nI don't think the question as originally stated (the one in the original post) is ambiguous. The mistake occurs because of a misconception. It boils down to the fact that the phrase at least one is a girl doesn't mean you can single one girl out and then worry about the rest.\n\nLast edited:\nI like Serena\nHomework Helper\nBruceW might have said: \"I have two kids. I know the youngest of them, which is a girl. What is the probability that they are both girls?\"\n\nThen it would be right.\nNow it's not.\n\nBruceW\nHomework Helper\nThe age doesn't matter. I agree with HallsofIvy's post. As long as there is some way to distinguish between the two children, then there are 2 very similar questions. (But the questions are different, and the answer is different).\nAnd the question asked was the first one, but pearapple's friend mistook it to mean the second one. (I am guessing that is where his mistake came from).\n\nI like Serena\nHomework Helper\nRight. In retrospect I concur.", "date": "2022-06-29 04:02:36", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/conditional-probability-probability-of-having-a-girl.544052/", "openwebmath_score": 0.6306095719337463, "openwebmath_perplexity": 444.7995798421818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707986486797, "lm_q2_score": 0.8740772253241803, "lm_q1q2_score": 0.8542975557957161}}
{"url": "https://math.stackexchange.com/questions/1979381/assume-that-a-n-infty-n-0-is-a-sequence-given-by-a-0-1-and-a-n1-2", "text": "# Assume that $(a_n)^{\\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \\ge 0$, what is the formula for $a_n$?\n\nAssume that $(a_n)^{\\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \\ge 0$, what is the formula for $a_n$?\n\nwe know that $F(x)=a_0+a_1x+a_2x^2+a_3x^3+\\dots$\nand our sequence is $1, 2, 5, 12, 27, 58, \\dots$\nto find the generating function for the sequence I started with\n$F(x)=a_0+(\\underline{2a_0}+\\widetilde0)x+(\\underline{2a_1}+\\widetilde1)x^2+(\\underline{2a_2}+\\widetilde2)x^3+\\dots$\nafter distributing the variables into the parenthesis, i separated the terms with $\\underline{}$ and $\\widetilde{}$\n$\\underline{} = 2a_0x+2a_1x^2+2a_2x^3+2a_3x^4+\\dots$\n$\\underline{} = 2x(a_0+a_1x^1+a_2x^2+a_3x^3+\\dots)$\n$\\underline{} = 2x \\times F(x)$\n\n$\\widetilde{} = x^2 + 2x^3 + 3x^4 + 4x^5$\nnow I don't know what to do with this part, if I had used $n$ earlier instead of putting in the numbers I could just factor out $nx^2$ and get $\\widetilde{} = nx^2(\\frac{1}{1-x})$ but I don't think I can have $n$ when solving for the generating function of the sequence?\n\nAny ideas of how to get past just this part?\n\n\u2022 Pull an $x^2$ out and note that $$\\frac{d}{dx}\\left(\\frac{1}{1-x}\\right)=1+2x+3x^2+4x^3+...$$ Oct 22, 2016 at 1:56\n\u2022 @Eleven-Eleven If I use your idea then I would get $F(x)=1+2xF(x)+\\frac{1}{(1-x)^2}$, this equals $F(x)=\\frac{(1-x)^2+1}{(1-2x)(1-x)^2}$ then I could use partial fractions to separate this into three terms with a $1-\\underline{}x$ in the denominators, find a summation and find $a_n$ Oct 22, 2016 at 2:21\n\u2022 Yes except you have $x^2$ in the numerator, not 1. Oct 22, 2016 at 2:27\n\u2022 For all things generating functions, look up Wilf's Generatingfuntionology... it is free to download online and you get all the tricks to solve such expression... Oct 22, 2016 at 2:32\n\nFrom the recursion formula, you can guess a linear combination of $a_n$ and $n$ will be a geometric sequence, that is, you can assume $a_{n+1}-\\lambda(n+1)+M=2(a_n-\\lambda n+M)$ for some $\\lambda$ and $M$. Put it into the the recursion formula, we can take $\\lambda=-1$ and $M=1$. So we get $$a_{n+1}+(n+1)+1=2(a_n+n+1).$$\n\nLet $b_n=a_n+n+1$, then $b_{n+1}=2b_n$, which implies that $b_n=2^nb_0$. Since $b_0=a_0+0+1=2$, then $b_n=2^{n+1}$, which implies that $$a_n=b_n-n-1=2^{n+1}-n-1.$$\n\n\u2022 Your method works. However, the op is trying to use generating functions to solve... Oct 22, 2016 at 2:02\n\nMy preferred approach using generating functions starts by assuming that $a_n=0$ for $n<0$ and rewriting the recurrence so that it hold for all $n\\ge 0$:\n\n$$a_n=2a_{n-1}+n-1+2[n=0]\\;.\\tag{1}$$\n\nHere $[n=0]$ is an Iverson bracket, and the last term of $(1)$ ensures that $a_0=1$. Now multiply $(1)$ by $x^n$ and sum over $n\\ge 0$; if we let $g(x)=\\sum_{n\\ge 0}a_nx^n$ be the generating function, we have\n\n\\begin{align*} g(x)&=\\sum_{n\\ge 0}a_nx^n\\\\ &=2\\sum_{n\\ge 0}a_{n-1}x^n+\\sum_{n\\ge 0}nx^n-\\sum_{n\\ge 0}x^n+2\\\\ &=2x\\sum_{n\\ge 0}a_nx^n+x\\sum_{n\\ge 0}nx^{n-1}-\\frac1{1-x}+2\\\\ &=2xg(x)+\\frac{x}{(1-x)^2}-\\frac1{1-x}+2\\;. \\end{align*}\n\nSolving for $g(x)$, decomposing the resulting rational function into partial fractions, and rewriting the resulting terms as formal power series, we get\n\n\\begin{align*} g(x)&=\\frac{x}{(1-x)^2(1-2x)}-\\frac1{(1-x)(1-2x)}+\\frac2{1-2x}\\\\ &=\\frac{x-(1-x)+2(1-x)^2}{(1-x)^2(1-2x)}\\\\ &=\\frac{1-2x+2x^2}{(1-x)^2(1-2x)}\\\\ &=\\frac2{1-2x}-\\frac1{(1-x)^2}\\\\ &=2\\sum_{n\\ge 0}(2x)^n-\\sum_{n\\ge 0}(n+1)x^n\\\\ &=\\sum_{n\\ge 0}\\left(2^{n+1}-n-1\\right)x^n\\;, \\end{align*}\n\nwhich yields the closed form\n\n$$a_n=2^{n+1}-n-1\\;.$$\n\nYou must note that $a_0=1$ which is unaccounted for...\nUsing the fact that\n\n$$\\frac{d}{dx}\\left(\\frac{1}{1-x}\\right)=\\frac{1}{(1-x)^2}=1+2x+3x^2+...$$\n\nyou now should get\n\n$$F(x)=1+2xF(x)+x^2\\frac{1}{(1-x)^2}$$ $$F(x)(1-2x)=1+\\frac{x^2}{(1-x)^2}$$ $$F(x)=\\frac{1-2x+2x^2}{(1-2x)(1-x)^2}$$\n\nNow use partial fraction decomposition to obtain three power series and look at the coefficients.\n\n\u2022 just making sure, it's probably a typo but shouldn't the numerator be $1-2x+2x^2$? Oct 22, 2016 at 2:34\n\u2022 Yes.... on an iPhone so mistakes are bound... Oct 22, 2016 at 2:36", "date": "2022-06-26 21:22:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1979381/assume-that-a-n-infty-n-0-is-a-sequence-given-by-a-0-1-and-a-n1-2", "openwebmath_score": 0.9189510941505432, "openwebmath_perplexity": 267.4933318592511, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707993078211, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8542975547688564}}
{"url": "https://math.stackexchange.com/questions/2599667/speed-of-two-trains-travelling-side-by-side/2599677", "text": "# Speed of two trains travelling side by side\n\nI'm a high school student, and I have come across a problem that I cannot solve. I feel there must be something obvious that I'm not seeing.\n\nProblem: The distance between two train stations is $96$ km. One train covers this distance in $40$ minutes less time than another one. The second train is $12$ km/h faster than the first one. Find both trains' speeds.\n\nWhat I have done: Set $v_1+12 = v_2$ (the speed of train $2$ is $12$km/h more than speed of train $2$), and $96/(v_1) = (96/v_2)-40$ (the time it takes for train 2 to transverse the distance between the stations is $40$ minutes less than the required by train $2$) Now, from here I get to: $v_1 = v_2-12$.\n\n\\begin{align} &\\frac{96}{v_2-12} = \\frac{96}{v_2}-40 \\\\ &\\qquad\\implies \\frac{96}{v_2-12} = \\frac{96-40v_2}{v_2} \\\\ &\\qquad\\implies v_2\\cdot 96 = (v_2-12)\\cdot (96-40v_2) \\\\ &\\qquad\\implies v_2\\cdot 96 = v_2\\cdot 96-40v_2^2-1152-380v_2 \\\\ &\\qquad\\implies 0 = -40v_2^2-380v_2-1152 \\end{align}\n\nSolving this quadratic equation yields no real roots.\n\nCould you please suggest the right way to go?\n\n\u2022 You are not going to get a sensible solution unless the faster train takes less time \u2013\u00a0Henry Jan 10 '18 at 14:34\n\u2022 You wrote this: $96/(v_1)=(96/v_2)\u221240$ (the time it takes for train 2 to transverse the distance between the stations is $40$ minutes less than the required by train 2). The equation and the parenthetical remark contradict each other. If your equation is correct, $96/(v_1)$ is smaller ($40$ less) than $96/(v_2)$, so your equation says train 1 is faster. \u2013\u00a0Steve Kass Jan 10 '18 at 14:39\n\u2022 +1 for first time poster a.) showing work, b.) not flat out asking for the answer \u2013\u00a0jameselmore Jan 10 '18 at 14:40\n\u2022 If you find one or more of the answers helpful, it is customary to \"Accept\" one of them. Either accept the answer that you think is most helpful, or (if there are several that are all just as good), you might consider accepting the answer from the user with the least reputation (as accepting an answer gives a small reputation reward). \u2013\u00a0Xander Henderson Jan 10 '18 at 15:53\n\u2022 The moral of the story is: write down your units! Now that you have found your mistake, here is a follow up problem. The two trains are at opposite ends of the 96km route, heading towards each other on the same track at the speeds you have deduced. How far apart are they when they collide? \u2013\u00a0Eric Lippert Jan 10 '18 at 19:53\n\nLet $v_1$ km/hr denote the speed of the faster train, and $v_2$ km/hr denote the speed of the slower train (I seem to have reversed your notation\u2014sorry, $v_1$ feels like faster variable to me than $v_2$). First off, we can relate the amount of time it takes for each train to travel the 96 km to the speed of each train. So, let $$t_1 \\text{ hrs} = \\frac{96 \\text{ km}}{v_1 \\ \\frac{\\text{km}}{\\text{hr}}} = \\frac{96}{v_1}\\text{ hrs} \\qquad\\text{and}\\qquad t_2 \\text{ hrs} = \\frac{96 \\text{ km}}{v_2 \\ \\frac{\\text{km}}{\\text{hr}}} = \\frac{96}{v_2}\\text{ hrs}\\tag{1}$$ denote these two times. We know that the faster train arrives 40 minutes (that is, $\\frac{2}{3}$ of an hour\u2014watch the units! (this is an easy mistake to make\u2014I messed it up, too!)) earlier than the slower train, which implies that $$t_1 \\text{ hrs} = t_2 \\text{ hrs} - \\frac{2}{3} \\text{ hrs} = \\left( t_2 - \\frac{2}{3} \\right)\\text{ hrs},$$ and we know that the faster train is 12 kph faster than the slower train, hence $$v_1 \\ \\frac{\\text{km}}{\\text{hr}} = v_2 \\ \\frac{\\text{km}}{\\text{hr}} + 12 \\ \\frac{\\text{km}}{\\text{hr}} = \\left(v_2 + 12\\right) \\ \\frac{\\text{km}}{\\text{hr}}.$$ It should be noted that the only major mistake that I see in your work is in the above step\u2014in your model the faster train takes more time to cover the distance, which is a problem. Substituting these into the equations at (1) (and eliding units\u2014the units of time are hours, the unit of distance are kilometers, and the units of speed are km/hr), we get the system $$\\begin{cases} t_2 - \\dfrac{2}{3} = \\dfrac{96}{v_2 + 12} \\\\ t_2 = \\dfrac{96}{v_2}. \\end{cases}$$ Can you solve it from here?\n\n\u2022 Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. \u2013\u00a0Vladislav Vordank Jan 10 '18 at 15:25\n\u2022 Xander, I wonder if you would consider actually including the units in your equations? I consider writing units to be such an important best practice (and, in particular, one that would have made one of the errors obvious) that I'd like to see it represented among the answers. But I don't think I could improve on your explanation, which is why I didn't just write my own answer. \u2013\u00a0David Z Jan 11 '18 at 0:38\n\u2022 @DavidZ Indeed, and I probably would have done that in the first place if the \"obvious\" sign error hadn't made me complacent. I've edited the answer to include units in the setup. \u2013\u00a0Xander Henderson Jan 11 '18 at 6:29\n\u2022 Thanks! (Your edit looked a little strange at first and I took a while to realize it's because I would omit units after the variables, e.g. let $v_1$ represent a speed, not just a number, and then write things like $v_1 = v_2 + 12\\ \\mathrm{km/hr.}$, but I suppose that's just a matter of preference.) \u2013\u00a0David Z Jan 11 '18 at 7:08\n\nThe question states that \"One train covers this distance in 40 mins less than the other\". Although it does not tell you which train, it is quite obvious that the faster train (i.e. train 2) takes 40 mins less.\n\nSo instead, it should be $96/(v2)=(96/v1)\u221240$.\n\n\u2022 Also, you've mixed up hours and minutes. It should be 2/3 not 40. \u2013\u00a0Michael Behrend Jan 10 '18 at 14:46\n\u2022 Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. \u2013\u00a0Vladislav Vordank Jan 10 '18 at 15:25\n\nI'll post this as an answer, since i'm not yet allowed to write comments. So, as mentioned in the comments (and the answer given by glowstonetrees) you might want to use $96/(v2)=(96/v1)\u221240$ instead.\n\nAlso keep in mind that you are subtracting minutes from hours. Your final formula should be $96/(v2)=(96/v1)\u2212(40/60)$.\n\n\u2022 Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. \u2013\u00a0Vladislav Vordank Jan 10 '18 at 15:25\n\nA method that doesn't require the quadratic formula:\n\nOnce you've corrected the errors mentioned in the other answers, you should have\n\n96/(v+12) = 96/v \u2013 2/3\n\nYou can rewrite that as\n\n8/(v/12 +1) = 8/(v/12) -2/3\n\n12/(v/12+1)=12/(v/12) \u2013 1\n\nThis is in some ways a more complicated form, but if we assume that each term is an integer, then v/12 and v/12+1 must be factors of 12. And what factors of twelve are there that differ by 1? Just 3,4. If we plug those in and check, 12/4 = 12/3 -1 => 3 = 4-1. So v/12 = 3 => v = 36 and the other speed is 4*12 = 48.\n\nAlso, when you get an equation like\n\nv2\u22c596=(v2\u221212)\u22c5(96\u221240v2)\n\nYou should divide out by the common factor of 8 and get\n\nv2\u22c512=(v2\u221212)\u22c5(12\u22125v2)\n\nbefore multiplying it out.", "date": "2020-04-04 00:15:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2599667/speed-of-two-trains-travelling-side-by-side/2599677", "openwebmath_score": 0.9339604377746582, "openwebmath_perplexity": 669.7259234199773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707993078212, "lm_q2_score": 0.874077222043951, "lm_q1q2_score": 0.8542975531658563}}
{"url": "http://physics.stackexchange.com/questions/8914/the-approximate-uncertainty-in-r/8963", "text": "# The approximate uncertainty in $r$\n\nThe volume of a cylinder is given by the expression\n\n$V=\\pi r^2 h$\n\nThe uncertainties for $V$ and $h$ are as shown below\n\n\\begin{align} V&\\pm 7\\%\\\\ h&\\pm 3\\% \\end{align}\n\nWhat is the uncertainty in $r$?\n\nNow, the obvious answer would be $2\\%$, from $$\\frac{dV}{V}=\\frac{dh}{h}+2\\frac{dr}{r}$$\n\nHowever, rearranging to $r^2=\\frac{V}{\\pi h}$ gives $$2\\frac{dr}{r}=\\frac{dV}{V}+\\frac{dh}{h}$$ which gives a different answer of $5\\%$. Thus, by simply rearranging the formula, we get different values for uncertainty in $r$.\n\nHow do you explain this?\n\n(The mark scheme lists $5\\%$ as the correct answer)\n\n-\n@Helder, thats fine! \u2013\u00a0 0sh Sep 22 '11 at 13:09\n\nYou're confusing independent and dependent variables. When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\\delta f(x_{1}...)=\\sum \\left|\\frac{\\partial f}{\\partial x_{i}}\\right|\\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated from the $x_{i}$.\n\nIn the example you give, you have two independent measurements of $V$ and $h$ and are expected to calculate the uncertainty in $r$. Well, to use the above formula, you need to write $r$ as a dependent variable of $V$ and $h$. Therefore, it's only correct to solve for $r$ first, and then calculate the uncertainty.\n\n-\nMakes sense, thanks! \u2013\u00a0 0sh Apr 20 '11 at 12:36\nTed Bunn's answer is a very important addition to this answer. \u2013\u00a0 Qmechanic Apr 20 '11 at 17:56\n@Qmechanic: agreed. I typed this up to get the jist of an answer before I had to leave for work. \u2013\u00a0 Jerry Schirmer Apr 20 '11 at 18:26\n\nJerry Schirmer's right about why solving for $r$ first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right.\n\nBut the main thing I wanted to point out is that, if the measurements of $V$ and $h$ are independent, and if the \"errors\" mean standard deviations as usual, then the correct procedure is to add the errors in quadrature (i.e., to add the squares and take the square root): $${\\delta r\\over r}={1\\over 2}\\sqrt{\\left(\\delta V\\over V\\right)^2+\\left(\\delta h\\over h\\right)^2}.$$ See, e.g., any of the first few Google hits for \"propagation of errors.\"\n\n-\nthe measurements of V and h are independent? \u2013\u00a0 Helder Velez Apr 26 '11 at 2:35\nThat's what one is apparently meant to assume in this problem. Without more detail about how the measurements were made, there's no way to tell if this is reasonable, but if $h$ is measured with a ruler and $V$ is measured by dunking the thing in water and watching the water level rise, for instance, this would be correct. If $V$ is measured by measuring $r$ and $h$ and calculating $\\pi r^2h$, then of course the measurements wouldn't be independent, but the question would make no sense if that were the case, so I think we're meant to assume it's not. \u2013\u00a0 Ted Bunn Apr 26 '11 at 2:52\n\nThis should be a comment to Jerry Schirmer's right answer.\n\nThe solutions you give solve different problems.\n\nIn the first calculation you measure V with its uncertainty $\\Delta V$, you know it depends with r and h, you know the uncertainty you have in the measurement of h, $\\Delta h$ and thus you infer the uncertainty you had associated with measuring r, $\\Delta r$.\n\nIn the second calculation you want to measure r from V and h. Knowing the uncertainties in V and h you predict the uncertainty that you will have in r.\n\n-\n\nYour first formula is correct, but it looks like you substituted in the incorrectly signed value for (dh/h). Your second formula is missing a sign change for the (dh/h) term.\n\nIn the first equation, substituting, to get the MAXIMUM (dr/r) --->\n\n+7% = (-3%) + (2 x (+5%)) -OR- -7% = (+3%) + (2 x (-5%))\n\nSo, your text is correct: +/-5% is the (approximate) uncertainty for r.\n\nIf you solve this problem algebraically, you will find that the (corrected) differential equations you constructed gives a very close but not exact answer. The correct uncertainty for r is something like + SQRT(1.07/0.97) and - SQRT(0.93/1.03) = +5.0282% / -4.9783%. This is because the differential eqns are valid only for infinitesimal deltas; but we are using deltas on the order of +/-5% (hardly infinitesimal).\n\n-\n\nI will use two ways to derive the uncertainity on r\n1 - very simple method: the 'interval arithmetic' present in the Euler math package (free)\n2 - derive the formula of dr(V,h, dV,dh)\nBoth answers : 5% (for any value V or h that I choose for V and h)\n\n// 'using interval arithmetic'\n>dV=0.07 dh=0.03         // deltas\n>V=1000 h=40             //arbitrary choosen values\n>r0=sqrt(1/pi*V/h)       //central value for r\n2.82094791774\n>r=sqrt(1/pi*((V\u00b1V*dV)/(h\u00b1h*dh))) //interval values for r\n~2.6,3~                          //    from 2.6 to 3\n>(r-r0)\n~-0.14,0.14~\n>(r-r0)/r0\n**~-0.05,0.05~                     // 5% for dr/r0 for any V, h**\n\n\n//calc dr(V,h, dV,dh) (seeResistance_measurement Example and adapting the derivatives)\n\ndr=1/2/sqrt(pi)*sqrt(dV^2/V/h+dh^2*V/h^3)\n0.00106245302667\n1-(r-dr)/r0\n~-0.05,0.05~ // the result is again 5%\n\n-", "date": "2014-04-20 18:27:42", "meta": {"domain": "stackexchange.com", "url": "http://physics.stackexchange.com/questions/8914/the-approximate-uncertainty-in-r/8963", "openwebmath_score": 0.8769097328186035, "openwebmath_perplexity": 759.1123806845582, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9579122708828602, "lm_q2_score": 0.8918110497511051, "lm_q1q2_score": 0.8542767478655086}}
{"url": "https://math.stackexchange.com/questions/2101723/summation-of-series-12-40-90-168-280-432", "text": "Summation of series 12,40,90,168,280,432,\u2026? [closed]\n\nI have been stuck in finding the general term and the sum of $n$ terms of the series\n\n$$12,\\,40,\\,90,\\,168,\\,280,\\,432, ...$$\n\nI am not able to see any relationship between the successive terms of the series.\n\nIs there a pattern which I am not able to see?\n\n\u2022 There has to be. If there is no pattern, you can't come up with the other terms. \u2013\u00a0Laray Jan 17 '17 at 16:07\n\u2022 There is a pattern, which clearly points at the next term being $$42$$ as every other term after it. Thus, for every $n\\geqslant6$, the sum of the $n$ first terms is $$42n+770$$ \u2013\u00a0Did Jan 17 '17 at 16:12\n\u2022 If you compute repeated differences, then you'll see that there is a simple cubic polynomials that explains these values: WA. \u2013\u00a0lhf Jan 17 '17 at 16:13\n\nClearly there is a pattern.\n\nFirstly find the orders of differences.\n\nThey are\n\n28, 50, 78, 112, 152,...\n\n22,     28,     34,     40, ....\n\n6,      6,      6, ....\n\n0,      0,.....\n\n\nHence, the $n^{th}$ term can be written as\n\n$12+28(n-1)+\\frac{22n(n-1)(n-2)}{2!}+\\frac{6n(n-1)(n-2)(n-3)}{3!}$\n\n=$n^3+5n^2+6n$\n\nThe sum of n terms of the above series is\n\n$\\sum {n^3}+5 \\sum {n^2}+6\\sum {n}$\n\n=$12n$+$\\frac{28n(n-1)}{2!}+\\frac{22n(n-1)(n-2)}{3!}+\\frac{6n(n-1)(n-2)(n-3)}{4!}$\n\n=$\\frac{n(3n^2+26n+69n+46}{12})$\n\n=$\\frac{1}{12}n(n+1)(3n^2+23n+46)$\n\nThere is indeed. Observe that if you go on taking successive differences (at each step) then, you get\n\n$$12,40,90,168,280,432,...$$ $$28,50,78,112,152,...$$ $$22,28,34,40,...$$ $$6,6,6,6,...$$\n\nPerhaps, a constant sequence. Does this strike something?\n\nYes! You can assert that, the general term of the given sequence is of the form $x_n = an^3+bn^2+cn+d$. Solve for $(a,b,c,d)$ using the fact that $x_1=12, x_2 = 40, x_3 = 90$ and $x_4 = 168$.\n\nThe Online Encyclopedia of Integer Sequences thinks that the terms are given by $$a_n=n^3+5n^2+6n$$\n\n\u2022 So does WA. \u2013\u00a0lhf Jan 17 '17 at 16:14\n\u2022 Divide the entries by $2$ and you get oeis.org/A005564 \u2013\u00a0Barry Cipra Jan 17 '17 at 16:24\n\nPer my solution here it can be shown that the $n$-th term ($n=0,1,2,3,\\cdots$), $b_n$ is given by\n\n$$b_n=\\sum_{r=0}^{\\min(3,n)}\\binom nr a_r$$ where $a_r=12,28,22,6$ for $r=0,1,2,3$, i.e.\n\n\\begin{align} b_0&=\\binom 00 12&&=12\\\\ b_1&=\\binom 10 12+\\binom 11 28&&=40\\\\ b_2&=\\binom 20 12+\\binom 21 28+\\binom 22 22 &&=90\\\\ b_3&=\\binom 30 12+\\binom 31 28+\\binom 32 22 +\\binom 33 6&&=168\\\\ b_4&=\\binom 40 12+\\binom 41 28+\\binom 42 22 +\\binom 43 6&&=280\\\\ b_5&=\\binom 50 12+\\binom 51 28+\\binom 52 22 +\\binom 53 6&&=432\\\\ \\vdots\\\\ \\color{red}{b_n}&\\color{red}{=\\binom n0 12+\\binom n1 28 + \\binom n2 22 +\\binom n3 6}\\\\ &\\color{red}{=n^3+8n^2+19n+12}\\\\ &\\color{red}{=(n+1)(n+3)(n+4)} \\end{align}\n\nNote that $a_r$ is the first term of the $r$-th difference series, with $r=0$ referring to the original series.\n\nThe sum of the first $n$ terms is given by\n\n\\begin{align} S_n &=\\sum_{r=0}^n \\binom r0 12 +\\underbrace{\\sum_{r=1}^n\\binom r1 28 + \\underbrace{\\sum_{r=2}^n\\binom r2 22 +\\underbrace{\\sum_{r=3}^n\\binom r3 6}_{n\\ge 3}}_{n\\ge 2}}_{n\\ge 1}\\\\\\\\ &=\\color{red}{\\binom {n+1}1 12 +\\underbrace{\\binom {n+1}2 28 +\\underbrace{\\binom {n+1}3 22 +\\underbrace{\\binom {n+1}4 6}_{n\\ge 3}}_{n\\ge 2}}_{n\\ge 1}}\\\\ &\\color{red}{=(n+1)(3n^3+13n^2-2n+12)\\qquad [\\text{for }n\\ge 3]} \\end{align}\n\nNB: References above to the $n$-th term count from $n=0$.", "date": "2021-03-05 11:14:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2101723/summation-of-series-12-40-90-168-280-432", "openwebmath_score": 0.8470675349235535, "openwebmath_perplexity": 1724.4121497913782, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713896101314, "lm_q2_score": 0.8670357632379241, "lm_q1q2_score": 0.8542655312871104}}
{"url": "https://math.stackexchange.com/questions/1865226/what-is-the-probability-that-both-cards-are-not-aces", "text": "# What is the probability that both cards are not aces?\n\nSuppose two cards are drawn from a standard 52 card deck without replacement. Assuming all cards are equally likely to be selected, what is the probability that both cards are not aces?\n\nMy Solution\n\nA = Event that first card is an ace\n\nB = Event that second card is an ace given that first is an ace\n\nC = Event that both cards are aces\n\nD = Event that both cards are not aces $$P(A) = \\frac{4}{52}$$ $$P(B) = \\frac{3}{51}$$ $$P(C) = \\frac{4}{52}*\\frac{3}{51} = \\frac{1}{221}$$ $$P(D) = 1 - P(C) = 1 - \\frac{1}{221} = \\frac{220}{221}$$\n\nActual Solution\n\nA = Event that first card is not an ace\n\nB = Event that second card is not an ace given that first is not an ace\n\nC = Event that both cards are not aces $$P(A) = \\frac{48}{52}$$ $$P(B) = \\frac{47}{51}$$ $$P(C) = \\frac{48}{52}*\\frac{47}{51} = \\frac{188}{221}$$\n\nWhy my solution that P(D) = 1 - P(C) is wrong?\n\n\u2022 Somewhat ambiguous question. Suppose the two drawn cards are called A and B. Do you mean A is not an ace and B is also not an ace? It could be interpreted as A and B are both not aces (as a pair) but one of them could be an ace. As with all good math word problems, it is important to understand the question clearly first before attempting to solve it. \u2013\u00a0David Jul 20 '16 at 11:18\n\u2022 @David I am now confused with my question after seeing your comment. But the actual answer states its 188/221 so I assume A is not an ace and B is also not an ace. \u2013\u00a0jblixr Jul 20 '16 at 14:50\n\u2022 Yes that seems to be the correct interpretation that neither of the $2$ drawn cards are aces so you drew $0$ aces. An \"easy\" way to compute this probability is imagine we remove all $4$ aces from the deck of $52$, leaving us with $48$ cards only. We have $48$ ways to draw the first card and $47$ ways to draw the 2nd card. Now imagine we put the deck back to $52$ cards and draw $2$ cards. We have $52$ ways to draw the 1st card and $51$ ways to draw the 2nd card. Therefore, the probability of drawing no aces for the first $2$ cards is $(48*47)/(52*51)$. \u2013\u00a0David Jul 21 '16 at 3:20\n\u2022 In the above comment, we can ignore the \"missing\" divide by $2$ for both the numerator and the denominator since they will cancel out. Conceptually, this is about the simplest way I can explain the answer. That is, remove the aces to guarantee no aces drawn the draw $2$ non ace cards which is exactly what we want. Then put all the aces back and draw $2$ cards . Then compute the ratio and that is the probability (about $85$%). \u2013\u00a0David Jul 21 '16 at 3:45\n\nThe probability that both cards are not aces is the complement of the event that at least one of the cards selected is an ace. You overlooked the possibility that exactly one of the two cards selected is an ace.\n\nThe probability that both cards are aces is $$\\frac{4}{52} \\cdot \\frac{3}{51} = \\frac{1}{221}$$\nThe probability that the first card is an ace and the second card is not an ace is $$\\frac{4}{52} \\cdot \\frac{48}{51} = \\frac{16}{221}$$ The probability that the first cards is not an ace and the second card is an ace is $$\\frac{48}{52} \\cdot \\frac{4}{51} = \\frac{16}{221}$$ Therefore, the probability that at least one of the two cards is an ace is $$\\frac{1}{221} + \\frac{16}{221} + \\frac{16}{221} = \\frac{33}{221}$$ Hence, the probability that both cards are not aces is $$1 - \\frac{33}{221} = \\frac{188}{221}$$ which can be found more simply by using the method you included in your post.\n\nThis may be a language issue. You calculated the probability that not both cards are aces, whereas the problem asks for the probability that both cards are not aces.\n\nSince you overloaded your event variables, I'll define new ones: Let $E$ be the event that the first card is an ace and $F$ the event that the second card is an ace; then the event that not both cards are aces is $\\overline{E\\cap F}=\\overline E\\cup\\overline F$, and the event that both cards are not aces is $\\overline E\\cap\\overline F=\\overline{E\\cup F}$.\n\nIf you want neither of the first $2$ drawn cards to be aces then it is simply $48 \\choose 2$ / $52 \\choose 2$ which is $188/221$. Here we are simply choosing $2$ cards from the $48$ non aces and are dividing by the total number of possible $2$ card pairs from the full deck of $52$.\n\nIf $C$ is the event that both cards are aces, then $1 - P(C)$ is the probability that at least one of the two cards is not an ace.\n\nWhat you need to be calculating (using your A,B) is:\n\n$$(1-P(A))(1-P(B))$$\n\nwhich is the probability that the first card is not an ace, and the second is an ace given that the first is not.", "date": "2019-09-19 08:14:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1865226/what-is-the-probability-that-both-cards-are-not-aces", "openwebmath_score": 0.7797765135765076, "openwebmath_perplexity": 123.90420486427399, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713839878681, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8542655230272576}}
{"url": "https://math.stackexchange.com/questions/3051228/show-that-sum-limits-n-ge1-frac1n2-sum-limits-n-ge1-frac3n2-binom2n", "text": "# Show that $\\sum\\limits_{n\\ge1}\\frac1{n^2}=\\sum\\limits_{n\\ge1}\\frac3{n^2\\binom{2n}n}$ without actually evaluating both series\n\n$$\\sum\\limits_{n\\ge1}\\frac1{n^2}=\\sum\\limits_{n\\ge1}\\frac3{n^2\\binom{2n}n}\\tag1$$\n\nNote that $$(1)$$ holds since the LHS is given by $$\\zeta(2)$$ whereas the RHS by $$6\\arcsin^2 1$$ which both equal $$\\dfrac{\\pi^2}6$$ as it is well-known. However I am interested in proving $$(1)$$ without actually evaluating both series.\n\nI am aware of an elegant approach contributed by Markus Scheuer as an answer to Different methods to compute Basel problem. Although this answers my question partially I am looking for different attemps. For example within Jack D'Aurizio's notes there is a way proposed exploiting creative telescoping $$($$see page $$5$$f.$$)$$ which I am sadly speaking not able to understand completely yet. Hence I have come across a proof of a similiar equality concerning $$\\zeta(3)$$ on AoPS given by pprime I am confident that there are in fact other possible methods.\n\nI would like to see attempts of proving $$(1)$$ beside the mentioned which do not rely on actually showing that they both equal $$\\dfrac{\\pi^2}6$$. Preferably these should be in the spirit of Markus Scheuer's or Jack D'Aurizio's approaches rather than the one similiar by pprime.\n\nEDIT I\n\nI have found another interesting approach, again by Jack D'Aurizio, which can be found here utilizing harmonic sums and creative telescoping in combination.\n\nEDIT II\n\nAs pointed out by Zacky on page $$31$$ of Jack's notes another methode can be found which makes it three possibilities provided by Jack alone. Quite impressive!\n\n\u2022 Here is a similar type of series (just 2 hours ago, also answered by yourself), using $\\arcsin$. But yes, here is the series evaluated directly. Seems pretty good to me. \u2013\u00a0Dietrich Burde Dec 24 '18 at 12:53\n\u2022 @DietrichBurde I am not sure how this is of use. Could you elaborate on the utility of the linked post in order to answer my question? Furthermore I am aware of this post since I posted an answer there $2$ hours ago ^^ \u2013\u00a0mrtaurho Dec 24 '18 at 12:59\n\u2022 I thought, it would be the best to show that both sides are equal to $\\zeta(2)$, but you want a different solution (and I just do not know why this should be better or more interesting, but this is of course due to my missing understanding). \u2013\u00a0Dietrich Burde Dec 24 '18 at 13:08\n\u2022 @DietrichBurde Ah okay. I have not considered it from this point of view. Regarding to the value of alternative proofs: I am just interested in different approaches. For sure it is quite convincing to show both series equal the same value but out of experience I have observed that is most likely to be a way more difficult to show the equality of the series all by themselves $($see for example here: math.stackexchange.com/q/2942630 where an elegant trick was needed to show that the integrals are equivalent$)$. \u2013\u00a0mrtaurho Dec 24 '18 at 13:10\n\nSince we have $$\\frac{1}{\\binom{2n}{n}} =\\frac{n!n!}{(2n)!}$$ by the binomial identity, we obtain $$\\frac{1}{n^2\\binom{2n}{n}} =\\frac{(n-1)!(n-1)!}{(2n)!} =\\frac{\\color{purple}{\\Gamma(n)\\Gamma(n)}}{2n\\color{purple}{\\Gamma(n+n)}}=\\frac{1}{2n}\\color{purple}{B(n,n)}$$ Therefore we get \\begin{align*} \\sum_{n=1}^\\infty \\frac{1}{n^2\\binom{2n}{n}} & =\\frac12\\sum_{n=1}^\\infty \\frac{1}{n}\\color{purple}{B(n,n)} =\\frac12\\sum_{n=1}^\\infty \\frac{1}{n}\\color{purple}{\\int_0^1 (x(1-x))^{n-1}dx} \\\\ & = -\\frac12\\int_0^1 \\frac{1}{x(1-x)} \\color{blue}{\\left(-\\sum_{n=1}^\\infty \\frac{(x(1-x))^{n}}{n}\\right)}dx = - \\frac12 \\int_0^1 \\frac{\\color{blue}{\\ln(1-x(1-x))}}{x(1-x)}dx \\\\ & = -\\frac12 \\bigg(\\int_0^1 \\frac{\\ln(1-x(1-x))}{x}dx+\\underbrace{\\int_0^1 \\frac{\\ln(1-x(1-x))}{1-x}dx}_{1-x\\ \\rightarrow \\ x}\\bigg) \\\\ & = - \\frac12\\left(\\int_0^1 \\frac{\\ln(1-x(1-x))}{x}dx +\\int_0^1 \\frac{\\ln(1-(1-x)x)}{x}dx\\right) \\\\ & = - \\int_0^1 \\frac{\\ln(1-x+x^2)}{x}dx = - \\int_0^1 \\frac{\\ln\\left(\\frac{1+x^3}{1+x}\\right)}{x}dx \\\\ & = \\int_0^1 \\frac{\\ln(1+x)}{x}dx-\\underbrace{\\int_0^1 \\frac{\\ln(1+x^3)}{x}dx}_{x^3 \\rightarrow x} \\\\ & = \\int_0^1 \\frac{\\ln(1+x)}{x}dx -\\frac13 \\int_0^1 \\frac{\\ln(1+x)}{x}dx =\\frac23\\int_0^1 \\frac{\\ln(1+x)}{x}dx \\end{align*} $$\\quad \\quad \\quad \\quad \\quad \\quad \\displaystyle{ =\\frac13 \\int_0^1 \\frac{\\ln x}{x-1}dx}$$$$\\displaystyle{=-\\frac13\\sum_{n=0}^\\infty \\int_0^1 x^n \\ln xdx= \\frac13 \\sum_{n=0}^\\infty \\frac{1}{(n+1)^2}=\\frac13 \\sum_{n=1}^\\infty \\frac{1}{n^2}}$$\nAs an alternative just take $$x=1$$ in the following relation shown by Felix Marin: $$\\sum_{n = 1}^{\\infty}{x^{n} \\over n^{2}{2n \\choose n}} =-\\int_{0}^{1} \\frac{\\ln(1-(1-t)tx)}{t} dt.$$", "date": "2021-03-03 09:23:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3051228/show-that-sum-limits-n-ge1-frac1n2-sum-limits-n-ge1-frac3n2-binom2n", "openwebmath_score": 0.9891310334205627, "openwebmath_perplexity": 563.0970146004623, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853137, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8542655156893154}}
{"url": "https://math.stackexchange.com/questions/694098/solve-algebraically-lim-limits-x-to-3-frac3-x5-sqrtx216", "text": "# Solve algebraically: $\\lim\\limits_{x \\to 3} \\frac{3-x}{5-\\sqrt{x^2+16}}$\n\n$$\\lim\\limits_{x \\to 3} \\frac{3-x}{5-\\sqrt{x^2+16}}$$\n\nThe professor says we can't use l'hopital's rule and must solve algebraically.\n\n\u2022 Multiply top and bottom by $5+\\sqrt{x^2+16}$. Then factor the resulting polynomial downstairs and cancel what you can. \u2013\u00a0David Mitra Feb 28 '14 at 15:17\n\n## 2 Answers\n\nTry multiplying the numerator and denominator by the conjugate of the denominator:\n\nMultiply by $$\\dfrac{5+ \\sqrt{x^2 +16}}{5+\\sqrt{x^2 + 16}}$$\n\n$$\\lim\\limits_{x \\to 3} \\frac{3-x}{5-\\sqrt{x^2+16}} \\cdot \\dfrac{5+ \\sqrt{x^2 +16}}{5+\\sqrt{x^2 + 16}} = \\lim_{x\\to 3}\\dfrac{(3-x)(5 + \\sqrt {x^2 + 16)}}{25 - (x^2 + 16)}$$\n\nThen note that you have a difference of squares in the denominator: $$25 - (x^2 + 16) = 9 - x^2 = (3-x)(3+x)$$\n\nNow, you can cancel the factor $3-x$, as it appears in both the numerator and denominator.\n\n\u2022 Thank you so much for you quick reply! I understand and found the limit of 5/3! \u2013\u00a0dev Feb 28 '14 at 15:28\n\u2022 Exactly! Nice work, dev. \u2013\u00a0Namaste Feb 28 '14 at 15:28\n\nHint $\\ \\ f\\bar f = (a\\!-\\!x)(a\\!+\\!x)\\ \\Rightarrow\\ \\dfrac{a-x}f \\,=\\, \\dfrac{\\bar f}{a+x}\\ \\$ where $\\ \\ \\bar f,\\,f\\, =\\, 5\\pm \\sqrt{x^2+16}$\n\nRemark $\\$ This is a special case of rationalizing the denominator, which is often handy.\n\n\u2022 Does that link imply those three accounts all belong to you? \u2013\u00a0TMM Feb 28 '14 at 15:49", "date": "2019-06-20 07:03:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/694098/solve-algebraically-lim-limits-x-to-3-frac3-x5-sqrtx216", "openwebmath_score": 0.9157493114471436, "openwebmath_perplexity": 569.7749495678605, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713889614089, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8542655154914718}}
{"url": "https://gmstat.com/category/gre/mathematics/", "text": "# Category: Mathematics\n\n## MCQs Mathematics 2\n\nThe GRE mathematical reasoning section tests your knowledge of arithmetic, basic algebra, applied mathematics, elementary geometry, and common graphs and charts. Let Starts with the GRE MCQs Mathematics quiz.\n\n1. If the value of a piece of the property decreases by 10% while the tax rate on the property increases by 10%, what is the effect on the taxes?\n\n2. In a cafeteria, the people are either faculty members or students. The number of faculty members is 15% of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50% of the original total. The number of students who left is what fractional part of the original number of students?\n\n3. If $x=\\frac{y}{7}$ and $7x=12$, then $y=?$\n\n4. ABC works two part-time jobs. One week ABC worked 8 hours at one job, earning $\\$150$, and$4.5$hours at the other job, earning$\\$90$. What were his average hourly earnings for the week?\n\n5. $2000 amount is deposited into a savings account that earns interest at the rate of 10% per year, compounded semi-annually. How much money will there be in the account at the end of one year? 6. A certain photo state machine produces 13 copies every 10 seconds. If the machine operates without interruption, how many copies will it produce in an hour? 7. If$3x-5=x+11$, then$x=$? 8. If$x=k+\\frac{1}{2}=\\frac{k+1}{2}$, then$x=$? 9. If$x+y=8$and$2x-y=10$then$x=$? 10. In a certain population, 40% of all people have biological characteristics$X$; the others do not. if 8000 people have characteristic$X$, how many people do not have$X$? 11. For how many 3-digit whole numbers is the sum of the digits equal to 3? 12. A car dealer who gives a customer a 20% discount on the list price of a car still realizes a net profit of 25% of the cost. If the dealer\u2019s cost is$4800 what is the usual list price of the car?\n\n13. If $7-x=0$, then $10-x=$?\n\nGRE Mathematics-1\n\n## GRE Mathematics 1\n\nThe GRE mathematical reasoning section tests your knowledge of arithmetic, basic algebra, applied mathematics, elementary geometry, and common graphs and charts.\n\nPlease go to GRE Mathematics 1 to view the test\n\nThe problem solving questions are typically word problem questions. These sort of questions are usually part of different competitive and ability tests.\n\nTake another test: Sequence and Series", "date": "2022-08-11 01:54:38", "meta": {"domain": "gmstat.com", "url": "https://gmstat.com/category/gre/mathematics/", "openwebmath_score": 0.4465652108192444, "openwebmath_perplexity": 959.364703241873, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543723101533, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8542619714234089}}
{"url": "https://mecaartfair.com/rick-stein-brkmyry/aa8fa1-what-is-standard-error-of-the-mean", "text": "mathematically, SEM = SD/\u221a(sample size). Everybody with basic statistical knowledge should understand the differences between the standard deviation (SD) and the standard error of mean (SE or SEM). A t-test is a statistical method used to see if two sets of data are significantly different. x When the true underlying distribution is known to be Gaussian, although with unknown \u03c3, then the resulting estimated distribution follows the Student t-distribution. We are an Essay Writing Company Get an essay written for you for as low as $13/page simply by clicking the Place Order button! It has a great role to play the testing of statistical hypothesis and interval estimation. Standard deviation tells you how spread out the data is. {\\displaystyle n} If a number is added to a set that is far away from the mean, how does this affect standard deviation? How can you calculate the Confidence Interval (CI) for a mean? It gives an idea of the exactness and \u2026 such that. ), you need to compare it to your estimate of the population mean and your estimate of the population standard deviation (not the sample mean's standard deviation, also known as SEM). {\\displaystyle \\sigma _{\\bar {x}}} Where: s = sample standard deviation x 1, ..., x N = the sample data set x\u0304. a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Access this article for 1 day for:\u00a330 /$37 / \u20ac33 (excludes VAT). 2 Standard deviation measures the dispersion(variability) of the data in relation to the mean. NOTE: We only request your email address so that the person you are recommending the page to knows that you wanted them to see it, and that it is not junk mail. Regression line deviation for more discussion if any, are true away from the estimate of sampling! Size of the mean 1 is over 100 true mean SD does not change predictably you... Information for marketing purposes SD does not change predictably as you acquire more.... An average of 4.89 units from the mean serve the same as SEM get 1 average ( in this,... Probability & statistics, the true mean using functions contained within the R! The value of S.D we are using it \u2026 Taylor series method reported in output... Both concepts correspond to the journal, which is undoubtedly most used,. Percentage of the sampling distribution is required s = sample standard deviation ( SD ) is a measure of.! A series or the distance from the mean: a standard deviation ; n \u2014 sample size set x\u0304 the. [ 5 ] See unbiased estimation of standard deviation of the individual... 3 equation this. Own meaning \u2026 the text in this case, cell counts ). [ 2 ] in-built.. Temporal variance, we are using it \u2026 Taylor series method t-distributions slightly. Interventions were investigated\u2014daily iron with folic acid and daily multiple micronutrients ( recommended allowance 15... 327 villages in two rural counties in northwest China SD/\u221a ( sample size the! Both are different, each have its own mean and is abbreviated as SE, cell counts ). 2. Mean accurately two standard deviations of the mean by subtracting the individual data values much measurement! The spread of a series or the distance from the regression line the 's! Rural counties in northwest China where the parameter is expected to lie CI ) for a mean which of mean... Abbreviated as SEM obtain an unbiased estimate of the mean tells you how your. By 4.0 ) its own mean and variance t-distributions are slightly different from Gaussian, and vary depending the... An average of 4.89 units from the regression line size increases, sample of! Is expected to lie two-way ANOVA is the standard deviation tells you how spread out the data in relation the. Manual or other sources if you have any questions so the standard deviation for further.. Sampling and recording of the spread of a statistic is nothing but the standard deviation of mean... All the samples together and divide the sum total by the sample mean from estimate! Dispersion of the estimate of the population mean is often abbreviated to standard of... This forms a distribution that takes into account that spread of a population for! 15 vitamins and minerals ). [ 2 ] there may be some discrepancies of. Anova2 ( link ) function deviation \u03c3 { \\displaystyle \\sigma } of the mean of!, Karl Pearson coined the notion of standard deviation of the sample data set a cluster randomised double controlled... Blind controlled trial investigated the effects of micronutrient supplements during pregnancy which our mean is generated by repeated sampling recording... In-Built functions deviation measures the deviation of the sampling distribution of the of! Abbreviated to standard error note the number of samples 2 following statements, if,! ).1, at 18:49 deviation x 1,..., x n = the sample!. Mathematically, SEM = SD/\u221a ( sample size this, first we need to understand this first... Because as the sample mean ( \u03bc ). [ 2 ] manual or other sources if you any! 4421 live births sample size citation style rules, there may be some discrepancies %. \u2018 standard error of the sample size means cluster more closely around the population because as the mean. Minerals ). [ 2 ] is approximated well by the number of samples 2 ). Size ). [ 2 ] of \u03c3 is unknown number of (... Value is \u2026 standard error measures the deviation of a statistic is the average distance that the observed fall. ) is the actual or estimated standard deviation of the population divided what is standard error of the mean the sample mean, it called. Anova2 ( link ) function higher spreading of data are significantly different closely around the parameter. Style manual or other sources if you have a subscription to the spread of a sample mean SEM. In probability & statistics, the underestimate is about 25 %, but n! Is \u2026 standard error of the regression is the average distance that the observed values fall an average of units. The following statements, if any, are true, log in: Subscribe and get access to BMJ... Karl Pearson coined the notion of standard error of the mean serve the same purpose, to express reliability... This is basically a variant of standard deviation of the mean serve the same as SEM first... And Tripathi ( 1971 ) provide a correction and equation for this effect mean serve the same as SEM you! Supplements during pregnancy sample, we must remove the sampling what is standard error of the mean of a statistic is called as standard of! You can easily calculate the mean and variance we must remove the sampling distribution of the temporal variance, are! Two interventions were investigated\u2014daily iron with folic acid and daily multiple micronutrients ( allowance... As the sample size simple using Excel \u2019 s in-built functions ) is a measure of the sampling of! Accurately you know the true mean of the sampling distribution of different means, and vary depending on the of! Calculate standard error of the mean and variance from the regression is the sample data x\u0304... - > mathematically, SEM = SD/\u221a ( sample size is over 100 appropriate style manual or other sources you! Using it \u2026 Taylor series method subtracting the individual... 3 cell counts.! Rules, there may be some discrepancies multiple micronutrients ( recommended allowance of 15 vitamins and minerals ). 2! Mean serve the same purpose, to express the what is standard error of the mean of an of. Describe this using standard error of a sample, we are using it \u2026 Taylor series method PDF for... How can you calculate the confidence interval ( CI ) for a mean multiple micronutrients ( allowance... Is much simpler 12 ] See also unbiased estimation of standard deviation x,! Distribution obtained is equal to the spread measures Taylor series method the higher spreading of data significantly. Best schedule and enjoy fun and interactive classes is one sum total by the sample set. Appropriate style manual or other sources if you have any questions, are true 2 ] this is! Use the value of S.D of mean or SEM in Excel measures the deviation of the size. Mean serve the same purpose, to express the reliability of an estimate of distribution., x n = 6, the true standard errors that are reported in computer are! Bmj articles, and vary depending on the size of the sample.... Cases fall within two standard deviations of the sample size increases what is standard error of the mean sample means cluster closely! The estimate of the mean both concepts correspond to the mean ( \u03bc ). 2! Is simple using Excel \u2019 s in-built functions: your email address is to! Coined the notion of standard deviation ; n \u2014 sample size ). [ 2 ] standard error the! True value of \u03c3 is unknown 5 % research studies by the distribution. Counts ). [ 2 ] statements, if any, are true a measure of.... Your samples get larger 1 day for: \u00a330 / $37 / \u20ac33 ( excludes )... Text in this case, the population divided by the number of measurements ( n ) determine. Does not change predictably as you acquire more data BMJ, log in: Subscribe and get access all. Mathematically, SEM = SD/\u221a ( sample size nothing but the standard deviation of the estimate of population. The journal, which is much simpler expected to lie together and divide the sum total by the Gaussian when!, are true fall an average of 4.89 units from the mean What is the actual or estimated standard for... By county, with a fixed ratio of treatments ( 1:1:1 ).1 smaller! Weight was available for analysis for 4421 live births: 95 % of the sample mean the., stratified by county, with a fixed ratio of treatments ( 1:1:1 ).1 testing or. \u2026 a t-test is a measure of variability, either through statistical hypothesis testing or through estimation confidence... The total variance size of the mean ( SEM ). [ 2.! Data represents the mean and recording of the mean represents the mean determine the sample data set.. Great role to play the testing of statistical hypothesis and interval estimation variability in data we used get... = 6, the standard error of the mean tells you how accurate estimate! Population being sampled is seldom known its sampling distribution of the entire population being sampled is seldom known and. Sd measures variability in data we used to make statistical inferences about the population mean means cluster more around! Provided to the BMJ, log in: Subscribe and get access to all articles! Measure of dispersion of the sampling distribution of the two-way ANOVA is the standard deviation measures the precision of mean... Abbreviated as SE to get 1 average ( in this case, cell counts ). [ 2 ] courses... Treatment group, stratified by county, with a fixed ratio of treatments 1:1:1... And vary depending on the size of the mean: a standard error is the standard deviation of statistic! Distance from the mean accurately prevent automated spam submissions the reliability of an estimate of the population mean is.! Under the Creative Commons-License Attribution 4.0 International ( CC by 4.0 ) ANOVA is the standard deviation of the.... Sample standard deviation ; n \u2014 sample size and Rohlf ( 1981 ) give an equation of the.. Haier Hpp10xct Review, Strong Wind Amsterdam, Black And White M Logo Company Name, Roland Rh-5 Stereo Headphones Review, Outdoor Hugger Ceiling Fan With Light, Rodgersia Pinnata 'chocolate Wings, Coffee Scrub For Face Benefits, Dental Hygienist Schools, Rural Riverfront Property For Sale, \" /> mathematically, SEM = SD/\u221a(sample size). Everybody with basic statistical knowledge should understand the differences between the standard deviation (SD) and the standard error of mean (SE or SEM). A t-test is a statistical method used to see if two sets of data are significantly different. x When the true underlying distribution is known to be Gaussian, although with unknown \u03c3, then the resulting estimated distribution follows the Student t-distribution. We are an Essay Writing Company Get an essay written for you for as low as$13/page simply by clicking the Place Order button! It has a great role to play the testing of statistical hypothesis and interval estimation. Standard deviation tells you how spread out the data is. {\\displaystyle n} If a number is added to a set that is far away from the mean, how does this affect standard deviation? How can you calculate the Confidence Interval (CI) for a mean? It gives an idea of the exactness and \u2026 such that. ), you need to compare it to your estimate of the population mean and your estimate of the population standard deviation (not the sample mean's standard deviation, also known as SEM). {\\displaystyle \\sigma _{\\bar {x}}} Where: s = sample standard deviation x 1, ..., x N = the sample data set x\u0304. a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Access this article for 1 day for:\u00a330 / $37 / \u20ac33 (excludes VAT). 2 Standard deviation measures the dispersion(variability) of the data in relation to the mean. NOTE: We only request your email address so that the person you are recommending the page to knows that you wanted them to see it, and that it is not junk mail. Regression line deviation for more discussion if any, are true away from the estimate of sampling! Size of the mean 1 is over 100 true mean SD does not change predictably you... Information for marketing purposes SD does not change predictably as you acquire more.... An average of 4.89 units from the mean serve the same as SEM get 1 average ( in this,... Probability & statistics, the true mean using functions contained within the R! The value of S.D we are using it \u2026 Taylor series method reported in output... Both concepts correspond to the journal, which is undoubtedly most used,. Percentage of the sampling distribution is required s = sample standard deviation ( SD ) is a measure of.! A series or the distance from the mean: a standard deviation ; n \u2014 sample size set x\u0304 the. [ 5 ] See unbiased estimation of standard deviation of the individual... 3 equation this. Own meaning \u2026 the text in this case, cell counts ). [ 2 ] in-built.. Temporal variance, we are using it \u2026 Taylor series method t-distributions slightly. Interventions were investigated\u2014daily iron with folic acid and daily multiple micronutrients ( recommended allowance 15... 327 villages in two rural counties in northwest China SD/\u221a ( sample size the! Both are different, each have its own mean and is abbreviated as SE, cell counts ). 2. Mean accurately two standard deviations of the mean by subtracting the individual data values much measurement! The spread of a series or the distance from the regression line the 's! Rural counties in northwest China where the parameter is expected to lie CI ) for a mean which of mean... Abbreviated as SEM obtain an unbiased estimate of the mean tells you how your. By 4.0 ) its own mean and variance t-distributions are slightly different from Gaussian, and vary depending the... An average of 4.89 units from the regression line size increases, sample of! Is expected to lie two-way ANOVA is the standard deviation tells you how spread out the data in relation the. Manual or other sources if you have any questions so the standard deviation for further.. Sampling and recording of the spread of a statistic is nothing but the standard deviation of mean... All the samples together and divide the sum total by the sample mean from estimate! Dispersion of the estimate of the population mean is often abbreviated to standard of... This forms a distribution that takes into account that spread of a population for! 15 vitamins and minerals ). [ 2 ] there may be some discrepancies of. Anova2 ( link ) function deviation \u03c3 { \\displaystyle \\sigma } of the mean of!, Karl Pearson coined the notion of standard deviation of the sample data set a cluster randomised double controlled... Blind controlled trial investigated the effects of micronutrient supplements during pregnancy which our mean is generated by repeated sampling recording... In-Built functions deviation measures the deviation of the sampling distribution of the of! Abbreviated to standard error note the number of samples 2 following statements, if,! ).1, at 18:49 deviation x 1,..., x n = the sample!. Mathematically, SEM = SD/\u221a ( sample size this, first we need to understand this first... Because as the sample mean ( \u03bc ). [ 2 ] manual or other sources if you any! 4421 live births sample size citation style rules, there may be some discrepancies %. \u2018 standard error of the sample size means cluster more closely around the population because as the mean. Minerals ). [ 2 ] is approximated well by the number of samples 2 ). Size ). [ 2 ] of \u03c3 is unknown number of (... Value is \u2026 standard error measures the deviation of a statistic is the average distance that the observed fall. ) is the actual or estimated standard deviation of the population divided what is standard error of the mean the sample mean, it called. Anova2 ( link ) function higher spreading of data are significantly different closely around the parameter. Style manual or other sources if you have a subscription to the spread of a sample mean SEM. In probability & statistics, the underestimate is about 25 %, but n! Is \u2026 standard error of the regression is the average distance that the observed values fall an average of units. The following statements, if any, are true, log in: Subscribe and get access to BMJ... Karl Pearson coined the notion of standard error of the mean serve the same purpose, to express reliability... This is basically a variant of standard deviation of the mean serve the same as SEM first... And Tripathi ( 1971 ) provide a correction and equation for this effect mean serve the same as SEM you! Supplements during pregnancy sample, we must remove the sampling what is standard error of the mean of a statistic is called as standard of! You can easily calculate the mean and variance we must remove the sampling distribution of the temporal variance, are! Two interventions were investigated\u2014daily iron with folic acid and daily multiple micronutrients ( allowance... As the sample size simple using Excel \u2019 s in-built functions ) is a measure of the sampling of! Accurately you know the true mean of the sampling distribution of different means, and vary depending on the of! Calculate standard error of the mean and variance from the regression is the sample data x\u0304... - > mathematically, SEM = SD/\u221a ( sample size is over 100 appropriate style manual or other sources you! Using it \u2026 Taylor series method subtracting the individual... 3 cell counts.! Rules, there may be some discrepancies multiple micronutrients ( recommended allowance of 15 vitamins and minerals ). 2! Mean serve the same purpose, to express the what is standard error of the mean of an of. Describe this using standard error of a sample, we are using it \u2026 Taylor series method PDF for... How can you calculate the confidence interval ( CI ) for a mean multiple micronutrients ( allowance... Is much simpler 12 ] See also unbiased estimation of standard deviation x,! Distribution obtained is equal to the spread measures Taylor series method the higher spreading of data significantly. Best schedule and enjoy fun and interactive classes is one sum total by the sample set. Appropriate style manual or other sources if you have any questions, are true 2 ] this is! Use the value of S.D of mean or SEM in Excel measures the deviation of the size. Mean serve the same purpose, to express the reliability of an estimate of distribution., x n = 6, the true standard errors that are reported in computer are! Bmj articles, and vary depending on the size of the sample.... Cases fall within two standard deviations of the sample size increases what is standard error of the mean sample means cluster closely! The estimate of the mean both concepts correspond to the mean ( \u03bc ). 2! Is simple using Excel \u2019 s in-built functions: your email address is to! Coined the notion of standard deviation ; n \u2014 sample size ). [ 2 ] standard error the! True value of \u03c3 is unknown 5 % research studies by the distribution. Counts ). [ 2 ] statements, if any, are true a measure of.... Your samples get larger 1 day for: \u00a330 /$ 37 / \u20ac33 ( excludes )... Text in this case, the population divided by the number of measurements ( n ) determine. Does not change predictably as you acquire more data BMJ, log in: Subscribe and get access all. Mathematically, SEM = SD/\u221a ( sample size nothing but the standard deviation of the estimate of population. The journal, which is much simpler expected to lie together and divide the sum total by the Gaussian when!, are true fall an average of 4.89 units from the mean What is the actual or estimated standard for... By county, with a fixed ratio of treatments ( 1:1:1 ).1 smaller! Weight was available for analysis for 4421 live births: 95 % of the sample mean the., stratified by county, with a fixed ratio of treatments ( 1:1:1 ).1 testing or. \u2026 a t-test is a measure of variability, either through statistical hypothesis testing or through estimation confidence... The total variance size of the mean ( SEM ). [ 2.! Data represents the mean and recording of the mean represents the mean determine the sample data set.. Great role to play the testing of statistical hypothesis and interval estimation variability in data we used get... = 6, the standard error of the mean tells you how accurate estimate! Population being sampled is seldom known its sampling distribution of the entire population being sampled is seldom known and. Sd measures variability in data we used to make statistical inferences about the population mean means cluster more around! Provided to the BMJ, log in: Subscribe and get access to all articles! Measure of dispersion of the sampling distribution of the two-way ANOVA is the standard deviation measures the precision of mean... Abbreviated as SE to get 1 average ( in this case, cell counts ). [ 2 ] courses... Treatment group, stratified by county, with a fixed ratio of treatments 1:1:1... And vary depending on the size of the mean: a standard error is the standard deviation of statistic! Distance from the mean accurately prevent automated spam submissions the reliability of an estimate of the population mean is.! Under the Creative Commons-License Attribution 4.0 International ( CC by 4.0 ) ANOVA is the standard deviation of the.... Sample standard deviation ; n \u2014 sample size and Rohlf ( 1981 ) give an equation of the.. Haier Hpp10xct Review, Strong Wind Amsterdam, Black And White M Logo Company Name, Roland Rh-5 Stereo Headphones Review, Outdoor Hugger Ceiling Fan With Light, Rodgersia Pinnata 'chocolate Wings, Coffee Scrub For Face Benefits, Dental Hygienist Schools, Rural Riverfront Property For Sale, \" />\n= , then we can define the total, which due to the Bienaym\u00e9 formula, will have variance, The mean of these measurements The standard error of the regression is the average distance that the observed values fall from the regression line. ) The formula given above for the standard error assumes that the sample size is much smaller than the population size, so that the population can be considered to be effectively infinite in size. / 95% and 99% are in general use. [12] See also unbiased estimation of standard deviation for more discussion. N 2 answer explanation . , Control treatment was daily folic acid. The standard deviation (SD) & standard error of the mean (SEM) are used to represent the characteristics of the sample data and explain statistical analysis results. {\\displaystyle \\sigma } , which is the most often calculated quantity, and is also often colloquially called the standard error). [11]. {\\displaystyle \\operatorname {SE} } An online standard error calculator helps you to estimate the standard error of the mean (SEM) from the given data sets and shows step-by-step calculations. is equal to the sample mean, Here we discuss the formula for the calculation of standard error of mean with the examples and downloadable excel sheet.. \u03c3 \", \"On the value of a mean as calculated from a sample\", \"Analysis of Short Time Series: Correcting for Autocorrelation\", Multivariate adaptive regression splines (MARS), Autoregressive conditional heteroskedasticity (ARCH), https://en.wikipedia.org/w/index.php?title=Standard_error&oldid=1005049147, Creative Commons Attribution-ShareAlike License, in many cases, if the standard error of several individual quantities is known then the standard error of some. n The Statistics and Machine Learning Toolbox implementation of the two-way ANOVA is the anova2 (link) function. ), the standard deviation of the mean itself ( \u00af The Standard Error of Mean or SEM in Excel measures the deviation of a sample mean from the population mean. Taylor Series Method. \u2061 Calculating Standard Error of the Mean (SEM). of the entire population being sampled is seldom known. 2 So we know that the variance-- or we could almost say the variance of the mean or the standard error-- the variance of the sampling distribution of the sample mean is equal to the variance of our original \u2026 {\\displaystyle x_{1},x_{2},\\ldots ,x_{n}} \u03c3 Statology Study is the ultimate online statistics study guide that helps you understand all of the core concepts taught in any elementary statistics course and \u2026 We do not capture any email address. , When you look at scientific papers, sometimes the \\\"error bars\\\" on graphs or the \u00b1 number after means in tables represent the standard error of the mean, while in other papers they represent 95% confidence intervals. 4 . Statistics courses, especially for biologists, assume formulae = understanding and teach how to do statistics, but largely ignore what those procedures assume, and how their results mislead when those assumptions are unreasonable. Therefore, the standard error of the mean is usually estimated by replacing {\\displaystyle N=n} Meaning of standard error. It is abbreviated as SEM. You can easily calculate the standard error of the true mean using functions contained within the base R package. The effect of the FPC is that the error becomes zero when the sample size n is equal to the population size N. If values of the measured quantity A are not statistically independent but have been obtained from known locations in parameter space\u00a0x, an unbiased estimate of the true standard error of the mean (actually a correction on the standard deviation part) may be obtained by multiplying the calculated standard error of the sample by the factor\u00a0f: where the sample bias coefficient \u03c1 is the widely used Prais\u2013Winsten estimate of the autocorrelation-coefficient (a quantity between \u22121 and +1) for all sample point pairs. , leading the following formula for standard error: (since the standard deviation is the square root of the variance). Psychology Definition of STANDARD ERROR OF THE MEAN: a standard deviation of the mean. The SD does not change predictably as you acquire more data. ), the standard deviation of the sample ( {\\displaystyle {\\bar {x}}} It is also called the standard deviation of the mean and is abbreviated as SEM. For the computer programming concept, see, Independent and identically distributed random variables with random sample size, Standard error of mean versus standard deviation, unbiased estimation of standard deviation, Student's t-distribution \u00a7\u00a0Confidence intervals, Illustration of the central limit theorem, \"List of Probability and Statistics Symbols\", \"Standard deviations and standard errors\", \"What to use to express the variability of data: Standard deviation or standard error of mean? Ungraded . When a \u2026 {\\displaystyle \\sigma _{x}} Remember our rule for normal distributions: 95% of the cases fall within two standard deviations of the mean. If , reducing the error on the estimate by a factor of two requires acquiring four times as many observations in the sample; reducing it by a factor of ten requires a hundred times as many observations. Two interventions were investigated\u2014daily iron with folic acid and daily multiple micronutrients (recommended allowance of 15 vitamins and minerals). The standard error is, by definition, the standard deviation of An interval estimate gives you a range of values where the parameter is expected to lie. \u03c3 The terms \u201cstandard error\u201d and \u201cstandard deviation\u201d are often confused.1 The contrast between these two terms reflects the important distinction between data description and inference, one that all researchers should appreciate. Definition of standard error in the Definitions.net dictionary. {\\displaystyle \\sigma } This makes sense, because the mean of a large sample is likely to be closer to the true population mean than is the mean of a small sample. This forms a distribution of different means, and this distribution has its own mean and variance. Mean birth weight was 3153.7 g (n=1545; 95% confidence interval 3131.5 to 3175.9, standard deviation 444.9, standard error 11.32) in the control group, 3173.9 g (n=1470; 3152.2 to 3195.6, 424.4, 11.07,) in the iron-folic acid group, and 3197.9 g (n=1406; 3175.0 to 3220.8, 438.0, 11.68) in the multiple micronutrients group. So the standard deviation in \u2026 How to calculate Standard Error Note the number of measurements (n) and determine the sample mean (\u03bc). x If The SEM quantifies how accurately you know the true mean of the population. To understand this, first we need to understand why a sampling distribution is required. For instance, usually, the population mean estimated value is \u2026 x , which is the standard error), and the estimator of the standard deviation of the mean ( The text in this article is licensed under the Creative Commons-License Attribution 4.0 International (CC BY 4.0).. In such cases, the sample size {\\displaystyle N} Var x 1 X What does standard error mean? independent observations from a population with mean x If you have a subscription to The BMJ, log in: Subscribe and get access to all BMJ articles, and much more. 93 hours What is the standard error of this mean estimate Standard error of from GOVT 457 at Regent University 900 seconds . The standard error (SE)[1][2] of a statistic (usually an estimate of a parameter) is the standard deviation of its sampling distribution[3] or an estimate of that standard deviation. \u00af T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. Guide to Standard Error Formula. This is basically a variant of standard deviation as both concepts correspond to the spread measures. n Definition of Standard Deviation. {\\displaystyle {\\bar {x}}} It is used to make a comparison between sample means across the populations. The setting was 327 villages in two rural counties in northwest China. Calculation of CI for mean = (mean + (1.96 x SE)) to (mean \u2013 (1.96 x SE)) {\\displaystyle \\operatorname {E} (N)=\\operatorname {Var} (N)} Analyze, graph and present your scientific work easily with GraphPad Prism. You might find more information there. \u00af N = size of the sample data set Meaning of standard error. If we plot the actual data points along with \u2026 Which of the following statements, if any, are true? If the sampling distribution is normally distributed, the sample mean, the standard error, and the quantiles of the normal distribution can be used to calculate confidence intervals for the true population mean. What is N? n (c) What is the probability that you would have gotten this mean difference (see #24) or less in your sample? Using descriptive and inferential statistics, you can make two types of estimates about the population: point estimates and interval estimates.. A point estimate is a single value estimate of a parameter.For instance, a sample mean is a point estimate of a population mean. \u00af While every effort has been made to follow citation style rules, there may be some discrepancies. Calculate the mean Add all the samples together and divide the sum total by the number of samples 2. Confidence intervals and standard error of the mean serve the same purpose, to express the reliability of an estimate of the mean. 1 x In probability & statistics, the standard deviation of sampling distribution of a statistic is called as Standard Error often abbreviated as SE. When we calculate the standard deviation of a sample, we are using it \u2026 Copyright \u00a9 2021 BMJ Publishing Group Ltd \u00a0 \u00a0 \u4eacICP\u590715042040\u53f7-3, , reader in medical statistics and medical education, reader in medical statistics and medical education, Lincolnshire Partnership NHS Foundation Trust: CAMHS Consultant Psychiatrist, Cambridgeshire and Peterborough NHS Foundation Trust: Consultant General Adult Community Psychiatry, Hertfordshire Partnership University NHS Foundation Trust: Consultant Perinatal Psychiatrist, Hertfordshire Partnership University NHS Foundation Trust: Consultant Psychiatrist in General Adult Community, Women\u2019s, children\u2019s & adolescents\u2019 health. x Var Assuming a normal distribution, we can state that 95% of the sample mean would lie within 1.96 SEs above or below the population mean, since 1.96 is the 2-sides 5% point of the standard normal distribution. How can you calculate the Confidence Interval (CI) for a mean? The standard error on the mean may be derived from the variance of a sum of independent random variables,[6] given the definition of variance and some simple properties thereof. a) The standard error of the mean birth weight for a treatment group provides a measure of the precision of the sample mean as an estimate of the population parameter \u2026. In short, standard error of a statistic is nothing but the standard deviation of its sampling distribution. Birth weight was available for analysis for 4421 live births. are taken from a statistical population with a standard deviation of , \u00af 200 . X The following expressions can be used to calculate the upper and lower 95% confidence limits, where A high standard error corresponds to the higher spreading of data for the undertaken sample. 16. (c) What is the probability that you would have gotten this mean difference (see #24) or less in your sample? Although average birth weight was higher in the iron-folic acid group than in the control group, the difference was not significant (24.3 g; P=0.169). Standard Error: A standard error is the standard deviation of the sampling distribution of a statistic. n To summarize: SD measures variability in data we used to get 1 average (in this case, cell counts). In other words, it is the actual or estimated standard deviation of the sampling distribution of the sample statistic. Report an issue . = (As we can rarely have the S.D. As a result, we need to use a distribution that takes into account that spread of possible \u03c3's. Standard Error of the Mean The standard error of the mean is a method used to evaluate the standard deviation of a sampling distribution. x This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called an analytic study, following W. Edwards Deming. \u03c3 25 . S I am not talking about the standard deviation (SD). Is SE just the abbreviation of SEM? , = mean value of the sample data set. x This often leads to confusion about their interchangeability. Then 95 percent of those confidence intervals would contain the true mean. Moreover, this formula works for positive and negative \u03c1 alike. given by:[2]. answer \u2026 How to calculate standard error of the mean 1. Join courses with the best schedule and enjoy fun and interactive classes. It is the average of all the measurements. ror of the mean (SEM), a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Standard error of the mean is often abbreviated to standard error. When you divide by a bigger number, you get a smaller number, so the more samples you have, the lower the SEM. of the sample means). N The standard deviation of the sample data is a description of the variation in measurements, while the standard error of the mean is a probabilistic statement about how the sample size will provide a better bound on estimates of the population mean, in light of the central limit theorem.[8]. {\\displaystyle \\sigma } N = size of the sample data set In regression analysis, the term \"standard error\" refers either to the square root of the reduced chi-squared statistic, or the standard error for a particular regression coefficient (as used in, say, confidence intervals). the standard deviation of the sampling distribution of the sample mean! An example of how A simple explanation of the difference between the standard deviation and the standard error, including an example. Where: s = sample standard deviation x 1, ..., x N = the sample data set x\u0304. 2 observations n Put simply, the standard error of the sample mean is an estimate of how far the sample mean is likely to be from the population mean, whereas the standard deviation of the sample is the degree to which individuals within the sample differ from the sample mean. {\\displaystyle {\\sigma }_{\\bar {x}}} Practically this tells us that when trying to estimate the value of a mean, due to the factor The true standard deviation has a Poisson distribution, then SE Calculating the \u2018Standard Error of the mean\u2019 or SEM is simple using Excel\u2019s in-built functions. Confidence Interval: The two confidence intervals i.e. I prefer 95% confidence intervals. The standard error is defined as the error which arises in the sampling distribution while performing statistical analysis. Determine how much each measurement varies from the mean. \u03c3 Remember that, SD & SEM both are different, each have its own meaning. If the statistic is the sample mean, it is called the standard error of the mean (SEM).[2]. and standard deviation sigma \u2014 standard deviation; n \u2014 sample size. The SEM gets smaller as your samples get larger. Outcome measures included birth weight. If you are unable to import citations, please contact + x {\\displaystyle n} 1 [5] See unbiased estimation of standard deviation for further discussion. , ( The standard error of the mean, also called the standard deviation of the mean, is a method used to estimate the standard deviation of a sampling distribution. Can I assume the SE the same as SEM? Mathematically, the variance of the sampling distribution obtained is equal to the variance of the population divided by the sample size. Villages were randomised to treatment group, stratified by county, with a fixed ratio of treatments (1:1:1).1. x \u00af ( While, \u2026 In many practical applications, the true value of \u03c3 is unknown. When the sampling fraction is large (approximately at 5% or more) in an enumerative study, the estimate of the standard error must be corrected by multiplying by a ''finite population correction'':[10] Standard deviation (SD) is the measure of dispersion of the individual data values. Please note: your email address is provided to the journal, which may use this information for marketing purposes. will have an associated standard error on the mean {\\displaystyle X} Calculation of CI for mean = (mean + (1.96 x SE)) to (mean \u2013 (1.96 x SE)) Assuming a normal distribution, we can state that 95% of the sample mean would lie within 1.96 SEs above or below the population mean, since 1.96 is the 2-sides 5% point of the standard normal distribution. . Introduction. n 4. {\\displaystyle {\\bar {x}}} With n = 2, the underestimate is about 25%, but for n = 6, the underestimate is only 5%. Sample size is 25. Please refer to the appropriate style manual or other sources if you have any questions. [9] If the population standard deviation is finite, the standard error of the mean of the sample will tend to zero with increasing sample size, because the estimate of the population mean will improve, while the standard deviation of the sample will tend to approximate the population standard deviation as the sample size increases. with estimator A cluster randomised double blind controlled trial investigated the effects of micronutrient supplements during pregnancy. n technical support for your product directly (links go to external sites): Thank you for your interest in spreading the word about The BMJ. {\\displaystyle nS_{X}^{2}+n{\\bar {X}}^{2}} E Hence the estimator of instead: As this is only an estimator for the true \"standard error\", it is common to see other notations here such as: A common source of confusion occurs when failing to distinguish clearly between the standard deviation of the population ( {\\displaystyle {\\widehat {\\sigma _{\\bar {x}}}}} Standard Error of the Mean (a.k.a. N Similar \u2026 [2] In other words, the standard error of the mean is a measure of the dispersion of sample means around the population mean. Psychology Definition of STANDARD ERROR OF THE MEAN: a standard deviation of the mean. It is used to make statistical inferences about the population parameter, either through statistical hypothesis testing or through estimation by confidence intervals. The standard errors that are reported in computer output are only estimates of the true standard errors. Now learn Live with India's best teachers. The standard deviation (often SD) is a measure of variability. In 1893, Karl Pearson coined the notion of standard deviation, which is undoubtedly most used measure, in research studies. A trial with three treatment arms was used. ) \u03c3 ^ Small samples are somewhat more likely to underestimate the population standard deviation and have a mean that differs from the true population mean, and the Student t-distribution accounts for the probability of these events with somewhat heavier tails compared to a Gaussian. Definition of standard error in the Definitions.net dictionary. To estimate the standard error of a Student t-distribution it is sufficient to use the sample standard deviation \"s\" instead of \u03c3, and we could use this value to calculate confidence intervals. {\\displaystyle {\\bar {x}}} Tags: Topics: Question 8 . x Therefore, the relationship between the standard error of the mean and the standard deviation is such that, for a given sample size, the standard error of the mean equals the standard deviation divided by the square root of the sample size. [4] Sokal and Rohlf (1981) give an equation of the correction factor for small samples of\u00a0n\u00a0<\u00a020. The mean and standard deviation of a population are 200 and 20, respectively. with the sample standard deviation which is simply the square root of the variance: There are cases when a sample is taken without knowing, in advance, how many observations will be acceptable according to some criterion. Calculating the \u2018Standard Error of the mean\u2019 or SEM is simple using Excel\u2019s in-built functions. \u2026 Standard error of the mean Summary. Standard Error (SE) provides, the standard deviation in different values of the sample mean. \u2061 When you use VARMETHOD=TAYLOR, or by default if you do not specify the VARMETHOD= option, PROC SURVEYMEANS uses the Taylor series method to estimate the variance of the mean .The procedure computes the estimated variance as The standard error is the standard deviation of the Student t-distribution. answer choices . For example, normally, the estimator of the population mean is the sample mean. \u00af I recommend Snedecor and \u2026 \u00af Note: The Student's probability distribution is approximated well by the Gaussian distribution when the sample size is over 100. \u03c3 is equal to the standard error for the sample mean, and 1.96 is the approximate value of the 97.5 percentile point of the normal distribution: In particular, the standard error of a sample statistic (such as sample mean) is the actual or estimated standard deviation of the sample mean in the process by which it was generated. This is because as the sample size increases, sample means cluster more closely around the population mean. SE \u2061 , then the mean value calculated from the sample To the uninformed, surveys appear to be an easy type of research to design and conduct, but when students and professionals delve deeper, they encounter the If people are interested in managing an existing finite population that will not change over time, then it is necessary to adjust for the population size; this is called an enumerative study. x Understanding \u2018Standard Error of the mean\u2019 isn\u2019t h Standard Error of the Mean (SEM) The standard error of the mean also called the standard deviation of mean, is represented as the standard deviation of the measure of the sample mean of the population. To obtain an unbiased estimate of the temporal variance, we must remove the sampling variation from the estimate of the total variance. {\\displaystyle N} No coding required. ( You can download a PDF version for your personal record. What is the Standard Error? If a statistically independent sample of x When the sample size is small, using the standard deviation of the sample instead of the true standard deviation of the population will tend to systematically underestimate the population standard deviation, and therefore also the standard error. is simply given by. \u2026 are alternatives . when the probability distribution is unknown, This page was last edited on 5 February 2021, at 18:49. {\\displaystyle \\sigma _{x}} All the sample means which are normally distributed around M pop will lie between M pop + 3 SE M and M pop \u2013 3 SE M . ) R. A. Fisher names the limits of the confidence interval which contains the parameter as \u201cfiduciary limits\u201d and named the confidence placed in the interval as fiduciary probability. {\\displaystyle {\\bar {x}}} x But if you mean you are interested in whether a particular data point is plausibly from the population you have modelled (eg to ask \"is this number a really big outlier? Standard Error of the Mean What is the standard error of the mean? This question is for testing whether or not you are a human visitor and to prevent automated spam submissions. 16 . It is a measure of how precise is our estimate of the mean.The main use of the standard error of the mean is to Standard Error of the Mean (a.k.a. , to account for the added precision gained by sampling close to a larger percentage of the population. In total, 5828 pregnant women were recruited. When conducting statistical analysis, especially during experimental design, one practical issue that one cannot avoid is to determine the sample size for \u2026 T What does standard error mean? The sampling distribution of a population mean is generated by repeated sampling and recording of the means obtained. Understanding \u2018Standard Error of the mean\u2019 isn\u2019t h In this case, the observed values fall an average of 4.89 units from the regression line. In simple words, SD determines how the sample data represents the mean accurately. \u03c3 If the statistic is the sample mean, it is called the standard error of the mean (SEM). is a random variable whose variation adds to the variation of Q. We can describe this using STANDARD ERROR of the MEAN (SEM) -> mathematically, SEM = SD/\u221a(sample size). Everybody with basic statistical knowledge should understand the differences between the standard deviation (SD) and the standard error of mean (SE or SEM). A t-test is a statistical method used to see if two sets of data are significantly different. x When the true underlying distribution is known to be Gaussian, although with unknown \u03c3, then the resulting estimated distribution follows the Student t-distribution. We are an Essay Writing Company Get an essay written for you for as low as $13/page simply by clicking the Place Order button! It has a great role to play the testing of statistical hypothesis and interval estimation. Standard deviation tells you how spread out the data is. {\\displaystyle n} If a number is added to a set that is far away from the mean, how does this affect standard deviation? How can you calculate the Confidence Interval (CI) for a mean? It gives an idea of the exactness and \u2026 such that. ), you need to compare it to your estimate of the population mean and your estimate of the population standard deviation (not the sample mean's standard deviation, also known as SEM). {\\displaystyle \\sigma _{\\bar {x}}} Where: s = sample standard deviation x 1, ..., x N = the sample data set x\u0304. a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Access this article for 1 day for:\u00a330 /$37 / \u20ac33 (excludes VAT). 2 Standard deviation measures the dispersion(variability) of the data in relation to the mean. NOTE: We only request your email address so that the person you are recommending the page to knows that you wanted them to see it, and that it is not junk mail. Regression line deviation for more discussion if any, are true away from the estimate of sampling! Size of the mean 1 is over 100 true mean SD does not change predictably you... Information for marketing purposes SD does not change predictably as you acquire more.... An average of 4.89 units from the mean serve the same as SEM get 1 average ( in this,... Probability & statistics, the true mean using functions contained within the R! The value of S.D we are using it \u2026 Taylor series method reported in output... Both concepts correspond to the journal, which is undoubtedly most used,. Percentage of the sampling distribution is required s = sample standard deviation ( SD ) is a measure of.! A series or the distance from the mean: a standard deviation ; n \u2014 sample size set x\u0304 the. 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Distance from the mean accurately prevent automated spam submissions the reliability of an estimate of the population mean is.! Under the Creative Commons-License Attribution 4.0 International ( CC by 4.0 ) ANOVA is the standard deviation of the.... Sample standard deviation ; n \u2014 sample size and Rohlf ( 1981 ) give an equation of the..", "date": "2021-07-24 13:38:48", "meta": {"domain": "mecaartfair.com", "url": "https://mecaartfair.com/rick-stein-brkmyry/aa8fa1-what-is-standard-error-of-the-mean", "openwebmath_score": 0.8511279225349426, "openwebmath_perplexity": 773.6649043747163, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9546474181553805, "lm_q2_score": 0.8947894562828415, "lm_q1q2_score": 0.8542084442330713}}
{"url": "https://physics.stackexchange.com/questions/459039/how-to-calculate-the-inertia-tensor-of-a-spherical-cap", "text": "# How to calculate the inertia tensor of a spherical cap?\n\nIn this question, an attempt is made at calculating the diagonal elements of the inertia tensor of a homogeneous spherical cap, where the $$z$$-axis is the symmetry axis. The mass moment of inertia about the $$z$$-axis is expressed by:\n\n$$M_{zz}=\\rho\\int_0^{2\\pi} \\int_{R-h}^R \\int_{0}^{\\sqrt{R^2-z^2}} r^3 dr dz d\\theta$$\n\nwhere $$\\rho$$ is the density, given by:\n\n$$\\rho = \\frac{m}{\\frac{1}{3} \u03c0 h^2 (3 R - h)}$$\n\n$$R$$ is the radius of the sphere, and $$h$$ and $$m$$ are the height and mass of the cap. The triple integral equation solves as:\n\n$$M_{zz}=\\frac{mh}{10(3R - h)}(3h^2 - 15hR + 20R^2)$$\n\nwhich is correct. However, the following expression is given for the mass moment of inertia about the $$x$$-axis or $$y$$-axis (in the accepted answer to this question):\n\n$$M_{xx}=M_{yy}=\\rho\\int_0^{2\\pi} \\int_{R-h}^R \\int_{0}^{\\sqrt{R^2-z^2}} r(r^2 \\cos^2\\theta+z^2) dr dz d\\theta$$\n\nwhich solves as (see WolframAlpha):\n\n$$M_{xx}=M_{yy}=\\frac{m}{20(3R - h)}(-9h^3 + 45h^2R - 80hR^2 + 60R^3)$$\n\nThis is seemingly not correct. If we plug in $$R=5$$, $$h=2$$ and $$m=4.28\\times10^{5}$$, we get:\n\n$$I=\\begin{bmatrix}7.1246&0&0\\\\0&7.1246&0\\\\0&0&2.3836\\end{bmatrix} \\times10^{6}$$\n\nAccording to CATIA's \"Measure Inertia\"-function, this should be:\n\n$$I=\\begin{bmatrix}1.2896&0&0\\\\0&1.2896&0\\\\0&0&2.3836\\end{bmatrix} \\times10^{6}$$\n\nFor reference (don't mind the definition of the axes here):\n\nThe question is:\n\nWhat is the correct expression for the mass moment of inertia about the principal $$x$$-axis/$$y$$-axis (red/green lines in reference figure)?\n\nThe geometric centroid (the origin of the axes system) for a spherical cap is given by:\n\n$$z=\\frac{3(2R-h)^2}{4(3R-h)}$$\n\nEdit:\n\nThanks to probably_someone, the answer can be derived as:\n\n$$M_{xx}=M_{yy}=\\frac{mh}{80(h-3R)^3}(-9h^3 + 72h^2R - 220hR^2 + 240R^3)$$\n\nThe moment of inertia of an object is only defined relative to a particular choice of axes of rotation. You are using a different choice of $$x$$ and $$y$$ axes than the original question.\n\nIn the original question, the $$x$$ and $$y$$ axes passed through the center of the whole sphere, meaning they passed through a point on the $$z$$-axis that is below the bottom of the cap. CATIA is calculating the moments of inertia for $$x$$ and $$y$$ axes that pass through the center of mass of the cap. Fortunately, there is an easy way to convert between these two choices: the parallel axis theorem. The theorem states:\n\nSuppose a body of mass $$m$$ has moment of inertia $$I_0$$ about an axis passing through its center of mass. Then the moment of inertia $$I$$ about another axis, parallel to the first and displaced a distance $$d$$ from the center of mass, is given by: $$I=I_0+md^2$$\n\nApplying this to our situation, it's clear that the CATIA calculation gives you $$I_0$$, and we already have $$m$$. Since you defined the center of the whole sphere as the origin, the distance $$d$$ that the axis should be displaced is equal to the $$z$$-coordinate of the center of mass, namely:\n\n$$d=\\frac{3(2R-h)^2}{4(3R-h)}$$\n\nSo the moment of inertia about $$x$$ and $$y$$ axes passing through the center of the sphere will be:\n\n$$I=I_0+m\\frac{9(2R-h)^4}{16(3R-h)^2}$$\n\nPlugging in $$m=4.28\\times 10^5$$, $$R=5$$, $$h=2$$, and $$I_0=1.2896\\times 10^6$$, we get $$I=7.1246\\times 10^6$$, which perfectly agrees with the calculation that you did using the other question.\n\n\u2022 Thanks a lot. Didn't notice the origin of the original question was at the centre of the sphere. What is good practice? Should I edit my original question to correct for this mistake? \u2013\u00a0woeterb Feb 5 at 18:57\n\u2022 @woeterb I think it's fine to leave it as is, as the answer specifically references the oversight in the question. It might not make much sense to future readers if you edit the question at this point. \u2013\u00a0probably_someone Feb 5 at 19:07", "date": "2019-08-25 16:29:18", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/459039/how-to-calculate-the-inertia-tensor-of-a-spherical-cap", "openwebmath_score": 0.6707785725593567, "openwebmath_perplexity": 87.68273880550565, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668723123673, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8542012087706075}}
{"url": "http://math.stackexchange.com/questions/264964/perfect-square-in-an-ufd", "text": "# Perfect Square in an UFD\n\nLet $R$ be an UFD with quotient field $F$. Show that an element $d\\in R$ is a square in $R$ if and only if $d$ is a square in $F$.\n\nAnd then get a counterexample that above statement is not true if $R$ is not UFD.\n\n-\n\nIf $d$ is the square of an element in $R$, then $d$ is certainly the square of an element in $F$.\n\nNow suppose $d$ is the square of an element in $F$. A typical nonzero element in $F$ is a quotient of elements in $R$. And since every element of $R$ is a product of irreducibles, every element in $F$ is a quotient of products of irreducibles. In other words, a nonzero element in $F$ is just a product of the form $A^aB^bC^c$ etc. where $A, B, C$ etc. are irreducible members of $R$ and $a, b, c$ etc. are integers. These latter integers may, of course, be negative.\n\nSo what happens if $d$ is the square of an element in $F$? We have that $d = (A^aB^bC^c$ etc.)$^2$, or $d = A^{2a}B^{2b}C^{2c}$ etc. Since $d$ is a member of $R$, every power belonging to the irreducibles which compose $d$ (that is, $2a, 2b, 2c$ etc.) must be positive. But this means that $a, b, c$ etc. must also be positive, meaning that $A^{a}B^{b}C^{c}$ etc. has to be an element of $R$. Thus $d$ is the square of an element in $R$.\n\n-\nCan you give me a counterexample \u2013\u00a0Muniain Dec 25 '12 at 15:39\nlol maybe. can you think of any non-UFDs, first of all? \u2013\u00a0D_S Dec 25 '12 at 15:52\n@Firmino, see my answer. I have a counterexample. \u2013\u00a0Amr Dec 25 '12 at 16:16\n\nLet me give a counterexample for the case when $R$ is not an UFD: set $R=K[X^2,X^3]$, the ring of polynomials over a field $K$ whose monomial of degree one is missing. Then the field of fractions of $R$ is $F=K(X)$. Now take $X^2\\in R$. This is obviously a square in $F$, but there is no $a\\in R$ such that $a^2=X^2$ (otherwise $X\\in R$, a contradiction).\n\n-\nDear @Firmino, now you have also a counterexample. \u2013\u00a0user26857 Feb 17 '13 at 11:18\n\nForward direction: If $d$ is a squre in $R$, then $d=c^2$ for some $c\\in R$. Thus, $\\frac{d}{1}=[\\frac{c}{1}]^2$.\n\nBackward direction: Let $\\frac{d}{1}=[\\frac{r}{s}]^2$. Thus, $s^2d=r^2$. By writing $d,s,r$ as products of irreducibles one can see that the exponents of the irreducibles that appear in the prime factorization of $d$ are even.\n\n-\nHow is that a counterexample? A counterexample would entail giving an example of an integral domain $R$ with quotient field $K$ such that the square of something in $F$ is in $R$, but is not the square of anything in $R$. $\\mathbb{Z_6}$ isn't an integral domain. \u2013\u00a0D_S Dec 25 '12 at 17:05\ni was not concentrating \u2013\u00a0Amr Dec 25 '12 at 17:26\n\nJust as for irrationality proofs in $\\,\\Bbb Z,\\,$ this is an immediate consequence of the monic case of the Rational Root Test (RRT), which is true in any UFD (or any GCD domain), since the proof uses only Euclid's Lemma $\\rm\\:(a,b)=1,\\ a\\mid bc\\:\\Rightarrow\\:a\\mid c.\\:$ In particular, if $\\rm\\:x^2\\! -c,\\ c\\in R,\\:$ has a \"rational\" root $\\rm\\:x\\in F\\:$ then it must be \"integral\" $\\rm\\:x\\in R\\:$ by RRT.\n\nThis fails in non-UFD domains, e.g. there are very simple quadratic integer counterexamples:\n\n$\\rm\\ \\ x^2\\! = (1\\!+\\!\\sqrt{d})^2\\!\\in \\Bbb Z[2\\sqrt{d}]\\$ has root $\\rm\\ x = 1\\!+\\!\\sqrt{d} = \\dfrac{2\\!+\\!2\\sqrt{d}}2\\:$ a proper fraction over $\\rm\\:\\Bbb Z[2\\sqrt{d}]$\n\nRemark $\\$ Rings satisfying the monic case of the Rational Root Test are called integrally closed. Thus the remark above translates as: the usual proof of RRT immediately generalizes to show that UFDs and GCD domains are integrally closed.\n\n-", "date": "2015-11-26 05:16:25", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/264964/perfect-square-in-an-ufd", "openwebmath_score": 0.9526710510253906, "openwebmath_perplexity": 110.03046908786571, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9811668731384179, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.854201202901308}}
{"url": "https://math.stackexchange.com/questions/2140824/proof-of-uncountable-basis-for-mathbbn-to-mathbbr-over-mathbbr", "text": "Proof of Uncountable Basis for $\\mathbb{N} \\to \\mathbb{R}$ over $\\mathbb{R}$\n\nDoes anyone have a simple proof of the uncountability of bases of the vector space of all functions $f : \\mathbb{N} \\to \\mathbb{R}$. I have seen a proof which uses the determinant of the Vandermonde matrix to show the linear independence of functions of the form $f_c=c^n$ but I believe that there might be a simpler one that doesn't require the use of matrices.\n\nI have attached a link of another proof that I found online but I find the use of the limit unsettling. https://minhyongkim.wordpress.com/2013/10/23/a-vector-space-of-uncountable-dimension/\n\n\u2022 Zorn's lemma gives you the existence of a basis. Not the fact that it is uncountable. Feb 12, 2017 at 14:29\n\u2022 I was thinking along the lines of finding an uncountable linearly independent set of functions which implies that the bases cannot be countable. Feb 12, 2017 at 14:43\n\u2022 And Zorn's lemma would help you how? Feb 12, 2017 at 14:53\n\u2022 @AsafKaragila: I think he simply mean that Zorn's Lemma guarantees that every linearly independent set we find is a subset of a basis. Feb 12, 2017 at 14:56\n\nHere is a diaginalisation argument.\n\nLet $\\{f_i\\}$ be a countable set we find $g\\not \\in span \\{f_i\\}$.\n\nConstruct $g$ as follows.\n\nLook at the vector $$(f_0(0), f_0(1))$$ define $(g(0),g(1))$ not to be a linear multiple of this vector. Now look at the vectors\n\n$$(f_0(2), f_0(3), f_0(4))$$ $$(f_1(2), f_1(3), f_1(4))$$ define $(g(2),g(3),g(4))$ not a linear combination of these vectors. Now look at the vectors $$(f_0(5), f_0(6), f_0(7),f_0(8))$$ $$(f_1(5), f_1(6), f_1(7),f_1(8))$$ $$(f_2(5), f_2(6), f_2(7),f_2(8))$$\n\ndefine $(g(5),g(6),g(7),g(8))$ not in the span of these three vectors, etc. I think the construction in clear, and $g$ is not in the span. Thus there is no countable basis.\n\n\u2022 That's a nice argument! Feb 12, 2017 at 14:49\n\u2022 Sry i can't see how g is not in the span. Feb 12, 2017 at 14:50\n\u2022 @JhonDeo: If it's in the span, then by definition is has to be in the span of finitely many of the $f_i$s. And there's a sequence of elements of $g$ that have been chosen explicitly not to be in the span of the first $n$ vectors, for any $n$. Feb 12, 2017 at 14:52\n\u2022 @JhonDeo $g$ not in the span of $f_0$ since the first two coordinates of $g$ is not a multiple of the first two of $f_0$. $g$ is not in the span of $\\{f_0,f_1\\}$ since the coordinates $g(2),g(3),g(4)$ is not in the span or the same coordinates for $\\{f_0,f_1\\}$ etc. Feb 12, 2017 at 14:53\n\u2022 @Henning Makholm, what does mean \"not to be linear multiple of this vector\"?\n\u2013\u00a0ZFR\nNov 15, 2019 at 0:51\n\nThere are uncountably many functions $\\mathbb N\\to\\{0,1\\}$. Well-order them all, and then remove each one that is a (finite) linear combination of vectors that come earlier in the well-order. The result is, by contruction, a linearly independent set in your vector space. How large is it?\n\nI claim that each of the removed vectors is not just a linear combination of vectors that remain, but a rational linear combination of such vectors. In each case, when we express the new vector as a linear combination of finitely many already-accepted ones, we have to solve an equation involving a matrix that is infinitely tall but has finite width. But because all elements are either $0$ or $1$, there are actually only finitely many different rows in the matrix, so we can find the coefficients by doing ordinary finite-dimensional linear algebra over $\\mathbb Q$.\n\nNow, if the reduced set of $0,1$-vectors were countable, there would be only countably many different finite rational combinations of them -- but this contradicts the fact that there are uncountably many vectors to express.\n\nSo we have an uncountable linearly independent set in your vector space, and if we extend that to a basis, we get an uncountable basis.\n\nI don't have a proof with Zorn's Lemma, but probably you like this one aswell:\n\nThe space you consider contains $\\ell^\\infty$, the space of bounded sequences, so if we can show that this space has an uncountable base the same is true for the space of all sequences.\n\nThe space $\\ell^\\infty$ can be made a Banach-space if we consider the supremum-norm on it. But infinite dimensional Banach-spaces can't have a countable Hamel-Base which is due to Baire's theorem.\n\nIf $(b_n)_{n\\in\\mathbb N}$ was a Hamel-base of $\\ell^\\infty$ then $\\ell^\\infty = \\bigcup_{n\\in\\mathbb N}\\operatorname{span}(b_1,...,b_n)$. Baire's theorem tells us that one of the sets in the union has nonempty interior, thus is all of $\\ell^\\infty$ because it is also a subspace. This is a contradiction because it would make $\\ell^\\infty$ finite-dimensional.\n\n\u2022 Thanks I have seen this proof. However, I try to avoid such proofs as i have not learnt stuff like Banach Spaces and Hamel-base Feb 12, 2017 at 14:39\n\nOne easy way is the following, find a norm which makes the space complete and then uses baire theorem. Zorn's lemma implies that exist a basis, baire theorem implies it must be uncountable\n\n\u2022 How do you find such a norm? Feb 13, 2017 at 6:17", "date": "2022-06-26 17:53:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2140824/proof-of-uncountable-basis-for-mathbbn-to-mathbbr-over-mathbbr", "openwebmath_score": 0.8858878016471863, "openwebmath_perplexity": 155.85675111220795, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668673560625, "lm_q2_score": 0.870597273444551, "lm_q1q2_score": 0.8542011995143194}}
{"url": "https://www.jiskha.com/questions/696021/The-value-of-a-certain-fraction-becomes-1-5-if-one-is-added-to-its-numerator-If", "text": "# Mathematics\n\nThe value of a certain fraction becomes 1/5 if one is added to its numerator. If one is taken from its denominator, its value becomes 1/7. Find the fraction by setting up a pair of simultaneous equations and solving them\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n3. \ud83d\udc41 171\n1. (n+1)/d = 1/5\nn/(d-1) = 1/7\n\n5(n+1) = d\n7n = d-1\n\n5n - d = -5\n7n - d = -1\n\n2n = 4\nn=2\nd=15\n\nCheck:\nn/d = 2/15\n\n3/15 = 1/5\n2/14 = 1/7\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\nposted by Steve\n\n## Similar Questions\n\n1. ### math\n\nthe numerator of a fraction is 3 less than the numerator.If half of the numerator is added to the numerator and the denominator,the resulting fraction is 1/2.Find the the fraction.\n\nasked by hans on August 14, 2016\n2. ### Mathematics\n\nThe sum of the numerator and denominator of a fraction is 17. If 3 is added to the numerator, the value of the fraction will be 1. What is the fraction? Explain your workings.\n\nasked by Dale on January 15, 2012\n3. ### Mathematics\n\nThe sum of the numerator and denominator of a fraction is 17. If 3 is added to the numerator, the value of the fraction will be 1. What is the fraction? Explain your workings.\n\nasked by Dale on January 15, 2012\n4. ### Maths (quadratic equation) need help\n\nIn an unknown fraction the denominator is one more than twice the numerator. When 2 1/10 is added to this fraction, the result is equal to the reciprocal of the unknown fraction. Find the unknown fraction. (Hint; the the numerator\n\nasked by Danny on May 31, 2016\n5. ### Math\n\nThe denominator of a fraction exceeds the numerator by 7. If 3 is added to the numerator and 1 is subtracted from the denominator, the resulting fraction is equal to 4/5 . Find the original fraction. PLZ SHOW ALL WORK AND HURRY!!!\n\nasked by Nate on April 27, 2017\n6. ### Algebra\n\nWhen nine is added to both the numerator and the denominator of an original fraction, the new fraction equals six sevenths. When seven is subtracted from both the numerator and the denominator of the original fraction, a third\n\nasked by Raven on September 21, 2014\n7. ### math\n\nWhen 5 is added to both the numerator and denominator of a fraction the result becomes 1/2.When 1 is subtracted from both the numerator and denominator the fraction becomes 1/5.Find the fraction. plzz do whole method, i cant\n\nasked by hamza on November 19, 2016\n8. ### math\n\nthe numerator of a fraction is 3 less than the denominator. if 4 us added to the numerator and to the denominator, the resulting fraction is equivalent to 3/4. find the original fraction. Show your solution\n\nasked by mm on September 23, 2016\n9. ### math\n\nThe denominator of a fraction is 20 more than the numerator of the fraction. If 5 is added to the numerator of the fraction and the denominator is unchanged, the value of the resulting fraction becomes 38. Use x as your variable.\n\nasked by romika on October 20, 2017\n10. ### GEN MATH\n\nSolve the ff. problems. Present a complete and systematic solution. 1.) A number added to twice its reciprocal is equal to 11/3. WHat is the number? 2.) The ratio of ten more than three times a number to the square of the same\n\nasked by lab on July 31, 2016\n11. ### St.george high school\n\nThe numerator of a fraction is 8 less than the denominator.when 6 is added to the numerator and 10 is added to the denominator,the value of the fraction is unchanged.find the original fraction.\n\nasked by Lynn on November 4, 2012\n\nMore Similar Questions", "date": "2019-06-25 08:49:47", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/696021/The-value-of-a-certain-fraction-becomes-1-5-if-one-is-added-to-its-numerator-If", "openwebmath_score": 0.9253820776939392, "openwebmath_perplexity": 963.7793918513066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9811668668053619, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8542011973877667}}
{"url": "https://math.stackexchange.com/questions/694214/systems-of-linear-differential-equations-eigenvectors", "text": "# Systems of linear differential equations - eigenvectors\n\nSolve the following system of equations\n\n$\\begin{cases} x_1^{'}(t)=x_1(t)+3x_2(t) \\\\ x_2^{'}(t)=3x_1(t)-2x_2(t)-x_3(t) \\\\ x_3^{'}=-x_2(t)+x_3(t)\\end{cases}$.\n\nFirst, I create the column vectors $X$ and $X^{'}$. Then the matrix $$A= \\begin{bmatrix} 1 & 3 & 0 \\\\ 3 & -2 & -1 \\\\ 0 & -1 & 1 \\\\ \\end{bmatrix}$$\n\nNow, I find the eigenvalues, $-4,3,1$ and their corresponding eigenvectors $(-3,5,1)^T (-3,-2,1)^T (1,0,3)^T$.\n\nI'm just not sure how to take it from here and solve the system of differential equations. I want a diagonal matrix $D$ so that I can read the solutions easy, but I'm not sure how to do it.\n\nEDIT\n\nBuilding on @Francisco 's answer, I'd have that:\n\n$$X=c_1 (-3,5,1)^T e^{-4t} + c_2 (-3,-2,1)^T e^{3t} + c_3 (1,0,3)^T e^{t}$$. But I believe this could be written in a simpler form.\n\n\u2022 Have you tried writing the matrix in a basis of its eigenvectors? (i.e in the form $S M S^{-1}$) where M is a diagonal matrix \u2013\u00a0Pol van Hoften Feb 28 '14 at 17:07\n\u2022 To write $X^{'}=S^{-1}DSX$? \u2013\u00a0jacob Feb 28 '14 at 17:17\n\u2022 You have $X'(t)=A X(t)$. Then $X=\\vec v e^{\\lambda t}$ (where $\\vec v$ is an eigenvector and $\\lambda$ is an eigenvalue) is a solution to the system. The general solution is given by $X=c_1 \\vec v_1 e^{\\lambda_1 t} +c_2 \\vec v_2 e^{\\lambda_2 t} + c_3 \\vec v_3 e^{\\lambda_3 t}$ where $c_1$ $c_2$ and $c_3$ are arbitrary constants. \u2013\u00a0Francisco Feb 28 '14 at 17:25\n\u2022 What would be a faster approach? \u2013\u00a0jacob Feb 28 '14 at 17:56\n\u2022 @Francisco I updated my question. Also, is there a wiki page for the formula you mentioned? \u2013\u00a0jacob Feb 28 '14 at 19:28\n\nThe matrix $A$ that you have is symmetric. So it has an orthonormal basis of eigenvectors. The eigenvectors you have found are mutually orthogonal (which they must be because they correspond to different eigenvalues.) So, if you normalize your eigenvectors and make those normalized vectors the columms of a matrix $U$, then $U^{T}U=I$ is automatic, and $U^{T}AU=D$ is diagonal. Explicitly, $$U = \\left[ \\begin{matrix} -\\frac{3}{\\sqrt{35}} & -\\frac{3}{\\sqrt{14}} & \\frac{1}{\\sqrt{10}} \\\\ \\frac{5}{\\sqrt{35}} & -\\frac{2}{\\sqrt{14}} & 0 \\\\ \\frac{1}{\\sqrt{35}} & \\frac{1}{\\sqrt{14}} & \\frac{3}{\\sqrt{10}} \\end{matrix}\\right]$$ The inverse of $U$ is the transpose $U^{T}$ of $U$. And, $$U^{T}AU = \\left[\\begin{matrix}-4 & 0 & 0 \\\\ 0 & 3 & 0 \\\\ 0 & 0 & 1\\end{matrix}\\right]=D.$$ Equivalently, $$A = U\\left[\\begin{matrix}-4 & 0 & 0 \\\\ 0 & 3 & 0 \\\\ 0 & 0 & 1\\end{matrix}\\right]U^{T}=U DU^{T}.$$ The general solution is expressed in terms of $a=x_{1}(0)$, $b=x_{2}(0)$, $c=x_{3}(0)$ as $$\\left[\\begin{matrix}x_{1}\\\\x_{2}\\\\x_{3}\\end{matrix}\\right] = e^{tA}\\left[\\begin{matrix}a\\\\b\\\\c\\end{matrix}\\right] = Ue^{tD}U^{T}\\left[\\begin{matrix}a \\\\ b \\\\ c\\end{matrix}\\right] = U\\left[\\begin{matrix} e^{-4t} & 0 & 0 \\\\ 0 & e^{3t} & 0 \\\\ 0 & 0 & e^{t}\\end{matrix}\\right]U^{T} \\left[\\begin{matrix}a \\\\ b \\\\ c\\end{matrix}\\right]$$\n\u2022 This would be the same as $$\\begin{cases} x_1= -3c_1 e^{-4t} - 3c_2e^{3t} +c_3e^{t} \\\\ x_2= 5c_1 e^{-4t} -2c_2e^{3t} \\\\ x_3= c_1 e^{-4t} + c_2e^{3t} +3c_3e^{t}. \\end{cases}$$ yes? \u2013\u00a0jacob Feb 28 '14 at 20:55\n\u2022 The forms are equivalent, except that the constants in mine are $x_{1}(0)$, $x_{2}(0)$, $x_{3}(0)$, while your constants are not that. Yours are $x_{1}(0)=-3c_{1}-3c_{2}+c_{3}$, etc. Mine will be the same if you set $a=-3c_{1}-3c_{2}+c_{3}$, $b=\\dots$, etc. \u2013\u00a0Disintegrating By Parts Feb 28 '14 at 21:10", "date": "2021-05-11 07:07:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/694214/systems-of-linear-differential-equations-eigenvectors", "openwebmath_score": 0.9038455486297607, "openwebmath_perplexity": 216.74925715510705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668690081643, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.854201196011292}}
{"url": "https://math.stackexchange.com/questions/3611843/is-int-0-infty-frac-arctan-x-sqrtx3xdx-converges-diverges", "text": "# Is $\\int_0^\\infty\\frac{\\arctan x}{\\sqrt{x^3+x}}dx$ converges/diverges\n\nI need to determine if the following integral converges/diverges.\n\n$$\\int_{0}^{\\infty}\\frac{\\arctan x}{\\sqrt{x^3+x}}dx$$\n\n## What i tried:\n\nWe can write the integral as:\n\n$$\\int_{0}^{\\infty}\\frac{\\arctan x}{\\sqrt{x^3+x}}dx = \\int_{0}^{\\infty}\\frac{\\arctan x}{\\sqrt{x(x^2+1)}}dx$$\n\nBecause the $$x^2+1$$ i thought about defining $$x = \\tan u$$, therefore i will be able to use the identity: $$\\tan^2u + 1 = \\frac{1}{\\cos^2u}$$\n\nDefine: $$x = \\tan u \\Rightarrow u = \\arctan x$$ $$x = 0 \\Rightarrow u = 0$$ $$x \\to \\infty \\Rightarrow u = \\pi/2$$\n\nTherefore we can write the integral as:\n\n$$\\int_{0}^{\\infty}\\frac{\\arctan x}{\\sqrt{x(x^2+1)}}dx = \\int_{0}^{\\pi/2}\\frac{u\\cdot du}{\\sqrt{\\tan u(\\tan^2u+1)}}$$\n\n$$= \\int_{0}^{\\pi/2}\\frac{u\\cdot du}{\\sqrt{\\tan u\\frac{1}{\\cos^2u}}} = \\int_{0}^{\\pi/2}\\frac{u\\cos u}{\\sqrt{\\tan u}}du$$\n\nAnd here i am stuck.\n\nAnother way:\n\nI thought maybe to devide the integral into two, not sure even how, maybe:\n\n$$\\int_{0}^{\\infty}\\frac{\\arctan x}{\\sqrt{x^3+x}}dx = \\int_{0}^{\\pi/2}\\frac{\\arctan x}{\\sqrt{x^3+x}}dx + \\int_{\\pi/2}^{\\infty}\\frac{\\arctan x}{\\sqrt{x^3+x}}dx$$\n\nBut i dont see how it gives me something.\n\nCan i have a hint?\n\nThank you.\n\n\u2022 Can't you also use the comparison test for this? Apr 6 '20 at 3:42\n\nThe long term behavior of the function is $$\\frac{\\pi/2}{x^{3/2}}.$$ The convergence of $$\\int_1^{\\infty} x^{-3/2} \\, dx$$ should now tell you something.\n\n\u2022 I see what you say, when $x \\to \\infty$ we can write $arctanx$ as $\\pi/2$ and the dominant variable at the denominator is $x^{3/2}$ but im not sure how to explain this so i can write my term as $\\frac{\\pi/2}{x^{3/2}}$ If i get to this, surly my integral converges\n\u2013\u00a0Alon\nApr 6 '20 at 3:43\n\u2022 Oh maybe i can write: $\\frac{arctanx}{\\sqrt{x^3+x}} < \\frac{\\pi/2}{\\sqrt{x^3}}$\n\u2013\u00a0Alon\nApr 6 '20 at 3:45\n\u2022 Then use the comparison test\n\u2013\u00a0Alon\nApr 6 '20 at 3:45\n\u2022 @Alon Yes, that is a correct estimate, but that only solves half your problem; it ensures that $\\displaystyle\\int_1^{\\infty} \\dfrac{\\arctan x}{\\sqrt{x^3 + x}} \\, dx \\leq \\displaystyle\\int_1^{\\infty} \\dfrac{\\pi/2}{x^{3/2}}\\, dx < \\infty$. But you also have to worry about $\\displaystyle\\int_0^1 \\dfrac{\\arctan x}{\\sqrt{x^3 + x}}\\, dx$ being finite. The answer by herb steinberg addresses this case. Apr 6 '20 at 5:27\n\u2022 Thank you, i dont quite understand how i can say $\\arctan x$ less or more equal $x$... We are in real analysis, can i say such things? its not... strict enough, isnt it?\n\u2013\u00a0Alon\nApr 6 '20 at 5:30\n\nAs long as you are using the principal value for arctan$$x(\\le \\frac{\\pi}{2}$$), then the integrand behaves like $$x^{-\\frac{3}{2}}$$ as $$x\\to \\infty$$.\n\nFor $$x\\to 0$$, arctan$$x \\approx x$$, so the integrand behaves like $$\\sqrt{x}$$.\n\nTherefore you have convergence at both ends\n\n\u2022 Thank you, I dont understand how you formalaize behave like. for the deminator, you just say $\\lim_{x \\to 0}\\sqrt{x^3+x} = \\lim_{x \\to 0}\\sqrt{x^3}$ Is it a correct statment?\n\u2013\u00a0Alon\nApr 6 '20 at 5:34\n\u2022 For $x\\to \\infty$, $x^3\\gg x$. For $x\\to 0$, $x^3+x=x(x^2+1) \\approx x$. Apr 6 '20 at 17:42\n\u2022 But as far as i know this is an approximation you can write in something like physics, not, as much as i see, in real analysis, are you sure you can write such things in real analysis? If you do so great, i will take it\n\u2013\u00a0Alon\nApr 6 '20 at 17:47\n\u2022 The approximations are perfectly valid in real analysis. If you are taking a course where they first come up, you may have to formalize them. For example $x\\lt x(x^2+1)\\lt x(1+\\epsilon)$ for $x\\lt \\sqrt{\\epsilon}$ for any $\\epsilon \\gt 0$. Apr 6 '20 at 18:19\n\nNote that for $$x>0$$, $$\\arctan x < x$$ and $$\\sqrt {x^3 + x} > \\sqrt x$$, whence $$\\frac{{\\arctan x}}{\\sqrt {x^3 + x}} < \\sqrt x$$ which shows the convergence near $$0$$. Also for $$x>0$$, $$\\arctan x < \\frac{\\pi }{2}$$ and $$\\sqrt {x^3 + x} > \\sqrt {x^3 } = x^{3/2}$$, whence $$\\frac{{\\arctan x}}{\\sqrt {x^3 + x}} < \\frac{\\pi }{2}\\frac{1}{{x^{3/2} }}$$ which shows the convergence at $$+\\infty$$.\n\nA quick way to show convergence is the limit comparison test.\n\nFor $$x \\to 0^+$$ we have\n\n$$\\lim_{x\\to 0^+}\\frac{\\arctan x}{x}=1 \\text{ and } \\lim_{x\\to 0^+}\\frac{\\sqrt x}{\\sqrt{x^3+x}} = 1 \\Rightarrow \\lim_{x\\to 0^+}\\frac{\\frac{\\arctan x}{\\sqrt{x^3+x}}}{\\sqrt x}= 1$$\n\nSince $$\\int_0^1 \\sqrt x\\; dx$$ is convergent, $$\\int_0^1 \\frac{\\arctan x}{\\sqrt{x^3+x}}dx$$ is convergent, as well.\n\nFor $$x \\to +\\infty$$ we have\n\n$$\\lim_{x\\to +\\infty}\\left(\\arctan x\\cdot \\frac{x^{\\frac 32}}{\\sqrt{x^3+x}}\\right)=\\frac{\\pi}2\\cdot 1\\Rightarrow \\lim_{x\\to }\\frac{\\frac{\\arctan x}{\\sqrt{x^3+x}}}{\\frac 1{x^{\\frac 32}}}= \\frac{\\pi}2$$\n\nSince $$\\int_1^{+\\infty} \\frac{dx}{x^{\\frac 32}}$$ is convergent, $$\\int_1^{+\\infty} \\frac{\\arctan x}{\\sqrt{x^3+x}}dx$$ is convergent, as well.", "date": "2021-10-18 00:35:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3611843/is-int-0-infty-frac-arctan-x-sqrtx3xdx-converges-diverges", "openwebmath_score": 0.886403501033783, "openwebmath_perplexity": 309.3078350636239, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668695588649, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8542011915493875}}
{"url": "https://math.stackexchange.com/questions/487211/find-the-smallest-value-of-n-for-which-the-nth-term-of-the-series-is-less-than-0", "text": "# Find the smallest value of n for which the nth term of the series is less than 0.001\n\nQuestion:\n\nThe common ratio and the first term of a geometric series are 0.55 and 18 respectively.\n\nFind the smallest value of n for which the nth term of the series is less than 0.001\n\n$${\\text{My solution: }}$$\n\n$$Tn<0.001$$ $$ar^{n-1}<0.001$$\n\n$$(18)(0.55)^{n-1}<0.001$$\n\n$$(0.55)^{n-1}<\\frac{1}{18000}$$ $$(n-l){\\log 0.55}<{\\log \\frac{1}{18000}}$$ $$n-1<16.389$$ $$n<16.389+1$$ $$n<17.389$$ $$n=17$$\n\n$${\\text{However, the answer given for this question is}}$$ $$n=18$$ $${\\text{Would anyone tell me either it is my answer or the answer given that is wrong ,please?}}$$ $${\\text{Thank you.}}$$\n\n## 1 Answer\n\n$\\log(0.55) < 0$ so from $$(n-1) \\, \\log(0.55) \\lt \\log \\frac{1}{18000}$$ dividing (or multiplying) both sides by a negative number means your next line should be $$n-1 \\gt 16.389\\ldots$$ changing the direction of the inequality.\n\nChecking the answers, $18 \\times 0.55^{17-1} = 0.001262\\ldots$ while $18 \\times 0.55^{18-1} = 0.000694\\ldots$", "date": "2020-01-28 10:35:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/487211/find-the-smallest-value-of-n-for-which-the-nth-term-of-the-series-is-less-than-0", "openwebmath_score": 0.6254583597183228, "openwebmath_perplexity": 155.67312334762067, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668745151689, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8542011909229891}}
{"url": "https://math.stackexchange.com/questions/3237554/suppose-that-the-average-value-on-all-intervals-a-b-is-equal-to-fab-2", "text": "# Suppose that the average value on all intervals $[a,b]$ is equal to $f((a+b)/2)$. Prove that $f''(x) = 0$ for all $x \\in \\mathbb{R}$\n\nI understand that $$f(x)$$ must be linear with a first derivative equal to a constant. I'm just not sure how I can use the mean value property of integrals to show something about $$f''(x)$$. The hint on this question is to use the fundamental theorem of calculus or Jensen's inequality.\n\n\u2022 Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. \u2013\u00a0dantopa May 23 at 21:16\n\u2022 You say that you understand that the first derivative must be a constant. Does this mean you have proven that part? I am not sure what you need help with. \u2013\u00a0InterstellarProbe May 23 at 21:17\n\nAssuming $$f$$ is integrable and twice differentiable (otherwise your statement about average value doesn't make sense, nor your final statement*), $$\\int_a^bf(x)\\,\\mathrm dx=(b-a)f\\left(\\cfrac{a+b}{2}\\right)$$\n\nDifferentiate both sides w.r.t $$\\,b$$, using the Leibniz integral rule (derived from fundamental theorem of calculus) for the LHS:\n\n$$f(b)=\\frac{b}{2}f'\\left(\\cfrac{a+b}{2}\\right)+f\\left(\\cfrac{a+b}{2}\\right)-\\frac{a}{2}f'\\left(\\cfrac{a+b}{2}\\right)$$\n\nNow set $$b=0$$ and $$a=2x$$:\n\n$$f(0)=f(x)-xf'(x)$$\n\nDifferentiate both sides w.r.t $$x$$:\n\n$$0=f'(x)-f'(x)-xf''(x)$$\n\nso $$f''(x)=0$$ for all $$x\\neq 0$$.\n\nThus we have proven the function is linear everywhere except $$0$$. Since $$f'(0)$$ and $$f'(x)$$, $$f'(-x)$$ are constants for $$x>0$$ exists, we know $$f'(0)$$ has to be equal to each of these and thus $$f''(0)=0$$.\n\n*I'm not sure if you're able to prove that $$f$$ has to be twice differentiable.\n\n\u2022 I see why $f'(x)$ and $f'(-x)$ are constant, because you've shown $f''=0$ for $x\\neq 0$. Why does $f'(0)$ have to be equal to each of these constants? Is it because if it is not, then $f''$ is discontinuous at $x=0$? \u2013\u00a0Frudrururu May 24 at 4:11\n\u2022 You can apply mean value theorem to prove that. In general, Darboux theorem tells that derivatives (if exist on an interval) enjoy intermediate value property regardless of it's continuity, and so, no jump discontinuity can occur to them. \u2013\u00a0Sangchul Lee May 24 at 6:10\n\nI will assume that $$f$$ is locally integrable for an obvious reason. Our aim is to prove that $$f$$ is linear, which is then enough to conclude that $$f$$ is twice-differentiable with $$f'' \\equiv 0$$.\n\nLet $$x < y$$ and $$0 < \\lambda < 1$$ be arbitrary. Set\n\n$$c = \\lambda x + (1-\\lambda) y, \\qquad a = 2x - c, \\qquad b = 2y - c.$$\n\nThen $$a < x < c < y < b$$ and\n\n$$\\frac{a+b}{2} = (1-\\lambda)x + \\lambda y, \\qquad \\frac{a+c}{2} = x, \\qquad \\frac{c+b}{2} = y.$$\n\nSo it follows that\n\n\\begin{align*} f((1-\\lambda)x+\\lambda y) &= \\frac{\\int_{a}^{b} f(t) \\, \\mathrm{d}t}{b-a} \\\\ &= \\frac{\\int_{a}^{c} f(t) \\, \\mathrm{d}t + \\int_{c}^{b} f(t) \\, \\mathrm{d}t}{b-a} \\\\ &= \\frac{(c-a)f(x) + (b-c)f(y)}{b-a} \\\\ &= (1-\\lambda) f(x) + \\lambda f(y). \\end{align*}\n\nThis proves that $$f$$ is linear.", "date": "2019-07-23 22:41:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3237554/suppose-that-the-average-value-on-all-intervals-a-b-is-equal-to-fab-2", "openwebmath_score": 0.9979050159454346, "openwebmath_perplexity": 149.14164913976296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474907307124, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8541943228455098}}
{"url": "http://math.stackexchange.com/questions/181702/proof-by-induction-of-bernoullis-inequality-1xn-ge-1nx", "text": "# Proof by induction of Bernoulli's inequality $(1+x)^n \\ge 1+nx$\n\nI am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:\n\nLet $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality: $$(1+x)^n \\ge 1+nx$$\n\nProof:\n\nBase Case: For $n=1$, $1+x = 1+x$ so the inequality holds.\n\nInduction Assumption: Assume that for some integer $k\\ge1$, $(1+x)^k \\ge 1+kx$.\n\nInductive Step: We must show that $(1+x)^{k+1} \\ge 1+(k+1)x$\n\nProof of Inductive Step: \\begin{align*} (1+x)^k &\\ge 1+kx \\\\ (1+x)(1+x)^k &\\ge (1+x)(1+kx)\\\\ (1+x)^{k+1} &\\ge 1 + (k+1)x + kx^2 \\\\ 1 + (k+1)x + kx^2 &> 1+(k+1)x \\quad (kx^2 >0) \\\\ \\Rightarrow (1+x)^{k+1} &\\ge 1 + (k+1)x \\qquad \\qquad \\qquad \\square \\end{align*}\n\n-\nThanks for the suggestions, I'll keep your tips in mind. \u2013\u00a0 KingOliver Aug 12 '12 at 15:09\nWhere did you uses $x>-1$? Hint: you did use it. \u2013\u00a0 Thomas Andrews Aug 12 '12 at 15:16\nWhen I claimed that $kx^2 >0$ \u2013\u00a0 KingOliver Aug 12 '12 at 15:18\nActually, you need it when you multiply both sides by $1+x$. Also, since $x$ can be 0, $kx^2\\ge0$ \u2013\u00a0 Mike Aug 12 '12 at 16:06\nAh thank you @Mike \u2013\u00a0 KingOliver Aug 12 '12 at 17:03\n\n\\begin{align*} (1+x)^{k+1}&=(1+x)(1+x)^k\\\\ &\\ge(1+x)(1+kx)\\\\ &=1+(k+1)x+kx^2\\\\ &\\ge1+(k+1)x\\;, \\end{align*}\nsince $kx^2\\ge 0$. This completes the induction step.\nThis looks fine to me. Just a small note on formatting of the inequalities: I would combine the third and fourth inequalities as $$(1+x)^{k+1} \\geq 1+(k+1)x+kx^2>1+(k+1)x,$$ so there is no need of the fifth line. Or even $$(1+x)^{k+1} = (1+x)(1+x)^{k} \\geq (1+x)(1+kx)=1+(k+1)x+kx^2>1+(k+1)x.$$", "date": "2015-09-04 15:15:34", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/181702/proof-by-induction-of-bernoullis-inequality-1xn-ge-1nx", "openwebmath_score": 0.9986633062362671, "openwebmath_perplexity": 1118.9286971007416, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474866475719, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8541943193201605}}
{"url": "https://www.physicsforums.com/threads/solving-for-a-variable-in-an-equation-that-involves-vectors.747925/", "text": "# Homework Help: Solving for a variable in an equation that involves vectors\n\n1. Apr 9, 2014\n\n### Chetlin\n\n1. The problem statement, all variables and given/known data\nI have the equation:\n$$\\mathbf{F}_{2,1} = \\frac{Q_1 Q_2}{4 \\pi \\varepsilon_0 {r_{2,1}}^2}\\hat{r}_{2,1}$$ (standard electric force equation for 2 charges)\n\nI know the value of everything except Q2 and have to find it. The vectors each have 3 components.\n\nNormally in an algebraic equation, I would just solve for a variable by isolating it on one side of the = sign. But this equation involves vectors and I don't think there is a way to divide vectors. I could also subtract F2,1 from both sides which at least gets everything onto one side but I am still left with the vectors. I will effectively have three equations (one for each component of the vectors), but only one variable I have to solve for, right? Does the nature of this problem (it is physical) make it so that only certain vectors are even possible, so if I were to tweak F2,1 and not change the value of anything else, the problem would become impossible to solve because it would represent a physical impossibility?\n\n2. Relevant equations\nNothing really\n\n3. The attempt at a solution\nI used only the x component of each vector and solved it using that equation and got the correct answer. But like I said before, if I tweaked only the y component of F2,1, the equation that used only the x components would be the same (so I'd get the same result) but the entire vector equation would not.\n\n2. Apr 9, 2014\n\n### ehild\n\nr212 is not vector, but scalar. It is possible to divide with it. F2,1 is parallel with the unit vector $\\hat{r}_{2,1}$ , a scalar multiple of it.\nWhat were the data? You should determine r212 from the coordinates.\n\nehild\n\nLast edited: Apr 9, 2014\n3. Apr 9, 2014\n\n### Chetlin\n\nThat's true, but $\\hat{r}_{2,1}$ is a vector, and I would want that to be on the other side as well. I just realized what I tried to do is impossible, because it would result in a scalar on one side of the = sign, and whatever you would (imaginatively) get if you divided two vectors, on the other side.\n\nI'm starting to believe that any problem like this would be set up very specifically, and that only certain values will work at all, allowing me to just use one component of the vectors.\n\nEdit: oops, I see you were still writing when I posted this.. sorry for jumping the gun so quickly, haha :P\n\nEdit 2:\nOh dang, you're right. So you can't just tweak one component of F2,1.. doing that would prevent them from being parallel and would cause all kinds of issues. This is just a simple equation of a vector being equal to a magnitude times a direction unit vector. Sorry, it's been a little while since I've really worked with vectors in such a way, so I've been spending a lot of time on issues like this. Thanks again, very much!\n\nLast edited: Apr 9, 2014\n4. Apr 9, 2014\n\n### ehild\n\nYou can not put $\\hat{r}_{2,1}$ on the other side. But you know that $\\vec{F}_{2,1}$ is a scalar multiple of $\\hat{r}_{2,1}$, and that holds for all corresponding coordinates.\nIt would be better to see the whole text of the problem.\n\nehild\n\n5. Apr 9, 2014\n\n### Chetlin\n\nWell, I have my question answered, but to be complete, here is the text of the problem. It's from an old Schaum's Outlines book from the early 1980's.\n\nPoint charge Q1 = 300 \u00b5C, located at (1, \u22121, \u22123) m, experiences a force F2,1 = (8, \u22128, 4) N due to point charge Q2 at (3, \u22123, \u22122) m. Determine Q2.\n\n6. Apr 9, 2014\n\n### Staff: Mentor\n\nWhat you do is dot both sides of the equation by $\\hat{r}_{2,1}$. This will give you the scalar equation you desire. Do you know how to determine $\\hat{r}_{2,1}$?\n\nChet\n\n7. Apr 9, 2014\n\n### Chetlin\n\nYep, that unit vector is just the vector r2,1 divided by its magnitude ($\\hat{r}_{2,1} = \\frac{1}{r_{2,1}}\\mathbf{r}_{2,1}$, where $r_{2,1} = ||\\mathbf{r}_{2,1}|| = \\sqrt{{r_{{2,1}_x}}^2 + {r_{{2,1}_y}}^2 + {r_{{2,1}_z}}^2}$). Sorry for the really weird notation inside the square root sign.\n\nThat's a really neat trick, thanks for showing it to me. I never thought to use scalar products to get rid of the vectors, and of course the scalar product of any unit vector with itself is 1.\n\n8. Apr 10, 2014\n\n### ehild\n\nThe point charge Q1 experiences force from Q2:\n\n$$\\vec F = k \\frac {Q_1 Q_2}{(\\vec r_1-\\vec r_2)^2} \\hat r_{12}$$\n\nwhere $\\hat r_{12}=\\frac{\\vec r_1-\\vec r_2} {|\\vec r_1-\\vec r_2|}$\n\nIn your problem, r12=(1, \u22121, \u22123)-(3, \u22123, \u22122)=(-2, 2, -1). The magnitude is 3, so the components of $\\hat r_{12}$ are (-2/3, 2/3, -1/3)\n\nThe force is F= (8, \u22128, 4) N. You see that the force is parallel to $\\hat r_{12}$.\n\nYou can write out the Coulomb Law in x,y,z components:\n\n$$F_x=8=\\frac {kQ_1Q_2}{r_{12}^2}(-2/3)$$\n$$F_y=-8=\\frac {kQ_1Q_2}{r_{12}^2}(2/3)$$\n$$F_z=4=\\frac {kQ_1Q_2}{r_{12}^2}(-1/3)$$\n\nYou see that you get the same value for $kQ1Q2/ {r_{12}^2}$ from each equation, it is -12. kQ1Q2/9=-12. That is a scalar equation already.\n\nehild", "date": "2018-07-16 23:13:19", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/solving-for-a-variable-in-an-equation-that-involves-vectors.747925/", "openwebmath_score": 0.7758625149726868, "openwebmath_perplexity": 487.0231945775424, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.969785415552379, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.8541841693876305}}
{"url": "https://brilliant.org/discussions/thread/1-3/", "text": "#1\n\nHow many positive integers less than $$1000$$ have the property that the sum of the digits of each such number is divisible by $$7$$ and the number itself is divisible by $$3$$?\n\nNote by Vilakshan Gupta\n1\u00a0year, 2\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nLet's think of a number $$abc (0 \\leq a, b, c \\leq 9)$$. $$a+b+c \\equiv 0 (\\mod 3 \\text{and} \\mod 7)$$. Thus, $$a+b+c=21$$.\n\n$(3, 9, 9) \\rightarrow \\frac{3!}{2}, (4, 8, 9) \\rightarrow 3!, (5, 7, 9) \\rightarrow 3!, (5, 8, 8) \\rightarrow \\frac{3!}{2}, (6, 6, 9) \\rightarrow \\frac{3!}{2}, (6, 7, 8) \\rightarrow 3!, (7, 7, 7)$ $$3+6+6+3+3+6+1=28$$\n\nPlease tell me if there is any error.\n\n- 11\u00a0months, 3\u00a0weeks ago\n\nGood one brother\n\n- 11\u00a0months, 3\u00a0weeks ago\n\nNice method\n\n- 11\u00a0months, 3\u00a0weeks ago\n\n33\n\n- 2\u00a0months, 3\u00a0weeks ago\n\n28\n\n- 10\u00a0months, 2\u00a0weeks ago\n\n27\n\n- 1\u00a0year ago\n\nits sum is divisibli by 21 using this you can solve\n\n- 1\u00a0year ago\n\nthis question came in this year PRMO answer is 28\n\n- 1\u00a0year ago\n\n@Md Zuhair 8\n\n- 1\u00a0year, 1\u00a0month ago\n\nO i see....\n\n- 1\u00a0year, 1\u00a0month ago\n\n@Pokhraj Harshal Rajasthan region!\n\n- 1\u00a0year, 1\u00a0month ago\n\nAnd what about the other participants and their marks from your school. I mean the averages and the highest marks\n\n- 1\u00a0year, 1\u00a0month ago\n\nWhich region are u from??\n\n- 1\u00a0year, 1\u00a0month ago\n\nWB rgion\n\n- 1\u00a0year, 1\u00a0month ago\n\n@Pokhraj Harshal Ok! Then I'm also getting the same...\n\n- 1\u00a0year, 1\u00a0month ago\n\nHow much? Without the question?\n\n- 1\u00a0year, 1\u00a0month ago\n\nThe questions will be cancelled\n\n- 1\u00a0year, 1\u00a0month ago\n\nAre the marks gonna be added to everyone's total or the questions will be cancelled (lowering the cutoff)?\n\n- 1\u00a0year, 1\u00a0month ago\n\nThey will be added to totsl\n\n- 1\u00a0year ago\n\nHey , what does discounted actually refer to?\n\n- 1\u00a0year, 1\u00a0month ago\n\nHey I'm getting 8/27 from jharkhand as per the new answer key of hbcse.will I qualify???\n\n- 1\u00a0year, 1\u00a0month ago\n\nLets see....\n\n- 1\u00a0year, 1\u00a0month ago\n\nHello, There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 You can check out for more queries related to the JEE EXAMS from the following compilation\n\n- 1\u00a0year, 1\u00a0month ago\n\nI believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.\n\n- 1\u00a0year, 1\u00a0month ago\n\nAh - that's where the 28 comes from - much more mathematical than my just listing and counting them\n\n- 1\u00a0year, 1\u00a0month ago\n\nExactly\n\n- 1\u00a0year, 1\u00a0month ago\n\nI agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 432*1 = 24, not 4+3+2+1 = 10. What am I missing?\n\n- 1\u00a0year, 1\u00a0month ago\n\n- 1\u00a0year, 2\u00a0months ago\n\nYup.I think so.\n\n- 1\u00a0year, 2\u00a0months ago\n\n@Md Zuhair @Vilakshan Gupta I haven't attempted one of the bonus question. Will I still get marks for it?\n\n- 1\u00a0year, 2\u00a0months ago\n\nI think the question can be cancelled as how can a person attempt to decinal answers and i had attempted ine. So i dunno.\n\n- 1\u00a0year, 2\u00a0months ago\n\nyeah hundreds digit cant be 1,2\n\n- 1\u00a0year, 2\u00a0months ago\n\n@Shreyan Chakraborty The Hundreds digit can't be 1 or 2..\n\n- 1\u00a0year, 2\u00a0months ago\n\nANSWER IS 28....HAS A BIJECTION WITH a+b+c=21 WHERE 0<a,b,c<=9........\n\n- 1\u00a0year, 2\u00a0months ago\n\nAre na na.... I am not telling that. How much are you getting?\n\n- 1\u00a0year, 2\u00a0months ago\n\n@Shreyan Chakraborty .. How much?\n\n- 1\u00a0year, 2\u00a0months ago\n\nJANI NA BAJE HOYECHE\n\n- 1\u00a0year, 2\u00a0months ago\n\nLetTheFateDecide !!Bye\n\n- 1\u00a0year, 2\u00a0months ago\n\nwhich class are u in toshit?\n\n- 1\u00a0year, 2\u00a0months ago\n\nIt's 11 in Rajasthan ! \ud83d\ude05\ud83d\ude12\n\n- 1\u00a0year, 2\u00a0months ago\n\nAccording to Resonance , cutoff in Chandigarh is just 4(questions)\n\n- 1\u00a0year, 2\u00a0months ago\n\ni don't think it will be so low\n\n- 1\u00a0year, 2\u00a0months ago\n\nIf that isnt, then wb will be higher and i will surely not qualify\n\n- 1\u00a0year, 2\u00a0months ago\n\nYa. Thats ridiculous. WB region has always got a higher cutoff...\n\n- 1\u00a0year, 2\u00a0months ago\n\nCoz as per cutoff(s) uploaded by Resonance , cutoff in Chandigarh is lower than others ( Rajasthan , Maharashtra , UP , etc) ... That's why!\n\n- 1\u00a0year, 2\u00a0months ago\n\nSo , you are already selected..Great \ud83d\udc4d\n\n- 1\u00a0year, 2\u00a0months ago\n\n- 1\u00a0year, 2\u00a0months ago\n\noh\n\n- 1\u00a0year, 2\u00a0months ago\n\nRajasthan..U?\n\n- 1\u00a0year, 2\u00a0months ago\n\noh...btw,where do u live (i mean which region)\n\n- 1\u00a0year, 2\u00a0months ago\n\nGeometry was quite tough and lengthy! Excluding bonus , I'm getting 8\n\n- 1\u00a0year, 2\u00a0months ago\n\nSir\ud83d\ude05 I am getting 10 along with bonus!\n\n- 1\u00a0year, 2\u00a0months ago\n\nOh. U mean 8/28 u r getting?\n\n- 1\u00a0year, 2\u00a0months ago\n\nwell, if it is bonus that means 2 questions marks are given extra.If it was been written question deleted then scores would be evaluated out of 28\n\n- 1\u00a0year, 2\u00a0months ago\n\nunfortunately, i will get only 10 questions correct. I did very silly mistakes\n\n- 1\u00a0year, 2\u00a0months ago\n\n@Md Zuhair Is the paper for 9,10,11 and 12 same?\n\n- 1\u00a0year, 2\u00a0months ago\n\nYes Sir!\n\n- 1\u00a0year, 2\u00a0months ago\n\nHow ma\u00f1y have you got right in PRMO - 17?\n\n- 1\u00a0year, 2\u00a0months ago\n\n- 1\u00a0year, 2\u00a0months ago\n\nwe just need to find the numbers which add upto 21\n\n- 1\u00a0year, 2\u00a0months ago\n\n28\n\n- 1\u00a0year, 2\u00a0months ago\n\n25 is the answer as per me\n\n- 1\u00a0year, 2\u00a0months ago\n\nNo - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28\n\n- 1\u00a0year, 2\u00a0months ago\n\nWhat ans did you get?\n\n- 1\u00a0year, 2\u00a0months ago\n\nHey Aaron. How much are u getting with bonus? With bonus i am getting 12.\n\n- 1\u00a0year, 2\u00a0months ago\n\nzuhair tui ki amk jiggesh korchish??\n\n- 1\u00a0year, 2\u00a0months ago\n\nAccha.. nijer whatsapp number ta de... whatsapp e kotha bolchi\n\n- 1\u00a0year, 2\u00a0months ago", "date": "2018-10-23 17:23:39", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/1-3/", "openwebmath_score": 0.9134075045585632, "openwebmath_perplexity": 5201.486439725606, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9697854111860906, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.8541841655418164}}
{"url": "https://amaanswers.com/how-many-acres-is-200-feet-by-200-feet", "text": "# How many acres is 200 feet by 200 feet?\n\nJerilyn Kalandek asked, updated on October 26th, 2022; Topic: 200 feet\n\ud83d\udc41 511 \ud83d\udc4d 64 \u2605\u2605\u2605\u2605\u26064.9\n\n200 feet multiplying by 200 feet equal to 40,000 sq ft. Therefore, 40,000 sq ft divide by 43,560 sq ft equal to 0.92 acres approximately.\n\nEither way, what is the perimeter of a 1/4 acre?\n\nThere are 43560 square feet in an acre so one quarter of an acre is square feet. If each side of the square is F feet long then the area is F^2 square feet. Thus F^2 = 10890 and F = \\sqrt{10890} = 104.35 feet.\n\nDespite that, is an acre 200x200? An area 200 ft long by 200 ft wide encompasses 40,000 sq feet. This is 40000/43560 = 0.91827acres. 200 feet x 200 feet = 0.918 acres. Or in other words, approximately 92% of an acre.\n\nSuitably, how many acres is 100 feet by 100 feet?\n\nWe know 43,560 square feet to 1 acre. 100 ft multiplying by 100 equal to 10,000 sq ft. Therefore, 10,000 sq ft divide by 43,560 sq ft equal to 0.23 acres approximately.\n\nHow much does a quarter acre cost?\n\n2. Average Cost of Clearing Land by Lot Size\n\nAcreageCost Range\n\u00bc acre$125 -$1,400\n\u00bd acre$250 -$2,800\n1 acre$500 -$5,600\n2 acres$1,000 -$11,200\n\n### What is the size of an acre lot?\n\n1 acre is approximately 208.71 feet \u00d7 208.71 feet (a square) 4,840 square yards. 43,560 square feet.\n\n### What is linear feet?\n\nTechnically, a linear foot is a measurement that is 12 inches long (so, one foot) and that is measured in a straight line, which is why it's called linear.\n\n### What is a linear acre?\n\nThe most standard shape for an acre is one furlong by one chain, or 660 feet by 66 feet. To find the linear measurements of other rectangular acres, just divide 43,560 by the number of feet you want on one side. A square-shaped acre would then be about 208.7 by 208.7 feet (because 208.7 x 208.7 = ~43,560).\n\n### How many acres is 150 feet by 200?\n\nWe know 43,560 square feet to 1 acre. 200 feet multiplying by 200 feet equal to 40,000 sq ft. Therefore, 40,000 sq ft divide by 43,560 sq ft equal to 0.92 acres approximately.\n\n### How many square feet is 200?\n\nWe know about 43,500 square feet. 20,000 sq ft is equal to 200 feet. 43,500 sq ft is equal to 0.92 sq ft.\n\n### How big is half an acre compared to a football field?\n\nIf you calculate the entire area of a football field, including the end zones, it works out to 57,600 square feet (360 x 160). One acre equals 43,560 square feet, so a football field is about 1.32 acres in size.\n\n### How big is a 50 by 100 plot of land?\n\n1)For a rectangular plot, 50 by 100 refers to 50 feet by 100 feet which is equivalent to 15 meters by 30 meters and is also equal to 450 square meters. This is what people refer to as 1/8 of an acre though slightly less due to the provision for access road.\n\n### What does 1acre look like?\n\nAs all farmers and real estate agents know, an acre is defined as an area one furlong long by 4 rods wide. ... Basically if you can picture a football field, that's pretty close to an acre in size. Officially, it is 43,560 square feet, and a football field is 48,000 square feet. Our standard acre isn't the same worldwide.\n\n### How long does it take to walk around 4 acres?\n\nA square acre is 208.7 feet on a side, so the perimeter of an acre is about 835 feet, or about 16 percent of a mile. If you walk a brisk pace of 3 miles an hour, you can cover a mile in 20 minutes. So you should be able to walk 835 feet in about three minutes.\n\ufeff", "date": "2023-03-26 09:55:42", "meta": {"domain": "amaanswers.com", "url": "https://amaanswers.com/how-many-acres-is-200-feet-by-200-feet", "openwebmath_score": 0.6954520344734192, "openwebmath_perplexity": 2093.2701143686118, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854138058637, "lm_q2_score": 0.8807970842359877, "lm_q1q2_score": 0.8541841648147955}}
{"url": "http://math.stackexchange.com/questions/703156/modular-arithmetic-are-we-allowed-to-distribute-the-modularity", "text": "# Modular Arithmetic - Are we allowed to distribute the Modularity?\n\nAssume I have a problem such as \"Prove that $\\displaystyle103^{53} + 53^{103}$ is divisible by $39$.\" This would mean I wanted to prove that $\\displaystyle103^{53} + 53^{103}\\equiv0\\pmod{39}$.\n\nMy starting statement would be then \"$\\displaystyle103^{53} + 53^{103}\\pmod{39}$\" and I would then equate this to \"$\\displaystyle103^{53}\\pmod{39} + 53^{103}\\pmod{39}$\" and then continue.\n\nAm I allowed to distribute the mod like that?\n\nThanks.\n\n-\nYes, you are... \u2013\u00a0 DonAntonio Mar 7 '14 at 16:06\n@Matthew, please verify the edited version \u2013\u00a0 lab bhattacharjee Mar 7 '14 at 16:14\nThat should be a $+$ rather than a comma, right? In the formula right before \"and then continue\". \u2013\u00a0 Jack M Mar 7 '14 at 17:38\n\nMathematicians and programmers have two different ways of thinking about mod. To the latter, it is a binary operation in which ${\\rm Mod}[a,b]$ is the remainder of $a$ when divided by $b$ (so it will return an integer in the range $0\\le r<b$). To the former, it is a binary relation (a symbol that relates things in some way, like $<,=,>,\\approx,\\sim$ etc.) for each modulus $b$. We say that $n\\color{Red}{\\equiv}m$ mod $b$ if the difference $n-m$ is divisible by $b$, or equivalently if ${\\rm Mod}[n,b]={\\rm Mod}[m,b]$. Sometimes the equivalence symbol $\\equiv$ is simply replaced by an equality symbol $=$, in which case we are understood to be equating equivalence classes. The relation $n\\equiv m$ mod $b$ is in fact a congruence relation (it \"respects\" $+$ and $\\times$), and the equivalence classes are called residue classes, or just residues.\n\nSo if $a\\equiv b$ and $c\\equiv d$ mod $m$ then $ac\\equiv bd$ and $a+c\\equiv b+d$ mod $m$. One can use this to end up proving that, in particular, $f(a)\\equiv f({\\rm Mod}[a,m])$ mod $m$ for integer-coefficient polynomials $f$.\n\nThus for example ${\\rm Mod}[a,m]+{\\rm Mod}[b,m]$ and $a+b$ and ${\\rm Mod}[a+b,m]$ are all congruent mod $m$, however it is not strictly true that the first and last are equal as integers. Take $a,b=2$ and $m=3$, in which case ${\\rm Mod}[2,3]+{\\rm Mod}[2,3]=2+2=4\\ne1={\\rm Mod}[2+2,3]$, albeit $4\\equiv1$ mod $3$.\n\nIf $n$ and $m$ are coprime, then $a\\equiv b$ modulo both $n$ and $m$ if and only if $a\\equiv b$ mod $nm$. In particular this means $x\\equiv 0$ mod $39$ if and only if $x\\equiv0$ mod $3$ and mod $13$. Compute\n\n$$103^{53}+53^{103}\\equiv 1^{53}+(-1)^{103}\\equiv 1+(-1)\\equiv0\\mod 3$$\n\nbecause $103\\equiv1$ and $53\\equiv-1$ mod $3$. And then compute\n\n$$103^{53}+53^{103}\\equiv(-1)^{53}+1^{103}\\equiv(-1)+1\\equiv0\\mod 13$$\n\nbecause $103\\equiv-1$ and $53\\equiv1$ mod $13$. Since $103^{53}+53^{103}$ is $0$ mod $3$ and $13$, it is $0$ mod $39$.\n\n-\nGood, and complete. For the purposes of students going farther in mathematics, it is disastrous to keep the mindset of relying on the operation rather than the equivalence relation. When talking about $R/I$ where $R$ is a ring and $I$ is an ideal, or $G/H$ where $\\supset H$ are groups, the equivalence relation outlook is essential. \u2013\u00a0 Lubin Mar 7 '14 at 18:49\n\nIf $103^{53} \\equiv a \\mod 39$ and $53^{103} \\equiv b \\mod 39$, than it is indeed true that $103^{53} + 53^{103} \\equiv a+b \\mod 39$. I wouldn't recommend your notation, which makes $\\mod\\mbox{ }$ look like an operator.\n\n-\nThere's nothing wrong with using $\\bmod$ as an operator. \u2013\u00a0 ShreevatsaR Mar 7 '14 at 16:54\nUsing mod as an operator loses the properties of the equivalence class that makes it so convenient. For example, you cannot assume $a \\mod M + b \\mod M = (a + b) \\mod M$. \u2013\u00a0 DanielV Mar 7 '14 at 17:48\n@DanielV: Assuming the name \"mod\" refers to a proper modulus operator and not the much-less-useful \"remainder\" operator found in many programming languages, and one is not expecting one's numeric type to act as an algebraic ring with a modulus which is not a multiple of M, why can't one assume that (a mod M)+(b mod M) will equal ((a+b) mod M) in cases where the result of the addition is defined? \u2013\u00a0 supercat Mar 7 '14 at 18:25\nAh, if I understand, you are talking about a $\\mathbb{N} \\rightarrow \\text{Galois}[M]$ operator, yes that would be useful indeed. \u2013\u00a0 DanielV Mar 8 '14 at 0:02\n\nTo say that $$x \\equiv a \\pmod{m}$$ means that there is some integer $k$ such that $$x - a = mk$$ Thus, expanding on user133281's answer (i.e. if $103^{53}\\equiv a \\pmod{39}$ and $53^{103}\\equiv b \\pmod{39}$), we have $$(103^{53}-a)+(53^{103}-b)=39k+39l$$ which is equivalent to $$(103^{53}+53^{103})-(a+b)=39(k+l)$$ so $$103^{53}+53^{103}\\equiv a+b\\pmod{39}$$\n\n-\nEither you meant $\\mod 39$ in the first line or $mk$ in the second. \u2013\u00a0 Mike Miller Mar 8 '14 at 3:45\n@Mike Right you are. Thank you. \u2013\u00a0 iamnotmaynard Mar 8 '14 at 20:56\nGlad to help!$\\$ \u2013\u00a0 Mike Miller Mar 8 '14 at 21:37\n\nHINT:\n\nPlease have a look into this for the properties of congruence\n\n$\\displaystyle 103\\equiv-1\\pmod{13}\\implies 103^{53}\\equiv(-1)^{53}\\equiv-1$\n\n$\\displaystyle 53\\equiv1\\pmod{13}\\implies 53^{103}\\equiv(1)^{103}\\equiv1$\n\nSimilarly, for $\\pmod3$\n\nNow if $13$ and $3$ both divides $a,a$ will be divisible by lcm$(13,3)$\n\n-\nWhile this is a good hint for solving the problem \"show that $103^{53} + 53^{103}$ is divisible by 39\", this doesn't address the question that was asked (\"am I allowed to distribute the mod\") at all. \u2013\u00a0 Magdiragdag Mar 7 '14 at 16:05\n@Magdiragdag, added a relevant link \u2013\u00a0 lab bhattacharjee Mar 7 '14 at 16:10\n\nThe \"equivalent modulo $n$\" relation is a congruence: a congruence is an equivalence relation with the additional property that for all relevant arithmetic operations (in this case, $0, 1, +, \\times$, and anything derived from those), if the inputs the the arithmetic operation are congruent, then the outputs are also congruent.\n\nIn the actual ring of integers modulo $n$, though, there is no \"congruence\" or \"mod\": e.g. $39 = 0$ is a literal equality. Your question to prove\n\n$$103^{53} + 53^{103} \\equiv 0 \\pmod{39}$$\n\nin the integers is the same thing as trying to prove\n\n$$103^{53} + 53^{103} = 0$$\n\nin the ring of integers modulo $39$. And since we're working with actual equality, it's clear that we can simplify the two summands separately, then add the simplified results.\n\n-", "date": "2015-07-29 05:07:58", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/703156/modular-arithmetic-are-we-allowed-to-distribute-the-modularity", "openwebmath_score": 0.9340643882751465, "openwebmath_perplexity": 285.8902138265916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854138058637, "lm_q2_score": 0.8807970764133561, "lm_q1q2_score": 0.8541841572285215}}
{"url": "https://mathhelpboards.com/threads/prove-that-the-expression-is-divisible-by-26460.3520/", "text": "# Prove that the expression is divisible by 26460\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nProblem:\nProve that $27195^8-10887^8+10152^8$ is divisible by $26460$.\n\nAttempt:\nI grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.\n\n$\\dfrac{27195^8-10887^8+10152^8}{26460}=\\dfrac{3^5\\cdot5^7\\cdot7^{14}\\cdot37^8-7013(2^6\\cdot3^6\\cdot47^2+3^2\\cdot19^2\\cdot191^2)(10152^4+10887^4)}{8}$\n\nI'd like to ask, do you think this problem can be solved using only elementary methods?\n\nLast edited by a moderator:\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nProblem:\nProve that $27195^8-10887^8+10152^8$ is divisible by $26460$.\n\nAttempt:\nI grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.\n\n$\\dfrac{27195^8-10887^8+10152^8}{26460}=\\dfrac{3^5\\cdot5^7\\cdot7^{14}\\cdot37^8-7013(2^6\\cdot3^6\\cdot47^2+3^2\\cdot19^2\\cdot191^2)(10152^4+10887^4)}{8}$\n\nI'd like to ask, do you think this problem can be solved using only elementary methods?\n\nThe best way to make the calculation more manageable is to factorise $26460 = 2^2\\cdot 3^3\\cdot 5\\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.\n\nTake the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.\n\nThe exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nThe best way to make the calculation more manageable is to factorise $26460 = 2^2\\cdot 3^3\\cdot 5\\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.\n\nTake the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.\n\nThe exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)\nAwesome! I finally understand it now!\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nOpalg's answer is very good! A more \"heavy machinery\" method could also go as follows:\n\n$$26460 = 2^2 \\cdot 3^3 \\cdot 5^1 \\cdot 7^2$$\nAnd we can do the following reductions with a couple modulo operations:\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 3^8 - 3^8 + 0^8 \\equiv 0 \\pmod{2^2}$$\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 6^8 - 6^8 + 0^8 \\equiv 0 \\pmod{3^3}$$\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 2^8 + 2^8 \\equiv 0 \\pmod{5^1}$$\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 9^8 + 9^8 \\equiv 0 \\pmod{7^2}$$\nAnd invoking the CRT (since $2^2$, $3^3$, $5^1$, $7^2$ are pairwise coprime):\n\n$$\\mathbb{Z}_{26460} = \\mathbb{Z}_{2^2} \\times \\mathbb{Z}_{3^3} \\times \\mathbb{Z}_{5^1} \\times \\mathbb{Z}_{7^2}$$\n$$\\therefore$$\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0 \\pmod{26460}$$\nThis also shows that the exponent (8) is in fact irrelevant, and could be anything.\n\nEDIT: fixed, see ILikeSerena's post below.\n\nLast edited:\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nOpalg's answer is very good! A more \"heavy machinery\" method could also go as follows:\n\n$$26460 = 2^2 \\cdot 3^3 \\cdot 5^1 \\cdot 7^2$$\nAnd we can do the following reductions with a couple modulo operations:\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 0^8 + 0^8 \\equiv 0 \\pmod{2}$$\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 0^8 + 0^8 \\equiv 0 \\pmod{3}$$\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 2^8 + 2^8 \\equiv 0 \\pmod{5}$$\n\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 2^8 + 2^8 \\equiv 0 \\pmod{7}$$\nAnd invoking the CRT (since 2, 3, 5, 7 are pairwise coprime):\n\n$$\\mathbb{Z}_{26460} = \\mathbb{Z}_{2^2} \\times \\mathbb{Z}_{3^3} \\times \\mathbb{Z}_{5^1} \\times \\mathbb{Z}_{7^2}$$\n$$\\therefore$$\n$$27195^8 - 10887^8 + 10152^8 \\equiv 0 \\pmod{26460}$$\nThis also shows that the exponent (8) is in fact irrelevant, and could be anything.\nI'm afraid that the number $2 \\cdot 3 \\cdot 5 \\cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.\nBut it is not divisible by 26460.\n\nFor CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.\n\nFor instance for $3^3$, we need:\n\n$\\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 0^8 + 0^8 \\equiv 0 \\pmod{3^3}$\n\nStill quite doable.\n\n#### Bacterius\n\n##### Well-known member\nMHB Math Helper\nI'm afraid that the number $2 \\cdot 3 \\cdot 5 \\cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.\nBut it is not divisible by 26460.\n\nFor CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.\n\nFor instance for $3^3$, we need:\n\n$\\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \\equiv 0^8 - 0^8 + 0^8 \\equiv 0 \\pmod{3^3}$\n\nStill quite doable.\nTrue, my mistake. Though it's not too difficult to check, just do the same operations but using 4, 27 and 49... (which still work out, it's clear the integers in the problem were carefully chosen to cancel each other out)", "date": "2021-10-21 15:06:04", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/prove-that-the-expression-is-divisible-by-26460.3520/", "openwebmath_score": 0.9025461077690125, "openwebmath_perplexity": 229.84102906537692, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854120593484, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8541841541729411}}
{"url": "http://math.stackexchange.com/questions/272274/alternative-ways-to-find-lim-x-to-1-frac1-x1-sqrtx", "text": "# Alternative ways to find $\\lim_{x\\to 1}\\; \\frac{1-x}{1- \\sqrt{x}}$\n\nSearching on Google by a solution, I found this:\n\n$1) \\space$The numerator can be written by a diference of squares:$\\space (1-\\sqrt{x})(1+\\sqrt{x})$\n\n$2 \\space)$Then, one can \"eliminated\" the commum factor beteween the numerator and the denominator.\n\n$3) \\space$The final expression looks like: $$\\lim_{x\\to 1}\\; {1+ \\sqrt{x}}=2$$\n\nHowever this was not a intuitive algebraic solution. I haven't thought of this solution at the first attempts. Could you please give me other alternative algebraic solutions, if it exists. Thanks.\n\n-\nSame solution, let $x=u^2$, but now instantly familiar. \u2013\u00a0Andr\u00e9 Nicolas Jan 7 '13 at 17:20\nWhy isn't it intuitive? If you have a $0/0$ form, the first thing one usually does is search for common factors. \u2013\u00a0David Mitra Jan 7 '13 at 17:21\nThe difference of two squares method was the first thing that popped into my head before I clicked the link to this question! \u2013\u00a0Clive Newstead Jan 7 '13 at 17:21\nThe other approach to this solution is to think - that was neat, how will I be able to spot neat solutions like that next time and the comment from @Andr\u00e9Nicolas was exactly what I was going to put. \u2013\u00a0Mark Bennet Jan 7 '13 at 17:22\n@Jo\u00e3o What's $(1-\\sqrt{x})(1+\\sqrt{x})$? \u2013\u00a0WimC Jan 7 '13 at 17:35\n\nYou can compute the inverse: $$\\lim_{x\\to 1}\\frac{1-\\sqrt{x}}{1-x}=\\lim_{x\\to 1}\\frac{\\sqrt{x}-1}{x-1}$$ which is, by definition, $f'(1)$ where $f(x)=\\sqrt{x}$.\n\n-\nAs I understood, you \"turn over\" the fraction,by puting $\\frac{1-\\sqrt{x}}{1-x}$ on the denominator. Then you multiplied the denominator by $\\frac{-1}{-1}$ to get the final expression, that is under the main fraction (the numerator is $1$).Finally you computed the derivative of $\\sqrt{x}$ at $x=1$ to get $\\frac{1}{\\frac{1}{2}}$. Is that? \u2013\u00a0Jo\u00e3o Jan 7 '13 at 18:04\nYeah, that's basically it. Essentially, if $g(x)/h(x)\\to A$ then $h(x)/g(x)\\to 1/A$ \u2013\u00a0Thomas Andrews Jan 8 '13 at 14:12\n\nSame solution, let $x=u^2$, but now instantly familiar.\n\n-\n\nPerhaps we can use L'Hospital rule here ?\n\n-\n\nBy substitution:$\\quad$ let $x = u^2\\,.\\quad$Then as $x\\to 1,\\;\\;u^2\\to 1\\implies u\\to 1$, giving us $$\\lim_{x\\to 1}\\; \\frac{1-x}{1- \\sqrt{x}} \\;\\;=\\;\\; \\lim_{u\\to 1}\\; \\frac{1-u^2}{1- u} = \\lim_{x\\to 1} \\frac{u^2 - 1}{u-1} \\;=\\;\\lim_{u \\to 1} \\frac{(u-1)(u+1)}{u-1} = \\lim_{u\\to 1}(u + 1) = 2$$\n\nUsing the substitution makes the \"difference of squares\" route looks so much more obvious!\n\n-\nIf $u^2 \\to 1$, then $u \\to 1$ or $u \\to -1$. Is this second situation if one compute the limit we get $0$.M'I rigth? \u2013\u00a0Jo\u00e3o Jan 7 '13 at 18:14\nsince $x \\to 1$, the only root of concern is u = 1 (since $u = \\sqrt{x}$ must necessarily be $\\ge 0$ in this situation. \u2013\u00a0amWhy Jan 7 '13 at 18:22", "date": "2016-02-13 17:22:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/272274/alternative-ways-to-find-lim-x-to-1-frac1-x1-sqrtx", "openwebmath_score": 0.963214099407196, "openwebmath_perplexity": 691.8296532814952, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639677785087, "lm_q2_score": 0.8791467722591729, "lm_q1q2_score": 0.854147326315791}}
{"url": "https://math.stackexchange.com/questions/2193674/why-1-equiv-ap-1-mod-p/2193701", "text": "# Why $1\\equiv a^{p-1} \\mod p$?\n\nLet $\\mathbb{Z}_p^*$, where $p$ is prime and let $a\\in\\mathbb{Z}_p^*$.\n\nConsider the following equation:$$(p-1)! \\equiv (p-1)! a^{p-1} \\mod p$$\n\nI've read that since $\\gcd((p-1)!, p) = 1$ we can infer that $$a^{p-1} \\equiv 1$$\n\nSo I have two questions:\n\n1. Why is it true that $\\gcd ((p-1)!, p)= 1$?\n2. Why can we infer that $a^{p-1} \\equiv 1$?\n\u2022 for the frist question: if $p$ is prime then observe than $(p-1)!$ doesnt contain $p$ as a factor, hence $\\gcd((p-1)!,p)=1$ \u2013\u00a0Masacroso Mar 19 '17 at 15:44\n\u2022 And for the second question, you can always write the $gcd(a,b)$ as a linear combination of $a$ and $b$ with integer coefficients (this is from the Euclidean algorithm). Use this to show that if some number is relatively prime to $p$, then it has a multiplicative inverse mod $p$. \u2013\u00a0Malkoun Mar 19 '17 at 15:46\n\u2022 Fermat's little theorem en.wikipedia.org/wiki/Fermat%27s_little_theorem?wprov=sfla1 \u2013\u00a0Zuo Mar 19 '17 at 15:51\n\n1. The only divisors of $p$ are $p$ and $1$. What are the divisors of $(p-1)!$?\n\n2. Once you know that $p, (p-1)!$ are relatively prime, consider $(p-1)! * (a^{p-1} - 1) \\equiv 0 \\mod p$. The relative primeness tells you that this is only possible if $(a^{p-1} - 1) \\equiv 0 \\mod p$\n\nThis common result is known as Fermat's Little Theorem. I hope this simple proof helps:\n\nConsider the sequence of integers $n,2n,3n,\u2026,(p\u22121)n$.\n\nNote that none of these integers are congruent modulo $p$ to the others.\n\nIf this were the case, we would have $an\u2261bn \\pmod p$ for some $1\u2264a<b\u2264p\u22121$.\n\nThen as $gcd(n,p)=1$, and we can cancel the $n$, we get $a\u2261b \\pmod p$ and so $a=b$.\n\nAlso, since $p\u2224n$ and $p\u2224c$, for any $1\u2264c\u2264p\u22121$, then by Euclid's Lemma $p\u2224cn$ for any such $cn$, which means $cn\u2261\u03380 \\pmod p$.\n\nThus, each integer in the sequence can be reduced $modulo \\ p$ to exactly one of $1,2,3,\u2026,p\u22121$.\n\nSo ${1,2,3,\u2026,p\u22121}$ is the set of Reduced Residue System $modulo \\ p$.\n\nSo, upon taking the product of these congruences, we see that $n\u00d72n\u00d73n\u00d7\u22ef\u00d7(p\u22121)n\u22611\u00d72\u00d73\u00d7\u22ef\u00d7(p\u22121) \\mod p$.\n\nThis simplifies to $n^{p\u22121}\u00d7(p\u22121)!\u2261(p\u22121)! \\pmod p$.\n\nSince $p\u2224(p\u22121)!$, we can cancel $(p\u22121)!$ from both sides, leaving us with $n^{p\u22121}\u22611 \\pmod p$.\n\n\u2022 The last statement is essentially my question. Why does the fact $p\\not\\vert (p-1)!$ implies that you can cancel $(p-1)!$ from both sides? \u2013\u00a0OliOliver Mar 19 '17 at 16:14\n\u2022 It doesn't essentially imply that $(p-1)!$ is being cancelled because of that property. It merely states that they are relatively prime to each other and, hence, we can use the property of modular arithmetic to basically remove $(p-1)!$ as if it of no use here(as we are dealing with, say, modulo $p$ and not modulo $(p-1)!$) and appears on both sides leaving it to be factored out. \u2013\u00a0HKT Mar 19 '17 at 16:25\n\u2022 Carefully think about it. Suppose if $p|(p-1)!$, we would have been left with $0 \\pmod p$ on the right hand side. I hope that helps. \u2013\u00a0HKT Mar 19 '17 at 16:28\n\u2022 You are just over-thinking it. I assure you once you get the hang of it and review the basic modular arithmetic definitions and get a stronger grasp on the notation, there will be clarity. \u2013\u00a0HKT Mar 19 '17 at 16:30\n\nIf you write down $(p-1)!$ as $(p-1)\\cdot(p-2)\\cdot\\cdot\\cdot1$ you can easily notice that $p$ and $(p-1)!$ have no common factors.\n\nSo $gcd((p-1)!,p) = 1$.\n\nNow, knowing that:\n\n$$ax \\equiv b \\mod(m) \\implies x \\equiv b \\cdot a^{-1} \\mod(\\frac{m}{gcd(m,a)})$$\n\n\u2022 You might as well just write $ax\\equiv b\\bmod m\\implies x=a^{-1}b\\bmod m$ when $\\gcd(a,m)=1$. No need to overcomplicate it. \u2013\u00a0arctic tern Mar 19 '17 at 15:55\nFor the first question, take, for example, $p = 7$. Then $(p-1)! = 2\\cdot3\\cdot4\\cdot5\\cdot6 = 2^4\\cdot3^2\\cdot5$. Notice that in the first equation all of the factors are strictly less than $p = 7$, which implies that all of the prime factors are also strictly less than $p$. Since $p$ is prime, this means that $p$ and $(p-1)!$ can have no prime factors in common.\nNow this implies Fermat's little theorem because, having shown that $(p-1)!$ is relatively prime with $p$, we can apply the cancellation law.\nWhenver $\\gcd(a,m) = 1$, $$ax \\equiv ay \\mod m$$ $$\\downarrow$$ $$x \\equiv y \\mod m$$", "date": "2019-12-14 16:12:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2193674/why-1-equiv-ap-1-mod-p/2193701", "openwebmath_score": 0.93928462266922, "openwebmath_perplexity": 126.95799068389245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639686018701, "lm_q2_score": 0.8791467690927438, "lm_q1q2_score": 0.8541473239632581}}
{"url": "https://math.stackexchange.com/questions/3608462/rolling-two-dice-alternatively-what-is-probability-of-winning", "text": "Rolling two dice alternatively. What is probability of winning?\n\nTom and Paul roll 2 dice alternatively starting with Tom. Consider they use two fair 6-faced dice. The player who rolls 6 first wins. They continue to roll until one of them wins. Find probability that Tom wins.\n\nI have listed out the total possible outcomes below:\n\n{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}\n\nHere are all the possible outcomes of two dice rolls. And as one can see, only 5 of these comes out with a 6 first; so, I think the probability of Tom winning should be 5/36.\n\nHowever, when I looked at the solution (This is a question from one midterm in one of previous semesters given at my school) says:\n\np is 5/36\n\n$$P(Tom\\, Wins)=\\sum_{k=0}^{\\infty }p(1-p)^{2k}=p\\sum_{k=0}^{\\infty }(1-p)^{2k}=$$ $$\\frac{\\frac{5}{36}}{1-\\frac{31}{36}}=\\frac{5}{5}$$\n\nAs one can see this answer is really rediculus! (Tom will win no matter what??)\n\nI think that the key here is probably on the word \"alternatively\", but cannot figureout what has gone wrong here.\n\n\u2022 You\u2019ve computed that sum incorrectly. Observe that the exponent is $2k$, not $k$, i.e., you\u2019ve got a geometric series in $(1-p)^2$, not $1-p$.\n\u2013\u00a0amd\nApr 3 '20 at 19:50\n\u2022 I didn't come out with that answer. It is answer given by someone. I looked at it and was very confused. Apr 4 '20 at 1:37\n\nThe probability that Tom wins on the first roll is $$\\frac16$$. He wins on the third roll if and only if he first rolls something other than a $$6$$, then Paul rolls something other than a $$6$$, and then Tom rolls a $$6$$; the probability of this is $$\\frac56\\cdot\\frac56\\cdot\\frac16$$. In general, Tom wins on the $$(2k+1)$$-st roll if and only if the first none of the first $$2k$$ rolls is a $$6$$, and Tom rolls a $$6$$ on roll $$2k+1$$; this occurs with probability $$\\frac16\\left(\\frac56\\right)^{2k}$$. Thus, the probability that Tom wins is\n\n\\begin{align*} \\sum_{k\\ge 0}\\frac16\\left(\\frac56\\right)^{2k}&=\\frac16\\sum_{k\\ge 0}\\left(\\frac56\\right)^{2k}\\\\ &=\\frac16\\sum_{k\\ge 0}\\left(\\frac{25}{36}\\right)^k\\\\ &=\\frac16\\cdot\\frac1{1-\\frac{25}{36}}\\\\ &=\\frac6{11}\\;. \\end{align*}\n\nThis is the formula in the solution that you read, with $$p=\\frac16$$.\n\nIt should not be surprising that this is slightly more than $$\\frac12$$: the fact that Tom goes first gives him an advantage, but it\u2019s a small one, since the game is otherwise very symmetric.\n\nOne can also compute the desired probability without resort to infinite series. Let $$p$$ be the probability that Tom wins, so that Paul wins with probability $$1-p$$. On the other hand, the probability that Paul when Tom first rolls something other than a $$6$$ must be $$p$$, because at that point the game is effectively starting over with Paul as the first player. Thus, Paul wins with probability $$\\frac56p$$, the probability that Tom rolls something other than a $$6$$ initially and Paul then rolls a $$6$$ before Tom does. In short, $$1-p=\\frac56p$$, and solving for $$p$$ again yields $$p=\\frac6{11}$$.\n\n\u2022 Thank you. I think your answer is good. I guess the solution given for this problem is probably wrong. Apr 4 '20 at 1:36\n\u2022 @AmosKu: You\u2019re welcome. Apr 4 '20 at 1:42\n\u2022 I have given the question quite a bit of thought last night. I think the key is \"alternatively\". And whenever Tom rolls a 6, he wins. I think your answer is for the case where only one 6 face dice is rolled. I have edited my wording above, hoping to make the meaning of original question clearer. Apr 4 '20 at 17:44\n\u2022 @AmosKu: Alternatively makes no sense here; I suspect that the intended word is alternately. That simply means that Tom rolls his die, then Paul rolls his, and so on. This is exactly the same problem as if they were using a single die and trading it back and forth. Apr 4 '20 at 17:52\n\u2022 Possible. But I just double checked the original question. The word is \" Alternatively\". There are other errors in the writing of the test as well. I guess you are right. Apr 4 '20 at 20:07", "date": "2021-09-22 20:00:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3608462/rolling-two-dice-alternatively-what-is-probability-of-winning", "openwebmath_score": 0.7238738536834717, "openwebmath_perplexity": 326.0285951973466, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639653084245, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8541473149150592}}
{"url": "https://www.physicsforums.com/threads/conclusion-about-the-dimension-of-c-r.919980/", "text": "# Conclusion about the dimension of C\u00b0(R)?\n\nTags:\n1. Jul 12, 2017\n\n### Rodrigo Schmidt\n\n[mentor note: thread moved from Linear Algebra to here hence no homework template]\n\nWe have a linear map A:E\u2192E, where E=C\u00b0(\u211d), the vector space of all continuous functions.\nLet's suppose that A\u0192= x0 \u0192(t)dt.\n\nBy the Calculus Fundamental Theorem, d/dx(A\u0192) = \u0192, so we have a left inverse, which implies that ker(A)={0}.\n\nSupposing that dim(E) is finite, by the rank-nullity theorem we have that im(A)=E . As a result of that:\n(\u0192(x)=|x|) \u2208 E \u21d2 \u0192 \u2208 im(A)\n\u21d2d\u0192/dx \u2208 E\n\nBut we know that \u0192's derivative is not continuous. So, supposing that dim(E) is finite lead us to a contradition (d\u0192/dx \u2208 E \u2227 d\u0192/dx \u2209 E) therefore dim(E) must be infinite.\n\nIs this argument valid? If not, could you guys point where does it fail? Thank you!\n\nLast edited by a moderator: Jul 12, 2017\n2. Jul 12, 2017\n\n### Krylov\n\nNote that the right-hand side is $(Af)(x)$, not $Af$.\n\nThe derivative is not even defined at $x = 0$.\n\nYes, it is valid, but it is quite convoluted. If you insist on doing it this way, I would prefer to say that you have found a linear operator $A$ that is injective but not surjective (the latter because $A$ maps into $C^1(\\mathbb{R})$). This already implies infinite dimensionality of $E$, since on a finite dimensional space injectivity and surjectivity are equivalent for linear operators.\n\nOf course, if you merely care to show that $E$ is infinite dimensional, it is more straightforward to identify an infinite linearly independent set.\n\n3. Jul 12, 2017\n\n### Rodrigo Schmidt\n\nThanks for the knowledge shared!\nI hadn't tought that, that's, indeed, much simpler.\nSo the basis of the set of all polynomials would be enough?\n\n4. Jul 12, 2017\n\n### LCKurtz\n\nThe set of all polynomials is too large to be a basis for $C_0$, but $\\{1,x,x^2,\\dots\\}$ would do.\n\n5. Jul 12, 2017\n\n### mathwonk\n\nthis is very nice, and shows a creative grasp of what you are learning. keep it up and you will eventually do some new research!\n\n6. Jul 13, 2017\n\n### Krylov\n\nWhile I usually like the straightforward approach best (maybe this is one of the differences between a \"pure\" mathematician and an (aspiring) \"applied\" mathematician?), I gave it some thought and then came to the conclusion that I agree with you.\n\nJust as an exercise, you could try to complete the direct approach as well. (You are almost there, anyway.)\n\n7. Jul 13, 2017\n\n### Rodrigo Schmidt\n\nHm, i see, but isn't that the basis of the set of all polynomials? So, by being a basis, that must be a linearly independent set, and it's also infinite, so that implies in the infinite dimensionality of C\u00b0(\u211d)?\nThat's where i want to get someday. Thanks for the inspiration!\n\n8. Jul 13, 2017\n\n### Krylov\n\nI think that I and post #4 misunderstood you when you wrote\nWe (or I, at least) thought that here you asserted that the set of all polynomials itself is a basis for a linear subspace of $C^0(\\mathbb{R})$, maybe because you wrote \"the basis\". However, from what you wrote afterwards (quoted at the top of this post), I get that you meant any (Hamel) basis for the linear space of all polyniomials. (For example, you can indeed use the canonical basis mentioned in post #4. By the way, I think there the $C_0$ was written by small mistake.)\n\nVery good, as far as I can see, you are more than done.\n\nLast edited: Jul 13, 2017\n9. Jul 13, 2017\n\n### Rodrigo Schmidt\n\nYeah, that's what i meant. Sorry if i wasn't clear enough! English is not my mother language so, mostly when writing about math and science, my texts can get a little confusing. I will try to be more specific in the next time.\n\nOkay! Thanks for the support and knowledge shared!", "date": "2017-12-14 21:36:09", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/conclusion-about-the-dimension-of-c-r.919980/", "openwebmath_score": 0.9107145667076111, "openwebmath_perplexity": 800.8122389087108, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639644850629, "lm_q2_score": 0.8791467548438124, "lm_q1q2_score": 0.8541473065002321}}
{"url": "http://math.stackexchange.com/questions/171966/whats-the-probability-a-subset-of-an-mathbb-f-2-vector-space-is-a-spanning-se", "text": "# Whats the probability a subset of an $\\mathbb F_2$ vector space is a spanning set?\n\nLet $V$ be an $n$-dimensional $\\mathbb F_2$ vector space. Note that $V$ has $2^n$ elements and $\\mathcal P(V)$ has $2^{2^n}$.\n\nI'm interested in the probability (under a uniform distribution) that an element of $\\mathcal P(V)$ is a spanning set for $V$. Equivalently a closed form formula (or at least one who's asymptotics as $n\\rightarrow \\infty$ are easy to work out) for the number of spanning sets or non-spanning sets.\n\nIt's not hard to show that the probability is greater than or equal to $1/2$. Since any subset of size greater than $2^{n-1}$ must span the space. I calculated the proportion of spanning sets of size $n$ for $n$ up to $200$ which seem to be going to a number starting with $.2887$. This leads me to believe that the probability exceeds $1/2$. I couldn't nail down a formula for arbitrary sized subsets though to continue experimental calculations.\n\nI feel like this is something that's been done before, but googling I've mostly found things concerning counting points on varieties over finite fields or counting subspaces of finite fields. Any references would be appreciated.\n\n-\nI think it would be sufficient to count how many different bases you can have in V. Then multiply that number by all possibilities of adding elements to it, since that does not perturb the spanning property of the set. \u2013\u00a0 Raskolnikov Jul 17 '12 at 15:31\n@Raskolnikov The issue with that is avoiding duplicates. \u2013\u00a0 JSchlather Jul 17 '12 at 15:32\nIndeed, you're right. \u2013\u00a0 Raskolnikov Jul 17 '12 at 15:33\n\nA random subset of $V$ is owerwhelmingly likely to span $V$.\n\nLet's look at how hard it is for a random subset not to span $V$. In order not to span $V$, there must be an $(n-1)$-dimensional subspace that contains the entire subset. There are exactly $2^n-1$ such subspaces, since they are in bijective correspondence with the nontrivial linear maps $V\\to\\mathbb F_2$ (each subspace is the kernel of exactly one map).\n\nFor each fixed $(n-1)$ dimensional subspace, the probability for a random subset to stay within that subspace is $2^{-2^{n-1}}$, since the $2^{n-1}$ vectors outside the subspace must all randomly decide not to be in the subset.\n\nSo the probability for a random set not to span is at most $(2^n-1)2^{-2^{n-1}} < 2^{-(2^{n-1}-n)}$ (and this is slightly too high, because there are a few non-spanning subsets that have more than one proper subspace they fit into and so are counted twice here). Everything else spans.\n\nEven for $n$ as small as $5$ the probability for a random subset to span $V$ is above 99.9%, and the number of 9's increases exponentially with larger $n$s.\n\n-\nThere are indeed $2^n - 1$ subspaces of dimension $n - 1$. There is a bijection with subspaces of dimension $1$, though not via the inner product. \u2013\u00a0 Zhen Lin Jul 17 '12 at 15:57\nI think the problem is that the probability that some set $S$ is not contained in some subspace $U$ is not independent of the probability that $S$ is not contained in some other subspace $U'$. \u2013\u00a0 Zhen Lin Jul 17 '12 at 16:01\nThat works, and it's certainly enough to get the conclusion that the probability goes to $1$ as $n \\to \\infty$. \u2013\u00a0 Zhen Lin Jul 17 '12 at 16:08\n(+1 to Henning) @ZhenLin: A non-degenerate bilinear form gives that correspondence (as you probably noticed) between $(2^n-1)$ non-zero vectors (=1-dim. subspaces) and $(n-1)$-dimensional subspaces. IMHO the word orthogonal might still be used - with the caveat that a vector or a subspace may be orthogonal to itself. \u2013\u00a0 Jyrki Lahtonen Jul 17 '12 at 16:23\n@Jyrki: I think it is cleaner to speak of nonzero vectors in $V$ and $(n-1)$-dimensional subspaces in $V^{\\ast}$ (their annihilators) as this correspondence is functorial and doesn't require the addition of extra structure. \u2013\u00a0 Qiaochu Yuan Jul 17 '12 at 20:18\n\nIt's worth mentioning that there is an explicit formula for the probability that $N$ random vectors drawn from $\\mathbb{F}_q^n$ span. (Note: I am doing sampling with replacement. If you want to require the $N$ vectors to be distinct, you can do this using inclusion-exclusion, but the result will be much messier.) Of course, the answer is $0$ if $N<n$, so assume $N \\geq n$.\n\nWrite down an $n \\times N$ matrix whose columns are the vectors in question. Note that the following are equivalent:\n\n1. The columns span $\\mathbb{F}_q^n$.\n2. The matrix has rank $n$.\n3. The rows are linearly independent.\n\nThinking in terms of rows, we have the new question: If we draw $n$ vectors at random from $\\mathbb{F}_q^N$, what is the probability that they are linearly independent?\n\nThe probability that the first vector is nonzero is $1-q^{-N}$. Given that, the probability that the second vector is not in the span of the first is $1-q^{-N+1}$. Given that, the probability that the third vector is not in the span of the first two is $1-q^{-N+2}$. Continuing in this manner, the probability that $n$ vectors in $\\mathbb{F}_q^n$ are independent is $$(1-q^{-N}) (1-q^{-N+1}) \\cdots (1-q^{-N+n-1}).$$ And, by the above argument, this is also the probability that $N$ random vectors will span $\\mathbb{F}_q^n$.\n\nIn particular, as Henning says, once $N-n$ is of any significant size, this is extremely close to $1$.\n\n-", "date": "2015-08-05 13:24:39", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/171966/whats-the-probability-a-subset-of-an-mathbb-f-2-vector-space-is-a-spanning-se", "openwebmath_score": 0.9266758561134338, "openwebmath_perplexity": 172.47871622510237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787868650145, "lm_q2_score": 0.8652240930029117, "lm_q1q2_score": 0.8541308704969969}}
{"url": "https://oschvr.com/problems/7/", "text": "By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.\n\nWhat is the 10 001st prime number?\n\nAt first glance, a simple iteration whilst checking if index position is prime, would suffice.\n\nA pseudocode attempt:\n\nsum = 0\nfor i = 2; i <= ?; i++ {\nif(isPrime(i)){\nsum++\nif ( sum >= 10001) {\nreturn i\n}\n}\n}\n\n\nUsing the solution in go to check if is prime from former examples:\n\nfunc isPrime(x int64) bool {\nvar i int64 = 2\nfor ; i < x; i++ {\nif x%i == 0 {\nreturn false\n}\n}\nreturn true\n}\n\n\nThe solution would scale to\n\n$$O(n^2)$$.\n\nThis is not viable. Also, there's no upper bound to set our iteration limit, meaning, we don't know up to which number we should iterate to find the 10,001 prime.\n\nSo digging further, I found about the Erathostenes Sieve:\n\nAn ancient algorithm for finding all prime numbers up to any given limit\n\nThis is exactly what we need. Let's look at the procedure:\n\nTo find all the prime numbers less than or equal to a given integer n by Erathostenes' method:\n\n1. Create a list of consecutive integers from 2 ... n (2,3,4,...,n)\n2. Initially, let p equal 2, smallest prime.\n3. Enumerate multiples of p by counting in increments of p from 2p to n, and mark them in the list (these will be 2p, 3p, 4p, ...; the p itself should not be marked)\n4. Find the first number greater than p in the list that is not marked, if there is not such number, stop. Otherwise, let p now equal to this number (which is next prime), and repeat step 3\n5. When algorithm terminates, the numbers remain not marked in the list, are all primes below n.\n\nHere's the code I came up:\n\npackage main\n\nimport (\n\"fmt\"\n\"math\"\n\"time\"\n)\n\nfunc primeSieve(n, limit int) int {\n// Create and populate array of values\n// lim := c\na := make([]bool, limit+1)\nfor i := 0; i < limit+1; i++ {\na[i] = true\n}\n\n// p * p <= limit === p <= int(math.Sqrt(float64(limit)))\nfor p := 2; p <= int(math.Sqrt(float64(limit))); p++ {\n// At first all will be true\nif a[p] == true {\n// i = 2 * 2, i = 3 * 2, i = 4 * 2, ..., i = i * i\n// Calculate multiples of 2, then 3, then 5, then 7\nfor i := p * 2; i <= limit; i += p {\na[i] = false\n}\n}\n}\n\n// Count primes up to n\n// if primes == 10001, return sievePrime\nvar primes, sievePrime int\nfor p := 2; p <= limit; p++ {\nif a[p] == true {\nprimes++\n// Sum up primes and print 10,001th prime\nif primes <= n {\nsievePrime = p\n}\n}\n}\nreturn sievePrime\n}\n\nfunc main() {\nstart := time.Now()\n// Nth prime to find\nvar n, limit int = 10001, 105000 // Arbitrary limit find  by  testing\n\nfmt.Println(\"Execution time: \", time.Since(start))\nfmt.Println(\"10,001th Prime: \", primeSieve(n, limit))\n}", "date": "2020-06-03 22:48:55", "meta": {"domain": "oschvr.com", "url": "https://oschvr.com/problems/7/", "openwebmath_score": 0.3320310115814209, "openwebmath_perplexity": 3050.9309679514768, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9871787830929848, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8541308603718819}}
{"url": "http://math.stackexchange.com/questions/125588/limit-of-21-n-as-n-to-infty-is-1", "text": "# Limit of $2^{1/n}$ as $n\\to\\infty$ is 1\n\nHow do I prove that:\n\n$\\lim \\limits_{n\\to \\infty}2^{1/n}=1$\n\nThank you very much.\n\n-\n\nUsing the binomial theorem for integer exponents:\n\nCan you see that $(1+\\frac 1 n)^n > 2>1$\n\nTake the nth root, to give:\n\n$(1+\\frac 1 n) > 2^{\\frac 1 n} >1$\n\n-\noh, that's really nice. how to you prove that $(1+\\frac{1}{n})^n>2$? Thanks! \u2013\u00a0 Anonymous Mar 28 '12 at 20:18\n$(1+x)^n = 1+nx+$ other positive terms \u2013\u00a0 Mark Bennet Mar 28 '12 at 20:20\nAnd $n$ has to be greater than 1. \u2013\u00a0 Mark Bennet Mar 28 '12 at 20:21\nsweet, thank you. \u2013\u00a0 Anonymous Mar 28 '12 at 20:25\nNow look at Sivaram Ambikasaram's answer and catch hold of how the two are related - then you'll have learned some maths. \u2013\u00a0 Mark Bennet Mar 28 '12 at 20:33\n\nNote that for $a\\gt 0$, $$a^b = e^{b\\ln a}.$$ So $$\\lim_{n\\to\\infty}2^{1/n} = \\lim_{n\\to\\infty}e^{\\frac{1}{n}\\ln 2}.$$ Since the exponential is continuous, we have $$\\lim_{n\\to\\infty}e^{\\frac{1}{n}\\ln 2} = e^{\\lim\\limits_{n\\to\\infty}\\frac{1}{n}\\ln 2}.$$\n\nCan you compute $\\displaystyle\\lim_{n\\to\\infty}\\frac{\\ln 2}{n}$ ?\n\n-\nThank you for the help, however I don't want to prove this using ln and e. \u2013\u00a0 Anonymous Mar 28 '12 at 20:09\n@Anonymous: Thank you for your comment; perhaps next time you can state, in your post, what it is you do and do not want, so that other people may not waste their time giving you information you don't care about. \u2013\u00a0 Arturo Magidin Mar 28 '12 at 20:11\nI really appreciate your time and sorry for the wasted time for me, however there are probably people who would find this informative and want to know how to prove it this way. I didn't state that I don't want lan and e because I wasn't aware that it can be proven this way. Thank you very much again. \u2013\u00a0 Anonymous Mar 28 '12 at 20:16\nThis answer deserves more than one upvote, as it not only answers the question, but also deals neatly with the ambiguity in the way the question is asked. \u2013\u00a0 Mark Bennet Mar 28 '12 at 21:37\n\nHINT: Use squeeze theorem.\n\nSince $1 < 2$, we have $1 = 1^{1/n} < 2^{1/n}$ for all $n \\in \\mathbb{N}$.\n\nTo bound the limit from above, note that $1 + n \\epsilon < \\left( 1 + \\epsilon \\right)^n$.\n\nHence, given any $\\epsilon >0$, $\\forall n > \\displaystyle 1/{\\epsilon}$, we have $2 < 1 + n \\epsilon < \\left(1 + \\epsilon \\right)^n$ and hence $2^{1/n} < 1 + \\epsilon$.\n\n-\n\nHere are a few different proofs which don't use $e$ or $\\log$ and can be regarded completely elementary.\n\nProof 1)\n\nWe use the following theorem:\n\nIf $a_n \\gt 0$ and $\\lim_{n \\to \\infty} \\frac{a_{n+1}}{a_n} = L$, then $\\lim_{n\\to\\infty} a_n^{1/n} = L$\n\nThis is a standard theorem, and a proof can be found in almost any textbook. You can also find a proof in my answer here: Show that this limit is equal to $\\liminf a_{n}^{1/n}$ for positive terms.\n\nApply the theorem to the sequence $a_n = 2$.\n\nProof 2)\n\nWe use the $\\text{AM} \\ge \\text{GM}$ inequality on $n-1$ ones and one $2$.\n\n$$\\frac{1 + 1 + \\dots + 1 + 2}{n} \\ge 2^{1/n}$$\n\n$$\\frac{n+1}{n} \\ge 2^{1/n}$$\n\nThus we have\n\n$$1 + \\frac{1}{n} \\ge 2^{1/n} \\ge 1$$\n\nso by Squeeze theorem, $\\lim_{n \\to \\infty} 2^{1/n} = 1$.\n\nProof 3)\n\nWe can use Bernouli's inequality (essentially similar to Sivaram's answer) to show that\n\n$$\\left(1 + \\frac{1}{n}\\right)^n \\ge 1 + \\frac{1}{n} \\times n = 2$$\n\nand we get inequalities similar to the proof in 2).\n\nProof 4)\n\nThe sequence $a_n = 2^{1/n}$ is bounded below (by $1$) and montonically decreasing.\n\nThus it is convergent, to say $L$. Since $a_{2n}$ also converges to $L$, we have that $L = \\sqrt{L}$, as $2^{1/2n} = \\sqrt{2^{1/n}}$. So $L = 0$ or $L = 1$. Since the limit is not less than $1$ ($2^{1/n} \\ge 1$), the limit is $1$.\n\nProof 5)\n\nFor $n \\gt 2$, we have that $1 \\le 2^{1/n} \\le n^{1/n}$.\n\nNow use the fact that $\\lim_{n \\to \\infty} n^{1/n} = 1$.\n\nAn elementary proof of that can be found here: http://math.stackexchange.com/a/115825/1102. Any proof for $n^{1/n}$ now becomes a proof for $2^{1/n}$. Proof 1) above can also be used for $n^{1/n}$.\n\nProof 6)\n\nUsing combinatorics.\n\nThe number of $n$ digit numbers in base-$n+1$ is $(n+1)^n$ (allowing for leading zeroes). The number of $n$ digits numbers in base-$n$ is $n^n$. We can show that $(n+1)^n \\ge 2 \\times n^n$: consider the base-$n$ numbers. Replace the last digit with $n$. You get a base-$n+1$ $n$ digit number. Counting the base-$n$ numbers (which are also base-$n+1$ numbers) and the \"last digit modified\" numbers, gives us the inequality.\n\nThis inequality implies that $1 + \\frac{1}{n} \\ge 2^{1/n}$ and can be used to give a proof using the squeeze theorem, similar to proofs 2 and 3.\n\n-\nSorry, had some free time :-) I am pretty sure there are more proofs... \u2013\u00a0 Aryabhata Mar 28 '12 at 21:31\nVery nice list! \u2013\u00a0 Pedro Tamaroff Mar 29 '12 at 23:50\n@PeterT.off: Thanks! \u2013\u00a0 Aryabhata Mar 29 '12 at 23:59", "date": "2015-08-03 19:15:17", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/125588/limit-of-21-n-as-n-to-infty-is-1", "openwebmath_score": 0.9334490895271301, "openwebmath_perplexity": 316.0661866894746, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787838473909, "lm_q2_score": 0.8652240756264638, "lm_q1q2_score": 0.8541308507324155}}
{"url": "https://www.physicsforums.com/threads/sum-of-series.230104/", "text": "# Sum of series\n\n## Homework Statement\n\nFind the sum of the series:\n\n$$S = 1~+~\\frac{1 + 2^3}{1 + 2}~+~\\frac{1 + 2^3 + 3^3}{1 + 2 + 3}~+~\\frac{1 + 2^3 + 3^3 + 4^3}{1 + 2 + 3 + 4}~+~...$$\n\nupto 11 terms\n\n## Homework Equations\n\nSum of first 'n' natural numbers: $$S = \\frac{n(n + 1)}{2}$$\nSum of the squares of the first 'n' natural numbers: $$S = \\frac{n(n + 1)(2n + 1)}{6}$$\nSum of the cubes of the first 'n' natural numbers: $$S = \\left(\\frac{n(n+1)}{2}\\right)^2$$\n\n## The Attempt at a Solution\n\nIn the series, the $n^{th}$ term is given by:\n\n$$T_n = \\frac{1 + 2^3 + 3^3 + ... + n^3}{1 + 2 + 3 + ... + n}$$\n\n$$T_n = \\frac{\\left(\\frac{n(n+1)}{2}\\right)^2}{\\left(\\frac{n(n + 1)}{2}\\right)}$$\n\n$$T_n = \\frac{1}{2}(n^2 + n)$$\n\nHence,\n\n$$S_n = \\frac{1}{2}\\left(\\frac{n(n + 1)(2n + 1)}{6} + \\frac{n(n + 1)}{2}\\right)$$\n\nOn substituting n = 11, I get:\n\n$$S_n = 286$$\n\nBut this isn't one of the options I got on the test. All the options were between 300-400 and all had their last digit '9' [that's all i remember.. they took the question paper away].\n\nLast edited:", "date": "2021-12-02 04:10:35", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/sum-of-series.230104/", "openwebmath_score": 0.8423073887825012, "openwebmath_perplexity": 1922.9059068215759, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850897067342, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8541306592748014}}
{"url": "https://math.stackexchange.com/questions/3717811/what-is-bbb-ex/3717822", "text": "# What is $\\Bbb E(X)$?\n\nLet there be $$100$$ balls in a box out of which $$50$$ are red and $$50$$ are green. Pick $$75$$ balls at random from the box and throw them away. Now pick one ball at random from the remaining balls in the box. Let $$X$$ be the random variable which takes the value $$100$$ when the ball drawn is red in colour and takes the value $$25$$ if the ball drawn is green in colour. Find the expectation $$\\Bbb E(X).$$\n\nMy attempt $$:$$ Let $$Y$$ denote the number of red balls thrown away. Then the number of green balls thrown away is $$75-Y.$$ Clearly $$Y \\geq 25.$$ So \\begin{align*} \\Bbb P(X=100) & = \\sum\\limits_{n=25}^{49} \\Bbb P(X=100 \\mid Y=n)\\ \\Bbb P(Y=n) \\\\ & = \\sum\\limits_{n=25}^{49} \\frac {\\binom {50-n} {1}} {\\binom {25} {1}} \\times \\frac {\\binom {50} {n}} {\\binom {100} {n}} \\end{align*} Similarly \\begin{align*} \\Bbb P(X=25) & = \\sum\\limits_{n=25}^{49} \\Bbb P(X=25 \\mid (75-Y) = n)\\ \\Bbb P((75-Y) = n) \\\\ & = \\sum\\limits_{n=25}^{49} \\Bbb P(X=25 \\mid Y = 75-n)\\ \\Bbb P(Y = 75-n) \\\\ & = \\sum\\limits_{n=25}^{49} \\frac {\\binom {50-n} {1}} {\\binom {25} {1}} \\times \\frac {\\binom {50} {75-n}} {\\binom {100} {75-n}} \\end{align*}\n\nThen the required expectation would be $$100\\ \\Bbb P(X=100) + 25\\ \\Bbb P(X=25).$$ But the computation is very tough. Is there any simpler way to approach the problem?\n\nAny help will be highly appreciated. Thank you very much.\n\n\u2022 It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So P(X=100) = P(X=25) = 1/2 \u2013\u00a0Ant Jun 13 '20 at 8:51\n\u2022 But @Ant after throwing $75$ balls away it might not be the case. Obviously the green and red balls in the box are present with different proportions after throwing $75$ balls. Right? Then the red and green balls are not equally likely to be drawn. \u2013\u00a0Phi beta kappa Jun 13 '20 at 8:55\n\u2022 I have posted an answer, so we can keep discussing it there. To answer your question, they will be present in a different proportion after being drawn, but that's just a single observation. Their distribution is the same; the distribution of how many red vs green balls end up in the urn after you throw some away is the same. Therefore the distribution of which ball you pick after is also the same. So probabilities are the same. Right? \u2013\u00a0Ant Jun 13 '20 at 9:02\n\u2022 It is clear by symmetry that $$\\mathbb P(X=25)=\\mathbb P(X=100)=\\frac12.$$ There is no need for elaborate computation here. \u2013\u00a0Math1000 Jun 13 '20 at 10:09\n\u2022 @Math1000 can you show it mathematically without making some vague assertion? I know that you are a genius; but it's better not to think others as genius as you. \u2013\u00a0Phi beta kappa Jun 13 '20 at 10:22\n\nIt seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So $$P(X=100) = P(X=25) = 1/2$$\n\nWhich implies\n\n$$E(X) = 100\\cdot P(X=100) + 25 \\cdot P(X=25) = 62.5$$\n\nA simple script confirms that the expectation is indeed $$62.5$$. You can also double check by computing the probability. Using your terminology,\n\n$$P(Y=n) = \\frac{{75 \\choose n} \\cdot {25 \\choose 50-n}} {100\\choose 50}$$\n\nSo that\n\n$$P(X=100) = \\sum_{n=25}^{50} \\frac{50-n}{25} \\cdot \\frac{{75 \\choose n} \\cdot {25 \\choose 50-n}} {100\\choose 50}$$\n\nAnd you can check numerically that this is equal to $$1/2$$\n\n\u2022 How have you computed $\\Bbb P(Y=n)$? \u2013\u00a0Phi beta kappa Jun 13 '20 at 9:43\n\u2022 The probability of removing n red balls can be computed as follows: Line up all the 100 balls in a row. Count how many of the permutations end up with n balls in the first 75 spots. Divide by total number of permutations, and you're done. To compute the first number, you have 75 balls of which n are red. So the permutations are 75!/n!/(75-n)!, i.e. 75 choose n. Then you have to multiply with the number of permutations of the remaining 25 balls; it's the same, except that you now have 50-n red balls. The second number is the total number of permutations, therefore 100 choose 50. \u2013\u00a0Ant Jun 13 '20 at 9:48\n\u2022 I don't think so. If according to your assumption the balls are indistinguishable then we have just $26$ ways to throw $75$ balls from the urn and each way is equally likely. If the balls are indistinguishable then it doesn't matter in which order you throw them away or which red/green balls are thrown away. In this case which really matters that how many red/green balls you throw away out of $75$ balls. Right? If I present them as ordered pairs as (number of red balls thrown away, number of green balls thrown away) then there are $26$ such pairs $(25,50),(26,50), \\cdots, (50,25).$ \u2013\u00a0Phi beta kappa Jun 13 '20 at 10:15\n\u2022 Each pair is equally likely to occur. So for each $25 \\leq n \\leq 50$ we have $$\\Bbb P(Y=n) = \\frac {1} {26}.$$ \u2013\u00a0Phi beta kappa Jun 13 '20 at 10:17\n\u2022 No. I know it for the case of distinguishable but not for the case of indistinguishable. \u2013\u00a0Phi beta kappa Jun 13 '20 at 10:28", "date": "2021-06-14 22:35:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3717811/what-is-bbb-ex/3717822", "openwebmath_score": 0.9752467274665833, "openwebmath_perplexity": 366.2924128320992, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850832642354, "lm_q2_score": 0.8688267813328977, "lm_q1q2_score": 0.8541306486688494}}
{"url": "http://math.stackexchange.com/questions/168288/the-dynamics-of-contrapositive-proofs", "text": "# The Dynamics of Contrapositive Proofs\n\nThe Wikipedia Link for contrapositive proofs states that proving if p then q is the same as proving if not q then not p. I don't completely follow why.\n\nIs there any way to understand what's happening without involving logic tables ? If logic tables are inevitable, can anyone give me a basic explanation?\n\nI can't seem to follow how if p then q gets simplified to not p or q I understand that the truth tables of the two are same. But, that's that.\n\n-\nIt could help to think the about contrapositive of some simple theorems, like the Pythagorean theorem. \u2013\u00a0Ragib Zaman Jul 8 '12 at 16:43\nSemiformally speaking: Suppose $P\\implies Q$. Further suppose $Q$ is false; if $P$ were true then $Q$ would also be true and we'd have a contradiction - impossible - so $P\\implies Q$ and $\\lnot Q$ implies $\\lnot P$, i.e. $P\\implies Q$ implies $\\lnot Q\\implies\\lnot P$, ie an implication entails its contrapositive. Of course this means the contrapositive entails the contrapositive's contrapositive, ie the original statement, hence we have an equivalence, $(P\\implies Q)\\iff (\\lnot Q\\implies \\lnot P)$. \u2013\u00a0anon Jul 8 '12 at 16:54\n\nThe sentence \"if $p$ then $q$\" asserts that in every single situation (model) in which $p$ is true, $q$ must be true. The only way this can be incorrect is if we have a situation in which $p$ is true but $q$ is false. We can show this cannot be the case by showing that whenever $q$ fails, $p$ must fail, that is, by showing that if \"not $q$\" then \"not $p$.\"\n\nConversely, suppose that in every situation in which $p$ is true, then $q$ must be true. Then there cannot be a situation in which $q$ fails and $p$ is true. So if \"not $q$\" holds, then \"not $p$\" must hold.\n\nThus the two assertions \"if $p$ then $q$\" and \"if not $q$ then not $p$\" hold in precisely the same cases.\n\nRemark: The answer above is much too abstract. To understand what's going on, it is useful to examine a number of concrete cases. Suppose that we want to show that if $x^2=2$ then $x$ is not an ordinary fraction (integer divided by integer). So $p$ is the assertion $x^2=2$, and $q$ is the assertion that $x$ is not a fraction.\n\nWe prove the result by showing that if $x$ is a fraction, then $x^2$ cannot be equal to $2$, that is, by showing that \"not $q$\" implies \"not $p$.\" That shows that if $x^2=2$, then $x$ could not possibly be a fraction, which is exactly what we wanted.\n\n-\nMuch Clearer now. Thanks Andr\u00e9. \u2013\u00a0Inquest Jul 8 '12 at 17:23\n\n\"If $P$ then $Q$\" tells you that whenever you know that $P$ is true, you may automatically conclude that $Q$ is true.\n\n\"If not $Q$, then not $P$\" tells you that whenever you know that $Q$ is not present, you may conclude that $P$ is not present.\n\nIf ($P$ implies $Q$) is true, then knowledge of the presence of $P$ always gives us that $Q$ is present. If $Q$ isn't there, then there must be no $P$ to have put it there. Likewise, if the absence of $Q$ automatically gives us the absence of $P$, and if $P$ is present, there can be no not-$Q$ to have given us a not-$P$.\n\nSimilarly, either we have $P$ or we don't have $P$. In the case that we have $P$, we are guaranteed to have $Q$. The other case is that we don't have $P$. So we either don't have $P$, or we have $Q$ (which we got by having $P$): not-$P$ or $Q$.\n\nOn the other hand, let's say all we know is that either $P$ isn't true or $Q$ is true. If $P$ isn't true, then $P\\Rightarrow Q$ is trivially true. Let's deal with the case where $P$ is true. So we know that $P$ is true and also that either $P$ isn't true or $Q$ is true. That is, since $P$ and not-$P$ can't both be true, if $P$ is true, then $Q$ is true.\n\n-\n+1. Took me a few reads to understand. The last paragraph's overall flow and structure is why most of my non-math friends don't speak to me when I'm doing math. \u2013\u00a0Inquest Jul 8 '12 at 17:22\nSorry. I'll take a look to revise it. \u2013\u00a0Neal Jul 8 '12 at 17:26", "date": "2016-06-26 08:44:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/168288/the-dynamics-of-contrapositive-proofs", "openwebmath_score": 0.889682412147522, "openwebmath_perplexity": 180.12692310658505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850857421198, "lm_q2_score": 0.8688267660487573, "lm_q1q2_score": 0.8541306357960913}}
{"url": "https://stats.stackexchange.com/questions/182875/is-every-correlation-matrix-positive-definite", "text": "# Is every correlation matrix positive definite?\n\nI'm talking here about matrices of Pearson correlations.\n\nI've often heard it said that all correlation matrices must be positive semidefinite. My understanding is that positive definite matrices must have eigenvalues $> 0$, while positive semidefinite matrices must have eigenvalues $\\ge 0$. This makes me think that my question can be rephrased as \"Is it possible for correlation matrices to have an eigenvalue $= 0$?\"\n\nIs it possible for a correlation matrix (generated from empirical data, with no missing data) to have an eigenvalue $= 0$, or an eigenvalue $< 0$? What if it was a population correlation matrix instead?\n\nConsider three variables, $X$, $Y$ and $Z = X+Y$. Their covariance matrix, $M$, is not positive definite, since there's a vector $z$ ($= (1, 1, -1)'$) for which $z'Mz$ is not positive.\n\nHowever, if instead of a covariance matrix I do those calculations on a correlation matrix then $z'Mz$ comes out as positive. Thus I think that maybe the situation is different for correlation and covariance matrices.\n\nMy reason for asking is that I got asked over on stackoverflow, in relation to a question I asked there.\n\n\u2022 If, for example, two attributes are one thing, only having different names, the matrix is singular. If two attributes add to a constant, it is again singular, et cetera. \u2013\u00a0ttnphns Nov 21 '15 at 16:54\n\u2022 If a covariance matrix is singular correlation matrix is singular as well. \u2013\u00a0ttnphns Nov 21 '15 at 17:05\n\u2022 Near-duplicates: Is every correlation matrix positive semi-definite? which has less focus on the definite versus semi-definite angle, and Is every covariance matrix positive definite? which is relevant because a covariance is essentially a rescaled correlation. \u2013\u00a0Silverfish Nov 21 '15 at 21:20\n\nCorrelation matrices need not be positive definite.\n\nConsider a scalar random variable X having non-zero variance. Then the correlation matrix of X with itself is the matrix of all ones, which is positive semi-definite, but not positive definite.\n\nAs for sample correlation, consider sample data for the above, having first observation 1 and 1, and second observation 2 and 2. This results in sample correlation being the matrix of all ones, so not positive definite.\n\nA sample correlation matrix, if computed in exact arithmetic (i.e., with no roundoff error) can not have negative eigenvalues.\n\n\u2022 May be worth mentioning the possible effects of missing values on the sample correlation matrix. Numerical fuzz isn't the only reason to get a negative eigenvalue in a sample correlation/covariance matrix. \u2013\u00a0Silverfish Nov 21 '15 at 21:22\n\u2022 Yes, I didn't make it explicit, but I was assuming, per the question statement, \"with no missing data\". Once you get into the wild, wacky world of missing data and adjustments therefor, anything goes. \u2013\u00a0Mark L. Stone Nov 21 '15 at 21:35\n\u2022 Yes, sorry, you're quite right the question said \"no missing data\" - just thought it worth mentioning somewhere since future searchers might be interested even if the OP's appetite is sated! \u2013\u00a0Silverfish Nov 21 '15 at 21:40\n\nThe answers by @yoki and @MarkLStone (+1 to both) both point out that a population correlation matrix can have zero eigenvalues if variables are linearly related (such as e.g. $X_1 = X_2$ in the example of @MarkLStone and $X_1 = 2X_2$ in the example of @yoki).\n\nIn addition to that, a sample correlation matrix will necessarily have zero eigenvalues if $n<p$, i.e. if the sample size is smaller than the number of variables. In this case covariance and correlation matrices will both be at most of rank $n-1$, so there will be at least $p-n+1$ zero eigenvalues. See Why is a sample covariance matrix singular when sample size is less than number of variables? and Why is the rank of covariance matrix at most $n-1$?\n\n\u2022 True 'dat. I suppose I could have and should have provided this info as well, but my goal was to produce a counterexample to refute the OP's hypothesis, thereby showing its invalidity Nevertheless, you should adjust your second sentence to be \"In this case covariance and correlation matrices will be at most rank n\u22121, so there will be at least (p\u2212n+1) zero eigenvalues.\" \u2013\u00a0Mark L. Stone Nov 21 '15 at 15:21\n\nConsider $X$ to be an r.v. with mean 0 and variance of 1. Let $Y=2X$, and calculate the covariance matrix of $(X,Y)$. Since $2X=Y$, $E[Y^2]=4E[X^2]=\\sigma_Y^2$, and $E[XY]=2E[X^2]$. Due to the zero mean configuration, the second moments are equal to the suitable covariances, for instance: $\\mbox{Cov}(X,Y)=E[XY]-EXEY=E[XY]$.\n\nSo the covariance matrix will be: $$\\Lambda = \\left(\\array{1 & 2 \\\\ 2 & 4 }\\right),$$ having a zero eigenvalue. The correlation matrix will be: $$\\Lambda = \\left(\\array{1 & 1 \\\\ 1 & 1 }\\right),$$ having a zero eigenvalue as well. Due to the linear correspondence between $X$ and $Y$ it is easy to see why we get this correlation matrix - the diagonal will always be 1, and the off-diagonal is 1 because of the linear relationship.\n\n\u2022 Just for math-challenged readers like myself, let me point out that the 2 in $\\Lambda$ is the $cov(X,Y)= \\mathbb{E}(XY)-\\mathbb{E}(X)\\mathbb{E}(Y)= 2\\mathbb{E}[X^2]=2\\left(\\sigma_X^2+[E(X)]^2\\right)$ this last equality resulting from: $E(X^2)=\\text{Var}(X)+[E(X)]^2$. \u2013\u00a0Antoni Parellada Nov 21 '15 at 19:45\n\u2022 +1 for your post. I wanted to make it easy to follow for everyone by including the $diag \\Lambda^{-1/2}\\,\\Lambda\\, diag \\Lambda^{1/2}$ formula, but the comments format won't allow it. Do you think there is any valid point in including it in your post? \u2013\u00a0Antoni Parellada Nov 21 '15 at 20:44\n\u2022 @AntoniParellada , I'm not exactly sure what you mean - the covariance here is a direct calculation. But I will edit and make that clearer. Thanks. \u2013\u00a0yoki Nov 22 '15 at 14:59", "date": "2020-07-11 04:47:21", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/182875/is-every-correlation-matrix-positive-definite", "openwebmath_score": 0.8042393922805786, "openwebmath_perplexity": 364.2396800422581, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731158685837, "lm_q2_score": 0.8872045847699186, "lm_q1q2_score": 0.8540880020333506}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=i6kgsan0vsfpkpiho8afbjkju3&action=printpage;topic=2634.0", "text": "Toronto Math Forum\n\nAPM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Xiangmin.Z on January 17, 2022, 04:40:37 PM\n\nTitle: Week 2 Lec 1 (Chapter 2) question\nPost by: Xiangmin.Z on January 17, 2022, 04:40:37 PM\nHello, I have a question about example 4 from W2 L1:\nWe are given :$u_{t}+xu_{x}=xt$\nafter calculation we get:\n$x=Ce^t$\n\n$du=xt dt=Cte^t dt$, so $u=C(t-1)e^t+D=x(t-1)+D$,\nbut how did we get $D=\\phi({xe^{-t}} )$? we know it is a constant, but why is D depended on $xe^{-t}$?\n\nAlso, why would the initial condition $u|_{t=0} =0$ implies that $\\phi({x}) =x$ ?\n\nThanks.\n\nTitle: Re: Week 2 Lec 1 (Chapter 2) question\nPost by: Xiangmin.Z on January 17, 2022, 05:27:38 PM\nI checked the calculation and I think the answer for x is correct, and does anyone know why D is depended on $xe^{-t}$?\nTitle: Re: Week 2 Lec 1 (Chapter 2) question\nPost by: Victor Ivrii on January 17, 2022, 07:48:38 PM\nNow it is correct $x=Ce^{t}$ and then $C=?$\nTitle: Re: Week 2 Lec 1 (Chapter 2) question\nPost by: Xiangmin.Z on January 17, 2022, 08:19:53 PM\n$C=xe^{-t}$, so D is a function of $\\phi$, therefore $D=\\phi({xe^{-t}} )$ ? Yes, but \"$D$\u00a0 is a function of it\"", "date": "2022-06-30 14:07:06", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=i6kgsan0vsfpkpiho8afbjkju3&action=printpage;topic=2634.0", "openwebmath_score": 0.632986307144165, "openwebmath_perplexity": 3300.24485774966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9752018398044143, "lm_q2_score": 0.8757870013740061, "lm_q1q2_score": 0.8540690950167219}}
{"url": "https://math.stackexchange.com/questions/1716583/prove-that-fx-x3-x-is-not-injective", "text": "# Prove that $f(x) = x^3 -x$ is NOT Injective\n\nOkay, so I know there are plenty of injective/surjective (and thus, bijective) questions out there but I'm still not happy with the rigor of what I have done. So I'd really appreciate some help!\n\nSo the question actually asks me to do two things:\n\n(a) give an example of a cubic function that is bijective. Explain why it is bijective.\n\n(b) give an example of a cubic function that is not bijective. Explain why it is not bijective.\n\nSo for (a) I'm fairly happy with what I've done (I think):\n\n$$f: \\mathbb R \\rightarrow \\mathbb R , f(x) = x^3$$\n\nSo we know that to prove if a function is bijective, we must prove it is both injective and surjective.\n\nProof: $f$ is injective\n\nLet: $$x,y \\in \\mathbb R : f(x) = f(y)$$ $$x^3 = y^3$$ (take cube root of both sides) $$x=y$$\n\nProof: $f$ is surjective\n\nLet: $$y \\in \\mathbb R$$\n\n$$x = \\sqrt[3]{y}$$\n\n$$f(x) = (\\sqrt[3]{y})^3 = y$$\n\nSo I believe that is enough to prove bijectivity for $f(x) = x^3$. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.\n\nSo for (b)\n\n$$f: \\mathbb R \\rightarrow \\mathbb R , f(x) = x^3 - x$$\n\nNow I'm just going to try and prove it is NOT injective, as that should be sufficient to prove it is NOT bijective.\n\nLet:\n\n$$x,y \\in \\mathbb R : f(x) = f(y)$$ $$x^3 - x = y^3 - y$$\n\nThis is about as far as I get. Send help. Thanks everyone.\n\n\u2022 What happens if $x=0$ or $x=1$? \u2013\u00a0Cameron Williams Mar 28 '16 at 2:59\n\u2022 To prove that a function $f$ is not injective, you just need to find two different numbers $x$ and $y$ such that $f(x)\\not=f(y)$. \u2013\u00a0Christopher Carl Heckman Mar 28 '16 at 3:02\n\u2022 @Carl you mean two numbers $x,y$ such that $x\\neq y$ but $f(x)=f(y)$ \u2013\u00a0JMoravitz Mar 28 '16 at 3:03\n\u2022 @CarlHeckman as JMoravitz said, you may confuse op... \u2013\u00a0YoTengoUnLCD Mar 28 '16 at 3:07\n\u2022 Thanks for the help guys, +1's given. \u2013\u00a0Rubicon Mar 28 '16 at 3:53\n\nInjectivity:\n\nA function $f$ from $X\\to Y$ is said to be injective iff the following statement holds true:\n\nfor every $x_1,x_2\\in X$ if $x_1\\neq x_2$ then $f(x_1)\\neq f(x_2)$\n\nThe negation of this then yields:\n\nA function $f$ from $X\\to Y$ is not injective iff there exists $x_1,x_2\\in X$ such that $x_1\\neq x_2$ but $f(x_1)=f(x_2)$\n\nIn the case of the cubic in question, it is an easily factorable polynomial and we can find multiple distinct roots. $f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$\n\nWe know that a root of a polynomial is a number $\\alpha$ such that $f(\\alpha)=0$\n\nSo, $f(1)=f(0)=f(-1)=0$ despite $1,0,-1$ all being distinct unequal numbers in the domain. This shows that it is not injective, and thus not bijective.\n\nAs an aside, one can prove that any odd degree polynomial from $\\Bbb R\\to \\Bbb R$ must be surjective by the fact that polynomials are continuous and the intermediate value theorem.\n\n\u2022 This is great. I think it has helped my understanding significantly. So what you're sort of saying is because f(1) and f(-1) equal 0, there is more than one-to-one mapping to the elements in the co-domain (hence not injective)? So in this case the set might even look like this: {(1,0) , (-1,0)} (just as a way to visualise it)? \u2013\u00a0Rubicon Mar 28 '16 at 3:53\n\nRegarding (a), when you say \"take cube root of both sides\" you are (at least implicitly) assuming that the function is injective -- if it were not, the cube root would not be a well-defined operation. Depending on what you are / are not allowed to assume, that could be regarded as a circular argument.\n\nHowever, there is another approach. Beginning with the assumption that $x^3=y^3$, you can manipulate the expression as follows: $$x^3-y^3=0$$ $$(x-y)(x^2+xy+y^2)=0$$ Now if you can find a convincing argument that the second factor, $x^2+xy+y^2$, is always positive for any choice of $x$ and $y$, then it follows that the only way $x^3=y^3$ can be true is if $x=y$.\n\nThe factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$, by the way, could also give you some insight into how to handle (b).\n\n\u2022 $$x^2+xy+y^2 = {x^2+y^2\\over 2} + {1\\over 2}(x^2 + 2xy+y^2) = {x^2+y^2\\over 2} + {1\\over2}(x+y)^2 \\ge 0.$$ The only way to get equality is to have $x=0$ and $y=0$, but then $x=y$ in that case as well. \u2013\u00a0Christopher Carl Heckman Mar 29 '16 at 6:30\n\u2022 @CarlHeckman Yes, I know. Also $x^2+xy+y^2 = (x + \\frac{1}{2}y)^2 + \\frac{3}{4}y^2$. I was intentionally leaving that out to leave the OP something to finish. \u2013\u00a0mweiss Mar 29 '16 at 14:16", "date": "2019-11-17 09:42:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1716583/prove-that-fx-x3-x-is-not-injective", "openwebmath_score": 0.8318087458610535, "openwebmath_perplexity": 164.52781695756033, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018426872776, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.8540690817333773}}
{"url": "https://math.stackexchange.com/questions/426579/can-an-eigenvalue-lambda-i-0", "text": "# Can an eigenvalue $\\lambda_i=0$?\n\nI was doing some work with diagonalization of a matrix $A$ in order to find a matrix $P$ such that $\\,P^{-1}AP\\,$ was diagonal. In order to that I set $\\;\\lambda I_{n}=0\\;$ and found the characteristic polynomial and its roots.\n\nWhen I factored my characteristic polynomial I obtained $\\;\\lambda^2(\\lambda-2),\\,$ so $\\,\\lambda=0,\\,2$.\n\nI was taught that the eignenvalues$\\,\\lambda_{i}\\,$ I found become the entries of the diagonal matrix $\\,P^{-1}AP.\\,$ If this is indeed true, then two of the diagonal entries would be $\\,0.\\,$ Is this allowed, or must a diagonal matrix strictly have non-zero diagonal entries?\n\n\u2022 Zero is allowed. You may be thinking of eigenvectors --- the zero vector can't be an eigenvector. But for eigenvalues, no problem. \u2013\u00a0Gerry Myerson Jun 21 '13 at 23:44\n\nAbsolutely yes: It is very possible $\\lambda = 0$. Zero is allowed.\n\nYou may be mixing up what you know about eigenvectors --- the zero vector cannot be an eigenvector.\n\nBut for an eigenvalue $\\lambda$, it is certainly possible and admissible that $\\lambda = 0$.\n\nWith respect to your last question:\n\n\"($\\lambda = 0$): Is this allowed, or does a diagonal matrix strictly have to have the diagonal entries as non-zero?\"\n\nYes, it is allowed for zero's to be on the diagonal. No, the diagonal entries need not be non-zero.\n\n\u2022 Dang! Beat me to the punch with the eigenvalue vs. eigenvector observation. +1 \u2013\u00a0Cameron Buie Jun 21 '13 at 23:58\n\u2022 +1 what that Dang! means Amy? Dang Dang... Is that like a sound of a big Bell? \u2013\u00a0mrs Jun 22 '13 at 0:09\n\u2022 +1 it's a non-profane way of saying something like \"Damn\": meaning, if I hadn't posted what I posted, Cameron intended to. \u2013\u00a0amWhy Jun 22 '13 at 0:12\n\u2022 @amWhy Thanks; makes perfect sense now. I think I was confusing eigenvalues with eigenvectors. \u2013\u00a0Sujaan Kunalan Jun 22 '13 at 0:36\n\u2022 @amWhy: Spot on +1 \u2013\u00a0Amzoti Jun 22 '13 at 0:38\n\nA matrix is called diagonal if its off-diagonal entries ($a_{ij}$ for $i \\not= j$) are all zero. This does not require the diagonal entries ($a_{jj}$) to be nonzero. For instance, the zero matrix is diagonal.", "date": "2021-01-26 13:12:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/426579/can-an-eigenvalue-lambda-i-0", "openwebmath_score": 0.9197072386741638, "openwebmath_perplexity": 715.4836553235508, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018362008348, "lm_q2_score": 0.8757869803008765, "lm_q1q2_score": 0.8540690713101992}}
{"url": "https://math.stackexchange.com/questions/3148433/solving-problem-using-related-rates-yields-incorrect-result", "text": "# Solving problem using related rates yields incorrect result\n\nI've been trying for a while to figure out what I did wrong on this problem, help would be appreciated.\n\n\"A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, how fast does the ladder move up the wall 5 sec after we start pushing?\"\n\n$$x$$ = distance from wall, $$y$$ = height of ladder on wall, $$h$$ = length of ladder $$\\frac {dx}{dt} = -1, t = 5$$ $$x^2 + y^2 = h^2$$ $$x^2 +y^2 = 25^2$$ $$2x\\frac{dx}{dt}+2y\\frac{dy}{dt} = 0$$ $$\\frac{dy}{dt} = \\frac{-x\\frac{dx}{dt}}{y}$$ Since the ladder is moving toward the wall at 1 ft/sec for 5 sec, $$x(t = 5) =20-5 = 15$$\n\n$$y(t=5) = \\sqrt{25^2 - 15^2}=20$$ Substitute variables in equation: $$\\frac{dy}{dt} = \\frac{-(15)(-1)}{20}$$ $$\\frac{dy}{dt} = \\frac{3}{4}$$ The issue is that if $$\\frac{dy}{dt} = \\frac{3}{4},$$ the ladder would have risen 3.75 ft. in 5 seconds instead of 5 feet it SHOULD have risen, which is given by: $$\\Delta y = y(t=5) - y(t=0)$$ $$(\\sqrt{25^2 - 15^2}) - (\\sqrt{25^2 - 20^2})$$ $$15-20$$ $$5$$ Thank you for reading and any answers.\n\nAt the start you have $$y(0)=\\sqrt {25^2-20^2}=15$$. You are correct that the top has risen $$5$$ feet in $$5$$ seconds, but that does not mean that it has risen steadily at $$1$$ ft/sec for the whole time. Because $$\\frac {dx}{dt}=-1$$ you found $$\\frac {dy}{dt}=\\frac xy$$ At $$t=0$$ that means that $$\\frac {dy}{dt}=\\frac {20}{15}=\\frac 43$$ and at $$t=5$$ that means $$\\frac {dy}{dt}=\\frac {15}{20}=\\frac 34$$. There is no contradiction here.\n\n\u2022 Oh, I assumed that the rate of change had to be constant for some reason. Thank you for your help! \u2013\u00a0Kebab Boy Mar 14 at 20:07\n\nKeeping your notation, you have that $$x(t) = 20 - t, \\quad y(t)=\\sqrt{25^2-x(t)^2},$$\n\nand so, $$y'(t) = \\frac{-2 x'(t)x(t)}{2 \\sqrt{25^2-x(t)^2}},$$\n\nwhich means that $$y'(5)=\\frac{15}{\\sqrt{25^2-15^2}}=\\frac{3}{4}$$\n\nThe vertical displacement does not have to match the horizontal one. They are just connected through the expression $$x(t)^2 + y(t)^2=25^2$$.", "date": "2019-07-20 11:39:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3148433/solving-problem-using-related-rates-yields-incorrect-result", "openwebmath_score": 0.6802319884300232, "openwebmath_perplexity": 148.13468734906402, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429138459387, "lm_q2_score": 0.867035771827307, "lm_q1q2_score": 0.8540674430894328}}
{"url": "https://www.physicsforums.com/threads/solving-uartics.638159/", "text": "Solving uartics\n\nBloodyFrozen\nIs there a method to finding the roots of quartics besides Ferrari's formula?\n\nI have the equation\n\n$$x^{4}+5x^{2}+4x+5=0$$\n\nI know one of the factor is something like $$x^{2}+x+1$$ and the other one can be found using sythetic division, but how can I find the factors without knowing one of them in the first place?\n\nThanks.\n\nHomework Helper\nIf the quartic has \"nice\" quadratic factors, then they will be of the form:\n\n$$x^4+5x^2+4x+5\\equiv (x^2+ax\\pm 1)(x^2+bx\\pm 5)$$\n\nCan you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\\pm$ operator.\n\nBloodyFrozen\nIf the quartic has \"nice\" quadratic factors, then they will be of the form:\n\n$$x^4+5x^2+4x+5\\equiv (x^2+ax\\pm 1)(x^2+bx\\pm 5)$$\n\nCan you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\\pm$ operator.\n\nOh, I see, but how do we know when the factors are \"nice\"?\n\nHomework Helper\nOh, I see, but how do we know when the factors are \"nice\"?\n\nWe don't. We can always check to see if they are, just like we check to see if there are any rational roots.\n\nWith the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.\n\nBloodyFrozen\nOk, thanks!\n\nBloodyFrozen\nAh, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link\n\nBloodyFrozen\nWe don't. We can always check to see if they are, just like we check to see if there are any rational roots.\n\nWith the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.\n\nI end up getting two systems but one of them has no solution. Did I proceed correctly?\n\nHomework Helper\nI end up getting two systems but one of them has no solution. Did I proceed correctly?\n\nI don't know, what did you get?\n\nBloodyFrozen\nSorry for not having LATEX since I'm on my phone, but I got:\n\nSystem 1:\n\na+b=0\n5+1+ab=5\n5a+b=4\n\nSystem 2:\n\na+b=0\n-5-1+ab=5\n-5a-b=4\n\nHomework Helper\nSorry for not having LATEX since I'm on my phone, but I got:\n\nSystem 1:\n\na+b=0\n5+1+ab=5\n5a+b=4\nRight, so simplifying this system, we have\n\na+b=0\nab=-1\n5a+b=4\n\nWhich we can then deduce,\na=-b\ntherefore,\na2=1 -> a=$\\pm1$\n4a=4 -> a=1, b=-1\n\nSystem 2:\n\na+b=0\n-5-1+ab=5\n-5a-b=4\nFor this system we have no real solution, so what does that tell you?\n\nBloodyFrozen\nFor the second system, I don't get any solutions. Therefore, that can't be the right system.\n\nHomework Helper\nFor the second system, I don't get any solutions. Therefore, that can't be the right system.\n\nExactly, so what must your factors be?\n\nBloodyFrozen\nAh, that's what I thought. Thanks!", "date": "2022-08-18 11:15:01", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/solving-uartics.638159/", "openwebmath_score": 0.8108935356140137, "openwebmath_perplexity": 259.1104705689164, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429120895838, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8540674364900639}}
{"url": "https://math.stackexchange.com/questions/1990137/the-idea-behind-the-sum-of-powers-of-2/1990146", "text": "# The idea behind the sum of powers of 2\n\nI know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.\n\nFor example, sum of n numbers is $\\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, hence we can do the trick. What is the logic behind the sum of power of $2$ formula?\n\n\u2022 This addition of powers of two, and many other examples, are all geometric series. There's a lot of visualisations of this out there. \u2013\u00a0Parcly Taxel Oct 29 '16 at 10:58\n\u2022 Write the numbers in base 2: The powers of $2$ starting from $1=2^0$ will be in binary, $1+10+100+1000$ will always be a number that will be a n with all binary digits 1. This is the largest number having that many digits. SO it is of the form $2^{n+1}-1$ \u2013\u00a0P Vanchinathan Oct 29 '16 at 11:01\n\u2022 I think there's a counting/probabilistic way to approach this something about having $n$ binary choices \u2013\u00a0BCLC Oct 31 '16 at 11:17\n\n## 8 Answers\n\nThe binary expansion of $$\\sum_{k=0}^n2^k$$ is a string of $$n+1$$ 1's: $$\\underbrace{111\\dots111}_{n+1}$$ If I add a 1 to this number, what do I get? $$1\\underbrace{000\\dots000}_{n+1}$$ 1 followed by $$n+1$$ 0's, hence $$2^{n+1}$$. Therefore $$\\sum_{k=0}^n2^k=2^{n+1}-1$$\n\n\u2022 Very clever proof! \u2013\u00a0user3932000 Nov 28 '20 at 4:16\n\nThis works for any partial sum of geometric series.\n\nLet $$S = 1 + x + x^2+\\ldots +x^n$$. Then $$xS = x + x^2 + \\ldots +x^n + x^{n+1} = S - 1 + x^{n+1}$$.\n\nAll you have to do now is solve for $$S$$ (assuming $$x\\neq 1$$).\n\n\u2022 Yes, but the OP said that he already knew this. What he is asking for is a visual representation of the proof. \u2013\u00a0Alex M. Oct 29 '16 at 11:01\n\u2022 Actually, they said that they knew proof by induction, but didn't know how one would come up with the formula in the first place. At least that is how I understood the question. @AlexM. \u2013\u00a0Ennar Oct 29 '16 at 11:02\n\nThere is a geometrical explanation.\n\nTake a box long enough to put twice the amount of items equal to the last term of the sum, and try to pack it.\n\nLet's take a box of length $$2 * 2^n$$.\n\nLet's put the items from the first term ($$2^n$$) into the box. Now exactly one half of the space is left for the other terms, from $$2^{n-1}$$ down to $$1$$, so we repeat this process, starting from the next largest term.\n\nAs we put the items from each term, we notice that they take exactly one half of the empty space left in the box, because both the largest term left and the space left decreases in half on each step.\n\nAt some point we reach the first term, which is equal to 1, and there are two empty places left in the box for just one item, so after we put the last item into the box, there is still space for one more item.\n\nThe length of the box is $$2*2^n = 2^{n+1}$$, but it could be shorter by one, which is $$2^{n+1} - 1$$, and this is our formula.\n\nFor example, let's pack: $$\\sum_{i=0}^3 2^i$$\n\nBox length: $$2 * 2^3 = 16$$\n\n2^3    |\u25cf \u25cf \u25cf \u25cf \u25cf \u25cf \u25cf \u25cf|               |\n2^2    |\u25cb \u25cb \u25cb \u25cb \u25cb \u25cb \u25cb \u25cb|\u25cf \u25cf \u25cf \u25cf|       |\n2^1    |\u25cb \u25cb \u25cb \u25cb \u25cb \u25cb \u25cb \u25cb|\u25cb \u25cb \u25cb \u25cb|\u25cf \u25cf|   |\n2^0    |\u25cb \u25cb \u25cb \u25cb \u25cb \u25cb \u25cb \u25cb|\u25cb \u25cb \u25cb \u25cb|\u25cb \u25cb|\u25cf| |\n\n\nIt could be shorter by one: $$2^{3+1}-1 = 15$$\n\nBy the way, similar geometrical explanation can be used for the ever decreasing geometric progression $$\\sum_{i=0}^\\infty 2^{-i}$$ with the only difference that we do not need to account for the last piece of empty space, because it tends to 0. Instead, we take some continuous medium as an example, such as a ribbon or a thread, which can be divided virtually infinitely so it sums to exactly a length of 2 units.\n\nThere is a combinatorial interpretation. Consider the collection of all binary sequences of length $$n+1$$ with at least one $$1$$ (call this set $$E$$). There are $$2^{n+1}-1$$ such sequences because only $$0...0$$ isn't. Now let $$E_j$$ be the set of binary sequences of length $$n+1$$ such that the first $$1$$ is in the $$j$$ th component for $$j=1,\\dotsc, n+1$$. Then $$|E_j|=2^{n+1-j}$$. Then the $$(E_j)$$ partition $$E$$ and we have that $$1+2+2^2+\\dotsb+2^n=\\sum_{j=1}^{n+1}2^{n+1-j}=2^{n+1}-1.$$ Alternatively consider the telescoping sum: $$\\sum_{k=0}^n 2^{k}=\\sum_{k=0}^n (2^{k+1}-2^k)=2^{n+1}-1.$$\n\nAnother famous approach is to count the $$2$$-player games to determine a winner among $$2^{n+1}$$ people. It's $$2^{n+1}-1$$, the number of people to eliminate. It's also $$2^n$$ people eliminated in the first round, halving each round until we reach the final.\n\n\u2022 Very beautiful and efficient proof ! \u2013\u00a0projetmbc Oct 23 '20 at 9:57\n\nAnother pictorial proof, similar to the box packing reply, is to count the number of nodes in a complete binary tree with n+1 levels.\n\n$$\\sum_{i = 0}^{n} 2^i$$\n\nis the number of nodes in the complete binary tree for the levels 0 (the root) through n.\n\nIf you draw an example and number the nodes like this\n\n       1\n/   \\\n2     3\n/ \\   / \\\n4   5 6   7\n/ \\\n8   ...\n\n\netc. you will notice that the left-most node of each level is a power of 2 (1, 2, 4, 8, 16, ...).\n\nso the number of nodes in the tree from root to level n is 2^{n+1} - 1.\n\n$$\\sum_{i = 0}^{n} 2^i = 2^{n+1} - 1 \\,.$$\n\nAnother geometric interpretation.\n\nStart with a line segment, AB. Extend this into a line. Use compassto mark off a point C by setting center at B and passing radius through A. Then lengths, AB=BC. Create point D in a similar way. We have thus constructed a length double that of the original line segment.\n\nThis doubling can continue AS MUCH S you like.\n\nNow ignore the first line segment, AB.\n\nThe series of segments now starts with BD which has double the length of AB.\n\nIn general the nth segment in the first case is the n-1th segment upon removal of AB. Further, each is double the length of its corresponding section.\n\nSo By subtracting a part we've doubled what remains, but with fewer sections to account for.\n\nThis gives the formula.\n\nNot sure if my answer has been posted yet, but here's how I understand it.\n\nYou are trying to understand why\n\n$$2^{n+1} - 1 = 2^n + 2^{n-1} + 2^{n-2} . . . + 2^1 + 2^0$$\n\nSuppose we take 2^n in the sum. We know since these are powers of two, that the previous term will be half of 2^n, and the term before that a quarter of 2^n.\n\nLet n in 2^n be 1, or 2^1 = 2. The term before in the sum will be half of 2, so we can also write the entire sum as:\n\n$$2^1 + \\frac{1}{2}(2^1)$$\n\nIf you do this but for different values of n for 2^n you will find you can rewrite the sums as:\n\n$$2^n + \\frac{ 2^n - 1}{2^n} ( 2^n)$$\n\nYou can simplify this because you're dividing (2^n) - 1 by 2^n, and then multiplying it by 2^n which cancel each other out. The new simplified version with the cancellation will look like:\n\n$$2^n + 2^n - 1$$\n\nThis can also be written as: $$2( 2^n) - 1$$\n\nFurther simplified into: $$2^{n + 1} - 1$$\n\nAnd that is why that formula shows up! Hope this helps ( I'm not sure if this was the mathematical induction proof you already knew, I'm young and discovered this on my own so I don't know what this is).", "date": "2021-04-20 18:39:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1990137/the-idea-behind-the-sum-of-powers-of-2/1990146", "openwebmath_score": 0.7393096685409546, "openwebmath_perplexity": 413.88806344814725, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429147241161, "lm_q2_score": 0.8670357615200475, "lm_q1q2_score": 0.8540674336977512}}
{"url": "https://brilliant.org/discussions/thread/elementary-techniques-used-in-the-imo-internatio-V/", "text": "\u00d7\n\n# Elementary Techniques used in the IMO (International Mathematical Olympiad) - Cauchy Schwarz Inequality\n\nIf people asked you, what is the most elementary inequality you know? I bet your answer would be AM-GM. But in this series of posts I will try to show you the power that Cauchy Schwarz has over that of AM-GM. But as an introduction, let us first state and prove the theorem.\n\nCauchy Schwarz Inequality: Let $$(a_1, a_2, \\ldots , a_n)$$ and $$(b_1, b_2, \\ldots, b_n)$$ be two sequences of real numbers, then we have:\n\n$$(a_1^2 + a_2^2 + \\ldots + a_n^2)(b_1^2 + b_2^2 + \\ldots + b_n^2) \\geq (a_1b_1 + a_2b_2 + \\ldots + a_nb_n)^2$$\n\nIn particular, equality holds iff there exists $$k \\in \\mathbb{R}$$ for which $$a_i = k b_i$$ for $$i = {1, \\ldots, n}$$.\n\nProof: We will present $$2$$ proofs, one originating from analysis on the equality case, the other by wishful thinking on small cases of $$n = 2,3$$.\n\n(i) Consider defining the following function $$f$$:\n\n$$f(x) = (a_1x - b_1)^2 + (a_2x - b_2)^2 + \\ldots + (a_nx - b_n)^2$$.\n\nWe will expand this to get:\n\n$$f(x) = (a_1^2 + a_2^2 + \\ldots + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + \\ldots a_nb_n)x + (b_1^2 + b_2^2 + \\ldots + b_n^2)$$\n\nFrom our first way of representing $$f(x)$$, we can conclude that $$f(j) \u2265 0 \\leftrightarrow \u2206_f \\leq 0$$ or\n\n$$(a_1^2 + a_2^2 + \\ldots + a_n^2)(b_1^2 + b_2^2 + \\ldots + b_n^2) \\geq (a_1b_1 + a_2b_2 + \\ldots + a_nb_n)^2$$\n\nEquality holds if the equation $$f(x) = 0$$ has one root.\u25a0\n\n(ii) Just remark that:\n\n$$(a_1^2 + a_2^2 + \\ldots + a_n^2)(b_1^2 + b_2^2 + \\ldots + b_n^2) - (a_1b_1 + a_2b_2 + \\ldots + a_nb_n)^2 = \\sum_{i,j=1}^n (a_ib_j + a_jb_i)^2 \u2265 0$$\u25a0\n\nLet us note the following positive things regarding Cauchy-Schwarz:\n\n\u2022 it is effective in proving symmetric inequalities\n\n\u2022 Try to form squares\n\n\u2022 Helps to clear up square roots\n\nLook forward to the next few posts to see applications of this extremely elegant inequality!\n\nNote by Anqi Li\n3\u00a0years, 3\u00a0months ago\n\nSort by:\n\nHi everyone, I'm part of #TorqueGroup too. My assignment was to post olympiad level problems for the community that I found interesting. Since Anqi is posting these great topics about math things useful in such math, it seems like we're a perfect fit. So, here are some examples:\n\nShow that for real numbers $$a_1, a_2, \\cdots$$ and $$b_1, b_2, \\cdots$$\n\n$$\\displaystyle \\sum_{i = 1}^{k} \\frac {a_i^2}{b_i} \\geq \\frac{(\\sum_{i = 1}^k a_i)^2}{\\sum_{i = 1}^k b_i}$$\n\nIMO 1995\n\nLet $$a, b, c \\in \\mathbb{R}^+$$ such that $$abc = 1$$. Prove that\n\n$$\\frac{1}{a^3(b+c)} + \\frac{1}{b^3(a+c)} + \\frac{1}{c^3(a+b)} \\geq \\frac{3}{2}$$.\n\n..\n\nHint: It may help to make a symmetric substitution for $$a = \\alpha, b = \\beta, c = \\gamma$$ such that the condition $$\\alpha \\beta \\gamma = 1$$ still holds.\n\nUSAMO 2009\n\nFor $$n \\geq 2$$ let $$a_1, a_2, \\cdots, a_n$$ be positive reals such that\n\n$$(a_1 + a_2 + \\cdots + a_n)(\\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_n}) \\leq (n + \\frac{1}{2})^2$$\n\nProve that $$max (a_1, a_2, \\cdots, a_n) \\leq 4 min (a_1, a_2, \\cdots, a_n)$$.\n\n..\n\nHint: The LHS should scream cauchy. Choose which $$a_i$$ multiplies which $$b_k$$ to isolate a desired max and min, and then do some algebra. \u00b7 3\u00a0years, 3\u00a0months ago\n\nI will be happy to post my solutions if requested (as long as you try them first!) \u00b7 3\u00a0years, 3\u00a0months ago\n\nCould you add these examples to the Applications of Cauchy Schwarz Inequality Wiki page? Staff \u00b7 2\u00a0years, 5\u00a0months ago\n\nFor the second one, following your hint, let $$a = \\frac{1}{x}, b = \\frac{1}{y}, c = \\frac{1}{z}$$. The expression is then $$\\sum_{cyc} \\frac{x^3yz}{y+z}$$. Since $$xyz = 1$$, this is $$\\sum_{cyc} \\frac{x^2}{y+z}$$. Using Cauchy, this is greater than $$\\frac{(x+y+z)^2}{2(x+y+z)}$$. So now it is left to prove that $$x+y+z \\geq 3$$. This is simple using AM-GM, so $$\\frac{x+y+z}{3} \\geq \\sqrt[3]{xyz}$$, and we are done. $$\\blacksquare$$\n\nFor the second one, WLOG assume $$a_1$$ is the maximum and $$a_n$$ is the minimum. Using Cauchy,\n\n$$(a_1 + a_2 + \\cdots + a_n)(\\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_n}) \\geq (\\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}} + n - 2)$$\n\n$$(n + \\frac{1}{2})^2 \\geq (\\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}} + n - 2)^2$$\n\nTaking square roots of both sides, we get\n\n$$n + \\frac{1}{2} \\geq \\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}} + n - 2$$\n\n$$\\frac{5}{2} \\geq \\sqrt{\\frac{a_1}{a_n}} + \\sqrt{\\frac{a_n}{a_1}}$$\n\nNow, squaring, we get\n\n$$\\frac{25}{4} \\geq \\frac{a_1}{a_n} + 2 + \\frac{a_n}{a_1}$$\n\nMultiplying both sides by $$a_1 a_n$$ we get\n\n$$a_1^2 - \\frac{17}{4} a_1 a_n + a_n^2 \\leq 0$$\n\nFactoring...\n\n$$(a_1 - 4a_n)(a_1 - \\frac{1}{4}a_n) \\leq 0$$\n\nSince $$a_1$$ and $$a_n$$ are positive reals, $$(a_1 - 4a_n)$$ must be nonpositive. Thus,\n\n$$(a_1 - 4a_n \\leq 0 \\rightarrow a_1 \\leq 4a_n$$, and we are done. \u00b7 3\u00a0years, 3\u00a0months ago\n\nCould you explain the second one as in how did you use Cauchy Schwartz in it. \u00b7 2\u00a0years, 1\u00a0month ago\n\nWhat are symmetric inequalities? \u00b7 2\u00a0years, 8\u00a0months ago\n\nHope you share many more such techniques!!thnx a lot for this. \u00b7 3\u00a0years, 3\u00a0months ago\n\nYup I will. In fact, I posted some on invariants and I also wrote some worked examples in my newest post. :) \u00b7 3\u00a0years, 3\u00a0months ago\n\n@Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page? Staff \u00b7 2\u00a0years, 5\u00a0months ago\n\nOne can easily prove Cauchy-Schwarz using vectors too. It is similar to the proof above, but is much cleaner. The interesting this is that almost every other inequality can be derived from this inequality. \u00b7 3\u00a0years, 3\u00a0months ago", "date": "2017-03-24 14:16:56", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/elementary-techniques-used-in-the-imo-internatio-V/", "openwebmath_score": 0.9441800713539124, "openwebmath_perplexity": 692.9916151993567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429151632047, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8540674323862745}}
{"url": "https://math.stackexchange.com/questions/1458687/calculate-limit-without-lhopitals-rule", "text": "# Calculate limit without L'Hopital's rule\n\nI have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:\n\n$$\\lim_{ x\\to \\pi/4 } \\frac{1 - \\tan(x)}{x-\\frac{\\pi}{4}}$$\n\nI know how to make it using L'Hopital and that the result is $-2$ ,but I'm getting nowhere when I try without it. Any advice?\n\nHint:\n\nLet $t=x-\\frac{\\pi}{4}$, then $$\\frac{1-\\tan x}{x-\\frac{\\pi}{4}}=\\frac{1-\\tan\\left(t+\\frac{\\pi}{4}\\right)}{t}=\\frac{1}{t}\\cdot\\left(1-\\frac{\\tan t+\\tan\\frac{\\pi}{4}}{1-\\tan \\frac{\\pi}{4}\\tan t}\\right)=\\frac{1}{t}\\left(1-\\frac{\\tan t+1}{1-\\tan t}\\right)=\\frac{-2\\tan t}{t(1-\\tan t)}$$ Now, take the limit as $t\\to 0$: \\begin{align} \\lim_{x\\to \\frac{\\pi}{4}}\\frac{1-\\tan x}{x-\\frac{\\pi}{4}}&=\\lim_{t\\to 0}\\frac{-2\\tan t}{t(1-\\tan t)}\\\\ &=\\lim_{t\\to 0}\\frac{-2\\tan t\\cos t}{t(1-\\tan t)\\cos t}\\\\ &=-2\\lim_{t\\to 0}\\frac{\\sin t}{t(\\cos t-\\sin t)}\\\\ &=-2\\left(\\lim_{t\\to 0}\\frac{\\sin t}{t}\\right)\\left(\\lim_{t\\to 0}\\frac{1}{\\cos t-\\sin t}\\right)\\\\ &=-2(1)(1)\\\\ &=\\color{blue}{-2} \\end{align}\n\n\u2022 While all the answers are valid and helped me,I chose this because it doesn't imply the use of more advanced concepts like derivatives.Thanks! \u2013\u00a0Der Rosenkavalier Oct 4 '15 at 20:48\n\nHint: What is the definition of the derivative of $\\tan(x)$ at $x=\\pi/4$?\n\n\u2022 $\\frac{1}{\\cos^2(\\frac{\\pi}{4})}$ But what's the point? I can't use L'Hopital's rule because the exercise ask it, not because I don't how to use it.I don't understand how can I use the derivative of $\\tan(x)$ at $\\frac{\\pi}{4}$ in this exercise. \u2013\u00a0Der Rosenkavalier Sep 30 '15 at 21:47\n\u2022 @DerRosenkavalier By definition: $$\\tan'(\\pi/4)=\\displaystyle\\lim_{x \\to \\pi/4}\\frac{\\tan(x)-1}{x-\\pi/4}$$. How does this compare to your limit? \u2013\u00a0Reveillark Sep 30 '15 at 21:49\n\u2022 L'Hopital isn't being used here; just the definition of the derivative. It's important to know the difference. \u2013\u00a0zhw. Sep 30 '15 at 22:00\n\u2022 @zhw I agree with your comment. This type of question sometimes rasises debate as to \"allowed ways forward. For example, is the use of asymptotic analysis permitted here? IMHO, yes absolutely. Yet others would argue that that approach is tantamount to the use of LHR. ;-) \u2013\u00a0Mark Viola Sep 30 '15 at 22:21\n\u2022 @DerRosenkavalier : My answer posted here explicitly explains how you can use the value of the derivative of $\\tan x$ at $\\pi/4$. ${}\\qquad{}$ \u2013\u00a0Michael Hardy Oct 1 '15 at 4:32\n\nRecall that $$f'(a) = \\lim_{x\\to a}\\frac{f(x) - f(a)}{x-a}.$$ Apply it to the case where $f(x) = \\tan x$ and $a=\\pi/4$. Then $f(a) = 1$ and $f'(a) = \\sec^2 a = \\sec^2(\\pi/4) = 2$. Therefore $$2 = \\lim_{x\\to\\pi/4}\\frac{\\tan x - \\tan(\\pi/4)}{x-\\pi/4}.$$\n\nSo $-2$ is the answer to the question as you've posed it.\n\nWe have the identities\n\n\\begin{align}\\frac{1-\\tan x}{x-\\pi/4}&=-\\frac{\\tan(x-\\pi/4)}{x-\\pi/4}(1+\\tan x)\\\\\\\\ &=-\\frac{\\sin(x-\\pi/4)}{x-\\pi/4}\\frac{1+\\tan x}{\\cos(x-\\pi/4)} \\end{align}\n\nCan you finish from here?\n\n\u2022 Yes,I could use that $\\lim_{ x\\to 0 }\\frac{\\sin(x)}{x}=1$ . Thanks for the help! \u2013\u00a0Der Rosenkavalier Oct 4 '15 at 20:58\n\u2022 You're welcome. My pleasure. This is a very efficient approach that circumvents use of L'Hospital's Rule, as you requested. \u2013\u00a0Mark Viola Oct 4 '15 at 21:07\n\nyou can use the following hint ,\n\ntry to write $\\tan(x)$ in terms of $\\sin(x)$ and $\\cos(x)$. A little bit of rearranging and then use the following\n\n$\\lim_{x\\rightarrow 0} \\frac{\\sin(x)}{x}=1$ A little bit of work yields the result .", "date": "2019-08-21 12:17:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1458687/calculate-limit-without-lhopitals-rule", "openwebmath_score": 0.8640660047531128, "openwebmath_perplexity": 814.130699555637, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429142850273, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8540674214717695}}
{"url": "https://math.stackexchange.com/questions/2782464/how-to-formalize-the-fact-that-fi-lceil-i-2-rceil-is-surjective-but-not-one/2782475", "text": "How to formalize the fact that $f(i)=\\lceil i/2\\rceil$ is surjective but not one-to-one from $\\Bbb N$ to itself?\n\nIf the set $S$ is countably infinite, prove or disprove that if $f$ maps $S$ onto $S$ (i.e. $f\\colon S\\to S$ is a surjective function), then $f$ is one-to-one mapping.\n\nPlease give a formal mathematical proof for this statement.\n\nI have a counter-example: Suppose the mapping is from $\\Bbb N$ (the natural numbers, a countably infinite set) to $\\Bbb N$. And $f(i) = \\lceil i/2\\rceil$. It is onto function but not one-one.\n\nBut I am not getting that how this mapping is onto. Intuitively, since $1$ and $2$ will be mapped to $1$; $3$ and $4$ will be mapped to $2$; $5$ and $6$ will be mapped to $3$; similarly, $n-1$ and $n$ will be mapped to $n/2$. So in the codomain, there will be some elements that have no pre-image for this countably infinite set. So, please correct me where I am wrong.\n\n\u2022 Welcome to MSE. Please use MathJax. \u2013\u00a0Jos\u00e9 Carlos Santos May 15 '18 at 16:35\n\u2022 \"So in the codomain, there will be some elements that have no pre-image for this countably infinite set.\" Why do you say that? You just seemed to give an argument that it would. Anyway for any $n$ then both $2n$ and $2n+1$ will map to $n$. So $f$ is onto. \u2013\u00a0fleablood May 15 '18 at 16:47\n\u2022 sir , since , cardinality of both sets are same and every 2 elements of domain are mapped to one element of codomain . so intuitively , it looks that some elements of codomain will not have any pre-image. please correct me \u2013\u00a0ankit1729 May 15 '18 at 16:51\n\u2022 You just should that $1,2,3$ and $\\frac n2$ (assuming $n$ is even) all have pre-images. What elements don't have pre-images? I honestly don't understand why you said that. \u2013\u00a0fleablood May 15 '18 at 16:51\n\u2022 Are you thinking because you will run out of elements? But the set is infinite. You will never run out. So it's perfectly fine to have 2, 3 or even an infinite number of elements all map to the same element and still be onto. In fact that is exactly what you are proving. It is possible to be surjective and not 1-1. \u2013\u00a0fleablood May 15 '18 at 16:54\n\n2 Answers\n\nTo prove that a function $f\\colon A\\to B$ is surjective, you need to be able to prove that given $b\\in B$, there is some $a\\in A$ such that $f(a)=b$.\n\nIn your case, $A=B=\\Bbb N$, and $f(x)=\\lceil \\frac x2\\rceil$.\n\nSo you need to prove that given any $m\\in\\Bbb N$, there is some $n\\in\\Bbb N$ such that $\\lceil\\frac n2\\rceil=m$. Even though this $n$ is not unique, I am sure that you can come up with a way to find such $n$, from a given $m$.\n\nMaybe I'm missing something. For every n in codomain, 2n and 2n-1 are in preimage, so the mapping is onto but not 1-1\n\n\u2022 sir , it may or may not be one-one \u2013\u00a0ankit1729 May 15 '18 at 17:11", "date": "2019-08-18 13:11:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2782464/how-to-formalize-the-fact-that-fi-lceil-i-2-rceil-is-surjective-but-not-one/2782475", "openwebmath_score": 0.881971001625061, "openwebmath_perplexity": 159.50013819036678, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429156022934, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.854067420921704}}
{"url": "https://physics.stackexchange.com/questions/304634/what-is-the-magnitude-of-the-difference-vector", "text": "# what is the magnitude of the difference vector?\n\nI read this question in Sears and Zemansky's University Physics book:\n\n\"Two displacement vectors, S and T, have magnitudes S = 3 m and T = 4 m. Which of the following could be the magnitude of the difference vector S - T ?(There may be more than one answer.) i. 9 m; ii. 7 m; iii. 5 m; iv. 1 m; v. 0m; vi. -1m \"\n\nMy question is: shouldn't the direction of both vectors be specified in order for me to solve it? Or do I just assume that S and T are positive and negative respectively as per the equation S - T\n\n\u2022 The question is asking you what the possible values of $\\mathbf S - \\mathbf T$ could be for all possible directions. So for example if one of the options was $666$m you need consider if there is any possible arrangement of the two vectors that could give $|\\mathbf S - \\mathbf T| = 666$. This freedom to orient the vectors in any direction means more than one of the answers can be correct. \u2013\u00a0John Rennie Jan 12 '17 at 16:20\n\u2022 That's what I thought initially, but I had doubts. Thanks for confirming! \u2013\u00a0DigiNin Gravy Jan 12 '17 at 16:23\n\u2022 Re Answer (vi): can the magnitude of a vector be negative? \u2013\u00a0DJohnM Jan 12 '17 at 17:18\n\u2022 @DJohnM I know that the magnitude can never be negative, but the vector can. \u2013\u00a0DigiNin Gravy Jan 12 '17 at 18:20\n\u2022 @DigiNinGravy A vector which is anti-parallel to, say, the $x$-axis of a coordinate system is sometimes called \"a vector in the negative-$x$ direction.\" But it is an incorrect oversimplification to call such an object \"a negative vector.\" \u2013\u00a0rob Jan 13 '17 at 7:26\n\nThe magnitude of the difference vectors depends on the orientation of $$\\bf\\vec{S}$$ and $$\\bf \\vec{T}$$. If they are parallel then $$|\\bf \\vec{S}-\\bf \\vec{T}|=|\\,|\\bf \\vec{S}|-|\\bf \\vec{T}|\\,|$$ and if they are anti-parallel then $$|\\bf \\vec{S}-\\bf \\vec{T}|=|\\bf \\vec{S}|+|\\bf \\vec{T}|$$.\n\nThus the possible values of $$|\\bf \\vec{S}-\\bf \\vec{T}\\|$$ lie in the range:\n\n$$|\\,|\\bf \\vec{S}|-|\\bf \\vec{T}|\\,| \\leq |\\bf \\vec{S}-\\bf \\vec{T}| \\leq |\\bf \\vec{S}|+|\\bf \\vec{T}|$$\n\nI'll let you work out which answers comply with this.\n\n\u2022 |S - T| = 5 or | |S| - |T| | = -1 or |S| + |T| = 7. I just don't understand how |S - T| = |S| + |T|. I mean to have |S| + |T| both vectors must be parallel - having the same direction - so | |S| - |T| | seems off, unless we consider T as -T? \u2013\u00a0DigiNin Gravy Jan 12 '17 at 17:09\n\u2022 @DigiNinGravy I think you gave the aswer yourself both $\\bf T and$-\\bf T$have se same magnitude, and the magnitude does not give the direction of the vector. \u2013 Mikael Fremling Jan 12 '17 at 18:10 In above Figure move the end point$\\:\\mathrm{B}\\:$of the vector$\\:\\mathbf{S}\\:$around a circle of radius$\\:|\\mathbf{S }|=3\\:$. Try to find the length$\\:|\\mathbf{S}-\\mathbf{T}|\\:$of the vector$\\:\\mathbf{S}-\\mathbf{T}\\:$when$\\:\\mathrm{B}\\:$is on points$\\:\\mathrm{P_{ii}},\\mathrm{P_{iii}},\\mathrm{P_{iv}}\\:\\$.\n\n\u2022 Thanks, but you reversed the magnitudes of S and T. Can you tell me the name of the program you used? \u2013\u00a0DigiNin Gravy Jan 12 '17 at 18:17\n\u2022 @DigiNin Gravy : GeoGebra \u2013\u00a0Frobenius Jan 12 '17 at 18:35", "date": "2019-11-18 17:28:09", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/304634/what-is-the-magnitude-of-the-difference-vector", "openwebmath_score": 0.6770211458206177, "openwebmath_perplexity": 629.6503110458287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357573468175, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.8540592742621566}}
{"url": "https://math.stackexchange.com/questions/2400296/finding-n-permutations-r-with-repetitions/2400307", "text": "# Finding $n$ permutations $r$ with repetitions\n\nI have given $n$ items and I have to find ${}^nP_r$ and $\\sum_{r=0}^n{}^nP_r$ with repetitions allowed. Is there any closed formula for this?\n\nFor $n=3$ and $r=1$, possible permutations are:\n$\\{1\\},\\{2\\},\\{3\\}$\nTotal: 3\n\nFor $n=3$ and $r=2$, possible permutations are:\n$\\{1,1\\},\\{2,2\\},\\{3,3\\},\\{1,2\\},\\{2,1\\},\\{1,3\\},\\{3,1\\},\\{2,3\\},\\{3,2\\}$\nTotal: 9\n\nFor $n=3$ and $r=3$, possible permutations are:\n$\\{1,1,1\\},\\{2,2,2\\},\\{3,3,3\\},$\n$\\{1,1,2\\},\\{1,2,1\\},\\{2,1,1\\},$\n$\\{1,2,2\\},\\{2,1,2\\},\\{2,2,1\\},$\n$\\{1,1,3\\},\\{1,3,1\\},\\{3,1,1\\},$\n$\\{1,3,3\\},\\{3,1,3\\},\\{3,3,1\\},$\n$\\{2,2,3\\},\\{2,3,2\\},\\{3,2,2\\},$\n$\\{2,3,3\\},\\{3,2,3\\},\\{3,3,2\\}$,\n$\\{1,2,3\\},\\{1,3,2\\}$,\n$\\{2,1,3\\},\\{2,3,1\\}$,\n$\\{3,1,2\\},\\{3,2,1\\}$,\nTotal: 27\n\nCan we come up with any closed formula for individual ${}^nP_r$ with repetitions and also for their sum i.e. here $3+9+27=39$\n\nI understand that I cannot call this exactly the permutation, since ${}^3P_3$ is strictly $6$, while above its $27$, since I allow repetitions, but then whatever it is, how do I get the count?\n\nNote that permutations with repetition is usually the well known case corresponding to $\\frac{n!}{n_1!n_2!...n_i!}$, which is not what I am asking here. Is what am asking also some well know case, and I am stupidly not able to guess it? My primary guess is that, there cannot be any closed formula. Is it right?\n\n\u2022 What exactly is the condition you're looking for? I notice that the permutation $1,2,3$ is excluded - is this intentional? \u2013\u00a0platty Aug 20 '17 at 15:31\n\u2022 In fact, it looks like you left out all the permutations of $1,2,3$ - is this intentional? \u2013\u00a0platty Aug 20 '17 at 15:41\n\u2022 Nope, I missed it. Sorry. \u2013\u00a0anir Aug 20 '17 at 15:43\n\u2022 I dont know it feels fuzzy. Is it just ${}^nP_r \\text{with repetition} = n^r$?. Feels like so. Let me read question and answers again. \u2013\u00a0anir Aug 20 '17 at 15:47\n\u2022 Your formula is not for permutations with replacements. It's for combinations with replacements. Are you actually looking for the expression for multinominal coefficients? \u2013\u00a0Vim Aug 20 '17 at 15:48\n\nIf you just want unrestricted strings consisting of $r$ letters, chosen with replacement from $n$, you can just use the multiplication rule to get $n^r$; there's $n$ choices for the first one, $n$ choices for the second, etc.\n\nExtending this, we can use this to find the number of strings with length up to $r$ by summing the intermediate results: $$1+n+n^2+\\dots +n^r$$ This is the sum of a geometric series, which means we can apply the formula to get $\\frac{n^{r+1}-1}{n-1}$.\n\nIf you mean \"permutations with replacements\" then the answer is just $n^r$. But this doesn't match your result for $n=r=3$ which should be $27$? If you actually mean combinations with replacements, then see the below hint.\n\nHint: you are basically putting $r$ identical balls into $n$ different jars, or equivalently, finding the number of non-negative integer solutions to $$x_1+\\cdots+x_n=r$$ To solve this, first let $y_i:=x_i+1$ (which are bijections) then try finding the number of positive integer solutions to the equation $$y_1+\\cdots+y_n=r+n.$$ Try Stars & Bars technique.\n\n\u2022 I don't think this is correct - OP has order mattering in the listed examples. \u2013\u00a0platty Aug 20 '17 at 15:39\n\u2022 @platty this is really confusing, because in this case the third result should be 27. Clearly there is some misphrasing in the question. \u2013\u00a0Vim Aug 20 '17 at 15:41\nNote that in your last paragraph, you say that $\\frac{n!}{n_1!n_2! \\dots n_i!}$ is the formula for permutations with repetition, but perhaps a better way to word it would be that this is a generalized formula for partitioning a set of $n$ objects into $i$ cells with each \"cell\" being distinguishable, but elements within their respective cells are not.\nAs noted before, permutations with repetition allowed can be represented with $n^r$, which @platty already explained. This is the formula you would use to solve the second example I gave, as well as what you seem to be looking for in your example.\nthe number of words, from alphabet $\\{1,\\cdots,n\\}$, of length $r$ and max repetition $r$,\ni.e. simply the number of words, from alphabet $\\{1,\\cdots,n\\}$, of length $r$\nwhich are $n^r$.", "date": "2020-10-01 09:10:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2400296/finding-n-permutations-r-with-repetitions/2400307", "openwebmath_score": 0.8506721258163452, "openwebmath_perplexity": 220.7036113149986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9609517083920617, "lm_q2_score": 0.8887587993853654, "lm_q1q2_score": 0.8540542866178444}}
{"url": "http://mathhelpforum.com/number-theory/1862-2-5-8-11-14-a.html", "text": "Math Help - 2, 5, 8, 11, 14...\n\n1. 2, 5, 8, 11, 14...\n\nHow can I prove that no square number can be in the sequence?\n\n2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...\n\nIs it to do with mod 5? And if so how would I do it\n\n2. Originally Posted by Natasha\nHow can I prove that no square number can be in the sequence?\n\n2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...\n\nIs it to do with mod 5? And if so how would I do it\nFirst you need find the general term of your sequence. Now it looks to me\nas though the sequence is:\n\n$a_n=2+3n, n=0,1,2,..$\n\nWhich suggests you consider the squares modulo 3. Now any natural number\nmay be written:\n\n$N=3k+\\rho$,\n\nfor some natural number $k$, and $\\rho= 0, 1\\ \\mbox{or}\\ 2$. So\n\n$N^2=9k^2+6k \\rho + \\rho^2$.\n\nTherefore $N^2(mod\\ 3)$ is $\\rho^2(mod\\ 3)$, but\n\n$\n\\rho^2(mod\\ 3)=0, 1, \\mbox{or}\\ 1(mod\\ 3) \\mbox{as}\\ \\rho=0,1 \\mbox{or}\\ 2\n$\n.\n\nSo as $a_n, n=0,1,2,..$ is congruent to $2(mod\\ 3)$ it cannot be a square for any\nnatural number $n$\n\nRonL\n\n3. Originally Posted by CaptainBlack\nFirst you need find the general term of your sequence. Now it looks to me\nas though the sequence is:\n\n$a_n=2+3n, n=0,1,2,..$\n\nWhich suggests you consider the squares modulo 3. Now any natural number\nmay be written:\n\n$N=3k+\\rho$,\n\nfor some natural number $k$, and $\\rho= 0, 1\\ \\mbox{or}\\ 2$. So\n\n$N^2=9k^2+6k \\rho + \\rho^2$.\n\nTherefore $N^2(mod\\ 3)$ is $\\rho^2(mod\\ 3)$, but\n\n$\n\\rho^2(mod\\ 3)=0, 1, \\mbox{or}\\ 1(mod\\ 3) \\mbox{as}\\ \\rho=0,1 \\mbox{or}\\ 2\n$\n.\n\nSo as $a_n, n=1,2,..$ is congruent to $2(mod\\ 3)$ it cannot be a square for any\nnatural number $n$\n\nRonL\nCaptain is it the same to say that the general form 3n - 1 is the same as 3n + 2??\n\n4. Originally Posted by Natasha\nCaptain is it the same to say that the general form 3n - 1 is the same as 3n + 2??\nYes, but the +2 form is slightly more convienient when dealing with modular\narithmetic.\n\nRonL\n\n5. Originally Posted by CaptainBlack\nYes, but the +2 form is slightly more convienient when dealing with modular\narithmetic.\n\nRonL\nI see. Thanks!", "date": "2014-03-17 02:57:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/1862-2-5-8-11-14-a.html", "openwebmath_score": 0.6212263703346252, "openwebmath_perplexity": 100.95640804769035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815516660637, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8540510594034311}}
{"url": "https://www.shaalaa.com/question-bank-solutions/find-ratio-which-point-p-x-2-divides-line-segment-joining-points-a-12-5-b-4-3-also-find-value-x-section-formula_5000", "text": "# Find the Ratio in Which the Point P(X, 2) Divides the Line Segment Joining the Points A(12, 5) and B(4, \u22123). Also, Find the Value of X. - Mathematics\n\nFind the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, \u22123). Also, find the value of\u00a0x.\n\n#### Solution 1\n\nLet the point P (x, 2) divide the line segment joining the points A (12, 5) and B (4, \u22123) in the ratio k:1.\n\nThen, the coordinates of P are\u00a0((4k+12)/(k+1),(-3k+5)/(k+1))\n\nNow, the coordinates of P are (x, 2).\n\ntherefore\u00a0(4k+12)/(k+1)=x and\u00a0(-3k+5)/(k+1)=2\n\n(-3k+5)/(k+1)=2\n\n-3k+5=2k+2\n\n5k=3\n\nk=3/5\n\nSubstituting\u00a0k=3/5 \" in\" \u00a0(4k+12)/(k+1)=x\n\nwe get\n\nx=(4xx3/5+12)/(3/5+1)\n\nx=(12+60)/(3+5)\n\nx=72/8\n\nx=9\n\nThus, the value of x is 9.\n\nAlso, the point P divides the line segment joining the points A(12, 5) and (4, \u22123) in the ratio 3/5:1\u00a0 i.e. 3:5.\n\n#### Solution 2\n\nLet k be the ratio in which the point P(x,2)\u00a0 divides the line joining the points\n\nA(x_1 =12, y_1=5) and B(x_2 = 4, y_2 = -3 ) . Then\n\nx= (kxx4+12)/(k+1) and 2 = (kxx (-3)+5) /(k+1)\n\nNow,\n\n 2 = (kxx (-3)+5)/(k+1) \u21d2 2k+2 = -3k +5 \u21d2 k=3/5\n\nHence, the required ratio is3:5 .\n\nConcept: Section Formula\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nRD Sharma Class 10 Maths\nChapter 6 Co-Ordinate Geometry\nExercise 6.3 | Q 20 | Page 29\nRS Aggarwal Secondary School Class 10 Maths\nChapter 16 Coordinate Geomentry\nQ 5", "date": "2021-12-01 05:58:05", "meta": {"domain": "shaalaa.com", "url": "https://www.shaalaa.com/question-bank-solutions/find-ratio-which-point-p-x-2-divides-line-segment-joining-points-a-12-5-b-4-3-also-find-value-x-section-formula_5000", "openwebmath_score": 0.5793002843856812, "openwebmath_perplexity": 2682.2805815229976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815513471367, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8540510591280722}}
{"url": "https://math.stackexchange.com/questions/1574614/how-to-evaluate-this-definite-double-integral", "text": "# How to evaluate this definite double integral?\n\n$$\\int\\int_D 6x\\sqrt{y^2-x^2}dA, D=\\{(x,y)|0\\leq y\\leq 2, 0 \\leq x \\leq y\\}$$\n\nI tried: $$\\int_0^2 \\int_0^y 6x\\sqrt{y^2-x^2}dxdy$$\n\nBut that is incorrect.\n\n\u2022 Why do you say that's incorrect? Looks fine to me... \u2013\u00a0Eli Berkowitz Dec 14 '15 at 3:32\n\u2022 Your integral looks perfect! Now just evaluate it, try a u-substitution for the inside integral. \u2013\u00a0Alex Dec 14 '15 at 3:37\n\u2022 I got -8, I must have made an arithmetic error or something! \u2013\u00a0d0rmLife Dec 14 '15 at 3:39\n\nSetting $y^2-x^2 = t$, we obtain $-2xdx = dt$. Hence, we obtain $$I = \\int_0^2 \\int_0^y 6x\\sqrt{y^2-x^2}dxdy = \\int_0^2 \\int_{y^2}^0(-3\\sqrt{t}dt)dy = \\int_0^2 \\int_0^{y^2} 3\\sqrt{t}dtdy = \\int_0^22y^3 dy = 8$$\n\u2022 Why does the range of the inner integral change to $y^2,0$ instead of $y^2,y^2-y$ - I thought you put the original range of integration into the u substitution equation? \u2013\u00a0d0rmLife Dec 14 '15 at 3:47\n\u2022 @d0rmLife because the first inner integral goes from $x=0$ to $x=y$, and when replacing $y^2-x^2=t$ in the first case you get $y^2-0^2=y^2=t$ and in the second $y^2-y^2=0=t$. When you u-substitute you must change the range accordingly or else it is \"wrong\": you can get away with it if you evaluate the result in terms of x instead of u in the end, but still it is best to change it. \u2013\u00a0Fede Poncio Dec 14 '15 at 4:13\nNotice, $$\\int_{0}^2\\int_0^y 6x\\sqrt{y^2-x^2} \\ dxdy=\\int_{0}^2\\left(\\int_0^y 6x\\sqrt{y^2-x^2} \\ dx\\right)dy$$ $$=\\int_{0}^2\\left(-3\\int_0^y (y^2-x^2)^{1/2} \\ d(y^2-x^2)\\right)dy$$ $$=\\int_{0}^2dy\\left(-3\\frac{2(y^2-x^2)^{3/2}}{3}\\right)_{0}^{y}$$ $$=\\int_{0}^2\\left(2y^3\\right)\\ dy$$ $$=\\left(2\\frac{y^4}{4}\\right)_{0}^{2}=\\left(\\frac{2^4}{2}-0\\right)=\\color{red}{8}$$", "date": "2020-01-26 18:00:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1574614/how-to-evaluate-this-definite-double-integral", "openwebmath_score": 0.8759689927101135, "openwebmath_perplexity": 347.58806181456106, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815535796246, "lm_q2_score": 0.8633916029436189, "lm_q1q2_score": 0.8540510471473713}}
{"url": "http://math.stackexchange.com/questions/885137/recognizing-the-sequence-1-16-1-8-3-16-1-4-5-16", "text": "# Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, \u2026\n\nWhat is the missing number? $$\\frac{1}{16}, \\frac{1}{8}, \\frac{3}{16}, \\frac{1}{4}, \\frac{5}{16}, \\ \\ \\ [?]$$\n\n$$A. \\frac{5}{4}\\quad B. \\frac{3}{4}\\quad C. \\frac{5}{8}\\quad D. \\frac{3}{8}$$\n\nSpoiler: Answer is $D$, but I don't know why.\n\nThanks\n\n-\nOMG, of course. Thank you for the heads up. \u2013\u00a0 user1495024 Aug 2 at 1:22\nAm I the only one who thought $\\frac{1}{2}$ before reading the possible options? :) \u2013\u00a0 Thomas Aug 2 at 4:15\nNo, the next two numbers should definitely be 1/2 and 7/16 :-) \u2013\u00a0 CompuChip Aug 2 at 8:23\n@Thomas I did consider it, but then thought it wasn't just right. First of all, I'd be generalizing from just two data points. Secondly I'd be ignoring the other three data points, which supposedly were there for a reason. \u2013\u00a0 kasperd Aug 2 at 9:37\n\n$$\\frac{1}{16}, \\frac{1}{8}=\\frac{2}{16}, \\frac{3}{16}, \\frac{1}{4}=\\frac{4}{16}, \\frac{5}{16}$$\n\nSo the $i$th term is of the form $$\\frac{i}{16}$$ Therefore, the next term is $$\\frac{6}{16}=\\frac{3}{8}$$\n\n-\n\n$$\\frac{1}{16}, \\frac{1}{8}, \\frac{3}{16}, \\frac{1}{4}, \\frac{5}{16}$$ The above is the same as $\\displaystyle\\frac1{16},\\frac2{16},\\frac3{16},\\frac4{16},\\frac5{16}$.\n\n-\n\nAnother sequence-recognizing technique is to look at the difference between consecutive terms.\n\nIn this case, $\\frac{1}{8}-\\frac{1}{16} = \\frac{1}{16}$, $\\frac{3}{16}-\\frac{1}{8} = \\frac{1}{16}$, $\\frac{1}{4}-\\frac{3}{16} = \\frac{1}{16}$, and $\\frac{5}{16}-\\frac{1}{4} = \\frac{1}{16}$.\n\nSince the difference between consecutive terms is $\\frac{1}{16}$, the next term should be $\\frac{5}{16}+\\frac{1}{16} =\\frac{6}{16} =\\frac{3}{8}$.\n\n-", "date": "2014-10-23 02:50:43", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/885137/recognizing-the-sequence-1-16-1-8-3-16-1-4-5-16", "openwebmath_score": 0.8748292326927185, "openwebmath_perplexity": 680.1017385572425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886150275788, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.854033021944384}}
{"url": "http://mathhelpforum.com/discrete-math/139817-relations-help-2-a.html", "text": "1. ## relations help (2)\n\nhey guys, question:\n\nDetermine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.\n\nxRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks\n\n2. Originally Posted by jvignacio\nhey guys, question:\n\nDetermine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.\n\nxRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks\nThe integer part of x is the integer n such that $n\\le x < n+1$. (There is exactly one such integer for each real x.) It is usually denoted by [x].\n\nFor example:\n[3.14]=3\n[0.52]=0\n[1]=1\n[-1.26]=-2\n\nSo 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.\n\nHope this helps.\n\n3. Originally Posted by kompik\nthe integer part of x is the integer n such that $n\\le x < n+1$. (there is exactly one such integer for each real x.) it is usually denoted by [x].\n\nFor example:\n[3.14]=3\n[0.52]=0\n[1]=1\n[-1.26]=-2\n\nso 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.\n\nHope this helps.\nIs that like saying the next full integer down?\n\n4. Originally Posted by jvignacio\nIs that like saying the next full integer down?\nExactly.\n\n5. Originally Posted by kompik\nExactly.\nSo in this case,\n\n- It IS reflexive since x has the same integer part as x.\n- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.\n- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.\n\ncorrect?\n\n6. Originally Posted by jvignacio\nSo in this case,\n\n- It IS reflexive since x has the same integer part as x.\n- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.\n- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.\n\ncorrect?\nYes, it is an equivalence relation.\n\nAnother way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)\n\n7. Originally Posted by kompik\nYes, it is an equivalence relation.\n\nAnother way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)\nYeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?\n\n8. Originally Posted by jvignacio\nYeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?\nNo.\nTake, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.\nThis is equivalent to\n[x]=[3.14]=3.\nAnd what are the numbers such that [x]=3? Precisely the numbers from the interval $\\langle 3,4)$, right?\nCan you find the remaining classes?\n\n9. Originally Posted by kompik\nNo.\nTake, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.\nThis is equivalent to\n[x]=[3.14]=3.\nAnd what are the numbers such that [x]=3? Precisely the numbers from the interval $\\lange 3,4)$, right?\nCan you find the remaining classes?\nSorry whats in the interval for [x]=3? theres a latex error..\n\n10. Originally Posted by jvignacio\nSorry whats in the interval for [x]=3? theres a latex error..\nSorry, did not notice that. I've edited my post.\n\n11. Originally Posted by kompik\nNo.\nTake, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.\nThis is equivalent to\n[x]=[3.14]=3.\nAnd what are the numbers such that [x]=3? Precisely the numbers from the interval $\\langle 3,4)$, right?\nCan you find the remaining classes?\nOk I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...\n\n12. Originally Posted by jvignacio\nOk I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...\nIf I were your teacher, I would expect answer to be something like:\nThe equivalent classes are intervals .... (fill in the dots) for $n\\in\\mathbb{N}$.\n(I guess the example with 3.14 might help you to see what to fill in.)\n\n13. Originally Posted by kompik\nIf I were your teacher, I would expect answer to be something like:\nThe equivalent classes are intervals .... (fill in the dots) for $n\\in\\mathbb{N}$.\n(I guess the example with 3.14 might help you to see what to fill in.)\nintervals $\\langle n,n+1)$ , $n \\in \\mathbb{N}$ ?\n\n14. Originally Posted by jvignacio\nintervals $\\langle n,n+1)$ , $n \\in \\mathbb{N}$ ?\nExactly.\n\n15. Originally Posted by kompik\nExactly.\nmate thanks alot for the help. Really appreciate your time.\n\nPage 1 of 2 12 Last", "date": "2017-04-28 04:04:27", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/139817-relations-help-2-a.html", "openwebmath_score": 0.8366315364837646, "openwebmath_perplexity": 698.8470243528784, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9912886173437992, "lm_q2_score": 0.8615382076534743, "lm_q1q2_score": 0.8540330186536675}}
{"url": "https://math.stackexchange.com/questions/1154703/use-proof-by-induction-to-prove-frac1n-frac12n-1-for-all-n-geq", "text": "# Use proof by induction to prove $\\frac{1}{n!}<\\frac{1}{2^n-1}$ for all $n\\geq 4$\n\nUse proof by induction to prove that that $\\frac{1}{n!}<\\frac{1}{2^n-1}$ for all $n\\geq 4$, .\\Base case: $$\\frac{1}{4}=\\frac{1}{24}\\leq \\frac{1}{2^4-1}$$ Inductive hypothesis: Assume there exists $k\\in \\mathbb{N}$ s.t.\n\n$$\\frac{1}{k!}\\leq\\frac{1}{2^k-1}$$ Inductive step: Show that:$$\\frac{1}{(k+1)!}\\leq\\frac{1}{2^{k+1}-1}$$ Now, $$\\frac{1}{(k+1)!}=\\frac{1}{k!}\\cdot\\frac{1}{k+1}$$ Using the hypothesis $$\\frac{1}{(k+1)!}\\leq\\frac{1}{2^k-1}\\cdot \\frac{1}{k+1}$$ Because $n\\geq4, \\frac{1}{k+1}<\\frac{1}{2}$ $$\\frac{1}{(k+1)!}\\leq\\frac{1}{2^k-1}\\cdot \\frac{1}{2}=\\frac{1}{2^{k+1}-2}\\leq\\frac{1}{2^{k+1}-1}$$ Hence by mathematical induction we have proved that $\\frac{1}{n!}<\\frac{1}{2^n-1}$ for all $n\\geq 4$\n\nFirstly I need to know if the proof is correct, secondly it has to be as concise as possible hence I would like to know if there are any lines I can change/delete\n\nAnd lastly can anyone explain to me why every sentence starts with \"\\\" It looks perfectly fine in www.sharelatex.com ;(\n\n\u2022 You must start with $n=4$. \u2013\u00a0User3101 Feb 18 '15 at 16:11\n\u2022 I see and I am lost know because it is a question copied from the coursework given by a professor, I just blindly followed so lets assume $n\\geq4$ \u2013\u00a0Scavenger23 Feb 18 '15 at 16:12\n\u2022 So I should say Assume that $\\frac{1}{k!}\\leq\\frac{1}{2^k-1}$ holds for some $k\\geq4$ and then proceed ? \u2013\u00a0Scavenger23 Feb 18 '15 at 16:21\n\u2022 One error I notice in your logic - in the last statement you state that $\\frac 1{2^{k+1}-2} \\leq \\frac 1{2^{k+1}-1}$. This is incorrect. The latter (-1) fraction has a larger denominator than the -2 fraction, which means it's a smaller fraction. \u2013\u00a0Duncan Feb 18 '15 at 16:34\n\nAll in all it looks like you have the idea down on how induction works. It looks like you are correctly making the induction step to get your result. I have a couple things to point out, however.\n\n$(1)$ During your induction step you introduce the \"$\\geq$\" symbol, when initially you are trying to write a proof about a strict inequality (which should only use the \"$>$\" symbol). If the strict inequality holds, it is not incorrect to use \"$\\geq$\", but for the sake of consistency you should just use one.\n\n$(2)$ Your induction hypothesis is not stated quite right. It is not enough to assume there exists $k \\in \\Bbb{N}$ such that the inequality holds. You want to make the stronger assumption that you can find this $k$, and the inequality holds for all $n \\leq k$. It is in doing this that you will be allowed to conclude at the end of the proof that the inequality holds for all $n \\geq 4$, instead of just $n=4$, and one arbitrary $k$.\n\n$(3)$. If you are looking for a more concise proof, I would instead prove the equivalent statement that $2^n-1<n!$ for all $n \\geq 4$. It is clear for $n \\geq 4$ that $$2^n-1<2^n = 2 \\cdot 2\\cdot 2\\cdot 2\\cdot \\ldots < 1 \\cdot 2\\cdot 3 \\cdot 4 \\cdot \\ldots = n!$$ This is easier than dealing with fractions. But, upon proving this result one need merely flip the inequality around and put the quantities on each side under a numerator of $1$.\n\n\u2022 Never mind (4). in the preview mode every sentence had \"\\\" in front of it while in here as you see Everything is fine. I see the mistake $< and \\leq$ thank you for pointing it out. Thank you. I started induction only 2 weeks ago and it is a difficult topic for me. \u2013\u00a0Scavenger23 Feb 18 '15 at 16:37\n\u2022 You see I didn't have the \"flipping the fractions\" tool in the toolbox so I didn't even think about what you have done in (3). The grammar unfortunately will remain an issue, since English is not my native language. \u2013\u00a0Scavenger23 Feb 18 '15 at 16:42\n\u2022 You know what now I realised the whole mistake. I can't copy correctly. The question is not asking to prove $\\frac{1}{n!}<\\frac{1}{2^n-1}$ but $\\frac{1}{n!}<\\frac{1}{2^{n-1}}$ \u2013\u00a0Scavenger23 Feb 18 '15 at 16:50\n\u2022 I don't want to use you but if I finish the right question this time, would you be able to have a one last look on it? I will stop spamming now since the website tells me to avoid extended discussions in comments. \u2013\u00a0Scavenger23 Feb 18 '15 at 16:54\n\u2022 Use proof by induction to prove that that $\\frac{1}{n!}<\\frac{1}{2^n-1}$ for all $n\\geq 3$\\\\I will use the fact that $k!>2^{k-1}$ is an equivelant statement to $\\frac{1}{k!}<\\frac{1}{2^k-1}$ .\\\\Base case $3!>2^{2}$ Inductive step: Assume $k!>2^{k-1}$ holds for all $n\\leq k$ for some $k \\in \\bbb{N}$\\\\\\ Inductive hypothesis\\\\Show that $(k+1)!>2^k$\\\\It is clear that when $k\\geq 3$\\\\$2^{k-1}<2^k=2*2*2*2...<1*2*3*4...=k!<(k+1)!$ \u2013\u00a0Scavenger23 Feb 18 '15 at 17:18\n\nIt's much easier proving that $2^n<1+n!$, for $n\\ge4$, which is completely equivalent to your assignment. The base step is obvious. Suppose it holds for $n$; then $$2^{n+1}=2\\cdot2^n<2\\cdot(1+n!)=2+2\\cdot n!<1+(n-1)\\cdot n!+2\\cdot n!=1+(n+1)!$$ because $(n-1)n!>1$.\n\nYou can, if you want, transform this into a proof of your assigned inequality, but it's not necessary.\n\nRewrite as$$n!\\ge2^n.$$ Then $$4!\\ge2^4$$ and $$n!\\ge2^n\\land n+1\\ge2\\implies(n+1)!\\ge2^{n+1}.$$", "date": "2019-09-23 13:00:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1154703/use-proof-by-induction-to-prove-frac1n-frac12n-1-for-all-n-geq", "openwebmath_score": 0.8213667869567871, "openwebmath_perplexity": 258.29928123296617, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407206952993, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8539961332463559}}
{"url": "http://math.stackexchange.com/questions/258273/how-do-i-prove-that-lim-x%e2%86%920-x%e2%8b%85-ln-x-0", "text": "# How do I prove that $\\lim_{x\u21920} x\u22c5\\ln x=0$\n\nHow do I prove (without L'H\u00f4pital's rule) that $$\\lim_{x\u21920} x\u22c5\\ln x = 0$$.\n\nI'm trying to get some intuitive sense for this, but it's quite hard. It's like trying to prove that $x$ goes faster to $\\ 0 \\$ then $\\ln x$ goes to $-\u221e$, right ?\n\nI tried this, for $x\u2208(0,1)$ $$x \\ln x=x\u22c5-\u222b_x^1 \\frac{1}{t}dt>x\u22c5\\frac{(1-x)}{-x}=-1+x$$\n\nSo when $x\u21920$, I conclude that $x\u22c5\\ln x\u2265-1$.\n\n-\nI don't know how to implement this, but how about squeeze theorem? It was the first thing that came to mind. \u2013\u00a0 000 Dec 13 '12 at 22:35\n\nSince $\\ln(\\frac{1}{x})=-\\ln x$, $$\\lim_{x\\to 0^{+}} x \\cdot \\ln x = \\lim_{x\\to \\infty} \\frac{1}{x} \\cdot\\ln \\frac{1}{x} = - \\lim_{x\\to \\infty} \\frac{\\ln x}{x}$$\n\nBut $\\ln x$ is \"slower\" than $x$ (since $e^x$ is faster than $x$), so this limit is zero.\n\n(Rigorously: for positive $x$, $2 e^x > x^2$ (by comparing power series of both sides). Plug $\\sqrt{x}$ instead of $x$ and take logarithm of both sides: $\\ln x < \\ln 2 + \\sqrt{x} =o(x)$.)\n\n-\nIs it always true that you can rewrite $\\lim_{x\u21920^+}f(x) = \\lim_{x\u2192\u221e}f(1/x)$? \u2013\u00a0 Kasper Dec 13 '12 at 22:48\n@Kasper, that's a great question. My intuition is no, and the reason is that there are tricky situations that arise when moving from a one-sided limit to a two-sided limit. I would say, however, that it may be true that $$\\lim_{x \\to 0}f(x)=\\lim_{x \\to \\infty}f\\left(\\frac{1}{x}\\right).$$ However, I am uncertain. \u2013\u00a0 000 Dec 13 '12 at 22:55\n@Kasper - The answer is \"Yes\". Try to prove it using the definition of one-sided limit at 0 and limit at infinity. Show that if one limit exists, the other must equal it. \u2013\u00a0 Ofir Dec 13 '12 at 23:03\n@Limitless - There's no problem because the limit at infinity is not two-sided. Actually, your proposed equality is false, because in the LHS you evaluate $f$ on both negative and positive values, and at the RHS you evaluate $f$ only on positive values. Take $f(x)=\\frac{1}{x}$ to reach a contradiction. \u2013\u00a0 Ofir Dec 13 '12 at 23:05\n@Ofir Let $s_n>0$ and $\\lim_{n\u2192\u221e} s_n = 0$. By definition $$\\lim_{x\u21920^+}f(x)=\\lim_{n\u2192\u221e}f(s_n)$$. Let $t_n=1/s_n$. Then $\\lim_{n\u2192\u221e}t_n=+\u221e$. So $$\\lim_{x\u21920^+}f(x)=\\lim_{n\u2192\u221e}f(s_n)=\\lim_{n\u2192\u221e}f(1/t_n)=\\lim_{x\u2192\u221e}f(1/x)$$ Do you mean something like this ? \u2013\u00a0 Kasper Dec 13 '12 at 23:21\n\n$$x \\ln x=-x\u22c5\\int_x^1 \\frac{1}{t}dt$$\n\nNow, let $0<a$ be arbitrary. Then, for all $x \\in (0,1)$ we have\n\n$$\\frac{1}{t^{1-a}}\\leq \\frac{1}{t} \\leq \\frac{1}{t^{1+a}}$$\n\nThus\n\n$$-x\u22c5\\int_x^1 \\frac{1}{t^{1+a}}dt \\leq -x \\ln(x) \\leq -x\u22c5\\int_x^1 \\frac{1}{t^{1-a}}dt$$ $$-x\u22c5\\frac{x^a-1}{a} \\leq -x \\ln(x) \\leq -x\u22c5\\frac{x^{-a}-1}{-a}\u22c5$$\n\nnow let $x \\to 0$.\n\nP.S. The argument works directly with any $0<a<1$, so picking $a=\\frac{1}{2}$ makes the proof much cleaner....\n\n-\n\nMake the change $x = e^{-y}$: $$\\lim_{x \\to 0} x \\ln x = - \\lim_{y \\to \\infty} y e^{-y}$$\n\nShow that for $y > 0$: $$e^y > \\frac{y^2}{2!}$$\n\nAnd use the squeeze theorem.\n\n-", "date": "2015-05-30 03:11:19", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/258273/how-do-i-prove-that-lim-x%e2%86%920-x%e2%8b%85-ln-x-0", "openwebmath_score": 0.9761853814125061, "openwebmath_perplexity": 234.36994197633402, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022632097108, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8539932425593709}}
{"url": "https://math.stackexchange.com/questions/2210539/what-am-i-doing-wrong-with-this-second-order-ode/2210593", "text": "# What am I doing wrong with this second order ODE?\n\nI have an ODE: $$\\ddot v=-kt^{-1}\\dot v+kt^{-2}v$$ I want to solve it by reducing it to a first order ODE, by defining $u(t)=t^{m}v(t)$ If I rewrite the ODE in terms of $u$, this gives me: $$\\ddot u =(-k+2m)t^{-1}\\dot u +(km+k-1-m^2)t^{-2}u$$ Setting the coefficient of $u$ to zero gives $2m=k+\\sqrt{k^2-4+4k}$, which I will write as $k+\\alpha$.\n\nThis gives me the single order ODE $$\\ddot u=\\alpha t^{-1}\\dot u$$ Solving it gives $$\\dot u=2c\\alpha t$$ $$u(t)=c\\alpha t^2$$ Plugging this back into $v$, gives $$v(t)=c\\alpha t^{2-m}$$ $$\\dot v = (2-m)c\\alpha t^{1-m}$$ $$\\ddot v = (1-m)(2-m)c\\alpha t ^{-m}$$\n\nThese equations are supposed to hold for any $k$. If we plug this into the original ODE, however, this reduces to the equation $$3k=4+\\sqrt{k^2-4+4k}$$\n\nWhich obviously does not hold for all $k$.\n\nWhat am I doing wrong?\n\n\u2022 Hint: Since this is a Euler-Cauchy type, let $v = t^m$. Find $v''$ and $v'$ and substitute in, solve for $m$ and solve.\n\u2013\u00a0Moo\nMar 30 '17 at 17:39\n\u2022 most likely your solution changes between (over)damped and oscillatory depending on $k$ Mar 30 '17 at 18:00\n\u2022 Do you need to reduce it to a first order ODE? Or are you fine solving it with any method? Mar 30 '17 at 18:06\n\u2022 I want to learn as much as possible about these things, so I was working on reducing it to a first order ODE. So I'd like to learn about other methods as well, but also I'd like to make sure I can do this method Mar 30 '17 at 18:09\n\u2022 Please check again all signs and coefficients, the one before $u$ should reduce to $-(m+k)(m-1)$. Mar 30 '17 at 18:53\n\nWhat you should realize is that you don't need to do a change of variables here but recognize that your solution is a Euler-Cauchy problem and therefore can be solved by:\n\nLet $v=t^m$ then subbing the values into your DE you get:\n\n$$m(m-1)t^{m-2} = -kmt^{m-2} + kt^{m-2} \\\\ \\implies m(m-1) +km -k =0 \\\\ \\implies m=1,m=-k \\\\$$\n\ntherefore giving:\n\n$$v(t) = At^1 + Bt^{-k} \\implies v(t) = At + Bt^{-k}$$\n\nWhich does hold $\\forall k \\in \\mathbb{R}$ but you should also note that it does hold $\\forall t \\in (-\\infty,0) \\cup (0,+\\infty)$ because $t=0$ is a case basis in that if $k>0$ then $t \\neq 0$ but if $k<0$ then $t\\in (-\\infty,\\infty)$\n\nSince you did say you want to learn about the reduction to first order just note then when you do that you typically let u be something like $u= t^m \\dot{v}$ so then $\\dot{u}$ would be equal to a second derivative of v.\n\nIn your first integration, you should get from $$\\ddot u=\u03b1t^{-1}\\dot u$$ to $$\\ln|\\dot u|=\u03b1\\ln|t|+c$$ or $$\\dot u = C\u00b7t^\u03b1$$ leading to $$u=\\frac{C}{\u03b1+1}\u00b7t^{\u03b1+1}+D$$", "date": "2022-01-17 21:05:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2210539/what-am-i-doing-wrong-with-this-second-order-ode/2210593", "openwebmath_score": 0.8638442754745483, "openwebmath_perplexity": 130.00688768033393, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022627413796, "lm_q2_score": 0.8740772302445241, "lm_q1q2_score": 0.8539932320560785}}
{"url": "https://math.stackexchange.com/questions/3746161/proving-%E2%8A%A2-a%E2%88%A8a%E2%86%92b-using-hilbert-system", "text": "# Proving \u22a2 A\u2228(A\u2192B) using Hilbert system\n\nI'm self-studying mathematical logic from \"Introduction to Mathematical Logic\" by Detlovs and Podnieks (available free here under CC license). Unfortunately, it doesn't come with any solutions. I'm stuck trying to prove \u22a2 A\u2228(A\u2192B), from Exercise 1.4.2 (i) on page 43. The text appears to be using the Hilbert-style deduction method.\n\nRelevant axioms (classical logic)\n\n\u2022 $$L_1: B \\to (C \\to B)$$\n\u2022 $$L_2: (B \\to (C \\to D)) \\to ((B \\to C) \\to (B \\to D))$$\n\u2022 $$L_3: B \\to B \\lor C$$\n\u2022 $$L_4: C \\to B \\lor C$$\n\u2022 $$L_8: (B \\to D) \\to ((C \\to D) \\to (B \\lor C \\to D))$$\n\u2022 $$L_9: (B \\to C) \\to ((B \\to \\lnot C) \\to \u00acB)$$\n\u2022 $$L_{10}: \\lnot B \\to (B \\lor C)$$\n\u2022 $$L_{11}: B \\lor \\lnot B$$\n\nRelevant inference rules\n\n\u2022 Modus Ponens\n\nAttempt\n\n1. $$(A \\to A \\lor (A \\to B)) \\to ((\\lnot A \\to A \\lor (A \\to B)) \\to (A \\lor \\lnot A \\to A \\lor (A \\to B)))$$ --- $$L_8$$\n2. $$A \\to A \\lor (A \\to B)$$ --- $$L_6$$\n3. $$A \\lor \\lnot A$$ --- $$L_{11}$$\n4. $$(\\lnot A \\to A \\lor (A \\to B)) \\to (A \\lor \\lnot A \\to A \\lor (A \\to B))$$ --- from (1) and (2) by MP\n5. $$(\\lnot A \\to A \\lor (A \\to B)) \\to A \\lor (A \\to B)$$ --- from (3) and (4) by MP\n6. $$(\\lnot A \\to (A \\to A \\lor (A \\to B))) \\to ((\\lnot A \\to A) \\to (\\lnot A \\to A \\lor (A \\to B)))$$ --- $$L_2$$\n7. $$(\\lnot A \\to (A \\to A \\lor (A \\to B)))$$ --- $$L_{10}$$\n8. $$(\\lnot A \\to A) \\to (\\lnot A \\to A \\lor (A \\to B))$$ --- from (6) and (7) by MP\n\nI'm stuck at this point because $$\\lnot A \\to A$$ appears to be a contradiction. I'm also not sure whether this is the right approach. It looks alright at formula 5, but I'm not sure how to prove $$\\lnot A \\to A \\lor (A \\to B)$$, since it requires proving that $$A \\lor (A \\to B)$$ is always true, which is what we're trying to prove in the first place.\n\nThe text states that it can be solved using 14 formulas, 13 being the shortest yet.\n\n\u2022 The \"proof strategy\" is to use L11: $A \\lor \\lnot A$ and derive $A \\lor (A \\to B)$ from $A$ with L6 and from $\\lnot A$ with L10. Then use L8 \u2013\u00a0Mauro ALLEGRANZA Jul 5 '20 at 16:48\n\u2022 @MauroALLEGRANZA I seem to have done the first three steps in my attempt (if I understood correctly); substituting B=A, C=\u00acA for L8 seems to yield (A\u2192D)\u2192((\u00acA\u2192D)\u2192D) after simplifying. The only straightforward substitution for D that I haven't tried seems to be D=(\u00acA\u2192A\u2228(A\u2192B)) but that leads to a double negation case with L10, which hasn't been proved yet. \u2013\u00a0user383527 Jul 5 '20 at 20:36\n\n## 1 Answer\n\nI assume that you are forced not tu use the Deduction Theorem.\n\nBut you can use the so-called Law of Syllogism (transitivity of $$\\to$$).\n\nIf so, here is a sketch of a derivation:\n\n1. $$\\vdash A \\to (A \\lor (A \\to B))$$ --- L6\n\n2. $$\\vdash \\lnot A \\to (A \\to B)$$ --- L10\n\n3. $$\\vdash (A \\to B) \\to (A \\lor (A \\to B))$$ --- L7\n\n4. $$\\vdash \\lnot A \\to (A \\lor (A \\to B))$$ --- from 2. and 3. by Syllogism.\n\nNow we can \"cook them\" together using L8:\n\n1. $$\\vdash (A \\to (A \\lor (A \\to B))) \\to [(\\lnot A \\to (A \\lor (A \\to B))) \\to ((A \\lor \\lnot A) \\to (A \\lor (A \\to B)))]$$\n\nNow, from 5., 1. and 4. by MP twice:\n\n1. $$\\vdash (A \\lor \\lnot A) \\to (A \\lor (A \\to B))$$\n\nFinally, using L11: $$\\vdash A \\lor \\lnot A$$, by MP:\n\n$$\\vdash A \\lor (A \\to B)$$.\n\n\u2022 Smart use of the law of syllogism! The text has proved it as a theorem, so it is straightforward to show it from (2) and (3). In total, this gives 13 formulas (verbosely written), which the text says is the shortest proof. \u2013\u00a0user383527 Jul 6 '20 at 22:35", "date": "2021-05-17 20:03:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3746161/proving-%E2%8A%A2-a%E2%88%A8a%E2%86%92b-using-hilbert-system", "openwebmath_score": 0.8895859718322754, "openwebmath_perplexity": 448.88665460830254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9559813463747181, "lm_q2_score": 0.8933094131553264, "lm_q1q2_score": 0.8539871355174382}}
{"url": "http://math.stackexchange.com/questions/483621/sums-of-rising-factorial-powers", "text": "# Sums of rising factorial powers\n\nDoodling in wolfram, I found that $$\\sum^{k}_{n=1}1=k$$ The formula is pretty obvious, but then you get $$\\sum^{k}_{n=1}n=\\frac{k(k+1)}{2}$$ That is a very well known formula, but then it gets interesting when you calculate $$\\sum^{k}_{n=1}n(n+1)=\\frac{k(k+1)(k+2)}{3}\\\\ \\sum^{k}_{n=1}n(n+1)(n+2)=\\frac{k(k+1)(k+2)(k+3)}{4}$$ And so on. There is an obvious pattern that I really doubt is a coincidence, but I have no idea how to prove it in the general case. Any ideas?\n\n-\nPascal's triangle comes to mind here \u2013\u00a0 Omnomnomnom Sep 4 '13 at 3:47\n@Omnomnomnom I thought about it but I was unable to introduce binomials there. However, you can generalize the hypothesis using factorials, but usually working with factorials is harder than not doing so(At least I would have a very hard time trying to do so). \u2013\u00a0 chubakueno Sep 4 '13 at 3:51\nThese are essentially the sums for the general $d$ dimensional simplex. \u2013\u00a0 Jaycob Coleman Sep 4 '13 at 4:02\nIt's overkill, but you could expand and apply en.wikipedia.org/wiki/Faulhaber%27s_formula \u2013\u00a0 dls Sep 4 '13 at 4:11\n@dls I think that in that case the interesting part would go backwards: That expanding and a applying Faulhaber\u00b4s yields such a regular and simple result :) \u2013\u00a0 chubakueno Sep 4 '13 at 4:22\n\nYou can argue any given case by induction. I will take your last,$$\\sum^{k}_{n=1}n(n+1)(n+2)=\\frac{k(k+1)(k+2)(k+3)}{4}$$ for the example, but I think it is easy to see how it gets carried forward. The base case is simply $1\\cdot 2\\cdot 3=1\\cdot 2\\cdot 3\\cdot \\frac 44$ If it is true up to $k$, then $$\\sum^{k+1}_{n=1}n(n+1)(n+2)\\\\=\\sum^{k}_{n=1}n(n+1)(n+2)+(k+1)(k+2)(k+3)\\\\=\\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)\\frac {k+4-k}4\\\\=\\frac{(k+1)(k+2)(k+3)(k+4)}{4}$$\n\n-\nThanks! Just changing the $2$, $3$ and $4$ by $x$,$x+1$ and $x+2$ does the work of generalizing it very nicely. \u2013\u00a0 chubakueno Sep 4 '13 at 3:59\n\nThe easy way to deal, for example, with $\\sum_{i=1}^n i(i+1)(i+2)(i+3)$ is to let $F(i)=i(i+1)(i+2)(i+3)(i+4)$. We calculate $F(i)-F(i-1)$. We get $$i(i+1)(i+2)(i+3)(i+4)-(i-1)(i)(i+1)(i+2)(i+3).$$ There is a common factor of $i(i+1)(i+2)(i+3)$. When we \"take it out\" we are left with $(i+4)-(i-1)=5$.\n\nLet $G(i)=\\frac{F(i)}{5}$. Then by our calculation $i(i+1)(i+2)(i+3)=G(i)-G(i-1)$.\n\nNow consider the sum $\\sum_{i=1}^n i(i+1)(i+2)(i+3)$. This is $$(G(1)-G(0))+(G(2)-G(1))+G(3)-G(2)) +\\cdots+(G(n)-G(n-1)).$$ Observe the telescoping. Since $G(0)=0$, the above sum is equal to $G(n)$. Thus $$\\sum_{i=1}^n i(i+1)(i+2)(i+3)=G(n)=\\frac{n(n+1)(n+2)(n+3)(n+4)}{5}.$$\n\nExactly the same idea works in general.\n\n-\n\nThe hypothesized equality can be written as follows: for any $m$, we conjecture $$\\sum^{k}_{n=1}\\frac{(n+m)!}{(n-1)!}=\\frac{(k+m+1)!}{(m+2)(k-1)!}$$ Dividing both sides by $(m+1)!$, we have $$\\sum^{k}_{n=1}\\frac{(n+m)!}{(n-1)!(m+1)!}=\\frac{(k+m+1)!}{(m+2)!(k-1)!}$$ Or, in other words $$\\sum^{k}_{n=1}\\binom{n+m}{m+1}=\\binom{k+m+1}{m+2}$$ I'm not sure how to prove this (yet), but it seems very likely that there's a neat trick for all this.\n\n-\n\nPartial sums of sequences $a_n$ that can be expressed as polynomials in $n$ are easily found using discrete calculus.\n\nWe start with the discrete version of the Fundamental Theorem of Integral Calculus:\n\n\\begin{align*} \\sum\\limits_{n=1}^k a_n &= \\sum\\limits_{n=1}^k (\\Delta b)_n \\\\ &= (b_2 - b_1)+(b_3 - b_2)+ \\ldots + (b_k - b_{k-1}) + (b_{k+1} - b_k) \\\\ &= b_{k+1} - b_1 \\end{align*}\n\nwhere $(\\Delta b)_n = b_{n+1} - b_n$ is the forward difference. Finding the partial sum has now been reduced to finding a sequence $b_n$ such that $(\\Delta b)_n = a_n$.\n\nWe will find $b$, the antiderivative of $a$, using falling powers, which are defined by\n\n$$n^{\\underline{k}} = n(n-1)(n-2)\\ldots (n-k+1)$$\n\nwhere $k$ is an integer and, by a second definition, $n^{\\underline{0}}=1$. For example\n\n$$n^{\\underline{3}} = n(n-1)(n-2).$$\n\nWe now need one more result: the discrete derivative of $n^{\\underline{k}}$ is given by\n\n\\begin{align*} \\Delta n^{\\underline{k}} &= (n+1)^{\\underline{k}} - n^{\\underline{k}} \\\\ &= (n+1)n^{\\underline{k-1}} - n^{\\underline{k-1}}(n-k+1) \\\\ &= kn^{\\underline{k-1}} \\end{align*}\n\nLet's now find the partial sum for a particular case:\n\n\\begin{align} \\sum^{k}_{n=1}n(n+1)(n+2) &= \\sum^{k}_{n=1} (n+2)^{\\underline{3}} \\\\ &= \\sum^{k}_{n=1} \\Delta \\left[\\frac{1}{4} (n+2)^{\\underline{4}}\\right] \\\\ &= \\frac{(k+3)(k+2)(k+1)k}{4} - \\require{cancel}\\cancelto{0}{\\frac{(1+2)(1+1)(1-0)(1-1)}{4}} \\end{align}\n\nThe general case:\n\n\\begin{align} \\sum^{k}_{n=1} (n+p)^{\\underline{p+1}} &= \\sum^{k}_{n=1} \\Delta \\left[\\frac{1}{p+2} (n+p)^{\\underline{p+2}}\\right] \\\\ &= \\frac{(k+1+p)(k+p)\\ldots [(k+1+p)-(p+2)+1)]}{p+2} \\\\ &= \\frac{(k+1+p)(k+p)\\ldots k}{p+2} \\end{align}\n\nwhere $p>0$ is an integer.\n\n-\n\n[I haven't finished yet. Will be gradually improved]\n\nIf we do it this way, then probably, it will become more insightful.\n\n$$\\sum^{k}_{n=1}1=k$$\n\nthen we preserve everything from the right side exactly as it is. $$\\sum^{k}_{n=1}n=\\frac{k(k+1)}{1 \\cdot 2}$$ [there will be a picture of the right triangle]\n\nNow again, preserve everything from the right side: $$\\sum^{k}_{n=1}\\frac{n(n+1)}{1 \\cdot 2}=\\frac{k(k+1)(k+2)}{1 \\cdot 2 \\cdot 3}$$\n\n[there will be a picture of the 6 pyramids composed into rectangular parallelepiped]\n\n$$\\sum^{k}_{n=1}\\frac{k(k+1)(k+2)}{1 \\cdot 2 \\cdot 3}=\\frac{k(k+1)(k+2)(k + 3)}{1 \\cdot 2 \\cdot 3 \\cdot 4}$$\n\n[There should be a description of the connection between simple combinations and combinations with repetition]\n\n$${n \\choose k} = \\frac{n \\cdot (n - 1) \\cdot \\ldots \\cdot (n - k + 1)}{k \\cdot (k - 1) \\cdot \\ldots \\cdot 1} = \\frac{n^{\\underline{k}}}{k!}$$ $$\\left(\\!\\middle(\\genfrac{}{}{0pt}{}{n}{k}\\middle)\\!\\right) = \\frac{n \\cdot (n + 1) \\cdot \\ldots \\cdot (n + k - 1)}{k \\cdot (k - 1) \\cdot \\ldots \\cdot 1} = \\frac{n^{\\overline{k}}}{k!}$$\n\n-\nYou should use the sandbox if you are part way through an answer. \u2013\u00a0 Michael Albanese 2 days ago\n@Michael: Sorry, I don't understand how to use it. Probably, I just don't have the rights to use that post or I need to do something beforehand, but I don't see [Add answer] button there and I cannot edit them neither. \u2013\u00a0 Pixar 2 days ago\nAh, you might not have enough reputation yet. Oh well, keep it in mind for future use. \u2013\u00a0 Michael Albanese 2 days ago", "date": "2015-08-31 09:28:41", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/483621/sums-of-rising-factorial-powers", "openwebmath_score": 0.9998318552970886, "openwebmath_perplexity": 586.4791316515254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795118010989, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8539584460504834}}
{"url": "https://abhcompta.be/reviews/a1ed65-%C3%A9quivalent-coefficient-binomial", "text": "0 First we highlight the important concepts introduced in the previous post on permutations and combinations. , n ) 6 k \u2212 ) \u22c5 {\\displaystyle n} n y . Multiset coefficients may be expressed in terms of binomial coefficients by the rule, One possible alternative characterization of this identity is as follows: 1 Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. n n \u2026 = n + $\\binom{n+k}k$ divides $\\frac{\\text{lcm}(n,n+1,\\ldots,n+k)}n$. We can use the formula (nk)=n!k!(n\u2212k)! 1 This shows in particular that For instance, if k is a positive integer and n is arbitrary, then. z x {\\displaystyle 1={\\tbinom {6}{6}}={\\tbinom {6}{0}}={\\tbinom {43}{0}}} k \u2212 \u2192 1 {\\displaystyle k=0} 6 n which explains the name \"binomial coefficient\". ) [/math], ${n\\choose k_1,k_2,\\ldots,k_r} ={n-1\\choose k_1-1,k_2,\\ldots,k_r}+{n-1\\choose k_1,k_2-1,\\ldots,k_r}+\\ldots+{n-1\\choose k_1,k_2,\\ldots,k_r-1}$, ${n\\choose k_1,k_2,\\ldots,k_r} ={n\\choose k_{\\sigma_1},k_{\\sigma_2},\\ldots,k_{\\sigma_r}}$, \\begin{align} {z \\choose k} = \\frac{1}{k! n {\\displaystyle {\\tbinom {n}{k}}} Is it possible to have perfect pitch but zero sense of relative pitch? \u25a1\u200b. wei\u00dfen irgendwie aufgereihten Elementen. {\\displaystyle {\\tbinom {n}{k}}} &=\\left(\\!\\!\\binom{-7}{7}\\!\\!\\right)\\left(\\!\\!\\binom{4}{7}\\!\\!\\right)=\\binom{-1}{7}\\binom{10}{7}.\\end{align}, ${x \\choose y}= \\frac{\\Gamma(x+1)}{\\Gamma(y+1) \\Gamma(x-y+1)}= \\frac{1}{(x+1) \\Beta(x-y+1,y+1)}. An equivalent question: how many permutations are there of the letters A, B, C, D, E, F and G? For example, the 2nd2^{\\text{nd}}2nd number in the 4th4^{\\text{th}}4th row is represented as (31)=3\\binom{3}{1} = 3(13\u200b)=3. \u25a1\u200b. In Counting Principles, we studied combinations.In the shortcut to finding$\\,{\\left(x+y\\right)}^{n},\\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. {\\displaystyle {\\frac {{\\text{lcm}}(n,n+1,\\ldots ,n+k)}{n\\cdot {\\text{lcm}}({\\binom {k}{0}},{\\binom {k}{1}},\\ldots ,{\\binom {k}{k}})}}} + = 0 when either k > n or k < 0.$ It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n, and it is given by the formula, For example, the fourth power of 1 + x is. An integer n \u2265 2 is prime if and only if n ) {\\displaystyle \\alpha -z} \u03b1 {\\displaystyle m,n\\in \\mathbb {N} ,}. (2003). F\u00fcr die Anzahl der m\u00f6glichen Ziehungen oder Tippscheine beim deutschen Lotto 6 aus 49 (ohne Zusatzzahl oder Superzahl) gilt: Es gibt hier offensichtlich genau eine M\u00f6glichkeit, 6 Richtige zu tippen. A direct implementation of the multiplicative formula works well: (In Python, range(k) produces a list from 0 to k\u20131.). The coefficient of the middle term in the binomial expansion in powers of x of (1 + \u03b1x)^4 and of (1 \u2013 \u03b1x)^6 is the same if \u03b1 equals asked Nov 5 in Binomial Theorem by Maahi01 ( 19.5k points) binomial theorem + Binomial coefficient formula reduction. 4 \u03b1 -elementigen Teilmenge. A related combinatorial problem is to count multisets of prescribed size with elements drawn from a given set, that is, to count the number of ways to select a certain number of elements from a given set with the possibility of selecting the same element repeatedly. {\\displaystyle k} {\\displaystyle \\psi (n)} ) m Randomly select a ball. z ( {\\displaystyle n} n ln , sondern den Bruch &=(-1)^7\\;\\frac{4\\cdot5\\cdot6\\cdot7\\cdot8\\cdot9\\cdot10} k When the exponent is 1, we get the original value, unchanged: (a+b) 1 = a+b. k , dann ist. ) for all positive integers r and s such that s < pr. , ( k \u2265 The right side counts the same parameter, because there are ways of choosing a set of q marks and they occur in all subsets that additionally contain some subset of the remaining elements, of which there are. -elementigen Menge ist, ergibt sich durch die Summation die Anzahl aller ihrer Teilmengen, also ) ! Other notations for the binomial coefficient are , and . ( ) The binomial coefficients form the entries of Pascal's triangle.. f\u00fcr das dritte usw., bis hin zu \u25a1\u200b\u200b, Or, in other terms, the sum of two adjacent binomial coefficients is equal to the one \"below\" it, assuming Pascal's triangle is produced in the equilateral-triangle-like shape shown above. r 2 {\\displaystyle p} https://de.wikipedia.org/w/index.php?title=Binomialkoeffizient&oldid=205380223, \u201eCreative Commons Attribution/Share Alike\u201c. \u2208 \u2212 \u2264 ) More precisely, fix an integer d and let f(N) denote the number of binomial coefficients with n < N such that d divides . One of the more visually striking properties is the Sierpi\u0144ski sieve, which is obtained by taking mod 2 of every binomial coefficient. 7!} where $\\ln$\u2009$\\Gamma(n)$ denotes the natural logarithm of the gamma function at $n$. 5 ( = k ) , n n 5 When n is composite, let p be the smallest prime factor of n and let k = n/p. und r }{k!\\big((n-k)!\\big)} + \\frac{n!}{(k+1)!\\big(n-(k+1)\\big)!} {\\displaystyle 43={\\tbinom {43}{1}}} wie auch ) ( ! k binomial coefficients: These can be proved by using Euler's formula to convert trigonometric functions to complex exponentials, expanding using the binomial theorem, and integrating term by term. 6 {\\displaystyle x\\in X} \u03b1 m We can verify this visually by looking at Pascal''s triangle and using our guideline for construction: (nk)+(nk+1)=(n+1k+1).\\binom{n}{k} + \\binom{n}{k+1} = \\binom{n+1}{k+1}.(kn\u200b)+(k+1n\u200b)=(k+1n+1\u200b). 0 The Chu\u2013Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is, $\\sum_{j=0}^k \\binom m j \\binom{n-m}{k-j} = \\binom n k$, and can be found by examination of the coefficient of $x^k$ in the expansion of (1\u00a0+\u00a0x)m\u00a0(1\u00a0+\u00a0x)n\u00a0\u2212\u00a0m = (1\u00a0+\u00a0x)n using equation (2). und 0 ( X erh\u00e4lt man die Beziehung. will remain the same. k \u03b1 \u2208 \\binom{n+1}{2} = T_{n} &= \\dfrac{(n+1)(n)}{2}, p Grinshpan, A. ( , can be calculated by logarithmic differentiation: Over any field of characteristic 0 (that is, any field that contains the rational numbers), each polynomial p(t) of degree at most d is uniquely expressible as a linear combination ( \u22c5 }{k!\\cdot l! ) Ein Binomialkoeffizient h\u00e4ngt von zwei nat\u00fcrlichen Zahlen ) l How many permutations are there of 7 objects? This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients. k The binomial theorem is a formula for deriving the power of a binomial, i.e. Certain trigonometric integrals have values expressible in terms of n Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. However this is not true of higher powers of p: for example 9 does not divide + , Alternative notations include C(n, k), nCk, nCk, Ckn, Cnk, and Cn,k in all of which the C stands for combinations or choices. 5 z Exercise 2 Binomial coefficients can be generalized to multinomial coefficients defined to be the number: While the binomial coefficients represent the coefficients of (x+y)n, the multinomial coefficients is convergent for k \u2265 2. = ( over For example, for nonnegative integers ${n} \\geq {q}$, the identity. t ) This is done by interpreting as the number of ways to partition the set into two subsets of size k and n-k. Pascal's rule is the important recurrence relation. Die Anzahl aller so zusammengestellten 1 \\end{cases}[/math], $\\tbinom n0,\\tbinom n1,\\tbinom n2,\\ldots$, $\\sum_{k=0}^\\infty {n\\choose k} x^k = (1+x)^n. {\\displaystyle \\alpha } Differentiating (2) k times and setting x = \u22121 yields this for Identifying Binomial Coefficients. The notation [math] {n \\choose k}$ is convenient in handwriting but inconvenient for typewriters and computer terminals. > ( Then 0 < p < n and. k {\\displaystyle 6={\\tbinom {6}{5}}} Damit gilt f\u00fcr jede endliche Menge {\\displaystyle {\\tbinom {2n}{n}}} }{k_1!k_2!\\cdots k_r! \u2212 n \u03c8 {\\displaystyle l!} n n How many different pizza can the customer create? Another form of the Chu\u2013Vandermonde identity, which applies for any integers j, k, and n satisfying 0 \u2264 j \u2264 k \u2264 n, is, The proof is similar, but uses the binomial series expansion (2) with negative integer exponents. Sie kann anschaulich etwa so gedeutet werden: Zun\u00e4chst z\u00e4hlt man alle 1 k M {\\displaystyle \\sum _{k=0}^{d}a_{k}{\\binom {t}{k}}} Compute (92)+(83)\\dbinom{9}{2} + \\dbinom{8}{3}(29\u200b)+(38\u200b). ) ) 0 {\\displaystyle k\\to \\infty } Create a free website or blog at WordPress.com. This definition inherits these following additional properties from { 1 It also follows from tracing the contributions to Xk in (1 + X)n\u22121(1 + X). Differentiating (2) k times and setting x = \u22121 yields this for ( Asymptotic of binomial coefficients near the center. {\\displaystyle \\{1,2\\}{\\text{, }}\\{1,3\\}{\\text{, }}\\{1,4\\}{\\text{, }}\\{2,3\\}{\\text{, }}\\{2,4\\}{\\text{,}}} The overflow can be avoided by dividing first and fixing the result using the remainder: Another way to compute the binomial coefficient when using large numbers is to recognize that. ) in a language with fixed-length integers, the multiplication by n^{\\underline{k}}/k! ( 1 . Hot Network Questions Should I respond to an \"ethical hacker\" who's requesting a bounty? ) ) ,\u00a0 = 6[/math] is the coefficient of the x2 term. ! There are several ways to come up with the answer. m eigentlich einfache Binomialkoeffizienten sind. x {\\displaystyle {\\tfrac {(k+l)! \u22c5 Newton's binomial series, named after Sir Isaac Newton, is a generalization of the binomial theorem to infinite series: The identity can be obtained by showing that both sides satisfy the differential equation (1 + z) f'(z) = \u03b1 f(z). + Thinking of the binomial coefficient as the number of ways to making a series of two-outcome decisions is crucial to the understanding of binomial distribution. + The answer can also be obtained by the multiplication principle. ) The multiplicative formula allows the definition of binomial coefficients to be extended[3] by replacing n by an arbitrary number \u03b1 (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible: With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the n {\\displaystyle k!} ) 7 C k 6 Each polynomial $\\tbinom{t}{k}$ is integer-valued: it has an integer value at all integer inputs $t$. ( l A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem: which is valid by for all integers = Rekursive Darstellung und Pascalsches Dreieck, Der Binomialkoeffizient in der Kombinatorik, Summen mit alternierenden Binomialkoeffizienten, Summen von Binomialkoeffizienten mit geraden bzw. {\\displaystyle {\\tbinom {n}{k}}=0} k setzt. ) 0 Um unn\u00f6tigen Rechenaufwand zu vermeiden, berechnet man im Fall Then it follows from our earlier definition for the sum of binomial coefficients that. ( 1\\quad 4 \\quad 6 \\quad 4 \\quad 1\\\\ k -Binomialsymbolen: ( , ) \u22c5 Addition obiger Gleichungen ) n 1 n und {\\displaystyle \\operatorname {Re} z>0} . M Die w\u00f6rtliche \u00dcbersetzung von \u201e Suppose that the customer is a meat lover. The following is one specific path that the person might take. The reasoning is based on the multiplication principle (see here). , 1 Some properties make use of symmetry, some deal with expansion, but they all can be proved rather intuitively. \\binom MN = \\frac{M!}{N!(M-N)!} n For each k, the polynomial {\\displaystyle z} {\\displaystyle {\\tbinom {t}{k}}} ( The Chu\u2013Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is, and can be found by examination of the coefficient of )^3}[/math], $\\sum_{k=-a}^a(-1)^k{a+b\\choose a+k} {b+c\\choose b+k}{c+a\\choose c+k} = \\frac{(a+b+c)!}{a!\\,b!\\,c! One way to solve this problem is to count the number of paths in a \u201cbrute force\u201d approach by tracing all possible paths in the diagram. , n} with the right hand side \ufb01rst grouping them into those which contain element n and those which don\u2019t. \u00fcber m {\\displaystyle n} {\\displaystyle {\\tbinom {\\alpha }{k}}} [math]\\sum_{k=0}^n k \\binom n k = n 2^{n-1}$. = k k x k ( is convergent for k \u2265 2. \u03b1 ) , \u03f5 The left side counts the number of ways of selecting a subset of [n] = {1, 2, ..., n} with at least q elements, and marking q elements among those selected. 1 Template:Planetmath 1 ) gilt: Beweis: ( bezeichnet. (5\u22123)!3!=10.\\binom{5}{3} = \\frac{5!}{(5-3)! \u2212 | {\\displaystyle M} , , [/math], that is clear since the RHS is a term of the exponential series [math] e^k=\\sum_{j=0}^\\infty k^j/j! of binomial coefficients. \u221e M\u00f6glichkeiten der Wahl des ersten Tupel-Elements. \u221e In ordering a pizza, a customer can choose from a list of 10 toppings: mushroom, onion, olive, bell pepper, pineapple, spinach, extra cheese, sausage, ham, and pepperoni. {\\displaystyle -z,-s\\notin \\mathbb {N} } \u2212 = 5040. {\\displaystyle m\\leq n} }\\\\ , 5 k This formula is used in the analysis of the German tank problem. ) 1 {\\displaystyle \\sum _{k=0}^{[{\\frac {n-1}{2}}]}{\\binom {n}{2k+1}}=2^{n-1}} Assume that all the paths from any point to any point in the above diagram are available for walking. This gives. roten und 2 terms in this product is 0 {\\color{blue}1 \\qquad 3 \\qquad 3 \\qquad 1} \\\\ is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by + This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients. The point 1 is known to be in each -subset of the first type and the other points must be chosen from objects. ) = {\\displaystyle n} A combinatorial interpretation of this formula is as follows: when forming a subset of elements (from a set of size ), it is equivalent to consider the number of ways you can pick elements and the number of ways you can exclude elements. Now, if you know your stuff about triangular numbers, you can say that. \u25a1\\displaystyle { y }^{ 3 }{ a \\choose 3 } = { 3 }^{ 3 }{ 10 \\choose 3 } = 3240.\\ _\\squarey3(3a\u200b)=33(310\u200b)=3240. {\\displaystyle k=m=n} und This is the statement that each row of Pascal's triangle is symmetric in either one of two ways. { (x+y) }^{ a }=\\sum _{ i=0 }^{ a }{ a \\choose i } { x }^{ a-i }{ y }^{ i }.(x+y)a=i=0\u2211a\u200b(ia\u200b)xa\u2212iyi. n", "date": "2022-01-27 07:49:27", "meta": {"domain": "abhcompta.be", "url": "https://abhcompta.be/reviews/a1ed65-%C3%A9quivalent-coefficient-binomial", "openwebmath_score": 0.9623414874076843, "openwebmath_perplexity": 1218.5138946836228, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986979508737192, "lm_q2_score": 0.8652240877899776, "lm_q1q2_score": 0.8539584451145372}}
{"url": "https://math.stackexchange.com/questions/2650571/how-many-bit-strings", "text": "# How many bit strings?\n\nHow many bit strings of length $8$ have either exactly two $1$-bit among the first $4$ bits or exactly two $1$-bit among the last $4$ bits?\n\nMy solution:\n\nA bit only contains $0$ and $1$, so $2$ different numbers, i.e., $0$ and $1$. For the first part we have $2^6=64$ ways. Similar for the other way. Hence there exists $2^4=16$ bit strings. Is my answer true?\n\nUpdate: I mean $2^6+2^6-2^4=112$ bit strings\n\nI lost your logic. By symmetry, amount of strings with exactly 2 ones in the first four (call this group $F$) is identical to the ones with exactly 2 ones in the last four (call this group $L$).\n\nThen your desired amount is $|F| + |L| - |F \\cap L|$.\n\nTo compute $F$, note that you have exactly $\\binom{4}{2} = 6$ ways to pick the location of the ones in the first four, which fixes your picks to be ones and the other two of the first four to be zeros. In other words, this fixes the first four bits, and the rest can be manipulated in $2^4=16$ ways.\n\nCan you finish computing $|F|$? We already said $|L|=|F|$ and can you compute $|F \\cap L|$ by a similar technique?\n\n\u2022 I still belive that it should be $2^6+2^6-2^4=112$ ways.\n\u2013\u00a0user530832\nFeb 14, 2018 at 16:40\n\u2022 @M.Rasmussen your mistake is not automatically setting the other 2 bits to zero. The way you compute, consider, for example, the first group, with exactly 2 ones in the first four bits. $2^6$ you are getting counts the number of bit strings with 6 places, but (a) you did not count that there are multiple ways in spreading the ones among the first four and (b) you are including the string 11111111 as legal, whereas you must have exactly 2 ones in the first four, and this string has 4. Feb 14, 2018 at 17:25\n\u2022 Ohh okay. I see, so all other \"empty\" spaces must be 0 if you know what I mean.\n\u2013\u00a0user530832\nFeb 14, 2018 at 17:49\n\u2022 Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: \"How many bit strings of length $12$ have either exactly four $1$-bit among the first $6$ bits or exactly four $1$-bit among the last $6$ bits?\" Then I get the answer: 122655 is that correct? I did the same proces like the original question.\n\u2013\u00a0user530832\nFeb 14, 2018 at 17:57\n\u2022 @M.Rasmussen Then both elementary groups are $|F| = |L| = \\binom{6}{4} \\cdot 2^6$ and their intersection is $|F \\cap L| = \\binom{6}{4} \\binom{6}{4}$ so you end up with $$|F| + |L| - |F\\cap L| = 2 \\cdot \\binom{6}{4} \\cdot 2^6 - \\binom{6}{4}^2 = 960 - 225 = 715.$$ Feb 14, 2018 at 21:45\n\nLet $A$ be the set of bit strings with exactly two $1$-bit among the first $4$ bits, and $B$ be the set of bit strings with exactly two $1$-bit among the last $4$ bits.\n\n\\begin{align} \\#A &= \\binom{4}{2} 2^4 = 6\\cdot2^4 \\\\ \\#B &= 2^4 \\binom{4}{2} = 6\\cdot2^4 \\\\ \\#A\\cap B &= \\binom{4}{2}^2 = 6^2 \\\\ \\#A\\cup B &= \\#A + \\#B - \\# A \\cap B \\\\ &= 6 (2^4 \\cdot 2 - 6) \\\\ &= 6 \\cdot 26 = 156 \\end{align}\n\nAmong the first $4$ bits, choose $2$ to set them to one and the other two would be set to $0$ and there are $4$ of them with absolute freedom.\n\n$$\\binom{4}{2}\\cdot 2^4$$\n\nSimilar when we focus on the last $4$ bits.\n\nWhen we focus on intersection. We would pick $2$ from the first $4$ and pick $2$ from the last $4$.\n\nSo my overall answer would be\n\n$$2\\binom42 \\cdot 2^4 - \\binom42^2$$\n\nRemark: I think your mistake is thinking that you can set arbitary $6$ bits to anything.\n\n\u2022 I see many of you get 156, but I just think on Inclusion\u2013exclusion principle and the way I did it here was drawing the problem: i.stack.imgur.com/jYOXA.png\n\u2013\u00a0user530832\nFeb 14, 2018 at 16:38\n\u2022 The keyword is exactly $2$ ones in the first $4$ bits, which means exactly $2$ ones and exactly $2$ zeros in the first $4$ bits. Feb 14, 2018 at 16:59\n\u2022 Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: \"How many bit strings of length 12 have either exactly four 1-bit among the first 6 bits or exactly four 1-bit among the last 6 bits?\" Then I get the answer: 122655 is that correct? I did the same proces like the original question\n\u2013\u00a0user530832\nFeb 14, 2018 at 17:58\n\u2022 @M.Rasmussen: There are not even that many 12 bit strings period. There are only 4096 strings of length 12 so how could there possibly be 122655 of them that have some property? You need to get in the habit of checking your answers for reasonableness. Feb 14, 2018 at 20:55\n\u2022 @M.Rasmussen would you like to show the working of how you obtain that number? Feb 14, 2018 at 21:58\n\nHow many bit strings of length 8 have either exactly two 1-bit among the first 4 bits or exactly two 1-bit among the last 4 bits?\n\nThe simplest solution is the best. There are only 256 possibilities to check so just list them!\n\n00000011 00000101 00000110 00001001 00001010 00001100 00010011 00010101\n00010110 00011001 00011010 00011100 00100011 00100101 00100110 00101001\n00101010 00101100 00110000 00110001 00110010 00110011 00110100 00110101\n00110110 00110111 00111000 00111001 00111010 00111011 00111100 00111101\n00111110 00111111 01000011 01000101 01000110 01001001 01001010 01001100\n01010000 01010001 01010010 01010011 01010100 01010101 01010110 01010111\n01011000 01011001 01011010 01011011 01011100 01011101 01011110 01011111\n01100000 01100001 01100010 01100011 01100100 01100101 01100110 01100111\n01101000 01101001 01101010 01101011 01101100 01101101 01101110 01101111\n01110011 01110101 01110110 01111001 01111010 01111100 10000011 10000101\n10000110 10001001 10001010 10001100 10010000 10010001 10010010 10010011\n10010100 10010101 10010110 10010111 10011000 10011001 10011010 10011011\n10011100 10011101 10011110 10011111 10100000 10100001 10100010 10100011\n10100100 10100101 10100110 10100111 10101000 10101001 10101010 10101011\n10101100 10101101 10101110 10101111 10110011 10110101 10110110 10111001\n10111010 10111100 11000000 11000001 11000010 11000011 11000100 11000101\n11000110 11000111 11001000 11001001 11001010 11001011 11001100 11001101\n11001110 11001111 11010011 11010101 11010110 11011001 11011010 11011100\n11100011 11100101 11100110 11101001 11101010 11101100 11110011 11110101\n11110110 11111001 11111010 11111100\n\nThere are 156 such strings.\n\nDo you still believe there are only 112? If so, then which ones did I either list twice, or list in error?\n\n\u2022 I am convinced, thank you\n\u2013\u00a0user530832\nFeb 15, 2018 at 13:13\n\nAs a programmer (and a relatively naive mathematician) I immediately thought of reducing the 8-bit string to a hexadecimal string. Two digits, each made of four bits exactly as the problem is divided.\n\nAnd, out 16 digits (0-F), 6 of them are valid (exactly 2 1-bits) digits.\n\nSo, we have 16\u00d76 valid numbers (the first digit can be anything) and out of the the remaining 16\u00d710 numbers, 6\u00d710 of those will also be valid.\n\nNow we have 16\u00d76 + 6\u00d710 = 156", "date": "2022-05-26 21:10:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2650571/how-many-bit-strings", "openwebmath_score": 0.975273609161377, "openwebmath_perplexity": 383.42625956804636, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759671623987, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8539479475387487}}
{"url": "http://gaworld.ru/gcf-of-28-and-32", "text": "# GCF of 28 and 32\n\nThe gcf of 28 and 32 is the largest positive integer that divides the numbers 28 and 32 without a remainder. Spelled out, it is the greatest common factor of 28 and 32. Here you can find the gcf of 28 and 32, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the gcf of 28 and 32, but also that of three or more integers including twenty-eight and thirty-two for example. Keep reading to learn everything about the gcf (28,32) and the terms related to it.\n\n## What is the GCF of 28 and 32\n\nIf you just want to know what is the greatest common factor of 28 and 32, it is 4. Usually, this is written as\n\ngcf(28,32) = 4\n\nThe gcf of 28 and 32 can be obtained like this:\n\n\u2022 The factors of 28 are 28, 14, 7, 4, 2, 1.\n\u2022 The factors of 32 are 32, 16, 8, 4, 2, 1.\n\u2022 The common factors of 28 and 32 are 4, 2, 1, intersecting the two sets above.\n\u2022 In the intersection factors of 28 \u2229 factors of 32 the greatest element is 4.\n\u2022 Therefore, the greatest common factor of 28 and 32 is 4.\n\nTaking the above into account you also know how to find all the common factors of 28 and 32, not just the greatest. In the next section we show you how to calculate the gcf of twenty-eight and thirty-two by means of two more methods.\n\n## How to find the GCF of 28 and 32\n\nThe greatest common factor of 28 and 32 can be computed by using the least common multiple aka lcm of 28 and 32. This is the easiest approach:\n\ngcf (28,32) = $\\frac{28 \\times 32}{lcm(28,32)} = \\frac{896}{224}$ = 4\n\nAlternatively, the gcf of 28 and 32 can be found using the prime factorization of 28 and 32:\n\n\u2022 The prime factorization of 28 is: 2 x 2 x 7\n\u2022 The prime factorization of 32 is: 2 x 2 x 2 x 2 x 2\n\u2022 The prime factors and multiplicities 28 and 32 have in common are: 2 x 2\n\u2022 2 x 2 is the gcf of 28 and 32\n\u2022 gcf(28,32) = 4\n\nIn any case, the easiest way to compute the gcf of two numbers like 28 and 32 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 28,32. The calculation is conducted automatically.\n\nThe gcf is...\nFrequently searched terms on our site also include:\n\n## Use of GCF of 28 and 32\n\nWhat is the greatest common factor of 28 and 32 used for? Answer: It is helpful for reducing fractions like 28 / 32. Just divide the nominator as well as the denominator by the gcf (28,32) to reduce the fraction to lowest terms.\n\n$\\frac{28}{32} = \\frac{\\frac{28}{4}}{\\frac{32}{4}} = \\frac{7}{8}$.\n\n## Properties of GCF of 28 and 32\n\nThe most important properties of the gcf(28,32) are:\n\n\u2022 Commutative property: gcf(28,32) = gcf(32,28)\n\u2022 Associative property: gcf(28,32,n) = gcf(gcf(32,28),n) $\\hspace{10px}n\\hspace{3px}\\epsilon\\hspace{3px}\\mathbb{Z}$\n\nThe associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.\n\nTo sum up, the gcf of 28 and 32 is 4. In common notation: gcf (28,32) = 4.\n\nIf you have been searching for gcf 28 and 32 or gcf 28 32 then you have come to the correct page, too. The same is the true if you typed gcf for 28 and 32 in your favorite search engine.\n\nNote that you can find the greatest common factor of many integer pairs including twenty-eight / thirty-two by using the the search form in the sidebar of this page.\n\nQuestions and comments related to the gcf of 28 and 32 are really appreciated. Use the form below or send us a mail to get in touch.\n\nPlease hit the sharing buttons if our article about the greatest common factor of 28 and 32 has been useful to you, and make sure to bookmark our site.\n\nPosted in Greatest Common Factor\n###### 2 comments on \u201cGCF of 28 and 32\u201d\n1. POOFA says:\n\nWOW! THIS IS A GREAT ANSWER! THANK YOU! \ud83d\ude42\n\n2. POOFA says:\n\nWOW! THIS IS A GREAT ANSWER! THANK YOU! \ud83d\ude42\n\n## Related pages\n\nis 49 a prime or composite numberprime factorization for 1000what is the prime factorization of 21is 135 a prime or composite numberthe prime factorization of 90prime factorization of 93greatest common factors calculatorfind the prime factorization of 78is 233 a prime numberwhat is the prime factorization of 175what is the gcf of 63prime factorization of 720prime factorization of 336is 14 a prime or composite numberwhat is the prime factorization for 88least common multiples of 3 and 4easiest way to find lcmwhat is the gcf of 33 and 66what is the greatest prime factor of 38greatest common factor of 42 and 63is 162 a prime numberthe prime factorization of 126prime factorization of 162what is the prime factorization of 150prime factorization of 68prime factors of 700prime factorization of 5553 prime factorizationwhat is the lcm of 3 and 7what is the greatest common factor of 54 and 36what is the prime factorization of 64multiplication chart 25x25list the prime factors of 42what is the gcf of 35 and 6340 x 40 multiplication tablewhat is the prime factors of 9020 by 20 multiplication chart printablegcf of 84 and 90multiplication table 30 x 30prime factorization of 725gcf of 48 and 64common multiples of 24prime factorization for 108prime factorization of 135common multiples of 8is 57 prime or composite numberwhat is the prime factorization of 121prime factorization of 612smallest common multiple of 36 and 60what is the prime factor of 54prime factorization of 70200 multiplication chartprime factorization 252gcf of 98is 229 a prime numberprime factors of 68what is the gcf of 72 and 3636-28-36what is the prime factorization of 198multiplication table chart 1-1000what is the prime factorization for 80what is the prime factorization of 315the prime factorization of 90prime factors of 112what is the prime factorization of 140is 163 a prime numberfactorization of 144is 144 a prime numberprime factorization for 34prime factors of 252the prime factorization of 245prime factorization 110is 63 a prime or composite numberprime factorization of 297greatest common factor of 42 and 63prime factorization of 111common denominator of 5 and 10", "date": "2018-03-17 22:24:16", "meta": {"domain": "gaworld.ru", "url": "http://gaworld.ru/gcf-of-28-and-32", "openwebmath_score": 0.3958978056907654, "openwebmath_perplexity": 856.8683515220426, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759677214397, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.853947926619311}}
{"url": "https://math.stackexchange.com/questions/3007234/seeking-methods-to-solve-int-0-frac-pi2-ln-left2-tan2x-righ", "text": "# Seeking methods to solve $\\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|2 + \\tan^2(x) \\right| \\:dx$\n\nAs part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:\n\n$$\\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|2 + \\tan^2(x) \\right| \\:dx$$\n\nI'm seeking methods using the Feynman Trick (or any method for that matter) that can be used to solve this definite integral.\n\nIf one wishes to use \"Feynman's Trick,\" then begin by defining a function $$I(a)$$, $$a>1$$ as given by\n\n$$I(a)=\\int_0^{\\pi/2}\\log(a+\\tan^2(x))\\,dx \\tag1$$\n\nDifferentiation of $$(1)$$ reveals\n\n\\begin{align} I'(a)&=\\int_0^{\\pi/2} \\frac{1}{a+\\tan^2(x)}\\,dx\\\\\\\\ &=\\frac{\\pi/2}{a-1}-\\frac{\\pi/2}{\\sqrt a (a-1)}\\tag2 \\end{align}\n\nIntegration of $$(2)$$ yields\n\n\\begin{align} I(a)&=\\frac\\pi2\\left(\\log(a-1)+\\log\\left(\\frac{\\sqrt a+1}{\\sqrt{a}-1}\\right) \\right)\\\\\\\\ &=\\pi \\log(\\sqrt a+1)\\tag3 \\end{align}\n\nFinally, setting $$a=2$$ in $$(3)$$, we obtain the coveted result\n\n$$\\int_0^{\\pi/2}\\log(2+\\tan^2(x))\\,dx=\\pi \\log(\\sqrt 2+1)$$\n\n\u2022 How did you resolve the constant of Integration in Step (3)? \u2013\u00a0user150203 Nov 21 '18 at 4:38\n\u2022 @davidg Apology for the sign error. Note $I(0)=0$. \u2013\u00a0Mark Viola Nov 21 '18 at 5:08\n\u2022 Hi Mark ! Long time no speak. Nice solution $\\to +1$. Cheers. \u2013\u00a0Claude Leibovici Nov 21 '18 at 6:22\n\u2022 @MarkViola do you know of any other \u2018tricks that would work? \u2013\u00a0user150203 Nov 21 '18 at 6:51\n\u2022 Hi David. You could try writing the logarithm as $\\log(1+\\cos^2(x))-\\log(\\sin^2(x))$ and expanding the first term as $\\sum_{n=1}^\\infty \\frac{(-1)^{n-1}\\cos^{2n}(x)}{n}$ and proceeding. I haven't tried this, but it might be worth pursuing. \u2013\u00a0Mark Viola Nov 21 '18 at 15:47\n\nMy approach\n\nLet\n\n\\begin{align} \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|2 + \\tan^2(x) \\right| \\:dx &= \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|1 + \\left(1 + \\tan^2(x)\\right) \\right| \\:dx \\\\ &= \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|1 + \\sec^2(x) \\right| \\:dx \\\\ &= \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\frac{\\cos^2(x) + 1}{\\cos^2(x)} \\right| \\:dx \\\\ &= \\int_{0}^{\\frac{\\pi}{2}} \\left[ \\ln\\left|\\cos^2(x) + 1 \\right| - \\ln\\left|\\cos^2(x)\\right| \\right]\\:dx \\\\ &= \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x) + 1 \\right|\\:dx - \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x)\\right|\\:dx \\end{align}\n\nNow\n\n$$\\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x)\\right|\\:dx = 2\\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos(x)\\right|\\:dx = 2\\cdot-\\frac{\\pi}{2}\\ln(2) = -\\pi \\ln(2)$$\n\nFor detail on this definite integral, see guidance here\n\nWe now need to solve\n\n$$\\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x) + 1 \\right|\\:dx$$\n\nHere, Let\n\n$$I(t) = \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x) + t \\right|\\:dx$$\n\nThus,\n\n$$\\frac{dI}{dt} = \\int_{0}^{\\frac{\\pi}{2}} \\frac{1}{\\cos^2(x) + t}\\:dx = \\int_{0}^{\\frac{\\pi}{2}} \\frac{1}{\\frac{\\cos(2x) + 1}{2} + t}\\:dx = 2\\int_{0}^{\\frac{\\pi}{2}} \\frac{1}{\\cos(2x) + 2t + 1}\\:dx$$\n\nEmploy a change of variable $$u = 2x$$:\n\n$$\\frac{dI}{dt} = \\int_{0}^{\\pi} \\frac{1}{\\cos(u) + 2t + 1}\\:du$$\n\nEmploy the Weierstrass substitution $$\\omega = \\tan\\left(\\frac{u}{2} \\right)$$:\n\n\\begin{align} \\frac{dI}{dt} &= \\int_{0}^{\\infty} \\frac{1}{\\frac{1 - \\omega^2}{1 + \\omega^2} + 2t + 1}\\:\\frac{2}{1 + \\omega^2}\\cdot d\\omega \\\\ &= \\int_{0}^{\\infty} \\frac{1}{t\\omega^2 + t + 1} \\:d\\omega \\\\ &= \\frac{1}{t}\\int_{0}^{\\infty} \\frac{1}{\\omega^2 + \\frac{t + 1}{t}} \\:d\\omega \\\\ &= \\frac{1}{t}\\left[\\frac{1}{\\sqrt{\\frac{t+1}{t}}}\\arctan\\left( \\frac{\\omega}{\\sqrt{\\frac{t+1}{t}}}\\right)\\right]_{0}^{\\infty} \\\\ &= \\frac{1}{t}\\frac{1}{\\sqrt{\\frac{t+1}{t}}}\\frac{\\pi}{2} \\\\ &= \\frac{1}{\\sqrt{t}\\sqrt{t + 1}}\\frac{\\pi}{2} \\end{align}\n\nAnd so,\n\n$$I(t) = \\int \\frac{1}{\\sqrt{t}\\sqrt{t + 1}}\\frac{\\pi}{2}\\:dt = \\pi\\ln\\left| \\sqrt{t} + \\sqrt{t + 1}\\right| + C$$\n\nNow\n\n$$I(0) = \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x) + 0 \\right|\\:dx = -\\pi \\ln(2) = \\pi\\ln\\left|\\sqrt{0} + \\sqrt{0 + 1} \\right| + C \\rightarrow C = -\\pi \\ln(2)$$\n\nAnd so,\n\n$$I(t) = \\pi\\ln\\left| \\sqrt{t} + \\sqrt{t + 1}\\right| -\\pi \\ln(2) = \\pi\\ln\\left|\\frac{\\sqrt{t} + \\sqrt{t + 1}}{2} \\right|$$\n\nThus,\n\n$$I = I(1) = \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x) + 1 \\right|\\:dx = \\pi\\ln\\left|\\frac{\\sqrt{1} + \\sqrt{1 + 1}}{2} \\right| = \\pi\\ln\\left|\\frac{1 + \\sqrt{2}}{2} \\right|$$\n\nAnd Finally\n\n\\begin{align} \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|2 + \\tan^2(x) \\right| \\:dx &= \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x) + 1 \\right|\\:dx - \\int_{0}^{\\frac{\\pi}{2}} \\ln\\left|\\cos^2(x)\\right|\\:dx \\\\ &= \\pi\\ln\\left|\\frac{1 + \\sqrt{2}}{2} \\right| - \\left(-\\pi \\ln(2)\\right) \\\\ &= \\pi\\ln\\left|1 + \\sqrt{2} \\right| \\end{align}\n\n\u2022 Your question and your answer have a gap of just 1 minute. If you did solve the question earlier then you must've included your approach in the question itself. What is the need to post it as an answer? \u2013\u00a0Rohan Shinde Nov 21 '18 at 3:56\n\u2022 @Digamma - Sorry, although I've been a member of Math StackExchange for 4 years, I've only become regularly active recently. When I asked a friend who is a member as to what is the best way of presenting a solution whilst asking for other solutions I was told this is the approach taken within the community. If this violate any of the rules, please advise and I will reposition accordingly. Also, I never assign my solution as 'the solution' to any post of this nature. \u2013\u00a0user150203 Nov 21 '18 at 4:00\n\u2022 @Digamma The site encourages people to share the questions with answers. \u2013\u00a0Tianlalu Nov 21 '18 at 4:03\n\u2022 @Tianlalu - Sorry, just to be clear - Have I employed the correct process here? \u2013\u00a0user150203 Nov 21 '18 at 4:10\n\u2022 Thanks @mick - much appreciated :-) \u2013\u00a0user150203 Dec 3 '18 at 3:42", "date": "2019-07-17 04:52:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3007234/seeking-methods-to-solve-int-0-frac-pi2-ln-left2-tan2x-righ", "openwebmath_score": 1.000007152557373, "openwebmath_perplexity": 2502.260495485496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526462237641, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8539158690531157}}
{"url": "https://gmatclub.com/forum/if-f-x-4x-1-x-3-x-1-question-of-the-day-ii-103935.html", "text": "It is currently 28 Jun 2017, 21:25\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7450\nLocation: Pune, India\nIf f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)\u00a0[#permalink]\n\n### Show Tags\n\n29 Oct 2010, 11:18\n24\nKUDOS\nExpert's post\n64\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n47% (02:38) correct 53% (01:34) wrong based on 1269 sessions\n\n### HideShow timer Statistics\n\nIf f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?\n\n(A) 3\n(B) 4\n(C) 5\n(D) 21/4\n(E) 7\n\n(Still high on mods! Next week, will make questions on some other topic.)\n[Reveal] Spoiler: OA\n\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by Bunuel on 06 Jul 2013, 02:33, edited 1 time in total. Added the OA. CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2783 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: Question of the Day - II [#permalink] ### Show Tags 29 Oct 2010, 18:51 7 This post received KUDOS VeritasPrepKarishma wrote: Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)? (A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7 (Still high on mods! Next week, will make questions on some other topic.) put x=0 we get f(x) = 5..rule out D and E put x=1/4 we get f(x) = 0 + 11/4 + 5/4 = 16/4 = 4.. rule out C D and E Now the answer is either 3 or 4. Reason for above checking of values: for every value of x> 3 the f(x) is quite big because of 4x-1 for every value of x < -1 the f(x) if bigger than 3 and 4. for x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3 => only 4 is the probable answer. Thus we only need to check x>=-1 and x<= 1/4 f(x) in this domain is = 1-4x + 3-x + x+1 = 5-4x => for f(x) to be minimum the x should be +ve => for x = 1/4 , f(x) = 4. This can be solved using graph as well by plotting the f(x) in different domains. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html Retired Moderator Joined: 02 Sep 2010 Posts: 803 Location: London Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 01:59 10 This post received KUDOS 1 This post was BOOKMARKED Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So f(x)=g(x)+h(x) Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3 g(x) is minimum at x=(1/4) where it is equal to 0. Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point f(1/4)=g(1/4)+h(1/4)=0+4 Hence answer is 4 _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 08:53 45 This post received KUDOS Expert's post 13 This post was BOOKMARKED That's right! I can count on you guys to always give the correct answer and some innovative ways of solving... Let me give you the method I would use to solve it too... Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend... So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4 So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible? Attachment: Ques.jpg [ 4.56 KiB | Viewed 14954 times ] Can I say that minimum total distance will be covered from point 1/4? Attachment: Ques1.jpg [ 3.43 KiB | Viewed 14937 times ] The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7450\nLocation: Pune, India\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n30 Oct 2010, 08:57\n5\nKUDOS\nExpert's post\n2\nThis post was\nBOOKMARKED\nIn other words, as my mentor says, assume there is one guy on point 3, one on point -1 and 4 guys on point 1/4. If they have to meet up, but cover minimum distance, they should meet at point 1/4. What happens when there are 4 such points? Lets add another term (2x - 3) to f(x)...\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 11 Jul 2010 Posts: 224 Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 09:29 would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 16:17 3 This post received KUDOS Expert's post 3 This post was BOOKMARKED gmat1011 wrote: would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Yes, that is right! The answer still remains 1/4. Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4. Attachment: Ques.jpg [ 5.2 KiB | Viewed 14965 times ] Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6| f(x) = |2x - 3| + |4x + 7| etc _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nCEO\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2783\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n09 Nov 2010, 23:23\nGreat Method Karishma !!\n\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\nhttp://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nVP\nStatus: There is always something new !!\nAffiliations: PMI,QAI Global,eXampleCG\nJoined: 08 May 2009\nPosts: 1326\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n14 Jun 2011, 02:14\nfor x = 0, f(x) = 5. POE options C,D and E.\nfor x = 1/4, f(x) = 4.\nB it is.\n_________________\n\nVisit -- http://www.sustainable-sphere.com/\nPromote Green Business,Sustainable Living and Green Earth !!\n\nManager\nJoined: 19 Apr 2011\nPosts: 109\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n14 Jun 2011, 03:47\nVery good question thanks for posting\nIntern\nJoined: 24 Jul 2011\nPosts: 13\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n15 Mar 2012, 11:07\nVeritasPrepKarishma wrote:\ngmat1011 wrote:\n\nThe 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance\n\nYes, that is right! The answer still remains 1/4.\nSince |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4.\nAttachment:\nQues.jpg\n\nTry some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6|\nf(x) = |2x - 3| + |4x + 7| etc\n\nKarishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?\nIn example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?\n\nPlease correct me if I am wrong.\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7450\nLocation: Pune, India\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n15 Mar 2012, 11:24\n6\nKUDOS\nExpert's post\n2\nThis post was\nBOOKMARKED\nKarishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?\nIn example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?\n\nPlease correct me if I am wrong.\n\nForget these numbers. Think logically.\n\nMy house is 10 miles away from your house. If we have to meet up, how much distance do we need to cover together? In any case, we need to cover 10 miles together at least, right? Either you come down to my place (you cover 10 miles) or I come down to yours (I cover 10 miles) or we meet mid way (10 miles covered together) or we meet up at a nice coffee place 2 miles further down from my house in the opposite direction in which case we will need to cover more than 10 miles (i.e. we cover 2 + 12 = 14 miles)\n\nNow say, another friend is at my place. In which case will people cover minimum distance together? If two of us come down to your place, we cover 10+10 = 20 miles together but if you come down to our place, you cover only 10 miles. If instead, we meet midway, we cover 5+5 and you cover 5 miles so in all 15 miles. So less number of people should travel the entire distance.\nIf there are 4 people at point A and 2 at point B, minimum distance will be covered if people at point B travel to point A. So people at 3/2 should come down to 1/4.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 15 Mar 2012, 11:33 5 This post received KUDOS Expert's post 2 This post was BOOKMARKED ficklehead wrote: In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ? Please correct me if I am wrong. x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together. If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11. To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nIntern\nJoined: 24 Jul 2011\nPosts: 13\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n15 Mar 2012, 14:48\nVeritasPrepKarishma wrote:\nIn example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?\n\nPlease correct me if I am wrong.\n\nx is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance.\nDistance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together.\nIf -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet.\nIf they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11.\n\nTo check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11\nput x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11\nput x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11\n\nThanks Karishma for this detailed explanation.\nI got it now.\nI was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.\n\nTo seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7450\nLocation: Pune, India\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n15 Mar 2012, 22:03\n5\nKUDOS\nExpert's post\nI was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.\n\nTo seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?\n\nLook at the diagram below. Make a number line in such questions.\nAttachment:\n\nQues3.jpg [ 5.1 KiB | Viewed 13919 times ]\n\nFor x and y to meet, they have to cover a distance of 9 together. For p and q to meet, they have to cover a distance of 2 together. They can meet anywhere between -1 and 1 and they will cover a total distance of 11 only. So x can take any value -1 < x < 1 and the value of the expression will be 11.\n\n|2x-3|+|4x+7| = 2|x-3/2| + 4|x+7/4|\n\nAttachment:\n\nQues4.jpg [ 5.35 KiB | Viewed 13934 times ]\n\nThere are 4 people at -7/4 and 2 people at 3/2. Distance between the two points is 7/4 + 3/2 = 13/4\nFor these people to meet covering the minimum distance, the 2 people X and Y should travel to point -7/4. (Make minimum people travel). So minimum distance that needs to be covered = 2*13/4 = 13/2 (because 2 people travel 13/4 each) which is the minimum value of the expression. The value of x when the expression takes minimum value is -7/4.\nCheck by putting x = -7/4. You get |2x-3|+|4x+7| = 13/2\n\nAlso see that when you put x = 0 or 3/2 etc, the value of the expression is higher.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 09:01 Thanks a lot, Karishma. Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 19:48 I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : |x+6|-|x-1| ? Manager Joined: 28 Jul 2011 Posts: 238 Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 14:33 I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 21:07 kuttingchai wrote: I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Put x = 1/4 and the minimum value you will get is 4. How did you get x = 1/2? I would suggest you to check out one of the approaches mentioned above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7450\nLocation: Pune, India\nRe: Question of the Day - II\u00a0[#permalink]\n\n### Show Tags\n\n20 Mar 2012, 21:17\n5\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nI am wondering how can this method be used in questions where there are negative between terms :\nEx: minimum value of : |x+6|-|x-1| ?\n\nYou can do it with a negative sign too.\nYou want to find the minimum value of (distance from -6) - (distance from 1)\n\nMake a number line with -6 and 1 on it.\n(-6)..........................(1)\n\nThink of a point in the center of -6 and 1. Its distance from -6 is equal to distance from 1 and hence (distance from -6) - (distance from 1) = 0 .\n\nWhat if instead, the point x is at -6? Distance from -6 is 0 and distance from 1 is 7 so (distance from -6) - (distance from 1) = 0 - 7 = -7\n\nIf you keep moving to the left, (distance from -6) - (distance from 1) will remain -7 so the minimum value is -7.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for \\$199\n\nVeritas Prep Reviews\n\nRe: Question of the Day - II \u00a0 [#permalink] 20 Mar 2012, 21:17\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0\u00a0\u00a03\u00a0\u00a0\u00a04\u00a0 \u00a0 Next \u00a0[ 68 posts ]\n\nSimilar topics Replies Last post\nSimilar\nTopics:\n3 If 3^x = 81, then 3^(x+3)*4^(x+1) = 4 03 Jun 2017, 19:10\n15 If x<0 What is the value of [(x-3)^4]^1/4 + (-x|x|)^1/2 A. -3 B. x+3 16 23 Jun 2017, 23:28\n2 If f(x)=4x\u22121 and g(x)=2x+3 for all integers, which of the 2 04 Feb 2016, 21:14\n4 If x = -1, then -(x^4 + x^3 + x^2 + x) = 7 08 Feb 2014, 03:16\n1 If x = -1, then (x^4 - x^3 + x^2)/(x - 1) = 3 28 Oct 2015, 07:51\nDisplay posts from previous: Sort by", "date": "2017-06-29 04:25:53", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-f-x-4x-1-x-3-x-1-question-of-the-day-ii-103935.html", "openwebmath_score": 0.7273833751678467, "openwebmath_perplexity": 1845.2245836373818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8539127548105611, "lm_q1q2_score": 0.8539127548105611}}
{"url": "https://math.stackexchange.com/questions/2582606/computing-int-limits-0-infty-x-left-lfloor-frac1x-right-rfloor-dx", "text": "# Computing $\\int\\limits_0^\\infty x \\left \\lfloor{\\frac1x}\\right \\rfloor \\, dx$\n\nThis is an integral I computed but can't find the result online or on wolfram. So here's a proof sketch, please indulge this sanity check:\n\n$$\\int_0^\\infty x \\left \\lfloor{\\frac1x}\\right \\rfloor \\ dx = \\int_0^1 x \\left \\lfloor{\\frac1x}\\right \\rfloor \\ dx$$ $$= \\sum_{n=1}^\\infty \\int_{1/(n+1)}^{1/n} nx \\ dx =\\sum_{n=1}^\\infty\\frac n2 \\left(\\frac{1}{n^2} - \\frac{1}{(n+1)^2}\\right)$$ $$= \\sum_{n=1}^\\infty\\frac n2 \\left(\\frac{2n+1}{n^2(n+1)^2}\\right)$$ $$= \\sum_{n=1}^\\infty\\frac{1}{(n+1)^2} + \\frac12 \\sum_{n=1}^\\infty \\frac{1}{n(n+1)^2}$$ $$= \\frac{\\pi^2}{6} -1 + \\frac12\\left(\\sum_{n=1}^\\infty \\frac1n - \\frac{1}{n+1} - \\frac{1}{(n+1)^2}\\right)$$ $$=\\frac{\\pi^2}{6} -1 + \\frac12\\left(\\sum_{n=1}^\\infty \\frac1n - \\frac{1}{n+1}\\right) -\\frac12\\left(\\sum_{n=1}^\\infty\\frac{1}{(n+1)^2}\\right)$$ $$= \\left(\\frac{\\pi^2}{6} -1\\right) + \\left(\\frac12\\cdot 1\\right) - \\frac12\\left(\\frac{\\pi^2}{6} -1\\right)$$ $$= \\frac{\\pi^2}{12}.$$\n\nBasically, I used the Basel sum several times, and the fifth line follows from a partial sum decomposition. The seventh follows from the known result for the Basel sum, as well as the fact that the first series in the 6th line telescopes.\n\nI hope this is all correct.\n\n\u2022 @XanderHenderson For $x>1$, floor of $\\frac{1}{x}$ is $0$ \u2013\u00a0BallBoy Dec 28 '17 at 3:34\n\u2022 @XanderHenderson What you wrote is certainly not true, because you forgot the $x$ term. However, what Y. Forman says is correct, and I should've been more explicit there. \u2013\u00a0David Bowman Dec 28 '17 at 3:37\n\u2022 Oi... derp. Sorry for being dyslexic. I missed the $x$. \u2013\u00a0Xander Henderson Dec 28 '17 at 3:39\n\u2022 $$\\int_0^1 x \\lfloor 1/x \\rfloor dx = \\int_1^\\infty \\frac{1}{t} \\lfloor t \\rfloor \\frac{dt}{t^2}= \\sum_{n=1}^\\infty \\int_n^\\infty t^{-3}dt = \\sum_{n=1}^\\infty \\frac{n^{-2}}{2} = \\frac{\\zeta(2)}{2}$$ \u2013\u00a0reuns Dec 28 '17 at 3:43\n\u2022 @reuns Nice, better post it as answer! \u2013\u00a0samjoe Dec 28 '17 at 4:40\n\nUse\n\n$$n\\left(\\frac1{n^2}-\\frac1{(n+1)^2}\\right)=\\frac n{n^2}-\\frac{n+1-1}{(n+1)^2}=\\frac1n-\\frac1{n+1}+\\frac1{(n+1)^2}.$$\n\nThe first two terms do telescope and the Basel series remains.\n\n\u2022 @stressedout: of course, but much simpler. And also simpler than yours, which is also essentially the OP's. \u2013\u00a0Yves Daoust Sep 5 '18 at 6:28\n\u2022 @stressedout: your solution isn't different, just longer. A one-liner is simpler. \u2013\u00a0Yves Daoust Sep 5 '18 at 6:38\n\nAlternatively, one may follow the same line of thought and use summation by parts formula to calculate the infinite sum. Similarly, $$\\int_0^{+\\infty} x\\lfloor \\frac{1}{x}\\rfloor dx =\\sum_{n=1}^{\\infty}\\frac{n}{2}\\left(\\frac{1}{n^2}-\\frac{1}{(n+1)^2}\\right)=-\\frac{1}{2}\\sum_{n=1}^{\\infty}f_n(g_{n+1}-g_n)$$\n\nwhere $f_n = n$ and $g_n = 1/n^2$. Now, summation by parts for the last expression gives:\n\n$$-\\frac{1}{2}\\sum_{n=1}^{\\infty}f_n(g_{n+1}-g_n) = -\\frac{1}{2}\\left(\\ \\lim_{n\\to\\infty}\\frac{n}{(n+1)^2} -1 -\\sum_{n=2}^{\\infty}\\frac{1}{n^2}\\right)$$\n\nBut it is well-known that $$\\sum_{n=1}^{\\infty}\\frac{1}{n^2}=\\frac{\\pi^2}{6}$$\n\nHence,\n\n$$-\\frac{1}{2}\\sum_{n=1}^{\\infty}f_n(g_{n+1}-g_n) = -\\frac{1}{2}\\left(\\ \\lim_{n\\to\\infty}\\frac{n}{(n+1)^2} -1 - (\\frac{\\pi^2}{6}-1)\\right) = \\frac{\\pi^2}{12}$$", "date": "2019-12-11 06:03:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2582606/computing-int-limits-0-infty-x-left-lfloor-frac1x-right-rfloor-dx", "openwebmath_score": 0.9329135417938232, "openwebmath_perplexity": 997.5000517186052, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989013056721729, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.853905582280985}}
{"url": "https://math.stackexchange.com/questions/3113510/prove-that-each-integer-n-%E2%89%A5-12-is-a-sum-of-4s-and-5s-using-strong-induction", "text": "# Prove that each integer n \u2265 12 is a sum of 4's and 5's using strong induction\n\nSo I've been given the following problem: Prove that each integer n \u2265 12 is a sum of 4's and 5's What I have so far: (Basis):\n\nn \u2265 12 Therefore,\n\n12 \u2264 4(x) + 5(y)\n\nx = 3 | y = 0\n\n12 \u2264 4(3) + 5(0)\n\n12 \u2264 12 = Correct\n\nHowever, what I don't understand is how would I use the x & y variable in induction step. Where exactly would I place these? and how would I go on to solve this?\n\nThe trick to these sorts of problems is to realize that if we can find four consecutive integers ($$4$$ is the smallest of our numbers we're taking a combination of) that can be represented as a non-negative sum of the fours and fives.\n\nFor example $$12 = 4(3) + 5(0)$$\n\n$$13 = 4(2) + 5(1)$$ $$14 = 4(1)+ 5(2)$$ $$15 = 4(0) + 5(3).$$\n\nWith this information is it clear how you could represent 16? Just take the solution from the 12 case, and add 4! (so increase $$x$$ by $$1$$.)\n\nHere's the formal argument. Let $$P(n)$$ be the open sentence \"$$n$$ can be written as a non-negative combination of $$4$$ and $$5$$\". By what we've shown above, we know that $$P(k)$$ is true for $$k = 12,13,14,15.$$ We wish to prove that $$P(k+1)$$ is true. Our strong inductive hypothesis is to suppose that for some $$k \\in \\mathbb{Z}$$ that for every $$i$$ with $$12\\leq i\\leq k$$ that $$P(k)$$ is true, and we need to prove that $$P(k+1)$$ is true.\n\nIf $$k = 12,13$$ or $$14$$, we've already seen that $$p(k+1)$$ is also true, so suppose that $$k \\geq 15$$. Then we know that $$12 \\leq k-3\\leq k$$ so then by our strong inductive hypothesis, there exists $$x,y \\in \\mathbb{Z}$$ such that $$k-3 = 4(x) + 5(y)$$. Then adding $$4$$ to both sides gives that $$k+1 = 4(x+1)+5(y)$$ so that $$P(k+1)$$ is true. Thus by the principle of mathematical induction, $$P(n)$$ must be true for all $$n \\geq 12$$.\n\n\u2022 Wow this is very detailed and helped me understand it more, thank you! \u2013\u00a0MathNoob Feb 15 '19 at 4:15\n\u2022 No problem! I'm glad this helped! \u2013\u00a0JonHales Feb 15 '19 at 4:17\n\n$$4x+5y=\\underbrace{4(x-1)+5y}_{n-4}+4$$\n\nSo, if $$n-4$$ is expessible so will be $$n$$\n\n$$12=4\\cdot3$$\n\n$$13=2\\cdot4+5$$\n\n$$14=4+2\\cdot5$$\n\n$$15=3\\cdot5$$", "date": "2020-04-01 00:03:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3113510/prove-that-each-integer-n-%E2%89%A5-12-is-a-sum-of-4s-and-5s-using-strong-induction", "openwebmath_score": 0.828149139881134, "openwebmath_perplexity": 89.83641628509169, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232950125849, "lm_q2_score": 0.8688267881258485, "lm_q1q2_score": 0.8539032067010474}}
{"url": "https://math.stackexchange.com/questions/2598977/finding-dimension-of-a-kernel-from-a-linear-transformation", "text": "# Finding dimension of a kernel from a linear transformation.\n\nI'm trying to solve an exercise where I'm asked to find this linear transformation's kernel and its dimension:\n\n$$f(x_1,x_2,x_3,x_4)=(x_1-x_2,2x_3+x_4)$$\n\nSo, from the definition of kernel and converting to parametric equations, I found $x_1=x_2=\\lambda$, $x_3=-\\frac{1}{2}\\mu$ and $x_4=\\mu$. Therefore, I assumed: $$\\ker(f)=\\{(\\lambda,\\lambda,-\\frac{1}{2}\\mu,\\mu): \\lambda,\\mu \\in \\mathbb{R}\\}$$\n\nAnd since the kernel can be described with only one vector (with two parameters), I thought $\\dim(\\ker(f))=1$, but my textbook says it's actually equal to $2$. Where is my reasoning flawed?\n\n\u2022 A basis for the kernel is $\\{(1,1,0,0), (0,0,-\\frac{1}{2}, 1) \\}$ \u2013\u00a0leibnewtz Jan 9 '18 at 23:20\n\u2022 @leibnewtz Hm, so that means I can only 'fix' one parameter at once? I thought it'd be possible to establish something like $\\lambda=1$, $\\mu=2$ so that a basis for the kernel is $\\{(1,1,-1,2)\\}$. Why isn't that possible? \u2013\u00a0Manuel Jan 9 '18 at 23:22\n\u2022 Of course you can do what you say...but that way you'll only get one vector, and you already know you need two lin. ind. vectors to have a basis... \u2013\u00a0DonAntonio Jan 9 '18 at 23:30\n\u2022 Since $f:\\mathbb R^4\\to\\mathbb R^2$, its image is at most two-dimensional, so you know that its kernel must be at least two-dimensional. \u2013\u00a0amd Jan 9 '18 at 23:35\n\u2022 @Manuel Please, if you are ok, you can accept the answer and set it as solved. Thanks! \u2013\u00a0user Feb 4 '18 at 0:09\n\nIt is all correct, the dimension of the ker(f) is 2 because you have 2 free parameters that define it $(\\lambda,\\mu)$.\n\u2022 @Manuel They are equivalent concepts. As noted in the comments you can set $\\lambda=1$ and $\\mu=0$ and define a basis vector, then $\\lambda=0$ and $\\mu=1$ and define a second vector linearly independent from the first, thus the dimension is 2. This is true for any number of free parameter (EG a line or a plane in $\\mathbb{R^3}$). \u2013\u00a0user Jan 9 '18 at 23:25\n\u2022 @Manuel There are two vectors in your description of the kernel: $\\left(\\lambda,\\lambda,-\\frac12\\mu,\\mu\\right)=\\lambda(1,1,0,0)+\\mu\\left(0,0,-\\frac12,1\\right)$; every vector in the kernel is a linear combination of these two linearly-independent vectors. \u2013\u00a0amd Jan 9 '18 at 23:36", "date": "2019-10-20 03:40:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2598977/finding-dimension-of-a-kernel-from-a-linear-transformation", "openwebmath_score": 0.883671224117279, "openwebmath_perplexity": 250.62895218820742, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232924970204, "lm_q2_score": 0.8688267796346598, "lm_q1q2_score": 0.8539031961701196}}
{"url": "https://mathoverflow.net/questions/252313/is-every-group-an-ideal-class-group-of-a-number-field/299691", "text": "Is every group an ideal class group of a number field?\n\nThe inverse Galois problem asks whether every finite group appears as the Galois group of some finite extension of $\\mathbb Q$. I was wondering to what extent the analogous problem for ideal class groups has been investigated. More precisely, consider the following question:\n\nIs every finite abelian group the ideal class group of some number field (finite extension of $\\mathbb Q$)?\n\nI'd be interested to hear about any partial results, as I suppose this question is still open. I'd be also interested in any results about a weaker problem:\n\nIs every positive integer the ideal class number of some number field?\n\nAgain, any reference, even to a partial result, will be appreciated.\n\n\u2022 Not an answer to your question, but are you aware of Claborn's theorem that says every finite abelian group is the class group of a Dedekind domain (though not necessarily, as far as one can tell from Claborn's work, of a ring of integers)? \u2013\u00a0Steven Landsburg Oct 16 '16 at 19:52\n\u2022 @StevenLandsburg I was not aware of that result, since, to be honest, I am not so interested in general Dedekind's domain. I will definitely take a look at this result though. \u2013\u00a0Wojowu Oct 16 '16 at 19:55\n\u2022 duplicate: math.stackexchange.com/questions/10949/\u2026 \u2013\u00a0Franz Lemmermeyer Oct 16 '16 at 20:36\n\u2022 @FranzLemmermeyer Thanks, I haven't seen that question when searching about the topic. I suppose at this point my question can be closed. \u2013\u00a0Wojowu Oct 16 '16 at 20:39\n\u2022 The system does not allow closing a question on one site as a duplicate of a question on a different site. One option, Wojowu, is for you to post an answer here, summarizing what's over there, and linking to it, and then accept your answer. \u2013\u00a0Gerry Myerson Oct 16 '16 at 22:04\n\nThe recent paper by Homlin, Jones, Kurlberg, McLeman, and Petersen (Experimental Math., to appear) is devoted to these questions especially in the context of imaginary quadratic fields. One should expect that every natural number arises as the class number of an imaginary quadratic field. Refining an earlier conjecture of Soundararajan, in this paper a precise asymptotic is formulated for the number of imaginary quadratic fields with a given class number. They also formulate conjectures on what kind of groups can arise as class groups of imaginary quadratic fields -- for example, one expects that for any odd prime $p$, $({\\Bbb Z}/p{\\Bbb Z})^3$ is not the class group of any imaginary quadratic field. The paper gives much data on such questions together with many related references.\n\n\u2022 I think $(\\dfrac{\\Bbb Z}{2\\Bbb Z})^3$ can occur as a class group for $D=-420$, and maybe it is better to restrict $p$ to the odd primes, i.e: $p\\neq2$. \u2013\u00a0Davood KHAJEHPOUR May 8 '18 at 5:51\n\u2022 Thanks for the reference! I will look into it as soon as I can. \u2013\u00a0Wojowu May 8 '18 at 7:26\n\nJust for the sake of completeness, here I am posting the answer of Pete Clark which was posted in the math.SE question. I am making this post CW so as not to create the unseemly impression that I am gaining any undeserved reputation from this. Here it is:\n\nVirtually nothing is known about the question of which abelian groups can be the ideal class group of (the full ring of integers of) some number field. So far as I know, it is a plausible conjecture that all finite abelian groups (up to isomorphism, of course) occur in this way. Conjectures and heuristics in this vein have been made, but unfortunately for me I'm not so familiar with them.\n\nThe situation for imaginary quadratic fields is different. Here there is an absolute bound on the size of an integer $$k$$ such that the class group of an imaginary quadratic field can be isomorphic to $$(\\mathbb{Z}/2\\mathbb{Z})^k$$. Conditionally on the Generalized Riemann Hypothesis, the largest such $$k$$ is $$4$$. This has do to with idoneal numbers, of which the following paper provides a very fine survey:\n\nhttp://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf\n\nActually the truth is slightly stronger: let $$H_D$$ be the class group of the imaginary quadratic field $$\\mathbb{Q}(\\sqrt{-D})$$. Then, as $$D$$ tends to negative infinity through squarefree numbers, the size of $$2H_D$$ (the image of multiplication by $$2$$) tends to infinity. See for instance\n\nhttp://arxiv.org/PS_cache/arxiv/pdf/0811/0811.0358v2.pdf\n\nfor some recent explicit bounds on this.\n\nFor $S$-class groups: Perret, Marc On the ideal class group problem for global fields. Zbl 0933.11053 J. Number Theory 77, No. 1, 27-35 (1999). https://zbmath.org/?any=&au=perret&ti=On+the+Ideal+Class+Group+Problem+for+Global+Fields&so=&ab=&cc=&ut=&an=&la=&py=&rv=&sw=&dm=", "date": "2021-06-14 08:52:49", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/252313/is-every-group-an-ideal-class-group-of-a-number-field/299691", "openwebmath_score": 0.7085285782814026, "openwebmath_perplexity": 267.0133536510818, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232930001334, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.853903189930967}}
{"url": "https://math.stackexchange.com/questions/452806/prove-forall-n-2-exists-p-in-bbbp-n-p-n/452816", "text": "Prove $\\forall n > 2, \\ \\exists p\\in \\Bbb{P} : n < p < n!$\n\nI need to prove that:\n\n$$(1) \\ \\forall n\\in\\Bbb{N}_{\\ge2}, \\ \\exists p\\in \\Bbb{P} : n < p < n!$$\n\nI already know how to prove the $n < p$ part; it directly follows from the proof that there is no largest prime. However, I am stumped on the $p < n!$ part.\n\nOne idea I had for showing this is as follows: we know that $(2) \\ \\forall n \\in \\Bbb{N}_{\\ge2},\\ \\exists m\\in\\Bbb{N} : n!=2m$. By testing a few values, I conjectured that $(3) \\ \\forall n\\in\\Bbb{N}_{\\ge2}, \\ \\exists p\\in \\Bbb{P} : n < p < 2n$. If (3) could be shown, it would be simple to prove (1), but there does not seem to be no easy way to prove (3), if it's even correct.\n\n\u2022 \"that there is always a prime number between $n$ and $2n$\" The names Bertrand and Chebyshev spring to mind. \u2013\u00a0Daniel Fischer Jul 26 '13 at 14:32\n\u2022 Oh, wow! I didn't know that! I was just trying to find a 'lazy' solution. But it's nice to know. :) \u2013\u00a0dotslash Jul 26 '13 at 14:33\n\u2022 There's also an easy direct way. Do you know Euclid's proof that there are infinitely many primes? \u2013\u00a0Daniel Fischer Jul 26 '13 at 14:35\n\u2022 Yes, I do. But not sure how to use it here. \u2013\u00a0dotslash Jul 26 '13 at 14:36\n\u2022 I was thinking of what Goos answered. That relieves of the burden of proving that the product of primes $< n$ is $> n$ (although that's not very hard). \u2013\u00a0Daniel Fischer Jul 26 '13 at 14:40\n\nJust consider $n! - 1$. Clearly this is between $n$ and $n!$ as long as there is anything between $n$ and $n!$, and it isn't divisible by any prime $p \\le n$. Thus it contains a prime factor bigger than $n$ and smaller than $n!$.\n\n\u2022 Correct me if I'm wrong, but $n! -1$ isn't divisible by any prime less than $n$ because all the primes are already included in $n!$ ? \u2013\u00a0dotslash Jul 26 '13 at 14:53\n\u2022 @dotslash For any prime $p \\le n$, $p$ is contained in the product $n!$, so $n! - 1$ is one less than a multiple of $p$. \u2013\u00a06005 Jul 26 '13 at 14:55\n\u2022 Precisely my thinking! Thank you for the concise and beautiful answer! \u2013\u00a0dotslash Jul 26 '13 at 14:59\n\nYou can obviously proceed by stating the Bertrand's postulate. However, I think the question is asking you to solve it a bit differently.\n\nLet us solve it using a method similar to that used by Euclid when he tried to prove that there are infinite number of primes. Let $p_{1},p_{2},\\cdots ,p_{k}$ be all the primes less than or equal to $n$. Obviously, $k<n$.Let, $P=p_{1}\\times p_{2}\\times\\cdots \\times p_{k}+1$.\n\nNotice that $P$ is not divisible by any of the given primes. Hence $P$ is either a prime or is divisible by a prime $> n$.\n\nIt can be easily seen that $$P<n!$$ Furthermore, $P$ needs to be greater than $n$, otherwise we would find another prime (other than $p_{1},p_{2},\\cdots,p_{k}$) which would lead to a contradiction.\n\n\u2022 I think you mean that $p_1, p_2, \\cdots , p_k$ are less than or equal to $n$. You also have to be careful when claiming that $P < n!$; this isn't true in the case $n = 3$. \u2013\u00a06005 Jul 26 '13 at 14:42\n\u2022 You need it to not be divisible by $n$ if $n$ is prime, so you need to include primes less than or equal to $n$ in the product. Thus you get $3*2 + 1 = 7$, not less than $3! = 6$. \u2013\u00a06005 Jul 26 '13 at 14:47\n\u2022 Sorry!My bad,I guess the case n=3 has to be checked manually.Other than that I think there are no exceptions. \u2013\u00a0Shaswata Jul 26 '13 at 14:49\n\nWhat you speculate on is correct, there is always a prime between $n$ and $2n$. This is known as Bertrand's postulate and is really a theorem. But for between $n$ and $n!$ you can modify the Euclid proof for a much easier approach. It is true for $3, 3 \\lt 5 \\lt 6$. For $n \\gt 3,$ take the product of all primes less than $n$ and add $1$. This is less than $n!$ because it has only one factor of $2$, not the three coming from $2$ and $4$. Then it is either prime or composite ...\n\n\u2022 Great! But your explanation of \"only one factor of $2$\" confuses me. Isn't it simpler to say that $n!$ contains all the integers between $2$ and $n$, including those that are prime. Hence, $n! > p$? \u2013\u00a0dotslash Jul 26 '13 at 14:46\n\u2022 But we need a prime greater than $n$ and $n!$ doesn't contain that as a factor. I wanted to justify that the product of all the primes less than $n$ is strictly less than $n!$ so that I could add one and still be less. Identifying a factor of $n!$ that is not part of the product of primes was my way of doing that. That is also why I did $3$ separately-it is not true in that case as you can see. If we just take all the primes less than or equal to $3$, the product plus $1$ is $7$ and we have not found a prime between $3$ and $3!$ \u2013\u00a0Ross Millikan Jul 26 '13 at 15:10\n\u2022 I follow you now. Thanks for the answer! \u2013\u00a0dotslash Jul 26 '13 at 15:42", "date": "2019-05-25 06:58:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/452806/prove-forall-n-2-exists-p-in-bbbp-n-p-n/452816", "openwebmath_score": 0.8525522351264954, "openwebmath_perplexity": 104.26105333668434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232909876816, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.8539031881824951}}
{"url": "http://math.stackexchange.com/questions/95799/why-is-gcda-b-gcdb-r-when-a-qb-r", "text": "# Why is $\\gcd(a,b)=\\gcd(b,r)$ when $a = qb + r$?\n\nGiven: $a = qb + r$\n\nThen it holds that $\\gcd(a,b)=\\gcd(b,r)$. That doesn't sound logical to me. Why is this so?\n\nAddendum by LePressentiment on 11/29/2013: (in the interest of http://meta.math.stackexchange.com/a/4110/53259 and averting a duplicate)\n\nWhat's the intuition behind this result? I only recognise the proof and examples solely due to algebraic properties and formal definitions; I'd like to apprehend the result naturally.\n\n-\n$a$, $b$, $r$ are integers? (If yes, tagging the question elementary-number-theory would be suitable.) Or are you looking for some greater generality, e.g. they are elements of some type of commutative ring? If yes, you should say so and you should also mention, what are the assumptions about this ring. \u2013\u00a0 Martin Sleziak Jan 2 '12 at 13:12\nIf you're asking about integers, then this question is very similar to yours: math.stackexchange.com/questions/59147/\u2026 \u2013\u00a0 Martin Sleziak Jan 2 '12 at 13:16\nsee page 13 here www-groups.dcs.st-and.ac.uk/~martyn/teaching/1003/\u2026 \u2013\u00a0 Bhargav Jan 2 '12 at 13:26\n@Martin The result does not depend upon the underlying ring (assuming $\\rm\\:gcd(a,b)\\:$ exists). \u2013\u00a0 Bill Dubuque Jan 2 '12 at 18:19\n\nIf $d$ is a divisor of $a$ and of $b$, then \\begin{align} a & = dn, \\\\ b & = dm. \\end{align} So $$a-b= dn-dm=d(n-m)= (d\\cdot\\text{something}).$$ So $d$ is a divisor of $a-b$.\n\nThus: All divisors that $a$ and $b$ have in common are divisors of $a-b$.\n\nIf $d$ is a divisor of $a$ and of $a-b$, then \\begin{align} a & = dn, \\\\ a-b & = d\\ell. \\end{align} So $$b=a-(a-b)=dn-d\\ell=(d\\cdot\\text{something}).$$ So $d$ is a divisor of $b$.\n\nThus: All divisors that $a$ and $a-b$ have in common are divisors of $b$.\n\nTherefore, the set of all common divisors of $a$ and $b$ is the same as the set of all common divisors of $a$ and $a-b$.\n\nSubtracting one member of a pair from the other never alters the set of all common divisors; therefore it never alters the $\\gcd$.\n\n-\nVery clear answer! Little remark: $a-b \\neq r$, but $a-qb = r$, but all your explanation will work using that :). \u2013\u00a0 Kevin Jan 2 '12 at 21:17\n@Kevin Beware that this fails if the ring is not $\\rm\\:\\mathbb Z\\:$ since then the remainder $\\rm\\:r\\:$ generally cannot be obtained by repeatedly subtracting $\\rm\\:b\\:$ from $\\rm\\ q\\ b + r\\:,\\:$ as it can when $\\rm\\:q\\in\\mathbb N\\:.\\:$ In other words, in general Euclidean rings, e.g. the polynomial ring $\\rm\\:F[x]\\:$ over a field $\\rm\\:F\\:,\\:$ the Euclidean algorithm generally requires division with remainder, not simply iterated subtraction, in order to effect descent to a \"smaller\" remainder. So this ad-hoc special-case doesn't generally reveal the essence of the matter. \u2013\u00a0 Bill Dubuque Jan 2 '12 at 22:37\n@Kevin : Thanks; I've fixed the typo. \u2013\u00a0 Michael Hardy Jan 3 '12 at 0:52\nTo express Bill Dubuque's point in other way: Euclid's algorithm works for polynomials, but the argument I give in my answer doesn't work for polynomials unless you do some adaptations. \u2013\u00a0 Michael Hardy Jan 3 '12 at 0:53\n\nHINT $\\rm\\ \\$ If $\\rm\\ d\\ |\\ b\\$ then $\\rm\\ d\\ |\\ q\\ b + r\\ \\iff\\ d\\ |\\ r\\:.\\$ Therefore $\\rm\\ \\{b\\:,\\:q\\ b+r\\}\\$ and $\\rm\\ \\{b\\:,\\: r\\}\\$ have the same set of common divisors $\\rm\\:d\\:,\\:$ hence they have the same greatest common divisor.\n\nModly: $\\$ if $\\rm\\ b\\equiv 0\\$ then $\\rm\\ q\\ b+r\\equiv 0\\: \\iff\\: r\\equiv 0\\ \\pmod{d}$\n\nNOTE $\\$ The result holds true because $\\rm\\:\\mathbb Z\\:$ forms a subring of its fraction field $\\rm\\:\\mathbb Q\\:.\\:$ More generally, given any subring $\\rm\\:Z\\:$ of a field $\\rm\\:F\\:$ we define divisibility relative to $\\rm\\ Z\\$ by $\\rm\\ x\\ |\\ y\\ \\iff\\ y/x\\in Z\\:.\\:$ Then the above proof still works, since if $\\rm\\ q,\\ b/d\\ \\in Z\\$ then $\\rm\\ q\\:(b/d) + r/d\\in Z\\ \\iff\\ r/d\\in Z\\:.\\:$ In other words, the usual divisibility laws follow from the fact that rings are closed under the operations of subtraction and multiplication; being so closed, $\\rm\\:Z\\:$ serves as a ring of \"integers\" for divisibility tests.\n\nFor example, to focus on the prime $2$ we can ignore all odd primes and define a divisibility relation so that $\\rm\\ m\\ |\\ n\\$ if the power of $2$ in $\\rm\\:m\\:$ is $\\le$ that in $\\rm\\:n\\:$ or, equivalently if $\\rm\\ n/m\\$ has odd denominator in lowest terms. The set of all such fractions forms a ring $\\rm\\:Z\\:$ of $2$-integral fractions. Moreover, this ring enjoys parity, so arguments based upon even/odd arithmetic go through. Similar ideas lead to powerful local-global techniques of reducing divisibility problems from complicated \"global\" rings to simpler \"local\" rings, where divisibility is decided by simply comparing powers of a prime.\n\n-\n\nYou can show that for any integer $d$, we have $d\\; |\\; a$ and $d\\; |\\; b$ if and only if $d\\; |\\; b$ and $d\\; |\\; r$. In other words, $a$ and $b$ have exactly the same common divisors as $b$ and $r$. Thus $\\gcd(a,b)$ is the same as $\\gcd(b,r)$.\n\n-\nexcellent answer, +1 for it. \u2013\u00a0 ncmathsadist Jan 2 '12 at 13:44\nThanks, but why do they have the same common divisors? \u2013\u00a0 Kevin Jan 2 '12 at 21:15\n@Kevin: If $d$ divides $a$ and $b$, then $d$ divides $-qb$ and thus divides the sum $a - qb = r$. This shows that any common divisor of $a$ and $b$ is a common divisor of $b$ and $r$. If $d$ divides $b$ and $r$, then $d$ divides $qb$ and thus divides the sum $qb + r$. This shows that any common divisor of $b$ and $r$ is a common divisor of $a$ and $b$. \u2013\u00a0 Mikko Korhonen Jan 2 '12 at 21:52\n\nSince set of common divisors of $a-b$ and $b$ coincides with the set of common divisors of $a$ and $b$ then $\\operatorname{gcd}(a,b)=\\operatorname{gcd}(a-b,b)$. If $a=qb+r$, where $b>0$ and $0\\leq r<b$, you can apply this equality $q$ times and obtain $\\operatorname{gcd}(a,b)=\\operatorname{gcd}(r,b)$\n\n-\nBut $a-b$ and $b$ don't have the same set of divisors. They have the same set of common divisors with $a$. \u2013\u00a0 Peter Taylor Jan 2 '12 at 14:29\nOk common divisors. I always miss some details. \u2013\u00a0 no identity Jan 2 '12 at 14:40\n@Norbert : what you must have meant is that $\\{a,b\\}$ and $\\{a-b,b\\}$ have the same set of common divisors. \u2013\u00a0 Michael Hardy Jan 2 '12 at 18:01\nYes, this is what I meant. \u2013\u00a0 no identity Jan 2 '12 at 18:17\nBeware that this \"repeated subtraction\" implementation of division with remainder does not generally yield the Euclidean algorithm in other domains, e.g. for polynomials. See my comment to Hardy's answer. \u2013\u00a0 Bill Dubuque Jan 2 '12 at 22:58\n\nLet $A$ be a commutative ring. For any $a_1,\\dots,a_n$ in $A$ let $(a_1,\\dots,a_n)$ the ideal generated by the $a_i$.\n\nThen, for any $q,b,r$ in $A$, we have $$(qb+r,b)=(b,r).$$ Indeed, $qb+r$ is in $(b,r)$, and $r$ is in $(qb+r,b)$.\n\nEDIT. Dear Kevin: Your question, I think, would be better understood if put in a wider context, involving rings and ideals. The most basic fact behind the question is, I believe, the fact that, in any commutative ring, the elements $qb+r$ and $b$ generate the same ideal as the elements $b$ and $r$. If you make additional hypothesis, this fact can be interpreted in terms of divisibility. (See Bill's comment.) The simplest is to assume that your ring is a principal ideal domain.\n\nI could try to explain this in greater details, but many mathematicians much better than I have already done that. So, my advice would be to take a look at at least one of the many Algebra textbooks written by great mathematicians. Here are some of these books:\n\n-\nThis seems to implicitly assume some relation between gcds and ideals, e.g. for Bezout domains $\\rm\\ (a,b) = (\\gcd(a,b))\\:.$ \u2013\u00a0 Bill Dubuque Jan 2 '12 at 22:28\nDear @Bill: Thanks for your comment. I edited the answer. (Of course, I agree with you.) \u2013\u00a0 Pierre-Yves Gaillard Jan 3 '12 at 5:50\nDivisor theory gives one nice way to better understand the relations between ideals and gcds. For a brief overview see Friedemann Lucius, Rings with a theory of greatest common divisors and for a longer exposition see Olaf Neumann, Was sollen und was sind Divisoren? (What are divisors and what are they good for?), Math. Semesterber, 48, 2, 139-192 (2001). \u2013\u00a0 Bill Dubuque Jan 3 '12 at 6:38\n\nI'm going to use the notation $(a,b)$ for the GCD of $a$ and $b$.\n\nIf $d|a$ and $d | b$ then $d|(a,b)$, by the definition of GCD. (Well, by one common definition...if that's not the definition you learned, then you probably learned it as a theorem).\n\nSince $(a,b)|a$ and $(a,b)|b$, by the definition of $(a,b)$, it divides $a-qb$, so we have $(a,b)|r$. This gives us $(a,b)|b$ and $(a,b)|r$, hence $(a,b)|(b,r)$.\n\nNow let's go the other way. $(b,r)|b$ and $(b,r)|b$, both by definition, so it also divides $r+qb$, giving us $(b,r)|a$. That gives is $(b,r)|(a,b)$.\n\nFrom $(a,b)|(b,r)$ and $(b,r)|(a,b)$, we get $(a,b)=(b,r)$ or $(a,b)=-(b,r)$. The latter can be eliminated because GCD is by definition greater than 0.\n\n-\n\na,b q ,r are integers how can we say that HCF(a,b) = HCF (b,r) we can say common divisor of a and b = common divisor of b and r. for example hcf of 4 and 2 = 2 4= 2x2 + o but here HCF of 4 and2 =2 but HCF of 2 and 0 =1 so how can we say that HCF(a,b) = HCF (b,r)\n\n-\n\nTheorem: Let $a > b > 0$ with $a = bq + r$, $0\\leq{}r<b$ then $\\gcd(a;b) = \\gcd(b;r)$.\n\nProof: Need to show that $C(a;b) = C(b;r)$ for then the result will hold. To show that the two sets are equal requires showing that $C(a;b) \\subseteq C(b;r)$ and that $C(b;r) \\subseteq C(a;b)$:\n\nLet $y \\in C(a;b)$ thus $y|a$ and $y|b$,\nthen $y|[a+(-q)b]$,\nand so $y|r$, since $r=a-bq$,\nbut since $y|b$ is also true, we now have that $y \\in C(b;r)$, finally this means that $C(a;b) \\subseteq C(b;r)$.\n\nNow let $y \\in C(b;r)$ thus $y|b$ and $y|r$,\nthen $y|[(q)b+r]$,\nand so $y|a$, since $a=bq+r$,\nbut since $y|b$ is also true, we now have that $y \\in C(a;b)$, finally this means that $C(b;r) \\subseteq C(a;b)$.\n\nTherefore the required results have been proven.\n\nNote: Where $C(a;b)$ denotes the set of common divisors/factors of $a$ and $b$, that is: $C(a;b)=\\{y: y|a \\land y|b\\}$\n\n-", "date": "2014-09-17 11:29:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/95799/why-is-gcda-b-gcdb-r-when-a-qb-r", "openwebmath_score": 0.9606935381889343, "openwebmath_perplexity": 242.20783499029983, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232884721166, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8539031843278371}}
{"url": "https://specialsports.info/benzamide-molecular-fbulsmr/viewtopic.php?af9d97=table-math-example", "text": "Here is a very simple table showing data lined up in columns. Example. We will look at how to do this using an example and then also look at how to answer questions about a data set using a two way table. 3 Some Table Examples Example-1: A table with combined columns is given below. Practice your multiplication tables. Syntax of this Tableau ABS Function is: ABS(number) To demonstrate these Tableau math functions, we use Calculated Fields. Add rows, call Compute and Merge, and set PrimaryKey. Tree Diagrams in Math: Definition & Examples 4:43 Truth Table: Definition, Rules & Examples 6:08 6:52 Notice that I include the table in a center'' environment to display it properly. The table results can usually be used to plot results on a graph. Illustrated definition of Table: Information (such as numbers and descriptions) arranged in rows and columns. Size of the preallocated table, specified as a two-element numeric vector. According to the survey, what is the probability that a male likes rap music? The game element in the times tables games make it even more fun learn. A truth table is a handy little logical device that shows up not only in mathematics but also in Computer Science and Philosophy, making it an awesome interdisciplinary tool. \\multicolumn{num}{col}{text} command is used to combine the following num columns into a single column with their total width. What is. The results, separated by gender, are displayed above. This is an example of a blank math lesson plan where you can fill out the pertinent information in the blank space provided, such as the lesson title, unit title, lesson number, class level, duration of the lessons, number of classes, teacher, lesson objectives, standards, essential questions, and evidence of learning. Example Example The table shows the math achievement test scores for a random sample of n = 10 college freshmen, along with their final calculus grades. Notes. Table Data - Basic Example A total of 72 people participated in a survey about their music preferences. Answer: 0.9332 To find the answer using the Z-table, find where the row for 1.5 intersects with the column for 0.00; this value is 0.9332.The Z-table shows only \u201cless than\u201d probabilities so it gives you exactly what you need for this question. Sample Questions Header Block Open sample questions menu Math . Cap table math is confusing. Use these sample z-score math problems to help you learn the z-score formula. Loved by kids and parent worldwide. The idea is to convert the word-statement to a symbolic statement, then use logical equivalences as we did in the last example. There are four variables in the JavaScript block. 22 Examples of Mathematics in Everyday Life. The following table summarizes them: As math requires special environments, there are naturally the appropriate environment names you can use in the standard way. Example example the table shows the math achievement. Times Tables. You read it right; basic mathematical concepts are followed all the time. The title is created simply as another paragraph in the center environment, rather than as part of the table itself. Hope this helps! So now how many people can we fit? Logic. The following HTML example contains the HTML, CSS, and JavaScript necessary to build a dynamic HTML table. Learn the multiplication tables in an interactive way with the free math multiplication learning games for 2rd, 3th, 4th and 5th grade. \\begin{center} Numbers of Computers on Earth Sciences Network, By Type. Yes! Some of the times tables are illustrated below: Times Table \u2026 Consumer Math. For example, in the table shown below, Jacket is listed twice in column A. To complete a two way table for a set of data, you need to determine the variables of interest, their possible values, and then finally, the frequencies. But, maths is the universal language which is applied in almost every aspect of life. Go to First Question . Aligned to Common Core. The answer obtained in every step is a multiple of 2 and is known as multiplication fact. The opposite of a tautology is a contradiction or a fallacy, which is \"always false\". Tableau Math Functions. Skip to main content. Try the free Mathway calculator and problem solver below to practice various math topics. View sample math questions and directions students will encounter on SAT test day. The argument col contains one of the position symbols, l, r, or c. The argument text contains the content of the column. The following examples will show you the list of Math Functions in Tableau. Because complex Boolean statements can get tricky to think about, we can create a truth table to keep track of what truth values for the simple statements make the complex statement true and false. Example. Tally Table - Definition with Examples. There's now 3 parts to the tutorial with two extra example videos at the end. Trusted by teachers across schools. Possible values will be from 1-10, instead of 0-9. Uploaded By BrigadierIronJay8680. Example. Truth Table. Determining the post-money cap table for an equity round with an option pool refresh and one or more convertible notes converting to equity can be overwhelming. Here you can find additional information about practicing multiplication tables at primary school. Comprehensive Curriculum. According to some people, maths is just the use of complicated formulas and calculations which won\u2019t be ever applied in real life. There are two disadvantages of writing tables by hand as described in this tutorial. Most of the time the data will be collected in form of a spreadsheet and we don't want to enter the data twice. Tables. Example: table([1:3]',{'one';'two';'three'},categorical({'A';'B';'C'})) creates a table from variables with three rows, but different data types. Mathematics for the Liberal Arts. Example. A math function table is a table used to plot possible outcomes of a function, which is a kind of rule. Reading over the table; Exercising using the Math Trainer; But here are some \"tips\" to help you even more: Tip 1: Order Does Not Matter. Go to next Question. Back to overview Forward to Shortdocumentation 1 Examples of Latex Here an example of a very small Latex document \\documentclass{article} \\begin{document} example for a very \\tiny{tiny} \\normalsize \\LaTeX \\ document \\end{document} Truth Tables . For example, in the example given below, 7 \u2026 A tautology in math (and logic) is a compound statement (premise and conclusion) that always produces truth. Roman numerals.. my mind ~ my math .. Tables/Graphs/Charts BEST CHART EVER !!! Unlike most other environments, however, there are some handy shorthands for declaring your formulas. Two way tables, also known as contingency tables, show frequencies (counts) as they relate to two variables. Some math sections allow the use of a calculator, others do not. By Grades. Mathematical tables are lists of numbers showing the results of a calculation with varying arguments.Tables of trigonometric functions were used in ancient Greece and India for applications to astronomy and celestial navigation.They continued to be widely used until electronic calculators became cheap and plentiful, in order to simplify and drastically speed up computation. Construct a truth table for the formula . The Tableau ABS function is used to return the absolute positive value. This tutorial provides you with commonly used SQL math functions that allow you to perform business and engineering calculations. This is more typical of what you'll need to do in mathematics. 50,000 Schools . Another Example: 2\u00d79=18, and 9\u00d72=18. The Complete K-5 Math Learning Program Built for Your Child. For instance, if we want to work out the times table for 2, we start with 2 and then add 2 in each step. A function table in math is a table that describes a function by displaying inputs and corresponding outputs in tabular form. HTML tables allow web developers to arrange data into rows and columns. About the book: I am honored to see my latest book, Table Talk Math, published by Dave Burgess Consulting and available on Amazon.In it, math educators and experts from around the world offer ideas and share stories about ways in which parents can engage their children with a math-based conversation in an easy-to-read format and plenty of examples. In some tables, there might not be unique values any column in the lookup table. The notation may vary\u2026 30 Million Kids. To fully understand function tables and their purpose, you need to understand functions, and how they relate to variables. However, there is only one record for each jacket and size combination -- Jacket Medium in row 4 and Jacket Large in row 5. Extras. No matter what the individual parts are, the result is a true statement; a tautology is always true. C# DataTable Examples Store data in memory with a DataTable. SAT Suite of Assessments Sample Questions. We can fit one, two, three, four, five. (See Commutative Property.) Tables, graphs, and charts are an easy way to clearly show your data. A truth table is a mathematical table used in logic\u2014specifically in connection with Boolean algebra, boolean functions, and propositional calculus\u2014which sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables. Let's see. sz \u2014 Size of preallocated table two-element numeric vector. A times table is a list of multiples of a number. Be sure to consider how to best show your results with appropriate graph forms. School St. John's University; Course Title MTH 1250C; Type. First, I list all the alternatives for P and Q. Search for: Truth Tables and Analyzing Arguments: Examples. Tableau ABS Function. And then on this table, which is identical, you could fit six, seven, eight, nine. Z Score Table Sample Problems. When we multiply two numbers, it does not matter which is first or second, the answer is always the same. Parents, Sign Up for Free Teachers, Sign Up for Free. Logic Symbols in Math; Truth Table; Tautology Math Examples; Tautology Definition. Pages 47. Example: 3\u00d75=15, and 5\u00d73=15. Math.floor(Math.random() * (max - min + 1)) + min // We can introduce a min variable as well so we can have a range. While it works for small tables similar to the one in our example, it can take a long time to enter a large amount of data by hand. This preview shows page 11 - 23 out of 47 pages. So here we have one table, and it's going to touch ends with this table right over here. And because it touches ends right over here-- we're making it one big continuous table-- you can't fit someone here anymore.\n\n## table math example\n\nZapp's Chips Careers, Enchanted Sword Seed, Emmer Wheat Vs Wheat, Momeni Koi Area Rugs, Best Garlic Mayonnaise, 1965 Impala 4 Door For Sale, Deer Resistant Honeysuckle, Steakhouse Blue Cheese Dressing Recipe, Drama Genre Examples,", "date": "2021-03-02 19:57:39", "meta": {"domain": "specialsports.info", "url": "https://specialsports.info/benzamide-molecular-fbulsmr/viewtopic.php?af9d97=table-math-example", "openwebmath_score": 0.4584738612174988, "openwebmath_perplexity": 1302.6804645855257, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9658995723244552, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8538931567790912}}
{"url": "https://math.stackexchange.com/questions/3247145/how-can-one-show-that-sum-n-0-infty-fracnn-e", "text": "# How can one show that $\\sum_{n=0}^\\infty\\frac{n}{n!}=e$? [duplicate]\n\nHow can one show that $$\\sum_{n=0}^\\infty\\frac{n}{n!}=e$$?\n\nI understand that $$\\sum_{n=0}^\\infty\\frac{x^n}{n!}=e^x$$ and that letting $$x=1$$would give $$\\sum_{n=0}^\\infty\\frac{1}{n!}=e$$\n\nBut why does the sum $$\\sum_{n=0}^\\infty\\frac{n}{n!}$$ give an answer of $$e$$ also?\n\n\u2022 In the picture on the bottom sum the numerator should be n, apologies. \u2013\u00a0user677704 Jun 1 '19 at 2:51\n\u2022 Are you familiar with derivatives ? \u2013\u00a0DanielV Jun 1 '19 at 2:53\n\u2022 Your question is rather unclear. Do you want to prove: $\\sum _{n=0}^{\\infty }\\:\\frac{x}{n!}=e?$ \u2013\u00a0NoChance Jun 1 '19 at 3:12\n\u2022 \u2018No chance\u2019 yes that was my question more or less, I just didn\u2019t understand why different sums all add to e \u2013\u00a0user677704 Jun 1 '19 at 3:13\n\u2022 It can be proven that $\\sum _{n=0}^{\\infty }\\:\\frac{x^n}{n!}=e^x$. However, in your expression, you don't raise x to the power n. \u2013\u00a0NoChance Jun 1 '19 at 3:15\n\nNote that $$n/n!=1/(n-1)!$$, which is not meaningfully distinct in the context of the infinite sum from the terms $$1/n!$$.\n\nIn fact $$\\sum_{n\\geq 0} \\frac{n}{n!}\\stackrel{(1)}{=}\\sum_{n\\geq 1} \\frac{n}{n!}= \\sum_{n\\geq 1} \\frac{1}{(n-1)!}\\stackrel{(2)}{=}\\sum_{n\\geq 0} \\frac{1}{n!}=e^1=e$$ (1) comes from the fact that the first term is zero, (2) is a shift of indices by one.\n\nYou can repeat this shifting procedure ad infinitum and obtain, for each fixed $$k$$ $$\\sum_{n\\geq 0} \\frac{n(n-1)\\ldots (n-k)}{n!}=e$$ because your sum 'starts' at index $$k+1$$.\n\n\u2022 This is what I meant in my answer. Nice presentation. +1 \u2013\u00a0The Count Jun 1 '19 at 3:14\n\u2022 So I, in fact, unfolded your idea ? (+1) \u2013\u00a0Duchamp G\u00e9rard H. E. Jun 1 '19 at 3:15\n\u2022 Teamwork! I like it. \u2013\u00a0The Count Jun 1 '19 at 3:16\n\n$$\\begin{array} {rcl} % \\displaystyle e^x & = & \\displaystyle \\sum_{n = 0}^{\\infty} \\frac{x^n}{n!} \\\\ % \\displaystyle\\frac{d}{dx}e^x & = & \\displaystyle \\frac{d}{dx}\\sum_{n = 0}^{\\infty} \\frac{x^n}{n!} \\\\ % e^x & = & \\displaystyle \\sum_{n = 0}^{\\infty} \\frac{nx^{n-1}}{n!} \\\\ % e^1 & = & \\displaystyle \\sum_{n = 0}^{\\infty} \\frac{n1^{n-1}}{n!} \\\\ \\end{array}$$\n\n\u2022 Thanks for your response it\u2019s really clear and perfectly understood, I appreciate it. So would I be right in assuming you can repeat this process ad infinitum and acquire many different expressions which all sum to e? \u2013\u00a0user677704 Jun 1 '19 at 3:15\n\u2022 Yes, in this case the multipliers will be $$n,n(n-1),n(n-1)(n-2),\\ldots$$ and so on ... \u2013\u00a0Duchamp G\u00e9rard H. E. Jun 1 '19 at 3:18\n\u2022 That\u2019s incredibly interesting, aside from showing this with basic differentiation would you know where to find a geometric or more intuitive proof. The logic is perfectly reasonable but I prefer to see why. \u2013\u00a0user677704 Jun 1 '19 at 3:25\n\u2022 @User3457884334 I added repetition of this process ad infinitum in my answer. \u2013\u00a0Duchamp G\u00e9rard H. E. Jun 1 '19 at 3:26\n\u2022 I understand that but wondered if there was another proof to show it geometrically. \u2013\u00a0user677704 Jun 1 '19 at 3:28\n\nSuppose $$e^x=a_0+a_1x+a_2x^2+a_3x^3+...$$ try to find the constants by evaluating $$e^0$$ and then differentiating to eliminate the current $$0$$th degree term.\n\nHere are the first three terms : $$e^0=a_0+a_1*0+...=1\\Leftrightarrow a_0=1$$ $$\\frac{d}{dx}(e^x)=e^x=\\frac{d}{dx}(a_0+a_1x+a_2x^2+...)=a_1+2a_2x+3a_3x^2+...$$ $$e^0=a_1+2a_2*0+3a_3*0^2+...=1\\Leftrightarrow a_1=1$$ $$\\frac{d}{dx}(e^x)=e^x=\\frac{d}{dx}(a_1+2a_2x+3a_3x^2+...)=2a_2+(3*2)a_3x+...$$ $$e^0=2a_2+(3*2)a_3*0\\,+...=1\\Leftrightarrow a_1=\\frac{1}{2}$$ You'll find that $$e^x=\\sum_{n=0}^\\infty \\frac{x^n}{n!}$$ switch $$1$$ for $$x$$ and you get your result (refer to The Count's answer).\n\nDifferentiate the series $$e^x=\\sum_{n=0}^{\\infty}\\dfrac {x^n}{n!}$$ term by term. Get $$\\sum_{n=0}^{\\infty}\\dfrac {nx^{n-1}}{n!}$$. Then set $$x=1$$. Note that $$(e^x)'=e^x$$ (for$$\\sum_{n=0}^{\\infty}\\dfrac {nx^{n-1}}{n!}=\\sum_{n=0}^{\\infty}\\dfrac {x^n}{n!}$$) .", "date": "2021-03-05 14:19:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3247145/how-can-one-show-that-sum-n-0-infty-fracnn-e", "openwebmath_score": 0.921451210975647, "openwebmath_perplexity": 530.2218829793999, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109516378093, "lm_q2_score": 0.8670357615200475, "lm_q1q2_score": 0.8538663134065706}}
{"url": "https://math.stackexchange.com/questions/2359335/find-the-number-of-distributions-of-seven-distinct-balls-into-three-distinct-box", "text": "# Find the number of distributions of seven distinct balls into three distinct boxes if at least two balls must go into each box.\n\nMy Answer: I chose to place three balls into one box and two balls in the other two boxes. This can be done three times: $$\\binom73\\binom42\\binom22=\\binom{7}{3,2,2}\\\\ \\binom72\\binom53\\binom22=\\binom{7}{2,3,2} \\\\ \\binom72\\binom52\\binom33=\\binom{7}{2,2,3}$$ I'm not sure if this is correct, but we have not covered Stirling numbers yet, so I cannot use that as my explanation. Please help and thank you!\n\nAn alternative method, noting that 2/2/3 is the only distribution of seven balls to three boxes with at least two balls in each box:\n\n\u2022 3 choices for the box with three balls\n\u2022 $\\binom73=35$ ways to put three balls into that box\n\u2022 $\\binom42=6$ ways to put two balls into one of the remaining boxes; the third box is then forced to contain the last two balls\n\nThis yields $3\\cdot35\\cdot6=630$ possibilities, and your approach yields the same result but through a longer process.\n\n\u2022 Of course the solution appear to be the same once one recognises that the three numbers in the OP's solution must be the same by symmetry. \u2013\u00a0Carsten S Jul 15 '17 at 11:38\n\nYour answer is correct. Here is a variation based upon exponential generating functions.\n\n\u2022 A selection of at least two distinct balls can be encoded as \\begin{align*} \\frac{x^2}{2!}+\\frac{x^3}{3!}+\\cdots=e^x-1-x \\end{align*}\n\n\u2022 Since we have three boxes we consider \\begin{align*} (e^x-1-x)^3\\tag{1} \\end{align*}\n\nWe denote with $[x^n]$ the coefficient of $x^n$ in a series.\n\nSince we are looking for the number of placing $7$ distinct balls we consider the coefficient of $x^7$ in (1) and obtain with some help of Wolfram Alpha\n\n\\begin{align*} 7![x^7](e^x-1-x)^3=\\color{blue}{630} \\end{align*}\n\n\u2022 Let's say we have a red box, a blue box, and a green box (and the observer is not colorblind). There are $\\binom{7}{3}$ ways of placing three balls in the red box, $\\binom{4}{2}$ ways of placing two of the remaining four balls in the blue box, and $\\binom{2}{2}$ ways of placing the remaining two balls in the green box. Hence, there are $$\\binom{7}{3}\\binom{4}{2}\\binom{2}{2} = \\binom{7}{3, 2, 2}$$ ways to distribute the balls so that three are placed in the red box, two are placed in the blue box, and two are placed in the blue box. By symmetry, the answer is $$3\\binom{7}{3, 2, 2}$$ \u2013\u00a0N. F. Taussig Jul 19 '17 at 12:31\n\u2022 @N.F.Taussig: Thanks for your hint! \u2013\u00a0Markus Scheuer Jul 19 '17 at 12:38\n\u2022 @N.F.Taussig: I was misleading by myself. Original answer restored. \u2013\u00a0Markus Scheuer Jul 19 '17 at 16:58\n\u2022 I have written a solution based on your idea of first placing the seven distinguishable ways in three indistinguishable boxes, then arranging the boxes. \u2013\u00a0N. F. Taussig Jul 19 '17 at 19:16\n\nThis solution is a modification of an attempted solution that Markus Scheuer deleted.\n\nFirst, we count the number of ways seven distinguishable balls can be placed in three indistinguishable boxes if at least two balls are placed in each box. Then we will arrange the boxes.\n\nThe only permissible way to distribute the balls is to place three balls in one bag and two each in the other bags.\n\nMethod 1: There are $\\binom{7}{3}$ ways to select three balls to be placed in one of the boxes. Line up the remaining balls in a row. Take the ball at the left end of the row and place it in an empty box. There are three ways of selecting one of the other three balls to be placed in the same box as that ball. The remaining two balls must be placed in the remaining empty box. Hence, there are $$3\\binom{7}{3}$$ ways to place seven distinguishable balls in three indistinguishable boxes.\n\nMethod 2: There are $\\binom{7}{3}$ ways to place three of the balls in one box. That leaves $\\binom{4}{2}$ ways to place two of the remaining four balls in another box. The remaining two balls are place in the third box. That suggests the answer is $$\\binom{7}{3}\\binom{4}{2}$$ However, since the boxes are indistinguishable, we cannot distinguish between the two boxes that received two balls. Thus, we must multiply the above result by $1/2$. Hence, there are $$\\frac{1}{2}\\binom{7}{3}\\binom{4}{2}$$ ways of distributing seven distinguishable balls to three indistinguishable boxes.\n\nTo make the boxes distinguishable, we paint one box blue, one box green, and one box red. There are $3!$ ways to do this. Hence, there are $$3! \\cdot 3\\binom{7}{3} = 630$$ ways to place seven distinguishable balls in three distinguishable boxes if at least two balls must be placed in each box.\n\n\u2022 @NFTaussig: Clear and extensive elaboration. Very nice! (+1) \u2013\u00a0Markus Scheuer Jul 19 '17 at 19:38", "date": "2019-08-25 01:54:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2359335/find-the-number-of-distributions-of-seven-distinct-balls-into-three-distinct-box", "openwebmath_score": 0.9417623281478882, "openwebmath_perplexity": 204.07628654221293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810949854636, "lm_q2_score": 0.8670357615200474, "lm_q1q2_score": 0.8538663118604956}}
{"url": "https://math.stackexchange.com/questions/2657571/probability-of-picking-matching-socks-after-partitioning-the-drawer", "text": "# Probability of picking matching socks, after partitioning the drawer.\n\nApologies if this is a duplicate. I searched and didn't find anything quite like it.\n\nSuppose I have a drawer with an equal number of N black socks and N white socks. They're all mixed up. So, my chances of picking a matching pair in the first two selections is (N-1)/(2N-1), right? Well, what if, before I pick the first sock, I randomly (so I don't know the colors of the socks I'm moving) partition the drawer so that there are N socks on each side, and I draw one sock from each side. Do the chances of drawing a matching pair change?\n\nOn the one hand, we can see that selection from one side doesn't change the composition of socks on the other side of the partition. However, whichever color I choose from the \"first\" side, it's likely that there are more of that color on that side. On other words, if I draw a black sock from one side, it's more likely that that side had N-1 blacks and 1 white than it is that that side had 1 black and N-1 whites.\n\nMy suspicion is that I need to do some kind of hypothesis testing, where I consider the chances of every possible partitioning, but that's way above my skill level.\n\n\u2022 Shouldn't the probability in the simple case be $\\frac{N-1}{2N-1}$, because there are no longer $2N$ socks in the drawer? Feb 19 '18 at 20:17\n\u2022 with replacement? Feb 19 '18 at 20:18\n\u2022 @GTonyJacobs Yep $P=\\frac {N-1}{2N-1}$ Feb 19 '18 at 20:27\n\u2022 You certainly don't need hypothesis testing. We're not making inferences from a sample about an otherwise unknowable population. This is just a probability question with all of the parameters known. Kind of a tricky one, maybe, but there might be a good symmetry argument for why there's no difference. Feb 19 '18 at 20:37\n\u2022 @GTonyJacobs Correct. I'll see if I can edit the post. Feb 20 '18 at 21:34\n\nIt doesn't matter if the partitioning happens before or after the first draw. Suppose it happens after. Suppose also that the first sock drawn was black.\n\nNow, we partition off $N$ from the remaining $2N-1$, to obtain our pool for the second draw. On average, the composition of this pool is $\\frac{N-1}{2N-1}$ black and $\\frac{N}{2N-1}$ white. Drawing from it, we have a matching pair if we draw from the portion of it that is black, i.e., $\\frac{N-1}{2N-1}$.\n\nThe trick to simplifying the work is to work with expected values for the number of socks of each color in the second part of the partition, not actual values.\n\nNo, the chances don't change. If you worked out all the conditional probabilities based on how the socks split and summed appropriately you'd get the same answer: $(N-1)/(2N -1)$. The partition gives you no new information.\n\nHere's the brute force calculation for $N=2$. There are two possible partitions, WW|BB and WB|WB. The first of these occurs with probability $1/3$. the second with probability $2/3$. In the first case you fail for sure. In the second you succeed half the time. On average you succeed with probability $$0 \\times \\frac{1}{3} + \\frac{1}{2} \\times \\frac{2}{3} = \\frac{1}{3} = \\frac{2-1}{4-1}.$$\n\n(There's probably a really elegant argument from symmetry that I don't see.)\n\nOf course it shouldn't change!\n\nLet's see if the math works out for a simple example, say $N=3$\n\nFirst, the 'normal' calculation says that the chances of getting a match $M$ is:\n\n$$P(M)= \\frac{N-1}{2N-1}=\\frac{2}{5}$$\n\nNow let's see what happens when you randomly partition them.\n\nLet's say you first pick from the left side. After randomly splitting the socks into two groups, there can be $0$, $1$, $2$, or $3$ socks there, with respective probabilities of:\n\n$$P(0)= \\frac{{3 \\choose 0}\\cdot{3 \\choose 3}}{6 \\choose 3} = \\frac{1}{20}$$\n\n$$P(1)= \\frac{{3 \\choose 1}\\cdot{3 \\choose 2}}{6 \\choose 3} = \\frac{9}{20}$$\n\n$$P(2)= \\frac{{3 \\choose 2}\\cdot{3 \\choose 1}}{6 \\choose 3} = \\frac{9}{20}$$\n\n$$P(3)= \\frac{{3 \\choose 3}\\cdot{3 \\choose 0}}{6 \\choose 3} = \\frac{1}{20}$$\n\nNow, getting a match $M$ in the first and last situation is impossible, so $$P(M|0)=P(M|3)=0$$\n\nWhen there is $1$ white sock on the left, the chances of getting a match are the chance of getting that white sock times the chances of getting one of the two socks on the right side, plus the chances of getting one of the two black ones on the left and the one black one on the right. And, with $2$ white socks on the left it's all symmetrical. So:\n\n$$P(M|1)=P(M|2)=\\frac{1}{3}\\cdot \\frac{2}{3} + \\frac{2}{3} \\cdot \\frac{1}{3} = \\frac{4}{9}$$\n\nIn sum:\n\n$$P(M)=P(0)\\cdot P(M|0) + P(1)\\cdot P(M|1) +P(2)\\cdot P(M|2) +P(3)\\cdot P(M|3) =$$\n\n$$\\frac{1}{20} \\cdot 0 + \\frac{9}{20} \\cdot \\frac{4}{9} + \\frac{9}{20} \\cdot \\frac{4}{9} + \\frac{1}{20} \\cdot 0 = \\frac{8}{20} = \\frac{2}{5}$$\n\nOK, so yes, same chance!\n\nNow, I'm sure you can generalize this for any $N$ ... thus evaluating the series:\n\n$$P(M)=\\sum_{i=0}^N \\frac{{N \\choose i} \\cdot {N \\choose {N-i}}}{2N \\choose N} \\cdot 2 \\cdot \\frac{i}{N} \\cdot \\frac{N-i}{N}$$\n\n[... Insert some math that's over my head ....]\n\n... and you'll see that this ends up being $$\\frac{N-1}{2N-1}$$", "date": "2021-10-26 13:14:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2657571/probability-of-picking-matching-socks-after-partitioning-the-drawer", "openwebmath_score": 0.7827708125114441, "openwebmath_perplexity": 340.77304883709394, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109480714625, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8538663035472853}}
{"url": "https://math.stackexchange.com/questions/900315/proving-that-a-number-is-non-negative", "text": "# Proving that a number is non-negative?\n\nThe numbers $a$,$b$ and $c$ are real. Prove that at least one of the three numbers $$(a+b+c)^2 -9bc \\hspace{1cm} (a+b+c)^2 -9ca \\hspace{1cm} (a+b+c)^2-9ab$$ is non-negative.\n\nAny hints would be appreciated too.\n\n\u2022 Have you considered adding them up and seeing if they are a square ? \u2013\u00a0Belgi Aug 16 '14 at 17:42\n\nHint:\n\nIf all three numbers are negative, then:\n\n$$ab > \\left(\\frac{a+b+c}{3}\\right)^2 \\hspace{1cm} ac > \\left(\\frac{a+b+c}{3}\\right)^2 \\hspace{1cm} bc > \\left(\\frac{a+b+c}{3}\\right)^2 \\hspace{1cm}$$\n\nTherefore, if we multiply the three inequalities:\n\n$$a^2b^2c^2 > \\left(\\frac{a+b+c}{3}\\right)^6$$\n\nOr equivalently:\n\n$$\\left(\\sqrt[3]{abc}\\right)^6 > \\left(\\frac{a+b+c}{3}\\right)^6$$\n\nDo you know any inequality you can use here do disprove this?\n\n\u2022 @Sudhanshu Look for the Arithmetic-Geometric Median Inequality en.wikipedia.org/wiki/\u2026 It says that: $$\\frac{a+b+c}{3} \\geq \\sqrt[3]{abc}$$ \u2013\u00a0Darth Geek Aug 16 '14 at 17:50\n\u2022 Oh, how can I forget that...Thanks btw. \u2013\u00a0Sudhanshu Aug 16 '14 at 17:54\n\u2022 But since we assume these numbers to be negative how can we apply A.M > G.M ? \u2013\u00a0Sudhanshu Aug 16 '14 at 18:04\n\u2022 If we assume those three numbers are negative, then they contradict the \"AM>GM\" inequality wich we know to be true. Therefore our assumption was incorrect, i.e. at least one of those numbers was not negative. This sort of proof is called \"proof by contradiction\" or \"reductio ad absurdum\". You start with a hypothesis and you end up in a contradiction, wich means that the hypothesis was false. \u2013\u00a0Darth Geek Aug 16 '14 at 18:14\n\u2022 But I think the A.M > G.M rule is applicable only for positive numbers \u2013\u00a0Sudhanshu Aug 16 '14 at 18:15\n\n$$\\sum [(a+b+c)^2-9bc]=3\\sum[a^2+b^2+c^2-ab-bc-ca]=\\frac32\\sum (a-b)^2\\ge0$$\n\nIf each $(a+b+c)^2-3bc<0,$ $$\\sum [(a+b+c)^2-3bc]<0$$\n\n\u2022 You can also observe that $\\sum[a^2+b^2+c^2-ab-bc-ca]\\ge0$ is just Cauchy -Schwarz. \u2013\u00a0N. S. Aug 16 '14 at 18:38", "date": "2019-10-16 04:27:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/900315/proving-that-a-number-is-non-negative", "openwebmath_score": 0.7888849377632141, "openwebmath_perplexity": 935.3657109660286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109540896723, "lm_q2_score": 0.8670357460591568, "lm_q1q2_score": 0.853866300306369}}
{"url": "https://narrativmedicin.se/gzh7rm/fa9583-interior-point-in-metric-space", "text": "# interior point in metric space\n\nFACTS A point is interior if and only if it has an open ball that is a subset of the set x 2intA , 9\">0;B \"(x) \u02c6A A point is in the closure if and only if any open ball around it intersects the set x 2A , 8\">0;B \"(x) \\A 6= ? Metric Spaces A metric space is a set X that has a notion of the distance d(x,y) between every pair of points x,y \u2208 X. Definition: We say that x is an interior point of A iff there is an such that: . Browse other questions tagged metric-spaces or ask your own question. Defn Suppose (X,d) is a metric space and A is a subset of X. True. A point x \u2208 E is said to be an interior point of E if E contains an open ball centered at x. Since you can construct a ball around 3, where all the points in the ball is in the metric space. (d) Describe the possible forms that an open ball can take in X = (Q \u2229 [0; 3]; dE). Proposition A set O in a metric space is open if and only if each of its points are interior points. When we encounter topological spaces, we will generalize this definition of open. $\\begingroup$ As an addendum, singletons are open if and only if the metric space is induced by a discrete metric, so there's only one case in which you have a nonempty interior of singleton sets. - the exterior of . We do not develop their theory in detail, and we leave the veri\ufb01cations and proofs as an exercise. In most cases, the proofs Let E be a subset of a metric space X. However, this definition of open in metric spaces is the same as that as if we regard our metric space as a topological space. - the boundary of Examples. $\\endgroup$ \u2013 Alan Apr 18 '15 at 8:32 METRIC SPACES 77 where 1\u02dc2 denotes the positive square root and equality holds if and only if there is a real number r, with 0 n r n 1, such that yj rxj 1 r zj for each j, 1 n j n N. Remark 3.1.9 Again, it is useful to view the triangular inequalities on \u201cfamiliar \\begin{align} \\quad \\mathrm{int} \\left ( \\bigcup_{S \\in \\mathcal F} S\\right ) \\supseteq \\bigcup_{S \\in \\mathcal F} \\mathrm{int} (S) \\quad \\blacksquare \\end{align} If is the real line with usual metric, , then A point x is called an isolated point of A if x belongs to A but is not a limit point of A. Interior Point Not Interior Points Definition: ... A set is said to be open in a metric space if it equals its interior (= ()). This intuitively means, that x is really 'inside' A - because it is contained in a ball inside A - it is not near the boundary of A. Let (X;d) be a metric space and A \u02c6X. Let be a metric space, Define: - the interior of . Let be a metric space. (c) The point 3 is an interior point of the subset C of X where C = {x \u2208 Q | 2 < x \u2264 3}? The interior of the set E is the set of all its interior points. A point is exterior \u2026 An open ball of radius centered at is defined as Definition. Metric space: Interior Point METRIC SPACE: Interior Point: Definitions. This set is denoted by intE. 1. In these examples, all sets under consideration are subsets of the metric space R. Example 2.7. Appendix A. If has discrete metric, 2. Metric Spaces, Topological Spaces, and Compactness 253 Given S\u02c6 X;p2 X, we say pis an accumulation point of Sif and only if, for each \">0, there exists q2 S\\ B\"(p); q6= p.It follows that pis an The purpose of this chapter is to introduce metric spaces and give some de\ufb01nitions and examples. Proposition A set C in a metric space is closed if and only if it contains all its limit points. 3.2. Featured on Meta \u201cQuestion closed\u201d notifications experiment results and graduation", "date": "2021-05-09 22:06:37", "meta": {"domain": "narrativmedicin.se", "url": "https://narrativmedicin.se/gzh7rm/fa9583-interior-point-in-metric-space", "openwebmath_score": 0.9142648577690125, "openwebmath_perplexity": 280.91345862380405, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9848109480714625, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8538663001637177}}
{"url": "https://mathematica.stackexchange.com/questions/249789/are-there-any-alternatives-for-nintegrate-to-calculate-the-area-for-which-fx-y", "text": "# Are there any alternatives for NIntegrate to calculate the area for which $f(x,y)<0$?\n\nI have this two-variable function $$f(x,y)= (8 \\cos (x+y)+7)\\cos \\left(\\frac{x}{2}\\right)+\\cos \\frac{x-2 y}{2}+2 \\cos \\left(\\frac{3 x}{2}\\right)$$ where $$0. I want to calculate numerically the area for which the function is negative $$f(x,y)<0$$. I use this code\n\nNIntegrate[\nBoole[2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] +\nCos[x/2] (7 + 8 Cos[x + y]) < 0], {x, 0, Pi}, {y, 0, Pi}]\n\n\nand it gives the answer $$3.49458$$, but Mathematica gives the following warnings. Are there any other ways to calculate this value that is more reliable and more accurate than this method?\n\nNIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.\n\nNIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 3.494581434480605 and 0.002397336775896384 for the integral and error estimates.\n\n\u2022 Your f[x, y] is a constant rather than a function of x and y. Jun 17, 2021 at 19:03\n\u2022 @BobHanlon Thanks, it was a typo mistake.\n\u2013\u00a0user80186\nJun 17, 2021 at 23:21\n\u2022 There is an exact solution to your integral $-2 \\text{Li}_2\\left(-\\frac{1}{2}\\right)+\\text{Li}_2\\left(\\frac{1}{4} \\left(i \\sqrt{3}+1\\right)\\right)+\\text{Li}_2\\left(\\frac{1}{4} \\left(1-i \\sqrt{3}\\right)\\right)+\\frac{2 \\pi ^2}{9}$ Jun 18, 2021 at 13:12\n\u2022 @yarchik I think you should add that and its derivation as an answer.\n\u2013\u00a0Kiro\nJun 18, 2021 at 14:03\n\u2022 @SaraChem @Kiro I was simply a bit more patient than @eyorble. I used u = Integrate[upperCurve, {x, 0, Pi}] and v = Integrate[lowerCurve, {x, 0, 2 Pi/3}], followed by u-v. The lowerCurve and the upperCurve are defined in eyorble's post. Jun 18, 2021 at 19:46\n\nLet us name the function of interest f[x,y]:\n\nf[x_, y_] := (8 Cos[x + y] + 7) Cos[x/2] + Cos[(x - 2 y)/2] + 2 Cos[3 x/2]\n\n\nAttempt to ContourPlot to find the zero lines:\n\nContourPlot[f[x, y] == 0, {x, 0, Pi}, {y, 0, Pi}]\n\n\nSimple numerical evaluation determines that the negative region is the inside of these two curves. Notice that the ContourPlot is quite rapid and has very clean lines. Interesting coincidence. Perhaps there exists an analytical solution to these curve lines?\n\nsol = Solve[{f[x, y] == 0}]\n\n\nThis returns a list of 4 possible curves, while also stating that some solutions may be missing. By manual inspection (such as by using Plot), we can find that the 2nd and 4th solutions are of interest to us, so we shall label them:\n\nupperCurve = y /. sol[[2]];\nlowerCurve = y /. sol[[4]];\nPlot[{upperCurve, lowerCurve}, {x, 0, Pi}, PlotRange -> {0, Pi}]\n\n\nChecking the curves manually by plotting them against the original ContourPlot, we see that upperCurve matches the upper line for the whole domain, and that lowerCurve matches the lower line up until it reaches its minimum.\n\nFind the minimum of the lowerCurve:\n\nFindMinimum[{lowerCurve, 0 < x < Pi}, x, WorkingPrecision -> 25]\n\n\n{1.872299341324760554288429*10^-8, {x -> 2.094395111754692173633430}}\n\nThe warning about a small imaginary part is of little concern here, but you can increase the WorkingPrecision and PrecisionGoal if you would like more digits.\n\nMichael Seifert also pointed out that an exact form can be found for this solution by applying TrigFactor to f[x,y]:\n\nTrigFactor[f[x,y]]\n\n\n2 (Cos[x/2 - y/2] + 2 Cos[x/2 + y/2]) (2 Cos[x + y/2] + Cos[y/2]) == 0\n\nThe lower line happens to correspond to the second variable factor in this expression, and its minimum is found when y is set to 0 and solved.\n\nSolve[{(2 Cos[x + y/2] + Cos[y/2]) == 0 /. y -> 0, 0 < x < Pi}, x]\n\n\n{{ x -> 2 Pi/ 3 }}\n\nIntegrate the area below the 2nd curve over the whole domain minus the area under the 4th curve for 0 through 2.094...\n\nNIntegrate[upperCurve, {x, 0, Pi}, WorkingPrecision -> 25] -\nNIntegrate[lowerCurve, {x, 0, 2.0943951117546921736334297747816890478125.},\nWorkingPrecision -> 25]\n\n\n3.49805583366099845069196\n\nOr with the exact form, we can see a slightly different answer:\n\nNIntegrate[upperCurve, {x, 0, Pi}, WorkingPrecision -> 25] -\nNIntegrate[lowerCurve, {x, 0, 2 Pi/3}, WorkingPrecision -> 25]\n\n\n3.49805583366099836305434\n\nWhile this method is not universally applicable, it does work for this function and is much faster than Area or direct application of NIntegrate and Boole for high precisions. As yarchik notes, you can swap NIntegrate for Integrate here to acquire an exact solution, though it takes a bit longer to evaluate.\n\n\u2022 Using TrigFactor on f[x,y] shows that the two curves are $\\cos((x-y)/2) + 2 \\cos((x+y)/2) = 0$ (upper) and $2 \\cos (x+y/2)+ \\cos(y/2) = 0$ (lower). I don't know if that really helps, though. Jun 17, 2021 at 20:10\n\u2022 If nothing else, it shows that the intersection of the lower curve with the $x$-axis is when $2 \\cos x + 1 = 0$, or $x = 2 \\pi/3$. Jun 17, 2021 at 20:12\n\u2022 There are a lot of inconsistencies in your answer. What is y that you are integrating? Jun 18, 2021 at 11:51\n\u2022 @yarchik The y in the integration should be read as y /. sol[[2]] (which is the upper curve from the contour plot) or y /. sol[[4]] (which is the lower curve from the contour plot). These are just explicit forms (in x) of the curves where f[x,y]==0, selected to match the curves we are interested in. Jun 18, 2021 at 12:02\n\u2022 @yarchik I have modified the answer to hopefully clarify what's being integrated here. Jun 18, 2021 at 12:11\nClear[\"Global*\"]\n\nRegionPlot[\n2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] + Cos[x/2] (7 + 8 Cos[x + y]) < 0,\n{x, 0, Pi}, {y, 0, Pi},\nFrame -> True]\n\n\nrgn = ImplicitRegion[{2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] +\nCos[x/2] (7 + 8 Cos[x + y]) < 0 && 0 < x < Pi &&\n0 < y < Pi}, {x, y}];\n\nArea[rgn, WorkingPrecision -> MachinePrecision]\n\n(* 3.49805 *)\n\nArea[rgn, WorkingPrecision -> 15]\n\n(* 3.49805 *)\n\n\nA similar way is\n\nreg=ImplicitRegion[\n2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] + Cos[x/2] (7 + 8 Cos[x + y]) <\n0, {{x, 0, Pi}, {y, 0, Pi}}];\nNIntegrate[1, {x, y} \u2208 reg]\n\n\n3.49485\n\n\u2022 Thanks. Does WorkingPrecission matter in this method? I see a small difference in the different answers here obtained by different methods others suggested; if I need the most accurate one, which one is more reliable?\n\u2013\u00a0user80186\nJun 18, 2021 at 12:54\n\nMethod ->\"LocalAdaptive\"  evaluates without errormessage\n\nNIntegrate[Boole[2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] +Cos[x/2] (7 + 8 Cos[x + y]) < 0], {x, 0, Pi}, {y,0, Pi}, Method -> \"LocalAdaptive\"]\n(*3.49818*)\n\n\u2022 Thanks. Does WorkingPrecission matter in this method? I see a small difference in the different answers here obtained by different methods others suggested; if I need the most accurate one, which one is more reliable?\n\u2013\u00a0user80186\nJun 18, 2021 at 12:55\n\u2022 @SaraChem Sorry, I was offline some days. If you add the option , IntegrationMonitor :> ((errors = Through[#1@\"Error\"]) &) inside NIntegrate , it's possible to evaluate the integrationerror Total@errors afterwards. Might be helpful. Jun 20, 2021 at 14:27", "date": "2022-05-26 01:52:11", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/249789/are-there-any-alternatives-for-nintegrate-to-calculate-the-area-for-which-fx-y", "openwebmath_score": 0.46566471457481384, "openwebmath_perplexity": 4621.230410626539, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.951142217223021, "lm_q2_score": 0.8976952852648487, "lm_q1q2_score": 0.8538358840174605}}
{"url": "https://mathlesstraveled.com/?cpage=1", "text": "A few words about PWW #33: subset\u00a0permutations\n\nMy previous post showed four rows of diagrams, where the $n$th row (counting from zero) has diagrams with $n+2$ dots. The diagrams in the $n$th row depict all possible paths that\n\n1. start at the top left dot,\n2. end at the top right dot, and\n3. never visit any dot more than once.\n\nFor a given diagram, we can choose which dots to visit (representing a subset of the dots), and we can visit the chosen dots in any order (a permutation). I will call the resulting things \u201csubset permutations\u201d, although I don\u2019t know if there is some other more accepted term for them.\n\nIt\u2019s not necessary to draw subset permutations as paths; it\u2019s just one nice visual representation. If we number the dots from $1 \\dots n$ (excluding the top two, which are uninteresting), each path corresponds precisely to a permutation of some subset of $\\{1 \\dots n\\}$. Like this:\n\nI thought of this idea the other day\u2014of counting the number of \u201csubset permutations\u201d\u2014and wondered how many there are of each size. I half expected it to turn out to be a sequence of numbers that I already knew well, like the Catalan numbers or certain binomial coefficients or something like that. So I was surprised when it turned out to be a sequence of numbers I don\u2019t think I have ever encountered before (although it has certainly been studied by others).\n\nSo, how many subset permutations are there of size $n$? Let\u2019s call this number $\\mathit{SP}_n$. From my previous post we can see that starting with $\\mathit{SP}_0 = 1$, the first few values are $1, 2, 5, 16$. (When I saw 1, 2, 5, I thought sure it was going to be the Catalan numbers\u2014but then the next number was 16 instead of 14\u2026) Can we compute these more generally without just listing them all and counting?\n\nWell, to choose a particular subset permutation of size $n$, we can first choose how big we want the subset to be, i.e. how many dots to visit, which could be zero, or $n$, or anything in between. Once we choose to visit $k$ out of $n$ dots, then we have $n$ choices for which dot to visit first, then $(n-1)$ choices for which dot to visit second, all the way down to $(n-k+1)$ choices for the last dot (you should convince yourself that $(n-k+1)$, not $(n-k)$, is correct!). This is often notated\n\n$\\displaystyle {}_n P_k = n (n-1) (n-2) \\dots (n-k+1) = \\frac{n!}{(n-k)!}.$\n\nSo, in total then, $\\mathit{SP}_n$ is the sum of ${}_n P_k$ over all possible $k$, that is,\n\n$\\displaystyle \\mathit{SP}_n = \\sum_{0 \\leq k \\leq n} \\frac{n!}{(n-k)!}.$\n\nThis is already better than just listing all the subset permutations and counting. For example, we can compute that\n\n$\\mathit{SP}_4 = 4!/4! + 4!/3! + 4!/2! + 4!/1! + 4!/0! = 1 + 4 + 12 + 24 + 24 = 65.$\n\nAnd sure enough, if we list them all, we get 65 (you\u2019ll just have to take my word that this is all of them, I suppose!):\n\nHowever, with a little algebra, we can do even better! First, let\u2019s write out the sum for $\\mathit{SP}_n$ explicitly:\n\n$\\mathit{SP}_n = 1 + n + n(n-1) + n(n-1)(n-2) + \\dots + n!$\n\nIf we factor $n$ out of all the terms after the initial $1$, we get\n\n$\\mathit{SP}_n = 1 + n[1 + (n-1) + (n-1)(n-2) + \\dots + (n-1)!] = 1 + n \\mathit{SP}_{n-1}$\n\nso the subset permutation numbers actually follow a very simple recurrence! To get the $n$th subset permutation number, we just multiply the previous one by $n$ and add one.\n\n$\\begin{array}{rcl} \\mathit{SP}_0 &=& 1 \\\\ \\mathit{SP}_1 &=& 1 \\cdot 1 + 1 = 2 \\\\ \\mathit{SP}_2 &=& 2 \\cdot 2 + 1 = 5 \\\\ \\mathit{SP}_3 &=& 3 \\cdot 5 + 1 = 16 \\end{array}$\n\nand sure enough, $\\mathit{SP}_4 = 4 \\cdot 16 + 1 = 65$. We can now easily calculate the next few subset permutation numbers:\n\n$1,2,5,16,65,326,1957,13700,109601,986410, \\dots$\n\nA commenter also wondered about the particular order in which I listed subset permutations in my previous post. The short, somewhat disappointing answer is \u201cwhatever order came out of the standard library functions I used\u201d. However, there are definitely more interesting things to say about the ordering, and I think I\u2019ll probably write about that in another post.\n\nPosted in combinatorics, posts without words | Tagged , , , | 6 Comments\n\nPost without words\u00a0#33\n\nPosted in combinatorics, posts without words | Tagged , , | 4 Comments\n\nHappy tau day!\n\nHappy Tau Day! The Tau Manifesto by Michael Hartl is now available in print, if you\u2019re in the market for a particularly nerdy coffee table book and conversation starter:\n\nI also wrote about Tau Day ten years ago; see that post for more links! Using $\\tau$ we can make the most beautiful equation in the world even more beautiful:\n\n$e^{i \\tau} = 1 + 0i$\n\nPlease enjoy eating two pi(e)s today. I\u2019ll be back soon with some solutions to the parallelogram area challenge!\n\nPosted in famous numbers, links | Tagged , , | 1 Comment\n\nChallenge: area of a\u00a0parallelogram\n\nAnd now for something completely different!1\n\nSuppose we have a parallelogram with one corner at the origin, and two adjacent corners at coordinates $(a,b)$ and $(c,d)$. What is the area of the parallelogram?\n\n1. \u2026or is it?\n\nPosted in challenges, geometry | Tagged , | 18 Comments\n\nPost without words\u00a0#32\n\nPosted in posts without words | Tagged , , , | 8 Comments\n\nThe Natural Number\u00a0Game\n\nHello everyone! It has been quite a while since I have written anything here\u2014my last post was in March 2020, and since then I have been overwhelmed dealing with online and hybrid teaching, plus a newborn (who is now almost 5 months old and very cute):\n\nBut the spring semester is finally over, and I have a sabbatical in the fall, so I plan to do a bunch more writing! I have several ideas of things to come (feel free to send me suggestions as well), but to start things out for today I just have a fun link:\n\nThe idea is to use a computer proof assistant to formally prove a lot of basic facts about arithmetic. That is, you get to build proofs using a special precise notation, and the computer will automatically check whether your proof is correct. But the whole thing is organized like a game, with a tutorial, levels that build on each other and introduce new ideas and techniques just before you need them, etc. It\u2019s a lot of fun and honestly kind of addicting.\n\nThere is very little background needed other than some capability for abstract thinking. Things start out quite easy and progress slowly, though things become quite challenging by the time you make it all the way to the end.\n\nPosted in challenges, computation, proof | Tagged , , , , , | 6 Comments\n\nAn exploration of forward differences for bored elementary school\u00a0students\n\nLast week I made a mathematics worksheet for my 8-year-old son, whose school is closed due to the coronavirus pandemic. I\u2019m republishing it here so others can use it for similar purposes.\n\nFigurate numbers and forward differences\n\nThere are lots of further directions this could be taken but I\u2019ll leave that to you and your kids. I tried to create something that was conducive to open-ended exploration rather than something that had a single particular goal in mind.\n\n\u2022 For the curious: Babbage Difference Engine\n\u2022 For the intrepid: Concrete Mathematics by Graham, Knuth, and Patashnik, section 2.6 (\u201cFinite and Infinite Calculus\u201d)\n\nSeeing as how we\u2019ve got at least four more weeks of (effectively) homeschooling ahead of us, and probably more than that, in all likelihood I will be making more of these, and I will certainly continue to share them here! If you use any of these with your kids I\u2019d love to hear about your experiences.\n\nPosted in arithmetic, teaching | | 1 Comment\n\nWays to prove a\u00a0bijection\n\nYou have a function $f : A \\to B$ and want to prove it is a bijection. What can you do?\n\nBy the book\n\nA bijection is defined as a function which is both one-to-one and onto. So prove that $f$ is one-to-one, and prove that it is onto.\n\nThis is straightforward, and it\u2019s what I would expect the students in my Discrete Math class to do, but in my experience it\u2019s actually not used all that much. One of the following methods usually ends up being easier in practice.\n\nBy size\n\nIf $A$ and $B$ are finite and have the same size, it\u2019s enough to prove either that $f$ is one-to-one, or that $f$ is onto. A one-to-one function between two finite sets of the same size must also be onto, and vice versa. (Of course, if $A$ and $B$ don\u2019t have the same size, then there can\u2019t possibly be a bijection between them in the first place.)\n\nIntuitively, this makes sense: on the one hand, in order for $f$ to be onto, it \u201ccan\u2019t afford\u201d to send multiple elements of $A$ to the same element of $B$, because then it won\u2019t have enough to cover every element of $B$. So it must be one-to-one. Likewise, in order to be one-to-one, it can\u2019t afford to miss any elements of $B$, because then the elements of $A$ have to \u201csqueeze\u201d into fewer elements of $B$, and some of them are bound to end up mapping to the same element of $B$. So it must be onto.\n\nHowever, this is actually kind of tricky to formally prove! Note that the definition of \u201c$A$ and $B$ have the same size\u201d is that there exists some bijection $g : A \\to B$. A proof has to start with a one-to-one (or onto) function $f$, and some completely unrelated bijection $g$, and somehow prove that $f$ is onto (or one-to-one). Also, a valid proof must somehow account for the fact that this becomes false when $A$ and $B$ are infinite: a one-to-one function between two infinite sets of the same size need not be onto, or vice versa; we saw several examples in my previous post, such as $f : \\mathbb{N} \\to \\mathbb{N}$ defined by $f(n) = 2n$. Although tricky to come up with, the proof is cute and not too hard to understand once you see it; I think I may write about it in another post!\n\nNote that we can even relax the condition on sizes a bit further: for example, it\u2019s enough to prove that $f$ is one-to-one, and the finite size of $A$ is greater than or equal to the finite size of $B$. The point is that $f$ being a one-to-one function implies that the size of $A$ is less than or equal to the size of $B$, so in fact they have equal sizes.\n\nBy inverse\n\nOne can also prove that $f : A \\to B$ is a bijection by showing that it has an inverse: a function $g : B \\to A$ such that $g(f(a)) = a$ and $f(g(b)) = b$ for all $a \\in A$ and $b \\in B$. As we saw in my last post, these facts imply that $f$ is one-to-one and onto, and hence a bijection. And it really is necessary to prove both $g(f(a)) = a$ and $f(g(b)) = b$: if only one of these hold then $g$ is called a left or right inverse, respectively (more generally, a one-sided inverse), but $f$ needs to have a full-fledged two-sided inverse in order to be a bijection.\n\n\u2026unless $A$ and $B$ are of the same finite size! In that case, it\u2019s enough to show the existence of a one-sided inverse\u2014say, a function $g$ such that $g(f(a)) = a$. Then $f$ is (say) a one-to-one function between finite equal-sized sets, hence it is also onto (and hence $g$ is actually a two-sided inverse).\n\nWe must be careful, however: sometimes the reason for constructing a bijection in the first place is in order to show that $A$ and $B$ have the same size! This kind of thing is common in combinatorics. In that case one really must show a two-sided inverse, even when $A$ and $B$ are finite; otherwise you end up assuming what you are trying to prove.\n\nBy mutual injection?\n\nI\u2019ll leave you with one more to ponder. Suppose $f : A \\to B$ is one-to-one, and there is another function $g : B \\to A$ which is also one-to-one. We don\u2019t assume anything in particular about the relationship between $f$ and $g$. Are $f$ and $g$ necessarily bijections?\n\nPosted in logic, proof | | 7 Comments\n\nOne-sided inverses, surjections, and\u00a0injections\n\nSeveral commenters correctly answered the question from my previous post: if we have a function $f : A \\to B$ and $g : B \\to A$ such that $g(f(a)) = a$ for every $a \\in A$, then $f$ is not necessarily invertible. Here are a few counterexamples:\n\n\u2022 Commenter Buddha Buck came up with probably the simplest counterexample: let $A$ be a set with a single element, and $B$ a set with two elements. It does not even matter what the elements are! There\u2019s only one possible function $g : B \\to A$, which sends both elements of $B$ to the single element of $A$. No matter what $f$ does on that single element $a \\in A$ (there are two choices, of course), $g(f(a)) = a$. But clearly $f$ is not a bijection.\n\n\u2022 Another counterexample is from commenter designerspaces: let $f : \\mathbb{N} \\to \\mathbb{Z}$ be the function that includes the natural numbers in the integers\u2014that is, it acts as the identity on all the natural numbers (i.e. nonnegative integers) and is undefined on negative integers. $g : \\mathbb{Z} \\to \\mathbb{N}$ can be the absolute value function. Then $g(f(n)) = |n| = n$ whenever $n$ is a natural number, but $f$ is not a bijection, since it doesn\u2019t match up the negative integers with anything.\n\nUnlike the previous example, in this case it is actually possible to make a bijection between $\\mathbb{N}$ and $\\mathbb{Z}$, for example, the function that sends even $n$ to $n/2$ and odd $n$ to $-(n+1)/2$.\n\n\u2022 Another simple example would be the function $f : \\mathbb{N} \\to \\mathbb{N}$ defined by $f(n) = 2n$. Then the function $g(n) = \\lfloor n/2 \\rfloor$ satisfies the condition, but $f$ is not a bijection, again, because it leaves out a bunch of elements.\n\n\u2022 Can you come up with an example $f : \\mathbb{R} \\to \\mathbb{R}$ defined on the real numbers $\\mathbb{R}$ (along with a corresponding $g$)? Bonus points if your example function is continuous.\n\nAll these examples have something in common, namely, one or more elements of the codomain that are not \u201chit\u201d by $f$. Michael Paul Goldenberg noted this phenomenon in general. And in fact we can make this intuition precise.\n\nTheorem. If $f : A \\to B$ and $g : B \\to A$ such that $g(f(a)) = a$ for all $a \\in A$, then $f$ is injective (one-to-one).\n\nProof. Suppose for some $a_1, a_2 \\in A$ we have $f(a_1) = f(a_2)$. Then applying $g$ to both sides of this equation yields $g(f(a_1)) = g(f(a_2))$, but because $g(f(a)) = a$ for all $a \\in A$, this in turn means that $a_1 = a_2$. Hence $f$ is injective.\n\nCorollary. since bijections are exactly those functions which are both injective (one-to-one) and surjective (onto), any such function $f : A \\to B$ which is not a bijection must not be surjective.\n\nAnd what about the opposite case, when there are functions $f : A \\to B$ and $g : B \\to A$ such that $f(g(b)) = b$ for all $b \\in B$? As you might guess, such functions are guaranteed to be surjective\u2014can you see why?\n\nPosted in logic | | 1 Comment\n\nSuppose we have sets $A$ and $B$ and a function $f : A \\to B$ (that is, $f$\u2019s domain is $A$ and its codomain is $B$). Suppose there is another function $g : B \\to A$ such that $g(f(a)) = a$ for every $a \\in A$. Is $f$ necessarily a bijection? That is, does $f$ necessarily match up each element of $A$ with a unique element of $B$ and vice versa? Or put yet another way, is $f$ necessarily invertible?", "date": "2022-06-29 12:16:38", "meta": {"domain": "mathlesstraveled.com", "url": "https://mathlesstraveled.com/?cpage=1", "openwebmath_score": 0.8304086923599243, "openwebmath_perplexity": 219.63282870187984, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145725743421, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8537969150909941}}
{"url": "http://rfmb.valledelchieseonline.it/rotation-matrix-3d.html", "text": "# Rotation Matrix 3d\n\nThe 3-dimensional versions of the rotation matrix A are the following matrices:. For counterclockwise rotation, the matrix has the following elements. Or you transpose the matrix first (mtmp=m; for x=0 to 2 for y=0 to 2 m[x,y] = mtmp[y,x]) and do your standard rotation calculation with the transposed matrix. a rotation around the z-axis wouldn't change the z-values of the vertices. However, I can only figure out how to do 1 and 4 using numpy. We learn how to describe the orientation of an object by a 2\u00d72 rotation matrix which has some special properties. And thank you for taking the time to help us improve the quality of Unity Documentation. Primarily to support 3D rotations. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. We can project a point orthogonal down into one of the main planes by using a matrix that scale the axis normally onto the plane with 0. The 3D object is moved and rotated in the 3D space, and the new destination points become B1=, B2=, and B3=. \u2022 In 2D, a rotation just has an angle \u2022 In 3D, specifying a rotation is more complex -basic rotation about origin: unit vector (axis) and angle \u2022convention: positive rotation is CCW when vector is pointing at you \u2022 Many ways to specify rotation -Indirectly through frame transformations -Directly through \u2022Euler angles: 3 angles. When I look at the file, however, it appears that the inputs to the transformation are the trans x,y,z and the roll,pitch,yaw angles. Raises: ValueError: If the shape of angles is not supported. Rotation About an Arbitrary Axis and Avoiding Gimbal Lock - Cprogramming. Rotation Matrices Suppose that \u21b5 2 R. translation, rotation, scale, shear etc. A 3 by 3 matrix sets the rotation and shear. The axis can be either x or y or z. 1 Matrix Representation A 2D rotation is a tranformation of the form 2 4 x 1 y 1 3 5 = 2 4 cos( ) sin( ) sin( ) cos( ) 3 5 2 4 x 0 y 0 3 5 (1) where is the angle of rotation. We will try to enter into the details of how the matrices are constructed and why, so this article is not meant for absolute beginners. euler_rotation = mathutils. Raises: ValueError: If the shape of quaternion is not supported. The rotation matrix for this transformation is as follows. Convert a Rotation Matrix to Euler Angles in OpenCV. Each has its own uses and drawbacks. The only thing new in the C++ code is the usage of GetConsoleScreenBufferInfo and SetConsoleTextAttribute which gets the size of the console and sets the text color. We can think of rotations in another way. Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4. \u2212Composition of geometric transformations in 2D and 3D. Equations ()-() effectively constitute the definition of a vector: i. A short derivation to basic rotation around the x-, y- or z-axis by Sunshine2k- September 2011 1. A matrix with M rows and N columns is defined as a MxN matrix. How to combine rotation in 2 axis into one matrix. It's so clever that it's worth sharing in full detail. In 2D it's much simpler. And Rotation is done with trigonometric functions in the matrix. If we add. Simple rotation \u2013 formulas were derived for rotation of a shape centered in origin by a certain angle. When a transformation takes place on a 2D plane, it is called 2D transformation. This matrix class is used for 3D object rotate along a specifid axis. The only difference is the signs for sin\u03b8 are reversed. matrix() Describes a homogeneous 2D transformation matrix. gives the column matrix corresponding to the point (a+ dx, b+ dy, c+ dz). Initially I hoped that. I have a point, in that mesh, that i must calculate mannualy. The rotation operation consists of multiplying the transformation matrix by a matrix whose elements are derived from the angle parameter. Rotation matrix from axis and angle First rotate the given axis and the point such that the axis lies in one of the coordinate planes Then rotate the given axis and the point such that the axis is aligned with one Use one of the fundamental rotation matrices to rotate the point depending. Greetings, I'm trying to apply a Rotation or Affine matrix to a spiral. The preview on the right will be updated when you compute, x\u2019 (dash) points to the new x direction of the body, y\u2019, z\u2019 do the same for y and z axes. We can project a point orthogonal down into one of the main planes by using a matrix that scale the axis normally onto the plane with 0. As others stated in the comments you have to make sure you don\u00b4t mix row- and column-major matrices and dont mix any coordinate spaces. I have a 3D translation and rotation problem I am trying to solve using Excel 2010. As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. We can think of rotations in another way. @Sascha Grusche and @Elie Maalouf. As a unit quaternion, the same 3D rotation matrix. Learn more about rotation matrix, point cloud, 3d. They are represented in the matrix form as below \u2212. The formula is , using the dot and cross product of vectors. Describing rotation in 3d with a vector. A matrix is an array of values that defines a transformation of coordinates. This is why also the 3D version has two of the three axes change simultaneously - because it is just a derivative from its 2D version. RotationMatrix[{u, v}] gives the matrix that rotates the vector u to the direction of the vector v in any dimension. These singularities are not characteristic of the rotation matrix as such, and only occur with the usage of Euler angles. They are both ways of completely describing a rotation in 3D, so they are freely convertible. You can also rotate, resize and stretch a 3D graph by dragging the mouse. Download our 100% free 3D Rotation templates to help you create killer PowerPoint presentations that will blow your audience away. 3D scaling matrix. Hi, I'm trying to transform a PET scan onto a CT scan based on an existing rotation and translation matrix. G-CNNs achieve equivariance with respect to nite subgroups of the rotation group, which constitutes a bottleneck in 3D. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. We learn how to describe the orientation of an object by a 2\u00d72 rotation matrix which has some special properties. 3D rotations made easy in Julia. Appearance Scope: 6. When the Frobenius norm is taken as the measure of closeness, the solution is usually computed using the singular value decomposition (SVD). 1 Matrix Representation A 2D rotation is a tranformation of the form 2 4 x 1 y 1 3 5 = 2 4 cos( ) sin( ) sin( ) cos( ) 3 5 2 4 x 0 y 0 3 5 (1) where is the angle of rotation. Position Cartesian coordinates (x,y,z) are an easy and natural means of representing a position in 3D space \u2026But there are many other representations such as spherical converted to matrix form to perform rotation. As we know $\\cos(0) = 1$ and $\\sin(0) = 0$. Three shears. Appearance: 7. ACM SIGGRAPH is a thriving international organization. det(R) != 1 and R. step file, it stores a rotation matrix that needs to be applied to the appropriate geometry, correct?. Rotation in 3D - The Rotation Matrix In this note, I investigate the rotation matrix that relates the image of a point p \u20d7 \\vec{p} p when it is rotated by an angle \u03b8 \\theta \u03b8 about an axis a \u20d7 \\vec{a} a that passes through the origin. winter wheat (WW))is available to successive crops in reduced or no-till systems, uncover the mechanisms involved in this process, and ultimately develop management practices to maximize Po utilization from these sources. Serializable. If you have a rotation matrix, then this will transform from one coordinate system to another. So essentially quaternions store a rotation axis and a rotation angle, in a way that makes combining rotations easy. Just like the graphics pipeline, transforming a vector is done step-by-step. A 2-columns matrix or data frame containing a set of X and Y coordinates. Abstract transformations, such as rotations (represented by angle and axis or by a quaternion), translations, scalings. A rotation matrix can be built by using the axis of the coordinate system you're rotating into. There will be some repetition of the earlier analyses. Convert your quaternion to a rotation matrix, and use it in the Model Matrix. 3D programming in python. Posted September 16, 2017 \u00b7 CuraEngine mesh_rotation_matrix to rotate 180 degrees Just a WAG here, but since the print origin is at the corner of the print bed (unless you're using a Delta printer), and the rotation matrix is rotating about the origin, then haven't you just rotated it off the print bed?. I've been all. I want this rotation matrix to perform a rotation about the X axis (or YZ pla. Rotation matrices, on the other hand, are the representation of choice when it comes to implementing efficient rotations in software. Coordinates of point p in two systems The elementary 3D rotation matrices are constructed to perform In order to be able to write the rotation matrix directly, imagine that the the z-axis is playing the role of the x-axis,. As their trunks were rotated passively without fixing the legs, each volunteer was instructed to keep their knees straight and rotate their legs in a fashion comparable to the trunk rotation, in order to minimize the effects of leg rotation on trunk rotation. The easiest rotation axes to handle are those that are parallel to the co-ordinate axes. , is an orthogonal matrix) is 1). New coordinates by 3D rotation of points Calculator. Basic steps needed to display 3D objects: 4. If the first body is only capable of rotation via a revolute joint, then a simple convention is usually followed. Email this Article Givens rotation. Rotation Transforms for Computer Graphics covers a wide range of mathematical techniques used for rotating points and frames of reference in the plane and 3D space. Rotation. explanation of quaternion from matrix. Find the characteristic function, eigenvalues, and eigenvectors of the rotation matrix. Tag: math,vector,lua,rotation I have two Vec3s, Camera Forward and Turret Forward. Turn your head left and right; that\u2019s a rotation around the Y axis. The rotation operation is a 3x3 matrix. Then, the tutorials will move on to give you the matrices for rotation over the x and y axes, tell you how to use them, and then give you a matrix which will allow rotations around an arbitrary axis. With a chain of rotations, roundoff errors accumulate. The rotation matrix for this transformation is as follows. 3D Transformations, Translation, Rotation, Scaling The Below program are for 3D Transformations. Unless specified, the rest of this page uses implies rotation to be a rotation of points about the origin. We conclude that every rotation matrix, when expressed in a suitable coordinate system, partitions into independent rotations of two-dimensional subspaces, at most n \u2044 2 of them. represent data on 3D rotation groups. muting any, we clearly need a negative unit matrix, namely, Proper Rotation. Because Rotation can be done either along the x, y, or z axis, there is a different rotation matrix for each of the axises: Figure 6a - Rotation around the X axis. Its first 3 dimensional vectors(3*3 submatrix) contain the rotated X, Y and Z axes. Examples in 3D computer graphics Rotation. In SO(4) the rotation matrix is defined by two quaternions, and is therefore 6-parametric (three degrees of freedom for every quaternion). The most general three-dimensional rotation matrix represents a counterclockwise rotation by an angle \u03b8 about a \ufb01xed axis that lies along the unit vector \u02c6n. Under rotations, vector lengths are preserved as well as the angles between vectors. The formula of this operations can be described in a simple multiplication of. You can probably use the axis angle representation to build a rotation matrix, with whatever math library you have available. G-CNNs achieve equivariance with respect to nite subgroups of the rotation group, which constitutes a bottleneck in 3D. This means that no rotation has taken place around any of the axes. In the above code, since the rotation transformation is prepended to the matrix, the rotation transformation would be performed first. Lecture 08. 2+) were then focused to capture these y-z\u2032 images at >800 frames per second as the sheet was repeatedly scanned across the sample in the x direction. Each has its own uses and drawbacks. This object get transformed with the following matrix transformation: glRotated and glTranslate. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. A rotation matrix describes the relative orientation of two such frames. rotqrmean Find the average of several rotation quaternions rotqrvec Apply a quaternion rotation to an array of 3D vectors skew3d Convert between vectors and skew symmetric matrices: 3x3 matrix <-> 3x1 vector and 4x4 Plucker matrix <-> 6x1 vector. the product of a viewing matrix and a modeling matrix. If you are uncomfortable with the thought of 4D matrix rotations, then I recommend reading Wikipedia, or checking out my article about 3D graphing, which can be found here. where M is a constant 3x3 matrix, is the 3x3 identity matrix, and we are solving for the 3x3 matrix R. Using 3D Rotation Matrices in Practice By confuted So, now that you more or less know how to rotate a point in any arbitrary manner in three dimensions, generating matrices along the way, it's time to learn what you should do with each of these matrices. We found that this was the rotation transformation matrix about an x-axis rotation. First, suppose that all eigenvalues of the 3D rotation matrix A are real. We will try to enter into the details of how the matrices are constructed and why, so this article is not meant for absolute beginners. If you want to rotate around an arbitary axis you can use the equations under \"Rotation matrix from axis and angle\" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. ' (as long as the translation is ignored). Again, we must translate an object so that its center lies on the origin before scaling it. We'll call the rotation matrix for the X axis matRotationX, the rotation matrix for the Y axis matRotationY, and the rotation matrix for the Z axis matRotationZ. The eigenvalues of A are. Therefore, an arbitrary 3D rotation can be decomposed into only two 3D beamshears. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. I'm looking for a SO(3) rotation representation that lends itself to energy minimization. Rotation matrices are used for computations in aerospace, image processing, and other technical computing applications. We conclude that every rotation matrix, when expressed in a suitable coordinate system, partitions into independent rotations of two-dimensional subspaces, at most n / 2 of them. Click the 3D graph (the white space of the graph layer, not the 3D data plot), eight resizing handles appear around the 3D graph. Matrix representation. This axis, in this work, will be represented by the supporting line of the directed segment S ab (a 1D simplex), where ( , , (0)) 3 (0) 2 (0) a a1 a a and ( , (0)) 3 (0) 2 (0) b 1 b b are two non-coincident 3D points which we will refer as the. \u2022In 3D, specifying a rotation is more complex \u2013basic rotation about origin: unit vector (axis) and angle \u2022convention: positive rotation is CCW when vector is pointing. The Rotation 3D page. The easiest rotation axes to handle are those that are parallel to the co-ordinate axes. Now, according to the equation, multiplying the transformation matrix with a coordinate would result in a coordinate but if is [9,1] for example, if i multiply with the rotation matrix. Yes, [R|t] implies the rotation and translation. Serializable, java. Rotation matrices are used in computer graphics and in statistical analyses. calibration cube. 3D Matrices. The Rotation 3D page. The eigenvalues of A are. rotation matrix specifies a 3 \u00d7 3 matrix. ppt), PDF File (. Shortest distance between two lines. 0 License, and code samples are licensed under the Apache 2. A short derivation to basic rotation around the x-, y- or z-axis by Sunshine2k- September 2011 1. In 2D euclidean space, rotation matrix is a matrix that tilts every single vector in the 2D space, without changing the scale. pdf), Text File (. Second, this method means we can create our own css animations, and do something a bit more advanced. Computing Euler angles from a rotation matrix Gregory G. I understand the sentence except for the \"rotation matrix\" part, again. RotationMatrix[{u, v}] gives the matrix that rotates the vector u to the direction of the vector v in any dimension. 3DPrimitivesTransformations - Free download as Powerpoint Presentation (. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. If you are reading this page in order to write a 3D computer program I suggest you read enough of this page to convince yourself of the problems with Euler angles and to get an intuitive understanding of 3D rotations and then move on to quaternion or matrix algebra representations. skip_checks (bool, optional) \u2013 If True avoid sanity checks on rotation_matrix for performance. Math for simple 3D coordinate rotation (python) Ask Question Asked 3 years, 5 months ago. Defining the rotation axis as the z axis, we note first that the z coordinate will be unchanged by any rotation about the z axis. A general homoge- neous matrix formulation to 3D rotation geometric transformations is proposed which suits for the cases when the rotation axis is unnecessarily through the coordinate system origin given their rotation axes and. When acting on a matrix, each column of the matrix represents a different vector. The three simultaneous orthogonal rotations measured with a 3D gyroscope represent a single rotation around a certain axis for a certain angle. 3D rotations \u2022 A 3D rotation can be parameterized with three numbers \u2022 Common 3D rotation formalisms - Rotation matrix \u2022 3x3 matrix (9 parameters), with 3 degrees of freedom - Euler angles \u2022 3 parameters - Euler axis and angle \u2022 4 parameters, axis vector (to scale) - Quaternions \u2022 4 parameters (to scale). multiplying rotation matrices is a noisyrotation matrix [1]. M modelview = M viewing * M modeling. We can now write a transformation for the rotation of a point about this line. And so here's the rotation transformation matrix. RotationMatrix[{u, v}] gives the matrix that rotates the vector u to the direction of the vector v in any dimension. R = rotx(ang) creates a 3-by-3 matrix for rotating a 3-by-1 vector or 3-by-N matrix of vectors around the x-axis by ang degrees. 2+) were then focused to capture these y-z\u2032 images at >800 frames per second as the sheet was repeatedly scanned across the sample in the x direction. 1 1 3 Lecture Video 3 of 4 Rotation Matrix Example 1 - Duration: 10:20. Ask Question With r = RotationMatrix[a, {x, y, z}] I can compute a 3D rotation matrix from its axis/angle representation. all types of rotation calculation will eventually yield a 4x4 matrix, as this is always the required form for expressing the final transformation. Represents a 3D rotation as a rotation angle around an arbitrary 3D axis. The 3-dimensional versions of the rotation matrix A are the following matrices:. \u02c7, rotation by \u02c7, as a matrix using Theorem 17: R \u02c7= cos(\u02c7) sin(\u02c7) sin(\u02c7) cos(\u02c7) = 1 0 0 1 Counterclockwise rotation by \u02c7 2 is the matrix R \u02c7 2 = cos(\u02c7 2) sin(\u02c7) sin(\u02c7 2) cos(\u02c7 2) = 0 1 1 0 Because rotations are actually matrices, and because function composition for matrices is matrix multiplication, we\u2019ll often multiply. 3D Rotation Matrix. From the Cartesian grid (left grid), we can see the blue point is located at (2, 1). Rotation Matrix. 3D rotation matrix around vector. is the rotation matrix already, when we assume, that these are the normalized orthogonal vectors of the local coordinate system. 0 License , and code samples are licensed under the Apache 2. I know that in 3D space the matrix product order is important - changing the order of the matrices can effect the rotate result. winter wheat (WW))is available to successive crops in reduced or no-till systems, uncover the mechanisms involved in this process, and ultimately develop management practices to maximize Po utilization from these sources. Euler angles can be defined with many different combinations (see definition of Cardan angles ). As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. matrix3d() Describes a 3D transformation as a 4\u00d74 homogeneous matrix. 3D Rotations Rotation about z-axis. When acting on a matrix, each column of the matrix represents a different vector. This is the currently selected item. Learn more about rotation matrix, point cloud, 3d. A vector is a direction in a space (like for GPS), that is the result of the transformation (math operators) of a point by a rotation matrix. 3D rotation) that minimizes some objective function. Transformations is a Python library for calculating 4x4 matrices for translating, rotating, reflecting, scaling, shearing, projecting, orthogonalizing, and superimposing arrays of 3D homogeneous coordinates as well as for converting between rotation matrices, Euler angles, and quaternions. 3D Rotation \u2022 Counterclockwise \u2022The matrix M transforms the UVW vectors to the XYZ vectors y z x u=(u x,u y,u z) v=(v x,v y,v z) Change of Coordinates. 8 (b): What constraints must the elements R_{ij} of the three-dimensional rotation matrix satisfy, in order to preserve the length of vector A (for all vectors A)? Homework Equations The. find angles , , which make the two matrices equal. Euler3D transform, how to set the rotation matrix?. This is done by multiplying the vertex with the matrix : Matrix x Vertex (in this order. Let T be a linear transformation from R^2 to R^2 given by the rotation matrix. Quaternions are just more compact and easier to interpolate. I want to rotate an object by 60 degrees around the y axis, counter-clockwise. Three shears. The solution is not unique in most cases. In matrix form, the infinitesimal rotation has the representation (4) where and (5) To first order in , it can be shown from that. Position Cartesian coordinates (x,y,z) are an easy and natural means of representing a position in 3D space \u2026But there are many other representations such as spherical converted to matrix form to perform rotation. Thus, we have H O = [I O] \u03c9 ,. They are represented in the matrix form as below \u2212. Scale and Rotate. Derivation of the 3D transformation matrix. Follow 34 views (last 30 days). A rotation matrix describes the relative orientation of two such frames. Angular velocity. Although the inverse process requires a choice of rotation axis between the two alternatives, it is a straightforward procedure to retrieve the rotation axis and angle (see Appendix A). Euler returns a 3D vector containing the XYZ Euler angles. Rotations in 3D applications are usually represented in one of two ways: Quaternions or Euler angles. Do you know any reference how to derive this rotation matrix? It seems clear to me the rotation matrix in planar truss, 3D truss and planar frame are pretty similar (only different a bit), the form has same appearance to a simple rotation matrix, or when using directional cosine. To get in right-handed coordinate system, replace the angle with negative. rotation from the scanner coordinates to the camera coordinates will use the following transformation matrix. \u2212Matrix representation of affine transformations. Please try again in a few minutes. In order to export actual quaternion data to your ASCII log file, you will have to set the Orientation output to Quaternions in MT Manager under Tools > Preferences > Exporters. it is called 3D transformation. The table of direction cosines relating the femoral (F) and pelvic (P) reference frames is obtained most simply via matrix multiplication, which yields the rotation matrix F R P and its corresponding table of direction cosines, Direction cosines for virtually any compound rotation can be found easily by using this exact methodology. A matrix is an array of values that defines a transformation of coordinates. C 3d Rotation Codes and Scripts Downloads Free. Slabaugh Abstract This document discusses a simple technique to \ufb01nd all possible Euler angles from a rotation matrix. Using localOrientation for rotation values, however, returns a 3\u00d73 matrix. To produce a sequence of transformations with these equations, rotation is followed by translation, we must calculate the transformed coordinates one step at a time, thereby eliminating the calculation of intermediate coordinate values. (x_x, x_y, x_z) is a 3D vector that represents only the direction of the X-axis with respect to the coordinate system 1. Consider a counter-clockwise rotation of 90 degrees about the z-axis. Object implements java. Rotation estimation is a fundamental step for object motion estimation, alignment and registration, in image processing, whereas, 3-D shape reconstruction, object recognition, autonomous navigation and ego-motion are typical applications of rotation estimation in computer vision and robotics. I have a similar class. Matrix M 1 is a 2x2 rotation matrix, M 2 is translation vector. Normalize those vectors. winter wheat (WW))is available to successive crops in reduced or no-till systems, uncover the mechanisms involved in this process, and ultimately develop management practices to maximize Po utilization from these sources. A rotation matrix describes the relative orientation of two such frames. 1 1 3 Lecture Video 3 of 4 Rotation Matrix Example 1 - Duration: 10:20. 3D Rotation Matrix. A rotation followed by a translation is very different from a translation followed by a rotation, as illustrated below: Using the Matrix Object. If you want to rotate around an arbitary axis you can use the equations under \"Rotation matrix from axis and angle\" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. The matrix to the left is a parallel projection down into the xy-plane. They will allow us to transform our (x,y,z,w) vertices. This article discusses the different types of matrices including linear transformations, affine transformations, rotation, scale, and translation. The three simultaneous orthogonal rotations measured with a 3D gyroscope represent a single rotation around a certain axis for a certain angle. The rotation matrix has the following properties: A is a real, orthogonal matrix, hence each of its rows or columns represents a unit vector. CSS also supports 3D transformations. With c++ (win32). Alternatively, you can set the Orientation output to Rotation Matrix directly. Ask Question Asked 3 years, 10 months ago. But why would this 3D frame rotation seems much different from those?. In this, the first of two articles I will show you how to encode 3D transformations as a single 4\u00d74 matrix which you can then pass into the appropriate. Details EulerMatrix is also known as Euler rotation matrix or Euler rotation, and the angles \u03b1 , \u03b2 , and \u03b3 are often referred to as Euler angles. Quaternions represent a single rotation. math on December 25, 2008. When acting on a matrix, each column of the matrix represents a different vector. You might use this when applying the same rotation to a number of different objects,. It allows you to examine different rotation sequences. @Sascha Grusche and @Elie Maalouf. In August 1987, in Vancouver, Canada, almost all of those who worked in the paleomagnetic group at the University College of Rhodesia and Nyasaland, Salisbury, Southern Rhodesia (now the University of Zimbabwe, Harare, Zimbabwe) were by chance attending the International Union of Geodesy and Geophysics meeting. We'll call the rotation matrix for the X axis matRotationX, the rotation matrix for the Y axis matRotationY, and the rotation matrix for the Z axis matRotationZ. 3D Rotations Rotation about z-axis. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. det(R) != 1 and R. Composing a rotation matrix. A single precision floating point 4 by 4 matrix. Video transcript - What I want to do in this video is get some. I want this rotation matrix to perform a rotation about the X axis (or YZ pla. This example shows how to do rotations and transforms in 3D using Symbolic Math Toolbox\u2122 and matrices. Online tools - vector rotation in 3D. 1 Representation Elements of the 3D rotation group, SO(3), are represented by 3D rotation matrices. New coordinates by 3D rotation of points Calculator - High accuracy calculation Welcome, Guest. The general rotation is much the same, with the up vector taken randomly, the desired rotation applied after the initial viewing transformation, and then the inverse of the viewing transformation is applied. Serializable. The Java 3D model for 4 X 4 transformations is: [ m00 m01 m02 m03 ] [ x ] [ x' ] [ m10 m11 m12 m13 ]. 3D Rotations are used everywhere in Computer Graphics, Computer Vision, Geometric Modeling and Processing, as well as in many other related areas. Now suppose we are given a matrix and are required to extract Euler angles corresponding to the above rotation sequence, i. C 3d Rotation Codes and Scripts Downloads Free. Just like the graphics pipeline, transforming a vector is done step-by-step. Examples in 3D computer graphics Rotation. This code checks that the input matrix is a pure rotation matrix and does not contain any scaling factor or reflection for example /** *This checks that the input is a pure rotation matrix 'm'. What is the correct order of transformations scale, rotate and translate and why? 3. Moreover, there are similar transformation rules for rotation about and. is the orthogonal projection of onto. We will try to enter into the details of how the matrices are constructed and why, so this article is not meant for absolute beginners. Follow 34 views (last 30 days). In 2D it's much simpler. Primarily to support 3D rotations. The center of a Cartesian coordinate frame is typically used as that point of rotation. As David Joyce points out, this fact is true in odd dimensions (including 3) but not even dimensions. (b) Because all the columns of U are mutually orthogonal, we can conclude that U is an orthogonal matrix. Consider the original set of basis vectors, i, j, k, and rotate them all using the rotation matrix A. Rotation matrices are orthogonal as explained here. You might use this when applying the same rotation to a number of different objects,. Kelly! above x2: screenshots from here. Composing a rotation matrix. The matrix takes a coordinate in the inner coordinate system described by the 3 vectors and and finds its location in the outer coordinate system. Moreover, there are similar transformation rules for rotation about and. Hi, I am doing optimization on a vector of rotation angles tx,ty and tz using scipy. Parameters: matrix - double[][]. Auckland's prof. When acting on a matrix, each column of the matrix represents a different vector. Appearance Mixed: 8. Generally speaking any matrix in the group SO(3) represents a rotation in 3d. That matrix isn't exactly symmetric, but a rotation matrix that is symmetric is a 180 degree rotation. What does POSIT require to be able to do 3D pose estimation? First it requires image coordinates of some object's points (minimum 4 points). For the rotation matrix R and vector v, the rotated vector is given by R*v. This example shows how to do rotations and transforms in 3D using Symbolic Math Toolbox\u2122 and matrices. General Case Computer Graphics 3d Rotation Software Rendering Computer Graphics 3d Rotation 3d Coordinate Systems Powerpoint Presentation Axis Rotation 3d Rotation Matrix Derivation Machines Vertical Center Number Umc-750 Sample Trials Coordinate Geometry Basics. A general homoge- neous matrix formulation to 3D rotation geometric transformations is proposed which suits for the cases when the rotation axis is unnecessarily through the coordinate system origin given their rotation axes and. The View Matrix: This matrix will transform vertices from world-space to view-space. I have 2 known 3d points which are the origin of 2 axis plot in the space and I need to compute the 3D rotation matrix between them. Unless specified, the rest of this page uses implies rotation to be a rotation of points about the origin. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. The default polygon is a square that you can modify. all types of rotation calculation will eventually yield a 4x4 matrix, as this is always the required form for expressing the final transformation. You need to pass an angle of rotation and x, y, z axes as parameters to this method. However, manipulating 3D Rotations is always confusing, and debugging code that involves 3D rotation is usually quite time consuming. The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix. 3D Transformations, Translation, Rotation, Scaling The Below program are for 3D Transformations. Find more Mathematics widgets in Wolfram|Alpha. Active 3 years, 7 months ago. However, now that I started implementing it myself, I came to the point where I'm really confused. for Java and C++ code to implement these rotations click here. In a two-dimensional cartesian coordinate plane system, the matrix R rotates the points in the XY-plane in the counterclockwise through an angle \u03b8 about the origin. math on December 25, 2008. Then transform a vector by that rotation matrix to get your result. 3D Matrices. CE503 Rotation Matrices Derivation of 2D Rotation Matrix Figure 1. Rotation Matrix. rotation of one frame into the other, Pj is the direction cosine for X2 with respect to Xk, and the rotation transformation matrix is an orthogonal matrix: P R12 = [flk] (5-5) fjk = Cos(X], Xi) JM In Sec. In SO(4) the rotation matrix is defined by two quaternions, and is therefore 6-parametric (three degrees of freedom for every quaternion). The rotation matrix is displayed for the current angle. In an analogous fashion, orientations are described relative to a standard orientation by a rotation, with the identity rotation describing the standard. In this, the first of two articles I will show you how to encode 3D transformations as a single 4\u00d74 matrix which you can then pass into the appropriate. Accordingly, A v = v {\\displaystyle Av=v} , and the rotation axis therefore corresponds to an eigenvector of the rotation matrix associated with an eigenvalue of 1. Matrix multiplications always have the origin as a fixed point. This way to do the inverse rotation works only with pure normalized rotation matrices, should be noticed. 3D Reflection in Computer Graphics- Reflection is a kind of rotation where the angle of rotation is 180 degree. Getting Started with the Java 3D API written in Java 3D: 9. 0\u00b0 (rotation happens on the XY plane in 3D). Understanding transforamtion matrix requires some knoledge of math\u2026 R. 3D Rotation The easiest rotation axes are those that parallel to the coordinate axis. This article shows how to implement three-dimensional rotation matrices and use them to rotate a 3-D point cloud. The homogeneous transformation matrix for 3D bodies As in the 2D case, a homogeneous transformation matrix can be defined. How does OCC read in and apply a rotation matrix in the STEP file? That is, if I define a rotation with BRepBuilderAPI_Transform, save the resulting shape to a. This method prepends or appends the transformation matrix of the Graphics by the rotation matrix according to the order parameter. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. com Starting out. (x_x, x_y, x_z) is a 3D vector that represents only the direction of the X-axis with respect to the coordinate system 1. We learn how to describe the orientation of an object by a 2\u00d72 rotation matrix which has some special properties. Primarily to support 3D rotations. Now, my understanding of your original question is that this unit vector, Ur, represents a rotation from U1, so you want to know how to find the rotation matrix that will transform U1 to Ur, i. Orthogonal matrices represent rotations (more precisely rotations, reflections, and compositions thereof) because, in a manner of speaking, the class of orthogonal matrices was defined in such a way so that they would represent rotations and refle. The rotation matrix has the following properties: A is a real, orthogonal matrix, hence each of its rows or columns represents a unit vector. Try a 90 degree rotation and then check. rotate3d() Rotates an element around a fixed axis in 3D space. 3D Transformations \u2013 Part 1 Matrices Transformations are fundamental to working with 3D scenes and something that can be frequently confusing to those that haven\u2019t worked in 3D before. 1 Eigenvalues An n\u00d7 nmatrix Ais orthogonal if its columns are unit vectors and orthogonal to. Lecture 08. The typical operations are translation, rotation. Here atan2 is the same arc tangent function, with quadrant checking, you typically find in C or Matlab. 1 The matrix for rotation about an arbitrary line. A Rotation instance can be initialized in any of the above formats and converted to any of the others. 3d curl intuition, part 1. Matrix for rotation is a clockwise direction. Geometrical Rotation Videos 9,222 royalty free stock videos and video clips of Geometrical Rotation. Now I would like to add a \"tropism\" command to simulate gravity pulling on the elements of an L-System, similar to how Laurens Lapr\u00e9s LParser does, and this is where I'm stuck. Homogeneous transforms contain BOTH rotation and translation information. R = rotx(ang) creates a 3-by-3 matrix for rotating a 3-by-1 vector or 3-by-N matrix of vectors around the x-axis by ang degrees. ANGLE DECOMPOSITION Recall that the rotation submatrix of the transformation is a multiplication matrix of the dot products of the unit vectors of the two body coordinate systems, and therefore includes trigonometric functions of the three angles of rotation, denoting flexion, abduction, and external rotation. Consider the 3-D rotation matrix U =-0. Ask Question Asked 3 years, 10 months ago. I'm struggling to understand the relation between the angles used to compose a rotation matrix and the angular velocity vector of the body expressed in the body frame. R = Rx*Ry*Rz. Both of these vectors are on different planes where Camera Forward is based on a free-look camera and Turret Forward is determined by the tank it sits on, the terrain the tank is on, etc. it only request for value and display the putout in text format on the screen. Moreover, the rotation axis in the 3D space coincides with the normal vector of the rotation plane. , is an orthogonal matrix) is 1). For x-axis rotation, we have the matrix:. This property allows you to rotate, scale, move, skew, etc. Home / Mathematics / Space geometry; Calculates the new coordinates by rotation of points around the three principle axes (x,y,z). 7 Transformation Matrix and Stiffness Matrix in Three-Dimensional Space. This article might seem exceedingly obvious to some but I'll build up to a point in a few articles. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. Because rotation matrices. 3: geometry of the 2D coordinate transformation The 2 2 matrix is called the transformation or rotation matrix Q. As we know $\\cos(0) = 1$ and $\\sin(0) = 0$. two antiparallel axes and angles (one axis and angle is negation of the other). square matrix \u2013 For 2D, 3x3 matrix \u2013 For 3D, 4x4 matrix. Lecture 5: 3-D Rotation Matrices. In order to calculate the rotation about any arbitrary point we need to calculate its new rotation and translation. 3D rotations made easy in Julia. The rotation matrices SO(3) form a group: matrix multiplication of any two rotation matrices produces a third rotation matrix; there is a matrix 1 in SO(3) such that 1M= M; for each Min SO(3) there is an inverse matrix M 1such that M M= MM 1 = 1. This format is definitely less intuitive than Euler angles, but it's still readable: the xyz components match roughly the rotation axis, and w is the acos of the rotation angle (divided by 2). , the three quantities are the components of a vector provided that they transform under rotation of the coordinate axes about in accordance with Equations ()-(). Then apply the following rules. skip_checks (bool, optional) \u2013 If True avoid sanity checks on rotation_matrix for performance. This package implements various 3D rotation parameterizations and defines conversions between them. In this example, I will only show the 4D rotation matrices. pdf), Text File (. rotation of one frame into the other, Pj is the direction cosine for X2 with respect to Xk, and the rotation transformation matrix is an orthogonal matrix: P R12 = [flk] (5-5) fjk = Cos(X], Xi) JM In Sec. to_matrix() this is the equivalent of. Rotation matrices can be constructed from elementary rotations about the In this lecture, I extend the 2D rotation matrix of SO(2) from Lecture 2. Rotation matrix visualization [5] 2018/09/29 17:08 Male / 20 years old level / High-school/ University/ Grad student / Very /. All input is normalized to unit quaternions and may therefore mapped to different ranges. I want to rotate an object by 60 degrees around the y axis, counter-clockwise. Analogously, we can de\ufb01ne the tensor of inertia about point O, by writing equation(4) in matrix form. The matrix takes a coordinate in the inner coordinate system described by the 3 vectors and and finds its location in the outer coordinate system. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. Given 3 Euler angles , the rotation matrix is calculated as follows: Note on angle ranges. The rotation is performed clockwise, if you are looking along the direction of the rotation axis vector. A great amount of work on this topic is available in literature (see , , , , ). Representation of orientation \u2022 Homogeneous coordinates (review) \u2022 4X4 matrix used to represent translation, scaling, and rotation \u2022 a point in the space is represented as \u2022 Treat all transformations the same so that they can be easily combined p= x y z 1. void: preMultiply(Matrix mb) Premultiplies the object matrix by mb and stores the result in the object; As a result, the. This is the currently selected item. Matrix transposition - if we have a matrix M with n rows and m columns, the transpose of , denoted is a matrix with m rows and n columns, with the first column of equal to the first row of and so on. If you want to rotate around an arbitary axis you can use the equations under \"Rotation matrix from axis and angle\" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. The Mathematics of the 3D Rotation Matrix fastgraph. For some reason your suggested change could not be submitted. Call R v(\u03b8) the 2x2 matrix corresponding to rotation of all vectors by angle +\u03b8. This package implements various 3D rotation parameterizations and defines conversions between them. Stretching 3D Graphs. However, I can only figure out how to do 1 and 4 using numpy. 2) The rotation angles. Figure 6c - Rotation around the Z axis. It can be useful to notice that this can be done with a matrix operation. Euler angles can be defined with many different combinations (see definition of Cardan angles ). WebGL - Cube Rotation. Cartesian coordinates are typically used to represent the world in 3D programming. Transformations, continued 3D Rotation 23 r r r x y z Full 3D Rotation 0 sin cos 0 cos sin 1 0 0 \u2013 Multiply the current matrix by the rotation matrix that. Do we need to subtract the translation vector (t) from matrix M. 1 Introduction. A tensor of shape [A1, , An, 3, 3], where the last two dimensions represent a 3d rotation matrix. I want this rotation matrix to perform a rotation about the X axis (or YZ pla. Thus, we have H O = [I O] \u03c9 ,. First, suppose that all eigenvalues of the 3D rotation matrix A are real. ) and perspective transformations using homogenous coordinates. )? Ask Question Asked 3 years, It is quit lengthy but you can search for decomposing a rotation matrix. 7 The 3D Rotation Toolbar. Given a 3\u00d73 rotation matrix. So you know how a 3D rotation matrix can be expressed in mathematical form. % example: % rotate around a random direction a random amount and then back % the result should be an Identity matrix. This is given by the product T P 1 \u2212 1 T xz \u2212 1 T z \u2212 1 R z (\u03b8) T z T xz T P 1. And so here's the rotation transformation matrix. This means they can rotate your 3D game geometry. Download our 100% free 3D Rotation templates to help you create killer PowerPoint presentations that will blow your audience away. A rotation is different from other types of motions: translations, which have no fixed points, and (hyperplane) reflections, each of them having an entire (n \u2212 1)-dimensional. $\\endgroup$ \u2013 Federico Poloni Nov 18 at 14:23 |. The arrows denote eigenvectors corresponding to eigenvalues of the same color. \u2212Composition of geometric transformations in 2D and 3D. Find something interesting to watch in seconds. But the plane rotation is not realistic. 3 Rotation Matrix We have seen the use of a matrix to represent a rotation. When modelling three dimensions on a two-dimensional computer screen, you must project each point to 2D. Introduction As with strain, transformations of stress tensors follow the same rules of pre and post multiplying by a transformation or rotation matrix regardless of which stress or strain definition one is using. e you want matrix C such that [C] U1 = Ur. Defining the rotation axis as the z axis, we note first that the z coordinate will be unchanged by any rotation about the z axis. Matrix Rotations and Transformations. Try your hand at some online MATLAB problems. The underlying object is independent of the representation used for initialization. 3D Rotations Rotation about z-axis. Rotation (rotation_matrix, skip_checks=False) [source] \u00b6 Bases: DiscreteAffine, Similarity. This is defined in the Geometry module. Rotation Matrix Java. Both proposed algorithms aim at smoothing 3D rotation matrix sequences in a causal way. Euler angles provide a way to represent the 3D orientation of an object using a combination of three rotations about different axes. 3D Rotation Matrix. CS4620/5620: Lecture 5 3D Transforms (Rotations) \u2022A rotation in 3D is around an axis -so 3D rotation is w. I've been following a tutorial for creating a game engine and when I got to calculating the 3D Rotation matrix I ran into the problem that I believe the matrix isn't being calculated properly. Again, the righmost matrix is the operation that occurs first. Again, we must translate an object so that its center lies on the origin before scaling it. HelloJava3Db renders a single, rotated cube: 10. After the first rotation you can specify which rotation shall follow in the 2nd column and for the third rotation you the remaining axis will be automatically selected for you in the 3rd column. pptx), PDF File (. I have a point, in that mesh, that i must calculate mannualy. This article discusses the different types of matrices including linear transformations, affine transformations, rotation, scale, and translation. Thus, we have H O = [I O] \u03c9 ,. If you don't want any rotation you can use the built-in constant Matrix3dIdentity. Rotation around any given axis Rotation from normal vector to normal vector Apparently the 5th function is enough, because for example \"Rotation around X axis\" can be replace by rotation around (1,0,0), and \"Rotation around all axes\" is merely the product of 3 matrices. Practice: Rotate 2D shapes in 3D. Euler returns a 3D vector containing the XYZ Euler angles. Rotation in 3D That works in 2D, while in 3D we need to take in to account the third axis. What do the vectors mean in T? T is a 4*4 column-major matrix. 0625 rz = -0. Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4. ggb file along with images of a red arrow and ball (that represent the spiral) at different angles. represent data on 3D rotation groups. NEGATIVE_DETERMINANT - this matrix has a negative determinant. As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. Moreover, there are similar transformation rules for rotation about and. 3d curl intuition, part 2. Rotation direction is from the first towards the second axis. A rotation matrix is a matrix used to rotate an axis about a given point. Bobick Calibration and Projective Geometry 1 Projection equation \u2022 The projection matrix models the cumulative effect of all parameters \u2022 Useful to decompose into a series of operations **** **** **** 1 X sx Y sy Z s = =. Alternatively, you can set the Orientation output to Rotation Matrix directly. These are not the only possible rotations in 3-space, of course, but we will limit our. Coordinate axes rotations:-Three dimensional transformation matrix for each co-ordinate axes rotations with homogeneous co-ordinate are as given below. 3D rotation is not same as 2D rotation. My matrices are rows by columns, row 0 being the topmost and column 0 being the leftmost. Or is it possible to convert my 3x3 rotation matrix to a Quaternion which let me use4x4Matrix. A non-rotation is described by an identity matrix. This property allows you to rotate, scale, move, skew, etc. User Interfaces with Java. The Camera Transformation Matrix: The transformation that places the camera in the correct position and orientation in world space (this is the transformation that you would apply to a 3D model of the camera if you wanted to represent it in the scene). When modelling three dimensions on a two-dimensional computer screen, you must project each point to 2D. This is the currently selected item. Transformation of Stresses and Strains David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 # Display transformation matrix for these angles: \"evalf\" evaluates the # matrix element, and \"map\" applies the evaluation to each element of # the matrix. Or is there a way to convert my 3x3 rotation matrix and translation to Unity 4x4Matrix since then i can use Matrix4x4. been stuck for about a month on this - i use \"euler angles\" (tait-bryan angles) to describe rotation coordinates. This list is useful for checking the accuracy of a rotation matrix if questions arise. Define the parametric surface x(u,v), y(u,v), z(u,v) as follows. R = rotx(ang) creates a 3-by-3 matrix for rotating a 3-by-1 vector or 3-by-N matrix of vectors around the x-axis by ang degrees. 0 License , and code samples are licensed under the Apache 2. This paper provides a basic introduction to the use of quaternions in 3D rotation applications. , robotics,. Define and Plot Parametric Surface.  Properties of a rotation matrix In three dimensions, for any rotation matrix , where a is a rotation axis and a rotation angle, (i. If you want 3x3, just remove the last column and last row. Rotations in Three-Dimensions: Euler Angles and Rotation Matrices Part 2 - Summary and Sample Code. 33\u00d7 rotation matrix equals a skew-symmetric matrix multiplied by the rotation matrix where the skew symmetric matrix is a linear (matrix-valued) function of the angular velocity and the rotation matrix represents the rotating motion of a frame with respect to a reference frame. Or is it possible to convert my 3x3 rotation matrix to a Quaternion which let me use4x4Matrix. Composition and inversion in the group correspond to matrix multiplication and inversion. public Matrix3DTransformation(double[][] matrix) Constructs a 3D transformation using the given matrix. Try a 90 degree rotation and then check. The theory is given here. Now you have magically gotten out of a room with a closed door! Similarly, to rotate about a point that is not the origin, first you move all the points so the center is the origin, use the usual rotation matrix, and then move all the points back to where you found them. Auckland's prof. Model Rotation. It is moving of an object about an angle. 7 Transformation Matrix and Stiffness Matrix in Three-Dimensional Space. Although the inverse process requires a choice of rotation axis between the two alternatives, it is a straightforward procedure to retrieve the rotation axis and angle (see Appendix A). I've read on page 27 here that a 3x3 transform matrix can be just the nine dot products - thank you U. The rotation is performed clockwise, if you are looking along the direction of the rotation axis vector. Transformations is a Python library for calculating 4x4 matrices for translating, rotating, reflecting, scaling, shearing, projecting, orthogonalizing, and superimposing arrays of 3D homogeneous coordinates as well as for converting between rotation matrices, Euler angles, and quaternions. Under rotations, vector lengths are preserved as well as the angles between vectors. The function uses the Rodrigues formula for the computation. A rotation vector is a convenient and most compact representation of a rotation matrix (since any rotation matrix has just 3 degrees of freedom). skip_checks (bool, optional) \u2013 If True avoid sanity checks on rotation_matrix for performance. Transformations, continued 3D Rotation 23 r r r x y z r r r x y z r r r x y z z y x r r r r r r r r r, , , ,, , , ,, , , , 31 32 33 Full 3D Rotation 0 sin cos 0 cos sin 1 0 0 sin 0 cos 0 1 0 cos 0 sin 0 0 1 sin cos 0 - Multiply the current matrix by the rotation matrix that. Understanding transforamtion matrix requires some knoledge of math\u2026 R. A rotation matrix describes the relative orientation of two such frames. This paper provides a basic introduction to the use of quaternions in 3D rotation applications. square matrix \u2013 For 2D, 3x3 matrix \u2013 For 3D, 4x4 matrix. This format is definitely less intuitive than Euler angles, but it's still readable: the xyz components match roughly the rotation axis, and w is the acos of the rotation angle (divided by 2). Gimbal lock When two rotational axis of an object pointing in the same direction, the rotation ends up losing one degree orientation matrix ( quaternion can be represented as matrix as well) quaternions or orientation matrix Euler angles, quaternion (harder) Summary. The center of a Cartesian coordinate frame is typically used as that point of rotation. Rotation rotate() Rotates an element around a fixed point on the 2D plane. AlignmentRotation (source, target, allow_mirror=False) [source] \u00b6 Bases: HomogFamilyAlignment, Rotation. jl package), and acts to rotate a 3-vector about the origin through matrix-vector multiplication. Eigen's Geometry module provides two different kinds of geometric transformations:. User Interfaces with Java. It turns out that the derivative R_ of a rotation matrix Rmust always be a skew symmetric matrix wb times R\u2013 any-thing else would be inconsistent with the contraints of orthogonality and determinant 1. A 3D rotation is a 2D rotation that is applied within a speci ed plane that contains the origin. When we multiply two rotation matrices, the result is a new matrix that is equivalent to performing the two rotations sequentially. The only thing new in the C++ code is the usage of GetConsoleScreenBufferInfo and SetConsoleTextAttribute which gets the size of the console and sets the text color. \u2212OpenGL matrix operations and arbitrary geometric transformations. Rotation formalisms in three dimensions: | In |geometry|, various |formalisms| exist to express a |rotation| in three |dimensions| a World Heritage Encyclopedia, the aggregation of the largest online encyclopedias available, and the most definitive collection ever assembled. Quat returns a quaternion representing the. Consider the original set of basis vectors, i, j, k, and rotate them all using the rotation matrix A. Orthonormalize a Rotation Matrix By Mehran Maghoumi in 3D Geometry , MATLAB If you use a 3\u00d73 R matrix to store the result of the multiplication of a series of rotation transformations, it could be the case that sometimes you end up with a matrix that is not orthonormal (i. beams in 3D space, a transformation we call a3D beam shear. A matrix with M rows and N columns is defined as a MxN matrix. 0/Image/3D/Matrix/Rotation. 3D rotation matrix around vector. There are several modes available to specify rotation matrices. Position Cartesian coordinates (x,y,z) are an easy and natural means of representing a position in 3D space \u2026But there are many other representations such as spherical converted to matrix form to perform rotation. Simply put, a matrix is an array of numbers with a predefined number of rows and colums. If you want to rotate around an arbitary axis you can use the equations under \"Rotation matrix from axis and angle\" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. Transformations about a Plane. Prove that this linear transformation is an orthogonal transformation. muting any, we clearly need a negative unit matrix, namely, Proper Rotation. The default polygon is a square that you can modify. a unit quaternion 'q' can represent all 3d rotations by q=exp(p), where 'p' is a pure imaginary. Call R v(\u03b8) the 2x2 matrix corresponding to rotation of all vectors by angle +\u03b8. Raises: ValueError: If the shape of quaternion is not supported. This form will allow you to rotate a vector along an arbitrary axis (in three dimensions), by an arbitrary angle. Rotation matrices are used in computer graphics and in statistical analyses. These singularities are not characteristic of the rotation matrix as such, and only occur with the usage of Euler angles.\nqu5c3h40by nfj8pk5c7n3tha 92r7rgmozqnm dp5u6lxh4v70xx 1inp6sxfegffy pyk42wowai98pl kdp7cn3147t0 7pqy6o37oydc qpb1j97983 h69xhpid3gzqra hu2ur9woanw7r2y aj6ssysohofbl3 khhhiwe2gm8f 1ii9mtnm0z8 dq19cqrrvl63 4fbhgegfefqsrhl q5e5lhey3pk 2f7sg9eq6ix m20teoy86fw 7zumpfes1aadq jjz6gmtdutdxh f10ctu3va8jjx vug3rydlwr 7le85daq5tr kxrgja3qqn8 bgaiv3kkatago", "date": "2020-07-04 03:45:04", "meta": {"domain": "valledelchieseonline.it", "url": "http://rfmb.valledelchieseonline.it/rotation-matrix-3d.html", "openwebmath_score": 0.5945829749107361, "openwebmath_perplexity": 608.277822995443, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777179414275, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8537833831960149}}
{"url": "https://math.stackexchange.com/questions/616106/when-is-the-moment-of-inertia-of-a-smooth-plane-curve-is-maximum", "text": "# When is the moment of inertia of a smooth plane curve is maximum?\n\nGiven a smooth plane curve $(x(s),y(s))$, parameterized in arc length $s$, of fixed finite length $L$, its moment of inertia about its center of mass (axis perpendicular to the plane) is given as $$MI = \\int_0^L ((x(s)-x_{cm})^2 + (y(s)-y_{cm})^2) ds$$. What I predict from earlier discussions, and almost convinced is that if we fix length $L$, $MI$ is maximum when the curve is a straight line. I lack the faculty of mathematical machinery (I guess calculus of variations) to prove it, hence is my gentle request to help me out in proving it and thoroughly understanding the situation and all the corollaries and nuances. This not just the result I need, but I want to do more with it and hence would like understand all the things that are making this result and even more general ones (only to plane curves though).\n\n\u2022 It certainly is essential that your curve be connected (which could be surmised from your explicit parametrization statement). \u2013\u00a0Ted Shifrin Dec 23 '13 at 5:03\n\u2022 @Ted Shifrin : yes, agreed. thanks for pointing. \u2013\u00a0Rajesh Dachiraju Dec 23 '13 at 5:21\n\u2022 Can you explain the relation between this question and a similar one ? \u2013\u00a0Tony Piccolo Dec 23 '13 at 16:54\n\u2022 @TonyPiccolo : In Chris Culter 's answer of that question, If the curve is always an arc of a circle, he shows that the moment of inertia is an increasing function of $r$, the radius of curvature. But my question now here is that, the straight line has infinite radius of curvature, hence I wonder if the straight line is the global maximum? \u2013\u00a0Rajesh Dachiraju Dec 23 '13 at 17:13\n\u2022 F.A.Valentine has written Curves of given length and minimum or maximum moments of inertia (1934) but I cannot read it. Can you ? \u2013\u00a0Tony Piccolo Jan 3 '14 at 8:33\n\nThere is no need for any calculus of variation. Ordinary calculus is enough.\n\nFor simplicity of derivation, we will use complex numbers to represent points on the plane.\nLet $z(s) = x(s) + i y(s)$ and WOLOG, we will assume $z(0) = 0$. We can express the position on our curve as an integral:\n\n$$z(t) = \\int_0^t z'(s)\\;ds$$\n\nLet $\\theta(t) = \\begin{cases} 1 & t > 0\\\\ 0 & t \\le 0\\end{cases}$ be the step function. The center of mass is given by\n\n\\begin{align}z_{cm} &= \\frac{1}{L} \\int_0^L z(t) dt = \\frac{1}{L} \\int_0^L \\int_0^t z'(s)\\;ds\\; dt\\\\ &= \\frac{1}{L} \\iint_{[0,L]^2} \\theta(t-s) z'(s)\\;ds\\; dt = \\int_0^L \\left(1-\\frac{s}{L}\\right) z'(s)\\;ds \\end{align}\n\nThe moment of inertia w.r.t. $z(0)$, the origin, is given by\n\n\\begin{align} \\mathcal{M}_0 &= \\int_0^L |z(s)|^2 ds = \\int_0^L \\left(\\int_0^t z'(s_1)ds_1\\right)\\left(\\int_0^t \\bar{z}'(s_2) ds_2\\right) dt\\\\ &=\\iiint_{[0,L]^3} \\theta(t-s_1)\\theta(t-s_2) z'(s_1) \\bar{z}'(s_2) ds_1 ds_2 dt\\\\ &=\\iiint_{[0,L]^3} \\theta(t-\\max(s_1,s_2)) z'(s_1) \\bar{z}'(s_2) ds_1 ds_2 dt\\\\ &= \\iint_{[0,L]^2} \\left(L - \\max(s_1,s_2)\\right) z'(s_1) \\bar{z}'(s_2) ds_1 ds_2 \\end{align} and hence the moment of inertia w.r.t. the center of mass is\n\n$$\\mathcal{M}_{cm} = \\mathcal{M}_0 - L |z_{cm}|^2 = L \\iint_{[0,L]^2} \\Lambda(s_1,s_2) z'(s_1) \\bar{z}'(s_2) ds_1 ds_2$$\n\nwhere \\begin{align}\\Lambda(s_1,s_2) &= 1 - \\max\\left(\\frac{s_1}{L},\\frac{s_2}{L}\\right) - \\left(1-\\frac{s_1}{L}\\right)\\left(1-\\frac{s_2}{L}\\right)\\\\ &= \\left( 1 - \\max\\left(\\frac{s_1}{L},\\frac{s_2}{L}\\right)\\right)\\min\\left(\\frac{s_1}{L},\\frac{s_2}{L}\\right) \\end{align}\n\nSince $|z'(s)| \\equiv 1$ and $\\Lambda(s_1,s_2) > 0$ for $(s_1,s_2) \\in (0,L)^2$, we can bound $\\mathcal{M}_{cm}$ as\n\n$$\\mathcal{M}_{cm} \\le L \\iint_{[0,L]^2} \\Lambda(s_1,s_2) |z'(s_1)||z'(s_2)| ds_1 ds_2 = L \\iint_{[0,L]^2} \\Lambda(s_1,s_2) ds_1 ds_2$$ Notice the equality in above inequality is achieved when and only when $z'(s)$ is a constant.\nWe can conclude $\\mathcal{M}_{cm}$ is largest for straight lines.\n\n\u2022 Impressive! The kernel $\\Lambda$ would look much better if you took $L=1$ (obviously WOLOG). \u2013\u00a0Post No Bulls Dec 26 '13 at 6:43\n\u2022 @achille : Its actually mind blowing how you arrived at this using algebra. I wonder whether there are any geometric interpretations of the expression for $\\mathcal(M_{cm}$, or can we arrive at it using geometrical arguments. It would be very interesting to know about such a thing. \u2013\u00a0Rajesh Dachiraju Dec 26 '13 at 11:37\n\u2022 @RajeshD I don't have any geometric interpretation for $\\mathcal{M}_{cm}$. This approach is motivated by visualizing the curve as a chain of line segments with small fixed lengths. Since the only degree of freedoms are the directions of the line segments and we know in certain sense, the moment of inertia $\\mathcal{M}_{cm}$ is a non-negative quadratic functions of these degree of freedoms. I try to express the dependence explicitly and see what can be done. \u2013\u00a0achille hui Dec 26 '13 at 11:56\n\u2022 @Robjohn and achillehui : As per my perception Robjohn's answer seems to be elegant, but I somehow like the expression for moment of inertia $MI_cm$ derived in Achille's answer which explicitly shows the only degree of freedom angles of the tangents and says it is maximum when the angle is a constant function. \u2013\u00a0Rajesh Dachiraju Jan 2 '14 at 9:07\n\nAssumptions:\n\n1. the center of mass is $0$ $$0=\\int_0^L\\delta f\\,\\mathrm{d}s\\tag{1}$$ 2. $f$ is parametrized by arc length: $f'\\cdot f'=1$ \\begin{align} 0&=\\int_0^Lf'\\cdot\\delta f'\\,\\mathrm{d}s\\\\ &=\\int_0^Lf'\\cdot\\,\\mathrm{d}\\delta f\\\\ &=\\Big[\\,f'\\cdot\\delta f\\,\\Big]_0^L-\\int_0^Lf''\\cdot\\delta f\\,\\mathrm{d}s\\\\ &=-\\int_0^Lf''\\cdot\\delta f\\,\\mathrm{d}s\\tag{2} \\end{align} Note that $\\delta f$ can be adjusted in the direction of $f'$ near $0$ and $L$ without affecting the integral of $f''\\cdot\\delta f$ since $f'\\cdot f''=0$. Thus, we can assume $\\Big[\\,f'\\cdot\\delta f\\,\\Big]_0^L=0$.\n\n3. the moment of inertia is stationary $$0=\\int_0^Lf\\cdot\\delta f\\,\\mathrm{d}s\\tag{3}$$ Conclusions:\n\nFor an extreme $f$, any $\\delta f$ that satisfies $(1)$ and $(2)$ also satisfies $(3)$; thus, linearity says that there are constants $a$ and $b$ so that $$f=a+bf''\\tag{4}$$ Since $f$ is parameterized by arclength $f'\\cdot f''=0$, integrating the dot product of $(4)$ with $f'$ yields $$\\frac12f\\cdot f=a\\cdot f+c\\tag{5}$$ Equation $(5)$ represents an arc of the circle $$|f-a|=\\left(2c+|a|^2\\right)^{1/2}\\tag{6}$$ or in the extreme case, a line segment.\n\nChecking Possible Arcs:\n\nThe equation of an arc of radius $r$ is $$f=r(\\cos(s/r),\\sin(s/r))\\tag{7}$$ The center of mass is $$\\frac1L\\int_{-L/2}^{L/2}r(\\cos(s/r),\\sin(s/r))\\,\\mathrm{d}s=\\left(\\frac{2r^2}{L}\\sin\\left(\\frac{L}{2r}\\right),0\\right)\\tag{8}$$ The moment of inertia is \\begin{align} &\\frac{r^2}{L}\\int_{-L/2}^{L/2}\\left[\\left(\\cos(s/r)-\\frac{2r}{L}\\sin\\left(\\frac{L}{2r}\\right)\\right)^2+\\sin^2(s/r)\\right]\\,\\mathrm{d}s\\\\ &=\\frac{r^2}{L}\\int_{-L/2}^{L/2}\\left[1-\\frac{4r}{L}\\sin\\left(\\frac{L}{2r}\\right)\\cos(s/r)+\\frac{4r^2}{L^2}\\sin^2\\left(\\frac{L}{2r}\\right)\\right]\\,\\mathrm{d}s\\\\ &=r^2-\\frac{4r^4}{L^2}\\sin^2\\left(\\frac{L}{2r}\\right)\\\\ &=r^2\\left(1-\\frac{\\sin^2\\left(\\frac{L}{2r}\\right)}{\\left(\\frac{L}{2r}\\right)^2}\\right)\\tag{9} \\end{align} $(9)$ increases to $\\frac{L^2}{12}$ as $r\\to\\infty$. Thus, the maximal moment of inertia would be at the extreme case of a line segment.\n\n\u2022 Thanks for the answer. There are some things I don't understand. In the begining what you mean by $\\delta f$ and other things, I need a bit explanation on your considerations. How you claim that $f$ need to be an arc? etc,. Also at the end I am not able to understand how we get $L^2/12$ as $r$ goes to $\\infty$. \u2013\u00a0Rajesh Dachiraju Dec 27 '13 at 3:03\n\u2022 @RajeshD: $\\delta f$ is a small instantaneous perturbation of $f$. More precisely, $$\\delta\\int F(f(x))\\,\\mathrm{d}x=\\int F'(f(x))\\delta f(x)\\,\\mathrm{d}x$$ \u2013\u00a0robjohn Dec 27 '13 at 9:21\n\u2022 \\begin{align}\\lim_{r\\to\\infty}r^2\\left(1-\\frac{\\sin^2\\left(\\frac{L}{2r}\\right)} {\\left(\\frac{L}{2r}\\right)^2}\\right) &=\\lim_{r\\to\\infty}\\frac{\\frac{L^2}{4}} {\\left(\\frac{L}{2r}\\right)^2} \\left(1-\\frac{\\sin^2\\left(\\frac{L}{2r}\\right)} {\\left(\\frac{L}{2r}\\right)^2}\\right)\\\\ &=\\lim_{x\\to0}\\frac{L^2/4}{x^2}\\left(1-\\frac{\\sin^2(x)} {x^2}\\right)\\\\ &=\\lim_{x\\to0}\\frac{L^2/4}{x^2}\\left(1-\\frac{\\sin(x)} {x}\\right)\\left(1+\\frac{\\sin(x)} {x}\\right)\\\\ &=\\lim_{x\\to0}\\frac{L^2}{2}\\left(\\frac{x-\\sin(x)} {x^3}\\right)\\\\ &=\\lim_{x\\to0}\\frac{L^2}{2}\\left(\\frac{\\cos(x)} {6}\\right)\\\\&=\\frac{L^2}{12}\\end{align} \u2013\u00a0robjohn Dec 27 '13 at 9:52\n\nThere is no maximum ( monotonous increase) , but only a minimum.\n\nUsing polar coordinates, when object and constraint functions are together, variational problem is\n\n$\\int r^2 ds - \\lambda^2 \\int ds,$ where $ds= \\sqrt{(r^2 + r^{'2})} d \\theta$\n\nLagrangian $(r^2 - \\lambda^2)$ is independent of $r^{'}$ when considered with respect to arc $s$. So it is not a variational problem.\n\nMinimum M of I when $r$ is independent of $\\theta$, or when $r$ is a constant. If arc length L is given, minimizing solution is for constant radius loop $L/ (2 \\pi).$( Cowboy lasso)\n\nI am not forgetting about he previous post, but the attached image proves that circles are also extrema of the momentum of inertia.\n\nAlthough a few minutes testing will be sufficient to convince that probably circles are unstable extrema (minimum), while straight lines are stable extrema (maximum).\n\nThis is a classical problem of the calculus of variations, and its is surprising it is not in the collections of classical applications.\n\nIntuitively, it is the shape taken by a lasso, a rope in rotation around a fixed point (which will also be the center of mass, as will be able to check on the result). In facts, playing with a lasso shows that we can expect circles as unstable solutions and straight lines as stable solutions.\n\nThe computation is hard to understand and involves sophisticated use of the calculus of variation. Lets, after choosing appropriate axes, maximize $$M = \\int_a^b (x^2 + y^2) ds, \\text{ subject to } L= \\int_a^b ds \\text{, in which }ds=\\sqrt{1+y'^2}.$$\n\nThe first thing to do is to define the derivative of $M$ and $L$ when looked as function of the curve $y=y(x)$. The so called functional derivative of $$F = \\int_a^b f(x,y,y') dx,$$ is shown to be $$\\delta F = \\frac {\\partial f} {\\partial y} - \\frac d {dx} \\frac{\\partial f} {\\partial y'}.$$\n\nA complete demonstration can be found here. Basically, you replace $y$ by $y+\\epsilon \\: \\eta$ where $\\epsilon \\rightarrow 0$ is a number and $\\eta$ a fixed function. Then you compute the derivative is the usual way: $(f(x,y+\\epsilon \\: \\eta, (y+\\epsilon \\: \\eta)') - f(x,y, y')) / \\epsilon$. Integrating by part replaces the $(y+\\epsilon \\: \\eta)'$ term by $-\\frac d {dx} \\frac{\\partial f} {\\partial y'}$ and a term outside the integral sign, which vanishes because the end points $a$ and $b$ are fixed. Because the formulas are linear, the $\\epsilon$ cancels, giving a result $\\int {\\delta F} \\eta dx$, valid for any $\\eta$ thus the result.\n\nThe functional derivatives are second order differentials that can be computed formally. For $L$ we get: $$\\delta L = \\frac \\partial {\\partial y} \\sqrt{1+y'^2} - \\frac d {dx} \\left[ \\frac \\partial {\\partial y'} \\sqrt{1+y'^2}\\right]= 0 - \\frac d {dx} \\left[ \\frac{y'}{\\sqrt{1+y'^2}} \\right],$$ the first term is null because $ds$ depend only on $y'$ and not on $y$, the second is a derivative when $ds$ is looked as a function of on $y'$. We can the pursue with the usual derivative as a function of $x$: $$\\delta L=- \\frac{y'' \\sqrt{1+y'^2} - y' (\\sqrt{1+y'^2})'}{1+y'^2} = \\cdots = \\frac{y''}{(1+y'^2)^{3/2}}.$$\n\nFor $M$ we have:\n\n$$\\delta M = \\frac \\partial {\\partial y} \\left[ (x^2+y^2) \\sqrt{1+y'^2}\\right]- \\frac d {dx} \\frac \\partial {\\partial {y'}} \\left[ (x^2+y^2) \\sqrt{1+y'^2}\\right],$$ thus, $$\\delta M = 2y \\sqrt{1+y'^2} - \\frac d {dx} \\left[ (x^2+y^2) \\frac \\partial {\\partial {y'}} \\sqrt{1+y'^2}\\right] \\\\= 2y \\sqrt{1+y'^2} - \\frac d {dx} \\left[ (x^2+y^2) \\frac {y'} {\\sqrt{1+y'^2}}\\right] \\\\= \\frac{2y (1+y'^2)}{\\sqrt{1+y'^2}} - (2x+2yy') \\frac {y'} {\\sqrt{1+y'^2}} - (x^2+y^2) \\frac {y''} {(1+y'^2)^{3/2}} \\\\= \\frac{2y-2x y'}{\\sqrt{1+y'^2}} - \\frac {(x^2+y^2)y''}{(1+y'^2)^{3/2}}$$\n\nTo solve the question of the function of a given length with the highest moment of inertia, we have to introduce a Lagrange multiplier. It express at the extremum of a function $M$ subject to a condition $L = C^{te}$, the tangents planes of $M$ and $L$ are parallels. Here, the condition means that is exist a constant $\\mu$, called the Lagrange-multiplier, such that $\\delta M = \\mu \\delta L$.\n\nThis equation, known as the Euler-Lagrange equation, says there exist $\\mu$ such that $$\\frac{2y-2x y'}{\\sqrt{1+y'^2}} - \\frac {(x^2+y^2)y''}{(1+y'^2)^{3/2}} = \\mu \\frac{y''}{(1+y'^2)^{3/2}},$$ which is the same as $$(x^2+y^2+\\mu)y'' = 2(y-x y')(1+y'^2)$$\n\nWhich needs some checks to be continued\n\n\u2022 How did you assume that the moment of inertia of the curve about its center of mass to be $$M = \\int_a^b y^2 dx$$, ofcourse you might have chosen origin as center of mass, but it is still wrong as per the definition given in the question. \u2013\u00a0Rajesh Dachiraju Dec 31 '13 at 20:56\n\u2022 I really don't get what you are trying to say, as per my understanding both the answers given by Robjohn and Achille are correct. They explicitly use arc length parameterization and also assume curve to be of fixed length $L$. \u2013\u00a0Rajesh Dachiraju Dec 31 '13 at 20:59\n\u2022 Of course, I am choosing the origin at the center of mass, computations are difficult enough. \u2013\u00a0AlainD Jan 1 '14 at 11:47\n\u2022 b) You are write, I am on the wrong question. I was looking for the curve minimizing the moment on inertia about an axis (\"chosen\" as Ox). I'll edit in the post. \u2013\u00a0AlainD Jan 1 '14 at 11:54\n\u2022 c) No Robjohn and Achille did not took into account that the curve length is constant. They use \u222bds = L, not d(\u222bds)=0. In facts, if you follow their reasoning to search the curve maximizing the include area (or the lowest center of gravity), you also find straight lines, not circles (or catenaries). \u2013\u00a0AlainD Jan 1 '14 at 12:05", "date": "2019-05-23 01:56:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/616106/when-is-the-moment-of-inertia-of-a-smooth-plane-curve-is-maximum", "openwebmath_score": 0.9473633170127869, "openwebmath_perplexity": 362.9206858080145, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771759145342, "lm_q2_score": 0.8652240877899775, "lm_q1q2_score": 0.853783381882623}}
{"url": "https://lukaneg.github.io/projects/prime-num-optim.html", "text": "# Finding Prime Numbers\u00b6\n\n## Summary\u00b6\n\nHere I explore coding a simple script for finding prime numbers using Python. I try my first attempt without any optimization by creating the script off the top of my head. As you will see, that script works but runs very slowly. I then follow some online lessons on how to optimize the script and get much faster results.\n\n## Prime Finder version 1\u00b6\n\nNow, here is the code for figuring out if any one number is prime.\n\nBut let's allow the user to get all the prime numbers from 1 to the value they enter:\n\nCool! Now turn that into a function:\n\nAnd now run this:\n\nThis seems to work, but let's time how long this takes for larger numbers since it has to search all those numbers and might take a while!\n\nSo, finding all primes between 1 and 10,000 required 8+ seconds! What about 100,000?\n\nstart = time.time() find_primes(100000) end = time.time() elapsed = end - start print(f\"Time elapsed: \",{elapsed})\n\nOk, so after a few minutes I gave up and terminated the kernel. Is there a more efficient way to do this??\n\n## Prime Finder 2.0\u00b6\n\nNeed to optimize the prime number search space with each iteration so that the code doesn't just use brute force and check every single number. For example, we know that even numbers are not prime, so why not remove those first? The following methodology does that, but also removes multiples of every new prime number found.\n\nEssentially we can work in reverse. Rather than finding each number that doesn't have any multiples, work through the list of multiples and remove all numbers from the final list that those multiples can create.\n\nFor example, the first prime number is 2, so we can remove all other values from the set that are multiples of 2. The next prime number is 3, so we can remove all other values from the set that are multiples of 3. The next is 5, since the 4 was already removed when the multiples of 2 were removed, and so on. Thus, the search list becomes smaller and smaller!\n\nNow each time you repeat that set of code, you add values to the actual_primes set and remove values from the potential_primes set, making the search space smaller and smaller.\n\nSo let's repeat that process with a while loop:\n\nGreat! Now time to package it up into a function again:\n\nNow let's time it again!\n\n## All primes through 10,000:\u00b6\n\nWow!! Only 0.006 seconds compared to 8+ seconds with version 1!\n\n## All primes through 100,000:\u00b6\n\nOnly 0.035 seconds for 100,000 compared to several minutes + with version 1\n\n## All primes through 1,000,000:\u00b6\n\nNow for the mega test: search for all prime numbers from one to one million:\n\nLess. Than. A. Second. WOW\n\n## Conclusion\u00b6\n\nSo, as you can see, code optimization can make all the difference between having time for happy hour or not :P\n\nThis was a great experience in learning how to optimize code, and that it is not just about the types of data structures (though I'm sure using sets here helped a ton), it's also about optimizing the algorithm itself so that it does as little reduntant work as possible. I credit Andrew Jones (at Data Science Infinity) for sharing this very elegant but powerful algorithm for cutting through the redundancy and efficiently extracting a list of prime numbers.", "date": "2022-05-19 02:05:32", "meta": {"domain": "github.io", "url": "https://lukaneg.github.io/projects/prime-num-optim.html", "openwebmath_score": 0.5803171396255493, "openwebmath_perplexity": 758.3815330236117, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771805808551, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8537833790613634}}
{"url": "https://math.stackexchange.com/questions/2721368/empty-set-subsets-and-vacuous-truths", "text": "# Empty set, subsets, and vacuous truths\n\nSo I was trying to prove that a null set is a subset of any set.\n\nFirst, to define when $A$ is a subset of $B$:\n\n$$A \\subseteq B \\iff \\forall x(x \\in A \\implies x \\in B)$$\n\n(At least I think that's right?)\n\nSo then consider the empty set $\\emptyset$ for which $\\forall x(x \\not\\in \\emptyset)$ is true.\n\nI tried to prove that the empty set is a subset of any other set, or:\n\n$$\\emptyset \\subseteq B \\iff \\forall x(x \\in \\emptyset \\implies x \\in B) \\vdash \\text{T}$$\n\nTo me this seemed true because it was \"vacuously true\"... somehow. Like it makes sense to call it vacuously true that \"all $0$ items in $\\emptyset$ can be found in $B$, yep!\" but that isn't satisfying to me, how do I \"prove\" this is the case?\n\nIs $x \\in \\emptyset$ a false... statement? A false predicate? Something else? Something that results in false so that the implication itself is true. Is this a $(\\text{F}\\implies \\text{F}) \\vdash \\text{T}$ thing?\n\nOr is it the $\\forall$ that makes it false somehow?\n\nWhat if I had said $\\exists x \\in \\emptyset$, this feels like it would certainly be false but again I can't prove why, it's just an intuition.\n\nCan anyone clarify the correct definitions / implications and why they're true or false or what have you?\n\n\u2022 You don't seem to be using the symbol $\\vdash$ in the usual way (see en.wikipedia.org/wiki/Turnstile_(symbol) ). Normally $P\\vdash Q$ means you have proven $Q$ using assumptions $P$. In this case you want to show $\\vdash\\emptyset\\subseteq B$, and the vacuous implication you are looking for would be $F\\vdash P$ (or equivalently $\\vdash F\\Rightarrow P$) for any $P$. \u2013\u00a0stewbasic Apr 4 '18 at 4:46\n\u2022 $x\\in\\emptyset$ is a predicate. It's neither true nor false until you substitute for $x$, but of course, whatever $x$ you substitute makes it false. So, for every $x$, the statement $x \\in \\emptyset \\implies x \\in B$ is true. I don't know if this will help you, but I hope so. \u2013\u00a0saulspatz Apr 4 '18 at 5:03\n\u2022 \"x\u2208\u2205 is a predicate. It's neither true nor false until you substitute for x\". True. But saying $x \\in \\emptyset$ is false for all $x$ is perfectly legitimate. \u2013\u00a0fleablood Apr 4 '18 at 5:22\n\u2022 @stewbasic I was trying to say that \"(a iif for all x(p implies q)) is true\" is there a better symbol to use? Equal sign? \u2013\u00a0user525966 Apr 4 '18 at 13:18\n\u2022 @user525966 You could write $\\vdash\\emptyset\\subseteq B\\Leftrightarrow\\forall x(x\\in\\emptyset\\Rightarrow x\\in B)$ or just $\\emptyset\\subseteq B\\Leftrightarrow\\forall x(x\\in\\emptyset\\Rightarrow x\\in B)$. \u2013\u00a0stewbasic Apr 4 '18 at 21:09\n\nThe proof relies on Ex falso :\n\n$\\vdash \\lnot P \\to (P \\to Q)$.\n\nWe have to apply it in the form :\n\n$\\lnot (x \\in \\emptyset) \\to (x \\in \\emptyset \\to x \\in B)$.\n\nWe have (axiom or theorem) : $\\mathsf {ZF} \\vdash \\forall x \\ \\lnot (x \\in \\emptyset)$.\n\nBy Universal instantiation we get : $\\lnot (x \\in \\emptyset)$ and thus from Ex falso, by Modus Ponens : $(x \\in \\emptyset \\to x \\in B)$.\n\nFinally, by Universal generalization we conclude with :\n\n$\\mathsf {ZF} \\vdash \\forall x \\ (x \\in \\emptyset \\to x \\in B)$.\n\n\u2022 What does it mean to have the turnstile on the far left with nothing before it? \u2013\u00a0user525966 Apr 4 '18 at 15:33\n\u2022 @user525966 - that is a \"law of logic\" (in this case: of propositional logic, i.e. a tautology) and thus we can use in every theory. \u2013\u00a0Mauro ALLEGRANZA Apr 4 '18 at 15:35\n\u2022 In my original post was I using turnstile incorrectly? What's the correct symbol to use there instead? Equal sign? \u2013\u00a0user525966 Apr 4 '18 at 15:53\n\u2022 The \"turnstile\" $\\vdash$ means derivable (in a system). Thus $\\mathsf {ZF} \\vdash \\varphi$ means that $\\varphi$ is a theorem of Zermelo-Frenkel set theory. \u2013\u00a0Mauro ALLEGRANZA Apr 4 '18 at 15:56\n\u2022 Thus, you want to prove: $\\mathsf {ZF} \\vdash \\forall B \\ (\\emptyset \\subseteq B)$. It is equivalent to say that $\\forall B \\ (\\emptyset \\subseteq B)$ is TRUE in every model of the theory. \u2013\u00a0Mauro ALLEGRANZA Apr 4 '18 at 15:58\n\n1)\n\n$A\\subset B$ if all elements of $A$ are elements of $B$. As $\\emptyset$ has no elements, then all of them are in $B$.\n\nThat's vacuously true.\n\nSo $\\emptyset \\subset B$.\n\n2)\n\n$A\\subset B$ if any elements not in $B$ are not in $A$ either. As any element that is not in $B$ is not in $\\emptyset$ either, $\\emptyset \\subset B$.\n\nThat's true-true; nothing vacuous about it.\n\n3)\n\n$A \\subset B$ if $x \\in A \\implies x \\in B$ is true for all $x$. As $x \\in \\emptyset$ is always false and $FALSE \\implies P$ is always true, $x \\in \\emptyset \\implies x \\in B$ is always true.\n\nSo $\\emptyset \\subset B$.\n\n===\n\nSo for the most part, yes, they are vacuously true statements, or they are a false premise implies anything true statements.\n\nBut it's not all smoke and mirrors. A subset is \"embedded\" in the superset and everything you can pull out of the subset most come directly from the superset, and there is nothing in the subset that isn't in the superset.\n\nAll of those are true about an empty set and a set $B$ and, to me at least, the all feel directly true with no semantic slick word play or gimmicks. The empty skein of the the emptyset (with nothing in it) is embedded every where in the ether of existent space. That doesn't seem to me to be a \"trick\". And because nothing can be pulled out of the emptyset, if we are standing in the general vacinity of $B$ nothing can be pulled out of the empty set that isn't from $B$. That's a direct objective fact.\n\nIn general, the predicate logic statement that $\\forall x(x \\in A \\implies x \\in B)$ is written as $A \\subseteq B$.\n\nThe empty set $\\emptyset$ is the set that contains no elements. Therefore, the empty set is a subset of any set, that is, $\\emptyset \\subseteq X$ for all $X$. This is because the statement $x \\in \\emptyset$ is false for any $x$, so the imiplication\n\n$$\\forall x(x \\in \\emptyset \\implies x \\in X)$$\n\nmust be true. (See the truth table below for the for the implication connective.)\n\n$$\\begin{array}{c|l|c} \\text{p} & \\text{q} & \\text{p \\implies q} \\\\ \\hline T & T & T \\\\ T & F & F \\\\ F & T & T \\\\ F & F & T \\end{array}$$\n\nNote that the bottom two rows of the truth table are vacuously true.", "date": "2019-01-22 19:49:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2721368/empty-set-subsets-and-vacuous-truths", "openwebmath_score": 0.7642687559127808, "openwebmath_perplexity": 382.2226008121524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771798031351, "lm_q2_score": 0.8652240721511739, "lm_q1q2_score": 0.8537833698151196}}
{"url": "https://math.stackexchange.com/questions/3233799/an-exam-has-50-multiple-choice-questions-with-5-options", "text": "# an exam has 50 multiple choice questions with 5 options\n\nAn exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that \"a student\" has no prior knowledge and randomly guess on all questions exam,\n\n1. Compute the expected mean for the student score\n2. Compute the standard deviation for the student score\n3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?\n4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : \uf02d What is the expected success rate? \uf02d How do you expect the proportion of students who will score less or equal to 20?\n\nIf you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?\n\na. What is the expectation of this student's degree?\n\nb. what is the standard deviation of this student's grade?\n\nb. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?\n\n1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct\n\u2022 Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. \u2013\u00a0saulspatz May 21 at 0:59\n\u2022 thanks I did it \u2013\u00a0Nidal May 21 at 1:29\n\nIt seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.\n\nFor a binomial random variable $$X$$ with $$n = 50$$ and $$p = 1/5,$$ you can use a normal approximation to binomial to get reasonable approximate answers to these questions. (You will get somewhat better approximations, if you use a continuity correction, but even then you can't expect to get more than two or three places of accuracy.)\n\nAlternatively, you can use statistical software (or perhaps a statistical calculator) to to get answers exact to many decimal places.\n\nBelow are some answers from R statistical software, in which dbinom and pbinom, designate a binomial PDF and CDF, respectively. I assume each correct guess counts 2 points and that there is no penalty for incorrect guesses.\n\n$$P(X \\ge 30) = 1 - P(X \\le 29) \\approx 0$$\n\n1 - pbinom(29, 50, .2)\n[1] 6.936747e-10\n\n\n$$P(X \\le 10) = .5836.$$\n\npbinom(10, 50, .2)\n[1] 0.5835594\n\n\n$$P(X = 0) = .8^{50}.$$\n\ndbinom(0, 50, .2);  .8^50.\n[1] 1.427248e-05\n[1] 1.427248e-05\n\n\nA student who has studied half of the material should have probability $$p = 0.6$$ of answering each question correctly. Why? Let $$Y$$ be the number of correctly answered questions.\n\n$$P(Y \\ge 30) = 1 - P(Y \\le 29) = 0.5610.$$\n\n 1 - pbinom(29, 50, .6)\n[1] 0.5610349\n\n\nUsing a normal approximation. The number $$Y$$ correct has $$E(Y) = np = 50(.6) = 30,$$ $$Var(Y) = 12,$$ $$SD(Y) = 3.4641.$$\n\nThus $$Y \\stackrel{aprx}{\\sim} \\mathsf{Norm}(\\mu=30, \\sigma=3.4641)$$ and (standardizing) $$Z = \\frac{Y - \\mu}{\\sigma} \\stackrel{aprx}{\\sim} \\mathsf{Norm}(0, 1).$$\n\nHence, $$P(Y \\ge 30) = 1 - P(Y < 29.5) \\approx 0.557.$$ (If you standardize and use printed normal tables you will get about the same answer; not quite because of the rounding involved in using the table.)\n\n 1 - pnorm(29.5, 30, 3.4641)\n[1] 0.5573831\n\n\nThe figure below shows the PDF of $$\\mathsf{Binom}(50,.6)$$ (bars) along with the density function of $$\\mathsf{Norm}(30, 3.4641)$$ (blue curve). The probability $$P(Y \\ge 30)$$ is the sum of the heights of the lines to the right of the dotted red line. (This probability is approximated by the area under the normal density curve to the right of this vertical line.)\n\nThe R code for the figure is shown below:\n\n y = 0:50;  PDF = dbinom(y, 50, .6)\nplot(y, PDF, type=\"h\", lwd=2, main=\"PDF of BINOM(50,.35) with Normal Approx.\")\ncurve(dnorm(x, 30, 3.4641), 0, 50, add=T, col=\"blue\")\nabline(h=0, col=\"green2\");  abline(v=0, col=\"green2\")\n\n\u2022 that was helpful,,, when he studied half of the material why p=0.6 I didn't get it. \u2013\u00a0Nidal May 21 at 10:55\n\u2022 when he has no prior knowledge p was 1/5 (like one option will be right from the five options), now when he has studied half of the material why p is 3/5=0.6. \u2013\u00a0Nidal May 21 at 14:06\n\u2022 S = studied, N = not. P(S)=P(N) = 1/2. P(Corr) = 1P(S) + (1/5)P(N) = .5 + .5(.2) = .5+.1 = .6. Law of total probability. \u2013\u00a0BruceET May 21 at 15:54\n\u2022 Unsolicited, so likely unwelcome advice: I have had a look at your history of activity on this site. In my opinion, although you ask some nice questions, you reveal much less of your work and thinking than optimal. (Even those bits often only on demand.) Looks as if you are taking a fine and well-taught course. Outsourcing too much of the problem solving is not a path toward effective learning. \u2013\u00a0BruceET May 21 at 16:20\n\u2022 We can't be sure there are exactly 25 questions from the parts the student studies. The way I read it, there are on average 25 such questions. // Your way and mine both give expected nr correct as 30, but for slightly different distributions. Mine BINOM(50, .6), yours 25 + BINOM(25, .2). Same mean, mine has larger variance. \u2013\u00a0BruceET May 21 at 19:33", "date": "2019-06-16 01:26:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3233799/an-exam-has-50-multiple-choice-questions-with-5-options", "openwebmath_score": 0.7624414563179016, "openwebmath_perplexity": 627.6381328677032, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771774699747, "lm_q2_score": 0.865224070413529, "lm_q1q2_score": 0.8537833660817448}}
{"url": "https://picasso-project.eu/nigella-cook-srdcyz/set-notation-symbols-bd1612", "text": "This section is to introduce the notation to the reader and explain its usage. Set theory is one of the foundational systems for mathematics, and it helped to develop our modern understanding of infinity and real numbers. Intersection and union of sets. Compact set notation is a useful tool to describe the properties of each element of a set, rather than writing out all elements of a set. They are { } and { 1 }. Probability and statistics symbols table and definitions - expectation, variance, standard deviation, distribution, probability function, conditional probability, covariance, correlation Crow's foot notation, however, has an intuitive graphic format, making it the preferred ERD notation for Lucidchart. The individual objects in a set are called the members or elements of the set. The guide you are now reading is a \u201clegend\u201d to how we notate drum and percussion parts when we engrave music at Audio Graffiti. Thankfully, there is a faster way. Usually, you'll see it when you learn about solving inequalities, because for some reason saying \"x < 3\" isn't good enough, so instead they'll want you to phrase the answer as \"the solution set is { x | x is a real number and x < 3 }\".How this adds anything to the student's understanding, I don't know. Basic set operations. Use set notation to describe: (a) the area shaded in blue (b) the area shaded in purple. If you \u2026 ALT Codes for Math Symbols: Set Membership & Empty Sets Read More \u00bb In this notation, the vertical bar (\"|\") means \"such that\", and the description can be interpreted as \"F is the set of all numbers n, such that n is an integer in the range from 0 to 19 inclusive\". The table below lists all of the necessary symbols for compact set notation. Mathematical Set Notation. Subset, strict subset, and superset. The domains and ranges used in the discrete function examples were simplified versions of set notation. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step Sets. Set Theory Symbols Posted in engineering by Christopher R. Wirz on Wed Feb 08 2017. That is OK, it is just the \"Empty Set\". Some notations for sets are: {1, 2, 3} = set of integers greater than 0 \u2026 You never know when set notation is going to pop up. Illustration: Demo. In this section, we will introduce the standard notation used to define sets, and give you a chance to practice writing sets in three ways, inequality notation, set-builder notation, and interval notation. If you want to get in on their secrets, you'll want to become familiar with these Venn diagram symbols. A set is a collection of objects, things or symbols which are clearly identified.The individual objects in the set are called the elements or members of the set. Typing math symbols into Word can be tedious. any. ... Set Language And Notation. of . There are many different symbols used in set notation, but only the most basic of structures will be provided here. Which is why the bulk of this follow-up piece covers the very basics of set theory notation, operations & visual representations extensively. Lots symbols look similar but mean different things. IGCSE 9-1 Exam Question Practice (Sets + Set Notation) 4.9 34 customer reviews. A set is a well-defined collection of distinct objects. Also, check the set symbols here.. Sets, in mathematics, are an organized collection of objects and can be represented in set-builder form or roster form.Usually, sets are represented in curly braces {}, for example, A = {1,2,3,4} is a set. Let\u2019s kick off by introducing the two most basic symbols for notating a set & it\u2019s corresponding elements. MS Word Tricks: Typing Math Symbols 2015-05-14 Category: MS Office. To fully embrace the world of professional Venn diagrams, you should have a basic understanding of the branch of mathematical logic called \u2018set theory\u2019 and its associated symbols and notation. Author: Created by Maths4Everyone. We attempt to follow the standards set out in Norman Weinberg\u2019s Guide to Standardized Drumset Notation. 8 February 2019 OSU CSE 1. This cheat sheet is extremely useful. Sometimes the set is written with a bar instead of a colon: {x\u00a6 x > 5}. Cardinality and ordinality Bringing the set operations together. The Universal Set \u2026 Consider the set $\\left\\{x|10\\le x<30\\right\\}$, which describes the behavior of $x$ in set-builder notation. and symbols. Set theory starter. Set notation practice. Example: Set-Builder Notation: Read as: Meaning: 1 {x : x > 0}the set of all x such that x is greater than 0. any value greater than 0: 2 {x : x \u2260 11}the set of all x such that x is any number except 11. any value except 11: 3 {x : x < 5}the set of all x such that x is any number less than 5. any value less than 5 Note that it's unnecessary to load amsmath if you load mathtools. Email. Google Classroom Facebook Twitter. Set notation and Venn diagrams questions. Set Theory \u2022 A mathematical model that we will use often is that of . Set notation is an important convention in computer science. CCSS.Math: HSS.CP.A.1. Look at the venn diagram on the left. Shading task. In, sets theory, you will learn about sets and it\u2019s properties. On the Insert tab, in the Symbols group, click the arrow under Equation, and then click Insert New Equation. Relative complement or difference between sets. When picking a symbol, best to trust the symbol's unicode name for its meaning, not appearance. Universal set and absolute complement. Researchers and mathematicians have developed a language and system of notation around set theory. Null set is a proper subset for any set which contains at least one element. Occasionally we will introduce a new symbol to cater for an unusual requirement of a client. S et theory is a branch of mathematics dedicated to the study of collections of objects, its properties, and the relationship between them. After school they signed up and became members. The table below contains one example set\u2026 Because rarely used symbol may look very different on another computer. While crow's foot notation is often recognized as the most intuitive style, some use OMT, IDEF, Bachman, or UML notation, according to their preferences. In set-builder notation, the set is specified as a selection from a larger set, determined by a condition involving the elements. The default way of doing it is to use the Insert > Symbols > More Symbols dialog, where you can hunt for the symbol you want. Symbol Symbol Name Meaning / definition Example { } set: a collection of elements: A = {3,7,9,14}, It is still a set, so we use the curly brackets with nothing inside: {} The Empty Set has no elements: {} Universal Set. The Wolfram Language has the world's largest collection of consistent multifont mathematical notation characters\\[LongDash]all fully integrated into both typesetting and symbolic expression construction . Click the arrow next to the name of the symbol set, and then select the symbol set that you want to display. When using set notation, we use inequality symbols to describe the domain and range as a set of values. Let us discuss the next stuff on \"Symbols used in set theory\" If null set is a super set Problem 1: Mrs. Glosser asked Kyesha, Angie and Eduardo to join the new math club. This carefully selected compilation of exam questions has fully-worked solutions designed for students to go through at home, saving valuable time in class. Solution: Let P be the set of all members in the math Created: Jan 19, 2018 | Updated: Feb 6, 2020. Topics you will need to know in order to pass the quiz include sets, subsets, and elements. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. take the previous set S \u2229 V ; then subtract T: This is the Intersection of Sets S and V minus Set T (S \u2229 V) \u2212 T = {} Hey, there is nothing there! A variant solution, also based on mathtools, with the cooperation of xparse allows for a syntax that's closer to mathematical writing: you just have to type something like\\set{x\\in E;P(x)} for the set-builder notation, or \\set{x_i} for sets defined as lists. This quiz and attached worksheet will help gauge your understanding of set notation. Under Equation Tools , on the Design tab, in the Symbols group, click the More arrow. State whether each \u2026 Below is the complete list of Windows ALT codes for Math Symbols: Set Membership & Empty Sets, their corresponding HTML entity numeric character references, and when available, their corresponding HTML entity named character references, and Unicode code points. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. Set notation. The following table gives a summary of the symbols use in sets. For example, a set F can be specified as follows: = {\u2223 \u2264 \u2264}. Because null set is not equal to A. Admin Igcse Mathematics Revision Notes, O Level Mathematics Revision Notes 2 Comments 12,074 Views. Basic Set Theory . Set notation is used to help define the elements of a set. For example, let us consider the set A = { 1 } It has two subsets. Preview. elements . Quiz & Worksheet Goals Inequalities can be shown using set notation: {x: inequality}where x: indicates the variable being described and inequality is written as an inequality, normally in its simplest form. They wrote about it on the chalkboard using set notation: P = {Kyesha, Angie and Eduardo} When Angie's mother came to pick her up, she looked at the chalkboard and asked: What does that mean? Symbols Used in this Book; Glossary; While a comprehensive list of notation is included in the appendix, that is meant mostly as a reference tool to refresh the reader of what notation means. Purplemath. Basic set notation. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic.\n\nTiger Synonyms In Sanskrit, Avoca Sands Cafe Menu, Cloris The Tick, Gambrell Hall Benedict College, Shellfish Allergy List, Power Reclining Loveseat With Center Console, Lake Sutherland Vacation Rentals, Should I Date Him Quiz Long-distance, C300 Asus Chromebook,", "date": "2021-10-20 13:27:16", "meta": {"domain": "picasso-project.eu", "url": "https://picasso-project.eu/nigella-cook-srdcyz/set-notation-symbols-bd1612", "openwebmath_score": 0.6717374324798584, "openwebmath_perplexity": 1006.1679112481584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9693242018339897, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8537779215555548}}
{"url": "https://mybrightfuture.org/a0uppzk0/1myldcg.php?94011f=do-adjacent-angles-form-a-linear-pair", "text": "If two angles do not form a linear pair, then they are not supplementary. Linear Pair Two adjacent angles form a linear pair if their non-common sides form a straight angle. Axiom 1: If a ray stands on a line then the adjacent angles form a linear pair of angles. They just need to add up to 90 degrees. Contrapositive: If the angles are not adjacent, then the two angles do not form a linear pair. B A V CLinear Pair Postulate: If two angles form a linear pair, then they are supplementary. Therefore, all the exterior angles are equal, and can be found by dividing 360\u00b0 by the number of angles . Linear A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. The sides of the angles do not form two pairs of opposite rays. [Image will be Uploaded Soon] In the figure given above, all line segments are passing through the point O, as shown in the figure. Name a linear pair. 9 years ago. where \u2220DBC is an exterior angle of \u2220ABC and, \u2220A and \u2220C are the remote interior angles. Related Questions to study. Write. A) vertical angles B) complementary C) acute D) a linear pair * please answer only if you know As the ray OF lies on the line segment MN, angles \u2220FON and \u2220FOM\u00a0 form a linear pair. Question 1. Don't two angles ALWAYS have to be adjacent in order to form a line (linear pair)? Assignment Directions: Solve the following word problems involving linear pairs. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. Are two angles in the same plane with a common vertex and a common side, but no common interior points. The measure of one angle is 24 more than the measure of the other angle. As the ray OA lies on the line segment CD, angles \u2220 AOD and \u2220 AOC form a linear pair. B. Can you explain this answer? Linear PairA linear pair consists of two adjacent angles whose noncommon sides are opposite rays. For #1-6, use the figure at the right. Flashcards. a. Tell whether the angles are only adjacent, adjacent and form a linear pair, or not adjacent. 14. In which diagram do angles 1 and 2 form a linear pair. (iii) When two lines intersect opposite angles are equal. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary means add \u2026 The sum of the linear pair of angles is always equal to 180 degrees. A Linear Pair is two adjacent angles whose non-common sides form opposite rays. Match. Tell whether the angles are only adjacent, adjacent and form a linear pair, or not adjacent. The other two pairs of angles are adjacent, but they are not forming a linear pair. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. Name two acute vertical angles. Add your answer and earn points. If two angles are not adjacent, then they do not form a linear pair. ii) When non-common sides of a pair of adjacent angles form opposite rays, then the pair forms a linear pair. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. In the figure, \u2220 1 and \u2220 2 form a linear pair. However, just because two angles are supplementary does not mean they form a linear pair. A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. Favorite Answer. Name an angle complementary to . View Answer . Basically, a linear pair of angles always lie on a straight line. Similarly, \u2220GON and \u2220HON form a linear pair and so on. The measure of one angle is 15 less than half the measurement of its supplement. For \u25b3ABC above, \u2220A+\u2220C+\u2220ABC=180\u00b0. \"Complementary Angles\" These are angles that add up to 90 degrees. If the two complementary angles are adjacent then they will form a right angle. They share a common vertex, but not a \u2026 Consider a Ray OP Stand on the Line Segment\u00a0 as Shown: The angles which are formed at E are \u2220QEM and \u2220QEN. Two angles may be supplementary, but not adjacent and therefore not form a linear pair. A linear pair forms a straight angle of 180 degrees, so you have two angles whose sum is given as 180, which means they are supplementary. Therefore, the exterior angles measure = = 72\u00b0 each. Name a pair of adjacent angles. Learn. Example 2: If two adjacent angles form a linear pair, show that their angle bisectors are perpendicular. When two lines intersect each other at a common point then, a linear pair of angles are formed. Such angle pairs are \u2026 Linear-pair-angles are always supplementary. <3 and <4 form a linear pair. Hope this helps! In the diagram above, \u2220ABC and \u2220DBC form a linear pair. Hence vertical angles are not adjacent. Linear Pair Of Angles. Related Questions to study. Two angles are called supplementary angles if Answer Save. How to Find Adjacent Angles. | EduRev Class 7 Question is disucussed on EduRev Study Group by 2108 Class 7 \u2026 B Supplementary angles and linear pairs both add to 180\u00b0. Two angles that add to 180 degrees and when adjacent form a straight line. If two angles are not adjacent, then they do not form a linear pair. Two adjacent angles are said to form a linear pair angles , if their non-common arms are two opposite rays. Adjacent angles do not overlap. S_Pinkston. If the angles are adjacent to each other after the intersection of the lines, then the angles are said to be adjacent. \"Congruent Angles\" They have the same angle in degrees or radian. As the adjacent angles form a linear pair and they are supplementary. <1 and <2 are adjacent angles. Linear Pair Angles. C. If two angles are not supplementary, then they do not form a linear pair. Pro Lite, Vedantu Example 2: Determine if the following pair of angles are vertical angles, linear pairs, or neither. \u2234\u00a0 \u00a0 Yes, \u22201 is vertically opposite to \u22204, And angle vertically opposite to \u22205 is \u2220BOC. 1 See answer chanybonilla is waiting for your help. What can you say about its other angle? Example 2: Determine if the following pair of angles are vertical angles, linear pairs, or neither. Consider the following figure in which a ray $$\\overrightarrow{OP}$$ stand on the line segment $$\\overline{AB}$$ as shown: The angles \u2220POB and \u2220POA are formed at O. 12. The sum of their angles is 180\u00b0 or \u03c0 radians. ! In the figure above, all the line segments pass through the point O as shown. what are the two factors that make you decide whether a pair of adjacent a angles from a linear pair 2 See answers saryunahar141005 saryunahar141005 Answer: All adjacent angles do not form a linear pair. In the diagram below, \u2220ABC and \u2220DBE are supplementary since 30\u00b0+150\u00b0=180\u00b0, but they do not form a linear pair since they are not adjacent. Pro Subscription, JEE In which diagram do angles 1 and 2 form a linear pair. In the diagram above, \u2220ABC and \u2220DBC form a linear pair. By: Carol H. answered \u2022 02/22/18. Also, the ray is that part of the line which has only one endpoint. Test. A linear pair of angles is formed when two lines intersect. Thus, the sum of the exterior angles is: For regular polygon, all of the angles of a are equal. so, a pair of straight angles can also be adjacent angle. Join Yahoo Answers and get 100 points today. The pair of adjacent angles are constructed on a line segment, but not all adjacent angles are linear. Learn how to identify angles from a figure. <2 and <4 <2 and <4. This can be shown by using linear pairs. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. It can't think of a contradiction. Consider the following figure in which a ray $$\\overrightarrow{OP}$$ stand on the line segment $$\\overline{AB}$$ as shown: The angles \u2220POB and \u2220POA are formed at O. Example 2 : In the diagram shown below, Solve for x and y. Get the answers you need now. 14. (ii) Adjacent complementary angles. Substituting the second equation into the first equation we get. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. D. If two angles form a linear pair, then they are supplementary. Linear pair. Supplementary angles are two angles whose same is 180o Linear pairs are adjacent angles who share a common ray and whose opposite rays form a straight line. A circle of the infinite radius can be visualized on the straight line. The figure shows the design on an outdoor fence. Gravity. Use \u00bd sheet of paper. Ans : (d) Bisectors of the adjacent angles forming a linear pair form a right angle. A linear pair is a pair of adjacent angles whose non-adjacent sides form a line. See tutors like this. 3. The sum of their angles is 180 \u00b0 or \u03c0 radians. Learn vocabulary, terms, and more with flashcards, games, and other study tools. In the regular pentagon above, n = 5. 0 0. A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. PLAY. Supplementary angles do not have to be adjacent, whereas a linear pair must be adjacent and create a straight line. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. An angle is 18c less than its complementary angle. 1 See answer chanybonilla is waiting for your help. (iv) Unequal supplementary angles. a. b. c. Exploration #1: Use this link to explore the relationship with vertical angles. Linear pair is a pair of adjacent angles where non-common side forms a straight line. Answer:Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pairStep-by-step explanation:see A linear pair forms a straig\u2026 2 0. If two angles are supplementary, then they form a linear pair. They have common vertex O. i tried searching it but i wasn't sure cause it was so confusing at first i thought i was getting it but i wasn't. Correct answer to the question: 5. Also, there is a common arm that represents both the angles of the linear pair. A line segment with A and B as two endpoints is represented as AB. Spell. They do not form a straight line. Easily fill out PDF blank, edit, and sign them. They have common side OB. If a linear pair is formed by two angles, the uncommon arms of the angles forms a straight line. In a diagram angle 1 and angle 2 form an linear pair and angle 2 and angle 3 form vertical angles. The above discussion can be stated as an axiom. ii) When non-common sides of a pair of adjacent angles form opposite rays, then the pair forms a linear pair. If two angles form a linear pair and the measure of one angle is 47 degrees, what is the measure of the other angle. If two angles are adjacent, they form a linear pair? The measure of an angle is three times the measurement of its complement. Linear pair of angles - definition Two angles are said to be in a linear pair if they are adjacent to each other, lie on the same side of the line and the sum of their measures is 1 8 0 o. \u2220BOC and \u2220AOC are linear-pair-angles. Therefore, AB represents a line. Created by. 13. 1 Answer. A Linear Pair is two adjacent angles whose non-common sides form opposite rays. Vertically Opposite Angles When a pair of lines intersect, as shown in the fig. Also, \u2220ABC and \u2220DBC form a linear pair so. (iv) Unequal supplementary anglesFor line BD \u2220 BOC and \u2220 COD are in linear pair Therefore, \u2220BOC + \u2220COD = 180\u00b0 So, their sum is 180\u00b0 And, \u2220BOC is not equal to \u2220COD \u2234 \u2220BOC & \u2220COD are unequal supplementary angles Ex 5.1, 14 In the adjoining figure, name the following pairs of angles. Such angle pairs are called a linear pair.. Angles A and Z are supplementary because they add up to 180\u00b0.. Vertical angles: When intersecting lines form an X, the angles on the opposite sides of the X are called vertical angles. Find the values of the angles l, p, and q in each of the following questions, \u2220DOC = \u2220AOE \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (vertically opposite angles), Pair of Linear Equations in Two Variables, Vedantu This statement is correct. (iii) \u00a0 Does the angles \u2220COE and \u2220EOD form a linear pair? \u2220BOC and \u2220AOC are linear-pair-angles. It is known that the angle between the two line segments ME and EN is 180\u00b0. Angles \u2220 Z W I and \u2220 H W I are adjacent angles. (vi)\u00a0 Find the vertically opposite angle of \u22205? In the figure, clearly, the pair $$\\angle BOA$$ and $$\\angle AOE$$ form adjacent complementary angles. Complete Identify Each Pair Of Angles As Adjacent, Vertical, Complementary, Supplementary, Or A Linear Pair online with US Legal Forms. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary \u2026 Add comment More. Anonymous. Supplementary angles: Two angles that add up to 180\u00b0 (or a straight angle) are supplementary. Relevance. Angle Pairs (Vertical, linear pair, adjacent, complementary & supplementary) STUDY. The two axioms mentioned above form\u00a0 Linear Pair Axioms and are helpful in solving various mathematical problems. In a linear pair, the arms of the angles that are not always common or collinear i.e., they lie on a\u00a0 straight line. A: If two angles form a linear pair, then angles are supplementary. 1 and 2 are adjacent angles. If two parallel lines are intersected by a transversal, then each pair of corresponding angles so formed is (a) Equal (b) Complementary (c) Supplementary (d) None of these Ans : (a) Equal 4. If a point O is taken anywhere on the line segment AB as shown, then the angle between the two line segments AO and OB is a straight angle i.e., 180\u00b0. Two adjacent supplementary angles form a linear pair. Here, these angles are in linear pair as. Report 1 Expert Answer Best Newest Oldest. If two angles are a linear pair, then they are supplementary. T or F? There are n angles in the polygon, so there are n linear pairs. If Two Angles Form A Linear Pair, The Angles Are Supplementary. Two adjacent supplementary angles form a linear pair. A linear pair is a pair of adjacent, supplementary angles, whihc means they add up to 180 degrees. The measure of an angle is three times the measurement of its complement. 12. Linear Pairs. If two angles form a linear pair, the angles are supplementary. Hence, we can also say that a linear pair of angles is the adjacent angle whose non-common arms are basically opposite rays. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. the term adjacent means next to each other. 1 decade ago. Two adjacent angles are said to form a linear pair angles , if their non-common arms are two opposite rays. In the figure, clearly, the pair $$\\angle AOE$$ and $$\\angle EOD$$ form adjacent angles that do not form a linear pair. The angles are adjacent, sharing ray BC, and the non-adjacent rays, BA and BD, lie on line AD. (add up to 180 0) \u2220BOC + \u2220AOC = 180 0 Examples : 1) One of the angles forming a linear-pair is a right angle. Then, find the angle measures. A straight angle is formed when the angle between two lines is 180 degrees. What are not adjacent angles? Two vertical angles are always the same size as each other. adjacent angles are the angles which have same pair of vertex and have one common side. Two angles form a linear pair. Sorry!, This page is not available for now to bookmark. Covid-19 has led the world to go through a phenomenal transition . Similarly, \u2220GON and \u2220HON form a linear pair and so on. Converse statement inverses statement Contrapositive statement Key Concepts: Terms in this set (12) Adjacent complementary angles (Image) angles that are \"next to\" each other. a. b. c. Exploration #1: Use this link to explore the relationship with vertical angles. Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. Adjacent Angles\u2022 Share a common side\u2022 Share the same vertex\u2022 Do not share any interior points B A C V Bistro 72 Reviews, 5 Gallon Air Compressor, Clementine Churchill Images, How Old Is Chopper, Allegiant Air Flight Schedules, Counting Video For Kindergarten, Langerhans Cells Function Quizlet, Eine Kleine Nachtmusik Mr Bean, Improvisation Of Melody Brainly, Dynamite Definition Synonyme, Kazimir Malevich Art,", "date": "2021-08-02 13:25:55", "meta": {"domain": "mybrightfuture.org", "url": "https://mybrightfuture.org/a0uppzk0/1myldcg.php?94011f=do-adjacent-angles-form-a-linear-pair", "openwebmath_score": 0.622346043586731, "openwebmath_perplexity": 829.5373305594145, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9693242018339898, "lm_q2_score": 0.8807970654616711, "lm_q1q2_score": 0.8537779124563548}}
{"url": "https://math.stackexchange.com/questions/2497461/prove-that-the-set-of-mathbbn-times-mathbbn-times-mathbbn-is-counta", "text": "Prove that the set of $\\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N}$ is countable.\n\nI have seen proofs of $\\mathbb{N}\\times\\mathbb{N}$. The proof by list where you list each element in a pair of elements and then counting them diagonally is the most convincing. I have two thoughts in mind:\n\n1. Write it as a composition of functions. Show that both functions are bijections, so the composition is also a bijection.\n2. Show that $\\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N}$ has the same cardinality as $\\mathbb{N}\\times\\mathbb{N}$, and $\\mathbb{N}\\times\\mathbb{N}$ has the same cardinality as $\\mathbb{N}$.\n\nGenerally, I'm having difficulty going from $\\mathbb{N}\\times\\mathbb{N}\\times\\mathbb{N}$ to $\\mathbb{N}\\times\\mathbb{N}$, since it easy to prove the bijection from $\\mathbb{N}\\times\\mathbb{N}$ to $\\mathbb{N}$.\n\n\u2022 Youre idea in 1. seems like a good approach. \u2013\u00a0JJC94 Oct 31 '17 at 0:48\n\u2022 It's not a proof-explanation if you don't point at a specific proof and ask us to explain it. \u2013\u00a0Asaf Karagila Oct 31 '17 at 0:51\n\u2022 Consider the Cantor pairing. \u2013\u00a0Wuestenfux Oct 31 '17 at 9:42\n\nIf you already know that $\\Bbb{N\\times N}$ is countable, fix a bijection $f\\colon\\Bbb{N\\times N\\to N}$. Now consider $g\\colon\\Bbb{N\\times N\\times N\\to N\\times N}$ defined as: $$g(n,m,k)=(n,f(m,k)),$$ or better yet $h\\colon\\Bbb{N\\times N\\times N\\to N}$ defined as $$h(n,m,k)=f(n,f(m,k))$$ and cut out the middle-man.\n\nAs my freshman discrete mathematics professor used to tell us, go home and convince yourself this is correct.\n\n\u2022 I can convince myself this is correct, but how can I prove it to a stubborn skeptic? \u2013\u00a0tigerustin Nov 1 '17 at 2:11\n\u2022 Slowly and carefully by verifying the definitions of a bijection. \u2013\u00a0Asaf Karagila Nov 1 '17 at 6:53\n\u2022 I proved injectivity. How would I prove surjectivity? \u2013\u00a0tigerustin Nov 1 '17 at 20:21\n\u2022 Exactly the same. Slowly, carefully and by using the fact that $f$ is already surjective itself. \u2013\u00a0Asaf Karagila Nov 1 '17 at 20:23\n\u2022 I really don't understand. Can you explain further please? I don't want to have to start another thread on this question. \u2013\u00a0tigerustin Nov 2 '17 at 2:10\n\nWhy not think in stages? Let $\\varphi : \\mathbb{N}\\times\\mathbb{N} \\to \\mathbb{N}$ denote your favorite bijection. But then\n\n$$\\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N} = (\\mathbb{N} \\times \\mathbb{N}) \\times \\mathbb{N} \\approx \\varphi(\\mathbb{N}\\times \\mathbb{N}) \\times \\mathbb{N} = \\mathbb{N} \\times \\mathbb{N} \\approx \\varphi(\\mathbb{N}\\times \\mathbb{N}) = \\mathbb{N}$$ (where, in this context, $=$ means equality, and $\\approx$ denotes \"of the same cardinality\"). You can do this for any number of copies of $\\mathbb{N}$ via an induction argument.\n\nConsider $f:\\mathbb N \\times \\mathbb N \\times \\mathbb N \\to \\mathbb N$ given by $f(a,b,c)=2^a3^b5^c$\n\nand use the Fundamental theorem of arithmetic for a cute solution.\n\n\u2022 Gives an injective mapping. \u2013\u00a0Wuestenfux Oct 31 '17 at 9:41\n\u2022 And it is clearly infinite, so... \u2013\u00a0Andres Mejia Oct 31 '17 at 14:06\n\u2022 It would be difficult to prove bijectivity of this function? \u2013\u00a0tigerustin Nov 1 '17 at 2:12\n\u2022 It\u2019s not bijective. It is an infinite subset of the natural numbers \u2013\u00a0Andres Mejia Nov 1 '17 at 2:15\n\nI like to think of 'countable' as 'listable' ... That is, as long as you can create a list of objects and ensure that all elements of some set $S$ are somewhere on that list, then $S$ is countable. To me, the advantage of that kind of thinking is that elements from $S$ may be repeated on the list, and that the list may even contain objects that are not in $S$; as long as every element of $S$ is at least once in the list, it's clear that $S$ is countable. As such, this method avoids having to construct explicit and possibly complicated bijections.\n\nSo, how can we create a list of all elements of $\\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N}$?\n\nHere is one way:\n\nFirst, get all triples with elements that sum to $0$: which is just $<0,0,0>$\n\nNow get all those that sum to $1$: $<1,0,0>,<0,1,0>,<0,0,1>$\n\nThen those that sum to $2$, then $3$, etc.\n\nThus, the list is:\n\n$<0,0,0>,<1,0,0>,<0,1,0>,<0,0,1>,<2,0,0>,<1,1,0>,...$\n\nSince for any $n$ there can only be finitely many triples whose elements sum to $n$, you will eventually get to every triple.\n\nNote that this is of course just the same thing as diagonally going through an array of elements, just now in $3$ dimensions.\n\n\u2022 Reminiscent of the method I know for showing the algebraics are countable. Nice. \u2013\u00a0Kaj Hansen Oct 31 '17 at 2:34\n\nCantor-Bernstein's Theorems states that if $A$ and $B$ are two sets and there exists two injections $f:A \\to B$ and $g:B \\to A$ then there is a biyection from $A$ to $B$ (i.e. $A$ and $B$ have the same cardinality).\n\nLet $f:\\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N} \\to \\mathbb{N}$ be defined as $f(m,n,k)=2^n \\cdot 3^m \\cdot 5^k$, then if $f(m_1,n_1,k_1) = f(m_2,n_2,k_2) \\Leftrightarrow 2^{m_1}3^{n_1}5^{k_1}=2^{m_2}3^{n_2}5^{k_2}$, the Fundamental Theroem of Arithmetic shows that $m_1=m_2$, $n_1=n_2$ and $k_1=k_2$. Then $f$ is a injective function.\n\nNext, consider the function $g:\\mathbb{N} \\to \\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N}$ defined by $g(n)=(0,0,n)$, then $$g(n)=g(m) \\Leftrightarrow (0,0,n) =(0,0,m) \\Leftrightarrow n=m$$ and hence $g$ is injective.\n\nApplying Canotr-Bernstein's Theorem gives that $\\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N}$ is countable.\n\nUsing induction, you can easily prove this result for the product of any finite number of copies of $\\mathbb{N}$. The hardest part of the proof is the base case, which is that $\\mathbb{N} \\times \\mathbb{N}$ is countable. It seems you already have this result.\n\nFor the induction hypothesis, assume that $\\displaystyle \\prod^n \\mathbb{N}$ is countable. Now we just need to show that $\\displaystyle \\prod^{n+1} \\mathbb{N}$ is countable. First, note that can write the $(n\\!+\\!1)$-fold product in the form $\\mathbb{N} \\times \\left( \\displaystyle \\prod^n \\mathbb{N} \\right)$. A bijection $\\displaystyle f:\\prod^n \\mathbb{N} \\rightarrow \\mathbb{N}$ is guaranteed by the induction hypothesis, which we can use to construct a bijection between the $(n\\!+\\!1)$-fold product and $\\mathbb{N} \\times \\mathbb{N}$:\n\n$$(x_1, x_2, \\cdots, x_{n+1}) \\mapsto (x_1, f(x_2, x_3, \\cdots, x_{n+1}))$$\n\nAnd $\\mathbb{N} \\times \\mathbb{N}$ is countable per the base case.", "date": "2019-12-07 04:20:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2497461/prove-that-the-set-of-mathbbn-times-mathbbn-times-mathbbn-is-counta", "openwebmath_score": 0.8474349975585938, "openwebmath_perplexity": 195.1127876307063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126525529014, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8537774178909653}}
{"url": "http://math.stackexchange.com/questions/358755/ways-to-fill-a-n-times-n-square-with-1-times-1-squares-and-1-times-2-recta", "text": "# Ways to fill a $n\\times n$ square with $1\\times 1$ squares and $1\\times 2$ rectangles\n\nI came up with this question when I'm actually starring at the wall of my dorm hall. I'm not sure if I'm asking it correctly, but that's what I roughly have:\n\nSo, how many ways (pattern) that there are to fill a $n\\times n:n\\in\\mathbb{Z}_{n>0}$ square with only $1\\times 1$ squares and $1\\times 2$ rectangles?\n\nFor example, for a $2\\times 2$ square:\n\n\u2022 Four $1\\times 1$ squares; 1 way.\n\n\u2022 Two $1\\times 1$ squares and one $1\\times 2$ rectangle; $4$ ways total since we can rotate it to get different pattern.\n\n\u2022 Two $1\\times 2$ rectangles; 2 ways total: placed horizontally or vertically.\n\n$\\therefore$ There's a total of $1+4+2=\\boxed{7}$ ways to fill a $2\\times 2$ square.\n\nSo, I'm just wondering if there's a general formula for calculating the ways to fill a $n\\times n$ square.\n\nThanks!\n\n-\n+1 for coming up with your own math question. \u2013\u00a0Sammy Black Apr 11 '13 at 20:58\nFor $1\\times n$ boards, you get the Fibonacci numbers. For $2\\times n$ boards it's already a bit harder. You get the recurrence relation $a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$, see oeis.org/A030186 \u2013\u00a0azimut Apr 11 '13 at 21:49\n\nA hand count for $n=3$ yields $131$ ways, and a search for $1,7,131$ yields OEIS sequence A028420, the \"number of monomer-dimer tilings of an $n\\times n$ chessboard\". The entry doesn't provide a formula (so it's likely that none is known), but some references.\n\n-\nso it's like a program? \u2013\u00a0user67258 Apr 11 '13 at 21:18\nSee also oeis.org/A210662 which gives all the rectangles, not just the squares. \u2013\u00a0Charles Apr 11 '13 at 21:32\n@herderp: Sorry, I don't understand that question. What's like a program? \u2013\u00a0joriki Apr 11 '13 at 22:12\nI thought the sequence is computed by a computer program \u2013\u00a0user67258 Apr 11 '13 at 22:24\n\nIf you just used $1\\times2$ rectangles, then this is same as finding the number of matchings in the $m \\times n$ rectangle graph, and a formula for that has been given by Kastelyn:\n\n$$\\sqrt{\\left|\\prod_{j=1}^{m}\\prod_{k=1}^{n} \\left(2 \\cos \\frac{\\pi j}{m+1} + 2i\\cos \\frac{\\pi k}{n+1}\\right)\\right|}$$\n\nThis was done, by mapping the number to the square root of the determinant of a weighted adjacency matrix of the graph.\n\nIf you include $1 \\times 1$ squares, you essentially need to find the sum over all subgraphs (think of placing a $1\\times 1$ square as deleting that vertex) of the $m \\times n$ graph, and for each subgraph, the determinant approach still works, I believe, but might not have a nice closed form.\n\nYou can find a nice exposition for the $1\\times 2$ case in the first chapter here: http://users.ictp.it/~pub_off/lectures/lns017/Kenyon/Kenyon.pdf\n\n-\n\nWe can probably give some upper and lower bounds though. Let $t_n$ be the possible ways to tile an $n\\times n$ in the manner you described. At each square, we may have $5$ possibilities: either a $1\\times 1$ square, or $4$ kinds of $1\\times 2$ rectangles going up, right, down, or left. This gives you the upper bound $t_n \\leq 5^{n^2}$.\n\nFor the lower bound, consider a $2n\\times 2n$ rectangle, and divide it to $n^2$ $2\\times 2$ blocks, starting from the top left and putting a $2\\times 2$ square, putting another $2\\times 2$ square to its right and so on... For each of these $2\\times 2$ squares, we have $5$ possible distinct ways of tiling. This gives the lower bound $t_{2n} \\geq 7^{n^2}$. Obviously, $t_{2n+1} \\geq t_{2n},\\,n \\geq 1$, and therefore $t_n \\geq 7^{\\lfloor \\frac{n}{2}\\rfloor ^2}$.\n\nHence, \\begin{align} 7^{\\lfloor \\frac{n}{2}\\rfloor ^2} \\leq t_n \\leq 5^{n^2}, \\end{align} or roughly (if $n$ is even), \\begin{align} (7^{1/4})^{n^2} \\leq t_n \\leq 5^{n^2}. \\end{align} BTW, $7^{1/4} \\geq 1.6$. So, at least we know $\\log t_n \\in \\Theta(n^2)$.\n\nNote: Doing the $3\\times 3$ case for the lower bound, we get $(131)^{1/9} \\geq 1.7$ which is slightly better.\n\n-", "date": "2016-06-25 01:52:31", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/358755/ways-to-fill-a-n-times-n-square-with-1-times-1-squares-and-1-times-2-recta", "openwebmath_score": 0.9462553858757019, "openwebmath_perplexity": 358.1395556863249, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126457229185, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8537774119328476}}
{"url": "http://math.stackexchange.com/questions/23596/why-is-the-eigenvector-of-a-covariance-matrix-equal-to-a-principal-component?answertab=oldest", "text": "# Why is the eigenvector of a covariance matrix equal to a principal component?\n\nIf I have a covariance matrix for a data set and I multiply it times one of it's eigenvectors. Let's say the eigenvector with the highest eigenvalue. The result is the eigenvector or a scaled version of the eigenvector.\n\nWhat does this really tell me? Why is this the principal component? What property makes it a principal component? Geometrically, I understand that the principal component (eigenvector) will be sloped at the general slope of the data (loosely speaking). Again, can someone help understand why this happens?\n\n-\n\nShort answer: The eigenvector with the largest eigenvalue is the direction along which the data set has the maximum variance. Meditate upon this.\n\nLong answer: Let's say you want to reduce the dimensionality of your data set, say down to just one dimension. In general, this means picking a unit vector $u$, and replacing each data point, $x_i$, with its projection along this vector, $u^T x_i$. Of course, you should choose $u$ so that you retain as much of the variation of the data points as possible: if your data points lay along a line and you picked $u$ orthogonal to that line, all the data points would project onto the same value, and you would lose almost all the information in the data set! So you would like to maximize the variance of the new data values $u^T x_i$. It's not hard to show that if the covariance matrix of the original data points $x_i$ was $\\Sigma$, the variance of the new data points is just $u^T \\Sigma u$. As $\\Sigma$ is symmetric, the unit vector $u$ which maximizes $u^T \\Sigma u$ is nothing but the eigenvector with the largest eigenvalue.\n\nIf you want to retain more than one dimension of your data set, in principle what you can do is first find the largest principal component, call it $u_1$, then subtract that out from all the data points to get a \"flattened\" data set that has no variance along $u_1$. Find the principal component of this flattened data set, call it $u_2$. If you stopped here, $u_1$ and $u_2$ would be a basis of the two-dimensional subspace which retains the most variance of the original data; or, you can repeat the process and get as many dimensions as you want. As it turns out, all the vectors $u_1, u_2, \\ldots$ you get from this process are just the eigenvectors of $\\Sigma$ in decreasing order of eigenvalue. That's why these are the principal components of the data set.\n\n-\nPhenomenal answer. Thank you. \u2013\u00a0Ryan Mar 28 '11 at 11:55\n\nSome informal explanation:\n\nCovariance matrix $C_y$ (it is symmetric) encodes the correlations between variables of a vector. In general a covariance matrix is non-diagonal (i.e. have non zero correlations with respect to different variables).\n\nBut it's interesting to ask, is it possible to diagonalize the covariance matrix by changing basis of the vector?. In this case there will be no (i.e. zero) correlations between different variables of the vector.\n\nDiagonalization of this symmetric matrix is possible with eigen value decomposition. You may read 'A Tutorial on Principal Component Analysis'by Jonathon Shlens to get a good understanding. (pages 6-7)\n\n-", "date": "2013-06-19 04:50:15", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/23596/why-is-the-eigenvector-of-a-covariance-matrix-equal-to-a-principal-component?answertab=oldest", "openwebmath_score": 0.8532469868659973, "openwebmath_perplexity": 206.79197546820686, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126438601956, "lm_q2_score": 0.8723473730188543, "lm_q1q2_score": 0.8537774038117792}}
{"url": "https://math.stackexchange.com/questions/2115633/using-mathematical-induction-prove-that-11-cdot-3n-3-cdot-7n-6-is-div", "text": "# Using mathematical induction prove that $11 \\cdot 3^n + 3 \\cdot 7^n - 6$ is divisible by 8\n\nProve that $\\phi(n) =11 \\cdot 3^n + 3 \\cdot 7^n - 6$ is divisible by 8 for all $n \\in N$.\n\nBase: $n = 0$\n\n$8 | 11 + 3 - 6$ is obvious.\n\nNow let $\\phi(n)$ be true we now prove that is also true for $\\phi(n+1)$.\n\nSo we get $11 \\cdot 3^{n+1} + 3 \\cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8.\n\nFor instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here.\n\nSuppose $11*3^n + 3*7^n - 6 = 8k$\n\nThe $11*3^{n+1} + 3*7^{n+1} - 6 = 11*3^n*3 + 3*7^n*7 - 6$\n\n$=3(11*3^n + 3*7^n-2) + 4*3*7^n$\n\n$= 3(11*3^n + 3*7^n - 6) + 4*3*7^n + 12$\n\n$= 3(8k) + 4(3*7^n + 3)$; $3*7^n$ is odd and $3$ is odd so $(3*7^n + 3)$ is even.\n\n$= 3(8k) + 8(\\frac{3*7^n + 3}2) = 8(3k + \\frac{3*7^n + 3}2)$.\n\n======\n\nActually I like and am inspired by Bill Dubuques answer.\n\nWe want to prove $\\phi(n) = 11*3^n + 3*7^n - 6 \\equiv 0 \\mod 8$\n\nAnd we know $\\phi(n) = 11*3^n + 3*7^n - 6 \\equiv 3*3^n + 3*(-1)^n -6 = 3^{n+1} + 3*(-1)^n - 6 \\mod 8$.\n\nSo it's a matter of showing $f(n) = 3^{n+1} + 3(-1)^n \\equiv 6 \\mod 8$.\n\nAnd if we notice $f(n+2) = 3^{n+3} + 3(-1)^{n+2} = 3^{n+1}*9 + 3(-1)^{n} \\equiv 3^n + 3(-1)^{n}= f(n) \\mod 8$.\n\nSo it's now just a matter of showing for $f(0) \\equiv f(1) \\equiv 6 \\mod 8$.\n\nWhich is easily verified $3^1 + 3*(-1)^0 =3+3= 6$ and $3^2 + 3*(-1)^1 = 9 -3 = 6$\n\n\u2022 Yap, that's it. Thank you, just what I was looking for. \u2013\u00a0stigma6 Jan 27 '17 at 7:15\n\u2022 You want a secret confession? When I began answering I didn't know how it would turn out. But I know I had to factor $f(n+1) = manipulate(f(n)) = manipulate(8k)$ and I had faith that $manipulate(8k) = 8j$ and I just chewed on it to see what would happen. That's how a lot of figuring induction proofs. You know $f(n) = Property(t)$ and $f(n+1) = manipulate(f(n))=manipulate(Property(t))$ and know you want $manipulate(Property(t))=Property(s)$. Then you just gum and chew away until you get it. \u2013\u00a0fleablood Jan 27 '17 at 7:36\n\n${\\rm mod}\\ 8\\!:\\ f(n\\!+\\!2)\\equiv f(n)\\,$ by $\\,a\\equiv 3,7\\,\\Rightarrow\\,a^{\\large 2}\\equiv 1\\,\\Rightarrow\\,a^{\\large n+2}\\equiv a^{\\large n}.\\,$ Thus $\\,8\\mid f(n)\\iff 8\\mid f(n\\!+\\!2),\\,$ hence by (strong/parity) induction, it is true for all $n$ $\\iff$ it is true for the base cases $\\,n=0,1.$\n\n\u2022 @Downvoter The downvote is puzzling. If something is not clear then please feel welcome to ask questions and I will be happy to elaborate. This is one of the easiest ways to prove it, \u2013\u00a0Bill Dubuque Jan 26 '17 at 22:40\n\u2022 If congruences are unknown then we can eliminate them, viz. show that $8$ divides $\\,f(n\\!+\\!2)-f(n)\\,$ because $8$ divides $\\,3^{\\large n+2}-3^{\\large n} = (3^{\\large 2}-1)3^{\\large n} = 8\\cdot 3^n,\\,$ and similarly $8$ divides $\\,7^{\\large n+2}-7^{\\large n}.\\ \\$ \u2013\u00a0Bill Dubuque Jan 26 '17 at 23:04\n\nSetup the same as your current work:\n\n$\\dots$\n\n$\\dots = 11\\cdot 3^{n+1}+3\\cdot 7^{n+1}-6 = 11\\cdot 3\\cdot 3^{n}+ 3\\cdot 7\\cdot 7^n - 6$\n\n$=33\\cdot 3^n + 21\\cdot 7^n - 6 = (11+22)\\cdot 3^n + (3 + 18)\\cdot 7^n - 6$\n\n$=\\underbrace{11\\cdot 3^n + 3\\cdot 7^n - 6}_{\\text{should be familiar}} + \\underbrace{22\\cdot 3^n + 18\\cdot 7^n}_{\\text{unknown}}$\n\nNow, what can we say about $22\\cdot 3^n+18\\cdot 7^n$? Anything? You say in a previous example, you had to run a second induction proof to finish, might that be useful here?\n\n\u2022 fyi: your answer was downvoted 1 second before mine, so it doesn't appear that the downvoter took the time to carefully consider the answers. \u2013\u00a0Bill Dubuque Jan 26 '17 at 22:50", "date": "2019-10-19 20:13:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2115633/using-mathematical-induction-prove-that-11-cdot-3n-3-cdot-7n-6-is-div", "openwebmath_score": 0.7768850326538086, "openwebmath_perplexity": 222.46163476834505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712645102011, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8537774032710418}}
{"url": "https://math.stackexchange.com/questions/1766653/why-does-z-mapsto-exp-z2-have-an-antiderivative-on-mathbb-c", "text": "Why does $z\\mapsto \\exp(-z^2)$ have an antiderivative on $\\mathbb C$?\n\nWhy does $z\\mapsto \\exp(-z^2)$ have an antiderivative on $\\mathbb C$?\n\nSo far I have seen the following results:\n\n1. If $f\\colon U\\to\\mathbb C$ has an antiderivative $F$ on $U$ then $\\displaystyle\\int_\\gamma f(z)~\\mathrm dz=F(\\gamma(b))-F(\\gamma(a))$ along a smooth curve $\\gamma\\colon[a,b]\\to\\mathbb C$.\n2. If $f$ has an antiderivative then $\\displaystyle\\int_\\gamma f(z)~\\mathrm dz$ depends only on the start and end point of the curve $\\gamma$.\n3. If $f$ has an antiderivative, then $\\displaystyle\\oint_\\gamma f(z)~\\mathrm dz=0$ for all closed curves $\\gamma$.\n\nMy problem is that I could use those results to show a function has no antiderivative (like $1/z$ yielding $2\\pi\\mathrm i\\neq 0$ on $\\partial B_r(0)$) but I am unsure what might work the other way round. What might be useful here in terms of complex analysis?\n\nEDIT\n\nThere is a follow-up question based on the problem above.\n\nUse the result from above to calculate\n\n$$\\int_{-\\infty}^\\infty\\exp(-x^2-\\mathrm ikx)\\,\\mathrm dx, k\\in\\mathbb R.$$\n\nI guess something constructive like GEdgar's suggestion might be useful here, isn't it?\n\nEDIT 2\n\nHere is my solution for the follow-up question\n\n\\begin{align*} \\int_{-\\infty}^\\infty \\exp(-x^2-\\mathrm ikx)\\,\\mathrm dx &= \\int_{-\\infty}^\\infty \\exp(-(x^2+\\mathrm ikx))\\,\\mathrm dx \\\\ &= \\int_{-\\infty}^\\infty \\exp\\left(-\\left(x^2+\\mathrm ikx-\\frac{k^2}{4}+\\frac{k^2}{4}\\right)\\right)\\,\\mathrm dx \\\\ &= \\int_{-\\infty}^\\infty \\exp\\left(-\\left(\\left(x+\\frac{1}{2} \\mathrm ik\\right)^2+\\frac{k^2}{4}\\right)\\right)\\,\\mathrm dx \\\\ &= \\exp\\left(-\\frac{k^2}{4}\\right)\\int_{-\\infty}^\\infty \\exp\\left(-\\left(x+\\frac{1}{2} \\mathrm ik\\right)^2\\right)\\,\\mathrm dx \\\\ &= \\exp\\left(-\\frac{k^2}{4}\\right)\\int_{-\\infty}^\\infty \\exp(-t^2)\\,\\mathrm dt \\\\ &= \\exp\\left(-\\frac{k^2}{4}\\right)\\sqrt{\\pi} \\end{align*}\n\nAnother way.\n\n$\\exp(-z^2)$ is an entire function. Its power series at the origin converges for all $z$.\nTake the anti-derivative of that series term-by-term. It still converges for all $z$. (Use the formula for radius of convergence.) So the series of antiderivatives is an antiderivative for $\\exp(-z^2)$.\n\nThat argument works for all entire functions. In this case: $$f(z) = \\sum_{n=0}^\\infty \\frac{(-1)^n}{n!} z^{2n} \\\\ F(z) :=\\sum_{n=0}^\\infty \\frac{(-1)^n}{n!}\\;\\frac{z^{2n+1}}{2n+1} \\\\ F'(z) = f(z)$$\n\nThe composition of functions with an antiderivative has an antiderivative.\n\n$\\exp$ has an antiderivative because it is its own derivative (and hence its own antiderivative) everywhere.\n\n$z \\mapsto -z$ has an antiderivative $z \\mapsto \\frac{-z^2}{2}$.\n\n$z \\mapsto z^2$ has antiderivative $z \\mapsto \\frac{z^3}{3}$.\n\n\u2022 Really? $\\exp(\\exp(x))$ doesn't have an antiderivative though, does it? \u2013\u00a0Lundborg May 1 '16 at 12:17\n\u2022 @Neutronic Not in closed form, certainly, but there is a function which differentiates to $\\exp \\circ \\exp$ everywhere. \u2013\u00a0Patrick Stevens May 1 '16 at 12:18\n\u2022 @PatrickStevens ... Still, we need a proof of that fact. \u2013\u00a0GEdgar May 1 '16 at 13:38\n\nDefine, for some $z_0\\in \\mathbb{C}$ $$g(z)=\\int_{z_0}^z e^{-t^2}dt$$ Now for any $2$ homotopic (within the domain where $e^{-z^2}$ is analytic) curves the value of this integral will be the same. Since $e^{-z^2}$ is analytic and $\\mathbb{C}$ is simply connected, we find that the value of $g(z)$ is independent of the chosen contours. Hence this is a well defined function of $z$, and clearly $g'=f$.\n\nmore details: We know that holomorphic differentials are closed, i.e. $d(fdz)=0$. We also know that on a simply connected domain closed forms are exact. This means we can write, using stokes theorem, $$\\int_{\\gamma}f dz = \\int_{\\gamma}dg=\\int_{\\partial\\gamma}g=g(\\gamma(1))-g(\\gamma(0))=g(z)-g(z_0)$$ which shows that the integral does not depend on the path $\\gamma$ we chose.\n\n\u2022 So far we haven't been talking about homotopy so I find it difficult to fully understand your proof. I have a feeling you are trying to pick arbitrary curves which start and end point are fixed and showing that the integrals along those curves is always the same. Can you confirm this? I am struggling to understand how you deduce this exactly. \u2013\u00a0Christian Ivicevic May 1 '16 at 13:04\n\u2022 Yes that is what I am doing. I'll include a bit more detail in my answer \u2013\u00a0user2520938 May 1 '16 at 13:07", "date": "2020-04-06 13:02:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1766653/why-does-z-mapsto-exp-z2-have-an-antiderivative-on-mathbb-c", "openwebmath_score": 0.9993290901184082, "openwebmath_perplexity": 265.2539088837637, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126451020108, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8537773935268504}}
{"url": "http://karamarine.com/be-born-dtmwpm/solve-linear-congruence-bd4eea", "text": "# solve linear congruence\n\nWe need now aplly the above recursive relation: Finally, solutions to the given congruence are, $$x \\equiv 61, 61 + 211, 61 \\pmod{422} \\equiv 61, 272 \\pmod{422}.$$. Necessary cookies are absolutely essential for the website to function properly. This says we can take x = (105*7 + 65)/50 = 16. Solving linear congruences is analogous to solving linear equations in calculus. To the above congruence\u00a0 we add the following congruence, By dividing the congruence by $7$, we have. The solution to the congruence $ax \\equiv b \\pmod m$ is now given with: $$x \\equiv v + t \\cdot m\u2019 \\pmod m, \\quad t= 0, 1, \\ldots, d-1.$$. So if g does divide b and there are solutions, how do we find them? stated modulo 90, and so the most satisfying answer is given in terms of congruence classes modulo 90. The solution of a linear congruence can be found in the Wolfram Language using Reduce[a*x == b, x, Modulus -> m]. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. We must now see how many distinct solutions are there. solve the linear congruence step by step. Let x 0 be any concrete solution to the above equation. However, if we divide both sides of the congru- Solve the linear system sa+ tm= 1: Then sba+ tbm= b: So sba b (mod m) gives the solution x= sb. We find y = 4. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. This problem has been solved! Linear Congruence Video. The algorithm says we should solve 100y \u00e2\u0089\u00a1 -13(mod 7). A Linear Congruence is a congruence mod p of the form where,,, and are constants and is the variable to be solved for. most likely will be coming back here in the future, Thank you! Here, \"=\" means the congruence symbol, i.e., the equality sign with three lines. The congruence $ax \\equiv b \\pmod m$ has solutions if and only if $d = \\gcd(a, m)$ divides $b$. If (a;m) = 1, then the congruence ax b mod mphas exactly one solution modulo m. Constructive. Then the solutions to ax \u00e2\u0089\u00a1 b (mod m) are x = y + tm/g where t = 0, 1, 2, \u00e2\u0080\u00a6,\u00c2\u00a0g-1. solutions of a linear congruence (1) by looking at solutions of Diophantine equation (2). We can repeat this process recursively until we get to a congruence that is trivial to solve. Solve x^11 + x^8 + 5 mod(49) I have a lot of non-linear congruence questions, so I need an example of the procedure. If u 1 and u 2 are solutions, then au 1 b (mod m) and au 2 b (mod m) =)au 1 au 2 (mod m) =)u 1 u Solve The Linear Congruence Step By Step ; Question: Solve The Linear Congruence Step By Step . Solve Linear Congruences Added May 29, 2011 by NegativeB+or- in Mathematics This widget will solve linear congruences for you. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. $3x \\equiv 8 \\pmod 2$ means that $3x-8$ must be divisible by $2$, that is, there must be an integer $y$ such that. Find all solutions to the linear congruence $5x \\equiv 12 \\pmod {23}$. Solve the following system of linear congruences: From the first linear congruence there exists a such that: Substituting this into the second linear congruence gives us: Notice that , and so there exists a solution. Then x 0 \u2261 \u2026 Let d = gcd(c,m), and choose q, r 2Z such that c = dq and m = d r. If b is a solution to (1), then it is also a If $d \\nmid b$, then the linear congruence $ax \\equiv b \\pmod m$ has no solutions. 1 point Solve the linear congruence 2x = 5 (mod 9). Example 1. Existence of solutions to a linear congruence. // Example: To solve \u20ac \u2026 Since 7 and 100 are relatively prime, there is a unique solution. Thanks a bunch, Your email address will not be published. Therefore, solution to the congruence\u00a0$3x \\equiv 8 \\pmod 2$ is, $$x = x_0 + 2t, \\quad t \\in \\mathbb{Z},$$. A linear congruence is the problem of finding an integer x satisfying, for specified integers a, b, and m. This problem could be restated as finding x such that, Two solutions are considered the same if they differ by a multiple of m. (It\u2019s easy to see that x is a solution if and only if x + km is a solution for all integers k.). Linear Congruence Calculator. First, suppose a and m are relatively prime. Thus: Hence our solution in least residue is 7 (mod 23). The algorithm can be formalized into a procedure suitable for programming. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Solving the congruence a x \u2261 b (mod m) is equivalent to solving the linear Diophantine equation a x \u2013 m y = b. See the answer. Your email address will not be published. The result is closely related to the Euclidean algorithm. One or two coding examples would\u2019ve been great, though =P, this really helpful for my project. Since $\\gcd(6,8) = 2$ and $2 \\nmid 7$, there are no solutions. We can calculate this using the division algorithm. The calculations are somewhat involved. This is a linear Diophantine equation and it has a solution if and only if $d = \\gcd(a, m)$ divides $b$. Theorem 1. If y solves this new congruence, then x = (my + b)/ a solves the original congruence. In particular, (1) can be rewritten as This field is denoted by $\\mathbb{Z}_p$. This is progress because this new problem is solving a congruence with a smaller modulus since a < m. If y solves this new congruence, then x = (my + b)/a solves the original congruence. Example. By the Euler\u2019s theorem, $$a^{\\varphi (m)} \\cdot b \\equiv b \\pmod m.$$, By comparing the above congruence with the initial congruence, we can show that, $$x \\equiv a^{\\varphi (m) -1} \\cdot b \\pmod m$$. The linear congruence The result is closely related to the Euclidean algorithm. Solving Linear Congruence by Finding an Inverse. Although Bill Cook's answer is completely, 100% correct (and based on the proof of the Chinese Remainder Theorem), one can also work with the congruences successively; we know from the CRT that a solution exists. This category only includes cookies that ensures basic functionalities and security features of the website. Recall that since $(31,24)=1$ and $1|12$ there is exactly one incongruent solution modulo $24.$ To find this solution let\u2019s use the definition of congruence, \u2026 Multiply the rst congruence by 2 1 mod 7 = 4 to get 4 2x 4 5 (mod 7). The CRT is used solve systems of congruences of the form $\\rm x\\equiv a_i\\bmod m_{\\,i}$ for distinct moduli $\\rm m_{\\,i}$; in our situation, there is only one variable and only one moduli, but different linear congruences, so this is not the sort of problem where CRT applies. Adding and subtracting rational expressions, Addition and subtraction of decimal numbers, Conversion of decimals, fractions and percents, Multiplying and dividing rational expressions, Cardano\u2019s formula for solving cubic equations, Integer solutions of a polynomial function, Inequality of arithmetic and geometric means, Mutual relations between line and ellipse, Unit circle definition of trigonometric functions, Solving word problems using integers and decimals. The linear congruence equation ax = b (mod n) may be rewritten as ax1 = b - nx2 where x1, x2 -E- Z. That works in theory, but it is impractical for large m. Cryptography applications, for example, require solving congruences where m is extremely large and brute force solutions are impossible. Gauss illustrates the Chinese remainder theorem on a problem involving calendars, namely, \"to find the years that have a certain period number with respect to the solar and lunar cycle and the Roman indiction.\" We first put the congruence ax \u00e2\u0089\u00a1 b (mod m) in a standard form. To the solution to the congruence $a\u2019v \\equiv b\u2019 \\pmod{m\u2019}$, where $a\u2019 = \\frac{a}{d}, b\u2019 = \\frac{b}{d}$ and $m\u2019 = \\frac{m}{d}$, can be reached by applying a simple recursive relation: $$v_{-1}= 0, \\quad v_0 = 1, \\quad v_i = v_{i-2} \u2013 q_{i-1}, \\quad i= 1, \\ldots, k,$$. This means that a linear congruence also has infinitely many solutions which are given in the form: $$x = x_0 + \\left( \\frac{m}{d}\\right) \\cdot t, \\quad t \\in \\mathbb{Z}.$$. With the increase in the number of congruences, the process becomes more complicated. Hence -9 can be used as an inverse to our linear congruence $5x \\equiv 12 \\pmod {23}$. Finally, again using the CRT, we can solve the remaining system and obtain a unique solution modulo \u20ac [m 1,m 2]. Example 4. If b is divisible by g, there are g solutions. Update: Here are the posts I intended to write: systems of congruences, quadratic congruences. If d does divide b, and if x 0is any solution, then the general solution is given by x = x We look forward to exploring the opportunity to help your company too. Menu. Also, we assume a < m. If not, subtract multiples of m from a until a < m. Now solve my \u00e2\u0089\u00a1 \u2013b (mod a). Linear Congruences ax b mod m Theorem 1. That help us the \u2026 Let's use the division algorithm to find the inverse of modulo : Hence we can use as our inverse. The equation 3x==75 mod 100 (== means congruence), input 3x into Variable and Coeffecient, input 100 into modulus, and input 75 into the last box. Required fields are marked *. 10 15 20 25 30 None of the above 1 point Using the binary modular exponentiation algorithm (as shown in lecture, Algorithm 5 in Section 4.2) to \u2026 So the solutions are 16, 37, 58, 79, and 100. Given the congruence, Suppose that $\\gcd(a, m) =1$. That is, assume g = gcd(a, m) = 1. We first note that $(5, 23) = 1$, hence we this linear congruence has 1 solution (mod 23). For another example, 8x \u00e2\u0089\u00a1 2 (mod 10) has two solutions, x = 4 and x = 9. Linear Congruence Calculator. Therefore, $x_1$ and $x_2$ are congruent modulo $m$ if and only if $k_1$ and $k_2$ are congruent modulo $d$. Thus: Hence for some , . Then first solve the congruence (a/g)y \u00e2\u0089\u00a1 (b/g) (mod (m/g)) using the algorithm above. Theorem 2. The algorithm above says we can solve this by first solving 21y \u00e2\u0089\u00a1 -13 (mod 10), which reduces immediately to y \u00e2\u0089\u00a1 7 (mod 10), and so we take y = 7. If we need to solve the congruence $ax \\equiv b \\pmod p$, we must first find the greatest common divisor $d= \\gcd(a,m)$ by using the Euclidean algorithm. So, we restrict ourselves to the context of Diophantine equations. Rather, this is linear algebra. The brute force solution would be to try each of the numbers 0, 1, 2, \u00e2\u0080\u00a6,\u00c2\u00a0m-1 and keep track of the ones that work. Solve the following congruence: We must first find $\\gcd(422, 186)$ by using the Euclidean algorithm: Therefore, $\\gcd(422, 186) = 2$. Proposition 5.1.1. This entails that a set of remainders $\\{0, 1, \\ldots, p-1 \\}$ by dividing by $p$, whit addition and multiplication $\\pmod p$, makes the field. If this condition is satisfied, then the above congruence has exactly $d$ solutions modulo $m$, and that, $$x = x_0 + \\frac{m}{d} \\cdot t, \\quad t = 0, 1, \\ldots, d-1.$$. Previous question Next question Get more help from Chegg. Thanks :) If the number $m =p$ is a prime number, and if $a$ is not divisible by $p$, then the congruence $ax \\equiv b \\pmod p$ always has a solution, and that solution is unique. Now substitute for x in the second congruence: 3(6+7t) 4 (mod 8). First, let\u2019s solve 7x \u00e2\u0089\u00a1 13 (mod 100). Example 2. In the table below, I have written x k first, because its coefficient is greater than that of y. In this case, $\\overline{v} \\equiv v_k \\pmod m\u2019$ is a solution to the congruence $a\u2019 \\overline{v} \\equiv 1 \\pmod{m\u2019}$, so $v \\equiv b\u2019 v_k \\pmod{m\u2019}$ is the solution to the congruence $a\u2019v \\equiv b\u2019 \\pmod{m\u2019}$. You can verify that 7*59 = 413 so 7*59 \u00e2\u0089\u00a1 13 (mod 100). By finding an inverse, solve the linear congruence $31 x\\equiv 12 \\pmod{24}.$ Solution. It is possible to solve the equation by judiciously adding variables and equations, considering the original equation plus the new equations as a system of linear \u2026 But opting out of some of these cookies may affect your browsing experience. 1/15 15 22 31 47 Fermat's Little Theorem is often used in computing large powers modulo n, 1 point under some conditions. By subtracting obtained equations we have: It follows: $x \u2013 x_0 = 2t, t \\in \\mathbb{Z}$. I intend to write posts in the future about how to solve simultaneous systems of linear congruences and how to solve quadratic congruences. For example, 8x \u00e2\u0089\u00a1 3 (mod 10) has no solution; 8x is always an even integer and so it can never end in 3 in base 10. Let $a$ and $m$ be natural numbers, and $b$ an integer. We also use third-party cookies that help us analyze and understand how you use this website. Let , and consider the equation (a) If , there are no solutions. Email: donsevcik@gmail.com Tel: 800-234-2933; The most important fact for solving them is as follows. Then x = (100*4 + 13)/7 = 59. My colleagues and I have decades of consulting experience helping companies solve complex problems involving data privacy, math, statistics, and computing. It turns out x = 9 will do, and in fact that is the only solution. Which of the following is a solution for x? We assume a > 0. Since gcd(50, 105) = 5 and 65 is divisible by 5, there are 5 solutions. So we first solve 10x \u00e2\u0089\u00a1 13 (mod 21). Then $x_0 \\equiv b \\pmod m$ is valid. For example, we may want to solve 7x \u00e2\u0089\u00a1 3 (mod 10). Solution to a linear congruence equation is equivalent to finding the value of a fractional congruence, for which a greedy-type algorithm exists. the congruences whose moduli are the larger of the two powers. These cookies do not store any personal information. Browse other questions tagged linear-algebra congruences or ask your own question. Solve the following congruence: $$x \\equiv 5^{\\varphi(13) -1} \\cdot 8 \\pmod{13}.$$, Since $\\varphi (13) =12$, that it follows, By substituting it in $x \\equiv 3^{11} \\cdot 8 \\pmod{13}$ we obtain. Since $\\frac{m}{d}$ divides $m$, that by the theorem 6. Proof. Example 1. For example 25x = 15 (mod 29) may be rewritten as 25x1 = 15 - 29x2. Our rst goal is to solve the linear congruence ax b pmod mqfor x. Unfortu-nately we cannot always divide both sides by a to solve for x. A linear congruence is an equation of the form. This simpli es to 5t 2 (mod 8), which we solve by multiplying both sides by The answer to the first question depends on the greatest common divisor of a and m. Let g = gcd(a, m). We have $a\u2019 = \\frac{186}{2} = 93$, $b\u2019 = \\frac{374}{2} = 187$ and $m\u2019 = \\frac{422}{2} = 211$. In this way we obtain the congruence which also specifies the class that is the solution. The given congruence we write in the form of a linear Diophantine equation, on the way described above. This website uses cookies to improve your experience while you navigate through the website. Now what if the numbers a and m are not relatively prime? The method of\u00a0\u00a0transformation of coefficients consist in the fact that to the given equation we add or subtract a well selected true congruence. Solutions we can write in the equivalent form: $$x_1 = 61 + 422t, \\quad x_2 = 272 + 422t, \\quad t \\in \\mathbb{Z}.$$, The Euler\u2019s method consist in the fact that we use the Euler\u2019s theorem. In the second example, the order is reversed because the coefficient of the x k is smaller than the coefficient of the y. and that is the solution to the given congruence. However, linear congruences don\u2019t always have a unique solution. If b is not divisible by g, there are no solutions. Solve the following congruence: Since $\\gcd(3, 2) = 1$, that, by the theorem 1., the congruence has a unique solution. With modulo, rather than talking about equality, it is customary to speak of congruence. 1 point In order to solve the linear congruence 15x = 31 (mod 47) given that the inverse of 15 modulo 47 is 22, what number should be multiplied to both sides in the given congruence? Get 1:1 help now from expert Advanced Math tutors (b) If , there are exactly d distinct solutions mod m.. Observe that Hence, (a) follows immediately from the corresponding result on linear \u2026 Solution: We have gcd(42,90) = 6, so there is a solution since 6 is a factor of 12. Here we use the algorithm to solve: 5x\u22123y=1 (5x\u22611 (mod 3), which is easily solved by testing. first place that I\u2019ve understood it, after looking through my book and all over the internet This means that there are exactly $d$ distinct solutions. The one particular solution to the equation above is $x_0 = 0, y_0 = -4$, so $3x_0 \u2013 2y_0 = 8$ is valid. The complete set of solutions to our original congruence can be found by adding multiples of 105/5 = 21. This was really helpful. You also have the option to opt-out of these cookies. There are several methods for solving linear congruences; connection with\u00a0 linear Diophantine equations, the method of transformation of coefficients, the Euler\u2019s method, and a method that uses the Euclidean algorithm\u2026, Connection with\u00a0 linear Diophantine equations. It is mandatory to procure user consent prior to running these cookies on your website. Theorem. Linear CongruencesSimultaneous Linear CongruencesSimultaneous Non-linear CongruencesChinese Remainder Theorem - An Extension Theorem (5.6) If d = gcd(a;n), then the linear congruence ax b mod (n) has a solution if and only if d jb. If this condition is met, then all solutions are given with the formula: $$x = x_0 + \\left (\\frac{m}{d} \\right) \\cdot t, \\quad y= y_0 \\left (\\frac{a}{d} \\right) \\cdot t,$$. Solving the congruence 42x \u2261 12 (mod 90) is equivalent to solving the equation 42x= 12+90qfor integers xand q. Suppose a solution exists. If we need to solve a system of three linear congruences with one unknown, then we need first solve a system of two linear congruences, and then see which of the obtained solutions also satisfy the third congruence. linear congruences (in one variable x). Linear Congruences In ordinary algebra, an equation of the form ax = b (where a and b are given real numbers) is called a linear equation, and its solution x = b=a is obtained by multiplying both sides of the equation by a1= 1=a. I enjoyed your article but impore you to give more examples in simpler forms, thank you for explaining this thoroughly and easy to understand Expert Answer . For this purpose, we take any two solutions from that set: $$x_1 = x_0 + \\left( \\frac{m}{d}\\right) \\cdot k_1,$$, $$x_2 = x_0 + \\left (\\frac{m}{d}\\right) \\cdot k_2.$$, $$x_0 + \\left( \\frac{m}{d} \\right) \\cdot k_1 \\equiv x_0 + \\left( \\frac{m}{d} \\right) \\cdot k_2 \\pmod m$$, $$\\left( \\frac{m}{d} \\right) \\cdot k_1 \\equiv \\left( \\frac{m}{d} \\right) \\cdot k_2 \\pmod m.$$. 24 8 pmod 16q. This website uses cookies to ensure you get the best experience on our website. Example 3. Substituting this into our equation for yields: Thus it follows that , so is the solution t\u2026 Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Now let\u2019s find all solutions to 50x \u00e2\u0089\u00a1 65 (mod 105). For daily tweets on algebra and other math, follow @AlgebraFact on Twitter. If it is now $x_1$ any number from the equivalence class determined with $x_0$, then from $x_1 \\equiv x_0 \\pmod m$ follows that $ax_1 \\equiv ax_0 \\pmod m$, so $ax_1 \\equiv b \\pmod m$, which means that $x_1$ is also the solution to $ax \\equiv \\pmod m$. How do I solve a linear congruence equation manually? This reduces to 7x= 2+15q, or 7x\u2261 \u2026 Let $x_0$ be any concrete solution to the above equation. Therefore, if $ax \\equiv b \\pmod m$ has a solution, then there is infinitely many solutions. The algorithm can be formalized into a procedure suitable for programming. A linear congruence\u00a0\u00a0$ax \\equiv b \\pmod m$ is equivalent to. In an equation a x \u2261 b (mod m) the first step is to reduce a and b mod m. For example, if we start off with a = 28, b = 14 and m = 6 the reduced equation would have a = 4 and b = 2. The notion of congruences was first introduced and used by Gauss in his Disquisitiones Arithmeticae of 1801. Lemma. In case the modulus is prime, everything you know from linear algebra goes over to systems of linear congruences. Systems of linear congruences can be solved using methods from linear algebra: Matrix inversion, Cramer's rule, or row reduction. x \u2261 (mod )--- Enter a mod b statement . Construction of number systems \u2013 rational numbers. is the solution to the initial congruence. This simpli es to x 6 (mod 7), so x = [6] 7 or x = 6 + 7t, where t 2Z. Linear Congruences. Since $2 \\mid 422$, that the given congruence has solutions ( it has exactly two solutions). If not, replace ax \u00e2\u0089\u00a1 b (mod m) with \u2013ax \u00e2\u0089\u00a1 \u2013b (mod m). For instance, solve the congruence $6x \\equiv 7 \\pmod 8$. Featured on Meta \u201cQuestion closed\u201d \u2026 The proof for r > 2 congruences consists of iterating the proof for two congruences r \u2013 1 times (since, e.g., \u20ac ([m 1,m 2],m 3)=1). Section 5.1 Solving Linear Congruences \u00b6 Our first goal to completely solve all linear congruences $$ax\\equiv b$$ (mod $$n$$). A modular equation is an equation (or a system of equation, with at least one unknown variable) valid according to a linear congruence (modulo/modulus). In general, we may have to apply the algorithm multiple times until we get down to a problem small enough to solve easily. These cookies will be stored in your browser only with your consent. We can repeat this process recursively until we get to a congruence that is trivial to solve. Find more at https://www.andyborne.com/math See how to solve Linear Congruences using modular arithmetic. Solve the following congruence: Since $\\gcd(7, 15) = 1$, that the given congruence has a unique solution. Linear Congruence Calculator. where $k$ is the least non-zero remainder and $q_i$ are quotients in the Euclidean algorithm. Since 100 \u00e2\u0089\u00a1 2 (mod 7) and -13 \u00e2\u0089\u00a1 1 (mod 7), this problem reduces to solving 2y \u00e2\u0089\u00a1 1 (mod 7), which is small enough to solve by simply sticking in numbers. Solving the congruence $ax \\equiv b \\pmod m$ is equivalent to solving the linear Diophantine equation $ax \u2013 my = b$. Let\u2019s talk. Adding multiples of 105/5 = 21 is mandatory to procure user consent prior to running these on! B ( mod 23 ) procedure suitable for programming systems of linear Added. G = gcd ( a ; m ) with three lines least non-zero remainder and b. Can apply that knowledge to solve \u20ac \u2026 linear congruences ( in one variable )! But opting out of some of these cookies on your website 42,90 ) = and... Of congruences, the order is reversed because the coefficient of the two powers exploring the to... First solve the congruence by 2 1 mod 7 = 4 and =. Complete set of solutions to 50x \u00e2\u0089\u00a1 65 ( mod ( m/g ) ) using the algorithm can be into. Value of a fractional congruence, for which a greedy-type algorithm exists ) 1 under... X in the solve linear congruence example, 8x \u00e2\u0089\u00a1 2 ( mod 10 ) has two solutions, how do solve... \u2261 ( mod 100 ) necessary cookies are absolutely essential for the website to function properly to... 2 $and$ m $is equivalent to solving the congruence which also specifies the class that the. Most important fact for solving them is as follows congruences don \u2019 t have! \\Pmod m$ be any concrete solution to the given congruence we write in the of... For daily tweets on algebra and other Math, statistics, and fact! T always have a unique solution 37, 58, 79, and computing d. A standard form. $solution because the coefficient of the website to help your company too the is! Write posts in the future about how to solve linear Diophantine equations g, there are solutions. Whose moduli are the larger of the y 2 1 mod 7 ) )! Equality, it is customary to speak of congruence this widget will solve linear (. Are quotients in the Euclidean algorithm not, replace ax \u00e2\u0089\u00a1 b ( mod m ) congruence can used... Point under some conditions non-zero remainder and$ q_i $are quotients in the second congruence: (., 1 point solve the linear congruence$ ax \\equiv b \\pmod m $has no.! \u00c2\u0089\u00a1 13 ( mod 7 ), though =P, this really for... 7 ( mod ( m/g ) ) using the algorithm can be formalized into a procedure for...:$ x \u2013 x_0 = 2t, t \\in \\mathbb { }! To a linear congruence $6x \\equiv 7 \\pmod 8$ = 2t, t \\in \\mathbb Z... We also use third-party cookies that ensures basic functionalities and security features of the website congruence equation manually congruence 1. 23 } $422$, there are no solutions you use this website uses to. An inverse, solve the linear congruence is an equation of the website 7 $, then the$!, by dividing the congruence 42x \u2261 12 ( mod m ) in a standard solve linear congruence. 'S use the division algorithm to find the inverse of modulo: Hence solution... Out x = ( 100 * 4 + 13 ) /7 = 59 4 ( mod 100 ) 7 8... How you use this website uses cookies to ensure you get the experience! And 100 consulting experience helping companies solve complex problems involving data privacy, Math follow. \\Nmid b $an integer solution in least residue is 7 ( mod )! Of coefficients consist in the fact that to the given congruence has solutions ( has! And 100 some of these cookies on your website we find them daily tweets on algebra other. 58, 79, and computing \u00e2\u0089\u00a1 2 ( mod m ) know how solve! Integers xand q specifies the class that is the only solution consider the equation 42x= 12+90qfor integers xand q,...$ \\gcd ( 6,8 ) = 1, then the congruence, for which greedy-type! Will be stored in your browser only with your consent } { }. G solutions // example: to solve simultaneous systems of linear congruences 5x... Great, though =P, this really helpful for my project 7 $, that the! To systems of linear congruences start Here ; our Story ; Hire a Tutor ; to. Result is closely related to the Euclidean algorithm that ensures basic functionalities and security of! In computing large powers modulo n, 1 point solve the congruence ( 1 ) by at! Solutions, x = 9 will do, and$ 2 \\nmid 7 $, then there a! Helping companies solve complex problems involving data privacy, Math, statistics, and$ q_i $are in... The number of solve linear congruence, the process becomes more complicated, x = 9, there. Quotients in the future about how to solve easily ask your own.! G does divide b and there are no solutions solutions ) solve linear congruence.! To systems of congruences, quadratic congruences by NegativeB+or- in Mathematics this widget will solve linear congruences ( one! Though =P, this really helpful for my project is smaller than the coefficient of following... Fermat 's Little Theorem is often used in computing large powers modulo n, 1 point under some conditions (. -- - Enter a mod b statement * 59 = 413 so 7 * 59 = so! 50, 105 ) = 1 how many distinct solutions are there Math... Here ; our Story ; Hire a Tutor ; Upgrade to Math Mastery though... 65 is divisible by 5, there are no solutions equation we add the following is factor., and computing b ) if, there are g solutions point under some.. Our solution in least residue is 7 ( mod 7 ): 3 ( mod m )$... 100 * 4 + 13 ) /7 = 59 * 59 = 413 so 7 * 59 \u00e2\u0089\u00a1 13 mod... In terms of congruence 105/5 = 21 concrete solution to a congruence that is to... 9 solve linear congruence do, and 100 are relatively prime start Here ; our Story ; Hire a Tutor Upgrade. Companies solve complex problems involving data privacy, Math, statistics, and consider the equation 42x= 12+90qfor xand! The numbers a and m are relatively prime ) 1 point solve the linear congruence by. The y equality, it is customary to speak of congruence solve 7x \u00e2\u0089\u00a1 3 ( mod )... - Enter a mod b statement your own question in two variables, we can apply that to... Forward to exploring the opportunity to help your company too that $\\gcd ( a, )! The solution to the above congruence we write in the second example, have. ( 100 * 4 + 13 ) /7 = 59 given in terms of congruence classes 90! Is greater than that of y your company too sign with three lines ) if, there infinitely... A unique solution for solving them is as follows - Enter a mod b statement repeat this recursively! Advanced Math tutors the congruences whose moduli are the larger of the congru- other. Tagged linear-algebra congruences or ask your own question multiples of 105/5 = 21 the value a! Example 25x = 15 - 29x2 navigate through the website ; question: solve the congruence \u2261... 105 ) = 1 now see how many distinct solutions are 16, 37,,. Suitable for programming this category only includes cookies that ensures basic functionalities and security features of form. Example, 8x \u00e2\u0089\u00a1 2 ( mod 100 )$ x \u2013 x_0 = 2t, t \\mathbb..., t \\in \\mathbb { Z } _p $statistics, and$ 2 \\mid $! About how to solve easily stated modulo 90 1/15 15 22 31 47 Fermat 's Theorem... To Math Mastery q_i$ are quotients in the second example, the equality sign with three lines but out. Your experience while you navigate through the website to function properly is as follows x \u2261 ( mod 10 has... 50, 105 ) = 1 13 ) /7 = 59 function properly reversed because the coefficient of two... Analyze and understand how you use this website uses cookies to improve your experience while you navigate the... Be rewritten as 25x1 = 15 - 29x2: systems of linear congruences and how to solve to... 4 to get 4 2x 4 5 ( mod ( m/g ) ) using the algorithm above knowledge... Ve been great, though =P, this really helpful for my project original congruence can be used an!, 58, 79, and consider the equation ( 2 ) 6x. = 4 and x = ( solve linear congruence * 4 + 13 ) /7 = 59 Here ; our ;. This means that there are no solutions 4 + 13 ) /7 = 59 since gcd ( a, )... Means the congruence 42x \u2261 12 ( mod 100 ) 4 2x 4 5 ( mod )... Theorem 6 you can verify that 7 * 59 \u00e2\u0089\u00a1 13 ( mod )... \\Equiv 7 \\pmod 8 \\$ posts in the second congruence: 3 ( 6+7t 4... In a standard form 1, then there is a unique solution x x_0! Y \u00e2\u0089\u00a1 ( b/g ) ( mod 7 ) is not divisible by,. The future about how to solve 7x \u00e2\u0089\u00a1 3 ( 6+7t ) 4 ( mod 90 ) is equivalent solving! Exactly two solutions ) means the congruence ax b mod mphas exactly one solution modulo m... 1 ) by looking at solutions of Diophantine equation ( 2 ) the congruences whose moduli are posts..., 105 ) that there are no solutions 22 31 47 Fermat Little!", "date": "2022-09-30 17:02:13", "meta": {"domain": "karamarine.com", "url": "http://karamarine.com/be-born-dtmwpm/solve-linear-congruence-bd4eea", "openwebmath_score": 0.7182279825210571, "openwebmath_perplexity": 413.5897726235827, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9711290930537121, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8537650137551713}}
{"url": "https://www.physicsforums.com/threads/monotonically-decreasing-increasing.775576/", "text": "# Monotonically decreasing/increasing\n\n1. Oct 11, 2014\n\n### BOAS\n\n1. The problem statement, all variables and given/known data\n\nPlot each function and decide, based on the plot whether or not it is monotonically increasing/decreasing or strictly monotonically increasing/decreasing.\n\nf(x) = -5\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nI can plot this function no problem, and show algebraically why it fits the definition of monotonically increasing and monotonically decreasing.\n\nMy question is, how do I justify this 'based on the plot'?\n\nThe next part of the question is to show it algebraically, but graphically i'm not really sure why my answer is correct...\n\n2. Oct 11, 2014\n\n### Ray Vickson\n\nYou really need to show more: what exact definitions are you using for \"monotonically increasing/decreasing\"? You seem to be making a distinction between monotonically increasing and strictly monotonically increasing, etc. Your answer could be right or wrong, depending on these details.\n\n3. Oct 11, 2014\n\n### Staff: Mentor\n\nThe question distinguishes between monotonically increasing/decreasing and strictly monotonically increasing/decreasing. In the latter, the graph of the function would have to be trending up (mon. increasing) or down (mon. decreasing). From your plot, which should show a horizontal line, which you could characterize as both monotically decreasing and monotonically increasing.\n\n4. Oct 11, 2014\n\n### BOAS\n\nI have attached the definitions I am using as a picture. I will explain how i'm using them to come to my result.\n\nFrom this, I say that the function $f(x) = -5$ is monotonically increasing and monotonically decreasing because for all $x_{1}, x_{2} \\in \\mathbb{R}$ where $x_{1} < x_{2}, f(x_{1}) \\leq f(x_{2})$.\n\nLikewise, for all $x_{1}, x_{2} \\in \\mathbb{R}$ where $x_{1} < x_{2}, f(x_{1}) \\geq f(x_{2})$.\n\n5. Oct 11, 2014\n\n### Staff: Mentor\n\nI figured that these were the definitions you were using. I agree with your answer.\n\n6. Oct 11, 2014\n\n### vela\n\nStaff Emeritus\nIs your question essentially \"How can a flat graph be considered increasing or decreasing?\"\n\n7. Oct 11, 2014\n\n### BOAS\n\nthat is at the heart of it.\n\n8. Oct 11, 2014\n\n### vela\n\nStaff Emeritus\nIt's just how it is. As long as the function satisfies the definition, we say it's monotonically increasing or decreasing, even if it may run counter to our everyday sense of the words. Your intuition aligns with strictly monotonically increasing or decreasing.\n\n9. Oct 11, 2014\n\n### Staff: Mentor\n\nNote that the definitions for monotonically increasing and decreasing include the possibility that f(x1) = f(x2). IOW, they include the possiblity of graphs that are flat. The \"strictly\" definitions don't include this.", "date": "2017-08-22 09:38:18", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/monotonically-decreasing-increasing.775576/", "openwebmath_score": 0.5020829439163208, "openwebmath_perplexity": 994.0720642106213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129092218133, "lm_q2_score": 0.8791467770088162, "lm_q1q2_score": 0.853765011483069}}
{"url": "https://math.stackexchange.com/questions/1721513/area-of-region-bounded-by-two-parabolas-using-double-integrals", "text": "# Area of region bounded by two parabolas using double integrals\n\n\"Let R be the region in the first quadrant bounded by the graphs of the parabolas $y=2x^2$, $y=9-x^2$ and the line x=0. Express the area of region R:\n(i) Integrating first with respect to y, and then with respect to x\n(ii) Integrating first with respect to x, and then with respect to y \"\n\nI have tried sketching the two curves, and expressing the double integral. However, my sketch and the double integral itself is quite different from the solution. Please have a look at my attempt and the solution below:\n\nThe Solution:\n\nCould someone please help me understand this solution? I feel like I am missing a very important concept about double integrals. Any help would be highly appreciated.\n\n\u2022 The area that you drew is not correct. It should extend upward until the $9-x^2$ parabola. And that is only one of two solutions. Mar 31 '16 at 9:01\n\u2022 Thanks for the reply. But after extending the graph, I would still get a similar shape to the one I currently have. The solution's graph is different. Could it be incorrect? Mar 31 '16 at 19:30\n\nIf I well understand your question maybe that this intuitive interpretation can help.\n\nWhen we calculate an area by a double integral, we subdivide this area in a sum of little ''area elements''. If we chose these elements as $dy dx$ or $dx dy$, this means that the elements are little rectangles of sides $dy$ and $dx$.\n\nIf we chose the order $dydx$ this means that we want to count these elements starting from ''vertical'' strips in which (in your case) the side $dy$ start from $2x^2$ and goes to $9-x^2$, so these values are the limits of the sum, and becomes the limits of the integration in $dy$, than we sum all these strips summing for the oter side $dx$ has limits $o$ and $\\sqrt{3}$, and this gives the double integral: $$\\int_0^{\\sqrt{3}}\\int_{2x^2}^{9-x^2}dydx$$\n\nIf we change the order to $dx dy$ we count the area elements starting from horizontal stripes, and in this case the side $dx$ start from $x=0$ and goes to $\\sqrt{y/2 }$ if $y\\le 6$, and to $\\sqrt{9-y }$ if $6<y\\le 9$. So you have the other double integral that is divided in two parts.\n\n\u2022 Thank you very much for this explanation; it makes perfect sense to me. But how do I go about sketching the projection? I have a habit of always sketching the region before setting up the double integrals, Apr 1 '16 at 13:34\n\u2022 I've added a figure. I hope it's useful. Apr 1 '16 at 13:50\n\u2022 Perfect! Thanks you :) Apr 3 '16 at 8:26\n\nThe first graph you draw is good, but you paint incorrectly. You should have painted from left: $x=0$, from up: $y = 9-x^2$, and from down: $y = 2x^2$.\n\nThe graph in the solution is not correct, i think, but the double integrals are true.\n\n\u2022 I thought so, the solution graph is incorrect. Thanks for responding Mar 31 '16 at 19:31", "date": "2021-10-26 06:40:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1721513/area-of-region-bounded-by-two-parabolas-using-double-integrals", "openwebmath_score": 0.8568595051765442, "openwebmath_perplexity": 150.274623287278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290905469752, "lm_q2_score": 0.8791467675095292, "lm_q1q2_score": 0.8537650007888421}}
{"url": "https://math.stackexchange.com/questions/2029326/rock-drawing-possibility/2029337", "text": "# Rock drawing possibility [closed]\n\nIn the latest episode of Survivor there were 6 people and they had to draw rocks from a bag. In the bag were 6 rocks, 5 white rocks and 1 black rock. First person drew a rock and kept it in his hand. Then the second person drew a rock. Last person took the only rock that left.\n\nThen they all showed what they got at the same time.\n\nWas the possibility to draw a black rock for each person 1/6 or what was it?\n\nFirst point of view is they can all draw the rocks at the same time without changing the outcome.\n\nAnother way of seeing this is to compute the probability that the $k$th person picks the black stone. The $k-1$ persons before must have picked a white stone, and the $k$th picks the black. The probability is $$\\frac56\\times\\frac45\\times\\dots\\times \\frac{6-(k-1)}{6-(k-1)+1}\\times \\frac{1}{6-(k-1)} = \\frac16$$ (telescopic product).\n\nYes.\n\nFor the first person it's obviously $1/6$.\n\nFor the second it is either $1/5$ if the first person drew a white ($5/6$), or $0$ if the first person drew black already ($1/6$). So $p($Person 2 draws black$)=5/6 \\times 1/5+1/6 \\times 0=1/6$.\n\nLogic continues and gives $1/6$ for each person.\n\nYes, the probability of drawing the black rock is 1/6 for each person. This is obvious for the first person. Intuitively \"by symmetry,\" it should be clear that everyone has the same probability 1/6 of drawing the black rock.\n\nOne more formal way to look at it is that there are $6! = 720$ orders in which the six rocks can be drawn. Let's number the rocks 1 through 6, with #1 being the black rock. Some examples among the 720 ways are 123456, 321456, 135246, 654132, and so on.\n\nSuppose you are third in line to draw. Then you would get the black rock for outcomes such as 321456 and 541326.\n\nHow many ways are there to arrange the rocks so that the black rock is third? Put the black rock in the 3rd position, and then there are 5! ways to arrange the other five rocks. So the number of ways for you to get the rock are 5! = 120.\n\nThen the probability you get the black rock is $5!/6! = 120/720 = 1/6.$\n\nAddendum: While I have been typing this and taking holiday phone calls, I see that two other answers have appeared. I have up-voted them both, as being reasonable alternative methods to show the same thing.", "date": "2022-05-19 02:57:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2029326/rock-drawing-possibility/2029337", "openwebmath_score": 0.6926605105400085, "openwebmath_perplexity": 295.82811814572585, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9888419694095655, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8537578669298872}}
{"url": "https://math.stackexchange.com/questions/661425/rate-of-flow-question/661430", "text": "# Rate of flow Question\n\nHow would you solve a general problem of a steady stream of leakage:\n\nWater leaks at a rate\n\n$$r(t)= 20 \\sqrt{3} \\sec^2 (2t) \\, \\frac{\\text{gallons}}{\\text{hour}}.$$\n\nAt time 0, there are 50 gallons of water.\n\nSo how should I find a function that represents the amount remaining at a certain time, say $\\pi/6$ hours?\n\nI understand to take the integral of $r(t)$, but how would you proceed then?\n\nAs a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity...\n\n\u2022 Hi! Welcome to MSE. I've editted your post for clarity without (hopefully) altering meaning. Best wishes. \u2013\u00a0Mark Fantini Feb 3 '14 at 0:32\n\u2022 Thanks. This makes it better. \u2013\u00a0user235059 Feb 3 '14 at 3:40\n\u2022 You are welcome. :) \u2013\u00a0Mark Fantini Feb 3 '14 at 9:29\n\u2022 As a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity... \u2013\u00a0user235059 Feb 5 '14 at 17:55\n\nHint: You are given the if $W(t)$ is the amount of water in the tank, your function $r(t)$ is $W'(t)$. So if you integrate $$\\int W'(t) dt=\\int r(t) dt$$ that will give you $W(t)$, the amount of water at time $t$. Use the fact that $W(0)=50$. Then the question is just asking what is $W(\\frac{\\pi}{6})$.\n\n\u2022 Not quite: since $\\ r(t) \\$ is a volume rate of change, the integral represents the amount of water added to or lost from the volume in the tank. This is why we must include an \"initial condition\" for the actual amount of water in the tank in order to obtain a specific function for the tank's contents. \u2013\u00a0colormegone Feb 3 '14 at 0:42\n\u2022 That's why I said to include $W(0)=50$! \u2013\u00a0Kyle L Feb 4 '14 at 7:23\n\u2022 The rate function is a way of writing a differential equation, $\\ \\frac{dV}{dt} \\ = \\ r(t) \\ .$ This does not give us exclusively a function for the rate of water volume change in the tank, but only the rate at which water is flowing. What I was taking issue with is that the integral can also describe the amount of water accumulating elsewhere, such as another tank, the outside environment, etc. Since the rate function given by OP is described as leakage, you need to write your integral as $\\int - r(t) \\ dt \\$ in order to produce the function you call $\\ W(t) \\ .$ \u2013\u00a0colormegone Feb 4 '14 at 8:29\n\nThe \"Net Change Theorem\" tells us that the volume of water lost from the container is\n\n$$\\Delta V (T) \\ = \\ \\int_0^{T} \\ - r(t) \\ \\ dt \\ ,$$\n\nthe negative sign being inserted since we are told this is \"leakage\", and thus a reduction of water volume in the container. What remains in the container at time $\\ T \\$ is then\n\n$$V(T) \\ = \\ 50 \\ + \\ \\Delta V(T) \\ .$$\n\n[We add the change in volume, which is a negative change for this problem.]", "date": "2020-12-05 15:35:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/661425/rate-of-flow-question/661430", "openwebmath_score": 0.8909657001495361, "openwebmath_perplexity": 317.1884998664008, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419677654415, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8537578602965756}}
{"url": "https://math.stackexchange.com/questions/2809429/example-sequence-b-n-with-lim-n-rightarrow-infty-nb-n-0-but-s", "text": "Example Sequence $\\{ b_n \\}$ with $\\lim_{n \\rightarrow \\infty} nb_n = 0$ but $\\sum_{n=1}^{\\infty} b_n$ diverges.\n\nAs the title states, I'm looking for an example of a strictly decreasing sequence of positive numbers with the properties that $$\\lim_{n \\rightarrow \\infty} nb_n = 0$$ but $$\\sum_{n=1}^{\\infty} b_n$$ diverges.\n\nMy efforts have been unsuccessful so far. I know that nothing of the form\n\n$$b_n = \\frac{1}{n^p}$$\n\nworks, as $\\lim_{n \\rightarrow \\infty} nb_n = 0$ if $p>1$, also implying that the series will converge via p-test. I've also tried more creative sequences like $$b_n = \\frac{\\sin(\\frac{1}{n})}{n}$$ but still no luck.\n\nMore than a specific example, is there a certain strategy I should employ to find such an example? I was thinking the sequence must go to zero must faster than $n$ goes to infinity, but not fast enough for the series to converge.\n\n\u2022 Check out $1/(n \\log(n))$. Unfortunately there is not a lot of strategy with some of these questions. It requires knowing a good handful of examples that you can assemble for future use. This is one of the most classic examples in calculus / analysis. \u2013\u00a0Cameron Williams Jun 5 '18 at 21:47\n\nHint :\n\nLook at Bertrand series.\n\nBertrand series are the series of the form : $$\\frac{1}{n^{\\alpha}(\\log n)^{\\beta}}$$ and you know that this serie converges only if : $\\alpha > 1$ or ($\\alpha = 1$ and $\\beta > 1$).\n\nWith $$s_n:=\\sum_{k=1}^n b_k,$$ try to achieve that $s_n\\to \\infty$, but sloooowly.\n\nIf $s_n= n$, then $b_n=1$ which is too large.\n\nIf $s_n=\\sqrt n$, then $b_n\\sim \\frac1{\\sqrt n}$, which makes $nb_n\\sim\\sqrt n$, still too large.\n\nIf $s_n=\\ln n$, then $b_n\\sim \\frac 1n$, still too large: $nb_n\\sim 1$.\n\nIn general, if $s_n=f(n)$ then $b_n\\sim f'(n)$.\n\nNow what if $s_n=\\ln\\ln n$? Then $b_n\\sim \\frac1{n\\ln n}$ and $nb_n\\sim \\frac1{\\ln n}\\to 0$!\n\nLet $p_n$ denote the $n$-th prime number. Then $$\\sum_{n\\in\\Bbb N}\\frac{1}{p_n}=\\infty\\tag{1}$$\n\nBut $$\\lim_{n\\to\\infty}\\frac{n}{p_n}=0\\tag{2}$$\n\nFor $(1)$, see Divergence of the sum of the reciprocals of the primes.\n\nFor $(2)$, this is the Prime Number Theorem: $p_n\\sim n\\log n$. See for instance this discussion. This is also another way to prove $(1)$.\n\nUsing the idea in this answer: Is there a slowest rate of divergence of a series?\n\nWe set $D_n = 1/n$, so $\\sum_{n = 1}^{\\infty}D_n$ converges. Write $H_n = \\sum_{k = 1}^{n}D_n$, the $n$th harmonic number. Finally, let $d_n = D_n/H_{n-1} = \\frac{1}{nH_{n-1}}$. We can easily show that $d_n$ is positive and decreasing and $nd_n \\rightarrow 0$, and that answer shows that $\\sum_{n = 2}^{\\infty}d_n$ diverges.\n\nAsymptotically this answer is no different from $b_n = \\frac{1}{n\\ln(n)}$ given in other responses, but I thought this would be a nice fact to share.", "date": "2020-01-27 19:45:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2809429/example-sequence-b-n-with-lim-n-rightarrow-infty-nb-n-0-but-s", "openwebmath_score": 0.9650533199310303, "openwebmath_perplexity": 136.84542803771276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419707248646, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8537578593758572}}
{"url": "https://math.stackexchange.com/questions/2528538/fourier-transform-integration", "text": "# Fourier Transform Integration\n\nI have a Fourier transform to complete with the definition of the Fourier Transform.\n\nLet $$\\phi$$ be defined as follows. $$\\tag{1} \\phi(x) = Ne^{-\\frac{(x-x_0)^2}{a^2}}e^{ik_0x}$$\n\nI must complete the Fourier transform of the function.\n\nThe definition of a Fourier transform is as follows.\n\n$$\\tag{2} \\hat f(k) = \\frac{1}{\\sqrt{2\\pi}}\\int f(x) e^{-ikx}dx$$\n\nTo compute the Fourier transform we must evaluate the following integral with $$\\phi$$ substituted into (2).\n\n$$\\tag{3} \\hat\\phi(k) = \\frac{1}{\\sqrt{2\\pi}}\\int \\phi(x) e^{-ikx}dx$$\n\n$$\\tag{4} \\hat\\phi(k) = \\frac{1}{\\sqrt{2\\pi}}\\int Ne^{-\\frac{(x-x_0)^2}{a^2}}e^{ik_0x} e^{-ikx}dx$$\n\nI have tried completing this integral with completion of squares. I cannot find a way to finish this integral. How can I solve this integral?\n\n\u2022 You were correct to pursue completing the square. You will need to evaluate a Gaussian integral thereafter. And to be rigorous, you need to deform the contour back to the real line by appealing to Cauchy's Integral Theorem. And that is about it. \u2013\u00a0Mark Viola Nov 20 '17 at 2:00\n\u2022 After completing the square, $\\hat\\phi$ will become the following integral. $$\\frac{1}{\\sqrt{2\\pi}}\\int Ne^{-(x-\\frac{2x_0 + ik_0a^2}{2}}e^{\\frac{(2x_0+ik_0a^2)^2}{2}} e^{\\frac{x_0^2}{a^2}}dx$$ \u2013\u00a0Jeremy Nov 20 '17 at 2:33\n\u2022 There is an error in the previous comment. $\\hat\\phi$ will be defined as follows. $$\\frac{1}{\\sqrt{2\\pi}}\\int Ne^{-(x-\\frac{2x_0 + ik_0a^2}{2})^2}e^{\\frac{(2x_0+ik_0a^2)^2}{2}} e^{\\frac{x_0^2}{a^2}}dx$$ \u2013\u00a0Jeremy Nov 20 '17 at 2:39\n\u2022 After evaluating the Gaussian integral $\\hat\\phi$ is defined as follows. $$\\hat\\phi(k) = \\frac{1}{\\sqrt{2}}\\int e^{\\frac{(2x_0+ik_0a^2)^2}{2}} e^{\\frac{x_0^2}{a^2}}dx$$ \u2013\u00a0Jeremy Nov 20 '17 at 2:41\n\u2022 I substituted $u = x-\\frac{(2x_0 + ik_0a^2)}{2}$ in. After squaring both sides and changing the integral to polar coordinates, I found that the integral evaluated to $\\sqrt{\\pi}$. This equation I have arrived at does not match the provided solution. I may have made an error in evaluating the Gaussian integral. Where did I go wrong? \u2013\u00a0Jeremy Nov 20 '17 at 2:46\n\nFirst, enforce the substitution $x-x_0\\to x$ so that\n\n\\begin{align} \\int_{-\\infty }^\\infty e^{-\\frac1{\\alpha^2}(x-x_0)^2-i(k-k_0)x}\\,dx&=e^{-i(k-k_0)x_0}\\int_{-\\infty }^\\infty e^{-\\frac1{\\alpha^2}x^2-i(k-k_0)x}\\,dx \\end{align}\n\nThen, enforce the substitution $x/\\alpha\\to x$ so that\n\n$$e^{-i(k-k_0)x_0}\\int_{-\\infty }^\\infty e^{-\\frac1{\\alpha^2}x^2-ikx}\\,dx=\\alpha e^{-i(k-k_0)x_0}\\int_{-\\infty }^\\infty e^{-x^2-i(k-k_0)\\alpha x}\\,dx$$\n\nCompleting the square reveals\n\n$$\\alpha e^{-i(k-k_0)x_0}\\int_{-\\infty }^\\infty e^{-x^2-i(k-k_0)\\alpha x}\\,dx=\\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\\alpha/2)^2}\\int_{-\\infty }^\\infty e^{-(x-i(k-k_0)\\alpha /2)^2}\\,dx$$\n\nEnforcing the substitution $x-i(k-k_0)\\alpha/2\\to x$ yields\n\n\\begin{align}\\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\\alpha/2)^2}\\int_{-\\infty }^\\infty e^{-(x-i(k-k_0)\\alpha /2)^2}\\,dx&=\\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\\alpha/2)^2}\\\\\\\\ &\\times \\int_{-\\infty-i(k-k_0)\\alpha/2 }^{\\infty-i(k-k_0)\\alpha/2} e^{-x^2}\\,dx\\end{align}\n\nApplying Cauchy's Integral Theorem, we can deform the contour back onto the real line to obtain\n\n$$\\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\\alpha/2)^2}\\int_{-\\infty-i(k-k_0)\\alpha/2 }^{\\infty-i(k-k_0)\\alpha/2} e^{-x^2}\\,dx=\\alpha e^{-((k-k_0)\\alpha/2)^2}\\underbrace{\\int_{-\\infty}^{\\infty} e^{-x^2}\\,dx}_{=\\sqrt\\pi}$$\n\nPutting it all together, we find that\n\n$$\\int_{-\\infty }^\\infty e^{-\\frac1{\\alpha^2}(x-x_0)^2-i(k-k_0)x}\\,dx=\\alpha e^{-i(k-k_0)x_0}\\sqrt\\pi e^{-((k-k_0)\\alpha/2)^2}$$\n\n\u2022 I think the same techniques you've applied in your solution can be applied to the original problem. I think you have misread the problem statement. You have two exponentials with $(x-x_0)$ in both. Please note in the original problem statement that one exponential contains $(k-k_0)$ \u2013\u00a0Jeremy Nov 20 '17 at 3:45\n\u2022 Jeremy, I edited on 20 November in response to your posted comment herein. Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to both up vote and accept an answer as you see fit. Happy holidays! -Mark \u2013\u00a0Mark Viola Dec 18 '17 at 18:40\n\u2022 Please feel free to up vote and accept an answer as you see fit. And Happy Holidays! \u2013\u00a0Mark Viola Dec 24 '17 at 17:29\n\u2022 Your solution is incorrect by a factor of $\\frac1{\\sqrt{2}}$. \u2013\u00a0Jeremy Dec 29 '17 at 22:48\n\u2022 @Jeremy Why do you say that? \u2013\u00a0Mark Viola Dec 29 '17 at 22:50", "date": "2021-04-15 23:27:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2528538/fourier-transform-integration", "openwebmath_score": 0.8833812475204468, "openwebmath_perplexity": 355.25525227469467, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.993096162133108, "lm_q2_score": 0.8596637541053281, "lm_q1q2_score": 0.8537287749269412}}
{"url": "http://mathoverflow.net/questions/71736/number-of-closed-walks-on-an-n-cube/71737", "text": "# Number of closed walks on an $n$-cube\n\nIs there a known formula for the number of closed walks of length (exactly) $r$ on the $n$-cube? If not, what are the best known upper and lower bounds?\n\n Note: the walk can repeat vertices.\n\n-\n\nYes (assuming a closed walk can repeat vertices). For any finite graph $G$ with adjacency matrix $A$, the total number of closed walks of length $r$ is given by\n\n$$\\text{tr } A^r = \\sum_i \\lambda_i^r$$\n\nwhere $\\lambda_i$ runs over all the eigenvalues of $A$. So it suffices to compute the eigenvalues of the adjacency matrix of the $n$-cube. But the $n$-cube is just the Cayley graph of $(\\mathbb{Z}/2\\mathbb{Z})^n$ with the standard generators, and the eigenvalues of a Cayley graph of any finite abelian group can be computed using the discrete Fourier transform (since the characters of the group automatically given eigenvectors of the adjacency matrix). We find that the eigenvalue $n - 2j$ occurs with multiplicity ${n \\choose j}$, hence\n\n$$\\text{tr } A^r = \\sum_{j=0}^n {n \\choose j} (n - 2j)^r.$$\n\nFor fixed $n$ as $r \\to \\infty$ the dominant term is given by $n^r + (-n)^r$.\n\n-\nThanks! that's what I needed. \u2013\u00a0 Lev Reyzin Jul 31 '11 at 19:24\nI'm guessing not, but is there any chance this expression has a closed form? \u2013\u00a0 Lev Reyzin Aug 2 '11 at 17:12\n@Lev: you mean without a summation over $n$? I doubt it. Is fixed $n$ as $r \\to \\infty$ not the regime you're interested in? \u2013\u00a0 Qiaochu Yuan Aug 2 '11 at 17:36\nIn some sense, I'm more interested in fixed r as n gets large. The summation is certainly quite helpful, but of course if a closed form existed, it would even be nicer :) \u2013\u00a0 Lev Reyzin Aug 2 '11 at 18:06\n\nThe number of such walks is $2^n$ (the number of vertices of the $n$-cube) times the number of walks that start (and end) at the origin. We may encode such a walk as a word in the letters $1, -1, \\dots, n, -n$ where $i$ represents a positive step in the $i$th coordinate direction and $-i$ represents a negative step in the $i$th coordinate direction. The words that encode walks that start and end at the origin are encoded as shuffles of words of the form $i\\ -i \\ \\ i \\ -i \\ \\cdots\\ i \\ -i$, for $i$ from 1 to $n$. Since for each $i$ there is exactly one word of this form for each even length, the number of shuffles of these words of total length $m$ is the coefficient of $x^m/m!$ in $$\\biggl(\\sum_{k=0}^\\infty \\frac{x^{2k}}{(2k)!}\\biggr)^{n} = \\left(\\frac{e^x + e^{-x}}{2}\\right)^n.$$ Expanding by the binomial theorem, extracting the coefficient of $x^r/r!$, and multiplying by $2^n$ gives Qiaochu's formula.\n\nLet $W(n,r)$ be the coefficient of $x^r/r!$ in $\\cosh^n x$, so that $$W(n,r) = \\frac{1}{2^n}\\sum_{j=0}^n\\binom{n}{j} (n-2j)^r.$$ Then we have the continued fraction, due originally to L. J. Rogers, $$\\sum_{r=0}^\\infty W(n,r) x^r = \\cfrac{1}{1- \\cfrac{1\\cdot nx^2}{ 1- \\cfrac{2(n-1)x^2}{1- \\cfrac{3(n-2)x^2}{\\frac{\\ddots\\strut} {\\displaystyle 1-n\\cdot 1 x^2} }}}}$$ A combinatorial proof of this formula, using paths that are essentially the same as walks on the $n$-cube, was given by I. P. Goulden and D. M. Jackson, Distributions, continued fractions, and the Ehrenfest urn model, J. Combin. Theory Ser. A 41 (1986), 21\u2013-31.\n\nIncidentally, the formula given above for $W(n,r)$ (equivalent to Qiaochu's formula) is given in Exercise 33b of Chapter 1 of the second edition of Richard Stanley's Enumerative Combinatorics, Volume 1 (not published yet, but available from his web page). Curiously, I had this page sitting on my desk for the past month (because I wanted to look at Exercise 35) but didn't notice until just now that this formula was on it.\n\n-\nthanks - this is nice. \u2013\u00a0 Lev Reyzin Aug 2 '11 at 23:08\n\nAssuming a \"closed walk\" can repeat vertices, we can count closed walks starting at $0$ by counting the $r$-sequences of $[n]$ so that each number appears an even number of times. The bijection is given by labeling edges by the coordinate that is toggled between the vertices. You can probably count these sequences by inclusion/exclusion and then multiply by $2^n/r$ to account for the choice of start position.\n\n-\nIf we assume the path moves in each dimension 0 or 2 times, you can select ${n \\choose r/2}$ dimensions and then permute them $r^1/2^r$ ways. This is a lower bound on the number of walks and is likely the right asymptotics. \u2013\u00a0 Derrick Stolee Jul 31 '11 at 17:03\nThat should be $r!/2^r$. \u2013\u00a0 Derrick Stolee Jul 31 '11 at 17:03", "date": "2015-05-25 07:46:45", "meta": {"domain": "mathoverflow.net", "url": "http://mathoverflow.net/questions/71736/number-of-closed-walks-on-an-n-cube/71737", "openwebmath_score": 0.9168342351913452, "openwebmath_perplexity": 193.67619611022087, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806552225684, "lm_q2_score": 0.8705972784807406, "lm_q1q2_score": 0.8536908497676295}}
{"url": "http://math.stackexchange.com/questions/217462/question-about-summation-notation-index?answertab=active", "text": "# Question about summation notation index\n\nI'm working through a proof involving the sum of covariances but the notation is tripping me up. What does it mean when you are taking a summation over the index $i < j$? For instance $\\sum_{i < j}\\mathrm{Cov}(X_{i},X_{j})$ where Cov is covariance.\n\nI'm sure it's nothing complicated I just want to make sure I understand.\n\n-\nYou run over all $j$ and for each $j$ you run over all $i$ such that $i < j$. For instance is $\\sum_{i < j} i = 1+2+3+\\cdots+n$ if $j$ can be maximal $n$. In particular this means here that you have $\\sum_{i<j} Cov(x_i,x_j) = \\sum_{i<1} Cov(x_i,x_1) + \\sum_{i<2} Cov(x_i,x_2) + \\sum_{i<3} Cov(x_i,x_3)+ \\cdots +\\sum_{i<n} Cov(x_i,x_n)$ if $j$ is maximal $n$. \u2013\u00a0 Andr\u00e9 Oct 20 '12 at 16:04\nIn general $\\sum_{i<j} = \\sum_{j=1}^n \\sum_{i=1}^j$ \u2013\u00a0 Andr\u00e9 Oct 20 '12 at 16:13\nThis is an old post. If your question is solved, maybe you'd like to pick an answer you'd like best. Regards. \u2013\u00a0 FrenzY DT. Dec 4 '12 at 17:03\n\n### An Illustration\n\nSuppose you have $i\\in I=\\{1,2,\\cdots,5\\}$ and $j\\in J=\\{1,2,\\cdots,6\\}$. The following picture shows you which of the terms $\\sum\\limits_{i<j} a_{i,j}$ include.\n\nAccording to the picture, if you choose a certain value of $i$, then $j$ should at least be $i+1$, otherwise the condition $i<j$ isn't met (falling on or below the line $i=j$).\n\nIn conclusion, $$\\sum\\limits_{i<j} a_{i,j} = \\sum_{i} \\sum_{\\text{All j's that's >i}} a_{i,j}.$$\n\nIf you sum $j$ last, you get $$\\sum\\limits_{i<j} a_{i,j} = \\sum_{j} \\sum_{\\text{All i's that's <j}} a_{i,j}.$$\n\n-\n\nThe notation means that you sum over all allowed $i$ and $j$ such that $i < j$. The notation is compact and protects you from errors that may come from doing the end-points incorrectly.\n\nHere is an example. Suppose that $1\\leq i \\leq m$ and $1\\leq j \\leq n$ where $m$ and $n$ are not the same. If $m < n$, the sum becomes\n\n$$\\sum_{i < j} a_{ij} = \\sum_{j = i+1}^n \\sum_{i=1}^{m}a_{ij}.$$\n\nIf $n < m$ we have\n\n$$\\sum_{i < j} a_{ij} = \\sum_{j = i+1}^n \\sum_{i=1}^{n-1}a_{ij}.$$\n\nFormally, these are two different expressions and to determine which one you are using can take a few seconds here but it could take a lot more time for more complicated sums.\n\n-", "date": "2014-10-25 20:38:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/217462/question-about-summation-notation-index?answertab=active", "openwebmath_score": 0.9213523268699646, "openwebmath_perplexity": 276.30872809639965, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806563575741, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8536908491096323}}
{"url": "https://math.stackexchange.com/questions/2776463/mentally-finding-factors-of-an-integer-of-up-to-three-digits", "text": "# Mentally finding factors of an integer of up to three-digits.\n\nI\u2019m looking for the easiest methods to find factors, not necessarily prime, of positive integers $<1000$, preferably mentally, but using pen and blank paper when required.\n\nAlthough these techniques might be useful for a certain televised Numbers Game, I\u2019m just looking for methods faster than trial division.\n\nI take for granted full knowledge of the twelve by twelve multiplication tables, and recognition of primes up to $101$\n\nThis, with the addition of $7\\cdot13=91$ should cover much of the factorisation up to two digits.\n\nNotation Using the idea of hundreds, tens and units, so $n=100H+10T+U$ is the number, I\u2019ll use $H,T,U$ for the individual digits, $HT$ for the first two digits and $TU$ for the last two.\n\nFor example, $n=456$ gives $H=4$, $T=5$, $U=6$, $HT=45$ and $TU=56$\n\nSome easy published divisibility tests\n\nIf $U=0$ then $10|n$\n\nIf $(TU)\\equiv 0{\\pmod{25}}$ then ${25}|n$\n\nIf $U=5$ then $5|n$\n\nIf $U\\equiv 0{\\pmod 2}$ then $2|n$\n\nIf $(H+T+U)\\equiv 0{\\pmod 9}$ then $9|n$\n\nIf $(H+T+U)\\equiv 0{\\pmod 3}$ then $3|n$\n\nMerged published divisibility tests for primes $p=7$ to $47$\n\nWith $z$ taken from the following, if $(HT+zU)\\equiv 0{\\pmod p}$ then $p|n$\n\n$$(p,z)$$ $$(7,-2)$$ $$(11,-1)$$ $$(13,4)$$ $$(17,-5)$$ $$(19,2)$$ $$(23,7)$$ $$(29,3)$$ $$(31,-3)$$ $$(37,-11)$$ $$(41,-4)$$ $$(43,13)$$ $$(43,-30)$$ $$(47,-14)$$\n\nFor some examples in the range, the test may need to be repeated on the result.\n\nAs $31^2=961$, the last few test are of limited use.\n\nAlternative divisibility test for $11$\n\nIf $(H+U-T)=0$ or $11$ then $11|n$\n\nDifference of two squares\n\nUsing $a^2-b^2\\equiv(a+b)(a-b)$ helps in a few cases. For example,\n\n$$899=900-1=(30+1)(30-1)=31\\cdot29$$\n\n$$91=100-9=(10+3)(10-3)=13\\cdot7$$\n\n$$391=400-9=(20+3)(20-3)=23\\cdot17$$\n\nMy question\n\nI was tempted to ask what other divisibility or factorisation techniques exist that are potential easier than trial division, but \u201ceasier\u201d is subjective; so I questions is:.\n\nWhat other mental divisibility or factorisation techniques exist for an integer up to three digits?\n\nClarification update 11 May 2018\n\nSorry, I don\u2019t feel I\u2019ve sufficiently emphasised the importance of time, just assuming that easier was faster.\n\nFull factorisation is ideal; partial factorisation is much better than nothing, so $663=3\\cdot13\\cdot17$ is best, but $663=3\\cdot221$ is better than nothing.\n\nObtaining more than one small prime factor with one test is desirable, where the prime factors are obvious, as per the tests for $10,25$ and $9$, but a full list of all possible factors of $n$ is not needed.\n\n\u2022 For three digit numbers you only have to do 11 trial divisions to find factors (for the primes from 2 up to 31). If none of them divides your number, no number does (except 1 and the number itself, of course). Trial divisions for 2,3,5,7 and 11 are easy. \u2013\u00a0YukiJ May 11 '18 at 10:58\n\u2022 @YukiJ The OP wants to find factors, not necessarily prime. For instance, 6 is a factor. \u2013\u00a0scaaahu May 11 '18 at 11:35\n\u2022 @scaaahu Yes, but if we observe that 2 and 3 are factors in a number then their product $2\\cdot 3 =6$ is necessarily also a factor. Hence, if you find the prime factors, you also found the remaining other factors. \u2013\u00a0YukiJ May 11 '18 at 11:37\n\u2022 @YukiJ Having a factor 9 implies having a factor 3, not the other way around. \u2013\u00a0scaaahu May 11 '18 at 11:52\n\u2022 @scaaahu True but if you know that 3 divides $n$ then you can also check if 3 also divides $n/3$. If so, 9 divides $n$. In the particular case with 9 you could even check if the sum of digits of $n$ is divisible by 9. \u2013\u00a0YukiJ May 11 '18 at 11:55\n\nAs you have stated:\n\nChecking a number is even will automatically give you the factor of $2$.\n\nSumming the digits, if the digit sum is a multiple of $3$ or $9$, the number is also a multiple of $3$ or $9$.\n\nMultiples of $5$ end in $5$, multiples of $10$ end in $0$.\n\nFor my extra method, let's consider $473$. It has no obvious factors based on those rules, so here I would use prime multiple subtraction. Start with $7$, the multiple of $7$ that ends in $3$ is $7\\cdot9=63$, so subtract $63$ to get $410$. $41$ is not a multiple of $7$ or $9$, so neither are factors. Then consider $11$. $3\\cdot 11=33$ so subtract $33$ to get $440$. $440=40\\cdot 11$, hence we can see that $473=43\\cdot 11$.\n\nNote for $11$ that if you have $ab\\cdot11$ , where $a,b$ represent the digits, the result is $a(a+b)b$, again to mean digits: e.g. $35\\cdot11=385$. When $a+b>10$, it is harder to spot but still possible. What you should look for then is: $a(a+b-11)b$, and this is $(a-1)b \\cdot 11$. E.g. for $528$, $5+8-11=2, \\therefore 528=11*48$\n\nIf you can identify numbers as concatenations of multiples of the same number, they will be a multiple of that number. E.g. $654$ contains $6$ and $54$, both of which are multiples of $6$, and so we can do $6\\div6=1$, and $54\\div6=09$, therefore $654\\div6=109$\n\n\u2022 Thanks for your answer, which I\u2019ve up-voted. I\u2019ve used a similar method with numbers such as $721$. \u2013\u00a0Old Peter May 11 '18 at 14:42\n\u2022 Thanks for the upvote. Note that $721$ is $7\\cdot103$, you can see this by the subtraction method, or notice that $7$ and $21$ are both multiples of $7$, and so we can simply divide both by $7$ for $1$ and $03$ and then put these back together. A different example might be $726$, we see $72=12\\cdot 6$ and $6=1\\cdot 6$, putting back together we get $121$. \u2013\u00a0Rhys Hughes May 11 '18 at 14:56", "date": "2019-12-06 23:45:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2776463/mentally-finding-factors-of-an-integer-of-up-to-three-digits", "openwebmath_score": 0.8441216945648193, "openwebmath_perplexity": 270.63407062838064, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806484125338, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.853690843838832}}
{"url": "https://mathhelpboards.com/threads/d-e-word-problem.9047/", "text": "# D.E word problem\n\n#### bergausstein\n\n##### Active member\n1. radium decompose at the rate proportional to the amount itself. if the half life is 1600, find the percentage remaining after at the end of 200 years.\n\ncan you me go about solving this. thanks!\n\n#### MarkFL\n\nStaff member\nLet $R(t)$ be the mass of radium in a given sample at time $t$. How can we mathematically state how this mass changes with time in general? Just look at the first sentence and use that to try to model this change with an initial value problem.\n\n#### bergausstein\n\n##### Active member\n$R(t)=R_oe^{kt}$ where $k<0$\n\nnow the half-life is 1600 so,\n\n$1600=3200e^{kt}$\n\n#### MarkFL\n\nStaff member\n$R(t)=R_oe^{kt}$ where $k<0$\n\nnow the half-life is 1600 so,\n\n$1600=3200e^{kt}$\n\nYou have the correct equation for the mass of a sample, but the half-life being 1600 years means that at time t=1600 then $R(t)=\\dfrac{1}{2}R_0$. That is (I like to always use a positive constant):\n\n$$\\displaystyle R(1600)=R_0e^{-1600k}=\\frac{1}{2}R_0$$\n\nDivide through by $$\\displaystyle R_0$$ and the convert from exponential to logarithmic form to solve for $k$. Another way to look at it is:\n\n$$\\displaystyle R(t)=R_02^{-\\frac{t}{1600}}$$\n\nAnd then to find the percentage remaining after $t$ years, use:\n\n$$\\displaystyle \\frac{100R(t)}{R_0}=100\\cdot2^{-\\frac{t}{1600}}$$\n\n#### bergausstein\n\n##### Active member\nfrom here solving for k\n\n$\\displaystyle R(1600)=R_0e^{-1600k}=\\frac{1}{2}R_0$\n\ndividing both sides by $R_o$\n\n$\\displaystyle e^{-1600k}=\\frac{1}{2}$\n\ntaking the $\\ln$ of both sides\n\n$-1600k=\\ln\\frac{1}{2}$\n\ndividing both sides by -1600\n\n$\\displaystyle k=-\\frac{-\\ln2}{1600}$ or $k=0.0004332$\n\nnow\n\n$\\displaystyle R(t)=R_0e^{-(0.0004332)t}$\n\nhow can I find the initial $R_o$?\n\nLast edited:\n\n#### MarkFL\n\nStaff member\nYou don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.\n\n#### bergausstein\n\n##### Active member\nYou don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.\nhow can I do that if there's no given initial amount? can you tell how to go about it?\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nfrom here solving for k\n\n$\\displaystyle R(1600)=R_0e^{-1600k}=\\frac{1}{2}R_0$\n\ndividing both sides by $R_o$\n\n$\\displaystyle e^{-1600k}=\\frac{1}{2}$\n\ntaking the $\\ln$ of both sides\n\n$-1600k=\\ln\\frac{1}{2}$\n\ndividing both sides by -1600\n\n$\\displaystyle k=-\\frac{-\\ln2}{1600}$ or $k=0.0004332$\n\nnow\n\n$\\displaystyle R(t)=R_0e^{-(0.0004332)t}$\n\nhow can I find the initial $R_o$?\nDo you think it's a good idea to give a decimal approximation when the function simplifies greatly if kept exact?\n\n\\displaystyle \\begin{align*} k &= \\frac{1}{1600} \\ln{ (2) } \\end{align*}\n\nand so\n\n\\displaystyle \\begin{align*} R(t) &= R_0 \\exp { \\left[ \\frac{t}{1600} \\ln{(2)} \\right] } \\\\ &= R_0 \\exp{ \\left[ \\ln{ \\left( 2 ^{ \\frac{t}{1600} } \\right) } \\right] } \\\\ &= R_0 \\cdot 2^{ \\frac{t}{1600} } \\end{align*}\n\nand so after 200 years, what proportion of the initial amount do you have?\n\n#### MarkFL\n\nStaff member\nhow can I do that if there's no given initial amount? can you tell how to go about it?\nThis is what I posted before:\n\n...to find the percentage remaining after $t$ years, use:\n\n$$\\displaystyle \\frac{100R(t)}{R_0}$$\n\n#### bergausstein\n\n##### Active member\n$\\displaystyle \\frac{100R(200)}{R_0}=100\\cdot2^{\\frac{200}{1600}}$\n\ndo you mean like this? how can I find the percentage of remaining here?\n\n#### MarkFL\n\nStaff member\n$\\displaystyle \\frac{100R(200)}{R_0}=100\\cdot2^{\\frac{200}{1600}}$\n\ndo you mean like this? how can I find the percentage of remaining here?\nReduce the exponent, and then that is the exact percentage remaining, and then you can use a calculator to obtain a decimal approximation if you like.\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nIt appears there was something wrong with your initial model.\n\nYou know that your model involves exponential decay, so after each unit of time passes, there is a certain constant amount multiplying through. In other words\n\n\\displaystyle \\begin{align*} R_1 &= C\\,R_0 \\\\ R_2 &= C^2\\,R_0 \\\\ R_3 &= C^3\\,R_0 \\\\ \\vdots \\\\ R_t &= C^t\\,R_0 \\end{align*}\n\nSince you know that when \\displaystyle \\begin{align*} t = 1600, R_{1600} = \\frac{1}{2}R_0 \\end{align*}, that means\n\n\\displaystyle \\begin{align*} \\frac{1}{2}R_0 &= C^{1600}\\,R_0 \\\\ \\frac{1}{2} &= C^{1600} \\\\ \\ln{ \\left( \\frac{1}{2} \\right) } &= \\ln{ \\left( C^{1600} \\right) } \\\\ \\ln{ \\left( \\frac{1}{2} \\right) } &= 1600 \\ln{(C)} \\\\ \\frac{1}{1600} \\ln{ \\left( \\frac{1}{2} \\right) } &= \\ln{(C)} \\\\ \\ln{ \\left[ \\left( \\frac{1}{2} \\right) ^{\\frac{1}{1600}} \\right] } &= \\ln{(C)} \\\\ \\left( \\frac{1}{2} \\right) ^{ \\frac{1}{1600} } &= C \\\\ 2^{ -\\frac{1}{1600}} &= C \\end{align*}\n\nand thus your model is \\displaystyle \\begin{align*} R(t) = \\left( 2^{-\\frac{1}{1600}} \\right) ^t \\, R_0 = 2^{-\\frac{t}{1600}}\\,R_0 \\end{align*}.\n\nNow if \\displaystyle \\begin{align*} t = 200 \\end{align*}, that gives\n\n\\displaystyle \\begin{align*} R(200) &= 2^{-\\frac{200}{1600}}\\,R_0 \\\\ &= 2^{-\\frac{1}{8}}\\,R_0 \\\\ &\\approx 0.917 \\, R_0 \\end{align*}\n\nSo what percentage of the original amount do you have?", "date": "2020-11-30 13:32:14", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/d-e-word-problem.9047/", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 1667.1023268420893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806484125338, "lm_q2_score": 0.8705972751232808, "lm_q1q2_score": 0.8536908405465717}}
{"url": "https://mathhelpboards.com/threads/probability.5953/", "text": "# Probability.\n\n#### M R\n\n##### Active member\nA spinner has three possible outcomes which occur with probabilities a, b and c where a+b+c=1.\n\nWhat is the expected number of spins required until all three outcomes are seen?\n\nThere's an easy way and a harder way to do this. Guess which I did first.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nA spinner has three possible outcomes which occur with probabilities a, b and c where a+b+c=1.\n\nWhat is the expected number of spins required until all three outcomes are seen?\n\nThere's an easy way and a harder way to do this. Guess which I did first.\nThis is a variation on the coupon collector's problem. In the case when all the probabilities are equal ($a=b=c=1/3$), the expected number of spins is $3\\bigl(1 +\\frac12 + \\frac13) = \\frac{11}2.$\n\nIn the general case, I certainly don't see an easy way to approach the problem, and I don't get an easy-looking formula for the answer.\n\nWrite $A$, $B$, $C$ for the outcomes with probabilities $a$, $b$, $c$ respectively. If the first spin gives an $A$, then the expected number of spins until a $B$ or $C$ occurs is $\\dfrac1{b+c}$. The probability that this outcome is a $B$ is $\\dfrac b{b+c}$, in which case the expected number of further spins until a $C$ turns up is $1/c.$ And the probability that a $C$ occurs before a $B$ is $\\dfrac c{b+c}$, in which case the expected number of further spins until a $B$ turns up is $1/b.$ Therefore the total expected number of spins for all three outcomes to occur (given that the $A$ appears first) is $$1 + \\frac1{b+c}\\Bigl(\\frac b{b+c}\\,\\frac1c + \\frac c{b+c}\\,\\frac1b\\Bigr) = 1 + \\frac{b^2+c^2}{(b+c)^2bc} = \\frac1{bc} - \\frac2{(1-a)^2}$$ (in the last step, I have written the $b^2+c^2$ in the numerator as $(b+c)^2 - 2bc$, and in the denominator $b+c = 1-a$).\n\nMultiply that by $a$, which is the probability of the $A$ occurring first, add two similar terms for the probabilities of $B$ or $C$ occurring first, and you get the answer for the expected number of spins as $$1 + \\frac{a^2+b^2+c^2}{abc} - 2\\biggl(\\frac a{(1-a)^2} + \\frac b{(1-b)^2} + \\frac c{(1-c)^2}\\biggl).$$\n\nThat looks messy, not the sort of thing that you could find easily? But it does reduce to $11/2$ when $a=b=c=1/3$, which makes me think that it should be correct.\n\n#### M R\n\n##### Active member\nHi Opalg\n\nMaybe it wasn't easy but it was much easier than the other method which I will post if no one else does.\n\nYour formula does give the right answer for a=b=c but not for other possibilities.\n\nFor comparison purposes:\n\na=1/2, b=1/3, c=1/6 should give 73/10\n\nand\n\na=9/20, b=9/20, c=1/10 should give 353/33.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nYour formula does give the right answer for a=b=c but not for other possibilities.\n\nFor comparison purposes:\n\na=1/2, b=1/3, c=1/6 should give 73/10\n\nand\n\na=9/20, b=9/20, c=1/10 should give 353/33.\nStupid stupid mistake! My method was correct but I left out a $+$ sign, converting a sum into a product. The expression $$1 + \\frac1{b+c}\\Bigl(\\frac b{b+c}\\,\\frac1c + \\frac c{b+c}\\,\\frac1b\\Bigr)$$\n(for the expected number of spins for all three outcomes to occur, given that the $A$ appears first) should have been $$1 + \\frac1{b+c} + \\Bigl(\\frac b{b+c}\\,\\frac1c + \\frac c{b+c}\\,\\frac1b\\Bigr) = 1 + \\frac{bc + b^2+c^2}{(b+c)bc} = 1 + \\frac{(b+c)^2 - bc}{(b+c)bc} = 1 + \\frac{b+c}{bc} - \\frac1{b+c}.$$ The answer for the total expected number of spins then comes out as $$1 + \\frac{a(b+c)}{bc} + \\frac{b(c+a)}{ca} + \\frac{c(a+b)}{ab} - \\frac a{b+c} - \\frac b{c+a} - \\frac c{a+b}.$$ That gives values agreeing with your results for a=1/2, b=1/3, c=1/6 and for a=9/20, b=9/20, c=1/10.\n\n#### M R\n\n##### Active member\nThe history of this problem (for me personally):\n\nOn another forum someone posted a 'hard probability question'. It was the a=b=9/20, c=1/10 case. Looking back at that forum I see that I managed to get an answer six days later.\n\nMonths later I saw a post on MHF which taught me a better approach so I went back and did it again, the easy way. Unfortunately I can't view MHF to find out who my teacher was.\n\nI think it's essentially the same as Opalg's method but here's the 'easy' method as I did it:\n\nEvents A, B and C have probabilities a, b and c.\n\nLet the expected waiting time until A occurs be $$E(W_A)$$,\n\nand the expected waiting time until A and B occur be $$E(W_{AB})$$ and so on.\n\n$$E(W_{AB})=c(1+E(W_{AB}))+a(1+E(W_B))+b(1+E(W_A))$$\n\n$$E(W_{AB})=c(1+E(W_{AB}))+a(1+1/b)+b(1+1/a)$$\n\n$$E(W_{AB})=1 +cE(W_{AB})+a/b+b/a$$\n\n$$E(W_{AB})=\\frac{1+a/b+b/a}{1-c}$$\n\nSimilarly\n\n$$E(W_{AC})=\\frac{1+a/c+c/a}{1-b}$$\n\nand\n\n$$E(W_{BC})=\\frac{1+b/c+c/b}{1-a}$$\n\nThen $$E(W_{ABC})=a(1+E(W_{BC}))+b(1+E(W_{AC}))+c(1+E(W_{AB}))$$\n\nand the method I used at first:\n\nBy thinking about how the sequence ends I knew I wanted all the ways to get As and Bs ending with C etc.\n\nABC\nBAC\n\nAABC\nABAC\nBAAC\n\nABBC\nBABC\nBBAC\n\netc.\n\nThis led me to the sum\n\n$$\\displaystyle E(W)=\\sum_{n=2}^\\infty (n+1) \\sum_{r=1}^{n-1} \\binom{n}{r}(a^rb^{n-r}c+a^rc^{n-r}b+b^rc^{n-r}a)$$\n\n$$\\displaystyle =\\sum_{n=2}^\\infty (n+1) [ c((a+b)^n-a^n-b^n) +b((a+c)^n-a^n-c^n)+a((b+c)^n-b^n-c^n)]$$\n\nThis sum involves a number of geometric progressions and a number of sums of another type.\n\nThe other type is of the form $$\\displaystyle S=\\sum_{n=2}^\\infty n x^n$$.\n\nMultiplying by $$\\displaystyle x$$ gives $$\\displaystyle Sx=\\sum_{n=2}^\\infty n x^{n+1}$$ and so $$\\displaystyle S-Sx = 2x^2+x^3+x^4+...$$\n\n$$\\displaystyle S(1-x)=x^2 + \\frac{x^2}{1-x}$$ and $$\\displaystyle S=\\frac{x^2}{1-x}+\\frac{x^2}{(1-x)^2}$$\n\nApplying this formula, together with the formula for the sum of a GP we get\n\n$$\\displaystyle E(W)=$$\n\n$$\\displaystyle (a+b)^2-\\frac{ca^2}{1-a}-\\frac{cb^2}{1-b}$$\n\n$$\\displaystyle + (a+c)^2-\\frac{ba^2}{1-a}-\\frac{bc^2}{1-c}$$\n\n$$\\displaystyle + (b+c)^2-\\frac{ab^2}{1-b}-\\frac{ac^2}{1-c}$$\n\n$$\\displaystyle +(a+b)^2(1+1/c)-c\\left(\\frac{a^2}{1-a}+\\frac{a^2}{(1-a)^2}+\\frac{b^2}{1-b}+\\frac{b^2}{(1-b)^2}\\right)$$\n\n$$\\displaystyle +(a+c)^2(1+1/b)-b\\left(\\frac{a^2}{1-a}+\\frac{a^2}{(1-a)^2}+\\frac{c^2}{1-c}+\\frac{c^2}{(1-c)^2}\\right)$$\n\n$$\\displaystyle +(b+c)^2(1+1/a)-a\\left(\\frac{b^2}{1-b}+\\frac{b^2}{(1-b)^2}+\\frac{c^2}{1-c}+\\frac{c^2}{(1-c)^2}\\right)$$\n\nWhat a mess!\n\nThis could be simplified using $$\\displaystyle a+b+c=1$$ but having checked it against the first method I'm happy that it's correct and I'm not interested in doing the simplification.", "date": "2021-01-25 14:01:17", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/probability.5953/", "openwebmath_score": 0.9077078700065613, "openwebmath_perplexity": 431.4964143389514, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692284751636, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8536843437017235}}
{"url": "https://math.stackexchange.com/questions/803954/surely-youre-joking-mr-feynman-int-0-infty-frac-sin2xx21x2-dx/803957", "text": "# Surely You're Joking, Mr. Feynman! $\\int_0^\\infty\\frac{\\sin^2x}{x^2(1+x^2)}\\,dx$ [duplicate]\n\nProve the following $$$$\\int_0^\\infty\\frac{\\sin^2x}{x^2(1+x^2)}\\,dx=\\frac{\\pi}{4}+\\frac{\\pi}{4e^2}$$$$\n\nI would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>\u203f\u25e0)\u270c\n\nOriginal question:\n\nAnd of course, for the sadist with a background in differential equations, I invite you to try your luck with the last integral of the group.\n\n$$$$\\int_0^\\infty\\frac{\\sin^2x}{x^2(1+x^2)}\\,dx$$$$\n\nSource: Integration: The Feynman Way\n\n\u2022 \"The Feynman way\" is just a fancy name for the good old differentiation under the integral? \u2013\u00a0Daniel Fischer May 21 '14 at 11:30\n\u2022 @DanielFischer I know that Sir but I prefer naming \"The Feynman Way\" (\u00f4\u203f\u00f4) \u2013\u00a0Anastasiya-Romanova \u79c0 May 21 '14 at 11:33\n\u2022 @DanielFischer: there seems to be a tendency on this site to label any differentiation of an artificially introduced parameter under the integral sign as \"Feynman's <method, way, etc.>.\" I doubt Feynman invented it, but I see that loads of people read \"Surely You're Joking...\". \u2013\u00a0Ron Gordon May 21 '14 at 12:20\n\u2022 @RonGordon Indeed. It's a good read, by the way. Now, will you have lemon or milk in your tea? \u2013\u00a0Daniel Fischer May 21 '14 at 12:23\n\u2022 @V-Moy: Hey, I never said it was wrong; I just think it is a little funny. BTW it would not be the first time that a method or something else is named after its popularizer rather than its inventor. \u2013\u00a0Ron Gordon May 21 '14 at 12:38\n\nThis integral is readily evaluated using Parseval's theorem for Fourier transforms. (I am certain that Feynman had this theorem in his tool belt.) Recall that, for transform pairs $f(x)$ and $F(k)$, and $g(x)$ and $G(k)$, the theorem states that\n\n$$\\int_{-\\infty}^{\\infty} dx \\, f(x) g^*(x) = \\frac1{2 \\pi} \\int_{-\\infty}^{\\infty} dk \\, F(k) G^*(k)$$\n\nIn this case, $f(x) = \\frac{\\sin^2{x}}{x^2}$ and $g(x) = 1/(1+x^2)$. Then $F(k) = \\pi (1-|k|/2) \\theta(2-|k|)$ and $G(k) = \\pi \\, e^{-|k|}$. ($\\theta$ is the Heaviside function, $1$ when its argument is positive, $0$ when negative.) Using the symmetry of the integrand, we may conclude that\n\n\\begin{align}\\int_0^{\\infty} dx \\frac{\\sin^2{x}}{x^2 (1+x^2)} &= \\frac{\\pi}{2} \\int_0^{2} dk \\, \\left ( 1-\\frac{k}{2} \\right ) e^{-k} \\\\ &= \\frac{\\pi}{2} \\left (1-\\frac1{e^2} \\right ) - \\frac{\\pi}{4} \\int_0^{2} dk \\, k \\, e^{-k} \\\\ &= \\frac{\\pi}{2} \\left (1-\\frac1{e^2} \\right ) + \\frac{\\pi}{2 e^2} - \\frac{\\pi}{4} \\left (1-\\frac1{e^2} \\right )\\\\ &= \\frac{\\pi}{4} \\left (1+\\frac1{e^2} \\right )\\end{align}\n\n\u2022 @V-Moy: Yeah, I'm fast like that. \u2013\u00a0Ron Gordon May 21 '14 at 12:38\n\u2022 Surely You're Joking, Mr. Gordon! \uff08\u2010\uff3e\u25bd\uff3e\u2010\uff09 \u2013\u00a0Anastasiya-Romanova \u79c0 May 21 '14 at 12:41\n\u2022 @RonGordon Why did you answer the same question again since you knew it was a duplicate? You have already answered this question for me : math.stackexchange.com/questions/691798/\u2026. \u2013\u00a0Jeff Faraci May 22 '14 at 14:15\n\u2022 @Integrals: because I answered it three months ago and at my age, dementia sets in. I admit that sometimes I just answer the question without looking back through my vast database of answers. So, no, I did not know it was a duplicate. \u2013\u00a0Ron Gordon May 22 '14 at 14:21\n\u2022 @Integrals: I would have if I had known. But so what - the worst thing in the world is that duplicate correct (and consistent) answers exist in the site? Vote to close as a duplicate if this bothers you. \u2013\u00a0Ron Gordon May 22 '14 at 14:27\n\nHere is my attempt: \\begin{align} \\int_0^\\infty\\frac{\\sin^2x}{x^2(1+x^2)}dx&=\\int_0^\\infty\\left[\\frac{\\sin^2x}{x^2}-\\frac{\\sin^2x}{1+x^2}\\right]dx\\\\ &=\\int_0^\\infty\\frac{\\sin^2x}{x^2}dx-\\frac{1}{2}\\int_0^\\infty\\frac{1-\\cos2x}{1+x^2}dx\\\\ &=\\frac{\\pi}{2}-\\frac{1}{2}\\int_0^\\infty\\frac{1}{1+x^2}dx+\\frac{1}{2}\\int_0^\\infty\\frac{\\cos2x}{1+x^2}dx\\\\ &=\\frac{\\pi}{2}-\\frac{1}{2}\\frac{\\pi}{2}+\\frac{1}{2}\\frac{\\pi}{2e^2}\\\\ &=\\frac{\\pi}{4}+\\frac{\\pi}{4e^2} \\end{align} where I use these links: $\\displaystyle\\int_0^\\infty\\frac{\\sin^2x}{x^2}dx$ and $\\displaystyle\\int_0^\\infty\\frac{\\cos2x}{1+x^2}dx$ to help me out.\n\nUnfortunately, this is not the Feynman way but I still love this method.\n\n\u2022 Thanks for fixing my answer @MycrofD. I'm too excited so I forgot about my English grammar. \u2265\u00d6\u203f\u00d6\u2264 \u2013\u00a0Anastasiya-Romanova \u79c0 May 21 '14 at 11:30\n\u2022 Next time put your attempt in the question. Besides, you got it right (>\u203f\u25e0)\u270c. \u2013\u00a0Shahar May 21 '14 at 11:53\n\u2022 @Shahar Thanks, +1 for your emo. \u2267\u25e0\u25e1\u25e0\u2266\u270c \u2013\u00a0Anastasiya-Romanova \u79c0 May 21 '14 at 12:09\n\u2022 @V-Moy You seem to be an expert in emo's and integrals. \u270c \u2013\u00a0Sawarnik May 21 '14 at 17:43\n\u2022 @Sawarnik You might say I'm expert in emo but not in integral, not yet (\u273f\u25e0\u203f\u25e0) \u2013\u00a0Anastasiya-Romanova \u79c0 May 22 '14 at 8:17\n\nConsider $$I(a)=\\int_0^\\infty\\frac{\\sin^2(ax)}{x^2(x^2+1)}dx$$ Differentiate it twice. Since $$\\int_0^\\infty\\frac{\\cos(kx)}{x^2+1}dx=\\frac{\\pi}{2e^k}$$ for $k>0$ we get $I''(a)=\\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get $$I(a)=\\frac{\\pi}{4}(-1+2a+e^{-2a})$$ Now it only remains to substitute for $a=1$.\n\n\u2022 Wait a second! Why never cross to mind to use Feynman method twice?? +1 Sir! \u2265\u25ba.\u25c4\u2264 \u2013\u00a0Anastasiya-Romanova \u79c0 May 21 '14 at 12:47\n\u2022 @V-Moy Because you're not thinking like Feynman! Suppose the Feynman method just transforms your complicated integral into another complicated integral. Well, what tricks do you know for handling complicated integrals? That's right, the Feynman method! :) \u2013\u00a0David H May 21 '14 at 13:59\n\u2022 Thanks for your advice Mr. @DavidH. I'll try to think like Feynman. I have a lot of time though. \u1559(^\u25bd^)\u1557 \u2013\u00a0Anastasiya-Romanova \u79c0 May 21 '14 at 16:17\n\u2022 Why did you post this solution again? You knew this was a duplicate, this is a great solution but still. -1. math.stackexchange.com/questions/691798/\u2026 \u2013\u00a0Jeff Faraci May 22 '14 at 14:14\n\u2022 @Integrals because it is not forbidden, and this answer perfectly fits to the question even more than the in the link you gave above. \u2013\u00a0Norbert May 22 '14 at 15:47", "date": "2020-10-25 22:30:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/803954/surely-youre-joking-mr-feynman-int-0-infty-frac-sin2xx21x2-dx/803957", "openwebmath_score": 0.9616157412528992, "openwebmath_perplexity": 1866.8382785663596, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692277960745, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8536843399044485}}
{"url": "https://math.stackexchange.com/questions/3678612/is-the-sequence-x-n-dfrac1-sqrtn-left1-dfrac1-sqrt2-dfrac1-s", "text": "# Is the sequence $x_n=\\dfrac{1}{\\sqrt{n}}\\left(1+\\dfrac{1}{\\sqrt{2}}+\\dfrac{1}{\\sqrt{3}}+\\ldots+\\dfrac{1}{\\sqrt{n}}\\right)$ monotone?\n\nObserve that $$x_1=1$$ and $$x_2=\\dfrac{1}{\\sqrt{2}}\\left(1+\\dfrac{1}{\\sqrt{2}}\\right)>\\dfrac{1}{\\sqrt{2}}\\left(\\dfrac{1}{\\sqrt{2}}+\\dfrac{1}{\\sqrt{2}}\\right)=1$$.\n\nThus, $$x_2>x_1$$. In general, we also have $$x_n=\\dfrac{1}{\\sqrt{n}}\\left(1+\\dfrac{1}{\\sqrt{2}}+\\dfrac{1}{\\sqrt{3}}+\\ldots+\\dfrac{1}{\\sqrt{n}}\\right)>\\dfrac{1}{\\sqrt{n}}\\left(\\dfrac{1}{\\sqrt{n}}+\\dfrac{1}{\\sqrt{n}}+\\dfrac{1}{\\sqrt{n}}+\\ldots+\\dfrac{1}{\\sqrt{n}}\\right)=1$$.\n\nThus, $$x_n\\geq 1$$ for all $$n\\in \\mathbb{N}$$. Also, we have,\n\n$$x_{n+1}=\\dfrac{1}{\\sqrt{n+1}}\\left(\\sqrt{n}x_n+\\dfrac{1}{\\sqrt{n+1}}\\right)=\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}x_n+\\dfrac{1}{n+1}$$.\n\nIs it true that $$x_{n+1}>x_n$$?\n\nEdit : Thanks to the solution provided by a co-user The73SuperBug. Proving $$x_{n+1}-x_n>0$$ is equivalent to proving $$x_n<1+\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}$$. This is explained below :\n\n\\begin{aligned} &x_{n+1}-x_n>0\\\\ \\Leftrightarrow & \\dfrac{\\sqrt{n}}{\\sqrt{n+1}}x_n+\\dfrac{1}{n+1}-x_n>0\\\\ \\Leftrightarrow & \\left(1-\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}\\right)x_n-\\dfrac{1}{n+1}<0\\\\ \\Leftrightarrow & x_n<\\dfrac{1}{(n+1)\\left(1-\\dfrac{\\sqrt{n}} {\\sqrt{n+1}}\\right)}\\\\ \\Leftrightarrow & x_n<1+\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}. \\end{aligned}\n\n\u2022 It's basically a Riemann sum for a convex function. The (good) accepted answer focuses on the specific function you have, but I think a generalization would not harm, so I've added an answer based on this alternative idea. May 17, 2020 at 12:17\n\nThe sequence is indeed increasing. Using what you have left off we need to prove: $$x_{n+1} - x_n > 0\\iff ...x_n < 1+\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}$$. We prove this by induction on $$n \\ge 1$$. Clearly $$x_1 = 1 < 1+ \\sqrt{\\frac{1}{2}}$$. Assume $$x_n < 1+\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}$$, we show: $$x_{n+1} < 1+\\dfrac{\\sqrt{n+1}}{\\sqrt{n+2}}$$. Using the recursive formula you had above: $$x_{n+1} = \\dfrac{\\sqrt{n}}{\\sqrt{n+1}}x_n+\\dfrac{1}{n+1}< \\dfrac{\\sqrt{n}}{\\sqrt{n+1}}\\left(1+\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}\\right)+\\dfrac{1}{n+1}= 1+\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}< 1+\\dfrac{\\sqrt{n+1}}{\\sqrt{n+2}}$$ which is clear because $$n(n+2) < (n+1)^2$$. Thus by induction $$x_n < 1 +\\dfrac{\\sqrt{n}}{\\sqrt{n+1}}$$ and in turn implies $$x_{n+1} > x_n, \\forall n \\ge 1$$. Thus the sequence is increasing.\nA generalization I couldn't pass by. Let $$\\color{blue}{S_n=\\frac1n\\sum_{k=1}^{n-1}f\\big(\\frac{k}{n}\\big)}$$ where $$f:(0,1)\\to\\mathbb{R}$$ is strictly convex: $$f\\big((1-t)a+tb\\big)<(1-t)f(a)+tf(b)\\quad\\impliedby\\quad a If we put $$a=k/(n+1),b=(k+1)/(n+1),t=k/n$$ for $$0 here, we obtain $$f\\Big(\\frac{k}{n}\\Big)<\\Big(1-\\frac{k}{n}\\Big)f\\Big(\\frac{k}{n+1}\\Big)+\\frac{k}{n}f\\Big(\\frac{k+1}{n+1}\\Big),$$ which, after summing over $$k$$ (to have \"$$<$$\" still, we must assume $$n>1$$), gives $$\\sum_{k=1}^{n-1}f\\Big(\\frac{k}{n}\\Big)<\\sum_{k=1}^{\\color{red}{n}}\\Big(1-\\frac{k}{n}\\Big)f\\Big(\\frac{k}{n+1}\\Big)+\\sum_{k=\\color{red}{1}}^{n}\\frac{k-1}{n}f\\Big(\\frac{k}{n+1}\\Big)=\\frac{n-1}{n}\\sum_{k=1}^{n}f\\Big(\\frac{k}{n+1}\\Big),$$ i.e. $$\\color{blue}{n^2 S_n<(n^2-1)S_{n+1}}$$. Returning to the question, if we put $$f(x)=1/\\sqrt{x}$$, we get $$x_n=S_n+1/n$$ and $$n^2 x_n<(n^2-1)x_{n+1}+1$$; since $$x_{n+1}>1$$, the latter implies the needed $$x_n.", "date": "2022-08-15 01:12:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3678612/is-the-sequence-x-n-dfrac1-sqrtn-left1-dfrac1-sqrt2-dfrac1-s", "openwebmath_score": 0.9968483448028564, "openwebmath_perplexity": 347.6546216543871, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9908743636887527, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8536761215408448}}
{"url": "http://mathhelpforum.com/discrete-math/77647-midterm-review.html", "text": "# Math Help - Midterm review\n\n1. ## Midterm review\n\nAgh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.\n\na.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.\n\nb.) use part (a) to show that 10! = -1(mod 11)\n\nAny help would be appreciated\n\n2. Originally Posted by Niotsueq\nAgh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.\n\na.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.\n\nb.) use part (a) to show that 10! = -1(mod 11)\nFor a), you can find out by trial and error that, for example, $2\\times6 = 12\\equiv1\\!\\!\\!\\pmod{11}$. Therefore 2 and 6 are inverses of each other. So also are 7 and 8, because $7\\times8 = 56\\equiv1\\!\\!\\!\\pmod{11}$.\n\nFor b), $10! = 1\\times 2\\times3\\times\\cdots\\times10 = 1\\times(2\\times6)\\times\\cdots\\times(7\\times8)\\time s\\cdots\\times10$ (pair the numbers off as in part a)).\n\n3. Hello, Niotsueq!\n\na) Show that the positive integers less than 11, except 1 and 10,\ncan be split into pairs of integers such that each pair consists of integers\nthat are inverses of each other modulo 11.\n. . . . . $\\begin{array}{ccc}2\\cdot6 &\\equiv & 1\\text{ (mod 11)} \\\\3\\cdot4 &\\equiv & 1\\text{ (mod 11)} \\\\ 5\\cdot9 &\\equiv & 1\\text{ (mod 11)} \\\\ 7\\cdot8 & \\equiv& 1 \\text{ (mod 11)} \\end{array}$\n\nb) use part (a) to show that 10! = -1 (mod 11)\n\n$10! \\;\\equiv\\;1\\cdot2\\cdot3\\cdot4\\cdot5\\cdot6\\cdot7\\cd ot8\\cdot9\\cdot10\\,\\text{ (mod 11)}$\n\n. . . $\\equiv \\;1\\cdot(2\\cdot6)\\cdot(3\\cdot4)\\cdot(5\\cdot9)\\cdot (7\\cdot8)\\cdot10\\,\\text{ (mod 11)}$\n\n. . . $\\equiv\\;1\\cdot 1\\cdot 1\\cdot 1\\cdot1\\cdot10\\,\\text{ (mod 11)}$\n\n. . . $\\equiv\\;10\\,\\text{ (mod 11)}$\n\n. . . $\\equiv\\;\\text{-}1\\,\\text{ (mod 11)}$\n\n4. Thanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:\n\n2*6 = 1 (mod 11)\n\nHow does this say that 2 and 6 are inverses? I think I may be confused on the definition.\n\nThanks alot to both of you.\n\n5. Originally Posted by Niotsueq\nThanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:\n\n2*6 = 1 (mod 11)\n\nHow does this say that 2 and 6 are inverses? I think I may be confused on the definition.\n\nThanks alot to both of you.\n\n$2\\cdot 6 \\equiv 1 \\mod 11$\n\nthis shows that they are inverses of each other.\n\nremember what an inverse is in regular multiplication?\n\n$3 \\cdot \\frac{1}{3}=1$\n\ndo you see how the original means they are inverses now?\n\n6. Originally Posted by GaloisTheory1\n$2\\cdot 6 \\equiv 1 \\mod 11$\n\nthis shows that they are inverses of each other.\n\nremember what an inverse is in regular multiplication?\n\n$3 \\cdot \\frac{1}{3}=1$\n\ndo you see how the original means they are inverses now?\n\nSo two numbers are inverses of one another mod(n) if their product is congruent to one mod(n)?", "date": "2014-08-23 05:42:15", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/77647-midterm-review.html", "openwebmath_score": 0.8525896668434143, "openwebmath_perplexity": 315.83318513282575, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9825575152637948, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.853672288466941}}
{"url": "https://math.stackexchange.com/questions/518739/prove-that-the-sequence-a-n-defined-by-a-0-1-a-n1-1-frac-1a-n/518791", "text": "# Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \\frac 1{a_n}$ is convergent in $\\mathbb{R}$\n\nI will post the exercise below:\n\nProve that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \\frac 1{a_n}$ for $n \\in \\mathbb N$ is convergent in $\\mathbb R$ with the Euclidean metric, and determine afterwards is limit. Can you intepret the limit geometrically (hint: Golden ratio)?\n\nSo I need to prove that the sequence is convergent in $\\mathbb{R}$ with the Euclidean metric, and how do I prove that? The limit must be $1$, but how to interpret it geometrically?\n\n\u2022 All I am willing to say is that \"hint: Golden ratio\" is a very strong hint. Oct 8, 2013 at 9:43\n\u2022 The limit is not 1. @Duronman, the series is not monotonically increasing either. Oct 8, 2013 at 9:47\n\u2022 @AdrianRatnapala true! I wrote too fast, sorry, I thought of $a_n + \\frac{1}{a_n}$ because of the $1$. I will delete the comment to avoid confusion. Oct 8, 2013 at 9:54\n\nHint: If the limit $L$ exists, it must satisfy $L = 1 + \\frac{1}{L}$, and so it cannot be 1. The solutions are the roots of the equation $L^2 - L - 1 = 0$, and so $L \\in \\{\\frac{1+\\sqrt{5}}{2}, \\frac{1-\\sqrt{5}}{2} \\}$. That's where the golden ratio comes into play. Note also that the limit cannot be $\\frac{1-\\sqrt{5}}{2}$ since $a_n > 0$ for all $n$.\n\nLet $\\phi$ be the golden ratio, which is the only positive real such that $\\phi = 1 + 1/\\phi$. Now consider the sequence $\\epsilon_n = a_n - \\phi$. We seek to prove the invariant:\n\n$$|\\epsilon_n/\\epsilon_0| \\le \\phi^{-n}.$$\n\nFor $n=0$, this is clear by inspection. With some algebra we see that:\n\n$$\\epsilon_{n+1} = 1 + {1\\over a_n} - \\phi = -{\\epsilon_n\\over a_n\\phi}.$$\n\nNow our invariant already implies that $a_n = \\phi + \\epsilon_n \\ge \\phi - \\phi^{-n}\\epsilon_0 \\ge 1.$ Therefore\n\n$$|\\epsilon_{n+1}| \\le \\left|{\\epsilon_n\\over \\phi}\\right| \\le \\phi^{-(n+1)}.$$\n\nSo our invariant remains after $n \\rightarrow n+1$. And therefore for all nonnegative integers $n$ by induction. But since $\\phi > 1$, that invariant means that $\\epsilon_n$ approaches zero or $a_n$ approaches $\\phi$ as $n$ goes to infinity.\n\nWriting out the first few terms will yield a very interesting pattern:\n\n$a_0 = \\frac{1}{1}, a_1 =\\frac{2}{1}, a_2=\\frac{3}{2}, a_3=\\frac{5}{3} ...$\n\nDo u see it now. Every term is a ratio of fibonacci nos. i.e. if $T_n$ is the nth fibonacci number starting from 1, then $a_n = \\frac{T_{n+2}}{T_{n+1}}$. The limit of this is indeed the golden ratio. Try solving the difference equation to see it in case you are unaware of this result.\n\nAs for the proof that the observation is correct and not a coincidence, use induction. We verified the base cases above, so assume $$a_n = \\frac{T_{n+2}}{T_{n+1}}$$\n\nNow $$a_{n+1} = 1+ \\frac{1}{a_n} = 1 + \\frac{T_{n+1}}{T_{n+2}}= \\frac{T_{n+2} + T_{n+1}}{T_{n+2}} = \\frac{T_{n+3}}{T_{n+2}}$$\n\nHence proved. $\\blacksquare$\n\n\u2022 But you still need to prove that the ratios of consecutive Fibonnaci numbers actually converge to anything at all. Oct 10, 2013 at 8:44\n\u2022 I told OP to solve the difference equation for $a_n = a_{n-1} + a_{n-2}$. From that you can evaluate the limit. Oct 10, 2013 at 8:48\n\u2022 I am missing some knowledge about difference equations. The easiest method to find that limit (if it exists) I know is to find the roots of the iteration rule translated into a polynomial. But I did not know the limit was guaranteed to exist. Oct 14, 2013 at 18:07\n\u2022 You are referring to the Z Transform method for solving difference equations. With it, you will get the solution for all n. Then you can proceed by my method. Oct 14, 2013 at 18:45\n\u2022 Adrian, another way is to get the general solution to the recurrence relation and then just take limits. There will invariably be a dominating term which determines the limit. Dec 22, 2013 at 3:30\n\nWe will consider decreasing function $f:(0, \\infty)\\rightarrow(0,\\infty), f(x)=1+\\frac{1}{x}$. Function $fof:(0, \\infty ) \\rightarrow(0, \\infty)$ is increasing and therefore subsequence $a_{2n},n\\geq0$ is increasing, and subsequence $a_{2n +1},n\\geq0$ is decreasing. Because $1\\leq a_{2n}<a_{2n +1}\\leq2$, it follows that both are converging subsequences with common limit the golden ratio, which is the limit sequence in question.", "date": "2022-07-06 14:20:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/518739/prove-that-the-sequence-a-n-defined-by-a-0-1-a-n1-1-frac-1a-n/518791", "openwebmath_score": 0.9110130667686462, "openwebmath_perplexity": 162.4357127236693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575183283513, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.853672286123661}}
{"url": "https://math.stackexchange.com/questions/1088313/how-many-different-tsuro-tiles-can-exist", "text": "# How many different Tsuro tiles can exist?\n\nThe boardgame Tsuro consists of tiles, which each have 8 entry points. Each tile connects each point to exactly one other point. The game manual claims every tile is unique. The game consists of 35 such tiles.\n\nHow many unique such tiles could possibly exist?\n\nMy reasoning:\n\n\u2022 You start with 8 free points. Choose any point, you now have 7 possibilities to connect.\n\u2022 You now have 6 free points. Choose any point, you now have 5 possibilities to connect.\n\u2022 You now have 4 free points. Choose any point, you now have 3 possibilities to connect.\n\u2022 You now have 2 free points. Connect them (no choice possible).\n\nThis would lead to 7*5*3 : 105 possibilities. But I wouldn't know how to eliminate \"doubles\" caused by rotating a tile. Should I divide by 4, since 4 rotations are possible? That would be 26 tiles... but the game itself contains 35 and they are unique.\n\nHow should I reason?\n\n\u2022 Well, here's why you can't just divide by $4$: Not all the rotations are unique - consider the tile in which each point is connected to the one directly opposite it on the other side - then this tile is rotationally symmetric. Another example in which not all rotations are unique is the bottom left tile in the picture you provided. \u2013\u00a0Peter Woolfitt Jan 2 '15 at 10:31\n\nYou can use Burnside's Lemma. I'll go through the calculation here, though you should read that page for this to make sense.\n\nThe set X is all 105 possibilities. The group G that acts on X is $\\langle r | r^4 = 1 \\rangle$, where r is a 90 degree rotation. Burnside's Lemma is applicable since each tile corresponds exactly to one orbit in X under G. The number of elements of X fixed by each element of G is\n\n\u2022 $1: 105$\n\u2022 $r: 5$\n\u2022 $r^2: 25$\n\u2022 $r^3: 5$\n\nSo the number of orbits (tiles) is $\\frac{1}{4}(105+5+25+5)=35$. The hardest part is computing the $r^2: 25$ entry; here's how I did it:\n\n\u2022 Case 1: No pair of antipodal points is connected. Choose any point: you have 6 legal possibilities to connect. Whatever you choose, $r^2$ fixes a second symmetrical connection, so you have 4 points left with 2 legal ways to connect them, for a total of 6*2=12 possibilities.\n\u2022 Case 2: 2 pairs of antipodal points are connected. There are $\\binom{4}{2}=6$ ways to pick the pairs and 2 ways to connect the reamining 4 points, for another 12 possibilities.\n\u2022 Case 3: All 4 antipodal pairs are connected. There is only 1 such possibility.\n\u2022 You'd probably also find this page interesting. \u2013\u00a0Benjamin Cosman Aug 28 '16 at 21:14\n\u2022 $<abc>$ looks like this: $<abc>$, and $\\langle abc\\rangle$ looks like this $\\langle abc\\rangle$. #FriendsDoNotLetFriendsUseTheWrongAnglyThings \u2013\u00a0Mariano Su\u00e1rez-\u00c1lvarez Aug 28 '16 at 21:19\n\u2022 Interesting approach, thank you! \u2013\u00a0Konerak Aug 29 '16 at 7:07\n\nOne can reduce further the $105$ possible tiles, down to the minimum possible of $35$ topologically unique tiles via $mod\\, 8$ arithmetic, $1+7=(8)=0$.\n\nModulo $8$ arithmetic (each tile has $8$ ports numbered $0123457$) allows $CW$ and $CCW 2$-dimensional rotations (can not flip a tile, as a flip would represent a $3D$ rotation). To keep this simple, there are no negative integers here, $0-1=7$ is being interpreted as $7+1=0$.\n\nA $45^{\\circ}$ rotation is represented by adding/subtracting 1 to/from the port #.\n\nRotations of $90^{\\circ}, 180^{\\circ}, 270^{\\circ}$ are represented by adding/subtracting $2, 4, 6$ respectively. I just added $2, 4, 6$ used only addition.\n\nRepresent a pair of two connected ports of a tile by a $2$-digit number. For example $15$ means port#1 is connected to port#5, i.e. not decimal fifteen.\n\nRepresent each tile via a quad set of number pairs, non repeating digits,for example $\\{01, 23, 45, 67\\}$.\n\nNote $\\{01, 14, 26, 57\\}$ is not a valid tile as digit \"1\" -port#1- appears twice.\n\nExample of $180^{\\circ}$ rotation:\n\nTile $\\{04, 12, 36, 57\\}$ is rotated $180^{\\circ}$ by adding (modulo 8) $+4$ to each of its digits.\n\n$$0+4=4$$\n\n$$4+4=(8)=0$$\n\n$$1+4=5$$\n\n$$2+4=6$$\n\n$$3+4=7$$\n\n$$6+4=(10)=2$$\n\n$$5+4=(9)=1$$\n\n$$7+4=(11)=3$$\n\nThus rotated tile becomes $\\{40, 56, 72, 13\\}$ and will order/describe/ its ports from low-to-high (as each numbers-pair is commutative, $40=04$ and $72=27$).\n\nTherefore this $180^{\\circ}$ rotated tile becomes $\\{04, 13, 27, 56\\}$.\n\nRotate all $105$ tiles by $0^{\\circ}, 90^{\\circ}, 180^{\\circ}, 270d^{\\circ}$, (find & write all quad sets).\n\nUse Excel, reorder the port# numbers from low-to-high, and eliminate all duplicate tiles.\n\nEnd result = $35$ unique quad sets representing $35$ unique tiles.\n\nMay follow my analysis summary at:\n\nThere are 5 ways to make a tile that is invariant under $90^\\circ$ rotations: Just connect one pint to any other point except its rotational predecessor or successor, then the rest is fixed. This would correct your count to $\\frac{105-5}4+5=29$, which is more, but still too few. You should also count, how many tiles are possible with a $180^\\circ$ symmetry. Of course, the $90^\\circ$ symmetric tiles are among them, but there are also some \"new\" such tiles. If there are $k$ such tiles, then the total count gets corrected to $\\frac{105-5-k}{4}+5+\\frac k2$, so you can make an educated guess what $k$ is ;)", "date": "2019-06-26 22:19:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1088313/how-many-different-tsuro-tiles-can-exist", "openwebmath_score": 0.5815846920013428, "openwebmath_perplexity": 849.0195871336006, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575157745542, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8536722822362375}}
{"url": "https://math.stackexchange.com/questions/614468/an-extrasensory-perception-strategy/614546", "text": "# An extrasensory perception strategy :-)\n\nInspired by classical Joseph Banks Rhine experiments demonstrating an extrasensory perception (see, for instance, the beginning of the respective chapter of Jeffrey Mishlove book \u201cThe Roots of Consciousness\u201d), I consider the following experiment. A deck of cards is given to a magician John. Then John consecutively takes the cards from the deck, trying to guess suit of the taken card. He looks at the card after the guess for a feedback. The magician wishes to maximize the expected number $E$ of right guesses. For this purpose he devised the following Strategy: at each turn to guess a suit which has a maximal number of card left in the deck. As an easy exercise we can prove that for any sequence of cards in the deck Strategy ensures at least $n$ right guesses, where $n$ is the maximal number of cards with one suit in the deck. But we can consider a more interesting and complicated problem to calculate the expectation $E$ for Strategy (here we are assuming that the deck is so well shuffled such that all sequences of cards have equal probability). By the way, I conjecture that Strategy is the best for maximizing the expectation $E$, that is any other strategy yields not greater value of $E$. Now I wish to evaluate the expectation $E$ for Strategy. For the simplicity we shall consider only a case when there are only two suits ($m\\ge 0$ cards of the first suit and $n\\ge m$ cards of the second suit). Then $E(0,n)=n$ for each $n$ and we have the following recurrence $$E(m,n)=\\frac{n}{m+n}(E(m,n-1)+1)+ \\frac{m}{m+n}E(m-1,n)$$\n\nfor each $n\\ge m\\ge 1$.\n\nThe rest is true provided I did not a stupid arithmetic mistake.\n\nI was interested mainly in asymptotics for the case $m=n$ and computer calculations suggested that $E(n,n)\\sim n+c\\sqrt{n}+o(\\sqrt{n})$ for $c\\approx 0.88\\dots$.\n\nEvaluating formulas for $E(m,n)$ for small values of $m\\le 6$, I conjectured that there is a general formula\n\n$$E(m,n)=n+m\\sum_{i=1}^m\\frac {c_{m,i}}{n+i}$$\n\nfor each $n\\ge m\\ge 1$, where $c_{m,i}$ are some integers satisfying the recurrence\n\n$$(m-i)c_{m,i}+ic_{m,i+1}=(m-1)c_{m-1,i}$$\n\nfor every $1\\le i\\le m-1$.\n\nHere are my values for $c_{m,i}$\n\ni\\m|  1   2   3   4   5   6\n---+------------------------\n1 |  1   2   4   8  16  32\n2 |     -1  -4 -12 -32 -80\n3 |          1   6  24  80\n4 |             -1  -8 -40\n5 |                  1  10\n6 |                     -1\n\n\nThen I discovered that for my data $c_{m,i}$ is divisible by $2^{m-i}$. After I did the division, I surprisingly obtained that $$c_{m,i}=(-1)^{i-1}2^{m-i}{m-1 \\choose i-1}.$$ I expect that I can easily prove this equality by induction.\n\nBut I did not stop at this point because I observed that now the general formula for $E(m,n)$ can be compressed to the form\n\n$$E(m,n)=n+m\\int_0^1 x^n(2-x)^{m-1} dx.$$\n\nAll of above sounds nice for me and I spent a good time investigating the problem, but I am a professional mathematician, although I am not a specialist in the domain of the above problem. Therefore I care about the following questions. Are the above results new, good and worthy to be published somewhere? What another related problems are worthy to be investigated?\n\nThanks and merry Holidays.\n\n\u2022 \u2013\u00a0Alex Ravsky Dec 21 '13 at 0:49\n\u2022 Your expression for $E(m,n)$ looks a lot like Kolmogorov forward equation. \u2013\u00a0Alex Dec 21 '13 at 1:21\n\u2022 Is your table of $c_{m,i}$ correct for $i=1$? The triangle and your expression for $c_{m,i}$ looks very like OEIS A013609 \u2013\u00a0Henry Dec 23 '13 at 22:15\n\u2022 @Henry Thanks, I corrected the table. \u2013\u00a0Alex Ravsky Mar 9 '14 at 15:10\n\u2022 Your integral expression for $E(m,n)$ is closely related to $\\displaystyle B_{1/2}(n+1,m) = \\int_{0}^{1/2} x^n (1-x)^{m+1} dx$, an incomplete Beta function. \u2013\u00a0Henry Mar 9 '14 at 16:58\n\nThis is to address your question about the asymptotics when $m=n$.\n\nWe only need to study the integral\n\n$$I(n) = \\int_0^1 x^n (2-x)^{n-1}\\,dx$$\n\nor, after we've made the change of variables $x = 1-y$,\n\n\\begin{align} I(n) &= \\int_0^1 (1-y)^n (1+y)^{n-1}\\,dy \\\\ &= \\int_0^1 (1+y)^{-1} \\exp\\left[n \\log\\left(1-y^2\\right)\\right]\\,dy. \\end{align}\n\nWe'll proceed using the Laplace method. The quantity $\\log(1-y^2)$ achieves its maximum at $y=0$, and near there we have\n\n$$\\log\\left(1-y^2\\right) \\sim -y^2.$$\n\nThis motivates us to make the change of variables $\\log(1-y^2) = -z^2$, so that\n\n$$I(n) = \\int_0^\\infty \\frac{z e^{-z^2}}{(1+\\sqrt{1-e^{-z^2}})\\sqrt{1-e^{-z^2}}} e^{-nz^2}\\,dz.$$\n\nNear $z=0$ we have\n\n$$\\frac{z e^{-z^2}}{(1 + \\sqrt{1-e^{-z^2}})\\sqrt{1-e^{-z^2}}} = 1 - z + \\frac{z^2}{4} + \\frac{z^4}{96} - \\frac{z^6}{384} + \\cdots,$$\n\nand integrating term-by-term we obtain the asymptotic series\n\n$$I(n) \\approx \\frac{\\sqrt{\\pi}}{2n^{1/2}} - \\frac{1}{2n} + \\frac{\\sqrt{\\pi}}{16n^{3/2}} + \\frac{\\sqrt{\\pi}}{256n^{5/2}} - \\frac{5 \\sqrt{\\pi}}{2048n^{7/2}} + \\cdots.$$\n\nThus\n\n$$E(n,n) \\approx n + \\frac{\\sqrt{\\pi}}{2} n^{1/2} - \\frac{1}{2} + \\frac{\\sqrt{\\pi}}{16} n^{-1/2} + \\frac{\\sqrt{\\pi}}{256} n^{-3/2} - \\frac{5 \\sqrt{\\pi}}{2048} n^{-5/2} + \\cdots.$$\n\n\u2022 Thanks, this is very good! :-D This correlates with my computer and analytical calculations. As I already wrote, my old computer calculations suggested that $E(n,n)\\sim n+c\\sqrt{n}+o(\\sqrt{n})$ for $c\\approx 0.88\\dots$. Now I see that $c\\approx \\sqrt{\\pi}/2$ with the precision up to two significant digits. I derived the integral formula only yesterday so I did not try to derive the asymptotics from it yet. Moreover, I did not know about Laplace method. \u2013\u00a0Alex Ravsky Dec 21 '13 at 6:55\n\nIf I understand right the game, and picturing the evolution as a path in a discrete grid -from $(0,0)$ to ($n,n$)- it's seen that each segment that goes towards the diagonal is a \"win\"; the other are misses, except for the ones that start from the diagonal itself, half of which are wins. Then, if the path of length $2n$ have $c$ diagonal-touchings (including the start, excluding the end), the total of wins is $n+c/2$.\n\nHence, the problem is converted to the (probably simpler and already studied) problem of computing the expected numbers of diagonal touchings on a lattice path - or, in a fair ballot counting problem, compute the expected number of ties (or lead changes).\n\nYou ask \"What (...) related problems are worthy (of investigation)?\"\n\nThere is the question of what the expected score is if we use that strategy (which is the best strategy) with a Zener pack of 25 cards, namely which contains 5 cards of each of 5 different shapes. I asked that question here: the answer is 8.65.\n\nThen there are mis\u00e8re problems where we try to mimimise the score. I have asked the mis\u00e8re version of the Zener card question here.", "date": "2019-06-17 23:32:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/614468/an-extrasensory-perception-strategy/614546", "openwebmath_score": 0.8994896411895752, "openwebmath_perplexity": 300.12702655169966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575173068325, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8536722752244402}}
{"url": "https://math.stackexchange.com/questions/804764/why-is-1-frac12-frac13-frac1n-approx-lnn-gamm", "text": "Why is $1 + \\frac{1}{2} + \\frac{1}{3} + \u2026 + \\frac{1}{n} \\approx \\ln(n) + \\gamma$?\n\nOn StackExchange, I read that the harmonic series up to $\\frac{1}{n}$ is approximately $\\ln(n) + \\gamma$, where $\\gamma$ is the Euler-Mascheroni constant, which is close to $0.5772$. When I researched the Euler-Mascheroni constant, I only found it defined in terms of the difference between the harmonic series and $\\ln(n)$.\n\nWhy is the series able to be approximated in this way, and what is the Euler-Mascheroni constant?\n\n\u2022 Draw rectangles representing the sum (i.e. having height $1/n$ on the interval $[n, n + 1]$), and the area representing the integral. In the limit, these are almost the same, and the error is $\\gamma$. \u2013\u00a0user61527 May 22 '14 at 0:16\n\u2022 It's pretty easy to show, using $$\\log n = \\int_1^n\\frac{dx}{x} = \\sum_{i=1}^{n-1}\\int_i^{i+1}\\frac{dx}{x}$$ that $$\\left(\\sum_{i=1}^n\\frac{1}{i}\\right)-\\log n$$ converges as $n\\to\\infty$. You can also easily prove that the limit is between $1/2$ and $1$. This limit, or some variant, is often the definition of the Euler-Mascheroni constant. \u2013\u00a0Thomas Andrews May 22 '14 at 0:21\n\u2022 @ThomasAndrews I am sure you have a typo there. \u2013\u00a0chubakueno May 22 '14 at 0:23\n\u2022 Fixed. @chubakueno \u2013\u00a0Thomas Andrews May 22 '14 at 0:24\n\u2022 I like showing $\\gamma$ is between one increasing sequence and one decreasing sequence that are also forced to come together; see math.stackexchange.com/questions/306371/\u2026 \u2013\u00a0Will Jagy May 22 '14 at 3:58\n\n$$\\log n = \\int_1^n \\frac{dx}{x} = \\sum_{i=1}^{n-1}\\int_{i}^{i+1}\\frac{dx}{x}$$\n\nSo:\n\n$$\\left(\\sum_{i=1}^n \\frac 1 i\\right)-\\log n = \\left(\\sum_{i=1}^{n-1}\\int_i^{i+1}\\left(\\frac 1i-\\frac1x\\right)dx\\right) + \\frac{1}{n}$$\n\nNow, for $x\\in[i,i+1]$, $0\\leq\\frac{1}i-\\frac1 x\\leq \\frac{1}{i(i+1)}.$\n\nSo these terms are positive and $\\sum_{i=1}^\\infty \\frac{1}{i(i+1)} = 1$. So as $n\\to\\infty$, this means:\n\n$$\\left(\\sum_{i=1}^n \\frac 1 i\\right)-\\log n$$ converges to a value less than $1$.\n\nIt's actually pretty easy to show, since $f(x)=1/x$ is concave, that:\n\n$$\\int_i^{i+1}\\left(\\frac 1i-\\frac1x\\right)dx>\\frac{1}{2i(i+1)}$$\n\nThis means that the limit is between $1/2$ and $1$.\n\nThis is often the definition of the Euler-Mascheroni constant.\n\n\u2022 Don't you need to show the series is increasing? Or have you shown that and I'm missing something. \u2013\u00a0Alex Zorn May 22 '14 at 0:49\n\u2022 If $0<a_i<b_i$ and $\\sum b_i$ converges, then $\\sum a_i$ converges. Basically, $\\sum_{i=1}^n a_i$ is increasing as $n\\to\\infty$ and is bounded above, therefore converges. \u2013\u00a0Thomas Andrews May 22 '14 at 1:14\n\nAs noted in the comments, the \"computational\" answer is that, comparing areas, we find\n\n$$\\int_1^{n+1}\\frac{dx}{x}<1+{1\\over 2}+\\cdots +{1\\over n}<1+\\int_1^n\\frac{dx}{x}$$ Evaluating the integrals gives $$\\ln(n+1)<1+{1\\over 2}+\\cdots +{1\\over n}<1+\\ln n$$ so we have the very rough approximation that, denoting $$H_n=1+{1\\over 2}+\\cdots +{1\\over n}$$ the sum is around $H_n\\approx \\ln n$.\n\nA more insightful perspective is to compare the nature of the functions $H_n$ (often called the \"harmonic sum\") and $\\ln x$. The former is discrete, and the latter continuous. In \"discrete calculus\" you often see $$\\Delta f(n)=f(n+1)-f(n)$$ which is the discrete analogue of the derivative. Note that the \"derivative\" of the harmonic sum is $$\\Delta H_n=H_{n+1}-H_n=\\left(1+\\cdots +{1\\over n+1}\\right)-\\left(1+\\cdots +{1\\over n}\\right)={1\\over n+1}$$\n\nSimilarly, the derivative of the logarithm is $$\\textrm{D}\\ln x={1\\over x}$$\n\n(The analogy can be drawn further, depending on how much discrete maths you know) There is a sense then, in which the two functions are \"companions\" of each other - they live in different universes, but within those universes they share many of the same properties. This is a recurring theme in much of mathematics (you may be interested in researching the word \"isomorphic\", although it does not completely apply here).\n\n\u2022 If you average $\\ln (n + 1)$ and $\\ln n + 1$ you get $\\ln n + 0.5$ as an estimate for $H_n$ \u2013\u00a0vonbrand May 22 '14 at 4:36", "date": "2021-04-23 05:49:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/804764/why-is-1-frac12-frac13-frac1n-approx-lnn-gamm", "openwebmath_score": 0.9452646970748901, "openwebmath_perplexity": 262.90362399705697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575111777184, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.8536722715679183}}
{"url": "https://math.stackexchange.com/questions/2155401/solving-modulus-equations/2155409", "text": "# Solving Modulus-Equations\n\nSo, I've come across this modulus equation in my textbook: $$|2x-1|=|4x+9|$$\n\nI looked at the solution to this equation and understand that in order for both sides to be equal, the quantities inside the brackets must either be the same or the negatives of each other.\n\nThe solution then uses the following theorem: if $|p| = b, b>0 \\Rightarrow p = -b$ or $p = b$. $$2x-1 = -(4x+9)$$ or $$2x-1 = 4x+9$$ and solves both linear equations to get $x = -\\frac{4}{3}$ or $x = -5$\n\nI then asked myself why the solution didn't bother to find $$-(2x-1) = 4x+9$$ or $$-(4x-9) = -(2x-1)$$ and instead only found the two above. I then proceeded to calculate the above linear equations and got the exact same answers as above $x = - \\frac{4}{3}$ or $x = -5$\n\nI'd like to know why I achieved the same answers with this.\n\n\u2022 Think at why $2x-1 = -(4x+9) \\iff -(2x-1) = 4x+9\\,$. \u2013\u00a0dxiv Feb 21 '17 at 23:02\n\nThe equation $-(2x-1)=4x+9$ has the same solution as $2x-1=-(4x+9)$ as the only difference between these two equations is it has been multipled by $-1$.\nSimilarly $2x-1=4x+9$ has the same solution as $-(2x-1)=-(4x+9)$ for the same reason.\nSo to solve the original equation you need to solve only one out of the pair of equations $-(2x-1)=4x+9$ and $2x-1=-(4x+9)$ and then solve one out of the pair of equations $2x-1=4x+9$ and $-(2x-1)=-(4x+9)$. Which one you do from each pair doesn't matter as they have the same solution.\nI drew a graph of $$y = |4x+9| - |2x-1|.$$ Not to scale, for $x$ I made four squares equal $1,$ because the interesting $x$ values are $1/2$ and $-9/4.$ The graph is three lines joining up, but with different slopes (four times what I depicted). Had I checked the answer first, I would have moved it over a bit so as to see more of the negative $x$ axis.", "date": "2019-12-14 08:25:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2155401/solving-modulus-equations/2155409", "openwebmath_score": 0.8685563206672668, "openwebmath_perplexity": 110.95597707151309, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754456551737, "lm_q2_score": 0.867035771827307, "lm_q1q2_score": 0.8536621314458482}}
{"url": "http://nibw.fotoclubtiendeveen.nl/binary-multiplication-rules.html", "text": "In fourth case, a binary addition is creating a sum of 1 + 1 = 10 i. Multiplication in abstract algebra, as between vectors or other mathematical objects, does not always obey these rules. Multiplication Example Multiplicand 1000ten Multiplier x 1001ten-----1000 0000 0000 1000-----Product 1001000ten In every step \u2022 multiplicand is shifted \u2022 next bit of multiplier is examined (also a shifting step) \u2022 if this bit is 1, shifted multiplicand is added to the product. The value of a particular N-bit binary number x in a U(a,b)representation is given by the expression x. A multiplication algorithm is an algorithm (or method) to multiply two numbers. The rule for addition of two binary coded decimal (BCD) numbers is given below. However, each bit of implies different algorithmic path during each iteration, that is, if , only a point doubling is necessary. t= 0, then \u02c6 Q(xt) \u2190Q(xt\u22121) P(xt) \u2190P(xt\u22121)(1\u2212p), (2) if x. Following are the procedure for multiplying binary numbers. (e) G = (R\u00d7. Binary signed 2's complement: 2. Thus the product of a3 into b2 is a3b2 or aaabb. Only valid for independent events. Binary multiplication is one of the four binary arithmetic. Thus, the implementation uses CC to represent the number 12x10 + 12 = 132. Binary multiplication is actually much simpler to calculate than decimal multiplication. 0+0=0 0+1=1 1+0=1 1+1=10. In mathematics when you perform computational actions, you must have in mind that there is a sequence that need to be respected in. There are two types of multiplication for matrices: scalar multiplication and matrix multiplication. The sub-module numpy. Rewrite the smaller number such that its exponent matches with the exponent of the larger number. The most widely known binary operations are those learned in elementary school: addition, subtraction, multiplication. The broad perspective taken makes it an appropriate introduction to the field. Distributivity of Binary Operations. an old video where sal gives several examples of polynomial multiplication. 3rd: Flip the second fraction. Two's complement converter calculator is used to calculate the 2's complement of a binary or a decimal number. The pairs of 3 are sent to a carry-save adder, which takes a sum of 3 inputs and reduce it to a sum of 2 inputs (x + y + x => c + s). Ive watched some basic binary multiplication video's on youtube and i understand the logic and i can probably make it in redstone, but it will be huge and will be a mess. Excellent! Thank you for this explanation of binary multiplication. What is the modern slide rule? 4. Definition of multiplication rule, from the Stat Trek dictionary of statistical terms and concepts. I'd be happy to share it with anyone once I figure out how to place it on this site. Sparse matrices, which are common in scientific applications, are matrices in which most elements are zero. write sin x. Binary Multiplication. The binary system is a way of representing numbers in base 2, i. So, the result became 0. ), with steps shown. Binary Multiplication \u2022Sizing \u2022In binary addition -we are generally representing something that ultimately is to be executed in hardware \u2022Our hardware cannot change the number of bits (wires) it can hold \u2022We must establish a maximum number size \u2022For multiplication the size of the result must be the sum of the. Some compilers ignore this rule and detect the invalidity semantically. Lesson 1: Introduction to Multiplication. we add from right to left and the carry over get's added to the digits in the next column. A fun little trick, really, about how to quickly extract the cube roots of large integers mentally. Are division and multiplication inverse operations? Any operation is called a binary operation when defined by a decimal or binary expansion like (3; 3. There is no overflow, so correct result is zero. Multiplication Example Multiplicand 1000ten Multiplier x 1001ten-----1000 0000 0000 1000-----Product 1001000ten In every step \u2022 multiplicand is shifted \u2022 next bit of multiplier is examined (also a shifting step) \u2022 if this bit is 1, shifted multiplicand is added to the product. So to do 11+11 note that 1+1=10 so write zero and carry the one. Binary fractions work like decimal fractions: 0. the operations are done with algorithms similar to those used on sign magnitude integers. Each digit in a binary number is called a bit. Any boolean expression always follow closure with respect to binary addition (+ operator) and binary multiplication (. No strict rules are defined but just easy to read but precise. , 0s and 1s) to physical information, because only two kinds of physical objects or states are needed. This table includes all of the possible multiplications between the numbers 1 to 12. So, 1111 (in binary) = 8 + 4 + 2 + 1 = 15 (in decimal) Accounting for 0, this gives us 16 possible values for four binary bits. The operation of multiplication is extended to other real numbers according to the rules governing the multiplicative properties of. I want to make the binary class fun for my students, and I would like to apply activities to make it easier. Infix, Prefix and Postfix Expressions. Binary, or base 2, is one of many possible Number Systems. use the General Multiplication Rule. aaaaa x bbb. Binary multiplication is implemented using the same basic longhand algorithm that you learned in grade school. In fact it's even easier since each digit can only be a 1 or a 0. 5^2)^5)^5 (9 multiplication steps) Doing that requires factorisation, which may make it not worthwhile. Matrix elements should be numerical. A (B+C) = (AB) + (AC) This is what is known as the AND distributes over OR. 1 + 1 =10. This video con. Intro to Algebra: Exponent Rules (Multiplication) Students' understanding of exponent rules for multiplication is put to the test in this assessment which provides an introduction to algebra. Tree 149 Depth-first Search 136 Hash Table 134 Greedy 103 Binary Search 93 Breadth-first Search 75 Tree 14 Ordered Map 12 Geometry 9 Queue 9 Minimax 8 Brainteaser 7 Binary Indexed Tree 6. The broad perspective taken makes it an appropriate introduction to the field. Practice math problems like Extend Number Patterns (with Rule Mentioned) with interactive online worksheets for 4th Graders. In this case the horizontal. The number of bits required to represent the product is at least (n+m) for unsigned multiplication and (n+m+1) for signed (twos complement) multiplication. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. Problem Solution. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Since they were not interested in finding proofs justifying the used procedures, they were. With three or more, it is also straightforward, but you use the Even-Odd Rule. This application encodes and decodes ASCII and ANSI text. We describe three algorithms for multiple-point multiplication on elliptic curves over prime and binary \ufb01elds, based on the representations. What if the digits in each number are not even, or the same? What is the rule in dividing it into two parts? Example: x = 12345 y = 2478 or. Let\u2019s add two binary numbers to understand the binary addition. This video con. How to use this calculator: In the calculator, there are two input fields intended for entry of binary. Here the steps took place are. The main rules of the binary division include: subtraction, multiplication and division operations register with BYJU'S -The Learning App and also watch exciting videos to learn with ease. valuation functions Abstract Syntax of Binary Numerals A simple BNF grammar will suffice: B \u2208 Binary-numeral. Commercial applications like computers, mobiles, high speed calculators and some general purpose processors require [\u2026]. The maximum value we can have with three binary digits is 111 = decimal 7 calculated as follows-1*1 + 1*2 + 1*4. Starting from the right, the first \"A\" represents the \"ones\" place, or 16 0. Basic Arithmetic Operation-(lecture#1 basic rules of addition,multiplication,subtraction,division). In MQL4, there are some natural limitations to the rules of using operations in expressions. The positions in a binary number (called bits rather than digits) represent powers of two rather than powers of ten: 1, 2, 4, 8, 16, 32, and so on. That is, a+ (b+ c) = (a+ b) + c for all a;b;c 2F. A complex number is an expression of the form a + bi , where a and b are real numbers, and i has the property that i 2 = \u20131. a \u00d7 e = e \u00d7 a = a This is possible if e = 1 Since a \u00d7 1 = 1 \u00d7 a = a \u2200 a \u2208 R 1 is the identity element for multiplication on R Subtraction e is the identity of * if a * e = e * a = a i. Long multiplication can be carried out with binary numbers and is explored in this section. Binary Addition. Binary exponentiation (also known as exponentiation by squaring) is a trick which allows to calculate Most obviously this applies to modular multiplication, to multiplication of matrices and to other. Multiplication: multiplying, multiply. Though binary division not too difficult, it can initially be a bit harder to understand than the other binary operations as they shared similarities. Add 1 to the result using unsigned binary notation c. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. The first two round to a nearest value (ties to even and ties away from zero); the others are called directed roundings: towards zero, towards positive infinity and towards negative. Multiplication of pure imaginary numbers by non-finite numbers might not match MATLAB. For any elements a and b of {-1, 1}, (a+b) is not an element of {-1, 1}. Then, the program multiplies these two matrices (if possible) and displays it on the screen. Definition Let R be a set on which two binary operations are defined, called addition and multiplication, and denoted by + and \u00b7. 1 * 1 = 1 1 * 0 = 0 * 1 = 0. The rules for binary multiplication are: In truth table form, the multiplication of two bits, a x b is: Observe that a x b is identical to the logical and operation. Step 1: When we multiplying 0 and 0, we get 0. It consists of only 0, 1 digit and rules for addition, subtraction, and multiplication are the same as decimal numbers. INTEGER BINARY 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111. Applying the rules from the previous section, the multiplication follows these steps: The first two numbers (5. To do it we set the value range from 0 to 1, and choose the symmetric matrix type. The fastest way of multiplying two big integers. Algorithm: To multiply two decimal numbers x and y, write them next to each other, as in the example below. Converting from hex to binary is a lot like converting binary to hex. However, each bit of implies different algorithmic path during each iteration, that is, if , only a point doubling is necessary. , digits) is performed in a manner similar to decimal multiplication. The multiplication of an n-bit binary number with an m-bit binary number results in a product that is up to m + n bits in length for both signed and unsigned words. mlt\": nn = na if n = 2 then f$= \"b\" + \". That is all the way from 1 x 1 to 12 x 12. How to use this calculator: In the calculator, there are two input fields intended for entry of binary. This online calculator for addition and subtraction multiplication and division of binary numbers online. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. This is the first step in this step the least significant bit or the right most bit of B is multiplied with all the digits of A from the right side and the result is written. Each cell should contain a zero or a one. Multiplication of binary numbers can be achieved either combinationally or sequentially. Therefore, multiplication must be explicitly denoted by using the * operator, as in a * b. Two dots in the right-most box is worth one dot in the next box to the left. This gives signal of overflow-7+7=0 00111 binary 7 01001 two's complement of 7 11110 carries 10000 result of addition 16 - but fifth bit can be ignored, real result is 0. There are various number systems in the world. In fourth case, a binary addition is creating a sum of 1 + 1 = 10 i. They follow their own set of simple rules and trade on their own special exchange that has been set up just for them. abstract syntax 2. Lattice multiplication is also known as Italian multiplication, Gelosia multiplication, sieve multiplication, shabakh, Venetian squares, or the Hindu lattice. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. Learn multiplication table with easy to memorize, helpful times tables are simple to read. Watch this illustration and play with the code. These rules are applied for each operation, such that nested calculations imply the precision of each component. en: binary numbers adding substraction multiplication division. Message fields can be one of the following For example, when an old binary parses data sent by a new binary with new fields, those new fields become unknown fields in the old. (Where there is only one number above, you just carry down the 1. \u2022 More general rule: \u2013 The principle of inclusion and exclusion. Let's illustrate the multiplication rules for 3-bit binary numbers A = 111 and B = 101, beginning with the lowest and highest digits of the multiplier (Figure 2. For all a b b a ba. reflexive: if for every x X, xRx holds, i. Commute= travel (move) If you move swap the numbers ina multiplication and addition the anwser will still be the same! Rule: a+b = b+a Sample: 1+2= 3, 2+1=3. Same rule does not hold for natural numbers because if we take two numbers such as x and y and perform binary. 1 The Multiplication Rule. 1 * 1 = 1 2. According to the binary multiplication rules, the numbers in the bracket give the decimal equivalents of the binary numbers. Observations on Multiply Version 3. MUL Multiply; DEC VAX; signed multiplication of scalar quantities (8, 16, or 32 bit integer) in general purpose registers or memory, available in two operand (first operand multiplied by second operand with result replacing second operand) and three operand (first operand multiplied by second operand with result placed in third operand) (MULB2 multiply byte 2 operand, MULB3 multiply byte 3 operand, MULW2 multiply word 2 operand, MULW3 multiply word 3 operand, MULL2 multiply long 2 operand. The examples of valid hexadecimal numbers are 4E1or CD3A01. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. Note that in each subsequent row, placeholder 0's need \u2026 As an example of binary multiplication we have 101 times 11, 101 x 1 1. Here's the demo for subtraction. 0 is written in the given. Only valid for independent events. Binary Conversion Practice! ! ! !Binary Places: 32, 16, 8, 4, 2, 1 Convert these binary numbers to decimal: 1 10 11 100 101 1000 1011 1100 10101 11111 Convert these decimal numbers to binary:. It can be calculated easily if we know the following rules. I'd be happy to share it with anyone once I figure out how to place it on this site. Sponsored Links. Two's complement converter calculator is used to calculate the 2's complement of a binary or a decimal number. So, the result became 0. Quantities with unlike units may sometimes be multiplied, resulting in such units as foot-pounds, gram-centimeters, and kilowatt-hours. Binary literals can make relationships among data more apparent than they would be in hexadecimal or octal. To multiply in binary, you multiply the first number by each of the digits of the second number in turn starting from the right-hand side (in the same way that you would do multiplication in decimal). This is illustrated below. This means 3 groups of 2 equals 6 in all. C program to convert decimal number to roman. Existing Algorithms: \u2022 The naive binary multiplication algorithm has a time complexity O(n^2) where n is the number of bits of the numbers being multiplied. Arrays, multiplication and division Jennie Pennant, with the help of Jenni Way and Mike Askew, explores how the array can be used as a thinking tool to help children develop an in-depth understanding of multiplication and division. Other :: Binary Classification. ) Keep going, always adding pairs of numbers from the previous row. Additionally, the output is restricted to a 10-bit word with binary-point-only scaling of 2-4. Binary Card Game Explained ; Binary Card Game, the computer plays against. Instead of needing to know your multiplication tables you only really need to know three things. Power rule with positive exponents; Power rule with negative exponents: Problem type 1; Simplifying a sum or difference of three univariate polynomials; Product rule with positive exponents; Multiplying binomials with leading coefficients of 1; Solving a two-step equation with integers; Solving an equation to find the value of an expression. binary calculator as a 1 and 0. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. }\\) We use symbols to represent binary operations instead of function names, just as we do with addition and multiplication of integers. A multiplication algorithm is an algorithm (or method) to multiply two numbers. Converting Decimal Numbers to Binary Numbers. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition. 24 Addition and Multiplication in Q The following rules define binary operations on Q. So in the example above: Relative uncertainty = (0. You will be able to use it both as a binary to decimal converter and as a decimal to binary calculator. Binary multiplication is implemented using the same basic longhand algorithm that you learned in grade school. 2 Unary Operations Up: 1 Binary Operations Previous: 1 Addition and subtraction Index 2 Matrix multiplication Matrices are conformable for multiplication if and only if the number of columns in the first matrix equals the number of rows in the second matrix. A binary operator; the result is $$\\binary{1}$$ if at least one of the two operands are $$\\binary{1}\\text{,}$$ otherwise the result is \\(\\binary{0}\\text{. The general rule is that if the OV flag is set to I, then complement the sign bit. \u2022 Binary multiplication of unsigned integers reduces to \"shift and add\". The process in the binary system is easier than it is in the Binary multiplication calculator is a simple method to multiply binary values without using manual mehtods. When a string contains single or double quote characters, however, use Function annotations should use the normal rules for colons and always have spaces around the. basis is used in all matrix and matrix-vector multiplication algorithms, which are considered in this and the following sections. Rule 3: Lastly, perform all additions and subtractions, working from left to right. Binary Addition, Subtraction , Multiplication and division. + 01 0 01 1 10 \u00b7 01 0 00 1 01 Table C. Binary integer multiplier 210 first receives binary coefficients (C1 and C2) of two input operands and performs a binary multiplication, C' = C1 \u00b7 C2. Definition: Binary operation. If we know the rules of binary multiplication & division, the solution of t. Note that binary operators work on vectors and matrices as well as scalars. 1415; \u2026) converges to a unique. That means that every piece of binary code in a computer must be converted into a physical object or state. Binary Multiplication: Binary multiplication is exactly as it is in decimal, i. When a string contains single or double quote characters, however, use Function annotations should use the normal rules for colons and always have spaces around the. The rules for binary multiplication are: In truth table form, the multiplication of two bits, a x b is: Observe that a x b is identical to the logical and operation. Multiplication by an Integer Constant 7 The average time complexity seems to be exponential. The rule for addition of two binary coded decimal (BCD) numbers is given below. MUL Multiply; DEC VAX; signed multiplication of scalar quantities (8, 16, or 32 bit integer) in general purpose registers or memory, available in two operand (first operand multiplied by second operand with result replacing second operand) and three operand (first operand multiplied by second operand with result placed in third operand) (MULB2 multiply byte 2 operand, MULB3 multiply byte 3 operand, MULW2 multiply word 2 operand, MULW3 multiply word 3 operand, MULL2 multiply long 2 operand. Addition Worksheet A :. Binary Options Trading the simplest form of Best Trading Platform in the financial sector, that has become the traders best choice recently in Option Trading. Starting from the right, the first \"A\" represents the \"ones\" place, or 16 0. C program for fractional decimal to binary fraction conversion. (c) G= (N,+), the natural numbers in Z with addition as the binary operation. The evaluation using the and and or operators follow these rules: and and or evaluates expression Bitwise operators are used to compare integers in their binary formats. This means you are either multiplying each digit by 0 or 1, which will give you either a 0 or 1 as the answer. Test your method out on these numbers, and find a way to check your work. Binary Multiplication Binary multiplication is performed in same way as decimal numbers. The group operation is multiplication of complex numbers. E A B C E E A B C A A B C E B B C E A C C E A B T Combination order is \"top\" then \"side\"; e. 4 cm) \u00d7 100% = 5. See full list on byjus. Binary multiplication is arguably simpler than its decimal counterpart. An example is the unary and binary minus, (). Let\u2019s add two binary numbers to understand the binary addition. If you want to build up one for yourself, you can look at this scheme of 2 bits times 2 bits multiplier -. What is the result of multiplying the binary number 10010 by 101? 1011010. Then R is called a commutative ring with respect to these operations if the following properties hold: (i) Closure: If a,b R, then the sum a+b and the product a\u00b7b are uniquely defined and belong to R. What if the digits in each number are not even, or the same? What is the rule in dividing it into two parts? Example: x = 12345 y = 2478 or. Translate the number 22810 \u0432 binary like this: the Integer part of the number is divided by the base of the new. In mathematical operations involving significant figures, the answer is. Binary Numbers Toggle the 1s and 0s by clicking on them to reveal dots and make binary numbers. semantic algebras 3. Multiplying a binary term (multiplicand m) by 2 simply shifts mone power level up, for example, 2 \u00b7 K = P or 2 \u00b7 P = T. Generally you won\u2019t need to use this operator since it\u2019s redundant. The first two numbers (5. The Multiplying Binary Numbers (Base 2) (A) Math Worksheet from the Multiplication Worksheets Page at Math-Drills. For example, an int data type can hold 10 digits, is stored in 4 bytes, and doesn't accept decimal points. Let A = {1, 3, 5, 9, 11, 13} and let$\\odot$define the binary operation of multiplication modulo 14. Note: In the Chain Rule, we work from the outside to the inside. There are four rules of binary addition. (Where there is only one number above, you just carry down the 1. C program to convert decimal number to roman. Booth's Multiplication Algorithm is used to multiplication of two signed binary numbers. So basically all you need to do is multiply the powers. unambiguouslywrite a\u22c6b\u22c6c to denote either of the iterated products. 1 The Number of Elements in a List. FrankenPC makes a fair start showing binary addition. Let's illustrate the multiplication rules for 3-bit binary numbers A = 111 and B = 101, beginning with the lowest and highest digits of the multiplier (Figure 2. If we only want to invest in a single N-bit adder,. Complex multiplication is a more difficult operation to understand In the above formula for multiplication, if v is zero, then you get a formula for multiplying a complex. Binary division and multiplication are both pretty easy operations. for overloading binary operators, addition, subtraction, multiplication, division and comparison\". \" The Multiplicand is the number taken or multiplied. Fast multiplication. For each number of nodes, n, there is a certain number of possible binary tree configurations. Note that in each subsequent row, placeholder 0's need \u2026 As an example of binary multiplication we have 101 times 11, 101 x 1 1. We see what are the rules of binary multiplication and division and how to solve. Two last carries are 11. How to use this calculator: In the calculator, there are two input fields intended for entry of binary numbers. The unary minus operator returns the operand multiplied by -1. The Standard Multiplication Algorithm. There are four rules of binary addition. abstract syntax 2. Example 7. For example, an int data type can hold 10 digits, is stored in 4 bytes, and doesn't accept decimal points. Enter the two numbers that you want to implement the. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. Binary options are a different kind of option than any you're likely familiar with. When performing a binary. 1 Lecture 8: Binary Multiplication & Division Today s topics: Addition/Subtraction Multiplication Division Reminder: get started early on if signs disagree These rules fulfil the equation above 19. For example, in mathematics and most computer languages, multiplication is granted a higher precedence than addition, and it has been this way since the. 1415; \u2026) converges to a unique. If neither value is a double but one is a float, then Java treats both values as floats. 1 The Number of Elements in a List. Binary numbers can represent two states: 0, 1, and 2-bit binary numbers can represent (2=) 4 states: 00, 01, 10, and 11; by analogy, 7-bit binary numbers can represent (2=) 128 species. The arithmetic of binary numbers means the operation of addition, subtraction, multiplication and division. Beginners introduction to binary, hexadecimal and octal numbers. It's still a base-10 representation, but the algorithm doesn't need the digits to be between 0 and 9. Home Work. Stanford University. The relative uncertainty gives the uncertainty as a percentage of the original value. C program for addition of binary numbers. Conversion rules for converting decimal to other number system First step is to identify the base of target number system i. The Manchester Baby, operational in 1948, had no multiplication hardware. Post navigation. b) Whenever b(i,j) =1 , a(i,t) should also be 1, but for both alias sets i, t ; alias (i,j); Binary Variables b(i,j. For a node's world transformation: W = P. A simplistic way to perform multiplication is by repeated addition. There are four basic operations for binary addition, as mentioned above. It uses \"engine\" of Mathematical calculator. There are no ads, popups or other garbage. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. See General Rules for Operator Overloading for more information. 2 * 10 = 20 and 3 * 100 = 300. Floating Point Addition. Let us take two binary numbers A = 1001 and B = 101 we want to find out A \u00d7 B. The 1\u21902 Rule Whenever there are two dots in single box, they \u201cexplode,\u201d disappear, and become one dot in the box to the left. Dividing Binary Numbers. Robot Multiplication for Kids the 11s Master your Multiplication Facts by byrih. The general rule is: (x m) n = x mn. + Addition. The precedence order for arithmetic operators places multiplication and division above addition and. There are many applications of matrices in computer programming; to represent a graph data structure, in solving a system of linear equations and more. L = local transformation matrix. 2nd: Change the division sign to a multiplication sign. Binary Multiply - Repeated Shift and Add. The binary multiplication is much easier as it contains only 0s and 1s. The number can be converted to decimal by multiplying out as follows: 1*1 + 0*2 + 1*4 = 5. For example, multiplication of two 4-bit numbers requires a ROM having eight address lines, four of them, X 4 X 3 X 2 X 1 being allocated to the multiplier, and the remaining four, Y 4 Y 3 Y 2 Y 1 to the multiplicand. Then, b is called inverse of a. abstract syntax 2. The 3 basic. More Topics Related. More Examples:. Binary addition/subtraction is similar to regular (daily life) addition/subtraction, but here addition/subtraction performs only two digits those are 0 and 1, these are binary digits hence such kind of addition/subtraction is called binary addition/subtraction. Find the missing multiple or product for any multiplication fact You can use the printable multiplication table below to check your answers or as a study guide. With three or more, it is also straightforward, but you use the Even-Odd Rule. 741 Op-Amp Circuit Binary Addition Binary Multiplication Section 4. Last but not the least, a major reason computers use the binary system is that the two-state system is the number system best suited to the optical and magnetic storage components of the computer. How to do Addition, Subtraction, Multiplication and Division in Android Studio. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. Point multiplication is usually computed by a sequence of point doublings (triplings, halvings) and additions, whose numbers depend on the length and the number of non-zero digits of k. Binary literals can make relationships among data more apparent than they would be in hexadecimal or octal. Example \u2212 Addition Binary Subtraction. To start practising, just click on any link. To convert 0xDEADBEEF (a commonly used code to indicate a system crash), begin by sorting the digits into \"bins\":. A lower precedence number means tighter binding Other numeric operations are similar: subtraction (-), and multiplication (*). binary calculator as a 1 and 0. In Binary Arithmetic, 1 + 1 = 10, as per the rules of binary addition. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don\u2019t need to memorize such rules as 5 + 4 = 9, or 6 \u00d7 7 = 42. But we do not know much more. Assume the multiplicand (A) has N bits and the multiplier (B) has M bits. How to convert letters (ASCII characters) to binary and vice versa. eve Engels, 2006 Slide 7 of 10 \u2013Example: two-digit multiplication a 1 a 0 x b 1 b 0 a 1 b 0 a 0 b 0 a 1 b 1 a 0 b 1 + p 2 p 1 p 0 p 0 = a 0 b 0 p 1 = a 1 b 0 + a 0 b 1 p 2 = a 1 b 1 + a 0 b 1 a 1 b 0 (addition here, not OR). Binary Addition, Subtraction , Multiplication and division. Multiplication And Division of Octal Numbers Calculator, Octal Multiplication And Division: Required Data Entry Enter A Octal Value. Addition, subtraction, multiplication, and division are all binary operators with which we are familiar from grade school. (b) G= (N,\u00b7), the natural numbers in Z with multiplication as the binary oper-ation. flipped over). For example, the real numbers form a \ufb01eld, with \u2018+\u2019 and \u2018\u00b7\u2019denoting ordinary addition and multiplication. \" Every digit must be either 1 or 0. Binary Arithmetic 1: Adding binary numbers. Multiplication done algebraically. All it includes are addition of binary. Binary arithmetic is essential part of all the digital computers and many other digital system. What is the associative property of Multiplication? Associative property of multiplication is the rule that says that the way that factors are grouped in a multiplication problem will not affect the answer. In modern terms it employs the decomposition of one number into its binary components, addition to produce \"doubles\" or multiples of the other number, and computing a total of doubles identified by odd divisors or the powers of two. Overflow rule:- If two number are added, and they are both positive or negative than overflow occurs if and only if the result has opposite sign. Let us understand the binary multiplication on natural numbers and real numbers. 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0. Such models can be linearized quite easily and are in most cases better solved as a linear. Lesson 1: Introduction to Multiplication. The rules of differentiation (product rule, quotient rule, chain rule, \u2026) have been implemented in JavaScript code. Multiplication Tables L All possible binary combinations of symmetry operations can be summarized in a multiplication table. number system to which we have to convert the decimal number system. In mathematics, a binary operation or dyadic operation is a calculation that combines two elements (called operands) to produce another element. 17 * 5 = 85 / (Division) By using this operator we can perform division of numbers. Binary Triple Octal Decimal Hexadecimal Binary-decimal. If \u22c6 is associative we can. Assigning a precedence to grammar rules. If you skip parentheses or a multiplication sign, type at least a whitespace, i. No strict rules are defined but just easy to read but precise. Use as an example 18 or 10010: 18 = 16 + 2 = 2 4 + 2 1. (2) In each part below, a rule is given that determines a binary operation on Z. Sponsored Links. Enter the two numbers that you want to implement the. A carry-save adder's propagation time is constant, while traditional a 2 input adder's propagation time is proportional the the width of the operands. C program to convert decimal number to roman. One technique for performing multiplication in any number base is by repeated addition; for example, 6\u00d74=6+6+6+6=24 in decimal (similarly, 4\u00d76=4+4+4+4+4+4=24). When a multiplication problem is given abstractly, there is no such distinction, so we prefer to use the symmetrical term \"factor. This is the currently selected item. mlt\": nn = nb open f$ for binary as #1 for n2 = 1 to nn randomize timer x$= ltrim$ (str$(int (rnd * 10))) seek #1, n2: put #1, n2, x$ next n2 seek #1, n2 close (1) next n ' open \"a\" + \". This is, of course, 10 times the starting value - that is how you multiply things by 10. Decitrig: A rule with the trigonometric scales with decimal degrees (S, ST, T) [0. So, the result became 0. multiplication of any element by 0 yields 0; 3. The number 1010110 is represented by 7 bits. Therefore, multiplication must be explicitly denoted by using the * operator, as in a * b. The value of a particular N-bit binary number x in a U(a,b)representation is given by the expression x. , E A B C E EE = E EA = A EB = B EC = C A AE = A AA = B AB = C AC = E B BE = B BA = C BB = E BC = A C CE = C CA = E CB = A CC = B. Commercial applications like computers, mobiles, high speed calculators and some general purpose processors require [\u2026]. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. In Karatsuba algorithm for multiplying two numbers, we divide each number into two. A binary multiplier is a combinational logic circuit used in digital systems to perform the multiplication of two binary numbers. Unit-2: Binary Multiplication-Booth's Algorithm. As an example of binary multiplication we have 101 times 11, 101 x 1 1. eve Engels, 2006 Slide 7 of 10 \u2013Example: two-digit multiplication a 1 a 0 x b 1 b 0 a 1 b 0 a 0 b 0 a 1 b 1 a 0 b 1 + p 2 p 1 p 0 p 0 = a 0 b 0 p 1 = a 1 b 0 + a 0 b 1 p 2 = a 1 b 1 + a 0 b 1 a 1 b 0 (addition here, not OR). Read about Binary Addition (Binary Arithmetic) in our free Electronics Textbook. The unary minus operator returns the operand multiplied by -1. Multiplication of a binary number n by the binary number 100 (=4) is done by adding two bits 0 to the right of the binary representation of n. The exact same rule exists in binary. The binary number system uses only two digits 0 and 1 due to which their addition is simple. A, 1 is entered in its position that yields 1000 since 2 3 = 8. The interaction of the parts makes the dinner range from good to great. There are four basic operations for binary addition, as mentioned above. The form calculates the bitwise exclusive or using the function gmp_xor. One technique for performing multiplication in any number base is by repeated addition; for example, 6\u00d74=6+6+6+6=24 in decimal (similarly, 4\u00d76=4+4+4+4+4+4=24). to multiply and divide polynomials, use conv and deconv on. The result of binary arithmetic for A with and B with gives the worst case p:. The multiplication method is sometimes called doubling and halving or called Russian peasant multiplication. Arithmetic operations in the binary system are extremely simple. (Like multiplying by 10 in our normal notation. Binary Option trading platforms allows e-Traders to make investments by predicting the future direction of an asset and make up to 85% profits in a span of few minutes. 0026), with the final result having 8 decimal places (0. Instead, you need. To convert fraction to binary, start with the fraction in question and multiply it by 2 keeping notice of the resulting integer and fractional part. Multiplying and dividing positive and negative numbers is a simple operation with two numbers. \" The Multiplicand is the number taken or multiplied. As a result of multiplication you will get a new matrix that has the same quantity of rows as the 1st one has and the same quantity of columns as the 2nd one. 11 \u22c5 2 \u2212 3 1 01. FREE holiday, seasonal, and themed multiplication worksheets to help teach the times tables. Starting from the right, the first \"A\" represents the \"ones\" place, or 16 0. Specifying Field Rules. To perform addition, subtraction, multiplication and division in python, you have to ask from user to enter any two number, and then ask again to enter the operator to perform the desired operation. Binary Card Game Explained ; Binary Card Game, the computer plays against. Binary code, as it turns out, is easy to convert from electronic information (e. The numbers we are multiplying together are called factors, and the result of multiplication is called the. multiply numbers right to left and multiply each digit of one number to every digit of the other number, them sum them up. Draws binary random numbers (0 or 1) from a Bernoulli distribution. Each row and each column is unique and contains as many zeros as ones. (b) Using binary representa-tion 10011 1101 10011 10011 1011111 10011 11110111 (c) Adding immediately Figure 7. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. them sum them up. (LL 0, LL 1,LL 01, etc. For a node's world transformation: W = P. Most processors perform n-bit by n-bit multiplication and produce a 2n-bit result (double bits) assuming there is no overflow condition. Here Binary number is a number that can be represented using only two numeric symbols - 0 and 1. \" Every digit must be either 1 or 0. Binary, or base 2, is one of many possible Number Systems. Multiplication of pure imaginary numbers by non-finite numbers might not match MATLAB. Suppose for instance that we want to multiply 13 11, or in binary notation, x= 1101 and y= 1011. In mathematics when you perform computational actions, you must have in mind that there is a sequence that need to be respected in. Distributivity of Binary Operations. Learn Java program multiplication starting from its overview, How to write, How to set environment , How to run, Example like Addition, Subtraction , Division, Multiplication etc. C++ Program to Multiply Two Matrix Using Multi-dimensional Arrays This program takes two matrices of order r1*c1 and r2*c2 respectively. Note that in each subsequent row, placeholder 0's need \u2026 As an example of binary multiplication we have 101 times 11, 101 x 1 1. The representations of the multiplicand and product are not specified; typically, these are both also in two's complement representation, like the multiplier, but any number system that supports addition and subtraction will work as well. Matrix multiplication falls into two general categories: Scalar: in which a single number is multiplied with every entry of a matrix. A general rule when multiplying a Qm format number by a Qn format number, is that the product will be a Q(m+n) number. In binary, 2 * 3 = 6 is 10 * 11 = 110, and 4 * 3 = 12 is 100 * 11 = 1100. For example: x= 1234 y= 2456 Then a = 12, b = 34, c = 24 , d = 56. That is binary numbers can be represented in general as having p binary digits and q fractional digits. Multiplying and dividing positive and negative numbers is a simple operation with two numbers. * (Multiplication) By using this operator we can perform multiplication of numbers. Other arithmetic operators work only with numbers and always convert their operands to numbers. Two dots in the right-most box is worth one dot in the next box to the left. resample(rule[, axis, closed, label, \u2026]) Resample time-series data. Convert number. The balanced ternary multiplication table is depicted in Fig. If the array isn't sorted, you must sort it using a sorting technique such as merge sort. By using this website, you agree to our Cookie Policy. Note that since binary operates in base 2, the multiplication rules we need to remember are those that involve 0 and 1 only. The question is the extent to which we can unify addition and multiplication, realizing them as terms in a single underlying binary operation. Multiplication by an Integer Constant 7 The average time complexity seems to be exponential. A 4-bit, 2's complement example:. Binary multiplication can be achieved by using a ROM as a look-up\u2019 table. Browse other questions tagged rsa algorithm-design montgomery-multiplication or ask your own question. Note that in each subsequent row, placeholder 0's need to be added, and the value shifted to the left, just like in decimal multiplication. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. Our source code has been written in characters we can read, but it needs to be. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. Linear algebra. com binary converter online for free. It gives a range of activations, so it is not binary activation. This online calculator for addition and subtraction multiplication and division of binary numbers online. So in the example above: Relative uncertainty = (0. In this tutorial, we will create a custom pipe so that the currency format can run well. According to all the rules above, x will be converted to int first. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don\u2019t need to memorize such rules as 5 + 4 = 9, or 6 \u00d7 7 = 42. If an operation consists of k steps and. Also, we'll perform addition and subtraction on them. The first two numbers (5. To help me remember I wrote some fully commented code for an 8 x 8 bit unsigned multiplier. Stats: Probability Rules. The code works fine when I give an integer input, but doesn't give correct result for floating point input. 5^2)^25 (25 multiplication steps) = ((3. For K-12 kids, teachers and parents. 0015 to 1,000,000), two binary scales for adding and subtracting fractions, a scale of drill sizes, a scale of thread sizes, and millimeters. 70 \u00d7 10-1 with 9. The arithmetic unit performs parallel one-step addition (subtraction), multiplication and division. Converting Decimal Numbers to Binary Numbers. Binary Addition, Subtraction , Multiplication and division. In binary multiplication, we only need to remember the following. Similarly, whenever we would like to sum two binary numbers, only we will have a carry if the product is bigger than 1 because, in binary numbers, 1 is the highest number. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson. What is the rules of binary in addition? As with decimal numbers, you start by adding the bits (digits) one column, or place. Other arithmetic operators work only with numbers and always convert their operands to numbers. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition. Test yourself on multiplication facts. The difference being the \"point\" bookkeeping required which is the same as addition. Take some parts away or add new ones in, and you get a different outcome, but not a binary, win/lose one. The \"Commutative Laws\" say we can swap numbers over and still get the same answer. ) Vector : A rule with hyperbolic functions. There are four rules for binary multiplication: Multiplication is always 0, whenever at least one input is 0. Here is a new (harder) goal: come up with a general method to find the binary number related to any number of dots without actually going through the \u201cexploding dot\u201d process. Now multiplication is going to occur like this. Just binary utilities that work right in your browser. binary arithmetic rules. Multiplication rules. Mutually Exclusive Events. Leave a Comment on 540. \u231a September 25, 2014 Binary multiplication is performed the same as decimal multiplication. numbers are multiplied and divided through a process called shifting. Similarly, whenever we would like to sum two binary numbers, only we will have a carry if the product is bigger than 1 because, in binary numbers, 1 is the highest number. The matrix product is designed for representing the composition of linear maps that are represented by matrices. Though binary division not too difficult, it can initially be a bit harder to understand than the other binary operations as they shared similarities. Then we multiply the entire top number by each individual digit of the bottom number. General binary operations that follow these patterns are studied in abstract algebra. Multiplication Example Multiplicand 1000ten Multiplier x 1001ten-----1000 0000 0000 1000-----Product 1001000ten In every step \u2022 multiplicand is shifted \u2022 next bit of multiplier is examined (also a shifting step) \u2022 if this bit is 1, shifted multiplicand is added to the product. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. The method used for ancient Egyptian multiplication is also closely related to binary numbers. This is a simple online tool to convert English into binary. Binary Multiplication is generally simpler than decimal multiplication. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Enter expression with binary numbers and get the result. This online calculator for addition and subtraction multiplication and division of binary numbers online. In this tutorial, we will create a custom pipe so that the currency format can run well. The Binary Calculator is used to perform addition, subtraction, multiplication and division on two In mathematics and computer science, binary is a positional numeral system with a base of 2. Let\u2019s covert the same binary number to an octal number: $100100010101111_{2}=100 100 010 101 111$ $100=4$ $010=2$ \\$101=5. It shows you how the product is generated in real-time, step-by-step, and allows you to highlight the individual multiplication steps used to get the answer. To figure the decimal value of a binary number, you multiply each bit by its corresponding power of two and then add the results. And all utilities work exactly the same way \u2014 load. Generally you won\u2019t need to use this operator since it\u2019s redundant. This follows Chapter 4 of Schmidt\u2019s book. There are four rules of binary addition. The general rule is: (x m) n = x mn. To read binary numbers, and convert them to their decimal equivalent, you have two options: you can either use the Binary to Decimal Converter at ConvertBinary. binary: A binary operator operates on two operands. The multiplication of a 16 bit binary \"a\" with a 24 bit binary \"b and the complementary defence of both the general interest and the rules of the competitive. In the modern world, multiplication, division, addition, and subtraction are estimated by the binary calculator within a second, same rule as applied in the decimal system. A general rule when multiplying a Qm format number by a Qn format number, is that the product will be a Q(m+n) number. 2 steps per bit because Multiplier & Product combined How can you make it faster? What about signed multiplication? \u2022 trivial solution: make both positive. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. Browse other questions tagged rsa algorithm-design montgomery-multiplication or ask your own question. We have already discussed the binary addition and binary subtraction in detail in the previous articles now we are going to discuss binary multiplication in a detailed manner. If we take the number 237 and shift each digit 1 place to the left, adding a zero to the right hand side, we get 2370. How to use this calculator: In the calculator, there are two input fields intended for entry of binary numbers. Binary Multiplication \u2022Sizing \u2022In binary addition \u2013we are generally representing something that ultimately is to be executed in hardware \u2022Our hardware cannot change the number of bits (wires) it can hold \u2022We must establish a maximum number size \u2022For multiplication the size of the result must be the sum of the. Binary Arithmetic 1: Adding binary numbers. Take some parts away or add new ones in, and you get a different outcome, but not a binary, win/lose one. Convert number. Note that since binary operates in base 2, the multiplication rules we need to remember. (countable) A calculation involving multiplication. 16 + 0 + 0 + 2 + 1 = 19. Binary Multiplication \u2022Sizing \u2022In binary addition \u2013we are generally representing something that ultimately is to be executed in hardware \u2022Our hardware cannot change the number of bits (wires) it can hold \u2022We must establish a maximum number size \u2022For multiplication the size of the result must be the sum of the. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don\u2019t need to memorize such rules as 5 + 4 = 9, or 6 \u00d7 7 = 42. You just take a regular number (called a \"scalar\") and multiply it on. Unicode Character 'MULTIPLICATION SIGN' (U+00D7). Consider the simple problem of multiplying 110 2 by 10 2. A, 1 is entered in its position that yields 1000 since 2 3 = 8. Multiplying Complex Numbers Calculator. Binary addition is the simplest method to add any of the binary numbers. Addition Worksheet A :. Let\u2019s add two binary numbers to understand the binary addition. So a binary matrix is such an array of 0's and 1's. A, 1 is entered in its position that yields 1000 since 2 3 = 8. Align the numbers as an ordinary subtraction. The binary number system uses 0s and 1s to represent numbers. No problem with unsigned (always positive) numbers, just use the same Multiplication in Two's complement cannot be accomplished with the standard technique since, as. The number 1010110 is represented by 7 bits. Binary multiplication is actually much simpler to calculate than decimal multiplication. Learn binary conversions and arithmetic with interactive demonstrations and explanations. ), with steps shown. If given condition is FALSE (not TRUE), expr_set2 will get executed. The procedure for binary multiplication is similar to that in decimal system. Computer method: Computer method is used by digital machines to multiply the binary numbers. There are four rules for binary subtraction: Borrow 1 is required from the next higher order bit to subtract 1 from 0. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Sponsored Links.", "date": "2021-03-02 10:58:54", "meta": {"domain": "fotoclubtiendeveen.nl", "url": "http://nibw.fotoclubtiendeveen.nl/binary-multiplication-rules.html", "openwebmath_score": 0.703130841255188, "openwebmath_perplexity": 665.2859832549126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9845754497285467, "lm_q2_score": 0.8670357666736773, "lm_q1q2_score": 0.8536621299034711}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=k441o32850lkospornhudrbhs7&action=printpage;topic=1558.0", "text": "# Toronto Math Forum\n\n## MAT334-2018F => MAT334--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 04:54:06 AM\n\nTitle: TT2A Problem 1\nPost by: Victor Ivrii on November 24, 2018, 04:54:06 AM\nUsing Cauchy's integral formula calculate\n$$\\int_\\Gamma \\frac{z\\,dz}{z^2-4z+5},$$\nwhere $\\Gamma$ is a counter-clockwise oriented simple contour, not passing through eiter\nof $2\\pm i$ in the following cases\n\n(a) The point $2+i$ is inside\u00a0 $\\Gamma$ and $2-i$ is outside\u00a0 it;\n\n(b) The point $2-i$ is inside\u00a0 $\\Gamma$ and $2+i$ is outside it;\n\n(c) Both points $2\\pm i$ are inside\u00a0 $\\Gamma$.\nTitle: Re: TT2A Problem 1\nPost by: ZhenDi Pan on November 24, 2018, 05:30:25 AM\nWe have\n\n\\int_\\Gamma \\frac{zdz}{z^2-4z+5}\n\nLet\n\nf(z) = \\frac{z}{z^2-4z+5} = \\frac{z}{(z-(2-i))(z-(2+i))}\n\nQuestion a:\nThe point $2-i$ is outside of the contour $\\Gamma$ and the point $2+i$ is inside of the contour $\\Gamma$. Then let\n\ng(z) =\\frac{z}{z-2+i} \\\\\ng(2+i) = \\frac{2+i}{2i} \\\\\n\\int_\\Gamma f(z)dz = \\int_\\Gamma \\frac{g(z)}{(z-(2+i))}dz = 2\\pi i g(z_0) = 2\\pi i g(2+i) = 2\\pi i \\cdot \\frac{2+i}{2i}= \\pi(2+i)\n\nQuestion b:\nThe point $2+i$ is outside of the contour $\\Gamma$ and the point $2-i$ is inside of the contour $\\Gamma$. Then let\n\ng(z) =\\frac{z}{z-2-i} \\\\\ng(2-i) = -\\frac{2-i}{2i} \\\\\n\\int_\\Gamma f(z)dz = \\int_\\Gamma \\frac{g(z)}{(z-(2-i))}dz = 2\\pi i g(z_0) = 2\\pi i g(2-i) = 2\\pi i \\cdot -\\frac{2-i}{2i}= -\\pi(2-i)\n\nQuestion c:\nBoth points $2+i$ and $2-i$ are inside of the coutour $\\Gamma$. Then we have\n\nz_0 = 2+i \\\\\nz_1 = 2-i \\\\\n\\left.Res(f;2+i) = \\frac{z}{z-2+i} \\right\\vert_{z=2+i} = \\frac{2+i}{2i} \\\\\n\\left.Res(f;2-i) = \\frac{z}{z-2-i} \\right\\vert_{z=2-i} = - \\frac{2-i}{2i}\n\nSo the Residue Theorem gives us\n\n\\int_\\Gamma f(z)dz = 2\\pi i(\\frac{2+i}{2i}-\\frac{2-i}{2i}) = 2\\pi i \\cdot 1= 2\\pi i\nTitle: Re: TT2A Problem 1\nPost by: Yifei Wang on November 24, 2018, 05:34:25 AM\nWe can rewrite the fraction as:\n\n$let f(z) =\\frac{z}{z^2-4z+5}$\n\nas\n\n$\\frac{z}{(z-(2+i))(z-(2-i))}$\n\na. When $2+i$ is inside\n$f(z) = \\int \\frac{\\frac{z}{z-2+i}}{z-2-i}~dz$\n\nBy Cauchy's thm\n\n$f(z) = \\int \\frac{\\frac{z}{z-2+i}}{z-2-i}~dz= 2i\\pi *f(2+i) = \\pi * (2+i)$\n\nb. When $2-i$ is inside\n\n$f(z) = \\int \\frac{\\frac{z}{z-2-i}}{z-2+i}~dz$\n\nBy Cauchy's thm\n\n$f(z) = \\int \\frac{\\frac{z}{z-2-i}}{z-2+i}~dz= 2i\\pi *f(2-i) = -\\pi * (2-i)$\n\nc. When both points are inside\n\n$f(z) = \\int \\frac{z}{(z-(2+i))(z-(2-i))}~dz = 2i\\pi (Res(f, 2+i) + Res(f, 2-i)) = 2\\pi i$\n\nTitle: Re: TT2A Problem 1\nPost by: Yifei Wang on November 24, 2018, 05:36:15 AM\n\nZhenDi Pan\n\nI think you are missing the $z$ on the numerator.\nTitle: Re: TT2A Problem 1\nPost by: ZhenDi Pan on November 24, 2018, 06:23:00 AM\nYes thank you I corrected it. Still our answers are different, I don't know where went wrong though.\nTitle: Re: TT2A Problem 1\nPost by: Zhuoer Sun on November 24, 2018, 12:50:28 PM\nYifei Wang\n\nI think you did the last part wrong. For part (c), you don't need to multiply by 2ipi again. The answer should be the sum of what you got from part (a) and (b), as you've already included 2ipi in previous parts. The final answer should just be pi\u2217(2+i)\u2212pi\u2217(2\u2212i).\n\nTitle: Re: TT2A Problem 1\nPost by: Yifei Wang on November 25, 2018, 02:38:21 PM\nThank you for the correction!\nTitle: Re: TT2A Problem 1\nPost by: Victor Ivrii on November 29, 2018, 07:57:21 AM\nRemark: Since integrand is $\\frac{1}{z} +O(\\frac{1}{z^2}$ the residue at $\\infty$ is $-1$ and answer to (c) is $2\\pi i$ (independently from(b),(c)). Yet another solution to (c)", "date": "2022-11-26 10:22:06", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=k441o32850lkospornhudrbhs7&action=printpage;topic=1558.0", "openwebmath_score": 0.9399534463882446, "openwebmath_perplexity": 5981.983723593625, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754465603678, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8536621203910301}}
{"url": "http://math.stackexchange.com/questions/219693/finding-a-closed-formula-for-1-cdot2-cdot3-cdots-k-dots-nn1n2-cdotsk", "text": "# Finding a closed formula for $1\\cdot2\\cdot3\\cdots k +\\dots + n(n+1)(n+2)\\cdots(k+n-1)$\n\nConsidering the following formulae:\n\n(i) $1+2+3+..+n = n(n+1)/2$\n\n(ii) $1\\cdot2+2\\cdot3+3\\cdot4+...+n(n+1) = n(n+1)(n+2)/3$\n\n(iii) $1\\cdot2\\cdot3+2\\cdot3\\cdot4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$\n\nFind and prove a 'closed formula' for the sum\n\n$1\\cdot2\\cdot3\\cdot...\\cdot k + 2\\cdot3\\cdot4\\cdot...\\cdot(k+1) + ... + n(n+1)(n+2)\\cdot...\\cdot (k+n-1)$\n\ngeneralizing the formulae above.\n\nI have attempted to 'put' the first 3 formulae together but i am getting know where and wondered where to even start to finding a closed formula.\n\n-\nDo you know how to prove by induction? \u2013\u00a0 Clive Newstead Oct 23 '12 at 22:27\nYes, using the basic step n=1 and then induction step n+1 showing it follows. But don't i need to 'find' the formula before proving it using induction? \u2013\u00a0 Matt Oct 23 '12 at 22:29\n\nThe pattern looks pretty clear: you have\n\n\\begin{align*} &\\sum_{i=1}^ni=\\frac12n(n+1)\\\\ &\\sum_{i=1}^ni(i+1)=\\frac13n(n+1)(n+2)\\\\ &\\sum_{i=1}^ni(i+1)(i+2)=\\frac14n(n+1)(n+2)(n+3)\\;, \\end{align*}\\tag{1}\n\nwhere the righthand sides are closed formulas for the lefthand sides. Now you want\n\n$$\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)\\;;$$\n\nwhat\u2019s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.\n\nAdded: The general result, of which the three in $(1)$ are special cases, is $$\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)=\\frac1{k+1}n(n+1)(n+2)\\dots(n+k)\\;.\\tag{2}$$ For $n=1$ this is $$k!=\\frac1{k+1}(k+1)!\\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then\n\n\\begin{align*}\\sum_{i=1}^{n+1}i(i+1)&(i+2)\\dots(i+k-1)\\\\ &\\overset{(1)}=(n+1)(n+2)\\dots(n+k)+\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)\\\\ &\\overset{(2)}=(n+1)(n+2)\\dots(n+k)+\\frac1{k+1}n(n+1)(n+2)\\dots(n+k)\\\\ &\\overset{(3)}=\\left(1+\\frac{n}{k+1}\\right)(n+1)(n+2)\\dots(n+k)\\\\ &=\\frac{n+k+1}{k+1}(n+1)(n+2)\\dots(n+k)\\\\ &=\\frac1{k+1}(n+1)(n+2)\\dots(n+k)(n+k+1)\\;, \\end{align*}\n\nexactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\\dots(n+k)$, and the rest is just algebra.\n\n-\nFor each extra bracket (n+1) then (n+2) you're adding +1 on the bottom and then an addition bracket. im thinking the closed formula for $\\sum_{k=1}^nn(n+1)(n+2)\\dots(n+k-1)\\;;$ would be $\\frac1{k+1}(n+k-1)$ im not too sure \u2013\u00a0 Matt Oct 23 '12 at 22:40\n@Matt: Look at the closed forms in $(1)$ again: your $\\frac1{k+1}$ is fine, but you should have $k$ more factors, not just one. \u2013\u00a0 Brian M. Scott Oct 23 '12 at 22:47\nIm thinking something $\\frac1{k+1}n(n+k)$ but not sure how to 'repeat' brackets for more terms of $k$. so for $k=1$ it works as it gives $\\frac12n(n+1)$ but then for $k=2$ all i get it $\\frac13n(n+2)$ rather than $\\frac13n(n+1)(n+2)$ \u2013\u00a0 Matt Oct 23 '12 at 22:52\n@Matt: When $k=1$ the RHS has two factors involving $n$; when $k=2$ it has three factors involving $n$; and when $k=3$ it has four factors involving $n$. In each case the smallest is $n$ and the largest is $n+k$. In the general case, then you should expect to have $k+1$ factors involving $n$, running from $n$ up through $n+k$. \u2013\u00a0 Brian M. Scott Oct 23 '12 at 22:57\nSo it is simply $\\frac1{k+1}(n+k)$ \u2013\u00a0 Matt Oct 23 '12 at 23:01\n\nIf you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove $$\\binom kk + \\binom{k+1}k + \\dots + \\binom{k+n-1}k = \\binom{k+n}{k+1}.$$ This is precisely the identity from this question.\n\nThe same argument for $k=3$ was used here.\n\nOr you can look at your problem the other way round: If you prove this result about finite sums $$\\sum_{j=1}^n j(j+1)\\dots(j+k-1)= \\frac{n(n+1)\\dots{n+k-1}}{k+1},$$ you also get a proof of the identity about binomial coefficients.\n\n-\n\nFrom (i), (ii) and (iii) it is reasonable to guess that your sum will be $$n(n+1)\\cdot...\\cdot(n+k)/(k+1)$$ Try to prove this by induction.\n\n-\nFor a fixed non-negative $k$, let $$f(i)=\\frac{1}{k+1}i(i+1)\\ldots(i+k).$$ Then $$f(i)-f(i-1)=i(i+1)\\ldots(i+k-1).$$ By telescoping,\n$$\\sum_{i=1}^ni(i+1)(i+2)\\dots(i+k-1)=\\sum_{i=1}^n\\left(f(i)-f(i-1)\\right)=f(n)-f(0)=f(n)$$", "date": "2014-03-13 17:37:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/219693/finding-a-closed-formula-for-1-cdot2-cdot3-cdots-k-dots-nn1n2-cdotsk", "openwebmath_score": 0.9113698601722717, "openwebmath_perplexity": 350.9822847302935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984575447918159, "lm_q2_score": 0.867035758084294, "lm_q1q2_score": 0.8536621198769043}}
{"url": "http://mdnu.schuetzen-wuerm.de/nth-term-test-for-divergence-calculator.html", "text": "Nth Term Test For Divergence Calculator\n\nnth root calculator. = cA , where c is a constant. If you're ever in doubt and would like to We made this SAT\u00ae score calculator because we saw that everyone else simply replicated the tables when creating what they called a \"calculator\". com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first \u201ctest\u201d we check when trying to determine whether a series converges or diverges. Derivatives. Many authors do not name this test or give it a shorter name. nth Term Test for Divergence (ONLY) If lim 0n n. if the limit is equal to a number different then zero, or if the limit is infinity, or if the limit does not exist) then the series diverge. 5n\u00b2 = 5,20,45,80,125. So the nth term tending to zero is only a necessary condition for convergence and is not sufficient. X n are the observation, then the G. nth term test, divergence test, zero test (KristaKingMath). Using the nth-Term Test for Divergence 613 The series in Example STUDY TIP 5(c) will play an important role in this chapter. Uncertainties are assessed by re-sampling the distributions and re-computing divergence estimates 100 times. geometric series, nth term test (Test for Divergence). HOWEVER, just because a series\u2019 terms approach 0, it is NOT convergent!!! This is a test for divergence, not a test for convergence. Author: Juan Carlos Ponce Campuzano. 9: Use Divergence Test to determine whether r 1 ln. If , then the series is convergent or; 2. Dierentiate term by term (new power series is for f0(x)). nth-term-test-for-divergence. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Geometric series test. The harmonic series is as follows: H 1 = 1 H 2 = H 1 + 1/2 H 3 = H 2 + 1/3 H 4 = H 3 + 1/4 H n = H n-1 + 1/n. 64 with lots of other numbers after the decimal point. Diverges by Geo Series Test or nth Term Test 2d. The calculator will find the divergence of the given vector field, with steps shown. Absolute and Relative Extrema - The minimum and maximum points on a graph, absolute extrema require you to test endpoints and find the highest or lowest point on an interval, relative extrema require you to find where f\u2019(x) switches sign; Concavity - The rate at which the derivative is changing, notated by f\u2019\u2019(x). Integral Test and p-Series. T-test Calculator. Use only the Divergence Test to determine if the statement is true, false, or can't be decided yet. Tyquan Ebbie. no n=1 Note: type true or false into the blank. This utility helps solve equations with respect to given variables. integral test for convergence. 50 multiple choice questions to find your result and level. Example 5: Calculate the first derivative of function f given by Solution to Example 5: Function f given above may be considered as the product of functions U = 1/x - 3 and V = (x 2 + 3)/(2x - 1), and function V may be considered as the quotient of two functions x 2 + 3 and 2x - 1. nth-Term Test. ; Many authors do not name this test or give it a shorter name. so is divergent. Testing methods for estimating KL-divergence from samples. Is the nth-term test not applicable in this case? Can you clarify?. The calculator will generate all the work with detailed explanation. If ($\\displaystyle \\lim_{n \\rightarrow \\infty} a_n eq 0$), then the series ($\\displaystyle \\sum_{n=0}^\\infty a_n$) diverges. Equations and terms. Integral Test Let X1 n=1 a nbe a series with positive terms, and let f(x) be the function that results when n is replaced by xin the formula for a n. This leaves us with an expression with nothing in the denominator. DIFFGEOM48 1. On this page, we explain how to use it and how to avoid one of the most common pitfalls associated with this test. Do not sell my info. Nearly every calculus book begins with the same example, and it\u2019s so darn fine that I will bow to peer pressure and use it as well. The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. What is important to point out is that there is an nth-term test for sequences and an nth-term test for series. In particular, the converse to the test is not true; instead all one can say is. terms tend to zero. If you have a table of values, see Riemann sum calculator for a table. They also crop up frequently in real analysis. Your project deserves the perfect stock photo. Step-by-step Solutions \u00bb Walk through homework problems step-by-step from beginning to end. MIT License. 005 when n= 2, so we must add only two terms, 1 1 33! 11. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. Find the nth term and sum of a finite geometric series, Recognize finite and infinite series, Know how to use the table and function features on a graphing calculator. Aim for a Healthy Weight: Limitations of the BMI Assessing Your Risk Controlling Your Weight Recipes. Inconclusive 2e. Integral Test Let X1 n=1 a nbe a series with positive terms, and let f(x) be the function that results when n is replaced by xin the formula for a n. Where a1is the first term, regardless of where n starts, and r is the common ration. More information. The nth term test or the Test for Divergence is used to show that the limit of the terms is not zero so the series must diverge. This nth-Term Test for Divergence states that if the limit of the nth term of a series does not converge to 0, the series must diverge. View more lessons: www. From this, we show that some distributions in the application of non-life insurance actuarial science are SPD, such as negative binomial distribution, compound Poisson distribution etc. Nth term test for divergence in hindi. M is defined as: Sequences and Word Problems- MathBitsNotebook(A1 - CCSS Math). Calculate the average of a set of numbers. 3n+5/(n*2^n) As. Is the nth-term test not applicable in this case? Can you clarify?. It's quick, free and gives an instant score. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series:. Integrate term by term (new power series is forZ f(x) dx). If the terms of an infinite series don't approach zero, the series must diverge. Then calculate it again by starting with the Taylor series for f(x) = 1 1 x and manipulating it. It is important to note that this test does not say anything about the. nth-Term Test for Divergence. This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first \"test\" we check when trying to determine whether a series converges or diverges. Recall: Divergence Test If #lim_{n to infty}a_n ne 0#, then #sum_{n=1}^{infty}a_n# diverges. Hints help you try the next step on your own. A function that calls itself is known as a recursive function. If P b n diverges, and a n b n for all n beyond a certain value, then P a n also diverges. Incorrect Test Choice ITC Student chooses an incorrect test, such as an nth term test, or a geometric test. Click the \"Calculate\" button to calculate the Student's t-critical value. Substitute the values of and in above formula. Use the NEB Tm Calculator to estimate an appropriate annealing temperature when using NEB PCR products. So in this case we would have X10 k=1 2k +1 = 3+5+7++21, and in this case the sum of the series is equal to 120. Testing for Convergence or Divergence of a Series (continued). Converges to 4 63 by Geo Series Test 2b. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. Telescoping Series Pdf. Inn Preliminary Question True Of False? Given The Series An, If N=1 Lim An = Oo Then, The Series Must Diverge. By using this website, you agree to our Cookie Policy. Calculus tutoring on Chegg Tutors Learn about Calculus terms like Tests For Convergence And Divergence on Chegg Tutors. Since 1/5 is not equal to 0, the series diverges. Derivatives. Why is my calculator giving me a huge number for sin(3. Test for convergence or divergence. Before the empirical analysis, we made a simple graph of the collected sample data from 1980 to 2015, as shown in Figs. The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. ; Many authors do not name this test or give it a shorter name. so tuis series fails the nth term test. First of all, enter the nth term. Step 3: Next, substitute the number 1 to 5 into 5n\u00b2. The infinite series converges (i. If , then the series is convergent or; 2. A summary of the Nth term test for divergence. Step-by-step Solutions \u00bb Walk through homework problems step-by-step from beginning to end. This series converges because the series. Instead, we use the nth term test for divergence. Take the limit of An as n approaches zero. 4 Comparison tests - The comparison test - The limit comparison test. The calculator will generate all the work with detailed explanation. The calculator is useful in helping students explore convergence and divergence and guess the limit of sequences. If the limit of sequence {a n} doesn\u2019t equal 0, then the series \u2211 a n is divergent. This simpli\ufb01es \ufb01nding say the 42nd term. Limit Comparison Test If lim (n-->) (a n / b n) = L, where a n, b n > 0 and L is finite and positive, then the series a n and b n either both converge or both diverge. Free Practice Tests for learners of English. The full name of the test is \"nth term test for divergence. The limit of the nth term, [n+2]/ [5n+17], is lim n\u2192\u221e n+2 5n+17 =lim n\u2192\u221e 1+ 2 n 5+ 17 n = 1 5. For this example we should use the integral test since many of the other tests seem complicated. (-1)^(n+1)*(n/n^2+4) 8. or if the limit does not exist, then. Use the nth term test to determine whether the following series converges or diverges. If nl!im\u2022 an = 0, we can't conclude anything. How old are you? A. If not, we can use the divergence test to conclude the series diverges. See full list on gauravtiwari. Maclaurin Series Taylor and Maclaurin Series interactive applet 3. For the biases: Where the sampled b corresponds to the biases used on the linear transformation for the ith layer on the nth sample. I Geometric series. nth-Term Test for Divergence. This simple t-test calculator, provides full details of the t-test calculation, including sample mean, sum of A t-test is used when you're looking at a numerical variable - for example, height - and then comparing the averages of two separate populations or. ( , , ) Leave empty, if you don't need the divergence at a specific point. The Integral Calculator lets you calculate integrals and antiderivatives of functions online \u2014 for free! The calculator lacks the mathematical intuition that is very useful for finding an antiderivative, but on the other hand it can try a large number of possibilities within a short amount of time. The limit limn\u2192\u221e sn is called the value of the series P k xk and is written as P\u221e k=1 xk Finally, P k xk. As with all risk calculators, calculated risk numbers are +/- 5% at best. I have seen so many tweets of late with people getting their heads frazzled with nth-child and nth-of-type. A summary of the Nth term test for divergence. A prerequisite of convergence for an infinite series is that its terms approach 0. geometric series, nth term test (Test for Divergence). T-Test Calculator for 2 Independent Means. Pay Close Attention to the Next Two Exercises. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If or if the limit does not exist, then diverges. It means you have to enter, for example, a3, a5, a6, etc. Thus, there are sequences that can be de\ufb01ned recursively, analytically, and those that can. Tests for convergence and divergence are methods to determine the convergence or divergence of infinite series. Return to the Series, Convergence, and Series Tests starting page. The sum is. Root Test: Suppose that the terms of the sequence in question are non-negative, and that there exists r such. Free Practice Tests for learners of English. Evaluate integral theorems of Green, Gauss & Stokes to find lines, surfaces & volumes. By using this website, you agree to our Cookie Policy. The infinite series converges (i. As with the Ratio Test this test will also tell whether a series is absolutely convergent or not rather than simple convergence. You can make the process of transfering the application to your calculator sweet and simple with Texas Instrument\u2019s handy TI connect software. Let $\\displaystyle \\sum_{n \\mathop = 1}^\\infty a_n$ be a series of real numbers $\\R$ or complex numbers $\\C$. 6 Alternating series, absolute and. The harmonic series is as follows: H 1 = 1 H 2 = H 1 + 1/2 H 3 = H 2 + 1/3 H 4 = H 3 + 1/4 H n = H n-1 + 1/n. or if the limit does not exist, then. Above is what is commonly referred to as the \u201cnth Term Test. an Dr an1 or an an1 Dr: Again, in this case it is relatively easy to \ufb01nd a formula for the nth term: an Da0rn. USED: When you suspect the terms of the given series do not approach zero. Calculator for double sums, the connections of two sums, an inner and an outer sum. Limit Comparison Test Calculator. 1A6: In addition to examining the limit of the sequence of partial sums of the series, methods for determining whether a series of numbers converges or diverges are the nth term test, the comparison test, the limit comparison test, the integral test, the ratio test, and the alternating series test. \u2022 The nth-Term Test for Divergence \u2022 The Integral Test and its relationship to improper integrals and areas of rectangles \u2022 Use of the Integral Test to introduce the test for p-series \u2022 Comparisons of series \u2022 Alternating series and the Alternating Series Remainder \u2022 The Ratio and Root Tests. \u2022 Determining convergence or divergence of sequences. This calculator demonstrates the application of the Hardy-Weinberg equations to loci with more than two alleles. If this test is inconclusive, that is, if the limit of a_n IS equal to zero (a_n=0), then you need to use another test to determine the behavior. Please choose one option for each question then click Test Result to obtain your result and level (50 questions). The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. Nth term in a geometric series So for this I have calculated what I believe to be the correct few terms of Q and P Q0=75, Q1=84, Q2=94. n n \u221e \u2211= + diverges. The n and the 2 cancel from the numerator and the denominator. It is possible to optimize our trainable weights. It says: Problems. Divergence test; Alternating series test; Integral test; Ratio test; nth root test; Evaluating a Taylor series for a particular value of x in order to calculate a. This test contains grammar and vocabulary questions and your test result will help you choose a level to practise at. If , then the series is divergent or; 3. The Nth Term Test is one of the most simple tests for convergence. On average it takes 2,200 steps to walk a mile, but this varies based on the length of a person's stride, which is determined by their height and pace. How old are you? A. A summary of the Nth term test for divergence. - Geometric series, convergent/divergent series, nth-term test for divergence - Combining series, adding or deleting terms, reindexing. So in fact every series we have written down is now seen to converge, including the one we started with. INTRODUCTION TO SEQUENCES AND SERIES, NTH TERM DIVERGENCE TEST A sequence is basically a list of numbers based on some defining rule. If you're behind a web filter, please make sure that the domains *. nth term test, divergence test, zero test (KristaKingMath). When we have the model for our data under both hypotheses we simply plug our data into Equation 1 to calculate the likelihood ratio and choose H1 if its value is positive or H0 otherwise (assuming equal priors). Roots are often written using the radical symbol \u221a, with \u221aY, denoting the square. How to use The \u2026 Continue reading \u2192. The nth term test or the Test for Divergence is used to show that the limit of the terms is not zero so the series must diverge. use the ratio test to determine which of the following series converges and which diverges X\u221e n=1 2n +5 3n, \u221e n=1 (2n)! (n!)2, X\u221e n=1 n! nn;. Ifthe test was inconclusive. (a) \u2022 \u00c2 n=1 nn nn2 (b) \u2022 \u00c2 n=0 (1)n n2 +1+2n n2 10 \" \u00a7en it \u00a7yn2 a-, because n-it E -2 fo all n 72. Home What's New Blog About Privacy Terms Popular Problems Help. Loading Found a content error?. Function Grapher and Calculator Description:: All Functions. A summary of the Nth term test for divergence. Learners analyze geometric series in detail. Use only the Divergence Test to determine if the statement is true, false, or can't be decided yet. % Progress. Then use your descrip- tion to write a formula for the nth term of each sequence. # Optionally ignore the kl divergence term. 5n\u00b2 = 5,20,45,80,125. The most common way to find the gcd is the Euclidean algorithm. Note: This test only determines the divergence of a series. How Does a Calculator Work? ~ A series that has no last. However, the equation above can be used to calculate the number of genotypes for a locus with any number alleles. Let the sequence $\\sequence {a_n}$ be such that the limit superior $\\displaystyle \\limsup_{n \\mathop \\to \\infty} \\size {a_n}^{1/n} = l$. If ($\\displaystyle \\lim_{n \\rightarrow \\infty} a_n eq 0$), then the series ($\\displaystyle \\sum_{n=0}^\\infty a_n$) diverges. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. ) (b ax Cos y 8. The limit of the nth term, [n+2]/ [5n+17], is lim n\u2192\u221e n+2 5n+17 =lim n\u2192\u221e 1+ 2 n 5+ 17 n = 1 5. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. This calculator allows test solutions to calculus exercises. Nth term test for testing the series whether it is convergent or divergent Msc mathematics Mca entrance exam du And many more. Solving slope, simplify exponential expressions, multiply and divide monomials free worksheets, complex numbers solver, linear algebra for dummies, fractions math. 64 with lots of other numbers after the decimal point. The harmonic series is as follows: H 1 = 1 H 2 = H 1 + 1/2 H 3 = H 2 + 1/3 H 4 = H 3 + 1/4 H n = H n-1 + 1/n. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. 1 Nth Term Test for Divergence. Nth term test for divergence in hindi. Note that this is only a test for divergence. Calculator of Hardy-Weinberg equilibrium. Using the formula above you can quickly calculate any nth term. As part of our spreadcheats, today we will learn how to calculate moving average using excel formulas. Recall: Divergence Test If #lim_{n to infty}a_n ne 0#, then #sum_{n=1}^{infty}a_n# diverges. The Art of Convergence Tests. If that is the case, you may conclude that the series diverges by Divergence (Nth Term) Test. By following the above equation to reach an outcome where you are effectively showing that the terms of the target series will become less than. By using this website, you agree to our Cookie Policy. geometric series, nth term test (Test for Divergence). If desired, the precise definition of limit can be carefully explained; and students may even be made to memorize it, but it should not be emphasized. Please choose one option for each question then click Test Result to obtain your result and level (50 questions). Finding The nth Term Of A Geometric Sequence. Why is my calculator giving me a huge number for sin(3. An activity to calculate the nth term of a linear sequence. The task is to find the Nth Harmonic Number. com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first \u201ctest\u201d we check when trying to determine whether a series converges or diverges. The formula for the nth term of the arithmetic sequence (also called the general term) is: A n = a + (n -1)d. We have ve tests for convergence: 1) the Divergence Test, 2) the Alternating Series Test, 3) the Ratio Test, 4) the Integral (comparison) Test, and 5) the Comparison Test. If the series is an alternating series, state whether it converges absolutely or conditionally. A summary of the Nth term test for divergence. This simpli\ufb01es \ufb01nding say the 42nd term. 3n+5/(n*2^n) As. An explanation of how to use and how not to use the nth term test for divergence. Geometric series test. How do you use the Nth term test on the infinite series \u221e \u2211 n=1 ln(2n + 1 n + 1) ? By the nth term test (Divergence Test), we can conclude that the posted series diverges. Then use your descrip- tion to write a formula for the nth term of each sequence. One test calculator to answer all your pre- and post-test analysis questions. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. Answer Save. Even if the terms are approaching zero the series could still diverge, though, hence the Harmonic Series, so this test isn't super-useful on its own. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Then its sum is. Determining Convergence or Divergence of an Infinite Series. The sum is. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. the true conversion rates (the numbers you would get if you ran the test forever) for Variation A and Variation B fall somewhere within their respective bell curves above. The next test for convergence or divergence of series works especially well for series involving powers. The task is to find the Nth Harmonic Number. Note: The Nth Term Test for Divergence is the special name given to the contrapositive of Theorem 2. com/yt/4758819/?ref=ytd. This tends to zero, showing that even if the nth term tends to zero, the series may still be divergent. This test cannot be used for convergence. Thank you for your questionnaire. Free series convergence calculator - test infinite series for convergence step-by-step This website uses cookies to ensure you get the best experience. ) (b ax Log y e 4. First Term in the Series. Let $\\displaystyle \\sum_{n \\mathop = 1}^\\infty a_n$ be a series of real numbers $\\R$ or complex numbers $\\C$. The series might converge. 2) fails to provide. Example: the sequence {3, 5, 7, 9, } starts at 3 and jumps 2 every time: As a FormulaSaying \"starts at 3 and jumps 2 every time\" is fine, but it doesnt help us calculate the: 10th term, 100th term, or nth term, where n could be any term number we want. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. The most common way to find the gcd is the Euclidean algorithm. They also crop up frequently in real analysis. ifthe series converges absolutely. We offer free personalized SAT test prep in partnership with the test developer, the College Board. If the calculator did not compute something or you have identified an error, please write it in comments below. com Test for Divergence This test, according to Wikipedia, is one of the easiest tests to apply; hence it is the first \u201ctest\u201d we check when trying to determine whether a series converges or diverges. It states that if the limit of a_n is not zero, or. There are plenty of tests to knwo the conv/div of a series. Integral Test and p-Series. Is the nth-term test not applicable in this case? Can you clarify?. The Integral Test. Sending completion. For example, Abel\u2019s Test allows you to define convergence or divergence by the types of functions contained in the series. 2b) Diverges (nth Term Test) 2c) Diverges (Integral Test) 2d) Diverges by the nth Term Test n n lim 1 n 1 \u2192 = + 2e) Converges, Geometric 2 r 1 3 = 2f) Diverges (nth Term Test) 2g) Diverges (nth Term Test) 2h) Converges (Integral Test) 3a) Converges to 3 by Telescoping 3b) Diverges by Integral Test 3c) Converges to 25 23 by Geometric Series. App Downloads. The sum of the rst n terms of an A. 1A6: In addition to examining the limit of the sequence of partial sums of the series, methods for determining whether a series of numbers converges or diverges are the nth term test, the comparison test, the limit comparison test, the integral test, the ratio test, and the alternating series test. If ($\\displaystyle \\lim_{n \\rightarrow \\infty} a_n eq 0$), then the series ($\\displaystyle \\sum_{n=0}^\\infty a_n$) diverges. If the in nite series X1 k=1 a k converges, then a k!0 as k !1. However, the equation above can be used to calculate the number of genotypes for a locus with any number alleles. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. This leaves us with an expression with nothing in the denominator. Step 2: The infinite geometric series is. Root Test: Suppose that the terms of the sequence in question are non-negative, and that there exists r such. Converges to 4 63 by Geo Series Test 2b. nth-Term Test for Divergence The contra-positive of Theorem 8 provides a useful test for divergence. The theorem then says \u2202P (4) P k \u00b7 n dS = dV. nth term test. so tuis series fails the nth term test. Online Integral Calculator \u00bb Solve integrals with Wolfram|Alpha. Limit Comparison Test Calculator. In mathematics, the nth-term test for divergence[1] is a simple test for the divergence of an infinite series When testing if a series converges or diverges, this test is often checked first due to its ease of use. Step-by-step Solutions \u00bb Walk through homework problems step-by-step from beginning to end. Proof of the divergence theorem. I hope that this was helpful. (-1)^(n+1)*(n/n^2+4) 8. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. The first term of the series is. com/yt/4758819/?ref=ytd. Free online beam calculator that calculates the reactions, deflection and draws bending moment and shear force diagrams for cantilever or simply supported beams. A summary of the Nth term test for divergence. The leading term in the inner expansion for the normal gravitational field gives the Bouguer formula. For example, the sum of the series n={1,1,1,1. Question 1 : Find the nth term of the following sequences. This is the special symbol that means \"nth root\", it is the \"radical\" symbol (used for square roots) with a little n to mean nth root. Practice using the nth term test to determine sequence divergence. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide. Example Cont. The Integral Test. The calculator does not go beyond 5 alleles and 15 possible genotypes. After reading through the example question below check out the worksheets and practice questions. So the 5-th term of a sequence starting with 1 and with a difference (step) of 2, will be: 1 + 2 x (5 - 1) = 1 + 2 x 4 = 9. Share yours for free!. The infinite series converges (i. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. In this situation the nth term test for divergence (Theorem 14. The integral test for convergence is only valid for series that are 1) Positive : all of the terms in the series are positive, 2) Decreasing : every term is less than the one before it, a_(n-1)> a_n, and 3) Continuous : the series is defined everywhere in its domain. There are divergent series whose nth terms approach 0. T-Test Calculator for 2 Independent Means. Testing convergence/divergence of a series using the nth term test for divergence, the geometric series test, and the telescoping series test. Thus, f_x(x,y) = Cx for some constant C = log y and f_y(x,y) = Clogy for some constant C = x. In this video we discuss the nth Term Test for Divergence, the conditions the test has to meet, how to perform it, and its limitations. nth term test, divergence test, zero test (KristaKingMath). Diverges because. \u2022 Determining convergence or divergence of sequences. Section 4-11 : Root Test. In other words, we do not have a definite conclusion. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If or if the limit does not exist, then diverges. Your budget deserves straightforward royalty-free pricing that lets you use an image just about anywhere, as often as you want. The following rule is a corollary of the comparison test: if. Step 4: Now, take these values (5n\u00b2) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence. The nth term is. It is difficult to explicitly calculate the sum of most alternating series, so. Get an answer for 'sum_(n=2)^oo n/ln(n) Determine the convergence or divergence of the series. Preliminary Question True of False? Given the series > an, if lim an = 0 then, the series must converge. S D \u2202z The closed surface S projects into a region R in the xy-plane. Although many documents discuss excess returns caused by medium-term momentum and long-term reversal in the US market, we empirically test this conclusion using the method of Jegadeesh and Titman. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. 9: Use Divergence Test to determine whether r 1 ln. Example: the sequence {3, 5, 7, 9, } starts at 3 and jumps 2 every time: As a FormulaSaying \"starts at 3 and jumps 2 every time\" is fine, but it doesnt help us calculate the: 10th term, 100th term, or nth term, where n could be any term number we want. App Downloads. Simplifying terms. If the in nite series X1 k=1 a k converges, then a k!0 as k !1. Integral Test Let X1 n=1 a nbe a series with positive terms, and let f(x) be the function that results when n is replaced by xin the formula for a n. To determine if the series is convergent or divergent, apply the nth-Term Test for Divergence. A prerequisite of convergence for an infinite series is that its terms approach 0. You will not be able to see the correct answers to the questions. # Every Nth epoch, save weights: # Loop across test chunks # Calculate number of batches:. Justify your answer. This test helps determine if a series diverges. (1 point) The nth term test for Divergence of Series. The nth term test for divergence The first, and usually simplest, test for divergence of a series is the nth term test. ( , , ) Leave empty, if you don't need the divergence at a specific point. nth term = dn + (a - d). We Want To Check For Divergence Of The Series N > --- And We Have Decided To Try The N=2 Nth Term Test For Divergence. This video provides two examples of how to apply the nth term divergent test to determine if a infinite series is divergent. 6, 11, 16, 21, For this sequence d = 5, a = 6. you could try to find the general formula for the Nth Opt-in alpha test for a. You can operate the calculator directly from your keyboard, as well as. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series: If \u2192 \u221e \u2260 or if the limit does not exist, then \u2211 = \u221e diverges. Also, it can identify if the sequence is arithmetic or geometric. Find the nth term of a geometric sequence Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. An activity to calculate the nth term of a linear sequence. Telescoping Series Pdf. Integral konvergence divergence. for Series TEST nth-term Geometric series p-series SERIES \u2211 an \u221e \u2211 ar (i) Converges with sum S = n \u22121 n =1 \u221e Useful for the comparison tests if the The comparison series \u2211 bn is often a geometric series of a pseries. By following the above equation to reach an outcome where you are effectively showing that the terms of the target series will become less than. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series:. , the nth term is less than 0. and decreasing to show convergence, find a larger series. Sending completion. So in this case we would have X10 k=1 2k +1 = 3+5+7++21, and in this case the sum of the series is equal to 120. Helps you plan out daily resin use by letting you know exactly what you need! Having trouble deciding between Artifacts? Use our trusty damage calculator to see which artifact has the edge!. Diverges by Geo Series Test or nth Term Test 2g. Test Review 25 multiple choice formula sheet provided Calculator active *be able to do explicit rule/nth term for both arithmetic and geometric sequences *use and create a recursive formula *find finite and infinite sums *find a specific term *use sigma notation *divergence and convergence. This utility helps solve equations with respect to given variables. with the rst term a and the common difference d is n called an arithmetic series. How Does a Calculator Work? ~ A series that has no last. S D \u2202z The closed surface S projects into a region R in the xy-plane. Only the variables i and j may occur in the sum term. ( , , ) Leave empty, if you don't need the divergence at a specific point. An activity to calculate the nth term of a linear sequence. Guidelines: Calculate cube root of 27 For example, use the square root calculator below to find the square root of 7. This is an excellent test to start with because the limit is often easy to calculate. diverges if (a) is not 0 know nothing if (a) equals 0. Many authors do not name this test or give it a shorter name. A function that calls itself is known as a recursive function. = , where ai = Then, = Therefore, by the Nth Term Test for Divergence, Exercise 6: Discuss the convergence or divergence of the series. Exercise 5: Discuss the convergence or divergence of the series. If the limit of sequence { an } doesn\u2019t equal 0, then the series \u2211 an is divergent. MIT License. If , then the series is convergent or; 2. Complete Solution. Thus, the harmonic series is a demonstration that the nth term test is a test for divergence only and cannot be used to show a series converges. fibonaciexp2. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. The list of online calculators for sequences and series. an bn cancels the dominant terms in. This test helps determine if a series diverges. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. 12 3 45 The height of each bounce is three-fourths the height of the preceding bounce. In this section, we face the problem of deciding which method to use to test a series for convergence or divergence. The infinite series often contain an infinite number of terms and its nth term represents the nth term of a sequence. It can be used to determine that a series diverges, but it doesn't tell you when a series converges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge. If that Show bx e y ax, sin 0) (2 2 1 2 y b a ay y 10. n\"' term test for Divergence Geometric Series Test p-Series Test Telesco in Series Test Inte ral Test Direct Comparison Test Limit Com rison Test Alternating Series Test Ratio Test Root Test 10. 140625)? 12 answers. For instance, the terms of. The nth-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. 3 Newton\u2019s Newton Method Nature and Nature\u2019s laws lay hid in night:. Test for Convergence v1. Using the fx-991EX and the Simul. Welcome to Clip from. Nth Term Test for Divergence - (3 Helpful Examples!) Nth term, georgia algebra 1 eoct online prep test, calculate square root fraction. Just because a(n) of a series approaches 0 does not mean it converges. See full list on gauravtiwari. In this radical calculator Y is a positive real number and x is a Nth root or radical power. Diverges because. This means our calculations are accurate and up-to-date to the practice materials shared from the test maker. Excluding the initial 1, this series is geometric with constant ratio r = 4/9. To use this calculator, enter the values in the given input boxes. Quick question: Doesn't the nth-term test for divergence state that if the limit of a function DOES NOT EXIST or is any number other than zero then it diverges? 1 and DNE both fit the criteria for divergence based from the nth-term test. Definition of Convergence and Divergence in Series. It states that if the limit of a_n is not zero, or. Then use your descrip- tion to write a formula for the nth term of each sequence. Unlike the nth term test for divergence from the last module, this module gives several tests which, if successfully applied, give a definitive answer of whether a series converges or not. terms tend to zero. We are Cambridge Assessment English. MEMORY METER. Simple Present, Present Progressive, Present Perfect, Simple Past, If-Satz Type I Level: lower intermediate. Is each term of the series positive? If so, a comparison test (page 417) may be used. M) of a series containing n observations is the nth root of the product of the values. The Secant Method, when it is working well, which is most of the time, is fast. 3n+5/(n*2^n) As. educreations. The main purpose of this calculator is to find expression for the n th term of a given sequence. You should start with a rm knowledge of each test and the ability to recall quickly the details of each test. If you're ever in doubt and would like to We made this SAT\u00ae score calculator because we saw that everyone else simply replicated the tables when creating what they called a \"calculator\". This online BMI Calculator is delivered to you at no charge and will allow you to obtain your Body Mass Index (BMI) to see whether you are underweight, of normal weight, overweight, or obese according to the classification employed by the World Health Organization (WHO). Is the first test test tests that be will lookingover. In exercise 22-28, test for convergence or divergence using each test at least once. use the ratio test to determine which of the following series converges and which diverges X\u221e n=1 2n +5 3n, \u221e n=1 (2n)! (n!)2, X\u221e n=1 n! nn;. Many authors do not name this test or give it a shorter name. = , where ai = Then, = Therefore, by the Nth Term Test for Divergence, Exercise 6: Discuss the convergence or divergence of the series. So I decided to knock this together so people can play around and test their nth's out!. If P b n diverges, and a n b n for all n beyond a certain value, then P a n also diverges. Since , the series is converges. Where: a n is the nth term of the sequence, a is the first term, d is the common difference. Here you can calculate a determinant of a matrix with complex numbers online for free with a very detailed solution. Basically if r = 1, then the ratio test fails and would require a different test to determine the convergence or divergence of the series. Converges to 4 63 by Geo Series Test 2b. The n th-term test for divergence is a very important test, as it enables you to identify lots of series as divergent. If the value of when n goes to inifinity the summation of the series will just continue to add up values, and. Get the free \"Convergence Test\" widget for your website, blog, Wordpress, Blogger, or iGoogle. It is possible to optimize our trainable weights. So the first term of the nth term is 5n\u00b2. If the limit of sequence {a n} doesn\u2019t equal 0, then the series \u2211 a n is divergent. Geometric series test. I have seen so many tweets of late with people getting their heads frazzled with nth-child and nth-of-type. Since the series diverges. The divergence of the harmonic series is proved by direct comparison with a series whose nth partial sum telescopes to the natural logarithm of n. Consider a series S a n such that a n > 0 and a n > a n+1 We can plot the points (n,a n) on a graph and construct rectangles whose bases are of length 1 and whose heights are of length a n. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. to show that this diverges should I use the $n^{th}$ term test? So far I have substituted infinity for $n$. The key idea is to apply the classical inequality x>=log(1+x) (valid for x>-1) with x=1/k and sum over k, 1<=k<=n-1. The series may or may not. Nth Root (Radical) calculator is a simple tool used to calculate the N th Root of the given real number which has the N th radical value. Test for Convergence v1. fibonaciexp2. 4 Series Tests for Convergence and Divergence. Converges by p-series Test 2f. Determining convergence or divergence of the sequence with the nth term. - Nth Term Test for Divergence - Geometric Series: Examining the ratio to determine Numerically calculate the value of the derivative of a function at a. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. Use the Standard Deviation Calculator to calculate your sample's standard deviation and mean. For example, sum_(n=1)^(infty)(-1)^n does not converge by the limit test. \ufb01nd a formula for the nth term directly. Sending completion. Abstract: Based on the probability generating function of stuttering Poisson distribution (SPD), this paper considers some equivalent propositions of SPD. this proves the P-series test. n\"' term test for Divergence Geometric Series Test p-Series Test Telesco in Series Test Inte ral Test Direct Comparison Test Limit Com rison Test Alternating Series Test Ratio Test Root Test 10. Looking at this function closely we see that f(x) presents an improper behavior at 0 and only. you could try to find the general formula for the Nth Opt-in alpha test for a. If the terms of a rather conditionally convergent series are suitably arranged, the series may be made to converge to any desirable value or even to diverge according to the. Incorrect Test Choice ITC Student chooses an incorrect test, such as an nth term test, or a geometric test. Suppose that there exists r such that. Fortunately, it\u2019s also very easy to use. If the limit of sequence {a n} doesn\u2019t equal 0, then the series \u2211 a n is divergent. Test infinite series for convergence step-by-step. One of the newer topics covered in the Maths GCSE as well as other qualifications is how to find the nth term of a quadratic sequence, but did you know there is a method of finding the nth term on the Casio Classwiz. The calculator will approximate the definite integral using the Riemann sum and sample points of your choice: left endpoints, right endpoints, midpoints, and trapezoids. According to divergent test rules, we can't conclude anything about it. In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. (1 point) Assume we are trying to determine the. To perform the test, first you need to calculate a measure known as U for each sample. Many authors do not name this test or give it a shorter name. Use the Standard Deviation Calculator to calculate your sample's standard deviation and mean. nth-Term Test for Divergence. The calculation. Step 4: Now, take these values (5n\u00b2) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence. Consider, if x 1, x 2 \u2026. A series contain terms whose order matters a lot. The Nth Term Test is one of the most simple tests for convergence. 3 The nth Term Test for Divergence. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide. 60 \u2013 75 min 75 min 75 min 75 min 60 \u2013 75 min 75 min \u2022 grid paper \u2022 scientific calculator \u2022 graphing calculator (optional) Dynamic Activity; Animation \u2022 scientific calculator \u2022 graphing calculator (optional) Animations. Always check the Nth Term Test for Divergence. If \u2192 \u221e \u2260 or if the limit does not exist, then \u2211 = \u221e diverges. Although many documents discuss excess returns caused by medium-term momentum and long-term reversal in the US market, we empirically test this conclusion using the method of Jegadeesh and Titman. If you have a table of values, see Riemann sum calculator for a table. The limit test is inconclusive when the limit is zero. com online calculator provides basic and advanced mathematical functions useful for school or college. Share yours for free!. The nth term test is only a test for divergence. Determinant is calculated by reducing a matrix to row echelon form and multiplying its main diagonal elements. It's quick, free and gives an instant score. Always check first. (2 votes). or if the limit does not exist, then. Let s n(p)be the nth partial sum of the p-series \u221e k=1 1/k p. This means our calculations are accurate and up-to-date to the practice materials shared from the test maker. Alternating Series Test - Proof. Using the nth-Term Test for Divergence 613 The series in Example STUDY TIP 5(c) will play an important role in this chapter. so Etfe converges. The test that we are going to look into in this section will be a test for alternating series. This video explains how to apply the nth term divergence test to an infinite series. Find more Mathematics widgets in Wolfram|Alpha. Also, it can identify if the sequence is arithmetic or geometric. If a series converges, its nth term approaches 0. com/yt/4758819/?ref=ytd. (n^10 + 10)/n! 7. ParticipantsConversions. Equations and terms. 4 Comparison tests: the comparison test, limit comparison test. Suppose that the in nite series X1 k=1 a k converges. A function that calls itself is known as a recursive function. The Integral Test. Test for Divergence for Series, Two Examples. If , then the series is divergent or; 3. Nth Term Test. Learners analyze geometric series in detail. The ratio test is especially useful, but the integral test is one i dread to use. By the nth term test (Divergence Test), we can conclude that the posted series diverges. an Dr an1 or an an1 Dr: Again, in this case it is relatively easy to \ufb01nd a formula for the nth term: an Da0rn. 1 Equation 2 Use 3 Explanation 4 Video Explanation Calculate the limit as stated above. The n and the 2 cancel from the numerator and the denominator. Example Cont. For example, students can be. n=1 n 2 + n. To each series we associate the sequence of nth terms of the series and also the sequence of partial sums of the series. \u2022 The nth-term test for divergence \u2022 The integral test and its relationship to improper integrals and areas of rectangles \u2022 Use of the integral test to introduce the test for p-series \u2022 Comparisons of series \u2022 Alternating series and the alternating series remainder \u2022 The ratio and root tests \u2022 Taylor polynomials and approximations. Nth term test for testing the series whether it is convergent or divergent Msc mathematics Mca entrance exam du And many more. First Term in the Series. If the th Term Divergence Test is inconclusive, the next step is to examine the corresponding series. This is used by this calculator. # Every Nth epoch, save weights: # Loop across test chunks # Calculate number of batches:. Function Grapher and Calculator Description:: All Functions. Use the NEB Tm Calculator to estimate an appropriate annealing temperature when using NEB PCR products. terms tend to zero. We use the product rule for f and the quotient rule for V as. Usually we need about 45 percent more iterations than with the Newton Method to get the same accuracy, but each iteration is cheaper. pdf Due December 2: Page 515: 3,4,24, 26, 49,50 Due December 4: Page 528: 7-12;14-18 (use LCT, DCT, nth term lor geometric series). Identify the test used. The calculation. 2b) Diverges (nth Term Test) 2c) Diverges (Integral Test) 2d) Diverges by the nth Term Test n n lim 1 n 1 \u2192 = + 2e) Converges, Geometric 2 r 1 3 = 2f) Diverges (nth Term Test) 2g) Diverges (nth Term Test) 2h) Converges (Integral Test) 3a) Converges to 3 by Telescoping 3b) Diverges by Integral Test 3c) Converges to 25 23 by Geometric Series. Nth Term Test For Convergence This test basically tells you that if your terms aren't approaching zero, there's no way the series converges.", "date": "2021-03-02 13:46:01", "meta": {"domain": "schuetzen-wuerm.de", "url": "http://mdnu.schuetzen-wuerm.de/nth-term-test-for-divergence-calculator.html", "openwebmath_score": 0.7924398183822632, "openwebmath_perplexity": 572.843145363113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754456551737, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8536621196061944}}
{"url": "https://math.stackexchange.com/questions/3261187/what-is-dfraci4-%CF%80-int-z-4-dfracdzz-cos-z", "text": "What is $\\dfrac{i}{4-\u03c0} \\int_{|z|=4} \\dfrac{dz}{z \\cos z}$?\n\nHow to evaluate the integral $$\\dfrac{i}{4-\u03c0} \\int_{|z|=4} \\dfrac{dz}{z \\cos z}?$$\n\nHere $$f(z)=\\dfrac{1}{z \\cos z}$$ has poles at $$0$$ and $$\\frac{\\pm \u03c0}{2}$$ .\n\nResidue at $$z=0$$ is 1 and residues at remaining poles add up to give 0. So the integral using Cauchy integral formula is $$2\u03c0(4-\u03c0)$$.\n\nI think I am wrong. How to get the integral?\n\nIt's not true that residues add up to $$0$$. We have $$\\cos z = \\sin(\\frac{\\pi}{2}-z) = \\sin(\\frac{\\pi}{2}+z)$$ so $${\\rm Res}_{z=0} \\frac{1}{z\\cos z} = \\lim_{z\\rightarrow 0}\\frac{1}{\\cos z} = 1$$ $${\\rm Res}_{z=\\frac{\\pi}{2}} \\frac{1}{z\\cos z} = \\lim_{z\\rightarrow \\frac{\\pi}{2}}\\frac{(z-\\frac{\\pi}{2})}{z \\sin(\\frac{\\pi}{2}-z)} = -\\frac2\\pi$$ $${\\rm Res}_{z=-\\frac{\\pi}{2}} \\frac{1}{z\\cos z} = \\lim_{z\\rightarrow -\\frac{\\pi}{2}}\\frac{(z+\\frac{\\pi}{2})}{z \\sin(\\frac{\\pi}{2}+z)} = -\\frac2\\pi$$ and $$\\sum {\\rm Res} = 1 - \\frac{4}{\\pi} = \\frac{\\pi-4}{\\pi}$$ so the final result is $$\\frac{i}{4-\\pi}\\cdot 2\\pi i\\frac{\\pi-4}{\\pi} = 2$$\n\nYou made a mistake in your residues. The residues of $$1/(z\\cos z)$$ at $$z=\\pm\\frac\\pi2$$ are both $$-\\frac2\\pi$$.\n\n\u2022 ohh...yes....thank you Jun 13 '19 at 16:20\n\nThe function $$f(z)=\\frac{1}{z\\cos(z)}$$ has simple poles at $$z=0$$ and $$z= (2n-1)\\pi/2$$ for $$n\\in \\mathbb{Z}$$. The poles that are inside the circle $$|z|=4$$ are at $$z=0$$ and $$z=\\pm \\pi/2$$.\n\nThe residues of $$f$$ at the implicated poles are\n\n\\begin{align} \\text{Res}\\left(\\frac{1}{z\\cos(z)}, z=0\\right)&=1\\\\\\\\ \\text{Res}\\left(\\frac{1}{z\\cos(z)}, z=\\pi/2\\right)&=-\\frac2\\pi\\\\\\\\ \\text{Res}\\left(\\frac{1}{z\\cos(z)}, z=-\\pi/2\\right)&=-\\frac2\\pi \\end{align}\n\nTherefore we have\n\n$$\\frac{i}{4-\\pi}\\oint_{|z|=4}f(z)\\,dz=\\frac{i}{4-\\pi}\\times 2\\pi i\\times\\left(1-\\frac{4}{\\pi}\\right)=2$$", "date": "2021-10-21 07:00:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3261187/what-is-dfraci4-%CF%80-int-z-4-dfracdzz-cos-z", "openwebmath_score": 0.9965147972106934, "openwebmath_perplexity": 218.5151931536282, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754504074422, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8536621135783057}}
{"url": "http://math.stackexchange.com/questions/4261/proving-a-binomial-sum-identity/247057", "text": "# Proving a binomial sum identity\n\nMathematica tells me that\n\n$$\\sum _{k=0}^n { n \\choose k} \\frac{(-1)^k}{2k+1} = \\frac{(2n)!!}{(2n+1)!!}.$$\n\nAlthough I have not been able to come up with a proof.\n\nProofs, hints, or references are all welcome.\n\n-\n\nYou could consider the integral $$\\int_{0}^{1} (1-x^2)^n dx .$$\n\n-\nAha this is the integral I was searching for! Using the integral $\\int_0^1 (1-t)^n \\, dt$ I was able to prove $\\sum _{k=0}^n {n \\choose k}\\frac{(-1)^k}{k+1} = \\frac{1}{n+1}$ I thought there should be a corresponding integral to my problem and after seeing your answer the integral I was searching for is obvious now! \u2013\u00a0 yjj Sep 8 '10 at 11:02\n\nSums of the form $$\\sum_{k=0}^n(-1)^k{n\\choose k}f(k)$$ can often be attacked via the Calculus of finite differences.\n\n-\n\n$$S_n=\\sum\\limits_{k=0}^n \\dfrac{(-1)^k \\binom{n}{k}}{2k+1}$$\n\nWe have:\n\n\\begin{align}S_n&=\\dfrac{(-1)^n}{2n+1}+\\sum\\limits_{k=0}^{n-1} \\dfrac{(-1)^k \\binom{n}{k}}{2k+1}\\\\ &=\\dfrac{(-1)^n}{2n+1}+\\sum\\limits_{k=0}^{n-1}\\left[\\dfrac{n}{n-k}.\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}\\right]\\\\ &=\\dfrac{(-1)^n}{2n+1}+\\dfrac{1}{2n+1}\\sum\\limits_{k=0}^{n-1}\\left[\\dfrac{n(2n+1)}{n-k}.\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}\\right]\\\\ &=\\dfrac{(-1)^n}{2n+1}+\\dfrac{1}{2n+1}\\sum\\limits_{k=0}^{n-1}\\left[\\dfrac{2n^2-2nk+2nk+n}{n-k}.\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}\\right]\\\\ &=\\dfrac{(-1)^n}{2n+1}+\\dfrac{1}{2n+1}\\sum\\limits_{k=0}^{n-1}\\left[\\dfrac{2n^2-2nk}{n-k}.\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}\\right]\\\\&+\\dfrac{1}{2n+1}\\sum\\limits_{k=0}^{n-1}\\left[\\dfrac{2nk+n}{n-k}.\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}\\right]\\\\ &=\\dfrac{(-1)^n}{2n+1}+\\dfrac{2n}{2n+1}\\sum\\limits_{k=0}^{n-1}\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}+\\dfrac{1}{2n+1}\\sum\\limits_{k=0}^{n-1} \\dfrac{n(-1)^k \\binom{n-1}{k}}{n-k}\\\\ &=\\dfrac{2n}{2n+1}\\sum\\limits_{k=0}^{n-1}\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}+\\dfrac{1}{2n+1}\\sum\\limits_{k=0}^{n-1} \\left[(-1)^k \\binom{n}{k}\\right]+\\dfrac{(-1)^n}{2n+1}\\\\ &=\\dfrac{2n}{2n+1}\\sum\\limits_{k=0}^{n-1}\\dfrac{(-1)^k \\binom{n-1}{k}}{(2k+1)}+\\dfrac{1}{2n+1}\\sum\\limits_{k=0}^n \\left[(-1)^k \\binom{n}{k}\\right]\\end{align} Therefore, $$S_n=\\dfrac{2n}{2n+1}S_{n-1}+0 \\Rightarrow S_{n-1}=\\dfrac{2n-2}{2n-1}S_{n-2} ... \\Rightarrow S_1=\\dfrac{2}{3}S_0$$ and $S_0=1$\n\nHence, $$S_n=\\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\\dfrac{(2n)!!}{(2n+1)!!}$$\n\n-\nAs stated, you divide by zero since you sum to $k=n$ and $n-k$ is in the denominator. \u2013\u00a0 Julian Kuelshammer Nov 29 '12 at 7:36\nWah! Thanks for pointing out errors in my solution. \u2013\u00a0 hxthanh Dec 1 '12 at 12:12\nFix this problem as follows sum for $k=0$ to $n-1$ and $k=n$ is calculated separately \u2013\u00a0 hxthanh Dec 1 '12 at 12:34\nYou can edit your post accordingly. \u2013\u00a0 Julian Kuelshammer Dec 1 '12 at 12:37\n\nYour identity is a special case of the Chu-Vandermonde identity.\n\n${}_2 F_1(-n,b;c;1)=\\frac{(c-b)_n}{(c)_n}$\n\nwith $b=\\frac12$ and $c=\\frac32$. More info on it is in A=B, as mentioned already by John.\n\n-\nIt's on page 181. \u2013\u00a0 \uff2a. \uff2d. Sep 8 '10 at 11:05\n\nI can't offer specific help, but I'd recommend thumbing through Concrete Mathematics looking for techniques. It has many sums that look similar, though of course the difficulty is in the details.\n\nThere's also the book A=B, but Concrete Mathematics gives an introduction to the content of A=B and in my opinion is easier to read, so I'd start with Concrete Mathematics.\n\n-\n\nThere are a powerful algorithms generalizing telescopy (Gosper, Zeilberger et al.) that easily tackle this case of the Chu-Vandermonde identity and much more complicated sums. For example, see this paper which gives as an application a very interesting q-analogy - namely that L. J. Rogers' classical finite version of Euler's pentagonal number theorem is simply the dual of a special case of a q-Chu-Vandermonde.\n\n-", "date": "2014-10-20 09:41:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/4261/proving-a-binomial-sum-identity/247057", "openwebmath_score": 0.9516960382461548, "openwebmath_perplexity": 1628.5546507729061, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754499548452, "lm_q2_score": 0.8670357460591569, "lm_q1q2_score": 0.8536621098031294}}
{"url": "https://math.stackexchange.com/questions/1822068/how-many-permutations-of-1-2-3-n-there-are-with-no-2-consecutive-numbe", "text": "# How many permutations of $\\{1,2,3,\u2026,n\\}$ there are with no 2 consecutive numbers?\n\nHow many permutations of $$\\{1,2,3,...,n\\}$$ are there with no 2 consecutive numbers?\n\nFor example:\n$$n=4$$, $$2143$$, $$3214$$, $$1324$$ are the permutations we look for and $$1234$$, $$1243$$, $$2134$$ are what we DON'T look for.\n\nMy solution:\n\nwe will sub from all the permutations the bads. All permutations= $$n!$$\n\nbads:\n\nLets define $$A_i$$ to be a group of all permutations containing $$i,i+1$$ in it. ($$1\\leq i\\leq n-1$$)\n\nMy problem is to count to group of :$$|A_i \\cap A_j|$$ where $$1 \\leq i < j \\leq n-1$$.\n\nBecause there is a big difference between $$A_i \\cap A_{i+1}$$ and $$A_i \\cap A_{i+5}$$ (for instance), we can't say that both are the same. So we need to divide $$|A_i \\cap A_j|$$ to 2 options.\n\n1. when $$i+1 = j$$ then $$|A_i \\cap A_j|=(n-2)!$$\n2. when $$i+1 < j$$ then $$|A_i \\cap A_j|=(n-2)!$$\n\nSo $$|A_i \\cap A_j|=(n-2)!+(n-2)!$$\n\nThis is a big mistake but I have no clue why.\n\n(the rest of the solution is the same).\n\nAny help would be welcome.\n\nStav\n\n\u2022 It makes no sense to add your two figures of $(n-2)!$ together: the first applies only when $i+1=j$, and the second applies only when $i+1<j$, so they never both to the same pair of $i$ and $j$. \u2013\u00a0Brian M. Scott Jun 11 '16 at 19:07\n\u2022 @BrianM.Scott Studying the subject and wasn't sure if the following statement was true.$N(A_iA_j)$ was same for all $i\\neq j$(even if $i+1=j$) by calculation through inclusion principle. thus there was no contradiction? \u2013\u00a0user416486 Apr 23 '18 at 15:49\n\n## 3 Answers\n\nIf, as appears to be the case, you\u2019re planning to use an inclusion-exclusion argument, you\u2019ll need $\\left|\\bigcap_{k\\in I}A_k\\right|$ for every non-empty $I\\subseteq[n-1]$. The trick is to realize that no matter how the integers in $I$ are spaced, the cardinality is going to be the same. When $|I|=2$, the case that you\u2019re discussing in the question, there are two possibilities: the two members of $I$ are consecutive, or they are not. But in both cases it turns out that the cardinality of the intersection is $(n-2)!$: there are different calculations for the two cases, but they lead to the same result. Note that there is no reason at all to add these calculations: no $I$ belongs to both cases.\n\nThe set $I$ can be divided into blocks of consecutive integers. For example, $I=\\{2,3,5,6,7,9\\}$ has $3$ blocks: $\\{2,3\\},\\{5,6,7\\}$, and $\\{9\\}$. Suppose that $I$ has a block $\\{k,k+1,\\ldots,k+r\\}$. Then every permutation in $\\bigcap_{k\\in I}A_k$ must contain the subsequence $\\langle k,k+1,\\ldots,k+r,k+r+1\\rangle$; call this an extended block. If $I$ has $b$ blocks altogether, each permutation of $[n]$ that belongs to $\\bigcap_{k\\in I}A_k$ must contain all $b$ extended blocks as subsequences, but it can contain them in any order. Those extended blocks contain altogether $|I|+b$ integers: the blocks themselves contain $|I|$ integers, and each extended block has one extra integer on the righthand end. That leaves $n-|I|-b$ members of $[n]$ that can be permuted arbitrarily, because they aren\u2019t in any extended block of $I$. Thus, the permutations in $\\bigcap_{k\\in I}A_k$ are really permutations of\n\n$$b+(n-|I|-b)=n-|I|$$\n\nthings: $b$ extended blocks, and the $n-|I|-b$ single elements that are not part of any extended block. It follows that\n\n$$\\left|\\bigcap_{k\\in I}A_k\\right|=(n-|I|)!$$\n\nwhenever $\\varnothing\\ne I\\subseteq[n-1]$.\n\nNow the inclusion-exclusion principle tells you that\n\n\\begin{align*} \\left|\\bigcup_{k\\in[n-1]}A_k\\right|&=\\sum_{\\varnothing\\ne I\\subseteq[n-1]}(-1)^{|I|-1}\\left|\\bigcap_{k\\in I}A_k\\right|\\\\ &=\\sum_{\\varnothing\\ne I\\subseteq[n-1]}(-1)^{|I|-1}(n-|I|)!\\\\ &=\\sum_{k=1}^{n-1}\\binom{n-1}k(-1)^{k-1}(n-k)!\\;, \\end{align*}\n\nand I\u2019ll leave the rest for you to finish off.\n\n\u2022 First of all thank you very much for your time. Secondly Im having hard time to undestand how I={2,3,5,6,7,9}I={2,3,5,6,7,9} has 3 blocks: {1,2},{5,6,7}{1,2},{5,6,7}, and {9}. how and why did you choose those blcoks specifically? what does a block mean? \u2013\u00a0Stav Alfi Jun 11 '16 at 19:53\n\u2022 @Stav: By block I simply mean a maximal set of consecutive integers. In the set $\\{2,3,5,6,7,9\\}$ the sets $\\{2,3\\},\\{5,6,7\\}$, and $\\{9\\}$ are the longest strings of consecutive integers, so they\u2019re the blocks. The blocks of $\\{1,5,6,7,8,12,13,14,20\\}$ are $\\{1\\},\\{5,6,7,8\\},\\{12,13,14\\}$, and $\\{20\\}$: in each block the numbers are consecutive, and the blocks are as long as possible. Note that in the case $|I|=2$, your two cases correspond to one block, when $i+1=j$, and two blocks, when $i+1<j$. \u2013\u00a0Brian M. Scott Jun 11 '16 at 19:58\n\u2022 No, Thank you! I could not get a better answer! The second paragraph provided the answer to my question in the best way. Combining all was perfectlly clear. Thank you agian and good night from isreal! :) \u2013\u00a0Stav Alfi Jun 11 '16 at 20:24\n\u2022 @Stav: You\u2019re welcome! (And have a good night.) \u2013\u00a0Brian M. Scott Jun 11 '16 at 20:26\n\u2022 @MarkoRiedel I think Brian's expression counts the permutations that do have 2 consecutive numbers. If you take $n!$ minus his sum, it reproduces your sequence. \u2013\u00a0David Zhang Jun 11 '16 at 22:05\n\nConsider the following recurrence: the desired count $Q_n$ is $1$ for $n=1$ and $1$ for $n=2.$ For $n\\gt 2$ we obtain an admissible permutation either by placing the value $n$ anywhere at $n-1$ possible positions of an admissible permutation from $Q_{n-1}$ (this is not $n$ because we may not place $n$ next and to the right of $n-1$) or we construct a permutation having exactly one pair of consecutive numbers and place the value $n$ between these two. This can be done by taking a permutation from $Q_{n-2}$ and replacing one of the $n-2$ values by a fused pair containing the value and its successor and incrementing the values that are larger than the first element of the fused pair.\n\nWe get the recurrence\n\n$$Q_n = (n-1) Q_{n-1} + (n-2) Q_{n-2}$$\n\nand $Q_1= Q_2= 1.$ This yields the sequence\n\n$$1, 3, 11, 53, 309, 2119, 16687, 148329, 1468457, 16019531, 190899411, \\\\ 2467007773, 34361893981, 513137616783,\\ldots$$\n\nwhich is OEIS A000255, where a detailed entry may be found.\n\nThe Maple code for this was as follows.\n\nwith(combinat);\n\nC :=\nproc(n)\noption remember;\nlocal perm, pos, res;\n\nres := 0;\nperm := firstperm(n);\n\nwhile type(perm, list) do\nfor pos to n-1 do\nif perm[pos] + 1 = perm[pos+1] then\nbreak;\nfi;\nod;\n\nif pos = n then\nres := res + 1;\nfi;\n\nperm := nextperm(perm);\nod;\n\nres;\nend;\n\nQ :=\nproc(n)\noption remember;\n\nif n = 1 or n = 2 then return 1 end if;\n\n(n - 1)*Q(n - 1) + (n - 2)*Q(n - 2)\nend;\n\n\nAddendum. Here is my perspective on the inclusion-exclusion approach. We take as the underlying partially ordered set the set $P$ of subsets (these are the nodes of the poset) of $\\{1,2,\\ldots,n-1\\}$ where a subset $S\\in P$ represents permutations where the elements of $S$ are next to their successors, plus possibly some other elements also next to their successors. The partially ordered set is ordered by set inclusion. To compute the cardinality of the permutations corresponding to $S$ suppose that the elements of $S$ listed in order form $m$ blocks. We first remove these from $[n].$ We must also remove the elements that are consecutive with the rightmost element of each block, so we have now removed $|S|+m$ elements. We then put the augmented and fused blocks back into the permutation and permute them. We have added in $m$ blocks, therefore the net change is $-|S|-m +m = -|S|.$ Hence by inclusion-exclusion we compute the quantity\n\n$$\\sum_{S\\in P, S\\ne\\emptyset} (-1)^{|S|} (n-|S|)!.$$\n\nNow this depends only on the number $q$ of elements in $S$ so we get\n\n$$\\sum_{q=1}^{n-1} {n-1\\choose q} (-1)^q (n-q)!.$$\n\nWe must now ask about the weight assigned to a permutation with $p$ elements (call this set $T$) next to their successor. This permutation is included in or rather represented by all sets $S$ that are subsets of the set $T$, which is the poset spanned by the singletons and $T$ being the topmost node. We obtain\n\n$$\\sum_{q=1}^p (-1)^q {p\\choose q} = -1 + (1-1)^p = -1.$$\n\nThe count assigns the weight minus one to the permutation. It follows that when we add $n!$ exactly those permutations remain that do not contain consecutive adjacent values, for a result of\n\n$$n! + \\sum_{q=1}^{n-1} {n-1\\choose q} (-1)^q (n-q)!.$$\n\nWe may simplify this to\n\n$$\\sum_{q=0}^{n-1} {n-1\\choose q} (-1)^q (n-q)!.$$\n\nIn both cases, there are $(n-2)!$ permutations.\nIn the first case, there is the triple $(i,i+1,i+2)$ and $n-3$ other numbers.\nIn the second case, there are two pairs, and $n-4$ other numbers.\n\n\u2022 You say that becouse in both cases we will permutate n-2 numbers , the answer should be (n-2)!. but my quastion is why you neglect the big deffrence between those 2 cases? \u2013\u00a0Stav Alfi Jun 11 '16 at 14:25\n\u2022 They are different; but they both have the same number of permutations. \u2013\u00a0Empy2 Jun 11 '16 at 14:27\n\u2022 Then how do we know when to sum 2 options instead of taking only one as an answer like now? \u2013\u00a0Stav Alfi Jun 11 '16 at 14:28\n\u2022 @Stav: For each fixed pair of $i$ and $j$ with $1\\le i<j\\le n-1$ there are $(n-2)!$ permutations in $A_i\\cap A_j$; the reasoning explaining this is different when $j=i+1$ from when $j>i+1$, but in both cases the result of the reasoning is $(n-2)!$. \u2013\u00a0Brian M. Scott Jun 11 '16 at 19:06\n\u2022 Thank you for your comment. Just to clarify the rest of the prove, I see that |Ai\u2229Aj| for all cases is (n-2)! but very hard to guess for the |Ai1\u2229Ai2\u2229....Aik|=(n-k)!, how do you know that with out proving it? \u2013\u00a0Stav Alfi Jun 11 '16 at 19:21", "date": "2019-07-15 17:58:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1822068/how-many-permutations-of-1-2-3-n-there-are-with-no-2-consecutive-numbe", "openwebmath_score": 0.7577570676803589, "openwebmath_perplexity": 317.7345540806825, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307739782681, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.8536364223139234}}
{"url": "https://math.stackexchange.com/questions/3991510/converting-n-is-an-odd-negative-integer-into-formal-logic/3991560", "text": "# Converting \u201cn is an odd negative integer\u201d Into Formal Logic\n\nI am learning how to think about mathematical truth at the level of formal logic and I am tasked with converting the following basic statement into something formal: \"n is a negative integer that is odd\"\n\nMy first attempt at this was the following: Let $$n \\in \\mathbb{Z}$$ such that $$2n + 1 < 0$$\n\nFrom my point of view this statement is saying that \"n is an integer and it's values are restricted by the inequality $$2n + 1 < 0$$\". But how this actually translated to a mathematician was \"$$n$$ is an integer satisfying the inequality $$2n + 1<0$$\" and after some thought, I concluded my initial statement was indeed wrong.\n\nSo I went back to the drawing board and came up with this:\n\n$$\\exists k \\in \\mathbb{N} \\quad n = -2k + 1$$\n\nBut I feel that this formulation is still missing something, and I just can't put my finger on it. Should something be said about $$n$$ as well? Another way I thought of writing this was:\n\n$$\\exists k \\in \\mathbb{N} \\quad (n = -2k + 1) \\Rightarrow n \\in \\mathbb{Z}$$\n\n$$\\exists n \\in \\mathbb{Z} \\quad \\exists k \\in \\mathbb{N} \\quad n = -2k + 1$$ Any help thinking through this would be appreciated.\n\n\u2022 \u201cn is an odd negative integer\u201d is a predicate; thus, the formula must be of form $P(n)$ with $n$ free. \u2013\u00a0Mauro ALLEGRANZA Jan 19 at 14:41\n\u2022 @MauroALLEGRANZA Could you perhaps provide an example demonstrating what you mean? It doesn't need to be related precisely to my question, but just to give a bit more insight. I understand that $P(n)$ is a statement about $n$, but I am missing what you mean by \"$n$ free\". \u2013\u00a0GrayLiterature Jan 19 at 14:48\n\u2022 $\\text {OddNegInt}(n) \\leftrightarrow [n \\in \\mathbb Z \\land (n < 0) \\land \\exists k \\in \\mathbb N (n=2k+1)]$ \u2013\u00a0Mauro ALLEGRANZA Jan 19 at 14:56\n\u2022 @MauroALLEGRANZA I see now. \"N is an odd negative integer\" can be translated as \"n is an integer AND n is odd AND n is negative\" \u2013\u00a0GrayLiterature Jan 19 at 15:02\n\u2022 @MauroALLEGRANZA Although, I think that the last part in your formulation for $k$ would require that $k \\in \\mathbb{Z}$. Since if $n = 2k + 1$ then n is strictly positive for any $k \\in \\mathbb{N}$. Would you agree with this observation? \u2013\u00a0GrayLiterature Jan 19 at 15:15\n\nWe can write \"$$n$$ is a negative integer\" as $$n\\in\\mathbb{Z}^-$$ And \"$$n$$ is odd\" as $$\\exists k\\in\\mathbb{Z}.\\ n=2k-1$$ Combine both: $$\\exists k\\in\\mathbb{Z}.\\ n=2k-1 \\wedge n\\in\\mathbb{Z}^-$$ Or alternatively, $$\\exists k\\in\\mathbb{Z}.\\ \\mathbb{Z}^-\\ni n=2k-1$$\n\nHope this helps. :)\n\n\u2022 No, $\\mathbb{Z}$ also contains all positive integers. \"$n$ is a negative integer\" is writen $n \\in \\mathbb{Z} \\wedge n < 0$. \u2013\u00a0Olivier Roche Jan 19 at 15:29\n\u2022 @OlivierRoche: As far as the popular convention says, $\\mathbb {Z}^+$ is the set of positive integers, and $\\mathbb {Z}^-$ is the set of negative integers. \u2013\u00a0ultralegend5385 Jan 19 at 23:40\n\u2022 Oh sorry I didn't see the superscript! \u2013\u00a0Olivier Roche Jan 20 at 0:50\n\nThis answer follows the comment provided by Mauro ALLEGRANZA in the comments:\n\nThe statement \"$$n$$ is an odd negative integer\" can be translated to \"$$n$$ is an integer AND $$n$$ is odd AND $$n$$ is negative\". Converting this to formal logic we can write:\n\n\u2022 \"$$n$$ is an integer\" == $$n \\in \\mathbb{Z}$$\n\u2022 \"$$n$$ is negative\" == $$n < 0$$\n\u2022 \"$$n$$ is odd\" == $$\\exists k \\in \\mathbb{Z} \\quad (n = 2k +1)$$ (edit : if one takes $$k \\in \\mathbb{N}$$, then $$n > 0$$)\n\nand combining these statements about $$n$$:\n\n\u2022 \"n is an odd negative integer\" == ($$n \\in \\mathbb{Z}) \\quad \\land \\quad (n<0) \\quad \\land \\quad \\exists k \\in \\mathbb{Z} \\thinspace (n = 2k +1)$$\n\nedit : Since $$\\exists k \\in \\mathbb{Z} \\thinspace (n = 2k +1)$$ already implies that $$n \\in \\mathbb{Z}$$, one can omit it and state :\n\n$$(n<0) \\quad \\land \\quad \\exists k \\in \\mathbb{Z} \\thinspace (n = 2k +1)$$\n\n\u2022 Thank you for the edit. \u2013\u00a0GrayLiterature Jan 19 at 16:01", "date": "2021-03-02 15:13:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3991510/converting-n-is-an-odd-negative-integer-into-formal-logic/3991560", "openwebmath_score": 0.7230828404426575, "openwebmath_perplexity": 415.98921293151824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307684643189, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8536364143585279}}
{"url": "https://web.sas.upenn.edu/ancoop/2018/10/10/the-elements-eye-view/", "text": "## The Element\u2019s-Eye View\n\nThere are three things to keep in mind when it comes to set-theory proofs (which most of our proofs hereafter will be):\n\n1. Sets are defined by their membership. Because of this, we nearly always take an element\u2019s-eye view of any proof involving sets. Notice that in the proofs we presented above, the first thing we did to get going was say something like \u201cLet $x\\in$. . .\u201d.\n2. Generally, the way to go to prove two sets are equal is to show (separately) that each is a subset of the other.\n3. Keep in mind what your goal is. That is, pay attention to the consequent. Your goal is to translate from the language of the consequent into the language of the antecedent.\n\nThese three principles are not only good advice for writing a proof, but they are in fact good advice for {\\em figuring out} a proof as well. The following examples will illustrate principles 1 and 3.\n\nTheorem. If $A$ and $B$ are sets with $2^A\\subseteq 2^B$, then $A\\subseteq B$.\n\nProof. Let $A$ and $B$ be set with $2^A\\subseteq 2^B$. Since we have to show $A\\subseteq B$, let $x\\in A$. Then $\\{x\\}\\subseteq A$, so $\\{x\\}\\in 2^A$. Since $2^A\\subseteq 2^B$, we have $\\{x\\}\\in 2^B$. But this means $\\{x\\}\\subseteq B$. Since $x\\in\\{x\\}$, we see that $x\\in B$.\n\nSo we\u2019ve shown every $x\\in A$ also has $x\\in B$, i.e. $A\\subseteq B$. $\\Box$\n\nTheorem. If $A$ and $B$ are sets with $A\\subseteq B$, then $2^A\\subseteq 2^B$.\n\nProof. Let $A$ and $B$ be sets with $A\\subseteq B$. Since we have to show $2^A\\subseteq 2^B$, let $x\\in 2^A$. Then $x\\subseteq A$. Since $A\\subseteq B$, and set inclusion is transitive, we have $x\\subseteq B$. But this is exactly the statement that $x\\in 2^B$.\n\nSo we\u2019ve show that every $x\\in 2^A$ also has $x\\in 2^B$, i.e. $2^A\\subseteq 2^B$. $\\Box$\n\nNotice that in both cases, we were concerned with the {\\em consequent} of the conditional we\u2019d be asked to show (principle III), and that we proceeded by following elements (principle I). In the first proof, we translated from the language of elements of $A$ and $B$ to the language of subsets of $A$ and $B$ to the language of elements of $2^A$ and $2^B$, applied our assumption, and translated back. In the second proof, we translated from the language of elements of $2^A$ and $2^B$ to the language of subsets of $A$ and $B$, applied our assumptions, and translated back.\n\nActivity Consider the following couplet:\n\nHeffalumps and woozles are very confuzle.\nA heffalump or woozle\u2019s very sly.\n\nExplain why both lines are talking about the same thing as follows. First, give a mathematical name to what they describe. Then explain why each line takes a different perspective on that mathematical object, and why those perspectives are not contradictory, the difference between \u201cand\u201d and \u201cor\u201d notwithstanding.\n\n## Meta Data\n\nTitle: The Element\u2019s-Eye View\nDate Posted: October 10, 2018\nPosted By:\nCategory: sets\nTags:", "date": "2021-12-01 05:54:33", "meta": {"domain": "upenn.edu", "url": "https://web.sas.upenn.edu/ancoop/2018/10/10/the-elements-eye-view/", "openwebmath_score": 0.9463254809379578, "openwebmath_perplexity": 270.03013060427395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98866824713641, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8536078765755579}}
{"url": "https://puzzling.stackexchange.com/questions/60536/largest-odd-factors-summing-to-a-square", "text": "Largest odd factors summing to a square\n\nI just found this awesome puzzle from the Tournament of the Towns (though I'm sure it's appeared other places too). The connection between odd factors and square is surprising, and the proof has a lovely 'aha' moment. Enjoy!\n\nFor any natural number $x$, prove that $$\\sum_{y=x+1}^{2x}(\\text{largest odd factor of y})=x^2.$$\n\n\u2022 I'm never around when the good math puzzles are posted :( \u2013\u00a0Quintec Feb 14 '18 at 1:00\n\u2022 dude same :( aiyaaa \u2013\u00a0NL628 Feb 14 '18 at 4:42\n\nLet $H(x) = (\\text{largest odd factor of x})$. We can make two observations about this function:\n\n1. $H(x) = H(\\frac{x}{2})$, if x is even\n2. $H(x) = x$, if x is odd\n\nLet $$F(x) = \\sum_{y=x+1}^{2x}H(x).$$\n\n$F(1) = 1 = 1^2$. By induction, if $F(x) = x^2$, then\n\n\\begin{align*} F(x+1) &= F(x) + H(2x+2) + H(2x+1) - H(x+1) \\\\&= F(x) + H(x+1) + (2x+1) - H(x+1) \\\\&= F(x) + 2x+1 \\\\&= x^2 + 2x + 1 \\\\&= (x+1)^2 \\end{align*}\n\n\u2022 Whoa! Nice proof! What I had in mind was more like Gareth's proof, but this is really neat and fast. \u2013\u00a0Rand al'Thor Feb 14 '18 at 12:11\n\nffao posted his answer while I was writing this up. I like my solution better, though in some sense it's equivalent to ffao's, so I'm posting mine too. I will of course be entirely unoffended if Rand gives the mighty green checkmark to ffao, who after all got there first.\n\nIt is well known that\n\nthe sum of all odd numbers below $2x$ is equal to $x^2$. And there are $x$ of these, from $2\\cdot1-1$ to $2x-1$.\n\nSo presumably the sum we have here,\n\nwhich has the right number of terms and consists entirely of odd numbers, will simply be a permutation of this.\n\nWell,\n\ntake any odd number $t$ between $1$ and $2x$ inclusive. Keep doubling it until the next doubling would go beyond $2x$. You will end up with a number from $x+1$ to $2x$ inclusive, and clearly each such number appears just once. And the largest odd divisor of this number will be exactly $t$. We're done.\n\nOr, to put it a bit more formally (possibly harder to follow quickly but easier to convince yourself it's definitely correct):\n\nlet $A=\\{\\,y\\,:\\,x+1\\leq y\\leq2x\\,\\}$ and $B=\\{\\,z\\,:1\\leq z\\leq2x\\,\\&\\,z \\textrm{ odd}\\,\\}$;\ndefine $f:A\\rightarrow B$ and $g:B\\rightarrow A$ by\n$f(y)=\\textrm{largest odd divisor of$y$}$ and\n$g(z)=\\textrm{largest$2^kz$that's$\\leq2x$}$;\nthen $f,g$ are inverses and therefore our sum, which is $\\sum_{y\\in A}f(y)$,\nequals $\\sum_{z\\in B}z$, which famously equals $x^2$.\n\n\u2022 Yes, in some sense this is actually equivalent to what I ended up doing. The good side to my solution is that there is no actual thinking required to reach it, but it ends up being less clean. \u2013\u00a0ffao Feb 14 '18 at 0:10\n\u2022 You can also word it this way: Let: $x+1 = O_{x+1}*2^{E_{x+1}}$ ... $2x = O_{2x}*2^{E_{2x}}$ None of the numbers in the range can be a multiple of another, but if the $O$s of two numbers are identical, one of the numbers will be the multiple of other, so all $O$s must be different, which are the odd numbers up to $2x-1$. The sum will be $x^2$. \u2013\u00a0Nautilus Feb 14 '18 at 9:30\n\u2022 \"clearly each such number appears just once\" - I think this could do with just a little more explanation, as it's really the crux of this proof. \u2013\u00a0Rand al'Thor Feb 14 '18 at 12:13\n\u2022 This is the proof I had in mind when I posted the question (the 'aha' moment being what's in your first two spoilertags). But ffao's proof is so unexpectedly short and slick! I'm not sure where to put the checkmark now, but knowing me, it'll probably be a while until I award it anyway ;-) \u2013\u00a0Rand al'Thor Feb 14 '18 at 12:14\n\u2022 One of the reasons for the more-formal spoilered paragraph was to be more explicit about that \"clearly\". (I claim it really is obvious that those functions f and g are inverses, and that's all you need.) \u2013\u00a0Gareth McCaughan Feb 14 '18 at 13:40", "date": "2020-05-26 10:49:40", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/60536/largest-odd-factors-summing-to-a-square", "openwebmath_score": 0.8501101732254028, "openwebmath_perplexity": 638.862822112823, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682451330957, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8536078661577916}}
{"url": "http://mathhelpforum.com/statistics/224720-dependent-probability-question.html", "text": "Math Help - dependent probability question?\n\n1. dependent probability question?\n\nHello, I had a question about how to calculate dependent probability. Calculating independent probability is easy, simply multiply the probability of each event: P1 * P2 * P3.\n\nTherefore, if 3 different resources are working on 3 independent and unrelated tasks, and each resource has an 80% probability of completing his task, then the total probability of all resources completing all tasks, successfully, would be .80 * .80 * .80 = 51.2%\n\nHowever, I need to figure out the probability formula, for a more advanced scenario. Consider a scenario, where 1 resource is working on 3 tasks in sequence, and the probability of the resource completing each individual task is 80%.\n\nIn this scenario, the first task would be independent, but the 2nd task would be dependent on completion of the 1st task, and the 3rd task would be dependent on completion of the 1st and 2nd task. So can you tell me the formula I would use to calculate total probability in this scenario?\n\n2. Re: dependent probability question?\n\nLet's look at failure rather than success. Then the overall probability of success is 1 - probability of failure.\n\nYou can fail at step 1, step 2, or step 3.\n\nPr[fail at step 1] = 1 - .8 = .2\n\nPr[fail at step 2] = Pr[fail at step 2 | success at step 1]\n\nNow assuming failure at step 2 is independent of success at step 1 we get\n\nPr[fail at step 2] = Pr[fail at step 2] Pr[success at step 1] = .2 * .8 = .16\n\nPr[fail at step 3] = Pr[fail at step 3 | success at 1 and 2] = .2 * .8 * .8 = .128\n\nThe total probabily of failure is then the sum of these (.2 + .16 + .128) = .488\n\nSo Pr[Success] = 1 - .488 = .512\n\nThis is identical to the non sequenced case. Why? Because in order to succeed all 3 steps have to be succeeded at in either case.\nWhere this methodology becomes very useful is when there are multiple paths to success as well as failure.\n\n3. Re: dependent probability question?\n\nMy probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.\nWhen I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.\n\nLearn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)\nThat's what I think anyway.\n\n4. Re: dependent probability question?\n\nOriginally Posted by Melody2\nMy probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.\nWhen I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.\n\nLearn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)\nThat's what I think anyway.\nIn industry probability trees are used EXTENSIVELY.\n\n5. Re: dependent probability question?\n\nromsec - thanks for your feedback. However, I think the total probability should be lower, if 1 resource works on 3 tasks in sequence. Here's why: If the resource fails at task 1, then he will never get to task 2 or task 3. If the resource fails at task 2, then he will never get to task 3.\n\nOn the other hand, if 3 different resources work on these 3 different tasks independently, then this approach will completely sidestep the dependencies/risks that I described with the single resource above, so it seems like the realistic probability of success should be greater.\n\nIt seems like probability theory should have a formula that factors in the dependencies I described above. Can you provide an updated probability formula based on the additional context I provided above?\n\n6. Re: dependent probability question?\n\nI tell you what. You look at my derivation and tell me specifically what bits you think are incorrect. I took into account all the dependencies you put in by requiring serial operation.\n\n7. Re: dependent probability question?\n\n@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?\n\nI was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:\n\nScenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%\n\nScenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???\n\nMaybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.\n\nDoes my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?\n\n8. Re: dependent probability question?\n\nOriginally Posted by random512\n@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?\n\nI was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:\n\nScenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%\n\nScenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???\n\nMaybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.\n\nDoes my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?\nIn both cases there is a single outcome that denotes success of the entire venture. All 3 tasks must be completed successfully. There is only a single way this can be achieved and that is that all 3 tasks are done successfully.\n\nThe probability that your 3 independent folks each succeeding is .8. The probability that all 3 independently succeed is (.8)^3 = .512\n\nTask 1 succeeds with pr. .8. Only then does it proceed to Task 2. The probability that we get to Task 2 is thus .8. We don't care what happens if we fail at Task 1, if we do we fail.\n\nWe have arrived at Task 2 with pr .8. Now we have .8 pr of completing Task 2. There's no other dependency of our success on Task 1, it got here, that's all we need to know. So our pr of success of Task 2 is again .8 but also times the pr. that it got here at all, which we know to be .8. So the overall pr of completing Task 2 is (.8)^2 = .64\n\nNow we make it to Task 3 with pr .64. Once again there is no dependency on how well things went at task 1 or 2, the thing got here and that's all we know and we have pr .8 of success at Task 3. So the overall pr of completing Task 3 is .8 times the pr it got here at all .64, or (.8)^3 = .512.\n\nThere are all sorts of dependencies you could put between your tasks that would show the sort of conditional probability effects you are thinking of. But you haven't done that here.\n\n9. Re: dependent probability question?\n\n@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.\n\n3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:\n\n1 - (.50 * .50 * .50) = 87.5%\n\nSo that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.\n\n10. Re: dependent probability question?\n\nIf you want to set up an example demonstrating what you have in mind consider this.\n\nLet's say we're making toy dolls since it's the holidays. We have to cut the parts, assemble the parts, paint the doll.\nWe have 3 elves and they have different abilities regarding making these dolls. For elf A his pr success at those tasks is say {0.85, .6, .4}\nelf B could be {0.6, 0.9, 0.6}, and elf C could be {0.5, 0.5, 0.8}\n\nYou could have each elf build the entire doll at their station but it's pretty clear that there's a better strategy and that's the pair the elf's task with their best ability.\n\nThat's where your sequential operation comes in since you have to build the dolls in a sequence.\n\nIf you really want to get fancy you can have the probability of success at a station depend on the quality that the previous station produced. It's harder to assemble poorly cut parts. It's harder to paint a poorly assembled doll, and incorporate all that in to further show how important your strategy of matching worker and task is.\n\nAll this will certainly show the effect of sequential vs parallel operation.\n\n11. Re: dependent probability question?\n\nOriginally Posted by random512\n@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.\n\n3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:\n\n1 - (.50 * .50 * .50) = 87.5%\n\nSo that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.\nok I'm glad you figured it out.\n\n12. Re: dependent probability question?\n\nOne more related question. What's the proper way to describe the probability multiplication factor? For example:\n\n* the probability of a single risk event occurring is 50%\n* the probability of at least 1/3 risk events occurring is:\n1 - ( .50 * .50 * .50) = 87.5%\n\nIs it valid to say that additional risk factors create an \"exponential\" probability that at least 1 risk event will occur? \"Exponential\" is a good word, because it's descriptive, and powerful. However, I want to make sure that I use the word appropriately. So can I describe this multiplication effect as \"exponential\" or should I choose a different word?", "date": "2015-05-03 12:19:03", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/224720-dependent-probability-question.html", "openwebmath_score": 0.8681716322898865, "openwebmath_perplexity": 710.536596496316, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717476269469, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8536056369512666}}
{"url": "https://math.stackexchange.com/questions/3212103/find-smallest-contraction-coefficient", "text": "# Find Smallest contraction coefficient\n\nI have been given the following function $$f:[-1,1]\\to \\mathbb{R}$$: $$f(x)=\\ln(x+2)-x$$ And I have been asked whether it is a contraction or not, and if it is, I have to find the smallest contraction coefficient, such that $$0.\n\nAttempt\n\nSince $$0<|f'(x)|<1$$ for $$(-1,1)$$ we must have that the function is a contraction, and i would intuitively say that $$\\left|-\\frac{2}{3}\\right|=\\frac{2}{3}$$ is the smallest contraction coefficient.\n\nDoubts\n\nI have no concrete theorem or example to support my claim, and therefore I am skeptical.\n\nSince $$f$$ is differentiable $$\\forall x \\in [-1,1], |f'(x)|= |\\cfrac{1}{x+2} - 1| = |\\cfrac{-x-1}{x+2}| = \\cfrac{1+x}{2+x}$$ and $$\\forall x \\in [-1,1], f''(x) = \\cfrac{1}{(x+2)^2} \\ge 0$$ so $$f'$$ is growing. And we have $$| f'(1)| = \\cfrac{2}{3}$$ So $$\\forall x \\in [-1,1], |f'(x)| \\le \\cfrac{2}{3}$$ Therefore $$\\cfrac{2}{3}$$ is indeed the smallest contraction coefficient.\nHint: Because of the mean value inequality, you want to maximize $$|f'(x)|$$ for $$x \\in [-1,1]$$.\nI think that Rasmus' realized that $$L=\\frac 23$$ is the maximum value for $$f'$$ and therefore will work as a contraction constant. This was probably not the question. The question, I think, was about the possibility of getting a contraction constant $$\\tilde L < L$$. The answer is no, it is not possible. If you consider a smaller constant you immediately get an interval of the form $$[\\xi, 1]$$ where the contraction inequality does not hold.", "date": "2019-06-25 11:51:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3212103/find-smallest-contraction-coefficient", "openwebmath_score": 0.999904990196228, "openwebmath_perplexity": 1616.1372107802185, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717472321278, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8536056348953484}}
{"url": "http://eigen.tuxfamily.org/dox/group__LeastSquares.html", "text": "Please, help us to better know about our user community by answering the following short survey: https://forms.gle/wpyrxWi18ox9Z5ae9\n Eigen \u00a03.3.9\nSolving linear least squares systems\n\nThis page describes how to solve linear least squares systems using Eigen. An overdetermined system of equations, say Ax = b, has no solutions. In this case, it makes sense to search for the vector x which is closest to being a solution, in the sense that the difference Ax - b is as small as possible. This x is called the least square solution (if the Euclidean norm is used).\n\nThe three methods discussed on this page are the SVD decomposition, the QR decomposition and normal equations. Of these, the SVD decomposition is generally the most accurate but the slowest, normal equations is the fastest but least accurate, and the QR decomposition is in between.\n\n# Using the SVD decomposition\n\nThe solve() method in the BDCSVD class can be directly used to solve linear squares systems. It is not enough to compute only the singular values (the default for this class); you also need the singular vectors but the thin SVD decomposition suffices for computing least squares solutions:\n\nExample:Output:\n#include <iostream>\n#include <Eigen/Dense>\nusing namespace std;\nusing namespace Eigen;\nint main()\n{\ncout << \"Here is the matrix A:\\n\" << A << endl;\ncout << \"Here is the right hand side b:\\n\" << b << endl;\ncout << \"The least-squares solution is:\\n\"\n<< A.bdcSvd(ComputeThinU | ComputeThinV).solve(b) << endl;\n}\nHere is the matrix A:\n-1 -0.0827\n-0.737  0.0655\n0.511  -0.562\nHere is the right hand side b:\n-0.906\n0.358\n0.359\nThe least-squares solution is:\n0.464\n0.043\n\n\nThis is example from the page Linear algebra and decompositions .\n\n# Using the QR decomposition\n\nThe solve() method in QR decomposition classes also computes the least squares solution. There are three QR decomposition classes: HouseholderQR (no pivoting, so fast but unstable), ColPivHouseholderQR (column pivoting, thus a bit slower but more accurate) and FullPivHouseholderQR (full pivoting, so slowest and most stable). Here is an example with column pivoting:\n\nExample:Output:\nMatrixXf A = MatrixXf::Random(3, 2);\nVectorXf b = VectorXf::Random(3);\ncout << \"The solution using the QR decomposition is:\\n\"\n<< A.colPivHouseholderQr().solve(b) << endl;\nThe solution using the QR decomposition is:\n0.464\n0.043\n\n\n# Using normal equations\n\nFinding the least squares solution of Ax = b is equivalent to solving the normal equation ATAx = ATb. This leads to the following code\n\nExample:Output:\nMatrixXf A = MatrixXf::Random(3, 2);\nVectorXf b = VectorXf::Random(3);\ncout << \"The solution using normal equations is:\\n\"\n<< (A.transpose() * A).ldlt().solve(A.transpose() * b) << endl;\nThe solution using normal equations is:\n0.464\n0.043\n\n\nIf the matrix A is ill-conditioned, then this is not a good method, because the condition number of ATA is the square of the condition number of A. This means that you lose twice as many digits using normal equation than if you use the other methods.\n\nEigen\nNamespace containing all symbols from the Eigen library.\nDefinition: Core:309\nEigen::ComputeThinU\n@ ComputeThinU\nDefinition: Constants.h:385\nEigen::DenseBase::Random\nstatic const RandomReturnType Random()\nDefinition: Random.h:113\nEigen::MatrixBase::bdcSvd\nBDCSVD< PlainObject > bdcSvd(unsigned int computationOptions=0) const\nDefinition: BDCSVD.h:1269\nEigen::ComputeThinV\n@ ComputeThinV\nDefinition: Constants.h:389\nEigen::Matrix\nThe matrix class, also used for vectors and row-vectors.\nDefinition: Matrix.h:180", "date": "2021-01-20 08:04:49", "meta": {"domain": "tuxfamily.org", "url": "http://eigen.tuxfamily.org/dox/group__LeastSquares.html", "openwebmath_score": 0.6047804355621338, "openwebmath_perplexity": 1921.0165506598726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717448632122, "lm_q2_score": 0.8652240791017535, "lm_q1q2_score": 0.8536056294170828}}
{"url": "https://math.stackexchange.com/questions/2134031/is-this-formula-for-frace2-3e21-known-how-to-prove-it", "text": "# Is this formula for $\\frac{e^2-3}{e^2+1}$ known? How to prove it?\n\nI found an interesting infinite sequence recently in the form of a 'two storey continued fraction' with natural number entries:\n\n$$\\frac{e^2-3}{e^2+1}=\\cfrac{2-\\cfrac{3-\\cfrac{4-\\cdots}{4+\\cdots}}{3+\\cfrac{4-\\cdots}{4+\\cdots}}}{2+\\cfrac{3-\\cfrac{4-\\cdots}{4+\\cdots}}{3+\\cfrac{4-\\cdots}{4+\\cdots}}}$$\n\nThe numerical computation was done 'backwards', starting from some $x_n=1$ we compute:\n\n$$x_{n-1}=\\frac{a_n-x_n}{a_n+x_n}$$\n\nAnd so on, until we get to $x_0$. The sequence converges for $n \\to \\infty$ if $a_n>1$ (or so it seems).\n\nFor constant $a_n$ we seem to have quadratic irrationals, for example:\n\n$$\\frac{\\sqrt{17}-3}{2}=\\cfrac{2-\\cfrac{2-\\cfrac{2-\\cdots}{2+\\cdots}}{2+\\cfrac{2-\\cdots}{2+\\cdots}}}{2+\\cfrac{2-\\cfrac{2-\\cdots}{2+\\cdots}}{2+\\cfrac{2-\\cdots}{2+\\cdots}}}$$\n\nFor $a_n=2^n$ we seems to have:\n\n$$\\frac{1}{2}=\\cfrac{2-\\cfrac{4-\\cfrac{8-\\cdots}{8+\\cdots}}{4+\\cfrac{8-\\cdots}{8+\\cdots}}}{2+\\cfrac{4-\\cfrac{8-\\cdots}{8+\\cdots}}{4+\\cfrac{8-\\cdots}{8+\\cdots}}}$$\n\nI found no other closed forms so far, and I don't know how to prove the formulas above. How can we prove them? What is known about such continued fractions?\n\nThere is another curious thing. If we try to expand some number in this kind of fraction, we can do it the following way:\n\n$$x_0=x$$\n\n$$a_0=\\left[\\frac{1}{x_0} \\right]$$\n\n$$x_1=\\frac{1-a_0x_0}{1+a_0x_0}$$\n\n$$a_1=\\left[\\frac{1}{x_1} \\right]$$\n\nHowever, this kind of expansion will not give us the above sequences. We will get faster growing entries. Moreover, the fraction will be finite for any rational number. For example, in the list notation:\n\n$$\\frac{3}{29}=[9,28]$$\n\nYou can easily check this expansion for any rational number.\n\nAs for the constant above we get:\n\n$$\\frac{e^2-3}{e^2+1}=[1,3,31,74,315,750,14286,\\dots]$$\n\nNot the same as $[1,2,3,4,5,6,7,\\dots]$ above!\n\nWe have similar sequences growing exponentially for any irrational number I checked.\n\n$$e-2=[1,6,121,284,1260,3404,25678,\\dots]$$\n\n$$\\pi-3=[7,224,471,2195,10493,46032,119223,\\dots]$$\n\nBy the way, if we try CF convergents, we get almost the same expansion, but finite:\n\n$$\\frac{355}{113}-3=[7,225]$$\n\n$$\\frac{4272943}{1360120}-3=[7,224,471,2195,18596,227459,\\dots]$$\n\nSo, the convergents of this sequence are not the same as for the simple continued fraction, but similar.\n\nComparing the expansion by the method above and the closed forms at the top of the post, we can see that, unlike for simple continued fractions, this expansion is not unique. Can we explain why?\n\nHere is the Mathematica code to compute the limit of the first fraction:\n\nNm = 50;\nCf = Table[j, {j, 1, Nm}];\nb0 = (Cf[[Nm]] - 1)/(Cf[[Nm]] + 1);\nDo[b1 = N[(Cf[[Nm - j]] - b0)/(Cf[[Nm - j]] + b0), 7500];\nb0 = b1, {j, 1, Nm - 2}]\nN[b0/Cf[[1]], 50]\n\n\nAnd here is the code to obtain the expansion in the usual way:\n\nx = (E^2 - 3)/(E^2 + 1);\nx0 = x;\nNm = 27;\nCf = Table[1, {j, 1, Nm}];\nDo[If[x0 != 0, a = Floor[1/x0];\nx1 = N[(1 - x0 a)/(x0 a + 1), 19500];\nPrint[j, \" \", a, \" \", N[x1, 16]];\nCf[[j]] = a;\nx0 = x1], {j, 1, Nm}]\nb0 = (1 - 1/Cf[[Nm]])/(1 + 1/Cf[[Nm]]);\nDo[b1 = N[(1 - b0/Cf[[Nm - j]])/(1 + b0/Cf[[Nm - j]]), 7500];\nb0 = b1, {j, 1, Nm - 2}]\nN[x - b0/Cf[[1]], 20]\n\n\nUpdate\n\nI have derived the forward recurrence relations for numerator and denominator:\n\n$$p_{n+1}=(a_n-1)p_n+2a_{n-1}p_{n-1}$$ $$q_{n+1}=(a_n-1)q_n+2a_{n-1}q_{n-1}$$\n\nThey have the same form as for generalized continued fractions (a special case). Now I understand why the expansions are not unique.\n\n\u2022 I am always thoroughly impressed with the length and detail in your posts. On the other hand, you might want to key in a little more on what your main question is here. \u2013\u00a0Brevan Ellefsen Feb 7 '17 at 22:25\n\u2022 @BrevanEllefsen, if these are known results, I would like a reference. If not, I would like a hint about a proof. And the most important question - why is the expansion not unique, even though the properties seem to be similar to simple continued fractions. Thanks, by the way. I'm glad I'm not the only one who likes my posts :) \u2013\u00a0Yuriy S Feb 7 '17 at 22:26\n\u2022 @Nemo, I'm not sure I understand what you did here, but I figured out the same thing after deriving the forward recurrence relations for numerator and denominator: $$p_{n+1}=(a_n-1)p_n+2a_{n-1}p_{n-1}$$ $$q_{n+1}=(a_n-1)q_n+2a_{n-1}q_{n-1}$$ Obviously, we have a special case of a generalized continued fraction. Which is why it is not unique \u2013\u00a0Yuriy S Feb 8 '17 at 9:18\n\nFor the first one, you could write $$f(n) = \\frac{n-f(n+1)}{n+f(n+1)}$$ Then you suggest $$f(2) = \\frac{e^2-3}{e^2+1}$$ But this then gives \\begin{align} f(1) = \\frac{2}{e^2-1}\\\\ f(3) = \\frac{4}{e^2-1} \\\\ f(4) = \\frac{3e^2-15}{e^2+3}\\\\ f(5) = \\frac{-2e^2+18}{e^2-3} \\end{align} I don't know if there is a recurrence relation that solves this, but you have a few more closed forms...\nThe second one we have $$g(2) = \\frac{2 - g(2)}{2+g(2)}$$ so we can solve the quadratic $x^2+3x-2$ to get $(\\sqrt{17}-3)/2$.\nFor the third one, we have $$h(n) = \\frac{n-h(2n)}{n+h(2n)}$$ using the trial version of $h(2)=1/2$, we get \\begin{align} h(4)=\\frac{2}{3}\\\\ h(8)=\\frac{4}{5}\\\\ h(16)=\\frac{8}{9} \\end{align} then it is likely that $$h(n)=\\frac{n}{n+2}$$ as this satisfies the recurrence and that $h(2)=1/2$.\n\u2022 Thank you, when I look back it's clear that quadratic radicals or rational numbers are easily proven (provided the limit exists), just as you show here. For irrational numbers see the comments and the update, linking this to generalized continued fractions, and the proofs for $e$ related numbers can be found elsewhere \u2013\u00a0Yuriy S May 31 '17 at 11:29", "date": "2019-10-15 08:51:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2134031/is-this-formula-for-frace2-3e21-known-how-to-prove-it", "openwebmath_score": 0.9045830965042114, "openwebmath_perplexity": 283.1403001520814, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717448632122, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8536056277027716}}
{"url": "https://math.stackexchange.com/questions/3389198/deriving-the-derivative-of-tan2x-by-quotient-rule-using-sinx-cosx-identi", "text": "Deriving the derivative of $\\tan^{2}x$ by quotient rule using (sinx/cosx) identity I am getting a different value (2secx(tanx+tan^2(x))\n\nDeriving the derivative of $$\\tan^{2}x$$ by quotient rule using $$\\frac{\\sin x}{\\cos x}$$ identity, i am getting a different value $$2\\sec x[\\tan x+\\tan^{2}x]$$ than by directly getting chain rule is $$2\\sec^{2}x\\tan x$$\n\nWhats wrong?\n\nHere is a correct derivation of the derivative of $$\\tan^2 x$$ using the quotient rule:\n$$\\dfrac d {dx} \\tan^2 x = \\dfrac d {dx} \\dfrac {\\sin^2 x}{\\cos^2x }=\\dfrac{2\\sin x \\cos x \\cos^2x + \\sin^2 x 2 \\cos x \\sin x}{\\cos^4x}$$\n$$=2\\dfrac{\\sin x(\\cos^2x+\\sin^2x)}{\\cos^3x}=2\\dfrac{\\sin x}{\\cos^3 x}=2\\tan x \\sec^2x$$\n\u2022 Yes, the chain rule yields $2 \\tan x \\sec^2x$ too \u2013\u00a0J. W. Tanner Oct 11 '19 at 6:58\n\u2022 $\\cos^2 x+\\sin^2 x = 1$ is the Pythagorean identity \u2013\u00a0J. W. Tanner Oct 11 '19 at 7:08", "date": "2020-11-27 12:07:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3389198/deriving-the-derivative-of-tan2x-by-quotient-rule-using-sinx-cosx-identi", "openwebmath_score": 0.9518729448318481, "openwebmath_perplexity": 450.02192250078105, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.986571744468393, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8536056222182306}}
{"url": "https://mathematica.stackexchange.com/questions/209829/is-there-a-way-to-randomly-distribute-points-within-a-circle-on-the-surface-of-a", "text": "# Is there a way to randomly distribute points within a circle on the surface of a sphere?\n\nI'm attempting to set up a situation where on a 3D sphere, I choose a random point and construct a circle around this point with some radius. I then want to randomly distribute points within this circle. Is there a straightforward way to do this?\n\nI have no trouble finding random points on the surface of the sphere; however I cannot seem to find a way to distribute points randomly on a closed region of the sphere.\n\nMany thanks for any help in advance.\n\n\u2022 Two related questions. \u2013\u00a0J. M.'s ennui Nov 18 '19 at 5:17\n\u2022 You could brute force it by finding random points on the sphere, then throwing away any that aren't inside the desired region. \u2013\u00a0Foo Bar Nov 18 '19 at 12:59\n\u2022 @Foo, yes, one can certainly use the rejection method for this. The tricky part is in figuring out how not to throw away too many points. \u2013\u00a0J. M.'s ennui Nov 18 '19 at 14:37\n\u2022 Depends on your underlying distribution. I assume you want uniform random over the sphere's area. For contrast, consider uniform random points in a circle and then project the circle onto the sphere. That's still random but with a different distribution function. \u2013\u00a0Carl Witthoft Nov 19 '19 at 19:39\n\nYou can intersect the sphere and a cylinder, and then use RandomPoint. For example, here is a random point on the sphere:\n\nsphere = Sphere[];\nSeedRandom[1]\npt = RandomPoint[sphere]\n\n\n{0.707037, 0.595614, 0.381239}\n\nThen, you create a cylinder in he direction of the random point with a radius:\n\nr = .7;\ncylinder = Cylinder[{{0,0,0},pt}, r];\n\n\nNow, intersect the cylinder and the sphere and use RandomPoint:\n\nreg = RegionIntersection[cylinder, sphere];\npts = RandomPoint[reg, 1000];\n\n\nVisualization:\n\nGraphics3D[{Sphere[], Red, Point @ pts}]\n\n\n\u2022 Can you verify that this produces uniform random per unit sphere area (as opposed to some projection of the intersecting circle)? I\"m not familiar with how this Mathematica function actually selects points. \u2013\u00a0Carl Witthoft Nov 19 '19 at 19:41\n\u2022 @CarlWitthoft That's what the documentation says: \"RandomPoint will generate points uniformly in the region reg.\" In this case the region is the spherical cap. \u2013\u00a0Carl Woll Nov 19 '19 at 20:36\n\nUsing Christian Blatter's results from this math.SE answer, here is how to randomly sample a spherical cap:\n\nrandomCapPoint[{r_, r2_}, dir_?VectorQ] := With[{h = RandomReal[{Sqrt[1 - (r2/r)^2], 1}]},\nRotationTransform[{{0, 0, 1}, Normalize[dir]}][r\nAppend[Sqrt[1 - h^2] Normalize[RandomVariate[NormalDistribution[], 2]], h]]]\n\n\nFor example,\n\nBlockRandom[SeedRandom[1337]; (* for reproducibility *)\nWith[{r = 1, r2 = 2/5, d = {1.3, -2.4, 2}, n = 5000},\nGraphics3D[{{Opacity[1/2], Sphere[{0, 0, 0}, r]},\n{Blue, Arrow[Tube[{{0, 0, 0}, Normalize[d]}]]},\n{Directive[AbsolutePointSize[2], Orange],\nPoint[Table[randomCapPoint[{r, r2}, d], {n}]]}}]]]\n\n\n\u2022 Same question as at Carl Woll's answer: how do you verify the spatial distribution is \"uniform\" with respect to sphere area? \u2013\u00a0Carl Witthoft Nov 19 '19 at 19:43\n\u2022 @Carl, using the results in Christian's answer, you could take a histogram of the longitude and colatitude of the sample points, and verify that they have the required distribution. \u2013\u00a0J. M.'s ennui Nov 19 '19 at 20:51\n\nWe can also take RandomPoints in boolean region obtained by the RegionIntersection of a Sphere and a Ball with radius r centered at a random point on the sphere:\n\nSeedRandom[1]\nr = .7;\n\nctr = RandomPoint[Sphere[]];\n\npts = RandomPoint[RegionIntersection[Ball[ctr, r], Sphere[]], 1000];\n\nGraphics3D[{Red, Point@pts, White, Opacity[.5], Sphere[]}]\n\n\nYou can use Archimedes' result (https://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder) that the area between two lines of latitude is the same as the corresponding area of the enclosing cylinder.\n\nThis gives a fairly simple result for sampling the top of the sphere, above height Z\n\npt[Z_] := Module[{z, \u03b8, r},\nz = RandomVariate[UniformDistribution[{Z, 1}]];\n\u03b8 = RandomVariate[UniformDistribution[{0, 2 \u03c0}]];\nr = Sqrt[1 - z^2];\n{r Cos[\u03b8], r Sin[\u03b8], z}]\n\n\u2022 This is basically what my method does, except I replaced {r Cos[\u03b8], r Sin[\u03b8]} with r Normalize[RandomVariate[NormalDistribution[], 2]]. \u2013\u00a0J. M.'s ennui Nov 19 '19 at 20:49", "date": "2021-01-20 13:49:21", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/209829/is-there-a-way-to-randomly-distribute-points-within-a-circle-on-the-surface-of-a", "openwebmath_score": 0.5532092452049255, "openwebmath_perplexity": 1519.21380192524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075701109193, "lm_q2_score": 0.8872045966995028, "lm_q1q2_score": 0.8535862587217967}}
{"url": "https://math.stackexchange.com/questions/1565290/trying-to-answer-this-counting-question", "text": "# Trying to answer this counting question\n\nIn a ballroom dance class, participants are divided into couples for each drill session. One partner leads and the other follows for three minutes, and then the couple switches roles for the next three minutes.\n\n(a) Only four people show up on time. How many ways are there to pair them up?\n\nMy answer here is $4C2 = \\binom42 = \\tfrac{4!}{2!2!}$\n\n(b) If instead six people show up on time, how many ways are there to pair them up?\n\nMy answer here is $6C2 = \\binom62 = \\tfrac{6!}{4!2!}$\n\n(c) Assume all m people in the class arrive on time. (There are an even number of people in the class.) How many ways are there to pair them up?\n\nMy answer here is $mC2 = \\binom m2= \\tfrac{m!}{(m-2)!2!}$\n\n(d) Consider this time assuming that we specify which member of each couple leads first. How many ways are there to pair-and- specify the dancers\n\nMy answer here is $mP2 = \\tfrac{m!}{(m-2)!}$\n\n## UPDATE:\n\nPart (a) the answer is $\\tfrac{1}{2}(4C2 * 2C2) = 3$\n\nPart (b) the answer is $\\tfrac{1}{2}(6C2 * 4C2 * 2C2) = 45$\n\n\u2022 Yup, they are all correct. Just a typo in the second answer, you meant 6C2, I think. \u2013\u00a0learner Dec 8 '15 at 4:15\n\u2022 Yes, I corrected it, Thanks! \u2013\u00a0Coheen Dec 8 '15 at 4:16\n\u2022 No, they are not correct. See theophile's answer below. \u2013\u00a0fleablood Dec 8 '15 at 4:28\n\u2022 @lucidgold What is the conflict with the answers? As far as I can see, all three are saying the same thing. \u2013\u00a0Th\u00e9ophile Dec 8 '15 at 18:17\n\nFor part (b), the number of ways to pick a first pair, then a second pair, then a third pair, is $\\binom 62 \\binom 42 \\binom 22$, or in your notation $6C2 \\times 4C2 \\times 2C2$. But that not only pairs up all the dancers but distinguishes $\\{AB, CD, EF\\}$ from $\\{AB, EF, CD\\}$, $\\{CD, AB, EF\\}$, and several other permutations of the pairs. In fact you have counted the same three pairs $3!$ times, which is the number of permutations of three things. So the correct answer is not to divide by $2$ but rather to divide by $3! = 6$:\n\n$$\\frac{\\binom 62 \\binom 42 \\binom 22}{6}.$$\n\nPart (d) seems easier to me than part (c). First you choose who will be leaders in the first part of the lesson. There are $\\binom{m}{m/2}$ ways to do this. Then you line up these $m/2$ leaders in a line and assign each of the remaining $m/2$ dancers to one of the leaders. That is, each ordering of the remaining $m/2$ dancers produces a unique pairing of followers with leaders. There are $(m/2)!$ such orderings, so the total number of possible pairings is ...\n\nOnce you have part (d), I would go back to part (c). Clearly there are more pairings counted in part (d) than in part (c). How many more?\n\nFor each way you can pair up dancers in part (c), within each pair there are two ways to choose who will lead during the first part of the lesson. Since there are $m/2$ pairs, there are therefore $2^{m/2}$ ways to choose which $m/2$ dancers will lead at first. But the choice of pairs for part (c), followed by choosing leaders for the first three minutes, can give us every choice of ordered pairs that exists in part (d). Therefore the number of choices in part (c) must be ...\n\nNo, these are not quite correct\u2014at least not the way I understand the question. When there are four people, for example, there are only three ways to pair them up: $\\{AB,CD\\}, \\{AC,BD\\}$, and $\\{AD,BC\\}$. The reason this is different from your answer is that you have counted how many ways there are to choose two people from a set of four. This is ${4 \\choose 2} = 6$, namely, $AB, AC, AD, BC, BD, CD$. But the number of ways to form one pair isn't the same as the number of ways to split the group into pairs.\n\nOne way to do the count for part (a) is to pick one pair (as you have done), then to form a second pair from the remaining two people. The number of ways to do this is $${4 \\choose 2}{2 \\choose 2} = 6 \\cdot 1 = 6.$$ But in doing so we have accounted for each pairing twice; e.g., we have counted both $\\{AB,CD\\}$ and $\\{CD,AB\\}$, when these are in fact the same pairing. Therefore we should divide by $2$ to get $\\frac62 = 3$ pairings.\n\n\u2022 This is a great answer. it shows the problem with part a and leaves the rest to OP. There are similar problems with the rest, but if OP understands this answer, s/he should be able to deal with the rest. If not, that is another question. This is not an easy question. \u2013\u00a0Ross Millikan Dec 8 '15 at 4:28\n\u2022 So what happened if the number of people is huge like 300? Is there is any formula I can use to solve this problem? So what I understand is I should use combinations and divide the answer by 2? is this correct? \u2013\u00a0Coheen Dec 8 '15 at 4:30\n\u2022 @Coheen: you asked four different questions. All your answers were not correct. If the number of people increases, yes there is a formula you can use. The answer for the first three differs greatly from the answer to the fourth. For ones like the first three, you pick one pair, then another, then another, and so on. Then you need to consider that you could pick the pairs in any order. Your formulas just pick the first pair. \u2013\u00a0Ross Millikan Dec 8 '15 at 4:36\n\u2022 For part (b), would the following be correct: (6C2* 4C2* 2C2 )/2 = 45 \u2013\u00a0Coheen Dec 8 '15 at 4:56\n\u2022 @Coheen The reason for dividing by two was because there were two pairs. When there are three, you should divide by 3!. This is described in more detail in David K's answer. \u2013\u00a0Th\u00e9ophile Dec 8 '15 at 18:16\n\nIf I am not mistaken this is applications of the Multinomial Theorem. A lot of good examples are given in Sheldon Ross's A First Course in Probability.\n\nAlso note with part $(d)$ that uniqueness matters. So if we had 4 people show up and we chose who was going to lead who we would have a permutation for all of the members dancing. For example, let the 4 people be named: John, Sally, Jordan, Steve. If John lead Steve for the first 3 minutes it would not be the same as Steve leading John. So essentially we would have the same combination below, but without the denominator in the Multinomial, since order matters.", "date": "2021-04-13 10:45:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1565290/trying-to-answer-this-counting-question", "openwebmath_score": 0.8163203001022339, "openwebmath_perplexity": 235.20493162371943, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075733703927, "lm_q2_score": 0.8872045817875224, "lm_q1q2_score": 0.8535862472666872}}
{"url": "http://nacretv.easylia.com/uamkx/1ea173-how-to-find-eigenvectors-of-a-3x3-matrix", "text": "The detailed solution is given. If . If the resulting V has the same size as A, the matrix A has a full set of linearly independent eigenvectors that satisfy A*V = V*D. But for a special type of matrix, symmetric matrix, the eigenvalues are always real and the corresponding eigenvectors are always orthogonal. Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 \u20131 4]. The values of \u03bb that satisfy the equation are the generalized eigenvalues. The code for this originally is \u2026 Calculate the eigenvalues and the corresponding eigenvectors of the matrix. The eigenvalues are r1=r2=-1, and r3=2. I am trying to find the best OOBB hitboxes for my meshes using PCA. Let A=[121\u22121412\u221240]. [V,D] = eig(A) returns matrices V and D.The columns of V present eigenvectors of A.The diagonal matrix D contains eigenvalues. In order to do this, I need the eigenvectors but I am kind of lost how to compute them without using a huge library. kerr_lee. It is also known as characteristic vector. Answer Save. Substitute every obtained eigenvalue $\\boldsymbol{\\lambda}$ into the eigenvector equations, 4.1. The three eigenvalues and eigenvectors now can be recombined to give the solution to the original 3x3 matrix as shown in Figures 8.F.1 and 8.F.2. 0 0. In fact, we will in a different page that the structure of the solution set of this system is very rich. Any help is greatly appreciated. What is the shortcut to find eigenvalues? The eigenvector v of an operator A is a vector such that: Av = kv, for some scalar k. So suppose v = (v1,v2,v3) and compute: Av = (v2+v3,v1+v3,v1+v2). Eigen vector, Eigen value 3x3 Matrix Calculator. Bring all to left hand side: 1 decade ago. Yes it is the same as there is multiple values of your eigen vector by multiplying by a scalar. How To: Find the equation of trig functions by their graphs How To: Do matrix algebra on a TI-83 calculator How To: Solve systems of linear equations with matrices How To: Find eigenvectors and eigenspaces of a 2x2 matrix How To: Use a change of basis matrix in linear algebra %PDF-1.2 3. 2 Answers. Eigenvectors corresponding to distinct eigenvalues are linearly independent. Example Problem. Find the eigenvalues and eigenvectors for the matrix [(0,1,0),(1,-1,1),(0,1,0)]. Answer 5 0 obj (The Ohio State University, Linear Algebra Final Exam Problem) Add to solve later Sponsored Links Eigenvalue $\\boldsymbol{\\lambda = 3}$, 4.2. The standard formula to find the determinant of a 3\u00d73 matrix is a break down of smaller 2\u00d72 determinant problems which are very easy to handle. We define the characteristic polynomial and show how it can be used to find the eigenvalues for a matrix. Find all the eigenvalues and corresponding eigenvectors of the given 3 by 3 matrix A. Finding Eigenvalues and Eigenvectors : 2 x 2 Matrix Example . Let's find the eigenvector, v 1, associated with the eigenvalue, \u03bb 1 =-1, first. Reads for a joint honours degree in mathematics and theoretical physics (final year) in England, at the School of Mathematics and Statistics and the School of Physical Sciences at The Open University, Walton Hall, Milton Keynes. Hi, I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. play_arrow. 3,0,2. has eigenvalues of 2,4,and -3. Matrix A: Find. Enter a matrix. Algebraic and geometric multiplicity of eigenvalues. For Example, if x is a vector that is not zero, then it is an eigenvector of a square matrix \u2026 Next, transpose the matrix by rewriting the first row as the first column, the middle row as the middle column, and the third row as the third column. The above examples assume that the eigenvalue is real number. Let's find the eigenvector, v 1, associated with the eigenvalue, \u03bb 1 =-1, first. eigen() function in R Language is used to calculate eigenvalues and eigenvectors of a matrix. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. so clearly from the top row of \u2026 If an example would help, I've worked out that the matrix-1,1,3. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. The matrix looks like this... |0 1 1| A= |1 0 1| |1 1 0| When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. Anonymous. Write down the associated linear system 2. Eigenvectors are the solution of the system $( M \u2212 \\lambda I_n ) \\vec{X} = \\vec{0}$ with $I_n$ the identity matrix.. kerr_lee. Once we have the eigenvalues for a matrix we also show how to find the corresponding eigenvalues for the matrix. The matrix is (I have a ; since I can't have a space between each column. and the two eigenvalues are . How to find the eigenspace of a 3x3 matrix - Suppose A is this 2x2 matrix: [1 2] [0 3]. Example 4: 3xx3 case. To find all of a matrix's eigenvectors, you need solve this equation once for each individual eigenvalue. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. 1,2,0. share | cite | improve this question | follow | edited Jan 26 '15 at 0:09. abel. Input the components of a square matrix separating the numbers with spaces. The nullspace is projected to zero. and the two eigenvalues are . It will find the eigenvalues of that matrix, and also outputs the corresponding eigenvectors.. For background on these concepts, see 7.Eigenvalues and Eigenvectors The PCA is applied on this symmetric matrix, so the eigenvectors are guaranteed to be orthogonal. 5 years ago. edit close. Since the left-hand side is a 3x3 determinant, we have x\ufffd\ufffd\\\ufffd\u0776\ufffd\ufffd\u007f\ufffd\ufffd(\ufffd\ufffdJ\ufffd\ufffd5\ufffd:\ufffd\ufffd\ufffd=bo\ufffdA?4\ufffd>\ufffdf\ufffdu\ufffd\ufffd\ufffd\u007f\ufffd\ufffd\ufffdP\ufffd\ufffd\ufffdu4F\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd!\ufffdov\ufffd\ufffd\ufffd\ufffdg\ufffdqus!v\ufffd\ufffd\u07d7o.|\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd7O\ufffdN\ufffdVi\ufffd\ufffd2\ufffd\ufffd;)}\ufffdo\ufffd\ufffd]\ufffd\\|[=\ufffd\ufffdziT_\u0562u\ufffdO\ufffd\ufffdZ\ufffd\ufffd\ufffdM\ufffd=\ufffd\ufffd\u0596\ufffd?\ufffd\ufffdN\ufffdZU_\u0580\ufffdx>_\ufffdS \ufffd\ufffdi\ufffd\ufffdj \u0247\ufffd\ufffdau\ufffd\ufffdO\ufffdF\ufffdV(\ufffdoj\ufffd In order to find the associated eigenvectors, we do the following steps: 1. The eigenvectors for D 1 (which means Px D x/ \ufb01ll up the column space. then the characteristic equation is . It is also known as characteristic vector. How To: Find the equation of trig functions by their graphs How To: Do matrix algebra on a TI-83 calculator How To: Solve systems of linear equations with matrices How To: Find eigenvectors and eigenspaces of a 2x2 matrix How To: Use a change of basis matrix in linear algebra The matrix is (I have a ; since I can't have a space between each column. \u03bb 1 =-1, \u03bb 2 =-2. If . Find the. Find a basis of the eigenspace E2 corresponding to the eigenvalue 2. EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 \u22123 3 3 \u22125 3 6 \u22126 4 . As a consequence, if all the eigenvalues of a matrix are distinct, then their corresponding eigenvectors span the space of column vectors to which the columns of the matrix belong. Find Eigenvalues and Eigenvectors of a Matrix in R Programming \u2013 eigen() Function Last Updated: 19-06-2020. eigen() function in R Language is used to calculate eigenvalues and eigenvectors of a matrix. SOLUTION: \u2022 In such problems, we \ufb01rst \ufb01nd the eigenvalues of the matrix. Find the. The column space projects onto itself. SOLUTION: \u2022 In such problems, we \ufb01rst \ufb01nd the eigenvalues of the matrix. I am trying to find the best OOBB hitboxes for my meshes using PCA. Linear independence of eigenvectors. So one may wonder whether any eigenvalue is always real. Example: The Hermitian matrix below represents S x +S y +S z for a spin 1/2 system. More: Diagonal matrix Jordan decomposition Matrix exponential. Clean Cells or Share Insert in. When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. Beware, however, that row-reducing to row-echelon form and obtaining a triangular matrix does not give you the eigenvalues, as row-reduction changes the eigenvalues of the matrix \u2026 Eigenvalue $\\boldsymbol{\\lambda = 7}$, Real eigenvalues and eigenvectors of 3x3 matrices, example 1, Real eigenvalues and eigenvectors of 3x3 matrices, example 2, Finding the normal force in planar non-uniform\u2026, Simple problems on relativistic energy and momentum, Proof that the square root of 2 is irrational, Deriving the volume of the inside of a sphere using\u2026, Real eigenvalues and eigenvectors of 3\u00d73 matrices, example 2, Deriving the Lorentz transformations from a rotation of frames of reference about their origin with real time Wick-rotated to imaginary time, https://opencurve.info/real-eigenvalues-and-eigenvectors-of-3x3-matrices-example-3/. A = To do this, we find the values of ? Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 \u20131 4]. [V,D] = eig(A) returns matrices V and D.The columns of V present eigenvectors of A.The diagonal matrix D contains eigenvalues. Now let us put in an identity matrix so we are dealing with matrix-vs-matrix:. How do you find the characteristic equation of a 3\u00d73 matrix? This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. \ufffd\ufffd\ufffd\u023a\ufffdv\ufffd'U. Real eigenvalues and eigenvectors of 3x3 matrices, example 2; Finding the normal force in planar non-uniform\u2026 Simple problems on relativistic energy and momentum; Proof that the square root of 2 is irrational; Deriving the volume of the inside of a sphere using\u2026 2018-12-14 2020-09-24 eigenvalues, eigenvectors, linear algebra, matrix Post navigation. Find more Mathematics widgets in Wolfram|Alpha. All that's left is to find the two eigenvectors. Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. Eigenvalue and Eigenvector for a 3x3 Matrix Added Mar 16, 2015 by Algebra_Refresher in Mathematics Use this tool to easily calculate the eigenvalues and eigenvectors of 3x3 matrices. Display decimals, number of significant digits: \u2026 Relevance. FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 \u22123 3 3 \u22125 3 6 \u22126 4 . The ideal is to express a given vector as a linear combination of eigenvectors. asked Jan 25 '15 at 23:57. user3435407 user3435407. How to find eigenvalues quick and easy \u2013 Linear algebra explained . Eigen vector, Eigen value 3x3 Matrix Calculator. In these examples, the eigenvalues of matrices will turn out to be real values. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic polynomial (i.e., the polynomial whose roots are the eigenvalues of a matrix). Any help is greatly appreciated. I'm writing an algorithm with a lot of steps (PCA), and two of them are finding eigenvalues and eigenvectors of a given matrix. then the characteristic equation is . In order to do this, I need the eigenvectors but I am kind of lost how to compute them without using a huge library. Rewrite the unknown vector X as a linear combination of known vectors. Check the determinant of the matrix. It will find the eigenvalues of that matrix, and also outputs the corresponding eigenvectors.. For background on these concepts, see 7.Eigenvalues and Eigenvectors by Marco Taboga, PhD. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. If the resulting V has the same size as A, the matrix A has a full set of linearly independent eigenvectors that satisfy A*V = V*D. FINDING EIGENVALUES \u2022 To do this, we \ufb01nd the values of \u03bb which satisfy the characteristic equation of the matrix A, namely those values of \u03bb for which det(A \u2212\u03bbI) = 0, stream That example demonstrates a very important concept in engineering and science - eigenvalues and eigenvectors- which is used widely in many applications, including calculus, search engines, population studies, aeronautics \u2026 EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 \u22123 3 3 \u22125 3 6 \u22126 4 . which satisfy the characteristic equation of the. Any help is greatly appreciated. Since the zero-vector is a solution, the system is consistent. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. 3xx3 matrices and their eigenvalues and eigenvectors. The eigenvectors for D 0 (which means Px D 0x/ \ufb01ll up the nullspace. Finding of eigenvalues and eigenvectors. 2 Answers. To find all of a matrix's eigenvectors, you need solve this equation once for each individual eigenvalue. We compute a power of a matrix if its eigenvalues and eigenvectors are given. Favorite Answer. In linear algebra, the Eigenvector does not change its direction under the associated linear transformation. In general, for any matrix, the eigenvectors are NOT always orthogonal. <> The result is a 3x1 (column) vector. FINDING EIGENVALUES \u2022 To do this, we \ufb01nd the values of \u2026 The eigenvalues are immediately found, and finding eigenvectors for these matrices then becomes much easier. Notice how we multiply a matrix by a vector and get the same result as when we multiply a scalar (just a number) by that vector.. How do we find these eigen things?. How do you find the eigenvectors of a matrix? Find 2 linearly independent Eigenvectors for the Eigenvalue 0 c.) The e-value 0 has both geometric and algebraic multiplicity 2. If $\\theta \\neq 0, \\pi$, then the eigenvectors corresponding to the eigenvalue $\\cos \\theta +i\\sin \\theta$ are In this section we will introduce the concept of eigenvalues and eigenvectors of a matrix. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. Visit http://ilectureonline.com for more math and science lectures!In this video I will find eigenvector=? Solve the system. The eigenvector v of an operator A is a vector such that: A = To do this, we find the values of ? What is the trace of a matrix? We start by finding the eigenvalue: we know this equation must be true:. This is a linear system for which the matrix coefficient is . If you love it, our example of the solution to eigenvalues and eigenvectors of 3\u00d73 matrix will help you get a better understanding of it. To find eigenvectors, take $M$ a square matrix of size $n$ and $\\lambda_i$ its eigenvalues. matrices eigenvalues-eigenvectors. The generalized eigenvalue problem is to determine the solution to the equation Av = \u03bbBv, where A and B are n-by-n matrices, v is a column vector of length n, and \u03bb is a scalar. \u01b2\ufffd\u03c8\ufffdo\ufffd\ufffd|\ufffd\u07db\ufffdz?cI\ufffd\ufffd\ufffd4\ufffd\ufffd^?\ufffd\ufffdR9\ufffd\ufffd\ufffd(/k\ufffd\ufffd\ufffd\ufffdk The projection keeps the column space and destroys the nullspace: Relevance. 1 decade ago. Set the characteristic determinant equal to zero and solve the quadratic. Find the eigenvalues and eigenvectors. Yes, finding the eigenvectors should be straightforward. Remark. The result is a 3x1 (column) vector. Notice, however, that you have x=1 as a double root. The matrix A has an eigenvalue 2. If the determinant is 0, the matrix has no inverse. This pages describes in detail how to diagonalize a 3x3 matrix througe an example. Find the eigenvalues and bases for each eigenspace. For example, say you need to solve the following equation: First, you can rewrite this equation as the following: I represents the identity matrix, with 1s along its diagonal and 0s otherwise: Remember that the solution to [\u2026] Free Matrix Eigenvectors calculator - calculate matrix eigenvectors step-by-step This website uses cookies to ensure you get the best experience. \ufffd\ufffd~\ufffd?.\ufffd\ufffd\ufffd\ufffd(x\ufffd$\u05c4\ufffd\ufffd;\ufffdoE|Ik\ufffd\ufffd\ufffd\ufffd\ufffd$P\ufffd\ufffd\ufffd?\ufffdIha\ufffd\ufffd\u05a6\ufffdBB')\ufffd\ufffd\ufffdq\ufffd\ufffd\ufffd\ufffd\ufffdd\ufffdz\ufffd\ufffdI;E\ufffd\ufffd\ufffdk\ufffd\ufffdy\ufffd \ufffd@\ufffd\ufffd\ufffd9P}\ufffd\ufffd\ufffd\ufffdT\ufffd\ufffd\ufffd3\ufffdT\u05f8\ufffd2q\ufffdw8\ufffd{\ufffdT\ufffd*\ufffd\u007fN\ufffdmk\ufffd\u01dfJBZ\ufffdem\ufffd\ufffd\ufffd58j\ufffd\ufffdk\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd~\ufffd\ufffd\ufffd-lQ9i\ufffd[$aT$A\ufffd_\ufffd1#sv;q\u543a\ufffd\ufffdzz{5\ufffd\ufffdiB\ufffdnq\ufffd\ufffd()\ufffd\ufffd\ufffd6\ufffdau\ufffd\u07bc \ufffd\ufffd\ufffd)\ufffd\ufffdF\ufffd\u0710QXk\ufffdjhi8[=\ufffd\ufffd\ufffdn\ufffdB\ufffdF\ufffd\ufffd$.\ufffdCFZ\u041d.\ufffdP\u04b7D\ufffd\ufffd\ufffd\ufffd\u007fG\u0585KZ\ufffd\ufffd\ufffd\ufffdv\ufffd\ufffdv\ufffd\ufffd\u0280~\ufffd\ufffd|rq\ufffd\u0677\ufffd\ufffd\ufffd\ufffd3B\ufffdf\ufffd\ufffd\u0672\ufffd\ufffdl More: Diagonal matrix Jordan decomposition Matrix exponential. EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix. By using this website, you agree to our Cookie Policy. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. View all posts by KJ Runia, 4. The generalized eigenvalue problem is to determine the solution to the equation Av = \u03bbBv, where A and B are n-by-n matrices, v is a column vector of length n, and \u03bb is a scalar. If$ \\mathbf{I} $is the identity matrix of$ \\mathbf{A} $and$ \\lambda $is the unknown eigenvalue (represent the unknown eigenvalues), then the characteristic equation is \\begin{equation*} \\det(\\mathbf{A}-\\lambda \\mathbf{I})=0. Find the characteristic polynomial of a matrix \u2013 What is the fastest way to find eigenvalues? On the previous page, Eigenvalues and eigenvectors - physical meaning and geometric interpretation appletwe saw the example of an elastic membrane being stretched, and how this was represented by a matrix multiplication, and in special cases equivalently by a scalar multiplication. There may be two independent eigenvectors corresponding to that. If you need a refresher, check out my other lesson on how to find the determinant of a 2\u00d72.Suppose we are given a square matrix A where, Eigenvalue is the factor by which a eigenvector is scaled. $$\\tag{1}$$ , which is a polynomial equation in the variable$\\lambda$. I tried to find the inverse of the eigenvectors, but it brought a wrong matrix. Without having to make extensive calculations explain why 0 is an eigenvalue of A b.) I implemented an algorithm that computes three eigenvalues given a 3x3 Matrix. In quantum physics, if you\u2019re given an operator in matrix form, you can find its eigenvectors and eigenvalues. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. How do you find the eigenvectors of a 3x3 matrix? so clearly from the top row of the equations we get. Enter a matrix. Do you know how to solve it? [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. Calculate eigenvalues and eigenvectors. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons). This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. SOLUTION: \u2022 In such problems, we \ufb01rst \ufb01nd the eigenvalues of the matrix. Syntax: eigen(x) Parameters: x: Matrix Example 1: filter_none. by Marco Taboga, PhD. Av = \u03bbIv. For Example, if x is a vector that is not zero, then it is an eigenvector of a square matrix \u2026 When I try to solve for the eigenvectors I end up with a 3x3 matrix containing all 1's and I get stumped there. Eigenvalue is the factor by which a eigenvector is scaled. Favorite Answer. Eigenvalues and eigenvectors calculator. Finding of eigenvalues and eigenvectors. Matrix A: Find. Note that if we took the second row we would get . Eigenvalues and eigenvectors calculator. /\ufffd7P=\u0161\ufffd EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix. Eigenvectors for: Now we must solve the following equation: First let\u2019s reduce the matrix: This reduces to the equation: There are two kinds of students: those who love math and those who hate it. Syntax: eigen(x) Parameters: x: Matrix \u2026 Source(s): eigenvectors 3x3 matric: https://tinyurl.im/fNPuM. All that's left is to find the two eigenvectors. The code for this originally is \u2026 Please check my work in finding an eigenbasis (eigenvectors) for the following problem. The process for finding the eigenvalues and eigenvectors of a 3xx3 matrix is similar to that for the 2xx2` case. \u03bb 1 =-1, \u03bb 2 =-2. https://www.khanacademy.org/.../v/linear-algebra-eigenvalues-of-a-3x3-matrix The determinant of matrix M can be represented symbolically as det(M). The Formula of the Determinant of 3\u00d73 Matrix. If the determinant is 0, then your work is finished, because the matrix has no inverse. I do not wish to write the whole code for it because I know it is a long job, so I searched for some adhoc code for that but just found 1 or 2 libraries and at first I prefer not to include libraries and I don't want to move to matlab. In summary, when$\\theta=0, \\pi$, the eigenvalues are$1, -1$, respectively, and every nonzero vector of$\\R^2$is an eigenvector. I'm having a problem finding the eigenvectors of a 3x3 matrix with given eigenvalues. Calculate the eigenvalues and the corresponding eigenvectors of the matrix. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Display decimals, number of significant digits: Clean. In other words, the eigenvalues and eigenvectors are in$\\mathbb{R}^n$. Eigenvalues and Eigenvectors Consider multiplying a square 3x3 matrix by a 3x1 (column) vector. \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdlMOK\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\u007f\ufffd n\ufffd\ufffdh vx{Vb\ufffdHL\ufffd\ufffd\ufffd\ufffd%f;bz\\5\ufffd This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Find the eigenvalues and bases for each eigenspace. Eigenvalue$ \\boldsymbol{\\lambda = 6} \\$, 4.3. FINDING EIGENVALUES \u2022 To do this, we \ufb01nd the values of \u03bb which satisfy the characteristic equation of the matrix A, namely those values of \u03bb for which det(A \u2212\u03bbI) = 0, where I is the 3\u00d73 identity matrix. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. I implemented an algorithm that computes three eigenvalues given a 3x3 Matrix. \u2192Below is a calculator to determine matrices for given Eigensystems. Thanks! Answer Save. The determinant of a triangular matrix is easy to find - it is simply the product of the diagonal elements. I am trying to find the eigenvectors for the following 3x3 matrix and are having trouble with it. 27.7k 1 1 gold badge 25 25 silver badges 52 52 bronze badges. You need to calculate the determinant of the matrix as an initial step. The values of \u03bb that satisfy the equation are the generalized eigenvalues. Av = \u03bbv. Illustrate the process of finding eigenvalues and corresponding eigenvectors of a 3x3 matrix. Notice, however, that you have x=1 as a double root. To find the inverse of a 3x3 matrix, first calculate the determinant of the matrix. Get the free \"Eigenvalue and Eigenvector for a 3x3 Matrix \" widget for your website, blog, Wordpress, Blogger, or iGoogle. The only eigenvalues of a projection matrix are 0 and 1. I have to find 4 things for the Matrix A which is a 3x3 matrix with all values equal to 1 A= 1 1 1 1 1 1 1 1 1 a.) In this page, we will basically discuss how to find the solutions. How to find the eigenspace of a 3x3 matrix - Suppose A is this 2x2 matrix: [1 2] [0 3]. which satisfy the characteristic equation of the.\n\n## how to find eigenvectors of a 3x3 matrix\n\nFacts About African Wild Dogs, Family Member In Vegetative State, Black-eyed Susan Climber, How To Bleach Hair Men, Brie Baguette Sandwich, Lumber Liquidators Stock, Ubuntu Emacs 27, Barbados Hurricane Risk, Where To Get Vodka, Bbq Gas Valve Replacement, Cool Youth Batting Gloves,", "date": "2021-03-03 13:59:29", "meta": {"domain": "easylia.com", "url": "http://nacretv.easylia.com/uamkx/1ea173-how-to-find-eigenvectors-of-a-3x3-matrix", "openwebmath_score": 0.8721694946289062, "openwebmath_perplexity": 350.64668699753804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.963779946215714, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8535538542034362}}
{"url": "http://bibliotekacmolas.pl/3bhqq/4s7hgw.php?page=6cdafa-simple-cryptography-examples", "text": "# simple cryptography examples\n\nIn the examples above, statement (A) is plaintext, while statement (B) is a reverse cipher text. Thus, cryptography is defined as the art of manipulating or scrambling plaintext into ciphertext. An example of primitive cryptography is an encrypted message in which it takes over from letters with other characters. I will discuss a simple method of enciphering and deciphering a message using matrix transformations and modular arithmetic, and show how elementary row operations can sometimes be used to break an opponent's code. Cryptography is the technique of protecting information by transforming it into a secure format. Introduction. Simple substitution cipher is the most commonly used cipher and includes an algorithm of substituting every plain text character for every cipher text character. Select primes p=11, q=3. Cryptography is the science of keeping information secret and safe by transforming it into form that unintended recipients cannot understand. Solved Examples 1) A very simple example of RSA encryption This is an extremely simple example using numbers you can work out on a pocket calculator (those of you over the age of 35 45 can probably even do it by hand). Final Example: RSA From Scratch This is the part that everyone has been waiting for: an example of RSA from the ground up. Hence the modulus is $$n = p \\times q = 143$$. 2. n = pq \u00e2\u0080\u00a6 In this process, alphabets are jumbled in comparison with Caesar cipher algorithm. I am first going to give an academic example, and then a real world example. A simple example of an encryption algorithm would be changing all Ns to a 3, or all Zs to a 1. Classic Encryption - The Caesar Cipher. Background: Many of the ideas we use to keep secrets in the digital age are far older than the Internet. Cryptography, or cryptology (from Ancient Greek: \u00ce\u00ba\u00cf\u0081\u00cf\u0085\u00cf\u0080\u00cf\u0084\u00cf\u008c\u00cf\u0082, romanized: krypt\u00f3s \"hidden, secret\"; and \u00ce\u00b3\u00cf\u0081\u00ce\u00ac\u00cf\u0086\u00ce\u00b5\u00ce\u00b9\u00ce\u00bd graphein, \"to write\", or -\u00ce\u00bb\u00ce\u00bf\u00ce\u00b3\u00ce\u00af\u00ce\u00b1-logia, \"study\", respectively), is the practice and study of techniques for secure communication in the presence of third parties called adversaries. Cryptography originated approximately 4000 years ago in Egypt. Keys for a simple substitution cipher usually consists of 26 letters. The word cryptography comes from the word: KRYPTOS and GRAPHEIN. The process of encoding a plain text message in some secret way is called Encryption. Simple ciphers Simple encryption algorithms, which were invented long before first computers, are based on substitution and transposition of single plaintext characters. The following are common examples of encryption. Encryption is the conversion of information into an cryptographic encoding that can't be read without a key.Encrypted data looks meaningless and is extremely difficult for unauthorized parties to decrypt without the correct key. 1. The RSA encryption system is the earliest implementation of public key cryptography. Quick Background. This project for my Linear Algebra class is about cryptography. Meanwhile, the operations performed in modern encryption algorithms are usually similar but they affect single bits and bytes. The routine may perform several passes and changes, called permutations, on the plaintext. An example key is \u00e2\u0088\u0092 Calculation of Modulus And Totient Lets choose two primes: $$p=11$$ and $$q=13$$. Example. The purpose of this note is to give an example of the method using numbers so small that the computations can easily be carried through by mental arithmetic or with a simple calculator. 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Character for every cipher text character for every cipher text character for every cipher text character for every text! Algorithms are usually similar but they affect single bits and bytes text message which. Called encryption the plaintext keep secrets in the digital age are far than! System is the most commonly used cipher and includes an algorithm of substituting plain! Cipher usually consists of 26 letters background: Many of the ideas we use to keep secrets in the age! Secret and safe by transforming it into a secure format RSA encryption system is the technique protecting...\n\nTen wpis zosta\u0142 opublikowany w kategorii Aktualno\u015bci. Dodaj zak\u0142adk\u0119 do bezpo\u015bredniego odno\u015bnika.\n\nMo\u017cliwo\u015b\u0107 komentowania jest wy\u0142\u0105czona.", "date": "2021-03-01 10:49:37", "meta": {"domain": "bibliotekacmolas.pl", "url": "http://bibliotekacmolas.pl/3bhqq/4s7hgw.php?page=6cdafa-simple-cryptography-examples", "openwebmath_score": 0.4021863639354706, "openwebmath_perplexity": 2478.223593725322, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9637799404938199, "lm_q2_score": 0.8856314768368161, "lm_q1q2_score": 0.8535538520452405}}
{"url": "https://math.stackexchange.com/questions/2327748/probability-or-rolling-1-1-2-with-six-dice", "text": "# probability or rolling 1 1 2 with six dice\n\nWhat is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?\n\nEquivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.\n\nThere are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478\n\nI'm not sure how to get this without the help of a computer though.\n\n\u2022 What have you tried? Also, since you have six dice, is unclear whether, for instance, $(1,1,2,3,1,2)$ is valid. Is it valid? \u2013\u00a0Arthur Jun 18 '17 at 20:29\n\u2022 yes, 1,1,2,x,x,x where x could be any value \u2013\u00a0mjmdavis Jun 18 '17 at 20:38\n\u2022 Common practice is to say at least two 1's and one 2 in your case and exactly \" ... \" otherwise. \u2013\u00a0infinitylord Jun 18 '17 at 20:44\n\u2022 @infinitylord like that? \u2013\u00a0mjmdavis Jun 18 '17 at 20:48\n\u2022 If you want positive responses, I recommend stating the question as: \"What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die\" and then (and this part is really huge for this site) show your attempt at the problem. \u2013\u00a0infinitylord Jun 18 '17 at 21:01\n\nApproach via inclusion-exclusion and De Morgan's laws.\n\nLet $A$ be the event that you roll at least two $1$'s. Let $B$ be the event that you roll at least one $2$.\n\nWe are attempting to calculate then $Pr(A\\cap B)$\n\n$Pr(A\\cap B)=1-Pr((A\\cap B)^c)=1-Pr(A^c\\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\\cap B^c)$\n\nWe calculate each term on the right now that they are in simpler form.\n\n$Pr(A^c)$ refers to the probability of strictly fewer than two $1$'s occurring, i.e. at most one $1$ occurring. Either exactly one $1$ occurs or no $1$'s occur. In the case of exactly one $1$, pick its location and then pick each remaining digit for a total of $6\\cdot 5^5$ possibilities. In the case of no $1$'s, pick each digit for a total of $5^6$ possibilities. Taking the ratio of this compared to the $6^6$ equally likely dice rolls, we calculate $Pr(A^c)=\\frac{6\\cdot 5^5+5^6}{6^6}$.\n\n$Pr(B^c)$ refers to the probability of strictly fewer than one $2$ occurring, i.e. no $2$'s. This occurs with probability $Pr(B^c)=\\frac{5^6}{6^6}$\n\n$Pr(A^c\\cap B^c)$ refers now to the probability of at most one $1$ and no $2$'s, which similarly to before we break into cases for either exactly one $1$ or no $1$'s. For exactly one $1$, first pick the location, then pick each remaining digit for $6\\cdot 4^5$ possibilities and for no $1$'s pick each digit for $4^6$ possibilities for a probability of $Pr(A^c\\cap B^c)=\\frac{6\\cdot 4^5+4^6}{6^6}$\n\nThis gives a final probability of:\n\n$$Pr(A\\cap B) = 1-\\frac{6\\cdot 5^5+5^6}{6^6}-\\frac{5^6}{6^6}+\\frac{6\\cdot 4^5+4^6}{6^6}$$\n\n$$=1-\\frac{39760}{46656} = \\frac{431}{2916}\\approx 0.147805$$\n\nYou can consider the possibility to find probability directly without inclusion-exclusion principle. This is a longer way, but there are those who like it. Consider all possibilities to roll at least two ones and at least one two with six 6-sided die. We can have exactly two $1$ and one $2$. Denote this event by $A_{21}$. Or we can have exactly two $1$ and two $2$: event $A_{22}$. There are $10$ disjoint events total: $A_{21}$, $A_{22}$, $A_{23}$, $A_{24}$, $A_{31}$, $A_{32}$, $A_{33}$, $A_{41}$, $A_{42}$, $A_{51}$.\n\nFind and then add their probabilities. The probability of $A_{21}$ is $$\\mathbb P(A_{21})=\\frac{\\binom{6}{2}\\binom{6-2}{1}4^{6-2-1}}{6^6}=\\frac{15\\cdot 4\\cdot 4^3}{6^6}.$$ By first multiplier $\\binom{6}{2}=15$ we choose the rolls which result in $1$, next multiplier $\\binom{6-2}{1}=4$ we choose the roll among the rest which results in $2$, and the last multiplier $4^3$ is number of possible outcomes in the rest $6-2-1=3$ rolls.\n\nSimilar, $$\\mathbb P(A_{22})=\\frac{\\binom{6}{2}\\binom{4}{2}4^{2}}{6^6}=\\frac{15\\cdot 6\\cdot 16}{6^6},$$ $$\\mathbb P(A_{23})=\\frac{15\\cdot 4\\cdot 4}{6^6},\\quad \\mathbb P(A_{24})=\\frac{15}{6^6},\\quad \\mathbb P(A_{31})=\\frac{20\\cdot 3\\cdot 16}{6^6},$$ $$\\mathbb P(A_{32})=\\frac{20\\cdot 3\\cdot 4}{6^6}, \\quad \\mathbb P(A_{33})=\\frac{20}{6^6}, \\quad \\mathbb P(A_{41})=\\frac{15\\cdot 2\\cdot 4}{6^6},$$ $$\\mathbb P(A_{42})=\\frac{15}{6^6}, \\quad \\mathbb P(A_{51})=\\frac{6}{6^6}.$$ The total probability of rolling at least two ones and at least one two with six 6-sided die is equal to the sum of probabilities above: $$\\frac{6896}{6^6}=0,147805213.$$\n\nAs I understood your question, you are asking about probability that between $6$ values will be at least two $1$'s and one $2$ (don't looking at other values and order).\n\nWell, so the probability of $(1,1,2,x,x,x)$ (excluding order and remarking that there may be other $1$'s and $2$'s between $x$'s) is exactly the number of sets of six digits from $1$ to $6$ that start with $1,1,2$ divided by total number of sets of six digits from $1$ to $6$. There are as many (disordered) sets of six digits from $1$ to $6$ that start with $1,1,2$ as just (disordered) sets of three digits from $1$ to $6$.\n\nThus the answer is $P = \\frac{(3^3)/3!}{(6^6)/6!} = \\frac{5}{72}$.\n\n\u2022 Hasek I'm not sure if this result is correct. \u2013\u00a0mjmdavis Jun 18 '17 at 23:07\n\u2022 The line \"There are as many (disordered) sets of six digits from 1 to 6 that start with 1,1,2 as just (disordered) sets of three digits from 1 to 6.\" is incorrect. After all, what does it mean for a \"disordered set to start with something\"? Further, your calculations don't appear to make much sense. What would $3^3/3!$, your supposed numerator, represent? Notice that $3!$ is even but $3^3$ is odd and so $3^3/3!$ is not an integer, so it could not possibly represent the answer to a counting problem. Similarly for the denominator. \u2013\u00a0JMoravitz Jun 18 '17 at 23:28\n\u2022 Yes, I'm definitely wrong, sorry. \u2013\u00a0Hasek Jun 19 '17 at 6:44", "date": "2019-02-19 06:57:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2327748/probability-or-rolling-1-1-2-with-six-dice", "openwebmath_score": 0.9045918583869934, "openwebmath_perplexity": 173.6843580484299, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319877136793, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.853553464481981}}
{"url": "http://clandiw.it/umtr/inverse-fourier-transform-of-unit-step-function.html", "text": "In reality there is no such thing as a unit step, and the Fourier transform is not 'real'. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. The Fourier transform we\u2019ll be int erested in signals de\ufb01ned for all t the Four ier transform of a signal f is the function F (\u03c9)= \u221e \u2212\u221e f (t) e \u2212 j\u03c9t dt \u2022 F is a function of a real variable \u03c9;thef unction value F (\u03c9) is (in general) a complex number F (\u03c9)= \u221e \u2212\u221e f (t)cos \u03c9tdt \u2212 j \u221e \u2212\u221e f (t)sin \u03c9tdt \u2022| F (\u03c9) | is called the amplitude spectrum of f; F (\u03c9) is the phase spectrum of f \u2022 notation: F = F (f) means F is the Fourier transform of f. Remark 6 The definition of the Fourier transform on implies that whenever , we have that. If any argument is an array, then fourier acts element-wise on all elements of the array. $\\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. How about going back? Recall our formula for the Fourier Series of f(t) : Now transform the sums to integrals from \u2013\u221eto \u221e, and again replace F m with F(\u03c9). 2 Fourier Series Consider a periodic function f = f (x),de\ufb01ned on the interval \u22121 2 L \u2264 x \u2264 1 2 L and having f (x + L)= f (x)for all. Because the original function and its inverse Laplace transform are only valid for t\u201a 0, some people introduce a Heaviside step function H ( t ) (see Section B. Fourier Transform Pairs (contd). 4M subscribers. Laplace transform of the unit step function | Laplace transform | Khan Academy - YouTube. \u2022 Sometimes we want to use one-dimensional Fourier transforms or inverse transforms. Explain briefly below. The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain. Unlike the inverse Fourier transform, the inverse Laplace transform in Eq. Step 7: Check the \u201cInverse\u201d box only if you have results from a prior analysis and you want to find the original function. The book is divided into four major parts: periodic functions and Fourier series, non-periodic functions and the Fourier integral, switched-on signals and the Laplace transform, and finally the discrete versions of these transforms, in particular the Dis-crete Fourier Transform together with its fast implementation, and the z-transform. Unit analysis, algebra solver free step by step, how fast can one learn algebra, mathsiequalities, online direction field. Solution: Here, =0 for <2 , then \u02dd =1 for \u22652. Fourier Transform \u0432\u0402\u201d Theoretical Physics Reference 0. Before proceeding into solving differential equations we should take a look at one more function. It cannot be said that time information is lost because it is possible to recover the original time domain observation using the Inverse Fourier. The calculator will find the Inverse Laplace Transform of the given function. The usual Fourier transform tables found online don't have many functional relationship rules. we can get the Fourier transform of a unit impulse as the time derivative of a unit step function: Alternatively, by definition, the forward Fourier transform of an impulse function is and the inverse transform is. The Fourier Transform (used in signal processing) The Laplace Transform (used in linear control systems) The Fourier Transform is a particular case of the Laplace Transform, so the properties of Laplace transforms are inherited by Fourier transforms. Interestingly, these transformations are very similar. Together with a great variety, the subject also has a great coherence, and the hope is students come to appreciate both. You take the Fourier transform fft of f. Suppose that the Fourier transform of f and its inverse exist. This is specifically due to its property that it is neither absolutely summable nor square summable. dft() and cv2. How can you create a delta function using some other function, the Fourier transform of which you already know. Fourier transform pair The function ! X(j\") is the Fourier transform of the signal x(t) and conversely x(t) is the inverse Fourier transform of! X(j\"). If any argument is an array, then fourier acts element-wise on all elements of the array. 2 Fourier Series Consider a periodic function f = f (x),de\ufb01ned on the interval \u22121 2 L \u2264 x \u2264 1 2 L and having f (x + L)= f (x)for all. Mathematicians have developed tables of commonly used Laplace transforms. PLOTTING STEP RESPONSE OF TRANSFER FUNCTION Learn more about fourier transform. The notation is introduced in Trott (2004, p. Discrete-Time Fourier Transform (DTFT) inverse DTFT. text orientation finding) where the Fourier Transform is used to gain information about the geometric structure of the. We can solve the integral by contour integration. Denoted , it is a linear operator of a function f(t) with a real argument t (t \u2265 0) that transforms it to a function F(s) with a complex argument s. The book only states a limited form of the Heavyside expansion theorem in problem 5 of section 53. The Z transform of the discrete time unit ramp function 42. (iii) Comment the time domain expression of the filter. 1 Dirac Delta Function 1 2 Fourier Transform 5 3 Laplace Transform 11 3. Visualizing Pole-Zero plot: Since the z-transform is a function of a complex variable, it is convenient to describe and interpret it using the complex z-plane. 12 tri is the triangular function 13 Dual of rule 12. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time t. calculating the Fourier transform of a signal, then exactly the same procedure with only minor modification can be used to implement the inverse Fourier transform. As such, the restriction of the Fourier transform of an L 2 (R n) function cannot be defined on sets of. Consider the Fourier transforms of the functions in Example 9. 1) into the integral in the de\ufb02nition of the inverse transform in (F. The special characteristics of the Fast Fourier Transform implementation are described. Inverse Transform 6. In this section we introduce the step or Heaviside function. is the Fourier Transform of f(t). Fourier transforms take the process a step further, to a continuum of n-values. For instance, the inverse continuous Fourier transform of both sides of Eq. Correlation, autocorrelation. For the Laplace transform, the Fourier transform existed if the ROC included the j!axis. I have to find the inverse fourier transform for: \\frac{e^{i 6\\omega}}{\\omega} So I'm using a table, then. In this section we introduce the Fourier transform and then we illustrate the fast Fourier transform algorithm, applied to the projection of unit-step i. Continuous Fourier Transform A general Fourier Transform for Spectrum Representation \u2022With the unit-impulse function incorporated, the continuous Fourier transform can represent a broad range of continuous-time signals. Basic Fourier Transform Theory: Relationship to Chap 3 on Fourier Series Interpretation of Inverse Fourier Transform Frequency Ranges of Biological, E&M, and other Signals. In reality there is no such thing as a unit step, and the Fourier transform is not 'real'. It is \"off\" (0) when < , the \"on\" (1) when \u2265. For math, science, nutrition, history. The Fourier Transform and its Inverse The Fourier Transform and its Inverse: So we can transform to the frequency domain and back. 3 Complex form of Fourier series, Fourier integral representation, Fourier Transform and Inverse Fourier transform of constant and exponential function. 1) into the integral in the de\ufb02nition of the inverse transform in (F. Laplace transform to solve a differential equation. Fourier Transform - Free download as Powerpoint Presentation (. Fourier transform of the unit step function and of the signum function: The signum function sgn( t) is a function that is related to the unit step function. In this study, a new inversion method is presented for performing two-dimensional (2D) Fourier transform. Consider the equation f(x) + A(f(x \u2212 1) + f(x + 1)) = u(x) where u(x) is a known function, absolutely integrable, on R and A is a constant. Evaluate one transform on data from step 3. The Laplace transform is similar to the Fourier transform. 12-2 Circuit Analysis Using the Fourier Transform Determine and plot the spectrum of the response Vo(co) of the circuit of Figure 15. \u2022 The unit step function (1 class) \u2022 The Dirac delta function (1 class) \u2022 Applications of step and impulse functions (1 class) \u2022 Periodic functions and their applications (2 classes) \u2022 Convolution and applications (2 classes) \u2022 Solving integral equations (1 class) \u2022 Fourier series (3 classes) \u2022 Fourier integral representation (1. Inverse Laplace Transform with unit step function, sect7. 1998 We start in the continuous world; then we get discrete. The Fourier transform is only valid for a periodic function, and a unit step is not periodic. So the Fourier transform X (jw) of x (t) is the convolution of X (jw) and sinc (w). Fast Fourier Transforms Phase factors There are functions that produce roots-of-one as a function of time (t) or place (x). e\ufb01ne the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f (t)= 0 t< 0 1 t \u2265 0? the Laplace transform is 1 /s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1 /j\u03c9 in fact, the integral \u221e \u2212\u221e f (t) e \u2212 j\u03c9t dt = \u221e 0 e \u2212 j\u03c9t dt = \u221e 0 cos. In this section we introduce the step or Heaviside function. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. The unit pulse function can be defined with the help of the Heaviside unit step function ( ) ( ) ( ) 0 x a f t Ht a Ht a 1 x a 0 x a <\u2212 = +\u2212 \u2212= < > a0 > The Fourier transform of this function can be determined as. of a second over a period of 10 seconds. Uniqueness of Fourier transforms, proof of Theorem 3. 2) The DTFT X. 3 Complex form of Fourier series, Fourier integral representation, Fourier Transform and Inverse Fourier transform of constant and exponential function. 3 Properties of The Continuous -Time Fourier Transform 4. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. So essentially, in the decomposition of x of t as a linear combination of complex exponentials, the complex amplitudes of those are, in effect, the Fourier transform scaled by the differential and scaled by 1 over 2 pi. 6 ) and obtain. There are different definitions of these transforms. Remembering the fact that we introduced a factor of i (and including a factor of 2 that just crops up. UNIT3: FOURIER TRANSFORM - Complex form of Fourier Transform and its inverse, Fourier sine and cosine transform and their inversion. Remembering the fact that we introduced a factor of i (and including a factor of 2 that just crops up. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Inverse Fourier Transform F [f] Step 3) Find the inverse transform. I think you should have to consider the Laplace Transform of f(x) as the Fourier Transform of Gamma(x)f(x)e^(bx), in which Gamma is a step function that delete the negative part of the integral and e^(bx) constitute the real part of the complex exponential. xxxiv), and and are sometimes also used to denote the Fourier transform and inverse Fourier transform, respectively (Krantz 1999, p. Dirac defined the delta function as shown below. If n is less than the length of the signal, then ifft ignores the remaining signal values past the nth entry and. Express f under an integral form. Overlapping in real time fourier transform? Hot Network Questions Numbers by Position a name for a boy, boy's name, boy name Mother milk of 6 Corona-positive (COVID-19) women does not contain the virus - can we make a confidence. How can you create a delta function using some other function, the Fourier transform of which you already know. The convergence criteria of the Fourier transform (namely, that the function be absolutely integrable on the real line) are quite severe due to the lack of the exponential decay term as seen in the Laplace transform, and it means that functions like polynomials, exponentials, and trigonometric functions all do not have Fourier transforms in the. Its value is not trivial to calculate, and ends up being. Any good reference to more detailed tables would be very helpful! My attempt: $\\mathcal F[f\\times u] = (\\mathcal Ff)*(\\mathcal Fu)$ where * denotes convolution. The usual Fourier transform tables found online don't have many functional relationship rules. As with the Laplace transform, calculating the Fourier transform of a function can be done directly by using the definition. Magnitude and phase representation of the Fourier transform and frequency response of LTI systems; Applications of the. ej!O /that results from the de\ufb01nition is a function of frequency !O. 2 Contents 2. The more general statement can be found in standard texts devoted to Laplace transforms. Determine the Fourier transform of the non-periodic signals shown in the figures below: (b) 8(1) -2 -1 0 1 2. Introduction to Inverse Problems Guillaume Bal 1 July 2, 2019 1University of Chicago, Chicago, IL 60637; [email\u00a0protected] Fourier transforms take the process a step further, to a continuum of n-values. 2 The Fourier transform Given a function f(x) de ned for all real x, we can give an alternative representation to it as an integral rather than as an in nite series, as follows f(x) = Z eikxg(k)dk Here g(x) is called the Fourier transform of f(x), and f(x) is the inverse Fourier transform of g(x). 8 Fourier transforms. Second Implicit Derivative (new) Derivative using Definition (new) Derivative Applications. In this section we introduce the step or Heaviside function. The unit step function, also known as the Heaviside function, is defined as such:. 2 Fourier Series Consider a periodic function f = f (x),de\ufb01ned on the interval \u22121 2 L \u2264 x \u2264 1 2 L and having f (x + L)= f (x)for all. The Fourier transform is a particular case of z-transform, i. INTRODUCTION AND FOURIER TRANSFORM OF A DERIVATIVE One can show that, for the Fourier transform g(k) = Z 1 1 f(x)eikx dx (1) to converge as the limits of integration tend to 1 , we must have f(x) ! 0 as. The derivation can be found by selecting the image or the text below. Solved examples of the Laplace transform of a unit step function. We also work a variety of examples showing how to take Laplace transforms and inverse Laplace transforms that involve Heaviside functions. It is clearly desirable that there should be a canonical definition of the Fourier Transform, consistent with classical definitions, which is applicable to all distributions - or, at least to some. The Discrete Fourier Transform the two transforms and then \ufb01look up\ufb02 the inverse transform to get the convolution. The excel fourier analysis tool. 4142*j]; x_n=ifft(X_K) Example 2: X_K=[10,-2+2*j,-2,-2-2*j]; x_n=. Discrete Time Fourier Transforms The discrete-time Fourier transform or the Fourier transform of a discrete-time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence. fast fourier transform. This transformation is essentially bijective for the majority of practical. !! If you apply the Fourier transform to function f(t), you get a new function F(w). Learn more about Chapter 8: The Fourier Transform on GlobalSpec. Since the transform of a lattice in real space is a reciprocal lattice, the diffraction pattern of the crystal samples the diffraction pattern of a single unit cell at the points of the reciprocal. When you have worked through this unit you should:. 2) factor (1/2\u03c0 )2 must be replaced by (1/2\u03c0 ) To avoid confusion, we shall indicate one-dimensional Fourier transforms by Fx, Fx-1 or Fky. We experi-ment here to see if Mathematica knows these functions, and if it can deal with their Fourier transforms. The Fast Fourier Transform for polynomials works in an analogous way to a slide rule. The Fourier Transform Saravanan Vijayakumaran [email\u00a0protected] 1) which is now called Heaviside step function. The fourier transform uses the assumption that any finite time-domain signal can be broken into an infinite sum of sinusoidal (sine and cosine waves) signals. I don't know where you got G(f), but it only a mathematical expression to \"give\" the value of the Fourier transform of a unit step. Let tqptqu. 1,791,367 views. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t \u2032 s c o r r e c t. Properties of the Fourier Series 51. The key step in the proof of (1. The properties are useful in determining the Fourier transform or inverse Fourier transform They help to represent a given signal in term of operations (e. Especially important among these properties is Parseval's Theorem, which states that power computed in either domain equals the power in the other. CT Fourier Transform Pairs signal (function of t) $\\longrightarrow$ Fourier transform (function of $\\omega$) : 1 CTFT of a unit impulse $\\delta (t)\\$ $1 \\$. The Fourier transform 45. Fourier transform. Inverse Fourier transform \u2013 be able to compute this from definition as well as from looking up the transform for elementary signals. This MATLAB function returns the Fourier Transform of f. UNIT STEP FUNCTIONS AND PERIODIC FUNCTIONS 157 Which implies that y(t) = t2 solves the DE. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. If Y is a matrix, then ifft (Y) returns the inverse transform of each column of the matrix. Q5(a) is given to be: 2 1 Feje() ( 1)\u03c9\u03c9jj\u03c9\u03c9 \u03c9 = \u2212\u2212 Use this information and the time-shifting and time-scaling properties, find the Fourier transforms of the signals. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. Introduction to Hilbert Transform. In this video tutorial, the tutor covers a range of topics from from basic signals and systems to signal analysis, properties of continuous-time Fourier transforms including Fourier transforms of standard signals, signal transmission through linear systems, relation between convolution and correlation of signals, and sampling theorems and techniques. Ada version of General N Point Fast Fourier Transform. Discrete-Time Fourier Transform (DTFT) inverse DTFT. Solved examples of the Laplace transform of a unit step function. Note that the usual results for Fourier transforms of even and odd functions still hold. IQ v(t) vo(t) FIGURE 15. The Fourier transform of controlled-source time-domain electromagnetic data by smooth spectrum inversion Yuji Mitsuhata. Problems at x!+1are removed by multiplying by e cx, where cis a positive real number. The Fourier transform we\u2019ll be int erested in signals de\ufb01ned for all t the Four ier transform of a signal f is the function F (\u03c9)= \u221e \u2212\u221e f (t) e \u2212 j\u03c9t dt \u2022 F is a function of a real variable \u03c9;thef unction value F (\u03c9) is (in general) a complex number F (\u03c9)= \u221e \u2212\u221e f (t)cos \u03c9tdt \u2212 j \u221e \u2212\u221e f (t)sin \u03c9tdt \u2022| F (\u03c9) | is called the amplitude spectrum of f; F (\u03c9) is the phase spectrum of f \u2022 notation: F = F (f) means F is the Fourier transform of f. Thus the Fourier transform on tempered distributions is an extension of the classical definition of the Fourier transform. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. he Fourier and Laplace transforms can be rectangular pulse: f (t)= 1 e\u043f\u00ac\u0403ne the Fourier transform of a step function or a constant signal, The aim of this post is to properly understand Numerical Fourier Transform on Python or Matlab with an example in fourier transform of the. study how a piecewise continuous function can be constructed using step functions. As in the FDK analysis, the s direction transforms require a total of 2 NMP 2 log(2 P ) operations for the forward direction and 2 NP 2 log(2 P ) for the inverse direction. 8 Filters 2. Inverse Z Transform: Part 2. of a unit step can be inferred, but it's natural with the Laplace. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)\u2026. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The step response is the convolution. Fourier Transform of Unit Step Function Guess Fourier Transform of Unit Step Function F ( ) 0 |F(j )| 0 t 1 f(t) Fourier Transforms of Special Functions Fourier Transform vs. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The usual Fourier transform tables found online don't have many functional relationship rules. Have these ideas in mind as we go through the examples in the rest of this section. This chapter introduces the Fourier Transform, also known as the Fourier Integral. Linearity and the result for the unit step, above. Fourier transform of unit step signal u(t). MAXIMA Quick Reference Labels. The Fourier transform of the unit step function is not any of those things. It's basically a set of Sine waves with amplitudes and phases. Laplace transform to solve a differential equation. Once the transformation has been applied, time information is hidden and cannot be easily observed. 10 Band-Pass Systems. Together with a great variety, the subject also has a great coherence, and the hope is students come to appreciate both. Unit analysis, algebra solver free step by step, how fast can one learn algebra, mathsiequalities, online direction field. The Fourier transform of an integrable function is continuous and the restriction of this function to any set is defined. Try to integrate them? Cite. That is, given the Fourier transform of an function, when can we recover the original function from ? We begin with a simple case where the recovery is quite easy. Table of Fourier Transform Pairs of Energy Signals Function name Time Domain x(t) Frequency Domain X Unit step () 10 00 t ut 2. Visualizing Pole-Zero plot: Since the z-transform is a function of a complex variable, it is convenient to describe and interpret it using the complex z-plane. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series Inverse Fourier transform: The Fourier integral theorem Example: the rect and sinc functions Cosine and Sine Transforms Symmetry properties Periodic signals and functions Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 22. The special characteristics of the Fast Fourier Transform implementation are described. From this block diagram we can find overall transfer function which is nonlinear in nature. Example 2-2 SPECTRUM OF AN EXPONENTILA PULSE By means of direct integration find the Fourier transform of ) ( t w < = - 0 , 0 0 , ) ( t t e t w t Properties of Fourier Transforms. These reviews did not try 44 to minimize Laplace-space function evaluations, since their functions were simple closed-form expressions, 45 not simulations. Time Reversal and Frequency Response By Clay S. Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform f t F( )ej td 2 1 ( ) Definition of Fourier Transform F() f (t)e j tdt Trigonometric Fourier Series 1 ( ) 0 cos( 0 ) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)\u2026. Fourier Transform We will use the convention that a time function, g(t), and the Fourier Transform (FT) of that function, g(!), are in the time or frequency domain as indicated by the argument list rather than some variation on the function symbol. transforms on pairs from step 1. 7 Transmission of Signals Through Linear Systems 2. Tags: EMML, inner product, probability density functions, likelihood function, linear functional, orthonormal basis, linear transformation, vector, Linear Algebra. Fourier Transforms and the Dirac Delta Function A. The usual Fourier transform tables found online don't have many functional relationship rules. If any argument is an array, then fourier acts element-wise on all elements of the array. 16) Several important transforms are listed in the following table: f(t) F( ) a. Last time, we saw the equations that calculate the Fourier Transform and its inverse. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. If you really want to understand the Fourier and Laplace transforms , how they work and why they work then this is the course for you. Characteristics of the Continuous Fourier Transform The plots in Figures 1-1 and 1-2 demonstrate two characteristics of the Fourier transforms of real time history functions: 1. 1 Dirac delta function The delta function \u2013(x) studied in this section is a function that takes on zero values at all x 6= 0, and is in\ufb02nite at x = 0, so that its integral +R1 \u00a11 \u2013(x)dx = 1. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. How to solve a basic math equation, foerester's algebra 1 suggested timelien, how to answe algebra problems, free algebra word problem solver, algebra1 answer keys g. 3 If f (x) is a good function with its Fourier transform g( y), then the Fourier transform of f (x) is 2\u03c0iyg( y), and the Fourier transform of f (ax + b) is |a|\u22121 e2\u03c0iby/a g( y/a). Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform f t F( )ej td 2 1 ( ) Definition of Fourier Transform F() f (t)e j tdt Trigonometric Fourier Series 1 ( ) 0 cos( 0 ) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)\u2026. So we can write S2+S as S(S+1) now we can rewrite the equation as (S+2. Fessler,May27,2004,13:11(studentversion) Subtleties in dening the ROC (optional reading!) We elaborate here on why the two possible denitions of the ROC are not equivalent, contrary to to the book's claim on p. As such, it transforms one function into another, which is called the frequency domain representation of the original function (where the original function is often a function in the time-domain). \" The full name of the function is \"sine cardinal,\" but it is commonly referred to by its abbreviation, \"sinc. Notice the minus sign! Usually, to get rid of that, the inverse transform is written with a minus sign inside the exponential. computation of the Z transform with contour integration 43. Trigonometric Polynomials 58. 2 Transforms of Derivatives and Integrals 6. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. 11-9) give sketches of possible Fourier transform magni- tudes. The second channel for the imaginary part of the result. To see how the Fourier transform works, we will begin with a one-dimensional signal and consider a simple step function. We experi-ment here to see if Mathematica knows these functions, and if it can deal with their Fourier transforms. The sinc function sinc(x) is a function that arises frequently in signal processing and the theory of Fourier transforms. I used overlapping windows of 1024 points. Learn more about Chapter 8: The Fourier Transform on GlobalSpec. FTIR spectrometers (Fourier Transform Infrared Spectrometer) are widely used in organic synthesis, polymer science, petrochemical engineering, pharmaceutical industry and food analysis. Analyzing the frequency components of a signal with a Fast Fourier Transform. Third Derivative. Example 21 Find the Fourier transform of the function where represents unit step function Solution: Fourier transform of is given by = = or Result: Note: If Fourier transform of is taken as , then Example 22 Find the inverse transform of the following functions: i. The expression in (7), called the Fourier Integral, is the analogy for a non-periodic f (t) to the Fourier series for a periodic f (t). Let tqptqu. Modular graph functions associate to a graph an SL(2,Z)-invariant function on the upper half plane. How to do a fast fourier transform fft in microsoft excel 1. Any good reference to more detailed tables would be very helpful! My attempt: $\\mathcal F[f\\times u] = (\\mathcal Ff)*(\\mathcal Fu)$ where * denotes convolution. We denote by Sn,k the set of all k-SIIRVs of order n. Usually, to find the Inverse Laplace Transform of a function, we use the property of linearity of the Laplace Transform. The Fourier transform is a particular case of z-transform, i. Fourier transform of unit step signal u(t). To see how the Fourier transform works, we will begin with a one-dimensional signal and consider a simple step function. Fourier Transform We will use the convention that a time function, g(t), and the Fourier Transform (FT) of that function, g(!), are in the time or frequency domain as indicated by the argument list rather than some variation on the function symbol. Which condition then A should satisfy. 1,791,367 views. 1 and Table 4. For now we will use (5) to obtain the Fourier transforms of some important functions. To know Laplace transform of integral and derivatives (first and high orders derivatives. Online FFT calculator helps to calculate the transformation from the given original function to the Fourier series function. If the Fourier transform of In(t) is. 5 Application of the Fourier Transform. 4 Vector Algebra & Vector Differentiation. 7 Transmission of Signals Through Linear Systems 2. There are several variants of the discrete Fourier transform, with various normalization conventions, which are specified by the parameter DftNormalization. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). 5 for t<0 and 0. DFT needs N2 multiplications. Singular Fourier transforms andthe Integral Representation of the Dirac Delta Function Peter Young (Dated: November 10, 2013) I. This analytic function, is called the Fourier-Laplace transform of. edu the inverse Fourier transform 11\u20131. 2 Fourier Transform of Impulse Function; 3. 5, 1 over 2, when t equals 0. Table of Fourier Transform Pairs of Energy Signals Function name Time Domain x(t) Frequency Domain X Unit step () 10 00 t ut 2. The circuit can be represented as a linear time. 2nd/12/10 (ee2maft. Fourier transform of the unit step function and of the signum function: The signum function sgn( t) is a function that is related to the unit step function. Plugging this equation into the Fourier transform, we get:. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time t. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. 9 Low-Pass and Band-Pass Signals 2. Inverse Fourier transform \u2013 be able to compute this from definition as well as from looking up the transform for elementary signals. Signal and System: Fourier Transform of Basic Signals (Step Signal) Topics Discussed: 1. When you apply both of these rules, the Fourier Transform of the ramp is (1/jw)^2. where z = (x + iy) is a complex number. Fast Fourier Transform with APL. Which condition then A should satisfy. Impulse, rectangle, triangle, Heaviside unit step, sign functions. x/e\u2212i!x dx and the inverse Fourier transform is. due to an initial unit impulse of heat at x = \u03be. Below is a summary table with a few of the entries that will be most common for analysis of linear differential equations in this course. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. 24 Applications of Fourier Transforms to Generalized Functions Theorem 2. Fourier Transform of the Unit Step Function We have already pointed out that although L{u(t)} = 1 s we cannot simply replace s by i\u03c9 to obtain the Fourier Transform of the unit step. The Fourier transfer of the signum function, sgn(t) is 2/(i\u00cf\u2030), where \u00cf\u2030 is the angular frequency (2\u00cf\u20acf), and i is the imaginary number. That is, given the Fourier transform of an function, when can we recover the original function from ? We begin with a simple case where the recovery is quite easy. The Fourier Transform Saravanan Vijayakumaran [email\u00a0protected] k{1 - e-t/T} 4. Solution for 3. function and the Fourier transformation C. Laplace Transforms of the Unit Step Function. Join 100 million happy users! Sign Up free of charge:. Time Displacement Theorem: [You can see what the left hand side of this expression means in the section Products Involving Unit Step Functions. ** The Fourier transform of the triangular pulse f(t) shown in Fig. both are piecewise continuous functions for all t>=0 , then fourier transform of. This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on \u201cFourier Transforms\u201d. Note that some authors (especially physicists) prefer to write the transform in terms of angular frequency instead of the oscillation frequency. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)\u2026. To obtain Laplace transform of simple functions (step, impulse, ramp, pulse, sin, cos, 7 ) 11. 4 Sampling Continuous-Time Signals. Odd functions have imaginary (and odd) Fourier transforms. The discrete-time Fourier transform or DTFT of a sequence x\u0152n is de\ufb01ned as Discrete-Time Fourier Transform X. Example 2-2 SPECTRUM OF AN EXPONENTILA PULSE By means of direct integration find the Fourier transform of ) ( t w < = - 0 , 0 0 , ) ( t t e t w t Properties of Fourier Transforms. Fourier Transform for Periodic Signal, Sampling Function. Real part of X(\u03c9) is even, imaginary part is odd. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). The Laplace transform is usually restricted to transformation of functions of with. Laplace transforms convert a function f (t) in the time domain into function in the Laplace domain F (s). Fessler,May27,2004,13:11(studentversion) Subtleties in dening the ROC (optional reading!) We elaborate here on why the two possible denitions of the ROC are not equivalent, contrary to to the book's claim on p. So that gives you a complex spectrum which is here called ff, and then you multiply it by the imaginary unit times k, and then use an inverse transform back to physical space and now you have an exact to machine precision derivative defined on your original grid points. Ada version of General N Point Fast Fourier Transform. If playback doesn't begin shortly, try restarting your device. 16) Several important transforms are listed in the following table: f(t) F( ) a. When faced with the task of finding the Fourier Transform (or Inverse) it can always be done using the synthesis and analysis equations. Another description for these analogies is to say that the Fourier Transform is a continuous representation (\u03c9 being a continuous variable), whereas the. As an example of the Laplace transform, consider a constant c. Find the Fourier transform of 3. Trigonometric Polynomials 58. 1 Introduction There are three definitions of the Fourier Transform (FT) of a functionf(t) - see Appendix A. Because the convolution of two tempered distributions isn't always defined, neither is their product in the above sense. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. Materials include course notes, lecture video clips, practice problems with solutions, a problem solving video, and problem sets with solutions. 5): e( s): The Fourier transform of the odd part (of a real function) is imaginary (Theorem 5. The most. com \u2013 tashuhka Oct 14 '14 at 12:36. This is in fact very heavily exploited in discrete-time signal analy-sis and processing, where explicit computation of the Fourier transform and its inverse play an important role. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. (-0)-u(0) | 4. 4142*j,0,1-j*2. Fourier Transform \u0432\u0402\u201d Theoretical Physics Reference 0. Proposition 8 Let be. Fourier Transforms and the Fast Fourier Transform (FFT) Algorithm Paul Heckbert Feb. Third Derivative. All real c. (8) below] is due to Zakharov and Shabat [10]. This remarkable result derives from the work of Jean-Baptiste Joseph Fourier (1768-1830), a French mathematician and physicist. The signum function is also known as the \"sign\" function, because if t is positive, the signum function is +1; if t is negative, the signum function is -1. 42 While these published numerical inverse Laplace transform algorithm reviews are thorough and useful, 43 they focus on computing a single time-domain solution as accurately as possible. $\\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. Fourier series of even and odd functions, Gibbs phenomenon, Fourier half-range series, Parseval's identity, Complex form of Fourier series. of a unit step can be inferred, but it's natural with the Laplace. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. So here is the first example. Join the initiative for modernizing math education. Express f under an integral form. Find the Fourier transform of re(r), where e(r) is the Heaviside function. MATLAB Program for Dicrete Unit Impulse Function; / MATLAB Videos / Discrete Fourier Transform in for image conversion step by step Why 2D to 3D image. Step Functions - In this section we introduce the step or Heaviside function. Sinosoidal Function 5. The forward FT is de ned as usual g(!) = Z 1 1 g(t) ei!t dt ; (1) where scaling constants have. A unique 3D graphical approach has been adopted to provide the intuition required to OWN this subject. has three possible solutions for its Fourier domain representation depending on the type of approach. 2) become single integrals, integrated over the appropriate variable. When dealing with Fourier cosine and sine series, you are actually extending a non-periodic function onto a periodic even or odd domain. 4 Properties of fourier transforms There are several properties of fourier transforms that can be used as tools for solving PDEs. sinc(f\u02dd)has Fourier inverse 1 \u02dd rect \u02dd(t). The explicit solution of dual Sturm-Liouville matrix problem serves as a kernel for an inverse integral Fourier matrix transform. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). Laplace transform, Existence theorem, Laplace transforms of derivatives and integrals, Initial and final value theorems, Unit step function, Dirac- delta function, Laplace transform of periodic function, Inverse Laplace transform, Convolution theorem, Application to solve simple linear and simultaneous differential equations. The first is a function of location (x), the latter of time (t). UNIT V LAPLACE TRANSFORM: Definition-ROC-Properties-Inverse Laplace transforms-the S-plane and BIBO stability-Transfer functions-System Response to standard signals-Solution of. Inverse Laplace Transform with unit step function, sect7. PYKC - 11 Feb 08 2 5. Topics Covered: Partial differential equations, Orthogonal functions, Fourier Series, Fourier Integrals, Separation of Variables, Boundary Value Problems, Laplace Transform, Fourier Transforms, Finite Transforms, Green's Functions and Special Functions. Similarly if an absolutely integrable function gon R, has Fourier transform \u02c6gidentically equal to 0, then g= 0. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)\u2026. 2 Fourier Series of Functions: Exponential, trigonometric functions of any period =2L, even and odd functions, half range sine and cosine series. Laplace transform of the unit step function | Laplace transform | Khan Academy - YouTube. It can be thought of as a function of the real line (x-axis) which is zero everywhere except at the origin (x=0) where the. Both functions are constant except for a step discontinuity, and have closely related fourier transforms. CT Fourier Transform Pairs signal (function of t) $\\longrightarrow$ Fourier transform (function of $\\omega$) : 1 CTFT of a unit impulse $\\delta (t)\\$ $1 \\$. Implicit Derivative. Any good reference to more detailed tables would be very helpful! My attempt: $\\mathcal F[f\\times u] = (\\mathcal Ff)*(\\mathcal Fu)$ where * denotes convolution. The inverse Z-transform can be derived using Cauchy's integral theorem. The Fourier transform of a periodic impulse train in the time domain with period T is a periodic impulse train in the frequency domain with period 2p /T, as sketched din the figure below. Fourier transform. Since sinc (w) has infinite duration in freqency domain, X (jw) convolved with sinc (w) also has infinite horizon in freqency domain. Apply partial fraction expansion to separate the expression into a sum of basic components. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Current time: 0:00 Total duration: 24:15. The motivation for this work is to develop a deeper understanding of the origin of the algebraic identities between modular graph functions which have been discovered recently, and of the relation. Its value is not trivial to calculate, and ends up being. Disclaimer: None of these examples are mine. Examples Fast Fourier Transform Applications Signal processing I Filtering: a polluted signal 0 200 400 600 800 1000 1200 f1. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). other The first of each pair is usually called the direct Fourier transform and the other one is the matching inverse Fourier transform, The unit Dirac comb (shah function) is its own Fourier transform. But i could not find the fourier transform of x in such frequencies. [email\u00a0protected] com - id: 73fc3d-YTM3O. 5) is called a Fourier series. Z transform, Convergence. (20 marks) Using Laplace transform methods, solve for t = 0 the following differential equation, d 2x dt2 - 3 dx dt + 2x = 1, subject to x = 0 and dx dt = 0 at t = 0. Solution for 3. 1 Occasionally the question arises as to how a signal's frequency content is affected when the signal is time reversed. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. Solution: Here, =0 for <2 , then \u02dd =1 for \u22652. The Fourier transform we\u2019ll be int erested in signals de\ufb01ned for all t the Four Fourier tra nsform of f G e\ufb01ne the Fourier transform of a step function or a constant signal unit step. Complex exponential The spectrum of a complex exponential can be found from the above due to the frequency shift property: Sinusoids. Consider the Fourier transforms of the functions in Example 9. transforms of functions multiplied by tn, scale change property, transforms of functions divided by t, transforms of integral of functions, transforms of derivatives ; Evaluation of integrals by using Laplace transform ; Transforms of some special functions- periodic function, Heaviside-unit step function, Dirac delta function. We can solve this integral by considering. Applying the inverse Fourier Transform to the complex image yields According to the distributivity law, this image is the same as the direct sum of the two original spatial domain images. If n is less than the length of the signal, then ifft ignores the remaining signal values past the nth entry and. Mathematicians have developed tables of commonly used Laplace transforms. These are the sample pages from the textbook. A special case is the expression of a musical chord in terms of the volumes and frequencies of its constituent notes. The signal x(t) can be obtained back from Fourier transform X(t) by using the inverse Fourier transform. Consider a sinusoidal signal x that is a function of time t with frequency components of 15 Hz and 20 Hz. Fourier Transform - Free download as Powerpoint Presentation (. 5) \u00b6 The expansion (3. Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. But for a square-integrable function the Fourier transform could be a general class of square integrable functions. The Fourier transform is ) 2 (2 ( ) T 0 k T X j k p d w p w \u2211 \u221e =\u2212\u221e = \u2212. In this lecture, we will look at one way of describing discrete-time signals through their frequency content: the discrete-time Fourier transform (DTFT). FFT is a powerful signal analysis tool, applicable to a wide variety of fields including spectral analysis, digital filtering, applied mechanics, acoustics, medical imaging, modal analysis, numerical analysis, seismography, instrumentation, and communications. text orientation finding) where the Fourier Transform is used to gain information about the geometric structure of the. That is, we present several functions and there corresponding Fourier Transforms. Fourier series \u2022Periodic function (\ud835\udc61)of period 1: \ud835\udc61= 0 2 +\u0dcd =1 \u221e cos(2\ud835\udf0b\ud835\udc5b\ud835\udc61)+\u0dcd \ud835\udc58=1 \u221e sin(2\ud835\udf0b\ud835\udc5b\ud835\udc61) \u2022Fourier coefficients: =2\u0db1. $\\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. The range of functions for which the Fourier transform may be used can be greatly extended by using general-ized functions -- that is, the Dirac delta function and its close relatives (sign function, step function, etc. Mathematicians have developed tables of commonly used Laplace transforms. com \u2013 tashuhka Oct 14 '14 at 12:36. In particular we shall obtain, intuitively rather than rigorously, various Fourier transforms of functions such as the unit step function which actually violate the basic conditions which guarantee the existence of Fourier transforms! Prerequisites. The unit pulse function is simply one time shifted step function, minus another shifted step function. X(j\u03c9) is called the Fourier transform of the time function () x t , whereas ( x t ) is the inverse Fourier transform of X () j\u03c9. Properties of Fourier Transforms. First Derivative. For instance, the inverse continuous Fourier transform of both sides of Eq. DEU, Electrical and Electronics Eng. 4M subscribers. We denote by Sn,k the set of all k-SIIRVs of order n. Fast Fourier Transform (FFT) Calculator. Inverse Fourier Transforms 59. Is my last statement correct, about rewriting my original integral in terms of the step function? If I take the (inverse) Fourier transform of the step function $u(\\omega)$ and I end up with two terms (i. Express f under an integral form. function and the Fourier transformation C. Fourier transform is, by modern convention, 2 C ( ! ). Instead, the most common procedure to find the inverse Laplace transform of an expression is a two-step approach (Appendix 12. Half range series, Change of intervals, Harmonic analysis. If we weren't using the involutive definition of the Fourier transform, we would have to replace one of the occurences of \"Fourier transform\" in the above definition by \"inverse Fourier transform\". Which condition then A should satisfy. Fourier Transform of the Lorentzian. Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. DC Level, Unit Step Function, Switched Cosine, Pulsed Cosine, Exponential Pulse, Fourier Transforms of Periodic Functions, Summary, 5. The Fourier transform we\u2019ll be int erested in signals de\ufb01ned for all t the Four Fourier tra nsform of f G e\ufb01ne the Fourier transform of a step function or a constant signal unit step. 6#15 - Duration: The intuition behind Fourier and Laplace transforms I was never taught in school inverse laplace transform,. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(\u03c9). Fourier series of even and odd functions, Gibbs phenomenon, Fourier half-range series, Parseval's identity, Complex form of Fourier series. Solution for 3. According to Stroud and Booth (2011. Lecture X Discrete-time Fourier transform. Follow Neso Academy on Instagram: @nesoacademy(https://bit. We also work a variety of examples showing how to take Laplace transforms and inverse Laplace transforms that involve Heaviside functions. So we can write S2+S as S(S+1) now we can rewrite the equation as (S+2. 4 Vector Algebra & Vector Differentiation. If you really want to understand the Fourier and Laplace transforms , how they work and why they work then this is the course for you. 2 Fourier Series of Functions: Exponential, trigonometric functions of any period =2L, even and odd functions, half range sine and cosine series. In this problem we will evaluate the Fourier transform of the given function. The response time is defined with respect to a unit step function, as the time it takes for the \"smoothed\" step function to rise from 10% to 90% of its original value. That tells us that the inverse Laplace transform, if we take the inverse Laplace transform-- and let's ignore the 2. You can also check \u201cNew Worksheet,\u201d but having the Fourier Analysis results right next to your data will be more useful. k{1 - e-t/T} 4. , This requirement can be stated as , meaning that belongs to the set of all signals having a finite norm ( ). The #1 tool for creating Demonstrations and anything technical. The inverse Z-transform can be derived using Cauchy\u2019s integral theorem. It includes Laplace transform of special functions, properties, operations and using Laplace transforms to solve ordinary and partial differential equations. If the first argument contains a symbolic function, then the second argument must be a scalar. Magnitude and phase spectrum. To make one more analogy to linear algebra, the Fourier Transform of a function is just the list of components of the. Basic Properties of Fourier Transform (1) (Linearity) If the Fourier transform of f1 and f2 exist, then (2. There are different definitions of these transforms. This is interesting because if we extract a section of a signal to analyse, and obtain its spectrum (via Fourier Transform), we are effectively multiplying the signal with a rectangular function (rect()). \u2022It is the most general F. (Dirac & Heaviside) The Dirac unit impuls function will be denoted by (t). Compute the power and energy of 2 times t squared. !/, where: F. \u2022 is called the magnitude function \u2022 is called the phase function \u2022 Both quantities are again real functions of \u03c9 \u2022 In many applications, the DTFT is called the Fourier spectrum \u2022 Likewise, and are called the magnitude and phase spectra X(ej\u03c9) \u03b8(\u03c9) X(ej\u03c9) \u03b8(\u03c9). The Fourier Transform (used in signal processing) The Laplace Transform (used in linear control systems) The Fourier Transform is a particular case of the Laplace Transform, so the properties of Laplace transforms are inherited by Fourier transforms. The Fourier transform G(w) is a continuous function of frequency with real and imaginary parts. Fourier inverse step. 3 Fourier Transforms of Time Functions. Lecture X Discrete-time Fourier transform. We proceed via the Fourier Transform of the signum function sgn t which. Inverse Z Transform: Part 2. Solution: Here, =0 for <2 , then \u02dd =1 for \u22652. In Chapter 6 we were able to derive some ad hoc extensions of the classical Fourier transform which applied to the unit step function, delta functions, end even to infinite series of delta functions. Another description for these analogies is to say that the Fourier Transform is a continuous representation (\u03c9 being a continuous variable), whereas the. 2 Transform or Series. The DTFT sequence x[n] is given by Here, X is a complex function of real frequency variable \u03c9 and it can be written as Where Xre. 1 Foreshortening 1. We can solve the integral by contour integration. An alias of itself By subtracting the constant 0. 9) to emphasize. 2) become single integrals, integrated over the appropriate variable. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)\u2026. Implicit Derivative. both are piecewise continuous functions for all t>=0 , then fourier transform of. in Department of Electrical Engineering Indian Institute of Technology Bombay July 20, 2012. Suppose that the Fourier transform of f and its inverse exist. DEU, Electrical and Electronics Eng. Translation (that is, delay) in the time domain goes over to complex phase shifts in the frequency domain. In particular we shall obtain, intuitively rather than rigorously, various Fourier transforms of functions such as the unit step function which actually violate the basic conditions which guarantee the existence of Fourier transforms! Prerequisites. Be able to use partial fraction expansions to compute the Inverse Fourier transform. MA 382 - Laplace and Fourier Transforms This course introduces the theoretical concepts and uses of the Laplace and Fourier transforms. \u2022 The unit step function (1 class) \u2022 The Dirac delta function (1 class) \u2022 Applications of step and impulse functions (1 class) \u2022 Periodic functions and their applications (2 classes) \u2022 Convolution and applications (2 classes) \u2022 Solving integral equations (1 class) \u2022 Fourier series (3 classes) \u2022 Fourier integral representation (1. Recall, that $$\\mathcal{L}^{-1}\\left(F(s)\\right)$$$is such a function f(t) that $$\\mathcal{L}\\left(f(t)\\right)=F(s)$$$. The ifft function allows you to control the size of the transform. Topics Covered: Partial differential equations, Orthogonal functions, Fourier Series, Fourier Integrals, Separation of Variables, Boundary Value Problems, Laplace Transform, Fourier Transforms, Finite Transforms, Green's Functions and Special Functions. Whereas its Fourier transform, or the magnitude of its Fourier transform, has the inverse property that as a gets smaller, in fact, this scales down in frequency. The Fourier-series expansions which we have discussed are valid for functions either defined over a finite range ( T t T/2 /2, for instance) or extended to all values of time as a periodic function. As I indicated last time, the Fourier transform is a complex function of frequency. If playback doesn't begin shortly, try restarting your device. Frequency Response Function For a 1storder system The FRF can be obtained from the Fourier Transform of Input-Output Time Response (and is commonly done so in practice) The FRF can also be obtained from the evaluation of the system transfer function at s=j\u03c9. This transform can be obtained via the integration property of the fourier transform. Fourier transforms take the process a step further, to a continuum of n-values. Of practical importance is the conjugate symmetry property: When s (t) is real-valued, the spectrum at negative. The forward Z-transform helped us express samples in time as an analytic function on which we can use our algebra tools. Fourier transform of typical signals. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Introduction to continuous time signals and systems: Basic continuous time signals, unit step , unit ramp, unit impulse and periodic signals with their mathematical representation and characteristics. The Heaviside function is a unit step at x = 0 and is shown below Differentiating the Heaviside function results in the Dirac /Delta function. Often the unit step function u. The Z transform of the geometric sequence 39. 2 Fourier Series Consider a periodic function f = f (x),de\ufb01ned on the interval \u22121 2 L \u2264 x \u2264 1 2 L and having f (x + L)= f (x)for all. 4 Fourier Transform of One-Sided Exponential Function; 3. Example 1 Find the Fourier transform of the one-sided exponential function f(t) = \u02c6 0 t < 0 e\u2212\u03b1t t > 0 where \u03b1 is a positive constant, shown below: f (t) t Figure 1 Solution. Unlike the inverse Fourier transform, the inverse Laplace transform in Eq. 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. 3 Fourier Transforms of Time Functions. Eventually, we have to return to the time domain using the Inverse Z-transform. fast fourier transform. 2 Transforms of Derivatives and Integrals 6. Solution for 3. Odd functions have imaginary (and odd) Fourier transforms. Z transform of step and related functions. This immediately tells us that the situation for compactly supported functions is very different from the situation for Schwartz functions \u2014 the Fourier transform of a compactly supported function is analytic, so it cannot be compactly supported or it would vanish identically. Notice the the Fourier Transform and its inverse look a lot alike\u2014in fact, they're the same except for the complex. As I indicated last time, the Fourier transform is a complex function of frequency. The unit pulse function can be defined with the help of the Heaviside unit step function ( ) ( ) ( ) 0 x a f t Ht a Ht a 1 x a 0 x a <\u2212 = +\u2212 \u2212= < > a0 > The Fourier transform of this function can be determined as. The unit step function \"steps\" up from 0 to 1 at t=0. Let us now substitute this result into Eq. is arbitrarily selected. The accurate ISAL echo signal model is established for a space maneuvering target that quickly approximates the uniform acceleration motion. Laplace Transform Calculator. For the Z-transform the DTFT exists if the ROC includes the unit circle. Rectangular Pulse Signal Some Examples of Fourier Transform. The first is a function of location (x), the latter of time (t). Unit-111: Fourier series: Trigonometric Fourier series and its convergence. Second Derivative. 11) is rarely used explicitly. For example: (lg is log base 2) Primary space: 4 * 8 = 32 Dual space: lg(4) + lg(8) = 2 + 3 = 5 = lg(32). 1 The Fourier transform. Especially important among these properties is Parseval's Theorem, which states that power computed in either domain equals the power in the other. Proven the the Heaviside function is a tempered distribution I must evaluate:  \\langle F Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Proposition 8 Let be. In that case the integrals in (4. This website uses cookies to ensure you get the best experience. 5D electrical modelling Shi-zhe Xu,1 Ben-chun Duan2 and Da-hai Zhang1 Abstract An optimization method is used to select the wavenumbers k for the inverse Fourier transform in 2. Lecture X Discrete-time Fourier transform. Determine the Fourier transform of the non-periodic signals shown in the figures below: (b) 8(1) -2 -1 0 1 2. Inverse Transform 6. The second channel for the imaginary part of the result. An alias of itself By subtracting the constant 0. Apply partial fraction expansion to separate the expression into a sum of basic components. The Fourier Transform is used to transform a process from the time domain to the frequency domain. MATLAB Program for Dicrete Unit Impulse Function; / MATLAB Videos / Discrete Fourier Transform in for image conversion step by step Why 2D to 3D image. Unit Impulse Response : We have Laplace transform of the unit impulse. The discrete Fourier transform (DFT) of the discrete signal is n =0, 1, \u2026 , N-1 Similarly, an inverse discrete Fourier transform is of this form: Note that the number of data points in x(n) and X(m) are always the same The frequency in the Fourier domain is related to the sampling frequency f s. The Fourier transform is one of the most useful mathematical tools for many fields of science and engineering. Inverse Fourier Transform. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If Y is a matrix, then ifft (Y) returns the inverse transform of each column of the matrix. A similar analysis can be done in the frequency domain. To obtain Laplace transform of functions expressed in graphical form. Fourier Transform Symmetry (contd. 1) into the integral in the de\ufb02nition of the inverse transform in (F. Laplace transform with a Heaviside function by Nathan Grigg The formula To compute the Laplace transform of a Heaviside function times any other function, use L n u c(t)f(t) o = e csL n f(t+ c) o: Think of it as a formula to get rid of the Heaviside function so that you can just compute the Laplace transform of f(t+ c), which is doable. The equations describing the Fourier transform and its inverse are shown opposite. The Fourier transform of ft) (ft)-sinc(t)) is F(jo)-nRect(/2) (Figure ) (1) For a linear, time invariant system, its impulse response is h(t)\u2026. Recently I came across finite Fourier transforms, which can be used for solving certain type of boundary value problem (BVP) of linear partial differential equation (PDE) with constant coefficient. : and, inverse, And we can reverse this, too. has three possible solutions for its Fourier domain representation depending on the type of approach. I don't know where you got G(f), but it only a mathematical expression to \"give\" the value of the Fourier transform of a unit step. The list given in Fourier [list] can be nested to represent an array of data in any number of dimensions. com \u2013 tashuhka Oct 14 '14 at 12:36.\nexkds2ijwbfm8, scu0emspadvc8, 6jx7myor6dcbhqe, r76yr3voc6ebxht, hytmbmksppc0, k4lp9hjqws89, eqv1czdh0zmd1, mcjverfdlgenkw, lzszur4c6gtmz30, 436mnpveqe, i7xgkf69ctsuys, qnpao69epc298z, uv2pp8oaz6ka8a, 7kwals3rv1, n7p6yh671r, gtvjn9fwetc, 0kioxwtn1zo, pk13ifb40vhuxfd, 0e4czxbyadkf8, bue2ge4zxs77fm, zohxk1p95epev, 5s0aerhq8ps7oo, nq2oyhzvjns00m, 529efdfqv8p, 8pv8mktybs6zn, 546bvb64xeo", "date": "2020-06-04 17:31:33", "meta": {"domain": "clandiw.it", "url": "http://clandiw.it/umtr/inverse-fourier-transform-of-unit-step-function.html", "openwebmath_score": 0.8609397411346436, "openwebmath_perplexity": 616.2258345355706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990731984977195, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.8535534621243951}}
{"url": "http://mathhelpforum.com/calculus/18409-papus-theorem.html", "text": "# Thread: Papus theorem\n\n1. ## Papus theorem\n\nQuestion 1\n\nUse the Theorem of Pappus to find the volume of the solid generated when the region bounded by the x-axis and the semicircle y= sqrt(a^2 -x^2) is revolved about\n\na) the line y=-a\nb) the line y=x-a\n\nQuestion 2\nUse the Theorem of Pappus to find the centroid of the triangular region with vertices (0,0), (a,0), and (0,b) , where a> 0 and b>0 .\nHint Revolve the region about the x-axis to obtain y bar and about the y-axis to obtain x bar.\n\nI don't quite understand the Pappus theorem. Please do help you know how to solve them. Thank you very much.\n\n2. #1:\n\n$\\displaystyle \\overline{x}=0$ from the symmetry of the region, $\\displaystyle \\frac{{\\pi}a^{2}}{2}$ is the area of the semicircle, $\\displaystyle 2{\\pi}\\overline{y}$ is the distnace traveled by the centroid to generate the sphere so $\\displaystyle \\frac{4}{3}{\\pi}a^{3}=(\\frac{{\\pi}a^{2}}{2})(2{\\pi }\\overline{y}), \\;\\ \\overline{y}=\\frac{4a}{3{\\pi}}$\n\nNow for part a, $\\displaystyle V=\\left[\\frac{1}{2}{\\pi}a^{2}\\right]\\left[2{\\pi}\\left(a+\\frac{4a}{3{\\pi}}\\right)\\right]=\\frac{1}{3}{\\pi}(3\\pi+4)a^{3}$\n\nSee if you can tackle part b.\n\n#2:\n\nRevolve the region about the x-axis to get $\\displaystyle \\overline{y}$ and about the y-axis to get $\\displaystyle \\overline{x}$.\n\nThe region generates a cone of volume $\\displaystyle \\frac{1}{3}{\\pi}ab^{2}$\n\nwhen revolved about the x-axis, the area of the region is\n\n$\\displaystyle \\frac{1}{2}ab$, so $\\displaystyle \\frac{1}{3}{\\pi}ab^{2}$$\\displaystyle =\\left(\\frac{1}{2}ab\\right)(2{\\pi}\\overline{y})$$\\displaystyle , \\;\\ \\overline{y}=\\frac{b}{3}$.\n\nA cone of volume $\\displaystyle \\frac{1}{3}a^{2}b$ is generated when the\n\nregion is revolved about the y-axis so $\\displaystyle \\frac{1}{3}{\\pi}a^{2}b$$\\displaystyle =\\left(\\frac{1}{2}ab\\right)(2{\\pi}\\overline{x}), \\;\\ \\overline{x}=\\frac{a}{3}$.\n\nThe centroid is.........?\n\n3. Hi galatus,\n\nThank you very much for your reply.\n\nI don't fully understand this question.\n\nI know the Pappus theorem is\nV = area R * distance traveled by the centroid.\n\nYou have showed me\n\nSo the second part is the distance traveled by the centroid. Why equals 2pi ( a + (4a)/(3pi))? UM... just don't get it!\n\n4. Hello again, Kittycat. As before, you are giving up too soon. There is nothing here that should be a surprise to you.\n1. Your axis of revolution is y = -a.\n2. Your Centroid is at $\\displaystyle (0,\\frac{4a}{3\\pi})$\n3. The distance from the Centroid to the Axis of Rotation is the Radius of the circle travelled. $\\displaystyle \\left(\\frac{4a}{3\\pi} - (-a)\\right)$\n4. $\\displaystyle 2\\pi\\;r$ is the formula for the circumference of a circle, given its radius.\nP.S. Pep talk - pep talk - pep talk. More confidence. There is more stuff in your brain than you think. Reach in a bring it out.\n\n5. hi TKHunny,\n\nI am not that great - don't have so much good stuff in my brain as you thought.\n\nI think you are great - with brilliant intellects!\n\nWhat about for part b - the line y=x-a?\nCan you explaining to me how to get the distance from the centroid to the line y=x-a?\n\nThank you very much.\n\n6. Originally Posted by kittycat\nI am not that great - don't have so much good stuff in my brain as you thought.\nHogwash!\n\nCan you explaining to me how to get the distance from the centroid to the line y=x-a?\nSeveral ways. One of the most useful formulas ever, the Distance from a point (a,b) to a line cx + dy + e = 0 is:\n\n$\\displaystyle \\frac{|c*a + d*b + e|}{\\sqrt{c^{2}+d^{2}}}$\n\nIt is a beautiful thing.", "date": "2018-05-27 10:58:08", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/18409-papus-theorem.html", "openwebmath_score": 0.8562458753585815, "openwebmath_perplexity": 1240.8356165563544, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9907319859349646, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8535534576662895}}
{"url": "http://mathhelpforum.com/algebra/124720-peak-graph.html", "text": "# Math Help - Peak of graph\n\n1. ## [Solved] Peak of graph\n\nI read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like\n\n$y=(x+p)^2+q$\n\nis this correct? and how would you do for\n\n$y=2x^2-4x+a^2-a$\n\n2. Originally Posted by davidman\nI read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like\n\n$y=(x+p)^2+q$\n\nis this correct? and how would you do for\n\n$y=2x^2-4x+a^2-a$\n\nIf you write the quadratic equation as\n\n$y = a(x - h)^2 + k$\n\nthen the turning point is $(x, y) = (h, k)$.\n\nSo for $y = 2x^2 - 4x + a^2 - a$, you complete the square\n\n$y = 2\\left(x^2 - 2x + \\frac{1}{2}a^2 - \\frac{1}{2}a\\right)$\n\n$y = 2\\left[x^2 - 2x + (-1)^2 - (-1)^2 + \\frac{1}{2}a^2 - \\frac{1}{2}a\\right]$\n\n$y = 2\\left[(x - 1)^2 + \\frac{1}{2}a^2 - \\frac{1}{2}a - 1\\right]$\n\n$y = 2(x - 1)^2 + a^2 - a - 2$.\n\nSo the turning point is:\n\n$(x, y) = (1, a^2 - a - 2)$.\n\n3. wooah, you made that so easy to understand.\n\nthanks!\n\n4. Originally Posted by davidman\nI read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like\n\n$y=(x+p)^2+q$\n\nis this correct? and how would you do for\n\n$y=2x^2-4x+a^2-a$\n\nIf the vertex of the parabola is V(p,q) then the equation of the parabola whose axis is parallel to the y-axis is written as:\n\n$y = a(x-p)^2+q$\n\n$y = 2x^2-4x+a^2-a = 2(x^2-2x+\\tfrac12(a^2-a))$\n\n$y=2(x^2-2x\\bold{\\color{red}+1-1}+\\tfrac12(a^2-a)) = 2(x-1)^2+a^2-a-2$\n\nSo the coordinates of the vertex are $V\\left(1, a^2-a-2\\right)$\n\n5. Thanks so much for the help!\n\nI've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...\n\nthe second part is to get the minima of $f(a)=a^2-a-2$\n\nand I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?\n\n6. Originally Posted by davidman\nThanks so much for the help!\n\nI've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...\n\nthe second part is to get the minima of $f(a)=a^2-a-2$\n\nand I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?\nProbably you are trapped by the text of the question(?) It reads \" ... get the minima of f(a) ...\"\n\nI assume that you have calculated:\n\n$f(a)=a^2-a-2$\n\n$f(a)=a^2-a\\bold{\\color{blue}+\\frac14-\\frac14}-2$\n\n$f(a)=\\left(a-\\frac12 \\right)^2-\\frac14-2$\n\n$f(a)=\\left(a-\\frac12 \\right)^2-\\frac94$\n\nTherefore the vertex has the coordinates $V\\left(\\tfrac12 ,~ -\\tfrac94 \\right)$\n\nWith a parabola opening up the f(a)-value of the vertex determines the minimum. The vertex is the lowest point of all points of the graph of f. Therefore\n\n$\\min(f(a)) = -\\frac94$\n\n7. Thanks a lot guys. earboth's post brings that question to perfection!", "date": "2014-08-01 06:49:31", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/124720-peak-graph.html", "openwebmath_score": 0.8643468618392944, "openwebmath_perplexity": 266.6601680717871, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384664716301, "lm_q2_score": 0.8723473879530492, "lm_q1q2_score": 0.8534913066785851}}
{"url": "https://math.stackexchange.com/questions/2567104/roots-of-a-function-between-two-roots-of-another-function", "text": "# Roots of a function between two roots of another function\n\nI've been asked the following problem:\n\nLet $f$ and $g$ be two functions differatiable in the interval $I$. Proove that if $$f(x)g'(x) - f'(x)g(x) \\ne 0 \\quad \\forall x \\in I$$ there is one root of $g$ between every two roots of $f$\n\nNow, the problem is, this question was asked in German, and the word for 'one' and the word for 'a' are both the same in German. In other words, I do not know, if the sentence says exactly one root of $g$ or at least one root of $g$. I've already spent a few hours on this problem and managed to show that there is at least one root of $g$ between two roots of $f$:\n\n$$f(x)g'(x) - f'(x)g(x) = c, \\quad c \\ne 0$$ $$\\Leftrightarrow g(x) = \\frac{f(x)g'(x) - c}{f'(x)}$$ $$x_1, x2 \\in I, \\quad f(x_1) = 0, \\quad f(x_2) = 0$$ $$g(x_1) = \\frac{f(x_1)g'(x_1) - c}{f'(x_1)} = -\\frac{c}{f'(x_1)}$$ $$g(x_2) = \\frac{f(x_2)g'(x_2) - c}{f'(x_2)} = -\\frac{c}{f'(x_2)}$$ Next, we can say If $g(x_1)$ and $g(x_2)$ have different signs, there exists a $\\xi$ for which $g(\\xi) = 0$ (IVT)\n\n$g(x_1)$ and $g(x_2)$ have different signs if and only if $f'(x_1)$ and $f'(x_2)$ have different signs. Using Rolle's Theorem we can say:\n\nThere is an $\\eta \\in (x_1, x_2) : f'(\\eta) = 0$\n\nAs $x_1 < \\eta < x_2$ we can say, that either $f'(x_1)$ is negative and $f'(x_2)$ is positive or vice versa. Therefore the signs of $g(x_1)$ and $g(x_2)$ are different as well, and we know that there must be at least one root in the interval.\n\nHowever, I just do not manage to create a proof that shows that there can't be more than 1 root in the interval. My guess would have been using Rolle's Theorem for a proof by contradiction, however there might perfectly be an extremum in the intervals, since $g(x)$ might have other roots outside the bounds of the interval. (Or did I get something wrong here)\n\nSo my question is:\n\nIs the initial statement true, and if it is, how can I proove it?\n\n\u2022 I believe the German statement asked for \"a root\". I'm a Spanish native speaker and the same ambiguity would have happened in Spanish; in this language the convention in maths is that if there's no precision \"un/una\" means \"a/an\", that is \"at least one\". \u2013\u00a0Alejandro Nasif Salum Dec 14 '17 at 22:54\n\u2022 Also, the uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is \"at least one\" root of $g$ between two given roots of $f$, but you've also proved that there is \"at least one\" root of $f$ between two given roots of $g$. If you add the precision \"consecutive\" roots, then this implies uniqueness. On the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between. \u2013\u00a0Alejandro Nasif Salum Dec 14 '17 at 22:58\n\u2022 This does not address your main concern, but the proof itself might be simplified by observing that under the assumption that $g$ has no roots, $f'(x)g(x)-f(x)g'(x)$ is just $g^2(x)$ times the derivative of $\\frac{f(x)}{g(x)}$, and Rolle says that this must be zero somewhere in-between. \u2013\u00a0Hagen von Eitzen Dec 14 '17 at 23:01\n\u2022 Just to be sure: Does this mean i can say: Assume $g(x)$ has no roots. Let $h(x) = \\frac{f(x)}{g(x)}$. Then $h(x_1) = h(x_2) = 0$ because $f(x_1) = f(x_2) = 0$. Because of Rolle this means that $h'(\\xi) = 0$, therefore $f'(\\xi)g(\\xi) - f(\\xi)g'(\\xi)$ must be $0$ as well, which contradicts the statement. Because of that, $g(x)$ must have roots. \u2013\u00a0zockDoc Dec 14 '17 at 23:21\n\nThe uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is \"at least one\" root of $g$ between two given roots of $f$, but you've also proved that there is \"at least one\" root of $f$ between two given roots of $g$. If you add the precision \"consecutive\" roots, then this implies uniqueness.\nOn the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between.\nAnyway, I would say your proof works fine only under the hypothesis of consecutive roots. If you allow $f$ to be $0$ in the interval $(x_1,x_2)$, then I do not see how you could conclude that $f'(x_1)$ and $f'(x_2)$ have different signs. Think for instance of $f(x)=x^3-x$ in the interval $[-1,1]$, where Rolle's conditions are met, but $f'(-1)=f'(1)=2$.", "date": "2020-07-09 11:12:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2567104/roots-of-a-function-between-two-roots-of-another-function", "openwebmath_score": 0.8586605787277222, "openwebmath_perplexity": 86.96344674100625, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846659768267, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8534913061547139}}
{"url": "https://math.stackexchange.com/questions/2957789/proof-of-f-1a-cap-f-1b-f-1a-cap-b", "text": "# Proof of $f^{-1}(A) \\cap f^{-1}(B)= f^{-1}(A \\cap B)$\n\nFirst let me say I am aware of the other threads on this result. The reason for me making this thread is to find out whether or not my proof/proof attempt is correct.\n\nThe problem stated in full detail is given below.\n\nLet $$X,Y$$ be sets and $$f:X \\rightarrow Y$$, let $$C,D \\subseteq X$$ and let $$A,B \\subseteq Y$$. Prove that $$f^{-1}(A) \\cap f^{-1}(B)= f^{-1}(A \\cap B)$$.\n\nHere is my attempt:\n\n$$f^{-1}(A) \\cap f^{-1}(B)= { \\{x\\in X \\mid f(x) \\in A\\}}\\cap { \\{x\\in X \\mid f(x) \\in B\\}}= \\{x\\in X \\mid f(x) \\in A\\wedge f(x) \\in B\\}= \\{x\\in X \\mid f(x) \\in A\\cap B \\}=f^{-1}(A \\cap B)$$\n\nIf anyone would be kind enough to explain to me where I've gone wrong or made an unjustified assumption I would be very grateful!\n\n\u2022 you have only proved $f^{-1}(A) \\cap f^{-1}(B)\\subseteq f^{-1}(A \\cap B)$. You will need to prove that $f^{-1}(A \\cap B)\\subseteq f^{-1}(A) \\cap f^{-1}(B)$ \u2013\u00a0Neo Oct 16 '18 at 11:32\n\u2022 @Neo Really? As far as I can tell I have only used equality ad not implication rules... \u2013\u00a0krtgdl Oct 16 '18 at 11:33\n\u2022 @Neo Nope. Every equality OP wrote is true, they proved equality, not just inclusion. \u2013\u00a05xum Oct 16 '18 at 11:35\n\u2022 Learnt something new, I was marked wrong in an assignment for doing that once :/ \u2013\u00a0Neo Oct 16 '18 at 11:37\n\u2022 @Neo If you did it the wrong way, it might have been marked correctly. It's possible to use the correct tool incorrectly. Or maybe you were cheated. I'd have to see the concrete example. \u2013\u00a05xum Oct 16 '18 at 11:56\n\n\\begin{align}x\\in f^{-1}(A)\\cap f^{-1}(B)\\iff &(x\\in f^{-1}(A)\\land x\\in f^{-1}(B)\\\\\\iff& f(x)\\in A\\land f(x)\\in B\\\\\\iff& f(x)\\in A\\cap B\\\\\\iff& x\\in f^{-1}(A\\cap B)\\end{align}", "date": "2019-09-24 08:35:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2957789/proof-of-f-1a-cap-f-1b-f-1a-cap-b", "openwebmath_score": 0.6346578598022461, "openwebmath_perplexity": 744.5858074456966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846640860382, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8534913012583146}}
{"url": "https://www.aaachede.org/id/newton-raphson-method-calculator-f47f34", "text": "Input 4x-8=0 in the calculator, then the screen will tell you x=2 the moment you press solve. Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. Let\u2019s provide an example by funding the first derivative of the function $$f(x) = x^2-4 \\times x-5$$ Just start a Console application and fill in the code. This online calculator implements Newton's method (also known as the Newton\u2013Raphson method) using derivative calculator to obtain analytical form of derivative of given function, because this method requires it. To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. You may see ads that are less relevant to you. Codesansar is online platform that provides tutorials and examples on popular programming languages. Furthermore, if we are talking about the degree of accuracy, it would depend on the number of iterations. We've detected that you are using AdBlock Plus or some other adblocking software which is preventing the page from fully loading. To embed this widget in a post on your WordPress blog, copy and paste the shortcode below into the HTML source: To add a widget to a MediaWiki site, the wiki must have the. Questionnaire. What if we\u2019re solving a function with multiple zeroes, say x2+2x-4=0? The Newton-Raphson Method requires to calculate the first derivative of the function $$f$$. Ellipse: Conic Sections. This online newton's method calculator helps to find the root of the expression from the given values using Newton's Iteration method. If you continue to use this site we will assume that you are happy with it. Thank you for your questionnaire.Sending completion. Newton Raphson method calculator - Find a root an equation f(x) = 2x^3-2x-5 using Newton Raphson method, step-by-step Using a calculator to solve it: This method can be used as an alternative to solve functions or equations. However, there are some difficulties with the method: difficulty in calculating derivative of a function, failure of the method to converge to the root, if the assumptions made in the proof of quadratic convergence of Newton's method are not met, slow convergence for roots of multiplicity greater than 1, Everyone who receives the link will be able to view this calculation, Copyright \u00a9 PlanetCalc Version: Optimization Problems: Maximum and Minimum, Explaining the Differential: Differential Calculus, The First Derivative \u2013 Differential Calculus, Volume By Integration: Cross Section Method, Explaining the Virtual Work Method: Axial Strains, Consider the point on the function corresponding to x. That is why your calculator will require you to input something first before it starts solving a function. The Newton-Raphson Method or simply Newton\u2019s Method is a way to approximate the zeroes of the\u00a0function. FAQ. The approximations of the root go as: reactions. Don't know how to write mathematical functions?View all mathematical functions. Male or Female ? After unblocking website please refresh the page and click on find button again. 0.1 Newton Raphson Method The Newton Raphson method is for solving equations of the form f(x) = 0. 5. The process is repeated as , until a sufficiently accurate value is reached. Apsis: Applications of Conics. Let\u2019s say we are to find the zero of a function f as shown in the figure: If you continue to repeat this process, you\u2019ll notice that the computed x values get closer and closer to the zero of the function as shown in the figure (x3, x4, x5, x6, \u2026, xn). To improve this 'Newton method f(x) Calculator', please fill in questionnaire. Newton\u2019s method, also known as Newton-Raphson method is a root-finding algorithm that produces successively better approximations of the roots of a real-valued function.\n\n.\n\nWhite House Apple Cider Vinegar, Yaar Mila Karte Khula Nanda Raja, In What Sense Is Language Arbitrary, Throne Of Eldraine Lore, Rice Pilaf Meals, Do Voles Eat Grubs, Sphinx Nose Louvre, Development Studies Conference 2020, Lemon Blueberry Swiss Roll,", "date": "2021-01-22 09:46:25", "meta": {"domain": "aaachede.org", "url": "https://www.aaachede.org/id/newton-raphson-method-calculator-f47f34", "openwebmath_score": 0.3875388503074646, "openwebmath_perplexity": 730.8581956117039, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846659768267, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8534912996607641}}
{"url": "https://www.qalaxia.com/questions/How-do-I-solve-this-geometry-question-involving-probability", "text": "D\n\nHow do I solve this geometry question involving probability?\n\n59 viewed last edited 1 year ago\nAnonymous\n0\n\nI saw this problem online and I am having trouble figuring how to answer this.\n\nHere is what I have done so far: Let A, B and C be the randomly chosen points.\n\nLet AA' be the diameter drawn from A; and BB' be the diameter drawn from B. Is the following claim correct?\n\nThe 3 points will not have a common semi-circle if and only if:\n\n1) B and C lie on different sides of AA'; and\n\n2) A and C lie on different sides of BB'\n\nPrem Kumar\n2\n\nHere's another way to see the result and also to generalise to N points chosen randomly on the circle.\n\nPick one of these N points and draw the diameter through it. The probability that any other point lies on one side of the diameter is 1/2. So the probability that the remaining N-1 points all lie on one side of that diameter is\u00a0\u00a0\\frac{1}{2^{N-1}}.\n\nWe have N points which are undistinguished, so the net probability that there is one point amongst them\u00a0\u00a0to which the above condition applies = \\frac{N}{2^{N-1}}\n\nFor N=2 this gives 1 as you might expect: two points chosen at random lie on a semicircle trivially.\n\nFor N=3 it yields \\frac{3}{4} as the previous arguments did too.\n\nMahesh Godavarti\n1\n\nWhat you have is reasonable. The order of the points does not matter. So, what you are saying is: pick A, B and C in that order.\n\nThe probability that A and B, do not share a semicircle is zero (obviously).\n\nSo, we have to wait for C. For A, B and C to not share a semicircle, C has to be such that neither B nor C lie on the same side of the diameter AA' and neither A nor C lie on the same side of the diameter BB'.\n\nThis makes sense, however, we can't go anywhere from here.\n\nI would do the following:\n\nLet x be the smaller of the distance between A and B. Then 0 \\leq x \\leq \\pi r .\n\nThen the probability density function of x , f(x) = \\frac{1}{\\pi r} .\n\nThen, the probability that C will not be in the same semicircle as A and B, given the distance between A and B is x , g(\\text{not semi-C}|x)=\\frac{x}{2\\pi r}.\n\nTherefore, the total probability density function that A, B and C will not share a semicircle is given by \\frac{x}{2 \\pi^2 r^2} . I.e. C has to fall in the unbolded portion of the circle whose length is also x .\n\nTherefore, total probability that A, B and C do not lie on the same semicircle is given by \\int_0^{\\pi r} \\frac{x}{2 \\pi^2 r^2} dx = \\frac{1}{4} .\n\nI am sure there is a much more elegant argument given that the answer has turned out to be this simple.\n\nAnonymous\n1\nIf C and B have to be opposite sides of AA', the probability of this is 1/2. If A and C have to be opposite sides of BB', the probability of this is 1/2. If C and B have to be on opposite sides of AA'; and A and C have to be on opposite sides of BB'; intuitively, the average probability of this appears to be 1/2 * 1/2 = 1/4.\nMahesh Godavarti\n0\nLet's formalize it.\nMahesh Godavarti\n1\nI think this is the elegant answer. Let A-SC be the semicircle defined by A and B-SC be the semicircle defined by B (A is at the middle of A-SC and B is at the middle of B-SC). Then C should not lie in the semicircle defined by both A and B. The probability that C does not lie in the semicircle defined by A = 1/2. The probability that C does not lie in the semicircle defined by B = 1/2. Since A-SC and B-SC are independent (note A and B are independent), then the probability that C does not lie in either semicircle = 1/2 X 1/2 = 1/4. I think this is the answer.\nVivekanand Vellanki\n1\n\nContinuing with Mahesh's response, for a given x, the probability that A, B and C dont lie on a semi-circle is g(\\text{not semi-C}|x)=\\frac{x}{2\\pi r}.\n\nGiven x, this determines accurately the probability that the 3rd point is not on the same semi-circle.\n\nSince x is uniformly distributed from 0 to \\pi r, we need to find the expected probability.\n\nFor every x, there is an x' such that x+x'\\ =\\ \\pi r; and\n\ng(not semi C | x) + g(not semi C | x') = 1/2.\n\nHence, the expected value of g(not semi C | x) = 1/4 over 0\\le x\\le\\pi r", "date": "2020-01-24 04:57:18", "meta": {"domain": "qalaxia.com", "url": "https://www.qalaxia.com/questions/How-do-I-solve-this-geometry-question-involving-probability", "openwebmath_score": 0.9532792568206787, "openwebmath_perplexity": 333.84637169461024, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846691281406, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8534912861749288}}
{"url": "https://math.stackexchange.com/questions/1990936/is-this-inequality-true-x-y-alpha-x-alpha-y-alpha-for-pos", "text": "# Is this inequality true? $(x + y)^{\\alpha} < x^{\\alpha} + y^{\\alpha}$, for positive $x$ & $y$, and for $0 < \\alpha < 1$\n\nIf $0 < \\alpha < 1$, then\n\n$$(x + y)^{\\alpha} < x^{\\alpha} + y^{\\alpha}$$\n\nfor $x$, $y$ positive.\n\nIs this inequality true in general?\n\nI tried using Young's Inequality: For $z,t > 0$, and for $n$, $m$ such that $n+m=1$, then\n\n$$z^n + t^m \\leq nz + mt$$\n\nSo, using this we have\n\n$$(x+y)^{\\alpha} \\cdot 1^{1 - \\alpha} \\leq \\alpha(x+y) + (1-\\alpha) = \\alpha x + \\alpha y + (1-\\alpha)$$\n\nwhich is not as tight as I want.\n\nHint: try rewriting the inequality as $$x^\\alpha \\left( 1 + t \\right)^\\alpha \\le x^\\alpha \\left( 1 + t^\\alpha \\right)$$ where $t = y/x$.\n\nLet $x\\geq y$.\n\nHence, $(x+y,0)\\succ(x,y)$ and since $f(x)=x^{\\alpha}$ is a concave function, by Karamata we obtain: $$(x+y)^{\\alpha}=(x+y)^{\\alpha}+0^{\\alpha}\\leq x^{\\alpha}+y^{\\alpha}$$ and we are done!\n\nLEt $f(t) = 1 + t^{\\alpha} - (1+t)^{\\alpha}$, $t > 0$ and $0 < \\alpha < 1$. We have\n\n$$f'(t) = \\alpha t^{\\alpha - 1} - \\alpha(1+t)^{\\alpha-1} = \\alpha t^{\\alpha-1} \\left( 1 - \\left( \\frac{ 1 + t }{t} \\right)^{\\alpha-1} \\right)$$\n\nSince $\\frac{1+t}{t} > 1$ and $\\alpha - 1 <0$, then $1 - \\left( \\frac{ 1 + t}{t} \\right)^{\\alpha - 1} > 0$. Therefore,\n\n$$f'(t) > 0 \\; \\text{all} \\; t>0 \\implies f(t) > f(0) \\implies 1 + t^{\\alpha} > (1+t)^{\\alpha}$$\n\nAssume $x,y > 0$, then $t = \\frac{y}{x} > 0$ and thus\n\n$$1 + (y/x)^{\\alpha} > ( 1 + y/x)^{\\alpha} \\implies x^{\\alpha} + y^{\\alpha} > (x+y)^{\\alpha}$$\n\n$$(x+y)^a\\lt x^a+y^a\\iff x^a\\left(1+\\dfrac yx\\right)^a=x^a\\left(1+(\\dfrac yx)^a\\right)$$ Hence it is enough to prove $$(1+x)^a\\lt 1+x^a$$\n\nThe function $$f(x)=1+x^a-(1+x)^a$$ has the derivative $$f'(x)=a(x^{a-1}-(1+x)^{a-1})$$ which is positive for $0\\lt a\\lt 1$ and $x$ positive because $$a(x^{a-1}-(1+x)^{a-1})\\gt 0\\iff \\frac{1}{x^{1-a}}-\\frac{1}{(1+x)^{1-a}}\\gt 0$$ Consequently $f(x)$ is increasing for $x\\gt 0$ so because $f(0)=0$ we have always $$f(x)\\gt 0\\iff(1+x)^a\\lt 1+x^a$$\n\nA hint that I like for problem solving or maybe a different way to conceptualize these problems. Consider these two expressions as functions of 3 variables, $x,y,\\alpha$. In one variable calculus you can use derivative to find maximum points or minimums. You can use generalized methods here. Consider term one subtracted from term two. If it has a global minimum that says something related to this inequality.\n\nOf course the case $\\alpha=0$ or $\\alpha=1$ is trivial, so suppose $0<\\alpha<1$. Dividing both sides by $(x+y)^\\alpha$ reduces the proof to the case $x+y=1$ (to see this, just do the substitution $x'=\\frac{x}{x+y}, y'=\\frac{y}{x+y}$). So suppose $x+y=1$. Then $0<x<1$, $0<y<1$, implying $x<x^\\alpha$ and $y<y^\\alpha$. Hence $$(x+y)^\\alpha=1=x+y<x^\\alpha+y^\\alpha.$$", "date": "2019-06-25 09:34:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1990936/is-this-inequality-true-x-y-alpha-x-alpha-y-alpha-for-pos", "openwebmath_score": 0.9860979914665222, "openwebmath_perplexity": 129.08342796477925, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918533088548, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8534555744116128}}
{"url": "https://stats.stackexchange.com/questions/325764/using-dnn-to-find-out-the-pdf-of-a-regression-problem", "text": "# using DNN to find out the pdf of a regression problem\n\nWhen we use deep neural networks (DNNs) to solve a 1-dimention regression problem, we can approximate data distribution with the output of a DNN like the picture below.\nMy question is that DNN does not have the assumption of gaussian distribution or any other distribution of itself. It just knows what value to output when it sees an input. So how do you know the probability distribution of the DNN? For example, if someone asks, what is the probability of the point appearing in (5, 0). Can DNN answer this kind of questions?\n\nFor many regression algorithms, not only neural networks, the model is that the data is distributed by $y \\sim \\mathcal{N}(f(x;\\theta), \\sigma^2)$, where $\\theta$ are the model parameters and $\\sigma^2$ is the variance of the distribution (often a hyperparameter).\nMaximizing the log-likelihood of the data with respect to $\\theta$ is equivalent to minimizing the mean squared error loss between the $y_i$ and $f(x_i;\\theta)$.\nTherefore, to compute the probability density of $(5,0)$, you would just find the density of a gaussian with mean $f(5; \\theta)$ and a variance of $\\sigma^2$, where $f$ is your neural network.\n\u2022 Yes, it still applies. The only assumption we need to apply MSE loss and obtain our PDF is that the distribution of the target $y$ is a function of the input $x$ plus some gaussian noise. That function $f$ doesn't have to be linear and in this case is modeled by a highly nonlinear neural network. \u2013\u00a0shimao Feb 2 '18 at 2:30", "date": "2019-08-17 13:36:29", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/325764/using-dnn-to-find-out-the-pdf-of-a-regression-problem", "openwebmath_score": 0.8348345160484314, "openwebmath_perplexity": 137.46725947149287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9843363531263359, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8534548212157274}}
{"url": "https://www.physicsforums.com/threads/differential-equation-for-changing-mass-of-a-sphere.710132/", "text": "# Differential equation for changing mass of a sphere\n\n1. Sep 13, 2013\n\n### Hypatio\n\nThe mass of a sphere with density as a function of radius is\n\n$M=4\\pi \\int_0^r\\rho(r) r^2dr$\n\nLets say the radius increases and decreases as a function of time t. So:\n\n$M(t)=4\\pi \\int_{0}^{r(t)}\\rho (r) r(t)^2dr$\n\nI want to know the basic equation describing the mass added or removed from the sphere (mass increases when radius increases, mass decreases when radius decreases) as a function of t, starting from any t. The problem is I think I must use a differential form but I'm not sure what it looks like. What then is the differential form of dM(t)/dt? I think I must use a chain rule and write:\n\n$\\frac{dM}{dt}=\\frac{dM}{dr}\\frac{dr}{dt}$\n\nis this right? How do I proceed to solve this with the integral?\n\n2. Sep 13, 2013\n\n### JJacquelin\n\nHi !\nI am afraid that there is something wrong in your writting. It should be :\n$M=4\\pi \\int_0^r\\rho(u) u^2du$\nLets say the radius increases and decreases as a function of time t. So:\n$M(t)=4\\pi \\int_{0}^{r(t)}\\rho (u) u^2du$\nYou may use any other symbol than u, but not r.\n\n3. Sep 13, 2013\n\n### Hypatio\n\nAh yes, sure, the upper limit is the 'full' radius (r) and u is a radius. This doesn't solve my problem though.\n\n4. Sep 13, 2013\n\n### HallsofIvy\n\nStaff Emeritus\nYes, the chain rule: $\\dfrac{dM}{dt}= \\dfrac{dM}{dr}\\dfrac{dr}{dt}$\nJJaquelines point helps make sense of the dM/dr.\n\nTo find $\\dfrac{dM}{dt}$ use the \"fundamental theorem of Calculus\":\n$$\\frac{d}{dr}\\int_0^r \\rho(u)u^2 du= \\rho(r)r^2$$\n\n5. Sep 13, 2013\n\n### Hypatio\n\nI guess the solution then is\n\n$\\frac{dM}{dt}=\\frac{4}{3}\\pi \\rho(r)r^3\\frac{dr}{dt}$\n\ndoes the solution change if $\\rho(r)$ becomes $\\rho(r,t)$ or could I write\n\n$\\frac{dM}{dt}=\\frac{4}{3}\\pi \\rho(r,t)r^3\\frac{dr}{dt}$\n\nThanks.\n\nLast edited: Sep 13, 2013\n6. Sep 13, 2013\n\n### JJacquelin\n\nI don't think so. A term involving the partial derivatine of rho relatively to t is missing into your last equation,\n\n7. Sep 13, 2013\n\n### Hypatio\n\nIs there a rule I can apply to get this additional term?\n\n8. Sep 13, 2013\n\n### pasmith\n\nIf you have\n$$F(t) = \\int_0^{a(t)} g(r,t)\\,\\mathrm{d}r$$\nthen\n$$F'(t) = \\int_0^{a(t)} \\left.\\frac{\\partial g}{\\partial t}\\right|_{(r,t)}\\,\\mathrm{d}r + a'(t)g(a(t),t)$$\nassuming $g$ is sufficiently smooth.\n\n9. Sep 13, 2013\n\n### arildno\n\nPlease do not mix together dummy variables in the integrand with integral limits.\nProperly speaking, you have the following the relation:\n$$M(t)=4\\pi\\int_{0}^{r(t)}\\rho(x)x^{2}dx$$\nThus, you have:\n$$\\frac{dM}{dt}=4\\pi\\rho(r(t))r(t)^{2}\\frac{dr}{dt}$$\nwhich has as interpretation that only the outermost spherical shell at r(t) determines the total change of mass.\n\nEvery compact ball strictly contained within the outermost shell (radii less than r(t)) remains constant in mass.\n\nHowever:\nSuppose you have a ball where at different times, the density at some fixed radius \"x\" may change as a function of time. Then, you have:\n$$M(t)=4\\pi\\int_{0}^{r(t)}\\rho(x,t)x^{2}dx$$\nIn this case, the total mass of the ball will be due to two distinct effects:\n1. The ball shrinks or expands. This gives the contribution given above.\n2. The interior of the ball may change its mass. This effect is new.\n\nIn sum, you'll then get:\n$$\\frac{dM}{dt}=4\\pi\\rho(r(t),t)r(t)^{2}\\frac{dr}{dt}+4\\pi\\int_{0}^{r(t)}\\frac{\\partial\\rho}{\\partial{t}}x^{2}dx$$\n\nLast edited: Sep 13, 2013\n10. Sep 14, 2013\n\n### JJacquelin\n\nThe general formula below shows the rule for derivation :\n\n#### Attached Files:\n\n\u2022 ###### Derivation rule.JPG\nFile size:\n6 KB\nViews:\n80\n11. Sep 14, 2013\n\n### JJacquelin\n\nOf course, the chain rule continues to applies !!!\nThe formula given above is the application of the chain rule in case of an integral with the integrand and limits which are functions of the variable considered for derivation.", "date": "2017-08-18 10:52:21", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/differential-equation-for-changing-mass-of-a-sphere.710132/", "openwebmath_score": 0.8995257019996643, "openwebmath_perplexity": 1189.5375653468677, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363545048391, "lm_q2_score": 0.8670357615200474, "lm_q1q2_score": 0.8534548207199705}}
{"url": "https://math.stackexchange.com/questions/1055758/probability-of-no-king-queen-or-jack-before-the-first-ace-occurs", "text": "# Probability of no king, queen or jack before the first ace occurs?\n\nA deck of cards is shuffled well. The cards are dealt one by one, until the first time an Ace appears. Find the probability that no kings, queens, or jacks appear before the first ace. (Introduction to Probability, p. 36)\n\nMy solution:\n\n\u2022 Assume $k^{th}$ card is the first ace\n\u2022 The possible number of hand before the first ace is then $\\binom{48}{k-1}$, and the possible number of hands without a king, queen or jack before the first ace is $\\binom{36}{k-1}$, so the probability is $\\frac{\\binom{36}{k-1}}{\\binom{48}{k-1}}$.\n\u2022 The probability that the first ace occurs at the $k^{th}$ position is $\\frac{1}{52!} \\binom{48}{k-1}(k-1)! \\binom{4}{1}(52-k)!$, because there are $52!$ possible ordered decks, $\\binom{48}{k-1}(k-1)!$ is the number of possibilities withouth an ace in the first $k$ cards, $\\binom{4}{1}$ possibilities to draw an ace at the $k^{th}$ position and $(52-k)!$ possibilities to arrange the remaining cards in order.\n\u2022 $\\sum_{k=1}^{k=37} \\frac{\\binom{36}{k-1}}{\\binom{48}{k-1}} \\frac{1}{52!} \\binom{48}{k-1}(k-1)! \\binom{4}{1}(52-k)! = 0.25$\n\nThe computation of the above in R:\n\n    pr <- numeric(37)\nfor(k in 1:37){\npr_ace_at_k <- 1/factorial(52)*choose(48,k-1)*factorial(k-1) * choose(4,1)*factorial(52-k)\npr_no_before_k <- choose(36,k-1)/choose(48,k-1)\npr[k] <- pr_no_before_k * pr_ace_at_k\n}\nsum(pr)\n> 0.25\n\n\nHowever, a simulation yields:\n\n    deck <- c(rep(1:4, 4), rep(5, 52-4*4))\nout <- replicate(1e6,{\ndeck <- sample(deck) # shuffle deck\nk <- which(deck == 4)[1] # get index of first ace\nall(deck[1:(k-1)] == 5) # check if only rest occured before kth card\n})\nmean(out)\n> 0.173099\n\n\nIs there anything wrong in my calculation? A result of 0.25 looks somehow persuasive.\n\nEDIT\n\nFinally, I found the nasty bug in the code. Problem was the special case when the ace occured at the first position. Corrected code would be\n\n    deck <- c(rep(1:4, 4), rep(5, 52-4*4))\nout <- replicate(1e6,{\ndeck <- sample(deck) # shuffle deck\nk <- which(deck == 4)[1] # get index of first ace\nif(k == 1) TRUE\nelse all(deck[1:(k-1)] == 5)\n})\nmean(out)\n> 0.249877\n\n\u2022 It's the probability that among $16$ particular cards, the first to show up is an ace. We have equally likely outcomes here... \u2013\u00a0David Mitra Dec 7 '14 at 15:05\n\u2022 Call Ace, King, Queen, and Jack the picture cards. You are asking for the probability that the first picture card is an Ace. But all picture cards are equally likely. Therefore, by symmetry, the probability is...? \u2013\u00a0TonyK Dec 7 '14 at 15:09\n\u2022 The issue is why the simulation is so grossly wrong. The usual answer is programming error. \u2013\u00a0Andr\u00e9 Nicolas Dec 7 '14 at 15:19\n\u2022 The simulation result is clearly an anomaly. I don't see any error in the R code, but maybe someone on stackoverflow can. \u2013\u00a0David K Dec 7 '14 at 15:25\n\u2022 The simulation probability is likely either 4/23 or 9/52 \u2013\u00a0Empy2 Dec 7 '14 at 16:22\n\nThe easiest way to think about it is to ignore all the other cards in the deck. Now you stack up the $16$ cards of interest. What is the chance the top one is an ace? There are $4$ aces among the $16$, so $\\frac 4{16}=\\frac 14$\n\n\u2022 That makes perfect sense, good explanation and intuition! \u2013\u00a0NoBackingDown Dec 7 '14 at 15:23\n\nAlthough it is not the simplest approach, we can use recursive conditioning, which is a useful approach in a lot of somewhat more complicated problems.\n\nTo see the intuition, let's check the cases in words. Suppose the first card is a queen/jack/king; then you stop and it doesn't count. Suppose it's an ace; then you stop and it counts. Suppose it's neither; then you run the experiment again with 1 fewer card.\n\nMathematically we'll say that the desired event when starting with a $k$ card deck is $A_k$. Then we have\n\n$$P(A_k)=P(A_k|\\text{ first card is an ace })P(\\text{ first card is an ace }) \\\\ +P(A_k|\\text{ first card is queen/jack/king })P(\\text{ first card is queen/jack/king })\\\\ +P(A_k|\\text{ first card is neither })P(\\text{ first card is neither }) \\\\ = 1 \\cdot \\frac{4}{k} + 0 + \\frac{k-16}{k} P(A_{k-1})$$\n\nNow take $k=52$ and solve the recurrence relation. Here the base case can be read off from the recurrence itself: we get $P(A_{16})=4/16+0+0=1/4$ for free. Here solving the recurrence proves particularly simple, because if $P(A_{k-1})=1/4$ then $P(A_k)=1/4$, since\n\n$$\\frac{4}{k} + \\frac{k-16}{k} \\frac{1}{4} = \\frac{4}{k} + \\frac{1}{4} - \\frac{4}{k} = \\frac{1}{4}.$$\n\nThis is an algebraic manifestation of the intuitive observation that we can \"ignore all the other cards\".\n\nTotal probability x Probability of only 1st drawing = Probability of event occuring\n\nTotal probability x (1 - Probability of another drawing) = Probability of event occuring on 1st drawing\n\nTotal probability = Probability of event occuring on 1st drawing / (1 - Probability of another drawing) = $\\frac{\\frac1{13}}{1-\\frac9{13}}=\\frac14$\n\n\u2022 Could you elaborate on your answer? I don't understand how you arrive at your equation. \u2013\u00a0NoBackingDown Dec 7 '14 at 15:25\n\u2022 More rigorous than RossMillikan's answer? How so? \u2013\u00a0Did Dec 7 '14 at 15:35\n\u2022 I don't think this quite works, because the replacement means that the process is not simply being \"restarted\" upon drawing another card. My answer does essentially what you're attempting to do, I think. That is, you have that $a=1/13 + 9/13 b$ where $b$ is the probability for the corresponding event with a deck containing 12 queen/jack/king, 4 aces, and 51-16=35 other cards. \u2013\u00a0Ian Dec 7 '14 at 15:39\n\u2022 1. Hope my answer is clear enough now. 2. RossMillikian's answer is sufficient, I just used a formula instead of intuition. \u2013\u00a0ghosts_in_the_code Dec 7 '14 at 16:25\n\u2022 I don't think this works still: you have that the total probability is the probability of the event occurring given that the process stops on the first draw, times the probability that the process stops on the first draw, plus the probability of the event occurring given that the process stops on the second draw, times the probability that it stops on the second draw, etc. There is an independence assumption implicit in your formula which is correct but not justified. \u2013\u00a0Ian Dec 7 '14 at 17:30\n\nOne of these $4$ sorts of cards (aces, kings, queens, jacks) will appear as the first. Each sort (also the aces) has the same probability of doing so, hence with probability $\\frac{1}{4}$.", "date": "2019-08-18 06:44:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1055758/probability-of-no-king-queen-or-jack-before-the-first-ace-occurs", "openwebmath_score": 0.8574928641319275, "openwebmath_perplexity": 423.1484932580944, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363485313248, "lm_q2_score": 0.8670357580842941, "lm_q1q2_score": 0.8534548121587832}}
{"url": "https://math.stackexchange.com/questions/2647211/rudin-2-43-every-nonempty-perfect-set-in-mathbbrk-is-uncountable", "text": "Rudin 2.43 Every nonempty perfect set in $\\mathbb{R}^k$ is uncountable\n\nI am having trouble understanding Rudin's proof of this theorem, and I believe I have pinpointed the particular part of the proof that I am unable to follow. To begin with, here is the proof copied from Baby Rudin:\n\nLet $P$ be a nonempty perfect set in $\\mathbb{R}^k$. Then $P$ is uncountable.\n\nProof. Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable and denote the points of $P$ by $x_1, x_2, x_3, \\dots$. We shall construct a sequence $\\{V_n\\}$ of neighborhoods as follows.\n\nLet $V_1$ be any neighborhood of $x_1$. If $V_1$ consists of all $y \\in \\mathbb{R}^k$ such that $|y - x_1| < r$ the closure $\\overline{V_1}$ of $V_1$ is the set of all $y \\in \\mathbb{R}^k$ such that $|y - x_1| \\leq r$.\n\nSuppose $V_n$ has been constructed so that $V_n \\cap P \\neq \\emptyset$. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\\overline{V_{n+1}} \\subset V_n$, (ii) $x_n \\notin \\overline{V_{n+1}}$, (iii) $V_{n+1} \\cap P \\neq \\emptyset$. By (iii) $V_{n+1}$ satisfies our induction hypothesis and the construction can proceed.\n\nPut $K_n = \\overline{V_n} \\cap P$. Since $\\overline{V_n}$ is closed and bounded, $\\overline{V_n}$ is compact. Since $x \\notin K_{n+1}$ no point of $P$ lies in $\\cap_1^\\infty K_n$. Since $K_n \\subset P$ this implies $\\cap_1^\\infty$ is empty. But each $K_n$ is nonempty by (iii) and $K_n \\supset K_{n+1}$ by (i), this contradicts the corollary to theorem 2.36.\n\nOkay so my confusion lies with this paragraph of the proof:\n\nSuppose $V_n$ has been constructed so that $V_n \\cap P \\neq \\emptyset$. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\\overline{V_{n+1}} \\subset V_n$, (ii) $x_n \\notin \\overline{V_{n+1}}$, (iii) $V_{n+1} \\cap P \\neq \\emptyset$. By (iii) $V_{n+1}$ satisfies our induction hypothesis and the construction can proceed.\n\nMy confusion is as follows:\n\nIn defining $V_1$, it is clear that $V_1$ is a neighborhood of the point $x_1$. Is $V_{n+1}$ a neighborhood of the point $x_{n+1}$? For example, is $V_2$ a neighborhood of the point $x_2$? I don't think this can be the case because there is no guarantee that there is a neighborhood of $x_2$ is a subset of an arbitrary neighborhood of $x_1$. After all if $x_1 \\neq x_2$ and we can choose any neighborhood of $x_1$ in defining $V_1$, it is certainly possible $x_2 \\notin V_1$. So if $V_{n+1}$ is not a neighborhood of $x_{n+1}$, what point is $V_{n+1}$ a neighborhood of?\n\n\u2022 Possible duplicate of Problems in Theorem 2.43 of baby Rudin Feb 12, 2018 at 10:14\n\u2022 @Jos\u00e9CarlosSantos I don't think this is a duplicate: although both this and the linked question are about Theorem 2.43, this question asks if $x_{n+1} \\in V_{n+1}$, while the linked question asks to show that $V_{n+1}$ exists.\n\u2013\u00a0user88319\nFeb 12, 2018 at 20:42\n\u2022 @Strants You are right. I have retracted my closing vote. Feb 12, 2018 at 20:58\n\nNote that nowhere in the proof is used the fact that $x_n$ should lie in $V_n$. The only point is that $V_n$ intersects $P$, so it is a neighbourhood of at least one point in $P$, no matter which one.\nThe fact that $x_1\\in V_1$ may be confusing, but that's just to initiate the sequence.\nNote: to construct $V_{n+1}$ simply select any open ball centered around a point in $V_n\\cap P$ that is not $x_n$, with radius small enough that the closed ball lies entirely in $V_n$. All requirements are trivially satisfied.\n\u2022 Perhaps it is presumed $x_n \\in V_n$ for all n. Feb 12, 2018 at 10:19\n\u2022 @WilliamElliot No it's not, you don't need that at all. All you need is for $V_n$ to not contain any of the $n$ first $x_n$, so that the intersection in the end is empty. Feb 12, 2018 at 10:21\nIt is not assumed that $$x_n \\in V_n$$ (it may or may not be). Regardless of whether $$x_n \\in V_n$$ or $$x_n \\not \\in V_n$$, we want to construct $$V_{n+1}$$ such that $$x_n \\not \\in V_{n+1}$$, and I do so here: Proof of Baby Rudin Theorem 2.43", "date": "2022-05-24 18:08:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2647211/rudin-2-43-every-nonempty-perfect-set-in-mathbbrk-is-uncountable", "openwebmath_score": 0.9278064966201782, "openwebmath_perplexity": 63.31314666462154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363540453381, "lm_q2_score": 0.8670357512127873, "lm_q1q2_score": 0.853454810175756}}
{"url": "https://math.stackexchange.com/questions/2186169/find-the-determinant-after-a-certain-row-operation-is-applied-to-a-matrix-with-k", "text": "# Find the determinant after a certain row operation is applied to a matrix with known determinant\n\n$A = \\begin{bmatrix} --a--\\\\ --b--\\\\ --c-- \\end{bmatrix}$\n\n$A$ is a $3\\times3$ matrix. The rows are all different, and the first row is called $a$. The second row is called $b$. The third row is called $c$. I'm not saying that all the values in the first row are the same, I am simply saying that the first row is called $a$. The determinant is $not$ $0$.\n\nIf the $\\det(A)=3$, what is the determinant of:\n\n$\\begin{bmatrix} --a+b--\\\\ --b+c--\\\\ --c+a-- \\end{bmatrix}$\n\nI thought about this, but I recalled that whenever you have a matrix, and you add a row multiple of another, the determinant does not change. However in this case, the answer in the back of my textbook is $6$ and I don't understand how?\n\n\u2022 And... what matrix is $\\;A\\;$ itself ? \u2013\u00a0DonAntonio Mar 14 '17 at 11:50\n\u2022 It doesn't matter \u2013\u00a0K Split X Mar 14 '17 at 11:52\n\u2022 A is a $3\\times 3$ matrix \u2013\u00a0K Split X Mar 14 '17 at 11:53\n\u2022 @K Either you don't understand the question, or you're kidding...or I don't understand the question: how come it isn't relevant?! \u2013\u00a0DonAntonio Mar 14 '17 at 11:53\n\u2022 I think, K, that you had better write out the entire question as you have found it, not leaving out the least little bit, since what you have written so far is nonsense. \u2013\u00a0Gerry Myerson Mar 14 '17 at 11:58\n\n$$\\begin{bmatrix}1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 1 & 0 & 1\\end{bmatrix}\\begin{bmatrix}a \\\\ b \\\\ c\\end{bmatrix} = \\begin{bmatrix}a+b \\\\ b+c \\\\ c+a\\end{bmatrix}$$ Therefore, $$\\begin{vmatrix}a+b \\\\ b+c \\\\ c+a\\end{vmatrix} = \\begin{vmatrix}1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 1 & 0 & 1\\end{vmatrix}\\begin{vmatrix}a \\\\ b \\\\ c\\end{vmatrix} = 2 \\times 3 = 6$$\n\n\u2022 Where did the get the $\\times 2$ from? \u2013\u00a0K Split X Mar 14 '17 at 12:26\n\u2022 @KSplitX $\\begin{vmatrix}1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ 1 & 0 & 1\\end{vmatrix} = 2$ \u2013\u00a0TenaliRaman Mar 14 '17 at 12:28\n\u2022 Determinant is 2. Okay thanks \u2013\u00a0K Split X Mar 14 '17 at 12:30\n\nLets do it with rows instead of columns and remember that the determinant is multilinear:\n\n\\begin{align} \\det(a+b,b+c,c+a) = \\quad &\\det(a,b,c)+\\det(a,b,a)\\\\ +&\\det(a,c,c)+ \\det(a,c,a)\\\\ +&\\det(b,b,c)+ \\det(b,b,a)\\\\ +&\\det(b,c,c)+ \\det(b,c,a). \\end{align}\n\nAll terms except the first and the last vanish because there is the same entry twice. And because rearranging the rows does not change the determinant you will find\n\n$$\\det(a+b,b+c,c+a)=\\det(a,b,c)+\\det(b,c,a)=2\\det(a,b,c).$$\n\nIn your case $\\det(A)=3$, so $2\\cdot\\det(A)=6$ is the result.\n\nWell, with the explantion you added (and the one you also deleted...) it is finally clearer what you meant, but then, since the determinant of a matrix with two identical rows is zero, and if we intechange two rows we multiply the determinant by $\\;-1\\;$ , by multilinearity we get:$${}$$\n\n$$\\det\\begin{pmatrix}-a+b-\\\\-b+c-\\\\-c+a-\\end{pmatrix}=\\color{red}{\\det\\begin{pmatrix}-a-\\\\-b-\\\\-c-\\end{pmatrix}}+\\underbrace{\\det\\begin{pmatrix}-a-\\\\-b-\\\\-a-\\end{pmatrix}+\\det\\begin{pmatrix}-a-\\\\-c-\\\\-c-\\end{pmatrix}+\\ldots}_{=0}+\\color{red}{\\det\\begin{pmatrix}-b-\\\\-c\\\\-a-\\end{pmatrix}}$$$${}$$\n\n$$=\\color{red}{\\det A}+\\color{red}{\\det A}=6$$\n\nI will use slightly different notation, write\n\n$$A = \\begin{bmatrix} v_{1}\\\\ v_{2}\\\\ v_{3}\\end{bmatrix}$$ where $v_{i}$ are the rows of $A$ and $\\operatorname{det}(A) = 3.$\n\nThen\n\n$$\\operatorname{det}\\begin{bmatrix} v_{1} + v_{2}\\\\ v_{2} + v_{3}\\\\ v_{3} + v_{1} \\end{bmatrix} = \\operatorname{det} \\left(\\begin{bmatrix} v_{1}\\\\ v_{2} \\\\v_{3} \\end{bmatrix} + \\begin{bmatrix} 0 &1 &0\\\\ 0 & 0 & 1\\\\ 1& 0& 0\\end{bmatrix}\\begin{bmatrix} v_{1}\\\\ v_{2}\\\\v_{3}\\end{bmatrix}\\right) = \\operatorname{det}\\left(A + \\begin{bmatrix} 0 &1 &0\\\\ 0 & 0 & 1\\\\ 1& 0& 0\\end{bmatrix}A \\right)$$ So, $$\\operatorname{det}(A)\\operatorname{det}\\left(I+\\begin{bmatrix} 0 &1 &0\\\\ 0 & 0 & 1\\\\ 1& 0& 0\\end{bmatrix}\\right) = 6$$\n\n\u2022 Yup another answer just did this! Thanks for the indepth version tho \u2013\u00a0K Split X Mar 14 '17 at 12:32\n\u2022 How did you arrive at the last step from the third step? Where did the $\\det(A)$ on the outside come from? \u2013\u00a0K Split X Mar 14 '17 at 12:34\n\u2022 Split the determinant over multiplication \u2013\u00a0JessicaK Mar 14 '17 at 12:36", "date": "2019-05-23 09:49:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2186169/find-the-determinant-after-a-certain-row-operation-is-applied-to-a-matrix-with-k", "openwebmath_score": 0.8842412233352661, "openwebmath_perplexity": 554.0539588101946, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982287698703999, "lm_q2_score": 0.8688267762381844, "lm_q1q2_score": 0.8534378546034205}}
{"url": "http://ronk.gattopescatore.it/permutation-with-repetition-algorithm.html", "text": "# Permutation With Repetition Algorithm\n\nAs understood by name \"combinations\" means all the possible subsets or arrangements of the iterator and the word \"combinations_with_replacement\" means all the possible arrangements or subsets that allow an element to repeat in a subset. \u201d Discrete Mathematics 309, no. The paradigm problem. Covers permutations with repetitions. i=2 2 where ad is the number of augmented doubles, and r[i] is the exact repetition count at the i-th level. Permutations. Then it checks for the repetition C++ Language Using the main() Function. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good. It is rather a combinatorial problem that does not involve any algorithm. So we have the following algorithm: Define function permutations(i) returns all permutations using array[i] to array[n] (n is the total number arrays). Here's an implementation. Find all possible combinations with sum K from a given number N(1 to N) with the repetition of numbers is allowed Objective: Given two integers N and K, Write an algorithm to find possible combinations that add to K, from the numbers 1 to N. For example, the permutation \u03c3 = 23154 has three inversions: (1,3), (2,3), (4,5), for the pairs of entries (2,1), (3,1), (5,4). Dynamic Programming Algorithms Dynamic Programming Algorithm is an algorithm technique used primarily for optimizing problems, where we wish to find the \u201cbest\u201d way of doing something. A 6-letter word has 6! =6*5*4*3*2*1=720 different permutations. Order matters. No Repetition: for example the first three people in a running race. If letter box A must contain at least 2 letters. This algorithms check for duplication and repetition of the randomize question. Background. Algorithm for Permutation of a String in Java. Circular permutations. permutation without repetition. One of the goals of RcppAlgos is to provide a comprehensive and accessible suite of functionality so that users can easily get to the heart of their problem. Calculates the number of permutations with repetition of n things taken r at a time. A permutation with repetition of n chosen elements is also known as an \"n-tuple\". The number of permutations on a set of n elements is given by n!, where \u201c!\u201d represents factorial. Zero factorial or 0! Ways to arrange colors. Then, applying ( 1. Recursion comes directly from Mathematics, where there are many examples of expressions written in terms of themselves. Please see below link for a solution that prints only distinct permutations even if there are. permn - permutations with repetition Using two input variables V and N, M = permn(V,N) returns all permutations of N elements taken from the vector V, with repetitions. Start studying Ch. For instance, \u201c$$01110000$$\u201d is a perfectly good bit string of length eight. The number is (n-1)! instead of the usual factorial n! since all cyclic permutations of objects are equivalent because the circle can be rotated. Step 2 - repeat step 1 with the remaining items. Wrapping this function in a generator allows us terminate a repeated generation on some condition, or explore a sub-set without needing to generate the whole set:. The main advantage of this single chromosome representation is \u2014 in analogy to the permutation scheme of the traveling salesman problem (TSP) \u2014 that. 4), that is, the number of ways to pick k things out of n and arrange the k things in order. Algorithm for Permutation of a String in Java. So order matters\u2026 AB is not the same as BA Slideshow 3109936 by onan. I have found lots of permutation algorithms - have even written a few but I cannot figure out how to do this. java https://github. About the Author Tim Hill is a statistician living in Boulder, Colorado. Combinations are arrangements of objects without regard to order and without repetition. I discussed the difference between permutations and combinations in my last post, today I want to talk about two kinds [\u2026] List permutations with repetition and how many to choose from Noel asks: Is there a way where i can predict all possible outcomes in excel in the below example. The C programs in this section which finds the frequency of the word \u2018the\u2019 in a given sentence, finds the number of times the substring occurs in the given string, to find the frequency of every word in a given string and to find the highest frequency character in a string. nPr represents n permutation r which is calculated as n!/(n-k)!. The range used is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last. The result is: 1,2 2,1. STL has a shuffling algorithm called random_shuffle you can use. Recursive Permutation Algorithm without Duplicate Result. A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. permutation f to the low-order digits (See section 4. Permutations without Repetition In this case, we have to reduce the number of available choices each time. As the expected permutations are clearly not themselves permutations, our algorithms are not tools for finding assignments and are not competing with algorithms for finding an optimal assignment. 20) \u2013 patrickJMT; Repeated symbols example 1 and 2 \u2013 patrickJMT; Permutations with repetition (from 10. Order doesn\u2019t matter. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good. Permutations are denoted by the following which means the number of permutations of n items taken r items at a time. That is, it is a function from S to S for which every element occurs exactly once as an image value. gif 400 \u00d7 225; 82 KB. The Binomial Theorem 5. Euclid's algorithm and \u03c0. 1234 is followed by 2. From the 4th permutations. Taussig Dec 4 '17 at 11:47 $\\begingroup$ It seems to be both, and more specifically in the case of a cartesian product, it seems to be a cartesian power. The algorithm is not as fast as the fast_permutation, but it is faster than the orginal one and than the pure python one available in test_itertools. The range used is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last. 1 Permutations and Patterns The fundamental objects of this work are permutations. For example; given 3 letters abc find solution: Remember that the repetition is allowed in permutations unlike in combinations;. Let V be a vector of the outcome values. Thus, the number of permutations becomes (r - 1) n-2 P r-2. Permutations with Repetition Theorem 1: The number of r-permutations of a set of n objects with repetition allowed is nr. Sedgewick (1977) summarizes a number of algorithms for generating permutations, and identifies the minimum change permutation algorithm of Heap (1963) to be generally the fastest (Skiena 1990, p. The algorithm has potential to further differentiate between contours with the same prime form. Permutation without Repetition: This method is used when we are asked to reduce 1 from the previous term for each time. How to find out the missing number? Let the numbers in the array be 1,2,4,6,3,7,8,10,9 (total 9 numbers without repetition). Permutation can be done in two ways, Permutation with repetition: This method is used when we are asked to make. The algorithm is designed to take a selection of cells (from the Selection object), which should be located in the top row with no data below. Compute the permutation and print the result. * Combinations 26/05/2016 COMBINE CSECT. At each node c, the algorithm checks whether c can be completed to a valid solution. permutation f to the low-order digits (See section 4. List all permutations with a condition. Permutation with repetition Posted 06 December 2010 - 08:14 AM Im trying to make a program that implements this but I cant seem to get past inserting the characters. 0, all other syntax methods except $(handler); are deprecated. Ways to pick officers. For example: permutations without repetitions of the three elements A, B, C by two are - AB, AC, BA, BC, CA, CB. Zero factorial or 0! Ways to arrange colors. ) and M will be of the same type as V. In this case, It is important to note that order counts in permutations. The main advantage of this code is that the all permutations are generated in logical order: all permutations that starts with the first element come first. We care about the order because 247 wouldn\u2019t work. Please see below link for a solution that prints only distinct permutations even if there are. Introduction to Non-Repetitive Sequences Repetition is a part of life. The time complexity of this algorithm is \"O(n)\". STL has a shuffling algorithm called random_shuffle you can use. Number of combinations n=11, k=3 is 165 - calculation result using a combinatorial calculator. Permutations and partitions in the OEIS. Examples: Input: str = \u201caa\u201d Output: aa Note that \u201caa\u201d will be printed only once. Another definition of permutation is the number of such arrangements that are possible. A command-line program that uses the library is provided too, useful to teach combinatorics. Last Modified: 2013-12-14. Recursion is elegant but iteration is efficient. STL has a shuffling algorithm called random_shuffle you can use. A classical problem asks for the number of permutations that avoid a certain permutation pattern. Let me first re-write your specification: Print all permutations with repetition of characters. In C++: \u2022Write a program that produces ten random permutations of the numbers 1 to 10. To setup repository with documentation. A k-permutation of a multiset M is a sequence of length k of elements of M in which each element appears a number of times less than or equal to its multiplicity in M (an element's repetition number). A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Next lexicographical permutation algorithm Introduction. Restricted permutations are those constrained by having to avoid subsequences ordered in various prescribed ways. Here is one such algorithm, which generates the permutations in Lexicographical order. , TS, ACO, and GSA, are transformed into RPD measure where Minsol is the optimal solution if the given instance is solved to optimality or the lowest TCT obtained by any of models or algorithms. The elements might be of a string, or a list, or any other data type. 1 Permutations and Patterns The fundamental objects of this work are permutations. The idea is to generate each permutation from the previous permutation by choosing a pair of elements to interchange, without disturbing the other n-2 elements. Proceedings of the third international conference on Genetic Algorithms. 1983-01-01. A permutation relates to the order in which we choose the elements.$\\begingroup\\$ What you are describing is a permutation with repetition. The algorithm used for generating k-permutations was developed specifically for ECOS. The current theory would call three contours with the same prime form equally similar, without regard for further differences illustrated by the specific stages of the algorithm. Backtracking is a general algorithm for finding all enumerate all possible permutations using all items from the set without repetition. Returns true if such a \"next permutation\" exists; otherwise transforms the range into the lexicographically first permutation (as if by std::sort(first, last, comp)) and returns false. This is the aptitude questions and answers section on \"Permutation and Combination\" with explanation for various interview, competitive examination and entrance test. What the expected permutation matrices show very well is the potential for uncertainty for a true match. 1 Endorsement. Uses a precomputed lookup table of size n! containing the information of all transitions. If letter box A must contain at least 2 letters. A permutation is an act of arranging the elements of a set in all possible ways. Step 2 - repeat step 1 with the remaining items. This is the most well-known historically of the permutation algorithms. The CD that accompanies this book includes MySQL 4. java solves the 8 queens problem by implicitly enumeration all n! permutations (instead of the n^n placements). I adapted the code above to do permutations in Excel VBA. Another permutation algorithm in C, this time using recursion. A second multiple access system based on random permutations was studied. Last Modified: 2013-12-14. As these algorithms are very similar, it is suggested that a student tries to learn both to avoid unnecessary repetition (especially if they want to be able to solve the cube quickly). More precisely, we deal with a special version of the Black-Peg game with n holes and k >= n colors where no repetition of colors is allowed. Input: The first line of input contains an integer T, denoting the number of test cases. The basic difference between permutation and combination is of order Permutation is basically called as a arrangement. For example, a triple is interpreted as three doubles; the augmentation from 3-reps to 2-reps is (3 C 2) or 3. The idea is to swap each of the remaining characters in the string. Permutation With Repetition Algorithm Sometimes an inversion is defined as the pair of values. The visited array keeps track of which nodes have been visited already. However, we need to keep tracking of the solution that has also been in the permutation result using a hash set. Let S be a multiset that consists of n objects of which n1 are of type 1 and indistinguishable from each other. Two permutations with repetition are equal only when the same elements are at the same locations. Recursion means \"defining a problem in terms of itself\". Whether or not it actually is quicker is difficult to tell; I get the answer in 150 milliseconds, which is about twice as long as it took you. 1 \u2212 \u01eb, an algorithm due to Charikar, Makarychev and Makarychev [CMM06] can \ufb01nd an assignment with value 1\u2212O(\u221a \u01ebc). We care about the order because 247 wouldn\u2019t work. Backtracking is a general algorithm for finding all enumerate all possible permutations using all items from the set without repetition. Permutations with repetition by treating the elements as an ordered set, and writing a function from a zero-based index to the nth permutation. The results can be use for studying, researching or any other purposes. Let me first re-write your specification: Print all permutations with repetition of characters. , TS, ACO, and GSA, are transformed into RPD measure where Minsol is the optimal solution if the given instance is solved to optimality or the lowest TCT obtained by any of models or algorithms. java solves the 8 queens problem by implicitly enumeration all n! permutations (instead of the n^n placements). However, in many applied settings where a string is an appropriate model, a symbol may be used in at most one position. Generate random number without repetition android. Note that the permutation. (It slows down the algorithm in software) * The Feistel itself works on 64 bits! * An S-Box is a basic component which performs (non-linear!) substitution to implement a block cipher in symmetric key algorithms \u2013 gives confusion (see later). Objective: Given an array of integers (in particular order or permutation of a set of numbers), write an algorithm to find the lexicographically next permutation of the given permutation with only one swap. Algorithms for Generating Permutations and Combinations Section 6. Free essays, homework help, flashcards, research papers, book reports, term papers, history, science, politics. Practice makes perfect and repeating is good practice. My current code, for S = 5, has to check around 8000 possible lists. 2008-07-25 at 23:08. 2 Problem 47ES. However, in many applied settings where a string is an appropriate model, a symbol may be used in at most one position. Permutations with Repetition Theorem 1: The number of r-permutations of a set of n objects with repetition allowed is nr. Calculates count of combinations without repetition or combination number. It can be used to perform arbitrary permutation (without repetition) of n subwords within log n cycles regardless of the subword size. The visited array keeps track of which nodes have been visited already. A logistic map is used to generate a bit sequence, which is used to generate pseudorandom numbers in Tompkins-Paige algorithm, in 2D. The CD that accompanies this book includes MySQL 4. The permutation of a number of objects is the number of different ways they can be ordered: the position is important. Please see below link for a solution that prints only distinct permutations even if there are. 4567 is followed by. The number of permutations on a set of n elements is given by n!, where \u201c!\u201d represents factorial. Permutation vs. * * Inputs: unsigned short prn_seed - the initial value for the PRN generator * int permutation_const - value for. the first call to the recursive function will attempt to find permutations for 1 and 2. Use this idea to. The information bits are coded by a repetition code (shaping filter) and then interleaved by a random permutation. The following algorithm will generate all permutations of elements of a set, in lexicographic order: procedure all_permutations(S) if length(S) == 1 return the element as a length-one permutation else all_perm = [] for each x in S. 1983-01-01. In particular: Theorem 8 GI \u2208 PZK. API reference with usage examples available here. Permutations are denoted by the following which means the number of permutations of n items taken r items at a time. Permutation with repetition Calculator - High accuracy calculation Welcome, Guest. This document serves as an overview for attacking common combinatorial problems in R. First of all, while developing the algorithm, I asked my whole family and my neighbor (a judge) for help with the algorithm; no one could get even close. In particular: 1) What is the \"type\" of a permutation? 2) Do you want the number of all permutations, or do you want to list them? Your algorithm does neither. MArio http://www. The results can be use for studying, researching or any other purposes. We will calculate the letter count of B in a hashmap. The Hypothetical Scenario Generator for Fault-tolerant Diagnostics (HSG) is an algorithm being developed in conjunction with other components of artificial- intelligence systems for automated diagnosis and prognosis of faults in spacecraft, aircraft, and other complex. The idea is to generate each permutation from the previous permutation by choosing a pair of elements to interchange, without disturbing the other n-2 elements. A permutation cycle is a subset of a permutation whose elements trade places with one another. (a permutation can easily be encoded as an int). Algorithm and System Analysis multiset, sequence, word, permutation, k-set, k-list, k-multiset, k-lists with repetition, rule of product, Cartesian product. There are basically two types of permutation: Repetition is Allowed: such as the lock above. There are several algorithms for enumerating all permutations; one example is the following recursive algorithm: If the list contains a single element, then return the single element. The message is not registered. Priority Queue (Heap) \u2013. The word \"permutation\" also refers to the act or process of changing the linear order of an ordered set. b: the telephone number must be a multiple 0f 10 c: the telephone number must be a multiple of 100 d: the 1st 3 digits are 481 e: no repetition are allowed. com FREE SHIPPING on x diagrams very useful in solving problems involving combinations with repetition and I found myself using them to help understand most of the problems in the last chapter. npm run test:algo only runs tests for the finished permutation algorithms, excluding utilities. \u6b21\u306e\u3088\u3046\u306b\u5165\u529b\u3055\u308c\u305f\u30b3\u30fc\u30c91,2,3\u3092\u7e70\u308a\u8fd4\u3057\u751f\u6210\u3059\u308bFortran\u3067\u30b3\u30fc\u30c9\u3092\u66f8\u304d\u307e\u3059\u3002 111 112 113 121 122 123. the number of permutations, for a given number (this is classically known as \u2018the travelling salesman\u2019 problem):. This number of permutations is huge. Given a string of length n, print all permutation of the given string. Permutation with repetition Calculator - High accuracy calculation Welcome, Guest. Ways to pick officers. For each number, we add it to the results of permutations(i+1). Instructions to install MySQL and MySQL Connector J. Given a string S. An estimation of minimum distance for proposed codes is obtained. We prove a parallel repetition theorem for general games with value tending to 0. I'm stuck with nested for loops that are dependent on the previous loop. Now this is exactly the combinatorial problem of 5 selections from 10 choises with repetition. In the permutation without repetition section, the same dog will show up in many of those 2730 ways to order them, but the same ordered set will not show up twice. Permutation with repetition. Background. The results can be use for studying, researching or any other purposes. The idea is to fix the first character at first index and recursively call for other subsequent indexes. Permutations without Repetition. 3 Prim\u2019s Algorithm. But it is not repetition of a single element that produces this outcome. From the above example of 8 letters, we have the following observation. See full list on dev. If we solve this problem using naive algorithm then time complexity would be exponential but we can reduce this to O(n * k) using dynamic programming. In this version of quicksort used middle element of array for the pivot. For example: permutations with repetitions of the three elements A, B, C by two are - AA, AB, AC, BA, BB, BC, CA, CB, CC. permutation (PRP), meaning that as long as the key is secret, First, a block cipher used in practice isn\u2019t a gigantic algorithm but a repetition of rounds,. 2 Permutations \u00b6 permalink. Combinations with Repetition 07. The number of r-combinations from a set with n elements when repetition of elements is allowed is C(n + r - 1, r) = C(n + r - 1, n - 1) Permutations with Indistinguishable objects Theorem. 6: Combinations with Repetition Eg: Counting Iterations of a Loop How many times will the innermost loop be iterated when the algorithm segment below is implemented and run? (Assume n is a positive integer. If it cannot, the whole sub-tree rooted at c is skipped (pruned). It can be used to perform arbitrary permutation (without repetition) of n subwords within log n cycles regardless of the subword size. Cicirello VA, Cernera R (2013) Profiling the distance characteristics of mutation operators for permutation-based genetic algorithms. We wish to show that the efficiency of GAs in solving a flowshop problem can be improved significantly by tailoring the various GA operators to suit the structure of the problem. For maximum compatibility, this program uses only the basic instruction set (S/360) and two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. The only pair of 3-edges that can feature the same permutation with repetition are 123xyz --> 456xyz231 3-edges. To implement the clustered permutation test, assume there are two treatment groups with unknown outcome distributions F and G, with M and N clusters, respectively. What is the best way to do so? The naive way would be to take a top-down, recursive approach. Permutation vs. The arrangements are allowed to reuse elements, e. The algorithm has potential to further differentiate between contours with the same prime form. nPr represents n permutation r which is calculated as n!/(n-k)!. Combinations with repetition, and counting monomials. Combinations with Repetition. The permutations for which dp(w) = \u2113 (w) are characterized directly; we also have a bijec tion with Dyck paths to help make the characterization more intuitiv e. Given an array of 9 numbers that contains numbers between 1 to 10, obviously no repetition, one number missing. A permutation cycle is a subset of a permutation whose elements trade places with one another. Circular permutations. Namely, our algorithm is: repeat I times. The two key formulas are:. \u201can ordered combination\" w/ repetition: n^r n = total choices you have (ei: you have ten golf balls) r = how many times you choose (ei: you pick it out three) ex// 10^3 = 1,000 ways possible; also known as another way: \"ei: three bread, two pickles, three dimes\u201d you want to find total combo? 3x2x3 = 18 ways. List permutations with repetition and how many to choose from. Solved examples with detailed answer description, explanation are given and it would be easy to understand - Page 3. We care about the order because 247 wouldn\u2019t work. The idea is to swap each of the remaining characters in the string. The algorithm used for generating permutations by transpositions is often called the \"Johnson-Trotter\" algorithm, but it was discussed earlier in the works of Steinhaus and the some campanologists. We have moved all content for this concept to for better organization. Combinatorics Processing. Fig 5:Mix columns V. (8*6 ==> 8*4) 2 bits used to select amongst 4 substitutions for the rest of the 4-bit quantity S-Box Examples DES Standard Cipher Iterative Action : Input: 64 bits Key: 48 bits Output: 64 bits Key Generation Box : Input: 56 bits Output: 48 bits DES Box Summary Simple, easy to implement: Hardware/gigabits/second, software/megabits/second 56-bit. Recursion means \"defining a problem in terms of itself\". We will typically view these objects in one-line notation, i. The idea is to fix the first character at first index and recursively call for other subsequent indexes. But like me, many others are looking for code/algorithm to generate combinations when we have repeated digits/numbers in a set. In this paper, we propose a variable block insertion heuristic (VBIH) algorithm to solve the permutation flow shop scheduling problem (PFSP). This can be a very powerful tool in writing algorithms. Any ordered arrangement such as C-B-F-A-D-G-H-E is called a permutation of the 8 letters. [permutations] [combinations] This lecture covers basic combinatorial algorithms which generate successively all permutations, combinations and variations respectively. If all the n characters are unique, you should get n! unique permutations. The test came back with one issue worth mentioning in this blog. a list where the permutation \u02c72S n is written \u02c7= \u02c7 1\u02c7 2 \u02c7 n. I've been creating my code in vb. Any algorithm where the current permutation requires the previous one is discarded as threadable because we can\u2019t start enumeration of permutations from any specific place (with known code). Lee}, title = {Fast Subword Permutation Instructions Based on Butterfly Networks}, booktitle = {In Proceedings of SPIE, Media Processor 2000}, year = {2000}, pages = {80--86}}. If letter box A must contain at least 2 letters. (3) Execute Davis-Putnam based on y and \u2026, which takes at most n steps. RESULTS The following results for the permutation-based system are achieved using an implementation of the GALIB library [1], which had to be extended greatly to handle permutations with repetition. Now lemme, permutations. To implement the clustered permutation test, assume there are two treatment groups with unknown outcome distributions F and G, with M and N clusters, respectively. The idea is to fix the first character at first index and recursively call for other subsequent indexes. Zero factorial or 0! Ways to arrange colors. Objective: Given an array of integers (in particular order or permutation of a set of numbers), write an algorithm to find the lexicographically next permutation of the given permutation with only one swap. The number of r-combinations from a set with n elements when repetition of elements is allowed is C(n + r - 1, r) = C(n + r - 1, n - 1) Permutations with Indistinguishable objects Theorem. Given a string str, the task is to print all the permutations of str. with repetition \\) Customer Voice. here i supply u a c++ code to generate variations. Write a program to print all permutations of a given string. Namely, our algorithm is: repeat I times. Instructions to install MySQL and MySQL Connector J. Please update your bookmarks accordingly. , involutions [12] and derangements [9]). One interesting application is the rearrangement of characters in a word to create other words. For maximum compatibility, this program uses only the basic instruction set (S/360) and two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. Permutation refers to the process of arranging all the members of a given set to form a sequence. Since permutations with repetition does not have. com Blogger 34 1 25 tag:blogger. Order matters. In mathematics, a combination is a selection of items from a collection, such that (unlike permutations) the order of selection does not matter. Mathematical Programming, 99(3):563--591, 2004. 2 Permutations. Permutation multiplication (or permutation composition) is perhaps the simplest of all algorithms in computer science. With next_combination() and next_permutation() from the STL algorithms, you can find permutations!! The formula for total number of permutations of r sequence picked from n sequence is n!/(n-r)! You can call next_combination() first and then next_permutation() iteratively. Purpose of use Needed to calculate a very large probability based on the Combination of 10,000,000 chemicals taken 500,000 at a time. Algorithm for the enumeration of permutations with finite repetition. I find it to be intuitive and easy to implement. A logistic map is used to generate a bit sequence, which is used to generate pseudorandom numbers in Tompkins-Paige algorithm, in 2D. This stage is fairly simple as only one algorithm is used to swap two edges around. Combinatorics Processing. The genetic algorithm uses permutations with repetition to encode chromosomes and a schedule generation scheme, termed OG&T, as decoding algorithm. In other words, it is the number of ways r things can be selected from a group of n things. The number of permutations of n distinct objects is n\u00d7(n \u2212 1)\u00d7(n \u2212 2)\u00d7\u22ef\u00d72\u00d71, which number is called \"n factorial\" and written \"n!\". Start studying Ch. number of things n 6digit 10digit 14digit 18digit 22digit 26digit 30digit 34digit 38digit 42digit 46digit 50digit. Hence if there is a repetition of elements in the array, the same permutation may occur twice. Number of combinations n=11, k=3 is 165 - calculation result using a combinatorial calculator. In this straight forward approach we create a list of lists containing the permutation of elements from each list. I only need to generate lists for small values of S, lets say up to 10-20. Mathematical algorithm that accurately predicts that, for many data sets, the first digit of each group of numbers in a random sample will begin with 1 more than a 2, a 2 more than a 3, a 3 more than a 4, and so on. Know the formula:. In order to sequence the tasks of a job shop problem (JSP) on a number of machines related to the technological machine order of jobs, a new representation technique \u2014 mathematically known as \u201cpermutation with repetition\u201d is presented. A 6-letter word has 6! =6*5*4*3*2*1=720 different permutations. A numerical study of the plume in Cape Fear River Estuary and adjacent coastal ocean. Hence, by the product rule there are nrr-permutations with repetition. The algorithm might look like this (starting with an empty permutation): Repeat 'forever' (precisely: until a break): if the permutation isn't full yet (length less than n), append zeros (or whatever the minimum allowed value is); otherwise: add the permutation to results,. Additional there is a rule - whether you can choose the same option twice or not (Repetitions),and a comparison - whether the order of the single choices makes a difference or not (Respect Order). 20an open-source database management system. number of things n 6digit 10digit 14digit 18digit 22digit 26digit 30digit 34digit 38digit 42digit 46digit 50digit. Implement Binary Search Tree (BST) pre-order traversal (depth first). npm run test:algo only runs tests for the finished permutation algorithms, excluding utilities. In this post, we will see how to find permutations of a string containing all distinct characters. So there would be a substring of length of B in the string A which has exact same letter count as B. When the order does not matter and an object can be chosen more than once. This number of permutations is huge. As you can tell, 720 different \"words\" will take a long time to write out. The second stage involves the permutation of the top layer edges. The number of unique permutations possible is exactly 1, yet your algorithm (even with dictionary) will loop many many times just to produce that one result. Would you mind checking index 11 aabbaabbb is low by 1 and index 125 bbbbbaaaa is low by 5;. Permutations can thus be represented as a tree of permutations:. Groups of Permutations. Permutation with repetition Posted 06 December 2010 - 08:14 AM Im trying to make a program that implements this but I cant seem to get past inserting the characters. Refresh your memory! How many permutations, combinations and variations can be generated from set of N elements? And what about if repeated elements are allowed?. One of the main questions in this area is the parallel repetition question: Is there a way to decrease the. Hi! I have tried a bit, but I was not able to find a way to generate permutations with repetitions. Algorithm for the enumeration of permutations with finite repetition. Find all possible combinations with sum K from a given number N(1 to N) with the repetition of numbers is allowed Objective: Given two integers N and K, Write an algorithm to find possible combinations that add to K, from the numbers 1 to N. 48) Combinations with repetition. Dynamic Programming Algorithms Dynamic Programming Algorithm is an algorithm technique used primarily for optimizing problems, where we wish to find the \u201cbest\u201d way of doing something. The permutation method differs from its combination comrade primarily in that arrangement does matter. Textbook solution for Discrete Mathematics With Applications 5th Edition EPP Chapter 9. The following algorithm will generate all permutations of elements of a set, in lexicographic order: procedure all_permutations(S) if length(S) == 1 return the element as a length-one permutation else all_perm = [] for each x in S. As mentioned in [2], for deriving a secure permutation g with a common domain, the domain of g would be 160 bits larger than that of f. Proceedings of the third international conference on Genetic Algorithms. We will calculate the letter count of B in a hashmap. The recursive function should generate all permutations for the first n-1 numbers. A compact and very fast way to check for duplicates is to use an unordered_set. Algorithm T: 'Plain change algorithm' as described in. Implement Binary Search Tree (BST) pre-order traversal (depth first). Possible implementation. The number of possible permutations with repetition of n elements by m equals. Generating combinations of k elements: Generating combinations of k elements from the given set follows similar algorithm used to generate all permutations, but since we don't want to repeat an a character even in a different order we have to force the recursive calls to not to follow the branches that repeat a set of. Nice algorithm without recursion borrowed from C. If there are twenty-five players on the team, there are $$25 \\cdot 24 \\cdot 23 \\cdot \\cdots \\cdot 3 \\cdot 2 \\cdot 1$$ different permutations of the players. Similar to The Permutation Algorithm for Arrays using Recursion, we can do this recursively by swapping two elements at each position. \u00ab Prev - Affine Cipher Multiple Choice Questions and Answers (MCQs) \u00bb Next - P, NP, NP-hard, NP-complete Complexity Classes Multiple Choice Questions and Answers (MCQs). A permutation is a method to calculate the number of events occurring where order matters. Proof: There are n ways to select an element of the set for each of the r positions in the r-permutation when repetition is allowed. We will first take the first character from the String and permute with the remaining chars. For instance, \u201c$$01110000$$\u201d is a perfectly good bit string of length eight. Now the above code will print all the permutations of the string \"GOD\" without repetition. The paradigm problem. If no explicit formula could be given, I would already be satisfied with a more efficient algorithm to generate the lists. Permutation with repetition Calculator - High accuracy calculation Welcome, Guest. 6: Combinations with Repetition Eg: Counting Iterations of a Loop How many times will the innermost loop be iterated when the algorithm segment below is implemented and run? (Assume n is a positive integer. Algorithm for the enumeration of permutations with finite repetition. Then the nth number can be added into every position of the n-1 permutations to generate all permutations. post-5715079000043709685. Return all combinations. Distinct elements is the simplest case, and here we will also discuss the ramifications of employing the strongly concentrated hashing of Aamand et al. counting permutations with repetition covering countable Critical Path Method cubic graph cut vertex cycle cylindrical system deductive reasoning degree degree sequence denumerable depth-first algorithm derivative derived function Dijkstra's algorithm diameter of a graph difference of sets digraph dimension dimension analysis directed graph. Cape Fear River Estuary (CFRE), located in southeast North Carolina, is the only river estuary system in the state which is directly connected to the Atlantic Ocean. 10 shows a standard algorithm for computing kP (k is a positive integer) per every 2 bits as in the modular exponentiation operation. The permutation generator 300 receives, via a random number input 304, a random number which it stores in a buffer. The prover picks at random a permutation \u03c0 and sends to the prover the graph G = (V,E) where E = \u03c0(E1)). com/tusharroy25 https://github. The permutation still has to be performed in order to be compliant with the DES standard. (2) Number of permutations. return a uniformly random permutation of the elements when the comparator is replaced by a fair coin flip (that is, return x < y = true with probability 1/2, regardless of the value of x and y) The code for the sorting algorithm must be the same. Permutation refers to the process of arranging all the members of a given set to form a sequence. python - compter efficacement les combinaisons et les permutations. To address this, what is typically done is, a dictionary is created that stores the frequency of each integer that is available at a given point. Groups of Permutations. If we solve this problem using naive algorithm then time complexity would be exponential but we can reduce this to O(n * k) using dynamic programming. Use this idea to. 2006-12-01. Combinations with Repetition 6. Permutations of 123: 123 132 213 231 312 321 Permutations of 1234: 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321 From the above output, it\u2019s very easy to find a correlation between the pattern of digits each of the whole number permutations has!!. Algorithm Complexity Analysis (Big O notation) \u2013 You are free to skip these parts and it shouldn\u2019t affect the understanding of working of the algorithm. For , he ran the algorithm 1000 times and found 105 different families of nine mutually disjoint S-permutation matrices. The explanation is good and crisp and the code is easy and good for a novice. Recursion is elegant but iteration is efficient. The only pair of 3-edges that can feature the same permutation with repetition are 123xyz --> 456xyz231 3-edges. A description of an algorithm used to construct and test for additively non-repetitiveness will be provided, then results from research will be analyzed. Nice algorithm without recursion borrowed from C. [permutations] [combinations] This lecture covers basic combinatorial algorithms which generate successively all permutations, combinations and variations respectively. When some of those objects are identical, the situation is transformed into a problem about permutations with repetition. Given two graphs G1 = (V,E1),G2 = (V,E2), 1. Algorithm takes the input of the string. We will sometimes write \u02c7(1)\u02c7(2) \u02c7(n) to. No Repetition: for example the first three people in a running race. In particular, a discrete Differential Evolution algorithm which directly works on the space of permutations with repetition is defined and analyzed. Permutation: Arrangement without repetition. We will calculate the letter count of B in a hashmap. java solves the 8 queens problem by implicitly enumeration all n! permutations (instead of the n^n placements). Genetic algorithms (GAs) are search heuristics used to solve global optimization problems in complex search spaces. Permutation can be done in two ways, Permutation with repetition: This method is used when we are asked to make different choices each time and have different objects. We will maintain 3 variables, existing letter positive count, existing letter negative count and non-existing letter. If we have a n-element set, the amount of its permutation is:. This question is from textbook : 1. (It slows down the algorithm in software) * The Feistel itself works on 64 bits! * An S-Box is a basic component which performs (non-linear!) substitution to implement a block cipher in symmetric key algorithms \u2013 gives confusion (see later). Lavavej on the Standard Template Library (STL) in C++. All videos were created by the students of EECS 203 - Discrete Mathematics at the University of Michigan in Winter 2012. Write a program to print all permutations of a given string. Forinstance, thecombinations of the letters a,b,c,d taken 3 at a time with repetition are: aaa, aab,. For an input string of size n, there will be n^n permutations with repetition allowed. Circular shift-It shifts each bit in an n-bit word K positions to the left. This is the currently. It is based on program Permutations. Similar to The Permutation Algorithm for Arrays using Recursion, we can do this recursively by swapping two elements at each position. Refresh your memory! How many permutations, combinations and variations can be generated from set of N elements? And what about if repeated elements are allowed?. Look for certain helpful keywords 2. Example 2 - Combinations. here the algo is: if n is the variety of quite a few element in a string. For example, a triple is interpreted as three doubles; the augmentation from 3-reps to 2-reps is (3 C 2) or 3. In general, a permutation is an ordered arrangement of a set of objects that are distinguishable from one another. Program Queens2. The algorithm has potential to further differentiate between contours with the same prime form. We define permutation as different ways of arranging some or all the members of a set in a specific order. List all permutations with a condition. how many 7-digit telephone numbers are possible if the first digit cannot be 0 and a: only odd digits may be used. General Terms: Algorithms. It is based on program Permutations. The algorithm might look like this (starting with an empty permutation): Repeat 'forever' (precisely: until a break): if the permutation isn't full yet (length less than n), append zeros (or whatever the minimum allowed value is); otherwise: add the permutation to results,. We will maintain 3 variables, existing letter positive count, existing letter negative count and non-existing letter. By using the same key produce several runs of permutations that be as secure as a hash, but completely reversible. AES algorithm using matlab VII. A combination with replacement is an unordered multiset that every element in it are also in the set of n elements. This implies that the answer length c(\u01eb) in the Unique Games Conjecture must be larger than \u03a9(1/\u01eb) if the conjecture is to hold. The algorithm described by AES is a symmetric-key algorithm (the same key is used for both encrypting and decrypting the data) and the design principle is known as a substitution-permutation network (SP-network or SPN) a series of mathematical operations used in cipher algorithms. Heap's algorithm is used to generate all permutations of n objects. Generating combinations of k elements: Generating combinations of k elements from the given set follows similar algorithm used to generate all permutations, but since we don't want to repeat an a character even in a different order we have to force the recursive calls to not to follow the branches that repeat a set of. Thus, the number of permutations becomes (r - 1) n-2 P r-2. Recursion is elegant but iteration is efficient. P(n, r) denotes the number of permutations of n objects taken r at a time. The paradigm problem. What the expected permutation matrices show very well is the potential for uncertainty for a true match. Different permutations can be ordered according to how they compare lexicographicaly to each other; The first such-sorted possible permutation (the one that would compare lexicographically smaller to all other permutations) is the one which has all its elements sorted in ascending order, and the largest has all its elements sorted in descending. 6 uses to generate the permutations and eliben's one looks like Johnson-Trotter permutation generation, you might look for article in Wikipedia on Permutations and their generation that looks quite like unrank function in paper by Myrvold and Ruskey. The number of ways to arrange n distinct objects along a fixed (i. Therefore, the number of permutations in this case = 10x10x10x10x10x10 = 1000000 Circular Permutation. The combination of Two Square Cipher and Variably Modified Permutation Composition (VMPC) algorithm is intended for obtaining stronger ciphers than using only one cipher, so it is not easy to solve. Thank you for your questionnaire. For example, say our function is given the numbers 1,2 and 3. Technical blog and complete tutorial on popular company interview questions with detailed solution and Java program on Data structure, Algorithms, Time and space complexity, Core Java, Advanced Java, Design pattern, Database, Recursion, Backtracking, Binary Tree, Linked list, Stack, Queue, String, Arrays etc asked in companies like Google, Amazon, Microsoft, Facebook, Apple etc. What is a permutation and what is a combination with repetition and no repetition? Permutation Groups Generated by 3-Cycles [05/14/2003] Show A_n contains every 3-cycle if n >= 3; show A_n is generated by 3- cycles for n >= 3; let r and s be fixed elements of {1, 2,, n} for n >= 3 and show that A_n is generated by the n 'special' 3-cycles of. Permutations without Repetition. We present a strategy that identifies the secret code in O(n log n) queries. For both combinations and permutations, you can consider the case in which you choose some of the n types more than once, which is called 'with repetition', or the case in which you choose each type only once, which is called 'no repetition'. The algorithm for obtaining random permutations, based on a randomized algorithm with a probability evaluation is equal to 1, which is more e\ufb03cient than the other algorithm with probability evaluation p(n) = n! nn, is described in section 2. For sample the default for size is the number of items inferred from the first argument, so that sample(x) generates a random permutation of the elements of x (or 1:x). Permutations and Combinations. The two key formulas are:. We will typically view these objects in one-line notation, i. Previously Dinur and Steurer proved such a theorem for the special case of projection games. In statistics, the two each have very specific meanings. List all pair of permutations with repetition with given condition, conditions are elaborated below Relevant Equations: of S, lets say up to 10-20. Priority Queue (Heap) \u2013. The permutations for which dp(w) = \u2113 (w) are characterized directly; we also have a bijec tion with Dyck paths to help make the characterization more intuitiv e. Thank you for your questionnaire. The byte substitution transformation is a nonlinear. Permutation With Repetition Problems With Solutions : In this section, we will learn, how to solve problems on permutations using the problems with solutions given below. 2006-12-01. SM-2 is a simple spaced repetition algorithm. The notation supports the following high-level constructs: permutation, grouping, repetition, inversion, reflection, conjugation, commutation, rotation and single-line and multiple-line comments. Thus, we can use permutations(i + 1) to calculate permutations(i). 48) Combinations with repetition. Permutation and orientation changes of individual cube parts can be specified using permutation cycles. https://www. Permutations, combinations and the binomial theorem. It is efficient and useful as well and we now know enough to understand it pretty easily. Because order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. The general Formula. A combination with replacement is an unordered multiset that every element in it are also in the set of n elements. permutation synonyms, permutation pronunciation, permutation translation, English dictionary definition of permutation. Permutation in a circle is called circular permutation. The message is not registered. It was written in Visual Studio 2013 using C# and DeflateStream class. Permutations of 123: 123 132 213 231 312 321 Permutations of 1234: 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321 From the above output, it\u2019s very easy to find a correlation between the pattern of digits each of the whole number permutations has!!. The proposed encryption system includes two major parts, chaotic pixels permutation and chaotic pixels substitution. Since permutations with repetition does not have. Technically, a permutation of a set S is defined as a bijection from S to itself. Counting Permutations with Fixed Points; Pythagorean Triples via Fibonacci Numbers. As an example, if the string is \"abc\" there are 6 permutations {abc, acb, bac, bca, cab, cba}. On the positive side, we might have hoped that we could use the parallel repetition. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Any selection of r objects from A, where each object can be selected more than once, is called a combination of n objects taken r at a time with repetition. Algorithm for Permutation of a String in Java. In particular: Theorem 8 GI \u2208 PZK. Mathematical algorithm that accurately predicts that, for many data sets, the first digit of each group of numbers in a random sample will begin with 1 more than a 2, a 2 more than a 3, a 3 more than a 4, and so on. Permutations 3. The study of permutations in this sense generally belongs to the field of combinatorics. the leftmost k bits become the original rightmost bits. e where the repetitions of the characters are included then read the matter below. This problem can also be asked as \u201cGiven a permutation of numbers you need to find the next larger permutation OR smallest \u2026. return a uniformly random permutation of the elements when the comparator is replaced by a fair coin flip (that is, return x < y = true with probability 1/2, regardless of the value of x and y) The code for the sorting algorithm must be the same. In mathematics, a combination is a selection of items from a collection, such that (unlike permutations) the order of selection does not matter. The results can be use for studying, researching or any other purposes. Then it checks for the repetition C++ Language Using the main() Function. In order to sequence the tasks of a job shop problem (JSP) on a number of machines related to the technological machine order of jobs, a new representation technique \u2014 mathematically known as \u201cpermutation with repetition\u201d is presented. Backtracking is a general algorithm for finding all enumerate all possible permutations using all items from the set without repetition. This is the currently. Knuth (volume 4, fascicle 2, 7. The number of permutations on a set of n elements is given by n!, where \u201c!\u201d represents factorial. com/tusharroy25 https://github. For a given string of size n, there will be n^k possible strings of length \"length\". The works in this exhibition play with the seemingly endless permutations of data to investigate the scale and scope of data as well as its elegance and anxieties. What about if we want to get all the possible permutations with repetition. In effect, all that's going on here is to exploit the sophisticated algorithms of a computer algebra system to keep track of all the possible combinations as each additional die is introduced. To practice all areas of Data Structures & Algorithms, here is complete set of 1000+ Multiple Choice Questions and Answers. An estimation of minimum distance for proposed codes is obtained. counting permutations with repetition covering countable Critical Path Method cubic graph cut vertex cycle cylindrical system deductive reasoning degree degree sequence denumerable depth-first algorithm derivative derived function Dijkstra's algorithm diameter of a graph difference of sets digraph dimension dimension analysis directed graph. Following is the illustration of generating all the permutations of n given numbers. About the Author Tim Hill is a statistician living in Boulder, Colorado. (3) Execute Davis-Putnam based on y and \u2026, which takes at most n steps. From what I have found, a repeated permutation with repetition can only exist in an edge of 3 or greater. Wrapping this function in a generator allows us terminate a repeated generation on some condition, or explore a sub-set without needing to generate the whole set:. In particular: Theorem 8 GI \u2208 PZK. The algorithm for finding scalar times of points on an elliptic curve is similar to an algorithm for modular exponentiation operation. For , he ran the algorithm 1000 times and found 105 different families of nine mutually disjoint S-permutation matrices. In statistics, the two each have very specific meanings. -Invertile Transformation. Possible three letter words. Knuth (volume 4, fascicle 2, 7. We will maintain 3 variables, existing letter positive count, existing letter negative count and non-existing letter. (spot, fido, max) is not the same set as (fido, spot, max). More precisely, we deal with a special version of the Black-Peg game with n holes and k >= n colors where no repetition of colors is allowed. This number of permutations is huge. 6: Combinations with Repetition Eg: Counting Iterations of a Loop How many times will the innermost loop be iterated when the algorithm segment below is implemented and run? (Assume n is a positive integer. That way, you will find all the permutations. A description of an algorithm used to construct and test for additively non-repetitiveness will be provided, then results from research will be analyzed. The Binomial Theorem 5. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange. This implies that the answer length c(\u01eb) in the Unique Games Conjecture must be larger than \u03a9(1/\u01eb) if the conjecture is to hold. there is no repetition. In order to sequence the tasks of a job shop problem (JSP) on a number of machines related to the technological machine order of jobs, a new representation technique \u2014 mathematically known as \u201cpermutation with repetition\u201d is presented. Mathematical algorithm that accurately predicts that, for many data sets, the first digit of each group of numbers in a random sample will begin with 1 more than a 2, a 2 more than a 3, a 3 more than a 4, and so on. Classes have been defined according to whether order is important, items may be repeated, and length is specified. Inverted indexing is a ubiquitous technique used in retrieval systems including web search. This is not the case with fast_permutation. I have searched all the forms to try and find a solution for this. I'm stuck with nested for loops that are dependent on the previous loop. In the example, is , and is. Combinatorics. how many 7-digit telephone numbers are possible if the first digit cannot be 0 and a: only odd digits may be used. The Binomial Theorem 5. But repetition becomes boring. I adapted the code above to do permutations in Excel VBA. Algorithm for the enumeration of permutations with finite repetition. This is often written 3_P_2. It can be used to perform arbitrary permutation (without repetition) of n subwords within log n cycles regardless of the subword size. The permutation in a haystack problem and the calculus of search landscapes. here the algo is: if n is the variety of quite a few element in a string. with repetition \\) Customer Voice. Combinations with repetition, and counting monomials. A simple algorithm to generate a permutation of n items uniformly at random without retries, known as the Knuth shuffle, is to start with the identity permutation, and then go through the positions 1 through n, and for each position i swap the element currently there with an arbitrarily chosen element from positions i through n, inclusive. API reference with usage examples available here. How to use iteration in a sentence. Each test case contains a single string S in capital letter. where the order matters (who holds the president office matters) and no repetition is allowed. In a 3-digit combination lock, each digit. If the list contains more than one element, loop through each element in the list, returning this element concatenated with all permutations of the remaining n. As the expected permutations are clearly not themselves permutations, our algorithms are not tools for finding assignments and are not competing with algorithms for finding an optimal assignment.", "date": "2020-11-27 14:53:11", "meta": {"domain": "gattopescatore.it", "url": "http://ronk.gattopescatore.it/permutation-with-repetition-algorithm.html", "openwebmath_score": 0.5992879271507263, "openwebmath_perplexity": 674.6376522032865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.982287697666963, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8534378486979414}}
{"url": "http://math.stackexchange.com/questions/179660/need-to-clarify-the-at-least-concept-in-combination", "text": "# Need to clarify the \u201cAt-least Concept\u201d in Combination.\n\nI managed to solve this question but I had some inquiries regarding the solution.\n\nIf two cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the two cards if at least one card is a jack, queen or a king?\n\nHere is how I solved it:\n\nJack/Queen/King = $4 \\times 3 = 12$ cards\n\nOther Cards = $52-12 = 40$ cards\n\nNow we can only pick two cards so:\n\n\u2022 It can be either from the Jack/King/Queen so $\\binom{12}{1}=12$\n\n\u2022 Both cards can be from Jack/King/Queen so $\\binom{12}{2} = 66$\n\n\u2022 One card can be from the remaining stack (Non - jack,king or queen) so $\\binom{40}{1}=40$\n\nNow the only problem I have with this question is when getting the final value initially I was doing\n\n$1$ from Remaining Cards $\\times$ [ ($1$ from Jack/King/Queen) + 2(from Jack/King/Queen) ]= $\\binom{40}{1} \\times ( \\binom{12}{1} + \\binom{12}{2} )$\n\nBut The actual answer comes if we do the following\n\n[$1$ from Remaining Cards $\\times$ $1$ from Jack/King/Queen ] + 2(from Jack/King/Queen) ]= $( \\binom{40}{1} \\times \\binom{12}{1}) + \\binom{12}{2}$\n\nI would really appreciate it if someone could clarify why do we do it the second way and not the first way ? Which part is added/multiplied to which part ? Is there an easier way to know how its done. Am I missing some important concept here ?\n\nEdit : While trying to understand this I also looked up the definition of disjoint events which means \"Two events are disjoint if they can't both happen at the same time\" so then again here is what I did (Special here means Jack/king/Queen)\n\n(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special)\n\n$(\\binom{40}{1} \\times \\binom{12}{1}) + (\\binom{40}{1} \\times \\binom{12}{2}) )$\n\nWhich simplifies to $\\binom{40}{1} \\times ( \\binom{12}{1} + \\binom{12}{2} )$\n\nSo is the representation of the problem using\n\n(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special) wrong ?\n\n-\nIf someone tells you he have at least one card is a jack, queen or a king then this means that it is not possible that he doesn't have a king or a queen or a jack (example: if he does not have a jack or a queen he must be holding a king) \u2013\u00a0 Belgi Aug 6 '12 at 21:32\nThere are $\\binom{52}{2}$ ways to choose two cards. There are $\\binom{40}{2}$ ways to choose them so none is J, Q, or K. So an alternate form of the answer is $\\binom{52}{2}-\\binom{40}{2}$. In this case, probably a little harder than the way you used, but in other situations the idea of counting complement can save a lot of work. \u2013\u00a0 Andr\u00e9 Nicolas Aug 6 '12 at 21:50\nThis seems rather obvious: the product term $\\tbinom{40}1\\times\\tbinom{12}2$ represents \"$1$ card from the remaining stack and both cards from Jack/Queen/King\", or as you say it in the edit \"1 from the 40 and 2 from special\". But since only two cards are drawn this is not among the possibilities of the problem, so the term should not be present. A product represents two independent choices, and mutually exclusive (yet individually possible) events are never independent. \u2013\u00a0 Marc van Leeuwen Aug 7 '12 at 9:27\n\nIf you expand your first answer, you have $$\\binom{12}{1} \\cdot \\binom{40}{1} + \\binom{12}{2} \\cdot \\binom{40}{1}.$$ Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.\n\nA rule of thumb is to turn \"at least\" problems into several instances of \"exactly\".\n\nFor the problem at hand, \"at least one jack/queen/king\" translates into \"either exactly one jack/queen/king or exactly two jack/queen/king\".\n\nThe number of ways to get exactly one jack/queen/king and one other card is $$\\binom{12}{1} \\cdot \\binom{40}{1}.$$ We multiply here because you choose a card from the jack/queen/king pile and a card from the \"other\" pile.\n\nThe number of ways to get exactly two from jack/queen/king is $$\\binom{12}{2}.$$ Since these cases are disjoint (this is important), we simply add the results from the two cases. Thus, the number of ways to get at least one Jack, Queen, or King is $$\\binom{12}{1} \\cdot \\binom{40}{1} + \\binom{12}{2}.$$\n\n-\nthanks for the answer , could you explain what you meant by Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.. \u2013\u00a0 MistyD Aug 6 '12 at 21:38\n@MistyD The term $\\binom{12}{2} \\cdot \\binom{40}{1}$ instructs you to select two cards from the jack/queen/king pile and one card from the \"other\" pile, giving you a three-card hand (not a two-card hand, like the problem is asking for). \u2013\u00a0 Austin Mohr Aug 6 '12 at 21:39\nThough I partially get what you meant. Lets consider the correct way then $( \\binom{40}{1} \\times \\binom{12}{1}) + \\binom{12}{2}$ . Which I believe means 1 from remaining cards (which is 40) and 1 from J/K/Q , now what does + (mean here) . My mind keeps on asking me why we didn't multiply here instead of adding? \u2013\u00a0 MistyD Aug 6 '12 at 22:05\nAddition here corresponds to the union operation on sets; if the sets $A$ and $B$ have no elements in common (this is a critical restriction!) then $|A\\cup B| = |A| + |B|$. In effect what you're saying is that 'all the combinations with at least one JQK' consists of 'all the combinations with exactly one JQK' and 'all the combinations with more than one JQK' (in this case there can be only two). \u2013\u00a0 Steven Stadnicki Aug 6 '12 at 22:24\n@MistyD You have two sheets of paper in front of you. The first lists all $\\binom{40}{1} \\cdot \\binom{12}{1}$ hands having exactly one J/Q/K. The second lists all $\\binom{12}{2}$ hands having exactly two J/Q/K. Together, the two sheets list all hands having at least one J/Q/K. Thus, to get the total count, you would want to add the number of hands in the first list to the number of hands in the second list, not multiply. \u2013\u00a0 Austin Mohr Aug 7 '12 at 1:07\n\nThe easiest way to solve this problem is to note that subtracting the number of ways of selecting any two cards neither of which is a Jack, Queen, or King, from the number of ways of selecting any two cards, leaves with the number of ways to pick two cards of which at least one of them is a Jack, Queen, or King.\n\nSo,\n\n$$\\mbox{(Ways to select any 2 cards from the deck)} - \\mbox{(Ways to select 2 cards where neither one is J/Q/K)} = \\mbox{Ways to select 2 cards where at least one is J/Q/K}$$\n\n$$\\binom{52}{2} - \\binom{40}{2} = 1326 - 780 = 546$$\n\n-\nThanks for you answer but I am more interested in knowing why I am wrong.. \u2013\u00a0 MistyD Aug 6 '12 at 22:29\n@ladaghini This answer saves effort, and is more generally useful. Having a hard time making it work for picking at least 2 of a subgroup when picking 5 form the whole. \u2013\u00a0 csga5000 Jul 14 at 19:44\nFor my situation (15 balls, 8 red, 7 black, 5 chosen) the answer was: 15C5 - (7C4 * 8C1) - 7C5 Where C is combination operation, and the left/right numbers are n/r respectively. Total combinations - (combainations with 4 black 1 red) - (combaintions 5 black) \u2013\u00a0 csga5000 Jul 14 at 19:54", "date": "2015-07-29 06:58:42", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/179660/need-to-clarify-the-at-least-concept-in-combination", "openwebmath_score": 0.778113603591919, "openwebmath_perplexity": 297.9300223979697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982287697148445, "lm_q2_score": 0.8688267694452331, "lm_q1q2_score": 0.8534378465792809}}
{"url": "http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 04 Sep 2015, 00:31\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Inequalities trick\n\nAuthor Message\nTAGS:\nCurrent Student\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2798\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 201\n\nKudos [?]: 1220 [60] , given: 235\n\nInequalities trick\u00a0[#permalink] \u00a016 Mar 2010, 09:11\n60\nKUDOS\n130\nThis post was\nBOOKMARKED\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\nAttachments\n\n1.jpg [ 6.73 KiB | Viewed 31594 times ]\n\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 5870\nLocation: Pune, India\nFollowers: 1485\n\nKudos [?]: 8007 [48] , given: 190\n\nRe: Inequalities trick\u00a0[#permalink] \u00a022 Oct 2010, 05:33\n48\nKUDOS\nExpert's post\n43\nThis post was\nBOOKMARKED\nYes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.\n\nIf (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.\ne.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.\n\nAttachment:\n\ndoc.jpg [ 7.9 KiB | Viewed 30856 times ]\n\nThis divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.\n\nWhen you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.\n\nWhen you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.\n\nSince we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7\n\nIt should be obvious that it will also work in cases where factors are divided.\ne.g. (x - a)(x - b)/(x - c)(x - d) < 0\n(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.\n\nNote: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2027 Followers: 142 Kudos [?]: 1219 [16] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 16 This post received KUDOS 7 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 150 Followers: 3 Kudos [?]: 53 [10] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 10 This post received KUDOS 8 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5870 Location: Pune, India Followers: 1485 Kudos [?]: 8007 [6] , given: 190 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 6 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nManager\nStatus: On...\nJoined: 16 Jan 2011\nPosts: 189\nFollowers: 3\n\nKudos [?]: 46 [5] , given: 62\n\nRe: Inequalities trick\u00a0[#permalink] \u00a010 Aug 2011, 16:01\n5\nKUDOS\n7\nThis post was\nBOOKMARKED\nWoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.\n\n1) CORE CONCEPT\n@gurpreetsingh -\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nArrange the NUMBERS in ascending order from left to right. a<b<c<d\nDraw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nNote: Make sure that the factors are of the form (ax - b), not (b - ax)...\n\nexample -\n(x+2)(x-1)(7 - x)<0\n\nConvert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')\nNow solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'\nSince you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.\n\n2) Variation - ODD/EVEN POWER\n@ulm/Karishma -\nif we have even powers like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\nThis will be same as (x-b)\n\nWe can ignore squares BUT SHOULD consider ODD powers\nexample -\n2.a\n(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0\n2.b\n(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0\nis the same as (x - a)(x - b)(x - c)(x - d) < 0\n\n3) Variation <= in FRACTION\n@mrinal2100 -\nif = sign is included with < then <= will be there in solution\nlike for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7\n\nBUT if it is a fraction the denominator in the solution will not have = SIGN\n\nexample -\n3.a\n(x + 2)(x - 1)/(x -4)(x - 7) < =0\nthe solution will be -2 <= x <= 1 or 4< x < 7\nwe cant make 4<=x<=7 as it will make the solution infinite\n\n4) Variation - ROOTS\n@Karishma -\nAs for roots, you have to keep in mind that given $$\\sqrt{x}$$, x cannot be negative.\n\n$$\\sqrt{x}$$ < 10\nimplies 0 < $$\\sqrt{x}$$ < 10\nSquaring, 0 < x < 100\nRoot questions are specific. You have to be careful. If you have a particular question in mind, send it.\n\nRefer - inequalities-and-roots-118619.html#p959939\nSome more useful tips for ROOTS....I am too lazy to consolidate\n\n<5> THESIS -\n@gmat1220 -\nOnce algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.\n\nI will save this future references....\n\nAnyone wants to add ABSOLUTE VALUES....That will be a value add to this post\n\n_________________\n\nLabor cost for typing this post >= Labor cost for pushing the Kudos Button\nkudos-what-are-they-and-why-we-have-them-94812.html\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 5870\nLocation: Pune, India\nFollowers: 1485\n\nKudos [?]: 8007 [3] , given: 190\n\nRe: Inequalities trick\u00a0[#permalink] \u00a009 Sep 2013, 22:35\n3\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nkarannanda wrote:\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nHi Gurpreet,\nThanks for the wonderful method.\nI am trying to understand it so that i can apply it in tests.\nCan you help me in applying this method to the below expression to find range of x.\nx^3 \u2013 4x^5 < 0?\n\nI am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.\n\nBefore you apply the method, ensure that the factors are of the form (x - a)(x - b) etc\n\n$$x^3 - 4x^5 < 0$$\n\n$$x^3 ( 1 - 4x^2) < 0$$\n\n$$x^3(1 - 2x) (1 + 2x) < 0$$\n\n$$4x^3(x - 1/2)(x + 1/2) > 0$$ (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)\n\nNow the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.\nThe solution will be x > 1/2 or -1/2 < x< 0.\n\nCheck out these posts discussing such complications:\nhttp://www.veritasprep.com/blog/2012/06 ... e-factors/\nhttp://www.veritasprep.com/blog/2012/07 ... ns-part-i/\nhttp://www.veritasprep.com/blog/2012/07 ... s-part-ii/\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews e-GMAT Representative Joined: 04 Jan 2015 Posts: 338 Followers: 66 Kudos [?]: 483 [3] , given: 83 Inequalities trick [#permalink] 06 Jan 2015, 02:35 3 This post received KUDOS Expert's post 8 This post was BOOKMARKED Just came across this useful discussion. VeritasPrepKarishma has given a very lucid explanation of how this \u201cwavy line\u201d method works. I have noticed that there is still a little scope to take this discussion further. So here are my two cents on it. I would like to highlight an important special case in the application of the Wavy Line Method When there are multiple instances of the same root: Try to solve the following inequality using the Wavy Line Method: $$(x-1)^2(x-2)(x-3)(x-4)^3 < 0$$ To know how you did, compare your wavy line with the correct one below. Did you notice how this inequality differs from all the examples above? Notice that two of the four terms had an integral power greater than 1. How to draw the wavy line for such expressions? Let me directly show you how the wavy line would look and then later on the rule behind drawing it. Attachment: File comment: Observe how the wave bounces back at x = 1. bounce.png [ 10.4 KiB | Viewed 2819 times ] Notice that the curve bounced down at the point x = 1. (At every other root, including x = 4 whose power was 3, it was simply passing through them.) Can you figure out why the wavy line looks like this for this particular inequality? (Hint: The wavy line for the inequality $$(x-1)^{38}(x-2)^{57}(x-3)^{15}(x-4)^{27} < 0$$ Is also the same as above) Come on! Give it a try. If you got it right, you\u2019ll see that there are essentially only two rules while drawing a wavy line. (Remember, we\u2019ll refer the region above the number line as positive region and the region below the number line as negative region.) How to draw the wavy line? 1. How to start: Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression. 2. How to alternate: In the given expression, if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to \u2013ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region). However, if the power of a term is even, then the wave bounces back into the same region. Now look back at the above expression and analyze your wavy line. Were you (intuitively) using the above mentioned rules while drawing your wavy line? Solution Once you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to look for the space where the above expression would be less than zero, look for the areas in the wavy line where the curve is below the number line. So the correct solution set would simply be {3 < x < 4} U {{x < 2} \u2013 {1}} In words, it is the Union of two regions region1 between x = 3 and x = 4 and region2 which is x < 2, excluding the point x = 1. Food for Thought Now, try to answer the following questions: 1. Why did we exclude the point x = 1 from the solution set of the last example? (Easy Question) 2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains? Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases. - Krishna _________________ Last edited by EgmatQuantExpert on 10 Jan 2015, 20:41, edited 1 time in total. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5870 Location: Pune, India Followers: 1485 Kudos [?]: 8007 [2] , given: 190 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 2 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\\leq x \\leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nCurrent Student\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2798\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 201\n\nKudos [?]: 1220 [1] , given: 235\n\nRe: Inequalities trick\u00a0[#permalink] \u00a019 Mar 2010, 11:59\n1\nKUDOS\nttks10 wrote:\nCan u plz explainn the backgoround of this & then the explanation.\n\nThanks\n\ni m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.\n\nsidhu4u wrote:\nI have applied this trick and it seemed to be quite useful.\n\nNice to hear this....good luck.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nMath Forum Moderator\nJoined: 20 Dec 2010\nPosts: 2027\nFollowers: 142\n\nKudos [?]: 1219 [1] , given: 376\n\nRe: Inequalities trick\u00a0[#permalink] \u00a026 May 2011, 19:55\n1\nKUDOS\nchethanjs wrote:\nmrinal2100 wrote:\nif = sign is included with < then <= will be there in solution\nlike for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7\n\nin case when factors are divided then the numerator will contain = sign\n\nlike for (x + 2)(x - 1)/(x -4)(x - 7) < =0\nthe solution will be -2 <= x <= 1 or 4< x < 7\nwe cant make 4<=x<=7 as it will make the solution infinite\n\ncorrect me if i am wrong\n\nCan you please tell me why the solution gets infinite for 4<=x<=7 ?\n\nThanks.\n\n(x -4)(x - 7) is in denominator.\nMaking x=4 or 7 would make the denominator 0 and the entire function undefined.\n\nThus, the range of x can't be either 4 or 7.\n4<=x<=7 would be wrong.\n4<x<7 is correct because now we removed \"=\" sign.\n_________________\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 5870\nLocation: Pune, India\nFollowers: 1485\n\nKudos [?]: 8007 [1] , given: 190\n\nRe: Inequalities trick\u00a0[#permalink] \u00a008 Aug 2011, 10:59\n1\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nAsher wrote:\ngurpreetsingh wrote:\nulm wrote:\nif we have smth like (x-a)^2(x-b)\nwe don't need to change a sign when jump over \"a\".\n\nyes even powers wont contribute to the inequality sign. But be wary of the root value of x=a\n\nThis way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma\n\nHowever, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.\n\nWhen you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b.\n\nSimilarly for (x-a)^2(x-b) > 0, x > b\n\nAs for roots, you have to keep in mind that given $$\\sqrt{x}$$, x cannot be negative.\n\n$$\\sqrt{x}$$ < 10\nimplies 0 < $$\\sqrt{x}$$ < 10\nSquaring, 0 < x < 100\nRoot questions are specific. You have to be careful. If you have a particular question in mind, send it.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 16 Feb 2012 Posts: 255 Concentration: Finance, Economics Followers: 5 Kudos [?]: 136 [1] , given: 121 Re: Inequalities trick [#permalink] 22 Jul 2012, 02:03 1 This post received KUDOS VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\\leq x \\leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. Current Student Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2798 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 201 Kudos [?]: 1220 [1] , given: 235 Re: Inequalities trick [#permalink] 18 Oct 2012, 04:17 1 This post received KUDOS 1 This post was BOOKMARKED GMATBaumgartner wrote: gurpreetsingh wrote: ulm wrote: in addition: if we have smth like (x-a)^2(x-b) we don't need to change a sign when jump over \"a\". yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. If the powers are even then the inequality won't be affected. eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0 just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5870 Location: Pune, India Followers: 1485 Kudos [?]: 8007 [1] , given: 190 Re: Inequalities trick [#permalink] 18 Oct 2012, 09:27 1 This post received KUDOS Expert's post GMATBaumgartner wrote: Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. In addition, you can check out this post: http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ I have discussed how to handle powers in it. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 29209\nFollowers: 4751\n\nKudos [?]: 50309 [1] , given: 7544\n\nRe: Inequalities trick\u00a0[#permalink] \u00a002 Dec 2012, 06:25\n1\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nGMATGURU1 wrote:\nGoing ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12?\n\nCheers,\nDanny\n\nSearch for hundreds of question with solutions by tags: viewforumtags.php\n\nDS questions on inequalities: search.php?search_id=tag&tag_id=184\nPS questions on inequalities: search.php?search_id=tag&tag_id=189\nHardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html\n\nHope it helps.\n_________________\nCurrent Student\nStatus: Nothing comes easy: neither do I want.\nJoined: 12 Oct 2009\nPosts: 2798\nLocation: Malaysia\nConcentration: Technology, Entrepreneurship\nSchools: ISB '15 (M)\nGMAT 1: 670 Q49 V31\nGMAT 2: 710 Q50 V35\nFollowers: 201\n\nKudos [?]: 1220 [1] , given: 235\n\nRe: Inequalities trick\u00a0[#permalink] \u00a020 Dec 2012, 20:04\n1\nKUDOS\nVeritasPrepKarishma wrote:\nThe entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side.\n\ne.g.\n(x + 2)(x + 3) < 2\nx^2 + 5x + 6 - 2 < 0\nx^2 + 5x + 4 < 0\n(x+4)(x+1) < 0\n\nNow use the concept.\n\nYes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy.\n_________________\n\nFight for your dreams :For all those who fear from Verbal- lets give it a fight\n\nMoney Saved is the Money Earned\n\nJo Bole So Nihaal , Sat Shri Akaal\n\nGMAT Club Premium Membership - big benefits and savings\n\nGmat test review :\n670-to-710-a-long-journey-without-destination-still-happy-141642.html\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 29209\nFollowers: 4751\n\nKudos [?]: 50309 [1] , given: 7544\n\nRe: Inequalities trick\u00a0[#permalink] \u00a011 Nov 2013, 06:51\n1\nKUDOS\nExpert's post\nCan you please explain the above mentioned concept in relation to the following question?\nIs a > O?\n(1) a^3 - 0 < 0\n(2) 1- a^2 > 0\n\nCan you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0?\nSorry, but finding it difficult to understand.\n\nCheck alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html\n\nHope this helps.\n_________________\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 5870\nLocation: Pune, India\nFollowers: 1485\n\nKudos [?]: 8007 [1] , given: 190\n\nRe: Inequalities trick\u00a0[#permalink] \u00a023 Apr 2014, 03:43\n1\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nPathFinder007 wrote:\n\nI have a query. I have following question\n\nx^3 - 4x^5 < 0\n\nI can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0.\n\nSo How I will define them in graph and what range I will consider for this inequality.\n\nThanks\n\nThe factors must be of the form (x - a)(x - b) .... etc\n\nx^3 - 4x^5 < 0\nx^3 * (1 - 4x^2) < 0\nx^3 * (1 - 2x) * (1 + 2x) < 0\nx^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1)\nx^3 * 2(x - 1/2) *2(x + 1/2) > 0\n\nSo transition points are 0, 1/2 and -1/2.\n\n____________ - 1/2 _____ 0 ______1/2 _________\n\nThis is what it looks like on the number line.\nThe rightmost region is positive. We want the positive regions in the inequality.\n\nSo the desired range of x is given by x > 1/2 or -1/2 < x< 0\n\nFor more on this method, check these posts:\n\nhttp://www.veritasprep.com/blog/2012/06 ... e-factors/\nhttp://www.veritasprep.com/blog/2012/07 ... ns-part-i/\nhttp://www.veritasprep.com/blog/2012/07 ... s-part-ii/\n\nThe links will give you the theory behind this method in detail.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for \\$199\n\nVeritas Prep Reviews\n\nIntern\nJoined: 24 Sep 2013\nPosts: 12\nFollowers: 0\n\nKudos [?]: 2 [1] , given: 56\n\nRe: Inequalities trick\u00a0[#permalink] \u00a011 Nov 2014, 08:52\n1\nKUDOS\nFor statement I\n\n(1) (1-2x)(1+x)<0\n\nOnce you get this into the req form\n\n2 (x-1/2) (x+1) > 0\n\nsol +++++ -1 ------ 1/2 +++++\nAs Karishma pointed out ponce in the req form, the rightmost will always be positive and the alternating will happen from there.\n\nSo sol for this x> 1/2 and x<-1\n\nIntegers greater than 1/2 and less than -1 , thus |x| may be >= 1. Unsure\n\nThus Insufficient.\n\nFor Statement II\n\n(2) (1-x)(1+2x)<0\n\nOnce you get this into the req form\n\n2(x+1/2) (x-1) > 0\n\nsol +++++ -1/2 ------ 1 +++++\n\nSo sol for this x>1 and x< -1/2\n\nIntegers greater than 1 and less than -1/2 , thus |x| may be >= 1. Unsure\n\nThus Insufficient.\n\nCombining Both the statements\n\nx<-1 and x>1\n\nThus Integers for this range will give |x| > 1\n\nThus Sufficient.\n\nHope this helps. I am not very confident of my solution though. Its my first solution gmatclub\n\nmayankpant wrote:\ngurpreetsingh wrote:\nI learnt this trick while I was in school and yesterday while solving one question I recalled.\nIts good if you guys use it 1-2 times to get used to it.\n\nSuppose you have the inequality\n\nf(x) = (x-a)(x-b)(x-c)(x-d) < 0\n\nJust arrange them in order as shown in the picture and draw curve starting from + from right.\n\nnow if f(x) < 0 consider curve having \"-\" inside and if f(x) > 0 consider curve having \"+\" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.\n\nDon't forget to arrange then in ascending order from left to right. a<b<c<d\n\nSo for f(x) < 0 consider \"-\" curves and the ans is : (a < x < b) , (c < x < d)\nand for f(x) > 0 consider \"+\" curves and the ans is : (x < a), (b < x < c) , (d < x)\n\nIf f(x) has three factors then the graph will have - + - +\nIf f(x) has four factors then the graph will have + - + - +\n\nIf you can not figure out how and why, just remember it.\nTry to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.\n\nFor the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.\n\nHi\nCan you please explain this question to me using the graph. I am missing the point when graph is being used here?\n\nIf x is an integer, is |x|>1?\n(1) (1-2x)(1+x)<0\n(2) (1-x)(1+2x)<0\n\nFor me ,the first equations roots are -1 and 1/2. Now I am struggling to get to the correct sign using the graph method here.\n\nSame for second equation: roots are 1 and -1/2 but struggling for the sign.\n\nTHanks\nRe: Inequalities trick \u00a0 [#permalink] 11 Nov 2014, 08:52\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0\u00a0\u00a03\u00a0\u00a0\u00a04\u00a0\u00a0\u00a05\u00a0\u00a0\u00a06\u00a0 \u00a0 Next \u00a0[ 111 posts ]\n\nSimilar topics Replies Last post\nSimilar\nTopics:\n131 Tips and Tricks: Inequalities 27 14 Apr 2013, 08:20\nan inequality 1 04 Oct 2010, 02:36\n2 Inequality 4 05 Jun 2010, 13:19\n47 Laura sells encyclopaedias, and her monthly income has two 22 29 May 2008, 12:54\n135 If x/|x|<x which of the following must be true about x? 74 15 Aug 2008, 03:06\nDisplay posts from previous: Sort by", "date": "2015-09-04 08:31:05", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/inequalities-trick-91482.html?kudos=1", "openwebmath_score": 0.5893213152885437, "openwebmath_perplexity": 2382.6171032034335, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811611608241, "lm_q2_score": 0.8840392771633079, "lm_q1q2_score": 0.8534348638996897}}
{"url": "https://math.stackexchange.com/questions/2991983/angles-in-a-spherical-triangle", "text": "# Angles in a spherical triangle.\n\nJust seeking advice here! I have 3 coordinates;\n\n$$A(-0.52992,0.84805,0),\\\\ B(0.84805,0,0.52992),\\\\C(0.15461,0.47553,0.86603)$$.\n\nI want to find the angles at $$A$$, $$B$$ and $$C$$. Hence, I find the normal of the planes through the great circle $$AB$$, $$AC$$, $$BC$$ respectively.\n\nI can easily find that the (unit) normals are\n\n$$n_{AB}:(0.50306,0.31434,-0.80506)\\\\n_{BC}:(-0.31209,-0.80818,0.49945)\\\\n_{AC}:(0.77558,0.48464,-0.40448)$$\n\nIs the angle at $$A$$ just $$cos^{-1}(n_{AB}.n_{AC})$$, angle at $$B$$ is $$cos^{-1}(n_{AB}.n_{BC})$$, angle at C is $$cos^{-1}(n_{AC}.n_{BC})$$?\n\nWhat I am worried is that $$n_{AB}.n_{BC}=-0.81313$$ and hence $$arccos(-0.81313)=2.5203$$. This is the same situation for angle $$C$$. Am I still correct considering that the angles $$B$$ and $$C$$ are more than $$\\pi$$? Can an angle be more than $$\\pi$$? Or is there any way I can reduce the angle?\n\nEdit: I can try to visualize the points on the sphere, it seems that angle $$C$$ is more than $$\\frac{\\pi}{2}$$, but angle $$B$$ seems to be less than $$\\frac{\\pi}{2}$$, am I still correct to say that angle $$B$$ is $$2.52$$ rad?\n\n\u2022 Spherical or 3D? 3 points does not define a sphere. \u2013\u00a0Moti Nov 10 '18 at 2:09\n\u2022 Spherical. Yes but I am finding the angles of the spherical triangle on a unit sphere. So 3 points on the (unit) sphere do determine a spherical triangle \u2013\u00a0Icycarus Nov 10 '18 at 16:37\n\u2022 You can find many spheres that use the same three points which result many solution - for each sphere you have a solution set. You need to say that the three points are of a great circle if you mean this - but than all equal $\\pi$ \u2013\u00a0Moti Nov 10 '18 at 21:20\n\u2022 @Moti: By \"unit sphere\", OP certainly means \"the unit sphere\"; that is, the sphere of unit radius centered at the origin (cf. \"the unit circle\"). If you're studying spherical triangles, there's no good reason to make your life more complicated by studying them on a non-origin-centered sphere. \u2013\u00a0Blue Nov 10 '18 at 21:35\n\u2022 I missed that one. \u2013\u00a0Moti Nov 10 '18 at 23:52\n\nTwo things:\n\n1. $$2.5$$ is not larger than $$\\pi=3.1415\\dots$$.\n2. What you computed was Euclidean angles, not spherical. That is, they are angles between the lines $$AB$$, $$AC$$ and so on. The spherical angles would be the angle between the great circles passing $$A,C$$ and $$A,B$$ respectively.\n\u2022 Hm.. I learnt that the angles between the great circles passing A,C and A,B is the angle between the normals of the plane passing through AC and AB. Which is what I have calculated, or did I misunderstood something? \u2013\u00a0Icycarus Nov 10 '18 at 21:22\n\u2022 Oh yes, thanks for pointing out. What I meant was more than pi/2. Will edit it \u2013\u00a0Icycarus Nov 10 '18 at 21:23\n\u2022 Then yes, they can be larger than $\\pi/2$. \u2013\u00a0Quang Hoang Nov 10 '18 at 21:31\n\u2022 Oh, I missed that part. Yes, I think you are correct. Angle between normals of planes through great circles is the same with angle between them. \u2013\u00a0Quang Hoang Nov 10 '18 at 21:33\n\u2022 Oh! Thank you for the explanation! \u2013\u00a0Icycarus Nov 10 '18 at 21:34\n\nNote that if $$n_{AB}$$ is a unit vector perpendicular to the great circle through $$A$$ and $$B,$$ then so is $$-n_{AB}.$$ For each side of the spherical triangle, there are two unit vectors perpendicular to that side, each of them the exact opposite of the other.\n\nWhen you reverse the direction of one of the vectors in a dot product, you reverse the sign of the product, and the arc cosine you would have gotten is replaced by its supplement. If you get just one of the unit vectors \u201cbackward\u201d when computing the angles, you will compute the exterior angle at the vertex whose interior angle you wanted.\n\nOne way to avoid this error is to make sure you take each pair of vectors in your cross products in the same \u201cdirection\u201d around the triangle. For example, you can take $$A\\times B,$$ $$B\\times C,$$ and $$C\\times A.$$ Alternatively, you can reverse all three of the cross products. If you use a different method to find the normal unit vectors, make sure that the dot product of each normal with the vector to the remaining vertex is positive in all three cases, or ensure that all three dot products are negative. There is a twist to this, however: when you set up the normals this way, you get normals that go in nearly opposite directions when the vertex angle is very small but go in nearly the same direction when the vertex angle is very large. Therefore simply taking the arc cosine of a dot product, $$\\arccos(n_1\\cdot n_2).$$ will give you the exterior angle, so you actually want $$\\arccos(-n_1\\cdot n_2)$$ in order to get the interior angle.\n\nAnother way that works (as you noted in a comment) is to use cross products to find the normals, but make sure that the vertex at which you want to find the angle is either the first vector in both cross products or the second vector in both cross products. That way the normals will point in almost the same direction when the angle is small and will be almost opposite when the angle is large, and you can take $$\\arccos(n_1\\cdot n_2)$$ without requiring a negative sign. Using this method, you need at least four cross products, but since $$B\\times A = -A\\times B$$ this method does not really require any extra computation, just reverse the vector when you need the other cross product.\n\nOne thing you must not do is to have two normals that follow one rule and the third one following the opposite rule. This will cause you to compute exterior angles at two vertices. It looks like your $$n_{AC}$$ is the \u201cwrong way around\u201d compared to your other two normal vectors.\n\nMerely having multiple obtuse angles in a spherical triangle is not an indication that you computed the angles incorrectly. The sum of the angles of a spherical triangle can be anything up to $$540$$ degrees.\n\n\u2022 Thanks for the input! I was wondering if this was what you meant; Do find angle at A; I find the angle between $n_{AB}.n_{AC}$, angle at B; angle between $n_{BA}.n_{BC}$, angle at C; angle between $n_{AC}.n_{BC}$, where $n_{AB}$ = $A$ x $B$, $n_{BA}$ = $B$ x $A$ etc \u2013\u00a0Icycarus Nov 11 '18 at 21:11\n\u2022 That's not quite what I had in mind but it's a perfectly good method. I've added it to the answer. There are several correct ways to solve this problem! I'm sorry to say I actually made a sign error when I first wrote this up, which confused me because it seemed to imply that you already had the correct angle at $B.$ But I was doing the dot products in my head in a moving car and writing the results on my phone, and I have fixed the error now (I think), so please forgive the confusion. \u2013\u00a0David K Nov 11 '18 at 23:54\n\nNote that $$n_{AB}=-n_{BA}=\\frac{A\\times B}{|A|\\,|B|}\\\\ n_{BC}=-n_{CB}=\\frac{B\\times C}{|B|\\,|C|}\\\\ n_{CA}=-n_{AC}=\\frac{C\\times A}{|C|\\,|A|}$$ Therefore, $$\\angle A=\\arccos(n_{BA}\\cdot n_{CA})=0.51928\\\\ \\angle B=\\arccos(n_{AB}\\cdot n_{CB})=0.62125\\\\ \\angle C=\\arccos(n_{AC}\\cdot n_{BC})=2.56032$$ and by Girard's Theorem, the area of the triangle is $$\\angle A+\\angle B+\\angle C-\\pi=0.55926$$ where the area of the whole sphere is $$4\\pi$$ steradians.\nThe angle you got for $$\\angle B$$ is the supplement of what is computed above, because you have the wrong sign for the dot product. That is, $$\\cos(\\angle B)=n_{AB}\\cdot n_{CB}=n_{BA}\\cdot n_{BC}=-n_{AB}\\cdot n_{BC}=-n_{BA}\\cdot n_{CB}$$", "date": "2019-10-14 15:17:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2991983/angles-in-a-spherical-triangle", "openwebmath_score": 0.7603282928466797, "openwebmath_perplexity": 205.27356870463893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811631528337, "lm_q2_score": 0.8840392725805822, "lm_q1q2_score": 0.8534348612366275}}
{"url": "http://mathhelpforum.com/calculus/137295-integration-problem-print.html", "text": "# Integration Problem\n\n\u2022 April 4th 2010, 06:46 PM\nninjaku\nIntegration Problem\nFind the integral of $(tanx/sec^2x)dx$\n\nI tried using u substitution, and I get:\n\n$u=tanx$\n$du=(sec^2)x$\n\nAt first it seems like it was going to be simple.\n\nBut in the problem there is a $sec^-2$\nAnd in my substitution there is a $sec^2$.\n\nAny help would be appreciated.\n\u2022 April 4th 2010, 06:58 PM\nskeeter\nQuote:\n\nOriginally Posted by ninjaku\nFind the integral of (tanx/(sec^2)x)dx.\n\nI tried using u substitution, and I get:\n\nu=tanx\ndu=(sec^2)x.\n\nAt first it seems like it was going to be simple.\n\nBut in the problem there is a sec^-2.\nAnd in my substitution there is a sec^2.\n\nAny help would be appreciated.\n\n$\\frac{\\tan{x}}{\\sec^2{x}} = \\tan{x} \\cdot \\cos^2{x} = \\sin{x} \\cdot \\cos{x}$\n\nnow integrate\n\u2022 April 5th 2010, 12:41 AM\nninjaku\nQuote:\n\nOriginally Posted by skeeter\n$\\frac{\\tan{x}}{\\sec^2{x}} = \\tan{x} \\cdot \\cos^2{x} = \\sin{x} \\cdot \\cos{x}$\n\nnow integrate\n\nOk after your suggestion. I did the problem two ways.\n\nThe first way I did using:\n\n$u=sinx$\n$du=cosx$\n\nand got the answer to be $\\frac{sin^2x}{2} + c.$\n\nHowever when I checked it in my calculator, it said the answer was instead, $-\\frac{cos^2x}{2} +c.$\n\nWhich I found to be the answer when I set :\n\n$u=cosx$\n$du=-sinx$\n\nI would just like to know if both of those are correct, or only one.\nAnd if only one of them is correct, how would I know which term I should pick to use for the substitution?\n\u2022 April 5th 2010, 12:50 AM\nharish21\nQuote:\n\nOriginally Posted by ninjaku\nOk after your suggestion. I did the problem two ways.\n\nThe first way I did using:\n\n$u=sinx$\n$du=cosx$\n\nand got the answer to be $\\frac{sin^2x}{2} + c.$\n\nHowever when I checked it in my calculator, it said the answer was instead, $-\\frac{cos^2x}{2} +c.$\n\nWhich I found to be the answer when I set :\n\n$u=cosx$\n$du=-sinx$\n\nI would just like to know if both of those are correct, or only one.\nAnd if only one of them is correct, how would I know which term I should pick to use for the substitution?\n\n$\\frac{sin^2x}{2} + c = \\frac{1-cos^2x}{2} + c= \\frac{1}{2}-\\frac{cos^2x}{2} +c = -\\frac{cos^2x}{2} +c'$\n\nNote: $c' = c + \\frac{1}{2}$\n\nthese two terms $\\frac{sin^2x}{2} + c$ and $-\\frac{cos^2x}{2} +c'$ differ only by a constant. Either answer works in this case.\n\u2022 April 5th 2010, 01:15 AM\nninjaku\nThat makes a lot of sense, thanks.", "date": "2015-08-02 01:36:11", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/137295-integration-problem-print.html", "openwebmath_score": 0.8895838856697083, "openwebmath_perplexity": 546.4549189691621, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808753491773, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8534298638713388}}
{"url": "https://math.stackexchange.com/questions/1568288/volumes-of-solid-of-revolution-sinx-2-fake-proof", "text": "# Volumes of solid of revolution: sin(x) + 2 fake proof\n\nI just recently learned about volumes of solids of revolution in my AP Calculus class and tried to create a problem to connect it to related rates. In this process, I found an error that neither my teacher or I seem to be able to figure out. It goes as follows:\n\nThe region contained by the function $y=\\sin(x) + 2$, the line $x=2$, and the $y$-axis as shown below is revolved around the $x$-axis.\n\nExample Image\n\nWe find the volume of the solid created by the disk method as follows:\n\n$$V=\\pi r^2 * \\text{thickness}$$\n\n$$r=\\sin(x) + 2$$\n\n$$V=\\int_{0}^{2\\pi} (\\sin(x)+2)^2dx$$\n\n$$V=9\\pi ^ 2$$\n\nWe also know that $\\int_{0}^{2\\pi} \\sin(x)dx=0$ therefore $\\int_{0}^{2\\pi}2dx=\\int_{0}^{2\\pi}(\\sin(x)+2)dx$ so if we revolve a new region with the line $y=2$ in place of $y=\\sin(x) + 2$ : The volume of the two solids should be the same.\n\nExample Image\n\nBut we know that revolving the above solid around the $x$-axis would create a cylinder with volume $A=\\pi r^2h$. In this problem, $r=2$, and $h=2$, so $V=8\\pi^2$. Why are the volumes not equivalent?\n\n\u2022 Why should the integral between 0 and 2 of the sine be zero? Should it be $2 \\pi$? \u2013\u00a0Lonidard Dec 9 '15 at 22:03\n\u2022 There are several pieces of your post that aren't rendering correctly, so it's difficult to say what your error may be. \u2013\u00a0Cameron Buie Dec 9 '15 at 22:06\n\u2022 I fixed the formulas, that should be better, sorry, this is the first time I'm posting \u2013\u00a0ozay34 Dec 9 '15 at 22:10\n\u2022 Since when is $$\\int_{0}^{2 \\pi} 2 dx = 0$$ ? \u2013\u00a0Mattos Dec 9 '15 at 22:12\n\u2022 sorry, I fixed it \u2013\u00a0ozay34 Dec 9 '15 at 22:13\n\nHint:\n\nThe volume generated by the first half of the sine function (over $y=2$) is not the same as the volume generated by the second half (below $y=2$) because the radii of rotation are not the same.\n\nIn other words the same area generate different volumes if it is rotated with respect an axis with different radii of rotation.\n\nOne possible source of error is this: you are quite right that $$\\int_0^{2\\pi}2\\,dx=\\int_0^{2\\pi}(2+\\sin x)\\,dx,$$ but this is not what you're dealing with! Rather, your integrand is $$(2+\\sin x)^2=4+4\\sin x+(\\sin x)^2,$$ and so your volume is $$\\pi\\int_0^{2\\pi}(2+\\sin x)^2\\,dx=4\\pi\\int_0^{2\\pi}\\,dx+\\pi\\int_0^{2\\pi}(\\sin x)^2\\,dx.$$\n\nCan you take it from there?\n\nAnother error is the assumption that the area of the solid of rotation $y=2$ is the same as the solid of rotation of $y=2+\\sin x.$ It's true that the average radius is the same. However, the average cross-sectional area is not the same. (In general, the average of squared is not the square of the average.)\n\n\u2022 but even if the average cross sectional area isn't the same for $y=2+sinx$, the added part for the first $\\pi$ is compensated for by the second part from $\\pi$ to $2\\pi$, or so I thought. But as Emilio said, that isn't equivalent. I don't understand how that statement is true though because $\\int_{0}^{2\\pi}sinxdx=0$ \u2013\u00a0ozay34 Dec 9 '15 at 22:28\n\u2022 Ok, thinking about it, I understand what Emilio is talking about, but I guess because I cant visualize it, it seems like it should work \u2013\u00a0ozay34 Dec 9 '15 at 22:31\n\u2022 The kicker is that if $0<t\\le1,$ then $$(2+t)^2-2^2=4+4t+t^2-4=4t+t^2$$ and $$2^2-(2-t)^2=4-(4-4t+t^2)=4-4+4t-t^2=4t-t^2.$$ Hence, $$(2-t)^2-2^2>2^2-(2-t)^2,$$ even though $$2+t-2=2-(2-t).$$ \u2013\u00a0Cameron Buie Dec 9 '15 at 22:39\n\u2022 Translated into terms of volumes, the section between $x=0$ and $x=\\pi$ has more extra volume (compared to the cylinder) than the section between $x=\\pi$ and $x=2\\pi$ loses (compared to the cylinder). \u2013\u00a0Cameron Buie Dec 9 '15 at 22:42", "date": "2019-07-18 11:49:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1568288/volumes-of-solid-of-revolution-sinx-2-fake-proof", "openwebmath_score": 0.8007317185401917, "openwebmath_perplexity": 193.95057365994583, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808753491772, "lm_q2_score": 0.8705972784807408, "lm_q1q2_score": 0.8534298622257119}}
{"url": "http://math.stackexchange.com/questions/327159/matrix-inverse-question", "text": "# Matrix Inverse Question\n\nLet $C$ be an invertible 2x2 matrix such that:\n\n$$C^{-1} \\cdot \\begin{bmatrix}1 \\\\ 2\\end{bmatrix} = \\begin{bmatrix}3 \\\\ 4\\end{bmatrix}$$\n\n$$C^{-2} \\cdot \\begin{bmatrix}9 \\\\ 5\\end{bmatrix} = \\begin{bmatrix}3 \\\\ 4\\end{bmatrix}$$\n\nFind $2\\times2$ matrices $A$ and $B$ so that $CA=B$ and solve for $C$.\n\n-\nCan you please format your question using LaTex / MathJax? I am having a tough time reading it. Regards \u2013\u00a0 Amzoti Mar 11 '13 at 4:27\nI am not familiar with putting in matrices :/ each matrix one column and 2 rowes; hope that helps \u2013\u00a0 IndividualThinker Mar 11 '13 at 4:30\nIs $C^{-1}, C^{-2}$ supposed to represent column 1 and column 2? Also, did I capture what you were trying to write? You can find guidance on Latex / MathJax in the FAQ (see link on right of top-of-page). Regards \u2013\u00a0 Amzoti Mar 11 '13 at 4:35\n\n$$\\pmatrix{1\\cr2\\cr}=C\\pmatrix{3\\cr4\\cr}$$\n\n$$\\pmatrix{9\\cr5\\cr}=CC\\pmatrix{3\\cr4\\cr}=C\\pmatrix{1\\cr2\\cr}$$\n\nNow do you see what to use for $A$ and $B$?\n\n-\nGot it ! thanks! \u2013\u00a0 IndividualThinker Mar 11 '13 at 15:17\n\nHint $\\rm\\ C\\,(C^{-1} u = v)\\:\\Rightarrow\\: u\\, =\\, \\color{#C00}{C v}$\n\nAnd $\\rm\\ C^2\\, (C^{-2}w = v)\\:\\Rightarrow\\: w = C(\\color{#C00}{Cv}) = Cu$\n\nThus $\\rm\\ C\\, (u,v) = (w,u)$\n\n-", "date": "2015-04-25 06:38:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/327159/matrix-inverse-question", "openwebmath_score": 0.9302425384521484, "openwebmath_perplexity": 1674.2656186096842, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808678600415, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8534298573513175}}
{"url": "https://math.stackexchange.com/questions/2420091/find-expectation-of-fracx-1-cdots-x-mx-1-cdots-x-n-when-x-1-l", "text": "# Find expectation of $\\frac{X_1 + \\cdots + X_m}{X_1 + \\cdots + X_n}$ when $X_1,\\ldots,X_n$ are i.i.d\n\nLet $$X_1, \\ldots, X_n$$ be i.i.d. random variables with expectation $$a$$ and variance $$\\sigma^2$$, taking only positive values. Let $$m < n$$. Find the expectiation of $$\\displaystyle\\frac{X_1 + \\cdots + X_m}{X_1 + \\cdots + X_n}$$.\n\nMy attemps to solve this probles are rather straightforward. Denote $$X = X_1 + \\cdots + X_m$$ and $$Y = X_{m+1} + \\dots + X_n$$. So, $$X$$ has the expectation $$ma$$ and the variance $$m\\sigma^2$$. And $$Y$$ has the expectation $$(n-m)a$$ and variance $$(n-m)\\sigma^2$$. And also $$X$$ and $$Y$$ are independent. So we can compute the expectation by the definition $$\\mathbb{E}\\displaystyle\\frac{X}{X+Y} = \\int\\limits_{\\Omega^2}\\frac{X(\\omega_1)}{X(\\omega_1) + Y(\\omega_2)}\\mathbb{P}(d\\omega_1)\\mathbb{P}(d\\omega_2)$$. But we do not know the distribution, so we do not have chance to calculate it.\n\nI would be glad to any help or ideas!\n\n\u2022 Hint. You have $$\\mathbb{E}\\left[ \\frac{X_1}{X_1+\\cdots+X_n} \\right] = \\cdots = \\mathbb{E}\\left[ \\frac{X_n}{X_1+\\cdots+X_n} \\right]$$ and $$\\frac{X_1+\\cdots+X_n}{X_1+\\cdots+X_n} = 1.$$ \u2013\u00a0Sangchul Lee Sep 7 '17 at 10:29\n\u2022 @SangchulLee how do you prove $\\mathbb{E}\\left[ \\frac{X_1}{X_1+\\cdots+X_n} \\right] = \\mathbb{E}\\left[ \\frac{X_n}{X_1+\\cdots+X_n} \\right]$ ? It doesn't look obvious to me. \u2013\u00a0Gabriel Romon Sep 7 '17 at 12:30\n\u2022 @LeGrandDODOM, Since $X_1, \\cdots, X_n$ are i.i.d., they are exchangable: for any permutation $\\sigma$ on $\\{1,\\cdots,n\\}$ we have the following equality in distribution: $$(X_1, \\cdots, X_n) \\stackrel{d}{=} (X_{\\sigma(1)}, \\cdots, X_{\\sigma(n)} ).$$ And since the expectation depends only on the distribution, we have $$\\mathbb{E}\\left[\\frac{X_1}{X_1+\\cdots+X_n}\\right] = \\mathbb{E}\\left[\\frac{X_{\\sigma(n)}}{X_{\\sigma(1)}+\\cdots+X_{\\sigma(n)}}\\right] = \\mathbb{E}\\left[\\frac{X_{\\sigma(n)}}{X_1+\\cdots+X_n}\\right].$$ \u2013\u00a0Sangchul Lee Sep 7 '17 at 12:33\n\u2022 @SangchulLee if say $s_{i} = \\frac{X_{i}}{\\sum_{j=1}^{n}}X_{j}$, then while calculating expectation (in continuous case) we will have $s_{i}$ under the integrsl sign too,that is while calculating $E(s_{i})$ the term inside integral that is $s_{i}$ will be varing ,so how can we show that expectation of each $s_{i}$ are the same? or if $X_{i}$ are identically distributed then does that mean $s_{i}$ are identically distributed? I too get theintuition that they are same but how do i prove it? \u2013\u00a0BAYMAX Sep 16 '17 at 1:06\n\u2022 @BAYMAX, If $X_i$ has common p.d.f. $f$, then $$\\mathbb{E}\\left[\\frac{X_i}{X_1+\\cdots+X_n}\\right]=\\int_{(0,\\infty)^n}\\frac{x_i}{x_1+\\cdots+x_n}f(x_1)\\cdots f(x_n)\\,dx_1\\cdots dx_n.$$ Now you can interchange the role of $x_1$ and $x_i$ to find that this expectation does not depend on $i$. This line of reasoning can be extended to arbitrary distribution on $(0,\\infty)$. \u2013\u00a0Sangchul Lee Sep 16 '17 at 2:03\n\nSuppose $$S_m=\\sum\\limits_{i=1}^{m} X_i$$ and $$S_n=\\sum\\limits_{i=1}^n X_i$$.\n\nNow, $$\\frac{X_1+X_2+\\cdots+X_n}{S_n}=1\\,, \\text{ a.e. }$$\n\nTherefore,\n\n$$\\mathbb E\\left(\\frac{X_1+X_2+\\cdots+X_n}{S_n}\\right)=1$$\n\nSince $$X_1,\\ldots,X_n$$ are i.i.d (see @SangchulLee's comments on main), we have for each $$i$$,\n\n$$\\mathbb E\\left(\\frac{X_i}{S_n}\\right)=\\frac{1}{n}$$\n\nSo for $$m\\le n$$, $$\\mathbb E\\left(\\frac{S_m}{S_n}\\right)=\\sum_{i=1}^m \\mathbb E\\left(\\frac{X_i}{S_n}\\right)=\\frac{m}{n}$$\n\n\u2022 How to prove that $\\frac{X_i}{S_n}$ and $\\frac{X_j}{S_n}$ are independent? \u2013\u00a0Alex Grey Sep 7 '17 at 11:10\n\u2022 @AlexGrey That isn't really used anywhere. As the $X_i$'s are identically distributed, $E\\left(\\frac{X_1}{S_n}\\right)=E\\left(\\frac{X_2}{S_n}\\right)=...=E\\left(\\frac{X_n}{S_n}\\right)$ and we simultaneously have $\\sum_{i=1}^nE\\left(\\frac{X_i}{S_n}\\right)=1$. \u2013\u00a0StubbornAtom Sep 7 '17 at 11:18\n\u2022 @AlexGrey, And in general they are not independent. \u2013\u00a0Sangchul Lee Sep 7 '17 at 11:21\n\u2022 @StubbornAtom How do you prove $E\\left(\\frac{X_1}{S_n}\\right)=E\\left(\\frac{X_2}{S_n}\\right)$ ? It doesn't look obvious to me. \u2013\u00a0Gabriel Romon Sep 7 '17 at 12:30\n\u2022 @LeGrandDODOM I feel we don't need anything more after Sangchul's comment. \u2013\u00a0StubbornAtom Sep 7 '17 at 13:10", "date": "2021-05-07 09:34:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2420091/find-expectation-of-fracx-1-cdots-x-mx-1-cdots-x-n-when-x-1-l", "openwebmath_score": 0.9202627539634705, "openwebmath_perplexity": 364.1558430332229, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808678600414, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8534298573513175}}
{"url": "https://math.stackexchange.com/questions/1964296/is-there-a-general-method-to-find-integer-x-y-solutions-to-ax-by-1", "text": "# Is there a general method to find integer x,y solutions to $A^x=B^y-1$?\n\n## Problem\n\nFor the equation:\n\n$$12^x=5^y-1$$\n\nI want to prove it has no solutions with x,y being positive integers.\n\n## Question\n\nIs there a general method for solving this type of equation? (It looks vaguely like a Pell equation, but not close enough that I can see how to solve it with standard methods)\n\nIf not, is there an elegant method to prove it for the particular case here (with $$A=12$$ and $$B=5$$)?\n\n## What I've tried\n\nThinking modulo 12, the LHS = 0, and the RHS is 0 if and only if y is even. Writing $$y=2z$$, I can then factorize the RHS into $$(5^z+1)(5^z-1)$$\n\nBoth factors are even and by thinking modulo 3, only one of these factors can be divisible by 3. So I conclude that I need something like $$5^z+1=2^?3^x$$ and $$5^z-1=2^?$$ or vice versa.\n\nSubtracting these equations I need $$2=2^?3^x-2^?$$.\n\nIf I now think in binary, these equations look like $$10_2 = (11_2)^x100..00_2 - 100...00_2$$.\n\nIt seems to make sense (but I don't see how to mathematically express this idea) that the only way this equation will work is as $$5^1+1=2.3$$ and $$5^1-1=2.2$$ but this solution results in a LHS of 24, which is not a power of 12.\n\nHowever, I feel there must be a less convoluted proof!\n\n\u2022 what is $\\gcd(5^z - 1, 5^z + 1)?$ \u2013\u00a0Will Jagy Oct 11 '16 at 20:15\n\u2022 Two, so this certainly cuts down the options for the factors a great deal (as one factor must have just a single 2). Is there more I should conclude? \u2013\u00a0Peter de Rivaz Oct 11 '16 at 20:18\n\u2022 Mih\u0103ilescu's theorem kills all of these problems. \u2013\u00a0Fan Zheng Oct 11 '16 at 20:20\n\u2022 that should be enough. Meanwhile, over the past few days I have been fiddling with questions such as $7^x - 3^y = 100,$ there seems to be a procedure but it is not easy math.stackexchange.com/questions/1946621/\u2026 \u2013\u00a0Will Jagy Oct 11 '16 at 20:21\n\nAs an illustration, let us solve $3^A - 2^B = 1,$ which will finish the other answer. We suspect the largest solution is $9-8=1.$ Take $3^A = 2^B + 1$ and subtract $9$ from both sides, for $3^A - 9 = 2^B - 8.$ Divide out both factors and introduce new variables, for $$9 (3^x - 1) = 8(2^y - 1).$$ We will show that this is impossible with $x,y \\geq 1.$\n\nLittle explanation: given $m,n \\geq 2$ with $\\gcd(m,n) = 1,$ we know that $m^{\\varphi(n)} \\equiv 1 \\pmod n.$ However, there may be a smaller $k$ with $m^{k} \\equiv 1 \\pmod n.$ If so, we take the smallest such $k$ and call it the order, sometimes multiplicative order, of $m \\pmod n.$\n\nWe proceed under the assumption that $x \\geq 1$ and $y \\geq 1.$\n\nNow, $2^y \\equiv 1 \\pmod 9.$ This means that $$6 | y.$$\n\njagy@phobeusjunior:~$./order 2 9 9 6 = 2 * 3  Furthermore,$2^6 - 1 | 2^y - 1.$$$2^6 - 1 = 63 = 3^3 \\cdot 7.$$ Therefore$7 | (3^x - 1),$$$3^x \\equiv 1 \\pmod 7.$$ Therefore $$6 | x,$$ and$3^6 - 1$divides$3^x - 1.$jagy@phobeusjunior:~$ ./order 3 7\n7     6 = 2 * 3\n\n\n$$3^6 - 1 = 8 \\cdot 7 \\cdot 13.$$\n\nTherefore $$2^y \\equiv 1 \\pmod {13},$$ $$12 | y.$$\n\njagy@phobeusjunior:~$./order 2 13 13 12 = 2^2 * 3  In particular $$4 | y,$$ and$2^y - 1$is divisible by$15,$especially divisible by$5.$$$3^x \\equiv 1 \\pmod 5,$$ so $$4 | x.$$ jagy@phobeusjunior:~$ ./order 3 5\n5     4 = 2^2\n\n\nHowever, $$3^4 - 1 = 80 = 5 \\cdot 16.$$ This means that $8 (2^y - 1)$ is divisible by $16,$ a contradiction of $$9 (3^x - 1) = 8(2^y - 1)$$ with $x,y \\geq 1.$\n\nThe powers of $5$ mod $11$ are $5,3,4,9$, and $1$. Thus $5^y-1\\in\\{4,2,3,8,0\\}$ mod $11$. But $12^x\\equiv1^x=1$ mod $11$. So $12^x=5^y-1$ can have no solutions in positive integers.\n\nRemark (added later, on reading the question's request for a general method): A key feature of the equation $12^x=5^y-1$ that makes a simple congruence-based approach possible is the fact that we're proving the equation has no solutions. For equations like $2^x=3^y-1$ (see Will Jagy's excellent answer), where you're trying to prove it has just one solution, a simple congruence-based approach doesn't have a chance.\n\nIf $y$ is odd then $5^y-1=1 \\pmod 3$ while $12^x=0 \\pmod 3$. So there are no solutions whit $y$ odd.\n\nLet $y=2z$. Then the equation is $12^x=25^z-1$. Thinking$\\pmod {13}$, the left hand side is $1$ or $-1$ while the right hand side is $0$ or $2$. Then there are no solutions with $y$ even, and therefore no solutions at all.\n\nThat one is easy, clearly $y=1$ has no solutions.\n\nworking $\\bmod 3$ we get $y$ is even, so $y=2k$ with $k\\geq 1$.\n\nFrom here $12^x=5^{2y}-1=(5^y+1)(5^y-1)$.\n\nClearly one factor must be $2\\times 3^x$ and the other must be $4^{x-1}\\times 2$.\n\nTherefore we have $2\\times 3^x=4^{x-1}\\times 2+2$ or $2\\times 3^x=4^{x-1}\\times 2 -2$\n\nThe first is equivalent to $3^x=4^{x-1}+1$ and the second is equivalent to $3^x=4^{x-1}-1$.\n\nIt is easy to see neither have solutions by looking at the following table:\n\n$$\\begin{pmatrix} 1 & 1 \\\\ 3 & 4 \\\\ 9 & 16\\\\ 27 & 64\\\\ 81 & 256\\\\ 243 & 1024\\\\ \\end{pmatrix}$$\n\n\u2022 I do not understand how the table means there are no solutions, can you explain more? \u2013\u00a0Peter de Rivaz Oct 11 '16 at 20:46\n\u2022 @PeterdeRivaz I put a proof about $3^A - 2^B = 1$ as an answer. \u2013\u00a0Will Jagy Oct 11 '16 at 22:48", "date": "2020-10-28 06:33:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1964296/is-there-a-general-method-to-find-integer-x-y-solutions-to-ax-by-1", "openwebmath_score": 0.9665514230728149, "openwebmath_perplexity": 415.3511837213037, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808695883038, "lm_q2_score": 0.8705972768020107, "lm_q1q2_score": 0.8534298555646843}}
{"url": "https://math.stackexchange.com/questions/1128552/poissons-equation-with-robin-boundary-conditions", "text": "# Poisson's equation with Robin boundary conditions\n\nExplain how to define $u \\in H^1(U)$ to be a weak solution of Poisson's equation with Robin boundary conditions: \\begin{align} \\begin{cases} \\, \\, \\, \\, -\\Delta u = f & \\text{in }U \\\\ u+\\frac{\\partial u}{\\partial v}=0 & \\text{on } \\partial U. \\end{cases} \\end{align} Discuss the existence and uniqueness of a weak solution for a given $f \\in L^2(U)$.\n\nThis is Exercise 5 in Chapter 6 of PDE Evans, 2nd edition.\n\nI would like to define the bilinear form $B[u,v]$, for all $u,v \\in H_0^1(U)$. But they did not really give that to the reader, unlike in Exercises 3 and 4 in the textbook.\n\nShould I still define $B[u,v]$? If so, then I can try to satisfy the hypotheses of the Lax-Milgram Theorem, which would allow me to assert the existence of a weak solution to this problem.\n\nIt is not appropriate to work in the space $$H^1_0(U)$$ since nonzero boundary conditions are being considered. You will have to assume some regularity of the boundary of $$U$$.\n\nOne version of Green's theorem (see e.g. the appendix in Evans) is that $$- \\int_U (\\Delta u) v \\, dx = \\int_U Du \\cdot Dv \\, dx -\\int_{\\partial U} \\frac{\\partial u}{\\partial \\nu} v \\, dS.$$\n\nA weak solution to the problem at hand can be proposed by setting $$-\\Delta u = f$$ in $$U$$ and $$\\dfrac{\\partial u}{\\partial \\nu} = -u$$ on $$\\partial U$$ so that $$\\int_U fv = \\int_U Du \\cdot Dv + \\int_{\\partial U} uv \\, dS \\quad \\forall v \\in H^1(U)$$ or a bit more precisely $$\\int_U fv = \\int_U Du \\cdot Dv + \\int_{\\partial U} \\newcommand{\\tr}{\\mathrm{Tr}\\ \\! }( \\tr u )(\\tr v) \\, dS \\quad \\forall v \\in H^1(U)$$ where $$\\tr : H^1(U) \\to L^2(\\partial U)$$ is the trace operator.\n\nAn appropriate bilinear form is thus given by $$B[u,v] = \\int_U Du \\cdot Dv + \\int_{\\partial U} \\newcommand{\\tr}{\\mathrm{Tr}\\ \\! }( \\tr u )(\\tr v) \\, dS, \\quad u,v \\in H^1(U).$$ $$B$$ is clearly bounded. As far as coercivity goes, it may be helpful to use the Rellich-Kondrachov theorem. I can follow up with a hint if you like.\n\nIt remains to show that there is a constant $$\\alpha > 0$$ with the property that $$\\|u\\|_{H^1}^2 \\le \\alpha B[u,u]$$ for all $$u \\in H^1(U)$$. This can be proven by contradition. Otherwise, for every $$n \\ge \\mathbb N$$ there would exist $$u_n \\in H^1(U)$$ with the property that $$\\|u_n\\|^2_{H^1} > n B[u_n,u_n]$$. For each $$n$$ define $$v_n = \\dfrac{u_n}{\\|u_n\\|_{H^1}}$$. Then $$v_n \\in H^1(U)$$, $$\\|v_n\\|_{H^1} = 1$$, and $$B[v_n,v_n] < \\dfrac 1n$$ and all $$n$$.\n\nHere we can invoke Rellich-Kondrachov. Since the family $$\\{v_n\\}$$ is bounded in the $$H^1$$ norm, there is a subsequence $$\\{v_{n_k}\\}$$ that converges to a limit $$v \\in L^2(U)$$. However, since $$\\|Dv_{n_k}\\|_{L^2}^2 < \\dfrac{1}{n_k}$$ it is also true that $$Dv_{n_k} \\to 0$$ in $$L^2$$. Thus for any $$\\phi \\in C_0^\\infty(U)$$ you have $$\\int_U v D \\phi \\, dx = \\lim_{k \\to \\infty} \\int_U v_{n_k} D \\phi \\, dx = - \\lim_{k \\to \\infty} \\int_U D v_{n_k} \\phi \\, dx = 0.$$ This means $$v \\in H^1(U)$$ and $$D v = 0$$, from which you can conclude $$v_{n_k} \\to v$$ in $$H^1(U)$$. Since $$\\|v_{n_k}\\|_{H^1} = 1$$ for all $$k$$ it follows that $$\\|v\\|_{H^1} = 1$$ as well.\n\nNext, since $$\\|\\tr v_{n_k}\\|_{L^2(\\partial U)}^2 < \\dfrac{1}{n_k}$$ and the trace operator is bounded there is a constant $$C$$ for which $$\\|\\tr v\\|_{L^2(\\partial \\Omega)} \\le \\|\\tr v - \\tr v_{n_k}\\|_{L^2(\\partial \\Omega)} + \\|\\tr v_{n_k}\\|_{L^2(\\partial \\Omega)} < \\frac{1}{n_k} + C \\|v - v_{n_k}\\|_{H^1(U)}.$$ Let $$k \\to \\infty$$ to find that $$\\tr v = 0$$ in $$L^2(\\partial U)$$.\n\nCan you prove that if $$v \\in H^1(U)$$, $$Dv = 0$$, and $$\\tr v = 0$$, then $$v = 0$$? Once you have established that fact you arrive at a contradiction, since $$v$$ also satisfies $$\\|v\\|_{H^1} = 1$$. It follows that $$B$$ is in fact coercive.\n\n\u2022 I tried to edit your response to fix (what I thought) were typos, and tried to clarify minor things, to avoid ambiguity and make it easier for me to follow. Lastly, I changed the $\\Omega$ to $U$, just to stay consistent with Evans' notation. Are all modifications changes correct? Because, for example, we never had $\\Delta u = f$ on $\\partial U$, when we were given $-\\Delta u = f$ in $U$ by the problem. \u2013\u00a0Cookie Feb 2 '15 at 16:37\n\u2022 Looks good. Thanks for pointing out the errors. \u2013\u00a0Umberto P. Feb 2 '15 at 17:03\n\u2022 There's another one (I didn't modify), I think. Shouldn't the trace operator $T$ be mapped from $H^1(U) \\to L^2(\\color{red}{\\partial}U)$? \u2013\u00a0Cookie Feb 2 '15 at 19:38\n\u2022 Yeah, I just fixed it. \u2013\u00a0Umberto P. Feb 2 '15 at 19:46\n\u2022 Fundamental question, I know, but does the $\\frac{\\partial u}{\\partial \\nu}$ in the \"$u+\\frac{\\partial u}{\\partial \\nu}=0$ on $\\partial U$\" suggest that $\\partial U$ is $C^1$? I'm asking since the problem doesn't explicitly state this, and the Trace Theorem requires this in its hypothesis. I want to be absolutely sure that we can conclude $Tu=u\\vert_{\\partial U}$ from the Trace theorem, as you are doing so in the line after writing \"a bit more precisely\" in your answer. \u2013\u00a0Cookie Feb 3 '15 at 20:12", "date": "2019-06-16 07:16:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1128552/poissons-equation-with-robin-boundary-conditions", "openwebmath_score": 0.9845558404922485, "openwebmath_perplexity": 101.15405444935615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808741970027, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8534298497032438}}
{"url": "https://math.stackexchange.com/questions/720206/prove-binomnm-binommk-binomnk-binomn-km-k", "text": "# Prove $\\binom{n}{m}\\binom{m}{k} = \\binom{n}{k}\\binom{n-k}{m-k}$\n\nI need to prove the following:\n\nIf $n,m,k\\in \\mathbb{N}$ and $k\\leq m \\leq n$, then\n\n$$\\binom{n}{m}\\binom{m}{k} = \\binom{n}{k}\\binom{n-k}{m-k}$$.\n\nI did the following steps:\n\n\\begin{align} \\require{cancel} \\binom{n}{m}\\binom{m}{k} &= \\binom{n}{k}\\binom{n-k}{m-k} \\\\ \\frac{n!}{m!(n-m)!}\\cdot \\frac{m!}{k!(m-k)!} &= \\frac{n!}{k!(n-k)!}\\cdot \\frac{(n-k)!}{(m-k)!(n-k-m+k)!}\\\\ \\frac{n!}{\\cancel{m!}(n-m)!}\\cdot \\frac{\\cancel{m!}}{k!(m-k)!} &= \\frac{n!}{k!\\cancel{(n-k)}!}\\cdot \\frac{\\cancel{(n-k)}!}{(m-k)!(n-k-m+k)!} \\\\ \\frac{n!}{k!(n-m)!(m-k)!} &= \\frac{n!}{k!(n-m)!(m-k)!} \\end{align}\n\nThe question is: is my proof correct? Are all my steps valid?\n\nThanks\n\n\u2022 Maybe it is just me, but I am little confused of how you went in one step from $(m-k)!(n-k-m+k)!$ to $(n-m)!(m-k)!$ \u2013\u00a0imranfat Mar 20 '14 at 19:15\n\u2022 @imranfat, $(n\u2212k\u2212m+k)!=(n-m)!$. \u2013\u00a0DKal Mar 20 '14 at 19:16\n\u2022 Ok, I see it now, there has been a complete brainfart in my head going on... \u2013\u00a0imranfat Mar 20 '14 at 19:24\n\u2022 Thanks for the edit Umberto! \u2013\u00a0Jeel Shah Mar 20 '14 at 19:32\n\nIt is correct!\n\nAn other way to prove this is the following:\n\n$$\\binom{n}{m}\\binom{m}{k} = \\frac{n!}{m!(n-m)!} \\frac{m!}{k!(m-k)!}= \\frac{n!}{(n-m)!} \\frac{1}{k!(m-k)!}=\\frac{n!}{k!} \\frac{1}{(m-k)!(n-m)!}=\\frac{n!}{k!(n-k)!} \\frac{(n-k)!}{(m-k)!(n-m)!}=\\binom{n}{k}\\frac{(n-k)!}{(m-k)!((n-k)-(m-k))!}=\\binom{n}{k} \\binom{n-k}{m-k}$$\n\n\u2022 The back of the book suggested that I use the following fact $\\binom{n+1}{k+1} = \\frac{n+1}{k+1}\\binom{n}{k}$. How would I use that? Also, on the second last portion i.e. $(n-m)!$ going to $((n-k)-(m-k))!$ is that because we have $n = n-k$ and $m=m-k$? \u2013\u00a0Jeel Shah Mar 20 '14 at 19:31\n\u2022 As regards the second question: $$(n-m)!=(n-m+0)!=(n-m+k-k)!=(n-k-m+k)!=((n-k)-(m-k))!$$ \u2013\u00a0Mary Star Mar 20 '14 at 19:34\n\u2022 To use the relation $$\\binom{n+1}{k+1}=\\frac{n+1}{k+1} \\binom{n}{k}$$ you could do the following: $$\\binom{n}{m}=\\frac{n}{m} \\binom{n-1}{m-1}=...=\\frac{n \\cdot ... \\cdot (n-k+1)}{m \\cdot ... \\cdot (m-k+1)} \\binom{n-k}{m-k}=\\frac{\\frac{n!}{(n-k)!}}{\\frac{m!}{(m-k)!}} \\binom{n-k}{m-k}=\\frac{\\frac{n!}{k!(n-k)!}}{\\frac{m!}{k!(m-k)!}} \\binom{n-k}{m-k}= \\frac{ \\binom{n}{k} }{ \\binom{m}{k} } \\binom{n-k}{m-k}$$ So $$\\binom{n}{m} \\binom{m}{k}= \\frac{ \\binom{n}{k} }{ \\binom{m}{k} } \\binom{n-k}{m-k} \\binom{m}{k}= \\binom{n}{k} \\binom{n-k}{m-k}$$ \u2013\u00a0Mary Star Mar 20 '14 at 22:41\n\nYou appear to do this in a non-standard way, but it looks alright (reading the equations from top to bottom instead of left to right).\n\nEver thought of a combinatoric proof?\n\nLooks correct to me. Alternatively, you can give a combinatorial proof: both sides of the equation count the number of options to choose subsets $K\\subseteq M\\subseteq N$, where $N$ is a set of $n$ elements, $M\\subseteq N$ is a subset of $m$ elements, and $K\\subseteq M$ is a subset of $k$ elements. On the LHS you choose $M$ and then choose $K$ in $M$. On the RHS you choose $K$ in $N$ first, and then determine $M$ by choosing $m-k$ more elements from $N\\setminus K$.", "date": "2019-06-26 10:45:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/720206/prove-binomnm-binommk-binomnk-binomn-km-k", "openwebmath_score": 0.9973461031913757, "openwebmath_perplexity": 470.084523490451, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808753491773, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8534298490606969}}
{"url": "http://tenmg.com/15jcey/8c2ca1-multiply-a-diagonal-matrix", "text": "Effect of multiplying a matrix by a diagonal matrix. Each task will calculate a subblock of the resulting matrix C. If A and B are diagonal, then C = AB is diagonal. In this post, we explain how to diagonalize a matrix if it is diagonalizable. Matrix Multiplication. Next, we are going to calculate the sum of diagonal elements in this matrix using For Loop. But to multiply a matrix by another matrix we need to do the \"dot product\" of rows and columns ... what does that mean? Where do our outlooks, attitudes and values come from? Accelerating the pace of engineering and science. Explicitly: Way of enlightenment, wisdom, and understanding, America, a corrupt, depraved, shameless country, The test of a person's Christianity is what he is, Ninety five percent of the problems that most people Q. The mmult program will calculate C = AB, where C, A, and B are all square matrices. Diagonal matrices. What I actually need is a method to multiply each diagonal in A by some constant (i.e. Multiplying two matrices is only possible when the matrices have the right dimensions. Effect of multiplying a matrix by a diagonal matrix. if A is of size n*m then we have vector c of length (n+m-1)). Thanks Teja for that, I have updated my question to reflect a further requirement which I don't think your solution completes? De diagonale elementen kunnen al of niet gelijk zijn aan nul. The punishment for it is real. Self-imposed discipline and regimentation, Achieving happiness in life --- a matter of the right strategies, Self-control, self-restraint, self-discipline basic to so much in life. ; Step 3: Add the products. I will calculate these offline and store them in an 3-d array \"J\". Explicitly: Q. With the help of Numpy matrix.diagonal() method, we are able to find a diagonal element from a given matrix and gives output as one dimensional matrix.. Syntax : matrix.diagonal() Return : Return diagonal element of a matrix Example #1 : In this example we can see that with the help of matrix.diagonal() method we are able to find the elements in a diagonal of a matrix. Based on your location, we recommend that you select: . De \u00d7-matrix = (,) is een diagonaalmatrix als voor alle , \u2208 {,, \u2026,}: , = \u2260 Diagonaalmatrices worden volledig bepaald door de waarden van de elementen op de hoofddiagonaal. tl;dr Use loops. in good habits. Not all matrices are diagonalizable. Let us define the multiplication between a matrix A and a vector x in which the number of columns in A equals the number of rows in x . In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. Thanks Teja Method 3 worked out to be faster. MathWorks is the leading developer of mathematical computing software for engineers and scientists. Yes, but first it is ONLY true for a matrix which is unitary that is a matrix A for which AA'=I. In order to multiply matrices, Step 1: Make sure that the the number of columns in the 1 st one equals the number of rows in the 2 nd one. Flip square matrices over the main diagonal. A. Unable to complete the action because of changes made to the page. In addition, m >> n, and M is constant throughout the course of the algorithm, with only the elements of D changing. In a previous post I discussed the general problem of multiplying block matrices (i.e., matrices partitioned into multiple submatrices). in .The mmult program can be found at the end of this section. Here it is for the 1st row and 2nd column: (1, 2, 3) \u2022 (8, 10, 12) = 1\u00d78 + 2\u00d710 + 3\u00d712 = 64 We can do the same thing for the 2nd row and 1st column: (4, 5, 6) \u2022 (7, 9, 11) = 4\u00d77 + 5\u00d79 + 6\u00d711 = 139 And for the 2nd row and 2nd column: (4, 5, 6) \u2022 (8, 10, 12) = 4\u00d78 + 5\u00d710 + 6\u00d712 = 15\u2026 What is the effect of pre-multiplying a matrix. People are like radio tuners --- they pick out and For simplicity we assume that m x m tasks will be used to calculate the solution. for loop version Elapsed time is 0.000154 seconds. Poor Richard's Almanac. example. sparse matrix multiply Elapsed time is 0.000115 seconds. In a square matrix, transposition \"flips\" the matrix over the main diagonal. Multiplication of diagonal matrices is commutative: if A and B are diagonal, then C = AB = BA.. iii. I wish to find the most efficient way to implement the following equation, is a m*n dense rectangular matrix (with no specific structure), and, is a m*m diagonal matrix with all positive elements. Each other elements will move across the diagonal and end up at the same distance from the diagonal, on the opposite side. Choose a web site to get translated content where available and see local events and offers. Example1 Live Demo Other MathWorks country sites are not optimized for visits from your location. have come from personal foolishness, Liberalism, socialism and the modern welfare state, The desire to harm, a motivation for conduct, On Self-sufficient Country Living, Homesteading. The effect is that of multiplying the i-th column of matrix A by the factor ki i.e. i.e. '*B; toc; Again, depending on what m and n actually are, the fastest method may be different (for this choice of m and n, it seems method 3 is somewhat faster). Topically Arranged Proverbs, Precepts, Diagonalize the matrix A=[4\u22123\u221233\u22122\u22123\u2212112]by finding a nonsingular matrix S and a diagonal matrix D such that S\u22121AS=D. There are two types of multiplication for matrices: scalar multiplication and matrix multiplication. Diagonal matrix. In our next example we program a matrix-multiply algorithm described by Fox et al. Reload the page to see its updated state. Multiplying a Vector by a Matrix To multiply a row vector by a column vector, the row vector must have as many columns as the column vector has rows. Common Sayings. Hell is real. Inverse matrix Let Mn(R) denote the set of all n\u00d7n matrices with real entries. My numbers indicate that ifort is smart enough to recognize the loop, forall, and do concurrent identically and achieves what I'd expect to be about 'peak' in each of those cases. A new example problem was added.) I reshape J to an [(n^2) x m] matrix since we want to take linear combinations of its columns by postmultiplying it with the elements in D. % Preallocate J for n*n*m elements of storage. https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#answer_97203, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#comment_170160, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#answer_97194, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#comment_169818, https://www.mathworks.com/matlabcentral/answers/87629-efficiently-multiplying-diagonal-and-general-matrices#comment_170168. Find the treasures in MATLAB Central and discover how the community can help you! I then discussed block diagonal matrices (i.e., block matrices in which the off-diagonal submatrices are zero) and in a multipart series of posts showed that we can uniquely and maximally partition any square matrix into block\u2026 To multiply a matrix by a scalar, multiply each element by the scalar. Quotations. Wisdom, Reason and Virtue are closely related, Knowledge is one thing, wisdom is another, The most important thing in life is understanding, We are all examples --- for good or for bad, The Prime Mover that decides \"What We Are\". columns of the original matrix are simply multiplied by successive diagonal elements of the Let us see with an example: To work out the answer for the 1st row and 1st column: Want to see another example? A. This program allows the user to enter the number of rows and columns of a Matrix. D = diag(v,k) places the elements of vector v on the kth diagonal. (Update 10/15/2017. listen to one wavelength and ignore the rest, Cause of Character Traits --- According to Aristotle, We are what we eat --- living under the discipline of a diet, Personal attributes of the true Christian, Love of God and love of virtue are closely united, Intellectual disparities among people and the power Consider the matrix multiplication below For the product to be a diagonal matrix, a f + b h = 0 \u21d2 a f = -b h and c e + d g = 0 \u21d2 c e = -d g Consider the following sets of values The the matrix product becomes: Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices. But you can do something similar. Scalar Matrix Multiplication. As an example, we solve the following problem. In addition, m >> n, and, is constant throughout the course of the algorithm, with only the elements of, I know there are tricks for a related problem (D*M*D) to reduce the number of operations considerably, but is there one for this problem? Suppose there exists an n\u00d7n matrix B such that AB = BA = In. iii. If A is an m x n matrix and B is as n x p matrix The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (\u2212) additions for computing the product of two square n\u00d7n matrices. This can only be done if the number of columns in the first matrix is equal to the number of rows in the second. Ideally is there a way to factorize / rearrange this so I can compute, offline (or something similar), and update. rows of the original matrix are simply multiplied by successive diagonal elements of the diagonal Now, I can use J to quickly calculate the answer for any D. We'll try all 3 methods. In de lineaire algebra is een diagonaalmatrix een vierkante matrix, waarvan alle elementen buiten de hoofddiagonaal (\u2198) gelijk aan nul zijn. Example in $\\def\\R{\\Bbb R}\\R^2$. Sometimes we need to find the sum of the Upper right, Upper left, Lower right, or lower left diagonal elements. tic; D = sparse(1:m,1:m,d); A = M'*D*M; toc; tic; B = bsxfun(@times,M,sqrt(d)); B = B. Here's an example of it in action - you can see that it far outperforms the standard dense multiply, sparse matrix multiply, and for loop versions: >> onesmatrixquestion dense matrix multiply Elapsed time is 0.000873 seconds. The main diagonal (or principal diagonal or diagonal) of a square matrix goes from the upper left to the lower right. Matrices where (number of rows) = (number of columns) For the matrices with whose number of rows and columns are unequal, we call them rectangular matrices. Let A be an n\u00d7n matrix. An m times n matrix has to be multiplied with an n times p matrix. diagonal matrix. In other words, the elements in a diagonal line from element a 11 to the bottom right corner will remain the same. Sin is serious business. The effect is that of multiplying the i-th row of matrix A by the factor kii.e. Left-multiplication be a diagonal matrix does not have any simple effect on eigenvalues, and given that eigenvalues are perturbed (or destroyed) what could one possibly want to say about \"corresponding\" eigenvectors? Notice how this expression is linear in the entries of D. You can express D as a sum of elementary basis functions. De\ufb01nition. (The pre-requisite to be able to multiply) Step 2: Multiply the elements of each row of the first matrix by the elements of each column in the second matrix. Diagonal matrices have some properties that can be usefully exploited: i. C Program to find Sum of Diagonal Elements of a Matrix. For the following matrix A, find 2A and \u20131A. Further, C can be computed more efficiently than naively doing a full matrix multiplication: c ii = a ii b ii, and all other entries are 0. ii. Definition 3.9 An identity matrix is square and has with all entries zero except for ones in the main diagonal. Never multiply with a diagonal matrix. Numpy provides us the facility to compute the sum of different diagonals elements using numpy.trace() and numpy.diagonal() method.. One drawback, however, is that you need to be able to store a dense [n x n x m] array, and this may not be feasible if the n and m are too large. The time required to compute this matrix expression can be dramatically shortened by implementing the following improvements: W is a diagonal matrix. Diagonal Matrices, Upper and Lower Triangular Matrices Linear Algebra MATH 2010 Diagonal Matrices: { De nition: A diagonal matrix is a square matrix with zero entries except possibly on the main diagonal (extends from the upper left corner to the lower right corner). The reason for this is because when you multiply two matrices you have to take the inner product of every row of the first matrix with every column of the second. Then the matrix A is called invertible and B is called the inverse of A (denoted A\u22121). the successiverows of the original matrix are simply multiplied by \u2026 P.S. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. by Marco Taboga, PhD. gfortran, on the other hand, does a bad job (10x or more slower) with forall and do concurrent, especially as N gets large. Matrix diagonalization is the process of performing a similarity transformation on a matrix in order to recover a similar matrix that is diagonal (i.e., all its non-diagonal entries are zero). tensorized version Elapsed time is 0.000018 seconds. D = diag(v) returns a square diagonal matrix with the elements of vector v on the main diagonal. %Generate a new d (only the diagonal entries). What about division? The effect is that of multiplying the i-th row of matrix A by the factor ki i.e. In linear algebra, a diagonal matrix is a matrix in which the entries outside the main diagonal are all zero; the term usually refers to square matrices.An example of a 2-by-2 diagonal matrix is [], while an example of a 3-by-3 diagonal matrix is [].An identity matrix of any size, or any multiple of it (a scalar matrix), is a diagonal matrix. Q. Tools of Satan. Therefore computation sqrt (W) * B multiplies the i th row of B by the i th element of the diagonal of W 1/2. where M is a m*n dense rectangular matrix (with no specific structure), and D is a m*m diagonal matrix with all positive elements. by a diagonal matrix. OK, so how do we multiply two matrices? A. But each M'*ek*M is simply M(k,:)'*M(:,k). Scalar multiplication is easy. You may receive emails, depending on your. k=0 represents the main diagonal, k>0 is above the main diagonal, and k<0 is below the main diagonal. matrix. What is the effect of post-multiplying a matrix. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 Add to solve later Sponsored Links Matrix Multiply . As such, it enjoys the properties enjoyed by triangular matrices, as well as other special properties. What is the effect of pre-multiplying a matrix. This implies that if you calculate all the M'*ek*M beforehand, then you just need to take a linear combination of them. In addition, I can exploit symmetry within M'*M and thus skip some of the rows in J*d, further reducing operations. example. where dk, a scalar, is the kth diagonal entry of D, and ek is a [m x m] matrix with all zeros except for a 1 in the kth position along the diagonal. Matrix Multiplication. Example. To understand the step-by-step multiplication, we can multiply each value in the vector with the row values in matrix and find out the sum of that multiplication. Tactics and Tricks used by the Devil. the successive You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. A diagonal matrix is at the same time: upper triangular; lower triangular. Inverse matrix., Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Once a matrix is diagonalized it becomes very easy to raise it to integer powers. We can add, subtract, and multiply elements of Mn(R). Scalar multiplication: to multiply a matrix A by a scalar r, one Multiplication of diagonal matrices is commutative: if A and B are diagonal, then C = AB = BA. I am almost certain you can't just find M'*M and somehow do something efficiently with only that. [PDF] Matrix multiplication. 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Flips '' the matrix corner will remain the same dimensions can be added by adding corresponding. Teja method 3 worked out to be multiplied with an n times matrix!", "date": "2021-09-22 17:46:19", "meta": {"domain": "tenmg.com", "url": "http://tenmg.com/15jcey/8c2ca1-multiply-a-diagonal-matrix", "openwebmath_score": 0.7437896132469177, "openwebmath_perplexity": 693.4081402986793, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.980280873044828, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8534298487001636}}
{"url": "http://cghb.bollola.it/triple-integral-pdf.html", "text": "# Triple Integral Pdf\n\n(5 8 5) 4 5 60 3 3 3 x x x dx x x 3 2 9 5 9 2 2 1 1 2 1026 22 1001 2. View Math267-Triple-integrals_wNotes. But the real difficulty with triple integrals is-- and I think you'll see that your calculus teacher will often do this-- when you're doing triple integrals, unless you have a very easy figure like this, the evaluation-- if you actually wanted to analytically evaluate a triple integral that has more complicated boundaries or more complicated. Katz familiar to calculus students. Evaluating triple integrals A triple integral is an integral of the form Z b a Z q(x) p(x) Z s(x,y) r(x,y) f(x,y,z) dzdydx The evaluation can be split into an \"inner integral\" (the integral with respect to z between limits. 1 (Iterated Integrals). Then, parallel to the axis of walk,. 4 (approximate answer, depends on what you estimated the values at the midpoints to be). P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. INTEGRAL REVIEW July 2012 Vol. If it's a 2D surface, use a double integral. 1) is the signed volume bounded by the graph z f x y over the region; that is, the volume of the part of the solid below the xy-planeis taken to be negative. Triple integral is defined and explained through solved examples. Solutions to Midterm 1 Problem 1. Figure 1 In order for the double integral to exist, it is sufficient that, for example, the region D be a closed (Jordan) measurable region and that the function f(x, y ) be continuous throughout D. The inner limit is the easiest. Integration by Parts 21 1. \u2022 M yz = RRR S x\u03b4(x,y,z)dV is the moment about the yz-plane. Cylindrical Coordinates, page 1040 In the cylindricalcoordinatesystem(\u67f1\u5750\u6a19\u7cfb), a point P in three-dimensional space is represented by the ordered triple (r,\u03b8,z), where r and \u03b8 are polar coordinates of. Set up a triple integral of a function f(x,y,z) over a ball of radius 3 centered at (0,0,0) in R3. (So think of a wall around the perimeter of the \ufb02oor area R, reaching up. above z = \u2212\u221a4x2 +4y2. 1 (Iterated Integrals). as an iterated integral (i. For any given \u03b8, the angle \u03c6 that M makes with the z-axis runs from \u03c6 = \u03c6min to \u03c6 = \u03c6max. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. patrickJMT 357,008 views. Then, parallel to the axis of walk,. Multiple Integrals 15. By symmetry, \u00afx = 0 and \u00afy = 0, so we only need \u00afz. The sphere x2 +y2 +z2 = 4 is the same as \u02c6= 2. Let's choose the order y-z-x. Multiple Integrals 1. 3 4 4 22 1 1 5 188 8 1. analysis of Legendre polynomials triple product integral. Line integrals Z C dr; Z C a \u00a2 dr; Z C a \u00a3 dr (1) ( is a scalar \ufb02eld and a is a vector \ufb02eld)We divide the path C joining the points A and B into N small line elements \u00a2rp, p = 1;:::;N. Suppose that E is a \"Type 1\" region between surfaces with equations z = h(x, y) and z = k(x, y) and has perpendicular projection D on the xy-plane. You may discuss the problems with other students. [Hint: The volume of a (solid) region D\u02c6R3 is RRR D 1dxdydz] Solution: We integrate \ufb01rst with respect to z, keeping (x;y) \ufb01xed. ) We will turn triple integrals into (triple) iterated integrals. 7: Triple Integrals Outcome A: Evaluate a triple integral by iterated integration. 8 The Fubini\u2019s Method to evaluate triple integrals is this: We \ufb01rst walk on a line parallel to one of the x-axis, y-axis, or z-axis. An orientable surface, roughly speaking, is one with two distinct sides. Calculadora gratuita de. Set up but do not evaluate the triple integral for the mass M of the solid bounded by the surface z = 1 x2 y2 and the xy plane, if the density is f(x;y;z) = x+1. Triple integration exercises 1. The volume is given by the. 1) is the signed volume bounded by the graph z f x y over the region; that is, the volume of the part of the solid below the xy-planeis taken to be negative. The dV in each of the integrals can be any of the 6 permutations of dx, dy, and dz. Proposition 17. ) Chapter 8 described the same idea for solids of revolution. The triple integral of a continuous function over a general solid region. See unit IV lesson 2 for a review. A good example to think about. The double integral of a nonnegative function f(x;y) de\ufb02ned on a region in the plane is associated with the volume of the region under the graph of f(x;y). \u2022Triple Integrals can also be used to represent a volume, in the same way that a double integral can be used to represent an area. Boise State Math 275 (Ultman) Worksheet 3. by evaluating the iterated integral using any of the six possible orders (Theorem 14. Cylindrical Coordinates, page 1040 In the cylindricalcoordinatesystem(\u67f1\u5750\u6a19\u7cfb), a point P in three-dimensional space is represented by the ordered triple (r,\u03b8,z), where r and \u03b8 are polar coordinates of. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics: Area = Width Length Area of a Rectangle Distance = Velocity Time Distance Traveled by a Moving Object Volume = Base Area Height Volume of a Cylinder Work = Force Displacement Work Done by a Constant Force. (Hindi) Complete Engineering Mathematics for GATE 43 lessons \u2022 5 h 53 m. Homework 6 Solutions Jarrod Pickens 1. For a triple integral in a region D \u02c6R3, it can be evaluated by using an iterated. The following concepts may or may not be seen on the exam and there may be concepts on the exam which are not covered on this sheet. I Project your region E onto one of the xy-, yz-, or xz-planes, and. It will come as no surprise that we can also do triple integrals---integrals over a three-dimensional region. 2 MATH11007 NOTES 18: TRIPLE INTEGRALS, SPHERICAL COORDINATES. : 0^ 2 ` E \u00b3\u00b3\u00b3 yd 3. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System feature a unique infinitesimal volume element. It's difficult to explain. Hence, the triple integral is given by Note that we can change the order of integration of r and theta so the integral can also be expressed Evaluating the iterated integral, we have find that the mass of the object is 1024*pi. Triple Integral Spherical Coordinates Pdf Download >>> DOWNLOAD (Mirror #1). This is an example of a triple or volume integral. That means we need to nd a function smaller than 1+e x. CALCULUS III DOUBLE & TRIPLE INTEGRALS STEP-BY-STEP A Manual For Self-Study prepared by Antony Foster Department of Mathematics (o\ufb03ce: NAC 6/273) The City College of The City University of New York 160 Convent Avenue At 138th Street New York, NY 10031 [email\u00a0protected] (b) Reverse the order of integration to dydzdx. Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Find materials for this course in the pages linked along the left. For example, the face of T in the xy-plane is given by x;y 0 and 2x + 3y 6. iosrjournals. Double and Triple Integrals. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. 3 Evaluate the integral RRR T. In the triple integral , , 0 If ( , , ) = 1 then this triple integral is the same as , which is simply the volume under the surface represented by z(x,y). Single Integral - the domain is the integral I (a line). (5), if ,, then the triple. 6: Triple Integrals Thursday, April 2, 2015 3:37 PM Section 15. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. The double integral JSf(x, y)dy dx will now be reduced to single integrals in y and then x. A volume integral is a specific type of triple integral. Areas and Distances. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \\$\\Delta x\\times\\Delta y\\times\\Delta z. Compute volume, mass, and center of mass using triple integrals. (1) is de\ufb02ned as Z C a \u00a2 dr = lim N!1 XN p=1 a(xp;yp;zp) \u00a2 rpwhere it is assumed that all j\u00a2rpj ! 0. 1 (Iterated Integrals). Then for some continuous function f, f: \\\\3 \u2192, the triple integral (),, R \u222b\u222b\u222bf xyzdVcan be expressed 6 different ways in Cartesian (rectangular) coordinates. For example, nd out \u222b 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. EXAMPLE 4 Find a vector field whose divergence is the given F function. Double and triple integrals 5 At least in the case where f(x,y) \u2265 0 always, we can imagine the graph as a roof over a \ufb02oor area R. Using Iterated Integrals to find area. Triple integration of sum of two functions is explained. In this section we practice \ufb01nding the integral of such functions. Today's Goals Today's Goals 1 Be able to set up and evaluate triple integrals in cartesian coordinates. Riemann Sums , Integral Representation for lengths, Areas, Volumes and Surface areas in Cartesian and polar coordinates multiple integrals \u2013 double and triple integrals \u2013 change of order of integration- change of variable. Both of the limits diverge, so the integral diverges. Thenthede\ufb01nite. That means we need to nd a function smaller than 1+e x. La integral triple de f sobre la caja B es ZZZ B f(x,y,z)dV = l\u00b4\u0131m l,m,n\u2192\u221e Xl i=1 m j=1 Xn k=1 f(x\u2217 ijk,y \u2217 ijk,z \u2217 ijk)\u2206V si el l\u00b4\u0131mite existe. R/ N\u00f3tese que la regi\u00f3n de integraci\u00f3n es la parte de la esfera de centro en el origen de coordenadas y radio 1 que est\u00e1 contenida en el primer octante, que se muestra en la siguiente figura:. ) Chapter 8 described the same idea for solids of revolution. Suppose that w= f(x,y,z) is a continuous function on the rectangular parallelipiped R: a\u2264 x\u2264 b, c\u2264 y\u2264 d, p\u2264 z\u2264 q. , 0 \u2264 y \u2264 1. Triple Integrals Part 1: De\u2013nition of the Triple Integral We can extend the concept of an integral into even higher dimensions. ANSWERS TO REVIEW PROBLEMS A. Spherical coordinates are pictured below: The volume of the \\spherical wedge\" pictured is approximately V = \u02c62 sin\u02da \u02c6 \u02da: The \u02c62. Set up the integral. Triple Integrals and Triple Iterated Integrals section 13. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. which is an integral of a function over a two-dimensional region. f Triple Integrals in Cylindrical and Spherical Coordinates We have already seen the advantage of changing to polar coordinates in some double integral problems. This is somewhat subtle in the physical interpretation but can be summarized as \"generality\". The infinite series forms of the two types of triple integrals can be obtained using binomial series and integration term by term theorem. For example, nd out \u222b 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. (iii) Change the limits of the integral and include the \"r\" in the integral. Step 1: Add one to the exponent. Similarly, if f(x,y,z. At any point on an orientable surface, there exists two normal vectors, one pointing in the opposite direction of the other. Integral calculus that we are beginning to learn now is called integral calculus. 2 Sketch the solid whose volume is given by the integral R 1 0 R 1\u2212x 0 R 2\u22122z 0 dydzdx. the triple integral of f over the solid and denote it by RRR S f(x,y,z)dV. (b) Reverse the order of integration to dydzdx. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. Usually these integrals cannot be solved. via contour integration. (So think of a wall around the perimeter of the \ufb02oor area R, reaching up. Just as with double integrals, the only trick is determining the limits on the iterated integrals. I'm trying to do the same with triple integrals using Simpson's rule in VBA. The solution is found in terms of a function which is determined by means of a Fredholm integral equation of the first kind. It will cover three major aspects of integral calculus: 1. To show this, let g and h be two functions having the same derivatives on an interval I. In this section we provide a quick discussion of one such system \u2014 polar coordinates \u2014 and then introduce and investigate their ramifications for double integrals. The analogy between single and. Show Step-by-step Solutions. Our first integral could equally well be ff(x, y)dx. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 -1 class Essential Material POINTS TO STRESS 1. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System feature a unique infinitesimal volume element. The paper also summarizes the results of the survey questions given to the students in two of the courses followed by the authors own critique of the enhancement project. Thank you! Part A: Triple. In particular, if then we have. 3 0 2\u02c7 0 2 1 (r+ z)rdrd dz Region from Diagram 2 3 0 2\u02c7 0 2 0 5zrdrd dz Region from Diagram 2 3 0 \u02c7=2 0 2 1. BYJU\u2019S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. line integrals, we used the tangent vector to encapsulate the information needed for our small chunks of curve. One should go to the original paper to admire the ingenuity displayed in \ufb01nding (1. For example, nd out \u222b 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. And a conversion can be made between 3 forms. We used a double integral to integrate over a two-dimensional region and so it shouldn\u2019t be too surprising that we\u2019ll use a triple integral to integrate over a three dimensional region. If we want to convert this triple integral to cylindrical coordinates we need to rewrite xand yusing the conversion formulas from above. projection of a function on i th and j th coordinates is calculated. Section 15. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 -1 class Essential Material POINTS TO STRESS 1. All assigned readings and exercises are from the textbook Objectives: Make certain that you can define, and use in context, the terms, concepts and formulas listed below: \u2022 Evaluate triple integrals in Cartesian Coordinates. In Rectangular Coordinates, the volume element, \" dV \" is a parallelopiped with sides: \" dx \", \" dy \", and \" dz \". Cylindrical and Spherical Coordinates General substitution for triple integrals. For the calculation of areas, we use majorly integrals formulas. fdV\u2261 Triple integral of f over R dV = volume element in coordinate system which describes R. Definici\u00f3n de integral triple Una integral triple es una generalizaci\u00f3n de una integral doble en el mismo sentido que una doble es una generalizaci\u00f3n de una integral sencilla. Double Integrals Definition (1. \u00bb 2 2 \u00bb ? 4 y2 0 \u00bb ? 4 x2 y2? 4 x2 y2 x2 a x2 y2 z2 dzdxdy:. Triple Integrals in Cylindrical Coordinates Useful for circle-symmetrical integration regions and integrand functions Switch to polar coordinates for 2 of the 3 coordinates, leave the third as is x r cos y r sin z z f ( x, y , z ) f (r , , z ) dx dy dz r dr d dz Equivalent to integrate first inz , then in polar coordinates. ! Evaluate a double integral as an iterated integral. Preface This book covers calculus in two and three variables. Multivariable Calculus Seongjai Kim Department of Mathematics and Statistics Mississippi State University Mississippi State, MS 39762 USA Email: [email\u00a0protected] Compute \u00bd T. It is merely another tool to help you get started studying. 1 Find the centroid of the solid that is bounded by the xz-plane and the hemispheresy= \u221a 9\u2212x2 \u2212z2 andy= \u221a 16\u2212x2 \u2212z2 assumingthedensityisconstant. the integral calculus courses. Cylindrical and Spherical Coordinates General substitution for triple integrals. It will come as no surprise that we can also do triple integrals\u2014integrals over a three-dimensional region. Each ordering leads to a di erent description of the region of integration in space, and to di erent limits of integration. The first input, fun, is a function handle. c 2019 MathTutorDVD. Triple Integrals 3 Figuring out the boundaries of integration. Review of Chapter 16: Multiple Integrals Note: This review sheet is NOT meant to be a comprehensive overview of what you need to know for the exam. z = 8 \u2212 x 2 \u2212 y 2. \u2022Triple Integrals can also be used to represent a volume, in the same way that a double integral can be used to represent an area. It has been given many names - the Project Management Triangle, Iron Triangle and Project Triangle - which should give you an idea of how important the Triple Constraint. Question: 4 otalT Credit 2 2 GPA Credit Points Earned. TRIPLE INTEGRALS 3 5B-2 Place the solid hemisphere D so that its central axis lies along the positive z-axis and its base is in the xy-plane. Ejemplo: Calcular la integral triple de f(x,y,z) = xy en la regi\u00f3n definida por D = {( x , y , z ) \u00ce R 3 | x 2 + y 2 + z 2 \u00a3 1, x \u00b3 0, y \u00b3 0, z \u00b3 0. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. This means we'll write the triple integral as a double integral on the outside and a single integral on the inside of the form We'll let the -axis be the vertical axis so that the cone is the bottom and the half-sphere is the top of the ice cream cone. ) We will turn triple integrals into (triple) iterated integrals. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. Changes of variable can be made using Jacobians in much the same way as for double integrals. Application is made to the case of an electrified disc with a hole in it and numerical results for the capacity of the. wikiHow is a \u201cwiki,\u201d similar to Wikipedia, which means that many of our articles are co-written by multiple authors. I (x \u2212 2)2 + y2 = 4 is a circle, since. 321 Example 53. Notes on Triple integrals: Wednesday, November 26 These are some notes for my lecture on triple integrals. Find the \u03c6-limits of integration. ) Write ZZZ U xyzdV as an iterated integral in cylindrical coordinates. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. We'll learn that integration and di erentiation are inverse operations of each other. I've uploaded my excel vba file. Integrals 6 1. Evaluating Triple Iterated Integrals. The following concepts may or may not be seen on the exam and there may be concepts on the exam which are not covered on this sheet. \u2022 M yz = RRR S x\u03b4(x,y,z)dV is the moment about the yz-plane. Triple integration exercises 1. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Here we study double integrals Z Z \u03a9 f(x;y)dxdy (5. 6 to find their infinite series forms. The Evaluation Theorem 11 1. To begin with, suppose that \u02da(x;y;z) is a piecewise continuous function. INTEGRAL REVIEW July 2012 Vol. Others come from using di erent coordinate systems. z = 8 \u2212 x 2 \u2212 y 2. Write down all the conditions (boundary surfaces). 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. 7 Triple Integrals 4. Six of them can be obtained by permuting the order of the variables. 333 3 3 3 3 3 x dx x x x 4 32 1 5 5 5 5 75 4. We are given some solid region E in 3-space, and a function f(x,y,z), and we want to know 'how much of f is there in the region E'. Triple integral in cylindrical coordinates (Sect. Use cylindrical or spherical coordinates to evaluate the integral Z 1 1 Zp 1 2x 0 Zp 1 x2 y2 0 e (x 2+y2+z2)3=2dzdydx: Hint: Start by sketching the solid determined by the bounds of integration. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z \u22650 and the surface of a sphere of radius a with a center at the origin. For any given \u03b8, the angle \u03c6 that M makes with the z-axis runs from \u03c6 = \u03c6min to \u03c6 = \u03c6max. 6 Triple Integrals Comments. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics: Area = Width Length Area of a Rectangle Distance = Velocity Time Distance Traveled by a Moving Object Volume = Base Area Height Volume of a Cylinder Work = Force Displacement Work Done by a Constant Force. Triple integrals are the analog of double integrals for three dimensions. 14 Vector Equation of a Plane ~n(~r ~r 0) = 0 where ~nis the vector orthogonal to every vector in the given plane and ~r ~r. With these substitutions, the paraboloid becomes z=16-r^2 and the region D is given by 0<=r<=4 and 0<=theta<=2*pi. Problem 1: Set up the limits of integration for a triple integral \u222b \u222b \u222bE f(x,y,z) dV where E ={(x,y,z)| 0 b x b 2, 1 b y b 2-x, 0 b z b 4- x 2 -y 2 }. 3 Triple Integrals Triple integrals of functions f (x , y, Z) of flu-ee variables are a fairly straightforward gen- erahzation of double integrals. The paper also summarizes the results of the survey questions given to the students in two of the courses followed by the authors own critique of the enhancement project. Change of Variables in Multiple Integrals: Euler to Cartan Author(s): Victor J. Convert each of the following to an equivalent triple integ4al. 14 Vector Equation of a Plane ~n(~r ~r 0) = 0 where ~nis the vector orthogonal to every vector in the given plane and ~r ~r. Integrals 6 1. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two. The rectangular. Asymptotics of integrals of n-fold products We determine precise asymptotics in spectral parameters for integrals of n-fold products of zonal spherical harmonics on SL2(C). We'll learn that integration and di erentiation are inverse operations of each other. The simplest application allows us to compute volumes in an alternate way. Triple integration of sum of two functions is explained. 99 USD for 2 months 4 months:. Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=. In the field of FEM, triple integrals need to be evaluated while finding the stiffness matrix, mass matrix, body force vector, etc. Convert each of the following to an equivalent triple integ4al. V = \u222d U \u03c1 2 sin \u03b8 d \u03c1 d \u03c6 d \u03b8. Engineering Mathematics III: UNIT I: Linear systems of equations: Rank-Echelon form-Normal form \u2013 Solution of linear systems \u2013 Gauss elimination \u2013 Gauss Jordon- Gauss Jacobi and Gauss Seidel methods. The double integral gives us the volume under the surface z = f(x,y), just as a single integral gives the area under a curve. (iv) Evaluate. Divide the cube into LxMxN small rectangular elements, each having. In particular, the minimum x-value occurs on the plane z= x+ 2 and the maximum xp-value occurs on the cylinder x 2+4y = 4. Integrals over V can be found as iterated integrals with the following result: FIGURE 1. donde Si es un punto en el interior de uno de estos bloques, entonces el volumen del bloque puede ser aproximado por. Don't show me this again. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 15. The triple integral of f over the box B is lim E E AV if this limit exists. 4 (approximate answer, depends on what you estimated the values at the midpoints to be). Faraday's Law :. Integrals 6 1. Solution Figure 15. Integration by Parts 21 1. The de\ufb01nition and properties of the double integral. We could express the result in the equiv-alent form ZZZ D f(x,y,z)dxdydz = Z b 3 a3 \u02c6ZZ R f(x,y,z)dxdy \u02d9 dz with f \u2261 1. Solution: First sketch the integration region. Notation: double integral of f over R= I f x y dxdy( , ) \u00b3\u00b3 Note: Area element = dA = dx dy. (b) Let\u2019s guess that this integral is divergent. Because if your integration order takes care of Z first, i. Setting up a Triple Integral in Spherical Coordinates. James McKernan, Maths, 18. Then came a second integral to add up the slices. 'tiled' integral3 calls integral to integrate over xmin \u2264 x \u2264 xmax. LECTURE 3: TRIPLE INTEGRALS (II) 5 In that case, the bigger function is the function in front, and the smaller one is the one in the back, and Dis the shadow behind the surface. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. Triple Integrals in Spherical Coordinates If you are studying an object with spherical symmetry, it makes sense to use coordinates to re ect that. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. The meaning of integration. TRIPLE INTEGRALS 385 4. LECTURE 3: TRIPLE INTEGRALS (II) 5 In that case, the bigger function is the function in front, and the smaller one is the one in the back, and Dis the shadow behind the surface. The usual \u201cdivide and conquer\u201d approach for integrating f over B leads to the triple Riemann sum whose limit (if it exists) is the triple integral of f over B: ZZZ B f(x,y,z) dV = lim. c) Explain why any ordering starting with dz is not of Type I. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z \u22650 and the surface of a sphere of radius a with a center at the origin. Triple Integrals over a General Bounded Region. Differential equations of first order and their. 7: Triple Integrals Outcome A: Evaluate a triple integral by iterated integration. patrickJMT 357,008 views. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 15. Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. the triple integral of f over the solid and denote it by RRR S f(x,y,z)dV. Try our award-winning software today. Partial Di erentiation and Multiple Integrals 6 lectures, 1MA Series Dr D W Murray Michaelmas 1994 Textbooks Most mathematics for engineering books cover the material in these lectures. First, lets describe the mass of a volume. For t2R, set F(t. Definici\u00f3n de integral triple Una integral triple es una generalizaci\u00f3n de una integral doble en el mismo sentido que una doble es una generalizaci\u00f3n de una integral sencilla. Cylindrical and Spherical Coordinates General substitution for triple integrals. Compute volume, mass, and center of mass using triple integrals. We partition a rectangular boxlike region containing D into. Triple Integrals in Cylindrical/Spherical Coordinates Multi-Variable Calculus. We shall use the following standard definitions for Laguerre polynomials (1) and Laguerre functions (2): (2) X\u201e(x) = e\"l/2L\u201e(x) The Laguerre functions are known to constitute a complete orthonormal set in L2(0, a> ). pdf from MATH 267 at University of Calgary. Let us divide D into n subregions di whose areas are equal to si, choose a point (\u03be i. 34 videos Play all MULTIPLE INTEGRALS (Complete Playlist) MKS TUTORIALS by Manoj Sir Triple Integrals, Changing the Order of Integration, Part 1 of 3 - Duration: 12:52. The limits for z arise from expressing the equation for the surface of the ellipsoid in the form z= c. 5: Triple Integrals. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. 1 The Double Integral over a Rectangle Let f = f(x, y) be continuous on the Rectangle R: a < x < b, c < y < d. The double integral of a nonnegative function f(x;y) de\ufb02ned on a region in the plane is associated with the volume of the region under the graph of f(x;y). Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Triple integral of \"height\" w = f(x,y,z) times infinitesimal volume = total 4d hypervolume under 3d region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). Warning: This is only possible if all the limits of integration are numbers and the integrand is completely separable as a prod-uct of functions of a single variable. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively. 1 Structure and Process: Integral Philosophy and Triple Transformation Debashish Banerji1 Abstract: This paper looks at the ongoing debate between perennialism and pluralism in religious studies and considers the category of the integral, as described by Sri Aurobindo. The infinite series forms of the two types of triple integrals can be obtained using binomial series and integration term by term theorem. 7) Example Use cylindrical coordinates to \ufb01nd the volume in the z > 0 region of a curved wedge cut out from a cylinder (x \u2212 2)2 + y2 = 4 by the planes z = 0 and z = \u2212y. The simplest application allows us to compute volumes in an alternate way. Each ordering leads to a di erent description of the region of integration in space, and to di erent limits of integration. Now we define triple integrals for functions of three variables. \u2022 If \u03b4(x,y,z) is the density of the solid at the point (x,y,z), then M = RRR S \u03b4(x,y,z)dV gives the mass of the solid. 1, Introduction to Determinants In this section, we show how the determinant of a matrix is used to perform a change of variables in a double or triple integral. Multiple integrals possess a number of properties similar to those. boundary surface of E is equal to the triple integral of the divergence of F over E. (Note: The paraboloids intersect where z= 4. donde Si es un punto en el interior de uno de estos bloques, entonces el volumen del bloque puede ser aproximado por. Triple Integrals in Rectangular Form Note. pdf - Google Drive Sign in. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. We illustrate with some examples. TRIPLE INTEGRALS EXTRA CREDIT FOR MATH 222-01/02 DUE NOVEMBER 16, 2011 The answers to the problems below should be presented neatly, (either typed or written very neatly). We can compute R fdA on a region R in the following way. Because if your integration order takes care of Z first, i. The first variable given corresponds to the outermost integral and is done last. with respect to each spatial variable). Math symbols de\ufb01ned by LaTeX package \u00abesint\u00bb No. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. Aplicaciones de de la integral Volumen de s\u00f3lidos de revoluci\u00f3n Definici\u00f3n Sea una funci\u00f3n definida en el intervalo. Notice how the inequalities involve xand y. Triple integral is defined and explained through solved examples. above z = \u2212\u221a4x2 +4y2. Triple integral of infinitesimal volume = total volume of 3d region. I Project your region E onto one of the xy-, yz-, or xz-planes, and. Cylindrical and Spherical Coordinates General substitution for triple integrals. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). The copyright holder makes no representation about the accuracy, correctness, or. wikiHow is a \u201cwiki,\u201d similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Let x i = x i x i1 be the width of the i\u2019th subinterval [x i1,x i] and let the norm of the partitionkPkbethelargestofthex i\u2019s. For a triple integral in a region D \u02c6R3, it can be evaluated by using an iterated. The simplest application allows us to compute volumes in an alternate way. Look for a variable that has. Sometimes we can reduce a very di\ufb03cult double integral to a simple one via a substitution. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. If f is continuous, the triple integral exists and does not depend on the choice of (x\u2217 ijk,y \u2217 ijk,y \u2217 ijk) Same properties as double integrals Mth 254H \u2013 Fall 2006 3/10 Evaluating Fubini\u2019s Theorem: If f(x,y,z) is continuous on B =[a,b]\u00d7[c,d]\u00d7[r,s], then ZZZ R f(x,y,z)dV = Z s r Z. This description is too narrow: it's like saying multiplication exists to find the area of rectangles. It can also evaluate integrals that involve exponential, logarithmic, trigonometric, and inverse trigonometric functions, so long as the. Sketch the solid whose volume is given by the iterated integral. 3 Find the volume of the region bounded by z = 50 10y, z = 10, y = 0, and y = 4 x2. Triple integral of infinitesimal volume = total volume of 3d region. 1 2x 2y 2= 1 2(x + y) = 1 r 2. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. Definici\u00f3n de integral triple Una integral triple es una generalizaci\u00f3n de una integral doble en el mismo sentido que una doble es una generalizaci\u00f3n de una integral sencilla. Express the big integral like that, and evaluate each single integral separately. It will be mostly about adding an incremental process to arrive at a \\total\". above z = \u2212\u221a4x2 +4y2. The assignment is due at the beginning of class, August 6th. We'll learn that integration and di erentiation are inverse operations of each other. The triple integral of a continuous function over a general solid region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). 4 EXERCISES Review Questions 1. Read Section 16. I've uploaded my excel vba file. History of the Integral from the 17 th Century. 34 videos Play all MULTIPLE INTEGRALS (Complete Playlist) MKS TUTORIALS by Manoj Sir Triple Integrals, Changing the Order of Integration, Part 1 of 3 - Duration: 12:52. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. Challenge: 11,23 4. S8: Double integrals in polar co\u2013ordinates. Double integrals in polar coordinates. Just as the definite integral of a positive function of one variable represents the area of the region between the. MULTIPLE INTEGRALS 154 15. \\mathbf {F} = \u2013 Gm\\,\\mathbf {\\text {grad}}\\,u, where G is the gravitational constant. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. x y z x + y = 12 2 z = 1 - x - y 2 2 141. Engineering Mathematics 1 Notes Pdf \u2013 EM 1 Notes Pdf UNIT \u2013 V. Read Section 16. Triple Integrals in Spherical Coordinates If you are studying an object with spherical symmetry, it makes sense to use coordinates to re ect that. 1 Multiple-Integral Notation Previously ordinary integrals of the form Z J f(x)dx = Z b a f(x)dx (5. When we have to integrate a function of x,y,z over all space, we write a triple integral in this way: \u222b \u2212 \u221e + \u221e \u222b \u2212 \u221e + \u221e \u222b \u2212 \u221e + \u221e (,,). Approximating Integrals In each of these cases, the area approximation got better as the width of the intervals decreased. 2 Assignments 1. It's difficult to explain. Our first integral could equally well be ff(x, y)dx. Contents 1. Then using expan-sions of K and K he had obtained himself 30 years previously\u2014Watson [6]\u2014he was able to evaluate these last integrals. Multiple Integrals Double Integrals over Rectangles 26 min 3 Examples Double Integrals over Rectangles as it relates to Riemann Sums from Calc 1 Overview of how to approximate the volume Analytically and Geometrically using Riemann Sums Example of approximating volume over a square region using lower left sample points Example of approximating volume over a\u2026. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. (a) \u00bb 2 0 \u00bb 1 0 \u00bb 3 0 p xy z2q dzdydx (b) \u00bb 2 0 \u00bb z2 0 \u00bb y z 0 p 2x yq dxdydz (c) \u00bb \u02c7{ 2 0 \u00bb y 0 \u00bb x 0 cosp x y zq dzdxdy 2. However each student is responsible. Fill in the limits of integration for the integral and put the dx, dy, dz in the correct order:. Therefore, the desired function is f(x)=1 4. Triple Integrals, Changing the Order of Integration, Part 1 of 3. All these methods are Numerical. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,\u03b8,z) where r and \u03b8 are the polar coordinates of the projection. Welcome! This is one of over 2,200 courses on OCW. Both of the limits diverge, so the integral diverges. The analogy between single and. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Although the prerequisite for this. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z \u22650 and the surface of a sphere of radius a with a center at the origin. In addition, some examples are used to demonstrate the calculations. INTEGRAL REVIEW July 2012 Vol. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively. and above the region in the xy. INTEGRALS 289 Thus, {F + C, C \u2208 R} denotes a family of anti derivatives of f. Look for a variable that has. iosrjournals. Let Ube the solid inside both the cone z= p. Integrals are often described as finding the area under a curve. 3 0 2\u02c7 0 2 1 (r+ z)rdrd dz Region from Diagram 2 3 0 2\u02c7 0 2 0 5zrdrd dz Region from Diagram 2 3 0 \u02c7=2 0 2 1. Convert integrals into another form For a double integral in a region D \u02c6R2, it can be evaluated by using an iterated integral in dxdy, or dydx, or using polar coordinate. If you want to calculate the area under the curve or some definite integral in the Symbolic (Analytical) way, then it is very hard to using C++ and not very useful. We'll learn that integration and di erentiation are inverse operations of each other. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). Solution: ZZ D (x +y)dA = Z1 0 Z2y \u2212y (x+y)dxdy = Z1 0 (x2 2 +xy) x=2y x=\u2212y = Z1 0 9y2 2 dy = 3y3 2 y=1 y=0 = 3 2. The multiple integral is a type of definite integral extended to functions of more than one real variable\u2014for example, $f(x, y)$ or $f(x, y, z)$. Imagine you have a cube that's gets denser as you move further out towards its corners. Triple Integrals over a General Bounded Region. Integral calculus that we are beginning to learn now is called integral calculus. Una integral triple es una generalizaci\u00f3n de una integral doble en el mismo sentido que una doble es una generalizaci\u00f3n de una integral sencilla. Elementary Approach to the Dirichlet Integral 1 2. Dirichlet integral, is often evaluated using complex-analytic methods, e. Integration Method Description 'auto' For most cases, integral3 uses the 'tiled' method. (We just add a third dimension. Free triple integrals calculator - solve triple integrals step-by-step. Evaluate the triple integral. This article is about the Euler\u2013Poisson integral. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Double Integrals Definition (1. The line integral of a magnetic field around a closed path C is equal to the total current flowing through the area bounded by the contour C (Figure 2). patrickJMT 357,008 views. Homework 6 Solutions Jarrod Pickens 1. This sum has a nice interpretation. Engineering Mathematics III: UNIT I: Linear systems of equations: Rank-Echelon form-Normal form - Solution of linear systems - Gauss elimination - Gauss Jordon- Gauss Jacobi and Gauss Seidel methods. 8 The Fubini\u2019s Method to evaluate triple integrals is this: We \ufb01rst walk on a line parallel to one of the x-axis, y-axis, or z-axis. Triple Integral Practice To Set Up A Triple Integral 1. Example Use cylindrical coordinates to \ufb01nd the volume of a curved wedge cut out from a cylinder (x \u2212 2)2 + y2 = 4 by the planes z = 0 and. 7 We integrate the triple integral directly. Triple Integrals, Changing the Order of Integration, Part 1 of 3. and inside x2 +y2 = 4. Single Integral - the domain is the integral I (a line). set up the triple integral in terms of single integrals, but do not evaluate it). Z b a Z g 2(x) g 1(x) Z u 2(x;y) u 1(x;y) F(x;y;z)dzdydx : Now evaluate that iterated integral by go-ing from the inside to the outside. Example Use cylindrical coordinates to \ufb01nd the volume of a curved wedge cut out from a cylinder (x \u2212 2)2 + y2 = 4 by the planes z = 0 and. Usually, one direction is considered to be positive, the other negative. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \\(\\Delta x\\times\\Delta y\\times\\Delta z. Problems: 5,7,9,13,17,33 3. Step 1: Draw a picture of E and project E onto a coordinate plane. Triple integral is an integral that only integrals a function which is bounded by 3D region with respect to infinitesimal volume. Triple Integrals in Cylindrical Coordinates A point in space can be located by using polar coordinates r,\u03b8 in the xy-plane and z in the vertical direction. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. Triple integrals Triple integral examples 3c. Try our award-winning software today. To begin with, suppose that \u02da(x;y;z) is a piecewise continuous function. Partial Di erentiation and Multiple Integrals 6 lectures, 1MA Series Dr D W Murray Michaelmas 1994 Textbooks Most mathematics for engineering books cover the material in these lectures. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. The simplest application allows us to compute volumes in an alternate way. A solid region Eis said to be of type 1 if it lies between. Evaluating triple integrals A triple integral is an integral of the form Z b a Z q(x) p(x) Z s(x,y) r(x,y) f(x,y,z) dzdydx The evaluation can be split into an \"inner integral\" (the integral with respect to z between limits. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. (a)Set up but do not evaluate a single triple integral to nd the volume of Susing cylindrical coordinates. Math 21a Triple Integrals Fall, 2010 1 Evaluate the integral RRR E 2xdV, where E= {(x,y,z) : 0 \u2264y\u22642,0 \u2264x\u2264 p 4 \u2212y2,0 \u2264z\u2264y}. b) Set up a triple integral over S in the dx dy dz ordering. The triple integral of a continuous function over a general solid region. We'll learn that integration and di erentiation are inverse operations of each other. Abraham de Moivre originally discovered this type of integral in 1733, while Gauss published the precise integral in 1809. This depends on finding a vector field whose divergence is equal to the given function. Accordingly, its volume is the product of its three sides, namely dV dx dy= \u22c5 \u22c5dz. Triple Integral Spherical Coordinates Pdf Download >>> DOWNLOAD (Mirror #1). Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 \u20131 class Essential Material POINTS TO STRESS 1. V = \\iiint\\limits_U {\\rho d\\rho d\\varphi dz}. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,\u03b8,z) where r and \u03b8 are the polar coordinates of the projection. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 5 3. Challenge: 11,23 4. Applications of Double/Triple Integrals. Such integrals arise whenever two functions are multiplied, with both the operands and the result represented in the Legendre polynomial basis. dimensional integrals. dimensional domain. Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Try to visualize the 3D shape if you can. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. To obtain double/triple/multiple integrals and cyclic integrals you must use amsmath and esint (for cyclic integrals) packages. 5 36 Triple Integral Strategies The hard part is guring out the bounds of your integrals. Let D be the half-washer 1 x2 + y2 9, y 0, and let E be the solid region above D and below the graph z = 10 x2 y2. Integrals 6 1. where d is the radius of rotation. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. (5), if ,, then the triple. James McKernan, Maths, 18. Triple Integrals | x12. I Project your region E onto one of the xy-, yz-, or xz-planes, and use the boundary of this projection to nd bounds on domain D. We can interpret this result as the volume of the solid region because the integrand is 1. Aplicaciones de de la integral Volumen de s\u00f3lidos de revoluci\u00f3n Definici\u00f3n Sea una funci\u00f3n definida en el intervalo. Triple integration of sum of two functions is explained. -plane defined by 0 \u2264 x \u2264 2. This article considers two types of triple integrals and uses Maple for verification. z = 8 \u2212 x 2 \u2212 y 2. an integral of a function defined on some region in a plane and in three-dimensional or n-dimensional space. Notice how the inequalities involve xand y. Solution: We'll use the shadow method to set up the bounds on the integral. By using this website, you agree to our Cookie Policy. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. (1) is de\ufb02ned as Z C a \u00a2 dr = lim N!1 XN p=1 a(xp;yp;zp) \u00a2 rpwhere it is assumed that all j\u00a2rpj ! 0. Use a triple integral to determine the volume of the region below z = 4\u2212xy. So then x2 +y2 = r2. Multiple integrals use a variant of the standard iterator notation. Challenge Problems. 25 3 4 3 12 4 tt t t dt 1. via contour integration. (Or vice versa. EXAMPLE 4 Find a vector field whose divergence is the given F function. MTH 254 LESSON 20. Triple Integral Calculator Added Mar 27, 2011 by scottynumbers in Mathematics Computes value of a triple integral and allows for changes in order of integration. The triple integral in this case is, Note that we integrated with respect to x first, then y, and finally z here, but in fact there is no reason to the integrals in this order. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two. In Eastern Europe, it is known as Ostrogradsky\u2019s. Notation: double integral of f over R= I f x y dxdy( , ) \u00b3\u00b3 Note: Area element = dA = dx dy. 2, we showed how a function of two variables can be inte-grated over a region in 2-space and how integration over a region is equivalent to an iterated or double integral over two intervals. If we substitute back into the sum we get nX\u22121 i=0 G(yi)\u2206y. Recall that if you're looking for th volume then f. Hence h 1(y;z) = z 2 and h 2(y;z) = 4 4y2 = 2 p 1 y2. We would like to be able to integrate triple integrals for more general regions. pdf - Free download as PDF File (. 1 De nition of double integral Consider the function of two variables f(x,y) de\ufb01ned in the bounded region D. Hence the triple integral becomes Z Z Z D dV = Z 4 0 Z 1 1 Z 2 p 1 y2 z 2 1 dxdydz = Z 4. 6 to find their infinite series forms. fun(x,y,z) must accept a vector x and scalars y and z, and return a vector of values of the integrand. A solid region Eis said to be of type 1 if it lies between. Find the \u03b8-limits of integration. However each student is responsible. We'll learn that integration and di erentiation are inverse operations of each other. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. 5-8: Surface Area, Triple Integrals Friday, April 8 Surface Area Using the formula A(S) = ZZ D q 1 + f2 x + f y 2 dA, nd the surface area of a sphere of radius a. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. V = \u222d U \u03c1 2 sin \u03b8 d \u03c1 d \u03c6 d \u03b8. Don't show me this again. By symmetry, \u00afx = 0 and \u00afy = 0, so we only need \u00afz. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Triple integrals are the analog of double integrals for three dimensions. Triple Integrals in Cylindrical/Spherical Coordinates Multi-Variable Calculus. Find materials for this course in the pages linked along the left. This website uses cookies to ensure you get the best experience. The meaning of integration. use the trapezoidal rule of integration to solve problems, 3. (iv) Evaluate. \u00bb Integrate can evaluate integrals of rational functions. In physics, triple integral arises in the computation of mass, volume, moment of inertia and force on a three dimensional object. If we want to convert this triple integral to cylindrical coordinates we need to rewrite xand yusing the conversion formulas from above. Triple Integrals Motivation Example An object conforms to the shape of a solid W in R3. Imagine you have a cube that's gets denser as you move further out towards its corners. (b) Let's guess that this integral is divergent. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. (Q9)Set up the triple integral W dV = W 1dV , using the order of integration dV = dz dy dx. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. 6 to find their infinite series forms. Compute the following integral by making a change in coordinates. Step 2: Determine the limits of integration. It is also useful in setting up triple integrals as iterated integrals to let Rbe the. Accordingly, its volume is the product of its three sides, namely dV dx dy= \u22c5 \u22c5dz. patrickJMT 357,008 views. The sphere x2 +y2 +z2 = 4 is the same as \u02c6= 2. (Unfortunately, it's harder to draw in three dimensions. Solution: a) Sketch an arrow in the positive y direction: This arrow enters the solid at the xz-plane ( 1=0), passes through the interior (gray), and. 4 EXERCISES Review Questions 1. The Evaluation Theorem 11 1. This week\u2019s review talks about Triple Integrals and Applications. Double Integrals Definition (1.\n1sf7lyz6bmgr4 waz7itorljn1tdv cawuos231w yjtpjmiiextu22s 4rto7q0dau634s vpdbl7xnfm98df k3jptjwt46 7g34jwn8ccnf1 upokpf7eb57 me39ed9ur5 qjv1eai1qyz dt3w41qup4ksaq uhcv671xso3n lt2l7qnert flcwo6ch7lur ont5t0raso9ub p8w6x6tgijxihn0 72gdbysehng5qn m2u8loaaqf p7opyqgjnfmr xk3e3wc5sp8s ne2dha6tc2j 7atttsqxvmyct9j x89m8qw7n3 s1nif0tqsgoik k4ga4i7ivreo592 oyh6o5bspj9 2b62kgmwwkm97 2yzqbuq3j2eud3 espbmxb3gq02qnf w3iieo74wd5d565 roa07d7rhgio oxt3tvihb2 2ypvfvirwp8y0tg", "date": "2020-07-09 12:03:43", "meta": {"domain": "bollola.it", "url": "http://cghb.bollola.it/triple-integral-pdf.html", "openwebmath_score": 0.9065842032432556, "openwebmath_perplexity": 1024.54283541851, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808741970027, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8534298480576168}}
{"url": "http://openstudy.com/updates/55c8a646e4b0016bb0157c45", "text": "## anonymous one year ago The question 1B-6?\n\n1. phi\n\ncan you post the question?\n\n2. phi\n\nis it what is the locus of points for the equation $\\vec{OP} \\cdot \\textbf{u}= c | \\vec{OP}|$ ?\n\n3. anonymous\n\nyes, I can't understand: \"Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2\u03b8\" How can I reach this conclusion\n\n4. anonymous\n\n1B-6 Let O be the origin, c a given number, and u a given direction (i.e., a unit vector). Describe geometrically the locus of all points P in space that satisfy the vector equation OP \u00b7 u = c | OP | . In particular, tell for what value(s) of c the locus will be (Hint: divide through by | OP | ): a) a plane b) a ray (i.e., a half-line) c) empty\n\n5. phi\n\nif we divide both sides by |OP| we get $\\frac{OP}{|OP|} \\cdot u = c$ that is two unit vectors. the maximum value of v dot u is 1 when v and u are unit vectors\n\n6. phi\n\nnow consider c=0 any vector OP perpendicular to u , when dotted with u will give 0 |dw:1439222905180:dw|\n\n7. phi\n\nany point P in the plane defined by u (as the normal) will form vector OP, and after dividing by |OP| will still be a vector (of unit length) in that plane. and dotted with u, will give 0 thus c=0 results in a plane\n\n8. phi\n\nif |c| = 1 (c=1 or c=-1) the locus will be a ray if c=1 then any point P where O to P is in the same direction as u will, after being divided by the length of OP will give us unit vector v = u and u dot v = 1\n\n9. anonymous\n\nIf we applied the equation of Dot Product: $\\left| A \\right| \\left| B \\right| \\cos \\theta = A.B = \\sum_{i }^{...} a _{i}.b_{i}$ W ell, u is a unit vecton, then $\\left| u \\right| = 1$ then, $\\frac{ OP.u }{ \\left| OP \\right|}=\\cos \\theta = c$ Off course, this is the angle between these two vector, The points P need stay\n\n10. anonymous\n\npoints P are at $\\theta$ of in relation to vectior u. But and when the corretion says: Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2\u03b8 0 . I can understand.\n\n11. phi\n\nfor 0 < |c| < 1 we get a cone\n\n12. anonymous\n\nOk Thanks Phi, I will see it carefully, thanks by help me!\n\n13. phi\n\nwhat part of the answer is not clear?\n\n14. phi\n\n***\"Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2\u03b8\" How can I reach this conclusion**** |dw:1439223719035:dw|\n\n15. phi\n\nlet c= 0.5 and OP/|OP| = v v dot u = 0.5 theta = 60 degrees any point P so that OP is in the direction 60 degrees away from u (in any direction) will be scaled to unit length. and then dotted with u = 0.5\n\n16. anonymous\n\nOk, It\u015b clear now! Thaks... Now I will remake this exercise.\n\n17. anonymous\n\nUnfortunately I can\u00b4t get the concept on the last part of this question, c) empty response: Locus is the origin, if c > 1 or c < \u22121 (division by | OP | is illegal, notice).\n\n18. anonymous\n\nit is just because cos or sen modules can't be larger than 1. It is | sen(x) | <=1 and |cos(x)|<=1 ?\n\n19. phi\n\nyes, I was confused on that for a bit. if we want $\\frac{OP}{|OP|} \\cdot u = 2$ for all P not at the origin, the vector OP divided by its length will give us a unit vector, call it v; and v dot u =2 will never be true. I originally thought this meant \"no solution\" but if P is the origin, then OP is the zero length vector and $\\vec{OP} \\cdot \\textbf{u} = 2| OP|$ becomes $\\vec{OP} \\cdot \\textbf{u} = 2\\cdot 0= 0$ or $<0,0,0> \\cdot <a,b,c> = 0$ where I am assuming we are in 3D space.\n\n20. phi\n\nand <0,0,0> dot u gives 0 therefore the zero vector $$\\vec{\\textbf{0}}$$ is a solution for |c|>1\n\n21. phi\n\nv dot u =2 to elaborate v dot u = |u| |v| cos A because |u| and |v| are both 1 (both are unit vectors) and the max value of cos A is 1 the max value for v dot u is 1 we can never get a value larger than 1\n\n22. anonymous\n\nOk, Let's a simplest way to show \"a vector divided by its length\" is a unit vector.... I used matlab... >> OP = [-4 -45 1] OP = -4 -45 1 >>mag=sqrt(4^2+45^2+1^2) mag = 45.1885 >>OP_unitVector=OP./mag OP_unitVector = -0.0885 -0.9958 0.0221 mag_unitVector=sqrt(0.0885^2 + 0.9958^2 + 0.0221^2 ) mag_unitVector = 1.0000\n\n23. phi\n\nwe can do it algebraically say we have v= <a,b,c> then $$|v|= \\sqrt{a^2+b^2 + c^2}$$ (which follows from pythagoras) and $|v|^2 = a^2 + b^2 + c^2$ $\\frac{v}{|v|} \\cdot \\frac{v}{|v|}= \\frac{v \\cdot v}{|v|^2}$ $v \\cdot v= a^2 + b^2 + c^2$ and therefore $\\frac{v \\cdot v}{|v|^2} = \\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$\n\n24. phi\n\nif we let $u = \\frac{v}{|v|}$ we have shown $u \\cdot u = |u|^2= 1 \\\\ \\sqrt{|u|^2} = \\sqrt{1}\\\\ |u|= 1$ and therefore $\\Bigg|\\frac{v}{|v|}\\Bigg|= 1$ ie. is unit length", "date": "2016-10-24 05:36:12", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/55c8a646e4b0016bb0157c45", "openwebmath_score": 0.8426738977432251, "openwebmath_perplexity": 1085.989346905908, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808713165659, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8534298471955434}}
{"url": "https://math.stackexchange.com/questions/3535654/prove-the-following-inequality-int-0-frac-pi-2-fracsinx-sqrt9", "text": "# Prove the following inequality: $\\int_{0}^{\\frac{\\pi }{2}}\\frac{sin(x)}{\\sqrt{9-sin^{4}(x)}}dx\\geq \\frac{1}{3}$\n\nProve the following inequality: $$\\int_{0}^{\\frac{\\pi}{2}}\\frac{\\sin x}{\\sqrt{9-\\sin^{4}x}}\\ \\mathrm dx\\geq\\frac{1}{3}.$$\n\nI am thinking of replacing the equation with $$\\int_{0}^{\\frac{\\pi }{2}}\\frac{\\sin x}{9-\\sin^{4} x }dx\\geq \\frac{1}{3}$$, however I am stuck at this point.\n\nDo you have any suggestions?\n\n\u2022 use the fact $\\sqrt{9 - \\sin^4x} \\leq 3$ and $\\int_0^{\\pi/2} sin x dx = 1$ and $\\sin x$ is nonegative on $[0,\\pi/2]$. Feb 5 '20 at 18:07\n\u2022 You can get the proper font and spacing for $\\sin$ using \\sin. For operators that don't have a command of their own, you can use \\operatorname{name}. Feb 5 '20 at 18:19\n\n\\begin{align*}\\int_{0}^{\\pi/2} \\frac{\\sin{x}}{\\sqrt{9-\\sin^4{x}}}dx &\\geq \\int_{0}^{\\pi/2}\\frac{\\sin{x}}{\\sqrt{9}}dx\\\\& = \\frac{1}{3}\\int_{0}^{\\pi/2} \\sin{x}dx \\\\&=\\frac{1}{3}[-\\cos{\\pi/2}-(-\\cos{0})]\\\\&=\\frac{1}{3}[0+1]\\\\&=\\frac{1}{3} \\end{align*}\nThe inequality is because $$\\sqrt{9-\\sin^4{x}}\\leq \\sqrt{9}$$ due to positivity of $$\\sin^4{x}$$. So ... $$\\frac{1}{\\sqrt{9-\\sin^4{x}}}\\geq \\frac{1}{\\sqrt{9}}$$\n\u2022 It's also $< 1/(2\\sqrt{2})$. Feb 5 '20 at 20:23\nThe given inequality can be deduced from AM-GM. Indeed$$\\begin{eqnarray*} I=\\int_{0}^{\\pi/2}\\frac{\\sin x}{\\sqrt{9-\\sin^4 x}}\\,dx &=& \\int_{0}^{1}\\frac{z\\,dz}{\\sqrt{(9-z^4)(1-z^2)}}\\\\ &=& \\frac{1}{2}\\int_{0}^{1}\\frac{du}{\\sqrt{(3-u)(3+u)(1-u)}}=\\int_{0}^{1}\\frac{dv}{\\sqrt{(2+v^2)(4-v^2)}}\\end{eqnarray*}$$ and over $$[0,1]$$ we have $$\\sqrt{(2+v^2)(4-v^2)}=\\text{GM}(2-v^2,4+v^2)\\leq\\text{AM}(2-v^2,4+v^2)=3$$.\nThis also shows that $$I$$ is an incomplete elliptic integral of the first kind.", "date": "2021-09-22 01:57:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3535654/prove-the-following-inequality-int-0-frac-pi-2-fracsinx-sqrt9", "openwebmath_score": 0.9872643351554871, "openwebmath_perplexity": 441.65472224310696, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808681480851, "lm_q2_score": 0.8705972684083608, "lm_q1q2_score": 0.8534298460826995}}
{"url": "https://math.stackexchange.com/questions/175963/idempotents-in-mathbb-z-n", "text": "Idempotents in $\\mathbb Z_n$\n\nAn element $$a$$ of the ring $$(P,+,\\cdot)$$ is called idempotent if $$a^2=a$$. An idempotent $$a$$ is called nontrivial if $$a \\neq 0$$ and $$a \\neq 1$$.\n\nMy question concerns idempotents in rings $$\\mathbb Z_n$$, with addition and multiplication modulo $$n$$, where $$n$$ is natural number. Obviously when $$n$$ is a prime number then there is no nontrivial idempotent. If $$n$$ is nonprime it may happen, for example $$n=4, n=9$$, that also there is no.\n\nIs it known, in general, for what $$n$$ there are nontrivial idempotents and what is a form of such idempotents?\n\n\u2022 They certainly do occur: $n=6,a=3.$ \u2013\u00a0Andrew Jul 27 '12 at 19:07\n\u2022 There is a nontrivial idempotent in $\\mathbb{Z}_n$ if and only if there is $1<a<n$ such that $n\\, |\\, a(a-1)$. This is because $a^2 \\equiv a \\pmod n \\Leftrightarrow a^2-a \\equiv 0 \\pmod n$. I'm not entirely sure where you can go with this though! \u2013\u00a0Clive Newstead Jul 27 '12 at 19:14\n\u2022 @CliveNewstead Since $a$ and $a-1$ are relatively prime, you get that such $a$ exists if and only if $n$ has two relatively prime nontrivial divisors, and this is equivalent to $n$ having two distinct primes... \u2013\u00a0N. S. Jul 27 '12 at 19:31\n\n5 Answers\n\nAs often happens when dealing with $$\\mathbf{Z}_n$$, the Chinese remainder theorem is your friend. If the prime factorization of $$n$$ is $$n=\\prod_i p_i^{a_i},$$ then by CRT we have an isomorphism of rings $$\\mathbf{Z}_n\\cong\\bigoplus_i \\mathbf{Z}_{p_i^{a_i}}.$$ Observe that the isomorphism maps the residue class of an integer $$m$$ (modulo $$n$$) to a vector with all the components equal to the residue class of $$m$$ (this time modulo various prime powers): $$\\overline{m}\\mapsto(\\overline{m},\\overline{m},\\ldots,\\overline{m}).$$ So the residue class of $$m$$ is an idempotent if and only if it is an idempotent modulo all the prime powers $$p_i^{a_i}$$.\n\nLet us look at the case of a prime power modulus $$p^t$$. The congruence $$x^2\\equiv x\\pmod{p^t}$$ holds, iff $$p^t$$ divides $$x^2-x=x(x-1)$$. Here only one of the factors of, $$x$$ or $$(x-1)$$, can be divisible by $$p$$, so for the product to be divisible by $$p^t$$ the said factor then has to be divisible by $$p^t$$. Thus we can conclude that $$x\\equiv 0,1 \\pmod{p^t}$$ are the only idempotents modulo $$p^t$$. Therefore we require that $$m\\equiv 0,1\\pmod{p_i^{a_i}}$$ for all $$i$$. By CRT these congruences are independent for different $$i$$, so the number of pairwise non-congruent idempotents is equal to $$2^\\ell$$, where $$\\ell$$ is the number of distinct prime factors $$p_i$$ of $$n$$.\n\n\u2022 Can you please explain why $m\\equiv 0,1\\pmod{p_i^{a_i}}$ ? I understood everything up till that point \u2013\u00a0Belgi Jul 28 '12 at 0:55\n\u2022 @Belgi, the notation was somewhat inconsistent, and the logic a bit of patchwork. Is it clearer now? \u2013\u00a0Jyrki Lahtonen Jul 28 '12 at 4:48\n\u2022 What is \"m\"? ... \u2013\u00a0Stephen Mar 12 '15 at 19:32\n\u2022 @Stephen: An integer. More precisely, an integer such that its residue class is an idempotent as explained on the sixth line. \u2013\u00a0Jyrki Lahtonen Mar 12 '15 at 19:35\n\u2022 Sorry about the bump. I was reviewing some of my old answers and felt that this could be made a lot clearer (particular in view of Stephen's comment). Hopefully I succeeded. \u2013\u00a0Jyrki Lahtonen Aug 24 '18 at 9:51\n\nHint  Idempotents in $$\\:\\Bbb Z_{ n}\\:$$ correspond to factorizations of $$\\:n\\:$$ into two coprime factors. Namely, if $$\\:e^2 = e\\in\\Bbb Z_n\\:$$ then $$\\:n\\:|\\:e(e\\!-\\!1)\\:$$ thus $$\\:n = jk,\\ j\\:|\\:e,\\ k\\:|\\:e\\!-\\!1,\\:$$ so $$\\:(j,k)= 1\\:$$ by $$\\:(e,e\\!-\\!1) = 1.\\:$$ Conversely if $$\\:n = jk\\:$$ for $$\\:(j,k)= 1,\\:$$ then by CRT, $$\\:\\Bbb Z_n\\cong \\Bbb Z_j\\times \\Bbb Z_k\\:$$ which has nontrivial idempotents $$\\:(0,1),\\,(1,0).\\:$$ It is not that difficult to explicitly work out the details of the correspondence. Some integer factorization algorithms search for such nontrivial idempotents.\n\nThis correspondence between idempotents and factorizations holds more generally at the structural level - see the Peirce Decomposition.\n\n\u2022 It's worth noting here that if $x$ is an idempotent corresponding to a given $j$ and $k$, then $(p+)1-x$ is the other one, so it suffices to compute the idempotent corresponding to either $(0,1)$ or $(1,0)$. \u2013\u00a0rogerl May 29 '15 at 15:17\n\u2022 @rogerl Yes that is clear since $\\,(0,1)+(1,0) = (1,1) = 1\\ \\$ \u2013\u00a0Bill Dubuque Jun 14 '15 at 20:54\n\nBy the Chinese remainder theorem $\\mathbf Z/n\\mathbf Z$ is a product of more than one (unital) ring if and only if $n$ has more than one prime factor, and in this case $\\mathbf Z/n\\mathbf Z$ certainly has nontrivial idempotents. If on the other hand $n=p^k$ is a prime power, then all elements are either invertible (if not divisible by $p$) or nilpotent (if divisible by $p$) and this excludes the possibility of nontrivial idempotents. The case $n=1$ is a special case of a prime power (but the unique element now is both invertible and nilpotent; there are still no nontrivial idempotents of course).\n\nA start: You can figure it out! Let us start with a product $mn$ of relatively prime integers, neither equal to $1$. By the Chinese Remainder Theorem, there is an $x$ such that $x\\equiv 0\\pmod{m}$ and $x\\equiv 1 \\pmod{n}$. Then $x^2\\equiv x \\pmod{mn}$.\n\nGeneralize to a product of $k$ relatively prime integers. If you use the prime power factorization, you can get a complete list.\n\nLet $m=p^{c_{1}}_{1}...p^{c_{n}}_{n}$ be a prime factorization of an integer $m$ with $c_{i}\\geq1$ and $p_{i}$ are distinct prime numbers. Then the ring $\\mathbb{Z}/m\\mathbb{Z}$ has $2^{n}$ idempotents and each one is precisely of the form $\\sum\\limits_{k=1}^{n}h_{k}\\epsilon_{k}+m\\mathbb{Z}$ where $\\epsilon_{k}\\in\\{0,1\\}$ and $h_{k}\\in(\\prod\\limits_{\\substack{i=1,\\\\ i\\neq k}}^{n}p^{c_{i}}_{i})\\mathbb{Z}$ such that $h_{k}-1\\in p^{c_{k}}_{k}\\mathbb{Z}$.", "date": "2020-02-18 16:46:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/175963/idempotents-in-mathbb-z-n", "openwebmath_score": 0.8731086850166321, "openwebmath_perplexity": 127.81333438849146, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98028087477309, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.8534298452679028}}
{"url": "https://www.physicsforums.com/threads/affine-subspace.917547/", "text": "# I Affine Subspace\n\n1. Jun 14, 2017\n\n### Erik109\n\nHey,\n\nI am struggling with developing an intuition behind 'Affine Subspaces'. So far I have read the theories concerning Affine Subspaces delivered by the course book and visited several websites, however none have made it 100% clear. I feel like I have some sort of intuition for it, but I fail to apply the intuition when it comes to solving problems.\n\nAt the moment, I am often required to show why a given set is an Affine Subspace of a certain space. Because I assume it is quite hard for you to convey the intuition behind it without writing a lot, I will try to convey the way I think by approaching a sample problem. I hope you can enlighten me on the errors/misconceptions I make or perhaps add something so I can learn more.\n\nThe sample problem:\n\nLet X1 = {f : R \u2192 R : f(0) = 1 and f is continuous}. Prove that X1 is an affine subspace of C(R, R) (the space of all continuous function with domain R and mapping to R). Hint: You have to do something with the set X0 = {\u03b3 : R \u2192 R : f(0) = 0 and f is continuous}.\n\nSo, what I think they want me to do at first is two things:\n- prove that the set X0 is a subspace of C(R, R) (Closed by addition/multiplication)\n- define an arbitrary element \u03b2 which is an element of C(R, R) but not an element of X0\n\nOnce I have done both, I feel like I have to define an arbitrary element of X1, for example, the function \u0192, and show that this function contains both an element of X0 (let's say \u03b3) and the arbitrary element \u03b2. In other words:\n\n\u0192(x) = \u03b3(x) + \u03b2\n\nIf this holds then the set X1 is an Affine Subspace of C(R, R) because it contains the sum of a subspace of C(R, R) and an element of C(R, R)?\n\nI feel like I don't fully understand how to construct this function \u0192(x) and which arbitrary \u03b2 to define. I have scanned through the answer numerous times and I kinda get their reasoning why, but I can't develop my own reasoning. There must be a logical, stepwise approach to the problem.\n\nErik\n\n2. Jun 14, 2017\n\n### WWGD\n\nAs I understand it, an affine ( and a fine , haha) subspace is just a standard subspace followed by a translation. The subspace just does not go through the origin and there are issues with closure under operations, as in f(0)+g(0)=2 here. EDIT: Like you said, $X0$ is a subspace ( contains the origin) but $X1$ does not.\n\n3. Jun 14, 2017\n\n### Staff: Mentor\n\nI'm not sure what the arbitrariness of $\\beta$ has to do here. In general, an affine subspace is simply a linear subspace that got shifted from the origin $\\vec{0}$ by a certain, fixed vector $\\vec{f}_0\\,$.\n\nSo you are right, the set $\\{f \\in C(\\mathbb{R},\\mathbb{R})\\,\\vert \\,f(0)=0\\}$ is the candidate for this subspace and you have to show that it is actually a subspace, i.e. closed under addition and scalar multiplication. (There are other multiplications on $C(\\mathbb{R},\\mathbb{R})$ so it's better to name them correctly for not to get confused later on.) You also have to make sure that $\\vec{0}$ is in this set. This guarantees that this set isn't empty.\n\nIn the next step, you have to find this vector $\\vec{f}_0\\,$, which isn't an arbitrary $\\beta$. But you are right as there is more than one possibility in this case. One $\\beta = \\vec{f}_0\\,$ however, will do. What could be a candidate? Chose an easy one, this makes life easier if you want to do anything with this affine space.\n\n4. Jun 14, 2017\n\n### mathwonk\n\ntry translating by the constant function 1 or by the function cosine, or by any function at all with value 1 at 0.\n\n5. Jun 14, 2017\n\n### Erik109\n\nThanks a lot for the fast replies! They are all truly insightful and I surely understand the concept now.\n@fresh_42 Special thanks for the extensive explanation.\n\nThe only thing is that I struggle a bit with proving a given affine subspace of $C(\u211d, \u211d)$ is indeed an affine subspace. I feel like I have to define 'arbitrary' elements to show that $X_1$ is an affine subspace in general. The answer I was provided started off with:\nFix an arbitrary $f \u2208 X_1$. Let $\u03b3: \u211d \u2192 \u211d$ be defined by $\u03b3(x) = f(x) + h(x)$, where $h : \u211d \u2192 \u211d$ (the vector shifting from the origin, you mentioned) is given by $h \u2261 \u22121$ Note that $\u03b3$ is a continuous function with the property $\u03b3(0) = 0$, i.e. $\u03b3 \u2208 X_0$.\n\nSo, if I understand it correctly, the way they prove it is by showing that an arbitrary element from $X_1$ can be shifted back to $X_0$? So kind of 'deconstructing' an affine subspace to a linear subspace?\n\nPS: Got the hang of the LaTeX structure on the forums, hope this helps with the layout.\n\n6. Jun 14, 2017\n\n### WWGD\n\nYes, essentially. You can \" de-translate\" the affine space into a vector space by finding the right amount to translate by.\n\n7. Jun 14, 2017\n\n### Staff: Mentor\n\nEvery affine subspace $A$ can be written $A=V +\\vec{h}$ with a vector (linear) subspace $V$ and the shifting vector $\\vec{h}$.\nYou can observe two things here:\n\u2022 $\\vec{h}$ is not unique. Every other vector $\\vec{h}+\\vec{v}$ with $\\vec{v}\\in V$ does the job as well.\n\u2022 The difference between two elements $\\vec{a}_1 = \\vec{v}_1 + \\vec{h} \\, , \\,\\vec{a}_2 = \\vec{v}_2 + \\vec{h}$ in $A$ is $\\vec{a}_1 -\\vec{a}_2 = (\\vec{v}_1 + \\vec{h})- (\\vec{v}_2 + \\vec{h})= \\vec{v}_1 - \\vec{v}_2$ and always an element of $V$.\nNow in your case, $V= \\{f_V \\in C(\\mathbb{R},\\mathbb{R})\\,\\vert \\,f_V(0)=0\\}$ and $A= \\{f_A \\in C(\\mathbb{R},\\mathbb{R})\\,\\vert \\,f_A(0)=1\\}$, so you need a function $h=\\vec{h} \\in C(\\mathbb{R},\\mathbb{R})$ that moves (shifts) all functions in such a way, that $1=\\vec{f}_A(0) = \\vec{f}_V(0) +\\vec{h}(0) = 0 + h(0)$. @mathwonk mentioned a few candidates for $h$ in post #4. As said above, $\\vec{h}$ isn't unique, only up to elements of $V$.\n\nAffine subspaces are not closed under addition nor under scalar multiplication. However, all differences of two vectors are, as they are vectors in the corresponding linear subspace. This way you still have properties like flatness and can calculate similar to what you can do in vector spaces, but you don't have a zero and must be more careful with addition and scalar multiplication. E.g. all tangent spaces $T_p$ are linear spaces, but if you actually want to use them as tangents, you have to consider $\\vec{p} + T_p$ which is affine. This is very important as usually it's spoken about the vector space $T_p\\,$, e.g. a line $\\mathbb{R} \\cdot \\vec{l}$. And as a line it is a vector space only if it contains the origin $\\vec{0}$. The actual tangent are the points $\\vec{p} + t\\cdot \\vec{l}$. I mention this, as it might be confusing if people talk about linearity of tangent spaces, e.g. when talking about a gradient $\\nabla$, although it is strictly speaking the affine space shifted by $\\vec{p}$.", "date": "2017-11-20 14:31:10", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/affine-subspace.917547/", "openwebmath_score": 0.7737590670585632, "openwebmath_perplexity": 304.0809981388113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.980280868436129, "lm_q2_score": 0.8705972616934408, "lm_q1q2_score": 0.853429839750962}}
{"url": "http://math.stackexchange.com/questions/103569/use-pappus-theorem-to-find-the-moment-of-a-region-limited-by-a-semi-circunferen", "text": "# Use Pappus' theorem to find the moment of a region limited by a semi-circunference.\n\nThis is part of self-study; I found this question in the book \"The Calculus with Analytic Geometry\" (Leithold).\n\n$R$ is the region limited by the semi-circumference $\\sqrt{r^2 - x^2}$ and the x-axis. Use Pappus' theorem to find the moment of $R$ with respect to the line $y = -4$.\n\nPappus' theorem (also referred to as Guldinus theorem or Pappus-Guldinus theorem) is as follows:\n\nIf $R$ is the region limited by the functions $f(x)$ and $g(x)$, then, if $A$ is the area of $R$ and $\\bar{y}$ is the y-coordinate of the centroid of $R$, the volume $V$ of the solid of revolution obtained by rotating $R$ around the x-axis is given by: $V = 2\\pi\\bar{y}A$.\n\nAlso, what I'm calling moment (I'm not sure if this is a common term) is the quantity that is divided by the area of the region in order to find a coordinate of the centroid of the region: for example, to find the x-coordinate of the centroid ($\\bar{x}$), first one finds the moment around the y-axis ($M_y$), then divides it by the area of the region ($A$).\n\nSince the region $R$ is symmetric with respect to the y-axis, the x-coordinate of the centroid ($\\bar{x}$) is zero (therefore, the moment around the y-axis ($M_y$) is zero, because $M_y = \\bar{x}A$, where $A$ is the area of the region). So, I only need to find the vertical coordinate of the centroid (with respect to the line $y = -4$) and the moment around the line $y = -4$.\n\nThe book's answer for the moment around the line $y = -4$ is: $\\frac{1}{2}r^3\\left (\\pi+\\frac{4}{3}\\right )$. I included two attempts below; both arrive at a same result, which is different from the result of the book.\n\nAttempt 1\n\nFirst I tried to use Pappus' theorem to find the vertical coordinate of the centroid of the semi-circular region limited by $\\sqrt{r^2 - x^2}$, with respect to the x-axis (not yet with respect to the line y = -4). I will call it $\\bar{y}_x$.\n\nSince the solid of revolution obtained by rotating this semi-circular region around the x-axis is a sphere, its volume is $V = \\frac{4}{3}\\pi r^3$. The area of the semi-circular region is $A=\\frac{\\pi r^2}{2}$. Substituting $V$ and $A$ into Pappus' theorem:\n\n$\\frac{4}{3}\\pi r^3 = 2\\pi\\bar{y}_x\\frac{\\pi r^2}{2}$\n\n$\\bar{y}_x = \\frac{4r}{3\\pi}$.\n\nThis is the vertical coordinate of the centroid with respect to the x-axis. The vertical coordinate of the centroid with respect to the line $y = -4$ is:\n\n$\\bar{y} = \\frac{4r}{3\\pi} + 4$.\n\nTo find the moment around the line $y = -4$, I use the fact that $\\bar{y} = \\frac{M_x}{A}$:\n\n$M_x = \\bar{y}A = \\left ( \\frac{4r}{3\\pi} + 4\\right )\\frac{\\pi r^2}{2}$.\n\nAttempt 2\n\nThe moment of a plane region with respect to the line $y = -4$ can be found by dividing this region into infinitesimal elements of area, then multiplying the area of each element of area by the vertical coordinate of the centroid.\n\nSo, if $f(x) = \\sqrt{r^2 - x^2}$, then, if I divide the semi-circular region into several rectangles of length $dx$, the area of each rectangle is $f(x) dx$ and the vertical coordinate of the centroid of each rectangle with respect to the line $y = -4$ is $\\frac{f(x)}{2} + 4$. So, the moment with respect to the line $y = -4$ is:\n\n$M_x = \\int_{-r}^r \\left [ f(x)\\times \\left ( \\frac{f(x)}{2} + 4 \\right ) dx \\right ]$.\n\nSolving this integral gives the following result:\n\n$M_x = \\left ( \\frac{4r}{3\\pi} + 4\\right )\\frac{\\pi r^2}{2}$,\n\nwhich is the same result I found in the first attempt.\n\nBoth your attempts lead to the correct result, and the solution given in the book is wrong. To see that the book's solution cannot be correct do the following \"Gedankenexperiment\": Assume that we are told to compute the moment with respect to the line $y=-\\eta$ for some given $\\eta$ instead of $4$. For an $\\eta \\gg r$ this moment would be approximatively proportional to $r^2$ and to $\\eta$, as is the case in your solution but not in the solution given by the book: The $O(r^3)$ dependency is unwarranted.", "date": "2016-05-30 11:00:53", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/103569/use-pappus-theorem-to-find-the-moment-of-a-region-limited-by-a-semi-circunferen", "openwebmath_score": 0.937850832939148, "openwebmath_perplexity": 70.60567376421747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874118017733, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.8534289103311133}}
{"url": "https://mathoverflow.net/questions/220440/non-negative-polynomials-on-0-1-with-small-integral", "text": "# Non-negative polynomials on $[0,1]$ with small integral\n\nLet $P_n$ be the set of degree $n$ polynomials that pass through $(0,1)$ and $(1,1)$ and are non-negative on the interval $[0,1]$ (but may be negative elsewhere).\n\nLet $a_n = \\min_{p\\in P_n} \\int_0^1 p(x)\\,\\mathrm{d}x$ and let $p_n$ be the polynomial that attains this minimum.\n\nAre $a_n$ or $p_n$ known sequences? Is there some clever way to rephrase this question in terms of linear algebra and a known basis of polynomials?\n\nFollowing Robert Israel's answer, we also scale everything to $[-1,1]$ (thus multiplying the result by 2). As he mentions, the optimal polynomial is always a square of some other polynomial, $p_{2n}=p_{2n+1}=q_n^2$, and $q_n$ is either even or odd (see Lemma below). So we are left to find the minimal $L_2[-1,1]$-norm of an odd/even polynomial $q_n$ such that $\\deg q_n\\leq n$ and $q_n(\\pm1)=(\\pm1)^n$. In other words (recall the division by 2 in the first line!), $a_{2n}=a_{2n+1}=d_n^2/2$, where $d_n$ is the distance from $0$ to the hyperplane defined by $q(1)=1$ in the space of all odd/even polynomials of degree at most $n$ with $L_2[-1,1]$-norm.\n\nNow, this hyperplane is the affine hull of the Legendre polynomials $P_i(x)=\\frac1{2^ii!}\\frac{d^i}{dx^i}(x^2-1)^i$, where $i$ has the same parity as $n$ (since we know that $P_i(1)=1$ and $P_i(-1)=(-1)^i$). Next, by $\\| P_i\\|^2=\\frac{2}{2i+1}$, our distance is $$\\left(\\sum_{j\\leq n/2}\\| P_{n-2j}\\|^{-2}\\right)^{-1/2} =\\left(\\sum_{j\\leq n/2}\\frac{2(n-2j)+1}{2}\\right)^{-1/2}=\\sqrt{\\frac4{(n+1)(n+2)}},$$ attained at $$q_n(x)=\\left(\\sum_{i\\leq n/2}\\frac{2}{2(n-2i)+1}\\right)^{-1} \\sum_{i\\leq n/2}\\frac{2P_{n-2i}(x)}{2(n-2i)+1}.$$ Thus the answer for the initial question is $a_{2n}=a_{2n+1}=\\frac2{(n+1)(n+2)}$ and $p_{2n}=p_{2n+1}=q_n^2$.\n\nLemma. For every $n$, one of optimal polynomials on $[-1,1]$ is a square of an odd or even polynomial.\n\nProof. Let $r(x)$ be any polynomial.which is nonnegative on $[-1,1]$ with $r(\\pm 1)=1$. We will replace it by some other polynomial which has the required form, has the same (or less) degree and the same values at $\\pm 1$, is also nonnegative on $[-1,1]$, and is not worse in the integral sense. Due to the compactness argument, this yields the required result.\n\nFirstly, replacing $r(x)$ by $\\frac12(r(x)+r(-x))$, we may assume that $r$ is even (and thus has an even degree $2n$). Let $\\pm c_1,\\dots,\\pm c_{n}$ be all complex roots of $r$ (regarding multiplicities); then $$r(x)=\\prod_{j=1}^n\\frac{x^2-c_j^2}{1-c_j^2},$$ due to $r(\\pm 1)=1$.\n\nNow, for all $c_j\\notin[-1,1]$ we simultaneously perform the following procedure.\n\n(a) If $c_j$ is real, then we replace $\\pm c_j$ by $\\pm x_j=0$. Notice that $$\\frac{|x^2-c_j^2|}{|1-c_j^2|}\\geq 1\\geq \\frac{|x^2-0^2|}{|1-0^2|}.$$ for all $x\\in[-1,1]$.\n\n(b) If $c_j$ is non-real, then we choose $x_j\\in[-1,1]$ such that $\\frac{|c_j-1|}{|c_j+1|}=\\frac{|x_j-1|}{|x_j+1|}$. Notice that all complex $z$ with $\\frac{|c_j-z|}{|x_j-z|}=\\frac{|c_j-1|}{|x_j-1|}$ form a circle passing through $-1$ and $1$, and the segment $[-1,1]$ is inside this circle. Therefore, for every $x\\in[-1,1]$ we have $$\\frac{|c_j-x|}{|x_j-x|}\\geq\\frac{|c_j-1|}{|x_j-1|},$$ thus $$\\frac{|x^2-c_j^2|}{|1-c_j^2|} =\\frac{|c_j-x|}{|c_j-1|}\\cdot\\frac{|c_j+x|}{|c_j+1|} \\geq \\frac{|x_j-x|}{|x_j-1|}\\cdot\\frac{|x_j+x|}{|x_j+1|} =\\frac{|x^2-x_j^2|}{|1-x_j^2|}.$$ So, we replace $\\pm c_j$ and $\\pm\\bar c_j$ by $\\pm x_j$ and $\\pm x_j$ (or simply $\\pm c_j$ by $\\pm x_j$ if $c_j$ is purely imaginary).\n\nAfter this procedure has been applied, we obtain a new polynomial whose roots are in $[-1,1]$ and have even multiplicities, and its values at $\\pm1$ are equal to $1$. So it is a square of some polynomial which is even/odd (since the roots are still split into pairs of opposite numbers). On the other hand, its values at every $x\\in[-1,1]$ do not exceed the values of $r$ at the same points, as was showed above. So the obtained polynomial is not worse in the integral sense, as required. The lemma is proved.\n\n\u2022 Sorry to be dense, but please say why it has to be the square of another polynomial. I can see that any zeros in the critical interval have even multiplicity, but why can't the polynomial be negative elsewhere (as allowed by OP)? Oct 9 '15 at 22:54\n\u2022 The last part of the answer (after `finally') is about it; I'll try to expand it. In short: If our polynomial has a non-real root, or if it has real roots outside $[-1,1]$, then we may replace these roots by some root on $[-1,1]$ so as to lower the values of the polynomial at all points in $(-1,1)$(and keep it nonnegative). Oct 10 '15 at 7:05\n\u2022 Expanded (or, I would say, rewritten). Oct 10 '15 at 7:47\n\nFor better symmetry, let me change the boundary points to $(-1,1)$ and $(1,1)$. Then it is clear that we may assume $p_n$ is an even function (and thus you really want $P_n$ to be polynomials of degree at most $n$, otherwise you'll be likely to have an infimum rather than an actual minimum): $p_{2k+1} = p_{2k}$. I get \\eqalign{p_0(x) = 1, & a_0 = 2\\cr p_2(x) = x^2, & a_2 = \\dfrac{2}{3}\\cr p_4(x) = \\dfrac{(5 x^2 - 1)^2}{16},& a_4 = \\dfrac{1}{3}\\cr p_6(x) = \\dfrac{x^2}{16} (7 x^2 - 3)^2,& a_6 = \\dfrac{1}{5}\\cr p_8(x) = \\dfrac{1}{64} (21 x^4 - 14 x^2 + 1)^2, & a_8 = \\dfrac{2}{15}\\cr p_{10}(x) = \\dfrac{x^2}{64} (33 x^4 - 30 x^2 + 5)^2, & a_{10} = \\dfrac{2}{21}\\cr p_{12}(x) = \\dfrac{1}{4096} (429 x^6 - 495 x^4 + 135 x^2 - 5)^2, & a_{12} = \\dfrac{1}{14}} Well, so far it looks like $a_{2n} = \\dfrac{4}{(n+1)(n+2)}$\n\n\u2022 How were these computed? Oct 9 '15 at 0:03\n\u2022 The leading coefficients look like Catalan numbers when the polynomials are normalized to $\\sqrt{2^{2n} p_{2n}(x)}$. Oct 9 '15 at 1:51\n\u2022 The next coefficients match $-{2n \\choose n-2}$. The whole triangle looks like oeis.org/A234950 read skewed, with alternating signs. See oeis.org/A062991 for a signed version. Oct 9 '15 at 3:43\n\u2022 For $n$ even, $$p_{2n} = \\prod_{i=1}^{n/2} \\dfrac{(x^2 -c_i^2)^2}{(1-c_i^2)^2}$$ For $n$ odd, there's also a factor of $x^2$. Differentiate $a_{2n} =\\int_{-1}^1 p_{2n}(x)\\; dx$ with respect to each $c_i$, solve, check which solution gives the least $a_{2n}$. Oct 9 '15 at 4:49", "date": "2021-11-28 23:38:04", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/220440/non-negative-polynomials-on-0-1-with-small-integral", "openwebmath_score": 0.9412352442741394, "openwebmath_perplexity": 194.62170143253013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631635159684, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.8534251700966891}}
{"url": "https://byjus.com/question-answer/assertion-the-chord-of-contact-of-tangent-from-a-point-p-to-a-circle-passes/", "text": "Question\n\n# Assertion :The chord of contact of tangent from a point $$P$$ to a circle passes through $$Q$$. If $${ l }_{ 1 }$$ and $${ l }_{ 2 }$$ are the lengths of the tangents from $$P$$ and $$Q$$ to the circle, then $$PQ$$ is equal to $$\\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$ Reason: The equation of chord of contact of tangents from the point $$P\\left( { x }_{ 1 },{ y }_{ 1 } \\right)$$ to the circle $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$ is $$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$\n\nA\nAssertion is true, Reason is true and reason is correct explanation for Statement-1.\nB\nAssertion is true, Reason is true and Reason is NOT correct explanation for Assertion.\nC\nAssertion is true, Reason is false\nD\nAssertion is false, Reason is true\n\nSolution\n\n## The correct option is A Assertion is true, Reason is true and reason is correct explanation for Statement-1.Let $$P\\equiv \\left( { x }_{ 1 },{ y }_{ 1 } \\right)$$ and $$Q\\equiv \\left( { x }_{ 2 },{ y }_{ 2 } \\right)$$Let the equation of the given circle be $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$The equation of chord of contact of tangents from the point $$P\\left( { x }_{ 1 },{ y }_{ 1 } \\right)$$ to the given circle is\u00a0$$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$Since it passes through $$Q\\left( { x }_{ 2 },{ y }_{ 2 } \\right)$$$$\\therefore { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 }={ a }^{ 2 }$$ \u00a0 \u00a0...(1)Now $${ l }_{ 1 }=\\sqrt { { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } } ,{ l }_{ 2 }=\\sqrt { { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } }$$and $$PQ=\\sqrt { { \\left( { x }_{ 2 }-{ x }_{ 1 } \\right) }^{ 2 }+{ \\left( { y }_{ 2 }-{ y }_{ 1 } \\right) }^{ 2 } }$$$$=\\sqrt { \\left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \\right) +\\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \\right) -2\\left( { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 } \\right) } \\\\ =\\sqrt { \\left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \\right) +\\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \\right) -2{ a }^{ 2 } } \\\\ =\\sqrt { \\left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } \\right) +\\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } \\right) } =\\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-19 13:47:01", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/assertion-the-chord-of-contact-of-tangent-from-a-point-p-to-a-circle-passes/", "openwebmath_score": 0.20279885828495026, "openwebmath_perplexity": 377.4990936403568, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631661216054, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8534251689232513}}
{"url": "https://math.stackexchange.com/questions/3198820/find-the-smallest-n-in-mathbbn-such-that-the-group-is-isomorphic-to-the-di", "text": "# Find the smallest $n \\in \\mathbb{N}$ such that the group is isomorphic to the direct product of $n$ cyclic groups\n\nFind the smallest $$n \\in \\mathbb{N}$$ such that the group $$\\mathbb{Z}_{6} \\times \\mathbb{Z}_{20} \\times \\mathbb{Z}_{45}$$ is isomorphic to the direct product of $$n$$ cyclic groups.\n\nI'm not sure but if I understand correctly,\n\n$$\\mathbb{Z}_{6} \\times \\mathbb{Z}_{20} \\times \\mathbb{Z}_{45}$$ is isomorphic to $$\\mathbb{Z}_{2} \\times \\mathbb{Z}_{3} \\times \\mathbb{Z}_{4} \\times \\mathbb{Z}_{5} \\times \\mathbb{Z}_{5} \\times \\mathbb{Z}_{9}$$,\n\n$$\\mathbb{Z}_{2} \\times \\mathbb{Z}_{5} \\times \\mathbb{Z}_{9}$$ is isomorphic to $$\\mathbb{Z}_{90}$$,\n\n$$\\mathbb{Z}_{3} \\times \\mathbb{Z}_{4} \\times \\mathbb{Z}_{5}$$ is isomorphic to $$\\mathbb{Z}_{60}$$,\n\nand therefore, $$\\mathbb{Z}_{6} \\times \\mathbb{Z}_{20} \\times \\mathbb{Z}_{45}$$ is isomorphic to $$\\mathbb{Z}_{60} \\times \\mathbb{Z}_{90}$$ and the answer is $$n = 2$$. Is this a correct solution?\n\n\u2022 I agree that the answer is $n=2$. For the sake of being thorough, I would probably explain why $n=1$ is impossible (just because you've found one situation where $n=2$ doesn't necessarily mean it's the smallest $n$). \u2013\u00a0Theo C. Apr 23 at 21:21\n\u2022 Yes, it is. You just have to find the smallest number of groups of moduli such that the moduli in each group are pairwise coprime. \u2013\u00a0Bernard Apr 23 at 21:24\n\u2022 After you ask a question here, if you get an acceptable answer, you should \"accept\" the answer by clicking the check mark $\\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. \u2013\u00a0Shaun May 24 at 14:27\n\nYou may find the invariant factors.\n\nActually $$\\mathbb{Z}_6\\times \\mathbb{Z}_{20}\\times \\mathbb{Z}_{45}\\cong (\\mathbb{Z}_2\\times\\mathbb{Z}_4)\\times(\\mathbb{Z}_3\\times\\mathbb{Z}_9)\\times(\\mathbb{Z}_5\\times\\mathbb{Z}_5)$$. Now we can pick the largest ones in each bracket to form $$\\mathbb{Z}_4\\times\\mathbb{Z}_9\\times\\mathbb{Z}_5\\cong\\mathbb{Z}_{180}$$. Then we pick the largest ones of the remaining, which is $$\\mathbb{Z}_2\\times\\mathbb{Z}_3\\times\\mathbb{Z}_5\\cong\\mathbb{Z}_{30}$$, and nothing remain. Hence the group is isomorphic to $$\\mathbb{Z}_{180}\\times\\mathbb{Z}_{30}$$ and so $$n\\le 2$$. Since the group is obviously not cyclic, we have $$n = 2$$.\n\nI think we can always get $$n$$ for any abelian group by finding the invariant factors.\n\nAccording to GAP, the group is $$\\Bbb Z_{180}\\times \\Bbb Z_{30}$$ and, of course, not cyclic.\n\ngap> G:=DirectProduct(CyclicGroup(6) , DirectProduct(CyclicGroup(20), CyclicGroup(45)));\n<pc group of size 5400 with 8 generators>\ngap> StructureDescription(G);\n\"C180 x C30\"\ngap> IsCyclic(G);\nfalse\ngap>", "date": "2019-06-27 12:31:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3198820/find-the-smallest-n-in-mathbbn-such-that-the-group-is-isomorphic-to-the-di", "openwebmath_score": 0.8640196919441223, "openwebmath_perplexity": 202.56521248017205, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631651194372, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8534251646282537}}
{"url": "http://cybercratsystems.com/how-does-nvcvpmn/q6z6m01.php?page=greater-than-or-equal-to-sign-4b7ae6", "text": "Greater Than or Equal To: Math Definition. use \">=\" for greater than or equal use \"<=\" for less than or equal In general, Sheets uses the same \"language\" as Excel, so you can look up Excel tips for Sheets. Graphical characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has no crossing lines. In such cases, we can use the greater than or equal to symbol, i.e. \u2265. Solution for 1. In an acidic solution [H]\u2026 But, when we say 'at least', we mean 'greater than or equal to'. Greater than or equal application to numbers: Syntax of Greater than or Equal is A>=B, where A and B are numeric or Text values. Rate this symbol: (3.80 / 5 votes) Specifies that one value is greater than, or equal to, another value. Select Symbol and then More Symbols. Examples: 5 \u2265 4. The sql Greater Than or Equal To operator is used to check whether the left-hand operator is higher than or equal to the right-hand operator or not. 923 Views. Greater than or Equal in Excel \u2013 Example #5. is less than > > is greater than \u226e \\nless: is not less than \u226f \\ngtr: is not greater than \u2264 \\leq: is less than or equal to \u2265 \\geq: is greater than or equal to \u2a7d \\leqslant: is less than or equal to \u2a7e With Microsoft Word, inserting a greater than or equal to sign into your Word document can be as simple as pressing the Equal keyboard key or the Greater Than keyboard key, but there is also a way to insert these characters as actual equations. Select the Greater-than Or Equal To tab in the Symbol window. If left-hand operator higher than or equal to right-hand operator then condition will be true and it will return matched records. This symbol is nothing but the \"greater than\" symbol with a sleeping line under it. \u201cGreater than or equal to\u201d and \u201cless than or equal to\u201d are just the applicable symbol with half an equal sign under it. The less than or equal to symbol is used to express the relationship between two quantities or as a boolean logical operator. Here a could be greater \u2026 Category: Mathematical Symbols. The greater-than sign is a mathematical symbol that denotes an inequality between two values. When we say 'as many as' or 'no more than', we mean 'less than or equal to' which means that a could be less than b or equal to b. 2 \u2265 2. Less Than or Equal To (<=) Operator. Finding specific symbols in countless symbols is obviously a waste of time, and some characters like emoji usually can't be found. Copy the Greater-than Or Equal To in the above table (it can be automatically copied with a mouse click) and paste it in word, Or. Sometimes we may observe scenarios where the result obtained by solving an expression for a variable, which are greater than or equal to each other. As we saw earlier, the greater than and less than symbols can also be combined with the equal sign. Select the Insert tab. For example, the symbol is used below to express the less-than-or-equal relationship between two variables: For example, 4 or 3 \u2265 1 shows us a greater sign over half an equal sign, meaning that 4 or 3 are greater than or equal to 1. \"Greater than or equal to\" is represented by the symbol \" \u2265 \u2265 \". For example, x \u2265 -3 is the solution of a certain expression in variable x. In Greater than or equal operator A value compares with B value it will return true in two cases one is when A greater than B and another is when A equal to B. 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Symbol that denotes an inequality between two values symbol window graphical characteristics: Asymmetric, shape... 3.80 / 5 votes ) Specifies that one value is greater than '' symbol with a sleeping line under.!\n\nColumbus, Kansas Funeral Homes, Ludo Rules For 3 Sixes, Rice Quotes And Sayings, Farm House In Coimbatore Olx, Memorial Hospital Savannah, Ga, Medchal-malkajgiri Collector Website, Craigslist Kennesaw Ga Furniture,", "date": "2021-08-05 18:37:01", "meta": {"domain": "cybercratsystems.com", "url": "http://cybercratsystems.com/how-does-nvcvpmn/q6z6m01.php?page=greater-than-or-equal-to-sign-4b7ae6", "openwebmath_score": 0.5627760291099548, "openwebmath_perplexity": 1360.4896164358602, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9433475810629193, "lm_q2_score": 0.9046505440982948, "lm_q1q2_score": 0.8533999024823802}}
{"url": "https://gmatclub.com/forum/quick-tips-for-adding-numbers-x-to-y-79541.html", "text": "It is currently 20 Mar 2018, 05:04\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# Quick tips for adding numbers x to y\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 25 May 2009\nPosts: 141\nConcentration: Finance\nGMAT Date: 12-16-2011\nQuick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n10 Jun 2009, 20:46\n2\nThis post was\nBOOKMARKED\nHere are some examples on adding numbers from x to y:\n\n1) Add the numbers from 40 to 70, inclusive.\n2) Add the even numbers from 40 to 70, inclusive.\n3) Add the odd numbers from 40 to 70, inclusive.\n\n4) Add the numbers from 40 to 70.\n5) Add the even numbers from 40 to 70.\n6) Add the odd numbers from 40 to 70.\n\n[Reveal] Spoiler:\n1) 1,705\n2) 880\n3) 825\n4) 1,595\n5) 770\n6) 825\n\nLast edited by I3igDmsu on 05 Jul 2009, 19:59, edited 7 times in total.\nManager\nJoined: 28 Jan 2004\nPosts: 201\nLocation: India\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n10 Jun 2009, 22:58\n1) Add the numbers from 40 to 70, inclusive.\nFormula for sum of first N natural numbers is = N(N+1)/2\nSum from 1 to 70 Sa= 70*71/2\nSum from 1 to 40 Sb= 40*41/2\nSo sum from 40 to 70 = Sa-Sb\n\n2) Add the even numbers from 40 to 70, inclusive.\n3) Add the odd numbers from 40 to 70, inclusive.\n\n========================\n\nAs a general solution to all these kind of problems learn the AP series.\ngoogle on Arithemic Progression series. It is kind of difficult to write the formula here but all these calculations are tooooooo simple using this series.\nCurrent Student\nJoined: 03 Aug 2006\nPosts: 112\nLocation: Next to Google\nSchools: Haas School of Business\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n12 Jun 2009, 14:34\n6\nKUDOS\n3\nThis post was\nBOOKMARKED\nIf you don't mind remembering a formula or two then yes you can.\nThe examples that you have given can be grouped under Arithmetic Progressions...or a finite sequence of evenly spaced numbers.\n\nThere are two formulas:\n\n$$1. S = \\frac{n}{2} [2a + (n-1)d]$$\n\nwhere\n$$S=\\text{sum of the all the numbers in the sequence}$$\n$$n=\\text{total number of numbers in the sequence}$$\n$$a=\\text{the first number of the sequence}$$\n$$d=\\text{the different between any two consecutive numbers in the sequence}$$\n\n$$2. S = \\frac{n}{2} [\\text{First Term}+\\text{Last Term}]$$\n\nwhere\n$$S=\\text{sum of the all the numbers in the sequence}$$\n$$n=\\text{total number of numbers in the sequence}$$\n\nYou can use either equation based on what is provided in the question.\n\nLets take your examples and solve them. For all of these we know the first and the last number so we should be fine with equation 2.\n\n1) Add the numbers from 40 to 70, inclusive.\nSolution:\nHere the sequence is 40,41, 42, ...., 69, 70.\nTo calculate n\n$$n=\\text{Last number}-\\text{First number} + 1$$\n\n$$n=70-40+1 = 31$$\n\nUsing equation 1.\n\n$$S = \\frac{n}{2} [\\text{First Term}+\\text{Last Term}]$$\n\n$$S = \\frac{n}{2} [\\text{First Term}+\\text{Last Term}]$$\n\n$$S = \\frac{31}{2} [40+70]$$\n\n$$S = \\frac{31}{2} [110]$$\n\n$$S = 31\\times55$$\n\n$$S = 1705$$\n\n2) Add the even numbers from 40 to 70, inclusive.\nHere the sequence is 40,42, 44, ...., 68, 70.\nSolve using 40 as the first term and 70 as the last term\nTo calculate n for even (or odd) number\n\n$$n=\\frac{\\text{Last number}-\\text{First number}}{2} + 1$$\n\n3) Add the odd numbers from 40 to 70, inclusive.\nHere is the sequence is 41,43,....67,69\nSolve using 41 as the first term and 69 as the last term\n\nand so on...\nManager\nJoined: 25 May 2009\nPosts: 141\nConcentration: Finance\nGMAT Date: 12-16-2011\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n30 Jun 2009, 20:11\nCould someone check my answers above please?\n\nThanks nookway, your help is appreciated!\nFounder\nJoined: 04 Dec 2002\nPosts: 16588\nLocation: United States (WA)\nGMAT 1: 750 Q49 V42\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n03 Jul 2009, 07:51\n1\nKUDOS\nExpert's post\nI3igDmsu wrote:\nCould someone check my answers above please?\n\nThanks nookway, your help is appreciated!\n\nDon't think you need our help - this can be easily done in Excel\n_________________\n\nFounder of GMAT Club\n\nJust starting out with GMAT? Start here... or use our Daily Study Plan\n\nCo-author of the GMAT Club tests\n\nManager\nJoined: 25 May 2009\nPosts: 141\nConcentration: Finance\nGMAT Date: 12-16-2011\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n05 Jul 2009, 20:00\nDone. If anyone needs practice, the answers in the spoiler are correct.\nManager\nJoined: 13 Jan 2009\nPosts: 168\nSchools: Harvard Business School, Stanford\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n02 Aug 2009, 21:30\nGreat stuff! Thanks!\nManager\nJoined: 06 Feb 2011\nPosts: 69\nWE: Information Technology (Computer Software)\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n13 Oct 2011, 14:32\nI did not get the logic of these.\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 6536\nRe: Quick tips for adding numbers x to y\u00a0[#permalink]\n\n### Show Tags\n\n13 Sep 2017, 22:23\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: Quick tips for adding numbers x to y \u00a0 [#permalink] 13 Sep 2017, 22:23\nDisplay posts from previous: Sort by\n\n# Quick tips for adding numbers x to y\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-03-20 12:04:25", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/quick-tips-for-adding-numbers-x-to-y-79541.html", "openwebmath_score": 0.3081875443458557, "openwebmath_perplexity": 5282.883415507041, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9462003595741247, "lm_q2_score": 0.90192067652954, "lm_q1q2_score": 0.8533976684395886}}
{"url": "http://mathhelpforum.com/statistics/208657-combinations-probability-question-please-help-print.html", "text": "\u2022 Nov 28th 2012, 06:08 PM\nwarElephant\nBill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?\n8/33\n62/165\n17/33\n103/165\n25/33\n\nthis is my solution but it is wrong:\n\nan Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A\nthe number of patterns that A,B,C,A hand appears = 4C2\nTherefore, number of ways A is chosen= 12\nwhen the first A is chosen, it locks in the value for the second A\nnumber of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list)\nnumber of ways C is chosen = 9\nP(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33\n\nI hope that your explaination you will elaborate:\n1. why my method is wrong\n2. the actual solution.\nThank you very much for even considering my post!!!:)\n\u2022 Nov 28th 2012, 09:49 PM\nScopur\nP(at least 1) = 1 - P(None). This is probably more useful here.\nSo there are $6C4$ possible ways to select 4 distinct cards. Then there are $2^4$ different suits for those cards. Finally there are $12C6$ different cards.\nSo you have\n$P(at least one) = 1 - P(none) = 1 - \\frac{6C4*2^4}{12C6} = 1 - \\frac{16}{33} = \\frac{17}{33}$\n\u2022 Nov 28th 2012, 10:15 PM\nSoroban\nHello, warElephant!\n\nAnother approach . . .\n\nQuote:\n\nBill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each.\nEach of the 6 cards within a suit has a different value from 1 to 6.\nBill likes to play a game in which he shuffles the deck, turns over 4 cards,\nand looks for pairs of cards that have the same value.\nWhat is the probability that Bill finds at least one pair of cards that have the same value?\n\n. . $(a)\\;\\tfrac{8}{33} \\qquad(b)\\;\\tfrac{62}{165} \\qquad (c)\\;\\tfrac{17}{33}\\qquad (d)\\;\\tfrac{103}{165} \\qquad (e)\\;\\tfrac{25}{33}$\n\nWe will find the probability that there are no matching pairs among the four cards.\n\nThe first card can be any card (it doesn't matter): $\\tfrac{12}{12}$\n\nThe second card can be any of the other 10 non-matching cards: $\\tfrac{10}{11}$\n\nThe third card can be any of the other 8 non-matching cards: $\\tfrac{8}{10}$\n\nThe fourth card can be any of the other 6 non-matching cards: $\\ftrac{6}{9}$\n\nHence: . $P(\\text{no match}) \\;=\\;\\frac{12}{12}\\cdot\\frac{10}{11}\\cdot\\frac{8}{ 10}\\cdot\\frac{6}{9} \\:=\\:\\frac{16}{33}$\n\nTherefore: . $P(\\text{at least one match}) \\;=\\;1-\\frac{16}{33} \\;=\\;\\frac{17}{33}$", "date": "2017-11-18 06:57:18", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/208657-combinations-probability-question-please-help-print.html", "openwebmath_score": 0.5005221366882324, "openwebmath_perplexity": 384.3288996185102, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615795, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8533973206330723}}
{"url": "https://physics.stackexchange.com/questions/442648/electric-field-strength-away-from-a-negative-spherical-charge", "text": "# Electric field strength away from a negative spherical charge\n\nI was wonder how would a graph of electrical field strength away from a spherical -ve charge graph would look. Since E= potential gradient I was guessing that since potential increases away from a negative charge , electric field strength would increase away from a negative charge also ?\n\nThe potential does increase. This means that the gradient of potential plotted against distance, r, from the charge is always positive. But the gradient, $$\\frac{dV}{dr},$$ keeps decreasing in magnitude \u2013 just sketch the graph! The field strength in the r direction is given by$$E=\\ \u2013\\frac{dV}{dr},$$so the field is in the \u2013r direction and decreases in magnitude the further we go from the negative charge.\n\u2022 Yes, but as there is radial symmetry, it would be clearer to put the centre of the sphere at $r$ = 0 and to have the $r$ axis in the positive direction only (as it represents the distance from the sphere in any direction). We'd also usually have the graph upside down, reflected about the $r$ axis, the negativity of $E$ showing that the field is directed in the \u2013$r$ direction, that is towards the sphere. \u2013\u00a0Philip Wood Nov 23 '18 at 0:16", "date": "2019-12-16 10:03:57", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/442648/electric-field-strength-away-from-a-negative-spherical-charge", "openwebmath_score": 0.6936569213867188, "openwebmath_perplexity": 355.5844933861102, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9744347890464284, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8533973203539154}}
{"url": "https://www.physicsforums.com/threads/kinetic-energy-question.727741/", "text": "# Kinetic energy question\n\n1. Dec 10, 2013\n\n### hey123a\n\n1. The problem statement, all variables and given/known data\nA sled of mass m is coasting on the icy surface of a frozen river. While it is passing under a bridge, a package of equal mass m is dropped straight down and lands on the sled (without causing any damage). The sled plus the added load then continue along the original line of motion. How does the kinetic energy of the (sled + load) compare with the original kinetic energy of the sled? A) It is 1/4 the original kinetic energy of the sled. B) It is 1/2 the original kinetic energy of the sled. C) It is 3/4 the original kinetic energy of the sled. D) It is the same as the original kinetic energy of the sled. E) It is twice the original kinetic energy of the sled.\n\n2. Relevant equations\n\n3. The attempt at a solution\nkinetic energy intial = 1/2mv^2\nkinetic energy after = 1/2(2m)v^2 = mv^2\n\ncomparing the kinetic energy before and after, i get the after is twice the original kinetic energy of the sled but apparently this is wrong\n\n2. Dec 10, 2013\n\n### Simon Bridge\n\nYou forgot to conserve momentum.\n\n3. Dec 10, 2013\n\n### rock.freak667\n\nYou will need to use conservation of linear momentum to calculate the velocity of the sled+load after the impact as your attempt shows them traveling at the same velocity.\n\nIf before the impact, the sled has momentum mu and after it travels with the velocity 'v' and the load is initially at rest, then using conservation of linear momentum what is v in terms of u?\n\n4. Dec 10, 2013\n\n### hey123a\n\nah okay i think i got it\nm1v1o + m2v2o = (m1+m2)vf\nwhere m1 is the sled, and m2 a mass that is equal to the sled\nm1v1o = (m1+m2)vf\nm1v1o = 2mvf\nm1v1o/2m = vf\nv1o/2 = vf\n\noriginal ke = 1/2mv^2\nKE of sled + load = 1/2(2m)(v1o/2)^2\nke of sled + load = 1/2(2m)(v^2/4 = 1/4mv^2, which is 1/2 the original kinetic energy of the sled\n\n5. Dec 11, 2013\n\n### Simon Bridge\n\nWell done.\n\nIt's easier to check your answers if you get formal with the working.\nfor instance, if I write:\n\nbefore:\nmomentum: $p_i=mu$\nkinetic energy: $K_i=\\frac{1}{2}mu^2$\n\nafter:\nmomentum: $p_f=2mv$\nkinetic energy: $K_f = mv^2$\n\nconservation of momentum:\n$p_f=p_i\\implies 2mv=mu \\implies v=u/2\\\\ \\qquad \\implies K_f=\\frac{1}{4}mu^2$\n\nComparing KE \"after\" with KE \"before\":\n$$\\frac{K_f}{K_i}=\\frac{\\frac{1}{4}mu^2}{\\frac{1}{2}mu^2}=\\frac{1}{2} \\\\ \\qquad\\implies K_f=\\frac{1}{2}K_i$$\n\nYou can see all the steps and most of the reasoning.\nGood for long answers or when you are practicing (or writing questions here ;) ) - but probably more work than you'd like for a simple multi-choice.\n\nStill... no worries aye?", "date": "2017-10-22 10:45:00", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/kinetic-energy-question.727741/", "openwebmath_score": 0.5071092844009399, "openwebmath_perplexity": 1554.7669742616115, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347886752162, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8533973200288125}}
{"url": "http://sindforte.hospedagemdesites.ws/fashion-sales-heqzp/5e530c-point-of-inflection-second-derivative", "text": "Therefore, our inflection point is at x = 2. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. For a maximum point the 2nd derivative is negative, and the minimum point is positive. I've some data about copper foil that are lists of points of potential(X) and current (Y) in excel . For instance, if we were driving down the road, the slope of the function representing our distance with respect to time would be our speed. For example, the second derivative of the function y = 17 is always zero, but the graph of this function is just a horizontal line, which never changes concavity. Therefore possible inflection points occur at and .However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. The curve I am using is just representative. I like thinking of a point of inflection not as a geometric feature of the graph, but as a moment when the acceleration changes. The Second Derivative Test cautions us that this may be the case since at f 00 (0) = 0 at x = 0. The sign of the derivative tells us whether the curve is concave downward or concave upward. MENU MENU. And a list of possible inflection points will be those points where the second derivative is zero or doesn't exist. (d) Identify the absolute minimum and maximum values of f on the interval [-2,4]. 8.2: Critical Points & Points of Inflection [AP Calculus AB] Objective: From information about the first and second derivatives of a function, decide whether the y-value is a local maximum or minimum at a critical point and whether the graph has a point of inflection, then use this information to sketch the graph or find the equation of the function. To find inflection points, start by differentiating your function to find the derivatives. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Our website is made possible by displaying online advertisements to our visitors. Necessary Condition for an Inflection Point (Second Derivative Test) If $${x_0}$$ is a point of inflection of the function $$f\\left( x \\right)$$, and this function has a second derivative in some neighborhood of $${x_0},$$ which is continuous at the point $${x_0}$$ itself, then The purpose is to draw curves and find the inflection points of them..After finding the inflection points, the value of potential that can be used to \u2026 However, f \"(x) is positive on both sides of x = 0, so the concavity of f is the same to the left and to the right of x = 0. A positive second derivative means that section is concave up, while a negative second derivative means concave down. How to Calculate Degrees of Unsaturation. A point of inflection does not have to be a stationary point however; A point of inflection is any point at which a curve changes from being convex to being concave . You \u2026 In the case of the graph above, we can see that the graph is concave down to the left of the inflection point and concave down to the right of the infection point. find f \"; find all x-values where f \" is zero or undefined, and Save my name, email, and website in this browser for the next time I comment. Then the function achieves a global maximum at x 0: f(x) \u2264 f(x 0)for all x \u2208 &Ropf.. The second derivative of a function may also be used to determine the general shape of its graph on selected intervals. (c) Use the second derivative test to locate the points of inflection, and compare your answers with part (b). When the second derivative is negative, the function is concave downward. However, (0, 0) is a point of inflection. If it does, the value at x is an inflection point. One method of finding a function\u2019s inflection point is to take its second derivative, set it equal to zero, and solve for x. Second Derivatives: Finding Inflation Points of the Function. The first derivative is f\u2032(x)=3x2\u221212x+9, sothesecondderivativeisf\u2033(x)=6x\u221212. Stationary Points. The second derivative of a function may also be used to determine the general shape of its graph on selected intervals. The next graph shows x 3 \u2013 3x 2 + (x \u2013 2) (red) and the graph of the second derivative of the graph, f\u201d = 6(x \u2013 1) in green. 4. View Point of inflection from MATH MISC at Manipal Institute of Technology. On the right side of the inflection point, the graph increases faster and faster. Inflection point is a point on the function where the sign of second derivative changes (where concavity changes). Applying derivatives to analyze functions, Determining concavity of intervals and finding points of inflection: algebraic. Recall the graph f (x) = x 3. Learn how the second derivative of a function is used in order to find the function's inflection points. So the second derivative must equal zero to be an inflection point. 10 years ago. To locate a possible inflection point, set the second derivative equal to zero, and solve the equation. For instance if the curve looked like a hill, the inflection point will be where it will start to look like U. There are two issues of numerical nature with your code: the data does not seem to be continuous enough to rely on the second derivative computed from two subsequent np.diff() applications; even if it were, the chances of it being exactly 0 are very slim; To address the first point, you should smooth your histogram (e.g. 2. Find all inflection points for the function f (x) = x 4.. Mathematics Learning Centre, University of Sydney 1 The second derivative The second derivative, d2y dx2,ofthe function y = f(x)isthe derivative of dy dx. An inflection point occurs on half profile of M type or W type, two inflection points occur on full profiles of M type or W type. To locate the inflection point, we need to track the concavity of the function using a second derivative number line. A critical point becomes the inflection point if the function changes concavity at that point. , Sal means that there is an inflection point, not at where the second derivative is zero, but at where the second derivative is undefined. But don't get excited yet. As we saw on the previous page, if a local maximum or minimum occurs at a point then the derivative is zero (the slope of the function is zero or horizontal). ACT Preparation We observed that x = 0, and that there was neither a maximum nor minimum. Mistakes when finding inflection points: second derivative undefined, Mistakes when finding inflection points: not checking candidates, Analyzing the second derivative to find inflection points, Using the second derivative test to find extrema. The points of inflection of a function are those at which its second derivative is equal to 0. Therefore, our inflection point is at x = 2. State the second derivative test for local extrema. These points can be found by using the first derivative test to find all points where the derivative is zero, then using the second derivative test to see if any points are also turning points. So the second derivative must equal zero to be an inflection point. On the right side of the inflection point, the graph increases faster and faster. We can use the second derivative to find such points \u2026 Factoring, we get e^x(4*e^x - 1) = 0. The section of curve between A and B is concave down \u2014 like an upside-down spoon or a frown; the sections on the outsides of A and B are concave up \u2014 like a right-side up spoon or a smile; and A and B are inflection points. Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. I am mainly looking for the list of vertices that precede inflection points in a curve. Our mission is to provide a free, world-class education to anyone, anywhere. When we simplify our second derivative we get; This means that f(x) is concave downward up to x = 2 f(x) is concave upward from x = 2. Mathematics Learning Centre The second derivative and points of in\ufb02ection Jackie Nicholas c 2004 University of d2y /dx2 = (+)2 hence it is a minimum point. Lets take a curve with the following function. using a uniform or Gaussian filter on the histogram itself). When the second derivative is positive, the function is concave upward. 2. What is the difference between inflection point and critical point? (this is not the same as saying that f has an extremum). The critical points of inflection of a function are the points at which the concavity changes and the tangent line is horizontal. Mind that this is the graph of f''(x), which is the Second derivative. Setting the second derivative of a function to zero sometimes . This means that a point of inflection is a point where the second derivative changes sign (from positive to negative or vice versa) Inflection Points: The inflection points of a function of an independent variable are related to the second derivative of the function. By using this website, you agree to our Cookie Policy. But if continuity is required in order for a point to be an inflection point, how can we consider points where the second derivative doesn't exist as inflection points? Call Us Today: 312-210-2261. There is a third possibility. Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function\u2019s graph. Mathematics Learning Centre, University of Sydney 1 The second derivative The second derivative, d2y dx2,ofthe function y = f(x)isthe derivative of dy dx. Even the first derivative exists in certain points of inflection, the second derivative may not exist at these points. The second derivative can be used as an easier way of determining the nature of stationary points (whether they are maximum points, minimum points or points of inflection). Test Preparation. The second derivative is 4*e^2x - e^x. Not every zero value in this method will be an inflection point, so it is necessary to test values on either side of x = 0 to make sure that the sign of the second derivative actually does change. The following figure shows the graphs of f, Home; About; Services. Computing the first derivative of an expression helps you find local minima and maxima of that expression. Candidates for inflection points include points whose second derivatives are 0 or undefined. Solution To determine concavity, we need to find the second derivative f\u2033(x). Inflection points can only occur when the second derivative is zero or undefined. If f 00 (c) = 0, then the test is inconclusive and x = c may be a point of inflection. Since it is an inflection point, shouldn't even the second derivative be zero? When x = ln 1/4, y = (1/4)^2 - 1/4 = 1/16 - 1/4 = -3/16. The second derivative is written d 2 y/dx 2, pronounced \"dee two y by d x squared\". Then, find the second derivative, or the derivative of the derivative, by differentiating again. exists but f \u201d(0) does not exist. Definition. h (x) = simplify (diff (f, x, 2)) Concavity may change anywhere the second derivative is zero. Recognizing inflection points of function from the graph of its second derivative ''. Let us consider a function f defined in the interval I and let $$c\\in I$$.Let the function be twice differentiable at c. A stationary point on a curve occurs when dy/dx = 0. How to Calculate Income Elasticity of Demand. When we simplify our second derivative we get; 6x = 12. x = 2. f \"(x) = 12x 2. Definition by Derivatives. 0 0? The second derivative is never undefined, and the only root of the second derivative is x = 0. Home > Highlights for High School > Mathematics > Calculus Exam Preparation > Second Derivatives > Points of Inflection - Concavity Changes Points of Inflection - Concavity Changes Exam Prep: Biology Stationary Points. By using this website, you agree to our Cookie Policy. This results in the graph being concave up on the right side of the inflection point. There might just be a point of inflection. The second derivative of the curve at the max/nib points confirms whether it is max/min. One way is to use the second derivative and look for change in the sign from +ve to -ve or viceversa. cannot. Explain the relationship between a function and its first and second derivatives. The second derivative tells us if the slope increases or decreases. How to obtain maximums, minimums and inflection points with derivatives. AP\u00ae is a registered trademark of the College Board, which has not reviewed this resource. List all inflection points forf.Use a graphing utility to confirm your results. The second derivative is written d 2 y/dx 2, pronounced \"dee two y by d x squared\". Learn which common mistakes to avoid in the process. For there to be a point of inflection at (x 0, y 0), the function has to change concavity from concave up to concave \u2026 Let us consider a function f defined in the interval I and let $$c\\in I$$.Let the function be twice differentiable at c. A function is said to be concave upward on an interval if f\u2033(x) > 0 at each point in the interval and concave downward on an interval if f\u2033(x) < 0 at each point in the interval. The second derivative test is used to find out the Maxima and Minima where the first derivative test fails to give the same for the given function.. Second Derivative Test To Find Maxima & Minima. In other words, the graph gets steeper and steeper. Anyway, fun definitional question. If you're seeing this message, it means we're having trouble loading external resources on our website. Khan Academy is a 501(c)(3) nonprofit organization. This results in the graph being concave up on the right side of the inflection point. 2x = 0 . The second derivative can be used as an easier way of determining the nature of stationary points (whether they are maximum points, minimum points or points of inflection). Inflection points are where the function changes concavity. dy/dx = 2x = 0 . Taking y = x^2 . Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. I just dont know how to do it. (c) Use the second derivative test to locate the points of inflection, and compare your answers with part (b). The Second Derivative Test cautions us that this may be the case since at f 00 (0) = 0 at x = 0. The concavity of this function would let us know when the slope of our function is increasing or decreasing, so it would tell us when we are speeding up or slowing down. The usual way to look for inflection points of f is to . Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. Thanks @xdze2! If x >0, f\u201d(x) > 0 ( concave upward. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points: Since e^x is never 0, the only possible inflection point is where 4*e^x = 1, which is ln 1/4. We find the inflection by finding the second derivative of the curve\u2019s function. Donate or volunteer today! Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. Inflection point is a point on the function where the sign of second derivative changes (where concavity changes). (d) Identify the absolute minimum and maximum values of f on the interval [-2,4]. This means that f (x) is concave downward up to x = 2 f (x) is concave upward from x = 2. The following figure shows the graphs of f, Points of Inflection are locations on a graph where the concavity changes. We can define variance as a measure of how far\u00a0\u2026, Income elasticity of demand (IED) refers to the sensitivity of\u00a0\u2026. The concavityof a function lets us know when the slope of the function is increasing or decreasing. I'm very new to Matlab. Example 3, If x < 0, f\u201d(x) < 0 ( concave downward. A function is said to be concave upward on an interval if f\u2033(x) > 0 at each point in the interval and concave downward on an interval if f\u2033(x) < 0 at each point in the interval. A common mistake is to ignore points whose second derivative are undefined, and miss a possible inflection point. It is not, however, true that when the derivative is zero we necessarily have a local maximum or minimum. dy dx is a function of x which describes the slope of the curve. If you're seeing this message, it means we're having trouble loading external resources on our website. dy dx is a function of x which describes the slope of the curve. Note: You have to be careful when the second derivative is zero. The second derivative has a very clear physical interpretation (as acceleration). Learn which common mistakes to avoid in the process. A point where the second derivative vanishes but does not change its sign is sometimes called a point of undulation or undulation point. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If y = e^2x - e^x . Points of Inflection. Candidates for inflection points are where the second derivative is 0. In other words, the graph gets steeper and steeper. And for that, we don\u2019t need smoothness, just continuity. And where the concavity switches from up to down or down \u2026 Second Derivatives: Finding Inflection Points Computing the second derivative lets you find inflection points of the expression. We observed that x = 0, and that there was neither a maximum nor minimum. Solution To determine concavity, we need to find the second derivative f\u2033(x). The second derivative test uses that information to make assumptions about inflection points. The second derivative at an inflection point vanishes. The first derivative is f\u2032(x)=3x2\u221212x+9, sothesecondderivativeisf\u2033(x)=6x\u221212. A point of inflection or inflection point, abbreviated IP, is an x-value at which the concavity of the function changes.In other words, an IP is an x-value where the sign of the second derivative changes.It might also be how we'd describe Peter Brady's voice.. x = 0 , but is it a max/or min. The only critical point in town test can also be defined in terms of derivatives: Suppose f: \u211d \u2192 \u211d has two continuous derivatives, has a single critical point x 0 and the second derivative f\u2032\u2032 x 0 < 0. First Derivatives: Finding Local Minima and Maxima. Also, an inflection point is like a critical point except it isn't an extremum, correct? Please consider supporting us by disabling your ad blocker. Lets begin by finding our first derivative. An inflection point is associated with a complex root in its neighborhood. This results in the graph being concave up on the right side of the inflection point. Recall the graph f (x) = x 3. Lv 6. Limits: Functions with Suprema. Then find our second derivative. The usual way to look for inflection points of f is to . \u2013 pyPN Aug 28 '19 at 13:51 For example, the second derivative of the function $$y = 17$$ is always zero, but the graph of this function is just a horizontal line, which never changes concavity. Inflection points are where the function changes concavity. Using the Second Derivatives. List all inflection points forf.Use a graphing utility to confirm your results. For ##x=-1## to be an *horizontal* inflection point, the first derivative ##y'## in ##-1## must be zero; and this gives the first condition: ##a=\\frac{2}{3}b##. A critical point is a point on the graph where the function's rate of change is altered wither from increasing to decreasing or in some unpredictable fashion. An inflection point is a point on a curve at which a change in the direction of curvature occurs. A stationary point on a curve occurs when dy/dx = 0. And the inflection point is where it goes from concave upward to concave downward \u2026 A point of inflection or inflection point, abbreviated IP, is an x-value at which the concavity of the function changes.In other words, an IP is an x-value where the sign of the second derivative changes.It might also be how we'd describe Peter Brady's voice.. The concavity of a function r\u2026 y\u2019 = 3x\u00b2 \u2013 12x. Here we have. In algebraic geometry an inflection point is defined slightly more generally, as a regular point where the tangent meets the curve to order at least 3, and an undulation point or hyperflex is defined as a point where the tangent meets the curve to order at least 4. then y' = e^2x 2 -e^x. The second derivative and points of in\ufb02ection Jackie Nicholas c 2004 University of Sydney . Inflection points in differential geometry are the points of the curve where the curvature changes its sign.. For example, the graph of the differentiable function has an inflection point at (x, f(x)) if and only if its first derivative f' has an isolated extremum at x. find f \"; find all x-values where f \" is zero or undefined, and In other words, the graph gets steeper and steeper. If f 00 (c) = 0, then the test is inconclusive and x = c may be a point of inflection. The second derivative and points of in\ufb02ection Jackie Nicholas c 2004 University of Sydney . Explain the concavity test for a function over an open interval. Now, I believe I should \"use\" the second derivative to obtain the second condition to solve the two-variables-system, but how? y\u201d = 6x -12. The first derivative is f '(x) = 4x 3 and the second derivative is. The second derivative test is used to find out the Maxima and Minima where the first derivative test fails to give the same for the given function.. Second Derivative Test To Find Maxima & Minima. Learn how the second derivative of a function is used in order to find the function's inflection points. A point of inflection is any point at which a curve changes from being convex to being concave This means that a point of inflection is a point where the second derivative changes sign (from positive to negative or vice versa) To find the points of inflection of a curve with equation y = f (x): Explanation: . 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External resources on our website is made possible by displaying online advertisements to our Cookie.!, if x < 0 ( concave upward on the right side of the inflection point is a point the. Recall the graph being concave up on the right side of the curve local minima and maxima of that.. Is f\u2032 ( x ) > 0 ( concave upward 0, but is a! On a graph where the concavity changes and the minimum point is positive, the graph f ( x =... Inflection: algebraic point of inflection second derivative of Sydney slope of the function is concave.! We get ; 6x = 12. x = 2 the list of point of inflection second derivative that inflection! Negative, and that there was neither a maximum nor minimum x is an point... You find inflection points step-by-step this website uses cookies to ensure you get best... Step-By-Step this website, you agree to our Cookie Policy ) nonprofit organization means down. Aug 28 '19 at 13:51 I 'm very new to Matlab to provide a free, world-class to. 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Sothesecondderivativeisf\u2033 ( x ) = simplify ( diff ( f, x, 2 ) Call! Is a point on a curve ap\u00ae is a minimum point is a point of inflection from MATH MISC Manipal. X which describes the slope of the derivative tells us whether the curve concave! Equal zero to be careful when the second derivative and points of inflection get! Y ) in excel of an expression helps you find local minima and maxima of expression... And current ( y ) in excel complex root in its neighborhood maximums, and... Computing the second derivative of a function over an open interval recall the graph faster... That section is concave downward or concave upward disabling your ad blocker very... F \u201d ( 0, 0 ) is a 501 ( c ) use second... Save my name, email, and the minimum point is where 4 e^x. You get the best experience =3x2\u221212x+9, sothesecondderivativeisf\u2033 ( x ) = simplify ( diff f. Not reviewed this resource ) use the second derivative test to locate the inflection point is at x =,! The histogram itself ) certain points of in\ufb02ection Jackie Nicholas c 2004 University of Sydney f\u2032 ( x =. Not reviewed this resource points can only occur when the slope of the curve looked like a hill the. [ -2,4 ] '19 at 13:51 I 'm very new to Matlab tangent line is horizontal the process equal to! And compare your answers with part ( b ) uses cookies to ensure you get the best experience the changes. + ) 2 hence it is n't an extremum ) inflection are locations on a curve you! Also be used to determine the general shape of its graph on selected intervals expression. 'S no point of inflection in and use all the features of Khan Academy, please make sure the! You have to be an inflection point, however, ( 0, 0 ) is a point... Sometimes this can happen even if there 's no point of inflection resources on our is! Max/Or min c 2004 University of Sydney point if the function the concavityof a function over an interval... True that when the second condition to solve the equation and for that, we need to track the test... To down or down \u2026 list all inflection points of the College Board, which is ln,... The value at x = 0 cookies to ensure you get the best.! E^2X - e^x ap\u00ae is a minimum point the only possible inflection point, set the second is. A local maximum or minimum general shape of its second derivative must equal to! Message, it means we 're having trouble loading external resources on our website is made possible displaying! Differentiating your function to find the function is concave downward or concave upward 501 ( c ) 3. Are where the second derivative of a function of x which describes the slope of the.! ) use the second derivative changes ( where concavity changes ) its derivative! Occur when the slope of the curve is concave upward of that expression also be used to determine the shape! Advertisements to our Cookie Policy, by differentiating your function to zero sometimes x < 0 ( concave or... Please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked minima and maxima of that expression for... Is an inflection point and critical point becomes the inflection point is a registered trademark of function. Lists of points of inflection: algebraic selected intervals 3 ) nonprofit organization whether it is an inflection,...", "date": "2021-10-23 17:01:48", "meta": {"domain": "hospedagemdesites.ws", "url": "http://sindforte.hospedagemdesites.ws/fashion-sales-heqzp/5e530c-point-of-inflection-second-derivative", "openwebmath_score": 0.5898116230964661, "openwebmath_perplexity": 406.405938920952, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347831070321, "lm_q2_score": 0.8757869997529961, "lm_q1q2_score": 0.8533973151522691}}
{"url": "http://math.stackexchange.com/questions/336097/finding-a-vector-orthogonal-to-a-subspace", "text": "# Finding a vector orthogonal to a subspace\n\nSuppose you were given vectors $a_1,\\dots,a_n \\in \\mathbb{R}^m$ then how would you compute some vector orthogonal to the given list of vectors? Note that you are allowed to return the zero vector only if the vectors span $\\mathbb{R}^m$.\n\nI thought about it for a while and the best I could do was to form the matrix with these vectors as rows and pick a vector from the nullspace. Is there an easier/faster way? Perhaps something geometric like Gram-Schmidt could be possible.\n\nThanks!\n\n-\nTo computer \"some vector\", you could just return $\\vec{0}$ but I guess you want another one. \u2013\u00a0 xavierm02 Mar 20 '13 at 19:16\nLet $V=\\operatorname{Vect}(\\{a_1,\\dots,a_n\\})$ Let $(e_1,\\dots,e_p)$ be an orthonormal basis of $V$ Then take any vector $x$ that is not in $V$ And then take $x-\\sum\\limits_{k=1}^p<e_p\\mid x>e_p$ \u2013\u00a0 xavierm02 Mar 20 '13 at 19:16\nHow do I pick a vector that is not in $V$? I could always pick a random vector in $\\mathbb{R}^m$ and the remove its components that lie in the subspace. But is there a deterministic way of doing this? \u2013\u00a0 anon Mar 21 '13 at 10:24\nWhat is wrong with your finding x such that Ax = 0 idea? I see it as completely correct way of going about this. \u2013\u00a0 TenaliRaman Mar 21 '13 at 10:29\nYou have a basis $(\\varepsilon_1,\\dots,\\varepsilon_m)$ of $\\Bbb R^m$, to pick a vector that is not in $V$, you can compute $\\varepsilon_i-\\sum\\limits_{k=1}^p<e_p\\mid \\varepsilon_i>e_p$ until you get a non-zero result. \u2013\u00a0 xavierm02 Mar 21 '13 at 12:18\n\nHere is the deterministic algorithm.\n\nLet $A$ be the $m \\times n$ matrix of your vectors $$A = \\pmatrix{a_0 & a_1 & \\cdots & a_n}$$ Use the QR factorization of it $$A = QR$$ so that the Q matrix will contain the entire null space you are looking for: $$A = \\pmatrix{Q_1 & Q_2}\\pmatrix{R_1 \\\\ 0}$$ Since $Q$ is orthonormal $$\\pmatrix{Q_1^\\top \\\\ Q_2^\\top}A = \\pmatrix{R_1 \\\\ 0}$$ To completely specify the null space, you can see here that it is $Q_2$ $$Q_2^\\top A = 0$$\n\n-\nAny deterministic algorithm will take as much computation as a row reduction, even to find a single vector in the null space. \u2013\u00a0 adam W Mar 21 '13 at 16:33\n\nThis is a very good occasion to apply the fundamental theorem of linear algebra regarding the four fundamental subspaces of a matrix. (The question is correctly answered by muzzlator already, I am just adding a more detailed explanation below and hope it is helpful for some people.)\n\nFirst, use $a_1,\\dots,a_n \\in \\mathbb{R}^m$ to form an $m\\times n$ matrix $A$, with each vector as a column of the matrix. The dimension of the column space of $A$ is designated as $r$, the number of linearly independent vectors among vectors $a_j, (j=1\\dots n)$, and $r$ is also the rank of $A$. Then the problem of finding some vector orthogonal to $a_1,\\dots, a_n$ is equivalent to finding the solution in the \"left nullspace\" of $A$, designated as $N(A^T)$, by solving the following equation:\n\n$$A^T x = 0.$$\n\n$A^T$ is $n\\times m$, and the dimension of $N(A^T) = m-r$. (Actually what we find here is the orthogonal complement of the original subspace. This is much better than finding just some orthogonal vectors.) If the original list of vectors span $\\mathbb{R}^m$, it means the rank of $A$ equals $m$, and the dimension of the left nullspace is $m-r = m-m =0$. So in this case the only solution (the orthogonal vector) is the zero vector, $(0,\\dots, 0)$.\n\nGauss elimination (to get the echelon matrix) is the method to find the solution to $A^Tx=0$. On the other hand, Gram-Schmidt is the process to build a normalized orthogonal (\"orthonormal\") basis after you have found the vectors in $N(A^T)$.\n\nIn your post, you are correct to use the vectors as rows in a matrix, so you don't need to transpose the matrix to find the answer. (Personally I prefer to keep the matrix as an $m\\times n$ matrix as much as possible.)\n\n-\n\nThere should indeed be a faster way. This is because Elementary row operations do not change the row space when solving $A^T x = 0$. You've described yourself as having already done this but we don't need to solve the whole nullspace to find something in it (unless you're already not doing this?).\n\nHere is a thought, start with any non-zero vector and find a non-zero component. Eliminate that component in every other vector (theoretically). Choose any other component, if every other vector (after the row reduction) has a $0$ there, then $(-a_2, a_1, \\dots, 0)$ is orthogonal to your vectors. Otherwise repeat this process on the smaller matrix produced by the first non-zero entry you see. If $m \\leq n$ and this process doesn't terminate after $m$ components are checked, there is only the trivial solution. If $n < m$ and the algorithm doesn't terminate after $n$ components are checked, then compute $e_{n+1} \\cdot a_i$ for each $i$ and use this to produce an orthogonal vector by choosing the appropriate coordinates in the first $n+1$ items.\n\nThis means a total of $\\frac{\\min\\{m,n\\}^2}{2} + m \\max\\{n-m, 0\\}$ operations are more or less are all you need at worst.\n\n-\nHm let me just check that, just realised this process wouldnt work if they span the space as it never gives a $0$ answer \u2013\u00a0 muzzlator Mar 21 '13 at 11:10\nAfter a gazillion edits, I think I have it finally right. As expected, if $m >> n$, the problem is a lot easier. If $n$ is very large, theres the stupid chance a lot of them could be linearly dependent and you'd have to go through almost everything to see if this is true. \u2013\u00a0 muzzlator Mar 21 '13 at 12:03\nThat being said, there's probably a random algorithm which solves this in far fewer steps and if it fails to produce an answer, there will be a sufficiently small probability that there actually was an orthogonal vector. \u2013\u00a0 muzzlator Mar 21 '13 at 12:09", "date": "2015-07-02 03:39:16", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/336097/finding-a-vector-orthogonal-to-a-subspace", "openwebmath_score": 0.9126025438308716, "openwebmath_perplexity": 195.76094254990858, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347838494567, "lm_q2_score": 0.8757869965109765, "lm_q1q2_score": 0.8533973126433383}}
{"url": "https://math.stackexchange.com/questions/4080968/rolling-a-die-twice-in-either-order-with-not-mutual-exclusive-events", "text": "# Rolling a die twice in either order with not mutual exclusive events\n\nI am reading the following problem:\n\nIf a single die is rolled twice, find the probability of rolling an odd number and a number greater than $$4$$ in any either order\n\nMy solution:\nProbability of rolling an odd number = $$\\frac{3}{6}$$\nProbability of rolling a number greater than $$4$$ = $$\\frac{2}{6}$$\nSince the events are not mutually exclusive ($$5$$ is counted twice) the probabilty of throwing an odd number followed by a number greater than $$4$$:\n$$\\frac{3}{6}\\cdot \\frac{2}{6} - \\frac{1}{36} = \\frac{5}{36}$$\n\nThe probablity of throwing a number greater than $$4$$ followed by an odd number is: $$\\frac{2}{6}\\cdot \\frac{3}{6} - \\frac{1}{36} = \\frac{5}{36}$$\n\nTherefore the my final solution is $$\\frac{5}{36} + \\frac{5}{36} = \\frac{10}{36}$$\n\nBut it is wrong, as the solution states that it should be $$\\frac{11}{36}$$\n\nWhat am I doing wrong here?\n\n\u2022 What is 1/36, and why do you think it corrects for double counting? Mar 28, 2021 at 21:35\n\u2022 @AleksejsFomins: The $5$ is odd and greater than $4$ so to avoid counting that twice I subtract this\n\u2013\u00a0Jim\nMar 28, 2021 at 21:47\n\u2022 As stated by @Karl below, the mistake seems to be in subtracting it from both orders Mar 28, 2021 at 21:50\n\u2022 @AleksejsFomins: I thought that if events are not mutually independent we subtract the common event occurence. E.g. if the question was about the probability of odd followed by number greater than $4$, should I subtract the $1$ occurence of $5$? I apply this logic for the reverse order too\n\u2013\u00a0Jim\nMar 28, 2021 at 21:54\n\nThe outcome potentially counted twice is $$(5,5)$$, not $$5$$.\n\nThe number of pairs $$(a,b)$$ where $$a$$ is odd and $$b>4$$ is $$3\\times2=6$$. This includes $$(5,5)$$.\n\nThe number of pairs $$(a,b)$$ where $$a>4$$ and $$b$$ is odd is $$2\\times3=6$$. This includes $$(5,5)$$.\n\nThe sum of these is $$6+6=12$$, but $$(5,5)$$ was counted twice; there are actually only $$11$$ pairs in the union of these two sets.\n\n\u2022 I thought that when having events that are not mutually exclusive we always subtract the $\\bigcap$. So for the number of pairs (\ud835\udc4e,\ud835\udc4f) where \ud835\udc4e is odd and \ud835\udc4f>4 is 3\u00d72=6. I should not include (5,5)\n\u2013\u00a0Jim\nMar 28, 2021 at 21:51\n\u2022 I'd suggest just trying to think clearly about the outcomes you want to count instead of focusing on formulaic rules. If we wanted the probability of getting odd or $>4$ in one roll of the die, your approach would be applicable: $$|\\{odd\\}\\cup\\{>4\\}|=|\\{odd\\}|+|\\{>4\\}|-|\\{odd\\}\\cap\\{>4\\}|=|\\{1,3,5\\}|+|\\{5,6\\}|-|\\{5\\}|=3+2-1=4.$$ But for the given problem we instead have $$|\\{(odd,>4)\\}\\cup\\{(>4,odd)\\}|=|\\{(odd,>4)\\}|+|\\{(>4,odd)\\}|-|\\{(odd,>4)\\}\\cap\\{(>4,odd)\\}|=6+6-1=11.$$\n\u2013\u00a0Karl\nMar 28, 2021 at 22:20\n\nIn saying that \"5 is counted twice\", meaning (presumably) you're removing the duplicate event $$(5, 5)$$, what you should be doing is saying \"The event $$(5, 5)$$ is part of both of the cases I've considered, so I need to only count it once, so I will remove it once from my calculation.\"\n\nInstead, what it looks like you've done is removed it from both of your cases, each of which assumes the other case has already counted it.\n\n\u2022 Since an odd number and a number greater than $4$ are not mutually exclusive, my understanding is that since $5$ appears in both events it should be counted once. Hence if we focus on odd followed by a number greater than $4$ we should subtract $1$ case. Similar logic for the reverse order.\n\u2013\u00a0Jim\nMar 28, 2021 at 21:52\n\u2022 In rolling an odd number followed by a number greater than 4, rolling the 5 first isn't the same as rolling the 5 second so you don't remove it. Mar 29, 2021 at 21:45\n\u2022 There are only 36 possibilities anyway, so try writing them out explicitly and counting them all? That might help you see what you missed. Mar 29, 2021 at 21:46", "date": "2022-05-19 01:48:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4080968/rolling-a-die-twice-in-either-order-with-not-mutual-exclusive-events", "openwebmath_score": 0.5227314233779907, "openwebmath_perplexity": 264.7678170061913, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347912737016, "lm_q2_score": 0.8757869867849167, "lm_q1q2_score": 0.8533973096679844}}
{"url": "https://math.stackexchange.com/questions/921243/combinatorics-question-why-doesnt-this-method-work", "text": "# Combinatorics question. Why doesn't this method work?\n\nIf I have the following set {$1,2,3,4$}, and I want to know the number of possible lists of size 3 that contain $1$ and $2$ in them, I tried the following, and think of it as the number of ways to select 3 elements:\n\n$2\u00b71\u00b72=4$ as in \"number of choices for the first number, then the second... \" But the actual number is 12.\n\nThe only way I can do it is by observing that I have two sets {$1,2,3$} and {$1,2,4$} And each one has 6 permutations, and therefore 12 total permutations or lists.\n\nHow can I find the number of lists and subsets that satisfy the property of containing 1 and 2?\n\n\u2022 Your first method works too : \"total permutations\" - (\"permutations without 1\" + \"permutations without 2\") $$4.3.2 - (3.2.1 + 3.2.1)$$ Sep 6 '14 at 10:08\n\u2022 By list you mean n-tuple? A subset isnt a ordered thing, it completely different the case for a 3-tuple and a subset of cardinality 3. Can you repeat elements on the lists/3-tuple? Sep 6 '14 at 10:36\n\u2022 Yeah, that's what I meant. But I want a robust method to find both lists and subsets given such criteria. Sep 6 '14 at 10:41\n\u2022 Apparently you want neither $(1,2,1)$, nor $(2,6,1)$, but I could not tell the first and hardly the second from your description. Sep 6 '14 at 11:22\n\nIn fact, $2*1*2=4$ isn't that far away from your other solution. What you did is calculate the number of possibilities to have either $2$ or $1$ in the first place of your set and the other one on the second spot, with the remaining $3$ and $4$ taking the last one.\n\nYou did not, however, keep in mind, that $2$ and $1$ could also be in the second and third place or in the first and the third place. In the end, you have essentially three possibilities to form your set concerning placement of $2$ and $1$, and each possibility has four different ways of fulfilling the condition.\n\nSo you get: $3*2*1*2 = 12$ , which is the correct solution.\n\nI hope that helped!\n\nSDV\n\nLet's say that you are looking for the number of sets of $\\left\\{ 1,\\dots,n\\right\\}$ of size $m$ that contain $\\left\\{ 1,\\dots,k\\right\\}$.\n\nThis comes to choosing $m-k$ members of this set from set $\\left\\{ k+1,\\dots,n\\right\\}$ so there are $$\\binom{n-k}{m-k}$$ possibilities.\n\nIf it comes to lists then each permution on one of these sets induces a new list, so there are $$\\binom{n-k}{m-k}m!$$ possibilities.\n\n\u2022 But it doesn't work for $m=1$, containing {1}: (4-1)!/(1-1)!= 6 Sep 6 '14 at 11:02\n\u2022 @Hex4869 It also workes for $m=1=k$. Note that $\\binom{n-1}{0}=1$ for each $n\\geq1$. Do not confuse it with $\\frac{(n-1)!}{0!}$ Sep 6 '14 at 11:44", "date": "2021-12-06 06:40:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/921243/combinatorics-question-why-doesnt-this-method-work", "openwebmath_score": 0.8240166306495667, "openwebmath_perplexity": 238.16383950202163, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347860767304, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.853397303536976}}
{"url": "https://stats.stackexchange.com/questions/317297/summation-of-expectation", "text": "# Summation of expectation\n\nLets suppose we are in a casino, having two slot machines. Our strategy is to flip a coin to decide which machine to start on. (Without loss of generality, we say that we play the slot one first with probability $c_1$ and play the slot two first with probability $1-c_1$)\n\nThen, we play on that machine until losing one game, then we switch to another slot machine to continue playing, we play a total n games (n is finite here).\n\nSuppose the probability to win in slot one and slot two is i.i.d $Bernoulli(p_1)$ and $Bernoulli(p_2)$, respectively, and each win just gain 1 dollar, and each loss just get nothing.\n\nThen, what should be the expected earning using this strategy???\n\nI approached it by iteration (drawing a tree), hoping that there will be some pattern and if we find the expected winning in each turn, then we can sum the n rounds up because the trials are independent.\n\nThe expected winning in round 1 is: $$c_1\\cdot p_1 + (1-c_1) \\cdot p_2$$\n\nThe expected winning in round 2 is: $$c_1 \\cdot[p_1^2+(1-p_1)\\cdot p_2]+ (1-c_1)\\cdot[p_2^2+(1-p_2)\\cdot p_1]$$\n\nThe expected winning in round 3 is: $$c_1 \\cdot \\Big[p_1^3 + p_1(1-p_1) p_2 +(1-p_1)p_2^2 + (1-p_1)(1-p_2)p_1\\Big] + (1-c_1) \\Big[p_2^3 + p_2(1-p_2) \\cdot p_1 + (1-p_2)p_1^2+(1-p_2)(1-p_1)p_2\\Big]$$\n\nI counted the expected winning up till round 4, which is: $$c_1 \\cdot \\Big[p_1^4 + p_1^2(1-p_1)p_2 + p_1(1-p_1) p_2^2 + p_1(1-p_1)(1-p_2)p_1 +(1-p_1)p_2^3 + (1-p_1)p_2(1-p_2)p_1 + (1-p_1)(1-p_2)p_1^2 + (1-p_1)^2(1-p_2)p_2\\Big]$$ $$+ (1-c_1) \\cdot \\Big[p_2^4 + p_2^2(1-p_2)p_1 + p_2(1-p_2) p_1^2 + p_2(1-p_2)(1-p_1)p_2 +(1-p_2)p_1^3 + (1-p_2)p_1(1-p_1)p_2 + (1-p_2)(1-p_1)p_2^2 + (1-p_2)^2(1-p_1)p_1\\Big]$$\n\nThis calculation seems endless so it does not work well. How could I find the expected winning in this case?\n\n\u2022 Oh it is the former, n is the number of times I pull a slot machine's lever~~ We switch only if we lose one game \u2013\u00a0son520804 Dec 5 '17 at 20:54\n\u2022 Let $X_1$ be the number of winning pulls on machine 1 before a losing pull. So $X_1 - 1$ is your payoff if you were to play machine 1. What is the expectation of $X_1$? And do any of these formulas for sums of a series help you compute the expectation of $X_1$? \u2013\u00a0Matthew Gunn Dec 5 '17 at 22:36\n\u2022 I could not easily compute the number of winning on each slot. We choose to play at slot 1 first by probability $c_1$ and play at slot 2 first by probability $c_2$. Then we play at the slot continuously, until lose 1 game and then play the another slot. (And the slot 1 and slot 2 each has winning probability $p_1$ and $p_2$) \u2013\u00a0son520804 Dec 5 '17 at 22:42\n\nConsider $d$ machines numbered $1,2,\\ldots, d$, each with chance $p(d)$ of winning. Let's find the expectation of beginning with machine $1$, playing until a loss, proceeding to machine $2$, etc, and cycling around from machine $d$ to machine $1$ if necessary until $n$ games have been played. The question concerns the case $d=2$, but the analysis (and computation) isn't any harder for larger $d$.\n\nLet $e_j(k)$ denote the expected winnings from playing $k$ games beginning with machine $j$. Because each game is either a win (gaining $1$ dollar) with probability $p(j)$ or a loss with probability $1-p(j)$, leaving $k-1$ games to go, the rules of conditional expectation imply\n\n$$e_j(k) = p(j)(e_j(k-1) + 1) + (1-p(j))(e_{(j \\operatorname{mod} d)+1}(k-1))$$\n\nwhen $k \\ge 1$ and otherwise $e_j(k)=0$.\n\nThis is a simple dynamic program requiring $O(dn)$ computational effort and $O(dn)$ storage: in other words, it's a quick easy computation. You can even do it by hand.\n\nAs an example, here is an R implementation. The argument p is an array of probabilities $p(j)$. It returns a $d\\times n+1$ array indexed by machine and number of games (from $0$ through $n$).\n\nexpectation <- function(p, n=10) {\nd <- length(p)\nP <- matrix(0, d, n+1, dimnames=list(Machine=names(p), Plays=0:n))\nif (n > 0)\nfor (i in 1:n) {\nfor (j in 1:d) {\nP[j, i+1] <- p[j] * (P[j, i] + 1) + (1-p[j]) * P[j%%d + 1, i]\n}\n}\nreturn(P)\n}\n\n\nGiven an initial probability distribution $c(1), c(2),\\ldots, c(d)$ for the choice of which machine to start with, the rules of conditional expectation state that the expected winnings will be the sum of $c(j)e_j(n)$. This answers the question.\n\nIntuitively, machines that tend to fail will rarely be operated and those that tend to win are allowed to run. The $e_j(n)$ therefore will rapidly tend to a common value approximately equal to the $p$-weighted average of the $e_j(n)$ as $n$ grows, and the differences among the $e_j(n)$ will be no greater than the differences among the raw probabilities $p(j)$. Here are some examples for $d=2$.\n\nThe panel headings list the probabilities for machine \"a\" and machine \"b\" in order.\n\nIt's always good to check probability calculations. I carried out simulations as shown in this R program:\n\np <- c(a=0.5, b=0.2)\nset.seed(17) # Comment this out when repeating simulations!\nsim <- replicate(1e4, n - sum(cumsum(rgeom(n, 1-p) + 1) <= n))\n\n\nIt exploits the fact that the number of wins obtained from a machine has a geometric distribution, allowing the simulation to proceed fairly quickly. (The computation actually amounts to counting how many machines were tried in turn, rather than counting the wins directly.)\n\nThe output is an array of winnings, one for each iteration of the scenario. Because this dataset is large and not too skewed, we may compare its mean to the theoretical calculation using a Z-test.\n\nm <- mean(sim)\ns <- sd(sim)\nmu <- expectation(p, n)[1,n+1]\nc(Simulation=m, Calculation=mu, Z=(m - mu)/s * sqrt(length(sim)))\n\n\nThe output in this (reproducible) case is\n\n Simulation Calculation           Z\n3.950       3.930       0.889\n\n\nIt says the average among the (10,000) simulations was 3.95, the calculation of the expectation came out to 3.93, and the Z-score is 0.889 (which is not a significant difference). Repeated simulations for different $p$ and $n$ continue to agree with the calculations, providing assurance the calculations are correct.", "date": "2020-02-26 06:49:54", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/317297/summation-of-expectation", "openwebmath_score": 0.776028037071228, "openwebmath_perplexity": 769.3784822603061, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347823646075, "lm_q2_score": 0.8757869835428965, "lm_q1q2_score": 0.8533972987063784}}
{"url": "https://mathematica.stackexchange.com/questions/137900/applying-the-vec-operator", "text": "# Applying the vec operator\n\nHow to apply the $\\operatorname{vec}$ operator in Mathematica? For example, how can I transform a $2 \\times 2$ matrix into a $1 \\times 4$ matrix as follows? $$\\operatorname{vec}\\left( \\begin{bmatrix} a_{1,1} & a_{1,2} \\\\ a_{2,1} & a_{2,2} \\end{bmatrix} \\right) = \\begin{bmatrix} a_{1,1} \\\\ a_{2,1} \\\\ a_{1,2} \\\\ a_{2,2} \\end{bmatrix}$$\n\n\u2022 vector and matrix are vague terms in Mathematica so please add input and expected output in terms of Mathematica code in order to make the question clear. \u2013\u00a0Kuba Feb 16 '17 at 10:31\n\u2022 Thanks for the Accept. Please see the additional example of Flatten that I added afterward. \u2013\u00a0Mr.Wizard Feb 16 '17 at 13:59\n\nUse Flatten to do this in a single operation.\n\nin = Array[a, {2, 2}]\n\nFlatten[in, {2, 1}]\n\n{{a[1, 1], a[1, 2]}, {a[2, 1], a[2, 2]}}\n\n{a[1, 1], a[2, 1], a[1, 2], a[2, 2]}\n\n\nIn Mathematica there are only vectors (lists), not column vectors and row vectors. However if you wish to convert a vector into an array with rows of length one for output the computationally fastest method is typically Partition:\n\nPartition[{a[1, 1], a[2, 1], a[1, 2], a[2, 2]}, 1]\n\n{{a[1, 1]}, {a[2, 1]}, {a[1, 2]}, {a[2, 2]}}\n\n\nIf that is your goal from the start you can also do that in a single operation using Flatten:\n\nFlatten[{in}, {3, 2}]     (* note the extra {} around in *)\n\n{{a[1, 1]}, {a[2, 1]}, {a[1, 2]}, {a[2, 2]}}\n\n\n\u2022 Unless OP sees column vector as: List /@ {a[1, 1], a[2, 1], a[1, 2], a[2, 2]} \u2013\u00a0Kuba Feb 16 '17 at 13:34\n\u2022 @Kuba Good point. There is no such thing as a column vector in Mathematica as you know, but that doesn't mean the OP doesn't potentially want a series of one element rows... \u2013\u00a0Mr.Wizard Feb 16 '17 at 13:37\n{{1, 2}, {4, 5}} // MatrixForm\n\n\n\\begin{bmatrix} 1 & 2 \\\\ 4 & 5 \\end{bmatrix}\n\nArrayReshape[Transpose[%], {4, 1}] // MatrixForm\n\n\n\\begin{bmatrix} 1 \\\\ 4\\\\ 2\\\\ 5 \\end{bmatrix}\n\nThanks to @cyrille.piatecki for the use of Transpose[].\n\nI propose this simple module\n\nvec[mat_] :=\nModule[{a = mat},\nArrayReshape[Transpose[\na], {Dimensions[mat][[1]] Dimensions[mat][[2]], 1}]]\n\n\napply with\n\naa = Table[Subscript[a, i, j], {i, 1, 2}, {j, 1, 2}]\n\n\nthis gives the expected result\n\nvec[aa] // MatrixForm", "date": "2020-04-02 16:05:05", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/137900/applying-the-vec-operator", "openwebmath_score": 0.18426832556724548, "openwebmath_perplexity": 3556.4582373555704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9585377237352756, "lm_q2_score": 0.8902942348544447, "lm_q1q2_score": 0.8533806093320183}}
{"url": "https://math.stackexchange.com/questions/3382832/does-minimal-ideal-always-imply-principal-ideal", "text": "# Does minimal ideal always imply principal ideal?\n\nFirst, let me specify two definitions i will use.\n\n$$[1.]$$ A (right/ left/ both) ideal $$I$$ of a ring $$R$$ (unity not assumed) is minimal if $$(1.) \\; I\\neq (0)$$ and $$(2.)$$ If $$J$$ is any nonzero (right/ left/ both) ideal of $$R$$ containied in $$I$$, then $$J=I$$\n\n$$[2.]$$ If $$x \\in R$$, then $$(x)$$ is the intersection of all (left/ right/ both) ideals of $$R$$ containing $$x$$.\n\nConsider the following propostition and its proof:\n\n$$\\textbf{A [right / left / both] ideal I of a ring R is minimal iff}$$\n\n$$\\textbf{ I is generated by any of its nonzero elements x \\in I }$$\n\nProof:\n\n$$1.(\\Rightarrow)$$ Suppose $$I$$ is minimal and and $$x \\in I$$ is nonzero. Consider the ideal $$J:=(x)$$ generated by $$x$$. By construction, $$J \\neq (0)$$ since $$x \\in J$$. Now, $$J \\subseteq I$$, since by definition $$J$$ is the smallest ideal containing $$x$$. But then , by minimality of $$I$$ we must have $$I=J$$, so $$I$$ is generated by $$x$$.\n\n$$2.(\\Leftarrow)$$ Suppose $$I=(x)$$ for any $$x \\in I$$, and that $$J$$ is any nonzero ideal of $$R$$ with $$J \\subseteq I$$. Let $$y$$ be any nonzero element of $$J$$. Then $$y \\in I$$, and by hypothesis we have $$I=(y)$$. But then we must have $$J=I$$, because $$(y)$$ is the smallest ideal of $$R$$ containing $$y$$.\n\nMy question is:\n\n$$\\textbf{Does this also prove that any minimal ideal is principal? }$$\n\n\u2022 (2) is not correct. That $I$ is principal does not mean $I = (y)$ for every $y \\in I$, it means $I = (y)$ from some $y \\in I$.\n\u2013\u00a0Jim\nOct 6, 2019 at 13:36\n\u2022 (1) is correct though, all minimal ideals are principal and for minimal ideals it actually is true that $I = (y)$ for every $y \\in I$. That condition is actually equivalent to the principal ideal being minimal.\n\u2013\u00a0Jim\nOct 6, 2019 at 13:37\n\u2022 Then you have read incorrectly. I'm not saying that I is principal. I'm saying that I is an ideal which has the property that it is generated by any of its elements. Oct 6, 2019 at 13:37\n\u2022 Oh, actually you're right! I did read what you were proving incorrectly, my bad.\n\u2013\u00a0Jim\nOct 6, 2019 at 13:39\n\u2022 In that case both (1) and (2) are correct.\n\u2013\u00a0Jim\nOct 6, 2019 at 13:39\n\nYes essentially, although I think it is safest to word it as \u201ca left (resp, right/twosided) ideal is minimal if and only if it is generated by any of its nonzero elements as a left (resp, right/twosided) ideal.\u201d\n\nThe fact that minimals are generated by a single element follows a fortiori from the $$\\implies$$ direction.\n\nLet $$I$$ be a minimal right ideal of a ring $$R$$. By definition, $$I$$ being a principal right ideal means that $$\\exists x \\in R \\, (I=xR)$$.\n\nIn fact, $$\\forall x \\in I \\setminus \\{0\\} \\, (I=xR)$$. Indeed, for any nonzero element $$x$$ of $$I$$, $$xR \\subseteq I$$ (because $$I$$ is a right ideal of $$R$$) and $$0 \\neq x \\in xR$$, so $$xR=I$$ because $$I$$ is assumed to be a minimal right ideal.\n\nSimilarly, if $$I$$ is a minimal left (resp. two-sided) ideal of $$R$$, then $$\\forall x \\in I \\setminus \\{0\\} \\, (I=Rx)$$ (resp. $$\\forall x \\in I \\setminus \\{0\\} \\, (I=RxR)$$).\n\nMore generally, any simple module is cyclic. Two-sided ideals of $$R$$ are the same as the right $$R^{op} \\otimes_{\\mathbb{Z}} R$$-submodules of $$R$$.\n\nI agree with the former part of your proof, but I do not with the latter part. (I missed the word ANY.)\n\nTo be specific, consider $$R = \\mathbb Z$$ and an ideal $$I = (n)$$ with $$n \\neq 0$$. Obviously, it contains a non-zero proper ideal $$(n) \\supset (2n) \\supset (0).$$ Hence, a principal ideal generated by a non-zero element needs not to be minimal. (Indeed, this argument shows that $$\\mathbb Z$$ has no minimal ideals.)\n\n\u2022 Yes, it does. (You've proved it, right?)\n\u2013\u00a0Orat\nOct 6, 2019 at 13:42\n\u2022 Well, I'm not sure, because I can't find the statement mentioned on the web. On the wikipedia page of minimal ideal, en.wikipedia.org/wiki/Minimal_ideal , they mention briefly that for a ring R with unity, this in neccesarily true for right ideals .. . Oct 6, 2019 at 13:47\n\u2022 Well, what you've done is basically the same as the argument shown on that wikipedia page. Depending on what type of ideals you're considering (right/left/two-sided), considering an ideal of the form ($xR$/$Rx$/$RxR$) is the key, anyway. BTW, many people may reserve the symbol $(x)$ to denote $RxR$ only.\n\u2013\u00a0Orat\nOct 6, 2019 at 13:53\n\u2022 Well, the fact that I=xR / I=Rx / I=RxR for principal ideals is only the case if R has unity, which isn't assumed... Oct 6, 2019 at 13:56\n\u2022 You're right; as I usually deal with unital rings, I didn't pay much attention to that. Anyway the former part of your proof is correct as it only uses the minimality, and does not use something like $xR$.\n\u2013\u00a0Orat\nOct 6, 2019 at 14:02", "date": "2022-05-27 23:38:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3382832/does-minimal-ideal-always-imply-principal-ideal", "openwebmath_score": 0.9307032227516174, "openwebmath_perplexity": 166.7801105031896, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9585377272885904, "lm_q2_score": 0.89029422102812, "lm_q1q2_score": 0.8533805992424601}}
{"url": "http://math.stackexchange.com/questions/233760/calculating-e/233768", "text": "# Calculating $e$\n\nIf I calculate $e$ using the following formula.\n\n$$e = \\sum_{k=0}^{\\infty}{\\frac{1}{k!}}$$\n\nIs it possible to predict how many correct decimal places I get when I stop summing at $n$ terms?\n\n-\nThere is a subtlety in the contrast between \"correct decimal places\" and accuracy. Most of us had interpreted \"$n$ correct decimal places as within $10^{-n}$, but as Dan Brumleve points out, you could be very close. If the correct answer is $1.9999$, an error of $10^{-4}$ can change the ones digit. Having a string of $9$'s is rare, but if you care about it you need to check. \u2013\u00a0Ross Millikan Nov 10 '12 at 4:37\n\nIf we use $n$ terms, the last term used is $\\dfrac{1}{(n-1)!}$. The missing \"tail\" is therefore $$\\frac{1}{n!}+\\frac{1}{(n+1)!}+\\frac{1}{(n+2)!}+\\frac{1}{(n+3)!}\\cdots.\\tag{1}$$ Note that $(n+1)!=n!(n+1)$ and $(n+2)!\\gt n!(n+1)^2$, and $(n+3)!\\gt n!(n+1)^3$ and so on. So our tail $(1)$ is less than $$\\frac{1}{n!}\\left(1+\\frac{1}{n+1}+\\frac{1}{(n+1)^2}+\\frac{1}{(n+1)^3}+\\cdots \\right).$$ Summing the geometric series, we find that the approximation error is less than $$\\frac{1}{n!}\\left(1+\\frac{1}{n}\\right).$$\n\n-\nThis is a good answer, why was it downvoted? \u2013\u00a0robjohn Nov 9 '12 at 19:45\nTo end the error vs correct digits discussion: If you calculate $e_n:=\\sum_{k=0}^n\\frac1{n!}$ and $e_n+\\frac1{n!}(1+\\frac1n)$ and both agree on $d$ decimals, then both estimates are correct to (at least) $d$ decimals. \u2013\u00a0Hagen von Eitzen Nov 9 '12 at 20:39\n@DanBrumleve How about instead of just downvoting (something that can't be undone after a few minutes) 10 (a slight exaggeration) different answers without a single comment (until after someone asked why), leave a comment and clarify. Then, if it turns out you misunderstood, you haven't just downvoted 10 different answers incorrectly. \u2013\u00a0Graphth Nov 9 '12 at 22:41\n@DanBrumleve Well, obviously many other people (everyone who answered?) believe otherwise. So, is it possible that at the very least it's just your preference? And, if so, is it really worth downvoting? \u2013\u00a0Graphth Nov 9 '12 at 23:31\n@DanBrumleve While this is more of a meta issue, IMHO there is a difference between 'this is not how I would have answered the question' or even 'this is not what I consider a complete answer to the question' and 'this is an unhelpful answer to the question' - I would consider the latter to be a minimal standard for downvoting (as opposed to simply not voting), and the FAQ seems to agree: 'Use your downvotes whenever you encounter an egregiously sloppy, no-effort-expended post, or an answer that is clearly and perhaps dangerously incorrect.' This answer is hardly egregious or dangerous. \u2013\u00a0Steven Stadnicki Nov 10 '12 at 1:08\n\nYou can use the remainder term in Taylor's expansion\n\n-\nTo what function are we applying Taylor's theorem? \u2013\u00a0robjohn Nov 9 '12 at 19:49\n@robjohn the definition of $e$ given by OP is actually the Taylor series for $e^x$ centered at $x=0$, evaluated at $x=1$. \u2013\u00a0Logan Stokols Nov 9 '12 at 23:48\n@Logan: As I mentioned in a comment to glebovg, since $f$ was not specified in the question, it would be good to mention it in the answer. \u2013\u00a0robjohn Nov 10 '12 at 0:04\n\nIn this answer, it is shown, by comparison to a geometric series, that $$0\\le n!\\left(e-\\sum_{k=0}^n\\frac1{k!}\\right)\\le\\frac1n$$ Therefore, the error after $n+1$ terms is at most $\\frac1{nn!}$ .\n\nTo $n$ decimal places:\n\nWhen asking for a number to $n$ decimal places, there are two common meanings\n\n1. the error is less than $\\frac12\\times10^{-n}$.\n\n2. the value is correct when rounded to $n$ decimal places. As has been pointed out, if a number is very close to $10^{-n}\\left(\\mathbb{Z}+\\frac12\\right)$, rounding to $n$ decimal places might require computing more decimal places to know the actual $n^{\\mathrm{th}}$ digit of the rounded number. This is not as easy to use as meaning 1, so it is not as commonly used.\n\n-\nThat was a fast downvote. Care to comment? \u2013\u00a0robjohn Nov 9 '12 at 20:21\nIs someone just downvoting for free? \u2013\u00a0Pedro Tamaroff Nov 10 '12 at 0:24\n@PeterTamaroff: nah... it still costs 1 rep to downvote an answer (afaik). \u2013\u00a0robjohn Nov 10 '12 at 0:26\n\nThe series converges rapidly. If you stop at $\\frac 1{ k!}$ you can bound the error by $\\frac 1{k(k!)}$ by bounding the remaining terms with a geometric series.\n\n-\nWhy was this downvoted? It seems a bit scant, but valid. \u2013\u00a0robjohn Nov 9 '12 at 19:43\nI downvoted for the same reason as most of the others: the question asks about the number of correct leading digits in the partial sum and this upper bound on the error term doesn't lead to any obvious answer to that question. \u2013\u00a0Dan Brumleve Nov 10 '12 at 4:29\n\nThe $n$-th Taylor polynomial is $${P_n}(x) = f(0) + \\frac{{f'(0)}}{{1!}}x + \\frac{{f''(0)}}{{2!}}{x^2} + \\cdots + \\frac{{{f^{(n)}}(0)}}{{n!}}{x^n}$$ (in this case $f(x)$ is simply $e$) and the error we incur in approximating the value of $f(x)$ by $n$-th Taylor polynomial is exactly $$f(x) - {P_n}(x) + \\frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}$$ where $0 < c < x$. This form of the remainder can be used to find an upper bound on the error. If the difference above is positive, then the approximation is too low, and likewise if the error is negative, then the approximation is too high. We only need to find an appropriate $c$.\n\n-\nTo what function are we applying Taylor's theorem? \u2013\u00a0robjohn Nov 9 '12 at 19:50\n@robjohn In this case $e^x$ for $x = 1$, but it works in general. \u2013\u00a0glebovg Nov 9 '12 at 20:17\nWhy downvote? Please comment. \u2013\u00a0glebovg Nov 9 '12 at 20:17\nI downvoted because it doesn't address the distinction between the error term and the number of correct digits in the partial sum. (Also consider simplifying or clarifying: the question asks only about $e$ not $e^x$.) \u2013\u00a0Dan Brumleve Nov 9 '12 at 20:21\n@glebovg: Since $f$ was not specified in the question, it would be good to mention it in the answer. \u2013\u00a0robjohn Nov 9 '12 at 22:49", "date": "2016-05-24 19:40:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/233760/calculating-e/233768", "openwebmath_score": 0.8965113759040833, "openwebmath_perplexity": 472.60097472746503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877717925422, "lm_q2_score": 0.8791467706759583, "lm_q1q2_score": 0.853377019906055}}
{"url": "https://math.stackexchange.com/questions/1235009/what-distribution-models-number-of-trials-needed-for-given-number-of-successes-a/1235601", "text": "# What distribution models number of trials needed for given number of successes and success rate?\n\nCase scenario: a retro-virus infects a healthy cell. The virus programs the cell to brew little viruses, at a rate of 0.5 per-sec, until finally the cell bursts when the number of virus inside it is 5. How to model this?\n\nIn Binomial, the random variable represents the number of successful trials obtained when throwing a coin a certain number of trials, at a certain probability of success per trial.\n\nI want a distribution whose random variable is the number of trials (coin tosses) that were necessary to perform, given a certain number of successful trials and a certain probability per trial.\n\nI am not even sure how I would write down the probability mass function.\n\nIs there such a distribution? Nothing rings a bell here: https://en.wikipedia.org/wiki/Category:Discrete_distributions\n\nThere are related questions to this one --- such as this one: How many trials until I each my desired outcome --- but no-one mentioned a distribution, or if any exists.\n\nJust to make it clear, the random generator for a random variable of such a distribution would look like this in R:\n\nrmy <- function(s, p) {\ni <- n <- 0\nwhile(i != s) {\ni <- i+rbinom(1,1,p)\nn <- n+1\n}\nn\n}\n\n\nThank you ! ps: sorry if the text was a little flowery, but it helps me think, since I am a junior mathematician hehe.\n\n\u2022 Let me see if I understand the process. Each second I flip a (fair) coin to see if I produce a virus. Since I have a $1/2$ probability of producing a single new virus, the average virus production rate is $1/2$. I stop when I produce five viruses. And you want to know the expected number of coin flips, is that right? \u2013\u00a0Brian Tung Apr 14 '15 at 22:23\n\u2022 Have you looked at Poisson Distribution? \u2013\u00a0rightskewed Apr 14 '15 at 22:46\n\u2022 @BrianTung, I want to know if there is a theoretical distribution. Expected value is easy, it is E=m/p=5/0.5=10, in this case, where m is number of successful trials and p is probability of success. \u2013\u00a0rpmcruz Apr 15 '15 at 8:05\n\u2022 @rightskewed, hmm what do you have in mind exactly? Using $\\lambda=m/p$; testing in R rpois(10, 5/0.5) shows it sampling values below <5, which is impossible in this model. \u2013\u00a0rpmcruz Apr 15 '15 at 8:08\n\u2022 Just to make it clear: I want to know if there is already a popular theoretical distribution for what I want. If anyone works to work out the probability mass function that would be superb too. :P Might make sense to migrate it to: stats.stackexchange.com . I did not remember about the statistics forum ... \u2013\u00a0rpmcruz Apr 15 '15 at 8:12\n\nI figure this one out. :)\n\nI can model it using a Negative Binomial: https://en.wikipedia.org/wiki/Negative_binomial_distribution\n\nFirst, let us change the values of my case scenario, just to make it clearer. \"Case scenario: a retro-virus infects a healthy cell. The virus programs the cell to brew little viruses, at a rate of 0.2 per-sec, until finally the cell bursts when the number of virus inside it is 5. How to model this?\"\n\nWe can model number of failures $Y$ as $Y\\sim\\mathcal{NB}(5,0.2)$. That answers the question, how many failed trails do we have, when we need 5 successful at a probability rate of 0.2. But we do not want failed trials, we want total trials, and total trials = failed trials + successful trials. We know successful trials, which is 5, so our random variable $X$ is such that $X\\sim5+\\mathcal{NB}(5,0.2)$.\n\nIn fact, comparing the random generator function I proposed in the question with the negative binomial random generator (with this adjustment):\n\npar(mfrow=c(1,2))\nhist(sapply(1:1e5, function(x) rmy(5, 0.2)))\nhist(5+rnbinom(1e5, 5, 0.2))\n\n\nAll functions mean, sd and summary are consistent as well.\n\nMy 2 cents here :\n\nUsing a Negative Binomial distribution can be a very good approximation. However :\n\n1) You totally remove the case r=0 (No failures).\n\n2) Also, you're not answering your initial question. You're computing the distribution of the #trials before achieving the known #fails (or #success). It's different than #trials knowing the #fails (or #success), as you can have fails following the last success !\n\nI believe the right answer involves Bayesian probabilities :\n\n\u2022 We consider that N is a random variable describing the # of trials (# of Bernoulli trials). Let's assume that N is following an Exponential Distribution $$N \\thicksim Exp(\\lambda)$$\n\u2022 We then consider that X is a random variable describing the number of successes after N=n trials, following a Binomial distribution $$N \\thicksim Binom(p, N)$$\n\nLet's compute $$P(n|k)$$ thanks to the Bayes formula :\n\n$$\\frac{1}{P(N=n|X=k)} = \\frac{P(X=k)}{P(X=k|N=n)P(N=n)}$$\n\nAnd by writing $$P(k) = \\sum_{m=k}^{\\infty}{P(X=k|N=m)P(N=m)}$$\n\n$$= \\sum_{m=k}^{\\infty}\\frac{P(X=k|N=m)P(N=m)}{P(X=k|N=n)P(N=n)}$$\n\nAnd because $$P(X=k|N=n)$$ and $$P(X=k|N=m)$$ are just Binomial distributions :\n\n$$= \\sum_{m=k}^{\\infty}\\frac{m!k!(n-k)!}{n!k!(m-k)!} \\frac{p^k}{p^k} \\frac{(1-p)^{m-k}}{(1-p)^{n-k}} \\frac{e^{-\\lambda.m}}{e^{-\\lambda.n}}$$\n\n$$= \\sum_{m=k}^{\\infty}{[e^{-\\lambda}(1-p)]^{m-n}}$$\n\nAnd by re-indexing the Serie with $$l=m-k <=> m=l+k$$\n\n$$= \\sum_{l=0}^{\\infty}{[e^{-\\lambda}(1-p)]^{l+k-n}}$$\n\n$$\\frac{1}{P(N=n|X=k)} = \\frac{[e^{-\\lambda}(1-p)]^{k-n}}{1 - e^{-\\lambda}(1-p)}$$\n\n$$P(N=n|X=k) = \\frac{1 - e^{-\\lambda}(1-p)}{[e^{-\\lambda}(1-p)]^{k-n}}$$\n\nAnd now you have a probability $$P>0$$ for $$k=0$$ ;-)\n\nEdit: The choice of the prior (Exponential) is totally arbitrary and leads to a lot of simplifications. If anyone wants to attempt the calculation with another prior, feel free :-)", "date": "2020-01-27 15:44:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1235009/what-distribution-models-number-of-trials-needed-for-given-number-of-successes-a/1235601", "openwebmath_score": 0.7430022358894348, "openwebmath_perplexity": 471.68785779457136, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.970687766704745, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.8533770154331346}}
{"url": "https://stats.stackexchange.com/questions/604226/why-are-x-sim-u-1-1-and-y-x2-dependent", "text": "# Why are $X \\sim U(-1,1)$ and $Y=X^2$ dependent?\n\nSuppose we have two continuous random variables $$X \\sim U(-1,1)$$ and $$Y=X^2$$. I don't understand why they are dependent.\n\n$$E[X] = 0$$ $$E[Y] = \\int_{-1}^{1} x^2 dx = 2/3$$ $$E[XY] = \\int_{-1}^{1} x^3 dx = 0$$\n\nThey are independent to me because $$E[XY]=E[X]E[Y]$$, so why they are dependent?\n\n\u2022 Plotting simulated values of $X$ and $Y$ will be enlightening, but you can also work out the density of $Y$ via a change of variables from $X$ in order to compare the probabilities in terms of definition of statistical independence. Feb 4 at 4:15\n\u2022 I find it challenging to find any function $h$ for which $X$ and $h(X)$ are independent.\n\u2013\u00a0whuber\nFeb 4 at 15:55\n\u2022 @whuber: one such function $h$ is $h(x)=0$. Another one is $h(x)=5$. Feb 4 at 17:25\n\u2022 @Kjetil Certainly. Not very interesting, are they? ;-)\n\u2013\u00a0whuber\nFeb 4 at 18:00\n\u2022 For specific distributions more interesting functions can be possible. For instance, $Y = \\sin(2\\pi X)$ is independent from $X$ if the domain of $X$ is integer values only. But yes, it will still be the not so interesting map where all values in the domain of $X$ need to be mapped to a single point. It is because $f(X)$ given $X=x$ is a singular distribution in the point $f(x)$. Independence requires that the distribution of $f(X)$ is the same distribution for every value of $X$, $f(x)$ must be a single value for every $x$. Feb 4 at 20:55\n\nLet me write this in a generic manner:\n\nTwo random variables for which the expectation of the product is equal to the product of expectations need not be independent.\n\nLet $$X\\sim \\mathcal U(-a, a)$$ where $$a\\in\\mathbb R_+$$ and let $$Y\\sim X^2.$$ Then as noted $$\\mathbb E[XY]=\\mathbb E[X^3] =0=\\mathbb E[X]\\mathbb E[Y].$$ But they are definitely not independent.\n\n## Reference:\n\n$$\\rm [I]$$ Counterexamples in Probability and Statistics, Joseph P. Romano, Andrew F. Siegel, Wadsworth, $$1986,$$ ex. $$4.15.$$\n\nSince $$Y = X^2$$, $$[-1/2 \\leq X \\leq 0] \\cap [Y > 1/4] = \\varnothing$$, hence $$P[-1/2 \\leq X \\leq 0, Y > 1/4] = 0$$. On the other hand, $$P[-1/2 \\leq X \\leq 0] = 1/4 > 0, P[Y > 1/4] = 1/2 > 0$$. Hence the independence defining relation $$P(A \\cap B) = P(A)P(B)$$ for all $$A \\in \\sigma(X), B \\in \\sigma(Y)$$ fails to hold for $$A = [-1/2 \\leq X \\leq 0]$$ and $$B = [Y > 1/4]$$, i.e., $$X$$ and $$Y$$ are not independent.\n\nAbove is a formal proof. Intuitively, since $$Y$$ is a deterministic function of $$X$$, knowing the value of $$X$$ means knowing the value of $$Y$$, hence $$Y$$ and $$X$$ of course cannot be independent (the heuristic definition of independence between $$X$$ and $$Y$$ requires that observing the information provided by $$X$$ does not increase the information of $$Y$$).\n\nIt is worth emphasizing that $$E[XY] = E[X]E[Y]$$ is a necessary condition, rather than a sufficient condition for the independence of $$X$$ and $$Y$$. If you really want to express the independence of $$X$$ and $$Y$$ in terms of expectations, it should be stated as\n\n$$X$$ and $$Y$$ are independent $$\\iff$$ $$E[f(X)g(Y)] = E[f(X)]E[g(Y)]$$ for all Borel measurable functions $$f$$ and $$g$$.\n\nEvidently, the condition $$E[f(X)g(Y)] = E[f(X)]E[g(Y)]$$ is much stronger than the condition $$E[XY] = E[X]E[Y]$$. The former is a system of infinitely many equations, while the latter is a single equation.\n\nYou need to distinguish between the concepts of uncorrelatedness, which involves the decomposition of the expectation of a product, and independence, which involves the decomposition of a joint probability distribution. What you're calling independence is actually uncorrelatedness. What happens, as the other answers show, is that independence is a stronger condition than uncorrelatedness, that is, independence implies uncorrelatedness but the vice versa is not true.\n\nAnd your example is exactly one example that is used to show that uncorrelatedness does not imply independence.\n\nThey are clearly not independent: if you know the value $$X$$ then this gives you information about the value of $$Y$$ and indeed allows you to calculate it precisely.\n\nYou have correctly said independence means $$P(X\\le a, Y\\le b)=P(X\\le a) P(Y\\le b)$$, so for example consider $$a=\\frac13$$ and $$b=\\frac 14$$.\n\n\u2022 $$X \\le \\frac 13$$, i.e. when $$-1 \\le X \\le \\frac13$$, has probability $$\\frac23$$\n\u2022 $$Y \\le \\frac 14$$, i.e. when $$-\\frac12 \\le X \\le \\frac12$$, has probability $$\\frac12$$\n\u2022 $$X \\le \\frac 13, Y \\le \\frac 14$$ together, i.e. when $$-\\frac12 \\le X \\le \\frac13$$, has probability $$\\frac5{12}$$\n\nbut $$\\frac23 \\times \\frac12 \\not=\\frac5{12}$$ so they are not independent. Try other examples with $$-1 < a < 1$$ and $$0 < b < 1$$.\n\n$$E[XY] = E[X]E[Y]$$ is a condition for zero covariance and so zero correlation, but this on its own does not imply independence. The bottom row of Wikipedia's chart illustrating correaltion has other examples of this; in your question you would get a parabolic U-shaped chart.\n\n\u2022 Note that zero covariance is sufficient for independence if $X$ and $Y$ are jointly Gaussian, but not necessarily (or maybe ever) for other distributions. That it works for multivariate Gaussian distributions might be the source of the confusion.\n\u2013\u00a0Dave\nFeb 4 at 14:03\n\nHere is a simulation of how the joint distribution looks like:\n\nIt shows clearly that the two variables are not independent.\n\nYour formula is more generally (including dependent variables).\n\n$$\\text{E}[XY] = \\text{E}[X]\\text{E}[Y] + \\text{COV}(X,Y)$$\n\nSo when you compute $$\\text{E}[XY] = \\text{E}[X]\\text{E}[Y]$$ then it must mean that $$\\text{COV}(X,Y) = 0$$. But that doesn't imply independence. This makes your case an example of: Why zero correlation does not necessarily imply independence\n\n\u2022 It is not clear where you got the idea for the product $\\text{E}[XY] = \\text{E}[X]\\text{E}[Y]$. Possibly you are confusing it for a description of independence as $P(A \\text{ and } B) = P(A) \\cdot P(B)$ Feb 4 at 19:27", "date": "2023-03-21 17:00:15", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/604226/why-are-x-sim-u-1-1-and-y-x2-dependent", "openwebmath_score": 0.9404534697532654, "openwebmath_perplexity": 169.8070347222628, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.970687770944576, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.8533770053293052}}
{"url": "https://www.physicsforums.com/threads/component-of-a-vector-along-another-vector.777615/", "text": "# Component of a vector along another vector.\n\nTags:\n1. Oct 22, 2014\n\n### Hijaz Aslam\n\n1. The problem statement, all variables and given/known data\nGiven $\\vec{A}=2\\hat{i}+3\\hat{j}$ and $\\vec{B}=\\hat{i}+\\hat{j}$.Find the component of $\\vec{A}$ along $\\vec{B}$.\n\n2. Relevant equations\n$\\vec{A}.\\vec{B}=ABcos\u03b8$ where \u03b8 is the angle between both the vectors.\n\n3. The attempt at a solution\nI attempted the question as follows:\nLet the angle between $\\vec{A}$ and $\\vec{B}$ be '\u03b8'. So the component of $\\vec{A}$ along $\\vec{B}$ is given by $Acos\u03b8\\hat{B}$ => $Acos\u03b8(\\frac{\\vec{B}}{B})$\n\nAs $\\vec{A}.\\vec{B}=ABcos\u03b8$ => $[( 2\\hat{i}+3\\hat{j})(\\hat{i}+\\hat{j})]/B=Acos\u03b8$ => $\\frac{5}{\\sqrt{2}}=Acos\u03b8$\n\nTherefore the component is : $\\frac{5}{\\sqrt{2}}(\\frac{\\hat{i}+\\hat{j}}{\\sqrt{2}})$ => $\\frac{5}{2}({\\hat{i}+\\hat{j}})$\n\nBut my text produces the solution as follows:\n$A_B=(\\vec{A}.\\vec{B})\\hat{B}=\\frac{5}{\\sqrt{2}}(\\hat{i}+\\hat{j})$.\n\n2. Oct 22, 2014\n\n### RUber\n\nI usually see this process broken down into basis components.\nThat is $\\hat B =\\sqrt{2}/2 \\hat i + \\sqrt{2}/2 \\hat j.$\nThen the component is $A\\cdot \\hat B_i \\hat i + A\\cdot \\hat B_j \\hat j$.\nSomewhere in your process, you divided by the magnitude of B twice.\n\n3. Oct 23, 2014\n\n### willem2\n\nYou're right and the book is wrong. The book answer as well as the formula for AB they use.\nThe length of your answer is smaller than the length of A as it should be. The book answer is larger.\nThe projection of A on B should only depend on the direction of B, not the magnitude. The formula used for AB in the book does depend on the magnitude of B.\n\n4. Oct 23, 2014\n\n### Delta\u00b2\n\nBook is wrong . We can verify this by standard euclidean geometry easily because by the definition of cosine, it will be cos(\u03b8)=(component of A along B)/A hence Acos(\u03b8)=(component of A along B). And we have to multiply this by the unit vector of B to get the required result.", "date": "2017-08-22 14:53:53", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/component-of-a-vector-along-another-vector.777615/", "openwebmath_score": 0.8723196983337402, "openwebmath_perplexity": 397.49518757889416, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561721629776, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8533656823903848}}
{"url": "https://www.physicsforums.com/threads/different-results-with-quotient-rule.693718/", "text": "# Different results with quotient rule\n\n1. May 26, 2013\n\n### mike82140\n\nI have been trying to figure out the derivative of (X\u00b2-1)/X. When I use the quotient rule, the result I get is 1-1/X\u00b2. However, when I simplify the expression first, then take the derivative, I get 1+1/X\u00b2\n\nWhy are the results different?\n\n2. May 26, 2013\n\n### Hypersphere\n\n3. May 26, 2013\n\n### mike82140\n\nCan you show how you evaluate it? I tried both ways, but get different results.\n\n4. May 26, 2013\n\n### Hypersphere\n\nAlright, for a function\n$$f(x) = \\frac{g(x)}{h(x)}$$\nthe quotient rule says\n$$f'(x) = \\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$\n\nIn your case, $g(x)=x^2-1$ and $h(x)=x$. Thus $g'(x)=2x$ and $h'(x)=1$. Then you get\n$$\\frac{\\mathrm{d}}{\\mathrm{dx}}\\frac{x^2-1}{x}=\\frac{2x\\cdot x - \\left( x^2 -1 \\right) \\cdot 1}{x^2}=\\frac{2x^2-x^2+1}{x^2}=1+\\frac{1}{x^2}$$\nAre you sure you included the $g'(x)$ term?\n\nSimplifying first,\n$$\\frac{\\mathrm{d}}{\\mathrm{dx}}\\frac{x^2-1}{x} = \\frac{\\mathrm{d}}{\\mathrm{dx}} \\left( x - \\frac{1}{x} \\right) = 1- \\frac{-1}{x^2}=1+\\frac{1}{x^2}$$\n\n5. May 26, 2013\n\n### mike82140\n\nWhere did +1 come from when you were multiplying x\u00b2-1 with 1?\n\n6. May 26, 2013\n\n### Hypersphere\n\nIt was also multiplied by -1, from the quotient rule formula.\n\n7. May 26, 2013\n\n### mike82140\n\nI'm sorry for asking so many questions, and this may sound stupid, but, what -1?\n\nEdit: It appears as though I have made a mistake. The -1 is the minus part that is in front of the x\u00b2-1, so the negative, or minus, distributes itself.\n\nThank you for the help, I appreciate it.\n\n8. May 26, 2013\n\n### Hypersphere\n\nYou see the minus sign in front of $\\left( x^2 -1 \\right) \\cdot 1$, right? That is just short notation for\n\n$$- \\left( x^2 -1 \\right) \\cdot 1=\\left( -1 \\right) \\cdot \\left( x^2 -1 \\right)= (-1) \\cdot x^2 + (-1) \\cdot (-1) = -x^2+1$$\n\nAgreed? Or are you pulling my leg here?\n\nEDIT: Ah, good. I was almost giving up for a while there.\n\n9. May 26, 2013\n\n### SteamKing\n\nStaff Emeritus\nLook at the formula for the quotient rule.\n\nIf h(x) = x, what is h'(x)?", "date": "2017-09-24 18:04:30", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/different-results-with-quotient-rule.693718/", "openwebmath_score": 0.788221538066864, "openwebmath_perplexity": 1672.8934159116618, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561712637256, "lm_q2_score": 0.8807970732843033, "lm_q1q2_score": 0.8533656800825252}}
{"url": "https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring", "text": "# Order is Preserved on Positive Reals by Squaring\n\n## Theorem\n\nLet $x, y \\in \\R: x > 0, y >0$.\n\nThen:\n\n$x < y \\iff x^2 < y^2$\n\n## Proof 1\n\n### Necessary Condition\n\nAssume $x < y$.\n\nThen:\n\n $\\ds x < y$ $\\implies$ $\\ds x \\times x < x \\times y$ Real Number Ordering is Compatible with Multiplication $\\ds x < y$ $\\implies$ $\\ds x \\times y < y \\times y$ Real Number Ordering is Compatible with Multiplication $\\ds$ $\\leadsto$ $\\ds x^2 < y^2$ Real Number Ordering is Transitive\n\nSo:\n\n$x < y \\implies x^2 < y^2$\n\n$\\Box$\n\n### Sufficient Condition\n\nAssume $x^2 < y^2$.\n\nThen:\n\n $\\ds x^2$ $<$ $\\ds y^2$ $\\ds \\leadsto \\ \\$ $\\ds 0$ $<$ $\\ds y^2 - x^2$ Real Number Ordering is Compatible with Addition $\\ds \\leadsto \\ \\$ $\\ds \\paren {y - x} \\paren {y + x}$ $>$ $\\ds 0$ Difference of Two Squares $\\ds \\leadsto \\ \\$ $\\ds \\paren {y - x} \\paren {y + x} \\paren {y + x}^{-1}$ $>$ $\\ds 0 \\times \\paren {y + x}^{-1}$ as $x + y > 0$ $\\ds \\leadsto \\ \\$ $\\ds y - x$ $>$ $\\ds 0$ $\\ds \\leadsto \\ \\$ $\\ds x$ $<$ $\\ds y$\n\nSo:\n\n$x^2 < y^2 \\implies x < y$\n\n$\\blacksquare$\n\n## Proof 2\n\nFrom Real Numbers form Ordered Field, the real numbers form an ordered field.\n\nThe result follows from Order of Squares in Ordered Field.\n\n$\\blacksquare$\n\n## Proof 3\n\nFrom Real Numbers form Ordered Field, the real numbers form an ordered field.\n\nBy definition, an ordered field is a totally ordered ring without proper zero divisors.\n\nThe result follows from Order of Squares in Totally Ordered Ring without Proper Zero Divisors.\n\n$\\blacksquare$\n\n## Proof 4\n\n### Necessary Condition\n\nLet $x < y$.\n\nThen:\n\n $\\ds x < y$ $\\implies$ $\\ds x \\times x < x \\times y$ Real Number Ordering is Compatible with Multiplication $\\ds x < y$ $\\implies$ $\\ds x \\times y < y \\times y$ Real Number Ordering is Compatible with Multiplication $\\ds$ $\\leadsto$ $\\ds x^2 < y^2$ Real Number Ordering is Transitive\n\nSo:\n\n$x < y \\implies x^2 < y^2$\n\n$\\Box$\n\n### Sufficient Condition\n\nLet $x^2 < y^2$.\n\nAiming for\u00a0a contradiction, suppose $x \\ge y$.\n\nThen:\n\n $\\ds x \\ge y$ $\\implies$ $\\ds x \\times x \\ge x \\times y$ Real Number Ordering is Compatible with Multiplication $\\ds x \\ge y$ $\\implies$ $\\ds x \\times y \\ge y \\times y$ Real Number Ordering is Compatible with Multiplication $\\ds$ $\\leadsto$ $\\ds x^2 \\ge y^2$ Real Number Ordering is Transitive\n\nBut this contradicts our assertion that $x^2 < y^2$.\n\nHence by Proof by Contradiction it follows that:\n\n$x < y$\n\n$\\blacksquare$", "date": "2021-08-01 00:42:17", "meta": {"domain": "proofwiki.org", "url": "https://proofwiki.org/wiki/Order_is_Preserved_on_Positive_Reals_by_Squaring", "openwebmath_score": 0.9143046140670776, "openwebmath_perplexity": 411.2855690758946, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904406026280905, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8533024268571389}}
{"url": "http://math.stackexchange.com/questions/78533/prove-that-2n-n2-is-even-if-n-is-a-positive-integer/114520", "text": "# prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer\n\nProve that $(2n)!/(n!)^2$ is even if $n$ is a positive integer. For clarity: the denominator is the only part being squared.\n\nMy thought process: The numerator is the product of the first n even numbers and the product of the first n odd numbers. That is, $(2n!) = (2n)(2n-2)(2n-4)\\cdots(2n-1)(2n-3)(2n-5).$ In effect, the product of even numbers can be cancelled out with n! resulting in the following quotient: $$(2^n)(2n-1)(2n-3)/(n!).$$ To me this looks even thanks to the powers of 2. But I am not convinced for some reason.\n\nSorry for the poor notation, I don't know any coding languages, my apologies.\n\n-\nTry counting the powers of 2 in the numerator and denominator. After a quick bit of experimentation, it looks like it's probably sufficient to consider $n=4m+k$ for $k=0,1,2,3$ (i.e. there are four cases) and you could demonstrate that there are always more powers of 2 in the numerator than in the denominator. \u2013\u00a0 Chris Taylor Nov 3 '11 at 9:40\nCould you give me more insight as to how you came up with the four cases for n? I don't see the derivation for n as you have shown. \u2013\u00a0 hdtv1104 Nov 3 '11 at 9:50\nDid you try using induction? It seems like that shouldn't be too difficult... \u2013\u00a0 Braindead Nov 3 '11 at 13:59\n\nWe have $$\\displaystyle \\frac{(2n)!}{(n!)^2} = \\binom{2n}{n}$$ and of course for every way to choose a combination of $n$ objects from a total of $2n$ objects, there exists a complementary choice (the left over $n$ objects). Since the ways to pick $n$ objects from $2n$ comes in pairs, it follows that the total number is even.\n\nAnother proof: $$\\frac{(2n)!}{(n!)^2} = \\frac{ 2n \\cdot (2n-1)! }{(n!)^2 } = 2 \\cdot \\frac{ n \\cdot (2n-1)! }{ n\\cdot (n-1)! \\cdot n!} = 2 \\cdot \\frac{(2n-1)!}{n! (n-1)!} = 2 \\binom{2n-1}{n}$$\n\nThis one can be written in another flavor (using $\\binom{n}{k} = \\binom{n-1}{k-1} + \\binom{n-1}{k}$ and the symmetric property): $$\\frac{(2n)!}{(n!)^2} = \\binom{2n}{n} = \\binom{2n-1}{n-1} + \\binom{2n-1}{n} = 2\\binom{2n-1}{n}.$$ Yet another: The sum of the $2n$-th row of Pascal's triangle is $2^{2n}$ which is even, and the sum of all the entries excluding the central coefficient is also even, because of the symmetric property $\\displaystyle \\binom{n}{k} = \\binom{n}{n-k}$. Thus, the central coefficient must also be even.\n\nWe can squeeze this problem a bit further and learn something interesting with this fourth proof. This one does not require any knowledge of binomial coefficients. To investigate the prime factors of factorials we use the following identity: $$l = \\sum_{k=1}^{\\infty} \\bigg\\lfloor \\frac{n}{p^k} \\bigg\\rfloor$$\n\nwhere $p$ is a prime number and $l$ is the unique natural number such that $p^l \\mid n!$ but $p^{l+1} \\nmid n!.$ Basically this counts the multiplicity of the prime factor $p$ in $n!$.\n\nThe idea of the proof is this: Write $n! = 1\\cdot 2 \\cdot 3 \\cdots (n-1) \\cdot n$, and ask yourself where the factors of $p$ come from. Clearly you get one factor of $p$ from $p, 2p, 3p, \\cdots$ so there are $\\lfloor n/p \\rfloor$ factors of $p$ from these. But that's not the whole story - when we checked over the multiples $p,2p,3p\\cdots$ we didn't count the second factor of $p$ every time there was $p^2, 2p^2, 3p^2 \\cdots$ so there comes another $\\lfloor n/p^2 \\rfloor$ to add. Wait, we didn't count another factor every time we missed cubes in $p^3, 2p^3,\\cdots$, so we add another $\\lfloor n/p^3 \\rfloor$ and so on. Also, note that this may look like an infinite series, but eventually $p^k > n$ so all the terms eventually become $0.$\n\nAnyway, back to the main goal. Using our identity, we see that there are $$\\sum_{k=1}^{\\infty} \\bigg \\lfloor \\frac{2n}{2^k} \\bigg \\rfloor - 2 \\bigg \\lfloor \\frac{n}{2^k} \\bigg \\rfloor$$ factors of $2$ in $\\displaystyle \\frac{(2n)!}{ (n!)^2}$, and we wish to show that this number is at least $1.$\n\nWith some easy casework we can see that $$\\lfloor 2x \\rfloor - 2 \\lfloor x \\rfloor = \\left\\{ \\begin{array}{lr} 1 & : \\{ x\\} \\geq \\frac{1}{2} \\\\ 0 & : \\{ x\\} < \\frac{1}{2} \\end{array} \\right.$$ where $\\{ x \\}$ denotes the fractional part of $x.$ Thus our problem is reduced to showing that there exists some $k\\in \\mathbb{N}$ such that $\\displaystyle \\frac{n}{2^k}$ has fractional part greater than or equal to $1/2.$ This is of course true, since if $m$ is the largest positive integer such that $\\displaystyle \\frac{n}{2^m} \\geq 1$ then $\\displaystyle \\frac{1}{2} \\leq \\frac{n}{2^{m+1} } < 1.$ Hence, $\\displaystyle \\frac{(2n)!}{(n!)^2}$ is even.\n\n-\nThanks for the combinatorics response to this proof. I see that it is easier to invoke the combinatorial interpretation but I myself could not see this interpretation as readily. What inspired the use of binomial? \u2013\u00a0 hdtv1104 Nov 3 '11 at 9:53\nJust a bit of practice I guess. Every time I see a ratio of factorials, I check whether it fits the form $\\frac{n!}{k! (n-k)!} = \\binom{n}{k}.$ Also, it was particularly easy for me here because this particular coefficient, $\\binom{2n}{n}$, is quite special and comes up relatively often. \u2013\u00a0 Ragib Zaman Nov 3 '11 at 9:55\nThanks a lot Ragib. \u2013\u00a0 hdtv1104 Nov 3 '11 at 9:58\nYou're welcome! \u2013\u00a0 Ragib Zaman Nov 3 '11 at 10:00\nIn fact, the number of factors of $2$ in $\\binom{2n}{n}$ is equal to the number of $1$ bits in the binary representation of $n$. \u2013\u00a0 robjohn Nov 14 '11 at 21:24\n\nIt is often fun to try to solve such problems using Lagrange's Theorem from finite group theory. That is, to prove that $a$ divides $b$ you try to find a group of cardinality $b$ with a subgroup of cardinality $a$.\n\nIn this case, the symmetric group on $2n$ letters, $S_{2n}$, is an obvious choice for a group of cardinality $(2n)!$. Let $G\\subset S_{2n}$ be the set of all permutations which map the set $\\{1,\\ldots, n\\}$ onto $\\{1,\\ldots, n\\}$ or $\\{n+1,\\ldots, 2n\\}$. You can check that $G$ is a subgroup of $S_{2n}$.\n\nTo compute the cardinality $\\lvert G \\rvert$, one option is to observe that $G\\cong S_n\\wr \\mathbb{Z}_2$, so $\\lvert G\\rvert = \\lvert S_n\\rvert^{\\lvert \\mathbb{Z}_2\\rvert} \\lvert \\mathbb{Z}_2\\rvert = 2(n!)^2$. Alternatively, choosing an element of $G$ corresponds to choosing two permutations of a set of size $n$ and a choice of whether to swap $\\{1,\\ldots,n\\}$ with $\\{n+1,\\ldots, 2n\\}$ or not, so $\\lvert G\\rvert = 2(n!)^2$.\n\nThus by Lagrange's Theorem $2(n!)^2$ divides $(2n)!$, or in other words, $\\frac{(2n)!}{(n!)^2}$ is even.\n\n-\n\nBy the binomial theorem, $$4^n=(1+1)^{2n}=\\sum\\limits_{k=0}^{2n}{2n\\choose k}={2n\\choose n}+\\sum\\limits_{k=0}^{n-1}{2n\\choose k}+\\sum\\limits_{k=n+1}^{2n}{2n\\choose k}.$$ Since Pascal's triangle is symmetrical, each term in the sum from $k=0$ to $n-1$ coincides with one term in the sum from $k=n+1$ to $2n$, hence these two sums are equal and $$\\frac{(2n)!}{(n!)^2}={2n\\choose n}=4^n-2\\cdot\\sum\\limits_{k=0}^{n-1}{2n\\choose k}$$ is even for every $n\\geqslant1$.\n\n-\nI had this one in my answer =) But not as fleshed out. \u2013\u00a0 Ragib Zaman Nov 14 '11 at 21:02\nOh yes, I missed it but I see it now. Sorry about that. \u2013\u00a0 Did Nov 14 '11 at 21:08\n\nLet's observe an example:\n\n$(2\\cdot 5)!=10\\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5!$ , we may conclude that:\n\n$(2n)!=2n\\cdot(2n-1)\\cdot(2n-2)....(n+1)n!$ so we get following expression :\n\n$$\\frac{2n\\cdot(2n-1)\\cdot(2n-2)....(n+1)}{n!}=\\frac{2n\\cdot(2n-1)\\cdot2\\cdot(n-1)\\cdot(2n-3)\\cdot 2 \\cdot (n-2)....(n+1)}{n!}$$\n\n$$=\\frac{2\\cdot2\\cdot2...\\cdot2\\cdot n\\cdot(n-1)\\cdot(n-2)\\cdot......\\cdot 1\\cdot(2n-1)\\cdot(2n-3)\\cdot......(n+1)}{n!}$$\n\n$$=\\frac{2\\cdot2\\cdot2...\\cdot2 \\cdot n!\\cdot(2n-1)\\cdot(2n-3)\\cdot......\\cdot (n+1)}{n!}=$$\n\n$$=2\\cdot2\\cdot2...\\cdot2 \\cdot(2n-1)\\cdot(2n-3)\\cdot......\\cdot (n+1)$$ so number is even.\n\n-\nBut I'm sorry, this proof is false (still got 11 upvotes). \u2013\u00a0 user144248 May 15 '14 at 21:32\nThe error here is at the third equality: The terms that have been divided by $2$ don't go all the way down to $1$, since there are only (about) $n/2$ such terms. In fact, for $n$ even, the last term that is of this form is $n + 2$, so the correct expression would be $$= \\frac{2 \\cdots 2 \\cdot n \\cdot (n - 1) \\cdot ... \\cdot \\frac{n + 2}{2} \\cdot (2n - 1) \\cdots (n + 1)}{n!}$$ \u2013\u00a0 user61527 May 16 '14 at 4:49\n\nThis answer has been moved from this question, which was closed because it was judged to be a duplicate of this question.\n\nThe number of factors of a prime $p$ in $n!$ is $$\\nu_p(n)=\\left\\lfloor\\frac{n}{p}\\right\\rfloor+\\left\\lfloor\\frac{n}{p^2}\\right\\rfloor+\\left\\lfloor\\frac{n}{p^3}\\right\\rfloor+\\dots\\tag{1}$$ $\\left\\lfloor\\frac{n}{p}\\right\\rfloor$ counts the number of positive multiples of $p$ no greater than $n$, but only once. $\\left\\lfloor\\frac{n}{p^2}\\right\\rfloor$ counts the multiples of $p^2$ a second time, and $\\left\\lfloor\\frac{n}{p^3}\\right\\rfloor$ the positive multiples of $p^3$ a third time, etc. Applying $(1)$ to the base-$p$ representation of $n$ $$n=\\sum_{i=0}^kn_ip^i\\tag{2}$$ where $0\\le n_i<p$, yields \\begin{align} \\nu_p(n) &=\\sum_{i=0}^kn_i\\left(1+p+p^2+\\dots+p^{i-1}\\right)\\\\ &=\\sum_{i=0}^kn_i\\left(\\frac{p^i-1}{p-1}\\right)\\\\ &=\\frac{1}{p-1}\\left(n-\\sum_{i=0}^kn_i\\right)\\\\ &=\\frac{n-\\sigma_p(n)}{p-1}\\tag{3} \\end{align} where $\\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.\n\nTherefore, the number of factors $p$ in $\\displaystyle\\binom{n}{k}=\\frac{n!}{k!(n-k)!}$ is \\begin{align} &\\nu_p(n)-\\nu_p(k)-\\nu_p(n-k)\\\\ &=\\frac{1}{p-1}\\left(n-\\sigma_p(n)-k+\\sigma_p(k)-(n-k)+\\sigma(n-k)\\right)\\\\ &=\\frac{\\sigma_p(k)+\\sigma_p(n-k)-\\sigma_p(n)}{p-1}\\tag{4} \\end{align} Using $(4)$ with $p=2$ says that the number of factors of $2$ in $\\displaystyle\\binom{2n}{n}$ is $\\sigma_2(n)+\\sigma_2(n)-\\sigma_2(2n)$ which is simply $\\sigma_2(n)$ since $\\sigma_2(n)=\\sigma_2(2n)$. Therefore, the number of factors of $2$ in $\\displaystyle\\binom{2n}{n}$ is the number of $1$-bits in the binary representation of $n$.\n\nThus, if $n>0$, then there is at least one $1$-bit in $n$ in binary, and so $2$ divides $\\displaystyle\\binom{2n}{n}$.\n\n-\n\nThe exact power of $2$ dividing $\\binom{2n}{n}$ is $2$^(sum of digits in base $2$ representation of $n$).\n\nThe exact power of $3$ dividing the trinomial coefficient $\\binom{3n}{n,n,n} = (3n)!/(n!)^3$ is $3$^(sum of digits in base $3$ representation of $n$).\n\nThe exact power of $p$ dividing the $p$-nomial coefficient $\\binom{pn}{n,n,\\dots , n} = (pn)!/(n!)^p$ is $p$^(sum of digits in base $p$ representation of $n$) for any prime $p$.\n\nThis is from the formula for the power of $p$ dividing $n!$, which is ($n$ - sum of digits of $n$ when written in base $p$)/($p-1$). There also are combinatorial and generating function proofs but they are more complicated.\n\n-", "date": "2015-08-30 08:46:05", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/78533/prove-that-2n-n2-is-even-if-n-is-a-positive-integer/114520", "openwebmath_score": 0.9431523680686951, "openwebmath_perplexity": 154.04376730357018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990440601781584, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8533024261278411}}
{"url": "http://mathhelpforum.com/pre-calculus/8982-finding-tangent-line-without-calculus.html", "text": "Math Help - Finding a Tangent Line without Calculus\n\n1. Finding a Tangent Line without Calculus\n\nI hope this is in the right place, I'm not in a hurry, just curious.\n\nHow can I find an equation for a line tangent to a point on a parabola without using calculus?\n\nI just started playing with this this morning\n\nThe equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)\n\nUsing calculus I found the equation to be y = 4x -18\n\nHow can I do this without calculus?\n\n2. Hello, Ranger SVO!\n\nYou're curious . . . good for you!\n\nWhile playing with parabolas in college (centuries ago),\n. . I \"discovered\" some facts that might help you.\n\nThe equation I'm using is $y \\:= \\:x^2 - 4x - 2$\nand I'm looking for the equation of the tangent line at point $(4,\\,-2)$\n\nUsing calculus I found the equation to be: $y \\:= \\:4x -18$\n\nHow can I do this without calculus?\nCode:\n        *       |       */\n|       /\n*      |      *P(x,y)\n*    |    */\n---------***---o----------\n|   /\u2191\n|  /(\u00bdx,0)\n| /\n|/\no \u2190(0,-y)\n/|\n\nGiven the parabola: . $y \\:=\\:ax^2$, the tangent at point $P(x,y)$\n. . has an x-intercept of $\\left(\\frac{1}{2}x,\\,0\\right)$ and a y-intercept of $(0,\\,-y)$\n\nThe tangent at $P(x,y)$ always has a point directly below the vertex\n. . with the same vertical displacement (only downward).\n\nYour parabola $y \\:=\\:x^2-4x-2$ has its vertex at $(2,-6).$\n\nYour point $P(4,-2)$ is 4 units above the level of the vertex.\nHence, it has a point 4 units directly below the vertex: $Q(2,-10).$\n\nNow you can write the equation through $P$ and $Q.$\n\n3. Originally Posted by Ranger SVO\nI hope this is in the right place, I'm not in a hurry, just curious.\n\nHow can I find an equation for a line tangent to a point on a parabola without using calculus?\n\nI just started playing with this this morning\n\nThe equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)\n\nUsing calculus I found the equation to be y = 4x -18\n\nHow can I do this without calculus?\nI have another approach.\nIt does not always work.\nBut it does sometimes.\n\nIn your example, the tangent line is the line that only intersects the parabola at one point. Thus, you need to find an equation of a line that one intersects at one point, i.e. solution set is only one element. That means you need the discrimanant to be zero.\n\n4. Soroban, I like your explination. But first, at my age curiousity is the only thing that keeps me from vegetating.\n\nHere is what I've been working on. If I draw a rectangle using the vertex and the point where I want the tagent line, I can see the secant line bisecting the rectangle. I can easily notice that the slope of the tangent is twice that of the secant.\n\nIs this always true? If so why?\n\nPerfect Hacker, does that mean that I can use the quadratic equation to find the slope of a line tangent to some point.\n\n5. Hello again, Ranger SVO!\n\nHere is what I've been working on.\nIf I draw a rectangle using the vertex and the point where I want the tangent line,\nI can see the secant line bisecting the rectangle.\nI can easily notice that the slope of the tangent is twice that of the secant.\n\nIs this always true? If so why?\n\nGiven the parabola: $y \\:=\\:ax^2$ and any point on it: $P(p,\\,ap^2)$\n\nThe slope of the secant $OP$ is: . $m_s \\:=\\:\\frac{ap^2 - 0}{p - 0} \\:=\\:ap$\n\nThe derivative is: . $y' \\:=\\:2ax$\n. . Hence, the slope of the tangent at $P$ is: $m_t\\:=\\:2ap$\n\nTherefore, the slope of the tangent is always twice the slope of the secant.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nIf you're really bored, verify these claims.\n\n1) On the parabola $y \\:=\\:ax^2$, pick any two points: . $P(p,\\,ap^2)$ and $Q(q,\\,aq^2)$\n\nDraw the chord $PQ.$\n\nLocate the tangent parallel to $PQ.$\n\nIt is found at: . $x \\:=\\:\\frac{p+q}{2}$ . . . halfway between $p$ and $q.$\n\n2) On the parabola $y \\:=\\:ax^2$, pick any three points: . $P(p,\\,ap^2),\\;Q(q,\\,aq^2),\\;R(r,\\,ar^2)$\nFind the area of $\\Delta PQR.$\n\nConsider the tangents to the parabola at $P,\\,Q,\\,R.$\n. . They will intersect at $A,\\,B,\\,C.$\nFind the area of $\\Delta ABC.$\n\nWe find that: . $(\\text{area }\\Delta ABC) \\:=\\:\\frac{1}{2}(\\text{area }\\Delta PQR)$\n\n6. y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)\nWe want a tangent line at $(4,-2)$.\nTo do that I assume you are familar with the point-slope formula:\n$y-y_0=m(x-x_0)$\nIn this case,\n$(x_0,y_0)=(4,-2)$.\n\nThus, the equation of the line (tangent) at that point is determined by,\n$y+3=m(x-4)$ that means ( $y=m(x-4)-3$)\nWhere,\n$m$\nIs the slope to be determined.\n\nAs I said we need to intersection to be a single point.\nThat is,\n$\\left\\{ \\begin{array}{c}y=x^2-4x-2\\\\y=m(x-4)-3 \\end{array} \\right\\}$\nWe want this to has a single solution.\nEquate,\n$x^2-4x-2=m(x-4)-3$\n$x^2-4x-2=mx-4m-3$\n$x^2+x(-4-m)+(1+4m)=0$\nTo have a single real solution we need the discrimanant to be zero,\n$(-4-m)^2-4(1)(1+4m)=0$\nSolve to get the value of $m$.", "date": "2014-10-22 07:27:46", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/8982-finding-tangent-line-without-calculus.html", "openwebmath_score": 0.7115028500556946, "openwebmath_perplexity": 287.79511760039446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990440598395557, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8533024232106494}}
{"url": "http://dzdb.whynotimmobiliare.it/cambridge-vector-calculus-lecture-notes.html", "text": "# Cambridge Vector Calculus Lecture Notes\n\nPublisher: Dalhousie University 2007 Number of pages: 106. 1 Vector calculus According to classical physics, The vector triple product,. Contents 1. These can be proven with a calculation, but what was more interesting is that the converse. They can be found here. Actually, this text also discusses integration and vector calculus (Chapter 10), but I personally found Rudin's treatment of such hard to follow when I was first learning the subject. Therefore the normal vector we look for is. For example we will see that the existence of partial derivatives does not guarantee that the function itself is continuous (as it is the case for a function of one variable). Unfortunatelythenumberof. Scalar and vector \ufb01elds. Description. Math 53 - Multivariable Calculus -- [4 units] Course Format: Three hours of lecture and three hours of discussion per week. Lecture Notes: 178 kb: Module-5 Linear and Quadratic Approximations,Newton and Picard Methods: Lecture 15 : Newton\\\\\\'s method: Lecture Notes: 184 kb: Module-6 Definition of Integral: Lecture 16 : Integral from upper and lower sums: Lecture Notes: 330 kb: Module-6 Definition of Integral: Lecture 17 : Fundamental theorem of calculus: Lecture. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. Vector Calculus; Line Integrals; Computational Vector Analysis Topics; Stoke\u2019s and Green\u2019s Theorems; The Joy of X\" by Steven Strogatz; Laplace Transforms; Series: Power and Fourier; Partial Differential Equations; Numerical Methods; Topics in Engineering Mathematics; Computational Fluid Dynamics. pdf Lecture notes. From E\u2192 to \u03a6. Index of Math Terms. Vector calculus identities explained. in Calculus. \u2022 how vector fields relate to certain physical phenomena, and how to carry out calculations using vector operations and vector identities, \u2022 how to evaluate line and surface integrals and to interpret these concepts into physical applications, \u2022 how to use the Divergence Theorem, Green, and Stokes to evaluate. The main focus of this module is on multivariable calculus in 2 and 3 dimensions, and vector calculus. Cambridge Notes Below are the notes I took during lectures in Cambridge, as well as the example sheets. , please let me know by e-mail. Higher derivatives and product rules 128 54. https://ocw. Vector geometry / Gilbert de B. Course Format: There are four 50-minute lectures each week. That weight function is commonly the arc length of the curve, or\u2014if you\u2019re integrating over a vector field\u2014the scalar product with a vector differential in the curve. A fraction of the price of most calculus books. than 10 dimensions. The way we did in class was to observe that this rectangular region is vertical to the -plane. Notes are applicalicable for both 1st and 2nd sem students of CBCS scheme. Operator notation Gradient. Andrew Steane's Lecture Courses. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Scalar Product: The \"dot\" product of two vectors is a scalar. MATH 226 (Calculus III) Fall 2016 39571R (MWF at noon in SOS B44) Final exam is Wednesday, December 7, 2-4 pm. Applied Geometry for Computer Graphics and ms excel 2007 user guide pdf CAD D. These notes, in my view, can be used as a short reference for an introductory course on tensor algebra and calculus. May 17, 2020 - Lecture Notes - Review of Vector Calculus Notes | EduRev is made by best teachers of. They consist largely of the material presented during the lectures, though we have taken the liberty of eshing them out in some places and of being more cursory here than in the lectures in other places. Applied Advanced Calculus Lecture Notes by Jan Vrbik. The first lecture of Cosmology II is on Monday, Oct 28th, and the first exercise session is on Friday, Nov 8th. 2) F (x; y z) = P x y z I + Q x y z J R x y z K: For example, the vector \ufb01eld (18. Chapter 18 Vector Calculus 282 x 18. called the Euclidean metric. In this lesson, we will use the Calculus Shell Method to find the volume of a solid of revolution. The notes are adapted to the structure of the course, which stretches over 9 weeks. Advanced Integer Programming, Advanced Linear Programming, Advanced Nonlinear Programming, Calculus I, Vector Calculus, and Differential Equations, Calculus II. Vector Calculus: This lecture Contain problem of GREEN\u2019S THEOREM PROBLEM-1 (pdf link notes is available in this description) Find Online Engineering Mathematics Online Solutions Here. MATH 1920 Lecture Notes - Lecture 31: Vector Calculus, Divergence Theorem, 2D Computer Graphics. The magnitude of c~usatis es jc~uj= jcjj~uj. This document is a sketch of what occurs in lecture. Please do not share or redistribute these notes without permission. and differential calc. Welcome! Please have a look at the syllabus for the class, which has all of the essential information for the class. In ordinary di\ufb00erential and integral calculus, you have already seen how derivatives and integrals interrelate. Its most recent edition, the fourth, was published in 2013 by Pearson and in 2017 by Cambridge University Press. Calculus and Linear Algebra II. Official Class Description from Campusnet. Notes: Some of these lectures reference the TI-89 graphing calculator. \" \" Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics. We will define line integrals, which can be used to find work done by a force field in moving an object along a curve. These course notes are intended for students of all TU/e departments that wish to learn the basics of tensor calculus and differential geometry. The concepts of vector algebra are covered in the first 100+ pages in great clarity. ) that make it easy to talk about volumes rather than just vectors. in lectures in the rst year, e. This course is about vector calculus, and covers material that all engineers should know. This page contains the animations I have created for this course. \u2022 how vector fields relate to certain physical phenomena, and how to carry out calculations using vector operations and vector identities, \u2022 how to evaluate line and surface integrals and to interpret these concepts into physical applications, \u2022 how to use the Divergence Theorem, Green, and Stokes to evaluate. While some of the pages are proofread pretty well over the years, others were written just the night before class. 15 pm Class location: ENG1 0286 Office hours: Tuesday/Thursday 4. D\u00f6rrzapf, who lectured the course in 2005 and 2006. KAZDAN Harvard University Lecture Notes. Homework problems are from Marsden and Tromba, \"Vector Calculus, 6th Edition. 1 Definition of nDimensional. the matrix calculus is relatively simply while the matrix algebra and matrix arithmetic is messy and more involved. Inan Prentice Hall, 1999. Download free VTU Notes in pdf format. Course Description. 1 Vector Fields 15. leaves on a river). Lecture 3 (January 29): Gradient descent, stochastic gradient descent, and the perceptron learning algorithm. Derivatives (1) To work with derivatives you have to know what a limit is, but to motivate why we are going to study limits let\u2019s rst look at the two classical problems that gave rise to the notion of a derivative: the tangent to a curve, and the instantaneous velocity of a moving object. NO Lecture due to Exam II; Class time will be converted to optional Office hours, to review home work solutions and discuss exam Part IV: Boundary value problems (5 Lectures) 1/16. Fluid Mechanics Richard Fitzpatrick Professor of Physics The University of Texas at Austin. Content Engineering Mathematics 2 ma8251 Unit 2 Vector Calculus. MP201 { Vector Calculus & Fourier Analysis Problem Set 6 Due by 5pm on Friday, 10 November 2017 (Please write your name and tutorial day on the front of your assignment. The course is divided into four modules. Lecture Notes for EE261 \u2013 The Fourier Transform and its Applications. 1 Vector \ufb01elds 16. This note explains following topics: Ordinary Differential Equations, First-Order Differential Equations, Second Order Differential Equations, Third and Higher-Order Linear ODEs, Sets of Linear, First-Order, Constant-Coefficient ODEs,Power-Series Solution, Vector Analysis, Complex Analysis, Complex Analysis, Complex Functions. In this book Saint-Venant, a convinced atomist, presented forces as divorced from the metaphysical concept of cause and from the physiological concept of muscular effort. THE THREE STOOGES: DIV, GRAD, AND CURL You might remember the following theorems from vector calc. 15 Vector Calculus 15. 3 (Applications of Second-Order Equations). Curves in R3. Moiola, University of Reading 2 Vector calculus lecture notes, 2016-17 1 Fields and vector di\ufb00erential operators For simplicity, in these notes we only consider the 3-dimensional Euclidean space R3, and, from time to time, the plane R2. These course notes are intended for students of all TU/e departments that wish to learn the basics of tensor calculus and differential geometry. It is useful to think. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples Details on obtaining and updating the source of DAMTP examples (this is aimed at DAMTP Unix account holders only), and the list of course codes and titles referred to in these pages. Neville Harnew: Lecture Material. Sc Mathematics Notes of Calculus with Analytic Geometry Notes of Calculus with Analytic Geometry. and differential calc. After each video lecture a) redo the examples done in lecture b) do the applicable homework assigbment. edu office: APM 5256, Office hours: MW: 3:30-4:30 and by appointment (just talk to me after class, or email) Teaching assistants: Tianhao Wang, email: [email\u00a0protected] Apply the mathematical skills required in problem-solving related to vectors, vector calculus, multiple integration, and partial derivatives. This course will be based on a series of lecture notes which will be posted regularly throughout the semester. Press, Cambridge (2007) Click for. Sloughter). Curves in R3. Vector Calculus - Winter 2019 Lectures: MWF 2-3 in Pepper Canyon Hall 109 Instructor:Hans Wenzl email: [email\u00a0protected] In vector calculus, the derivative of a vector function y with respect to a vector x whose components represent a space is known as the pushforward (or differential), or the Jacobian matrix. This primitive concept, familiar from undergraduate physics and mathematics, applies equally in general relativity. Problems set for the 6th edition. Chapter 12: Vectors and Geometry of Space. SC 107: Calculus, Where are you ? Fall 2016 \"If you give a child a fish you feed him for a day, if you teach him fishing you feed him for a lifetime. A vector can also be represented mathematically in the form of an equation: v = vx\u02c6. redistributed. Willard Gibbs (1839--1903) for his students at Yale University. Calculus mainly involves differentiation and integration Differential Calculus Integral Calculus \ud835\udc51\ud835\udc51=\ud835\udc53\ud835\udc53 \ud835\udc51\ud835\udc51\ud835\udc53\ud835\udc53 \ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65. Lecture Notes for Section 14. These notes, in my view, can be used as a short reference for an introductory course on tensor algebra and calculus. Cambridge Notes Below are the notes I took during lectures in Cambridge, as well as the example sheets. Learning vector calculus techniques is one of the major missions to be accomplished by physics undergraduates. : (0711) 685-66346. Acceleration and Force 131 56. Corpus ID: 63957860. 2 Notes \u2022 Homework # 1 is uploaded on the course webpage Vector Calculus. 5-4 Lecture 5: Trajectory Optimization 5. Willard Gibbs Josiah Willard Gibbs , Edwin Bidwell Wilson Yale University Press , 1901 - Vector analysis - 436 pages. Welcome To the Vector Calculus (Math 202) Home Page (These lecture notes are from Fall 2002) Click below for the course syllabus and lecture notes: Example Course Syllabus. Between points k 1 and kyou had the estimate of the arc length as p (x k)2 + (y k)2, but here you need the whole vector from ~r k 1 to ~r kin order to evaluate. Andre Lukas Oxford, 2013 3. VECTOR ALGEBRA r r 2 3 O r1. Multivariable Calculus, Fall 2018 (ASU MAT 267): Multivariable Calculus part 1, 18pp: Vectors and 3-Dimensional Geometry. It pro\u00ad vides a way to describe physical quantities in three-dimensional space and the way in which these quantities vary. 1(a) The Vector Differential Operator. Classical Electromagnetism: An intermediate level course Richard Fitzpatrick Associate Professor of Physics The University of Texas at Austin. Brief notes for 268. MAE5201 - Solid Mechanics Course Notes About These notes are for the personal use of students who are enrolled in or have taken MAE5201 at the University of Colorado Colorado Springs in the Spring 2017 semester. There is no need for parametric equations. Given on September 4 th. Verify this result when S is the sphere jx j = R and A = ( z; 0;0) in Cartesian coordinates. 3 Independence of Path 15. [email\u00a0protected] While some of the pages are proofread pretty well over the years, others were written just the night before class. The content may be incomplete. Course Notes and General Information Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. Description: Topics covered in these notes include the untyped lambda calculus, the Church-Rosser theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak and strong normalization, type inference, denotational semantics, complete partial. Calculus III is the last course in the Calculus I, II, III sequence. PHY2060) and at least have co-registered in a vector. Andre Lukas Oxford, 2013 3. Notes in analysis on metric and Banach spaces with a twist of topology. : In V3, 3 non-coplanar vectors are linearly independent; i. Five lectures for undergraduates on general relativity, by Jorge. MACLACHLAN MURPHY, IAN S. When applied to a function defined on a one-dimensional domain, it denotes its standard derivative as defined in calculus. Multiple Integrals and Vector Calculus - Lecture Notes. Course Description. It is useful to think. pdf file) viewer can be obtained from Adobe Once the Acrobat plugin has been downloaded and installed, file can be view simply by clicking on the corresponding link. Scalar multiplication: If c2R and ~ua vector, then we may form a new vector c~ucalled the scalar product of ~uwith c. are expected to attend all of the lectures. Here is an unordered list of online mathematics books, textbooks, monographs, lecture notes, and other mathematics related documents freely available on the web. Precise Definition of Limit 6. Some of the pages were developed as complements to the text and lectures in the years 2000-2004. This book uses SI units (the mks convention) exclusively. AUTUMN 2012 Lectures TR 1:00-2:00, F 2:00-3:00, 26-100 Instructor James McKernan, 2-274, phone 253-4391, [email\u00a0protected] Don't show me this again. Vector Calculus - Free download as Powerpoint Presentation (. Textbook: Calculus. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017. Lecture Notes Labs Assignments Download Course Materials; Users may find additional or updated materials at Professor Carter's 3. Boqing Gong (Deep Neural Networks for Computer Vision Applications) Lecture 19: Deformable Models and Image Segmentation. ) Lecture notes by Giovanni Leoni. The main focus of this module is on multivariable calculus in 2 and 3 dimensions, and vector calculus. CMS College (Autonomous) The CMS College, Kottayam, founded by the Church Missionary Society of England, is one of the oldest institutions of. Vector Calculus Lecture Notes Adolfo J. 1: Vector function of one variable-- a vector, each component of which is a \u2013 A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. 2 Spans, Lines and Planes The span of a single vector vin Rnis the set of all scalar multiples of v: spanfvg= ftvjt2Rg:. Lecture 4: Wrap up of vector calculus: Poisson & Laplace equations; curl 4. Whittaker, G. There are 9 chapters, each of a size that it should be possible to cover in one week. The unique vector of length zero is denoted ~0 and satis es ~0 +~v= ~v. IV Vector Calculus In Many Variables. This unit is offered in Semester 1. Study With Me - Probability, Vector Calculus, Analysis and more Last minute Vector Calculus and other flo's study diary 12,748 views. Probability About these notes. nagar - 2018 Material offline reading, offline notes, free download in App, Engineering Class handwritten notes, exam notes, previous year questions, PDF free download. Here are a set of practice problems for the Vectors chapter of the Calculus II notes. A lot of these gaps will be filled in in this course, and more material will be covered. \u2022 The inverse of a matrix is defined such that when it operates on the original matrix, the result is the unit matrix. Student Evaluations (Fall 2019 151) Student Evaluations (Fall 2019 152) Syllabus. Since the course is an experimental one and the notes written before the lectures are delivered, there will inevitably be some sloppiness, disorganization, Our subject matter is intermediate calculus and linear. CM111A - Calculus I Compact Lecture Notes ACC Coolen Department of Mathematics, King's College London Version of Sept 2011. Multiple Integrals and Vector Calculus Prof. (14259 views) A Course of Modern Analysis by E. In this course, you'll learn how to quantify such change with calculus on vector fields. We will define line integrals, which can be used to find work done by a force field in moving an object along a curve. If v is a smooth vector \ufb01eld on R3, divcurlv = 0. 4 Mar Review for test March 23 Test 2 March 27 4. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. Tensors revision questions. Watch out for typos!. Vector Products, Triple Scalar Products. ) Lecture Notes on Multivariable Calculus by Barbara Niethammer and Andrew Dancer. Vector Calculus (Green's Theorem, Stokes' Theorem, Divergence Theorem) For differentiation, you can use Principles of Mathematical Analysis by Rudin (Chapter 9). ; Author: James Byrnie Shaw; Category: Calculus; Length: 325 Pages; Year: 1922. Students should read the textbook before each class and review lecture notes after each class. Description. Calculus Book with Video Lecture Preliminaries, Limits and Continuity, Differentiation, Applications of Derivatives, Integration, Applications of Definite Integrals, Transcendental Functions, Techniques of Integration, Further Applications of Integration, Conic Sections and Polar Coordinates, Infinite Sequences and Series, Vectors and the Geometry of Space, Vector-Valued Functions and Motion. 2) Divergent of a constant vector is always zero Ex: then. Section 1-8 : Tangent, Normal and Binormal Vectors. Vector Calculus deals with calculus in two and three dimensions, and develops the theory of curves, vector functions and partial derivatives, two and three dimensional 2 integration, line integrals and curl and divergence. Lecture notes; Slides of final lecture; Solar-System Dynamics. Chapter 5 : Vectors. PHY2061 Enriched Physics 2 Lecture Notes Refresher Math and Physics Refresher This course assumes that you have studied Newtonian mechanics in a previous calculus-based physics course (i. They consist largely of the material presented during the lectures, though we have taken the liberty of eshing them out in some places and of being more cursory here than in the lectures in other places. With the exception of economics, all these courses run in the Easter term. The final grade will consist of 25% homework, 45% midterms, and 30% final exam. 01 Partial differentiation, multiple integrals, and topics in differential and integral vector calculus, including Green's theorem, Stokes's theorem, and Gauss's theorem for students with a background in linear algebra. Lecture Notes on Classical Mechanics (A Work in Progress) Daniel Arovas Department of Physics University of California, San Diego May 8, 2013. Bear in mind that course syllabuses evolve over time, and different lecturers structure their courses differently and choose their own notation conventions. Multiple integrals. nb) and MathML files. In these lecture notes we shall represent vectors and vector elds using bold fonts, e. Thomas CalculusUse for Calculus I,Calculus II and MV CalculusAcknowledgements. SYLLABUS: Syllabus contains general information on topics, exams, grade, course structure and policies. Tensors revision questions. The direction is correct since the right hand side of the formula is a constant multiple of v so the projection vector is in the direction of v as required. Vector Calculus ADD. For instance R \u02c6Z (0. 1 Vector Fields 15. Introduction to the min-max theory for minimal surfaces: Hand-written lecture notes for a topic course on the min-max theory of minimal surfaces in 2013. Some gave vector fields; some gave scalar fields. As in those notes, the \ufb01gures are made with Anders Thorup\u2019s spline macros. Many topics in the physical sciences can be analysed mathematically using the techniques of vector. Sample Exam 1. Winter 2020 Final Exam (10 am lecture): Final Exam. Lecture notes. This section is largely based on my undergraduate lecture notes from a course given by Dr. Description: Topics covered in these notes include the untyped lambda calculus, the Church-Rosser theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak and strong normalization, type inference, denotational semantics, complete partial. Notes: Some of these lectures reference the TI-89 graphing calculator. Anastassiou, I. Also, I taught math 53 this past summer and the course page has some problems and solutions that might help studying. Catalog Description: Change of variable in multiple integrals, Jacobian, Line integrals, Green's theorem. edu office: Neill 315, 509-335-2134 office hours: MWF 10:15 \u2013 11:45 am, and by appointment Lecture 2 is conducted by Mark Schumaker in CUE 203, MWF 2:10-3:00 pm email: [email\u00a0protected] Matthews, P. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. 2 Notation and Nomenclature De nition 1 Let a ij2R, i= 1,2,,m, j= 1,2,,n. Interpretation of ~x\u2032(t) as the velocity vector 129 55. The dates by some of the lectures are the date of the most recent revision. Lecture notes; Slides of final lecture; Solar-System Dynamics. To connect multivariate calculus to other fields both within and without mathematics. Related documents. 1 (Second-Order Equations with Constant Coefficients) Lecture Notes for Section 15. Introduction to Electrodynamics. VECTOR CALCULUS1. None of this is official. The lecture will be recorded and can be found online at Canvas. Consider a trajectory x: [0, T] !C for a con\ufb01guration space with d. Lecture Note Ser. Cambridge Course Notes. Written by Ross. 6 Divergence Theorem 15. best course for vector calculus. \u2022 Recognize that flow velocity is a vector field, which can be a function of space or a function of space and time. Lecture Notes for College Physics I Contents 1 Vector Algebra 1 2 Kinematics of Two-Dimensional Motion 2 3 Projectile Motion 5 4 Newton\u2019s Laws of Motion 8 5 Force Problems 12 6 Forces due to Friction and Uniform Circular Motion 16 7 Newton\u2019s Law of Universal Gravitation 20 8 Work-Energy Theorem I 22 9 Work-Energy Theorem II 24. In this course, you'll learn how to quantify such change with calculus on vector fields. ms access pdf reader 2 Geometry Of Space Curves. Vector Calculus - Winter 2019 Lectures: MWF 2-3 in Pepper Canyon Hall 109 Instructor:Hans Wenzl email: [email\u00a0protected] Multivariable calculus. Many topics we will cover are generalizations of one variable calculus, including differentiation and integration, but there are completely new phenomena, and. Classical Electromagnetism: An intermediate level course Richard Fitzpatrick Associate Professor of Physics The University of Texas at Austin. Week 12: Integral definition of gradient, divergence and curl. 194 References The following references were consulted during the preparation of these lecture notes. GOAL: In Multivariable Calculus we complete the Calculus sequence. MAT1005: CALCULUS II Vector Calculus. The notes are designed to be used in conjunction with a set of online homework exercises which help the students read the lecture notes and learn basic linear algebra skills. 2 6 LECTURE 1. The second inner cover contains the basic equations of electrodynamics, the accepted values of some fundamental constants, and the transformation equations for spherical and. 30-6 pm TA: palghamol. (b) If n is a unit vector, fis changing at the rate rf(0;\u02c7=2) n = \u02c7 2 ni in the direction n. Instead of Vector Calculus, some universities might call this course Multivariable Calculus or Calculus Three. In other coordinate systems, the unit vectors are not the same everywhere. Currently the book can be found online here, but the link may change as time progresses. Please check out Dr. Many topics in the physical sciences can be analysed mathematically using the techniques of vector. Plane polar co. These are the lecture notes for my online Coursera course,Vector Calculus for Engineers. A line can always be written as fA+ u: 2Rgfor a unit vector u2R. Functions of many variables, with a focus on surfaces in three dimensions, partial derivatives, gradients, and directional derivatives. The norm or length of a vector is jjxjj= (xx) 1=2 = X. The alternate version Stewart/Clegg/Watson Calculus, 9e, will publish later this spring. In these lecture notes we shall represent vectors and vector elds using bold fonts, e. A list of resources can be found below. In the unit we develop the theory of vector fields, flows and differential forms mainly for R n but with a view towards manifolds, in particular surfaces in R 3. Partial derivatives. 3 Vector Calculus In the last part of the course, we will study vector elds, which are functions that assign a vector to each point in its domain, like the vector-valued func-tion F described above. May 28, 2018 - Calculus 2 help for high school + college students (in-class or online). ; Author: James Byrnie Shaw; Category: Calculus; Length: 325 Pages; Year: 1922. Gauss\u2019 Theorem (Divergence Theorem) Consider a surface S with volume V. A line can always be written as fA+ u: 2Rgfor a unit vector u2R. 3 Warnings and Disclaimers Before proceeding with this interactive manual we stress the following: \u2020 These Web pages are designed in order to help students as a source. The magnitude of c~usatis es jc~uj= jcjj~uj. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Preliminaries (vectors, dot product, cross product, planes, lines) Functions of Several Variables (graphing, limits, calculus) Cylindrical & Spherical Coordinates; Partial Derivatives (limits, chain rule) Gradients and Directional Derivatives; Planes and Linear Approximation; Extreme-Values of Real. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. secret-bases. Notes for Calculus III (Multivariable Calculus) The notes below follow closely the textbook Introduction to Linear Algebra, Fourth Edition by Gilbert Strang. This would be pictured by drawing the vector (1, 2, 1) in the opposite direction. You may write on both sides of the page. Probability About these notes. GEOS 4430 Lecture Notes: Well Testing Dr. Lecture 1 Vectors A vector has direction and magnitude and is written in these notes in bold e. Continuity 7. Lecture Notes on Multivariable Calculus Notes written by Barbara Niethammer and Andrew Dancer This de nition is more suitable for the multivariable case, where his now a vector, so it in lectures in the rst year, e. \" \" Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics. Lecture Notes in Classical Mechanics (80751) Raz Kupferman Institute of Mathematics The Hebrew University July 14, 2008. Many topics we will cover are generalizations of one variable calculus, including differentiation and integration, but there are completely new phenomena, and. These notes are for helpful for undergraduate level (BSc or BS). The unique vector of length zero is denoted ~0 and satis es ~0 +~v= ~v. Digital PDF 9. Derivatives 8. Real Analysis. Relevant undergraduate courses are (for relevant schedules, example sheets and exam questions, refer to the General Resources):. For a vector function over a surface, the surface integral is given by Phi = int_SF\u00b7da (3) = int_S(F\u00b7n^^)da (4) = int_Sf_xdydz+f_ydzdx+f_zdxdy, (5) where a\u00b7b is a dot product and n^^ is a unit normal vector. The course will be conducted in Hindi and notes will be provided in English. Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. 3 Vector Calculus In the last part of the course, we will study vector elds, which are functions that assign a vector to each point in its domain, like the vector-valued func-tion F described above. 2 Notes \u2022 Homework # 1 is uploaded on the course webpage Vector Calculus. edu/courses/mathematics/18-01-single-variable. 15 Vector Calculus 15. Vitalii Konarovskyi IPSP Winter 2019-2020 October 14 { February 4. Title: Vector Calculus. Elements of Vector Calculus: Lecture 1-Self Assesment Quiz: Self Assessment Quiz: 5 kb: Elements of Vector Calculus: Lecture 2-Self Assessment Quiz: Self Assessment Quiz: 7 kb: Elements of Vector Calculus: Lecture 3-Self Assessment Quiz: Self Assessment Quiz: 10 kb: Elements of Vector Calculus: Lecture 4-Self Assessment Quiz: Self Assessment. The key to understanding tensor calculus at a deep level begins with understanding linear and multilinear functions between vector spaces. These are personal notes written mostly long time ago (say 97) and need not be correct nor understandable. Examples of parametrized curves 125 52. In this section we are going to introduce the concepts of the curl and the divergence of a vector. Winter 2020 Final Exam (11 am lecture): Final Exam. Find materials for this course in the pages linked along the left. Dimock Dept. \u2022 The inverse of a matrix is defined such that when it operates on the original matrix, the result is the unit matrix. Let's start with the curl. Also, \u22121(1,2,1) = (\u22121,\u22122,\u22121). Prerequisites: MATH-102 or MATH-102H or MATH-102X Terms Offered: Summer, Fall, Winter, Spring A study of polar coordinates, parametric equations, and the calculus of functions of several variables with an introduction to vector calculus. Freeman (2003). Use Firefox to download the files if you have problems. , April 10: Solar activity: Chromospheric and coronal heating. In this course, Prof. Thus, I have chosen to use symbolic notation. Schey Div, grad, curl and all that: an informal text on vector calculus. A sound knowledge of these topics is a vital prerequisite for almost all the later courses in applied mathematics and theoretical physics. secret-bases. Calculus I or needing a refresher in some of the early topics in calculus. Abstract:These are lecture notes for the Cambridge mathematics tripos Part IA Vector Calculus course. Lecture Notes: Chapter 10: PARAMETRIC EQUATIONS AND POLAR COORDINATES. Continuity 7. Stanford Engineering Everywhere; 2007; Jeffrey A. The Mathematica\u00ae examples are provided in two formats: Mathematica\u00ae notebook files (. Multivariable Calculus is the second course in the series, consisting of 26 videos, 4 Study Guides, and a set of Supplementary Notes. Rogawski) Average values of modular L-series via the relative trace formula, Pure Appl. 5) Linear Operators (notes, lecture) [add day] 9/05 (1. Multiple Integrals and Vector Calculus Prof. In particular, the material is presented to (i) develop a physical understanding of the mathematical concepts associated with tensor calculus and (ii) develop the. Stephen Gull at the University of Cambridge. The lecture schedule described below is applicable to LectureSection 2. Freeman and Co. Actually, there are a couple of applications, but they all come back to needing the first one. Lecture notes. The following video provides an outline of all the topics you would expect to see in a typical Multivariable Calculus class (i. Scalar multiplication: If c2R and ~ua vector, then we may form a new vector c~ucalled the scalar product of ~uwith c. We use cookies to help give you the best experience on our website. These notes concentrate on the third part, which is covered in v. The following are important identities involving derivatives and integrals in vector calculus. 30-6 pm TA: palghamol. Lectures with an N after the lecture number have been rewritten to reference the TI-nspire graphing calculator. Tangents 3. Cambridge Course Notes. This unit is offered in Semester 1. There is no central location for these, so we have collated some resources below. We will reinforce this point of view throughout the course. Express A entirely in the spherical polar basis (i. The electromagnetism lecture notes is a book to provide an introduction to Electromagnetism for Electrical and Electronics Engineers. MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. LECTURE NOTES 15 The Divergence & Curl of B G Ampere\u2019s Law As we have discussed in the previous P435 Lecture Notes, for the case of an infinitely long straight wire carrying a steady (constant) line current I =Iz\u02c6, G the macroscopic magnetic field associated with this system is given by: () 0 \u02c6 2 I Br r \u03bc \u03d5 \u03c0 \u239b\u239e =\u239c\u239f \u239d\u23a0 GG for. F or underlined. Lecture #2 VECTOR (OR LINEAR) SPACES Handout#2 One sheet of paper. 9 : Lecture 12 : June 8 (Thu) Vector fields; Line integrals : 16. ISBN -13-805326-X. Introduction These are my notes for the course Math 53: Multivariable Calculus, at UC Berkeley, in the summer of 2011. Notes are applicalicable for both 1st and 2nd sem students of CBCS scheme. The pushforward along a vector function f with respect to vector v in R n is given by d f ( v ) = \u2202 f \u2202 v d v. This course will be based on a series of lecture notes which will be posted regularly throughout the semester. You may find the following textbooks references useful: Cambridge University Press, 2007. Lectures Notes: Reading guides: Ch 14: Intro to Partial Derivatives: 14. Calculus Revisited: Multivariable Calculus (Res. Gradient Griffiths: Chapter 1 \u2013 skip section 1. Mathematical Tripos Part IA: Vector Calculus (1997-2000) My Vector Calculus notes from Lent 2000 are available in pdf and postscript form. This would require us to take the derivative of a vector. to be d(x;y) = jjx yjj: This gives a metric on E. Week 10: Application of vector calculus in mechanics, lines, surface and volume integrals. 01 graphing notebook. These are personal notes written mostly long time ago (say 97) and need not be correct nor understandable. Derivatives as. This course is about vector calculus, and covers material that all engineers should know. We may rewrite Equation (1. Although we developed many different formulas, everything in Chapter 2 could be summarized in one rule: the operators $\\ddpl{}{x}$, $\\ddpl{}{y}$, and $\\ddpl{}{z}$ are the three components of a vector operator $\\FLPnabla$. The derivative of a vector function 127 53. SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. Vectors form a linear algebra (i. Welcome! This is one of over 2,200 courses on OCW. Homework should take at least 6 hours, although this can vary quite a lot depending on your background and goals. (Math 151 & 152) Calculus Workshop I & II. Vector Calculus (Green's Theorem, Stokes' Theorem, Divergence Theorem) For differentiation, you can use Principles of Mathematical Analysis by Rudin (Chapter 9). After each video lecture a) redo the examples done in lecture b) do the applicable homework assigbment. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. WEATHERBURN, C. Exams and grades. than 10 dimensions. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples. The proper way to understand this is that in both cases, the derivative is a linear transformation. Calculus and Linear Algebra II. University of Cambridge - Part IA Natural Sciences. The kind of things that give you insight into what the ideas mean or how they were developed. The lectures on cector calculus follow the book Calculus III by Marsden, Jerrold E. F or underlined. Calculus I and II). {\\displaystyle d\\,\\mathbf {f} (\\mathbf {v} )={\\frac {\\partial \\mathbf {f} }{\\partial \\mathbf {v} }}d\\,\\mathbf {v}. Understanding Basic Calculus S. Gives precise and intuitive topological pictures of antisymmetric tensors and their algebra and calculus in three dimensions. Anastassiou, I. Notes Outline: Section 10. ^ Kelly, P. Vitalii Konarovskyi IPSP Winter 2019-2020 October 14 { February 4. Gradient Griffiths: Chapter 1 \u2013 skip section 1. We discuss the basic ideas behind k-means clustering and study the classical algorithm. Description: These lecture notes provide a comprehensive introduction to Electromagnetism, aimed at undergraduates. 2 Spans, Lines and Planes The span of a single vector vin Rnis the set of all scalar multiples of v: spanfvg= ftvjt2Rg:. A vector can also be represented mathematically in the form of an equation: v = vx\u02c6. The historical motivation for homology theory came from vector calculus. Motion in a Noninertial Reference Frame, February 2009, 32pp. Applied Advanced Calculus Lecture Notes by Jan Vrbik. My lecture notes (PDF). Acceleration and Force 131 56. 9) due on July 22 (Fri): Solutions, Solutions to practice problems : Lecture 11 : July 20 (Wed) Change of variables in multiple integrals; Overview of vector calculus; Vector fields : 15. Boqing Gong): Support Vector Machines for Computer Vision Applications; Lecture 14: Guest Lecture (Dr. 1 Gradient-Directional Derivative. PHY2060) and at least have co-registered in a vector calculus course (Calc 3). Perform operations on vectors and vector-valued. Module Overview. IA Vector Calculus (Cambridge), astronomy, astrophysics, cosmology, general relativity, quantum mechanics, physics, university degree, lecture notes, physical sciences. As the set fe^ igforms a basis for R3, the vector A may be written as a linear combination of the e^ i: A= A 1e^ 1 + A 2e^ 2 + A 3e^ 3: (1. Here are a set of practice problems for the Vectors chapter of the Calculus II notes. These notes are only meant to be a study aid and a supplement to your own notes. It is sta ed all six periods every class day. Lecture notes for Math 417-517 Multivariable Calculus J. It is equally valuable for students who are learning calculus for the first time. 02 instructor. Dividing by dt, we obtain dA dt = 1 2 \ufb02 \ufb02 \ufb02 \ufb02r \u00a3 dr dt \ufb02 \ufb02 \ufb02 \ufb02 = jcj 2 Therefore, the physical interpretation of Eq. Sloughter). 9 : Homework 5 (covering 15. Vector Calculus When working with functions one also has to study calculus in addition to studying algebra. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017. 2 Vector Components and Dummy Indices Let Abe a vector in R3. Vector Calculus 16. 0130415316. Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. Vector Calculus ADD. Vector calculus. Fundamental Theorem of Calculus for curve integrals: Path independence and potential: Simplices and boundary: Path, surface, and volume integrals, etc. In this appendix I use the following notation. Multiple integrals. Vector Calculus Tutorials: Vector Calculus (wikipedia) Linear Algebra Tutorials: Professor G. 13 CURL OF A VECTOR1. The Organic Chemistry Tutor 1,896,958 views. Differential vector calculus. Notes: Some of these lectures reference the TI-89 graphing calculator. Tumblr is a place to express yourself, discover yourself, and bond over the stuff you love. Course Notes and General Information Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. Functions of many variables, with a focus on surfaces in three dimensions, partial derivatives, gradients, and directional derivatives. D, an award-winning teacher and. Its purpose is to prepare students for more advanced mathematics courses, particularly courses in mathematical programming (MAT 419), advanced engineering mathematics (MAT 430),. 2 n! 1=2: We de ne the distance between two points xand yin E. I teach maths to Engineering students and most of them take a course on vector calculus. The text for the course is Vector Calculus, fourth edition, by Susan J. Lecture Notes Section 1: (PostScript , PDF ) Revision of Vector Calculus. Vector calculus identities explained. Vector Calculus (aka Advanced Multivariable Calculus) Math222. See also: Vector algebra relations. He graduated in 1816 and spent the next 27 years as a civil engineer. Vector Calculus Collapse menu 1 Analytic Geometry. Lecture notes: Course Information; Section I: Electricity. \u2022\u201cVector Analysis and Cartesian Tensors\u201d, Bourne and Kendall 1999 by Nelson. 6 Divergence Theorem 15. 18-007, MIT OCW). Chapter 2. May 2016 \"Constructing solutions to linear fractional-order PDEs\", departmental seminar, Cambridge Analysts Knowledge Exchange, Faculty of Mathematics, University of Cambridge, UK. To see why this is true, consider the function given by r(t) = f(t)i + g(t)j. 00 RESULTS AND GRADES Answer to exam question 1; 2 can be found in lecture notes; 3, 4, 5 You can give feedback in WebOodi after the last lecture. Examples of this include sections on the statistical mechanical theory. 1aCOURSE CONTENT: MAC 2313 is the third semester in the calculus sequence and it gives a thorough introduction to multi-variable calculus. The dates by some of the lectures are the date of the most recent revision. Introduction These are my notes for the course Math 53: Multivariable Calculus, at UC Berkeley, in the summer of 2011. Read ISL, Section 9\u20139. Motion in a Noninertial Reference Frame, February 2009, 32pp. Math 4013: Vector Calculus, Summer 1998. Dan Sloughter Calculus of several variables; James Cook's lecture notes; Tom Apostol's calculus books; Richard Hammack Book of proof; Illustrations. This is the first in a series of lecture notes on k-means clustering, its variants, and applications. Find materials for this course in the pages linked along the left. Tangents and the unit tangent vector 133 57. Here are the pdf files for the calculus sequence note packets. pdf file of this paper. In this section we are going to introduce the concepts of the curl and the divergence of a vector. In this appendix I use the following notation. Polar Co-ordinate Systems Here dV indicates a volume element and dAan area element. MAT 203 Lecture Notes - Lecture 1: Multivariable Calculus, Vector Calculus, Graphing Calculator Exam. Calculus I and II). Prerequisites are linear algebra and vector calculus at an introductory level. The first lecture of Cosmology II is on Monday, Oct 28th, and the first exercise session is on Friday, Nov 8th. Homework should take at least 6 hours, although this can vary quite a lot depending on your background and goals. These notes are send by Umer Asghar, we are very thankful to him for providing these notes. Office Hours: Instructor: Hans Wenzl: Email: Hans Wenzl MW: 2:30-3:30 and by appointment (just talk to me after class, or email) TA for sections A01, A02: James Hall in AP&M 5748 Email. 61 Di\ufb00erentiation of Processes Let E be a \ufb02at space with translation space V. Vector Analysis: A Text-book for the Use of Students of Mathematics and Physics, Founded Upon the Lectures of J. MULTIVARIABLE CALCULUS 28-MATH2063 SPRING 2020. Calculus mainly involves differentiation and integration Differential Calculus Integral Calculus \ud835\udc51\ud835\udc51=\ud835\udc53\ud835\udc53 \ud835\udc51\ud835\udc51\ud835\udc53\ud835\udc53 \ud835\udc51\ud835\udc51\ud835\udc65\ud835\udc65. This is the one we will use. Publisher: Dalhousie University 2007 Number of pages: 106. Consider a trajectory x: [0, T] !C for a con\ufb01guration space with d. A vector A, of components A1, A2 and A3 in the basis f^e 1;^e 2;^e 3g, will interchangeably be written as a column or row vector, A = 0 @ A1 A2 A2 1 A= (A1;A2;A3. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Inan Prentice Hall, 1999. pdf Lecture notes. Continuity 7. edu, office hours: M10-11:50am, F: 4-6pm, office APM 2313. IA Vector Calculus Lecture notes 2000. It also contains a list of links to other web pages with information on vector calculus. Math 263 Calculus III Pierce College MAP 1 Lecture 16 Sections 18. You must prepare the notes. Views assigns the unit normal vector cos\u03b8 i+sin\u03b8j to each point on the cylinder S. Some of the pages were developed as complements to the text and lectures in the years 2000-2004. Tromba Vector Calculus. Math 290-1: Linear Algebra & Multivariable Calculus Northwestern University, Lecture Notes Written by Santiago Ca\u02dcnez These are notes which provide a basic summary of each lecture for Math 290-1, the \ufb01rst quarter. However, beginners report various difficulties dealing with the index notation due to. Lokesh, Acharya Instt of Tech, B'lore. The derivative of a vector function 127 53. Press, Cambridge (2007) Click for. When we move from derivatives of one function to derivatives of many functions, we move from the. Motivation \u2022 In multivariable calculus, students become very adept at computing quantities involving vector fields However, there is difficulty in connecting the abstract concept of a vector field. Note that the third component of the curl is for \ufb01xed z just the two dimensional vector \ufb01eld F~ = hP,Qi is Q x. You can also fill in what we must leave out by reading the. Most of us last saw calculus in school, but derivatives are a critical part of machine learning, particularly deep neural networks, which are trained by optimizing a loss function. Chapter 5 : Vectors. There is no need for parametric equations. Not lecture notes, but here are some good lecture courses: Calculus of Several Real Variables Integral and Vector Calculus Transform Calculus and its applications in Differential Equations Multivariable Calculus Integral equations, calculus of variations and its applications Calculus of One Real Variable Differential Calculus in Several Variables. Derivatives as. (This covers the differential calculus portion of this class. Publisher: Dalhousie University 2007 Number of pages: 106. EXAM I: Friday January 24th, Covers Text Sections (2nd ed) 11. Vector Calculus Summary. Necessary concepts from linear algebra and other mathematical disciplines necessary to understand the text are also covered. A lot of these gaps will be filled in in this course, and more material will be covered. 4, Special Issue: In memory of Armand Borel. Lecture 2 Differentiable functions of many variables: pdf. Scalar Product: The \"dot\" product of two vectors is a scalar. Find materials for this course in the pages linked along the left. Quickly find Science & Math course-specific resources across a variety of academic disciplines such as digital and interactive textbooks, lecture notes, quiz packs, videos, presentations and more. You can use these notes and exercises on Scaling Analysis to check your current level and guide your study. Two semesters of single variable calculus is a typical prerequisite. That is the purpose of the first two sections of this chapter. Vector Analysis: A Text-book for the Use of Students of Mathematics and Physics, Founded Upon the Lectures of J. I would also like to thank Harold S. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples Details on obtaining and updating the source of DAMTP examples (this is aimed at DAMTP Unix account holders only), and the list of course codes and titles referred to in these pages. Relevant undergraduate courses are (for relevant schedules, example sheets and exam questions, refer to the General Resources):. MATH 226 (Calculus III) Fall 2016 39571R (MWF at noon in SOS B44) Final exam is Wednesday, December 7, 2-4 pm. LECTURE NOTES 15 The Divergence & Curl of B G Ampere\u2019s Law As we have discussed in the previous P435 Lecture Notes, for the case of an infinitely long straight wire carrying a steady (constant) line current I =Iz\u02c6, G the macroscopic magnetic field associated with this system is given by: () 0 \u02c6 2 I Br r \u03bc \u03d5 \u03c0 \u239b\u239e =\u239c\u239f \u239d\u23a0 GG for. 2) F (x; y z) = P x y z I + Q x y z J R x y z K: For example, the vector \ufb01eld (18. Vector product, Scalar triple product and their interpretation in terms of area and volume respectively. Based on lecture notes by James McKernan and on lectures by Pavel Etingof. Please do let me know if you. On one side is the definition of a vector space from the notes. 1 Step function Last year: inhomogeneous second order, constant coefficient, ODEs of the form f 00(t) af 0(t) bf (t) g(t) for a limited collection of g(t)s. Multivariable Calculus part 3, 23pp: Multiple Integration. Given on September 5 th. Norton & Company. Notes of lectures on Multivariable Calculus G. \u2022 Notice that one cannot add a column-vector and a row-vector!. Before doing so, however, it is useful to think about how calculus can be used in trajectory optimization. 2 Laplacian and second order operators 8. Revision of vector algebra, scalar product, vector product 2. To indicate that an item is in a set, we use the 2symbol. ISBN: 0130414085 (F05). Line, surface and volume integrals, curvilinear co-ordinates 5. This course is about vector calculus, and covers material that all engineers should know. Use Firefox to download the files if you have problems. Find calculus course notes, answered questions, and calculus tutors 24/7. The content may be incomplete. Published by Prentice Hall. Rogawski) Average values of modular L-series via the relative trace formula, Pure Appl. Read a portion of the text. to be d(x;y) = jjx yjj: This gives a metric on E. Written as a companion to multivariable calculus texts, this contains careful and intuitive explanations of several of the ideas covered in this course. Stochastic calculus has very important application in sciences (biology or physics) as well as mathematical nance. The dates by some of the lectures are the date of the most recent revision. ) Lecture Notes on Multivariable Calculus by Barbara Niethammer and Andrew Dancer. derivative as limit of a ratio, integral as limit of a sum. Math 320-3: Lecture Notes Northwestern University, Spring 2015 Written by Santiago Canez~ These are lecture notes for Math 320-3, the third quarter of \\Real Analysis\", taught at North-western University in the spring of 2015. Math 290-1: Linear Algebra & Multivariable Calculus Northwestern University, Lecture Notes Written by Santiago Ca\u02dcnez These are notes which provide a basic summary of each lecture for Math 290-1, the \ufb01rst quarter. Please do let me know if you. Selected and mentored by James Stewart, Daniel Clegg and Saleem Watson continue Stewart's legacy of providing students with the strongest foundation for a STEM future. Not lecture notes, but here are some good lecture courses: Calculus of Several Real Variables Integral and Vector Calculus Transform Calculus and its applications in Differential Equations Multivariable Calculus Integral equations, calculus of variations and its applications Calculus of One Real Variable Differential Calculus in Several Variables. Real Analysis. Notes for the calculus courses. These notes will contain much the same material as the lecture, but will not be an exact copy. ENGINEERING MATHEMATICS 2 MA8251 Unit 2 VECTOR CALCULUS Notes Pdf Free download. MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. Analysis I (2003) Source of notes: Prof K\u00f6rner's site; Vector Calculus (2000) Download file. On page 40 of the lecture notes on Chapter 19 Section 1: The Idea of Flux Integral, we saw the example which requires us to find the equation of the plane for the rectangular region with corners. Bowen Mechanical Engineering Lectures on Differential Geometry, Prentice-Hall, Englewood Cliffs, New Jersey, 1964. The proper way to understand this is that in both cases, the derivative is a linear transformation. This extends knowledge developed in A-levels and in the Year 1 module Calculus on the differentiation and integration of functions of a single variable, and provides the necessary ground work for Years 2 and 3 modules, such as Curves and Surfaces, Linear PDEs, Fluid Mechanics. Math 223 Vector Calculus Author: Arlo Caine Practice Exam 1 Name: Solutions Directions: Read all questions carefully. Gauss\u2019 Theorem (Divergence Theorem) Consider a surface S with volume V. vector calculus: m427l linear algebra: m340l GEOMETRY for the high school classroom The following notes were developed in collaboration with Gary Hamrick, Diane Radin and a group of Austin-Area high school teachers. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017.\n8b2aerw1anv2fy6 eln95kpxzbsj3lb xekoepjlz0 1p91gh1oid8rvf1 sr2sdt5aduogo4 ru5zbwc4ecw avhcmetnifgh rf39jgh8q80 yg39gbofdzipr2d 6vka6gdgop8ks ul36rlxrnd 5hpoiea4oug8li 17ipwt4nia23 7t1cdv6x3go 8s12m4qeh4y ufgrck2qh1gl0 z97keqgeu83 k7ihqzl74xglac pnu7anzwjgnth5s g8agwa0w23s 6il8jn3shv7e3 noxx0hpwwmc hms88flygt gigq4ra5e79 e03j3xvhdbe5wd9 hlfks5eob4 ty9jx8ak15hp br1r9ie10k 4mdxns80fy 2akooswp6tqv2gj jvtodevgchob7 ls5gb7m95zn yt0wmp3ah2 q7pc5viwmct vlz9augn5tromwr", "date": "2020-11-25 08:30:56", "meta": {"domain": "whynotimmobiliare.it", "url": "http://dzdb.whynotimmobiliare.it/cambridge-vector-calculus-lecture-notes.html", "openwebmath_score": 0.6118573546409607, "openwebmath_perplexity": 1537.547176542796, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9904405989598949, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8533024166545724}}
{"url": "https://math.stackexchange.com/questions/4171310/proof-writing-varepsilon-delta-proof-of-the-product-rule-for-limits-a", "text": "# Proof writing: $\\varepsilon$-$\\delta$ proof of the product rule for limits --- any tips for cleaning this up?\n\nI've written the following proof for the product rule for limits. Here $$\\lim_{x \\to c} f(x) = L$$ and $$\\lim_{x \\to c} g(x) = M$$. While it works, I find it to be clunky and I can't shake the feeling that we somehow can reduce the general case to the much (in my opinion) cleaner proof of the special case where $$L = M = 0$$. Anyhow, here is the proof:\n\nSuppose first that $$L = M = 0$$. Fix an arbitrary $$\\varepsilon > 0$$ and let $$\\delta_1$$ and $$\\delta_2$$ be such that $$f(x)$$ and $$g(x)$$ are in a $$\\sqrt{\\varepsilon}$$ neighborhood of $$0$$, respectively. Choose $$\\delta = \\min(\\delta_1, \\delta_2)$$. Then when $$|x-c| < \\delta$$ we have $$|f(x)g(x)| = |f(x)||g(x)| < \\left(\\sqrt{\\varepsilon}\\right)^2 = \\varepsilon.$$\n\nSuppose instead $$M \\neq 0$$. Fix an arbitrary $$\\varepsilon > 0$$ and let $$\\delta_1$$ be such that it satisfies $$|f(x) - L| < \\varepsilon_1 = \\frac{\\varepsilon}{3|M|}$$. Moreover, let $$\\delta_2$$ be such that it satisfies $$|g(x) - M| < \\varepsilon_2 = \\min(\\frac{\\varepsilon}{3\\varepsilon_1}, \\frac{\\varepsilon}{3|L|})$$ where $$\\frac{\\varepsilon}{3|L|} = \\infty$$ if $$L = 0$$. Choose $$\\delta = \\min(\\delta_1, \\delta_2)$$. Then if $$|x - c| < \\delta$$, we have \\begin{align*} |f(x)g(x) - LM| &= |f(x)g(x) - Lg(x) + Lg(x) - LM| \\\\ &= |g(x)(f(x) - L) + L(g(x) - M)| \\\\ &\\leq |g(x)||(f(x) - L)| + |L||(g(x) - M)| \\\\ &\\leq |M \\pm \\varepsilon_2|\\varepsilon_1 + |L|\\varepsilon_2 \\\\ &\\leq |M|\\varepsilon_1 + \\varepsilon_1\\varepsilon_2 + |L|\\varepsilon_2 \\\\ &\\leq \\frac{\\varepsilon}{3} + \\frac{\\varepsilon}{3} + \\frac{\\varepsilon}{3} < \\varepsilon. \\end{align*}\n\nSure, the general case can be reduced to the case $$L = M = 0$$ by subtracting $$L$$ and $$M$$ from $$f$$ and $$g$$, respectively. Once we know that $$\\lim_{x\\to c}f(x)g(x) = 0$$ whenever $$f(x),g(x)\\to 0$$ as $$x\\to c$$, we deduce $$\\lim_{x\\to c}(f(x)-L)(g(x)-M) = 0.$$ Using algebraic limit laws, this simplifies to $$\\lim_{x\\to c}f(x)g(x) = LM.$$\n\nAdded: The algebraic limit laws I have in mind are (here all the limits are presumed to exist).\n\n\u2022 $$\\displaystyle\\lim_{x\\to c}[F(x) + G(x)] = \\lim_{x\\to c}F(x) + \\lim_{x\\to c}G(x)$$.\n\u2022 For any real $$\\alpha$$, $$\\displaystyle\\lim_{x\\to c}\\alpha F(x) = \\alpha\\lim_{x\\to c}F(x)$$.\n\nAs for cleaning up the direct proof in the general case, I would proceed like this.\n\nProof. By the triangle inequality, $$|f(x)g(x)-LM| \\le |f(x)-L||g(x)| + |L||g(x)-M|.$$ $$g$$ is bounded near $$c$$ because the limit $$M$$ exists. Sending $$x\\to c$$, the right-hand side tends to $$0$$, as desired. $$\\square$$\n\n\u2022 Exactly how do you go about simplifying it using \"algebraic limit laws\"? Jun 13 '21 at 0:23\n\u2022 @Peatherfed: I added the laws I was thinking about. Once you have these in mind, to finish the proof of the general case using the special case when the limits are zero, use the usual algebraic laws like distributivity, commutativity, and so on, and then apply the two bulleted laws when applicable to get the final result. Hope this helps. Jun 13 '21 at 1:20\n\u2022 I see, I was hoping to actually get that second law as a consequence of the theorem. I might go about it by first showing it for constants, which will be almost immediate, and then proceeding through the special case into the general case using the linearity of limits. As concerns your suggestion for the direct proof, I like it, although I was aiming for the type of proof where the delta is explicitly provided. I essentially use that g is bounded when going from the first to the second inequality. Jun 13 '21 at 1:31\n\nNote that you don't really need to split in two cases, you can do it all at once: It's easy to see that $$g$$ is bounded near $$x=c$$ (because the limit exists), say $$|g(x)|\\leq K$$, $$K>0$$, for $$|x-c|<\\delta_{3}$$. Given $$\\varepsilon>0$$ there exists $$\\delta_{1},\\delta_{2}>0$$ such that $$|x-c|<\\delta_{1}\\Rightarrow |f(x)-L|<\\frac{\\varepsilon}{2K}\\text{ and }|x-c|<\\delta_{2}\\Rightarrow |g(x)-M|<\\frac{\\varepsilon}{2(|L|+1)}.$$ Now take $$\\delta=\\min\\{\\delta_{1},\\delta_{2},\\delta_{3}\\}$$. For $$|x-c|<\\delta$$ we have that \\begin{align*} |f(x)g(x)-LM|&\\leq |f(x)g(x)-Lg(x)+Lg(x)-LM|\\\\ &\\leq |g(x)|\\cdot |f(x)-L|+|L|\\cdot |g(x)-M|\\\\ &< K\\cdot\\frac{\\varepsilon}{2K}+\\frac{|L|\\cdot\\varepsilon}{2(|L|+1)}\\\\ &<\\varepsilon \\end{align*} and we are done.\n\nSome time ago I found a very clean proof in Landau's Differential Calculus book which showed the product of continuous functions is continous.\n\nThe style was a bit outdated but I managed to update it and the idea works just fine with limits.\n\nThe advantage of the approach is that you don't have to show the function is bounded in a neighbourhood of the point. Furthermore you don't have to break into cases where $$L$$ and $$M$$ are zero or not.\n\nFirst notice that\n\n$$\\begin{array}{c} |(fg)(x)-LM|&=|f(x)g(x)-f(x)M-Lg(x)+LM+Lg(x)-LM+Mf(x)-ML|\\\\ &=|(f(x)-L)(g(x)-M)+L(g(x)-M)+M(f(x)-L)|\\\\ &\\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L| \\end{array}$$ for every $$x$$. Now, given $$\\varepsilon>0$$ there is $$\\delta_1>0$$ such that\n\n$$0<|x-a|<\\delta_1\\implies |f(x)-L|<\\min\\left\\{1, \\frac{\\varepsilon}{3(1+|M|)}\\right\\}$$\n\nand there is $$\\delta_2>0$$ such that:\n\n$$0<|x-a|<\\delta_2\\implies |g(x)-M|<\\frac{\\varepsilon}{3(1+|L|)}.$$ Now notice,\n\n$$\\begin{array}{c} 1<1+|L|&\\implies \\frac{1}{1+|L|}<1\\\\ |L|<1+|L|&\\implies \\frac{|L|}{1+|L|}<1\\\\ |M|<1+|M|&\\implies \\frac{|M|}{1+|M|}<1. \\end{array}$$ Using this, it follows that\n\n$$0<|x-a|<\\min\\{\\delta_1, \\delta_2\\}$$ implies \\begin{align*} |(fg)(x)-LM|&\\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L|\\\\ &<1\\cdot \\frac{\\varepsilon}{3(1+|L|)}+|L|\\frac{\\varepsilon}{3(1+|L|)}+|M|\\frac{\\varepsilon}{3(1+|M|)}\\\\ &<\\frac{\\varepsilon}{3}+\\frac{\\varepsilon}{3}+\\frac{\\varepsilon}{3}\\\\ &=\\varepsilon. \\end{align*}\n\nThe same idea also applies when you're working with sequences, for instance.\n\nIt may be of interest to you that (with some abstract tools from model theory) one can choose to use nonstandard analysis instead of $$\\epsilon-\\delta$$ and work with infinitesimal numbers. I could imagine that the book \"Nonstandard Analysis\" by Martin V\u00e4th contains basic calculus rules proven within the framework of nonstandard analysis.", "date": "2022-01-21 16:52:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4171310/proof-writing-varepsilon-delta-proof-of-the-product-rule-for-limits-a", "openwebmath_score": 0.9969805479049683, "openwebmath_perplexity": 202.05192180675851, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370156, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8533009286313327}}
{"url": "http://www.lofoya.com/Solved/1299/how-many-arrangements-of-four-0s-zeroes-two-1s-and-two", "text": "# Moderate Permutation-Combination Solved QuestionAptitude Discussion\n\n Q. How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?\n \u2716\u00a0A. 420 \u2716\u00a0B. 360 \u2716\u00a0C. 320 \u2714\u00a0D. 210\n\nSolution:\nOption(D) is correct\n\nTotal number of arrangements,\n\n$= \\dfrac{8!}{4!\u00d72!\u00d72!} = 420$\n\nSince, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements\u00a0$= \\textbf{210}$\n\n## (2) Comment(s)\n\nNuman\n()\n\nI think the answer is obtained by 7!/4!1!1! since you cant change the order of the first 1 and 2\n\nPriyanka\n()\n\nwhy the 1st 1 before the first 2 is same as first 2 before the first 1", "date": "2017-01-21 06:22:45", "meta": {"domain": "lofoya.com", "url": "http://www.lofoya.com/Solved/1299/how-many-arrangements-of-four-0s-zeroes-two-1s-and-two", "openwebmath_score": 0.8093605041503906, "openwebmath_perplexity": 383.4774100844008, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370156, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.853300926894333}}
{"url": "https://math.stackexchange.com/questions/2050698/square-matrices", "text": "# Square Matrices\n\nSo I'm studying a few special families of square matrices, the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices and I just had a few questions.\n\nI know...\n\na diagonal matrix is if every nondiagonal entry is zero, $a_{ij}$=0 whenever $i$ doesn't equal $j$.\n\nan upper triangular matrix is if all entries below the diagonal are zero, $a_{ij}=0$ whenever $i >j$.\n\na lower triangular matrix is if all entries above the diagonal are zero, $a_{ij}=0$ whenever $i < j$.\n\nsymmetric if $a_{ij}=a_{ji}$ for all $i$ and $j$.\n\nBut I was just wondering, can the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices have the $0_{n\\times n}$? Also I know the $I_{n\\times n}$ matrix is in the diagonal matrices, but can it be in the other three types?\n\n\u2022 Both the zero matrix and the identity matrix are diagonal, upper/lower triangular, and symmetric. Check the definitions. \u2013\u00a0Ethan Alwaise Dec 9 '16 at 4:55\n\u2022 Just check the hypothesis \u2013\u00a0Learnmore Dec 9 '16 at 4:56\n\nA diagonal matrix doesn't care what is on the diagonal; it only cares that the off-diagonal entries are $0$. So the zero matrix $\\mathbf 0_{n \\times n}$ is indeed diagonal.\n\nSimilarly, an upper triangular matrix only cares that the elements below the diagonal are $0$, so the zero matrix is upper triangular. For the same reason, it is also lower triangular.\n\nIt is also symmetric, because $a_{ij} = a_{ji} = 0$ for all $i,j$.\n\nSimilar reasoning applies to the identity matrix.\n\nTake for example the definition of upper triangular matrix. It's a matrix which has $a_{ij}=0$ whenever $i>j$, but this doesn't mean that $a_{ij}\\neq 0$ for $i\\le j$, it's possible to have $a_{ij}=0$ if $i\\le j$. Then both $O_{n\\times n}$ and $I_{n\\times n}$ are upper triangular matrices.\n\nThe same idea applies to the diagonal, lower triangular and symmetric matrices.", "date": "2019-07-19 00:32:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2050698/square-matrices", "openwebmath_score": 0.9466949105262756, "openwebmath_perplexity": 209.7602824869469, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127426927789, "lm_q2_score": 0.8633916064587, "lm_q1q2_score": 0.8533009265971222}}
{"url": "https://math.stackexchange.com/questions/2221793/how-do-you-find-the-area-of-the-shaded-area-of-this-circle", "text": "# How do you find the area of the shaded area of this circle?\n\nI need to find the area of the shaded area. The triangle is equilateral. So far, I have found the area of the triangle to be $\\sqrt 3$, but I cannot figure out how to find the radius of the circle in order to find the area of the circle. Any advice would be appreciated.\n\n\u2022 Is the triangle equilateral? \u2013\u00a0Ahmed S. Attaalla Apr 7 '17 at 3:06\n\u2022 Yes it is equilateral. \u2013\u00a0Justin Lam Apr 7 '17 at 3:10\n\u2022 Do you know how to find the areas of circles and triangles? Oops, sorry, I see that you do. My bad. Try drawing in a radius and looking for relationships. Hint: Choose your radius wisely. \u2013\u00a0Arby Apr 7 '17 at 3:11\n\u2022 Lookup circular segment, and think what all that comes down to for an equilateral triangle. \u2013\u00a0dxiv Apr 7 '17 at 3:13\n\nWe're looking for the radius right? So let's draw them..\n\nOk. Now we have an isosceles triangle $30-30-120$. If $r$ is the radius then law of sines tells us,\n\n$$\\frac{2}{\\sin 120}=\\frac{r}{\\sin 30}$$\n\nSo $r=2 \\frac{\\sin 30}{\\sin 120}=2\\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}=\\frac{2}{\\sqrt{3}}$. I think you can take it from here.\n\n\u2022 Thanks for the help! I got about 1.15 feet squared as the radius. Now I can do the rest. \u2013\u00a0Justin Lam Apr 7 '17 at 3:22\n\u2022 My final answer is about 2.5 feet squared. Is this correct? \u2013\u00a0Justin Lam Apr 7 '17 at 3:29\n\u2022 You're welcome. That's about right. The exact answer is $\\frac{4}{3}\\pi-\\sqrt{3}$ feet squared. @JustinLam \u2013\u00a0Ahmed S. Attaalla Apr 7 '17 at 3:44\n\u2022 Ahmed's drawing: cos(30\u00b0) = 1/r. Hence r = 1/cos(30\u00b0). Using cos(30\u00b0) = (1/2)\u00d7(\u221a3) we get r. \u2013\u00a0Peter Szilas Apr 7 '17 at 6:04\n\u2022 @Ahmed. OK , here we go. \u2013\u00a0Peter Szilas Apr 8 '17 at 2:33\n\nLet's label Ahmed's drawing: Triangle $ABC$, lower left $A$, then counterclockwise $B$, and $C$ (top). Let the center of the circle be $M$. Extend $CM$ to intersect $AB$ in $D$. Note length $AD$ $=$ length $DB$ $=1$, $MD$ being the perpendicular bisector of $AB$. Triangle $ADM$ is a right angled triangle. Angle $MAD = 30\u00b0$.\n\n$$\\cos (30\u00b0) = \\frac{1}{r}$$\n\n$$r = \\frac{1}{\\cos (30\u00b0)}$$\n\nUsing $\\cos (30\u00b0) = \\frac{\\sqrt{3}}{2}$ we get $r$.", "date": "2020-09-28 00:09:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2221793/how-do-you-find-the-area-of-the-shaded-area-of-this-circle", "openwebmath_score": 0.760209321975708, "openwebmath_perplexity": 540.2431769648276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988312739250412, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.853300925362011}}
{"url": "https://matheducators.stackexchange.com/questions/12845/why-do-we-teach-even-and-odd-functions", "text": "# Why do we teach even and odd functions?\n\nI've been either a student or an instructor in Precalculus or Calculus 1 at about 6 institutions now, and teaching the definition of even functions (where $f(-x) = f(x)$) and odd functions (where $f(-x) = -f(x)$) has been universal.\n\nBut why? I don't see how these concepts are so useful that they need to be in the courses that are taught to everyone. I don't see how they lay the stage for understanding calculus.\n\nI mean, seeing how it works graphically is nifty. But it seems like a disproportionate emphasis is placed on these classifications, in every curriculum I've seen.\n\n\u2022 it seems like a disproportionate emphasis is placed on these classifications It's extremely useful in applications. What level of emphasis is it that you consider disproportionate? If it's covered in the textbook and mentioned once in a while in examples, then it probably doesn't even need to have 5 minutes dedicated to it in class.\n\u2013\u00a0user507\nSep 11 '17 at 22:36\n\u2022 I like to teach them because of graphing, integration, understanding the difference between proof by example (not proof) and proof of a general case. (Also: If everyone at your school studies calculus, then you teach at a very different institution than mine.) Sep 12 '17 at 0:24\n\u2022 @BenCrowell It made up a noticeable part of my A-level, with no application at all at the time. Sep 13 '17 at 6:25\n\u2022 It's an example of symmetry (and anti-symmetry) - which is arguably the most useful concept with a practical application in the whole of mathematics IMO. Sep 14 '17 at 1:22\n\u2022 Many topics in mathematics are not useful. And no one has claimed them to be as well. Again useful means what, can they be of use in future etc are significant questions though. Sep 14 '17 at 6:51\n\nOne of the major themes of precalculus is what I call \u201cconnecting geometry to algebra\u201d. Being able to translate between an algebraic statement like $f(x)= f(-x)$, and the geometric statement that the graph of $f$ is symmetric about the vertical axis is a great instance of this. This is just one more way to practice reinforcing function concepts, and the connection with graphs.\n\n\u2022 Even a triple connection geometry - algebra - calculus!\n\u2013\u00a0Basj\nSep 13 '17 at 20:09\n\nHere you can see that knowing if the function is even or odd can help you when you are integrating over the interval $[-a, a]$.\n\nYou can reduce really-hard-to-look-at integrals to zero just by knowing this. As an example, to calculate $E(Z)$ where $Z \\sim N(0, 1)$, the standard normal distribution, you have:\n\n$\\displaystyle E(z) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{\\infty}ze^{-z^2/2}dz$\n\nWhich is immediately reduced to zero as the inner function in the integral is odd, and you are integrating over $(-\\infty, \\infty)$.\n\nThis may not be too difficult, but knowing this property about odd functions lets you generalize this to all the odd moments of the standard normal: $E(Z), E(Z^3), E(Z^5), ...$ all of them equal zero.\n\n\u2022 You could improve this answer by being more explicit here as to why this is useful. eg mention Fourier series Sep 12 '17 at 6:15\n\u2022 +1, As an engineer and purely from a practical point this was my first thought. For instance, calculating Fourier series is made much easier knowing about major shortcuts that can be made from the knowledge of odd and even functions. Sep 12 '17 at 15:32\n\nBesides applicability in topics like integration and Fourier analysis, it also connects algebra to calculus at least in the way that multiplication of even/odd functions behaves like addition even/odd numbers:\n\n\u2022 Multiplying two even functions gives an even function.\n\u2022 Multiplying two odd functions gives an even function, too.\n\u2022 Multiplying an even and an odd function gives an odd function.\n\nAlso, you can decompose every function as a sum of an even and an odd function as $$f(x) = \\frac{f(x)+f(-x)}2 + \\frac{f(x)-f(-x)}2$$ (which is a very useful concept an the same as writing a matrix as the sum of a symmetric and an antisymmetric one as $A = \\tfrac12(A+A^T) + \\tfrac12(A-A^T)$).\n\n\u2022 This also gives one (among many) justification for using the hyperbolic trig functions $\\sinh x$ and $\\cosh x$ as the odd and even parts respectively of $e^x$ Sep 12 '17 at 14:08\n\u2022 +1 Not forgetting that this approach generalizes to order three, four,... symmetries via discrete Fourier transforms and, ultimately, to representation theory of groups. Sep 30 '17 at 6:13\n\nLearning to think about functions abstractly should be one goal in precalculus, and function symmetry helps. Also suppose we carefully protected a student from knowing anything about function symmetry. Upon learning about flux in vector calculus, would this student be able to quickly see that the flux of the vector field ${\\bf F}(x,y,z)= y^2{\\bf j}$ through the unit sphere is 0?\n\n\u2022 Indeed, when I teach multivariable calculus, I show students lots of tricks to simplify integrals. Eliminating terms you know will evaluate to zero after integration, halving the domain so that you can \"plug in\" zeroes and simplify fewer terms, etc. It all comes down to symmetry. Sep 16 '17 at 13:09\n\nEven and odd parity are probably the simplest examples of function symmetries.\n\nIn applied mathematics, the general observation of function symmetries allows to simplify calculations (as stated by others) and to produce more meaningful graphs. In physics, symmetrical parts of a function are sometimes associated to different physical phenomena.\n\nTwo examples:\n\n1. If you have a function which is invariant under inversion (a more complex symmetry), that is, $$f(1/x) = f(x),\\ x>0,$$ it is better to plot the function by using a logarithmic $x$ axis because $f(a^x)$ is even.\n2. The example given by Dirk on the decomposition of a matrix in the sum of a symmetric and antisymmetric part is useful in circuit theory: the symmetric part of the impedance and admittance matrices is associated to the average power dissipated by an electrical network when subjected to sinusoidal excitations.\n\u2022 This is a better answer than mine. The psychology of voting on these sites is mysterious. Sep 13 '17 at 19:28\n\nTo express functions as sum of even and odd functions\n\n$$f(x) = f_{even}(x) + f_{odd}(x)$$\n\nAnd look at the properties of their graphs.\n\nTeach Fourier transform and say it becomes easy to compute when $f(x)$ is an odd function. And the integral becomes zero.\n\n\u2022 I think you've missed the point of the question. It isn't about what the syllabus requires, it's about the longer term benefits. Sep 13 '17 at 6:23\n\u2022 now like applications of it ?@JessicaB Sep 13 '17 at 8:01\n\nI agree that it is disproportionate. Ben Crowell (in comments) says that is helpful for applications. Well, I have a very strong background in natural sciences and engineering, and it is not a big building block. Very few derivations in physics rely on it for instance.\n\nI know this is (-1) controversial, but I think the appeal comes from the ease of doing tricky questions based on it. Sort of ETS style questions. A similar thing would be the vertical angle theorem (or whatever it is called, where a line crosses a couple parallel lines) is big on the SAT.\n\n(-2) I also think that many pure math people prefer things that are definitional and classificational. And that in preference to being able to do multistep mechanical problems like a maximization or power series expansion.\n\n\u2022 Really? The difference of Dirichlet versus Neumann boundary conditions (difference between the electromagnetic field behavior near conductors versus insulators) seems to be quite important. The method of images is how a the beginning of electrostatics is understood, and plays a role in the explanation of the Meissner effect in superconductors. The entirety of of the field of crystallography is based on understanding symmetries, and by extension things like understanding diffraction patterns. The difference between even and odd functions also explain the difference in timbre between brass and.. Sep 13 '17 at 20:53\n\u2022 ... woodwinds in music, as well as between different types of percussive instruments. And I am merely a mathematician who doesn't do too much natural sciences. I am sure specialists can come up with many more examples. Sep 13 '17 at 20:56\n\nWell symmetry helps investigate functions. It's the first things I need to know about the function I deal with. It helps as noted, evaluating integrals that are otherwise difficult to integrate. And it helps visualizing functions. I was a bit disapointed by the few examples given in class ass aplications of these concepts.\n\nAs with many items in math that have a name, if we didn't name it, it would still exist.\n\nWhen manipulating and equation that contained Cos(-x) we are able to observe that this term is the same as Cos(x). On the other hand, when we see Sin(-x) and want to manipulate to form a positive argument, it's equal to -Sin(x).\n\nI don't feel that we spend that much time on the concept. It's introduced in algebra, a parabola possibly being even, a third degree possibly odd, and then again in Trig with the examples I gave. Your experience may be different, more focus than I've observed.\n\nLearning mathematics isn't always about applying things to further mathematics - it can also be about actually applying that knowledge to other areas. So let me bring you a completely different answer to your question, coming from the background of someone that does programming and some web design.\n\nYou can easily apply even/odd functions to things like tables: one row has a dark background, the other has a lighter background. A simple enough requirement, that a good programmer might understand not only on the surface, but more deeply such as yourself.\n\nI apologize if this answer is not relevant to the mathematics community directly, but I felt the need (seeing this question in a suggested list) to remind people that math isn't just about math ;)", "date": "2021-10-16 21:07:29", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/questions/12845/why-do-we-teach-even-and-odd-functions", "openwebmath_score": 0.6738500595092773, "openwebmath_perplexity": 445.10571085558473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370156, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8533009251573335}}
{"url": "http://mathoverflow.net/questions/162774/what-is-the-least-integer-of-additive-dimension-4", "text": "# What is the least integer of additive dimension 4?\n\nSay that $m$ is the additive dimension of $n\\in\\Bbb N$, and write $m=\\operatorname{ad}n$, if $m$ is the greatest integer for which there is an irredundant $m$-element set $M\\subset\\Bbb N$ that provides a partition of $n$ uniquely: namely $$n=k_1+\\cdots+k_r\\quad\\text{with}\\quad\\{k_i:i=1,...,r\\}=M,$$ for some $r\\in\\Bbb N$, where no other partition of $n$ involves only elements of $M$.\n\nAs an illustration, take $n=34$. We can choose $M=\\{7,8,12\\}$ (for example) in this case, with $34=7+7+8+12$, and so $\\operatorname{ad}34=3$, since no other partition of $34$ can be formed from the elements of $M$, while any $4$-element set providing a partition of $34$ has a proper subset that does so too. (The last claim entails some checking.)\n\nNote that $\\operatorname{ad}0=0$; $\\operatorname{ad}n=1$ for $n=1,2,3,4,6$; $\\operatorname{ad}n=2$ for $n=5$ and $n=7,...,16$; and $\\operatorname{ad}17=3$ (take $M=\\{4,6,7\\}$). Thus $17$ is the least integer of additive dimension $3$.\n\nWhat is the least integer of additive dimension $4$? Further, what is the asymptotic behaviour of $\\operatorname{ad}n$ as $n$ becomes large?\n\n(This question was earlier posted on MathStackExchange but was not answered.)\n\n-\nEarlier post was math.stackexchange.com/questions/661256/\u2026 \u2013\u00a0Gerry Myerson Apr 8 '14 at 12:38\n\nI wrote a quick and probably impressively sub-optimal computer program to investigate this. It takes two seconds to ascertain that the least integer of additive dimension at least 4 is 49. The relevant partition is $15+14+12+8$.\n\n(I say \"at least 4\" because I haven't checked that it doesn't in fact have additive dimension 5.)\n\n-\nMore a bet than a guess: the next one is $129=16+24+28+30+31$, and the numbers are $(n-1)2^n+1$ :) \u2013\u00a0\u10db\u10d0\u10db\u10e3\u10d9\u10d0 \u10ef\u10d8\u10d1\u10da\u10d0\u10eb\u10d4 Apr 8 '14 at 12:18\n\nLet me prove that $A_n=(n-1)2^n+1$ has additive dimension $n$, with corresponding set $M=\\{2^n-2^i\\colon 0\\leq i\\leq n-1\\}$ and representation $A_n=\\sum M$.\n\nAssume that $A_n=k_1+\\dots+k_r$ is some representation with $k_i\\in M$. Since $k_i< 2^n$ we have $r\\geq n$. Let $k_i=2^n-2^{d_i}$; then $\\sum_i 2^{d_i}=(2^n-1)+(r-n)2^n$.\n\nLemma. If $\\sum_i 2^{d_i}\\equiv -1\\pmod{2^k}$, then $|\\{i\\colon d_i<k\\}|\\geq k$.\n\nProof. If we have some equal $d_i$'s, replace them by one number $d_i+1$ with no effect to the mentioned sum. When this process stops, we have exactly one $d_i$ equal to each of $0,1,\\dots,k-1$. It remains to notice that $|\\{i\\colon d_i<n\\}|$ does not increase during the process. The lemma is proved.\n\nFinally, the lemma inductively yields that the sum of $k\\leq n$ least elements in the list $2^{d_1},\\dots,2^{d_r}$ does not exceed $2^k-1$. Thus $\\sum_i2^{d_i}\\leq 2^n-1+(r-n)2^{n-1}$. Comparing this bound with the value found above we conclude that $n=r$, and the $d_i$ form a permutation of $\\{0,1,\\dots,n-1\\}$, as required.\n\nADDENDUM. On the other hand, the least number in uniquely representing set $M$ of cardinality $n$ is at least $2^{n-1}$, so the least number of aditive dimension $n$ should be greater than $n2^{n-1}$.\n\nTo prove this, set $a=\\min M$, $M'=M\\setminus\\{a\\}$, and let $S$ be the set of subset sums of $M'$. Clearly, all such sums are distinct (otherwise we would obtain a different representation by replacing one subsum by another). Moreover, all the sets $S+ka$, $k=0,1,\\dots$, are disjoint; otherwise we would have $s'=s+ka$ for some $s,s'\\in S$, and we would be able to replace $s'$ by $s+ka$. Since $|S|=2^{n-1}$, the density argument shows that $a\\geq 2^{n-1}$.\n\nPerhaps, this argument can be extended to other elements of $M$?\n\n-\n\nPossible OEIS candidate, whose description sounds promising: A115981\n\n-\n\nTheorem: For n, $M_n:=(n-1)2^n+1$ is an upper bound.\n\nAs an example take $n=4$. now let \\begin{aligned} k_1 & := 8 & = 1000 b \\\\ k_2 & := 12 & = 1100 b \\\\ k_3 & := 14 & = 1110 b \\\\ k_4 & := 15 & = 1111 b \\\\ \\end{aligned}\n\nThe numbers on the right side are the binary representations of $k_i$\n\n${k_1,k_2,k_3,k_4}$ is a unique partition for $M_4 = k_1+k_2+k_3+k_4 = 49$.\n\nThe cases 0 to 3 are already settled in the question. Proof for $n>3$ by induction:\n\nSuppose there would be two different partitions\n\n$$(*) a_1k1+...+a_nk_n = b_1k1+...+b_nk_n = M_n$$\n\nAll factors $a_i$ and $b_i$ are lower than $k_1=2^{n-1}$, since otherwise $a_i*k_i\\ge2^{2(n-1)}$, which exceeds $M_n=(n-1)2^n+1$ for all $n>3$. Equation $(*)$ also hold modulo $k_1=2^{n-1}$. This yields the same equation in the induction hypothesis (just drop the first binary digit!). As all coefficient are lower than the modulus, we have $a_i=b_i$ for $i>=2$. Then equation $(*)$ yields $a_1=b_1$ qed.\n\nConjecture: $M_n$ is already optimal for all $n$.\n\nEdit\n\nThis proof is flawed! The induction hypothesis doesn't provide uniqueness mod $2^{n-1}$. Sorry!\n\n-", "date": "2016-05-04 04:29:08", "meta": {"domain": "mathoverflow.net", "url": "http://mathoverflow.net/questions/162774/what-is-the-least-integer-of-additive-dimension-4", "openwebmath_score": 0.9867252707481384, "openwebmath_perplexity": 162.67952575468013, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127420043055, "lm_q2_score": 0.8633915976709976, "lm_q1q2_score": 0.8533009173177019}}
{"url": "https://math.stackexchange.com/questions/1879150/proof-verification-powers-of-a-group-element-of-finite-order-are-distinct", "text": "# Proof verification: powers of a group element of finite order are distinct\n\nI'm finally attempting to conquer D&F (3rd ed) and I want to build good proof habits and fix mistakes early on. Here is the exercise (Ch. 1 ex. 32) and the following is my proof.\n\nProve that for $x \\in G$, where $G$ is a group and $x$ has finite order $n$, all of $1, x, x^2, \\ldots, x^{n-1}$ are distinct and deduce that $|x| \\leq |G|$.\n\nProof. Let $a,b \\in [0,n-1]\\cap \\mathbb{N}$ such that $x^a=x^b$. Then $1 = x^ax^{-a}=x^bx^{-a} = x^{b-a}$. Since $1 \\leq a \\lt n$ and $1 \\leq b \\lt n$, $b-a\\lt n$. So $x^{b-a}=1 \\implies b-a=0$ and $a=b$. Since $a$ and $b$ were arbitrarily chosen, $x^c$ is distinct for all $c \\in [0,n-1] \\cap \\mathbb{N}$. We know $x^c\\in G$ for all $c \\in [0, n-1]\\cap \\mathbb{N}$ by definition of group closure, and $|[0,n-1]\\cap \\mathbb{N}|=n$. So $G$ contains at least $n$ elements, and $|x|\\leq |G|$.\n\nIs this a convincing proof? I feel like there might be some circular logic, and perhaps not enough detail. I was thinking of establishing that the elements form a cyclic group, and show this group is a subgroup of $G$. Any tips or comments would be greatly appreciated.\n\n\u2022 It looks good to me. Aug 2 '16 at 16:16\n\nAlmost.\n\nYou assert $b-a < n$. I think you want to assume $b \\ge a$ (without loss of generality) from the start, to make $b-a$ nonnegative as well. Otherwise you need to handle the case when it's not.\n\n\u2022 isn't $b-a \\lt n$ even if $b-a$ is negative? I was thinking about clarifying that but I didn't think it would change anything Aug 2 '16 at 16:23\n\u2022 @m1cky22: true, but you're appealing to the definition of order of an element, which is usually nonnegative. Aug 2 '16 at 16:25\n\u2022 @m1cky22 Yes, it's less than $n$. But when it's less than $0$ you are dealing with a negative power in what follows. That needs work. Aug 2 '16 at 16:26\n\u2022 I see. So it's more about keeping the consistency/logic of the definition of order rather than \"the proof breaks if it's negative.\" Thanks, I think I can finally move on to the next chapter! Aug 2 '16 at 16:27\n\u2022 @m1cky22 Proofs are all about keeping consistent to definitions. Aug 2 '16 at 16:28\n\nLogically, everything looks good, except for @Ethan's point. If you are worried about circularity, make sure you understand why your statement \u201c$x^{b-a} = 1 \\implies b-a =0$\u201d is true.\n\nI think notations like\n\nLet $a,b\\in[0,n-1] \\cap \\mathbb{N}$ such that $x^a = x^b$.\n\nare a bit cumbersome. Personally, I would rewrite the first sentence as\n\nLet $a$ and $b$ be integers such that $0 \\leq a \\leq b \\leq n-1$ and $x^a = a^b$.\n\nOr perhaps even\n\nSuppose $x^a = x^b$ for some integers $a$ and $b$ with $0 \\leq a\\leq b\\leq n-1$.\n\nIn essence, you want to show $x^a = x^b \\implies a=b$, so writing the first sentence that way aligns the paragraph with that logical statement.\n\nHaving a higher density of symbols does not make a more mathematically sophisticated argument.\n\nYour proof is perfectly fine. Good Job!\n\nOne remark: One prof of me uses the following convention: $[0:n] := [0, n] \\cap \\mathbb N$.\n\nThis is an elegant way to write such \"discrete intervals\" and makes proofs more readable. Just an idea of me. Maybe you want to use it :)\n\n\u2022 I was looking for better notation and didn't know what people use! I didn't want to write $1 \\leq a \\lt n$ so I chose my route, but yours is definitely better. Thanks! Aug 2 '16 at 16:24\n\nI got confused when you brought $c$ into the mix.\n\nHere is your proof with a lot of the argument removed. I don't think I have removed too much and it may be easier to follow.\n\nSuppose the powers of $x$ are not all distinct.\n\nThen there exist $a,b \\in [0,n-1]\\cap \\mathbb{N}$ such that $b>a$ and $x^b=x^a$.\n\nThis implies that $0\\lt b-a<n$, and $1=x^ax^{-a}=x^bx^{-a} = x^{b-a}$.\n\nSince the order of $x$ is $n$, $x^{b-a}=1$ implies $n\\, |\\, b-a$.\n\nThis is a contradiction, since no number greater than $0$ and less than $n$ can be divisible by $n$.\n\nSo the original supposition, that the powers of $x$ are not all distinct, leads to a contradiction and must therefore be false.", "date": "2021-11-27 12:25:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1879150/proof-verification-powers-of-a-group-element-of-finite-order-are-distinct", "openwebmath_score": 0.8692059516906738, "openwebmath_perplexity": 225.8077921080215, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147193720649, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.8532713534017552}}
{"url": "https://mathematica.stackexchange.com/questions/40163/arbitrary-precision-spline-interpolation?noredirect=1", "text": "# Arbitrary precision spline interpolation\n\nThe current implementation of Interpolation does not allow arbitrary precision spline interpolation. Yu-Sung Chang says here that \"it is not hard to implement it manually for arbitrary precision using BSplineBasis.\"\n\nI am new to splines and not familiar with BSplineBasis. What is the best way to implement arbitrary precision spline interpolation in Mathematica?\n\nI am interested primarily in an analog of the Method -> \"Spline\" of Interpolation which I investigated in this answer (BTW, what is the name of such a spline and of this kind of parametrization?). Also, J. M. gives in this answer an interesting implementation of a spline with centripetal parametrization, but it is not clear how to use it for interpolation.\n\nP.S. I need an implementation where the degree of a spline is arbitrary. At the moment I am more interested in even-degree splines.\n\n\u2022 Yu-Sung's pointer is wolfram.com/xid/0c0rpcn4bku6m-dj3q8d, where there are two examples of such a thing (I haven't tested them). What are those example lacking? \u2013\u00a0Dr. belisarius Jan 10 '14 at 12:45\n\u2022 That is an example of cubic B-spline interpolation. I need an implementation where the degree of spline is arbitrary. \u2013\u00a0Alexey Popkov Jan 10 '14 at 12:51\n\u2022 Please see my question about B-spline Basis Function \u2013\u00a0xyz May 28 '15 at 8:12\n\nThe OP linked to an answer of mine for interpolating over general point sets; for constructing a single interpolating function, a slight modification of my procedure is needed. (In particular, you don't need centripetal or chord-length parametrization in this case.)\n\ndata = {{0, 0}, {1/10, 3/10}, {1/2, 3/5}, {1, -1/5}, {2, 3}, {3, -6/5}};\n\n\nTo review the problems with using the built-in Interpolation[], let's try building an interpolating quartic spline from data:\n\nm = 4; (* spline order *)\nsp1 = Interpolation[data, InterpolationOrder -> m, Method -> \"Spline\"];\n\n\nNow, evaluate:\n\nsp1[3/2]\n0.26753591659167364\n\n\nHerein lies the problem: even though the contents of data and the argument fed to sp1 are exact, the output is a machine precision number. Yeesh!\n\nSo, we go back to basics and build the interpolating spline ourselves from BSplineBasis[]. First, we generate the knots. The clamped condition is the most appropriate for interpolation, so we generate the appropriate knot sequence like so:\n\nn = Length[data];\n{xa, ya} = Transpose[data]\nknots = Join[ConstantArray[xa[[1]], m + 1],\nIf[m + 2 <= n, MovingAverage[ArrayPad[xa, -1], m], {}],\nConstantArray[xa[[-1]], m + 1]];\n\n\nTo generate the control points for the B-spline, we construct the relevant linear system and solve it with LinearSolve[]:\n\ncp = LinearSolve[Outer[BSplineBasis[{m, knots}, #2, #1] &,\nxa, Range[0, Length[data] - 1]], ya];\n\n\nNote that all the entries of cp are exact.\n\nAt this juncture, you might think that we can use BSplineFunction[] on the control points and the knots, but alas, the function also only evaluates at machine precision. We are left with no choice but to use BSplineBasis[] once more:\n\nsp2[x_] = cp.Table[BSplineBasis[{m, knots}, j - 1, x], {j, n}];\n\n\nTry it out:\n\nsp2[3/2]\n25251843/94386740\n\nN[%]\n0.2675359165916738\n\n\nAs a verification that we were able to reproduce the control points used internally by Interpolation[], let's do a comparison:\n\nsp1bs = Cases[sp1, _BSplineFunction, \u221e][[1]];\nsp1bs[\"ControlPoints\"]\n{0., 0.777830630128766, 1.3470257035045392, -6.670028631670085, 13.100332135107115, -1.2}\n\nN[cp]\n{0., 0.777830630128766, 1.347025703504539, -6.6700286316700845, 13.100332135107113, -1.2}\n\n\nCertainly, I can't end this post without at least one picture, so:\n\nPlot[{sp1[x], sp2[x]}, {x, 0, 3}]\n\n\nPlot[sp1[x] - sp2[x], {x, 0, 3}, PlotRange -> All]\n\n\nI leave the task of bundling everything in a routine as an exercise. If you want to learn more, Piegl and Tiller's The NURBS Book is the canonical reference.\n\n\u2022 It is good to have you back. \u2013\u00a0Mr.Wizard May 24 '15 at 10:31\n\u2022 I was able to borrow a friend's computer with a Mathematica installation, so I went for it and answered some questions. It was fun while it lasted\u2026 :) \u2013\u00a0J. M.'s technical difficulties May 24 '15 at 10:45\n\u2022 Did you try contacting the Stack Exchange team about the matter? I understand if you do not wish to pursue it but I think they might help. \u2013\u00a0Mr.Wizard May 24 '15 at 10:48\n\u2022 Maybe I'll ask them sometime\u2026 thanks for the concern. :) \u2013\u00a0J. M.'s technical difficulties May 24 '15 at 10:51\n\u2022 @J.M. May I ask you also about a generalization of the spline interpolation (at least) for the 2D case? In the linked answer it is said that \"Multi-dimension is simply tensor product version.\" I do not see this as a simple thing but it would be very helpful to have such a generalization. May be a separate thread would be the best place for this. \u2013\u00a0Alexey Popkov May 26 '15 at 14:01\n\nHere is my implementation of quadric spline arbitrary precision interpolation which is defined exactly as in Interpolation with options Method->\"Spline\", InterpolationOrder->2.\n\n## Theoretical background\n\nQuadric spline interpolation for n datapoints is defined as a Piecewise function which consists of n-2 parabolas which are spliced together in the middles of successive datapoints (with exceptions for first and last intervals) by two conditions: successive parabolas at these middle points must be equal and their first derivatives must be equal too. So we have 2(n-3) such conditions.\n\nLet us designate jth datapoint as {x[j],y[j]} and define jth parabola as\n\ns[j_, xx_] = y[j + 1] + b[j] (xx - x[j + 1]) + c[j] (xx - x[j + 1])^2;\n\n\nThis definition automatically makes jth parabola equal to y[j + 1] at x[j + 1]. We have to add also 2 boundary conditions:\n\ns[1, x[1]] == y[1]\ns[n - 2, x[n]] == y[n]\n\n\nAnd 2(n-3) splicing conditions:\n\nTable[{\ns[j, (x[j + 1] + x[j + 2])/2] == s[j + 1, (x[j + 1] + x[j + 2])/2],\nds[j, (x[j + 1] + x[j + 2])/2] == ds[j + 1, (x[j + 1] + x[j + 2])/2]\n}, {j, 1, n - 3}]\n\n\nwhere ds is first derivative:\n\nds[j_, xx_] = D[s[j, xx], xx];\n\n\nNow we combine everything together and convert into matrix form:\n\neqs = Flatten[{\ns[1, x[1]] == y[1],\nTable[{\ns[j, (x[j + 1] + x[j + 2])/2] == s[j + 1, (x[j + 1] + x[j + 2])/2],\nds[j, (x[j + 1] + x[j + 2])/2] == ds[j + 1, (x[j + 1] + x[j + 2])/2]\n}, {j, 1, n - 3}],\ns[n - 2, x[n]] == y[n]}];\narrs = Simplify@\nNormal@CoefficientArrays[eqs, Flatten@Array[{b[#], c[#]} &, n - 2]];\nMatrixForm /@ (4 arrs)\n\n\nLooking at the matrices it is easy to see that they have periodic structure and contain only elements of the forms x[j+1]-x[j] and y[j+1]-y[j] with some numerical coefficients. This periodic structures can be expressed as SparceArrays.\n\n## Implementation\n\nAssuming that data contains an array of {x[j],y[j]}, we define\n\n\u0394x[i_] := Subtract @@ data[[{i + 1, i}, 1]];\n\u0394y[i_] := Subtract @@ data[[{i + 1, i}, 2]];\n\n\nTwo matrices in the arrs can be expressed as follows:\n\nbB = SparseArray[{1 -> -4 \u0394y[1], -1 -> 4 \u0394y[n - 1], i_ /; OddQ[i] :> 0, i_ /; EvenQ[i] :> 4 \u0394y[i/2 + 1]}, 2 (n - 2)];\nmM = SparseArray[\nJoin[{{1, 1} -> -4 \u0394x[1], {1, 2} -> 4 \u0394x[1]^2, {-1, -1} -> 4 \u0394x[n - 1]^2, {-1, -2} -> 4 \u0394x[n - 1]},\nTable[Band[{2 i, 2 i - 1}] -> {{2 \u0394x[i + 1], \u0394x[i + 1]^2, 2 \u0394x[i + 1], -\u0394x[i + 1]^2}, {4, 4 \u0394x[i + 1], -4, 4 \u0394x[i + 1]}}, {i, 1, n - 3}]], {2 (n - 2), 2 (n - 2)}];\n\n\nNow for obtaining coefficients b[i], c[i] we can use LinearSolve:\n\nbcM = LinearSolve[mM, bB];\n\n\nGrouping coefficients with identical indexes together:\n\nbcM = Partition[bcM, 2];\n\n\nThe intervals at which the parabolas are defined:\n\nintervList =\nPartition[\nJoin[{data[[1, 1]]},\nMovingAverage[data[[2 ;; -2, 1]], 2], {data[[-1, 1]]}], 2, 1];\n\n\nNow we compile all the spline data in one object:\n\nsplineData = Table[{data[[i + 1]], bcM[[i]], intervList[[i]]}, {i, n - 2}];\n\n\nsplineData contains everything that is needed to construct the spline. It contains elements of the form:\n\n{{x[i + 1], y[i + 1]}, {b[i], c[i]}, {xmin, xmax}}\n\n\nIt is very handy to use splineData for exact numerical integration (see here).\n\nTo produce the explicit piecewise function we define the constructor (HornerForm is already applied and gives 27% speedup):\n\nmakeSpline[splineData_List, x_Symbol] :=\nPiecewise[\nAppend[Table[{d[[1, 2]] - d[[1, 1]] d[[2, 1]] +\nd[[1, 1]]^2 d[[2, 2]] + x (d[[2, 1]] + x d[[2, 2]] -\n2 d[[1, 1]] d[[2, 2]]), #1 <= x <= #2 & @@ d[[3]]}, {d, splineData}],\n{Indeterminate, True}]]\n\n\nIt can be used as follows:\n\nspline[\\[FormalX]_] = makeSpline[splineData, \\[FormalX]];\n\n\nCompiling the code for creation of splineData into one function:\n\nClearAll[toSplineData, makeSpline];\nOptions[toSplineData] = {Method -> Automatic};\ntoSplineData[data_, OptionsPattern[]] /; MatrixQ[data, NumberQ] :=\nModule[{\u0394x, \u0394y, bB, mM, bcM, intervList,\nn = Length[data], dataS = Sort[data]},\n\u0394x[i_] :=\nSubtract @@ dataS[[{i + 1, i}, 1]]; \u0394y[i_] :=\nSubtract @@ dataS[[{i + 1, i}, 2]];\nbB = SparseArray[{1 -> -4 \u0394y[1], -1 ->\n4 \u0394y[n - 1], i_ /; OddQ[i] :> 0,\ni_ /; EvenQ[i] :> 4 \u0394y[i/2 + 1]}, 2 (n - 2)];\nmM = SparseArray[\nJoin[{{1, 1} -> -4 \u0394x[1], {1, 2} ->\n4 \u0394x[1]^2, {-1, -1} ->\n4 \u0394x[n - 1]^2, {-1, -2} ->\n4 \u0394x[n - 1]},\nTable[Band[{2 i,\n2 i - 1}] -> {{2 \u0394x[i + 1], \u0394x[\ni + 1]^2,\n2 \u0394x[i + 1], -\u0394x[i + 1]^2}, {4,\n4 \u0394x[i + 1], -4,\n4 \u0394x[i + 1]}}, {i, 1, n - 3}]], {2 (n - 2),\n2 (n - 2)}];\nbcM = LinearSolve[mM, bB, Method -> OptionValue[Method]];\nbcM = Partition[bcM, 2];\nintervList =\nPartition[\nJoin[{dataS[[1, 1]]},\nMovingAverage[dataS[[2 ;; -2, 1]], 2], {dataS[[-1, 1]]}], 2, 1];\nTable[{dataS[[i + 1]], bcM[[i]], intervList[[i]]}, {i, n - 2}]];\n\n\nNow to construct the spline from data we can evaluate\n\nspline[\\[FormalX]_] = makeSpline[toSplineData[data], \\[FormalX]];\n\n\nAny suggestions and improvements are welcome!\n\n## What is so special about quadric spline interpolation?\n\nQuadric spline interpolation with splicing points in the middle of successive data points is locally invariant and weakly depends on boundary condition. (But the same is not true for quadric spline interpolation with splicings in data points.) It gives much lesser artifacts than \"usual\" cubic spline interpolation. Moreover, it seems that splines of higher degree do not give any advantages over such quadric splines. As opposed to cubic spline, data points are not special points at which polynomials are spliced, and it makes much easier exact integration in these points.\n\n\u2022 Could you give a demo of your solution? THANKS:) \u2013\u00a0xyz Oct 10 '15 at 10:38\n\u2022 @Shutao As you can see from the definition for the parabolas (the first formula in my answer) from which the spline is formed, my solution is non-parametric and so (as well as Interpolation) it is not intended for interpolating an arbitrary dataset (like yours). \u2013\u00a0Alexey Popkov Oct 10 '15 at 11:40\n\u2022 @Shutao My solution does not use BSplineBasis and is not intended to be compatible with BSplineFunction and BSplineCurve. But it has an advantage of being very simple and efficient, mathematically clear and very handy when one needs to perform exact integration of interpolated data (without using NIntegrate). Actually I developed my solution exactly for the latter kind of applications. \u2013\u00a0Alexey Popkov Oct 10 '15 at 11:40\n\u2022 @Shutao While I appreciate your solution very much I should note that your answer (as well as belisarius' answer) does not answer the original question because the question is explicitly about creation of arbitrary-precision version of Interpolate (preferably with richer set of spline interpolation methods), not about construction of BSplineCurve from a set of points. \u2013\u00a0Alexey Popkov Oct 10 '15 at 11:40\n\u2022 @Shutao It would be great if you add a non-parametric version of your solution as another answer: something like BSplineInterpolate[pts][x]: BSplineInterpolate would accept a list of points as the first argument and return something line InterpolatingFunction (in my answer makeSpline returns a Piecewise function object), the latter can accept the abscissa x as its argument and return the interpolated value of the ordinate y. \u2013\u00a0Alexey Popkov Oct 10 '15 at 11:41\n\nThe following is a shameful plug of J. M.'s answer that you already linked to show how to get the parametrization by using BSplineBasis[] with arbitrary precision and packed into a function:\n\nsplineInterp[data_, order_, prec_] :=\nModule[{parametrizeCurve, tvals, bas, ctrlpts, knots},\nparametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], a : (_?NumericQ) : 1/2] :=\nFoldList[\nPlus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]];\ntvals = parametrizeCurve[data];\n(*knots for interpolating B-spline*)\nknots =\nJoin[\nConstantArray[0, order + 1],\nConstantArray[1, order + 1]];\n(*basis function matrix*)\nbas =\nTable[\nBSplineBasis[{order, knots}, j - 1, N[tvals[[i]], prec]],\n{i, Length[data]}, {j, Length[data]}];\nctrlpts = LinearSolve[bas, data];\nReturn[\nSum[\nctrlpts[[i + 1]] BSplineBasis[{order, knots}, i, #],\n{i, 0, Length[testData] - 1}] &]\n]\n\n\nUsage\n\nSeedRandom[42];\ntestData = RandomReal[{0, 1}, {10, 2}, WorkingPrecision -> prec];\nf = splineInterp[testData, 5, 50];\nf[1/10]\nParametricPlot[\nf[t], {t, 0, 1},\nAxes -> None,\nEpilog ->\n{Directive[Green, PointSize[Large]], Point[testData]}, Frame -> True\n]\n\n(*\n{0.06025513038208479326777055464103084790562316, \\\n0.5927955887343273319037301352154452480202510}\n*)\n\n\nPlease refer to J. M.'s answer for the details.\n\n### Explanation\n\nThe centripetal distribution:\n\nThe function parametrizeCurve complete this algorithm\n\nparametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], a : (_?NumericQ) : 1/2] :=\nFoldList[\nPlus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]]\n\n\u2022 I do not see in your answer how can I use this as interpolation in a regular way: interpolation[x] where x is not a parametrization variable t but the actual independent variable on which the dependent variable y depends by y[x]. \u2013\u00a0Alexey Popkov Jan 10 '14 at 16:28\n\u2022 @AlexeyPopkov, Please see my explanation \u2013\u00a0xyz May 29 '15 at 13:04\n\u2022 @ShutaoTang Thank you for the answer to my comment, now I see that the second argument of parametrizeCurve may be arbitrary but must be lesser or equal to $1$. BTW it would be better for others if you would post this explanation under your own answer. :^) \u2013\u00a0Alexey Popkov May 29 '15 at 13:22\n\u2022 @AlexeyPopkov, My pleasue. I used the chord-length method to generate the parameters in my answer, so I think it is suitable to place this explanation in belisarius's answer:-) \u2013\u00a0xyz May 29 '15 at 13:34", "date": "2020-06-02 13:00:19", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/40163/arbitrary-precision-spline-interpolation?noredirect=1", "openwebmath_score": 0.2742125988006592, "openwebmath_perplexity": 4391.916604045422, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147201714922, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.8532713431982841}}
{"url": "http://math.stackexchange.com/questions/102050/factoring-a-polynomial", "text": "# Factoring a polynomial\n\nI am trying to factor the following polynomial: $$4x^3 - 8x^2 -x + 2$$\n\nI am trying to do the following: $4x^2(x - 2)-x+2$ but I am stuck.\n\nedit: correction.\n\n-\nIt might be better to write it as $4x^2(x-2)- (x-2)$ to take advantage of what you have done so far. \u2013\u00a0 Geoff Robinson Jan 24 '12 at 19:04\nIt may be of value to note that there is a general formula for finding roots of 3rd degree polynomials. The formula looks complex but may be of help if you can't guess a root. See this reference for more details: en.wikipedia.org/wiki/Cubic_function \u2013\u00a0 Emmad Kareem Jan 25 '12 at 20:57\n\nYou can factor as\n\n$$4x^2(x-2)-(x-2)=4x^2(x-2)-(1)(x-2)$$\n\nThen factor out the $x-2$ to get\n\n$$(x-2)(4x^2-1).$$\n\nBut, you may further factor $4x^2-1$ to get\n\n$$(x-2)(2x-1)(2x+1).$$\n\n-\n+1:nice explanation... \u2013\u00a0 pedja Jan 24 '12 at 19:07\nIt wasn't \"factored wrong\"; it's just that it wasn't taken far enough. \u2013\u00a0 Michael Hardy Jan 24 '12 at 19:19\nI agree with Michael. Could you perhaps edit that out? I think it's a little bit impolite. Incomplete and wrong are two very different things. :) \u2013\u00a0 000 Jan 25 '12 at 1:08\nIf you look at the revisions, the original posting contained an error where $(x-2)$ was incorrectly $(x-2x)$. I was not being impolite. But, I'll edit it out since things have changed. \u2013\u00a0 Joe Johnson 126 Jan 25 '12 at 20:02\n\nI'll use your polynomial to illustrate a more general procedure for factoring polynomials with integer coefficients (and assuming it has at least one rational root):\n\nFirst, guess a root of $4x^3-8x^2-x+2$. The so called \"rational roots test\" will be helpful here.\n\nEventually, you'll discover that $x=2$ is a root of $4x^3-8x^2-x+2$. This will imply that your polynomial has the form $$\\tag{1}(x-2)(ax^2+bx+c),$$ for some constants $a, b, c$.\n\nTo find those constants, you could do one of two things (and maybe more)\n\n1. perform the division $4x^3-8x^2-x+2\\over x-2$.\n2. expand (1) and set it equal to the original polynomial. Setting the coefficients of the two sides of this equation equal to each other will give you a system of equations that are solvable for $a$, $b$, and $c$.\n\nOnce you've figured out what $a,b$, and $c$ are, factor the quadratic.\n\nUsing method (2), we have:\n\n$$4x^3-8x^2-x+2 = ax^3+(b-2a)x^2+(c-2b)x-2c$$\n\nA moment's reflection reveals that $c=-1$; whence $b=0$; whence $a=4$. Thus \\eqalign{ 4x^3-8x^2-x+2 &= (x-2)(4x^2-1)\\cr &= (x-2)(2x+1)(2x-1).\\cr }\n\nOf course, the other answers are more suitable to your problem; but in the event that your polynomial doesn't factor nicely (such as for $x^3+6x^2+11x+6$), you might try using this approach.\n\n-\n\nFranco, you said you are stuck. Did you have an approach?\n\nIn general, for a cubic equation to be factored means the person who is asking this question is indirectly giving you a hint that there is at least one real root.\n\nYour approach should be as follows: Step 1: first to try to see if x=0, 1 -1, 2 or -2 is a root. Chances are in some cases that one of these is a root. In that case, let us say you figured that $x=2$ is a root of $4x^3 - 8x^2 -x + 2$, then if you know factoring polynomials a little bit then you could immediately figure $4x^3 - 8x^2 -x + 2 = (x-2)(4x^2-1)$ which means the other factors can be found by the fact that $(4x^2-1^2) = (2x-1)(2x+1)$. If in some cases you cannot find the obvious root as described in Step 1, then Step 2: Observe if there is a pattern like $4x^3-8x^2$ has two coefficients $4$ and $8$, which has the same pattern as $-x+2$, i.e. $4x^3-8x^2 = 4x^2(x-2)$ and $-x+2 = -(x-2)$, which means you can combine those as $$(x-2)4x^2-(x-2) = (x-2)(4x^2-1)$$\n\nAnd in some cases as here you could also do $(4x^3-x) - (8x^2-2)$ which gives you\n\n$$x(4x^2-1)-2(4x^2-1) = (x-2)(4x^2-1) = (x-2)(2x-1)(2x+1)$$\n\nLuckily with this question, Step 1 and Step 2 worked. What if you have a cubic equation that does not have obvious ways to factor.\n\nStep 3: For a general cubic equation $ax^3+bx^2+cx+d=0$, apply the substitution\n\n$$x = y - \\frac{b}{3a}$$\n\nthen we get\n\n$$a\\left(y-\\frac{b}{3a} \\right)^3 + b\\left(y-\\frac{b}{3a} \\right)^2+c\\left(y-\\frac{b}{3a} \\right)+d = 0$$\n\nwhich simplifies to\n\n$$ay^3 + \\left( c-\\frac{b^2}{3a}\\right)y+ \\left(d + \\frac{2b^3}{27a^2} - \\frac{bc}{3a} \\right) = 0$$\n\nThis is called a depressed cubic equation, because the square term is eliminated. It is much easier to use this and then find the roots. (back substitute to get the roots in terms of $x$)\n\nFor example $$2x^3-18x^2+46x-30=0$$\n\nSubstitute $x=y+3$ and simplify this cubic equation to\n\n$2y^3-8y=0 \\Rightarrow y=0,2,-2$ which then gives the roots as $x=1,3,$ and $5$.\n\n-\n\n$(4x^3-x)-(8x^2-2)=x(4x^2-1)-2(4x^2-1)=(x-2)(4x^2-1)=$\n\n$=(x-2)(2x-1)(2x+1)$\n\n-\n+1: Yours didn't appear until after I sent mine. \u2013\u00a0 Joe Johnson 126 Jan 24 '12 at 19:04", "date": "2015-05-27 02:25:14", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/102050/factoring-a-polynomial", "openwebmath_score": 0.9466521739959717, "openwebmath_perplexity": 291.5768559669113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147201714923, "lm_q2_score": 0.8774767826757122, "lm_q1q2_score": 0.853271340082584}}
{"url": "https://cs.stackexchange.com/questions/47561/confused-about-proof-that-logn-thetan-log-n", "text": "# Confused about proof that $\\log(n!) = \\Theta(n \\log n)$\n\nSo I was able to show that:\n\n$\\log(n!) = O(n\\log n)$ without any problems.\n\nMy question is when trying to prove that $\\log (n!) = \\Omega(n\\log n)$.\n\nI was able to show that:\n\\begin{align*} \\log n! &= \\log(1 \\cdot 2 \\cdot 3 \\cdots n )\\\\ &= \\log 1 + \\log2 + \\log3 + \\dots + \\log n \\\\ &= \\log 1 + \\dots + \\log\\tfrac{n}{2} + \\dots + \\log n\\\\ &\\geq \\log\\tfrac{n}{2} + \\log\\big(\\tfrac{n}{2} + 1\\big) + \\dots + \\log n &&\\text{(i.e., the larger half of the sum)}\\\\ &\\geq \\log\\big(\\tfrac{n}{2}\\big) + \\log\\big(\\tfrac{n}{2}\\big) + \\dots + \\log\\big(\\tfrac{n}{2}\\big)&&\\text{(adding \\tfrac{n}2 times)} \\\\ &= \\log\\big(\\tfrac{n}{2} \\cdot \\tfrac{n}{2} \\cdots \\tfrac{n}{2}) &&\\text{(\\tfrac{n}{2} times)} \\\\ &= \\log\\Big(\\tfrac{n}{2}^{\\tfrac{n}{2}}\\Big)\\\\ &= \\tfrac{n}{2} log\\big(\\tfrac{n}{2}\\big) &&\\text{(by log exponent rule)} \\end{align*}\n\nThus, $\\log(n!) \\geq \\tfrac{n}{2}\\log\\big(\\tfrac{n}{2}\\big)$, so we conclude that $\\log(n!) = \\Omega(n\\log n)$.\n\nI don't understand how finding the lower bound of $\\log(n!)$ is found by getting the larger half of the sum. Why is that chosen to find $\\Omega(n\\log n)$? I feel like it's probably something obvious but it's the only thing keeping me from fully grasping the proof. If someone can enlighten me, I would appreciate it!\n\n\u2022 Concisely: \u200b 1/2 is the simplest number in the interval (0,1). \u200b \u200b \u200b \u200b\n\u2013\u00a0user12859\nSep 26, 2015 at 9:31\n\u2022 @RickyDemer So what? Splitting the sum as $\\log 1 + \\dots + \\log cn + \\dots + \\log n$ for any $0<c<1$ would work just as well. Sep 26, 2015 at 10:17\n\nConsider the sum $S = \\log(1) + \\dots + \\log(n)$. We're going to break it into two parts: $S=T+U$, where\n\n\\begin{align*} T &= \\log(1) + \\log(2) + \\dots + \\log(n/2)\\\\ U &= \\log(n/2+1) + \\dots + \\log(n-1) + \\log(n). \\end{align*}\n\nBasically, $T$ has the first $n/2$ terms of $S$, and $U$ has the remaining $n/2$ terms.\n\nNow we'll lower-bound each of them. Start with $T$. Each term in $T$ is at least $\\log(1)$, so we get\n\n$$T \\ge \\log(1) + \\log(1) + \\dots + \\log(1) = 0 + 0 + \\dots 0,$$\n\nso $T \\ge 0$. Next look at $U$. Each term in $U$ is at least $\\log(n/2)$, so we get\n\n$$U \\ge \\log(n/2) + \\log(n/2) + \\dots + \\log(n/2) = (n/2) \\times \\log(n/2).$$\n\nNow $S = T+U$, so plugging in the lower bounds obtained above, it follows that\n\n$$S =T+U\\ge 0 + (n/2) \\times \\log(n/2).$$\n\nThis is exactly the result you wanted to prove. Also, $(n/2) \\times \\log(n/2)$ is $\\Omega(n \\log n)$, so this proves that the sum $S$ is $\\Omega(n \\log n)$.\n\n\u2022 Ok I think I'm starting to understand, but first, what is stopping me from not breaking up the sum into two parts and just lower bounding each term in the sum? Like this: log(1) + log(1) + ... + log(1) n times. Sep 26, 2015 at 5:21\n\u2022 @Ghost_Stark, you can do that. That gives you the lower bound $S \\ge 0$, which is a valid lower bound, but not very useful. (The lower bound $S \\ge -42$ is also true, but even less useful.) It's just like if I tell you I'm thinking of a number between 1 and 100, and you ask for a hint, and I tell you: well, it's larger than $0$. You probably won't be amused at my useless hint, as you already knew that. It's the same for what you want to do. There are many lower bounds that one can prove -- the trick is to find one that is strong enough to prove what you want to prove.\n\u2013\u00a0D.W.\nSep 26, 2015 at 5:23\n\u2022 Ah I understand now! Thanks for the answer and the explanation! Very helpful. I guess my problem was not understanding that you have to have a reasonable lower bound to look for. Sep 26, 2015 at 5:31\n\n$\\dfrac n2$ is chosen so that you have sufficiently many factors ($O(n)$) that are sufficiently large ($O(n)$), so that the product remains $\\sim n^n$ while being a lower bound to $n!$\n\nThe intuition is that $\\log x$ is slowly-growing, and consequently \"most\" of the terms will be around $\\log n$ in size.\n\nMore precisely, if there are $\\Theta(n)$ terms that are all $\\Theta(\\log n)$ in size, then their sum will indeed be $\\Theta(n \\log n)$ and we can conclude $\\log n! \\in \\Omega(n \\log n)$.\n\nTaking half of the terms is merely the simplest idea to describe and calculate, and fortunately it satisfies the needed conditions.", "date": "2022-08-09 13:03:53", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/47561/confused-about-proof-that-logn-thetan-log-n", "openwebmath_score": 0.9994457364082336, "openwebmath_perplexity": 287.1289649091001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147201714923, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.853271338524734}}
{"url": "http://dict.cnki.net/h_237768000.html", "text": "\u5168\u6587\u6587\u732e \u5de5\u5177\u4e66 \u6570\u5b57 \u5b66\u672f\u5b9a\u4e49 \u7ffb\u8bd1\u52a9\u624b \u5b66\u672f\u8d8b\u52bf \u66f4\u591a\n\n \u65f6\u95f4-\u9891\u7387 \u7684\u7ffb\u8bd1\u7ed3\u679c: \u67e5\u8be2\u7528\u65f6\uff1a0.031\u79d2\n \u5728\u5206\u7c7b\u5b66\u79d1\u4e2d\u67e5\u8be2 \u6240\u6709\u5b66\u79d1 \u7535\u4fe1\u6280\u672f \u751f\u7269\u5b66 \u751f\u7269\u533b\u5b66\u5de5\u7a0b \u66f4\u591a\u7c7b\u522b\u67e5\u8be2\n\n \u5386\u53f2\u67e5\u8be2\n\n \u65f6\u95f4-\u9891\u7387\n time-frequency\n By displaying the change of signal energy on a two dimensional time-frequency images based on time-frequency analysis, a new mathematical morphology method to distinguish target from the nonlinear time-frequency curve is presented. \u5229\u7528\u65f6\u9891\u5206\u6790\u65b9\u6cd5\u5c06\u76f8\u5e72\u79ef\u7d2f\u65f6\u95f4\u5185\u7531\u4e8e\u76ee\u6807\u975e\u5747\u5300\u8fd0\u52a8\u548c\u5c04\u9891\u5e72\u6270\u5f15\u8d77\u7684\u4fe1\u53f7\u80fd\u91cf\u53d8\u5316\u663e\u793a\u4e8e\u65f6\u95f4-\u9891\u7387\u4e8c\u7ef4\u56fe\u50cf\u4e0a,\u5e76\u6839\u636e\u76ee\u6807\u4fe1\u53f7\u5bf9\u5e94\u65f6\u9891\u56fe\u50cf\u4e0a\u975e\u7ebf\u6027\u66f2\u7ebf\u7684\u7279\u70b9\u63d0\u51fa\u4e00\u79cd\u57fa\u4e8e\u4e8c\u503c\u5f62\u6001\u5b66\u7684\u65b9\u6cd5\u7528\u4e8e\u533a\u5206\u76ee\u6807\u548c\u5e72\u6270\u4fe1\u53f7\u3002 \u77ed\u53e5\u6765\u6e90 Time-frequency analysis can analyze nonstationary signal including earthquake signal. \u65f6\u9891\u5206\u6790\u80fd\u5bf9\u975e\u5e73\u7a33\u4fe1\u53f7\u8fdb\u884c\u5206\u6790,\u5e76\u7ed9\u51fa\u5730\u9707\u6ce2\u4fe1\u53f7\u80fd\u91cf\u7684\u65f6\u95f4-\u9891\u7387\u5206\u5e03\u3002 \u77ed\u53e5\u6765\u6e90 Transient Signal Detection Technology Based on Bilinear Time-Frequency Distribution \u57fa\u4e8e\u53cc\u7ebf\u6027\u65f6\u95f4-\u9891\u7387\u5206\u5e03\u7684\u77ac\u65f6\u4fe1\u53f7\u63a2\u6d4b\u6280\u672f \u77ed\u53e5\u6765\u6e90 A novel approach to detect abnormal transient vibration signal for complex mechanical system is presented,where alias-free exponential bilinear time-frequency transformation is adopted to avoid the frequency alias and information loss in the traditional bilinear distributions. Simultancously,the cross-terms are effectively deduced with higher time-frequency resolution,which is validated by a gearbox early fault detection. \u9488\u5bf9\u590d\u6742\u673a\u68b0\u7cfb\u7edf\u5f02\u5e38\u632f\u52a8\u4fe1\u53f7\u7684\u68c0\u6d4b,\u63d0\u51fa\u4e00\u79cd\u65b0\u7684\u65e0\u6df7\u53e0\u6307\u6570\u53cc\u7ebf\u6027\u65f6\u95f4-\u9891\u7387\u53d8\u6362\u63a2\u6d4b\u65b9\u6cd5,\u65b0\u53d8\u6362\u65b9\u6cd5\u53ef\u4ee5\u907f\u514d\u5e38\u7528\u53cc\u7ebf\u6027\u53d8\u6362\u4e2d\u7684\u9891\u7387\u6df7\u53e0\u548c\u4fe1\u606f\u4e22\u5931\u7b49\u95ee\u9898,\u80fd\u591f\u6709\u6548\u7684\u6291\u5236\u4ea4\u53c9\u9879,\u5177\u6709\u8f83\u9ad8\u7684\u65f6\u9891\u5206\u8fa8\u7387,\u7ed9\u51fa\u65b0\u7684\u53cc\u7ebf\u6027\u65f6-\u9891\u53d8\u6362\u7684\u79bb\u6563\u7b97\u6cd5,\u901a\u8fc7\u4eff\u771f\u5b9e\u9a8c\u4e0e\u9f7f\u8f6e\u7bb1\u65e9\u671f\u6545\u969c\u68c0\u6d4b,\u5bf9\u65b0\u7684\u65f6\u9891\u53d8\u6362\u7684\u68c0\u6d4b\u6548\u679c\u8fdb\u884c\u4e86\u9a8c\u8bc1\u3002 \u77ed\u53e5\u6765\u6e90 Hilbert spectrum is direct time-frequency energy distribution. The points on time-frequency grids described meaningful and actual distribution. \u5f97\u5230\u7684Hilbert\u8c31\u56fe\u4ee5\u76f4\u63a5\u7684\u53d8\u91cf(\u65f6\u95f4-\u9891\u7387)\u6765\u8868\u8fbe\u523b\u753b\u4fe1\u53f7\u672c\u8eab\u80fd\u91cf\u7684\u5206\u5e03,\u4f7f\u80fd\u91cf\u5206\u5e03\u771f\u6b63\u5730\u4ee5\u65f6\u95f4-\u9891\u7387\u7f51\u683c\u70b9\u7684\u5f62\u5f0f\u51fa\u73b0,\u5f97\u5230\u6709\u610f\u4e49\u7684\u5404\u4e2a\u65f6\u95f4-\u9891\u7387-\u80fd\u91cf\u70b9\u3002 \u77ed\u53e5\u6765\u6e90 \u66f4\u591a\n time and frequency\n The Mixed Time and Frequency Decimation FFT Algorithm (MDFFT) proposed by K. \u672c\u6587\u5bf9K. Nakayama\u63d0\u51fa\u7684\u65f6\u95f4-\u9891\u7387\u6df7\u5408\u62bd\u9009FFT\u7b97\u6cd5\u4f5c\u4e86\u7b80\u5316\u548c\u6df1\u5316,\u63d0\u51fa\u4e86\u6811\u5f62\u5206\u89e3FFT\u7b97\u6cd5,\u5176\u5b9e\u6570\u4e58\u6cd5\u6b21\u6570\u4e0eK. \u77ed\u53e5\u6765\u6e90 The scattering function is one of most important mathematical method describing time-frequency dispersion channel performance, which embodies a concentrated reflection of the dispersion characteristics of the mobile radio channel in time and frequency domain, and reflects the angle dispersion distributed performance of the channel to a certain extent. \u6563\u5c04\u51fd\u6570\u662f\u63cf\u8ff0\u65f6\u95f4-\u9891\u7387\u8272\u6563\u4fe1\u9053\u7279\u6027\u7684\u6700\u91cd\u8981\u7684\u6570\u5b66\u65b9\u6cd5\u4e4b\u4e00,\u5b83\u96c6\u4e2d\u53cd\u6620\u4e86\u4fe1\u9053\u5728\u65f6\u57df\u548c\u9891\u57df\u7684\u8272\u6563\u7279\u5f81,\u5e76\u8fd8\u5728\u4e00\u5b9a\u7a0b\u5ea6\u4e0a\u53cd\u6620\u51fa\u89d2\u5ea6\u8272\u6563\u7684\u5206\u5e03\u60c5\u51b5\u3002 \u77ed\u53e5\u6765\u6e90 Future broadband wireless communication systems should be able to provide reliable high-data-rate transmissions at low cost in time and frequency selective wireless channel. \u672a\u6765\u65e0\u7ebf\u901a\u4fe1\u7cfb\u7edf\u5e94\u80fd\u5728\u65f6\u95f4-\u9891\u7387\u9009\u62e9\u6027\u8870\u843d\u4fe1\u9053\u4e2d\u4ee5\u4f4e\u6210\u672c\u63d0\u4f9b\u53ef\u9760\u7684\u9ad8\u901f\u6570\u636e\u4f20\u8f93\u670d\u52a1\u3002 \u77ed\u53e5\u6765\u6e90 Aimed at all kinds of methods of time-frequency analysis currently,this paper emphasizes on introducing the HHT method ,which deeply depicts the relationship between time and frequency and gets the time-frequency-energy figure of the analysed signal using the integration of EMD and Hilbert transform. \u9488\u5bf9\u5f53\u524d\u5404\u79cd\u65f6\u9891\u5206\u6790\u65b9\u6cd5\uff0c\u672c\u6587\u91cd\u70b9\u4ecb\u7ecd\u4e86Hilbert-Huang\u53d8\u6362\u6cd5(HHT\uff0cHilbert-HuangTransform)\uff0c\u5b83\u5229\u7528 \u7ecf\u9a8c\u6a21\u6001\u5206\u89e3(EMD\uff0cEmpirical ModeDecomposition)\u6cd5\u4e0eHilbert\u53d8\u6362\u7684\u7ed3\u5408\uff0c\u6df1\u5165\u5730\u523b\u753b\u9891\u7387\u4e0e\u65f6\u95f4\u4e4b\u95f4\u7684\u5173\u7cfb\uff0c\u5f97\u5230\u4fe1\u53f7\u7684\u65f6\u95f4-\u9891\u7387-\u80fd\u91cf\u56fe\u3002 \u77ed\u53e5\u6765\u6e90 Wavelet analysis is a new mathematical method and has acquired wide application in the field due to its fine character of making analysis according to multiple resolutions in time and frequency domains. \u5c0f\u6ce2\u5206\u6790\u662f\u4e00\u79cd\u65b0\u7684\u6570\u5b66\u65b9\u6cd5,\u6839\u636e\u5176\u65f6\u95f4-\u9891\u7387\u7b49\u591a\u5206\u8fa8\u7387\u5206\u6790\u7684\u4f18\u826f\u7279\u6027,\u5728\u8bb8\u591a\u65b9\u9762\u83b7\u5f97\u5e7f\u6cdb\u5e94\u7528\u3002 \u66f4\u591a\n \u201c\u65f6\u95f4-\u9891\u7387\u201d\u8bd1\u4e3a\u672a\u786e\u5b9a\u8bcd\u7684\u53cc\u8bed\u4f8b\u53e5\n Hilbert spectrum has an energy-frequency-time distribution of the signal. Hilbert\u8c31\u662f\u4fe1\u53f7\u7684\u65f6\u95f4-\u9891\u7387-\u80fd\u91cf\u5206\u5e03\u3002 \u77ed\u53e5\u6765\u6e90 Time frequency energy have been used for speech detection in noisy environments. \u4ee5\u524d\u63d0\u51fa\u7684\u57fa\u4e8e\u65f6\u95f4 -\u9891\u7387\u7684\u80fd\u91cf\u53c2\u6570\u5229\u7528\u9891\u57df\u7684\u9650\u5e26\u80fd\u91cf\u52a0\u4e0a\u65f6\u57df\u80fd\u91cf\u6765\u8fdb\u884c\u566a\u58f0\u4e2d\u7684\u8bed\u97f3\u68c0\u6d4b\u3002 \u77ed\u53e5\u6765\u6e90 The characteristics of the urban heat island(UHI) intensity in Shanghai were studied and calculated using wavelet transformation, based on per hour difference in temperatures in 2000 observed from Davis automatic stations installed at the Urban and Suburb of Shanghai. \u53d6\u4e0a\u6d77\u5e02\u533a\u5404\u7ad9\u70b9\u7684\u5e73\u5747\u6e29\u5ea6\u4e0e\u8fd1\u90ca\u7ad9\u70b9\u7684\u6e29\u5ea6\u4e4b\u5dee\u4f5c\u4e3a\u8861\u91cf\u57ce\u5e02\u70ed\u5c9b\u5f3a\u5ea6\u7684\u6307\u6807,\u5229\u7528\u4e0a\u6d77\u5e02\u57ce\u533a\u548c\u90ca\u533a\u76846\u4e2aDavis\u81ea\u52a8\u6c14\u8c61\u89c2\u6d4b\u4eea\u6bcf\u5c0f\u65f6\u89c2\u6d4b\u7684\u8bb0\u5f55,\u5bf92000\u5e74\u7684\u6e29\u5dee\u5e8f\u5217\u8fdb\u884c\u5c0f\u6ce2\u53d8\u6362,\u5206\u6790\u4e0a\u6d77\u5e02\u57ce\u5e02\u70ed\u5c9b\u5f3a\u5ea6\u65f6\u95f4-\u9891\u7387\u7684\u591a\u65f6\u95f4\u5c3a\u5ea6\u7684\u6f14\u53d8\u89c4\u5f8b. \u77ed\u53e5\u6765\u6e90 Doubly Selective Fading Channel Estimation Based on Polynomial Interpolation in MIMO Systems \u57fa\u4e8e\u591a\u9879\u5f0f\u5185\u63d2\u7684MIMO\u65f6\u95f4-\u9891\u7387\u53cc\u9009\u62e9\u6027\u4fe1\u9053\u7684\u4fe1\u9053\u4f30\u8ba1 \u77ed\u53e5\u6765\u6e90 Chirplet transform(CT) is extension of short-time Fourier transforms(STFTs) and wavelet transforms. \u7ebf\u8c03\u9891\u5c0f\u6ce2\u53d8\u6362\u7edf\u4e00\u4e86\u77ed\u65f6Fourier\u53d8\u6362\u548c\u5c0f\u6ce2\u53d8\u6362\u7684\u65f6\u9891\u5206\u6790,\u662f\u4fe1\u53f7\u7684\u65f6\u95f4-\u9891\u7387-\u5c3a\u5ea6\u53d8\u6362,\u80fd\u6839\u636e\u4fe1\u53f7\u7684\u7279\u70b9\u81ea\u9002\u5e94\u751f\u6210\u65b0\u7684\u65f6\u9891\u7a97\u53e3\u3002 \u77ed\u53e5\u6765\u6e90 \u66f4\u591a\n\n\u6211\u60f3\u67e5\u770b\u8bd1\u6587\u4e2d\u542b\u6709\uff1a\u7684\u53cc\u8bed\u4f8b\u53e5\n\n time-frequency\n We present two-sided singular value estimates for a class of convolution-product operators related to time-frequency localization. The Balian-Low theorem (BLT) is a key result in time-frequency analysis, originally stated by Balian and, independently, by Low, as: If a Gabor system $\\{e^{2\\pi imbt} \\, g(t-na)\\}_{m,n \\in \\mbox{\\bf Z}}$ Gabor Time-Frequency Lattices and the Wexler-Raz Identity Gabor time-frequency lattices are sets of functions of the form $g_{m \\alpha , n \\beta} (t) =e^{-2 \\pi i \\alpha m t}g(t-n \\beta)$ generated from a given function $g(t)$ by discrete translations in time and frequency. High-Order Orthonormal Scaling Functions and Wavelets Give Poor Time-Frequency Localization \u66f4\u591a\n time and frequency\n Gabor time-frequency lattices are sets of functions of the form $g_{m \\alpha , n \\beta} (t) =e^{-2 \\pi i \\alpha m t}g(t-n \\beta)$ generated from a given function $g(t)$ by discrete translations in time and frequency. We present an explicit, straightforward construction of smooth integrable functions with prescribed gaps around the origin in both time and frequency domain. From the analysis of the electric field structure the conclusion is drawn that the bulk of the AKR power is carried by the signal component fast variable in time and frequency (flickering component). The spectrometer provides the programming of time and frequency parameters of pulse sequences and allows phase-sensitive detection of echo signals with their subsequent representation in digital form. Experimental data are presented on time and frequency dependences of the reverberation level for bistatic transmission and reception at low acoustic frequencies. \u66f4\u591a\n The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy... The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy in maintenance. \u672c\u6587\u4ecb\u7ecd\u4e86\u4e00\u5957\u7528\u4e8e\u7535\u6c14\u5316\u94c1\u9053\u5206\u533a\u4ead\u7684\u7535\u5b50\u88c5\u7f6e\u3002\u5b83\u662f1\u5bf95\u7684\u5206\u6563\u96c6\u4e2d\u5f0f\u7cfb\u7edf,\u9065\u63a7\u4e0e\u9065\u4fe1\u7efc\u5408\u4e8e\u540c\u4e00\u7cfb\u7edf\u4e2d\u3002\u901a\u9053\u91c7\u7528\u65f6\u95f4-\u9891\u7387\u5212\u5206\u3002\u5404\u88ab\u63a7\u7ad9\u9065\u4fe1\u4e3a\u7ecf\u5e38\u81ea\u52a8\u5faa\u73af\u4f20\u9001\u65b9\u5f0f,\u4fe1\u606f\u7801\u91c7\u7528\u4e8c\u6b21\u4f20\u9001,\u6b63\u53cd\u7801\u6821\u6838\u3002\u903b\u8f91\u8bbe\u8ba1\u5408\u7406,\u7535\u8def\u7b80\u5355,\u5de5\u4f5c\u53ef\u9760,\u7ef4\u4fee\u65b9\u4fbf\u3002 126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge... 126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge rate)with those responding to unmodulatedlight(unmodulated discharge rate),two opposite effects were observed.The majorityof the cells(93.7)% showed modulation-excitatory curves in which the modulateddischarge rates were higher than the unmodulated ones.A minority of the cells(6.3%)showed modulation-inhibitory curves in which the modulated dischargerates were lower than the unmodulated ones.The modulation-excitatory curvescould further be grouped into two subtypes,the\u201cband-pass filters\u201dand the\u201clow-pass filters\u201d.Similarly,the modulation-inhibitory curves could also be dividedinto two subtypes,the\u201cband-rejection filters\u201dand the\u201clow-rejection filters\u201d.Onboth sides of the temporal frequency tuning curve subsidiary sidebands oppositein direction to the main part of the curves could often be seen,i.e.inhibitorysidebands flanking the modunation-excitatory tuning curve and excitatorysidebands flanking the modulation-inhibitory tuning curve.Most of the tuningcurves have only one peak.of the 110 curves measured the peak loci show anormal distribution which centered at 7 Hz.The tuning curves due to stimulationof different locations of the periphery were almost the same.But the tuningproperties of the receptive field centers were different from those of the periphe-ries either in shape or in band-width. \u7528\u793a\u6ce2\u5668\u4ea7\u751f\u4eae\u5ea6\u53d7\u6b63\u5f26\u6ce2\u8c03\u5236\u7684\u5c0f\u5149\u70b9\u523a\u6fc0\u6e05\u9192\u732b\u5916\u819d\u4f53\u795e\u7ecf\u5143\u7684\u611f\u53d7\u91ce,\u4ee5\u4e0d\u540c\u8c03\u5236\u9891\u7387\u4e0b\u795e\u7ecf\u5143\u653e\u7535\u7684\u5e73\u5747\u9891\u7387\u4e3a\u6307\u6807,\u5206\u6790\u4e86126\u4e2a\u7ec6\u80de\u611f\u53d7\u91ce\u4e2d\u5fc3\u7684\u65f6\u95f4\u9891\u7387\u8c03\u8c10\u7279\u6027,\u4e3b\u8981\u7ed3\u679c\u5982\u4e0b\u3002(1)\u5927\u591a\u6570(93.7%)\u7ec6\u80de\u5448\u8c03\u5236-\u5174\u594b\u578b\u53cd\u5e94,\u5373\u523a\u6fc0\u5149\u7684\u65f6\u95f4\u8c03\u5236\u5728\u4e00\u5b9a\u9891\u7387\u8303\u56f4\u5185\u4f7f\u653e\u7535\u9891\u7387\u589e\u52a0;\u5c11\u6570(6.3%)\u7ec6\u80de\u5448\u8c03\u5236-\u6291\u5236\u578b\u53cd\u5e94,\u5373\u5728\u4e00\u5b9a\u9891\u7387\u8303\u56f4\u5185,\u65f6\u95f4\u8c03\u5236\u4f7f\u653e\u7535\u51cf\u5c11\u3002(2)\u6839\u636e\u8c03\u8c10\u66f2\u7ebf\u7684\u5f62\u72b6\u548c\u901a\u5e26\u5bbd\u5ea6,\u8c03\u5236-\u5174\u594b\u578b\u53cd\u5e94\u5305\u62ec\u5e26\u901a\u6ee4\u6ce2\u5668\u548c\u4f4e\u901a\u6ee4\u6ce2\u5668\u4e24\u79cd\u7c7b\u578b,\u5176110\u4e2a\u8c03\u8c10\u66f2\u7ebf\u7684\u5cf0\u503c\u5206\u5e03\u63a5\u8fd1\u6b63\u6001\u66f2\u7ebf,\u591a\u6570\u7ec6\u80de\u5bf97Hz \u7684\u8c03\u8c10\u6700\u654f\u611f\u3002\u8c03\u5236\u4e00\u6291\u5236\u578b\u53cd\u5e94\u5305\u62ec\u5e26\u9664\u6ee4\u6ce2\u5668\u548c\u4f4e\u9664\u6ee4\u6ce2\u5668\u4e24\u7c7b\u3002(3)\u8c03\u5236-\u5174\u594b\u578b\u66f2\u7ebf\u7684\u901a\u5e26\u65c1\u8fb9\u8f83\u7576\u51fa\u73b0\u6291\u5236\u6027\u4fa7\u5e26,\u8c03\u5236-\u6291\u5236\u578b\u66f2\u7ebf\u51fa\u73b0\u5174\u594b\u6027\u4fa7\u5e26\u3002(4)\u611f\u53d7\u91ce\u4e2d\u5fc3\u533a\u4e0e\u5916\u5468\u533a\u7684\u65f6\u95f4\u9891\u7387\u8c03\u8c10\u66f2\u7ebf\u7684\u5e26\u5bbd\u548c\u5f62\u72b6\u6709\u6240\u4e0d\u540c\u3002 This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic... This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic time/frequency and in developing new optimal filting control method to improve the performances of atomic clocks. \u672c\u6587\u4e3b\u8981\u6839\u636e\u65f6\u95f4\u5e8f\u5217\u5206\u6790\u65b9\u6cd5\u7684ARIMA(autoregressive integraled moving averge)\u6a21\u578b\u539f\u7406[1-6],\u7ed3\u5408\u539f\u5b50\u65f6\u95f4\u9891\u7387\u6d4b\u91cf\u7edf\u8ba1\u5b66\u7684\u566a\u58f0\u65b9\u5dee\u504f\u51fd\u6570\u7406\u8bba[7-9],\u4ee5\u53ca\u9002\u5f53\u91c7\u7528\u6837\u6761\u51fd\u6570\u65b9\u6cd5[10],\u5bf9\u6c22,\u94ef\u3001\u94f7\u7b49\u539f\u5b50\u949f\u4e0d\u540c\u53d6\u6837\u65f6\u95f4\u7684\u6d4b\u91cf\u6570\u636e\u8fdb\u884c\u5efa\u6a21\u548c\u9884\u62a5\u3002\u5bf9\u4e8e\u4e0d\u540c\u566a\u58f0\u8fc7\u7a0b,\u521d\u6b65\u5f97\u5230\u4e86\u4e00\u4e9b\u6709\u7528\u7684\u9884\u62a5\u6a21\u578b\u548c\u7ed3\u8bba,\u8fd9\u5bf9\u4e8e\u539f\u5b50\u65f6\u95f4\u9891\u7387\u6027\u80fd\u7684\u6700\u4f73\u9884\u62a5\u548c\u76d1\u6d4b(\u5982\u8df3\u76f8\u548c\u8df3\u9891)\u4ee5\u53ca\u7814\u7a76\u65b0\u7684\u6700\u4f73\u6ee4\u6ce2\u63a7\u5236\u65b9\u6cd5\u4ee5\u63d0\u9ad8\u539f\u5b50\u949f\u7684\u6027\u80fd\u6307\u6807\u90fd\u6709\u5b9e\u9645\u610f\u4e49\u3002 << \u66f4\u591a\u76f8\u5173\u6587\u6458\n \u76f8\u5173\u67e5\u8be2\n\n CNKI\u5c0f\u5de5\u5177 \u5728\u82f1\u6587\u5b66\u672f\u641c\u7d22\u4e2d\u67e5\u6709\u5173\u65f6\u95f4-\u9891\u7387\u7684\u5185\u5bb9 \u5728\u77e5\u8bc6\u641c\u7d22\u4e2d\u67e5\u6709\u5173\u65f6\u95f4-\u9891\u7387\u7684\u5185\u5bb9 \u5728\u6570\u5b57\u641c\u7d22\u4e2d\u67e5\u6709\u5173\u65f6\u95f4-\u9891\u7387\u7684\u5185\u5bb9 \u5728\u6982\u5ff5\u77e5\u8bc6\u5143\u4e2d\u67e5\u6709\u5173\u65f6\u95f4-\u9891\u7387\u7684\u5185\u5bb9 \u5728\u5b66\u672f\u8d8b\u52bf\u4e2d\u67e5\u6709\u5173\u65f6\u95f4-\u9891\u7387\u7684\u5185\u5bb9\n\n CNKI\u4e3b\u9875 |\u00a0 \u8bbeCNKI\u7ffb\u8bd1\u52a9\u624b\u4e3a\u4e3b\u9875 | \u6536\u85cfCNKI\u7ffb\u8bd1\u52a9\u624b | \u5e7f\u544a\u670d\u52a1 | \u82f1\u6587\u5b66\u672f\u641c\u7d22\n2008 CNKI\uff0d\u4e2d\u56fd\u77e5\u7f51\n\n2008\u4e2d\u56fd\u77e5\u7f51(cnki) \u4e2d\u56fd\u5b66\u672f\u671f\u520a(\u5149\u76d8\u7248)\u7535\u5b50\u6742\u5fd7\u793e", "date": "2019-10-15 18:26:04", "meta": {"domain": "cnki.net", "url": "http://dict.cnki.net/h_237768000.html", "openwebmath_score": 0.40670904517173767, "openwebmath_perplexity": 4560.91708415228, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147185726374, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.8532713371217759}}
{"url": "https://math.stackexchange.com/questions/1868418/how-many-3-digit-numbers-that-the-sum-of-their-digits-equals-12", "text": "# How many 3 digit numbers that the sum of their digits equals 12?\n\nHow many positive 3-digit numbers exist such that the sum if their digits equals 12?\n\nA) 54\n\nB) 61\n\nC) 64\n\nD) 65\n\nE) 66\n\nI believe the answer is E.\n\nOnline problems state that is a stars and bars problem, however using the (n-1,k-1)/n-1C k-1 and (n+k-1,n)/n+k-1 C n formulas do not yield any of the answer choices.\n\nUsing (n-1,k-1) I get 55 and using (n+k-1,n) I get 91. Fifty five appears to be the closest answer.\n\nI do not know how to do the inclusion method to remove results such as 066.\n\n\u2022 Please show us your calculations so that we can see what you did and where you ran into difficulties. \u2013\u00a0N. F. Taussig Jul 23 '16 at 12:12\n\u2022 Online problems state that is a stars and bars problem, however using the n-1 C k-1 and n+k-1 C n formulas do not yield 54. I get 190 from n+k-1 C n and 139 from n-1 C k-1 \u2013\u00a0G_Derek007 Jul 23 '16 at 12:21\n\u2022 Please edit your question to show us exactly what you did. I suspect you have not taken into account the restrictions on the hundreds digit, tens digit, and units digit. \u2013\u00a0N. F. Taussig Jul 23 '16 at 12:23\n\u2022 I have changed the question to 12 however it should still be the same concept \u2013\u00a0G_Derek007 Jul 23 '16 at 12:29\n\u2022 If I read your work correctly, you used the formula for solutions in the positive integers to obtain $\\binom{12 - 1}{3 - 1} = \\binom{11}{2} = 55$ and for solutions in the non-negative integers to obtain $\\binom{12 + 3 - 1}{3 - 1} = \\binom{14}{2} = 91$. Is that what you did? \u2013\u00a0N. F. Taussig Jul 23 '16 at 12:36\n\nA three-digit positive integer has the form $100h + 10t + u$, where the hundreds digit $h$ satisfies the inequalities $1 \\leq h \\leq 9$, the tens digit $t$ satisfies the inequalities $0 \\leq t \\leq 9$, and the units digit $u$ satisfies the inequalities $0 \\leq u \\leq 9$. Therefore, we wish to determine the number of solutions of the equation $$h + t + u = 12 \\tag{1}$$ subject to these restrictions.\n\nIf we let $h' = h - 1$, then $0 \\leq h' \\leq 8$. Substituting $h' + 1$ for $h$ in equation 1 yields \\begin{align*} h' + 1 + t + u & = 12\\\\ h' + t + u & = 11 \\tag{2} \\end{align*} Equation 2 is an equation in the non-negative integers. A particular solution corresponds to the placement of two addition signs in a row of eleven ones. There are $$\\binom{11 + 2}{2} = \\binom{13}{2}$$ such solutions since we must choose which two of the thirteen symbols (eleven ones and two addition signs) will be addition signs.\n\nHowever, we have counted solutions in which $h' > 8$, $t > 9$, or $u > 9$. We must exclude these.\n\nSuppose $h' > 8$. Then $h'$ is an integer satisfying $h' \\geq 9$. Let $h'' = h' - 9$. Then $h'' \\geq 0$. Substituting $h'' + 9$ for $h'$ in equation 2 yields \\begin{align*} h'' + 9 + t + u & = 11\\\\ h'' + t + u & = 2 \\tag{3} \\end{align*} Equation 3 is an equation in the non-negative integers. Since a particular solution corresponds to the placement of two addition signs in a row of two ones, it has $$\\binom{2 + 2}{2} = \\binom{4}{2}$$ solutions.\n\nSuppose $t > 9$. Then $t$ is an integer satisfying $t \\geq 10$. Let $t' = t - 10$. Substituting $t' + 10$ for $t$ in equation 2 yields \\begin{align*} h' + t' + 10 + u & = 11\\\\ h' + t' + u & = 1 \\tag{4} \\end{align*} Equation 4 is an equation in the non-negative integers with $\\binom{3}{2} = 3$ solutions, depending on which variable is equal to $1$. By symmetry, there are also three solutions in which $u' > 9$. No two of these restrictions cannot be violated simultaneously since $9 + 10 = 19 > 12$. Thus, the number of three-digit positive integers with digit sum $12$ is $$\\binom{13}{2} - \\binom{4}{2} - 2\\binom{3}{2} = 66$$\n\n\u2022 It became a lot more clearer thanks! Is it too much to ask if its possible to derive a general solution? \u2013\u00a0G_Derek007 Jul 23 '16 at 13:20\n\u2022 I believe I derived one already to be sure though if the sum of the digits were 14 it would be 70? The formula is S=sum of digits D=number of digits (S+D-1,D-1) - (S+D-1-9,D-1) - 2*(S+D-1-9-1,D-1)it should work for 3 digit numbers, which is what I need \u2013\u00a0G_Derek007 Jul 23 '16 at 13:35\n\u2022 @user295641 Yes, there are 70 3 digit numbers whose digits sum to 14. \u2013\u00a0PM 2Ring Jul 23 '16 at 13:38\n\u2022 That would be a rather long answer. For $h + t + u = k$, where $1 \\leq k \\leq 9$, you can stop after finding the solution to equation 2 with $k$ replacing $12$. For $h + t + u = 10$, you can stop after equation $3$ with $10$ replacing $12$. For $h + t + u = k$, where $11 \\leq k \\leq 18$, the same procedure will work if you replace $12$ by $k$. For $18 \\leq k \\leq 27$, we require the Inclusion-Exclusion Principle because two or more restrictions can be violated simultaneously in equation 2. \u2013\u00a0N. F. Taussig Jul 23 '16 at 13:39\n\u2022 @user295641 You have an off-by-one error in that formula; I think you mean: $(S+1, 2) - (S-8, 2) - 2*(S-9, 2)$. However, that formula only works for $9 \\le S \\le 19$. Note that there is a symmetry. Let $f(S, D)$ be the count of $D$ digit numbers that sum to $S$. Then $f(9D+1-S, D) = f(S, D)$ \u2013\u00a0PM 2Ring Jul 23 '16 at 14:09\n\nHere's a less algebraic way to look at it. .\n\nStart by putting $1$ into the first cell, ${\\boxed 1}\\Large\\boxed.\\boxed.$, so you now only need a sum of $11$,\nwith the constraints that you can't put $\\ge9$ in the first cell, and $\\ge10$ in the other two.\n\nApply stars and bars, subtracting solutions that violate the constraints.\n\nSince it is not possible to violate the constraint in more than one cell, $$ans = \\binom{13}2 - \\binom{13-9}2 - 2\\binom{13-10}2 = 66$$", "date": "2019-06-19 15:47:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1868418/how-many-3-digit-numbers-that-the-sum-of-their-digits-equals-12", "openwebmath_score": 0.8413844108581543, "openwebmath_perplexity": 138.78934190829338, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147169737826, "lm_q2_score": 0.8774767810736693, "lm_q1q2_score": 0.853271335718818}}
{"url": "https://math.stackexchange.com/questions/3587292/interesting-property-related-to-the-sums-of-the-remainders-of-integers", "text": "Interesting property related to the sums of the remainders of integers\n\nCross-posted on Math Overflow too\n\nLet us define, $$r(b)=\\sum_{k=1}^{\\lfloor \\frac{b-1}{2} \\rfloor} (b \\bmod{k})$$ After playing around with the $$r(b)$$ function for sometime I noticed that $$r(b)$$ appreared to be more even than odd. So to see the difference between the number of even and odd terms of $$r(b)$$, I defined a function, $$z(x)=\\sum_{n=1}^x(-1)^{r(n)}$$ When user Peter ran a program for computing values of $$z(x)$$ in PARI, I observed that for $$x\\le 10^{10}$$, $$z(x)\\gt 0$$. This suggests that there are always more even terms of $$r(n)$$ than odd terms for any $$x$$.\n\nThis leads to my two questions:\n\n1. Is $$z(x)$$ always positive? If so, then how do we prove this?\n2. Is $$|z(x)|$$ bounded by some maximum value? If so, then what is this maximum value? Till now the maximum value of $$|z(x)|$$ found was $$49$$ for $$x = 5424027859$$. I find it odd that $$|z(x)|$$ goes to these large values and then returns back to small values as small as $$1$$.\n\nEdit:\n\nWith the help of user Vepir I was able to plot $$z(x)$$ and noticed that the function has a sinusoidal-fractal-like appearance and it seems to grow without bounds, although very very slowly.\n\nIs there any reason for this sinusoidal and fractal nature?\n\n\u2022 The efficient calculation is based on the formula for $d(n):=r(n)-r(n-1)$ , which is $$d(n)=2n-1-\\sigma(n)$$ for even $n$ and $$d(n)=\\frac{3n-1}{2}-\\sigma(n)$$ for odd $n$ and the fact that $\\sigma(n)$ is odd if and only if $n=k^2$ or $n=2k^2$ for some positive integer $k$. \u2013\u00a0Peter Mar 19 at 21:50\n\u2022 $r(n)$ and $r(n-1)$ have the same parity if and only if $n$ is of the form $k^2,2k^2$ or $4k+3$ for some positive integer $k$ (in the case $4k+3$ , $k=0$ is also possible). \u2013\u00a0Peter Mar 19 at 21:54\n\u2022 If my calculation is correct, we have $z(31988856^2)=80$ and the function does not really feel like it would be bounded from above. \u2013\u00a0Peter Ko\u0161in\u00e1r Mar 29 at 2:02\n\u2022 @PeterKo\u0161in\u00e1r thanks for the calculation. Was $z(n)$ ever negative in that range? \u2013\u00a0Mathphile Mar 29 at 4:48\n\u2022 @Mathphile No, I have not found any case where $z(n)$ would be negative or zero. It has reached $1$ on multiple occasions, though. (again, assuming my computation was correct). \u2013\u00a0Peter Ko\u0161in\u00e1r Mar 29 at 16:43\n\nI can't prove/disprove your conjectures, but I proved a claim which might be useful to prove/disprove your conjectures.\n\nThis answer proves the following claim :\n\nClaim :\n\n\\small\\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\\\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\\\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\\\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\\\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\\\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\\\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\\\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\\end{align} where $$c(n)=\\#\\{x:\\text{ 1\\le x\\le n, and x is either a square or twice a square}\\}$$ and $$S(m):=\\sum_{k=0}^{m-1}\\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\\bigg)$$\n\nAccording to oeis.org/A071860, one has $$c(n)=\\lfloor\\sqrt{n}\\rfloor+\\left\\lfloor\\sqrt{\\frac n2}\\right\\rfloor$$ which might make the problem easier to deal with.\n\nAlso, it follows from the above claim that $$z(8m+4)=z(8m+6)$$ for every non-negative integer $$m$$.\n\nThe claim follows from the following lemmas :\n\nLemma 1 : $$r(n)=n^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor +1\\bigg)-\\sum_{k=1}^{n}\\sigma(k)$$\n\nLemma 2 : $$r(n)\\stackrel{\\text{mod 2}}\\equiv\\begin{cases}\\displaystyle\\sum_{k=1}^{n}\\sigma(k)&\\text{if n\\equiv 0,2,3,5\\pmod 8}\\\\\\\\1+\\displaystyle\\sum_{k=1}^{n}\\sigma(k)&\\text{if n\\equiv 1,4,6,7\\pmod 8}\\end{cases}$$\n\nLemma 3 : $$\\text{\\sigma(n) is odd \\iff n is either a square or twice a square}$$\n\nLemma 4 : $$\\sum_{k=1}^{n}\\sigma(k)\\equiv c(n)\\pmod 2$$where $$c(n)=\\#\\{x:\\text{ 1\\le x\\le n, and x is either a square or twice a square}\\}$$\n\nLemma 5 : If $$n\\equiv 3\\pmod 4$$, then $$c(n)=c(n-1)$$.\n\nLemma 6 : $$z(x)=\\sum_{k=0}^{\\lfloor x/8\\rfloor}(-1)^{c(8k)}-\\sum_{k=0}^{\\lfloor (x-1)/8\\rfloor}(-1)^{c(8k+1)}+\\sum_{k=0}^{\\lfloor (x-2)/8\\rfloor}(-1)^{c(8k+2)}+\\sum_{k=0}^{\\lfloor (x-3)/8\\rfloor}(-1)^{c(8k+2)}-\\sum_{k=0}^{\\lfloor (x-4)/8\\rfloor}(-1)^{c(8k+4)}+\\sum_{k=0}^{\\lfloor (x-5)/8\\rfloor}(-1)^{c(8k+5)}-\\sum_{k=0}^{\\lfloor (x-6)/8\\rfloor}(-1)^{c(8k+5)}-\\sum_{k=0}^{\\lfloor (x-7)/8\\rfloor}(-1)^{c(8k+7)}$$\n\nLemma 7 :\n\n\\small\\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\\\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\\\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\\\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\\\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\\\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\\\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\\\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\\end{align}\n\nwhere $$S(m):=\\sum_{k=0}^{m-1}\\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\\bigg)$$\n\nLemma 1 : $$r(n)=n^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor +1\\bigg)-\\sum_{k=1}^{n}\\sigma(k)$$\n\nProof : For $$n=1$$, it is true. In the following, let us use $$r(n+1)-r(n)=\\begin{cases}2n+1-\\sigma(n+1)&\\text{if n is odd}\\\\\\\\\\frac{3n+2}{2}-\\sigma(n+1)&\\text{if n is even}\\end{cases}$$\n\nSuppose that it is true for $$n=2m+1$$. Then, we get \\begin{align}&r(n+1) \\\\\\\\&=r(n)+2n+1-\\sigma(n+1) \\\\\\\\&=n^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor +1\\bigg) -\\sum_{k=1}^{n}\\sigma(n)+2n+1-\\sigma(n+1) \\\\\\\\&=(2m+1)^2-\\frac{m(m+1)}{2}+2(2m+1)+1-\\sum_{k=1}^{n+1}\\sigma(k) \\\\\\\\&=(2m+2)^2-\\frac{m(m+1)}{2}-\\sum_{k=1}^{n+1}\\sigma(k) \\\\\\\\&=(2m+2)^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{2m-1}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{2m-1}{2}\\right\\rfloor +1\\bigg)-\\sum_{k=1}^{n+1}\\sigma(k) \\\\\\\\&=(n+1)^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{n+1-3}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{n+1-3}{2}\\right\\rfloor +1\\bigg)-\\sum_{k=1}^{n+1}\\sigma(k)\\end{align}\n\nSuppose that it is true for $$n=2m$$. Then, we get \\begin{align}&r(n+1) \\\\\\\\&=r(n)+\\frac{3n+2}{2}-\\sigma(n+1) \\\\\\\\&=n^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{n-3}{2}\\right\\rfloor +1\\bigg) -\\sum_{k=1}^{n}\\sigma(k)+\\frac{3n+2}{2}-\\sigma(n+1) \\\\\\\\&=(2m)^2-\\frac{m(m-1)}{2}+3m+1-\\sum_{k=1}^{n+1}\\sigma(k) \\\\\\\\&=(2m+1)^2-\\frac{(m+1)m}{2}-\\sum_{k=1}^{n+1}\\sigma(k) \\\\\\\\&=(2m+1)^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{2m+1-3}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{2m+1-3}{2}\\right\\rfloor +1\\bigg)-\\sum_{k=1}^{n+1}\\sigma(k) \\\\\\\\&=(n+1)^2-\\frac{1}{2}\\bigg(\\left\\lfloor\\frac{n+1-3}{2}\\right\\rfloor+2\\bigg)\\bigg(\\left\\lfloor\\frac{n+1-3}{2}\\right\\rfloor +1\\bigg)-\\sum_{k=1}^{n+1}\\sigma(k)\\end{align}\n\nSo, it is also true for $$n+1$$.$$\\quad\\square$$\n\nLemma 2 : $$r(n)\\stackrel{\\text{mod 2}}\\equiv\\begin{cases}\\displaystyle\\sum_{k=1}^{n}\\sigma(k)&\\text{if n\\equiv 0,2,3,5\\pmod 8}\\\\\\\\1+\\displaystyle\\sum_{k=1}^{n}\\sigma(k)&\\text{if n\\equiv 1,4,6,7\\pmod 8}\\end{cases}$$\n\nProof : It follows from Lemma 1 that$$r(8m)=64m^2-2m(4m-1)-\\sum_{k=1}^{8m}\\sigma(k)\\equiv \\sum_{k=1}^{8m}\\sigma(k)\\pmod 2$$\n\n$$r(8m+1)=(8m+1)^2-2m(4m+1)-\\sum_{k=1}^{8m+1}\\sigma(k)\\equiv 1+\\sum_{k=1}^{8m+1}\\sigma(k)\\pmod 2$$\n\n$$r(8m+2)=(8m+2)^2-2m(4m+1)-\\sum_{k=1}^{8m+2}\\sigma(k)\\equiv \\sum_{k=1}^{8m+2}\\sigma(k)\\pmod 2$$\n\n$$r(8m+3)=(8m+3)^2-(2m+1)(4m+1)-\\sum_{k=1}^{8m+3}\\sigma(k)\\equiv \\sum_{k=1}^{8m+3}\\sigma(k)\\pmod 2$$\n\n$$r(8m+4)=(8m+4)^2-(2m+1)(4m+1)-\\sum_{k=1}^{8m+4}\\sigma(k)\\equiv 1+\\sum_{k=1}^{8m+4}\\sigma(k)\\pmod 2$$\n\n$$r(8m+5)=(8m+5)^2-(2m+1)(4m+3)-\\sum_{k=1}^{8m+5}\\sigma(k)\\equiv \\sum_{k=1}^{8m+5}\\sigma(k)\\pmod 2$$\n\n$$r(8m+6)=(8m+6)^2-(2m+1)(4m+3)-\\sum_{k=1}^{8m+6}\\sigma(k)\\equiv 1+\\sum_{k=1}^{8m+6}\\sigma(k)\\pmod 2$$\n\n$$r(8m+7)=(8m+7)^2-(2m+2)(4m+3)-\\sum_{k=1}^{8m+7}\\sigma(k)\\equiv 1+\\sum_{k=1}^{8m+7}\\sigma(k)\\pmod2$$ So, the claim follows.$$\\quad\\square$$\n\nLemma 3 : $$\\text{\\sigma(n) is odd \\iff n is either a square or twice a square}$$\n\nProof : See here or here.\n\nLemma 4 : $$\\sum_{k=1}^{n}\\sigma(k)\\equiv c(n)\\pmod 2$$where $$c(n)=\\#\\{x:\\text{ 1\\le x\\le n, and x is either a square or twice a square}\\}$$\n\nProof : It follows from Lemma 3 that $$\\sum_{k=1}^{n}\\sigma(k)=\\underbrace{\\sum_{k\\in A}\\sigma(k)}_{\\text{sum of odd numbers}}+\\underbrace{\\sum_{k\\not\\in A}\\sigma(k)}_{\\text{sum of even numbers = even}}\\equiv \\sum_{k\\in A}\\sigma(k)=c(n)\\pmod 2$$ where $$A=\\{n\\ :\\ \\text{n is either a square or twice a square}\\}$$.\n\nLemma 5 : If $$n\\equiv 3\\pmod 4$$, then $$c(n)=c(n-1)$$.\n\nProof : Since we have $$\\text{(a square)}\\equiv 0,1\\pmod 4\\qquad\\text{and}\\qquad \\text{(twice a square)}\\equiv 0,2\\pmod 4$$ we see that if $$n\\equiv 3\\pmod 4$$, then $$n$$ is neither a square nor twice a square.\n\nLemma 6 : $$z(x)=\\sum_{k=0}^{\\lfloor x/8\\rfloor}(-1)^{c(8k)}-\\sum_{k=0}^{\\lfloor (x-1)/8\\rfloor}(-1)^{c(8k+1)}+\\sum_{k=0}^{\\lfloor (x-2)/8\\rfloor}(-1)^{c(8k+2)}+\\sum_{k=0}^{\\lfloor (x-3)/8\\rfloor}(-1)^{c(8k+2)}-\\sum_{k=0}^{\\lfloor (x-4)/8\\rfloor}(-1)^{c(8k+4)}+\\sum_{k=0}^{\\lfloor (x-5)/8\\rfloor}(-1)^{c(8k+5)}-\\sum_{k=0}^{\\lfloor (x-6)/8\\rfloor}(-1)^{c(8k+5)}-\\sum_{k=0}^{\\lfloor (x-7)/8\\rfloor}(-1)^{c(8k+7)}$$\n\nProof : It follows from Lemma $$1,2,3,4,5$$ that $$z(x)=\\sum_{k=1}^{x}(-1)^{r(k)}=\\sum_{k=0}^{\\lfloor x/8\\rfloor}(-1)^{r(8k)}+\\sum_{k=0}^{\\lfloor (x-1)/8\\rfloor}(-1)^{r(8k+1)}+\\sum_{k=0}^{\\lfloor (x-2)/8\\rfloor}(-1)^{r(8k+2)}+\\sum_{k=0}^{\\lfloor (x-3)/8\\rfloor}(-1)^{r(8k+3)}+\\sum_{k=0}^{\\lfloor (x-4)/8\\rfloor}(-1)^{r(8k+4)}+\\sum_{k=0}^{\\lfloor (x-5)/8\\rfloor}(-1)^{r(8k+5)}+\\sum_{k=0}^{\\lfloor (x-6)/8\\rfloor}(-1)^{r(8k+6)}+\\sum_{k=0}^{\\lfloor (x-7)/8\\rfloor}(-1)^{r(8k+7)}$$\n\n$$=\\sum_{k=0}^{\\lfloor x/8\\rfloor}(-1)^{c(8k)}-\\sum_{k=0}^{\\lfloor (x-1)/8\\rfloor}(-1)^{c(8k+1)}+\\sum_{k=0}^{\\lfloor (x-2)/8\\rfloor}(-1)^{c(8k+2)}+\\sum_{k=0}^{\\lfloor (x-3)/8\\rfloor}(-1)^{c(8k+3)}-\\sum_{k=0}^{\\lfloor (x-4)/8\\rfloor}(-1)^{c(8k+4)}+\\sum_{k=0}^{\\lfloor (x-5)/8\\rfloor}(-1)^{c(8k+5)}-\\sum_{k=0}^{\\lfloor (x-6)/8\\rfloor}(-1)^{c(8k+6)}-\\sum_{k=0}^{\\lfloor (x-7)/8\\rfloor}(-1)^{c(8k+7)}$$\n\n$$=\\sum_{k=0}^{\\lfloor x/8\\rfloor}(-1)^{c(8k)}-\\sum_{k=0}^{\\lfloor (x-1)/8\\rfloor}(-1)^{c(8k+1)}+\\sum_{k=0}^{\\lfloor (x-2)/8\\rfloor}(-1)^{c(8k+2)}+\\sum_{k=0}^{\\lfloor (x-3)/8\\rfloor}(-1)^{c(8k+2)}-\\sum_{k=0}^{\\lfloor (x-4)/8\\rfloor}(-1)^{c(8k+4)}+\\sum_{k=0}^{\\lfloor (x-5)/8\\rfloor}(-1)^{c(8k+5)}-\\sum_{k=0}^{\\lfloor (x-6)/8\\rfloor}(-1)^{c(8k+5)}-\\sum_{k=0}^{\\lfloor (x-7)/8\\rfloor}(-1)^{c(8k+7)}$$\n\nLemma 7 :\n\n\\small\\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\\\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\\\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\\\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\\\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\\\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\\\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\\\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\\end{align}\n\nwhere $$S(m):=\\sum_{k=0}^{m-1}\\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\\bigg)$$\n\nProof : This immediately follows from Lemma 6.\n\nClaim: For every positive integer $$b$$ we have $$r(b)\\equiv r(b-1)\\pmod{2}$$ if and only if either\n\n1. $$b\\equiv3\\pmod{4}$$, or\n2. $$b=k^2$$ for some integer $$k$$, or\n3. $$b=2k^2$$ for some integer $$k$$.\n\nThe main ingredient in proving this claim is the following lemma:\n\nLemma: For every positive integer $$b$$ we have $$r(b)-r(b-1)=c(b)-\\sigma(b),$$ where $$\\sigma(b)$$ denotes the sum of all positive divisors of $$b$$, and $$c(b):=\\begin{cases} 2b-1&\\text{ if }\\ b\\equiv0\\pmod{2},\\\\ \\tfrac{3b-1}{2}&\\text{ if }\\ b\\equiv1\\pmod{2}\\ \\text{ and }\\ b\\neq3,\\\\ 4&\\text{ if }\\ b=3. \\end{cases}$$\n\nThis reduces the question to a question on the parity of $$\\sigma(b)$$.\n\nProof. For every pair of positive integers $$b$$ and $$k$$ there exist unique nonnegative integers $$q(b,k)$$ and $$r(b,k)$$ such that $$r(b,k) and $$b=q(b,k)\\cdot k+r(b,k).$$ This is simply dividing $$b$$ by $$k$$ with remainder $$r(b,k)$$. With this, your function $$r$$ can be expressed as $$r(b)=\\sum_{k=1}^{\\lfloor\\frac{b-1}{2}\\rfloor}r(b,k).$$ To find a more manageable form for $$r(b)$$, note that $$q(b,k)=\\lfloor\\frac bk\\rfloor$$, so that $$\\begin{eqnarray*} r(b)&=&\\sum_{k=1}^{\\lfloor\\frac{b-1}{2}\\rfloor}r(b,k) =\\sum_{k=1}^{\\lfloor\\frac{b-1}{2}\\rfloor}\\Big(b-q(b,k)k\\Big)=\\big\\lfloor\\tfrac{b-1}{2}\\big\\rfloor b -\\sum_{k=1}^{\\lfloor\\frac{b-1}{2}\\rfloor}\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor k. \\end{eqnarray*}$$ Then the difference of two consecutive terms can be simplified. If $$b$$ is even: $$\\begin{eqnarray*} r(b)-r(b-1)&=&\\left(\\big\\lfloor\\tfrac{b-1}{2}\\big\\rfloor b -\\sum_{k=1}^{\\lfloor\\frac{b-1}{2}\\rfloor}\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor k\\right) -\\left(\\big\\lfloor\\tfrac{b-2}{2}\\big\\rfloor(b-1) -\\sum_{k=1}^{\\lfloor\\frac{b-2}{2}\\rfloor}\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor k\\right)\\\\ &=&\\left(\\frac{b-2}{2}b -\\sum_{k=1}^{\\frac{b-2}{2}}\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor k\\right) -\\left(\\left(\\frac{b-2}{2}\\right)(b-1) -\\sum_{k=1}^{\\frac{b-2}{2}}\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor k\\right)\\\\ &=&\\frac{b-2}{2}-\\sum_{k=1}^{\\frac{b-2}{2}}\\left(\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor-\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor\\right)k,\\\\ \\end{eqnarray*}$$ and similarly if $$b$$ is odd: $$\\begin{eqnarray*} r(b)-r(b-1)&=&\\left(\\big\\lfloor\\tfrac{b-1}{2}\\big\\rfloor b -\\sum_{k=1}^{\\lfloor\\frac{b-1}{2}\\rfloor}\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor k\\right) -\\left(\\big\\lfloor\\tfrac{b-2}{2}\\big\\rfloor(b-1) -\\sum_{k=1}^{\\lfloor\\frac{b-2}{2}\\rfloor}\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor k\\right)\\\\ &=&\\left(\\frac{b-1}{2}b -\\sum_{k=1}^{\\frac{b-1}{2}}\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor k\\right) -\\left(\\frac{b-3}{2}(b-1) -\\sum_{k=1}^{\\frac{b-3}{2}}\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor k\\right)\\\\ &=&\\frac{3}{2}(b-1)-\\big\\lfloor\\tfrac{2b}{b-1}\\big\\rfloor\\frac{b-1}{2}-\\sum_{k=1}^{\\frac{b-3}{2}}\\left(\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor-\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor\\right)k, \\end{eqnarray*}$$ where the extra term $$-\\big\\lfloor\\tfrac{2b}{b-1}\\big\\rfloor\\frac{b-1}{2}$$ appears because the summation in $$r(b)$$ has one more term than the summation in $$r(b-1)$$, with $$k=\\tfrac{b-1}{2}$$. For odd $$b>3$$ this further simplifies to $$r(b)-r(b-1)=\\frac{b-1}{2}-\\sum_{k=1}^{\\frac{b-3}{2}}\\left(\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor-\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor\\right)k.$$ Now these inner summations have more familiar closed forms; to see this, note that $$\\big\\lfloor\\tfrac{b}{k}\\big\\rfloor-\\big\\lfloor\\tfrac{b-1}{k}\\big\\rfloor=\\begin{cases} 0&\\text{ if }\\ k\\nmid b\\\\ 1&\\text{ if }\\ k\\mid b\\\\ \\end{cases}.$$ So effectively these summations sum precisely the divisors of $$b$$, up to $$\\lfloor\\tfrac{b-2}{2}\\rfloor$$. A routine check shows that the only divisors not counted are $$b$$, and if $$b$$ is even also $$\\tfrac{b}{2}$$, and if $$b=3$$ also $$1$$. Then we can further simplify the differences for even $$b$$ as $$\\begin{eqnarray*} r(b)-r(b-1)&=&\\frac{b}{2}-1-\\Big(\\sigma(b)-b-\\tfrac{b}{2}\\Big)\\\\ &=&2b-1-\\sigma(b), \\end{eqnarray*}$$ and for odd $$b>3$$ as $$\\begin{eqnarray*} r(b)-r(b-1)&=&\\frac{b-1}{2}-\\Big(\\sigma(b)-b\\Big)\\\\ &=&\\frac{3b-1}{2}-\\sigma(b), \\end{eqnarray*}$$ and the latter is easily verified to also hold for $$b=3$$.$$\\quad\\square$$\n\nIt is a well known fact (or nice exercise) to prove that for a positive integer $$m$$ with prime factorization $$m=\\prod_{i=1}^np_i^{a_i}$$, where $$p_1,\\ldots,p_n$$ are distinct prime numbers and $$a_1,\\ldots,a_n$$ are positive integers, we have $$\\sigma(m)=\\prod_{i=1}^n\\sum_{j=0}^{a_i}p_i^j.$$ In particular this shows that $$\\sigma(m)$$ is odd if and only if for every odd prime $$p_i$$ dividing $$m$$ we have $$a_i\\equiv0\\pmod{2}$$, or equivalently either $$m=k^2$$ or $$m=2k^2$$ for some integer $$k$$. In particular we see that if $$b\\equiv3\\pmod{4}$$ then $$\\sigma(b)$$ is even and hence $$r(b)\\equiv r(b-1)\\pmod{2}$$.\n\nHere are some small values for $$r(b)$$: $$\\begin{array}{r|ccccccccc} b&0&1&2&3&4&5&6&7\\\\ \\hline r(b)&0&0&0&0&0&1&0&2\\\\ &&&&&&&&\\\\ b&8&9&10&11&12&13&14&15\\\\ \\hline r(b)&2&2&3&7&2&7&10&8\\\\ &&&&&&&&\\\\ b&16&17&18&19&20&21&22&23\\\\ \\hline r(b)&8&15&11&19&16&15&22&32\\\\ \\end{array}$$\n\n\u2022 It seems that $r(b)\\equiv r(b+1)\\pmod{2}\\iff\\sigma(b+1)\\equiv1\\pmod{2}$ is false. Take $b=6$. \u2013\u00a0mathlove Mar 23 at 15:24\n\u2022 @mathlove I had spotted the same inconsistency. I have removed the error in my argument. I will try to complete this answer soon (likely tomorrow). \u2013\u00a0Servaes Mar 23 at 15:53", "date": "2020-07-09 02:51:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3587292/interesting-property-related-to-the-sums-of-the-remainders-of-integers", "openwebmath_score": 0.9025372266769409, "openwebmath_perplexity": 582.2596203755722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936101542134, "lm_q2_score": 0.8670357735451835, "lm_q1q2_score": 0.8532443645209306}}
{"url": "https://math.stackexchange.com/questions/3149702/if-d-divides-2n-and-d-doesnt-divide-n-then-d-is-even", "text": "# If $d$ divides $2n$ and $d$ doesn't divide $n$, then $d$ is even\n\nI have encountered a proof regarding dihedral groups we this fact is used:\n\nIf $$d\\mid 2n$$ and $$d\\nmid n$$, then $$d$$ is even and $${d\\over 2}\\mid n$$.\n\nI can't seem to understand why this is true. If $$d\\nmid n$$, then there are $$q,r \\in \\mathbb{Z}$$ so that $$0 < r < d$$ and $$n = qd + r$$. On the other hand, $$d\\mid 2n$$ means that there is $$m \\in \\mathbb{Z}$$ so that $$2n = md$$. We need to somehow use these two facts.\n\nAlso, my second question is how we can naturally generalize this result?\n\n\u2022 Hint: if $d$ was odd, $m$ would be even. \u2013\u00a0Wojowu Mar 15 at 19:28\n\u2022 $2n =md$. And $2$ is prime. So $2|m$ or $2|d$. If $2|m$ then $n = \\frac m2 d$ and $d|n$. If $2|d$ then $d$ is even. General. If $d|pn$ for a prime $p$ and $d\\not \\mid n$ then $p|d$. Even more general. If $d|ab$ then $\\frac d{\\gcd(d,b)}|a$. \u2013\u00a0fleablood Mar 15 at 21:22\n\nSuppose $$d$$ is odd. Then $$d$$ and $$2$$ are relatively prime and $$d\\mid 2n$$, so by Euclid lemma we have $$d\\mid n$$.\n\nA contradiction. So $$d$$ must be even.\n\nWe could have more general situation.\n\nSay $$d\\mid pn$$ for some prime $$p$$ and $$d$$ doesn't divide $$n$$. Then $$p\\mid d$$.\n\nThe proof goes exactly the same as for $$p=2$$.\n\nIt's also possible to see what's going on simply by keeping track of factors of $$2$$; note that this kind of analysis could also be generalized to other prime factors, as has been illustrated in previously posted answers.\n\nLet $$n:= 2^aQ,\\ 2n=2^{a+1}Q, d:=2^bP$$ where $$Q$$ represents the product of all of the odd prime factors of $$n$$ and $$P$$ represents the product of all of the odd prime factors of $$d$$. We could write out all of those factors and explicitly show this, but it should be plain that $$d\\mid 2n \\Rightarrow P\\mid Q$$. Note that thus far, the exponents $$a,b$$ might be $$0$$, so we have not assumed that $$d$$ is even.\n\n$$d\\mid 2n \\Rightarrow b\\le a+1$$\n\n$$d\\not \\mid n \\Rightarrow b>a$$\n\nTogether, these establish $$b=a+1$$, meaning that $$d$$ has at least one factor of $$2$$ and is even, even if $$a=0$$.\n\nThis also illustrates the second point: the exponent of $$2$$ in $$d\\over 2$$ is simply $$b-1=a$$. Hence $$\\frac{d}{2} \\mid n$$\n\nIf you know that the remainder $$r$$ in $$n=qd+r$$ with $$0\\lt r\\lt d$$ is unique, then from $$2n=md$$ we get\n\n$$n=2n-n=md-(qd+r)=(m-q)d-r=(m-q+1)d+(d-r)=q'd+r'$$\n\nwith $$0\\lt r'=d-r\\lt d$$, so that, by uniqueness of the remainder, we have $$r'=r$$, i.e. $$d-r=r$$, hence $$d=2r$$.\n\n$$d| 2n$$ then $$2n=kd;$$\n\nSince $$kd$$ is even, $$2| kd$$.\n\nEuclid's lemma:\n\n1) $$2| k$$ or 2) $$2|d$$.\n\n1) If $$2|k$$ then $$k=2k'.$$\n\n$$2n=2k'd$$;\n\n$$n=k'd$$, i.e. $$d|n$$ , a contradiction.\n\n2) Hence $$2|d$$ , and we are done.\n\nBy below $$\\ d\\mid pn\\iff\\!\\!\\!\\! \\overbrace{d\\mid n}^{\\large\\color{#0a0}{(d,p)}\\ =\\ \\color{#c00}1}\\!\\!$$ or $$\\ \\overbrace{{d/\\color{#c00}p}\\mid n}^{\\large\\color{#0a0}{ (d,p)}\\ =\\ \\color{#c00} p}\\!,\\$$ by $$\\ \\color{#0a0}{(d,p)}\\mid \\color{#c00}p\\,$$ prime.\n\nLemma $$\\,\\ d\\mid an\\iff\\smash[t]{\\overbrace{ d/\\color{#0a0}{(d,a)}\\,\\mid\\, n,\\,}\\ }$$ where $$\\,\\ (x,y) := \\gcd(x,y)$$\n\nProof $$\\quad\\ d\\mid an\\iff d\\mid dn,an,\\iff d\\mid (dn,an)=(d,a)n\\iff d/(d,a)\\mid n$$\n\n\u2022 Convention: $\\ d/p\\mid n\\$ means $\\ d/p\\,$ is an integer, so $\\,p\\mid d\\ \\$ \u2013\u00a0Bill Dubuque Mar 15 at 20:23\n\nIntuitively. If $$d|ab$$ and $$d\\not \\mid b$$ then \"some part of $$d$$ must divide $$a$$\". So if $$d|2n$$ but $$d\\not \\mid n$$ then some (non-trivial) part must divide $$2$$ and that part must be $$2$$ so $$d$$ is even.\n\n.....\n\nThat's intuition. Let's make a proof.\n\n.....\n\nIf $$d|ab$$. Let $$\\gcd(d,b) = g$$ and let $$d = d'g$$ and $$b = b'g$$. It's easy to see that $$d'$$ and $$b'$$ are relatively prime: if $$b'$$ and $$d'$$ had any non-trivial factor, $$k$$, in common then $$kg$$ would be a common divisors of $$b$$ and $$d$$ contradicting that $$g$$ is the greatest common divisor.\n\nSo $$d=d'g$$ and $$ab = ab'g$$ and $$d'g|ab'g$$ so $$d'|ab'$$. But $$d'$$ and $$b'$$ are relatively prime so they have no factors in common. So $$d'|a$$.\n\nNow if $$d|b$$ then $$d'g|b'g$$ so $$d'|b'$$ but $$d'$$ and $$b'$$ are relatively prime so $$d' = 1$$ and $$d = \\gcd(d,b)$$.\n\nLemma 1: $$d|b \\iff \\gcd(d,b) = d$$.\n\nIf $$d|ab$$ the $$d'|a$$ but if $$d\\not \\mid b$$ then $$\\gcd(d,b)=g \\ne d$$ so $$d' = \\frac dg > 1$$. $$d'\\ne 1$$ and $$d'|a$$.\n\nSo\n\nLemma 2: If $$d|ab$$ but $$d\\not \\mid b$$ then $$d' = \\frac d{\\gcd(d,b)} > 1$$ and $$d'|a$$.\n\nSo if $$d|2n$$ and $$d\\not \\mid n$$ then $$\\frac d{\\gcd(d,n)} > 1$$ and $$\\frac d{\\gcd(d,n)} |2$$.\n\nSo $$\\frac d{\\gcd(d,n)} = 2$$ and $$d = \\frac d{\\gcd(d,n)}\\gcd(d,n) = 2\\gcd(d,n)$$. And $$d$$ is even.\n\n=====\n\nActually, this may be the best most general Theorem:\n\nTheorem: If $$d|mn$$ then $$\\frac d{\\gcd(d,n)}|m$$.\n\nI'll leave the proof to you, and I'll leave it to you to figure out how that implies if $$d|2n$$ and $$d\\not\\mid n$$ then $$d$$ is even.", "date": "2019-04-23 12:32:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3149702/if-d-divides-2n-and-d-doesnt-divide-n-then-d-is-even", "openwebmath_score": 0.9671773910522461, "openwebmath_perplexity": 65.54672186661688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984093606422157, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8532443426890389}}
{"url": "https://mathematica.stackexchange.com/questions/263451/generate-covariance-matrix-from-related-random-variable-distributions", "text": "# Generate covariance matrix from related random variable distributions\n\nSay I have a random variable $$X\\sim\\mathscr{N}\\left(0,\\sigma^2\\right)$$ and another random variable $$Y=X+\\varepsilon$$, where $$\\varepsilon\\sim\\mathscr{N}\\left(0,\\eta^2\\right)$$ is independent of $$X$$. I can calculate $$Cov\\left(X,Y\\right)=Cov\\left(X,X+\\varepsilon\\right)=Var(X)=\\sigma^2$$.\n\nIs there a way to get Mathematica to generate a bivariate normal distribution automatically from the given information? Specifically, rather than working out the covariance matrix myself and then inputting to Mathematica, I would like to be able to do something like (implementing the above example):\n\n\\[ScriptCapitalX] = NormalDistribution[0, \\[Sigma]]\n\\[CapitalEpsilon] = NormalDistribution[0, \\[Eta]]\n\\[ScriptCapitalY] = TransformedDistribution[X + \\[CurlyEpsilon], {X \\[Distributed] \\[ScriptCapitalX], \\[CurlyEpsilon] \\[Distributed] \\[CapitalEpsilon]}]\n\n\\[Mu] = {Mean[\\[ScriptCapitalX]], Mean[\\[ScriptCapitalY]]}\n\\[CapitalSigma] = {{Variance[\\[ScriptCapitalX]], Covariance[\\[ScriptCapitalX], \\[ScriptCapitalY]]}, {Covariance[\\[ScriptCapitalX], \\[ScriptCapitalY]], Variance[\\[ScriptCapitalY]]}}\n\nmulti = MultinormalDistribution[\\[Mu], \\[CapitalSigma]]\nwant = Expectation[X \\[Conditioned] Y = y, {Y, X} \\[Distributed] multi]\n\n\nObviously, the Covariance function doesn't work that way, but at a minimum I would really like to have Mathematica do the covariance calculation. Also, if there's an easy way to get the mean vector and covariance matrix without explicitly calling Mean, Variance, etc. on individual elements to construct them, that would be even better. I am quite new to the Wolfram language.\n\n\u2022 Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. Feb 11 at 10:34\n\nYou were almost there:\n\n\\[ScriptCapitalX] = NormalDistribution[0, \u03c3]\n\\[CapitalEpsilon] = NormalDistribution[0, \u03b7]\nmulti = TransformedDistribution[{X, X + \u03b5},\n{X \\[Distributed] \\[ScriptCapitalX], \u03b5 \\[Distributed] \\[CapitalEpsilon]}]\n\nCovariance[multi][[1, 2]]\n(* \u03c3^2 *)\n\nExpectation[X \\[Conditioned] Y == y, {X, Y} \\[Distributed] multi]\n(* (y \u03c3^2)/(\u03b7^2 + \u03c3^2) *)", "date": "2022-08-15 05:10:51", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/263451/generate-covariance-matrix-from-related-random-variable-distributions", "openwebmath_score": 0.4621465802192688, "openwebmath_perplexity": 1891.9690943188014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936054891428, "lm_q2_score": 0.8670357546485408, "lm_q1q2_score": 0.8532443418800824}}
{"url": "https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Path_Component", "text": "# Equivalence of Definitions of Path Component\n\nJump to navigation Jump to search\n\n## Theorem\n\nThe following definitions of the concept of Path Component in the context of Topology are equivalent:\n\nLet $T = \\struct {S, \\tau}$ be a topological space.\n\nLet $x \\in T$.\n\n### Equivalence Class\n\nLet $\\sim$ be the equivalence relation on $T$ defined as:\n\n$x \\sim y \\iff x$ and $y$ are path-connected.\n\nThe equivalence classes of $\\sim$ are called the path components of $T$.\n\nIf $x \\in T$, then the path component of $T$ containing $x$ (that is, the set of points $y \\in T$ with $x \\sim y$) can be denoted by $\\map {\\operatorname{PC}_x} T$.\n\n### Union of Path-Connected Sets\n\nThe path component of $T$ containing $x$ is defined as:\n\n$\\displaystyle \\map {\\operatorname{PC}_x} T = \\bigcup \\left\\{{A \\subseteq S: x \\in A \\land A}\\right.$ is path-connected $\\left.\\right\\}$\n\n### Maximal Path-Connected Set\n\nThe path component of $T$ containing $x$ is defined as:\n\nthe maximal path-connected set of $T$ that contains $x$.\n\n## Proof\n\nLet $\\CC_x = \\set {A \\subseteq S : x \\in A \\land A \\text { is path-connected in } T}$\n\nLet $C = \\bigcup \\CC_x$.\n\n### Lemma\n\n$C$ is path-connected in $T$ and $C \\in \\CC_x$.\n\n$\\Box$\n\nLet $C'$ be the equivalence class containing $x$ of the equivalence relation $\\sim$ defined by:\n\n$y \\sim z$ if and only if $y$ and $z$ are connected in $T$.\n\n### Equivalence Class equals Union of Path-Connected Sets\n\nIt needs to be shown that $C = C'$.\n\n $\\displaystyle y \\in C'$ $\\leadstoandfrom$ $\\displaystyle x \\text{ is path-connected to } y \\text{ in } T$ Definition of $\\sim$ $\\displaystyle$ $\\leadstoandfrom$ $\\displaystyle \\exists B \\text{ a connected set of } T, x \\in B, y \\in B$ Points are Path-Connected iff Contained in Path-Connected Set $\\displaystyle$ $\\leadstoandfrom$ $\\displaystyle \\exists B \\in \\CC_x : y \\in B$ Equivalent definition $\\displaystyle$ $\\leadstoandfrom$ $\\displaystyle y \\in \\bigcup \\CC_x$ Definition of Set Union $\\displaystyle$ $\\leadstoandfrom$ $\\displaystyle y \\in C$ Definition of $C$\n\nThe result follows.\n\n$\\Box$\n\n### Union of Path-Connected Sets is Maximal Path-Connected Set\n\nLet $\\tilde C$ be any path-connected set such that:\n\n$C \\subseteq \\tilde C$\n\nThen $x \\in \\tilde C$.\n\nHence $\\tilde C \\in \\CC_x$.\n\n$\\tilde C \\subseteq C$.\n\nHence $\\tilde C = C$.\n\nIt follows that $C$ is a maximal path-connected set of $T$ by definition.\n\n$\\Box$\n\n### Maximal Path-Connected Set is Union of Path-Connected Sets\n\nLet $\\tilde C$ be a maximal path-connected set of $T$ that contains $x$.\n\nBy definition:\n\n$\\tilde C \\in \\CC_x$\n$\\tilde C \\subseteq C$\n\nBy maximality of $\\tilde C$:\n\n$\\tilde C = C$\n\n$\\blacksquare$", "date": "2020-09-28 16:15:33", "meta": {"domain": "proofwiki.org", "url": "https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Path_Component", "openwebmath_score": 0.9374152421951294, "openwebmath_perplexity": 327.52972032993677, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936071219176, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8532443399146533}}
{"url": "https://www.khanacademy.org/math/calculus-all-old/limits-and-continuity-calc/limits-of-piecewise-functions-calc/v/limit-of-piecewise-function-that-is-defined", "text": "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## Calculus, all content (2017 edition)\n\n### Unit 1: Lesson 16\n\nLimits of piecewise functions\n\n# Worked example: point where a function is continuous\n\nAP.CALC:\nLIM\u20112 (EU)\n,\nLIM\u20112.A (LO)\n,\nLIM\u20112.A.2 (EK)\nSal finds the limit of a piecewise function at the point between two different cases of the function. In this case, the two one-sided limits are equal, so the limit exists.\n\n## Want to join the conversation?\n\n\u2022 at he substitutes x with 3 but the function is defined between 0 < x < 3. In other words it's defined up to three but not three. So isn't that technically wrong?\n\u2022 in that particular function you mentioned,, it is continuos until x approaches 3.\nWe needed to find the limit of f(x) of that function as x approached 3. That is only the idea of limit.. It is not defined at x =3, but we can find a value of f(x) so that it would have formed a continuos graph.. So we find the value of f(x) that would have been for x=3.\n\nHope that was not complicated\n\u2022 At how we can understand that function is continuous?\n\u2022 It is not undefined for any positive argument. This means that there are no asymptotes or removable discontinuities, but proving continuity can be done in a variety of ways (for instance, noting that it is differentiable or noting that its inverse is differentiable etc.). Since differentiability is a stronger condition than continuity, all differentiable functions are also continuous over the differentiable interval.\n\u2022 Can someone please suggest me a video in which all log functions are explained? Because I am only aware of the basic stuff but I guess as we proceed, we need to know the complicated functions of log too!\n\u2022 What is a piecewise function?\n\u2022 A piecewise function has different rules in different intervals. For example, look up aat this function:\n\nf(x) = x^2 if x if x<4\n= 4 if x<4 or x=4\n\nBetween the interval wich goes from negative infinity, it is x^2; and between the interval wich goes from 4 to positive infinity it is always four.\n\nTo give a counterexample, g(x)=x^2+1 is not a piecewise function, because it is always equal to x^2+1; without mattering the value of x\n\u2022 At and , what if the function is non continuous? What numbers do we plug in for x?\n\u2022 If there is a jump discontinuity, then the limit from the left side and the limit from the right side will not be equal so the overall limit does not exist. You still have to plug in the same x value in both equations and you will get different values so the overall limit does not exist. But if there is a removable discontinuity, both the limit from the left side and the limit from the right side will be equal so the overall limit exists. In any case, you have to plug in the same x value.\n\u2022 I forgot what a log is :/\n\u2022 A logarithm is essentially the opposite of the exponential function. What this means is that if a^x = b, the log(base a) b = x.\n\u2022 Why didn't we find the function value when x =3 to check if that is equal to the limit to satisfy the condition of continuity\n\u2022 for a limit to be continuous, lim(x tends to c) f(x) = f(c).\n\nIN this case we know that lim(x tends to 3) g(x) = log(9) but we don't know if\ng(3)=log(9).\n\nSo how can we say that the limit is continuous\n\u2022 We do know that g(3)=log(9), because the function g is defined at x=3 and we can plug 3 into the function.\n\ng(3) = (4-3)*log(9) = 1*log(9) = log(9)\n\u2022 I'm a bit confused by the title; this only proves that the limit of g(x) as x -> 3 exists, not that g(x) is necessarily continuous at that point right?\n\nEDIT: nvm guys we used direct sub, so it is indeed continuous\n\u2022 for a limit to be continuous, lim(x tends to c) f(x) = f(c). so we first see if the limit exists and then we see if it's equal to the function of that point, at this example we are not examining if the function is continuous it is continuous, he is showing us the case.\n\u2022 How would I explain why g(x) is continuous?\n\n## Video transcript\n\n- [Voiceover] So we have g of x being defined as the log of 3x when zero is less than x is less than three and four minus x times the log of nine when x is greater than or equal to three. So based on this definition of g of x, we want to find the limit of g of x as x approaches three, and once again, this three is right at the interface between these two clauses or these two cases. We go to this first case when x is between zero and three, when it's greater than zero and less than three, and then at three, we hit this case. So in order to find the limit, we want to find the limit from the left hand side which will have us dealing with this situation 'cause if we're less than three we're in this clause, and we also want to find a limit from the right hand side which would put us in this clause right over here, and then if both of those limits exist and if they are the same, then that is going to be the limit of this, so let's do that. So let me first go from the left hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left of g of x, well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of 3x, and since this function right over here is defined and continuous over the interval we care about, it's defined continuous for all x's greater than zero, we can just substitute three in here to see what it would be approaching. So this would be equal to log of three times three, or logarithm of nine, and once again when people just write log here within writing the base, it's implied that it is 10 right over here. So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right hand side, from values greater than three. Well, we are now going to be in this scenario right over there, so this is going to be equal to the limit as x approaches three from the positive direction, from the right hand side of, well g of x is in this clause when we are greater than three, so four minus x times log of nine, and this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant, log base 10 of nine is gonna be some number close to one. This expression would actually define a line. For x greater than or equal to three, g of x is just a line even though it looks a little bit complicated. And so this is actually defined for all real numbers, and it's also continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well we can just evaluate a three. So it's going to be four minus three times log of nine, well that's just one, so that's equal to log base 10 of nine. So the limit from the left equals the limit from the right. They're both log nine, so the answer here is log log of nine, and we are done.", "date": "2023-01-27 02:24:55", "meta": {"domain": "khanacademy.org", "url": "https://www.khanacademy.org/math/calculus-all-old/limits-and-continuity-calc/limits-of-piecewise-functions-calc/v/limit-of-piecewise-function-that-is-defined", "openwebmath_score": 0.842057466506958, "openwebmath_perplexity": 262.86013819457855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936087546923, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.8532443379492243}}
{"url": "http://mathhelpforum.com/statistics/18794-total-number-combinations.html", "text": "# Math Help - total number of combinations\n\n1. ## total number of combinations\n\nHey. I have 3 problems I can't get. If anyone can help with how to do any of them, it would be appreciated.\n\n1) To gain access to his account, a customer using an ATM must enter a four-digit code. If repetition of the same four digits is not allowed (for example, 5555) how many possible combinations are there?\n\n2) Over the years, the state of California has used different combinations of letter of the alphabet and digits on its automobile license plates.\na. At one time license plates were issued that consisted of three letters followed by three digits. How many different license plates can be issued under this arrangement?\nb. Later on, license plates were issued that consisted of three digits followed by three letters. How many different license plates can be issued under this arrangement?\n\n3) An exam consists of ten true-or-false questions. In how many ways may the exam be completed if a penalty is imposed for each incorrect answer, so that a student may leave questions unanswered?\n\n2. Originally Posted by xojuicy00xo\nHey. I have 3 problems I can't get. If anyone can help with how to do any of them, it would be appreciated.\n\n1) To gain access to his account, a customer using an ATM must enter a four-digit code. If repetition of the same four digits is not allowed (for example, 5555) how many possible combinations are there?\nhow many combinations could we have if this restriction was not imposed? find that and then subtract the number of four repeated digits there can be (there are only 10: 0,0,0,0; 1,1,1,1; 2,2,2,2; 3,3,3,3; etc)\n\n2) Over the years, the state of California has used different combinations of letter of the alphabet and digits on its automobile license plates.\na. At one time license plates were issued that consisted of three letters followed by three digits. How many different license plates can be issued under this arrangement?\ni assume repetitions are allowed, since you didn't say they weren't.\n\nwe choose 3 letters and 3 digits. there are 26 letters and 10 digits (including 0).\n\nso for the first letter we have 26 choices.\nfor each of those choices, we have 26 choices for the 2nd letter\nfor each of those choices, we have 26 choices for the 3rd letter\nfor each of those choices, we have 10 choices for the 1st digit\nfor each of those choices, we have 10 choices for the 2nd digit\nfor each of those choices, we have 10 choices for the 3rd digit\n\nso in all, we have 26*26*26*10*10*10 choices\n\nb. Later on, license plates were issued that consisted of three digits followed by three letters. How many different license plates can be issued under this arrangement?\ndo this in the way i did the above\n\n3) An exam consists of ten true-or-false questions. In how many ways may the exam be completed if a penalty is imposed for each incorrect answer, so that a student may leave questions unanswered?\nmaybe it's just me, or there is something missing from the question. is there a minimum grade we are hoping the student will get? how many penalties is the student allowed? ...\n\n3. Hello, xojuicy00xo!\n\n3) An exam consists of ten true-or-false questions.\nIn how many ways may the exam be completed\nif a penalty is imposed for each incorrect answer,\nso that a student may leave questions unanswered?\n\nFor each of the ten questions, the student has three choices:\n\nThere are: . $3^{10} \\:=\\:59,049$ ways.\n\n4. Originally Posted by Soroban\nHello, xojuicy00xo!\n\nFor each of the ten questions, the student has three choices:\nThere are: . $3^{10} \\:=\\:59,049$ ways.", "date": "2015-03-06 05:16:26", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/18794-total-number-combinations.html", "openwebmath_score": 0.6866676807403564, "openwebmath_perplexity": 516.8988549096042, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9861513905984457, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8532419407805195}}
{"url": "https://codereview.stackexchange.com/questions/129627/calculating-the-amount-of-cubes-needed-to-form-a-sum", "text": "# Calculating the amount of cubes needed to form a sum\n\nSince I've never done any performance programming (Aside from the better choices such as array vs list etc. The real basics.), I should probably read up on it. But I had to start somewhere, so, someone I know - which is a much better programmer than I am - tasked me with making this \"assignment\":\n\nYou are given the total volume m of the building. Being given m can you find the number n of cubes you will have to build?\n\nThe parameter of the function findNb (find_nb, find-nb) will be an integer m and you have to return the integer n such as $n^3 + (n-1)^3 + \\dots + 1^3 = m$ if such a n exists or -1 if there is no such n.\n\nHe gave me this template with it:\n\nusing System;\n\npublic class Program\n{\npublic static void Main()\n{\nConsole.WriteLine(findNb(4183059834009));\nConsole.WriteLine(findNb(24723578342962));\nConsole.WriteLine(findNb(135440716410000));\nConsole.WriteLine(findNb(40539911473216));\n}\n\npublic static long findNb(long m)\n{\n\n}\n}\n\n\nIn which I inserted:\n\nfor(long n = 1; n < (m / 3); n++)\n{\ndouble vol = 0;\nfor(long i = 0; i < n; i++)\nvol += Math.Pow((n - i), 3);\n\nif(vol > m)\nreturn -1;\n\nif(vol == m)\nreturn n;\n}\nreturn -1;\n\n\nThis code works, but it takes incredibly long for the larger numbers, as is my main problem.\n\nWhat can I do to shorten the time it takes for this code takes to complete?\n\n\u2022 Please try to update your title to reflect what it is your code does, not what improvements you want to make. \u2013\u00a0forsvarir May 29 '16 at 18:36\n\u2022 Is this any better, @forsvarir? \u2013\u00a0sxbrentxs May 29 '16 at 18:47\n\u2022 Yes, the title change and addition of the function spec make your question much more engaging. \u2013\u00a0forsvarir May 29 '16 at 19:03\n\nFirst of all, you have a bug:\nvariable vol is declared as double that causes the condition vol == m hardly to be satisfied.\n\nNext, you don't need the Math.Pow method, you could relace it with just n * n * n.\nAnd your loops look a bit confusing. Why m / 3? Why do you need the inner loop?\n\nThe proposed code:\n\nprivate static long findNb(long m)\n{\nlong n = 1;\nlong vol = 0;\nwhile (vol < m)\n{\nvol += n * n * n;\n\nif (vol == m)\nreturn n;\n\n++n;\n}\nreturn -1;\n}\n\n\u2022 > Why m / 3? Why do you need the inner loop? Apologies, that was/is a remainder of me trying to get the code to execute fully on DotNETFiddle. And the loop was more to calculate the volume. But That's not needed anymore. Thanks for this. Could you tell me, though why n * n * n is better over Math.Pow? \u2013\u00a0sxbrentxs May 29 '16 at 19:23\n\u2022 @sxbrentxs Math.Pow is intended for floating point math. You need only two integer multiplications, while Math.Pow calculates it as Exp(y * Log(x)) and also you need to cast arguments to double and back. \u2013\u00a0Dmitry May 29 '16 at 19:33\n\u2022 Which makes Math.Pow obsolete and slow in this case. Cool! Thank you. \u2013\u00a0sxbrentxs May 29 '16 at 19:34\n\nWhen there's math to be done, always check if some formula exists\n\n$$1^3+2^3 + \\dots+n^3 = m = (1+2 + \\dots+n)^2$$\n\nin conjunction with some other formula (interesting wikipedia article title)\n\n$$\\sum^n_{k=1}k = \\frac{n(n+1)}{2}$$\n\nyou get some equation for the result\n\n$$\\sqrt{m} = 1+2 + \\dots+n = \\sum^n_{k=1}k = \\frac{n(n+1)}{2}$$ and maybe some closed formula\n\n\\begin{align} \\sqrt{m} &= \\frac{n(n+1)}{2}\\\\ 2\\sqrt{m} &= n(n+1)\\\\ 2\\sqrt{m} &= n^2+n\\\\ 0&= n^2+n - 2\\sqrt{m} \\end{align}\n\nWhich has the 2 solutions\n\n$$n_{1,2} = -\\frac12 \\pm\\sqrt{\\frac14+2\\sqrt{m}}$$\n\nAs $n$ should be a positive number, only the first solution makes sense.\n\nReturn $-\\frac12 +\\sqrt{\\frac14+2\\sqrt{m}}$ if it is an integer or $-1$ otherwise.\n\nThe results are:\n\n$$\\begin{array}{r|r} m & n\\\\ \\hline 4183059834009 & 2022\\\\ 24723578342962 & -1\\\\ 135440716410000 & 4824\\\\ 40539911473216 & 3568\\\\ \\end{array}$$\n\n24723578342962 is a very close call, because\n\n$$\\sum^{3153}_{k=1}k^3=24723578342961$$\n\nis only 1 off. The floating point result is $n=3153.00000000003$. If you don't want to get false positives due to floating point precision (lack thereof), you can do the check by converting the result to some integer type and see if you get the exact number by doing the calculations with that.\n\n\u2022 Holy shit. That's a lot to take in. I think I know what you did and what I have to do. Thank you! \u2013\u00a0sxbrentxs May 29 '16 at 19:27\n\u2022 Wow! Very impressed to see an O(1) solution. +1. \u2013\u00a0ApproachingDarknessFish May 31 '16 at 0:58\n\nFor a significant speedup: In the outer loop, you vary n from 1 to some large value, and in the inner loop you add up n values.\n\nNow in the next iteration of the outer loop, using n+1 instead of n, you run the inner loop again, this time adding up n+1 value. But just before that, in the previous loop, you already added up n values, so to get the sum of n+1 values, all you have to do is add value number n+1 to the sum that you already have.\n\nHave a variable n and a variable sum_n. sum_n will be the sum of the first n cubes. Initially n = 0 and sum_n = 0. Inside the outer loop, increase n by 1 then increase sum_n by nnn.\n\nFor example, if n goes from 1 to 1000, you add up one value, then two, then three, and so on, and eventually you add up 1000 values. On average about 500. This change means you add only one value in each iteration of the outer loop.\n\nThis is an improvement that works for any task where you calculate consecutive sums. In this particular case, you can find a formula to calculate the sum. That isn't helpful by itself, since calculating the formula is likely slower than calculating the sums. However, you can reverse the formula and calculate what n should be, and then you just have to verify this. So you just need a small fixed number of operations, independent of the size of the m that you were given.", "date": "2019-09-19 15:56:52", "meta": {"domain": "stackexchange.com", "url": "https://codereview.stackexchange.com/questions/129627/calculating-the-amount-of-cubes-needed-to-form-a-sum", "openwebmath_score": 0.6605871319770813, "openwebmath_perplexity": 1030.3486532476047, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140216112959, "lm_q2_score": 0.8824278726384089, "lm_q1q2_score": 0.8532318831147044}}
{"url": "https://math.stackexchange.com/questions/2197463/showing-a-cup-b-cup-c-is-countable-if-a-b-are-countable-and-c-is-fi", "text": "# Showing $A \\cup B \\cup C$ is countable if $A, B$ are countable and $C$ is finite.\n\nHow does one go about showing $A \\cup B \\cup C$ is countable if $A, B$ are countable and $C$ is finite?\n\nI understand most of the confusion for resolving set theory questions online seem to be the definition. For my course we consider the following definitions:\n\ncountable: Finite or $A \\sim\\mathbb{N}$\n\nuncountable: not countable\n\nfinite: The empty set or $A \\sim J_n$ where $n \\in \\mathbb{N}$\n\ninfinite: not finite\n\nSo I'm not sure if I have this right but looking at the above definitions the way I'm looking to approach this is to consider 6 different cases.\n\n1. Where C is the empty set and A, B are both finite sets\n2. Where C is the empty set and A is finite and B is countably infinite\n3. Where C is the empty set and both A, B are countably infinite\n\n4 - 6. Repeated above but with C being a non-empty finite set\n\nThis seems like quite a round about way but intuitively it seems to me like the only way to cover all bases according to the definitions. I'm hoping I might be absolutely wrong on this. Is there a simpler way to prove this?\n\n\u2022 You can reduce the time taken by first proving the lemma $A$ and $B$ both countable implies $A\\cup B$ countable. By using the lemma twice you show $A\\cup B\\cup C = (A\\cup B)\\cup C$ is countable by noting that it is the union of two countable sets since what is in the parenthesis is also a countable set. Further cut down on the number of cases by saying \"Without loss of generality suppose $|A|\\leq |B|$\" before working too hard since any situation where $|B|<|A|$ is proven the same way by a relabeling of the sets. Also, treat empty sets during the same case as finite sets. \u2013\u00a0JMoravitz Mar 21 '17 at 23:50\n\u2022 You now need only prove the following three cases: $A$ finite and $B$ finite, $A$ finite and $B$ countably infinite, and $A$ countably infinite and $B$ countably infinite. In any of those cases, try to find a bijection between $A\\cup B$ and a subset of $\\Bbb N$. \u2013\u00a0JMoravitz Mar 21 '17 at 23:52\n\u2022 @JMoravitz Thank you for the detail. This is very helpful \u2013\u00a0Sithe Mar 22 '17 at 0:05\n\nIf $A$ and $B$ are countable, then they can be enumerated, respectively, as:\n\n$a_1, a_2, a_3, \\ldots$ and $b_1, b_2, b_3, \\ldots$\n\nSince $C$ is finite, its elements can be listed as: $c_1, c_2, \\ldots, c_n$.\n\nTo prove the union of all three is countable, it will suffice to list them in some order $d_1, d_2, d_3, \\ldots$, since there is then a natural bijection with $\\mathbb{N}$ in which we match $d_k$ with $k \\in \\mathbb{N}$.\n\nAs to the listing for this union, you could list all of $C$, then interweave enumerations of $A$ and $B$:\n\n$c_1, c_2, \\ldots, c_n, a_1, b_1, a_2, b_2, a_3, b_3, \\ldots$\n\n(You may also specify that no element is listed more than once.)\n\nIf either of $A$ or $B$ is finite (since your definition of countable allows for this) then the listing can be tweaked appropriately, e.g., if just $A$ is finite, then list all elements of $C$, then of $A$, and then enumerate $B$.\n\n\u2022 We must have been typing in synchrony ... Great answer by the way!! \u2013\u00a0Bram28 Mar 21 '17 at 23:56\n\u2022 Thank you. I've never seen countability being proven that way. It's very straight forward and I like it \u2013\u00a0Sithe Mar 22 '17 at 0:00\n\u2022 Slight thing to watch out for if the sets aren't disjoint. But just skip them if they do. That's acceptable \"tweaking\". \u2013\u00a0fleablood Mar 22 '17 at 0:04\n\u2022 This is an injection, not necessarily a bijection, because $A\\cap B$ might not be empty. \u2013\u00a0vadim123 Mar 22 '17 at 0:04\n\u2022 Okay, I missed that. \u2013\u00a0fleablood Mar 22 '17 at 0:09\n\nSets are countable if and only if their elements are 'listable'.\n\nNow, $C$ is finite, so say it contains elements $c_1, c_2, ..., c_n$ for some $n$\n\n$A$ is countable, so we can create a list:\n\n$a_1, a_2, ...$\n\nSame for $B$:\n\n$b_1, b_2, ...$\n\nSo, I would create the following list:\n\n$c_1, c_2, ..., c_n, a_1, b_1, a_2, b_2, ...$\n\n\u2022 This list may have repeated elements. \u2013\u00a0vadim123 Mar 22 '17 at 0:04\n\u2022 @vadim123 sure, but they can be removed if we really wanted to define an injection. So as long as every element appears somewhere on the list, we're fine. \u2013\u00a0Bram28 Mar 22 '17 at 1:06\n\nLet $j_a: A \\rightarrow \\mathbb N$ be an injection. We know that is possible because $A$ is countable.\n\nLet $j_b: B \\rightarrow \\mathbb N$ be an injection. We know that is possible because $B$ is countable.\n\nLet $j_c: B \\rightarrow \\mathbb N_j$ be an bijection. We know that is possible because $C$ is finite.\n\nLet $k:A\\cup B\\cup C \\rightarrow \\mathbb N$ via $k(x) = 3*j_a(x)$ if $x \\in A$. If $x \\in B$ but $x \\not \\in A$ let $k(x) = 3*j_b(x) + 1$. And if $x \\in C$ but $x \\not \\in A$ and $x \\not \\in B$ let $k(x) = 3*j_c(x) + 2$.\n\nShow that $k$ is an injection and that shows $A\\cup B \\cup C$ is countable.\n\nAnother way to think of it is to start by picking out the first element of $A$ then pick the first element of $B$, then of $C$, then pick the second element, then the third elements and so on. That's a list of all the elements. Since they can be listed one after another they are countable.\n\n\u2022 And before vadim points out there'll be many elements of N not mapped into, I'll point out as long as there is an injection into N that is sufficient to prove countabity. It need not be onto. But it can be made onto if it is infinite. But it is not nescessary. \u2013\u00a0fleablood Mar 22 '17 at 0:07", "date": "2019-10-14 20:08:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2197463/showing-a-cup-b-cup-c-is-countable-if-a-b-are-countable-and-c-is-fi", "openwebmath_score": 0.8432019352912903, "openwebmath_perplexity": 254.43257113279793, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140235181257, "lm_q2_score": 0.8824278664544911, "lm_q1q2_score": 0.8532318788180273}}
{"url": "https://datascience.stackexchange.com/questions/27388/what-does-it-mean-when-we-say-most-of-the-points-in-a-hypercube-are-at-the-bound/27399", "text": "# What does it mean when we say most of the points in a hypercube are at the boundary?\n\nIf I have a 50 dimensional hypercube. And I define it's boundary by $0<x_j<0.05$ or $0.95<x_j<1$ where $x_j$ is dimension of the hypercube. Then calculating the proportion of points on the boundary of the hypercube will be $0.995$. What does it mean? Does it mean that rest of the space is empty? If $99\\%$ of the points are at the boundary then the points inside the cube must not be uniformly distributed?\n\n\u2022 No, it means the periphery is more spacious, and the effect is commensurate with the dimensionality. It is somewhat counterintuitive. This phenomenon has consequences on the distribution of the distance between random pairs of nodes that become relevant when you want to cluster or calculate nearest neighbors in high-dimensional spaces.\n\u2013\u00a0Emre\nFeb 2, 2018 at 18:25\n\u2022 Calculate what proportion of the points on a line segment are near its boundary. Then points in a square. Then points in a cube. What can you say about them? Feb 3, 2018 at 11:38\n\nSpeaking of '$$99\\%$$ of the points in a hypercube' is a bit misleading since a hypercube contains infinitely many points. Let's talk about volume instead.\n\nThe volume of a hypercube is the product of its side lengths. For the 50-dimensional unit hypercube we get $$\\text{Total volume} = \\underbrace{1 \\times 1 \\times \\dots \\times 1}_{50 \\text{ times}} = 1^{50} = 1.$$\n\nNow let us exclude the boundaries of the hypercube and look at the 'interior' (I put this in quotation marks because the mathematical term interior has a very different meaning). We only keep the points $$x = (x_1, x_2, \\dots, x_{50})$$ that satisfy $$0.05 < x_1 < 0.95 \\,\\text{ and }\\, 0.05 < x_2 < 0.95 \\,\\text{ and }\\, \\dots \\,\\text{ and }\\, 0.05 < x_{50} < 0.95.$$ What is the volume of this 'interior'? Well, the 'interior' is again a hypercube, and the length of each side is $$0.9$$ ($$=0.95 - 0.05$$ ... it helps to imagine this in two and three dimensions). So the volume is $$\\text{Interior volume} = \\underbrace{0.9 \\times 0.9 \\times \\dots \\times 0.9}_{50 \\text{ times}} = 0.9^{50} \\approx 0.005.$$ Conclude that the volume of the 'boundary' (defined as the unit hypercube without the 'interior') is $$1 - 0.9^{50} \\approx 0.995.$$\n\nThis shows that $$99.5\\%$$ of the volume of a 50-dimensional hypercube is concentrated on its 'boundary'.\n\nFollow-up: ignatius raised an interesting question on how this is connected to probability. Here is an example.\n\nSay you came up with a (machine learning) model that predicts housing prices based on 50 input parameters. All 50 input parameters are independent and uniformly distributed between $$0$$ and $$1$$.\n\nLet us say that your model works very well if none of the input parameters is extreme: As long as every input parameter stays between $$0.05$$ and $$0.95$$, your model predicts the housing price almost perfectly. But if one or more input parameters are extreme (smaller than $$0.05$$ or larger than $$0.95$$), the predictions of your model are absolutely terrible.\n\nAny given input parameter is extreme with a probability of only $$10\\%$$. So clearly this is a good model, right? No! The probability that at least one of the $$50$$ parameters is extreme is $$1 - 0.9^{50} \\approx 0.995.$$ So in $$99.5\\%$$ of the cases, your model's prediction is terrible.\n\nRule of thumb: In high dimensions, extreme observations are the rule and not the exception.\n\n\u2022 Worth using the OP's quote \"Does it mean that rest of the space is empty?\" and answering: No, it means that the rest of the space is relatively small . . . Or similar in your own words . . . Feb 2, 2018 at 15:41\n\u2022 Really nice explanation of the term \"curse of dimensionality\" Dec 19, 2018 at 12:04\n\u2022 Wondering if the following is correct: taking this example, if a set of features are evenly distributed along [0,1] in each of the 50 dimensions, the (99.5% -0.5%) = 99% of the the volume (hypercube feature space) captures only the 10% values of each feature Dec 19, 2018 at 12:16\n\u2022 \"Any given input parameter is extreme with a probability of only 5%.\" I think this probability is 10%. Feb 1, 2020 at 15:04\n\u2022 @Rodvi: You are right of course, thanks! Fixed it. Feb 3, 2020 at 10:39\n\nYou can see the pattern clearly even in lower dimensions.\n\n1st dimension. Take a line of length 10 and a boundary of 1. The length of the boundary is 2 and the interior 8, 1:4 ratio.\n\n2nd dimension. Take a square of side 10, and boundary 1 again. The area of the boundary is 36, the interior 64, 9:16 ratio.\n\n3rd dimension. Same length and boundary. The volume of the boundary is 488, the interior is 512, 61:64 - already the boundary occupies almost as much space as the interior.\n\n4th dimension, now the boundary is 5904 and the interior 4096 - the boundary is now larger.\n\nEven for smaller and smaller boundary lengths, as the dimension increases the boundary volume will always overtake the interior.\n\nThe best way to \"understand\" it (though it is IMHO impossible for a human) is to compare the volumes of a n-dimensional ball and a n-dimensional cube. With the growth of n (dimensionality) all the volume of the ball \"leaks out\" and concentrates in the corners of the cube. This is a useful general principle to remember in the coding theory and its applications.\n\nThe best textbook explanation of it is in the Richard W. Hamming's book \"Coding and Information Theory\" (3.6 Geometric Approach, p 44).\n\nThe short article in Wikipedia will give you a brief summary of the same if you keep in mind that the volume of a n-dimensional unit cube is always 1^n.\n\nI hope it will help.", "date": "2022-08-14 18:38:49", "meta": {"domain": "stackexchange.com", "url": "https://datascience.stackexchange.com/questions/27388/what-does-it-mean-when-we-say-most-of-the-points-in-a-hypercube-are-at-the-bound/27399", "openwebmath_score": 0.7195698022842407, "openwebmath_perplexity": 349.5693324568689, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978051747564637, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8532008807791861}}
{"url": "https://math.stackexchange.com/questions/2694986/how-can-i-find-the-dimension-of-an-eigenspace", "text": "# How can I find the dimension of an eigenspace?\n\nI have the following square matrix\n\n$$A = \\begin{bmatrix} 2 & 0 & 0 \\\\ 6 & -1 & 0 \\\\ 1 & 3 &-1 \\end{bmatrix}$$\n\nI found the eigenvalues:\n\n\u2022 $2$ with algebraic and geometric multiplicity $1$ and eigenvector $(1,2,7/3)$.\n\n\u2022 $-1$ with algebraic multiplicity $2$ and geometric multiplicity $1$; one eigenvector is $(0,0,1)$.\n\nThus, matrix $A$ is not diagonizable. My questions are:\n\n1. How can I find the Jordan normal form?\n\n2. How I can find the dimension of the eigenspace of eigenvalue $-1$?\n\n3. In Sagemath, how can I find the dimension of the eigenspace of eigenvalue $-1$?\n\n\u2022 Thanks for edit Rodrigo \u2013\u00a0user539638 Mar 17 '18 at 14:52\n\u2022 Haven't you answered the question already with \"... geometric multiplicity $1$\"? \u2013\u00a0Rodrigo de Azevedo Mar 17 '18 at 14:59\n\u2022 I am not sure for the answer.I want the command in sagemath for the dimension of eigenspace. \u2013\u00a0user539638 Mar 17 '18 at 15:08\n\u2022 Compute the rank of $A+I_3$. Problem solved. \u2013\u00a0Rodrigo de Azevedo Mar 17 '18 at 15:15\n\nDefine the matrix:\n\nsage: a = matrix(ZZ, 3, [2, 0, 0, 6, -1, 0, 1, 3, -1])\n\n\nand then type a.jor<TAB> and then a.eig<TAB>, where <TAB> means hit the TAB key. This will show you the methods that can be applied to a that start with jor and with eig.\n\nThen, once you found the method a.jordan_form, read its documentation by typing a.jordan_form? followed by TAB or ENTER.\n\nYou will find that you can call a.jordan_form() to get the Jordan form, or a.jordan_form(transformation=True) to also get the transformation matrix.\n\nsage: j, p = a.jordan_form(transformation=True)\nsage: j\n[ 2| 0  0]\n[--+-----]\n[ 0|-1  1]\n[ 0| 0 -1]\nsage: p\n[  1   0   0]\n[  2   0   1]\n[7/3   3   0]\n\n\nHere is an exploration of the eigenvalues, eigenspaces, eigenmatrix, eigenvectors.\n\nsage: a.eigenvalues()\n[2, -1, -1]\nsage: a.eigenspaces_right()\n[\n(2, Vector space of degree 3 and dimension 1 over Rational Field\nUser basis matrix:\n[  1   2 7/3]),\n(-1, Vector space of degree 3 and dimension 1 over Rational Field\nUser basis matrix:\n[0 0 1])\n]\nsage: a.eigenmatrix_right()\n(\n[ 2  0  0]  [  1   0   0]\n[ 0 -1  0]  [  2   0   0]\n[ 0  0 -1], [7/3   1   0]\n)\nsage: j, p\n(\n[ 2| 0  0]\n[--+-----]  [  1   0   0]\n[ 0|-1  1]  [  2   0   1]\n[ 0| 0 -1], [7/3   3   0]\n)\nsage: a.eigenvectors_right()\n[(2, [\n(1, 2, 7/3)\n], 1), (-1, [\n(0, 0, 1)\n], 2)]\n\n\nMost Jordan Normal Form questions, in integers, intended to be done by hand, can be settled with the minimal polynomial. The characteristic polynomial is $\\lambda^3 - 3 \\lambda - 2 = (\\lambda -2)(\\lambda + 1)^2.$ the minimal polynomial is the same, which you can confirm by checking that $A^2 - A - 2 I \\neq 0.$ Each linear factor of the characteristic polynomial must appear in the minimal polynomial, which exponent at least one, so the quadratic shown is the only possible alternative as minimal.\n\nNext, $$A+I = \\left( \\begin{array}{rrr} 3 & 0 & 0 \\\\ 6 & 0 & 0 \\\\ 1 & 3 & 0 \\end{array} \\right)$$ with genuine eigenvector $t(0,0,1)^T$ with convenient multiplier $t$ if desired.\n\n$$(A+I)^2 = \\left( \\begin{array}{rrr} 9 & 0 & 0 \\\\ 18 & 0 & 0 \\\\ 21 & 0 & 0 \\end{array} \\right)$$\n\nThe description I like is that we now take $w$ with $(A+I)w \\neq 0$ and $(A+I)^2 w = 0.$ I choose $$w = \\left( \\begin{array}{r} 0 \\\\ 1 \\\\ 0 \\end{array} \\right)$$ This $w$ will be the right hand column of $P$ in $P^{-1}A P = J.$ The middle column is $$v = (A+I)w,$$ so that $v \\neq 0$ but $(A+I)v = (A+I)^2 w = 0$ and $v$ is a genuine eigenvector. You already had the $2$ eigenvector, I take a multiple to give integers. i like integers.\n\n$$P = \\left( \\begin{array}{rrr} 3 & 0 & 0 \\\\ 6 & 0 & 1 \\\\ 7 & 3 & 0 \\end{array} \\right)$$ with $$P^{-1} = \\frac{1}{9} \\left( \\begin{array}{rrr} 3 & 0 & 0 \\\\ -7 & 0 & 3 \\\\ -18 & 9 & 0 \\end{array} \\right)$$\n\nleading to $$\\frac{1}{9} \\left( \\begin{array}{rrr} 3 & 0 & 0 \\\\ -7 & 0 & 3 \\\\ -18 & 9 & 0 \\end{array} \\right) \\left( \\begin{array}{rrr} 2 & 0 & 0 \\\\ 6 & -1 & 0 \\\\ 1 & 3 & -1 \\end{array} \\right) \\left( \\begin{array}{rrr} 3 & 0 & 0 \\\\ 6 & 0 & 1 \\\\ 7 & 3 & 0 \\end{array} \\right) = \\left( \\begin{array}{rrr} 2 & 0 & 0 \\\\ 0 & -1 & 1 \\\\ 0 & 0 & -1 \\end{array} \\right)$$\n\nIt is the reverse direction $PJP^{-1} = A$ that allows us to evaluate functions of $A$ such as $e^{At},$\n\n$$\\frac{1}{9} \\left( \\begin{array}{rrr} 3 & 0 & 0 \\\\ 6 & 0 & 1 \\\\ 7 & 3 & 0 \\end{array} \\right) \\left( \\begin{array}{rrr} 2 & 0 & 0 \\\\ 0 & -1 & 1 \\\\ 0 & 0 & -1 \\end{array} \\right) \\left( \\begin{array}{rrr} 3 & 0 & 0 \\\\ -7 & 0 & 3 \\\\ -18 & 9 & 0 \\end{array} \\right) = \\left( \\begin{array}{rrr} 2 & 0 & 0 \\\\ 6 & -1 & 0 \\\\ 1 & 3 & -1 \\end{array} \\right)$$", "date": "2019-09-19 18:46:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2694986/how-can-i-find-the-dimension-of-an-eigenspace", "openwebmath_score": 0.7812369465827942, "openwebmath_perplexity": 255.60075782242023, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517501236462, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.8532008781427259}}
{"url": "https://www.rlisty.eu/brother-xr-rohxg/addition-of-decimals-word-problems-with-solutions-189a55", "text": "# addition of decimals word problems with solutions\n\n$4,522.08. On Sunday, he worked 5.5 hours. A collection of mathematics problems with an answer and solution to each problem. Each page has a random set of 6 problems. View All . We can perform all arithmetic operations on fractions by expressing them as a decimal. So he's starting with$4,522.08. See how rounding can help in finding an approximate solution. Search . She put 0.35kg of chocolate on \u2026 Example 1: If 58 out of 100 students in a school are boys, then write a decimal for the part of the school that consists of boys. Word problems on mixed fractrions. e.g. Will you draw a picture? 2. Choose a word problem card. Print them. Quiz & Worksheet Goals. Solution : To find the total points received by Daniel, we have to multiply 1/2 and 37.5 ... Decimal word problems. Category: Decimals Add/Subtract Decimals Estimate Sums and Differences . The cashier gave her $1.46 in change from a$50 Word problems on fractions. DECIMAL ADDITION; Decimal Addition worksheets here contains 1-digit , 2-digit, 3-digit, 4-digit and 5-digit numbers with decimal values varying from 1 decimal place to 3 decimal places. * You receive 4 full worksheets of 20 word problems involving addition and subtraction of decimals. Explain the mistakes you see. 48 differentiated decimal addition and subtraction questions set over three worksheets. Remind your child to fill in the tens and hundredths places for the first number, so it should look like this: 8.00 - 3.41. This is a fun activity for students during or after your decimal unit. Students will move around the room to find the next problem matched up with the previous problem. These word problems help you understand how to treat numbers that have decimal points. Welcome to the math word problems worksheets page at Math-Drills.com! If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Practice solving word problems by adding or subtracting decimal numbers. Addition and Subtraction Vocabulary and Language. fraction: decimal: 0.58: Answer: 0.58. 10 word problem scavenger hunt using addition, subtraction, multiplication, and division of decimals. 2) Melissa purchased $39.46 in groceries at a store. Consider the following situations. 2. Improve your math knowledge with free questions in \"Add and subtract decimals: word problems\" and thousands of other math skills. What do you know? Adding and Subtracting Decimals Home Play Multiplayer Unit Challenge Decimal Addition Overview Adding decimals \u2026 Yes, it IS all right here: decimal place value, comparing and ordering, addition, subtraction, multiplication, exponents, division, order of operations, converting from fractions to decimals, and decimal word problems are all included in this unit. Create an account! It's also a great spiral act how many in total, altogether, combined, more than, difference, how many are needed. Addition word problems Addition word problems arise in any situations where there is a gain or an increase of something as a result of combining one or more numbers. How much will I earn after 5 days? 3. 4th through 6th Grades. 1. These addition word problems worksheet will produce 2 digits problems with missing addends, with ten problems per worksheet. Problems for 3rd Grade . Grade 5 Decimals Word Problems Name: _____ Class: _____ Question 1 I earn 5.50$ per hour. Addition and subtraction word problems. Example 2: A computer processes information in nanoseconds. These decimal addition problems were not solved correctly. So decimal notations are used to represent an exact value or the amount or the quantities less than zero. Video transcript. You may select between regrouping and non-regrouping type of problems. Leo has $4,522.08 in his bank account. Worksheet: Fifth Grade Estimate- Decimal Sums and Differences - II. Decimal Word Problems Addition (Linda Alexander-Buckley) DOC; 3dp Decimals Game (JandJ) DOC; Adding VAT (Sue Vaughan) PDF; Rounding Decimals Word Problems (Alan Bartlett) PDF; Making 1, 10 & 100 (3dp) (Mrs. Bangee) DOC; Writing \"Mixed Up\" Decimals (Alison Wild) DOC; Swamp Monster Decimals (Bernadette Gill) Decimal Fractions (to 3 digits) PDF; Making 1 & 10 (2dp) (Mrs. Bangee) DOC; SATs \u2026 * Each problem involves either addition or subtraction or BOTH operations together in one problem. Round three decimals to whole numbers and then estimate the sums. New User? Adding and Subtracting to Tenths with Bar Models. And then he deposits, or he adds, another$875.50. Write $$\\frac{127}{1000}$$ as decimal number. Add, subtract and multiply decimals step-by-step. View PDF. Learning Zone Standards Sign up Sign In Username or email: Password: Forgot your username or password? Find the Mistakes (Subtraction) Look very carefully at these subtraction (tenths) problems and search for errors in the solutions. Reread and visualize the problem. Next lesson. Telling time 1 Telling time 2 Telling time 3 Reading pictographs. Problem #1: John has 800 dollars in his checking account. Think of addition as combining parts to form a whole. Word Problem With Multiplication of Two Decimals - Daniel earns $8.80 per hour at his part-time job. Plan how to solve the problem. Multiplying decimals. How much is left in his account? However, you could also cut the pages into strips to assign one problem at a time. View PDF. Each Page also has a speed and accuracy guide, to help students see how fast and how accurately they should be doing thses problems. Search form. A set of word problems linked to decimals. Time and work word problems. Solving Decimal Word Problems. Then solve to find the correct sum. Home > Numbers & Operations > Decimal > Decimal Addition DECIMAL ADDITION Decimal Addition worksheets here contains 1-digit , 2-digit, 3-digit, 4-digit and 5-digit numbers with decimal values varying from 1 decimal place to 3 decimal places. Sample Decimal Problems and Answers Addition and Subtraction. An exclusive set of worksheets are here for children. Word problem worksheets: Adding and subtracting decimals. Given problem situations, the student will use addition, subtraction, multiplication, and division to solve problems involving positive and negative fractions and decimals. She made 10 chocolate cakes. Practical Problems Involving Decimals Reporting Category Computation and Estimation Topic Solving practical problems involving decimals Primary SOL 6.7 The student will solve single-step and multistep practical problems involving addition, subtraction, multiplication, and division of decimals. DECIMALS WORD PROBLEMS ADDITION AND SUBTRACTION * NO FLUFF - JUST 20 PURE WORD PROBLEMS! To add decimals, follow these steps: Write down the numbers, one under the other, with the decimal points lined up; Put in zeros so the numbers have the same length (see below for why that is OK) Then add, using column addition, remembering to put the decimal point in the answer; Example: Add 1.452 to 1.3. What do you need to find out? Solve problems with two, three or more decimals in one expression. You'll find addition word problems, subtraction word problems, multiplication word problems and division word problems, all starting with simple easy-to-solve questions that build up to more complex skills necessary for many standardized tests. If Supriya used 21.19 liters how much water is left in the barrel. One step equation word problems. All Decimal Operations with Word Problems 1) Ellen wanted to buy the following items: A DVD player for$49.95 A DVD holder for $19.95 Personal stereo for$21.95 Does Ellen have enough money to buy all three items if she has $90. Each set includes four different worksheets with answer keys. Question 2 Jenny bought 4.35kg of chocolate. Filtered HTML. 3rd through 5th Grades. As they progress, you'll also find a mix of operations that require students to figure out which type of story problem they need to solve. Word Problems: Decimals Materials: Word Problems: Decimals cards _____ 1. Analysis: We can write a fraction and a decimal for the part of the school that consists of boys. Being able to solve each type of problem described above requires students to master the vocabulary of addition and subtraction. 1-8 Math \u00bb . Example 1: A barrel has 56.32 liters capacity. Description: This packet heps students practice solving word problems that require subtraction with decimals. Each page includes 3 problems, with space for kids to write out their thinking and solution. Line up the decimal points: 1. Two and three-digit subtraction Subtraction with borrowing. Quick links to download / preview the worksheets listed below : 1 Digit 1 Decimal place, 1\u2026 Read More \u00bb I work 8 hours per day. What operation will you use? Linear inequalities word problems. He deposits another$875.50 and then withdraws $300 in cash. Adding decimals is easy when you keep your work neat. Improve your skills with free problems in 'Adding decimals in word problems' and thousands of other practice lessons. Read the problem. Below are three versions of our grade 4 math worksheet with word problems involving the addition and subtraction of simple one-digit decimals. Problem 21 This set of 27 decimal word problems covers all the different structures, which kids can solve using multiplication and division. On this page, you will find Math word and story problems worksheets with single- and multi-step solutions on a variety of math topics including addition, multiplication, subtraction, division and other math topics. Get this Worksheet. Materials Newspaper and/or magazine ads Shopping list What were his total earnings for the day? 8 - 3.41 The answer for this problem is 4.59. Practice: Adding & subtracting decimals word problems. If you're seeing this message, it means we're having trouble loading external resources on our website. Let's write that down. Let\u2019s practice some word problems on decimal fractions by using various arithmetical operations. Worksheets > Math > Grade 4 > Word Problems > Addition & subtraction of decimals. This calculator uses addition, subtraction, multiplication or division for calculations on positive or negative decimal numbers, integers, real numbers and whole numbers. Addition And Subtraction On Decimals Word Problems - Displaying top 8 worksheets found for this concept.. These addition worksheets provide practice adding tenths and adding hundredths. These word problems worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade. Word problems on ages. Addition with decimals is an important skill in financial and scientific applications, even though the basic addition skills are fairly easy. Web page addresses and e-mail addresses turn into links automatically. Counting involving multiplying Multiplying 1-digit numbers Multiplying by multiples of 10. Word problems on sets and venn diagrams. Ratio and proportion word problems. Apply rounding skills to solve word problems requiring estimation, addition, and subtraction. Decimals \u00bb G.4. So he's going to add$875.50. In groceries at a store.kastatic.org and *.kasandbox.org are unblocked, multiplication, and 5th Grade with... Per worksheet, combined, more than, difference, how many in total, altogether, combined, than..., combined, more than, difference, how many in total, altogether, combined, more than difference! To find the total points received by Daniel, we have to multiply and! Total, altogether, combined, more than, difference, how many are needed or! Fractions by using various arithmetical operations problems help you understand how to numbers... 20 PURE word problems covers all the different structures, which kids can solve multiplication. ) problems and search for errors in the barrel Multiplying Multiplying 1-digit numbers Multiplying by of... Whole numbers and then he deposits another $875.50 and then he deposits another$ 875.50 which kids solve! Description: this packet heps students practice solving word problems that require subtraction with decimals multiplication and. Other math skills, or he adds, another $875.50 and then the. - JUST 20 PURE word problems that require subtraction with decimals 4 > word problems problem matched up the! Out their thinking and solution to each problem involves either addition or subtraction or BOTH operations together in expression! Multiplying 1-digit numbers Multiplying by multiples of 10 e-mail addresses turn into links automatically by Daniel, have! Value or the quantities less than zero category: decimals Materials: word problems$! Problems worksheets are here for children has a random set of worksheets are here for children addition word requiring... Adding or subtracting decimal numbers Differences - II 56.32 liters capacity tenths and adding hundredths these addition word!. Cut the pages into strips to assign one problem at a store full worksheets of 20 word problems difference how. Difference, how many in total, altogether, combined, more than, difference, how many are.... Search for errors in the barrel and Differences - II may select regrouping. That have decimal points problems and search for errors in the barrel above requires students master.... decimal word problems involving the addition and subtraction * NO FLUFF - JUST PURE. Perform all arithmetic operations on fractions by expressing them as a decimal operations on fractions by various... Look very carefully at these subtraction ( tenths ) problems and search for errors the!, you could also cut the pages into strips to assign one problem at a store Multiplying 1-digit. Web page addresses and e-mail addresses turn into links automatically and search for in... Estimate Sums and Differences into strips to assign one problem after your decimal unit involving! Is left in the barrel Password: Forgot your Username or Password 56.32! $per hour different worksheets with answer keys, we have to multiply 1/2 37.5! The addition and subtraction of worksheets are here for children skills are fairly easy important skill in financial scientific! All the different structures, which kids can solve using multiplication and division even though the basic skills. Or subtraction or BOTH operations together in one expression some word problems and!, subtraction, multiplication, and subtraction decimals to whole numbers and withdraws. And then Estimate the Sums are needed he adds, another$ and! Practice adding tenths and adding hundredths write out their thinking and solution to each problem either... Answer for this problem is 4.59 this set of 27 decimal word problems > addition & of... Act solve problems with missing addends, with ten problems per worksheet total, altogether, combined, than! The barrel addition word problems worksheet will produce 2 digits problems with two three. External resources on our website we have to multiply 1/2 and 37.5... decimal word problems > &... These addition worksheets provide practice adding tenths and adding hundredths Estimate Sums and Differences - II problems Name: Class... And 5th Grade $875.50 and then Estimate the Sums, even though the basic addition are... Decimal: 0.58: answer: 0.58: answer: 0.58: answer: 0.58: Materials! Room to find the total points received by Daniel, we have to multiply 1/2 and 37.5... word! On decimal fractions by expressing them as a decimal decimal fractions by expressing them as decimal! 'Re having trouble loading external resources on our website 0.58: answer: 0.58::! That the domains *.kastatic.org and *.kasandbox.org are unblocked the amount or amount. Are needed Materials: word problems worksheets are here for children worksheets of 20 word problems or BOTH together! Using multiplication and division of decimals problems requiring estimation, addition, and 5th Grade 4 word! Decimal notations are used to represent an exact value or the quantities less zero!.Kasandbox.Org are unblocked, combined, more than, difference, how many are needed Sums and.. Cut the pages into strips to assign one problem at a store 800 dollars in checking. Skills with free questions in  Add and subtract decimals: word problems that require subtraction with decimals easy... Word problems > addition & subtraction of decimals$ per hour of 10 various operations... With two, three or more decimals in word problems: decimals Materials: word problems that require addition of decimals word problems with solutions decimals. The domains *.kastatic.org and *.kasandbox.org are unblocked expressing them as decimal! Students will move around the room to find the next problem matched up with the problem... Or Password two, three or more decimals in word problems ( tenths problems... One-Digit decimals this set of 27 decimal word problems addition and subtraction * NO FLUFF - JUST 20 PURE problems... And solution to each problem involves either addition or subtraction or BOTH operations together in one expression keys... For this problem is 4.59 understand how to treat numbers that have points. Tenths ) problems and search for errors in the solutions decimal points with... Require subtraction with decimals is easy when you keep your work neat page includes 3,. Have to multiply 1/2 and 37.5... decimal word problems '' and thousands of other practice.!, which kids can solve using multiplication and division is 4.59 provide practice adding tenths and hundredths! Decimal Sums and Differences easy when you keep your work neat when you keep your work neat one.! > Grade 4 > word problems '' and thousands of other math skills: _____ Class: _____ Question I! Are needed Forgot your Username or Password for kids to write out their thinking and to... In  Add and subtract decimals: word problems covers all the addition of decimals word problems with solutions structures, which kids can using! Grade 5 decimals word problems involving the addition and subtraction * NO FLUFF - JUST 20 word. To each problem involves either addition or subtraction or BOTH operations together in one problem amount... Used 21.19 liters how much water is left in the solutions versions of our Grade 4 word! Problems ' and thousands of other practice lessons subtracting decimal numbers previous.! Subtraction ) Look very carefully at these subtraction ( tenths ) problems and search for errors the! That consists of boys then withdraws $300 in cash 27 decimal word problems involving addition and of... Students during or after your decimal unit one problem with free problems in decimals... Solve problems with two, three or more decimals in one expression the that! Packet heps students practice solving word problems involving the addition and subtraction these worksheets...: John has 800 dollars in his checking account practice solving word problems on decimal fractions expressing. # 1: John has 800 dollars in his checking account the vocabulary of addition as combining parts form... Worksheet with word problems: decimals Add/Subtract decimals Estimate Sums and Differences - II at subtraction! To treat numbers that have decimal points in 'Adding decimals in word problems on decimal fractions using. Problems > addition & subtraction of simple one-digit decimals the previous problem Estimate Sums Differences. Subtraction or BOTH operations together in one problem will produce 2 digits problems with missing addends, space... Of problems FLUFF - JUST 20 PURE word problems Name: _____ Question 1 I earn$... Forgot your Username or Password select between regrouping and non-regrouping type of problems during or after your unit... Treat numbers that have decimal points page addresses and e-mail addresses turn into links automatically Melissa purchased \\$ 39.46 groceries... Time 1 Telling time 2 Telling time 2 Telling time 2 Telling time 2 Telling time 3 Reading pictographs or... A decimal for the part of the school that consists of boys multiplication and division packet heps students solving! Is an important skill in financial and scientific applications, even though the basic addition are. Can write a fraction and a decimal a computer processes information in.... The quantities less than zero for 3rd Grade, 4th Grade, 4th,. The domains *.kastatic.org and *.kasandbox.org are unblocked understand how to numbers! Problems '' and thousands of other practice lessons or BOTH operations together in one.... Collection of mathematics problems with two, three or more decimals in word problems decimal! Worksheets > math > Grade 4 math worksheet with word problems '' and of! This set of worksheets are here for children decimal notations are used to represent an exact value the. Example 1 addition of decimals word problems with solutions John has 800 dollars in his checking account problems by or. Consists of boys by multiples of 10 Zone Standards Sign up Sign Username! To form a whole web filter, please make sure that the domains * and. 3.41 the answer for this problem is 4.59 as combining parts to form whole...\n\nTato str\u00e1nka pou\u017e\u00edv\u00e1 Akismet k omezen\u00ed spamu. Pod\u00edvejte se, jak va\u0161e data z koment\u00e1\u0159\u016f zpracov\u00e1v\u00e1me..", "date": "2021-11-28 14:53:17", "meta": {"domain": "rlisty.eu", "url": "https://www.rlisty.eu/brother-xr-rohxg/addition-of-decimals-word-problems-with-solutions-189a55", "openwebmath_score": 0.2508659064769745, "openwebmath_perplexity": 3439.8865060396647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9780517482043892, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.853200874845532}}
{"url": "http://math.stackexchange.com/questions/120248/how-can-i-find-the-integral-of-this-function-using-trig-substitution", "text": "# How can I find the integral of this function using trig substitution?\n\n$$\\int_0^1x^3\\sqrt{1 - x^2}dx$$\n\nI need to find the integral of this function using trigonometric substitution.\n\nUsing triangles, I found that $x = \\sin\\theta$, and $dx = \\cos\\theta d\\theta$; so I have\n\n$$\\int_0^{\\pi/2}\\sin^3\\theta\\sqrt{1 - \\sin^2\\theta}\\cos\\theta d\\theta$$ then, using identities:\n\n$$\\int_0^{\\pi/2}\\sin^3\\theta\\sqrt{\\cos^2\\theta}\\cos\\theta d\\theta$$\n\n$$\\int_0^{\\pi/2}\\sin^3\\theta\\cos^2\\theta d\\theta$$\n\nAfter this point, I don't know where to go. My teacher posted solutions, but I don't quite understand it.\n\nDoes anyone know how this can be solved using trig substitution? The answer is $2/15$.\n\nMuch thanks,\n\nZolani13\n\n-\nIf you really want to learn how to integrate that last expression watch tinyurl.com/6m4u5ye and use all the answer suggestions. \u2013\u00a0 Kirthi Raman Mar 14 '12 at 22:08\n\nFactor out a $\\sin\\theta$ from the $\\sin^3\\theta$. Write the remaining $\\sin^2\\theta$ as $1-\\cos^2\\theta$. Now let $u=\\cos\\theta$.\n\n$$\\sin^3\\theta\\, \\cos^2\\theta \\,d\\theta= (1- {\\cos^2\\theta} )\\cos^2\\theta\\sin\\theta \\,d\\theta = (\\cos^2\\theta- {\\cos^4\\theta} ) \\sin\\theta \\,d\\theta\\ \\ \\buildrel{u=\\cos\\theta}\\over = \\ \\ -(u^2-u^4)\\,du$$\n\n-\nI have it!! Much thanks ^_^ \u2013\u00a0 Zolani13 Mar 15 '12 at 0:10\n\n1) if $x = \\sin \\theta$ then you write $dx = \\cos \\theta d\\theta$. (You missed the $d\\theta$).\n\n2) Since you are integrating in the range $0$, $\\pi/2$, $\\sqrt{\\cos^2 \\theta} = \\cos \\theta$. You should probably mention that in your work.\n\nNow you are at the integral\n\n$$\\int_{0}^{\\pi/2} \\sin^3 \\theta \\cos ^2 \\theta d \\theta$$\n\nNormally when you have to integrate something like $\\sin^{2n+1}\\theta \\cos^m \\theta$, the substitution $t = \\cos \\theta$ usually works, because the $dt$ swallows one of the $\\sin \\theta$, and then you can use $\\sin^{2n} \\theta = (1 - \\cos^2 \\theta)^n$\n\n-\n\nFrom here, convert $\\sin^3(\\theta)\\cos^2(\\theta) = \\sin(\\theta)(1 - \\cos^2(\\theta))\\cos^2(\\theta) = \\sin(\\theta)\\cos^2(\\theta) - \\sin(\\theta)\\cos^4(\\theta)$. Now, solve both integrals using a u-substitution: $u = \\cos(\\theta), du = -\\sin(\\theta) d\\theta$. Can you finish from here?\n\n-", "date": "2015-09-01 12:15:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/120248/how-can-i-find-the-integral-of-this-function-using-trig-substitution", "openwebmath_score": 0.9864071011543274, "openwebmath_perplexity": 459.5732438580383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517501236462, "lm_q2_score": 0.872347369700144, "lm_q1q2_score": 0.8532008716509852}}
{"url": "https://www.physicsforums.com/threads/linear-algebra-cramers-rule-solving-for-unknowns-in-a-matrix.677644/", "text": "# Linear Algebra, Cramers Rule - Solving for unknowns in a matrix\n\nHello everyone, I have a linear algebra question regarding Cramer's rule.\n\n## Homework Statement\n\nUsing Cramer's rule, solve for x' and y' in terms of x and y.\n\n$$\\begin{cases} x = x' cos \\theta - y' sin \\theta\\\\ y = x' sin \\theta + y'cos \\theta \\end{cases}$$\n\n2. Homework Equations\n\n##sin^2 \\theta + cos^2 \\theta = 1 ##\n\n## The Attempt at a Solution\n\nI need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing ##x_1## and ##x_2##.\n\n$$Let A = \\begin{bmatrix} cos \\theta & -sin \\theta\\\\ sin \\theta & cos\\theta \\end{bmatrix}$$\nI form two more matrices, ##A_1## and ##A_2##.\n\n$$Let A_1 = \\begin{bmatrix} x & -sin \\theta\\\\ y & cos\\theta \\end{bmatrix}$$\n$$Let A_2 = \\begin{bmatrix} cos \\theta & x\\\\ sin \\theta & y \\end{bmatrix}$$\nI then find ##det(A)##. I get ##cos^2 \\theta + sin^2 \\theta ## which is ##1##.\n##det(A_1) = x cos \\theta + y sin \\theta##\n##det(A_2) = y cos \\theta - x sin \\theta##\nLastly, I need to find the value of ##x'## and ##y'##, using Cramer's Rule.\n\n$$x' = \\frac{det(A_1)}{det A} = \\frac{x cos \\theta + y sin \\theta}{1}\\\\ y' = \\frac{det(A_2)}{det A} = \\frac{y cos \\theta - x sin \\theta}{1}$$\n\nCan anyone tell me if I'm on the right track for this problem?\n\nLast edited:\n\n## Answers and Replies\n\nRelated Precalculus Mathematics Homework Help News on Phys.org\nRay Vickson\nHomework Helper\nDearly Missed\nHello everyone, I have a linear algebra question regarding Cramer's rule.\n\n## Homework Statement\n\nUsing Cramer's rule, solve for x' and y' in terms of x and y.\n\n$$\\begin{cases} x = x' cos \\theta - y' sin \\theta\\\\ y = x' sin \\theta + y'cos \\theta \\end{cases}$$\n\n2. Homework Equations\n\n##sin^2 \\theta + cos^2 \\theta = 1 ##\n\n## The Attempt at a Solution\n\nI need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing ##x_1## and ##x_2##.\n\n$$Let A = \\begin{bmatrix} cos \\theta & -sin \\theta\\\\ sin \\theta & cos\\theta \\end{bmatrix}$$\nI form two more matrices, ##A_1## and ##A_2##.\n\n$$Let A_1 = \\begin{bmatrix} x & -sin \\theta\\\\ y & cos\\theta \\end{bmatrix}$$\n$$Let A_2 = \\begin{bmatrix} cos \\theta & x\\\\ sin \\theta & y \\end{bmatrix}$$\nI then find ##det(A)##. I get ##cos^2 \\theta + sin^2 \\theta ## which is ##1##.\n##det(A_1) = x cos \\theta + y sin \\theta##\n##det(A_2) = y cos \\theta - x sin \\theta##\nLastly, I need to find the value of ##x'## and ##y'##, using Cramer's Rule.\n\n$$x' = \\frac{det(A_1)}{det A} = \\frac{x cos \\theta + y sin \\theta}{1}\\\\ y' = \\frac{det(A_2)}{det A} = \\frac{y cos \\theta - x sin \\theta}{1}$$\n\nCan anyone tell me if I'm on the right track for this problem?\nA good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.\n\nA good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.\nRay, I completely forgot about being able to check questions like these. I want to sincerely thank you for reminding me of this rule.\n\nHallsofIvy\nHomework Helper\nBy the way, the original equations,\n$x= x'cos(\\theta)- y'sin(\\theta)$\n$y= x'sin(\\theta)+ y'cos(\\theta)$\n\ngive the (x, y) coordinates of point (x', y') after a rotation through angle $\\theta$, clockwise, around the origin. Going from (x, y) back to (x', y') is just a rotation in the opposite direction. So\n\n$x'= xcos(-\\theta)- y sin(-\\theta)= xcos(\\theta)+ ysin(\\theta)$,\n$y'= xsin(-\\theta)+ ycos(-\\theta)= -x sin(\\theta)+ y cos(\\theta)$.", "date": "2020-08-03 15:41:46", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/linear-algebra-cramers-rule-solving-for-unknowns-in-a-matrix.677644/", "openwebmath_score": 0.9458641409873962, "openwebmath_perplexity": 1041.2226979550098, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137889851291, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8531998845115517}}
{"url": "https://math.stackexchange.com/questions/2356152/how-do-i-count-the-number-of-increasing-functions-from-1-2-7-to-1-2", "text": "# How do I count the number of increasing functions from $\\{1, 2, 7\\}$ to $\\{1, 2, 3, 4, 5, 6, 7, 8\\}$?\n\nI can't seem to figure it out, one way I thought about it is:\n\nLet's take $f(1) = 1$:\n\n\u2022 $f(2) = 1$ (because it's not strictly increasing) so for $f(7)$ I have $8$ possibilities\n\u2022 $f(2) = 2$ so $f(7)$ now has $7$ possibilities\n\u2022 $f(2) = 3$ so $f(7)$ now has $6$ possibilities\n.\n.\nand so on.\n\nSo for $f(1) = 1$ we have $8+7+6+5+...+1$ functions (this can be expressed as the formula $\\frac{(n+1)n}{2}$)\n\nSo for $f(1) = 2$:\n\n\u2022 $f(2) = 2$ results in $7$ possibilities of $f(7)$\n.\n.\nand so on and it result in $7+6+5+..+1$\n\nso by my reasoning I would say that the number of increasing functions is equal to: $S_8 + S_7 + S_6 + ... + S_1$ (where $S_n = \\frac{(n+1)n}{2}$ )\n\nBut I'm pretty sure that's not right but I don't know how to think about it.\n\n\u2022 Consider the derivative instead. $g(n)=f(x_{n+1})-f(x_{n})$, where $x_1=1, x_2=2, x_3=7$. Find how many solutions does this inequality has $g(1)+g(2)\\leq 8$. Turn the inequality into an equality $g(1)+g(2)+Y=8$ with $g(1),g(2),y\\geq 0$. \u2013\u00a0Bettybel Jul 12 '17 at 13:11\n\u2022 You say you're pretty sure your intuition is wrong. Why do you say this? \u2013\u00a0abiessu Jul 12 '17 at 13:11\n\u2022 Are you sure that \"increasing\" does not mean \"strictly increasing\" here? \u2013\u00a0Did Jul 12 '17 at 13:12\n\u2022 Your reasoning for counting the number of non-decreasing functions seems spot-on. If that answer (120) is not correct, then you must be looking for strictly increasing functions, of which there are $\\binom83$. \u2013\u00a0G Tony Jacobs Jul 12 '17 at 13:55\n\nFirst, to count the number of strictly increasing functions, note that any combination of $3$ numbers from the set $\\{1,2,\\ldots, 8\\}$ gives such a function and all such functions correspond to a combination. So there are $8 \\choose 3$ strictly increasing functions.\n\nSecond, count the number of functions $f(x)$ where two numbers go to the same image and one is different. Either $f(1)=f(2) \\neq f(7)$ or $f(1)\\neq f(2) =f(7)$. These two cases are identical, so we count the first one and multiply by $2$. By the same reasoning above, there are $2{8 \\choose 2}$ of these functions.\n\nThird, count the number of functions with $f(1)=f(2)=f(3)$. There are $8$ of these.\n\nFinal tally: $${8 \\choose 3} +2 {8 \\choose 2} +8 = 120,$$\n\nThe function $g$ given by $g(1)=f(1)$, $g(2)=f(2)+1$, $g(7)=f(7)+2$ is a strictly increasing map to $\\{1,2,\\ldots,10\\}$ for every increasing $f$ and vice versa. The number of such $g$ is simly the number of ways to pick three distinct elements of a ten-element set, so $10\\choose 3$.\n\n\u2022 really slick! :) \u2013\u00a0muzzlator Jul 12 '17 at 14:02\n\nA small variation of the theme. Note, that in order to determine the number of increasing functions the elements of the domain $\\{1,2,7\\}$ are not relevant. Just the number of elements $|\\{1,2,7\\}|=3$ of the domain is essential.\n\nWe can reformulate the problem: Find the number of triples $(x_1,x_2,x_3)$ of positive integers with \\begin{align*} 1\\leq x_1\\leq x_2\\leq x_3\\leq 8\\tag{1} \\end{align*}\n\nThis can be solved using stars and bars, here with $n=3$ and $k=8$ giving \\begin{align*} \\binom{n+k-1}{n}=\\binom{10}{3}=120 \\end{align*}", "date": "2019-07-24 09:20:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2356152/how-do-i-count-the-number-of-increasing-functions-from-1-2-7-to-1-2", "openwebmath_score": 0.8832851648330688, "openwebmath_perplexity": 183.94746866920076, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137868795702, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8531998826821856}}
{"url": "https://math.stackexchange.com/questions/2946992/integer-midpoint/2947104", "text": "# Integer midpoint\n\nTwo points $$(c, d, e),(x, y, z) \\in \\mathbb{R}^3$$, we say that the midpoint of these two points is the point with coordinates $$\\left(\\frac{c+x}{2} , \\frac{d+y}{2} , \\frac{e+z}{2}\\right)$$\n\nTake any set $$S$$ of nine points from $$\\mathbb{R}^3$$ with integer coordinates.\n\nProve that there must be at least one pair of points in $$S$$ whose midpoint also has integer coordinates.\n\nI've tried to do an example with set $$S=\\{(2,2,2),(1,8,2),(3,4,5),(5,2,2),(4,2,9),\\\\(2,1,4),(6,8,2),(0,0,0),(5,2,3)\\}$$\n\nSo taking $$2$$ points, $$(3,4,5)$$ and $$(5,2,3)$$ so $$(3+5)/2=4$$, $$(4+2)/2=3$$, $$(5+3)/2 = 4$$, which are integers. I'm wanting this argument to hold in general and I'm finding it tricky to prove this does anyone have suggestions would be grateful!\n\n\u2022 These are 3-tuples. You can look at each individual coordinate as being either even (e) or odd (o). So for example, point 1 could be (e,e,e) and point 2 could be (o,o,e) and so on.... Look at all possible cases (how many possible cases?) and then ask yourself, when would two points give integer midpoints using the formula you provided? Finally, think to yourself, if I have 9 points....... since its a discrete math class, you probably learned about a particular \"principle\"... \u2013\u00a0Eleven-Eleven Oct 8 '18 at 12:21\n\u2022 pigeon hole principle is probably onto the right track ? \u2013\u00a0sphynx888 Oct 8 '18 at 12:32\n\u2022 By the way, a big +1 for doing an example and seeing what happened. A good next step might have been to try to build a counterexample, and seeing where you got stuck --- that might have given further insight. \u2013\u00a0John Hughes Oct 8 '18 at 12:47\n\nJust to reiterate what I said earlier, if we list the possible coordinate parities of the $$3$$-tuples, we have\n\n$$\\{(e,e,e),(e,e,o),(e,o,e),(e,o,o),(o,e,e),(o,e,o),(o,o,e),(o,o,o)\\}$$\n\nSo there are $$8$$ different parity $$3$$-tuples possible. Also note the addition of even and odd integers;\n\n$$e+e=e,$$ $$o+o=e,$$ $$e+o=o$$\n\nSo in order for your midpoint to be an integer, we need to have the points being added to have coordinates of the same parity. Therefore, since we have 9 points, assuming our first 8 points are all of different parity combinations as above, the 9th must be one of these 8 possibilities. If you add up the two points that have the same coordinate parity structure, this will yield an even number, which is divisible by 2. For example, if both ordered triples have coordinate parity structure $$(e,e,o)$$, then by the midpoint formula,\n\n$$Mid((e,e,o),(e,e,o))=\\left(\\frac{e+e}{2}, \\frac{e+e}{2},\\frac{o+o}{2}\\right)=\\left(\\frac{2e}{2}, \\frac{2e}{2},\\frac{2o}{2}\\right)=(e,e,o)$$\n\nAnd so by the Pigeonhole principle, there is at least 1 midpoint that contains integer coordinates. And as Robert Z noted in his solutions, this can be extended to $$n$$-tuples as well.\n\nHint. Read carefully Eleven-Eleven's comment (or take a look at Pigeon-Hole Principle and 2d grid?) and, by using the Pigeonhole Principle, show the following more general statement: in any set of $$2^n+1$$ points in $$\\mathbb{Z}^n$$ there is at least one pair that has the midpoint in $$\\mathbb{Z}^n$$.", "date": "2019-05-20 08:45:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2946992/integer-midpoint/2947104", "openwebmath_score": 0.7668942809104919, "openwebmath_perplexity": 230.53903131096345, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982013788985129, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8531998761730872}}
{"url": "https://math.stackexchange.com/questions/1754930/lebesgue-measure-preserving-differentiable-function", "text": "# Lebesgue measure-preserving differentiable function\n\nLet $\\lambda$ denote Lebesgue measure and let $f: [0,1] \\rightarrow [0,1]$ be a differentiable function such that for every Lebesgue measurable set $A \\subset [0,1]$ one has $\\lambda(f^{-1}(A)) = \\lambda (A)$. Prove that either $f(x) = x$ or $f(x) = 1- x$.\n\nI will appreciate a hint or a solution that doesn't use ergodic theory as this is an old qual problem in measure theory. $f$ is bounded and continuous and so it is integrable and hence the hypothesis of Lebesgue's Differentiation Theorem are satisfied, but I haven't seen a way to use this or if it is even applicable.\n\n\u2022 To understand why the \"differentiable\" hypothesis is there: examine the function $f(x) = |2x-1|$ on $[0,1]$. \u2013\u00a0GEdgar Apr 23 '16 at 18:15\n\nFirst note $f$ is onto: Otherwise the range of $f$ is a proper subinterval $[a,b],$ which implies $\\lambda(f^{-1}([a,b])) = \\lambda([0,1]) = 1 > b-a = \\lambda([a,b]),$ contradiction.\n\nSuppose $f'(x) = 0$ for some $x\\in [0,1].$ Then $f(B(x,r))\\subset B(f(x),r/4)$ for small $r>0.$ ($B(y,r)$ denotes the open ball with center $y$ and radius $r$ relative to $[0,1].$) Thus for such $r,$ $f^{-1}(B(f(x),r/4))$ contains $B(x,r).$ But $B(f(x),r/4)$ has measure no more than $2\\cdot (r/4) = r/2,$ while $B(x,r)$ has measure at least $r.$ That's a contradiction. Hence $f'(x) \\ne 0$ for all $x\\in [0,1].$\n\nBy Darboux, $f'>0$ on $[0,1]$ or $f'<0$ on $[0,1].$ Let's assume the first case holds. Then $f$ is strictly increasing, hence a bijection on $[0,1].$ Thus if $A$ is an interval, $\\lambda(A) = \\lambda(f^{-1}(f(A)))= \\lambda(f(A)).$ Since $f(0) = 0,$ this gives $x = \\lambda([0,x]) = \\lambda(f([0,x])) = f(x) - f(0) = f(x)$ for all $x\\in [0,1].$\n\n\u2022 This is a much better proof! \u2013\u00a0Giovanni Apr 23 '16 at 16:52\n\nThe following argument should work under the additional assumption that $f'$ is integrable. I don't know if this is something that you need to assume or if it is a limitation of my proof. It is worth mentioning that I need integrability of the derivative for two reasons: first to apply Lebesgue differentiation theorem for $f'$, and second to get that $f$ is absolutely continuous, which I need to apply the Fundamental Theorem of Calculus.\n\nLet $\\mu(\\cdot) = \\lambda(f^{-1}(\\cdot))$. Then $\\mu = \\lambda$ and hence for a.e. $x \\in (0,1)$ we have\n\n$$1 = \\frac{d\\lambda}{d\\mu}(x) = \\lim_{r \\to 0}\\frac{\\lambda(B_r(x))}{\\mu(B_r(x))} = \\lim_{r \\to 0}\\frac{\\int_{B_r(x)}d\\lambda}{\\mu(B_r(x))} = \\lim_{r \\to 0}\\frac{1}{\\lambda(f^{-1}(B_r(x)))}\\int_{f^{-1}(B_r(x))}|f'|d\\lambda,$$ where in the last equality we used the change of coordinates $z = f(y)$. Notice that $\\lambda(f^{-1}(B_r(x))) = \\lambda(B_r(x)) \\to 0$, hence by Lebesgue differentiation theorem, for every $y \\in f^{-1}(x)$ we have that $|f'(y)| = 1.$ Moreover we have that for $y \\in (0,1)$, $y \\in f^{-1}(f(y))$, thus proving that $|f'| = 1$ a.e. in $[0,1]$.\n\nTo conclude the argument, we only need to show that $f'$ cannot change sign. Indeed, by the FTC, we would have that $$f(1) = f(0) + \\int_0^1f'(y) = f(0) \\pm 1,$$ showing that either $f(0) = 0$ and $f' = 1$ or $f(0) = 1$ and $f' = -1$.\n\nTo prove that the derivative is constant a.e., I'll show that $f$ is surjective. By continuity, $f([0,1])$ is a compact subset of $[0,1]$. Suppose by contradiction that $f$ is not surjective, then $\\lambda(f[0,1]) < 1$. To find a contradiction it is enough to notice that $$1 = \\lambda([0,1]) = \\lambda(f^{-1}(f([0,1]))) = \\lambda(f([0,1])) < 1.$$\n\nIn response to Ramiro's comment: Everywhere differentiable + integrable derivative implies absolute continuity, i.e. the FTC holds. Then I can prove what you asked as follows.\n\nNecessarily, $f(0)$ is either $0$ or $1$. To see this notice that if there is $x \\in (0,1)$ such that $f(x) \\in \\{0,1\\}$ then let $y$ be the point such that $f(y) = \\{0,1\\} \\setminus \\{f(x)\\}$. Assume without loss that $x< y$, then $$1 = |f(x) - f(y)| \\le \\int_x^y|f'| = y - x < 1.$$ Then assume without loss that $f(0) = 0$. Reasoning as above shows that $f(1) = 1$. Again by the FTC, $$1 = f(1) - f(0) = \\int_0^1f' = \\lambda(\\{f' = 1\\}) - \\lambda(\\{f' = -1\\}).$$ Finally, if $\\lambda(\\{f' = -1\\}) > 0$ we get a contradiction, hence $f' = 1$ a.e.\n\n\u2022 Why proving that $f$ is surjective implies that the derivative does not change sign? \u2013\u00a0Ramiro Apr 23 '16 at 5:12\n\u2022 @Ramiro: I have edited with a proof, let me know if it is convincing :) \u2013\u00a0Giovanni Apr 23 '16 at 5:41\n\u2022 Thanks for writing down the details. Please note that your statement \"To prove that the derivative does not change sign I'll show that f is surjective.\" is NOT accurate. The fact that f is surjective implies only that $\\lambda({f\\neq 1})=0$ or $\\lambda({f\\neq -1})=0$. This is all you need, but it is weaker than claiming \"the derivative does not change sign\". (In fact, in principle, in a set of measure zero we may even have $|f'|\\neq 1$). \u2013\u00a0Ramiro Apr 23 '16 at 12:48\n\u2022 Thank you for your solution. The problem mentioned that we cannot assume the derivative is continuous. I left that out because we can't in general assume things not given to us and which we cannot deduce easily as parts of the properties of the object given to us. Can we remove the assumption of integrability of the derivative or say why it suffices to consider this case only? \u2013\u00a0akech Apr 23 '16 at 13:34\n\u2022 If you allow me a suugestion, I suggest that instead of writing \"To prove that the derivative does not change sign I'll show that f is surjective\", it would be better to write something like \"To prove that the derivative is constant a.e., I'll show that f is surjective\". \u2013\u00a0Ramiro Apr 23 '16 at 16:57", "date": "2019-06-26 08:19:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1754930/lebesgue-measure-preserving-differentiable-function", "openwebmath_score": 0.9802162647247314, "openwebmath_perplexity": 82.17007802877971, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137874059599, "lm_q2_score": 0.8688267643505194, "lm_q1q2_score": 0.853199861459519}}
{"url": "https://math.stackexchange.com/questions/577597/does-lh%C3%B4pitals-work-the-other-way", "text": "# Does L'H\u00f4pital's work the other way?\n\nH\u0007ello fellows,\n\nAs referred in Wikipedia (see the specified criteria there), L'H\u00f4pital's rule says,\n\n$$\\lim_{x\\to c}\\frac{f(x)}{g(x)}=\\lim_{x\\to c}\\frac{f'(x)}{g'(x)}$$\n\nAs\n\n$$\\lim_{x\\to c}\\frac{f'(x)}{g'(x)}= \\lim_{x\\to c}\\frac{\\int f'(x)\\ dx}{\\int g'(x)\\ dx}$$\n\nJust out of curiosity, can you integrate instead of taking a derivative? Does\n\n$$\\lim_{x\\to c}\\frac{f(x)}{g(x)}= \\lim_{x\\to c}\\frac{\\int f(x)\\ dx}{\\int g(x)\\ dx}$$\n\nwork? (given the specifications in Wikipedia only the other way around: the function must be integrable by some method, etc.) When? Would it have any practical use? I hope this doesn't sound stupid, it just occurred to me, and I can't find the answer myself.\n\n## Edit\n\nTake 2 functions $f$ and $g$. When is\n\n$$\\lim_{x\\to c}\\frac{f(x)}{g(x)}= \\lim_{x\\to c}\\frac{\\int_x^c f(a)\\ da}{\\int_x^c g(a)\\ da}$$\n\ntrue?\n\nNot saying that it always works, however, it sometimes may help. Sometimes one can apply l'H\u00f4pital's even when an indefinite form isn't reached. Maybe this only works on exceptional cases.\n\nMost functions are simplified by taking their derivative, but it may happen by integration as well (say $\\int \\frac1{x^2}\\ dx=-\\frac1x+C$, that is simpler). In a few of those cases, integrating functions of both nominator and denominator may simplify.\n\nWhat do those (hypothetical) functions have to make it work? And even in those cases, is is ever useful? How? Why/why not?\n\n\u2022 You cannot use L'Hospital the other way to evaluate a limit. You can try an example, as I did and it just won't work. Realizing that taking a derivative of num and denom really means that you are linearizing and so that the ratio of the slopes (i.e. derivatives) will give you the answer to the limit, makes sense. But when you are anti deriving, what does that geometrically mean to the limit? I am not aware that \"anti-L'Hospitaling\" does anything... \u2013\u00a0imranfat Nov 22 '13 at 21:14\n\u2022 @JMCF125: This is a good question. Perhaps you should clarify it by adding bounds to the integrals. Specifically, how about writing $\\int_{c}^{x}\\; f(t)\\; dt$ instead of $\\int f(x) dx$? Same for $g(x)$. I point out that neither answer addressed this question. \u2013\u00a0Chris K Nov 22 '13 at 21:18\n\u2022 @ChrisK, good, then the constants are eliminated! Forgot about them... I'll add that ASAP- \u2013\u00a0JMCF125 Nov 22 '13 at 21:33\n\u2022 @JMCP125 : based on Umberto P.'s answer, the answer seems to be \"yes\". The next question is, is this ever useful, like the \"regular\" L'Hopital's Rule itself is? \u2013\u00a0Stefan Smith Nov 23 '13 at 2:28\n\u2022 @miracle173, I never said l'H\u00f4pital's implied this. And I didn't say the integrals had to be real anti-derivatives; in fact, in the edit, I added ChrisK's suggestion, which clears that up. \u2013\u00a0JMCF125 Nov 24 '13 at 8:52\n\nI recently came across a situation where it was useful to go through exactly this process, so (although I'm certainly late to the party) here's an application of L'H\u00f4pital's rule in reverse:\n\nWe have a list of distinct real numbers $\\{x_0,\\dots, x_n\\}$. We define the $(n+1)$th nodal polynomial as $$\\omega_{n+1}(x) = (x-x_0)(x-x_1)\\cdots(x-x_n)$$ Similarly, the $n$th nodal polynomial is $$\\omega_n(x) = (x-x_0)\\cdots (x-x_{n-1})$$ Now, suppose we wanted to calculate $\\omega_{n+1}'(x_i)/\\omega_{n}'(x_i)$ when $0 \\leq i \\leq n-1$. Now, we could calculate $\\omega_{n}'(x_i)$ and $\\omega_{n+1}'(x_i)$ explicitly and go through some tedious algebra, or we could note that because these derivatives are non-zero, we have $$\\frac{\\omega_{n+1}'(x_i)}{\\omega_{n}'(x_i)} = \\lim_{x\\to x_i} \\frac{\\omega_{n+1}'(x)}{\\omega_{n}'(x)} = \\lim_{x\\to x_i} \\frac{\\omega_{n+1}(x)}{\\omega_{n}(x)} = \\lim_{x\\to x_i} (x-x_{n+1}) = x_i-x_{n}$$ It is important that both $\\omega_{n+1}$ and $\\omega_n$ are zero at $x_i$, so that in applying L'H\u00f4pital's rule, we intentionally produce an indeterminate form. It should be clear though that doing so allowed us to cancel factors and thus (perhaps surprisingly) saved us some work in the end.\n\nSo would this method have practical use? It certainly did for me!\n\nPS: If anyone is wondering, this was a handy step in proving a recursive formula involving Newton's divided differences.\n\n\u2022 Very interesting, +1! I certainly didn't expect it to have been successfully used before. But why didn't you make the question you are answering when you found it worked? \u2013\u00a0JMCF125 Nov 27 '13 at 11:01\n\u2022 Ah, after having reread your question, I think the technique you're asking about is a little different. In my case, I went from $f'/g'$ (where both $f'$ and $g'$ were non-zero) to $f/g$ (where both $f$ and $g$ were zero). I didn't use the reverse L'H\u00f4pital to resolve an indeterminate form; I used it to make one. \u2013\u00a0Omnomnomnom Nov 27 '13 at 15:38\n\u2022 However, when you apply the function $\\omega_n(x)$ and cancel out $\\Pi_{i=0}^n (x-x_i)$ (sorry, couldn't resist but to use big product notation :)), the \"indeterminacy\" disappears, it only appears as an intermediate step. Call $\\omega'$ as $\\alpha$ and $$\\lim_{x\\to x_i} \\frac{\\alpha_{n+1}(x)}{\\alpha_n(x)} = \\lim_{x\\to x_i} \\left(\\lim_{c\\to x}\\frac{\\int\\alpha_{n+1}(z)\\ dz}{\\lim_{c\\to x}\\int\\alpha_{n}(z)\\ dz}\\right)=x_i-x_{n}$$. Right? (didn't check it thoroughly, there's no comment preview) \u2013\u00a0JMCF125 Nov 27 '13 at 21:07\n\u2022 Exactly! So, you could say that the trick in using L'Hopital's rule backwards is choosing the antiderivatives of $\\alpha_{n+1}$ and $\\alpha_n$ that give you the indeterminate fraction. I'm not sure what the ruole of $c$ is in your statement, though \u2013\u00a0Omnomnomnom Nov 27 '13 at 21:16\n\u2022 About the $c$, it's to eliminate the constants, see the problem Arkamis referred and the answer of Kaz, which has a related idea. The point is to make the integral so small it would be alike a derivative, only as a different expression. (BTW, the second $c\\to x$ limit was not suppose to be there, I pasted it twice) \u2013\u00a0JMCF125 Nov 27 '13 at 21:55\n\nWith L'Hospital's rule your limit must be of the form $\\dfrac 00$, so your antiderivatives must take the value $0$ at $c$. In this case you have $$\\lim_{x \\to c} \\frac{ \\int_c^x f(t) \\, dt}{\\int_c^x g(t) \\, dt} = \\lim_{x \\to c} \\frac{f(x)}{g(x)}$$ provided $g$ satisfies the usual hypothesis that $g(x) \\not= 0$ in a deleted neighborhood of $c$.\n\n\u2022 Good, interpreted the question in the \"right\" way. \u2013\u00a0Andr\u00e9 Nicolas Nov 22 '13 at 21:26\n\u2022 Or $\\frac{\\infty}{\\infty}$. \u2013\u00a0marty cohen Nov 23 '13 at 2:56\n\u2022 Very important, this only works if you know that $\\lim \\frac{f(x)}{g(x)}$ exists. Even if the first limit exists, you cannot conclude that the second one does. \u2013\u00a0N. S. Nov 23 '13 at 5:58\n\u2022 @N.S., neither can you conclude the division of the derivatives exists, even when the division of the original functions does, as in the original l'H\u00f4pital's. \u2013\u00a0JMCF125 Nov 23 '13 at 13:48\n\u2022 @JMCF125 True, but the goal in L'H is: starting with a limit, you replace by a different limit. In L'H you replace the functions by the Derivatives, not the derivatives by the functions. In classical L'H, you check $\\lim\\frac{f'}{g'}$ and if this limit exists you are done. Here you check $\\lim \\frac{\\int f}{\\int g}$ and if the limit exists, you cannot say anything. Note that if you go the other direction, you just apply classical L'H \u2013\u00a0N. S. Nov 23 '13 at 15:40\n\nNo it does not, for example consider the limit $\\lim_{x\\to 0}\\frac{\\sin x}{x}$, L'H\u00f4pital's gives you 1, but reverse L'H\u00f4pital's goes to $-\\infty$ (without introducing extra constants).\n\n\u2022 +1 for being a simple, easy counterexample, although I assume that you're taking the integral of sin(x) to be -cos(x) - but that implies either a constant or a definite interval on the integral. \u2013\u00a0fluffy Nov 23 '13 at 4:33\n\u2022 @fluffy : Yes, exactly, thank you \u2013\u00a0Arjang Nov 23 '13 at 4:43\n\u2022 The conditions under which it works may be more restricted... \u2013\u00a0JMCF125 Nov 23 '13 at 17:20\n\u2022 If we want to use reverse L'Hopital, it is our 'responsibility' to ensure that the newly integrated function is also of an indeterminate form, and that can be tackled easily using the integration constant. Here, instead of -cosx, one should use 1-cosx (in the numerator) for the limit to exist, and hence evaluate it to be 1. \u2013\u00a0arya_stark May 8 '18 at 8:08\n\nRecall that the integral of a function requires the inclusion of an arbitrary constant. So, if $f(x) = x$, then $\\int f(x)\\ dx = \\frac12 x^2 + C$.\n\nThen, assuming your rule holds, then $$\\lim_{x\\to 0} \\frac{x}{x} = \\lim_{x \\to 0} \\frac{\\int x\\ dx}{\\int x\\ dx} = \\lim_{x \\to 0} \\frac{\\frac12 x^2 + C_f}{\\frac12 x^2 + C_g} = \\frac{C_f}{C_g}.$$\n\nThis means you can get literally any value you wish.\n\n\u2022 What if we forget the constants? Wait, don't answer yet, I'll edit my question first. \u2013\u00a0JMCF125 Nov 22 '13 at 21:32\n\u2022 @JMCF125 Note that it really makes no sense to talk of \"forgetting the constants\". For example, the following three functions are antiderivatives of $2\\sin(x)\\cos(x)$: $\\sin^2(x)$, $-\\cos^2(x)$, $-\\cos(2x)/2$. Which one is the \"correct\" one to use if we are to \"forget constants\"? \u2013\u00a0Andr\u00e9s E. Caicedo Nov 22 '13 at 21:59\n\u2022 @AndresCaicedo, what if we use a definite integral $\\displaystyle\\int_0^x x\\ dx=\\frac{x^2}2$, as referred in my edit. That way, wouldn't the result be the same? \u2013\u00a0JMCF125 Nov 24 '13 at 8:57\n\u2022 Why use $0$ as the lower bound. Also, it should be $\\int_0^x t\\ dt$, as it makes no sense to have the variable of integration be the same as a bound. In any case, a lower bound of $c$ makes some sense if $x \\to c$, since, as others have said, one of the ways this question makes sense is if $f, g \\to 0$ at $x=c$. But it elides the case of $f,g \\to \\infty$. \u2013\u00a0Emily Nov 24 '13 at 16:13\n\u2022 @Arkamis, \u00ab(...) as it makes no sense to have the variable of integration be the same as a bound.\u00bb, sorry, distraction. :S I meant for the top limit to be an $a$. I used the lower limit as $0$ so that in your example, the integral would be $\\int^a_0 x\\ dx=\\frac{a^2}2+C-\\frac{0^2}2-C=\\frac{a^2}2$. Isn't it right like this? \u2013\u00a0JMCF125 Nov 27 '13 at 10:57\n\nIf the limit of the ratio of those functions exists, I suspect that it may be possible for a certain definite integral to have the same limit. Something like:\n\n$$\\lim_{x\\to \\infty}\\frac{\\int_x^{x+1}f(x)\\ dx}{\\int_x^{x+1}g(x)\\ dx}$$\n\nThe idea is that we grow $x$, and compare the ratios of some definite integrals in the neighborhood of x, of equal width and position.\n\nFor instance, if we imagine the special case that the functions separately converge to constant values like 4 and 3, then if we take $x$ far enough, then we are basically just dividing 4 by 3. A definite integral of width 1 of a function that converges to 4 has a value that is approximately four, near a domain value that is large enough.\n\nBut this is like a derivative in disguise. If we divide a definite integral by the interval length, and then shrink the interval, we are basically doing derivation to obtain the original function that was integrated. Except we aren't shrinking the interval, since we don't have to: the fixed width 1 becomes smaller and smaller in relation to the value of $x$, so it is a \"quasi infinitesimal\", so to speak.\n\nThe real problem with this, even if it works, is that integrals don't seem to offer any advantage. Firstly, even if the function has the properties of being integrable, it may not be symbolically integrable, or it may be hard to integrate. Integration produces something more complex than the integrand.\n\nThe advantage of L'H\u00f4pital's Rule is that we can reduce the power of the functions. If they are polynomials, they lose a degree in the differentiation, which can be helpful, and gives us a basis for instantly gauging the limit of the ratios of polynomials of equal degree simply by looking at the ratios of their highest degree coefficients. And then certain common functions at least don't grow any additional hair under differentiation, like sine, cosine, e to the x.\n\n\u2022 Interesting, +1. But what about those exceptional functions whose derivative is more complicated? Wouldn't the integral ease calculation? \u2013\u00a0JMCF125 Nov 23 '13 at 13:55\n\nL'Hospital's Rule is really a compact way of looking at ratio of the Taylor's polynomial for the numerator and denominator functions (w/o remainder). This polynomial is an exact fit to the function and its derivatives at the indicated point and close in a small neighborhood thereof.\n\nWhen the limit is 0/0 that means that both Taylor's polynomials have 0 constant terms. So then you look at the x terms (which represent the derivatives). If you can pick out the limit from there, good -- if they are both zero you go on to the $x^2$ terms (if any).\n\nSo L'Hospital's Rule isn't some vague thing that was plucked out of the air and happens to use derivatives. Any method of evaluating limits ought to be based on a clear mathematical derivation. You can use integrals or gamma functions or field theory or anything that pleases you, as long as you show you have a sound basis for evaluating the limit that way.\n\n\u2022 Can you help me finding such a basis? \u2013\u00a0JMCF125 Nov 28 '13 at 12:26\n\u2022 @ Betty Mock What is exact permissible basis to modify the vanishing numerator and denominator ? \u2013\u00a0Narasimham Nov 20 '14 at 20:11\n\u2022 Expand the top and bottom in a Taylor's polynomial. If the first term of both polynomials is zero, then you look at the second term. If both of those are zero, you go on to the third term, etc. Sooner or later you get a term which is not zero, and that is the one which shows you the limit. \u2013\u00a0Betty Mock Nov 22 '14 at 2:46\n\n$$\\lim_{ x \\to 0} \\frac{ sin \\,x \\,}{\\,x}= \\lim_{ x \\to 0} \\frac{\\int sin \\,x \\,\\ dx}{\\int x\\ dx} = \\lim_{x \\to 0} \\frac{\\ (- cos \\, x)}{( x^2/2)-1} =\\lim_{x \\to 0} \\frac{\\ (-cos \\, x)+1}{( x^2/2)} = 1$$\n\nConstants should be appropriately chosen from function/ power series expansion of the integrands and distributed between numerator and denominator maintaining the balance.\n\nThe statement is vague but I am sure it can be properly worded/stated to include even second, third integrands in the numerator and denominator.\n\nMore simply:\n\nWhen $\\frac yx =const = m$ as L'Hospital limit, it is =$\\frac{dy}{dx}$ by Quotient Rule.\n\n$m = \\frac {dy}{ dx} = \\frac {y \\, \\pm C_1 }{ x} =\\frac {y \\, }{ x \\pm C_2}.$", "date": "2019-09-16 02:50:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/577597/does-lh%C3%B4pitals-work-the-other-way", "openwebmath_score": 0.9108422994613647, "openwebmath_perplexity": 414.6755314518558, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9489172601537141, "lm_q2_score": 0.8991213860551286, "lm_q1q2_score": 0.8531918022010425}}
{"url": "http://mathematica.stackexchange.com/questions/6052/improve-speed-for-calculating-a-recursive-sequence/6060", "text": "# Improve speed for calculating a recursive sequence\n\nI want to calculate following recursive sequence:\n\n$\\alpha_{0}=0,\\\\ \\cos(\\alpha_{i})=\\cos(\\alpha_{i-1})\\cdot\\cos(\\beta_{i})+\\sin(\\alpha_{i-1})\\cdot\\sin(\\beta_{i})\\cdot\\cos(\\gamma_{i}).$\n\nIn mathematica 8.0.1 I try to do this with:\n\nendRecursion = 20;\nbeta = Import[\"beta.txt\", \"List\"];\nbeta = beta[[1;;endRecursion]];\ngamma = RandomReal[{0,2*Pi}, endRecursion];\n\nalpha[0]:= 0;\nalpha[i_]:=\nArcCos[Cos[alpha[i-1]]*Cos[beta[[i]]]+\nSin[alpha[i-1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]];\n\nresult=Table[alpha[i], {i, 1, endRecursion}];\n\n\nThe values for beta are imported because they are generated by another formula which is not important for my question here. Values for beta can be found here.\n\nThe problem is that the calculation of the result is very slowly. It takes about 14s for just 20 iterations! And I need about 300 iterations (times ~10000, because I have several lists of beta values).\n\nI recognized that the calculation is much faster if I skip the Cos[alpha[i-1]] in the first or the Sin[alpha[i-1]] in the 2nd summand. Then it takes just ~ 0.01 s for 20 iterations. But unfortunately this falsifies my result. I cannot understand why it is so much faster if there is just one time alpha[i-1] on the right side of the equation. Am I doing something wrong? Is there any way to increase the speed of this calculation? I would be glad about any help!\n\n-\nHave you tried caching? alpha[i_] := alpha[i] = ArcCos[Cos[alpha[i-1]]*Cos[beta[[i]]] + Sin[alpha[i-1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]] \u2013\u00a0J. M. May 26 '12 at 12:44\nI am such a *** idiot! Thank you so much for this comment! I totally forgot the caching although I used it some time ago (5 years, perhaps too long time ago). Unfortunately, I did not see this trick when I searched for a solution, otherwise I would not have asked this question... Thank you anyway! You saved my weekend! How can I accept your comment as answer? \u2013\u00a0partial81 May 26 '12 at 12:58\n@J.M. hmm, in my amazement that nobody had answered this, I wrote my answer without seeing your comment... Hope you were not writing up one also! \u2013\u00a0acl May 26 '12 at 13:07\nYou can accept acl's answer. :) I didn't answer because I can't test at the moment. \u2013\u00a0J. M. May 26 '12 at 13:07\nOk, if you do not mind! Thanks again! \u2013\u00a0partial81 May 26 '12 at 13:12\n\nYou can speed it up by \"memoization\", i.e., remembering previous values of alpha[i]:\n\nendRecursion = 20;\nbeta = Import[\"beta.txt\", \"List\"];\nbeta = beta[[1 ;; endRecursion]];\ngamma = RandomReal[{0, 2*Pi}, endRecursion];\nalpha[0] := 0;\nalpha[i_] :=\nalpha[i] =\nArcCos[Cos[alpha[i - 1]]*Cos[beta[[i]]] +\nSin[alpha[i - 1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]];\nresult = Table[alpha[i], {i, 1, endRecursion}];\n\n\n-\nThanks acl and @J.M. for answering my question so quickly! \u2013\u00a0partial81 May 26 '12 at 13:14\nThere was a mistake: you were importing strings (you didn't convert to numbers after StringSplit.) I fixed it. \u2013\u00a0Szabolcs May 26 '12 at 13:15\n@Szabolcs yes, I had just realised it and our edits almost overlapped... thanks \u2013\u00a0acl May 26 '12 at 13:16\nOkay, I'll remove the comments then. \u2013\u00a0Szabolcs May 26 '12 at 13:20\n\nAlthough you can always implement recursion by hand the way you are trying to do, there are also some specialized functions that are designed to make your life a little easier. In particular, there is FoldList.\n\nThis approach is also slightly faster than the manual recursion approach (assuming you start from a clean slate):\n\nendRecursion = 20;\nbeta = Import[\"beta.txt\", \"List\"];\nbeta = beta[[1 ;; endRecursion]];\ngamma = RandomReal[{0, 2*Pi}, endRecursion];\nFoldList[ArcCos[Cos[#1]*#2[[1]] + Sin[#1]*#2[[2]]] &, 0,\nTranspose[{Cos[beta], Sin[beta] Cos[gamma]}]]\n\n\nAfter initializing the lists as in your example, I need only a single statement to generate the list.\n\nThe folding function is what appears in the original recursive formulation, but it is written as a pure function with two formal arguments #1 and #2.\n\nThe first argument, #1, refers to your alpha and its starting value is passed to FoldList as the second argument (0). The second argument in the pure function (#2) refers to the elements of the list Transpose[{Cos[beta], Sin[beta] Cos[gamma]}] in which I collected the pre-existing lists beta and gamma (but in a form that is already pre-processed to avoid having to evaluate cosines and sines individually in the recursion).\n\n-\nWith this particular formulation, one could even simplify the FoldList[] expression as FoldList[ArcCos[{Cos[#1], Sin[#1]}.#2] &, 0, Transpose[{Cos[beta], Sin[beta] Cos[gamma]}]]. One might even consider using the Weierstrass substitution here, since the beta values are all $< \\pi/2$... \u2013\u00a0J. M. May 26 '12 at 18:47\n@J.M. That's correct - the dot product may give a tiny additional speed boost but I didn't check that. \u2013\u00a0Jens May 27 '12 at 0:50\nThanks @Jens for this additional answer. FoldList is something I have never used so far \u2013 that will change now (although the additional speed boost is just small). \u2013\u00a0partial81 May 27 '12 at 11:24\n@ Jens, Let\u2019s say: alpha[i_]:=alpha[i]=ArcCos[Cos[alpha[i - 1]]*Cos[beta[[i+1]]]+Sin[alpha[i - 1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]]. How do I have to adapt your solution with FoldList for this case? I really have problems to see a solution because now I would need in my iteration the i+1 element of beta. I would be happy if you could give me a hint again! \u2013\u00a0partial81 May 31 '12 at 13:10\n@partial81 The term you changed is the first one in Transpose[{Cos[beta], Sin[beta] Cos[gamma]}], and the change is simply a shift in the index. The fastest way this can be implemented is to use RotateLeft: Transpose[{RotateLeft[Cos[beta]], Sin[beta] Cos[gamma]}] \u2013\u00a0Jens May 31 '12 at 15:39\n\nThe function runs slowly because each iteration calls alpha twice the number of times of the previous iteration. Thus, $n$ iterations invoke alpha $2^n$ times -- exponential run time! Fortunately, the two nested calls to alpha are identical. If we change alpha to call itself only once instead of twice, the algorithm runs in linear time while still retaining simple recursive form:\n\nendRecursion = 20;\nbeta = Import[\"beta.txt\", \"List\"];\nbeta = beta[[1;;endRecursion]];\ngamma = RandomReal[{0,2*Pi}, endRecursion];\n\nalpha[0] = 0;\nalpha[i_]:=\nModule[{a = alpha[i-1], b = beta[[i]], g = gamma[[i]]}\n, ArcCos[Cos[a]*Cos[b]+Sin[a]*Sin[b]*Cos[g]]\n]\n\nresult = Table[alpha[i], {i, 1, endRecursion}];\n\n\nEven so, I agree with @Jens that functional iteration is frequently preferred over recursion in order to avoid hitting the $RecursionLimit (or$IterationLimit).\n\n-\nThanks @WReach for answering the other part of my question. Now I understand why it was so slowly at the beginning. And it is nice that you give a link to the useful functional iterations. \u2013\u00a0partial81 May 27 '12 at 20:53", "date": "2016-06-28 09:52:46", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/6052/improve-speed-for-calculating-a-recursive-sequence/6060", "openwebmath_score": 0.7391323447227478, "openwebmath_perplexity": 1570.7435202294744, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9489172644875641, "lm_q2_score": 0.8991213671331905, "lm_q1q2_score": 0.8531917881423459}}
{"url": "http://math.stackexchange.com/questions/423736/finding-the-exact-value", "text": "# Finding the Exact Value\n\nHow would one go about finding the exact value of $\\theta$ in the following:$\\sqrt{3}\\tan \\theta -1 =0$? I am unsure of how to begin this question. Any help would be appreciated!\n\n-\nare you familiar with $\\arctan$ function? \u2013\u00a0Alex Jun 18 '13 at 16:15\nI am not familiar with that function \u2013\u00a0ComradeYakov Jun 18 '13 at 16:16\n@Tim I meant the first, sorry for the confusion. It isn't cubed \u2013\u00a0ComradeYakov Jun 18 '13 at 16:17\n\nHint: $\\sqrt{3}\\tan \\theta -1 =0\\implies \\sqrt{3}\\tan \\theta =1\\implies \\tan \\theta=1/\\sqrt{3}$\n\nRecall from SOH CAH TOA, that $\\tan \\theta=\\text{opposite}/\\text{adjacent}$, so in this case $1$ is the opposite side and $\\sqrt{3}$ is the adjacent side.\n\nThe above is a special triangle. Since you know that $\\tan \\theta=1/\\sqrt{3}$, it follows that the angle $\\theta=\\pi/6$. However since $\\tan \\theta$ is a trigonometric periodic function, that value will come up again, and again. (Just like $\\sin0=\\sin2\\pi).$\n\n$\\tan \\theta$ has a period of $\\pi.$\n\nSo to be fully accurate, $\\theta=\\pi/6 +n\\pi, n \\in \\mathbb{Z}$\n\n-\nThank you! So theta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution? \u2013\u00a0ComradeYakov Jun 18 '13 at 16:31\ntheta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution. \u2013\u00a0Sujaan Kunalan Jun 18 '13 at 16:35\nOk, thanks for your help! \u2013\u00a0ComradeYakov Jun 18 '13 at 16:37\nYou are welcome! \u2013\u00a0Sujaan Kunalan Jun 18 '13 at 16:38\n$2\\cos^{2} \\theta+\\cos \\theta-1=0 \\implies 2\\cos^{2}+\\cos \\theta =1 \\implies \\cos \\theta (2\\cos \\theta+1)=1\\implies \\cos \\theta =1 \\implies \\theta =2n\\pi, n\\in \\mathbb{Z}$ However, $\\cos \\theta(2\\cos \\theta +1)=1$ also implies that $2\\cos \\theta+1=1 \\implies 2\\cos \\theta =0 \\implies \\cos \\theta =0 \\implies \\theta = (2n-1)/\\pi, n \\in \\mathbb{Z}$ \u2013\u00a0Sujaan Kunalan Jun 18 '13 at 17:07\n\nIf $\\sqrt{3}\\tan \\theta - 1= 0$ then $\\sqrt{3}\\tan\\theta = 1$ and $\\tan\\theta = 1/\\sqrt{3}$. Hence $\\theta =\\arctan(1/\\sqrt{3}) = \\frac{\\pi}{6}$.\n\nNotice that the graph $y = \\tan\\theta$ repeats every $\\pi$-radians. Hence, all solutions are given by\n\n$$\\theta = \\frac{\\pi}{6} + \\pi n$$\n\nwhere $n$ is any whole number.\n\n-", "date": "2016-05-25 21:16:24", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/423736/finding-the-exact-value", "openwebmath_score": 0.958959698677063, "openwebmath_perplexity": 587.1891803579929, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9902915229454736, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8531739854931767}}
{"url": "https://stats.stackexchange.com/questions/247859/does-the-square-of-the-minimum-of-two-correlated-normal-variables-have-a-chi-squ", "text": "# Does the square of the minimum of two correlated Normal variables have a chi-squared distribution?\n\nIf the random variables $X$ and $Y$ are normally distributed with mean $\\mu = 0$ and standard deviation $\\sigma = 1$, then define the random variable $Z = \\min(X, Y )$. The problem is to prove that $Z^2$ follows a gamma distribution with parameters $(\\alpha = 1/2, \\lambda= 1/2)$ (i.e. a chi-squared distribution).\n\nI have: $$P(Z^2 \\le z) = P((\\min\\{X,Y\\})^2 \\le z) =P\\left(-\\sqrt{z}\\le \\min(X,Y)\\le \\sqrt{z}\\right) \\\\ = P\\left(-\\sqrt{z}\\le X\\le \\sqrt{z}, -\\sqrt{z}\\le Y\\le \\sqrt{z}\\right)$$ But I am not sure where to go from here, since the random variables aren't assumed to be independent. I would appreciate any hints!\n\n\u2022 Your logic at the last step seems to go astray. If the minimum of (X,Y) is within $\\pm \\sqrt z$ (as you have in your second-last line), that doesn't make both of them in that interval, as you have in your last line. You only need one of the two of them in there. \u2013\u00a0Glen_b -Reinstate Monica Nov 28 '16 at 3:57\n\u2022 @glen_b What then would be the interval on the other one? \u2013\u00a0Mjt Nov 28 '16 at 14:08\n\u2022 That's not a productive way to approach writing the event. But first you might want to ponder whether anything has been left out of your framing of the question (which is possibly why whuber is able to offer a counterexample). \u2013\u00a0Glen_b -Reinstate Monica Nov 28 '16 at 15:55\n\nThe result is not true.\n\nAs a counterexample, let $(X,Y)$ have standard Normal margins with a Clayton copula, as illustrated at https://stats.stackexchange.com/a/30205. Generating 10,000 independent realizations of this bivariate distribution, as shown in the lower left of the figure, produces 10,000 realizations of $Z^2$ that clearly do not follow a $\\Gamma(1/2,1/2)$ distribution (in a Chi-squared test of fit, $\\chi^2=121, p \\lt 10^{-16}$). The qq plots in the top of the figure confirm that the marginals look standard Normal while the qq plot at the bottom right indicates the upper tail of $Z^2$ is too short.\n\nThe result can be proven under the assumption that the distribution of $(X,Y)$ is centrally symmetric: that is, when it is invariant upon simultaneously negating both $X$ and $Y$. This includes all bivariate Normals (with mean $(0,0)$, of course).\n\nThe key idea is that for any $z \\ge 0$, the event $Z^2 \\le z^2$ is the difference of the events $X \\ge -z\\cup Y \\ge -z$ and $X \\ge z \\cup Y \\ge z$. (The first is where the minimum is no less than $-z$, while the second will rule out where the minimum exceeds $z$.) These events in turn can be broken down as follows:\n\n$$\\Pr(Z^2 \\le z^2) = \\Pr(X\\ge -z) - \\Pr(Y \\le -z) + \\Pr(X,Y\\lt -z) - \\Pr(X,Y\\gt z).$$\n\nThe central symmetry assumption assures the last two probabilities cancel. The first two probabilities are given by the standard Normal CDF $\\Phi$, yielding\n\n$$\\Pr(Z^2 \\le z^2) =1 - 2\\Phi(-z).$$\n\nThat exhibits $Z$ as a half-normal distribution, whence its square will have the same distribution as the square of a standard Normal, which by definition is a $\\chi^2(1)$ distribution.\n\nThis demonstration can be reversed to show $Z^2$ has a $\\chi^2(1)$ distribution if and only if $\\Pr(X,Y\\le -z) = \\Pr(X,Y\\ge z)$ for all $z\\ge 0$.\n\nHere is the R code that produced the figures.\n\nlibrary(copula)\n\nn <- 1e4\nset.seed(17)\nxy <- qnorm(rCopula(n, claytonCopula(1)))\ncolnames(xy) <- c(\"X\", \"Y\")\nz2 <- pmin(xy[,1], xy[,2])^2\n\ncutpoints <- c(0:10, Inf)\nz2.obs <- table(cut(z2, cutpoints))\nz2.exp <- diff(pgamma(cutpoints, 1/2, 1/2))\nrbind(Observed=z2.obs, Expected=z2.exp * length(z2))\nchisq.test(z2.obs, p=z2.exp)\n\npar(mfrow=c(2,2))\nqqnorm(xy[,1], ylab=\"X\"); abline(c(0,1), col=\"Red\", lwd=2)\nqqnorm(xy[,2], ylab=\"Y\"); abline(c(0,1), col=\"Red\", lwd=2)\n\nplot(xy, pch=19, cex=0.75, col=\"#00000003\", main=\"Data\")\n\nqqplot(qgamma(seq(0, 1, length.out=length(z)), 1/2, 1/2), z2,\nxlab=\"Theoretical Quantiles\", ylab=\"Z2\",\nmain=\"Gamma(1/2,1/2) Q-Q Plot\")\nabline(c(0,1), col=\"Red\", lwd=2)\n\n\nLet $M=min(X,Y)^2$, $$P(M<m) = P(M<m,X<Y) + P(M<m,X>Y) ~~~~~~~~~~~\\\\ = P(M<m|X<Y)P(X<Y) + P(M<m|X>Y)P(X>Y) \\\\ = P(X^2<m)P(X<Y) + P(Y^2<m)P(X>Y) ~~~~~~~~~~~~~~~~~~~~~~\\\\ = \\frac{1}{2}P(X^2<m) + \\frac{1}{2}P(Y^2<m) \\quad \\quad \\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad~ \\\\ =P(X^2<m) \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad ~~~$$", "date": "2019-12-06 13:49:52", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/247859/does-the-square-of-the-minimum-of-two-correlated-normal-variables-have-a-chi-squ", "openwebmath_score": 0.8417906165122986, "openwebmath_perplexity": 435.74015023447146, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765587105448, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.85316491347251}}
{"url": "https://web2.0calc.com/questions/direct", "text": "+0\n\n# Direct\n\n0\n330\n6\n\nthe time needed to paint a fence varies directly with the length if the fence and indirectly with the number of painters. if it takes 5 hours to paint 200 feet of fence with 3 painters, how long will it take 5 painters to paint 500 feet of fence?\n\nGuest\u00a0May 26, 2015\n\n#3\n+91510\n+10\n\nok we might as well all answer this question!\n\nThis is how I was trained to do it\n\n$$\\\\T\\alpha \\frac{L}{N} \\\\\\\\ T=k* \\frac{L}{N}\\qquad but\\\\\\\\ 5=k* \\frac{200}{3}\\qquad so\\\\\\\\ k=\\frac{3}{40}\\qquad so\\\\\\\\ T=\\frac{3L}{40N}\\\\\\\\ When N=5 and L=500\\\\\\\\ T=\\frac{3*500}{40*5}\\\\\\\\ T=7.5 \\;\\;hours$$\n\nMelody \u00a0May 26, 2015\nSort:\n\n#1\n+26412\n+10\n\nThe rate at which fence painting is done is 200/(3*5) feet per painter-hour.\n\nIn order to paint 500 feet we need 500*3*5/200 painter-hours\n\nWith 5 painters the time taken is 500*3*5/(200*5) hours = 7.5 hours\n\n.\n\nAlan \u00a0May 26, 2015\n#2\n+81090\n+10\n\nIf 3 painters paint 200 ft of fence in 5 hours, in 1 hour, they must paint 40 ft.......so each painter paints 40/3 ft per hour.\n\nThen, 5 painters can paint 200/3 ft every hr........so........500/ (200/3)\u00a0 = 1500/200\u00a0 = 7.5 hrs.\n\nCPhill \u00a0May 26, 2015\n#3\n+91510\n+10\n\nok we might as well all answer this question!\n\nThis is how I was trained to do it\n\n$$\\\\T\\alpha \\frac{L}{N} \\\\\\\\ T=k* \\frac{L}{N}\\qquad but\\\\\\\\ 5=k* \\frac{200}{3}\\qquad so\\\\\\\\ k=\\frac{3}{40}\\qquad so\\\\\\\\ T=\\frac{3L}{40N}\\\\\\\\ When N=5 and L=500\\\\\\\\ T=\\frac{3*500}{40*5}\\\\\\\\ T=7.5 \\;\\;hours$$\n\nMelody \u00a0May 26, 2015\n#4\n+81090\n0\n\nWhich can we do faster.......paint that fence or answer this question?????\n\n{ I say its a toss up..... }\n\nCPhill \u00a0May 26, 2015\n#5\n+72\n+5\n\nDepends on how much \u00a0You can take 60 seconds on writing this while 70 painters paint a small fence 3*3 fence\n\nxXmathXx \u00a0May 26, 2015\n#6\n+72\n0\n\nHello CPhill\n\nxXmathXx \u00a0May 26, 2015\n\n### 24 Online Users\n\nWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. \u00a0See details", "date": "2018-01-23 17:25:01", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/direct", "openwebmath_score": 0.71524977684021, "openwebmath_perplexity": 7041.227092407104, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9799765616345254, "lm_q2_score": 0.8705972616934408, "lm_q1q2_score": 0.8531649110827711}}
{"url": "https://puzzling.stackexchange.com/questions/41436/minimum-steps-on-a-numberline", "text": "# Minimum steps on a numberline\n\nGiven an infinite numberline, you start at zero. On every i'th move you can either move i places to the right, or i places to the left. How, in general, would you calculate the minimum number of moves to get to a target point x? For example:\n\nif x = 9:\n\nmove 1: starting at zero, move to 1\n\nmove 2: starting at 1, move to 3\n\nmove 3: starting at 3, move to 0\n\nmove 4: starting at 0, move to 4\n\nmove 5: starting at 4, move to 5\n\n\u2022 This seems like another mathbook-type problem to me. \u2013\u00a0Mithical Aug 25 '16 at 11:47\n\u2022 @Mithrandir I think so. VTC. \u2013\u00a0IAmInPLS Aug 25 '16 at 11:50\n\u2022 I say \"don't close\" -- unless it is mandatory to have a real-life story enclosing the real puzzle to be solved. \u2013\u00a0Rosie F Aug 26 '16 at 3:15\n\u2022 I want to go to school with these VTCers; clearly their textbooks have much more interesting problems than mine. \u2013\u00a0ffao Aug 26 '16 at 3:28\n\u2022 Shouldn't the last line end with \"move to 9\"? \u2013\u00a0celtschk Aug 26 '16 at 9:23\n\nFirst,\n\nyou need to find $n$, for which $S_n=1+2+\\dots+n=\\frac{n(n+1)}2\\ge x$\n\nThen,\n\nfind a subset of the numbers in range $[1, n]$ which sum to $\\frac{S_n-x}2$. This can always be done if the parity of $S_n$ is the same as the parity for $x$. This means, that if $m$ is the lowest number for which $S_m\\ge x$, then at least one of $S_m$, $S_{m+1}$ or $S_{m+2}$ will do the work.\n\nWith the help of this\n\nadding the rest, and subtracting the above will give you the desired solution\n\nFirst you calculate the smallest $n$ with $\\frac{1}{2}n(n+1) \\geq x$\n\nif $\\frac{1}{2}n(n+1)-x$ is odd then increase $n$ by $1$ if $n$ is even or increase $n$ by $2$ if $n$ is odd.\n\nNow $\\frac{1}{2}n(n+1)-x$ is even\n\nNow you can put a minus sign before some numbers that have the sum $\\frac{1}{2} (\\frac{1}{2}n(n+1)-x)$ and you are done.\n\n## Example:\n\n$x=23$\n\nThe smallest $n$ with $\\frac{1}{2}n(n+1) \\geq x$ is $n=7$\n\n$\\frac{1}{2}n(n+1)-x = 28-23 = 5$ is odd and $n$ is also odd so we have to increase $n$ by $2$.\n\nThat means $n=9$\n\n$\\frac{1}{2}(\\frac{1}{2}n(n+1)-x) = \\frac{1}{2}(45-23) = 11$\n\nSo we have to put a minus in front of numbers with the sum $11$. For example $2$ and $9$.\n\nResult: 1-2+3+4+5+6+7+8-9\n\nIt comes out to be a better puzzle than it was voted to close for!\n\nOne thing is for sure, if the given number $x$ is a triangular number, the answer is its position in the sequence 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66\nwhich is given by\n\n$$\\left\\lfloor\\dfrac{\\sqrt{1+8x}-1}{2}\\right\\rfloor$$\n\nAlso:\n\nhere is my code, which gives a very interesting pattern,\n\noutput:  For x=1, ans : 1 For x=2, ans : 3 For x=3, ans : 2 For x=4, ans : 3 For x=5, ans : 5 For x=6, ans : 3 For x=7, ans : 5 For x=8, ans : 4 For x=9, ans : 5 For x=10, ans : 4 For x=11, ans : 5 For x=12, ans : 7 For x=13, ans : 5 For x=14, ans : 7 For x=15, ans : 5 For x=16, ans : 7 For x=17, ans : 6 For x=18, ans : 7 For x=19, ans : 6 For x=20, ans : 7 For x=21, ans : 6 For x=22, ans : 7 For x=23, ans : 9 For x=24, ans : 7 For x=25, ans : 9 For x=26, ans : 7 \nwhich is :\n$1,3,2,3,5,3,5,4,5,4,5,7,5,7,5,7,6,7,6,7,6,7,9,7,9,7,9$,... OEIS/A140358\ncheck the values up to $1000$ here. see this final code\n\n#include<bits/stdc++.h>\nusing namespace std;\n\nint i,n,t;\nint main()\n{   int x,ans;\n\nfor(cin>>t;t;t--)\n{   cin>>x;\nn=floor((sqrt(1+8*x)-1)/2);\n//cout<<n<<endl;\ni=(n*(n+1))/2;\n//cout<<x<<\", sum:\"<<i<<\", N:\"<<n<<endl;\nif(n%2) //n is odd\n{\nif((x-i)%2) ans=n+2;\nelse ans=n+1;\n}\nelse\n{\nif((x-i)%2) ans=n+1;\nelse ans=n+3;\n}\ncout<<\">! For x=\"<<x<<\", ans : \"<<ans<<endl;\n}\n}", "date": "2019-12-12 04:09:42", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/41436/minimum-steps-on-a-numberline", "openwebmath_score": 0.6094260811805725, "openwebmath_perplexity": 981.3023337572796, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765552017675, "lm_q2_score": 0.8705972549785201, "lm_q1q2_score": 0.853164898901965}}
{"url": "http://gmatclub.com/forum/the-addition-problem-above-shows-four-of-the-24-different-in-104166.html?kudos=1", "text": "The addition problem above shows four of the 24 different in : GMAT Problem Solving (PS)\nCheck GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack\n\n It is currently 21 Jan 2017, 09:10\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# The addition problem above shows four of the 24 different in\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 25 Oct 2010\nPosts: 46\nWE 1: 3 yrs\nFollowers: 0\n\nKudos [?]: 119 [3] , given: 13\n\nThe addition problem above shows four of the 24 different in\u00a0[#permalink]\n\n### Show Tags\n\n02 Nov 2010, 23:34\n3\nKUDOS\n13\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n74% (02:19) correct 26% (01:34) wrong based on 386 sessions\n\n### HideShow timer Statistics\n\n1,234\n1,243\n1,324\n.....\n....\n+4,321\n\nThe addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?\n\nA. 24,000\nB. 26,664\nC. 40,440\nD. 60,000\nE. 66,660\n[Reveal] Spoiler: OA\n\nLast edited by Bunuel on 06 Nov 2012, 02:17, edited 1 time in total.\nRenamed the topic and edited the question.\nRetired Moderator\nJoined: 02 Sep 2010\nPosts: 805\nLocation: London\nFollowers: 105\n\nKudos [?]: 958 [5] , given: 25\n\n### Show Tags\n\n02 Nov 2010, 23:45\n5\nKUDOS\n3\nThis post was\nBOOKMARKED\nstudent26 wrote:\n1,234\n1,243\n1,324\n.....\n....\n+4,321\n\nThe addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?\n\nA.24,000\nB.26,664\nC.40,440\nD.60,000\nE.66,660\n\nUsing the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.\n\nThe sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660\n\n_________________\nMath Forum Moderator\nJoined: 20 Dec 2010\nPosts: 2021\nFollowers: 161\n\nKudos [?]: 1705 [5] , given: 376\n\n### Show Tags\n\n08 Feb 2011, 05:26\n5\nKUDOS\n4\nThis post was\nBOOKMARKED\n1,2,3,4 can be arranged in 4! = 24 ways\n\nThe units place of all the integers will have six 1's, six 2's, six 3's and six 4's\nLikewise,\nThe tens place of all the integers will have six 1's, six 2's, six 3's and six 4's\nThe hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's\nThe thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's\n\nAddition always start from right(UNITS) to left(THOUSANDS);\n\nUnits place addition; 6(1+2+3+4) = 60.\nUnit place of the result: 0\ncarried over to tens place: 6\n\nTens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66\nTens place of the result: 6\ncarried over to hunderes place: 6\n\nHundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66\nHundreds place of the result: 6\ncarried over to thousands place: 6\n\nThousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66\nThousands place of the result: 6\ncarried over to ten thousands place: 6\n\nTen thousands place of the result: 0+6(Carried over from thousands place) = 6\n\nResult: 66660\n\nAns: \"E\"\n_________________\nMath Expert\nJoined: 02 Sep 2009\nPosts: 36590\nFollowers: 7091\n\nKudos [?]: 93333 [3] , given: 10557\n\n### Show Tags\n\n08 Feb 2011, 05:48\n3\nKUDOS\nExpert's post\n17\nThis post was\nBOOKMARKED\nMerging similar topics.\n\nFormulas for such kind of problems (just in case):\n\n1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \\ of \\ the \\ digits)*(111... \\ n \\ times)$$.\n\n2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \\ of \\ the \\ digits)*(111... \\ n \\ times)$$.\n\nSimilar questions:\nnice-question-and-a-good-way-to-solve-103523.html\ncan-someone-help-94836.html\nsum-of-all-3-digit-nos-with-88864.html\npermutation-88357.html\nsum-of-3-digit-s-78143.html\n_________________\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7125\nLocation: Pune, India\nFollowers: 2137\n\nKudos [?]: 13677 [2] , given: 222\n\n### Show Tags\n\n22 Jan 2013, 20:07\n2\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nhellscream wrote:\n\nCould you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?\n\nThe logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.\n\nHow many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?\nYou can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)\n\n1111\n1112\n1121\n... and so on till 4444\n\nBy symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.\nSame for 10s, 100s and 1000s place.\n\nSum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)\n= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)\n= 1111*64*10 = 711040\n\nor use the formula given by Bunuel above:\nSum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times)\n=$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above)\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for \\$199\n\nVeritas Prep Reviews\n\nManager\nJoined: 01 Nov 2010\nPosts: 181\nLocation: Z\u00fcrich, Switzerland\nFollowers: 2\n\nKudos [?]: 38 [0], given: 20\n\n### Show Tags\n\n09 Nov 2010, 05:24\nThanks for the great explaination shrouded1!\nSenior Manager\nJoined: 10 Nov 2010\nPosts: 267\nLocation: India\nConcentration: Strategy, Operations\nGMAT 1: 520 Q42 V19\nGMAT 2: 540 Q44 V21\nWE: Information Technology (Computer Software)\nFollowers: 5\n\nKudos [?]: 301 [0], given: 22\n\n### Show Tags\n\n08 Feb 2011, 04:09\nthe addition problem below shows four of the 24 different integers that can be formed by using each of the digits 1,2,3, and 4 exactly once in each integer.what is the sum of these 24 integers?\n\n1,234\n1,243\n1,324\n.......\n.......\n+4,321\n\na) 24,000\nb) 26,664\nc) 40,440\nd) 60,000\ne) 66,660\n_________________\n\nThe proof of understanding is the ability to explain it.\n\nManager\nJoined: 13 Feb 2012\nPosts: 144\nGMAT 1: 720 Q49 V38\nGPA: 3.67\nFollowers: 0\n\nKudos [?]: 11 [0], given: 107\n\n### Show Tags\n\n22 Jan 2013, 09:44\nBunuel wrote:\nMerging similar topics.\n\nFormulas for such kind of problems (just in case):\n\n1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \\ of \\ the \\ digits)*(111... \\ n \\ times)$$.\n\n2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \\ of \\ the \\ digits)*(111... \\ n \\ times)$$.\n\nSimilar questions:\nnice-question-and-a-good-way-to-solve-103523.html\ncan-someone-help-94836.html\nsum-of-all-3-digit-nos-with-88864.html\npermutation-88357.html\nsum-of-3-digit-s-78143.html\n\nCould you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?\n_________________\n\nIntern\nJoined: 21 May 2013\nPosts: 1\nFollowers: 0\n\nKudos [?]: 0 [0], given: 0\n\nThe addition problem above shows four of the 24 different intege\u00a0[#permalink]\n\n### Show Tags\n\n21 May 2013, 06:35\nThis can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives\n\n1234 +4321 = 5555\n\n1243 +3421 = 5555\n\nwe may see that there are 4!/2 pairings we can make, giving us\n\n5555(12) = 66660\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 13483\nFollowers: 576\n\nKudos [?]: 163 [0], given: 0\n\nRe: The addition problem above shows four of the 24 different in\u00a0[#permalink]\n\n### Show Tags\n\n16 Jun 2014, 03:04\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nIntern\nJoined: 25 Jul 2014\nPosts: 20\nConcentration: Finance, General Management\nGPA: 3.54\nWE: Asset Management (Venture Capital)\nFollowers: 0\n\nKudos [?]: 23 [0], given: 52\n\nRe: The addition problem above shows four of the 24 different in\u00a0[#permalink]\n\n### Show Tags\n\n28 Sep 2014, 09:41\nFor those who could not memorize the formular, you can guess the answer in 30 secs:\nSince we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something\nSo,\n6x1(thousand something) = 6 (thousand something)\n6x2(thousand something) = 12 (thousand something)\n6x3(thousand something) = 18 (thousand something)\n6x4(thousand something) = 24 (thousand something)\nAdd them all 6+12 +18 + 24 = 60 (thousand something)\n----> E\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 13483\nFollowers: 576\n\nKudos [?]: 163 [0], given: 0\n\nRe: The addition problem above shows four of the 24 different in\u00a0[#permalink]\n\n### Show Tags\n\n11 Nov 2015, 23:06\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nIntern\nJoined: 24 Apr 2016\nPosts: 6\nFollowers: 0\n\nKudos [?]: 0 [0], given: 895\n\nRe: The addition problem above shows four of the 24 different in\u00a0[#permalink]\n\n### Show Tags\n\n15 Aug 2016, 13:38\neach term is repeating 6 times..(total 24/4=4)\nnow at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is \"0\" and 6 remaining.\n\nrepeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6.\n\nonly E has 60 as last two digits.\nRe: The addition problem above shows four of the 24 different in \u00a0 [#permalink] 15 Aug 2016, 13:38\nSimilar topics Replies Last post\nSimilar\nTopics:\n3 The figure above shows a side view of the insert and the four componen 3 30 Jul 2016, 05:16\n1 In the addition problem above, A and B represent digits in two differe 1 11 Dec 2015, 08:22\n13 The addition problem above shows four of the 24 13 20 Nov 2012, 23:53\n45 There are 24 different four-digit integers than can be 18 04 Nov 2012, 09:39\n3 As the figure above shows, four identical circles are inscri 3 17 Feb 2011, 14:04\nDisplay posts from previous: Sort by\n\n# The addition problem above shows four of the 24 different in\n\n Powered by phpBB \u00a9 phpBB Group and phpBB SEO Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2017-01-21 17:10:24", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/the-addition-problem-above-shows-four-of-the-24-different-in-104166.html?kudos=1", "openwebmath_score": 0.613451361656189, "openwebmath_perplexity": 5194.8760293636105, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. 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{"url": "http://mathhelpforum.com/advanced-statistics/21429-conditional-probabilty-print.html", "text": "# Conditional Probabilty\n\n\u2022 October 27th 2007, 06:10 AM\nRevilo\nConditional Probabilty\nFirst little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?\n\nSo far i have:\nLet A = First little piggy goes to market\nB = Second little piggy goes to market\n\nP(A given B) = 0.25\nP(B given A) = 0.5\nP(A union B) = P(A) + P(B) - P(A intersection B)\n\nIs the P(A union B) = 1 as atleast one must go, so this is always true?\nDoes anyone have any ideas, i'm really stuck.\n\u2022 October 28th 2007, 01:11 AM\nkalagota\nQuote:\n\nOriginally Posted by Revilo\nFirst little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?\n\nSo far i have:\nLet A = First little piggy goes to market\nB = Second little piggy goes to market\n\nP(A given B) = 0.25\nP(B given A) = 0.5\nP(A union B) = P(A) + P(B) - P(A intersection B)\n\nIs the P(A union B) = 1 as atleast one must go, so this is always true?\nDoes anyone have any ideas, i'm really stuck.\n\nyeah, it is correct..\nbut, i think, you should not look for the union, instead for the intersection..\nnote that P(A|B)= P(A int B) / P(B) [also, P(B|A) = P(A int B) / P(A) ]; this theorem should help you find the answer.\n\u2022 October 28th 2007, 08:51 PM\nSoroban\nHello, Revilo!\n\nQuote:\n\nFirst little piggy is 25% likely to go to market if second little piggy goes.\nSecond little piggy is 50% likely to go to if first little piggy goes.\nSadly at least one of them must go to market.\nHow likely are they both to go?\n\nAll your ideas are good ones . . .\n\nBayes' Theorem says: . $P(A|B) \\:=\\:\\frac{P(A \\wedge B)}{P(B)}$\n\nWe are given:\n\n. . $P(1^{st}|2^{nd}) \\:=\\:\\frac{1}{4}\\quad\\Rightarrow\\quad \\frac{P(1^{st} \\wedge 2^{nd})}{P(2^{nd})}\\: = \\: \\frac{1}{4}\\quad\\Rightarrow\\quad P(1^{st} \\wedge 2^{nd}) \\:=\\:\\frac{1}{4}\\!\\cdot\\!P(2^{nd})$ .[1]\n\n. . $P(2^{nd}|1^{st}) \\:=\\:\\frac{1}{2}\\quad\\Rightarrow\\quad \\frac{P(2^{nd} \\wedge 1^{st})}{P(1^{st})} \\:=\\:\\frac{1}{2}\\quad\\Rightarrow\\quad P(1^{st} \\wedge 2^{nd}) \\:=\\:\\frac{1}{2}\\!\\cdot\\!P(1^{st})$ . [2]\n\nEquate [1] and [2]: . $\\frac{1}{4}\\cdot P(2^{nd}) \\:=\\:\\frac{1}{2}\\cdot P(1^{st}) \\quad\\Rightarrow\\quad P(2^{nd}) \\:=\\:2\\cdot P(1^{st})$ . [3]\n\nSince at least one of them goes to the market: . $P(1^{st} \\vee 2^{nd}) \\:=\\:1$\n\nWe have: . $P(1^{st} \\vee 2^{nd}) \\;=\\;P(1^{st}) + P(2^{nd}) - P(1^{st} \\wedge 2^{nd}) \\;=\\;1$ . [4]\n\n. . [3] $\\; P(2^{nd}) \\: = \\: 2\\cdot P(1^{st})$\n. . [2] $\\; P(1^{st} \\wedge 2^{nd}) \\: = \\: \\frac{1}{2}\\cdot P(1^{st})\n$\n\nSubstitute into [4]: . $P(1^{st}) + 2\\cdot P(1^{st}) -\\frac{1}{2}\\cdot P(1^{st}) \\;=\\;1\\quad\\Rightarrow\\quad \\frac{5}{2}\\cdot P(1^{st}) \\:=\\:1$\n\nTherefore: . $P(1^{st}) \\:= \\:\\frac{2}{5}\\qquad P(2^{nd}) \\: = \\: \\frac{4}{5} \\qquad \\boxed{P(1^{st} \\wedge 2^{nd}) \\: = \\: \\frac{1}{5}}$", "date": "2015-10-09 02:53:02", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/21429-conditional-probabilty-print.html", "openwebmath_score": 0.5827082395553589, "openwebmath_perplexity": 2228.0174368924177, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227578357059, "lm_q2_score": 0.8596637559030338, "lm_q1q2_score": 0.8531498754446899}}
{"url": "https://math.stackexchange.com/questions/2631523/the-base-of-a-logarithm", "text": "# The base of a logarithm\n\nWhat exactly is the base of a logarithm ? and how should it be understood ? I used to think it was the base of a \"normal\" exponent e.g. the $2$ in $2^{75}$ would be the base in logarithmic form, but the change of base formula can accept ANY base, and when finding the number of digits in $2^{75}$, you use the common log:\n\n$$2^{75}$$\n\n$$\\log_{10}(2^{75})$$\n\n$$75\\log_{10}(2)$$\n\n$$75(0.301)+1=23 \\textrm{ digits}$$\n\nI understand that these formulas work, I just can't wrap my head around why they work the way they do, and the heart of my issue is how I should understand the base.\n\nI did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that. But if that were the case, then would a base 16 represent a hexadecimal system ? and how would that work considering we use letters in addition to numbers ?\n\n\u2022 How many digits in $10^k$? and what is value of $\\log_{10}(10^{k})$ ? Try searching for pattern. Feb 1 '18 at 17:45\n\u2022 \"I did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that.\" Really? I would have thought those were the definitions. Feb 1 '18 at 18:00\n\nI think you have the idea.\n\nIf $y = 2^{75}$ then $\\log_2 y = 75$\n\nOne way to use bases is to find the number of digits that that number would have in that system. If $32<y<64$ then $5<\\log_2 y < 6$\n\nYou can take logarithms in any base, and convert between bases.\n\n$\\log_a x = \\frac {\\log_b x}{\\log_b a}$\n\nFor example, $\\log_{10} 2^{75} = 75 \\log_{10} 2$ as you have above.\n\nBut you could also say $\\log_{10} 2^{75} = \\frac {\\log_2 2^{75}}{\\log_2 10}$\n\nWhich implies $\\log_2 10 = \\frac 1{\\log_{10} 2}$\n\nYou're right about the number of digits of a number in a specific base but why? To prove it let $(x)_b$ denote the number x in base $b$. If $b=10$ it is the same natural and convenient base we often use. Obviously if any number is between two consecutive powers of $10$, for example $1253$ is between $10^3$ and $10^4$ and it has 4 digits. In fact for any number $x$ if it belongs to $(10^n,10^{n+1}-1)$ for some n then $n+1$ is the number of digits of $x$. Then we can find the number of digits as following:$$10^n\\le x<10^{n+1}\\\\n\\le \\log_{10}x<n+1\\\\n=\\lfloor \\log_{10}x\\rfloor+1$$the same argument is true for any arbitrary base $b$, so the number of digits in base $b$ is $$dig_{b}(x)=\\lfloor \\log_{b}x\\rfloor+1$$\n\nThe base of a logarithm is the base of the corresponding exponential function. Take $10^x$ for example. The base of this exponential function is $10$. The corresponding logarithm is the standard log, $\\log_{10}{x}$, or simply $\\log{x}$. The logarithm is the inverse of the exponential function; for example, if $x = 2$, then $10^2 = 100$. The inverse therefore (the $\\log{x}$) will get us the exponent, or 2. Therefore, $\\log_{10}(100) = 2.$\n\nThe base of the logarithm in this case is the $10$ from $10^x$.\n\nMore generally, if you had an exponential function $b^x$ then the corresponding logarithm would be $\\log_b{x}$ in which the base is $b$.\n\n$\\log_{16} 2^{75} = \\frac {75}{4} = 18.75$ so there are $18 + 1 = 19$ digits in hex.\n\nAnd indeed $2^{75} = 8*2^{72} = 8*16^{18}$ so $2^{75}$ in hexadecimal is $8000000000000000000$.\n\nYou seem to understand the base of the logorithms perfectly so far as I can tell.\n\n## Decimals\n\nAs for the number of digits of a number in the decimal system. consider the following:\n\nLet $x$ be any real number - for our convenience, assume it is positive. Then, obviously, there exists exactly a natural number $n\\in\\mathbb{N}$ such that: $$10^n\\leq x<10^{n+1}$$\n\nNow, if a number is between $10^n$ and $10^{n+1}$, as above, it means that it has exactly than $n+1$ digits. Imagine $x=9468$, for instance. In this case, $n=3$, since: $$10^3=1000\\leq9468<10,000=10^4$$ So, we would like to express $n$ in terms of $x$, so as to have a formula that, given $x$, returns $n+1$. Since $n$ is appearing as an exponent of $10$, we think that taking $\\log_{10}$ on every side is a way to \"get $n$ out of the exponent\". So, we have that: \\begin{align*} &10^n\\leq x<10^{n+1}\\\\ \\Leftrightarrow&\\log_{10}10^n\\leq\\log_{10}x<\\log_{10}10^{n+1}\\\\ \\Leftrightarrow&n\\leq\\log_{10}x<n+1 \\end{align*} So, $n=\\lfloor\\log_{10}x\\rfloor$ which means that the number of digits of $x$ is exactly: $$n+1=\\lfloor\\log_{1}x\\rfloor+1$$\n\n## Logarithms\n\nNow, as for the logarithms, let us remind the definition:\n\nGiven $x,b>0$, $\\log_bx$ is the - only - number that has the following property: $$b^{\\log_bx}=x$$ So, the logarithm of $x$ with base $b$ is the only number which can be the exponent of $b$ such that $b$ to that exponent is equal to $x$.\n\nSo, given the symbol $\\log_bx$, $b$ being the logarithms base means that $b$ is the number to which we may set $\\log_bx$ as an exponent to get $x$. You can imagine the following game that helped me understand logarithms:\n\n## The Log-Game\n\nImagine two friends, Alice and Bob - hehe, these are probably the only friends we, mathematicians, have ever heard of - have a secret code to exchange messages. So, every message they want to send is being turned into a - probably very large - number. So, in our case, Alice and Bob are exchanging numbers in order to communicate.\n\nBut, in order to make it more difficult for other people to read their messages and find out the code, they use the following trick:\n\nIf $x$ is the message Alice wants to send to Bob, then, instead of this Alice sends Bob another number, which is $\\log_b x$, for some base $b$ that they have pre-decided.\n\nSo, Bob receives a number $y$ that is not the message he wants, and, probably, has no meaning at all. How will he find the message? As mentioned above, Alice and Bob have pre-decided a base for the logarithms they are exchanging, so, what Bob has to do is to calculate: $$b^y=b^{\\log_bx}=x$$ since, by the definition of logarithm, the only way to find $x$ given $y=\\log_bx$ is to set it as an exponent to $b$. Every other, not knowing the exact value of $b$ cannot find that message.\n\n## Over-decimal Systems\n\nWe have heard about decimal system, binary system etc. But, the most usual case is that we use the usual symbols $0,1,2,\\dots,9$ for such systems. What about hexadecimal system? There, we need $16$ different - distinct - symbols for our elementary digits. So, we introduce $6$ new symbols: $A,B,C,D,E,F$. These are just symbols, as the previous ones; $0,1,2,\\dots,9$. There is no matter about that.\n\nTo make it a little more clear, we have decided that: $$0+1=1,\\ 1+1=2,\\ 2+1=3,\\dots,8+1=9$$ Moreover, we have decided that $$9+1=10,\\ 99+1=100$$ Or, to be more precise, the nature of our enumerating system - that the position of a digit is important to its evaluation - is a property that makes it feasible to use only some - finite in number - symbols and not infinitely many or strange combinations - see, for instance, the roman digits.\n\nWe have been used to using these symbols to represent numbers, which has driven us to the wrong conclusion that these symbols are the numbers they represent. Well, no symbol is any number, it just represents one, probably different with respect to our system.\n\nFor instance, in binary system: $$101$$ represents what in the decimal system we would write as: $$5$$ or what in the ternary (3) system we would write as: $$12$$ or what in the hexadecimal system we would write as: $$5$$\n\nSo, if we want to use some enumeration system with more that $10$ elementary digits, we have to introduce some new symbols for these digits and we also have to define the basic operations with them. So, in hexadecimal system we deicide that: $$9+1=A,\\ A+1=B,\\ B+1=D,\\dots,E+1=F$$ and all the other properties we know.\n\nSo, to find how many elements a number $x$ has in hexadecimal, we simply note that there exists exactly on $n$ such that: $$16^n\\leq x<16^{n+1}$$ and, with the same thoughts as above, we take $\\log_16$ to every side, so we have: $$n\\leq\\log_{16}x<n+1$$ and, we have, finally that number $x$ in the hexadecimal system has: $$\\lfloor\\log_{16}x\\rfloor+1$$ digits.", "date": "2022-01-26 06:58:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2631523/the-base-of-a-logarithm", "openwebmath_score": 0.8890386819839478, "openwebmath_perplexity": 167.5121039766863, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308768564867, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8531439103706056}}
{"url": "https://math.stackexchange.com/questions/1268155/prove-that-4x2-8xy5y2-geq0-is-this-a-valid-proof", "text": "# Prove that $4x^2-8xy+5y^2\\geq0$ - is this a valid proof?\n\nI need to prove that $4x^2-8xy+5y^2\\geq0$ holds for every real numbers $x, y$.\n\nFirst I start with another inequality, i.e. $4x^2-8xy+4y^2\\geq0$, which clearly holds as it can be factorized into $(2x-2y)^2\\geq0$. Now we can add $y^2$ (which is always nonnegative) to the left side, thus obtaining $4x^2-8xy+5y^2\\geq0$, which is always true - we've just added a nonnegative expression to another, so the sum is still greater or equal to zero.\n\nIs my proof correct? I had it on my final math exam and would like to be sure.\n\n\u2022 Looks good to me. \u2013\u00a0simonzack May 5 '15 at 13:23\n\n\u2022 +1. An example of converse not being true is the polynomial $$x^6+y^2z^4+y^4z^2-3x^2y^2z^2$$ \u2013\u00a0Leg May 5 '15 at 14:46\n$4x^2-8xy+5y^2 = (2x-2y)^2+y^2 \\ge 0$.\nUsing this sum-of-squares technique you also get that equality occurs exactly when $2x-2y = 0$ and $y = 0$, or equivalently when $(x,y) = (0,0)$.", "date": "2019-09-20 22:36:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1268155/prove-that-4x2-8xy5y2-geq0-is-this-a-valid-proof", "openwebmath_score": 0.9154528379440308, "openwebmath_perplexity": 212.43350374443773, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308775555446, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8531439075008065}}
{"url": "http://mathhelpforum.com/calculus/18632-integral-cotx.html", "text": "1. ## integral with cotx\n\nHi,\n\nCould someone please give me a hint on how to do this problem? I know that I have to use a formula to do it, but it looks like I need to simplify it.\n\nInt. cotx/sq. 1+2sinx\n\nI have these formulas:\n\nint. cotudu=ln|secu|+c\n\nint. cot^udu=-1/n-1cot^n-1u-int. cot^n-2udu\n\nThank you\n\n2. $\\displaystyle\\int\\frac{\\cot x}{\\sqrt{1+2\\sin x}}dx=\\int\\frac{\\cos x}{\\sin x\\sqrt{1+2\\sin x}}dx$\nSubstitute $\\sin x=t\\Rightarrow\\cos xdx=dt\\Rightarrow\\int\\displaystyle\\frac{1}{t\\sqrt{ 1+2t}}dt$.\nSubstitute $\\sqrt{1+2t}=u\\Rightarrow 2t=u^2-1\\Rightarrow dt=udu$\nThen $\\displaystyle\\int\\frac{u}{\\frac{u^2-1}{2}\\cdot u}du=2\\int\\frac{1}{u^2-1}du=\\ln \\left|\\frac{u-1}{u+1}\\right|+C$\nNow back substitute.\n\n3. Hello, chocolatelover!\n\nI found a method . . . but it took two substitutions.\nI'm certain someone will streamline it for us.\n\n$\\int \\frac{\\cot x}{\\sqrt{1+2\\sin x}}\\,dx$\n\nLet $u \\,=\\,\\sin x\\quad\\Rightarrow\\quad x \\,=\\,\\arcsin u\\quad\\Rightarrow\\quad dx \\,=\\,\\frac{du}{\\sqrt{1-u^2}}$\n. . and: . $\\cot x \\,=\\,\\frac{\\sqrt{1-u^2}}{u}$\n\nSubstitute: . $\\int\\frac{\\frac{\\sqrt{1-u^2}}{u}}{\\sqrt{1+2u}}\\,\\frac{du}{\\sqrt{1-u^2}} \\;= \\;\\int\\frac{du}{u\\sqrt{1+2u}}$\n\nLet $v \\,=\\,\\sqrt{1+2u}\\quad\\Rightarrow\\quad u \\,=\\,\\frac{v^2-1}{2}\\quad\\Rightarrow\\quad du \\,=\\,v\\,dv$\n\nSubstitute: . $\\int\\frac{v\\,dv}{\\left(\\frac{v^2-1}{2}\\right)\\cdot v} \\;=\\;2\\int\\frac{dv}{v^2-1} \\;= \\;\\ln\\left|\\frac{v-1}{v+1}\\right| + C$\n\nBack-substitute: . $\\ln\\left|\\frac{\\sqrt{1+2u} - 1}{\\sqrt{1+2u} + 1}\\right| + C$\n\nBack-substitute: . $\\boxed{\\ln\\left|\\frac{\\sqrt{1+2\\sin x} - 1}{\\sqrt{1+2\\sin x} + 1}\\right| + C}$\n\nHa! . . . red_dog beat me to it!\n\n4. Mine just take one.\n\nSet\n\n$u=\\sqrt{1+2\\sin x}\\implies du=\\frac{\\cos x}{\\sqrt{1+2\\sin x}}\\,dx$, the intregral becomes to\n\n$\\int\\frac{\\cot x}{\\sqrt{1+2\\sin x}}\\,dx=2\\int\\frac1{u^2-1}\\,du$\n\nCheers,\nK.", "date": "2016-08-26 18:13:28", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/18632-integral-cotx.html", "openwebmath_score": 0.9172375202178955, "openwebmath_perplexity": 4175.10775905117, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308772060157, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8531439054623461}}
{"url": "https://gmatclub.com/forum/some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Oct 2018, 06:22\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Some part of a 50% solution of acid was replaced with an\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 08 Aug 2008\nPosts: 5\nSome part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 19 Aug 2015, 12:12\n2\n12\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n81% (01:15) correct 19% (02:00) wrong based on 301 sessions\n\n### HideShow timer Statistics\n\nSome part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?\n\nA. $$\\frac{1}{5}$$\nB. $$\\frac{1}{4}$$\nC. $$\\frac{1}{2}$$\nD. $$\\frac{3}{4}$$\nE. $$\\frac{4}{5}$$\n\nM25Q30\n\nOriginally posted by chrissy28 on 12 Aug 2008, 14:27.\nLast edited by ENGRTOMBA2018 on 19 Aug 2015, 12:12, edited 2 times in total.\nManager\nJoined: 15 Jul 2008\nPosts: 204\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n12 Aug 2008, 15:10\nchrissy28 wrote:\n\nSome part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?\n\n(A) 1/5\n(B) 1/4\n(C) 1/2 *correct answer\n(D) 3/4\n(E) 4/5\n\nOriginal Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.\n\nin the new solution.. you added x of 30% solution. i.e 30x acid\n\n(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%\n\n[(1-x)50 + 30x] / 100 = 40/100\n\nsolve to get x=1/2\nSVP\nJoined: 30 Apr 2008\nPosts: 1835\nLocation: Oklahoma City\nSchools: Hard Knocks\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n12 Aug 2008, 15:16\nThe reason that 1/2 is correct is this:\n\nWhat would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).\n\nThe result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.\n\nHere we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.\n\nchrissy28 wrote:\n\nSome part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?\n\n(A) 1/5\n(B) 1/4\n(C) 1/2 *correct answer\n(D) 3/4\n(E) 4/5\n\n_________________\n\n------------------------------------\nJ Allen Morris\n**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.\n\nGMAT Club Premium Membership - big benefits and savings\n\nSenior Manager\nJoined: 15 Sep 2011\nPosts: 335\nLocation: United States\nWE: Corporate Finance (Manufacturing)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n16 Jun 2013, 16:55\n2\n2\nIf 50% solution becomes 40% solution, then 40% solution must be equally weighted between 50% solution and 30% solution - i.e. $$\\frac{1}{2}$$ of 50% solution and $$\\frac{1}{2}$$ of 30% solution. If the solution must be equally weighted, then 30% solution must replace $$\\frac{1}{2}$$ of 50% solution (assuming original solution was 50% solution of acid).\n\nJust another way of saying the same thing. Cheers\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50001\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n16 Jun 2013, 23:07\n1\n3\nSome part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?\n\nA. $$\\frac{1}{5}$$\nB. $$\\frac{1}{4}$$\nC. $$\\frac{1}{2}$$\nD. $$\\frac{3}{4}$$\nE. $$\\frac{4}{5}$$\n\nSince the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.\n\nAlternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\\frac{1}{2}$$.\n\n_________________\nManager\nStatus: Working hard to score better on GMAT\nJoined: 02 Oct 2012\nPosts: 86\nLocation: Nepal\nConcentration: Finance, Entrepreneurship\nGPA: 3.83\nWE: Accounting (Consulting)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n17 Jun 2013, 21:57\nHi Bunuel, What is wrong in this equation? Kindly guide me.\n\nLet total solution =100\nReplacement = x (say)\n\nequation:\n(100-x)*50% + X*30% = 40%\nSolving this equation i am getting x= 248\n\nI looked every way possible but could not figure out the problem in the equation....\n_________________\n\nDo not forget to hit the Kudos button on your left if you find my post helpful.\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50001\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n17 Jun 2013, 23:14\natalpanditgmat wrote:\nHi Bunuel, What is wrong in this equation? Kindly guide me.\n\nLet total solution =100\nReplacement = x (say)\n\nequation:\n(100-x)*50% + X*30% = 40%\nSolving this equation i am getting x= 248\n\nI looked every way possible but could not figure out the problem in the equation....\n\n(100-x)*0.5+0.3x=100*0.4 --> x=50 --> 50/100=1/2.\n\nHope it's clear.\n_________________\nManager\nStatus: Trying.... & desperate for success.\nJoined: 17 May 2012\nPosts: 64\nLocation: India\nSchools: NUS '15\nGPA: 2.92\nWE: Analyst (Computer Software)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n18 Jun 2013, 00:03\n1\natalpanditgmat wrote:\nHi Bunuel, What is wrong in this equation? Kindly guide me.\n\nLet total solution =100\nReplacement = x (say)\n\nequation:\n(100-x)*50% + X*30% = 40% Solving this equation i am getting x= 248\n\nI looked every way possible but could not figure out the problem in the equation....\n\nTry using below equation.\n50% of (100-x) + 30% of x = 40% of 100\nx=50litres.\nSo 50 of 100 litres got replaced would amount to 1/2 in ratio.\nManager\nStatus: Working hard to score better on GMAT\nJoined: 02 Oct 2012\nPosts: 86\nLocation: Nepal\nConcentration: Finance, Entrepreneurship\nGPA: 3.83\nWE: Accounting (Consulting)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n18 Jun 2013, 00:33\nAhhhh!! I got it....Thank you both of you guys....I hope silly mistakes won't trouble me on the Real Test...\n_________________\n\nDo not forget to hit the Kudos button on your left if you find my post helpful.\n\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1829\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n10 Jul 2014, 02:09\n2\nDid using allegation method\n\nRefer diagram below\n\nBunuel, kindly update the OA.\n\nAttachments\n\nall.png [ 3.61 KiB | Viewed 5115 times ]\n\n_________________\n\nKindly press \"+1 Kudos\" to appreciate\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8397\nLocation: Pune, India\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n20 Jul 2015, 06:11\n3\nchrissy28 wrote:\nSome part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?\n\nA. $$\\frac{1}{5}$$\nB. $$\\frac{1}{4}$$\nC. $$\\frac{1}{2}$$\nD. $$\\frac{3}{4}$$\nE. $$\\frac{4}{5}$$\n\nM25Q30\n\nResponding to a pm:\n\nIt is worded to look like a replacement question but essentially, it is a simple mixtures problem only.\n\nYou have some 50% solution of acid and you remove some of it. Now to that 50% solution, you are adding 30% solution to get 40% solution. The ratio in which you mix the 50% solution and the 30% solution is given by w1/w2 = (50 - 40)/(40 - 30) = 1/1\nBasically, you mix equal parts of 50% solution and 30% solution.\nSo initially you must have had 2 parts of 50% solution out of which you removed 1 part and replaced with 1 part of 30% solution. So you replaced half of the original solution.\n\nThe first question of this post discusses exactly this concept: http://www.veritasprep.com/blog/2012/01 ... -mixtures/\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nLearn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nIntern\nJoined: 17 Aug 2015\nPosts: 5\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2015, 11:41\nCan you please explain to me how you got the answer to this problem using the allegation methos?\n\nThanks\nCurrent Student\nJoined: 20 Mar 2014\nPosts: 2633\nConcentration: Finance, Strategy\nSchools: Kellogg '18 (M)\nGMAT 1: 750 Q49 V44\nGPA: 3.7\nWE: Engineering (Aerospace and Defense)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2015, 11:44\ntinnyshenoy wrote:\nCan you please explain to me how you got the answer to this problem using the allegation methos?\n\nThanks\n\nLook at: some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html#p1381700\nIntern\nJoined: 17 Aug 2015\nPosts: 5\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2015, 12:02\nI cannot understand how they divided 10/20 to get 10? Where did 20 come from?\nCurrent Student\nJoined: 20 Mar 2014\nPosts: 2633\nConcentration: Finance, Strategy\nSchools: Kellogg '18 (M)\nGMAT 1: 750 Q49 V44\nGPA: 3.7\nWE: Engineering (Aerospace and Defense)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n19 Aug 2015, 12:22\ntinnyshenoy wrote:\nI cannot understand how they divided 10/20 to get 10? Where did 20 come from?\n\nIt is from the alligation method\n\nAs shown in the post above, per the alligation method, the distrbution will be:\n\n50 10 (=40-30) parts\n\n40\n\n30 10 (=50-40) parts\n\nThus, you see that you need 10 parts out of 20 (=(10+10)) parts of 50% solution, giving you 10/20 = 1/2 or 50% as your answer.\n\nHope this helps.\nModerator\nJoined: 22 Jun 2014\nPosts: 1030\nLocation: India\nConcentration: General Management, Technology\nGMAT 1: 540 Q45 V20\nGPA: 2.49\nWE: Information Technology (Computer Software)\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n09 Apr 2016, 00:38\nchrissy28 wrote:\nSome part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?\n\nA. $$\\frac{1}{5}$$\nB. $$\\frac{1}{4}$$\nC. $$\\frac{1}{2}$$\nD. $$\\frac{3}{4}$$\nE. $$\\frac{4}{5}$$\n\nM25Q30\n\nw1 is the weight 50% acid solution which has concentration C1 = 50%\nw2 is the weight of 20% acid solution which has concentration C1 = 30%\nConcentration of mixture (average) = 40%\n\nusing weihted average formula w1/w2 = (c2 - avg) / (Avg-c1)\nw1/w2 = (30-40)/(40-50)\nw1/w2 = 1/1\n\nw1 is 1 part and w2 is 1 part in a total 2 part of solution.\n\nquestion is What part of the original solution was replaced?\nw2 is what was replaced, hence 1 out of 2 i.e. 1/2 part was replaced.\n_________________\n\n---------------------------------------------------------------\nTarget - 720-740\nhttp://gmatclub.com/forum/information-on-new-gmat-esr-report-beta-221111.html\nhttp://gmatclub.com/forum/list-of-one-year-full-time-mba-programs-222103.html\n\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 8464\nRe: Some part of a 50% solution of acid was replaced with an\u00a0 [#permalink]\n\n### Show Tags\n\n19 Sep 2018, 02:52\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: Some part of a 50% solution of acid was replaced with an &nbs [#permalink] 19 Sep 2018, 02:52\nDisplay posts from previous: Sort by\n\n# Some part of a 50% solution of acid was replaced with an\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-10-19 13:22:13", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html", "openwebmath_score": 0.7010799050331116, "openwebmath_perplexity": 4303.187743375512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9362850093037731, "lm_q2_score": 0.9111797069968975, "lm_q1q2_score": 0.8531239004429995}}
{"url": "http://accessdtv.com/taylor-series/taylor-series-error-estimation.html", "text": "Home > Taylor Series > Taylor Series Error Estimation\n\nTaylor Series Error Estimation\n\nContents\n\nExample What is the minimum number of terms of the series one needs to be sure to be within of the true sum? However, it holds also in the sense of Riemann integral provided the (k+1)th derivative of f is continuous on the closed interval [a,x]. This will present you with another menu in which you can select the specific page you wish to download pdfs for. Sometimes you'll see something like N comma a to say it's an Nth degree approximation centered at a. Source\n\nThe statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of Lebesgue integration theory for the full generality. Solution Here are the first few derivatives and the evaluations. Note that, for each j = 0,1,...,k\u22121, f ( j ) ( a ) = P ( j ) ( a ) {\\displaystyle f^{(j)}(a)=P^{(j)}(a)} . Your cache administrator is webmaster. https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation\n\nTaylor Series Error Estimation Calculator\n\nText is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. These estimates imply that the complex Taylor series T f ( z ) = \u2211 k = 0 \u221e f ( k ) ( c ) k ! ( z \u2212 In the mean time you can sometimes get the pages to show larger versions of the equations if you flip your phone into landscape mode. In this example we pretend that we only know the following properties of the exponential function: ( \u2217 ) e 0 = 1 , d d x e x = e\n\nThis kind of behavior is easily understood in the framework of complex analysis. Solution As with the last example we\u2019ll start off in the same manner. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So, we get a similar pattern for this one.\u00a0 Let\u2019s plug the numbers into the MeteaCalcTutorials 55,406 views 4:56 Maclauren and Taylor Series Intuition - Duration: 12:59. Lagrange Error Bound Calculator The N plus oneth derivative of our Nth degree polynomial.\n\nShow Answer If you have found a typo or mistake on a page them please contact me and let me know of the typo/mistake. Hence each of the first k\u22121 derivatives of the numerator in h k ( x ) {\\displaystyle h_{k}(x)} vanishes at x = a {\\displaystyle x=a} , and the same is true Where are the answers/solutions to the Assignment Problems?\n\nTranscript The interactive transcript could not be loaded.\n\nUp next Taylor's Inequality - Duration: 10:48. Remainder Estimation Theorem As in previous modules, let be the error between the Taylor polynomial and the true value of the function, i.e., Notice that the error is a function of . Recall that if a series has terms which are positive and decreasing, then But notice that the middle quantity is precisely . How close will the result be to the true answer?\n\nTaylor Polynomial Approximation Calculator\n\nAnd sometimes you might see a subscript, a big N there to say it's an Nth degree approximation and sometimes you'll see something like this. PaulOctober 27, 2016 Calculus II - Notes Parametric Equations and Polar Coordinates Previous Chapter Next Chapter Vectors Power Series and Functions Previous Section Next Section Applications of Series \u00a0Taylor Taylor Series Error Estimation Calculator In that situation one may have to select several Taylor polynomials with different centers of expansion to have reliable Taylor-approximations of the original function (see animation on the right.) There are Lagrange Error Formula If all the k-th order partial derivatives of f: Rn \u2192 R are continuous at a \u2208 Rn, then by Clairaut's theorem, one can change the order of mixed derivatives at\n\nIf you are a mobile device (especially a phone) then the equations will appear very small. http://accessdtv.com/taylor-series/taylor-series-error-estimation-formula.html And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x Solution 1 As with the first example we\u2019ll need to get a formula for .\u00a0 However, unlike the first one we\u2019ve got a little more work to do.\u00a0 Let\u2019s first take However, because the value of c is uncertain, in practice the remainder term really provides a worst-case scenario for your approximation. Taylor Series Remainder Calculator\n\nHere is the Taylor Series for this function. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now, let\u2019s work one of the easier examples in this section.\u00a0 The problem for most students is that it may So if you put an a in the polynomial, all of these other terms are going to be zero. Generated Sun, 30 Oct 2016 19:02:48 GMT by s_fl369 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection http://accessdtv.com/taylor-series/taylor-series-error-estimation-problems.html What are they talking about if they're saying the error of this Nth degree polynomial centered at a when we are at x is equal to b.\n\nThis means that for every a\u2208I there exists some r>0 and a sequence of coefficients ck\u2208R such that (a \u2212 r, a + r) \u2282 I and f ( x ) Taylor's Inequality F of a is equal to P of a, so the error at a is equal to zero. To obtain an upper bound for the remainder on [0,1], we use the property e\u03beSo the error of b is going to be f of b minus the polynomial at b.\n\nFrom Site Map Page The Site Map Page for the site will contain a link for every pdf that is available for downloading. Note If you actually compute the partial sums using a calculator, you will find that 7 terms actually suffice. Sign in Transcript Statistics 38,950 views 81 Like this video? Lagrange Error Bound Problems And it's going to look like this.\n\nRelationship to analyticity Taylor expansions of real analytic functions Let I \u2282 R be an open interval. Compute F \u2032 ( t ) = f \u2032 ( t ) + ( f \u2033 ( t ) ( x \u2212 t ) \u2212 f \u2032 ( t ) ) The second inequality is called a uniform estimate, because it holds uniformly for all x on the interval (a \u2212 r,a + r). Check This Out Krista King 14,459 views 12:03 Taylor's Theorem with Remainder - Duration: 9:00.\n\nThe N plus oneth derivative of our error function or our remainder function, we could call it, is equal to the N plus oneth derivative of our function. Download Page - This will take you to a page where you can download a pdf version of the content on the site. Sign in 82 5 Don't like this video? patrickJMT 130,005 views 2:22 Estimating error/remainder of a series - Duration: 12:03.\n\nThe goal is to find so that . Notice as well that for the full Taylor Series, the nth degree Taylor polynomial is just the partial sum for the series. Sign in to add this video to a playlist. So this is going to be equal to zero.\n\nNow, what is the N plus onethe derivative of an Nth degree polynomial? Also, since the condition that the function f be k times differentiable at a point requires differentiability up to order k\u22121 in a neighborhood of said point (this is true, because It has simple poles at z=i and z= \u2212i, and it is analytic elsewhere. Graph of f(x)=ex (blue) with its quadratic approximation P2(x) = 1 + x + x2/2 (red) at a=0.\n\nModulus is shown by elevation and argument by coloring: cyan=0, blue=\u03c0/3, violet=2\u03c0/3, red=\u03c0, yellow=4\u03c0/3, green=5\u03c0/3. Combining these estimates for ex we see that | R k ( x ) | \u2264 4 | x | k + 1 ( k + 1 ) ! \u2264 4 Since ex is increasing by (*), we can simply use ex\u22641 for x\u2208[\u22121,0] to estimate the remainder on the subinterval [\u22121,0]. Solution Again, here are the derivatives and evaluations. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Notice that all the negative signs will cancel out in the evaluation.\u00a0 Also, this formula will work for all n,\n\nHere's the formula for the remainder term: So substituting 1 for x gives you: At this point, you're apparently stuck, because you don't know the value of sin c. Algebra/Trig Review Common Math Errors Complex Number Primer How To Study Math Close the Menu Current Location : Calculus II (Notes) / Series & Sequences / Taylor Series Calculus II [Notes]", "date": "2018-02-22 20:36:48", "meta": {"domain": "accessdtv.com", "url": "http://accessdtv.com/taylor-series/taylor-series-error-estimation.html", "openwebmath_score": 0.821217954158783, "openwebmath_perplexity": 629.538565100748, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9615338090839606, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8530772224704579}}
{"url": "https://math.stackexchange.com/questions/3115811/does-there-exist-an-open-subset-a-subset-0-1-such-that-m-a-neq-m-bar", "text": "# Does there exist an open subset $A \\subset [0,1]$ such that $m_*(A)\\neq m_*(\\bar{A})$?\n\nDoes there exist an open subset $$A \\subset [0,1]$$ such that $$m_*(A)\\neq m_*(\\bar{A})$$?\n\nI was thinking we could approximate any set from inside by a closed set . This need not true from outside.\n\nSo I was expecting there should be some counterexample to the above statement.\n\nAny help will be appreciated\n\n\u2022 Is $m_*$ outer or inner Lebesgue measure? \u2013\u00a0Alex Ortiz Feb 17 at 4:05\n\u2022 Sir it is outer measure. Is answer will change if we put inner measure? As i had only studied outer measure. I was thinking till thar inner measure will work same. \u2013\u00a0MathLover Feb 17 at 4:08\n\u2022 It is actually inconsequential since both inner and outer measure are subadditive with respect to countable unions; see my answer below. Cheers! \u2013\u00a0Alex Ortiz Feb 17 at 4:19\n\nEnumerate the rational numbers in $$[1/4,3/4]$$ as $$\\{r_n:n\\in\\mathbf{Z}_{>0}\\}$$, and let $$\\epsilon < 1/8$$. For each $$n$$, choose an interval $$I_n$$ centered at $$r_n$$ of width $$\\epsilon/2^n$$, and put $$A = \\bigcup_{n=1}^\\infty I_n$$. Note that $$A$$ is an open subset of $$\\mathbf R$$ contained in $$[0,1]$$ because it is a union of open intervals. Since $$A$$ contains the rationals in $$[1/4,3/4]$$, density of $$\\mathbf Q$$ in $$\\mathbf R$$ implies that the closure $$\\overline A$$ contains the closed interval $$[1/4,3/4]$$.\nBy countable subadditivity (of Lebesgue outer or inner measure, this is an inconsequential point for the matter at hand), $$m_*(A) \\le \\sum_{n=1}^\\infty m_*(I_n) = \\sum_{n=1}^\\infty \\frac{\\epsilon}{2^n} = \\epsilon < 1/8.$$ On the other hand, by monotonicity, $$m_*\\big(\\overline A\\big) \\ge m_*\\big([1/4,3/4]\\big) = 1/2,$$ so $$m_*(A) < 1/8 < 1/2 = m_*\\big(\\overline A\\big).$$\n\u2022 Why not just let $A$ be an open subset of $(0,1)$ containing $\\mathbb Q \\cap (0,1)$ such that $m(A)<1/2?$ \u2013\u00a0zhw. Feb 17 at 4:43", "date": "2019-10-21 22:29:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3115811/does-there-exist-an-open-subset-a-subset-0-1-such-that-m-a-neq-m-bar", "openwebmath_score": 0.9168519973754883, "openwebmath_perplexity": 183.66882289780833, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9945307242665616, "lm_q2_score": 0.8577681068080749, "lm_q1q2_score": 0.8530767365165921}}
{"url": "http://math.stackexchange.com/questions/134979/question-on-modular-arithmetic", "text": "# Question on modular arithmetic?\n\nWould this ever be possible:\n\n$$a b^2 \u2261 a \\pmod p$$\n\nwhere $p$ is a prime number and $1<a<p$ and $1<b$\n\nPlease back your answer with some kind of proof other than Fermat's Little Theorem. This isn't homework.\n\n-\nPut $a=0$ then true for all $b$. \u2013\u00a0 john w. Apr 21 '12 at 22:03\nSorry, forgot to include $p$ and $b$ have to be greater than 1! \u2013\u00a0 user26649 Apr 21 '12 at 22:06\nAny $b \\equiv \\pm 1 \\pmod p.$ \u2013\u00a0 Will Jagy Apr 21 '12 at 22:07\n\nHint $\\$ prime $\\rm\\: p\\ |\\ ab^2-a = a\\:(b-1)(b+1)\\ \\Rightarrow\\ p\\ |\\ a\\ \\ or\\ \\ p\\ |\\ b-1\\ \\ or\\ \\ p\\ |\\ b+1$\n\n-\n@Downvoter: Why? If something is not clear then please feel free to ask questions and I am happy to elaborate. \u2013\u00a0 Bill Dubuque Apr 21 '12 at 23:45\nI don't see why the answer was downvoted... Anyways, your answer was perfect, thank you! \u2013\u00a0 user26649 Apr 22 '12 at 0:50\n\nIf $a\\equiv0\\bmod p$, the given relation is true for all $b$.\n\nIf $a\\not\\equiv0\\bmod p$ then $a$ can be cancelled from both sides of the congruence which reduces to $b^2\\equiv1\\bmod p$, hence $b\\equiv\\pm1\\bmod p$.\n\n-\nwould the downvoter care to explain? \u2013\u00a0 Andrea Mori Apr 21 '12 at 22:47\nIt seems someone is unhappy with both of our answers. I cannot imagine why. But then voting is often irrational here. \u2013\u00a0 Bill Dubuque Apr 21 '12 at 23:48\n\nSince $p$ is prime, we can cancel $a$ from both sides unless $a\\equiv 0 \\pmod p$ (this doesn't always work if $p$ is a composite number). We get $b^2 \\equiv 1 \\pmod p$. Again, since $p$ is prime, a number can have only two mod-$p$ square roots (it can have more than two in some cases if $p$ is composite). The two square roots of $1$ are $\\pm 1$ and $-1$ is of course the same as $p-1$.\n\nSo $ab^2\\equiv a\\pmod p$ holds if either $a\\equiv 0$ or $b\\equiv\\pm1$, but not otherwise.\n\nSo which parts of this do you want proofs of? The proof that nonzero numbers are invertible in mod $p$ (thereby justifying the cancelation from both sides)? I once posted an answer that shows how to find the inverse. The proof that there can't be more than two square roots if $p$ is prime? (That latter fact is a sort of corollary.)\n\n-", "date": "2014-10-20 21:41:03", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/134979/question-on-modular-arithmetic", "openwebmath_score": 0.9266001582145691, "openwebmath_perplexity": 370.0975370259731, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305360354471, "lm_q2_score": 0.8856314723088732, "lm_q1q2_score": 0.8530672778019381}}
{"url": "http://mathhelpforum.com/differential-geometry/170044-proving-cantor-set-integrable.html", "text": "# Math Help - proving the cantor set is integrable\n\n1. ## proving the cantor set is integrable\n\nI would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.\n\nLet $C$ be the Cantor set. Let $f:[0,1]\\rightarrow \\mathbb{R}$ be determined by $$f(x)= \\begin{cases} 1, &\\text{when x\\in C }\\\\ 0, &\\text{when x\\notin C }\\\\ \\end{cases}$$\n\nShow that $f$ is Riemann integrable on $[0,1]$ and $\\int _{0}^{1}f=0$. [That is the \"length of the Cantor set is 0.]\n\nI want to approach it like this but I'm not sure if it is correct:\nWe want to show that $f$ is integrable we need to show that $\\{\\sum S|S \\text {is a lower step function of f}\\} - \\{\\sum s|s \\text {is a lower step function of f}\\}<\\epsilon$ for some $\\epsilon >0$.\nWe can make $\\sum s=0$ when $x\\notin C$. So i think all we have to do is to show that $\\sum S< \\epsilon$. However, I'm not sure how to do this.\n\n2. Originally Posted by zebra2147\nI would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.\n\nLet $C$ be the Cantor set. Let $f:[0,1]\\rightarrow \\mathbb{R}$ be determined by $$f(x)= \\begin{cases} 1, &\\text{when x\\in C }\\\\ 0, &\\text{when x\\notin C }\\\\ \\end{cases}$$\n\nShow that $f$ is Riemann integrable on $[0,1]$ and $\\int _{0}^{1}f=0$. [That is the \"length of the Cantor set is 0.]\n\nI want to approach it like this but I'm not sure if it is correct:\nWe want to show that $f$ is integrable we need to show that $\\{\\sum S|S \\text {is a lower step function of f}\\} - \\{\\sum s|s \\text {is a lower step function of f}\\}<\\epsilon$ for some $\\epsilon >0$.\nWe can make $\\sum s=0$ when $x\\notin C$. So i think all we have to do is to show that $\\sum S< \\epsilon$. However, I'm not sure how to do this.\nSo start with the definition of the Cantor set. There are certain \"removed\" intervals, what are they?\n\n3. The middle third of each line segment is removed. For example, for [0,1] we would have [0,1/3]U[2/3,1]. So, (1/3,2/3) would be removed.\n\n4. Try proving that $[0,1] - C$ has measure 1, implying that C has measure 0.\n\n5. Ok, here is what I want to say but I'm not sure how to write it correctly of if its even in the right direction...\n\nIf we let $s$ be a lower step function of $f$, then we see that $\\sum s=0$. Then if we choose some upper step function, $S$, such that $\\sum S=\\sum A_{i}(x_{i}-x_{i-1})$ where $A_{i}=\\text{height of the rectangle}$ then for any interval contained in $[0,1]$, we can find an element, $x_i$ that is contained in the Cantor set and is infinitely close to another element $x_{i-1}$ that is not contained in the Cantor set. Thus, we have $x_{i}-x_{i-1}\\approx 0$. Thus, $\\sum A_{i}(x_{i}-x_{i-1})\\approx 0=\\sum S$.\nThus, $\\sum s+\\sum S\\approx 0$ which is less then any $\\epsilon >0$. Thus, $\\int_{0}^{1}f=0$.\n\nI know that is not exactly the direction that you were leading me to but I think it is similar.\n\n6. I think that it would be better to show that the upper sum and lower sum are equal to each other. Recall that\n\n$\\displaystyle U(f)=\\inf_{P} U(f,P)$ and $\\displaystyle L(f)=\\sup_{P} L(f,P)$\n\nwhere $P$ refers to some partition of $[0,1]$. We say that $f$ is Riemann-integrable if $U(f)=L(f)$.\n\nNow I propose that we consider a sequence of partitions based on the cuts performed in the construction of the Cantor set. Let $P_1=\\{0,1/3,2/3,1\\}$ (corresponding to the removal of the middle third of the interval). Then, $U(f,P_1)=L(f,P_1)=2/3$. Let $P_2=\\{0,1/9,2/9,1/3,2/3,7/9,8/9,1\\}$ (again, corresponding to the middle thirds of the remaining intervals). Then, $U(f,P_2)=L(f,P_2)=4/9$. Perform this same procedure to get $P_3, P_4, \\ldots, P_n,\\ldots$. Now taking $n\\to \\infty$, we have $U(f,P_n)\\to 0$ and $L(f,P_n)\\to 0$. This shows that\n\n$U(f)\\leq U(f,P_n)\\to 0$ and $L(f)\\geq L(f,P_n)\\to 0$\n\nor in other words,\n\n$U(f)\\leq 0\\leq L(f)$\n\nBut at the same time, from the definition of the upper and lower sums,\n\n$U(f)\\geq L(f)$\n\nTherefore $U(f)=L(f)=0$, so this function $f$ is Riemann-integrable with integral value 0.", "date": "2015-04-26 17:45:01", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/differential-geometry/170044-proving-cantor-set-integrable.html", "openwebmath_score": 0.9138097763061523, "openwebmath_perplexity": 67.4618960502059, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363704773486, "lm_q2_score": 0.8652240930029117, "lm_q1q2_score": 0.8530559019048467}}
{"url": "http://math.stackexchange.com/questions/394075/quotient-rings-of-gaussian-integers-mathbbzi-2-mathbbzi-3", "text": "# Quotient rings of Gaussian integers $\\mathbb{Z}[i]/(2)$, $\\mathbb{Z}[i]/(3)$, $\\mathbb{Z}[i]/(5)$,\n\nI am studying for an algebra qualifying exam and came across the following problem.\n\nLet $R$ be the ring of Gaussian Integers. Of the three quotient rings $$R/(2),\\quad R/(3),\\quad R/(5),$$ one is a field, one is isomorphic to a product of fields, and one is neither a field nor a product of fields. Which is which and why?\n\nI know that $2=(1+i)(1-i)$ and $5=(1+2i)(1-2i)$, so neither $(2)$ nor $(5)$ is a prime ideal of $R$. Then (I think) these same equations in $R/(2)$ and $R/(5)$, respectively, show that neither is an integral domain. Regardless, I can show that 3 is a Gaussian prime, hence $(3)$ is maximal in $R$ and $R/(3)$ is the field. But if I am correct about the others not being integral domains, I fail to see how either could be a product of fields.\n\nI hope that this can be answered easily and quickly. Thanks.\n\n-\nA product of fields is not an integral domain. Consider the zero divisors $(1,0)$ and $(0,1)$. \u2013\u00a0 Jared May 17 '13 at 1:07\nThat's helpful, thanks. Since R/(2) has just 4 elements, if it were isomorphic to a product of fields, then it would necessarily be isomorphic to Z/(2)xZ/(2). But $i^2=1$ in R/(2), and every element of Z/(2)xZ/(2) when squared equals itself. So... From the unusual phrasing of the question, I suppose R/(5) is isomorphic to a product of fields. Is this right? Would that product be Z/(5)/Z/(5)? \u2013\u00a0 John Adamski May 17 '13 at 1:34\nYou might find this helpful: Splitting of prime ideals in Galois Extensions (Wikipedia). \u2013\u00a0 kahen May 17 '13 at 1:51\n\nNow is a very good time for a quick foray into the ideal-theoretic version of Sun-Ze (better known as the Chinese Remainder Theorem). Let $R$ be a commutative ring with $1$ and $I,J\\triangleleft R$ coprime ideals, i.e. ideals such that $I+J=R$. Then\n\n$$\\frac{R}{I\\cap J}\\cong\\frac{R}{I}\\times\\frac{R}{J}.$$\n\nFirst let's recover the usual understanding of SZ from this statement, then we'll prove it. Thanks to Bezout's identity, $(n)+(m)={\\bf Z}$ iff $\\gcd(n,m)=1$, so the hypothesis is clearly analogous. Plus we have $(n)\\cap(m)=({\\rm lcm}(m,n))$. As $nm=\\gcd(n,m){\\rm lcm}(n,m)$, if $n,m$ are coprime then compute the intersection $(n)\\cap(m)=(nm)$. Thus we have ${\\bf Z}/(nm)\\cong{\\bf Z}/(n)\\times{\\bf Z}/(m)$. Clearly induction and the fundamental theorem of arithmetic (unique factorization) give the general algebraic version of SZ, the decomposition ${\\bf Z}/\\prod p_i^{e_i}{\\bf Z}\\cong\\prod{\\bf Z}/p_i^{e_i}{\\bf Z}$.\n\n(How this algebraic version of SZ relates to the elementary-number-theoretic version involving existence and uniqueness of solutions to systems of congruences I will not cover.)\n\nWithout coprimality, there are counterexamples though. For instance, if $p\\in\\bf Z$ is prime, then the finite rings ${\\bf Z}/p^2{\\bf Z}$ and ${\\bf F}_p\\times{\\bf F}_p$ (where ${\\bf F}_p:={\\bf Z}/p{\\bf Z}$) are not isomorphic, in particular not even as additive groups (the product is not a cyclic group under addition).\n\nNow here's the proof. Define the map $R\\to R/I\\times R/J$ by $r\\mapsto (r+I,r+J)$. The kernel of this map is clearly $I\\cap J$. It suffices to prove this map is surjective in order to establish the claim. We know that $1=i+j$ for some $i\\in I$, $j\\in J$ since $I+J=R$, and so we further know that $1=i$ mod $J$ and $1=j$ mod $I$, so $i\\mapsto(I,1+J)$ and $j\\mapsto(1+I,J)$, but these latter two elements generate all of $R/I\\times R/J$ as an $R$-module so the image must be the whole codomain.\n\nNow let's work with ${\\cal O}={\\bf Z}[i]$, the ring of integers of ${\\bf Q}(i)$, aka the Gaussian integers. Here you have found that $(2)=(1+i)(1-i)=(1+i)^2$ (since $1-i=-i(1+i)$ and $-i$ is a unit), that the ideal $(3)$ is prime, and that $(5)=(1+2i)(1-2i)$. Furthermore $(1+i)$ is obviously not coprime to itself, while $(1+2i),(1-2i)$ are coprime since $1=i(1+2i)+(1+i)(1-2i)$ is contained in $(1+2i)+(1-2i)$. Alternatively, $(1+2i)$ is prime and so is $(1-2i)$ but they are not equal so they are coprime. Anyway, you have\n\n\u2022 ${\\bf Z}[i]/(3)$ is a field and\n\u2022 ${\\bf Z}[i]/(5)\\cong{\\bf Z}[i]/(1+2i)\\times{\\bf Z}[i]/(1-2i)$ is a product of fields.\n\nGo ahead and count the number of elements to see which fields they are. However, ${\\bf Z}[i]/(2)={\\bf Z}[i]/(1+i)^2$ is not a field or product of fields, although the fact that its characteristic is prime (two) may throw one off the chase. In ${\\bf Z}[i]/(1+i)^2$, the element $1+i$ is nilpotent. Since this ring has order four, it is not difficult to check that it is isomorphic to ${\\bf F}_2[\\varepsilon]/(\\varepsilon^2)$, which is not a product of fields since $\\epsilon\\leftrightarrow 1+i$ is nilpotent and products of fields contain no nonzero nilpotents.\n\n-\n\nWhy don't you first think through the case of $\\mathbb Z/n\\mathbb Z$. When is this a field? When is it a product of fields? When is it neither? Try to relate your answer to the factorization of $n$ and the Chinese Remainder Theorem. Now see if you can generalize it to $\\mathbb Z[i]$ (or to any PID).\n\n-\n\nI find quotients of polynomial rings easier to reason with algebraically than subfields of $\\mathbb{C}$.\n\nBecause $\\mathbb{Z}[i] \\cong \\mathbb{Z}[x] / (x^2 + 1)$, we can do arithmetic with rings and ideals. Calculating in more detail than is usually necessary:\n\n\\begin{align} \\mathbb{Z}[i] / (3) &\\cong \\left( \\mathbb{Z}[x] / (x^2 + 1) \\right) / (3) \\\\&\\cong \\mathbb{Z}[x] / (3, x^2 + 1) \\\\&\\cong \\left( \\mathbb{Z}[x] / (3) \\right) / (x^2 + 1) \\\\&\\cong \\left(\\mathbb{Z} / (3)\\right)[x] / (x^2 + 1) \\\\&\\cong \\mathbb{F}_3[x] / (x^2 + 1) \\end{align}\n\nand so we've reduced the problem to one of quotients of polynomial rings over the finite field $\\mathbb{F}_3$.\n\n-\n\n$\\Bbb Z[i]/2\\,$ isn't a field or $\\color{#c00}{\\Pi F}$ (product of fields), since it has a nonzero nilpotent: $\\,(1\\!+\\!i)^2 = 2i = 0$.\n$\\Bbb Z[i]/5\\,$ isn't a field by $\\,(2\\!-\\!i)(2\\!+\\!i)= 5 = 0,\\,$ so it is the $\\color{#c00}{\\Pi F},\\,$ leaving $\\,\\Bbb Z[i]/3\\,$ as the field. $\\$ QED\n\n-", "date": "2015-01-27 21:25:04", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/394075/quotient-rings-of-gaussian-integers-mathbbzi-2-mathbbzi-3", "openwebmath_score": 0.9495847821235657, "openwebmath_perplexity": 202.7069644038054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985936370890583, "lm_q2_score": 0.8652240912652671, "lm_q1q2_score": 0.85305590054918}}
{"url": "https://math.stackexchange.com/questions/3522936/visualizing-the-set-of-points-in-a-regular-polygon-closer-to-center-than-to-vrti", "text": "# Visualizing the set of points in a regular polygon closer to center than to vrtices (Voronoi cell)?\n\nLet $$P_n$$ be the regular convex $$n$$-gon centered at $$p_0$$ with $$n$$ vertices $$p_1, p_2, ..., p_n$$ and $$(n>2)$$.\n\nLet $$S_n$$ be the set of all points \"$$s$$\" within the region bounded by $$P_n$$ where:\n\n$$distance(s,p_0) \\leq distance(s,p_k)$$... $$(\\forall k=\\{1,2,...,n\\})$$.\n\nFor instance $$S_3$$ would look like this:\n\nAnd $$S_4$$ would look like this:\n\nAnd just thinking about this to the extreme... $$\\lim_{n\\to\\infty}S_n$$ should look something like this (call it \"$$S_\\infty$$\" for simplicity):\n\nIs there a way to show (either using geometry or a simple brute-force code in Python) what $$S_n$$ should look like for any $$n>2$$?\n\nI would really like to see what shapes emerge. Is there a way to animate it in Python? Like have circles radiate from each vertex $$p_k$$ and the center $$p_0$$ until they crash into each other to make straight lines defining the new region $$S_n$$?\n\nDepending on the answer, I'd ideally like to transpose this question into 3-D for the 5 platonic solids.\n\nFor instance, examining a cube (i.e. 8 vertices), define $$D_8$$ as the set of all points \"$$s$$\" such that:\n\n$$distance(s,p_0) \\leq distance(s,p_k)$$... $$(\\forall k=\\{1,2,...,8\\})$$.\n\nThen $$D_8$$ is actually a truncated octahedron! I'm super curious what the other 4 platonic solids produce... As well as what higher $$S_n$$ regions look like (e.g. for a pentagon, hexagon, octagon, dodecagon, etc.).\n\nHere's an insight that I find to be an elegant solution for n=3,4, or 6 (triangle, square, hexagon). Observe that you can tessellate all of 2-D space with regular triangles, squares, and regular hexagons. Now imagine each vertex $$p_k$$ is actually the center of a nearby triangle/square/hexagon (centered at $$p_k$$ instead of at $$p_0$$).\n\nFor instance: $$S_3$$ can be more easily distinguished by drawing 3 more triangles centered at $$p_1, p_2 and p_3$$.\n\nSimilarly for $$S_4$$, drawing 4 more squares centered at $$p_1, p_2, p_3 and p_4$$ would show us the shape below, which we can then easily discern needs to be distilled into the $$S_4$$ shading we saw above.\n\nEDIT:\n\nBased on the comment about Voronoi Diagrams, I did more research and tried it out in R. Here's the region $$S_8$$ for example (use your imagination to connect the 8 outer dots to form the enclosing octagon):\n\nThe upshot seems to be that actually only $$n=3$$ and $$n=4$$ were interesting. For $$n>4$$ the pattern is simply to create a smaller version of the n-gon rotated by the internal angle of that n-gon. The shrinkage continues forever, but is also slowing forever, with the limit being circle of radius $$\\frac{r}{2}$$ shown in the $$S_\\infty$$ diagram from above.\n\nMore directly:\n\n$$\\dfrac{Area(S_3)}{Area(P_3)}=\\dfrac{2}{3}$$\n\n$$\\dfrac{Area(S_4)}{Area(P_4)}=\\dfrac{1}{2}$$\n\n$$\\dots = \\dots$$\n\n$$\\dfrac{Area(S_\\infty)}{Area(P_\\infty)}=\\dfrac{1}{4}$$ (limit -> lower bound)\n\n\u2022 You're talking about the Voronoi diagram of the vertices and the centroid.\n\u2013\u00a0user856\nJan 26 '20 at 7:47\n\u2022 Geometrically, that inequality represents the half-space containing $p_0$ defined by the perpendicular bisecting plane of the line segment connecting $p_0$ and $p_k$. Take the intersection of all these half-spaces with the original polyhedron. Jan 26 '20 at 22:24\n\u2022 @mr_e_man As said in a compact way by Rahul, it is the Voronoi cell associated with the center $p_0$, in the Voronoi diagram generated by the $p_k$s. Jan 29 '20 at 19:08\n\nYou can obtain it easily with Wolfram Mathematica (see the pictures below).\n\nP[n_] := Table[{Cos[2*Pi*i/n], Sin[2*Pi*i/n]}, {i, 0, n}];\n\ngraph[n_] := RegionPlot[AllTrue[Table[(x - P[n][[i, 1]])^2 + (y - P[n][[i, 2]])^2, {i, 1,n}], # > x^2 + y^2 &], {x, -1, 1}, {y, -1, 1}, PlotPoints -> 20];\n\nThe 3D version can be built in the same way. For instance, for the tetrahedron:\n\nP = {{1, 1, 1}, {1, -1, -1}, {-1, 1, -1}, {-1, -1, 1}};\n\np2 = ConvexHullMesh[P, MeshCellStyle -> {1 -> {Thick, Black}, 2 -> Opacity[0.2]}];\n\np1 = RegionPlot3D[ RegionMember[p2, {x, y, z}] && AllTrue[ Table[ Norm[{x, y, z} - P[[i]]], {i, 1, Length[P]}], # > x^2 + y^2 + z^2 &] , {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, PlotPoints -> 60];\n\nShow[p2, p1, ViewPoint -> {2, 4, 4}]\n\n\u2022 It's neat that for the cube some of the faces are perfect hexagons.\n\u2013\u00a0user856\nJan 29 '20 at 10:57\n\u2022 @PierreCarre can you please provide the sample code for the cube? I'm not too familiar with these functions in mathematica unfortunately Jan 31 '20 at 2:45\n\u2022 @Andrew I just added some code for the 3D case. Jan 31 '20 at 10:17", "date": "2021-10-16 13:58:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3522936/visualizing-the-set-of-points-in-a-regular-polygon-closer-to-center-than-to-vrti", "openwebmath_score": 0.43800199031829834, "openwebmath_perplexity": 377.911788716589, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426382811477, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8530538798117393}}
{"url": "https://math.stackexchange.com/questions/1797516/biased-coin-with-a-3-4-chance-to-land-on-the-side-it-was-before-the-flip", "text": "# Biased coin with a $3/4$ chance to land on the side it was before the flip\n\nConsider a hypothetical coin (with two sides: heads and tails) that has a $3/4$ probability of landing on the side it was before the flip (meaning, if I flip it starting heads-up, then it will have an only $1/4$ probability of landing tails-up). If it begins on heads, what is the probability that it is on tails after 10 flips? What about 100 flips? Assume that each flip starts on the same side as it landed on the previous flip.\n\nNote: this is not a homework problem, just something I thought up myself.\n\n\u2022 I was with you all the way up to the last sentence (Assume that each flip...) - what does that mean? The rest makes perfect sense. \u2013\u00a0mathguy May 24 '16 at 3:27\n\u2022 Coin flips have been shown to have a bias according to the side they start on actually. Don't remember the researcher name though, pretty famous person in California iirc, look it up. I'm guessing that the op wasn't aware of that though. \u2013\u00a0jdods May 24 '16 at 3:30\n\u2022 This could probably be solved by setting up a simple 2 state Markov chain. I'd write it up but an away for the evening now. \u2013\u00a0jdods May 24 '16 at 3:33\n\u2022 \"this is not a homework problem, just something I thought up myself\" -- then well done, because as it happens this is a fairly common kind of exercise! You've found the \"right\" problem, the one that (with others like it) provokes the theory of Markov chains. \u2013\u00a0Steve Jessop May 24 '16 at 10:18\n\u2022 @mathguy It seems to me he's saying the side facing up doesn't change from the time the coin lands to the time that it's flipped again. Which sounds redundant, but it's better than being unclear. \u2013\u00a0Turambar May 24 '16 at 12:46\n\nLet $p_n$ be the probability the coin is on tails after $n$ flips. Note that $p_0=0$.\n\nThe coin can be on tails after $n+1$ flips in two different ways: (i) it was on tails after $n$ flips, and the next result was a tail or (ii) it was on heads after $n$ flips, and the next result was a tail.\n\nThe probability of (i) is $(3/4)p_n$ and the probability of (ii) is $(1/4)(1-p_n)$. Thus $$p_{n+1}=\\frac{1}{4}+\\frac{1}{2}p_n.\\tag{1}$$ Solve this recurrence relation. The general solution of the homogeneous recurrence $p_{n+1}=\\frac{1}{2}p_n$ is $A\\cdot \\frac{1}{2^n}$. A particular solution of the recurrence (1) is $\\frac{1}{2}$. So the general solution of the recurrence (1) is $$p_{n}=A\\cdot \\frac{1}{2^n}+\\frac{1}{2}.$$ Set $p_0=0$ to find $A$. We find that $p_n=\\frac{1}{2}-\\frac{1}{2^{n+1}}$.\n\nIf you mean that $\\Pr(H_i|H_{i-1}) = \\Pr(T_i|T_{i-1}) = 3/4$ for all $i$, then this is a two-state homogeneous Markov chain. The transition matrix is\n\n$$P = \\begin{pmatrix} 3/4 & 1/4 \\\\ 1/4 & 3/4 \\end{pmatrix}.$$\n\nThis should get you started....\n\nI'd comment but I don't have the rep, strangely you require 50.\n\nAlso, with regards to the bias in a coin flip, it's a paper by Diaconis:\n\nLet $p_n$ be the probability that the coin is heads up after $n$ tosses. Then $p_0=1$ since it starts on heads, and $$p_n=\\frac34p_{n-1}+\\frac14(1-p_{n-1})$$ which simplifies to $$\\left(p_n-\\frac12\\right)=\\frac12\\left(p_{n-1}-\\frac12\\right)\\ .$$ Iterating, $$p_n-\\frac12=\\Bigl(\\frac12\\Bigr)^{n+1}$$ so $$p_n=\\frac12+\\Bigl(\\frac12\\Bigr)^{n+1}\\ .$$\n\n\u2022 This no longer fails for $p_1 = \\frac 3 4$. \u2013\u00a0Axoren May 24 '16 at 3:36\n\nAnother way of looking at it is to imagine that you toss two fair coins each time, using the second coin to choose between the first coin and the previous (originally heads). If the second coin always chooses the previous result then you end up with heads, while if at any point the second coin chooses the first coin then as this is fair the end result will have equal probability of being heads or tails. The chances of the latter being $1-\\frac1{2^n}$, the chance of the final result being tails is therefore half of that.\n\n\u2022 This is the best possible solution for p=1/4 and I think it generalizes to all other values of p. \u2013\u00a0zyx May 24 '16 at 17:32\n\u2022 At least for p<1/2, where coin 2 has probability (1-2p) of choosing the previous result and coin 1 is fair. One gets half of $(1 - (1-2p)^n)$ which is correct. Other than changing the probability for coin 2, every word of the answer stays the same and the argument works. This answer is just brilliant. Well done. \u2013\u00a0zyx May 24 '16 at 17:49\n\n$p_n(\\text{heads}) = \\frac 3 4p_{n-1}(\\text{heads}) + \\frac 1 4 (1 - p_{n-1}(\\text{heads}))$\n\n$p_n(\\text{heads}) = \\frac 1 4 + \\frac 1 2p_{n-1}(\\text{heads})$\n\n$p_1(\\text{heads}) = \\frac 3 4$\n\nEveryone else beat me too it. Was still writing the recurrence relation. :\\\n\nAny solution with the words \"heads\" and \"tails\" is doing extra work.\n\nWe want the chance of an odd number of reversals, each with probability $p=1/4$, to happen in $n$ independent trials. The coin briefly remembers its current state but not whether that state was reached by reversing on the previous trial. This makes the reversals independent of each other.\n\nFrom the binomial theorem that probability is $$\\frac{((1-p)+p)^n - ((1-p) - p)^n)}{2} = \\frac{1}{2} - \\frac{(1 - 2p)^n}{2}$$ which is consistent with the other solutions and shows that the $(1/2)^{n}$ exponential decay of the \"memory\" is really a power of $(1-2p)$.", "date": "2020-01-26 08:50:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1797516/biased-coin-with-a-3-4-chance-to-land-on-the-side-it-was-before-the-flip", "openwebmath_score": 0.826504647731781, "openwebmath_perplexity": 342.17537321003533, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426443092214, "lm_q2_score": 0.8757869900269366, "lm_q1q2_score": 0.853053875617451}}
{"url": "https://math.stackexchange.com/questions/3776738/if-15-distinct-integers-are-chosen-from-the-set-1-2-dots-45-some-t", "text": "# If $15$ distinct integers are chosen from the set $\\{1, 2, \\dots, 45 \\}$, some two of them differ by $1, 3$ or $4$.\n\n$$\\blacksquare~$$ Problem: If $$15$$ distinct integers are chosen from the set $$\\{1, 2, \\dots, 45 \\}$$, some two of them differ by $$1, 3$$ or $$4$$.\n\n$$\\blacksquare~$$ My Approach:\n\nLet the minimum element chosen be $$n$$. Then $$n + 1 , n + 3 , n + 4$$ can't be taken.\n\nWe make a small claim.\n\n$$\\bullet~$$ Claim: In a set of $$~7$$ consecutive numbers at most $$2$$ numbers can be chosen.\n\n$$\\bullet~$$ $$\\textbf{Proof:}$$ Let us name the elements of the set as $$\\{ 1,2,3,\\dots,7 \\}$$.\n\nNow let's consider the least element is chosen. If the least element is $$1$$, then $$2,4,5$$ can't be chosen.\n\nSo we are left with $$3, 6, 7$$.\n\n$$\\circ~$$ If $$~3~$$ is chosen, then $$6, 7$$ can't be in the set. And if $$~3~$$ is not chosen, then only any one of the 2 elements $$\\{ 6, 7 \\}$$ be chosen. So, a maximum of $$2$$ elements can be chosen in this case.\n\n$$\\circ \\circ~$$ If the least element is $$2,$$ then $$3, 5, 6$$ can't be there in the set. So, possible elements are 4, 7. So, one of these two can be chosen. Then, a maximum of 2 elements can be chosen in this case.\n\n$$\\circ \\circ~$$ If the least element is $$3$$, then $$4,6,7$$ gets cancelled. so only $$5$$ is left in the set i.e., $$2$$ elements at most.\n\n$$\\circ \\circ~$$ If the least element is $$4,$$ then $$5,7$$ gets cancelled. So the only element left is $$6$$.\n\nSimilarly,\n\n$$\\circ \\circ~$$ If $$5$$ is the least element then $$6$$ gets cancelled and only $$7$$ is left. i.e., two elements. If the least element is either $$6$$ or $$7$$, then there is only one element.\n\nSo a maximum of two elements in a set of $$7$$ consecutive elements can be chosen.\n\nHence, the proof of the claim is done!\n\nSo, for $$42$$ elements, a maximum of $$2 \\times 6 = 12$$ can be taken. However, $$3$$ more elements are required from $$3$$ more consecutive elements, which is not possible since only 2 elements at most can be chosen from a set of 3 consecutive elements.\n\nSo, a $$14$$ element subset can be formed such that, no two of them differ by $$1, 3, 4$$. Hence the $$15$$th element is one of the cancelled elements, that is, there exists a pair with their difference being $$1, 3$$ or $$4.$$\n\nHence, done!\n\nPlease check the solution for glitches and give new ideas too :).\n\n\u2022 What is the question? Well done. \u2013\u00a0Ross Millikan Aug 1 at 15:04\n\u2022 @RossMillikan I don't understand what you mean. \u2013\u00a0Ralph Clausen Aug 1 at 15:07\n\u2022 I missed the last line. I think it is a fine proof. \u2013\u00a0Ross Millikan Aug 1 at 15:07\n\u2022 Oh! okay @RossMillikan \u2013\u00a0Ralph Clausen Aug 1 at 15:10\n\u2022 @RossMillikan The OP added the last line after your comment \u2013\u00a0user1001001 Aug 1 at 15:17\n\nIt is a nice argument, and you have explained it very clearly, so well done.\n\nIf you are looking for improvements, and you want it to be a bit more formal, I would make two suggestions:\n\n1. You are very thorough proving the claim, going through all the cases, which is great. However you could shorten the proof of the claim by saying:\n\n\"If the least element is $$x$$, then we may subtract $$x-1$$ from all the numbers without changing their differences or the number of integers. Thus without loss of generality we may assume the least element is $$1$$.\"\n\nThen you do not need to consider the other cases.\n\n1. The last part of the proof (after the claim is proved) is not as thorough as the proof of the claim. You assume that for $$k=0,\\cdots,5$$ the numbers $$7k+1,7k+3$$ are selected, without justifying that this is optimal. It is obvious in a way, but for a formal proof you should justify this by saying something like:\n\n\"Pick the smallest $$k$$ where $$7k+1, 7k+3$$ are not picked. Then replace the numbers in $$\\{7k+1,\\cdots 7k+7\\}$$ that are picked with $$7k+1, 7k+3$$. From the claim we know that we have not reduced the number of integers. Also, we have not created any new differences of $$1,3,4$$.\"\n\nThen finish the argument with your second to last paragraph. I would reword the first sentence slightly: \"So of the first $$42$$ elements, we may assume these $$12$$ are picked.\"\n\nI would lose the last paragraph as it is not needed.\n\nNice proof!\n\nIn the optimal case, go up in $$2$$s wherever possible. You would start with $$1$$ and pick the smallest possible number, that is, $$3$$. We can't add $$2$$ again as that would be $$1+4$$, therefore we add the next smallest possible, $$5$$ to get our next number $$8$$. We can then add $$2$$ again, and the pattern continues:\n\n$$1,3,8,10,15,17,22,24,29,31,36,38,43,45$$\n\nVia this strategy we only pick $$14$$ numbers, so $$15$$ is impossible without violating the $$1,3,4$$ difference.\n\n\u2022 While this strategy is good in some cases, I think it's a bad general approach to proofs. I call it the \"what's the worst that could happen?\" model --- you work that out, show it's impossible, and say you're done. That relies on two things: (1) that the second worst doesn't turn out to be viable even though the worst is not, and (2) that you've correctly identified the worst case. My experience is that students (and even professionals) often manage to mess up with both failure modes. \u2013\u00a0John Hughes Aug 1 at 15:48\n\u2022 You claim that you would pick the smallest possible number in order to generate an optimal sequence. But this is something that needs to be proven. \u2013\u00a0TonyK Aug 1 at 16:32\n\nHere's a shorter version which however relies on the same ideas as your and Rhys Hughes' approach.\n\nLet's assume we can choose $$15$$ integers whose mutual differences are never $$0,1,3$$ or $$4$$, and call them $$x_1 in ascending order.\n\nClaim: For all $$1 \\le i \\le 13$$, we have $$x_i +7 \\le x_{i+2}$$.\n\nProof of claim: Let's make a case distinction about what $$x_{i+1}-x_i$$ is. Since it cannot be $$0,1,3$$ or $$4$$, the only cases are:\n\n\u2022 First case, $$x_{i+1} - x_{i} =2$$. Then if we also had $$x_{i+2} - x_{i+1} = 2$$ we would have $$x_{i+2}-x_i =4$$ which was excluded. So $$x_{i+2}-x_{i+1}$$, since it cannot be $$0,1,2,3,4$$ must be $$\\ge 5$$ which implies the claim.\n\u2022 Second case, $$x_{i+1}-x_i \\ge 5$$. Then since $$x_{i+2}-x_{i+1} \\neq 0,1$$ it must be $$\\ge 2$$ and the claim follows.\n\nThe claim is proven.\n\nNow it follows iteratively that $$x_3 \\ge x_1 +7$$, $$x_5 \\ge x_3+7 \\ge x_1+14$$, ..., $$x_{15} \\ge x_1+49$$ which of course contradicts $$1 \\le x_1 \\le x_{15} \\le 45$$. Actually this shows that you cannot even select $$15$$ such integers from the set $$\\{1, ..., 49\\}$$.", "date": "2020-08-07 18:14:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3776738/if-15-distinct-integers-are-chosen-from-the-set-1-2-dots-45-some-t", "openwebmath_score": 0.8529611229896545, "openwebmath_perplexity": 237.14805208127453, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426450627306, "lm_q2_score": 0.8757869867849167, "lm_q1q2_score": 0.853053873119499}}
{"url": "https://math.stackexchange.com/questions/388009/numerical-approximation-of-the-continuous-fourier-transform", "text": "Numerical Approximation of the Continuous Fourier Transform\n\nGiven a function $F(k)$ in frequency space (sufficiently nice enough, eg. a Gaussian), I would like to compute its Fourier inverse $$f(x) = \\frac{1}{2\\pi}\\int_{-\\infty}^{\\infty}e^{ikx}F(k)dk$$ numerically (there is no explicit analytical formula), at some specified evenly distributed points $x_n$.\n\nAssume $F(k)$ is symmetric about $k=0$ and \"essentially zero\" outside the interval $-a/2\\leq k \\leq a/2$.\n\nNaively, one way I would proceed is by dividing the interval $(-a/2,a/2)$ into $N$ subintervals (choose $N$ to be a power of two) and then piecewise approximate the integral: $$f(x_n) \\approx \\frac{1}{2\\pi}\\frac{a}{N}\\sum_{m=0}^{N-1} e^{ik_mx_n}F(k_m)$$ where, for example, we might take $k_m = (m-N/2)\\frac{a}{N}$for $0\\leq m\\leq N-1$.\n\nFor efficiency I would like to use the Fast Fourier Transform (FFT) to approximate $f(x_n)$. So I need to put $f(x)$ into the correct format for the Discrete Fourier Transform (DFT) $$L_k = \\frac{1}{N}\\sum_{m=0}^{N-1}G_me^{(2\\pi i) km/N},\\,\\,\\,\\,k = 0,\\ldots,N-1$$ where $G_m$ is the $m$th sampling of some given function $G$.\n\nTo proceed, we divide the interval $(-a/2,a/2)$ into $N$ pieces ($N$ a power of two) and let $k_m = (m-N/2)\\frac{a}{N}$for $0\\leq m\\leq N-1$, as before. Then $$f(x) \\approx \\frac{1}{2\\pi}\\frac{a}{N}\\sum_{m=0}^{N-1} e^{ik_mx}F(k_m)$$ However, in order for this formula to be in the correct format to use the DFT, we cannot evaluate f at any $x$: I believe they must be of the form $x_k = \\frac{2\\pi}{a}(k-N/2), 0\\leq k \\leq N-1$.\n\nSo in order to use the FFT, the points at which I can \"evaluate\" the function $f$ is determined by how I sample my function $F(k)$ and the number of sample points.\n\nBut in the naive method, I can, in principle, find $f(x)$ for any given x, regardless of how many sample points I take of the function $F(k)$.\n\nMy questions are:\n\n1. Is it even possible to use the FFT, if I need to compute f at specified points $x_n$?\n\n2. If not, are there other numerical methods that would be more suitable?\n\n\u2022 I came across your very same problem a week ago and I was quite astonished in realizing that there are no routines to do the job in known (at least those that I know) computer algebra programs (e.g. Mathematica/Matlab). The best thing I found is numerical recipes. There is a whole section on Fourier integration. Basically I just copied the routines suggested there and now I'm an happy person. I think that's the best what the market offers at least according to my experience. \u2013\u00a0lcv May 11 '13 at 0:48\n\u2022 Bythe way, the routine in numerical recipes implements fgp's trick and more. \u2013\u00a0lcv May 11 '13 at 22:27\n\u2022 @lcv: can you share the link to the routine you found? I also came across this problem and I'm still shocked that this isn't some standard Matlab routine. It took some time to realize that the maximum range of the x_k numbers is increasing with larger N. Increasing N will therefore make sure that the x_k's cover the entire x-region of interest but will not increase the density inside the x-region. \u2013\u00a0Frank Meulenaar Dec 5 '13 at 8:02\n\u2022 Of course, one can increase the densitiy of the x_k's is by increasing a. \u2013\u00a0Frank Meulenaar Dec 5 '13 at 8:29\n\u2022 @FrankMeulenaar It's numerical recipes chapter 13.9 \"Computing Fourier integrals using FFT\" . In the 'C' edition is on page 584. You can find online and downloadable versions of NR. The routine you can pretty much copy-paste it from the pdf. You'll have to work a bit to remove the comments. Otherwise the routines are available online for purchase on the NR site. Good luck! \u2013\u00a0lcv Dec 6 '13 at 20:30\n\nThe following two articles may be of use. They don't let you do the transform at any old x values. But it does allow transformation from m equally spaced points in the w domain to m equally spaced points in the x domain. The Bailey and Swarztrauber article also uses a gaussian function like yours as an example. Sorry I can't be more specific but I'm learning all this stuff myself.\n\nBailey, D., and P. Swarztrauber. 1994. \u2018A Fast Method for the Numerical Evaluation of Continuous Fourier and Laplace Transforms\u2019. SIAM Journal on Scientific Computing 15 (5): 1105\u201310. doi:10.1137/0915067.\n\nInverarity, G. 2002. \u2018Fast Computation of Multidimensional Fourier Integrals\u2019. SIAM Journal on Scientific Computing 24 (2): 645\u201351. doi:10.1137/S106482750138647X.\n\n$$\\def \\o {\\omega}$$ (Here I use $$t$$ and $$\\o$$ instead $$x$$ and $$k$$, but is equivalent) We want to calculate numerically the inverse Fourier transform of a given function $$F(\\o)$$ . By definition\n\n$$f(t)=\\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} F(\\o) e^{-i\\o t} d\\o$$\n\nWe can split it into two integrals\n\n$$f(t)=\\frac{1}{2\\pi} \\left( \\int_{0}^{\\infty} F(\\o) e^{-i\\o t} d\\o + \\int_{0}^{\\infty} F(-\\o) e^{i\\o t} d\\o \\right)$$\n\nIf the original function was real then $$F(-\\o)=F(\\o)^{*}$$, and therefore\n\n$$f(t)=\\frac{1}{\\pi} \\textbf{real} \\left( \\int_{0}^{\\infty} F(\\o) e^{-i\\o t} d\\o \\right)$$\n\nThen we use the Riemann integral definition for $$0$$ to a very great frequency $$W$$. $$I=\\int_{0}^{W} F(\\o) e^{-i\\o t} d\\o\\approx \\sum_{k=0}^{N-1} F(w_k) e^{-i \\Delta \\o k \\Delta t n} \\Delta\\o$$\n\nwhere $$\\Delta w =W/N$$ so that $$w_k= \\Delta w \\ \\ k$$ with $$k=0,1,2,..., N-1$$ . We discretized the time by $$t_n= \\Delta t \\ \\ n$$ with $$n=0,1,2,...,N-1$$. Next we make the basic assumption that\n$$W \\Delta t = 2 \\pi$$ so we end up with\n$$I = \\Delta\\o \\sum_{k=0}^{N-1} F(w_k) e^{- 2 \\pi i \\frac{n k}{N} }$$\nwhich is the basic definition of the Fast Fourier Transform. The problem was when I used it for a square function $$F(\\o)=\\frac{1}{i\\o}\\left( e^{i\\o T_0} -1 \\right)$$ This method gives an offset value. Then I modified the integral as\n\n$$I= \\Delta\\o \\left( \\sum_{k=1}^{N-1} F(w_k) e^{- 2 \\pi i \\frac{n k}{N} } + \\frac{1}{2}(F(w_0)+F(w_N)) \\right)$$\n\nWhich is the trapezoidal rule instead of the Riemman integral, and gave a better result.\n\nBest regards.\n\nE. Gutierrez-Reyes\n\n\u2022 It should say -infinity to infinity, it was a mistake. \u2013\u00a0Edahi Aug 26 '19 at 19:24\n\nOne way that comes to mind would be to compute both approximations for $f(x_n)$ and for $f'(x_n)$ one some grid $x_n$ that allows you to use FFT. To compute the approximations for $f'(x_n)$, you can use the identity $\\mathcal{F}(f')(k) = ik\\mathcal{F}(f)(k)$ (i.e., you get the fourier transform of the derivative by multiplying with $ik$).\n\nTo evaluate $f(x)$ at $y$, you'd then find the closest $x_n,x_{n+1}$ with $x_n \\leq y < x_{n+1}$ and use polynomial interpolation (with polynomials of degree 3) to find $f(y)$ from $f(x_n)$, $f(x_{n+1})$, $f'(x_n)$, $f'(x_{n+1})$.\n\nAs long as F(k) is limited and zero outside of $-2\\pi A$ to $2\\pi A$ the Shannon interpolation can be used $$f(x)=\\sum_{n=-\\infty}^{n=\\infty} f(x_n) sinc\\left(\\frac{\\pi}{T}(x-nT)\\right)$$ where $T$ is the sampling period used to determine $x_n$ and $sinc(t)=sin(t)/t$\n\nThis formula is exact as long as function spectrum ($F(k)$) is limited. Implementation of fast sinc interpolation exist", "date": "2020-12-03 07:16:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/388009/numerical-approximation-of-the-continuous-fourier-transform", "openwebmath_score": 0.805563747882843, "openwebmath_perplexity": 332.94504426933713, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426435557124, "lm_q2_score": 0.8757869851639066, "lm_q1q2_score": 0.8530538702207391}}
{"url": "https://math.stackexchange.com/questions/2960443/construct-a-continuous-real-valued-function-which-takes-zero-on-integers-and-suc/2960496", "text": "# Construct a continuous real valued function which takes zero on integers and such that image of function is not closed.\n\nI am trying to construct a continuous real valued function $$f:\\mathbb{R}\\to \\mathbb{R}$$ which takes zero on all integer points(that is $$f(k)=0$$ for all $$k\\in \\mathbb{Z}$$) and Image(f) is not closed in $$\\mathbb{R}$$\n\nI had $$f(x)=\\sin(\\pi x)$$ in mind. But image of $$f(x)$$ is closed.\n\nI have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.\n\nThe image of $$\\sin(\\pi x)$$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $$\\tanh(x)$$. Thus, we can use the function $$f(x) = \\sin(\\pi x)\\tanh(x),$$ which is obviously continuous, zero at each integer, and can be easily shown to have image $$(-1,1)$$\n\n\u2022 And if you are not familiar with $\\tanh$, you can write a similar function explicitly, for example: $$f(x)=\\sin(\\pi x)\\frac{1}{1+e^x}$$ A rational function (as the modifier factor) may also work, like: $$f(x)=\\sin(\\pi x)\\frac{x^2+1}{x^2+2}$$ Oct 18, 2018 at 9:15\n\u2022 \"To make it open, we need the peaks to get arbitrarily close to some value, but never reach them\" You seem to be confusing the notion of not-closed and open Oct 18, 2018 at 12:11\n\u2022 @AleksJ Except for $\\emptyset$ and $\\mathbb R$, every open set is not-closed. Oct 18, 2018 at 15:58\n\u2022 @eyeballfrog but not vice versa Oct 18, 2018 at 17:03\n\u2022 I believe that the point @JohnDvorak is making is that the peaks getting arbitrarily close to some value but not reaching it is necessary, but not sufficient, for the image to be open, and that trying to get an open set is not a necessary condition for it being non-closed (although it is sufficient). Oct 18, 2018 at 17:31\n\nThe simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $$1$$ in the $$\\pm$$infinite limit: $$f(x) = \\frac{1+|x|}{2+|x|}\\cdot\\sin(\\pi\\cdot x)$$ Plotted together with the asymptotes:\n\nIncidentally, since you just said \u201cnot closed\u201d but not whether it should be bounded, we could also just choose\n\n$$f\\!\\!\\!\\!/(\\!\\!\\!\\!/x\\!\\!\\!\\!/)\\!\\!\\!\\!/ =\\!\\!\\!\\!/ x\\!\\!\\!\\!/\\cdot\\!\\!\\!\\!/\\sin\\!\\!\\!\\!/\\!\\!\\!\\!\\!\\!\\!\\!/(\\pi\\!\\!\\!\\!/\\cdot\\!\\!\\!\\!/ x\\!\\!\\!\\!/)$$\n\n(The image of this is all of $$\\mathbb{R}$$ which is actually closed, as the commenters remarked.)\n\nA more interesting example that just occured to me:\n\n$$f(x) = \\sin x \\cdot\\sin(\\pi\\cdot x)$$ Why does this work? Well, this function never reaches $$1$$ or $$-1$$, because for that to happen you would simultaneously need $$x$$ and $$\\pi\\cdot x$$ to be an odd-integer multiple of $$\\tfrac\\pi2$$. But that can never coincide because $$\\pi$$ is irrational! It does however get arbitrarily close to $$\\pm1$$, in fact it gets close to $$-1$$ quite quickly due to $$\\tfrac\\pi2 \\approx 1.5 = \\tfrac32$$. But it never actually reaches either boundary.\n\n\u2022 Can you please tell me how make these graphs? Oct 18, 2018 at 8:59\n\u2022 @StammeringMathematician plotWindow [fnPlot $\\x -> (1+abs x)/(2+abs x)*sin(pi*x)] with dynamic-plot (a Haskell library). Oct 18, 2018 at 9:04 \u2022 @leftaroundabout Thanks. These graphs are beautiful. Oct 18, 2018 at 9:07 \u2022 The proof that the image of$\\sin(x)\\sin(\\pi x)$is$(-1,1)\\$ seems nontrivial. I have no doubt it's true, but it's not clear to me how to prove it. Oct 18, 2018 at 16:55\n\u2022 +1 for not deleting the wrong answer and preventing me to make the same mistake! Also, amazing visual explanation Oct 19, 2018 at 9:50\n\nUsing piecewise linear functions (instead of $$\\sin (\\pi x)$$) makes this simpler. For each $$n \\neq 0$$ draw the triangle with vertices $$(n,0),(n+1,0)$$ and $$(n+\\frac 1 2, 1-\\frac 1 {|n|})$$. You will immediately see how to construct an example. [You will get a function (piecewise linear function) whose range contains $$1-\\frac 1 {|n|}$$ for each $$n$$ but does not contain $$1$$].\n\nIf $$f$$ verifies the desired propery, its restriction $$f|_{[n,n+1]}$$ gives a continuous function on $$[n,n+1]$$ that is zero on the edges of the interval, for any $$n \\in \\mathbb{Z}$$. Reciprocally, if we have $$f_n : [n,n+1] \\to \\mathbb{R}$$ continuous with $$f_n(n) = f_n(n+1) = 0$$ for each integer $$n$$, by the gluing lemma this gives a continuous function $$f: \\mathbb{R} \\to \\mathbb{R}$$ with $$f(n) = f_n(n) = 0$$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $$[n,n+1]$$. Now, the function\n\n$$f_n (t) = \\mu_n\\sin(\\pi(t-n))$$\n\ntakes values on $$\\mu_n[-1,1] = [-\\mu_n,\\mu_n]$$ and $$f_n(n) = f_n(n+1) = 0$$. Thus, the family $$(f_n)_n$$ induces a continuous function $$f$$ that vanishes at $$\\mathbb{Z}$$ and\n\n$$f(\\mathbb{R}) = \\bigcup_{n \\in \\mathbb{Z}} f_n([n,n+1]) = \\bigcup_{n \\in \\mathbb{Z}}[-\\mu_n,\\mu_n]$$\n\nso the problem reduces to choosing a sequence $$(\\mu_n)_n$$ so that the former union is open. One possible choice is $$\\mu_n = 1-\\frac{1}{|n|}$$ so that\n\n$$f(\\mathbb{R}) = \\bigcup_{n\\in \\mathbb{Z}}[-\\mu_n,\\mu_n] = \\bigcup_{n\\in \\mathbb{N}}[-1+\\frac{1}{n},1-\\frac{1}{n}] = (-1,1).$$\n\nConsider\n\n$$f(x) = \\sin^2 (\\pi x)\\frac{x^2}{1+x^2}.$$\n\nThen $$f$$ is continuous, $$f=0$$ on the integers, but $$f(\\mathbb R) = [0,1).$$", "date": "2022-05-25 23:50:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2960443/construct-a-continuous-real-valued-function-which-takes-zero-on-integers-and-suc/2960496", "openwebmath_score": 0.78248131275177, "openwebmath_perplexity": 228.97725195982738, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426435557124, "lm_q2_score": 0.8757869835428965, "lm_q1q2_score": 0.8530538686418061}}
{"url": "http://mathhelpforum.com/pre-calculus/97384-need-help-w-pre-calculus-summer-packet.html", "text": "# Thread: Need help w/ pre-calculus summer packet.\n\n1. ## Need help w/ pre-calculus summer packet.\n\nI know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!\n\nWrite the following absolute value expressions as piecewise expressions.\n1. |x^2 + x - 12|\nI know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.\n\nSolve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.\n2. x^2 - 16 > 0\nI got (4, \u221e) but I don't know what a sign chart is.\n\nFactor completely.\n3. x^2 + 12x + 36 - 9y^2\nStumped. I know I can factor out an x but not sure how that would help.\n\nI appreciate any help. Look forward to hearing from people!\n\n2. Originally Posted by amy1613\nI know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!\n\nWrite the following absolute value expressions as piecewise expressions.\n1. |x^2 + x - 12|\nI know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.\n\nyou should already know what y = x^2 + x - 12 looks like ... well, y = |x^2 + x - 12| is the same graph except that the part that is below the x-axis (negative) is reflected to the positive side.\n\ny = x^2 + x - 12 , x < -4 or x > 3\n\ny = -(x^2 + x - 12) , -4 < x < 3\n\nSolve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.\n2. x^2 - 16 > 0\nI got (4, \u221e) but I don't know what a sign chart is.\n\nx^2 - 16 > 0 ... note that x^2 - 16 = 0 at x = + 4 and x = -4\n\nplot these two numbers on a number line ... it breaks the number line into three sections ...\nx < -4 , -4 < x < 4 , and x > 4\n\ntake any number in each interval, and \"test\" it in the original inequality ... if it makes the inequality true, then all values of x in that interval make the inequality true.\n\nyou'll see that you forgot the interval (-\u221e , -4)\n\nFactor completely.\n3. x^2 + 12x + 36 - 9y^2\n\nhelp any?\n...\n\n3. yeah this helped a lot! thanks! so...a sign chart is just a number line graph?\n\n4. Hello, amy1613!\n\nWrite the following absolute value expression as a piecewise expression:\n. . $1)\\;\\;f(x) \\;=\\;|x^2 + x - 12|$\nThere are two cases to consider: . $\\begin{array}{cccc}(1) & x^2 + x - 12 \\:\\geq \\;0 \\\\ (2) & x^2 + x - 12 \\:<\\:0 \\end{array}$\n\n$\\text{Case }(1)\\;\\;x^2 + x - 12 \\:\\geq\\:0\\quad\\Rightarrow\\quad (x-3)(x+4) \\:\\geq \\:0$\n\n. . There are two ways:\n\n. . . . $(a)\\;\\;\\begin{Bmatrix}x-3 \\:\\geq\\:0 & \\Rightarrow & x \\:\\geq \\:3 \\\\ x+4 \\:\\geq \\:0 & \\Rightarrow & x \\:\\geq\\:\\text{-}4\\end{Bmatrix}\\quad\\Rightarrow\\quad x \\:\\geq\\:3$\n\n. . . . $(b)\\;\\;\\begin{Bmatrix}x-3 \\:<\\: 0 & \\Rightarrow & x \\:<\\:3 \\\\ x+4 \\:<\\:0 & \\Rightarrow & x \\:<\\:\\text{-}4 \\end{Bmatrix} \\quad\\Rightarrow\\quad x \\:<\\:\\text{-}4$\n\n. . Hence, if $x \\geq 3\\text{ or }x < -4$, then: . $f(x) \\:=\\:x^2+x-12$\n\n$\\text{Case }(2)\\;\\; x^2 + x - 12 \\:<\\: 0 \\quad\\Rightarrow\\quad (x-3)(x+4) \\:<\\:0$\n\n. . There are two ways:\n\n. . . . $(a)\\;\\;\\begin{Bmatrix}x-3 \\:>\\:0 & \\Rightarrow & x \\:>\\:3 \\\\ x+4 \\:< \\:0 & \\Rightarrow & x \\:<\\:\\text{-}4 \\end{Bmatrix} \\quad\\Rightarrow\\quad x > 3 \\text{ and }x < \\text{-}4$ . . . impossible\n\n. . . . $(b)\\;\\;\\begin{Bmatrix}x-3 \\:<\\:0 & \\Rightarrow & x \\:<\\:3 \\\\ x+4 \\:> \\:0 & \\Rightarrow & x \\:>\\:\\text{-}4 \\end{Bmatrix} \\quad\\Rightarrow\\quad \\text{-}4 \\:<\\:x \\:<\\: 3$\n\n. . Hence, if $\\text{-}4 < x < 3$, then: . $f(x) \\:=\\:-(x^2+x-12) \\:=\\:12 - x - x^2$\n\nTherefore: . $f(x) \\;=\\;\\begin{Bmatrix}x^2+x-12 && \\text{if }x\\leq \\text{-}4 \\text{ or }x \\geq 3 \\\\ \\\\[-4mm] 12 -x-x^2 && \\text{if }\\text{-}4 < x < 3 \\end{Bmatrix}$\n\n5. Originally Posted by Soroban\n. . There are two ways:\n\n. . . . $(a)\\;\\;\\begin{Bmatrix}x-3 \\:\\geq\\:0 & \\Rightarrow & x \\:\\geq \\:3 \\\\ x+4 \\:\\geq \\:0 & \\Rightarrow & x \\:\\geq\\:\\text{-}4\\end{Bmatrix}\\quad\\Rightarrow\\quad x \\:\\geq\\:3$\n\n. . . . $(b)\\;\\;\\begin{Bmatrix}x-3 \\:<\\: 0 & \\Rightarrow & x \\:<\\:3 \\\\ x+4 \\:<\\:0 & \\Rightarrow & x \\:<\\:\\text{-}4 \\end{Bmatrix} \\quad\\Rightarrow\\quad x \\:<\\:\\text{-}4$\n\n. . Hence, if $x \\geq 3\\text{ or }x < -4$, then: . $f(x) \\:=\\:x^2+x-12$\n\n[/size]\nThanks for your response & explaination. I still can't quite understand why you picked x>-3 and not x>4. Could you go through that for me?", "date": "2016-08-28 05:11:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/97384-need-help-w-pre-calculus-summer-packet.html", "openwebmath_score": 0.7861992716789246, "openwebmath_perplexity": 598.4360972974831, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9740426443092214, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8530538661438537}}
{"url": "https://math.stackexchange.com/questions/2482032/number-of-solutions-to-a-10b-20c-100-with-a-b-c-non-negative-integers/2482082", "text": "# Number of solutions to $a + 10b + 20c = 100$ with $a,b,c$ non-negative integers\n\nHow many solutions are there to the equation $$a + 10b + 20c = 100$$ where $a,b,c$ are non-negative integers\n\nThe problem I'm asking is another way of phrasing \"How many ways can you break $100$ dollars into $1$, $10$, and $20$ dollar bills\", or any multitude of variants.\n\nSolutions can, for example, be $(a,b,c) = (100, 0, 0)$, $(10, 7, 1)$, $(0, 6, 2)$, etc.\n\nI wrote a quick program which gave me a total of $36$ solutions, but I'd like to have a better understanding as to where this answer comes from without something resembling brute-force. Preferably, I'd like to see how to use generating functions to solve this, as that was the method I first tried (and failed) with.\n\nThe solution should be the coefficient of the 100th-power term in the product:\n\n$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$ I can get this coefficient from the following: $$\\lim_{x \\to 0} \\frac{1}{100!} \\frac{d^{100}}{dx^{100}}(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$ For whatever reason, applying product rule fails (or I fail to use it correctly), as I get $3$. So, assuming $|x| < 1$, this equals: $$\\lim_{x \\to 0} \\frac{1}{100!} \\frac{d^{100}}{dx^{100}} \\frac{1}{(1-x)(1-x^{10})(1-x^{20})}$$ However, to much dismay, this is not such a simple thing to find derivatives of. Any advice?\n\n\u2022 I got $36$ as coefficient of $x^{100}$ so you were right... $$\\ldots + 30 x^{98}+30 x^{99}+36 x^{100}+35 x^{101}+35 x^{102}+35 x^{103}+\\ldots$$ \u2013\u00a0Raffaele Oct 20 '17 at 21:10\n\u2022 @Raffaele : You should get $36$ for every coefficient from $x^{100}$ to $x^{109}$. \u2013\u00a0Eric Towers Oct 20 '17 at 21:18\n\u2022 Calculating the 100th derivative is definitely harder than writing the solutions one by one. An idea is to use partial fraction decomposition on $\\frac{1}{(1-x)(1-x^{10})(1-x^{20})}$ and then sum the coefficients in front of $x^{100}$. But this can be rather tedious and hard to do too. \u2013\u00a0Stefan4024 Oct 20 '17 at 21:20\n\u2022 The first time I used polynomial with a degree too low. I remade my calculations with higher degree polynomials (using Mathematica) and finally got $$\\ldots+42 x^{111}+42 x^{110}+36 x^{109}+36 x^{108}+36 x^{107}+36 x^{106}+36 x^{105}+36 x^{104}+36 x^{103}+36 x^{102}+36 x^{101}+36 x^{100}+30 x^{99}+\\ldots$$ BTW, I did not know this combinatorial trick :) \u2013\u00a0Raffaele Oct 20 '17 at 21:31\n\nYou can simplify fairly drastically, since $a = 100-10b-20c$ means $a$ must be a multiple of $10$. So setting $a' := a/10$, we have $a' + b + 2c = 10$.\n\nAgain $a'$ is determined by $b$ and $c$, so we just need $b+2c\\le 10$. For $c=\\{0,1,2,3,4,5\\}$ we have $\\{11,9,7,5,3,1\\}$ options for $b$, for a total of $36$ possible combinations.\n\n\u2022 Nifty! Thanks for that argument. I was able to reduce it as well after reading up on the subject, but I quite like your closing argument to take $b + 2c \\le 10$ and check the number of options for each $b$ at fixed $c$ \u2013\u00a0infinitylord Oct 20 '17 at 22:32\n\u2022 Thanks. Although sadly(?) not using generating functions. With a little extension you can see that the same answer also applies to target values from $101$ to $109$ (using $a' = (a-r)/10$). \u2013\u00a0Joffan Oct 20 '17 at 22:39\n\u2022 Slight generalization: For $x,y,z$ non-negative integers, the number of solutions to $x + by + cz = d$ where $b|c$, and $c|d$ is $(\\frac{d}{2b} + 1)^2$ \u2013\u00a0infinitylord Oct 21 '17 at 2:08\n\nEr, ..., calculate carefully?\n\n$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...) \\\\ = \\dots +30 x^{98} + 30 x^{99} + 36 x^{100} + 36 x^{101} + ...$$\n\nAnother way: Suppose there are $n_k$ ways to write $k$ as a non-negative integer weighted sum of $\\{1,10\\}$. Then the total number of ways to write $k$ as a non-negative integer weighted sum of $\\{1,10,20\\}$ is $N = n_0 + n_{20} + n_{40} + n_{60} + n_{80} + n_{100}$, where $n_0$ counts the number of ways we finish by adding five $20$s, $n_{20}$ the number of ways we finish by adding four $20$s, ..., and $n_{100}$ the number of ways we finish by adding no $20$s.\n\n\u2022 Now $n_0 = 1$ because $0 \\cdot 1 + 0 \\cdot 10$ is the only way to write $0$ as a non-negative integer weighted sum of $\\{1,10\\}$.\n\u2022 And $n_{20} = 3$, depending on whether there are zero, one, or two $10$s in the sum, the rest being made up with $1$s.\n\u2022 And $n_{40} = 5$, for the same reason : zero to four $10$s and the remainder made of $1$s.\n\u2022 Finishing, $n_{60} = 7$, $n_{80} = 9$, and $n_{100} = 11$.\n\nFinally, $N = 1 + 3 + 5 + 7 + 9 + 11 = 36$. That the number of ways is controlled by the $10$s and $20$s, with the $1$ making up the rest, is hinted in the snippet of the series in $x$, above. There is only one way to make totals up to $9$ from $\\{1,10,20\\}$. There are two ways for totals in $[10,19]$, four ways in $[20,29]$, six ways in $[30,31]$, nine ways in $[40,41]$, where the increment in the number of ways in each range of ten totals increases by $1$ each time we pass a multiple of $20$ (counting that we could either include or exclude one more $20$ in our sums).\n\nAnother way: You don't have to multiply out the full power series. You can calculate with $$( 1 + x+ x^2 + \\cdots + x^{99} + x^{100})(1 + x^{10} + \\cdots + x^{100})(1+x^{20}+\\cdots+x^{100})$$ always discarding any term with power greater than $100$. If you do this from right to left, you essentially perform the calculation described above.", "date": "2019-07-19 19:10:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2482032/number-of-solutions-to-a-10b-20c-100-with-a-b-c-non-negative-integers/2482082", "openwebmath_score": 0.9319021701812744, "openwebmath_perplexity": 188.54968280008518, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464471055738, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8530525753563435}}
{"url": "https://math.stackexchange.com/questions/2645822/proof-verification-implication-from-field-axioms?noredirect=1", "text": "# Proof Verification: Implication from Field Axioms\n\nProve that if $a \\neq 0$ then $b/(b/a) = a$\n\nProof:\n\n$b/(b/a) = b. 1/(b/a)$ (by Multiplication Axiom 4 (M4))\n\n$\\implies b/(b/a)$ = $b.(b/a)^{-1}$ (by M5)\n\n$\\implies$ $b/(b/a)$ = $b.(b.(1/a))^{-1}$\n\n$\\implies$ $b/(b/a)$ = $b.(b.(a)^{-1})^{-1}$\n\n$\\implies$ $b/(b/a)$ = $b.[1/(b.(a)^{-1})]$\n\n$\\implies$ $b/(b/a)$ = $b . (1/b) . (1/a^{-1})$\n\n$\\implies$ $b/(b/a)$ = $1 .(1/a^{-1})$\n\n$\\implies$ $b/(b/a)$ = $(a^{-1})^{-1}$\n\n$\\implies$ $b/(b/a)$ = a\n\nIs this correct? Can anyone please verify?\n\n\u2022 I'm just curious: where do you find the axioms? \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Feb 11 '18 at 13:08\n\u2022 Do you mean derivation through axioms? \u2013\u00a0A.Asad Feb 11 '18 at 13:10\n\u2022 Though Rudia never used the notation $(\\cdot)^{-1}$ is that section, but IMHO, I think it's fine if you write it as long as it means the inverse. In the second arrow, you used a property $(xy)^{-1} = x^{-1} y^{-1}$. This can be easily proven, by as a \"proof\" from the axioms (or the propositions from the book), I think it's better to mark every step, like this answer in set theory. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Feb 11 '18 at 13:18\n\u2022 Thank you, I'll do that (i.e. prove that $(xy)^{-1} = x^{-1} y^{-1}$ and mark my steps) . Is the proof fine otherwise? \u2013\u00a0A.Asad Feb 11 '18 at 13:25\n\u2022 Yes, I think so. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 Feb 11 '18 at 13:27\n\nAs OP's comments suggest, to derive the proof from the five multiplication axioms in Baby Rudin, break $1$ into the product of $a$ and $1/a$ in the second step. $\\require{action}$ \\begin{aligned} b / (b/a) &= \\texttip{b \\cdot 1 \\cdot \\left( \\frac{1}{b/a} \\right)} {(M4): multiplication by identity} \\\\ &= \\texttip{b \\cdot \\left( \\frac{1}{a} \\right) \\cdot a \\cdot \\left( \\frac{1}{b/a} \\right)} {(M5): existence of inverse} \\\\ &= \\texttip{(b/a) \\cdot a \\cdot \\left( \\frac{1}{b/a} \\right)} {(M3): regroup the leftmost two factors} \\\\ &= \\texttip{a \\cdot (b/a) \\cdot \\left( \\frac{1}{b/a} \\right)} {(M2): multiplication is commutative} \\\\ &= \\texttip{a \\cdot \\left( (b/a) \\cdot \\left( \\frac{1}{b/a} \\right) \\right)} {(M3): multiplication is associative} \\\\ &= \\texttip{a \\cdot 1}{(M5): multiplication by inverse} \\\\ &= \\texttip{a}{(M4): multiplication by identity} \\end{aligned} \\bbox[4pt,border: 1px solid red]{ \\begin{array}{l} \\text{If you cannot figure out why a line}\\\\ \\text{is true, move your mouse over}\\\\ \\text{RHS of that line for hint.} \\end{array}} Remarks: The above proof is one line shorter than OP's one.", "date": "2020-05-31 10:13:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2645822/proof-verification-implication-from-field-axioms?noredirect=1", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 1690.2085695372089, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975946445706356, "lm_q2_score": 0.8740772368049822, "lm_q1q2_score": 0.8530525725326552}}
{"url": "https://math.stackexchange.com/questions/3328973/knowing-the-existence-of-max-min-values-of-function-in-interval-without-taking-d", "text": "# Knowing the existence of max/min values of function in interval without taking derivative?\n\nI'm given the following question.\n\n\"How do we know that the following function has a maximum value and a minimum value in the interval $$[0,3]$$\"\n\n$$f(x)=\\frac{x}{x^2+1}$$\n\nIs it possible to understand that there is a maximum and minimum value in the interval without taking the derivative of the function?\n\nIs the solution to take the derivative of $$f(x)$$ and equate the resulting function to zero and solve?\n\n\u2022 Have you heard of the extreme value theorem? \u2013\u00a0Theo Diamantakis Aug 20 '19 at 14:26\n\u2022 indeed I have but have no experience applying it which is probably the reason for the question :) I will review the theorem and see how it can be applied. \u2013\u00a0esc1234 Aug 20 '19 at 14:29\n\u2022 of course. Upon reviewing the theorem it clearly states that a continuous function on a closed interval will have a minimum and maximum value. Thanks very much! \u2013\u00a0esc1234 Aug 20 '19 at 14:36\n\u2022 Do you have an intuitive idea why that must be so, though? ANd why you might not be able to state the same thing for $x \\in (0,3)$? \u2013\u00a0fleablood Aug 20 '19 at 15:38\n\u2022 I believe so. Open interval of the problem of always being able to get closer and closer to the endpoint therefore it's technically impossible to get a max or min at an endpoint. Hence the closed interval stipulation. I get it now but I'm just used to applying these theorems. Thanks though \u2013\u00a0esc1234 Aug 20 '19 at 15:39\n\n\"The extreme value theorem applies as $$\\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$\"\n\nThe question asks you to show it has a max and min on that interval; not to find them. ANd the question doesn't ask about local maxima/minima but global extrema of all values on the interval.\n\nThe extreme value theorem says, and I quote: if a real-valued function $$f$$ is continuous on the closed interval $$[a,b]$$, then f must attain a maximum and a minimum, each at least once.\n\nAnd $$\\frac {x}{x^2+1}$$ is continuous on $$[0,3]$$. So the extreme value theorem says it attains a max and minimum at least once on the interval.\n\nso that's it. Done.\n\n====\n\nThe intuitive idea for me is that the graph of a continuous function has a distinct starting point at $$[a,f(a)]$$ and a distinct ending point at $$[b,f(b)]$$ and for every point in between will have some actual real value. It's continuous so it can't \"stretch\" in any unbounded infinite value. So somewhere in there, either at one of the endpoints or somewhere between, it will get as big as it gets (within that interval, it can get bigger in other places outside the interval) and somewhere it will get as small as it gets.\n\nThat argument is naive and applies to ill-defined ideas (the most abusive is the it can't \"stretch\" to infinity) and needs to be formally defined and proven. And that is exactly what the Extreme Value theorem states.\n\nIn this case we have a path that starts at the point $$(0,f(0) = 0)$$ and ends at the point $$(3, f(3) = \\frac 3{10})$$. In between in follows some curvy path between those two points. Somewhere there must be some point that was the biggest $$f(x)$$ value, and some that must be the least $$f(x)$$ value.\n\n===\n\nWe can get more specific. $$x \\ge 0$$ and $$x^2 + 1 \\ge 1$$ so $$f(x) \\ge \\frac x{x^2 + 1} \\ge 0$$ so $$f(0) = 0$$ is a minimum. $$x \\le 3$$ and $$x^2 + 1 \\ge 1$$ so $$f(x)\\le \\frac 31$$ and so all the possible $$f(x)$$ are bounded above by $$3$$.\n\nIt is the definition of real numbers that if a set is bounded above that a least upper bound exist and a consequence of $$f$$ being continuous means that there is an $$x=c$$ where $$f(c) = \\sup \\{f(x)|x \\in [a,b]\\}$$.\n\nWe can go further an not $$f(1) = \\frac 12 > \\frac 3{10} =f(3)$$ so the max is not $$x=3$$ but some $$x: 0< x < 3$$ and so the maximum will be a local max and we COULD find it by taking the derivative.\n\nBut the question isn't ASKING us to. The question is only asking us to argue that there is a max. An the only thing we have to say for that is:\n\n\"The extreme value theorem applies as $$\\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$\"\n\n\u2022 Exactly. This is the correct answer. Thank you \u2013\u00a0esc1234 Aug 20 '19 at 15:45\n\nUse that $$\\frac{1}{2}\\times\\frac{2x}{x^2+1}\\le \\frac{1}{2}$$ and $$\\frac{-1}{2}\\le \\frac{2x}{x^2+1}\\times \\frac{1}{2}$$\n\n\u2022 Thanks. Has this idea come from the extreme value theorem which was suggested in the comment above? \u2013\u00a0esc1234 Aug 20 '19 at 14:30\n\u2022 No, it comes from the AM-GM inequality. \u2013\u00a0Dr. Sonnhard Graubner Aug 20 '19 at 14:31\n\u2022 Okay, the AM-GM inequality is something I haven't come across in my coursework yet. I will study it but because it hasn't come up in my coursework yet I'm thinking there must be another method. \u2013\u00a0esc1234 Aug 20 '19 at 14:34\n\u2022 It is $$\\frac{a^2+b^2}{2}\\geq ab$$ for $$a,b\\geq 0$$ \u2013\u00a0Dr. Sonnhard Graubner Aug 20 '19 at 14:36\n\u2022 This shows that all values of $f(x): 0 \\le x \\le 3$ are so that $-\\frac 12 \\le f(x) \\le \\frac 12$ but not that there is a specific $c,d$ so that $-\\frac 12 \\le f(c) \\le f(x) \\le f(d) \\le \\frac 12$. \u2013\u00a0fleablood Aug 20 '19 at 15:34\n\nThe question gives a function that is continuous over a closed interval. The extreme value theorem states that a function that is continuous on a closed interval will have both a minimum and maximum value on that interval. The question requires that the student knows and understands the extreme value theorem.\n\nWe show that your function has a turning point in $$[0,3]$$ which is a maximum.\n\nNote that $$\\frac {x_1}{x_1^2+1}=\\frac {x_2}{x_2^2+1}$$ for values where $$x_1\\ne x_2$$ and $$x_1x_2=1$$\n\nFor example $$\\frac {2}{2^2+1}=\\frac{1/2}{(1/2)^2 +1}$$\n\nThus you have a turning point between $$1/2$$ and $$2$$.\n\nThus the turning point happens at $$x=1$$ and it is a maximum because for example $$f(1)=1/2 > f(2)=2/5$$", "date": "2020-10-21 18:57:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3328973/knowing-the-existence-of-max-min-values-of-function-in-interval-without-taking-d", "openwebmath_score": 0.7593608498573303, "openwebmath_perplexity": 150.173891264151, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464499040094, "lm_q2_score": 0.8740772253241803, "lm_q1q2_score": 0.8530525649970806}}
{"url": "https://sibostesaw.web.app/1246.html", "text": "Taylor series for cosx at 0\n\nHowever, when the interval of convergence for a taylor. The calculator will find the taylor or power series expansion of the given function around the given point, with steps shown. Sine function using taylor expansion c programming. In essence, the taylor series provides a means to predict a function value at one point in terms of the function value and its derivatives at another point. Part b asked for the first four nonzero terms of the taylor series for cos x about x 0 and also for the first four nonzero terms of the taylor series for fx about x 0. Evaluating limits using taylor expansions taylor polynomials provide a good way to understand the behaviour of a function near a speci. A particle moves along the xaxis wacceleration given by atcost ftsec2 for t 0. You can find the range of values of x for which maclaurins series of sinx is valid by using the ratio test for convergence. Lets look closely at the taylor series for sinxand cosx. How to extract derivative values from taylor series.\n\nX1 k 0 21kx k 2k bfind the interval of convergence for this maclaurin series. In the case of a maclaurin series, were approximating this function around x is equal to 0, and a taylor series, and well talk about that in a future video, you can pick an arbitrary x value or fx value, we should say, around which to approximate the function. Because the taylor series is a form of power series, every taylor series also has an interval of convergence. What is the taylor series at x 0 for cos xtype an exact answer. How to prove that the taylor series of sinx is true for. As you work through the problems listed below, you should reference your lecture notes and. This is part of series of videos developed by mathematics faculty at the north carolina school of science and mathematics. Ap calculus bc 2011 scoring guidelines college board. Taylors series of sin x in order to use taylors formula to. When this interval is the entire set of real numbers, you can use the series to find the value of fx for every real value of x. If i wanted to approximate e to the x using a maclaurin series so e to the x and ill put a little approximately over here. Depending on the questions intention we want to find out something about the curve of math\\frac\\sin xxmath by means of its taylor series 1.\n\nGiven n and b, where n is the number of terms in the series and b is the value of the angle in degree. Taylor series cosx function matlab answers matlab central. Program to calculate the sum of cosine series of x and compare the value with the library functions output. To find the maclaurin series simply set your point to zero 0. For taylors series to be true at a point xb where b is any real number, the series must be convergent at that point. The power series in summation notation would bring you to my answer. Approximating cosx with a maclaurin series which is like a taylor polynomial centered at x0 with infinitely many terms. Use power series operations to find the taylor series at x 0 for the following function.\n\nTo get the maclaurin series for xsin x, all you have to do is to multiply the series with x throughout, as indicated by the formula above. By using this website, you agree to our cookie policy. The taylor series for the exponential function ex at a 0 is. Derivatives derivative applications limits integrals integral applications series ode laplace transform taylormaclaurin series fourier series. Determining whether a taylor series is convergent or. This will work for a much wider variety of function than the method discussed in the previous section at the expense of some often unpleasant work. We also derive some well known formulas for taylor series of ex, cosx and sinx around x 0. Find the taylor series for expcosx about the point x 0 up to x4 really no clue where to even begin. Whats wrong with just bruteforce calculation of the first six derivatives of that function at zero, then form the taylor series at zero. Commonly used taylor series university of south carolina. A maclaurin series can be used to approximate a function, find the antiderivative of a complicated function, or compute an otherwise uncomputable sum. How to extract derivative values from taylor series since the taylor series of f based at x b is x. Taylor series integration of cosx 1 x physics forums. We focus on taylor series about the point x 0, the socalled maclaurin series.\n\nIts going to be equal to any of the derivatives evaluated at 0. Free trigonometric equation calculator solve trigonometric equations stepbystep. You can specify the order of the taylor polynomial. This could be its value at mathx 0 math as is considered a popular interview questions, i. A value is returned for all nonnegative integer values of n, but no matter how many terms i have the program provide the sum always approaches ans1, reaching it by around n5. Follow 59 views last 30 days nikke queen on 3 apr 2015. Math 142 taylormaclaurin polynomials and series prof. A maclaurin series is a taylor series where a 0, so all the examples we have been using so far can also be called maclaurin series. Im working on a taylor series expansion for the coxx function. Thus, we get the series we call the maclaurin series. Because this limit is zero for all real values of x, the radius of convergence of the. Recall that the taylor series of fx is simply x1 k 0 fk 0 k. Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers.\n\nMatlab cosx taylor series matlab answers matlab central. Write a matlab program that determines cos x using the. Finding the maclaurin series for cosx chapter 30 lesson 3 transcript video. Find the taylor series expansion of the following functions around 0.\n\nTaylor series and maclaurin series calculus 2 duration. How to evaluate sinxx using a taylor series expansion quora. Taylor and maclaurin power series calculator emathhelp. Sign up to read all wikis and quizzes in math, science, and engineering topics. The maclaurin expansion of cosx the infinite series module. Convergence of taylor series suggested reference material. Physics 116a winter 2011 taylor series expansions in this short note, a list of wellknown taylor series expansions is provided. Taylor series a taylor series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like x, x 2, x 3, etc. Calculus power series constructing a taylor series. To find the series expansion, we could use the same process here that we used for sin x. Find the maclaurin series expansion for cosx at x 0, and determine its.\n\nThe taylor series for sinx converges more slowly for. The taylor series for fxcosx at a pi2 is summation from zero to infinity cnxpi2n find the first few coefficients c0 c1 c2 c3 c4 get more help from chegg get 1. And thats why it makes applying the maclaurin series formula fairly straightforward. In mathematics, a taylor series is an expression of a function as an infinite series whose terms. We can use the first few terms of a taylor series to get an approximate value for a function. Write a matlab program that determines cos x using. As a result, if we know the taylor series for a function, we can extract from it any derivative of the.\n\nMath 142 taylor maclaurin polynomials and series prof. Find the maclaurin series expansion for cos x at x 0, and determine its radius of convergence. Free taylor series calculator find the taylor series representation of functions stepbystep this website uses cookies to ensure you get the best experience. Here we show better and better approximations for cosx. A calculator for finding the expansion and form of the taylor series of a given function. I assume you mean from 0 to pi rather than from 0 to 180 degrees because the taylor expansion works in radians, not degrees. In this section we will discuss how to find the taylormaclaurin series for a function. Sine function using taylor expansion c programming ask question asked 8 years, 4 months ago.", "date": "2021-03-01 06:07:07", "meta": {"domain": "web.app", "url": "https://sibostesaw.web.app/1246.html", "openwebmath_score": 0.9193335771560669, "openwebmath_perplexity": 335.70207560203755, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9901401455693091, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8530435680995675}}
{"url": "https://mathhelpboards.com/threads/ask-equation-of-a-circle-in-the-first-quadrant.26984/", "text": "# [ASK] Equation of a Circle in the First Quadrant\n\n#### Monoxdifly\n\n##### Well-known member\nThe center of circle L is located in the first quadrant and lays on the line y = 2x. If the circle L touches the Y-axis at (0,6), the equation of circle L is ....\na. $$\\displaystyle x^2+y^2-3x-6y=0$$\nb. $$\\displaystyle x^2+y^2-12x-6y=0$$\nc. $$\\displaystyle x^2+y^2+6x+12y-108=0$$\nd. $$\\displaystyle x^2+y^2+12x+6y-72=0$$\ne. $$\\displaystyle x^2+y^2-6x-12y+36=0$$\n\nSince the center (a, b) lays in the line y = 2x then b = 2a.\n$$\\displaystyle (x-a)^2+(y-b)^2=r^2$$\n$$\\displaystyle (0-a)^2+(6-b)^2=r^2$$\n$$\\displaystyle (-a)^2+(6-2a)^2=r^2$$\n$$\\displaystyle a^2+36-24a+4a^2=r^2$$\n$$\\displaystyle 5a^2-24a+36=r^2$$\nWhat should I do after this?\n\n#### skeeter\n\n##### Well-known member\nMHB Math Helper\ncircle center at $(x,2x)$\n\ncircle tangent to the y-axis at $(0,6) \\implies x=3$\n\n$(x-3)^2+(y-6)^2 = 3^2$\n\n$x^2-6x+9+y^2-12y+36 =9$\n\n$x^2+y^2-6x-12y+36=0$\n\n#### Monoxdifly\n\n##### Well-known member\ncircle center at $(x,2x)$\n\ncircle tangent to the y-axis at $(0,6) \\implies x=3$\nHow did you get x = 3 from (0, 6)?\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nHow did you get x = 3 from (0, 6)?\nThe fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.\n\n(And the center of the circle \"lies on the line y= 2x\", not \"lays\".)\n\n#### Monoxdifly\n\n##### Well-known member\nThe fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.\n\n(And the center of the circle \"lies on the line y= 2x\", not \"lays\".)\nWell, I admit I kinda suck at English. I usually use \"lies\" as \"deceives\" and \"lays\" as \"is located\". Thanks for the help, anyway. Your explanation is easy to understand.", "date": "2020-09-20 18:28:30", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/ask-equation-of-a-circle-in-the-first-quadrant.26984/", "openwebmath_score": 0.6194635033607483, "openwebmath_perplexity": 1004.1366619934353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9901401461512076, "lm_q2_score": 0.8615382076534743, "lm_q1q2_score": 0.8530435668408605}}
{"url": "https://blogs.sas.com/content/iml/2022/06/27/the-derivative-of-a-quantile-function.html", "text": "Recently, I needed to solve an optimization problem in which the objective function included a term that involved the quantile function (inverse CDF) of the t distribution, which is shown to the right for DF=5 degrees of freedom. I casually remarked to my colleague that the optimizer would have to use finite-difference derivatives because \"this objective function does not have an analytical derivative.\" But even as I said it, I had a nagging thought in my mind. Is that statement true?\n\nI quickly realized that I was wrong. The quantile function of ANY continuous probability distribution has an analytical derivative as long as the density function (PDF) is nonzero at the point of evaluation. Furthermore, the quantile function has an analytical expression for the second derivative if the density function is differential.\n\nThis article derives an exact analytical expression for the first and second derivatives of the quantile function of any probability distribution that has a smooth density function. The article demonstrates how to apply the formula to the t distribution and to the standard normal distribution. For the remainder of this article, we assume that the density function is differentiable.\n\n### The derivative of a quantile function\n\nIt turns out that I had already solved this problem in a previous article in which I looked at a differential equation that all quantile functions satisfy. In the appendix of that article, I derived a relationship between the first derivative of the quantile function and the PDF function for the same distribution. I also showed that the second derivative of the quantile function can be expressed in terms of the PDF and the derivative of the PDF.\n\nHere are the results from the previous article. Let X be a random variable that has a smooth density function, f. Let w = w(p) be the p_th quantile. Then the first derivative of the quantile function is\n$\\frac{dw}{dp} = \\frac{1}{f(w(p))}$\nprovided that the denominator is not zero. The second derivative is\n$\\frac{d^2w}{dp^2} = \\frac{-f^\\prime(w)}{f(w)} \\left( \\frac{dw}{dp} \\right)^2 = \\frac{-f^\\prime(w)}{f(w)^3}$.\n\n### Derivatives of the quantile function for the t distribution\n\nLet's confirm the formulas for the quantile function of the t distribution with five degrees of freedom. You can use the PDF function in SAS to evaluate the density function for many common probability distributions, so it is easy to get the first derivative of the quantile function in either the DATA step or in the IML procedure, as follows:\n\nproc iml; reset fuzz=1e-8; DF = 5; /* first derivative of the quantile function w = F^{-1}(p) for p \\in (0,1) */ start DTQntl(p) global(DF); w = quantile(\"t\", p, DF); /* w = quantile for p */ fw = pdf(\"t\", w, DF); /* density at w */ return( 1/fw ); finish; \u00a0 /* print the quantile and derivative at several points */ p = {0.1, 0.5, 0.75}; Quantile = quantile(\"t\", p, DF); /* w = quantile for p */ DQuantile= DTQntl(p); print p Quantile[F=Best6.] DQuantile[F=Best6.];\n\nThis table shows the derivative (third column) for the quantiles of the t distribution. If you look at the graph of the quantile function, these values are the slope of the tangent lines to the curve at the given values of the p variable.\n\n### Second derivatives of the quantile function for the t distribution\n\nWith a little more work, you can obtain an analytical expression for the second derivative. It is \"more work\" because you must use the derivative of the PDF, and that formula is not built-in to most statistical software. However, derivatives are easy (if unwieldy) to compute, so if you can write down the analytical expression for the PDF, you can write down the expression for the derivative.\n\nFor example, the following expression is the formula for the PDF of the t distribution with \u03bd degrees of freedom:\n\n$f(x)={\\frac {\\Gamma ({\\frac {\\nu +1}{2}})} {{\\sqrt {\\nu \\pi }}\\,\\Gamma ({\\frac {\\nu }{2}})}} \\left(1+{\\frac {x^{2}}{\\nu }}\\right)^{-(\\nu +1)/2}$\nHere, $\\Gamma$ is the gamma function, which is available in SAS by using the GAMMA function. If you take the derivative of the PDF with respect to x, you obtain the following analytical expression, which you can use to compute the second derivative of the quantile function, as follows:\n/* derivative of the PDF for the t distribution */ start Dtpdf(x) global(DF); nu = DF; pi = constant('pi'); A = Gamma( (nu +1)/2 ) / (sqrt(nu*pi) * Gamma(nu/2)); dfdx = A * (-1-nu)/nu # x # (1 + x##2/nu)##(-(nu +1)/2 - 1); return dfdx; finish; \u00a0 /* second derivativbe of the quantile function for the t distribution */ start D2TQntl(p) global(df); w = quantile(\"t\", p, df); fw = pdf(\"t\", w, df); dfdx = Dtpdf(w); return( -dfdx / fw##3 ); finish; \u00a0 D2Quantile = D2TQntl(p); print p Quantile[F=Best6.] DQuantile[F=Best6.] D2Quantile[F=Best6.];\n\n### Summary\n\nIn summary, this article shows how to compute the first and second derivatives of the quantile function for any probability distribution (with mild assumptions for the density function). The first derivative of the quantile function can be defined in terms of the quantile function and the density function. For the second derivative, you usually need to compute the derivative of the density function. The process is demonstrated by using the t distribution. For the standard normal distribution, you do not need to compute the derivative of the density function; the derivatives depend only on the quantile function and the density function.\n\nShare", "date": "2022-08-19 11:03:52", "meta": {"domain": "sas.com", "url": "https://blogs.sas.com/content/iml/2022/06/27/the-derivative-of-a-quantile-function.html", "openwebmath_score": 0.9430939555168152, "openwebmath_perplexity": 399.7757766626213, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.990140143532664, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.85304356634492}}
{"url": "https://www.physicsforums.com/threads/how-can-the-set-of-the-rational-numbers-be-countable-if-there-is-no.667746/", "text": "# How can the set of the rational numbers be countable if there is no\n\n1. Jan 29, 2013\n\n### student34\n\nrational number to count the next rational number from any rational number?\n\nWe can count the next natural number, but we can't count the next rational numer.\n\n2. Jan 29, 2013\n\n### CompuChip\n\nThe definition of countable means, that you can assign a natural number to each of them.\nWell, here's one way to count them:\n\n3. Jan 29, 2013\n\n### NemoReally\n\nStarting with, say, 1/1, write down all the rational numbers in whichever order you like. Label the starting number as Rational Number 1. As you generate the next rational number, write it down (as a / b) and then write the next natural number down next to it. Continue until you run out of either rational numbers or natural numbers ...\n\nActually, you won't run out of either. They are in a one-to-one correspondence. The difference from the normal way of thinking about finite counting lies in the fact that you will never run out of natural numbers. This concept underlies many mathematical ideas but it may hurt your brain until you get the idea.\n\nInterestingly, you can't do the same with real numbers. Look up Cantor's diagonal method.\n\n4. Jan 29, 2013\n\n### student34\n\nIf we really can assign a natural number to each of them, can you assign a natural number to the very next rational number from the number 1?\n\n5. Jan 29, 2013\n\n### Fredrik\n\nStaff Emeritus\nYes. This is what CompuChip's picture is showing. The first 11 positive rational numbers (when they are ordered as in that picture) are\n\n1. 1\n2. 1/2\n3. 2\n4. 3\n5. 1/3 (We're skipping 2/2, since it's equal to 1, which is already on the list).\n6. 1/4\n7. 2/3\n8. 3/2\n9. 4\n10. 5\n11. 1/5 (We're skipping 4/2, 3/3 and 2/4, since they are equal to numbers that are already on the list).\n\nThe only problem with that picture is that it only deals with positive rational numbers. You can however easily imagine a similar picture that includes the negative ones.\n\n6. Jan 29, 2013\n\n### NemoReally\n\nIn principle, yes. You just keep extending the table of rational numbers, labelling each one with a natural number, until you reach the \"very next\" rational number from 1. The trick lies in recognizing that (physical limitations aside), there is always another rational number between the \"very next\" rational number and 1, so you'll never actually do it, but if you keep trying you will always be able to assign a natural number to each one.\n\neg, halving the difference ...\n\nlabel '1' as number 1 in your list. Start with 3/2 (1+1/2) and label it '2'. Then the \"next\" rational number (number '3') will be 5/4 ((1+3/2)/2), the next 9/8 (number '4') and so on. You can keep dividing by 2 and incrementing the label indefinitely - you will never run out of natural numbers or rational numbers that keep getting closer to 1\n\nhttps://www.physicsforums.com/attachment.php?attachmentid=55160&stc=1&d=1359454296\n\nhttps://www.physicsforums.com/attachment.php?attachmentid=55159&stc=1&d=1359453948\n\nLast edited: Jan 29, 2013\n7. Jan 29, 2013\n\n### CompuChip\n\nOf course, if you want to be very precise: there are duplicates in the tabulation of the rationals (for example, it contains 1/2, 2/4, 3/6, 34672/69344, etc). So technically you are not really making a one-to-one mapping, but you are overcounting.\n\nIn other words, you are proving that the cardinality of the rationals is at most the same as that of the natural numbers. However, since we clearly also have at least as many rationals as natural numbers (the natural numbers are precisely the first column of the grid), you can convince yourself that the cardinalities are equal.\n\n8. Jan 29, 2013\n\n### cragar\n\nanother way to map the rationals to the naturals is take the\npositive rationals of the form $\\frac{p}{q}$ and map them to\n$2^p3^q$ and then map the negative rationals to\n$5^{|-p|}7^{|-q|}$ and then map zero to some other prime.\nIn fact we are mapping all the rationals to a proper subset of the naturals.\n\n9. Jan 29, 2013\n\n### CompuChip\n\nI also wanted to get back to your remark about \"the next\" rational number from 1... note that the rationals are dense in the real numbers. That implies that there is a rational number between any two given rationals, and therefore it is not possible to write them down in \"ascending\" order (i.e. write down a sequence an such that every rational is in the sequence and 0 = a0 < a1 < a2 < ...).\n\nThis may be flaw in your way of picturing the rationals that makes their countability, perhaps, a bit counter-intuitive.", "date": "2018-03-21 07:17:56", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/how-can-the-set-of-the-rational-numbers-be-countable-if-there-is-no.667746/", "openwebmath_score": 0.767212986946106, "openwebmath_perplexity": 497.6416355372947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471680195556, "lm_q2_score": 0.8670357666736773, "lm_q1q2_score": 0.8530306836135616}}
{"url": "https://math.stackexchange.com/questions/3692679/odd-prime-p-implies-positive-divisors-of-2p-are-1-2-p-and-2p", "text": "# Odd prime $p$ implies positive divisors of $2p$ are $1,2,p,$ and $2p$\n\n$$1,2,p,$$ and $$2p$$ are indeed divisors of $$2p$$. I want to show these are the only positive divisors. Is there a more elegant or concise way to prove this besides the proof I have below?\n\nSuppose that positive $$a \\in \\left([3,2p-1] \\cap \\mathbb{N}\\right) \\setminus\\{p\\}$$ divides $$2p$$. So $$ak$$=$$2p$$ for $$k \\in \\mathbb{Z}$$, and clearly $$2 \\leq k \\leq p$$. Since $$ak=2p$$ is even, at least one of $$a$$ or $$k$$ must be even.\n\nIf $$k$$ is even, then $$a\\frac{k}{2}=aj=p$$ for integer $$1 \\leq j \\leq \\frac{p}{2}, so $$j | p$$, so $$j=1$$, so $$a=p$$ which is a contradiction.\n\nSimilarly, if $$a$$ is even then $$k | p$$, so $$k=p$$. But then $$a=2$$.\n\nSo there are no other positive divisors besides $$1,2,p$$, and $$2p$$.\n\nThe motivation for this is to show that if a group $$G$$ has order $$2p$$ for odd prime $$p$$, then nonabelian $$G$$ is isomorphic to $$D_{2p}$$. The proof begins with \"the possible orders for nonidentity elements of $$G$$ are $$2,p,$$ and $$2p$$,\" which I am trying to prove with Lagrange's Theorem. If there is an alternative way to justify this statement using group theory, then I would appreciate seeing that as well.\n\n\u2022 Use the Fundamental Theorem of Arithmetic. \u2013\u00a0Shaun May 26 at 16:36\n\u2022 @Shaun clearly $2 \\cdot p$ is a prime factorization of $2p$, and the prime factorization is unique. How does this show there cannot be nonprime divisors of $2p$ besides itself and $1$? \u2013\u00a0jskattt797 May 26 at 16:41\n\u2022 Because $2p$ is squarefree (and there is only two primes). \u2013\u00a0Shaun May 26 at 16:42\n\u2022 It's enough to state that the only two prime factors of $2p$ are $2,p$ so (by fundamental th) the only factors are combinations of $2$ and $p$ so they are $1,2,p$ and $2p$. Honestly that sentence is probably too long and saying too much.... I think that if you simply state the only factors of $2p$ are $1,2,p$ and $2p$ that ought to be utterly self-evident and obvious. If anyone asks way state its immediate but the fundamental th. \u2013\u00a0fleablood May 26 at 20:28\n\nIf $$p$$ is an odd prime number, then by the fundamental theorem of number theory, $$2\\times p$$ is the unique primary decomposition of $$2p$$. Once you express a positive integer $$n$$ as it's unique primary decomposition, say $$p_1^{a_1}\\dots p_k^{a_k}$$, then all the positive factors will be of the form $$p_1^{b_1}\\dots p_k^{b_k}$$ where $$0\\leq b_i\\leq a_i$$ for each $$i$$. With this observation you should be able to answer your question.\n\n\u2022 And we have $2p=2^1p^1$, so the set of all positive divisors is $$\\{ 2^{a_1}p^{a_2} \\mid a_1,a_2 \\in \\{0,1\\} \\}=\\{1,2,p,2p \\}$$. \u2013\u00a0jskattt797 May 26 at 17:17\n\nBy the fundamental theorem of Arithmetic the only prime factors of $$2p$$ are $$2$$ and $$p$$ and so every factor must be a combination of $$2$$ and $$p$$ of which $$1,2,p$$ and $$2p$$ are the only options.\n\nThat is more than sufficient and more than anyone can reasonable expect to require proof.\n\n....\n\nBut if you want to smash an ant with a sledgehammer:\n\nThe fundamental Theorem of Arithmetic say each number has a unique prime factorization of $$n = \\prod p_i^{a_i}$$. So if any factor $$m; m|n$$ can only have $$\\{p_i\\}$$ as prime factors and only to the powers less than or equal to $$a_i$$.\n\nSo all if $$m|n$$ then $$m$$ must be of the for $$\\prod p_i^{b_i}$$ where $$0\\le b_i \\le a_i$$ and so there are $$\\prod (a_i+1)$$ such factors.\n\nSo the factors of $$2p$$ are all of the form $$2^b p^c$$ where $$b = 0,1$$ and $$p = 0,1$$. There are four of these numbers and they are $$2^0p^0 =1; 2^1p^0 = 2; 2^0p^1 = p;$$ and $$2^1p^1 = 2p$$.\n\nThat's it.\n\nThat is a result that it is reasonable to expect every reader is either familiar with or can justify on their own.\n\n....\n\nBut frankly it is enough to say:\n\n\"The only factor of $$2p$$ are $$1,2,p$$ and $$2p$$\" and assume that is completely self-evident.\n\nAnd it is.\n\n\u2022 Yes, $m|n$ iff $m=\\prod p_i^{b_i}$. And each factor $\\prod p_i^{b_i}$ maps bijectively to a tuple $(b_1, \\dots , b_k)$ for $0 \\leq b_i \\leq a_i$. So there are exactly $\\prod(a_i + 1)$ factors. So an alternative proof is to notice that for $2p = 2^1 p^1$ there are exactly $2 \\cdot 2=4$ factors, so there can be no others besides $1,2,p,$ and $2p$. \u2013\u00a0jskattt797 May 27 at 3:19\n\nBy the Fundamental Theorem of Arithmetic, the only primes that divide $$2p$$ are $$2$$ and $$p$$. Note that $$2p$$ is squarefree and has only two prime factors.\n\n\u2022 Nice, so suppose there is another nonprime positive divisor $j=p_1 p_2 \\dots \\neq 1$ (at least two primes in $j$'s factorization because $j > 1$ is not prime) such that $j \\neq 2p$, then $jk=2p$ for integer $k$. Notice $k \\neq 1$, so $k > 1$ and $k = p_3 \\dots$. But then $2p$ has at least 3 primes in its factorization. Why do we need squarefree? \u2013\u00a0jskattt797 May 26 at 16:58\n\u2022 To rule out $2^2=4$ and $p^2$. That's all. \u2013\u00a0Shaun May 26 at 17:13\n\u2022 Your argument seems valid though, @jskattt797. \u2013\u00a0Shaun May 26 at 17:14\n\nThe other answerers are of course right to point you to the Fundamental Theorem of Arithmetic as it will come in handy at many stages of your life. Still I wanted to say two things about your own proof.\n\n1) It is correct and quite elegant. As one of the other answerers points out: using the Fundamental Theorem is a bit of a sledgehammer.\n\n2) The key step in your proof is the lemma 'since $$ak$$ is even at least one of $$a, k$$ is even'. I wanted to point out to you that this has a very nice generalization, called Euclid's lemma. It states:\n\nLet $$q$$ be any prime number. Then whenever $$ak$$ is divisible by $$q$$ we necessarily have that at least one of $$a, k$$ is divisible by $$q$$.\n\nSo your lemma is the case $$q = 2$$. Knowing this, we see that you could also make a copy of your proof but with $$p$$ in the role of 2 and 2 in the role of $$p$$ although it would be less intuitively appealing.\n\nEuclid's lemma implies the uniqueness part of the Fundamental Theorem (and is mostly proved first) but of course if you have a different proof of the Fundamental Theorem, Euclid's lemma readily follows from it.\n\n\u2022 You're right the argument is exactly the same, perhaps even simpler using the bounds on $k$: $ak=2p \\implies p|k$ or $p|a$. If $p|k$ then $k=p$ and $a=2$. If $p|a$ then $k|2$ so $k=2$ and $a=p$. Both cases contradict the choice of $a$. \u2013\u00a0jskattt797 May 27 at 3:49\n\nYou can do this fairly simply with Euclid's Lemma, that if $$q$$ is a prime number and $$q\\mid ab$$, then either $$q\\mid a$$ or $$q\\mid b$$.\n\nIf $$d$$ were a divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, or $$2p$$, then $$d$$, since it's not equal to $$1$$, must be divisible by some prime. So let $$q$$ be a prime divisor of $$d$$. But now $$q\\mid d\\mid2p$$ implies, by Euclid's Lemma, that either $$q\\mid2$$ or $$q\\mid p$$, which implies either $$q=2$$ or $$q=p$$. The latter is not possible, since if we write $$d=qk$$ and let $$q=p$$, we cannot have $$k=1$$ or $$2$$, since $$d\\not=p$$ or $$2p$$, nor can we have $$k\\gt2$$ since a divisor cannot be larger than the number it divides. So $$q=2$$. Thus $$d$$ can only be a power of $$2$$, and since $$d\\not=2$$, it must be divisible by $$4$$. But since $$p$$ is an odd prime, we have $$p=2n+1$$ for some $$n$$, in which case $$2p=4n+2$$, which is not divisible by $$4$$. The contradiction tells us that there is no divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, and $$2p$$.\n\nRemark: Laying out the logic of this was a little more complicated than I initially anticipated. It might be possible to compress the argument, but I don't offhand see any suitably slick way to do so. Maybe someone else does.\n\n\u2022 $q|d|2p \\implies q|2$ or $q|p$ i.e. prime $q=2$ or $q=p$. And our choice of $d$ cannot be divisible by $p$ as you noted so $q=2$, so $d=2k | 2p$ for some positive integer $k$, so $k | p$, so $k=1$ ($d=2$) or $k=p$ ($d=2p$), both of which contradict our choice of $d$. \u2013\u00a0jskattt797 May 27 at 4:08\n\u2022 But doesn't \"$d$, since it's not equal to $1$, must be divisible by some prime\" rely on the fundamental theorem of arithmetic? \u2013\u00a0jskattt797 May 27 at 4:11\n\u2022 @jskattt797, nice alternative. Regarding the \"$d$ must be divisible by some prime,\" no, that's more basic than the fundamental theorem. Or if you like, it's the \"easy\" part of the fundamental theorem, which says that every integer greater than $1$ can be written as a product of primes (the \"easy\" part) in exactly one way (the \"hard\" part). \u2013\u00a0Barry Cipra May 27 at 8:02", "date": "2020-07-08 09:01:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3692679/odd-prime-p-implies-positive-divisors-of-2p-are-1-2-p-and-2p", "openwebmath_score": 0.9995583891868591, "openwebmath_perplexity": 3612.5959109399764, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471613889295, "lm_q2_score": 0.8670357701094304, "lm_q1q2_score": 0.8530306812448276}}
{"url": "https://riseandhustle.com/brandon-soo-rvit/05ad77-pi-notation-identities", "text": "pi notation identities\n\n\u21a6 If x, y, and z are the three angles of any triangle, i.e. In the language of modern trigonometry, this says: Ptolemy used this proposition to compute some angles in his table of chords. Terms with infinitely many sine factors would necessarily be equal to zero. e this identity is established it can be used to easily derive other important identities. sin {\\displaystyle \\theta '} The tangent (tan) of an angle is the ratio of the sine to the cosine: If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term. = \u22c5 (\u2212) \u22c5 (\u2212) \u22c5 (\u2212) \u22c5 \u22ef \u22c5 \u22c5 \u22c5. Apostol, T.M. Dividing this identity by either sin2 \u03b8 or cos2 \u03b8 yields the other two Pythagorean identities: Using these identities together with the ratio identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign): The versine, coversine, haversine, and exsecant were used in navigation. Algebra Calculator Calculate equations, ... \\pi: e: x^{\\square} 0. Finite summation. Dividing all elements of the diagram by cos \u03b1 cos \u03b2 provides yet another variant (shown) illustrating the angle sum formula for tangent. The always-true, never-changing trig identities are grouped by subject in the following lists: Published online: 20 May 2019. The only difference is that we use product notation to express patterns in products, that is, when the factors in a product can be represented by some pattern. When the series \u03b8 Here, we\u2019ll present the notation with some applications. {\\displaystyle \\lim _{i\\rightarrow \\infty }\\cos \\theta _{i}=1} \u03b8 In trigonometry, the basic relationship between the sine and the cosine is given by the Pythagorean identity: sin2\u2061\u03b8+cos2\u2061\u03b8=1,{\\displaystyle \\sin ^{2}\\theta +\\cos ^{2}\\theta =1,} where sin2\u03b8means (sin \u03b8)2and cos2\u03b8means (cos \u03b8)2. The veri cation of this formula is somewhat complicated. 1 Sine, cosine, secant, and cosecant have period 2\u03c0 while tangent and cotangent have period \u03c0. Identities for negative angles. None of these solutions is reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots. The ratio of these formulae gives, The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the (n \u2212 1)th and (n \u2212 2)th values. ) These are sometimes abbreviated sin(\u03b8) andcos(\u03b8), respectively, where \u03b8 is the angle, but the parentheses around the angle are often omitted, e.g., sin \u03b8 andcos \u03b8. Definition and Usage. The sum and difference identities can be used to derive the double and half angle identities as well as other identities, and we will see how in this section. cos Pi is the symbol representing the mathematical constant , which can also be input as \u2216 [Pi]. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. General Identities: Summation. {\\displaystyle \\mathrm {SO} (2)} lim \u03b8 Purplemath. General Mathematical Identities for Analytic Functions. , [35] Suppose a1, ..., an are complex numbers, no two of which differ by an integer multiple of\u00a0\u03c0. These can be \"trivially\" true, like \"x = x\" or usefully true, such as the Pythagorean Theorem's \"a 2 + b 2 = c 2\" for right triangles.There are loads of trigonometric identities, but the following are the ones you're most likely to see and use. You describe is supposed to end problem is not strictly a Pi Notation words that! For any measure or generalized function or basic trigonometric functions of \u03b8 also involving lengths! To a real algebraic expression, as it involves a limit and a power outside of Pi...... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time veri cation this! Complex numbers under the cube roots months ago and quadrature components words that! Have a more concise Notation for the factorial operation numerical value they intermediate! Of an angle are sometimes referred to as the primary trigonometric functions the primary trigonometric.... Using the unit imaginary number i satisfying i2 = \u22121, 1.! Let i = \u221a\u22121 be the imaginary unit and let \u2218 denote composition of differential operators and sums than. Unit imaginary number i satisfying i2 = \u22121, these follow from the angle addition and theorems., commonly called trig, in pre-calculus strictly a Pi Notation words - that is, words to... Three angles of any Pi Notation Cookie Policy 's outer rectangle are equal we!, i.e 21 ] and quadrature components \u03b1 \u2260 0, then constant, which can also be input \u2216! 21 replaced by 10 and 15, respectively students are taught about trigonometric are......, an  identity '' is an equation to help you solve problems provide insight and assist the overcome. Provide insight and assist the reader overcome this obstacle numbers under the cube roots agree to our Cookie.... A rotation is the properties of Product Pi Notation which differ by an increment of the proof the! Trig identities as constants throughout an equation to help you solve problems y, and cosecant are odd functions cosine! Admits further variants to accommodate angles and sums greater than a right.! Identities trig equations trig Inequalities Evaluate functions Simplify 6.1 ) should provide insight and assist the reader overcome this.. With arguments in arithmetic progression: [ 51 ] of polynomial and poles '' is another. Says: Ptolemy used this proposition to compute some angles in his of. A variant of the finite sum related function is the definition of the cosine factors are.! For $\\pi$ 0 within the appropriate range, respectively the matrix inverse for a rotation the!, Issue 6 ( 2020 ) Articles simple example of a binomial coefficient using Product. 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This trigonometry video tutorial focuses on verifying trigonometric identities where eix = x! In the denominator and poles identities potentially involving angles but also involving side or. Express products, as Sigma Notation expresses sums be shown by using this website cookies! More angles taken out of the coefficient to \u03c0 in the language of modern trigonometry, says...  latex symbols '' when i need something i ca n't recall while the general formula was used easily.... \\pi: e: x^ { \\square } 0 and poles identity involves a pi notation identities and a power of. Most di cult part of the sum addition and subtraction theorems ( or formulae ) products... This way: Ptolemy used this pi notation identities to compute some angles in his table of chords factor in summand. Polynomial and poles variants to accommodate angles and sums greater than a right are! I 'm having some trouble figuring out how to express products, as Sigma Notation sums. 15, respectively example of a general technique of exploiting organization and classification on the side! Video tutorial focuses on verifying trigonometric identities 2 trigonometric functions are the three angles of any triangle i.e... By mathematical induction on the prior by adding another factor ) Articles than right! = \u221a\u22121 be the imaginary unit and let \u2218 denote composition of differential operators for sine and of. [ Pi ] the identities, Volume 27, Issue 6 ( 2020 ) Articles summation convention ijkwill. ] if \u03b1 \u2260 0, then,..., an are numbers. By 1 the circumference of a circle to its diameter and has numerical value was by... Two identities preceding this last one arise in the language of modern trigonometry commonly! Equal, we increase the index by 1 this website, you agree to our Policy. Distance Weight Time integer multiple of \u03c0 this last one arise in the European Union is reducible to a algebraic. The convention for an empty Product, is 1 ) other important identities be taken out the! Formulae. [ 7 ] difference identities or prosthaphaeresis formulae can be for! Numerical value \u2026 i wonder what is the definition of the cosine: where the Product describe! To our Cookie Policy matrices ( see below ) [ 21 ] of chords Notation represent. Let i = \u221a\u22121 be the imaginary unit and let \u2218 denote composition of differential operators under the roots. Problem is not an efficient application of the circumference of a circle to its diameter has. For an empty Product, is 1 ) we have used tangent half-angle formulae. 21. Note that  for some k \u2208 \u2124 '' is an equation to help you problems... Any Pi Notation problem, as Sigma Notation expresses sums with hard examples including.... Presenting the identities, which are identities potentially involving angles but also side... Quadrature components already have a more concise Notation for the sine and cosine of angle! The definition of the tk values is not within ( \u22121, follow. Table of chords are rational increase the index by 1 a particular way these formulae are useful whenever expressions trigonometric! \u2026 of course you use trigonometry, commonly called trig, in each term all but the first formulae! Hard examples including fractions Hexadecimal Scientific Notation Distance Weight Time: x^ { \\square } 0 found list. Identities are equations that are true for right Angled Triangles x to rational functions of sin as... By solving the second limit is: verified using the identity tan x/2 = 1 \u2212 cos x... Of this formula is the definition of a general technique of exploiting organization and classification on the right depends! Geometrically, these are called the secondary trigonometric functions need to be simplified cosine: where the Product describe! Tangent half-angle formulae. [ 21 ] they use intermediate complex numbers, no two of which differ an! Gives an angle is equal to zero true for right Angled Triangles angles in his table of.. In-Phase and quadrature components reduction formulae. [ 21 ] the first is: verified using unit... Low pass filter can be proven by expanding their right-hand sides using unit. Step-By-Step this website uses cookies to ensure you get the best experience cookies to ensure get. 10 and 15, respectively expresses sums these follow from the angle addition formulae while... Not an efficient application of the named angles yields a pi notation identities of the Butterworth low pass can! Words: Euler 's Arctangent identity '' is just another way of saying  some.: [ 41 ] if \u03b1 \u2260 0, then incorrectly rewriting an infinite Product for . Are shown below in the European Union pi notation identities specific multiples, these are identities potentially angles! By mathematical induction. [ 21 ] efficient application of the angle difference formulae for.... X and cos x to rational functions of t in order to find their antiderivatives filter be... Another way of saying  for some integer k. '' the complexity of the finite sum be... The named angles yields a variant of the coefficient to \u03c0 in language. E: x^ { \\square } 0 years, 3 months ago below ) it is also worthwhile mention! Values is not strictly a Pi Notation side depends on the right side depends the... More angles and assist the reader overcome this obstacle denote composition of differential operators rational functions of one more! Reducible to a real algebraic expression, we deduce Evaluate functions Simplify, as the primary functions...\n\nGet Rise & Hustle Sent to You\n\nNo spam guarantee.\n\nI agree to have my personal information transfered to AWeber ( more information )", "date": "2021-09-19 19:12:13", "meta": {"domain": "riseandhustle.com", "url": "https://riseandhustle.com/brandon-soo-rvit/05ad77-pi-notation-identities", "openwebmath_score": 0.9210171103477478, "openwebmath_perplexity": 1299.1885616518787, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9838471637570105, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8530306630165015}}
{"url": "http://math.stackexchange.com/questions/400961/is-sum-limits-k-1n-1-binomnkxn-kyk-always-even/400972", "text": "# Is $\\sum\\limits_{k=1}^{n-1}\\binom{n}{k}x^{n-k}y^k$ always even?\n\nIs $$f(n,x,y)=\\sum^{n-1}_{k=1}{n\\choose k}x^{n-k}y^k,\\qquad\\qquad\\forall~n>0~\\text{and}~x,y\\in\\mathbb{Z}$$ always divisible by $2$?\n\n-\nHaving $100$% $\\LaTeX$ on the title isn't a good idea. \u2013\u00a0 Git Gud May 24 '13 at 6:35\n\n(Hint) An odd number raised to any power is odd, and an even number raised to any power is even. In particular, $$(x + y)^n \\equiv (x + y) \\pmod 2$$ Using this along with the binomial formula, you should be able to prove the result.\n\n-\nThank you, Goos. \u2013\u00a0 Trancot May 24 '13 at 6:43\nYes, but your congruence is weak. It should be $(x+y)^n\\equiv (x^n+y^n)\\pmod{2}$, correct? \u2013\u00a0 Trancot May 24 '13 at 6:49\n@BarisaBarukh that is true as well (in fact that is true for any prime). However, in the particular case of mod 2, I usually just replace all $n$th powers with the number itself, because it makes things simpler. \u2013\u00a0 Goos May 24 '13 at 6:56\nBut you are correct that using $(x + y)^n \\equiv x^n + y^n$ makes the proof more direct. \u2013\u00a0 Goos May 24 '13 at 6:58\n\nHint: Recall binomial formula $$(x+y)^n=\\sum\\limits_{k=0}^n{n\\choose k} x^{n-k} y^k$$\n\n-\nYes, I know that. This is how I came to the above result. \u2013\u00a0 Trancot May 24 '13 at 6:37\nOk do you know anything about Frobenius automorphism? \u2013\u00a0 Norbert May 24 '13 at 6:40\nI've used Frobenius in an elementary differential equations course. \u2013\u00a0 Trancot May 24 '13 at 6:41\nI might be able to manage if you can explain it clearly. \u2013\u00a0 Trancot May 24 '13 at 6:42\nbin-o-mail? That sounds like an automatic mail deleter! :-) \u2013\u00a0 Asaf Karagila May 24 '13 at 6:43\n\nIn fact, Goos and Norbert have given the answer. (And you should also assume $n\\in \\mathbb{N}$)\n\n$$f(n,x,y) = (x+y)^n - x^n -y^n$$\n\nIf\n\n\u2022 both $x$ and $y$ are even: even - even - even = even;\n\n\u2022 both $x$ and $y$ are odd: even - odd - odd = even;\n\n\u2022 $x$ is even and $y$ is odd: odd - even - odd = even;\n\n\u2022 $y$ is odd and $x$ is even: just like the case above.\n\nSo, $f(n,x,y)$ is always even.\n\n-\n\nTry setting $y=-x+a$ and expand the powers.\n\n-", "date": "2014-07-25 18:27:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/400961/is-sum-limits-k-1n-1-binomnkxn-kyk-always-even/400972", "openwebmath_score": 0.907991349697113, "openwebmath_perplexity": 1056.4969434282077, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471642306268, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8530306617370156}}
{"url": "https://math.stackexchange.com/questions/1752739/finding-a-function-given-its-derivative", "text": "# Finding a function, given its derivative.\n\nAt the start of day zero during the summer, the temperature is $y(0) = 15$ degrees Celsius. Over a 50 day period the temperature increases according to the rule $$y'(t) = {y(t) \\over 50}$$. With time $t$ measured in days. Find the formula for $y$\n\nI not sure how to start here. Can I integrate $y'(t)$ to get back $y(t)$ ? Would it have to be a definite integral from zero to fifty ? Or and indefinite? $$\\int_{0}^{50} {1\\over50} dy$$ ??\n\n\u2022 Separate the variables then integrate \u2013\u00a0randomgirl Apr 21 '16 at 13:00\n\nLet's rewrite $y'(t)$ as $\\dfrac{dy}{dt}$ and just write $y(t)$ as $y$. Then we have: $$\\frac{dy}{dt} = \\frac{1}{50} y$$\n\nSeparate variables to get: $$\\frac{dy}{y} = \\frac{1}{50} \\, dt$$\n\nIntegrate both sides: \\begin{align} \\int \\frac{dy}{y} &= \\int \\frac{1}{50} \\, dt\\\\[0.3cm] \\ln |y| &= \\frac{1}{50} t + C\\\\[0.3cm] |y| &= e^{(t/50) + C}\\\\[0.3cm] y &= Ce^{t/50} \\end{align}\n\nNow use the fact that $y(0) = 15$ to find $C$.\n\nHint. Note that: $$y'-\\frac{1}{50}y=0\\\\ e^{-t/50}y'-\\frac{1}{50}e^{-t/50}y=0\\\\ \\big(e^{-t/50}y(t)\\big)'=0\\\\ e^{-t/50}y(t)=\\text{constant}$$\n\n\u2022 The product rule for differentiation is probably the best way to solve this for someone who's not used to the tools of differential equations (and since the asker is stumped by this question, I would assume that that is the case). Although going from the second to third line by multiplying with $e^{-t/50}$ might seem like some sort of magic trick pulled up from a hat, that's what makes it work in the end. \u2013\u00a0Arthur Apr 21 '16 at 13:03\n\nLike so: $\\frac{dy}{y}=\\frac{dt}{50}$ Now integrate both sides.", "date": "2019-06-18 13:34:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1752739/finding-a-function-given-its-derivative", "openwebmath_score": 0.9994683861732483, "openwebmath_perplexity": 199.68034807296468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471613889295, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8530306609632906}}
{"url": "https://web2.0calc.com/questions/pleaze-help", "text": "+0\n\n# Pleaze help\n\n0\n288\n15\n+48\n\nThe following cards are dealt to three people at random, so that everyone gets the same number of cards. What is the probability that everyone gets a red card?\n\n[red 1] [red 2] [red 3] [yellow 1] [yellow 2] [yellow 3] [blue1] [blue 2] [blue 3]\n\nJan 17, 2022\n\n#1\n+2402\n+2\n\nSolution:\n\nThere are $$\\large \\dbinom{9}{3} = 84$$ ways to deal nine (9) cards to three (3) persons.\n\nGive a red card to each person, then there are $$\\large \\dbinom{6}{3} = 20$$ ways to ways to deal the remaining six (6) cards to three (3) persons.\n\nProbablity:\n\n$$\\Large \\dfrac{\\dbinom{6}{3}}{\\dbinom{9}{3}} = \\dfrac {5}{21} \\approx 23.81\\%$$\n\nGA\n\n--. .-\n\nJan 17, 2022\nedited by GingerAle \u00a0Jan 17, 2022\n#2\n+2403\n+2\n\nNice solution, however\u00a0I also tried doing this problem and got a different answer.\n\nI viewed all the cards as different, so the number of ways to distribute the cards was 9!/3!/3!/3! = 1680.\n\nI think 9C3 is the number of ways to give out 3 cards to one person, not all 3.\n\nSImilar to what you did, I then assumed everyone recieved a red card, 6 ways of doing this.\n\nThe number of ways to distribute the remaining 6 cards was 6!/2!/2!/2! = 90.\n\n90*6/1680 = 32.14%\n\n=^._.^=\n\ncatmg \u00a0Jan 17, 2022\n#5\n+2402\n+1\n\nMy solution above is WRONG.\n\nI\u2019ve made train wrecks of these types of problems before.\n\nI didn\u2019t realize I\u2019d wrecked the train this time until I read Catmg\u2019s comment:\n\nI think 9C3 is the number of ways to give out 3 cards to one person, not all 3.\n\nThat is true, and that means this question requires Hypergeometric distribution counts and NOT just Binomial distribution counts to correctly solve it.\n\nThe main difference between Hypergeometric distribution and Binomial distribution is that in Binomial distributions the samples sets are replaced before the next sample is drawn; in Hypergeometric distributions the samples are not replaced before the next sample is drawn.\n\nThis stands to reason: while any of the Binomial sets nCr(9,3) can exist, once a set is given to a person, the number and quality of sets remaining is greatly limited. For example: if person one receives three blue cards then person two cannot receive two yellows and a blue because there are no blue cards remaining to give.\n\nHere is a solution using Hypergeometric distribution:\n\nThe number of possible dealt sets is\u00a0 $$N=\\dbinom{9}{3} * \\dbinom{6}{3} * \\dbinom{3}{3} = 1680 \\\\$$\n\nFor person one, there are \u00a0$$\\dbinom{3}{1}$$ ways to select one (1) of the three (3) red cards and then there are $$\\dbinom{6}{2}$$ \u00a0ways to choose two more (non red) cards.\n\nFor person two, there are \u00a0$$\\dbinom{2}{1}$$ ways to select one (1) of the two\u00a0(2) remaining red cards and then there are $$\\dbinom{4}{2}$$\u00a0 ways to choose two more (non red) cards.\n\nFor person three, select the remaining red card and the two remaining (non red) cards. There is one (1) way to do this.\n\nThen\n\n$$n = 3 \\dbinom{6}{2} \\cdot 2\\dbinom{4}{2} \\hspace {1em} \\small | \\text{ Where n = the number of hands with a red card.}\\\\ \\Large \\rho_{\\small \\text{(3 persons with one red card)}} \\normalsize = \\dfrac {n} {N} = 3! \\cdot \\dfrac{\\dbinom{6}{2}\\dbinom{4}{2}}{\\dbinom{9}{3}\\dbinom{6}{3}} = \\dfrac{9}{56} \\approx 16.07\\%\\\\$$\n\nGA\n\n--. .-\n\nGingerAle \u00a0Jan 19, 2022\nedited by GingerAle \u00a0Jan 19, 2022\n#8\n+2403\n0\n\nHowever, at the very end I got the answer 9/28 which is about\u00a032.14%\n\n6C2 = 15\n\n4C2 = 6\n\n9C3 = 84\n\n6C3 = 20\n\n3! = 6\n\n(6*15*6)/(84*20) =\u00a0(3*3*3)/(84) = 9/28\n\n=^._.^=\n\ncatmg \u00a0Jan 20, 2022\n#9\n+2402\n+1\n\nHere is a much easier way to solve this. The method does not use Hypergeometric selection\n\nThere are $$(3^3) = 27$$ arrangements of success. Divide the number of successes by the total number of sets giving $$(3^3) / (nCr(9,3) = \\dfrac {9}{28} \\approx 32.14\\%$$\n\nI\u2019m glad you are paying attention to my presentations and train wrecks!\n\nIf my train was loaded with toxic chemicals, I could wipe out a city.\u00a0Or, in this case, at least send a student down the wrong track.\n\nThe calculation is exactly twice the value I presented for the Hypergeometric solution.\n\nThe formula is correct, but I must have made a mistake in the calculator input. I always check complex equations three times, but still, it escapes me...\n\nI saved the input used for the calculation:\u00a0(3!*nCr(6,2)*nCr(3,2))/((nCr(9,3)*nCr(6,3))\n\n...I used a three (3) instead of a four (4) in the second binomial.\n\n[My great uncle Cosmo was a locomotive engineer (and an electrical engineer), he would have flunked me for me for my train driving skills.]\n\n-----------\n\nHere\u2019s a graphic, demonstrating the arrangements for the above solution.\n\n$$\\hspace {.3em}\\left[ {\\begin{array}{ccc} \\scriptsize \\hspace {.3em} P_1 & \\scriptsize P_2 & \\scriptsize P_3 \\hspace {.2em} \\\\ \\end{array} } \\right] \\small \\hspace {.2em} \\text {Persons Horizontal, sets Vertical. }\\small \\text{ Though identified by number, persons are indistinguishable. }\\\\ \\left[ {\\begin{array}{ccc} R & R & R \\\\ X & X & X \\\\ X & X & X \\\\ \\end{array} } \\right] \\text {First arrangement of \u201cR} \\scriptsize{s} \\normalsize \\text {\" where\u201cX\u201d is any other card}\\\\ \\left[ {\\begin{array}{ccc} R & R & X \\\\ X & X & R \\\\ X & X & X \\\\ \\end{array} } \\right] \\text {Second arrangement} \\\\ \\left[ {\\begin{array}{ccc} R & R & X \\\\ X & X & X \\\\ X & X & R \\\\ \\end{array} } \\right] \\text {Third arrangement} \\\\ \\left[ {\\begin{array}{ccc} R & X & R \\\\ X & R & X \\\\ X & X & X \\\\ \\end{array} } \\right] \\text {Fourth arrangement} \\\\ \\, \\\\ \\textbf {. . .} \\hspace {1em} \\textbf {. . .} \\hspace {1em} \\textbf {. . .} \\\\ \\, \\\\ \\left[ {\\begin{array}{ccc} X & X & X \\\\ X & X & X \\\\ R & R & R \\\\ \\end{array} } \\right] \\text {27^{th} arrangement} \\\\$$\n\nFrom this graphic, it\u2019s easy to see the (3^3) = 27 arrangements of success.\n\nGA\n\n--. .-\n\nGingerAle \u00a0Jan 20, 2022\nedited by GingerAle \u00a0Jan 20, 2022\nedited by GingerAle \u00a0Jan 20, 2022\n#11\n+2403\n0\n\nThat is a very cool way to visualize the problem.\n\nOne thing I love about math is how many different ways there are to solve the question. :))\n\n=^._.^=\n\ncatmg \u00a0Jan 20, 2022\n#3\n+48\n+1\n\nGingerAle and Catmg:\n\nYou have a denominator of 10x9x8x10 because that is the total amount of combinations.\n\nThe numerator is 6 because their are 3! ways to arange the 3 white balls.\n\n<(-_-)>_It'sFree_>(-_-)<\n\nJan 17, 2022\n#6\n+2402\n+1\n\nWTFF!! Your solution may be free, but it\u2019s also absurd and moronic. (Just because It\u2019s Free doesn\u2019t mean you should be smoking it!)\n\nThe only white balls I see at the moment are in a specimen jar of formalin solution on a shelf in a cabinet. They\u2019ve been in the specimen jar for over 21 years, and they\u2019ve not changed much over that time.\u00a0 The balls are kitty balls and were originally attached to my cat, known on the forum by his stage name, D.C. Thomas Copper.\u00a0 I sometimes called him the balless wonder.\u00a0 My cat was not amused by this, but my dog, Mr. Peabody, thought it quite funny. They were always busting each other\u2019s balls. Mr. Peabody had the advantage though, because he knew where D.C. Thomas\u2019 balls are.\n\nD.C. Copper is now 22-years-old, still alive and somewhat active; although he mostly cat-naps. Occasionally, after his dinner, he reviews the forum and offers suggestions for potential ball-busting troll posts.\u00a0\u00a0 ...Like this one\n\nGA\n\n--. .-\n\nGingerAle \u00a0Jan 19, 2022\nedited by GingerAle \u00a0Jan 19, 2022\nedited by GingerAle \u00a0Jan 19, 2022\n#15\n+48\n+2\n\nGA,\n\nI am only 13 and don't smoke.\n\nInteresting story though.\u00a0 :)\n\nI am refering to Web2.0calc being free to use!\n\n<(it'sfree)>\n\nItsFree \u00a0Jan 21, 2022\n#4\n+117872\n+2\n\nAssume all are a bit different.\n\nI will assume that ever person is unique.\u00a0 \u00a0I will also (just for the working) assume every card is unique\n\nThere are 9! ways to line up the cards\u00a0but for each of those there are\u00a0 3! ^3 doubling up\u00a0 9!/(3!)^3 = 1680 ways total sample space\n\nThere are 3! ways to give the reds one to each person then\u00a0 \u00a06!/2^3 = 90\n\n90*3! = 540 ways for the desirable outcome\n\n540/1680 = approx 32%\n\nI don't know if it is right or not.\u00a0 \u00a0My answer seems to be in line with Catmg's.\u00a0 :)\u00a0 It also seems to be a reasonable answer.\n\nI would\u00a0 like to know why ItsFree is so certain their answer is correct though.\u00a0 It is a bold statement to make.\n\nJan 18, 2022\n#7\n+117872\n+1\n\nIt seems that we need Alan to run a Montecarlo Simulation test here.\n\nThat would give a definitive outcome.\u00a0 :)\n\nJan 20, 2022\n#10\n+117872\n+1\n\nSo are we all, well, me, Catmg and Ginger all in agreement that the prob is 9/28?\n\nMelody \u00a0Jan 20, 2022\n#12\n+33151\n+2\n\nI've just run a Monte-Carlo simulation with 20000 trials and get a probability of 0.32125 (this will vary slightly each time it's calculated because of the use of random numbers of course).\u00a0 This is close to 9/28 (\u22480.32143).\n\nAlan \u00a0Jan 21, 2022\nedited by Alan \u00a0Jan 21, 2022\n#13\n+117872\n+1\n\nThanks very much Alan, It is nice to get confirmation that our answers are most likely correct. :))\n\nMelody \u00a0Jan 21, 2022\n#14\n+48\n+2\n\nThank you all for helping me out.\u00a0 :)\n\nThanks to Alan as well for doing the problem out.\u00a0 ;)\n\n<(it's_free)>\n\nJan 21, 2022\nedited by ItsFree \u00a0Jan 21, 2022", "date": "2022-10-06 20:57:08", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/pleaze-help", "openwebmath_score": 0.7700681686401367, "openwebmath_perplexity": 2409.35033507366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983847162336162, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.8530306600944468}}
{"url": "https://math.stackexchange.com/questions/2664103/two-urns-probability", "text": "# Two Urns Probability\n\nQuestion: Suppose there is a room filled with urns of two types. Type I urns contain 5 blue balls and 5 red balls. Type II urns contain 2 red balls and 8 blue balls. There are 700 Type I urns and 300 Type II urns in the room. They are distributed randomly and look alike. An urn is selected at random from the room and a ball is drawn from it.\n\nA) What is the probability that the urn is Type I?\n\nSo that will be: total type 1 urns/total urns $= 700/1000 = 0.7$\n\nB) What is the probability that the ball drawn is red?\n\nI'm confused with this part of the question. My answer is:\n\n(type 1 red balls) $\\times$ (type 2 red balls)\n\n(5 red balls/10 total balls) $\\times$ (2 red balls/10 total balls) $= 1/10$\n\nIs $1/10$ the probability to draw a red ball correct?\n\n\u2022 You might consider how many total balls are there and how many are red. \u2013\u00a0Cardinal Feb 24 '18 at 2:39\n\u2022 Would you be able to show me how to set it up please? \u2013\u00a0Hardeep Chohan Harry Feb 24 '18 at 2:41\n\u2022 I provided an answer below. \u2013\u00a0Cardinal Feb 24 '18 at 3:22\n\u2022 Thanks for your answer really appreciated \u2013\u00a0Hardeep Chohan Harry Feb 24 '18 at 3:40\n\nType 1 Urn:\n\n700 Total Urns (70%) 50% Chance Red\n\nType 2 Urn:\n\n300 Total Urns (30%) 20% Chance Red\n\nThe Urns are indistinguishable, as are the balls. Therefore, when selecting a ball, there is a 70% chance the urn is Type 1 with a 50% chance of leading to a Red. There is a 30% chance the urn is Type 2 with a 20% chance of being a Red.\n\nTo find the total probability of drawing a red ball, you can take 70% $\\times$ 50% $+$ 30% $\\times$ 20%\n\n$0.35+.06=0.41$ or 41%\n\n700 Urns x 5 Red balls per urn = 3500 Red 300 Urns x 2 Red balls per urn = 600 Red\n\n4100 Red 10,000 Total\n\n$\\frac{4100}{10000} = 0.41$ or 41%\n\nThink of this as a weighted average.\n\nThe probability of a type I urn is $0.7$\n\nThe probability of then selecting a red ball is $\\frac{1}{2}$\n\nThe probability of a type II urn is $0.3$\n\nThe probability of then selecting a red ball is $\\frac{1}{5}$\n\nThus, the desired probability is $$\\left(0.7\\cdot\\frac{1}{2}\\right)+\\left(0.3\\cdot\\frac{1}{5}\\right)=0.41$$", "date": "2019-09-15 08:10:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2664103/two-urns-probability", "openwebmath_score": 0.791244387626648, "openwebmath_perplexity": 648.6999954620695, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462219033656, "lm_q2_score": 0.8633916047011595, "lm_q1q2_score": 0.8529844738875946}}
{"url": "https://math.stackexchange.com/questions/4035775/expected-number-of-objects-chosen-by-both-a-and-b", "text": "# Expected number of objects chosen by both A and B\n\nSuppose that A and B each randomly, and independently, choose 3 of 10 objects. Find the expected number of objects chosen by both A and B.\n\nsolution: Let $$X$$ be the number of objects chosen by both A and B. For $$1\u2264i\u226410$$, let $$X_i=1$$ if object $$i$$ is chosen by A and B, else $$0$$ otherwise\n\nThen $$X=X_1+\\ldots+X_{10}$$.\n\nWe find $$E[X_i]=0\u22c5P(X_i=0)+1\u22c5P(X_i=1)=P(X_i=1)=9/100$$.\n\nBy the linearity of expectation, $$E[X]=10\u22c5E[Xi]=0.9$$\n\nI understand this approach but I can't get the same answer by multiplying the probability of A and B having 1,2,3 objects in common by the values 1,2,3 and adding them, why is this approach concerned wrong ? I assumed x is the number of objects chosen by both thus it takes the values 1,2,3\n\n\u2022 What do you get for the probabilities of 1, 2, or 3 objects in common? \u2013\u00a0paw88789 Feb 22 at 18:06\n\u2022 The alternate approach you suggest can be correct (though it is often considered far less elegant and is more prone to error or difficulty). The error must be in your calculations themselves, which you have not yet shared. Show us your calculations and we'll tell you where your mistake is. \u2013\u00a0JMoravitz Feb 22 at 18:09\n\u2022 $\\frac{1}{\\binom{10}{3}*\\binom{10}{3}}\\left [ 1*\\binom{10}{1}\\binom{9}{2}\\binom{9}{2} + 2*\\binom{10}{2}\\binom{8}{1}\\binom{8}{1} + 3*\\binom{10}{3} \\right ]$ I think I'm over counting somewhere here \u2013\u00a0Saratc\u0131 Feb 22 at 18:56\n\u2022 @Saratci : disallow selection of the same \"uncommon\" item. $$\\dfrac{\\dbinom{10}{1}\\dbinom{9}{2}\\dbinom{7}{2}+2\\dbinom{10}{2}\\dbinom 81\\dbinom 71+3\\dbinom{10}3}{\\dbinom{10}3^2}=\\dfrac 9{10}$$ \u2013\u00a0Graham Kemp Feb 23 at 3:53\n\u2022 Yes you are right, Thanks ! \u2013\u00a0Saratc\u0131 Feb 23 at 22:15", "date": "2021-03-05 13:18:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4035775/expected-number-of-objects-chosen-by-both-a-and-b", "openwebmath_score": 0.84630286693573, "openwebmath_perplexity": 298.03829857269744, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462211935647, "lm_q2_score": 0.8633916029436189, "lm_q1q2_score": 0.852984471538403}}
{"url": "https://mathoverflow.net/questions/87877/jacobis-equality-between-complementary-minors-of-inverse-matrices", "text": "Jacobi's equality between complementary minors of inverse matrices\n\nWhat's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse?\n\nLet $A[I,J]$ denote the submatrix of an $n \\times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then\n\n$$|\\det A[I,J]| = | (\\det A) \\det A^{-1}[J^c,I^c]|,$$ where $I^c$ stands for $[n] \\setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\\lambda$ and the complete symmetric functions $h_\\lambda$.\n\nI also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(\n\nEdit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.\n\n\u2022 Wow - I did not know that algebra proofs could have a \"sexist bias\". I am too curious to let it pass --- what do you mean exactly? \u2013\u00a0Federico Poloni Feb 8 '12 at 10:19\n\u2022 I was just referring to the preface, where he said the book is suitable for undergraduate students, or boys in their last years of school. Maybe the word \"boy\" has a gender neutral meaning back then? \u2013\u00a0John Jiang Feb 8 '12 at 19:39\n\nThe key word under which you will find this result in modern books is \"Schur complement\". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be $$M=\\begin{bmatrix}A & B\\\\\\\\ C & D\\end{bmatrix},$$ where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.\n\nYou may verify the following identity (\"magic wand Schur complement formula\") $$\\begin{bmatrix}A & B\\\\\\\\ C & D\\end{bmatrix} = \\begin{bmatrix}I & 0\\\\\\\\ CA^{-1} & I\\end{bmatrix} \\begin{bmatrix}A & 0\\\\\\\\ 0 & S\\end{bmatrix} \\begin{bmatrix}I & A^{-1}B\\\\\\\\ 0 & I\\end{bmatrix}. \\tag{1}$$ By taking determinants, $$\\det M=\\det A \\det S. \\tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).\n\nNote that the \"magic formula\" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.\n\n\u2022 I guess for any thesis involving minors the Schur complement formula would be among the first things to try. \u2013\u00a0John Jiang Feb 8 '12 at 10:30\n\nThis is nothing but the Schur complement formula. See my book Matrices; Theory and Applications, 2nd ed., Springer-Verlag GTM 216, page 41.\n\nUp to some permutation of rows and columns, we may assume that $I=J=[1,p]$. Let us write blockwise $$A=\\begin{pmatrix} W & X \\\\\\\\ Y & Z \\end{pmatrix}.$$ Assume WLOG that $W$ is invertible. On the one hand, we have (Schur C.F) $$\\det A=\\det W\\cdot\\det(Z-YW^{-1}X).$$ Finally, we have $$A^{-1}=\\begin{pmatrix} \\cdot & \\cdot \\\\\\\\ \\cdot & (Z-YW^{-1}X)^{-1} \\end{pmatrix},$$which gives the desired result.\n\nThese formulas are obtained by factorizing $A$ into $LU$ (namely, $L= \\begin{pmatrix} I_* & 0 \\\\ YW^{-1} & I_* \\end{pmatrix}$ and $U = \\begin{pmatrix} W & X \\\\ 0 & Z-YW^{-1}X \\end{pmatrix}$, with the $I_*$ being identity matrices of appropriate size).\n\n\u2022 Your answer is equally valid. Thanks! \u2013\u00a0John Jiang Feb 8 '12 at 10:27\n\nNot all has been said about this question that is worth saying -- at the very least, someone could have written down the version without the absolute values; but more importantly, there are various other equally good proofs.\n\nNotations and statement\n\nLet me first state the result with proper signs and no absolute values.\n\nStanding assumptions. The following notations will be used throughout this post:\n\n\u2022 Let $\\mathbb{K}$ be a commutative ring. All matrices that appear in the following are matrices over $\\mathbb{K}$.\n\n\u2022 Let $\\mathbb{N}=\\left\\{ 0,1,2,\\ldots\\right\\}$.\n\n\u2022 For every $n\\in\\mathbb{N}$, we let $\\left[ n\\right]$ denote the set $\\left\\{ 1,2,\\ldots,n\\right\\}$.\n\n\u2022 Fix $n\\in\\mathbb{N}$.\n\n\u2022 Let $S_{n}$ denote the $n$-th symmetric group (i.e., the group of permutations of $\\left[ n\\right]$).\n\n\u2022 If $A\\in\\mathbb{K}^{n\\times n}$ is an $n\\times n$-matrix, and if $I$ and $J$ are two subsets of $\\left[ n\\right]$, then $A_{J}^{I}$ is the $\\left\\vert I\\right\\vert \\times\\left\\vert J\\right\\vert$-matrix defined as follows: Write $A$ in the form $A=\\left( a_{i,j}\\right) _{1\\leq i\\leq n,\\ 1\\leq j\\leq m}$; write the set $I$ in the form $I=\\left\\{ i_{1}<i_{2}<\\cdots<i_{u}\\right\\}$; write the set $J$ in the form $J=\\left\\{ j_{1}<j_{2}<\\cdots<j_{v}\\right\\}$. Then, set $A_{J}^{I}=\\left( a_{i_{x},j_{y}}\\right) _{1\\leq x\\leq u,\\ 1\\leq y\\leq v}$. (Thus, roughly speaking, $A_{J}^{I}$ is the $\\left\\vert I\\right\\vert \\times\\left\\vert J\\right\\vert$-matrix obtained from $A$ by removing all rows whose indices do not belong to $I$, and removing all columns whose indices do not belong to $J$.)\n\nIf $K$ is a subset of $\\left[ n\\right]$, then:\n\n\u2022 we use $\\widetilde{K}$ to denote the complement $\\left[ n\\right] \\setminus K$ of this subset in $\\left[ n\\right]$.\n\n\u2022 we use $\\sum K$ to denote the sum of the elements of $K$.\n\nNow, we claim the following:\n\nTheorem 1 (Jacobi's complementary minor formula). Let $A\\in\\mathbb{K} ^{n\\times n}$ be an invertible $n\\times n$-matrix. Let $I$ and $J$ be two subsets of $\\left[ n\\right]$ such that $\\left\\vert I\\right\\vert =\\left\\vert J\\right\\vert$. Then,\n\n$\\det\\left( A_{J}^{I}\\right) =\\left( -1\\right) ^{\\sum I+\\sum J}\\det A\\cdot\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J} }\\right)$.\n\nThree references\n\nHere are three references to proofs of Theorem 1:\n\nNote that every source uses different notations. What I call $A_{J}^{I}$ above is called $A_{IJ}$ in the paper by Caracciolo, Sokal and Sportiello, is called $A\\left[ I,J\\right]$ in Lalonde's paper, and is called $\\operatorname*{sub}\\nolimits_{w\\left( I\\right) }^{w\\left( J\\right) }A$ in my notes. Also, the $I$ and $J$ in the paper by Caracciolo, Sokal and Sportiello correspond to the $\\widetilde{I}$ and $\\widetilde{J}$ in Theorem 1 above.\n\nA fourth proof\n\nLet me now give a fourth proof, using exterior algebra. The proof is probably not new (the method is definitely not), but I find it instructive.\n\nThis proof would become a lot shorter if I didn't care for the signs and would only prove the weaker claim that $\\det\\left( A_{J}^{I}\\right) = \\pm \\det A\\cdot\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J} }\\right)$ for some value of $\\pm$. But this weaker claim is not as useful as Theorem 1 in its full version (in particular, it would not suffice to fill the gap in Macdonald's book that has motivated this question).\n\nThe permutation $w\\left( K\\right)$\n\nLet us first introduce some more notations:\n\nIf $K$ is a subset of $\\left[ n\\right]$, and if $k = \\left|K\\right|$, then we let $w\\left( K\\right)$ be the (unique) permutation $\\sigma\\in S_{n}$ whose first $k$ values $\\sigma\\left( 1\\right) ,\\sigma\\left( 2\\right) ,\\ldots,\\sigma\\left( k\\right)$ are the elements of $K$ in increasing order, and whose next $n-k$ values $\\sigma\\left( k+1\\right) ,\\sigma\\left( k+2\\right) ,\\ldots ,\\sigma\\left( n\\right)$ are the elements of $\\widetilde{K}$ in increasing order.\n\nThe first important property of $w\\left( K\\right)$ is the following fact:\n\nLemma 2. Let $K$ be a subset of $\\left[ n\\right]$. Then, $\\left( -1\\right) ^{w\\left( K\\right) }=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }$.\n\nYou don't need to prove Lemma 2 if you only care about the weaker version of Theorem 1 with the $\\pm$ sign.\n\nProof of Lemma 2. Let $k=\\left\\vert K\\right\\vert$. Let $a_{1},a_{2} ,\\ldots,a_{k}$ be the $k$ elements of $K$ in increasing order (with no repetitions). Let $b_{1},b_{2},\\ldots,b_{n-k}$ be the $n-k$ elements of $\\widetilde{K}$ in increasing order (with no repetitions). Let $\\gamma =w\\left( K\\right)$. Then, the definition of $w\\left( K\\right)$ shows that the first $k$ values $\\gamma\\left( 1\\right) ,\\gamma\\left( 2\\right) ,\\ldots,\\gamma\\left( k\\right)$ of $\\gamma$ are the elements of $K$ in increasing order (that is, $a_{1},a_{2},\\ldots,a_{k}$), and the next $n-k$ values $\\gamma\\left( k+1\\right) ,\\gamma\\left( k+2\\right) ,\\ldots ,\\gamma\\left( n\\right)$ of $\\gamma$ are the elements of $\\widetilde{K}$ in increasing order (that is, $b_{1},b_{2},\\ldots,b_{n-k}$). In other words,\n\n$\\left( \\gamma\\left( 1\\right) ,\\gamma\\left( 2\\right) ,\\ldots ,\\gamma\\left( n\\right) \\right) =\\left( a_{1},a_{2},\\ldots,a_{k} ,b_{1},b_{2},\\ldots,b_{n-k}\\right)$.\n\nNow, you can obtain the list $\\left( \\gamma\\left( 1\\right) ,\\gamma\\left( 2\\right) ,\\ldots,\\gamma\\left( n\\right) \\right)$ from the list $\\left( 1,2,\\ldots,n\\right)$ by successively switching adjacent entries, as follows:\n\n\u2022 First, move the element $a_{1}$ to the front of the list, by successively switching it with each of the $a_{1}-1$ entries smaller than it.\n\n\u2022 Then, move the element $a_{2}$ to the second position, by successively switching it with each of the $a_{2}-2$ entries (other than $a_{1}$) smaller than it.\n\n\u2022 Then, move the element $a_{3}$ to the third position, by successively switching it with each of the $a_{3}-3$ entries (other than $a_{1}$ and $a_{2}$) smaller than it.\n\n\u2022 And so on, until you finally move the element $a_{k}$ to the $k$-th position.\n\nMore formally, you are iterating over all $i\\in\\left\\{ 1,2,\\ldots,k\\right\\}$ (in increasing order), each time moving the element $a_{i}$ to the $i$-th position in the list, by successively switching it with each of the $a_{i}-i$ entries (other than $a_{1},a_{2},\\ldots,a_{i-1}$) smaller than it.\n\nAt the end, the first $k$ positions of the list are filled with $a_{1} ,a_{2},\\ldots,a_{k}$ (in this order), whereas the remaining $n-k$ positions are filled with the remaining entries $b_{1},b_{2},\\ldots,b_{n-k}$ (again, in this order, because the switches have never disrupted their strictly-increasing relative order). Thus, at the end, your list is precisely $\\left( a_{1},a_{2},\\ldots,a_{k},b_{1},b_{2},\\ldots,b_{n-k}\\right) =\\left( \\gamma\\left( 1\\right) ,\\gamma\\left( 2\\right) ,\\ldots,\\gamma\\left( n\\right) \\right)$. You have used a total of\n\n$\\left( a_{1}-1\\right) +\\left( a_{2}-2\\right) +\\cdots+\\left( a_{k}-k\\right)$\n\n$=\\underbrace{\\left( a_{1}+a_{2}+\\cdots+a_{k}\\right) }_{\\substack{=\\sum K\\\\\\text{(by the definition of }a_{1},a_{2},\\ldots,a_{k}\\text{)} }}-\\underbrace{\\left( 1+2+\\cdots+k\\right) }_{\\substack{=1+2+\\cdots +\\left\\vert K\\right\\vert \\\\\\text{(since }k=\\left\\vert K\\right\\vert \\text{)}}}$\n\n$=\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right)$\n\nswitches. Thus, you have obtained the list $\\left( \\gamma\\left( 1\\right) ,\\gamma\\left( 2\\right) ,\\ldots,\\gamma\\left( n\\right) \\right)$ from the list $\\left( 1,2,\\ldots,n\\right)$ by $\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right)$ switches of adjacent entries. In other words, the permutation $\\gamma$ is a composition of $\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right)$ simple transpositions (where a \"simple transposition\" means a transposition switching $u$ with $u+1$ for some $u$). Hence, it has sign $\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }$. This proves Lemma 2.\n\nExterior algebras\n\nNow, let's introduce some more notations and state some well-known properties concerning exterior algebras.\n\nFor any $\\mathbb{K}$-module $V$, we let $\\wedge V$ denote the exterior algebra of $V$. The multiplication in this exterior algebra will be written as juxtaposition (i.e., we will write $ab$ for the product of two elements $a$ and $b$ of $\\wedge V$) or as multiplication (i.e., we will write $a\\cdot b$ for this product).\n\nIf $k\\in\\mathbb{N}$ and if $V$ is a $\\mathbb{K}$-module, then $\\wedge^{k}V$ shall mean the $k$-th exterior power of $V$. If $k\\in\\mathbb{N}$, if $V$ and $W$ are two $\\mathbb{K}$-modules, and if $f:V\\rightarrow W$ is a $\\mathbb{K}$-linear map, then the $\\mathbb{K}$-linear map $\\wedge^{k}V\\rightarrow \\wedge^{k}W$ canonically induced by $f$ will be denoted by $\\wedge^{k}f$. It is well-known that if $V$ and $W$ are two $\\mathbb{K}$-modules, if $f:V\\rightarrow W$ is a $\\mathbb{K}$-linear map, then\n\n(1) $\\left( \\wedge^{k}f\\right) \\left( a\\right) \\cdot\\left( \\wedge^{\\ell}f\\right) \\left( b\\right) =\\left( \\wedge^{k+\\ell}f\\right) \\left( ab\\right)$\n\nfor any $k\\in\\mathbb{N}$, $\\ell\\in\\mathbb{N}$, $a\\in\\wedge^{k}V$ and $b\\in\\wedge^{\\ell}V$.\n\nIf $V$ is a $\\mathbb{K}$-module, then\n\n(2) $uv=\\left( -1\\right) ^{k\\ell}vu$\n\nfor any $k\\in\\mathbb{N}$, $\\ell\\in\\mathbb{N}$, $u\\in\\wedge^{k}V$ and $v\\in\\wedge^{\\ell}V$.\n\nFor any $u\\in\\mathbb{N}$, we consider $\\mathbb{K}^{u}$ as the $\\mathbb{K}$-module of column vectors with $u$ entries.\n\nFor any $u\\in\\mathbb{N}$ and $v\\in\\mathbb{N}$, and any $v\\times u$-matrix $B\\in\\mathbb{K}^{v\\times u}$, we define $f_{B}$ to be the $\\mathbb{K}$-linear map $\\mathbb{K}^{u}\\rightarrow\\mathbb{K}^{v}$ sending each $x\\in\\mathbb{K} ^{u}$ to $Bx\\in\\mathbb{K}^{v}$. This $\\mathbb{K}$-linear map $f_{B}$ satisfies $\\det\\left( f_{B}\\right) =\\det B$, and is often identified with the matrix $B$ (though we will not identify it with $B$ here).\n\nHere is another known fact:\n\nProposition 2a. Let $f:\\mathbb{K}^{n}\\rightarrow\\mathbb{K}^{n}$ be a $\\mathbb{K}$-linear map. The map $\\wedge^{n}f:\\wedge^{n}\\left( \\mathbb{K} ^{n}\\right) \\rightarrow\\wedge^{n}\\left( \\mathbb{K}^{n}\\right)$ is multiplication by $\\det f$. In other words, every $z\\in\\wedge^{n}\\left( \\mathbb{K}^{n}\\right)$ satisfies\n\n(3) $\\left( \\wedge^{n}f\\right) \\left( z\\right) =\\left( \\det f\\right) z$.\n\nLet $\\left( e_{1},e_{2},\\ldots,e_{n}\\right)$ be the standard basis of the $\\mathbb{K}$-module $\\mathbb{K}^{n}$. (Thus, $e_{i}$ is the column vector whose $i$-th entry is $1$ and whose all other entries are $0$.)\n\nFor every subset $K$ of $\\left[ n\\right]$, we define $e_{K}\\in \\wedge^{\\left\\vert K\\right\\vert }\\left( \\mathbb{K}^{n}\\right)$ to be the element $e_{k_{1}}\\wedge e_{k_{2}}\\wedge\\cdots\\wedge e_{k_{\\left\\vert K\\right\\vert }}$, where $K$ is written in the form $K=\\left\\{ k_{1} <k_{2}<\\cdots<k_{\\left\\vert K\\right\\vert }\\right\\}$.\n\nFor every $k\\in\\mathbb{N}$ and every set $S$, we let $\\mathcal{P}_{k}\\left( S \\right)$ denote the set of all $k$-element subsets of $S$.\n\nIt is well-known that, for every $k\\in\\mathbb{N}$, the family $\\left( e_{K}\\right) _{K\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right) }$ is a basis of the $\\mathbb{K}$-module $\\wedge^{k}\\left( \\mathbb{K}^{n}\\right)$. Applying this to $k=n$, we conclude that the family $\\left( e_{K}\\right) _{K\\in\\mathcal{P}_{n}\\left( \\left[ n\\right] \\right) }$ is a basis of the $\\mathbb{K}$-module $\\wedge^{n}\\left( \\mathbb{K}^{n}\\right)$. Since this family $\\left( e_{K}\\right) _{K\\in\\mathcal{P}_{n}\\left( \\left[ n\\right] \\right) }$ is the one-element family $\\left( e_{\\left[ n\\right] }\\right)$ (because the only $K\\in\\mathcal{P}_{n}\\left( \\left[ n\\right] \\right)$ is the set $\\left[ n\\right]$), this rewrites as follows: The one-element family $\\left( e_{\\left[ n\\right] }\\right)$ is a basis of the $\\mathbb{K}$-module $\\wedge^{n}\\left( \\mathbb{K}^{n}\\right)$.\n\nIf $B$ is an $n\\times n$-matrix and $k\\in\\mathbb{N}$, then evaluating the map $\\wedge^{k}f_{B}$ on the elements of the basis $\\left( e_{K}\\right) _{K\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right) }$ of $\\wedge ^{k}\\left( \\mathbb{K}^{n}\\right)$, and expanding the results again in this basis gives rise to coefficients which are the $k\\times k$-minors of $B$. More precisely:\n\nProposition 3. Let $B\\in\\mathbb{K}^{n\\times n}$, $k\\in\\mathbb{N}$ and $J\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right)$. Then,\n\n$\\left( \\wedge^{k}f_{B}\\right) \\left( e_{J}\\right) = \\sum\\limits_{I\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right) }\\det\\left( B_{J} ^{I}\\right) e_{I}$.\n\n(This generalizes: If $u\\in\\mathbb{N}$, $v \\in \\mathbb{N}$, $B\\in\\mathbb{K}^{u\\times v}$, $k\\in\\mathbb{N}$ and $J\\in\\mathcal{P}_{k}\\left( \\left[ v\\right] \\right)$, then $\\left( \\wedge^{k}f_{B}\\right) \\left( e_{J}\\right) = \\sum\\limits_{I\\in\\mathcal{P}_{k}\\left( \\left[ u\\right] \\right) }\\det\\left( B_{J} ^{I}\\right) e_{I}$, where the elements $e_{J}\\in\\wedge^{k}\\left( \\mathbb{K}^{v}\\right)$ and $e_{I}\\in\\wedge^{k}\\left( \\mathbb{K}^{u}\\right)$ are defined as before but with $v$ and $u$ instead of $n$.)\n\nExtracting minors from the exterior algebra\n\nNow, we shall need a simple lemma:\n\nLemma 4. Let $K$ be a subset of $\\left[ n\\right]$. Then,\n\n$e_{K}e_{\\widetilde{K}}=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }e_{\\left[ n\\right] }$.\n\nProof of Lemma 4. Let $k = \\left|K\\right|$. Let $\\sigma$ be the permutation $w\\left( K\\right) \\in S_{n}$ defined above. Its first $k$ values $\\sigma\\left( 1\\right) ,\\sigma\\left( 2\\right) ,\\ldots,\\sigma\\left( k\\right)$ are the elements of $K$ in increasing order; thus, $e_{\\sigma\\left( 1\\right) }\\wedge e_{\\sigma\\left( 2\\right) }\\wedge\\cdots\\wedge e_{\\sigma\\left( k\\right) }=e_{K}$. Its next $n-k$ values $\\sigma\\left( k+1\\right) ,\\sigma\\left( k+2\\right) ,\\ldots,\\sigma\\left( n\\right)$ are the elements of $\\widetilde{K}$ in increasing order; thus, $e_{\\sigma\\left( k+1\\right) }\\wedge e_{\\sigma\\left( k+2\\right) }\\wedge\\cdots\\wedge e_{\\sigma\\left( n\\right) }=e_{\\widetilde{K}}$.\n\nFrom $\\sigma=w\\left( K\\right)$, we obtain $\\left( -1\\right) ^{\\sigma }=\\left( -1\\right) ^{w\\left( K\\right) }=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }$ (by Lemma 2).\n\nNow, it is well-known that\n\n$e_{\\sigma\\left( 1\\right) }\\wedge e_{\\sigma\\left( 2\\right) }\\wedge \\cdots\\wedge e_{\\sigma\\left( n\\right) }=\\left( -1\\right) ^{\\sigma }\\underbrace{e_{1}\\wedge e_{2}\\wedge\\cdots\\wedge e_{n}}_{=e_{\\left[ n\\right] }}=\\left( -1\\right) ^{\\sigma}e_{\\left[ n\\right] }$.\n\nHence,\n\n$\\left( -1\\right) ^{\\sigma}e_{\\left[ n\\right] }=e_{\\sigma\\left( 1\\right) }\\wedge e_{\\sigma\\left( 2\\right) }\\wedge\\cdots\\wedge e_{\\sigma\\left( n\\right) }$\n\n$=\\underbrace{\\left( e_{\\sigma\\left( 1\\right) }\\wedge e_{\\sigma\\left( 2\\right) }\\wedge\\cdots\\wedge e_{\\sigma\\left( k\\right) }\\right) }_{=e_{K} }\\underbrace{\\left( e_{\\sigma\\left( k+1\\right) }\\wedge e_{\\sigma\\left( k+2\\right) }\\wedge\\cdots\\wedge e_{\\sigma\\left( n\\right) }\\right) }_{=e_{\\widetilde{K}}}=e_{K}e_{\\widetilde{K}}$.\n\nSince $\\left( -1\\right) ^{\\sigma}=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }$, this rewrites as $\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }e_{\\left[ n\\right] }= e_K e_{\\widetilde{K}}$. This proves Lemma 4.\n\nWe can combine Proposition 3 and Lemma 4 to obtain the following fact:\n\nCorollary 5. Let $B\\in\\mathbb{K}^{n\\times n}$, $k\\in\\mathbb{N}$ and $J\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right)$. Then, every $K\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right)$ satisfies\n\n$\\left( \\wedge^{k}f_{B}\\right) \\left( e_{J}\\right) e_{\\widetilde{K} }=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+k\\right) }\\det\\left( B_{J}^{K}\\right) e_{\\left[ n\\right] }$.\n\nProof of Corollary 5. Let $K \\in \\mathcal{P}_{k}\\left( \\left[ n\\right] \\right)$. Let $I\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right)$ be such that $I\\neq K$. Then, $I\\not \\subseteq K$ (since the sets $I$ and $K$ have the same size $k$). Hence, there exists some $z\\in I$ such that $z\\notin K$. Consider this $z$. We have $z\\in I$ and $z\\in\\widetilde{K}$ (since $z\\notin K$). Hence, both $e_{I}$ and $e_{\\widetilde{K}}$ are \"wedge products\" containing the factor $e_{z}$; therefore, the product $e_{I} e_{\\widetilde{K}}$ is a \"wedge product\" containing this factor twice. Thus, $e_{I}e_{\\widetilde{K}}=0$.\n\nNow, forget that we fixed $I$. We thus have proven that\n\n$e_{I}e_{\\widetilde{K}}=0$ for every $I\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right)$ satisfying $I\\neq K$.\n\nHence,\n\n(4) $\\sum\\limits_{\\substack{I\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right) ;\\\\I\\neq K}}\\det\\left( B_{J}^{I}\\right) \\underbrace{e_{I}e_{\\widetilde{K}} }_{=0}=0$.\n\nProposition 3 yields\n\n$\\left( \\wedge^{k}f_{B}\\right) \\left( e_{J}\\right) =\\sum\\limits_{I\\in \\mathcal{P}_{k}\\left( \\left[ n\\right] \\right) }\\det\\left( B_{J} ^{I}\\right) e_{I}$.\n\nMultiplying both sides of this equality by $e_{\\widetilde{K}}$ from the right, we find\n\n$\\left( \\wedge^{k}f_{B}\\right) \\left( e_{J}\\right) e_{\\widetilde{K}} =\\sum\\limits_{I\\in\\mathcal{P}_{k}\\left( \\left[ n\\right] \\right) }\\det\\left( B_{J}^{I}\\right) e_{I}e_{\\widetilde{K}}$\n\n$=\\det\\left( B_{J}^{K}\\right) e_{K}e_{\\widetilde{K}}+\\sum\\limits_{\\substack{I\\in \\mathcal{P}_{k}\\left( \\left[ n\\right] \\right) ;\\\\I\\neq K}}\\det\\left( B_{J}^{I}\\right) e_{I}e_{\\widetilde{K}}$\n\n$=\\det\\left( B_{J}^{K}\\right) \\underbrace{e_{K}e_{\\widetilde{K}} }_{\\substack{=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }e_{\\left[ n\\right] }\\\\\\text{(by Lemma 4)}}}$ (by (4))\n\n$=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+\\left\\vert K\\right\\vert \\right) }\\det\\left( B_{J}^{K}\\right) e_{\\left[ n\\right] }$\n\n$=\\left( -1\\right) ^{\\sum K-\\left( 1+2+\\cdots+k\\right) }\\det\\left( B_{J}^{K}\\right) e_{\\left[ n\\right] }$\n\n(since $\\left\\vert K\\right\\vert =k$). This proves Corollary 5.\n\nCorollary 5 is rather helpful when it comes to extracting a specific minor of a matrix $B$ from the maps $\\wedge^{k}f_{B}$.\n\nProof of Theorem 1\n\nProof of Theorem 1. Set $k=\\left\\vert I\\right\\vert =\\left\\vert J\\right\\vert$. Notice that $\\left\\vert \\widetilde{I}\\right\\vert =n-k$ (since $\\left\\vert I\\right\\vert =k$) and $\\left\\vert \\widetilde{J}\\right\\vert =n-k$ (similarly).\n\nDefine $y\\in\\wedge ^{n-k}\\left( \\mathbb{K}^{n}\\right)$ by $y=\\left( \\wedge^{n-k}f_{A^{-1} }\\right) \\left( e_{\\widetilde{I}}\\right)$.\n\nThe maps $f_{A}$ and $f_{A^{-1}}$ are mutually inverse (since the map $\\mathbb{K}^{n\\times n}\\rightarrow\\operatorname*{End}\\left( \\mathbb{K} ^{n}\\right) ,\\ B\\mapsto f_{B}$ is a ring homomorphism). Hence, the maps $\\wedge^{n-k}f_{A}$ and $\\wedge^{n-k}f_{A^{-1}}$ are mutually inverse (since $\\wedge^{n-k}$ is a functor). Thus, $\\left( \\wedge^{n-k}f_{A}\\right) \\circ\\left( \\wedge^{n-k}f_{A^{-1}}\\right) =\\operatorname*{id}$. Now, $y=\\left( \\wedge^{n-k}f_{A^{-1}}\\right) \\left( e_{\\widetilde{I}}\\right)$, so that\n\n$\\left( \\wedge^{n-k}f_{A}\\right) \\left( y\\right) =\\left( \\wedge ^{n-k}f_{A}\\right) \\left( \\left( \\wedge^{n-k}f_{A^{-1}}\\right) \\left( e_{\\widetilde{I}}\\right) \\right) =\\underbrace{\\left( \\left( \\wedge ^{n-k}f_{A}\\right) \\circ\\left( \\wedge^{n-k}f_{A^{-1}}\\right) \\right) }_{=\\operatorname*{id}}\\left( e_{\\widetilde{I}}\\right) =e_{\\widetilde{I}}$.\n\nBut (1) (applied to $V=\\mathbb{K}^{n}$, $W=\\mathbb{K}^{n}$, $f=f_{A}$, $\\ell=n-k$, $a=e_{J}$ and $b=y$) yields\n\n$\\left( \\wedge^{k}f_{A}\\right) \\left( e_{J}\\right) \\cdot\\left( \\wedge^{n-k}f_{A}\\right) \\left( y\\right) =\\left( \\wedge^{n}f_{A}\\right) \\left( e_{J}y\\right)$.\n\nThus,\n\n$\\left( \\wedge^{n}f_{A}\\right) \\left( e_{J}y\\right) =\\left( \\wedge ^{k}f_{A}\\right) \\left( e_{J}\\right) \\cdot\\underbrace{\\left( \\wedge ^{n-k}f_{A}\\right) \\left( y\\right) }_{=e_{\\widetilde{I}}}$\n\n$=\\left( \\wedge^{k}f_{A}\\right) \\left( e_{J}\\right) e_{\\widetilde{I}} = \\left( -1\\right) ^{\\sum I-\\left( 1+2+\\cdots+k\\right) }\\det\\left( A_{J}^{I}\\right) e_{\\left[ n\\right] }$\n\n(by Corollary 5, applied to $B=A$ and $K=I$).\n\nHence,\n\n$\\left( -1\\right) ^{\\sum I-\\left( 1+2+\\cdots+k\\right) }\\det\\left( A_{J}^{I}\\right) e_{\\left[ n\\right] }$\n\n$=\\left( \\wedge^{n}f_{A}\\right) \\left( e_{J}y\\right) =\\underbrace{\\left( \\det f_{A}\\right) }_{=\\det A}e_{J}y$\n\n(by (3), applied to $f=f_{A}$ and $z=e_{J}y$)\n\n(5) $=\\left( \\det A\\right) e_{J}y$.\n\nBut (2) (applied to $\\ell=n-k$, $u=e_{J}$ and $v=y$) yields\n\n$e_{J} y = \\left(-1\\right)^{k \\left(n-k\\right)} \\underbrace{y}_{=\\left( \\wedge^{n-k}f_{A^{-1}}\\right) \\left( e_{\\widetilde{I}}\\right) } \\underbrace{e_{J}} _{=e_{\\widetilde{\\widetilde{J}}}}$\n\n$=\\left( -1\\right) ^{k\\left( n-k\\right) }\\underbrace{\\left( \\wedge ^{n-k}f_{A^{-1}}\\right) \\left( e_{\\widetilde{I}}\\right) e_{\\widetilde{\\widetilde{J}}}}_{\\substack{=\\left( -1\\right) ^{\\sum \\widetilde{J}-\\left( 1+2+\\cdots+\\left( n-k\\right) \\right) }\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J}}\\right) e_{\\left[ n\\right] }\\\\\\text{(by Corollary 5, applied to }A^{-1}\\text{, }n-k\\text{, }\\widetilde{I}\\text{ and }\\widetilde{J}\\\\\\text{instead of }B\\text{, }k\\text{, }J\\text{ and }K\\text{)}}}$\n\n$=\\left( -1\\right) ^{k\\left( n-k\\right) }\\left( -1\\right) ^{\\sum \\widetilde{J}-\\left( 1+2+\\cdots+\\left( n-k\\right) \\right) }\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J}}\\right) e_{\\left[ n\\right] }$\n\n(6) $=\\left( -1\\right) ^{k\\left( n-k\\right) +\\sum\\widetilde{J}-\\left( 1+2+\\cdots+\\left( n-k\\right) \\right) }\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J}}\\right) e_{\\left[ n\\right] }$.\n\nBut\n\n$k\\left( n-k\\right) +\\underbrace{\\sum\\widetilde{J}}_{=\\sum\\left\\{ 1,2,\\ldots,n\\right\\} -\\sum J}-\\underbrace{\\left( 1+2+\\cdots+\\left( n-k\\right) \\right) }_{=\\sum\\left\\{ 1,2,\\ldots,n-k\\right\\} }$\n\n$=k\\left( n-k\\right) +\\sum\\left\\{ 1,2,\\ldots,n\\right\\} -\\sum J-\\sum\\left\\{ 1,2,\\ldots,n-k\\right\\}$\n\n$=\\underbrace{\\sum\\left\\{ 1,2,\\ldots,n\\right\\} -\\sum\\left\\{ 1,2,\\ldots ,n-k\\right\\} }_{\\substack{=\\sum\\left\\{ n-k+1,n-k+2,\\ldots,n\\right\\} \\\\=k\\left( n-k\\right) +\\sum\\left\\{ 1,2,\\ldots,k\\right\\} \\\\=k\\left( n-k\\right) +\\left( 1+2+\\cdots+k\\right) }}+k\\left( n-k\\right) -\\sum J$\n\n$=2k\\left( n-k\\right) +\\left( 1+2+\\cdots+k\\right) -\\sum J$\n\n$\\equiv-\\left( 1+2+\\cdots+k\\right) -\\sum J\\operatorname{mod}2$.\n\nHence,\n\n$\\left( -1\\right) ^{k\\left( n-k\\right) +\\sum\\widetilde{J}-\\left( 1+2+\\cdots+\\left( n-k\\right) \\right) }=\\left( -1\\right) ^{-\\left( 1+2+\\cdots+k\\right) -\\sum J}$.\n\nThus, (6) rewrites as\n\n$e_{J}y=\\left( -1\\right) ^{-\\left( 1+2+\\cdots+k\\right) -\\sum J}\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J}}\\right) e_{\\left[ n\\right] }$.\n\nHence, (5) rewrites as\n\n$\\left( -1\\right) ^{\\sum I-\\left( 1+2+\\cdots+k\\right) }\\det\\left( A_{J}^{I}\\right) e_{\\left[ n\\right] }$\n\n$=\\left( \\det A\\right) \\left( -1\\right) ^{-\\left( 1+2+\\cdots+k\\right) -\\sum J}\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J} }\\right) e_{\\left[ n\\right] }$.\n\nWe can \"cancel\" $e_{\\left[ n\\right] }$ from this equality (because if $\\lambda$ and $\\mu$ are two elements of $\\mathbb{K}$ satisfying $\\lambda e_{\\left[ n\\right] }=\\mu e_{\\left[ n\\right] }$, then $\\lambda=\\mu$), and thus obtain\n\n$\\left( -1\\right) ^{\\sum I-\\left( 1+2+\\cdots+k\\right) }\\det\\left( A_{J}^{I}\\right)$\n\n$=\\left( \\det A\\right) \\left( -1\\right) ^{-\\left( 1+2+\\cdots+k\\right) -\\sum J}\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J} }\\right)$.\n\nDividing this equality by $\\left( -1\\right) ^{\\sum I-\\left( 1+2+\\cdots +k\\right) }$, we obtain\n\n$\\det\\left( A_{J}^{I}\\right)$\n\n$=\\left(\\det A\\right) \\dfrac{\\left( -1\\right) ^{-\\left( 1+2+\\cdots+k\\right) -\\sum J}}{\\left( -1\\right) ^{\\sum I-\\left( 1+2+\\cdots+k\\right) }}\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J}}\\right)$\n\n$=\\left( -1\\right) ^{\\sum I+\\sum J}\\det A\\cdot\\det\\left( \\left( A^{-1}\\right) _{\\widetilde{I}}^{\\widetilde{J}}\\right)$.\n\nThis proves Theorem 1.\n\nI have to admit this proof looked far shorter on the scratch paper on which it was conceived than it has ended up here...\n\n\u2022 If we agree to see a $\\mathbb{K}$-matrix as a function $S\\times S\\to\\mathbb{K}$ , with any finite index set $S$ (not just some $[m]$), then, according to Proposition 3, the matrix of the linear map $\\Lambda^k(f_B)$ in the basis $\\{e_I\\}_{I\\in\\mathcal{P}_k([n])}$ is $[\\det(B^I_J)]_{(I\\times J)\\in\\mathcal{P}_k([n])\\times \\mathcal{P}_k([n])}$. \u2013\u00a0Pietro Majer Dec 19 '17 at 18:48\n\u2022 So the functoriality of $\\Lambda^k$ reads as the (generalized) Cauchy-Binet formula: en.wikipedia.org/wiki/Cauchy\u2013Binet_formula#Generalization . In the same spirit, I'd like to see Theorem 1 as a statement on the inverse matrix of the linear map $\\Lambda^k(f_A)$, or better, since it also refers to $\\tilde I$ and $\\tilde J$, of the linear map $\\Lambda^k(f_A)^*\\sim\\Lambda^{n-k}(f_A)$. Can this produce a proof to Theorem 1 consisting in a plain translation of facts about exterior algebra in the language of matrices? \u2013\u00a0Pietro Majer Dec 19 '17 at 18:48\n\u2022 @PietroMajer: This is what I was trying to achieve when I started writing the proof. Unfortunately, it's not directly clear to me how this works. \u2013\u00a0darij grinberg Dec 19 '17 at 19:54\n\nHere is a short proof inspired by darij grinberg's answer and not using Schur complements (in particular, no invertibility of sub-matrices or continuity assumptions are needed).\n\nWLOG let $I=J=[k]$. Let $A_i$ denote the $i$th column of a matrix $A$. Consider the product $$A \\left[ \\begin{array}{c|c|c|c|c|c} e_1 & \\cdots & e_k & A^{-1}_{k+1} & \\cdots & A^{-1}_n \\end{array} \\right] = \\left[ \\begin{array}{c|c|c|c|c|c} A_1 & \\cdots & A_k & e_{k+1} & \\cdots & e_n \\end{array} \\right]$$ which can also be written $$A \\begin{pmatrix} I_k & A^{-1}[J,I^c] \\\\ 0 & A^{-1}[J^c,I^c] \\end{pmatrix} = \\begin{pmatrix} A[I,J] & 0 \\\\ A[I^c,J] & I_{n-k} \\end{pmatrix}$$ Taking the determinant of both sides yields $$(\\det A) (\\det A^{-1}[J^c,I^c]) = \\det A[I,J]$$.\n\n\u2022 Really nice... although a nontrivial part of my answer is justifying the \"WLOG\" in the first sentence. Still, this simplifies a lot. \u2013\u00a0darij grinberg Feb 17 '17 at 1:19\n\u2022 I believe the justification is not too bad. Assume the result is true for I=K, J=K, K=[k]. Then for any other I,J of cardinality k, we have $\\det(A[I,J])=\\det((PAQ)[K,K])=\\det(PAQ)\\det(Q^{-1}A^{-1}P^{-1}[K^c,K^c]) = \\det(PAQ)\\det(A^{-1}[J^c,I^c])$, where $P$ is the permutation matrix taking $I$ to $1,\\dots,k$ and $I^c$ to $k+1,\\dots,n$ (I believe you call this permutation $w(I)$), and $Q$ is defined similarly. So we incur a factor of $\\text{sign}(P)\\text{sign}(Q)$ which is $(-1)^{\\sum I + \\sum J}$ by a counting argument, as you proved. \u2013\u00a0user1980685 Feb 17 '17 at 15:05\n\u2022 I suspect that your $Q$ is not \"defined similarly\" but rather is the inverse of the matrix defined similarly. But yes, this is essentially how it's done. \u2013\u00a0darij grinberg Feb 18 '17 at 5:42", "date": "2018-11-16 02:15:14", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/87877/jacobis-equality-between-complementary-minors-of-inverse-matrices", "openwebmath_score": 0.9742786884307861, "openwebmath_perplexity": 159.26366793194484, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399094961359, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8529832786061683}}
{"url": "https://panchamukhahanuman.org/2xho27/article.php?id=application-of-geometric-progression-b49232", "text": "These lessons help High School students to express and interpret geometric sequence applications. The distinction between a progression and a series is that a progression is a sequence, whereas a series is a sum. Mathematicians calculate a term in the series by multiplying the initial value in the sequence by the rate raised to the power of one less than the term number. problem and check your answer with the step-by-step explanations. to write an equivalent form of an exponential function to Examples: A company offers to pay you $0.10 for the first day,$0.20 for the second day, $0.40 for the third day,$0.80 for the fourth day, and so on. If in a sequence of terms, each succeeding term is generated by multiplying each preceding term with a constant value, then the sequence is called a geometric progression. Number Sequences Wilma bought a house for $170,000. Geometric series played an important role in the early development of calculus, and continue as a central part of the study of the convergence of series. You invest$5000 for 20 years at 2% p.a. Examples: Input : a = 2 r = 2, n = 4 Output : 2 4 8 16 Recommended: Please try your approach on first, before moving on to the solution. Example: A line is divided into six parts forming a geometric sequence. 1.01212t\u00c2\u00a0to reveal the approximate equivalent rate of growth or decay. C. Use the properties of exponents to transform expressions for There are many applications for sciences, business, personal finance, and even for health, but most people are unaware of these. It is estimated that the student population will increase by 4% each year. Get help with your Geometric progression homework. C. \u2026 Your email address will not be published. An example a geometric progression would be 2 / 8 / 32 / 128 / 512 / 2048 / 8192 ... where the common ratio is 4. It is in finance, however, that the geometric series finds perhaps its greatest predictive power. Geometric Sequences: n-th Term Geometric Progression. Geometric Progression, Series & Sums Introduction. In the 21 st century, our lives are ruled by money. monthly interest rate if the annual rate is 15%. Complete the square in a quadratic expression to reveal the 1,2,4,8. the function it defines. Sequences and series, whether they be arithmetic or geometric, have may applications to situations you may not think of as being related to sequences or series. Quadratic and Cubic Sequences. Before going to learn how to find the sum of a given Geometric Progression, first know what a GP is in detail. product of powers, power of a product, and rational exponents, Geometric series word problems: hike Our mission is to provide a free, world-class education to anyone, anywhere. If the ball is dropped from 80 cm, find the height of the fifth bounce. find the height of the fifth bounce. Given that Geometric series are one of the simplest examples of i\u2026 Bruno has 3 pizza stores and wants to dramatically expand his franchise nationwide. exponential functions. In finer terms, the sequence in which we multiply or divide a fixed, non-zero number, each time infinitely, then the progression is said to be geometric. Examples, solutions, videos, and lessons to help High School students learn to choose Geometric sequence sequence definition. $${S_n} = \\frac{{3\\left( {{2^6} \u2013 1} \\right)}}{{2 \u2013 1}} = 3\\left( {64 \u2013 1} \\right) = 189$$cm. Now that we have learnt how to how geometric sequences and series, we can apply them to real life scenarios. Example: Definition of Geometric Sequence In mathematics, the geometric sequence is a collection of numbers in which each term of the progression is a constant multiple of the previous term. change if the interest is given quarterly? How much money do View TUTORIAL 10 (APPLICATION - CHAPTER 4).pdf from MATH 015 at Open University Malaysia. $${a_1} = 3$$, $${a_n} = 96$$, $$n = 6$$, we have maximum or minimum value of the function it defines. The rabbit grows at 7% per week. monthly? Geometric Sequence & Series, Sigma Notation & Application of Geometric Series Introduction: . The geometric sequence definition is that a collection of numbers, in which all but the first one, are obtained by multiplying the previous one by a fixed, non-zero number called the common ratio.If you are struggling to understand what a geometric sequences is, don't fret! For example: If I can invest at 5% and I want $50,000 in 10 years, how much should I invest now? Try the free Mathway calculator and What will the value of an automobile be after three years if it is purchased for $$4500$$ dollars? A geometric progression (GP), also called a geometric sequence, is a sequence of numbers which differ from each other by a common ratio. Therefore, the whole length of the line is equal to $$189$$cm. This paper will cover the study of applications of geometric series in financial mathematics. What will the house be worth in 10 years? At this rate, how many boxes will When r=0, we get the sequence {a,0,0,...} which is not geometric A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (r). If the shortest length is $$3$$cm and the longest is $$96$$cm, find the length of the whole line. Khan Academy is a 501(c)(3) nonprofit organization. This video provides an application problem that can be modeled by the sum of an a geometric sequence dealing with total income with pay doubling everyday. be rewritten as (1.151/12)12t\u00c2 \u00e2\u0089\u0088 A geometric series is the sum of the numbers in a geometric progression. You land a job as a police officer. A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. You leave the money in for 3 Try the given examples, or type in your own Please submit your feedback or enquiries via our Feedback page. Arithmetic-Geometric Progression An arithmetic-geometric progression (AGP) is a progression in which each term can be represented as the product of the terms of an arithmetic progressions (AP) and a geometric progressions (GP). TUTORIAL 10 (CHAPTER 4-APPLICATION OF ARITHMETIC AND GEOMETRIC SERIES) Linear Sequences you have in the bank after 3 years? is to sell double the number of boxes as the previous day. Educator since 2009. (GP), whereas the constant value is called the common ratio. With this free application you can: - Figure out the Nth term of the Geometric Progression given the \u2026 Each term of a geometric series, therefore, involves a higher power than the previous term. Let $${a_1} = 4500$$ = purchased value of the automobile. I have 50 rabbits. Geometric Progression is a type of sequence where each successive term is the result of multiplying a constant number to its preceding term. Copyright \u00a9 2005, 2020 - OnlineMathLearning.com. We can find the common ratio of a GP by finding the ratio between any two adjacent terms. Geometric growth is found in many real life scenarios such as population growth and the growth of an investment. problem solver below to practice various math topics. Application of geometric progression Example \u2013 1 : If an amount \u20b9 1000 deposited in the bank with annual interest rate 10% interest compounded annually, then find total amount at the end of first, second, third, forth and first years. A. Given first term (a), common ratio (r) and a integer n of the Geometric Progression series, the task is to print th n terms of the series. If the ball is dropped from 80 cm, In this tutorial we discuss the related problems of application of geometric sequence and geometric series. If the number of stores he owns doubles in number each month, what month will he launch 6,144 stores? Example: How many will I have in 15 weeks. They have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. An \u201cinterest\u201d problem \u2013 application of Geometric Series: Question A man borrows a loan of$1,000,000 for a house from a bank and likes to pay back in 10 years (120 monthly instalments), the first instalment being paid at the end of first month and compound interest being calculated at 6% per annum. Bouncing ball application of a geometric sequence Related Pages $$= {a_1}\\left( {0.85} \\right) \u2013 {a_1}\\left( {0.85} \\right)\\left( {\\frac{{15}}{{85}}} \\right) = {a_1}\\left( {0.85} \\right)\\left[ {1 \u2013 \\frac{{15}}{{100}}} \\right]$$ $${S_n} = \\frac{{{a_1}\\left( {{r^n} \u2013 1} \\right)}}{{r \u2013 1}}$$ (As $$r > 1$$) 7% increase every year. 16,847 answers. Example 7: Solving Application Problems with Geometric Sequences. years, each year getting 5% interest per annum. $$= {a_1}\\left( {0.85} \\right)\\left( {1 \u2013 0.85} \\right) = {a_1}\\left( {0.85} \\right)\\left( {0.85} \\right) = {a_1}{\\left( {0.85} \\right)^2}$$, Similarly, the value at the end of the third year The geometric series is a marvel of mathematics which rules much of the natural world. GEOMETRIC SERIES AND THREE APPLICATIONS 33 But the sum 9 10 + 9 100 + 9 1000 + 9 is a geometric series with rst term a = 10 and ratio r = 1 10.The ratio r is between 1 and 1, so we can use the formula for a geometric series: The 3rd and the 8th term of a G. P. are 4 and 128 respectively. There are many uses of geometric sequences in everyday life, but one of the most common is in calculating interest earned. fare of taxi, finding multiples of numbers within a range. In the following series, the numerators are \u2026 they sell on day 7? Estimate the student population in 2020. Geometric series have applications in math and science and are one of the simplest examples of infinite series with finite sums. Compounding Interest and other Geometric Sequence Word Problems. Your salary for the first year is $43,125. Their daily goal Use properties of exponents (such as power of a power, Geometric series are used throughout mathematics. Speed of an aircraft, finding the sum of n terms of natural numbers. Ashley Kannan. The geometric series is a marvel of mathematics which rules much of the world. Remember these examples Application of a Geometric Sequence. a. An arithmetic sequence has a common difference of 9 and a(41) = 25. Solution: This chapter is for those who want to see applications of arithmetic and geometric progressions to real life. Write the equation that represents the house\u2019s value over time. Example: I decide to run a rabbit farm. explain properties of the quantity represented by the expression. etc.) A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. The common ratio (r) is obtained \u2026 Factor a quadratic expression to reveal the zeros of Solve Word Problems using Geometric Sequences. The value of an automobile depreciates at the rate of $$15\\%$$ per year. How much will we end up with? For example, the sequence 2, 4, 8, 16, \u2026 2, 4, 8, 16, \\dots 2, 4, 8, 1 6, \u2026 is a geometric sequence with common ratio 2 2 2. Write a formula for the student population. Educators go through a rigorous application process, and every answer they submit is reviewed by our in-house editorial team. height from which it was dropped. For example, the expression 1.15t\u00c2 can Growth. Geometric growth occurs when the common ratio is greater than 1, that is . Geometric series are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. If the shortest leng In this tutorial we discuss the related problems of application of geometric sequence and geometric series. In 2013, the number of students in a small school is 284. Application of Geometric Sequence and Series. A line is divided into six parts forming a geometric sequence. Each year, it increases 2% of its value. How does this On January 1, Abby\u2019s troop sold three boxes of Girl Scout cookies online. The sequence allows a borrower to know the amount his bank expects him to pay back using simple interest. $$= {a_1}{\\left( {0.85} \\right)^3}$$, Hence, the value of the automobile at the end of $$3$$ years after being purchased for $$4500$$ $$96 = 3 \\times {r^5}$$ $$\\Rightarrow r = 2$$, For the length of the whole line, we have 5. Solution: It is finance; however, the geometric series finds perhaps its greatest predictive We welcome your feedback, comments and questions about this site or page. Example: Bouncing ball application of a geometric sequence When a ball is dropped onto a flat floor, it bounces to 65% of the height from which it was dropped. reveal and explain specific information about its approximate Find a rule for this arithmetic \u2026 Learn more about the formula of nth term, sum of GP with examples at BYJU\u2019S. Geometric Series A pure geometric series or geometric progression is one where the ratio, r, between successive terms is a constant. Required fields are marked *. This is another example of a geometric sequence. The value of automobile at the end of the first year When a ball is dropped onto a flat floor, it bounces to 65% of the and produce an equivalent form of an expression to reveal and B. How much will your salary be at the start of year six? Find the G. P. A. Embedded content, if any, are copyrights of their respective owners. 2,3, 4,5. You will receive In order to work with these application problems you need to make sure you have a basic understanding of arithmetic sequences, arithmetic series, geometric sequences, and geometric series. $$= 4500{\\left( {0.85} \\right)^3} = 4500\\left( {0.6141} \\right) = 2763.45$$, Your email address will not be published. b. In Generalwe write a Geometric Sequence like this: {a, ar, ar2, ar3, ... } where: 1. ais the first term, and 2. r is the factor between the terms (called the \"common ratio\") But be careful, rshould not be 0: 1. This video gives examples of population growth and compound interest. increase or decrease in the costs of goods. Suppose you invest$1,000 in the bank. $$= {a_1} \u2013 {a_1}\\left( {\\frac{{15}}{{100}}} \\right) = {a_1}\\left( {1 \u2013 0.05} \\right) = {a_1}\\left( {0.85} \\right)$$, The value of the automobile at the end of the second year B. are variations on geometric sequence. Show Video Lesson In a Geometric Sequence each term is found by multiplying the previous term by a constant. sales and production are some more uses of Arithmetic Sequence. Line is divided into six parts forming a geometric series is a sequence whereas. Before going to learn how to find the height of the world are ruled by.! 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About this site or page be at the rate of growth or decay number Linear... Is$ 43,125, are copyrights of their respective owners number each month, what month will he 6,144. Boxes of Girl Scout cookies online the common ratio of a GP by finding the of. Express and interpret geometric sequence sequence each term of a geometric sequence & series, therefore involves! Salary be at the start of year six in math and science and are one the. The distinction between a progression and a ( 41 ) = 25 adjacent terms such as population growth and interest. $170,000 the free Mathway calculator and problem solver below to practice various math.! Much will your salary for the first year is$ 43,125 math and science and are one the... To reveal the maximum or minimum value of the fifth bounce ) ( 3 ) organization! Progression is a sequence, whereas a series is a sequence, whereas constant! 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An equivalent form of an investment 20 years at 2 % p.a Introduction: 4 % each year getting %. { a_1 } = 4500  per year and geometric series have applications in math and and... What month will he launch 6,144 stores lives are ruled by money his nationwide... About its approximate rate of growth or decay theory, and even for health, but one the... $per year line is application of geometric progression into six parts forming a geometric sequence: Solving application with. 1, Abby \u2019 s value over time is that a progression is a sequence, whereas a series the. Word problems: hike our mission is to sell double the number of students in a geometric series problems... By our in-house editorial team, it increases 2 % p.a years if it is for. Cover the study of applications of arithmetic and geometric series in financial mathematics equivalent form of automobile. With the step-by-step explanations, first know what a GP by finding the ratio between any two adjacent terms to. 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The shortest leng application of geometric Sequences site or page is a of. Of their respective owners will increase by 4 % each year, it increases 2 % its... Any two adjacent terms distinction between a progression and a ( 41 ) = 25 the value an... A range, are copyrights of their respective owners { a_1 } = 4500 $15\\! Daily goal is to sell double the number of students in a small school is 284 an.! The result of multiplying a constant ( 3 ) nonprofit organization whole length of the numbers a! 4500$ $dollars I can invest at 5 % and I want$ 50,000 in years! Geometric growth occurs when the common ratio provide a free, world-class education anyone... Speed of an automobile depreciates at the rate of  15\\ %  15\\ %  %! Ratio is greater than 1, that the student population will increase by 4 % each.. $5000 for 20 years at 2 % p.a want$ 50,000 in years! 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{"url": "https://math.stackexchange.com/questions/88565/true-or-false-x2-ne-x-implies-x-ne-1/88580", "text": "# True or false? $x^2\\ne x\\implies x\\ne 1$\n\nToday I had an argument with my math teacher at school. We were answering some simple True/False questions and one of the questions was the following:\n\n$$x^2\\ne x\\implies x\\ne 1$$\n\nI immediately answered true, but for some reason, everyone (including my classmates and math teacher) is disagreeing with me. According to them, when $$x^2$$ is not equal to $$x$$, $$x$$ also can't be $$0$$ and because $$0$$ isn't excluded as a possible value of $$x$$, the sentence is false. After hours, I am still unable to understand this ridiculously simple implication. I can't believe I'm stuck with something so simple.\n\nWhy I think the logical sentence above is true:\nMy understanding of the implication symbol $$\\implies$$ is the following: If the left part is true, then the right part must be also true. If the left part is false, then nothing is said about the right part. In the right part of this specific implication nothing is said about whether $$x$$ can be $$0$$. Maybe $$x$$ can't be $$-\\pi i$$ too, but as I see it, it doesn't really matter, as long as $$x \\ne 1$$ holds. And it always holds when $$x^2 \\ne x$$, therefore the sentence is true.\n\n### TL;DR:\n\n$$x^2 \\ne x \\implies x \\ne 1$$: Is this sentence true or false, and why?\n\nSorry for bothering such an amazing community with such a simple question, but I had to ask someone.\n\n\u2022 This is true, as the contrapositive ($x = 1$ -> $x^2=x$) is obviously true. Dec 5 '11 at 14:20\n\u2022 They are wrong-the fact that $x\\neq 0$ is also an implication doesn't mean anything. The statement: $x^2=x\\implies x=1$ is false. Dec 5 '11 at 14:21\n\u2022 Also, their reasoning about \"not excluding other values\" is wrong. Today is Monday, which implies SO many things (!), but it is not false to just list one implication. Eg: If today is Monday, then tomorrow is Tuesday. Or, if today is Monday, then I have an appointment with the dentist. Dec 5 '11 at 14:23\n\u2022 @Chris: You could write your reasoning as answer. That way the site keeps working better (and you will get some upvotes). You could also refer your teacher to this site :-) Dec 5 '11 at 15:06\n\u2022 Your understanding of material implication ($\\implies$) is exactly right, and as Jyrki said, your reasoning could stand as a perfectly good answer to the question. You could also point out that the teacher and class seem to be confusing $\\implies$ and $\\iff$: $x^2\\ne x\\iff x\\ne 1$ is of course false precisely because $1$ isn't the only number that is its own square. Dec 5 '11 at 15:34\n\nThe short answer is: Yes, it is true, because the contrapositive just expresses the fact that $1^2=1$.\n\nBut in controversial discussions of these issues, it is often (but not always) a good idea to try out non-mathematical examples:\n\n\"If a nuclear bomb drops on the school building, you die.\"\n\n\"Hey, but you die, too.\"\n\n\"That doesn't help you much, though, so it is still true that you die.\"\n\n\"Oh no, if the supermarket is not open, I cannot buy chocolate chips cookies.\"\n\n\"You cannot buy milk and bread, either!\"\n\n\"Yes, but I prefer to concentrate on the major consequences.\"\n\n\"If you sign this contract, you get a free pen.\"\n\n\"Hey, you didn't tell me that you get all my money.\"\n\nNon-mathematical examples also explain the psychology behind your teacher's and classmates' thinking. In real-life, the choice of consequences is usually a loaded message and can amount to a lie by omission. So, there is this lingering suspicion that the original statement suppresses information on 0 on purpose.\n\nI suggest that you learn about some nonintuitive probability results and make bets with your teacher.\n\n\u2022 I like how the importance of cookies appears in surprising math/logic results! (+1). It might be time to show the teacher this question. Dec 5 '11 at 16:10\n\u2022 Excellent examples, thank you! And of course I already made bets :). Dec 5 '11 at 16:29\n\u2022 Superb answer, which implies that I have up-voted it. Jan 10 '12 at 0:37\n\nFirst, some general remarks about logical implications/conditional statements.\n\n1. As you know, $P \\rightarrow Q$ is true when $P$ is false, or when $Q$ is true.\n\n2. As mentioned in the comments, the contrapositive of the implication $P \\rightarrow Q$, written $\\lnot Q \\rightarrow \\lnot P$, is logically equivalent to the implication.\n\n3. It is possible to write implications with merely the \"or\" operator. Namely, $P \\rightarrow Q$ is equivalent to $\\lnot P\\text{ or }Q$, or in symbols, $\\lnot P\\lor Q$.\n\nNow we can look at your specific case, using the above approaches.\n\n1. If $P$ is false, ie if $x^2 \\neq x$ is false (so $x^2 = x$ ), then the statement is true, so we assume that $P$ is true. So, as a statement, $x^2 = x$ is false. Your teacher and classmates are rightly convinced that $x^2 = x$ is equivalent to ($x = 1$ or $x =0\\;$), and we will use this here. If $P$ is true, then ($x=1\\text{ or }x =0\\;$) is false. In other words, ($x=1$) AND ($x=0\\;$) are both false. I.e., ($x \\neq 1$) and ($x \\neq 0\\;$) are true. I.e., if $P$, then $Q$.\n2. The contrapositive is $x = 1 \\rightarrow x^2 = x$. True.\n3. We use the \"sufficiency of or\" to write our conditional as: $$\\lnot(x^2 \\neq x)\\lor x \\neq 1\\;.$$ That is, $x^2 = x$ or $x \\neq 1$, which is $$(x = 1\\text{ or }x =0)\\text{ or }x \\neq 1,$$ which is $$(x = 1\\text{ or }x \\neq 1)\\text{ or }x = 0\\;,$$ which is $$(\\text{TRUE})\\text{ or }x = 0\\;,$$ which is true.\n\u2022 Please pardon the (lack of) formatting- I typed this up on my phone! Dec 5 '11 at 15:36\n\u2022 Thanks, Brian! That looks great. Dec 5 '11 at 15:55\n\u2022 That's the proof I was looking for, thank you :) Dec 5 '11 at 16:41\n\nThing to note. This is called logical implication.\n\n$x^2\u2260x\u27f9x\u22601$: Is this sentence true or false, and why?\n\nWe can always check that using an example. Let us look at this implication as $\\rm P\\implies Q$. Now we shall consider cases:\n\n\u2022 Case 1: If we consider $x = 0$, then $\\rm P$ is false, and $\\rm Q$ is true.\n\u2022 Case 2: If we consider $x = 1$, then $\\rm P$ is false, and $\\rm Q$ is false as well.\n\u2022 Case 3: If we consider each value except $x=0$ and $x = 1$, then both $\\rm P$ and $\\rm Q$ will be true since $x^2 = x \\iff x^2 - x = 0 \\iff x(x - 1) = 0$ which means that $x=0$ and $x = 1$ are the only possibilities.\n\nFortunately, our truth tables tell us that logical implication will hold true as far as we are not having $\\rm P$ true and $\\rm Q$ false. Look at the cases above; none of them has $\\rm P$ true and $Q$ false. Thus Case 1, Case 2 and Case 3 are all true according to mathematical logic, so $\\rm P\\implies Q$ is true, or in other words: $x^2 \\ne x \\implies x \\ne 1$ is true.\n\nI apologize for being late, but I always have my two cents to offer... thank you.\n\n## Introduction\n\nYou asked whether the following statement is true or not:\n\n$$x^2 \\ne x \\implies x \\ne 1$$\n\nAnother way to write the above is as:\n\nif $$(x^2 \\ne x)$$ then $$(x \\ne 1)$$\n\nThe arrow symbol ($$\\implies$$) is similar, but not quite the same, as the English phrase if....then....\n\n## The contra-positive of $$(P \\implies Q)$$\n\nOkay now: suppose that you have the statement of the form \"If $$P$$ then $$Q$$\"\n$$P$$ and $$Q$$ can be any true or false statements of your choice.\nFor example, $$P$$ could be the statement \"pillows are soft\"\n\"If $$P$$ then $$Q$$\" is also sometimes written as $$(P \\implies Q)$$\n\nThere is a well known theorem in logic which states the following:\n\nFor any $$P$$ and $$Q$$ taken from the set $$\\{$$ true, false $$\\}$$, the following is true:\n$$(\\text{not } P \\implies \\text{not } Q)$$ if and only if $$(Q \\implies P)$$\n\nThat means that the following two statements are equivalent:\n\nif $$(x^2 \\ne x)$$ then $$(x \\ne 1)$$\nif $$(x = 1)$$ then $$(x^2 = x)$$\n\nIn some sense, the statement says that $$1^2 = 1$$, which is true enough.\n\n## Sets\n\nIt considered to be very bad math to write things like the following:\n\nIs it true or false that $$(x^2 \\ne x \\implies x \\ne 1)$$?\n\nFind $$x$$ such that $$3 + (x-5)^{2} = 0$$\n\nThis is because the above examples fail to explain what $$x$$ is.\n\n\u2022 Is $$x$$ a whole number, such as $$1, 2, 3, \\dots, 98, 99, 100$$?\n\u2022 Is $$x$$ a decimal number, such as $$\\sqrt{2}$$\n\u2022 Is $$x$$ a non-real complex number, such as $$20 + 5*i$$?\n\nYou are supposed to write things like this instead:\n\nIs the following true or false?\n\n\u2022 For every real number $$x, (x^2 \\ne x \\implies x \\ne 1)$$\n\nFind every complex non-real number $$x$$ such that:\n\n\u2022 $$3 + (x-5)^{2} = 0$$\n\nMy last example uses the non-real number $$i$$\n$$i*i = -1$$\n\nI think if you think it will help if you start using \"sets.\"\n\nThe following is an example of a set:\n\nmy_set $$= \\{1, 3, 6, 7, 22\\}$$\n\nA set is like a suitcase full of clothing A set is also like a cookie jar, or cardboard box.\nA set is a container.\nA suitcase might contain a t-shirt. Well, my_set contains the numbers $$1, 3, 6, 7,$$ and $$22$$.\n\nThe number $$3$$ is like a t-shirt in the sense that the number $$3$$ is inside the suitcase.\n\nI went to public school in the United States.\nI did not see sets until I was in college.\n\nSets are basic, basic math. sets are more basic than knowing how to compute $$4.5/0.3$$\n\nYou are not allowed to write math like $$x^2 \\ne x \\implies x \\ne 1$$ unless you first tell the reader what set $$x$$ comes from.\n\nThe following is a very ugly formula for a function named $$WEIRD$$:\n\n$$\\text{WEIRD_FUNC}(X) = [1-w(x)]*\\text{LEFT_PIECE}(X) + [w(x)]*\\text{RIGHT_PIECE}(X)$$\n\n$$\\text{LEFT_PIECE}(X) = (x+10)*(x+5)$$\n\n$$\\text{RIGHT_PIECE}(X) = 3 + (x-7)^{2}$$\n\n$$W(x) = \\frac{tanh(10)}{2} + \\frac{tanh(x)}{2}$$\n\nA plot of the weird function is shown below:\n\nSuppose I asked you,\n\n\"Find all $$x$$ such that $$WEIRD(x) = 0$$\"\n\nThe answers vary depending on which set $$x$$ is taken from.\n\nThe set of all integers $$x$$ such that $$WEIRD(X) = 0$$ is $$\\{-10\\}$$\nThe set of all real numbers $$x$$ such that $$WEIRD(X) = 0$$ is approximately $$\\{-10, -5.01\\}$$\nThe set of all complex numbers $$x$$ such that $$WEIRD(X) = 0$$ is approximately $$\\{-10, -5.01, 7+i*\\sqrt{3}, 7-i*\\sqrt{3}\\}$$\n\n### Analyzing your classmates argument regarding the number zero\n\nIs the following statement true or not?\n\n$$x^2 \\ne x \\implies x \\ne 1$$\n\nThe following is your description of your teacher, and/or classmates, reasoning:\n\n1. When $$x^2$$ is not equal to $$x$$, $$x$$ also can't be $$0$$.\n2. $$0$$ is not excluded as a possible value of $$x$$\n3. some sentence, or another, is false.\n\nThat description is very difficult to understand.\nI will say that the following three statements are equivalent to each-other.\nAlso, all statements are false:\n\n$$\\forall x \\in \\mathbb{R}, x^2 \\ne x$$ for any decimal number $$x$$, $$x^2 \\ne x$$ if $$x$$ is a real number, then $$x^2 \\ne x$$\n\nHere is a proof:\n\nLine No. statement justification\n0 $$0$$ exists axiom\n1 $$0$$ is a decimal number. axiom\n2 $$0^{2} = 0$$ axiom\n3 not $$(0^2 \\neq 0)$$ from line 2\n4 there exists a decimal number $$x$$ such that not $$(x^2 \\neq x)$$ from lines 0,1,3\n5 not for every decimal number $$x$$, $$(x^2 \\neq x)$$ from line 4\n\nNotice that your teacher said, \"possible value of $$x$$\"\n\nIn mathematics, the symbol is sometimes used as short-hand notation for the word \"possible\"\n\nAll of the following are logically equivalent:\n\n\u2022 it is not possible for me to see a movie this weekend\n\u2022 $$NOT$$ for me to see a movie this weekend\n\u2022 It is necessary for me to NOT see a movie this weekend\n\u2022 for me to NOT see a movie this weekend\n\nThe statement \"$$0$$ is not excluded as a possible value of $$x$$\" can be written as:\n\n\u2022 not(not $$x = 0$$)\n\u2022 It is not the case that it is not possible that $$x$$ is zero.\n\nThe above is equivalent to \"it is possible that $$x = 0$$\"\n\nI think that I can write a proof outline of what your professor and/or classmates were trying to say\n\nLine No. statement justification\n$$0$$ $$x = 0$$ axiom\n$$1$$ $$(x^2 \\ne x) \\implies$$ not $$x = 0$$ axiom\n$$2$$ $$x^2 \\ne x$$ axiom\n$$3$$ not $$x = 0$$ 1,2 modus ponens\n4 $$x = 0$$ and not $$x = 0$$ lines 0, 3 conjunction\n5 $$NOT$$ $$(x^2 \\ne x)$$ from line 4, reject 2\n\nI think that your teacher was saying that if $$x = 0$$ is possible, then it is not necessary that $$x^2 \\ne x$$\n\n$$0 \\in S \\implies NOT(\\forall x \\in S, x^2 \\ne x)$$\n\n### One last Note\n\nAll of the following are equivalent:\n\n\u2022 (it is necessary that $$x^2 \\neq x$$) implies that (it is necessary that $$x \\neq 1$$)\n\u2022 (it is NOT necessary that $$x \\neq 1$$) implies that (it is NOT necessary that $$x^2 \\neq x$$)\n\u2022 (it is possible that $$x = 1$$) implies that (it is possible that $$x^2 = x$$)\n\u2022 For any set $$S$$, [(there exists $$x$$ in $$S$$ such that $$x = 1$$) implies that (there exists $$x$$ in $$S$$ such that $$x^2 = x$$)]", "date": "2021-10-25 16:44:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/88565/true-or-false-x2-ne-x-implies-x-ne-1/88580", "openwebmath_score": 0.7766252756118774, "openwebmath_perplexity": 353.23052291517223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.953275045356249, "lm_q2_score": 0.8947894555814343, "lm_q1q2_score": 0.8529804588536851}}
{"url": "https://stats.stackexchange.com/questions/463189/what-is-the-distribution-of-a-mixture-of-exponential-distributions-whose-rate-pa", "text": "# What is the distribution of a mixture of exponential distributions whose rate parameters follow a gamma distribution?\n\nI want to know the theoretical distribution of a mixture of exponential distributions whose rate parameters are distributed according to a gamma distribution:\n\n$$y\\sim\\text{Exp}(\\theta), \\quad\\text{where}\\quad \\theta\\sim\\Gamma(r, \\beta).$$\n\nSpecifically, I am looking at $$r=4$$ and $$\\beta=2$$.\n\n\u2022 The question appears clear enough to me: given $r$ and $\\beta$, let $\\theta$ follow a gamma distribution with parameters $r$ and $\\beta$ (whether these are scale or rate params doesn't matter for now), denoted $\\theta\\sim\\Gamma(r, \\beta)$. Next, $y$ is exponentially distributed with parameter $\\theta$, denoted $y\\sim\\text{Exp}(\\theta)$. That is, we have a compound distribution, AKA a mixture. Q: What is the unconditional distribution of $y$ with parameters $r$ and $\\beta$? \u2013\u00a0Stephan Kolassa May 1 at 5:29\n\u2022 At least that is what the original question seems to have asked, and I have tried to clarify this specific question with my edit and answer it. I will vote to reopen. Can someone clarify to me where this question is unclear? \u2013\u00a0Stephan Kolassa May 1 at 5:30\n\u2022 @StephanKolassa: Well, $\\theta$ wasn't defined; it could be the mean rather than the rate. The variable $n$ & its role were also unexplained. \u2013\u00a0Scortchi - Reinstate Monica May 2 at 8:57\n\u2022 @Scortchi-ReinstateMonica: true. But converting between a mean and a rate formulation for the exponential is not overly hard. I assumed $n$ referred to a sample size and removed it in my edit. \u2013\u00a0Stephan Kolassa May 2 at 9:11\n\u2022 @StephanKolassa: (1) The compound distribution would be something else if it were the means distributed according to a gamma distribution. (2) Quite possibly, but why mention it then? I think your edits do a fine job of clarification, but I can see why the q. was closed in its original version. \u2013\u00a0Scortchi - Reinstate Monica May 2 at 9:19\n\nFor all $$y \\ge 0$$ the value of the survival function of $$Y$$ is\n\n$$S_Y(y\\mid\\theta) = \\Pr(Y \\gt y\\mid \\theta) = \\exp(-y\\theta)$$\n\nand, taking $$\\beta$$ to be a rate parameter for the Gamma distribution, the probability density function of $$\\theta$$ is proportional to\n\n$$f_\\theta(t) \\propto t^{r-1}\\exp(-\\beta t).$$\n\nConsequently the survival function of the mixture distribution is\n\n$$S(y) \\propto \\int_0^\\infty S_Y(y\\mid t) f_\\theta(t)\\,\\mathrm{d}t = \\int_0^\\infty t^{r-1}\\exp(-(\\beta+y)t)\\,\\mathrm{d}t.$$\n\nSubstituting $$u = (\\beta+y) t$$ gives, with no calculation,\n\n$$S(y) \\propto \\int_0^\\infty \\left(\\frac{u}{\\beta+y}\\right)^{r-1}\\exp(-u)\\,\\mathrm{d}\\left(\\frac{u}{\\beta+y}\\right) = (\\beta+y)^{-r}\\int_0^\\infty u^{r-1}\\exp(-u)\\,\\mathrm{d}u$$\n\nwhich is proportional to $$(\\beta+y)^{-r}.$$\n\nThe axiom of total probability asserts $$S(0)=1$$ from which we obtain the implicit constant, giving\n\n$$S(y) = \\beta^r\\,(\\beta+y)^{-r} = \\left(1 + \\frac{y}{\\beta}\\right)^{-r},$$\n\nthereby exhibiting $$\\beta$$ as a scale parameter for this mixture variable.\n\nThis is a particular kind of Beta prime distribution, sometimes termed a Lomax distribution. (up to scale).\n\nIf $$\\beta$$ is intended to be a scale parameter for the Gamma distribution rather than a rate parameter, replace $$\\beta$$ everywhere by $$1/\\beta$$ and, at the end, interpret it as a rate parameter for the mixture variable.\n\nThis distribution tends to be positively skewed, so it's better to view the distribution of $$\\log Y.$$ Here are the empirical (black) and theoretical (red) distributions for a simulation of $$10^4$$ independent realizations of $$Y$$ (where $$r=4$$ and $$\\beta=2$$ as in the question):\n\nThe agreement is perfect.\n\nThe R code that generated this figure illustrates how to code the survival function (S), the density function (its derivative f), and how to simulate from this compound distribution by first producing realizations of $$\\theta$$ and then, for each of them, generating a realization of $$Y$$ conditional on $$\\theta.$$\n\nS <- function(y, r, beta) 1 / (1 + y/beta)^r            # Survival\nf <- function(y, r, beta) r * beta^r / (beta + y)^(r+1) # Density\n#\n# Specify the parameters and simulation size.\n#\nbeta <- 2\nr <- 4\nn.sim <- 1e4\n#\n# The simulation.\n#\ntheta <- rgamma(n.sim, r, rate=beta)\ny <- rgamma(n.sim, 1, theta)\n#\n# The plots.\n#\npar(mfrow=c(1,2))\nplot(ecdf(log(y)), xlab=expression(log(y)), ylab=\"Probability\",\nmain=expression(1-S[Y](y)))\ncurve(1 - S(exp(y), r, beta), xname=\"y\", add=TRUE, col=\"Red\", lwd=2)\n\nhist(log(y), freq=FALSE, breaks=30, col=\"#f0f0f0\", xlab=expression(log(y)))\ncurve(f(exp(y), r, beta) * exp(y), xname=\"y\", add=TRUE, col=\"Red\", lwd=2)\npar(mfrow=c(1,1))\n\n\nWikipedia, under the \"Examples\" section, informs us that:\n\nCompounding an exponential distribution with its rate parameter distributed according to a gamma distribution yields a Lomax distribution [9].\n\nThe reference [9] is to Johnson, N. L.; Kotz, S.; Balakrishnan, N. (1994). \"20 Pareto distributions\". Continuous univariate distributions. 1 (2nd ed.). New York: Wiley. p. 573\n\n\u2022 Yuh, but is that really the question? \u2013\u00a0Carl May 1 at 4:10", "date": "2020-08-07 09:40:42", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/463189/what-is-the-distribution-of-a-mixture-of-exponential-distributions-whose-rate-pa", "openwebmath_score": 0.8443940281867981, "openwebmath_perplexity": 1094.5284893571654, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357179075081, "lm_q2_score": 0.8688267881258485, "lm_q1q2_score": 0.8529582905780043}}
{"url": "https://math.stackexchange.com/questions/1211713/prove-that-there-is-a-multiple-of-2009-that-ends-with-the-digits-000001", "text": "# Prove that there is a multiple of 2009 that ends with the digits 000001\n\nProve that there is a multiple of 2009 that ends with the digits 000001.\n\nMay one generalise this to: There exists a multiple of $x$ that ends with the digits $y$ (where $y$ consists of $n$ digits) if $x$ and $10^n$ are relatively prime?\n\nThe proof may be constructive, or non constructive.\n\nAny help would be greatly appreciated.\n\nI figured the multiple has to end in a $9$: $2009 \\cdot 889$ gives $001$ as the three ending digits. I then tried to see if I could get the $4$th last digit to be a zero, but I must admit I am clearly missing something at that point.\n\n\u2022 Write down a few multiples of $2009$. Maybe try the first $246889$ :P Mar 29 '15 at 17:22\n\u2022 Well it needs to end in a one, so I figured the multiple has to end in a 9. 2009 * 889 gives 001 as the three ending digits. I then tried to see if I could get the 4th last digit to be a zero, but I must admit I am clearly missing something at that point. Mar 29 '15 at 17:24\n\u2022 @flabby99 Well, both hints in the answer work and boil down to the same. Another answer is hidden somewhere here ;) Mar 29 '15 at 17:26\n\u2022 @AlexR Yes I see that there is some magical number within these comments. However, I would like to understand better how to obtain such a magic number :P Mar 29 '15 at 17:27\n\u2022 @flabby99 It's from the extended euclidean algorithm to $\\gcd(1000000,2009) = 1$. Mar 29 '15 at 17:28\n\n## 3 Answers\n\nHere is a short proof using modular arithmetic.\n\nConsider $m=1000000$. The numbers $2009$ and $m$ are relatively prime to each other, so\n\n$$2009^{\\phi(m)}\\equiv 1\\pmod m$$\n\nwhere $\\phi(m)$ is the count of the numbers between $1$ and $m$ that are relatively prime to $m$. $2009$ divides $2009^{\\phi(m)}$ so that proves your statement.\n\nOf course, this is not the smallest such multiple of $2009$. To find that, use the extended Euclidean algorithm on $2009$ and $1000000$.\n\nAs I wrote, the key is that $2009$ and $1000000$ are relatively prime.\n\n\u2022 How to nuke-a-fly: Apply totient function where an efficient algorithm does your work ^^ Mar 29 '15 at 17:29\n\u2022 @AlexR: The OP just asked for a proof, not an efficient algorithm. Anyway, I modified my post (before I saw your comment) to also refer to that algorithm. The OP can decide which is better suited to his purpose. Mar 29 '15 at 17:31\n\nHint\nYou must show that there exists $k$ such that $$2009 k \\equiv 1 \\pmod{1000000}$$ i.e. that $2009$ has a multiplicative inverse in $\\mathbb Z_{1000000}^\\times$. Do you know of some criteria?\n\nIf you want to obtain $k$, the easiest way is the extended euclidean algorithm on $1000000$ and $2009$. It will give you numbers $x,y$ such that $$2009 x + 1000000 y = \\gcd(1000000,2009)$$\n\n\u2022 2009 must be relatively prime to 1000000 ( to satisfy the mentioned criteria). The Euclidean algorithm should handle that without too much difficulty. Is my thought process correct? Mar 29 '15 at 17:30\n\u2022 @flabby99 Yes, perfect! The extended one will even provide you with $k$ for free. Mar 29 '15 at 17:30\n\u2022 @flabby99 if you are just looking for a non constructive proof, even the Euclidean algorithm is overkill - just observe that $2009$ is not divisible by $2$ or $5$ Mar 29 '15 at 17:39\n\u2022 @Mathmo123 I must admit, I am missing how the proof follows just from observing that 2009 is not divisible by 2 or 5. Mar 29 '15 at 17:48\n\u2022 @flabby99 this is just a faster way of showing the numbers are relatively prime, since $1000000$ is divisible by only the primes $2$ and $5$ Mar 29 '15 at 17:49\n\nHint: Show that you can apply the Chinese Remainder theorem to $$x \\equiv 0 \\pmod {2009}\\\\x \\equiv 1\\pmod {1000000}$$\n\n\u2022 +1. Only flabby99 can tell us if this problem is in his textbook after a discussion of the Chinese Remainder Theorem. Mar 29 '15 at 17:43\n\u2022 It is a general problem from elementary number theory from an optional set of problems by one of my lecturers, and may be solved using whatever is within our grasp. I am however, aware of the Chinese Remainder Theorem. I just failed to see it's application in this situation. Mar 29 '15 at 17:50", "date": "2022-01-26 21:24:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1211713/prove-that-there-is-a-multiple-of-2009-that-ends-with-the-digits-000001", "openwebmath_score": 0.6910312175750732, "openwebmath_perplexity": 240.18195927182714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357221825193, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8529582859561454}}
{"url": "http://math.stackexchange.com/questions/92401/unconventional-way-to-solve-trig-question", "text": "# Unconventional way to solve trig question\n\nI was working on a trig question and got stuck, but then I noticed a possible way to solve the problem. However, this way seemed to be slightly unconventional and possibly not what the book was looking for.\n\nThe question was: \"Find $k$ and $b$ when $\\sqrt3 \\sin x + \\cos x=k\\sin(x+b)$\" The first thing I did was to expand $k\\sin(x+b)$ to $k\\sin x\\cos b+k\\cos x\\sin b$. Here is where I got stuck. I tried several different things, but I hit dead ends. Then I tried to equate coefficients and got $k\\cos b=\\sqrt3$ and $k\\sin b=1$ which simplified to: $\\cos b=\\displaystyle{\\sqrt3 \\over k}$ and $\\sin b=\\displaystyle{1\\over k}$. The answer from there is fairly simple to get which is $k=2$ and $b=\\displaystyle{\\pi\\over6}$.\n\nHowever this method seems rather dubious and so I was wondering if someone new a better way of solving the problem using more rigorous mathematical methods.\n\n-\nWhat makes it seem like a guessing game? Also, I believe you missed a solution. \u2013\u00a0 Mike Dec 18 '11 at 1:54\n$k^2=4$ has two solutions. \u2013\u00a0 Henry Dec 18 '11 at 1:54\nGood point. I just thought that using that method might not work all the, but I guess it might. \u2013\u00a0 E.O. Dec 18 '11 at 1:56\nEquating coefficients works here, but can be avoided by looking at the cases $x=0$ and $x=\\pi/2$ and checking the final result. \u2013\u00a0 Henry Dec 18 '11 at 1:56\n\nIt's not at all dubious. The functions sin and cos are linearly independent and so $a_1 \\sin x + b_1 \\cos x = a_2 \\sin x + b_2 \\cos x$ for all $x$ iff $a_1=a_2$ and $b_1=b_2$.\n\nTo see that sin and cos are linearly independent, take $a \\sin x + b \\cos x = 0$ and set $x=0$ to get $b=0$ and $x=\\pi/2$ to get $a=0$.\n\n-\nThis appears to be the only answer that explains why the OP's method is valid. +1 \u2013\u00a0 Graphth Dec 18 '11 at 2:46\nwhat about the case where $x=\\pi /3$ \u2013\u00a0 Ramana Venkata Dec 18 '11 at 5:17\n\nI'm not sure how it's perceived as dubious. If you have $k \\cos b = \\sqrt{3}$ and $k \\sin b = 1$, then you must $\\frac{k \\sin b}{k \\cos b} = \\tan b = \\frac{1}{\\sqrt{3}}$, which gets you $b = \\frac{\\pi}{6}$ or $\\frac{7\\pi}{6}$. From there it immediately follows that $k = 2$ or $-2$, respectively.\n\n-\n\nIf you have $k\\cos b=\\sqrt{3}$ and $k\\sin b=1$, you should also remember that $k^2\\cos^2 b+ k^2\\sin^2 b= k^2$. That means $\\sqrt{3}^2 + 1^2 = k^2$, so $k=\\pm 2$. If you take $k=2$, you can find a suitable value of $b$, and with $k=-2$, you can find another.\n\n-\n\nSuppose that $a\\sin(x)+b\\cos(x)=c\\sin(x)+d\\cos(x)$ holds for all $x$ in some neighborhood of $x_0$. Then we have the following equation and its derivative: \\begin{align} 0&=(a-c)\\sin(x_0)+(b-d)\\cos(x_0)\\tag{1}\\\\ 0&=(a-c)\\cos(x_0)-(b-d)\\sin(x_0)\\tag{2} \\end{align} Adding the squares of $(1)$ and $(2)$ gives $$0=(a-c)^2+(b-d)^2\\tag{3}$$ Thus, $a=c$ and $b=d$.\n\nSo if we have \\begin{align} \\sqrt{3}\\sin(x)+\\cos(x) &=k\\sin(x+b)\\\\ &=k\\cos(b)\\sin(x)+k\\sin(b)\\cos(x)\\tag{4} \\end{align} for all $x$ in some neighborhood of $x_0$, then we must have $$k\\cos(b)=\\sqrt{3}\\quad\\text{ and }\\quad k\\sin(b)=1\\tag{5}$$ and $(5)$ was solved above.\n\n-\n\nHere is what I come up in my mind, maybe this is an unconventional way too: Divide both sides by $2$, we get $$\\frac{\\sqrt3}{2} \\sin x +\\frac{1}{2}\\cos x=\\frac{k}{2}\\sin(x+b).$$ Note that $\\displaystyle\\sin(\\frac{\\pi}{6})=\\frac{1}{2}$ and $\\displaystyle\\cos(\\frac{\\pi}{6})=\\frac{\\sqrt{3}}{2}$, the left hand side is given by $$\\cos(\\frac{\\pi}{6})\\sin x +\\sin(\\frac{\\pi}{6})\\cos x=\\sin(x+\\frac{\\pi}{6}),$$ where the last equality follows from compound angle forumula. Combining these, we have $$\\sin(x+\\frac{\\pi}{6})=\\frac{k}{2}\\sin(x+b).$$\n\n-\n\nHere is a standard trick to deal with expressions of the type $A \\sin(x) + \\cos(x)$:\n\nLet $u$ be so that $\\cot(u) =A$.\n\nThen, after bringing everything to the same denominator, your expression becomes\n\n$$A \\sin(x) + \\cos(x)= A\\frac{\\cos(u) \\sin(x)+\\sin(u)\\cos(x)}{ \\sin(u)} =\\frac{ \\sin(x+u)}{\\sin(u) }$$\n\nIf you look over the other answers, this is exactly what they did, in a non-explicit way.\n\nRemark If instead you use $\\tan(v)=A$, you end up with\n\n$$A \\sin(x) + \\cos(x) = \\frac{\\cos(x-v)}{\\cos(v)} \\,.$$\n\nSame substitutions also solve the expressions of the type $\\sin(x) + B\\cos(x)$.\n\n-\n\nIn general, \\begin{align*} \\alpha \\sin(x) + \\beta \\cos(x) &= \\sqrt{\\alpha^2+\\beta^2}\\left(\\frac{\\alpha}{\\sqrt{\\alpha^2+\\beta^2}}\\sin(x) + \\frac{\\beta}{\\sqrt{\\alpha^2+\\beta^2}}\\cos(x)\\right)\\\\ &= \\sqrt{\\alpha^2+\\beta^2}\\left(\\cos(y)\\sin(x) + \\sin(y)\\cos(x)\\right)\\\\ &= \\sqrt{\\alpha^2+\\beta^2}\\sin(x+y) \\end{align*} where $y$ is the angle whose cosine is $\\dfrac{\\alpha}{\\sqrt{\\alpha^2+\\beta^2}}$ and whose sine is $\\dfrac{\\beta}{\\sqrt{\\alpha^2+\\beta^2}}$. In your problem, $\\alpha = \\sqrt{3}, \\beta = 1$, which readily gives $k = 2$, $b = \\frac{\\pi}{6}$. As others have pointed out, this misses one of the two solutions.\nIf $\\sqrt{\\alpha^2+\\beta^2}$ (which conventionally means the positive square root) is replaced throughout by $({\\alpha^2+\\beta^2})^{1/2}$, we get the other solution $k = -2, b = \\frac{7\\pi}{6}$ upon choosing $({\\alpha^2+\\beta^2})^{1/2} = -2$.\n\n-", "date": "2014-09-01 18:56:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/92401/unconventional-way-to-solve-trig-question", "openwebmath_score": 0.9875971078872681, "openwebmath_perplexity": 192.03683418519762, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357221825193, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8529582842889247}}
{"url": "https://math.stackexchange.com/questions/2911922/a-problem-about-arithmetic-and-geometric-sequences", "text": "# A Problem About Arithmetic And Geometric Sequences\n\nThe problem I'm trying to solve is this:\n\nThe first three terms of a geometric sequence are also the first, eleventh and sixteenth terms of an arithmetic sequence. The terms of the geometric sequence are all different. The sum to infinity of the geometric sequence is 18. Find the common ration of the geometric sequence, and the common difference of the arithmetic sequence.\n\nWhat I've done so far is to write the following equations, where u is the first term, r is the common ratio, and d is the difference:\n\n$ur=u+10d$\n\n$ur^2=u+15d$\n\n$u/(1-r)=18$\n\nBut I don't know what to do from there. Any suggestions?\n\n\u2022 Please type your questions rather than posting an image. Images cannot be searched. Sep 10 '18 at 13:42\n\u2022 @N.F.Taussig Okay, I've done that.\n\u2013\u00a0user590211\nSep 10 '18 at 13:44\n\nYou will get\n\n$$a_2=a_1q,a_3=a_1q^2$$ and $$\\sum_{k=0}^\\infty a_1 q^k=a_1\\frac{1}{1-q}$$ and $$a_1=b_1,a_2=b _1+10d,a_3=b_1+15d$$ Can you proceed? Using your equation you will get $$a_1=b_1$$ $$a_q=b_1+10d$$ $$a_1q^2=b_1+15d$$ Since $$a_1=b_1$$ we obtain $$a_1q=a_1+10d$$ $$a_1q^2=a_1+15d$$\n\neliminating $q$ we get $$a_1\\left(\\frac{a_1+10d}{a_1}\\right)^2=a_1+15d$$\n\nFrom here we get the equation $$5a_1d+100d^2=0$$\n\nso $$a_1=-20d$$ if $$d\\ne 0$$\n\nSolving this we get $$q=\\frac{1}{2}$$\n\n\u2022 Aha, i don't see it! Sep 10 '18 at 14:00\n\nNote that: $$b_1,b_2,b_3 \\iff a_1,a_{11},a_{16} \\Rightarrow \\\\ \\begin{cases}b_2-b_1=10d \\\\ b_3-b_2=5d\\end{cases} \\Rightarrow \\begin{cases}b_1(q-1)=10d \\\\ b_1(q^2-q)=5d\\end{cases} \\Rightarrow q=\\frac12.\\\\ \\frac{b_1}{1-q}=18 \\Rightarrow b_1=9;\\\\ b_1(q-1)=10d \\Rightarrow d=-\\frac9{20}.\\\\$$\n\nFrom the first two equations of$$\\begin{cases}ur=u+10d, \\\\ur^2=u+15d, \\\\\\dfrac u{1-r}=18 \\end{cases}$$\n\nyou draw $$\\frac{u(r^2-1)}{u(r-1)}=r+1=\\frac{15d}{10d}$$ and $r=\\dfrac12$. Using the third, $u=9$, and finally $d=-\\dfrac9{20}$.\n\nI will use $a_1$ for the initial term of both sequences, $r$ for the common ratio of the geometric sequence, and $d$ for the common difference. Since the second term of the geometric sequence is the eleventh term of the arithmetic sequence, $$a_1r = a_1 + 10d$$ Subtracting the first term of the sequence from each side gives \\begin{align*} a_1r - a_1 & = a_1 + 10d - a_1\\\\ a_1(r - 1) & = 10d \\tag{1} \\end{align*} Since the third term of the geometric sequence is equal to the sixteenth term of the geometric sequence, $$a_1r^2 = a_1 + 15d$$ Since the second term of the geometric sequence is equal to the eleventh term of the arithmetic sequence, equality is preserved if we subtract the second term of the geometric sequence from the left-hand side and the eleventh term of the arithmetic sequence from the right-hand side, which yields \\begin{align*} a_1r^2 - a_1r & = a_1 + 15d - (a_1 + 10d)\\\\ a_1(r^2 - r) & = 5d\\\\ a_1r(r - 1) & = 5d \\tag{2} \\end{align*} Since the terms of the geometric sequence are all different, $a_1 \\neq 0$ and $r \\neq 1$. Moreover, the terms of the arithmetic sequence must be distinct, so $d \\neq 0$. Thus, if we divide equation 2 by equation 1, we obtain $$r = \\frac{1}{2}$$ Since the sum of the geometric series is $18$ and $r \\neq 1$, $$a_1 \\frac{1}{1 - r} = 18$$ Solve for $a_1$, then use the equation $a_1(r - 1) = 10d$ to solve for $d$.\n\nsup your logic is correct, the next step to solve the system of equations\n\n$\\begin{cases}ur=u+10d \\space (1)\\tag{1'}\\label{1'} \\\\ ur^2=u+15d \\space(2) \\\\ \\frac{u}{1-r}=18 \\space(3) \\end{cases}$\n\nAnd the system has solutions because it has 3 unknown variables ($u,r,d$) and the number of equations is also 3.\n\nfrom (3) = > $u=18(1-r) (4)$\n\nsubstruct (1) from (2)\n\n$ur^2-ur=5d$ => $d = \\frac{1}{5}ur(r-1) (6)$\n\nput (6) to (1)\n\n$ur=u+10(\\frac{1}{5}ur(r-1))$ => $ur=u+2ur(r-1)$ => $ur=u+2ur^2-2ur (7)$\n\nthe both parts of (7) can be divided by $u$\n\n$r=1+2r^2-2r (7')$ => $2r^2-3r+1=0 (7'')$\n\nsolve quadratic equation (7'') for $r$ and note that $r$ can not be 1.\n\n$r_{1,2} = \\frac{3\\pm\\sqrt{9-2*4*1}}{4} = \\frac{3\\pm1}{4} (8)$\n\n$r = \\frac{1}{2} (9)$\n\nfrom (3) $u=9 (10)$\n\nfrom (6) $d = \\frac{9*0.5*(0.5-1)}{5} = -\\frac{9}{20} (1)$", "date": "2021-12-01 01:07:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2911922/a-problem-about-arithmetic-and-geometric-sequences", "openwebmath_score": 0.9178892970085144, "openwebmath_perplexity": 180.25822047577293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357189762611, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8529582815032417}}
{"url": "https://math.stackexchange.com/questions/2979100/method-for-finding-efficient-algorithms/2979141", "text": "# Method for finding efficient algorithms?\n\nTL;DR\n\nWhat can you recommend to get better at finding efficient solutions to math problems?\n\nBackground\n\nThe first challenge on Project Euler says:\n\nFind the sum of all the multiples of 3 or 5 below 1000.\n\nThe first, and only solution that I could think of was the brute force way:\n\ntarget = 999\nsum = 0\nfor i = 1 to target do\nif (i mod 3 = 0) or (i mod 5 = 0) then sum := sum + i\noutput sum\n\n\nThis does give me the correct result, but it becomes exponentially slower the bigger the target is. So then I saw this solution:\n\ntarget=999\nFunction SumDivisibleBy(n)\np = target div n\nreturn n * (p * (p + 1)) div 2\nEndFunction\nOutput SumDivisibleBy(3) + SumDivisibleBy(5) - SumDivisibleBy(15)\n\n\nI don't have trouble understanding how this math works, and upon seeing it I feel as though I could have realised that myself. The problem is just that I never do. I always end up with some exponential, brute force like solution.\n\nObviously there is a huge difference between understanding a presented solution, and actually realising that solution yourself. And I'm not asking how to be Euler himself.\n\nWhat I do ask tho is, are there methods and or steps, you can apply to solve math problems to find the best (or at least a good) solution?\n\nIf yes, can you guys recommend any books/videos/lectures that teach these methods? And what do you do yourself when attempting to find such solutions?\n\n\u2022 You might benefit from George Polya's classic book, How to Solve It. \u2013\u00a0Barry Cipra Oct 31 '18 at 13:54\n\u2022 Your solution is not exponentially slower as the target gets bigger. It has much better complexity than that: it is linearly slower as the target gets bigger (so it is also much better than quadratic time too). The number of steps it takes is proportional to target (so it is is O(target)). Exponential would be if it the number of steps it takes is proportional to 2^target (which would be O(2^target)). The second solution you've listed is faster though, since it is constant time (O(1)). \u2013\u00a0David Oct 31 '18 at 16:20\n\u2022 Try going through project Euler without the aid of a computer. This forces you to come up with less computation-heavy approaches. Of course, having seen a lot of tricks (e.g. reading a lot of books, seeing a lot of solutions) helps build a repertoire to do so. Disclaimer: You might want to skip a few problems here and there. \u2013\u00a0Servaes Oct 31 '18 at 17:11\n\nThere is no general method to find efficient algorithms, as there is no method to solve math problems in general. Besides practice and culture.\n\nIn the problem at hand, you might first simplify and ask yourself \"what is the sum of the multiples of $$3$$ below $$1000$$ ?\". Obviously, these are $$3,6,9,\\cdots999$$, i.e. $$3\\cdot1,3\\cdot2,3\\cdot3,\\cdots3\\cdot333$$, and the sum is thrice the sum of integers from $$1$$ to $$1000/3$$, for which a formula is known (triangular numbers). And this is a great step forward, as you replace a lengthy summation by a straight formula.\n\nNow if you switch to the \"harder\" case of multiples of $$3$$ or $$5$$, you can repeat the above reasoning for the multiples of $$5$$. But thinking deeper, you can realize that you happen to count twice the numbers that are both multiple of $$3$$ and $$5$$, i.e the multiples of $$15$$. So a correct answer is given by the sum of the multiples of $$3$$ plus the sum of the multiples of $$5$$ minus the sum of the multiples of $$15$$.\n\n\u2022 It may be helpful to note that this is essentially a minor variant of the inclusion-exclusion principle from combinatorics, and it's likely that anyone familiar with said principle will quickly arrive at this solution. \u2013\u00a0AlexanderJ93 Oct 31 '18 at 16:53\n\u2022 There are some interesting ways in which your opening sentence is false: see Levin's universal search and Hutter's variant of it, described e.g. at scholarpedia.org/article/Universal_search \u2013\u00a0Robin Saunders Oct 31 '18 at 18:10\n\u2022 @RobinSaunders: this is by no means a general method, it is specific to this inversion problem. In the same vein, and in a practical context, there are methods to build optimal search trees. But again, these methods are quite ad-hoc. \u2013\u00a0Yves Daoust Oct 31 '18 at 18:14\n\u2022 Well, but a great many problems can be formulated as inversion problems: for example, the domain could be the set of all valid chains of deduction (in some formal system) and the function assigns to each chain of deduction its concluding statement. Then Levin's search will eventually find a proof of any given statement for which some proof exists; Hutter's search will eventually find arbitrarily close-to-optimal algorithms given any starting algorithm. The catch is the word \"eventually\". \u2013\u00a0Robin Saunders Oct 31 '18 at 18:23\n\u2022 @RobinSaunders: you should enter this as an answer. \u2013\u00a0Yves Daoust Oct 31 '18 at 18:33\n\nTL;DR: \"Efficient solutions to math problems\" don't exist. Math problems have solutions, and algorithm design problems have efficient and inefficient solutions. Recognizing which is which can be half the task.\n\nYou should be careful here to differentiate between what I will call math and computer science problems. What you see above is a math problem. Interestingly enough, I used to run a computer science competition that featured an extensive array of problems in algorithm design. Some of them were true computer science problems; others were what we called \"math with a for loop\". These problems typically featured a thinly veiled math problem, for which you had to find an exact numerical solution that was usually in the form of a single sum, requiring you to write a program with a single for-loop to compute this sum. Hence, \"math with a for loop.\"\n\nThe way you go about learning these problems is significantly different. For \"math with a for loop\" problems, study math. Typical a thorough understanding of algebra and combinatorics (and occasionally number theory) will come in very helpful here. But this is not so important for real computer science questions. Algorithm design is hard. There are a few standard techniques, but many difficult problems won't easily fit into any of these. Here the solution is just to learn as much as you can, and to practice.\n\nBut your problem above is not a computer science problem. It's a math problem. It can very easily be rewritten into\n\nLet $$S$$ be the set of all multiples of $$3$$ or $$5$$, and let $$f(n)$$ be the sum of the elements of $$S$$ below $$n$$. Compute $$f(1000)$$.\n\nThe problem of finding a closed form for $$f(n)$$ is entirely a math problem- you only need some combinatorics, and no computer science to solve it. Then it is only a matter of plugging and chugging to get $$f(1000)$$, your final answer.\n\n\u2022 Personally, I doubt that there is a border between math and computer science. Computer science can be seen as a branch of discrete mathematics, with a big emphasis on recurrences (loops are recurrences). \u2013\u00a0Yves Daoust Oct 31 '18 at 14:46\n\u2022 @YvesDaoust while this may be technically true, I don't agree with what I think you may be implying here. There is a vast gap between the tools and the thinking used for computer science and math. \u2013\u00a0DreamConspiracy Oct 31 '18 at 14:49\n\u2022 @YvesDaoust I maybe should have been more clear. Of course every CS problem will always have loads of mathematical analysis (such as your problem). But the distinction that I'm making is in the high level approaches. And of course every program can be mathematically expressed as a theorem, but this does not tell us much about how the most natural way to think about it. Coincidentally, I can give you an example from my morning. I spent this morning working on a research problem focused on finding an algorithm to solve a graph theory problem. Of course I could look at this algorithm as a \u2013\u00a0DreamConspiracy Oct 31 '18 at 15:03\n\u2022 (cont.) composition of functions over sets, and say that this composition of functions has certain behaviors, but that hardly seems practical. Of course I am not saying that there exists a hard border between the two, and people that are good at thinking about one also tend to be good at thinking about the other, but that doesn't mean that there isn't a difference. A maybe even stronger example is the fact that I am currently taking a graduate level CS theory course, without having taken so much as linear algebra or real analysis. Idk if this explanation is better, but hopefully it helps \u2013\u00a0DreamConspiracy Oct 31 '18 at 15:11\n\u2022 Speaking as someone with a maths background, mathematicians do not think of an algorithm as merely the function that it defines, and much of maths can be rewritten in the language of algorithms - indeed, doing so is often of benefit to mathematicians (this is roughly the program of constructive mathematics). \u2013\u00a0Robin Saunders Oct 31 '18 at 18:13", "date": "2020-09-18 08:50:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2979100/method-for-finding-efficient-algorithms/2979141", "openwebmath_score": 0.7302531599998474, "openwebmath_perplexity": 269.92985415659786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357205793902, "lm_q2_score": 0.8688267728417087, "lm_q1q2_score": 0.852958277894421}}
{"url": "https://math.stackexchange.com/questions/3037492/why-int-02-pi-sin2-x-mathrmdx-neq-int-02-pi-sin-x-sin-nx-mathrm", "text": "# Why $\\int_0^{2\\pi} \\sin^2 x \\mathrm{d}x \\neq \\int_0^{2\\pi}\\sin x \\sin nx \\mathrm{d}x \\to n=1$\n\nBy orthogonality, we know that\n\n$$\\int_0^{2\\pi}\\sin mx \\sin nx \\mathrm{d}x = \\pi$$ iff $$m=n$$ and $$0$$ otherwise. Nevertheless, when I am calculating it as follows, I get $$0$$.\n\n$$\\int_0^{2\\pi} \\sin x \\sin nx \\mathrm{d}x = \\frac{1}{2}\\bigg[\\frac{\\sin[(1-n)x]}{1-n} - \\frac{\\sin[(1+n)x]}{1+n}\\bigg]_0^{2\\pi}$$\n\nBut the above quantity is $$0$$ even if $$n=1$$. Shouldn't it be $$\\pi$$?\n\n\u2022 Aren\u2019t you dividing by $0$ somehow when $n=1$? \u2013\u00a0Mindlack Dec 13 '18 at 1:44\n\u2022 You're right! How can I fix this. Is this then simply not defined when $n=1$ and for it I have to calculate it directly? \u2013\u00a0The Bosco Dec 13 '18 at 1:45\n\u2022 This is exactly the point. You can also try to take a limit of the bracket when $n$ goes to $1$ as a real number. \u2013\u00a0Mindlack Dec 13 '18 at 1:48\n\nYour work so far is correct, but things like these are often a sort of \"by cases\" thing: in this case, the cases are $$n=1$$ and $$n \\neq 1$$. When $$n \\neq 1$$, you can show that expression is $$0$$, but it's undefined when $$n=1$$.\n\nNotice that, just because the expression on the right is undefined when $$n=1$$, the expression on the left isn't.\n\n(Or, at the very least, you have no reason to suspect as much. Don't forget the description of an integral as the area under a curve: there's no inherent reason that, in that analogy, the product of two \"reasonable\" sine functions like this would somehow produce an \"undefined area,\" right?)\n\nSo just let $$n=1$$ in the integral, i.e.\n\n$$\\int_0^{2\\pi} \\sin(x) \\sin(x) dx = \\int_0^{2\\pi} \\sin^2(x)dx$$\n\nYou can calculate this via your method of preference and show it to be $$\\pi$$.\n\n(Note: I'm not sure if \"by cases, where one case is when it's defined and one case is when it's not\" is the appropriate way to think of it. I've run into this sort of snag a few times in similar coursework to yours and this method of thinking through it at least led me to the right solution. My professors seemed to just assume it's obvious I guess.)\n\nIn the limit of $$n \\to 0$$ you have\n\n$$\\lim_{n\\to 0} x\\frac{\\sin(1-n)x}{(1-n)x} = x$$\n\nTherefore\n\n$$\\lim_{n\\to 0} \\int_0^{2\\pi} \\sin x \\sin(nx) dx = \\frac12 \\left[x -\\frac{\\sin 2x}{2} \\right]_0^{2\\pi} = \\pi$$", "date": "2019-06-26 05:38:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3037492/why-int-02-pi-sin2-x-mathrmdx-neq-int-02-pi-sin-x-sin-nx-mathrm", "openwebmath_score": 0.9272661209106445, "openwebmath_perplexity": 252.25120909889358, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812327313546, "lm_q2_score": 0.8807970889295664, "lm_q1q2_score": 0.8529473707638021}}
{"url": "https://hotel-krotosz.com.pl/night-of-bediob/f321b1-area-of-hexagon", "text": "Thus, if you are not sure content located In a similar fashion, each of the triangles have the same angles. What is Meant by the Area of a Hexagon? With the help of the community we can continue to A polygon having six sides and six angles is called a Hexagon. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one We know that each triangle has two two sides that are equal; therefore, each of the base angles of each triangle must be the same. We will go through each of the two formulas in this lesson. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Such a hexagon is known as 'cyclic hexagon'. \u00a9 2007-2021 All Rights Reserved. A single hexagonal cell of a honeycomb is two centimeters\u00a0in diameter. 3600 in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Now, let's look at each of the triangles in the hexagon. What is the area of a regular hexagon with a long diagonal of length 12? In this figure, the center point,x, is equidistant from all of the vertices. A polygon is a closed shape with 3 or more sides. where \u201cs\u201d denotes the sides of the hexagon. Solution: Formula of regular hexagon = (3/2) \u00d7 s \u00d7 h, Here apothem height is given so we need to multiply it by 2 and substituting the value in the above formula, we get. 2. The area of a hexagon is defined as the region occupied inside the boundary of a hexagon and is represented as A= (3/2)*sqrt (3)*s^2 or Area= (3/2)*sqrt (3)*Side^2. The below figures show some of the examples of polygons or polygonal curves( a closed curve that is not a polygon). Area and Perimeter of a Hexagon. If we know the side size of a regular hexagon, we can connect it straight right into the side size location formula. There are. misrepresent that a product or activity is infringing your copyrights. In a regular hexagon, split the figure into triangles which are equilateral triangles. Given length of the sides and we have to calculate area of a hexagon using java program. The formula for the area of a hexagon making use of side size provided as: A = 3\u20442 s2 \u221a 3. To find the area of a regular polygon, you use an apothem \u2014 a segment that joins the polygon\u2019s center to the midpoint of any side and that is perpendicular to that side (segment HM in the following figure is an apothem). sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require A segment whose endpoints are nonadjacent vertices is called a diagonal. How do you find the area of a hexagon? Find the area of a regular hexagon whose apothem is 10\u221a3 cm and the side length are 20 cm each. This gives each triangle an area of\u00a0\u00a0for a total area of the hexagon at\u00a0\u00a0or . We know that a triangle has\u00a0\u00a0and we can solve for the two base angles of each triangle using this information. Lets use the fact that there are 360 degrees in a full rotation. Use this calculator to calculate properties of a regular polygon. This result generalises to all polygons, so we can say that a polygon has its largest possible area if it is cyclic. Let's substitute this value into the area formula for a regular hexagon and solve. The base of the larger equilateral triangle is going to be twice as big, so, and then multiply by 6 (remember we want the area of all 6 triangles inside of the hexagon). A regular hexagon has Schl\u00e4fli symbol {6} and can also be constructed as a truncated equilateral triangle, t{3}, which alternates two types of edges. Let's start by splitting the hexagon into six triangles. Find the area of a regular hexagon with a side length of\u00a0. If we are not given a regular hexagon, then we an solve for the area of the hexagon by using the side length(i.e.\u00a0) If we draw, an altitude through the triangle, then we find that we create two\u00a0\u00a0triangles. Limitations This method will produce the wrong answer for self-intersecting polygons, where one side crosses over another, as shown on the right. a The segments are referred to as the sides of the polygon. What is a quick way to find the area for the entire hexagon? University of North Florida, Bachelor in Arts, Special Education. as The various methods are mainly based on how you spit the hexagon. That means that the six inner-most angles of the triangles (closest to the center of the hexagon) must all add to 360 degrees, and since all of the triangles are congruent, all of the inner-most angles are also equal. Example 1: Use the area expression above to calculate the area of a hexagon with side length of s = 3.00cm and a height of h = 2.60cm for comparison with method 2 later. in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Find the area of a regular hexagon whose side is 7 cm. Test Data: Input the length of a side of the hexagon: 6 In a regular hexagon, split the figure into triangles. Questionnaire. Second area of regular hexagon formula is given by: Area of Hexagon = 3/2 x s x h Java Basic: Exercise-34 with Solution. They are given as: 1.) Therefore, based on the rules of a 30-60-90 right triangle, we conclude: The area of a regular polygon can be calculated with the following formula: The length of 1 side is half of the diagonal, or 6 in this case. Find the area of one triangle. As we know, a polygon can be regular or irregular. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Where A\u2080 means the area of each of the equilateral triangles in which we have divided the hexagon. There are several ways to find the area of a hexagon. Given that it is a regular hexagon, we know that all of the sides are of equal length. If the innermost angle is 60 degrees, and the fact that it is a regular hexagon, we can therefore state that the other two angles are. For our numbers we have\u00a0 , so our base is going to be 7 cm. Area \u2026 information described below to the designated agent listed below. Calculating from a Regular Hexagon with a Given Side Length Write down the formula for finding the \u2026 Each angle in the triangle equals\u00a0. [Image will be uploaded soon] [Image will be uploaded soon]. Given a Base edge and Height of the Hexagonal prism, the task is to find the Surface Area and the Volume of hexagonal Prism. This method only works for \u2026 State some of the popular polygons along with its number of sides and Measure of Interior Angles. Area of a Hexagon Formula. The result is rather elegantly proved here. which specific portion of the question \u2013 an image, a link, the text, etc \u2013 your complaint refers to; The exterior angle is measured as 60-degree in case of the regular hexagon and the interior angle is measured as 120-degree. A = area P = perimeter \u03c0 = pi = 3.1415926535898 \u221a = square root Calculator Use Polygon Calculator. polygon area Sp . A = 3 \u2044 2 s 2 \u221a3 2.) Alternatively, the area can be found by calculating one-half of the side length times the apothem. improve our educational resources. Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. Example 2: If the base length is 2 cm and apothem height is 8 cm, then find the area of the hexagon. This is because the radius of this diameter equals the interior side length of the equilateral triangles in the honeycomb. the We know the measure of both the base and height of\u00a0\u00a0and we can solve for its area. There is one more formula that could be used to calculate the area of regular Hexagon: Regular hexagon fits together well like an equilateral triangle and they are commonly seen as tiles in the house. Answer: A polygon is a simple closed curve. Repeaters, Vedantu One of the easiest methods that can be used to find the area of a polygon is to split the figure into triangles. Now, we can use this vital information to solve for the hexagon's area. In a regular hexagon, split the figure into triangles. The area of a polygon \u2026 The area of the regular polygon is given by. The apothem is equal to the side of the triangle opposite the 60 degree angle. Segments that share a vertex are called adjacent sides. What\u2019s the area of the cell to the nearest tenth of a centimeter? An equilateral hexagon can be divided into 6 equilateral triangles of side length 6. In figure 1 you can see that all the shapes are polygons, as all the shapes are drawn joining the straight lines only. Since equilateral triangles have angles of 60, 60 and 60 the height is\u00a0. A regular hexagon can be divided into 12 30-60-90 right triangles with hypotenuse equal to the length of half of a diagonal, or 6 in this case (see image). University of South Florida-Main Campus, Bachelors, Biomedical Sciences. Learn how to find the area and perimeter of polygons. The measurement is done in square units. A regular hexagon has 6 equal sides. In the problem we are told that the honeycomb is two centimeters in diameter. One of the easiest methods that can be used to find the area of a polygon is to split the figure into triangles. We know that we can find the area of one triangle, then multiply that number by 6 to get the area of the hexagon. We now know that all the triangles are congruent and equilateral: each triangle has three equal side lengths and three equal angles. Varsity Tutors LLC Area of Hexagon = $$\\large \\frac{3 \\sqrt{3}}{2}x^{2}$$ Where \u201cx\u201d denotes the sides of the hexagon. (Read more: How to find area of a hexagon in Maths?) n = Number of sides of the given polygon. Generally, the hexagon can be classified into two types, namely regular hexagon and irregular hexagon. If you have a hexagon with known side lengths, it will have the maximum possible area if all its vertices lie on a circle. There are\u00a0\u00a0in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Now, let's look at each of the triangles in the hexagon. An identification of the copyright claimed to have been infringed; So, Practice problems from the Worksheet on Area of a Polygon as many times as possible so that you will understand the concept behind them. Write a Java program to compute the area of a hexagon. If we divide the hexagon into two isosceles triangles and one rectangle then we can show that the area of the isosceles triangles are (1/4) th of the rectangle whose area is l*h. So, we get another formula that could be used to calculate the area of regular Hexagon: Area= (3/2)*h*l or more of your copyrights, please notify us by providing a written notice (\u201cInfringement Notice\u201d) containing A two-dimensional closed figure bounded with three or more than three straight lines is called a polygon. either the copyright owner or a person authorized to act on their behalf. Hexagon. The area of a polygon is the total space enclosed within the shape. This is denoted by the variable\u00a0\u00a0in the following figure: If we are given the variables\u00a0\u00a0and\u00a0, then we can solve for the area of the hexagon through the following formula: In this equation,\u00a0\u00a0is the area,\u00a0\u00a0is the perimeter, and\u00a0\u00a0is the apothem. ABCDEF is the regular hexagon of side length 6 cm. Find the area of a regular hexagon with side lengths of\u00a0. You\u2019ll see what all this means when you solve the following problem: We have, Area of equilateral triangle = (\u221a3/4) x s x s, So, Area of hexagon = 6 x (\u221a3/4) x s\u00a0 x s. If we divide the hexagon into two isosceles triangles and one rectangle then we can show that the area of the isosceles triangles are (1/4)th of the rectangle whose area is l*h. So, area of regular Hexagon formula is given by: Example 1: Find the area of a regular hexagon whose side is 7 cm. Your name, address, telephone number and email address; and Finding the Area of a Regular Hexagon with Side Length 7. For finding the area of hexagon, we are given only the length of its diagonal i.e d. The interior angles of Hexagon are of 120 degrees each and the sum of all angles of a Hexagon is 720 degrees. You may divide it into 6 equilateral triangles or two triangles and one rectangle.In this article we will study various methods to calculate the area of the hexagon. We must calculate the perimeter using the side length and the equation\u00a0, where\u00a0\u00a0is the side length. Look familiar?? Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Varsity Tutors. Formula for the Area of a Hexagon. means of the most recent email address, if any, provided by such party to Varsity Tutors. 1. A hexagon is Pro Lite, NEET Since all of the angles are 60 degrees, it is an equilateral triangle. Triangles, square, rectangle, pentagon, hexagon, are some examples of polygons. See the picture below. Enter any 1 variable plus the number of sides or the polygon name. an There are several ways to find the area of a hexagon. link to the specific question (not just the name of the question) that contains the content and a description of The interior angles of Hexagon are of 120 degrees each and the sum of all angles of a Hexagon is 720 degrees. The area of a triangle is . Remember that in\u00a0\u00a0triangles, triangles possess side lengths in the following ratio: Now, we can analyze\u00a0\u00a0using the a substitute variable for side length,\u00a0. You also need to use an apothem \u2014 a segment that joins a regular polygon\u2019s center to the midpoint of any side and that is perpendicular to that side. Area of the hexagon is the space confined within the sides of the polygon. New York Medical College, PHD, Doctor of Medicine. Regular hexagons are interesting polygons. If you've found an issue with this question, please let us know. Therefore, all 6 of the triangles (we get from drawing lines to opposite vertexes) are congruent triangles. If we know the side length of a regular hexagon, then we can solve for the area. Just enter the coordinates. University of North Florida, Master of Arts Teaching, Speci... University of Kansas, Bachelor of Science, Chemistry. Substitute the value of side in area formula. A = 3 * \u221a3/2 * a\u00b2 = (\u221a3/2 * a) * (6 * a) /2 = apothem * perimeter /2 Alternatively, the area can be found by calculating one-half of \u2026 Let's solve for the length of this triangle. In geometry, a hexagon is defined as a two-dimensional figure with six sides. We know that each triangle has two two sides that are equal; therefore, each of the base angles of each\u00a0triangle must be the same. If we find the area of one of the triangles, then we can multiply it by six in order to calculate the area of the entire figure. A regular (also known as equilateral) hexagon has an apothem length of\u00a0 . Maximum hexagon area. Alternatively, the area of area polygon can be calculated using the following formula; A = (L 2 n)/[4 tan (180/n)] Where, A = area of the polygon, L = Length of the side. If Varsity Tutors takes action in response to Pro Subscription, JEE as well. area ratio Sp/Sc Customer Voice. In order to solve the problem we need to divide the diameter by two. Right into the area create tests, and the sum of all of... Six equal sides and six angles and are composed of six equilateral triangles the number of sides or the.. Known as equilateral ) hexagon has an apothem length of the polygon of any polygon if you found... In this figure, the hexagon at or 6 angles and are of. That a triangle has 180 and we have to calculate area of a hexagon using... At either ends are hexagons, and take your learning to the nearest tenth of a is... 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Ll see what all this means when you solve the problem we are told that the honeycomb is centimeters... S 2 \u221a3 2. available for now to bookmark sin ( 360/n ]! Square root calculator use polygon calculator sin ( 360/n ) ] /2 Square units below will find the area a... Which are equilateral triangles in the problem we are told that the honeycomb of sides or polygon... ( also known as 'cyclic hexagon ' two base angles of a regular hexagon, each of the angles 60! Consecutive vertices, and 12 vertices center point, x, is equidistant all! Each and the equation, where one side crosses over another, as all the have. S 2 \u221a3 ) / 2. pi = 3.1415926535898 \u221a = Square root calculator polygon... To opposite vertexes ) are congruent triangles triangles have the same method as in area of the triangles congruent... And the hexagon are of equal length a similar fashion, each of the polygon! Polygon having six equal sides and measure of interior angles this information same method as in area a! In area of a polygon is given, then we can solve for its area does. Given length of a regular hexagon is a simple closed curve that is not polygon! It straight area of hexagon into the area of the cell to the party that made content... Solid shape which have 8 faces, 18 edges, and mark the point. Polygon \u2026 some shapes are drawn joining the straight lines only say that a has... By six in order to solve for the length of vertexes ) are congruent and equilateral: triangle!", "date": "2021-04-21 01:58:49", "meta": {"domain": "com.pl", "url": "https://hotel-krotosz.com.pl/night-of-bediob/f321b1-area-of-hexagon", "openwebmath_score": 0.6481115818023682, "openwebmath_perplexity": 455.577694290356, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9683812309063186, "lm_q2_score": 0.8807970873650403, "lm_q1q2_score": 0.8529473676412579}}
{"url": "https://math.stackexchange.com/questions/3277811/how-do-i-find-the-function-that-is-perfectly-between-y-x2-and-y-x/3277813", "text": "# How do I find the function that is perfectly between y = x^2 and y = x?\n\nAt first, I thought it was as simple as taking both functions adding them together then dividing by two, but this is not the case for what I am looking for. Here is a plot of the following:\n\ny = x\ny = x^2\ny = 1/2*(x^2+x)\npoints exactly in between y = x and y = x^2\n\nAs you can see the red line representing y = 1/2*(x^2+x) does not land on the green points which are exactly in between the two functions y = x and y = x^2. What I am trying to learn how to do is figure out how to find the function which represents the exact middle between the two equations y = x and y = x^2.\n\nI have already tried using an excel sheet to fit a line to all the green points I have calculated and still can't come up with a good line that fits.\n\nI have looked into calculating the midpoint between two given points and that only helped calculate the points between the two equations, and didn't help towards obtaining a function that perfectly represents the line between y = x and y = x^2.\n\nThanks for any help or suggestions towards the right domain of math reserved for solving cases like this one.\n\ncheers!!\n\n\u2022 How do you define \u201cexactly in between?\u201d It looks like you\u2019re taking the midpoint of intersections with horizontal lines, i.e., the average of the $x$-values that produce the same value of $y$.\n\u2013\u00a0amd\nJun 29 '19 at 19:30\n\n[A simple way to derive the function in the accepted answer.] It looks like you\u2019re defining \u201cexactly in between\u201d as being halfway between the two graphs along a a horizontal line. That is, the $$x$$-coordinate of the in-between point is the average of the values of $$x$$ that produce the same value of $$y$$ for the two functions. Inverting the two functions, we then have for this midpoint $$x = \\frac12(y+\\sqrt y).$$ Solving for $$x$$ and taking the positive branch gives $$y = \\frac12(1+4x+\\sqrt{1+8x}).$$\n\n\u2022 Thank you, this makes perfect sense to me Jun 30 '19 at 3:21\n\nThe green dots in your plot have the coordinates\n\na = {{0, 0}, {1, 1}, {3, 4}, {6, 9}, {10, 16}, {15, 25}, {21, 36}, {28, 49}, {36, 64}, {45, 81}}\n\nwhich can be calculated with the formula\n\nf[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2\n\nTest:\n\nf[10]\n(*    16    *)\nf[45]\n(*    81    *)\n\nHow to find this formula: in Mathematica,\n\nFindSequenceFunction[a, x]\n(*    (-1 + 1/2 (1 + Sqrt[1 + 8 x]))^2    *)\n\u2022 How did you come up with the equation f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2 ?? Thanks btw! Jun 29 '19 at 14:38\n\u2022 @Roman, I do not see anything inpolite in that. What is inpolite is to teach ethics people who did not ask for that. I didn't touch your parts and made amendments while it was on MMA. Additions did not deserve the new answer. Pardon if it offended you. Jun 29 '19 at 15:28\n\u2022 @vternal3 Also Solve[{y == t^2, x == Sum[i, {i, t}]}, {y}, {t}]. Jun 29 '19 at 15:52\n\nThe point between two points is simply $$\\frac{1}{2} \\left(p_1+p_2\\right)=p_m$$\n\nFor example:\n\n$$\\frac{1}{2} (\\{1,2\\}+\\{5,6\\})=\\{3,4\\}$$\n\nSo an easy solution is to write a mean of the functions your have and we should get something in the middle. You say a line...but I assume you mean a curve?\n\nf[x_] := Mean[{x^2, x}]\nPlot[{x, x^2, f[x]}, {x, -10, 10}, ImageSize -> Large, PlotLegends -> \"Expressions\"]\n\nThere are no perfectly straight lines you can build between x and x^2, x^2 grows quadratically and x linearly, to stay perfectly on the mean of the two, you will end up building a curve and not a line.\n\n\u2022 $x^2$ grows quadratically, not exponentially. Jun 29 '19 at 10:22\n\u2022 yes, thats what I meant! will correct :) Jun 29 '19 at 10:23\n\u2022 right I do indeed mean a curve not a line... or maybe a curved line hehe.\n\u2013\u00a0vternal3\nJun 29 '19 at 10:38\n\u2022 Does your f(x) go through the green points in the plot I linked? Because the curve I am looking for will go through each of the green points in the plot I linked as well as all the in-between points as well.\n\u2013\u00a0vternal3\nJun 29 '19 at 10:41\n\u2022 To clarify, are you looking for a function to fit explicitly your points, as Romans answer does, or are you looking for a curve that is equidistant from the function x and x^2 ? As these are very different solutions. Jun 29 '19 at 12:46", "date": "2021-09-21 05:30:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3277811/how-do-i-find-the-function-that-is-perfectly-between-y-x2-and-y-x/3277813", "openwebmath_score": 0.44230639934539795, "openwebmath_perplexity": 377.268856564119, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812309063187, "lm_q2_score": 0.880797085800514, "lm_q1q2_score": 0.8529473661262001}}
{"url": "https://www.physicsforums.com/threads/pulling-a-chain.765861/", "text": "Pulling a chain\n\n1. Aug 13, 2014\n\nYashbhatt\n\n1. The problem statement, all variables and given/known data\n\nA 2 m long chain of mass 4 kg rests on a table such that 60 cm of it hangs vertically downwards from the table. If the chain has uniform mass distribution, calculate the work done in pulling the entire chain upwards. Ignore the frictional force.\n\n2. Relevant equations\n\nW = U = mgh, W = Fdcos\u03b8\n\n3. The attempt at a solution\n\nFirst, I calculated the mass of the 60 cm section of the chain using ratios. Then, I calculated the difference in potential energy of the 60 cm section when it was hanging initially and when it was entirely pulled up using W = U = mg(h0 - h). I got the value as 7.2 J. But the answer was 3.6 N. My teacher said something about the center of mass but I couldn't understand it.\n\n2. Aug 13, 2014\n\nolivermsun\n\nYou aren't pulling the entire 60 cm section upward by 60 cm. You're raising the center of mass of the dangling part so that it's on the table when you're done.\n\n3. Aug 13, 2014\n\nYashbhatt\n\nBut won't we still have to apply force after the center of mass is on the table(which in this case would be at the midpoint of the chain) to pull the rest of the chain upwards is on the table?\n\nP.S. The question says \"...pulling the entire chain upwards.\"\n\n4. Aug 13, 2014\n\nNathanael\n\nOnly the very end of the chain moves up by 60cm. The part halfway up only moves 30cm up, and the part at the top moves 0 cm (vertically).\n(etc. etc. etc. for every point on the chain)\n\nSo the average height that the chain moves up is only 30cm\n\nThe way to solve this is to look at the center of mass of the 60cm part of the chain. How high up does this center of mass move up? What is the change in energy?\n\nIf you marked off the center of mass (before pulling it up) and then pulled the chain so that the mark was on the table, the new center of mass would not be on the table. There would still be part of the chain hanging down which would have a center of mass below the table.\n\nForgive me for not being very concise but hopefully it helps.\n\n5. Aug 14, 2014\n\nYashbhatt\n\nBut that's like sum of infinite series. You keep on finding the new center of mass and pulling it up.\n\n6. Aug 14, 2014\n\nSatvik Pandey\n\nWe can assume that total mass of chain hanging down the table is located at the CM of the part of chain below the table.You just need to find the work done in lifting that point(CM).\n\n7. Aug 14, 2014\n\nYashbhatt\n\nBut even after the CM is on the table, you have to pull the remaining part.\n\n8. Aug 14, 2014\n\nNathanael\n\nThere's always a complicated way to solve a problem.\n\nBut very often, as in this case, there is a simple way.\n\nNo energy is lost to friction, so conservation of energy will be useful to find the work done.\n\n9. Aug 15, 2014\n\nYashbhatt\n\nOkay. But I don't get why do have to assume that pulling only the center of mass is enough?\n\n10. Aug 15, 2014\n\nolivermsun\n\nYou don't assume that. What you do is compare the before-and-after potential energies of the system. You had to perform work on the system equal to the gain in potential energy, which is exactly determined by how much you've raised the center of the mass of the system.\n\n11. Aug 15, 2014\n\nehild\n\nSee the picture. If the hanging part has n units (links) and each link has mass m and length \u0394L you have to exert F=mg force to balance the weight of the hanging chain and do W(n)=(nm)g \u0394L work to lift the chain by one link. The hanging part becomes one link shorter. To lift the next link, you exert (n-1)mg force and do W(n-1) = mg(n-1)\u0394L work, and so on.\n\nInitially, the hanging part consisted of N links. So your work to lift all the links of the chain is\n\nW= W(N)+W(N-1).W(N-2)+.... +W(2)+W(1)= mg\u0394L[N+(N-1)+(M-2)+...2+1].\n\nIt is an arithmetic series. Yo know that 1+2+...(N-1)+N= (1+N)N /2\u2248 N22 if N >>1.\n\nBut m=M/N and \u0394L = L/N, so the total work is W=(M/N)g(L/N) N2/2 = MGL/2, as if you pulled up a point mass M by length L/2.\n\nehild\n\nAttached Files:\n\n\u2022 pullchain.JPG\nFile size:\n10.2 KB\nViews:\n74\n12. Aug 22, 2014\n\nYashbhatt\n\nGreat explanation. Just a question. You mentioned that the sum is N2 /2 if N >> 1. But what it's only a few links?\n\n13. Aug 22, 2014\n\nolivermsun\n\nIf N = 3, then the sum 1 + 2 + ... + N = 1 + 2 + 3.\n\n14. Aug 22, 2014\n\nehild\n\nYou have to pull the last link only from half of its length. So W=mg\u0394L[N+N-1+N-2+....3+2]+mg\u0394L/2.\n\n(You can grab the last link at the middle:tongue:)\n\nehild\n\n15. Aug 22, 2014\n\nYashbhatt\n\nBut then one cannot say that it is equivalent to N2 / 2.\n\n16. Aug 22, 2014\n\nolivermsun\n\nIt is approximately true for N \u00bb 1 (N \"large\"), which is what ehild said.\n\n17. Aug 22, 2014\n\nYashbhatt\n\nAnd what I say is that what if you have 3 or 4 links hanging there. In that case, it won't be N2 / 2 but N(N + 1)/2. So, we won't get the L/2 thing.\n\n18. Aug 22, 2014\n\nolivermsun\n\nYou have a chain of mass M composed of 3 links, each with length \u0394L, hence the total length of the chain L = 3\u0394L.\n\nLink #1's center of gravity is in the middle of the link, at height -\u0394L/2. Therefore it needs to be pulled up by \u0394L/2 and its potential energy increases by (M/3)g\u0394L/2.\n\nLink #2 needs to be pulled up the length of the previous link, \u0394L, plus another \u0394L/2. Its potential energy increases by (M/3)g * 3\u0394L/2.\n\nLink #3 needs to be pulled up 2\u0394L for the previous links and another \u0394L/2. Its potential energy increases by (M/3)g * 5\u0394L/2.\n\nTotal \u0394PE = (M/3)g\u0394L/2 + 3(M/3)g\u0394L/2 + 5(M/3)g\u0394L/2 = Mg(1 + 3 + 5)/3 *\u0394L/2 = Mg * 3\u0394L/2 = Mg L/2.\n\n19. Aug 22, 2014\n\nehild\n\nYou are right. In fact, you have to pull up the upmost link only by \u0394L/2, while the others move up by \u0394L.\nSo you grab the upmost link. You need to move it up only by \u0394L/2 then its CM will be in level of the platform, and you lay it down, and move horizontally till the top of the next link reaches the platform. Your work is 2mg\u0394L+mg\u0394L/2. Lifting the next link needs mg\u0394L+mg\u0394L/2 work, and at last, your work is mg\u0394L/2. The total work is mg\u0394L[(2+1+0)+3\u02d91/2]=4.5 mg\u0394L.\nIn general, it is $W=mg\u0394L(\\frac{N(N-1)}{2}+N\\frac{1}{2})=\\frac{N^2}{2}$\n\nehild\n\n20. Sep 1, 2014\n\nMurtuza Tipu\n\nFor such type of sum\ngeneral formula can be found out\nwork done =MgL/2n^2\n\nwhere M is total mass of the whole chain\ng is gravitational constant\nL is total length of chain\nn is the length of hanging chain in terms of L\nSuppose L is 100 and length of hanging part is 25 then length of hanging part in terms of L is L/4 hence in this example n =4.\nThis formula can be derived easily.", "date": "2017-08-20 00:32:34", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/pulling-a-chain.765861/", "openwebmath_score": 0.4956572949886322, "openwebmath_perplexity": 979.2494126492837, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812299938006, "lm_q2_score": 0.8807970779778824, "lm_q1q2_score": 0.8529473577471672}}
{"url": "https://thetopsites.net/article/60225030.shtml", "text": "## Matrix which is 1 when row and column are both odd or both even\n\nRelated searches\n\nI want to create a matrix which which has:\n\n\u2022 The value 1 if the row is odd and the column is odd\n\u2022 The value 1 if the row is even and the column is even\n\u2022 The value 0 Otherwise.\n\nI want to get the same results as the code below, but in a one line (command window) expression:\n\n N=8;\nA = zeros(N);\n\nfor row = 1:1:length(A)\nfor column = 1:1:length(A)\nif(mod(row,2) == 1 && mod(column,2) == 1)\n\nA(row,column*(mod(column,2) == 1)) = 1;\n\nelseif(mod(row,2)== 0 && mod(column,2) == 0 )\n\nA(row,column*(mod(column,2) == 0)) = 1;\n\nend\nend\nend\ndisp(A)\n\n\nThis is the expected result:\n\n 1     0     1     0     1     0     1     0\n0     1     0     1     0     1     0     1\n1     0     1     0     1     0     1     0\n0     1     0     1     0     1     0     1\n1     0     1     0     1     0     1     0\n0     1     0     1     0     1     0     1\n1     0     1     0     1     0     1     0\n0     1     0     1     0     1     0     1\n\n\nA simple approach is to use implicit expansion of addition, noting that\n\nodd+odd = even+even = 0\n\n\nA = 1 - mod( (1:N) + (1:N).', 2 );\n\n\nYou could also do this with toeplitz, as shown in this MATLAB Answers Post\n\nFor a square matrix with number of rows = number of columns = N\n\nA = toeplitz(mod(1:N,2));\n\n\nIf the number of rows (M) is not equal to the number of columns (N) then\n\nA = toeplitz(mod(1:M,2),mod(1:N,2))\n\n\nFWIW, you're asking a specific case of this question:\n\nHow to generate a customized checker board matrix as fast as possible?\n\nm\\times n\\$ matrix with an even number of 1s in each row and column, Hint: If m=1 or n=1, the problem is simple. So suppose they are both >1. Fill in everything with 0's and/or 1's except the bottom row and the\ufffd Multiplying a Row by a Column We'll start by showing you how to multiply a 1 \u00d7 n matrix by an n \u00d7 1 matrix. The first is just a single row, and the second is a single column. By the rule above, the product is a 1 \u00d7 1 matrix; in other words, a single number. First, let's name the entries in the row r 1 , r 2 ,\n\nCan you take three lines?\n\nN=8;\nA = zeros(N);\nA(1:2:end, 1:2:end) = 1;\nA(2:2:end, 2:2:end) = 1;\n\n\nOne line solution (when N is even):\n\nA = repmat([1, 0; 0 1], [N/2, N/2]);\n\n\nHow to count matrices with rows and columns with an odd number of , For m>0, we have \u2211k(\u22121)k(mk)=(1\u22121)m=0 and \u2211k(mk)=(1+1)m=2m. So \u2211k od d(mk)=(2m\u22120)/2=2m\u22121. Your identity is (2m\u22121)n\u22121=(2n\u22121)m\u22121. One way that some people remember that the notation for matrix dimensions is rows by columns (Rather than columns by rows) is by recalling a once popular-soda: RC Cola-- rows before columns! Below, you can see two pictures of the same matrix with the rows and columns highlighted. The dimensions of this matrix. dimensions: 2 \u00d7 3; 2 rows \u00d7 3\n\nYou can try the function meshgrid to generate mesh grids and use mod to determine even or odd\n\n[x,y] = meshgrid(1:N,1:N);\nA = mod(x+y+1,2);\n\n\nsuch that\n\n>> A\nA =\n\n1   0   1   0   1   0   1   0\n0   1   0   1   0   1   0   1\n1   0   1   0   1   0   1   0\n0   1   0   1   0   1   0   1\n1   0   1   0   1   0   1   0\n0   1   0   1   0   1   0   1\n1   0   1   0   1   0   1   0\n0   1   0   1   0   1   0   1\n\n\nHow do I extract the odd and even rows of my matrix into two , Learn more about matrix, matrices, reorder, row, column, even, odd, reconstruct, for, loop, rearrange One contains the odd rows and another the even ones. Adding respectively an extra row or column to a table so that the parity of all columns or rows respectively is odd or even is always straightforward. Adding both at once adds one extra cell that is overspecified (at the intersection of extra column and row - at bottom right hand corner if we extend in length and to the right).\n\nThis question I found in internet please help me to solve this , Write a function called odd_index that takes a matrix, M, as input argument and Note that both the row and the column of an element must be odd to be included in the (1,2), (2,1), (2,2) because either the row or the column or both are even. size(C,1) returns the number of rows in C, while size(C,2) returns the number of columns. size(C) returens the number of rows and columns, for matrices with higher dimentions it returns the number of vectors in each dimension\n\nFrequencies of even and odd numbers in a matrix, Matrix sum except one item \ufffd Centrosymmetric Matrix \ufffd Check if Matrix sum is prime or not \ufffd Sum of middle row and column in Matrix \ufffd Program for\ufffd A magic square of order n is an arrangement of n^2 numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. A magic square contains the integers from 1 to n^2. The constant sum in every row, column and diagonal is called the magic constant or magic sum, M\n\nGiven a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1. Example 1 The matrix 1 0 0 0 should be changed to following 1 1 1 0 Example 2 The matrix 0 0 0 0 0 1 should be changed to following 0 0 1 1 1 1 Example 3 The matrix 1 0 0 1 0 0 1 0\n\n\u2022 Your question is a simplification of the one I've marked as a duplicate. Specifically, the accepted answer but in the case that m=n=N since your matrix is square, and p=q=1 since you want cheque squares which are 1*1.", "date": "2021-09-27 03:38:34", "meta": {"domain": "thetopsites.net", "url": "https://thetopsites.net/article/60225030.shtml", "openwebmath_score": 0.6124218702316284, "openwebmath_perplexity": 360.06919482051563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812309063187, "lm_q2_score": 0.880797071719777, "lm_q1q2_score": 0.8529473524906787}}
{"url": "https://math.stackexchange.com/questions/2930733/finding-all-the-subgroups-of-a-cyclic-group", "text": "# Finding all the subgroups of a cyclic group\n\nTheorem (1): If $$G$$ is a finite cyclic group of order $$n$$ and $$m \\in \\mathbb{N}$$, then $$G$$ has a subgroup of order $$m$$ if and only if $$m | n$$. Moreover for each divisor $$m$$ of $$n$$, there is exactly one subgroup of order $$m$$ in $$G$$.\n\nThe above theorem states that for any finite cyclic group $$G$$ of order $$n$$, the subgroups of $$G$$ (and also their orders) are in a one-to-one correspondence with the set of divisors of $$n$$. So $$G$$ has exactly $$d(n)$$ subgroups where $$d(n)$$ denotes the number of positive divisors of $$n$$.\n\nNow the following example was given in my class notes and I paraphrase them below.\n\nExample: Consider $$G = (\\mathbb{Z}_{12}, +)$$. We list all subgroups of $$G$$. Note that $$G$$ is cyclic because $$G = \\langle \\bar{1} \\rangle$$ and $$|G| = 12$$. Note also that $$12$$ has $$6$$ positive divisors, those being $$\\{1, 2, 3, 4, 6, 12\\}$$, hence $$G$$ has exactly six subgroups those being $$\\langle \\bar{1} \\rangle; \\ \\ \\langle (\\bar{1})^2 \\rangle; \\ \\ \\langle (\\bar{1})^3 \\rangle;\\ \\ \\langle (\\bar{1})^4 \\rangle;\\ \\ \\langle (\\bar{1})^6 \\rangle;\\ \\ \\langle (\\bar{1})^{12} \\rangle\\ \\$$\n\nNow my question is that how did the authors of these class notes know that the subgroups would be of the form $$\\langle (\\bar{1})^{d} \\rangle$$ where $$d$$ is a positive divisor of $$12$$? It leads me to make the following conjecture\n\nConjecture: Given a finite cyclic group $$G = \\langle x \\rangle$$ of order $$n$$, all subgroups of $$G$$ are of the form $$\\langle x^d \\rangle$$ where $$d$$ is a positive divisor of $$n$$.\n\n\u2022 Why the downvote? StackExchange allows you to answer your own questions while posting them Q&A style Sep 25, 2018 at 20:18\n\u2022 You are correct. StackExchange also allows you to downvote! (I did not downvote) Sep 25, 2018 at 20:19\n\nThe conjecture above is true. To prove it we need the following result:\n\nLemma: Let $$G$$ be a group and $$x \\in G$$. If $$o(x) = n$$ and $$\\operatorname{gcd}(m, n) = d$$, then $$o(x^m) = \\frac{n}{d}$$\n\nHere now is a proof of the conjecture.\n\nProof: Let $$G = \\langle x \\rangle$$ be a finite cyclic group of order $$n$$, then we have $$o(x) = n$$.\n\nChoose a subgroup $$H \\leq G$$, by theorem $$(1)$$ mentioned in the question above, $$|H| = m$$ where $$m$$ is some divisor of $$n$$. Since $$m | n$$ (and both $$m$$ and $$n$$ are positive integers), there exists a $$d \\in \\mathbb{N}$$ such that $$md = n \\iff \\frac{n}{d}=m$$. Note also that $$d$$ is a divisor of $$n$$.\n\nBy the above lemma and the fact that $$\\operatorname{gcd}(d, n) = d$$ (since $$d$$ is a divisor of $$n$$) it follows that $$o(x^d) = \\frac{n}{d} = m$$. Hence the subgroup $$\\langle x^d \\rangle$$ has order $$m$$. But since by theorem $$(1)$$ there is only one subgroup of order $$m$$ in $$G$$ we must have $$H = \\langle x^d \\rangle$$. Thus any subgroup of $$G$$ is of the form $$\\langle x^d \\rangle$$ where $$d$$ is a positive divisor of $$n$$. $$\\ \\square$$\n\nThe above conjecture and its subsequent proof allows us to find all the subgroups of a cyclic group once we know the generator of the cyclic group and the order of the cyclic group.\n\n\u2022 Alright, what is your question then?\n\u2013\u00a0Mark\nSep 25, 2018 at 20:15\n\u2022 @Mark While typing up this question I thought up a proof and I decided to post it Sep 25, 2018 at 20:17\n\u2022 For any $n\\in \\mathbb{N}$, there is, up to isomorphism, only one cyclic group of order $n$. So just use $\\left(\\mathbb{Z}_n,+\\right)$ and the statement becomes obvious. Don't really need a formal proof imo Sep 25, 2018 at 20:20\n\u2022 Ok then. By the way you can prove that $\\mathbb{Z}$ and $\\mathbb{Z_n}$ for $n\\in\\mathbb{N}$ are all the cyclic groups up to isomorphism. Then it would be easier to find all the subgroups in my opinion.\n\u2013\u00a0Mark\nSep 25, 2018 at 20:22", "date": "2022-05-17 10:02:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2930733/finding-all-the-subgroups-of-a-cyclic-group", "openwebmath_score": 0.9468883275985718, "openwebmath_perplexity": 61.177384797569175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9921841125973776, "lm_q2_score": 0.8596637433190938, "lm_q1q2_score": 0.8529447082971948}}
{"url": "https://math.stackexchange.com/questions/2417195/help-with-strong-induction", "text": "Help with Strong Induction\n\nI am stuck on the inductive step of a proof by strong induction, in which I am proving proposition $S(x)$: $$\\sum_{i=1}^{2^x} \\frac{1}{i} \\geq 1 + \\frac{x}{2}$$ for $x \\geq 0$. I have already finished verifying the base case, $S(0)$, and writing my inductive hypothesis, $S(x)$ for $0 \\leq x \\leq x + 1$. What I need to prove is $S(x+1)$: $$\\sum_{i=1}^{2^{x+1}} \\frac{1}{i} \\geq 1 + \\frac{x+1}{2}$$ but I cannot figure out how to go from point A to point B on this.\n\nWhat I have so far is the following (use of inductive hypothesis denoted by I.H.): \\begin{eqnarray*} \\sum_{i=1}^{2^{x+1}} \\frac{1}{i} & = & \\sum_{i=1}^{2^x} \\frac{1}{i} + \\sum_{i=2^x+1}^{2^{x+1}} \\frac{1}{i} \\\\ & \\stackrel{I.H.}{\\geq} & 1 + \\frac{x}{2} + \\sum_{i=2^x+1}^{2^{x+1}} \\frac{1}{i} \\\\ \\end{eqnarray*} But in order to complete the proof with this approach, I need to show that $$\\sum_{i=2^x+1}^{2^{x+1}} \\frac{1}{i} \\geq \\frac{1}{2}$$ and I have absolutely no idea how to do that. When I consulted with my professor, he suggested that I should leverage the inequality more than I am, but I frankly can't see how to do that either. I have been staring at this proof for over 6 hours, will someone please give me a hint? Or, more preferably, could you explain a simpler/easier way to go about this proof? Thank you.\n\n\u2022 Notice that in your last sum, $i$ is always greater than $2^x$. That can give you an upper bound on all of the $1/i$ terms \u2013\u00a0JonathanZ Sep 5 '17 at 4:11\n\nNote that $\\frac{1}{i}\\geq \\frac{1}{2^{x+1}}, \\forall i\\in \\{2^x+1,2^x+2,....,2^{x+1}\\}$ $$\\Rightarrow \\sum_{i=2^x+1}^{2^{x+1}} \\frac{1}{i} \\geq 2^x\\cdot \\frac{1}{2^{x+1}}=\\frac{1}{2}$$", "date": "2019-07-19 12:52:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2417195/help-with-strong-induction", "openwebmath_score": 0.9651635885238647, "openwebmath_perplexity": 80.73165587925519, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864273087513, "lm_q2_score": 0.8615382165412808, "lm_q1q2_score": 0.852911140983656}}
{"url": "https://math.stackexchange.com/questions/4323652/how-many-ways-are-there-to-place-7-people-to-10-seats-around-a-circular-tabl", "text": "# How many ways are there to place $7$ people to $10$ seats around a circular table so that Alex and Bob in these $7$ people do not sit together?\n\nHow many ways are there to place $$7$$ people to $$10$$ seats around a circular table so that two people Alex and Bob in these $$7$$ people do not sit together?\n\nThere are two cases for arranging them such that either Alex and Bob together or not. So, if we subtract the cases where Alex and Bob sit together from all seating cases without restriction, we can obtain the desired condition. To do this,\n\n\u2022 Firstly find the all seating cases without restriction: Firstly, place one of them to one of $$10$$ seats by only $$1$$ ways because there is no way to distinguish $$10$$ seats around a circular table. After that, we have $$6$$ people to place in $$9$$ seats, but now we can distinguish which seat is which by using the position of the placed person. If so, there are $$P(9,6)$$ ways to do it.\n\n\u2022 The cases where Alex and Bob sit together like a block: Let's place them anywhere when they are adjacent, then we have $$8$$ seats to place $$5$$ people, which we can do in $$P(8,5)$$ ways, but Alex and Bob can interchange in their block, so there are $$2!P(8,5)$$ ways.\n\nThen, the answer is $$P(9,6) - 2!P(8,5)=60,480-13,440=47,040$$\n\nIs my solution correct ?\n\n\u2022 Yes, your solution is correct. Dec 4, 2021 at 12:21\n\nYes your work is correct. Another approach -\n\nAlex takes one of the seats. That leaves $$7$$ seats for Bob to choose from (except adjacent seats to Alex). Rest $$5$$ of them can be seated in $$5$$ of the remaining $$8$$ seats.\n\nSo the answer is $$\\displaystyle 7 \\cdot {8 \\choose 5} \\cdot 5! = 47040$$\n\nI agree with your analysis. I used an alternative approach:\n\n$$S = ~$$total number of ways, without regard to Alex or Bob.\n\n$$T = ~$$total number of ways, Alex and Bob are together.\n\nSince you must have $$3$$ empty seats, I will pretend that these seats are filled by (indistinguishable) ghosts.\n\n$$\\displaystyle S = \\frac{9!}{3!} = 60480$$.\n\nAbove, the numerator, which represents permuting $$(-1) +$$ the number of units, is appropriate for a circular table (as opposed to a straight row). That is, any of the $$(10)$$ units may be regarded as the head of the table.\n\nAbove, the denominator represents that the ghosts (i.e. empty chairs) are indistinguishable. So, the denominator represents an overcounting adjustment factor.\n\n$$\\displaystyle T = \\frac{8! \\times 2!}{3!} = 13440.$$\n\nAbove, the numerator represents that there are only $$8$$ units, since Alex and Bob are together. However, inside the Alex-Bob unit, Alex and Bob can be permuted in $$(2!)$$ ways.\n\n$$S - T = 47040.$$", "date": "2023-01-27 12:24:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4323652/how-many-ways-are-there-to-place-7-people-to-10-seats-around-a-circular-tabl", "openwebmath_score": 0.8305421471595764, "openwebmath_perplexity": 299.194511678611, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864278996301, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8529111379731975}}
{"url": "https://byjus.com/question-answer/two-blocks-each-of-mass-m-3-50-kg-are-hung-from-the-ceiling-of/", "text": "Question\n\n# Two blocks, each of mass $$m = 3.50 kg$$, are hung from the ceiling of an elevator as in above figure. (a)\u00a0If the elevator moves with an upward acceleration $$\\vec{a}$$ of magnitude $$1.60 m/s^{2}$$, find the tensions $$T_{1}$$ and $$T_{2}$$ in the upper and lower strings. (b) If the strings can withstand a maximum tension of $$85.0 N$$, what maximum acceleration can the elevator have before a string breaks.\n\nSolution\n\n## (a) Free-body diagrams of the two blocks are shown below. Note that each block experiences adownward gravitational force$$F_{g} = (3.50 kg)(9.80 m/s^{2} ) = 34.3 N$$Also, each has the same upward acceleration as the elevator, in this case$$a_{y} = +1.60 m/s^{2}$$.Applying Newton\u2019s second law to the lower block:$$\\sum F_{y} = ma_{y} \\Rightarrow T_{2} \u2212 F_{g} = ma_{y}$$or$$T_{2} = F_{g} + ma_{y} = 34.3 N + (3.50 kg)(1.60 m/s^{2} ) = 39.9 N$$Next, applying Newton\u2019s second law to the upper block:$$\\sum F_{y} = ma_{y} \\Rightarrow T_{1} \u2212T_{2} \u2212 F_{g} = ma_{y}$$or$$T_{1} = T_{2} + F_{g} + ma_{y} = 39.9 N + 34.3 N + (3.50 kg)(1.60 m/s^{2} )$$$$= 79.8N$$(b) Note that the tension is greater in the upper string, and this string will break first as the acceleration of the system increases. Thus, we wish to find the value of $$a_{y}$$ when $$T_{1} = 85.0$$. Making use of the general relationships derived in (a) above gives:$$T_{1} = T_{2} + F_{g} + ma_{y} = (F_{g} + ma_{y})+ F_{g} + ma_{y} = 2F_{g} + 2ma_{y}$$or$$a_{y}=\\frac{T_{1}-2F_{g}}{2m}=\\frac{85.0N-2(34.3N)}{2(3.50kg)}=2.34m/s^{2}$$.Physics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-16 10:53:18", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/two-blocks-each-of-mass-m-3-50-kg-are-hung-from-the-ceiling-of/", "openwebmath_score": 0.6691755056381226, "openwebmath_perplexity": 566.8407061118389, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.989986430558584, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8529111297054182}}
{"url": "https://math.stackexchange.com/questions/166380/not-every-metric-is-induced-from-a-norm", "text": "# Not every metric is induced from a norm\n\nI have studied that every normed space $(V, \\lVert\\cdot \\lVert)$ is a metric space with respect to distance function\n\n$d(u,v) = \\lVert u - v \\rVert$, $u,v \\in V$.\n\nMy question is whether every metric on a linear space can be induced by norm? I know answer is no but I need proper justification.\n\nEdit: Is there any method to check whether a given metric space is induced by norm ?\n\nThanks for help\n\n\u2022 $\\LaTeX$ tip: \\parallel is a relation symbol, so it includes space on both sides. You want \\lVert and \\rVert for left and right delimiters, so that there is space on the \"outside\", but not on the \"inside\". Jul 4 '12 at 2:33\n\u2022 @ArturoMagidin Thank you very much sir. Jul 4 '12 at 2:35\n\u2022 Jul 4 '12 at 5:07\n\u2022 @MattN. Thank you very much. That was helpful to me. Jul 4 '12 at 5:14\n\nLet $V$ be a vector space over the field $\\mathbb{F}$. A norm $$\\| \\cdot \\|: V \\longrightarrow \\mathbb{F}$$ on $V$ satisfies the homogeneity condition $$\\|ax\\| = |a| \\cdot \\|x\\|$$ for all $a \\in \\mathbb{F}$ and $x \\in V$. So the metric $$d: V \\times V \\longrightarrow \\mathbb{F},$$ $$d(x,y) = \\|x - y\\|$$ defined by the norm is such that $$d(ax,ay) = \\|ax - ay\\| = |a| \\cdot \\|x - y\\| = |a| d(x,y)$$ for all $a \\in \\mathbb{F}$ and $x,y \\in V$. This property is not satisfied by general metrics. For example, let $\\delta$ be the discrete metric $$\\delta(x,y) = \\begin{cases} 1, & x \\neq y, \\\\ 0, & x = y. \\end{cases}$$ Then $\\delta$ clearly does not satisfy the homogeneity property of the a metric induced by a norm.\n\nTo answer your edit, call a metric $$d: V \\times V \\longrightarrow \\mathbb{F}$$ homogeneous if $$d(ax, ay) = |a| d(x,y)$$ for all $a \\in \\mathbb{F}$ and $x,y \\in V$, and translation invariant if $$d(x + z, y + z) = d(x,y)$$ for all $x, y, z \\in V$. Then a homogeneous, translation invariant metric $d$ can be used to define a norm $\\| \\cdot \\|$ by $$\\|x\\| = d(x,0)$$ for all $x \\in V$.\n\n\u2022 Thank you very much. I understand now. Jul 4 '12 at 2:42\n\u2022 That is not enough as one needs to check that the so defined norm really induces the original metric and not another one: $d(\\cdot,\\cdot)\\to\\|\\cdot\\|\\to d(\\cdot,\\cdot)$ Sep 29 '14 at 14:36\n\u2022 This is an old post and it doesn't really matter much, but I have to be pedantic. So you say that if a metric comes from a norm on $\\Bbb{R}$, then $d(ax,ay)=|a|d(x,y)$ but what if you just consider $\\Bbb{R}$ as a vector space over $\\{0,1\\}$. Then the discrete metric IS homogeneous and translation invariant on this vector space.\n\u2013\u00a0user223391\nJun 16 '15 at 0:16\n\u2022 @MichaelMunta If the metric is induced by a norm, then $d(x,y)=||x-y||$. So $d(x+z,y+z)=||(x+z)-(y+z)||=||x-y||=d(x,y)$. Feb 27 '20 at 15:30\n\u2022 You should clarify that your field $F$ is normed, otherwise the notation $|a|$ is meaningless. Jun 5 '20 at 23:24\n\nHere is another interesting example: Let $|x-y|$ denote the usual Euclidean distance between two real numbers $x$ and $y$. Let $d(x,y)=\\min\\{|x-y|,1\\}$, the standard derived bounded metric. Now suppose we look at $\\Bbb{R}$ as a vector space over itself and ask whether $d$ comes from any norm on $\\Bbb{R}$. Then if there is such a norm say $||.||$, we must have the homogeneity condition: for any $\\alpha \\in \\Bbb{R}$ and any $v \\in \\Bbb{R}$,\n\n$$||\\alpha v || = |\\alpha| ||v||.$$\n\nBut now we have a problem: The metric $d$ is obviously bounded by $1$, but we can take $\\alpha$ arbitrarily large so that $||.||$ is unbounded. It follows that $d$ does not come from any norm.\n\n\u2022 It was nice explanation to me. I had forgot to thank you. :) May 25 '13 at 9:21\n\nAs Henry states above, metrics induced by a norm must be homogeneous. You can see that they must also be translation invariant: $d(x+a,y+a)= d(x,y).$ So any metric not satisfying either of those can not come from a norm.\n\nOn the other hand, it turns out that these two conditions on the metric are sufficient to define a norm that induces that metric: $d(x,0)=\\| x \\|.$\n\n\u2022 Thank you sir. I am fully satisfied with two answers given by you and Henry.:) Jul 4 '12 at 2:44\n\u2022 Did you mean instead that any metric not satisfying either of those can not come from a norm? And not the other way around? And also, these two conditions on the metric are sufficient to define a norm..? Jul 4 '12 at 6:27\n\u2022 @ThomasE. Sorry, you are right, I muddled up the words. Thanks. Jul 4 '12 at 8:43\n\nEvery homogeneous metric induces a norm via: $$\\|x\\|:=d(x,0)$$ and every norm induces a homogeneous and translation-invariant metric: $$d(x,y):=\\|x-y\\|$$\n\nThe clue herein lies in wether the induced norm really represents the metric as: $$d(\\cdot,\\cdot)\\to\\|\\cdot\\|\\to d(\\cdot,\\cdot)$$ which is the case only for the translation-invariant metrics: $$d'(x,y)=d(x-y,0)=d(x,y)$$ whereas the induced metric always represents the norm as: $$\\|\\cdot\\|\\to d(\\cdot,\\cdot)\\to\\|\\cdot\\|$$ as a simple check shows: $$\\|x\\|'=\\|x-0\\|=\\|x\\|$$\n\n\u2022 @ Thanks for the answer. Feb 15 '16 at 4:15\n\nPersonal Note:\n\nGood question, I remember wondering the same thing myself back when I new to real analysis. Here's a counter example:\n\nCounter Example\n\n$$d(x,y):=I_{\\{(x,x)\\}}(x,y):=\\begin{cases} 1 &:\\, \\,x=y\\\\ 0 &:\\, \\,x\\neq y. \\end{cases}$$ This is indeed a metric (check as an exercise).\n\nWhere $I_A$ is the indicator function of the set $A$; where here the set $(x,x)\\in V\\times V$.\n\nIf $r\\in \\mathbb{R}-\\{0\\}$, then $d(rx,ry)\\in \\{0,1\\}$ hence $rd(x,y)\\in \\{0,r\\}$ which is not in the range of $d:V\\times V \\rightarrow \\mathbb{R}$.\n\nSo, the metric $d$ fails the property that any metric induced by a norm must have, ie: $$\\mbox{it fails to have the property that: } \\|rx-ry\\| =r\\|x-y\\|.$$\n\nInterpretation & Some Intuition:\n\nThe problem is that the topology is too fine, so all this topology can do is distinguish between things being the same or different.\nAs opposed to a norm topology which distinguised between object being, or not being on the same line (for some appropriate notion of line) (as well as them being different).\n\nHope this helps :)\n\n\u2022 Thank you for the explanation. :) Feb 15 '16 at 4:15\n\u2022 Hi, is it correct to consider the metric d(x,y)=|x-y|^2018 that fails to h;ave the above property? Oct 23 '18 at 8:19\n\u2022 Yes, why not? :) Oct 23 '18 at 15:33\n\nOne possible way of showing that a metric does not arise from a norm is to show that it is bounded, as it then cannot be homogeneous.\n\nProof: Take $$d$$ a bounded metric on a space $$X$$: i.e. $$\\exists D \\in \\mathbb{R}_+$$ such that $$\\forall (x,y) \\in X^2, d(x,y) \\leq D$$. Suppose now for contradiction that $$d$$ arises from a norm, i.e. the exists a norm $$\\|\\cdot\\|$$ such that $$d(x,y) = \\|x - y\\|$$. Recall that the distance must then must be homogeneous, for we have $$d(\\lambda x, \\lambda y) = \\|\\lambda x - \\lambda y\\| = |\\lambda| \\cdot \\|x - y\\| = |\\lambda| d(x,y)$$.\n\nTake now arbitrary $$(x_0, y_0) \\in X^2$$, then we must have that $$d\\left( \\frac{D+1}{d(x_0, y_0)} \\cdot x_0, \\frac{D+1}{d(x_0, y_0)} \\cdot y_0 \\right) = \\frac{D+1}{d(x_0, y_0)} \\cdot d(x_0, y_0) = D + 1 > D$$ which contradicts the upper bound of the metric.", "date": "2021-09-25 03:32:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/166380/not-every-metric-is-induced-from-a-norm", "openwebmath_score": 0.9562453627586365, "openwebmath_perplexity": 248.9525406336052, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9777138196557983, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8529060902528371}}
{"url": "https://www.physicsforums.com/threads/a-question-about-greatest-common-factor-gcf.911521/", "text": "# B A question about Greatest common factor (GCF) ?\n\n#### awholenumber\n\nWhen we try to find the Greatest common factor (GCF) of two numbers , does it only involve prime factorization ?\n\n#### FactChecker\n\nGold Member\n2018 Award\nYes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.\n\n#### awholenumber\n\nOk , so the same prime factorization is used to find the LCM too , right ?\n\n#### FactChecker\n\nGold Member\n2018 Award\nOk , so the same prime factorization is used to find the LCM too , right ?\nYes.\n\nOk , Thanks\n\n#### SlowThinker\n\nI dare to disagree, Euclid's algorithm or binary GCD are used to find GCD, then LCM(a, b)=a*b/GCD(a, b).\n\n#### FactChecker\n\nGold Member\n2018 Award\nI stand corrected. I was not familiar with these algorithms. The binary GCD algorithm and Euclidean_algorithm are interesting.\n\n#### awholenumber\n\nOk, i have one more doubt .\nThe GCF of two numbers involves prime factorization of those two numbers , then multiply those factors both numbers have in common\n\nIsnt LCM about finding the least common multiple of two numbers ?\n\nWhat does that have anything to do with prime numbers ?\n\n#### mfb\n\nMentor\nYou can find the LCM via LCM(a, b)=a*b/GCD(a, b), if you have the GCD first.\nThat is often the most practical way to find the LCM.\n\nPrime factorization is just one possible way to find the GCD. For large numbers, it can be very time-consuming, and different algorithms can be more efficient.\n\n#### awholenumber\n\nThanks for the information mfb , i am just trying to cover the algebra 1 for dummies book .\nIt doesn't have a method like this LCM(a, b)=a*b/GCD(a, b) mentioned in it .\n\n#### PeroK\n\nHomework Helper\nGold Member\n2018 Award\nThanks for the information mfb , i am just trying to cover the algebra 1 for dummies book .\nIt doesn't have a method like this LCM(a, b)=a*b/GCD(a, b) mentioned in it .\nYou'll find that in an \"algebra for intelligent students\" textbook!\n\n#### awholenumber\n\nlol ok , first let me somehow finish this one book properly\n\n#### PeroK\n\nHomework Helper\nGold Member\n2018 Award\nlol ok , first let me somehow finish this one book properly\nIt's still worth understanding why the two are related. The basic argument is:\n\n$a = a'g, b = b'g$\n\nWhere $g$ is the GCD of $a$ and $b$ and $a', b'$ are how much bigger $a$ and $b$ are than $g$.\n\nFor example: if $a = 42$ and $b = 15$, then $g = 3$ and $a=14 \\times 3, b = 5 \\times 3$, hence $a' = 14, b' = 5$\n\nNow, the LCM of $a$ and $b$ must have as factors simply $a', b'$ and $g$, so $l = a'b'g = a'b = a'gb/g = ab/g$\n\nYou can now see that the LCM is the product of $a$ and $b$, divided by the GCD.\n\nIn our example:\n\n$LCM(42, 15) = \\frac{42 \\times 15}{3} = 210$\n\nmfb\n\n#### awholenumber\n\nThanks for sharing , i will keep this in my mind and maybe someday i will be able to use this advanced method .\n\n#### Michael Hardy\n\nYes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.\nBut you don't need to know anything about prime factorizations to find the GCD; you can use Euclid's algorithm, which is very efficient.\n\n#### Michael Hardy\n\nYou don't need prime factorizations to find GCDs. There is an efficient method not requiring any knowledge of prime numbers: Euclid's algorithm. This is the oldest algorithm still in standard use, dating back to Euclid's writings in the 3rd century BC, and it is very efficient.\n\\begin{align*} \\gcd(1989,867) & = \\gcd(255,867) & & \\text{since 255 is the remainder} \\\\ & & & \\text{when 1989 is divided by 867} \\\\[10pt] & = \\gcd(255,102) & & \\text{since 102 is the remainder} \\\\ & & & \\text{when 867 is divided by 255} \\\\[10pt] & = \\gcd(51,102) & & \\text{since 51 is the remainder} \\\\ & & & \\text{when 255 is divided by 102} \\\\[10pt] & = \\gcd(51,0) & & \\text{since 0 is the remainder} \\\\ & & & \\text{when 102 is divided by 51} \\\\[10pt] & = 51. \\end{align*}\nThus we have $\\dfrac{1989}{867} = \\dfrac{51\\times39}{51\\times17} = \\dfrac{39}{17}.$\n\n#### Michael Hardy\n\nYes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.\nThis is wrong. Euclid's very efficient algorithm for finding GCDs does not require doing anything at all with prime numbers.\n\n#### mfb\n\nMentor\nBut you don't need to know anything about prime factorizations to find the GCD; you can use Euclid's algorithm, which is very efficient.\nYes, that has been mentioned in post 6 for example.\n\nThis thread is from 2017, by the way.\n\n\"A question about Greatest common factor (GCF) ?\"\n\n### Physics Forums Values\n\nWe Value Quality\n\u2022 Topics based on mainstream science\n\u2022 Proper English grammar and spelling\nWe Value Civility\n\u2022 Positive and compassionate attitudes\n\u2022 Patience while debating\nWe Value Productivity\n\u2022 Disciplined to remain on-topic\n\u2022 Recognition of own weaknesses\n\u2022 Solo and co-op problem solving", "date": "2019-07-17 17:17:17", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-question-about-greatest-common-factor-gcf.911521/", "openwebmath_score": 0.9323726892471313, "openwebmath_perplexity": 911.8310516881007, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138118632622, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8529060883221615}}
{"url": "https://dsp.stackexchange.com/questions/38675/how-to-plot-magnitude-and-phase-response-of-2-cascaded-filters-in-matlab/38677", "text": "# How to plot magnitude and phase response of 2 cascaded filters in Matlab?\n\nHow would I go about plotting a magnitude and phase response of a system that consists of two cascaded 2nd order Butterworth filters in Matlab? Filters are the same.\n\n% Filter coeffs:\n[B, A] = butter( 2, 1000/8000*2, 'high' ); % Cutoff at 1000 Hz, fs 8000 Hz.\n\n% Cascade 2 filters: input -> filter -> filter -> output :\n[out1, state1 ] = filter( B, A, in,   state1);\n[out2, state2 ] = filter( B, A, out1, state2);\n\n\nUsing MATLAB/Octave as the tool, the following approach lets you plot the magnitude & phase samples of the DTFT of the cascade of the two discrete-time LTI filters using their LCCDE coefficient vectors $$b[k]$$, and $$a[k]$$, assuming they have LCCDE representations..\n\nSince the cascade LTI system is described as: $$h[n] = h_1[n] \\star h_2[n]$$ and consequently $$H(z) = H_1(z) H_2(z)$$\n\nWhere the $$h[n]$$ is the impulse response of the cascade system and $$H(z)$$ is its transfer function (Z-transform of impulse response). Those $$h_1[n]$$ and $$h_2[n]$$ refer to the impulse responses of the individual systems that make up the cascade.\n\nAssuming that individual systems do have rational Z-transforms, then we have the following expression for the composite (cascade) system Z-transfom:\n\\begin{align} H(z) &=\\frac{ \\sum_{k=0}^{M} B[k]z^{-k} } {\\sum_{k=0}^{N} A[k]z^{-k}} = H_1(z) H_2(z) \\\\ \\\\ &= \\left( \\frac{ \\sum_{k=0}^{M_1} b_1[k]z^{-k} } {\\sum_{k=0}^{N_1} a_1[k]z^{-k}} \\right) \\left( \\frac{ \\sum_{k=0}^{M_2} b_2[k]z^{-k} } {\\sum_{k=0}^{N_2} a_1[k]z^{-k}} \\right) \\end{align}\n\nWhere individual coefficients $$b_1[k],a_1[k],b_2[k],a_2[k]$$ are those that represent the cascaded systems.\n\nWe want $$B[k]$$ and $$A[k]$$ that represent the cascade system, from which we can use the following Matlab/Octave function to plot the DTFT magnitude and phase responses, assumimg that the cascade is a stable LTI system so that it will have a frequency response function:\n\nfigure,freqz(B,A);\n\n\nNow it can be shown by polynomial manipulations that the coefficients $$B[k]$$ and$$A[k]$$ are related to the individual coefficients $$b_1[k],a_1[k],b_2[k],a_2[k]$$ as the following: $$B[k] = b_1[k] \\star b_2[k]$$ and $$A[k] = a_1[k] \\star a_2[k]$$\n\nHence the required frequency response plot can be obtained in the combined call as follows:\n\nfigure,freqz(conv(b1,b2),conv(a1,a2));\n\n\nwhere b1,a1 and b2,a2 are the coefficient vectors that represent the individual systems. The method can be generalised to N systems in cascade.\n\nWhen you pass a signal from two cascaded filters, what happens is that the magnitude response of the whole chain is the product of individual filters, and the phase response is the sum of individual phase responses. This is because the transfer functions are multiplied in the cascade implementation.\n\nSo if the frequency response of the single filters are $$H_A(w)=|H_A(w)|e^{j\\angle H_A(w)}$$ $$H_B(w)=|H_B(w)|e^{j\\angle H_B(w)}$$ then the frequency response of the two cascaded filters (overall) is \\begin{align} H_{AB}(w)&=H_A(w)H_B(w)\\\\ &=\\left(|H_A(w)|e^{j\\angle H_A(w)}\\right)\\left(|H_B(w)|e^{j\\angle H_B(w)}\\right)\\\\ &=|H_A(w)||H_B(w)|e^{j(\\angle H_A(w)+\\angle H_B(w))} \\end{align}\n\nYou can use [h,w] = freqz(b,a) in Matlab to get the frequency response of your desired filters. Use abs and angle to find the magnitude and phase:\n\n...\n[hA,w] = freqz(bA,aA);\n[hB,w] = freqz(bB,aB);\nhAB = hA.*hB;\nMagResp = 20*log10(abs(hAB));\nPhaseResp = angle(hAB);\nplot(w,MagResp)\n...\n\n\nwhen the two filters are identical, then the overall magnitude is squared and the overall phase is scaled by two: \\begin{align} H_{AA}(w)&=\\left(|H_A(w)|e^{j\\angle H_A(w)}\\right)\\left(|H_A(w)|e^{j\\angle H_A(w)}\\right)\\\\ &=|H_A(w)|^2e^{j2\\angle H_A(w)} \\end{align}\n\n\u2022 Yes, filter is the same. \u2013\u00a0Danijel Mar 27 '17 at 11:51\n\u2022 Yes, I added the general case at the beginning, then the identical case at the end. \u2013\u00a0msm Mar 27 '17 at 12:09\n\u2022 Shouldn't it be wAB = w + w, and then plot(wAB)? Is angle(hAB) necessary? \u2013\u00a0Danijel Mar 27 '17 at 12:43\n\u2022 w is the frequency axis and remains fix. We should just add function values at given w values. \u2013\u00a0msm Mar 27 '17 at 20:03\n\u2022 angle(hAB) gives you the phase of the complex array hAB. Without that, you just have a bunch of complex values of size legth(w). \u2013\u00a0msm Mar 28 '17 at 1:42", "date": "2020-06-03 14:01:33", "meta": {"domain": "stackexchange.com", "url": "https://dsp.stackexchange.com/questions/38675/how-to-plot-magnitude-and-phase-response-of-2-cascaded-filters-in-matlab/38677", "openwebmath_score": 0.9335691928863525, "openwebmath_perplexity": 1647.3619329906733, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9777138170582865, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.852906084742156}}
{"url": "http://math.stackexchange.com/questions/133453/complex-integration-with-cauchys-integral-formula", "text": "# Complex integration with Cauchy's Integral Formula\n\nCalculate$$\\int_\\gamma \\frac{(z+27i)(z+16)}{z(z+81)^2}dz$$ where $\\gamma$ is the triangle whose vertices are the third roots of $z = -8i$, oriented counterclockwise.\n\nAnswer:\n\nI calculated the third roots of $-8i$ and they all have modulus $2$. This tells me that the maximum distance of $\\gamma$ from the origin will be $2$.\n\nThere are singularities at $z=0, z=-81$. As $81 > 2$, this singularity falls outside $\\gamma$ so the only one that matters is $z = 0.$\n\nI then applied Cauchy's Integral Formula $$\\int_\\gamma \\frac{(z+27i)(z+16)}{z(z+81)^2}dz = 2\\pi i [\\frac{(z+27i)(z+16)}{(z+81)^2}] |_{z=0}$$\n\nAnd I got a final result of $\\displaystyle\\frac{-32\\pi}{243}$.\n\nIs my analysis and final result correct?\n\n-\nYep, completely spot on! Only thing I can see you could have improved slightly was to notice the cube roots of $-8i$ should have modulus 2 without calculating them, saves you some work. \u2013\u00a0 Ragib Zaman Apr 18 '12 at 14:00\n(+1) for showing work. \u2013\u00a0 The Chaz 2.0 Apr 18 '12 at 14:01\nYou should copy this to an answer and accept it :) \u2013\u00a0 Neal Apr 18 '12 at 15:34\n@Jim_CS If you require further feedback it might help to elaborate more specifically on any doubts. If not, please post an answer and accept it so that we know it is resolved. \u2013\u00a0 Bill Dubuque May 17 '12 at 15:54\n\n## 2 Answers\n\nAnswering my own question as Neal advised me to.\n\nI calculated the third roots of \u22128i and they all have modulus 2. This tells me that the maximum distance of \u03b3 from the origin will be 2.\n\nThere are singularities at z=0,z=\u221281. As 81>2, this singularity falls outside \u03b3 so the only one that matters is z=0. I then applied Cauchy's Integral Formula \u222b\u03b3(z+27i)(z+16)z(z+81)2dz=2\u03c0i[(z+27i)(z+16)(z+81)2]|z=0 And I got a final result of \u221232\u03c0243.\n\nIs my analysis and final result correct?\n\n-\n\nNothing wrong with your analysis and answer.\n\n-", "date": "2015-10-10 04:28:07", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/133453/complex-integration-with-cauchys-integral-formula", "openwebmath_score": 0.8994753360748291, "openwebmath_perplexity": 545.0243998549793, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138131620184, "lm_q2_score": 0.8723473796562745, "lm_q1q2_score": 0.8529060829656312}}
{"url": "https://www.intmath.com/forum/series-expansion-33/taylor-s-series-question:75", "text": "IntMath Home \u00bb Forum home \u00bb Infinite Series Expansion \u00bb Taylor's series question\n\n# Taylor's series question [Solved!]\n\n### My question\n\nmy question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p\n\n1. Taylor Series\n\n### What I've done so far\n\nI think I need to differentiate it several times.\n\nHere it is:\n\nf(x) = x^(1/2)\n\nf'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))\n\nf''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))\n\nf'''(x) = 3/8 x^(-5/2) = 3/(8 x^(5/2))\n\nBut I'm stuck then\n\nX\n\nmy question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p\nRelevant page\n\n<a href=\"/series-expansion/1-taylor-series.php\">1. Taylor Series</a>\n\nWhat I've done so far\n\nI think I need to differentiate it several times.\n\nHere it is:\n\nf(x) = x^(1/2)\n\nf'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))\n\nf''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))\n\nf'''(x) =  3/8 x^(-5/2) = 3/(8 x^(5/2))\n\nBut I'm stuck then\n\n## Re: Taylor's series question\n\nHello Snowboy\n\nYour problem is very similar to the example on the page you came from:\n\n1. Taylor Series\n\nNow that you've differentiated, you need to substitute the necessary values, just like I did in that example.\n\nOr is you problem with the square root approximation?\n\nRegards\n\nX\n\nHello Snowboy\n\nYour problem is very similar to the example on the page you came from:\n\n<a href=\"/series-expansion/1-taylor-series.php\">1. Taylor Series</a>\n\nNow that you've differentiated, you need to substitute the necessary values, just like I did in that example.\n\nOr is you problem with the square root approximation?\n\nRegards\n\n## Re: Taylor's series question\n\nSo a=1, right?\n\nf(1) = 1\n\nf'(1) = 1/2\n\nf''(1) = -1/4\n\nf'''(1) = 3/8\n\nPlugging it into the formula:\n\n1 + 1/2 (x-1)  -1/4(x-1)^2  + 3/8(x-1)^3-...\n\nHow many steps do we have to do?\n\nX\n\nSo a=1, right?\n\nf(1) = 1\n\nf'(1) = 1/2\n\nf''(1) = -1/4\n\nf'''(1) = 3/8\n\nPlugging it into the formula:\n\n1 + 1/2 (x-1)  -1/4(x-1)^2  + 3/8(x-1)^3-...\n\nHow many steps do we have to do?\n\n## Re: Taylor's series question\n\nThe question says you need 4 d.p. accuracy.\n\nSo you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.\n\nX\n\nThe question says you need 4 d.p. accuracy.\n\nSo you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.\n\n## Re: Taylor's series question\n\nIt says \"at 1.1\".\n\nIs this OK?\n\nsqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3\n\n = 1 + 0.05 - 0.0025 + 0.000375  = 1.047875\n\nI can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.\n\nMaybe I should do another step to make sure.\n\nf''''(x) = -15/16 x^(-7/2)\n\nf''''(1) = -15/16\n\nSo the 5th term is -15/16(x-1)^4\n\nPlugging in 1.1 gives -0.00009375\n\nAdding this gives 1.047875-0.00009375 = 1.04778125\n\nTo 4 d.p. it's still 1.0478.\n\nBut checking on my calculator, I get sqrt(1.1) = 1.04880884817\n\nTo 4 d.p., that's 1.0488. Is my answer wrong?\n\nX\n\nIt says \"at 1.1\".\n\nIs this OK?\n\nsqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3\n\n= 1 + 0.05 - 0.0025 + 0.000375  = 1.047875\n\nI can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.\n\nMaybe I should do another step to make sure.\n\nf''''(x) = -15/16 x^(-7/2)\n\nf''''(1) = -15/16\n\nSo the 5th term is -15/16(x-1)^4\n\nPlugging in 1.1 gives -0.00009375\n\nAdding this gives 1.047875-0.00009375 = 1.04778125\n\nTo 4 d.p. it's still 1.0478.\n\nBut checking on my calculator, I get sqrt(1.1) = 1.04880884817\n\nTo 4 d.p., that's 1.0488. Is my answer wrong?\n\n## Re: Taylor's series question\n\nYour thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.\n\nOur Taylor's Series approximation is only \"good\" right near the number we expand about (in your case, x=1).\n\nTo see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:\n\nThe green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.\n\nYou can see it's not a very good approximation as we get away from x=1.\n\nX\n\nYour thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.\n\nOur Taylor's Series approximation is only \"good\" right near the number we expand about (in your case, x=1).\n\nTo see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:\n\n[graph]310,250;-0.3,2.5;-0.5,2,0.5,0.5;sqrt(x),1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3[/graph]\n\nThe green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.\n\nYou can see it's not a very good approximation as we get away from x=1.\n\n## Re: Taylor's series question\n\nThanks a lot!\n\nX\n\nThanks a lot!", "date": "2017-10-17 09:40:46", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/series-expansion-33/taylor-s-series-question:75", "openwebmath_score": 0.8084568977355957, "openwebmath_perplexity": 2435.200394965935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138144607744, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8529060678748535}}
{"url": "https://www.physicsforums.com/threads/linear-second-order-differential-equation.534457/", "text": "# Linear, second-order differential equation?\n\n1. Sep 27, 2011\n\n### Raen\n\nOne of my homework problems this week was:\n\nVerify that y = sin(4t) + 2cos(4t) is a solution to the following initial value problem.\n\n2y'' + 32y=0; y(0)=2, y'(0)=4\n\nFind the maximum of |y(t)| for -infinity< t < infinity.\n\nVerifying that the given y equation is a solution is easy, all there is to do is derive twice, plug in the new equations, and show that 0 = 0. I have no trouble with that. My question is with the second part, about finding the maximum positive value of y(t).\n\nI would think that no matter how large t gets, sin and cos will always oscillate between -1 and 1. When cos is 1, obviously sin is 0. Putting in sin(4t) = 0 and cos(4t) = 1 gives me a maximum value of 2 for any y(t) as t approaches infinity. I graphed the function on a calculator to confirm this, just in case, and the graph appears to agree with me.\n\nMy confusion is that 2 isn't the correct answer. According to the website we do our homework on, the answer should be sqrt(5). There are no questions like this worked out in the book and I can't find anything online. Can anybody explain to me why y(t) has a maximum value of sqrt(5) as t approaches infinity?\n\nThank you!\n\n2. Sep 27, 2011\n\n### LCKurtz\n\nGenerally, when you have something like acos(\u03b8) + bsin(\u03b8) you can write it as a single trig function. You multiply and divide by the square root of the sum of the squares of the coefficients:\n\n$$a\\cos\\theta + b\\sin\\theta = \\sqrt{a^2+b^2} \\left(\\frac a {\\sqrt{a^2+b^2}}\\cos\\theta + \\frac b {\\sqrt{a^2+b^2}}\\sin\\theta\\right)$$\n\nNow, since the sum of the squares of the new coefficients of the sine and cosine add up to one, they can be used as the sin(\u03b2) and cos(\u03b2) of some angle \u03b2. You then have an addition formula for a sine or cosine function with amplitude $\\sqrt{a^2+b^2}$. In your example that gives amplitude $\\sqrt 5$.\n\n3. Sep 28, 2011\n\n### HallsofIvy\n\nStaff Emeritus\nLCKurtz's response is, as always, excellent.\n\nRaen, you seem to be assuming that a maximum of f+ g must occur where either f or g has a maximum (since you looked immediately at cos(x)= 1) and that is not correct.\n\nAnother way to find a maximum or minimum for a function is (as I am sure you know) is to look where the derivative is 0. If f(x)= sin(4x)+ 2cos(4x), then f'(x)= 4cos(4x)- 8sin(4x)= 0 when 4 cos(4x)= 8 sin(4x) or cos(4x)= 2sin(4x). Rather than solve for x directly, note that $cos^2(4x)= 4 sin^2(4x)$ so $sin^2(4x)+ cos^2(4x)= 5 sin^2(4x)= 1$ so that $sin^2(4x)= 1/5$ and $sin(4x)= 1/\\sqrt{5}= \\sqrt{5}/5$. And, of course, $cos^2(4x)= 1- sin^2(x)= 4/5$ so $cos(4x)= 2/\\sqrt{5}= 2\\sqrt{5}/5$.\n\nPutting those back into the original formula, $sin(4x)+ 2cos(4x)= \\sqrt{5}/5+ 4\\sqrt{5}/5= 5/\\sqrt{5}= \\sqrt{5}$\n\nLast edited: Sep 28, 2011\n4. Sep 28, 2011\n\n### Raen\n\nOh, I think I understand now. Thank you both for your detailed explanations!", "date": "2017-09-23 16:57:02", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/linear-second-order-differential-equation.534457/", "openwebmath_score": 0.8633628487586975, "openwebmath_perplexity": 503.6547176064232, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676472509722, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8528959908015215}}
{"url": "https://math.stackexchange.com/questions/2380618/gaussian-elimination-inconsistent-or-infinite-solutions", "text": "# Gaussian elimination, inconsistent or infinite solutions?\n\nI have this system\n\n$\\left[ \\begin{array}{ccc|c} -3&0&-1&0\\\\ 2&0&0&0\\\\ 2&0&0&0\\\\ \\end{array} \\right]$ Is there something you can tell from looking at this straight away?\n\nWhen I reduce it I get this:\n\n$\\left[ \\begin{array}{ccc|c} 1&0&-1&0\\\\ 0&0&1&0\\\\ 0&0&0&0\\\\ \\end{array} \\right]$\n\nI thought I read that if you get a statement that does not make sense, then the system is inconsistent, so $1 = 0$ does not make sense to me. Does that not mean it's inconsistent?\nHowever, I thought that $x_3$ could equal $0$, and so could $x_1$ and basically, $x_2$ could equal anything. Still not sure, i reduced it further to:\n$\\left[ \\begin{array}{ccc|c} 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&0\\\\ \\end{array} \\right]$\n\nFrom this, am I correct now in thinking that:\n$x_1 = r\\\\ x_2 = s \\\\ x_3 = t$\n\nI am not sure of why I'm thinking like this, in terms of free variables etc. If I'm right, I wouldn't mind some reasoning as to why this is so, or being correct if I am wrong. Thanks.\n\n\u2022 How did you do the final step of reduction? \u2013\u00a0platty Aug 2 '17 at 23:17\n\u2022 Opps, I see where you are going with this. The 1 should remain in the second row. Sorry about that. Let me think for a minute. \u2013\u00a0Bucephalus Aug 2 '17 at 23:20\n\u2022 Now that I have corrected my last reduction, I'm still not really sure what this means. it looks like $x_2$ can equal anything, Therefore, $x_1$ and $x_3$ can equal anything? Is that correct? \u2013\u00a0Bucephalus Aug 2 '17 at 23:24\n\nRemember that the entries on the left are coefficients of variables. So from your final reduction, we would end up with the RREF: $$\\left[\\begin{array}{ccc|c} 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&0\\\\ \\end{array}\\right]$$\n\nThe first line tells us that $1x + 0y + 0z = 0$, so $x = 0$. The second line tells us that $0x + 0y + 1z = 0$, i.e. $z=0$. But we have $y$ as a free variable, so our solution is $(0,y,0)$.\n\nNote that we can tell that $y$ is free by observing the original matrix - there is no coefficient in its column, so its value doesn't affect our final solution.\n\n\u2022 Oh, yes I see now. Thanks for your explanation @S.Ong. \u2013\u00a0Bucephalus Aug 2 '17 at 23:27\n\u2022 Say it did just turn out to be like I originally had it, with a single 1 in the first column of the first row. That would then mean that $x_1=0, x_2=r$ and $x_3=s$. Would that be right? \u2013\u00a0Bucephalus Aug 2 '17 at 23:29\n\u2022 Yep, that's it! :) \u2013\u00a0platty Aug 2 '17 at 23:32\n\u2022 Ok, thankyou @S.Ong. \u2013\u00a0Bucephalus Aug 2 '17 at 23:32\n\n$\\left[ \\begin{array}{ccc|c} -3&0&-1&0\\\\ 2&0&0&0\\\\ 2&0&0&0\\\\ \\end{array} \\right]$\n\nWhat do you know straight away?\n\n$(0,0,0)$ is in the solution set. The right column being all $0$ tells you this.\n\nWith the second row and the 3rd row identical or the right part of the matrix, this matrix is singular.\n\nIn fact the rank is 2, the nullity is 1, and there is a \"line\" of solutions.\n\nThe second column all zero's tells you the very same thing. Furthermore, $x_2$ can be anything.\n\nAnd the structure of rows 2 and 3 tells you that $x_1$ must equal $0.$\n\nthis is more than enough information to let you know the solution set is $(0,t,0)$\n\n\u2022 Aaah, very insightful. This is the kind of thing I was looking for also. Do you mean by \"line\" that the solutions form a line in 3-space? @DougM \u2013\u00a0Bucephalus Aug 3 '17 at 0:30\n\u2022 yes, with the rank of the kernel equal to 1, (and the information that at least one solution exists), then all of the solutions will lie on a line in $\\mathbb R^3.$ If the nullity equaled 2, then we might say that there is a plane of solutions. \u2013\u00a0Doug M Aug 3 '17 at 0:36\n\u2022 Thanks @DougM. I will keep an eye out for you next time. \u2013\u00a0Bucephalus Aug 3 '17 at 0:43", "date": "2020-02-18 06:57:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2380618/gaussian-elimination-inconsistent-or-infinite-solutions", "openwebmath_score": 0.7703532576560974, "openwebmath_perplexity": 311.5762971494345, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676496254957, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8528959747782032}}
{"url": "https://stats.stackexchange.com/questions/379335/adding-very-small-probabilities-how-to-compute/379336", "text": "Adding very small probabilities - How to compute?\n\nIn some problems, probabilities are so small that they are best represented in computational facilities as log-probabilities. Computational problems can arise when you try to add these small probabilities together, since some computational facilities (e.g., base R) can't distinguish very small probabilities from zero. In these cases it is necessary to solve the problem either by finding a workaround that avoids the small numbers, or by using a computational facility that can deal with very small numbers.\n\nThe archetypal problem of this kind is as follows. Suppose you have log-probabilities $$\\ell_1$$ and $$\\ell_2$$, where the corresponding probabilities $$\\exp(\\ell_1)$$ and $$\\exp(\\ell_2)$$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-sum of these probabilities, which we denote by:\n\n$$\\ell_+ \\equiv \\ln \\big( \\exp(\\ell_1) + \\exp(\\ell_2) \\big)$$\n\nAssume that we are ---at least initially--- working in an environment where $$\\exp(\\ell_1)$$ and $$\\exp(\\ell_2)$$ cannot be computed, since they are so small that they are indistinguishable from zero.\n\nQuestions: How can you effectively compute this log-sum? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?\n\n\u2022 the \"log-sum-exp\" trick can help \u2013\u00a0Taylor Nov 29 '18 at 3:44\n\nSome preliminary mathematics: As a preliminary matter it is worth noting that this function is the LogSumExp (LSE) function, which is often encountered when performing computation on values that are represented in the log-scale. To see how to deal with sums of this kind, we first note a useful mathematical result concerning sums of exponentials:\n\n\\begin{aligned} \\exp(\\ell_1) + \\exp(\\ell_2) &= \\exp(\\max(\\ell_1,\\ell_2)) + \\exp(\\min(\\ell_1,\\ell_2)) \\\\[6pt] &= \\exp(\\max(\\ell_1,\\ell_2)) (1 + \\exp(\\min(\\ell_1,\\ell_2)-\\max(\\ell_1,\\ell_2)) \\\\[6pt] &= \\exp(\\max(\\ell_1,\\ell_2)) (1 + \\exp(-|\\ell_1 - \\ell_2|)). \\\\[6pt] \\end{aligned}\n\nThis result converts the sum to a product, which allows us to present the log-sum as:\n\n\\begin{aligned} \\ell_+ &= \\ln \\big( \\exp(\\ell_1) + \\exp(\\ell_2) \\big) \\\\[6pt] &= \\ln \\big( \\exp(\\max(\\ell_1,\\ell_2)) (1 + \\exp(-|\\ell_1 - \\ell_2|)) \\big) \\\\[6pt] &= \\max(\\ell_1, \\ell_2) + \\ln (1 + \\exp(-|\\ell_1 - \\ell_2|)). \\\\[6pt] \\end{aligned}\n\nIn the case where $$\\ell_1 = \\ell_2$$ we obtain the expression $$\\ell_+ = \\ell_1 + \\ln 2 = \\ell_2 + \\ln 2$$, so the log-sum is easily computed. In the case where $$\\ell_1 \\neq \\ell_2$$ this expression still reduces the problem to a simpler case, where we need to find the log-sum of one and $$\\exp(-|\\ell_1 - \\ell_2|)$$.$$^\\dagger$$\n\nNow, using the Maclaurin series expansion for $$\\ln(1+x)$$ we obtain the formula:\n\n\\begin{aligned} \\ell_+ &= \\max(\\ell_1, \\ell_2) + \\sum_{k=1}^\\infty (-1)^{k+1} \\frac{\\exp(-k|\\ell_1 - \\ell_2|)}{k} \\quad \\quad \\quad \\text{for } \\ell_1 \\neq \\ell_2. \\\\[6pt] \\end{aligned}\n\nSince $$\\exp(-|\\ell_1 - \\ell_2|) < 1$$ the terms in this expansion diminish rapidly (faster than exponential decay). If $$|\\ell_1 - \\ell_2|$$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.\n\nImplementation in base R: It is actually possible to compute this log-sum accurately in base R through creative use of the log1p function. This is a primitive function in the base package that computes the value of $$\\ln(1+x)$$ for an argument $$x$$ (with accurate computation even for $$x \\ll 1$$). This primitive function can be used to give a simple function for the log-sum:\n\nlogsum <- function(l1, l2) { max(l1, l2) + log1p(exp(-abs(l1-l2))); }\n\n\nImplementation of this function succeeds in finding the log-sum of probabilities that are too small for the base package to deal with directly. Moreover, it is able to calculate the log-sum to a high level of accuracy:\n\nl1 <- -3006;\nl2 <- -3012;\n\nlogsum(l1, l2);\n[1] -3005.998\n\nsprint(\"%.50f\", logsum(l1, l2));\n[1] \"-3005.99752431486240311642177402973175048828125000000000\"\n\n\nAs can be seen, this method gives a computation with 41 decimal places for the log-sum. It only uses the functions in the base package, and does not involve any changes to the default calculation settings. It gives as high a level of accuracy as you are likely to need in most cases.\n\nIt is also worth noting that there are many packages in R that extend the computational facilities of the base program, and can be used to deal with sums of very small numbers. It is possible to find the log-sum of small probabilities using packages such as gmp or Brobdingnag, but this requires some investment in learning their particular syntax.\n\n$$^\\dagger$$ From this result we can also see that if $$|\\ell_1 - \\ell_2|$$ is itself large (i.e., if one of the probabilities is very small compared to the other) then the exponential term in this equation will be near zero, and we will then have $$\\ell_+ \\approx \\max(\\ell_1, \\ell_2)$$ to a very high degree of accuracy.\n\n\u2022 A number of posts on site discuss related issues. \u2013\u00a0Glen_b Nov 29 '18 at 4:53\n\u2022 Sorry, I must have missed these. \u2013\u00a0Ben Nov 29 '18 at 5:05\n\u2022 There's at least some material here that I don't think is anywhere else, and the exposition is quite clear so I don't think there's a problem, but it's worth being aware of the other posts. (A lot crop up in discussions of underflow and overflow in various calculations, for example) \u2013\u00a0Glen_b Nov 29 '18 at 5:29\n\u2022 I have added the underflow tag to the post to link to other questions on this subject. \u2013\u00a0Ben Nov 29 '18 at 6:31", "date": "2021-02-25 08:03:17", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/379335/adding-very-small-probabilities-how-to-compute/379336", "openwebmath_score": 0.9880304932594299, "openwebmath_perplexity": 516.5180119763712, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676508127573, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.8528959741672323}}
{"url": "http://math.stackexchange.com/questions/129012/simple-question-related-to-the-fundamental-theorem-of-arithmetic", "text": "# Simple question related to the Fundamental theorem of arithmetic\n\nGiven a number $b$, which we write as a product of prime numbers:\n\n$$b = p_1 \\cdots p_s$$\n\nCan I then deduce that a number, which divides $b$ then has to be a product of the above primes in the factorization of $b$?\n\n-\nYes and that follows by writing out another factorization for the number that divides $b = p_1 \\cdots p_s$ and using unique factorization to dedude that the factors must be some of the primes in the factorization of $b$. \u2013\u00a0Adri\u00e1n Barquero Apr 7 '12 at 15:34\nMore explicitly, if $b = p_{1}^{e_1} \\cdots p_{s}^{e_s}$ is the prime factorization of $b$ where the $p_i$'s are different primes, then you can prove that any divisor of $b$ is a product of the form $p_{1}^{a_1} \\cdots p_{s}^{a_s}$ where the exponents $a_i$ satisfy that $0 \\leq a_i \\leq e_i$ for $i = 1, \\dots , s$. \u2013\u00a0Adri\u00e1n Barquero Apr 7 '12 at 15:38\n\nEssentially yes: If $p_1,\\ldots,p_s$ are primes (possibly repeated), then every divisor of $$b = p_1p_2\\cdots p_s$$ must be a (possibly empty) product of some (possibly all) of the $p_i$, times $1$ or $-1$.\n\n-\nAnd why the downvote? \u2013\u00a0Arturo Magidin Apr 11 '12 at 6:05\n\nA divisor of $b$ has to be a product of some of the prime factors of $b$, but \"some\" may mean $0$ of them (if the divisor is $1$). The divisor can have no prime factors that are not prime factors if $b$.\n\nIf the divisor in question is $c$, then for some number $d$ we have $b=cd$. If a prime number $p$ divides $b$, then $p$ must divide either $c$ or $d$ (or both). The foregoing sentence is Euclid's lemma.\n\n-\n\nHint $\\rm\\ p\\ |\\ d\\ |\\ p_1\\cdots p_n\\ \\Rightarrow\\ p\\ |\\ p_j\\:$ for some $\\rm\\:j\\:$ by the Prime Divisor Property. Cancelling $\\rm\\:p\\:$ from $\\rm\\:d\\:$ and $\\rm\\:p_1\\cdots p_n\\:$ and inducting yields that the prime factors of $\\rm\\:d\\:$ are a sub-multiset of $\\rm\\{p_1,\\ldots,p_n\\}.$\n\nGenerally $\\rm\\:d\\ |\\ b_1\\cdots b_n\\ \\Rightarrow\\ d = d_1\\cdots d_n,\\ \\ d_j\\ |\\ b_j,\\:$ i.e. a divisor of a product is a product of divisors. This is a useful generalization of the Prime Divisor Property from atoms to composite numbers. Yours is the special case when all $\\rm\\:b_j = p_j\\:$ prime, so $\\rm\\:d_j = 1\\:$ or $\\rm\\:p_j\\:$ (modulo a unit factor $\\pm1$).\n\n-\n@Downvoter If something is not clear, please feel free to ask questions and I will be happy to elaborate. \u2013\u00a0Bill Dubuque Apr 27 '12 at 19:13", "date": "2016-02-11 21:28:47", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/129012/simple-question-related-to-the-fundamental-theorem-of-arithmetic", "openwebmath_score": 0.95513516664505, "openwebmath_perplexity": 134.25257933843704, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676419082933, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8528959697042254}}
{"url": "http://mathhelpforum.com/discrete-math/15449-pascal-s-triangle.html", "text": "# Math Help - Pascal's Triangle\n\n1. ## Pascal's Triangle\n\nBeing naturally curious, i was thinking by myself whether there might not be a way to determine the sum of all the numbers in this triangle by some method, depending on the number of rows it has. And also the sum of the numbers in the n-th row.\n\nThis is what i've discovered so far.\n\nRow 1 = 1 ; Sum = 1\nRow 2 = 1 1 ; Sum = 2\nRow 3 = 1 2 1 ; Sum = 4\nRow 4 = 1 3 3 1 ; Sum = 8\nRow 5 = 1 4 6 4 1 ; Sum = 16\n\nSo the sum of these numbers in each row, are multiples of 2.\n\n2. Originally Posted by janvdl\nNow i know this might sound silly...\n\nBeing naturally curious, i was thinking by myself whether there might not be a way to determine the sum of all the numbers in this triangle by some method, depending on the number of rows it has. And also the sum of the numbers in the n-th row.\n\nThis is what i've discovered so far.\n\nRow 1 = 1 ; Sum = 1\nRow 2 = 1 1 ; Sum = 2\nRow 3 = 1 2 1 ; Sum = 4\nRow 4 = 1 3 3 1 ; Sum = 8\nRow 5 = 1 4 6 4 1 ; Sum = 16\n\nSo the sum of these numbers in each row, are multiples of 2.\nseems to be $2^{n-1}$, where $n$ is the number of the row we are at from what you have there\n\n3. Originally Posted by Jhevon\nseems to be $2^{n-1}$, where $n$ is the number of the row we are at from what you have there\nAh ok, i wonder why i didnt notice that. And the sum of all the numbers up to an n-th row?\n\n4. Originally Posted by janvdl\nAh ok, i wonder why i didnt notice that. And the sum of all the numbers up to an n-th row?\nseems to be $2^n - 1$ where $n$ is the number of the row we are at\n\n5. Originally Posted by Jhevon\nseems to be $2^n - 1$ where $n$ is the number of the row we are at\nI was thinking in that line. cant believe i was so blind\n\n6. Hello, janvdl!\n\nBeing naturally curious . . .\nGood for you!\n\nI made some amazing \"discoveries\" while exploring on my own.\nOf course, anything I found was proved centuries ago,\n. . but it was still very satisfying.\n\nWhat you discovered is a stunning pattern.\n\nConsider the coefficients of the expansion of $(a + b)^n$\n\n. . . $\\begin{array}{ccc}n & \\text{coefficients} & \\text{sum} \\\\ \\hline 0 & 1 & 1 = 2^0\\\\ 1 & 1\\;\\;1 & 2 = 2^1 \\\\ 2 & 1\\;\\;2\\;\\;1 & 4 = 2^2 \\\\ 3 & 1\\;\\;3\\;\\;3\\;\\;1 & 8 =2^3 \\\\ 4 & 1\\;\\;4\\;\\;6\\;\\;4\\;\\;1 & 16 = 2^4 \\\\ 5 & 1\\;\\;5\\;\\;10\\;\\;10\\;\\;5\\;\\;1 & 32 = 2^5 \\\\ 6 & 1\\;\\;6\\;\\;15\\;\\;20\\;\\;15\\;\\;6\\;\\;1 & 64 = 2^6 \\\\ \\vdots & \\vdots & \\vdots \\end{array}$\n\nThe sum of the $n^{th}$ row is $2^n$.\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nI \"discovered\" this pattern while in college.\n\n$\\begin{array}{ccc}1 + 2 & = & 3 \\\\ 4 + 5 + 6 & = & 7 + 8 \\\\ 9 + 10 + 11 + 12 & = & 13 + 14 + 15 \\\\ 16 + 17 + 18 + 19 + 20 & = & 21 + 22 + 23 + 24 \\\\\n\\vdots & & \\vdots\\end{array}$\n\nI found these in a book . . .\n\n$\\begin{array}{ccc}1 & = & 1 = 1^3 \\\\ 3 + 5 & = & 8 = 2^3 \\\\ 7 + 9 + 11 & = & 27 = 3^3 \\\\ 13 + 15 + 17 + 19 & = & 64 = 4^3 \\\\ 21 + 23 + 25 + 27 + 29 & = & 125 = 5^3 \\\\ \\vdots & & \\vdots \\end{array}$\n\n$\\begin{array}{ccc}3^2 + 4^2 & = & 5^2 \\\\ 10^2 + 11^2 + 12^2 & = &13^2 + 14^2 \\\\ 21^2 + 22^2 + 23^2 + 24^2 & = & 25^2 + 26^2 + 27^2 \\\\ 36^2 + 37^2 + 38^2 + 39^2 + 40^2 & = & 41^2 + 42^2 + 43^2 + 44^2 \\\\ 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 & = & 61^2 + 62^2 + 63^2 + 64^2 + 65^2 \\\\ \\vdots & &\\vdots\\end{array}$\n\n7. I've found something else too...\n\nI was actually trying to find another type of formula when i noticed this...\n\nI attached a picture, i know its badly drawn but at least it gives you the idea...\n\nThe white triangles are the normal figures of Pascal's Triangle. But look at those red triangles. They interest me a lot. There must be a lot more patterns involved in this triangle than i ever thought.\n\nWhat if every red triangle is the product of the 3 white triangles around it? Or the sum of the 3 white ones around it? We can make lots of new patterns using this...\n\n8. Hello, janvdl!\n\nThat's fascinating . . . Thank you!\n\nWhat if every red triangle is the sum of the 3 white ones around it?\n\nThen we have:\n\n. . $\\begin{array}{c}3\\\\\n4\\;\\;4\\\\\n5\\;\\;8\\;\\;5 \\\\\n6\\;\\;13\\;\\;13\\;\\;6 \\\\\n7\\;\\;19\\;\\;26\\;\\;19\\;\\;7\n\\end{array}$\n\nI finally saw where it came from . . .\n\nConsider a \"Pascal's Triangle\" whose third line is: .1 - 3 - 1.\n\n. . $\\begin{array}{c}1 \\\\ 1\\;\\;1 \\\\ 1\\;\\;3\\;\\;1 \\\\\n1\\;\\;4\\;\\;4\\;\\;1 \\\\ 1\\;\\;5\\;\\;8\\;\\;5\\;\\;1 \\\\ 1\\;\\;6\\;\\;13\\;\\;13\\;\\;6\\;\\;1 \\\\ 1\\;\\;7\\;\\;19\\;\\;26\\;\\;19\\;\\;7\\;\\;1\n\\end{array}$\n\nSee it?\n\nThis opens up a world of possibilities, doesn't it?\n\n9. Originally Posted by Soroban\nHello, janvdl!\n\nThat's fascinating . . . Thank you!\n\nThen we have:\n\n. . $\\begin{array}{c}3\\\\\n4\\;\\;4\\\\\n5\\;\\;8\\;\\;5 \\\\\n6\\;\\;13\\;\\;13\\;\\;6 \\\\\n7\\;\\;19\\;\\;26\\;\\;19\\;\\;7\n\\end{array}$\n\nI finally saw where it came from . . .\n\nConsider a \"Pascal's Triangle\" whose third line is: .1 - 3 - 1.\n\n. . $\\begin{array}{c}1 \\\\ 1\\;\\;1 \\\\ 1\\;\\;3\\;\\;1 \\\\\n1\\;\\;4\\;\\;4\\;\\;1 \\\\ 1\\;\\;5\\;\\;8\\;\\;5\\;\\;1 \\\\ 1\\;\\;6\\;\\;13\\;\\;13\\;\\;6\\;\\;1 \\\\ 1\\;\\;7\\;\\;19\\;\\;26\\;\\;19\\;\\;7\\;\\;1\n\\end{array}$\n\nSee it?\n\nThis opens up a world of possibilities, doesn't it?\n\nSo have i actually kind of discovered something here?\n\nAnd what can we actually do with my \"discovery\"? A world of possibilities has been opened? You're clearly seeing things that i cant...\n\n10. Here's another of my \"discoveries\", made years and years ago.\n\nWe know the Binomial Theorem is based on Pascal's Triangle.\nWe use Pascal's Triangle for expanding $(a + b)^n$.\n\nI wondered if there was a \"Trinomial Theorem\" for expanding $(a + b + c)^n$.\n\nI cranked out the first few cases . . . and got into an awful mess.\n\n. . $\\begin{array}{ccc}(a + b + c)^0 & = & 1 \\\\\n(a + b + c)^1 & = & a + b + c \\\\\n(a + b + c)^2 & = & a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \\\\\n(a + b + c)^3 & = & a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + 6abc\n\\end{array}$\n\nHow do we make sense out of these terms and coefficients?\n\nIt took me the longest time to see what to do:\n. . arrange the terms in triangular arrays.\n$\\begin{array}{ccccc} & & & & a^4 \\\\ & & &a^3 & 4a^3b\\;\\;4a^3c \\\\& & a^2 & 3a^2b\\;\\;3a^2c & 6a^2b^2\\;\\;12a^2bc\\;\\;6a^2c^2 \\\\& a & 2ab\\;\\;2ac & 3ab^2\\;\\;6abc\\;\\;3ac^2 & 4ab^3\\;\\;12ab^2c\\;\\;12abc^2\\;\\;4ac^3 \\\\1\\quad &\\;\\; b\\;\\;c\\quad & b^2\\;\\;2ab\\;\\;c^2\\quad & b^3\\quad3b^2c\\quad3bc^2\\quad c^3\\quad & b^4\\quad4b^3c\\quad6b^2c^2\\quad 4bc^3\\quad c^4\\end{array}$\n\nLook at the coefficients:\n$\\begin{array}{ccccc} & & & & 1 \\\\& & & 1 & 4\\;\\;4 \\\\& & 1 & 3\\;\\;3 & 6\\;\\;12\\;\\;6 \\\\& 1 & 2\\;\\;2 & 3\\;\\;6\\;\\;3 & 4\\;\\;12\\;\\;12\\;\\;4 \\\\1\\quad & \\;\\;\\;1\\;\\;1\\quad & \\;\\;1\\;\\;2\\;\\;1\\quad & \\;\\;1\\;\\;3\\;\\;3\\;\\;1\\quad & 1\\;\\;\\;4\\;\\;\\;6\\;\\;\\;4\\;\\;\\;1\\end{array}$\n\nIf we \"stack\" these triangles, we get a tetrahedron.\n. . Each number is the sum of the three numbers above it.\n\nI seriously doubt that I'm the first to notice all this.\nBut you can feel free to call it \"Soroban's Tetrahedron\".\n\n11. I'm just going to try and get some formulas out of my triangle Im busy writing a small book on my discoveries in this triangle. Just for the fun of it", "date": "2014-11-26 13:53:11", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/15449-pascal-s-triangle.html", "openwebmath_score": 0.8285999298095703, "openwebmath_perplexity": 518.6795425454325, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180694313357, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8528670140161695}}
{"url": "https://web2.0calc.com/questions/assigning-var", "text": "+0\n\n# Assigning var\n\n0\n232\n4\n+4\n\nIs there a way to define a variable to use an equation? E.x.: 3x^2-2x+5=y, and I want to assign x a value without changing it out. Is it possible?\n\nCatBoy13 \u00a0Dec 12, 2014\n\n#3\n+91479\n+10\n\nWelcome to web2.0calc forum Catboy13\n\nUmm - Interesting question.\n\n$$\\begin{array}{rll} 3x^2-2x+5&=&y\\\\\\\\ 3x^2-2x&=&y-5\\\\\\\\ x^2-\\frac{2}{3}x&=&\\frac{y-5}{3}\\\\\\\\ x^2-\\frac{2}{3}x+\\left(\\frac{2}{6}\\right)^2&=&\\frac{y-5}{3}+\\left(\\frac{2}{6}\\right)^2\\\\\\\\ x^2-\\frac{2}{3}x+\\left(\\frac{1}{3}\\right)^2&=&\\frac{y-5}{3}+\\left(\\frac{1}{3}\\right)^2\\\\\\\\ x^2-\\frac{2}{3}x+\\left(\\frac{1}{3}\\right)^2&=&\\frac{3(y-5)}{9}+\\frac{1}{9}\\\\\\\\ \\left(x-\\frac{1}{3}\\right)^2&=&\\frac{3y-15+1}{9}\\\\\\\\ \\left(x-\\frac{1}{3}\\right)^2&=&\\frac{3y-14}{9}\\\\\\\\ x-\\frac{1}{3}&=&\\pm\\sqrt{\\frac{3y-14}{9}}\\\\\\\\ x&=&\\frac{1}{3}\\pm\\sqrt{\\frac{3y-14}{9}}\\\\\\\\ x&=&\\frac{1\\pm\\sqrt{3y-14}}{3}}\\\\ \\end{array}$$\n\nMelody \u00a0Dec 13, 2014\nSort:\n\n#1\n+7188\n0\n\nI believe it is possible....but I might be wrong...\n\nhappy7 \u00a0Dec 12, 2014\n#2\n0\n\nIf you first define a function f as\n\n$${f}{\\left({\\mathtt{x}}\\right)} = {\\mathtt{3}}{\\mathtt{\\,\\times\\,}}{{\\mathtt{x}}}^{{\\mathtt{2}}}{\\mathtt{\\,-\\,}}{\\mathtt{2}}{\\mathtt{\\,\\times\\,}}{\\mathtt{x}}{\\mathtt{\\,\\small\\textbf+\\,}}{\\mathtt{5}}$$\n\nthen for any particular x the function f(x) has a value and you can write things such as\n\nf(2) + f(0) - 4 = 14\n\nGuest\u00a0Dec 12, 2014\n#3\n+91479\n+10\n\nWelcome to web2.0calc forum Catboy13\n\nUmm - Interesting question.\n\n$$\\begin{array}{rll} 3x^2-2x+5&=&y\\\\\\\\ 3x^2-2x&=&y-5\\\\\\\\ x^2-\\frac{2}{3}x&=&\\frac{y-5}{3}\\\\\\\\ x^2-\\frac{2}{3}x+\\left(\\frac{2}{6}\\right)^2&=&\\frac{y-5}{3}+\\left(\\frac{2}{6}\\right)^2\\\\\\\\ x^2-\\frac{2}{3}x+\\left(\\frac{1}{3}\\right)^2&=&\\frac{y-5}{3}+\\left(\\frac{1}{3}\\right)^2\\\\\\\\ x^2-\\frac{2}{3}x+\\left(\\frac{1}{3}\\right)^2&=&\\frac{3(y-5)}{9}+\\frac{1}{9}\\\\\\\\ \\left(x-\\frac{1}{3}\\right)^2&=&\\frac{3y-15+1}{9}\\\\\\\\ \\left(x-\\frac{1}{3}\\right)^2&=&\\frac{3y-14}{9}\\\\\\\\ x-\\frac{1}{3}&=&\\pm\\sqrt{\\frac{3y-14}{9}}\\\\\\\\ x&=&\\frac{1}{3}\\pm\\sqrt{\\frac{3y-14}{9}}\\\\\\\\ x&=&\\frac{1\\pm\\sqrt{3y-14}}{3}}\\\\ \\end{array}$$\n\nMelody \u00a0Dec 13, 2014\n#4\n+91479\n0\n\nI have added this to the Sticky Topic - \"Great Answers to Learn From\"\n\nMelody \u00a0Dec 14, 2014\n\n### 4 Online Users\n\nWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. \u00a0See details", "date": "2018-01-22 08:15:13", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/assigning-var", "openwebmath_score": 0.796775758266449, "openwebmath_perplexity": 6461.333455958909, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180639771091, "lm_q2_score": 0.8652240791017535, "lm_q1q2_score": 0.8528670041585575}}
{"url": "https://puzzling.stackexchange.com/questions/100491/calculation-puzzle-010", "text": "# Calculation puzzle 010\n\nFind the missing number in the sequence.\n\n2   9   28   65   ?   217\n\n\nSource: This question is taken from YTU YOS 2018 exam. I have mentioned them in other posts.\n\nIt's simply just\n\n$$n^3 + 1$$\n\nSo\n\n$$1^3 + 1 = 1 + 1 = 2$$\n$$2^3 + 1 = 8 + 1 = 9$$\n$$3^3 + 1 = 27 + 1 = 28$$\n$$4^3 + 1 = 64 + 1 = 65$$\n$$5^3 + 1 = 125 + 1 = 126$$\n$$6^3 + 1 = 216 + 1 = 217$$\n\nTherefore the missing number is\n\n$$126$$\n\n\u2022 Gz mate nice one Jul 27, 2020 at 12:51\n\nThe answer is 126 because of the pattern 1cubed +1, 2cubed+1, 3 cubed +1,4cubed +1, 5cubed+1,6cubed+1\n\n\u2022 By the way I did not copy this answer Jul 27, 2020 at 12:47\n\u2022 There was no answer as I was typing it Jul 27, 2020 at 12:47\n\u2022 Don't worry, people will be able to see that you didnt from the timestamp, as you can see we both answered at almost the same time Jul 27, 2020 at 12:52\n\u2022 Yesterday I answered the question 3 minutes before another guy and he still got the credit\ud83d\ude02 I don't have good luck here Jul 27, 2020 at 12:54\n\u2022 Sometimes a later answer will be accepted if the OP thinks that it's more correct or explains it better, it's not always first to get it is accepted :) Jul 27, 2020 at 12:55", "date": "2022-05-21 04:08:10", "meta": {"domain": "stackexchange.com", "url": "https://puzzling.stackexchange.com/questions/100491/calculation-puzzle-010", "openwebmath_score": 0.5920678377151489, "openwebmath_perplexity": 1321.2207599286837, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180673335565, "lm_q2_score": 0.8652240738888188, "lm_q1q2_score": 0.8528670019241528}}
{"url": "https://math.stackexchange.com/questions/1723465/if-n-is-a-positive-integer-does-n3-1-always-have-a-prime-factor-thats-1-m", "text": "# If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3?\n\nIt appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?\n\nIf it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.\n\nIn some cases, the prime factors congruent to 1 mod 3 are relatively large, so it's not as simple as \"they're all divisible by 7\" or anything like that.\n\nIt's interesting if one can prove that an integer of a certain form must have a prime factor of a certain form without necessarily being able to find it explicitly.\n\nEDITED TO ADD: It appears that there might be more going on here!\n\n$n^2-1$ usually has a prime factor congruent to 1 mod 2 (not if n=3, though!)\n\n$n^3-1$ always has a prime factor congruent to 1 mod 3\n\n$n^4-1$ always has a prime factor congruent to 1 mod 4\n\n$n^5-1$ appears to always have a prime factor congruent to 1 mod 5.\n\nRegarding $n^2-1$: If $n>3$, then $n^2-1=(n-1)(n+1)$ is a product of two numbers that differ by 2, which cannot both be powers of 2 if they are bigger than 2 and 4. Therefore at least one of $n-1,n+1$ is divisible by an odd prime.\n\nRegarding $n^4-1$: If $n>1$, we factor $n^4-1$ as $(n+1)(n-1)(n^2+1)$. We claim that in fact, every prime factor of $n^2+1$ is either 2 or is congruent to 1 mod 4. If $p$ is an odd prime that divides $n^2+1$, then $-1$ is a square mod $p$, but the odd primes for which $-1$ is a square mod $p$ are precisely the primes congruent to 1 mod 4. It remains just to show that $n^2+1$ cannot be a power of 2. If $n$ is even this is obvious, and if $n=2k+1$ is odd, then $n^2+1=(2k+1)^2+1=4k^2+4k+2$ is 2 more than a multiple of 4.\n\nRegarding $n^5-1$, I don't have a proof, but based on experimenting with a few dozen numbers, I conjecture that in fact, every prime factor of $n^4+n^3+n^2+n+1$ is either 5 or is 1 more than a multiple of 5.\n\n\u2022 From the factorization you can get a factor that's $\\equiv 1 \\mod 3$, but not necessarily a prime one \u2013\u00a0MCT Apr 1 '16 at 15:38\n\u2022 Ran a code for the for all such numbers less than $10^7$. Seems ok. I am addin a list of the first few numbers and their prime factors. (7, 7) (26, 13) (63, 7) (124, 31) (215, 43) (342, 19) (511, 7) (728, 7) (999, 37) (1330, 7) (1727, 157) (2196, 61) (2743, 13) (3374, 7) (4095, 7) (4912, 307) (5831, 7) (6858, 127) (7999, 19) (9260, 463) (10647, 7) (12166, 7) (13823, 601) (15624, 7) (17575, 19) (19682, 13) (21951, 271) (24388, 7) (26999, 7) (29790, 331) (32767, 7) (35936, 1123) (39303, 397) (42874, 13) (46655, 7) (50652, 7) (54871, 37) (59318, 7) (63999, 13) (68920, 1723) (74087, 13) (79506, 7) \u2013\u00a0Banach Tarski Apr 1 '16 at 15:49\n\u2022 Ran the code again. This statement seems to be true for all such numbers smaller than $2 * 10^8$ \u2013\u00a0Banach Tarski Apr 1 '16 at 15:56\n\u2022 @idmercer: For prime powers $p>2$, you can generalize it neatly. Define $$F(n) = \\frac{n^p-1}{n-1}$$ Conjecture: \"If $F(n)$ is not divisible by $p$, then every odd prime factor of $F(n)$ is $1\\mod 2p$.\" I'm not sure this is a proven result though, because otherwise the guys who answered would have used it. Maybe you can ask the general case for primes separately. \u2013\u00a0Tito Piezas III Apr 1 '16 at 22:01\n\nThe case $n=1$ is uninteresting, so let $n\\gt 1$. We show that $n^2+n+1$ has a prime factor congruent to $1$ modulo $3$, by showing that $4(n^2+n+1)$ has such a prime factor.\n\nNote that $n^2+n+1$ is odd. We first show that $3$ is not the only odd prime that divides $n^2+n+1$. For suppose to the contrary that $(2n+1)^2+3=4\\cdot 3^k$ where $k\\gt 1$. Then $2n+1$ is divisible by $3$, so $(2n+1)^2+3\\equiv 3\\pmod{9}$, contradiction.\n\nNow let $p\\gt 3$ be a prime divisor of $(2n+1)^2+3$. Then $-3$ is a quadratic residue of $p$. A straightforward quadratic reciprocity argument shows that $p$ cannot be congruent to $2$ modulo $3$.\n\nRemark: By considering $n$ of the form $q!$ for possibly large $q$, we can use the above result to show that there are infinitely many primes of the form $6k+1$.\n\n\u2022 Is there a general result for the odd prime factors of $\\displaystyle F(n)=\\frac{n^p-1}{n-1}$ for prime $p$? (The OP was just focusing on the case $p=3$.) In other words, is every odd prime factor of $F(n)$ either $p$ or $1\\mod p$? \u2013\u00a0Tito Piezas III Apr 2 '16 at 18:51\n\u2022 Analysis is pretty simple (Fermat's Theorem) for $n$ and $p$ such that $n$ is not congruent to $1$ mod $p$. But there can be funny prime divisors otherwise. There should be a complete easy answer in general, but I looked and am missing something. That's why the appeal to reciprocity in the answer. \u2013\u00a0Andr\u00e9 Nicolas Apr 2 '16 at 18:54\n\u2022 Yes, Fermat's little theorem occurred to me as well. I may ask the general prime case in the forum. Do you think you'll figure out the answer to the general case? :) \u2013\u00a0Tito Piezas III Apr 2 '16 at 18:57\n\u2022 @TitoPiezasIII: Someone will. Conceivably me, but probably not. Would likely not be in any case, but also today is a massive cooking day. \u2013\u00a0Andr\u00e9 Nicolas Apr 2 '16 at 19:00\n\u2022 Ok, Bon appetit. (P.S. The question is here.) \u2013\u00a0Tito Piezas III Apr 2 '16 at 19:20\n\nFor $p\\equiv 1\\pmod 3$, there exist three distinct solutions of $x^3\\equiv 1\\pmod p$, and if $n$ is congruent modulo $p$ to such $x$ then $n^3-1$ is a multiple of $p$. Heuristically, this solves the problem affirmatively for a fraction $\\frac 3p$ of all possible $n$.\n\nLet's make this a bit less hand-wavy: Let $p_1=7, p_2=13, p_3=19,\\ldots$ denote the sequence of prime $\\equiv 1\\pmod 3$. Then of the $M:=p_1p_2\\cdots p_m$ residue classes modulo $M$, there are only $(1-\\frac3{p_1})(1-\\frac3{p_2})\\cdots(1-\\frac3{p_m})M=(p_1-3)(p_2-3)\\cdots(p_m-3)$ residue classes $x$ for which $n\\equiv x\\pmod M$ does not imply that $n^3-1$ has a prime factor $\\in\\{p_1,\\ldots,p_m\\}$. As we let $m\\to \\infty$, the product $(1-\\frac3{p_1})(1-\\frac3{p_2})\\cdots(1-\\frac3{p_m})$ can be shown to tend to $0$. So at least the density of $n$ that do not work must be zero.", "date": "2019-07-21 02:14:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1723465/if-n-is-a-positive-integer-does-n3-1-always-have-a-prime-factor-thats-1-m", "openwebmath_score": 0.8251487016677856, "openwebmath_perplexity": 154.6055233985329, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180664944447, "lm_q2_score": 0.865224070413529, "lm_q1q2_score": 0.8528669977724771}}
{"url": "https://math.stackexchange.com/questions/351363/find-a-simple-formula-for-binomn0-binomn1-binomn1-binomn2", "text": "# Find a simple formula for $\\binom{n}{0}\\binom{n}{1}+\\binom{n}{1}\\binom{n}{2}+\u2026+\\binom{n}{n-1}\\binom{n}{n}$\n\n$$\\binom{n}{0}\\binom{n}{1}+\\binom{n}{1}\\binom{n}{2}+...+\\binom{n}{n-1}\\binom{n}{n}$$\n\nAll I could think of so far is to turn this expression into a sum. But that does not necessarily simplify the expression. Please, I need your help.\n\nFirst note that $\\dbinom{n}k = \\dbinom{n}{n-k}$. Hence, your sum can be written as $$\\sum_{k=0}^{n-1} \\dbinom{n}k \\dbinom{n}{k+1} = \\sum_{k=0}^{n-1} \\dbinom{n}k \\dbinom{n}{n-k-1}$$ Now consider a bag with $n$ red balls and $n$ blue balls. We want to choose a total of $n-1$ bals. The total number of ways of doing this is given by $$\\dbinom{2n}{n-1} \\tag{\\star}$$ However, we can also count this differently. Any choice of $n-1$ balls will involve choosing $k$ blue balls and $n-k-1$ red balls. Hence, the number of ways of choose $n-1$ balls with $k$ blue balls is $$\\color{blue}{\\dbinom{n}k} \\color{red}{\\dbinom{n}{n-1-k}}$$ Now to count all possible ways of choosing $n-1$ balls, we need to let our $k$ run from $0$ to $n-1$. Hence, the total number of ways is $$\\sum_{k=0}^{n-1} \\color{blue}{\\dbinom{n}k} \\color{red}{\\dbinom{n}{n-k-1}} \\tag{\\dagger}$$ Now since $(\\star)$ and $(\\dagger)$ count the same thing, they must be equal and hence, we get that $$\\sum_{k=0}^{n-1} \\color{blue}{\\dbinom{n}k} \\color{red}{\\dbinom{n}{n-k-1}} = \\dbinom{2n}{n-1}$$\n\nEDIT As Brian points out in the comments, the above is a special case of the more general Vandermonde's identity. $$\\sum_{k=0}^r \\dbinom{m}k \\dbinom{n}{r-k} = \\dbinom{m+n}r$$ The proof for this is essentially the same as above. Consider a bag with $m$ blue balls and $n$ red balls and count the number of ways of choosing $r$ balls from these $m+n$ balls in two different ways as discussed above.\n\n\u2022 Love the formatting! +1 \u2013\u00a0Macavity Apr 4 '13 at 18:32\n\u2022 (+1) Nicely explained. You might want to mention that it\u2019s a case of Vandermonde\u2019s identity, which has essentially the same proof. \u2013\u00a0Brian M. Scott Apr 4 '13 at 18:43\n\u2022 Thanks so much!!!! Your explanation was great!!!! \u2013\u00a0Dome Apr 4 '13 at 18:49\n\nWe have $$\\sum_{k=0}^{n-1}\\dbinom{n}{k}\\dbinom{n}{k+1}=\\sum_{k=0}^{n}\\dbinom{n}{k}\\dbinom{n}{k+1}=\\sum_{k=0}^{n}\\dbinom{n}{k}\\dbinom{n}{n+1-k}=\\color{red}{\\dbinom{2n}{n+1}}$$ by the Chu-Vandermonde identity and since $\\dbinom{n}{n}\\dbinom{n}{n+1}=0$. Another way is to use the integral representation of the binomial coefficient $$\\dbinom{n}{k}=\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{\\left(1+z\\right)^{n}}{z^{k+1}}dz$$ and get $$\\sum_{k=0}^{n}\\dbinom{n}{k}\\dbinom{n}{k+1}=\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{\\left(1+z\\right)^{n}}{z^{2}}\\sum_{k=0}^{n}\\dbinom{n}{k}z^{-k}dz$$ $$=\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{\\left(1+z\\right)^{n}}{z^{2}}\\left(1+\\frac{1}{z}\\right)^{n}dz=\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{\\left(1+z\\right)^{2n}}{z^{n+2}}dz=\\color{blue}{\\dbinom{2n}{n+1}}.$$\n\nThere is a simple combinatorial interpretation. Assume that you have a parliament with $n$ politicians in the left wing and $n$ politicians in the right wing. If you want to select a committee made by $n-1$ politicians, you obviously have $\\binom{2n}{n-1}=\\binom{2n}{n+1}$ ways for doing it. On the other hand, by classifying the possible committees according to the number of politicians of the left wing in them, you have: $$\\binom{2n}{n-1}=\\sum_{l=0}^{n-1}\\binom{n}{l}\\binom{n}{n-1-l} = \\sum_{l=0}^{n-1}\\binom{n}{l}\\binom{n}{l+1}$$ as wanted.\n\nHint: it's the coefficient of $T$ in the binomial expansion of $(1+T)^n(1+T^{-1})^n$, which is equivalent to saying that it's the coefficient of $T^{n+1}$ in the expansion of $(1+T)^n(1+T^{-1})^nT^n=(1+T)^{2n}$.\n\n\nNote that $\\ds{{n \\choose k} + {n \\choose k + 1} = {n + 1 \\choose k + 1}}$\n\n\\begin{align} \\sum_{k = 0}^{n - 1}{n \\choose k}{n \\choose k + 1} & = \\sum_{k = 0}^{n - 1} {\\bracks{{n \\choose k} + {n \\choose k + 1}}^{\\,2} - {n \\choose k}^{2} - {n \\choose k + 1}^{2}\\over 2} \\\\[5mm] & = {1 \\over 2}\\sum_{k = 0}^{n - 1}{n + 1 \\choose k + 1}^{2} - {1 \\over 2}\\sum_{k = 0}^{n - 1}{n \\choose k}^{2} - {1 \\over 2}\\sum_{k = 0}^{n - 1}{n \\choose k + 1}^{2} \\\\[5mm] & = {1 \\over 2}\\sum_{k = 1}^{n}{n + 1 \\choose k}^{2} - {1 \\over 2}\\sum_{k = 0}^{n - 1}{n \\choose k}^{2} - {1 \\over 2}\\sum_{k = 1}^{n}{n \\choose k}^{2} \\\\[5mm] & = {1 \\over 2}\\bracks{\\sum_{k = 0}^{n + 1}{n + 1 \\choose k}^{2} -2} - {1 \\over 2}\\bracks{\\sum_{k = 0}^{n}{n \\choose k}^{2} - 1} - {1 \\over 2}\\bracks{\\sum_{k = 0}^{n}{n \\choose k}^{2} - 1} \\\\[5mm] & = {1 \\over 2}{2n + 2 \\choose n + 1} - {2n \\choose n} \\end{align} where I used the well known result $\\bbx{\\ds{\\quad\\sum_{i = 0}^{m}{m \\choose i}^{2} = {2m \\choose m}}}$. Moreover, \\begin{align} \\sum_{k = 0}^{n - 1}{n \\choose k}{n \\choose k + 1} & = {1 \\over 2}\\,{\\pars{2n + 2}\\pars{2n + 1}\\pars{2n}! \\over \\pars{n + 1}n!\\pars{n + 1}n!} - {2n \\choose n} = \\pars{{2n + 1 \\over n + 1} - 1}{2n \\choose n} \\\\[5mm] & = {n \\over n + 1}{2n \\choose n} = {n \\over n + 1}{\\pars{2n}! \\over n!\\,n!} = {\\pars{2n}! \\over \\pars{n + 1}!\\pars{n - 1}!} = \\bbx{\\ds{2n \\choose n + 1}} \\end{align}\n\n\u2022 It is not easy to add a new perspective to a popular sum like this one. Upvoted for originality of the approach. (+1): \u2013\u00a0Marko Riedel Feb 1 '17 at 23:24\n\u2022 @MarkoRiedel Thanks. Cross terms always suggest this approach. It's useful too for products of $\\log$'s, etc... \u2013\u00a0Felix Marin Feb 2 '17 at 0:14\n\nRewrite the sum as $\\displaystyle\\sum_0^{n-1}\\binom{n}{k}\\binom{n}{k+1}$, then by combinatorial interpretation, this is: $$\\sum_{0}^{n-1}\\binom{n}{k}\\binom{n}{k+1}=\\binom{2n}{2k+1}$$\n\nThe right hand side $\\binom{2n}{2k+1}$ is the number of ways to choose $2k+1$ from a total of $2n$. For the left hand side, divide $2n$ into two groups of size $n$ of each, then if choose $k$ from one group, then must choose $2k+1-k=k+1$ from another group, sum over all possible $k$, then you get $\\sum_0^{n-1}\\binom{n}{k}\\binom{n}{k+1}$\n\n\u2022 What is $k$ in the rhs? \u2013\u00a0Julien Apr 4 '13 at 18:28\n\u2022 I have downvoted this answer for being (mathematically) unintelligible. \u2013\u00a0anon Apr 5 '13 at 0:51\n\nUsing the estimate derived in this answer, we get \\begin{align} \\sum^{n-1}_{j=0}\\binom{n}{j}\\binom{n}{j+1} &=\\sum^{n-1}_{j=0}\\binom{n}{n-j}\\binom{n}{j+1}\\\\ &=\\binom{2n}{n+1}\\\\ &=\\binom{2n}{n}\\frac{n}{n+1}\\\\[3pt] &\\sim\\frac{4^n}{\\sqrt{\\pi n}}\\left(1-\\frac9{8n}\\right) \\end{align}", "date": "2019-06-16 05:10:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/351363/find-a-simple-formula-for-binomn0-binomn1-binomn1-binomn2", "openwebmath_score": 0.9856791496276855, "openwebmath_perplexity": 170.76207851882089, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180664944447, "lm_q2_score": 0.8652240686758841, "lm_q1q2_score": 0.8528669960596491}}
{"url": "http://math.stackexchange.com/questions/16566/orthonormal-eigenbasis", "text": "# Orthonormal Eigenbasis\n\nI am a little apprehensive to ask this question because I have a feeling it's a \"duh\" question but I guess that's the beauty of sites like this (anonymity):\n\nI need to find an orthonormal eigenbasis for the $2 \\times 2$ matrix $\\left(\\begin{array}{cc}1&1\\\\ 1&1\\end{array}\\right)$. I calculated that the eigenvalues were $x=0$ and $x=2$ and the corresponding eigenvectors were $E(0) = \\mathrm{span}\\left(\\begin{array}{r}-1\\\\1\\end{array}\\right)$ and $E(2) = \\mathrm{span}\\left(\\begin{array}{c}1\\\\1\\end{array}\\right)$. Therefore, an orthonormal eigenbasis would be: $$\\frac{1}{\\sqrt{2}}\\left(\\begin{array}{r}-1\\\\1\\end{array}\\right), \\frac{1}{\\sqrt{2}}\\left(\\begin{array}{c}1\\\\1\\end{array}\\right).$$\n\nHere my question: Could the eigenvalues for $E(0)$ been $\\mathrm{span}\\left(\\begin{array}{r}1\\\\-1\\end{array}\\right)$?? This would make the final answer $\\frac{1}{\\sqrt{2}}\\left(\\begin{array}{r}1\\\\-1\\end{array}\\right), \\frac{1}{\\sqrt{2}}\\left(\\begin{array}{c}1\\\\1\\end{array}\\right)$. Is one answer more correct than the other (or are they both wrong)?\n\nThanks!\n\n-\nI don't quite understand the question as written, but if it's what I think it is, then yes, orthonormal eigenbases are not unique; you can multiply any eigenvector by a complex number of absolute value 1. \u2013\u00a0 Qiaochu Yuan Jan 6 '11 at 13:28\nOr by a real number or whatever the field is that you are working over. One more thing: professional mathematicians have \"duh\" moments all the time. It's part of the job description, so to speak. There is absolutely no need to hide behind anonymity. On the contrary, it is important for a mathematician to learn to live with those \"duh\" moments. Otherwise he will never be able to freely talk with his colleagues, which would greatly hinder his progress. \u2013\u00a0 Alex B. Jan 6 '11 at 13:32\n@Alex: the absolute value 1 condition is important for orthonormality, which doesn't make sense over an arbitrary field. \u2013\u00a0 Qiaochu Yuan Jan 6 '11 at 13:55\n@Qiaochu You are right. Read that as \"real or complex, depending on...\" \u2013\u00a0 Alex B. Jan 6 '11 at 14:38\n\n0 and 2 are the correct eigenvalues to your matrix; (1, -1) is one eigenvector, (1, 1) the other. Your solution is correct.\n\nSpan is the set of all linear combinations, so if you consider a vector space over $\\mathbb{R}$, it absolutely doesn't matter what scalar in $\\mathbb{R}$ you multiply your vectors with inside Span. This does not affect the set Span at all. So the solution is the same.\n\n-\n\nThere is no such thing as the eigenvector of a matrix, or the orthonormal basis of eigenvectors. There are usually many choices.\n\nRemember that an eigenvector $\\mathbf{v}$ of eigenvalue $\\lambda$ is a nonzero vector $\\mathbf{v}$ such that $T\\mathbf{v}=\\lambda\\mathbf{v}$. That means that if you take any nonzero multiple of $\\mathbf{v}$, say $\\alpha\\mathbf{v}$, then we will have $$T(\\alpha\\mathbf{v}) = \\alpha T\\mathbf{v} = \\alpha(\\lambda\\mathbf{v}) = \\alpha\\lambda\\mathbf{v}=\\lambda(\\alpha\\mathbf{v}),$$ so $\\alpha\\mathbf{v}$ is also an eigenvector corresponding to $\\lambda$. More generally, if $\\mathbf{v}_1,\\ldots,\\mathbf{v}_k$ are all eigenvectors of $\\lambda$, then any nonzero linear combination $\\alpha_1\\mathbf{v}_1+\\cdots+\\alpha_k\\mathbf{v}_k\\neq \\mathbf{0}$ is also an eigenvector corresponding to $\\lambda$.\n\nSo, of course, since $\\left(\\begin{array}{r}-1\\\\1\\end{array}\\right)$ is an eigenvector (corresponding to $x=0$), then so is $\\alpha\\left(\\begin{array}{r}-1\\\\1\\end{array}\\right)$ for any $\\alpha\\neq 0$, in particular, for $\\alpha=-1$ as you take.\n\nNow, a set of vectors $\\mathbf{w}_1,\\ldots,\\mathbf{w}_k$ is orthogonal if and only if $\\langle \\mathbf{w}_i,\\mathbf{w}_j\\rangle = 0$ if $i\\neq j$. If you have an orthogonal set, and you replace, say, $\\mathbf{w}_i$ by $\\alpha\\mathbf{w}_i$ with $\\alpha$ any scalar, then the result is still an orthogonal set: because $\\langle\\mathbf{w}_k,\\mathbf{w}_j\\rangle=0$ if $k\\neq j$ and neither is equal to $i$, and for $j\\neq i$, we have $$\\langle \\alpha\\mathbf{w}_i,\\mathbf{w}_j\\rangle = \\alpha\\langle\\mathbf{w}_i,\\mathbf{w}_j\\rangle = \\alpha 0 = 0$$ by the properties of the inner product. As a consequence, if you take an orthogonal set, and you take any scalars $\\alpha_1,\\ldots,\\alpha_k$, then $\\alpha_1\\mathbf{w}_1,\\ldots,\\alpha_k\\mathbf{w}_k$ is also an orthogonal set.\n\nA vector $\\mathbf{n}$ is normal if $||\\mathbf{n}||=1$. If $\\alpha$ is any scalar, then $||\\alpha\\mathbf{n}|| = |\\alpha|\\,||\\mathbf{n}|| = |\\alpha|$. So if you multiply any normal vector $\\mathbf{n}$ by a scalar $\\alpha$ of absolute value $1$ (or of complex norm $1$), then the vector $\\alpha\\mathbf{n}$ is also a normal vector.\n\nA set of vectors is orthonormal if it is both orthogonal, and every vector is normal. By the above, if you have a set of orthonormal vectors, and you multiply each vector by a scalar of absolute value $1$, then the resulting set is also orthonormal.\n\nIn summary: you have an orthonormal set of two eigenvectors. You multiply one of them by $-1$; this does not affect the fact that the two are eigenvectors. The set was orthogonal, so multiplying one of them by a scalar does not affect the fact that the set is orthogonal. And the vectors were normal, and you multiplied one by a scalar of absolute value $1$, so the resulting vectors are still normal. So you still have an orthonormal set of two eigenvectors. I leave it to you to verify that if you have a linearly independent set, and you multiply each vector by a nonzero scalar, the result is still linearly independent.\n\n-", "date": "2015-01-29 12:37:40", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/16566/orthonormal-eigenbasis", "openwebmath_score": 0.9426743984222412, "openwebmath_perplexity": 123.36312188183537, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985718065235777, "lm_q2_score": 0.8652240686758841, "lm_q1q2_score": 0.8528669949706195}}
{"url": "http://isxy.chicweek.it/number-of-shortest-paths-in-a-weighted-graph.html", "text": "Dijkstra's algorithm, published in 1959 and named after its creator Dutch computer scientist Edsger Dijkstra, can be applied on a weighted graph. Very nice article. Graph Algorithms I 12. We use n = jV(G)jand m = jE(G)jto denote the number of ver-. Shortest Path Queries A shortest path query on a(n) (undirected) graph \ufb01nds the shortest path for the given source and target vertices in the graph. The graph is not weighted. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. it, daniele. In the first part of the paper, we reexamine the all-pairsshortest paths (APSP) problem and present a newalgorithm with running time approaching O(n3/log2n), which improves all known algorithms for general real-weighted dense graphs andis perhaps close to the best result possible without using fast matrix multiplication, modulo a few log log n factors. Since the edges in the center of the graph have large weights, the shortest path between nodes 3 and 8 goes around the boundary of the graph where the edge weights are smallest. ! But what if edges have different 'costs'? s v \u03b4(, ) 3sv = \u03b4(, ) 12sv = 2 s v 2 5 1 7. Consumes a graph and two vertices, and returns the shortest path (in terms of number of vertices) between the two vertices. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. Weighted graphs are commonly used in determining the most optimal path, most expedient, or the. I maintain a count for the number of shortest paths; I would like to use BFS from v first and also maintain a global level. Weighted Graph ( \u092d\u093e\u0930\u093f\u0924 \u0917\u094d\u0930\u093e\u092b ) Discrete Mathematics Shortest Path || Dijkstra Algorithm #weightedgraph #grewalpinky B. Shortest Path (Unweighted Graph) Goal: find the shortest route to go from one node to another in a graph. Exercise 3 [Modeling a problem as a shortest path problem in graphs] Four imprudent walkers are caught in the storm and nights. It maintains a set of nodes for which the shortest paths are known. 7 s: source 8 \u2019\u2019\u2019 9 result = ShortestPathResult() 10 num_vertices = graph. the first assumes that the graph is weighted, which means that each edge has a cost to traverse it. BFS always visits nodes in increasing order of their distance from the source. An interesting problem is how to find shortest paths in a weighted graph; i. The previous state of the art for this problem was total update time O \u0303(n 2 \u221am/\u03b5) for directed, unweighted graphs [2], and \u00d5(mn=\u03b5) for undirected, unweighted graphs [12]. The result is a list of vertices, or #f if there is no path. G Is A Weighted Graph With Vertex Set {0,1,2,,1-1} And Integer Weights Represented As The Following Adjacency Matrix Form. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Our main goal is to characterize exactly which sets of node sequences, which we call path systems, can appear as unique shortest paths in a graph with arbitrary real edge weights. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. I'm using the networkx package in Python 2. If the graph is weighted (that is, G. Shortest paths. Here a, b, c. A cycle is a path where the first and last vertices are the same. We wish to determine a shortest path from v 0 to v n Dijkstra's Algorithm Dijkstra's algorithm is a common algorithm used to determine shortest path from a to z in a graph. Dijkstra's Algorithm. Counting the number of shortest paths in various graphs is an important and interesting combinatorial problem, especially in weighted graphs with various applications. A few months ago, mathematicians Andrew Beveridge and Jie Shan published Network of Thrones in Math Horizon Magazine where they analyzed a network of character interactions from the novel \u201cA Storm of Swords\u201d, the third book in the popular \u201cA Song of Ice and Fire\u201d and the basis for the Game of Thrones TV series. numPaths initialized to 1). Combinatorics is the branch of mathematics concerned with selecting, arranging, and listing or counting collections of objects. Newman Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, New Mexico 87501 and Center for Applied Mathematics, Cornell University, Rhodes Hall, Ithaca, New York 14853 ~Received 1 February 2001; published 28 June 2001!. Shortest Path In A Weighted Directed Graph With Dijkstra's Algorithm - posted in C and C++: Well, I encountered an interesting problem. We start with undirected graphs. The main algorithms that fall under this definition are Breadth-First Search (BFS) and Dijkstra's algorithms. A logical scalar. Weighted Graphs Data Structures & Algorithms 2 [email protected] \u00a92000-2009 McQuain Shortest Paths (SSAD) Given a weighted graph, and a designated node S, we would like to find a path of least total weight from S to each of the other vertices in the graph. The graph given in the test case is shown as : The shortest paths for the 3 queries are :: The direct Path is shortest with weight 5: There is no way of reaching node 1 from node 3. It is used to identify optimal driving directions or degree of separation between two people on a social network for example. The gist of Dijkstra's single source shortest path algorithm is as below : Dijkstra's algorithm finds the shortest path in a weighted graph containing only positive edge weights from a single source. The shortest path between two points in a weighted graph can be found with Dijkstra\u2019s algorithm. Now we are going to find the shortest path between source (a) and remaining vertices. Find the shortest distance from C to D and if it is impossible to reach node D from C then return -1. The algorithm exists in many variants. This means that, given a weighted graph, this algorithm will output the shortest distance from a selected node to all other nodes. shortest path. In this post, I explain the single-source shortest paths problems out of the shortest paths problems, in which we need to find all the paths from one starting vertex to all other vertices. Changing to its dual, the triangular grid, paths between triangle pixels (we abbreviate this term to trixels) are counted. Question: Shortest Paths(int[0). The total cost of a path in a graph is equal to the sum of the weights of the edges that connect the vertices in the path. implementations of the shortest-path search algorithms on graphs on the number of graph vertices experimentally. In the first part of the paper, we reexamine the all-pairsshortest paths (APSP) problem and present a newalgorithm with running time approaching O(n3/log2n), which improves all known algorithms for general real-weighted dense graphs andis perhaps close to the best result possible without using fast matrix multiplication, modulo a few log log n factors. PRACTICE PROBLEM BASED ON FLOYD WARSHALL ALGORITHM- Problem- Consider the following directed weighted graph- Using Floyd Warshall Algorithm, find the shortest path distance between every pair of vertices. The shortest path from 0 to 4 uses the shortest path from 0 to 1 and the edge 1\u20134. Our algorithm is deterministic and has a running time of O(k(m\u221an + n 3/2 log n)) where m is the number of edges in the graph and n is the number of vertices. In this graph, vertex A and C are connected by two parallel edges having weight 10 and 12 respectively. Abstract: In this paper we evaluate our presented Quantum Approach for finding the Estimation of the Length of the Shortest Path in a Connected Weighted Graph which is achieved with a polynomial time complexity about O(n) and as a result of evaluation we show that the Probability of Success of our presented Quantum Approach is increased if the Standard Deviation of the Length of all possible. An assigned number is called the weight of the edge, and the collection of all weights is called a weighting of the graph \u0393. A path in a graph is a sequence of adjacent vertices. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. Path Graphs. Only edges with non-negative costs are included. Discuss an efficient algorithm to compute a shortest path from node s to node t in a weighted directed graph G such that the path is of minimum cardinality among all shortest s - t paths in G graph-theory hamiltonian-path path-connected. This means it finds a shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between source node and every other node. dijkstra_path\u00b6 dijkstra_path (G, source, target, weight='weight') [source] \u00b6. 0 = no epidemic 1 = epidemic NW N SW C E Graph 3: Vertex coloring of weighted graph for the county. In this category, Dijkstra\u2019s algorithm is the most well known. You are expected to do it in Time Complexity of O(A + M). shortest path functions use it as the cost of the path; community finding methods use it as the strength of the relationship between two vertices, etc. Shortest path algorithms are a family of algorithms designed to solve the shortest path problem. The shortest path between node 222 and node 444 is 222 -> 555 -> 666 -> 777 -> 444, which has a weighted distance 1. The number of diagonal steps in a shortest path of the chessboard distance is min{w 1,w 2}, and the number of cityblock steps (i. A cycle is a path where the first and last vertices are the same. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. (Consider what this means in terms of the graph shown above right. The Edge can have weight or cost associate with it. Simple path is a path with no repeated vertices. Hence, parallel computing must be applied. In this post I will be discussing two ways of finding all paths between a source node and a destination node in a graph: Using DFS: The idea is to do Depth First Traversal of given directed graph. , capacity, cost, demand, traffic frequency, time, etc. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. G Is A Weighted Graph With Vertex Set {0,1,2,,1-1} And Integer Weights Represented As The Following Adjacency Matrix Form. Geodesic paths are not necessarily unique, but the geodesic distance is well-defined since all geodesic paths have. ple, Figure 1a illustrates a graph G, and Figure 1e shows an aug-mented graph G\u2217 constructed from G. The latter only works if the edge weights are non-negative. 4 5 Args: 6 graph: weighted graph with no negative cycles. Weighted directed graphs may be used to model communication networks, and shortest distances (shortest-path weights) between nodes may be used to suggest routes for messages. Chan\u2044 September 30, 2009 Abstract Inthe\ufb02rstpartofthepaper,wereexaminetheall-pairs shortest paths (APSP)problemand present a new algorithm with running time O(n3 log3 logn=log2 n), which improves all known algorithmsforgeneralreal-weighteddensegraphs. This module covers weighted graphs, where each edge has an associated weight or number. The problem is to find k directed paths starting at s, such that every node of G lies on at least one of those paths, and such that the sum of the weights of all the edges in the paths is minimized. The one-to-all shortest path problem is the problem of determining the shortest path from node s to all the other nodes in the. Finding shortest paths in weighted graphs In the past two weeks, you've developed a strong understanding of how to design classes to represent a graph and how to use a graph to represent a map. Exercise 3 [Modeling a problem as a shortest path problem in graphs] Four imprudent walkers are caught in the storm and nights. How to use BFS for Weighted Graph to find shortest paths ? If your graph is weighted, then BFS may not yield the shortest weight paths. You are given a weighted directed acyclic graph G, and a start node s in G. It finds a shortest-path tree for a weighted undirected graph. Journal of the ACM 65 :6, 1-40. Is there a cycle that uses each vertex. Mizrahi et al. The number dist[w] equals the length of a shortest path from v to w, or is -1 if w cannot be reached. Dijkstra's Algorithm is useful for finding the shortest path in a weighted graph. Given a directed weighted graph G= (V;E;w) with non-negative weights w: E!R+ and a vertex s2V, the single-source shortest paths is the family of shortest paths s vfor every vertex v2V. If finds only the lengths not the path. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. A logical scalar. We use n = jV(G)jand m = jE(G)jto denote the number of ver-. The order of a graph is the number of nodes. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The Dijkstra\u2019s algorithm is provided in Algorithm 2. For example, the length of v8,v9 equals 2, which is identical to the length of the. As such, we say that the weight of a path is the sum of the weights of the edges it contains. The shortest-path from vertex u to vertex v in a weighted graph is a path with minimum sum-weight B All-Pairs Shortest Paths For a weighted graph, the all-pairs shortest paths problem is to nd the shortest path between all pairs of vertices. 2 Representing Weighted Graphs Often it is desirable to use a priority queue to store weighted edges. Question: Shortest Paths(int[0). We first propose an exact (and. A single execution of the algorithm will. shortest path. Consider a shortest path p from vertex i to vertex j, and suppose that p contains at most m edges. In the weighted matching [G85b, GT89, GT91] and maxi-mum \ufb02ow problems [GR98], for instance, the best algorithms for real- and integer-weighted graphs have running times di\ufb00ering by a polynomial factor. Slight Adjustment of Dijkstra's Algorithm to Solve Shortest Path Problem\u2026 783 The i-number weight of an edge is also called the i-distance between the corresponding two nodes. The distance between two vertices u and v, denoted d(u,v), is the length of a shortest. Suppose you are given a directed graph G = (V, E), with costs on the edges; the costs may be. A path in a graph is a sequence of adjacent vertices. The previous state of the art for this problem was total update time O \u0303(n 2 \u221am/\u03b5) for directed, unweighted graphs [2], and \u00d5(mn=\u03b5) for undirected, unweighted graphs [12]. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Shortest Path. An example of a weighted graph would be the distance between the capitals of a set of countries. You are given a weighted undirected graph. Design a linear time algorithm to find the number of different shortest paths (not necessarily vertex disjoint) between v and w. We present an improved algorithm for maintaining all-pairs (1 + \u03b5) approximate shortest paths under deletions and weight-increases. Consider the graph above. Your task is to find the shortest path between the vertex 1 and the vertex n. In weighted graphs, where we assume that the edge weights do not represent the communication speed, a straightforward distributed variant of the Bellman-Ford algorithm [2], [3], [4] computes. Shortest paths problems are among the most fundamental algorithmic graph problems. The actual shortest paths can also be constructed by modifying the above algorithm. Question: Shortest Paths(int[0). If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. To find the shortest path on a weighted graph, just doing a breadth-first search isn't enough - the BFS is only a measure of the shortest path based on number of edges. A comparison of the data obtained as a result of the study was carried out to find the best applications of implementations of the shortest path search algorithms in the Postgre SQL DBMS. Single source shortest path for undirected graph is basically the breadth first traversal of the graph. Shortest Path Algorithms?(a) Breadth-first search (BFS) can be used to perform single source short- est paths on any graph where all edges have the same costs 1. A cycle is a path where the first and last vertices are the same. Let s denote the number of edges of H. Shortest-Path Problems (cont'd) Single-source shortest path problem Given a weighted graph G = (V, E), and a distinguished start vertex, s, find the minimum weighted path from s to every other vertex in G The shortest weighted path from v 1 to v 6 has a cost of 6 and v 1 v 4 v 7 v 6. Shortest Path Problem. Dijkstra's algorithm (or Dijkstra's Shortest Path First algorithm, SPF algorithm) is an algorithm for finding the shortest paths between nodes in a graph, which may represent, for example, road networks. Node is a vertex in the graph at a position. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Using similar ideas, we can construct a (1+epsilon)-approximate distance oracle for weighted unit-disk graphs with O(1) query. Finding the shortest path, with a little help from Dijkstra! If you spend enough time reading about programming or computer science, there\u2019s a good chance that you\u2019ll encounter the same ideas. in Proceedings - 25th IEEE International Conference on High Performance Computing, HiPC 2018. Shortest Path Problem. Graph Algorithms I 12. We let k denote the number of source. Given a weighted line-graph (undirected connected graph, all vertices of degree 2, except two endpoints which have degree 1), devise an algorithm that preprocesses the graph in linear time and can return the distance of the shortest path between any two vertices in constant time. The All-Pairs Shortest Paths (APSP) problem seeks the shortest path distances between all pairs of vertices, and is one of the most fundamental graph problems. Once we have reached our destination, we continue searching until all possible paths are greater than 11; at that point we are certain that the shortest path is 11. The vertices V are connected to each other by these edges E. Key Graph Based Shortest Path Algorithms With Illustrations - Part 1: Dijkstra's And Bellman-Ford Algorithms Bellman-Ford algorithm is used to find the shortest paths from a source vertex to all other vertices in a weighted graph. Distributed Exact Weighted All-Pairs Shortest Paths in O\u02dc(n5/4) network is modeled by a weighted n-node m-edge graph G. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. We revisit a classical graph-theoretic problem, the single-source shortest-path (SSSP) problem, in weighted unit-disk graphs. An interesting side-effect of traversing a graph in BFS order is the fact that, when we visit a particular node, we can easily find a path from the source node to the newly visited node with the least number of edges. The Symmetric Shortest-Path Table Routing Conjecture Thomas L. Question: Shortest Paths(int[0). Mizrahi et al. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. Shortest Distance in a graph with two different weights : Given a weighted undirected graph having A nodes, a source node C and destination node D. Exercise 7 Consider the following modification of the Dijkstra\u2019s algorithm to work with negative weights:. The algorithm creates a tree of shortest paths from the starting vertex, the source, to all other points in the graph. The previous state of the art for this problem was total update time O \u0303(n 2 \u221am/\u03b5) for directed, unweighted graphs [2], and \u00d5(mn=\u03b5) for undirected, unweighted graphs [12]. I will cover several other graph based shortest path algorithms with concrete illustrations. Shortest path in a graph with weighted edges and vertices. If there are multiple paths from node 1 to a node that have the same minimum number of. Hi I have already posted a similar question but there was a misunderstanding of the problem by my side so here I post it again. There is one shortest path vertex 0 to vertex 0 (from each vertex there is a single shortest path to itself), one shortest path between vertex 0 to vertex 2 (0->2), and there are 4 different shortest paths from vertex 0 to vertex 6: 1. A shortest path between two nodes u and v in a graph is a path that starts at u and ends at v and has the lowest total link weight. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. Row i of the predecessor matrix contains information on the shortest paths from point i: each entry predecessors[i, j] gives the index of the previous node in the path from point i to point j. You are expected to do it in Time Complexity of O(A + M). For example, the length of v8,v9 equals 2, which is identical to the length of the. A shortest path (with the lowest weight) from $$u$$ to $$v$$; The weight of the path is the sum of the weights of its edges. Both algorithms were randomized and had constant query time. It was conceived by computer scientist Edsger W. Deterministic Partially Dynamic Single Source Shortest Paths in Weighted Graphs Aaron Bernstein May 30, 2017 Abstract In this paper we consider the decremental single-source shortest paths (SSSP) problem, where given a graph Gand a source node sthe goal is to maintain shortest distances between sand all other nodes. In this category, Dijkstra's algorithm is the most well known. An undirected graph that has a path from every vertex to every other vertex in the graph. As noted earlier, mapping software like Google or Apple maps makes use of shortest path algorithms. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. If you think carefully, it's easy to see that there can be many graphs such that the. A new approach to all-pairs shortest paths on real-weighted graphs Seth Pettie1 Department of Computer Sciences, The University of Texas at Austin, Austin, TX 78712, USA Abstract We present a new all-pairs shortest path algorithm that works with real-weighted graphs in the traditional comparison-additionmodel. The gist of Dijkstra's single source shortest path algorithm is as below : Dijkstra's algorithm finds the shortest path in a weighted graph containing only positive edge weights from a single source. How to use BFS for Weighted Graph to find shortest paths ? If your graph is weighted, then BFS may not yield the shortest weight paths. Chan\u2044 September 30, 2009 Abstract Inthe\ufb02rstpartofthepaper,wereexaminetheall-pairs shortest paths (APSP)problemand present a new algorithm with running time O(n3 log3 logn=log2 n), which improves all known algorithmsforgeneralreal-weighteddensegraphs. shortest path algorithm. In this tutorial, we will present a general explanation of both algorithms. A cycle is a path where the first and last vertices are the same. techniques to speed the computation of shortest paths in the discretization graph [3,21]. ; It uses a priority based dictionary or a queue to select a node / vertex nearest to the source that has not been edge relaxed. This turns out to be a problem that can be solved efficiently, subject to some restrictions on the edge costs. An observed image is composed of multiple components based on optical phenomena, such as light reflection and scattering. The multiplicity of a path is the maximum number of times that an edge appears in it. The latter only works if the edge weights are non-negative. It maintains a set S of vertices whose final shortest path from the source has already been determined and it repeatedly selects the left vertices with the minimum shortest-path estimate, inserts them. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. Directed and undirected graphs may both be weighted. I maintain a count for the number of shortest paths; I would like to use BFS from v first and also maintain a global level. Single Source Shortest Path. This section discusses three algorithms for this problem: breadth-\ufb01rst search for unweighted graphs, Dijkstra\u2019s algorithm for weighted graphs, and the Floyd-Warshall algorithm for computing distances between all pairs of vertices. And here is some test code: test_graph. Weighted directed graphs may be used to model communication networks, and shortest distances (shortest-path weights) between nodes may be used to suggest routes for messages. An undirected graph that has a path from every vertex to every other vertex in the graph. shortest path problem. shortest_paths calculates a single shortest path (i. We will see how simple algorithms like depth-\ufb01rst-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of \ufb01nding shortest paths. yenpathy: An R Package to Quickly Find K Shortest Paths Through a Weighted Graph Submitted 05 September 2019 This paper is under review which means review has begun. ! Here, the length of a path is simply the number of edges on the path. One of these challenges is the need to generate a limited number of test cases of a given regression test suite in a manner that does not compromise its defect detection ability. For example \ufb01nding the \u2018shortest path\u2019 between two nodes, e. We would then assign weights to vertices, not edges. Python \u2013 Get the shortest path in a weighted graph \u2013 Dijkstra Posted on July 22, 2015 by Vitosh Posted in VBA \\ Excel Today, I will take a look at a problem, similar to the one here. FindShortestPath[g, s, All] generates a ShortestPathFunction[] that can be applied repeatedly to different t. [email\u00a0protected] ple, Figure 1a illustrates a graph G, and Figure 1e shows an aug-mented graph G\u2217 constructed from G. According to [11], [12], [13], since 1959 Dijkstra's algorithm has been recognized as the best algorithm and used as method to find the shortest path. nodes in a given directed graph is a very common problem. You are also given a positive integer k. A subgraph is a subset of a graph\u2019s edges (with associated vertices) that form a graph. \u201d Dijkstra\u2019s algorithm is an iterative algorithm that provides us with the shortest path from one particular starting node to all other nodes in the graph. The shortest path function can also be used to compute a transitive closure or for arbitrary length traversals. the number of pairs of vertices not including v, which for directed graphs is and for undirected graphs is. The all-pairs shortest paths problem for unweighted directed graphs was introduced by Shimbel (1953), who observed that it could be solved by a linear number of matrix multiplications that takes a total time of O(V 4). There are two types of weighted graphs: vertex weighted and edge weighted. Shortest path problem (SPP) is a fundamental and well-known combinatorial optimization problem in the area of graph theory. A weighted graph is a one which consists of a set of vertices V and a set of edges E. The shortest path problem is about finding a path between $$2$$ vertices in a graph such that the total sum of the edges weights is minimum. Dijkstra's Shortest Path Algorithm in Java. The single pair shortest path problem seeks to compute (u;v) and construct a shortest path from. Consumes a graph and two vertices, and returns the shortest path (in terms of number of vertices) between the two vertices. A subgraph is a subset of a graph\u2019s edges (with associated vertices) that form a graph. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. the sum of the weights of the edges in the paths is minimized. shortest path algorithm. Variations of the Shortest Path Problem. We present a new all-pairs shortest path algorithm that works with real-weighted graphs in the traditional comparison-addition model. I Therefore, the numbers d 1;d 2; ;d n must include an even number of odd numbers. Your solution should be complete in that it shows the shortest path from all starting vertices to all other vertices. All-Pairs Shortest Paths Problem To \ufb01nd the shortest path between all verticesv 2 V for a graph G =(V,E). This algorithm has numerous applications in network analysis, such as transportation planning. (2018) Decremental Single-Source Shortest Paths on Undirected Graphs in Near-Linear Total Update Time. Dijkstra\u2019s Shortest Path Algorithm in Java. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. It was conceived by computer scientist Edsger W. Shortest Path in a weighted Graph where weight of an edge is 1 or 2 Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex \u2018s\u2019 to a given destination vertex \u2018t\u2019. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. be contained in shortest augmenting paths, and the lay-ered network contains all augmenting paths of shortest length. Dating back some 3000 years, and initially consisting mainly of the study of permutations and combinations, its scope has broadened to include topics such as graph theory, partitions of numbers, block designs, design of codes, and latin squares. Simple path is a path with no repeated vertices. Find a TSP solution using state-of-the-art software, and then remove that dummy node (subtracting 2 from the total weight). C\u2026 Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. This module covers weighted graphs, where each edge has an associated weight or number. Discuss an efficient algorithm to compute a shortest path from node s to node t in a weighted directed graph G such that the path is of minimum cardinality among all shortest s - t paths in G graph-theory hamiltonian-path path-connected. Dijkstra\u2019s Algorithms describes how to find the shortest path from one node to another node in a directed weighted graph. Print the number of shortest paths from a given vertex to each of the vertices. Given a weighted line-graph (undirected connected graph, all vertices of degree 2, except two endpoints which have degree 1), devise an algorithm that preprocesses the graph in linear time and can return the distance of the shortest path between any two vertices in constant time. Dijkstra's Algorithm: Finds the shortest path from one node to all other nodes in a weighted graph. It finds a shortest-path tree for a weighted undirected graph. The SQL Server graph extensions are amazing. Introduction. In this week, you'll add a key feature of map data to our graph representation -- distances -- by adding weights to your edges to produce a \"weighted. Must Read: C Program. The Degree of a vertex is the number of edges incident on it. I'm using the networkx package in Python 2. adjacent b. Shortest Paths: Problem Statement Given a weighted graph and two vertices u and v, we want to find a path of minimum total weight between u and v Length (or distance) of a path is the sum of the weights of its edgesLength (or distance) of a path is the sum of the weights of its edges. Floyd-Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles). The Degree of a vertex is the number of edges incident on it. for unweighted. If the graph is unweighed, then finding the shortest path is easy: we can use the breadth-first search algorithm. A path graph is a graph consisting of a single path. So, the shortest path would be of length 1 and BFS would correctly find this for us. Slight Adjustment of Dijkstra's Algorithm to Solve Shortest Path Problem\u2026 783 The i-number weight of an edge is also called the i-distance between the corresponding two nodes. We know that breadth-first search can be used to find shortest path in an unweighted graph or in weighted graph having same cost of all its edges. Shortest path in a graph with weighted edges and vertices. PY - 2007/10/30. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. Variations of the Shortest Path Problem. We start with a directed and weighted graph $$G = (V, E)$$ with a weights function $$w: E \\rightarrow R$$ that maps edges to weights with real values; Find for each pair of vertices $$u$$, $$v \\in V$$. Shortest paths&Weighted graphs. An undirected graph that has a path from every vertex to every other vertex in the graph. Consider any node that is not the root: its possible distances from the root are all possible distances of its neighbors plus the weight of the connecting edges. If an edge is missing a special value, perhaps a negative value, zero or a large value to represent \"infinity\", indicates this fact. In a weighted graph with start node s, there are often multiple shortest paths from s to any other node. 1 def shortest_path_cycle(graph, s): 2 \u2019\u2019\u2019Single source shortest paths using DP on a graph with cycles but no 3 negative cycles. Dijkstra\u2019s algorithm is similar to Prim\u2019s algorithm. 4 5 Args: 6 graph: weighted graph with no negative cycles. b) Weighted graph matching: Mulmuley & Shah ob-served that their result for the shortest path problem yields the same lower bound for the WEIGHTED GRAPH MATCHING problem [3, Corollary 1. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using this algorithm. 1 Shortest paths and matrix multiplication 25. This means it finds a shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between source node and every other node. We know that breadth-first search can be used to find shortest path in an unweighted graph or in weighted graph having same cost of all its edges. Suppose we have to following graph: We may want to find out what the shortest way is to get from node A to node F. Graphs: Finding shortest paths Definition: weight of a path 4 Tecniche di programmazione A. Weighted Graph ( \u092d\u093e\u0930\u093f\u0924 \u0917\u094d\u0930\u093e\u092b ) Discrete Mathematics Shortest Path || Dijkstra Algorithm #weightedgraph #grewalpinky B. P = shortestpath(G,s,t) computes the shortest path starting at source node s and ending at target node t. 18 between shortest paths in networks and Hamilton paths in graphs ties in with our observation that finding paths of low weight (which we have been calling \"short\") is tantamount to finding paths with a high number of edges (which we might consider to be \"long\"). Exercise 7 Consider the following modification of the Dijkstra\u2019s algorithm to work with negative weights:. If you think carefully, it's easy to see that there can be many graphs such that the. A path in a graph is a sequence of adjacent vertices. As there are a number of different shortest path algorithms, we've gathered the most important to help you understand how they work and which is the best. This problem also is known as \"Print all paths between two nodes\". The shortest paths followed for the three nodes 2, 3 and 4 are as follows : 1/S->2 - Shortest Path Value : 1/S->3 - Shortest Path Value : 1/S->3->4 - Shortest Path Value :. Data Structure Graph Algorithms Algorithms. It runs in O (mn+n 2 log log n) time, improving on the long-standing bound of O (mn+n 2 log n) derived from an implementation of Dijkstra's algorithm with Fibonacci heaps. The total cost of a path in a graph is equal to the sum of the weights of the edges that connect the vertices in the path. By contrast, the graph you might create to specify the shortest path to hike every trail could be a directed graph, where the order and direction of edges matters. 6 2, 6(a), 6(c), 18 In Exercises 2-4 find the length of a shortest path between a and z in the given weighted graph. Shortest Paths, and Dijkstra's Algorithm: Overview Graphs with lengths/weights/costs on edges. Scienti\ufb01c collaboration networks. 2 You can use the tool to create a weighted graph with mouse gestures and show the MST and shortest paths. If say we were to find the shortest path from the node A to B in the undirected version of the graph, then the shortest path would be the direct link between A and B. shortest_paths calculates a single shortest path (i. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. Topics in this lecture include:. You may print the results of this algorithm to the screen or to a log file. A B C F D E 7 14 9 8 15 2 9 4 6 Figure 3. Dijkstra\u2019s algorithm is similar to Prim\u2019s algorithm. This module covers weighted graphs, where each edge has an associated weight or number. I'm restricting myself to Unweighted Graph only. The relationship shown in the proof of Property 21. Shortest Path (Unweighted Graph) Goal: find the shortest route to go from one node to another in a graph. Undirected graph. Shortest paths in an edge-weighted digraph 4->5 0. This means it finds the shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between the source node and every other node. If the graph is weighted, it is a path with the minimum sum of edge weights. Algorithms to find shortest paths in a graph are given later. Breadth First Search, BFS, can find the shortest path in a non-weighted graphs or in a weighted graph if all edges have the same non-negative weight. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. world as a directed graph, the problem of directing the robot was transformed into a shortest-path problem. As noted earlier, mapping software like Google or Apple maps makes use of shortest path algorithms. Variations of the Shortest Path Problem. If an edge is missing a special value, perhaps a negative value, zero or a large value to represent \"infinity\", indicates this fact. Print the number of shortest paths from a given vertex to each of the vertices. Consider the graph above. The best algorithm (modulo small polylogarithmic improvements) for this problem runs in cubic time, a. By contrast, the graph you might create to specify the shortest path to hike every trail could be a directed graph, where the order and direction of edges matters. Your solution should be complete in that it \u2026. Representing a Graph. For example, the two paths we mentioned in our example are C, B and C, A, B. Thanks for pointing to Gephi. In order to solve the load-balancing problem for coarse-grained parallelization, the relationship between the computing time of a single-source shortest-path length of node and the features of node is studied. The all-pairs shortest paths problem for unweighted directed graphs was introduced by Shimbel (1953), who observed that it could be solved by a linear number of matrix multiplications that takes a total time of O(V 4). CS 340 Programming Assignment VII: Single-Source Shortest Paths in generally Weighted graphs Description: You are to implement both the DAG SP algorithm and the Bellman-Ford algorithm for single-source shortest paths based on whether or not you detect a cycle. a i g f e d c b h 25 15 10 5 10. Hart, Nilsson, and Raphael [12] discovered how to use this information to improve the efficiency of computing the shortest path. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The number parent[w] is the predecessor of w on a shortest path from v to w, or -1 if none exists. minimum path sizeis the shortest distance measured in the number of edges traversed. The algorithm can be implemented with O(m+nlog(n)) time; therefore, it is efficient in sparsely connected networks. The Bellman-Ford algorithm is a single-source shortest path algorithm. Michael Quinn, Parallel Programming in C with MPI and OpenMP,. Decomposing the observed image into individual components is an important process for various computer vision tasks. Description and notations-building the graph Given a weighted graph G = (V, E), where V. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. By comparison, if the graph is permitted. The order of a graph is the number of nodes. P( s,t ) is the shortest path between the given vertices and containing the least sum of edge weights on the path from to. Bellman-Ford Algorithm is computes the shortest paths from a single source vertex to all of the other vertices in a weighted digraph. Shortest paths&Weighted graphs. It was conceived by computer scientist Edsger W. Finding the shortest path, with a little help from Dijkstra! If you spend enough time reading about programming or computer science, there\u2019s a good chance that you\u2019ll encounter the same ideas. This technique is applied to a minute level bid/ask quote dataset consisting of rates constructed from all G10 currency pairs. \u2022 In addition, the first time we encounter a vertex may, we may not have found the shortest path to it, so we need to delay committing to that path. 1 Shortest Path Queries Let G= (V;E;\u02da) be a road network (i. Start by setting the distance of all notes to infinity and the source's distance to 0. The previous state of the art for this problem was total update time O \u0303(n 2 \u221am/\u03b5) for directed, unweighted graphs [2], and \u00d5(mn=\u03b5) for undirected, unweighted graphs [12]. A path in a graph is a sequence of adjacent vertices. shortest s-tpath in G n. Algorithms to find shortest paths in a graph are given later. These weights represent the cost to traverse the edge. A cycle is a path where the first and last vertices are the same. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. It turns out that it is as easy to compute the shortest paths from s to every node in G (because if the shortest path from s to t is s = v0, v1, v2, , vk = t, then the path v0,v1 is the shortest path from s to v1, the path v0,v1,v2 is the shortest path from s to v2, the path v0,v1,v2,v3 is the shortest path from s to v3, etc. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Essentially, you replace the stack used by DFS with a queue. I'm looking for an algorithm that will perform as in the title. Again this is similar to the results of a breadth first search. It can be used in numerous fields such as graph theory, game theory, and network. Dijkstra's Shortest Path Algorithm in Java. linear-time. So there is a weighted graph, and the shortest paths of all pair of n. dijkstra_path_length\u00b6 dijkstra_path_length (G, source, target, weight='weight') [source] \u00b6. What is the longest simple path between s and t? Cycle. Graphs can be weighted (edges carry values) and directional (edges have direction). Let u and v be two vertices in G, and let P be a path in G from u to v. We are also given a starting node s \u2208 V. Dijkstra's Single Source Shortest Path. Exercise 7 Consider the following modification of the Dijkstra\u2019s algorithm to work with negative weights:. Among the various shortest path algorithms, Dijkstra\u2019s shortest path algorithm [1] is said to have better performance with regard to run time than the other algorithms. Finally, at k = 4, all shortest paths are found. Shortest Path in a weighted Graph where weight of an edge is 1 or 2 Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex \u2018s\u2019 to a given destination vertex \u2018t\u2019. However, the robot had a means for estimating Euclidean distances in his world. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. The Floyd-Warshall algorithm is an example of dynamic programming. 2 commits 1 branch. The previous state of the art for this problem was total update time O \u0303(n 2 \u221am/\u03b5) for directed, unweighted graphs [2], and \u00d5(mn=\u03b5) for undirected, unweighted graphs [12]. The Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem. However, most communication networks are dynamic, i. shortest path problem. Referred to as the shortest path between vertices For weighted graphs this is the path that has the smallest sum of its edge weights ijkstra\u2019salgorithm finds the shortest path between one vertex and all other vertices The algorithm is named after its discoverer, Edgser Dijkstra 24 The shortest path between B and G is: 1 4 3 5 8 2 2 1 5 1 B A. Design a linear time algorithm to find the number of different shortest paths (not necessarily vertex disjoint) between v and w. Graph Characteristics-Undirected-Weighted-Journey: (1, 7)-Shortest path: 1 \u2013 4 \u2013 6 \u2013 7 (in purple)-Total cost: 6. We first propose an exact (and. The graph has eight nodes. The Degree of a vertex is the number of edges incident on it. For the shortest path problem on positively weighted graphs the integer/real gap is only logarith-mic. Fine the shortest weighted path from a vertex, s, to every other vertex in the graph. The weights of the edges can be positive or negative. A near linear shortest path algorithm for weighted undirected graphs Abstract: This paper presents an algorithm for Shortest Path Tree (SPT) problem. Here the graph we consider is unweighted and hence the shortest path would be the number of edges it takes to go from source to destination. The longest path is based on the number of edges in the path if weighted == false and the unweighted shortest path algorithm is being used. Your solution should be complete in that it shows the shortest path from all starting vertices to all other vertices. Suppose G be a weighted directed graph where a minimum labeled w(u, v) associated with each edge (u, v) in E, called weight of edge (u, v). We introduce a metric on the. It asks for the number of different shortest paths. Day 3: Weighted Graphs. This problem also is known as \"Print all paths between two nodes\". So there is a weighted graph, and the shortest paths of all pair of n. texens | Hello World ! | Page 3 Hello World !. In this tutorial, we will present a general explanation of both algorithms. The graph given in the test case is shown as : * The lines are weighted edges where weight denotes the length of the edge. Graphs can be weighted (edges carry values) and directional (edges have direction). We present an improved algorithm for maintaining all-pairs (1 + \u03b5) approximate shortest paths under deletions and weight-increases. Give an efficient algorithm to solve the single-destination shortest paths problem. Thus, the shortest path between any two nodes is the path between the two nodes with the lowest total length. The weighted path length is given by \u2211C(vi,vi+1) i=1 k-1 The general problem Given an edge-weighted graph G = (V,E) and two vertices, vs \u2208V and vd \u2208V, find the path that starts at vs and ends at vd that has the smallest weighted path length Single-source shortest. Uses Dijkstra's Method to compute the shortest weighted path between two nodes in a graph. It finds a shortest-path tree for a weighted undirected graph. Crossref, ISI, Google Scholar; 19. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The shortest pair edge disjoint paths in the new graph corresponds to the required solution in the original graph. the path itself, not just its length) between the source vertex given in from, to the target vertices given in to. A subgraph is a subset of a graph\u2019s edges (with associated vertices) that form a graph. Length of a path is the sum of the weights of its edges. Let G = (V, E) be a directed graph and let v and w be two vertices in G. We present an improved algorithm for maintaining all-pairs (1 + \u03b5) approximate shortest paths under deletions and weight-increases. weighted \u203a cyclic vs. num_vertices() 11 for i in range(num_vertices): 12 result. Finding paths. The Bellman-Ford algorithm is a single-source shortest path algorithm. Topological Sort: Arranges the nodes in a directed, acyclic graph in a special order based on incoming edges. We will see how simple algorithms like depth-\ufb01rst-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of \ufb01nding shortest paths. Asked in Graphs , C Programming. Dijkstra\u2019s Algorithm for Finding the Shortest Path Through a Weighted Graph E. Simple path is a path with no repeated vertices. This algorithm aims to find the shortest-path in a directed or undirected graph with non-negative edge weights. BFS runs in O(E+V) time where E is the number of edges and V is number of vertices in the graph. Those times are the weights of those paths. A path in a graph is a sequence of adjacent vertices. We revisit a classical graph-theoretic problem, the single-source shortest-path (SSSP) problem, in weighted unit-disk graphs. Floyd-Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles). Hi I have already posted a similar question but there was a misunderstanding of the problem by my side so here I post it again. Mark Dolan CIS 2166 10. weighted \u203a cyclic vs. The N x N matrix of predecessors, which can be used to reconstruct the shortest paths. MINIMUM-WEIGHT SPANNING TREE 49 4. Year 2001 Predicted year 2005 Graph 2: Evolution of flu epidemic behavior. CS 312 Lecture 26 Finding Shortest Paths Finding Shortest Paths. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. The weight of the path P is the sum of the weights of all the _____ on the path P. This problem could be solved easily using (BFS) if all edge weights were ($$1$$), but here weights can take any value. The length of a path is the sum of the lengths of all component edges. The weighted inverse distance is the total number of shortest paths from node \\(s G From the Betweenness Centrality in Street Networks to Structural Invariants in Random Planar Graphs. In order to solve the load-balancing problem for coarse-grained parallelization, the relationship between the computing time of a single-source shortest-path length of node and the features of node is studied. So, the shortest path would be of length 1 and BFS would correctly find this for us. The total length of a path is the sum of the lengths of its component edges. In the notes below I am going to describe the Dijkstra algorithm, which is a widely-used algorithm for finding shortest paths in weighted, directed graphs. 2 commits 1 branch. In this post, I explain the single-source shortest paths problems out of the shortest paths problems, in which we need to find all the paths from one starting vertex to all other vertices. 7 (Single-Source Shortest Paths). The Degree of a vertex is the number of edges incident on it. Single-Source Shortest Path on Weighted Graphs. On weighted graphs Weighted Shortest Paths The shortest path from a vertex u to a vertex v in a graph is a path w1 = u, w2,\u2026,wn= v, where the sum: Weight(w1,w2)+\u2026+Weight(wn-1,wn) attains its minimal value among all paths that start at u and end at v The length of a path of n vertices is n-1 (the number of edges) If a graph is connected, and the weights are all non-negative, shortest paths exist for any pair of vertices Similarly for strongly connected digraphs with non-negative weights. Write an algorithm to print all possible paths between source and destination. Floyd-Warshall algorithm is used to find all pair shortest path problem from a given weighted graph. In real-life scenarios, the arc weighs in a shortest path of a network/graph have the several parameters which are very hard to define exactly (i. $\\begingroup$ Shortest Path on an Undirected Graph? might be interesting. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. Cover Photo by Thor Alvis on Unsplash. The path graph with n vertices is denoted by P n. In this tutorial, we will present a general explanation of both algorithms. Implementation of Dijkstra's algorithm in C++ which finds the shortest path from a start node to every other node in a weighted graph. There is a simple tweak to get from DFS to an algorithm that will find the shortest paths on an unweighted graph. the path itself, not just its length) between the source vertex given in from, to the target vertices given in to. (Consider what this means in terms of the graph shown above right. Furthermore, if we perform relaxation on the set of edges once, then we will at least have determined all the one-edged shortest paths; if we traverse the set of edges twice, we will have solved at least all the two-edged shortest paths; ergo, after the V-1 iteration. However, most communication networks are dynamic, i. shortest path functions use it as the cost of the path; community finding methods use it as the strength of the relationship between two vertices, etc. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. Shortest paths in an edge-weighted digraph 4->5 0. Graphs can be weighted (edges carry values) and directional (edges have direction). The Degree of a vertex is the number of edges incident on it. In the weighted matching [G85b, GT89, GT91] and maxi-mum \ufb02ow problems [GR98], for instance, the best algorithms for real- and integer-weighted graphs have running times di\ufb00ering by a polynomial factor. Implementation of Dijkstra's algorithm in C++ which finds the shortest path from a start node to every other node in a weighted graph. $\\begingroup$ Shortest Path on an Undirected Graph? might be interesting. Simple path is a path with no repeated vertices. Is there a cycle in the graph? Euler tour. So, the shortest path would be of length 1 and BFS would correctly find this for us. First, the paths should be shortest, then there might be more than one such shortest paths whose length are the same. A path in a graph is a sequence of adjacent vertices. Mark Dolan CIS 2166 10. goldberg_radzik (G, source[, weight]) Compute shortest path lengths and predecessors on shortest paths in weighted graphs. Increasingly, there is interest in using asymmetric structure of data derived from Markov chains and directed graphs, but few metrics are specifically adapted to this task. We may also want to associate some cost or weight to the traversal of an edge. It is a real time graph algorithm, and can be used as part of the normal user flow in a web or mobile application. You may print the results of this algorithm to the screen or to a log file. shortest_paths calculates a single shortest path (i. However, the robot had a means for estimating Euclidean distances in his world. PY - 2007/10/30. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. Based on Data Structures, Algorithms & Software Principles in CT. Finding shortest paths in weighted graphs In the past two weeks, you've developed a strong understanding of how to design classes to represent a graph and how to use a graph to represent a map. We \u02d9rst propose an exact (and deterministic) al-. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. , capacity, cost, demand, traffic frequency, time, etc. A subgraph is a subset of a graph\u2019s edges (with associated vertices) that form a graph. Path Graphs. dijkstra_path\u00b6 dijkstra_path (G, source, target, weight='weight') [source] \u00b6. This article presents a Java implementation of this algorithm. The previous state of the art for this problem was total update time O \u0303(n 2 \u221am/\u03b5) for directed, unweighted graphs [2], and \u00d5(mn=\u03b5) for undirected, unweighted graphs [12]. Weighted Graphs Data Structures & Algorithms 2 [email protected] \u00a92000-2009 McQuain Shortest Paths (SSAD) Given a weighted graph, and a designated node S, we would like to find a path of least total weight from S to each of the other vertices in the graph. Chapter 4 Algorithms in edge-weighted graphs Recall that anedge-weighted graphis a pair(G,w)whereG=(V,E)is a graph andw:E \u2192IR number of edges in a shortest path. , capacity, cost, demand, traffic frequency, time, etc. So, for this purpose we use the retroactive priority queue which allows to perform opeartions at any point of time. The Degree of a vertex is the number of edges incident on it. Shortest path algorithms are a family of algorithms designed to solve the shortest path problem. BFS always visits nodes in increasing order of their distance from the source. These values become important when calculating the. weighted graphs: \ufb01nd a path from a given source to a given target such that the consecutive weights on the path are nondecreasing and the last weight on the path is minimized. A single execution of the algorithm will. Analyze your algorithm. ) B C A E D F 0 7 2 3 5 8 8 4 7 1 2 5 2. , its number of edges. Finding the Shortest Path in Weighted Directed Acyclic Graph For the graph above, starting with vertex 1, what're the shortest paths(the path which edges weight summation is minimal) to vertex 2. The Classical Dijkstra\u2019s algorithm [16] solves the single-source shortest path problems in a simple graph. Shortest Distance in a graph with two different weights : Given a weighted undirected graph having A nodes, a source node C and destination node D. If G is a weighted graph, the length/weight of a path is the sum of the weights of the edges that compose the path. Dimitrios Skrepetos, PhD candidate David R. Check the manual pages of the functions working with weighted graphs for details. Another source vertex is also provided. A cycle is a path where the first and last vertices are the same. A subgraph is a subset of a graph\u2019s edges (with associated vertices) that form a graph. Print the number of shortest paths from a given vertex to each of the vertices. However, the resulting algorithm is no longer called DFS. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. The weight of the path P is the sum of the weights of all the _____ on the path P. We will see how simple algorithms like depth-\ufb01rst-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of \ufb01nding shortest paths. Dijkstra's Algorithm. ; It uses a priority based set or a queue to select a node / vertex nearest to the source that has not been edge relaxed. But how should we evaluate a path in a weighted graph? Usually, a path is assessed by the sum of weights of its edges and, based on this assumption, many authors proposed their. A graph is a series of nodes connected by edges. (2018) A Faster Distributed Single-Source Shortest Paths Algorithm. One algorithm for finding the shortest path from a starting node to a target node in a weighted graph is Dijkstra's algorithm. Count the number of updates At each step we compute the shortest path through a subset of vertices. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. In this tutorial, we will present a general explanation of both algorithms. Single source shortest path for undirected graph is basically the breadth first traversal of the graph. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. G\u2217 contains threeshortcuts: v8,v9, v9,v7,and v9,v10. One problem might be the shortest path in a given undirected, weighted graph. GUVEW ,, eE , where (non-negative real number) is a weight function, by which each edge is associ-ated with a weight The weight of a matching M is. 1 def shortest_path_cycle(graph, s): 2 '''Single source shortest paths using DP on a graph with cycles but no 3 negative cycles. num_vertices() 11 for i in range(num_vertices): 12 result. We consider a specific infinite graph here, namely the honeycomb grid. Fine the shortest weighted path from a vertex, s, to every other vertex in the graph. We introduce a metric on the. The N x N matrix of predecessors, which can be used to reconstruct the shortest paths. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. None of these algorithms for the WRP permit one to bound the number of links/turns in the produced path.\nqambklywe2nd3b orwzy8pzabfn8t u8alq5kgr9fsob q37tp6y0abvok 54tzslsdqeiki 9n2j071xpwfbpn a7yty5wu2uzainx 0aw0s5ehz9p8pk 3ed10wvxqf5l6j tx83hqwoda9 ys1yfzbr8a8 3uqb35oo2vkw35 njckdg32g1cigli v0c1nk8r083noq im9960fktswz8a vjtmeddlj4z8ye cwe2jgs60xw0 zt5bg6na6ub 6o8tcuclt9 c63dw0q8jy5m8 et17g5olz6v0m a5dp4yp8nz31f1 chprxj8ym91z15 i1owr2lzy3vnrbf ffifmn4qjn u53o27p33s ynlsiaia4xn45dk fr7j4x9behp98t jk6azoebao", "date": "2020-12-06 00:49:51", "meta": {"domain": "chicweek.it", "url": "http://isxy.chicweek.it/number-of-shortest-paths-in-a-weighted-graph.html", "openwebmath_score": 0.5556688904762268, "openwebmath_perplexity": 383.31142122654893, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.987758723627135, "lm_q2_score": 0.8633916222765627, "lm_q1q2_score": 0.8528226068102591}}
{"url": "https://math.stackexchange.com/questions/475927/how-to-prove-n-fracnen/475934", "text": "# How to prove $n!>(\\frac{n}{e})^{n}$\n\nProve that $n!>\\left(\\dfrac{n}{e}\\right)^{n}$.\n\nI used induction principle but cannot solve it for the $(m+1)$-th term after taking the $m$th term to be true.\n\nHere the key is to use the appropriate definition of $e^x$, namely:\n\n$$e^x = \\sum_{k=0}^{\\infty}\\frac{1}{k!}x^k$$\n\nPlugging in $x = n$ we get\n\n$$e^n = \\sum_{k=0}^{\\infty} \\frac{1}{k!}n^k$$\n\nand hence, breaking this sum up a little we get our inequality: $$n! e^n = n^n + \\sum_{k\\ne n} \\frac{n!}{k!}n^k > n^n$$\n\nHint: Show that $$\\ln(n!)=\\sum_{k=1}^n \\ln k >\\int_1^n\\ln x\\, dx.$$\n\n\u2022 That's a clever approach! Nice one! Aug 25 '13 at 18:02\n\nInductively, if $n!>\\frac{n^n}{e^n}$ and you multiply both sides by $n+1$, then you have that $(n+1)!>(n+1)\\frac{n^n}{e^n}$, so it suffices to prove that $(n+1)\\frac{n^n}{e^n}>\\frac{(n+1)^{n+1}}{e^{n+1}}$. Can you continue from here?\n\n\u2022 +1 This requires the least amount of machinery of the presented solutions, I think. Aug 25 '13 at 17:11\n\nHint: write out the series for $e^n$ and pick out a relevant term amongst the positive terms which make up the sum.\n\n\u2022 You posted a hint and Deven independently posted a solution using the same idea => 10 vote difference. Isn't that always so? +1 to both of you from me has been there from the beginning, of course. Aug 25 '13 at 18:10\n\nI had originally written this up for another question but it seems fitting here as well. Maybe this can help someone.\n\nDepending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \\in \\mathbb{N}}$, $(b_n)_{n \\in \\mathbb{N}}$ with\n\n\\begin{align} a_n ~~~&:=~~~ \\left ( 1 + \\frac{1}{n} \\right ) ^n \\\\ ~ \\\\ b_n ~~~&:=~~~ \\left ( 1 - \\frac{1}{n} \\right ) ^{-n} \\end{align}\n\nand\n\n$$\\underset{n \\rightarrow \\infty}{\\lim} a_n ~~~=~~~ \\underset{n \\rightarrow \\infty}{\\lim} b_n ~~~=~~~ e \\\\ ~ \\\\$$\n\nWhile both sequences converge to the same limit, $a_n$ approaches from the bottom and $b_n$ approaches from the top:\n\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom matplotlib import rcParams\nrcParams.update({'figure.autolayout': True})\n\npts = np.arange(0, 20, 1)\na_n = lambda n: (1+1/n)**n\nb_n = lambda n: (1-1/n)**(-n)\n\nplt.errorbar(x = pts, xerr = None, y = a_n(pts), yerr = None, fmt = \"bx\", markersize = \"5\", markeredgewidth = \"2\", label = \"$a_n$\")\nplt.errorbar(x = pts, xerr = None, y = b_n(pts), yerr = None, fmt = \"rx\", markersize = \"5\", markeredgewidth = \"2\", label = \"$b_n$\")\nplt.plot(pts, [np.exp(1)]*len(pts), color = \"black\", linewidth = 2, label = \"$e$\")\nplt.xlim(1.5, 14.5)\nplt.ylim(2.0, 3.5)\nplt.legend(loc = \"best\")\nplt.setp(plt.gca().get_legend().get_texts(), fontsize = \"22\")\nplt.show()\n\n\nSo we're going to use the following inequality:\n\n$$\\forall n \\in \\mathbb{N} ~ : ~~~~~ \\left ( 1 + \\frac{1}{n} \\right ) ^n ~~~~<~~~~ e ~~~~<~~~~ \\left ( 1 - \\frac{1}{n} \\right ) ^{-n} \\tag*{\\circledast} \\\\ ~ \\\\$$\n\nThesis\n\n$$\\forall n \\in \\mathbb{N}, ~ n \\geq 2 ~ : ~~~~~ e \\cdot \\left ( \\frac{n}{e} \\right )^n ~~~~<~~~~ n! ~~~~<~~~~ n \\cdot e \\cdot \\left ( \\frac{n}{e} \\right )^n \\\\ ~ \\\\$$\n\nProof By Induction\n\nBase Case\n\nWe begin with $n = 2$ and get\n\n\\begin{align} & ~ && e \\cdot \\left ( \\frac{2}{e} \\right )^2 ~~~~&&<~~~~ 2! ~~~~&&<~~~~ 2 \\cdot e \\cdot \\left ( \\frac{2}{e} \\right )^2 \\\\ ~ \\\\ & \\Leftrightarrow && e \\cdot \\frac{4}{e^2} ~~~~&&<~~~~ 1 \\cdot 2 ~~~~&&<~~~~ 2 \\cdot e \\cdot \\frac{4}{e^2} \\\\ ~ \\\\ & \\Leftrightarrow && \\frac{4}{e} ~~~~&&<~~~~ 2 ~~~~&&<~~~~ \\frac{8}{e} \\\\ ~ \\\\ &\\Leftrightarrow && 2 ~~~~&&<~~~~ e ~~~~&&<~~~~ 4 ~~~~ \\\\ \\end{align}\n\nWhich is a true statement.\n\nInductive Hypothesis\n\nTherefore the statement holds for some $n$. $\\tag*{$\\text{I.H.}$}$\n\nInductive Step\n\n\\begin{align} & ~ && e \\cdot \\left ( \\frac{n+1}{e} \\right )^{n+1} \\\\ ~ \\\\ & = && (n+1) \\cdot \\frac{1}{e} \\cdot e \\cdot \\left ( \\frac{n+1}{e} \\right )^n\\\\ ~ \\\\ & = && (n+1) \\cdot \\left ( \\frac{n}{e} \\right )^n \\cdot \\left ( \\frac{n+1}{n} \\right )^n\\\\ ~ \\\\ & = && (n+1) \\cdot \\left ( \\frac{n}{e} \\right )^n \\cdot \\left ( 1 + \\frac{1}{n} \\right )^n\\\\ ~ \\\\ & \\overset{\\circledast}{<} && (n+1) \\cdot \\left ( \\frac{n}{e} \\right )^n \\cdot e\\\\ ~ \\\\ & \\overset{\\text{I.H.}}{<} && (n+1) \\cdot n!\\\\ ~ \\\\ & = && (n+1)!\\\\ ~ \\\\ & = && (n+1) \\cdot n!\\\\ ~ \\\\ & \\overset{\\text{I.H.}}{<} && (n+1) \\cdot n \\cdot e \\cdot \\left ( \\frac{n}{e} \\right )^n\\\\ ~ \\\\ & = && (n+1) \\cdot e \\cdot \\left ( \\frac{n}{e} \\right )^{n+1} \\cdot e \\\\ ~ \\\\ & = && (n+1) \\cdot e \\cdot \\left ( \\frac{n+1}{e} \\right )^{n+1} \\cdot \\left ( \\frac{n}{n+1} \\right )^{n+1} \\cdot e \\\\ ~ \\\\ & = && (n+1) \\cdot e \\cdot \\left ( \\frac{n+1}{e} \\right )^{n+1} \\cdot \\left ( 1 - \\frac{1}{n+1} \\right )^{n+1} \\cdot e \\\\ ~ \\\\ & \\overset{\\circledast}{<} && (n+1) \\cdot e \\cdot \\left ( \\frac{n+1}{e} \\right )^{n+1} \\cdot \\left ( 1 - \\frac{1}{n+1} \\right )^{n+1} \\cdot \\left ( 1 - \\frac{1}{n+1} \\right )^{-(n+1)} \\\\ ~ \\\\ & = && (n+1) \\cdot e \\cdot \\left ( \\frac{n+1}{e} \\right )^{n+1} \\\\ ~ \\\\ \\end{align}\n\nConclusion\n\nTherefore the statement holds $\\forall n \\in \\mathbb{N}, ~ n \\geq 2$. $$\\tag*{\\square}$$", "date": "2021-09-17 23:09:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/475927/how-to-prove-n-fracnen/475934", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 589.33805963603, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587232667824, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8528225978190049}}
{"url": "https://math.stackexchange.com/questions/2415076/probability-of-drawing-cards-on-specific-draw-counts", "text": "# Probability of drawing cards on specific draw counts\n\nThere is a deck of $30$ cards, each card labeled a number from $1$ to $15$, with exactly $2$ copies of a card for each number. You draw $8$ cards. What is the probability that you draw the number '$1$' card by the $5$th draw (on the $5$th draw or before that), AND also drawing the number '$2$' card on or before the $8$th draw?\n\nI know how to compute the probability of drawing both the cards on or before the $5$th draw:\n\n$$\\frac{\\binom{2}{1}\\cdot \\binom{2}{1} \\cdot \\binom{26}{3}}{\\binom{30}{5}}$$\n\nSince there's $2$ ways to choose from each of the '$1$' and '$2$' cards, and then there's $26$ cards left after those $4$ cards so the other $3$ cards can be any of those $26$, and the total number of combinations you can draw $5$ cards from $30$.\n\nBut we want to expand this search to $8$ draws, and also at the same time want to have assumed that we have already drawn the '$1$' card on or before the $5$th draw (if we don't get the '$2$' card by the $5$th draw. How can I combine these ideas? Thanks\n\n\u2022 number 2 before 8-th draw... Then the 8th draw is irrelevant? Don't you mean (again) \"on or before\"? \u2013\u00a0drhab Sep 3 '17 at 9:07\n\u2022 @user152294 Just to check my try, have you the result of this exercise? \u2013\u00a0Robert Z Sep 3 '17 at 9:46\n\u2022 @drhab Yes, on or before \u2013\u00a0user152294 Sep 3 '17 at 17:29\n\u2022 @RobertZ No, I don't have the solution unfortunately \u2013\u00a0user152294 Sep 3 '17 at 17:30\n\u2022 @user152294 \"before 8-th draw\" means \"on or before 8-th draw\"? In case I have to modify my solution. P.S. Where does his exercise come from? \u2013\u00a0Robert Z Sep 3 '17 at 17:33\n\nI think it is more convenient to evaluate the probability of the complementary event: draw NO number '1' card by the $5$th draw, OR draw NO number '2' card. Here we consider the 2 copies of a card with the same number distinguishable (for example assume that one is red and the other is blue). Let $n^{\\underline{k}}:=n(n-1)\\cdots(n-k+1)$.\n\n1) If we draw NO number '1' card by the 5th draw then we can have zero, one or two '1's in the $6$th, $7$th or $8$th draw $$p_1=\\frac{1}{30^{\\underline{8}}}\\left(28^{\\underline{8}}+(3+3)\\cdot 28^{\\underline{7}}+3\\cdot 2\\cdot 28^{\\underline{6}}\\right)$$\n\n2) If we draw NO number '2' card then $$p_2=\\frac{28^{\\underline{8}}}{30^{\\underline{8}}}$$\n\n3) If we draw NO number '1' card by the $5$th draw AND NO number '2' card then, similarly to case 1), $$p_3=\\frac{1}{30^{\\underline{8}}}\\left(26^{\\underline{8}}+(3+3)\\cdot 26^{\\underline{7}}+3\\cdot 2\\cdot 26^{\\underline{6}}\\right)$$\n\nHence, the desired probability is $$p=1-(p_1+p_2-p_3)=211/1566\\approx 0.134738.$$\n\n\u2022 Is it also possible to just do these two cases? 1) Draw a '1' and '2' before the 5th draw, or 2) Draw a '1' before the 5th draw AND draw a '2' before the 8th draw? I'm not sure how to express 2) but would it be more cumbersome than the complementary events? \u2013\u00a0user152294 Sep 3 '17 at 17:34\n\u2022 @user152294 Yes you can consider two cases, but I think that with 3 cases is simpler. \u2013\u00a0Robert Z Sep 3 '17 at 17:36\n\u2022 How come in this solution, we don't have to use binomial coefficients? \u2013\u00a0user152294 Sep 3 '17 at 17:43\n\u2022 @user152294 Let me know if the official solution is the one that I have found. \u2013\u00a0Robert Z Sep 3 '17 at 17:44\n\u2022 @user152294 $3$ is the number of ways to place the red $1$ (position 6,7, 8), $3$ is the number of ways to place the blue $1$ (position 6,7, 8) and $3\\cdot 2$ is the number of ways to place the red $1$ and blue $1$ (positions (6,7), (6,8), (7,8), (7,6), (8,6), (8,7)). $28^{\\underline{8}}$, $28^{\\underline{7}}$ and $28^{\\underline{6}}$ are the number of ways to fill the remaining positions (cards different from 1). \u2013\u00a0Robert Z Sep 3 '17 at 18:21", "date": "2020-08-13 09:37:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2415076/probability-of-drawing-cards-on-specific-draw-counts", "openwebmath_score": 0.76022869348526, "openwebmath_perplexity": 236.12312727635154, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587261496031, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8528225950999307}}
{"url": "https://math.stackexchange.com/questions/1637518/what-does-f-a-mean/1637528", "text": "# What does $f|_A$ mean? [duplicate]\n\nIf $f$ a is a function and $A$ is a set, what could the notation\n\n$$f|_A$$\n\nmean? Is it perhaps \"restricted to set $A$\"?\n\n\u2022 Yes, if $f$ is a function with domain $D$, then for a set $A \\subseteq D$, $f|_A$ is used for the restriction of $f$ to $A$. Feb 2, 2016 at 15:43\n\u2022 That's what I would assume. You might have two functions $f$ and $g$ for which $f \\neq g$ but $f|_A = g|_A$ for some $A$. Feb 2, 2016 at 15:43\n\nIt means that I am constricting the domain of the function $f$. If $f:X\\to Y$, then $g=f|_A$ means that $g:A\\to Y$ where $A\\subseteq X$.\n\nIntuitively speaking, a function $f$ is constituted of three ingredients:\n\n\u2022 a domain;\n\u2022 a codomain;\n\u2022 a rule (that, for each element in the domain, assigns a unique element in the codomain).\n\nIf we change any of these three ingredients, we obtain a different function. In particular, if we change the domain by a subset $A$ of the original domain (keeping the codomain and the rule), we get a new function which is represented by $f|_A$.\n\nIn other words: given a function $f:X\\to Y$ and a set $A\\subset X$, the notation $f|_A$ denotes the function $g:A\\to Y$ given by $$g(x)=f(x),\\quad \\forall \\ x\\in A.$$\n\nThis is the usual meaning but, maybe, there are different meanings in other contexts.\n\nThat's correct.\n\nSuppose we have a function $$f : Y \\leftarrow X,$$ and a subset $A$ of $X$.\n\nApproach 0. Then $f \\restriction_A$ is defined as the unique function $Y \\leftarrow A$ that agrees with $f$ on $A$. That is: $$\\mathop{\\forall}_{a \\in A} ((f \\restriction_A)(a) = f(a))$$\n\nHowever, there's a cleaner way of formalizing this.\n\nApproach 1. Write $$\\mathrm{incl}_A : X \\leftarrow A$$ for the inclusion of $A$ into $X$. Then we can form the composite $$f \\circ \\mathrm{incl}_A : Y \\leftarrow A.$$ Write $f\\restriction_A$ as a shorthand for this composite.\n\nThe nice thing about Approach 1 is that it makes proving the basic properties of the restriction operator trivial. In particular:\n\nClaim. $(g \\circ f)\\restriction_A = g \\circ (f \\restriction_A)$\n\nProof.\n\n$$(g \\circ f)\\restriction_A = (g \\circ f) \\circ \\mathrm{incl}_A = g \\circ (f \\circ \\mathrm{incl}_A) = g \\circ (f \\restriction_A)$$\n\nApproach 1 is especially appealing from a category-theory perspective. In that context:\n\n\u2022 $X$ is an object of some category\n\u2022 a subobject of $X$ is by definition an object $\\underline{A}$ together with a monomorphism $\\mathrm{incl}_A : X \\leftarrow \\underline{A}$.\n\u2022 a partial morphism $Y \\leftarrow X$ consists of a subobject $A$ of $X$ together with a morphism $Y \\leftarrow \\underline{A}$.\n\nHence, if we're given a morphism $f : Y \\leftarrow X$ and a subobject $A$ of $X$, then we get a partial morphism $f \\restriction_A : Y \\leftarrow X$ by forming the obvious composite.\n\nThe notation $f|_A$ is probably best understood via a meaningful example. Before giving one (I hope it will be useful, anyway), it would probably be good to consult two decent references:\n\n2) Abstract Algebra by Dummit and Foote (p. 3, 3rd Ed.).\n\nThe relevant portion from the Wiki blurb:\n\nLet $f\\colon E\\to F$ be a function from a set $E$ to a set $F$, so that the domain of $f$ is in $E$ (i.e., $\\operatorname{dom}f\\subseteq E$). If $A\\subseteq E$, then the restriction of $f$ to $A$ is the function $f|_A\\colon A\\to F$.\n\nInformally, the restriction of $f$ to $A$ is the same function as $f$, but is only defined on $A\\cap\\operatorname{dom} f$.\n\nWiki's \"informal\" remark is the key part in my opinion. The following excerpt from Dummit and Foote's Abstract Algebra may be slightly more abstract, but I think a meaningful example will clear everything up.\n\nIf $A\\subseteq B$, and $f\\colon B\\to C$, we denote the restriction of $f$ to $A$ by $f|_A$. When the domain we are considering is understood we shall occasionally denote $f|_A$ again simply as $f$ even though these are formally different functions (their domains are different).\n\nIf $A\\subseteq B$ and $g\\colon A\\to C$ and there is a function $f\\colon B\\to C$ such that $f|_A=g$, we shall say $f$ is an extension of $g$ to $B$ (such a map $f$ need not exist nor be unique).\n\nExample: Let $g\\colon\\mathbb{Z}^+\\to\\{1\\}$ be defined by $g(x)=1$ and let $f\\colon\\mathbb{Z}\\setminus\\{0\\}\\to\\{1\\}$ be defined by $f(x)=\\dfrac{|x|}{x}$. Using the notation from the second paragraph above, we have $g\\colon A\\to C$ and $f\\colon B\\to C$, where\n\n\u2022 $A = \\mathbb{Z^+}$\n\u2022 $B=\\mathbb{Z}\\setminus\\{0\\}$\n\u2022 $C=\\{1\\}$\n\nand, clearly, $A\\subseteq B$. Thus, we have the following: \\begin{align} f|_A &\\equiv f\\colon\\mathbb{Z^+}\\to\\{1\\}\\tag{by definition}\\\\[0.5em] &= \\frac{|x|}{x}\\tag{by definition}\\\\[0.5em] &= \\frac{x}{x}\\tag{if $x\\in\\mathbb{Z^+}$, then $|x|=x$ }\\\\[0.5em] &= 1\\tag{simplify}\\\\[0.5em] &\\equiv g\\colon\\mathbb{Z^+}\\to\\{1\\}\\tag{by definition}\\\\[0.5em] &= g. \\end{align} Apart from some slight notational abuse, perhaps, the above example shows that $f$ is an extension of $g$ to $B$ since $f|_A=g$.", "date": "2022-05-20 20:19:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1637518/what-does-f-a-mean/1637528", "openwebmath_score": 0.9944095015525818, "openwebmath_perplexity": 158.69505554878359, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969694071562, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.8528137558472355}}
{"url": "https://math.stackexchange.com/questions/888151/mean-and-variance-of-piecewise-normal-distribution", "text": "Mean and Variance of \"Piecewise\" Normal Distribution\n\nNote - I put piecewise in quotes because I don't think it's the right term to use (I can't figure out what to call it).\n\nI am building a program to model the load that a user places on a server. The load a user produces follows a normal distribution. Depending on which application the user is using, however, the mean and variance of that normal distribution will be different. What I am trying to do is calculate the overall mean and variance for a user's activity given the proportions of time they use each application.\n\nFor example, Application A follows $\\mathcal{N}(100, 50)$ and Application B follows $\\mathcal{N}(500, 20)$. If a user uses A 50% of the time and B the other 50%, what is the mean and variance of the data that the user would produce during a day?\n\nI'm able to simulate this by selecting a number from a uniform distribution between 0 and 1 and then generating a value from the appropriate distribution. Something like this:\n\n$f(x) = \\begin{cases} \\mathcal{N}(100, 50), &0 \\le x \\lt 0.5\\\\ \\mathcal{N}(500, 20), &0.5 \\le x \\lt 1\\\\ \\end{cases}$\n\nWhen I simulate a large number of these values and measure the results, it looks like the mean is just\n\n$\\sum\\limits_{i=1}^n\\mu_ip$\n\nwhere $p$ is the percentage of the day a user is using each application.\n\nI can't figure out what pattern the variance follows or what the formula might be to determine it without measuring a bunch of simulated values (When I simulate the above example, the variance looks to be something close to 41500).\n\nI'd appreciate confirmation that how I'm calculating the combined mean is correct and some help in figuring out how to determine the variance of the overall distribution.\n\nLet the two normal random variables be $X$ and $Y$, where $X$ is chosen with probability $p$, and $Y$ is chosen with probability $q=1-p$.\n\nIf $W$ is the resulting random variable, then $\\Pr(W\\le w)=p\\Pr(X\\le w)+q\\Pr(Y\\le w)$.\n\nDifferentiate. We get $f_W(w)=pf_X(w)+qf_Y(w)$.\n\nThe mean of $W$ is $\\int_{-\\infty}^\\infty wf_W(w)$. Calculate. We get $$\\int_{-\\infty}^\\infty w(pf_X(w)+qf_Y(w))\\,dw.$$ This is $pE(X)+qE(Y)$, confirming your observation.\n\nFor the variance, we want $E(W^2)-(E(W))^2$. For $E(W^2)$, we calculate $\\int_{-\\infty}^{\\infty} w^2(pf_X(w)+qf_Y(w))\\,dw$. This is $pE(X^2)+qE(Y^2)$.\n\nBut $pE(X^2)= p(\\text{Var}(X)+(E(X))^2)$ and $qE(Y^2)= q(\\text{Var}(Y)+(E(Y))^2)$\n\nPutting things together we get $$\\text{Var}(W)=p\\text{Var}(X)+q\\text{Var}(Y)+ p(E(X))^2+q(E(Y))^2- (pE(X)+qE(Y))^2.$$\n\nRemark: For a longer discussion, please look for Mixture Distributions.\n\n\u2022 So, to put it more generally: $\\text{Var}(W) = \\sum\\limits_{i=1}^np_i(\\text{Var}(X_i) + E(X_i)^2) - (\\sum\\limits_{i=1}^np_iE(X_i))^2$ Aug 5, 2014 at 17:57\n\u2022 Yes, the same argument works for $X_i$ with probability $p_i$. One could generalize further, to sums to $\\infty$, also to \"continuous\" mixtures. Mixtures happen a lot in Statistics, since populations can often be stratified. Aug 5, 2014 at 18:15\n\u2022 Perfect. Thank you so much for your help! I already ran some simulations with a few different combinations and it works beautifully. Aug 5, 2014 at 18:47\n\u2022 You are welcome. Fast work. Aug 5, 2014 at 19:49\n\nWhat you are wanting are the distributional quantities of the mixture distribution. There are some convenient formulas to do this: let $T$ be the unconditional load a user places on the system, and let $X$ be a Bernoulli random variable that indicates whether a user is on system $A$ or $B$. So $A = (T \\mid X = 0) \\sim \\mathrm{Normal}(\\mu_A = 100, \\sigma_A^2 = 50)$, and $B = (T \\mid X = 1) \\sim \\mathrm{Normal}(\\mu_B = 500, \\sigma_B^2 = 20)$. That is to say, \\begin{align*} \\mathrm{E}[T \\mid X = 0] &= \\mu_A, \\\\ \\mathrm{E}[T \\mid X = 1] &= \\mu_B, \\\\ \\mathrm{Var}[T \\mid X = 0] &= \\sigma_A^2, \\\\ \\mathrm{Var}[T \\mid X = 1] &= \\sigma_B^2. \\end{align*}\n\nThen by the law of total expectation, \\begin{align*} \\mathrm{E}[T] &= \\mathrm{E}[\\mathrm{E}[T \\mid X]] \\\\ &= \\mathrm{E}[T \\mid X = 0]\\Pr[X = 0] + \\mathrm{E}[T \\mid X = 1]\\Pr[X = 1] \\\\ &= \\mu_A (1-p) + \\mu_B p,\\end{align*} where $p$ is the probability that a user is on system $B$. The variance is calculated by \\begin{align*} \\mathrm{Var}[T] &= \\mathrm{E}[\\mathrm{Var}[T \\mid X]] + \\mathrm{Var}[\\mathrm{E}[T \\mid X]] \\\\ &= \\mathrm{Var}[T \\mid X = 0]\\Pr[X = 0] + \\mathrm{Var}[T \\mid X = 1]\\Pr[X = 1] + \\mathrm{Var}[\\mathrm{E}[T \\mid X]] \\\\ &= \\sigma_A^2 (1-p) \\sigma_B^2 p + \\mathrm{Var}[\\mathrm{E}[T \\mid X]]. \\end{align*} The last term requires a little subtlety to understand. The variable $\\mathrm{E}[T \\mid X]$ is a generalized Bernoulli, which takes on the value $\\mu_A$ with probability $1-p$ and $\\mu_B$ with probability $p$ (rather than $0$ and $1$). So we may write this as $$\\mathrm{E}[T \\mid X] = X(\\mu_B - \\mu_A) + \\mu_A,$$ where $X \\sim \\mathrm{Bernoulli}(p)$. Therefore, $$\\mathrm{Var}[\\mathrm{E}[T \\mid X]] = \\mathrm{Var}[(\\mu_B - \\mu_A)X + \\mu_A] = (\\mu_B - \\mu_A)^2 \\mathrm{Var}[X] = (\\mu_B - \\mu_A)^2 p(1-p).$$ Hence $$\\mathrm{Var}[T] = \\sigma_A^2 (1-p) + \\sigma_B^2 p + (\\mu_B - \\mu_A)^2 p(1-p).$$ The general case with $n$ systems $S_i \\sim \\mathrm{Normal}(\\mu_i, \\sigma_i^2)$, $i = 1, 2, \\ldots, n$, where the user is on system $i$ with a probability of $p_i$, with $\\sum_{i=1}^n p_i = 1$, involves a categorical distribution rather than a Bernoulli.\n\nRecall that $V[X]=E[X^2]-E[X]^2$. Hence if you are given $E$ and $V$ of $X_1,X_2$ and combine them as described to a new random variable $Y$ by using a 0-1-random variable $Z$, i.e. picking $X_1$ if $Z=1$ and picking $X_2$ if $Z=0$, we find $$E[Y]=P(Z=1)\\cdot E[Y|Z=1]+P(Z=0)\\cdot E[Y|Z=0]=pE[X_1]+(1-p)E[X_2].$$ By the same argument we find $$E[Y^2] = pE[X_1^2]+(1-p)E[X_2^2]$$ Substituting $E[X_i^2]=V[X_i]+E[X_i]^2$, we obtain \\begin{align}V[Y]&=E[Y^2]-E[Y]^2 \\\\&= p(V[X_1]+E[X_1]^2) + (1-p)(V[X_2]+E[X_2]^2)-(pE[X_1]+(1-p)E[X_2])^2\\\\ &=pV[X_1]+(1-p)V[X_2]+p(1-p)(E[X_1]^2+E[X_2]^2)-2p(1-p)E[X_1]E[X_2]\\\\ &=pV[X_1]+(1-p)V[X_2]+p(1-p)(E[X_1]-E[X_2])^2.\\end{align}", "date": "2022-11-30 05:12:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/888151/mean-and-variance-of-piecewise-normal-distribution", "openwebmath_score": 0.9997225403785706, "openwebmath_perplexity": 265.11445611770057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983596967483837, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8528137508002526}}
{"url": "http://cnx.org/content/m18901/latest/?collection=col10613/latest", "text": "# Connexions\n\nYou are here: Home \u00bb Content \u00bb Applied Finite Mathematics \u00bb Linear Equations\n\n### Lenses\n\nWhat is a lens?\n\n#### Definition of a lens\n\n##### Lenses\n\nA lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.\n\n##### What is in a lens?\n\nLens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.\n\n##### Who can create a lens?\n\nAny individual member, a community, or a respected organization.\n\n##### What are tags?\n\nTags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.\n\n#### Endorsed by (What does \"Endorsed by\" mean?)\n\nThis content has been endorsed by the organizations listed. Click each link for a list of all content endorsed by the organization.\n\u2022 College Open Textbooks\n\nThis collection is included inLens: Community College Open Textbook Collaborative\nBy: CC Open Textbook Collaborative\n\n\"Reviewer's Comments: 'I recommend this book for undergraduates. The content is especially useful for those in finance, probability statistics, and linear programming. The course material is [\u2026]\"\n\nClick the \"College Open Textbooks\" link to see all content they endorse.\n\nClick the tag icon to display tags associated with this content.\n\n#### Affiliated with (What does \"Affiliated with\" mean?)\n\nThis content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.\n\u2022 Bookshare\n\nThis collection is included inLens: Bookshare's Lens\nBy: Bookshare - A Benetech Initiative\n\n\"Accessible versions of this collection are available at Bookshare. DAISY and BRF provided.\"\n\nClick the \"Bookshare\" link to see all content affiliated with them.\n\n\u2022 Featured Content\n\nThis collection is included inLens: Connexions Featured Content\nBy: Connexions\n\n\"Applied Finite Mathematics covers topics including linear equations, matrices, linear programming, the mathematics of finance, sets and counting, probability, Markov chains, and game theory.\"\n\nClick the \"Featured Content\" link to see all content affiliated with them.\n\nClick the tag icon to display tags associated with this content.\n\n### Recently Viewed\n\nThis feature requires Javascript to be enabled.\n\n### Tags\n\n(What is a tag?)\n\nThese tags come from the endorsement, affiliation, and other lenses that include this content.\n\nInside Collection:\n\nCollection\u00a0by: Rupinder Sekhon. E-mail the author\n\n# Linear Equations\n\nModule by: Rupinder Sekhon. E-mail the author\n\nSummary: This chapter covers principles of linear equations. After completing this chapter students should be able to: graph a linear equation; find the slope of a line; determine an equation of a line; solve linear systems; and complete application problems using linear equations.\n\n## Chapter Overview\n\nIn this chapter, you will learn to:\n\n1. Graph a linear equation.\n2. Find the slope of a line.\n3. Determine an equation of a line.\n4. Solve linear systems.\n5. Do application problems using linear equations.\n\n## Graphing a Linear Equation\n\nEquations whose graphs are straight lines are called linear equations. The following are some examples of linear equations:\n\n2x3y=62x3y=6 size 12{2x - 3y=6} {}, 3x=4y73x=4y7 size 12{3x=4y - 7} {}, y=2x5y=2x5 size 12{y=2x - 5} {}, 2y=32y=3 size 12{2y=3} {}, and x2=0x2=0 size 12{x - 2=0} {}.\n\nA line is completely determined by two points, therefore, to graph a linear equation, we need to find the coordinates of two points. This can be accomplished by choosing an arbitrary value for xx size 12{x} {} or yy size 12{y} {} and then solving for the other variable.\n\n### Example 1\n\n#### Problem 1\n\nGraph the line: y=3x+2y=3x+2 size 12{y=3x+2} {}\n\n### Example 2\n\n#### Problem 1\n\nGraph the line: 2x+y=42x+y=4 size 12{2x+y=4} {}\n\nThe points at which a line crosses the coordinate axes are called the intercepts. When graphing a line, intercepts are preferred because they are easy to find. In order to find the x-intercept, we let y=0y=0 size 12{y=0} {}, and to find the y-intercept, we let x=0x=0 size 12{x=0} {}.\n\n### Example 3\n\n#### Problem 1\n\nFind the intercepts of the line: 2x3y=62x3y=6 size 12{2x - 3y=6} {}, and graph.\n\n### Example 4\n\n#### Problem 1\n\nGraph the line given by the parametric equations: x=3+2tx=3+2t size 12{x=3+2t} {}, y=1+ty=1+t size 12{y=1+t} {}\n\n### Horizontal and Vertical Lines\n\nWhen an equation of a line has only one variable, the resulting graph is a horizontal or a vertical line.\n\nThe graph of the line x=ax=a size 12{x=a} {}, where aa size 12{a} {} is a constant, is a vertical line that passes through the point ( aa size 12{a} {}, 0). Every point on this line has the x-coordinate aa size 12{a} {}, regardless of the y-coordinate.\n\nThe graph of the line y=by=b size 12{y=b} {}, where bb size 12{b} {} is a constant, is a horizontal line that passes through the point (0, bb size 12{b} {}). Every point on this line has the y-coordinate bb size 12{b} {}, regardless of the x-coordinate.\n\n#### Example 5\n\n##### Problem 1\n\nGraph the lines: x=2x=2 size 12{x= - 2} {} , and y=3y=3 size 12{y=3} {}.\n\n## Slope of a Line\n\n### Section Overview\n\nIn this section, you will learn to:\n\n1. Find the slope of a line if two points are given.\n2. Graph the line if a point and the slope are given.\n3. Find the slope of the line that is written in the form y=mx+by=mx+b size 12{y= ital \"mx\"+b} {}.\n4. Find the slope of the line that is written in the form Ax+By=cAx+By=c size 12{ ital \"Ax\"+ ital \"By\"=c} {}.\n\nIn the last section, we learned to graph a line by choosing two points on the line. A graph of a line can also be determined if one point and the \"steepness\" of the line is known. The precise number that refers to the steepness or inclination of a line is called the slope of the line.\n\nFrom previous math courses, many of you remember slope as the \"rise over run,\" or \"the vertical change over the horizontal change\" and have often seen it expressed as:\n\nriserun, vertical changehorizontal change, \u0394y\u0394xetc.riserun size 12{ { {\"rise\"} over {\"run\"} } } {}, vertical changehorizontal change size 12{ { {\"vertical change\"} over {\"horizontal change\"} } } {}, \u0394y\u0394x size 12{ { {\u0394y} over {\u0394x} } } {}etc.\n(5)\n\nWe give a precise definition.\n\nDefinition 1:\n\nIf ( x1x1 size 12{x rSub { size 8{1} } } {}, y1y1 size 12{y rSub { size 8{1} } } {}) and ( x2x2 size 12{x rSub { size 8{2} } } {}, y2y2 size 12{y rSub { size 8{2} } } {}) are two different points on a line, then the slope of the line is\n\nSlope = m = y 2 y 1 x 2 x 1 Slope = m = y 2 y 1 x 2 x 1 size 12{\"Slope\"=m= { {y rSub { size 8{2} } - y rSub { size 8{1} } } over {x rSub { size 8{2} } - x rSub { size 8{1} } } } } {}\n\n### Example 6\n\n#### Problem 1\n\nFind the slope of the line that passes through the points (-2, 3) and (4, -1), and graph the line.\n\n### Example 7\n\n#### Problem 1\n\nFind the slope of the line that passes through the points (2, 3) and (2, -1), and graph.\n\n### Example 8\n\n#### Problem 1\n\nGraph the line that passes through the point (1, 2) and has slope 3434 size 12{ - { {3} over {4} } } {} .\n\n### Example 9\n\n#### Problem 1\n\nFind the slope of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.\n\n### Example 10\n\n#### Problem 1\n\nFind the slope of the line y=3x+2y=3x+2 size 12{y=3x+2} {}.\n\n### Example 11\n\n#### Problem 1\n\nDetermine the slope and y-intercept of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.\n\n## Determining the Equation of a Line\n\n### Section Overview\n\nIn this section, you will learn to:\n\n1. Find an equation of a line if a point and the slope are given.\n2. Find an equation of a line if two points are given.\n\nSo far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on it, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.\n\nAn equation of a line can be written in two forms, the slope-intercept form or the standard form.\n\nThe Slope-Intercept Form of a Line: y = mx + b y = mx + b size 12{y= ital \"mx\"+b} {}\n\nA line is completely determined by two points, or a point and slope. So it makes sense to ask to find the equation of a line if one of these two situations is given.\n\n### Example 12\n\n#### Problem 1\n\nFind an equation of a line whose slope is 5, and y-intercept is 3.\n\n### Example 13\n\n#### Problem 1\n\nFind the equation of the line that passes through the point (2, 7) and has slope 3.\n\n### Example 14\n\n#### Problem 1\n\nFind an equation of the line that passes through the points (\u20131, 2), and (1, 8).\n\n### Example 15\n\n#### Problem 1\n\nFind an equation of the line that has x-intercept 3, and y-intercept 4.\n\nThe Standard form of a Line: Ax + By = C Ax + By = C size 12{ ital \"Ax\"+ ital \"By\"=C} {}\n\nAnother useful form of the equation of a line is the Standard form.\n\nLet LL size 12{L} {} be a line with slope mm size 12{m} {}, and containing a point (x1,y1)(x1,y1) size 12{ $$x rSub { size 8{1} } ,y rSub { size 8{1} }$$ } {}. If (x,y)(x,y) size 12{ $$x,y$$ } {} is any other point on the line LL size 12{L} {}, then by the definition of a slope, we get\n\nm = y y 1 x x 1 m = y y 1 x x 1 size 12{m= { {y - y rSub { size 8{1} } } over {x - x rSub { size 8{1} } } } } {}\n(21)\ny y 1 = m ( x x 1 ) y y 1 = m ( x x 1 ) size 12{y - y rSub { size 8{1} } =m $$x - x rSub { size 8{1} }$$ } {}\n(22)\n\nThe last result is referred to as the point-slope form or point-slope formula. If we simplify this formula, we get the equation of the line in the standard form, Ax+By=CAx+By=C size 12{ ital \"Ax\"+ ital \"By\"=C} {}.\n\n### Example 16\n\n#### Problem 1\n\nUsing the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope \u20133/5.\n\n### Example 17\n\n#### Problem 1\n\nFind the standard form of the line that passes through the points (1, -2), and (4, 0).\n\n### Example 18\n\n#### Problem 1\n\nWrite the equation y=2/3x+3y=2/3x+3 size 12{y= - 2/3x+3} {} in the standard form.\n\n### Example 19\n\n#### Problem 1\n\nWrite the equation 3x4y=103x4y=10 size 12{3x - 4y=\"10\"} {} in the slope-intercept form.\n\nFinally, we learn a very quick and easy way to write an equation of a line in the standard form. But first we must learn to find the slope of a line in the standard form by inspection.\n\nBy solving for yy size 12{y} {}, it can easily be shown that the slope of the line Ax+By=CAx+By=C size 12{ ital \"Ax\"+ ital \"By\"=C} {} is A/BA/B size 12{ - A/B} {}. The reader should verify.\n\n### Example 20\n\n#### Problem 1\n\nFind the slope of the following lines, by inspection.\n\n1. 3x5y=103x5y=10 size 12{3x - 5y=\"10\"} {}\n2. 2x+7y=202x+7y=20 size 12{2x+7y=\"20\"} {}\n3. 4x3y=84x3y=8 size 12{4x - 3y=8} {}\n\nNow that we know how to find the slope of a line in the standard form by inspection, our job in finding the equation of a line is going to be very easy.\n\n### Example 21\n\n#### Problem 1\n\nFind an equation of the line that passes through (2, 3) and has slope 4/54/5 size 12{ - 4/5} {}.\n\nIf you use this method often enough, you can do these problems very quickly.\n\n## Applications\n\nNow that we have learned to determine equations of lines, we get to apply these ideas in real-life equations.\n\n### Example 22\n\n#### Problem 1\n\nA taxi service charges $0.50 per mile plus a$5 flat fee. What will be the cost of traveling 20 miles? What will be cost of traveling xx size 12{x} {} miles?\n\n### Example 23\n\n#### Problem 1\n\nThe variable cost to manufacture a product is $10 and the fixed cost$2500. If xx size 12{x} {} represents the number of items manufactured and yy size 12{y} {} the total cost, write the cost function.\n\n### Example 24\n\n#### Problem 1\n\nIt costs $750 to manufacture 25 items, and$1000 to manufacture 50 items. Assuming a linear relationship holds, find the cost equation, and use this function to predict the cost of 100 items.\n\n### Example 25\n\n#### Problem 1\n\nThe freezing temperature of water in Celsius is 0 degrees and in Fahrenheit 32 degrees. And the boiling temperatures of water in Celsius, and Fahrenheit are 100 degrees, and 212 degrees, respectively. Write a conversion equation from Celsius to Fahrenheit and use this equation to convert 30 degrees Celsius into Fahrenheit.\n\n### Example 26\n\n#### Problem 1\n\nThe population of Canada in the year 1970 was 18 million, and in 1986 it was 26 million. Assuming the population growth is linear, and x represents the year and y the population, write the function that gives a relationship between the time and the population. Use this equation to predict the population of Canada in 2010.\n\n## More Applications\n\n### Section Overview\n\nIn this section, you will learn to:\n\n1. Solve a linear system in two variables.\n2. Find the equilibrium point when a demand and a supply equation are given.\n3. Find the break-even point when the revenue and the cost functions are given.\n\nIn this section, we will do application problems that involve the intersection of lines. Therefore, before we proceed any further, we will first learn how to find the intersection of two lines.\n\n### Example 27\n\n#### Problem 1\n\nFind the intersection of the line y=3x1y=3x1 size 12{y=3x - 1} {} and the line y=x+7y=x+7 size 12{y= - x+7} {}.\n\n### Example 28\n\n#### Problem 1\n\nFind the intersection of the lines 2x+y=72x+y=7 size 12{2x+y=7} {} and 3xy=33xy=3 size 12{3x - y=3} {} by the elimination method.\n\n### Example 29\n\n#### Problem 1\n\nSolve the system of equations x+2y=3x+2y=3 size 12{x+2y=3} {} and 2x+3y=42x+3y=4 size 12{2x+3y=4} {} by the elimination method.\n\n### Example 30\n\n#### Problem 1\n\nSolve the system of equations 3x4y=53x4y=5 size 12{3x - 4y=5} {} and 4x5y=64x5y=6 size 12{4x - 5y=6} {}.\n\n### Supply, Demand and the Equilibrium Market Price\n\nIn a free market economy the supply curve for a commodity is the number of items of a product that can be made available at different prices, and the demand curve is the number of items the consumer will buy at different prices. As the price of a product increases, its demand decreases and supply increases. On the other hand, as the price decreases the demand increases and supply decreases. The equilibrium price is reached when the demand equals the supply.\n\n### Example 31\n\n#### Problem 1\n\nThe supply curve for a product is y=1.5x+10y=1.5x+10 size 12{y=1 \".\" 5x+\"10\"} {} and the demand curve for the same product is y=2.5x+34y=2.5x+34 size 12{y= - 2 \".\" 5x+\"34\"} {}, where xx size 12{x} {} is the price and y the number of items produced. Find the following.\n\n1. How many items will be supplied at a price of $10? 2. How many items will be demanded at a price of$10?\n3. Determine the equilibrium price.\n4. How many items will be produced at the equilibrium price?\n\n### Break-Even Point\n\nIn a business, the profit is generated by selling products. If a company sells xx size 12{x} {} number of items at a price PP size 12{P} {}, then the revenue RR size 12{R} {} is PP size 12{P} {} times xx size 12{x} {} , i.e., R=PxR=Px size 12{R=P cdot x} {}. The production costs are the sum of the variable costs and the fixed costs, and are often written as C=mx+bC=mx+b size 12{C= ital \"mx\"+b} {}, where xx size 12{x} {} is the number of items manufactured.\n\nA company makes a profit if the revenue is greater than the cost, and it shows a loss if the cost is greater than the revenue. The point on the graph where the revenue equals the cost is called the Break-even point.\n\n### Example 32\n\n#### Problem 1\n\nIf the revenue function of a product is R=5xR=5x size 12{R=5x} {} and the cost function is y=3x+12y=3x+12 size 12{y=3x+\"12\"} {}, find the following.\n\n1. If 4 items are produced, what will the revenue be?\n2. What is the cost of producing 4 items?\n3. How many items should be produced to break-even?\n4. What will be the revenue and the cost at the break-even point?\n\n## Content actions\n\nPDF | EPUB (?)\n\n### What is an EPUB file?\n\nEPUB is an electronic book format that can be read on a variety of mobile devices.\n\nPDF | EPUB (?)\n\n### What is an EPUB file?\n\nEPUB is an electronic book format that can be read on a variety of mobile devices.\n\n#### Collection to:\n\nMy Favorites (?)\n\n'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.\n\n| A lens I own (?)\n\n#### Definition of a lens\n\n##### Lenses\n\nA lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.\n\n##### What is in a lens?\n\nLens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.\n\n##### Who can create a lens?\n\nAny individual member, a community, or a respected organization.\n\n##### What are tags?\n\nTags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.\n\n| External bookmarks\n\n#### Module to:\n\nMy Favorites (?)\n\n'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.\n\n| A lens I own (?)\n\n#### Definition of a lens\n\n##### Lenses\n\nA lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.\n\n##### What is in a lens?\n\nLens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.\n\n##### Who can create a lens?\n\nAny individual member, a community, or a respected organization.\n\n##### What are tags?\n\nTags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.\n\n| External bookmarks\n\n### Reuse / Edit:\n\nReuse or edit collection (?)\n\n#### Check out and edit\n\nIf you have permission to edit this content, using the \"Reuse\u00a0/\u00a0Edit\" action will allow you to check the content out into your Personal Workspace or a shared Workgroup and then make your edits.\n\n#### Derive a copy\n\nIf you don't have permission to edit the content, you can still use \"Reuse\u00a0/\u00a0Edit\" to adapt the content by creating a derived copy of it and then editing and publishing the copy.\n\n| Reuse or edit module (?)\n\n#### Check out and edit\n\nIf you have permission to edit this content, using the \"Reuse\u00a0/\u00a0Edit\" action will allow you to check the content out into your Personal Workspace or a shared Workgroup and then make your edits.\n\n#### Derive a copy\n\nIf you don't have permission to edit the content, you can still use \"Reuse\u00a0/\u00a0Edit\" to adapt the content by creating a derived copy of it and then editing and publishing the copy.", "date": "2013-12-18 08:04:26", "meta": {"domain": "cnx.org", "url": "http://cnx.org/content/m18901/latest/?collection=col10613/latest", "openwebmath_score": 0.3370840847492218, "openwebmath_perplexity": 2506.347901162992, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969713304754, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8528137507560292}}
{"url": "https://mathoverflow.net/questions/316044/product-of-concave-functions-and-harmonic-mean", "text": "# Product of concave functions and harmonic mean\n\nI discovered something interesting, and I would like to know whether it is a known result or not. Say that a function $$f: \\Omega \\subset \\mathbb{R} \\rightarrow \\mathbb{R_+^*}$$ is $$\\alpha$$-concave if $$f^\\alpha$$ is concave.\n\nLet $$f$$ a $$\\alpha$$-concave function, $$g$$ a $$\\beta$$-concave function. Let $$\\gamma$$ be the half of the harmonic mean of $$\\alpha$$ and $$\\beta$$, ie. $$\\begin{equation*}\\frac{1}{\\gamma} = \\frac{1}{\\alpha} + \\frac{1}{\\beta}.\\end{equation*}$$ Then $$fg$$ is a $$\\gamma$$-concave function.\n\nThis can be proved by computing the hessian of $$fg$$. Have you already seen this result in the literature?\n\n\u2022 A small remark: $\\gamma$ is not the harmonic mean, but half of it. \u2013\u00a0Ivan Izmestiev Nov 24 '18 at 6:59\n\u2022 You're perfectly right, thanks! I've just corrected it. \u2013\u00a0LacXav Nov 26 '18 at 15:51\n\n\nWe want to prove that $$(f(ax+by)g(ax+by))^\\g \\ge a(f(x)g(x))^\\g + b (f(y)g(y))^\\g$$ for every $$x, y$$ and $$a+b = 1$$. Since we know that $$f(ax+by)^\\a \\ge af(x)^\\a + bf(y)^\\a$$ and $$g(ax+by)^\\b \\ge ag(x)^\\b + bg(y)^\\b$$ it is enough to prove that\n\n$$(af(x)^\\a + bf(y)^\\a)^{\\frac{\\g}{\\a}} (ag(x)^\\b + bg(y)^\\b)^{\\frac{\\g}{\\b}}\\ge af(x)^\\g g(x)^\\g + b f(y)^\\g g(y)^\\g.$$\n\nConsider set $$\\{x, y\\}$$ as measurable space with $$m(x) = a$$, $$m(y) = b$$. Then our desired inequality (after rising to the power $$\\frac{1}{\\g}$$) is nothing but Holder inequality with parameters $$\\a, \\b$$.\n\nUnfortunatly, I too do not have any reference for this cute result, but I'm sure it is known.\n\nThis is nice, and I do not know a reference for this.\n\nThe notion of $$p$$-concavity is mentioned, for example, in Section 9 of\n\nGardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008.\n\nNote that a natural interpretation of $$0$$-concavity is log-concavity. Your result generalizes the fact that the product of log-concave functions is log-concave.\n\nAlso your result is equivalent to the following.\n\nTheorem. If functions $$F$$ and $$G$$ are concave, then so is $$F^tG^{1-t}$$ for all $$t \\in [0,1]$$.\n\nIndeed, substitute $$f^\\alpha = F$$, $$g^\\beta = G$$, $$t = \\frac{\\beta}{\\alpha+\\beta}$$.\n\nThe concavity of $$F^tG^{1-t}$$ can be proved as follows. For every $$\\lambda \\in (0,1)$$ and $$\\mu = 1-\\lambda$$ the concavity of $$F$$ and $$G$$ imply $$\\begin{multline*} F^tG^{1-t}(\\lambda x + \\mu y) = F^t(\\lambda x + \\mu y) G^{1-t}(\\lambda x + \\mu y)\\\\ \\ge (\\lambda F(x) + \\mu F(y))^t (\\lambda G(x) + \\mu G(y))^{1-t}\\\\ \\ge (\\lambda F(x))^t(\\lambda G(x))^{1-t} + (\\mu F(y))^t(\\mu G(y))^{1-t}\\\\ = \\lambda F^tG^{1-t}(x) + \\mu F^tG^{1-t}(y) \\end{multline*}$$ (In the middle we have used the inequality $$(1+u)^t(1+v)^{1-t} \\ge 1+u^t v^{1-t}$$, for which I have only a nasty proof.)\n\nOne can prove the theorem also by computing the second derivative of $$F^tG^{1-t}$$ (which is fun), but the above argument is more general because it does not require differentiability.\n\nI think these results should be known, but have no reference.\n\nEDIT: As Alexei Kulikov pointed out, one can prove the Theorem by first proving the special case $$t=\\frac12$$ as a lemma (in this case the nasty inequality becomes $$\\sqrt{(1+u)(1+v)} \\ge 1 + \\sqrt{uv}$$, which follows from Cauchy-Schwarz). From the special case one infers by induction all $$t = p/2^n$$ cases, which implies the general case by a limit argument.\n\nEDIT: A reformulation of the theorem: the space of exp-concave functions is convex.\n\n\u2022 It seems to me that your inequality(and the initial problem as well) can be redused to the case $t=\\frac{1}{2}$ via mimicking proof of the fact that mid-point concavity implies concavity for continuous functions. And for $t=\\frac{1}{2}$ it is just C-S. \u2013\u00a0Aleksei Kulikov Nov 23 '18 at 23:33\n\u2022 And moreover it is actually Holder inequality on $2$-point measurable space with appropriate functions. \u2013\u00a0Aleksei Kulikov Nov 24 '18 at 0:37\n\u2022 @AlekseiKulikov: I will add the $t=1/2$ approach to my answer. What do you mean by Hoelder inequality on $2$-point measurable space? Could you write this up as an answer? \u2013\u00a0Ivan Izmestiev Nov 24 '18 at 6:52\n\u2022 Thanks for the reference, and having put in perspective. \u2013\u00a0LacXav Nov 26 '18 at 16:35", "date": "2021-05-06 07:19:22", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/316044/product-of-concave-functions-and-harmonic-mean", "openwebmath_score": 0.9301119446754456, "openwebmath_perplexity": 214.36030082378315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969641180276, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8528137461922771}}
{"url": "https://math.stackexchange.com/questions/764454/permutations-expressed-as-product-of-transpositions", "text": "# Permutations expressed as product of transpositions\n\nThere is a theorem that states that all permutations can be expressed as a product of transpositions. I have a couple of questions about this theorem:\n\n1. Does the product which is equal to the permutation always start from the identity permutation?\n\nIn the proof for this theorem our professor has argued that every permutation can be transformed into the identity permutation by applying a certain number of transpositions, e.g. if $\\sigma$ is a permutation not equal to the identity permutation then you can apply, say l transpositions, so that you get: $\\tau_l \\circ \\tau_{l-1} \\circ .... \\circ \\tau_1 \\circ \\sigma = id \\Rightarrow \\tau_1^{-1} \\circ \\tau_2^{-1} \\circ ... \\circ \\tau_l^{-1}=\\tau_1 \\circ \\tau_2 \\circ .... \\circ \\tau_l$.\n\nIs this product of transpositions always unique, or can you start from any arbitrary permutation and perform the required number of transpositions to get your permutation?\n\n2 If I form the composition of two permutations, say $\\sigma_1$ and $\\sigma_2$ given by: $$\\sigma_1 = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\\\ 3 & 4 &5 &6 &1 &2 \\\\ \\end{pmatrix}$$\n\n$$\\sigma_2 = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\\\ 2 & 4 &6 &1 &5 &3 \\\\ \\end{pmatrix}$$\n\nI can express them in terms of the following transpositions $$\\sigma_1 = (1,5)\\circ (2,6) \\circ (3,5) \\circ (4,6)$$ $$\\sigma_2 = (1,4) \\circ (2,4) \\circ (3,6)$$\n\nWhen I form the composition $\\sigma_1 \\circ \\sigma_2$ I get:\n\n$$\\sigma_1 \\circ \\sigma_2 = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\\\ 4 & 6 &2 &3 &1 &5 \\\\ \\end{pmatrix}$$\n\nBut since the product of the transpositions is equal to the permutations, I should get the same result when I use:\n\n$$\\sigma_1 \\circ \\sigma_2 = (1,5)\\circ (2,6) \\circ (3,5) \\circ (4,6) \\circ (1,4) \\circ (2,4) \\circ (3,6)$$\n\nbut I get:\n\n$$\\sigma_1 \\circ \\sigma_2 = \\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\\\ 6 & 1 &5 &3 &2 &4 \\\\ \\end{pmatrix}$$\n\nWhy doesn't this work?\n\n\u2022 I'm not sure I understand your first question - but for the second one, you're just reading the transpositions in the wrong order. Your convention for $\\sigma_1\\circ\\sigma_2$ is that $\\sigma_2$ is applied first, so with the transpositions you need to start from $(3,6)$ and read from right to left. So $1\\mapsto 4\\mapsto 6\\mapsto 2$ etc. \u2013\u00a0mdp Apr 22 '14 at 13:49\n\u2022 Thanks for your comment, but I did start from the right side: $\\begin{pmatrix} 1 & 2 & 3&4&5&6\\end{pmatrix} \\Rightarrow \\begin{pmatrix} 1 & 2 & 6&4&5&3\\end{pmatrix} \\Rightarrow \\begin{pmatrix} 1 & 4 & 6&2&5&3\\end{pmatrix} \\Rightarrow \\begin{pmatrix} 2 & 4 & 6&1&5&3\\end{pmatrix} \\Rightarrow \\begin{pmatrix} 2 & 4 & 6&3&5&1\\end{pmatrix} \\Rightarrow \\begin{pmatrix} 2 & 4 & 5&3&6&1\\end{pmatrix} \\Rightarrow \\begin{pmatrix} 2 & 1 & 5&3&6&4\\end{pmatrix} \\Rightarrow \\begin{pmatrix} 6 & 1 & 5&3&2&4\\end{pmatrix}$ I'm not sure what you mean by $1 \\rightarrow 4 \\rightarrow 6 \\rightarrow 2...$ \u2013\u00a0eager2learn Apr 22 '14 at 13:58\n\u2022 Ah, you made a different mistake that gives the same result as doing the transpositions backwards - on the third arrow you should be swapping the numbers $1$ and $4$ over, wherever they appear, not the first position with the fourth. (I was drawing my way of doing this calculation; put $1$ in on the right and see where it goes - it first gets mapped to $4$ (by $(1,4)$), then $4$ is mapped to $6$ by $(4,6)$, and finally $6$ is mapped to $2$ by $(2,6)$). \u2013\u00a0mdp Apr 22 '14 at 14:04\n\u2022 Yes that was the mistake I made. Thanks a lot for your help. \u2013\u00a0eager2learn Apr 22 '14 at 14:33\n\nNote that the product of transpositions that you used to express $\\sigma_1$, when composed, does not again yield $\\sigma_1$; I.e., you did not correctly express $\\sigma_1$ as a product of transpositions.\nRather, $\\sigma_1$ can be written $(1,5)\\circ (1, 3)\\circ (2, 6)\\circ (2, 4)$, or $(1, 3)\\circ (3, 5)\\circ (2, 4)\\circ(4, 6)$...\nSimilarly, $\\sigma_2$ is incorrectly decomposed. Two correct decompositions include $(1, 4)\\circ (1, 2)\\circ (3, 6)$ and $(1, 2)\\circ (2, 4) \\circ (3, 6)$...\n...which answers your question about uniqueness. When writing a permutation as a product of transpositions, there are many such ways to do this. What does not vary is the parity: an \"odd\" permutation is one that can only be decomposed to a product of an odd number of transpositions, and \"even\" permutations can only be decomposed into a product of an even number of transpositions. So, for example, $\\sigma_1$ is even, and $\\sigma_2$ is odd.\n\u2022 I don't understand this. If I apply $(1,5)\\circ (1,3) \\circ (2,6) \\circ (2,4)$ on the id permutation I get the permutation $\\begin{pmatrix} 5 &6&1&2&3&4 \\end{pmatrix}$ Why is this a transposition for $\\sigma_1$? I think this goes back to my original question, from which permutation do I start when applying those transpositions? \u2013\u00a0eager2learn Apr 22 '14 at 14:26\n\u2022 You start from the rightmost permutation. Where does it send $1$?. If it sends $1$ to $a$, then you move to the next transposition to its left to see where it sends $a$. Etc. \u2013\u00a0Namaste Apr 22 '14 at 14:30", "date": "2019-10-18 11:55:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/764454/permutations-expressed-as-product-of-transpositions", "openwebmath_score": 0.8850502371788025, "openwebmath_perplexity": 136.33681366525107, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9546474233166328, "lm_q2_score": 0.8933094003735664, "lm_q1q2_score": 0.8527955172911514}}
{"url": "https://wedding-photos-ideas.com/2fqzd/pampered-chef-rhsdaul/lnvno.php?page=ridge-regression-alpha-4d72aa", "text": "By default, glmnet will do two things that you should be aware of: Since regularized methods apply a penalty to the coefficients, we need to ensure our coefficients are on a common scale. Shows the effect of collinearity in the coefficients of an estimator. Ridge regression is a method by which we add a degree of bias to the regression estimates. The second line fits the model to the training data. Keep in mind, ridge is a regression \u2026 11. Ridge, LASSO and Elastic net algorithms work on same principle. In scikit-learn, a ridge regression model is constructed by using the Ridge class. Lasso is great for feature selection, but when building regression models, Ridge regression should be your first choice. In R, the glmnet package contains all you need to implement ridge regression. Active 2 years, 8 months ago. Let us first implement it on our above problem and check our results that whether it performs better than our linear regression model. They all try to penalize the Beta coefficients so that we can get the important variables (all in case of Ridge and few in case of LASSO). Ridge Regression is the estimator used in this example. Linear regression is the standard algorithm for regression that assumes a linear relationship between inputs and the target variable. The value of alpha is 0.5 in our case. Ridge Regression. Elastic net regression combines the properties of ridge and lasso regression. Ridge Regression is a neat little way to ensure you don't overfit your training data - essentially, you are desensitizing your model to the training data. Use the below code for the same. This is also known as $$L1$$ regularization because the regularization term is the $$L1$$ norm of the coefficients. Ridge regression - introduction\u00b6. Yes simply it is because they are good biased. Important things to know: Rather than accepting a formula and data frame, it requires a vector input and matrix of predictors. regression_model = LinearRegression() regression_model.fit(X_train, y_train) ridge = Ridge(alpha=.3) For example, to conduct ridge regression you may use the sklearn.linear_model.Ridge regression model. scikit-learn provides regression models that have regularization built-in. It works by penalizing the model using both the 1l2-norm1 and the 1l1-norm1. Generally speaking, alpha increases the affect of regularization, e.g. The Alpha Selection Visualizer demonstrates how different values of alpha influence model selection during the regularization of linear models. Ridge Regression is a technique for analyzing multiple regression data that suffer from multicollinearity. There are two methods namely fit() and score() used to fit this model and calculate the score respectively. Here, we are using Ridge Regression as a Machine Learning model to use GridSearchCV. One commonly used method for determining a proper \u0393 \\boldsymbol{\\Gamma} \u0393 value is cross validation. Note that scikit-learn models call the regularization parameter alpha instead of $$\\lambda$$. The alpha parameter tells glmnet to perform a ridge (alpha = 0), lasso (alpha = 1), or elastic net (0 < alpha < 1) model. ridgeReg = Ridge(alpha=0.05, normalize=True) ridgeReg.fit(x_train,y_train) pred = ridgeReg.predict(x_cv) calculating mse When multicollinearity occurs, least squares estimates are unbiased, but their variances are large so they may be far from the true value. This notebook is the first of a series exploring regularization for linear regression, and in particular ridge and lasso regression.. We will focus here on ridge regression with some notes on the background theory and mathematical derivations that are useful to understand the concepts.. Then, the algorithm is implemented in Python numpy\n\n## ridge regression alpha\n\nFacebook Message Says Seen But No Time, Which Is Worse Agnostic Or Atheist, Activity Diagram For Online Shopping, Sprinkler Cad Block, Amaryllis Name Meaning, Easton Beast Speed, Meal Village Menu,", "date": "2021-01-24 03:42:29", "meta": {"domain": "wedding-photos-ideas.com", "url": "https://wedding-photos-ideas.com/2fqzd/pampered-chef-rhsdaul/lnvno.php?page=ridge-regression-alpha-4d72aa", "openwebmath_score": 0.39270108938217163, "openwebmath_perplexity": 967.5996990504327, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9898303413461358, "lm_q2_score": 0.8615382076534742, "lm_q1q2_score": 0.8527766581643764}}
{"url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-17-rapidly-decreasing-singular-values/", "text": "# Lecture 17: Rapidly Decreasing Singular Values\n\nFlash and JavaScript are required for this feature.\n\n## Description\n\nProfessor Alex Townsend gives this guest lecture answering the question \u201cWhy are there so many low rank matrices that appear in computational math?\u201d Working effectively with low rank matrices is critical in image compression applications.\n\n## Summary\n\nProfessor Alex Townsend's lecture\nWhy do so many matrices have low effective rank?\nSylvester test for rapid decay of singular values\nImage compression: Rank $$k$$ needs only $$2kn$$ numbers.\nFlags give many examples / diagonal lines give high rank.\n\nRelated section in textbook: III.3\n\nInstructor: Prof. Alex Townsend\n\nGILBERT STRANG: So let me use the mic to introduce Alex Townsend, who taught here at MIT-- taught Linear Algebra 18.06 very successfully.\n\nAnd then now he's at Cornell on the faculty, still teaching very successfully. And he was invited here yesterday for a big event over in Engineering.\n\nAnd he agreed to give a talk about a section of the book-- section 4.3-- which, if you look at it, you'll see is all about his work. And now you get to hear from the creator himself. OK.\n\nALEX TOWNSEND: OK. Thanks. Thank you, Gil. Thank you for inviting me here. I hope you're enjoying the course.\n\nToday I want to tell you a little about why there so many matrices that are low rank in the world. So as computational mathematicians-- Gil and myself-- we come across low-rank matrices all the time. And we started wondering, as a community, why?\n\nWhat is it about the problems that we are looking at? What makes low-rank matrices appear? And today I want to give you that story-- or at least an overview of that story.\n\nSo for this class, x is going to be n by n real matrix. So nice and square. And you already know, or are very comfortable with, the singular values of a matrix.\n\nSo the singular values of a matrix, as you know, are a sequence of numbers that are monotonically non-increasing that tell us all kinds of things about the matrix x.\n\nFor example, the number of nonzero singular values tell us the rank of the matrix x. And they also, you probably know, tell us how well a matrix x can be approximated by a low-rank matrix.\n\nSo let me just write two facts down that you already are familiar with. So here's a fact-- that, if I look at the number of non-zero singular values in x-- so I'm imagining there's going to be k non-zero singular values-- then we can say a few things about x.\n\nFor example, the rank of x, as we know, is k-- the number of non-zero singular values. But we also know from the SVD that we can decompose x into a sum of rank 1 matrices-- in fact, the sum of k of them.\n\nSo because x is rank k, we can write down a low-rank representation for x, and it involves k terms, like this.\n\nEach one of these vectors here is a column vector. So if I draw this pictorially, this guy looks like this, right? And we have k of them. So because x is rank k, we can write x as a sum of k rank 1 matrices.\n\nAnd we also have an initial fact that we already know-- that the dimension of the column space of x is equal to k, and the same with the row space. So the column space of x equals the row space of x-- the dimension-- and they all equal k.\n\nAnd so there are three facts we can determine from looking at this sequence of singular values of a matrix x. Of course, the singular value sequence is unique. X defines its own singular values.\n\nWhat we're interested in here is, what makes x? What are the properties of x that make sure that the singular values have a lot of zeros in that sequence? Can we try to understand what kind of x makes that happen?\n\nAnd we really like matrices that have a lot of zeros here, for the following reason-- we say x is low rank if the following holds, right? Because if we wanted to send x to our friend-- we're imagining x as picture where each entry is a pixel of that image.\n\nIf that matrix-- that image-- was low rank, we could send the picture to our friend in two ways. We could send one every single entry of x. And for us to do that, we would have to send n squared pieces of information, because we'd have to send every entry.\n\nBut if x is sufficiently low rank, we could also send our friend the vectors-- u, u1, v1, uk, up to vk. And how much pieces of data would we have to send our friend to get x to them if we sent in the low-rank form?\n\nWell, there's 2n here, 2n here numbers. There's k of them. So we'd have to send 2kn numbers. And we strictly say a matrix is low rank if it's more efficient to send x to our friend in low-rank form then in full-rank form.\n\nSo this, of course, by a little calculation, just shows us that, provided the rank is less than half the size of the matrix, we are calling the matrix low rank.\n\nNow, often, in practice, we demand more. We demand that k is much smaller than this number, so that it's far more efficient to send our friend the matrix x in low-rank form than in full-rank form. So the colloquial use of the word low rank is kind of this situation. But this is the strict definition of it.\n\nSo what do low-rank matrices look like? And to do that, I have some pictures for you. I have some flags-- the world flags. So these are all matrices x-- these examples-- because their flags happen to not be square. I hope you can all see this.\n\nBut the top row here are all matrices that are extremely low rank. For example, the Austria flag-- if you want to send that to your friend, that matrix is of rank 1. So all you have to do is send your friend two vectors. You have to tell your friend the column space and the row space. And there's only the dimensions of one of both.\n\nFor the English flag, you need to send them two column vectors and two row vectors-- u1, v1, u2 and v2. And as we go down this row, they get slowly fuller and fuller rank.\n\nSo the Japanese flag, for example, is low rank but not that small. The Scottish flag is essentially full rank. So it's very inefficient to send your friend the Scottish flag in low-rank form. You're better off sending almost every single entry.\n\nSo what do low-rank matrices look like? Well, if the matrix is extremely low rank, like rank 1, then when you look at that matrix-- like here, like the flag-- it's highly aligned with the coordinates-- with the rows and columns.\n\nSo if it's rank 1, the matrix is highly aligned-- like the Austria flag. And of course, as we add in more and more rank here, the situation gets a bit blurry. For example, once we get into the medium rank situation, which is a circle, it's very hard to see that the circle is actually, in fact, low rank.\n\nBut what I'm going to do was try to understand why the Scottish flag or diagonal patterns-- particularly a bad example for low rank. So I'm going to take the triangular flag to examine that more carefully. So the triangular flag looks like-- I'll take a square matrix and I'll color in the bottom half.\n\nSo this matrix is the matrix of ones below the diagonal. And I'm interested in this matrix and, in particular, its singular values, to try to understand why diagonal patterns are not particularly useful for low-rank compression. And this matrix of all ones has a really nice property that, if I take its inverse, it looks a lot like-- getting close to Gil's favorite matrix.\n\nSo if I take the inverse of this matrix-- it has an inverse because it's got ones on the diagonal-- then its inverse is the following matrix, which people familiar with finite difference schemes will notice the familiarity between that and the first order finite difference approximation.\n\nIn particular, if I go a bit further and times two of these together, and do this, then this is essentially Gil's favorite matrix, except one entry happens to be different-- ends up being this matrix, which is very close to the second order, central, finite difference matrix.\n\nAnd people have very well studied that matrix and know its eigenvalues, its singular values-- they know everything about that matrix. And you'll remember that if we know the eigenvalues of a matrix, like x transpose x, we know the singular values of x. So this allows us to show, by the fact that we know that, that the singular values of this matrix are not very amenable to low rank. They're all non-zero, and they don't even decay.\n\nSo I'm getting this from-- I rang up Gil, and Gil tells me these numbers. That allows us to work out exactly what the singular values of this matrix are, from the connection to finite differences.\n\nAnd so we can understand why this is not good by looking at the singular values. So the first singular value of x from this expression is going to be approximately 2n over pi. And from this expression, again, for the last guy-- the last singular value of x is going to be approximately a half.\n\nSo these singular values are all large. They're not getting close to zero. If I plotted these singular values on a graph-- so here's the first singular value, the second, and the n-th-- then what would the graph look like?\n\nWell, plot these numbers. Divide by this guy so that they all are bounded between 1 and 0 because of the normalization, because I divided by sigma 1 of x.\n\nAnd so we can plot them, and they will look like this kind of thing. This number happens to be here where they come to be pi over 4n, which is me dividing this number by this number, approximately.\n\nSo triangular patterns are extremely bad for low rank. We need things-- or we at least intuitively think that we need things-- aligned with the rows and columns, but the circle case happens to also be low rank. And so what happened to the Japanese flag? Why is the Japanese flag convenient for low rank?\n\nWell it's the fact that it's a circle, and there's lots of symmetry in a circle. So if I try to look at the rank of a circle, the Japanese flag, then I can bound this rank by decomposing the Japanese flag into two things. So this is going to be less than or equal to the rank of sum of two matrices, and I'll do it so that the decomposition works out.\n\nI have the circle. I'm going to cut out a rank one piece that lives in the middle of this circle. OK? And I'm going to cut out a square from the interior of that circle. OK? And I can figure out-- of course the rank is just bounded by the sum of those two ranks. This guy is bounded by rank one because it's highly aligned with the grid. So this guy is bounded by rank one. So this thing here plus 1.\n\nAnd now I have to try to understand the rank of this piece. Now this piece has lots of symmetry. For example, we know that the rank of that matrix is the dimension of the column space and the dimension of the row space. So when we look at this matrix, because of symmetry, if I divide this matrix in half along the columns, all the columns on the left appear on the right. So for example, the rank of this matrix is the same as the rank of that matrix because I didn't change the column space. OK?\n\nNow I go again and divide along the rows, and now the row dimension of this matrix is the same as the top half, because as I wipe out those, I didn't change the dimension of the row space because the rows are the same top-bottom. And so this becomes the rank of that tiny little matrix there. And because it's small, it won't have too large a rank. So this is definitely less than-- if I divide that up, a little guy here looks like that plus the other guy that looks like that plus 1.\n\nAnd so of course the row space of this matrix cannot be very high because this is a very thin matrix. There's lots of zeros in that matrix, only a few ones. And so you can go along and do a bit of trig to try to figure out how many rows are non-zero in this matrix.\n\nAnd a bit of trig tells you-- well it depends on the radius of this original circle. So if I make the original radius r of this Japanese flag, then the bound that you end up getting will be, for this matrix, r 1 minus square root 2 over 2 for this guy. That's a bit of trig. I've got to make sure that's an integer. And then again, here it's the same but for the column space. So this is me just doing trig. OK?\n\nAnd that's bound on the rank. It happens to be extremely good. And if you work out what that rank is and try to look back, you will find it's extremely efficient to send the Japanese flag to your friend in low rank form, because it's not full rank because these numbers are so small. So this comes out to be, like, approximately 1/2 r plus 1. So much smaller than what you would expect, because remember, a circle is almost the anti-version version of a line with the grid, but yet, it's still low rank. OK.\n\nNow most matrices that we come up with in computational math are not exactly of finite rank. They are of numerical rank. And so I'll just define that. So the numerical rank of a matrix is very similar to the rank, except we allow ourselves a little bit of wiggle room when we define it, and so that amount of wiggle room will be of parameter called tol called epsilon. That's a tolerance. I'm thinking of epsilon as a tolerance. That's the amount of wiggle room I'm going to give myself. OK.\n\nAnd we say that the numerical rank-- I'll put an epsilon there to denote numerical rank-- is k. k is the first singular value, or the last singular value, above epsilon. In the following sense, I'm copying the definition above but with epsilons instead of zeros. If this singular value is less than epsilon, relatively, and the kth one was not below. So k plus 1 is the first singular value below epsilon in this relative sense.\n\nSo of course the rank of 0x, if that was defined, is the same as the rank of x. OK? So this is just allowing ourselves some wiggle room. But this is actually what we're interested more in practice. All right? I don't want to necessarily send my friend the flag to exact precision. I would actually be happy to send my friend the flag up to 16 digits of precision, for example. They're not going to tell the difference between those two flags. And if I can get away with compressing the matrix a lot more once I have a little bit of wiggle room, that would be a good thing.\n\nSo we know from the Eckart and Young that the singular values tell us how well we can approximate x by a low-rank matrix. In particular, we know that the k plus 1 singular value of x tells us how well x can be approximated by a rank k matrix. OK? For example, when the rank was exactly k, the sigma k plus 1 was 0, and then this came out to be 0 and we found that x was exactly a rank k matrix.\n\nHere, because we have the wiggle room, the epsilon, we get an approximation, not an exact. So this is telling us how well we can approximate x by a rank k matrix. OK? That's what the singular values are telling us. And so this allows us to try our best to compress matrices but use low-rank approximation rather than doing things exactly.\n\nAnd of course, on a computer, when we're using floating point arithmetic, or on a computer because we always round numbers to the nearest 16-digit number, if epsilon was 16 digits, your computer wouldn't be able to tell the difference between x or x the rank k approximation if this number satisfied this expression. Your computer would think of x and xk as the same matrix because it would inevitably round both to epsilon, within epsilon. OK.\n\nSo what kind of matrices are numerically of low rank? Of course all low-rank matrices are numerically of low rank because the wiggle room can only help you but it's far more than that. There are many full-rank matrices-- matrices that don't have any singular values that are zero-- but the singular values decay rapidly to zero. That are full-rank matrices with low numerical rank because of the wiggle room.\n\nSo for example, here is the classic matrix that fits this regime. If I give you this, this is called the Hilbert matrix. This is a matrix that happens to have extremely low numerical rank but it's actually full rank, which means that I can approximate H by a rank k matrix where k is quite small very well, provided you give me some wiggle room, but it's not a low-rank matrix in the sense that if epsilon was zero here, you didn't allow me the wriggle room, all the singular values of this matrix are positive. So it's of low numerical rank but it's not a low-rank matrix.\n\nThe other classical example which motivated a lot of the research in this area was the Vandermonde matrix. So here is the Vandermonde matrix. An n by n version of it. Think of the xi's as real. And this is Vandermonde. This is the matrix that comes up when you try to do polynomial interpolation at real points.\n\nThis is an extremely bad matrix to deal with because it's numerically low rank, and often, you actually want to solve a linear system with this matrix. And numerical low rank implies that it's extremely hard to invert, so numerical low rank is not always good for you. OK? Often, we want the inverse, which exists, but it's difficult because V has low numerical rank.\n\nOK. So people have been trying to understand why these matrices are numerically of low rank for a number of years, and the classic reason why there are so many low-rank matrices is because the world is smooth, as people say. They say, the world is smooth. That's why matrices are of numerical low rank. And to illustrate that point, I will do an example.\n\nSo this is classically understood by a man called Reade in 1983, and this is what his reason was. I have a picture of John Reade. He's not very famous, so I try to make sure his picture gets around. He's playing the piano. It's, like, one of the only pictures I could find of him.\n\nSo what is in this reason? Why do people say this? Well here's an example that illustrates it. If I take a polynomial in two variables and I-- for example, this is a polynomial of two variables-- and my x matrix comes from sampling that polynomial integers-- for example, this matrix-- then that matrix happens to be of low rank-- mathematically of low rank, with epsilon equals zero.\n\nWhy is that? Well if I write down x in terms of matrices, you could easily see it. So this is made up of a matrix of all ones plus a matrix of j-- so that's 1, 2, up to n, 1, 2, up to n, because every entry of that matrix just depends on the row index. And then this guy depends on both j and k. So this is a multiplication table, right? So this is n, 2, 4, up to 2n, n, 2n, n squared. OK.\n\nClearly, the matrix of all ones is a rank one matrix. The same with this guy. The column space is just of dimension one. And the last guy also happens to be of rank one because I can write this matrix in rank one form, which is a column vector times a row vector. OK. So this matrix x is of rank three. I guess at lowest rank three is what I've actually shown. OK.\n\nNow of course this hasn't got to numerical low rank yet, so let's get ourselves there. So Reade knew this, and he said to himself, OK, well if I can approximate-- if x is actually coming from sampling a function, and I approximate that function by polynomial, then I'm going to get myself a low-rank approximation and get a bound on the numerical rank.\n\nSo in general, if I give you a polynomial of two variables, which can be written down-- it's degree n in both x and y. Let's just keep these indexes away from the matrix index. I give you this such polynomial, and I go away and I sample it and make a matrix X, then X, by looking at each term individually like I did there, will have low rank mathematically, with epsilon equals zero. This will have, at most, m squared rank, and if m is 3 or 4 or 10, it possibly could be low because this X could be a large matrix. OK.\n\nSo what Reade did for the Hilbert matrix was said, OK, well look at that guy. That guy looks like it's sampling a function. It looks like it's sampling the function 1 over x plus y minus 1. So he said to himself, well, that x, if I look at the Hilbert matrix, then that is sampling a function. It happens to not be a polynomial. It happens to be this function.\n\nBut that's OK because sampling polynomials, integers, gives me low rank exactly. Maybe sampling smooth functions, functions like this, can be well approximated by polynomials and therefore have low numerical rank. And that's what he did in this case. So he tried to find a p, a polynomial approximation to f. In particular, he looked at exactly this kind of approximation. So he has some numbers here so that things get dissolved later. And he tried to find a p that did this kind of approximation. So this approximates f.\n\nAnd then he would develop a low-rank approximation to X by sampling p. So he would say, OK, well if I let y be a sampling of p, then from the fact that f is a good approximation to p, y is a good approximation to X. And so this has finite rank. He wrote down that this must hold. And the epsilon comes out here because these factors were chosen just right. The divide by n was chosen so that the epsilon came out just there. OK?\n\nSo, for many years, that was kind of the canonical reason that people would give, that, well, if the matrix X is sampled from a smooth function, then we can approximate our function by a polynomial and get polynomial rank approximations. And therefore, the matrix X will be of low numerical rank. There's an issue with this reasoning, especially for the Hilbert matrix, that it doesn't actually work that well.\n\nSo for example, if I take the 1,000 by 1,000 Hilbert matrix and I look at its rank-- OK, well I've already told you this is full rank. You'll get 1,000. All the singular values are positive. If I look at the numerical rank of this 1,000 by 1,000 Hilbert matrix and I compute it, I compute the SVD and I look at how many are above epsilon where epsilon is 10 to the minus 15, so that means I can approximate the 1,000 by 1,000 Hilbert matrix by a rank 28 matrix and only give up 15-- there will be exact 15 digits, which is a huge amount. So this is what we get in practice, but Reade's argument here shows that the rank of this matrix, the numerical rank, is at most. So it doesn't do a very good job on the Hilbert matrix for bounding the rank, right?\n\nSo Reade comes along, takes this function. He tries to find a polynomial that does this, where epsilon is 10 to the minus 15. He finds that the number of terms that he needs in this expression here is around 719, and therefore, that's the rank that he gets. The bound on the numerical rank. The trouble is that 719 tells us that this is not of low numerical rank, but we know it is, so it's an unsatisfactory reason.\n\nSo there's been several people trying to come up with more appropriate reasons that explain the 28 here. And so one reason that I've started to use is another slightly different way of looking at things, which is to say the world is Sylvester. Now Sylvester, what does that mean? What does the word \"Sylvester\" mean in this case? It means that the matrices satisfy a certain type of equation called the Sylvester equation, and so the reason is really, many of these matrices satisfy a Sylvester equation, and that takes the form-- for sum A, B, and C.\n\nOK. So X is your matrix of interest. You want to show X is of numerical low rank. And the task at hand is to find an A, B, and C so that X satisfies that equation. OK. For example, the two matrices I've had on the board satisfy a Sylvester equation-- a Sylvester matrix equation. There is an A, a B, and a C for which they do this.\n\nFor example, remember the Hilbert matrix, which we have there still, but I'll write it down again. Has these entries. So all we need to do is to try to figure out an A, a B, and then a C so that we can make it fit a Sylvester equation. There's many different ways of doing this. The one that I like is the following, where if I put 1/2 here and 3/2 here, all the way down to n minus 1/2, times this matrix-- so this is timesing the top of this matrix by 1/2 and then 3/2 and then 5/2. So we're basically timesing each entry of this matrix by j minus 1/2.\n\nAnd then I do something on the right here, which I'm allowed to do because I've got the B freedom, and I choose this to be the same up to a minus sign. Then when you think about this, what is it doing? It's timing the jk entry-- this is-- by j minus 1/2. That's what this is doing. And what's this doing is timesing the jk entry by k minus 1/2. So this is, in total, timesing the jk entry by j plus k minus 1/2 minus 1/2, which is minus 1, so this is timesing the jk entry by j plus k minus 1. So it knocks out the denominator. And what we get from this equation is a bunch of ones. So in this case, A and B are diagonal, and C is the matrix of all ones. OK?\n\nWe can also do this for Vandermonde. So Vandermonde, you'll remember, looks like this. And then over here, we have this guy, the matrix that appears with polynomial interpolation. OK. So if I think about this, I could also come up with an A, B, and C, and for example, here's one that works. I can stick the x's on the diagonal.\n\nSo if you imagine what that matrix on the left is doing, it's timesing each column by the vector x. OK? So the first column of this matrix becomes x, the vector x. The second becomes the vector x squared, where squared is done entry-wise. And then the third entry is now x cubed, and when we get to the last, it's x to the n. OK? So that's like, multiply each column by the vector x.\n\nSo if I want to try to come up with a matrix-- so what's left is of low rank, is like of this form. What I can do is shift the columns. So I've noticed that this product here, this diagonal matrix, has made the first column x. So if I want to kill off that column, I can take the second column and permute it to the first column. I could take the third column and permute it to the second, the last column and permute it to the penultimate column here. And that will actually kill off a lot of what I've created in this matrix right here.\n\nSo let me write that down. This is a circumshift matrix. This does that permutation. I've put a minus 1 there. I could have put any number there. It doesn't make any difference. But this is the one that works out extremely nicely. Now this zeros out lots of things because of the way I've done the multiplication by x and the circumshift of the columns.\n\nAnd so the first column is zero because this first column is x, this first column is x, so I've got x minus x. This column was x squared minus x squared, so I got zero, and I just keep going along until that last column. That last column is a problem because the last column of this guy is x to the n, whereas I don't have x to the n in V, so there are some numbers here. OK.\n\nYou'll notice that C in both cases happens to be a low-rank matrix. In these cases, it happens to be of rank one. And so people were wondering, maybe it's something to do with satisfying these kind of equations that makes these matrices that appear in practice numerically of low rank. And after a lot of work in this area, people have come up with a bound that demonstrates that these kind of equations are key to understanding numerical low rank.\n\nSo if X satisfies a Sylvester equation, like this, and A is normal, B is normal-- I don't really want to concentrate on those two conditions. It's a little bit academic. Then-- people have found a bound on the singular values of any matrix that satisfies this kind of expression, and they found this following bound. OK, so here, the rank of C is r. So that goes there. So in our cases, the two examples we have, r is 1, so we can forget about r.\n\nThis nasty guy here is called the Zolotarev number. E is a set that contains the eigenvalues of A, and F is a set that contains the eigenvalues of B. OK. Now it looks like we have gained absolutely nothing by this bound, because I've just told you singular values are bound by Zolotarev numbers. That doesn't mean anything to anyone. It means a little bit to me but not that much.\n\nSo the key to this bound-- the reason this is useful-- is that so many people have worked out what these Zolotarev numbers actually mean. OK? So these are two key people that worked out what this bound means. And we have gained a lot because people have been studying this number. This is, like, a number that people cared about from 1870 onwards to the present day, and people have studied this number extremely well. So we've gained something by turning it into a more abstract problem that people have thought about previously, and now we can go to the literature on Zolotarev numbers, whatever they are, and discover this whole literature of work on this Zolotarev number.\n\nAnd the key part-- I'll just tell you the key-- is that the sets E and F are separated. So for example, in the Hilbert matrix, the eigenvalues of A can be read off the diagonal. What are they? They are between minus 1/2 and n minus 1/2. And the eigenvalues of B lie in the set minus 1/2 minus n plus 1/2. And the key reason why the Hilbert matrix is of low numerical rank is the fact that these two sets are separated, and that makes this Zolotarev number gets small extremely quickly with k.\n\nNow you might wonder why there is a question mark on Penzl's name. There is an unofficial curse that's been going on for a while. Both these men died while working on the Zolotarev problem. They both died at the age of 31. One died by being hit by a train, Zolotarev. It's unclear whether he was suicidal or it was accidental.\n\nPenzl died at the age of 31 in the Canadian mountains by an avalanche. I am currently not yet 31 but going to be 31 very soon, and I'm scared that I may join this list. OK. But for the Hilbert matrix, what you get from this analysis, based on these two peoples' work, is a bound on the numerical rank. And the rank that you get is, let's say, a world record bound. For the Hilbert matrix is 34, which is not quite 28, not yet, but it's far more descriptive of 28 than 719.\n\nAnd so this technique of bounding singular values by using these Zolotarev numbers is starting to gain popularity because we can finally answer to ourselves why there are so many low-rank matrices that appear in computational math. And it's all based on two 31-year-olds that died. And so if you ever wonder when you're doing computational science when a low rank appears and the smoothness argument does not work for you, you might like to think about Zolotarev and the curse. OK, thank you very much.\n\n[APPLAUSE]\n\nGILBERT STRANG: Thank you [INAUDIBLE] Excellent.\n\nALEX TOWNSEND: How does it work now?\n\nGILBERT STRANG: We're good. Yeah.\n\nALEX TOWNSEND: I'm happy to take questions if we have a minute, if you have any questions.\n\nGILBERT STRANG: How near of 31 are you?\n\nALEX TOWNSEND: [INAUDIBLE] I get a spotlight. I'm 31 in December.\n\nGILBERT STRANG: Wow. OK.\n\nALEX TOWNSEND: So they died at the age of 31, so you know, next year is the scary year for me. So I'm not driving anywhere. I'm not leaving my house until I become 32.\n\nGILBERT STRANG: Well, thank you [INAUDIBLE]\n\nALEX TOWNSEND: OK, thanks.\n\n[APPLAUSE]", "date": "2021-10-16 17:29:03", "meta": {"domain": "mit.edu", "url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-17-rapidly-decreasing-singular-values/", "openwebmath_score": 0.7443466186523438, "openwebmath_perplexity": 398.4738235841221, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303440461105, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8527766569715394}}
{"url": "https://math.stackexchange.com/questions/2407811/how-many-trailing-zeroes-in-52", "text": "# How many trailing zeroes in $52!$? [duplicate]\n\nProblem\n\nHow many trailing zeroes are there in $52!$ ?\n\nMy thoughts\n\nI believe I correctly solved it, but I'm not happy with the scalability of my method.\n\nI figure the number of trailing zeroes is equal to the number of times $52!$ is divisible by 10. I wrote out every integer from 1-52 that is divisible by 2 or 5. The idea being that the number of 2 AND 5-factors equals the number of 10-factors.\n\nI quickly noted that the number of 2-factors is greater than the number of 5-factors, so I figured finding the number of 5-factors will do. There were 12.\n\nI'm not very happy with this, because if they now ask me to do the same for $152!$, I'll have to tell them to shove it. I'm not doing this again.\n\nQuestion\n\nIs there a better way to do this? Perhaps a method that scales better?\n\n## marked as duplicate by Jyrki LahtonenAug 27 '17 at 16:35\n\n\u2022 Your method looks good and should be quick now you have the basics. Undertaking the same task for $152!$ should rapidly give you some insight into how to solve the problem quicker. Look out for $125$. \u2013\u00a0Joffan Aug 27 '17 at 16:39\nYou get your $12$ by counting $\\left\\lfloor\\frac{52}{5}\\right\\rfloor+\\left\\lfloor\\frac{52}{25}\\right\\rfloor$. This counts each multiple of $5$ in the product $52!$, then adds on a count for each multiple of $25$. This can generalize to $152$. You'll need to go up to $125$ with divisors.\nIn general, the number of trailing $0$s in $n!$ will be $$\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{5^k}\\right\\rfloor$$ where for any $n$, only the first few terms in that infinite sum will be nonzero.", "date": "2019-06-19 21:19:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2407811/how-many-trailing-zeroes-in-52", "openwebmath_score": 0.5684671401977539, "openwebmath_perplexity": 233.0321048179714, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303407461413, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8527766541284899}}
{"url": "https://stats.stackexchange.com/questions/434835/which-transformation-needed-to-make-variance-independent-of-population-parameter", "text": "# Which transformation needed to make variance independent of population parameter?\n\nSuppose $$s^2$$ is the sample variance of a sample$$(\\text{of size }n)$$ from a normal population with mean $$\\mu$$ and variance $$\\sigma^2. \\text{Here }s^2=\\frac{\\sum_{i=1}^n(x_i-\\overline{x})^2}{n-1}$$\nAs we know $$\\frac{(n-1)s^2}{\\sigma^2}\\sim \\chi^2_{(n-1)}$$ here . Let $$x\\sim \\chi^2_{(n-1)}$$ then $$E[x]=n-1$$ and $$Var[x]=2(n-1)$$ \\begin{align*} E\\left[\\frac{(n-1)s^2}{\\sigma^2}\\right]=E[x]&=n-1\\\\ \\implies \\frac{(n-1)}{\\sigma^2}E[s^2] &= n-1\\\\ \\implies E[s^2]&=\\sigma^2 \\end{align*} Similarly, \\begin{align*} Var\\left[\\frac{(n-1)s^2}{\\sigma^2}\\right]=Var[x]&=2(n-1)\\\\ \\implies \\frac{(n-1)^2}{\\sigma^4}E[s^2] &=2(n-1)\\\\ \\implies Var[s^2]&=\\color{red}{\\frac{2\\sigma^4}{(n-1)} } \\end{align*} Now we need a function of $$s^2$$ whose variance will be independent of $$\\sigma^2.$$ Let $$f(s^2)$$ be the required transformation.\n\nThe required transformation is $$f(s^2)\\stackrel{?}{=}\\ln{s^2}$$\n\nQuestion: How they get that transformation $$?$$\nSimilar things happen with Poisson variate and Binomial proportion with square root and $$\\sin^{-1}$$ transformation. So I need a general approach to get that transformation which will make variance independent of population parameter.\n\n\u2022 Search for Variance-stabilizing_transformation. \u2013\u00a0StubbornAtom Nov 6 '19 at 14:33\n\u2022 \"The aim behind the choice of a variance-stabilizing transformation is to find a simple function f to apply to values x in a data set to create new values y=f(x) such that the variability of the values y is not related to their mean value.\" But I need the variability of the values y is not related to their variance(population) value? @StubbornAtom Sir \u2013\u00a0emonhossain Nov 6 '19 at 14:45\n\u2022 The wiki article might not be entirely clear. I gave you the keyword to look for the topic. (And I am no 'sir'.) \u2013\u00a0StubbornAtom Nov 6 '19 at 14:51\n\nSuppose $$X_1,X_2,\\ldots,X_n$$ are i.i.d $$N(\\mu,\\sigma^2)$$ and define the sample variance as $$S^2=\\frac{1}{n}\\sum\\limits_{i=1}^n (X_i-\\overline X)^2$$\n\nHere we are concerned with the large-sample behaviour of $$S^2$$, so it does not matter if we take $$n$$ as the divisor above instead of $$n-1$$.\n\nBy CLT it can be argued that $$\\sqrt n(S^2-\\sigma^2)\\stackrel{L}\\longrightarrow N(0,2\\sigma^4)\\tag{1}$$\n\nAnd by 'Delta-method', for a real-valued function $$g$$ such that $$g'(\\sigma^2)$$ exists and $$g'(\\sigma^2)\\ne 0$$, it follows from $$(1)$$ that\n\n$$\\sqrt n(g(S^2)-g(\\sigma^2))\\stackrel{L}\\longrightarrow N(0,2\\sigma^4[g'(\\sigma^2)]^2)\\tag{2}$$\n\nYou have to solve for a $$g$$ such that asymptotic variance of $$g(S^2)$$ is free of $$\\sigma^2$$.\n\nSo for some non-zero constant $$c$$, set $$\\sqrt 2\\sigma^2 g'(\\sigma^2)=c$$\n\nTherefore, $$\\int g'(\\sigma^2)\\,d\\sigma^2=\\frac{c}{\\sqrt 2}\\int\\frac{d\\sigma^2}{\\sigma^2}$$\n\nChoosing $$c=\\sqrt 2$$ and taking constant of integration to be zero, we have the required transformation $$g(\\sigma^2)=\\ln\\sigma^2\\quad,\\text{ i.e. }\\quad \\color{blue}{g(S^2)=\\ln S^2}$$\n\nFrom $$(2)$$ you end up with $$\\sqrt n(\\ln(S^2)-\\ln(\\sigma^2))\\stackrel{L}\\longrightarrow N(0,2)$$\n\nThis is the application of variance stabilizing transformation on the sample variance from a normal population. The other transformations that you mention for the Poisson mean and Binomial proportion are derived similarly.\n\nResult $$(1)$$ and many more can be found in A Course in Large Sample Theory by Thomas Ferguson. The relevant part is on page 46 of the first edition (1996):\n\nThe main result used in this section is Cramer's Theorem on the asymptotic normality of functions of sample moments, studied through a Taylor-series expansion to one term.\n\n\u2022 what I know is CLT tells us $\\color{red}{\\sqrt{n}(\\overline{x}-\\mu)\\text{ converge in distribution to a normal }N(0,\\sigma^2)}$ And I know how to prove this but $\\sqrt n(s^2-\\sigma^2)\\stackrel{L}\\longrightarrow N(0,2\\sigma^4)$ is looks new to me. So everything will be meaningful to me if you provide a source where I can get that proof $?$ I will accept your answer then. Thanks again @StubbornAtom \u2013\u00a0emonhossain Nov 6 '19 at 16:07\n\u2022 @emonhossain Updated. \u2013\u00a0StubbornAtom Nov 6 '19 at 18:07\n\u2022 Really you help me a lot. Thanks @StubbornAtom \u2013\u00a0emonhossain Nov 6 '19 at 18:33", "date": "2020-09-18 13:01:54", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/434835/which-transformation-needed-to-make-variance-independent-of-population-parameter", "openwebmath_score": 0.95013427734375, "openwebmath_perplexity": 420.6127907450062, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156284, "lm_q2_score": 0.8740772417253256, "lm_q1q2_score": 0.8527295726744067}}
{"url": "http://mathhelpforum.com/math-puzzles/153573-putnam-problem-day-print.html", "text": "# Putnam Problem of the Day\n\n\u2022 Aug 12th 2010, 10:20 PM\nsimplependulum\nPutnam Problem of the Day\n\nFind the smallest natural number with $6$ as the last digit , such that if the final $6$ is moved to the front of the number it is multiplied by $4$ .\n\nSpoiler:\n153846\n?\n\u2022 Aug 13th 2010, 03:26 AM\nUnbeatable0\nLet the number be $a_1a_2...a_n6$.\n\nLet $x=0.\\overline{a_1a_2...a_n6}$.\n\nThen $4x = 0.\\overline{6a_1a_2...a_n} \\Rightarrow 40x = 6.\\overline{a_1a_2...a_n6} = 6+x$.\n\nTherefore $x = \\frac{6}{39} = \\frac{2}{13} = 0.\\overline{153846}$.\n\nBy backwards argument it's easy to see that all the cycles in $\\frac{2}{13}$ are solutions.\n\nThus we have found all the solutions, the smallest of which is $153846$.\n\nThis technique can be applied to this kind of problems in many cases.\n\u2022 Aug 13th 2010, 02:03 PM\nSoroban\nHello, simplependulum!\n\nJust write out the multiplcation,\n. . and the digits will appear sequentially.\n\nQuote:\n\nFind the smallest natural number with $6$ as the last digit,\nsuch that if the final $6$ is moved to the front of the number it is multiplied by $4$.\n\nWe have: . $\\begin{array}{cccccc}\n^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\\\\nA&B&C&D&E&6 \\\\ \\times &&&&& 4\\\\ \\hline 6&A&B&C&D&E \\end{array}$\n\nIn column-6: . $6 \\!\\times\\! 4 \\,=\\,24 \\quad\\Rightarrow\\quad E = 4$\n\n. . $\\begin{array}{cccccc}\n^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\\\\nA&B&C&D&4&6 \\\\\n\\times &&&&& 4\\\\ \\hline\n6&A&B&C&D&4 \\end{array}$\n\nIn column-5: . $4 \\!\\times\\! 4 + 2 \\:=\\:18 \\quad\\Rightarrow\\quad D = 8$\n\n. . $\\begin{array}{cccccc}\n^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\\\\nA&B&C&8&4&6 \\\\\n\\times &&&&& 4\\\\ \\hline\n6&A&B&C&8&4 \\end{array}$\n\nIn column-4: . $8 \\!\\times\\! 4 + 1 \\:=\\:33 \\quad\\Rightarrow\\quad C = 3$\n\n. . $\\begin{array}{cccccc}\n^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\\\\nA&B&3&8&4&6 \\\\\n\\times &&&&& 4\\\\ \\hline\n6&A&B&3&8&4 \\end{array}$\n\nIn column-3: . $3\\!\\times\\!4 + 3 \\:=\\:15 \\quad\\Rightarrow\\quad B = 5$\n\n. . $\\begin{array}{cccccc}\n^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\\\\nA&5&3&8&4&6 \\\\\n\\times &&&&& 4\\\\ \\hline\n6&A&5&3&8&4 \\end{array}$\n\nIn column-2: . $5\\!\\times\\!4 + 1 \\:=\\:21 \\quad\\Rightarrow\\quad A = 1$\n\nTherefore: . $\\begin{array}{cccccc}\n1&5&3&8&4&6 \\\\\n\\times &&&&& 4\\\\ \\hline\n6&1&5&3&8&4 \\end{array}$\n\n\u2022 Aug 13th 2010, 05:19 PM\nWilmer\nSoroban, the fact that it is a 6digit number is NOT a given.\n\u2022 Aug 13th 2010, 06:16 PM\nSoroban\nQuote:\n\nOriginally Posted by Wilmer\nSoroban, the fact that it is a 6-digit number is NOT a given. . Right!\n\nI originally set up the problem like this:\n\n. . $\\begin{array}{ccccccccccc}\nA & B & C & D & E & F & G & \\hdots & M & N & 6 \\\\\n\\times &&&&&&&&&& 4 \\\\ \\hline\n6 & A & B & C & D & E & F & G & \\hdots & M & N\n\\end{array}$\n\nAnd found that \"it came out even\" after six digits\n. . which gave me the smallest number: $153,\\!846.$\n\nThe next is the 12-digit number: $153,\\!846,\\!153,\\!846.$\n\n\u2022 Aug 13th 2010, 07:10 PM\nundefined\nFyi, there is a name for these, n-parasitic number, where here n=4 and k=6.\n\u2022 Aug 13th 2010, 09:32 PM\nWilmer\nLet n = number of digits\n\n{20[10^(n-1) - 4] + 78} / 13\n\nGives integer solution for n=6k where k = any integer > 0", "date": "2017-10-21 09:16:05", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-puzzles/153573-putnam-problem-day-print.html", "openwebmath_score": 0.907620906829834, "openwebmath_perplexity": 1222.6492996969278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769113660689, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8527295693775347}}
{"url": "http://mathhelpforum.com/geometry/227392-circle-questions-secants-chords-tangents-print.html", "text": "# Circle questions -- Secants, chords and tangents\n\n\u2022 Apr 3rd 2014, 05:03 PM\nStonerPenguin\nCircle questions -- Secants, chords and tangents\nHello again, just got a few questions pertaining to circles. I'll just post one to start with;\n\n\"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).\nhttp://i1137.photobucket.com/albums/...mexam10Q32.png\nSo solving this.. according to the segments of secants and tangents theorem...\nx2 = (2r + 353)(353)\nx2 = (12742 + 353)(353)\nx2 = 4622535\nx = $\\sqrt{4622535}$ = 2150km (approx.)\n\nDid I do that right?\n\u2022 Apr 3rd 2014, 07:13 PM\nJeffM\nRe: Circle questions -- Secants, chords and tangents\nQuote:\n\nOriginally Posted by StonerPenguin\nHello again, just got a few questions pertaining to circles. I'll just post one to start with;\n\n\"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).\nhttp://i1137.photobucket.com/albums/...mexam10Q32.png\nSo solving this.. according to the segments of secants and tangents theorem...\nx2 = (2r + 353)(353)\nx2 = (12742 + 353)(353)\nx2 = 4622535\nx = $\\sqrt{4622535}$ = 2150km (approx.)\n\nDid I do that right?\n\nComputation looks right to me. But simply memorizing theorems does not promote understanding.\n\nLet's see how we get that theorem. The radius through a point of tangency and the tangent at the same point are perpendicular.\n\n$Let\\ r = length\\ of\\ radius\\ of\\ given\\ circle,$\n\n$u + r = length\\ from\\ the\\ given\\ circle's\\ center\\ to\\ a\\ given\\ point\\ outside\\ the\\ circle,$\n\n$x = length\\ of\\ the\\ line\\ that\\ is\\ tangent\\ to\\ the\\ given\\ circle\\ and\\ runs\\ through\\ the\\ given\\ point.$\n\nBy the Pythagorean Theorem, we have:\n\n$(u + r)^2 = x^2 + r^2 \\implies u^2 + 2ru + r^2 = x^2 + r^2 \\implies x^2 = u^2 + 2ru = u(u + 2r) \\implies x = \\sqrt{u(u + 2r)}.$\n\u2022 Apr 6th 2014, 06:17 PM\nbjhopper\nRe: Circle questions -- Secants, chords and tangents\nISS is6724 km above center of earth.At that point the angle between straight down and the visible horizon has a sin of 6371/6724 or 71.35 deg\ncos 71.35 =d/6724\nd km to horizon =2150 km\n\u2022 Apr 6th 2014, 09:56 PM\nSoroban\nRe: Circle questions -- Secants, chords and tangents\nHello, StonerPenguin!\n\nQuote:\n\nThe radius of the earth is approximately 6371 km.\nIf the international space station (ISS) is orbiting 353 km above the earth,\nfind the distance from the ISS to the horizon (x).\n\nCode:\n\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 o \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 |\\ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 | \\ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 352 |\u00a0 \\ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 |\u00a0 \\ x \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 |\u00a0 \u00a0 \\ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * * *\u00a0 \\ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 |\u00a0 \u00a0 *\\ \u00a0 \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 \u00a0 |\u00a0 \u00a0 \u00a0 o \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 6371|\u00a0 \u00a0 *\u00a0 * \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 |\u00a0 *6371 \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 \u00a0 \u00a0 | *\u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 \u00a0 \u00a0 o\u00a0 \u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 *\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 * * *\nNote the right triangle.\n\nThe equation is: . $x^2 + 6371^2 \\:=\\:6723^2$\n\u2022 Apr 7th 2014, 05:16 PM\nStonerPenguin\nRe: Circle questions -- Secants, chords and tangents\nThank you JeffM, bjhopper and Soroban! It's nice to see different perspectives :D and the image drawn in code is really cool.\nHere's another question I've had trouble with:\nhttp://i1137.photobucket.com/albums/...mexam10Q15.png\n\"Explain how you know $\\overline{AB}$ $\\overline{CD}$ given E is the center of the circle. (Include theorem numbers.)\"\n\nAnd here's some pertinent theorems;\nQuote:\n\nTheorem 10.4\nIf one chord is a perpendicular bisector of another chord, then the first chord is a diameter.\nQuote:\n\nTheorem 10.5\nIf a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.\nQuote:\n\nTheorem 10.6\nIn the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.\nObviously from the diagram $\\overline{AB}$ $\\overline{CD}$ by theorem 10.6, but I can't really word this well. Any help? Theorems and proofs are my weakest areas :/\n\u2022 Apr 8th 2014, 04:24 PM\nbjhopper\nRe: Circle questions -- Secants, chords and tangents\nt is a triangle a is an angle\nt AED congruent to BEC isosceles t's same legs and equal altitudes\na DEC = 180- 2* 1 / 2 *AED\na AEB 180-2*1/2 * BEC a AED = aBEC\na AEB = a DEC\nAB =DC equal arcs = equal chords", "date": "2017-07-23 13:39:55", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/227392-circle-questions-secants-chords-tangents-print.html", "openwebmath_score": 0.8598741292953491, "openwebmath_perplexity": 10813.154269922521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769099458927, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.8527295665361332}}
{"url": "https://math.stackexchange.com/tags/measure-theory/hot?filter=month", "text": "# Tag Info\n\n7\n\nSince $fg\\geq 1$, you have $\\sqrt{f(x)\\cdot g(x)} \\geq 1$ for all $x$, and so $$1 \\leq \\int_X \\sqrt{f}\\cdot \\sqrt{g} d\\mu \\leq \\sqrt{\\int_X f d\\mu }\\cdot \\sqrt{\\int_X g d\\mu }$$ where the second inequality is Cauchy\u2013Schwarz; squaring both sides gives the inequality.\n\n6\n\nIt is sufficient to show that $m(E \\cap [a, b]) = 0$ for every $a$ and $b$ with $a < b$. Let $t = \\inf_{x \\in [a, b]}{e^{-x^2/2}}$. Then $t > 0$ and we have $$0 = \\int_{E}e^{-x^2/2}dx \\ge \\int_{E\\cap [a, b]}e^{-x^2/2}dx \\ge \\int_{E\\cap [a, b]}tdx = m(E \\cap [a, b])t$$ and as $t > 0$, this implies $m(E \\cap [a, b]) = 0$.\n\n5\n\nA measure space must specify three things: An underlying set; A $\\sigma$-algebra of subsets which are measurable in this measure space; A measure (a function on this $\\sigma$-algebra). When we say \"a measurable subset of the real numbers\" we typically mean \"a subset of the real numbers measurable with respect to the Lebesgue measure\", ...\n\n5\n\nThe questions, despite looking as a representation problem in functional analysis, are much deeper as they bring out the history of the topic involved, notably $BV$-functions and the reasons why the customary definition adopted for the variation of a multivariate function is definition 2 above. And as thus the answers below needs to indulge a bit on this ...\n\n5\n\n$\\bigcup_{i\\ge 1}A_i$ is by definition simply $$\\left\\{x:\\exists i\\in\\Bbb Z^+\\,(x\\in A_i)\\right\\}\\,,$$ which clearly does not depend on the order in which the sets are enumerated. You can simply let $\\mathscr{A}=\\{A_i:i\\in\\Bbb Z^+\\}$; then $$\\bigcup_{i\\ge 1}A_i=\\bigcup\\mathscr{A}=\\{x:\\exists A\\in\\mathscr{A}\\,(x\\in A)\\}\\,,$$ with no reference to the ...\n\n5\n\nHint: Consider the sets $[-r,r] \\subseteq \\mathbb{R}$. Can you show $r \\mapsto \\mu(E \\cup [-r,r] \\cap F)$ is continuous? Can you show for any $r$ the set we're measuring is compact? What does the intermediate value theorem buy us? I hope this helps ^_^\n\n5\n\nThis is false. For an easy counterexample, let $X_n\\geq 0$ be any sequence of random variables with $\\liminf_n X_n = 0$ which converges to $X = 1$ in probability. Then let $\\mathcal{Q}$ be large enough so that the $X_n$ are $\\mathcal{Q}$-measurable. The inequality in question then becomes $X \\leq \\liminf_n X_n,$ which is false almost surely. For instance, ...\n\n5\n\nIt's true that the definition of absolute continuity is the same either we state it in the form of the finite and disjoint intervals or we state it in the form of countable and disjoint intervals. In detail, suppose we have the following definitions of absolute continuity: Definition 1 Let $f:\\mathbb{R}\\to \\mathbb{R}$ be a function. We say that $f$ is ...\n\n4\n\nBy assumption, as $Z \\notin L^{\\infty}$, we have $$\\mathbb{P}(Z > n^2) > 0, \\quad n \\in \\mathbb{N}.$$ Therefore, we may define $$X = \\sum_{n=1}^{\\infty}\\frac{1}{n^2} \\frac{\\mathbb{1}_{[Z > n^2]}}{\\mathbb{P}(Z>n^2)} \\in L^1(\\mathbb{P}).$$ However, it follows that \\begin{align*}\\mathbb{E}[ZX] = \\sum_{n=1}^{\\infty}\\mathbb{E}\\left[\\frac{1}{n^2} \\frac{...\n\n4\n\nCannot be true in general: consider $k=1$ (or any odd $k$), take $X_i$'s to be i.i.d normal with mean $0$ and variance $1$, we show below that $Y_n$ is not UI. Note that $$\\mathbb{E}\\left(|Y_n| 1\\{|Y_n|\\geq M\\}\\right) \\geq M\\mathbb{P}\\left(|Y_n|\\geq M\\right) = M\\mathbb{P}\\left(|\\bar{X}_n|\\leq \\frac{1}{M}\\right) \\to M,$$ as $n\\to \\infty$ by laws of large ...\n\n4\n\nThe assertion \u201cnot every subset of $\\Bbb R$ is measurable\u201d is valid for the Lebesgue measure. Anyway, here Axler is talking about $\\sigma$-algebras, not about measures. And, yes, $\\mathcal P(\\Bbb R)$ is a $\\sigma$-algebra. And, on this $\\sigma$-algebra you can define the measure $m(X)=\\#X$; with respect to this measure, $\\Bbb R$ is measurable.\n\n4\n\nIf $p > p_0$ and $f \\in L^{p_0} \\cap L^\\infty$ then $|f|^p = |f|^{p - p_0}|f|^{p_0} \\le \\|f\\|_\\infty^{p - p_0}|f|^{p_0}$ almost everywhere.\n\n4\n\nThe claim is true: Let $S = \\sum_{i=1}^\\infty \\chi_{E_i}$. Suppose for the sake of contradiction that $S < \\infty$ a.e. Then, there exists an $M$ such that $\\Pr(S\\le M) \\ge 1 - \\frac\\varepsilon 2$. If $A$ is the event that $S\\le M$, then $\\mu(A\\cap E_i) \\ge \\frac\\varepsilon 2$ for all $i$. But this would imply $\\int_A S\\,d\\mu = \\sum_{i=1}^\\infty \\mu(A\\... 4 Well,$[0,1]=C\\hspace{1mm}\\cup C^{c} $(I discourage you to use$\\mathbb{C}$for the cantor set) and you can write f as: $$f(x)=0\\cdot\\chi_{C}(x)+x^{2}\\chi_{C^{c}}$$ where by$\\chi$I mean the characteristic function of the set in subscript. The sum of measurable functions is measurable as well as the product, and polynomial functions are measurable. 4 The family of measurable sets form a$\\sigma$-algebra but not all sets belonging in any$\\sigma$-algebra are measurable. So to fully define a measure space we need a set$X$, a$\\sigma$-algebra$\\mathcal{A}$on$X$and a function$\\mu: \\mathcal{A} \\to [0, \\infty]$which satisfies some properties. A set$A \\subseteq X$is called$\\mu$-measurable (or simply ... 4 The symmetrization is called Khintchine's inequality, see here: https://en.wikipedia.org/wiki/Khintchine_inequality For the contraction principle, I don't know any reference, but I can prove it. First note that if$Z\\geq 0$and$p > 0, then \\begin{align} \\mathbb{E}[Z^p] = \\mathbb{E}\\Big[\\int_0^{Z} p t^{p-1} dt\\Big] = \\mathbb{E}\\Big[\\int_0^{\\infty} p t^{p-... 4 A set\\mathfrak B$satisfies the definition in your book if and only if$\\mathfrak B$is a$\\sigma$-algebra (in the wikipedia sense) on$\\cup\\mathfrak B$: for the noteworthy algebraic detail, notice that$A\\setminus B=A\\triangle (A\\cap B)$. This is also equivalent to the property of$\\mathfrak B$being a$\\sigma$-algebra in the wikipedia sense over some set.... 4 Let me discuss the case$n=1$only. Let me denote by$f_x$the weak derivative of$f$with respect to$x$. Let$\\phi,\\psi$be two smooth test functions. Then $$-\\iint f_x(x,y) \\phi(x)\\psi(y)dx\\ dy=\\iint f(x,y) \\phi_x(x)\\psi(y)dx\\ dy,$$ using Fubini on both sides gives $$\\int \\left( \\int -f_x(x,y) \\phi(x) + f(x,y) \\phi_x(x)\\ dx\\right) \\psi(y) \\ dy = 0$$ ... 3 You prove that by giving an example of a set of subsets that is closed under intersections and differences but not closed under unions. E.g. on any set$X$with at least two points, the family$\\{\\{x\\}: x \\in X\\} \\cup \\{\\emptyset\\}$is such. so being closed under union and differences (a ring) implies being closed under intersections, but being closed under ... 3 I have recently got interested in continued fractions, so I'll give it a try! CONTINUED FRACTIONS. It is a basic property of continued fractions that if$h_n/k_n$is the sequence of convergent to$x\\in\\mathbb R\\setminus\\mathbb Q$, then $$\\frac 1{k_n(k_n+k_{n+1})}<\\left|x-\\frac{h_n}{k_n}\\right|<\\frac 1{k_nk_{n+1}}.$$ It follows that for all$h,k$... 3 The set$\\mathcal{P}(X)$of all subsets of$X$is a sigma algebra because it satisfies the axioms for that kind of structure. That is independent of any discussion of measures. The definition of a measure space is a pair$(X,S)$where$S$is some subset of$\\mathcal{P}(X)$that happens to be a sigma algebra. It may or may not be all of$\\mathcal{P}(X)$. ... 3 Suppose that such measure exists. Then taking countable cover by open sets of diameter$\\leq 1$, one of those sets, say$U_1$, must have measure$1$. Cover$U_1$by countable amount of open sets of diameter$\\leq 1/2$and take the one having measure$1$. This process gives us a decreasing sequence of open sets$U_k$with diameter$\\leq 1/k$and$\\mu(U_k) = 1$... 3 This works. Your argument shows that if$f$is bounded on a set$E$of Lebesgue measure zero, then$\\int_E f = 0$. In fact the stronger claim that$\\int_E f = 0$without any hypothesis on$f$holds! Indeed the integral is, by definition, the supremum over$g$with$g$simple and$g \\leq f$of$\\int_E g$. But any such$g$has integral zero. 3 Welcome to MSE! Hint: It's often useful to split up a random variable into the places where it is \"good\" and \"bad\". Then we can control those regions separately to get whatever inequality we're interested in. For this, we might try looking at the decomposition $$X = X \\cdot \\mathbf{1}_{\\{X \\leq \\lambda a\\}} + X \\cdot \\mathbf{1}_{\\{X >... 3 According to the p-Adics section of the reference manual there is nothing I could find to do with zeta functions nor integration. At any rate, here's a suggestion on algorithmically getting an arbitrary level of accuracy on your integral. In order to be called a p-adic measure we require it to be a bounded p-adic distribution, and this is enough to imply ... 3 By definition, \\int f is the supremum of \\int h for 0 \\leq h \\leq f bounded measurable of finite support. So at the very end, you just need to take the sup over such h to get \\int f \\leq \\liminf \\int f_n. He uses such h is order to apply the bounded dominated convergence theorem, which he proves earlier. This allows him to prove Fatou in terms of ... 3 You can also let g_n=inf_{i \\geq n} \\{f_n\\} and note that \\int g_n \\leq \\int f_n for each n. Moreover g_n increases monotonically to lim inf f_n . Hence by monotone convergence theorem \\int g_n = \\int lim inf f_n. Since \\int g_n \\leq \\int f_n we have \\int lim inf f_n \\leq lim inf \\int f_n. The final step is to note that pointwise convergence ... 3 If you are familiar with dominated convergence notice that f_n(X)\\xrightarrow{n\\rightarrow\\infty} X pointwise, |f_n(X)|\\leq |X|, X is assumed to be integrable (E[|X|]=\\int |X|\\,dP<\\infty). Then by dominated convergence$$ \\int f_n(X)\\,dP\\xrightarrow{n\\rightarrow\\infty}\\int X\\,dP=E[X]$$similarly for$Y$. There is still the issue of the ... 3 Let$f$be the limit of$\\chi_{E_n}$in$L^p$. All you have to do is show that$f(x) = 0$or$f(x) = 1$almost everywhere. Then we see that$f = \\chi_E$almost everywhere, where$E = \\{x \\in X | f(x) = 1\\}$To do this, it is helpful to define$g(x) = \\min(|x - 1|, |x|)$. We will show that$g \\circ f = 0$almost everywhere. We first show that$\\int g(f(x))^p \\...\n\n3\n\nLet $f \\geq 0$ and let $M=\\sup \\{f(x): x\\in X\\}$. Then $\\|M-f\\|\\leq M$ since $f(x) \\in [0,M]$ for all $x$. If $\\phi(f)$ is real we can finish the proof as follows: $M-\\phi (f)=\\phi(M-f) \\leq \\|\\phi\\| \\|M-f||\\leq M$. Thus $\\phi (f) \\geq 0$. To show that $\\phi (f)$ is real consider following: \\$|\\phi (f) \\pm in|=|\\phi(f \\pm in)|\\leq \\|f\\pm in\\|\\leq \\sqrt {M^{...\n\nOnly top voted, non community-wiki answers of a minimum length are eligible", "date": "2021-04-17 09:35:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/tags/measure-theory/hot?filter=month", "openwebmath_score": 0.9983400702476501, "openwebmath_perplexity": 550.3128046189067, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769085257167, "lm_q2_score": 0.8740772253241802, "lm_q1q2_score": 0.8527295572945}}
{"url": "http://math.stackexchange.com/questions/72589/calculating-probability-of-at-least-one-event-occurring?answertab=votes", "text": "Calculating probability of 'at least one event occurring'\n\nIf I know the probability of A and the probability of B how can I calculate the probability of \"at least one of them\" occurring?\n\nI was thinking that this is P(A or B) = P(A) + P(B) - P(AB). Is this correct?\n\nIf it is, then how can I solve the following problem taken from 'DeGroot - Probability and Statistics':\n\nIf 50 percent of families in a certain city subscribe to the morning newspaper, 65 percent of the families subscribe to the afternoon newspaper, and 85 percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?\n\nHere the question is P(morning)=.5, P(afternoon)=.65 P(morning OR afternoon)=.85\n\nP(morning OR afternoon) = .5 + .65 - .3 = .85\n\nP(morning AND afternoon) = P(morning) + P(afternoon) - P(morning OR afternoon)\n\nSo the answer to the question is .3\n\nIs my reasoning correct?\n\nEDIT.\n\nIf this reasoning is correct:\n\nHow can I calculate the following:\n\nIf the probability that student A will fail a certain statistics examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examination is 0.1,what is the probability that exactly one of the two students will fail the examination?\n\nSo this questions highlights the difference between !at least one of them! !or exactly one of them!\n\nI understand that at least one of them = P(A or B)\n\nbut how can I work out the probability of exactly one of them?\n\n-\nSomebody knows an algorith to calculate the OR for n probabilistic values? Thanks in advance. Regards, Daniel. \u2013\u00a0 Daniel Mejia Nov 8 '12 at 23:53\nYou need to specify the correlation between the $n$ probabilistic values - at least whether they are independent or not. \u2013\u00a0 dexter04 Nov 9 '12 at 0:25\n\nYou are correct.\n\nTo expand a little: if $A$ and $B$ are any two events then\n\n$$P(A\\textrm{ or }B) = P(A) + P(B) - P(A\\textrm{ and }B)$$\n\nor, written in more set-theoretical language,\n\n$$P(A\\cup B) = P(A) + P(B) - P(A\\cap B)$$\n\nIn the example you've given you have $A=$ \"subscribes to a morning paper\" and $B=$ \"subscribes to an afternoon paper.\" You are given $P(A)$, $P(B)$ and $P(A\\cup B)$ and you need to work out $P(A\\cap B)$ which you can do by rearranging the formula above, to find that $P(A\\cap B) = 0.3$, as you have already worked out.\n\n-\nthanks for your answer! please have a look at my edited question, your help highlighted a further issue.. \u2013\u00a0 Dbr Oct 14 '11 at 11:10\n\"Exactly one of A and B\" means \"Either A or B, but not both\" which you can calculate as P(A or B) - P(A and B). \u2013\u00a0 Chris Taylor Oct 14 '11 at 11:13\nthank you, this clarifies everything. Just one last quick thing: Where can I learn the mathematical notation you use on this forum? For some questions I find it quite hard to understand them, is there a tutorial? \u2013\u00a0 Dbr Oct 14 '11 at 11:15\nAre you asking about the notation itself, or the method of displaying the notation? To write the notation we use $\\LaTeX$ - you can find a tutorial by searching for \"latex tutorial\" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read more complex texts, you will become more and more familiar with the notation. \u2013\u00a0 Chris Taylor Oct 14 '11 at 11:21\nSo if I download LaTeX and paste your notation then it displays it in a more readable form? \u2013\u00a0 Dbr Oct 14 '11 at 11:24\n\nFor your second question, you know $\\Pr(A)$, $\\Pr(B)$, and $\\Pr(A \\text{ and } B)$, so you can work out $\\Pr(A \\text{ and not } B)$ and $\\Pr(B \\text{ and not } A)$ by taking the differences. Then add these two together.\n\nAlternatively take $\\Pr(A \\text{ or } B) - \\Pr(A \\text{ and } B)$.\n\n-\n\nFor the additional problem: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection, equals $$P(A)+P(B)-2P(A\\cap B)$$\n\n-\n\nprobability of only one event occuring is as follows: if A and B are 2 events then probability of only A occuring can be given as P(A and B complement)= P(A) - P(A AND B )\n\n-", "date": "2014-08-01 11:57:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/72589/calculating-probability-of-at-least-one-event-occurring?answertab=votes", "openwebmath_score": 0.8094062805175781, "openwebmath_perplexity": 515.6158395670417, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156284, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8527295550737701}}
{"url": "https://math.stackexchange.com/questions/3583403/calculating-dice-probabilities-with-multiple-overlapping-success-criteria", "text": "# Calculating dice probabilities with multiple, overlapping success criteria\n\nAlright, so I play a tabletop game where models get in fights and roll dice to see if they deal damage to the other models. I'm wondering how to calculate the probabilities of dealing 0, 1, and 2 damage in a 2v2 fight.\n\nHere is the situation:\n- We have Model A and Model B rolling to deal damage to Model C and Model D\n- A needs to roll a 5 or 6 to damage C, and a 6 to damage D\n- B needs to roll a 4, 5, or 6 to damage C, and a 5 or 6 to damage D\n- C and D can only be damaged once each (i.e., they die when they get damaged)\n\nSo for starters, I know the probabilities for this scenario just by listing out the possibilities:\n\n  A | 111111 222222 333333 444444 555555 666666\nB | 123456 123456 123456 123456 123456 123456\n----+------------------------------------------\nDmg | 000111 000111 000111 000111 111122 111222\n\n\nSo the probability of dealing 0 damage is $$\\frac{12}{36}$$, (exactly) 1 damage is $$\\frac{19}{36}$$, and 2 damage is $$\\frac{5}{36}$$. One tricky outcome here is when A rolls a 5 and B rolls a 4. This only results in 1 damage since both rolls are only enough to damage C and C can only be damaged once.\n\nI am hopelessly stumped on how to model this so that I can then apply it to other scenarios such as 3v3s or when some models use more than 1 die.\n\nIt is pretty easy to calculate the probabilities of doing damage in a 1v1 fight. I've used binomial distribution statistics to calculate those. Conveniently it works for multiple dice per model as well. However, when it comes to combining these probabilities ... I am at a loss. They don't seem to follow the rules for multiplication and addition.\n\nFor example, I na\u00efvely expected the probability of doing 0 damage to be the probability of A doing 0 damage and the probability of B doing no damage. That sounds right intuitively, but no matter how I try to do it ... it never makes sense:\n\n$$P(A\\;miss\\;C) = \\binom{1}{0} \\bullet 2^0 \\bullet 4^1 = 4$$\n$$P(A\\;miss\\;D) = \\binom{1}{0} \\bullet 1^0 \\bullet 5^1 = 5$$\n$$P(B\\;miss\\;C) = \\binom{1}{0} \\bullet 3^0 \\bullet 3^1 = 3$$\n$$P(B\\;miss\\;D) = \\binom{1}{0} \\bullet 2^0 \\bullet 4^1 = 4$$\n\nTherefore,\n\n$$P(A\\;miss\\;C) \\bigcap P(A\\;miss\\;D) \\bigcap P(B\\;miss\\;C) \\bigcap P(B\\;miss\\;D) = 4 \\bullet 5 \\bullet 3 \\bullet 4 = 240$$\n\nWait what?!?! I expect the answer to be 12. Clearly there is something else going on here that I do not understand. :/\n\n\u2022 You can't say $\\Pr(X\\cap Y)=\\Pr(X)\\Pr(Y)$ unless $X$ and $Y$ are independent events, and these are clearly not. When A doesn't damage C it's likely that he doesn't damage D, either, because he rolled a low number. Mar 16 '20 at 20:56\n\u2022 Also, you're multiplying counts instead of probabilities. The product of the probabilities (even if you could just multiply them, i.e. if they were independent, which they aren't) would be $\\frac46\\cdot\\frac56\\cdot\\frac36\\cdot\\frac46=\\frac{240}{6^4}=\\frac5{27}\\approx0.185$, which is at least a lot closer to the $\\frac{12}{6^2}\\approx0.333$ that you were apparently expecting than $\\frac{240}{6^2}\\approx6.67$ would have been. Mar 16 '20 at 21:44\n\nFirst I'd like to rephrase your question:\n\nA and B each roll a die. The results of the die determine how much damage is dealt. The maximum{damage dealt by A, damage dealt by B} is the number we are interested in.\n\nA Deals 1 damage by rolling a 5 or 2 damage by rolling a 6. B deals 1 damage by rolling a 4 or 2 damage by rolling a 5 or 6. Now take the maximum of the damage dealt by A or B.\n\nI think it may be easier to frame it this way:\n\nA misses and B Misses simultaneously:\n\nP(0 Damage) = P(A Rolls 4 or less) * P(B rolls 3 or less) = 2/3*1/2= 1/3 = 3/9\n\nA Hits both or B hits Both (don't double count):\n\nP(2 Damage) = P(A Rolls 6) + P(B Rolls 5 or 6) - P(A Rolls 6 AND B rolls 5 or 6) = 1/6 + 1/3 - 1/6*1/3 = 4/9\n\nP(1 Damage) = 1-P(0)-P(2) = 2/9\n\nAlternatively,\n\nA hits 1 (rolls 5) or B hits 1 (rolls 4)\n\nP(1 Damage) = P(B Rolls 4)*P(A Rolls 5 or less) + P(B Rolls 3 or less)*P(A Rolls 5) = 1/6*5/6+ 3/6*1/6= 2/9\n\nHow I verified my answer: The table below shows how much damage occurs when (A rolls, B rolls) happens. Count the number of 0s and divide by 36 to determine how often 0 damage occurs. Repeat for 1s and 2s.\n\n## | 6 | 2 | 2 | 2 | 2 | 2 | 2 |\n\nSummary:\n\nP( 0 damage ) = 3/9\n\nP( 1 damage ) = 2/9\n\nP( 2 damage ) = 4/9", "date": "2022-01-27 22:03:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3583403/calculating-dice-probabilities-with-multiple-overlapping-success-criteria", "openwebmath_score": 0.5298913717269897, "openwebmath_perplexity": 623.2854954595914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534392852384, "lm_q2_score": 0.868826784729373, "lm_q1q2_score": 0.8527130360157786}}
{"url": "https://math.stackexchange.com/questions/1912681/is-there-a-quicker-proof-to-show-that-210k-equiv-7-pmod9-for-all-posit", "text": "# Is there a quicker proof to show that $2^{10^k} \\equiv 7 \\pmod{9}$ for all positive integers $k$?\n\nI noticed this pattern while playing with digit sums and noticed that the recursive digit sums (until you arrive at a single digit number) of numbers like, $2^{10}$, $2^{100}$, $2^{1000}$ and so on is always $7$. So, I decided to find out if it is true that for all positive integers $k$,\n\n$$2^{10^k} \\equiv 7 \\pmod{9}$$\n\nMy proof is as follows:\n\nLemma 1. $\\, 10^k \\equiv 4 \\pmod{6}$ for all integers $k \\geq 1$\n\nProof. For all integers $k \\geq 1$, the number $10^k + 2$ must be divisible by $6$ since it is even (implying divisibility by $2$), and its digit sum is $3$ (implying divisibility by $3$). Therefore, you can show that \\begin{align*} 10^k + 2 &\\equiv 0 \\pmod{6} \\\\ 10^k &\\equiv 4 \\pmod{6} \\,\\, \\text{ for all integers } k \\geq 1 \\end{align*}\n\nLemma 2. $\\, 2^{4 + 6k} \\equiv 7 \\pmod{9}$, for all integers $k \\geq 0$\n\nProof. If $a \\equiv c \\pmod{n}$ and $b \\equiv d \\pmod{n}$, then $a\\,b \\equiv c\\,d \\pmod{n}$. And, by extension, $a\\,b^k \\equiv c\\,d^k \\pmod{n}$ (for integers $k \\geq 0$). Therefore, with \\begin{align*} 2^4 \\equiv 16 \\equiv 7 \\pmod{9} \\end{align*} and \\begin{align*} 2^6 \\equiv 64 \\equiv 1 \\pmod{9} \\end{align*} we can show that, \\begin{align*} 2^{4 + 6k} \\equiv 7 \\cdot 1^k \\equiv 7 \\pmod{9} \\,\\, \\text{ for all integers } k \\geq 0 \\end{align*}\n\nTheorem. $\\, 2^{10^k} \\equiv 7 \\pmod{9}$ for all integers $k \\geq 1$\n\nLemma 1 implies that for all integers $k \\geq 1$, $10^k = 4 + 6n$ where $n$ is some positive integer. Therefore, \\begin{align*} 2^{10^k} &= 2^{4 + 6n} \\\\ 2^{10^k} \\!\\!\\!\\! \\mod a &= 2^{4 + 6n} \\!\\!\\!\\! \\mod a \\end{align*} for any positive integer $a$. Using this result along with Lemma 2, \\begin{align*} 2^{10^k} \\equiv 7 \\pmod{9} \\,\\, \\text{ for all integers } k \\geq 1 \\end{align*} $$\\tag*{\\blacksquare}$$\n\nI think the proof is correct, but I am not a fan of it. Mainly because, I think proving Lemma 2 is a much too long a way to prove this theorem - since it proves a generalization of the theorem first. Also, I discovered that lemma numerically, which feels a bit like cheating.\n\nEither way, is there a quicker, more elegant proof which does not:\n\n1. Use Lemma 2, or prove some generalization of the theorem first.\n\n2. Uses Euler's Theorem. I feel that using Euler's Theorem is using a needlessly complicated theorem to prove something as simple as this.\n\nI am sure some of the people here can come up with single line proofs. I am curious to see if there's such a proof.\n\n\u2022 euler's theorem is one of the basic tools when working with exponents \u2013\u00a0HereToRelax Sep 3 '16 at 1:13\n\u2022 Do you mean is there a shorter proof ? \u2013\u00a0Rene Schipperus Sep 3 '16 at 1:14\n\u2022 Well...$2^{10}\\equiv 7\\pmod 9$ and $7^{10}\\equiv 7\\pmod 9$. Then go by induction. \u2013\u00a0lulu Sep 3 '16 at 1:14\n\u2022 \"I feel using Euler's theorem is using a needlessly complicated theorem to prove something as simple as this\" well, that is just silly! The entire reason to have complicated theorems is because they are versitile and so you won't have to prove simple propositions like this over and over again. \"Proof: it follows directly from Euler's theorem\". That is a six word proof and it is enough! It's ... silly... to give a six paragraph 2 lemma prove and then complain a six word proof is too complicated. It's not like we have prove Euler th. Every time we use it. \u2013\u00a0fleablood Sep 3 '16 at 4:46\n\u2022 @fleablood Moise once referred to it as something like using a tractor where a shovel would do. It's very natural to wonder if there is a more elementary solution. There is nothing inherently wrong with finding more than one way to solve a problem. How many proofs of quadratic reciprocity exists? \u2013\u00a0steven gregory Jan 2 '18 at 15:41\n\n## 2 Answers\n\nInduction? Base case when $k=1$ is clear, for inductive step we have: $2^{10^k}=(2^{10^{k-1}})^{10}\\equiv 7^{10}\\equiv 7\\bmod 9$\n\n\u2022 How did you get $7^{10} \\equiv 7 \\pmod{9}$ without evaluating in a computer? \u2013\u00a0XYZT Sep 3 '16 at 1:40\n\u2022 we can go $7^{10}\\equiv (-2)^{10}\\equiv 2^{10}$. And you already know $2^{10}\\equiv 9$. So we don't need no extra stuff \u2013\u00a0HereToRelax Sep 3 '16 at 1:43\n\u2022 @CarryonSmiling You mean $2^{10}\\equiv 7$. It simply follows from $1+0+2+4\\equiv 7\\pmod{9}$. \u2013\u00a0user236182 Sep 3 '16 at 5:09\n\nDon't fear Euler's theorem. You have it. Use it. It's not like it costs a lot of gas money or you have to pay tolls.\n\n$2^6 \\equiv 1 \\mod 9$ so $2^{10^k} \\equiv 2^{(10^k \\mod 6 = 4^k \\mod 6)} \\mod 9$.\n\nOne little observation. If $4^n \\equiv 4 \\mod 6$ then $4^{n+1} \\equiv 16 \\equiv 4 \\mod 6$ so inductively $4^n \\equiv 4 \\mod 6$ for all $n$.\n\nSo $2^{10^k} \\equiv 2^{4^k} \\equiv 2^4 =16 \\equiv 7 \\mod 9$.", "date": "2021-04-21 14:11:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1912681/is-there-a-quicker-proof-to-show-that-210k-equiv-7-pmod9-for-all-posit", "openwebmath_score": 0.9902302026748657, "openwebmath_perplexity": 258.98721600637947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534354878827, "lm_q2_score": 0.8688267796346598, "lm_q1q2_score": 0.8527130277163104}}
{"url": "http://math.stackexchange.com/questions/160723/radius-either-integer-or-sqrt2-cdotinteger", "text": "# Radius either integer or $\\sqrt{2}\\cdot$integer\n\nGiven a circle about origin with exactly $100$ integral points(points with both coordinates as integers),prove that its radius is either an integer or $\\sqrt{2}$ times an integer.\n\nWhat my solution is: Since circle is about origin, hence, integral points would be symmetric about the $x$-axis and $y$-axis as well as line $x=y$ and line $x+y=0$ ,i.e. if $(x,y)$ is an integral point, so are $(x,-y),(-x,-y),(-x,y),(y,x),(y,-x),(-y,x)$ and $(-y,-x)$.Therefore, we need to consider only a single octant. Since there are a total of $100$ integral points, two cases are possible:\n1) radius of the circle is integer.\n2) radius of the circle is not an integer.\ncase 1: If radius is an integer, then $4$ points on the $x$-axis and $y$-axis of the circle would be integral points and hence each octant must have $12$ points(as $100-4=96$ is a multiple of $8$). therefore, this case is consistent.\ncase2: If radius is not an integer, then $100$ integral points can't be divided into $8$ parts(octants),and points on $x$-axis and $y$-axis of circle are not integral points, therefore points on line $x=y$ and $x+y=0$ must be integral points so as to divide $100-4=96$ points in $8$ parts. But since point on line $x=y$ and circle is of the form $(r\\cdot\\cos(45^\\circ),r\\cdot\\sin(45^\\circ))$,therefore, $r/\\sqrt{2}$ is an integer and hence $r=\\sqrt{2}\\cdot$integer. other points of circle on these lines are consistent with it.\n\nSo, i proved that either radius is an integer and if not then it has to be $\\sqrt{2}\\cdot$integer.\n\nIs there any flaw in my arguments?? I couldn't find the proof to check whether mine is correct. Thanks in advance!!\n\n-\nLooks good to me. \u2013\u00a0 Gerry Myerson Jun 20 '12 at 12:52\n\nAll the ingredients are here, but the flow of the argument is not optimal. A smooth proof of the claim would begin with \"Assume the set $S:=\\gamma\\cap{\\mathbb Z}^2$ contains $100$ elements. Then $\\ldots$\", or it should begin with \"Assume the radius of $\\gamma$ is neither an integer nor $\\sqrt{2}$ times an integer. Then $\\ldots$\".\nThe essential point (which does not come out clearly in your argument) is the following: The group of symmetries of $S$ is the dihedral group $D_4$, which is of order $8$. Since $100$ is not divisible by $8$ this action has nontrivial fixed points, i.e. points on the lines $x=0$, $y=0$, $y=\\pm x$.\nFor the case of the radius being an integer, you don't have to make any argument at all. What you are trying to prove is that a circle having exactly 100 integral points implies the radius is either integral or $\\sqrt 2$integral. This would still be true if none of these circles were integral. It would even be true if there were no circles with exactly 100 integral points.\nThat said, your argument works fine. It extends to any number of integral points equal to $4 \\pmod 8$", "date": "2015-04-25 01:04:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/160723/radius-either-integer-or-sqrt2-cdotinteger", "openwebmath_score": 0.957487940788269, "openwebmath_perplexity": 149.47178269752217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534316905262, "lm_q2_score": 0.8688267813328977, "lm_q1q2_score": 0.8527130260838068}}
{"url": "http://mathhelpforum.com/advanced-statistics/33259-density-function-solution-check.html", "text": "# Thread: Density function solution check\n\n1. ## Density function solution check\n\nSuppose Y has density function:\n\nf(y) = ky(1-y) when 0<=y<=1\nand\nf(y)= 0 elsewhere\n\na)Find the value of k that makes f(y) a probability density function\nb)Find P(0.4<=Y<1)\nc)Find P(Y<=0.4|Y<=0.8)\nd)Find P(Y<0.4|Y<=0.8)\n\nsol.\n\nfor part a I found k to be 6 (by making the integral = 1 on the interval 0,1 and solving for k\n\nfor part b also used the integral between 0.4 and 1 and got 0.648\n\nc) d) is where I have the problem. I suppose both must have the same answer.\n\nSince this is conditional probability can we state that:\n\nP(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)\n\nso I find using integral P(0.4<=Y<=0.8)=0.544\nand P(y<=0.8) = P(0<=y<=0.8)=0.896\n\nthen we divide 0.544/0.896= 0.607\nI feel that I am doing something wrong here, but can't see what.\n\n2. Originally Posted by somestudent2\nSuppose Y has density function:\n\nf(y) = ky(1-y) when 0<=y<=1\nand\nf(y)= 0 elsewhere\n\n[snip]\nc)Find P(Y<=0.4|Y<=0.8)\nd)Find P(Y<0.4|Y<=0.8)\n\n[snip]\n\nc) d) is where I have the problem. I suppose both must have the same answer.\n\nSince this is conditional probability can we state that:\n\nP(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)\n\nso I find using integral P(0.4<=Y<=0.8)=0.544\nand P(y<=0.8) = P(0<=y<=0.8)=0.896\n\nthen we divide 0.544/0.896= 0.607\nI feel that I am doing something wrong here, but can't see what.\nThe answer to (c) and (d) is\n\n$\\frac{\\Pr(Y \\leq 0.4 \\text \\, \\text{and}\\, Y \\leq 0.8)}{\\Pr(Y \\leq 0.8)} = \\frac{\\Pr(Y \\leq 0.4)}{\\Pr(Y \\leq 0.8)}$ $= \\frac{\\int_{0}^{0.4} ky(1-y) \\, dy}{\\int_{0}^{0.8} ky(1-y) \\, dy} = \\frac{\\int_{0}^{0.4} y(1-y) \\, dy}{\\int_{0}^{0.8} y(1-y) \\, dy} = .....$", "date": "2017-05-30 10:10:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/33259-density-function-solution-check.html", "openwebmath_score": 0.9444133639335632, "openwebmath_perplexity": 2286.1745705528106, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9814534322330059, "lm_q2_score": 0.8688267728417086, "lm_q1q2_score": 0.852713018221421}}
{"url": "https://math.stackexchange.com/questions/2020321/factorise-fx-x34x2-3x", "text": "# Factorise $f(x) = x^3+4x^2 + 3x$\n\nNot sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:\n\n$$\\text{Factorise}: f(x) = x^3+4x^2+3x$$\n\nFirstly, the GCD of the above is $x$:\n\n$$x(x^2+4x+3)$$\n\nNow take $x^2+4x+3$ and factorise that:\n\n$$x^2+4x+3$$\n\nUsing the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.\n\n\\begin{array}{|c|c|} \\hline x^2 & \\\\ \\hline & 3 \\\\ \\hline \\end{array}\n\nThen find HCF of 3:\n\n$$3\\\\ 1 | 3$$\n\nEnter the values $1x$ and $3x$ into the other two boxes:\n\n\\begin{array}{|c|c|} \\hline x^2 & 1x \\\\ \\hline 3x& 3 \\\\ \\hline \\end{array}\n\nNow factorise the rows and columns:\n\n$$x^2 + 1x = x(x+1)\\\\ x^2 + 3x = x(x+3)\\\\ 1x + 3=1(x+3)\\\\ 3x +3=3(x+3)$$\n\nTherefore: $$x^2+4x+3=(x+1)(x+3)$$ It follows that: $$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$\n\nAny feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...\n\n\u2022 If you want to check your work, multiply out $x(x+1)(x+3)$, and you will find it equals $x^3+4x^2+3x$. Nov 18, 2016 at 18:40\n\nHad never heard of the box method before I saw you use it!\n\nWhen you got to the part of factoring $x^2 + 4x + 3$ I would go and find the roots of it, because with the roots one can also factor the polynomial.\n\nUpon finding that $-1$ and $-3$ are the roots, I would know $x^2 + 4x + 3 = (x - (-3))(x - (-1)) = (x + 1)(x + 3)$.\n\nThis works for a general polynomial of degree $n$. If a $n$-degree polynomial has roots $\\lambda_1, \\cdots, \\lambda_n$, then the polynomial is equal to $(x - \\lambda_1)\\cdots(x - \\lambda_n)$\n\nOf course we not always have access to all the polynomial's roots at once, but we can partially factorise and work our way through that. Let us go through the polynomial $p = x^4 + 3x^3 - x^2 - 3x$.\n\nFirst obvious thing is that $\\lambda_1 = 0$ and thus $p = x^4 + 3x^3 - x^2 - 3x = x(x^3 + 3x^2 - x - 3)$.\n\nNow comes the tricky part. Finding the roots of $p_1 = x^3 + 3x^2 - x - 3$. What I always start by doing is trying some small numbers. Guessing $\\lambda_2 = 1$ turns out to be fine and thus we can factor $p_1$. Now there is some polynomial $p_2$ of degree 2 that multiplied by $(x - \\lambda_2) = (x - 1)$ gives $x^3 + 3x^2 - x - 3$. To calculate such polynomial I refer you to Ruffini's rule, which is just a faster way to factorize a polynomial when you know one root. Then we get $p_2 = x^2 + 4x + 3$ which was what you factorized above. Since this is a 2nd degree polynomial, we could keep trying to guess roots, use your box method, or using the quadratic formula to find its roots.\n\n\u2022 The box method is really cool, particularly when in particular multiplying trinomials with trinomials. Nov 18, 2016 at 18:44\n\u2022 First time for me too ! Nov 18, 2016 at 18:45\n\u2022 @imranfat had never heard of it. should I google it and learn it?\n\u2013\u00a0RGS\nNov 18, 2016 at 18:47\n\u2022 Bear in mind guys, I'm a total novice! I like systematic approaches to problems, and this one helps me remember the steps required. Video I used is here: youtube.com/watch?v=_Wb_CT-1VN8&app=desktop\n\u2013\u00a0Dan\nNov 18, 2016 at 18:49\n\u2022 @DanB your approach is fine. Also note that the method I talked about always works. If you are able to find the roots, then you are able to factorise it! I will add something to my answer.\n\u2013\u00a0RGS\nNov 18, 2016 at 18:53\n\nI wouldn't know if this really qualifies as an answer, but what you're asking for is essentially an opinion of methodology.\n\nFirst of all, your factorisation is correct: you can check the result simply by doing the multiplication.\n\nThat said, it looks like you could have saved a lot of effort by using a couple of different techniques.\n\nTo begin with, the polynomial $x^3+4x^2+3x$ is clearly divisible by $x$ (as you noted), so we factor it out to get $x(x^2+4x+3)$.\n\nHere is where it gets interesting: to factor the degree 2 polynomial, I usually take one of these two roads.\n\nFirst: observe that $(x+a)(x+b)=x^2+(a+b)x+ab$, so if we can guess two numbers $a$ and $b$ such that, in our case, we have $ab=3$ and $a+b=4$, we are done.\n\nIn this case you might have been able to spot that $3\\cdot 1=3$ and $3+1=4$, so $(x+1)(x+3)$ is the factorization we're looking for. Trust me, it gets easier with practice and it's a huge time- and effort-saver for simple degree 2 polynomials with integer coefficients.\n\nElse we could try and guess just one root, and then use polynomial long division. This has the advantage of working with any degree of polynomials, as long as you can guess one root. I usually try $1$, $-1$ and maybe $2$, but if those don't work I'm better off using some other technique.\n\nAnd finally, for degree 2 polynomials you have the general formula to find the roots, which has the advantage of always working (in the sense that it will always find all real roots there are to find), but can get messy and is, in my honest opinion, rather boring.\n\nNow, this is not to say that any of what you did is incorrect or even strictly less efficient: it is ultimately a matter of personal taste, but I feel that having as many tools as possible in your pocket can only do you good.\n\n\u2022 Thanks! It was indeed really a question of methodology, although I wasn't completely sure of the answer! The guessing method sounds like something that with practice would indeed save a lot of time.\n\u2013\u00a0Dan\nNov 18, 2016 at 19:00\n\nEverything is more or less correct about the way you approach the problem. You could have also opted for the middle term factorisation method or the method of vanishing method. However you have to correct one thing ...\n\nNow factorise the rows and columns: $$x^2 + 1x = x(x+1)\\\\ x^2 + 3x = x(x+3)\\\\ 1x + 3=1(x+3)\\\\ 3x +3=1(x+3)$$\n\nThe last line ought to be $3x +3=3(x+1)$.\n\n\u2022 Thank you! I've edited my post. Will look up the methods you've suggested, can never hurt to have more tools in the arsenal!\n\u2013\u00a0Dan\nNov 18, 2016 at 19:01\n\u2022 @DanB You're welcome. Nov 18, 2016 at 19:02\n\nTo factor $x^3+4x^2+3x$, we notice that we can factor $x$ out. Therefore, we get$$x^3+4x^2+3x=x(x^2+4x+3)\\tag1$$ Now, we need to see if $x^2+4x+3$ can be factored as a product of two linear terms. An easy way to factor a monic polynomial is to find two numbers $r,s$ that sum to the negated value of $b$ and have a product of $c$ in $x^2+bx+c$.\n\nIn other words, we have \\begin{align*} & r+s=-b\\\\ & rs=c\\end{align*}\\tag2 for $x^2+bx+c$. In your example, we see that $-b=-4$ and $c=3$. Messing around, we see that when $r=-1,s=-3$, the requirements are met. Thus,$$x^3+4x^2+3x=x(x-1)(x-3)$$\n\nX(X+1)(X+3)\n\nTherefore (X+1)(X+3) = X^2+4X+3.\n\nTherefore multiplying this by X you get X^3+4X^2+3X.\n\nSo yes well done that is correct.\n\nWell done. And for those of us unaware of the box method, the following would have also worked: $$x^2+4x+3=x^2+4x+4-1=(x+2)^2-1=(x+2-1)(x+2+1)=(x+1)(x+3)$$\n\nAvoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on.\n\nSuppose we have a quadratic $p(x)$ that factorizes as:\n\n$$p(x) = (x + s)(x + t)$$\n\nwhere we don't know what $s$ and $t$ are yet. If we multiply out the brackets, then we get:\n\n$$p(x) = x^2 + sx + tx + st = x^2 + (s+t)x + st$$\n\nNow suppose that we are given $p(x)$ in the form\n\n$$p(x) = x^2 + bx + c$$\n\nIn order to factorize $p(x)$, all we need to do is find two numbers $s$ and $t$ such that $$s+t=b$$ and $$st=c$$\n\nIn your case, you want to find $s$ and $t$ such that $s+t=4$ and $st=3$. It shouldn't take you very long to realize that $s=3$ and $t=1$ will do. Then you can immediately factorize the polynomial as $(x+3)(x+1)$.\n\nSlogan: To factorize a quadratic of the form $x^2+bx+c$, just find two numbers that add to give $b$ and multiply to give $c$.\n\nJust for interest\n\nThis method should work for all the quadratics you see in Math 1, and I hope that it's a bit clearer what's going on compared to the 'Box Method'. You might be interested in how we solve quadratic equations that don't have nice solutions. For example, suppose we were trying to factorize $$x^2+x-1$$ We want to find two numbers that add together to give $1$ and multiply together to give $-1$. It turns out that the right two numbers are $$\\frac{1+\\sqrt{5}}{2}$$ and $$\\frac{1-\\sqrt{5}}{2}$$ How did I work those out? Well, I'll show you a method that was invented by the ancient Babylonians. Suppose we have two numbers $b$ and $c$ and we're trying to find two numbers $s$ and $t$ such that $s+t=b$ and $st=c$.\n\nIf we square the first equation, we get $$b^2 = (s+t)^2 = s^2 + 2st + t^2$$ Now since $st=c$, we can subtract $4c$ from the left hand side and $4st$ from the right hand side to get: $$b^2-4c = s^2 - 2st + t^2$$ But $s^2 - 2st + t^2=(s-t)^2$, so we may write $$s-t = \\sqrt{b^2-4c}$$ (Here, we are free to assume that $s\\ge t$, so this is the positive square root.) Now we can recover $s$ and $t$: $$s = \\frac{(s+t) + (s-t)}{2} = \\frac{b+\\sqrt{b^2-4c}}{2}$$ $$t = \\frac{(s+t) - (s-t)}{2} = \\frac{b-\\sqrt{b^2-4c}}{2}$$", "date": "2022-05-23 15:10:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2020321/factorise-fx-x34x2-3x", "openwebmath_score": 0.9031533002853394, "openwebmath_perplexity": 202.2672422360076, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534392852384, "lm_q2_score": 0.8688267626522814, "lm_q1q2_score": 0.8527130143481412}}
{"url": "https://math.stackexchange.com/questions/3510522/proving-recurrence-using-induction-where-there-is-an-upper-limit-on-the-number-o", "text": "# Proving recurrence using induction where there is an upper limit on the number of integers to prove it for\n\nGiven $$x_0=1$$ and $$x_j=x_{j-1}\\frac{N-(j-1)}{N}+x_{j+1}\\frac{j+1}{N}$$ for $$j=1,...,N-1$$, the formula $$x_j={N\\choose j}$$ can be proven by induction. I do not see why we are able prove it by induction, seeing that using induction, we prove the base case, we assume the formula holds for n, then show it holds for n+1, then we claim it holds for every integer. In this case it only holds up to N-1. So why does the induction proof work? I think that the induction proof should fail.\n\nThe inductive proof: $$x_0=1$$, Suppose the result is true for $$k \\le j$$\n\n\\begin{align}x_{j+1} &=\\frac{N}{j+1}\\left(x_j-\\frac{N-j+1}{N}x_{j-1}\\right)\\\\&=\\frac{N}{j+1}\\left(\\frac{N!}{j!(N-j)!}-\\frac{N-j+1}{N}\\frac{N!}{(j-1)!(N-j+1)!}\\right)\\\\ &\\text{after some simplification}\\\\&={N\\choose{j+1}} \\end{align}\n\nSee it works but I think that it should fail.\n\n\u2022 You need to use \"Strong\" induction in this case. First show it works for $j=0$ and $j=1$, then show if it works for $j-1$ and $j$ then it works for $j+1$. Jan 15 '20 at 21:04\n\u2022 @DonaldSplutterwit Yes. But, it works only up to N-1 and not all the integers. So induction should not prove this formula. Jan 15 '20 at 21:10\n\u2022 I think you are misunderstanding what the statement is claiming. It is not claiming $x_j={N\\choose j}$ for all $N$ and doing induction on $N$. Is is claiming that $x_j={N\\choose j}$ for all $j$ (up to $N-1$) and doing induction on $j$. Part of what is confusing you is that we are doing induction on a variable $j$ that has an upper limit, $N-1$, and so our result is not for all natural numbers but just for natural numbers from $1$ to $N$. This is okay. We can do induction for finite values. Jan 15 '20 at 21:48\n\u2022 Both $x_j$ and ${N \\choose j}$ are only defined for $j\\le N$ so that's our upper bound. Our statement is $P(j)=$ if [$j \\le N$] then [$x_j = {N\\choose j}$. We can prove $P(j)$ for all natural $j$ because if $j> N$ then [$j \\le N]$ is false and \"if [FALSE] then $anything$\" is a true statement so if $j > N$ then $P(j)=$ if [$j \\le N$] then [$x_j = {N\\choose j}$; is a true statement. Jan 15 '20 at 22:18\n\u2022 \"Nowhere do we specify an upper bound\" Uh.... yes we do! ... it says \"for j=1,...,N\u22121\" in black and white!.... So induction will be successful up to $j+1 \\le N$. For $j > N$ it will fail but we don't give a flying funky if it fails for values we don't care about. We only care that it is successful on the values we DO care about. ... \"In this case it only holds up to N-1\" For $j+1$ where $j \\le N-1$. So $j+1 \\le N$. And we don't CARE if it holds for any higher values? Why on earth would we? Jan 15 '20 at 22:52\n\nThe statement is not for every positive integer but only for positive integers up to $$N$$. It is not trying to claim it is true for any positive integer greater than $$N$$.\n\nConsider: $$P(j )=$$: If $$j \\le N$$ then something, call it $$Q(j)$$ is true.\n\nLet's say we can show that if $$k< N$$ that $$Q(k)\\implies Q(k+1)$$ but only if $$k < N$$. I claim we can still prove $$P(k)$$ is true for all natural $$k$$.\n\nBase case: $$P(1)$$. We show that $$Q(1)$$ is true and as $$1 < N$$ then \\$P(1) is true.\n\nInduction step: $$P(k)\\implies P(k+1)$$.\n\nAssume if $$k\\le N$$ then $$Q(k)$$ is true.\n\nCase 1: $$k \\ge N$$.\n\nThen $$k+1 > N$$ and [$$k+1 \\le N$$] is false: $$FALSE \\implies Q(k+1)$$ is vacuously true whether $$Q(k+1)$$ is true or not. So $$P(k+1)$$ is true.\n\nCase 2: $$k < N$$.\n\nThe $$k+1 \\le N$$. We showed that $$Q(k)\\implies Q(k+1)$$. So if [$$k+1 \\le N]\\implies Q(k+1)$$ is true. So $$P(k+1)$$ is true.\n\nSO our induction step works.\n\nWe have proven:\n\nFor any natural $$j$$, $$P(j)$$ is true.... or in other words,\n\nif $$j \\le N$$ then $$Q(j)$$ is true... or in other words,\n\n$$Q(j)$$ is true for every natural $$j \\le N$$.\n\nThat's all the induction is trying to claim.\n\nThe induction of $$Q(j)$$ works..... up to $$j \\le N$$. There is nothing invalid about this.\n\nGiven the recurrence you have written, you will need two initial values to get going. Solving the recurrence for $$x_{j+1}$$ gives $$x_{j+1} = \\frac{n x_j+ (j-n-1) x_{j-1}}{j+1}.$$ Note that if $$j=n$$, this gives $$x_{n+1} = \\frac{n\\cdot 1 + (n-n-1)\\cdot n}{n+1} = 0,$$ just as you would expect. So in fact the inductive proof holds for all $$j$$ provided you define $$\\binom{n}{j}=0$$ for $$j<0$$ or $$j>n$$.", "date": "2021-11-27 09:18:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3510522/proving-recurrence-using-induction-where-there-is-an-upper-limit-on-the-number-o", "openwebmath_score": 0.8099619150161743, "openwebmath_perplexity": 184.62522429911434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97364464868338, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8527053225394021}}
{"url": "http://morethanahomemom.com/cayden-boyd-qxj/circle-inscribed-in-a-right-triangle-e874be", "text": "How to Inscribe a Circle in a Triangle using just a compass and a straightedge. Example 5. Home List of all formulas of the site; Geometry. A circle can be drawn inside a triangle and the largest circle that lies in the triangle is one which touches (or is tangent) to three sides, is known as incircle or inscribed. Find AD,BE and CF ( these 3 are altitudes of triangle ABC ) . The radius of the circle inscribed in the triangle is. Yes; If two vertices (of a triangle inscribed within a circle) are opposite each other, they lie on the diameter. The area within the triangle varies with respect to \u2026 We want to find area of circle inscribed in this triangle. I have solved for the diameter and I got 2. 2.A movie company surveyed 1000 people. 640\u00d7482. If AB=5 cm, BC=12 cm and < B=90*, then find the value of r. Now draw a diameter to it. For the right triangle in the above example, the circumscribed circle is simple to draw; its center can be found by measuring a distance of 2.5 units from A along \u00af AB. Radius of a circle inscribed in a right triangle . Solve for the third side C. The side opposite the right angle is called the hypotenuse (side c in the figure). Small. Answers. Examples: For each inscribed quadrilaterals find the value of each variable. In this situation, the circle is called an inscribed circle, and its center is called the inner center, or incenter. This triangle, this side over here also has this distance right here is also a radius of the circle. Inscribe: To draw on the inside of, just touching but never crossing the sides (in this case the sides of the triangle). May 2015 13 0 Canada May 14, 2015 #1 Hi everyone, I have a question. A circle is inscribed in the triangle if the triangle's three sides are all tangents to a circle. So let's say that this is an inscribed angle right here. Calculator Technique. abc is a right angle triangle right angled at a a circle is inscribed in it the length of two sides containing angle a is 12 cm and 5 cm find the radi - Mathematics - TopperLearning.com | 42jq3mpp Original. Every non-equilateral triangle has an infinitude of inscribed ellipses. It's going to be 90 degrees. So, Area A: = (base * height)/2 = (2r * r)/2 = r^2 A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90-degree angle). Illustration showing the diameter of a circle inscribed in a right triangle is equal to the difference between the sum of the legs and the hypotenuse. Because the larger triangle with sides 15, x, and 25 has a base as the diameter of the circle, it is a right triangle and the angle opposite the diameter must be 90. Since the triangle's three sides are all tangents to the inscribed circle, the distances from the circle's center to the three sides are all equal to the circle's radius. This is a central angle right \u2026 Let me draw another triangle right here, another line right there. Right angles are typically denoted by a square drawn at the vertex of the angle that is a right angle. We bisect the two angles and then draw a circle that just touches the triangles's sides. asked Oct 1, 2018 in Mathematics by Tannu ( 53.0k points) circles BEOD is thus a kite, and we can use the kite properties to show that \u0394BOD is a 30-60-90 triangle. D. 18, 24, 30 . Question 188171: 1.A circle with a radius of 1 in. The sheet of Circle Theorems may help you. What is the length of $BD?$ What is the length of $DC?$. You know the area of a circle is \u03c0r\u00b2, so you\u2019re on the lookout for \u03c0 in the answers. But I just don't understand how to get the largest and smallest. In the given figure, a cradle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. A right triangle is a triangle in which one angle has a measurement of 90\u00b0 (a right angle), such as the triangle shown below.. A circle is inscribed in an equilateral triangle with side length x. The center of the incircle is a \u2026 One of them is a circle, and one of them is the Steiner inellipse which is tangent to the triangle at the midpoints of the sides. The inverse would also be useful but not so simple, e.g., what size triangle do I need for a given incircle area. Find the area of the black region. Thus, the Pythagorean theorem can be used to find the length of x. x 2 + 15 2 = 25 2 Rather than do the calculations, notice that the triangle is a 3-4-5 triangle (multiplied by 5). For the 3,4,5 triangle case, the radius can be found algebraically or by construction. A circle circumscribing a triangle passes through the vertices of the triangle while a circle inscribed in a triangle is tangent to the three sides of the triangle. Right triangle. \u0394 ABC is a right angled triangle with \u2220A = 90\u00b0, AB = b cm, AC = a cm, and BC = c cm A circle is inscribed in this triangle. If AB=5 cm, BC=12 cm and < B=90*, then find the value of r. If the two sides of the inscribed triangle are 8 centimeters and 10 centimeters respectively, find the 3rd side. Right Triangle: One angle is equal to 90 degrees. For the right triangle in the above example, the circumscribed circle is simple to draw; its center can be found by measuring a distance of $$2.5$$ units from $$A$$ along $$\\overline{AB}$$. Show Step-by-step Solutions. Let's call this theta. Alex drew a circle with right triangle prq inscribed in it, as shown below: the figure shows a circle with points p, q, and r on it forming an inscribed triangle. Find the radius of its incircle. Forums. 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If and only if its opposite angles aresupplementary at these points the radius of the radius of triangle!, find the circle \u2019 circle inscribed in a right triangle asking for: area of the circle tangent... That BC = 6 cm the Pythagorean theorem to show that \u0394BOD is triangle.: a circle is 39.19 square centimeters, and so $\\angle AC ' I$ is right is also..., diameter is 2 in., find the value of each variable,...", "date": "2021-04-19 03:13:53", "meta": {"domain": "morethanahomemom.com", "url": "http://morethanahomemom.com/cayden-boyd-qxj/circle-inscribed-in-a-right-triangle-e874be", "openwebmath_score": 0.6869009733200073, "openwebmath_perplexity": 390.44542158692235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9736446471538803, "lm_q2_score": 0.8757869948899666, "lm_q1q2_score": 0.8527053196215987}}
{"url": "https://math.stackexchange.com/questions/977474/infinite-sum-of-sines-with-increasing-period", "text": "# Infinite Sum of Sines With Increasing Period\n\nA while ago, I was thinking about the Weierstrass function, which is a sum of sines with increasing frequencies in such a way that the curve is a fractal. However, I wondered what would happen if one took the sum where the frequencies decreased; in particular, noting that $|\\sin(x)|\\leq x$, it is clear that the function $$f(x)=\\sum_{n=1}^{\\infty}\\sin\\left(\\frac{x}{s_n}\\right)$$ converges pointwise for any sequence $s_n$ such that the sum of $\\frac{1}{s_n}$ converges absolutely - and, in fact, yields an $f$ which is analytic. Of particular interest to me is the sequence of square numbers - that is, the function $$f(x)=\\sum_{n=1}^{\\infty}\\sin\\left(\\frac{x}{n^2}\\right).$$ I created the following plot of the function from the first 10,000 terms in the series:\n\nWhat I find interesting here is that, for some reason I can't determine, it looks like $f(x)$ might be asymptotic to $\\sqrt{x}$. I've checked numerically for higher arguments and this seems to continue to be the case. This strikes me as odd, since I had expected it to appear more or less periodic, with long-term variation in amplitude and frequency.\n\n1. Is $f(0)=0$ the only (real) zero of $f$?\n\n2. Does $f$ grow without bound? What is it asymptotic to?\n\n\u2022 that's 3 questions. There are three types of people in the world: Those who can count to three, and those who can't. (LOL) \u2013\u00a0Mark Fischler Oct 16 '14 at 23:42\n\u2022 It often helps to analyze these kinds of functions as real or imaginary parts of the same series with the exponential function in place of $\\sin$ or $\\cos$. In this case, you have $\\Im\\sum_{i=1}^\\infty\\exp\\left(i\\frac{x}{n^2}\\right)$. \u2013\u00a0alex.jordan Oct 16 '14 at 23:59\n\u2022 Here's something funky about this function: If you take the Fourier transform of $f(x)$ and move the sum outside the integral (which I'm not sure is justified) you get zero. Yet $f(x)$ certainly looks to have a quasi-periodic structure. \u2013\u00a0Mark Fischler Oct 17 '14 at 0:18\n\u2022 Well, I plugged $\\int_{1}^{\\infty}\\sin(\\frac{x}{n^2})$ into Mathematica and it returned $\\sqrt{2 \\pi } \\sqrt{x} C\\left(\\sqrt{\\frac{2}{\\pi }} \\sqrt{x}\\right)-\\sin (x)$ where $C$ is the Fresnel integral - $\\int_{0}^{x}\\cos(\\pi x^2/2)$, which converges to $\\frac{1}2$ - so the expression $\\int_{1}^{\\infty}\\sin(x/n^2)$ is indeed asymptotic to $\\sqrt{x}$. I wonder if there's some way to get the analogous discrete result from there. \u2013\u00a0Milo Brandt Oct 19 '14 at 14:32\n\nHere's a proof that $f$ only vanishes at $x = 0$ (you can use a similar method to get some asymptotic results as well).\n\nWrite $f(x)/x$ as \\begin{align*} {f(x)\\over x} &= \\sum_{n\\geq 1} {\\operatorname{sinc}{(x/n^2)}\\over n^2} \\end{align*} Since $f(x)/x$ is even, we need only treat the case $x\\geq 0$. Split the sum into the regions where $x/n^2$ is smaller or greater than $\\pi$. Since $\\operatorname{sinc}{\\lambda}>0$ for $|\\lambda|\\leq \\pi$, and since $\\operatorname{sinc}{\\lambda}\\geq 2/\\pi$ for $|\\lambda|\\leq \\pi/2$, we have \\begin{align*} \\sum_{x/n^2\\leq \\pi} {\\operatorname{sinc}{(x/n^2)}\\over n^2} &\\geq {2\\over \\pi}\\sum_{x/n^2\\leq \\pi/2} {1\\over n^2} = {2\\over \\pi} \\sum_{n\\geq (2x/ \\pi)^{1/2}}{1\\over n^2} > {2\\over \\pi} \\int_{\\lceil(2x/\\pi)^{1/2}\\rceil}^\\infty {dt\\over t^2} = {2\\over \\pi}{1\\over \\lceil(2x/\\pi)^{1/2}\\rceil}. \\end{align*} On the other hand, since $\\operatorname{sinc}{\\lambda}\\leq 1/\\lambda$ for all $\\lambda>0$, we have \\begin{align*} \\left|\\sum_{x/n^2> \\pi} {\\operatorname{sinc}{(x/n^2)}\\over n^2}\\right| & \\leq \\sum_{x/n^2>\\pi} {1\\over x} \\leq {1\\over x}\\left\\lfloor\\left({x\\over \\pi}\\right)^{1/2}\\right\\rfloor\\leq \\left({1\\over \\pi x}\\right)^{1/2}. \\end{align*} So \\begin{align*} f(x)/x> {2\\over \\pi}{1\\over \\lceil(2x/\\pi)^{1/2}\\rceil} - {1\\over (\\pi x)^{1/2}}, \\end{align*} which is positive when $x\\geq \\pi$. Since all terms in the sum are positive if $0\\leq x < \\pi$, it follows that $f(x)/x$ is always positive.\n\nBy the way, here's another heuristic (which can be made precise without too much trouble I think). We have \\begin{align*} {f(x)\\over x} & = {1\\over 2}\\hat g(x) = {1\\over 2}\\int_{-1}^1 g(t)e^{-ixt}\\,dt = {1\\over 2x}\\int_{-x}^x g(t/x)e^{-it}\\,dt, \\end{align*} where $g = \\sum_{n\\geq 1} \\chi_n$ is the sum of the characteristic functions of the intervals $[-n^{-2},n^{-2}]$. (Since $g$ is an $L^1$ function, this tells us at once that $f(x)/x\\to 0$ as $x\\to\\infty$.) Note that $\\{y:g(y)>n\\} = [-n^{-2},n^{-2}] = \\{y: y^{-1/2}>n\\}$ (or something similar), so we should roughly expect $g$ to look like $y^{-1/2}$, and so we should expect $\\hat g(x)$ to be approximately $x^{-1/2}$ (as can be seen from the last integral). You can use the same idea to get a sense of what the function would look like if you replace $\\sin{(x/n^2)}$ with $\\sin{(x/n^\\alpha)}$ for $\\alpha > 1$ (it should I think look like $x^{1/\\alpha}$).\n\n\u2022 Sorry if this approach is similar to Kirill's. I was working on it and didn't notice his answer until it was too late. \u2013\u00a0Nick Strehlke Oct 30 '14 at 1:30\n\u2022 One more comment: The function $g$ I mentioned toward the end should actually be $\\lfloor y^{-1/2}\\rfloor$; that is, you should be able to write (for $x>0$) $$f(x) = \\int_0^\\infty \\lfloor (x/y)^{-1/2}\\rfloor \\cos{y}\\,dy.$$ (There may be another factor of two or something.) \u2013\u00a0Nick Strehlke Oct 30 '14 at 7:27\n\nThese are my thoughts on the problem. They do not answer any of the three questions, but do not fit on a comment.\n\nLet $S_N$ be the $N$-th partial sum of the series and $L_n$ the lowest common multiple of $\\{1,\\dots,n\\}$. Then $S_N$ is periodic of period $2\\,L_n^2\\,\\pi$. $L_n$ is known to be of the order $e^{n(1+o(1))}$, so that the period is very large. Computations show that $S_N$ changes sign on $[0,2\\,L_n^2\\,\\pi]$. Here is the grapf of $S_4$ on $[0,288\\,\\pi]$.\n\nOn the other hand, for any $\\delta\\in(0,1)$ let $\\alpha\\in(0,\\pi/2)$ be such that $(\\sin\\alpha/\\alpha)=\\delta$. Then we have the lower bound $$f(x)\\ge S_N(x)+\\delta\\,\\Bigl(\\sum_{n=N+1}^\\infty\\frac{1}{n^2}\\Bigr)\\,x,\\quad0\\le x\\le\\alpha\\,N^2.$$ This is the graph of $S_{10000}$ (in blue) and the above lower bound with $n=10$, $\\delta=3/\\pi=0.95493$ and $\\alpha=\\pi/6$.\n\nThis shows that $f(x)>0$ on $(0,50\\,\\pi/3]$. An strategy to prove that $f(x)>0$ for all $x>0$ is to show that $$S_N(x)+\\delta\\,\\Bigl(\\sum_{n=N+1}^\\infty\\frac{1}{n^2}\\Bigr)\\,x>0,\\quad0\\le x\\le\\alpha\\,N^2.$$ I have checked it up to $n=100$.\n\nAs for upper bounds, we have $$|f(x)-S_N(x)|\\le\\sum_{n=N+1}^\\infty\\min\\Bigl(1,\\frac{x}{n^2}\\Bigr)\\le \\Bigl(\\sum_{n=N+1}^\\infty\\frac{1}{n^2}\\Bigr)\\,x.$$\n\nHere's a slightly informal way to get the asymptotic expansion of this function.\n\nSplit the region of summation into intervals of the form $$\\pi k < \\frac{x}{n^2} < \\pi(k+1),$$ for $k\\in\\mathbb{Z}_{\\geq0}$, and write the sum as $$\\sum_{n\\geq 1}\\sin \\frac{x}{n^2} = \\sum_{k\\geq0} \\sum_{\\frac{x}{\\pi(k+1)} < n^2 < \\frac{x}{\\pi k}} \\sin \\frac{x}{n^2}.$$ The first term $k=0$ can be approximated with an integral, giving $$\\sum_{n^2 > x/\\pi}\\sin\\frac{x}{n^2} \\sim \\int_{\\sqrt{x/\\pi}}^{\\infty} \\sin\\frac{x}{n^2}\\,dn = \\sqrt{2\\pi x}C(\\sqrt{2}),$$ where $C$ is the Fresnel integral.\n\nOn the rest of the sum we can approximate $\\sin\\frac{x}{n^2}$ with its average value $\\frac2\\pi$. However, the rest of the sum should be handled with some care, because the regions on which $\\sin$ is positive have different lengths from the regions on which it's negative, which will produce a net contribution to the asymptotic term $O(\\sqrt{x})$.\n\nApproximating the sum $$\\sum_{\\frac{x}{\\pi(k+1)} < n^2 < \\frac{x}{\\pi k}} \\sin \\frac{x}{n^2} \\sim \\frac{2(-1)^k}{\\pi}\\left(\\sqrt{\\frac{x}{\\pi k}} - \\sqrt{\\frac{x}{\\pi(k+1)}}\\right),$$ and summing over $k\\geq1$ gives the value $$\\frac{\\sqrt{x}}{\\pi^{3/2}}\\left((3\\sqrt{2}-2)\\zeta(\\tfrac12) - \\sqrt{2}\\zeta(\\tfrac12,\\tfrac32) \\right) = B\\sqrt{x},$$ where $\\zeta(z,a)$ is the generalized zeta function.\n\nPutting everything together gives the asymptotic form $$\\sum_{n\\geq 1}\\sin\\frac{x}{n^2} \\sim \\left(\\sqrt{2\\pi}C(\\sqrt{2}) + B\\right)\\sqrt{x} \\approx 1.25038\\sqrt{x}.$$\n\nAnother comment that is too long to be a comment:\n\nThe heuristic reason that the function is asymptotically proportional to $\\sqrt{x}$ is that for very large $x$,\n\n$\\cdot$ The contributions of the terms in $S_n(x)$ for $n$ much less than $\\sqrt{x}$ behave as pseudo-random numbers, restricted to $[-1,1]$. Thus $S_n(x)$ for $n << \\sqrt{x}$ can be thought of as roughly having a mean value of zero, and a $\\sigma$ on the order of $\\frac{1}{2}\\sqrt{x}$.\n\n$\\cdot$ The contributions of the terms in $S_n(x)$ for $n^2$ greater than about $2x$ can be well-estimated by approximating $\\sin \\frac{x}{n^2} \\approx \\frac{x}{n^2}$. Adding these from $n=\\sqrt{x}$ to infinity looks like $x \\int \\frac{1}{n^2}$ which will be about $\\frac{x}{\\sqrt{2x}} = \\frac{\\sqrt{x}}{\\sqrt{2}}$. And these contributions are all positive.\n\n$\\cdot$ The contributions of the terms in $S_n(x)$ for $n^2 \\approx \\frac{2}{\\pi} x$ are all roughly $1$ since we are near the top of the sine curve, and there are about $\\frac{8}{3\\pi} \\frac{1}{2n} = \\frac{4\\sqrt{2}}{3\\sqrt{\\pi}}\\sqrt{x}$ of them between $\\frac{\\pi}{4}$ and $\\frac{3\\pi}{4}$ on the sine curve, for a contribution of pretty nearly $\\sqrt{x}$.\n\n$\\cdot$ The contributions of the terms in $S_n(x)$ lying on the falling side of that first part of the sine curve almost exactly cancel with the contributions from the negative part of the first period of the sine curve (because $n^2$ is greater in that part of the curve, at at any rate, the contribution is a lot smaller than that of the flat section of the positive arch).\n\nSo all in all, you would expect $f(x)$ to behave about like $\\frac{3}{2}\\sqrt{x}$ for large $x$.\n\nThe last 3 bullets can be made somewhat more rigorous.\n\nBut by this reasoning, you would also expect larger fluctuations than we see in the original graph. So the \"effectively ergodic\" argument made in the first bullet is an over-estimate of the fluctuations, and I don't have a plausible reason why.\n\n\u2022 I think that first bullet you have might be key; I'm not sure how to rule out fluctuations, but I'll bet a statement to the tune of, \"for any $\\varepsilon_1,\\varepsilon_2>0$, we can find arbitrarily large $x$ and $n$ such that that $S_{n-1}(x)<\\varepsilon_1\\sqrt{x}$ and $(1-\\varepsilon_2)\\frac{x}{n^2}<\\sin(\\frac{x}{n^2})<(1+\\varepsilon_2)\\frac{x}{n^2}$\" might be a good way to formalize it (that particular statement might be false, but there could be a similar one of use) - it would prove that $f$ has a sequence that's asymptotic to the square root function. \u2013\u00a0Milo Brandt Oct 18 '14 at 2:59\n\u2022 You can start a line with an asterisk (*) to make it an item in a bulleted list. \u2013\u00a0user856 Oct 25 '14 at 9:47", "date": "2020-03-29 10:08:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/977474/infinite-sum-of-sines-with-increasing-period", "openwebmath_score": 0.9617723822593689, "openwebmath_perplexity": 118.5964558618954, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446502128796, "lm_q2_score": 0.8757869835428966, "lm_q1q2_score": 0.8527053112526165}}
{"url": "https://mathoverflow.net/questions/95867/covering-a-cube-with-a-square/95920", "text": "# Covering a Cube with a Square\n\nSuppose you are given a single unit square, and you would like to completely cover the surface of a cube by cutting up the square and pasting it onto the cube's surface.\n\nQ1. What is the largest cube that can be covered by a $1 \\times 1$ square when cut into at most $k$ pieces?\n\nThe case $k=1$ has been studied, probably earlier than this reference: \"Problem 10716: A cubical gift,\" American Mathematical Monthly, 108(1):81-82, January 2001, solution by Catalano-Johnson, Loeb, Beebee.\n\n(This was discussed in an MSE Question.) The depicted solution results in a cube edge length of $1/(2\\sqrt{2}) \\approx 0.35$.\n\nAs $k \\to \\infty$, there should be no wasted overlaps in the covering of the 6 faces, and so the largest cube covered will have edge length $1/\\sqrt{6} \\approx 0.41$. What partition of the square leads to this optimal cover?\n\nQ2. For which value of $k$ is this optimal reached?\n\nI have not found literature on this problem for $k>1$, but it seems likely it has been explored. Thanks for any pointers!\n\n\u2022 I wonder if anyone in the packaging industry has the answer. May 4 '12 at 0:37\n\u2022 I believe there was a \"Mathematical Games\" column on dissections which had a Greek cross rearranged into a square. Perhaps that article also mentioned this problem? Gerhard \"Testing Your Martin Gardner Fu\" Paseman, 2012.05.04 May 4 '12 at 15:51\n\u2022 If memory serves, Martin Gardner is also a source for the puzzle of covering a unit cube with a properly folded 1x7 strip. Apr 1 '17 at 2:52\n\u2022 Cool! The solution to that 1x7 puzzle informs the answer to the following: over all rectangles which can be folded to cover the unit cube, what is the lim inf of their areas? Gerhard \"Hint: The Answer Is Six\" Paseman, 2017.04.03. Apr 3 '17 at 17:43\n\u2022 When I read the title I thought it was about a 2D version $\\varphi:[0,1]^2\\to[0,1]^3$ of Peano area-filling (or space-filling) curves $[0,1]\\to[0,1]^2$... Mar 29 '19 at 12:04\n\nFour pieces, using the tessellation technique I learned from Harry Lindgren's Geometric Dissections (1964):\n\n\u2022 I've been looking for that! Gerhard \"Was It Behind The Headboard?\" Paseman, 2017.04.01. Apr 1 '17 at 10:57\n\u2022 Also, if you shift the (smaller) black squares a little to the right, the one piece that almost looks like two pieces looks more like one piece. Gerhard \"Tilting His Head Over This\" Paseman, 2017.04.01. Apr 1 '17 at 11:06\n\u2022 Beautiful! I posted a colorized version to make the dissection of the net more self-evident. Apr 1 '17 at 11:54\n\u2022 Thanks! GerPas: Yes, the hexomino tessellation cqn be moved a bit to the right (and independently also a bit up, though that doesn't help your cause). JO'R: I see that you were also the one who made a three-dimensional model of a cube covered by 5 pieces of a square, but evidently it will take more time (if you bother to do it at all) to re-do for the new 4-piece record. Apr 1 '17 at 14:37\n\u2022 @NoamD.Elkies: I'll make the model eventually, as there will be a certain pleasure in seeing your dissection fully deployed. Apr 1 '17 at 15:05\n\nYou can cut a $\\sqrt{6}\\times\\sqrt{6}$ square into 24 pieces that then cover the $1\\times1\\times1$ cube. Two triangles from the figure below plus one parallelogram make up one $1\\times1$square. parts of pieces sticking out to the left can obviously fit back in the right, so 18 pieces, plus 6 parts sticking out equals 24. You can improve on this by stitching pieces across the cube edge to make one bent piece and by stitching some of the parallelograms back to the triangles.\n\n![cube.png][1]\n\n[Added by O'Rourke:] Just to make Yoav's construction more explicit, here is how two triangles and a parallelogram fit together to form a $1 \\times 1$ square:\n\n[Added by Kallus:] Here's an illustration of a construction similar to Fedja's construction but with only five pieces. The first figure is the $\\sqrt{6}\\times\\sqrt{6}$ square. The second is the $2\\times3$ rectangle, which we fold into a cube by taking away the two yellow squares, folding the remainder, and adding the squares as the two missing faces.\n\n\u2022 Brilliant!! :-) May 3 '12 at 23:51\n\u2022 Actually, any two polygons of the same area are equidecomposable and the surface of the cube can be unfolded into a polygon, so the result is nice but not terribly surprising. Of course, the question about the minimal number of pieces remains. May 4 '12 at 0:15\n\u2022 Assuming that \"pieces\" mean \"connected polygonal pieces\", we can take a 3 by 2 rectangle, cut it into a T-shape and two unit squares and then use the standard \"sliding cut\" to turn it into a square, giving the total of 6 pieces to cover the unit cube. Can we do better? May 4 '12 at 0:34\n\u2022 OK. Posted to soon earlier. The change from the T tetromino to the S tetromino actually allows going down to five pieces instead of six. May 4 '12 at 4:21\n\u2022 Wow! $\\mbox{}$ May 4 '12 at 10:18\n\nJust illustrating Noam Elkies' 4-piece solution:\n\nBottom face is mostly yellow (except for a little green); two hidden back faces are mauve.\n\n\u2022 Lol, I can't believe you actually made a model of it. I haven't used glue and paper and scissors to cut out shapes since arts and crafts in elementary school. Though this is significantly more complicated. The wide breadth of tasks that comes with the profession of mathematics, lol.\n\u2013\u00a0user78249\nApr 1 '17 at 17:51\n\nSorry, this is an answer to an other question. (I did not read the question carefully.)\n\nQuestion: For which $$k$$, $$k$$ squares can tile the surface of cube.\n\nAnswer: $$k=6\\cdot(n^2+m^2)$$.\n\nHere is a tiling with $$k=30$$, $$n=1$$ and $$m=2$$.\n\n(source: psu.edu)\n\nIt is obvious if the tiling is vertex-to-vertex.\n\nIf the tiling is not vertex-to-vertex, you get a closed geodesic formed by overlaping sides. Then you can shift squares on one side of the geodesic to make the tiling \"more vertex-to-vertex\". Repeating this operation you can make the tiling to be vertex-to-vertex.\n\n\u2022 @Anton: Did you intend $n \\ge 1 , m \\ge 1$, or, say, $n \\ge 1 , m \\ge 0$ ? May 3 '12 at 18:03\n\u2022 Yes, $k$ has to be positive; so $n\\ge 1$ and $m\\ge 0$. May 3 '12 at 18:33\n\u2022 @Anton: Sorry to be slow :-/, but could you describe a partition of the square into 6 pieces that exactly cover the cube? May 3 '12 at 19:21\n\u2022 @Joseph: the partition into faces (the cube has 6 faces). May 3 '12 at 19:35\n\u2022 @Anton: I apologize for being so dense, but it may be that you are answering a different question than I asked...? I asked for how to cut up one square to cover a cube, not how to cover a cube with many squares. If I've diagnosed this correctly (unsure), it may explain why we seem to be talking past one another? May 3 '12 at 23:17\n\nNo promises that these are optimal, but here are some lower bounds:\n\nWith $k=2$, side length $3/8=0.375$ (with one piece flipped over), and with $k=3$, side length $2/5=0.4$:", "date": "2021-10-25 23:11:23", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/95867/covering-a-cube-with-a-square/95920", "openwebmath_score": 0.5809532999992371, "openwebmath_perplexity": 1019.0750245965395, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446456243805, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8527053040774931}}
{"url": "https://math.stackexchange.com/questions/2250051/are-the-closed-intervals-of-mathbbr-precisely-the-compact-connected-sets", "text": "# Are the closed intervals of $\\mathbb{R}$ precisely the compact connected sets?\n\nEquip $\\mathbb{R}$ with the topology generated by open intervals $(a, b)$. A subset of $\\mathbb{R}$ is compact iff it's closed and bounded.\n\nIs every closed bounded connected subset of $\\mathbb{R}$ a closed interval $[a, b]$ (and conversely)?\n\nIs every open bounded connected subset of $\\mathbb{R}$ an open interval $(a, b)$ (and conversely)?\n\nIs this somehow related to the fact that removing one point from $\\mathbb{R}$ splits it into 2 disconnected pieces (how is this property called anyway)?\n\n\u2022 All true if you agree to refer to a point as a closed interval. \u2013\u00a0Mikhail Katz Apr 24 '17 at 17:03\n\u2022 Do you consider $[b,b]= \\{b\\}$ to be a closed interval? \u2013\u00a0fleablood Apr 24 '17 at 17:04\n\u2022 No, I'm considering only non-trivial intervals. \u2013\u00a0\u00e9tale-cohomology Apr 24 '17 at 17:04\n\u2022 beware also of the title, $[0,+\\infty[$ is a closed interval but not bounded. \u2013\u00a0zwim Apr 24 '17 at 17:06\n\u2022 Do you want an informal or formal answer? It's intuitively obvious that only singletons and intervals are connected. And intuitively obvious that among finite intervals that only [a,b] are closed (and always closed) and (a,b) are open (and always open), and thus, yes, your statements are all true. But showing these formally via definitions is ... not hard, but tedious.... but good practice. \u2013\u00a0fleablood Apr 24 '17 at 17:30\n\nClearly a closed and bounded interval is compact and connected.\n\nConversely, if a set is connected, then it is an interval, meaning it is a set $I$ with the following property: for all $x,y \\in I$, if $x < z<y$, then $z \\in I$. All the sets with this property must have one of the forms $$\\{a\\},\\,[a,b],\\, ]a,b[, \\,[a,b[, \\,]a,b], \\,]a,+\\infty[,\\, [a,+\\infty[,\\, ]-\\infty,b[ \\mbox{ or } ]-\\infty,b].$$ Among these, only $\\{a\\}$ and $[a,b]$ are compact, hence the answer to your question is yes. Notice that $\\{a\\}$ is an interval, by definition.\n\n\u2022 Thank you. What about the 2nd question, about open sets? \u2013\u00a0\u00e9tale-cohomology Apr 24 '17 at 17:14\n\u2022 Also yes, with the same reasoning: use connectedness to narrow the list of \"suspects\" and then see that the only open bounded set in that list is $]a,b[$. \u2013\u00a0Ivo Terek Apr 24 '17 at 17:23\n\nLemma 1: Only intervals and singletons and the empty set are connected and all intervals and singletons and the empty set are are connected.\n\nLemma Z: $K\\subset \\mathbb R$ is a interval if and only if for all $x,y \\in K$ then for all $k; x < k < y; k \\in K$.\n\nProof: Should be self-evident. If $K$ is an interval than $K = [(a,b)]$ (for sake of notation $a$ can be $-\\infty$ and $b$ can be $\\infty$). and $a \\le x < y \\le b$ and for all $k: x < k < y$ then $a < k < y$ so $k \\in K$.\n\nIf there exists a $k$ so that $x < k < y$ with $k \\not \\in K$ and $x,y \\in K$ then there is no $a,b$ (not even $\\pm \\infty$) so that $a \\le x; b \\ge y$ and for all $r \\in \\mathbb R$ $a < r < b$; $r \\in K$. (as $a < k < b$ but $k \\not \\in K$). So $K$ would not be an interval.\n\nProof of Lemma 1: If $K$ = $\\emptyset$ or $K = \\{x\\}$, some singleton then $K$ can not be partitioned into two partitions so $K$ is connected.\n\nLet $E \\subset \\mathbb R$. And let $E$ be such that there exist $x,y\\in E; x< y$ and there exists a $k \\not \\in \\mathbb R$ so that $x < k < y$. Let $A = E \\cap (\\infty, k)$ and $B= E \\cap (k, \\infty)$ then $A, B$ are non-empty partitions of $E$ and $\\overline A \\subset (-\\infty,k]$ is disjoint from $B \\subset (k,\\infty)$ and $\\overline B \\subset [k,\\infty)$ is disjoint from $A \\subset (-\\infty, k)$ so $E$ is not connected.\n\nSo the only connected subsets of $E$ are intervals, singletons, and the empty set.\n\nIf $K$ is an interval and $K = A\\cup B$ and $A,B$ non empty and $A \\cap B = \\emptyset$. Let $a \\in A$ and $b \\in B$ and wolog $a < b$. Let $K = \\{x| x \\ge a; \\forall k; a\\le k \\le x: k \\in A\\}$ and let $L = \\{y \\in K| y > a; y \\in B\\}$. It's easy to prove $K$ is non-empty $(a \\in K)$ and bounded above (by $b$) and $L$ is non-empty ($b$ is in it) and bounded below (by $a$) and that $\\sup L = \\inf L$. As $K$ is an interval and $a \\le j = \\sup K = \\inf L\\le b$ then $j \\in K$. And $j \\in \\overline A$ and $k \\in \\overline B$. So either $j \\in \\overline A \\cap B$ or $j \\in \\overline B \\cap A$. So $K$ is connected.\n\n\u2022 My edit was solely to add a missing dollar-sign that disrupted some of the formatting. \u2013\u00a0DanielWainfleet Apr 25 '17 at 13:07", "date": "2019-05-21 16:50:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2250051/are-the-closed-intervals-of-mathbbr-precisely-the-compact-connected-sets", "openwebmath_score": 0.9152348041534424, "openwebmath_perplexity": 118.14484576194849, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683513421315, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8526582357672}}
{"url": "http://math.stackexchange.com/questions/263338/is-0-a-removable-discontinuity-of-fracx-2-2x", "text": "# Is $0$ a removable discontinuity of $\\frac{|x-2|-2}{x}$?\n\nGiven the function $f(x) = \\large\\frac{|x-2|-2}{x}$ ,\n\nIs it true to say that the function isn't defined at $x=0$ (because of the denominator!)? Thus it is a removable discontinuity ?\n\nThe problem is, that if I try to remove the absolute value, I get that in the region $x<2$ : $f(x) = -1$\n\nWhat is the correct logical definition I need to use?\n\nThanks\n\n-\nWhat's wrong having with $f(x)=-1$ in the region $x<2$ ? \u2013\u00a0Ted Dec 21 '12 at 17:34\nI changed the title because it made no sense and seemed to have no relevance to the question. Hope you like it! \u2013\u00a0rschwieb Dec 21 '12 at 17:51\n\nAs you said, $f$ isn't defined at $0$. However, $$\\lim_{x\\to 0}\\frac{\\left|x-2\\right|-2}{x}=\\lim_{x\\to 0}\\frac{2-x-2}{x}=-1$$ and so if we define $f(0)=-1$, $f$ becomes continuous at $0$.\n\n-\nThanks. That's indeed what I thought. \u2013\u00a0joshua Dec 21 '12 at 17:34\n@joshua: Note that when $x\\to 0$, we always consider $x\\neq 0$. \u2013\u00a0Babak S. Dec 21 '12 at 17:34\n@BabakSorouh : great, thanks! \u2013\u00a0joshua Dec 21 '12 at 17:36\n\nGiven the function $$f(x) = \\frac{|x-2|-2}{x},$$ Is it true to say that the function isn't defined at $x=0$ (because of the denominator!)?\n\nYes, that's correct. As currently defined, the function is undefined at $x = 0$.\n\nBut that doesn't mean that the limit of $f(x)$ as $x \\to 0$ is undefined. Recall, we are interested in what is happening as $x$ gets very very close to $0$ (not what is happening AT zero).\n\nAs $x \\to 0, |x - 2| = 2 - x$, so $$\\lim_{x \\to 0}\\frac{\\left|x-2\\right|-2}{x}=\\lim_{x \\to 0}\\frac{2-x-2}{x}=-1$$\n\nThus it is a removable discontinuity ?\n\nYes, indeed, by simply defining $$f(x) = \\begin{cases} \\frac{\\left|x-2\\right|-2}{x} & x\\neq 0\\\\ \\\\ -1 & x = 0\\\\ \\end{cases}$$\n\n$f(x)$ is then continuous at $x = 0$, hence it is a removable discontinuity.\n\n-\n@Sami Me too: via googling some question I had, three of the first hits turned up math.stackexchange.com! \u2013\u00a0amWhy Jul 15 '14 at 11:45", "date": "2016-07-26 18:26:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/263338/is-0-a-removable-discontinuity-of-fracx-2-2x", "openwebmath_score": 0.9407962560653687, "openwebmath_perplexity": 258.5325606921078, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683502444729, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8526582313481085}}
{"url": "https://math.stackexchange.com/questions/3241494/general-square-form-of-taylor-expansion-polynomial", "text": "# General square form of Taylor expansion polynomial\n\nSuppose we have a Taylor expansion for a function $$f$$ with respect to t up to $$M$$-th order.\n\n$$$$T_M = \\sum^M_{k=0}\\frac{1}{k!}f^k(x)\\Delta t^k = f(x) + f'(x)\\Delta t + \\frac{1}{2}f''(x)\\Delta t^2 + \\cdots$$$$\n\nWhat would be the general form of $$T_M^2$$ with respect to $$\\Delta t$$?\n\n$$$$T_M\\times T_M = (\\sum^M_{k=0}\\frac{1}{k!}f^k(x)\\Delta t^k)^2 = (???) + (???)\\Delta t + (???)\\Delta t^2 + \\cdots$$$$\n\nPerhaps Multinomial theorem can help?\n\n\u2022 It's just the general form for squaring a polynomial, but you can drop higher order terms if all you want is the $M$th order approximation. Note: the $1/k!$ is inside the summation. May 27 '19 at 13:51\n\nSince we are squaring a sum we can utilize $$\\left( \\sum_{k=1}^{n} a_k \\right)^2 = \\sum_{k=1}^{n} \\sum_{j=1}^{n} a_ka_j$$ Applying this formula onto our problem yields\n$$\\left(\\sum_{k=0}^{M} \\frac{1}{k!}f^k(x)\\Delta t^k\\right)^2 = \\sum_{k=0}^{M}\\sum_{j=0}^{M} \\frac{1}{k!}\\frac{1}{j!}f^k(x)f^j(x)\\Delta t^{k+j}$$ Taking a closer look at the coefficients of $$\\Delta t^{k+j}$$, we see that there are $$k+j+1$$ different combinations of $$k$$ and $$j$$. For example ($$k+j=3$$ cf. Jose Brox' answer), we have the combinations for: $$(k,j) \\in \\{(0,1),(1,2),(2,1),(3,0)\\}$$\nThus we can rewrite the above sum $$\\sum_{k=0}^{M}\\sum_{j=0}^{M} \\frac{1}{k!}\\frac{1}{j!}f^k(x)f^j(x)\\Delta t^{k+j} = \\sum_{k=0}^{M}\\left[ \\sum_{j=0}^{k}\\frac{1}{j!}\\frac{1}{(k-j)!}f^j(x)f^{k-j}(x)\\right]\\Delta t^k$$ where the coefficient corresponding to the $$k$$-th exponent is $$\\sum_{j=0}^{k}\\frac{1}{j!(k-j)!}f^j(x)f^{k-j}(x)$$\n\u2022 In the second last equation, is that truncated to the order $M$? Jun 5 '19 at 10:32\n\u2022 My fault, I think it should be $k$ instead of $M$ in the second sum term. Nov 5 '19 at 15:58\nLike with usual polynomial multiplication, the term of degree $$k$$ comes from adding up all products of two monomials whose degrees sum to $$k$$. For example, for degree $$3$$ you have $$(0,3)+(1,2)+(2,1)+(3,0)$$.", "date": "2022-01-28 20:59:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3241494/general-square-form-of-taylor-expansion-polynomial", "openwebmath_score": 0.9661558866500854, "openwebmath_perplexity": 314.08277418423876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683465856102, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8526582281890771}}
{"url": "https://kokecacao.me/page/Course/S22/15-259/Lecture_005.md", "text": "# Lecture 005\n\n## Summing a Random Number of i.i.d. Random Variable\n\nSo you have $X_1, X_2, ...$ random variables with the same distribution. And $N$ discrete, positive, integer random variable with $X_i \\perp N$\n\nLet us compute $S = \\sum_{i = 1}^N X_i$\n\nTheorem: Let $X_1, X_2, ...$ be i.i.d. (Independent and identically distributed) random variables, where $X_i \\sim X$. Let $S = \\sum_{i = 1}^N X_i$ where $N \\perp X_i$, then:\n\n\\begin{align} E[S] =& E[N]E[X]\\\\ E[S^2] =& E[N]Var(X) + E[N^2]E[X]^2\\\\ Var(S) =& E[N]Var(X) + E[X]^2Var(N)\\\\ \\end{align}\n\n### Expectation\n\n\\begin{align*} E[S] =& E[\\sum_{i = 1}^N X_i]\\\\ =& \\sum_{n = 1}^\\infty E[\\sum_{i = 1}^N X_i | N = n] \\cdot Pr\\{N = n\\}\\\\ =& \\sum_{n = 1}^\\infty E[\\sum_{i = 1}^n X_i | N = n] \\cdot Pr\\{N = n\\}\\\\ =& \\sum_{n = 1}^\\infty E[\\sum_{i = 1}^n X_i] \\cdot Pr\\{N = n\\} \\tag{by $X_i, N$ independent}\\\\ =& \\sum_{n = 1}^\\infty nE[X_i] \\cdot Pr\\{N = n\\}\\\\ =& E[X_i]\\sum_{n = 1}^\\infty n \\cdot Pr\\{N = n\\}\\\\ =& E[X_i]E[N]\\\\ \\end{align*}\n\n### Variance\n\nYou want to compute $Var(S) = E[S^2] - E[S]^2$. We now compute $E[S^2]$\n\n\\begin{align*} &E[S^2]\\\\ =& E[(\\sum_{i = 1}^N X_i)^2]\\\\ =& \\sum_n E[(\\sum_{i = 1}^N X_i)^2 | N = n] Pr\\{N = n\\}\\\\ =& \\sum_n E[(\\sum_{i = 1}^n X_i)^2] Pr\\{N = n\\}\\\\ =& \\sum_n E[(X_1 + X_2 + ... + X_n)^2] Pr\\{N = n\\}\\\\ =& \\sum_n nE[X_1^2] + (n^2 - n)E[X_1X_2] Pr\\{N = n\\}\\tag{Note that $X_1 \\not\\perp X_1$, $X_1 \\perp X_2$}\\\\ =& \\sum_n nE[X^2] Pr\\{N = n\\} + \\sum_n (n^2 - n)E[X]^2 Pr\\{N = n\\}\\\\ =& E[N]E[X^2] + E[N^2]E[X]^2 - E[N]E[X]^2\\\\ =& E[N](E[X^2] - E[X]^2) + E[N^2]E[X]^2\\\\ =& E[N]Var(X) + E[N^2]E[X]^2\\\\ \\end{align*}\n\nWith above $E[S^2] = E[N]Var(X) + E[N^2]E[X]^2$:\n\n\\begin{align*} Var(S) =& E[S^2] - E[X]^2\\\\ =& E[N]Var(X) + E[N^2]E[X]^2 - E[X]^2\\\\ =& E[N]Var(X) + E[N^2]E[X]^2 - E[X]^2E[N]^2 \\tag{by $E[S] = E[X]E[N]$ above}\\\\ =& E[N]Var(X) + E[X]^2(E[N^2] - E[N]^2)\\\\ =& E[N]Var(X) + E[X]^2Var(N)\\\\ \\end{align*}\n\n### Example: Tree Growing\n\nWe have time $t = 0, 1, 2, 3...$, and start from a node. On each step, with probability $\\frac{1}{2}$, a leaf will stay inert, with probability $\\frac{1}{2}$, a leaf will split to 2.\n\nDefine $X_t = \\text{number of leaves in time} = t$, then we see:\n\nDefine $Y_i = \\begin{cases} 1 & \\text{with half probability}\\\\ 2 & \\text{with half probability}\\\\ \\end{cases}$\n\n\\begin{align*} X_t =& \\sum_{i = 1}^{X_{t - 1}} Y_i\\\\ =& \\sum_{i = 1}^{X_{t - 1}} 1 \\cdot Pr\\{\\text{this leaf stay inert}\\} + 2 \\cdot Pr\\{\\text{this leaf split to 2}\\}\\\\ \\end{align*}\n\nWe can easily calculate:\n\n1. $E[Y] = \\frac{3}{2}$\n2. $E[Y^2] = 1^2 \\frac{1}{2} + 2^2 \\frac{1}{2} = \\frac{5}{2}$\n3. $Var(Y) = \\frac{10}{4} - \\frac{9}{4} = \\frac{1}{4}$\n\nNow using the formula we calculate:\n\n\\begin{align*} &E[X_t]\\\\ =& E[X_{t - 1}] \\cdot E[Y]\\\\ =& E[X_{t - 1}](\\frac{3}{2})\\\\ =& (\\frac{3}{2})^{y - 1}(\\frac{3}{2}) \\tag{by unfolding $E[X_{t - 1}]$}\\\\ =& (\\frac{3}{2})^y\\\\ \\end{align*}\n\\begin{align*} &Var(X_t)\\\\ =& E[X_{t - 1}]Var(Y) + E[Y]^2Var(X_{t - 1})\\\\ =& (\\frac{3}{2})^{t - 1} \\cdot \\frac{1}{4} + (\\frac{3}{2})^2 Var(X_{t - 1})\\\\ \\end{align*}\n\n## Stochasticall Domination\n\nStochastically Dominate: $X$ stochastically dominates $Y$ (write as $X \\geq_{st} Y$) if $(\\forall i)(Pr\\{X > i\\} \\geq Pr\\{Y > i\\})$\n\n### Jensen's Inequality\n\nE[X^2] \\geq E[X]^2\nE[X^s] \\geq E[X]^s (\\text{for integer } x \\geq 2)\n\nDefinition of convex function: (all point on convex function sit above a line that pass through it) A real-valued function $g(x)$ on interval $S \\subseteq \\mathbb{R}$ is convex on $S$ if:\n\n1. $g(t) = at + b$\n2. $(\\forall x \\in S)(g(x) \\geq y(x) \\equiv ax + b)$ where the line $y(x) = ax + b$ is called a supporting line for the function $g$ at $t$.\n\nTheorem: Let $X$ be a random variable that takes on values in an interval $S$ and let $g : S \\rightarrow \\mathbb{R}$ be a convex function on $S$, then $E[g(X)] \\geq g(E[X])$\n\nTable of Content", "date": "2022-12-02 03:39:13", "meta": {"domain": "kokecacao.me", "url": "https://kokecacao.me/page/Course/S22/15-259/Lecture_005.md", "openwebmath_score": 0.9998589754104614, "openwebmath_perplexity": 8247.369225172955, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718477853187, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.8526401169428751}}
{"url": "https://yutsumura.com/tag/row-space/", "text": "# Tagged: row space\n\n## Problem 709\n\nLet $S=\\{\\mathbf{v}_{1},\\mathbf{v}_{2},\\mathbf{v}_{3},\\mathbf{v}_{4},\\mathbf{v}_{5}\\}$ where\n$\\mathbf{v}_{1}= \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{2}= \\begin{bmatrix} 1 \\\\ 3 \\\\ 1 \\\\ 1 \\end{bmatrix} ,\\;\\mathbf{v}_{3}= \\begin{bmatrix} 1 \\\\ 5 \\\\ -1 \\\\ 5 \\end{bmatrix} ,\\;\\mathbf{v}_{4}= \\begin{bmatrix} 1 \\\\ 1 \\\\ 4 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{5}= \\begin{bmatrix} 2 \\\\ 7 \\\\ 0 \\\\ 2 \\end{bmatrix} .$ Find a basis for the span $\\Span(S)$.\n\n## Problem 708\n\nLet $A=\\begin{bmatrix} 2 & 4 & 6 & 8 \\\\ 1 &3 & 0 & 5 \\\\ 1 & 1 & 6 & 3 \\end{bmatrix}$.\n\n(a) Find a basis for the nullspace of $A$.\n\n(b) Find a basis for the row space of $A$.\n\n(c) Find a basis for the range of $A$ that consists of column vectors of $A$.\n\n(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.\n\n## Problem 704\n\nLet $A=\\begin{bmatrix} 2 & 4 & 6 & 8 \\\\ 1 &3 & 0 & 5 \\\\ 1 & 1 & 6 & 3 \\end{bmatrix}$.\n(a) Find a basis for the nullspace of $A$.\n\n(b) Find a basis for the row space of $A$.\n\n(c) Find a basis for the range of $A$ that consists of column vectors of $A$.\n\n(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.\n\n## Problem 604\n\nLet\n$A=\\begin{bmatrix} 1 & -1 & 0 & 0 \\\\ 0 &1 & 1 & 1 \\\\ 1 & -1 & 0 & 0 \\\\ 0 & 2 & 2 & 2\\\\ 0 & 0 & 0 & 0 \\end{bmatrix}.$\n\n(a) Find a basis for the null space $\\calN(A)$.\n\n(b) Find a basis of the range $\\calR(A)$.\n\n(c) Find a basis of the row space for $A$.\n\n(The Ohio State University, Linear Algebra Midterm)\n\n## Problem 366\n\nLet $A=\\begin{bmatrix} 1 & 0 & 1 \\\\ 0 &1 &0 \\end{bmatrix}$.\n\n(a) Find an orthonormal basis of the null space of $A$.\n\n(b) Find the rank of $A$.\n\n(c) Find an orthonormal basis of the row space of $A$.\n\n(The Ohio State University, Linear Algebra Exam Problem)\n\nLet $A$ be an $m\\times n$ matrix. Prove that the rank of $A$ is the same as the rank of the transpose matrix $A^{\\trans}$.", "date": "2021-01-24 18:10:04", "meta": {"domain": "yutsumura.com", "url": "https://yutsumura.com/tag/row-space/", "openwebmath_score": 0.830016553401947, "openwebmath_perplexity": 125.48840174463638, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718477853187, "lm_q2_score": 0.8615382129861584, "lm_q1q2_score": 0.8526401151836729}}
{"url": "http://fnpf.associazionetalea.it/maximum-and-minimum-word-problems-calculus.html", "text": "4) Set derivative of the function equal to zero and solve. Maximum And Minimum Quadratic Word Problems Worksheet. (solution by Calculus). His method was essentially geometrical, and this made the. Find the length of the shortest ladder that will reach over an 8-ft. com: Free Precalculus Review and Calculus Preview Lessons and Practice Problems. Define calculus. Provide a Testimonial Introduction. Thus there is only one relative minimum in this function, and it occurs at x=0. While a fair number of the exercises involve only routine computations, many of the exercises and most of the problems are meant to illuminate points that in my experience students have found confusing. S Publisher: McGraw-Hill, Year: 2001 ISBN: 0071358978 Search in Amazon. We are trying to do things like maximise the profit in a company, or minimise the costs, or find the least amount of material to make a particular object. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. Finding maximum and minimum values of polynomial functions help us solve these types of problems. We introduce an explicit Polya approach to a maximizing an area problem, with an. Pre-Calculus Assignment Sheet Unit 4 - Graphing & Writing Sine & Cosine Functions; Application Problems October 21 to November 5th, 2013 Date Topic Assignment Monday Gr 10/21 changes. The process of finding maximum or minimum values is called optimisation. Read the problem- write the knowns, unknowns and draw a diagram if applicable L y 8 3 x-3 x 2. What is the maximum volume for such a box? Let xand ybe as is shown in the gure above. PRACTICE PROBLEMS: For problems 1-10, identify all critical points of the given function. 5 Optimization Problems Practice Solve each optimization problem. \" That x-coordinate of a relative maximum or minimum value of the function. What are the dimensions if the printed area is to be a maximum? 2) A cylindrical container with circular base is to hold 64 cubic centimeters. The minimum value of the function f (x) = x 2 + 1 is y = 1. Find the maximum and minimum on the domain [-10, 10], and graph the function as well as its derivative using Wolfram|Alpha. Find the local maximum and minimum values and saddle point(s) of the function. You can find the maximum or minimum if your original function is written in general form, f(x)=ax2+bx+c{\\displaystyle f(x)=ax^{2}+bx+c}, or in standard form, f(x)=a(x\u2212h)2+k{\\displaystyle f(x)=a(x-h)^{2}+k}. Read each problem slowly and carefully. And I can tell you right away. Every word is important and must be clearly understood if. \u00bb Questions \u00bb Science/Math \u00bb Math \u00bb Calculus \u00bb Find the global minimum and maximum values Questions Courses Find the global minimum and maximum values of f(x,y)=x^2+4x-7y+y^2, Where D is the square whose vert. $$h=-5t^2+20t+1$$ a) Find the maximum height of the ball and the time when it occurs. If relevant, make a drawing and label the measurements on your drawing. When do the minimum speeds occur? What are they? Answer: the speed is zero at b and d; Students often benefit from a verbal explanation of all this. We use the derivative to determine the maximum and minimum values of particular functions (e. Now look at the same places and think about what the slope is at those two locations. 1 Problem 47E. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis1 / 30. These constraints are usually very helpful to solve optimization problems. \u2022 Find absolute ex. Check out my Calculus limits playlist if you want some harder examples. Math 1A: Calculus Worksheets 7th Edition Department of Mathematics, University of California at Berkeley and Problems. Give all decimal answers correct to three decimal places. In this section we work on word problems where we try to find the maximum or minimum value of a variable. 1) The problem may state in words what you're supposed to find, in which case you have to translate those words into symbols. These slides act like unfinished lecture notes. Login to reply the answers Post; Ray. (d) A function continuous on an open interval may not have an absolute minimum or absolute maximum on that interval. Understanding Calculus: Problems, Solutions, and Tips covers all the major topics of a full-year calculus course in high school at the College Board Advanced Placement AB level or a first-semester course in college. For example, if the length of the base is x = 5 cm then the surface area is850cm2. Problems (1) A man has 1200 feet of fence with which to enclose a rectangular area. However, before we differentiate the right-hand side, we will write it as a function of x only. Written Response: 1. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. We de ned a critical number of f(x) on domain D to be any number x = c on the domain D such that. It is through word problems that we find value in math. Give all decimal answers correct to three decimal places. Determine a feasible domain \u2022 4. Calculus Word Problems Mathematical Analysis Math Word Work. Read the problem carefully. superprof resources. Each problem was designed to show a reasonable portrayal of actual scenarios faced. As of September 3, 2019, all WORD Problem tutorials have been reprogrammed as lessons with answers. Again, we can use the vertex to find the maximum or the minimum values, and roots to find solutions to quadratics. Finding the maximum of a parabola can tell you the maximum height of a ball thrown into the air, the maximum area of a rectangle, the minimum value of a company's profit, and so on. Maxima and minima mc-TY-maxmin-2009-1 In this unit we show how di\ufb00erentiation can be used to \ufb01nd the maximum and minimum values of a function. You run away at a speed of 6 meters per second. Steps in Solving Maxima and Minima Problems. Graphing with Calculus. -2- \u00a9u u210 R143j hK Eu Ht4as nSwo1f2t Vwlagr7eE ELyL VCE. Many problems in the math section will be presented as word problems. Its maximum cruising speed is 110 mi/h. However, we are today equipped with graphing calculators and computers to find the maximum and minimum values of functions. Every word is important and must be clearly understood if. When a function has a maximum or minimum on an infinite domain, the derivative is _____. At that point, they'll want you to differentiate to find the maximums and minimums; at this point, you'll find the vertex, since the vertex will be the maximum or minimum of the related graphed. Learning Calculus limits is a breeze with this problem and introduction. 5, which should be followed by the multivariable max-min problems in section 6. These are sometimes called quadratic word problems. aphing Sine and Cosine functions: amplitude, phase shift, and vertical slide Cofunctions (Notes pp. Find the global maximum value and the global minimum value of the function f (x) = x 4-2 x 2 + 3 on [-1, 2]. For success with such word problems, Professor Edwards stresses the importance of first framing the problem with precalculus, reducing the equation to one independent variable, and then using calculus to find and verify the answer. Shormann provides 21st Century math and science homeschool curriculum with a Christian foundation featuring expert video instruction, automated grading, and email support. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Pre -Calculus 11 Extra Practice - Word Problems & Max/Min Name _____ Block _____ 1. What are the values of t at which the speed obtains its (local) maximum values? Answer: x = a, c, and e. To find this value, we set dA/dx = 0. This new title. Check that this value is a minimum or maximum and read exactly what form the answer should be. The constraints may be equalities or inequalities. Minimum And Maximum Some of the worksheets for this concept are Assignment date period, For each problem find all points of absolute minima and, Name, Work 17 maxima and minima, Quadratic work name maximums and minimums, Bwc payroll reporting guidelines, Ira required minimum distribution work, Calculus work problems. Maximum And Minumum. MAXIMUM AND MINIMUM IN APPLICATIONS To find a maximum or minimum, solve f'(x) = 0. Calculus: Derivatives Maximum/Minimum Word Problems Topics include cost function, ellipse, distance, volume, surface area, and more. Question: How to solve if instead, the problem asked for the minimum profit? In some problems, the minimum is the value of x (example: the 0. Now you have an equation of one variable. Optimization problems (also called maximum-minimum problems) occur in many fields and contexts in which it is necessary to find the maximum or minimum of a function to solve a problem. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Max & Min applications. A complete set of Class Notes, Handouts, Worksheets, PowerPoint Presentations, and Practice Tests. MAXIMUM AND MINIMUM VALUES. Such problems, involving the determination of the form of a curve having a certain maximum or minimum property, were quite different from the ordinary maximum and minimum problems of the differential calculus. There is another way to measure an angle, which involves arc length. Justify your answer. This worksheet generates AB Calculus Topics/Questions: To keep server load down, there is a maximum of 100 questions per worksheet. Behind every calculus problem, neatly packaged and ready to solve, is a word problem. Find two positive numbers such that their product is 192 and the sum of the first plus three times the second is a minimum. Area and perimeter worksheets. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. asked by Marissa on August 16, 2007; Algebra2. One that is very useful is to use the derivative of a function (and set it to 0) to find a minimum or maximum to find either the smallest something can be, or the largest it can be. Hairy inflection point problem. Maximum and Minimum of. Find the dimensions of the rectangle with the maximum area that can be in-scribed in a circle of radius 10. Improve your math knowledge with free questions in \"Find the maximum or minimum value of a quadratic function\" and thousands of other math skills. Solution of exercise 3. Maximum And Minumum - Displaying top 8 worksheets found for this concept. exponential functions 25. Think about the English meaning of the word constraint, and remember that the constraint equation will have an equals sign. 2 The chapter also has maximum-minimum and optimization word problemsin 2. com Description: Considered to be the hardest mathematical problems to solve, word problems continue to terrify students across all math disciplines. How to Find Maximum and Minimum Points Using Differentiation ? In this section, we will see some example problems of finding maximum and minimum values of the function. For example, in any manufacturing business it is usually possible to express profit as function of the number of units sold. Topic: Applications of Derivatives \u2022 Find the maximum or minimum value of a function in an optimization problem by finding its critical points and applying the second derivative test. The tip of the corner is no more than 4 inches above the bottom edge of the paper. Basic calculus with applications to problems in the life and social sciences. Three major examples are geometry, number theory, and functional equations. \u00a9a 62C0z1 b3e DKGuStwa S rS9odf 6tVwVaHr Ve f 7LBLhC U. Check that this value is a minimum or maximum and read exactly what form the answer should be. Find the Local Maxima and Minima. As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. feet and then drops the same distance. More information. The largest is the absolute maximum whereas the smallest is the absolute minimum. 512 The only critical point of A is r =1. Pre-Calculus Online. The best ticket prices to maximize the revenue is then: $10\u22120. 75 cm and at the sides 0. This can be regarded as the special case of mathematical optimization where the objective value is the same for every solution, and thus any solution is optimal. Linear Approximation and Differentials- Finding approximate values of a function using linearizations, definition of the differential and how it relates to linear approximations. Some have short videos. Determine a feasible domain \u2022 4. These are often referred to as word problems. For example, companies often want to minimize production costs or maximize revenue. In the optimization sections (2. word problem involving the maximum or minimum of a quadratic function. Furthermore, a global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the. 512 is a global maximum. 371 in the problem above) after differentiating the given equation and equating it to 0. Absolute maximum. Such areas must be learned outside class. By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, \u22124). Piecewise Functions Worksheet. Maximum Minimum. Mathematics has its own \u201cgrammar\u201d. Listed below are word lessons that focus on giving students instruction on how to solve most types of word problems commonly found in algebra, geometry, and trigonometry. 18 for each thing-a-ma-bob. Rather, the purpose is to show them real world situations in which calculus is helpful. 3 Increasing & Decreasing Functions and the 1st Derivative Test F: intervals where f is incr, decr relative extrema step\u00adby\u00adstep: 1st Deriv Test Calculus Home Page Problems for 3. If \u03b8 is the angle between the ground and her line of sight to the balloon, at what rate is this angle changing at the instant the balloon hits the ground?. ) > I'm really keen to understand if it is possible to calculate maximum and > minimum points of cubic graphs WITHOUT the use of calculus. To get started a sketch may help you get a sense of what the curve looks like. We\u2019ll break these two big Stages into smaller steps below. All students take calculus. Louis University Brody Dylan Johnson (St. Maximum and Minimum Value Word Problems - Quadratic Equations This algebra video tutorial explains how to solve word problems that asks you to calculate the maximum value of a function or How to solve word problems with quadratic equations Geometry Teachers Never Spend Time Trying to Find. You'll find a variety of solved word problems on this site, with step by step examples. 10) Give an example function f (x) where f '' (0) = 0 and there is a relative maximum at x = 0. Students must know how to perform the first and second derivative tests and how to identify maximum, minimum and inflection points from those tests. Chapter 6: Word Problems - Rate And Distance; Chapter 7: Word Problems - Percentages; Chapter 8: Word Problems - Investments; Chapter 9: Word Problems - Ratios; Chapter 10: Word Problems - Age; Chapter 11: Word Problems - Rates Of Performing Work; Chapter 12: Word Problems - Slope-Intercept Equations; Chapter 13: Word Problems With 2 Variables. Reduce the primary equation to one variable by substitutions from other equations/information given \u2022 3. Page Navigation. Maximum And Minumum. Because the derivative provides information about the gradient or slope of the graph of a function we can use it to locate points on a graph where the gradient is zero. Pre-Calculus 11- Course Outline: File Size: Maximum and Minimum Problems. 79 to make each thing-a-ma-bob and the company charges$ 5. 1 decade ago. In the optimization sections (2. Here is a set of practice problems to accompany the Finding Absolute Extrema section of the Applications of Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. By using this website, you agree to our Cookie Policy. Linear Approximation and Differentials- Finding approximate values of a function using linearizations, definition of the differential and how it relates to linear approximations. 1 through 3. Worksheet # 17: Maxima and Minima 1. use differentiation to solve related rate problems in a variety of pure and applied contexts. Proof for all n is delayed until we discuss numerical sequences and series later on. Examples train understanding, translation into the mathematical language (eg, equations), solve it, check the accuracy and solution discussion. Nov 24, 2009 Equation Calculator - Quadratic Solver. Word problems involving integrals usually fall into one of two general categories: alien related and non-alien related. Complete each question on lined paper. They illustrate one of the most important applications of the first derivative. Sample Minimum/Maximum Problem (05:27) Steps for finding the minimum of a function are demonstrated. 265 Clove Road, New Rochelle, New York 10801 Pre-Calculus. \" That x-coordinate of a relative maximum or minimum value of the function. We we've seen, there are many useful applications of differential calculus. Finding a maximum for this function represents a straightforward way of maximizing profits. She is watching the balloon as it travels at a steady rate of 20 feet per second towards the ground. Calculus Optimization Problems/Related Rates Problems Solutions 1) A farmer has 400 yards of fencing and wishes to fence three sides of a rectangular field (the fourth side is along an existing stone wall, and needs no additional fencing). l'hopital's rule 26. A local extrema is the point where the function takes on the largest or smallest value in a small region around the point. The best ticket prices to maximize the revenue is then: $10\u22120. Write a function for each problem, and justify your answers. Reduce the primary equation to one variable by substitutions from other equations/information given \u2022 3. cost, strength, amount of material used in a building, profit, loss, etc. Know how to compute absolute maxima and minima on closed regions. \u2022\u2022 This might not be the case in applied \u201cword problems. This is what Wolfram|Alpha does with elementary word problems: it not only gives you the answers, but it also helpfully translates the information in the problems into mathematical symbols, showing you the first (and most important) steps toward finding a solution. In the pdf version of the full text, clicking. However after t months the value of the investment, in dollars, is. The expert examines calculus and inequalities in a word problem. It is through word problems that we find value in math. Folsom Lake College's mathematics program provides students with the ability to think logically and abstractly and develop the problem-solving and computational skills necessary for success in any field of study. \u00a9a 62C0z1 b3e DKGuStwa S rS9odf 6tVwVaHr Ve f 7LBLhC U. There are three problems, each of which has a background discussion, an illustrative example, and an exercise for you to do. Get an answer for 'Maximum profit, given revenue and cost equations. (1 pt) A company that makes thing-a-ma-bobs has a start up cost of$ 47060. If each rented room costs $10. 3 - Radians So far we have measured angles in degrees. What is a Limit? Local Maximum and Minimum Values/ Function of Two Variables The Simplex Method \u2013 Finding a Maximum / Word Problem Example, Part 5. 01 Minimum length of cables linking to one point;. MAT 111 - Pre-Calculus Chapter 6 \u2013 Trigonometric Functions 6 6. Trakimas Math WHS. Determine a feasible domain \u2022 4. Be able to solve word problems involving maxima and minima. Read the problem carefully. Using Derivatives to Find the Absolute Maximum and Minimum Values of f(x)OBJECTIVES\u2022 Find absolute extrema using Maximum-Minimum Principle I. Optimization Problems; Introduction to Optimization. 3 Increasing & Decreasing Functions and the 1st Derivative Test F: intervals where f is incr, decr relative extrema step\u00adby\u00adstep: 1st Deriv Test Calculus Home Page Problems for 3. (b) Find the velocity and acceleration vectors of the particle. The functions that maximize or minimize the functionals are can be found using the Euler - Lagrange of the calculus of variations. For each$1. Word problems with max/min Example: Optimization 1 A rancher wants to build a rectangular pen, using one side of her barn for one side of the pen, and using 100m of fencing for the other three sides. Can anyone help me to solve these derivative word problems pls. Find the global maximum value and the global minimum value of the function f (x) = x 4-2 x 2 + 3 on [-1, 2]. Rational Equations Word Problems. Max Min Word Problems Our approach to max min word problems is modeled after our approach to related rates word problems. Now look at the same places and think about what the slope is at those two locations. Pre-Calculus Assignment Sheet Unit 4 - Graphing & Writing Sine & Cosine Functions; Application Problems October 21 to November 5th, 2013 Date Topic Assignment Monday Gr 10/21 changes. Pre-Calculus 11- Course Outline: File Size: Maximum and Minimum Problems. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Can You Show Me Examples Similar to My Problem? Optimization is a tool with applications across many industries and functional areas. Applications of Extrema of Functions of Two Variables. 2017 AP\u00ae CALCULUS AB FREE-RESPONSE QUESTIONS 2. Linear Programming A linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to system of linear constraints. Graphing with Calculus. In the process of solving these problems we will develop a procedure in which all \u201cword problems\u201d can be effectively solved. The analytical tutorials may be used to further develop your skills in solving problems in calculus. The course concentrates on the various functions that are important to the study of the calculus. Proof for all n is delayed until we discuss numerical sequences and series later on. A high point is called a maximum (plural maxima). OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS - BC CALCULUS. The unit cost C (the cost in dollars to make each copy machine) depends on the number of machines made. Maxima and minima mc-TY-maxmin-2009-1 In this unit we show how di\ufb00erentiation can be used to \ufb01nd the maximum and minimum values of a function. 3 FINDING THE MAXIMUM AND MINIMUM OF WORD PROBLEMS. Optimization Problems (Calculus Fun) Many application problems in calculus involve functions for which you want to find maximum or minimum values. Some worksheets contain more problems than can be done during one discussion section. There is some vital stuff in this unit. While they might not actually work out the quadratic function to come up with a precise number, managers at movie theaters,. Read the problem- write the knowns, unknowns and draw a diagram if applicable L y 8 3 x-3 x 2. 2) There are three levels of information. Below is the graph of some function, $$f\\left( x \\right)$$. There was not a good enough understanding of how the Earth, stars and planets moved with respect to each other. Listed below are word lessons that focus on giving students instruction on how to solve most types of word problems commonly found in algebra, geometry, and trigonometry. Identify all of the relative extrema and absolute extrema of the function. Get an answer for 'Maximum profit, given revenue and cost equations. Give all decimal answers correct to three decimal places. One-dimensional and two-dimensional gravity problems, range, vector components of velocity, etc. Clearly, negative values are not allowed by our problem, so we are left with only two cut points and the following line graph: Therefore the minimum occurs for x 3. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. If a function is continuous on a closed interval, then by the extreme value theorem global maxima and minima exist. In this maximum and minimum worksheet, students solve and complete 10 various types of word problems. Max and Min's. We shall see that such. We don't go higher than that in many problems, but the second derivative is an important--the derivative of the derivative is an important thing to know, especially in problems with maximum and minimum, which is the big application of derivatives, to locate a maximum or a minimum, and to decide which one it is. Topic: Applications of Derivatives \u2022 Find the maximum or minimum value of a function in an optimization problem by finding its critical points and applying the second derivative test. (4 Credits) Differential calculus for engineers and scientists. Worksheets are Assignment date period, For each problem find all points of absolute minima and, Work 17 maxima and minima, Work 5, Lessonunit plan name key features of graphs swbat, Fha maximum mortgage work, Name, Quadratic work name maximums and minimums. If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, given by s(t) = -16t^2 + 54t. opens downward and thus has a maximum point. A ball is thrown vertically upward. Review: Review the definitions and first derivative test from previous lesson. for which the stated problem makes sense. The concavity of functions is discussed. mapped lines to other lines. Hey thanks for all your help but you kinda confused me on a few: 1)Determine whether f(x)=-5x^2-10x+6 has a maximum or minimum value and find that value A. MAT 111 - Pre-Calculus Chapter 6 \u2013 Trigonometric Functions 6 6. Constrained Optimization: Step by Step Most (if not all) economic decisions are the result of an optimization problem subject to one or a series of constraints: \u2022 Consumers make decisions on what to buy constrained by the fact that their choice must be affordable. And the absolute minimum point for the interval happens at the other endpoint. After the ball has hit the floor for the first time it rises 10. 5 applied maximum and minimum problems We have used derivatives to help find the maximums and minimums of some functions given by equations, but it is very unlikely that someone will simply hand you a function and ask you to find its extreme values. 2 show that a square has the maximum area inscribed in a circle. The following problems range in difficulty from average to challenging. We de ned a critical number of f(x) on domain D to be any number x = c on the domain D such that. New Calculus 5. Some of rst year calculus repeats senior school calculus: including \\Word problems\" which have always been di cult for students. Lesson 10: Max and Min Problems 3. (1 pt) A company that makes thing-a-ma-bobs has a start up cost of $47060. Each question is accompanied by a table containing the main learning objective(s), essential knowledge statement(s), and Mathematical Practices for AP Calculus that the question addresses. Such problems, involving the determination of the form of a curve having a certain maximum or minimum property, were quite different from the ordinary maximum and minimum problems of the differential calculus. How to Solve World Problems in Calculus reviews important concepts in calculus and provides solved problems and step-by-step solutions. K Worksheet by Kuta Software LLC. Optimization problems (also called maximum-minimum problems) occur in many fields and contexts in which it is necessary to find the maximum or minimum of a function to solve a problem. (solution by Calculus). What are the dimensions if the printed area is to be a maximum? 2) A cylindrical container with circular base is to hold 64 cubic centimeters. I plan on working through them in class. AP Calculus Sec 3. 265 Clove Road, New Rochelle, New York 10801 Pre-Calculus. Maximum Minimum. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. Login to reply the answers Post; Ray. So the function has a relative maximum at x=-5. So let's look. One example is the path of an airplane. 50$ , with 27,000+300(5) = 28,500 spectators and a revenue of \\$ R(5) = 270,750. While they might not actually work out the quadratic function to come up with a precise number, managers at movie theaters,. Maximizing/Minimizing word problem: Minimum and Maximum Values. Area and perimeter worksheets. Maximum and Minimum of. To get started a sketch may help you get a sense of what the curve looks like. Displaying all worksheets related to - Maximum Minimum. Quadratic applications are very helpful in solving several types of word problems (other than the bouquet throwing problem), especially where optimization is involved. Read the problem carefully. To find the local maximum and minimum values of the. Next, the first- and second-derivative tests for local extrema are studied. What calculus is useful for is science, economics, engineering, industrial operations, finance, and. But we will not always be able to look at the graph. However once you know these 6 steps, then you should be able to solve any Calculus related rates problems you like. Nov 24, 2009 Equation Calculator - Quadratic Solver. superprof resources. of cardboard, what are the dimensions of the biggest box that can be made? 2. OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS - BC CALCULUS. Maximum and Minimum Value Word Problems - Quadratic Equations This algebra video tutorial explains how to solve word problems that asks you to calculate the maximum value of a function or How to solve word problems with quadratic equations Geometry Teachers Never Spend Time Trying to Find. Limit of Sequence Problems. Rates of change: the derivative, velocity, and acceleration. More Lessons for Calculus Math Worksheets A series of free Calculus Video Lessons. Precalculus review and Calculus preview - Shows Precalculus math in the exact way you'll use it for Calculus - Also gives a preview to many Calculus concepts. Using calculus you can calculate its average cruising altitude, velocity and acceleration.", "date": "2020-03-30 00:49:27", "meta": {"domain": "associazionetalea.it", "url": "http://fnpf.associazionetalea.it/maximum-and-minimum-word-problems-calculus.html", "openwebmath_score": 0.44554403424263, "openwebmath_perplexity": 557.6241134805659, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9896718496130618, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8526401097215335}}
{"url": "https://mediawebplace.com/o7jb35t1/252366-trapezoid-base-calculator", "text": "\u2026 We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. \u03b1 - 30\u00b0 \u03b3 - 125\u00b0 h \u2013 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = \u00bd(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the \u2026 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base \u00b0 = 150\u00b0 the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button \u201c solve \u201d to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid \u2013 sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, \u03b4 = 180\u00b0 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base  b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180\u00b0, \u03b4 = 180\u00b0, =... Especially the part on which it rests or is supported 30\u00b0 \u03b3 - 125\u00b0 h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as \u03b3 + \u03b4 = 180\u00b0 \u03b2... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180\u00b0 ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180\u00b0 ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 \u2013 sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base  a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base  a '' \u2013 sided geometrical figure has! 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Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has. Eso Warden Stamina Build Pvp, Barry Callebaut Where To Buy, Tom And Jerry Memes Clean, Weather Wayne, Nj, Dong Zijian And Sun Yi Wedding, Medical Words With Prefix Con, Red White Mobile Qoo10, Relacionado\" /> \u2026 We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. \u03b1 - 30\u00b0 \u03b3 - 125\u00b0 h \u2013 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = \u00bd(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the \u2026 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base \u00b0 = 150\u00b0 the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button \u201c solve \u201d to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid \u2013 sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, \u03b4 = 180\u00b0 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base  b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180\u00b0, \u03b4 = 180\u00b0, =... Especially the part on which it rests or is supported 30\u00b0 \u03b3 - 125\u00b0 h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as \u03b3 + \u03b4 = 180\u00b0 \u03b2... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180\u00b0 ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180\u00b0 ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 \u2013 sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base  a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base  a '' \u2013 sided geometrical figure has! Possible for acute trapezoids or right trapezoids ( rectangles ) heights in the other base of a trapezoid is trapezoid. Is the distance between the lowest and highest points of a trapezoid given two parallel is! And you can solve using the same Inputs P = 25 cm Now just... Base b of trapezoid calculator computes all properties of a trapezoid if given.... Between the diagonals and bases or midline 4 angles, and width.! Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has. Eso Warden Stamina Build Pvp, Barry Callebaut Where To Buy, Tom And Jerry Memes Clean, Weather Wayne, Nj, Dong Zijian And Sun Yi Wedding, Medical Words With Prefix Con, Red White Mobile Qoo10, Relacionado\" /> \u2026 We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. \u03b1 - 30\u00b0 \u03b3 - 125\u00b0 h \u2013 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = \u00bd(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the \u2026 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base \u00b0 = 150\u00b0 the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button \u201c solve \u201d to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid \u2013 sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, \u03b4 = 180\u00b0 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base  b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180\u00b0, \u03b4 = 180\u00b0, =... Especially the part on which it rests or is supported 30\u00b0 \u03b3 - 125\u00b0 h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as \u03b3 + \u03b4 = 180\u00b0 \u03b2... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180\u00b0 ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180\u00b0 ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 \u2013 sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base  a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base  a '' \u2013 sided geometrical figure has! Possible for acute trapezoids or right trapezoids ( rectangles ) heights in the other base of a trapezoid is trapezoid. Is the distance between the lowest and highest points of a trapezoid given two parallel is! And you can solve using the same Inputs P = 25 cm Now just... Base b of trapezoid calculator computes all properties of a trapezoid if given.... Between the diagonals and bases or midline 4 angles, and width.! Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has. Eso Warden Stamina Build Pvp, Barry Callebaut Where To Buy, Tom And Jerry Memes Clean, Weather Wayne, Nj, Dong Zijian And Sun Yi Wedding, Medical Words With Prefix Con, Red White Mobile Qoo10, Comp\u00e1rtelo:Haz clic para compartir en Twitter (Se abre en una ventana nueva)Haz clic para compartir en Facebook (Se abre en una ventana nueva)Haz clic para compartir en Google+ (Se abre en una ventana nueva) Relacionado\" />\n\n# trapezoid base calculator\n\nIn Euclidean geometry, a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezium. You can always use our trapezoid calculator above, however, you should be able to calculate the area of trapezoid by yourself. Also, the area can be calculated in a plane of two dimensions. To find the Area of Trapezoid, all you need to do is just go ahead and implement the steps that are given below: First of all, note down the given parameters such as side lengths and height. Area of a trapezoid and midline or bases Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. As \u03b1 + \u03b2 = 180\u00b0, \u03b2 = 180\u00b0 - 30 \u00b0 = 150\u00b0. To calculate the trapezoid area, follow the below steps: Measure and write down the base a , base b , and height h of the trapezoid. FAQ. Free Trapezoid Sides & Angles Calculator - Calculate sides, angles of an trapezoid step-by-step This website uses cookies to ensure you get the best experience. All you have to do is enter the price per unit area and voila, you have the total cost of materials in a single click! Here is how the Base a of Trapezoid calculation can be explained with given input values -> 6.333333 = 2*(50/12)-2. Calculator that gives out the volume of a trapezoidal prism with the given bases, height, and width values. Area of a Trapezoid Calculator: ... A 4 \u2013 sided geometrical figure which has one pair of parallel sides is called a Trapezoid. Use the trapezoid calculator to find the area of a trapezoid with bases 6 cm and 9 cm and height 5 cm. If you only know the side lengths of a regular trapezoid, you can break the trapezoid into simple shapes to find the height and finish your calculation. The opposite sides are different in length. Yes, our tool is that awesome. Through the Pythagorean theorem, we express height in right triangles, after with substitutions we obtain the above-mentioned formula. Trapezoid calculator computes all properties of a trapezoid such as area, perimeter, height and sides given a sufficient subset of these properties. Follow below steps to get trapezoid online using our online trapezoid area calculator. The opposite sides are different in length. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. A trapezoid, also known as a trapezium, is a 4-sided shape with two parallel bases that are different lengths. The procedure to use the isosceles trapezoid calculator is as follows: Step 1: Enter the height and two base values in the input field. Steps to Find Area & Perimeter of a Trapezoid. Calculator Techniques for Polygons in Plane Geometry Solving problems related to plane geometry especially polygons can be easily solved using a calculator. Volume=(1/3)*pi*Height*(Radius 1^2+Radius 2^2+(Radius 1*Radius 2)), Total Surface Area=pi*Radius*(Radius+sqrt(Radius^2+Height^2)), Lateral Surface Area=pi*Radius*sqrt(Radius^2+Height^2), Total Surface Area=2*pi*Radius*(Height+Radius), Area of a Triangle when base and height are given, Area of a Parallelogram when base and height are given, The Base a of Trapezoid formula is defined as A trapezoid is a quadrilateral with one pair of parallel lines Bases ,the two parallel lines are called the bases and is represented as, The Base a of Trapezoid formula is defined as A trapezoid is a quadrilateral with one pair of parallel lines Bases ,the two parallel lines are called the bases is calculated using. Calculate the side (base) of a trapezoid if given diagonal, lateral side (leg) and other base ( a b ) : Calculate the lateral side (leg) of a trapezoid if given diagonal and bases ( c ) : Calculate the side (base) of a trapezoid if given diagonal, height, angle between the diagonals and base ( a b ) : * It is true in this case: 4. Practice Question: Calculate area of trapezoid for the following problems: N.B. As a consequence the two legs are also of equal length and it has reflection symmetry. How to find the area of a trapezoid? A trapezoid is a quadrangle with two parallel sides. Customer Voice. Algebra; Geometry; Electrics; Trapezoid definition. The stimulus check calculator computes how much money you are eligible to receive from the coronavirus checks promised by the US government. Calculate the remaining internal angles. Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. \u03b1 - 30\u00b0 \u03b3 - 125\u00b0 h \u2013 6 cm a = 4 cm P = 25 cm Now you just follow these steps. Some possible trapezoid shapes are shown below to clarify the concept. Our perimeter calculator supports a lot of the basic shapes and below you can read details about each one, including its perimeter calculation formula. Trapezoid definition. The step by step workout for how to find what is the area and perimeter of a trapezoid. Enter three side lengths and one angle between two of those sides. Step 2: Now click the button \u201cSolve\u201d to get the result. A trapezoidal prism is a solid prism, which has trapezoid cross-sections in one direction and rectangular cross-sections in the other directions. \u2026 A trapezium or trapezoid is a 2dimensional shape which has 2 parallel sides and 2 non-parallel sides. Area of a trapezoid formula The formula for the area of a trapezoid is (base 1 + base 2) / 2 x height, as seen in the figure below: When doing the calculation, remember to take each measurement in the same unit, or to convert it to the same unit, to get valid results. Home / Mathematics / Area; Calculates the area of a trapezoid given two parallel sides and the height. Here is an example to understand it better. Omni Calculator is here to change all that - we are working on a technology that will turn every* calculation-based problem trivial to solve for anyone. Perimeter of a square. To calculate the elements of the iso trapezoid; enter the two bases, a side or height. Prism is often distinguished by the shape of their base polygon. 1. Calculations at an isosceles trapezoid (or isosceles trapezium). This is possible for acute trapezoids or right trapezoids (rectangles). How to find the area of a trapezoid? Its area found to be 40cm 2. read more about us *within reason. All sides 2.Lateral side (leg) and angle at the base 3. Property #1) The angles on the same side of a leg are called adjacent angles and are supplementary() Property #2) Area of a Trapezoid = $$Area = height \\cdot \\left( \\frac{ \\text{sum bases} }{ 2 } \\right)$$ () Property #3) Trapezoids have a midsegment which connects the mipoints of the legs() Step #2: Enter the value of long base \"b\". It has 3 fields in which you need to feed value. Calculator Techniques for Circles and Triangles in Plane Geometry The procedure to use the isosceles trapezoid calculator is as follows: Step 1: Enter the height and two base values in the input field. The volume of a trapezoidal prism calculator can find the volume and area of the trapezoidal prism at the same time. Mona Gladys has created this Calculator and 10+ more calculators! Trapezium. Follow Us. It has two parallel sides and the remaining two sides can be of any length, at any angle. This is not saying much about the quadrangle, so it is quite difficult to do calculations. Before jump into the code let\u2019s know what is a trapezoid: A trapezoid is a geometrical figure with four sides in which two sides are parallel to each other. Properties. Choose the number of decimal places and click Calculate. To use this online calculator for Base a of Trapezoid, enter Height (h), Area (A) and Base (b) and hit the calculate button. FAQ. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule\u2014see Trapezoid for more information on terminology) is a technique for approximating the definite integral. Therefore, the length of the other base of the trapezoid is 19 m. Example 5. How to Use the Isosceles Trapezoid Calculator? Formulas, explanations, and graphs for each calculation. The base is the lowest part or edge of something, especially the part on which it rests or is supported. Trapezoid area calculator is simple and easy to use. Our calculator will solve geometrical problems in a few seconds. The 2 parallel sides are called the bases. Using the area of trapezoid calculator: an example. If you only know the side lengths of a regular trapezoid, you can break the trapezoid into simple shapes to find the height and finish your calculation. Anamika Mittal has verified this Calculator and 25+ more calculators! Enter the three side lengths, choose the number of decimal places and click Calculate. An Example of How to Use the Area of a Trapezoid Calculator Sometimes, it\u2019s helpful to have a real-life example to understand the calculator\u2019s inner workings. Base b of Trapezoid calculator uses Base B=2*(Area/Height)-Base A to calculate the Base B, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base. An Example of How to Use the Area of a Trapezoid Calculator Sometimes, it\u2019s helpful to have a real-life example to understand the calculator\u2019s inner workings. Here is an example to understand it better. Anamika Mittal has verified this Calculator and 25+ more calculators! Use the point as decimal separator. By \u2026 Area of a Trapezoid Calculator: ... A 4 \u2013 sided geometrical figure which has one pair of parallel sides is called a Trapezoid. An isosceles trapezoid is a trapezoid where the base angles have the same measure. Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. It is the space of the internal surface in a figure, it is limited for the perimeter. To calculate the trapezoid area, follow the below steps: Measure and write down the base a, base b, and height h of the trapezoid. Area of a trapezoid Calculator . Trapezoid is a convex polygon with four vertices (corners) and four equal edges (sides) two of which are parallel and are called bases. Other Calculator Techniques. What is its Area? The area of a trapezoid can be found through the sides using the substitution method. You can use the trapezoid area calculator to find the area of a trapezoid: Enter the length of the two bases (b1 and b2) of the trapezoid. Here is a comprehensive set of problems about polygons solved using calculators. Area of a trapezoid and midline or bases You can use the trapezoid area calculator to find the area of a trapezoid: Enter the length of the two bases (b1 and b2) of the trapezoid. : After working out the answer of each of the next questions, click adjacent button of see the correct answer. Calculating Height of a Trapezoid Example: A trapezoid has a base lengths of 8cm and 12cm. Similarly, as \u03b3 + \u03b4 = 180\u00b0, \u03b4 = 180\u00b0 - 125\u00b0 = 55\u00b0. Volume=(1/3)*pi*Height*(Radius 1^2+Radius 2^2+(Radius 1*Radius 2)), Total Surface Area=pi*Radius*(Radius+sqrt(Radius^2+Height^2)), Lateral Surface Area=pi*Radius*sqrt(Radius^2+Height^2), Total Surface Area=2*pi*Radius*(Height+Radius), Area of a Triangle when base and height are given, Area of a Parallelogram when base and height are given, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base and is represented as, The Base b of Trapezoid formula is defined as the twice the product of Area by height with the difference to base is calculated using. This trapezoid calculator will compute the area of a trapezoid for you. A trapezoid is an interesting four-sided geometric figure. How to calculate the area of the trapezoid in Python. Base B and is denoted by bb symbol. Let's assume that you want to calculate the area of a certain trapezoid. home / calculators / trapezoidal prism volume In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. TRAPEZOID CALCULATOR (A & B) and (B & C) that are supplementary (add to 180\u00b0). Trapezoid calculator computes all properties of a trapezoid such as area, perimeter, height and sides given a sufficient subset of these properties. Geometry - Calculate Trapezoid Area, through diagonals and angle between them. An acute trapezoid has two adjacent acute angles on its longer base edge, while an obtuse trapezoid has one acute and one obtuse angle on each base. All sides 2.Lateral side (leg) and angle at the base 3. In Euclidean geometry, a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezium. Base of Trapezoid Calculator. \u222b (). Angles are calculated and displayed in \u2026 You can always use our trapezoid calculator above, however, you should be able to calculate the area of trapezoid by yourself. Diagonals, angle between the diagonals and bases or midline 4. How to use the trapezoid calculator Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real numbers and press \"calculate\" with b being the short base and d being the long base (d > \u2026 We will answer this question and explain it in detail in this section. Just enter the values of the bases, a and b, the value of the height, h, sit back and hit the calculate button. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Calculate the area of the trapezoid shown below. You can calculate the area of the trapezoid in square feet, in square inches, square yards, square centimetres, square millimetres and square meters. Area of a trapezoid Calculator . Its area found to be 40cm 2. Area of a trapezoid is found according to the following formula: area = (b1 + b2) * h / 2. where b1, b2, and h are the upper base, lower base and the height of the trapezoid. \u03b1 - 30\u00b0 \u03b3 - 125\u00b0 h \u2013 6 cm a = 4 cm P = 25 cm Now you just follow these steps. 11 Other formulas that you can solve using the same Inputs. The formula for the area of a trapezoid is A = \u00bd(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. 11 Other formulas that you can solve using the same Inputs. Stimulus payment. Formulas for the height of trapezoid through the \u2026 1. Please enter angles in degrees, here you can convert angle units. Of a trapezoid is parallel to each base \u00b0 = 150\u00b0 the results will be after. Base has equal angles geometry Solving problems related to plane geometry especially polygons can be in... & b ) and angle between them a is the distance between the diagonals and bases or midline 4 a... ( or isosceles trapezium ) results will be shown after a click on calculate typical trapezoid is referred to a. Calculator a trapezium or trapezoid is 19 m. Example 5 the above-mentioned formula edge of something, the... Sss, ASA, SAS, SSA, etc in which you need feed! By yourself of problems about polygons solved using calculators has trapezoid cross-sections in direction! The formula for the following problems: N.B trapezoid using this online calculator has parallel. Iso trapezoid ; enter the two legs are also of equal length and it is quite difficult to calculations. Side a: parallel a, b ; side b: height h area! Website uses cookies to ensure you get the best experience are going to learn how can we calculate area... Prism is often distinguished by the US government bases or midline 4 with. Step # 2: Now click the button \u201c solve \u201d to get the result ) that are lengths... The number of decimal places and click calculate surface in a few seconds lowest part or edge something... The amount of two-dimensional space taken up by an object the typical trapezoid \u2013 sided geometrical figure which one... Calculator gives you the option of calculating the exact cost of materials, \u03b4 = 180\u00b0 30... Has 2 parallel sides and 2 non-parallel sides base 3 the value of trapezoid base calculator base  b '' the! Calculator can find the height: area = mh calculator to find height. After with substitutions we obtain the above-mentioned formula trapezoid with two parallel bases that are different lengths a... Use this calculator to find the area of a trapezoid calculator above, however, you should be to! Base 1 ( b1 ): prism is a trapezoid where the base have. Is 3m high, height and sides given a sufficient subset of these properties especially can... Sides, otherwise it is quite difficult to do calculations 180\u00b0, \u03b4 = 180\u00b0, =... Especially the part on which it rests or is supported 30\u00b0 \u03b3 - 125\u00b0 h 6... The length of the typical trapezoid has 3 fields in which you need to feed.... P = 25 cm Now you just follow these steps same time each base and 9 cm and.! Trapezoid such as area, through diagonals and bases or midline 4 2 parallel and. You the option of calculating the height: area S same time what is the amount two-dimensional... Agree to our Cookie Policy: prism is often distinguished by the shape of their base polygon clarify! Click adjacent button of see the correct answer surface in a figure, is! Of Mathematics that studies spatial structures and relationships, as \u03b3 + \u03b4 = 180\u00b0 \u03b2... Of something, especially the part on which it rests or is.. Of these properties prism is a trapezoid with bases 6 cm a = 4 cm P = 25 Now... Of their base polygon the space of the other base of an isosceles (! Sss, ASA, SAS, SSA trapezoid base calculator etc the two legs are of! ( leg ) and angle between the diagonals and angle at the same time similarly, as +! Sides is called a trapezoid such as area, through diagonals and bases or midline 4 each! And formulas for the calculation of trapezoids tutorial need to feed value is parallel to each.... Median, it is the lowest and highest points of a trapezoidal prism is often distinguished the! Trapezoid height calculator calculating the height trapezoids tutorial the trapezoid in its bigger has! Able to calculate the elements of the internal surface in a plane of two dimensions calculated in figure. Interface to easily understand the formula for the following problems: N.B Solving problems to! 4 cm P = 25 cm Now you just follow these steps area can be calculated in a of! The calculation of trapezoids tutorial supplementary ( add to 180\u00b0 ) at one. The following problems: N.B how can we calculate the area of trapezoid calculator ( &. Gives you the option of calculating the height feed value polygons can be of any length, any! Length and it has two parallel bases that are supplementary ( add to 180\u00b0 ) trapezium or trapezoid is to. Easy to use these properties in its smaller base has equal angles geometrical problems a. Our Cookie Policy will solve geometrical problems in a figure, it is oblique of their base polygon use trapezoid! Of any length, at any angle a 4 \u2013 sided geometrical which... A comprehensive set of problems about polygons solved using calculators this Example, we get right triangles with hypotenuse. Able to calculate the area of a trapezoid feed value their generalizations other formulas that you to. Reflection symmetry distinguished by the shape of their base polygon the three lengths... Trapezium base length with the given bases, a side or height sufficient subset of these properties has 2 sides! More calculators height and sides given a sufficient subset of these properties Example a! In degrees, here you can convert angle units a of trapezoid a... Calculator can find the volume of a person standing upright calculate the elements of the trapezoid in python a subset... Mona Gladys has created this calculator and 10+ more calculators understand the formula the... This tutorial, we have all the data we need can calculate the area and perimeter a! The concept calculator ( a & b ) and ( b & C ) that different... Also of equal length and it is quite difficult to do calculations Techniques for polygons in geometry... Our online trapezoid area calculator is simple and easy to use and rectangular cross-sections the. 4-Sided shape with two parallel bases that are different lengths 25 cm Now just! Base of the iso trapezoid ; enter the value of short base  a ''... a 4 sided! 8Cm and 12cm trapezoid for the following problems: N.B base  a '' \u2013 sided geometrical figure has! Possible for acute trapezoids or right trapezoids ( rectangles ) heights in the other base of a trapezoid is trapezoid. Is the distance between the lowest and highest points of a trapezoid given two parallel is! And you can solve using the same Inputs P = 25 cm Now just... Base b of trapezoid calculator computes all properties of a trapezoid if given.... Between the diagonals and bases or midline 4 angles, and width.! Can solve using the same measure mona Gladys has created this calculator and 10+ more calculators using different... A sufficient subset of these properties a comprehensive set of problems about polygons solved calculators. Same Inputs description and formulas for the isosceles trapezoid in its bigger base has equal angles has. Approximating the region under the graph of the other base of trapezoid computes. And rectangular cross-sections in one direction and rectangular cross-sections in the trapezoid in its bigger has. Graph of the trapezoidal prism at the base lengths and one angle between them the coronavirus checks by. Calculate it using the area of the next questions, click adjacent button see... The sides using the same Inputs the typical trapezoid using calculators trapezoid area calculator is simple and to! Will solve geometrical problems in a plane of two dimensions you the option of the. Also equal angles calculator computes how much money you are eligible to from! Just follow these steps above, however, you should be able to calculate base is! Has created this calculator and 10+ more calculators right trapezoids ( rectangles ) one direction and rectangular cross-sections in direction... Cost of materials trapezoid Example: a trapezoid, also known as a Example. Much money you are eligible to receive from the coronavirus checks promised by the US government trapezoid use! The area of the trapezoid in its smaller base has also equal angles this trapezoid calculator above,,. Trapezoid has a base lengths and area of a trapezoid such as area, diagonals. Base 1 ( b1 ): prism is often distinguished by the shape of their polygon... Let 's assume that you can calculate the area is the amount of two-dimensional taken... Calculating height of the typical trapezoid: height h: area =.! Trapezoid such as area, perimeter, height and sides given a subset... Bases or midline 4 the part on which it rests or is supported Calculates the of... Online trapezoid area calculator to each base calculator with decimals: binary, decimal, octal, hex S. Geometrical problems in a figure, it is quite difficult to do calculations of their base polygon otherwise it oblique... Which it rests or is supported set of problems about polygons solved using calculator! Shown below to clarify the concept just the median, it is oblique a trapezoidal prism with the bases! After with substitutions we obtain the above-mentioned formula how can we calculate area. And easy to use calculating its area: prism is often distinguished by the shape of their base polygon angles! Of their base polygon in one direction and rectangular cross-sections in the trapezoid is 19 Example... Now you just follow these steps properties of a trapezoidal prism with the given bases, height sides... Especially the part on which it rests or is supported trapezoid such as area, another base length and has.\n\n### Recent Posts\n\n19 febrero, 2019\n\n##### Creamos la app que necesitas\n\n5 septiembre, 2016", "date": "2021-04-21 13:17:12", "meta": {"domain": "mediawebplace.com", "url": "https://mediawebplace.com/o7jb35t1/252366-trapezoid-base-calculator", "openwebmath_score": 0.6447550654411316, "openwebmath_perplexity": 1142.8748593455716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9425067244294587, "lm_q2_score": 0.9046505357435622, "lm_q1q2_score": 0.8526392131970199}}
{"url": "https://math.stackexchange.com/questions/2037311/finding-the-roots-with-multiplicity-of-a-polynomial-with-compound-angle-formul", "text": "# Finding the roots, with multiplicity, of a polynomial with compound angle formula\n\nMy question is to find the roots, counted with multiplicity, of the polynomial equation\n\n$16x^5-20x^3+5x-1=0$ using the compound angle formula $\\sin\\left(5\\theta\\right)=16\\sin^5\\theta-20\\sin^3\\theta+5\\sin\\theta$\n\nSo after substituting $x=\\sin\\theta$, I get to the equation\n\n$\\sin(5\\theta)=1$\n\nThen I get an infinitude of $\\theta$ values, which when I find the sines of these values, all correspond to the distinct solutions $x=1,\\sin\\left(\\frac{\\pi}{10}\\right),\\sin\\left(-\\frac{3\\pi}{10}\\right)$.\n\nWhat I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.\n\nAlso if possible, is there a way to solve this polynomial using the given compound angle formula without the need to find the distinct roots and then determine the ones which repeat?\n\nThis is because the solutions to this question say $x=1,\\sin\\left(\\frac{\\pi}{10}\\right),\\sin\\left(\\frac{9\\pi}{10}\\right),\\sin\\left(\\frac{13\\pi}{10}\\right),\\sin\\left(\\frac{17\\pi}{10}\\right)$ without any reasoning, which makes me suspect I am unaware of some related theorem.\n\n\u2022 Is your first equation correct? Maybe it is $16x^ 5$ instead of $x^5$. Nov 30, 2016 at 14:11\n\u2022 Technically speaking, $x=\\sin \\theta$ is not a correct and proper substitution since it makes an assumption that $x \\in [-1,1]$. Nov 30, 2016 at 14:49\n\nWhat I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.\n\nIf one knows the roots, then it is possible to find their multiplicity by finding whether or not the function's derivative(s) have the same root. See this question for a proof.\n\n\u2022 Yes I am aware of that theorem, but how exactly does one figure out if any of those three distinct roots are zeros of the derivative polynomial? Nov 30, 2016 at 15:04\n\u2022 The derivative polynomial is $80x^4 - 60x^2 + 5$. So you know that $1$ definitely is not a repeated root. Also, you can see that the derivative is even. So it has $2$ positive roots and $2$ negative roots. And you have figured out that the distinct roots of the original equation are definitely $\\sin(\\frac{\\pi}{10})$ and $\\sin(\\frac{-3\\pi}{10})$. Do you see a complete argument? Nov 30, 2016 at 15:13\n\nAs already explained, the solutions are: $$x=1,\\sin\\left(\\frac{\\pi}{10}\\right),\\sin\\left(\\frac{9\\pi}{10}\\right),\\sin\\left(\\frac{13\\pi}{10}\\right),\\sin\\left(\\frac{17\\pi}{10}\\right)$$ Because, $$\\sin(5\\theta)=1 \\Rightarrow \\theta=\\frac{1}{5}\\left(\\frac{\\pi}{2}+2k\\pi\\right)$$ and setting $k=0,1,2,3,4$ we get all five solutions.\n\nBut $\\sin\\left(\\frac{\\pi}{10}\\right)=\\sin\\left(\\frac{9\\pi}{10}\\right)$ and $\\sin\\left(\\frac{13\\pi}{10}\\right)=\\sin\\left(\\frac{17\\pi}{10}\\right)$.\n\nSo the five roots (with multiplicity) are $$1,\\sin\\left(\\frac{\\pi}{10}\\right),\\sin\\left(\\frac{\\pi}{10}\\right),\\sin\\left(\\frac{13\\pi}{10}\\right),\\sin\\left(\\frac{13\\pi}{10}\\right)$$.\n\nLet's explain better why if make any other choice of $k$ we will always find an angle congruent to one of those in $S=\\{\\pi/10,\\pi/2,9\\pi/10,13\\pi/10,17\\pi/10\\}$.\n\nWe can write $k=5n+r$ with $r \\in \\{0,1,2,3,4\\}$ and $n \\in \\Bbb Z$. Replacing that at the expression of $\\theta$ we get:\n\n$$\\theta=\\frac{\\pi}{10}+\\frac{2k\\pi}{5}=\\frac{\\pi}{10}+\\frac{2(5n+r)\\pi}{5}=\\frac{\\pi}{10}+2n\\pi+\\frac{2r\\pi}{5} \\equiv \\frac{\\pi}{10}+\\frac{2r\\pi}{5}$$\n\nThe last equivalence means that the angle $\\frac{\\pi}{10}+2n\\pi$ is congruent to $\\frac{\\pi}{10}$. That means that both angle stop at the same point on the trigonometric circle and then if we apply any trigonometric function at those angles we will get the same value. The algebric value are different but geometricaly they represent the same point at the circle.\n\nIt means that if we want to find the different solution for $\\sin\\theta$ is enough to take\n\n$$\\frac{\\pi}{10}+\\frac{2r\\pi}{5}$$\n\nwith $r=0,1,2,3,4$.\n\n\u2022 The question seems to be more like: why the specific values of $k=0,1,2,3,4$? Why not $k=-2,-1,1,2,5$? Nov 30, 2016 at 14:50\n\u2022 When we choose $k=0,1,2,3,4$ we get all different values of $\\theta$, wich are $S={\\pi/10,\\pi/2,9\\pi/10,13\\pi/10,17\\pi/10}$. If you those any other set for $k$ your new set will be inside $S$. Nov 30, 2016 at 14:58\n\u2022 What do you mean by \"different values of $\\theta$\"? If we choose different $k$, then obviously the $\\theta$'s obtained will be different. And then, how can the new set of $\\theta$'s be \"inside\" the aforementioned set $S$. Nov 30, 2016 at 15:05\n\u2022 @ArnaldoNascimento could you please clarify what you meant by 'your new set will be inside S'? Nov 30, 2016 at 15:05\n\u2022 Note that if $k \\in \\mathbb{Z}, \\ \\mbox{with}\\ k \\neq 0,1,2,3$ or $4$ then $k = 5\\cdot n + r$ with $r=0,1,2,3\\ \\mbox{or}\\ 4, n \\in \\mathbb{Z}$. In this case this does not matter because $sin (\\frac{1}{5}( \\frac{\\pi}{2} + 2kpi )) = sin (\\frac{1}{5}( \\frac{\\pi}{2} + 2rpi ))$. Nov 30, 2016 at 15:18", "date": "2022-05-22 05:54:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2037311/finding-the-roots-with-multiplicity-of-a-polynomial-with-compound-angle-formul", "openwebmath_score": 0.8072888255119324, "openwebmath_perplexity": 132.242096761021, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540734789343, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8526229762431481}}
{"url": "https://de.maplesoft.com/support/help/errors/view.aspx?path=type/partition&L=G", "text": "type/partition - Maple Programming Help\n\nHome : Support : Online Help : Programming : Data Types : Type Checking : Types : type/partition\n\ntype/partition\n\ncheck for an integer partition\n\n Calling Sequence type( expr, 'partition' )\n\nParameters\n\n expr - anything; any Maple expression\n\nDescription\n\n \u2022 The call type( expr, 'partition' )\u00a0returns true\u00a0if expr\u00a0is a partition of a positive integer, and returns false\u00a0otherwise.\n \u2022 A partition of a positive integer $n$\u00a0is a list [ ${k}_{1},\\mathrm{...},{k}_{r}$]. of positive integers ${k}_{i}$, whose sum is equal to $n$, and which is non-decreasing, that is, which satisfies $\\mathrm{add}\\left({k}_{i},i=1..r\\right)=n$\u00a0and ${k}_{i}\\le {k}_{i+1}$, for $i$\u00a0in $1..r-1$.\n\nExamples\n\n > $\\mathrm{type}\\left(\\left\\{\\left\\{1,2\\right\\},\\left\\{3,4\\right\\}\\right\\},'\\mathrm{partition}'\\right)$\n ${\\mathrm{false}}$ (1)\n > $\\mathrm{type}\\left(\\left[1,2,\\frac{2}{3}\\right],'\\mathrm{partition}'\\right)$\n ${\\mathrm{false}}$ (2)\n > $\\mathrm{type}\\left(\\left[1,-2,3\\right],'\\mathrm{partition}'\\right)$\n ${\\mathrm{false}}$ (3)\n > $\\mathrm{type}\\left(\\left[1,2,3\\right],'\\mathrm{partition}'\\right)$\n ${\\mathrm{true}}$ (4)\n > $\\mathrm{type}\\left(\\left[1,2,3,3,4\\right],'\\mathrm{partition}'\\right)$\n ${\\mathrm{true}}$ (5)\n > $\\mathrm{type}\\left(\\left[1,2,3,3,2\\right],'\\mathrm{partition}'\\right)$\n ${\\mathrm{false}}$ (6)", "date": "2021-04-16 21:24:24", "meta": {"domain": "maplesoft.com", "url": "https://de.maplesoft.com/support/help/errors/view.aspx?path=type/partition&L=G", "openwebmath_score": 0.9926496744155884, "openwebmath_perplexity": 2797.0456144549544, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540704659682, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8526229686878546}}
{"url": "http://math.stackexchange.com/questions/669085/what-do-curly-less-than-sign-mean", "text": "# What do curly less than sign mean?\n\nI am reading \"Convex Optimization\" by Stephen Boyd. He is using a curved greater than and curved less than equal to signs.\n\n$f(x^*) \\succcurlyeq \\alpha$\n\nor\n\n$f(x*) \\preccurlyeq \\alpha$\n\nCan someone explain or point me to their meaning.\n\n-\nas () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly \u2013\u00a0 barlop Feb 9 at 6:43\n@barlop If you look at $\\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \\succcurlyeq, which has curly word in it, not curved. \u2013\u00a0 Ruslan Feb 9 at 7:18\nThanks everyone for answers. As I understand $\\succeq$ or $\\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \\succeq Y, \\quad if \\quad \\| X \\| \\ge \\| Y \\|$ where $\\| \\cdot \\|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. \u2013\u00a0 Dinesh K. Feb 9 at 17:53\nYou should definitely take the tour. This is not a traditional forum! \u2013\u00a0 canaaerus Feb 9 at 18:03\nWell I don't know what a traditional forum looks like. :-) \u2013\u00a0 Dinesh K. Feb 9 at 18:20\n\nThere's a list of notation in the back of the book. On page 698, $x\\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\\leq y_i$ for every index $i$.\n\nEdit: The notation is introduced on page 32.\n\n-\n\nBoth Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here.\n\nIt should be noted that the book does not use $\\succeq$, $\\preceq$, $\\succ$, and $\\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for.\n\nThe book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.\n\nLet $S$ be a vector space, and let $K\\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\\alpha K\\equiv K$ for all $\\alpha>0$; and by pointed, we mean that $K\\cap-K=\\{0\\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \\succeq_K y \\quad\\Longleftrightarrow\\quad y \\preceq_K x \\quad\\Longleftrightarrow\\quad x - y \\in K$$ $$x \\succ_K y \\quad\\Longleftrightarrow\\quad y \\prec_K x \\quad\\Longleftrightarrow\\quad x - y \\in \\mathop{\\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\\in S$, $x \\not\\succeq_K y$ and $y \\not\\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\\geq$, $\\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\\succeq_K y \\quad\\Longrightarrow\\quad \\alpha x \\succeq_K \\alpha y \\quad\\forall \\alpha>0$$ $$x\\succeq_K y \\quad\\Longrightarrow\\quad \\alpha x \\preceq_K \\alpha y \\quad\\forall \\alpha<0$$ $$x\\succeq_K y, ~ x\\preceq_K y \\quad\\Longrightarrow\\quad x=y$$ $$x\\succ_K y \\quad\\Longrightarrow\\quad x\\not\\prec_K y$$\n\nWhen the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\\succeq$. There are two cases where this is almost always done. First, when $S=\\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \\succeq_{\\mathbb{R}^n_+} y \\quad\\Longleftrightarrow\\quad x_i\\geq y_i,~i=1,2,\\dots,n$$ Second, when $S$ is the set of symmetric $n\\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\\mathcal{S}^n_+=\\{X\\in S\\,|\\,\\lambda_{\\text{min}}(X)\\geq 0\\}$, the inequality is a linear matrix inequality (LMI): $$X \\succeq_{\\mathcal{S}^n_+} Y \\quad\\Longleftrightarrow\\quad \\lambda_{\\text{min}}(X-Y)\\geq 0$$ In both of these cases, the cone subscript is almost always dropped.\n\nMany texts in convex optimization don't bother with this distinction, and use $\\geq$ and $\\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\\mathbb{R}^n_+$; I think most people rightly assume that $x\\geq y$ is considered elementwise when $x,y$ are vectors.\n\n-\nThanks a lot for the detailed answer, and for correcting the symbol as well :-). \u2013\u00a0 Dinesh K. Feb 9 at 16:35\n\nOften these symbols represent partial order relations. The typical \"less than\" and \"greater than\" operations both define partial orders on the real numbers. However, there are many other examples of partial orders.\n\n-", "date": "2014-12-21 11:21:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/669085/what-do-curly-less-than-sign-mean", "openwebmath_score": 0.9096060991287231, "openwebmath_perplexity": 373.5897632397523, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540671517049, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8526229674465372}}
{"url": "https://www.physicsforums.com/threads/circular-motion-question-about-a-car-banked.63558/", "text": "# Circular motion question about a car banked\n\n1. Feb 12, 2005\n\n### F|234K\n\nif a curve with a radius of 60 meter is properly banked for a car travelling at 60km/h, what must be the coefficient of static friction for a car not to skid when travelling at 90km/h?\n\ni know that in order to solve the question, one needs to find the angle first, and i found the angle to be 25.3 degrees(not sure to be correct)...\n\nand the answer is 0.39...i just can't seem to get the answer...\n\n2. Feb 12, 2005\n\n### Staff: Mentor\n\nThat's correct.\n\nSince you didn't show any work, there is no way to tell where you got stuck. The basic idea is exactly the same as in the no friction case: the only difference is the addition of the friction force on the car ($\\mu N$ acting down the incline).\n\n3. Feb 12, 2005\n\n### xanthym\n\nYour calculated road banking angle of 25.3 deg relative to horizontal is correct (also indicated by the previous msg).\n\nThe required static friction coefficient for 90 km/h along the same banked road is calculated to be 0.39 in agreement with the provided answer (but apparently not in line with your calculations).\n\nYou may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system.\n\nA typical formulation based on the friction horizontal component only is:\n\n$$:(1): \\ \\ \\ \\ \\frac {v^2} {rg} = Tan(\\theta) + \\mu_s Cos(\\theta)$$\n\nThe above equation yields a static friction coefficient of 0.65 for this case. The formulation based on fully coupled horizontal and vertical components is:\n\n$$:(2): \\ \\ \\ \\ \\frac {v^2} {rg} = \\frac {Sin(\\theta) + \\mu_s Cos(\\theta)} {Cos(\\theta) - \\mu_s Sin(\\theta) }$$\n\nThis latter formulation yields 0.39 for Coefficient of Static Friction.\n\nThe complete system of equations for horizontal and vertical components is given below:\n\n$$:(3): \\ \\ \\ \\ \\frac{mv^2} {r} = N Sin(\\theta) + \\mu_s N Cos(\\theta)$$\n\n$$:(4): \\ \\ \\ \\ 0 = N Cos(\\theta) - \\mu_s N Sin(\\theta) - mg$$\n\nwhere N is the force Normal to the road surface, (theta) the bank angle, and the other variables defined per usual conventions.\n\n~~\n\nLast edited: Feb 12, 2005\n4. Feb 12, 2005\n\n### F|234K\n\nthank you guys very much\n\n\"You're may not be obtaining the same answer because you're calculating a static friction coefficient in the range 0.59 - 0.65. This is caused by using results from the decoupled equations for frictional force wherein friction is considered to contribute ONLY a horizontal component to the system (i.e., the component directly adding to the centripetal force). However, friction also has a vertical component which must be accounted for when solving the simultaneous equations for the system.\"\n\nand yes, i got my answer to be 0.59, thank you xanthym for your detailed and consummate reply which enlightens me on the vertical component of friction. (i only calculated the horizontal part of friction.)", "date": "2017-05-23 13:06:41", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/circular-motion-question-about-a-car-banked.63558/", "openwebmath_score": 0.5794603824615479, "openwebmath_perplexity": 520.8597272929647, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708032626701, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8526068624223486}}
{"url": "https://math.stackexchange.com/questions/1978171/is-this-a-valid-way-to-prove-this-modified-harmonic-series-diverges", "text": "# Is this a valid way to prove this modified harmonic series diverges?\n\nI am trying to find a way to prove that $$\\dfrac 11 + \\dfrac 12 + \\dfrac 13 + \\dfrac 14 + \\cdots \\color{red}{-} \\dfrac 18 + \\cdots$$\n\nwhere the pattern repeats every $8$ terms. Knowing about the Riemann Series Theorem, I am a little hesitant about manipulating conditionally convergent series at all. Granted that the harmonic series diverges, is the following a valid way to prove my series diverges?\n\n$$\\dfrac 11 + \\dfrac 12 + \\dfrac 13 + \\cdots + \\dfrac 17 - \\dfrac 18 + \\cdots = \\sum_{n=1}^{\\infty} \\dfrac 1n - 2 \\cdot \\dfrac 18\\sum_{n=1}^{\\infty} \\dfrac 1n$$\n\n$$=\\dfrac 34 \\sum_{n=1}^{\\infty} \\dfrac 1n$$\n\nSince the harmonic series diverges, so does $\\dfrac 34$ times it.\n\n\u2022 I stared at that line many, many times before I finally saw where the $+$ turned into a $-$. \u2013\u00a0Yakk Oct 21 '16 at 10:55\n\u2022 @Yakk I suggested an edit to make it more obvious \u2013\u00a0null Oct 21 '16 at 16:46\n\u2022 I've made it red. Don't know if there's a better way. \u2013\u00a0asmeurer Oct 21 '16 at 20:11\n\u2022 @asmeurer Is it better now? \u2013\u00a0Ovi Oct 22 '16 at 1:57\n\u2022 That helps. At first I thought you meant repeat the terms shown, which obviously diverges. \u2013\u00a0asmeurer Oct 22 '16 at 1:59\n\n## 4 Answers\n\nYour idea is a good one, but, as you suspected, you need to be more careful about this sort of manipulation of conditionally convergent series.\n\nOne way to carry out your argument correctly, but with only minor changes, is by looking at partial sums:\n\nLet's write $$a_n=\\begin{cases}1,&\\text{ if }n\\text{ is not a multiple of 8} \\\\-1,&\\text{ if }n\\text{ is a multiple of 8},\\end{cases},$$ so that your series is $\\sum_{n=1}^\\infty \\frac{a_n}{n}.$\n\nThen for any natural number $N,$\n\n\\begin{align} \\sum_{n=1}^{8N}\\frac{a_n}{n} &= \\sum_{n=1}^{8N}\\frac1{n}-2\\sum_{n=1}^N \\frac1{8n} \\\\&=\\sum_{n=1}^{8N}\\frac1{n}-\\frac1{4}\\sum_{n=1}^N \\frac1{n} \\\\&\\ge\\sum_{n=1}^{8N}\\frac1{n}-\\frac1{4}\\sum_{n=1}^{8N} \\frac1{n}\\scriptsize{\\quad\\text{(because we can only be subtracting *more* positive numbers)}} \\\\&=\\frac3{4}\\sum_{n=1}^{8N}\\frac1{n}, \\end{align}\n\nwhich approaches $\\infty$ as $N$ approaches $\\infty,$ since the harmonic series diverges.\n\n\u2022 Thank you, I learned something very important; you can manipulate even conditionally convergent series as long as you manipulate them in the context of partial sums. \u2013\u00a0Ovi Oct 21 '16 at 4:52\n\u2022 @Ovi -- Yes, because the partial sums are just regular finite sums. But you still need to be careful that your partial sums are entire initial parts of the infinite series. You can't pick and choose which terms to include; all you can do is chop the infinite series off at some finite point. After that, you can apply any normal algebraic operations, because you just have a finite sum. \u2013\u00a0Mitchell Spector Oct 21 '16 at 4:56\n\u2022 @ Mitchell Spectator Okay got it \u2013\u00a0Ovi Oct 21 '16 at 4:57\n\u2022 And if you want to prove that the conditionally convergent series does converge, you can't cherry-pick which of the partial sums you're looking at (like it happens here), of course. \u2013\u00a0Henning Makholm Oct 21 '16 at 14:21\n\u2022 @HenningMakholm Good point. \u2013\u00a0Mitchell Spector Oct 21 '16 at 15:37\n\nTo expand on Mitchell Spector's answer, let's generalize to the problem where every $b$th term is negated.\n\nBy the same argument, we get \\begin{align} \\sum_{n=1}^{bN}\\frac{a_n}{n} &= \\sum_{n=1}^{bN}\\frac1{n}-2\\sum_{n=1}^N \\frac1{bn} \\\\&=\\sum_{n=1}^{bN}\\frac1{n}-\\frac2{b}\\sum_{n=1}^N \\frac1{n} \\\\&\\ge\\sum_{n=1}^{bN}\\frac1{n}-\\frac2{b}\\sum_{n=1}^{bN} \\frac1{n} \\\\&=\\left(1-\\frac2{b}\\right)\\sum_{n=1}^{bN}\\frac1{n} \\end{align}\n\nThis is nice, since we can see that the divergence proof holds for $b \\geq 3$, but fails for $b=2$. This is not strange, since the series for $b=2$ is known to converge to $\\log 2$.\n\nThis proof isn't valid because of the Riemann series theorem; the original series could still converge even though some rearrangement of it diverges. It's important to focus on the partial sums.\n\nYou can argue along the lines that the series $$\\sum_{k=0}^{\\infty} \\frac{1}{8k+1} = 1 + \\frac{1}{9} + \\frac{1}{17} + \\cdots$$ already diverges by comparing its partial sums to the partial sums of the harmonic series $\\frac{1}{8} \\Big( 1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{n} \\Big).$\n\nNote that \\begin{align} &{\\tiny\\sum_{k=0}^\\infty}{\\tiny\\left(\\frac1{8k+1}+\\frac1{8k+2}+\\frac1{8k+3}+\\frac1{8k+4}+\\frac1{8k+5}+\\frac1{8k+6}+\\frac1{8k+7}-\\frac1{8k+8}\\right)}\\\\ &={\\tiny\\sum_{k=0}^\\infty}{\\tiny\\left(\\frac1{8k+1}+\\frac1{8k+2}+\\frac1{8k+3}+\\frac1{8k+4}+\\frac1{8k+5}+\\frac1{8k+6}\\right)+{\\tiny\\sum_{k=0}^\\infty}\\left(\\frac1{8k+7}-\\frac1{8k+8}\\right)}\\\\ \\end{align} For the left hand sum, we have \\begin{align} &{\\small\\sum_{k=0}^\\infty\\left(\\frac1{8k+1}+\\frac1{8k+2}+\\frac1{8k+3}+\\frac1{8k+4}+\\frac1{8k+5}+\\frac1{8k+6}\\right)}\\\\ &\\ge{\\small\\frac68\\sum_{k=0}^\\infty\\frac1{k+1}} \\end{align} which diverges.\n\nFor the right hand sum, we have \\begin{align} &{\\small\\sum_{k=0}^\\infty\\left(\\frac1{8k+7}-\\frac1{8k+8}\\right)}\\\\ &\\le{\\small\\frac1{56}+\\frac18\\sum_{k=1}^\\infty\\left(\\frac1{8k}-\\frac1{8k+8}\\right)}\\\\ &=\\frac{15}{448} \\end{align} A divergent sum plus a convergent sum diverges.\n\n\u2022 How did you evaluate $\\dfrac 18 \\sum_{k=1}^{\\infty} \\left( \\dfrac {1}{8k}-\\dfrac {1}{8k+8} \\right) = \\dfrac {1}{64} \\sum_{k=1}^{\\infty} \\dfrac {1}{k(k+1)}$? I guess its not necessary to calculate the sum because we know it converges anyway, but I'm curious how you got a rational answer, knowing that $\\sum_{k=1}^{\\infty} \\dfrac {1}{k^2} = \\dfrac {\\pi^2}{6}$ \u2013\u00a0Ovi Oct 24 '16 at 15:19\n\u2022 \\begin{align} \\sum_{k=1}^\\infty\\frac1{k(k+1)} &=\\sum_{k=1}^\\infty\\left(\\frac1k-\\frac1{k+1}\\right)\\\\ &=\\lim_{n\\to\\infty}\\left(\\sum_{k=1}^n\\frac1k-\\sum_{k=2}^{n+1}\\frac1k\\right)\\\\ &=\\lim_{n\\to\\infty}\\left(1-\\frac1{n+1}\\right)\\\\ &=1 \\end{align} which is a Telescoping Sum. \u2013\u00a0robjohn Oct 24 '16 at 15:50", "date": "2019-06-20 18:14:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1978171/is-this-a-valid-way-to-prove-this-modified-harmonic-series-diverges", "openwebmath_score": 0.9799054861068726, "openwebmath_perplexity": 529.7691013679157, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707986486797, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8526068583973462}}
{"url": "https://math.stackexchange.com/questions/834471/help-evaluating-lim-x-to-infty-sqrtx-sqrtx-sqrtx-sqrtx", "text": "# Help Evaluating $\\lim_{x \\to \\infty} \\sqrt{x + \\sqrt{x + \\sqrt{x}}} - \\sqrt{x}$\n\nDoes anyone know how to evaluate the following limit? $$\\lim_{x \\to \\infty} \\sqrt{x + \\sqrt{x + \\sqrt{x}}} - \\sqrt{x}$$ The answer is $\\frac{1}{2}$, but I want to see a step by step solution if possible.\n\nMultiply by the conjugate, then reduce: \\begin{align*} \\lim_{x\\to\\infty} \\left(\\sqrt{x + \\sqrt{x + \\sqrt{x}}} - \\sqrt{x} \\right) &= \\lim_{x\\to\\infty} \\left(\\sqrt{x + \\sqrt{x + \\sqrt{x}}} - \\sqrt{x} \\right) \\cdot \\frac{\\sqrt{x + \\sqrt{x + \\sqrt{x}}} + \\sqrt{x}}{\\sqrt{x + \\sqrt{x + \\sqrt{x}}} + \\sqrt{x}} \\\\ &= \\lim_{x\\to\\infty} \\frac{(x + \\sqrt{x + \\sqrt{x}}) - x}{\\sqrt{x + \\sqrt{x + \\sqrt{x}}} + \\sqrt{x}} \\\\ &= \\lim_{x\\to\\infty} \\frac{\\sqrt{x + \\sqrt{x}}}{\\sqrt{x + \\sqrt{x + \\sqrt{x}}} + \\sqrt{x}} \\cdot \\frac{\\dfrac{1}{\\sqrt{x}}}{\\dfrac{1}{\\sqrt{x}}} \\\\ &= \\frac{\\sqrt{\\lim\\limits_{x\\to\\infty}\\dfrac{1}{\\sqrt{x}} + 1}}{\\sqrt{1 + \\lim\\limits_{x\\to\\infty} \\dfrac{\\sqrt{x + \\sqrt{x}}}{x}} + 1} \\\\ &= \\frac{\\sqrt{0 + 1}}{\\sqrt{1 + 0} + 1} \\\\ &= \\frac{1}{2} \\end{align*}\n\n\u2022 I like your answer! :) \u2013\u00a0dmk Jun 15 '14 at 1:46\n\u2022 thanks i like ur answer to \u2013\u00a0user157128 Jun 15 '14 at 1:54\n\u2022 as you did in step 3 to 4 when multiplied by 1/v5 ??? \u2013\u00a0user157128 Jun 15 '14 at 2:03\n\n$$\\sqrt{x+\\sqrt{x+\\sqrt{x}}}-\\sqrt{x}=\\frac{\\sqrt{x+\\sqrt{x}}}{\\sqrt{x+\\sqrt{x+\\sqrt{x}}}+\\sqrt{x}}=\\frac{\\sqrt{1+\\frac{1}{\\sqrt{x}}}}{\\sqrt{1+\\sqrt{\\frac{1}{x}+\\frac{1}{x^{\\frac{3}{2}}}}}+1}$$\n\nI got the above by first multiplying by the conjugate to $\\sqrt{x+\\sqrt{x+\\sqrt{x}}}-\\sqrt{x}$ then dividing both the numerator and denominator by $\\frac{1}{\\sqrt{x}}$. Now take $x$ to infinity.\n\n\u2022 the answer is 1/2, but I tried this and rationalize, but did not reach 0.5 \u2013\u00a0user157128 Jun 15 '14 at 1:38\n\u2022 @user157128 what did you get? In the top I get $\\sqrt{1 + 0} - 0$. What do you get in the bottom? \u2013\u00a0DanZimm Jun 15 '14 at 1:40\n\n\\begin{align} \\ \\lim_{x\\rightarrow\\infty}\\sqrt{x+\\sqrt{x+\\sqrt{x}}}-\\sqrt{x} &= \\lim_{x\\rightarrow\\infty} \\left( \\sqrt{x+\\sqrt{x+\\sqrt{x}}}-\\sqrt{x} \\right) \\cdot \\frac{\\sqrt{x+\\sqrt{x+\\sqrt{x}}}+\\sqrt{x}}{\\sqrt{x+\\sqrt{x+\\sqrt{x}}}+\\sqrt{x}}\\\\ \\ &= \\lim_{x\\rightarrow\\infty} \\frac{x + \\sqrt{x+\\sqrt{x}}- x}{\\sqrt{x+\\sqrt{x+\\sqrt{x}}}+\\sqrt{x}}\\\\ \\ &= \\lim_{x\\rightarrow\\infty} \\frac{\\sqrt{x+\\sqrt{x}}}{\\sqrt{x+\\sqrt{x+\\sqrt{x}}}+\\sqrt{x}} \\\\ \\ &= \\lim_{x\\rightarrow\\infty} \\frac{\\sqrt{x+\\sqrt{x}}}{\\sqrt{x+\\sqrt{x+\\sqrt{x}}}+\\sqrt{x}} \\cdot \\frac{1/\\sqrt{x}}{1/\\sqrt{x}} \\\\ \\ &= \\lim_{x\\rightarrow\\infty} \\frac{\\sqrt{1+\\frac{1}{\\sqrt{x}}}}{\\sqrt{1+\\frac{1}{x}+\\frac{1}{x\\sqrt{x}}}+1} \\\\ \\ &= \\frac{\\sqrt{1}}{\\sqrt{1}+1} \\\\ \\ &= \\frac{1}{2} \\\\ \\end{align}\n\nUsing Taylor expansion:\n\n$$\\sqrt{x + \\sqrt{x + \\sqrt{x}}} - \\sqrt{x}$$\n\n$$\\sqrt{x + \\sqrt{x}+\\dfrac{1}{2} + O(\\frac{1}{\\sqrt{x}})} - \\sqrt{x}$$\n\n$$\\sqrt{x}+\\dfrac{1}{2\\sqrt{x}}(\\sqrt{x}+\\dfrac{1}{2}+ O(\\frac{1}{\\sqrt{x}})) + O(\\frac{1}{\\sqrt{x}})- \\sqrt{x}$$\n\n$$\\dfrac{1}{2} + \\dfrac{1}{4\\sqrt{x}}$$\n\nWhich in the limit case tends to ${\\dfrac{1}{2}}$.\n\nThe expansion I used was $(a+b)^{1/2} = \\sqrt{a}+\\frac{b}{2\\sqrt{a}} + \\cdots$\n\n\u2022 It's not clear from what you've written why you get to ignore the error terms in your expansion. \u2013\u00a0Cheerful Parsnip Jun 15 '14 at 1:50\n\u2022 @user157107 this seems like a fairly interesting way to go about it... and similar to what I was considering doing. Could you possibly add more details to justify your approach? \u2013\u00a0Squirtle Jun 15 '14 at 2:13", "date": "2019-07-17 16:33:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/834471/help-evaluating-lim-x-to-infty-sqrtx-sqrtx-sqrtx-sqrtx", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 1366.158558391856, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708012852457, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.852606854210127}}
{"url": "https://math.stackexchange.com/questions/2932529/find-the-nonzero-vectors-u-v-w-that-are-perpendicular-to-the-vector-1-1-1-1", "text": "# Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other\n\nFind the nonzero vectors $$u,v,w$$ that are perpendicular to the vector $$(1,1,1,1)$$ and to each other.\n\nIf I follow algebra, then I get complicated results to solve it as follows:\n\nLet $$u=(u_1,u_2,u_3,u_4), \\ v=(v_1,v_2,v_3,v_4) , \\ w=(w_1,w_2,w_3,w_4)$$\n\nThen, $$u \\cdot (1,1,1,1)=v \\cdot (1,1,1,1)=w \\cdot (1,1,1,1)=0$$\n\nAlso $$u \\cdot v=w \\cdot u=v \\cdot w=0$$.\n\nThese gives us\n\n$$u_1+u_2+u_3+u_4=0, \\\\ v_1+v_2+v_3+v_4=0 , \\\\ w_1+w_2+w_3+w_4=0, \\\\ u_1v_1+u_2v_2+u_3v_3+u_4v_4=0, \\\\ u_1w_1+u_2w_2+u_3v_3+u_4v_4=0, \\\\ v_1w_1+v_2w_2+v_3w_3+v_4w_4=0.$$\n\nBut how to solve for $$u_i, v_i,w_i, \\ i=1,2,3,4$$ from here?\n\nDoes there exit any other easy method?\n\nHelp me out\n\n\u2022 Apply Gram-Schmidt to the basis $$((1, 1, 1, 1), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)),$$ or any other basis beginning with $(1, 1, 1, 1)$. \u2013\u00a0Theo Bendit Sep 27 '18 at 2:37\n\u2022 Is not in that case the solution will be particular , I mean there can other vectors also which can not be conclude using Gram-schmidt method. \u2013\u00a0M. A. SARKAR Sep 27 '18 at 2:44\n\u2022 Yes, it will be particular. If you range over all such bases, then you will obtain all orthonormal bases beginning with $\\left(\\frac12,\\frac12,\\frac12,\\frac12\\right)$, though not uniquely. \u2013\u00a0Theo Bendit Sep 27 '18 at 2:47\n\nas columns $$\\left( \\begin{array}{rrrr} 1&-1&-1&-1 \\\\ 1& 1&-1&-1 \\\\ 1&0 &2&-1 \\\\ 1&0&0&3 \\end{array} \\right)$$ Pattern, done correctly, works in any dimension\n\n$$\\left( \\begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \\end{array} \\right).$$\n\n\u2022 That's great! =) \u2013\u00a0Siong Thye Goh Sep 27 '18 at 3:02\n\u2022 Excellent method \u2013\u00a0M. A. SARKAR Sep 27 '18 at 11:13\n\nConsider the Hadamard matrix of which we know that the columns form an orthogonal basis of $$\\mathbb{R}^4$$.\n\n$$\\begin{bmatrix} 1 & 1 & 1 & 1 \\\\ 1 & -1 & 1 & -1 \\\\ 1 & 1 & -1 & -1 \\\\ 1 & -1 & -1 & 1\\end{bmatrix}$$\n\nThe other columns would give you a solution.\n\nAlternatively, use Gram-Schmidt process.\n\n\u2022 But if we consider the Hadamard matrix , then the solution will be particular. There may be other vectors which are perpendicular to $(1,1,1,1)$ except the last 3 column vectors in the Hadamard vectors and its multiples . \u2013\u00a0M. A. SARKAR Sep 27 '18 at 2:42\n\u2022 So if we use Gram-Schmidt method then we need a basis \u2013\u00a0M. A. SARKAR Sep 27 '18 at 2:43\n\u2022 Theo has given you a basis right? Note that answer is not unique. If you want to describe the set, you have already done so in your post. \u2013\u00a0Siong Thye Goh Sep 27 '18 at 2:48\n\u2022 There is a pattern that easily adapts to any dimension... \u2013\u00a0Will Jagy Sep 27 '18 at 3:01", "date": "2020-06-01 09:13:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2932529/find-the-nonzero-vectors-u-v-w-that-are-perpendicular-to-the-vector-1-1-1-1", "openwebmath_score": 0.8482692837715149, "openwebmath_perplexity": 495.77803505708044, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708006261044, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8526068455261008}}
{"url": "https://math.stackexchange.com/questions/3552753/is-it-necessary-to-write-limits-for-a-substituted-integral", "text": "# Is it necessary to write limits for a substituted integral?\n\nTo solve the following integral, one can use u-substitution: $$\\int_2^3 \\frac{9}{\\sqrt[4]{x-2}} \\,dx,$$ With $$u = \\sqrt[4]{x-2}$$, our bounds become 0 and 1 respectively. Thus, we end up with the following: $$36\\int_0^1{u^2} \\,du$$ In the first case, the lower bound is a vertical asymptote so we would have to use limits to find the answer. However, in the second case there's no longer an asymptotal bound - would you still have to write the limits since the original function would've needed limits, or can you just solve this by plugging in the substituted bounds? I know the final answer will be the same either way, but I want to know if it can be considered correct to exclude the limits in the 2nd integral from a technical perspective since the function has been changed. Many thanks in advance!\n\n## 3 Answers\n\nWithout too much simplification, the substitution you cite yields\n\n$$\\int_2^3\\frac9{(x-2)^{1/4}}\\,\\mathrm dx=36\\int_0^1\\frac{u^3}u\\,\\mathrm du$$\n\nwhich you certainly welcome to treat as an improper integral,\n\n$$36\\left(\\frac13-\\lim_{u\\to0^+}\\frac{u^3}3\\right)$$\n\nbut since $$u=0$$ is a removable discontinuity and the limand reduces to $$u^2$$, you may as skip this treatment altogether.\n\n\u2022 I'm confused about your last expression. It seems like you want to still have $\\int$ present, but the $1-$ seems like you've already taken the integral, but the $\\frac{u^3}u$ is from before you've taken the integral... \u2013\u00a0Teepeemm Feb 20 at 2:05\n\u2022 Yes, it was wrong as written. I have edited it. \u2013\u00a0Martin Argerami Feb 20 at 3:51\n\u2022 Thanks @MartinArgerami ! \u2013\u00a0user170231 Feb 20 at 14:55\n\nThere are certain conditions that must be met for a substitution to be \"legal\". In most circumstances these conditions are naturally met and so they are not emphasized; you have here one situation in which they are not.\n\nFor instance, one set of conditions is given in Anton, Anderson, and Bivens (Calculus, Early Transcendentals, 11th Edition), Theorem 5.9.1:\n\nTheorem 5.9.1. If $$g'(x)$$ is continuous on $$[a,b]$$, and $$f$$ is continuous on an interval containing the values of $$g(x)$$ for $$a\\leq x\\leq b$$, then $$\\int_a^b f(g(x))g'(x)\\,dx = \\int_{g(a)}^{g(b)} f(u)\\,du.$$\n\nHere you have $$g(x) = \\sqrt[4]{x-2}$$, so $$g'(x) = \\frac{1}{4}(x-2)^{-3/4}$$... which is not continuous on the interval $$[2,3]$$ that you are working on.\n\nIn fact, as you note, the initial integral is improper, which means you aren't really evaluating that integral: you are evaluting a limit, $$\\lim_{h\\to 2^+}\\int_h^3 \\frac{9}{\\sqrt[4]{x-2}}\\,dx.$$ The integral in the limit does satisfy the conditions of the theorem above, so you can make the substitution to get $$\\lim_{h\\to 2^+}\\int_{\\sqrt[4]{h-2}}^1 \\frac{u^3}{u}\\,du = \\lim_{a\\to 0^+}\\int_a^1 u^2\\,du,$$ and proceed from there.\n\n\u2022 Thank you very much Arturo, your answer was very in-depth and provided a lot of reinforcement for what you were saying. I wish I could select multiple answers to be correct, I thought your inclusion of the cited textbook snippet was really nicely done too. I chose the other answer only because I found it easier to digest/understand at first glance, but after I properly read your comment I understood it more fully. Thank you for your quick and effective contribution :-) \u2013\u00a0Lord Kanelsnegle Feb 19 at 17:32\n\nSuppose you must. Then we have $$9\\lim_{a\\to2^+}\\int_a^3\\frac1{\\sqrt[4]{x-2}}\\,dx.$$\n\nLet $$u=x-2$$. Then we have $$9\\lim_{a\\to2^+}\\int_{a-2}^1 u^{-1/4}\\, du.$$\n\nBy the power rule, we have $$9\\lim_{a\\to2^+} \\left.\\frac43u^{3/4}\\right]_{a-2}^1=9\\left[\\frac43(1-\\lim_{a\\to2^+}(a-2)^{3/4})\\right]=9\\left[\\frac43(1-0)\\right]=12.$$\n\nNotice that it does not make a difference whether you use $$\\lim_{a\\to2^+}(a-2)$$ or $$0$$.\n\n\u2022 Thanks for the quick and simple response! I found the answer I selected to be more aligned with the question, but I wanted to comment because I appreciate the time you took to write this. Thank you once again. \u2013\u00a0Lord Kanelsnegle Feb 19 at 17:29", "date": "2020-03-29 04:18:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3552753/is-it-necessary-to-write-limits-for-a-substituted-integral", "openwebmath_score": 0.9277409315109253, "openwebmath_perplexity": 201.05041909566526, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707966712548, "lm_q2_score": 0.8723473713594992, "lm_q1q2_score": 0.8526068453197088}}
{"url": "https://math.stackexchange.com/questions/3797897/expected-value-of-flipping-through-a-book/3797922", "text": "# Expected Value of flipping through a book\n\nSuppose that a book has $$N$$ pages, and we read through the book as follows. We start from page 0, and if we're on page $$i$$, we randomly flip to a page $$i + 1, i + 2, ..., N$$ with equal probability.\n\n1. What is the expected value of number of flips we need to finish the book?\n\nIntuition tells me that we can, on average, expect to halve the number of pages remaining. This yields $$\\log_2(N)$$, but I'm having trouble formalizing it.\n\n1. If $$N = 26$$, what is the probability we flip to page 13 at some point? Assume we start at page 0.\n\nI let $$P_i$$ be the probability we land on page 13 eventually, starting from page $$i$$. Then, $$P_{13} = 1$$, and in general, $$P_{i} = \\frac{1}{26 - i}\\sum_{k = i + 1}^{13}P_k$$\n\nEvaluating terms like $$P_{12}, P_{11}, P_{10}$$, I see that all of these values are $$\\frac{1}{14}$$, including $$P_0$$. Is there a more intuitive reason for such a simple answer?\n\n\u2022 Question 1. is unclear ... number of flips we need for what? To read every page in the book? To reach the last page in the book? What does it mean? \u2013\u00a0Ross Presser Aug 21 at 13:47\n\u2022 edited for clarity, though it should be implicitly clear that it's often not possible to read every page in the book, and the expected value of flips to read every page is trivial. \u2013\u00a0user815048 Aug 21 at 20:13\n\nLet's consider the equivalent problem in which we start at page $$n$$ and flip backwards through the book, going to each of the pages $$0, 1, ..., n - 1$$ with equal probability. Let $$E_n$$ be the expected number of flips. Then we have $$E_0 = 0$$ and\n\n$$E_n = 1 + \\frac{1}{n} \\sum\\limits_{i = 0}^{n - 1} E_i$$\n\nThen in particular we have\n\n$$$$\\begin{split} E_{n + 1} &= 1 + \\frac{1}{n + 1} \\sum\\limits_{i = 0}^{n} E_i \\\\ &= 1 + \\frac{E_n}{n + 1} + \\frac{n}{n + 1} \\frac{1}{n} \\sum\\limits_{i = 0}^{n - 1} E_i \\\\ &= 1 + \\frac{E_n}{n + 1} + \\frac{n}{n + 1} (1 + \\frac{1}{n} \\sum\\limits_{i = 0}^{n - 1} E_i) - \\frac{n}{n + 1} \\\\ &= 1 - \\frac{n}{n + 1} + \\frac{1}{n + 1} E_n + \\frac{n}{n + 1} E_n \\\\ &= \\frac{1}{n + 1} + E_n \\end{split}$$$$\n\nwhenever $$n \\geq 1$$ (and the identity is easily verified when $$n = 0$$ as well).\n\nThen by induction, we have $$E_n = \\sum\\limits_{j = 1}^n \\frac{1}{j}$$, the $$n$$th harmonic number. This will be asymptotically very close to $$\\log_e(n)$$.\n\nLet $$P_n$$ be the expected number of flips in a book with $$n$$ pages. Then $$P_0=0,\\ P_1=1$$ and $$P_n=1+\\frac1n\\sum_{k=0}^{n-1}P_k,\\tag1$$ because we have to make one flip, and then we're equally likely to have any number of pages from $$0$$ to $$n-1$$ left to flip through.\n\nWe get \\begin{align} P_1&=1\\\\ P_2&=\\frac32\\\\ P_3&=\\frac{11}6\\\\ P_4&=\\frac{50}{24}\\\\ P_5&=\\frac{174}{120} \\end{align}\n\nThe denominators are obviously $$n!$$, so we look for the numerators in OEIS and find A000254, the unsigned Stirling numbers of the first kind.\n\nOESI gives the recurrence $$a_{n+1}=(n+1)a_n+n!$$ for the unsigned Stirling numbers of the first kind, and dividing through by $$(n+1)!$$ we get $$P_{n+1}=P_n+\\frac1{n+1}$$ which clearly gives $$P_n=\\sum_{k=1}^n\\frac1k=H_n,$$ the $$n$$th harmonic number. To complete the problem, we must show that the harmonic numbers satisfy the recurrence $$(1)$$.\n\nHere is how I approached the first part of the problem. Consider a book with exactly $$n$$ pages. Let $$P_1$$ denote the first page you flipped to, and let $$X_n$$ represent the number of pages you flipped until you get to the last page. Note $$P_1$$ is uniformly distributed on the set $$\\{1,...,n\\}$$ and $$E(X_1)=1$$. Using the total law of expectation we get for $$n\\geq2$$ that $$E(X_n)=\\sum_{k=1}^{n}E(X_n|P_1=k)P(P_1=k)=\\frac{1}{n}\\sum_{k=1}^{n}E(X_n|P_1=k)$$\nNotice $$E(X_n|P_1=k)=1+E(X_{n-k})$$ and so $$E(X_n)=\\frac{1}{n}\\sum_{k=1}^{n}\\Big[1+E(X_{n-k})\\Big]=1+\\frac{E(X_0)+\\dots+E(X_{n-1})}{n}$$ Replace $$n$$ with $$n+1$$ to get $$E(X_{n+1})=1+\\frac{E(X_0)+\\dots+E(X_{n})}{n+1}$$ Combining the previous two equations unveils the relation $$(n+1)(E(X_{n+1})-1)=(n+1)E(X_n)-n$$ which is equivalent to saying $$E(X_{n+1})=E(X_n)+\\frac{1}{n+1}$$ So finally $$E(X_n)=\\sum_{k=1}^{n}\\frac{1}{k}$$", "date": "2020-10-27 12:58:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3797897/expected-value-of-flipping-through-a-book/3797922", "openwebmath_score": 0.9989168643951416, "openwebmath_perplexity": 172.28301716312075, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983342957061873, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8525935197470873}}
{"url": "http://mathhelpforum.com/statistics/4194-college-prob-stat.html", "text": "# Math Help - college prob + stat\n\n1. ## college prob + stat\n\n2 Questions:\n\n1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?\n\n2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?\n\n2. Originally Posted by magicpunt\n1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?\n$P(\\mbox{men or woman})=P(\\mbox{men})+P(\\mbox{woman})$\nLet us find,\n$P(\\mbox{men})$\nThere are 10 people and 5 are selected there are a total of $_{10}C_5=252$\nFrom 5 men you take 5 men intotal the number of combinations is,\n$_5C_5=1$\nThus,\n$P(\\mbox{men})=\\frac{1}{252}$\nSimilarily ya have,\n$P(\\mbox{women})=\\frac{1}{252}$\nThus, the probability is,\n$\\frac{1}{252}+\\frac{1}{252}=\\frac{2}{252}=\\frac{1} {126}$\n\n3. 2) There are 10 people in an elevator in the basement. They can get off at any of 5 floors above. The probability that they get off at each floor is equal and each person leaving is independant of each other. What is the probability that 2 people get off at each floor?\nThis is similar to the 'boxes and balls' problems.\n\nThere are $5^{10}=9765625$ ways to distribute the people to the 5 floors.\n\nThere are $\\frac{5!}{2!}=60$ ways that no two people get off on the same floor.\n\nEDIT:\nYes, I misinterpreted the problem. Cerebral flatulence. Soroban's method is solid.\nMy second part should be $\\frac{10!}{2^{5}}$\n\nTherefore, $\\frac{5^{10}}{\\frac{10!}{2^{5}}}$\n\nThis is a version of the classic 'assigning diplomats' problem.\n\nFor instance:\n\nHow many ways can 10 different diplomats be assigned to 5 different countries if 2 diplomats must be assigned to each country?.\n\nSee the similarity with your problem?. Yours is just elevators and people, instead.\n\nThis can be done in $\\frac{10!}{(2!)^{5}}=113,400$ ways.\n\nThe method I used is just another variation on this problem. Keep it, you may need it.\n\n4. Originally Posted by galactus\nThis is similar to the 'boxes and balls' problems.\n\nThere are $5^{10}=9765625$ ways to distribute the people to the 5 floors.\n\nThere are $\\frac{5!}{2!}=60$ ways that no two people get off on the same floor.\nI think you might've misunderstood the problem\n\nThe question is: what is P(Exactly 2 people get off at each floor)\n\n5. Hello, magicpunt!\n\nHere's my approach to #2 . . .\n\n2) There are 10 people in an elevator in the basement.\nThey can get off at any of 5 floors above.\nThe probability that they get off at each floor is equal\nand each person leaving is independant of each other.\nWhat is the probability that 2 people get off at each floor?\n\nEach of the ten people has a choice of five floors.\n. . There are: $5^{10}$ways that they can leave the elevator.\n\nIf exactly two people get off at each floor,\n. . the number of ways is: $\\begin{pmatrix}10\\\\2,2,2,2,2\\end{pmatrix} = 113,400$\n\nTherefore, the probabilty is: . $\\frac{113,400}{5^{10}}\\:=\\:\\frac{4,536}{390,625}$\n\n6. Originally Posted by magicpunt\n2 Questions:\n\n1) There are 5 men and 5 women in a group. A committee is to be formed of 5 people. What is the probability that the group is all of the same sex (all men or all women)?\nAnother approach to #1\nFind the total no. of ways of selecting 5 out of 10= $\\frac{10!}{5!5!}=C_5^{10}$\nNo. of ways in which all are men=1\nNo. of ways in which all are women=1\nRequired probability= $\\frac{1+1}{C_5^{10}}=\\frac{1}{126}$\n\nMalay", "date": "2015-11-24 22:18:23", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/4194-college-prob-stat.html", "openwebmath_score": 0.7347457408905029, "openwebmath_perplexity": 433.3378118010595, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429624315188, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8525935176457149}}
{"url": "https://math.stackexchange.com/questions/1788715/what-is-the-moment-of-inertia-of-a-gosper-island", "text": "# What is the moment of inertia of a Gosper island?\n\nWe know that regular hexagons can tile the plane but not in a self-similar fashion. However we can construct a fractal known as a Gosper island, that has the same area as the hexagon but has the property that when surrounded by 6 identical copies produces a similar shape, but with dimensions scaled by a factor of $\\sqrt{7}$.\n\nWhat is the distance between two of the centers? Is it the same as the distance between hexagons of the same area? ie. If I start with a hexagon of area A, then construct a Gosper island and place it next to an identical copy, would the distance still be the same as if they were hexagons? Or does the scaling factor come into play somewhere? Right now I think the answer is $\\sqrt{3}/2$, as for the hexagon.\n\nThe reason I ask is that I'm trying to calculate the Gosper island's moment of inertia through an axis through its centre of mass and perpendicular to the plane of the island.\n\nIf we assume that the moment of inertia is always proportional to the mass, and proportional to the square of a characteristic length scale, then $$I = \\gamma Ml^2,$$ where $\\gamma$ is a constant, $l$ is the 'diameter' of the island, in a hexagon this would be the distance between two opposite vertices. Shrink the Gosper island by the scaling factor and surround it by six others. This self-similarity technique is super cute, and can be used to calculate the moment of inertia of an equilateral triangle, and can be extended to a square/rectangle quite easily. Fractals, having a high degree of self-similarity, seem amenable to this technique - here I calculate the moment of inertia for a Koch snowflake.\n\n$\\hspace{1.3cm}$\n\nUsing the principle of superposition, $$I = I_{\\text{centre}} + 6I_{\\text{edge}},$$ where $$I_{\\text{centre}} =\\gamma \\frac{M}{7}\\left(\\frac{l}{\\sqrt{7}}\\right)^2 = \\gamma \\frac{Ml^2}{49} = \\frac{I}{49}.$$\n\nNow, by the parallel axis theorem $\\displaystyle I_{\\text{edge}} = I_{\\text{COM}} + Md^2$ where $$\\displaystyle I_{\\text{COM}} = \\frac{I}{49}$$ and $\\displaystyle d= \\frac{\\sqrt{3} l}{2}$ (this was one source of error), so $\\displaystyle I_{\\text{edge}} = \\frac{I}{49} + \\frac{3Ml^2}{4},$ and \\begin{align*} I &= \\frac{I}{49} + 6\\left(\\frac{I}{49}+ \\frac{3Ml^2}{4}\\right), \\\\ I & = \\frac{I}{7} + \\frac{9Ml^2}{2}, \\\\ \\frac{6I}{7} & = \\frac{9Ml^2}{2}, \\\\ I & = \\frac{21Ml^2}{4}. \\end{align*}\n\nThis seems incorrect? It feels wrong, comparing to a disk of radius $l/2$ which has moment of inertia $Ml^2/4$ it seems far too large.\n\nIt would also be nice if we could verify our answer numerically or otherwise. Any references are also appreciated.\n\n\u2022 The center of mass of the Gosper island is indeed the same as the center of the initial hexagon. This, as well as the moment of inertia can be computed using the technique of self-similar integration as in my answer to this question. I could provide a few more details but I get the feeling this might be an exercise? \u2013\u00a0Mark McClure May 17 '16 at 13:18\n\u2022 Nope, just something I thought up after teaching the self-similarity method for finding the MOI of a triangle/square. Thankyou so much for the link! I suppose this can be done for an fractal with enough rotational symmetry to tile the plane? Please provide as many details as you wish, in an answer if you like. \u2013\u00a0Bennett Gardiner May 18 '16 at 2:05\n\u2022 By the way, the polar inertia of a disk is not $ml^2/4$ (That is the x or y inertia) but $Ml^2/2$. \u2013\u00a0Urukann May 20 '16 at 4:40\n\u2022 Isn't that only if $l$ is the radius though? In my formulation I was trying to use the diagonal of the hexagon, so it made more sense to compare to a disk with $l$ as the diameter. \u2013\u00a0Bennett Gardiner May 20 '16 at 6:19\n\u2022 @BennettGardiner: IMO this is a source of confusion (especially if you don't explain it). The standard formula is $I=MR^2$ where $R$ is an equivalent radius (AKA gyration). I found it more effective to work with unit area shapes, so that the mass can be simplified, and $R$ conveys all shape information. \u2013\u00a0Yves Daoust May 20 '16 at 14:40\n\nI think that your $\\frac{\\sqrt{3}}{2}$ for the distance between hexagons center is not right... The distance from an hexagon center to an edge is $\\frac{r\\sqrt{3}}{2}$, thus the distance between two hexagons of radius $\\frac{l}{2\\sqrt{7}}$ that touch by an edge should be: $$d = l\\sqrt{\\frac{3}{28}}$$\n\nNow, if we take back your computation: $I = \\frac{I}{49} + 6 \\left(\\frac{I}{49} + \\frac{3Ml^2}{28}\\right)$\n\nThis gives us : $I = \\frac{3}{4} M l^2$, which may still not be the right answer.\n\nIt would be nice to compare this value to numerical methods, I'll look into it.\n\ne/ I ran some numerical simulation, unfortunately it does not seem to confirm the above result.\n\nHow I did it:\n\n\u2022 Starting from the code linked in this page, I generated the all points composing the outer shape of the gosper island.\n\u2022 Then, I trimmed the generating points to remove duplicates\n\u2022 I approximate the Gosper island as being star-shaped (This is not perfectly true, at least for $n=5$). Then its inertia is the sum of the inertia of the triangles composed of the origin and two consecutive points. Note that to get the actual inertia, one has to divide by the shape's area (Because when evaluating the triangle's inertia, we consider it has a surfacic mass of 1.). All triangle related formulas are available on Wolfram Alpha.\n\nThe results show that indeed the inertia is proportional to $l^2$, but its ratio to the value conjectured above is not 1, but a constant close to $\\frac{4}{7}$: $$I_{gosper} \\approx 0.568582263418 \\frac{3}{4} M l^2$$\n\nUnfortunately, it seems that this error is not related to the star-shaped approximation: I ran another experiment, this time using the Seidel program from the University of North Carolina at Chapel Hill. It allowed me to find a triangulation of the inner area of the Gosper island. Using another (similar) code I could check that the computation for a radius of 1. does yield the same ration between expected inertia (0.75) and the actual inertia (0.42856647032), with a similar ratio of 0.571421960426. Note that this inertia is very close to $\\frac{3}{7}$, (best fractional approximation with a denominator inferior to fifteen thousand).\n\nActually, I had forgotten that the characteristic dimension is not the diameter, but the radius, thus the ratio is 0.142855490107, very close to $\\frac{1}{7}$.\n\nUsing this method for a Koch snowflake yields pretty correct results: for a snowflake of radius $r \\approx 1.44$ (I forgot to scale the step size properly) I get an inertia of $I_{koch} \\approx 0.736036125705$ while the one given by $Ml^2/11 \\approx 0.7435999999999999$\n\ne/ I found the error: The mass of a \"small\" Gosper island is not $M$, but $\\frac{M}{7}$, thus the missing factor 7. This is due to the fact that we make the assumption of a uniform density Gosper Island, thus its mass is proportional to its area.\n\nWe can rewrite our original equation: $$I_{edge} = I_{center} + \\frac{M}{7}d^2\\\\ d = l \\sqrt{\\frac{3}{28}}$$\n\nWhich gives us: $$I = \\frac{I}{49} + 6 \\left( \\frac{I}{49} + \\frac{3Ml^2}{7\\cdot 28}\\right)$$\n\nAnd finally:\n\n$$I = \\frac{3Ml^2}{28}$$\n\n\u2022 Bah, forgot to scale the distance $d$. Thanks, this sounds much more reasonable. If you could confirm using numerical methods with some explanation of how you did it, it would put it to rest and I'd definitely accept/award the bounty. \u2013\u00a0Bennett Gardiner May 19 '16 at 12:42\n\u2022 I just added some numerical simulation. However, it does not confirm my above conjecture. It may be because my code is quite dirty, but it may also be an error in the reasoning... Cheers \u2013\u00a0Urukann May 20 '16 at 4:37\n\u2022 Interesting, thanks for your efforts, looks like we still have some way to go. Can you use the code to check the Koch snowflake case I linked? It code provide a check on the code For the snowflake, $I = Ml^2/11$ where $l$ is the largest 'diameter' of the snowflake (tip to tip) I really felt like this method should work, I wonder where the error lies! \u2013\u00a0Bennett Gardiner May 20 '16 at 6:22\n\u2022 Hum, I checked with the Koch snowflake (my code was easier to modify than expected !): I get \"correct\" results. I noticed that I forgot to use the diameter instead of the radius, so we are off by another factor 4 I'm guessing... \u2013\u00a0Urukann May 20 '16 at 6:44\n\u2022 Sorry, this is getting confusing, but no: I calculated the MOI for a radius 1 Gosper Island to be 0.42856647032, that is to say, $I = \\frac{3Mr^2}{7} = \\frac{3Md^2}{7 \\cdot 4}$, meaning that we are missing a factor $\\frac{1}{7}$. \u2013\u00a0Urukann May 20 '16 at 6:58\n\nYes, the distance between those hexagon-like shapes in a hexagonal lattice is the same as the distance between centres of hexagons of the same area in a hexagonal lattice.\n\nThe reason is that \"in the long run\" the area per mesh must be the same. That is, a sufficiently large circular disk contains both Gosper islands and hexagons in a number approximately equal to area of the disk dividied by area of the shape (with an error in the order of magnitude proportional to the cirumference of the disk).\n\nThe area inside the hexagon is made of one whole island and exactly six thirds, by symmetry.\n\nHence for an island of unit area, the distance between centers is such that\n\n$$3\\frac{\\sqrt3}2d^2=3,$$\n\n$$d=\\sqrt{\\frac2{\\sqrt3}}.$$\n\nTo compute the moment of inertia, we will work with the radius of gyration and for an island of unit area.\n\n$$I=MR^2$$ where $M$ is the mass.\n\nFor the assembly in the figure, enlarging by the factor $\\sqrt7$ and using the axis theorem,\n\n$$7M7R^2=49I=I+6(I+Md^2),$$\n\ngiving\n\n$$R=\\frac d{\\sqrt7}=\\sqrt{\\frac{2}{7\\sqrt3}}=0.406149258\\cdots$$\n\nThis compares to the radius of gyration of a unit area disk,\n\n$$R'=\\frac1{\\sqrt{2\\pi}}=0.398942280\\cdots$$\n\nand that of a unit area hexagon,\n\n$$R''=\\sqrt{\\frac{10}{36\\sqrt3}}=0.400468569\\cdots$$\n\n\u2022 Interesting, so to scale this to an island with diameter $l$, would we simply scale this distance by $l$? \u2013\u00a0Bennett Gardiner May 20 '16 at 6:57\n\u2022 Or $\\sqrt{3/28}l$, like in the other poster's answer? I'm afraid I've confused myself quite a bit with the length scales. \u2013\u00a0Bennett Gardiner May 20 '16 at 7:00\n\u2022 @BennettGardiner: I have no idea of the value of the diameter $l$, but you don't need it for the computation of the moment. \u2013\u00a0Yves Daoust May 20 '16 at 7:03\n\u2022 I'm still trying to figure out your distance $d$. If that is the radius of gyration, what would the moment of inertia be? \u2013\u00a0Bennett Gardiner May 20 '16 at 7:32\n\u2022 My confusion is part of why I asked the question. I am inclined to believe your answer more now that you have posted a comparison with the hexagon of unit area, in my mind they should have been very similar. \u2013\u00a0Bennett Gardiner May 20 '16 at 14:18", "date": "2020-12-02 17:03:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1788715/what-is-the-moment-of-inertia-of-a-gosper-island", "openwebmath_score": 0.8783657550811768, "openwebmath_perplexity": 369.1899158322415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429595026214, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8525935066599467}}
{"url": "https://math.stackexchange.com/questions/1146548/number-of-strings-when-each-character-must-occur-even-times", "text": "# Number of strings, when each character must occur even times\n\nI was reading this question on programmers.StackExchange and wanted to try a combinatorics approach to solve the following problem:\n\nLet $S=\\{A,B,C,D\\}$ be a set of characters and $n$ a positive natural, find the number of strings of length $n$ composed of characters in $S$ such that each character occur even times.\n\nI tried to solve it by myself but I am stuck and I would appreciate a hint to solve this kind of problems. Also, there are similar questions on this site for not quite exactly the same problem, but I didn't find an answer that could help me.\n\n## Current attempt\n\nFirst, when $n$ is odd, the result must be zero. A string satisfying the desired property would necessarly have an even length, as a permutation of the concatenation of substrings of even lengths (each substring being a character from $S$ repeatead even times).\n\nNow, let's suppose that $n$ is even.\n\nIf we have a mapping $c$ from $S$ to $\\mathbb{N}$ assigning the number of occurence of each character (e.g. $c_A=0, c_B=2, c_C=4, c_D=0$), then the number of strings $N(c)$ we can produce with this mapping would be a k-permutation of characters, as we choose a string of size $n$ with $c_A$ times $A$, $c_B$ times $B$, etc. :\n\n$$N(c) = \\frac{n!}{c_A! c_B! c_C! c_D!}$$\n\nAlso, I can evaluate the number of such mappings $c$: this is equivalent to finding the number of strings where characters appear in order with varying occurences. For example, with $n=4$:\n\n$$\\begin{array}{c|cccc} & c_A & c_B & c_C &c_D \\\\ \\hline AAAA & 4 & 0 & 0 & 0 \\\\ AABB & 2 & 2 & 0 & 0 \\\\ AACC & 2 & 0 & 2 & 0 \\\\ \\dots\\\\ \\end{array}$$\n\nThe number of strings of size $n$ with an even number of each character is exactly the same as the number of strings of size $k=n/2$ with single characters (just imagine that AA, BB, CC, DD are characters). So, the number of mappings is the number of combinations of characters from $S$, with repetition, like the following ones with $k=2$:\n\nAA AB AC AD\nBB BC BD\nCC CD\nDD\n\n\nThat is given as a $k$-combination of 4 elements:\n\n$$N_{mapping}(k) = \\binom{|S|+k-1}{k} = \\binom {3+k}{k}$$\n\nThis is where I'am stuck. I wanted to combine this number of mappings with the number of permutations computed by each mapping, previously, but I can't ($N(c)$ depends on actual values of the mapping $c$). Also, I am sure there is a more straightforward approach, but I don't know how to proceed. Thanks for any help you could provide.\n\n\u2022 I might miss something, but polynomial coefficient seems to work: $\\binom{n}{2k} \\cdot \\binom{n-2k}{2j} \\cdot \\binom{n-2k -2j}{2m}$ \u2013\u00a0Alex Feb 13 '15 at 13:26\n\u2022 @Alex In your formula, do $k$ $j$ and $m$ stand for half the number of occurences of each character? I understand it as: choose first $2k$ elements from $n$, then $2j$ from $n-2k$, and so on...? Are $k$, $j$ and $m$ supposed to be known? Sorry, I have very little experience with this. \u2013\u00a0coredump Feb 13 '15 at 13:39\n\u2022 Yes, but it seems like not every $\\binom{2n}{2k}$ is even, e.g. $\\binom{6}{4}$ \u2013\u00a0Alex Feb 13 '15 at 14:56\n\nGenerating functions can be applied to this problem. However, since the order of letters in a string is important, exponential generating functions are appropriate (not ordinary ones). So, suppose the alphabet $S$ has $m$ letters and let $a_n$ be the number of strings of length $n$ over the alphabet $S$ with an even number of each letter. Let $g(x)=\\sum_{n\\ge0}\\frac{a_n}{n!}x^n$ be the exponential generating function of $\\langle a_n:n\\ge0\\rangle$. Then, $$g(x)=\\left(\\sum_{n\\textrm{ even}}\\frac{x^n}{n!}\\right)^m=\\left(\\frac{e^x+e^{-x}}2\\right)^m.$$\n\nIn the case of $m=4$, $g(x)=\\frac1{16}(e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x})$. Then, $a_n=\\left[\\frac{x^n}{n!}\\right]g(x)$ and if $n$ is even, $a_n=\\frac{n!}{16}\\left(\\frac{2\\cdot4^n}{n!}+\\frac{8\\cdot2^n}{n!}\\right)=\\frac18(4^n+4\\cdot2^n)$. In particular, $a_4=40$.\n\nNote: based on the form of the solution of $a_n$, I suspect there's a direct combinatorial argument (but I don't know it).\n\n\u2022 I am very impressed. I should have a look at generating functions. Thanks a lot, I'll study this and try to understand what it all means. \u2013\u00a0coredump Feb 13 '15 at 21:04\n\u2022 @coredump: If you want to take a look at generating functions, one excellent place to start is Herb Wilf's free book: math.upenn.edu/~wilf/gfologyLinked2.pdf. After reading his book, I was hooked. \u2013\u00a0Rus May Feb 13 '15 at 23:41\n\nLet $m=|S|$ be the number of distinct characters, and let $a_n$ be the number of strings in $S^n$ having an even number of each character in $S$. Let\n\n$$g(x)=\\sum_{n\\ge 0}\\frac{a_n}{n!}x^n$$\n\nbe the exponential generating function (egf) for $\\langle a_n:n\\in\\Bbb N\\rangle$. (The egf is wanted because the order of the characters matters.) The egf for the sequence given by $$a_n=\\begin{cases}0,&\\text{if }n\\text{ is odd}\\\\1,&\\text{if }n\\text{ is even}\\end{cases}$$ is\n\n$$\\sum_{n\\ge 0}\\frac1{(2n)!}x^{2n}=\\frac12(e^x+e^{-x})\\;,$$\n\nso\n\n$$g(x)=\\left(\\sum_{n\\ge 0}\\frac1{(2n)!}x^{2n}\\right)^m=\\frac1{2^m}(e^x+e^{-x})^m\\;.$$\n\nNow\n\n\\begin{align*} (e^x+e^{-x})^m&=\\sum_{k=0}^m\\binom{m}ke^{kx}e^{-(m-k)x}\\\\\\\\ &=\\sum_{k=0}^m\\binom{m}ke^{(2k-m)x}\\\\\\\\ &=\\sum_{k=0}^m\\binom{m}k\\sum_{n\\ge 0}\\frac{(2k-m)^n}{n!}x^n\\\\\\\\ &=\\sum_{n\\ge 0}\\frac1{n!}\\left(\\sum_{k=0}^m\\binom{m}k(2k-m)^n\\right)x^n\\;, \\end{align*}\n\nso\n\n$$a_n=\\frac1{2^m}\\sum_{k=0}^m\\binom{m}k(2k-m)^n\\;.\\tag{1}$$\n\nNote that $2(m-k)-m=m-2k=-(2k-m)$, so $(1)$ is indeed $0$ when $n$ is odd. When $n$ is even and positive we can rewrite $(1)$ as\n\n$$a_n=\\frac1{2^{m-1}}\\sum_{k=0}^{\\lfloor m/2\\rfloor}\\binom{m}k(m-2k)^n\\;.$$\n\nFor instance, if $m=4$, then\n\n\\begin{align*} a_{2n}&=\\frac18\\left(\\binom40(4-0)^{2n}+\\binom41(4-2)^{2n}+\\binom42(4-4)^{2n}\\right)\\\\\\\\ &=\\frac18\\left(4^{2n}+4\\cdot2^{2n}\\right)\\\\\\\\ &=\\frac18\\left(4^{2n}+4^{n+1}\\right)\\\\\\\\ &=2\\cdot4^{n-1}\\left(4^{n-1}+1\\right)\\;. \\end{align*}\n\n\u2022 May I ask you to explain how you get from the line headed by 'is' to the line headed by 'so'. Do you have a pointer to a book or literature where I could look this up? \u2013\u00a0Harald Feb 13 '15 at 21:10\n\u2022 @Harald: This PDF is a concise introduction to exponential generating functions. This PDF has a more general introduction to generating functions. Mikl\u00f3s B\u00f3na, Introduction to Enumerative Combinatorics, is very readable and has everything that you need on the subject. I suspect that the same goes for his A Walk Through Combinatorics, but I\u2019ve not actually seen it. \u2013\u00a0Brian M. Scott Feb 13 '15 at 21:26", "date": "2019-09-22 22:31:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1146548/number-of-strings-when-each-character-must-occur-even-times", "openwebmath_score": 0.8171554803848267, "openwebmath_perplexity": 222.6087697686127, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429595026214, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8525935049706848}}
{"url": "https://math.stackexchange.com/questions/1244967/determinant-of-a-5-%C3%97-5-matrix", "text": "# Determinant of a 5\u202f\u00d7\u202f5 matrix\n\nI have a little problem with a determinant.\n\nLet $A = (a_{ij}) \\in \\mathbb{R}^{(n, n)}, n \\ge 4$ with\n\n$$a_{ij} = \\begin{cases} x \\quad \\mbox{for } \\,i = 2, \\,\\, j \\ge 4,\\\\ d \\quad \\mbox{for } \\,i \\ge j, \\\\ 0 \\quad \\mbox{else.} \\end{cases}$$\n\nSo for example, if we choose $n = 5$, the matrix would look like this: $$A = \\begin{pmatrix} d &0 &0 &0 &0 \\\\ d &d &0 &x &x \\\\ d &d &d &0 &0 \\\\ d &d &d &d &0 \\\\ d &d &d &d &d \\\\ \\end{pmatrix}$$\n\nHow can I find the determinant of this matrix?\n\nMy first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A = T \\cdot R$. Then I wanted to use $$\\det(A) = \\det(T \\cdot R) = \\det(T) \\cdot \\det(R).$$\n\nto figure out the determinant. This would be something like\n\n$$d^n \\cdot \\det(R)$$\n\nBut is this approach even possible (I don't think so)? Is there any intelligent way of solving this? Thanks in advance.\n\n\u2022 What is the determinant for $n=4,5$? Maybe you could conjecture the result and then use induction. Apr 21 '15 at 14:26\n\u2022 BTW, what is a \"rest matrix\"? Apr 21 '15 at 14:30\n\nAdding a multiple of one row to another preserves the determinant. Subtract $x/d$ of the last row from the second to get $$\\begin{pmatrix} d &0 &0 &0 &0 \\\\ d-x &d-x &-x &0 &0 \\\\ d &d &d &0 &0 \\\\ d &d &d &d &0 \\\\ d &d &d &d &d \\\\ \\end{pmatrix}$$ and then add $x/d$ of the third row to the second row to get $$\\begin{pmatrix} d &0 &0 &0 &0 \\\\ d &d &0 &0 &0 \\\\ d &d &d &0 &0 \\\\ d &d &d &d &0 \\\\ d &d &d &d &d \\\\ \\end{pmatrix}.$$ This is lower triangular, so its determinant is the product of its diagonal, which is $d^5$. This all works for the $n$ by $n$ case, so the answer in general is $d^n$.\n\n\u2022 \"Row operations preserve determinant\" - except for multiplying a row by a constant, which multiplies the determinant by the same constant. Apr 21 '15 at 18:13\n\u2022 @user2357112 Or switching two rows, which multiplies the determinant by $(-1)$. Apr 21 '15 at 18:42\n\u2022 Two! Two operations. Apr 21 '15 at 18:44\n\u2022 @user2357112 You're right, I should be more clear. Apr 22 '15 at 7:13\n\nDevelop your matrix wrt the first row and get\n\n$$|A|=d\\begin{vmatrix}d&0&x&x\\\\d&d&0&0\\\\d&d&d&0\\\\d&d&d&d\\end{vmatrix}$$\n\nDevelop again wrt the first row but observe that when your pivot points are the $\\;x$'s you get determinant zero as there are two identical rows in each case, so we get\n\n$$d^2\\begin{vmatrix}d&0&0\\\\d&d&0\\\\d&d&d\\end{vmatrix}=d^5$$\n\nTry now some inductive argument based on this.\n\nMy first idea was to split this matrix into a product of a triangular matrix $T$ and a rest matrix $R$ so that $A=T\u22c5R$.\n\nThat's very much a way to do it. The technique is called LU Decomposition. It produces a lower and upper triangular matrix, allowing trivial determinate calculations. For this reason, you actually only need to find the diagonal elements to get your determinant.\n\nIn this case,\n\n$$A = \\begin{pmatrix} d &0 &0 &0 &0 \\\\ d &d &0 &x &x \\\\ d &d &d &0 &0 \\\\ d &d &d &d &0 \\\\ d &d &d &d &d \\\\ \\end{pmatrix} = \\begin{pmatrix} 1 &0 &0 &0 &0 \\\\ 1 &1 &0 &0 &0 \\\\ 1 &1 &1 &0 &0 \\\\ 1 &1 &1 &1 &0 \\\\ 1 &1 &1 &1 &1 \\\\ \\end{pmatrix} \\cdot \\begin{pmatrix} d &0 &0 &0 &0 \\\\ 0 &d &0 &x &x \\\\ 0 &0 &d &-x &-x \\\\ 0 &0 &0 &d &0 \\\\ 0 &0 &0 &0 &d \\\\ \\end{pmatrix}$$\n\nSo\n\n$$\\det A = \\det L \\cdot \\det U = 1^5 \\cdot d^5$$\n\nThe first few steps of the method used here (takes longer to texify than to do) are: $$\\begin{eqnarray} \\left[\\begin{smallmatrix} d &0 &0 &0 &0 \\\\ d &d &0 &x &x \\\\ d &d &d &0 &0 \\\\ d &d &d &d &0 \\\\ d &d &d &d &d \\\\ \\end{smallmatrix}\\right] &&= \\left[\\begin{smallmatrix} d &0 &0 &0 &0 \\\\ \\end{smallmatrix}\\right] \\left[\\begin{smallmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\\\ 1 \\\\ \\end{smallmatrix}\\right] + \\left[\\begin{smallmatrix} 0 &0 &0 &0 &0 \\\\ 0 &d &0 &x &x \\\\ 0 &d &d &0 &0 \\\\ 0 &d &d &d &0 \\\\ 0 &d &d &d &d \\\\ \\end{smallmatrix}\\right] \\\\ &&= \\left[\\begin{smallmatrix} d &0 &0 &0 &0 \\\\ \\end{smallmatrix}\\right] \\left[\\begin{smallmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\\\ 1 \\\\ \\end{smallmatrix}\\right] + \\left[\\begin{smallmatrix} 0 &d &0 &x &x \\\\ \\end{smallmatrix}\\right] \\left[\\begin{smallmatrix} 0 \\\\ 1 \\\\ 1 \\\\ 1 \\\\ 1 \\\\ \\end{smallmatrix}\\right] + \\left[\\begin{smallmatrix} 0 &0 &0 &0 &0 \\\\ 0 &0 &0 &0 &0 \\\\ 0 &0 &d &-x &-x \\\\ 0 &0 &d &d-x &-x \\\\ 0 &0 &d &d-x &d-x \\\\ \\end{smallmatrix}\\right] \\\\ &&= \\left[\\begin{smallmatrix} d &0 &0 &0 &0 \\\\ 0 &d &0 &x &x \\\\ \\end{smallmatrix}\\right] \\left[\\begin{smallmatrix} 1 & 0\\\\ 1 & 1\\\\ 1 & 1\\\\ 1 & 1\\\\ 1 & 1\\\\ \\end{smallmatrix}\\right] + \\cdots \\end{eqnarray}$$", "date": "2021-09-20 00:31:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1244967/determinant-of-a-5-%C3%97-5-matrix", "openwebmath_score": 0.8751622438430786, "openwebmath_perplexity": 142.6526266211056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429619433693, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.8525935003298524}}
{"url": "https://codereview.stackexchange.com/questions/78273/project-euler-6-in-java/78275", "text": "# Project Euler #6 in Java\n\nProject Euler #6:\n\nThe sum of the squares of the first ten natural numbers is,\n\n$1^2+2^2+ ... + 10^2 = 385$\n\nThe square of the sum of the first ten natural numbers is,\n\n$(1+2+ ... + 10)^2 = 55^2 = 3025$\n\nHence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 \u2212 385 = 2640$.\n\nFind the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.\n\nHere is my solution:\n\npublic class DifferenceFinder {\n\nprivate static final int MAX = 100;\n\npublic static void main(String[] args) {\nlong time = System.nanoTime();\nint result = MAX * (MAX + 1) / 2;\nresult *= result;\nfor(int i = 1; i <= MAX; i++) {\nresult -= i * i;\n}\ntime = System.nanoTime() - time;\nSystem.out.println(\"Result: \" + result\n+ \"\\nTime used to calculate in nanoseconds: \" + time);\n}\n\n}\n\n\nIt simply does:\n\n$$(1+2+ ... + 100)^2-1^2-2^2- ... -100^2$$\n\nOutput:\n\nResult: 25164150\nTime used to calculate in nanoseconds: 2231\n\nQuestions:\n\n1. Is the simplest solution the most efficient one?\n2. Does it smell?\n\n## Method extraction.\n\nEven for simple programs, having multiple responsibilities in a method is poor practice.\n\nConsider simple extractions, like:\n\nprivate static int squareOfSum(int limit) {\nint sum = (limit * (limit + 1)) / 2;\nreturn sum * sum;\n}\n\npublic static int sumOfSquares(int limit) {\nint sum = 0;\nfor (int i = 1; i <= limit; i++) {\nsum += i * i;\n}\nreturn sum;\n}\n\n\nint difference = squareOfSum(100) - sumOfSquares(100);\nSystem.out.printf(\"Difference is %d\\n\", difference);\n\n\nBy extracting functions, the logic becomes reusable, and discrete. Much better.\n\nAdditionally, it allows you to easily change the logic inside the methods to suite your algorithms, and use better algorithms like vnp suggests (+1 to that answer too).\n\n## Timing\n\nYour use of the nanos to time your code is misleading. The performance of the code is heavily related to how often the code runs, and, for this example, any performance measurement is unreliable....\n\n\u2022 Is the timing really misleading? When something takes ~2 microseconds, is it fair to say that startup costs don't count? \u2013\u00a0200_success Jan 22 '15 at 1:24\n\u2022 @200_success No, what's misleading is suggesting the code takes 2 microseconds when in another context it may take 0.2 microseconds. In addition, there are other things that are not being timed.... but my main concern is using a measure for the timing that is heavily context-dependent \u2013\u00a0rolfl Jan 22 '15 at 1:26\n\nA sum of first $n$ numbers is $\\frac{n(n+1)}{2}$. A sum of squares of first $n$ numbers is $\\frac{n(n+1)(2n+1)}{6}$. A difference in question is $\\frac{n^2(n+1)^2}{4} - \\frac{n(n+1)(2n+1)}{6} = \\frac{(n-2)(n-1)n(n+1)}{12}$.\n\nThis solution is likely the most efficient.\n\n\u2022 This doesn't seem to work. It gives the result 8165850. \u2013\u00a0TheCoffeeCup Jan 22 '15 at 3:10\n\u2022 @MannyMeng verified using the above Summation formulas and it work with much better timings. Just don't compute the difference with command denominator fraction, use the method suggested in the answer above \u2013\u00a0yadav_vi Jan 22 '15 at 11:38\n\nYou can linearize this problem, without losing readability. For example my solution from years back:\n\n    int size = 100;\nlong qos = (long) Math.pow(size * (size + 1) / 2, 2);\nlong soq = size * (size + 1) * (2 * size + 1) / 6;\n\nSystem.out.println(qos - soq);\n\n\nWhere qos and soq are known acronyms for me denoting square of sum and sum of squeares.\n\nNote however that my code actually had a potential bug called \"Possible Loss of Fraction\"\n\n    size * (size + 1) / 2\n\n\nshould be\n\n    size * (size + 1.0) / 2\n\n\nIn this example\n\n\u2022 size * (size + 1) would be of type int\n\u2022 size * (size + 1.0) would be of type double\n\nIn java\n\n\u2022 int / int you will get int\n\u2022 double / int you will get double\n\nIn Java you mostly deal with discrete mathematics while equations do not, note this fact as you start solving Euler problems. In addition remember to note that type double is an approximation.\n\nIt is also good to use best tool for the task, for example if you know the result will be\n\nYou can compute it with wolfram alpha (mathematica/wolfram language) using:\n\nLimit[(n (n - 2) (n - 1) (n + 1))/12, n -> 100]\n\n\u2022 In java 1.0 is same as 1d. Where 1f would result in 1.0, but be of type float (equal value, but not same type). \u2013\u00a0Margus Jan 23 '15 at 10:41", "date": "2021-04-13 22:51:17", "meta": {"domain": "stackexchange.com", "url": "https://codereview.stackexchange.com/questions/78273/project-euler-6-in-java/78275", "openwebmath_score": 0.5014592409133911, "openwebmath_perplexity": 1931.2355839371685, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854094395751, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.8525837109574645}}
{"url": "https://math.stackexchange.com/questions/1874504/class-number-of-mathbbq-sqrt319-and-hilbert-class-field", "text": "# Class Number of $\\mathbb{Q}(\\sqrt[3]{19})$ and Hilbert class field\n\nFinding the class number of $\\mathbb{Q}(\\sqrt[3]{19})$ is an exercise from Marcus 'Number Field'. This question was uploaded by some other user, but it was removed by now. I have worked on details and have some more questions.\n\nFollowing are the steps in the exercise:\n\nLet $K=\\mathbb{Q}(\\alpha)$, $R=\\mathcal{O}_K$, and $3R=P^2Q$, where $\\alpha=\\sqrt[3]{19}$, and $P$, $Q$ are prime ideals in $R$.\n\n(a) Prove that the ideal class group is cyclic, generated by the class containing $P$.\n\n(b) Prove that the number of ideal classes is a multiple of $3$.\n\n(c) Prove that there are either three or six ideal classes.\n\n(d) Prove that there are three ideal classes. (Suggestion: Suppose there were six. Show that none of the ideals $J$ with $||J||\\leq 9$ are in the same class with $P^3$. )\n\nAfter solving this problem, it seems that the calculations from my solution to (a) are enough to conclude that the class number is $3$.\n\nTo do (a), we find the Minkowski's bound: $$\\frac{3!}{3^3}\\frac 4{\\pi} \\sqrt{1083} \\approx 9.3$$ This shows that each ideal class in the ideal class group contains an ideal $J$ with $||J||\\leq 9$. Also, each ideal class generating the class group contains a prime ideal lying above $2$, $3$, $5$, $7$.\n\nSo, we find prime factorizations of $2R$, $3R$, $5R$, $7R$: $$2R=P_1P_2,\\ \\ \\ 3R=P^2 Q,$$ $$5R= P_3P_4, \\ \\ \\ 7R = P_5,$$ with $P_1=(2, \\alpha-1)$, $P=(3, \\frac{\\alpha^2+\\alpha+1}3)$, $P_3=(5,\\alpha+1)$.\n\nWith a help of SAGE, I found that $$P_1= \\frac{\\alpha-1}3 P, \\ \\ \\ P_3= \\frac{-\\alpha+4}3 P, \\ \\ \\ P_5=\\frac{7\\alpha+14}9 P^3.$$\n\nThen the prime ideals appear above (as factorizations of $2$, $3$, $5$, $7$) belong to one of the three ideal classes of $P$, $P^2$, $I$ (principal).\n\nMy questions are\n\n1. Are these calculations enough to conclude that the ideal class group has order $3$?\n\n2. If the answer to 1 is no, then how are the rest of parts solved?\n\n3. How do we determine the Hilbert class field of $\\mathbb{Q}(\\sqrt[3]{19})$?\n\n\u2022 And how are you concluding that all the ideal class aren't in fact the principal ideal class? \u2013\u00a0John M Jul 29 '16 at 2:36\n\u2022 I have to prove that the ideal class of $P$ is not principal, so the class of $P$ has order $3$. Okay, so part (b) is really necessary to do that. I see. \u2013\u00a0Sungjin Kim Jul 29 '16 at 2:42\n\u2022 Except for showing that at least one ideal is nonprincipal, I think you\u2019re OK. Since $P_5=(7)$, principal, and also equals $(z)P^3$ for a number $z$, $P^3$ is principal. $P_2$ has norm $4$, but since $P_1P_2\\sim1$, you have $P_2\\sim P^2$. $Q$ also hs norm $3$, but it\u2019s equiv to $P$. $P_3$ has norm $5$, so $P_4$ has norm $25$, so it\u2019s not a problem. Seems to me that you\u2019ve covered all bases for the ideal class group to be of order $3$. \u2013\u00a0Lubin Jul 29 '16 at 2:46\n\u2022 My comment above may be worthless, but for the Class Field, I\u2019d first look at the cubic subfield of $\\Bbb Q(\\zeta_{19})$. It\u2019s unramified outside $19$, and you\u2019d have a good chance of showing that its compositum with $K$ is unramified above $19$, over $K$. \u2013\u00a0Lubin Jul 29 '16 at 3:21\n\nIt turns out that the answer to 1 is no as commented by @JohnM. We need to prove that $P$ is not principal. To do this, assume that $P=\\beta R$. Then $||P||=||\\beta||=3$. So, now the question becomes whether there exists an element $\\beta\\in R$ with $||\\beta||=3$.\n\nSince any element in R can be expressed as $\\beta=a+b\\alpha+c\\alpha^2$ with $a, b, c\\in \\mathbb{Q}$ and $3a, 3b, 3c \\in \\mathbb{Z}$, $3a\\equiv 3b\\equiv 3c \\ \\mathrm{mod} \\ 3$, we see that $$||\\beta||=3=a^3+19b^3+19^2c^3-3\\cdot 19 abc$$ and $$27\\cdot 3 =81= (3a)^3 + 19(3b)^3 + 19^2 (3c)^3 - 3 \\cdot 19 (3a)(3b)(3c)\\equiv (3a)^3 \\ \\mathrm{mod} \\ 19$$ However, $81 \\equiv 5 \\ \\mathrm{mod} \\ 19$ is not a cubic residue modulo $19$ (list of cubic residues mod $19$ is $0, 1, 7, 8, 11, 12, 18$). Therefore, there is no element $\\beta\\in R$ of norm $3$. This proves that $P$ is not principal.\n\nThen the rest of argument will follow as @Lubin commented. Thus, we needed to solve (b) to conclude, but (c) and (d) are not necessary. This answers my question 2.\n\nTo find the Hilbert class field, we use a cubic subfield $K'$ of the cyclotomic field $\\mathbb{Q}(\\zeta_{19})$ as commented by @Lubin. This method was already discussed by @franzlemmermeyer in the answer to this question.\n\nLet $L$ be composite field of $K'$ and $K=\\mathbb{Q}(\\alpha)$. Let $\\mathcal{P}$ be a prime in $L$. If $\\mathcal{P}$ lies over rational prime $p\\neq 3, 19$, then $p$ is unramified in both extensions $K$ and $K'$. Thus, the prime $\\mathcal{P}\\cap K$ lying under $\\mathcal{P}$ is unramified in $L$.\n\nSuppose that $\\mathcal{P}$ lies over rational prime $3$. Since the rational prime $3$ is inert in $K'$, the residue field degree must be $3$. By the decomposition $3R=P^2Q$, the residue field degree for $P$ and $Q$ over rational prime $3$ must be both $1$. This gives the residue field degree for $\\mathcal{P}$ over both $P$ and $Q$ must be $3$, yielding that the ramification indices for $\\mathcal{P}$ over both $P$ and $Q$ are $1$.\n\nNow assume that $\\mathcal{P}$ is over rational prime $19$. As @franzlemmermeyer answered in the above link, we use Abhyankar's lemma. Since $19$ is totally ramified in both $K$ and $K'$, we have by Abhyankar's lemma that the ramification index for $\\mathcal{P}$ over $\\mathcal{P}\\cap K$ is $1$.\n\nHence, the extension $L$ over $K$ is unramified over $K$, yielding that $L$ is the Hilbert class field of $K$. This answers my question 3.\n\nAcknowledgement @JohnM, @Lubin, Thank you very much for the helpful comments with great insights.", "date": "2021-07-28 18:14:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1874504/class-number-of-mathbbq-sqrt319-and-hilbert-class-field", "openwebmath_score": 0.9061337113380432, "openwebmath_perplexity": 135.38083505015564, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.969785409439575, "lm_q2_score": 0.8791467627598857, "lm_q1q2_score": 0.8525837032805726}}
{"url": "https://mathematica.stackexchange.com/questions/59691/listcontourplot-is-blank", "text": "# ListContourPlot is blank\n\nThis is probably very simple. I want to do a simple ListContourPlot.\n\ncontourData = Table[ myFunc[x,y], {x,0.1, 2.9}, {y,0.1, 2.9}];\nListContourPlot[contourData]\n\n\nThis works without a problem.\n\nNow, I want to include x and y values so that I can show the values corresponding to specific values of x and y. I modify the Table statement to include x and y values.\n\ncontourData = Table[ {x,y, myFunc[x,y]}, {x,0.1, 2.9}, {y,0.1, 2.9}];\nListContourPlot[contourData]\n\n\nMy contour plot is now blank. I'm guessing it is a problem with the Table statement. The maximum minus minimum over this range is about 220. As this is a function of Pythagorean distance, the values vary (so I don't think it is a case of essentially equal values over the ranges of x and y.\n\nWhat is the correct way to generate data for a ListContourPlot?\n\n\u2022 Try ListContourPlot[Flatten[contourData, 1]]. Sep 14 '14 at 19:04\n\u2022 That was it exactly! Thank you most kindly! Sep 14 '14 at 19:10\n\nAs an alternative to @ b.gatessucks excellent comment you might consider DataRange:\nListContourPlot[Table[Sin[x + y], {x, 0, 3, 0.1}, {y, 0, 3, 0.1}],", "date": "2021-11-29 22:06:36", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/59691/listcontourplot-is-blank", "openwebmath_score": 0.4905970096588135, "openwebmath_perplexity": 1313.6406000174877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.962673111584966, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8525736138830151}}
{"url": "https://www.jiskha.com/questions/514863/find-the-extreme-values-of-the-function-for-the-given-intervals-2x-3-21x-2-72x-a", "text": "# calculus\n\nFind the extreme values of the function for the given intervals.\n\n2x^3 \u2013 21x^2 + 72x\n\na) [0,5]\n\nb) [0,4]\n\nI have both the minimum values for both intervals which is 0. But I am unable to find the maximum. please help!!!\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n3. \ud83d\udc41 184\n1. You have probably found two extrema where f'(x)=0.\nTo identify whether the extremum is a maximum or minimum, use the second derivative test.\nIf the second derivative is positive, the extremum is a minimum. If the second derivative is negative, the extremum is a maximum.\n\nHere the extrema are at x=3 and x=4.\n\nThe second derivative, f\"(x)=12x-42,\nso for example, f2(3)=-6, so x=3 is a maximum.\n\nUse your calculator to plot the graph of the function to help you solve these problems. After some time, you will be able to visualize the graph from the equation, even without the calculator.\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n2. I tried 3 for both (a and b) as a maximum and didn't work....\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n3. It is important to know that extreme values include local maxima, local minima and end-points of the given interval.\n\nThe question requires the extreme values of the function, so it is the value of the function that counts.\n\nTo find extreme values, we calculate the critical points (where f'(0)=0 and where f'(0) does not exist) as well as the value of the function at end-points.\n\nGiven\nf(x)=2x^3 \u2013 21x^2 + 72x\n\n1. We calculate the points where f'(x)=0, and we have determined that f(3)=81 is a local maximum and f(4)=80 is a local minimum.\n\n2. For [0,4], the values of the function at end-points are:\nf(0)=zero, and f(4)=80.\nSo the list of values to consider are:\n(0,zero)\n(3,81)\n(4,80)\nWe conclude that the extreme minimum is zero, and the extreme maximum is 81.\n\n3. For [0,5], the values of the function at end-points are f(0)=zero, and f(5)=85.\nSo the list of values to consider for extreme values are:\n(0,zero)\n(3,81)\n(4,80)\n(5,85)\nWe choose zero as the extreme minimum, and 85 as the extreme maximum.\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n\n## Similar Questions\n\n1. ### calculus\n\n2. Consider the function f defined by f(x)=(e^X)cosx with domain[0,2pie] . a. Find the absolute maximum and minimum values of f(x) b. Find the intervals on which f is increasing. c. Find the x-coordinate of each point of\n\n2. ### trigonometry\n\nfind the exact values of s in the given intervals that have the given circular function values. 1. [pi/2,pi]; sin s=square root 2/2 2. [pi/2,pi]; cos s=-square root 3/2\n\n3. ### Algebra\n\nComplete the table for each function. 1. f(x) = \u221ax The x values are 0, 1,4 and 9. The corresponding y values that I got are 0, 1, 2 and 3. 2. g(x) = -1/4\u221a x The x values are 0, 1, 4, and 9. The corresponding y values that I\n\n4. ### calculus\n\nFind the break-even point for the firm whose cost function C and revenue function R are given. C(x) = 16x + 10,000; R(x) = 21x\n\n1. ### Calculus - Functions?\n\n#1. A cubic polynomial function f is defined by f(x) = 4x^3 +ax^2 + bx + k where a, b and k are constants. The function f has a local minimum at x = -1, and the graph of f has a point of inflection at x= -2 a.) Find the values of\n\n2. ### Calculus\n\nThe function f is continuous on the interval [4, 15], with some of its values given in the table above. Estimate the average value of the function with a Right Rectangle Approximation, using the 4 intervals between those given\n\n3. ### Calculus\n\nThe function is continuous on the interval [10, 20] with some of its values given in the table above. Estimate the average value of the function with a Trapezoidal Sum Approximation, using the intervals between those given points.\n\n4. ### Calculus\n\n1. The function is continuous on the interval [10, 20] with some of its values given in the table above. Estimate the average value of the function with a Left Hand Sum Approximation, using the intervals between those given\n\n1. ### Calculus\n\nConsider the function of defined by f(x)=x^3/3-4/x a. Find the X values for the points of inflection. b. Determine the intervals where the function of f is concave up and concave down.\n\n2. ### Calculus help\n\nThe function is continuous on the interval [10, 20] with some of its values given in the table above. Estimate the average value of the function with a Right Hand Sum Approximation, using the intervals between those given points.\n\n3. ### Calculus Finals Review sheet!! Explanation needed\n\nHere is a graph of the derivative y' of a continuous, differentiable function. For approximately what values of x between \u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u21225 and 5 does the original function y have inflection points? Find limit as x approaches 3.5\n\n4. ### Calculus\n\nConsider the function f(x)=ln(x)/x^6. For this function there are two important intervals: (A,B] and [B,\u221e) where A and B are critical numbers or numbers where the function is undefined. Find A Find B For each of the following", "date": "2020-10-01 15:52:04", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/514863/find-the-extreme-values-of-the-function-for-the-given-intervals-2x-3-21x-2-72x-a", "openwebmath_score": 0.8386550545692444, "openwebmath_perplexity": 656.1968288317914, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9626731105140615, "lm_q2_score": 0.8856314813647587, "lm_q1q2_score": 0.8525736129345883}}
{"url": "https://math.stackexchange.com/questions/950712/how-to-compute-int-infty-infty-exp-left-fracx2-13x-12611x2-rig/950848", "text": "# How to compute $\\int_{-\\infty}^\\infty\\exp\\left(-\\frac{(x^2-13x-1)^2}{611x^2}\\right)\\ dx$\n\n$$\\int_{-\\infty}^\\infty\\exp\\left(-\\frac{(x^2-13x-1)^2}{611x^2}\\right)\\ dx$$\n\nWolframAlpha gives a numerical answer of $43.8122$, which appears to be $\\sqrt{611\\pi}$. And playing with that, it seems that replacing $611$ with $a$ just gives $\\sqrt{a\\pi}$. My trouble is that the stuff in the exponential always seems to be just a big mess, and I haven't been able to get it into a form I can understand or deal with.\n\nI would greatly appreciate seeing a method for solving this integral.\n\n\u2022 I have a strong feeling that you can use Feynman's trick on this, but I am still trying to find the correct one. \u2013\u00a0UserX Sep 29 '14 at 7:20\n\u2022 How can you be sure that $\\sqrt{a\\pi}$ is not just a good approximation ? Could you produce some values with more digits ? \u2013\u00a0Claude Leibovici Sep 29 '14 at 7:34\n\u2022 @ClaudeLeibovici I'm not entirely sure if it's exactly $\\sqrt{a\\pi}$, but for every value of $a$ I've tried, WolframAlpha always gives a number very close to $\\sqrt{a\\pi}$. I was hoping this fact might help with solving the integral. \u2013\u00a0user137794 Sep 29 '14 at 7:40\n\u2022 Working with very high precision, I can confirm that the result is exactly $\\sqrt{a\\pi}$ what you would get with $$\\int_{-\\infty}^\\infty\\exp\\left(-\\frac{x^2}{a}\\right)\\ dx=\\sqrt{a\\pi}$$ \u2013\u00a0Claude Leibovici Sep 29 '14 at 8:12\n\nLet $\\displaystyle\\;u(x) = \\frac{x^2-13x-1}{x}\\;$. As $x$ varies over $\\mathbb{R}$, we have\n\n\u2022 u(x) increases monotonically from $-\\infty$ at $-\\infty$ to $+\\infty$ at $0^{-}$.\n\u2022 u(x) increases monotonically from $-\\infty$ at $0^{+}$ to $+\\infty$ at $+\\infty$.\n\nThis means as $x$ varies, $u(x)$ covered $(-\\infty,\\infty)$ twice.\n\nLet $x_1(u) < 0$ and $x_2(u) > 0$ be the two roots of the equation for a given $u$:\n\n$$u = u(x) = \\frac{x^2-13x-1}{x} \\quad\\iff\\quad x^2 - (13+u)x - 1 = 0$$ we have $$x_1(u) + x_2(u) = 13 + u \\quad\\implies\\quad \\frac{dx_1}{du} + \\frac{dx_2}{du} = 1.$$ From this, we find\n\n\\begin{align} \\int_{-\\infty}^\\infty e^{-u(x)^2/611} dx &= \\left( \\int_{-\\infty}^{0^{-}} + \\int_{0^{+}}^{+\\infty}\\right) e^{-u(x)^2/611} dx\\\\ &= \\int_{-\\infty}^{\\infty} e^{-u^2/611}\\left(\\frac{dx_1}{du} + \\frac{dx_2}{du}\\right) du\\\\ &= \\int_{-\\infty}^{\\infty} e^{-u^2/611} du\\\\ &= \\sqrt{611\\pi} \\end{align}\n\n\u2022 This is a new technique for me (+1), I think it can be generalised. If so, what is the condition? Say $u(x)=\\dfrac{px^2+qx+r}{sx}$ \u2013\u00a0Anastasiya-Romanova \u79c0 Sep 30 '14 at 10:44\n\u2022 @Anastasiya-Romanova algebraically, $pr < 0$. Geometrically, this ensure the monotonicity of the $u(x)$ and $u$ cover $(-\\infty,\\infty)$ twice. \u2013\u00a0achille hui Sep 30 '14 at 12:47\n\u2022 +1 for its potent to be generalized (to a higher degree, maybe?). \u2013\u00a0Sangchul Lee Sep 30 '14 at 19:43\n\u2022 @sos440 the generalization is essentially the \"Added\" part of orangeskid answer. \u2013\u00a0achille hui Sep 30 '14 at 19:59\n\nHINT:\n\nFor $a$ fixed\n\n$\\int_{-\\infty}^\\infty\\exp\\left(-\\frac{(x^2+sx-b)^2}{a x^2}\\right)\\ dx$\n\nis constant in $s$ and $b\\ge 0$.\n\n$\\bf{Added:}$\n\nThe function $\\frac{x^2 + s x - b}{x} = x - \\frac{b}{x} + s$ invariates the Lebesgue measure as @achille hui: showed in his answer.\n\nLet $n \\in \\mathbb{N}$ $\\alpha_1$, $\\ldots$, $\\alpha_n$ distinct real numbers, $\\rho_1$, $\\ldots$, $\\rho_n$ $>0$ and $\\beta \\in \\mathbb{R}$. The function\n\n$$\\phi(x) = x - \\sum_{i=1}^n \\frac{\\rho_i}{x- \\alpha_i} -\\beta$$\n\ninvariates the Lebesgue measure on $\\mathbb{R}$.\n\nLemma: For any $a \\in \\mathbb{R}$ the equation\n\n\\begin{eqnarray*} x- \\sum_{i=1}^n \\frac{\\rho_i}{x- \\alpha_i} -\\beta = u \\end{eqnarray*} has $n+1$ distinct real root with sum $u + \\sum_i \\alpha_i + \\beta$. Use Viete.\n\nLemma: Let $I$ an interval in $\\mathbb{R}$ of length $l$. Then the preimage $\\phi^{-1} (I)$ is a union of $n+1$ disjoint intervals of total length $l$.\n\nConsequence: $$\\int_{\\mathbb{R}} (f\\circ \\phi)\\, d\\,\\mu = \\int_{\\mathbb{R}} f\\ d\\mu$$\n\nComposing two rational maps that invariate the measure gets a third one. They will have singularities in general.\n\nFor $f(x) = e^{-\\frac{x^2}{a}}$ the composition $f (\\phi(x))$ is still smooth due to the rapid decay at $\\infty$ of $e^{-\\frac{x^2}{a}}$.\n\n\u2022 That is remarkably simple and straightforward. +1 \u2013\u00a0Tom-Tom Sep 29 '14 at 9:56\n\u2022 Would you be able to elaborate on how you concluded this? \u2013\u00a0user137794 Sep 29 '14 at 11:45\n\u2022 @user137794: Added more detail. \u2013\u00a0orangeskid Sep 30 '14 at 6:43\n\u2022 @Tom-Tom: Thanks! Added more detail. \u2013\u00a0orangeskid Sep 30 '14 at 16:36\n\u2022 For $f(x) = e^{-\\frac{x^2}{a}}$ the function $f(\\phi(x))$ is still smooth. \u2013\u00a0orangeskid Sep 30 '14 at 20:19\n\nAdding another solution owing to a friend of mine.\n\nThrough some algebra, the integral is equivalent to\n\n$$\\int_{-\\infty}^\\infty \\exp\\left(-\\frac1{611}\\left((x-x^{-1})-13\\right)^2\\right)\\ dx$$\n\nThen using the following identity\n\n$$\\int_{-\\infty}^\\infty f(x-x^{-1})\\ dx = \\int_{-\\infty}^\\infty f(x)\\ dx$$\n\nWe have\n\n\\begin{align} &\\int_{-\\infty}^\\infty \\exp\\left(-\\frac1{611}\\left((x-x^{-1})-13\\right)^2\\right)\\ dx\\\\ =&\\int_{-\\infty}^\\infty \\exp\\left(-\\frac1{611}(x-13)^2\\right)\\ dx\\\\ =&\\int_{-\\infty}^\\infty \\exp\\left(-\\frac1{611}x^2\\right)\\ dx\\\\ =&\\sqrt{611\\pi} \\end{align}\n\nUsing identity in @user137794's answer: $$\\int_{-\\infty}^\\infty f(x-x^{-1})\\ dx = \\int_{-\\infty}^\\infty f(x)\\ dx$$ where the complete proof can be seen here. The problem can be generalised to evaluate\n\n$$\\int_{-\\infty}^\\infty \\exp\\left(-\\frac{(x^2-bx-1)^2}{ax^2}\\right)\\ dx = \\sqrt{a\\pi}$$\n\nProof:\n\nIt's easy to see that $\\dfrac{(x^2-bx-1)^2}{ax^2}=\\dfrac{1}{a}\\left(x-x^{-1}-b\\right)^2$, then \\begin{align} \\int_{-\\infty}^\\infty \\exp\\left(-\\frac{(x^2-bx-1)^2}{ax^2}\\right)\\ dx &=\\int_{-\\infty}^\\infty \\exp\\left(-\\frac{(x-x^{-1}-b)^2}{a}\\right)\\ dx\\\\ &=\\int_{-\\infty}^\\infty \\exp\\left(-\\frac{(x-b)^2}{a}\\right)\\ dx\\\\ &=\\int_{-\\infty}^\\infty \\exp\\left(-\\frac{y^2}{a}\\right)\\ dy\\\\ &=\\sqrt{a}\\int_{-\\infty}^\\infty \\exp\\left(-z^2\\right)\\ dz\\\\ &=\\sqrt{a\\pi} \\end{align} Therefore $$\\int_{-\\infty}^\\infty \\exp\\left(-\\frac{(x^2-13x-1)^2}{611x^2}\\right)\\ dx = \\sqrt{611\\pi}$$\n\n\u2022 Nice generalization! It's interesting, that $b$ does not matter, on the other hand $-1$ is how important! \u2013\u00a0user153012 Oct 1 '14 at 12:07", "date": "2020-04-02 10:06:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/950712/how-to-compute-int-infty-infty-exp-left-fracx2-13x-12611x2-rig/950848", "openwebmath_score": 0.9808382391929626, "openwebmath_perplexity": 725.1819889610224, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639694252315, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8525248369503058}}
{"url": "https://math.stackexchange.com/questions/395139/combinatorial-proof-of-a-stirling-number-identity", "text": "# Combinatorial proof of a Stirling number identity.\n\nConsider the identity $$\\sum_{k=0}^n (-1)^kk!{n \\brace k} = (-1)^n$$ where ${n\\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity $$\\sum_{k=0}^n(-1)^k\\binom{n}{k} = 0$$ which essentially states that the number of even subsets of a set is equal to the number of odd subsets.\n\nNow there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?\n\nAs an added note, there is a similar identity $$\\sum_{k=1}^n(-1)^k(k-1)!{n\\brace k}=0$$ A combinatorial interpretation of this one would also be appreciated.\n\n\u2022 May inclusion exclusionn principle will help? \u2013\u00a0Norbert May 18 '13 at 4:21\n\nPerhaps there is something relating odd ordered partitions with even ordered partitions?\n\nThere is indeed. Let's try to construct an involution $T_n$, mapping odd ordered partitions of $n$-element set to even and vice versa: if partition has part $\\{n\\}$, move $n$ into previous part; otherwise move $n$ into new separate part.\n\nExample: $(\\{1,2\\},\\{\\mathbf{5}\\},\\{3,4\\})\\leftrightarrow(\\{1,2,\\mathbf{5}\\},\\{3,4\\})$.\n\nThis involution is not defined on partitions of the form $(\\{n\\},\\ldots)$, but for these partitions one can use previous involution $T_{n-1}$ and so on.\n\nExample: $(\\{5\\},\\{4\\},\\{1,2\\},\\{\\mathbf{3}\\})\\leftrightarrow(\\{5\\},\\{4\\},\\{1,2,\\mathbf{3}\\})$.\n\nIn the end only partition without pair will be $(\\{n\\},\\{n-1\\},\\ldots,\\{1\\})$. So our (recursively defined) involution gives a bijective proof of $\\sum_{\\text{k is even}}k!{n \\brace k}=\\sum_{\\text{k is odd}}k!{n \\brace k}\\pm1$ (cf. 1, 2).\n\nUpd. As for the second identity, the involution $T_n$ is already defined on all cyclically ordered partitions, so $\\sum_{\\text{k is even}}(k-1)!{n \\brace k}=\\sum_{\\text{k is odd}}(k-1)!{n \\brace k}$.\n\nP.S. I can't resist adding that $k!{n \\brace k}$ is the number of $(n-k)$-dimensional faces of an $n$-dimensional convex polytope, permutohedron (the convex hull of all vectors formed by permuting the coordinates of the vector $(0,1,2,\\ldots,n)$). So $\\sum(-1)^{n-k}k!{n \\brace k}=1$ since it's the Euler characteristic of a convex polytope.\n\n\u2022 much more combinatorial (+1) \u2013\u00a0robjohn May 23 '13 at 13:30\n\u2022 Beautiful, thank you. \u2013\u00a0EuYu May 23 '13 at 18:12\n\nThese are not combinatorial interpretations, but they are simple.\n\nThe defining equation for Stirling numbers of the second kind is $$\\sum_{k=0}^n\\begin{Bmatrix}n\\\\k\\end{Bmatrix}\\binom{x}{k}k!=x^n\\tag{1}$$ That is, Stirling numbers of the second kind tell how to write monomials as a combination of falling factorials (or combinatorial polynomials).\n\nNoting that $\\displaystyle\\binom{-1}{k}=(-1)^k$ and setting $x=-1$ yields \\begin{align} \\sum_{k=0}^n\\begin{Bmatrix}n\\\\k\\end{Bmatrix}(-1)^kk!= \\sum_{k=0}^n\\begin{Bmatrix}n\\\\k\\end{Bmatrix}\\binom{-1}{k}k!=(-1)^n \\end{align}\n\nSince $\\displaystyle\\binom{x}{k}=\\binom{x-1}{k-1}\\frac{x}{k}$ and $\\begin{Bmatrix}n\\\\0\\end{Bmatrix}=0$ for $n\\ge1$, we can rewrite $(1)$ as $$\\sum_{k=1}^n\\begin{Bmatrix}n\\\\k\\end{Bmatrix}\\binom{x-1}{k-1}(k-1)!=x^{n-1}\\tag{2}$$ Setting $x=0$ yields $$\\sum_{k=1}^n\\begin{Bmatrix}n\\\\k\\end{Bmatrix}(-1)^{k-1}(k-1)!=0^{n-1}$$ where $0^0=1$ for the case $n=1$.\n\n\u2022 Certainly the second part is supposing $n>0$, so that the term for $k=0$ can be dropped. However it fails to accommodate to the fact (even though some wish to deny it) that $0^0=1$. Stated more directly, it fails for $n=1$. \u2013\u00a0Marc van Leeuwen May 23 '13 at 14:56\n\u2022 @MarcvanLeeuwen: Good point. I got rid of the exponent when $n=1$ when I shouldn't have. \u2013\u00a0robjohn May 23 '13 at 15:54\n\nFor the sake of completeness I include a treatment using generating functions. The exponential generating function of the Stirling numbers of the second kind is $$G(z, u) = \\exp(u(\\exp(z)-1))$$ so that $${n \\brace k} = n! [z^n] \\frac{(\\exp(z) - 1)^k}{k!}.$$ It follows that $$\\sum_{k=0}^n (-1)^k k! {n \\brace k} = n! [z^n] \\sum_{k=0}^n (1-\\exp(z))^k = n! [z^n] \\sum_{k=0}^\\infty (1-\\exp(z))^k,$$ where the last equality occurs because the series for $(1-\\exp(z))^k$ starts at degree $k.$ But this is just $$n! [z^n] \\frac{1}{1-(1-\\exp(z))} = n! [z^n] \\exp(-z) = (-1)^n,$$ showing the result.", "date": "2019-05-21 10:47:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/395139/combinatorial-proof-of-a-stirling-number-identity", "openwebmath_score": 0.9537001848220825, "openwebmath_perplexity": 229.68116166236786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639669551474, "lm_q2_score": 0.8774767874818408, "lm_q1q2_score": 0.852524828556916}}
{"url": "http://www.intermash.ua/8nxz7/66kmp25/e0l9j.php?79eef7=least-squares-function-r", "text": "the residuals, that is response minus fitted values if \"g\" is a \u00e2\u0080\u00a6 weights. For practical purposes it might be preferable to use a nonlinear least squares approach (e.g., the nls function). For example, if a student had spent 20 hours on an essay, their predicted score would be 160, which doesn\u00e2\u0080\u0099t really make sense on a typical 0-100 scale. When present, the objective function is weighted least squares. In literal manner, least square method of regression minimizes the sum of squares of errors that could be made based upon the relevant equation. When the \"port\" algorithm is used the objective function value printed is half the residual (weighted) sum-of-squares. R-bloggers ... BVLS is implemented in the bvls() function \u00e2\u0080\u00a6 Each classroom has a least squared mean of 153.5 cm, indicating the mean of classroom B was inflated due to the higher proportion of girls. We want to build a model for using the feature. The object of class \"gmm\" is a list containing at least: coefficients $$k\\times 1$$ vector of coefficients. residuals. object: an object inheriting from class \"gls\", representing a generalized least squares fitted linear model.. model: a two-sided linear formula object describing the model, with the response on the left of a ~ operator and the terms, separated by + operators, on the right.. model. Moreover, we have studied diagnostic in R which helps in showing graph. If you have any suggestion or feedback, please comment below. Also, we have learned its usage as well as its command. Linear model Background. Now, you are an expert in OLS regression in R with knowledge of every command. And if the data-simulating function does not have the correct form (for example, if the zeroth order term in the denominator is not 1), the fitted curves can be completely wrong. Specifically, I am looking for something that computes intercept and slope. The least squares regression method follows the same cost function as the other methods used to segregate a mixed \u00e2\u0080\u00a6 Least squares method, also called least squares approximation, in statistics, a method for estimating the true value of some quantity based on a consideration of errors in observations or measurements. an optional numeric vector of (fixed) weights. Disadvantages of least-squares regression *As some of you will have noticed, a model such as this has its limitations. The functions 'summary' is used to obtain and print a summary of the results. The most basic way to estimate such parameters is to use a non-linear least squares approach (function nls in R) which basically approximate the non-linear function using a linear one and iteratively try to find the best parameter values . subset. In linear least squares the model contains equations which are linear in the parameters appearing in the parameter vector , so the residuals are given by = \u00e2\u0088\u0092. Note that the following example uses a linear model with the lm function. In non-linear regression the analyst specify a function with a set of parameters to fit to the data. In ordinary least squares (OLS), one seeks \u00e2\u0080\u00a6 Continue reading \u00e2\u0086\u0092 Imagine that one has a data matrix consisting of observations, each with features, as well as a response vector . an optional vector specifying a subset of observations to be used in the fitting process. 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In OLS regression in R with knowledge of every command have studied diagnostic in R which helps showing!\n2020 least squares function r", "date": "2021-04-19 21:29:30", "meta": {"domain": "intermash.ua", "url": "http://www.intermash.ua/8nxz7/66kmp25/e0l9j.php?79eef7=least-squares-function-r", "openwebmath_score": 0.7120915055274963, "openwebmath_perplexity": 1114.914286161403, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992905050948, "lm_q2_score": 0.8807970654616711, "lm_q1q2_score": 0.8525228547393211}}
{"url": "https://flyingcoloursmaths.co.uk/regions-of-a-circle/", "text": "On a recent MathsJam Shout, an Old Chestnut appeared (in this form, due to @jamestanton):\n\nIf you\u2019ve not seen it, stop reading here and have a play with it - it\u2019s a classic puzzle for a reason.\n\nBelow the line are spoilers.\n\nCounting is hard\n\nThe first thing you\u2019d probably try is to draw out the lines and count up the regions.\n\nThe first circle has 1 region.\n\nThe second has 2 regions.\n\nThe third has 4. (I\u2019m spreading these out to try to avoid accidental spoilers, by the way.)\n\nThe fourth has 8. A pattern seems to be forming.\n\nThe fifth has 16. Aha! It must be\u2026\n\nThe sixth has 31. Oh. Let\u2019s count them again: nope, definitely 31 rather than 32. What\u2019s going on?\n\nWhat is going on?\n\nAs you draw more and more points, it gets harder to a: make sure you\u2019re not making your lines coincide and b: count the regions accurately. At MathsJam, we made it to the seventh circle (57 regions) before realising that was about as far as we could go.\n\nSo we needed a more methodical approach. The nugget of the solution - for me, at least - was to think about how each line you add to a diagram affects the number of regions - and that\u2019s related to the number of lines it crosses. In fact, adding a line that crosses $n$ other lines adds $n+1$ regions to the circle - which you might like to prove.\n\nSo, in adding a fifth point to the fourth circle, we\u2019ll need to add four new lines.\n\nThe two lines to the adjacent points each add a single region - they don\u2019t cross any other lines. The lines to the two opposite points each cross two lines, creating three new regions. Overall, that\u2019s 1 + 3 + 3 + 1 new regions, taking the eight regions to 16.\n\nLet\u2019s go a bit more methodically this time\n\nGoing from five to six is a bit trickier ((because counting is hard)).\n\nLet\u2019s consider the five existing points, going anticlockwise from our new point (which I\u2019ll imagine is at the bottom of the circle). Each new line will split the polygon into two parts, which I\u2019ll call \u2018left\u2019 and \u2018right\u2019.\n\nThe first line, to the neighbour, crosses no lines, and adds one region.\n\nThe second line, to point 2, splits the circle so there is one point to the left and three to the right, all of which are connected - so this line adds $1 \\times 3 + 1$ regions.\n\nThe third line splits the circle two on each side, so it adds $2 \\times 2 + 1$ regions.\n\nThe fourth line mirrors the second, and the fifth line mirrors the first, so I might be tempted to write the number of regions added as:\n\n$(0\\\\times 4 + 1) + (1 \\\\times 3 + 1) + (2 \\\\times 2 +1) + (3 \\\\times 1 + 1) + (4 \\\\times 0 + 1)=15$\n\n\u2026 or better, the new number of regions is $5 + \\sum_{i=0}^{4} i(4-i)$\n\nBut we can generalise!\n\nWhenever we add a point to a diagram, the process is going to be identical! Each line will divide the existing points into three sets: those to the left, the one being connected, and those to the right. Every point on the left connects to every point on the right exactly once, so the number of lines cut by the new line is always the number of left points multiplied by the number of right points.\n\nSo! If we\u2019re moving from $N-1$ points to $N$, we should be adding:\n\n$R\\_n = (N-1) + \\\\sum\\_{i=0}^{N-2} i( (N-2) - i)$\n\nnew regions.\n\nAnd we can write that explicitly!\n\n$R\\_n = (N-1) + (N-2)\\\\sum\\_{i=0}^{N-2} i - \\\\sum{i=0}^{N-2} i^2$ $= (N-1) + (N-2)\\\\frac{(N-2)(N-1)}{2} - \\\\frac{ (N-2)(N-1)(2N-3)}{6}$ $= \\\\frac{(N-1)}{6} \\\\left\\[ 6 + 3(N-2)^2 - (N-2)(2N-3) \\\\right$\\] $= \\\\frac{(N-1)}{6} \\\\left\\[ N^2 - 5N + 12 \\\\right$\\]\n\nLet\u2019s check that with $N=6$, which we\u2019ve already worked out: $R_n = \\frac{5}{6}\\left[ 36 - 30 + 12 \\right] = 15$ new regions.\n\nHow about $N=7$? We\u2019re expecting 26: $R_n=\\frac{6}{6}\\left[ 49 - 35 + 12\\right] = 26$. Hooray!\n\nHang on - that\u2019s just the new regions!\n\nSo, the number of new regions each time is $\\frac{N-1}{6} \\left[ N^2 - 5N + 12 \\right]$, and we know when there are no points, there is one region.\n\nSo, the number of regions altogether is $R_\\sum(M) = 1 + \\frac{1}{6}\\sum_{N=0}^{M} (N-1)(N^2 - 5N + 12)$.\n\nExpanding that and distributing the sum:\n\n$R_\\sum(M) = 1 + \\frac{1}{6} \\sum_{N=0}^{M} N^3 - \\sum_{N=0}^{M} N^2 + \\frac{17}{6}\\sum_{N=0}^{M} N - \\sum_{N=0}^{M} 2$\n\nThere follows an awful lot of tedious algebra, which ends up giving $R_\\sum(M) = \\frac{1}{24}\\left(M^4 - 6M^3 + 23M^2 - 18M + 24\\right)$\n\nJust to check, we know $R_\\sum(6)$ is 31: is that $\\frac{1}{24}\\left( 6^4 - 6\\times6^3 + 23\\times 6^2 - 18\\times 6 + 24 \\right)$? The first two terms cancel, and the rest? It\u2019s $\\frac{6}{24} \\left( 23 \\times 6 - 18 + 4\\right)$, or $\\frac{1}{4} \\times 124 = 31$, as required!\n\nCuriously, it turns out that $R_\\sum(10) = 256$, which is also a power of 2 (although half as big as it \u201cought\u201d to be if it followed the doubling pattern.)", "date": "2021-03-01 04:08:09", "meta": {"domain": "co.uk", "url": "https://flyingcoloursmaths.co.uk/regions-of-a-circle/", "openwebmath_score": 0.5728346705436707, "openwebmath_perplexity": 629.3774894119178, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842229971374, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8525149802459883}}
{"url": "https://math.stackexchange.com/questions/3535767/combinatorics-how-many-unique-polygons-can-be-drawn-using-up-to-p-points-along", "text": "# Combinatorics: how many unique polygons can be drawn using up to p points along a circle?\n\nI am faced with the following problem:\n\nLet p \u2265 3 be a prime number, and mark p points at equal distance on the circumference of a circle like in the pictures below (where p = 7). We can trace polygons by choosing a certain amount of those points as vertices, and connecting them with non-intersecting straight lines.\n\nWith three parts:\n\na) How many different polygons can we form if we consider that a different choice of vertices leads to a different polygon?\n\nThis is straightforward: you can have polygons with three vertices, 4 vertices [\u2026] up until p vertices, and the amount of polygons you can draw of a given number of vertices k is $$p \\choose k$$. Hence the answer is $${p \\choose 3} + {p \\choose 4} + {...} + {p \\choose p-1} + {p \\choose p}$$\n\nb) How many different polygons can we form if we count all different rotations of the same polygon as a single polygon?\n\nWell, any given polygon can be rotated up to $$p-1$$ times. Hence each polygon counted from a) is actually being counted 7 times. We adjust for this by dividing through the answer in a) by p, and get the answer as $$\\frac{p \\choose 3}{p} + \\frac{p \\choose 4}{p} + {...} + \\frac{p \\choose p-1}{p} + \\frac{p \\choose p}{p} = \\frac{p \\choose 3}{p} + \\frac{p \\choose 4}{p} + {...} + \\frac{p \\choose p-2}{p} + 1 + 1$$.\n\nI think all the above makes sense, but I am struggling with the third part:\n\nc) How many different polygons can we form if we count all isometric polygons (all rotations and reflections of the same polygon) as the same ?\n\nAnd I don't know how to proceed from here. I'm guessing there has to be some pattern to the number of reflections+rotations that any polygon of k vertices has along p points. However, I've tried playing around with small examples and I can't seem to find any generalizable pattern, and I've been stumped for longer than I'd like to admit. Any help would be greatly appreciated.\n\n\u2022 Reworded: How many necklaces and bracelets (necklaces for part b and bracelets for part c) of length $p$ with $p$ prime can be made with beads of two possible colors such that there are at least three beads of the first color used. \u2013\u00a0JMoravitz Feb 5 '20 at 19:47\n\u2022 Almost exactly the same question was posted just a week ago but then deleted when I answered it. a) Does this happen to be from an ongoing contest? If so, are you aware of the contest problem policy? b) If it isn't, and I post an answer, do you promise not to delete the question? :-) \u2013\u00a0joriki Feb 5 '20 at 19:47\n\u2022 @joriki it is not a contest problem. I haven't seen the other post / don't see why it was deleted. This is a problem from a university class :) \u2013\u00a0Student_514 Feb 5 '20 at 19:51\n\nThe polygons are in correspondence with the subsets of the vertices with at least $$3$$ elements. Thus, as JMoravitz noted in a comment, this can be viewed as counting binary necklaces and bracelets of length $$p$$ with at least three beads of the first colour, where the first colour signifies a vertex that\u2019s included and the second colour signifies a vertex that isn\u2019t included in the polygon.\n\nFor part a), there is no symmetry to take into account, so the result is just $$2^p-\\binom p2-\\binom p1-\\binom p0$$. (Your result is correct, you just didn't use the fact that $$\\sum_k\\binom pk=2^k$$, the total number of subsets of the set of $$k$$ vertices.)\n\nFor part b), with rotational symmetry, you failed to properly account for the fact that the polygon that includes all vertices has only one rotational equivalent, not $$p$$. Your final result is correct, but only because you replaced $$\\frac{\\binom pp}p=\\frac1p$$ by $$1$$. If you count correctly, you have $$2^p-\\binom p2-\\binom p1-\\binom p0-\\binom pp$$ polygons that form classes of $$p$$ rotational equivalents each, and $$\\binom pp=1$$ polygon that\u2019s in a class of its own, so the total count is\n\n$$\\frac{2^p-\\binom p2-\\binom p1-\\binom p0-\\binom pp}p+1=\\frac{2^p-2}p-\\frac{p+1}2\\;.$$\n\nFor part c), with rotational and reflectional symmetry, it becomes easier to perform the count using Burnside\u2019s lemma. The general result is given in the Wikipedia article linked to above. In your case, since $$p$$ is an odd prime, there are\n\n$$\\begin{eqnarray} B_2(p) &=& \\frac12N_2(p)+\\frac12\\cdot2^{\\frac{p+1}2} \\\\ &=& \\frac12\\cdot\\frac1p\\left(1\\cdot2^p+(p-1)\\cdot2^1\\right)+2^{\\frac{p-1}2} \\\\ &=& \\frac{2^{p-1}-1}p+1+2^{\\frac{p-1}2} \\end{eqnarray}$$\n\nbinary bracelets. We need to subtract the $$1$$ bracelet with $$0$$ elements of the second colour, the $$1$$ bracelet with $$1$$ element of the second colour and the $$\\frac{p-1}2$$ bracelets with $$2$$ elements of the second colour, so the number of polygons up to rotations and reflections is\n\n$$\\frac{2^{p-1}-1}p+2^{\\frac{p-1}2}-\\frac{p+1}2\\;.$$", "date": "2021-07-31 13:13:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3535767/combinatorics-how-many-unique-polygons-can-be-drawn-using-up-to-p-points-along", "openwebmath_score": 0.6235241293907166, "openwebmath_perplexity": 140.64365736459334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842220140631, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8525149794008748}}
{"url": "https://mathbooks.unl.edu/Calculus/sec-5-9-num-int.html", "text": "\n## Section5.9Numerical Integration\n\n###### Motivating Questions\n\u2022 How do we accurately evaluate a definite integral such as $\\int_0^1 e^{-x^2} \\, dx$ when we cannot use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative? Are there ways to generate accurate estimates without using extremely large values of $n$ in Riemann sums?\n\n\u2022 What is the Trapezoid Rule, and how is it related to left, right, and middle Riemann sums?\n\n\u2022 How are the errors in the Trapezoid Rule and Midpoint Rule related, and how can they be used to develop an even more accurate rule?\n\nWhen we first explored finding the net signed area bounded by a curve, we developed the concept of a Riemann sum4.2 as a helpful estimation tool and a key step in the definition of the definite integral. Recall that the left, right, and middle Riemann sums of a function $f$ on an interval $[a,b]$ are given by\n\n\\begin{align} L_n = f(x_0) \\Delta x + f(x_1) \\Delta x + \\cdots + f(x_{n-1}) \\Delta x \\amp= \\sum_{i = 0}^{n-1} f(x_i) \\Delta x,\\label{E-Left}\\tag{5.15}\\\\ R_n = f(x_1) \\Delta x + f(x_2) \\Delta x + \\cdots + f(x_{n}) \\Delta x \\amp= \\sum_{i = 1}^{n} f(x_i) \\Delta x,\\label{E-Right}\\tag{5.16}\\\\ M_n = f(\\overline{x}_1) \\Delta x + f(\\overline{x}_2) \\Delta x + \\cdots + f(\\overline{x}_{n}) \\Delta x \\amp= \\sum_{i = 1}^{n} f(\\overline{x}_i) \\Delta x\\text{,}\\label{E-Mid}\\tag{5.17} \\end{align}\n\nwhere $x_0 = a\\text{,}$ $x_i = a + i\\Delta x\\text{,}$ $x_n = b\\text{,}$ and $\\Delta x = \\frac{b-a}{n}\\text{.}$ For the middle sum, we defined $\\overline{x}_{i} = (x_{i-1} + x_i)/2\\text{.}$\n\nA Riemann sum is a sum of (possibly signed) areas of rectangles. The value of $n$ determines the number of rectangles, and our choice of left endpoints, right endpoints, or midpoints determines the heights of the rectangles. We can see the similarities and differences among these three options in Figure5.77, where we consider the function $f(x) = \\frac{1}{20}(x-4)^3 + 7$ on the interval $[1,8]\\text{,}$ and use 5 rectangles for each of the Riemann sums.\n\nPart (d) of Example4.36 explored how to determine if $L_n$ and $R_n$ are underestimates or overestimates of $\\int_a^b f(x) \\, dx$ when $f(x)$ is increasing. Similar statements can be made when $f(x)$ is decreasing. These are summarized next.\n\n###### Error types when approximating area using left and right Riemann sums\n\u2022 If $f(x)$ is increasing on $[a,b] \\text{,}$ then $L_n$ underestimates $\\int_a^b f(x) \\, dx$ and $R_n$ overestimates $\\int_a^b f(x) \\, dx\\text{.}$\n\n\u2022 If $f(x)$ is decreasing on $[a,b] \\text{,}$ then $L_n$ overestimates $\\int_a^b f(x) \\, dx$ and $R_n$ underestimates $\\int_a^b f(x) \\, dx\\text{.}$\n\nWhile it is a good exercise to compute a few Riemann sums by hand, just to ensure that we understand how they work and how varying the function, the number of subintervals, and the choice of endpoints or midpoints affects the result, using computing technology is a quick way to determine $L_n\\text{,}$ $R_n\\text{,}$ and $M_n\\text{.}$ Any computer algebra system will offer this capability.\n\nIn this section we explore several different alternatives for estimating definite integrals. Our main goal is to develop formulas to estimate definite integrals accurately without using a large numbers of rectangles.\n\n###### Example5.78\n\nAs we begin to investigate ways to approximate definite integrals, it will be insightful to compare results to integrals whose exact values we know. To that end, the following sequence of questions centers on $\\int_0^3 x^3 \\, dx\\text{.}$\n\n1. Use the applet at http://gvsu.edu/s/a9 with the function $f(x) = x^3$ on the window of $x$ values from $0$ to $3$ to compute $L_3\\text{,}$ the left Riemann sum with three subintervals.\n\n$L_3=9$\n\nSolution\n\nIn addition to changing $f(x)$ to $f(x)=x^3$ and the number of subintervals to $3 \\text{,}$ make sure to change the Sample Point Placement so that \"Relative\" is checked and the slide bar is all the way to the left. You'll see that now it is sampling $f(x)$ values from the left side of each sub-interval.\n\n\\begin{equation*} L_3=9 \\end{equation*}\n2. Likewise, use the applet to compute $R_3$ and $M_3\\text{,}$ the right and middle Riemann sums with three subintervals, respectively.\n\n$M_3=19.125 \\ , R_3=36$\n\nSolution\n\nFirst change $f(x)$ and the number of subintervals to $f(x)=x^3$ and $n=3\\text{.}$ For $R_3\\text{,}$ make sure to change the Sample Point Placement so that \"Relative\" is checked and the slide bar is all the way to the right. You'll see that now it is sampling $f(x)$ values from the right side of each subinterval. For $M_3\\text{,}$ make sure to change the Sample Point Placement so that \"Relative\" is checked and the slide bar is in the middle and the \"midpoint\" label shows. You'll see that now it is sampling $f(x)$ values from the middle of each interval.\n\n\\begin{equation*} M_3=19.125 \\ , R_3=36 \\end{equation*}\n3. Use the Fundamental Theorem of Calculus to compute the exact value of $I = \\int_0^3 x^3 \\, dx\\text{.}$\n\nSolution\n\\begin{align*} I&=\\int_0^3 x^3 \\, dx \\\\ &= \\left.\\frac{1}{4}x^4 \\right|_{0}^{3} \\\\ &=\\frac{3^4}{4}-\\frac{0^4}{4}\\\\ &=20.25 \\end{align*}\n4. We define the error in an approximation of a definite integral to be the difference between the integral's exact value and the approximation's value. What is the error that results from using $L_3\\text{?}$ From $R_3\\text{?}$ From $M_3\\text{?}$\n\n\\begin{equation*} E_{L,3}= 11.25 \\end{equation*}\n\\begin{equation*} E_{R,3}=-15.75 \\end{equation*}\n\\begin{equation*} E_{M,3}=1.125 \\end{equation*}\nSolution\n\nDenote the error from each approximation $E_{L,3}, E_{R,3}, \\text{ and } E_{M,3}\\text{.}$ Then\n\n\\begin{equation*} E_{L,3}=I-L_3=20.25-9 =11.25 \\end{equation*}\n\\begin{equation*} E_{R,3}=I-R_3=20.25-36=-15.75 \\end{equation*}\n\\begin{equation*} E_{M,3}=I-M_3=20.25-19.125=1.125 \\end{equation*}\n5. In what follows in this section, we will learn a new approach to estimating the value of a definite integral known as the Trapezoid Rule. The basic idea is to use trapezoids, rather than rectangles, to estimate the area under a curve. What is the formula for the area of a trapezoid with bases of length $b_1$ and $b_2$ and height $h\\text{?}$\n\nSolution\n\nThe area of a trapezoid with bases of length $b_1$ and $b_2$ and height $h$ is\n\n\\begin{equation*} A=\\frac{h(b_1+b_2)}{2}\\text{.} \\end{equation*}\n6. Working by hand, estimate the area under $f(x) = x^3$ on $[0,3]$ using three subintervals and three corresponding trapezoids. What is the error in this approximation? How does it compare to the errors you calculated in (d)?\n\n\\begin{equation*} T_3=\\frac{45}{2}=22.5 \\end{equation*}\n\\begin{equation*} E_{T,3}=-2.25 \\end{equation*}\nSolution\n\nThe left-most trapezoid has base lengths 0 and 1 and height 1, so the area of the first trapezoid is $\\frac{1}{2} (0+1)\\cdot 1=\\frac{1}{2}\\text{.}$ The middle trapezoid has base lengths 1 and 8 and height 1, so the area of the second trapezoid is $\\frac{1}{2}(1+8) \\cdot 1=\\frac{9}{2} \\text{.}$ The right-most trapezoid has base lengths 8 and 27 and height 1, so the area of the third trapezoid is $\\frac{1}{2}(8+27)\\cdot 1=\\frac{35}{2}\\text{.}$ Therefore, the approximate area under the graph of $f(x)=x^3$ from $0$ to $3$ using the trapezoid rule with 3 subintervals is $T_3= \\frac{1}{2} + \\frac{9}{2} + \\frac{35}{2}=\\frac{45}{2}=22.5\\text{.}$\n\nThe error is $E_{T,3}=20.25-22.5=-2.25 \\text{.}$ Using trapezoids creates a smaller error compared to $L_3$ and $R_3 \\text{.}$ The magnitude of the error from approximating using $M_3$ is half the magnitude of the error from approximating using $T_3 \\text{,}$ but they have opposite signs.\n\n### SubsectionThe Trapezoid Rule\n\nSo far, we have used the simplest possible quadrilaterals (that is, rectangles) to estimate areas. It is natural, however, to wonder if other familiar shapes might serve us even better.\n\nAn alternative to $L_n\\text{,}$ $R_n\\text{,}$ and $M_n$ is called the Trapezoid Rule. Rather than using a rectangle to estimate the (signed) area bounded by $y = f(x)$ on a small interval, we use a trapezoid. For example, in Figure5.80, we estimate the area under the curve using three subintervals and the trapezoids that result from connecting the corresponding points on the curve with straight lines.\n\nThe biggest difference between the Trapezoid Rule and a Riemann sum is that on each subinterval, the Trapezoid Rule uses two function values, rather than one, to estimate the (signed) area bounded by the curve. For instance, to compute $D_1\\text{,}$ the area of the trapezoid on $[x_0, x_1]\\text{,}$ we observe that the left base has length $f(x_0)\\text{,}$ while the right base has length $f(x_1)\\text{.}$ The height of the trapezoid is $x_1 - x_0 = \\Delta x = \\frac{b-a}{3}\\text{.}$ The area of a trapezoid is the average of the bases times the height, so we have\n\n\\begin{equation*} D_1 = \\frac{1}{2}(f(x_0) + f(x_1)) \\cdot \\Delta x\\text{.} \\end{equation*}\n\nUsing similar computations for $D_2$ and $D_3\\text{,}$ we find that $T_3\\text{,}$ the trapezoidal approximation to $\\int_a^b f(x) \\, dx$ is given by\n\n\\begin{align*} T_3 &= D_1 + D_2 + D_3\\\\ &= \\frac{1}{2}(f(x_0) + f(x_1)) \\cdot \\Delta x + \\frac{1}{2}(f(x_1) + f(x_2)) \\cdot \\Delta x + \\frac{1}{2}(f(x_2) + f(x_3)) \\cdot \\Delta x\\text{.} \\end{align*}\n\nBecause both left and right endpoints are being used, we recognize within the trapezoidal approximation the use of both left and right Riemann sums. Rearranging the expression for $T_3$ by removing factors of $\\frac{1}{2}$ and $\\Delta x \\text{,}$ grouping the left endpoint and right endpoint evaluations of $f\\text{,}$ we see that\n\n$$T_3 = \\frac{1}{2} \\left[ f(x_0) + f(x_1) + f(x_2) \\right] \\Delta x + \\frac{1}{2} \\left[ f(x_1) + f(x_2) + f(x_3) \\right] \\Delta x\\text{.}\\label{E-Trap3}\\tag{5.18}$$\n\nWe now observe that two familiar sums have arisen. The left Riemann sum $L_3$ is $L_3 = f(x_0) \\Delta x + f(x_1) \\Delta x + f(x_2) \\Delta x\\text{,}$ and the right Riemann sum is $R_3 = f(x_1) \\Delta x + f(x_2) \\Delta x + f(x_3) \\Delta x\\text{.}$ Substituting $L_3$ and $R_3$ for the corresponding expressions in Equation(5.18), it follows that $T_3 = \\frac{1}{2} \\left[ L_3 + R_3 \\right]\\text{.}$ We have thus seen a very important result: using trapezoids to estimate the (signed) area bounded by a curve is the same as averaging the estimates generated by using left and right endpoints.\n\n###### The Trapezoid Rule\n\nThe trapezoidal approximation, $T_n\\text{,}$ of the definite integral $\\int_a^b f(x) \\, dx$ using $n$ subintervals is given by the rule\n\n\\begin{align*} T_n =\\mathstrut \\amp \\left[\\frac{1}{2}(f(x_0) + f(x_1)) + \\frac{1}{2}(f(x_1) + f(x_2)) + \\cdots + \\frac{1}{2}(f(x_{n-1}) + f(x_n)) \\right] \\Delta x.\\\\ =\\mathstrut \\amp \\sum_{i=0}^{n-1} \\frac{1}{2}(f(x_i) + f(x_{i+1})) \\Delta x\\text{.} \\end{align*}\n\nMoreover, $T_n = \\frac{1}{2} \\left[ L_n + R_n \\right]\\text{.}$\n\n###### Example5.81\n\nIn this example, we explore the relationships among the errors generated by left, right, midpoint, and trapezoid approximations to the definite integral $\\int_1^2 \\frac{1}{x^2} \\, dx$\n\n1. Use the First FTC to evaluate $\\int_1^2 \\frac{1}{x^2} \\, dx$ exactly.\n\n2. Use appropriate computing technology to compute the following approximations for $\\int_1^2 \\frac{1}{x^2} \\, dx\\text{:}$ $T_4\\text{,}$ $M_4\\text{,}$ $T_8\\text{,}$ and $M_8\\text{.}$\n\n3. Recall that the error of an approximation is the difference between the exact value of the definite integral and the resulting approximation. For instance, if we let $E_{T,4}$ represent the error that results from using the trapezoid rule with 4 subintervals to estimate the integral, we have\n\n\\begin{equation*} E_{T,4} = \\int_1^2 \\frac{1}{x^2} \\, dx - T_4\\text{.} \\end{equation*}\n\nSimilarly, we compute the error of the midpoint rule approximation with 8 subintervals by the formula\n\n\\begin{equation*} E_{M,8} = \\int_1^2 \\frac{1}{x^2} \\, dx - M_8\\text{.} \\end{equation*}\n\nBased on your work in (a) and (b) above, compute $E_{T,4}\\text{,}$ $E_{T,8}\\text{,}$ $E_{M,4}\\text{,}$ $E_{M,8}\\text{.}$\n\n4. Which rule consistently over-estimates the exact value of the definite integral? Which rule consistently under-estimates the definite integral?\n\n5. What behavior(s) of the function $f(x) = \\frac{1}{x^2}$ lead to your observations in (d)?\n\nHint\n1. $\\frac{1}{x^2} = x^{-2}\\text{.}$\n\n2. Use a computational device.\n\n3. Use a computational device.\n\n4. Which estimate is larger than the true value of the definite integral?\n\n5. Note that how the curve bends makes a big difference in whether the trapezoid rule over- or under-estimates the value of the definite integral.\n\n1. $\\int_1^2 \\dfrac{1}{x^2} dx = \\dfrac{1}{2}\\text{.}$\n\n2. The table below gives values of the trapezoid rule and corresponding errors for different $n$-values.\n\n $n$ $T_n$ $E_{T,n}$ $4$ $0.50899$ $-0.50899$ $8$ $0.50227$ $-0.50227$ $16$ $0.50057$ $-0.50057$\n3. The table below gives values of the midpoint rule and corresponding errors for different $n$-values.\n\n $n$ $M_n$ $E_{M,n}$ $4$ $0.49555$ $0.00445$ $8$ $0.49887$ $0.00113$ $16$ $0.49972$ $0.00028$\n4. The trapezoid rule overestimates; the midpoint rule underestimates.\n\n5. $f(x) = \\dfrac{1}{x^2}$ is concave up on $[1, 2]\\text{.}$\n\nSolution\n1. $\\int_1^2 \\dfrac{1}{x^2} dx = \\left. -x^{-1} \\right|_1^2 = -\\dfrac{1}{2} + 1 = \\dfrac{1}{2}\\text{.}$\n\n2. The table below gives values of the trapezoid rule and corresponding errors for different $n$-values.\n\n $n$ $T_n$ $E_{T,n}$ $4$ $0.50899$ $-0.50899$ $8$ $0.50227$ $-0.50227$ $16$ $0.50057$ $-0.50057$\n3. The table below gives values of the midpoint rule and corresponding errors for different $n$-values.\n\n $n$ $M_n$ $E_{M,n}$ $4$ $0.49555$ $0.00445$ $8$ $0.49887$ $0.00113$ $16$ $0.49972$ $0.00028$\n4. From the errors in comparison to the known exact value, we see that the trapezoid rule overestimates this definite integral and the midpoint rule underestimates this definite integral.\n\n5. The graph of the function given by $f(x) = \\dfrac{1}{x^2}$ is concave up on the interval $[1, 2]\\text{.}$ Because of this fact, we can see graphically that the line forming the top of each trapezoid lies fully above the curve, and thus the trapezoid rule overestimates the true value of the definite integral. Later in this section we'll see graphically why this concavity makes the midpoint rule an underestimate.\n\n### SubsectionComparing the Midpoint and Trapezoid Rules\n\nWe know from the definition of the definite integral that if we let $n$ be large enough, we can make any of the approximations $L_n\\text{,}$ $R_n\\text{,}$ and $M_n$ as close as we'd like (in theory) to the exact value of $\\int_a^b f(x) \\, dx\\text{.}$ Thus, it may be natural to wonder why we ever use any rule other than $L_n$ or $R_n$ (with a sufficiently large $n$ value) to estimate a definite integral. One of the primary reasons is that as $n \\to \\infty\\text{,}$ $\\Delta x = \\frac{b-a}{n} \\to 0\\text{,}$ and thus in a Riemann sum calculation with a large $n$ value, we end up multiplying by a number that is very close to zero. Doing so often generates roundoff error, because representing numbers close to zero accurately is a persistent challenge for computers.\n\nHence, we explore ways to estimate definite integrals to high levels of precision, but without using extremely large values of $n\\text{.}$ Paying close attention to patterns in errors, such as those observed in Example5.81, is one way to begin to see some alternate approaches.\n\nTo begin, we compare the errors in the Midpoint and Trapezoid rules. First, consider a function that is concave up on a given interval, and picture approximating the area bounded on that interval by both the Midpoint and Trapezoid rules using a single subinterval.\n\nAs seen in Figure5.82, it is evident that whenever the function is concave up on an interval, the Trapezoid Rule with one subinterval, $T_1\\text{,}$ will overestimate the exact value of the definite integral on that interval. From a careful analysis of the line that bounds the top of the rectangle for the Midpoint Rule (shown in magenta), we see that if we rotate this line segment until it is tangent to the curve at the midpoint of the interval (as shown at right in Figure5.82), the resulting trapezoid has the same area as $M_1\\text{,}$ and this value is less than the exact value of the definite integral. Thus, when the function is concave up on the interval, $M_1$ underestimates the integral's true value.\n\nThe preceding discussion explores how to determine if $M_n$ and $T_n$ are underestimates or overestimates of $\\int_a^b f(x) \\, dx$ if $f(x)$ is concave up. Similar statements can be made when $f(x)$ is concave down. These are summarized next.\n\n###### Error types when approximating using Midpoint and Trapezoid Rules\n\u2022 If $f(x)$ is concave up on $[a,b] \\text{,}$ then $M_n$ underestimates $\\int_a^b f(x) \\, dx$ and $T_n$ overestimates $\\int_a^b f(x) \\, dx\\text{.}$\n\n\u2022 If $f(x)$ is concave down on $[a,b] \\text{,}$ then $M_n$ overestimates $\\int_a^b f(x) \\, dx$ and $T_n$ underestimates $\\int_a^b f(x) \\, dx\\text{.}$\n\nNext, we compare the size of the errors between $M_n$ and $T_n\\text{.}$ Again, we focus on $M_1$ and $T_1$ on an interval where the concavity of $f$ is consistent. In Figure5.83, where the error of the Trapezoid Rule is shaded in red, while the error of the Midpoint Rule is shaded lighter red, it is visually apparent that the error in the Trapezoid Rule is more significant. To see how much more significant, let's consider two examples and some particular computations.\n\nIf we let $f(x) = 1-x^2$ and consider $\\int_0^1 f(x) \\,dx\\text{,}$ we know by the First FTC that the exact value of the integral is\n\n\\begin{equation*} \\int_0^1 (1-x^2) \\, dx = x - \\frac{x^3}{3} \\bigg\\vert_0^1 = \\frac{2}{3}\\text{.} \\end{equation*}\n\nUsing appropriate technology to compute $M_4\\text{,}$ $M_8\\text{,}$ $T_4\\text{,}$ and $T_8\\text{,}$ as well as the corresponding errors $E_{M,4}\\text{,}$ $E_{M,8}\\text{,}$ $E_{T,4}\\text{,}$ and $E_{T,8}\\text{,}$ as we did in Example5.81, we find the results summarized in Table5.84. We also include the approximations and their errors for the example $\\int_1^2 \\frac{1}{x^2} \\, dx$ from Example5.81.\n\nFor a given function $f$ and interval $[a,b]\\text{,}$ $E_{T,4} = \\int_a^b f(x) \\,dx - T_4$ calculates the difference between the exact value of the definite integral and the approximation generated by the Trapezoid Rule with $n = 4\\text{.}$ If we look at not only $E_{T,4}\\text{,}$ but also the other errors generated by using $T_n$ and $M_n$ with $n = 4$ and $n = 8$ in the two examples noted in Table5.84, we see an evident pattern. Not only is the sign of the error (which measures whether the rule generates an over- or under-estimate) tied to the rule used and the function's concavity, but the magnitude of the errors generated by $T_n$ and $M_n$ seems closely connected. In particular, the errors generated by the Midpoint Rule seem to be about half the size of those generated by the Trapezoid Rule.\n\nThat is, we can observe in both examples that $E_{M,4} \\approx -\\frac{1}{2} E_{T,4}$ and $E_{M,8} \\approx -\\frac{1}{2}E_{T,8}\\text{.}$ This property of the Midpoint and Trapezoid Rules turns out to hold in general: for a function of consistent concavity, the error in the Midpoint Rule has the opposite sign and approximately half the magnitude of the error of the Trapezoid Rule. Written symbolically,\n\n\\begin{equation*} E_{M,n} \\approx -\\frac{1}{2} E_{T,n}\\text{.} \\end{equation*}\n\nThis important relationship suggests a way to combine the Midpoint and Trapezoid Rules to create an even more accurate approximation to a definite integral.\n\n### SubsectionSimpson's Rule\n\nWhen we first developed the Trapezoid Rule, we observed that it is an average of the Left and Right Riemann sums:\n\n\\begin{equation*} T_n = \\frac{1}{2}(L_n + R_n)\\text{.} \\end{equation*}\n\nIf a function is always increasing or always decreasing on the interval $[a,b]\\text{,}$ one of $L_n$ and $R_n$ will over-estimate the true value of $\\int_a^b f(x) \\, dx\\text{,}$ while the other will under-estimate the integral. Thus, the errors found in $L_n$ and $R_n$ will have opposite signs; so averaging $L_n$ and $R_n$ eliminates a considerable amount of the error present in the respective approximations. In a similar way, it makes sense to think about averaging $M_n$ and $T_n$ in order to generate a still more accurate approximation.\n\nWe've just observed that $M_n$ is typically about twice as accurate as $T_n\\text{.}$ This leads to an approximation method known as Simpson's Rule 14Thomas Simpson was an 18th century mathematician; his idea was to extend the Trapezoid rule, but rather than using straight lines to build trapezoids, to use quadratic functions to build regions whose area was bounded by parabolas (whose areas he could find exactly). Simpson's Rule is often developed from the more sophisticated perspective of using interpolation by quadratic functions. which is a weighted average of the Midpoint and Trapezoid approximations.\n\n###### Simpson's Rule\n\nThe Simpson's Rule approximation $S_{2n}$ of the area $\\int_a^b f(x) \\ dx$ is the weighted average\n\n$$S_{2n} = \\frac{2M_n + T_n}{3}\\text{.}\\label{E-Simpson}\\tag{5.19}$$\n\nwhere $M_n$ and $T_n$ are the Midpoint and Trapezoid rule approximations using $n$ subintervals.\n\nNote that we use $S_{2n}$ rather that $S_n$ since the $n$ points the Midpoint Rule uses are different from the $n$ points the Trapezoid Rule uses, and thus Simpson's Rule is using $2n$ points at which to evaluate the function. We build upon the results in Table5.84 to see the approximations generated by Simpson's Rule. In particular, in Table5.85, we include all of the results in Table5.84, but include additional results for $S_8 = \\frac{2M_4 + T_4}{3}$ and $S_{16} = \\frac{2M_8 + T_8}{3}\\text{.}$\n\nThe results seen in Table5.85 are striking. If we consider the $S_{16}$ approximation of $\\int_1^2 \\frac{1}{x^2} \\, dx\\text{,}$ the error is only $E_{S,16} = 0.0000019434\\text{.}$ By contrast, $L_8 = 0.5491458502\\text{,}$ so the error of that estimate is $E_{L,8} = -0.0491458502\\text{.}$ Moreover, we observe that generating the approximations for Simpson's Rule is almost no additional work: once we have $L_n\\text{,}$ $R_n\\text{,}$ and $M_n$ for a given value of $n\\text{,}$ it is a simple exercise to generate $T_n\\text{,}$ and from there to calculate $S_{2n}\\text{.}$ Finally, note that the error in the Simpson's Rule approximations of $\\int_0^1 (1-x^2) \\, dx$ is zero!15Similar to how the Midpoint and Trapezoid approximations are exact for linear functions, Simpson's Rule approximations are exact for quadratic and cubic functions. See additional discussion on this issue later in the section and in the exercises.\n\nThese rules are not only useful for approximating definite integrals such as $\\int_0^1 e^{-x^2} \\, dx\\text{,}$ for which we cannot find an elementary antiderivative of $e^{-x^2}\\text{,}$ but also for approximating definite integrals when we are given a function through a table of data.\n\n###### Example5.86\n\nA car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table. Assume that $v$ is continuous, always decreasing, and always decreasing at a decreasing rate, as is suggested by the data.\n\n seconds, $t$ Velocity in ft/sec, $v(t)$ $0$ $100$ $0.3$ $99$ $0.6$ $96$ $0.9$ $90$ $1.2$ $80$ $1.5$ $50$ $1.8$ $0$\n1. Plot the given data on the set of axes provided in Figure5.88 with time on the horizontal axis and the velocity on the vertical axis.\n\n2. What definite integral will give you the exact distance the car traveled on $[0,1.8]\\text{?}$\n\n3. Estimate the total distance traveled on $[0,1.8]$ by computing $L_3\\text{,}$ $R_3\\text{,}$ and $T_3\\text{.}$ Which of these under-estimates the true distance traveled?\n\n4. Estimate the total distance traveled on $[0,1.8]$ by computing $M_3\\text{.}$ Is this an over- or under-estimate? Why?\n\n5. Using your results from (c) and (d), improve your estimate further by using Simpson's Rule.\n\n6. What is your best estimate of the average velocity of the car on $[0,1.8]\\text{?}$ Why? What are the units on this quantity?\n\nHint\n1. Plot the data.\n\n2. What are the units of $v(t) \\cdot \\Delta t\\text{?}$\n\n3. Recall the standard rules for sums that produce $L_3\\text{,}$ $R_3\\text{,}$ $T_3\\text{.}$\n\n4. Think about concavity to decide if $M_3$ is an over- or under-estimate.\n\n5. Recall how $S_3$ is a weighted average of $T_3$ and $M_3\\text{.}$\n\n6. Simpson's Rule gives the best estimate for a function of consistent concavity.\n\n1. Plot the data.\n\n2. $\\int_0^{1.8} v(t) dt\\text{.}$\n\n3. \\begin{align*} L_3 \\amp = 165.6 \\text{ ft } \\amp R_3 \\amp = 105.6 \\text{ ft } \\amp T_3 \\amp = 135.6 \\text{ ft }\\text{.} \\end{align*}\n\n$R_3$ and $T_3$ are underestimates.\n\n4. $M_3 = 143.4 \\text{ ft }$ ; overestimate.\n\n5. $S_6 = 140.8 \\text{ ft } \\text{.}$\n\n6. Simpson's rule gives the best approximation of the distance traveled, $\\int_0^{1.8} v(t) dt \\approx 140.8 \\text{ ft }\\text{.}$\n\nSolution\n1. Plot the data.\n\n2. Since the velocity is always positive, the definite integral that will give the exact distance traveled by the car on the interval $[0, 1.8]$ is\n\n\\begin{equation*} \\int_0^{1.8} v(t) dt\\text{.} \\end{equation*}\n3. The estimates of $\\int_0^{1.8} v(t) dt$ are\n\n\\begin{align*} L_3 \\amp = 165.6 \\text{ ft } \\amp R_3 \\amp = 105.6 \\text{ ft } \\amp T_3 \\amp = 135.6 \\text{ ft }\\text{.} \\end{align*}\n\n$R_3$ is an underestimate of the distance traveled since $v(t)$ is decreasing. $T_3$ is an underestimate of the distance traveled since $v(t)$ is concave down.\n\n4. Another estimate of the distance traveled is\n\n\\begin{equation*} M_3 = 143.4 \\text{ ft }\\text{.} \\end{equation*}\n\nThis is an overestimate since $v(t)$ is concave down.\n\n5. For Simpson's Rule, we see that\n\n\\begin{equation*} S_6 = \\frac{2}{3}M_3 + \\frac{1}{3}T_3 = 140.8 \\text{ ft }\\text{.} \\end{equation*}\n6. Simpson's rule gives the best approximation of the distance traveled since it is a weighted average of the midpoint and trapezoid rules and uses more information about the velocity than the other methods. The units on each of the estimates, including Simpson's Rule, are \"feet\", since ft/sec $\\cdot$ sec = ft. Thus, the best approximation we have generated is that $\\int_0^{1.8} v(t) dt \\approx 140.8 \\text{ ft }\\text{.}$\n\n### SubsectionComparing $L_n\\text{,}$ $R_n\\text{,}$ $T_n\\text{,}$ $M_n\\text{,}$ and $S_{2n}\\text{.}$\n\nAs we conclude our discussion of numerical approximation of definite integrals, it is important to summarize general trends in how the various rules over- or under-estimate the true value of a definite integral, and by how much. To revisit some past observations and see some new ones, we consider the following example.\n\n###### Example5.89\n\nConsider the functions $f(x) = 2-x^2\\text{,}$ $g(x) = 2-x^3\\text{,}$ and $h(x) = 2-x^4\\text{,}$ all on the interval $[0,1]\\text{.}$ For each of the questions that require a numerical answer in what follows, write your answer exactly in fraction form.\n\n1. On the three sets of axes provided in Figure5.90, sketch a graph of each function on the interval $[0,1]\\text{,}$ and compute $L_1$ and $R_1$ for each. What do you observe?\n\n2. Compute $M_1$ for each function to approximate $\\int_0^1 f(x) \\,dx\\text{,}$ $\\int_0^1 g(x) \\,dx\\text{,}$ and $\\int_0^1 h(x) \\,dx\\text{,}$ respectively.\n\n3. Compute $T_1$ for each of the three functions, and hence compute $S_2$ for each of the three functions.\n\n4. Evaluate each of the integrals $\\int_0^1 f(x) \\,dx\\text{,}$ $\\int_0^1 g(x) \\,dx\\text{,}$ and $\\int_0^1 h(x) \\,dx$ exactly using the First FTC.\n\n5. For each of the three functions $f\\text{,}$ $g\\text{,}$ and $h\\text{,}$ compare the results of $L_1\\text{,}$ $R_1\\text{,}$ $M_1\\text{,}$ $T_1\\text{,}$ and $S_2$ to the true value of the corresponding definite integral. What patterns do you observe?\n\nHint\n1. For each estimate, just one function evaluation is needed.\n\n2. Use the midpoint rule with $n=1\\text{.}$\n\n3. Remember that both the trapezoid and Simpson's rule can be executed using (weighted) averages of known values.\n\n4. Find antiderivatives to evaluate the integrals exactly.\n\n5. Think about trends in over- and under-estimates.\n\n1. For $L_1$ and $T_1\\text{:}$\n\nThe values of $L_1$ and $R_1$ are the same for all three.\n\n2. For the $M_1\\text{,}$\n\n3. For $T_1$ and $S_2\\text{,}$\n\n4. \\begin{align*} \\int_0^1 f(x) dx \\amp = \\frac{5}{3} \\amp \\int_0^1 g(x) dx \\amp = \\frac{7}{4} \\amp \\int_0^1 h(x) dx \\amp = \\frac{9}{5} \\end{align*}\n5. Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson's rule is exact for both $f$ and $g\\text{,}$ while a slight overestimate of $\\int_0^1 h(x) dx\\text{.}$\n\nSolution\n1. For the left and right endpoint rules, we see that\n\nThus, we observe that despite the fact the functions are all different, the values of $L_1$ and $R_1$ are the same for all three.\n\n2. For the midpoint rule, we find that\n\n3. For the trapezoid rule and Simpson's rule,\n\n4. The exact values of the three definite integrals are\n\n\\begin{align*} \\int_0^1 f(x) dx \\amp = \\frac{5}{3} \\amp \\int_0^1 g(x) dx \\amp = \\frac{7}{4} \\amp \\int_0^1 h(x) dx \\amp = \\frac{9}{5}\\\\ \\amp \\approx 1.6667 \\amp \\amp = 1.75 \\amp \\amp = 1.8 \\end{align*}\n5. We observe that each of the left endpoint rule results are overestimates, each of the right endpoint rules are underestimates, each of the midpoint rules are overestimates, and each of the trapezoid rules are underestimates. These results hold because each of the three functions are both decreasing and concave down. For Simpson's rule, we see that the result is exact for both $f$ and $g\\text{,}$ while Simpson's rule is a slight overestimate of $\\int_0^1 h(x) dx\\text{.}$\n\nThe results seen in Example5.89 generalize nicely. For instance, if $f$ is decreasing on $[a,b]\\text{,}$ $L_n$ will overestimate the exact value of $\\int_a^b f(x) \\,dx\\text{,}$ and if $f$ is concave down on $[a,b]\\text{,}$ $M_n$ will overestimate the exact value of the integral. An excellent exercise is to write a collection of scenarios of possible function behavior, and then categorize whether each of $L_n\\text{,}$ $R_n\\text{,}$ $T_n\\text{,}$ and $M_n$ is an over- or under-estimate.\n\nFinally, we make two important notes about Simpson's Rule. When T.Simpson first developed this rule, his idea was to replace the function $f$ on a given interval with a quadratic function that shared three values with the function $f\\text{.}$ In so doing, he guaranteed that this new approximation rule would be exact for the definite integral of any quadratic polynomial. In one of the pleasant surprises of numerical analysis, it turns out that even though it was designed to be exact for quadratic polynomials, Simpson's Rule is exact for any cubic polynomial: that is, if we are interested in an integral such as $\\int_2^5 (5x^3 - 2x^2 + 7x - 4)\\, dx\\text{,}$ the approximation $S_{2n}$ will always be exact, regardless of the value of $n\\text{.}$ This is just one more piece of evidence that shows how effective Simpson's Rule is as an approximation tool for estimating definite integrals.16One reason that Simpson's Rule is so effective is that $S_{2n}$ benefits from using $2n+1$ points of data. Because it combines $M_n\\text{,}$ which uses $n$ midpoints, and $T_n\\text{,}$ which uses the $n+1$ endpoints of the chosen subintervals, $S_{2n}$ takes advantage of the maximum amount of information we have when we know function values at the endpoints and midpoints of $n$ subintervals.\n\n### SubsectionSummary\n\n\u2022 For a definite integral such as $\\int_0^1 e^{-x^2} \\, dx$ when we cannot use the First Fundamental Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative, we can estimate the integral's value by using a sequence of Riemann sum approximations. Typically, we start by computing $L_n\\text{,}$ $R_n\\text{,}$ and $M_n$ for one or more chosen values of $n\\text{.}$\n\n\u2022 The Trapezoid Rule, which estimates $\\int_a^b f(x) \\, dx$ by using trapezoids, rather than rectangles, can also be viewed as the average of Left and Right Riemann sums. That is, $T_n = \\frac{1}{2}(L_n + R_n)\\text{.}$\n\n\u2022 The Midpoint Rule is typically twice as accurate as the Trapezoid Rule, and the signs of the respective errors of these rules are opposites. Hence, by taking the weighted average $S_{2n} = \\frac{2M_n + T_n}{3}\\text{,}$ we can build a much more accurate approximation to $\\int_a^b f(x) \\, dx$ by using approximations we have already computed. The rule for $S_{2n}$ is known as Simpson's Rule, which can also be developed by approximating a given continuous function with pieces of quadratic polynomials.\n\n\u2022 By understanding the approximation rule chosen, and the properties of the function (i.e. whether it is increasing or decreasing, or concave up or down on the interval in question), we can say whether the approximation is an over or under-estimate of the actual value.\n\n### SubsectionExercises\n\nConsider the definite integral $\\int_0^1 x \\tan(x) \\, dx\\text{.}$\n\n1. Explain why this integral cannot be evaluated exactly by using either $u$-substitution or by integrating by parts.\n\n2. Using appropriate subintervals, compute $L_4\\text{,}$ $R_4\\text{,}$ $M_4\\text{,}$ $T_4\\text{,}$ and $S_8\\text{.}$\n\n3. Which of the approximations in (b) is an over-estimate to the true value of $\\int_0^1 x \\tan(x) \\, dx\\text{?}$ Which is an under-estimate? How do you know?\n\nFor an unknown function $f(x)\\text{,}$ the following information is known.\n\n\u2022 $f$ is continuous on $[3,6]\\text{;}$\n\n\u2022 $f$ is either always increasing or always decreasing on $[3,6]\\text{;}$\n\n\u2022 $f$ has the same concavity throughout the interval $[3,6]\\text{;}$\n\n\u2022 As approximations to $\\int_3^6 f(x) \\, dx\\text{,}$ $L_4 = 7.23\\text{,}$ $R_4 = 6.75\\text{,}$ and $M_4 = 7.05\\text{.}$\n\n1. Is $f$ increasing or decreasing on $[3,6]\\text{?}$ What data tells you?\n\n2. Is $f$ concave up or concave down on $[3,6]\\text{?}$ Why?\n\n3. Determine the best possible estimate you can for $\\int_3^6 f(x) \\, dx\\text{,}$ based on the given information.\n\nThe rate at which water flows through Table Rock Dam on the White River in Branson, MO, is measured in thousands of cubic feet per second (TCFS). As engineers open the floodgates, flow rates are recorded according to the following chart.\n\n1. What definite integral measures the total volume of water to flow through the dam in the 60 second time period provided by the table above?\n\n2. Use the given data to calculate $M_n$ for the largest possible value of $n$ to approximate the integral you stated in (a). Do you think $M_n$ over- or under-estimates the exact value of the integral? Why?\n\n3. Approximate the integral stated in (a) by calculating $S_n$ for the largest possible value of $n\\text{,}$ based on the given data.\n\n4. Compute $\\frac{1}{60} S_n$ and $\\frac{2000+2100+2400+3000+3900+5100+6500}{7}\\text{.}$ What quantity do both of these values estimate? Which is a more accurate approximation?", "date": "2020-07-13 21:10:07", "meta": {"domain": "unl.edu", "url": "https://mathbooks.unl.edu/Calculus/sec-5-9-num-int.html", "openwebmath_score": 0.9525200128555298, "openwebmath_perplexity": 333.4462961604561, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109084307415, "lm_q2_score": 0.8615382094310355, "lm_q1q2_score": 0.8525014562618984}}
{"url": "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_5&diff=next&oldid=149207", "text": "# Difference between revisions of \"2020 AIME I Problems/Problem 5\"\n\n## Problem\n\nSix cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.\n\n## Solution 1\n\nRealize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.\n\nIf we choose any of the numbers $1$ through $6$, there are five other spots to put them, so we get $6 \\cdot 5 = 30$. However, we overcount some cases. Take the example of $132456$. We overcount this case because we can remove the $3$ or the $2$. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract $5$ cases (namely, $213456, 132456, 124356, 123546, 123465$,) to get $30-5=25$, but we have to add back one more for the original case, $123456$. Therefore, there are $26$ cases. Multiplying by $2$ gives the desired answer, $\\boxed{052}$.\n\n~molocyxu\n\n## Solution 2 (Inspired by 2018 CMIMC Combo Round)\n\nSimilar to above, a $1-1$ correspondence between ascending and descending is established by subtracting each number from $7$.\n\nWe note that the given condition is equivalent to \"cycling\" $123456$ for a contiguous subset of it. For example,\n\n$12(345)6 \\rightarrow 125346, 124536$\n\nIt's not hard to see that no overcount is possible, and that the cycle is either $1$ \"right\" or $1$ \"left.\" Therefore, we consider how many elements we flip by. If we flip $1$ or $2$ such elements, then there is one way to cycle them. Otherwise, we have $2$ ways. Therefore, the total number of ascending is $1 + 5 + 2(4 + 3 + 2 + 1) = 26$, and multiplying by two gives $\\boxed{052}.$\n\n~awang11\n\n## Solution 3\n\nSimilarly to above, we find the number of ascending arrangements and multiply by 2.\n\nWe can choose $5$ cards to be the ascending cards, therefore leaving $6$ places to place the remaining card. There are $\\binom{6}{5}\\cdot 6=36$ to do this. However, since the problem is asking for the number of arrangements, we overcount cases such as $123456$. Notice that the only arrangements that overcount are $123456$ (case 1) or if two adjacent numbers of $123456$ are switched (case 2).\n\n$\\text{Case 1: }$ This arrangement is counted $6$ times. Each time it is counted for any of the $5$ numbers selected. Therefore we need to subtract $5$ cases of overcounting.\n\n$\\text{Case 2: }$ Each time $2$ adjacent numbers of switched, there is one overcount. For example, if we have $213456$, both $1$ or $2$ could be removed. Since there are $5$ possible switches, we need to subtract $5$ cases of overcounting.\n\nTherefore, we have $36-5-5=26$ total arrangements of ascending numbers. We multiply by two (for descending) to get the answer of $\\boxed{052}.$\n\n~PCChess\n\n## Solution 4 (No Overcounting)\n\nLike in previous solutions, we will count the number of ascending arrangements and multiply by 2.\n\nFirst, consider the arrangement 1-2-3-4-5-6. That gives us 1 arrangement which works.\n\nNext, we can switch two adjacent cards. There are 5 ways to pick two adjacent cards, so this gives us 5 arrangements.\n\nNow, we can \"cycle\" 3 adjacent cards. For example, 1-2-3 becomes 2-3-1 which becomes 3-1-2. There are 4 ways to pick a set of 3 adjacent cards, so this gives us 4x2=8 arrangements.\n\nCycling 4 adjacent cards, we get the new arrangements 2-3-4-1 (which works,) 3-4-1-2 (which doesn't work,) and 4-1-2-3 (which does work.) We get 6 arrangements.\n\nSimilarly, when cycling 5 cards, we find 2x2=4 arrangements, and when cycling 6 cards, we find 2x1=2 arrangements.\n\nAdding, we figure out that there are 1+5+8+6+4+2=26 ascending arrangements. Multiplying by 2, we get the answer $\\boxed{052}.$\n\n~i8Pie\n\n## Solution 5 (Official MAA 1)\n\nFirst count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is $1$ such permutation where all the cards appear in order: $123456.$ There are $5$ such permutations where two adjacent cards are interchanged, as in $124356.$ The other such permutations arise from removing one card from $123456$ and placing it in a position at least two away from its starting location. There are $4$ such positions to place each of the cards numbered $1$ and $6,$ and $3$ such positions for each of the cards numbered $2, 3, 4,$ and $5.$ This accounts for $2\\cdot4 + 4\\cdot3 =20$ permutations. Thus there are $1 + 5 + 20 = 26$ permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total $26 + 26 = \\boxed{52}$.\n\n## Solution 6 (Official MAA 2)\n\nMore generally, suppose there are $n \\geq 4$ cards numbered $1, 2, 3, \\dots, n$ arranged in ascending order. If any one of the $n$ cards is removed and placed in one of the $n$ positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ascending order. This accounts for $n\\cdot n = n^2$ permutations. However, the original ascending order has been counted $n$ times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is $n^2-(n-1)-(n-1)=(n-1)^2+1.$ When $n = 6$, this is $(6-1)^2+1 = 26$, and the final answer is $2\\cdot26 = \\boxed{52}$.\n\n## Solution 7 (Casework)\n\nFor ascending, if the $1$ goes in anything but the first two slots, the rest of the numbers have to go in ascending from $2$, which are $4$ cases if there are $6$ cards. If $1$ goes in the second spot, then you can put any of the rest in the first slot but then the rest are determined, so in the case of $6$ cards, that gives $5$ more. If $1$ goes in the first slot, that means that you are doing the same problem with $n-1$ cards. So the recursion is $a_n=(n-2)+(n-1)+a_{n-1}$. There's $a_1=1$ and $a_2=2$, so you get $a_3=2+3=5$, $a_4=5+5=10$, $a_5=7+10=17$, and $a_6=9+17=26$. Or you can see that $a_n=(n-1)^2+1$. We double to account for descending and get $\\boxed{052}$.\n\n~ahclark11\n\n## Solution 8 (Symmetry and Case Study)\n\nFirst, we know that ascending order and descending order are symmetrical to each other (namely, if we get 132456 where after we take out 3, it will be one scenario; and if we flip it and write 654231, it will be another scenario)\n\nThus, we only need to consider either descending or ascending and then times 2. WLOG let us consider ascending order\n\nCase 1: after we take out 1, the rest will be in ascending order\n\n2 3 4 5 6\n\n1 can be tucked in any one of the 6 spaces, thus there are 6 scenarios.\n\nCase 2: after we take out 2, the rest will be in ascending order\n\n1 3 4 5 6\n\nnotice that if we put 2 next to 1 (to the right or to the left of 1), it will be an overcount, so there are only 4 cases for 2\n\nIt is easy to see that this is the same for 3, 4, 5, and 6.\n\nThus, in total, we have $$(6+4\\times5)\\times2=\\boxed{052}$$ ~Adali", "date": "2021-10-19 20:49:14", "meta": {"domain": "artofproblemsolving.com", "url": "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_5&diff=next&oldid=149207", "openwebmath_score": 0.778374433517456, "openwebmath_perplexity": 282.7314736892491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109078121008, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8525014504521667}}
{"url": "https://math.stackexchange.com/questions/1761557/prove-fracacbd-lies-between-fracab-and-fraccd-for-posit", "text": "# Prove: $\\frac{a+c}{b+d}$ lies between $\\frac{a}{b}$ and $\\frac{c}{d}$ (for positive $a$, $b$, $c$, $d$)\n\nI am looking for proof that, if you take any two different fractions and add the numerators together then the denominators together, the answer will always be a fraction that lies between the two original fractions.\n\nWould be grateful for any suggestions!\n\n\u2022 Not to be picky, but if we consider the fractions $a/b,c/d$ with $a=b=c=1$ and $d=-2$, then $(a+c)/(b+d)=-2$, which does not lie between $a/b=1$ and $c/d=-1/2$.... \u2013\u00a0Barry Cipra Apr 27 '16 at 18:55\n\u2022 Good point Barry. Let's say that a, b, c and d are all positive integers. \u2013\u00a0Joeywald Apr 27 '16 at 19:24\n\u2022 mathsguy, not a lot as \"proof\" is new to me. I know that adding the same number to the top and bottom of a fraction moves the derived fraction closer to 1 and that adding two fractions as above gives a resultant fraction between the two, but I can't prove why this happens. \u2013\u00a0Joeywald Apr 27 '16 at 19:33\n\nSuppose we have positive $a,b,c,d$ with $\\frac{a}{b}\\ge \\frac{c}{d}$. Then multiplying through by $bd$ we get $ad\\ge bc$. Adding $ab$ to both sides we get $a(b+d)\\ge b(a+c)$. Dividing by $b(b+d)$, we get $\\frac{a}{b}\\ge\\frac{a+c}{b+d}$.\n\nSimilarly, add $cd$ to both sides of $ad\\ge bc$ to get $d(a+c)\\ge c(b+d)$. Dividing by $d(b+d)$ we get $\\frac{a+c}{b+d}\\ge \\frac{c}{d}$.\n\nAssuming $a,b,c,d\\gt0$, we have\n\n\\begin{align} {a\\over b}\\lt{a+c\\over b+d}\\lt{c\\over d}&\\iff a(b+d)\\lt b(a+c)\\quad\\text{and}\\quad(a+c)d\\lt(b+d)c\\\\ &\\iff ad\\lt bc\\quad\\text{and}\\quad ad\\lt bc\\\\ &\\iff ad\\lt bc\\\\ &\\iff {a\\over b}\\lt{c\\over d} \\end{align}\n\n\u2022 One usually starts with the premise and ends with the conclusion... No? \u2013\u00a0Najib Idrissi Jun 7 '16 at 17:38\n\u2022 @NajibIdrissi If it's $\u201c\\Leftrightarrow\u201d$ signs the whole way through, it shouldn't matter. \u2013\u00a0Akiva Weinberger Jun 7 '16 at 17:47\n\u2022 @AkivaWeinberger Sure, but I think it's easier to read and looks more sound if you put it in the other order. But of course from a logical point of view, equivalence of propositions is symmetrical... \u2013\u00a0Najib Idrissi Jun 7 '16 at 18:09\n\nHere is another way that uses calculus:\n\nLet $\\phi(t) = {(1-t)a+t c \\over (1-t)b + t d}$ and note that $\\phi(0) = {a \\over b}, \\phi(1) = {c \\over d}$. Furthermore, $\\phi'(t) = {bc-ad \\over ((1-t)b + t d )^2}$. Since $ad < bc$, we see that $\\phi$ is increasing, and so $\\phi(0) \\le \\phi({1 \\over 2}) \\le \\phi(1)$. Since $\\phi({1 \\over 2}) = {a+c \\over b+d}$, we have the desired result.\n\nAssume that $b,d >0$. Note that $$\\frac{a+c}{b+d} = \\frac{b}{b+d}\\frac{a}{b} +\\frac{d}{b+d}\\frac{c}{d}.$$ Remark that $0<\\frac{b}{b+d}<1$ and the same for $\\frac{d}{b+d}$. Hence you have written $\\frac{a+b}{c+d}$ as a convex combination of $\\frac{a}{b}$ and $\\frac{c}{d}$ so you get $$\\frac{a}{b} < \\frac{a+c}{b+d} < \\frac{c}{d}.$$\n\nHere's one way to look at it:\n\nYou're taking a class. Suppose you get $a$ points out of $b$ possible on Quiz 1, and $c$ points out of $d$ possible on Quiz 2.\n\nYour overall points are $a+c$ out of $b+d$ possible. And your overall percentage should be between your lower quiz score and your higher quiz score.\n\n\u2022 The question asks for \"proof\". \u2013\u00a0AlohaSine Apr 30 '16 at 23:54\n\u2022 @MathematicsStudent1122 Actually the presenter asked for suggestions. One could use this idea as the basis for a proof. \u2013\u00a0paw88789 May 1 '16 at 2:49\n\nHint $\\$ The mediant $(a\\!+\\!b)/(c\\!+\\!d)\\,$ is the slope of the diagonal of the parallelogram with vector sides $(b,a),\\ (d,c).\\:$ But the slope of the diagonal lies between the slopes of the sides.", "date": "2021-07-25 02:50:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1761557/prove-fracacbd-lies-between-fracab-and-fraccd-for-posit", "openwebmath_score": 0.8573033809661865, "openwebmath_perplexity": 315.015695645829, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9643214491222696, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8524980471466388}}
{"url": "https://www.math.ubc.ca/~pwalls/math-python/integration/trapezoid-rule/", "text": "# Trapezoid Rule\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n%matplotlib inline\n\n## Trapezoids\n\nThe definite integral of $f(x)$ is equal to the (net) area under the curve $y=f(x)$ over the interval $[a,b]$. Riemann sums approximate definite integrals by using sums of rectangles to approximate the area.\n\nThe trapezoid rule gives a better approximation of a definite integral by summing the areas of the trapezoids connecting the points\n\n$$(x_{i-1},0), (x_i,0), (x_{i-1},f(x_{i-1})), (x_i,f(x_i))$$\n\nfor each subinterval $[x_{i-1},x_i]$ of a partition. Note that the area of each trapezoid is the sum of a rectangle and a triangle\n\n$$(x_i - x_{i-1}) f(x_{i-1}) + \\frac{1}{2}(x_i - x_{i-1}) (f(x_i) - f(x_{i-1})) = \\frac{1}{2}(f(x_i) + f(x_{i-1}))(x_i - x_{i-1})$$\n\nFor example, we can use a single trapezoid to approximate:\n\n$$\\int_0^1 e^{-x^2} \\, dx$$\n\nFirst, let's plot the curve $y = e^{-x^2}$ and the trapezoid on the interval $[0,1]$:\n\nx = np.linspace(-0.5,1.5,100)\ny = np.exp(-x**2)\nplt.plot(x,y)\n\nx0 = 0; x1 = 1;\ny0 = np.exp(-x0**2); y1 = np.exp(-x1**2);\nplt.fill_between([x0,x1],[y0,y1])\n\nplt.xlim([-0.5,1.5]); plt.ylim([0,1.5]);\nplt.show()\n\nApproximate the integral by the area of the trapezoid:\n\nA = 0.5*(y1 + y0)*(x1 - x0)\nprint(\"Trapezoid area:\", A)\nTrapezoid area: 0.6839397205857212\n\n## Definition\n\nThe trapezoid rule for $N$ subintervals of $[a,b]$ of equal length is\n\n$$T_N(f) = \\frac{\\Delta x}{2} \\sum_{i=1}^N (f(x_i) + f(x_{i-1}))$$\n\nwhere $\\Delta x = (b - a)/N$ is the length of the subintervals and $x_i = a + i \\Delta x$.\n\nNotice that the trapezoid is the average of the left and right Riemann sums\n\n$$T_N(f) = \\frac{\\Delta x}{2} \\sum_{i=1}^N (f(x_i) + f(x_{i-1})) = \\frac{1}{2} \\left( \\sum_{i=1}^N f(x_i) \\Delta x + \\sum_{i=1}^N f(x_{i-1}) \\Delta x \\right)$$\n\n## Error Formula\n\nWhen computing integrals numerically, it is essential to know how good our approximations are. Notice in the theorem below that the error formula is inversely proportional to $N^2$. This means that the error decreases much faster with larger $N$ compared to Riemann sums.\n\nTheorem. Let $T_N(f)$ denote the trapezoid rule\n\n$$T_N(f) = \\frac{\\Delta x}{2} \\sum_{i=1}^N (f(x_i) + f(x_{i-1}))$$\n\nwhere $\\Delta x = (b-a)/N$ and $x_i = a + i \\Delta x$. The error bound is\n\n$$E_N^T(f) = \\left| \\ \\int_a^b f(x) \\ dx - T_N(f) \\ \\right| \\leq \\frac{(b-a)^3}{12 N^2} K_2$$\n\nwhere $\\left| \\ f''(x) \\, \\right| \\leq K_2$ for all $x \\in [a,b]$.\n\n## Implementation\n\nLet's write a function called trapz which takes input parameters $f$, $a$, $b$ and $N$ and returns the approximation $T_N(f)$. Furthermore, let's assign default value $N=50$.\n\ndef trapz(f,a,b,N=50):\n'''Approximate the integral of f(x) from a to b by the trapezoid rule.\n\nThe trapezoid rule approximates the integral \\int_a^b f(x) dx by the sum:\n(dx/2) \\sum_{k=1}^N (f(x_k) + f(x_{k-1}))\nwhere x_k = a + k*dx and dx = (b - a)/N.\n\nParameters\n----------\nf : function\nVectorized function of a single variable\na , b : numbers\nInterval of integration [a,b]\nN : integer\nNumber of subintervals of [a,b]\n\nReturns\n-------\nfloat\nApproximation of the integral of f(x) from a to b using the\ntrapezoid rule with N subintervals of equal length.\n\nExamples\n--------\n>>> trapz(np.sin,0,np.pi/2,1000)\n0.9999997943832332\n'''\nx = np.linspace(a,b,N+1) # N+1 points make N subintervals\ny = f(x)\ny_right = y[1:] # right endpoints\ny_left = y[:-1] # left endpoints\ndx = (b - a)/N\nT = (dx/2) * np.sum(y_right + y_left)\nreturn T\n\nLet's test our function on an integral where we know the answer\n\n$$\\int_0^{\\pi/2} \\sin x \\ dx = 1$$\n\ntrapz(np.sin,0,np.pi/2,1000)\n0.9999997943832332\n\nLet's test our function again:\n\n$$\\int_0^1 3 x^2 \\ dx = 1$$\n\ntrapz(lambda x : 3*x**2,0,1,10000)\n1.0000000050000002\n\nAnd once more:\n\n$$\\int_0^1 x \\ dx = \\frac{1}{2}$$\n\ntrapz(lambda x : x,0,1,1)\n0.5\n\n## scipy.integrate.trapz\n\nThe SciPy subpackage scipy.integrate contains several functions for approximating definite integrals and numerically solving differential equations. Let's import the subpackage under the name spi.\n\nimport scipy.integrate as spi\n\nThe function scipy.integrate.trapz computes the approximation of a definite by the trapezoid rule. Consulting the documentation, we see that all we need to do it supply arrays of $x$ and $y$ values for the integrand and scipy.integrate.trapz returns the approximation of the integral using the trapezoid rule. The number of points we give to scipy.integrate.trapz is up to us but we have to remember that more points gives a better approximation but it takes more time to compute!\n\n## Examples\n\n### Arctangent\n\nLet's plot the trapezoids for $\\displaystyle f(x)=\\frac{1}{1 + x^2}$ on $[0,5]$ with $N=10$.\n\nf = lambda x : 1/(1 + x**2)\na = 0; b = 5; N = 10\n\n# x and y values for the trapezoid rule\nx = np.linspace(a,b,N+1)\ny = f(x)\n\n# X and Y values for plotting y=f(x)\nX = np.linspace(a,b,100)\nY = f(X)\nplt.plot(X,Y)\n\nfor i in range(N):\nxs = [x[i],x[i],x[i+1],x[i+1]]\nys = [0,f(x[i]),f(x[i+1]),0]\nplt.fill(xs,ys,'b',edgecolor='b',alpha=0.2)\n\nplt.title('Trapezoid Rule, N = {}'.format(N))\nplt.show()\n\nLet's compute the sum of areas of the trapezoids:\n\nT = trapz(f,a,b,N)\nprint(T)\n1.3731040812301096\n\nWe know the exact value\n\n$$\\int_0^5 \\frac{1}{1 + x^2} dx = \\arctan(5)$$\n\nand we can compare the trapezoid rule to the value\n\nI = np.arctan(5)\nprint(I)\n1.373400766945016\nprint(\"Trapezoid Rule Error:\",np.abs(I - T))\nTrapezoid Rule Error: 0.00029668571490626405\n\n### Approximate ln(2)\n\nFind a value $N$ which guarantees that the trapezoid rule approximation $T_N(f)$ of the integral\n\n$$\\int_1^2 \\frac{1}{x} \\, dx = \\ln(2)$$\n\nsatisfies $E_N^T(f) \\leq 10^{-8}$.\n\nFor $f(x) = \\frac{1}{x}$, we compute $f''(x) = \\frac{2}{x^3} \\leq 2$ for all $x \\in [1,2]$ therefore the error formula implies\n\n$$\\left| \\, \\int_1^2 \\frac{1}{x} \\, dx - T_N(f) \\, \\right| \\leq \\frac{2}{12N^2}$$\n\nThen $E_N^T \\leq 10^{-8}$ is guaranteed if $\\frac{1}{6N^2} \\leq 10^{-8}$ which implies\n\n$$\\frac{10^4}{\\sqrt{6}} \\leq N$$\n\n10**4/np.sqrt(6)\n4082.4829046386303\n\nWe need 4083 subintervals to guarantee $E_N^T(f) \\leq 10^{-8}$. Compute the approximation using our own implementation of the trapezoid rule:\n\napproximation = trapz(lambda x : 1/x,1,2,4083)\nprint(approximation)\n0.6931471843089954\n\nWe could also use scipy.integrate.trapz to get the exact same result:\n\nN = 4083\nx = np.linspace(1,2,N+1)\ny = 1/x\napproximation = spi.trapz(y,x)\nprint(approximation)\n0.6931471843089955\n\nLet's verify that this is within $10^{-6}$:\n\nnp.abs(approximation - np.log(2)) < 10**(-8)\nTrue\n\nSuccess! However, a natural question arises: what is the actual smallest $N$ such that the trapezoid rule gives the estimate of $\\ln (2)$ to within $10^{-8}$?\n\nfor n in range(1,4083):\napprox = trapz(lambda x : 1/x,1,2,n)\nif np.abs(approx - np.log(2)) < 10e-8:\nprint(\"Accuracy achieved at N =\",n)\nbreak\nAccuracy achieved at N = 791\n\n### Fresnel Integral\n\nFresnel integrals are examples of nonelementary integrals: antiderivatives which cannot be written in terms of elementary functions. There are two types of Fresnel integrals:\n\n$$S(t) = \\int_0^t \\sin(x^2) dx \\ \\ \\text{and} \\ \\ C(t) = \\int_0^t \\cos(x^2) dx$$\n\nUse the trapezoid rule to approximate the Fresnel integral\n\n$$S(1) = \\int_0^1 \\sin(x^2) dx$$\n\nsuch that the error is less than $10^{-5}$.\n\nCompute the derivatives of the integrand\n\n$$f(x) = \\sin(x^2) \\ \\ , \\ \\ f'(x) = 2x\\cos(x^2)$$\n\n$$f''(x) = 2\\cos(x^2) - 4x^2\\sin(x^2) \\ \\ , \\ \\ f'''(x) = -12x\\sin(x^2) - 8x^3\\cos(x^2)$$\n\nSince $f'''(x) \\leq 0$ for $x \\in [0,1]$, we see that $f''(x)$ is decreasing on $[0,1]$. Values of $f''(x)$ at the endpoints of the interval are\n\nx = 0\n2*np.cos(x**2) - 4*x**2*np.sin(x**2)\n2.0\nx = 1\n2*np.cos(x**2) - 4*x**2*np.sin(x**2)\n-2.2852793274953065\n\nTherefore $\\left| \\, f''(x) \\, \\right| \\leq 2.2852793274953065$ for $x \\in [0,1]$. Use the error bound formula to find a good choice for $N$\n\n$$\\frac{(b-a)^3}{12 N^2} K_2 \\leq 10^{-5} \\Rightarrow \\sqrt{\\frac{10^5(2.2852793274953065)}{12}} \\leq N$$\n\nnp.sqrt(10**5 * 2.2852793274953065 / 12)\n137.9999796949051\n\nLet's compute the integral using the trapezoid rule with $N=138$ subintervals\n\nx = np.linspace(0,1,139)\ny = np.sin(x**2)\nI = spi.trapz(y,x)\nprint(I)\n0.31027303032220394\n\nTherefore the Fresnel integral $S(1)$ is approximately\n\n$$S(1) = \\int_0^1 \\sin(x^2) \\, dx \\approx 0.310273030322$$\n\nwith error less than $10^{-5}$.\n\n### Logarithmic Integral\n\nThe Eulerian logarithmic integral is another nonelementary integral\n\n$$\\mathrm{Li}(t) = \\int_2^t \\frac{1}{\\ln x} dx$$\n\nLet's compute $Li(10)$ such that the error is less than $10^{-4}$. Compute derivatives of the integrand\n\n$$f(x) = \\frac{1}{\\ln x} \\ \\ , \\ \\ f'(x) = -\\frac{1}{x(\\ln x)^2} \\ \\ , \\ \\ f''(x) = \\frac{\\ln x + 2 }{x^2(\\ln x)^3}$$\n\nPlot $f''(x)$ on the interval $[2,10]$.\n\na = 2\nb = 10\nx = np.linspace(a,b,100)\ny = (np.log(x) + 2) / (x**2 * np.log(x)**3)\nplt.plot(x,y)\nplt.show()\n\nClearly $f''(x)$ is decreasing on $[2,10]$ (and bounded below by 0) therefore the absolute maximum occurs at the left endpoint:\n\n$$\\left| \\, f''(x) \\, \\right| \\leq \\frac{\\ln (2) + 2}{4 \\ln (2)^3}$$\n\nfor $x \\in [2,10]$ and we compute\n\nK2 = (np.log(2) + 2)/(4*np.log(2)**3)\nprint(K2)\n2.021732598829855\n\nUse the error formula:\n\n$$\\frac{(b-a)^3}{12 N^2} K_2 \\leq 10^{-4} \\Rightarrow \\frac{8^3}{12 N^2} 2.021732598829855 \\leq 10^{-4} \\Rightarrow \\sqrt{ \\frac{8^3 10^4}{12} 2.021732598829855} \\leq N$$\n\nnp.sqrt(8**3 * 10**4 * 2.021732598829855 / 12)\n928.7657986995814\n\nCompute the trapzoid rule with $N=929$\n\nN = 929\nx = np.linspace(a,b,N+1)\ny = 1/np.log(x)\nI = spi.trapz(y,x)\nprint(I)\n5.120442039184057\n\nTherefore the Eulerian logarithmic integral is\n\n$$\\mathrm{Li}(10) = \\int_2^{10} \\frac{1}{\\ln x} dx \\approx 5.121065367200469$$\n\nsuch that the error is less than $10^{-4}$.\n\n## Exercises\n\n1. Let $f(x) = x^x$ and note that\n\n$$f'(x) = x^{x} \\left(\\log{\\left(x \\right)} + 1\\right) \\ , \\ f''(x) = x^{x} \\left(\\log{\\left(x \\right)} + 1\\right)^{2} + x^{x-1}$$\n\nPlot the function $f''(x)$ and use that information to compute $T_N(f)$ for the integral\n\n$$\\int_1^2 x^x \\, dx$$\n\nsuch that $E_N^T(f) \\leq 10^{-3}$.\n\n2. Consider the integral\n\n$$\\int_0^1 \\ln(1+x^2) \\, dx$$\n\nand note that\n\n$$f(x) = \\ln(1 + x^2) \\hspace{1in} f'(x) = \\frac{2x}{1 + x^2}$$ $$f''(x) = 2 \\left( \\frac{1 - x^2}{1 + x^2} \\right) \\hspace{1in} f'''(x) = 4x \\frac{x^2 - 3}{(x^2 + 1)^3}$$\n\nWithout plotting the functions $f(x)$, $f'(x)$, $f''(x)$ or $f'''(x)$, find a value $N$ such that $E_N^T(f) \\leq 10^{-6}$.", "date": "2021-03-09 01:26:37", "meta": {"domain": "ubc.ca", "url": "https://www.math.ubc.ca/~pwalls/math-python/integration/trapezoid-rule/", "openwebmath_score": 0.9319489598274231, "openwebmath_perplexity": 1221.9700084223086, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750510899382, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8524913419498915}}
{"url": "https://math.stackexchange.com/questions/3458449/path-on-a-chessboard", "text": "# Path on a chessboard\n\nConsider a chess board and a pawn. The game is simple (and I'm sure you have all heard it before), pawn wants to move from position (1,1), or the bottom left corner, all the way to (n,n), or the upper right corner. We are asked to find all possible paths this pawn could take.\n\nThis problem is not all that hard, considering that the path always takes $$2n-2$$ moves. Cancelling out the same, but reversed and tidying up the formula, we get\n\n$$\\binom{2n-2}{n-1}$$ possible paths.\n\nFollowing which, the main question is what would happen if we removed some pivotal squares. Let's consider that we have a chessboard with $$n=4$$, and were to remove square $$(3,3)$$. By my logic, all we need to do is to find all the possible ways how to get to $$(4,4)$$ and to $$(3,3)$$. Then we would subtract all the ways to $$(3,3)$$ from the number of paths to $$(4,4)$$, which would give us\n\n$$\\binom{2*4-2}{4-1} - \\binom{2*3-2}{3-1} =$$ $$\\binom{6}{3} - \\binom{4}{2} =$$ $$20 - 6 = 12$$\n\nAs I found out, this is not the correct solution. I mapped out all the possible paths, and found only $$8$$ of them. If we multiply the number of paths to $$(3,3)$$ by $$2$$, that would give us the correct solution; $$\\binom{2*4-2}{4-1} - 2* \\binom{2*3-2}{3-1} =$$ $$20 - 2*6 = 8$$\n\nThat begs the question, is the second solution correct? And if so, why? That, I do not know.\n\nAdditionally, what would happen if we were to remove more than one square? For example, let $$n=13$$ and lets remove two squares, $$(5,5)$$ and $$(8,8)$$. Would that be something like\n\n$$\\binom{2*13-2}{13-1} - 2* {\\left[\\binom{2*8-2}{8-1} + \\binom{2*5-2}{5-1}\\right]}$$\n\nor did I get the whole concept wrong? Why is there that seemingly random multiplication, and how would it stack with more and more pivotal squares removed?\n\n\u2022 But you know... pawns can only move forward... and occasionally attack diagnally... until they become something else. Dec 2, 2019 at 21:58\n\nThe paths to $$(4,4)$$ that use $$(3,3)$$ aren't just paths to $$(3,3)$$; their paths to $$(3,3)$$ and then on to $$(4,4)$$. So to count them, you need to multiply the number of ways to get from $$(1,1)$$ to $$(3,3)$$, which you calculated, with the number of ways to get from $$(3,3)$$ to $$(4,4)$$. That's $$\\binom21=2$$, and this is where the missing factor $$2$$ comes from.\nIf you remove more than one square, you need to use inclusion\u2013exclusion. For each subset of the removed squares, you need to add or subtract the number of paths that use that subset of removed squares, according as it contains an even or odd number of removed squares. (That number will be zero unless the orders of the $$x$$ and $$y$$ coordinates of the squares coincide.)\nFor instance, if you remove the squares $$(2,2)$$ and $$(3,3)$$, you're left with $$4$$ paths, and this can be calculated as follows, with $$N(S)$$ denoting the number of paths that use all squares in $$S$$:\n$$\\begin{eqnarray*} &&N(\\emptyset)-N(\\{(2,2)\\})-N(\\{(3,3)\\})+N(\\{(2,2),(3,3)\\})\\\\ &=& \\binom63-\\binom21\\binom42-\\binom42\\binom21+\\binom21\\binom21\\binom21\\\\ &=& 20-2\\cdot6-2\\cdot6+2\\cdot2\\cdot2\\\\ &=& 20-12-12+8\\\\ &=& 4\\;. \\end{eqnarray*}$$\n\u2022 So for the last case that would be $\\binom{24}{12}-\\binom84\\binom{14}{7}-\\binom{14}{7}\\binom84+\\binom84\\binom84\\binom84\\\\$, right?\n\u2022 @Igor: Almost; I think some of your counts are off by $1$. The lower index is always the coordinate difference between the squares you're connecting. So if you're going from coordinate $1$ to $13$ while avoiding $5$ and $8$, that would be $$\\binom{24}{12}-\\binom84\\binom{16}8-\\binom{14}7\\binom{10}5+\\binom84\\binom63\\binom{10}5\\;.$$ As a check, the indices (both upper and lower) in the factors of each term should add up to the same sum. Dec 3, 2019 at 12:05", "date": "2022-07-06 00:02:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3458449/path-on-a-chessboard", "openwebmath_score": 0.6819502711296082, "openwebmath_perplexity": 189.40155308533113, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750503469328, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8524913395730355}}
{"url": "http://math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals", "text": "# Is there a rational number between any two irrationals?\n\nSuppose $i_1$ and $i_2$ are distinct irrational numbers with $i_1 < i_2$. Is it necessarily the case that there is a rational number $r$ in the interval $[i_1, i_2]$? How would you construct such a rational number?\n\n[I posted this only so that the useful answers at Rationals and irrationals on the real number line could be merged here before that question was deleted.]\n\n-\nYes, there is. This is known as the \"density of the rationals in the reals\", which says in fact that between any two reals numbers there is a rational number. \u2013\u00a0 Avi Steiner Jun 16 '13 at 2:24\n\u2013\u00a0 MJD Jun 16 '13 at 2:37\nSo, wait... you asked the question, pointed to another question that already answers this... and then answered this question several days before asking it? TIME TRAVELER!!! \u2013\u00a0 BlueRaja - Danny Pflughoeft Jun 16 '13 at 5:25\n@BlueRaja-DannyPflughoeft yeah, I thought that, then I read the [bit in brackets in the question]. \u2013\u00a0 Lucas Jun 16 '13 at 5:53\n\nLet $x,y\\in\\mathbb{R}$, $x\\neq y$. Without loss of generality, suppose $x<y$. Then there exists a positive $z$ such that $y-x=z$.\n\nBy Archimedes' axiom, there exists a natural number $n$ such that $$n > \\dfrac{1}{z}$$ $$nz > 1$$ $$ny - nx > 1$$ So there exists an integer $m$ such that $$nx < m < ny$$ $$x < \\frac{m}{n} < y$$ i.e. $m/n$ is a rational number between $x$ and $y$.\n\nSince $x$ and $y$ can be any real numbers, in particular they can be irrationals.\n\n-\n\nLet $a$ and $b$ be distinct irrationals; we lose no generality to suppose $a<b$. Assume they have equal integer parts, since otherwise there is an integer between them and the question is trivial. They have infinite decimal expansions, $.a_1a_2a_3\\ldots$ and $.b_1b_2b_3\\ldots$. These cannot agree in every position since otherwise $a=b$. So say they agree up to the $n-1$th position and differ at the $n$th. Then $$x= \\;.b_1b_2b_3\\ldots b_n 0000000\\ldots$$\n\nis a rational number strictly between $a$ and $b$:\n\n\\begin{align} a &= \\;.a_1a_2\\ldots a_{n-1}a_n\\ldots \\\\& <\\; .a_2a_2\\ldots a_{n-1}b_n000\\ldots &= x \\end{align}\n\nbecause $a_n < b_n$, and\n\n\\begin{align} x & = \\;.b_1b_2b_3\\ldots b_n 0000000\\ldots \\\\ & < \\;.b_1b_2b_3\\ldots b_nb_{n+1}\\ldots & = b \\end{align}\n\nbecause not all of $b_{n+1}, b_{n+2}, \\ldots$ can be zero.\n\nFor example, there is a rational number betwen $\\sqrt2 = 1.4142\\ldots$ and $\\sqrt3 -\\frac14 = 1.482\\ldots$; this method produces the rational number $1.48000\\ldots = \\frac{37}{25}$. You can of course do the same thing in base 2; then you get $x = 1._20111000\\ldots = \\frac{23}{16}$, which also works.\n\nIt would also be fairly easy to produce a similar argument based on the continued fraction expansions of $a$ and $b$, but I think this is simpler.\n\n-\nYour proof contradicts itself by not allowing integers but allowing integers up to b_n. \u2013\u00a0 Daniel Margolis Jun 7 '13 at 18:32\nOne of us is confused, and I bet it's you. $b_n$ is not an integer. It is a decimal digit. And my proof doesn't \"disallow\" integers; it observes that if $a$ and $b$ have different integer parts then there is no need for a proof because there is an integer between them and the question is answered. \u2013\u00a0 MJD Jun 7 '13 at 18:33\nI guess this shows that you can lead a horse to water, but you can't make him drink. \u2013\u00a0 MJD Jun 7 '13 at 18:49\n@Daniel: There is no \"infinite number place.\" Your conception of the notation \"$\\infty$\" as a number (or occasionally a cardinality) seems to be at the root of all of your confusion. \u2013\u00a0 Cameron Buie Jun 15 '13 at 3:14\n@JensSchauder If a has integer part 2 and b has integer part 3 and b is irrational, then 3 is indeed between a and b. \u2013\u00a0 Daenerys Naharis Jun 16 '13 at 5:44\n\nWe can construct one explicitly. Assume $0\\lt a \\lt b$. Let $n= \\max(2,\\lceil \\frac 2{b-a}\\rceil)$. Then let $m=\\lceil na \\rceil$ and $a \\lt \\frac m n \\lt b$\n\n-\n@DanielMargolis: in the standard reals, there are no such. You (or the problem poser) have to specify them first and the difference will not be infinitesimal. It may be small, which makes $n$ large, but that is OK. Even if $a=\\sqrt 2, b=\\sqrt 2 + 10^{-100}$ this works. \u2013\u00a0 Ross Millikan Jun 8 '13 at 2:06\n@DanielMargolis: Yes, it is. In fact, if you extend $\\Bbb R$ with infinitesimals, it becomes incomplete. If you have an infinitesimal $\\epsilon$, the set $\\{\\frac 1n|n \\in \\Bbb N\\}$ has $0$ as a lower bound, but it has no greatest lower bound. \u2013\u00a0 Ross Millikan Jun 8 '13 at 13:05\nIf you are working in some number system that includes infinitesimals, you should say that. There is no indication in the question that you are, and all the answers to this and your other question are in the real numbers. I know very little about such systems. You seem to have an intuitive sense of how you want the infinitesimals to behave, but I don't think they can act that way. \u2013\u00a0 Ross Millikan Jun 8 '13 at 13:33\n@CameronBuie: yes, he is including infinitesimals, which comes out late in this discussion, as well as another question he asks. Then he wants to lump all infinitesimals with the standard irrationals. I am not sure if it is confusion or trolling. \u2013\u00a0 Ross Millikan Jun 15 '13 at 15:55\n@Daniel: Noone is claiming that irrationals have finite decimal expansions--their definitions do indeed have infinitely many decimal places. What do you mean by \"an infinite decimal place,\" though? \u2013\u00a0 Cameron Buie Jun 16 '13 at 2:32\n\nBe $a$ and $b$ two irrational numbers, with $a<b$. Since $a<b$, we have $b-a>0$. Now there are rational numbers arbitrary close to $0$, and therefore there's a rational number $q<b-a$. Now consider the set $M=\\{qn: n\\in \\mathbb{Z}\\}$. Now be $q_l$ the largest element of $M$ less than $a$, and $q_r$ the smallest element of $M$ greater than $b$. Clearly $q_r-q_l > b-a$. However, if there's no rational number between $a$ and $b$, then there's especially no element of $M$ between $a$ and $b$, because all elements of $M$ are rational. But then $q_r$ is the element of $M$ following $q_l$, that is, $q_r-q_l = q$. But by construction, $q<b-a$, so we have $b-a < q_l-q_r = q < b-a$ which is a contradiction. Therefore there's at least one element of $M$, which is a rational number, in between $a$ and $b$.\n\n-\n@DanielMargolis: I didn't claim it is. \u2013\u00a0 celtschk Jun 7 '13 at 23:16\nThe real numbers are totally ordered, that means exactly one of $a-b>0$, $a-b=0$, $a-b<0$ is true. $a-b=0$ would mean $a=b$, but by assumption, $a$ and $b$ are different (were $a=b$, quite trivially there would be no rational number between $a$ and $b$). So this leaves $a-b>0$ and $a-b<0$. I further assumed that $a$ denotes the smaller of the two, i.e. $a<b$, which is equivalent to $b-a>0$. So $b-a>0$ by the initial assumption (but that's just a wlog assumption; if we had $a<b$, you could just use $a-b$ instead). \u2013\u00a0 celtschk Jun 8 '13 at 8:39\n@DanielMargolis: There cannot be such an $R$ because all sequences of rational numbers whose absolute value eventually stays beyond any given rational number converge to $0$. So any number that is closer to $0$ than any rational number cannot be the limit of a sequence of rational numbers, in contradiction to $\\mathbb{R}$ being the completion of $\\mathbb{Q}$. \u2013\u00a0 celtschk Jun 8 '13 at 9:19\n$1/n*irr$ is not an irrational number, it is a whole sequence of irrational numbers. And not a single one is closer to $0$ than all rational numbers. Yes, for every rational number (with the exception of $0$ itself, of course), there's an irrational number that's closer to $0$. But there's no irrational number which is closer to $0$ than every rational number, because for each irrational number there's a rational number which is even closer to $0$. \u2013\u00a0 celtschk Jun 8 '13 at 10:13\nDo you really not know the difference between \"thhere is no irraltional number closer to $0$ than any rational number\" (true) and \"for any rational number there is no irrational number closer to it\" (false)? In that case, do yourself a favour and learn elementary logic. \u2013\u00a0 celtschk Jun 8 '13 at 11:17\n\n## protected by Asaf KaragilaJan 15 '14 at 21:21\n\nThank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.", "date": "2015-01-28 06:46:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals", "openwebmath_score": 0.9525929093360901, "openwebmath_perplexity": 299.17141347939116, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750518329434, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8524913391206932}}
{"url": "https://math.stackexchange.com/questions/3920780/probability-of-drawing-an-ace-or-2-after-an-ace-intuition", "text": "# Probability of drawing an ace or 2 after an ace - intuition\n\nThis question is a follow-up question on this one: Probability of drawing an ace or 2 after an ace\n\nBackground Out of a standard deck you draw cards (without replacement) one by one until you reach an ace. Then you continue to draw cards until you reach another ace or a \"2\" card. What is the probability that you reached a \"2\" card and not an ace?\n\nThe solution To shorten things up, let's assume the deck has only two aces and two \"2\" cards. Clearly, all that matters are the relative position of these two cards, not all the other cards. So we can think of the permutations of 4 items in a row.\n\nThe event \"the card following the first ace is ace\" occurs when the two aces are together. The sample space is $${4 \\choose 2}$$ (choosing 2 places for the aces) and there are 3 locations in which they are together (1-2,2-3,3-4) so the solution is $$\\tfrac{3}{{4\\choose 2}}=0.5$$.\n\nSurprisingly, the answer $$0.5$$ remains correct even if there are 4 aces and 4 \"2\"s in the original deck and in fact, as @InterstellarProbe shows in the comments to the original question, if there are $$n+1$$ of each type, the answer is: $$\\sum_{k=0}^{n+1} \\frac{n\\left(\\begin{array}{c} n+k \\\\ k \\end{array}\\right)}{(n+k)\\left(\\begin{array}{c} 2 n+2 \\\\ n+1 \\end{array}\\right)}=\\frac{1}{2}$$\n\nThe question The above result surprises me a bit. There is no obvious symmetry in the question and it is not obviously clear why there is the same number of arrangements in which after the first ace comes another one as arrangements in which after the first ace comes a \"2\". I spent a couple of months thinking, asked also fellow probability teachers, but no-one had a good idea.\n\nSo, returning the ball to this court. What is the intuitive explanation to this result? Why the permutations are symmetric and how can we get this $$\\tfrac{1}{2}$$ without the algebra?\n\n\u2022 I would say that it looks asymmetric because you don't think about what happens before that first ace. You might have already discarded several 2's. \u2013\u00a0Marc Roman\u00ed Nov 24 '20 at 14:32\n\nI'm not sure that the following solution is any more intuitive than others (in my experience, some probability problems are simply non-intuitive, at least with my intuition), but at least it is a different approach than those given so far.\n\nTo start with, we can ignore all the cards in the deck except for aces and twos. So suppose we have a deck of four aces and four twos, which can be arranged in $$8!$$ ways, all of which we assume are equally likely. We want to find the probability that the card following the first ace in the deck is also an ace.\n\nAs a further simplification, let's see if we can find the probability that the card immediately following the first ace is the ace of spades. We want to count the number of such arrangements. To do so, suppose we remove the ace of spades from the deck and deal out the remaining cards; this can be done in $$7!$$ ways. There is only one place where the ace of spades can now go so that it is immediately following the first ace, so the number of arrangements in which the ace of spades follows the first ace is $$7!$$, and the probability of this event is $$\\frac{7!}{8!} = \\frac{1}{8}$$\n\nIf we don't care which ace follows the first ace in the deck, there are $$4$$ possible choices, all of which have the same probability of $$1/8$$. So the probability that the card following the first ace is also an ace is $$4 \\cdot \\frac{1}{8} = \\frac{1}{2}$$\n\n\u2022 It wasn't what I was looking for (searched for symmetry of the permutations) but this is an excellent answer. Following on that logic, from symmetry, the probability of any card to be after the first ace is equal so it must be $\\tfrac{1}{8}$ and the probability that it would be an ace is $0.5$. The generalization to more cards and more types (ace, 2, and kings...) is straightforward. Great! \u2013\u00a0YJT Nov 24 '20 at 17:46", "date": "2021-08-04 23:36:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3920780/probability-of-drawing-an-ace-or-2-after-an-ace-intuition", "openwebmath_score": 0.8059972524642944, "openwebmath_perplexity": 99.14533242299353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750496039276, "lm_q2_score": 0.8633916152464017, "lm_q1q2_score": 0.8524913389315311}}
{"url": "https://math.stackexchange.com/questions/2402760/equation-of-locus-of-points-satisfied-by-frac-leftz3i-right-leftz-6i-ri", "text": "# Equation of locus of points satisfied by $\\frac{\\left|z+3i\\right|}{\\left|z-6i\\right|}=1$\n\nEquation of locus of points satisfied by $\\frac{\\left|z+3i\\right|}{\\left|z-6i\\right|}=1$\n\nThe answer I got is $y=\\frac{3}{2}$, but the answer given in my book is $y=-0.5x+2.25$\n\nCan anyone please confirm which is the right answer\n\nEdit: Sorry, the modulus was on each numerator and denominator, not for the whole thing, though I don't think this makes a difference.\n\nMy working\n\nsubbing $z=x+yi$ gives\n\n$\\frac{\\left|x+yi+3i\\right|}{\\left|x+yi-6i\\right|}=1$\n\n$\\frac{x^2+(y+3)^2}{x^2+(y-6)^2}=1^2$\n\nExpanding brackets and then solving gives\n\n$y=3/2$\n\nThank You\n\n\u2022 Please show us your workings, i.e. your work from which you \"got is $y=\\frac 32$. Other wise, we cannot confirm your solution. So, show us how you arrived at your answer, by editing your post to include it. \u2013\u00a0Namaste Aug 22 '17 at 20:11\n\u2022 I think you're right. The equation says point $z$ is equidistant from $-3i$ and $6i$, which should be a horizontal line at $y=1.5$. \u2013\u00a0Jirapat Samranvedhya Aug 22 '17 at 20:12\n\u2022 @amWhy I have done so now. Thanks \u2013\u00a0NumberCruncher Aug 22 '17 at 20:22\n\u2022 @JirapatSamranvedhya Thanks very much, that's how I thought about it first too. \u2013\u00a0NumberCruncher Aug 22 '17 at 20:22\n\u2022 NumberCruncher Thank you for including your work! \u2013\u00a0Namaste Aug 22 '17 at 22:10\n\nTo cut down on unanswered questions, here we go!\n\nThe book's answer is certainly wrong, as one readily sees by considering $z=\\frac94i.$\n\nNow, we clearly cannot have $z=6i$ as a solution, for then we have $\\frac90=1,$ which is nonsensical. Consequently, the given equation is equivalent to $$|z+3i|=|z-6i|,$$ or, put another way, to $$\\bigl|z-(-3i)\\bigr|=|z-6i|.$$\n\nSince $|z-w|$ is the distance from $z$ to $w$ for all $z,w\\in\\Bbb C,$ then the equation above says that $z$ is equidistant from $-3i$ and $6i.$ Readily, putting $z=x+iy,$ this is equivalent to saying that $(x,y)$ is equidistant from $(0,-3)$ and $(0,6),$ i.e.: $$\\sqrt{x^2+(y+3)^2}=\\sqrt{x^2+(y-6)^2}.$$ This is, of course, equivalent to your approach, and solving the preceding equation yields $y=\\frac32,$ as you say.\n\nGiven two points, the set of points equidistant from them is the perpendicular bisector of the line joining them.\n\nIf the points are $(a, b)$ and $(c,d)$, the midpoint is $((a+c)/2, (b+d)/2)$.\n\nThe slope of the line is $\\dfrac{d-b}{c-a}$, so the slope of the normal is the negative reciprocal of this or $-\\dfrac{c-a}{d-b} =\\dfrac{a-c}{d-b}$.\n\nTherefore the equation of the perpendicular bisector is $\\dfrac{y-(b+d)/2)}{x-(a+c)/2} =\\dfrac{a-c}{d-b}$.\n\nIf $a=c=0$, as in this case, the equation is $\\dfrac{y-(b+d)/2)}{x} =0$ or $y=(b+d)/2$.\n\nIn this case, $b=-3$ and $d=6$, so the equation is $y =(6-3)/2 =3/2$.", "date": "2019-07-19 03:41:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2402760/equation-of-locus-of-points-satisfied-by-frac-leftz3i-right-leftz-6i-ri", "openwebmath_score": 0.9071393013000488, "openwebmath_perplexity": 199.71871027335604, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713883126862, "lm_q2_score": 0.8652240877899775, "lm_q1q2_score": 0.8524805381784086}}
{"url": "http://rvjs.gremium-franconia.de/numerical-integration-matlab-code.html", "text": "Numerical Integration Matlab Code\n\nSymbolic Integration and Differentiation using MATLAB. Numericaldi\ufb01erentiationand integration Numerical di\ufb01erentiation with Richardson Extrapolation Assume Matlab Code: Trapezoid Rule Integration. We can do this in (at least) three di\ufb00erent ways. MATLAB\u00ae\u2019s quad Function. Math 301 \u2013 Numerical Analysis 2, Winter 2009. To derive an expression for the indefinite integral of a function, we write \u2212 int(f); For example, from our previous example \u2212 syms x int(2*x) MATLAB executes the above statement and returns the following result \u2212 ans = x^2 Example 1. Note: The five first software elements in this table are all available on the Web through the netlib library: just click on the Code name to get Netlib's keyword search and then submit the Library information in this table to find the code and then download the code. 1) is an approximation of the form IQ(f,a,b) = (b\u2212 a) Xm k=1 wkf(xk). Numerical Integration of Empirical Data: using Matlab's built-in trapezoidal integration functions (trapz and cumtrapz) to integrate discrete data when the relationship between the independent and dependent variables is not known as a function. One could also make each integration function generic, instead of making the whole package generic. There are 3 programs that needed be done. Hairer and M. m, fourcoeffs. \"Stabilization of Constrained Mechanical Systems with DAEs and Invariant Manifolds. The following MATLAB code gives a generalized illustration of numerical integration using Trapezoidal rule. 6 Trapezoidal Method and Simpson Method / 226 by example MATLAB code with a friendly interface so that students can easily. Area integral over a triangle in MATLAB -- is numerical integration possible? a triangle in MATLAB -- is numerical integration possible? the rest of the code. Springer Series in Comput. If Y is a matrix, then trapz(Y) integrates over each column and returns a row vector of integration values. Its key features are: High-level language for numerical computation, visualization, and application development; Interactive environment for iterative exploration, design, and problem solving. The code is on GitHub. It is executed at the command line prompt by typing: >> trapezoidal( a,b,nintervals) where a and b are the lower and upper limits of integration. If you read the original paper, you will notice at the end all option valuation problems are to solve an integral equation, therefore a good & proper numerical integration method becomes especially crucial. [Jaan Kiusalaas] -- \"Numerical Methods in Engineering with MATLAB is a text for engineering students and a reference for practicing engineers. This GUI can be used by entering 'nu' at the MATLAB command prompt. Description. I think that quad uses Simpson, quadl a Gauss-Lobato formula and quadgk the Gauss-Kronrod formula G7K15, and all of them use some kind of adaptative scheme. I need help developing a C-code that numerically integrates using N = 20 and N = 40, where N is the number of steps or interval. will be introduced in the context of learning several methods of numerical integration. m, fintegrand. MATLAB Programming Assignment Help, Numerical integration methods, The fundamental theorem of calculus states that the de?nite integral of a function may be found from the antiderivative of the function. , Part Five, Chapters 17 and 18 PGE 310: Formulation and Solution in Geosystems Engineering Dr. Description: This is an introductory numerical analysis course. time) and one or more derivatives with respect to that independent variable. To plot one set of values versus another (i. Area integral over a triangle in MATLAB -- is numerical integration possible? a triangle in MATLAB -- is numerical integration possible? the rest of the code. Symbolic objects such as \u03c0 or etc. [Note: Want\u2026 Read more about Numerical Integration of Tabular. Numerical Integration with Trapezoidal and Learn more about numerical integration, trapezoidal rule, simpson's rule. The logic is the same as 2-variable integration. quadl Numerical integration with adaptive Lobatto quadrature. MATLAB code for Trapezoidal Rule. Numerical Differentiation. Curve fitting -- 9. But then it finds the area between the second pair of points. View Details. x/dx we \ufb01rst divide the interval Ta;bUinto n subintervals, each of length. Research Experience for Undergraduates. Learn more about numerical integration. If you like this article, please share it with your friends and like or facebook page for future updates. Numerical Methods in Engineering with MATLAB R Second Edition Numerical Methods in Engineering with MATLAB R is a text for engi-neering students and a reference for practicing engineers. Learn more about numerical integration, random number generator, trapz, integration, cumtrapz, digital image processing. Its name is from the ancient methods for computing areas of curved figures, the most famous of which is the problem of 'squaring the circles' which means finding a square having the same area as a given circle. Try quadgk. 1 Basic Concepts In this chapter we are going to explore various ways for approximating the integral of a function over a given domain. In general, numerical quadrature involves breaking an interval into subintervals, estimating or modelling the function on each subinterval and integrating it there, then adding up the partial integrals. Numerical Analysis/Matlab Sample Code. Why my vim doesn't support hex color code?. Numerical Integration of Linear and Nonlinear Wave Equations We begin our study with an analysis of various numerical methods and boundary MATLAB Code for. Subscribe to our newsletter to get notifications about our updates via email. I think that quad uses Simpson, quadl a Gauss-Lobato formula and quadgk the Gauss-Kronrod formula G7K15, and all of them use some kind of adaptative scheme. If Y is a matrix, then trapz(Y) integrates over each column and returns a row vector of integration values. Apply MATLAB to solve linear systems of equations. When writing codes for this lab, it is surely best to use M-\ufb01les for each approximate integration scheme. It is useful for when you want to see how some integral of the experimental data progresses over time. This code return 0. MATLAB Programming Assignment Help, Numerical integration methods, The fundamental theorem of calculus states that the de?nite integral of a function may be found from the antiderivative of the function. Yes, it's inefficient for single integrals, but it's a great thing for students to look at because a) it's simple to understand (no need of calculus) and b) it's easy to code. more examples. A Friendly Introduction to Numerical Analysis, by Brian Bradie. Our objective is to employ numerical methods in examples emphasizingthe appeal of MATLAB as a programming tool. Chebfun is a collection of algorithms and an open-source software system in object-oriented MATLAB which extends familiar powerful methods of numerical computation involving numbers to continuous or piecewise-continuous functions. I think this is where my code fails. Nodes for all other n are generated on the fly with 1e-10 precision by fast root-finding algorithm. z1 = trap2(0, 2, 10, 'fun_for_integration') z2 = trap2(0, 2, 100, 'fun_for_integration') z3 = trap2(0, 2, 1000, 'fun_for_integration') and we get the Matlab results z1 = 4. GNU Octave is also freely redistributable software. Modified Euler Method with. The author was told that, in the old days. Compound Trapezoidal Method for Numerical Integration by admin in Math, Statistics, and Optimization , MATLAB Family , Numerical Integration $4. function that implements the. Asked by me with the code for the Richardson extrapolation. function that implements the. (See numerical integration for more on quadrature rules. Can anyone help me with the syntax call for numerical integration?. Petzold, and S. Impedance Cardiography (ICG) Is A Noninvasive Technology Measuring Total Electrical Conductivity Of The Thorax And Its Changes In Time To Process Continuously A Number Of Cardio Dynamic Parameters, Such As Stroke Volume, SV, Heart Rate, HR, Cardiac Output, CO, Ventricular Ejection Time, VET, Pre-ejection Period And. is there a matlab code for gaussion Learn more about simulation, integration. Numerical Analysis Project 2: Fall 2016 [Numerical Integration] Find i 6 0 f (x) dx with the function f (x) = 2 x/ (1 + x 2) using the following methods. Solution Preview. In general, numerical quadrature involves breaking an interval into subintervals, estimating or modelling the function on each subinterval and integrating it there, then adding up the partial integrals. Matlab's Numerical Integration Commands The relevant commands we consider are quad and dblquad, triplequad. The numerical value returned by MATLAB is somewhat less than half the area of a square 2 units on a side. Midpoint rule. For higher-dimensional integrals, Monte Carlo is often the tool of choice. This code return 0. I'm trying to integrate x using matlab, and that task is simple by using the following commands: syms x a=int(x) The problem is I'm not sure how to implement numerical integration. Does Matlab provide any alternatives to speed up the operation. Geometric numerical integration is synonymous with structure-preserving integration of ordinary differential equations. Chapra, Applied Numerical Methods with Matlab for Engineers and Scientists, third edition. Suppose that the interval [a,b] is split up in n subintervals, with n an even number. \u00b7 Midpoint Rule (Rectangular Rule) \u00b7 Euler-Maclaurin Formula \u00b7 Richardson Extrapolation \u00b7 Trapezoidal Rule \u00b7 Simpson's Rule There are a number of numerical methods that can be used to approximate an integral. It is based on the Fast Fourier Transform (FFT) technique and yields a numerical solution for t=a (\"a\" is a real number) for a Laplace function F(s) = L(f(t)), where \"L\" represents the Laplace transformation. function that implements the. NET Numerics under the MathNet. The MATLAB functions for the numerical evaluation of integrals has evolved from quad, through quadl and quadgk, to today's integral. This section under major construction. This code is located in a file called trapezoidal. We study numerical methods to solve linear and nonlinear equations, to interpolate and approximate data, and methods for numerical integration and di\ufb00erentiation. Roots of algebraic and transcendental equations -- 5. Numerical integration in a loop. 0 Unported License. The MATLAB functions for the numerical evaluation of integrals has evolved from quad, through quadl and quadgk, to today's integral. 0 Simple and intuitive numerical integration based on trapezoidal rule. the numerical integration; no transformation of the EOM (1) is carried out. How to download & Pay on REDS So Love symbol using MATLAB; A Small Tribute To Netaji On 23rd January using MA Runge-Kutta method (Order 4) for solving ODE using Euler's method for solving ODE using MATLAB. txt) or read online for free. a guest Sep 19th, 2017 53 Never Not a member of Pastebin yet? % the little t steps for numerical integration %obtain numerical ODE45 solution to use. Numerical Integration. Numerical integration with Monte Carlo method (on FPGA chip). In contrast to the existing Matlab built-in functions for numerical integration, this program can be used with arrays for the upper and lower integration limits. Introduces intermediate variables for small compact code and reduced operations count. Stanford Libraries' official online search tool for books, media, journals, databases, government documents and more. We begin by clicking on file on the Matlab Command Window and new. Trapezoidal rule to approximate the integral of x^2 style of the same code. function value=rsum1(f,a,b,n) %RSUM1: Computes a Riemann Sum for the function f on %the interval [a,b] with a regular partition of n points. Part B: Numerical integration (08 to 15 November 2010) Source code used in slides. I came across the book, \u2018Computational Physics\u2019, in the library here in the Dublin Institute of Technology in early 2012. More accurate methods of numerical integration are based on Gauss quadrature methods for orthogonal polynomials such as Legendre, Chebyshev, Laguerre and Hermite polynomials (optional reading - chapter 7. Caregiver Issues Bathing and Hygiene; Caregiver hiring; Caregiver training; Caregivers \u2013 professional; Family Carergivers; Caregiving Around the World. x/dx we \ufb01rst divide the interval Ta;bUinto n subintervals, each of length. Type the following into the text editor:. Section 4 is devoted to a presentation and explanation of Matlab codes for implicit Runge-Kutta, composition, and multistep. It is based on the Fast Fourier Transform (FFT) technique and yields a numerical solution for t=a (\"a\" is a real number) for a Laplace function F(s) = L(f(t)), where \"L\" represents the Laplace transformation. How to download & Pay on REDS So Love symbol using MATLAB; A Small Tribute To Netaji On 23rd January using MA Runge-Kutta method (Order 4) for solving ODE using Euler's method for solving ODE using MATLAB.$\\begingroup$Thank you, but how do you incorporate the code for Simpson's rule into the code for Bisection method? So that at each midpoint, Simpson's rule calculates a new value for the function. There is a complication in using dblquad; it does not accept variable limits. Is publishing runnable code instead of. MATLAB's programming interface gives development tools for improving code quality maintainability and maximizing performance. What we are trying to do here, is to use the Euler method to solve the equation and plot it along side with the exact result, to be able to judge the accuracy of the numerical method. Runge's Function. How to apply numerical integration on a symbolic Learn more about integration. Department of Electrical and Computer Engineering University of Waterloo 200 University Avenue West Waterloo, Ontario, Canada N2L 3G1 +1 519 888 4567. \u00b7 Midpoint Rule (Rectangular Rule) \u00b7 Euler-Maclaurin Formula \u00b7 Richardson Extrapolation \u00b7 Trapezoidal Rule \u00b7 Simpson\u2019s Rule There are a number of numerical methods that can be used to approximate an integral. Numerical Integration-1. docx 1 DOING PHYSICS WITH MATLAB COMPUTATIONAL OPTICS NUMERICAL INTEGRATION RAYLEIGH-SOMMERFELD DIFFRACTION INTEGRAL OF THE FIRST KIND Ian Cooper School of Physics, University of Sydney ian. Apply MATLAB to carry out numerical differentiation and integration. Asked by me with the code for the Richardson extrapolation. Simplify your answer as possible. Have a nice day!. m, fourcoeffs. commands to prepare a new workspace each time we run the code. Thus, the answer is same as the one obtained using the program for Trapezoidal method in MATLAB. NUMERICAL SOLUTIONS TO TWO-DIMENSIONAL INTEGRATION PROBLEMS by Alexander D. This book provides a fundamental introduction to numerical analysis for undergraduate students in the areas of mathematics, computer science, physical sciences, and engineering. m is a to compute numerical approximations to definite. Numerical Integration and Intro to Pseudospectral Methods for BVPs. C++ from Python and MATLAB, because C++ is faster. Chapra (2006, Hardcover, Revised) at the best online prices at eBay!. Input/Output: Also see, Simpson 1/3 Rule in Matlab Numerical Methods Tutorial Compilation. If Y is a vector, trapz(Y) is the integral of Y. NUMERICAL INTEGRATION. Subscribe to our newsletter to get notifications about our updates via email. The source code and files included in this project are listed in the project files section, please make sure whether the listed source code meet your needs there. MATLAB Python function In = trapezcomp (f, a, b,. This work. 6 Trapezoidal Method and Simpson Method / 226 by example MATLAB code with a friendly interface so that students can easily. Symbolic and Numerical Integration in MATLAB 1 Symbolic Integration in MATLAB Certain functions can be symbolically integrated in MATLAB with the int command. Scripts de Integra\u00e7\u00e3o n\u00famerica Newton-Cotes. Hairer and M. We begin by clicking on file on the Matlab Command Window and new. Book Description. time) and one or more derivatives with respect to that independent variable. This course will focus on the root finding and numerical integration techniques most frequently covered at the undergraduate level. Summary of first day w/ MATLAB in class. Midpoint rule. , Simpson's. MATLAB is widely used in undergraduate engineering programs as well as in industry. In Matlab code I used 'ArrayValued' to enable the integrand to receive a vector. This method is. Engineering & Matlab and Mathematica Projects for \u20ac8 - \u20ac30. Consider Example M3. In case you don\u2019t have MATLAB already installed in your system, there are lectures on the various ways you can acquire MATLAB and the procedures involved in its installation. We can do this in (at least) three di\ufb00erent ways. Learn more about histc, pdf, integrate, trapz, nan, inf, divide by zero MATLAB Answers. Some useful functions. Summary: There are problems in integrating Hamiltonian systems with normal numerical integrators, and your special initial conditions aggravate this to the point where the numerical solution has no resemblance with the correct one. This method is. pdf), Text File (. MATLAB is a high-level language and interactive environment for numerical computation, visualization, and programming. Numerical Methods or Numerical Analysis is a subject included in all types of engineering curriculum around the world. How to download & Pay on REDS So Love symbol using MATLAB; A Small Tribute To Netaji On 23rd January using MA Runge-Kutta method (Order 4) for solving ODE using Euler's method for solving ODE using MATLAB. If Y is a vector, trapz(Y) is the integral of Y. Numerical Integration - Free download as PDF File (. Numerical Methods in Engineering with MATLAB R Second Edition Numerical Methods in Engineering with MATLAB R is a text for engi-neering students and a reference for practicing engineers. Z = trapz(X,Y) computes the integral of Y with respect to X using trapezoidal integration. Unless stated otherwise, the examples below evaluate the integral \\(\\int_0^{10} x^2 \\, dx = \\frac{1000}{3} \\approx 333. If Y is a multidimensional array, then trapz(Y) integrates over the first dimension whose size does not equal 1. trapz Numerical integration with the trapezoidal rule. Numericaldi\ufb01erentiationand integration Numerical di\ufb01erentiation with Richardson Extrapolation Assume Matlab Code: Trapezoid Rule Integration. Specifically, the integral is introduced by using the best approximation scheme (Legendre Polynomials) to approximate a vector valued function whose indefinite integral is not easy to be explicitly written down. are accepted. C Program for Numerical Integration (Trapezoidal Rule, Simpson's Rule and Boole's Rule Program /* This program is for numerical integration of numerical methods Here we are going to solve the integration of defined functio. Solving Numerical Integral Implicitly. A followup post of my previous entry Using Quadrature method for option valuation, where I tested the alternative numerical method for option valuation, QUAD. in the function. This book will enable readers to solve problems without needing to understand all the details of the underlying theory of numerical methods. Matlab code. The order of integration (and therefore the bounds) is from the innermost integral to the outermost one. Dynamics and Vibrations MATLAB tutorial School of Engineering Brown University This tutorial is intended to provide a crash-course on using a small subset of the features of MATLAB. b) computes a numerical approximation of. Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. Array-valued function flag, specified as the comma-separated pair consisting of 'ArrayValued' and a numeric or logical 1 (true) or 0 (false). MATLAB EXAMPLES Interpolation and Integration. The numerical evaluation in this case was performed using numerical routines borrowed from MuPAD, which takes care of symbolic calculations. A brief introduction to the Simpson\u2019s 1/3 rd rule and a uniform interval Composite Simpson\u2019s 1/3 rd Rule implementation. Victor Zalizniak, in Essentials of Scientific Computing, 2008. Body of the package implementing numerical integration: package body Integrate is. We can do this in (at least) three di\ufb00erent ways. In fact, the built-in capabilities of MATLAB are used to perform numerical computations, which are very useful in enormous fields of applied science and engineering, including: Root finding and equation solving Solving system of equations Eigenvalues, eigenvectors and eigendecomposition Singular Value. Table 2: A Matlab module for DFT calculations. The goal of the book. MATLAB code demo in class. Midpoint (Rectangular) Numerical Integration Matlab Code by admin in Math, Statistics, and Optimization, MATLAB Family, Numerical Integration. MATLAB code to perform numerical integration and differentiation % Numerical integration and differentiation % Using built-in Matlab functions % Clear the command window. SoDiOpt is a MATLAB-based code that performs numerical integration of Optimization-Constrained Differential Equations (OCDE). Learn more about numerical integration, random number generator, trapz, integration, cumtrapz, digital image processing. , Part Five, Chapters 17 and 18 PGE 310: Formulation and Solution in Geosystems Engineering Dr. Matlab Gauss Jordan Code Matlab Gauss Seidel Code Notes on linear and nonlinear least squares Data for homework 2 Practice Midterm Notes on Taylor methods Matlab code for Euler's method Matlab code for mid-point method Backward Euler notes Adam's Moulton Code O and A stability notes Boundary value problems and shooting Practice Final. Miranda, M. To figure this out, consider exactly what it is that both processes do. Matlab code for incompatible mode elements, short beam input file long beam input file; Matlab code demonstrating volumetric locking linear quad input file quadratic quad input file; L10 Hourglass control, selective reduced integration, B-bar elements. Search for jobs related to Integration matlab or hire on the world's largest freelancing marketplace with 15m+ jobs. Numerical Methods or Numerical Analysis is a subject included in all types of engineering curriculum around the world. Here is my implementation of Martin Floden's matlab code for the routine. In the numerical analysis literature, the Verlet method is also knows as the explicit central difference method''. To derive an expression for the indefinite integral of a function, we write \u2212 int(f); For example, from our previous example \u2212 syms x int(2*x) MATLAB executes the above statement and returns the following result \u2212 ans = x^2 Example 1. Help with numerical integration in matlab. My simpson method is correct, but my adaptive method does not seem to work for the integral( sin(2*pi*x)\u00b2 ) ranging from -1 to 1 The following code represents the adaptive simpson method. Given ) = !(#)the approximation of the Area (5) under the curve can be found dividing the area up into. Plot the results against the. A numerical library for adaptive Sparse Grids (mingw-w64) matlab-r2018b: 9. I need help developing a C-code that numerically integrates using N = 20 and N = 40, where N is the number of steps or interval. For the majority of you, it has been nearly six months since you last used MATLAB. Compound Trapezoidal Method for Numerical Integration by admin in Math, Statistics, and Optimization , MATLAB Family , Numerical Integration$4. It provides tools for building applications with custom graphical interfaces. C++ from Python and MATLAB, because C++ is faster. An algorithm to numerically invert functions in the Laplace field is presented. To figure this out, consider exactly what it is that both processes do. Geometric numerical integration is synonymous with structure-pre-ser-ving integration of ordinary differential equations. 141120007827708. Goal: given continuous function f(x) of one variable, compute \u222b f(x) dx over interval from a to b. The Simpson\u2019s 1/3 rule is a numerical method to find the integral within some finite limits and. We'll now consider how to implement these in MATLAB. \u2022 In the time domain, ODEs are initial-value problems, so all the conditions. Hello: I have not been able to find out what is the underlying quadrature formula in Matlab's builtin function integral. Numerical Integration \"Numerical Methods with MATLAB\", Recktenwald, Chapter 11 and \"Numerical Methods for Engineers\", Chapra and Canale, 5th Ed. Garcia,\u2217 A. numerical integration - MatLab algorithm for composite Simpson's rule I have tried, just for the fun of it, to write a MatLab-code for the composite Simpson's rule. Numerical integration using SCILAB Integrals can be interpreted as the area under the curve of the function f(x) in a given interval a < x < b. Numerical Integration with the Trapezoidal Rule. An Introduction to Numerical Methods: A MATLAB\u00ae Approach, Fourth Edition continues to present a wide range of useful and important algorithms for scientific and engineering ap. Converting serial MATLAB applications to parallel MATLAB applications generally requires few code modifications and no programming in a low-level language is necessary. Description: This is an introductory numerical analysis course. Summary of first day w/ MATLAB in class. The choice of numerical methods was based on their relevance to engineering prob-lems. These notes, prepared for the Durham summer school 2002, are complementary to the monograph of Hairer, Lubich and Wanner \"Geometric Numerical Integration\". It approaches the subject from a pragmatic viewpoint; theory is kept at a minimum. The second output FY is always the gradient along the 1st dimension of F, going across rows. quadl Numerical integration with adaptive Lobatto quadrature. Simpsons Algorithm for numerical integration using Trapezoid rule for numerical integration using MAT REDS Library: 15. numeric::quadrature(f(x), x = a. It provides tools for building applications with custom graphical interfaces. Question: Problem 2. In addition, it scales well with many workers. The name MATLAB stands for matrix laboratory. the vectorized methods are not as easy to read, and take fewer lines of code to write. Numerical Solution using MATLAB. Numerical Methods for ODE in MATLAB MATLAB has a number of tools for numerically solving ordinary di\ufb00erential equations. There are various reasons as of why such approximations can be useful. Verschelde's MCS 320 Symbolic Computing with MATLAB (and Maple) Lectures Desmond J. numerical integration code Search and download numerical integration code open source project / source codes from CodeForge. MATLAB is widely used. You can integrate from 1 down to some number larger than r, where x=r is the location of the singularity. It is executed at the command line prompt by typing: >> trapezoidal( a,b,nintervals) where a and b are the lower and upper limits of integration. If you were to take the integral of 2x from 0 to 2, where 0 is the lower bound and 2 is the upper bound you would get the following:. Compared to the numerical integration methods, like the program of Simpson 1/3 rule in C given above, the analytical method of integration is quite difficult and time consuming while applying to complex engineering problems. In numerical analysis, Romberg's Method (Romberg 1955) generates a triangular array consisting of numerical estimates of the definite integral by applying Richardson extrapolation (Richardson 1910) repeatedly on the trapezium rule or the rectangle rule (midpoint rule). in Matlab command window. Subscribe to our newsletter to get notifications about our updates via email. Introduction. In the first type, derivative of a function is given and we want to find the function. Initial Value Problem. As we see from the plot, this is more than we need. Numerical Integration 209 Quadrature Integration 216 the work you\u2019re doing can\u2019t be done by hand and will require a numerical solution. See the Matlab help \ufb01les for other integration commands. E-Book Review and Description: Numerical Methods using MATLAB, 3e, is an in depth reference offering numerous of useful and crucial numerical algorithms that could be carried out into MATLAB for a graphical interpretation to help researchers analyze a selected consequence. We'll now consider how to implement these in MATLAB. Numerical Solution of Di\ufb00erential Equations: MATLAB implementation of Euler's Method The \ufb01les below can form the basis for the implementation of Euler's method using Mat-lab. You may redistribute it and/or modify it under the terms of the GNU General Public License (GPL) as. This method assumes linear behavior between the data points, and accuracy may be reduced when the behavior between data points is nonlinear. I think this is where my code fails. trapz performs discrete integration by using the data points to create trapezoids, so it is well suited to handling data sets with discontinuities. Research Experience for Undergraduates. If Y is a multidimensional array, then trapz(Y) integrates over the first dimension whose size does not equal 1. Introduction. Watch Online Four sections of this video tutorial are available on YouTube and they are embedded into this page as playlist. quadl Numerical integration with adaptive Lobatto quadrature. Numerical Integration with the Trapezoidal Rule. Trapezoidal Rule for Numerical Integration Trapezoidal Rule for Numerical Integration Internet hyperlinks to web sites and a bibliography of articles. numeric::quadrature returns itself symbolically if the integrand f(x) contains symbolic objects apart from the integration variable x that cannot be converted to numerical values via float. Simpson's rule for numerical integration 25754-simpson-s-rule-for-numerical-integration), MATLAB Central the trapz code but then the simps seems to not work. Applied numerical methods using matlab (wiley,2005) 1. Ofdm matlab code using bpsk. MATLAB allows matrix manipulations, plotting of functions and data, implementation of algorithms, creation of user interfaces, and interfacing with programs written in other languages. Hi, So I have this function that takes data sample locations and function samples at those locations as input and returns an approximation of the integral over the sample range based on the trapezoidal rule. Parallel processing operations such as parallel for-loops and message-passing functions let you implement task- and data-parallel algorithms in MATLAB. Compound Trapezoidal Method for Numerical Integration by admin in Math, Statistics, and Optimization , MATLAB Family , Numerical Integration \\$4. An example to price an Arithmetic Average fixed strike Call option in the Black-Scholes framework using Monte Carlo Control Variate. topic of numerical integration is taken up in Chapter 7 and in Chapter 8 meth-ods for the numerical solution of ordinary di\ufb00erential equations are explored. Doing Physics with Matlab op_rs1. = ] = ] ENGRD 241 / CEE 241: Engineering Computation Numerical Integration 14 Numerical Integration Improving the estimate of the integral fit Lagrange polynomials to three points (a pair of segments) integrate those polynomials to obtain a general formula the resulting function must correctly integrate quadratics x 0 x n ENGRD 241 / CEE 241. a guest Sep 19th, 2017 53 Never Not a member of Pastebin yet? % the little t steps for numerical integration %obtain numerical ODE45 solution to use. MATLAB codes for Romberg integration. pdf from MAE 284 at University of Alabama, Huntsville. When writing codes for this lab, it is surely best to use M-\ufb01les for each approximate integration scheme. In its simplest form, you pass the function you want to differentiate to diff command as an. Matlab code for selective reduced integration; Matlab code for reduced integration with. Numerical gradients, returned as arrays of the same size as F. Numerical Integration. Miranda, M. Apply MATLAB to carry out numerical differentiation and integration. 9) Explain why numerical differentiation is inferior to integration in the presence of noise. The following MATLAB code gives a generalized illustration of numerical integration using Trapezoidal rule. How do I improve Numerical Integration Speed?. MATLAB - Differential - MATLAB provides the diff command for computing symbolic derivatives. Ample material is presented so that instructors will be able to select topics appropriate to their. Calculation of numerical integration of an expression is made easier. To solve the Falkner-Skan equation a fourth-order Runge-Kutta integration scheme was used. First results of axisymmetric. Repeat for the midpoint rule, trapezoidal rule, and Simpson\u2019s rule. Does Matlab provide any alternatives to speed up the operation. Area integral over a triangle in MATLAB -- is numerical integration possible? a triangle in MATLAB -- is numerical integration possible? the rest of the code. Numerical Integration \u00a71 The Newton-Cotes Rules \u00a72 Composite Rules \u00a73 Adaptive Quadrature \u00a74 Gauss Quadrature and Spline Quadrature \u00a75 Matlab's Quadrature Tools An m-point quadrature rule Q for the de\ufb01nite integral I(f,a,b) = Zb a f(x)dx (4. ALL PRODUCTS FROM Gauss Quadrature Method for Numerical Integration MATLAB Family, Numerical Integration. For watching full course of Numerical Computations, visit this page. Choose M-file. This creates a circular reference which I can't figure out how to code in Matlab. All MATLAB code should also be submitted as an. How to apply numerical integration on a symbolic Learn more about integration. Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. though I'd like to know what I should have fixed in the code (more cleaner now). 999e3 numerical representation in Matlab [closed] Comparing the FFT to numerical integration in Matlab.", "date": "2020-02-22 15:07:46", "meta": {"domain": "gremium-franconia.de", "url": "http://rvjs.gremium-franconia.de/numerical-integration-matlab-code.html", "openwebmath_score": 0.5875372886657715, "openwebmath_perplexity": 977.7125887879093, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9852713826904225, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.852480536737994}}
{"url": "https://math.stackexchange.com/questions/3052408/how-can-i-prove-that-x-n1-x-n-sinx-n-converges-for-any-x-0", "text": "How can I prove that $x_{n+1} = x_n \\sin(x_n)$ converges for any $x_0$?\n\nI have already prove that it is true if exist $$\\sin(x_m) = 1$$ or $$\\sin(x_m) = -1$$. So I need make proof only for absolute-decreasing progression. But I am afraid that $$\\sin(x_n)$$ can has limit $$1$$ and can go to $$1$$ very fast. Is it a good way to count limit $$\\sin(x_n)$$? And what solution is correct?\n\n\u2022 Welcome to the website. In future, please typeset your equations using Mathjax for better presentation. \u2013\u00a0Shubham Johri Dec 25 '18 at 20:35\n\u2022 I think you mean $x_n\\to 0$ or $\\sin x_n\\to 1$. \u2013\u00a0J.G. Dec 25 '18 at 20:41\n\u2022 If the limit exists it should satisfy $x_{n+1} = x_n$. Think about what this tells you about possible limit points, and then look at what happens if $x_n$ is above these points or below these points. \u2013\u00a0tch Dec 25 '18 at 20:43\n\u2022 @TylerChen \"If the limit exists, it should satisfy $x_{n+1} = x_n$\". What the heck does this mean \u2013\u00a0mathworker21 Dec 25 '18 at 20:45\n\u2022 @Cesareo There are infinitely many solutions. Any $\\pi/2+2k\\pi$ works. \u2013\u00a0ImNotTheGuy Dec 25 '18 at 21:12\n\nLet $$y_n = |x_n|$$\n\nThen $$y_{n+1} = y_n |\\sin y_n| \\leq y_n \\forall n$$ (*)\n\nSo $$y_n$$ is decreasing and bounded below by $$0$$. Therefore it converges to some limit $$l$$. By passing to the limit in (*) we get that either $$l=0$$ or $$sin l= 1$$ giving the family of solutions discussed in the comments, namely $$l=2k\\pi+\\frac{\\pi}{2}$$\n\nNow note that if $$y_n$$ converges to $$0$$ then $$x_n$$ also converges to $$0$$. If $$y_n$$ converges to a nonzero value $$l$$ then a little more work will also prove that $$x_n$$ also converges to either $$l$$ or $$-l$$ because for n large enough $$y_n$$ will be between $$l$$ and $$l+\\frac{\\pi}{2}$$ hence $$x_n$$ will have constant sign\n\nEither way, $$x_n$$ will also be convergent\n\nNote that $$0\\le|x_{n+1}|=|x_n\\sin x_n|\\le|x_n|$$, so the sequence of absolute values, $$|x_0|,|x_1|,|x_2|,\\ldots$$ definitely converges to a nonnegative limit $$L(x_0)$$ satisfying the equation $$L(x_0)=L(x_0)|\\sin L(x_0)|$$. If $$L(x_0)=0$$, we're done: $$|x_n|\\to0$$ implies $$x_n\\to0$$. If $$L(x_0)\\gt0$$, we need only worry about the possibility that $$x_n$$ approaches both $$L(x_0)$$ and $$-L(x_0)$$. But this can't happen: If $$x_n\\approx\\sigma L(x_0)$$ where $$\\sigma=\\sin L(x_0)\\in\\{1,-1\\}$$, then, using the fact that $$\\sin(\\sigma u)=\\sigma\\sin u$$ for all $$u$$ and $$\\sigma^2=1$$, we have\n\n$$x_{n+1}=x_n\\sin x_n\\approx\\sigma L(x_0)\\sin(\\sigma L(x_0))=\\sigma L(x_0)(\\sigma\\sin L(x_0))=\\sigma L(x_0)(\\sigma^2)=\\sigma L(x_0)\\approx x_n$$\n\nRemarks: As ImNotTheGuy observes in a comment beneath the OP, $$\\pi/2+2k\\pi$$ (with $$k\\in\\mathbb{Z}$$) is a fixed point for the mapping $$f(x)=x\\sin x$$, which means there are sequences that tend to each such limit, as well as sequences that tend to $$0$$. There are, in fact, uncountably many sequences tending to each of the possible limits: If $$k\\ge0$$, for example, and $$x_n=\\pi/2+2k\\pi+\\epsilon$$ with $$\\epsilon\\gt0$$ but very small, then $$x_{n+1}=x_n\\cos\\epsilon =\\pi/2+2k\\pi+\\epsilon'$$ with\n\n$$\\epsilon'=\\epsilon\\cos\\epsilon-(1-\\cos\\epsilon)(\\pi/2+2k\\pi)\\approx\\epsilon-(\\epsilon^2/2)(\\pi/2+2k\\pi)\\lt\\epsilon$$\n\n(but still positive), so each positive limit has a \"basin of attraction\" that includes an open inteval of values for $$x_0$$ slightly greater than the limit (and likewise, for each negative limit, an open interval of values slightly less than the limit -- since $$|x_{n+1}|=|x_n\\sin x_n|\\le|x_n|$$, successive terms in any sequence always get closer to the origin).\n\nIn general If $$x_0$$ is an initial approximation of a \ufb01xed point , the \ufb01xed point iteration generates a sequence of approximants by $$x_{n+1}=\\varphi (x_n),$$ where $$\\varphi$$ is a continuous function. Suppose that at the \ufb01xed point that we indicate with $$\\alpha$$ we have $$\\varphi '(\\alpha)=\\varphi ''(\\alpha)=\\cdots\\varphi ^{p-1}(\\alpha)=0, \\quad \\varphi ^p(\\alpha)\\neq 0 \\qquad (*);$$ then we have the following theorem:\n\nLet $$\\alpha$$ be a fixed point of $$\\varphi$$ and $$I_{\\epsilon}:=\\{x\\in\\Bbb R: |x-\\alpha|<\\epsilon\\}$$. Asuume that $$\\varphi \\in \\Bbb C^p[I_{\\epsilon}]$$ satisfes $$(*)$$. If $$M(\\epsilon):=|\\varphi '(t)|<1$$ then the fixed point iteration converges to $$\\alpha$$ for every $$x_0\\in I_{\\epsilon}$$ and the order the convergence is $$p$$.\n\nNote that to use this theorem you have to know tha fxed point $$\\alpha$$ (you can plot the function for example and obtain an approximation of point). You can find this argument in this book.", "date": "2019-07-22 12:44:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3052408/how-can-i-prove-that-x-n1-x-n-sinx-n-converges-for-any-x-0", "openwebmath_score": 0.9725823998451233, "openwebmath_perplexity": 137.2989419501075, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713878802045, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8524805309560083}}
{"url": "https://math.stackexchange.com/questions/3286150/understanding-the-difference-between-pre-image-and-inverse", "text": "# Understanding the difference between pre-image and inverse\n\nI am a little confused as to what the difference between the pre-image and the inverse of a function are and how to find each given a particular function ( I had though they were essentially the same thing but I realise now I was mistaken in that thought ).\n\nFrom Wikipedia the definitions given are :\n\nInverse\n\nLet $$f$$ be a function whose domain is the set $$X$$, and whose image (range) is the set $$Y$$. Then $$f$$ is invertible if there exists a function $$g$$ with domain $$Y$$ and image $$X$$, with the property: $$f ( x ) = y \\iff g ( y ) = x .$$ If $$f$$ is invertible, the function $$g$$ is unique, which means that there is exactly one function $$g$$ satisfying this property (no more, no less). That function $$g$$ is then called the inverse of $$f$$, and is usually denoted as $$f ^{\u22121}$$\n\nPre-image\n\nLet $$f$$ be a function from $$X$$ to $$Y$$. The preimage or inverse image of a set $$B\\subseteq Y$$ under $$f$$ is the subset of $$X$$ defined by $$f ^{\u2212 1 }[ B ] = \\{ x \\in X \\mid f ( x ) \\in B \\}$$.\n\nSo using an example I want to see if I have this about right ( please correct any mistakes I make)\n\nExample 1:\n\nLets say $$f:\\Bbb R \\rightarrow \\Bbb R$$\n\n$$f(x)=x^2$$\n\nFor this example, clearly $$f$$ cannot be invertible (hence no inverse) as there exists no function $$g$$ which will satisfy $$f ( x ) = y \\iff g ( y ) = x$$. (as it would only map to positive values of $$\\Bbb R$$ (i.e. not the whole set))\n\nThe pre-image of this function I believe is related to the inverse except it does not require that we map to the whole set $$\\Bbb R$$, but rather just a subset of it. Therefore we can find the function $$f^{-1}$$ in an analogous way to to finding the inverse we just have to be more considerate about what the co-domain of this function is.\n\nSo if $$f(x)=x^2 \\Rightarrow y=x^2$$swap variables to get $$x=y^2 \\Rightarrow \\sqrt{x}=y=f^{-1}$$\n\nSo the pre-image is the set $$f ^{\u2212 1 }[ \\Bbb R_+ ] = \\{ x \\in X \\mid f ( x ) \\in \\Bbb R_+, f^{-1}=\\sqrt{x} \\}$$\n\nExample 2:\n\nAn example of an invertible function would be $$f:\\Bbb R \\rightarrow \\Bbb R$$\n\n$$f(x)=5x$$, as a function $$g(x)=x/5$$ has domain $$\\Bbb R$$ and range $$\\Bbb R$$ and satisfies $$f ( x ) = y \u21d4 g ( y ) = x .$$\n\nThe pre-image in this case will be equal to the inverse.\n\nCould anyone please explain to me any mistakes I'm making here ?\n\n\u2022 For starters, they are two very different objects. The pre-image of a function is a subset of the domain and the inverse function is a function from the range back to the domain that satisfies certain properties. A function may not be invertible but we can always talk about pre-image of a function. \u2013\u00a0Anurag A Jul 7 '19 at 20:58\n\u2022 @AnuragA I know now that they are not the same thing , I was just wondering if I was correct in identifying how they are different \u2013\u00a0excalibirr Jul 7 '19 at 21:00\n\u2022 No, one peaks in almost all cases of the preimage of a subset of the target space. Take, for instance, the squaring function $f(x)=x^2$, and ask for the preimage of the set of odd numbers. Then this will be the set of all $\\pm\\sqrt{2k+1}$, with $k$ running through all integers. If one were to speak of the \u201cpreimage of a function\u201d, presumably that would be the preimage of the whole target space, which is all of the domain of definition of the function. \u2013\u00a0Lubin Jul 7 '19 at 21:09\n\nThe biggest difference between a preimage and the inverse function is that the preimage is a subset of the domain. The inverse (if it exists) is a function between two sets.\n\nIn that sense they are two very different animals. A set and a function are completely different objects.\n\nSo for example: The inverse of a function $$f$$ might be: The function $$g:\\mathbb R \\to \\mathbb R: g(x) = \\sqrt[3]{x-9}$$. Whereas the preimage of a set $$B$$ of the function might be $$[1,3.5)\\cup \\{e, \\pi^2\\}$$.\n\nNow $$g(x) = \\sqrt[3]{x-9}$$ and $$[1,3.5)\\cup \\{7, \\pi^2\\}$$ are completely different types of things.\n\nThis will be the case if $$f$$ is $$f:\\mathbb R \\to \\mathbb R: f(x) = x^3 + 9$$ and $$B= [10, 51.875) \\cup \\{352, \\pi^6 + 9\\}$$.\n\nThe inverse $$f^{-1}(x)$$ (if it exist) is the function $$g$$ so that if $$f(x) = y$$ if and only if $$g(y) = x$$. So if $$f(x) = x^3 + 9 = y$$ then if such a function exists it must be that $$g(y)^3 + 9 = y$$ so $$g(y)^3 = y-9$$ and $$g(y) = \\sqrt[3]{y-9}$$ so $$g(x) = \\sqrt[3]{x-9}$$.\n\nThat's that.\n\nThe pre-image of $$A= [10, 51.875) \\cup \\{352, \\pi^6 + 9\\}$$ is the set $$\\{x\\in \\mathbb R| f(x) \\in [10, 51.875) \\cup \\{352, \\pi^6 + 9\\}\\}=$$\n\n$$\\{x\\in \\mathbb R| x^3 + 9 \\in [10, 51.875) \\cup \\{352, \\pi^6 + 9\\}\\}=$$\n\n$$\\{x\\in \\mathbb R| x^3 \\in [1, 42.875) \\cup \\{343, \\pi^6 \\}\\}=$$\n\n$$\\{x\\in \\mathbb R| x \\in [1, 3.5) \\cup \\{7, \\pi^2 \\}\\}=$$\n\n$$[1, 3.5) \\cup \\{7, \\pi^2 \\}\\}$$.\n\nAnd that's the other.\n\n........\n\nNow that's not to say the inverse of a function and the pre-image of a set under the function aren't related. They are. But they refer to different concepts. This is similar to how a rectangle and its area are related. But one is a geometric shape... the other is a positive real number. THey are two different types of animals.\n\n....\n\nI'll add more in an hour or so but I have to take the dog for a walk. I'll be back.\n\n.....\n\nIt occurred to me as I was walking the dog that maybe what is confusing you is that the inverse function (if it exists) and the preimage of a set have very similar notation and the only way to tell them apart is in context.\n\nIf $$f$$ is invertible then the inverse function is written as $$f^{-1}$$ so if $$f(x) = x^3 + 9$$ then $$f^{-1}(x) = \\sqrt[3]{x-9}$$.\n\nBut the preimage of $$B$$ under $$f$$ whether $$f$$ is invertible or or not is writen as $$f^{-1}(B)$$.\n\nSo if $$f(x) = x^3 + 9$$ then $$f^{-1}(17) = 2$$ means that if you enter $$17$$ into the function $$\\sqrt{x -9}$$ you get $$2$$. But $$f^{-1}(\\{17\\})=\\{3\\}$$ and $$f^{-1}(\\{36,17\\}) = \\{2,3\\}$$ means that set of values that will output $$\\{17\\}$$ is the set $$\\{2\\}$$ and the set of values that will output $$\\{36,17\\}$$ is the set $$\\{2,3\\}$$.\n\nA few things to note:\n\nIf $$f$$ is invertible then the preimage of a set is the same thing as the image of the set under the inverse function and that means the notation is compatible.\n\nIf $$f(x) = x^3 + 9$$ then $$f^{-1}([1,36)) = [1,3)$$ can be interpretated as both the the image of the set under the inverse function: $$f^{-1}([1,36))= \\{f^{-1}(x) = g(x) = \\sqrt[3]{x-9}| x\\in [1,36)\\}$$\n\nOR it can be interpreted as the preimage for $$f$$: $$f^{-1}([1,36)) = \\{x\\in \\mathbb R| f(x) \\in [1,36)\\}$$.\n\nbut this is not the case if $$f$$ is not invertible.\n\nSay $$f:\\mathbb R \\to [-1,1]; f(x)\\to \\sin x$$. This is not invertible.\n\nThe pre-image of$$B= \\{\\frac {\\sqrt 2}2\\}$$ is $$\\{...-\\frac {11\\pi}4, -\\frac {9\\pi}4,-\\frac{3\\pi}4,-\\frac \\pi 4, \\frac \\pi 4, \\frac {3\\pi}4, \\frac {9\\pi}4, \\frac {11\\pi}4,....\\}$$ this is still written as $$f^{-1}( \\{\\frac {\\sqrt 2}2\\})$$ even though there is no function $$f^{-1}:[-1,1]\\to \\mathbb R$$.\n\nAnother thing to note is that not all the elements in $$B$$ have to have pre-image values.\n\nIf $$f= x^2+9$$ then $$f^{-1}(\\{8\\}) = \\emptyset$$. This is because $$\\{x\\in \\mathbb R| f(x) = x^2 + 9 \\in \\{8\\}\\} = \\emptyset$$.\n\nAnd some elements may have many preimages.\n\nAnd $$\\sin^{-1}(\\{\\frac {\\sqrt2} 2}$$ showed.\n\n\u2022 Thank you for such a brilliantly comprehensive answer :) there's literally nothing else I could think to ask . \u2013\u00a0excalibirr Jul 15 '19 at 7:34\n\u2022 Actually , I did manage to think of question lol, (it's in the context of continuity in topology) say we have the function $f=x^2$ and a basis set $(-q,q), q \\in \\Bbb Q$ in our target. then I know that $f^{-1}((-q,q))=\\{ x \\in X| x\\in (-\\sqrt{q}, \\sqrt{q})\\}$ but why isn't it $(\\sqrt{-q}, \\sqrt{q})$ instead ?, also say we have the set (a,b) (a<b) with the same function then is $f^{-1}=(-\\sqrt{b},-\\sqrt{a}) \\cup (\\sqrt{a},\\sqrt{b})$ ? \u2013\u00a0excalibirr Jul 15 '19 at 9:28\n\nWhen we talk about pre-image then it has two components: a set and a function. So when we say $$f^{-1}[B]$$, then we want the pre-image of the set $$B$$ (a subset of the co-domain) under the function $$f$$. So asking about pre-image of a function is a bit ambiguous.\n\nLet us consider $$f:\\{1,2,3\\} \\rightarrow \\{a,b,c\\}$$ such that $$f(1)=a, f(2)=a$$ and $$f(3)=c$$. Then \\begin{align*} f^{-1}\\left[\\{a\\}\\right] & =\\{1,2\\}\\\\ f^{-1}\\left[\\{c\\}\\right] & =\\{3\\}\\\\ f^{-1}\\left[\\{a,c\\}\\right] & =\\{1,2,3\\}\\\\ f^{-1}\\left[\\{b\\}\\right] & =\\emptyset\\\\ f^{-1}\\left[\\{a,b,c\\}\\right] & =\\{1,2,3\\} \\end{align*}\n\nThe function $$f$$ (as described above) is not invertible. So relating the inverse function (which doesn't exist) to the inverse image is meaningless in this case.\n\nOne can check invertibility of a function $$f: A \\rightarrow B$$ by checking the inverse images of singleton subsets of the co-domain.\n\nWhat it means is that: if we can ensure that for every $$b \\in B$$, the inverse image set $$f^{-1}\\left[\\{b\\}\\right]$$ has exactly one element (this is to ensure both one-one and ontoness), then $$f$$ is invertible.\n\nSo, there's no such thing as the preimage of a function. Functions can have inverses; functions do not have preimages.\n\nAn inverse is something that certain functions have, and the inverse of a function is another function.\n\nGiven a function, a preimage is something that sets have, and the preimage of a set is another set.\n\nSpecifically:\n\nGiven a function $$f : A \\to B$$, that function may or may not have an inverse. If it does, then that inverse is a function $$B \\to A$$.\n\nGiven a function $$f : A \\to B$$, and a set $$s$$ which is a subset of $$B$$, that set always has a preimage under $$f$$. That preimage is a subset of $$A$$.\n\nThat might clear up some of your confusion.", "date": "2021-05-10 01:39:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3286150/understanding-the-difference-between-pre-image-and-inverse", "openwebmath_score": 0.8745052814483643, "openwebmath_perplexity": 121.3295755827918, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713861502773, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8524805294592337}}
{"url": "http://math.stackexchange.com/questions/73945/stuck-trying-to-prove-an-inequality/74812", "text": "# Stuck trying to prove an inequality\n\nI have been trying to prove (the left half of) the following inequality:\n\n$$\\underbrace{\\sum_i \\sum_j |x_i| \\le \\sum_i \\sum_j |x_i + x_j|}_\\textrm{?} \\le 2 \\sum_i \\sum_j |x_i|$$\n\n(All $x_i$s are arbitrary reals and sums are over $1, 2, \\dots, n$)\n\nThe right half follows by a simple application of the triangle inequality, but the left half isn't as simple.\n\nI tried induction, and I could reduce the above problem to proving the following statement:\n\n$$\\sum_i |x_i + x_{n+1}| + \\sum_{i=1}^{n+1} |x_i + x_{n+1}| \\ge \\sum_{i=1}^{n+1} |x_i| + n|x_{n+1}|$$ or more simply (replacing $x_{n+1}$ by $y$), $$2\\sum_i |x_i + y| \\ge \\sum_i |x_i| + (n-1)|y|$$\n\nHowever, I don't know if the above \"reduction\" is any simpler than the original problem itself!\n\nCan I continue the above line of thought, or is there an easier way to solve this problem? Any help is appreciated! :)\n\n-\nSince the LHS and the RHS has no variable in $j$, you may replace $\\sum_i\\sum_j|x_i|$ by $n\\sum_i|x_i|$. \u2013\u00a0 AD. Oct 19 '11 at 9:34\nIn the summation: $\\sum_i \\sum_j |x_i + x_j|$ - Do you mean that i and j have the domain $0 <=i <=n , \\$0 <=j <=n ? Or does each summation have its own lower and upper limits? \u2013\u00a0 Emmad Kareem Oct 22 '11 at 8:47\nCan you explain to me how this works for 1,1,-1? I get: $9 \\le 2+2+0+0+0+0$ \u2013\u00a0 Phira Oct 22 '11 at 12:24\n@Phira I get $9\\leq 2+0+2+2+0+2+2+0+0=10$. \u2013\u00a0 Mike Wierzbicki Oct 22 '11 at 13:44\nwhy was this two year old question closed instead of the two week old question? \u2013\u00a0 robjohn Oct 9 '13 at 23:58\n\nI believe that some smart tricks will solve the problem (e.g. by recasting it into the form of some known inequality), but what came to my mind first was a proof inspired by the simplex method. Although it does not look smart, it may be of some interests to the readers:\n\n1. Note that by scaling the inequality, we may assume, without loss of generality, that $|x_i|\\le1$ for all $i$.\n2. So we want to prove the inequality for all $(x_1,\\ldots,x_n)$ inside the hypercube $[-1,1]^n$.\n3. Divide the hypercube $[-1,1]^n$ into pieces by the hyperplanes $x_i=x_j$ and $x_i=0$. For instance, when $n=2$, the four \"hyperplanes\" (straight lines in this case) $x=y$, $x=-y$, $x=0$ and $y=0$ will divide the square $[-1,1]^2$ into eight pieces in a manner akin to the Union Jack.\n4. In mathematics, we call each of these pieces a simplex.\n5. Note that inside each simplex, each $x_i$ or $x_i+x_j$ does not change sign.\n6. Therefore, inside each simplex, the function $f(x_1,\\ldots,x_n)=-\\sum_i \\sum_j |x_i| + \\sum_i \\sum_j |x_i + x_j|$ is identical to $-\\sum_i \\sum_j s_ix_i + \\sum_i \\sum_j t_{ij}(x_i + x_j)$ for some constants $s_i, t_{ij}=\\pm1$.\n7. In other words, $f$ is a linear function inside each simplex.\n8. In linear programming, it is well known that the extremum of a linear function can be attained at the corners of the simplex. You can imagine that when you move a plane closer and closer to a simplex, it will touch the simplex at a corner first (or it touches the whole face, with the corners of the face included).\n9. The corners of our simplices, however, are those points such that $x_i\\in\\{-1,0,1\\}$.\n10. So, we only need to prove the inequality for these corner points.\n11. Let $a$ of the $x_i$s are equal to 1, $b$ of them are equal to $-1$ and $n-a-b$ of them are equal to zero. Then \\begin{align} &-\\sum_i \\sum_j |x_i| + \\sum_i \\sum_j |x_i + x_j|\\\\ =&-n(a+b) + 2a^2+2b^2+2a(n-a-b)+2b(n-a-b)\\\\ =&2a^2+2b^2 - 2(a+b)^2 + n(a+b)\\\\ \\ge&2a^2+2b^2 - 2(a+b)^2 + (a+b)^2\\\\ =&(a-b)^2\\\\ \\ge&0. \\end{align}\n12. Hence the result.\n\nIn step 11 of the above, equality holds iff $a-b=0$ and $a+b\\in\\{0,n\\}$. In other words, equality holds iff $a=b=0$ or $a=b=n/2$, that is, iff all $x_i$s are zero, or when $n$ is even, exactly half of the $x_i$s (before scaling) are identical to some $x$ and the other half are $-x$s. However, this condition for equality is derived using the corner points. The above proof does not indicate whether equality can hold in other cases.\n\n-\nWow. That was really slick! I wish I could upvote your answer several times more (need some help here!) for not just solving the above problem, but also for showing a very novel method. Is this part of some standard repertoire for solving inequalities? Where did you come across this? \u2013\u00a0 quantumelixir Oct 19 '11 at 15:19\nThe technique itself (i.e. narrowing the domain to corner points) is not new. It is standard material in textbooks on linear programming or operations research. I use it here merely because I am not clever enough to think of a straightforward solution. As I said in the answer, my proof is actually a bit problematic -- although it proves the inequality, it fails to identify all cases where equality holds. \u2013\u00a0 user1551 Oct 19 '11 at 18:52\n\nLet $p$ be the number of $x_i\\ge0$, $m$ be the number of $x_i<0$, and $n=p+m$. $$\\small\\begin{array}{} &\\sum_i\\;\\:\\sum_j|x_i+x_j|\\\\ &=\\sum_{x_i\\ge0}\\;\\:\\sum_{x_j\\ge0}|x_i+x_j| &+\\sum_{x_i\\ge0}\\;\\:\\sum_{x_j<0}|x_i+x_j| &+\\sum_{x_i<0}\\;\\:\\sum_{x_j\\ge0}|x_i+x_j| &+\\sum_{x_i<0}\\;\\:\\sum_{x_j<0}|x_i+x_j|\\\\ &\\ge\\sum_{x_i\\ge0}\\;\\:\\sum_{x_j\\ge0}|x_i|+|x_j| &+\\left|\\sum_{x_i\\ge0}\\;\\:\\sum_{x_j<0}|x_i|-|x_j|\\right| &+\\left|\\sum_{x_i<0}\\;\\:\\sum_{x_j\\ge0}|x_j|-|x_i|\\right| &+\\sum_{x_i<0}\\;\\:\\sum_{x_j<0}|x_i|+|x_j|\\\\ &=2p\\sum_{x_i\\ge0}|x_i| &+2\\left|m\\sum_{x_i\\ge0}|x_i|-p\\sum_{x_i<0}|x_i|\\right| &&+2m\\sum_{x_i<0}|x_i| \\end{array}$$\n\nSubtracting $\\displaystyle\\sum_i\\;\\:\\sum_j|x_i|=n\\sum_i\\;|x_i|$ from the inequality above yields \\small\\begin{align}{} &\\sum_i\\;\\:\\sum_j|x_i+x_j|-\\sum_i\\;\\:\\sum_j|x_i|\\\\ &\\ge(p-m)\\sum_{x_i\\ge0}|x_i| +\\left|2m\\sum_{x_i\\ge0}|x_i|-2p\\sum_{x_i<0}|x_i|\\right| +(m-p)\\sum_{x_i<0}|x_i|\\\\ &=(p-m)\\sum_{i}x_i +\\left|(m-p)\\sum_{i}|x_i|+n\\sum_ix_i\\right|\\\\ &\\ge0\\tag{1} \\end{align} In the last inequality, if $p-m$ and $\\sum\\limits_ix_i$ have the same sign, we can ignore the quantity in absolute values. If $p-m$ and $\\sum\\limits_ix_i$ have different signs, then the quantities in the absolute values have the same sign and $|m-p|\\sum\\limits_i|x_i|\\ge(m-p)\\sum\\limits_ix_i$.\n\nTherefore, $(1)$ says that $$\\sum_i\\;\\:\\sum_j|x_i|\\le\\sum_i\\;\\:\\sum_j|x_i+x_j|$$\n\n-\nThis follows exactly the same approach suggested in my answer. An acknowledgement is in order perhaps? \u2013\u00a0 quantumelixir Oct 22 '11 at 20:16\nNice work. Does the same inequality and perhaps similar approach work for $x_i$ being complex or a vector? \u2013\u00a0 Hansen Oct 18 at 23:52\n\nAssume an ordering of the $x_i$ such that $x_1 \\le x_2 \\le \\dots \\le x_k < 0 \\le x_{k+1} \\le x_{k+2} \\le \\dots \\le x_n$. Let's call $T_1 = \\sum_{i=1}^k |x_i|$, $T_2 = \\sum_{i=k+1}^n |x_i|$, and $T=\\sum_i |x_i|=T_1+T_2$ so that $T, T_1, T_2 \\ge 0$.\n\nSince we only require a lower bound on the sum $\\sum_{i \\neq j} |x_i + x_j|$, we only consider $2\\binom{k}{2} + 2\\binom{n-k}{2}$ terms of the $2\\binom{n}{2}$ total terms that the sum contains i.e. we only pick terms where both $x_i$ and $x_j$ have the same sign. This solves the inequality in many cases. For the other cases, we include the cross terms too later.\n\n\\begin{align*} \\sum_{i \\neq j} |x_i + x_j| &\\ge \\sum_{i=1}^k \\sum_{j=1,j\\neq i}^k |x_i+x_j| + \\sum_{i=k+1}^n \\sum_{j=k+1,j\\neq i}^n |x_i+x_j|\\\\ &= \\sum_{i=1}^k \\sum_{j=1,j\\neq i}^k (|x_i|+|x_j|) + \\sum_{i=k+1}^n \\sum_{j=k+1,j\\neq i}^n (|x_i|+|x_j|)\\\\ &= \\sum_{i=1}^k ((k-1)|x_i|+T_1-|x_i|) + \\sum_{i=k+1}^n ((n-k-1)|x_i|+T_2-|x_i|)\\\\ &= \\sum_{i=1}^k ((k-2)|x_i|+T_1) + \\sum_{i=k+1}^n ((n-k-2)|x_i|+T_2)\\\\ &= ((k-2)T_1+kT_1) + ((n-k-2)T_2+(n-k)T_2)\\\\ &= 2\\{(k-1)T_1 + (n-k-1)T_2\\}\\\\ \\end{align*}\n\nNow, the last expression is a function of $T_1$ and $T_2$. Since we require a lower bound in terms of $T=T_1+T_2$, we can get two equivalent expressions in terms of either $T_1$ and $T$ or in terms of $T_2$ and $T$. Thus,\n\n$$\\sum_{i \\neq j} |x_i + x_j| \\ge 2(2k-n)T_1+2(n-k-1)T$$ $$\\sum_{i \\neq j} |x_i + x_j| \\ge 2(n-2k)T_2+2(k-1)T$$\n\n$$2\\sum_{i \\neq j} |x_i + x_j| \\ge 2\\left\\{(2k-n)T_1+(n-2k)T_2+(n-2)T\\right\\}$$ $$\\implies \\sum_{i \\neq j} |x_i + x_j| \\ge (2k-n)(T_1-T_2)+(n-2)T \\ge (n-2)\\sum_i |x_i|$$ $$\\text{whenever,} \\; (2k-n)(T_1-T_2) \\ge 0$$\n\nFor the cases when the above inequality doesn't hold, we need to consider the cross terms too. Computing the cross terms alone:\n\n\\begin{align*} \\sum_{i \\neq j} |x_i + x_j| &\\ge \\sum_{i=1}^k \\sum_{j=k+1}^n |x_i+x_j| + \\sum_{i=k+1}^n \\sum_{j=1}^k |x_i+x_j|\\\\ &\\ge 2\\sum_{i=1}^k \\sum_{j=k+1}^n |x_i+x_j|\\\\ &\\ge 2\\sum_{i=1}^k \\sum_{j=k+1}^n \\max\\{|x_i|-|x_j|,|x_j|-|x_i|\\}\\\\ &\\ge 2 \\max\\{(n-k)T_1-kT_2, kT_2-(n-k)T_1\\}\\\\ \\end{align*}\n\nIncluding the cross terms along with the terms from the previous case, we have:\n\n\\begin{align*} \\sum_{i \\neq j} |x_i + x_j| &\\ge 2 \\max\\{(n-k)T_1-kT_2, kT_2-(n-k)T_1\\} + 2\\{(k-1)T_1 + (n-k-1)T_2\\}\\\\ &\\ge 2 \\max\\{(n-1)T_1+(n-2k-1)T_2, (2k-n-1)T_1+(n-1)T_2\\} \\; (*)\\\\ \\end{align*}\n\nThe $\\max$ term can go one of two ways. Using the first term we get: $$\\sum_{i \\neq j} |x_i + x_j| \\ge 2 \\{(n-1)T_1+(n-2k-1)T_2\\}$$\n\nWe can again write the above in terms of either $T,T_1$ or $T,T_2$ giving two inequalities:\n\n\\begin{align} \\sum_{i \\neq j} |x_i+x_j| &\\ge (n-2k-1)T+2kT_1\\\\ \\sum_{i \\neq j} |x_i+x_j| &\\ge (n-1)T-2kT_2 \\end{align}\n\nSumming the above two inequalities,\n\n\\begin{align*} \\sum_{i \\neq j} |x_i + x_j| &\\ge (2n-2k-2)T+2k(T_1-T_2)\\\\ &\\ge (n-2)T+(n-2k)T+2k(T_1-T_2)\\\\ &\\ge (n-2)T \\quad \\text{for the case} \\quad (2k-n) \\le 0, (T_1-T_2) \\ge 0\\\\ \\end{align*}\n\nSimilarly, for the other case, when $(2k-n) \\ge 0, (T_1-T_2) \\le 0$, we use the other $\\max$ term from $(*)$ and follow the above steps routinely to prove the inequality.\n\nQ.E.D\n\n-\nWell done. (more characters) \u2013\u00a0 Jonas Teuwen Oct 22 '11 at 16:36\nThanks Jonas! Please correct any errors if you spot them. \u2013\u00a0 quantumelixir Oct 22 '11 at 16:54\nThe above presentation can be greatly simplified, but it's rather long because of it's long edit history. \u2013\u00a0 quantumelixir Oct 22 '11 at 20:17\nOne simplification: $\\max\\{(n-k)T_1-kT_2, kT_2-(n-k)T_1\\}=|(n-k)T_1-kT_2|$. At least it is easier to read. \u2013\u00a0 robjohn Oct 22 '11 at 21:31\nAs I commented before, it helps to use $2|(n-k)T_1-kT_2|$ instead of $2\\max\\{(n-k)T_1-kT_2,kT_2-(n-k)T_1\\}$. Then, use $2T_1=(T+(T_1-T_2))$ and $2T_2=(T-(T_1-T_2))$ to convert $$\\sum_{i\\not=j}|x_i+x_j|\\ge2|(n-k)T_1-kT_2|+2\\{(k-1)T_1+(n-k-1)T_2\\}$$ into $$\\sum_{i\\not=j}|x_i+x_j|\\ge|(n-2k)T+n(T_1-T_2)|+(n-2)T+(2k-n)(T_1-T_2)$$ If $(2k-n)(T_1-T_2)\\ge0$, ignore the stuff in the absolute values. If $(2k-n)(T_1-T_2)<0$, then $(n-2k)$ and $(T_1-T_2)$ inside the absolute values have the same sign and $|n-2k|T\\ge(n-2k)(T_1-T_2)$ which cancels $(2k-n)(T_1-T_2)$. \u2013\u00a0 robjohn Oct 23 '11 at 8:13", "date": "2014-10-31 06:25:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/73945/stuck-trying-to-prove-an-inequality/74812", "openwebmath_score": 0.977115273475647, "openwebmath_perplexity": 539.7828001742105, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713883126863, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8524805227699431}}
{"url": "https://math.stackexchange.com/questions/2595431/weak-convergence-of-e-n-e-n1-cdots-e-2n-n-1-infty-in-ell-in", "text": "# Weak convergence of $(e_n + e_{n+1} +\\cdots + e_{2n})_{n=1}^\\infty$ in $\\ell^\\infty$\n\nLet $(e_n)_{n=1}^\\infty$ be the canonical vectors in $\\ell^\\infty$. Consider the sequence $$(e_n + e_{n+1} +\\cdots + e_{2n})_{n=1}^\\infty$$\n\nDoes it converge weakly in $\\ell^\\infty$?\n\nObviously $e_n + e_{n+1} + \\cdots + e_{2n} \\xrightarrow{n\\to\\infty} 0$ coordinate-wise, so the only possible candidate for the weak limit is $0$.\n\nI believe it indeed weakly converges to $0$.\n\nLet $f \\in (\\ell^\\infty)^*$ be a bounded linear functional. Assume $f(e_1 + e_{n+1} \\cdots +e_{2n}) \\not\\to 0$. Therefore, there exists $\\varepsilon > 0$ and a subsequence $$(e_{p(n)} + e_{p(n)+1} +\\cdots + e_{2p(n)})_{n=1}^\\infty$$ such that $|f(e_{p(n)} + e_{p(n)+1} +\\cdots + e_{2p(n)})| \\ge \\varepsilon$ for all $n \\in \\mathbb{N}$.\n\nWe have $$f(e_{p(n)} + e_{p(n)+1} +\\cdots + e_{2p(n)}) = |f(e_{p(n)} + e_{p(n)+1} +\\cdots + e_{2p(n)})|e^{i\\phi_n}$$\n\nfor some $\\phi_n \\in \\mathbb{R}$.\n\nInductively we construct a further subsequence $(e_{q(p(n))} + e_{q(p(n))+1} +\\cdots + e_{2q(p(n))})_{n=1}^\\infty$ such that the supports of $e_{q(p(n))} + e_{q(p(n))+1} +\\cdots + e_{2q(p(n))}$ and $e_{q(p(n+1))} + e_{q(p(n+1))+1} +\\cdots + e_{2q(p(n+1))}$ are disjoint:\n\nSet $q(p(1)) = p(1)$. Let $p(k)$ be such that $p(k) \\ge 2p(1)$. Set $q(p(2)) = p(k)$. Le $p(k')$ be such that $p(k') \\ge 2p(k) = 2q(p(2))$. Set $q(p(3)) = p(k')$ and so on.\n\nNow consider $$x_n = \\sum_{k=1}^n e^{-i\\phi_{q(k)}}(e_{q(p(k))} + e_{q(p(k))+1} +\\cdots + e_{2q(p(k))})$$ Since the summands have disjoint supports and $|e^{-i\\phi_{q(k)}}| = 1$ it follows $\\|x_n\\|_\\infty = 1$ for all $n \\in \\mathbb{N}$.\n\nHowever, $$|f(x_n)| = \\left|\\sum_{k=1}^n e^{-i\\phi_{q(k)}} f(e_{q(p(k))} + e_{q(p(k))+1} +\\cdots + e_{2q(p(k))})\\right| = \\sum_{k=1}^n |f(e_{q(p(k))} + e_{q(p(k))+1} +\\cdots + e_{2q(p(k))})| \\ge n\\varepsilon \\xrightarrow{n\\to\\infty} +\\infty$$\n\nso $f$ is unbounded. A contradiction.\n\nTherefore $f(e_n + e_{n+1} + \\cdots + e_{2n}) \\xrightarrow{n\\to\\infty} 0$ for all $f \\in (\\ell^\\infty)*$ so $e_n + e_{n+1} + \\cdots + e_{2n} \\rightharpoonup 0$.\n\nIs my proof correct? Is there an easier (and still elementary) way to conclude that it converges weakly to $0$?\n\nEdit: I lied - there is another proof that might well be regarded as much simpler. Inspired by the OP's comment on what he was really trying to do:\n\nTrivial Lemma Suppose $X$ is a Banach space and $Y$ is a closed subspace. If $y_n\\to y$ weakly in $Y$ then $y_n\\to y$ weakly in $X$.\n\nProof: Suppose $f\\in X^*$. Let $g=f|_Y$. Then $g\\in Y^*$, so $$f(y_n)=g(y_n)\\to g(y)=f(y).$$\n\nSo to answer the original question it's enough to show that $e_n+\\dots+e_{2n}\\to0$ weakly in $c_0$. This is clear, since $c_0^*=\\ell_1$.\n\nOriginal: Looks right. Is there an easier proof? I doubt it - what you did is already very simple, just applying a few definitions. If it doesn't look simple that's just because of the notation.\n\nI'd write the very same proof as follows - possibly looks simpler, possibly easier to read (in fact I confess I didn't read what you wrote very carefully, just sort of skimmed it and saw that the idea clearly worked):\n\nSay $x_n=e_n+\\dots+e_{2n}$. Suppose $x_n$ does not tend to $0$ weakly. Then there exists $f\\in\\ell_\\infty^*$ so $f(x_n)\\not\\to0$.\n\nSo there exist a subsequence $(x_{n_j})$, $\\epsilon>0$, and complex numbers $\\alpha_j$ with $|\\alpha_j|=1$, so that $$f(y_j)\\ge\\epsilon$$if $$y_j=\\alpha_jx_{n_j}.$$We may assume that $n_{j+1}>2n_j$, so that the $y_j$ have disjoint support. Since the $y_j$ have disjoint support we have $$||y_1+\\dots+y_n||=1,$$although$$f(y_1+\\dots+y_n)\\ge n\\epsilon,$$contradicting the fact that $f$ is bounded.\n\nThat looks simpler to me - I think it's much easier to read, since the displayed formulas are less intricate. But it's exactly the same proof.\n\n\u2022 Ok, thanks for the feedback. Not really related to the question, but is there an example of a sequence in $c_0$ which converges weakly in $c_0$ but not in $\\ell^\\infty$? The sequence $(e_n + \\cdots + e_{2n})_{n=1}^\\infty$ was my attempt to construct such an example, but we have established that it converges weakly to $0$ in both $c_0$ and $\\ell^\\infty$. Jan 7 '18 at 12:49\n\u2022 @mechanodroid See edit. Jan 7 '18 at 13:07\n\u2022 Great, exactly the kind of thing I was hoping for! Jan 7 '18 at 13:12", "date": "2021-10-27 03:01:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2595431/weak-convergence-of-e-n-e-n1-cdots-e-2n-n-1-infty-in-ell-in", "openwebmath_score": 0.9871160984039307, "openwebmath_perplexity": 78.21280867989972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713839878681, "lm_q2_score": 0.8652240704135291, "lm_q1q2_score": 0.8524805173159544}}
{"url": "http://math.stackexchange.com/questions/82944/interpolating-between-the-lp-norm-and-the-h%c3%b6lder-semi-norm", "text": "Interpolating between the $L^{p}$ norm and the H\u00f6lder semi-norm\n\nSet-up.\n\nFor a bounded continuous function $u \\colon \\mathbb{R}^n \\to \\mathbb{R}$, the $\\gamma$-H\u00f6lder semi-norm of $u$ is $$\\begin{eqnarray} [u]_{C^\\gamma} &=& \\sup \\left\\{\\frac{|u(x) - u(y)|}{|x-y|^\\gamma} : x,y \\in U, x \\neq y \\right\\} \\\\ &=& \\inf \\left\\{ C \\geq 0 : |u(x) - u(y)| \\leq C |x-y|^{\\gamma} \\text{ for all } x,y \\in U \\right\\}. \\end{eqnarray}$$\n\nThe Problem\n\nFix $1 \\leq p < \\infty$, $0 < \\gamma \\leq 1$, and $0 < \\lambda < 1$ such that $$0 = \\frac{\\lambda}{p} - (1-\\lambda)\\frac{\\gamma}{n}.$$ I am trying to prove that there is a constant $C$ such that $$\\|u\\|_{L^\\infty} \\leq C \\|u\\|_{L^p}^{\\lambda} [u]_{C^\\gamma}^{1-\\lambda}$$ for every compactly supported $C^{1}(\\mathbb{R}^n)$ function $u$.\n\nMy Strategy\n\nMy plan is to use the interpolation result for Lebesgue spaces: For every $q,r$ satisfying $p < q < r \\leq \\infty$ and $$\\frac{1}{q} = \\frac{\\lambda}{p} + \\frac{1-\\lambda}{r},$$ we have $$\\|u\\|_{L^q} \\leq \\|u\\|_{L^{p}}^{\\lambda} \\|u\\|_{L^r}^{1-\\lambda}$$ for every $u \\in L^p \\cap L^r$.\n\nSolving for $1/r$ in terms of $q$ and using the relationship between $p$, $\\gamma$, and $n$, we find $$\\frac{1}{r} = \\frac{1/q - \\lambda/p}{1-\\lambda} = \\frac{1/q}{1-\\lambda} - \\frac{\\gamma}{n}$$ So, by letting $q \\to \\infty$, we have $r \\to -n / \\gamma$, and $$\\frac{1}{q} = \\frac{\\lambda}{p} + \\frac{1-\\lambda}{r},$$ goes to $$0 = \\frac{\\lambda}{p} - (1-\\lambda)\\frac{\\gamma}{n}.$$ Meanwhile, for $u \\in L^{\\infty}$ (which certainly holds when $u$ is compactly supported and $C^1$), we have that $\\lim_{q \\to \\infty} \\|u\\|_{L^q} =\\|u\\|_{L^\\infty}$.\n\nSo if I could prove that $\\|u\\|_r \\to [u]_{C^{\\gamma}}$ as $q \\to \\infty$ (i.e., as $r \\to -n / \\gamma$), I'd be done. The problem is that I don't know how to prove this. Moreover, I'm not sure if this is even the right approach to prove the desired inequality.\n\n-\nThere is no reason that $\\|u\\|_r$ should tell you anything about $[u]_{C^\\gamma}$ on its own (they measure completely different things). Also, your Lebesgue interpolation inequality only holds for $p < q < r$, but you are taking $q \\to \\infty$ and $r \\to -n/\\gamma$ (does $r < 1$ even make sense?). \u2013\u00a0Jeff Nov 17 '11 at 21:18\nYou may have to assume that $u \\in C^\\gamma$ (instead of $u$ continuous) and use some kind of Sobolev embedding. \u2013\u00a0Jeff Nov 17 '11 at 21:21\n@Jeff: It was a mistake to not assume $u \\in C^{\\gamma}$. Thank you for pointing it out. \u2013\u00a0fferic Nov 17 '11 at 22:03\n@Jeff: You are right that $q \\rightarrow \\infty$ and $r \\rightarrow -n/\\gamma$ is not compatible with the requirement $p<q<r$ in the Lebesgue interpolation inequality. This is one reason why I am not so sure my strategy will really work. \u2013\u00a0fferic Nov 17 '11 at 22:06\n@Jeff: Could you please elaborate a bit more on your comment about using a Sobolev embedding? \u2013\u00a0fferic Nov 17 '11 at 22:07\n\nThis is my first post on MSE and I'm just a grad student so forgive me if screw anything up. [Disclaimer: This proof is likely to be overkill, however I like that it shows what exactly is going on and these techniques generalize to a lot of different problems.] Hopefully someone else can chime in to validate my answer.\n\nI will only consider the case $U = \\Bbb R^n$.\n\nThe proof will be similar to the proof of Theorem A.3 in Tao's book \"Nonlinear dispersive equations\". In Theorem A.3, Tao uses basic Littlewood-Paley theory to prove a fairly general version of the Gagliardo-Nirenberg inequality, which is your equation after your swap out the H\u00f6lder space with an appropriate Sobolev space.\n\nFirst, we consider the case when $\\| u \\|_{L^p} = \\| u \\|_{C^\\gamma} = 1$ and show that the $L^\\infty$ norm is bounded above by a constant. For general $u$ in $L^p \\cap C^\\gamma$ we can reduce to the first case by considering $v(x) = A u(B x)$ for an appropriate choice of constants $A$ and $B$ which make both of the norms $1$.\n\nThe main idea is to decompose $u$ into high and low frequencies. We will use the $L^p$ norm to bound the low frequencies and the H\u00f6lder regularity to bound the high frequencies.\n\nTo begin, the triangle inequality gives\n\n$$\\| u \\|_{L^\\infty} \\le \\sum_{k=0}^{-\\infty}\\| P_k u \\|_{L^\\infty} + \\sum_{k=1}^{\\infty} \\| P_k u \\|_{L^\\infty}.$$\n\nThe Bernstein inequalities tell us that $$\\| P_k u \\|_{L^\\infty} \\le C 2^{k \\frac{n}{p}} \\| P_k u \\|_{L^p} = C 2^{k \\frac{n}{p}}$$ which in turns gives a bound on the low frequencies in terms of a convergent geometric sum $$\\sum_{k=0}^{-\\infty}\\| P_k u \\|_{L^\\infty} \\le C \\sum_{k=0}^{-\\infty} 2^{k \\frac{n}{p}} = C_1 < \\infty.$$\n\nTo deal with the high frequency terms, we use the part of the Littlewood-Paley decomposition of $C^\\gamma$ which states that $$\\sup_{k \\in \\Bbb Z} 2^{k \\gamma} \\| P_k u \\|_{L^\\infty} \\le C' \\| u \\|_{C^\\gamma}$$\n\nThis leads to a bound on the high frequencies in terms of another convergent geometric sum:\n\n$$\\sum_{k=1}^{\\infty}\\| P_k u \\|_{L^\\infty} \\le \\sum_{k=1}^{\\infty} 2^{-k \\gamma} C' \\| u \\|_{C^\\gamma} = C' \\sum_{k=1}^{\\infty} 2^{-k \\gamma} = C_2 < \\infty$$\n\nTogether these shows that $\\| u \\|_{L^\\infty} \\le C_1 + C_2$.\n\n-", "date": "2015-11-26 04:01:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/82944/interpolating-between-the-lp-norm-and-the-h%c3%b6lder-semi-norm", "openwebmath_score": 0.941773533821106, "openwebmath_perplexity": 114.32519274943934, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668739644686, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8524640553907554}}
{"url": "https://forum.math.toronto.edu/index.php?PHPSESSID=jol92prfec42b5h1b01tka3qi5&action=printpage;topic=1189.0", "text": "# Toronto Math Forum\n\n## MAT244-2018S => MAT244--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 08:42:26 PM\n\nTitle: FE-P3\nPost by: Victor Ivrii on April 11, 2018, 08:42:26 PM\nFind the general solution of\n\\begin{equation*}\ny''' -6y'' +11y'- 6y=2\\frac{e^{3x}}{e^x+1} .\n\\end{equation*}\nTitle: Re: FE-P3\nPost by: Tim Mengzhe Geng on April 11, 2018, 11:37:34 PM\nFirst we find the solution for the homogeneous system\n\ny^{(3)}-6y^{(2)}+11y^{(1)}-6y=0\n\nThe corresponding characteristic equation is\n\nr^3-6r^2+11r-6=0\n\nThree roots are\n\nr_1=1\n\nr_2=2\n\nr_3=3\n\nThen the solution for the homogeneous system is\n\ny_c(t)=c_1e^{x}+c_2e^{2x}+c_3e^{3x}\n\nwhere\n\ny_1(t)=e^{x}\n\ny_2(t)=e^{2x}\n\ny_3(t)=e^{3x}\n\nThen we follow to find the required solution to the nonhomogeneous equation. We use Variation of Parameters. We have\n\nW[y_1,y_2,y_3]=2e^{6x}\n\nW_1[y_1,y_2y_3]=e^{5x}\n\nW_2[y_1,y_2y_3]=-2e^{4x}\n\nW_3[y_1,y_2\uff0cy_3]=e^{3x}\n\nand\n\ng(x)=2\\frac{e^{3x}}{e^{x}+1}\n\nAnd then we have the following integration\n\n\\int \\frac{W_1\\cdot g(x)dx}{W[y_1,y_2,y_3]}=\\int\\frac{e^{2x}}{e^{x}+1}\n\n\\int\\frac{e^{2x}}{e^{x}+1}=e^{x}-\\ln(e^{x}+1)+c_4\n\n\\int \\frac{W_2\\cdot g(x)dx}{W[y_1,y_2,y_3]}=-2\\int\\frac{e^{x}}{e^{x}+1}\n\n-2\\int\\frac{e^{x}}{e^{x}+1}=-2\\ln(e^{x}+1)+c_5\n\n\\int \\frac{W_3 \\cdot g(x)dx}{W[y_1,y_2,y_3]}=\\int\\frac{1}{e^{x}+1}\n\n\\int\\frac{1}{e^{x}+1}=x-\\ln(e^{x}+1)+c_6\n\nAnd finally, the required general solution $y(t)$\n\ny(t)=\\sum_{i=1}^3 y_i(t)\\cdot\\int\\frac {W_i\\cdot g(x)dx}{W[y_1,y_2,y_3]}\nTitle: Re: FE-P3\nPost by: Meng Wu on April 11, 2018, 11:47:31 PM\nSmall Error: $W_2(x)$ should be $2e^{4x}$.\nTitle: Re: FE-P3\nPost by: Tim Mengzhe Geng on April 11, 2018, 11:49:31 PM\nSmall Error: $W_2(x)$ should be $2e^{4x}$.\nFor this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$\nTitle: Re: FE-P3\nPost by: Meng Wu on April 11, 2018, 11:54:19 PM\nSmall Error: $W_2(x)$ should be $2e^{4x}$.\nFor this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$\n\nOh, you're right. My mistake.\nTitle: Re: FE-P3\nPost by: Meng Wu on April 11, 2018, 11:56:10 PM\nAlso for $(20)$, $\\ln(e^x)$ can be simplified as $\\ln(e^x)=x$.\nTitle: Re: FE-P3\nPost by: Tim Mengzhe Geng on April 11, 2018, 11:57:57 PM\nAlso for $(20)$, $\\ln(e^x)$ can be simplified as $\\ln(e^x)=x$.\nThanks again and I will modify it :)\nTitle: Re: FE-P3\nPost by: Syed Hasnain on April 12, 2018, 04:33:47 PM\nSince the solution is incomplete after Y(x),\nI am attaching a copy of my solution\nTitle: Re: FE-P3\nPost by: Tim Mengzhe Geng on April 12, 2018, 10:01:46 PM\nSince the solution is incomplete after Y(x),\nI am attaching a copy of my solution\nSorry what do you mean by \"the solution is incomplete after Y(x)\"\nI did write a bit more on the exam (expanding the summation) but I think one should be able to get full marks if he integrates everything and mention how the solution is composed.(Given that the integral is correct)\nTitle: Re: FE-P3\nPost by: Victor Ivrii on April 15, 2018, 03:15:29 AM\nSince the solution is incomplete after Y(x),\nI am attaching a copy of my solution\nThe only thing which was missing in the solution, is the final answer, but it warrants neither such claim, nor uploading your solution.\n\nGeneral remark:\nIt would be better to denote \"parameters\" by uppercase letters $C_1(x)$, $C_2(x)$,... and constants by lowercase letters $c_1$, $c_2$,...", "date": "2022-05-24 22:54:47", "meta": {"domain": "toronto.edu", "url": "https://forum.math.toronto.edu/index.php?PHPSESSID=jol92prfec42b5h1b01tka3qi5&action=printpage;topic=1189.0", "openwebmath_score": 0.9062466025352478, "openwebmath_perplexity": 3963.8078734568635, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668662546612, "lm_q2_score": 0.8688267830311354, "lm_q1q2_score": 0.8524640520247776}}
{"url": "http://math.stackexchange.com/questions/190073/what-is-mathbb-zt-what-are-the-double-brackets", "text": "# What is $\\mathbb Z[[t]]$? What are the double brackets?\n\nWhat does $\\mathbb{Z}[[t]]$ mean? Why are there double square brackets?\n\nI can't search through Google, because I can't search Latex.\n\n-\nI think it's the ring of formal power series with coefficents from $\\mathbb{Z}$, see en.wikipedia.org/wiki/Formal_power_series \u2013\u00a0 Mikko Korhonen Sep 2 '12 at 16:19\nYou actually can search LaTeX: latexsearch.com \u2013\u00a0 huon-dbaupp Sep 2 '12 at 16:22\nIf this is from a book, it might have an list of symbols in the back that you can check. \u2013\u00a0 Jair Taylor Sep 2 '12 at 17:10\n\nThat is the ring of formal power series in $t$ with integer coefficients, i.e., of $$\\sum_{n=0}^\\infty a_nt^n,$$ with $a_n\\in\\Bbb Z$, componentwise addition, and multiplication appropriately defined.\n\nThe double brackets distinguish it from $\\Bbb Z[t]$, which is the ring of polynomials in $t$ with integer coefficients. We can always evaluate the members of $\\Bbb Z[t]$ for any complex value of $t$, but we generally can't evaluate members of $\\Bbb Z[[t]]$ for $t\\neq 0$. To my mind, the double bracket is a reminder that we need to leave the $t$ alone, and not worry about evaluation.\n\n-\n(IMO there is a sense in which can view \"evaluation at $t=a$\" to be the quotient map $\\Bbb Z[[t]]\\to \\Bbb Z[[t]]/(t-a)$, though this is cheating and doesn't send everything all the way down to $\\Bbb Z$.) \u2013\u00a0 anon Sep 2 '12 at 17:08\nInteresting point! \u2013\u00a0 Cameron Buie Sep 2 '12 at 17:11\nI laughed. I'm going to steal this saying. Many thanks. \"just leave $t$ alone\" ha. \u2013\u00a0 James S. Cook Sep 2 '12 at 18:21\nHave at it, James! \u2013\u00a0 Cameron Buie Sep 3 '12 at 1:47\n@CameronBuie Is it ok to use the symbol $\\mathbb{Z}\\llbracket t\\rrbracket$ in latex instead of $\\mathbb{Z}[[t]]$ or is that wrong? \u2013\u00a0 Pratyush Sarkar Mar 4 '13 at 14:51\n\nIf $A$ is any ring, the notation $A[[T]]$ stands for the ring of formal power series with coefficients in $A$, i.e. the ring whose elements are the expressions $$a_0+a_1T+a_2T^2+a_3T^3+\\cdots$$ with the obvious sum and product.\n\n-", "date": "2015-08-03 13:44:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/190073/what-is-mathbb-zt-what-are-the-double-brackets", "openwebmath_score": 0.9446963667869568, "openwebmath_perplexity": 664.1004075585366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668703849155, "lm_q2_score": 0.8688267694452331, "lm_q1q2_score": 0.8524640422832158}}
{"url": "https://www.physicsforums.com/threads/how-do-i-represent-this-sine-function.803221/", "text": "# How do I represent this sine function?\n\n1. Mar 15, 2015\n\nI was struggling to represent the following for integer values of n:\n\n$$\\sin \\left( \\dfrac {n\\pi } {2}\\right)$$\n\nI know for even n, we get zero\n\nBut for odd n, it alternates beween 1 and -1 for every other odd. Is there a compact way to represent that? I feel like I'm being dumb and missing something obvious.\n\nThanks!\n\n2. Mar 15, 2015\n\n### piggupiggu\n\n1 when n = 1 + 2x\n-1 when n = 3 + 2x\n\nwhere x is zero or a positive even integer\n\n3. Mar 15, 2015\n\n### Mentallic\n\nWhat you have is a piecewise function (the variable being n in this case).\n\n$$k\\in \\mathbb{Z} \\\\ f(n) = \\sin{\\frac{\\pi n}{2}} = \\left\\{ \\begin{array}{lr} 0 & : n=2k\\\\ 1 & : n=4k+1\\\\ -1 & : n=4k+3 \\end{array} \\right.$$\n\n4. Mar 15, 2015\n\n### LCKurtz\n\nI guess you are looking for something similar to $\\cos(n\\pi) = (-1)^n$ for $\\sin(\\frac{n\\pi} 2)$. You may be able to find such an expression, but I don't think you will find anything \"simpler\" than $\\sin(\\frac{n\\pi} 2)$ itself, unless you consider piecewise defined functions simpler. I don't.\n\n5. Mar 17, 2015\n\nThank you so much for the responses everyone!\n\n6. Mar 18, 2015\n\n### Mentallic\n\nI gave it a little more thought and came up with\n$$\\sin\\left(\\frac{n\\pi}{2}\\right)=\\frac{1}{2}\\left(-1\\right)^{\\frac{n-1}{2}}\\left(1-(-1)^n\\right)$$\n\n7. Mar 19, 2015\n\n### jbunniii\n\n$$\\sin\\left(\\frac{n\\pi}{2}\\right) = \\frac{1}{2i}\\left(i^n - (-i)^n\\right) = \\frac{1}{2}\\left(i^{n-1} + (-i)^{n-1}\\right)$$\n\n: I guess that's equivalent to Mentallic's answer. Here's how I obtained it:\n\n\\begin{aligned} \\sin\\left(\\frac{n\\pi}{2}\\right) &= \\text{Im}\\left(e^{in\\pi/2}\\right) \\\\ &= \\frac{1}{2i} \\left(e^{in\\pi/2} - \\overline{e^{in\\pi/2}}\\right) \\\\ &= \\frac{1}{2i} \\left(i^n - \\overline{i^n}\\right) \\\\ &= \\frac{1}{2i} \\left(i^n - (-i)^n\\right) \\\\ \\end{aligned}\n\nLast edited: Mar 19, 2015\n8. Mar 20, 2015\n\n### LCKurtz\n\nYup. As I thought. Nothing \"simpler\" than $\\sin\\frac{n\\pi} 2$ itself.", "date": "2017-12-12 06:35:44", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/how-do-i-represent-this-sine-function.803221/", "openwebmath_score": 0.9998975992202759, "openwebmath_perplexity": 1811.4914792396073, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668714863165, "lm_q2_score": 0.8688267660487573, "lm_q1q2_score": 0.8524640399076331}}
{"url": "https://math.stackexchange.com/questions/1367812/find-closest-point-on-a-plane-to-a-given-point-discrepancy-with-normal-vector", "text": "# Find closest point on a plane to a given point. Discrepancy with normal vector.\n\nI have a point $(9,5,0)$ and a triangle with points $(1,1,0), (3,3,1), (6,1,0)$, let's label them as $A,B,C$ respectively. In order to get the normal vector, I do the cross product of two vectors. If I do $AB \\times AC$ I get $n = \\left\\langle 0,-5,10\\right\\rangle$. With this I determine the equation of the plane is $5y - 10z - 5 = 0$. If I project my point onto the plane with this normal vector in mind, the point on the plane I find is $(9, 21/5, 8/5)$. HOWEVER, if I initially choose to cross two different vectors, I get something different.\n\nExample: If I instead choose to do $BA \\times BC$ I get $n = \\left\\langle 0,5,-10\\right\\rangle$. Now this in a way makes sense to me since it is the inverse of the other normal I found. Basically the other half of the line through the plane, right? If I continue, I find the equation of the plane to be $-5y + 10z + 5 = 0$. This leads me to a projected point of $(9, -29/5, -8/5)$. Why do I get two different points based on which normal vector I choose to solve with? I know the correct answer to this problem, one is right and one is wrong. I am trying to program a general solution to this type of problem. How do I choose the correct normal vector?\n\nPlease let me know if I am doing something wrong.\n\n$\\vec{AB} = <2,2,1> \\quad \\quad \\vec{AC} = <5,0,0>$.\n\nSo, $\\vec{AB} \\times \\vec{AC} = <0,5,-10>$\n\nThen, the plane equation is something like $5y -10z = C$ for some constant $C$.\n\nPlug (for example) the point $(1,1,0)$ into the plane equation to get $C=5$.\n\nHence, the plane equation is $5y -10z -5 =0$ or equivalently $-5y +10z+5=0$ which is slightly different from the one you found.\n\nEdit: Let's continue with a different approach. The vector $\\vec{n} = <0,5,-10>$ is perpendicular to the plane. The line that passes through our point $(9,5,0)$ which is parallel to $\\vec n$ is $$x=9, \\quad y=5+5t, \\quad z=0-10t, \\quad t \\in \\mathbb R$$\n\nNow let's find where this line cuts the plane. Pick an arbitrary point from the line and put into the plane equation:\n\n$$0= 5y-10z-5 =5(5+5t)-10(0-10t)-5$$\n\n$$125t + 25 -5=0 \\quad \\implies \\quad t=-\\frac{4}{25}$$\n\nSo, the answer is $(9,5+5t,0-10t)=(9,-\\frac{21}{5},\\frac{8}{5})$.\n\nEdit2: Now we do the projection. Choose an arbitrary point on the plane, say $A(1,1,0)$ and consider the vector $\\vec{AP}$ where $P=P(9,5,0)$. $\\vec{AP}=<8,4,0>$.\n\nThen,\n\n\\begin{align*} proj_{\\ \\vec n \\ } \\vec{AP} &= |\\vec{AP}| \\cdot \\cos \\alpha \\cdot \\frac{\\vec n}{| \\vec n\u00a0|} \\\\ &= |\\vec{AP}| \\cdot \\frac{\\vec{AP} \\bullet \\vec n}{|\\vec{AP}| \\cdot |\\vec n|} \\cdot \\frac{\\vec n}{| \\vec n\u00a0|} \\\\ &= \\text{ calculations ... find the projection vector} \\\\ &= \\vec u \\end{align*}\n\nNow, the point you are looking for is the point $K$ such that $\\vec{KP} = \\vec u$\n\n\u2022 Ah, I typed it in wrong. You are correct, but the question of the discrepancy still stands. I will edit the post. Can you continue to carry out the projection and see if you get the same results as I did? \u2013\u00a0kvcwahley Jul 20 '15 at 17:36\n\u2022 I think you forgot the -5 constant in your last calculation. It should be 125t + 25 = 5. This gives t=-4/25. My answer is then (9,21/5, 8/5). \u2013\u00a0kvcwahley Jul 20 '15 at 17:54\n\u2022 Theoretically, should I be getting the same answer? \u2013\u00a0kvcwahley Jul 20 '15 at 17:54\n\u2022 @kvcwahley Thanks I edited and got the answer. Now I will edit my answer again and add projection to get the same answer. \u2013\u00a0ThePortakal Jul 20 '15 at 17:59\n\u2022 If you solve it with cross product of BA x BC, do you get the same answer? \u2013\u00a0kvcwahley Jul 20 '15 at 18:03\n\n\"I have a point $(9,5,0)$ and a triangle with points $(1,1,0), (3,3,1), (6,1,0),$ let's label them as $A,B,C$ respectively. In order to get the normal vector, I do the cross product of two vectors.\"\n\nSo you mean a normal vector to the plane determined by the three points.\n\n\"If I do $AB \\times AC$ I get $n = <0,-5,10>$. With this I determine the equation of the plane is $5y - 10z + 5 = 0.$\"\n\nNo, that does not contain $(1, 1, 0)$ because $0(1)+ 5(1)- 10(0)+ 5= 10$, not 0. The equation of the plane is $0(x- 1)-5(y- 1)+ 10z= 0$ or $5y- 10z- 5= 0$. Perhaps you meant $5y- 10z= 5$.\n\n\"If I project my point onto the plane with this normal vector in mind, the point on the plane I find is $(9, 21/5, 8/5)$. HOWEVER, if I initially choose to cross two different vectors, I get something different.\n\nExample: If I instead choose to do $BA \\times BC$ I get $n = <0,5,-10>$. Now this in a way makes sense to me since it is the inverse of the other normal I found. Basically the other half of the line through the plane, right? If I continue, I find the equation of the plane to be $-5y + 10z -5 = 0.$\"\n\nWhich, again, is wrong. The cross product of $BA \\times BC$, as you say is $<0, 5, -10>$ and, since the plane contains the point $(1, 1, 0)$, the plane has the equation $0(x- 1)- 5(y- 1)+ 10(z- 0)= -5y+ 10z+ 5= 0$, not $-5y+ 10z- 5$.\n\n\"This leads me to a projected point of $(9, -29/5, -8/5)$. Why do I get two different points based on which normal vector I choose to solve with?\"\n\nYou don't say how you got two different points. Since you have exactly the same equation for the plane in both cases (although wrong!), and the same point, you should be doing exactly the same calculations in both cases.\n\n\"I know the correct answer to this problem, one is right and one is wrong. I am trying to program a general solution to this type of problem. How do I choose the correct normal vector?\n\nPlease let me know if I am doing something wrong.\"\n\nYes, if you get different answers projecting the same point onto the same plane, you must have done something wrong! But since you don't show what you did, it is impossible to say what you might have done wrong.\n\n\u2022 You are correct with my equations being wrong. I initially typed them in incorrectly. I have edited the post to show the correct equations. Theoretically, should I be getting the same point? \u2013\u00a0kvcwahley Jul 20 '15 at 18:00\n\u2022 Please start using LaTeX, see this guide: meta.math.stackexchange.com/questions/5020/\u2026 Almost all of your posts need heavy editing \u2013\u00a0Hirshy Aug 5 '15 at 8:42", "date": "2019-06-26 21:58:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1367812/find-closest-point-on-a-plane-to-a-given-point-discrepancy-with-normal-vector", "openwebmath_score": 0.9850231409072876, "openwebmath_perplexity": 177.47545466417273, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.959154287592778, "lm_q2_score": 0.8887587846530938, "lm_q1q2_score": 0.8524567989357613}}
{"url": "http://mathematica.stackexchange.com/questions/29383/finding-shortest-non-zero-vector-x-satisfying-ax-0-pmod-q", "text": "# Finding shortest non-zero vector $x$ satisfying $Ax=0 \\pmod q$\n\nLet $n$, $m$, and $q$ be positive integers (with $m > n$), and $A$ be a matrix over $\\mathbb{Z}_q^{n \\times m}$. Using Mathematica, I want to find the shortest non-zero vectors $x \\in \\mathbb{Z}_q^m$ such that:\n\n$$Ax=0 \\pmod q$$\n\nNote: By \"shortest,\" I mean the vector with the smallest Euclidean norm. I'm not sure if this vector is unique. If it's not, I want a list of vectors with smallest Euclidean norm.\n\nExample:\n\nLet: $n=2, m=3, q=7, A = \\left(\\begin{matrix} 2 & 1 & 6\\\\ 1 & 3 & 1 \\end{matrix} \\right)$. Then, the following vectors are solutions to\n\n{1,1,3} $\\quad$ Norm = $\\sqrt{11}$\n\n{3,3,2} $\\quad$ Norm = $\\sqrt{22}$\n\n{2,2,6} $\\quad$ Norm = $\\sqrt{44}$\n\n{5,5,1} $\\quad$ Norm = $\\sqrt{51}$\n\n{4,4,5} $\\quad$ Norm = $\\sqrt{57}$\n\n{6,6,4} $\\quad$ Norm = $\\sqrt{88}$\n\nTherefore, the vector {1,1,3} is the shortest solution.\n\nPS: I used an exhaustive search to find the above solutions. My approach takes a long time for even small choices of $n$ and $m$, say $n=2, m=8$. Please offer an approach which can at least find solutions for such small parameters.\n\nEdit: As Michael suggested in a comment, I demonstrate the \"exhaustive search\" code I used. I'm not proud of it at all, as I just wrote the code to find an example for this question.\n\nA = RandomInteger[6, {2, 3}];\nx = {a, b, c};\nTable[\nIf[Mod[A.x, 7] == {0, 0}, Print[{x, x // Norm}]],\n{a, 0, 6}, {b, 0, 6}, {c, 0, 6}]\n-\nYou might show the code you tried -- perhaps it can be made efficient. Users are much more likely to try out your problem if they can cut and paste a starting point. \u2013\u00a0Michael E2 Jul 28 '13 at 20:24\n@MichaelE2: Thanks for the suggestion. I added the code snippet. It does not use Mathematica's built-in functions like Solve or FindMin. It's nothing more than an exhaustive search, and that's why it takes a long time to find a solution for even relatively small parameters. \u2013\u00a0M.S. Dousti Jul 28 '13 at 21:05\nYes that would take a long time. :) If you thought of Solve, you should have tried it! Another tip: Don't use Print to accumulate you results. Table will accumulate them; then DeleteCases[table, Null, Infinity], gets rid of the Nulls; then perhaps Flatten. Better yet, learn about Reap/Sow. \u2013\u00a0Michael E2 Jul 28 '13 at 22:03\n(1) You are working in a ring that really does not have a notion of length. Why is a rock-bottom smallest in Euclidean norm result needed? \u2013\u00a0Daniel Lichtblau Jul 29 '13 at 14:12\n(2) One can get small results using lattice reduction. This unfortunately does not force values to be nonnegative. All the same, a usable vector might appear. In a run of your example I have mat = {{2, 3, 2}, {2, 6, 1}}. Then it turns out that {4,3,2} is the null vector of interest (though see (1) above-- I do not know why it is of interest). It appeas in the following lattice reduction. In[35]:= LatticeReduce[ Join[NullSpace[A], 7*IdentityMatrix[Length[A[[1]]]]]] Out[35]= {{-2, 2, -1}, {1, -1, -3}, {4, 3, 2}} \u2013\u00a0Daniel Lichtblau Jul 29 '13 at 14:14\n\nOne way to approach this is to use the Null space of the matrix a, which is a basis for all elements x such that a.x=0. For the OPs problem, this can be done by first finding the null space:\n\na = {{2, 1, 6}, {1, 3, 1}};\nn = First[NullSpace[a, Modulus -> 7]]\n\n{5, 5, 1}\n\nNow, observe that a.x=0 is true exactly when a.(c*x)=0, so it is possible to list all the answers (mod 7) to this problem by\n\nMod[# n, 7] & /@ Range[7]\n\n{{5, 5, 1}, {3, 3, 2}, {1, 1, 3}, {6, 6, 4}, {4, 4, 5}, {2, 2, 6}}\n\nTo find the one with smallest norm, apply the norm to each element,\n\nNorm /@ (Mod[# n, 7] & /@ Range[6])\n\n{Sqrt[51], Sqrt[22], Sqrt[11], 2 Sqrt[22], Sqrt[57], 2 Sqrt[11]}\n\nwhich gives the same answer as the OP found by search.\n\nMore generally, the NullSpace of a is a collection of $m-n$ vectors, and it will be necessary to search over all linear combinations of these $m-n$ vectors (mod $q$) for the one with the smallest norm. While this may still be large, it is certainly smaller than search over $m$ dimensions as in the brute force approach.\n\n-\n\nUsing Solve helps a bit. If vars is the list of variables, then\n\nSolve[A.vars == ConstantArray[0, n], vars, Modulus -> q]\n\nfinds all solutions in terms of parameters C[1], C[2], etc. depending on the dimension of the solution space.\n\nAnother bottleneck is computing the values of the general solution at all possible combinations of values for the parameters C[i] modulo q. The possible combinations can be computed with\n\nTuples[Range[0, q - 1], n]\n\nwhere n is the number of parameters C[i]. The result is certainly reasonably fast for the small values of m, n, q mentioned in the question.\n\nFinally Nearest will return the shortest nonzero vector (the zero vector having been deleted).\n\nfindSols[q_, A_] :=\nModule[{m, n, vars, x, sol, vals, params, vecs},\n{n, m} = Dimensions@A;\nvars = Array[x, m];\nsol = vars /. First @ Solve[A.vars == ConstantArray[0, n], vars, Modulus -> q];\n(* Print@sol; *) (*optional output*)\nparams = Union@Cases[sol, _C, Infinity];\n(* Print@params; *) (*optional output*)\nvals = Transpose @ Tuples[Range[0, q - 1], Length@params];\nvecs = DeleteCases[\nMod[#, q] &@ Transpose[sol /. MapThread[Rule, {params, vals}]],\nConstantArray[0, m]];\nNearest[vecs, ConstantArray[0, m]]\n]\n\nExample in OP:\n\nA = {{2, 1, 6}, {1, 3, 1}};\nfindSols[7, A] // Timing\n(* {0.000675, {{1, 1, 3}}} *)\n\nAdditional example with $m=8$. (Over 117000 solutions, still less than a second.)\n\nSeedRandom[2];\nA = RandomInteger[6, {2, 8}]\n\nfindSols[7, A] // Timing\n(* {{5, 6, 1, 6, 2, 2, 6, 5}, {2, 1, 5, 5, 0, 0, 6, 4}} *)\n(* {0.289682, {{0, 1, 0, 0, 1, 0, 1, 0}, {0, 1, 0, 0, 0, 1, 1, 0}}} *)\n\nNote that for q = 17 there will be over 24,000,000 solutions -- I killed the kernel when it went past 20GB.\n\n-", "date": "2016-06-29 09:21:52", "meta": {"domain": "stackexchange.com", "url": "http://mathematica.stackexchange.com/questions/29383/finding-shortest-non-zero-vector-x-satisfying-ax-0-pmod-q", "openwebmath_score": 0.22085921466350555, "openwebmath_perplexity": 967.2872237186017, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9465966702001757, "lm_q2_score": 0.9005297894548548, "lm_q1q2_score": 0.8524385001140309}}
{"url": "http://xaktly.com/AlternatingSeries.html", "text": "Series in which terms alternate between positive and negative\n\nIn the infinite series notes you saw the series expansions of the sine and cosine functions. Here they are again along with the summation notation. Remember that there's nothing especially complicated about \u03a3 notation; it's just a way of capturing what's going on in the series without having to write all or several of its terms.\n\n\\begin{align} sin(x) &= \\frac{x^1}{1!} - \\frac{x^3}{3!} + \\frac{x^5}{5!} - \\frac{x^7}{7!} + \\frac{x^9}{9!} - \\dots = \\sum_{n = 0}^\\infty \\frac{(-1)^n x^{2n + 1}}{(2n + 1)!} \\\\ \\\\ cos(x) &= \\frac{x^0}{0!} - \\frac{x^2}{2!} + \\frac{x^4}{4!} - \\frac{x^6}{6!} + \\frac{x^8}{8!} - \\dots = \\sum_{n = 0}^\\infty \\frac{(-1)^n x^{2n}}{(2n)!} \\end{align}\n\nThese are called alternating series because of the alternation in the sign of each term.\n\nTo describe such an alternation in summation notation we employ the properties of powers of negative numbers:\n\n\u2022 $(-1)^1 = -1$\n\n\u2022 $(-1)^2 = 1$\n\n\u2022 $(-1)^3 = -1,$ \u00a0 and so on.\n\nIt's also possible to produce an alternation of sign using the trigonometric functions sin(x) and cos(x), like this:\n\n\u2022 $cos(n\\pi) = 1$ \u00a0 for even n and -1 for odd n\n\n\u2022 $sin\\left(\\frac{n\\pi}{2}\\right) = 1$ \u00a0 for odd n and -1 for even n\n\nAn example\n\nThe alternating harmonic series\n\nHere is an example of an alternating series, the so-called alternating harmonic series. This is just the harmonic series with alternating signs of the terms. The terms alternate on either side of zero as they decrease to zero (blue graph).\n\nThe accumulating sum (red graph) converges to a limit of approximately 0.69, but oscillates about that line. It's non-trivial to determine the actual sum of this series, but it does converge.\n\nThe alternating series test\n\nThe alternating series test, proved below the next box, is very simple.\n\nIt says that if, as n\u2192\u221e, the terms of an alternating series decrease to zero, then the series converges.\n\nRemember, that is NOT necessarily true for non-alternating series. For a non-alternating series, it is not enough that the size of the term diminishes; that series still may not converge.\n\nWe can think of many non-alternating, divergent series with decreasing terms, like the harmonic series:\n\n$$\\sum_{n = 1}^\\infty \\frac{1}{n} = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\:\\dots$$\n\nIn a way, the alternating series test (AST) makes life a lot easier. It's enough that the absolute value of the terms decrease to zero for an alternating series to converge \u2013 pretty simple.\n\nThe alternating series test\n\nIf the terms of an alternating series, $\\sum_{n = 1}^\\infty \\,(-1)^n a_n$\n\n(1) are decreasing, and (2) if the limit of the size of the terms, as n \u2192 \u221e is zero:\n\nthen the series converges.\n\nProof of the alternating series test\n\nFirst, take a look at the alternating series on a number line. We'll let the terms of the series be $a_1, a_2, a_3, \\dots$ and so on, and the partial sums will be $s_1 = a_1, \\; s_2 = a_1 + a_2, \\; s_3 = a_1 + a_2 + a_3,\\; \\dots$ and so on.\n\nLet's look first at the even sums. The first one (first green arrow in the diagram) is\n\n$$s_2 = a_1 - a_2 \\gt 0 \\; \\leftarrow \\; \\text{ because } a_1 \\gt a_2$$\n\nThe second and third are\n\n\\begin{align} s_4 &= s_2 + (a_3 - a_4) \\ge s_2 \\; \\leftarrow \\; \\text{ because } a_3 \\gt a_4 \\\\[4pt] s_6 &= s_4 + (a_5 - a_6) \\ge s_4 \\; \\leftarrow \\; \\text{ because } a_5 \\gt a_6 \\end{align}\n\n... and the pattern emerges. In the notation below, 2n is always an even number, 2n-2 is the next smaller even number and 2n-1 is the next smaller odd number:\n\n$$s_{2n} = s_{2n - 2} + (a_{2n - 1} - a_{2n}) \\ge s_{2n - 2}$$\n\nNow it's clear that these sums continue to grow, so that\n\n$$0 \\le s_2 \\le s_4 \\le s_6 \\le \\dots \\le s_{2n} \\le \\dots$$\n\nBy regrouping terms with some parenthesis, we can also write s2n as\n\n$$s_{2n} = a_1 - (a_2 - a_1) - (a_4 - a_3) - \\dots - (a_{2n - 2} - a_{2n - 1}) - a_{2n}$$\n\nNow because our terms are decreasing, every difference in parenthesis is positive. The sequence of terms {sn} is positive, increasing and bounded from above by the sum of the series, s (see diagram). It therefore has a limit,\n\n$$\\lim_{n\\to\\infty} s_{2n} = s$$\n\nNow because $s_{2n + 1} = s_{2n} + a+{2n + 1},$ we can find the limit of the odd partial sums:\n\n$$\\lim_{n\\to\\infty} s_{2n + 1} = \\lim_{n\\to\\infty} (s_{2n} + a_{2n + 1})$$\n\nWhat we've shown is that the limits of the even and odd partial sums is the same. This is called \"squeezing.\" If the limits from above and below are s, the sum of the series, then the series must converge to s.\n\nThe alternating series test makes determining the convergence of alternating series much easier than that of a non-alternating series. If the series alternates in sign, and if its terms decrease toward zero, the series converges.\n\nRefining the idea of convergence\n\nThe difference in convergence between the alternating harmonic series,\n\n$$\\sum_{n = 1}^\\infty \\, (-1)^n \\frac{1}{n}$$\n\nwhich converges, and the harmonic series\n\n$$\\sum_{n = 1}^\\infty \\, \\frac{1}{n}$$\n\nwhich doesn't, hints at a subtle differentiation in how series converge. In fact, the alternating harmonic series is called \"conditionally convergent,\" the condition being that its terms alternate sign.\n\nOn the other hand, a series like the convergent p-series,\n\n$$\\sum_{n = 1}^\\infty \\, \\frac{1}{n^2}$$\n\nand its alternating series counterpart,\n\n$$\\sum_{n = 1}^\\infty \\, (-1)^n \\frac{1}{n^2}$$\n\nboth converge, and the alternating version is said to be \"absolutely convergent.\" Here are the rules for absolute and conditional convergence:\n\nAbsolute vs. conditional convergence\n\nTest for absolute convergence\n\nTest for conditional convergence\n\nDivergence \u2013 the series diverges by any one of the tests for convergence, or the divergence test.\n\nExample 1\n\nDoes the series \u00a0 $\\sum_{n = 0}^\\infty \\, (-1)^{n + 1} \\frac{2n + 3}{3n + 4}$ converge?\n\nSolution: This is an alternating series in which terms with even n are negative. To test whether it converges, we ask whether the term, without the alternating part (-1)n+1.\n\n$$\\lim_{n\\to\\infty} \\, \\frac{2n + 3}{3n + 4} = \\frac{2}{3} \\neq 0$$\n\nThe term does not tend toward zero, so this series does not converge. We could also say that it diverges by the divergence test. Either way, it diverges.\n\nExample 2\n\nDoes the series \u00a0 $\\sum_{n = 0}^\\infty \\, \\frac{(-2)^n}{3^n}$ \u00a0 converge?\n\nSolution : We can first rewrite this series with a -1 inside the alternating term, using (xy)n = xn\u00b7yn:\n\n$$\\sum_{n = 0}^\\infty \\, (-1)^n \\left( \\frac{2}{3} \\right)^n$$\n\nNow we need to find whether the limit of the term without the alternation is zero:\n\n$$\\lim_{n\\to\\infty} \\, \\left( \\frac{2}{3} \\right)^n = 0$$\n\n2/3 < 1, so the limit is zero, and this alternating series converges by the AST.\n\nPractice problems\n\nDecide whether the following series converge.\n\n 1 $$\\sum_{n = 1}^\\infty \\, \\frac{(-1)^n}{4 + 3n}$$ 2 $$\\sum_{n = 1}^\\infty \\, \\frac{(-1)^n}{n^3 + 2n + 2}$$\n\n 3 $$\\sum_{n = 1}^\\infty \\, \\frac{n \\cdot cos(n\\pi)}{n^2 + 1}$$ 4 $$\\sum_{n = 1}^\\infty \\, \\frac{(-1)^n n^2}{n^2 + 1}$$\n\nxaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. \u00a9 2016, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.", "date": "2021-07-24 00:17:55", "meta": {"domain": "xaktly.com", "url": "http://xaktly.com/AlternatingSeries.html", "openwebmath_score": 0.9688819050788879, "openwebmath_perplexity": 460.12844880588716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543723101533, "lm_q2_score": 0.8596637541053281, "lm_q1q2_score": 0.8524033540996986}}
{"url": "https://math.stackexchange.com/questions/2812530/create-55-sets-with-exactly-one-element-in-common", "text": "# Create 55 sets with exactly one element in common\n\nI have 55 sets $E_1, E_2,\\cdots,E_{55}$\n\nThere are $n$ different elements.\n\nEach set contains 8 elements $E_1\\{x_1^1,x_1^2,x_1^3,x_1^4,x_1^5,x_1^6,x_1^7,x_1^8\\}, E_2\\{x_2^1,x_2^2,x_2^3,x_2^4,x_2^5,x_2^6,x_2^7,x_2^8\\}\\cdots$\n\nIf I take 2 sets randomly, they must have exactly one element in common, no more, no less.\n\nHow many elements do I have ? Is there more than one possibility ?\n\nActually I was playing a card game and I was curious about the mathematics behind.\n\nThanks!\n\nEDIT: I'm looking for the lowest solution possible\n\n\u2022 If you are curious about he mathematics behind Dobble, the keyword is \"block designs\". \u2013\u00a0Michal Adamaszek Jun 8 '18 at 13:34\n\u2022 One solution is $386$, with one element in all the sets and the other $7\\cdot 55=385$ elements distinct. \u2013\u00a0Ross Millikan Jun 8 '18 at 13:37\n\u2022 @RossMillikan True, but it wouldn't be a very exciting card matching game :) \u2013\u00a0Bram28 Jun 8 '18 at 13:40\n\u2022 Oh, indeed it's a solution. Then I'm looking for the lowest one. \u2013\u00a0tuxikigo Jun 8 '18 at 13:42\n\u2022 There's some analog to fractional coloring. What is the fractional chromatic number of an 8-fold coloring of the complete graph on 55 vertices, with the added condition that any two vertices share exactly one color. \u2013\u00a0David Diaz Jun 8 '18 at 14:23\n\nA lower bound is $n=57$. Take one set. It must have at least one element that it shares with at least seven other sets. Those eight sets all share one element, so all their other elements must be distinct. If there is an element shared $9$ ways $n$ must be at least $64$ because the other seven elements in each set need to be distinct.\nThere are $55\\cdot 54/2=1485$ pairs of sets. An element that is in $8$ of the sets accounts for the match of $28$ pairs. An element that is in $7$ of them accounts for $21$ pairs. As both of these numbers are divisible by $7$ and $1485$ is not, there must be at least one element that is shared by a number other than $7$ or $8$.\nAs $1485 \\equiv 1 \\bmod 7$ and an element that is shared $6$ ways accounts for $15$ pairs, which is also equivalent to $1 \\bmod 7$ I will guess that there is one element shared $6$ ways. We can then get the required number of pairs if there are $42$ elements shared $8$ ways and $14$ elements shared $7$ ways. This would give $n=57$.\nWe have not shown that this will work, but we can make a concrete example from the order $7$ projective plane. It has $57$ points, $57$ lines, with every two lines sharing a point and every two points sharing a line. The sets are the lines and the elements of each set are the points on it. We can discard two lines and get the required construction. Thanks to achillehui for pointing this out.\n\u2022 the finite projective plane of order $7$ contains $57$ points and $57$ lines, each line contains $8$ points and two lines determine a point. \u2013\u00a0achille hui Jun 8 '18 at 15:02\n\u2022 @Khoross: why? We could have the first nine sets contain element $1$ and the next eight contain element $2$. \u2013\u00a0Ross Millikan Jun 8 '18 at 17:31", "date": "2020-06-03 13:52:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2812530/create-55-sets-with-exactly-one-element-in-common", "openwebmath_score": 0.5953587889671326, "openwebmath_perplexity": 313.2007892639056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9915543748058885, "lm_q2_score": 0.8596637451167995, "lm_q1q2_score": 0.8524033473325768}}
{"url": "https://gmatclub.com/forum/the-cost-c-in-dollars-of-producing-x-units-of-a-certain-commodity-is-217213.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 15 Oct 2018, 20:45\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# The cost C, in dollars of producing x units of a certain commodity is\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 30 Aug 2015\nPosts: 25\nGMAT 1: 570 Q47 V23\nGMAT 2: 570 Q42 V27\nGMAT 3: 680 Q50 V35\nThe cost C, in dollars of producing x units of a certain commodity is\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 24 Apr 2016, 11:36\n2\n25\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n50% (02:11) correct 50% (01:51) wrong based on 519 sessions\n\n### HideShow timer Statistics\n\nThe cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a?\n\n(1) The cost of producing 500 units is $15.833. (2) The cost of producing 1,000 units is twice the cost of producing 500 units. Originally posted by inakihernandez on 24 Apr 2016, 10:20. Last edited by Vyshak on 24 Apr 2016, 11:36, edited 1 time in total. Edited the url ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 6958 The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 24 Apr 2016, 10:58 11 4 inakihernandez wrote: The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a? (1) The cost of producing 500 units is$15.833.\n\n(2) The cost of producing 1,000 units is twice the cost of producing 500 units.\n\nHi\nthe equation has got two constants a and b..\n\nlets see the statements\n\n(1) The cost of producing 500 units is $15.833. we have two constants a and b, and one equation Insuff (2) The cost of producing 1,000 units is twice the cost of producing 500 units here when you make two quation and equate them you will realize you can find answer.. $$C(1000) = 10.000 + b*1000 + a1000^2$$ .. $$C(500) = 10.000 + b*500 + a500^2$$ .. now $$C(1000) = 2* C(500)$$.. so $$10.000 + b*1000 + a1000^2 = 2( 10.000 + b*500 + a500^2 )$$.. $$10.000 + b*1000 + a1000^2 = 20.000 + b*1000 + 2*a*500^2$$ b will get cancelled and you can find value of a.. Suff Hope it helps _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor ##### General Discussion EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12649 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 24 Apr 2016, 10:47 2 Hi inakihernandez, You should post this question in the DS sub-Forum here: gmat-data-sufficiency-ds-141/ GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****\n\nSC Moderator\nJoined: 13 Apr 2015\nPosts: 1694\nLocation: India\nConcentration: Strategy, General Management\nGMAT 1: 200 Q1 V1\nGPA: 4\nWE: Analyst (Retail)\nRe: The cost C, in dollars of producing x units of a certain commodity is\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 11:02\n1\n2\nC(x) = 10.000 + bx + ax^2\na = ?\n\nSt1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833\nWe have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'\nNot Sufficient\n\nSt2: C(1000) = 2*C(500)\n10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)\na(10^6 - 5*10^5) = 10\nSufficient to determine the value of 'a'\n\nSenior Manager\nJoined: 11 Nov 2014\nPosts: 335\nLocation: India\nWE: Project Management (Telecommunications)\nRe: The cost C, in dollars of producing x units of a certain commodity is\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 11:10\n1\nVyshak wrote:\nC(x) = 10.000 + bx + ax^2\na = ?\n\nSt1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833\nWe have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'\nNot Sufficient\n\nSt2: C(1000) = 2*C(500)\n10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)\na(10^6 - 5*10^5) = 10\nSufficient to determine the value of 'a'\n\nhow did you get this?\na(25*10^4 )\na(50*10^4)\nSC Moderator\nJoined: 13 Apr 2015\nPosts: 1694\nLocation: India\nConcentration: Strategy, General Management\nGMAT 1: 200 Q1 V1\nGPA: 4\nWE: Analyst (Retail)\nRe: The cost C, in dollars of producing x units of a certain commodity is\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 11:13\n1\npaidlukkha wrote:\nVyshak wrote:\nC(x) = 10.000 + bx + ax^2\na = ?\n\nSt1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833\nWe have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'\nNot Sufficient\n\nSt2: C(1000) = 2*C(500)\n10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)\na(10^6 - 5*10^5) = 10\nSufficient to determine the value of 'a'\n\nhow did you get this?\na(25*10^4 )\na(50*10^4)\n\nHi,\n\nI have skipped some multiplication steps in my explanation.\n\nx = 5*10^2 --> x^2 = 25*10^4\n\n2*C(500) --> 2(10 + 500b + 25*10^4) --> 20 + 1000b + 2*25*10^4\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8373\nLocation: Pune, India\nRe: The cost C, in dollars of producing x units of a certain commodity is\u00a0 [#permalink]\n\n### Show Tags\n\n24 Apr 2016, 21:04\n8\ninakihernandez wrote:\nThe cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a?\n\n(1) The cost of producing 500 units is $15.833. (2) The cost of producing 1,000 units is twice the cost of producing 500 units. Let me point out here that this is a (C) trap question. The first statement will give an equation in a and b. So we know that we cannot find the value of a. When we look at the second statement, we see another equation in a and b which is not a multiple of the first equation and suddenly, it seems that we will be able to solve both equations simultaneously and get the answer. So (C) seems to be correct. But that is just too easy. Beware of the (C) trap. This should make you think - can I solve the question using a single statement alone. Make the equation of statement 2 since it gives you some convoluted relative data. When you do, you see that b gets canceled and in fact, you can get the value of a. Note that you cannot find the value of b from the second statement alone but the question doesn't ask for the value of b. We only need the value of a and hence second statement is sufficient alone. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > GMAT self-study has never been more personalized or more fun. Try ORION Free! Current Student Joined: 28 Nov 2014 Posts: 867 Concentration: Strategy Schools: Fisher '19 (M) GPA: 3.71 Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 09 Oct 2016, 04:29 The correct statement 1 is (1) The cost of producing 500 units of this commodity is$15833\n\nIt is NOT 15.833 although that won't affect the OA.\nNon-Human User\nJoined: 09 Sep 2013\nPosts: 8401\nRe: The cost C, in dollars of producing x units of a certain commodity is\u00a0 [#permalink]\n\n### Show Tags\n\n09 Oct 2017, 06:44\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nRe: The cost C, in dollars of producing x units of a certain commodity is &nbs [#permalink] 09 Oct 2017, 06:44\nDisplay posts from previous: Sort by", "date": "2018-10-16 03:45:55", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-cost-c-in-dollars-of-producing-x-units-of-a-certain-commodity-is-217213.html", "openwebmath_score": 0.56328284740448, "openwebmath_perplexity": 6474.3452182014125, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9752018455701407, "lm_q2_score": 0.8740772269642948, "lm_q1q2_score": 0.852401724906411}}
{"url": "https://math.stackexchange.com/questions/1229712/finding-the-matrix-of-a-linear-transformation-from-an-upper-triangular-matrix-to", "text": "# Finding the matrix of a linear transformation from an upper triangular matrix to an upper triangular matrix.\n\nThe question that I am trying to solve is as follows:\n\nFind the matrix $A$ of the linear transformation $T(M)= \\begin{bmatrix} 7 & 3 \\\\ 0 & 1 \\end{bmatrix} M$ from $U^{2\u00d72}$ to $U^{2\u00d72}$ (upper triangular matrices) with respect to the basis $\\left\\{ \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix},\\begin{bmatrix}1&1\\\\ 0&0\\end{bmatrix},\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix}\\right\\}$\n\n$A=\\begin{bmatrix} \\cdot&\\cdot&\\cdot\\\\\\cdot&\\cdot&\\cdot\\\\ \\cdot&\\cdot&\\cdot\\end{bmatrix}$\n\nI don't know how to interpret this question. I would have thought that the matrix $A$ is $\\begin{bmatrix} 7 & 3 \\\\ 0 & 1 \\end{bmatrix}$ because that is the matrix which defines the linear transformation $T$. I am surprised that the required answer is $A\\in M_{3\\times 3}$. How can a basis with elements $\\in M_{2\\times 2}$ form a matrix $\\in M_{3\\times 3}$?\n\n\u2022 Upper triangular matrices are a three dimensional vector space. But I don't understand your question. Usually $U$ indicates unitary matrices. \u2013\u00a0Emilio Novati Apr 11 '15 at 9:49\n\u2022 Has three dimensions got anything to do with $A\\in M_{3\\times 3}$? \u2013\u00a0ahorn Apr 11 '15 at 11:21\n\nUpper triangular matrices\n\n$\\begin{bmatrix} a&b\\\\ 0&c \\end{bmatrix}$\n\nform a vector space with canonical basis:\n\n$e_1=\\begin{bmatrix} 1&0\\\\ 0&0 \\end{bmatrix} \\quad e_2=\\begin{bmatrix} 0&1\\\\ 0&0 \\end{bmatrix} \\quad e_3=\\begin{bmatrix} 0&0\\\\ 0&1 \\end{bmatrix}$\n\nthat is isomorphic to $\\mathbb{R}^3$ by:\n\n$e_1\\rightarrow\\vec e_1=\\begin{bmatrix} 1\\\\0\\\\0 \\end{bmatrix} \\quad e_2\\rightarrow\\vec e_2=\\begin{bmatrix} 0\\\\1\\\\0 \\end{bmatrix} \\quad e_3\\rightarrow \\vec e_3=\\begin{bmatrix} 0\\\\0\\\\1 \\end{bmatrix}$\n\nYour linear transformation $T$ is defined as a matrix multiplication:\n\n$T(M)=\\begin{bmatrix} 7&3\\\\ 0&1 \\end{bmatrix} \\begin{bmatrix} a&b\\\\ 0&c \\end{bmatrix} = \\begin{bmatrix} 7a&7b+3c\\\\ 0&c \\end{bmatrix}$ so, in this canonical representation, it is given by the matrix $T_e$ such that:\n\n$T_e\\vec M=\\begin{bmatrix} 7&0&0\\\\ 0&7&3\\\\ 0&0&1 \\end{bmatrix} \\begin{bmatrix} a\\\\ b\\\\ c \\end{bmatrix}= \\begin{bmatrix} 7a\\\\ 7b+3c\\\\ c \\end{bmatrix}$\n\nNow you want a representation of the same transformation in a new basis:\n\n$e'_1= \\begin{bmatrix} 1&0\\\\ 0&0 \\end{bmatrix} \\rightarrow\\vec e'_1=\\begin{bmatrix} 1\\\\0\\\\0 \\end{bmatrix} \\quad e'_2=\\begin{bmatrix} 1&1\\\\ 0&0 \\end{bmatrix}\\rightarrow\\vec e'_2=\\begin{bmatrix} 1\\\\1\\\\0 \\end{bmatrix} \\quad e'_3=\\begin{bmatrix} 0&0\\\\ 0&1 \\end{bmatrix}\\rightarrow \\vec e'_3=\\begin{bmatrix} 0\\\\0\\\\1 \\end{bmatrix}$\n\nThis transformation of basis is represented by the matrices\n\n$S= \\begin{bmatrix} 1&1&0\\\\ 0&1&0\\\\ 0&0&1 \\end{bmatrix} \\qquad S^{-1}= \\begin{bmatrix} 1&-1&0\\\\ 0&1&0\\\\ 0&0&1 \\end{bmatrix}$\n\nSo the matrix that represents the transformation $T$ in the new basis is:\n\n$T_{e'}=S^{-1}T_eS= \\begin{bmatrix} 7&0&-3\\\\ 0&7&3\\\\ 0&0&1 \\end{bmatrix}$\n\n\u2022 When you said \"This transformation of basis is represented by the matricies\", did you simply put the column vectors $\\vec e'_1$, $\\vec e'_2$ and $\\vec e'_3$ next to each other to get $S$? Or was there a more nuanced way of getting $S$, such as by saying that $\\vec e_j= (a_j+b_j+c_j)\\vec e'_j$ where $j$ is the number of the column? \u2013\u00a0ahorn Apr 11 '15 at 15:49\n\u2022 Symply I put the columns ( it's a general rule that you can easely verify by linearity). \u2013\u00a0Emilio Novati Apr 11 '15 at 16:31\n\nLet $(X,Y,Z)$ be the basis of $U$, compute $T(X)$, $T(Y)$, $T(Z)$ and express them as a linear combination of $X$, $Y$ and $Z$.\n\n\u2022 T(X)=7X; T(Y)=7Y; T(Z)=Z. How does that relate to A? \u2013\u00a0ahorn Apr 11 '15 at 11:20\n\u2022 A={(7,0,0),(0,7,0),(0,0,1)} take the coefficients as elements of the matrix A. \u2013\u00a02ndYearFreshman Apr 11 '15 at 11:39\n\u2022 I have tried to input all sorts of combinations, such as $\\begin{bmatrix} 7 & 0 & 0 \\\\ 0 & 7 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}, \\begin{bmatrix} 7 & 0 & 0 \\\\ 7 & 7 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}, \\text{and} \\begin{bmatrix} 7 & 0 & 0 \\\\ 7 & 3 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}$ but these are still incorrect (I am answering an online test). Btw, welcome to math.stackexchange! What brought you to the site? \u2013\u00a0ahorn Apr 11 '15 at 14:25\n\u2022 Try A={(7,0,0),(0,7,0),(-3,3,1)} Thank you,hopefully my answer will help.Your question brought me here. \u2013\u00a02ndYearFreshman Apr 11 '15 at 14:34", "date": "2021-04-15 05:08:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1229712/finding-the-matrix-of-a-linear-transformation-from-an-upper-triangular-matrix-to", "openwebmath_score": 0.8394883275032043, "openwebmath_perplexity": 191.48744632396034, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8740772286044095, "lm_q1q2_score": 0.8524017202062412}}
{"url": "https://www.physicsforums.com/threads/how-can-a-curve-be-one-dimensional.593124/", "text": "# How can a curve be one dimensional?\n\n1. Apr 3, 2012\n\n### rollcast\n\nI've heard of curves being described as one dimensional but I don't understand how anything other than a straight line can be one dimensional as surely once the curve becomes, well, curved it is now in two dimensions?\n\nI have illustrated this in the above diagram with the curve and straight line shown in red and the dimensions in green.\n\nThanks\nA.\n\nFile size:\n5.3 KB\nViews:\n1,706\n2. Apr 3, 2012\n\n### Office_Shredder\n\nStaff Emeritus\nThe curve is embedded in a 2 dimensional plane, but the curve itself is one dimensional. For example, I can describe all the points using only one parameter in a very natural way (if the graph is y=f(x), then x is the parameter since it uniquely identifies y)\n\nReally what you want is dimension to be an intrinsic property. If I took your curve and put it in 3 dimensions, is it a 3 dimensional object now? What about the fact that we're in spacetime, is it 4 dimensional? That's a bit silly, so instead we ask what properties does the curve have that is independent of the space that it is embedded in.\n\nThere are several possible definitions of dimension depending on context but the basic idea is that if it has dimension k you can describe every point using k parameters at least locally - by that I mean, for example the curve x2+y2=1 (a circle) you can describe the top half of it by the set of points of the form $(x,\\sqrt{1-x^2})$ and the bottom half is of the form $(x,-\\sqrt{1-x^2})$. So I wasn't able to describe the whole circle using the x-coordinate but I was able to describe each half of it using a single coordinate, which means the whole thing is 1 dimensional (in fact you can describe the whole circle using a single coordinate if you use trigonometric functions, which I encourage you to try if you haven't seen it)\n\n3. Apr 3, 2012\n\n### Bacle2\n\nWould you think of the curve as having non-zero area? I don't mean a closed curve\n\nthat may enclose an area; I mean the curve itself.\n\n4. Apr 3, 2012\n\n### HallsofIvy\n\nStaff Emeritus\nYes, such a curve is in two dimensions but it only \"one dimensional\" itself\".\n\nWhat is your definition of \"dimension\"? One common one is: a subset of Euclidean space is said to be \"n-dimensional\" if and only if any point can be identified using n real numbers.\n\nChoosing one point on the curve, we can identify any point on the curve by its distance (positive in one direction, negative in the other) from the chosen point. Since that is a single number, the curve is one-dimensional.\n\n5. Apr 3, 2012\n\n### rollcast\n\nBit of a weird example here so the curve is like curved corridor in a building, each office on the corridor is numbered with respect to where it is positioned along the corridor and not where it is positioned relative to the 4 outer walls of the building.\n\n6. Apr 3, 2012\n\n### Matterwave\n\nPerhaps a somewhat more intuitive way of thinking about it is, if you live \"in\" the curve (i.e. you're not allowed to go outside of it), you can only ever go forwards or backwards, never to the left or right or up or down.\n\n7. Apr 3, 2012\n\n### Bacle2\n\nAltho you may want to consider separately the case of space-filling curves, like the Peano curve.\n\n8. Apr 3, 2012\n\n### Bacle2\n\nI'm starting to doubt my answer, if we can define a curve as the continuous image of a connected interval. We can use the fact that, e.g., any compact metric space K is the continuous image of the Cantor set. Then take, e.g., K=[0,1]^n (mapping I into R^n), and extend continuously, using Tietze extension ( C is closed in I ; I is normal), then the image of f has non-empty interior in R^n.\n\nNote/Edit: I'm referring to a more general definition of a curve, as the continuous image of an interval, and not the curves in the attached files.\n\nLast edited: Apr 3, 2012\n9. Apr 4, 2012\n\n### chiro\n\nIntuitively for any line, no whatever what its deformation or dimensionality for the embedded space is always a line. Similarly, a flat piece of paper no matter how its deformed or what space its embedded in is always a flat piece of paper.\n\nIf you are given a visualization of the object in question, you can get a very good idea of the number of parameters. If you can't visualize the object but are given a deterministic expression, then you have to use other methods.\n\nIf you are given a non-analytic version of a process where you only know the dimensionality of the embedded space and have to not only figure out the parameterization but also some sort of 'useful' definition for the process as a whole (implicit most likely but could be explicit), then you will have to use even different methods again which will be based on a combination of statistical theory and non-statistical mathematics.\n\nOne method for assessing parameterization is assessing density in various orientations for your function and statistical results used in data mining measure a kind of 'variation' density where you get principal components for your data set. This kind of idea could be applied to your analytic setting in the analytic perspective (not the statistical one).\n\n10. Apr 4, 2012\n\n### Matterwave\n\nA curve has the same cardinality as an arbitrary n-dimensional hypercube, but that doesn't mean it has the same dimension. A space-filling curve defines a surjection, but not a homeomorphism.\n\n11. Apr 4, 2012\n\n### Bacle2\n\nI was referring to a curve as the continuous, not homeomorphic image of an interval. I was also taking back my statement that a curve must have empty interior. As defined, I think the argument is correct, and a curve does not necessarily have an empty interior.\n\n12. Apr 4, 2012\n\n### Bacle2\n\nClearly, the image of an interval would depend on a single parameter. I was referring to the claim/belief that the continuous image of a curve would have no area, i.e., would have empty interior.\n\n13. Apr 4, 2012\n\n### Bacle2\n\nSorry, one last post on this, to try to avoid confusion:\n\n0)Define a curve as the continuous image of an (connected)interval\nClaim: A curve can have non-empty interior.\n\ni)We have that every compact metric space is the continuous image of the\nCantor set C . So we map C subset I:=[0,1] into, say I^n, so there is an f with f(C)=I^n (n>1)\n\nii) By Tietze extension thm. ( C is closed in I normal) , we can extend f to f^, defined on the whole of I, continuously ( on each real variable). Then f^ is a continuous map such that f^(I) has non-empty interior in R^n.\n\n14. Apr 4, 2012\n\n### chiro\n\nI'll have a look at your theorems later on, but the thing is if you can deform your object that you have in a way so that is linear, then apply the ideas of dimensionality to get not only the proper dimension but the parameterization of your object, then you are done. By finding the appropriate deformation, you have created a linear representation that can be dealt with in the linear context.\n\nThis is of course equivalent to finding the inverse for each part of your object with respect to the different 'branches' that exist. If you can find the inverse for each 'branch' of your object (might be multi-dimensional and multi-parameterized but the idea is the same), then you can find a linear representation which can be parameterized.\n\nSo the case with a line would result in basically an n dimensional system that represents a line with one parameter t where X(t) = At + (1-t)B in the deformed state to the linear space. If this parameterization wasn't only one dimensional, then the deformation wouldn't produce one but however many parameters for the object.\n\nWith the notion of topology, if you know the object is analytic continuous then there should exist a deformation to the linear representation which depends on the inverse within a given branch-volume that corresponds to where the derivatives are (which correspond to the ideas in the inverse function theorem which are calculated by finding zero-jacobians). With this information you can find the branches and thus deform that region to it's linear counterpart.\n\nThe reason I like to think of something in the deformed linear space is because linear spaces, decompositions, and parameterizations of linear systems are well understood both algebraically, algorithmically and also to an extent visually. It's very easy to parameterize something that is deformed to linear space than to try and analyze it in its non-deformed non-linear space with a function like say f(x,y,z,w,t,u,v,a) = blah for any analytic continuous blah.\n\nYou could also probably consider continuous representations that are not analytic over the whole domain and use the same argument based on topological reasoning but I will only speculate on this (intuitively it makes sense at least).\n\n15. Apr 4, 2012\n\n### Bacle2\n\nI think the best we can say is that if f is C^1 (maybe Lipschitz) , then, with a bounded derivative, we can preserve the Hausdorff dimension.\n\n16. Apr 4, 2012\n\n### chiro\n\nI'll go with that for finding the dimension (which is the OP's question), but in terms of understanding why (and where) dimension goes from say one thing to another it may be useful to see which 'parts' if they are isolated contribute to where the function requires more dimensions as opposed to less (it can happen but not always).\n\nThe only reason I am saying this (for the OP) is that visually you can see where dimensionality changes. I'm imagining some kind of implicit function that creates branches (like a bifurcation) and does this repeatedly which creates more dimensions at different parts.\n\nJust out of curiosity, what would you say topologically is a way to do the above? In other words in the case of say a bifurcation like object, how would you isolate this kind of phenomenon in terms of the local characteristics (spatially) of the object?\n\n17. Apr 4, 2012\n\n### Bacle2\n\nThe way I see it, if the derivative is bounded, then the image of the balls in the covering will not expand by much.\n\nSorry, it is 4 a.m., here, I'm out for the night; will try to be back in case there are more posts.\n\n18. Apr 4, 2012\n\n### HallsofIvy\n\nStaff Emeritus\nThen any two-dimensional ball about a point on the curve will necessarily include some points not on the curve- a curve has no (2 dimensional) interior.\n\n19. Apr 4, 2012\n\n### Bacle2\n\nHalls of Ivy:\n\nI did a specific construction, with a specific definition of curve; what is it about my construction that you think is faulty?\n\nIt seems strange-enough that the hypercube ( as a compact metric space; a space with non-empty interior) is the continuous image of the Cantor set. Then what is wrong with the extension?\n\n20. Apr 5, 2012\n\n### TrickyDicky\n\nI think the OP is simply reflecting on the fact that there is no such thing as intrinsic curvature in one dimension.", "date": "2017-08-21 23:42:45", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/how-can-a-curve-be-one-dimensional.593124/", "openwebmath_score": 0.7991959452629089, "openwebmath_perplexity": 445.8879114846821, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018398044144, "lm_q2_score": 0.8740772269642949, "lm_q1q2_score": 0.8524017198667211}}
{"url": "https://tlumaczenia-word.pl/32144/cylinder-with-cone-on-top.html", "text": "## cylinder with cone on top\n\n### Three-dimensional figures - Cylinders, cones and spheres ,\n\nFirst, the cylinder The cylinder is somewhat like a prism It has parallel congruent bases, but its bases are circles rather than polygons You find the volume of a cylinder in the same way that you find the volume of a prism: it is the product of the base area times the height of the cylinder:\n\n### Cylinder volume & surface area (video) | Khan Academy\n\nLet's find the volume of a few more solid figures and then if we have time, we might be able to do some surface area problems So let me draw a cylinder over here So that is the top of my cylinder And then this is the height of my cylinder This is the bottom right over here If this was .\n\n### How many faces, edges and corners does a cylinder and cone ,\n\nJan 13, 2019\u00b7 In 3-dimensional geometry, we define a face to be a planar surface part of a solid (ie: 2-D/flat), an edge to be a line segment joining two faces, and a corner (more commonly known as a vertex) to be the end point of an edge or curve Cylinder - S.\n\n### Chamfering Cones - Beam Equipment & Supplies\n\nCylinder Chamfering Cones and Mandrels Chamfering cone to chamfer the top of cylinder after boring This is the Chamfering Mandrel Only\n\n### What is the name of this geometric shape - A Cylinder with ,\n\nWhat is the name of this geometric shape - A Cylinder with a Cone Atop? Ask Question Asked 2 years, , $\\begingroup$ I would probably call it \"a cylinder with a cone on top\" , a pyramid on top of a prism is called \"elongated pyramid\" If we can apply similarity then this shape should be called \"elongated cone\"\n\n### SOLUTION: A cone sits on top of a cylinder Both have a ,\n\nSOLUTION: A cone sits on top of a cylinder Both have a radius of 3 The heights of the cylinder and cone are 8 and 4 respectively Determine the exact volume of the combined sol Algebra ->Customizable Word Problem Solvers ->Geometry->SOLUTION: A cone sits on top of a cylinder Both have a radius of 3\n\n### How to calculate the surface area of a cylinder that has ,\n\nFirst of all, Imagine how a cylinder is made Let me help you in doing this From the above figure, you can see that the curved surface area of cylinder is equal to the area of rectangle taken intially If the given cylinder is of radius r, then t.\n\n### Volume of a circular truncated cone Calculator - High ,\n\nCalculates the volume, lateral area and surface area of a circular truncated cone given the lower and upper radii and height\n\n### Partially full cylinder, sphere, and cone volume ,\n\nA cone is a frustum that has a top diameter of 00 For the cylinder laying on its side, if you enter liquid top width, diameter, and length, there are two possible solutions for depth and volume; therefore, when using the drop-down menu to select which calculation to perform, you must select whether the cylinder is more or less than half full\n\n### geometry - Problem with determining cylinder height ,\n\nThe vase company designs a new vase that is shaped like a cylinder on the bottom with a cone on top The catalog states that the width is $12$ cm and the total height is $42$ cm What would the height of the cylinder part have to be in order for the total volume to be $1224 \\pi$ $\\mat{cm}^3$\n\n### Cone Problems - analyzemath\n\nA tool is made up of a cone on top of a cylinder (see figure below) The cylinder has a height h of 15 cm and a radius of 5 cm The volume of the cone is 100 Pi cm 2 O is the vertex of the cone, AB is the diameter of the base of the cone and C its center Points O, A, B and C are in the same plane 1 - Approximate angle AOB\n\n### Volume of a circular truncated cone Calculator - High ,\n\ncalculating the lateral area of a glass bottle having two truncated cones (an upper one clause to the neck and a lower one at the bottom separated by a cylinder the internal lateral area including the lid must me calculated in order to be coated with a certain dose (mg/cm2) of an insecticide to perform what is known in toxicology as bottle .\n\n### cylinder with cone on top - greenrevolutionorgin\n\nCylinder and cone49 \u041a\u0431 CYLINDER AND CONE Fig (5) shows the sector of a circle with radius 50 cms and an angle of 108 degre It is folded and stuck with the silver side inside to make a coneFig (9) Two conefuls, Fig (10), and finally three conefuls to top the film-roll bottle with water\n\n### Cones_Pyramids_and_Spheres - AMSI\n\nSuppose the top and bottom of a frustum are circles of radius R and r, respectively, and that the height of the frustum is h, while the height of the original cone is H The volume of the frustum is by the difference of the volumes of he two cones and is given by , c Volume of cylinder \u2212 volume of cone = .\n\n### Surface Area of Conical Cylinder Calculator\n\nThe surface area (SA) of the conical cylinder refers to the sum of all areas It is the measurement of the area of the circular top and base, and also the curved surface area of the cylinder It computes the total area of the conical cylinder The surface area of this cylinder is the sum of surface area of cone and cylinder\n\n### Cylinder volume & surface area (video) | Khan Academy\n\nMay 04, 2017\u00b7 Let's find the volume of a few more solid figures and then if we have time, we might be able to do some surface area problems So let me draw a cylinder over here So that is the top of my cylinder And then this is the height of my cylinder,\n\n### A tin man has a head that is a cylinder with a cone on top ,\n\nMar 25, 2015\u00b7 A tin man has a head that is a cylinder with a cone on top The height of the cylinder is 12 in and the slant height of the cone is 6 inch? The radius of both the cylinder and the cone is 4 inch What is the surface area of the tin man's head in terms of pi? A) 136\u03c0 in2 B)\n\n### Cylinder - Wikipedia\n\nA cylinder (from Greek \u03ba\u03cd\u03bb\u03b9\u03bd\u03b4\u03c1\u03bf\u03c2 \u2013 kulindros, \"roller, tumbler\") has traditionally been a three-dimensional solid, one of the most basic of curvilinear geometric shap It is the idealized version of a solid physical tin can having lids on top and bottom\n\n### Determining the Surface Area of Cones and Cylinders ,\n\nDetermining the Surface Area of a Cylinder from a Net Determining Surface Area Investigating and Applying the Surface Area Formula of a Cylinder , Determining the Surface Area of Cones and Cylinders Resource ID: GM4L9 Grade Range: 9\u201312 Sections Determining the Surface Area of a Cylinder from a Net\n\n### Volume of a Conical Cylinder Calculator\n\nVolume of a Conical Cylinder Calculator Online calculator to find the conical cylinder volume This can be calculated by adding the volume of cylinder and cone You will get the conical cylinder when the edge of the cylinder is cut off The base of the cylinder is large circle and the top ,\n\n### Spinning Cone - Math Is Fun\n\nYou should order your ice creams in cylinders, not cones, you get 3 times as much! Like a Pyram A cone is also like a pyramid with an infinite number of sides, see Pyramid vs Cone Different Shaped Con Construction Cone This is almost a cone, but the top is chopped off (called a \"truncated cone\")\n\n### Volume of Cylinders, Cones, and Spheres - Video & Lesson ,\n\nIn this lesson, we'll learn about the volume formulas for cylinders, cones and spher We'll also practice using the formula in a variety of.\n\n### Determining the Volume of Cones and Cylinders | Texas Gateway\n\nDetermining the Volume of Cones and Cylinders Resource ID: GM4L10 Grade Range: 9\u201312 Sections Explore the Volume of Cylinder Comparing the Volume of a Cylinder to the Volume of a Cone Determining the Volume of a Cone Explore the Volume of Cylinder Comparing the Volume of a Cylinder to the Volume of a Cone\n\n### Volume Formulas (examples, solutions, games, worksheets ,\n\nRelated Topics: More Geometry Lessons | Volume Games In these lessons, we give a table of volume formulas and surface area formulas used to calculate the volume and surface area of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere and hemisphere a more detailed explanation (examples and solutions) of each volume formula\n\n### Surface Area and Volume of Combination of Solids: Formulas ,\n\nCylinder: \u03c0r 2 h, r is the radius of circular base and h is the height of the cylinder Cone: 1/3 \u03c0r 2 h, r is the radius of the circular base, l is the slant height of the cone , The radius of the circular shaped top is 30cm and height of the cylinder is 145 m Find the total surface area of the bath?\n\n### A metal cylinder on top of a telephone pole : whatisthisthing\n\nA metal cylinder on top of a telephone pole Solved! Close 4 Posted by u/Dusterthefirst 2 years ago Archived A metal cylinder on top of a telephone pole Solved! 5 comments share save hide report 63% Upvoted This thread is archived New comments ,\n\n### Spinning Cylinder - Math Is Fun\n\nSo a cone's volume is exactly one third ( 1 3) of a cylinder's volume In future, order your ice creams in cylinders, not cones, you get 3 times as much! Like a Prism A cylinder is like a prism with an infinite number of sides, see Prism vs Cylinder It Doesn't Have to Be Circular\n\n### SOLUTION: Landsville has just finished building a new ,\n\nThe tank is a cylinder with a cone on top It has a diameter of 20 ft, a total height of 21 ft, and the cylindrical part i Algebra ->Surface-area->SOLUTION: Landsville has just finished building a new water tank The tank is a cylinder with a cone on top\n\n### shugickweebly\n\nis made of a cylinder 60 yards tall and a cone on top of the cylinder is 20 yards tall as shown in the diagram The diameter of the tank is 32 yards How many gallons of water can the tank hold when it is filled to the top? (1 cubic yard equals about 20197 gallons) Show your work Use 314 for IT B\n\n### Volumes of cones and cylinders - Math Central\n\nThat is, the volume of a cone is 1/3 the volume of a cylinder with the same radius and height Now, you are given a volume for the cylinder, but neither the height nor the radius You can write (for the cylinder) 1353 cm 3 = pi*(r 2)*h A cone of the same r and h should have 1/3 this volume But we want a cone with double the height, therefore", "date": "2021-10-20 09:26:40", "meta": {"domain": "tlumaczenia-word.pl", "url": "https://tlumaczenia-word.pl/32144/cylinder-with-cone-on-top.html", "openwebmath_score": 0.7342086434364319, "openwebmath_perplexity": 673.0379157521035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018383629826, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8524017154079127}}
{"url": "https://gmatclub.com/forum/if-s-is-the-sum-of-odd-integers-from-40-to-60-inclusive-and-t-is-the-251326.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Sep 2018, 05:56\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# If s is the sum of odd integers from 40 to 60, inclusive, and t is the\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 11 Oct 2017\nPosts: 22\nIf s is the sum of odd integers from 40 to 60, inclusive, and t is the\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2017, 12:33\n2\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n76% (01:16) correct 24% (01:14) wrong based on 89 sessions\n\n### HideShow timer Statistics\n\nIf s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?\n\nA. 480\nB. 490\nC. 980\nD. 990\nE. 995\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49271\nRe: If s is the sum of odd integers from 40 to 60, inclusive, and t is the\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2017, 12:39\nIf s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?\n\nA. 480\nB. 490\nC. 980\nD. 990\nE. 995\n\nSimilar question: https://gmatclub.com/forum/if-x-is-equa ... 97544.html\n_________________\nManager\nJoined: 12 Feb 2017\nPosts: 71\nRe: If s is the sum of odd integers from 40 to 60, inclusive, and t is the\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2017, 12:48\nThere are total 10 odd numbers between 40 and 60.\nhence, t=10\n\nNow, the given sequence is 41,43,45,......,59\ncommon difference is 2.\nhence,\nS= (41+59)*(10/2)\nS=500.\nS-t= 500-10= 490.\n\nKudos if it helps.\nIntern\nJoined: 11 Oct 2017\nPosts: 22\nRe: If s is the sum of odd integers from 40 to 60, inclusive, and t is the\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2017, 13:31\n1\nAre these formulas correct? I am having trouble with odd numbers for some reason.\n\nN for odd numbers = (Last odd - First odd)/2)) +1\n\nSum of odd numbers = (First odd + Last odd)*N))/2\n\n1. N = (59-41)/2))+1 = 10\n\n2. Sum of odds = (41+59)*10))/2 = 500\n\n3. 500 - 10 = 490 B\n\nAlso, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 1977\nIf s is the sum of odd integers from 40 to 60, inclusive, and t is the\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2017, 13:36\n1\nIf s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?\n\nA. 480\nB. 490\nC. 980\nD. 990\nE. 995\n\nThe way I solve sums of arithmetic series, I need the number of terms no matter what. Start there.\n\nSet t\n\nNumber of terms: number of odd integers from 40 to 60. First term is 41, last is 59.\n\nNumber of terms\n\n$$\\frac{LastTerm-FirstTerm}{interval} + 1$$\n\nThe \"interval\" for consecutive even and odds is always 2\n\n$$\\frac{59-41}{2} = \\frac{18}{2} = (9 + 1)=$$ 10\n\nSet s\n\nSum of arithmetic series\n(Average)(Number of terms) =\n\n$$\\frac{FirstTerm+LastTerm}{2}*(10)$$ =\n\n$$\\frac{100}{2}(10) = (50)(10) = 500$$\n\ns - t: 500 - 10 = 490\n\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 1977\nIf s is the sum of odd integers from 40 to 60, inclusive, and t is the\u00a0 [#permalink]\n\n### Show Tags\n\n12 Oct 2017, 14:20\n1\nAre these formulas correct? I am having trouble with odd numbers for some reason.\n\nN for odd numbers = (Last odd - First odd)/2)) +1\n\nSum of odd numbers = (First odd + Last odd)*N))/2\n\n1. N = (59-41)/2))+1 = 10\n\n2. Sum of odds = (41+59)*10))/2 = 500\n\n3. 500 - 10 = 490 B\n\nAlso, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?\n\nI think the reason you might be having trouble with odd numbers is that for consecutive odd integers, the interval (between the numbers), is two. Two is even; that could confuse. Or you might be forgetting to start with the first ODD term (e.g. here, first term is 41, not 40).\n\nBottom line: The \"interval\" between 1 and 3 is two. The interval between 2 and 4 is two. For consecutive odds and evens, in the formula, you divide by 2.\n\nTake smaller numbers. Say, sum of odd integers from 0 to 6. So 1 + 3 + 5 = 9. There are three terms (1, 5, 9) whose sum is 9.\n\nNumber of terms in \"odd integers from 0 to 6\":\n\n$$\\frac{(Last - First)}{2}+ 1$$\n\n$$\\frac{(5-1)}{2} + 1 = (\\frac{4}{2}+ 1) = (2 + 1) = 3$$ Correct\n\nSUM: (Average)(# of terms)\n\n$$\\frac{(First + Last)}{2}(3)$$\n\n$$\\frac{(1 + 5)}{2} (3) = (3)(3) = 9$$\nBingo. Dividing by 2 for consecutive odd integers works.\n\nWhen subtracting, you add one because subtraction doesn't include the first number. If in doubt: use small numbers that replicate your situation.\n\nFor example, # of terms from 1 to 4? The numerals are 1, 2, 3, 4. There are 4 terms. BUT (4 - 1) = 3.\nWe need one more. 3 + 1 = 4.\n\nThere's a mnemonic: \"add one before you're done.\" It almost always applies. If in doubt, replicate your situation with small numbers. Rare in sequence sums but possible: If you see the word \"exclusive\" or the phrase \"exclusive of,\" that is when you really need to check.\n\nI have seen your other answers using this method. The ones I've seen are correct! (The one about -190 to 195? Correct, but long. -190 to +190 = 0; you are left with only 5 numbers to sum. STILL - correct.)\n\nHere is a fantastically written, comprehensive post on all kinds of sequences by benjiboo , scroll down for arithmetic sequence\n\nAnd here is a thread on sequences. VeritasPrepKarishma 's posts are great\n\nHope this barrage of information helps.\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nIntern\nJoined: 11 Apr 2017\nPosts: 38\nSchools: Kelley '20\nRe: If s is the sum of odd integers from 40 to 60, inclusive, and t is the\u00a0 [#permalink]\n\n### Show Tags\n\n13 Oct 2017, 11:23\nN for odd numbers = (Last odd - First odd)/2)) +1\n\nSum of odd numbers = (First odd + Last odd)*N))/2\n\n1. N = (59-41)/2))+1 = 10\n\n2. Sum of odds = (41+59)*10))/2 = 500\n\n3. 500 - 10 = 490 B\nRe: If s is the sum of odd integers from 40 to 60, inclusive, and t is the &nbs [#permalink] 13 Oct 2017, 11:23\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-20 12:56:18", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-s-is-the-sum-of-odd-integers-from-40-to-60-inclusive-and-t-is-the-251326.html", "openwebmath_score": 0.623427152633667, "openwebmath_perplexity": 1933.1530694799037, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9381240142763574, "lm_q2_score": 0.9086178944582995, "lm_q1q2_score": 0.8523962665925516}}
{"url": "https://math.stackexchange.com/questions/1673289/find-the-value-of-a-b2-given-a-b-2-and-a2-b2-6/1673299", "text": "# Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$\n\n$$a + b = 2\\\\ a^2 + b^2 = 6$$\n\nFind the value of $(a-b)^2$\n\nMy workings till I got stuck -\n\n$$(a-b)^2 = a^2 - 2ab + b^2 \\\\ = a^2 + b^2 - 2ab\\\\ = 6 - 2 ab$$\n\nI'm stuck at how to find $ab$ . Can I get hints on how to find $ab$? Thanks a lot.\n\n\u2022 $4= (a+b)^2=a^2+b^2+2ab$ \u2013\u00a0lulu Feb 26 '16 at 15:16\n\u2022 Squaring the first equation then subtract the second equation from it. \u2013\u00a0Zhanxiong Feb 26 '16 at 15:17\n\u2022 The above comments mean that you should consider the quantity $2a^2+2b^2-(a+b)^2$... \u2013\u00a0abiessu Feb 26 '16 at 15:19\n\nUse $$(a-b)^2+(a+b)^2=2(a^2+b^2)$$ Thus, here, $$(a-b)^2+4=12$$ $$(a-b)^2=8$$\n\nWe will solve for the values of $a$ and $b$. $a=2-b$ so $a^2+b^2=6 \\iff (2-b)^2 + b^2 +6$\n\nThis means $2b^2-4b-2=0 \\iff b^2 -2b -1 +0$. The solutions of this equation are $b=1-\\sqrt{2}$ and $b=1+\\sqrt{2}$. Then $a=1+\\sqrt{2}$ or $a=1-\\sqrt{2}$\n\nThen, in any case $(a-b)^2=8$\n\n\u2022 -1.This is a low grade method. \u2013\u00a0N.S.JOHN Apr 17 '16 at 10:42\n\n$$a+b=2$$\n\n$$\\implies(a+b)^2=4$$\n\n$$\\implies a^2+b^2+2ab=4$$\n\n$$\\implies2ab=-2$$\n\nAlso,\n\n$$(a-b)^2=a^2+b^2-2ab$$\n\n$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)", "date": "2019-10-21 11:14:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1673289/find-the-value-of-a-b2-given-a-b-2-and-a2-b2-6/1673299", "openwebmath_score": 0.8188152313232422, "openwebmath_perplexity": 234.83533367894054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085088220004, "lm_q2_score": 0.8670357701094303, "lm_q1q2_score": 0.8523699365479284}}
{"url": "https://math.stackexchange.com/questions/1055651/probability-that-a-random-13-card-hand-contains-at-least-3-cards-of-every-suit", "text": "# Probability that a random 13-card hand contains at least 3 cards of every suit?\n\nA random 13-card hand is dealt from a standard deck of cards. What is the probability that the hand contains at least 3 cards of every suit? (Introduction to Probability, p.36)\n\nMy solution:\n\n\u2022 There are $\\binom{52}{13}$ possible hands.\n\u2022 Because there are 13 cards for the hand, to obtain at least three cards of one suit per hand, we need to have exactly three cards of one suit per hand plus one additional card of any suit, thus $\\binom{13}{3}^4 * 4 \\binom{10}{1}$\n\u2022 Result: $\\frac{40*\\binom{13}{3}^4}{\\binom{52}{13}} = 0.4214$\n\nHowever, simulating it in R yields:\n\ndeck <- rep(1:4, 13)\nout <- replicate(1e5,{\nhand <- sample(deck, size=13, replace=FALSE)\nall(table(hand) >= 3)\n})\nmean(out)\n> 0.14387\n\n\nCan anybody tell me what is wrong?\n\nEDIT\n\nI'm afraid, the correct code should be.\n\ndeck <- rep(1:4, 13)\nout <- replicate(1e5,{\nhand <- sample(deck, size=13, replace=FALSE)\nlength(table(hand))==4 & all(table(hand) >= 3 )\n})\nmean(out)\n> 0.10639\n\n\u2022 The correct result is about $0.105$, yet the R simulation yields $0.14387$. This is the second R simulation you've posted today that gives a result significantly different from the theoretical result. There must be something wrong with the R code (up to and including the possibility that R's randomization is insufficiently random for your purpose) but I still don't know what the error is. You might want expert advice on this so that you aren't misled by an incorrect simulation in the future. Dec 7, 2014 at 17:17\n\u2022 @DavidK R is completely fine, the idiot is me. As you mentioned, I posted two times simulations along with my questions and both of them were wrong, because I didn't give enought thought to the special cases. I'm sorry for the confusion I caused. Motivation for the code was that I didn't just want to ask for whether correct or not, but give some justification for my doubt and demonstrate previous work on the problem. Dec 7, 2014 at 17:32\n\u2022 Good for you for finding that correction to the R code. The error was sufficiently subtle for me to miss it entirely, even knowing that something must be wrong. Dec 7, 2014 at 17:47\n\nDominik, your answer was off by a factor of 4. This happened because you counted a hand containing J,K,Q,A of spades (for example) 4 times: (JQK)(A), (QKA)(J), (KAJ)(Q), and (JAQ)(K)\n\n\u2022 The answer above is a correct description of the basic issue that led to the overcount. Dec 7, 2014 at 15:40\n\nWe count the \"favourables,\" the 4-3-3-3 hands. The suit in which we have $4$ cards can be chosen in $\\binom{4}{1}$ ways. For each of these ways, the actual $4$ cards in that suit can be chosen in $\\binom{13}{4}$ ways. For each of these ways, the cards in the other three suits can be chosen in $\\binom{13}{3}^3$ ways, for a total of $\\binom{4}{1}\\binom{13}{4}\\binom{13}{3}^3$.\n\nRemark: Your counting procedure results in multiple counting. Think, for example, of the 4-3-3-3 hand that has the K, Q, J, 10 of hearts, and some specific cards for the rest of the hand. Your calculation views, for example, K, Q, J of hearts, and later 10 of hearts, as different from K, J, 10 of hearts, and then Q of hearts.\n\n\u2022 As @judith pointed out above, it seems that $\\binom{4}{1} \\binom{13}{4} \\binom{13}{3}^3 = \\binom{13}{3}^4 \\binom{10}{1}$. However, I still don't understand what I am overcounting. What do you mean by \"the 4-3-3-3 hand that has the K, Q, J, 10 of hearts\"? Could you rephrase/eleborate on your remark please? Dec 7, 2014 at 15:18\n\u2022 You counted the $12$-card 3-3-3-3 hands and then added a card in one of the suits. Each 4-3-3-3 hand comes up in $4$ ways, so you are overcounting by a factor of $4$. Here is a simpler example. How many $3$ card hands have $2$ hearts and $1$ diamond? The obvious answer is $\\binom{13}{2}\\binom{13}{1}$. (Cont) Dec 7, 2014 at 15:33\n\u2022 (Cont) Here is a wrong way of counting. Pick a heart and a diamond, $\\binom{13}{1}^2$ ways. Then add a heart, The wrong way counts twice the hand that has K, Q of hearts, and 5 of diamonds, once as K of hearts, 5 of diamonds, then Q of hearts, and also as Q of hearts, 5 of diamonds, and then K of hearts. Dec 7, 2014 at 15:35\n\u2022 Thank you for elaborating on it. I think I understand my error now. Dec 7, 2014 at 16:18\n\u2022 You are welcome. This sort of inadvertent multiple counting happens fairly often, until one gets sensitized to it. Dec 7, 2014 at 16:21\n\nI doubt, that it is still interesting for you, but I want to check myself.I think either you or I misunderstand the question. Because it asks to find the probability of the hand contains AT LEAST 3 cards of every suit and you calculated the probability of EXACTLY 3 cards of every suit. From my point of view answer must be:\n\n1. for 0 cards of every suit there is no point to calculate, since nothing happened\n\n2. for 1 cards of every suit (choose from 13-1)^3(choose from 13-remainig 10)(choose from 4-1) and divide all this to (choose from 52-13)\n\n3. for 2 cards of every suit (choose from 13-2)^3(choose from 13-remainig 7)(choose from 4-1) and divide all this to (choose from 52-13)\n\nThen just substract from 1 the calculated probability and you will get P(>=3)\n\n\u2022 OP is definitely not calculating exactly three cards of every suit. Because there are 13 cards in a hand, the only distribution(s) that have at least three cards in every suit have three suits with three cards in them and one suit with four; this is what OP attempts to calculate (with the overcounting mentioned in other answers). The hand size is fixed, so the notion of '1 card of every suit' doesn't make sense; it's an impossible configuration. Jul 9, 2021 at 3:11\n\nAnother solution that's more mechanical and generalizable to other arrangements like two pair or full house:\n\nWe know we must have a 4-3-3-3 hand. You can pick any card for the first suit: 52. The next three cards must match the suit of the first so there are 12 x 11 x 10 possibilities for those. Now for the 4th card, we can pick any card not in the 1st suit: 39. Using the same reasoning as before, for the next two cards there are 12 x 11 possibilities. We continue until we have picked all cards:\n\n$$X = (52 \\times 12 \\times 11 \\times 10) \\times (39 \\times 12 \\times 11) \\times (26 \\times 12 \\times 11) \\times (13 \\times 12 \\times 11)$$\n\n$$X$$ gives us all arrangements of the pattern WWWW XXX YYY ZZZ, as such we overcount all the internal arrangements of W, X, Y, and Z ($$4!3!^3$$). We also overcount that X, Y, and Z can be interchanged ($$3!$$).\n\nSo the answer is $$X /(4!3!^4) / {52 \\choose 13}$$.", "date": "2022-08-11 00:16:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1055651/probability-that-a-random-13-card-hand-contains-at-least-3-cards-of-every-suit", "openwebmath_score": 0.6990851163864136, "openwebmath_perplexity": 609.8782367337674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850847509661, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8523699352289673}}
{"url": "https://www.physicsforums.com/threads/what-is-the-integral-of-x.520571/", "text": "# What is the integral of |x| ?\n\n1. Aug 10, 2011\n\n### agentredlum\n\nThe question is in the title.\n\nThis is not a homework question, just curiousity on my part.\n\n2. Aug 10, 2011\n\n### HallsofIvy\n\nStaff Emeritus\nIf x< 0, then |x|= -x and its integral is $-x^2/2+ C$\n\nIf $x\\ge 0$, then |x| x and its integral is $x^2/2+ C$\n\nLast edited: Aug 10, 2011\n3. Aug 10, 2011\n\n### disregardthat\n\nor |x|x/2 + C\n\n4. Aug 10, 2011\n\n### ReaverKS\n\nHow? I thought the whole purpose of the absolute value stemmed back to distances along a number line, therefore you couldn't get a negative value out of the absolute value of a number or a function?\n\n5. Aug 10, 2011\n\n### Citan Uzuki\n\nIf x<0, then -x>0.\n\n6. Aug 10, 2011\n\n### BloodyFrozen\n\n7. Aug 10, 2011\n\n### Bohrok\n\n$$\\int|x| dx = \\frac{1}{2}x|x| + c$$If you write |x| as \u221a(x2) and use integration by parts, you can find the integral.\n\n8. Aug 10, 2011\n\n### agentredlum\n\nI know HallsofIvy figured it out because calculations were shown, but did the rest of you look it up?\n\nHallsofIvy, can you combine your 2 solutions into a single solution to get wiki answer?\n\nI know wiki mentions integration by parts, but did anyone use a simpler way to get the answer? It doesn't have to be rigorous, it just has to work. Don't worry about rigor, just functionality.\n\nWhat is the integral of |x^3| ?\n\nWhat is the integral of |x^n| for odd n?\n\nEDIT* There is nothing wrong (IMHO) with looking up the answer, but there is nothing wrong with working it out for yourself.\n\n9. Aug 10, 2011\n\n### Bohrok\n\nI never knew wikipedia showed how to integrate it; just noticed it right now. I personally did it once because I wanted to derive it myself, and for higher odd powers of x such as |x3| and there is a pattern.$$\\int|x^3|dx = \\frac{1}{4}x^3|x| + c, \\int|x^5|dx = \\frac{1}{6}x^5|x| + c, \\text{etc.}$$So in general,$$\\int|x^n|dx = \\frac{1}{n + 1}x^n|x| + c \\text{ for odd n}$$\n\n10. Aug 10, 2011\n\n### agentredlum\n\nOh, this is interesting. Is there any difference to the answer if you represent it as |x^n|x/(n+1) + c for odd n ?\n\nWell done in deriving it for yourself.\n\n11. Aug 10, 2011\n\n### Mute\n\nWhen you have an absolute value, you just split it into cases:\n\nIf $x > 0,~|x^n| = x^n$, so the integral is $x^{n+1}/(n+1)$.\n\nIf $x < 0,~|x^n| = - x^n$, so the integral is $-x^{n+1}/(n+1)$.\n\n(for n odd, of course)\nThis can be written as a single results using the \"sign function\", $\\mbox{sgn}(x)$, which returns the sign of x. Hence, the integral may be written\n\n$$\\mbox{sgn}(x) \\frac{x^{n+1}}{n+1}.$$\n\nFor x not zero, the sign function is equivalent to |x|/x or x/|x| (as is hopefully obvious after thinking about it for a second), so you can insert the first form there to get the results people have been throwing around.\n\nNote that since the sign function to an even power is always 1, you can always insert sign functions of even powers to shift around the exponents, so you can write the integral as\n\n$$\\frac{|x|^ax^b}{n+1},$$\nwhere a + b = n+1, and a and b are odd numbers.\n\n12. Aug 10, 2011\n\n### Bohrok\n\nNo difference; it's the same way as how I have it. I should also add positive to the restriction on n: for positive odd n.\n\n13. Aug 10, 2011\n\n### agentredlum\n\nOf course, right you are. I haven't explored what happens when n is negative. It is interesting to me that the integrals we are considering, |x^n|, for positive odd n, to get the answer you just write it down, multiply by x, divide by the total number of factors of x and add a 'c'. Integration by parts is not necessary. It makes me wonder if there are other functions out there that work simply like this, besides the normal x^n without absolute value symbol.\n\nWhat is the integral of |x^3 + x| ?\n\nLast edited: Aug 11, 2011\n14. Aug 11, 2011\n\n### Bohrok\n\nI thought this would be difficult and involve integration by parts, but it turned out quite simple actually:\n|x3 + x| = |x(x2 + 1)| = |x||x2 + 1| = |x|(x2 + 1) = |x|x2 + |x| = |x3| + |x|, using the facts that x2 + 1 = |x2 + 1| and x2 = |x2| since the former in each case is never negative, and |ab| = |a||b|, so\n\u222b|x3 + x| dx = \u222b|x3| dx + \u222b|x| dx\nOnce you integrate those, you can combine them and make the result look more condensed.\n\nIf you had |x3 - x|, that would be a different story!\n\nLast edited: Aug 11, 2011\n15. Aug 11, 2011\n\n### agentredlum\n\nIt's very nice how you turned this problem into something we have a shortcut for already!\n\nSo the integral of |x^3 + x| is |x^3|x/4 + |x|x/2 + C\n\nI like your algebra manipulation of absolute value so let me use it too. Disregard the C for now...\n\n(|x^3| + 2|x|)x/4 common denominator 4 and factor out x...now work inside the parenthesis...\n\n|x|(|x^2| + 2)x/4 for the same reasons you gave above...\n\n|x^2| + 2 = |x^2 + 2|\n\n|x||x^2 + 2| = |x^3 + 2x|\n\n|x^3 + 2x|x/4 again for same reasoning as you above... now 1/4 = 1/|4| so you can insert 1/4 into the parenthesis without fear, so you get finally...\n\n|x^3/4 + x/2|x + C\n\nNow if you compare this answer to the question...what is integral of |x^3 + x|, you write it down, divide x^3 by 4 inside absolute value (n + 1 for n = 3), divide x by 2 inside absolute value (n + 1 for n = 1), multiply the whole thing by x and add C\n\nPutting it all together...\n\nintegral of |x^3 + x| = |x^3/4 + x/2|x + C\n\nI think the shortcut is obvious now in this case and should work for any integral of |ax^3 + bx| where both a,b are non negative.\n\nI'm still working on integral of |x^3 - x| and trying to make it co-operate with the shortcut, any ideas?\n\n16. Aug 11, 2011\n\n### disregardthat\n\nSplit the integral as such:\n\n$$\\int^t_{1} |x||x^2-1| dx = \\int^t_{1} |x|(x^2-1) dx$$ for t >= 1,\n\n$$\\int^t_{-1} |x||x^2-1| dx = \\int^t_{-1} |x|(x^2-1) dx$$ for t <= -1\n\n$$\\int^t_{0} |x||x^2-1| dx = \\int^t_{0} |x|(1-x^2) dx$$ for -1 < t < 1\n\nThese three integrals are solved by the method above. For each domain, scale by a constant to get an answer on the form F(x) + C, but it is not necessary.\n\n17. Aug 11, 2011\n\n### agentredlum\n\nThank you. This is all very nice and important but right now i'm working on my own calculations using pencil and notebook. My PS3 browser does not decode LaTeX so its very hard sometimes to decode it using my brain. Can you solve all 3 integrals above and combine them into a single expression that gives the correct area of |x^3 - x| between the curve and the x-axis from x= -5 to x= 3 for example?\n\nLet me give an simple example. The integral of x from x= -5 to x= 5 gives zero. I'm interested in the integral as AREA between 2 curves.\n\nWhen you take the absolute value of a function it has 2 important graphical effects for my interests concerning area.\n\n1) Portions below the x-axis are reflected by symmetry to portions above the x-axis with area unchanged.\n\n2) Portions above the x-axis remain unchanged\n\nThis can be helpful when you have a function with portions below the x-axis. If one is interested in AREA then you don't have to find roots, split up the integral, use symmetry, etc. Just integrate it's absolute value. If you can find the integral then a single expression will give you the right answer for the area.\n\nNow, I realise integrating |f(x)| is harder than integrating f(x) unless shortcuts can be found and we have shown in previous post's that at least some shortcuts exist. The question interesting to me is can the shortcuts be extended to include more general cases. Any ideas?\n\nThis can be helpful, for example, if you are writing a computer program designed to calculate areas. If the program recognises a particular expression as lending itself to the shortcut, then it doesn't have to do integration by parts or some other more exotic method.\n\nOr consider it as a contest between bohrok, who knows the shortcut, and someone who doesn't know the shortcut. The question is find integral of |x^n| for positive odd n. Bohrok gets the answer in 1523 milliseconds, the other person not nearly as fast\n\nLast edited: Aug 11, 2011\n18. Aug 11, 2011\n\n### Bohrok\n\n19. Aug 11, 2011\n\n### agentredlum\n\nOH...MY...GOD you have no idea how much i want to thank you for that link. For months i've been hearing about wolfram alpha but i thought it was a computer program like Maple or Mathematica. I didn't know it was on the net, and my PS3 browser decodes it!!!!!\n\nIf i was in your neighborhood right now, i'd buy you a beer, i'd buy you 2 beers!.\n\nAs far as integral of |x^3 - x| my shortcuts are not working unless i made a mistake.\n\nAs for integral of |x^3| wolfram gave answer involving sgn(x) but you and i both know we don't need that. Haaaaaaaa.... Haaaa!\n\nHumanity strikes back against the machines.\n\n20. Aug 12, 2011\n\n### agentredlum\n\nAccording to wolframalpha...sqrt(abs(x)) has the same graph as abs(sqrt(x)) but that can't be right since domains are different\n\nThe first one has domain all real x\n\nThe second one has domain all rea NON-NEGATIVE x\n\nWhat is going on here?\n\nhttp://www.wolframalpha.com/input/?...,abs(x^(1/2))&asynchronous=false&equal=Submit\n\nIn the input interpretation it makes the correct interpretation but in the plots it doesn't.\n\nLast edited: Aug 12, 2011", "date": "2017-09-26 11:39:50", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/what-is-the-integral-of-x.520571/", "openwebmath_score": 0.8494406342506409, "openwebmath_perplexity": 1077.1842131615492, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850832642354, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8523699339399186}}
{"url": "http://www.lidea.com.tw/yx58gj/e3002.php?xvx=probability-problems-on-balls-with-solutions-pdf", "text": "We denote the event that the rst n persons are born under di erent signs, exactly as in example 5 page 62. A ball is drawn at random from the box and not replaced. 18 The probability r of no success at a certain day is equal to a red ball is drawn on. So the desired probability is just 15/48 = 5/16 = 0. 2) Letting R be the event that all of the balls are red, B be the event that all of the balls are blue, and G. How many balls are left over____? 4. Find the probability of winning $250,000 in the lottery. There are two bins with black and white balls. Probability problem on Balls. Tech interviews, Banks, IAS and SSC exams probability practice questions. Learn to find the probability when there are balls of different colors are there in a bag. Example 31. 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To win the Ohio Super Lotto Plus lottery, you must correctly pick which six balls are chosen out of a collection of forty-nine numbered balls. Each of the 6 sectors of the prize wheel are equal sizes. (a) (probability that the total after rolling 4 fair dice is 21. What is the probability that the three pieces can be assembled into a triangle?. Both are distinguishable. Thursday is the 4th of July here are a few problems relating to Independence Day!! While walking around at their town's Independence Day Festival, Latrease and Hannah decide to buy lunch. The function f(x) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x-axis is equal to 1. Exercise 1. A Short Introduction to Probability Prof. Tossing a Coin. I take out a pen and lay it on the desk; each pen has the same chance of being selected. Goodman, David Famolari August 27, 2014 1. Solutions of Problems on Probability theory P. Question 15. 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If we assume that as we draw the each ball in the urn is equally likely to be drawn, what is the probability that both balls drawn are red? Let R1 and R2 denote respectively the events that the first and the second ball drawn is red. Instructions Use black ink or ball-point pen. The function f(x) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x-axis is equal to 1. Correspond-ingly, Rndenotes the event that the n'th person is the rst person born under the same. So the probability total number of balls >= 16 will be 1/9 * 1/9 = 1/81. , probability laws) used for solving probability problems. School Overview Mowbray Heights Primary School is an. What is the probability of winning the Ohio Lottery if you have one ticket? Answer: The number of possible outcomes (each equally likely) is 49C 6 = 13983816, but only one. 3 or Miscellaneous Exercises in English Medium to study online or in PDF form. Introduction to Probability and Statistics (PDF) Solutions to Exam 1. Find the probability that in a given year it will not snow on January 1st in that town. The first of these items is mandatory, for practice with the above Poisson explanation (you don't know it until you can do it). Prior probability is the probability you attribute to a certain event without further knowledge about it. Use the binomal expansion to calculate the probabilities of HH, HT, and TT. Probability Tricks. This probability is written P(B|A), notation for the probability of B given A. , 2000 Outline 5-1 Introduction 5-2 Sample Spaces and Probability 5-3 The Addition Rules for Probability 5-4 The Multiplication Rules and Conditional Probability \u00a9The McGraw-Hill Companies, Inc. In this PDF we have given probability Questions and Answers, we hope this will be helpful for your preparation. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. Suppose that one face of a regular tetrahedron has three colors: red, green, and blue. Voiculescu, Lectures on Probability Theory and Statistics, Ecole d\u2019Ete e de Probabilit es de Saint-Flour XXVII { 1998, Ed. Now, you are asked to fill the bowls in any manner you like (using all the balls). She also has 4 extra bars. Various problems/solutions of mathematics in linear algebra, abstract algebra, number theory. 11th Annual Johns Hopkins Math Tournament Sunday, April 11, 2010 Grab Bag-Lower Division: Solutions (1) (6) Below is a square, divided by several lines (not to scale). Probability problems for aptitude pdf download, probability problems and solutions for aptitude, probability problems, random variables and probability distributions problems and solutions, probability word problems with solutions and answers, probability distribution problems and solutions, probability problems on balls with solutions, basic. P(passing in second given he passed in the first one) = P(A\ua4f5B)/P(A) = 0. Probability distribution problems solutions pdf Random variables and their probability distributions can save us significant. ) are represented as colored balls in an urn or other container. if p EA implies p E B, then A is called a. (Total 4 marks). Your answers may vary slightly. com provides you Free PDF download of NCERT Exemplar of Class 10 Maths chapter 13 Statistics and Probability solved by expert teachers as per NCERT (CBSE) book guidelines. Of course, you will learn best if you rst attempt to solve the exercises on your own, and only consult this manual when you are really stuck (or to check your solution after you think you have it right). These types of questions test whether employees can have fun together, not solve real world problems together. Ex: The wandering mathematician in previous example is an ergodic Markov chain. How should we change the probabilities of the remaining events? We shall call the. Probability Homework Solutions 1. 1 Combinations, Permutations, and Elementary Probability Roughly speaking, Permutations are ways of grouping things where the order is important. There are two bins with black and white balls. It prescribes a set of mathematical rules for manipulat-ing and calculating probabilities and expectations. Probability 2. Visit us for practice questions and solved examples. problems (2003 - 2006). Data Analysis, Statistics and Probability gum balls Scoring Guide- 2 The gum ball machine has 100 gum balls; 20 are yellow, 30 are blue, and 50 are red. In the preface, Feller wrote about his treatment of uctuation in coin tossing: \\The results are so amazing and so at variance with common intuition that even sophisticated colleagues doubted that coins actually misbehave as theory predicts. They are placed in a bag and one is drawn at random. So the answer is Floor 1. probability problems, probability, probability examples, how to solve probability word problems, probability based on area, examples with step by step solutions and answers, How to use permutations and combinations to solve probability problems, How to find the probability of of simple events, multiple independent events, a union of two events. A second urn contains 16 red balls and an unknown number of blue balls. A biased coin (with probability of obtaining a Head equal to p > 0) is tossed repeatedly and independently until the \ufb01rst head is observed. Jenny gets 10 gum balls from this machine. Ma 162 Spring 2010 Ma 162 Spring 2010 April 21, 2010 Problem 1. As 5 of the balls are red, and there are 10 balls, the probability that a red ball is drawn from the box is Pr(X = Red) = 5/10 = 1/2. Toggle navigation Close Menu. Probability Example 1. Solution: (1a) Sample space has jSj= 10 2 = 45. Suppose for example that there are three people: Ann, Bob and Carol. Probability and Statistics Problems - Solutions 1. There are 10 red and 20 blue balls in a box. Our extensive online study community is made up of college and high school students, teachers, professors, parents and subject enthusiasts who contribute to our vast collection of study resources: textbook solutions, study guides, practice tests, practice problems, lecture notes, equation sheets and more. Yates, David J. \u2022 Answer the questions in the spaces provided \u2013 there may be more space than you need. Represent events as a In the same small groups, students are given sample solutions to analyze and. An event that is certain to occur has a probability of 1. Solutions to Problems in H. Aptitude Made Easy - Probability \u2013 7 Tricks to solve problems on Balls and bags \u2013 Part 1 7 Tricks to solve problems on Balls and bags \u2013 Part 2 - Duration: Probability Word Problems. 1 Solutions will be available on Blackboard on the 29th Oct. Event A has jAj= 6 2 = 6 5=2 = 15. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. ) An urn contains twenty red balls and \ufb01ve blue balls. (a) (probability that the total after rolling 4 fair dice is 21. [1] By Jerome Dancis [2] 1. Permutation & Combination Problems with Solutions for bank exams-: Today, I am going to share with you to solve \"permutation & combination questions\". You also had the choice of two metal rackets and one wooden one. And, let X denote the number of people he selects until he finds his first success. A container contains ve red balls. introduction to counting and probability Download introduction to counting and probability or read online here in PDF or EPUB. STAT/MATH 394, Probability I, covers the basic elements of probability theory. HOMEWORKS All homeworks are due Thursday at 5:00pm unless otherwise stated. \u2022 Answer all questions. As 5 of the balls are red, and there are 10 balls, the probability that a red ball is drawn from the box is Pr(X = Red) = 5/10 = 1/2. It follows simply from the axioms of conditional probability, but can be used to powerfully reason about a wide range of problems involving belief updates. red, blue, black. You have some trick coins that land heads 60% of the time and tails 40%. Mathematics (Linear) \u2013 1MA0 PROBABILITY Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. 45 which is around 55%. Suppose Z has a continuous distribution with pdf given by A bucket contains 2 green balls and. When the brakes of a car are applied, the road surface works on the car\u2019s tyres to bring it to stop. To find the number of ways to match five numbers and not the Mega Ball number, break the selection of balls into two. A ball is chosen at random and it is noted whether it is red. This new probability is referred to as a conditional probability, because we have some prior information. 882 \u2014 PS3: Practice problems: Revised \u2014 Fall 2010 2 7. There is one desired outcome and six possible outcomes. As nincreases, the proportion of heads gets closer to 1/2, but the di\ufb01erence between the number of heads and half the number of \u00b0ips tends to increase (although it will occasionally be 0). Solutions Manual for Introduction to Statistical Physics (draft) 8- Consider again problem 7, with a distribution w(s) of the is the probability of -nding N. What is the probability of getting at least one post? Solution: The probability the candidate does not get an offer from the first interview is 2/3. Balls and Urns \u201cBalls and urns\u201d problems are paradigmatic. (a) Find the probability that the \ufb01rst ball is red. 3 or Miscellaneous Exercises in English Medium to study online or in PDF form. pdf from IOE 202 at University of Michigan. You have some trick coins that land heads 60% of the time and tails 40%. A function f is said to be probability density function pdf of the. As this has 4 elements there are 24 = 16 subsets, namely. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. 9 CONVERGENCE IN PROBABILITY 117 (There is no di\ufb00erence between home and away games. Yates and David J. P(B1) (first ball selected is black). The notes on the Web tend to be aimed at treating the more complex versions, so I'm working on winnowing out some references for the basic situation. Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. We can also count these outcomes as follows. Get The Official GMAT Club's App - GMAT TOOLKIT 2. You pick two random shoes from the bag. Probability Puzzles - Interview question on probability, maths puzzles and aptitude. Calculate the number of blue balls in the second urn. Balls in Bins 9. Solution : There are six possible ways in which a die can fall, out of these only one is favourable to the event. 11th Annual Johns Hopkins Math Tournament Sunday, April 11, 2010 Grab Bag-Lower Division: Solutions (1) (6) Below is a square, divided by several lines (not to scale). The answers and explanations are given for the practice questions. The probability that both balls are the same color is 0. Because we are drawing with replacement, the outcome of the second draw does not depend on the outcome of the first. Hannah bought a hot dog, a soda, and a bag of chips for $5. 1 in Pitman This is another version of the birthday problem. Probability Class 10 Extra Questions Maths Chapter 15 Extra Questions for Class 10 Maths Chapter 15 Probability Extra Questions for Class 10 Maths NCERT Solutions for. We have made a substantial e ort to check the solution to every quiz. Joe Blitzstein (Department of Statistics, Harvard University) 1 Naive De\ufb01nition of Probability 1. \" Let X 1 denote the random variable that equals 0 when we observe tails and equals 1 when we observe heads. We have 2 opaque bags, each containing 2 balls. Probability is the chance or likelihood that an event will happen. NCERT Exemplar Problems Class 10 Maths - Statistics and Probability August 19, 2019 by Bhagya 3 Comments CBSETuts. (Balls and Bins) This problem involves a balls and bins experiment in which m balls are tossed independently into n bins with each ball equally likely to land in any bin. Let X denote the number of red balls. If no problem is clearly marked as skipped, only the rst four problems will be graded. A bag contains fifteen red balls, twelve pink balls, and eight yellow balls. Aptitude Questions - Probability Set 14 the other from Bag B at random then what is the probability for the Balls to be Red? Quiz Questions and Answers PDF;. Harvard & HBR Business Case Study Solution and Analysis Online - Buy Harvard Case Study Solution and Analysis done by MBA writers for homework and assignments. CSE 103 Homework 2: Solutions October 13, 2010 (1) There are 4! total ways to return the hats to the women, but only 1 way for all women to receive the correct hat back. Here in this page we give few examples on Probability shortcut tricks. Solutions to Problems in H. Professor Eli Upfal of Brown University's Department of Computer Science (Brown CS) and Michael Mitzenmacher of Harvard University have just released a significantly larger second edition of their widely-used textbook, Probability and Computing: Randomization and Probabilistic Techniques in Algorithms and Data Analysis. What is the probability that the car is behind Door A (as a function of p)? If you like this problem, you might also like the Blinky Monty Problem. Law of Total Probability: The \u201cLaw of Total Probability\u201d (also known as the \u201cMethod of C onditioning\u201d) allows one to compute the probability of an event E by conditioning on cases, according to a partition of the sample space. If you know time management then everything will be easier for you. 3 or Miscellaneous Exercises in English Medium to study online or in PDF form. Conditional Probability and Independence One of the most important concepts in the theory of probability is based on the question: How do we modify the probability of an event in light of the fact that something new is known? What is the chance that we will win the game now that we have taken the \ufb01rst point?. Its goal is to help the student of probability theory to master the theory more pro\u00ad foundly and to acquaint him with the application of probability theory methods to the solution of practical problems. Groups: solve application problems and present solutions Play door selection game several times (partners) Discuss optimal strategies, correct probability inference Simulate and analyze problem with many doors (two versions) Play Golden Balls with random partners Discuss results of simulation, incentives to split or steal. Prior probability is the probability you attribute to a certain event without further knowledge about it. An outcomes has the form (x,y,z), where each of x, y and z is an H or a T, independently. Solution We have already calculated the number of ways to select the numbers for the Mega Millions Lottery, 175,711,536 ways. The mathematical theorem on probability shows that the probability of the simultaneous occurrence of two events A and B is equal to the product of the probability of one of these events and the conditional probability of the other, given that the first one has occurred. Boston Office (Near MIT/Kendall 'T'): Cambridge Innovation Center, One Broadway, 14th Floor,. (Balls and Bins) This problem involves a balls and bins experiment in which m balls are tossed independently into n bins with each ball equally likely to land in any bin. Chapter 1 Probability spaces 1. 4) 5) If a person is randomly selected, \ufb01nd the probability that his or her birthday is not in May. One bag has 2 black balls and the other has a black ball and a white ball. Get The Official GMAT Club's App - GMAT TOOLKIT 2. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Set books The notes cover only material in the Probability I course. Probability 1) A bag contains a green ball, a white ball and a black ball all balls being of the same shape and size. Miles, Bogdan Petrenko,. Kymbrea's comic book collection currently has 30 comic books in it, and she is adding to her collection at the rate of 2 comic books per month. LaShawn's comic book collection currently has 10 comic books in it, and he is adding to his collection at the rate of 6 comic books per month. All that was left when it was your turn to choose were three white balls and one orange ball. Into a bag of 14 balls, some white and the remainder red, the person drawing the two balls at the same time might insert a hand containing a white and a red ball, and then exchange (unseen, inside the bag) one of these with one of. You need at most one of the three textbooks listed below, but you will need the statistical tables. the probability of choosing a red ball is$\\frac{30}{100}$, and we repeat this in$20. If p(two purple balls) = 1 16 then find how many purple balls there are. We are simultaneously and randomly drawing 2 balls out of the 10. five numbers, but not the Mega Ball number. Your score will appear next to your identification number. Both the chosen bulbs are non defective, and probability of that is 18C2 / 20C2 = 153/190 as you have already calculated. (This is called a Bernoulli random variable. Probability of an impossible event is 0 and that of a sure event is 1. [1] By Jerome Dancis [2] 1. We can generalize this result with the theorem or law of total probability. Probability that the 1st ball is red: 4/11 Probability the 2nd ball is green: 5/10 Combined probability is 4/11 * 5/10 = 20/110 = 2/11. Probability of an impossible event is 0 and that of a sure event is 1. Problem 34. Conditional Probability and Independence One of the most important concepts in the theory of probability is based on the question: How do we modify the probability of an event in light of the fact that something new is known? What is the chance that we will win the game now that we have taken the \ufb01rst point?. By multiplication principle the probability of getting all two outcomes either a 1 or a 3 is 2/6 2/6 4/36 b). We provide step by step Solutions for ICSE Mathematics Class 10 Solutions Pdf. Question 15. Many prob-lems can be recast as balls and urns problems, once we \ufb01gure out which are the balls and which are the urns. A function f is said to be probability density function pdf of the. Probability Homework Solutions 1. A common topic in introductory probability is solving problems involving coin flips. rar, link Errata For Pitman, Probability, 1993 Springerverlag. Suppose a batter has probability 1 3 to hit the ball. Maths in a crowd. 14 A stick is broken into three pieces by picking two points independently and uniformly along the stick, and breaking the stick at those two points. Tree diagrams are useful to answer probability questions where there is more than one event happening in succession. You need to solve any four and indicate which problem you skip. Rosa will toss a fair coin twice. Determine the value of x. We denote the event that the rst n persons are born under di erent signs, exactly as in example 5 page 62. Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Third Edition STUDENT'S SOLUTION MANUAL (Solutions to the odd-numbered problems) Roy D. Can You Solve This Intro Probability Problem? in the above problem it tells us the probability of the both balls being the same color is 0. Many events can't be predicted with total certainty. We also discuss some applications of probability theory to computing, including systems for making likely inferences from data and a class of useful algorithms that work \"with high probability\" but are not guaranteed to work all the time. Conditional Probability Based on the data that Bryant had a. If three balls are drawn out at random (without replacement), what is the probability that exactly one of them is blue? It\u2019s the probability of drawing two red balls and one blue ball, which is \u00b5 20 2 \u00b6\u00b5 5 1 \u00b6 \u00b5 25 3 \u00b6. Various problems/solutions of mathematics in linear algebra, abstract algebra, number theory. What is the probability that he takes out a black ball?? Solution: Since Rohan takes the ball out without looking. What is the probability of drawing the black marble from the bag? _____ 3. If you know how to manage time then you will surely do great in your exam. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions Problem 1 If a ball is drawn at random, from a bag containing 5 white and 3 black balls, then write the number of successes and failures for the ball to be a black one. NCERT (National Council of educational research and training) was established in the year 1961 and provides with quality books that are prescribed by the Central Board of Education to provide better quality academics. A FIRST COURSE IN PROBABILITY. Similar calculations for the other colours yields the probability density function given by the following table.", "date": "2019-10-17 23:53:27", "meta": {"domain": "com.tw", "url": "http://www.lidea.com.tw/yx58gj/e3002.php?xvx=probability-problems-on-balls-with-solutions-pdf", "openwebmath_score": 0.6983087658882141, "openwebmath_perplexity": 571.372607393667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9830850827686585, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8523699284437792}}
{"url": "http://summer.mruni.eu/sja9myq/73v7b.php?e021c1=properties-of-matrix-multiplication-proof", "text": "The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; In the next subsection, we will state and prove the relevant theorems. $$\\begin{pmatrix} e & f \\\\ g & h \\end{pmatrix} \\cdot \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} ae + cf & be + df \\\\ ag + ch & bg + dh \\end{pmatrix}$$ i.e., (AT) ij = A ji \u2200 i,j. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. De\ufb01nition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, De\ufb01nition A square matrix A is symmetric if AT = A. The proof of Equation \\ref{matrixproperties2} follows the same pattern and is \u2026 19 (2) We can have A 2 = 0 even though A \u2260 0. Example 1: Verify the associative property of matrix multiplication \u2026 (3) We can write linear systems of equations as matrix equations AX = B, where A is the m \u00d7 n matrix of coefficients, X is the n \u00d7 1 column matrix of unknowns, and B is the m \u00d7 1 column matrix of constants. But first, we need a theorem that provides an alternate means of multiplying two matrices. A matrix is an array of numbers arranged in the form of rows and columns. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. MATRIX MULTIPLICATION. The following are other important properties of matrix multiplication. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. Selecting row 1 of this matrix will simplify the process because it contains a zero. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. A matrix consisting of only zero elements is called a zero matrix or null matrix. For sums we have. Subsection MMEE Matrix Multiplication, Entry-by-Entry. Let us check linearity. proof of properties of trace of a matrix. The first element of row one is occupied by the number 1 \u2026 Properties of transpose Proof of Properties: 1. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Even though matrix multiplication is not commutative, it is associative in the following sense. Given the matrix D we select any row or column. Equality of matrices Matrix transpose AT = 15 33 52 \u221221 A = 135\u22122 532 1 Example Transpose operation can be viewed as \ufb02ipping entries about the diagonal. Associative law: (AB) C = A (BC) 4. The determinant of a 4\u00d74 matrix can be calculated by finding the determinants of a group of submatrices. While certain \u201cnatural\u201d properties of multiplication do not hold, many more do. Notice that these properties hold only when the size of matrices are such that the products are defined. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If $$A$$ is an $$m\\times p$$ matrix, $$B$$ is a $$p \\times q$$ matrix, and $$C$$ is a $$q \\times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. Zero matrix on multiplication If AB = O, then A \u2260 O, B \u2260 O is possible 3. 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The number 1 \u2026 Subsection MMEE matrix multiplication \u2026 matrix multiplication first element of row is..., we will state and prove the relevant theorems multiplication is not commutative, it is associative the! Matrix is an array of numbers arranged in the next Subsection, we need A theorem that an. \u201c natural \u201d properties of transpose even though matrix multiplication, Entry-by-Entry the next Subsection, we need A that. I, j 0 even though matrix multiplication, Entry-by-Entry B ) C A! When the size of matrices are such that IA = AI =A 2 0! Because it contains A zero matrix or null matrix Subsection, we have A 2 = 0 1 0.... Numbers arranged in the form of rows and columns row or column = AC + BC.!\nLodge Wildlife Series 12\", 1 Timothy 3 Tagalog, Recruitment Business For Sale In Canada, Dance Literature Definition, Where To Buy Butterfly Eggs Online, Oathbreaker Meaning In Malayalam, You Call This A Utopia, Brannan Manor A Local Haunt, The Book Of Tawheed English Pdf, Beef Farms In Maine, Cbse 10th Science Syllabus Pdf,", "date": "2021-10-26 14:26:05", "meta": {"domain": "mruni.eu", "url": "http://summer.mruni.eu/sja9myq/73v7b.php?e021c1=properties-of-matrix-multiplication-proof", "openwebmath_score": 0.8481342792510986, "openwebmath_perplexity": 423.47140363606474, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850887155806, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.852369921778241}}
{"url": "https://math.stackexchange.com/questions/2997942/probability-concepts-how-can-balls-of-same-colour-be-distinguishable", "text": "# Probability concepts - how can balls of same colour be distinguishable?\n\nAn urn contains $$6$$ white and $$4$$ black balls. A fair die is rolled and that number of balls are chosen from the urn. Find the probability that the balls selected are white.\n\nI know the basic way to go about solving the problem.\n\nLet $$W$$ be the event of finally drawing all white balls. Let $$P(n)$$ denote the probability of appearance of $$n$$ on the die.\n\nWe want: $$P(W) = P(1)P(W\\mid 1)+ P(2)P(W\\mid 2)+\\dots \\implies P(W) = \\dfrac{1}{6}\\left(\\sum_{i=1}^6P(W\\mid i)\\right)$$\n\nNow, I am actually facing trouble in computing $$P(W/i)$$. I saw author's method and in it he has used $$P(W\\mid i) = \\dfrac{^6C_i}{^{10}C_i}$$\n\nbut I fail to understand how that can be true when all white balls are identical and all black balls are identical.\n\nHere, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here?\n\n\u2022 Doesn't it look pretty much as a Hypergeometric distribution were the number of draws is a rv? $K=6$, $N=10$, $n=k$ is a random variable. Does it make any sense to you? \u2013\u00a0Ramiro Scorolli Nov 14 at 8:13\n\u2022 I don't know what's a hypergeometric distribution @RamiroScorolli \u2013\u00a0Abcd Nov 14 at 8:13\n\u2022 Basically you have $N$ balls(10 balls in total), $K$ represent success (6 white balls), you draw $n$ balls (where n is the number obtained with the dice) and you are expecting $k$ successes. (basically n cause you want all to be white). en.m.wikipedia.org/wiki/Hypergeometric_distribution \u2013\u00a0Ramiro Scorolli Nov 14 at 8:20\n\u2022 I'm not completely sure but I think that this could be the $P(W/I)$ you are looking for \u2013\u00a0Ramiro Scorolli Nov 14 at 8:20\n\u2022 How about something similar with a smaller amount of balls? If you have two identical black balls and an otherwise similar but white ball in a bag, and you pull out one in random, what's the probability of getting a black ball? Is it 1/2 because there are only two colors (and you couldn't tell the black ones apart)? \u2013\u00a0ilkkachu Nov 14 at 14:43\n\n\u2022 The probability of drawing $$i$$ white balls is the probability of drawing a white ball and then (without replacement) drawing a second white ball, and so on up to $$i$$\n\n\u2022 The first ball drawn has a probability of $$\\frac{6}{10}$$ of being white\n\n\u2022 Given that the first ball drawn is white, the second ball drawn has a probability of $$\\frac{5}{9}$$ of being white\n\n\u2022 An so on up to the $$i$$th ball having a probability of $$\\frac{6-i+1}{10-i+1}$$ of being white\n\n\u2022 So the probability all $$i$$ balls drawn are white is $$\\frac{6 \\times 5 \\times \\cdots \\times(6-i+1)}{10 \\times 9 \\times \\cdots \\times(10-i+1)} = \\dfrac{\\frac{6!}{(6-i)!}}{\\frac{10!}{(10-i)!} }= \\dfrac{\\frac{6!}{(6-i)!i!}}{\\frac{10!}{(10-i)!i!}} = \\dfrac{^6C_i}{^{10}C_i}$$\n\n\u2022 Wow, this is amazing. \u2013\u00a0Abcd Nov 14 at 8:18\n\u2022 So does this also imply that considering them as distinct is a sort of \"trick\" that is always going to work? \u2013\u00a0Abcd Nov 14 at 8:22\n\u2022 @Abcd If you are going to use a counting calculation, then you want each event to be of equal probability, and treating the balls as physically distinct even when they look the same does this. Otherwise you risk saying that the probability of all white when $i=6$ might be higher than when $i=4$, since when $i=6$ you can get $4$, $5$ or $6$ white balls while with $i=4$ you can get $0$, $1$, $2$, $3$ or $4$ white balls; you need to be able say what the differing probabilities of the various possible outcomes are \u2013\u00a0Henry Nov 14 at 8:55\n\nIf you draw the $$i$$ balls sequentially $$P(W\\mid i)=\\frac{6}{10}\\cdot\\frac{5}{9}\\cdots\\frac{6-i+1}{10-i+1}$$ If you draw the $$i$$ balls simultaneously $$P(W\\mid i)=\\frac{\\dbinom{6}{i}\\cdot\\dbinom{4}{0}}{\\dbinom{10}{i}}$$ Now, can you convince yourself that these two are equivalent?\n\nSince I don't see it anywhere, let me address this part of your question, rather than specific case:\n\nhow that can be true when all blue balls are identical and all black balls are identical.\n\nHere, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here?\n\nLet's make a simple thought experiment. Let's say that rather than having indistinguishable white and black balls you have 10 balls numbered 1 to 10. Balls numbered 1 to 6 are white and those numbered 7 to 10 are black.\n\nNow you have 6 white balls and 4 black balls that are distinguishable. How does that change your probability?\n\nThe numbers on balls change nothing in probability as long as the only thing you're concerned is ball colour. The resulting probability has to be the same as if there were no numbers. We may simply ignore the numbers on the balls and follow the original problem. We still have 6 white balls and 4 black balls so the old approach holds. Whatever other way you count the probability it has to provide the same result.\n\nBut now your balls are distinguishable so you can apply methods specific to distinguishable balls, specifically use combination. The results, as shown earlier will be the same.\n\nThis is why to indistinguishable balls you can always apply approach \"Let's assume the balls are distinguishable...\"\n\nLet me answer from a different direction.\n\nThe reason you must use the equations which treat the balls as distinguishable is that thousands of repeated experiments have shown that the statistical behavior of all macroscopic objects is that of distinguishable items. There is no \"proof\" of this, since it's a physical reality. (by comparison, Bosons often do act as indistinguishable, leading to cool stuff in the super-cold regime).\n\nNow, in all statistics, the wording of the question is critical (see arguments about the Monty Hall problem!). If your teacher says \"Assume all balls of a given color are truly indistinguishable,\" then it's a theoretical problem and you basically divide by the number of permutations of distinguishable objects. But unless indistinguishability is specifically given as a premise, objects are distinguishable.", "date": "2018-12-15 04:18:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2997942/probability-concepts-how-can-balls-of-same-colour-be-distinguishable", "openwebmath_score": 0.7628301978111267, "openwebmath_perplexity": 249.24326735334031, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9830850892111574, "lm_q2_score": 0.8670357512127873, "lm_q1q2_score": 0.8523699188302859}}
{"url": "http://mathhelpforum.com/calculus/161029-difficult-definite-integral.html", "text": "# Math Help - Difficult definite integral\n\n1. ## Difficult definite integral\n\nThe question:\n$\\int_{-1}^{0} \\sqrt{t^2 + t^4} dt$\n\nI'm stumped. Not sure how to attempt this.\n\nAny assistance would be great.\n\n2. How about first taking $t^2$ from within the square root, and move it outside as $|t|$:\n\n$\\sqrt{t^2+t^4} = \\sqrt{t^2(1+t^2)} = |t|\\sqrt{1+t^2} = -t\\sqrt{1+t^2},$\n\nwhere $|t|=-t$, because we find ourselves in the range $-1\\leq t\\leq 0$. This last expression can be more easily integrated. I suggest substitution.\n\n3. Originally Posted by Glitch\nThe question:\n$\\int_{-1}^{0} \\sqrt{t^2 + t^4} dt$\n\nI'm stumped. Not sure how to attempt this.\n\nAny assistance would be great.\n\nThe integrand function is even, so we can as well work with $\\int\\limits_0^1\\sqrt{t^2+t^4}dt=\\frac{1}{2}\\int\\li mits^1_0 2t\\sqrt{1+t^2}dt =$ ....take it from here.\n\nNote: the last one is an automatic integral!\n\nTonio\n\n4. Originally Posted by tonio\nThe integrand function is even, so we can as well work with $\\int\\limits_0^1\\sqrt{t^2+t^4}dt=\\frac{1}{2}\\int\\li mits^1_0 2t\\sqrt{1+t^2}dt =$ ....take it from here.\n\nNote: the last one is an automatic integral!\n\nTonio\nI wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...\n\n$\\displaystyle \\sqrt{t^{2} + t^{4}} = t\\ \\sqrt{1+t^{2}}$\n\nIt is quite obvious that the second term of the 'identity' (1) is an odd function ...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n5. Originally Posted by HappyJoe\nHow about first taking $t^2$ from within the square root, and move it outside as $|t|$:\n\n$\\sqrt{t^2+t^4} = \\sqrt{t^2(1+t^2)} = |t|\\sqrt{1+t^2} = -t\\sqrt{1+t^2},$\n\nwhere $|t|=-t$, because we find ourselves in the range $-1\\leq t\\leq 0$. This last expression can be more easily integrated. I suggest substitution.\nI'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?\n\n6. Originally Posted by chisigma\nI wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...\n\n$\\displaystyle \\sqrt{t^{2} + t^{4}} = t\\ \\sqrt{1+t^{2}}$\n\nIt is quite obvious that the second term of the 'identity' (1) is an odd function ...\n\nKind regards\n\n$\\chi$ $\\sigma$\nThis really took me by surprise, though I think it doesn't matter here since the even thing was just to change the integral's limits.\n\nIn fact I should have writtent $\\sqrt{t^2+t^4}=|t|\\sqrt{1+t^2}$ , but in this case is irrelevant since the\n\nvariable t is already positive in the new integration's limits.\n\nTonio\n\n7. Originally Posted by Glitch\nI'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?\nLet $-u = t$, then $dt = -du$, then:\n\n$\\displaystyle \\int_{-1}^{0} \\sqrt{t^2+t^4}\\;{dt} = -\\int_{1}^{0}\\sqrt{(-u)^2+(-u)^4}\\;{du} = -\\int_{1}^{0}\\sqrt{u^2+u^4}\\;{du} = -\\int_{1}^{0} \\sqrt{t^2+t^4}\\;{dt}.$\n\nNow we can write $\\sqrt{t^2+t^4} = t\\sqrt{1+t^2}$, as our limits are non-negative.", "date": "2015-05-05 13:41:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/161029-difficult-definite-integral.html", "openwebmath_score": 0.9384992122650146, "openwebmath_perplexity": 607.8304465891345, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850837598122, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8523699174814126}}
{"url": "https://math.stackexchange.com/questions/866320/determine-if-the-alternating-series-converges-absolutely-conditionally-or-diver/866322", "text": "Determine if the alternating series converges absolutely, conditionally or diverges\n\nTrying to determine if this alternating series converges absolutely or conditionally. ATS criteria has been met (terms are positive [ignoring signs] & decreasing, and the lim n->inf = 0, assuming I haven't made a mistake) so I know it's at least convergent but need to prove absolute convergence. However, I believe that if the absolute value of the series is convergent then it is impossible for the original series to be divergent and, thus, the series has to be absolutely convergent. Does that make sense? Thanks for taking a look at this.\n\nThe terms, at least after a while, are decreasing in absolute value. However, there is no need to show that. For $$\\frac{n+2^n}{n+3^n}\\lt \\frac{2\\cdot 2^n}{3^n},$$ so by comparison with a geometric series, our series converges absolutely, and hence converges.\n\n\u2022 This is great but am I correct in saying that if the absolute value of an alternating series is convergent then the original alternating series is also convergent? In other words, is it correct to say that it is impossible for the original series to be divergent if it's absolute value is convergent? \u2013\u00a0joe schmoe Jul 13 '14 at 22:02\n\u2022 Yes, you are correct. The answer above says so: \"our series converges absolutely, and hence converges.\" The signs need not actually alternate. For any sequence $(a_n)$, if $\\sum|a_n|$ converges, then $\\sum a_n$ converges. Of course, the converse need not hold. If $\\sum a_n$ converges, then $\\sum|a_n|$ need not converge. \u2013\u00a0Andr\u00e9 Nicolas Jul 13 '14 at 22:06\n\u2022 You are welcome. The terminology is initially a bit confusing, but once things get clear, they stay clear forever. \u2013\u00a0Andr\u00e9 Nicolas Jul 13 '14 at 22:47\n\nHint: $$\\left|(-1)^{n}\\frac{n+2^{n}}{n+3^{n}}\\right|=\\frac{1+\\frac{n}{2^{n}}}{1+\\frac{n}{3^{n}}}\\left(\\frac{2}{3}\\right)^{n}$$\n\n\u2022 isn't it $\\left(\\dfrac{3}{2}\\right)^{n}$? \u2013\u00a0Kamster Jul 13 '14 at 21:11\n\u2022 Perhaps I've made a factoring error. $\\frac{n+2^{n}}{n+3^{n}}=\\frac{(\\frac{n}{2^{n}}+1)2^{n}}{(\\frac{n}{3^{n}}+1)3^{n}}=\\frac{1+\\frac{n}{2^{n}}}{1+\\frac{n}{3^{n}}}(\\frac{2}{3})^{n}$? \u2013\u00a0user71352 Jul 13 '14 at 21:13\n\u2022 Oh no you are completely right, miss read it. I thought you took a $\\frac{1}{2^{n}}$ and $\\frac{1}{3^{n}}$ for some reason \u2013\u00a0Kamster Jul 13 '14 at 21:14\n\u2022 Yea if there is any most common mistake I make in proofs, its algebra mistakes \u2013\u00a0Kamster Jul 13 '14 at 21:16\n\u2022 I have the same problem. \u2013\u00a0user71352 Jul 13 '14 at 21:18", "date": "2019-08-21 12:05:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/866320/determine-if-the-alternating-series-converges-absolutely-conditionally-or-diver/866322", "openwebmath_score": 0.9130003452301025, "openwebmath_perplexity": 264.4723886294657, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9830850862376966, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.8523699128745509}}
{"url": "https://byjus.com/question-answer/if-a-1-a-2-a-3-are-in-ap-a-2-a-3-a/", "text": "Question\n\n# If\u00a0$$a_{1}$$, $$a_{2}$$, $$a_{3}$$ are in AP\u00a0$$a_{2}$$, $$a_{3}$$, $$a_{4}$$ are in GP and\u00a0$$a_{3}$$, $$a_{4}$$,\u00a0$$a_{5}$$ are in HP then\u00a0$$a_{1}$$, $$a_{3}$$, $$a_{5}$$ are in\n\nA\nAP\nB\nGP\nC\nHP\nD\nnone of these\n\nSolution\n\n## The correct option is A GPGiven $${ a }_{ 1 },{ a }_{ 2 },,{ a }_{ 3 }$$ are in A.PLet's assume that $${ a }_{ 1 }=A-d\\\\ { a }_{ 2 }=A\\\\ { a }_{ 3 }=A+d$$similarly $${ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }$$ are in G.P$$so\\quad { a }_{ 2 }=A\\\\ { a }_{ 3 }=Ar\\\\ { a }_{ 4 }=A{ r }^{ 2 }$$$$\\Rightarrow A+d=Ar\\\\ \\Rightarrow d=Ar-A$$and finally $${ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 }$$ are in H.P$$\\Rightarrow \\dfrac { 1 }{ { a }_{ 3 } } ,\\dfrac { 1 }{ { a }_{ 4 } } ,\\dfrac { 1 }{ { a }_{ 5 } }$$ are in A.Phence $$\\Rightarrow \\dfrac { 1 }{ { a }_{ 5 } } =\\dfrac { 1 }{ { a }_{ 4 } } +\\left( \\dfrac { 1 }{ { a }_{ 4 } } -\\dfrac { 1 }{ { a }_{ 3 } } \\right) \\\\ \\Rightarrow \\dfrac { 1 }{ { a }_{ 5 } } =\\dfrac { 2 }{ A{ r }^{ 2 } } -\\dfrac { 1 }{ Ar } =\\dfrac { 2-r }{ A{ r }^{ 2 } } \\\\ \\Rightarrow { a }_{ 5 }=\\dfrac { A{ r }^{ 2 } }{ 2-r }$$and $${ a }_{ 1 }=A-\\left( Ar-A \\right) =2A-Ar=A\\left( 2-r \\right) \\\\ { a }_{ 3 }=Ar$$clearly we can see that $${ a }_{ 1 }{ a }_{ 5 }={ { a }_{ 3 } }^{ 2 }$$Thus $${ a }_{ 1 },{ { a }_{ 3 } },{ a }_{ 5 }$$ are in G.PMaths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-18 19:06:40", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/if-a-1-a-2-a-3-are-in-ap-a-2-a-3-a/", "openwebmath_score": 0.9351657629013062, "openwebmath_perplexity": 4205.284702701892, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474883750545, "lm_q2_score": 0.8615382094310355, "lm_q1q2_score": 0.8523606636397367}}
{"url": "http://math.stackexchange.com/questions/75712/why-is-sum-limits-k-0n-1k-k-binomnk-0/75720", "text": "# Why is $\\sum \\limits_{k = 0}^{n} (-1)^{k} k\\binom{n}{k} = 0$?\n\nI know that the expansion of $\\sum \\limits_{k = 0}^{n} (-1)^{k} \\binom{n}{k}$ equals to zero. But why is $\\sum \\limits_{k = 0}^{n} (-1)^{k} k\\binom{n}{k}$ also equal to zero for $n \\geq 2$?\n\nI've been using the first to derive the second, but it ended with no clue at all. Anyone know about how to derive this formula?\n\n$$\\displaystyle \\sum_{k = 0}^{n} (-1)^{k} k\\binom{n}{k} = 0 .$$\n\n-\n\nNote that $(1-x)^n = \\sum_{k=0}^n (-1)^k x^k \\binom{n}{k}$. Thus the sum you are interested in is $\\left. \\frac{\\mathrm{d}}{\\mathrm{d} x} (1-x)^n \\right|_{x=1} = \\left. -n (1-x)^{n-1} \\right|_{x=1} = -n (1-1)^{n-1}$. Thus it is zero for $n > 1$.\n\nIndeed for $n=1$ is the sum is $-1$, which can be explicitly checked.\n\n-\n\nI would like to give another different proof of the problem the OP proposed. My solution is based on the identity\n\n$$k\\binom{n}{k}=n\\binom{n-1}{k-1}.$$\n\nLet us first prove this identity: suppose we are given a class of $n$ children and suppose we want to form a team of $k$ people from the class, and moreover we want to elect a captain for our team. We can count the possibilities of doing so in two ways:\n\nFirst select $k$ people from the class and then elect the captain. Then we have $k$ possibilities for any previously chosen team, so in total $$k\\binom{n}{k}$$ ways of proceeding along this path.\n\nBut we may also elect first the captain, which can be done in $n$ ways, then form the team, for which we need other $k-1$ children out of $n-1$ remaining. In this other way we count $$n\\binom{n-1}{k-1}$$ ways to fulfill our task.\n\nThis proves in a combinatorical way the identity which can be however verified by algebraic means.\n\nBut then our formula reduces to $$n \\sum_{k=0}^{n} (-1)^k\\binom{n-1}{k-1}=n\\sum_{k=1}^n(-1)^k\\binom{n-1}{k-1}=0.$$\n\n-\nThe first part of your proof can even be shorter. Remember the definition of the binomial by factorials: $\\small k \\binom{n}{k} = k { n! \\over k! (n-k)! } = {n! \\over (k-1)! (n-k)! } = n { (n-1)! \\over (k-1)! ((n-1)-(k-1))! } = n \\binom{n-1}{k-1}$ \u2013\u00a0Gottfried Helms Oct 25 '11 at 16:15\n@Gottfried: I know.. However i'm in deep love with combinatorical double counting proofs so that's why i've chosen this one. However if you see i mentioned the fact that the identity may be proven algebraically. \u2013\u00a0uforoboa Oct 25 '11 at 16:24\n\nThe identity you ask about has a direct algebraic proof using the identity you already know. Let $g(n) = \\sum_{k = 0}^{n} (-1)^{k} k\\binom{n}{k}$, and let $f(n) = \\sum_{k = 0}^{n} (-1)^{k} \\binom{n}{k}$. We will show that $g(n+1) = - f(n)$, and thus the fact that $f(n) = [n=0]$ implies $g(n) = -[n=1]$. (Here, [statement] evaluates to $1$ if statement is true and $0$ if statement is false. It's called the Iverson bracket.)\n\nWe have $$g(n+1) - g(n) = \\sum_k (-1)^{k} k\\left(\\binom{n+1}{k} - \\binom{n}{k}\\right) = \\sum_k (-1)^{k} k\\binom{n}{k-1}$$ $$= \\sum_k (-1)^{k+1} (k+1)\\binom{n}{k} = -g(n) - f(n).$$ Thus $g(n+1) = -f(n) \\Longrightarrow g(n) = - f(n-1) = - [n-1=0] = -[n=1]$.\n\nGeneralization. If $g(n) = \\sum_{k = 0}^{n} (-1)^{k} \\binom{n}{k} b_k$, and $f(n) = \\sum_{k = 0}^{n} (-1)^{k} \\binom{n}{k} \\Delta b_k$ (where $\\Delta b_k = b_{k+1} - b_k$), then $g(n) = -f(n-1) + b_0[n=0]$. This relationship can be applied iteratively, starting with the problem above, to show that\n\n$$\\sum_{k=0}^n \\binom{n}{k} (-1)^k k^{\\underline{m}} = (-1)^m m![n=m],$$ and from there to $$\\sum_{k=0}^n \\binom{n}{k}(-1)^k k^m = \\left\\{ m \\atop n \\right\\}(-1)^n n!,$$ where $\\left\\{ m \\atop n \\right\\}$ is a Stirling number of the second kind.\n\n(See, for example, Section 3 of my paper \"Combinatorial Sums and Finite Differences,\" Discrete Mathematics, 307 (24): 3130-3146, 2007.)\n\n-\n\nHere's a purely combinatorial proof that doesn't reduce the sum to the known identity $\\sum \\limits_{k = 0}^{n} (-1)^{k} \\binom{n}{k} = 0$.\n\nThe quantity $\\binom{n}{k}k$ counts the number of ways to partition people numbered $\\{1, 2, \\ldots, n\\}$ into a chaired committee $A$ of size $k$ and an unchaired committee $B$ of size $n-k$. Given a particular commmittee pair $(A,B)$, let $x$ be the highest-numbered person in either committee who is not the chair of $A$. Move $x$ to the other committee. This mapping is defined for all pairs of committees when $n >1$, is its own inverse (and so is one-to-one), and changes the parity on committee pairs. Thus, for $n > 1$, there are as many committee pairs with even parity as there are with odd parity. In other words, $$\\sum_{k = 0}^{n} (-1)^{k} \\binom{n}{k} k = 0$$ when $n > 1$.\n\n-", "date": "2016-02-07 01:32:48", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/75712/why-is-sum-limits-k-0n-1k-k-binomnk-0/75720", "openwebmath_score": 0.9462562203407288, "openwebmath_perplexity": 202.1560345903079, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9893474877468791, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.852360657822662}}
{"url": "https://mathhelpboards.com/threads/determine-the-position-using-an-iteration-method.27205/", "text": "# Determine the position using an iteration method\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nHey!!\n\nThe function \\begin{equation*}f(x)=\\frac{x}{e^{x/9}}\\cdot \\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}\\end{equation*} has at exactly one position $\\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\\overline{x}$ using an iteration method with accuracy of two decimal digits.\n\nI have done the following:\n\nWe have that \\begin{equation*}f(\\overline{x})=f(1)\\Rightarrow f(\\overline{x})-f(1)=0 \\Rightarrow g(x):=f(x)-f(1)\\end{equation*}\n\nFirst we have to calculate $f(1)$:\n\\begin{align*}f(1)&=\\lim_{x\\rightarrow 1}f(x)=\\lim_{x\\rightarrow 1}\\frac{x}{e^{x/9}}\\cdot \\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}=\\frac{1}{e^{1/9}}\\cdot \\lim_{x\\rightarrow 1}\\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}\\ \\overset{DLH}{ = } \\ \\frac{1}{e^{1/9}}\\cdot \\lim_{x\\rightarrow 1}\\frac{\\pi \\cos \\left (\\pi (x-1)\\right )}{1}\\\\ & =\\frac{1}{e^{1/9}}\\cdot \\pi \\cos \\left (\\pi \\cdot 0\\right )=\\frac{1}{e^{1/9}}\\cdot \\pi =\\frac{\\pi}{e^{1/9}}\\end{align*}\n\nTherefore we get the function \\begin{equation*}g(x)=\\frac{x}{e^{x/9}}\\cdot \\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}-\\frac{\\pi}{e^{1/9}}\\end{equation*}\n\nNow we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?\n\nWe don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nHey!!\n\nThe function \\begin{equation*}f(x)=\\frac{x}{e^{x/9}}\\cdot \\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}\\end{equation*} has at exactly one position $\\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\\overline{x}$ using an iteration method with accuracy of two decimal digits.\n\nI have done the following:\n\nWe have that \\begin{equation*}f(\\overline{x})=f(1)\\Rightarrow f(\\overline{x})-f(1)=0 \\Rightarrow g(x):=f(x)-f(1)\\end{equation*}\n\nFirst we have to calculate $f(1)$:\n\\begin{align*}f(1)&=\\lim_{x\\rightarrow 1}f(x)=\\lim_{x\\rightarrow 1}\\frac{x}{e^{x/9}}\\cdot \\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}=\\frac{1}{e^{1/9}}\\cdot \\lim_{x\\rightarrow 1}\\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}\\ \\overset{DLH}{ = } \\ \\frac{1}{e^{1/9}}\\cdot \\lim_{x\\rightarrow 1}\\frac{\\pi \\cos \\left (\\pi (x-1)\\right )}{1}\\\\ & =\\frac{1}{e^{1/9}}\\cdot \\pi \\cos \\left (\\pi \\cdot 0\\right )=\\frac{1}{e^{1/9}}\\cdot \\pi =\\frac{\\pi}{e^{1/9}}\\end{align*}\n\nTherefore we get the function \\begin{equation*}g(x)=\\frac{x}{e^{x/9}}\\cdot \\frac{\\sin \\left (\\pi (x-1)\\right )}{x-1}-\\frac{\\pi}{e^{1/9}}\\end{equation*}\n\nNow we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?\n\nWe don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?\n[DESMOS=-1,4,-1,4]\\frac{\\left(x\\sin\\left(\\pi\\left(x-1\\right)\\right)\\right)}{e^{x/9}(x-1)}[/DESMOS]\nThe above graph shows that the function repeats its value at $x=1$ when $x$ is somewhere near $1.4$ or $1.5$. So I would take $x_0=1.5$.\n\nLast edited:\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nNow we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?\n\nWe don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?\nHey mathmari !!\n\nThis is an example where Newton-Raphson can have problems if we are not careful.\nIf we pick a starting value that is too far from the zero, it will likely not converge.\nHowever, a starting value that starts slightly to the right of the zero, such as the 1.5 that Opalg pointed out, should do the job. And it will converge quadratically.\nStarting below 1.4 or above 2.1 will likely diverge though.\n\nAlternatively algorithms are bisection and regula falsi.\nFirst we might search for values that are on opposite sides of the x-axis.\nThat is, we can start with some initial interval, and then either double or half its size until we find both a positive and a negative function value.\nThe root must then in between those, after which both bisection and regula falsi will find it.\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nWe have \\begin{align*}&g(x)=\\frac{x\\sin \\left (\\pi(x-1)\\right )}{e^{x/9}(x-1)}-\\frac{\\pi}{e^{1/9}}\\\\ &g'(x)=\\frac{\\left [\\sin \\left (\\pi (x-1)\\right )+x\\pi \\cos \\left (\\pi (x-1)\\right )\\right ]\\left (x-1\\right )-x\\sin \\left (\\pi (x-1)\\right ) \\frac{8+x}{9}}{e^{x/9}(x-1)^2}\\end{align*}\n\nChoosing as initial value $x_0=1.5$ we get the following:\n\\begin{align*}x_1=x_0-\\frac{g(x_0)}{g'(x_0)}\\approx 1.4259 \\\\ x_2=x_1-\\frac{g(x_1)}{g'(x_1)}\\approx 1.4149 \\\\ x_3=x_2-\\frac{g(x_2)}{g'(x_2)}\\approx 1.4147\\end{align*}\nThe first two decimal digits are the same as in the previous step and so position that we are looking for is $1.41$.\n\nIs everything correct?\n\nStaff member\nYep.", "date": "2020-05-25 20:09:14", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/determine-the-position-using-an-iteration-method.27205/", "openwebmath_score": 0.9956327080726624, "openwebmath_perplexity": 1742.5253729472686, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407160384082, "lm_q2_score": 0.8757869965109764, "lm_q1q2_score": 0.8523515635814696}}
{"url": "https://math.stackexchange.com/questions/2640694/efficient-method-for-computing-the-properties-of-a-block-anti-diagonal-matrix", "text": "# Efficient method for computing the properties of a block anti-diagonal matrix\n\nEDIT: the title and content of this question were reformulated for the sake of clarity and to avoid diverting attention from the main issue.\n\nI was trying to answer a question about a system of linear differential equations and it was necessary to find the eigenvalues, eigenvectors and generalized eigenvectors for the following matrix: $$M=\\begin{bmatrix}0&E_{\\times}\\\\B_{\\times}&0\\end{bmatrix},$$ where $E=[E_1,E_2,E_3]^T,B=[B_1,B_2,B_3]$, $E^TB=0$ and $E_\\times,B_\\times$ are the representation of vectors through antisymmetric matrices: $$E_{\\times}= \\begin{pmatrix} 0 & -E_{3} & E_{2}\\\\ E_{3}& 0 & -E_{1}\\\\ -E_{2} & E_{1} & 0 \\\\ \\end{pmatrix},\\qquad B_{\\times}= \\begin{pmatrix} 0 & -B_{3} & B_{2}\\\\ B_{3}& 0 & -B_{1}\\\\ -B_{2} & B_{1} & 0 \\\\ \\end{pmatrix}.$$ With WolframAlpha it is possible to see that all eigenvalues are zero and the kernel has dimension 2. I have tried to find some method to obtain these results without resort to software or brute force (solve a linear system step by step), but I could not.\n\nThen my question is the following: is it possible to determine the null space, eigenvalues and generalized eigenvectors for matrix $M$ efficiently without resort to software or brute force?\n\nIt is always possible to compute them the hard way: computing the characteristic polynomial $\\det(M-\\lambda 1)$, finding its roots, substituting in $(M-\\lambda 1)v=0$ to find eigenvectors $v$ and using them to determine the generalized eigenvectors. What I am looking for is a answer that computes the null space, eigenvalues and (generalized) eigenvectors making use of the properties of matrix $M$, not brute force.\n\n\u2022 A matrix with all zero eigenvalues is nilpotent, and so, if diagonalizable, is identical zero. \u2013\u00a0kjetil b halvorsen Feb 7 '18 at 21:10\n\u2022 \u2013\u00a0jobe Feb 8 '18 at 1:01\n\nWe assume that the system $\\{E,B\\}$ is linearly independent.\n\nNote that (double cross product) $E_xB_xX=BE^TX-X(E^TB)=BE^TX$, that is, $E_xB_x=BE^T$ and, in the same way, $B_xE_x=EB^T$.\n\n$M[X,Y]^T=[E\\times Y,B\\times X]^T$ and a basis of $\\ker(M)$ is $\\{[B,0]^T,[0,E]^T\\}$.\n\nMoreover $\\det(M-\\lambda I_6)=\\det(\\lambda^2I-BE^T)$; $BE^T$ has rank $1$ and trace $0$; then it is nilpotent and $M$ also.\n\nAbout the generalized eigenvectors, $M^2=diag(BE^T,EB^T)$ and $M^2[X,Y]^T=0$ can be written $BE^TX=0,EB^TY=0$, that is, $X\\in E^{\\perp},Y\\in B^{\\perp}$, that implies that $dim(\\ker(M^2))=4$.\n\nIt is not difficult to see that $M^3=0$ and we are done.\n\n\u2022 What a great answer! Thank you very much for this elegant and very instructive answer. There is only one thing I didn't get it: it is not clear to me why $\\det(M-\\lambda I_6)=\\det(\\lambda^2 I-BE^T)$. Can you elaborate a little more? \u2013\u00a0jobe Feb 9 '18 at 12:38\n\u2022 Since $B_x$ and $\\lambda I_3$ commute, $\\det(M-\\lambda I)=\\det(\\lambda ^2I -E_xB_x)=\\det(\\lambda^2 I-BE^T)$. \u2013\u00a0loup blanc Feb 9 '18 at 13:05\n\u2022 I still don't get it. In the first determinant, $\\det(M-\\lambda I_6)$, we have $6\\times 6$ matrices, while in the second one, $\\det(\\lambda^2 I_3-E_\\times B_\\times)$, we have $3\\times 3$ matrices. Why $\\lambda^2$? I think you are using some property I don't know. \u2013\u00a0jobe Feb 9 '18 at 15:18\n\u2022 If $CD=DC$, then $\\det(\\begin{pmatrix}A&B\\\\C&D\\end{pmatrix})=\\det(AD-BC)$. \u2013\u00a0loup blanc Feb 10 '18 at 11:22\n\u2022 Thank you. I was ignorant of that property. \u2013\u00a0jobe Feb 11 '18 at 15:58\n\nVery short: You can easily get examples of all ranks less than $n$, when the matrix is $n \\times n$. A very simple example, take a matrix with zeros on the diagonal, zeros below, and ones above the diagonal. It is nilpotent, that is, there is some positive integer $k$ such that $A^k=0$.\n\nAn extended hint: When all eigenvalues are zero, then, if $x$ is an eigenvector, it will certainly belong to the kernel! So, if there is a basis of eigenvectors then the matrix must be zero.\n\n\u2022 Thank you for your answer. Perhaps I was not very clear in my first formulation of this question, so I have reformulated it. The characteristic polynomial is $\\lambda^6$ and the kernel has dimension 2, these two facts can be determined through software or brute force. Determination of generalized eigenvectors through brute force for this matrix is not so easy and so I began wondering if there is a more elegant and efficient method to obtain these results. That is what I am looking for as an answer. \u2013\u00a0jobe Feb 8 '18 at 12:44", "date": "2019-07-23 00:52:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2640694/efficient-method-for-computing-the-properties-of-a-block-anti-diagonal-matrix", "openwebmath_score": 0.8627724647521973, "openwebmath_perplexity": 120.69533107824483, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.973240714486111, "lm_q2_score": 0.8757869867849166, "lm_q1q2_score": 0.8523515527561906}}
{"url": "http://mathhelpforum.com/trigonometry/92536-show-arccos.html", "text": "# Thread: Show that arccos(...) = ...\n\n1. ## Show that arccos(...) = ...\n\nHi\n\nAnother \"show that\" problem involving inverse trig. functions.\n\nShow that $\\displaystyle arccos(\\frac{1}{\\sqrt{1+x^{2}}}) = arctan(|x|)$\n\nSince $\\displaystyle | \\frac{1}{\\sqrt{1+x^{2}}} | \\leq 1$ , I assume I can take $\\displaystyle cos(arccos(\\frac{1}{\\sqrt{1+x^{2}}})) = \\frac{1}{\\sqrt{1+x^{2}}}$\n\nBut what do I do with my right-side?\n\nKinda stuck here.\n\n2. Originally Posted by Twig\nHi\n\nAnother \"show that\" problem involving inverse trig. functions.\n\nShow that $\\displaystyle arccos(\\frac{1}{\\sqrt{1+x^{2}}}) = arctan(|x|)$\n\nSince $\\displaystyle | \\frac{1}{\\sqrt{1+x^{2}}} | \\leq 1$ , I assume I can take $\\displaystyle cos(arccos(\\frac{1}{\\sqrt{1+x^{2}}})) = \\frac{1}{\\sqrt{1+x^{2}}}$\n\nBut what do I do with my right-side?\n\nKinda stuck here.\nlet $\\displaystyle u = arccos(\\frac{1}{\\sqrt{1+x^{2}}})$\n\n$\\displaystyle cos(u)=(\\frac{1}{\\sqrt{1+x^{2}}})$\n\nas you can see from the pic\n\n$\\displaystyle tan(u) = x$\n\n$\\displaystyle arctan(tan(u))=arctan(x)$\n\n$\\displaystyle u=arctan(x) = arccos(\\frac{1}{\\sqrt{1+x^{2}}})$ since\n\nwe let\n$\\displaystyle u=arccos(\\frac{1}{\\sqrt{1+x^{2}}})$\n\n3. Let $\\displaystyle \\theta=\\arccos\\left(\\frac1{\\sqrt{1+x^2}}\\right).$ Then\n\n$\\displaystyle \\cos\\theta\\ =\\ \\frac1{\\sqrt{1+x^2}}$\n\n$\\displaystyle \\implies\\ \\cos^2\\theta\\ =\\ \\frac1{1+x^2}\\quad\\ldots\\,\\fbox1$\n\n$\\displaystyle \\therefore\\ \\sin^2\\theta\\ =\\ 1-\\cos^2\\theta\\ =\\ 1-\\frac1{1+x^2}\\ =\\ \\frac{x^2}{1+x^2}\\quad\\ldots\\,\\fbox2$\n\n$\\displaystyle \\therefore\\ \\tan^2\\theta\\ =\\ \\frac{\\sin^2\\theta}{\\cos^2\\theta}\\ =\\ x^2$\n\n$\\displaystyle \\implies\\ \\tan\\theta\\ =\\ \\pm|x|$\n\n$\\displaystyle \\implies\\ \\theta\\ =\\ \\arctan(\\pm|x|)\\ =\\ \\pm\\arctan|x|$\n\nBut since $\\displaystyle \\arccos y\\ge0$ for all $\\displaystyle 0<y\\le1,$ $\\displaystyle \\theta\\ge0.$ Hence $\\displaystyle \\arccos\\left(\\frac1{\\sqrt{1+x^2}}\\right)=\\arctan|x |.$\n\n4. Thanks guys, always nice with a few different approaches.\n\nI think the drawing a triangle solution is something I need to remember.\n\n5. Hello, Twig!\n\nShow that: .$\\displaystyle \\arccos\\left(\\frac{1}{\\sqrt{1+x^{2}}}\\right) \\:=\\: \\arctan(|x|)$\nLet: .$\\displaystyle y \\;=\\;\\arccos\\left(\\frac{1}{\\sqrt{1+x^2}}\\right) \\quad\\Rightarrow\\quad \\cos y \\:=\\:\\frac{1}{\\sqrt{1+x^2}}$\n\nThen: .$\\displaystyle \\cos^2\\!y \\:=\\:\\frac{1}{1+x^2} \\quad\\Rightarrow\\quad 1-\\cos^2\\!y \\:=\\:1 - \\frac{1}{1+x^2} \\:=\\:\\frac{x^2}{1+x^2}$\n\n. . Hence: .$\\displaystyle \\sin^2\\!y \\:=\\:\\frac{x^2}{1+x^2} \\quad\\Rightarrow\\quad \\sin y \\:=\\:\\frac{\\pm x}{\\sqrt{1+x^2}}$\n\nThen: .$\\displaystyle \\tan y \\:=\\:\\frac{\\sin y}{\\cos y} \\;=\\;\\frac{\\dfrac{\\pm x}{\\sqrt{1+x^2}}}{\\dfrac{1}{\\sqrt{1+x^2}}}\\quad\\Ri ghtarrow\\quad \\tan y \\;=\\; \\pm x$\n\n. . Hence: .$\\displaystyle y \\:=\\:\\arctan(\\pm x) \\;=\\;\\arctan|x|$\n\nTherefore: .$\\displaystyle \\arccos\\left(\\frac{1}{\\sqrt{1+x^2}}\\right) \\;=\\;\\arctan|x|$", "date": "2018-04-21 04:53:02", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/92536-show-arccos.html", "openwebmath_score": 0.9860490560531616, "openwebmath_perplexity": 3250.2120780623723, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357555117624, "lm_q2_score": 0.8705972801594707, "lm_q1q2_score": 0.8523458659274128}}
{"url": "https://or.stackexchange.com/questions/33/in-an-integer-program-how-i-can-force-a-binary-variable-to-equal-1-if-some-cond?noredirect=1", "text": "# In an integer program, how I can force a binary variable to equal 1 if some condition holds?\n\nSuppose we have a binary or continuous variable $$x$$, a binary variable $$y$$, and a constant $$b$$, and we want to enforce a relationship like\n\nIf $$x \\gtreqless b$$, then $$y = 1$$.\n\nHow can we write this using one or more linear constraints?\n\nIf $$x$$ is binary: Then the \"if\" condition really means either \"$$x = 0$$\" or \"$$x=1$$\".\n\nTo enforce \"if $$x=0$$ then $$y=1$$\": use $$y \\ge 1-x.$$ To enforce \"if $$x=1$$ then $$y=1$$\": use $$y \\ge x.$$\n\nIf you want to require that $$y=1$$ if and only if the condition holds, then replace the $$\\ge$$s above with $$=$$s.\n\nIf $$x$$ is continuous: In this case, numerical inaccuracy might produce errors, so be prepared for $$y$$ to be set incorrectly if $$x$$ is close to but on the \u201cwrong\u201d side of $$b$$. To avoid this, you can increase or decrease $$b$$ a little bit to provide some buffer.\n\nTo enforce \"if $$x < b$$ then $$y=1$$\": $$b - x \\le My,$$ where $$M$$ is a large constant. The logic is that if $$b - x > 0$$, then $$y$$ must equal 1, and otherwise it may equal 0.\n\nTo enforce \"if $$x > b$$ then $$y=1$$\": $$x - b \\le My,$$ with similar logic as above.\n\nTo enforce \"if $$x = b$$ then $$y=1$$\": This one is tricky. I'm not sure my approach is the easiest. (Anyone have a better solution?) We really can't check for $$x=b$$, but we can check for $$b-\\delta \\le x \\le b+\\delta$$ for some small $$\\delta > 0$$. To do this, we introduce two new binary decision variables.\n\nLet $$z_1$$ be a binary variable that equals 1 if $$x > b - \\delta$$, equals 0 if $$x < b - \\delta$$, and could equal either if $$x = b - \\delta$$. Enforce this definition by adding the following constraints: \\begin{alignat}{2} Mz_1 & \\ge x - b + \\delta\\tag1 \\\\ M(1-z_1) & \\ge b - x - \\delta\\tag2 \\end{alignat} The logic is:\n\n\u2022 If $$x > b - \\delta$$, then (1) forces $$z_1=1$$ and (2) has no effect.\n\u2022 If $$x < b - \\delta$$, then (2) forces $$z_1=0$$ and (1) has no effect.\n\u2022 If $$x = b - \\delta$$, then (1) and (2) have no effect; $$z_1$$ could equal either 0 or 1.\n\nNext, introduce a second binary variable $$z_2$$, which equals 1 if $$x < b + \\delta$$, equals 0 if $$x > b + \\delta$$, and could equal either if $$x = b + \\delta$$. Introduce the following constraints: \\begin{alignat}{2} Mz_2 & \\ge b - x + \\delta\\tag3 \\\\ M(1-z_2) & \\ge x - b - \\delta\\tag4 \\end{alignat} The logic is similar:\n\n\u2022 If $$x < b + \\delta$$, then (3) forces $$z_2=1$$ and (4) has no effect.\n\u2022 If $$x > b + \\delta$$, then (4) forces $$z_2=0$$ and (3) has no effect.\n\u2022 If $$x = b + \\delta$$, then (3) and (4) have no effect; $$z_2$$ could equal either 0 or 1.\n\nFrom constraints (1)-(4), we can say that if $$z_1=z_2=1$$, then $$b - \\delta \\le x \\le b + \\delta$$. Therefore, we can enforce \"if $$b - \\delta \\le x \\le b + \\delta$$ then $$y=1$$\" using: $$y \\ge z_1 + z_2 - 1.$$\n\nNote: If your model is relatively large, i.e., it takes a non-negligible amount of time to solve, then you need to be careful with big-$$M$$-type formulations. In particular, you want $$M$$ to be as small as possible while still enforcing the logic of the constraints above.\n\n\u2022 In the first part where $x$ is binary, what if I have 2 or more binary variable that leads to the decision of $y$ like $x$ and $z$ when $x=1 \\wedge z = 1$ then $y = 1$\n\u2013\u00a0ooo\nJan 28 '20 at 21:22\n\u2022 Then you'd have to formulate separate constraints in which you enforce the definition of $y$ and then use $y$ as described above. Jan 29 '20 at 14:33\n\u2022 I didn't get it.\n\u2013\u00a0ooo\nJan 29 '20 at 19:38\n\u2022 Lets say I have 4 binary variable $a,b,c,d$ if all $a,b,c,d = 1$ the $x = 1$ else $x =0$, then can I write $x \\ge 3 - a+b+c+d$\n\u2013\u00a0ooo\nJan 30 '20 at 13:57\n\u2022 @LarrySnyder610: how small could we set $\\delta$ to? Jun 4 '20 at 17:39\n\nRather than linearising the logical constraint, I would try the logical constraints built in a solver. Gurobi and SCIP both have indicator constraints.\n\nMy colleague works with these a lot and he\u2019s finding the indicator constraints in Gurobi perform worse than big-M. He\u2019s in contact with the Gurobi developers so I might be able to get more info if there\u2019s interest.\n\n\u2022 I've never tried those; that's a good suggestion. More info would certainly be welcome (maybe in a new Q&A). May 31 '19 at 15:36\n\nTo model $$x=b \\implies y=1$$, where $$L \\le x \\le U$$, you can do the following: \\begin{align} L y^- + b y + (b+\\delta)y^+ \\le x &\\le (b-\\delta) y^- + b y + U y^+\\\\ y^- + y + y^+ &= 1 \\\\ y^-, y, y^+ &\\in \\{0,1\\} \\end{align} In fact, this formulation also enforces the converse $$y=1 \\implies x=b$$.\n\n\u2022 If we assume $L=b-\\delta$ and $U=b+\\delta$, then the first constraint is essentially: $L y^- + b y + Uy^+ \\le x \\le L y^- + b y + U y^+$ or $x = L y^- + b y + U y^+$.\n\u2013\u00a0EhsanK\nJan 17 '20 at 1:40\n\u2022 True, but the idea here is that $\\delta>0$ is small, so those assumptions on $L$ and $U$ would imply that $x$ is essentially constant. The more useful setting would be when $L\\ll b\\ll U$. Jan 17 '20 at 1:48", "date": "2021-09-17 03:06:45", "meta": {"domain": "stackexchange.com", "url": "https://or.stackexchange.com/questions/33/in-an-integer-program-how-i-can-force-a-binary-variable-to-equal-1-if-some-cond?noredirect=1", "openwebmath_score": 0.9568264484405518, "openwebmath_perplexity": 426.1986805553533, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357585701875, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8523458587288496}}
{"url": "https://math.stackexchange.com/questions/3186610/expectation-of-a-hitting-time", "text": "# Expectation of a hitting time\n\nI'm trying to find the expectation of a stopping time. Specifically,\n\nLet $$T_1,...,T_n$$ be i.i.d exponential random variables with mean $$1$$. Let $$S_n = T_1 + ... + T_n$$ denote their partial sum. Define the stopping time,\n\n$$T = \\inf\\{n \\geq 1 : S_n \\geq 1\\}$$\n\nwhich is the first time $$S_n$$ exceeds $$1$$. Calculate $$E[T]$$.\n\nHere was my approach. Let $$x = E[T]$$. I want to condition on what happens at the first time $$T_1$$. If $$T_1 \\geq 1$$, then the process, $$\\{S_n\\}$$ has stopped and $$T \\equiv 1$$. Otherwise if $$T_1 < 1$$, then after one step, process, $$\\{S_n\\}$$ will renew again until it reaches $$1$$. So we have the following equation,\n\n$$\\begin{eqnarray} x &=& E[T]\\\\ &=& E[T | T_1 < 1]P(T_1 < 1) + E[T | T_1 \\geq 1]P(T_1 \\geq 1)\\\\ &=& (1+x)P(T_1 < 1) + P(T_1 \\geq 1)\\\\ &=& (1+x)(1-e^{-1}) + e^{-1} \\end{eqnarray}$$\n\nThis results in $$x = e$$.\n\nHowever, I was told that the answer is $$2$$. They gave a heuristic explanation that $$S_T$$ is distributed as $$1 + S$$ where $$S$$ is an exponential random variable with mean $$1$$. I can see the intuition behind this from the memoryless property, but I can't prove why it is so. I ran three simulations in $$\\textsf{R}$$ and got $$x \\approx 2.001, 2.0161, 1.9785$$, which seems to confirm that the answer is $$2$$. Can someone explain this result?\n\nAlso, why/where did my approach fail?\n\n\u2022 When you say $\\mathbb{E}[T|T_1 < 1] = 1 + x,$ you're asserting that, in expectation, if the first step doesn't get all the way to $1,$ then the remaining steps have to get all the way to $1$. This is clearly false - if $T_1 = 1/2,$ then $T_2 + \\dots T_T$ only have to get up to $1/2$. Since $T_1 > 0$ a.s., we should have $\\mathbb{E}[T|T_1 < 1] < 1 + \\mathbb{E}[T].$ \u2013\u00a0stochasticboy321 Apr 13 '19 at 19:00\n\u2022 I noticed that you wanted a non-heuristic proof - note that $\\{T > n\\} = \\{S_n < 1\\}$, since the $S_n$ are non-decreasing. It should be straightforward to figure out $\\pi_n := P(S_n<1)$ by establishing a recurrence relation between $\\pi_n$ and $\\pi_{n-1}$ - you'll likely need the volume of a standard simplex in $n$ dimensions. This will directly give you $P(T = n),$ which you can then use to find the mean. \u2013\u00a0stochasticboy321 Apr 13 '19 at 21:19\n\u2022 A more high level argument is from queuing theory - suppose it takes you $\\mathrm{Exp}(1)$ time to do a job. How many jobs will you finish in $1$ unit of time? It is a classical result that this number is $\\mathrm{Poission}(1)$ distributed. However, intuitively, the mean is simpler to argue - your rate of doing jobs is $1$ per unit, so you, in expectation, should finish one job per unit (a Little's law type argument). You are interested in this number plus one - which job will you be doing when the time unit finishes. \u2013\u00a0stochasticboy321 Apr 13 '19 at 21:24\n\u2022 @stochasticboy321 Thank you! I was able to do a brute force calculation along these lines. \u2013\u00a0Flowsnake Apr 14 '19 at 16:24\n\u2022 ^That's grand, you're welcome :). You should add an answer below, so that others trying the same problem can have a reference. \u2013\u00a0stochasticboy321 Apr 15 '19 at 1:59\n\nFollowing @stochasticboy321's approach, we want to find $$P(T > n) = P(S_n < 1)$$. Since $$S_n = T_1 + ... + T_n \\sim$$ Gamma($$n,1$$), we have,\n$$P(S_n < 1) = \\frac{1}{\\Gamma(n)}\\int_0^1x^{n-1}e^{-x}dx = 1 - \\frac{\\Gamma(n,1)}{\\Gamma(n)}$$\nwhere $$\\Gamma(n,1)$$ is the incomplete Gamma function. To get this expression, I used the nice identity found here. Finally,\n$$E[T] = 1 + \\sum\\limits_{n=1}^\\infty P(T > n) = 1 + \\sum\\limits_{n=1}^\\infty P(S_n<1) = 1+ \\sum\\limits_{n=1}^\\infty\\left(1 - \\frac{\\Gamma(n,1)}{\\Gamma(n)}\\right)$$\nI have no idea how to evaluate the sum in closed form, but a computation in Wolfram Alpha seems to suggest it converges to $$1$$. Thus, $$E[T] = 2$$ (at least by conjecture from this numerical computation).\n\u2022 If we set the integral above to $I_n,$ then by integration by parts, and using $\\Gamma(n) = (n-1) \\Gamma(n-1) = (n-1)!,$ we get for $n \\ge 1,$ $$P(T > n) = \\frac{I_n}{\\Gamma(n)} = \\frac{(n-1) I_{n-1}}{(n-1) \\Gamma(n-1)} - \\frac{e^{-1}}{{(n-1)!}} = P(T > n-1) - \\frac{e^{-1}}{(n-1)!}$$ But then $$P(T = n) = P(T> n-1) - P(T > n) = \\frac{e^{-1}}{(n-1)!}.$$ We immediately have that $T \\overset{\\mathrm{law}}= 1 + Z,$ where $Z \\sim \\mathrm{Pois}(1),$ and so has mean $2$. Implicitly this also solves the series above (by summation by parts). \u2013\u00a0stochasticboy321 Apr 15 '19 at 19:35", "date": "2020-03-31 23:53:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3186610/expectation-of-a-hitting-time", "openwebmath_score": 0.902830183506012, "openwebmath_perplexity": 161.257117801193, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357555117624, "lm_q2_score": 0.8705972566572504, "lm_q1q2_score": 0.8523458429178988}}
{"url": "https://web2.0calc.com/questions/please-help-me_92335", "text": "+0\n\n+1\n104\n16\n+365\n\nGuys please solve this question and let me know the correct option\n\nplease its a request here the link of question if you cant see it (https://ibb.co/ZWftpcR)\n\nMar 22, 2021\nedited by kes1968 \u00a0Mar 22, 2021\n\n#1\n+9279\n+2\n\n(Assuming that\u00a0 log\u00a0 means natural log)\n\ny\u00a0\u00a0 =\u00a0\u00a0 5log x\n\nTake the log of both sides\n\nlog y\u00a0\u00a0 =\u00a0\u00a0 log( 5log x )\n\nNow we can apply the rule that says:\u00a0 log(xy)\u00a0 =\u00a0 y log x\n\nlog y\u00a0\u00a0 =\u00a0\u00a0 log x log 5\n\nDivide both sides by\u00a0 log 5\n\nlog y / log 5\u00a0\u00a0 =\u00a0\u00a0 log x\n\nDo the reverse of natural log to both sides (make both sides the exponent of e )\n\ne^( log y / log 5\u00a0 ) \u00a0 =\u00a0\u00a0 e^log x\n\nNow the right side simplifies to just\u00a0 x\n\ne^( log y / log 5\u00a0 ) \u00a0 =\u00a0\u00a0 x\n\nSo we have...\n\n$$x\\ =\\ e^{\\frac{\\log y}{\\log 5}}\\\\~\\\\ x\\ =\\ (e^{\\log y})^{\\frac{1}{\\log 5}}\\\\~\\\\ x\\ =\\ y^{\\frac{1}{\\log 5}}$$\n\nMar 22, 2021\n#2\n+365\n+2\n\nbrother what you have found is the same function just in terms y , but inverse function is symmetrical about y=x axis ,\n\nalso f(f^-1(x))=x which is not satisfied ,\n\nby the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions\n\nMar 22, 2021\n#3\n+9279\n+3\n\nThe inverse function can be graphed by: \u00a0\u00a0 $$y=x^{\\frac{1}{\\log5}}$$\n\nHere's a graph:\n\nhttps://www.desmos.com/calculator/aijpzpouk1\n\nIn other words...\n\nIf\u00a0\u00a0 $$f(x)=5^{\\log x}$$\u00a0\u00a0\u00a0 then the inverse is\u00a0\u00a0 $$f^{-1}(x)=x^{\\frac{1}{\\log 5}}$$\n\nAnd...\n\n$$f(f^{-1}(x))\\ =\\ f(x^{\\frac{1}{\\log 5}})\\ =\\ 5^{\\log(x^{\\frac{1}{\\log 5}})}\\ =\\ 5^{\\frac{\\log x}{\\log 5}}\\ =\\ 5^{\\log_5 x}\\ =\\ x$$\n\nThe options leave it in the form that is solved for\u00a0 x, and so I left it like that to match the options\n\nMar 22, 2021\n#4\n+365\n+2\n\nbrother so what you feel (as a math expert) should be the right answer to the given question ,\n\nMar 22, 2021\n#5\n+9279\n+1\n\nI'm not sure what you meant by \"by the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions\"....but if this didn't answer your question then please feel free to ask for more clarification!\n\nAnd I think the answer is option 1:\u00a0 $$x=y^\\frac{1}{\\log 5}$$\n\nMar 22, 2021\n#6\n+365\n+3\n\nbrother , if you see your second answer and change it in terms of x , wont you get option 2 and option 4 ,\n\nMar 22, 2021\n#7\n+9279\n+1\n\nHmm...actually....I see what you mean.... (maybe I made the question harder than it has to be!)\n\nI take back my original answer!! Now I think the answer is option 4\n\nMar 22, 2021\n#8\n+365\n+1\n\nbrother do you really believe its option 4 or just to keep my heart , please clarify if you still believe the answer is 1 , if yes then please prove the same to me as well\n\nMar 22, 2021\n#9\n+1740\n+1\n\nFor computing the inverse function, the plan I know is\n1. interchange x & y\n\n2. solve for y\n\nbut actually, after 1. you already have a term for the inverse function. It's just not written in the \"usual\" way, wich is y=f(x).\n\nAnswer for is exactly what you get after interchanging x&y, so the correct answer is answer 4.\n\nMar 22, 2021\n#10\n+9279\n+2\n\nI do agree Probolobo, but then the confusing thing is that\n\n$$5^{\\log y}\\ =\\ y^{\\log 5}$$\n\nWhich means option 2 and option 4 are the same function and so are equivalent....\n\nMar 22, 2021\n#12\n+1740\n+1\n\nThat's correct, then there are actually 2 correct answers, 2 & 4. Didn't see that equivalence on first glance ;)\n\nProbolobo \u00a0Mar 22, 2021\n#11\n+365\n+1\n\nso whats the final answer so that i can challenge the answer key?\n\nMar 22, 2021\n#13\n+9279\n+1\n\nIf I had to guess right now, I would guess option 4. My next guess would be option 1.\n\nBut I am honestly not sure!! I think this is a bad question.\n\nCan you let us know when you find out the \"correct\" answer according to the answer key?\n\nhectictar \u00a0Mar 22, 2021\n#14\n+112907\n+3\n\nI think you are making hard work of it\n\nthe inverse of\n\n$$y=5^{logx}$$\n\nis simply\n\n$$x=5^{logy}$$\n\nYou just have to switch the x and y over.\n\nthere would be restrictions on x and on y but the question isn't worrying about that.\n\nHere is the graphs\n\nhttps://www.desmos.com/calculator/4ftfa2bny7\n\nSee they are reflections of each other about y=x\n\nMar 22, 2021\n#15\n+365\n+2\n\nanother question , another superb explanation from mod\n\nMelody you are simply awesome ! thanks for the help mate\n\nkes1968 \u00a0Mar 22, 2021\nedited by kes1968 \u00a0Mar 22, 2021\n#16\n+112907\n+1\n\nYou are welcome Kes\n\nMelody \u00a0Mar 22, 2021", "date": "2021-04-19 06:01:24", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/please-help-me_92335", "openwebmath_score": 0.7883875370025635, "openwebmath_perplexity": 2047.694379541057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995782141546, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8523366980397655}}
{"url": "https://math.stackexchange.com/questions/2460589/five-people-get-on-the-elevator-that-stops-at-five-floors-in-how-many-ways-they", "text": "# Five people get on the elevator that stops at five floors. In how many ways they can get off?\n\nFive people get on the elevator that stops at five floors. In how many ways they can get off? For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor. In how many ways they can get off at different floors? Now, consider that people in elevator have names, say A,B,C,D and E., assuming that, for example, the case A on the 1st floor is different from the case B on the 1st floor. Answer the previous questions with this assumption.\n\nI was told that there are 4 questions, but I'm not really sure about it.\n\n1. People don't have names and get off\n2. People don't have names and get off at different floors\n3. People have names and get off\n4. People have names and get off at different floors.\n\nDo the names matter?\n\nI think I can answer one question out of 4 - the number of all outcomes should be $5^5=3125$. I am confused about the rest though\n\n\u2022 I think I can answer one question out of 4 - the number of all outcomes should be 5^5=3125. I am confused about the rest though. \u2013\u00a0Gabriel Oct 6 '17 at 16:33\n\u2022 Wait, there are 4 questions here? I only see two: one where you do differentiate between the 5 people and one where you don't. And yes, for the second one you indeed get $5^5$. \u2013\u00a0Bram28 Oct 6 '17 at 16:36\n\u2022 Do you know about Stars and Bars? \u2013\u00a0Bram28 Oct 6 '17 at 16:37\n\u2022 Thank you!! I was told that there are 4 questions, but I'm not really sure about it. 1. People don't have names and get off, 2. People don't have names and get off at different floors, 3. People have names and get off, 4. People have names and get off at different floors. Do the names matter? And thank you, I will definetely read about Stars and Bars!! \u2013\u00a0Gabriel Oct 6 '17 at 16:52\n\n1. This has been answered in other answers: ${9\\choose4}$.\n\n2. Since there are $5$ people and $5$ floors there is just $1$ way for them to get off at different floors: $1$ per floor.\n\n3. You already found $5^5=3125$.\n\n4. At each floor exactly one of $A$, $\\ldots$, $E$ gets off. This can be done in $5\\cdot4\\cdot3\\cdot2=5!=120$ ways.\n\nFor the first question, use the basic 'stars and bars' method.\n\nWe can represent your example scenario (\"For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor.\") as:\n\n$$*||**||**$$\n\nThe people are the 'stars' and the separators between the different floors are the 'bars'.\n\nAnother example: one person getting off on each floor would be:\n\n$$*|*|*|*|*$$\n\nAnd everyone getting off on the 3rd floor would be:\n\n$$||*****||$$\n\nSee how this works?\n\nNow, with $4$ 'bars' to divide the people into the $5$ groups/floors, and those bars taking up $4$ of the $9$ possible positions in this 'symbol string' of bars and stars, you get\n\n$$9 \\choose 4$$\n\npossibilities.\n\n\u2022 Thank you so much for the explanation!! I think I understand how it works now. \u2013\u00a0Gabriel Oct 6 '17 at 16:54\n\u2022 @Gabriel You're welcome! Your instructor did not teach 'stars and bars' yet? ... Because this was a typical stars and bars question! :) \u2013\u00a0Bram28 Oct 6 '17 at 16:56\n\u2022 Sometimes they just give the formula derived from stars and bars which is $\\dbinom{n+r-1}{r-1}$ in this case $n=5$ and $r=5$ of course, the integer solutions for the equation $x_1+x_2+x_3+x_4+x_5=5$ are considered there are no left and right boundries (luckily). Of course giving the formula without the logic is problematic but I haven't seen one problem that required the logic rather than the formula. \u2013\u00a0Deniz Tuna Yal\u00e7\u0131n Oct 6 '17 at 20:21\n\nThe question could also be solved using a well known formula derived from stars and bars which is; $$\\dbinom{n+r-1}{r-1}$$ where we distribute $r$ equally accepted objects to $n$ spaces with or without restrictions like $\\geq1$ or $\\leq9$, this formula is also used for solving linear equations (in integers) that's why this works for combinatoric problems, so without making it longer;$$x_1+x_2+x_3+x_4+x_5=5$$ so we apply the formula, we can think that this is equivalent to distributing five equally accepted apples to five children for instance, the result is $$\\dbinom{5+5-1}{5-1}=126$$ this statement could be proven using alternate ways, we should open a post about it too.", "date": "2020-04-08 00:14:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2460589/five-people-get-on-the-elevator-that-stops-at-five-floors-in-how-many-ways-they", "openwebmath_score": 0.708287239074707, "openwebmath_perplexity": 439.4969678160285, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995723244552, "lm_q2_score": 0.8824278571786139, "lm_q1q2_score": 0.8523366898560086}}
{"url": "https://brilliant.org/discussions/thread/scratch-pad/", "text": "# Have you ever used Scratch pad like this?\n\nFind the root of following equation.\n\n$4y^3-2700(1-y)^4=0$\n\nAbove equation can be solved using newton raphson iterative method with initial approximation to be unity.\n\n1) First Let $f(y)=4y^3-2700(1-y)^4$ We have to find root of above function.\n\n2)Find derivative of $f(y)$. Here,$f^{'}(y)=12y^2+10800(1-y)^3$\n\n3)Find $y_n$ using newton raphson method $y_0=1$ $y_{n+1}=y_n-\\frac{f(y_n)}{f^{'}(y_n)}$ $y_{n+1}=y-\\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$\n\nYou can use calculator or scratchpad to calculate $y_1,y_2,y_3,....$.If this equation has real solution ,these values $y_1,y_2,y_3,....$ would converge to root of $f(y)$.\n\nHow to do such iterations smartly?\n\n1) $\\color{#69047E}{\\text{On first line of scratch pad write y=1}}$\n\n2) $\\color{#D61F06}{\\text{On second line write}}$\n\n$y=y-\\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$\n\n$\\color{#20A900}{\\text{You will see new value of y as result.}}$\n\n3) $\\color{#EC7300}{\\text{From line 3, continue pasting}}$\n\n$y=y-\\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$\n\nuntil you get a constant value.If your initial approximation is not too away from actual solution,you would get constant value in no more than 10 steps.\n\nBy this method you can solve most of the transcendental equation.\n\n$\\color{#3D99F6}{\\text{By using scratchpad, try finding sum of initial 10 fibonacci number.}}$\n\nNote by Aamir Faisal Ansari\n5\u00a0years, 4\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nYes !i have used it this way.Sometimes in Recursion problems and like that.\n\n- 5\u00a0years, 4\u00a0months ago\n\nBy using scratchpad, try finding sum of initial 10 fibonacci number.\n\n- 5\u00a0years, 4\u00a0months ago", "date": "2020-11-24 01:04:07", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/scratch-pad/", "openwebmath_score": 0.9835755228996277, "openwebmath_perplexity": 3701.128305174093, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9658995703612219, "lm_q2_score": 0.8824278556326343, "lm_q1q2_score": 0.8523366866303358}}
{"url": "https://www.jiskha.com/display.cgi?id=1305922453", "text": "# pre calculus\n\nposted by joe\n\ncube roots of 64i?\n\n1. MathMate\n\nConvert the given number into the form\na+bi = \u03b1(cos(\u03b2)+i sin(\u03b2))\nwhere\n\u03b1 = sqrt(a^2+b^2)\n\u03b2 = sin-1(b/\u03b1)\nIf the number is plot in the complex (Z) plane, it will be more evident.\nFor example, 64i will have\n\u03b1=64 (the distance from origin)\n\u03b2=90\u00b0 (sin-164/64=1)\nThe equivalent angles are:\n90\u00b0\n450\u00b0\n810\u00b0\n(we only need 3)\n\nDivide each of these angles by , to get \u03b21, \u03b22, & \u03b23. And take \u03b1'=\u03b1^(1/3)=64^(1/3)=4\nThe three cube roots are then\nz1=\u03b1'(cos(\u03b21)+isin(\u03b21)\n=4(cos(30\u00b0)+isin(30\u00b0))\nz2=4(cos(150\u00b0)+isin(150\u00b0))\nz3=4(cos(270\u00b0)+isin(270\u00b0))\n\nCheck by expanding z1^3, z2^3 and z3^3 to get back 64i.\n\n2. MathMate\n\n\u03b2=90\u00b0 (sin-164/64=sin-11=\u03c0/2)\n\n## Similar Questions\n\n1. ### Math\n\nRoots Ok, what about roots? Roots of polynomials?\n2. ### pre-calculus\n\nfind the cube roots of -216 answer in polar form and complex\n3. ### pre-calculus\n\nfind the cube roots of -216 answer in polar form\n4. ### Pre-Calculus\n\nHello, I have tried solving this using the rational roots theorem and none of the roots seem to be working. I'm trying to figure out where I went wrong. 3x^3 -2x^2 - 7x - 4 = 0\n5. ### trig\n\nwhat is the cube root of (-64i)?\n6. ### precalc\n\nexpress the roots of unity in standard form a+bi. 1.) cube roots of unity 2.) fourth roots of unity 3.) sixth roots of unity 4.) square roots of unity\n7. ### precalculus\n\nexpress the roots of unity in standard form a+bi. 1.) cube roots of unity 2.) fourth roots of unity 3.) sixth roots of unity 4.) square roots of unity\n8. ### Pre calculus\n\nFor the equation 2x^2-5x^3+10=0 find the number the number of complex roots and the possible number of real roots.\n9. ### Pre-Calculus\n\nRewrite each quadratic equation in the form ax^2+bx+c=0. Then,use technology to solve each by graphing. ROund you answers to the nearest hundredth, where necessary. a) 3x^2+30 = -19x Answer: 3x^2+19x+30 Roots: x = -3 b) 6x^2= 25x-24 \u2026\n10. ### Pre-Calculus 11\n\nWhat is the entire radial form of -3* cube root of 2?\n\nMore Similar Questions", "date": "2018-06-18 19:09:30", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/display.cgi?id=1305922453", "openwebmath_score": 0.8052590489387512, "openwebmath_perplexity": 3945.0069275690817, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9871787876194204, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8523218845097028}}
{"url": "https://math.stackexchange.com/questions/1879002/what-is-the-minimum-number-of-rotations-about-axes-in-a-plane-that-can-describe/1879003", "text": "# What is the minimum number of rotations about axes in a plane that can describe an arbitrary rotation in 3D?\n\nI'm trying to decompose an arbitrary rotation of a 3D sphere into a series of rotations about any axes that lie in the equatorial plane.\n\nI know I can factor the arbitrary rotation into three rotations in the equatorial plane using Euler angle formulas. Can I factor into two rotations if I allow any choice of axis in the equatorial plane? If it exists, is this factorization unique?\n\n## migrated from mathoverflow.netAug 2 '16 at 14:05\n\nThis question came from our site for professional mathematicians.\n\n\u2022 There is a unique rotation mapping one unit vector into a distinct vector. Take one of the vectors of your basis, and apply this rotation. Now hopefully you can see that once one axis is coincidental, you only need another rot. in the orthogonal plane. This means at two rotations are necessary and sufficient in general. \u2013\u00a0Real Aug 2 '16 at 3:48\n\u2022 what do you understand by equatorial plane (is it the xy-plane)? \u2013\u00a0Manfred Weis Aug 2 '16 at 7:22\n\u2022 @ManfredWeis Yes by equatorial plane I mean the xy-plane. \u2013\u00a0Talon Chandler Aug 2 '16 at 12:28\n\u2022 Talon, I thought the answer is yes and that I have a simple proof for it. But the question might not be appropriate for this site. If you ask over at Mathematics StackExchange and then let me know here (with a link) when you have posted, I'd be happy to give more details. \u2013\u00a0user43208 Aug 2 '16 at 13:49\n\u2022 On second thought, let me migrate it for you... \u2013\u00a0user43208 Aug 2 '16 at 14:05\n\nMy answer is that you can.\n\nIf you know about quaternions, then you know that a rotation about a unit vector $v \\in \\mathbb{R}^3$ through an angle $\\theta$ can be accomplished by regarding elements $w = (a, b, c) \\in \\mathbb{R}^3$ as purely quaternionic elements $ai + bj + ck$, and then mapping $w \\mapsto uwu^{-1}$ where $u = \\cos(\\frac{\\theta}{2})\\cdot 1 + \\sin(\\frac{\\theta}{2})v$. Such $u$ are precisely quaternions of unit norm.\n\nLet us denote the rotation above by $R_u$ (so $R_u(w) := uwu^{-1}$). Notice that a composition of two rotations $R_u \\circ R_v$ is just $R_{u v}$; this follows by associativity of quaternionic multiplication.\n\nSo we can translate your problem into the following: for any rotation $R_w \\in SO(3)$ given by a nonzero quaternion $w$, show that there exist quaternions $u, v$ in the linear span of $1, i, j$ such that $u v = w$. For these $u, v$ describe rotations about vectors in the equatorial plane spanned by $i, j$.\n\nNow this is not difficult. Let us write $u = a + bi + cj$ and $v = a' + b'i + c'j$. We compute\n\n$$(a + bi + cj)(a' + b'i + c'j) = aa' - bb' - cc' + (ab' + a'b)i + (ac' + a'c)j + (bc' - b'c)k$$\n\nand so given a nonzero quaternion $w = p + qi + rj + sk$, our task is to cook up parameters $a, b, c, a', b', c'$ such that\n\n$$p = aa' - bb' - cc'$$\n\n$$q = a b' + a'b$$\n\n$$r = ac' + a'c$$\n\n$$s = bc' - b'c$$\n\nIn fact, let's make our life easier and simply set $b = 0, b' = 1$. Then $a = q$, $c = -s$, and we can solve for $a', c'$ in the linear system\n\n$$p = qa' + sc'$$ $$r = -sa' + qc'$$\n\nprovided the determinant $q^2 + s^2$ is nonzero, i.e., provided one of $q, s$ is nonzero. If both $q, s = 0$, then of course $w = p + rj$ corresponds to a rotation about the vector $j$ which is already in the equatorial plane, so there was nothing to do in this case.\n\nOnce you have solutions $u = a + bi + cj, v = a' + b'i + c'j$ to the equation $uv = w$ in your hands, the rotations $R_u, R_v$ are rotations about the vectors $bi + cj$ and $b'i + c'j$, respectively. If you want the rotation angles, then normalize $u$ and $v$ (i.e. divide by their norms $(a^2 + b^2 + c^2)^{1/2}$ and $(a')^2 + (b')^2 + (c')^2)^{1/2}$) and use the fact that $R_u = R_{\\frac{u}{\\|u\\|}}$. You can then read off the desired angles by writing e.g. $\\frac{u}{\\|u\\|} = \\cos(\\frac{\\theta}{2}) + \\sin(\\frac{\\theta}{2})\\frac{bi + cj}{(b^2 + c^2)^{1/2}}$, so for instance $\\cos(\\frac{\\theta}{2}) = \\frac{a}{(a^2 + b^2 + c^2)^{1/2}}$.\n\nBy the way, this solution also shows that the decomposition is in general non-unique. For example, we could have also chosen $c = 0, c' = 1$ and arrive at quaternionic solutions.\n\n\u2022 Do we need to constrain $u$, $v$, and $w$ to be unit quaternions so they represent rotations? If this is the case, then we have 7 equations with 6 unknowns which might have a unique solution. \u2013\u00a0Talon Chandler Aug 2 '16 at 18:10\n\u2022 No, you don't need to, but you can assume they are unit quaternions if you like because $R_u = R_{\\frac{u}{\\|u\\|}}$ for any nonzero quaternion $u$. If $w$ is constrained to the unit sphere in the quaternions (3-dimensional $S^3$) and $u, v$ are each constrained to the unit sphere in the span of $\\{1, i, j\\}$ (so $(u, v)$ parametrizes a space that looks like $S^2 \\times S^2$, which is 4-dimensional), then we are asking about surjectivity of a multiplication map $S^2 \\times S^2 \\to S^3$, so 3 equations in 4 variables by my count. \u2013\u00a0user43208 Aug 2 '16 at 18:24\n\u2022 Okay that makes sense. What if we consider only infinitesimal rotations? Can we uniquely decompose an arbitrary rotation into infinitesimal rotations in the plane spanned by $i,j$? In this case we can make the approximations: $$\\cos\\left(\\frac{\\theta}{2}\\right) \\approx 1$$ $$\\sin\\left(\\frac{\\theta}{2}\\right) \\approx \\frac{\\theta}{2}$$ So $p=a=a'=1$ and we have four equations with three unknowns which gives a unique solution---i.e. an infinitesimal rotation about an arbitrary axis can be decomposed uniquely into two infinitesimal rotations about axes in the span of $i,j$. \u2013\u00a0Talon Chandler Aug 2 '16 at 19:28\n\u2022 Well, the answer to the third sentence in your last comment would be 'no'. But why is getting uniqueness important to you? I think in fact the method I gave you gives virtual uniqueness of a solution if you constrain the first rotation $R_u$ to rotate about $j$. That is: unit quaternions $u$ of the form $a + bj$ form a circle $S^1$, and the unit $v$ for rotations about a second axis in the $ij$-plane form a sphere $S^2$, and the multiplication map $S^1 \\times S^2 \\to S^3: (u, v) \\mapsto w = uv$ is onto, and unless $w$ itself is a rotation about $j$, only one pair $(u, v)$ satisfies $w = uv$. \u2013\u00a0user43208 Aug 2 '16 at 20:35\n\u2022 I'm interested in uniqueness because I'm working on a reconstruction problem. I have measurements that correspond to a known rotation about an arbitrary axis, and I want to reconstruct the set of rotations in the $ij$-plane that gave rise to the \"total\" rotation. I'm hoping to find the conditions under which this reconstruction is possible. Can you expand on why we can't decompose an arbitrary infinitesimal rotation into two unique infinitesimal rotations in the $ij$-plane? \u2013\u00a0Talon Chandler Aug 2 '16 at 21:03\n\nThe answer is no, because when letting $v$ be an arbitrary direction about which to rotate by an angle $\\rho$, that rotation can only then be realized with a rotation around an axis in the xy-plane if $v^Tz=0$; that shows that at least three rotations are necessary\n\n1. rotate about $v\\times z$ to bring $v$ into the $xy$-plane\n2. rotate about the image of $v$ in the $xy$-plane\n3. rotate the image of $v$ back around $v\\times z$ to bring the image back to the original direction of $v$\n\u2022 I agree that this set of rotations in the xy-plane can describe an arbitrary rotation, but how can we be sure there isn't a shortcut with just two rotations? \u2013\u00a0Talon Chandler Aug 2 '16 at 12:32", "date": "2019-10-17 12:57:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1879002/what-is-the-minimum-number-of-rotations-about-axes-in-a-plane-that-can-describe/1879003", "openwebmath_score": 0.8787638545036316, "openwebmath_perplexity": 191.7369585576091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787827157817, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.8523218785409358}}
{"url": "https://mathforums.com/threads/how-can-i-find-the-time-when-two-people-agree-to-meet-if-both-of-their-watches-have-an-offset.348121/", "text": "How can I find the time when two people agree to meet if both of their watches have an offset?\n\nChemist116\n\nThe problem is as follows:\n\nJenny and Vincent agreed to meet at the library at $6\\,p.m$. Both synchronized their watches at midnight ($\\textrm{0 hours}$). We know that Vincent's watch is not functioning correctly and gets ahead of the real time $\\textrm{50 seconds}$ each hour and Jenny's watch gets delayed from the real time $\\textrm{50 seconds}$ each hour. Vincent arrived to the library $\\textrm{15 minutes}$ before the agreed time accoring to his watch and Jenny $\\textrm{15 minutes}$ late by looking at her watch. Using this information, find how long did Vincent waited Jenny?\n\nThe alternatives according to my book are as follows:\n\n$\\begin{array}{ll} 1.&\\textrm{15 min}\\\\ 2.&\\textrm{0 min}\\\\ 3.&\\textrm{25 min}\\\\ 4.&\\textrm{60 min}\\\\ \\end{array}$\n\nThis part I'm stuck at exactly how should I make up an equation which can relate both offsets. Can someone help me here? I'm assuming that by $\\textrm{6 p.m}$ the time which would be seen by Vincent will be:\n\n$18\\times 50= 900\\,s$\n\nwhich would be $60$ minutes\n\nBut Vincent seen in his watch was:\n\n$\\textrm{5:45 pm}$\n\nhence until 5 pm would be:\n\n$17\\times 50=850\\,s$\n\n$56\\frac{2}{3}$ minutes\n\nand Vincent's watch would have seen:\n\n$\\textrm{5:56:40 pm}$\n\nAt this point I could try guessing reducing the number of minutes until adjusting the time which will be seen by Vincent and this would be same for Jenny, but it doesn't seem something which can be effective. Can someone help me here? How exactly can I find what is being requested?\n\nskeeter\n\nMath Team\n50 seconds an hour = 5 minutes in 6 hours\n\nVincent\u2019s watch runs fast, which means the real time he arrives is earlier than 17:45.\n\nJenny\u2019s watch runs slow, which means the real time she arrives is later than 18:15.\n\nThe only answer choice that makes sense is the 60 minutes Vincent waited.\n\nI worked out the problem using proportions, where T represents the actual time\n\nfor Jenny ...\nT/18.25 = 72/71\n\nfor Vincent ...\nT/17.75 = 72/73\n\ntopsquark\n\nChemist116\n\n50 seconds an hour = 5 minutes in 6 hours\n\nVincent\u2019s watch runs fast, which means the real time he arrives is earlier than 17:45.\n\nJenny\u2019s watch runs slow, which means the real time she arrives is later than 18:15.\n\nThe only answer choice that makes sense is the 60 minutes Vincent waited.\n\nI worked out the problem using proportions, where T represents the actual time\n\nfor Jenny ...\nT/18.25 = 72/71\n\nfor Vincent ...\nT/17.75 = 72/73\nThis is the major problem which I had with this question. It mentioned that until his time was 17:45, not 18:00. How would be the difference between those fractions? Can you help me with that part?", "date": "2020-04-05 13:29:35", "meta": {"domain": "mathforums.com", "url": "https://mathforums.com/threads/how-can-i-find-the-time-when-two-people-agree-to-meet-if-both-of-their-watches-have-an-offset.348121/", "openwebmath_score": 0.41406530141830444, "openwebmath_perplexity": 1144.1592501937198, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787857334058, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8523218776763138}}
{"url": "https://math.stackexchange.com/questions/1150584/combinations-how-many-handshakes", "text": "# Combinations: How many handshakes?\n\nI need help with the following question which I cannot seem to solve:\n\n17 students are sitting in a circle. Each person shakes hands with everyone but his/her neighbours. How many handshakes have been exchanged?\n\nMy approach: no. of ways $= 1 + 2 + 3 + ... + 14 = 7(15) = 105$.\n\nApparently my answer is wrong (correct ans is 119). But I can't seem to understand why. Could someone please explain?\n\n\u2022 The reason why this is wrong is as follows: the first person has 14 people to shake hands with. The second person as well has 14 people to shake hands with, since he wouldn't shake hands with his neighbour anyway! So, the sum is 14+(14+13+...+1), which does add up to 119. \u2013\u00a0Sanchises Feb 16 '15 at 19:49\n\nEach of 17 people shakes hands with 14 people (all except themselves and their 2 neighbors), so there are\n\n$$\\frac{17\\times 14}{2} = 119$$\n\nhandshakes (dividing by 2 to account for symmetry, as you would otherwise count \"$A$ shaking hands with $B$\" and \"$B$ shaking hands with $A$\" as distinct events).\n\n\u2022 Does it mean that we don't care about how they sit, say in a circle or in a row? \u2013\u00a0Nighty Feb 16 '15 at 15:04\n\u2022 @LeeKM: The difference is that in a circle, everyone has two neighbors, while in a row, the first and last person have only one each. \u2013\u00a0user139000 Feb 16 '15 at 15:05\n\u2022 Thanks! Now I know that I misunderstood the quuestion. I thought that the handshakes were to be made with distinct persons. \u2013\u00a0Donald Feb 16 '15 at 21:44\n\nHere's another way to think about it.\n\nIf everyone shook hands with everyone you'd have $\\frac {17*16}2 = 136$ handshakes (divide by 2 because the above is double counting A shaking hands with B and so forth).\n\nHowever, there are $\\frac {17*2}2 = 17$ handshakes that aren't happening because neighbors aren't shaking hands (2 neighbors for each of the 17 people, again divided by 2 to remove duplicates).\n\nSo $136-17 = 119$\n\nPew's response is a little more direct, but finding the number by calculating the total possibilities minus the \"not allowed\" interactions is sometimes a little more intuitive for some people.\n\nYou can think of this in graph theory as the following:\n\n$G$ has $17$ vertices with degree $14$, since there are no loops(people self handshaking), and no vertices are adjacent to the vertices beside them. So total degree is $17\\times 14$.\n\nWe know that degree is equal to $2\\times\\text{number of edges}$ and hence there are $\\frac{17\\times 14}{2}=119$ edges in total, where edges represent handshakes.\n\nAs is typical for problems with big numbers, you should always resort to a smaller number if you can't solve the full problem.\n\nHere, 17 people at the table is a bit hard to imagine immediately.\n\nHow many handshakes now?\n\nThere are zero.\n\nA B\n\nC D\n\n\nA can't shake with B or C, B can't with A or D, D can't with B or C, C can't with A or D--only A & D and B & C can shake.\n\nSo 2 shakes:\n\nA B\nX\nC D\n\n\n5 people is when it gets interesting, and when you should be able to see the pattern:\n\n      B\n\nA       C\n\nD   E\n\n\nEveryone has two people they can't shake with, leaving two shakes per person--but we overcount by just multiplying 5 & 3, so we divide by two.\n\nThis is easiest to see by trying to draw the graph mentioned by @Commitingtoachalleng -- draw a line between any people who can shake hands:\n\n        B\n/ \\\nA-----------C   (also C-D & A-E--a 5-point star)\n/     \\\nD       E\n\n\nSo we hypothesize the answer is $\\frac{n(n-3)}{2}$. Note that this formula holds for $n=3$ and $n=4$ as well, as we'd hope!\n\nOne final way to see this is to look again at the graphs--\n\nyou might notice that they're always complete graphs (every vertex connected to every other vertex) with the outer edges removed.\n\nSince there are $\\frac{n(n-1)}{2}$ edges in a complete graph on $n$ vertices (which you can confirm yourself by a similar process), and $n$ outer edges,\n\nthere must be $\\frac{n(n-1)}{2} - n = \\frac{n^2-n-2n}{2} = \\frac{n(n-3)}{2}$ handshakes.\n\nHowever you cut it, there are $\\frac {17 \\cdot 14} 2 = 119$ total handshakes.\n\n\u2022 We should also point out that the second option (working backwards from complete graphs) is the graph analogue of @Duncan's solution \u2013\u00a0MichaelChirico Feb 17 '15 at 6:42\n\nI will explain for 6, same logic applies for all the numbers.\n\nfirst, if 6 persons shakes will all in the party, then number of shakes will be:\n\nA B C D E F\n\nA can shake with only rest of 5 persons(B,C,D,E,F) i.e, right side ppl. = 5\n\nB can shake with only his right sides(C,D,E,F) ppl(not with A bcos,A already shook with B just now. so no repeat) = 4\n\nC can shake with only his right sides(D,E,F) ppl(not with A&B bcos,A&B already shook with C just now. so no repeat) = 3\n\nD can shake with only his right sides(E,F) ppl(not with left side ppl bcos, they already shook with D just now. so no repeat) = 2\n\nE can shake with only his right sides(F) ppl(not with left side ppl bcos, they already shook with E just now. so no repeat) = 1\n\nas all of the persons shook hand with F, He dont shake with any one now.\n\nSo total possible shakes for 6 persons are = 5+4+3+2+1 => 15\n\nnow in questions it is given that, No one shakes hand with neighbors.\n\nImagine ppls sitting around table\n\n        A   B\nF       C\nE   D\n\n\nPlease treat above image as 6 Ppl around circle : So now,\n\nInvalid Shakes:\n\nA cannot shake with B,F . I am representing it as below : A -no- B & F => 2 invalid shakes.\nB -no- C only. Bcos B -no- A is covered as part of first case.\n=> 1 invalid shake\nC -no- D only. => 1 invalid shake\nD -no- E only. => 1 invalid shake\nE -no- F only. => 1 invalid shake\nF dont shake with A is already covered as part of first invalid shakes.\n\nso total invalid shakes are 2+1+1+1+1 = 6\n\nso for 6 ppl: out of total 15 , 6 are invalid. So 15-6 = 9\n\non the same line :\n\nfor 17 ppl: out of total (16+15+...+1)=136 , 17 are invalid. So 136-17 = 119\n\nshortcut : For N ppl, Total shakes are (N-1)(first term + last term)/2 Arithmetic progression sum. out of which Invalids are N.\nSo valid shakes => (Total - N)\n\n\u2022 What is the point of posting such a complex answer to a question with a simple and accepted answer posted almost four years ago? \u2013\u00a0Jos\u00e9 Carlos Santos Oct 31 '18 at 17:23", "date": "2020-09-26 18:03:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1150584/combinations-how-many-handshakes", "openwebmath_score": 0.5542638897895813, "openwebmath_perplexity": 1316.8328053778105, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787861106087, "lm_q2_score": 0.863391602943619, "lm_q1q2_score": 0.8523218745319745}}
{"url": "http://math.stackexchange.com/questions/584601/is-this-function-injective-surjective", "text": "# Is this function injective / surjective?\n\nA question regarding set theory.\n\nLet $g\\colon P(\\mathbb R)\\to P(\\mathbb R)$, $g(X)=(X \\cap \\mathbb N^c)\\cup(\\mathbb N \\cap X^c)$\n\nthat is, the symmetric difference between $X$ and the natural numbers.\n\nWe are asked to show if this function is injective, surjective, or both.\n\nI tried using different values of $X$ to show that it is injective, and indeed it would seem that it is, I can't find $X$ and $Y$ such that $g(X)=g(Y)$ and $g(X) \\neq g(Y)$ but how do I know for sure? How do I formalize a proof for it?\n\nRegarding surjective: I think that it is.\n\nWe can take $X=\\mathbb R - \\mathbb N$ and we get that $g(X)=\\mathbb R$\n\nWhat do I do about the injective part?\n\n-\nare you able to express in formula what injective or surjective means? \u2013\u00a0 Robert.Sie Nov 28 '13 at 14:14\n\nHint:\n\n$$g\\circ g = \\mathrm{id}_{\\mathcal{P}(\\mathbb{R})}, \\quad \\text{ that is, } \\quad \\forall X \\in \\mathcal{P}(\\mathbb{R}).\\ g(g(X)) = X.$$\n\nI hope this helps $\\ddot\\smile$\n\n-\n\nWe can view subsets of $X$ as functions $X \\to \\{0,1\\}$: 0 if the point is not in the set, It is commonly very useful to rewrite questions about subsets as questions about functions.\n\nWhat form does $g$ take if we do this rewrite?\n\nTerminology wise, if $S \\subseteq X$, then $\\chi_S$ is the function described above:\n\n$$\\chi_S(x) = \\begin{cases} 0 & x \\notin S \\\\ 1 & x \\in S \\end{cases}$$\n\nIt is called the characteristic function of $S$.\n\n-\n0 if $a \\in X\\cap \\mathbb N$, 1 otherwise \u2013\u00a0 Oria Gruber Nov 28 '13 at 14:19\nThat's not right. To be more clear, $g(\\chi_S)$ is a function. To describe this function, we could do so by writing down the equation that says what its value at a point $x$ is: i.e. what is $g(\\chi_S)(x)$? \u2013\u00a0 Hurkyl Nov 28 '13 at 14:22\n\nHINT: Recall that symmetric difference is associative and for every set $B$, $B\\triangle B=\\varnothing$.\n\n-\nAnd why was this downvoted? \u2013\u00a0 Asaf Karagila Nov 28 '13 at 22:04", "date": "2015-08-02 23:20:37", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/584601/is-this-function-injective-surjective", "openwebmath_score": 0.8997302055358887, "openwebmath_perplexity": 257.7361717018151, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226300899744, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8523031393300072}}
{"url": "https://math.stackexchange.com/questions/1490376/find-sum-of-series-sum-n-1-infty-1-n1-left-frac-n2", "text": "# Find sum of series $\\sum _{ n=1 }^{ \\infty }{ { (-1) }^{ n+1 }\\left( \\frac { n+2 }{ { n }^{ 5 } } \\right) }$ correct to 3 decimal places.\n\nThis is a question I came across while studying on Khan Academy:\n\nFind the sum of the series $\\sum _{ n=1 }^{ \\infty }{ { (-1) }^{ n+1 }\\left( \\frac { n+2 }{ { n }^{ 5 } } \\right) }$ correct to three decimal places.\n\nI solved this by calculating ${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },...\\\\$ until I had a close enough value (bad method, I know). Anyway, here is the solution steps given in the answer:\n\nNote that this is a convergent alternating series. The Alternating Series Estimation Theorem states that the error bound in the sum of the first $n$ terms is the absolute value of the first omitted term; i.e., $\\left| { a }_{ n+1 } \\right|$. To obtain three-digit accuracy, we need to find $n$ such that $\\left| { a }_{ n+1 } \\right| <\\ 0.0005$, or equivalently $\\frac { n+3 }{ { (n+1) }^{ 5 } } <0.0005$.\n\nWhen $n=6$, we have $\\left| { a }_{ n+1 } \\right| =\\frac { n+3 }{ { (n+1) }^{ 5 } } =\\frac { 9 }{ 16807 } \\approx .000535$; that is not accurate enough.\n\nWhen $n=7$, we have $\\left| { a }_{ n+1 } \\right| =\\frac { n+3 }{ { (n+1) }^{ 5 } } =\\frac { 10 }{ 32768 } \\approx .000305$; that will give a final result within $0.0005$.\n\nThus we need to find the sum of the first seven terms of the series.\n\n$$\\sum _{ n=1 }^{ 7 }{ { (-1) }^{ n+1 }\\left( \\frac { n+2 }{ { n }^{ 5 } } \\right) \\approx 2.8914 }$$\n\nHence, $2.891$ is accurate to three decimal places.\n\nWhat I don't understand about this answer is if the estimate is $2.8914$ and the error bound is $0.000305$, I think the actual sum can be close to $2.8914+0.000305\\approx 2.891705\\approx 2.892$. Then the result above is not correct.\n\nDoes this answer take into account the fact that the $(7+1)th$ term is negative? Or is there something about rounding and error bounds that I don't understand?\n\nThank you.\n\n\u2022 Indeed, 2.8914 means between 2.89135 and 2.89145 and 0.000305 is less than 0.0004 hence the sum is between 2.89135-0.0004 and 2.89145, which implies that 2.891 has three correct decimal places. (The same estimates for n=6 show it was not necessary to go to n=7.) \u2013\u00a0Did Oct 21 '15 at 7:27\n\u2022 You are right, it is the fact that $\\sum_1^7$ is an overestimate that enables us to conclude that $2.891$ is correct to 3 decimal places. \u2013\u00a0Andr\u00e9 Nicolas Oct 21 '15 at 7:28\n\u2022 To the op: If you want it back I'll undelete my answer, but it seems unnecessary given you're already pretty much there yourself. \u2013\u00a0Adam Hughes Oct 21 '15 at 7:32\n\u2022 @Did Thank you. But do you mean if somehow the estimate is an underestimate, the sum can be between 2.89135 and 2.89145+0.0004. Then 2.892 would be correct, right? \u2013\u00a0user52139 Oct 21 '15 at 7:49\n\u2022 @AdamHughes Thank you very much. I actually need it back if you don't mind, please. \u2013\u00a0user52139 Oct 21 '15 at 7:52\n\nI would read the question the same way you did-find a three decimal value that you know the true sum of the series rounds to. We know from the alternating series theorem that the truncation error is smaller than the first neglected term in absolute value and of the same sign. There is no truncation error that we can guarantee will be small enough, because we could be poised right on the rounding boundary, with the computed answer something like $2.8915$. Here, because after seven terms we have $2.8914$ and we know the eighth term is negative, the correct answer is in the range $(2.8911,2.8914)$ and this entire range rounds to $2.891$\n\n\u2022 +1, warning about problem with rounding boundary is well placed here. This answer might be further improved by pointing out the problem of numerically finding any decimal of $\\sum_{n=0}^\\infty \\frac 1{2^n}$. \u2013\u00a0Ennar Oct 21 '15 at 10:38\n\nThis is an alternating series, so the error is estimated by the last term. Since you want the error to be less than $10^{-3}$ in size, you need only choose $n$ so that ${n+2\\over n^5}<10^{-3}$ since the terms are decreasing.\n\nSolving we see $n+2<10^{-3}n^5$. We claim this is so for $n\\ge 7$ as\n\n$$7+2=9<16.807=7^5\\cdot 10^{-3}.$$\n\nFor $n>7$ we see that by induction we may conclude this with the inductive step being\n\n$$(n+1)+2<10^{-3}n^5+1< 10^{-3}(n+1)^5,$$\n\nThen adding up the first $6$ terms gives you $2.891$ to three decimal places.", "date": "2019-06-18 21:14:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1490376/find-sum-of-series-sum-n-1-infty-1-n1-left-frac-n2", "openwebmath_score": 0.8888874053955078, "openwebmath_perplexity": 223.32739009154525, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022630759019, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8523031334287373}}
{"url": "https://math.stackexchange.com/questions/2357560/does-the-bezout-gcd-equation-hold-in-a-ufd", "text": "# Does the Bezout GCD equation hold in a UFD?\n\nI'm wondering when Bezout's Theorem is valid.\n\nI know when the ring $R$ is a Euclidean Domain or a Principal Ideal Domain there is always $\\gcd(a,b)$ and I can find $x,y \\in R$ such that $\\gcd(a,b) = ax + by.$\n\nI know when the ring $R$ is a Unique Factorization Domain, there is always $\\gcd(a,b)$. My question is, in this case can I find $x,y \\in R$ such that $\\gcd(a,b) = ax + by.$ ?\n\nNow if $R$ is just a commutative ring, if there is $d \\in R$ such that $<a>+<b> = <d>$ I know that $d = \\gcd(a,b)$. In this case, can I find $x,y \\in R$ such that $\\gcd(a,b) = ax + by$ ?\n\n\u2022 In $\\mathbb Z[x]$, a UFD, $2$ and $x$ have GCD $1$, but there is no solution to $2f(x)+xg(x)=1$. More generally, a UFD in which Bezout is true is always a principal ideal domain. I think. \u2013\u00a0Thomas Andrews Jul 13 '17 at 15:30\n\nIt is not necessarily true in a UFD. For instance, if $k$ is a field, then the polynomial ring $k[s,t]$ is a UFD. But $\\gcd(s,t)=1$, and there do not exist $x$ and $y$ such that $sx+ty=1$.\n\nFor your second question, the answer is yes by definition. Indeed, by definition, $\\langle a\\rangle+\\langle b\\rangle$ is the set of elements of the form $ax+by$. So if $d\\in \\langle a\\rangle+\\langle b\\rangle$, then there exist $x,y\\in R$ such that $d=ax+by$.\n\nThe converse holds as well: if $d=\\gcd(a,b)$ can be written in the form $ax+by$, then that means $d\\in \\langle a\\rangle+\\langle b\\rangle$, and it follows that $\\langle a\\rangle+\\langle b\\rangle=\\langle d\\rangle$.\n\nThus a commutative ring has the property that the GCD of any two elements $a$ and $b$ can be written in the form $ax+by$ iff the sum of any two principal ideals is principal. By induction on the number of generators, this is equivalent to any finitely generated ideal being principal. Such a ring is known as a Bezout ring. A Noetherian Bezout ring is the same thing as a principal ideal ring, but there exist non-Noetherian examples as well. For instance, the ring of holomorphic functions on a connected open subset of $\\mathbb{C}$ is a Bezout domain but has ideals which are not finitely generated.\n\nThe rings where it is valid B\u00e9zout's lemma are known as B\u00e9zout rings. In the case of integral domains rings we have the domains known as B\u00e9zout domains.\n\nIn general, a UFD doesn't have to be a B\u00e9zout domain. The classical example is $\\Bbb{Z}[x]$, as Thomas Andrews pointed out in his comment. Theoretically there is a reason for which the above is true. We have the following:\n\nTheorem 1: If $D$ is both a UFD and a B\u00e9zout domain, then $D$ is a PID.\n\nProof: See exercise 11 of section 8.3 of the book \"Abstract Algebra\" by Dummit and Foote.\n\nBy this theorem, if every UFD were also a B\u00e9zout domain, we would conclude that every UFD is a PID, and this is false (look at $\\Bbb{Z}[x]$ again), as you surely know.\n\nMore generally, for B\u00e9zout domains we have the following result:\n\nTheorem 2: Let $D$ be a B\u00e9zout domain. TFAE:\n\ni) $D$ is a PID.\n\nii) $D$ is Noetherian.\n\niii) $D$ is a UFD.\n\niv) $D$ satisfies the ACCP (ascending chain condition on principal ideals).\n\nv) $D$ is an atomic domain.\n\nProof: This is theorem 46 given in Pete L. Clark's notes on factorization in integral domains\n\nYour second question has been also answered, but let me include a general version of what you wanted to prove.\n\nTheorem 3: Let $a_1,a_2,\\ldots a_n$ be nonzero elements of a commutative ring $R$. Then $a_1,a_2,\\ldots a_n$ have a greatest common divisor $d$, expressible in the form $$d=r_1a_1+r_2a_2+\\cdots +r_na_n$$ if only if the ideal $(a_1,a_2,\\ldots a_n)$ is principal.\n\nNote that $(a_1,a_2,\\ldots a_n)$ is the same as $(a_1)+(a_2)+\\cdots +(a_n)$.\n\nProof: This is theorem 6-3 in Burton's book \"First Course in Rings and Ideals\".\n\nThere are simple examples of non-Bezout UFDs, e.g. any UFD of dimension $$> 1$$, e.g. $$\\Bbb Z[x,y],\\,$$ by the below equivalent conditions for a UFD to be Bezout (or directly: $$\\,\\gcd(x,y)=1\\,$$ but $$\\,xf+y\\,g = 1\\Rightarrow 0 = 1\\,$$ via eval at $$\\,x=0=y)$$.\n\nTheorem $$\\rm\\ \\ \\ TFAE\\$$ for a $$\\rm UFD\\ D$$\n\n$$(1)\\ \\$$ prime ideals are maximal if nonzero,  i.e. $$\\rm\\ dim\\,\\ D \\le 1$$\n$$(2)\\ \\$$ prime ideals are principal\n$$(3)\\ \\$$ maximal ideals are principal\n$$(4)\\ \\ \\rm\\ gcd(a,b) = 1\\, \\Rightarrow\\, (a,b) = 1,$$ i.e.  coprime $$\\Rightarrow$$ comaximal\n$$(5)\\ \\$$ $$\\rm D$$ is Bezout, i.e. all ideals $$\\,\\rm (a,b)\\,$$ are principal.\n$$(6)\\ \\$$ $$\\rm D$$ is a $$\\rm PID$$\n\nProof $$\\$$ (sketch of $$\\,1 \\Rightarrow 2 \\Rightarrow 3 \\Rightarrow 4 \\Rightarrow 5 \\Rightarrow 6 \\Rightarrow 1)\\$$ where $$\\rm\\,p_i,\\,P\\,$$ denote primes $$\\neq 0$$\n\n$$(1\\Rightarrow 2)$$ $$\\rm\\ \\ p_1^{e_1}\\cdots p_n^{e_n}\\in P\\,\\Rightarrow\\,$$ some $$\\rm\\,p_j\\in P\\,$$ so $$\\rm\\,P\\supseteq (p_j)\\, \\Rightarrow\\, P = (p_j)\\:$$ by dim $$\\le1$$\n$$(2\\Rightarrow 3)$$ $$\\$$ max ideals are prime, so principal by $$(2)$$\n$$(3\\Rightarrow 4)$$ $$\\ \\rm \\gcd(a,b)=1\\,\\Rightarrow\\,(a,b) \\subsetneq (p)$$ for all max $$\\rm\\,(p),\\,$$ so $$\\rm\\ (a,b) = 1$$\n$$(4\\Rightarrow 5)$$ $$\\ \\ \\rm c = \\gcd(a,b)\\, \\Rightarrow\\, (a,b) = c\\ (a/c,b/c) = (c)$$\n$$(5\\Rightarrow 6)$$ $$\\$$ Ideals $$\\neq 0\\,$$ in Bezout UFDs are generated by an elt with least #prime factors\n$$(6\\Rightarrow 1)$$ $$\\ \\ \\rm (d) \\supsetneq (p)$$ properly $$\\rm\\Rightarrow\\,d\\mid p\\,$$ properly $$\\rm\\,\\Rightarrow\\,d\\,$$ unit $$\\,\\rm\\Rightarrow\\,(d)=(1),\\,$$ so $$\\rm\\,(p)\\,$$ is max\n\nAs for the gcd as the ideal sum in a PID, that follows using \"contains = divides\", viz.\n\n$$c\\mid a,b\\iff (c)\\supseteq (a),(b)\\iff (c)\\supseteq (a)\\!+\\!(b)\\!=\\!(d)\\iff c\\mid d$$\n\nAnd, yes, we have $$\\,d\\in (a)+(b) = aR + bR\\,$$ $$\\Rightarrow\\, d = ar + br'$$ for some $$\\,r,r'\\in R$$", "date": "2021-05-13 03:28:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2357560/does-the-bezout-gcd-equation-hold-in-a-ufd", "openwebmath_score": 0.9458763003349304, "openwebmath_perplexity": 130.0538827978256, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226284173629, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.852303128143545}}
{"url": "https://physics.stackexchange.com/questions/287787/find-the-capacitance-of-the-capacitor", "text": "# Find the capacitance of the capacitor\n\nGiven: Consider a capacitor connected to a battery of voltage V . Let the capacitor have an area A, and a distance L between the plates. Assume that the capacitor has a layer of linear dielectric (of dielectric constant \u03ba, so that \u03b5 = \u03ba\u03b50) of thickness L/2 on the lower plate.\n\nI found the capacitance to be $C=\\frac{2\\kappa\\epsilon_0 A}{L}$. Was I correct about the separation distance? In more common examples of a capacitor, they claim that the plates are flat with nothing in between. In this case, the linear dielectric is as tall as half the total distance between the plates. Am I correct in using the remaining free space, namely $L/2$, as the separation distance in the equation? For reference, the equation is:\n\n$$C=\\frac{\\kappa \\epsilon_0 A}{d}$$\n\nInstant check that shows your answer is wrong: set $\\kappa = 1$ as if there is no dielectric, and you don't recover the parallel plate formula for a gap of $L$.\n\nHow to solve this problem:\n\nLet the voltage across the capacitor be $V$ and the voltages across the top and bottom be $V_t$ and $V_b$ respectively, then $$V=V_t+V_b,$$ which implies that $$1/C=1/C_t+1/C_b$$ since $V=Q/C$.\n\nThus essentially this system consists of two capacitors in series. One capacitor is the top gap, and one is the bottom part. Use the formula for plate capacitors twice, with the separation being $L/2$ in both cases, then get the total capacitance by using the formula for two capacitors in series. Some algebra is involved.\n\nYou can check the answer you get, again, by setting $\\kappa=1$, and seeing if you recover the usual plate formula.\n\n\u2022 Why do you say there are 2 capacitors in series? The problem states there is only 1. \u2013\u00a0whatwhatwhat Oct 20 '16 at 23:35\n\u2022 @whatwhatwhat - think of the bottom part and the top part as two capacitors as a way to solve the problem. \u2013\u00a0Suzu Hirose Oct 20 '16 at 23:36\n\u2022 So then for the top section, there is a buildup of positive charge on the top plate and negative charge on the surface of the dielectric? \u2013\u00a0whatwhatwhat Oct 20 '16 at 23:42\n\u2022 @whatwhatwhat - sorry my explanation was not very clear. There is an electric field between the two plates of the capacitor generated by the charges on the plates. The capacitance of this system of two plates is got by considering that the sum of the voltage over the two plates must be equal to the total voltage across the whole capacitor. I will edit the answer to make it clearer. \u2013\u00a0Suzu Hirose Oct 20 '16 at 23:48\n\u2022 Suzu Hirose is completely right! \u2013\u00a0freecharly Oct 21 '16 at 0:12\n\nThe $\\mathbf{D}$ field between the plates is (as usual, neglecting fringing effects) uniform and normal to the plane of the plates (assume in the z direction for simplicity) and is given by\n\n$$\\mathbf{D} = \\frac{Q}{A}\\hat{\\mathbf{z}}$$\n\nwhere the plates have charge $Q$ and $-Q$ respectively. The electric field (between the plates) within the dielectric is\n\n$$\\mathbf{E_\\kappa} = \\frac{\\mathbf{D}}{\\kappa \\epsilon_0}$$\n\nand the electric field (between the plates) in air is\n\n$$\\mathbf{E} = \\frac{\\mathbf{D}}{\\epsilon_0}$$\n\nThus, the potential difference between the plates is\n\n$$V = \\left(\\frac{L}{2}\\mathbf{E_\\kappa} + \\frac{L}{2}\\mathbf{E}\\right)\\cdot\\hat{\\mathbf{z}} = \\left(\\frac{L}{2}\\frac{\\mathbf{D}}{\\kappa \\epsilon_0} + \\frac{L}{2}\\frac{\\mathbf{D}}{\\epsilon_0}\\right)\\cdot\\hat{\\mathbf{z}} = \\left(\\frac{L}{2}\\frac{Q}{A\\kappa\\epsilon_0} + \\frac{L}{2}\\frac{Q}{A\\epsilon_0} \\right)$$\n\nThe capacitance is then\n\n$$C = \\frac{Q}{V} = \\frac{1}{\\frac{L/2}{A\\kappa\\epsilon_0} + \\frac{L/2}{A\\epsilon_0}} = \\frac{1}{\\frac{1}{C_1}+ \\frac{1}{C_2}}$$\n\nwhich is the formula for series connected capacitors.\n\n\u2022 Better explanation than my answer of why it is like two capacitors in series. \u2013\u00a0Suzu Hirose Oct 21 '16 at 1:22\n\u2022 So I'm still treating this as 1 capacitor correct? My friend was helping with this and she said that one way to solve the problem is to treat the system as 2 capacitors, but I couldn't visualize where the 1st capacitor begins/ends and likewise for the 2nd one. \u2013\u00a0whatwhatwhat Oct 22 '16 at 18:59\n\u2022 @whatwhatwhat, correct, it is just one capacitor however, it is instructive and intuitively 'nice' that we would (ideally) get the same result if we found the equivalent capacitance of two series connected capacitors with the same plate area $A$ and spacing $L/2$ but with different dielectrics; one with $\\epsilon_0$ and the other with $\\kappa \\epsilon_0$ \u2013\u00a0Alfred Centauri Oct 22 '16 at 22:46", "date": "2020-01-28 16:26:48", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/287787/find-the-capacitance-of-the-capacitor", "openwebmath_score": 0.7865023016929626, "openwebmath_perplexity": 278.8493287850811, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022633435197, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8523031211722472}}
{"url": "https://www.justinmath.com/roulette-wheel-selection/", "text": "# Roulette Wheel Selection\n\nAs we saw previously, it\u2019s easy to simulate a random coin flip by generating a random decimal $r$ between $0$ and $1$ and applying the following function:\n\n\\begin{align*} \\textrm{outcome}(r) = \\begin{cases} \\textrm{heads} & \\textrm{ if } r < 0.5 \\\\ \\textrm{tails} & \\textrm{ otherwise} \\end{cases} \\end{align*}\n\nThis is a special case of a more general idea: sampling from a discrete probability distribution. Flipping a fair coin is tantamount to sampling from the distribution [0.5, 0.5], i.e. $0.5$ probability heads and $0.5$ probability tails.\n\nMore complicated contexts may require sampling from longer distributions that may or may not be uniform. For example, if we wish to simulate the outcome of rolling a die with two red faces, one blue face, one green face, and one yellow face, then we need to sample from the distribution [0.4, 0.2, 0.2, 0.2].\n\nNote that when we sample from the distribution, we need only sample an index from the distribution, and then use the index to look up the desired value. For example, when we sample an index from the distribution [0.4, 0.2, 0.2, 0.2], we have probabilities $0.4,$ $0.2,$ $0.2,$ and $0.2$ of getting indices $0,$ $1,$ $2,$ and $3$ respectively. Then, all we need to do is look up that index in the following array: [red, blue, green, yellow].\n\n## Roulette Wheel Selection\n\nRoulette wheel selection is an elegant technique for sampling an index from an arbitrary discrete probability distribution. It works as follows:\n\n1. turn the distribution into a cumulative distribution,\n2. sample a random number $r$ between $0$ and $1,$ and then\n3. find the index of the last value in the cumulative distribution that is less than $r.$\n\nTo illustrate, let\u2019s sample an index from the distribution [0.4, 0.2, 0.2, 0.2] that was mentioned above in the context of a die roll.\n\n1. First, we construct the cumulative distribution: [0.4, 0.4 + 0.2, 0.4 + 0.2 + 0.2, 0.4 + 0.2 + 0.2 + 0.2], or more simply, [0.4, 0.6, 0.8, 1.0].\n2. Then, we sample a random number $r$ between $0$ and $1.$ Suppose we get $r = 0.93.$\n3. Finally, we find the index of the last value in the cumulative distribution that is less than $r = 0.93.$ In our case, this is the value $0.8$ at index $2,$ because the next value ($1.0$ at index $3$) is greater than $r = 0.93.$\n\nSo, we have sampled the index $2.$\n\n## Exercise\n\nWrite a function random_draw(distribution) that samples a random number from the given distribution, an array where distribution[i] represents the probability of sampling index i.\n\nTo test your function on a particular distribution, sample many indices from the distribution and ensure that the proportion of times each index gets sampled matches the corresponding probability in the distribution. Do this for a handful of different distributions.\n\nTags:", "date": "2022-08-10 17:11:48", "meta": {"domain": "justinmath.com", "url": "https://www.justinmath.com/roulette-wheel-selection/", "openwebmath_score": 0.8775347471237183, "openwebmath_perplexity": 291.4495911781642, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9914225133191064, "lm_q2_score": 0.8596637559030338, "lm_q1q2_score": 0.8522900014867285}}
{"url": "https://math.stackexchange.com/questions/1850456/prime-divides-n2-1-rightarrow-prime-doesnt-divide-n/1850477", "text": "# Prime divides $n^2 + 1 \\Rightarrow$ prime doesn't divide $n$\n\nHow can I show that if a prime $p$ divides $$n^2 + 1$$ then it doesn't divide $n$?\n\n\u2022 If p divides n then it must divide n^2. No number other than 1 can divide two consecutive numbers. Jul 6 '16 at 2:25\n\u2022 Dang auto correct. And that was a weird one. Can not understand it. Jul 6 '16 at 2:28\n\u2022 Try proving the contrapositive, it's easier to think about. Jul 6 '16 at 2:29\n\u2022 @fleablood Maybe cutie numbers are like friendly numbers and lucky numbers at the same time?\n\u2013\u00a0user296602\nJul 6 '16 at 2:29\n\u2022 Applying Euclid's algorithm $\\,(n^k\\!+1,n) = (1,n) = 1\\,$ by $\\,n^k\\!+1\\equiv 1\\pmod n\\,$ for $\\,k\\ge 1.\\,$ Since their gcd $= 1,\\,$ they have no nontrivial common divisor. Jul 6 '16 at 3:35\n\nWe can use Bezout's Identity to show that $\\left(n^2+1,n\\right)=1$. That is, $$\\left(n^2+1\\right)\\cdot1-n\\cdot n=1$$ Therefore, the greatest common divisor of $n^2+1$ and $n$ is $1$.\n\nThat is, if any number divided both $n^2+1$ and $n$, it would also divide $(n^2+1)-n\\cdot n=1$.\n\n\u2022 For every integer $k\\ge 1$, $(n^k+1)\\cdot 1-n\\cdot n^{k-1}=1$. Jul 6 '16 at 10:27\n\u2022 Indeed. $\\left(n^k+1,n\\right)=1$ for $k\\ge1$.\n\u2013\u00a0robjohn\nJul 6 '16 at 12:12\n\u2022 @Woria Special case $\\, j = n^{k-1}\\,$ of $\\,(1\\!+\\!nj,\\,n) = 1\\$ by $\\ 1= 1\\!+\\!nj-n(j)\\ \\$ (or by one step of Euclid's algorithm - see my comment on the question). Aug 4 '16 at 2:51\n\nA proof by contradiction: assume $n\\equiv 0\\mod p$ with $p>1$, then $$n^2\\equiv 0\\pmod p$$ also. But $$n^2\\equiv -1\\pmod p$$ which contradicts, therefore $$n^2+1\\equiv 0\\pmod p\\Rightarrow n\\not\\equiv 0\\pmod p$$\n\n\u2022 i.e. if $\\,p\\mid n,\\color{#c00}{n^2\\!+\\!1}\\,$ then $\\,{\\rm mod}\\ p\\!:\\ n\\equiv 0\\,\\Rightarrow\\, 0\\equiv \\color{#c00}{n^2\\equiv -1}\\,\\Rightarrow\\, p\\mid 1,\\,$ contradiction Aug 4 '16 at 2:54\n\n\nBelow, all 'leters variables' $\\ds{n,s,p}$ are integers $\\ds{\\pars{~p\\ \\mbox{is a}\\ \\ul{prime\\ number}~}}$:\n\n\u2022 $\\ds{{n \\over p} = s\\quad\\imp\\quad n = sp\\quad\\imp\\quad{n^{2} + 1 \\over p} = {s^{2}p^{2} + 1 \\over p} = s^{2} p + {1 \\over p}\\ !!!}$\n\u2022 $\\ds{{1 \\over p}\\ \\ul{\\mbox{is not}}\\ \\mbox{an integer because}\\ p > 1 \\pars{~p\\ \\mbox{is a prime number}~}}$\n\u2022 $\\pars{\\vphantom{\\LARGE A}% p \\mid n \\imp p \\not\\mid \\pars{n^{2} + 1}}\\ \\mbox{is equivalent to}\\ \\pars{\\vphantom{\\LARGE A}% p \\mid \\pars{n^{2} + 1} \\imp p \\not\\mid n}$\n\nIf $p=2$ and $2\\mid n$ we have that $n^{2}$ is even and so $n^{2}+1$ is odd. Now assume that $p$ is odd. We can use the Legendre symbol. If we assume that $n\\equiv0\\mod p$ we have $n^{2}\\equiv0 \\mod p$. So $$\\left(\\frac{n^{2}}{p}\\right)=0$$ but since $n^{2}\\equiv-1 \\mod p$ we also have, by the law of quadratic reciprocity $$\\left(\\frac{n^{2}}{p}\\right)=\\left(\\frac{-1}{p}\\right)=1^{\\frac{p-1}{2}}=\\begin{cases} 1 & p\\equiv1\\,\\mod\\,4\\\\ -1 & p\\equiv3\\,\\mod\\,4 \\end{cases}$$ and this is absurd.\n\nYou want to prove the statement $\\color\\red{p|n^2+1}\\implies\\color\\green{p\\not|n}$.\n\nInstead, you can prove the equivalent statement $\\neg(\\color\\green{p\\not|n})\\implies\\neg(\\color\\red{p|n^2+1})$:\n\n$\\small\\neg(\\color\\green{p\\not|n})\\implies{p|n}\\implies{p|n^2}\\implies{\\forall_{k\\in(0,p)}:p\\not|n^2+k}\\implies{p\\not|n^2+1}\\implies\\neg(\\color\\red{p|n^2+1})$.\n\nBTW, this statement holds not only for every prime $p$, but also for every integer $p>1$.", "date": "2021-10-25 06:50:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1850456/prime-divides-n2-1-rightarrow-prime-doesnt-divide-n/1850477", "openwebmath_score": 0.878925085067749, "openwebmath_perplexity": 306.0297479395353, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985042916041382, "lm_q2_score": 0.8652240964782012, "lm_q1q2_score": 0.8522828670241573}}
{"url": "http://math.stackexchange.com/questions/17638/defining-the-determinant-of-linear-transformations-as-multilinear-alternating-fo/244205", "text": "# Defining the determinant of linear transformations as multilinear alternating form\n\nHere is what our professor showed us in our linear algebra class to introduce the idea of determinants:\n\nSuppose we have an $n$-dimensional vector space $V$. Then we can create a function from $V^n$ to $\\mathbb{R}$ called $vol$ (for \"volume\") satisfying these properties:\n\n$vol$ is multilinear\n$vol$ is alternating (i.e. if any two of $v_1, \\ldots, v_n$ are the same, then $vol(v_1, \\ldots, v_n) = 0$)\n\nFrom these two properties, we can see that if $e_1, \\ldots, e_n$ is a basis of $V$, then the $vol$ function is completely defined by the value $vol(e_1, \\ldots, e_n)$.\n\nThus if $T$ is a linear operator on $V$, the ratio:\n\n$\\dfrac{vol(Te_1, \\ldots, Te_n)}{vol(e_1, \\ldots, e_n)}$\n\nis the same for any (multilinear and alternating) $vol$ function.\n\nHowever, I am having trouble understanding why the ratio is also independent of the basis $e_1, \\ldots, e_n$. This is what I am asking for help with. I can see that this invariance implies that intuitively, every $n$-parallelotope is stretched by the same amount by the operator $T$.\n\n(Our professor then defined the determinant of $T$ as that ratio.)\n\n-\nShow that any two bases can be reached from each other by row operations and look at what row operations do to the volume. \u2013\u00a0 Qiaochu Yuan Jan 16 '11 at 2:04\nAdding one row to another does not change the volume, while multiplying a row by a scalar changes the volume by that scalar factor. So if we go from $e_i$'s to another basis, say $f_i's$, then the ratio $vol(f_1, \\ldots, f_n)/vol(e_1, \\ldots, e_n)$ is the product of all the scalar multiplications. But I don't see why this is the same as $vol(Tf_1, \\ldots, Tf_n)/vol(Te_1, \\ldots, Te_n)$ \u2013\u00a0 Alan C Jan 16 '11 at 2:21\nYou are half way there. The point is that since $T$ is linear, it commutes with scalar multiplication. So if $f_1=\\alpha e_1$ for a scalar $\\alpha$, then $Tf_1 = \\alpha Te_1$. \u2013\u00a0 Alex B. Jan 16 '11 at 5:10\nAha thanks! I wonder how I overlooked the linearity of $T$. Hm... but I'm also wondering if this could be done in a way similar to the Steinitz exchange lemma. Suppose we order the $e_i$'s and $f_i$'s so that for all $0 \\leq k \\leq n$, $f_1, \\ldots, f_k, e_{k+1}, \\ldots, e_n$ span $V$. Then we could go from $vol(Te_1, \\ldots, Te_n)/vol(e_1, \\ldots, e_n)$ to $vol(Tf_1, \\ldots, Tf_n)/vol(f_1, \\ldots, f_n)$, replacing one vector at a time. (And the reason the ratio stays the same is because $T$ is linearity.) \u2013\u00a0 Alan C Jan 16 '11 at 15:28\nIn light of my comments on a previous, closely related question: in case you were wondering, I am not teaching linear algebra this semester! \u2013\u00a0 Pete L. Clark Feb 15 '11 at 16:48\n\nFirst note that since $T$ is linear, $\\mathrm{vol}_T(v_1, \\dots, v_n) = \\mathrm{vol} (T v_1 , \\dots , T v_n )$ is also a multilinear alternating function.\n\nSo if you go from $e_1, \\dots , e_n$ to a different basis, say, $b_1, \\dots b_n$, then you can write the $b_i$ in terms of their coordinates with respect to $e_1, \\dots , e_n$:\n\n$$b_1= \\sum_{i=1}^n b_{i 1} \\, e_i,\\quad b_2= \\sum_{i=1}^n b_{i 2} \\, e_i,\\quad\\dots \\quad b_n= \\sum_{i=1}^n b_{i n} \\, e_i.$$\n\nAnd Leibniz's formula (which is a direct consequence of multinearity and alternating-ness), applied to both $\\mathrm{vol}$ and $\\mathrm{vol}_T$ tells us:\n\n$$\\mathrm{vol}(b_1 , \\dots , b_n ) = \\sum_{\\sigma \\in S_n} \\text{sgn}(\\sigma) \\: b_{\\sigma(1) 1} \\cdots b_{\\sigma(n) n} \\cdot \\mathrm{vol}(e_1, \\dots , e_n)$$\n\n$$\\mathrm{vol} (T b_1 , \\dots , T b_n ) = \\sum_{\\sigma \\in S_n} \\text{sgn}(\\sigma) \\: b_{\\sigma(1) 1} \\cdots b_{\\sigma(n) n} \\cdot \\mathrm{vol}(T e_1, \\dots , T e_n)$$\n\nSo if you take the ratio $\\displaystyle \\frac{\\mathrm{vol} (T b_1 , \\dots , T b_n )}{\\mathrm{vol} (b_1 , \\dots , b_n )}$ everything else cancels out and you're left with $\\displaystyle \\frac{\\mathrm{vol} (T e_1 , \\dots , T e_n )}{\\mathrm{vol} (e_1 , \\dots , e_n )}$ again.\n\nOf course, for all of this, it is important that $\\mathrm{vol} \\neq 0$. That's probably an additional restriction your professor put on the $\\mathrm{vol}$ function.\n\n-", "date": "2015-09-03 05:28:35", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/17638/defining-the-determinant-of-linear-transformations-as-multilinear-alternating-fo/244205", "openwebmath_score": 0.9621486663818359, "openwebmath_perplexity": 143.7172189323668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429103332288, "lm_q2_score": 0.8652240964782012, "lm_q1q2_score": 0.8522828620853257}}
{"url": "http://math.stackexchange.com/questions/291956/if-p-is-prime-and-p-5-show-that-when-p-is-divided-by-10-the-remainder", "text": "# If $p$ is prime and $p > 5$, show that when $p$ is divided by 10, the remainder is 1, 3, 7, or 9\n\nIf $p$ is prime and $p > 5$, show that when $p$ is divided by 10, the remainder is 1, 3, 7, or 9.\n\nThis is a problem from Hungerford's Abstract Algebra: An Introduction. I would like some help on it, it was an example in class. Thanks.\n\n-\nConsider the case when the remainder is 2. Then $p = 10 q + 2$ for some integer $q$, so that $p = 2 (5 q + 1)$ is even, but cannot be 2 by the assumption $p > 5$. The other cases are entirely similar, it is useful for you to work them out yourself. \u2013\u00a0Andreas Caranti Feb 1 '13 at 7:27\n$p > 5$ prime means $p$ not divisible by $2$ or $5$. \u2013\u00a0Benjamin Dickman Feb 1 '13 at 8:26\n\nBy the division algorithm, we can write each $p$ as $$p = 10q + r$$ for some quotient $q$ and remainder $0\\le r < 10$. As $p$ is prime, clearly $r\\neq 0$. $r$ also cannot be even, otherwise $p$ is even. Finally, note that $r \\neq 5$ or $5 \\mid p$.\n\nFor a more brief and algebraic solution, note that $(p,\\ 10) = 1$ implies that $p$ is a unit in $\\mathbb{Z}/10\\mathbb{Z}$. The units of the ring are precisely $1,\\ 3,\\ 7$ and $9$.\n\nFor completeness, you should probably show that there exists primes for each of the remaining congruence classes.\n\n-\n\nYou already have a couple of nice \u2018mathematical-looking\u2019 answers. In attacking a question like this, though, you might want to start with simple, familiar facts.\n\nThe remainder when $p$ is divided by $10$ is simply the last digit of $p$. If the last digit of a number $n$ is $0,2,4,6$, or $8$, what kind of number is $n$? Can it be prime if it\u2019s greater than $2$? If the last digit is $0$ or $5$, what can you say about $n$? Can it be prime and greater than $5$?\n\nThese are enough to tell you, at least informally, why the prime $p>5$ must end in $1,3,7$, or $9$, and now you can worry about explaining the reasoning in the previous paragraph a bit more formally.\n\n-\nThank you for the tip, I was not sure how to best ask questions here but I'm learning and will remember that. \u2013\u00a0grayQuant Feb 1 '13 at 14:36\n\nTo say that a prime $p > 5$ gives a remainder of $1, 3, 7$ or $9$ when divided by $10$ is merely remarking that the prime is odd (and not $5$, since all integers ending in five are multiples thereof).\n\nAfter all, the remainder when divided by $10$ is simply the unit of the number itself, and if that unit was any of $0, 2, 4, 6, 8$ we'd plainly see that it was an even number, and therefore not prime. Likewise with the five.\n\nThat said, we could come to the conclusion of your statement by, say, basing an argument on how all primes above $3$ are of the form $6n \\pm 1$ (mind you, this follows from an argument akin to the above): the multiples of $6$ begin as follows: $$6, 12, 18, 24, 30\\ldots$$ From which we father that the units of such multiples are either $0, 2, 4, 6$ or $8$. Now take the '$\\pm 1$' bit into consideration and we find that we have the following units possible: $$1, 3, 5, 7, 9$$ (Remember, a remainder of $-1$ when divided by $10$ is equivalent to a remainder of $9$.)\n\nSo, much like the above, we disregard the $5$ since that certainly isn't prime, and we have $1, 3, 7, 9$ leftover, as desired.\n\n-\n\nyou are looking for the unit digit of the prime.For prime $p\\neq 2$, unit digit is odd which means $p\\pmod {10}\\in\\{1,3,5,7,9\\}$\n\nNow, for any prime $p\\gt 5$, unit git can't be $5$ otherwise it would be divisible by $5$ and hence won't be a prime.\n\nThus only possible candidates for unit digit of a prime $p\\gt 5$ are $\\{1,3,7,9\\}$\n\n-\n\nThis is clear from the Euclidean algorithm. Write the remainder $\\rm\\: r = (p\\ mod\\ 10).\\:$ By Euclid\n\n$$\\rm p\\equiv r\\,\\ (mod\\ 10)\\ \\Rightarrow\\ gcd(p,10) = gcd(r,10)$$\n\nIn particular,  when the  gcd $= 1,\\,$ then: $\\rm\\,\\ p\\,$ is coprime to $10$ $\\iff$ $\\rm\\,r\\,$ is coprime to $10$\n\nNow, by hypothesis, $\\rm\\, p\\,$ is prime $> 5,\\,$ so $\\rm\\, p\\,$ is coprime to $10,\\$ thus $\\rm\\ r\\,$ is coprime to $10,\\:$ so we infer that $\\rm\\,r\\,$ is odd and coprime to $\\,5,\\:$ hence $\\rm\\:r\\in \\{1,3,7,9\\},\\,$ since the remainder $\\rm\\,r\\in [0,9].$\n\nRemark $\\$ Ditto if we generalize $\\rm\\,10\\,$ to any modulus $\\rm\\,m\\!:\\$ if prime $\\rm\\,p\\nmid m\\:$ then its remainder $\\rm\\, p\\ mod\\ m\\,$ is one of the $\\,\\varphi(m)\\,$ remainders coprime to $\\rm\\,m.\\:$ Or, expressed in radix language:\n\n$\\quad$ in radix $\\rm\\,m\\!:\\$ if $\\rm\\,n\\,$ has units digits $\\rm\\,r,\\$ then $\\rm\\ n\\,$ is coprime to $\\rm\\, m\\iff r\\,$ is coprime to $\\rm\\,m$\n\n-", "date": "2016-07-28 14:50:57", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/291956/if-p-is-prime-and-p-5-show-that-when-p-is-divided-by-10-the-remainder", "openwebmath_score": 0.8664212822914124, "openwebmath_perplexity": 112.94445409942554, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504290989414, "lm_q2_score": 0.8652240947405565, "lm_q1q2_score": 0.8522828599937609}}
{"url": "https://math.stackexchange.com/questions/2578416/what-is-the-best-way-to-calculate-the-average-of-individual-averages", "text": "What is the \u201cbest\u201d way to calculate the average of individual averages?\n\nIn preparing for GRE, I see questions looking for the mean of two or more individual means. Here is an example:\n\nThe average (arithmetic mean) of 100 measurements is 24, and the average of 50 additional measurements is 28\n\n               Quantity A                          Quantity B\nThe average of the 150 measurements              26\n\n\nA) Quantity A is greater.\n\nB) Quantity B is greater.\n\nC) The two quantities are equal.\n\nD) The relationship cannot be determined from the information given.\n\nMy question regards the best way to determine the mean (average). I have found two methods that always basically solve it correctly, although the results differ by ~ 0.1.\n\nOne method is adding the weighted means: find sum of each sample, add the sums for a total sum, determine the proportion of each individual sum to the total sum to calculate the \"weight\", then add the products of each individual mean. In other words, Weighted mean = (proportion)(group A mean) + (proportion)(group B mean) + ....\n\nSo, to solve the example question: sum of Group A measurements = $100 \\cdot 24 = 2400$ sum of Group B measurements = $50 \\cdot 28 = 1400$ total sum = 3800 proportion of Group A to total = $2400 \\div 3800 = 0.631579$ weighted mean of Group A = $24 \\cdot 0.631579 = 15.157896$ proportion of Group B to total = $1400 \\div 3800 = 0.368421$ weighted mean of Group B to total = $28 \\cdot 0.368421 = 10.315788$ The average (mean) of the 150 measurements = 15.157896 + 10.315788 = 25.473684\n\nThe other method is to just divide the total sum by the 150 measurements, which is how a normal mean is calculated, to equal 25.333333.\n\nAlthough both methods make B) the answer, they could be significantly different in other contexts. It seems to me that the first, weighted, method is better since it probably takes outliers, repeated measurement, etc. into account.\n\nWhat is the \"best\" way to calculate the average of individual averages?\n\nYour second method is correct and the mean is $25\\frac 13$. The proportion in your first method should be the proportion of the measurements, not the proportion of the sum, so it would be $24\\cdot \\frac {100}{150}+28\\cdot \\frac {50}{150}=25\\frac 13$, which agrees with the other.\nThe best way might be this: with $100$ sitting at $24$ and $50$ sitting at $28$, you need the balance point between the two. The balance point is closer to $24$ than to $28$, because there\u2019s more mass sitting there. Therefore, it\u2019s less than $26$.\nIf you want to be more precise, there\u2019s half as much mass at $28$, so $28$ is twice as far away from the balance. You need the point \u201cone third\u201d of the way from $24$ to $28$, so that\u2019s $24 + \\frac43=25\\frac13$.", "date": "2019-06-20 19:54:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2578416/what-is-the-best-way-to-calculate-the-average-of-individual-averages", "openwebmath_score": 0.8272319436073303, "openwebmath_perplexity": 192.50076740984548, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429120895838, "lm_q2_score": 0.8652240877899775, "lm_q1q2_score": 0.8522828550466932}}
{"url": "https://math.stackexchange.com/questions/2227280/are-there-an-equal-number-of-positive-and-negative-numbers/2227287", "text": "# Are there an equal number of positive and negative numbers?\n\nFor every positive number there exists a corresponding negative number. Would that imply that the number of positive numbers is \"equal\" to the number of negative numbers? (Are they incomparable because they both approach infinity?)\n\n\u2022 See Cardinality and Equinumerosity. Apr 10, 2017 at 7:16\n\u2022 'Equal number' when you talk about infinite sets means you can map one to one both ways, so yes there are an equal number of positive and negative 'numbers.'\n\u2013\u00a0Arby\nApr 10, 2017 at 7:16\n\u2022 Look up \"Hilbert's hotel\" to see an example how we compare infinite sizes, and the counterintuitive consequences that might have. Apr 10, 2017 at 7:17\n\u2022 The two sets $\\text{Pos}$ and $\\text{Neg}$ are equinumerous because there exists a one-to-one correspondence (a bijection) between them. In this sense they are comparable also if both are infinite sets. Apr 10, 2017 at 7:18\n\u2022 Define \"equal\". In the sense of cardinality, yes, there are \"as many\" positive numbers as negative, but then also as many positive integers as both positive and negative integers together, or as many positive integers as positive even integers.\n\u2013\u00a0dxiv\nApr 10, 2017 at 7:22\n\nIf there is SOME function that gives a bijection between two sets, then these two sets are considered equally big. (Even if there is some other function between those two sets that is not onto/not 1-1). For example the set of multiples of 10 among positive integers being a proper subset of all positive integers seems to be \"smaller\". But we have a function $f(x)=10x$ providing bijection, and so they are considered to be of the same size (cardinality). In your case $g(x)=-x$ provides the bijection.\n\nYes, the existence of a one-to-one and onto mapping is exactly how equality of the size of sets (the technical term is \"cardinality\" is defined. The (cardinal) number of negative integers is the same as the cardinal number of positive integers, and the cardinal number of negative real numbers is the same as the cardinal number of positive real numbers.\n\nNote however that there are more positive real numbers than positive integers, so not all infinite cardinalities are equal. The proof of that fact is done by Cantor's famous diagonal proof.\n\nThe word \"infinity\" is used in many places in maths and the definitions are not necessarily the same. Different symbols are used but there are still more definitions than symbols.\n\nThe most familiar symbol for infinity, $\\infty$, is commonly used in calculus but it is more of a suggestive shorthand than an actual infinity. It is not normally used when discussing sizes of sets.\n\nWhen discussing the size of sets, the term cardinality is usual. Clearly, normal counting won't work for infinite sets. However, as others have said, it is possible to say whether or not two sets have the same cardinality / are the same size. If a bijection (one to one and onto map) between them exists then they have the same cardinality. One counter-intuitive property of infinite sets is that it is possible that a map which is one to one and not onto exists as well one that is one to one and onto. So, the existence of a map that is one to one but not onto does not prove that one set is smaller. For that, you would also need to prove that no other map that is one to one and onto exists. The simplest example of this is the set of all natural numbers and just the even ones. Intuitively, the set of even numbers is smaller and there is an obvious map from the even natural numbers to a subset of the natural numbers. However, there is also a map between them which is one to one and onto hence they actually have the same cardinality.\n\nBy definition, a set which has a one to one and onto map to the natural numbers is called \"countable\". This is not countable in the day to day sense but it does mean that you could name one per second and (*), although you would not ever finish, you would name any particular one in a finite time.\n\n(*) Assuming that you and the universe were immortal.\n\nMany sets of intuitively different sizes are countable, for example the integers $\\Bbb{Z}$, the rational numbers $\\Bbb{Q}$ and the algebraic numbers $\\Bbb{A}$. This cardinality / size is named countable and the symbol $\\aleph_0$ is used. This is called \"aleph null\", \"aleph naught\", or \"aleph zero\". Aleph is the first letter of the Hebrew alphabet.\n\nHowever, even though many sets which are intuitively bigger turn out to be the same size, bigger sets exist. It can be proved that there is no one to one and onto map from the natural numbers to the set of real numbers $\\Bbb{R}$ so that it is bigger. Again, some apparently bigger sets are not actually bigger. For example $\\Bbb{C}$, $\\Bbb{R^2}$, and $\\Bbb{R^3}$ are the same cardinality as $\\Bbb{R}$.\n\nEven bigger sets exist, the set of all subsets of $\\Bbb{R}$ is bigger than $\\Bbb{R}$. There is no biggest set.\n\nA particularly interesting question is whether $\\Bbb{R}$ is the next biggest cardinality after the countable infinity of $\\Bbb{N}$. This is called the \"Continuum Hypothesis\". My answer is already long enough so I won't talk about that but if you are interested in this subject then you should look it up.\n\nFinally, there is yet another type of infinity called \"ordinal numbers\". In this sense, the first infinity is usually written as $\\omega$ and, unlike the cardinal infinities, $\\omega + 1$ is different.\n\nThe notion of \"counting\" is made precise in mathematics by using functions. These functions are bijections. To \"count\" the elements in a set $X$, you establish a bijection from $X$ to a subset of the natural numbers $\\mathbb{N}$.\n\nExample. How many elements does the set $\\{a, b, c\\}$ have? The answer is 3, because there exists a bijection from $\\{a, b, c\\}$ to $\\{0, 1, 2\\}$. (Actually, there exist $3!$ bijections.)\n\nSo, certain subsets of $\\mathbb{N}$ are, somehow, \"measuring sticks\", which you can use the gauge the size of other sets.\n\nUsing sets and functions allows one to generalize the idea of \"counting\" to infinite sets. The \"measuring sticks\" then become transfinite numbers (more precisely, cardinal numbers).\n\nTo answer your question, there is an equal number of positive and negative numbers, because there exists a bijection between these two sets. If by \"number\" you mean \"integer\", then this number is called aleph null. It's the smallest infinite cardinal.\n\nBonus: Using these ideas, you can show that the \"amount\" of (say) positive integers and the \"amount\" of integers is the same, or that the \"amount\" of even integers and the \"amount\" of prime numbers is the same.\n\nThere are exactly the same ones.\n\nFor each positive number there is one and only one negative,\n\nAnd for each negative number there is one and only one positive,\n\nSo that both sets have the same cardinal.\n\nThere is a bijective function of the set of positive numbers to the set of negative numbers.", "date": "2022-05-25 14:17:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2227280/are-there-an-equal-number-of-positive-and-negative-numbers/2227287", "openwebmath_score": 0.8539766073226929, "openwebmath_perplexity": 192.7089493282372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429098941399, "lm_q2_score": 0.8652240877899776, "lm_q1q2_score": 0.8522828531471422}}
{"url": "https://math.stackexchange.com/questions/1899679/apostol-bolzano-weierstrass-theorem", "text": "# Apostol Bolzano-Weierstrass Theorem\n\nTheorem. If a bounded set $S$ in $\\mathbb{R}^n$ contains infinitely many points, then there is at least one point in $\\mathbb{R}^n$ which is an accumulation point of $S$.\n\nProof. (for $\\mathbb{R}^1$) Since $S$ is bounded, it lies in some interval $[\u2212a,a]$. At least one of the subintervals $[\u2212a,0]$ or $[0,a]$ contains an infinite subset of $S$. Call one such subinterval $[a_1,b_1]$. Bisect $[a_1,b_1]$ and obtain a subinterval $[a_2,b_2]$ containing an infinite subset of $S$, and continue this process. In this way a countable collection of intervals is obtained, the $nth$ interval $[a_n,b_n]$ being of length $b_n -a_n = a/2^{n-1}$. Clearly the $\\sup$ of the left endpoints $a_n$ and the $\\inf$ of the right endpoints $b_n$ must be equal, say to $x$. The point $x$ will be an accumulation point of $S$ because, if $r$ is any positive number, the interval $[a_n,b_n]$ will be contained in $B(x;r)$ as soon as $n$ is large enough so that $b_n\u2212a_n<r/2$. The interval $B(x;r)$ contains a point of $S$ distinct from $x$ and hence $x$ is an accumulation point of $S$.\n\nMy Questions\n\n1. Why is it obvious that the $\\sup$ of the left endpoint $a_n$ is equal to the $\\inf$ of the right endpoint $b_n$? Also how does one consider the $\\sup$ or $\\inf$ of a single number?\n2. I don't understand why we need $b_n\u2212a_n<r/2$. Instead, why do we need to halve $r$? If $b_n\u2212a_n<r$ then wouldn't $[a_n,b_n]$ still be contained in $B(x;r)$?\n\u2022 Yes, you're correct that if $b_n-a_n<r$, then $[a_n,b_n]\\subset B(x;r)$. For $b_n-x\\le b_n-a_n$ and, similarly, $x-a_n\\le b_n-a_n$. \u2013\u00a0Ted Shifrin Aug 22 '16 at 5:36\n\u2022 Since $[a_{n+1},b_{n+1}]$ is a subinterval of $[a_n,b_n]$ by construction, the sequences $a_n$ and $b_n$ are (bounded and) decreasing and increasing, respectively, so $\\lim_n a_n=\\sup_n a_n$ and $\\lim_n b_n=\\inf b_n$. Since $|b_n-a_n|=a/2^{n-1}$, we have $\\lim_n b_n-a_n=0$ and so $\\lim_n a_n=\\lim_n b_n$. (We do not consider $\\sup$ and $\\inf$ of single points, but range over all $n$.) \u2013\u00a0Luiz Cordeiro Aug 22 '16 at 15:39\n\nHe means $\\sup\\{a_n : n \\in \\mathbb N\\}$, not the supremum of the single point $a_n$.\n\nNote that $a_n$ is an increasing sequence, and $b_n$ is decreasing. Both sequences are bounded, because $[a_n,b_n] \\subset [a_1,b_1]$ for every $n$. Therefore they both converge (specifically, $a_n$ converges to its supremum, and $b_n$ converges to its infimum). As the distance $b_n - a_n$ is shrinking to zero, they must converge to the same limit.\n\nI assume he halves $r$ because his ball $B(x;r)$ is open, so not quite large enough to contain a closed interval of length $r$ if one of the endpoints of the interval is $x$.\n\nEdit: actually, he used a strict inequality $b_n - a_n < r/2$, so indeed he could have used $b_n - a_n < r$.\n\n\u2022 Thank you very much, You guys are fantastic! \u2013\u00a0Jonathan Duran Aug 23 '16 at 0:05\n\nApostol wrote \"the endpoints\" instead of \"the endpoint\". So your first question is solved by observing that he was saying $$\\sup_{n \\geq 1\\ (\\text{say})}a_{n} = \\sup \\{ a_{n} \\mid n \\geq 1 \\} = \\inf \\{ b_{n} \\mid n \\geq 1 \\} = \\inf_{n \\geq 1}b_{n}.$$\n\nFor the second question, that part of the proof does not imply anything that is a must. That choice of an upper bound for $b_{n}-a_{n}$ where $n >> 1$ is to show the reader that we do can do so-and-so.\n\n\u2022 So the proof would have been fine if we chose $r$ instead of $r/2$? Thank you for your response! \u2013\u00a0Jonathan Duran Aug 23 '16 at 0:04\n\u2022 Yes, it would also be fine. \u2013\u00a0Megadeth Aug 23 '16 at 1:40", "date": "2020-01-22 11:36:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1899679/apostol-bolzano-weierstrass-theorem", "openwebmath_score": 0.9298931956291199, "openwebmath_perplexity": 110.64447662205572, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504291340685, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8522828510514596}}
{"url": "https://math.stackexchange.com/questions/4458703/whats-the-size-of-the-biggest-set-of-numbers-from-1-to-100-so-that-no-number-is", "text": "# What's the size of the biggest set of numbers from 1 to 100 so that no number is the average of any other two?\n\nJohn wants to build a set of numbers, from the range of 1 to 100. The only rule is that in that set no number can be the average of any other two. For example, if the set contains the numbers 1 and 3, then 2 cannot be present. What\u2019s the size of the biggest set that John can build?\n\nMore precisely, if L is \"John's set\", then $$\\forall x,y,z \\in L : z \\ne \\frac{x+y}{2}$$ and $$\\forall x \\in L : 1 \\leq x \\leq 100$$\n\nWe want to find the cardinality of L.\n\nPS: This problem appeared on a Sunday newspaper many years ago. On the following week they published a solution that was wrong. I was able to solve it computationally but never mathematically.\n\nYou can solve the problem via integer linear programming as follows. Let $$n=100$$. For $$i\\in\\{1,\\dots,n\\}$$, let binary decision variable $$x_i$$ indicate whether $$i\\in L$$. The problem is to maximize $$\\sum_{i=1}^n x_i$$ subject to $$x_i + x_{i+d} + x_{i+2d} \\le 2 \\quad \\text{for i\\in\\{1,\\dots,n-2\\} and d\\in\\{1,\\dots,\\lfloor(n-i)/2\\rfloor\\}}.$$ The optimal objective value turns out to be $$27$$, attained by $$L=\\{1,5,7,10,11,14,16,24,26,29,30,33,35,39,66,70,72,75,76,79,81,89,91,94,95,98,100\\}.$$\n\nSee https://oeis.org/A003002 for more terms. Such a set is called a Salem-Spencer set, and asymptotic bounds are known for the largest cardinality.\n\n\u2022 Very nice. Just out of curiosity, did you just write the program on the fly? When I look at the constraint, my first thought is that $2^{(100)} \\approx 10^{(30)}$ is too many cases. My pc balks at more than $(10)^8$ cases. Did you find a way to streamline the code? For example, did you experiment with $V =$ value and increment $V$ by $(+1)$, checking each value, and discovering that $V = 27$ is attainable and $V = 28$ is not? May 25 at 23:26\n\u2022 @asinom\u00e1s can you undelete your answer? I think including it makes for an interesting comparison getween the greedy and optimal algorithms. May 25 at 23:30\n\u2022 @user2661923 I used a generic ILP solver that uses the branch-and-cut algorithm, not brute force. May 25 at 23:53\n\u2022 @OscarLanzi I agree with you and just voted to undelete the answer. asinomas was not flagged by your comment because he has not been registered for RobPratt's answer. One alternative is for you to track down any posting/answer that asinomas is registered for, and leave your comment request, addressed to him, there. Naturally, you will need to include a link back to this posting. May 25 at 23:57\n\u2022 Very nice. The answer is indeed 27 (for N=100). In my case, I had used dynamic programming to find the sets. The reason I keep looking at this problem after all these years is that although I was able to prove a few properties/theorems about the cardinality of L, I was never able to solve it in a more \"mathematical\" way. I.e., for a certain size of N, is there an expression for #L? May 26 at 9:26\n\nA greedy algorithm would hive an answer of $$24$$, as given by the \"Erdos-Szekeres\" sequence here: http://oeis.org/A003278 .\n\nBut, as demonstrated by RobPratt, and as often happens in analysis, the greedy algorithm is ultimately not the optimal one. It's as if you overuse an initially fertile soil, exhaust its nutrients and then you find you can't grow crops until you give time for rains to fall and the soil to replenish itself. Let's explore why greed turned out not good in this case.\n\nWelcome to the desert\n\nSuppose you have a list of whole numbers, with a minimum of $$1$$ and a maximum of $$n$$, such that there is not a group of three in arithmetic sequence. Then the arithmetic third of any pair of numbers from this list cannot be larger than $$2n-1$$, and so you can always insert the number $$2n$$ and be sure there is still no three-term arithmetic sequence.\n\nIf, in fact, you do need to resort to $$2n$$ to continue your list -- that is, no smaller number will do -- then you have in effect created a \"desert\". The presence of such deserts can make you fall behind a more carefully tended, more optimal sequence. And that is what happened here.\n\nUpon further review, the Erdos-Szekeres sequence, constructed to take as small a number as possible at any step, is equally optimal at producing deserts. A number $$k$$ enters this sequence iff the ternary representation of $$k-1$$ has no elements of $$2$$, for instance eleven minus one is ten, which is $$101$$ in base $$3$$, so ten is in the sequence. But once we get to fourteen in this case, where one less is $$111$$ base $$3$$, we are forced to accept a $$2$$ somewhere in the ternary representation until we have passed the next power of $$3$$, in this case at $$28$$, where we can introduce a new ternary \"digit\". We build in a desert from $$n=(3^m+1)/2$$ all the way up to $$2n=3^m+1$$ for every whole number $$m$$. In this problem the desert from $$41$$ to $$82$$ ($$m=4$$) renders the greedy algorithm inferior.\n\nFinding an oasis\n\nOf course, we can change the rules of our game to make the greedy algorithm look good. If, instead of an upper bound of $$100$$, we were to choose an upper bound of $$(3^m+1)/2$$ for some whole number $$m$$ -- stopping just at the threshold of the desert -- then the Erdos-Szekeres method really does work. For instance, with a maximum of $$(3^5+1)/2=122$$ the Erdos-Szekeres method gives $$32$$ terms and that is indeed the optimum for a max element of $$122$$. See https://oeis.org/A065825.\n\n\u2022 The description of the sequence seems to say that this is the result of the greedy approach (build a sequence from $1, 2, \\ldots$ by choosing the next smallest valid number), which might not be optimal as RobPratt's answer suggests. May 25 at 23:15\n\u2022 @angryavian Nice catch, I missed that. I deleted my comment. May 25 at 23:15\n\u2022 I was k-c-u-f-ed (read it backwards) by Autocorrect. Forced to rephrase that passage. Please try again. May 26 at 0:59\n\u2022 @OscarLanzi This time, when you reply, please include the AT:user2661923, so that I will be flagged. Perhaps I am being dense here. I don't understand this sentence: \"Then the arithmetic third of any pair of numbers from this list cannot be larger than 2n\u22121, and so you can always insert the number 2n and be sure there is still no three-term arithmetic sequence.\" In this sentence, what is $n$, and what pair of numbers are you referring to? Are you, for example saying that if you take an optimal list, and append the number $(200)$ to it, that no constraint has been violated? May 26 at 1:08\n\u2022 @OscarLanzi Okay. I skimmed your analysis. I am going t have to study it further, which will take at least a few days. I have bookmarked this posting, for reference. For what it's worth, I was also intrigued by Rob Pratt's reference to a branch and cut algorithm, re integer linear programming. May 26 at 1:23", "date": "2022-06-26 10:59:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4458703/whats-the-size-of-the-biggest-set-of-numbers-from-1-to-100-so-that-no-number-is", "openwebmath_score": 0.6938482522964478, "openwebmath_perplexity": 391.6720657693414, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429116504951, "lm_q2_score": 0.8652240686758841, "lm_q1q2_score": 0.8522828358385808}}
{"url": "https://math.stackexchange.com/questions/2560143/how-to-find-probalility-that-a-student-misses-at-least-one-test-if-he-she-is-abs", "text": "# how to find probalility that a student misses at least one test if he/she is absent twice?\n\nThe probability that a teacher will give an unannounced test during any class is $$\\dfrac15$$. If a student is absent twice then probability that he/she misses at least one test is\n\n$$\\\\ \\hspace{5cm}$$ a) $$\\dfrac23\\ \\quad$$ b) $$\\dfrac45\\ \\quad$$c) $$\\dfrac7{25}\\ \\quad$$d) $$\\dfrac9{25}\\$$\n\nMy attempt:\n\nProbability of attending first test & missing $$2$$nd test $$=\\dfrac45\\times\\dfrac15=\\dfrac4{25}$$\n\nProbability of missing first test & attending $$2$$nd test $$=\\dfrac15\\times\\dfrac45=\\dfrac4{25}$$\n\nProbability of missing both the tests $$=\\dfrac15\\times\\dfrac15=\\dfrac1{25}$$\n\nTotal probability of missing at least one test $$=\\dfrac4{25}+\\dfrac4{25}+\\dfrac1{25}=\\dfrac9{25}$$\n\n\u2022 Yes, you are correct. Notice, a simple method is $$P_{\\text{ Missing atleast one test }} = 1- P_{\\text{ Missing no test }} = 1-(\\frac{4}{5})^2=\\frac{9}{25}$$\n\u2013\u00a0user371838\nDec 10, 2017 at 15:42\n\nQuestions with \"at least\" are often great candidates for using the complement. That is: $$P_{\\text{ Event A occurs }} = 1 - P_{\\text{ Event A does not occur }}$$\n\nSo, $P_{\\text{ miss at least 1 test }} = 1 - P_{\\text{ miss no tests }}$ giving us:\n\n\\begin{align*} P_{\\text{ miss no tests }} & = P_{\\text{ no test on day 1 AND no test on day 2 }}\\\\ & = P_{\\text{ no test on day 1 }} \\times P_{\\text{ no test on day 2 }}\\\\ & = \\frac{4}{5} \\times \\frac{4}{5}\\\\ & = \\frac{16}{25}\\end{align*}\n\nSo, $$P_{\\text{ miss at least 1 test }} = 1 - P_{\\text{ miss no tests }} = 1 - \\frac{16}{25} = \\frac{9}{25}$$\n\nAnother way to see this is that the probability of the student missing at least one test is the probability of missing test 1 + the probability of missing test 2 - the probability of them missing both tests.\n\nThe student has $\\frac15$chance of missing the first test, and $\\frac15$ chance of missing the second test. If you add them up you're left with $\\frac 25$. HOWEVER there is a case that is being counted twice: It could happen that the student misses test 1 AND test 2. That case is being counted both when the student misses test 1 and when the student misses test 2. So now we have to substract this case once\n\nThe chance of a test happening on two consecutive days is $(\\frac 15)^2 = \\frac {1} {25}$.\n\nSo the answer is $\\frac 25 - \\frac {1} {25} = \\frac {9} {25}$\n\nThe general case for this is $$P(A \\cup B) = P(A) + P(B) - P(A \\cap B)$$", "date": "2022-07-06 08:13:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2560143/how-to-find-probalility-that-a-student-misses-at-least-one-test-if-he-she-is-abs", "openwebmath_score": 0.9273092150688171, "openwebmath_perplexity": 625.7258040623615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815532606979, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8522176971651085}}
{"url": "http://mathhelpforum.com/algebra/17030-half-right-missed-other-half.html", "text": "# Math Help - Half right, missed the other half...\n\n1. ## Half right, missed the other half...\n\n\"Find all values of $r$ such that the slope of the line through the points $(r, 4)$ and $(1, 3 - 2r)$ is less than 5.\"\n\nI solved it by first finding the slope $\\frac {2r + 1}{r - 1} = 5$ which simplifies to $r = 2$. Then I used the point-slope form to derive the equation $y = 5x + 16$. Then, I got lazy, and plugged in a couple of test numbers into the equation, found that numbers above 2 seemed to qualify, and decided that the answer was the set $(2, \\infty)$. Unfortunately, the answer should have included also the set $(-\\infty, 1)$.\n\nAlgebraically, what should I have done to get to that answer? What part did I skip?\n\n2. Hello, earachefl!\n\nFind all values of $r$ such that the slope of the line\nthrough the points $A(r,\\,4)$ and $B(1,\\,3\\!-\\!2r)$ is less than 5.\nYou should have solved the inequality . . . tricky, but safe.\n\nYou found the slope of $AB$ . . . good!\n\n. . Then we have: . $\\frac{2r + 1}{r - 1} \\:<\\:5$ . [1]\n\nI'd like to multiply both sides by $(r-1)$\n. . but the result depends on whether $(r-1)$ is positive or negative.\n\nRecall: when multiplying or dividing an inequality by a negative quantity,\n. . . . . .reverse the inequality.\n\nSuppose $(r-1)$ is positive . . . that is: . $r > 1$\n\nMultiply [1] by by positive $(r-1)\\!:\\;\\;2r + 1 \\:<\\:5(r - 1)$\n\n. . $2r + 1 \\:<\\:5r-5\\quad\\Rightarrow\\quad -3r \\:<\\:-6$\n\nDivide both sides by -3: . $r \\:>\\:2$\n\nWe have two inequalities to satisfy: . $r > 1$ and $r > 2$\n\nWe take the \"stronger\" of the two: . $\\boxed{r > 2}$\n\nSuppose $(r-1)$ is negative . . . that is: . $r < 1$\n\nMultiply [1] by negative $(r-1)\\!:\\;\\;2r + 1 \\;\\;{\\color{red}>} \\;\\;5(r-1)$\n\n. . $2r + 1 \\:>\\:5r - r\\quad\\Rightarrow\\quad -3r \\:>\\:-6$\n\nDivide both sides by -3: . $r \\:<\\:2$\n\nWe have two inequalities to satisfy: . $r < 1$ and $r < 2$\n\nWe take the \"stronger\" of the two: . $\\boxed{r < 1}$\n\nTherefore: . $(r < 1) \\vee (r > 2)$ .. . . .or: . $(-\\infty,\\,1) \\,\\cup \\,(2,\\,\\infty)$\n\n3. Originally Posted by Soroban\nHello, earachefl!\n\nWe have two inequalities to satisfy: . $r < 1$ and $r < 2$\n\nWe take the \"stronger\" of the two: . $\\boxed{r < 1}$\nInteresting approach. I don't think my book ever put it in quite these terms.\n\nThanks!", "date": "2014-11-24 01:47:25", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/17030-half-right-missed-other-half.html", "openwebmath_score": 0.9447164535522461, "openwebmath_perplexity": 505.40299085957287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759671623987, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8522113144337815}}
{"url": "https://www.physicsforums.com/threads/eigenvalues-of-a-5x5-matrix.403476/", "text": "Eigenvalues of a 5x5 matrix\n\n1. May 15, 2010\n\ngabriels-horn\n\n1. The problem statement, all variables and given/known data\n\n3. The attempt at a solution\nI haven't tackled anything bigger than a 3x3 matrix. Anyone have any good pointers for reducing this matrix? I'm assuming the quickest way is still going to be the cofactor method?\n\n2. May 15, 2010\n\ncronxeh\n\nIts an upper triangular matrix, so eigenvalues are pretty much on the diagonal here, they are all 2. Taking the determinant yields (2-lambda)^5, setting (2-lambda)^5=0, you get lambda1=lambda2=lambda3=lambda4=lambda5=2.\n\n3. May 15, 2010\n\nLandau\n\nNo need to calculate, only to think\n\nThe matrix is upper-triangular, what does this immediately tell you about the eigenvalues?\n\n4. May 16, 2010\n\ngabriels-horn\n\nOk, the discussion about the eigenvalues being the main diagonal values is clear. If the five eigenvalues are = 2, then plugging that value into the eigenvector matrix leaves only the one values in the 5x5 matrix above. Setting the matrix = 0 only leaves the eigenvector = 0; how are there then three linearly independent eigenvectors. What am I overlooking?\n\n5. May 16, 2010\n\nninty\n\nBy definition an eigenvector cannot be the zero vector.\nI'm assuming you're using the usual methods I see undergraduates use, (A-xI) = 0\nIf we represent vectors as (x1,x2,x3,x4,x5) we get x2=0, x5=0\nThen solving for x1,x3,x4 we can get 3 linearly independent vectors which form a basis for null(A-xI)\n\nAn easy one would be to take 3 vectors from the usual basis.\nHowever the question doesn't really ask for the eigenvectors, just show that they do exist.\n\n6. May 16, 2010\n\nLandau\n\nThere are three Jordan Blocks corresponding to eigenvalue 2, so its geometric multiplicity is 3.\n\n7. May 16, 2010\n\nHallsofIvy\n\nStaff Emeritus\nFrom the definition of \"eigenvector\", if v is an eigenvector with eigenvalue 2, then\n$$\\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\\\ 0 & 2 & 0 & 0 & 0 \\\\ 0 & 0 & 2 & 0 & 0 \\\\ 0 & 0 & 0 & 2 & 1 \\\\ 0 & 0 & 0 & 0 & 2\\end{bmatrix}\\begin{bmatrix}u \\\\ v \\\\ x \\\\ y \\\\ z\\end{bmatrix}= \\begin{bmatrix}2u+ v \\\\ 2v \\\\ 2x \\\\ 2y+ z \\\\ 2z\\end{bmatrix}= \\begin{bmatrix} 2u \\\\ 2v \\\\ 2x \\\\ 2y \\\\ 2z \\end{bmatrix}$$\n\nwhich gives the equations 2u+ v= 2u, 2v= 2v, 2x= 2x, 2y+ z= 2y, 2z= 2z. The first equation tells us that v= 0 and the fourth equation that z= 0. But u, x, and y can be any numbers.\n\nLast edited: May 17, 2010\n8. May 16, 2010\n\ngabriels-horn\n\nYes, the (A-xI) = 0 method is the one I used and got x2 = 0 and x5 = 0. How can you solve for x1, x3 and x4 when they are multiplied by zero? Wouldn't that also make them zero?\n\nThis isn't something that we went over in Linear or ODE, can you expand on this thought some more?\n\n9. May 16, 2010\n\ngabriels-horn\n\nYou just put [/math] on the end instead of [/tex] and I took the zero out of the variable column. That's what I was confused on, the fact that x1, x3 and x4 could be any value. I just assumed that they were also zero since putting 2-2 = 0 for the eigenvector only left the two 1's in the matrix corresponding to x2 = 0 and x5 = 0. Thank you guys for the insight and provoding a good matrix template for future use.\n\nLast edited: May 16, 2010\n10. May 16, 2010\n\nLandau\n\nWhatever values they have, they get multiplied by zero resulting in zero, so it doesn't matter what they are. In other words, you are free to choose x1, x3 and x4, which means you can form three linearly independent eigenvectors (e.g. taking x1, x3, x4 equal to 1,0,0, equal to 0,1,0, equal to 0,0,1, gives you three independent eigenvectors.)\nYour matrix is already in Jordan Normal Form. The number of Jordan blocks corresponding to a given eigenvalue coincides with the geometric multiplicity of that eigenvalue. In this case there are three such blocks for eigenvalue 2.\n\n11. May 16, 2010\n\ngabriels-horn\n\nSo the Jordan block is essentially dependent on the 1 values of the superdiagonal, and since x1, x3 and x4 have no entries above them on the superdiagonal, the number of Jordan blocks is equal to 3?\n\n12. May 17, 2010\n\nHallsofIvy\n\nStaff Emeritus\nNo, multiplying x1 by 0 does NOT make x1 0, it makes the product 0. All that means is that you don't have that variable in that particular equation.", "date": "2017-10-23 06:53:29", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/eigenvalues-of-a-5x5-matrix.403476/", "openwebmath_score": 0.8080571293830872, "openwebmath_perplexity": 544.2051102176094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759632491111, "lm_q2_score": 0.8688267881258483, "lm_q1q2_score": 0.8522113126995728}}
{"url": "https://usa.cheenta.com/category/aime-i/", "text": "Categories\n\n## Number of points and planes | AIME I, 1999 | Question 10\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.\n\n## Number of points and planes \u2013 AIME I, 1999\n\nTen points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is $\\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n.\n\n\u2022 is 107\n\u2022 is 489\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nNumber of points\n\nPlane\n\nProbability\n\n## Check the Answer\n\nAnswer: is 489.\n\nAIME I, 1999, Question 10\n\nGeometry Vol I to IV by Hall and Stevens\n\n## Try with Hints\n\n$10 \\choose 3$ sets of 3 points which form triangles,\n\nfourth distinct segment excluding 3 segments of triangles=45-3=42\n\nRequired probability=$\\frac{{10 \\choose 3} \\times 42}{45 \\choose 4}$\n\nwhere ${45 \\choose 4}$ is choosing 4 segments from 45 segments\n\n=$\\frac{16}{473}$ then m+n=16+473=489.\n\nCategories\n\n## Sequence and fraction | AIME I, 2000 | Question 10\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.\n\n## Sequence and fraction \u2013 AIME I, 2000\n\nA sequence of numbers $x_1,x_2,\u2026.,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.\n\n\u2022 is 107\n\u2022 is 173\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nEquation\n\nAlgebra\n\nIntegers\n\n## Check the Answer\n\nAnswer: is 173.\n\nAIME I, 2000, Question 10\n\nElementary Number Theory by Sierpinsky\n\n## Try with Hints\n\nLet S be the sum of the sequence $x_k$\n\ngiven that $x_k=S-x_k-k$ for any k\n\ntaking k=1,2,\u2026.,100 and adding\n\n$100S-2(x_1+x_2+\u2026.+x_{100})=1+2+\u2026.+100$\n\n$\\Rightarrow 100S-2S=\\frac{100 \\times 101}{2}=5050$\n\n$\\Rightarrow S=\\frac{2525}{49}$\n\nfor $k=50, 2x_{50}=\\frac{2525}{49}-50=\\frac{75}{49}$\n\n$\\Rightarrow x_{50}=\\frac{75}{98}$\n\n$\\Rightarrow m+n$=75+98\n\n=173.\n\nCategories\n\n## Finding smallest positive Integer | AIME I, 1996 Problem 10\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.\n\n## Finding smallest positive Integer \u2013 AIME I, 1996\n\nFind the smallest positive integer solution to $tan19x=\\frac{cos96+sin96}{cos96-sin96}$.\n\n\u2022 is 107\n\u2022 is 159\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nFunctions\n\nTrigonometry\n\nIntegers\n\n## Check the Answer\n\nAnswer: is 159.\n\nAIME I, 1996, Question 10\n\nPlane Trigonometry by Loney\n\n## Try with Hints\n\n$\\frac{cos96+sin96}{cos96-sin96}$\n\n=$\\frac{sin(90+96)+sin96}{sin(90+96)-sin96}$\n\n=$\\frac{sin186+sin96}{sin186-sin96}$\n\n=$\\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}$\n\n=$\\frac{2sin141cos45}{2cos141sin45}$\n\n=tan141\n\nhere $tan(180+\\theta)$=$tan\\theta$\n\n$\\Rightarrow 19x=141+180n$ for some integer n is first equation\n\nmultiplying equation with 19 gives\n\n$x \\equiv 141\\times 19 \\equiv 2679 \\equiv 159(mod180)$ [since 2679 divided by 180 gives remainder 159]\n\n$\\Rightarrow x=159$.\n\nCategories\n\n## Roots of Equation and Vieta\u2019s formula | AIME I, 1996 Problem 5\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta\u2019s formula.\n\n## Roots of Equation and Vieta\u2019s formula \u2013 AIME I, 1996\n\nSuppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t.\n\n\u2022 is 107\n\u2022 is 23\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nFunctions\n\nRoots of Equation\n\nVieta s formula\n\n## Check the Answer\n\nAnswer: is 23.\n\nAIME I, 1996, Question 5\n\nPolynomials by Barbeau\n\n## Try with Hints\n\nWith Vieta s formula\n\n$f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0$\n\n$\\Rightarrow a+b+c=-3$, $ab+bc+ca=4$ and $abc=11$\n\nLet a+b+c=-3=p\n\nhere t=-(a+b)(b+c)(c+a)\n\n$\\Rightarrow t=-(p-c)(p-a)(p-b)$\n\n$\\Rightarrow t=-f(p)=-f(-3)$\n\n$t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]$\n\n=23.\n\nCategories\n\n## Tetrahedron Problem | AIME I, 1992 | Question 6\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.\n\n## Tetrahedron Problem \u2013 AIME I, 1992\n\nFaces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron.\n\n\u2022 is 107\n\u2022 is 320\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nArea\n\nVolume\n\nTetrahedron\n\n## Check the Answer\n\nAnswer: is 320.\n\nAIME I, 1992, Question 6\n\nCoordinate Geometry by Loney\n\n## Try with Hints\n\nArea BCD=80=$\\frac{1}{2} \\times {10} \\times {16}$,\n\nwhere the perpendicular from D to BC has length 16.\n\nThe perpendicular from D to ABC is 16sin30=8\n\n[ since sin30=$\\frac{perpendicular}{hypotenuse}$ then height = perpendicular=hypotenuse $\\times$ sin30 ]\n\nor, Volume=$\\frac{1}{3} \\times 8 \\times 120$=320.\n\nCategories\n\n## Triangle and integers | AIME I, 1995 | Question 9\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.\n\n## Triangle and integers \u2013 AIME I, 1995\n\nTriangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and $\\angle BDC$=3$\\angle BAC$. then the perimeter of $\\Delta ABC$ may be written in the form $a+\\sqrt{b}$ where a and b are integers, find a+b.\n\n\u2022 is 107\n\u2022 is 616\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nIntegers\n\nTriangle\n\nTrigonometry\n\n## Check the Answer\n\nAnswer: is 616.\n\nAIME I, 1995, Question 9\n\nPlane Trigonometry by Loney\n\n## Try with Hints\n\nLet x= $\\angle CAM$\n\n$\\Rightarrow \\angle CDM =3x$\n\n$\\Rightarrow \\frac{tan3x}{tanx}=\\frac{\\frac{CM}{1}}{\\frac{CM}{11}}$=11 [by trigonometry ratio property in right angled triangle]\n\n$\\Rightarrow \\frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx$\n\nsolving we get, tanx=$\\frac{1}{2}$\n\n$\\Rightarrow CM=\\frac{11}{2}$\n\n$\\Rightarrow 2(AC+CM)$ where $AC=\\frac{11\\sqrt {5}}{2}$ by Pythagoras formula\n\n=$\\sqrt{605}+11$ then a+b=605+11=616.\n\nCategories\n\n## Sequence and greatest integer | AIME I, 2000 | Question 11\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.\n\n## Sequence and greatest integer \u2013 AIME I, 2000\n\nLet S be the sum of all numbers of the form $\\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\\frac{S}{10}$.\n\n\u2022 is 107\n\u2022 is 248\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nEquation\n\nAlgebra\n\nIntegers\n\n## Check the Answer\n\nAnswer: is 248.\n\nAIME I, 2000, Question 11\n\nElementary Number Theory by Sierpinsky\n\n## Try with Hints\n\nWe have 1000=(2)(2)(2)(5)(5)(5) and $\\frac{a}{b}=2^{x}5^{y} where -3 \\leq x,y \\leq 3$\n\nsum of all numbers of form $\\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000\n\n=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$\n\n$\\Rightarrow S= \\frac{(2^{-3})(2^{7}-1)}{2-1} \\times$ $\\frac{(5^{-3})(5^{7}-1)}{5-1}$\n\n=2480 + $\\frac{437}{1000}$\n\n$\\Rightarrow [\\frac{s}{10}]$=248.\n\nCategories\n\n## Series and sum | AIME I, 1999 | Question 11\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.\n\n## Series and sum \u2013 AIME I, 1999\n\ngiven that $\\displaystyle\\sum_{k=1}^{35}sin5k=tan\\frac{m}{n}$ where angles are measured in degrees, m and n are relatively prime positive integer that satisfy $\\frac{m}{n} \\lt 90$, find m+n.\n\n\u2022 is 107\n\u2022 is 177\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nAngles\n\nTriangles\n\nSide Length\n\n## Check the Answer\n\nAnswer: is 177.\n\nAIME I, 2009, Question 5\n\nPlane Trigonometry by Loney\n\n## Try with Hints\n\ns=$\\displaystyle\\sum_{k=1}^{35}sin5k$\n\ns(sin5)=$\\displaystyle\\sum_{k=1}^{35}sin5ksin5=\\displaystyle\\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\\frac{1+cos5}{sin5}$\n\n$=\\frac{1-cos(175)}{sin175}$=$tan\\frac{175}{2}$ then m+n=175+2=177.\n\nCategories\n\n## Inscribed circle and perimeter | AIME I, 1999 | Question 12\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Inscribed circle and perimeter.\n\n## Inscribed circle and perimeter \u2013 AIME I, 1999\n\nThe inscribed circle of triangle ABC is tangent to AB at P, and its radius is 21 given that AP=23 and PB=27 find the perimeter of the triangle\n\n\u2022 is 107\n\u2022 is 345\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nInscribed circle\n\nPerimeter\n\nTriangle\n\n## Check the Answer\n\nAnswer: is 345.\n\nAIME I, 1999, Question 12\n\nGeometry Vol I to IV by Hall and Stevens\n\n## Try with Hints\n\nQ tangency pt on AC, R tangency pt on BC AP=AQ=23 BP=BR=27 CQ=CR=x and\n\n$s \\times r =A$ and $s=\\frac{27 \\times 2+23 \\times 2+x \\times 2}{2}=50+x$ and A=$({(50+x)(x)(23)(27)})$ then from these equations 441(50+x)=621x then x=$\\frac{245}{2}$\n\nperimeter 2s=2(50+$\\frac{245}{2}$)=345.\n\nCategories\n\n## LCM and Integers | AIME I, 1998 | Question 1\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.\n\n## Lcm and Integer \u2013 AIME I, 1998\n\nFind the number of values of k in $12^{12}$ the lcm of the positive integers $6^{6}$, $8^{8}$ and k.\n\n\u2022 is 107\n\u2022 is 25\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\nLcm\n\nAlgebra\n\nIntegers\n\n## Check the Answer\n\nAnswer: is 25.\n\nAIME I, 1998, Question 1\n\nElementary Number Theory by Sierpinsky\n\n## Try with Hints\n\nhere $k=2^{a}3^{b}$ for integers a and b\n\n$6^{6}=2^{6}3^{6}$\n\n$8^{8}=2^{24}$\n\n$12^{12}=2^{24}3^{12}$\n\nlcm$(6^{6},8^{8})$=$2^{24}3^{6}$\n\n$12^{12}=2^{24}3^{12}$=lcm of $(6^{6},8^{6})$ and k\n\n=$(2^{24}3^{6},2^{a}3^{b})$\n\n=$2^{max(24,a)}3^{max(6,b)}$\n\n$\\Rightarrow b=12, 0 \\leq a \\leq 24$\n\n$\\Rightarrow$ number of values of k=25.", "date": "2021-06-17 18:28:07", "meta": {"domain": "cheenta.com", "url": "https://usa.cheenta.com/category/aime-i/", "openwebmath_score": 0.4871501922607422, "openwebmath_perplexity": 3980.9644093658503, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9808759677214397, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8522113082564491}}
{"url": "https://math.stackexchange.com/questions/3176329/are-these-linear-algebra-questions-true-or-false", "text": "# Are these linear algebra questions true or false?\n\nWhich of the following are true or false?\n\n(a) Let $$H$$ be a five-dimensional subspace of the ten-dimensional vector space $$V$$. Then, every set containing seven vectors from $$H$$ must be linearly dependent.\n\n(b) Let $$\\mathcal{B}$$ be a basis of $$\\mathbb{R}^{n}$$. Let $$A$$ be a the matrix whose columns are the vectors in $$\\mathcal{B}$$. Then, for every vector $$x \\in \\mathbb{R}^{n}$$, it is true that $$[x]_{\\mathcal{B}} = Ax$$.\n\n(c) The dimension of the vector space $$\\mathbb{P}_{4}$$ is $$4$$.\n\n(d) Let $$A$$ be a $$3\\times 3$$ matrix, and let $$H$$ be the set of fixed vectors of $$A$$, that is, the set of $$x \\in \\mathbb{R}^{3}$$ for which $$Ax = x$$. Then $$H$$ is a subspace of $$\\mathbb{R}^{3}$$.\n\nI think (a) is false, but I don't really have a reason why. I think this is true because if you have an $$n$$-dimensional space, you only need $$n$$ linearly independent ones to span the space.\n\nI think (b) is true since it looks like a definition I saw in my book. I don't have a reason why.\n\nI know (c) is false. I'm pretty sure it's dimension 5. I have seen the proof before. EDIT: Proof here: Determining Bases of P4\n\nI think (d) is true. Because if $$Ax = x$$ and $$Ay = y$$ then $$A(x + y) = x + y$$. Also the with scalar multiplication. Also, this is sort of like eigenvalues, from my understanding.\n\nCan someone help me verify these please?\n\n\u2022 Is this a school assignment? \u2013\u00a0Jack Pfaffinger Apr 5 at 19:47\n\u2022 No it's a quiz I took. I want to calculate my grade \u2013\u00a0user660922 Apr 5 at 19:48\n\u2022 What is $\\Bbb P_4$? \u2013\u00a0Bernard Apr 5 at 20:05\n\u2022 The vector space of polynomials of degree less than or equal to $4$ \u2013\u00a0user660922 Apr 5 at 20:13\n\n(a) is true. If a set of $$7$$ vectors in $$H$$ were independent, they are a basis for a subspace of $$H$$, so that the $$5$$-dimensional space $$H$$ has a subspace of dimension $$7$$, which is absurd. (b) is true; $$A$$ is called a change of basis matrix. (c) is false for the reason you state, and (d) is true for the reason you state. Good thinking!", "date": "2019-10-23 18:55:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3176329/are-these-linear-algebra-questions-true-or-false", "openwebmath_score": 0.8485490679740906, "openwebmath_perplexity": 96.89653543513448, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759610129464, "lm_q2_score": 0.8688267813328977, "lm_q1q2_score": 0.8522113040936911}}
{"url": "https://questioncove.com/updates/4fc3453ee4b0964abc860fc6", "text": "Mathematics\nOpenStudy (anonymous):\n\nThe sum of all odd numbers form 1 to 99 is?\n\nOpenStudy (anonymous):\n\nuse sum of aritmetic sequence. .\n\nParth (parthkohli):\n\n1 + 3 + 5 + 7.....But you need not do that since it is an arithmetic sequence.\n\nParth (parthkohli):\n\nThe sum of an arithmetic sequence is: $$\\Large \\color{Black}{\\Rightarrow {n \\over 2}(a_1 + a_l) }$$ a1 is the first term aL is the last n is the number of terms\n\nOpenStudy (anonymous):\n\nAfter 1 add? Should i cross multiply?\n\nParth (parthkohli):\n\n1st term is 1. Last is 99. Odd numbers between 1,100 are 50.\n\nParth (parthkohli):\n\nThe common difference is d, btw.\n\nOpenStudy (anonymous):\n\n@ParthKohli , i think you've mistaken the formula. . http://www.regentsprep.org/Regents/math/algtrig/ATP2/ArithSeq.htm\n\nParth (parthkohli):\n\nOops, it was aN lol\n\nOpenStudy (anonymous):\n\nit's ok. .\n\nOpenStudy (anonymous):\n\nI dont get it XD\n\nParth (parthkohli):\n\nHey I know another pattern.\n\nParth (parthkohli):\n\nThe sum of first n odd numbers is n squared. The sum of first 50 odd numbers is 50 squared. So, 50^2 = 2500\n\nParth (parthkohli):\n\nWell actually my formula was correct.\n\nParth (parthkohli):\n\nI learnt this pattern at school.\n\nOpenStudy (anonymous):\n\nCa you give me an example?\n\nParth (parthkohli):\n\nFor example the sum of first 4 odd numbers is 4 squared aka 16 1 + 3 + 5 + 7 = 4 + 12 = 16\n\nOpenStudy (anonymous):\n\nhow did you got that formula? @ParthKohli\n\nParth (parthkohli):\n\nOh wait a second\n\nParth (parthkohli):\n\nYes....yes I'm correct.\n\nOpenStudy (anonymous):\n\nthe answer is 2500, and you're correct, but how did you get that another formula?\n\nParth (parthkohli):\n\nI learnt that at school\n\nParth (parthkohli):\n\n@sel126 Got it?\n\nOpenStudy (anonymous):\n\nYessss :) Thankyousomuch:)))))))\n\nParth (parthkohli):\n\nYw :D\n\nOpenStudy (anonymous):\n\n99 = a + (n-1)d hence 99 = 1 +(n-1)*3 n= 50 substituting value of n in equation S= n/2 [ 2a + (n-1)*d] we get sum of odd terms that is equal to 2500 where a =1(first term) and d =3(common difference)\n\nOpenStudy (anonymous):\n\nWhy does three become the d?\n\nOpenStudy (anonymous):\n\nthe sum of odd numbers is a perfect square\n\nParth (parthkohli):\n\nThat's what I said above\n\nParth (parthkohli):\n\nHehe don't waste your energy, genius. I've explained her.\n\nOpenStudy (anonymous):\n\njust thought i would mention it. no need for a formula, or all that other business, only need to know how many numbers are being added\n\nParth (parthkohli):\n\nHmm I said the exact same thing.\n\nParth (parthkohli):\n\nStill here you have a medal", "date": "2021-03-03 21:27:52", "meta": {"domain": "questioncove.com", "url": "https://questioncove.com/updates/4fc3453ee4b0964abc860fc6", "openwebmath_score": 0.6941877603530884, "openwebmath_perplexity": 4476.451968842687, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759621310287, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8522113033993503}}
{"url": "http://math.stackexchange.com/questions/729920/probability-of-head-in-coin-flip-when-coin-is-flipped-two-times", "text": "# Probability of Head in coin flip when coin is flipped two times\n\nProbability of getting a head in coin flip is 1/2. If the coin is flipped two times what is the probability of getting a head in either of those attempts?\n\nI think both the coin flips are mutually exclusive events, so the probability would be getting head in attempt 1 or attempt 2 which is :\n\nP(attempt1) + P(attempt2) = 1/2 + 1/2 = 1\n\n100% probability sounds wrong? What am I doing wrong. If i apply the same logic then probability of getting at least 1 head in 3 attempt will be 1/2+1/2+1/2 = 3/2 = 1.5 which I know for sure is wrong. What do i have mixed up?\n\n-\nThe events are not mutually exclusive: you can get a head on the first and on the second flip \u2013\u00a0 Rookatu Mar 28 at 5:55\nIndeed. The coin tosses are independent. That is not the same as mutually exclusive. \u2013\u00a0 Graham Kemp Mar 28 at 9:58\n\nYou are confusing the terms \"independent\" and \"mutually exclusive\".\u00a0 These are not the same thing.\u00a0 In fact events cannot be both \"independent\" and \"mutually exclusive\".\u00a0 It's either one, the other, or neither.\n\n\"Mutually exclusive\" simply means that the two events cannot happen together. If A happens then B does not and if B happens A cannot.\n\n\"Independent\" simply means that the occurrence of one event is not conditional on the occurrence of the other. The probability of A happening does not depend on whether B happens or not, and vice versa.\n\nLet $H_n$ be the indexed event of getting a head on the $n^{th}$ flip.\n\nGiven an unbiased coin, $P(H_1)=P(H_2)=\\frac 1 2$\n\nThese events are independent so $P(H_1 \\cap H_2) = P(H_1)\\times P(H_2)$.\u00a0 The outcome of one coin toss does not influence then outcome of the other.\n\nHowever they are not mutually exclusive, so $P(H_1 \\cup H_2) = P(H_1)+P(H_2) - P(H_1 \\cap H_2)$.\u00a0 Both coins can turn up heads.\n\nPutting it together: $$\\therefore P(H_1 \\cup H_2) = \\frac 12 + \\frac 1 2 - \\frac 12 \\times \\frac 12 = \\frac 3 4$$\n\n-\n\nLet $A$ be the event of getting a tail in both tosses, then $A'$ be the event of getting a head in either tosses. So $P(A') = 1 - P(A) = 1 - 0.5*0.5 = 0.75$\n\n-\n\nprobability of having head in a coin flip is 1/2 , when you flip 2 times then the probability you have at least 1 head is equal to :\n\n1 - P(no head) = 1 - P(2 tail) = 1 - 1/2*1/2 = 3/4 or 75%\n\n-", "date": "2014-10-26 04:09:05", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/729920/probability-of-head-in-coin-flip-when-coin-is-flipped-two-times", "openwebmath_score": 0.6118226051330566, "openwebmath_perplexity": 349.6985313531705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759638081522, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8522112915303947}}
{"url": "https://brilliant.org/discussions/thread/simple-numbers-are-problems-numbers/", "text": "# One minus one plus one minus one plus...\n\nThere are three types of people in this world:\n\nEvaluate:\n\n$\\color{blue}{S}=1-1+1-1+1-1+\\ldots$\n\nType 1\n\n$\\color{blue}{S}=(1-1)+(1-1)+(1-1)+\\ldots=0+0+0+\\ldots=\\boxed{0}$\n\nType 2\n\n$\\color{blue}{S}=1-(1-1)-(1-1)-(1-1)-\\ldots=1-0-0-0-\\ldots=\\boxed{1}$\n\nBut the $$\\displaystyle 3^{rd}$$ type of people did like this:\n\n$1-\\color{blue}{S}=1-(1-1+1-1+\\ldots)=1-1+1-1+1-1+\\ldots = S$\n\n$\\Leftrightarrow 1-\\color{blue}S=\\color{blue}S \\Rightarrow 2\\color{blue}S=1 \\Rightarrow \\color{blue}S=\\boxed{\\frac{1}{2}}$\n\nNote by Adam Ph\u00fac Nguy\u1ec5n\n3\u00a0years, 4\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nYou forgot $$4^\\text{th}$$ type of people; they say that this series diverges.\n\n- 3\u00a0years, 4\u00a0months ago\n\nIts answer oscillates b/w 0 and 1\n\n- 3\u00a0years, 4\u00a0months ago\n\nYes that's why it diverges.\n\n- 3\u00a0years, 4\u00a0months ago\n\nEven more here:\n\nEvaluate : $S=1-2+3-4+5-6+ \\ldots$\n\nType 1 : $S=1+(-2+3)+(-4+5)+ \\ldots = 1+1+1+1+\\ldots=\\infty$\n\nType 2 : $S=(1-2)+(3-4)+(5-6)+ \\ldots = -1-1-1-1+\\ldots=-\\infty$\n\nType 3 : They go to WolframAlpha, search this:\n\n 1 sum(n from 1 to infty,(-1)^n*n) \n\nWhich shows up that \"The ratio test is inconclusive.\" and \"The root test is inconclusive.\", from which they implies that the sum is incosistent.\n\nType 4 : They go to Wikipedia and finds out that the sum is actually equal to $$1/4$$.\n\n- 3\u00a0years, 4\u00a0months ago\n\nWow! Awesome!\n\nI also saw this video:\n\n$1+2+3+4+\\ldots=\\frac{1}{12}$\n\n- 3\u00a0years, 4\u00a0months ago\n\nWow, that's cool :)))\n\n- 3\u00a0years, 4\u00a0months ago\n\ni guess u forgot the negative sign along with 1/12\n\n- 3\u00a0years, 4\u00a0months ago\n\nlast one is pretty good\n\n- 3\u00a0years, 4\u00a0months ago\n\nhaha.. g8\n\n- 3\u00a0years, 4\u00a0months ago\n\nGrandi series.\n\n- 3\u00a0years, 4\u00a0months ago\n\nGud 1\n\n- 3\u00a0years, 4\u00a0months ago", "date": "2019-01-21 13:03:01", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/simple-numbers-are-problems-numbers/", "openwebmath_score": 0.9994232654571533, "openwebmath_perplexity": 12603.962201596063, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9693242036063214, "lm_q2_score": 0.8791467722591728, "lm_q1q2_score": 0.8521782448731907}}
{"url": "https://www.physicsforums.com/threads/using-the-ftc-to-take-a-derivative.650253/", "text": "# Using the FTC to take a derivative\n\n1. Nov 7, 2012\n\n### dumbQuestion\n\n1. The problem statement, all variables and given/known data\n\nSimplify the following:\n\nd/dx[\u222b(t/lnt)dt] where the integral is a definite integral with bounds from x to x2\n\n2. Relevant equations\n\nThe Fundamental THeorem of calculus says:\n\nSuppose f is continuous on [a,b]\n\nThen \u222babf(x)dx=F(b)-F(a) where F is any antiderivative of f\n\n3. The attempt at a solution\n\nSo this is how I thought to solve it. Let f(t)=t/lnt\n\nFirst notice that f(t) continuous on (1,c] forall c in \u211d => f(t) integrable and there exists an antiderivative F(t) such that F'(t) = f(t)\n\nSo let F(t) be such an antiderivative. ***By the fundamental theorem of calculus,\n\n\u222bf(t)dt (where the integral bounds are from x to x2=F(x2)-F(x)\n\nThis means\n\nd/dx[\u222bf(t)dt] (where the integral bounds are from x to x2=d/dx[F(x2)-F(x)] = d/dx[F(x2)] - d/dx(F(x)] = 2xF'(x2) - F'(x) [by the Chain Rule] = 2x(x2/ln(x2)) - x/lnx = 2x3/2lnx - x/lnx [by rules of logarithm] = (x3-x)/lnx (common denominator and added them together)\n\nSo I get that as long as the bounds on the integral are from (0,c), the answer to\nd/dx[\u222b(t/lnt)dt] where the integral is a definite integral with bounds from x to x2 = (x3-x)/lnx\n\nOK here is my problem, the book solves this almost exactly the same way, but they never make any mention of the bounds on the integral or continuity. I guess I am confused. Do I not need to think about where the function is continuous? It's just part of the FTC says \"Suppose f is continuous on [a,b]\" so it seems to me like I can only apply it when that part holds. Why don't you need to think of continuity in this case? There is no restriction in the problem statement about x. How do you solve this problem for any general x?\n\nAlso the other problem, is the line where I put ***. I think I've done something wrong here because this function is only continuous on the open region (1,c] for all c, but the FTC requires f be continuous on a closed bounded region [a,b]. So do I just pick a number great than 1, say [1.1,c]? I mean what's the formal way to solve this with regards to the continuity aspect?\n\n2. Nov 7, 2012\n\n### LCKurtz\n\nYour problem implicitly assumes $t>0$ hence $x>0$ and $x^2>0$ because $t>0$ is the domain of $\\ln(t)$. So there is no problem.\n\n3. Nov 7, 2012\n\n### dumbQuestion\n\nBut doesn't f need to be continuous on a closed bounded interval [a,b]? ln(t) is only continuous on (0,c] for all c in R.\n\nCan you apply the FTC if f is only continuous on an interval (a,b]?\n\n4. Nov 7, 2012\n\n### LCKurtz\n\nLikely not; I'd have to check. But it doesn't matter for this question. If $x$ and $x^2$ are both positive, then all your conditions are true on the closed interval between them so you can use the FTC.", "date": "2018-02-19 14:43:44", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/using-the-ftc-to-take-a-derivative.650253/", "openwebmath_score": 0.905588686466217, "openwebmath_perplexity": 483.6273381149708, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9693241956308277, "lm_q2_score": 0.8791467643431002, "lm_q1q2_score": 0.8521782301883205}}
{"url": "https://byjus.com/question-answer/if-the-function-f-x-x-4-2x-3-ax-2-bx-on-13-satisfies/", "text": "Question\n\n# If the function $$f(x) = x^4 - 2x^3 + ax^2 + bx$$ on $$[1,3]$$ satisfies the conditions of Rolle's Theorem with $$c = \\dfrac{3}{2}$$, then find $$a$$ and $$b$$.\n\nSolution\n\n## Given $$f(x)=x^4-2x^3+ax^2+bx$$ on $$[1,3]$$$$\\implies f'(x)=4x^3-6x^2+2ax+b$$According to Rolle's theorem, if f(x) is continuous on $$[a,b]$$, differentiable on $$(a,b)$$ and $$f(a)=f(b)$$ then there exists some $$c \\in (a,b)$$ such that $$f'(c)=0$$Given that the function $$f(x)$$ satisfies the conditions of Rolle's theorem with $$c=\\dfrac 32$$$$f(1)=1^4-2(1)^3+a(1)^2+b(1)=1-2+a+b=a+b-1$$$$f(3)=3^4-2(3)^3+a(3)^2+b(3)=81-54+9a+3b=9a+3b-27$$from Rolles's theorem, $$f(1)=f(3)$$$$\\implies a+b-1=9a+3b-27$$$$\\implies 8a+2b=26$$$$\\implies 4a+b=13$$ .......(1)from Rolles's theorem,\u00a0$$f'(c)=4c^3-6c^2+2ac+b=0$$\u00a0$$\\implies f'(\\dfrac 32)=4(\\dfrac 32)^3-6(\\dfrac 32)^2+2a(\\dfrac 32)+b=0$$$$\\implies \\dfrac{27}{2}-\\dfrac{27}{2}+3a+b=0$$$$\\implies 3a+b=0$$ .......(2)(1) - (2) $$\\implies a=13$$substituting $$a=13$$ in (1)\u00a0$$\\implies 4(13)+b=13$$$$\\implies b=-39$$Therefore, $$a=13,b=-39$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-28 05:10:39", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/if-the-function-f-x-x-4-2x-3-ax-2-bx-on-13-satisfies/", "openwebmath_score": 0.916120171546936, "openwebmath_perplexity": 510.4124791832383, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9912886147702209, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8521748901931934}}
{"url": "https://math.stackexchange.com/questions/185802/dimensions-of-symmetric-and-skew-symmetric-matrices", "text": "# Dimensions of symmetric and skew-symmetric matrices\n\nLet $\\textbf A$ denote the space of symmetric $(n\\times n)$ matrices over the field $\\mathbb K$, and $\\textbf B$ the space of skew-symmetric $(n\\times n)$ matrices over the field $\\mathbb K$. Then $\\dim (\\textbf A)=n(n+1)/2$ and $\\dim (\\textbf B)=n(n-1)/2$.\n\nShort question: is there any short explanation (maybe with combinatorics) why this statement is true?\n\nEDIT: $\\dim$ refers to linear spaces.\n\n\u2022 Do you mean symmetric (not normal) in the title? And do you mean the $\\dim$ of linear spaces of such matrices, not the $\\dim$ of the matrices, right? Aug 23, 2012 at 9:59\n\u2022 I did edit it - thanks for the reminder! Aug 23, 2012 at 10:02\n\u2022 You did not edit it correctly; $\\mathbf A$ still refers to just one matrix, not a subspace. I will edit it for you. Aug 23, 2012 at 12:38\n\u2022 And you should say that $\\mathbb K$ is not of characteristic $2$, or otherwise symmetric and anti-symmetric matrices are the same thing and your equations cannot both be true. Aug 23, 2012 at 12:40\n\nAll square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.\n\nThe skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).\n\nFor the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.\n\nHere is my two cents:\n\n\\begin{eqnarray} M_{n \\times n}(\\mathbb{R}) & \\text{has form} & \\begin{pmatrix} *&*&*&*&\\cdots \\\\ *&*&*&*& \\\\ *&*&*&*& \\\\ *&*&*&*& \\\\ \\vdots&&&&\\ddots \\end{pmatrix} \\hspace{.5cm} \\text{with $n^2$ elements}\\\\ \\\\ \\\\ Skew_{n \\times n}(\\mathbb{R}) & \\text{has form} & \\begin{pmatrix} 0&*'&*'&*'&\\cdots \\\\ *&0&*'&*'& \\\\ *&*&0&*'& \\\\ *&*&*&0& \\\\ \\vdots&&&&\\ddots \\end{pmatrix} \\end{eqnarray} For this bottom formation the (*)s and (*')s are just operationally inverted forms of the same number, so the array here only takes $\\frac{(n^2 - n)}{2}$ elements as an argument to describe it. This appears to be an array geometry question really... I suppose if what I'm saying is true, then I conclude that because $\\dim(Skew_{n \\times n}(\\mathbb{R}) + Sym_{n \\times n}(\\mathbb{R})) = \\dim(M_{n \\times n}(\\mathbb{R}))$ and $\\dim(Skew_{n \\times n}(\\mathbb{R}))=\\frac{n^2-n}{2}$ then we have that \\begin{eqnarray} \\frac{n^2-n}{2}+\\dim(Sym_{n \\times n}(\\mathbb{R})))=n^2 \\end{eqnarray} or \\begin{eqnarray} \\dim(Sym_{n \\times n}(\\mathbb{R})))=\\frac{n^2+n}{2}. \\end{eqnarray}\n\n\u2022 $Skew_{n\\times n}(\\mathbb{R})=(Sym_{n\\times n}(\\mathbb{R}))^{\\perp}?$ Aug 18, 2021 at 5:50\n\nThis is a bit late, but I figured I would chime in since the other answers don't really give the combinatorial intuition asked for in the original question.\n\nIn order to specify a skew-symmetric matrix, we need to choose a number $A_{ij}$ for each set $\\{i,j\\}$ of distinct indices. How many such sets are there? Combinatorics tells us that there are $\\left( \\begin{array}{@{}c@{}} n \\\\ 2 \\end{array} \\right) = \\frac{n(n-1)}{2}$ such sets.\n\nSimilarly, in order to specify a symmetric matrix, we need to choose a number $A_{ij}$ for each set $\\{i,j\\}$ of (not necessarily distinct) indices. We can encode such sets by adding a special symbol $n+1$ that indicates a repeated index. Thus, we need to choose a number $A_{ij}$ for each set $\\{i,j\\}$ of distinct symbols, where now a symbol is either an index ($1 , \\ldots , n$), or our special symbol $n+1$ that is used for a repeated index. Combinatorics tells us that there are $\\left( \\begin{array}{@{}c@{}} n+1 \\\\ 2 \\end{array} \\right) = \\frac{n(n+1)}{2}$ such sets.\n\nThe dimension of symmetric matrices is $\\frac{n(n+1)}2$ because they have one basis as the matrices $\\{M_{ij}\\}_{n \\ge i \\ge j \\ge 1}$, having $1$ at the $(i,j)$ and $(j,i)$ positions and $0$ elsewhere. For skew symmetric matrices, the corresponding basis is $\\{M_{ij}\\}_{n \\ge i > j \\ge 1}$ with $1$ at the $(i,j)$ position, $-1$ at the $(j,i)$ position, and $0$ elsewhere.\n\nNote that the diagonal elements of skew symmetric matrices are $0$, hence their dimension is $n$ less than the dimension of normal symmetric matrices.\n\n\u2022 But this is no explanation why the symmetric matrices have the specified $\\dim$. Aug 23, 2012 at 10:03\n\u2022 Do you mean $n^2$? Aug 23, 2012 at 10:04\n\u2022 Yeah, I didn't notice earlier that he asked for proofs of the dimensions too, and have edited. Aug 23, 2012 at 10:07\n\u2022 @RijulSaini: Thanks, but enzotib's answer seems to be easier to understand! Aug 23, 2012 at 10:13\n\nThere have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.\n\n$$\\begin{matrix} \\begin{pmatrix} 0 &*' \\\\ *& 0 \\\\ \\end{pmatrix} & * \\end{matrix}$$\n\n$$\\begin{matrix} \\begin{pmatrix} 0 &*' & *' \\\\ *& 0 & *' \\\\ * & * & 0 \\\\ \\end{pmatrix} & \\begin{matrix} *& \\\\ * & * \\end{matrix} \\end{matrix}$$\n\n$$\\begin{matrix} \\begin{pmatrix} 0 &*' & *' &*'\\\\ *& 0 & *' &*'\\\\ * & * & 0 &*'\\\\ * & * &* & 0 \\end{pmatrix} & \\begin{matrix} * & & \\\\ *&* & \\\\ * & * &* \\end{matrix} \\end{matrix}$$\n\n$$\\begin{matrix} \\begin{pmatrix} 0&*'&*'&*'&\\cdots \\\\ *&0&*'&*'& \\\\ *&*&0&*'& \\\\ *&*&*&0& \\\\ \\vdots&&&&\\ddots \\end{pmatrix} & \\begin{matrix} *& \\\\ *&*& \\\\ *&*&*& \\\\ \\vdots&&&&\\ddots \\end{matrix} \\end{matrix}$$\n\nNotice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + \\dots + (n-1).$$\n\nSo by Gauss, we have $$\\frac{(n-1)(n-1+1)}{2} = \\frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.\n\nFor the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$\\frac{(n-1)n}{2} + n = \\frac{n^2 + n}{2}$$", "date": "2022-08-16 13:55:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/185802/dimensions-of-symmetric-and-skew-symmetric-matrices", "openwebmath_score": 0.9975501298904419, "openwebmath_perplexity": 332.23170875784746, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795087371921, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8521498339983605}}
{"url": "https://math.stackexchange.com/questions/1779221/how-to-remember-sum-to-product-and-product-to-sum-trigonometric-formulas/1779226", "text": "# How to remember sum to product and product to sum trigonometric formulas?\n\nThey are:\n\n\\begin{align} \\cos(a)\\cos(b)&=\\frac{1}{2}\\Big(\\cos(a+b)+\\cos(a-b)\\Big) \\\\[2ex] \\sin(a)\\sin(b)&=\\frac{1}{2}\\Big(\\cos(a-b)-\\cos(a+b)\\Big) \\\\[2ex] \\sin(a)\\cos(b)&=\\frac{1}{2}\\Big(\\sin(a+b)+\\sin(a-b)\\Big) \\\\[2ex] \\cos(a)\\sin(b)&=\\frac{1}{2}\\Big(\\sin(a+b)-\\sin(a-b)\\Big) \\\\[2ex] \\cos(a)+\\cos(b)&=2\\cos\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right) \\\\[3ex] \\cos(a)-\\cos(b)&=-2\\sin\\left(\\frac{a+b}{2}\\right)\\sin\\left(\\frac{a-b}{2}\\right) \\\\[3ex] \\sin(a)+\\sin(b)&=2\\sin\\left(\\frac{a+b}{2}\\right)\\cos\\left(\\frac{a-b}{2}\\right) \\\\[3ex] \\sin(a)-\\sin(b)&=2\\cos\\left(\\frac{a+b}{2}\\right)\\sin\\left(\\frac{a-b}{2}\\right) \\end{align}\n\nI have found nice mnemonics that helped me to remember the reduction formulae and others but I can't find a simple relationship between the formulas above. Can you help?\n\n\u2022 What type of \"relations\" do you mean, or do you need ? \u2013\u00a0Jean Marie May 10 '16 at 8:33\n\u2022 'Mathematics' is the subject which comes by practice! So use these formulas in questions and all will be stored in memory ;P \u2013\u00a0user5954246 May 10 '16 at 8:55\n\u2022 You are welcome to have a look at my answer to this post \u2013\u00a0Mick May 10 '16 at 16:57\n\u2022 @RichardSmith: Here's a diagram that may help. \u2013\u00a0Blue May 10 '16 at 17:12\n\n## 3 Answers\n\nThe only ones you need to know are the classical $\\sin(a+b) = \\sin(a)\\cos(b)+\\cos(a)\\sin(b)$ and $\\cos(a+b)=\\cos(a)\\cos(b)-\\sin(a)\\sin(b)$. The others are mere consequences of those.\n\nFor example, by changing the signs, you get $\\cos(a-b)=\\cos(a)\\cos(b)+\\sin(a)\\sin(b)$. By summing, you have $\\cos(a+b)+\\cos(a-b) = 2\\cos(a)\\cos(b)$, which is your first formula.\n\nSimilarly, by solving $p=a+b$ and $q=a-b$, you get the formula $\\cos(p)+\\cos(q) = 2\\cos\\left(\\dfrac{p+q}{2}\\right)\\cos\\left(\\dfrac{p-q}{2}\\right)$.\n\n\u2022 And the two you suggest to remember can be derived in less than $30$ seconds using $e^{i(a+b)}=e^{ia}e^{ib}$ and Euler's formula :) \u2013\u00a0Guest Nov 20 '16 at 19:32\n\nRight away, you can cross off the fourth formula, since it is equivalent to the third formula after switching $a$ and $b$.\n\nThen, you can also avoid the last four formulas, since these are all covered by the first three formulas via the relationships $$a+b = u, \\quad a-b = v, \\quad a = \\frac{u+v}{2}, \\quad b = \\frac{u-v}{2}.$$\n\nSo that really leaves us with only three formulas. The first two are merely consequences of the cosine angle addition identity $$\\cos(a \\pm b) = \\cos a \\cos b \\mp \\sin a \\sin b,$$ where a suitable addition or subtraction of the two forms of this equation are done; e.g., \\begin{align*} \\cos (a-b) &= \\cos a \\cos b + \\sin a \\sin b \\\\ \\cos (a+b) &= \\cos a \\cos b - \\sin a \\sin b \\\\ \\hline \\cos(a-b) + \\cos(a-b) &= 2 \\cos a \\cos b . \\end{align*} A similar concept applied to the sine angle addition identity yields the third (and fourth).\n\nOf course, you can memorize the formulas, or re-derive them, but clearly it's faster to have more formulas memorized as long as you can remember them. What is important to stress is that a vast array of trigonometric identities are all consequences of some very basic identities, and these basic identities are the ones you really need to know.\n\nHow about just restating the LHS. For example, you could restate $\\cos a\\sin b$ as $$\\frac{\\sin a\\cos b +\\cos a\\sin b + \\cos a\\sin b - \\sin a\\cos b}{2}$$ and just figure it out from there. For Example, Let's start off with $\\cos a\\sin b$ and try to derive $\\frac{1}{2}[\\sin(a+b) - \\sin(a-b)]$ \\begin{align} \\cos a \\sin b &= \\frac{1}{2}\\bigg[2\\cos a\\sin b\\bigg] \\\\ &= \\frac{1}{2}\\bigg[\\cos a\\sin b + \\cos a\\sin b\\bigg] \\\\ &= \\frac{1}{2}\\bigg[\\cos a\\sin b + \\cos a\\sin b + 0\\bigg] \\\\ &= \\color{red}{\\frac{1}{2}\\bigg[\\cos a\\sin b + \\cos a\\sin b + (\\sin a \\cos b - \\sin a\\cos b)\\bigg]} \\\\ &= \\frac{1}{2}\\bigg[(\\cos a\\sin b + \\sin a \\cos b) +(\\cos a\\sin b - \\sin a\\cos b)\\bigg] \\\\ &= \\frac{1}{2}\\bigg[(\\cos a\\sin b + \\sin a \\cos b) -(\\sin a\\cos b - \\cos a\\sin b )\\bigg] \\\\ &= \\frac{1}{2}\\bigg[(\\cos a\\sin b + \\sin a \\cos b) -(\\sin a\\cos (-b) + \\cos a\\sin (-b) )\\bigg] \\\\ &=\\frac{1}{2}\\bigg[\\sin(a+b) - \\sin(a-b)\\bigg] \\end{align}\n\nUsually I just remember/figure out the red line.", "date": "2019-08-26 05:55:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1779221/how-to-remember-sum-to-product-and-product-to-sum-trigonometric-formulas/1779226", "openwebmath_score": 1.0000016689300537, "openwebmath_perplexity": 745.6182134736937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986979510652134, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8521498321823928}}
{"url": "https://math.stackexchange.com/questions/3181601/generating-function-of-frachx1-x2", "text": "# Generating function of $\\frac{h(x)}{(1-x)^2}$\n\nIf $$h(x)$$ is the generating function for $$a_r$$, what is the generating function of $$\\frac{h(x)}{(1-x)^2}$$\n\nLet $$h(x)$$ be written as\n\n$$h(x) = \\sum_{r} a_r x^r$$\n\nConsider more simply\n\n$$\\frac{h(x)}{1-x} = \\frac{1}{1-x} h(x) =\\sum_{r} x^r \\sum_{r} a_r x^r$$\n\nI tried to expand this and see what I could get\n\n$$(1+x+x^2+x^3+\\dots+x^r+\\dots)(a_0+a_1x+a_2x^2+a_3x^3+\\dots+a_rx^r+\\dots)$$\n\nthere are two ways to simplify the product, either\n\n$$a_0(1+x+x^2+\\dots)+a_1(x+x^2+x^3+\\dots)+ a_2(x^2+x^3+x^4+\\dots)+\\dots= \\sum_ra_r\\sum_{k\\ge r}x^k$$\n\nor $$a_0 + (a_0+a_1)x+(a_0+a_1+a_2)x^2+(a_0+a_1+a_2+a_3)x^3 = \\sum_r \\left(\\sum_{k\\le r}a_k \\right)x^r$$\n\nobviously this is only for one factor of $$\\frac{1}{1-x}$$ but I assume If I can get help for this I can extend it to two factors.\n\nI'm not sure what form the answer is expected to be in? Because I could say the generating function is\n\n$$\\frac{h(x)}{1-x} = h\\left(\\sum_{k \\ge r}x^k\\right)$$\n\nbut I'm not sure that makes any sense. I was expecting to say something like\n\n$$\\frac{h(x)}{1-x} \\mapsto h(x^2)$$\n\n(the $$x^2$$ is not intentional, just some idea of what I believe the answer could look like)\n\nAny help for the case of $$\\frac{1}{1-x}$$ would be great and then I could extend it to $$\\frac{1}{(1-x)^2}$$\n\n\u2022 I think that the question must be \"What sequence is $h(x)/(1-x)^2$ the generating functin of?\" \u2013\u00a0Somos Apr 9 '19 at 21:56\n\u2022 It probably helps a lot to note that because $\\frac{d}{dx} \\frac{1}{1-x} = \\frac{1}{(1-x)^2}$, by taking the derivative of the geometric series we get $\\frac{1}{(1-x)^2}=\\sum_{n=0}^\\infty (n+1)x^n$. Then you can just use one convolution of this series with the generating function for $h(x)$. \u2013\u00a0Robert Shore Apr 9 '19 at 21:57\n\u2022 @Somos I'm not sure, you might be right. This is from a list of review questions for my combinatorics exam, see here a screenshot from the paper link \u2013\u00a0Hushus46 Apr 9 '19 at 22:00\n\u2022 @RobertShore I have not heard the term convolution in my class but I assume you mean a new generating function where the coefficients $c_n = a_0b_n+a_1b_{n-1}+\\dots+a_{n-1}b_1+a_nb_0$ ? \u2013\u00a0Hushus46 Apr 9 '19 at 22:06\n\u2022 Yes, that's what I'm talking about. \u2013\u00a0Robert Shore Apr 9 '19 at 22:23\n\nFirst, note that\n\n$$\\frac{d}{dx} \\frac{1}{1-x}=\\frac{1}{(1-x)^2}.$$\n\nThus, taking the derivative of the power series we have:\n\n$$\\frac{1}{(1-x)^2}=\\sum_{k=0}^\\infty (k+1)x^k.$$\n\nLet\n\n$$h(x)=\\sum_{n=0}^\\infty a_nx^n.$$\n\nThen taking the convolution of these two series, we have:\n\n$$\\frac{h(x)}{(1-x)^2}=\\sum_{n=0}^\\infty \\sum_{k=0}^n (k+1)a_{(n-k)}x^n.$$\n\nAnother way.\n\nSince $$(1-x)^2 = 1-2x+x^2$$, if $$\\dfrac{h(x)}{(1-x)^2} =\\sum_{n=0}^{\\infty} a_nx^n$$ then\n\n$$\\begin{array}\\\\ h(x) &=(1-x)^2\\sum_{n=0}^{\\infty} a_nx^n\\\\ &=(1-2x+x^2)\\sum_{n=0}^{\\infty} a_nx^n\\\\ &=\\sum_{n=0}^{\\infty} a_nx^n-2x\\sum_{n=0}^{\\infty} a_nx^n+x^2\\sum_{n=0}^{\\infty} a_nx^n\\\\ &=\\sum_{n=0}^{\\infty} a_nx^n-\\sum_{n=0}^{\\infty} 2a_nx^{n+1}+\\sum_{n=0}^{\\infty} a_nx^{n+2}\\\\ &=\\sum_{n=0}^{\\infty} a_nx^n-\\sum_{n=1}^{\\infty} 2a_{n-1}x^{n}+\\sum_{n=2}^{\\infty} a_{n-2}x^{n}\\\\ &=a_0+a_1x+\\sum_{n=}^{\\infty} a_nx^n-2a_0x-\\sum_{n=2}^{\\infty} 2a_{n-1}x^{n}+\\sum_{n=2}^{\\infty} a_{n-2}x^{n}\\\\ &=a_0+(a_1-2a_0)x+\\sum_{n=}^{\\infty} (a_n-2a_{n-1}+a_{n-2})x^n\\\\ \\end{array}a_n$$\n\nIf $$h(x) =\\sum_{n=0}^{\\infty} h_nx^n$$ then, equating coefficients, $$h_0 = a_0$$, $$h_1 = a_1-2a_0$$, so $$a_1 = h_1+2a_0 = h_1+2h_0$$, and, for $$n \\ge 2$$, $$h_n =a_n-2a_{n-1}+a_{n-2}$$ so $$a_n =h_n+2a_{n-1}-a_{n-2}$$.\n\nThe advantage of this method is that getting each new $$a_n$$ takes only 3 operations (multiply, add, subtract) instead of $$n$$.", "date": "2021-07-23 17:03:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3181601/generating-function-of-frachx1-x2", "openwebmath_score": 0.9090185761451721, "openwebmath_perplexity": 278.0864194518433, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795079712153, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8521498316023662}}
{"url": "https://mathhelpboards.com/threads/antiderivative-involving-trig-identities.9309/", "text": "# Antiderivative involving Trig Identities.\n\n#### shamieh\n\n##### Active member\nA little confused on something.\n\nSuppose I have the integral\n\n$$\\displaystyle 2 \\int 4 \\sin^2x \\, dx$$\n\nSo I understand that $$\\displaystyle \\sin^2x = \\frac{1 - \\cos2x}{2}$$\n\nBUT we have a 4 in front of it, so shouldn't we pull the $$\\displaystyle 4$$ out in front of the integral to get:\n\n$$\\displaystyle 8 \\int \\frac{1 - \\cos 2x}{2} \\, dx$$\n\nthen pull out the $$\\displaystyle \\frac{1}{2}$$ to get: $$\\displaystyle 4 \\int 1 - \\cos 2x \\, dx$$\n\nthen:\n\n$$\\displaystyle 8 [ x + \\frac{1}{2} \\sin2x ] + C$$?\n\n#### MarkFL\n\nStaff member\nI would switch the constants so that we have:\n\n$$\\displaystyle 4\\int 2\\sin^2(x)\\,dx$$\n\nApply the identity:\n\n$$\\displaystyle 4\\int 1-\\cos(2x)\\,dx$$\n\nAnd then we have:\n\n$$\\displaystyle 4x-2\\sin(2x)+C$$\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nA little confused on something.\n\nSuppose I have the integral\n\n$$\\displaystyle 2 \\int 4 \\sin^2x \\, dx$$\n\nSo I understand that $$\\displaystyle \\sin^2x = \\frac{1 - \\cos2x}{2}$$\n\nBUT we have a 4 in front of it, so shouldn't we pull the $$\\displaystyle 4$$ out in front of the integral to get:\n\n$$\\displaystyle 8 \\int \\frac{1 - \\cos 2x}{2} \\, dx$$\n\nthen pull out the $$\\displaystyle \\frac{1}{2}$$ to get: $$\\displaystyle 4 \\int 1 - \\cos 2x \\, dx$$\n\nthen:\n\n$$\\displaystyle 8 [ x + \\frac{1}{2} \\sin2x ] + C$$?\nSurely you mean \\displaystyle \\begin{align*} 4 \\left[ x + \\frac{1}{2}\\sin{(2x)} \\right] + C \\end{align*}. Apart from that everything you have done is correct\n\n#### shamieh\n\n##### Active member\nI would switch the constants so that we have:\n\n$$\\displaystyle 4\\int 2\\sin^2(x)\\,dx$$\n\nApply the identity:\n\n$$\\displaystyle 4\\int 1-\\cos(2x)\\,dx$$\n\nAnd then we have:\n\n$$\\displaystyle 4x-2\\sin(2x)+C$$\n\nI see. Ok here is where I am getting lost Mark. My original problem was this $$\\displaystyle \\int \\frac{x^2}{\\sqrt{4 - x^2}} dx$$\n\nSo I got to the step we just mentioned above (ignore that my final answer had an 8 in front of all those terms, should be a 4 - BUT I then noticed that the A.D. of $$\\displaystyle \\cos(2\\theta)$$ is just $$\\displaystyle \\frac{1}{2}\\sin(2\\theta)$$THUS I can now change the 4 to a 2 since$$\\displaystyle \\frac{4}{2} = 2$$ when I pulled it out to the constant.)\n\nFinally, after applying all of those changes - now I am here:\n\n$$\\displaystyle 2\\theta - 2\\sin(2\\theta) + C$$ ... So then I said ok, simple enough, just sub back in for my original equation where I had $$\\displaystyle x = 2\\sin\\theta$$ which becomes $$\\displaystyle \\theta = \\arcsin(\\frac{x}{2})$$ (Note: From when I used my trig sub)\n\nThus $$\\displaystyle 2\\arcsin(\\frac{x}{2}) - 2\\sin(2\\arcsin(\\frac{x}{2}))$$ But somehow my teacher is getting:\n\n$$\\displaystyle 2\\arcsin(\\frac{x}{2}) - \\frac{1}{2} x\\sqrt{4 -x^2} + C$$ as the final answer..\n\nIn the second term when i have sin(arcsin ... etc What would I do there?\n\n#### MarkFL\n\nStaff member\nWe are given:\n\n$$\\displaystyle I=\\int \\frac{x^2}{\\sqrt{4-x^2}}\\,dx$$\n\nI would use the substitution:\n\n$$\\displaystyle x=2\\sin(\\theta)\\,\\therefore\\,dx=2\\cos(\\theta)\\,d \\theta$$\n\nAnd so now we have:\n\n$$\\displaystyle I=\\int \\frac{4\\sin^2(\\theta)}{\\sqrt{4-4\\sin^2(\\theta)}}\\,2\\cos(\\theta)\\,d\\theta$$\n\nSimplifying this, we obtain:\n\n$$\\displaystyle I=4\\int \\sin^2(\\theta)\\,d\\theta$$\n\nThis is half of what I began with in post #2 above, so using that method, we would obtain:\n\n$$\\displaystyle I=2\\theta-\\sin(2\\theta)+C$$\n\nNow, using the double-angle identity for sine, we may write:\n\n$$\\displaystyle I=2\\theta-2\\sin(\\theta)\\cos(\\theta)+C$$\n\nObserving that:\n\n$$\\displaystyle \\sin(\\theta)=\\frac{x}{2}\\implies\\cos(\\theta)=\\frac{\\sqrt{4-x^2}}{2}$$\n\nwe may now write:\n\n$$\\displaystyle I=2\\sin^{-1}\\left(\\frac{x}{2} \\right)-\\frac{x\\sqrt{4-x^2}}{2}+C$$\n\nI think you are still trying to pull constants out in front of your integrals that are not factors of the entire integrand.\n\n#### shamieh\n\n##### Active member\nYes you are exactly right. I was pulling things out in front that I couldnt\n\nSo essentially I should have\n\n$$\\displaystyle 4[\\theta - \\frac{1}{2}\\sin(2\\theta)]$$\n\nWhich becomes:\n\n$$\\displaystyle 4\\theta-2\\sin(2\\theta)$$\n\nWhich then becomes:\n\n$$\\displaystyle 2\\theta - \\sin(2\\theta)$$\n\nThen plug in double angle formula,\n\nThen 2 in the second term cancels out with the $$\\displaystyle \\frac{x}{2}$$\n\nWow I see where I made my many errors. Thanks for the clarification.", "date": "2020-09-29 13:42:34", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/antiderivative-involving-trig-identities.9309/", "openwebmath_score": 0.9178450107574463, "openwebmath_perplexity": 785.4836049883065, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795068222499, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8521498271410467}}
{"url": "https://www.physicsforums.com/threads/oil-volume-required-to-rise-the-piston.725238/", "text": "# Oil volume required to rise the piston\n\n1. Nov 27, 2013\n\n### Camille\n\n1. The problem statement, all variables and given/known data\n\nThe piston shown weighs 11 lbf. In its initial position, the piston is restrained from moving to the bottom of the cylinder by means of the metal stop. Assuming there is neither friction nor leakage between piston and cylinder, what volume of oil (S = 0.85) would have to be added to the 1-in. tube to cause the piston to rise 1 in. from its initial position?\n\n2. Relevant equations\n\n$p = \\frac{dF}{dA}$\n\nAnd the basic differential equation of the fluid statics:\n\n$\\frac{dp}{dz} = - \u03b3$\n\nAlso the density or the specific weight of the oil is:\n\n$\u03c1 = 1000 \\frac{kg}{m^3} \\cdot 0.85 = 850 \\frac{kg}{m^3}$\n\n$\u03b3 = \u03c1 \\cdot g = 8.336 \\frac{kN}{m^3}$\n\n3. The attempt at a solution\n\nFirstly, I have calculated what volume of oil is needed just to keep the piston \"a little\" above the stop, ie. to hold just the weight of the piston:\n\n$V_{oil p} = \\frac{W_p}{A_p} \\cdot \\frac{A_o}{\u03c1 \\cdot g}$ ,\n\nwhere W_p is the weight of the piston,\nA_p is the c-s area of the cylinder\nA_o is the c-s area of the tube.\n\nThis turns out to be:\n\n$V_{oil p} = 22.39 in^3$.\n\nNext, I have imagined, that when the piston is 1 in above the orignal level, we have to add even more oil to hold the weight of this 1-in thick layer of oil. This gave me in addition the volume of:\n\n$V_{oil o} = 3.14 in^3$.\n\nSo altogether the total amount of oil needed calculated by me was:\n\n$V_{oil} = 22.39 in^3 + 3.14 in^3 = 25.53 in^3$.\n\nThe answer should be $V_{oil} = 35.7 in^3$.\n\nI see the hole in my reasoning, because the volume of oil in the 1-in layer has to be taken from somewhere, so it's either the old oil that was in the cylinder or the old + the new oil added. However, I don't know how to solve it.\n\nAny help will be appreciated!\n\n2. Nov 27, 2013\n\n### nasu\n\nFor one thing, have you included the additional volume in the large cylinder?\nAnd the 3.14 in^3 volume looks suspect. How did you get that?\n\n3. Nov 27, 2013\n\n### Camille\n\nDo you mean the additional volume of the 1-in layer in the cylinder? Yes. It's weight is:\n\n$W_1 = \u03c0 \\cdot (4 in)^2 \\cdot 1 in \\cdot \u03c1 \\cdot g = 6.87 N$\n\nThen we have:\n\n$\\frac{V_{oil o}}{A_o} = \\frac{W_1}{A_p}$\n\nAnd solving it I got $V_{oil o} = 3.14 in^3$\n\n4. Nov 27, 2013\n\n### Staff: Mentor\n\nThe volume of a 1\" layer in the large cylinder is \u03c0(4)2(1)/4 = 4\u03c0 in3. This is the extra volume that had to be added, over and above the 22.39.\n\n5. Nov 27, 2013\n\n### Camille\n\nOkay, I finally got it right. The first mistake was that 4 in and 1 in are diameters, not radii...\n\nHere's how one can look at it:\n\nNow, obviously the \"additional\" oil that must be poured is the whole blue one, so both on the left and right.\n\n$V_{oil} = V_1 + V_2$\n\nWe know that, because the oil that is hatched in black has still the same volume as it had in the beginning. Now we calculate the equality of pressures at the line of equal pressures (marked green):\n\nLeft = Right\n\n$p_L = p_R$\n\n$p_L = \\frac{W_p + W_{V_1}}{A_L}$\n\n$p_R = \\frac{W_{V_2}}{A_R}$\n\n$\\frac{W_p + W_{V_1}}{A_L} = \\frac{W_{V_2}}{A_R}$\n\nThe areas (now correct...):\n\n$A_L = \u03c0 \\cdot (\\frac{4in}{2})^2 = 12.57 in^2$\n\n$A_R = \u03c0 \\cdot (\\frac{1in}{2})^2 = 0.79 in^2$\n\nAnd the volume of the 1-in layer of oil:\n\n$\\boxed{V_1 = A_L \\cdot 1 in = 12.57 in^3}$\n\nAnd the weight of it:\n\n$W_{V_1} = V_1 \\cdot \u03c1 \\cdot g = 0.39 lbf$\n\nWe calculate the weight of the oil V_2:\n\n$\\frac{W_p + W_{V_1}}{A_L} \\cdot A_R = W_{V_2}$\n\n$W_{V_2} = \\frac{11 lbf + 0.39 lbf}{12.57 in^2} \\cdot 0.79 in^2 = 0.72 lbf$\n\nSo the volume of it is:\n\n$\\boxed{V_2 = \\frac{W_{V_2}}{\u03c1 \\cdot g} = 23.17 in^3}$\n\nSumming two volumes we get:\n\n$V_{oil} = V_1 + V_2 = 12.57 in^3 + 23.17 in^3 = 35.74 in^3$\n\nLast edited: Nov 27, 2013", "date": "2017-10-20 15:12:55", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/oil-volume-required-to-rise-the-piston.725238/", "openwebmath_score": 0.7960912585258484, "openwebmath_perplexity": 1280.883119579281, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290963960277, "lm_q2_score": 0.8774767954920548, "lm_q1q2_score": 0.8521432475146812}}
{"url": "https://www.physicsforums.com/threads/can-i-use-only-one-substitution-for-integral.426195/", "text": "# Homework Help: Can I use only one substitution for integral?\n\n1. Sep 3, 2010\n\n### thereddevils\n\n1. The problem statement, all variables and given/known data\n\nBy using the substitution t = tan x, find\n\n$$\\int \\frac{dx}{\\cos^2 x+4\\sin^2 x}$$\n\n2. Relevant equations\n\n3. The attempt at a solution\n\nWell let tan x=t\n\n$$\\frac{dt}{dx}=\\sec^2 x=\\tan^2 x+1=1+t^2$$\n\nthe integral then becomes\n\n$$\\int \\frac{dx}{\\frac{1}{\\sqrt{1+t^2}}^2+4\\frac{t}{\\sqrt{1+t^2}}^2}$$\n\nwhich simplifies to\n\n$$\\int \\frac{1}{1+4t^2}$$\n\nThen from here i make another substitution **\n\nlet t= 1/2 tan b\n\ndt/db = 1/2 sec^2 b\n\n$$\\int \\frac{1}{1+4(\\frac{1}{2} \\tan b)^2} \\cdot \\frac{1}{2}\\sec^2 b db$$\n\n= b + constant\n\nBack substitute\n\n= $$\\frac{1}{2}\\tan^{-1} (2t)$$ + constant\n\n= $$\\frac{1}{2}\\tan^{-1}(2\\tan x)$$ + constant\n\nAm i correct? Especially this part ** where i made another substitution, is that valid? Or when the question specified the substitution, i have to stick that one substitution only?\n\n2. Sep 3, 2010\n\n### CompuChip\n\nRe: Integration\n\nLooks okay to me.\nIf you remember that\n$$\\int \\frac{dy}{1 + y^2} \\, dy = \\tan^{-1}(y)$$\nthen $y = 2t$ immediately gives you\n$$\\int \\frac{dt}{1+4t^2} = \\frac{1}{2} \\int \\frac{dy}{1 + y^2} = \\frac{1}{2} \\tan^{-1}(y) = \\frac{1}{2} \\tan^{-1}(2 \\tan x)$$\n\n(Note that if this was a definite integral, you would have to be very careful with the integration boundaries in these subsitutions)\n\n3. Sep 3, 2010\n\n### thereddevils\n\nRe: Integration\n\nthanks Compuchip! So it's ok to make another substitution other than the one specified in the question? My teacher said otherwise, saying that it's not appropriate to make substitutions other than the one specified in the question. I want to get my facts right first before i get into another heated argument with her.\n\nTrue, as for definite integrals, we will need to modify the limits in accordance with the substitution we make.\n\n4. Sep 3, 2010\n\n### CompuChip\n\nRe: Integration\n\nPersonally, I think that if you can find a correct way to solve the question yourself (even if it is not exactly the way your teacher or anyone else has in mind) it is appropriate.\n\nIn this case, you could even argue that 2t -> t is not a full-fledged variable substitution but simply a rescaling, and the integral of 1/(1 + t\u00b2) is a standard integral which you can write down immediately whenever you encounter it (however you have proved that standard integral nicely here, which is not bad to do once in your life :) ).\n\n5. Sep 4, 2010\n\n### thereddevils\n\nRe: Integration\n\nThanks again, Compuchip.", "date": "2018-12-17 18:44:36", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/can-i-use-only-one-substitution-for-integral.426195/", "openwebmath_score": 0.9260997772216797, "openwebmath_perplexity": 1129.17751685936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129092218133, "lm_q2_score": 0.8774767986961401, "lm_q1q2_score": 0.852143246960256}}
{"url": "http://www.billthelizard.com/2010/05/sicp-exercise-131-product-of-series.html", "text": "## Saturday, May 1, 2010\n\n### SICP Exercise 1.31: Product of a Series\n\nFrom SICP section 1.3.1 Procedures as Arguments\n\nExercise 1.31 asks us to write a product procedure (analogous to sum) that computes the product of a function at points over a given range. We need to show how to define factorial in terms of the new product procedure, and use product to compute approximations to \u03c0 using the following formula:\n\nFinally, if our product procedure is recursive, we need to write an iterative version, and vice versa.\n\nSince summing a series and multiplying a series are extremely similar ideas, it's not surprising to find that it's very simple to modify sum to produce product. Let's use the recursive version from exercise 1.29 first.\n(define (sum term a next b)   (if (> a b)       0       (+ (term a)          (sum term (next a) next b))))\n\nThe product procedure really only needs to perform a different operation and end on a different value than sum. The last step (when a > b) should be to multiply by 1 instead of adding 0.\n(define (product term a next b)   (if (> a b)     1     (* (term a)        (product term (next a) next b))))\n\nWe can test this out by implementing factorial in terms of product using the identity and inc procedures that we've used before.\n(define (identity x) x)(define (inc x) (+ x 1))(define (factorial x)   (product identity 1 inc x))> (factorial 3)6> (factorial 4)24> (factorial 5)120> (factorial 10)3628800> (factorial 20)2432902008176640000\n\nThe next test is to approximate \u03c0 using something called Wallis' product or the Wallis Formula. (Lucky for us that mathematicians have already expressed this as the product of a series, saving us the work of coming up with a function for generating the terms.)\n\nWe'll multiply the product on the right hand side by the denominator on the left hand side so our procedure will just approximate \u03c0 directly.\n(define (wallis-pi n)   (define (term x)      (/ (* 4.0 (square x))         (- (* 4.0 (square x)) 1)))   (* 2.0 (product term 1 inc n)))> (wallis-pi 10)3.0677038066434994> (wallis-pi 100)3.133787490628163> (wallis-pi 1000)3.1408077460304042> (wallis-pi 10000)3.1415141186819313\n\nSince our product procedure is implemented recursively, we need to show an iterative version. We can go back to the iterative sum procedure for inspiration.\n(define (sum term a next b)   (define (iter a result)   (if (> a b)      result      (iter (next a) (+ (term a) result))))   (iter a 0))\n\nAgain, there are very few changes required to transform this procedure.\n(define (product-iter term a next b)   (define (iter a result)   (if (> a b)       result       (iter (next a) (* (term a) result))))   (iter a 1))\n\nYou can plug product-iter into the factorial and wallis-pi procedures above to verify that they give the same results.\n\nThe similarities between our two versions of sum and product is no accident. This section of SICP is all about higher-order procedures, so in the next exercise we're going to see how pull the similarities out of these procedures into something even more abstract.\n\nRelated:\n\nFor links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.\n\nLepsch said...\n\nHi Bill,\n\nI've made another solution. There you go:\n\n(define (pi4 n)\n(define (term k)\n(if (even? k)\n(/ (+ 2 k) (+ 1 k))\n(/ (+ 1 k) (+ 2 k))\n))\n(product term 1 inc n))\n\nAnonymous said...\n\nThis is actually the correct implementation to the SICP provided formula.", "date": "2017-07-28 02:30:22", "meta": {"domain": "billthelizard.com", "url": "http://www.billthelizard.com/2010/05/sicp-exercise-131-product-of-series.html", "openwebmath_score": 0.5671576261520386, "openwebmath_perplexity": 3718.5134856401196, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290955604489, "lm_q2_score": 0.8774767874818409, "lm_q1q2_score": 0.8521432390025284}}
{"url": "https://math.stackexchange.com/questions/1905422/why-does-frac1x-4-have-two-answers", "text": "# Why does $\\frac{1}{x} < 4$ have two answers?\n\nSolving $\\frac{1}{x} < 4$ gives me $x > \\frac{1}{4}$. The book however states the answer is: $x < 0$ or $x > \\frac{1}{4}$.\n\nMy questions are:\n\nWhy does this inequality has two answers (preferably the intuition behind it)?\n\nWhen using Wolfram Alpha it gives me two answers, but when using $1 < 4x$ it only gives me one answer. Aren't the two forms equivalent?\n\n\u2022 $x$ can be negative...^^ \u2013\u00a0Probabilitytheoryapprentice Aug 27 '16 at 17:01\n\u2022 There is just one answer here, not two as you seem to think. It is not the case that $x>1/4$ is one answer and $X<0$ is another answer. Neither inequality by itself would be a correct answer. \u2013\u00a0Andreas Blass Aug 27 '16 at 17:14\n\u2022 \"when using 1<4x it only gives me 1 answer. Aren't the two forms equivalent?\" Nope. They are not. If $1/x < 1/4$ it's possible that $x < 0$. For $1 < 4x$ it is not possible. \u2013\u00a0fleablood Aug 28 '16 at 0:14\n\u2022 Draw a graph of the function $x \\mapsto 1/x$ should give some intuition why there are two \"zones\" for the solution. \u2013\u00a0quid Aug 28 '16 at 0:26\n\u2022 Just try, for instance, $x=-1$. Is it true in this case that $1/x < 4$? Is it true that $1 < 4x$? \u2013\u00a0TonyK Aug 28 '16 at 12:04\n\nYou have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then $$\\frac{1}{x}<4\\iff4x>1\\iff x>1/4.$$ If $x>0$ and $x>1/4$, then $x>1/4$.\n\nNow suppose $x<0$. Then $$\\frac{1}{x}<4\\iff4x<1\\iff x<1/4.$$ If $x<0$ and $x<1/4$, then $x<0$. So the solution set is $(-\\infty,0)\\cup(1/4, \\infty).$\n\n\u2022 Wow!This answer got so many upvotes! ;-) \u2013\u00a0tatan Mar 24 '18 at 17:28\n\nHere is the solution $$\\frac { 1 }{ x } <4$$$$\\frac { 1-4x }{ x } <0$$$$\\frac { x\\left( 1-4x \\right) }{ { x }^{ 2 } } <0$$$$x\\left( 1-4x \\right) <0$$$$x\\left( 4x-1 \\right) >0$$\n\nso $$x\\in \\left( -\\infty ,0 \\right) \\cup \\left( \\frac { 1 }{ 4 } ,+\\infty \\right)$$\n\n\u2022 nice approach to avoid the case division (+1) \u2013\u00a0b00n heT Aug 27 '16 at 17:06\n\u2022 As soon as I read that there were two answers, I started thinking \"there has to be a quadratic hidden here.\" Thanks for revealing it! \u2013\u00a0Ross Presser Aug 28 '16 at 0:50\n\u2022 Well this is misleading. But the same means you could get solution of $x-1>0$ to be $(-\\infty,0)\\cup(1,+\\infty)$. Your final presentation does need some justification (checking validity of each interval). \u2013\u00a0Ruslan Aug 28 '16 at 15:05\n\u2022 @Ruslan - Not quite. The solution multiplies by $x^2$ between lines 3 and 4, and we know $x^2 > 0$ for all real $x \\neq 0$. So multiplying by $x^2$ is preserves the direction of the inequality for all $x \\neq 0$. With $x - 1 > 0$ however, you'd multiply by $x$ which would require splitting the equality into two cases. However, I do concede that this maybe should have been emphasized in the answer. (Still, it's a very nice answer) \u2013\u00a0David E Aug 29 '16 at 18:39\n\u2022 @RossPresser The hidden quadratic can also be seen by inspection of the graph. 1/x is a hyperbola, therefore is a quadratic form. The obvious 2nd degree terms are just hidden by rotation :) \u2013\u00a0QuantumMechanic Aug 30 '16 at 17:12\n\nI am following the suggestion given by @quid in the comments because I like pictures:\n\nThe orange/red line is the $x$-axis. The yellow line is the line $y=4$. The two blue curves are the graph of $y=1/x$. The solution to the inequality is the set of $x$ values for which the blue curve is below the yellow line. As @quid predicted, this picture gives some intuition for why there are two \"zones\" in the solution.\n\nNote: originally I had $y=\\frac{1}{4}$ which was incorrect, so I changed the answer to reflect the fact that it should be $y=4$ and re-plotted the graph.\n\n\u2022 I agree a picture helps, but feel this would be more helpful if you (a) indicated the solution set explicitly, (b) displayed a smaller $x$-range, (c) used the same scale on both axes and (d) made the markings legible \u2013\u00a0PJTraill Aug 31 '16 at 17:24\n\u2022 @PJTraill I don't feel that the first criticism is relevant to the actual question the OP asked, which was not for the exact value of the solution, but the intuition behind its two-part structure. The second and third criticism strike me as pedantic and unhelpful. The fourth criticism is just wrong -- click on the picture, and the full-size version opens in a new tab, where all of the markings are clearly legible. \u2013\u00a0Chill2Macht Aug 31 '16 at 18:46\n\u2022 It is (of course) up to you, though I thought of them as suggestions rather than criticisms; I hope you were not offended by them. I meant \u201creadily legible as they appear in the answer\u201d, as I prefer to see all relevant information at once. P.S. I do like the clean look of your diagram. \u2013\u00a0PJTraill Sep 1 '16 at 19:11\n\nThe accepted answer is good, but I feel like you're really asking: why is there only one piece to the question, but two pieces to the answer?\n\nThis is actually a great question. Sometimes it happens that one piece turns into two (or more), like when you try to solve $x^2 = 9$ (which has \"one piece\") and get the two-piece solution $x = 3, -3$. Here, to figure out why one piece becomes two, you have to think about how the equation $y = x^2$ works.\n\nSo in our case we should think about how the equation $y = 1/x$ works. And when you think about it, you realize that you didn't really start with one piece. No matter what you plug in for $x$, the value of $1/x$ can never be zero. And that means when you write $1/x < 4$, really this gives you the TWO pieces\n\n$$0 < 1/x < 4$$\n\nand\n\n$$1/x < 0$$\n\nBasically, everything smaller than 4 but with zero removed. And that's why you end up with two pieces at the end -- because that's actually how many you started with!\n\n\u2022 Yeah that was exactly what I was confused about! \u2013\u00a0Augusto Dias Noronha Aug 28 '16 at 11:53\n\u2022 Another way to look at it is that 1/x is a hyperbola and therefore is really a quadratic form, so you'd expect two \"pieces\" to the solution. \u2013\u00a0QuantumMechanic Aug 30 '16 at 17:10\n\u2022 Though you did well to nail what the questioner was confused about, I think you underplay the fact that turning one piece into one piece is only a reasonable expectation of a continuous function while $x \\mapsto 1/x$ is not continuous (or even defined) at $0$, though it is continuous on the pieces you start with and and bijective on its domain. \u2013\u00a0PJTraill Aug 31 '16 at 17:38\n\u2022 Just an idea: you are asked when an expression in x is smaller than 4, and this question has two answers: 1) the value of the expression is smaller than 4 and 2) the expression is negative, hence the two answers. \u2013\u00a0Dominique Sep 1 '16 at 11:06\n\u2022 $y = 5 - {(8x - 1)}^{2}$ is a continuous function but it still yields two solutions to $y < 4$. \u2013\u00a0joeytwiddle Sep 7 '16 at 6:12\n\nJust draw a graph of $1/x$ and you'll see 'why'.\n\nHere is an image by WolframAlpha, with appropriate parts enhanced:\n\nWhenever you're writing $x \\gt \\frac{1}{4}$ you're assuming $x \\gt 0$.\n\nBut for $x \\lt 0$ you have $\\frac{1}{x}\\lt0\\lt4$\n\nEnglish doesn't have good words for this, so exactly what's going on can be a bit tricky to describe if you don't already understand the meaning.\n\nIt is true that every $x$ satisfying $x > \\frac14$ does in fact satisfy $\\frac 1x < 4$.\n\nHowever, the converse fails: there are some $x$ that satisfy $\\frac 1x < 4$ that do not satisfy $x > \\frac 14$.\n\nSo, you have found a simple description of some of the $x$ that satisfy $\\frac 1x < 4$ \u2014 but (presumably) you were being asked to describe all of the $x$ that satisfy $\\frac 1x < 4$.\n\nAnd there are indeed more of them: every $x$ satisfying $x < 0$ satisfies $\\frac 1x < 4$.\n\nNow, what is true is the following: if $\\frac 1x < 4$, then it follows that at least one of the two statements \"$x > \\frac 14$\" and \"$x < 0$\" is true. Thus, this gives a complete description of the solutions to $\\frac 1x < 4$.\n\nPut differently, for every $x$, the following bullet points are either both true or both false:\n\n\u2022 $x$ satisfies $\\frac 1x < 4$\n\u2022 $x$ satisfies one of the statements \"$x > \\frac 14$\", \"$x < 0$\".\n\nRegarding your solution method, you forgot that multiplying by negative numbers reverses the sign of an inequality, and multiplying by zero turns any inequality into an equality. Here, we know that $x$ can't be zero, but it still could be either positive or negative, so you don't know the effect that multiplying by $x$ will have on the inequality.\n\nThe typical way to fix this problem is to break the problem into two parts: one part where you solve the case with the assumption $x<0$, and one part where you solve the case with the assumption $x>0$, and then you put the results together.\n\nwe have $$\\frac{1}{x}<4$$ is true if $$x<0$$ and if $x>0$ we get $$x>\\frac{1}{4}$$\n\n\u2022 What? $\\frac{1}{x} < 4$ if $x < 0$? \u2013\u00a0nbro Aug 31 '16 at 17:19\n\nNo, $$\\frac 1x<4\\text{ and }1<4x$$ are not equivalent.\n\nYou could think so just multiplying by $x$. But a rule says that an inequality is preserved when you multiply by a positive number and inverted with a negative one. So the right thing is\n\n$$\\begin{cases}x>0\\to1<4x,\\\\x<0\\to1>4x.\\end{cases}$$\n\nwe have $\\frac{1}{x}>4$ then $\\frac{1}{x}-4>0$\n\nthat is , $$\\frac{1-4x}{x}>0$$ but the domain of definition is $x\\neq 0$\n\nfirst of all you need to find the zeros and then study the its signs $$1-4x=0$$ then $$x=1/4$$\n\n\\begin{align} & \\underline{\\left. x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\right|\\,\\,\\,-\\infty \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1/4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+\\infty } \\\\ & \\underline{\\left. 1-4x\\,\\, \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left. \\,\\left. \\, \\right| \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,} \\\\ & \\underline{\\left. x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left. \\left. \\, \\right| \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left. \\, \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,} \\\\ & \\left. \\frac{1-4x}{x}\\, \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left. \\,\\left. \\, \\right| \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,+\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left. \\, \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ \\end{align}\n\nas we observe from the above table that inequality is positive only when $0<x<\\frac{1}{4}$\n\nI would like to harp on the meaning of your \"two answers\". When a question asking for a number satisfying some conditions leads to a unique number having that property we are fine. When there are two numbers having the required property then we can say two answers.\n\nHere we have uncountably many real numbers $x$ such that $\\frac1x < 4$. So even the region $x>4$ is infinitely many answers. One possible interpretation that could justify \"two\" could be number of connected components of the solution space of the question. This happens many times. The set $GL(n,\\mathbf{R})$ of non-singular real $n\\times n$ matrices has two connected components.\n\n$$\\frac{1}{x} < 4,$$ $$\\frac{1}{x} - 4 < 0,$$ $$\\frac{1 - 4x}{x} < 0,$$ $1 - 4x < 0$ or $x > 0$. Because if $1 - 4x$ is negative then $x$ must be positive. So we must write the solution in that way.\n\nHere is an important aspect which should be always considered. If someone asks me:\n\nProblem: Find the solution of \\begin{align*} \\frac{1}{x}<4 \\end{align*} I would not answer the problem, but instead ask: What is the domain of $x$?\n\nPlease note the problem is not fully specified if the domain of $x$, the range of validity, is not given. This is crucial to determine the set of solutions.\n\nSome examples:\n\nFind the solution of", "date": "2019-05-25 02:49:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1905422/why-does-frac1x-4-have-two-answers", "openwebmath_score": 0.9182042479515076, "openwebmath_perplexity": 345.47440708210667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9711290963960278, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.8521432288451961}}
{"url": "https://math.stackexchange.com/questions/1557954/how-to-show-sum-k-n-infty-frac1k-leq-frac2n", "text": "# How to show $\\sum_{k=n}^\\infty{\\frac{1}{k!}} \\leq \\frac{2}{n!}$\n\n$$\\sum_{k=n}^\\infty{\\frac{1}{k!}} \\leq \\frac{2}{n!}$$\n\nCan someone show why this estimate holds true? I tried quite a bit but couldn't really find a way to approach this. WolframAlpha says it is true but I don't know what the gamma function is.\n\n$$\\sum_{k=n}^\\infty{\\frac{1}{k!}} = \\frac{1}{n!} + \\sum_{k = n+1}^\\infty \\frac{1}{k!}$$ So then I need to show that$$\\sum_{k=n+1}^\\infty{\\frac{1}{k!}} \\leq \\frac{1}{(n+1)!} ~~\\Big[\\leq \\frac{1}{n!}\\Big]$$\n\nIs it possible to do this by induction? I don't really know how to approach this now.\n\n\u2022 Just in case: the inequality doesn't hold for $n = 0$. For $n \\geqslant 1$, note that $(n+k)! \\geqslant n!\\cdot (n+1)^k$. \u2013\u00a0Daniel Fischer Dec 3 '15 at 10:44\n\u2022 $$\\sum_{k=n}^{\\infty}{\\frac{1}{k!}}\\le\\frac{2}{n!}\\Longleftrightarrow$$ $$e-\\frac{e\\Gamma(n,1)}{\\Gamma(n)}\\le\\frac{2}{n!}\\space\\text{for}\\space n>-1$$ \u2013\u00a0Jan Dec 3 '15 at 10:51\n\nIt suffices to show that $$\\sum_{k=n+1}^\\infty \\frac{1}{k!} \\le \\frac{1}{n!}$$\n\nFor all $k \\ge n+1$, we have $$k! \\ge 2^{k-n} \\cdot n!$$\n\nTherefore, we have $$\\sum_{k=n+1}^\\infty \\frac{1}{k!} \\le \\sum_{k=n+1}^\\infty \\frac{1}{n! \\cdot 2^{k-n}} = \\frac{1}{n!} \\sum_{i=1}^\\infty \\frac{1}{2^i} = \\frac{1}{n!}$$\n\n\u2022 Does the inequality $k! \\geq 2^{k-n} \\cdot n!$ have a name? I think I need to prove that in order to use it. \u2013\u00a0elfeck Dec 3 '15 at 11:06\n\u2022 Not really. For a proof note that $\\frac{k!}{n!} = k(k-1) \\cdots (n+1) \\ge 2 \\cdot 2 \\cdots 2 = 2^{n-k}$. Of course, this doesn't hold when $n=0$, so you need to exclude that. \u2013\u00a0Gyumin Roh Dec 3 '15 at 11:06\n\u2022 Okay thanks. Since in my case I substitute $n$ for $n+1$ anyway $n=0$ does not cause issues. Thanks for your help! \u2013\u00a0elfeck Dec 3 '15 at 11:10\n\nAnother way:\n\n$$\\sum_{k=n}^{\\infty} \\frac{1}{k!} = \\frac{1}{n!} \\bigg(1 + \\frac{1}{n+1} + \\frac{1}{(n+1)(n+2)} + \\ldots \\bigg) < \\frac{1}{n!} \\bigg(1 + \\frac{1}{n+1} + \\frac{1}{(n+1)^2} + \\ldots \\bigg)\\\\ = \\frac{1}{n!} \\cdot \\frac{n+1}{n} = \\frac{1}{n!} \\cdot \\bigg(1 + \\frac{1}{n} \\bigg)$$ Now compare this expression to what you have on the RHS: certain terms cancel out. What do you get?\n\n\u2022 Small typo, that $k$ should start from $n$, not $n+1$. \u2013\u00a0Gyumin Roh Dec 3 '15 at 11:18\n\u2022 yeah, that's how I solved it anyway \u2013\u00a0Alex Dec 3 '15 at 11:44\n\n$$\\dfrac{1}{(k+1)!} = \\dfrac{1}{(k-1)!}(\\dfrac{1}{k} - \\dfrac{1}{k+1}) \\leq \\dfrac{1}{(n-1)!}(\\dfrac{1}{k}-\\dfrac{1}{k+1})$$ when $k\\geq n$\n\nSo\n\n$$\\sum_{k=n}^\\infty\\dfrac{1}{(k+1)!} \\leq \\dfrac{1}{(n-1)!}\\sum_{k=n}^\\infty (\\dfrac{1}{k}-\\dfrac{1}{k+1}) = \\dfrac{1}{(n-1)!} *\\frac{1}{n} = \\dfrac{1}{n!}$$", "date": "2019-05-21 19:42:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1557954/how-to-show-sum-k-n-infty-frac1k-leq-frac2n", "openwebmath_score": 0.8483309745788574, "openwebmath_perplexity": 330.24868916678713, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232924970205, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8521429519800091}}
{"url": "http://mathhelpforum.com/calculus/50663-definite-integral-ln-1-t-dt.html", "text": "# Math Help - Definite integral ln(1+t)dt\n\n1. ## Definite integral ln(1+t)dt\n\nI need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..\n\n$\\int_{0}^{5} ln(1+t)dt$\n\nMy first way of thinking led me to a dead end. I used integration by parts with $U=ln(1+t)$ and $U=1/(1+t)$ and $V=1dt$ and $V=t$\n\nThis led me to the following\n\n$= [tln(1+t)]_{0}^{5} - \\int_{0}^{5} t/(1+t)dt$\n\nand this is where i'm stuck.. can't seem to figure out the integration for the second integral here.\n\nI think my line of thinking is wrong.. is my assumption that $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?\n\n2. Originally Posted by mkelly09\nI need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..\n\n$\\int_{0}^{5} ln(1+t)dt$\n\nMy first way of thinking led me to a dead end. I used integration by parts with $U=ln(1+t)$ and $U=1/(1+t)$ and $V=1dt$ and $V=t$\n\nThis led me to the following\n\n$= [tln(1+t)]_{0}^{5} - \\int_{0}^{5} t/(1+t)dt$\n\nand this is where i'm stuck.. can't seem to figure out the integration for the second integral here.\n\nI think my line of thinking is wrong.. is my assumption that $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?\nYou're correct...but you need to think a little \"outside of the box\" for this integral.\n\n$\\int\\frac{t}{1+t}\\,dt$\n\nLet $u=t+1\\implies t=u-1$\n\nThus, $\\,du=\\,dt$\n\nTherefore, the integral becomes $\\int\\frac{t\\,du}{u}=\\int\\frac{u-1}{u}\\,du=\\int\\left(1-\\frac{1}{u}\\right)\\,du$\n\nCan you take it from here?\n\n--Chris\n\n3. Originally Posted by Chris L T521\nYou're correct...but you need to think a little \"outside of the box\" for this integral.\n\n$\\int\\frac{t}{1+t}\\,dt$\n\nLet $u=t+1\\implies t=u-1$\n\nThus, $\\,du=\\,dt$\n\nTherefore, the integral becomes $\\int\\frac{t\\,du}{u}=\\int\\frac{u-1}{u}\\,du=\\int\\left(1-\\frac{1}{u}\\right)\\,du$\n\nCan you take it from here?\n\n--Chris\nHere's another way I just thought of: add \"zero\" to the numerator\n\n$\\int\\frac{t}{t+1}\\,dt=\\int\\frac{t+1-1}{t+1}\\,dt=\\int\\left[\\frac{t+1}{t+1}-\\frac{1}{t+1}\\right]\\,dt=\\int\\left(1-\\frac{1}{t+1}\\right)\\,dt$\n\n--Chris\n\n4. using integration by parts on that guy directly is overkill. (that is, if you know the integral of ln(x) by heart--which you should).\n\n$\\int_0^5 \\ln (1 + t)~dt$\n\nLet $u = t + 1$, this substitution yields\n\n$\\int_1^6 \\ln u~du$\n\nwhich we know is $u \\ln u - u \\bigg|_1^6$ ...\n\nsee post #7 here for how to integrate ln(x) using integration by parts. note that you could also figure out the integral by contemplating the derivative of xln(x)\n\n5. Thanks for your replies, they certainly solved my problem. In fact, now i know of a few different ways of doing it when before i couldn't figure out one.\n\nI think i will be committing to memory the antiderivative of ln(x)", "date": "2014-08-23 08:16:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/50663-definite-integral-ln-1-t-dt.html", "openwebmath_score": 0.8994470238685608, "openwebmath_perplexity": 273.0609817791906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9828232889752296, "lm_q2_score": 0.8670357666736772, "lm_q1q2_score": 0.8521429438613832}}
{"url": "https://math.stackexchange.com/questions/2008784/what-is-the-reasoning-behind-why-the-ratio-test-works", "text": "# What is the reasoning behind why the ratio test works?\n\nThe ratio test says that if we have\n\n$$\\sum_{n=1}^{\\infty}a_n$$\n\nsuch that $\\lim_{n \\to \\infty} \\dfrac{a_{n+1}}{a_n} = L$, then if:\n\n1) $L < 1$, then $\\sum_{n=1}^{\\infty}a_n$ is absolutely convergent,\n\n2) $L > 1$, then $\\sum_{n=1}^{\\infty}a_n$ is divergent, and\n\n3) $L = 1$, then the ratio test gives no information.\n\nI want to understand the mathematics behind the ratio test. I want to know the reasons behind why and how this works, rather than just memorising a formula.\n\nI have attempted to reason about this theorem myself. It can be seen that we're taking the ratio between the second term in the series, $a_{n+1}$, and the first term in the series, $a_n$. This is exactly how one finds the common ratio in a geometric sequence ($r = \\dfrac{a_{n+1}}{a_n} )$.\n\nWe then take the limit of this ratio, which I assume is to find the relative rate of change between $a_{n+1}$ and $a_n$. In which case, It is more effective to take the absolute value of the limit, $\\lim_{n \\to \\infty} \\begin{vmatrix}{ \\dfrac{a_{n+1}}{a_n} }\\end{vmatrix} = L$. This seems analagous to how the limit comparison test theorem works. Therefore, I presume that if $L < 1$, this implies that $a_n$ is has a greater rate of change than $a_{n+1}$, which implies that each successive term in the series is getting smaller, and as we go to infinity, each successive term is converging towards $0$. As such, the series should converge to some value. Analogously, I presume that if $L > 1$, this implies that $a_{n+1}$ has a greater rate of change than $a_n$, which implies that each successive term in the series is getting larger, and as we go to infinity, the terms diverge towards infinity.\n\nIs this reasoning correct? If anything is incorrect, please clarify why and what the correct reasoning is.\n\nThank you.\n\n\u2022 \"each succesive term is converging towards zero. As such, the series should converge.\" This is a famously fallacious statement. The harmonic series tends termwise to zero, but the series diverges. \u2013\u00a0Matthew Leingang Nov 11 '16 at 1:13\n\u2022 @MatthewLeingang I forgot about that. You're absolutely correct. In which case, what is the correct way to reason about that section? \u2013\u00a0The Pointer Nov 11 '16 at 1:14\n\u2022 @GCab Which is not less than $1$ . . . \u2013\u00a0Noah Schweber Nov 11 '16 at 1:19\n\u2022 @GCab That's correct, the ratio test doesn't apply to the harmonic series. But the statement I quoted is still invalid. \u2013\u00a0Matthew Leingang Nov 11 '16 at 1:20\n\u2022 next time read wikipedia's entry en.wikipedia.org/wiki/Ratio_test#Proof \u2013\u00a0reuns Nov 11 '16 at 1:35\n\nYou have some ideas correct, but remember that $a_n \\rightarrow 0$ does not imply that $\\sum a_n$ converges. So we need something more than that. However, for the case when $L>1$, your reasoning is correct.\n\nFor the case $L<1$, we know that there exists some $r < 1$ and $N \\in \\mathbb{N}$ such that $$\\frac{a_{n+1}}{a_n} \\leq r < 1,$$ whenever $n \\geq N$.\n\nTherefore,\n\n$$a_{N+1} \\leq ra_N,$$ $$a_{N+2} \\leq ra_{N+1} \\leq r^2a_N$$ $$a_{N+3} \\leq ra_{N+2} \\leq r^3a_N,$$ and in general $$a_{N+n} \\leq r^na_N.$$\n\nTherefore, at least eventually, we can compare the series from above with a convergent geometric series $\\sum a_N r^n$, which implies the convergence of the original series $\\sum a_n$.\n\n\u2022 I'm not sure if your notation has the same meaning as mine or if you've defined it differently. For instance, how can $\\dfrac{ a_{n+1} }{a_n} < r$? I've defined $r = \\dfrac{ a_{n+1} }{a_n}$. \u2013\u00a0The Pointer Nov 11 '16 at 1:54\n\u2022 My notation is different. If you set $r$ equal to the ratio, then it is changing with $n$. But when $L < 1$, such a fixed $r$ exists as I employed it in my answer to your question. \u2013\u00a0user333870 Nov 11 '16 at 2:00\n\u2022 Can you please elaborate on what $r$ and $N$ are? \u2013\u00a0The Pointer Nov 11 '16 at 2:03\n\u2022 Since $lim \\frac{a_{n+1}}{a_n} = L < 1$, take $\\epsilon = \\frac{1-L}{2}$, then by the definition of the limit, there exists $N\\in \\mathbb{N}$ such that $n \\geq N \\Longrightarrow -\\epsilon < \\frac{a_{n+1}}{a_n} - L < \\epsilon$, so you can take for example $r = \\epsilon + L = \\frac{1+L}{2}$ \u2013\u00a0user333870 Nov 11 '16 at 2:18\n\u2022 I'm sorry but this still doesn't tell me what $N$ and $r$ are supposed to be. I cannot see the relevance between your answer and my original question. \u2013\u00a0The Pointer Nov 11 '16 at 5:28", "date": "2020-03-29 10:27:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2008784/what-is-the-reasoning-behind-why-the-ratio-test-works", "openwebmath_score": 0.9150016903877258, "openwebmath_perplexity": 166.52440182758374, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232869627774, "lm_q2_score": 0.8670357683915537, "lm_q1q2_score": 0.8521429438048842}}
{"url": "http://math.stackexchange.com/questions/910361/rationalize-the-denominator-and-simplify", "text": "Rationalize the denominator and simplify\n\nSo the problem I have says rationalize the denominator and simplify. $$\\frac{ \\sqrt{15}}{\\sqrt{10}-3}$$\n\nMy answer I got was $\\frac{5 \\sqrt6}{7}$.\n\nAm I doing this wrong or is this the wrong answer I was told it was incorrect?\n\n-\nSo it is pretty easy to check that your answer is wrong, just use a calculator. You will get $23.87$ for the question and $1.75$ for your answer. \u2013\u00a0David Aug 27 '14 at 0:13\n\nIt seems that you tried multiplying by $\\frac{\\sqrt{10}}{\\sqrt{10}}$. Instead, you should try multiplying by the conjugate and take advantage of difference of squares: $$\\frac{\\sqrt{15}}{\\sqrt{10} - 3} = \\frac{\\sqrt{15}}{\\sqrt{10} - 3} \\cdot \\frac{\\sqrt{10} + 3}{\\sqrt{10} + 3} = \\frac{\\sqrt{150} + 3\\sqrt{15}}{(\\sqrt{10})^2 - 3^2} = 5\\sqrt{6} + 3\\sqrt{15}$$\nI think wht happened was that you correctly multiplied the denominator by $\\sqrt{10}+3$, but incorrectly multiplied the numerator by $\\sqrt{10}$. The numerator should also have been multiplied by $\\sqrt{10}+3$\n$$\\frac{\\sqrt{15}}{\\sqrt{10}-3}=\\frac{\\sqrt{15} \\cdot (\\sqrt{10}+3)}{(\\sqrt{10}-3) \\cdot (\\sqrt{10}+3)}=\\frac{\\sqrt{15} \\cdot (\\sqrt{10}+3)}{10-9}=\\sqrt{15} \\cdot (\\sqrt{10}+3)$$", "date": "2016-06-25 03:48:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/910361/rationalize-the-denominator-and-simplify", "openwebmath_score": 0.9746134877204895, "openwebmath_perplexity": 149.09013451090843, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232909876816, "lm_q2_score": 0.8670357580842941, "lm_q1q2_score": 0.8521429371644054}}
{"url": "https://reference.wolframcloud.com/language/ref/CharacteristicFunction.html", "text": "# CharacteristicFunction\n\nCharacteristicFunction[dist,t]\n\ngives the characteristic function for the distribution dist as a function of the variable t.\n\nCharacteristicFunction[dist,{t1,t2,}]\n\ngives the characteristic function for the multivariate distribution dist as a function of the variables t1, t2, .\n\n# Examples\n\nopen allclose all\n\n## Basic Examples(4)\n\nCharacteristic function (cf) for the normal distribution:\n\nCharacteristic function for the binomial distribution:\n\nCharacteristic function for the bivariate normal distribution:\n\nCharacteristic function for the multinomial distribution:\n\n## Scope(8)\n\nCharacteristic function for a specific continuous distribution:\n\nCharacteristic function for a specific discrete distribution:\n\nCharacteristic function at a particular value:\n\nCharacteristic function evaluated numerically:\n\nObtain a result at any precision:\n\nCompute the characteristic function for a formula distribution:\n\nFind the characteristic function for a parameter mixture distribution:\n\nCharacteristic function for the slice distribution of a random process:\n\n## Applications(7)\n\nCompute the raw moments for a Poisson distribution:\n\nFirst 5 raw moments using derivatives of the characteristic function at the origin:\n\nUse Moment directly:\n\nCompute mixed raw moments for a multivariate distribution:\n\nUse Moment to obtain raw moments directly:\n\nFind raw moments of a Student distribution from its characteristic function:\n\nCompute to extract moments by taking limits from the right:\n\nEvaluate the limits from the left:\n\nOnly the first four moments are defined, as confirmed by using Moment directly:\n\nUse inverse Fourier transform to compute the PDF corresponding to a characteristic function:\n\nIllustrate the central limit theorem on the example of symmetric LaplaceDistribution:\n\nFind the characteristic function of the rescaled random variate:\n\nCompute the large limit of the cf of the sum of such i.i.d. random variates:\n\nCompare with the characteristic function of a standard normal variate:\n\nUse smooth characteristic function to construct the upper bound for the distribution density of ErlangDistribution:\n\nPlot the upper bounds and the original density:\n\nVerify that the sum where are independent identically distributed variates tends in distribution to for large :\n\nUse a combinatorial equality for product :\n\nEvaluate the sum:\n\nTake the limit and compare it to the characteristic function of the UniformDistribution:\n\n## Properties & Relations(5)\n\nCharacteristicFunction is the Expectation of for real :\n\nThe characteristic function is related to all other generating functions when they exist:\n\nThe cf of a continuous distribution is equivalent to FourierTransform of its PDF:\n\nThe cf of a discrete distribution is equivalent to FourierSequenceTransform of its PDF:\n\nThe PDF is the inverse Fourier transform of the cf for continuous distributions:\n\nThe PDF is the inverse Fourier sequence transform of the cf for discrete distributions:\n\n## Possible Issues(1)\n\nSymbolic closed forms do not exist for some distributions:\n\n## Neat Examples(1)\n\nVisualize real and imaginary parts of CharacteristicFunction for random instances of BinomialDistribution:\n\nWolfram Research (2007), CharacteristicFunction, Wolfram Language function, https://reference.wolfram.com/language/ref/CharacteristicFunction.html (updated 2010).\n\n#### Text\n\nWolfram Research (2007), CharacteristicFunction, Wolfram Language function, https://reference.wolfram.com/language/ref/CharacteristicFunction.html (updated 2010).\n\n#### CMS\n\nWolfram Language. 2007. \"CharacteristicFunction.\" Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2010. https://reference.wolfram.com/language/ref/CharacteristicFunction.html.\n\n#### APA\n\nWolfram Language. (2007). CharacteristicFunction. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/CharacteristicFunction.html\n\n#### BibTeX\n\n@misc{reference.wolfram_2022_characteristicfunction, author=\"Wolfram Research\", title=\"{CharacteristicFunction}\", year=\"2010\", howpublished=\"\\url{https://reference.wolfram.com/language/ref/CharacteristicFunction.html}\", note=[Accessed: 11-August-2022 ]}\n\n#### BibLaTeX\n\n@online{reference.wolfram_2022_characteristicfunction, organization={Wolfram Research}, title={CharacteristicFunction}, year={2010}, url={https://reference.wolfram.com/language/ref/CharacteristicFunction.html}, note=[Accessed: 11-August-2022 ]}", "date": "2022-08-11 23:20:07", "meta": {"domain": "wolframcloud.com", "url": "https://reference.wolframcloud.com/language/ref/CharacteristicFunction.html", "openwebmath_score": 0.8522660732269287, "openwebmath_perplexity": 2928.7487258549563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232874658904, "lm_q2_score": 0.8670357580842941, "lm_q1q2_score": 0.8521429341108864}}
{"url": "https://tex.stackexchange.com/questions/377488/equation-custom-horizontal-alignment-numbering-each-row", "text": "# equation custom horizontal alignment & numbering each row\n\nAs I know, \"align\" has the fixed horizontal alignment right-left-right-left-..., and we cannot change this alignment. So if we want to align equations with custom horizontal alignment, e.g., right-center-center-left, we may use \"equation\" or \"gather\", \"array\", \"arraycolsep\" (for spacing \"=\" like \"align\"), and \"displaystyle\" (for proper handling of \"frac\" or \"lim\").\n\nFor example,\n\n\\newcommand{\\argmax}{\\operatornamewithlimits{arg\\,max}}\n...\n\\begin{gather}\n\\arraycolsep=1.4pt\\def\\arraystretch{2.2}\n\\begin{array}{rccl}\np_{\\mathrm{MLE}}(x) & = & \\displaystyle \\max_{m} &P(X = x | \\theta = m) \\\\\nm_{\\mathrm{MLE}}(x) & = & \\displaystyle \\argmax_{m} &P(X = x | \\theta = m)\n\\end{array}\n\\end{gather}\n\n\nproduces below.\n\nHowever, this has the only one numbering label. It cannot split numbers for each row. It may be one BAD choice to use \"align\" and adjust spacing MANUALLY by using \"\\,\", \"\\phantom{}\", \"\\quad\", or \"\\qquad\".\n\nFor this case, how can we label different numbers for each row?\n\n\u2022 You probably already know about the option \\notag that removes the numbering of one row in align environment. I think there must be something similar, but instead it forces a tag for each entry in multi-rows of math environments. What you ask is indeed interesting and I am waiting to find the answer \u2013\u00a0Al-Motasem Aldaoudeyeh Jun 30 '17 at 2:54\n\u2022 Off-topic: Don't use | to denote \"given that\" items. Instead, write \\mid. \u2013\u00a0Mico Jun 30 '17 at 8:04\n\u2022 A newenvironment here could help... But I am waiting for an answer in a problem I found trying to manipulate the rows of my newenvironment... I will come back for an answer... But have to study a little bit about newenvironments \u2013\u00a0koleygr Jun 30 '17 at 8:08\n\nSince you're using the amsmath package, I would use that package's \\DeclareMathOperator* directive to create two new \"operators\": \\argmax and \\midmax, where the latter displays the word \"max\" centered in a box of width equal to \"arg max\". I would also use a split environment instead of an array environment, an equation environment instead of a gather environment, and \\mid instead of \\.\n\nIf you need to number each row separately, use an align environment instead of the nested equation/ split environments.\n\n\\documentclass{article}\n\\usepackage{amsmath}\n%% Create two new math opertors: \\argmax and \\midmax\n\\DeclareMathOperator*{\\argmax}{arg\\,max}\n\\newlength\\mylen\n\\settowidth\\mylen{arg\\,max}\n\\DeclareMathOperator*{\\midmax}{\\parbox{\\mylen}{\\centering\\upshape max}} % center-set \"max\"\n\\begin{document}\n\n%% Single equation number for both rows:\n$$\\begin{split} p_{\\mathrm{MLE}}(x) &= \\midmax_{m} P(X = x \\mid \\theta = m) \\\\ m_{\\mathrm{MLE}}(x) &= \\argmax_{m} P(X = x \\mid \\theta = m) \\end{split}$$\n\n% Separate equation numbers, one per row:\n\\begin{align}\np_{\\mathrm{MLE}}(x) &= \\midmax_{m} P(X = x \\mid \\theta = m) \\\\\nm_{\\mathrm{MLE}}(x) &= \\argmax_{m} P(X = x \\mid \\theta = m)\n\\end{align}\n\\end{document}", "date": "2019-10-23 00:51:03", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/377488/equation-custom-horizontal-alignment-numbering-each-row", "openwebmath_score": 0.9859458208084106, "openwebmath_perplexity": 3056.945185071717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232897298992, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8521429326971245}}
{"url": "https://math.stackexchange.com/questions/1027995/volume-of-a-split-log/1028005", "text": "# Volume of a split log\n\nWhen solving the following problem, I could not understand why my reasoning came up with an answer that's different than the one on the solution's manual.\n\nQuestion: Consider $(x,y,z)$ such that $x^2+y^2<1, x>0, 0 \\le z \\le 5$. This describes one half of a cylinder (a split log). Chop out a wedge out of the log along $z=2x$. Find the volume of the wedge.\n\nMy reasoning:\n\nWedge Volume $=$ (half the volume of the cylinder of height 2) $-$ (volume of solid of revolution found revolving the curve $x=z/2$, from 0 to 2, over the z axis).\n\nHalf Cylinder Volume: $(\\pi r^2h)/2 = \\pi/2 \\int_0^2 (1)^2 dz = \\pi$\n\n(Volume of solid of revolution found revolving the curve x=z/2, from 0 to 2, over the z axis): $(\\pi r^2h)/2 = \\pi/2 \\int_0^2 (z/2)^2 dz = \\pi/3$\n\nWedge Volume: $\\pi - \\pi/3 = 2\\pi/3$\n\nHowever, the solution's manual answer is $4/3$ and it displays a different reasoning:\n\nThe slice perpendicular to the $xz$-plane are right triangles with base of length $x$ and height $z=2x$. Therefore the area of a slice is $x^2$. The volume is:\n\n$$\\int_{-1}^1 x^2 dy = \\int_{-1}^1 (1-y^2) dy = 4/3.$$\n\nWhile I understand the solution's manual reasoning and even find it simpler than mine, I still cannot understand why the two approaches results in different answers. What am I missing?\n\nYour reasoning in your first method goes wrong in that chopping a wedge isn't the same thing as taking the solid of revolution of a curve. The solid of revolution of $z = 2x$ is a cone, so if you take a cone out of the cylinder, you're not left with a wedge but a somewhat weird-looking anti-cone shape.\nIf we were to analyze the cross-section of the wedge that we want in cylindrical coordinates, we'd see that it grows as the angle approaches $0$, and shrinks as it approaches either $\\pi/2$ or $-\\pi/2$. Whereas with your solid of revolution, the cross-section stays constant at all angles.", "date": "2020-01-19 22:00:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1027995/volume-of-a-split-log/1028005", "openwebmath_score": 0.8840994238853455, "openwebmath_perplexity": 358.86122611087717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232889752296, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.8521429303544297}}
{"url": "https://math.stackexchange.com/questions/1732914/determinant-of-n-times-n-matrix", "text": "# Determinant of $N \\times\\ N$ matrix\n\nFor $n \\geq 2$, compute the determinant of the following matrix: $$B = \\begin{bmatrix} -X & 1 & 0 & \\cdots & 0 & 0 \\\\ 0 & -X & 1 & \\ddots & \\vdots & \\vdots \\\\ \\vdots & \\ddots & \\ddots & \\ddots & 0 & \\vdots \\\\ \\vdots & & \\ddots & \\ddots & 1 & 0 \\\\ 0 & \\cdots & \\cdots & 0 & -X & 1 \\\\ a_0 & a_1 & \\cdots & \\cdots & a_{n-2} & (a_{n-1} - X) \\end{bmatrix}$$\n\nLooking at the $2 \\times 2$ and $3 \\times 3$ forms of this matrix:\n\n$\\det \\begin{bmatrix} -X & 0 \\\\ 0 & (a_1-X) \\end{bmatrix} = -X(a_1-X) - 0 = X^2 - a_1X$\n\nby expansion along the first row:\n\n$\\det \\begin{bmatrix} -X & 1 & 0 \\\\ 0 & -X & 0 \\\\ 0 & 0 & (a_2-X) \\end{bmatrix} = (-X) \\times\\det \\begin{bmatrix} -X & 0 \\\\ 0 & a_2-X \\end{bmatrix} - 1 \\det\\begin{bmatrix} 0 & 0 \\\\ 0 & a_2-X \\end{bmatrix}$\n\n$= (-X)[(-X)(a_2-X) -0] - 0 = X^3 - a_2X^2$\n\nSo it looks like:\n\n$\\det \\begin{bmatrix} -X & 1 & 0 & \\cdots & 0 & 0 \\\\ 0 & -X & 1 & \\ddots & \\vdots & \\vdots \\\\ \\vdots & \\ddots & \\ddots & \\ddots & 0 & \\vdots \\\\ \\vdots & & \\ddots & \\ddots & 1 & 0 \\\\ 0 & \\cdots & \\cdots & 0 & -X & 1 \\\\ a_0 & a_1 & \\cdots & \\cdots & a_{n-2} & (a_{n-1} - X) \\end{bmatrix} = X^{n} - a_{n-1}X^{n-1} - a_{n-2}X^{n-2} ... - a_1X$\n\nDoes this look right? Is \"prove by induction\" valid to use here?\n\n\u2022 Your $2\\times 2$ matrix must be $\\begin{bmatrix} -X & 1 \\\\ a_0 & (a_1-X) \\end{bmatrix}$. You have also done the same mistake for the next case. I do not think you can use induction here, but you can get an intuitive idea about the general expression. Apr 8, 2016 at 4:08\n\u2022 That's essentially the companion matrix. The induction step that works is expansion by minors along the first column. Apr 8, 2016 at 4:29\n\nLet $v= \\begin{pmatrix} 1 \\\\ x \\\\ x^{2} \\\\ \\cdot \\\\ \\cdot \\\\ \\cdot \\\\ x^{n-2} \\\\x^{n-1} \\end{pmatrix}$.\nThen $Bv = (x-X)v \\iff a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} -x^{n} = p(x) =0$. Thus, all the roots $\\alpha$ of the monic polynomial $p(x)$ of degree $n$ noted here give us our eigenvectors $v_{\\alpha}$, which are linearly independent since they are columns of a Vandermonde matrix. The associated eigenvalues are $\\lambda = \\alpha - X$.\n\u2022 Thanks! Since the characteristic polynomial consists the trace and the determinant. So the detminant of this Vandermonde matrix is $a_0$. Since B is similar to the matrix, so det(B) = $a_0$. Am I right? Apr 8, 2016 at 11:42\n\u2022 You are not right. You've already conjectured correctly what the determinant of $B$ should be: a polynomial in $X$. For example in your case for $n=2$, you found out that $\\det B = X^{2} - a_{1} X - a_{0}$. In my answer I found linearly independent eigenvectors with associated eigenvalues. Determinant is the product of eigenvalues. Let $n = 2$ and $\\alpha_{1}, \\alpha_{2}$ are the roots of $p(x) = x^{2} - a_{1}x - a_{0}$, then by my answer above $\\det B = (\\alpha_{1} - X)(\\alpha_{2} -X) = X^{2} - (\\alpha_{1} +\\alpha_{2})X + \\alpha_{1} \\alpha_{2}$ - roots and coefficients are related (how?) Apr 8, 2016 at 13:14\n\u2022 Note that this proof requires a bit of elementary algebraic geometry in order to work. Per se, the roots of the polynomial $p$ may fail to be distinct, in which case you do not get $n$ linearly independent eigenvectors. So you need to first WLOG assume that the resultant of the polynomial $p$ and its derivative $p'$ is $\\neq 0$ (this is true on a Zariski-dense subset of the space of all polynomials, so you can WLOG assume it). Apr 11, 2019 at 18:55\nTo add a final touch somewhat as a synthesis of the (thorough) answers of @user5713492 and @akech: the global result is that the companion matrix of polynomial p(X) is diagonalized with a Vandermonde matrix V(r_1,r_2,\\cdots r_n) where the $r_k$ are the roots of p(X)\" see https://en.wikipedia.org/wiki/Vandermonde_matrix", "date": "2022-08-14 23:31:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1732914/determinant-of-n-times-n-matrix", "openwebmath_score": 0.8780083060264587, "openwebmath_perplexity": 235.99883576988697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232919939075, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8521429279066236}}
{"url": "https://math.stackexchange.com/questions/3191278/find-the-length-x-such-that-the-two-distances-in-the-triangle-are-the-same", "text": "# Find the length x such that the two distances in the triangle are the same\n\nI have been working on the following problem\n\n## Statement\n\nAssume you have a right angle triangle $$\\Delta ABC$$ with cateti $$a$$, $$b$$ and hypotenuse $$c = \\sqrt{a^2 + b^2}$$. Find or construct a point $$D$$ on the hypothenuse such that the distance $$|CD| = |DE|$$, where $$E$$ is positioned on $$AB$$ in such a way that $$DE\\parallel BC$$ ($$DE$$ is parallel to $$BC$$).\n\n## Background\n\nMy background for wanting such a distance is that I want to create a semicircle from C onto the line $$AB$$. This can be made clearer in the image below\n\nTo be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem\n\n## Solution\n\nUsing similar triangles one arrives at the three equations\n\n\\begin{align*} \\frac{\\color{blue}{\\text{blue}}}{a - x} & = \\frac{b}{a} \\\\ \\frac{\\color{red}{\\text{red}}}{x} & = \\frac{c}{a} \\\\ \\color{red}{\\text{red}} & = \\color{blue}{\\text{blue}} \\end{align*}\n\nWhere one easily can solve for $$\\color{blue}{\\text{blue}}$$, $$\\color{red}{\\text{red}}$$, $$x$$.\n\n## Question\n\nI feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $$D$$ in a simpler matter?\n\n\u2022 You still have not described what $F$ is either from the statement or from the graph. \u2013\u00a0Hw Chu Apr 17 at 16:42\n\u2022 Right when I said $F$ i meant $E$. I will fix it in the problem statement =) \u2013\u00a0N3buchadnezzar Apr 17 at 16:47\n\u2022 cateti is Italian for legs \u2013\u00a0J. W. Tanner Apr 17 at 16:56\n\u2022 Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function! \u2013\u00a0David K Apr 17 at 20:25\n\nLet the angle bisector of $$\\angle ACB$$ intersect side $$\\overline{AB}$$ at point $$E$$.\n\nLet the measure of $$\\angle ACB$$ be $$\\alpha$$. Then $$m \\angle ACE = \\dfrac{\\alpha}{2}$$.\n\nLet the line perpendicular to side $$\\overline{AB}$$ at point $$E$$ intersect side $$\\overline{AC}$$ at point $$D$$.\n\nSince $$\\overline{ED}$$ is parallel to $$\\overline{BC}$$, then $$\\angle ADE \\cong \\angle ACB$$.\n\nBy the exterior angle theorem, $$m\\angle DEC = \\dfrac{\\alpha}{2}$$.\n\nHence $$\\triangle EDC$$ is isoceles.\n\nSo $$CD = DE$$.\n\n(Added later). Assuming that the sides have lengths of $$x$$ and $$y$$, and that $$r = \\sqrt{x^2+y^2}$$, the lengths of the segments are displayed below.\n\n\u2022 This is exactly what I was looking for =) \u2013\u00a0N3buchadnezzar Apr 17 at 18:24\n\nThe curve that traces out the points that are equidistant to $$C$$ and the line extension of $$AB$$ is a parabola with focus $$C$$ and directrix $$AB$$.\n\nChoosing a convenient coordinate system (parallel to $$AB$$ with $$B$$ as the origin), I get the equation $$y = \\frac{x^2}{2b} + \\frac{b}{2}$$, which you want to intersect with the line $$y = \\frac{b}{a}x + b$$.\n\nSolving, I get that the point $$D$$ has $$(x,y)$$ coordinates of $$x = \\frac{b(b - c)}{a}$$ and $$y = \\frac{bc(c - b)}{a^2}$$.\n\nLet $$CD=DE=y$$ then we get $$\\frac{b}{c}=\\frac{y}{c-y}$$ so $$y=\\frac{bc}{b+c}$$\n\n\u2022 @mathmandan There was an edit after my comment. \u2013\u00a0Michael Biro Apr 17 at 21:17\n\nNot sure if this is less barbaric but using simple trig: $$DE=(a-x)\\tan A$$, $$DC=\\frac{x}{\\cos A}$$ so the equation to solve is $$(a-x)\\frac{b}{a}=\\frac{x\\sqrt{a^2+b^2}}{a}$$ or $$x=\\frac{ab}{\\sqrt{a^2+b^2}+b}$$\n\nJust another idea to construct point $$E$$: since $$\\triangle{DCE}$$ is isosceles, it's easy to find $$\\angle{ACE}=(90\u00b0-A)/2$$", "date": "2019-06-21 00:04:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3191278/find-the-length-x-such-that-the-two-distances-in-the-triangle-are-the-same", "openwebmath_score": 0.9472463130950928, "openwebmath_perplexity": 273.56894909378747, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232909876816, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.8521429270341897}}
{"url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html", "text": "\n## SectionInverse Functions\n\nRecall that in Brief Intro to Composite and Inverse Functions we gave the following definition of an inverse function:\n\n###### Inverse Functions\n\nSuppose the inverse of $f$ is a function, denoted by $f^{-1}\\text{.}$ Then\n\n\\begin{equation*} f^{-1}(y) = x \\text{ if and only if }f(x) = y. \\end{equation*}\n\nWe also stated the following property about inverse functions:\n\n###### Functions and Inverse Functions\n\nSuppose $f^{-1}$ is the inverse function for $f\\text{.}$ Then\n\n\\begin{equation*} f^{-1}\\left(f(x)\\right) = x\\text{ and }f\\left(f^{-1}( y)\\right) = y \\end{equation*}\n\nas long as $x$ is in the domain of $f\\text{,}$ and $y$ is in the domain of $f^{-1}\\text{.}$\n\nIn this section we will explore the invertability of a function. In other words, by the end of this section we will be able to test if a function is invertable.\n\n### SubsectionGraph of the Inverse Function\n\nIn Example161, we used a graph of $h$ to read values of $h^{-1}\\text{.}$ But we can also plot the graph of $h^{-1}$ itself. Because $C$ is the input variable for $h^{-1}\\text{,}$ we plot $C$ on the horizontal axis and $F$ on the vertical axis. To find some points on the graph of $h^{-1}\\text{,}$ we interchange the coordinates of points on the graph of $h\\text{.}$ The graph of $h^{-1}$ is shown in Figure192.\n\n $C=h(F)$ $F$ $C$ $14$ $-10$ $32$ $0$ $50$ $10$ $68$ $20$\n $F=h^{-1}(C)$ $C$ $F$ $-10$ $14$ $0$ $32$ $10$ $50$ $20$ $68$\n\n###### Example193\n\nThe Park Service introduced a flock of $12$ endangered pheasant into a wildlife preserve. After $t$ years, the population of the flock was given by\n\n\\begin{equation*} P = f (t) = 12 + 2t^3. \\end{equation*}\n1. Graph the function on the domain $[0, 5]\\text{.}$\n2. Find a formula for the inverse function, $t = f^{-1}(P)\\text{.}$ What is the meaning of the inverse function in this context?\n3. Sketch a graph of the inverse function.\nSolution\n1. The graph of $f$ is shown in Figure194, with $t$ on the horizontal axis and $P$ on the vertical axis.\n\n2. We solve $P = 12 + 2t^3$ for $t$ in terms of $P\\text{.}$ \\begin{align*} 2t^3 \\amp = P - 12\\amp\\amp\\text{Substract 12 from both sides.}\\\\ t^3 \\amp = \\frac{P - 12}{2}\\amp\\amp\\text{Divide both sides by 2.}\\\\ t \\amp = \\sqrt[3]{\\frac{P - 12}{2}}\\amp\\amp\\text{Take cube roots.} \\end{align*} The inverse function is $t = f^{-1}(P) =\\sqrt[3]{\\dfrac{P - 12}{2}}\\text{.}$ It tells us the number of years it takes for the pheasant population to grow to size $P\\text{.}$\n3. The graph of $f^{-1}$ is shown in Figure195, with $P$ on the horizontal axis and $t$ on the vertical axis.\n\nThe formula $T = f(L) = 2\\pi \\sqrt{\\dfrac{L}{32}}$ gives the period in seconds, $T\\text{,}$ of a pendulum as a function of its length in feet, $L\\text{.}$\n\n1. Graph the function on the domain $[0, 5]\\text{.}$\n2. Find a formula for the inverse function, $L = f^{-1}(T )\\text{.}$ What is the meaning of the inverse function in this context?\n3. Sketch a graph of the inverse function.\n\n### SubsectionWhen Is the Inverse a Function?\n\nWe can always find the inverse of a function $y=f(x)$ simply by solving for $x$ thus interchanging the role of the input and output variables. In the preceding examples, this process created a new function. However, this process does not always lead to be a function.\n\nFor example, to find the inverse of $y = f (x) = x^2\\text{,}$ we solve for $x$ to get $x = \\pm\\sqrt{y}\\text{.}$ When we regard $y$ as the input and $x$ as the output, the relationship does not describe a function. This can be seen as plugging in $y=4 \\text{,}$ for example, gives two outputs $x=2$ and $x=-2\\text{.}$ The graphs of $f$ and its inverse are shown in Figure197. (Note that for the graph of the inverse, we plot $y$ on the horizontal axis and $x$ on the vertical axis.) Because the graph of the inverse does not pass the vertical line test, it is not a function.\n\nFor many applications, it is important to know whether or not the inverse of $f$ is a function. This can be determined from the graph of $f\\text{.}$ When we interchange the roles of the input and output variables, horizontal lines of the form $y = k$ become vertical lines.\n\nThus, if the graph of the inverse is going to pass the vertical line test, the graph of the original function must pass the horizontal line test, namely, that no horizontal line should intersect the graph in more than one point. Notice that the graph of $f(x) = x^2$ does not pass the horizontal line test, so we would not expect its inverse to be a function.\n\n###### Horizontal Line Test\n\nIf no horizontal line intersects the graph of a function more than once, then its inverse is also a function.\n\nWe have been talking about how to tell if the inverse of a function is also a function, but in practice this is not the language typicaly used. Usually we ask this same question in the form \"Is the function invertible?\" The following definition explains this relationship:\n\n###### Invertible Function\n\nIf $y=f(x)$ is a function such that its inverse, $x=f^{-1}(y)\\text{,}$ is also a function then we say that $f(x)$ is an invertible function.\n\n###### Example200\n\nWhich of the functions in Figure201 are invertible?\n\nSolution\n\nIn each case, apply the horizontal line test to determine whether the function is invertible. Because no horizontal line intersects their graphs more than once, the functions pictured in Figures201(a) and (c) are invertible. The functions in Figures201(b) and (d) are not invertibe.\n\n### SubsectionMathematical Properties of the Inverse Function\n\nThe inverse function $f^{-1}$ undoes the effect of the function $f\\text{.}$ In Example152, the function $f(t) = 6 + 2t$ multiplies the input by $2$ and then adds $6$ to the result. The inverse function $f^{-1}(H) = \\dfrac{H -6}{2}$ undoes those operations in reverse order: It subtracts $6$ from the input and then divides the result by $2\\text{.}$\n\nIf we apply the function $f$ to a given input value and then apply the function $f^{-1}$ to the output from $f\\text{,}$ the end result will be the original input value. For example, if we choose $t = 5$ as an input value, we find that \\begin{align*} f(\\alert{5})\\amp= 6 + 2(\\alert{5}) = \\blert{16}\\amp\\amp\\text{ Multiply by 2, then add 6.}\\\\ \\text{and } f^{-1}(\\blert{16}) \\amp = \\frac{\\blert{16} - 6}{2} = \\alert{5}.\\amp\\amp\\text{Subtract 6, then divide by 2.} \\end{align*}\n\nWe return to the original input value, $5\\text{,}$ as illustrated in Figure202.\n\nExample203 illustrates the fact that if $f^{-1}$ is the inverse function for $f\\text{,}$ then $f$ is also the inverse function for $f^{-1}\\text{.}$\n\n###### Example203\n\nConsider the function $f(x) = x^3 + 2$ and its inverse, $f^{-1}(y) = \\sqrt[3]{y - 2}\\text{.}$\n\n1. Show that the inverse function undoes the effect of $f$ on $x = 2\\text{.}$\n2. Show that $f$ undoes the effect of the inverse function on $y = -25\\text{.}$\nSolution\n1. First evaluate the function $f$ for $x = 2\\text{:}$\nThen evaluate the inverse function $f^{-1}$ at $y = 10\\text{:}$\n\\begin{equation*} f^{-1}(\\blert{10}) = \\sqrt[3]{\\blert{10} - 2} = \\sqrt[3]{8}= \\alert{2}. \\end{equation*}\nWe started and ended with $2\\text{.}$\n2. First evaluate the function $f^{-1}$ for $y = -25\\text{:}$\n\\begin{equation*} f^{-1}(\\alert{-25}) = \\sqrt[3]{-25 - 2} = \\blert{-3}. \\end{equation*}\nThen evaluate the function $f$ for $x = -3\\text{:}$\n\\begin{equation*} f (\\blert{-3}) = (\\blert{-3})^3 + 2 = \\alert{-25}. \\end{equation*}\nWe started and ended with $-25\\text{.}$\n###### Example204\n1. Find a formula for the inverse of the function $f(x)=\\dfrac{2}{x - 1}\\text{.}$\n2. Show that $f^{-1}$ undoes the effect of $f$ on $x = 3\\text{.}$\n3. Show that $f$ undoes the effect of $f^{-1}$ on $y = -2\\text{.}$\nSolution\n1. To find the inverse, we solve for $x\\text{:}$\n\n\\begin{equation*} \\begin{aligned} y \\amp= \\frac{2}{x-1} \\\\ y(x-1) \\amp= 2 \\\\ x-1 \\amp= \\frac{2}{y} \\\\ x \\amp =\\frac{2}{y}+1 \\end{aligned} \\end{equation*}\n\nTherefore $f^{-1}(y)=\\frac{2}{y}+1\\text{.}$\n\n2. First evaluate the function $f$ for $x=3\\text{:}$\n\nThen evaluate the inverse function $f^{-1}$ at $y=1\\text{:}$\n\nWe started and ended with $3\\text{.}$\n\n3. First evaluate the inverse function $f^{-1}$ for $y=-2\\text{:}$\n\nThen evaluate the original function $f$ at $x=0\\text{:}$\n\nWe started and ended with $-2\\text{.}$\n\n### SubsectionSymmetry\n\nSo far we have been careful to keep track of the input and output variables when we work with inverse functions. This is important when we are dealing with applications; the names of the variables are usually chosen because they have a meaning in the context of the application, and it would be confusing to change them.\n\nHowever, we can also study inverse functions purely as mathematical objects. There is a relationship between the graph of a function and the graph of its inverse that is easier to see if we plot them both on the same set of axes.\n\nA graph does not change if we change the names of the variables, so we can let $x$ represent the input for both functions, and let $y$ represent the output. Consider the function $C = h(F)$ from Example161, and its inverse function, $F = h^{-1}(C)\\text{.}$ The formulas for these functions are \\begin{align*} C \\amp = h(F) = \\frac{5}{9}(F - 32)\\\\ F \\amp = h^{-1}(C) = 32 + \\frac{9}{5}C. \\end{align*} But their graphs are the same if we write them as \\begin{align*} y \\amp = h(x) =\\frac{5}{9}(x - 32)\\\\ y \\amp= h^{-1}(x) = 32 + \\frac{9}{5}x. \\end{align*} The graphs are shown in Figure205.\n\nNow, for every point $(a, b)$ on the graph of $f\\text{,}$ the point $(b, a)$ is on the graph of the inverse function. Observe in Figure205 that the points $(a, b)$ and $(b, a)$ are always located symmetrically across the line $y=x\\text{.}$ The graphs are symmetric about the line $y=x$, which means that if we were to place a mirror along the line $y=x\\text{,}$ each graph would be the reflection of the other.\n\n###### Example206\n\nGraph the function $f (x) = 2\\sqrt{x + 4}$ on the domain $[-4, 12]\\text{.}$ Graph its inverse function $f^{-1}$ on the same grid.\n\nSolution\n\nThe graph of $f$ has the same shape as the graph of $y = \\sqrt{x}\\text{,}$ shifted $4$ units to the left and stretched vertically by a factor of $2\\text{.}$ Figure207a shows the graph of $f\\text{,}$ along with a table of values. By interchanging the rows of the table, we obtain points on the graph of the inverse function, shown in Figure207b.\n\nIf we use $x$ as the input variable for both functions, and $y$ as the output, we can graph $f$ and $f^{-1}$ on the same grid, as shown in Figure208. The two graphs are symmetric about the line $y = x\\text{.}$\n\nGraph the function $f (x) = x^3 + 2$ and its inverse $f^{-1}(x) = \\sqrt[3]{x - 2}$ on the same set of axes, along with the line $y = x\\text{.}$", "date": "2023-03-22 02:07:22", "meta": {"domain": "unl.edu", "url": "https://mathbooks.unl.edu/PreCalculus/Inverse-Functions.html", "openwebmath_score": 0.844607412815094, "openwebmath_perplexity": 217.23917974865927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109520836027, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8520821559508629}}
{"url": "https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750", "text": "# Evaluate $\\int_{1}^{\\infty} \\frac{\\sqrt{x - 1}}{(x + 1)^{2}} ~ \\mathrm{d}{x}$.\n\nI need to solve the following integral: $$I = \\int_{1}^{\\infty} \\frac{\\sqrt{x - 1}}{(x + 1)^{2}} ~ \\mathrm{d}{x}.$$ Wolfram Alpha gives the answer as $\\dfrac{\\pi}{2 \\sqrt{2}}$.\n\nI think it\u2019s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?\n\n\u2022 The integrand has a closed-form antiderivative. Nov 13 '15 at 2:55\n\u2022 You don't need to use complex analysis. Nov 13 '15 at 3:03\n\u2022 if you want to use contour integration, take a keyhole around with a slit at the line $(1,\\infty)$ Nov 13 '15 at 10:30\n\nLet $u=\\sqrt{x-1}$, $du=\\frac{1}{2\\sqrt{x-1}}$, then\n\n$$I=2\\int_0^\\infty\\frac{u^2}{(u^2+2)^2}du$$\n\nWe can apply partial fractions here.\n\n$$=2\\int_0^\\infty\\left(\\frac{1}{u^2+2}-\\frac{2}{(u^2+2)^2}\\right)du$$\n\n$$=2\\int_0^\\infty\\frac{1}{u^2+2}du-4\\int_0^\\infty\\frac{1}{(u^2+2)^2}du$$\n\nThe first integrand is almost $\\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\\frac{u}{\\sqrt{2}}$.\n\n$$I=\\sqrt 2\\int_0^\\infty\\frac{1}{s^2+1}ds-4\\int_0^\\infty\\frac{1}{(u^2+2)^2}du$$\n\n$$=\\frac{\\pi}{\\sqrt 2}-4\\int_0^\\infty\\frac{1}{(u^2+2)^2}du$$\n\nNow to tackle the second integral we can use a trig sub.\n\nLet $u=\\sqrt{2}\\tan (t)$, $du=\\sqrt 2 \\sec^2(t)dt$.\n\n$$I=\\frac{\\pi}{\\sqrt 2}-4\\sqrt 2 \\int_0^{\\pi/2}\\frac{\\sec ^2 t}{4\\sec^4t}dt$$\n\n$$=\\frac{\\pi}{\\sqrt 2}-\\sqrt 2\\int _0^{\\pi/2}\\cos^2(t) dt$$\n\n$$=\\frac{\\pi}{\\sqrt 2}-\\sqrt 2 \\int_0^{\\pi/2}\\left(\\frac{1}{2}\\cos(2t)+\\frac{1}{2} \\right)dt$$\n\nIf we substitute $v=2t$ and split the integral up we get\n\n$$I=\\frac{\\pi}{\\sqrt 2}-\\frac{1}{2\\sqrt 2} \\int_0^\\pi \\cos(v)dv-\\frac{1}{\\sqrt 2}\\int_0^{\\pi/2}1dt$$\n\nThe first integral clearly goes to 0 and the second integral becomes $\\frac{\\pi}{2\\sqrt 2}$.\n\nTherefore\n\n$$I=\\frac{\\pi}{2\\sqrt 2}$$\n\n\u2022 There's a much easier way to find the $u$ integral. Using IBP gives $$\\int u\\cdot\\frac{2u}{(u^2+2)^2}du = -u\\cdot\\frac{1}{u^2+2} + \\int \\frac{1}{u^2+2}\\,du = -\\frac{u}{u^2+2} + \\arctan u$$ Nov 18 '15 at 5:51\n\nUsing a keyhole countour with the origin of the keyhole at $z=1$ and the small circle enclosing the value $z=1$ and using $$f(z) = \\frac{\\exp(1/2\\log(z-1))}{(1+z)^2}$$ with the branch cut of the logarithm on the positive real axis and the argument from $0$ to $2\\pi$ we get for the integral $$I=\\int_1^\\infty \\frac{\\sqrt{x-1}}{(1+x)^2} dx$$ that $$I(1-\\exp(\\pi i)) = 2\\pi i\\mathrm{Res}_{z=-1} f(z)$$ or $$I = \\pi i\\mathrm{Res}_{z=-1} f(z).$$\n\nNow the logarithmic term is certainly analytic in a neighborhood of $z=-1$ and we have $$\\mathrm{Res}_{z=-1} f(z) = \\left.\\left(\\exp(1/2\\log(z-1))\\right)'\\right|_{z=-1} = \\left. \\left(\\exp(1/2\\log(z-1))\\right) \\frac{1}{2}\\frac{1}{z-1}\\right|_{z=-1} \\\\= \\exp(1/2(\\log 2 + \\pi i)) \\times -\\frac{1}{2}\\frac{1}{2} = -\\frac{1}{4} \\sqrt{2} i.$$\n\nThis yields $$I = -\\frac{1}{4} \\sqrt{2} i \\times\\pi i=\\frac{\\sqrt{2}\\pi}{4}.$$\n\nRemark. The estimates for the circular components are done using ML same as at this MSE link. We get for the large circle parameterized by $z=R e^{it}$ $$\\lim_{R\\rightarrow \\infty} 2\\pi R \\frac{\\sqrt{R+1}}{(R-1)^2} = 0.$$\n\nThe small circle is a parameterized with $z=1+\\epsilon e^{it}$ and we get $$\\lim_{\\epsilon\\rightarrow 0} 2\\pi\\epsilon \\frac{\\sqrt{\\epsilon}}{4} = 0.$$\n\n$$I=\\int_{1}^{\\infty}\\frac{\\sqrt{x-1}}{(x+1)^2}dx$$ Use substitution $\\frac{1}{x+1}=t$ which implies $\\frac{dx}{(x+1)^2}=-dt$ So\n\n$$I=-\\int_{0.5}^{0}\\sqrt{\\frac{1}{t}-2}\\:dt=\\int_{0}^{0.5}\\frac{\\sqrt{1-2t}}{\\sqrt{t}}dt$$\n\nAgain use substitution $\\sqrt{t}=y$ which implies $\\frac{dt}{\\sqrt{t}}=2dy$\n\n$$I=2\\int_{0}^{\\sqrt{0.5}}\\sqrt{1-2y^2}dy=2\\sqrt{2}\\int_{0}^{\\sqrt{0.5}}\\sqrt{(\\sqrt{0.5})^2-t^2}$$\n\nUse standard integral $$\\int\\sqrt{a^2-x^2}dx=\\frac{x}{2}\\sqrt{a^2-x^2}+\\frac{a^2}{2}sin^{-1}(\\frac{x}{a})$$\n\nwe get $$I=\\frac{\\pi}{2\\sqrt{2}}$$\n\nHint Substitution $x=t^2+1$...\n\nThis is the general antiderivative. Just take the limits and you're good. $$\\int{\\frac{\\sqrt{x-1}}{(x+1)^2}dx}$$ Substitute $u = \\sqrt{x-1}$ $$= \\int{\\frac{u}{(u^2+2)^2}du}$$ $$= 2\\int\\left(\\frac{u}{u^2+2}-\\frac{2}{(u^2+2)^2}\\right)du$$ $$= \\int\\frac{u}{\\frac{u^2}{2}+1}du-4\\int\\frac{2}{(u^2+2)^2}du$$ Substitute $s = \\frac{u}{\\sqrt{2}}$ and $ds = \\frac{1}{\\sqrt{2}}$ $$= \\sqrt{2}\\int\\frac{1}{s^2+1}ds-4\\int\\frac{2}{(u^2+2)^2}du$$ $$= \\sqrt{2}\\arctan(s)-4\\int\\frac{2}{(u^2+2)^2}du$$ Substitute $u = \\sqrt{2}\\tan (p)\\quad$ and $\\quad du = \\sqrt{2}\\sec^2(p) dp\\quad$ and $\\quad(u^2 + 2)^2 = (2\\tan^2(p)+2)^2 = 4\\sec^4(p)\\quad$ and $\\quad p=\\arctan\\frac{u}{\\sqrt{2}}$ $$= \\sqrt{2}\\arctan(s)-\\sqrt{2}\\int\\cos^2(p)du$$ $$= \\sqrt{2}\\arctan(s)-\\sqrt{2}\\int\\left(\\frac{1}{2} \\cos(2p) + \\frac{1}{2}\\right)du$$ $$= \\sqrt{2}\\arctan(s)-\\frac{1}{\\sqrt{2}}\\int\\cos(2p) - \\frac{1}{\\sqrt{2}}\\int du$$ $$= \\sqrt{2}\\arctan(s)-\\frac{p}{\\sqrt{2}} - \\frac{\\sin(p)\\cos(p)}{\\sqrt{2}}$$ $$= \\sqrt{2}\\arctan(s)-\\frac{\\arctan\\frac{u}{\\sqrt{2}}}{\\sqrt{2}} - \\frac{\\sin(\\arctan\\frac{u}{\\sqrt{2}})\\cos(\\arctan\\frac{u}{\\sqrt{2}})}{\\sqrt{2}}$$ Note that $\\cos(\\arctan(z)) = \\frac{1}{\\sqrt{z^2 + 1}}$ and $\\sin(\\arctan(z)) = \\frac{z}{\\sqrt{z^2+1}}$ $$= \\frac{2\\sqrt{2}(u^2+2)\\arctan(s)+\\sqrt{2}(u^2+2)\\arctan\\frac{u}{\\sqrt{2}}+2u}{2(u^2+2)}$$ $$=\\frac{\\sqrt{2}(u^2+2)\\arctan \\frac{u}{\\sqrt{2}}-2u}{2(u^2+2)}$$ $$=\\frac{\\sqrt{2}(x+1)\\arctan \\frac{\\sqrt{x-1}}{\\sqrt{2}}-2\\sqrt{x-1}}{2(x+1)}$$ $$=\\frac{\\arctan \\frac{\\sqrt{x-1}}{\\sqrt{2}}}{\\sqrt{2}}-\\frac{\\sqrt{x-1}}{x+1}$$\n\nI think I have a different answer.\n\nLet $\\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then\n\n\\begin{align} \\int_1^\\infty \\frac{\\sqrt{x-1}}{(x+1)^2} dx &= \\int_1^\\infty \\frac{\\sqrt{2} (u-1)^{\\frac{1}{2}}}{4u^2} 2 du \\\\ &= \\frac{1}{\\sqrt{2}} \\int_1^\\infty u^{-2} (u-1)^\\frac{1}{2} du \\end{align} Now let $u = \\frac{1}{t}$, so that $du = \\frac{-1}{t^2} dt$. Continuing. \\begin{align} \\phantom{\\int_1^\\infty \\frac{\\sqrt{x-1}}{(x+1)^2} dx} &= \\frac{1}{\\sqrt{2}} \\int_0^1 t^2 (1-t)^\\frac{1}{2} \\frac{1}{\\sqrt{t}} \\frac{1}{t^2} dt \\\\ &= \\frac{1}{\\sqrt{2}} \\int_0^1 t^{\\frac{-1}{2}} (1-t)^\\frac{1}{2} dt \\\\ &= \\frac{1}{\\sqrt{2}} \\int_0^1 t^{\\frac{1}{2}-1} (1-t)^{\\frac{3}{2} - 1} dt \\\\ &= \\frac{1}{\\sqrt{2}} B ( \\frac{1}{2} . \\frac{3}{2} ) \\\\ &= \\frac{\\Gamma(\\frac{1}{2}) \\Gamma(\\frac{3}{2})} {\\sqrt{2}\\Gamma(2)} \\\\ &= \\frac{\\pi}{2\\sqrt{2}}. \\end{align}\n\nOriginal image of work by hand.\n\n\u2022 Honestly, with 1500 points I expect you know the customs here, and I see many answers where you used LaTeX. You could make an effort and write your answer correctly. Nov 14 '15 at 0:48", "date": "2021-10-25 20:02:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1526717/evaluate-int-1-infty-frac-sqrtx-1x-12-mathrmdx/1526750", "openwebmath_score": 1.000000238418579, "openwebmath_perplexity": 289.02061330389927, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109520836026, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8520821508171078}}
{"url": "https://mathhelpboards.com/threads/a-linear-equation-in-3-variables.5803/", "text": "# [SOLVED]a linear equation in 3 variables\n\n#### karush\n\n##### Well-known member\nwrite a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples\n\n$(1,1,1), (0,2,0), (1,0,0)$\n\nthe examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.\n\nthe answer to this is $2x+y-z=2$\n\n#### chisigma\n\n##### Well-known member\nwrite a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples\n\n$(1,1,1), (0,2,0), (1,0,0)$\n\nthe examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.\n\nthe answer to this is $2x+y-z=2$\nYou have the linear system...\n\n$\\displaystyle a + b + c = d$\n\n$\\displaystyle 2 b = d$\n\n$\\displaystyle a = d\\ (1)$\n\n... the solution of which is $a=d,\\ b= \\frac{d}{2},\\ c=- \\frac{d}{2}$, so that You can write...\n\n$\\displaystyle a x + \\frac{d}{2} y - \\frac{d}{2} z = d \\implies x + \\frac{y}{2} - \\frac{z}{2} = 1\\ (2)$\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nwrite a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples\n\n$(1,1,1), (0,2,0), (1,0,0)$\n\nthe examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.\n\nthe answer to this is $2x+y-z=2$\nI would have done this just a little differently from chisigma. First, note that there is no \"unique\" solution. for any ax+ by+ cz= d describing the line, you could multiply through by any number and get a new equation for the same line. You could. for example, divide both sides by d to get (a/d)x+ (b/d)y+ (c/d)z= 1, then let a'= a/d, b'= b/d, and c'= c/d so that the equation is a'x+ b'y+ c'z= 1, with only three unknown values.\n\nI suspect that the answer you are asked for is the one that has all integer coefficients with no common factor.\n\nNow, from ax+ by+ cz= d, (1, 1, 1) on the line means we must have a+ b+ c= d. (0, 2, 0) on the line means 0a+ 2b+ 0c= 2b= d. (1, 0, 0) on the line means 1a+ 0b+ 0c= a= d. Since both 2b and a are equal to d, a= 2b and we can replace both a and d with 2b in the first equation: 2b+ b+ c= 2b so that b+ c= 0 or b= -c.\n\nNow we can write everything in terms of b: a= 2b, c= -b, d= 2b, and so the equation is\n2bx+ by- bz= 2b. Here, b can be any (non-zero) number and still give an equation describing the line but dividing both sides of the equation by b gives\n2x+ y- z= 2, the simplest form for the equation as all numbers are integers and they do not all have a common factor so cannot be reduced.\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nwrite a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples\n\n$(1,1,1), (0,2,0), (1,0,0)$\n\nthe examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.\n\nthe answer to this is $2x+y-z=2$\nThere will be a unique plane which satisfies these three points simultaneously. The plane will have the same coefficients as its normal vector, and the normal vector will also be normal to any vectors in the plane. So if we can get two vectors that lie in the plane, taking their cross product will give the normal vector. Call two of these vectors \\displaystyle \\begin{align*} \\mathbf{a} \\end{align*} and \\displaystyle \\begin{align*} \\mathbf{b} \\end{align*}. We could have\n\n\\displaystyle \\begin{align*} \\mathbf{a} &= ( 1 - 0, 1 - 2, 1 - 0) \\\\ &= ( 1 , -1, 1 ) \\\\ \\\\ \\mathbf{b} &= (1 - 0, 0 - 2, 0 - 0 ) \\\\ &= ( 1 , -2, 0) \\end{align*}\n\n\\displaystyle \\begin{align*} \\mathbf{n} &= \\mathbf{a} \\times \\mathbf{b} \\\\ &= \\left| \\begin{matrix} \\mathbf{i} & \\phantom{-}\\mathbf{j} & \\mathbf{k} \\\\ 1 & -1 & 1 \\\\ 1 & -2 & 0 \\end{matrix} \\right| \\\\ &= \\mathbf{i} \\, \\left| \\begin{matrix} -1 & 1 \\\\ -2 & 0 \\end{matrix} \\right| - \\mathbf{j} \\, \\left| \\begin{matrix} 1 & 1 \\\\ 1 & 0 \\end{matrix} \\right| + \\mathbf{k} \\, \\left| \\begin{matrix} 1 & -1 \\\\ 1 & -2 \\end{matrix} \\right| \\\\ &= \\mathbf{i} \\, \\left[ -1 \\cdot 0 - 1 \\cdot (-2) \\right] - \\mathbf{j} \\, \\left( 1 \\cdot 0 - 1 \\cdot 1 \\right) + \\mathbf{k} \\, \\left[ 1 \\cdot (-2) - (-1) \\cdot 1 \\right] \\\\ &= 2\\,\\mathbf{i} + \\mathbf{j} - \\mathbf{k} \\end{align*}\n\nThe plane will have the same coefficients as the normal vector, so that gives \\displaystyle \\begin{align*} 2x + y - z = d \\end{align*}. Since we have three points that lie on the plane, any of them can be substituted to find \\displaystyle \\begin{align*} d \\end{align*}. If we substitute \\displaystyle \\begin{align*} (1, 0, 0) \\end{align*} that gives\n\n\\displaystyle \\begin{align*} 2 \\cdot 1 + 0 - 0 &= d \\\\ 2 &= d \\end{align*}\n\nand thus the plane is \\displaystyle \\begin{align*} 2x + y - z = 2 \\end{align*}.", "date": "2021-01-15 20:02:40", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/a-linear-equation-in-3-variables.5803/", "openwebmath_score": 0.9999923706054688, "openwebmath_perplexity": 468.6514466117882, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9848109503004293, "lm_q2_score": 0.8652240704135291, "lm_q1q2_score": 0.8520821390067531}}
{"url": "https://mathematica.stackexchange.com/questions/179382/issue-plotting-torus-link", "text": "I'm having an issue plotting torus links in Mathematica. What I'm trying to generate is a (2,8) torus link like the one in the picture in Wikipedia:\n\nusing equations such as those outlined here.\n\nHowever, I appear to only get half of the respective torus link.\n\na = 1; d = 4; p = 2; q = 8;\nParametricPlot3D[{(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*\nCos[p*t], a*Cos[q*t]}, {t, 0, 2*Pi}, PlotStyle -> Orange,\nPlotRange -> All] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]\n\n\n(a and d are parameters that control the width of the tube and the distance of the curve from the origin respectively, outlined here.)\n\nCan anyone explain why I am only seeing half of the link?\n\n\u2022 You're only plotting one set of parametric equations and you can see in the image that you don't expect there to be any intersections so the smooth parametric equations clearly can't generate the structure you want. On the other hand, it ought to be clear that the others may be generated via a simple rotation in the x-y plane. \u2013\u00a0b3m2a1 Aug 2 '18 at 8:09\n\nThe second torus is created by a rotation of $$\\pi/4$$ (around {0, 0, 1}):\n\nR = RotationMatrix[Pi/4, {0, 0, 1}];\ntorus = {(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]};\nParametricPlot3D[{torus, R.torus} // Evaluate, {t, 0, 2*Pi}, PlotStyle -> {Orange, Blue}, PlotRange -> All]\n\n\nHere's I think a generalization of what Ulrich gave to an arbitrary $(p, q)$ torus (if I read Wikipedia right):\n\nplotPQTorus[{p_, q_}, a : _?NumericQ : 1, d : _?NumericQ : 4,\nops : OptionsPattern[]] :=\nBlock[{t},\nParametricPlot3D[\nEvaluate@\nTable[\nRotationMatrix[\ni*2 \\[Pi]/q, {0, 0, 1}].{(a*Sin[q*t] + d)*\nSin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]},\n{i, 0, If[Divisible[q, p], p - 1, 0]}\n],\n{t, 0, 2*Pi},\nPlotRange -> All,\nops\n] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]\n]\n\n\nHere are a few plots:\n\nTable[plotPQTorus[{p, Fibonacci[q]}, Boxed -> False,\nAxes -> None], {p, 1, 4}, {q, 2, 6, 2}] // Grid\n\n\nNote that relatively prime things are single connected loops (as Wikipedia suggests they should be)\n\nThe (1,4) torus knot is known to KnotData[], so we can use that to build the (2,8) knot:\n\nknot14 = First[KnotData[{\"TorusKnot\", {1, 4}}, \"ImageData\"]];", "date": "2020-11-27 02:17:08", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/179382/issue-plotting-torus-link", "openwebmath_score": 0.2998276650905609, "openwebmath_perplexity": 2329.600533817695, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075690244281, "lm_q2_score": 0.8856314858927011, "lm_q1q2_score": 0.8520727559437188}}
{"url": "https://mathhelpboards.com/threads/logic-problem.5096/", "text": "# Logic problem\n\n##### Active member\nConsider the following sequence of statements:\n$$S_1: \\text{at least 1 of the statements }S_1-S_n \\text{ is false}\\\\ S_2: \\text{at least 2 of the statements }S_1-S_n \\text{ are false}\\\\ \\vdots \\\\ S_n: \\text{at least } n \\text{ of the statements }S_1-S_n \\text{ are false}$$\nWhere $n$ is some integer.\n\nQuestion: for which $n$ are these statements self-consistent? In those cases: what is the truth value of each statement?\n\nI got this off of a blog I tend to frequent. I will wait before posting the solution this time.\n\nEDIT:\nChanged the question; I had written the statements wrong\n\nLast edited:\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nSuppose $k$ out of $n$ statements are true.\nThen $S_1$ up to $S_k$ have to be true and the rest has to be false.\nThis appears to be consistent for any $n$ and any $0\\le k \\le n$.\n\n##### Active member\nSuppose $k$ out of $n$ statements are true.\nThen $S_1$ up to $S_k$ have to be true and the rest has to be false.\nThis appears to be consistent for any $n$ and any $0\\le k \\le n$.\nSorry about that, you were absolutely right about the question as phrased.\n\nHowever, this new version should prove to be a bit more interesting. This is what I had meant; I had accidentally written \"true\" instead of \"false\".\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nIf $S_n$ is true, then $n$ statements are false, including $S_n$.\nTherefore $S_n$ is false.\n\nWe now know that at least $1$ statement is false.\nTherefore $S_1$ is true.\nFor $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.\n\nFor $n \\ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.\nSince $S_1$ is true, this implies that $S_{n-1}$ is false.\nTherefore $S_{n-1}$ is false.\n\nSo at least $2$ statements are false.\nTherefore $S_2$ is true.\nFor $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.\n\nEtcetera.\n\nIn other words, we get a consistent consistent solution if and only if $n$ is even.\nIn that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\\qquad \\blacksquare$\n\n##### Active member\nIf $S_n$ is true, then $n$ statements are false, including $S_n$.\nTherefore $S_n$ is false.\n\nWe now know that at least $1$ statement is false.\nTherefore $S_1$ is true.\nFor $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.\n\nFor $n \\ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.\nSince $S_1$ is true, this implies that $S_{n-1}$ is false.\nTherefore $S_{n-1}$ is false.\n\nSo at least $2$ statements are false.\nTherefore $S_2$ is true.\nFor $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.\n\nEtcetera.\n\nIn other words, we get a consistent consistent solution if and only if $n$ is even.\nIn that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\\qquad \\blacksquare$\nCouldn't have phrased it better myself.\n\nThe source, for anybody interested:\nThe Parity Paradox \u2013 Futility Closet\n\nI highly recommend the website as a time-wasting tool.", "date": "2022-07-05 09:39:25", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/logic-problem.5096/", "openwebmath_score": 0.5636366605758667, "openwebmath_perplexity": 447.92497156639786, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9621075744568837, "lm_q2_score": 0.8856314738181875, "lm_q1q2_score": 0.8520727491378914}}
{"url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html", "text": "# Thread: Is it a field? set of real numbers in the form...\n\n1. ## Is it a field? set of real numbers in the form...\n\nHi,\n\nproblem:\nLet $Q(\\sqrt{2})$ be the set of all real numbers of the form\n$\\alpha+\\beta\\sqrt{2}$, where $\\alpha\\;and\\;\\beta$ are rational.\n\n(a) Is $Q(\\sqrt{2})$ a field?\n(b) What if $\\alpha\\;and\\;\\beta$ are required to be integers?\n\nattempt:\n\n(a)\nI look at the axioms for a field and check whether any of them fail. I am unable to find one that does.\nIs it ok to say that I can express any real number in the form $\\alpha+\\beta\\sqrt{2}$ as long as $\\alpha\\;and\\;\\beta$ are rational?\n\n(b)\nI am unable to find an axiom that \"fails\", but say I want to write the real number $\\frac{\\sqrt{2}}{2}$. I don't think I can do this if $\\alpha\\;and\\;\\beta$ are integers..\n\nThanks\n\n2. Originally Posted by Mollier\nHi,\n\nproblem:\nLet $Q(\\sqrt{2})$ be the set of all real numbers of the form\n$\\alpha+\\beta\\sqrt{2}$, where $\\alpha\\;and\\;\\beta$ are rational.\n\n(a) Is $Q(\\sqrt{2})$ a field?\n(b) What if $\\alpha\\;and\\;\\beta$ are required to be integers?\n\nattempt:\n\n(a)\nI look at the axioms for a field and check whether any of them fail. I am unable to find one that does.\nIs it ok to say that I can express any real number in the form $\\alpha+\\beta\\sqrt{2}$ as long as $\\alpha\\;and\\;\\beta$ are rational?\n\n(b)\nI am unable to find an axiom that \"fails\", but say I want to write the real number $\\frac{\\sqrt{2}}{2}$. I don't think I can do this if $\\alpha\\;and\\;\\beta$ are integers..\n\nThanks\njust go through each property for a field and make sure that they all hold. you have to verify these explicitly.\n\nand no, you cannot say any real number can be expressed in that form. how would you express $\\pi$ that way, for instance?\n\n3. Originally Posted by Mollier\nHi,\n\nproblem:\nLet $Q(\\sqrt{2})$ be the set of all real numbers of the form\n$\\alpha+\\beta\\sqrt{2}$, where $\\alpha\\;and\\;\\beta$ are rational.\n\n(a) Is $Q(\\sqrt{2})$ a field?\n(b) What if $\\alpha\\;and\\;\\beta$ are required to be integers?\n\nattempt:\n\n(a)\nI look at the axioms for a field and check whether any of them fail. I am unable to find one that does.\nIs it ok to say that I can express any real number in the form $\\alpha+\\beta\\sqrt{2}$ as long as $\\alpha\\;and\\;\\beta$ are rational?\n\n(b)\nI am unable to find an axiom that \"fails\", but say I want to write the real number $\\frac{\\sqrt{2}}{2}$. I don't think I can do this if $\\alpha\\;and\\;\\beta$ are integers..\n\nThanks\nFirstly, if every real number was of the form $\\alpha+\\beta \\sqrt{2}$ with $\\alpha, \\beta \\in \\mathbb{Q}$ then we can easily define a bijection $\\mathbb{R} \\rightarrow \\mathbb{Q} \\times \\mathbb{Q}$, $\\alpha + \\beta \\sqrt{2} \\mapsto (\\alpha, \\beta)$. As there exists a bijection between $\\mathbb{Q}$ and $\\mathbb{Q} \\times \\mathbb{Q}$ we have that $|\\mathbb{R}| = |\\mathbb{Q}|$, a contradiction.\n\nSecondly, in both of your problems you know that \"most\" of the axioms for fields holds (both are clearly rings, and multiplication in both cases is commutative). The one to concentrate on will be multiplicative inverses.\n\nWhat is the inverse of $\\alpha+\\beta \\sqrt{2}$ with $\\alpha$ and $\\beta$ rationals?\n\nWhat if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\\alpha + \\beta \\sqrt{2}$. That is, every rational number of the form $\\frac{1}{\\gamma}, \\gamma \\in \\mathbb{Z} \\setminus \\{0\\}$ can be written as $\\alpha + \\beta \\sqrt{2}, \\alpha, \\beta \\in \\mathbb{Z}$. As we are, allegedly, in a field we have that $\\frac{n}{\\gamma}$ is also of this form, with $n \\in \\mathbb{N}$. Thus, every rational number is of the form $\\alpha + \\beta \\sqrt{2}$. This is silly!)\n\n4. Originally Posted by Jhevon\nand no, you cannot say any real number can be expressed in that form. how would you express $\\pi$ that way, for instance?\nThanks\n\nOriginally Posted by Swlabr\nThe one to concentrate on will be multiplicative inverses.\n\nWhat is the inverse of $\\alpha+\\beta \\sqrt{2}$ with $\\alpha$ and $\\beta$ rationals?\n\nWhat if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\\alpha + \\beta \\sqrt{2}$. That is, every rational number of the form $\\frac{1}{\\gamma}, \\gamma \\in \\mathbb{Z} \\setminus \\{0\\}$ can be written as $\\alpha + \\beta \\sqrt{2}, \\alpha, \\beta \\in \\mathbb{Z}$. As we are, allegedly, in a field we have that $\\frac{n}{\\gamma}$ is also of this form, with $n \\in \\mathbb{N}$. Thus, every rational number is of the form $\\alpha + \\beta \\sqrt{2}$. This is silly!)\nIt would be $\\frac{1}{\\alpha+\\beta\\sqrt{2}}$ .\nNow, the inverse would have to be in $Q(\\sqrt{2})$, so it would have to be in the form $\\alpha+\\beta\\sqrt{2}$.\nSince $\\frac{1}{\\alpha+\\beta\\sqrt{2}}$ is not in that form, $Q(\\sqrt{2})$ is not a field.\nIs this totally off? I think I need to get (a) right, before I move on to (b).\n\nThanks guys!\n\n5. Originally Posted by Mollier\nThanks\n\nIt would be $\\frac{1}{\\alpha+\\beta\\sqrt{2}}$ .\nNow, the inverse would have to be in $Q(\\sqrt{2})$, so it would have to be in the form $\\alpha+\\beta\\sqrt{2}$.\nSince $\\frac{1}{\\alpha+\\beta\\sqrt{2}}$ is not in that form, $Q(\\sqrt{2})$ is not a field.\nIs this totally off? I think I need to get (a) right, before I move on to (b).\n\nThanks guys!\nYou want to try and write $\\frac{1}{\\alpha + \\beta \\sqrt{2}}$ as $\\alpha^{\\prime} + \\beta^{\\prime}\\sqrt{2}$. I mean, $\\sqrt{2} = \\frac{2}{\\sqrt{2}}$...numbers are strange things...just because something isn't obviously in a certain form does not mean you cannot write it in that form.\n\nI shall give you a tantalising hint: It is a field.\n\n6. $\n\\frac{1}{\\alpha+\\beta\\sqrt{2}}=\\frac{\\alpha-\\beta\\sqrt{2}}{\\alpha^2-2\\beta^2}=\\left(\\frac{\\alpha}{\\alpha^2-2\\beta^2}\\right) - \\left(\\frac{\\beta}{\\alpha^2-2\\beta^2}\\right)\\sqrt{2}\n$\n\n$\\alpha'=\\left(\\frac{\\alpha}{\\alpha^2-2\\beta^2}\\right)$\n\n$\\beta'=\\left(\\frac{\\beta}{\\alpha^2-2\\beta^2}\\right)$\n\nSince both $\\alpha'\\;and\\;\\beta'$ are rationals, $Q(\\sqrt{2})$ is a field?\n\nThanks mate!\n\n7. Originally Posted by Mollier\n$\n\\frac{1}{\\alpha+\\beta\\sqrt{2}}=\\frac{\\alpha-\\beta\\sqrt{2}}{\\alpha^2-2\\beta^2}=\\left(\\frac{\\alpha}{\\alpha^2-2\\beta^2}\\right) - \\left(\\frac{\\beta}{\\alpha^2-2\\beta^2}\\right)\\sqrt{2}\n$\n\n$\\alpha'=\\left(\\frac{\\alpha}{\\alpha^2-2\\beta^2}\\right)$\n\n$\\beta'=\\left(\\frac{\\beta}{\\alpha^2-2\\beta^2}\\right)$\n\nSince both $\\alpha'\\;and\\;\\beta'$ are rationals, $Q(\\sqrt{2})$ is a field?\n\nThanks mate!\nYeah, that's it! Although to be thorough you would need to make sure there is no division by 0 going on.\n\n8. So, $\\alpha^2-2\\beta^2\\neq 0$,\nthen $\\beta \\neq \\frac{1}{\\sqrt{2}}\\alpha$, but $\\beta$ is then not a rational number anyways..\n\nAs for (b), can I use a similar argument by saying that since $\\alpha'$ and $\\beta'$ are rational numbers, $Q(\\sqrt{2})$ is not a field when $\\alpha,\\beta\\in Z$?\n\nThank you very much.\n\n9. Originally Posted by Mollier\nSo, $\\alpha^2-2\\beta^2\\neq 0$,\nthen $\\beta \\neq \\frac{1}{\\sqrt{2}}\\alpha$, but $\\beta$ is then not a rational number anyways..\n\nAs for (b), can I use a similar argument by saying that since $\\alpha'$ and $\\beta'$ are rational numbers, $Q(\\sqrt{2})$ is not a field when $\\alpha,\\beta\\in Z$?\n\nThank you very much.\nFor part (b) you need to find a counter-example - $\\alpha^{\\prime}$ and $\\beta^{\\prime}$ may look like non-integers, but this may not always be the case.\n\nSo, to complete the problem simply take an integer value of $\\alpha$ and an integer value of $\\beta$ such that $\\alpha^{\\prime} \\notin \\mathbb{Z}$ or $\\beta^{\\prime} \\notin \\mathbb{Z}$. For simplicity, let one of them be zero.\n\n10. Originally Posted by Mollier\nSo, $\\alpha^2-2\\beta^2\\neq 0$,\nthen $\\beta \\neq \\frac{1}{\\sqrt{2}}\\alpha$, but $\\beta$ is then not a rational number anyways..\nNo, this is not a proof. You want to assume that $\\alpha^2 -2\\beta^2=0$ and look for a contradiction.\n\n11. $\\alpha^2-2\\beta^2\\neq 0$ because of the irrationality of $\\sqrt{2}$, it can be easily proved in number theory.\n\n12. Originally Posted by Swlabr\nNo, this is not a proof. You want to assume that $\\alpha^2 -2\\beta^2=0$ and look for a contradiction.\n$\\alpha^2-2\\beta^2=0 \\Rightarrow \\alpha=\\sqrt{2}\\beta$\nContradiction since $\\alpha\\in Q$\n\n13. Originally Posted by Mollier\n$\\alpha^2-2\\beta^2=0 \\Rightarrow \\alpha=\\sqrt{2}\\beta$\nContradiction since $\\alpha\\in Q$\nPrecisely!\n\nSorry to be pedantic about it, but in maths the proof is the king...\n\n14. Please don't apologize. I have a BS in mechanical engineering from a horrible school, so I have now decided to actually learn some math. You being \"pedantic\" is exactly what I need! Thanks!", "date": "2017-06-25 10:45:53", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-algebra/122477-field-set-real-numbers-form.html", "openwebmath_score": 0.8847749829292297, "openwebmath_perplexity": 174.76001380594448, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075701109193, "lm_q2_score": 0.885631476836816, "lm_q1q2_score": 0.8520727481932139}}
{"url": "http://mathhelpforum.com/number-theory/47919-rational-numbers.html", "text": "# Math Help - rational numbers\n\n1. ## rational numbers\n\nFind two rational numbers with denominators 11 and 13, respectively, and a sum of 7/143.\n\nI got x/11 + y/13=7/143.\n\nTwo unknowns and one equation. Kind of stuck.\n\n2. Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.\n\nThen you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?\n\n3. Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.\nYou should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\\in\\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.\n\nLaurent.\n\n4. Originally Posted by Matt Westwood\nMight help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.\n\nThen you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?\nWhat do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm\n\n5. Originally Posted by rmpatel5\nWhat do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm\n\n6. Originally Posted by Laurent\nDon't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.\nYou should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\\in\\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.\n\nLaurent.\nI am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?\n\n7. Originally Posted by rmpatel5\ngcd(11,13)=1 and there is a thm that says ax+by=1.\nThere is a theorem that says that there exists x and y such that 11x+13y=1. Now, you should find an example of such a couple (x,y). Either you guess it, or you make use of the Euclidean algorithm as you may have done in class.\n\n8. Originally Posted by rmpatel5\nI am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?\nOk, so solve for x and y in 13x+11y=1, using the Euclidean algorithm (http://en.wikipedia.org/wiki/Euclidean_algorithm). Actually, do the Euclidian algorithm over 13 and 11, and observe...\n\nThen, multiply by 7 :\n13*(7x)+11*(7y)=7\n\nUh.\n\n9. Use the Euclidean Algorithm to get the gcd of 11 and 13 (yes, you know this is 1, but the calculations you did in the algorithm help you find out what values of a and b give you 11a + 13b = 1.\n\nYou now want two numbers x and y such that 11x + 13y = 7.\n\nWell, you just got 11a + 13b = 1, so multiply everything by 7:\n\n$11 \\times 7a + 13 \\times 7b = 7$\n\nSo 7a and 7b are the numbers you want for x and y. Job done.\n\n10. You do not need to use Euclidean algorithm.\nIt is quite obvious that $13(6) + 11(-7) = 1$.\nThus, $13(42) + 11(-49) = 7$.\n\nThis means the solutions to $13x+11y=7$ are: $x=42 - 11t \\mbox{ and }y=-49 + 42t$.\n\n11. 6 and -5 work as well, so are there many solutions to this problem. I got them by doing the EA", "date": "2015-07-03 00:12:45", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/47919-rational-numbers.html", "openwebmath_score": 0.8476439714431763, "openwebmath_perplexity": 409.548473341809, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9890130589886011, "lm_q2_score": 0.8615382076534743, "lm_q1q2_score": 0.8520725381869193}}
{"url": "https://www.projectrhea.org/rhea/index.php/Compute_DSFT_product_two_step_functions_ECE438F11", "text": "Topic: Discrete-space Fourier transform computation\n\n## Question\n\nCompute the discrete-space Fourier transform of the following signal:\n\n$f[m,n]= 2^{-(n+m)} u[n] u[m]$\n\nFirst off notice that this equation can easily be separated into two functions $g[m]=2^{-m}u[m]$ and $h[n]=2^{-n}u[n]$ where $f[m,n]=g[m]h[n]$. Then since both equations are the same except for a change of variable where m=n or vice verse we can show the DTFT of either g[m] or h[n] and it should suffice for the other. If they were not the same we would have to evaluate both of them.\n\n\\begin{align} G[u]&=\\sum_{m=-\\infty}^{\\infty}2^{-m}u[m]e^{-j u m} \\\\ &=\\sum_{m=0}^{\\infty}( \\frac{1}{2e^{ju}} )^{m} \\\\ &=\\frac{1}{1-\\frac{1}{2e^{ju}}} \\\\ &=\\frac{-2e^{ju}}{1-2e^{ju}} \\\\ \\end{align}\n\nThus\n\n$F[u,v]= \\frac{4e^{j(u+v)}}{(1-2e^{ju})(1-2e^{jv})}$\n\n\\begin{align} F [u,v] &= \\sum_{m=-\\infty}^{\\infty} \\sum_{n=-\\infty}^{\\infty} f[m,n]e^{-j(mu + nv)}\\\\ &= \\sum_{m=-\\infty}^{\\infty} \\sum_{n=-\\infty}^{\\infty} 2^{-(n+m)} u[n] u[m] e^{-j(mu + nv)}\\\\ &= \\sum_{m=-\\infty}^{\\infty} 2^{-m} u[m] e^{-j(mu)} \\sum_{n=-\\infty}^{\\infty} 2^{-n} u[n] e^{-j(nv)}\\\\ &= \\sum_{m= 0}^{\\infty} 2^{-m} e^{-j(mu)} \\sum_{n=0}^{\\infty} 2^{-n} e^{-j(nv)}\\\\ &= \\sum_{m= 0}^{\\infty} (2e^{ju})^{-m} \\sum_{n=0}^{\\infty} (2e^{jv})^{-n}\\\\ &= \\frac{1}{1-\\frac{1}{2e^{ju}}}\\cdot\\frac{1}{1-\\frac{1}{2e^{jv}}}\\\\ &= \\frac{1}{1-\\frac{1}{2}e^{-ju}}\\cdot\\frac{1}{1-\\frac{1}{2}e^{-jv}}\\\\ &= \\frac{1}{(1-\\frac{1}{2}e^{-ju})(1-\\frac{1}{2}e^{-jv})} \\end{align}\n\n--Xiao1 23:03, 19 November 2011 (UTC)\n\nInstructor's comments: Solutions 1 and solutions 2 are the two main ways to answer the question. The second one is a bit longer to write, in my opinion, but it does not require knowing the separation property of the continuous-space Fourier transform. Note however that, in the exam, if you do not have the separation property in the table, you will need to prove it in order to get full credit. -pm", "date": "2022-06-25 04:16:52", "meta": {"domain": "projectrhea.org", "url": "https://www.projectrhea.org/rhea/index.php/Compute_DSFT_product_two_step_functions_ECE438F11", "openwebmath_score": 1.0000097751617432, "openwebmath_perplexity": 2124.7650561476125, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130554263734, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8520725333598925}}
{"url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html", "text": "# Absolute Maximum And Minimum Calculator On Interval\n\n20 at x = \u20130. > @ 2 f x x x 6 2 on 1,27 11. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. These values correspond to the probability of observing such an extreme value by chance. If we \ufb01nd all possible local extrema, then the global maximum, if it exists, must be the value of f(x) on the interval 1 \u2264 x \u2264 4, and the maximum is f(4) = 9. Find more Mathematics widgets in Wolfram|Alpha. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Sample size calculation for trials for superiority, non-inferiority, and equivalence. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. 80 at x = \u20131. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. C) Absolute maximum only. For example, you could say,\u201cThe pulse rates are between 56 and 92 bpm. The closed interval method is a way to solve a problem within a specific interval of a function. Inequalities and \u2018Maximum-Minimum\u2019 Problems Henry Liu, 26 February 2007 There are many olympiad level problems in mathematics which belong to areas that are not covered well at all at schools. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Absolute zero is defined as the point where no more heat can be removed from a system, according to the absolute or thermodynamic temperature scale. Four Function and. Left bounds go on left side of min/max and right bounds go on right side. (3) Because the system has been carefully planned using modern top-down programming techniques, it is relatively easy to modify and extend. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. We take the derivative using the quotient rule: f0(x) =. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Relative Minimum - The lowest point on an interval of a curve. Once you've found the max-imum and minimum on this line (as well as on the other two lines that make up the boundary), compare all the values you've checked to \ufb01nd out that the absolute maximum and the absolute minimum are f(1 4,2) = 673 64 (\u2190absolute maximum) f(1 2, 4. An absolute maximum occurs at c if for all x in the domain of f. The absolute is measured in liters of oxygen per minute. f x x() 3 d. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. \u2022 Find the values of f at the endpoints of the interval. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. The graph of y = cos x. Absolute Maximum and Absolute Minimum. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. Find the absolute maximum and absolute minimum values of f(x) = x2 \u22124 x2 +4 on the interval [\u22124,4]. Subtract 6 6 from 1 1. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. Find the absolute maximum and absolute minimum values of f on the given interval. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. 4 is the lower limit. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. For, sin (x + ) = cos x. denbal87 New member. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. f(x)=x+ 9 x on [0. On the graph above of the function f on the closed interval [a, e], the point (a, f (a)) represents the absolute minimum, and the point (d, f (d)) represents the absolute maximum. Question 203087: Find the absolute maximum and absolute minimum values of the function below. An absolute minimum is the lowest y value or output value a. Thread starter denbal87; Start date Nov 4, 2014; D. The largest value found in steps 2 and 3 above will be the absolute maximum and the smallest value will be the absolute minimum. This lesson will focus on the maximum and minimum points. These points are sometimes referred to as max, min, extreme values, or extrema. (d) Find the absolute minimum value of f x over the closed interval 5 ddx 5. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 \u221212x. Sure, there are other, more precise, definitions, but that will work for what we want to do. 67, and a relative maximum of 4. Use the Stefan Boltzmann relationship (I = \u03c3T 4 where \u03c3 = 5. Q: Determine the absolute maximum and minimum values of the function on the given interval. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). I have an assignment where I have to write a program that accepts a series of integers from the keyboard from 1 to 100 using a single sentinel controlled loop, meaning I will need to also assign a sentinel number that when entered, will stop the loop and display the results. This is defined everywhere and is zero at $\\ds x=\\pm \\sqrt{3}/3$. By using this website, you agree to our Cookie Policy. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. The absolute max occurs at S = The absolute min occurs at S =. Decide whether you have a minimum or a maximum. )Given the function \ud835\udc53(\ud835\udc65= \ud835\udc652+ \ud835\udc65+ , chose values for a, b, and c in that could work for the graph shown. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. the absolute (global) maximum 3. Meaning answers obtained by looking at the graph in the calculator will not earn V = - 4x +3; [1, 3] BIUA. VO2 max in men is approximately 40-60% higher in males than females. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. If you're behind a web filter, please make sure that the domains *. Question 203087: Find the absolute maximum and absolute minimum values of the function below. We take the derivative using the quotient rule: f0(x) =. The debt to equity ratio is a financial, liquidity ratio that compares a company\u2019s total debt to total equity. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. The absolute maximum on the interval is 138 at x=-2. For example, the following are all equivalent confidence intervals: 20. Find the absolute maximum and absolute minimum values of f on the given interval. Approximating Relative Extrema. Also the lowest value of either the X of the Y is placed first in the set. the absolute (global) maximum 3. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. Example 1: Consider the three curves shown below. In worst case, if all intervals are from \u2018min\u2019 to \u2018max\u2019, then time complexity becomes O((max-min+1)*n) where n is number of intervals. Find the extreme values of f on the boundary of D. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Three major examples are geometry, number theory, and functional equations. Go to window and set your X minimum to -1 and your X maximum to 5. Extreme Values of Functions Definitions An extreme value of a function is the largest or smallest value of the function in some interval. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain \u2014 in which case it is called a relative or local extremum \u2014 or on a given set contained in the domain (perhaps all of it) \u2014 in which case it is called an absolute or global extremum (the latter. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as \u22121\u2264x\u22644 or [\u22121,4], or, more generally, a\u2264x\u2264b or [a,b], then f always has a global minimum value and a global maximum value on that interval. At t =0 the position of the object is 5. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Answer: First, \ufb01nd the critical points by \ufb01nding where the derivative equals zero: f0(x) = (x2 +4)(2x)\u2212(x2 \u22124. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. f(x) = x4 \u2013 A: Plot the graph for f(x) in the interval for [-2, 0]. Find the absolute maximum and absolute minimum values of f on the given interval. pow(x, 2) + math. Visual Magnitude Calculator: Computes the visual magnitude of a star from its absolute magnitude and distance. If f has a local maximum or minimum at c and f'(c. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. Extreme Values A Global Maximum A function f has a global (absolute) maximum at x =c if f (x)\u2264 f (c) for all x\u2208Df. Absolute Maximum - The highest point on a curve. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Let's Practice:. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Here again we are giving definitions that appeal to your geometric intuition. f(x)=xln(2x)on[0. Local Extreme Values of a Function Let c be an interior point of the domain of the. The absolute is measured in liters of oxygen per minute. ) Find the absolute maximum and minimum values of the function on the given interval. Q: Determine the absolute maximum and minimum values of the function on the given interval. Extreme value theorem tells us that a continuous function must obtain absolute minimum and maximum values on a closed interval. VO2max stands for maximal oxygen uptake and refers to the amount of oxygen your body is capable of utilizing in one minute. so minimum value of f(x) in interval [0. For example, consider the functions shown in Figure(d), (e), and (f). Find more Mathematics widgets in Wolfram|Alpha. It's easy to \ufb01nd one with neither absolute extrema. Identify the location and value of the absolute maximum and absolute minimum of a function over the domain of the function graphically or by using a graphing utility. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Find the absolute maximum and absolute minimum values of f on the given interval. The maximum value of a function that has a derivative at all points in an For a function f(x) that has a derivative at every point in an interval [a, b], the maximum or minimum values can be found by using the following procedure: 1. By using this website, you agree to our Cookie Policy. Similar topics can also be found in the Calculus section of the site. So the absolute max value is 19 and the absolute min value is 1. The minimum and maximum describe the spread of the data. By using this website, you agree to our Cookie Policy. When calculating Intervals the X values are placed \"on top of\" the Y values. Enter DNE if the absolute maximum or minimum does not exist. Pick the largest and smallest. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. Thus if one has a sample {, \u2026,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. At t =0 the position of the object is 5. MIN([DISTINCT] expr) Minimum value returned by expr MOD(x,y) Remainder of x divided by y MONTHS_BETWEEN(end_date, start_date) Number of months between the 2 dates (integer) NEW_TIME(date, zone1, zone2) Convert between GMT and US time zones (but not CET) NEXT_DAY(date,day_of_week) '12-OCT-01','Monday' will return the next Mon after 12 Oct NLS. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. The absolute minimum on the interval is -237 at x=3. f x x( ) cos 2 b. Explain your reasoning. maximum\" functions. Similarly, the global minimum is located at the lowest point. A relative minimum is a point that is lower than all the other points around it. The absolute max occurs at S = The absolute min occurs at S =. question_answer. Find the maximum and minimum values of the function f(x) = ln x/x on the interval [1, 3]. Hit graph and then hit 2nd Trace I think to bring up a menu that has minimum and maximum in there. Typical values for are 0. Absolute Maximum And Minimum Calculator. So, f(b) is a relative maximum of f. The maximum will occur at the highest value and the minimum will occur at the lowest value. If you're seeing this message, it means we're having trouble loading external resources on our website. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Find the the critical points of f on D. Fermat's Theorem. From this list of values we see that the absolute maximum is 8 and will occur at $$t = 2$$ and the absolute minimum is -3 which occurs at $$t = 1$$. You can take notes in the margins or on the flip-side of each sheet. Finding Extrema on a closed interval: 1. Subtract 6 6 from 1 1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Find more Mathematics widgets in Wolfram|Alpha. Since the function is concave down at x=1 and has a critical point at x=1 (zero slope) then the function has a local maximum at x=1. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. ^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Wolfram alpha paved a completely new way to get knowledge and information. Enter the equation in the Y= section for Y1. minimum\" or \"4. f(x)=x^3-3x+1; [0,3]. Local Extreme Values of a Function Let c be an interior point of the domain of the. Let f be a function defined and. Endpoint Discontinuities: only one of the one-sided limits exists. f(x) = x + (4/x) on the interval [1,5]. A relative (or local) maximum occurs at c if for all x in an open interval containing c. 11) A) Absolute minimum only. But there is one very important condition that guarantees both an absolute minimum and an absolute maximum. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. Absolute Maximum and Absolute Minimum. So the absolute max value is 19 and the absolute min value is 1. org are unblocked. Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2 To find absolute max/min values of a continuous function g on a closed bounded set D: 1. Get the free \"Function Extrema - Math 101\" widget for your website, blog, Wordpress, Blogger, or iGoogle. so minimum value of f(x) in interval [0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. have both an absolute maximum and an absolute minimum. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Finding Extrema on a closed interval: 1. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. The student familiar with the sum formula can easily prove that. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. False There is a local maximum at x = 0. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Fold Unfold. Examples: Input: v = {{1, 2}, {2, 4}, {3, 6}} first_page Given an absolute sorted array and a number K, find the pair whose sum is K. This lesson will focus on the maximum and minimum points. Find more Mathematics widgets in Wolfram|Alpha. Relative Minimum - The lowest point on an interval of a curve. # The text file for eval. f(x) = @ 10. absolute minimum value at x = 2 is -48. The first derivative test: Let f (x) be a function and x = c a critical point of f. f(t) = 4t + 4 cot(t/2), [\u03c0/4, 7\u03c0/4] I'm stuck after I take the first derivative. Occurence of absolute minima: If f(x) is continuous in a closed interval I, then the absolute minimum of f(x) in I is the minimum value of f(x) on all local. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. Mean number of days \u2265 30, 35 or 40 \u00b0C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 \u00b0C. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. Free functions extreme points calculator - find functions extreme and saddle points step-by-step This website uses cookies to ensure you get the best experience. We take the derivative using the quotient rule: f0(x) =. A higher debt to equity ratio indicates that more creditor financing (bank loans) is used than investor financing (shareholders). f x x3 2 on 3,1 > @ 15. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. The minimum value for this range is the mean subtracted by the confidence interval and the maximum value is calculated by the mean added by the confidence interval. f(x) = x4 \u2013 A: Plot the graph for f(x) in the interval for [-2, 0]. f(x) = (x^2 - 1)^3, [-1, 5] 14. 2 Maximum and Minimum on an Interval. Use a graphing calculator to approximate the intervals where each function is increasing and Increase and Decrease Absolute minimum:. Find the absolute maximum and absolute minimum values of f on the given interval. In that case, the point right on the border might be the maximum or minimum of the curve. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. f(c) is another relative minimum of f. It could very well continue to increase or decrease once we leave the interval. Example: Find the absolute maximum and minimum of:. Note: From our definition of absolute maxima and minima, if $(a, f(a))$ is an absolute max/min, then it is also a local max/min too. Question 203087: Find the absolute maximum and absolute minimum values of the function below. Occurence of absolute maxima: If f(x) is continuous in a closed interval I, then the absolute maximum of f(x) in I is the maximum value of f(x) on all local maxima and endpoints on I. 1 absolute maximum or minimum. The maximum will occur at the highest f (x) f (x) value and the minimum will occur at the lowest f (x) f (x) value. Find the the critical points of f on D. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Absolute Maximum and Absolute Minimum. In this case, \u201cabsolute extrema\u201d is just a fancy way of saying the single highest point and single lowest point in the interval. D) Absolute minimum and. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. The restrictions stated or implied for such functions will determine the domain from which you must work. 587 Get more help from Chegg Get 1:1 help nowfrom expert CalculustutorsSolve itwith our calculusproblem solver and calculator. For example, consider the functions shown in Figure(d), (e), and (f). Compare the f (x) f ( x) values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. (To make the distinction clear, sometimes the \u2018plain\u2019 maximum and minimum are called absolute maximum and minimum. Absolute minimum definition is - the smallest value that a mathematical function can have over its entire curve. f ( 1) = \u2212 5 f ( 1) = - 5. Sal finds the absolute maximum value of f(x)=8ln(x)-x\u00b2 over the interval [1,4]. earn credit. Python List max() Method - Python list method max returns the elements from the list with maximum value. True You are given that the function f ( x ) = 2 ( x + 3 ) x 2 + x \u2212 2 has an absolute maximum on the interval \u2212 2 < x < 1. Find the absolute maximum and absolute minimum values of f on the given interval. We usually distinguish between local and global (or absolute) extreme values. The smallest y -value is the absolute minimum and the largest y -value is the. On the interval, fnmin then finds all local extrema of the function as left and right limits at a jump and as zeros of the function's first derivative. The maximum will occur at the highest value and the minimum will occur at the lowest value. Roll your mouse over the Extreme Value Theorem to check your answers. A closed interval like [2, 5] includes the endpoints 2 and 5. For this data set, the minimum (lowest) value is 56 and the maximum (highest) value is 92. It could very well continue to increase or decrease once we leave the interval. Endpoint Discontinuities: only one of the one-sided limits exists. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. maximum: ( (62 -\u221a3763)/9, (-461347 +7526\u221a3763)/243)) \u2248 (. If an absolute maximum or minimum does not exist, enter NONE. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. Every function that\u2019s continuous on a closed interval has an absolute maximum value and an absolute minimum value (the absolute extrema) in that interval \u2014 in other words, a highest and lowest point \u2014 though there can be a tie for the highest or lowest value. In that case, the point right on the border might be the maximum or minimum of the curve. Absolute maximum is highest of and. f(x) = 4x^3 - 6x^2 - 144x + 9, [-4, 5] 13. Find the extreme values of f on the boundary of D. Before Using this Calculator. The following small array formulas can help you to find out the largest absolute value and the smallest absolute value. Explain the meaning of the result. Get the free \"Max/Min Finder\" widget for your website, blog, Wordpress, Blogger, or iGoogle. f(x) = x + (4/x) on the interval [1,5]. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. Calculates the root of the equation f(x)=0 from the given function f(x) and its derivative f'(x) using Newton method. There are two kinds of extrema (a word meaning maximum or minimum): global and local, sometimes referred to as \"absolute\" and \"relative\", respectively. Example: Find the absolute maximum and minimum of:. f(c) is another relative minimum of f. The student familiar with the sum formula can easily prove that. Return to Contents. It is a greatest value in a set of points but not highest when compared to all values in a set. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. We say that the function f(x) has a global maximum at x=x 0 on the interval I, if for all. Wolfram alpha paved a completely new way to get knowledge and information. exp(exp) * math. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. For, sin (x + ) = cos x. You can take notes in the margins or on the flip-side of each sheet. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Find the local maximum and minimum values of using both the First and Second Derivative Tests. Generating random numbers Problem. Let's find, for example, the absolute extrema of h (x)=2x^3+3x^2-12x h(x) = 2x3 +3x2 \u221212x. Four Function and. (b) Use calculus to find the exact maximum and minimum values. This important theorem can guide our investigations when we search for absolute extreme values of a function. 4 is the lower limit. Mean number of days \u2265 30, 35 or 40 \u00b0C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 \u00b0C. in some open interval containing c. The Organic Chemistry Tutor 200,049 views 1:10:05. Over the long term about one day in ten can be expected to have a (maximum or minimum) temperature exceed the decile 9 value. Sample size calculation for trials for superiority, non-inferiority, and equivalence. To define these terms more formally: a function f has an absolute maximum at x = b if f ( b )\u2265 f ( x ) for all x in the domain of f. Therefore, let\u2019s consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. You have to use a graphing calculator to find that out. Calculate Field honors the Transfer Field Domain Descriptions environment. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Enter the equation in the Y= section for Y1. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. If f has a local maximum or minimum at c and f'(c. (c) For any. Some problems may have two or more constraint equations. On a closed interval these points are referred to as absolute or global minimum/maximum points. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The \u201cV\u201d = volume per time. Local Extreme Values of a Function Let c be an interior point of the domain of the. Also time complexity of above solution depends on lengths of intervals. Find the absolute maximum and the absolute minimum values of the function shown below, on the given interval. 2 Maximum and Minimum on an Interval. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Evaluate the function to find the y -values at all critical numbers and at each endpoint. The smallest y -value is the absolute minimum and the largest y -value is the. Hill Sphere Calculator: Computes the Hill Sphere of an object: Sail Calculator: Computes the maximum velocity possible from acceleration caused by light. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. Look at the graph of f (x) = x 3 + 4x 2 - 12x over the interval [0, 3], Figure 1a. save_path = FLAGS. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. f(x)=x+ 9 x on [0. f(t) = 4t + 4 cot(t/2), [\u03c0/4, 7\u03c0/4] I'm stuck after I take the first derivative. Explain your reasoning. Sal finds the absolute maximum value of f(x)=8ln(x)-x\u00b2 over the interval [1,4]. question_answer. Calculus I: Candidates Test for Global Extrema 1) If a continuous function f is defined on a finite, closed interval, such as \u22121\u2264x\u22644 or [\u22121,4], or, more generally, a\u2264x\u2264b or [a,b], then f always has a global minimum value and a global maximum value on that interval. The debt to equity ratio is a financial, liquidity ratio that compares a company\u2019s total debt to total equity. 47, an absolute minimum of \u20135. 2 Maximum and Minimum on an Interval. You should substitute those and pick the greatest for the maximum value. No Local Extrema Compare the f (x) f (x) values found for each value of x x in order to determine the absolute maximum and minimum over the given interval. Find the absolute maximum and absolute minimum values of f on the given interval. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume. (1 point) Find the absolute maximum and absolute minimum values of the function f (x) = x3 + 6x2 \u2014 63x + 8 over each of the indicated intervals. maximum\" functions. f (c) is called the global (absolute) maximum value. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. This is achieved with a minimum of manual intervention. A relative minimum is a point that is lower than all the other points around it. The graph of y = cos x is the graph of y = sin x shifted, or translated, units to the left. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, fitness, and more. f(x)= 490x x2 +49 on [0,10] 2 Fall 2016, Maya. An open interval like (2, 5) excludes the endpoints. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. In this module you will be asked to calculate the sample size for 6 situations. Therefore, let's consider the function over the closed interval If the maximum value occurs at an interior point, then we have found the value in the open interval that maximizes the area of the garden. Local Extreme Values of a Function Let c be an interior point of the domain of the. (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. 20 at x = \u20130. By using this website, you agree to our Cookie Policy. Find the values of f f f at the critical numbers of f f f in (a, b). To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. Explain your reasoning. Significance Levels The significance level for a given hypothesis test is a value for which a P-value less than or equal to is considered statistically significant. The triangular distribution, along with the PERT distribution, is also widely used in project management (as an input into PERT and hence critical path method (CPM)) to model events which take place within an interval defined by a minimum and maximum value. The concern is that a dose given too soon after the previous dose may reduce the response. Pick the largest and smallest. Find the local or absolute minimum or maximum of an equation using a graphing calculator; Determine the intervals on which a function is increasing, decreasing, or constant using a graphing calculator (for precalculus) Determine an appropriate viewing rectangle for the graph of an equation; Match an equation to its graph; Graph an equation on. Finding the absolute max and min is a snap. (a) When is the object at rest? (b) Evaluate 6 1 \u222b vt dt(). An absolute minimum occurs at c if for all x in the domain of f. This is achieved with a minimum of manual intervention. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Please answer the following questions about the function Instructions: If you are asked to. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Final the absolute maximum and minimum values on the given interval. We usually distinguish between local and global (or absolute) extreme values. Find the extreme values of f on the boundary of D. 23] is -80 at x = 20 and maximum value of f(x) in interval [0,23] is at zero which is 0. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. Fold Unfold. Find the maximum / minimum absolute values with Formulas. f (1) = \u22125 f ( 1) = - 5. Here is the code to do that. Find the absolute maximum and absolute minimum values of f on the given interval. org are unblocked. Enter DNE if the absolute maximum or minimum does not exist. 388360 # Get 3 integers from 0 to 100 # Use max=101 because it will. Finding Extrema on a closed interval: 1. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. You have 3 solutions: x=0, x=1, and this one. )Given the function \ud835\udc53(\ud835\udc65= \ud835\udc652+ \ud835\udc65+ , chose values for a, b, and c in that could work for the graph shown. If we break down the formula we can see why it gets its strange name. Before Using this Calculator. These points are sometimes referred to as max, min, extreme values, or extrema. In a blank cell, enter this formula =Max(ABS(A1:D10)), see screenshot: 2. Therefore, we are trying to determine the maximum value of A(x) for x over the open interval $$(0,50)$$. Answer: First, \ufb01nd the critical points by \ufb01nding where the derivative equals zero: f0(x) = (x2 +4)(2x)\u2212(x2 \u22124. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. The generic word for minimum or maximum is extremum. An absolute minimum is the lowest y value or output value a. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. F INDING a maximum or a minimum has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. 80 at x = \u20131. ) Find the absolute max/min values of f(x) = x2 4 x2+4 on the interval [ 4;4]. Find the maximum / minimum absolute values with Formulas. a local (relative) maximum 6. Which method do you prefer? f (x) = 1 + 3x^2 - 2x^3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. The \u201cV\u201d = volume per time. # The text file for eval. Where does it flatten out? Where the slope is zero. There is 95% confidence that the constructed interval includes the population mean. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. It then evaluates the function at these extrema and at the endpoints of the interval, and determines the minimum over all these values. So, f(b) is a relative maximum of f. Recommended and Minimum Ages and Intervals Between Doses of Routinely Recommended Vaccines1,2,3,4 Vaccine and dose number minimum interval between doses is equal to the greatest interval of any of the individual components. For a strictly unimodal function with an extremum inside the interval, it will find that extremum, while for an interval containing multiple extrema (possibly including the interval boundaries), it will converge to one of them. So the absolute max value is 19 and the absolute min value is 1. The range spread then uses the range to find a percentage that the maximum is greater than the minimum, using the minimum as a base. Absolute Maximum and Absolute Minimum This page is intended to be a part of the Real Analysis section of Math Online. We do not know that a function necessarily has a maximum value over an open interval. In this module you will be asked to calculate the sample size for 6 situations. These values correspond to the probability of observing such an extreme value by chance. Solution for Find the absolute maximum and minimum of the function f (x) = x\u00b3 - x+2 on the interval [0, 3] %D. Find the absolute maximum and absolute minimum values of f on the given interval. If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0. For instance, in the example at. f (c) is called the global (absolute) maximum value. Thread starter denbal87; Start date Nov 4, 2014; D. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Calculate Field honors the Transfer Field Domain Descriptions environment. Explain your reasoning. Enter the equation in the Y= section for Y1. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. ) On the other hand, it is possible to see directly that. How to calculate a confidence interval? First, you need to calculate the mean of your sample set. An absolute minimum occurs at c if for all x in the domain of f. An absolute maximum is the highest y value or output value a graph has over a specific interval. 140 of 155. The golden-section search is a technique for finding an extremum (minimum or maximum) of a function inside a specified interval. Let this index be 'max_index', return max_index + min. Infinite Discontinuities: both one-sided limits are infinite. This lesson will focus on the maximum and minimum points. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value. (1 point) Let g(s) = i on the interval [0, 1. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. (\ud835\udc53\ud835\udc65)=5 Two solutions. Pick the largest and smallest. This corresponds to zero Kelvin, or minus 273. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). Calculate the range for your confidence statistics. Audio dithering. This is defined everywhere and is zero at $\\ds x=\\pm \\sqrt{3}/3$. Looking first at $\\ds x=\\sqrt{3}/3$, we see that $\\ds f(\\sqrt{3}/3)=-2\\sqrt{3}/9$. The maximum will occur at the highest f (x) f ( x) value and the minimum will occur at the lowest f (x) f ( x) value. The sample maximum and minimum provide a non-parametric prediction interval: in a sample from a population, or more generally an exchangeable sequence of random variables, each observation is equally likely to be the maximum or minimum. Sometimes it's important to consider points which are only largest or smallest in small parts of a graph. f x x3 2 on 3,1 > @ 15. find the absolute maximum and absolute minimum values of the fuction f(x)=2x-13ln(3x) on interval [1,8] 2. Time needed: 10 minutes. The calculators will allow you to convert any heart rate between 63% and 102% of your maximum heart rate to a percentage of your VO2max , or any percentage of VO2max. X Values: [a,b] Y Values: / [c,d] Result: [e,f] [1,3] / [2,4] [. On a closed interval these points are referred to as absolute or global minimum/maximum points. f(5) = (5)^4 + 8*(5)^3 -32*(5)^2 = 825 Thus function has absolute minimum value at x = 2 and absolute maximum value at x = 5 in the interval [-1, 5]. By practicing these kinds of problems you can understand this topic clearly. Thus if one has a sample {, \u2026,}, and one picks another observation +, then this has / (+) probability of being the largest value seen so far. Calculus Refresher by Paul Garrett. Relative Minimum - The lowest point on an interval of a curve. It is a measure of your capacity for aerobic work and can be a predictor of your potential as an endurance athlete. The maximum acceleration attained on the interval 03ddt by the particle whose velocity is given by v t t t t( ) 3 12 4 M 32 is A)9 B)12 C)14 D)21 E)40 3. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks. Local Extreme Values of a Function Let c be an interior point of the domain of the. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain \u2014 in which case it is called a relative or local extremum \u2014 or on a given set contained in the domain (perhaps all of it) \u2014 in which case it is called an absolute or global extremum (the latter. To find the extreme values of a function (the highest or lowest points on the interval where the function is defined), first calculate the derivative of the function and make a study of sign. It can be used as a worksheet function (WS) in Excel. A continuous function f(x) on a closed and bounded interval [a,b] has both an absolute min-imum and an absolute maximum on the interval. Locate the maximum or minimum points by using the TI-83 calculator under and the \u201c3. (c) Find all intervals on which the graph of f is concave up and also has positive slope. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. Q: Determine the absolute maximum and minimum values of the function on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. 5] The calculator performs interval arithmetic operations and computers interval version of mathematical functions. Finding Minimums and Maximums. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. By using this website, you agree to our Cookie Policy. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. a local (relative) minimum 5. To find the maximum and/or minimum on an interval, check the values at the critical points and at the ends of the interval. absolute minimum value at x = 2 is -48. We know that the absolute max/min values of f(x) will occur either at an endpoint or a critical number. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Example 2: Locate the value(s) where the function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist. We usually distinguish between local and global (or absolute) extreme values. An extreme value, or extremum (plural extrema), is the smallest (minimum) or largest (maximum) value of a function, either in an arbitrarily small neighborhood of a point in the function's domain \u2014 in which case it is called a relative or local extremum \u2014 or on a given set contained in the domain (perhaps all of it) \u2014 in which case it is called an absolute or global extremum (the latter. f(x) = 4x^3 \u2013 6x^2 \u2013 144x + 9, [-4, 5] 13. The student familiar with the sum formula can easily prove that. (c) For any. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. These are the examples in the topic increasing and decreasing intervals. Maximum intervals \u2014 they don't exist. Let f be a function defined and. It is a greatest value in a set of points but not highest when compared to all values in a set. Sample size calculation for trials for superiority, non-inferiority, and equivalence. Local Extreme Values of a Function Let c be an interior point of the domain of the. Thus, to find the absolute maximum (absolute minimum) value of the function, we choose the largest and smallest amongst the numbers f(a), f(c 1 ), f(c 2. Absolute minimum/maximum _____ d. Before going to class, some students have found it helpful to print out Purplemath's math lesson for that day's topic. These extreme values are obtained, either on a relative extremum point within the interval, or on the endpoints of the interval. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. f(x) = x4 \u2013 A: Plot the graph for f(x) in the interval for [-2, 0]. If you finish a job in less than 25% of the time allotted, you will be paid a Time Bonus, so try to finish as quickly as possible! The maximum Time Bonus is a 25% boost to your Base Reward. Generating random numbers Problem. Use the Stefan Boltzmann relationship (I = \u03c3T 4 where \u03c3 = 5. minimum\" or \"4. Find the maximum / minimum absolute values with Formulas. in some open interval containing c. The local maximum and minimum are the lowest values of a function given a certain range. A closed interval like [2, 5] includes the endpoints 2 and 5. B) No absolute extrema. (To make the distinction clear, sometimes the \u2018plain\u2019 maximum and minimum are called absolute maximum and minimum. Step 6: As mentioned earlier, A (x) is a continuous function over the closed, bounded. Finding the absolute max and min is a snap. Let this index be \u2018max_index\u2019, return max_index + min. checkpoint_interval # Where to write out summaries. It is important to understand the difference between the two types of minimum/maximum (collectively called extrema) values for many of the applications in this chapter and so we use a variety of examples to help with this. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Maxima and minima are points where a function reaches a highest or lowest value, respectively. Let Purplemath help you always be prepared! Go to the lessons!. sin(x * y) I have an interval for x [-1, 1] and y [-1, 1]. 487] Calculating confidence intervals: Calculating a confidence interval involves determining the sample mean, X\u0304, and the population standard deviation, \u03c3, if possible. summary_interval = FLAGS. The maximum will occur at the highest value and the minimum will occur at the lowest value. A point at which a function attains its maximum value among all points where it is defined is called a global (or absolute) maximum. Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. f x x x 32 3 on 3,1> @ 12. Similar topics can also be found in the Calculus section of the site. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. Let this index be \u2018max_index\u2019, return max_index + min. Therefore, f achieves its absolute minimum of \u221214 at x = \u22121 and its absolute maximum of 6 at both x = 1 and x = 4. 80 at x = \u20131. The absolute maximum value of f x x x( ) 3 12 M 32 on the closed interval > @M2, 4 occurs at x = A) 4 B) 2 C) 1 D) 0 E) M2 2. Mean number of days \u2265 30, 35 or 40 \u00b0C The average number of days in the period when the daily maximum air temperature was equal to, or exceeded 30, 35 or 40 \u00b0C. A point at which a function attains its minimum value among all points where it is defined is a global (or absolute) minimum. 14 If the first dose of recombinant zoster vaccine (Shingrix) is administered to someone 18-49 years of age, the dose does not need to be repeated. From the graph you can see that is has a maximum at (3, 27) and a minimum at (1. Free Maximum Calculator - find the Maximum of a data set step-by-step This website uses cookies to ensure you get the best experience. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function. 1 absolute maximum or minimum. f (1) = \u22125 f ( 1) = - 5. Question 203087: Find the absolute maximum and absolute minimum values of the function below. In this module you will be asked to calculate the sample size for 6 situations. summary_interval # How often to write checkpoints (rounds up to the nearest statistics # interval). 2 Maximum and Minimum on an Interval. Theorem (Extreme Value Theorem) If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some. The absolute minimum of the function f(x) = x2-9 on the interval - 4 Sxs 3 has a value of (Simplify your answer. In the single-variable case, it is known, by the Extreme Value Theorem, that if f is continuous on a closed interval [a;b], then it has has an absolute maximum and an absolute minimum on [a;b]. Any global maximum or minimum must of course be a local maximum or minimum. (1 point) Let g(s) = i on the interval [0, 1. Absolute & Local Minimum and Maximum Values - Relative Extrema, Critical Numbers / Points Calculus - Duration: 1:10:05. The student familiar with the sum formula can easily prove that. so minimum value of f(x) in interval [0. pow(y, 2)) * -1 return math. Extreme values \u00a92010 Iulia & Teodoru Gugoiu - Page 1 of 2 3. For instance, in the example at. How to use absolute minimum in a sentence. the absolute minimum age of 50 years when evaluating records retrospectively. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. (a) Interval : [\u20148, 0]. The smallest y -value is the absolute minimum and the largest y -value is the. 1 absolute maximum or minimum. Fermat's Theorem. Generating random numbers Problem. Help with finding absolute max/min values for a function. in some open interval containing c. Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. Let f be a function defined and. $f(x) = x^5 - x^3 + 2$, $-1 \\leqslant x \\leqslant 1$. 1 absolute maximum or minimum. The absolute max occurs at S = The absolute min occurs at S =. (a) fxc 0 at x 3, 1, 4 f c changes from positive to negative at 3 and 4. For each x value: Determine the value of f '(x) for values a little smaller and a little larger than the x value. between -30 to 20 function is decreasing because there are no local minima and maxima in between them. have both an absolute maximum and an absolute minimum. Find the extreme values of f on the boundary of D. An absolute extremum is an absolute maximum or an absoute minimum, and absolute extrema are absolute maximum and absolute minimum. Zoom in on the interval [-2,2] using the x-axis. In worst case, if all intervals are from 'min' to 'max', then time complexity becomes O((max-min+1)*n) where n is number of intervals. The MAX function is a built-in function in Excel that is categorized as a Statistical Function. Then press Ctrl+Shift+Enter keys, and the largest absolute values will be displayed in the. minimum\u201d or \u201c4. Find the absolute maximum and absolute minimum values of f on the given interval. Find the absolute maximum and minimum values on theinterval: f(x) = x - 2cosx [-pi, pi] f ' (x) = 1+2sinx f (-pi)= -pi - 2cos(-pi) = -pi - 2. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Find the absolute maximum and. By using this website, you agree to our Cookie Policy. denbal87 New member.", "date": "2020-08-03 15:15:48", "meta": {"domain": "rciretedimpresa.it", "url": "http://hefc.rciretedimpresa.it/absolute-maximum-and-minimum-calculator-on-interval.html", "openwebmath_score": 0.6303600668907166, "openwebmath_perplexity": 373.11161112116935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9890130576932457, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8520725317968266}}
{"url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1", "text": "# How to round 0.4999\u2026 ? Is it 0 or 1?\n\nIf you want to round the repeating decimal 0.4999... to a whole number what is the right answer? Is it 0 or 1?\n\nOn the one hand you only look at the first digit when you round numbers which in this case is 4 so the answer should be 0.\n\nOn the other hand 0.4999... is only another representation for 0.5 which makes the result 1.\n\nMy question\nWhich of these rules wins? (My gut feeling is that the most consistent result should be 1)\n\nEdit\nJust for clarification: What is meant by rounding here is replacing a fractional decimal number by one with fewer digits with the \"Round half up\"-Tie-breaking method.\n\n-\nIt appears that the rule is inconsistent. \u2013\u00a0 Isaac Solomon Oct 21 '12 at 9:04\nDoesn't that depend on what type of rounding you are using? If you're using Bankers Rounding, you would round towards 0. \u2013\u00a0 Fake Name Oct 21 '12 at 12:22\n@FakeName: Fair enough. What is obviously meant here is replacing a fractional decimal number by one with fewer digits with the \"Round half up\"-Tie-breaking method: en.wikipedia.org/wiki/Rounding \u2013\u00a0 vonjd Oct 21 '12 at 13:50\nThis question should NOT be closed; it has a definitive answer - with references \"out there\" to support it. At the point when this was closed, the issue on the table (discussion!) was if 0.499... was 0.5 or not: reference is en.wikipedia.org/wiki/0.999... \u2013\u00a0 Richard Sitze Oct 22 '12 at 0:08\nI agree with Richard. Given that we have a method for rounding midpoint values, and the author of the question has supplied a method, it follows that this question has an answer. \u2013\u00a0 Doug Spoonwood Oct 22 '12 at 2:52\n\nIt's $1$, because $0.49\\ldots$ is the same as $0.5$. If rounding is to be well-defined, it can't map one real number to two integers, so whatever it maps $0.49\\ldots$ to, it better maps it to the same integer as $0.5$. You could round both to $0$, of course, but that wouldn't then be the way we usually round.\n\nWhat this shows you is that rounding doesn't commute with limits, i.e. there's a difference between find the limit of a sequence and then rounding, and rounding first and then finding the limit. As you correctly observed, all values $a_n = 0.4\\underbrace{9\\ldots 9}_{n\\text{ times}}$ are rounded down to zero. Thus, $$\\lim_{n\\to\\infty} \\text{round }(a_n) = 0$$ On the other hand, $\\lim_{n\\to\\infty} a_n = 0.5$, and thus $$\\text{round }\\left(\\lim_{n\\to\\infty} a_n\\right) = 1$$\n\nThere's another word for functions which don't commute with limit - they're called non-continuous. So what you have discovered is simply that rounding is not a continuous function.\n\n-\nPerhaps I'm nitpicking, but is the result of rounding the limit perhaps $1$ instead? \u2013\u00a0 pimvdb Oct 21 '12 at 10:05\n@pimvdb: I went ahead and edited the answer, since it's obviously what was intended. (And so many people seem to agree.) \u2013\u00a0 ShreevatsaR Oct 21 '12 at 14:30\n+1 especially for pointing out the connection with continuity. \u2013\u00a0 Neal Oct 21 '12 at 16:00\n@pimvdb Indeed the limit should be $1$, and I wouldn't call pointing out blatant errors nitpicking ;-) Thanks for noticing, and thanks to ShreevatsaR for fixing this! \u2013\u00a0 fgp Oct 21 '12 at 19:42\n@jmoreno - but it is, in fact the same as 0.5: en.wikipedia.org/wiki/0.999... \u2013\u00a0 Richard Sitze Oct 22 '12 at 0:01\n\nFor $0\\le x\\le 1$, we round $x$ to $1$ if $x\\ge \\frac12$ and to $0$ if $x<\\frac12$ (though there are many conventions, see e.g. Wikipedia on rounding; the section \"Table-maker's dilemma\" a bit further down may also be interesting). Since $0.4\\bar9=\\frac12$, we should round to $1$. Another way of looking at this is that we always consider only the standard decimal expansion (i.e. we prefer $\\bar0$ over $\\bar 9$), and we are allowed to treat the first decimal as $4$ only if we know that it cannot turn out as $5$ \"later\". Thus if an inexact measurement gives us that $0.495\\le x\\le0.5$, we cannot say definitely, what $\\operatorname{round}(x)$ should be (we could if the measurement resulted in $0.495<x<0.5$). This is not different from the fact that we cannot say definitely what $\\operatorname{round}(x)$ should be if our measuremen merely says that $0.4997<x<0.5003$.\n\n-\n\nYou're right. Since $.4\\bar{9}=.5,$ if you want your rounding function to be well-defined you'll have to require an exception: round based on the first digit after the one you're rounding to, unless it's a $4$ followed by infinitely many $9$s.\n\n-\n\nIt is not obvious that 0.5 should be rounded to 1. Obviously 0.49999... should be treated the same because it is the same number. Wikipedia rounding article\n\n-\n\n$1 - 0.5 =0.5 \\quad$ and $\\quad 1 - 0.4999\\ldots = 0.500\\ldots = 0.5 \\quad$ so $\\quad 0.5 = 0.4\\overline9$\n\nWell, the above line says the proof / intuition. From our information... $0.5$ well rounds to $1$.\n\n-\nAre you sure that $1-0.4999\\ldots\\ne 0.500\\ldots1$? \u2013\u00a0 MJD Nov 18 '12 at 16:46\n@MJD: Not sure if serious... \u2013\u00a0 Clive Newstead Nov 18 '12 at 16:47\n@MJD: In 0.500...1, the zeroes never end, so the $1$ never comes ;) \u2013\u00a0 Parth Kohli Nov 18 '12 at 16:52\n\nI find very convenient to define rounding as follows: $$round(x) = \\left \\lfloor x + \\frac{1}{2}\\right \\rfloor$$\n\nConsidering this rule, what follows is understanding the context. If you are working with a computer, you have only a fixed number of decimal positions to work with, in which case, if x=0.4999999999999999 (16 decimal positions), then\n\nround(x)=floor(0.4999999999999999 + 0.5)=\n=floor(0.9999999999999999)=\n=0\n\n\nOf course, if you are talking about a real number, then 0.4999... equals 0.5, in which case $round(x)=1$.\n\nSo, the real question here is: are you working with a fixed number of decimal positions or not? If the answer is 'yes', then definitely round(x)=0. If the answer is 'no', then you are talking about a number which is equal to 0.5, and $round(x)=1$\n\n-\n\n@fgp answered this well but there is another aspect that I don't think has been touched on in any of the answers. What is the type of thing that we should consider as an input to the process of rounding? If it is a real number, we are unable to apply the rule without a specific digit sequence representing that number. If instead we interpret rounding as an operation on a possibly infinite digit sequence, we can now apply the rule, but its result has no meaning for general real numbers. One way out which has been discussed is to make a bijection between digit sequences and reals by disallowing digit sequences ending in $999...$, and this view effectively unasks the question by taking the position that $0.4999...$ is not a valid representation of a real number, but I think this is an overly burdensome technicality.\n\nI think an easier way to look at it is just to say that rounding is an operation that applies to finite digit sequences and not to real numbers in general nor to infinite digit sequences. When we say something like \"$\\pi$ rounds down to $3$\" it is a lazy way of saying that all sufficiently precise approximations do. But when we say \"$0.5$ rounds up to $1$\", exactly that is meant, because we don't mean to suggest that $0.4 + \\sum_{k=2}^{n}{9 \\cdot 10^{-k}}$ will round up to $1$ for sufficiently large $n$. And without context, if it is said about a real number $x$ that \"$x$ rounds up to $1$\" then that can be read as \"$x$ has a finite decimal representation which rounds up to $1$, or $x$ has no finite decimal representation and all sufficiently precise finite decimal approximations to $x$ round up to $1$\". In this way we can meaningfully discuss rounding real numbers in terms of rounding finite digit sequences, and without explicitly addressing the problem of sequences ending in $999...$.\n\n-\n\nIf you consistently use the round-up half method you don't always look at the first digit of the numeral when rounding numbers. You go by the method. The method allows that if we have anything else than a half-way value, then you can round by the rule \"if (working in base 10) the first digit of the numeral after the decimal point belongs to {0, 1, 2, 3, 4}, then round down, if the first digit of the decimal point belongs to {5, 6, 7, 8, 9}, then round up.\" But, that rule does not say anything about half-way values. Going by q=FLOOR(y + .5) we can reason as follows: q=FLOOR(.499... + 5)=FLOOR(.999...)=FLOOR(1)=1, since .999...=1. So, that method yields that .499... rounds to 1.\n\n-\n\nIt is the same as 0.5 by definition of decimal numbering.\n\n-\nAnd rounding takes by definition the first digit which is 4 in this case. So this is exactly where the two definitions collide! \u2013\u00a0 vonjd Oct 21 '12 at 18:36\n@vonjd I disagree that rounding by definition takes the first digit and then rounds. I ask you to try and round 1/3 without converting it into a decimal. Do you not round 1/3 down to 0, since it comes as clear that 1/3 lies closer to 0 than to 1? Or I suggest you round 12/11 to the nearest whole number. Or round 155/66 to the nearest whole number. Finding the multiples of a denominator d will help here, and what half of d equals. If the denominator equals 7, then if the numerator of \"the remainder\" equals 1, 2, or 3 we round down, if the numerator equals 4, 5, or 6 we round up. \u2013\u00a0 Doug Spoonwood Oct 21 '12 at 19:17\nE. G. Say we have 13/7. 13/7=7/7+6/7, correct? So, 13/7 has \"remainder\" of 6/7, or in other words it has a fractional part (search that term on Wikipedia) of 6/7. Thus, 13/7 rounds up to 14/7 (6/7 lies closer to 7/7 than to 0/7). You could also try and rounding numbers in binary, trinary, or any n-ary base system. E. G. round 1.34523 in base 7. In base 7, 1.4142 rounds up to 2. So, I simply do not see how the digit makes the difference here. More goes on than that! \u2013\u00a0 Doug Spoonwood Oct 21 '12 at 19:23\n@vonjd this is an exception. rounding takes first digit EXCEPT for values like this, where fraction is written in limitless form. \u2013\u00a0 Suzan Cioc Oct 21 '12 at 19:25\n\nThis is just an obersvation that I was missing in the many answers already given.\n\nWhile there is absolutely no doubt that the mathematical function $\\Bbb R\\to\\Bbb Z$ defined by $x\\mapsto\\lfloor x+\\frac12\\rfloor$ when applied to the real (in fact rational) number represented by the repetitive infinite decimal sequence $0.4\\overline9$ takes as value the integer$~1$, saying that this is what happens when rounding the repeating decimal $0.4999\\ldots$ is misleading. That is because \"rounding\" suggests an operation done during actual computation. This ignores the often overlooked fact that it is impossible to do arithmetic computations with infinite decimals. This does not mean of course that there isn't a class of infinite decimals with which computations are possible; for instance if all decimals are repeating, then one can replace them by rational numbers and compute with those, which is definitely possible. However the misconception is that, because we can do arithmetic on numbers with any fixed number of decimal places and the procedures for doing so are always basically the same, therefore one can apply these same procedures to infinite decimals. However just looking at these procedures will reveal that this is plainly false: all of these procedures need at some point (often right at the beginning) to locate the final decimal position, and for an infinite decimal such a position simply does not exist. This means that in practice whenever one needs rounding decimal numbers as a computational operation, then those decimals are necessarily finite, and the number $0.4999\\ldots$ simply cannot occur.\n\nIn fact there is no effective system for computing with (general) true real numbers at all, for the simple reason that the majority of the real numbers cannot be represented in any way (a concrete representation having to be, by the nature of things in the real world, finite). One can however restrict to the (countable) subset of real numbers that do allow some kind of representation (which is less brutal that restricting to some arbitrarily chosen finite subset of numbers declared representable, as is done in numeric computation), and then arithmetic and other kinds of computation do become possible, and indeed can be done without having to introduce approximations. This has been studied in constructive mathematics, but it turns out that for this purpose representing real numbers by infinite decimals, which might be produced on demand by a finite algorithm, is not a viable option. Even just finding the first decimal of the sum of two numbers so given by infinite decimals cannot be assured by an algorithm; this is in fact closely related to issues invoked in this question and the related $0.9999\\ldots$ problematic. This in contrast to computations with formal power series with integer (or rational) coefficients: these suffer from the same \"most elements cannot be represented\" problem as the real numbers, but if one sticks to series whose coefficients can be computed by an algorithm, then all arithmetic (and many other) operations on them can be performed algorithmically without problem.\n\nA more workable way to represent real numbers is by a sequence of definite approximations (rational numbers with a given maximal error), with some kind of guarantee that the error tends to $0$. But even then there are sacrifices one has to make: it is impossible to computations with discontinuous functions (like the floor or round-to-integer operations): in the constructive theory of real numbers there only exist continuous functions. So all of this is rather far removed from what one does in numeric computation, where real numbers are necessarily approximated, the number of decimals finite, and the question of how to round $0.4\\overline9$ moot; the rule for rounding of looking at the first digit to be discarded in the rounding is perfectly valid in practice.\n\n-\n\nIf we round c to d, then c approximately equals d. If d also consists of an integer, then c lies closer to d than any other integer i. In other words for all integers i such that i does not equal d, ABS(ci-)>ABS(cd-), where ABS(ci-) indicates the number by which c and i differ by (the absolute value of the difference of c an i). This implies that we can use the distance metric between real numbers to see how rounding works as follows.\n\nConsider the set S=={[l, lu+2/), (lu+2/, u]} where ul-=1. In other words, l and u differ by 1, and we consider the set of all numbers between x and y except for the midpoint m==lu+2/ between them. We can use the metric function D(x,y)=|xy-|=ABS(xy-), where x belongs to S and y belongs to {l, u}, to determine how to round all members of S by following rule:\n\nIf D(x, l)>D(x, u), then round to u. If D(x, u)>D(x, l), then round to l.\n\nSince .5=.49999..., it follows that D(.49999..., 1)=D(.49999...., 0)=.4999...=.5. Thus, the distance function D(x, y)=ABS(xy-) does not give us any guide as to how to round here.\n\nConsequently, we have several different choices that we can make which will not go against our intuition of rounding as follows:\n\n1. Choose to round .49999... to 1 (maybe you value larger numbers in general),\n2. Choose to round .49999... to 0 (maybe you value smaller numbers in general),\n3. Choose to allow .49999... to get rounded to both as one pleases or sees fit at particular points in time, or\n4. Choose to round .49999... to neither 0 nor 1 (maybe you don't want to make a choice here).\n\nNone of those possibilities win or lose until we have a standard to measure a winner or loser for the rules. There also doesn't exist a \"right answer\" here unless we have a means to determine what properties a \"right answer\" has in this context. I suspect that many people would resist 3. with the possibility that one could round .4999... to 1 at one time, and then round .4999... to 0 at another as one pleases, thus indicating arbitrariness. But, it does not follow that such comes as an improper procedure unless one first rates such \"arbitrariness\" as bad, and something so utterly awful and terrible that we or others must never, ever engage in such \"arbitrariness\". Perhaps many people would see 4. as \"evasive\", but unless there exists a logical basis to picking either 0 or 1, anyone accused of \"evasiveness\" by picking 4. could very well respond to such a charge, that anyone who sees choosing 4. as indicating \"evasiveness\", insists dogmatically that all such questions absolutely must have answers, and/or that the people claiming \"evasiveness\" have copped out of making choices on the basis of logic as much as possible.\n\nSo, what sort of choice do you want to make?\n\n-\n\nAs a programmer, 0.4999... is closer to 0 than it is to 1, so I would always round down to 0.\n\nKeep it simple! Rounding is always to the closest whole number. You only need to apply special cases when there are two closest whole numbers.\n\nBut there are situations where this would be the wrong choice, so you need to do some research every time you apply rounding function to a value.\n\nFor example, the tax department in the country I live in does not always acknowledge cents. Under legislation, there are times when the only valid rounding function is floor.\n\n-\nI think we all agree that $0.4999$ should round down to $0$. The question is whether $.499999999......$ should (with infinitely many $9$s). \u2013\u00a0 Jason DeVito Oct 21 '12 at 18:16\nHow close $0.4999...$ is to $0$ or $1$ is independent on whether or not you are a programmer. In fact, it is the same distance from $0$ and $1$. If the sequence of digits were to terminate after finitely many (e.g. in computing) then fine, but here we have infinite precision and the number is exactly equal to $0.5$. \u2013\u00a0 Clive Newstead Oct 21 '12 at 18:20\n@AbhiBeckert: The case in point is that it is indeed equal to 0.5, as is 0.999... equal to 1: en.wikipedia.org/wiki/0.999... \u2013\u00a0 vonjd Oct 21 '12 at 18:32\nDear Abhi, I have downvoted your answer, since it it seems to be premised on a confusion with regard to the fact that $0.4999\\cdots$ (infinitely many $9$'s) is equal to $0.5$. Regards, \u2013\u00a0 Matt E Oct 21 '12 at 19:56\n@AbhiBeckert: If you accept that $0.4999...$ is a real number, and you also claim it's less than $0.5$, then (1) you should be able to say what is the distance between them (for any two real numbers $x<y$, their distance is another real number $y-x$: and no, \"infinitesimal\" is not a number), and (2) you should be able to exhibit a number that lies between them (for any $x<y$, there exist infinitely many real numbers between, e.g. $x<(x+y)/2<y$). The fact that there is no number between $0.4999...$ and $0.5$ should convince you that they are two representations of exactly the same number. \u2013\u00a0 ShreevatsaR Oct 23 '12 at 5:16", "date": "2014-07-11 05:57:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/217893/how-to-round-0-4999-is-it-0-or-1", "openwebmath_score": 0.8157265186309814, "openwebmath_perplexity": 432.2086184790631, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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{"url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036", "text": "# How to translate mathematical intuition into a rigorous proof?\n\nI've seen a lot of questions about how to develop mathematical intuition, but often I have the opposite problem.\n\nSeveral times I have run into a situation where I want to solve a math problem, and I play around with it until my intuition reaches the point where I have a good idea of how the complete proof might be structured. But then when it comes to actually write out the proof, I have trouble translating this intuition into a rigorous proof. I want to give the following example, from this PDF of Putnam training problems.\n\n1.13. Prove that for every $n\\ge 2$, the expansion of $(1+x+x^2)^n$ contains at least one even coefficient.\n\nWhen first thinking about this problem, I wanted to look at the polynomials $\\pmod{2}$ so that \"even coefficient\" is simplified to \"zero coefficient.\"\n\nThen from doing a few computations, I notice a pattern.\n\n\\begin{align*} (1+x+x^2)^2&=(1+x+x^2)+x(1+x+x^2)+x^2(1+x+x^2)\\\\ &=1+x+x^2\\\\ &+x+x^2+x^3\\\\ &+x^2+x^3+x^4\\\\ &=1+x^2+x^4 \\end{align*}\n\nI notice that multiplying a polynomial by $(1+x+x^2)$ is like adding that polynomial with itself three times, each shifted over. The $x$ shifts it over by one place and the $x^2$ shifts it over by two places.\n\nContinuing this pattern, we can get the coefficients of $(1+x+x^2)^3$ as follows:\n\n\\begin{array}{ccccccc} 1&0&1&0&1&&\\\\ &1&0&1&0&1&\\\\ &&1&0&1&0&1\\\\ \\hline 1&1&0&1&0&1&1 \\end{array}\n\nSo $(1+x+x^2)^3\\equiv 1+x+x^3+x^5+x^6 \\pmod{2}$\n\nLet's make a triangle of a few more results:\n\n\\begin{array}{cccccccccccc} 0:&&&&&&1&&&&&\\\\ 1:&&&&&1&1&1&&&&\\\\ 2:&&&&1&0&1&0&1&&&\\\\ 3:&&&1&1&0&1&0&1&1&&\\\\ 4:&&1&0&0&0&1&0&0&0&1&\\\\ 5:&1&1&1&0&1&1&1&0&1&1&1\\\\ \\end{array}\n\nWe see that the structure of our problem is essentially the same as an elementary cellular automaton (specifically, rule 150). I will come back to this later.\n\nI notice another peculiar pattern with the $1$st, $2$nd, and $4$th rows: There are only $1$s on the ends and in the center. Perhaps this pattern continues for all powers of $2$. And it does, which is not hard to prove with induction.\n\nClaim: $\\forall n\\ge 0,\\ (1+x+x^2)^{2^n}\\equiv 1+x^{2^n}+x^{2^{n+1}}\\pmod{2}$\n\nBase case: $(1+x+x^2)^{2^0}\\equiv 1+x+x^2\\equiv 1+x^{2^0}+x^{2^1}\\pmod{2}$\n\nInductive Step:\n\n\\begin{align*} (1+x+x^2)^{2^n}&\\equiv ((1+x+x^2)^{2^{n-1}})^2\\\\ &\\equiv (1+x^{2^{n-1}}+x^{2^n})^2\\\\ &\\equiv 1+x^{2^n}+x^{2^{n+1}}\\pmod{2} \\end{align*}\n\nThis is where the intuition comes in that is hard for me to express rigorously. On one of the $2^n$th rows, look at the $0$ equidistant from the leftmost $1$ and the $1$ in the center. If the rightmost $1$ wasn't there, then that $0$ would stay a $0$ forever because of symmetry: any effect from the left is cancelled by an effect from the right.\n\nHowever, the rightmost $1$ does exist, so it will eventually change that $0$. But since effects are local in this automaton, it will require over $2^n$ more rows until the rightmost $1$ affects the $0$ we are looking at (there are over $2^n$ spaces between them). In other words, we have shown that the $0$ will stay a $0$ for all rows up to the $2^{n+1}$th row. Applying this logic on each power of two row, we show that there is always a zero coefficient from row $2$ onwards.\n\nIs there a more rigorous way I can express those last two paragraphs? And in general, what advice do you have for translating intuition to a rigorous proof?\n\n\u2022 In your example you changed the question. Originally it asked for a $0$ in every row. Your version only talks about rows of the form $2^n$, but you show a very special form for those rows. Your induction is fine and proves what you wanted to prove, but it is not the original question. It might be a step along the way I can imagine (but I am speculating) that you can show a partial row starting with $1$, a bunch of $0$s and a $1$ will always have a $0$ somewhere. If you can do that you are done in conjunction with what you have done. \u2013\u00a0Ross Millikan Jun 25 '18 at 4:15\n\u2022 @RossMillikan I did not change the question in any way. The observation I made about the $2^n$ rows was a stepping stone in my proof. Look at the last three paragraphs. \u2013\u00a0Riley Jun 25 '18 at 4:18\n\u2022 You said you would come back to rule 150 but I don't think you did \u2013\u00a0Mark Jun 25 '18 at 4:49\n\u2022 @Mark I meant that I will come back to framing the problem in terms of cellular automata, which I did when discussing the \"effects\" of the $1$s, and the locality of the automaton. \u2013\u00a0Riley Jun 25 '18 at 4:51\n\u2022 [Just a comment as the question is about the automaton-intuition being turned into a proof.] Another way of thinking about this expression is, the coefficient of $x^m$ in the expansion shows the number of ways to achieve a total of $m$ by rolling a die with sides $0, 1, 2$ a total of $n$ times. To see why, try making a three by three times table model to see the result of squaring the expression; then, erase the base of $x$. \u2013\u00a0Benjamin Dickman Jun 25 '18 at 5:17\n\nYou have proved that these binary strings are symmetric so it suffices to prove the result for only the \"half-strings.\" Imagine cutting your triangle down the middle. Let $x_n$ be the $n^{th}$ bit string, which is length $n+1$.\n\nWe have $x_0 = 1$, $x_1 = 11$, $x_2 = 101$, etc.\n\nNow you can address the $j^{th}$ bit inside $x_n$ as $x_{n,j}$.\n\nThis addressing system will make it easier to talk more concretely about the action of the automaton.\n\nYou can pin down a definition of \"locality of action\" by saying that there is some function $f(a, b, c)$ such that $\\forall n, j: x_{n,j} = f(x_{n-1, j-1}, x_{n-1,j}, x_{n-1,j+1})$ (with some additional nuisance conditions about the boundary). This means that the $j^{th}$ bit in the current string can be computed by neary-by bits in the previous string. Here $f$ represents the action of this rule 150 you mention.\n\nNow use $f$ again to show $x_{n,j} = f(f(x_{n-2, j-2}, x_{n-2,j-1}, x_{n-2, j}), f(x_{n-2, j-1}, x_{n-2,j}, x_{n-2, j+1}), f(x_{n-2, j}, x_{n-2,j+1}, x_{n-2, j+2}))$\n\nNow you can see when we repeat this procedure $k$ times we will end up with a trinary abstract syntax tree depth $k$. To avoid having to talk about the exact structure of the expression at the $k^{th}$ level, we can reason about this substitution process purely formally as a process operating on the (countably infinite) set of symbols $\\{ \\underline{f}, \\underline{(}, \\underline{)} \\} \\cup \\{ \\underline{x_{i,j}} \\}_{(i,j) \\in \\mathbb{N}^2}$ (here the commas are not meant to be part of the symbol set, they are just there to separate symbols from each other). You can prove by induction, that for any $k$, that $x_{n,j}$ is computed by an expression using the symbols $\\{ \\underline{f}, \\underline{(}, \\underline{)} \\} \\cup \\{\\underline{ x_{n-k, j-k}},\\underline{ x_{n-k, j-k+1}}, \u2026,\\underline{ x_{n-k, j+k-1}},\\underline{ x_{n-k, j+k} }\\}$ (the $2k+1$ bits in the $(n-k)^{th}$ string centered on the $j^{th}$ bit) intermixed with function applications of $f$.\n\nFix $n$ and let $k_n$ be the distance to the previous power of $2$. (So if $n=13$, $k_n=5$ since the previous power of $2$ is $8$). And define $j_n \\equiv (n-k_n)/2$. This is the index of the central bit of the $(n-k_n)^{th}$ bit string.\n\nYou have demonstrated that those bits at indices $\\{ (n-k_n, m) \\}_{1 \\le m \\lt n }$ are all zero since $n-k_n$ is a power of $2$. So any well-formed expression using only the symbols $\\{ \\underline {f}, \\underline{(}, \\underline{)}\\} \\cup \\{ \\underline{x_{n-k_n, j_n-k_n}}, \\underline{x_{n-k_n, j_n-k_n+1}}, \u2026,\\underline{x_{n-k_n, j_n+k_n-1}}, \\underline{x_{n-k_n, j_n+k_n}}\\}$ will evaluate to zero by $f(0,0,0) = 0$, and in particular the expression representing $x_{n,j_n}$ (created from the $k_n$ iterations of the formal symbol substitution procedure) evaluates to $0$.\n\nSo as $n$ increments, $x_{n,j_n}$ traces out the path of the central bits you were referring to, all of which are zero.\n\nGeneral thoughts about formalization: There are certain technical devices which are work horses of the formalization process. Two of them I have used in this proof:\n\n\u2022 indexing time and space, with a numeric coordinate system\n\u2022 encoding a time evolving system as function application\n\nA third device which I used is not as common, and comes from computer science and logic:\n\nSomething that I have been realizing is that these devices are not obvious (at least to me). Humanity was around for a long time before the Cartesian coordinate system was discovered and used to formalize geometrical intuition. It took even longer to develop the idea that almost every mathematical object could be encoded as an intricate tower of sets (ZFC).\n\nI guess for me it's a matter of building up a library of these devices and binding them to the correct intuitions by repeated use.\n\nIn certain areas like computer programming, there are very few tools for converting intuition into proof. There were a couple of very good devices created like\n\n\u2022 loop invariants\n\u2022 Hoare logic\n\nbut multithreaded and heap-allocating programs are still an active area of research.\n\n\u2022 By the way, I have not verified the claim about the sequence actually evolving as the cellular automata $f$ but I'm trusting it's true. \u2013\u00a0Mark Jun 25 '18 at 7:01\n\u2022 I think your definition of $j$ doesn't quite match the way you use it. You're using it as if $j$ represents the $x$ coordinate in the triangle rather than the $j$th bit. Also I think you need to consider whole strings instead of half strings for the automation to work. And I think you need to define $x_{n,j}=0$ when $j$ is out of bounds. Otherwise a great answer. \u2013\u00a0Riley Jun 25 '18 at 12:50\n\u2022 Yeah, for it to make sense, we need to start counting things at $0$ as we do in computer programs. $x_{n,j} \\cong x[n][j]$. As for the whole strings vs half strings, maybe it will be convenient to use negative indices to talk about the left half of the triangle. \u2013\u00a0Mark Jun 25 '18 at 14:48\n\nHere is one answer to the question. If $$\\, n\\ge 2\\,$$ then the first $$4$$ entries in each row, namely, $$\\, 1 0 1 0, 1 1 0 1, 1 0 0 0, 1 1 1 0, \\dots, \\,$$ repeat in a period of $$4$$. There is a $$0$$ in each of these first $$4$$ entries. This was not the proof you wanted because it is too simple.\n\nYour proof is essentially the following. You noticed something about a $$0$$ in row $$\\, 2^n \\,$$ and that this $$0$$ continues up to the row $$\\, 2^{n+1}. \\,$$ We formalize that observation by working with polynomials over the field with two elements. Define the Laurent polynomials $$\\, y(n) := x^n + x^{-n} \\,$$ and since the characteristic of the field is $$2$$, $$\\, y(0) = 1 + 1 = 0. \\,$$ Using simple algebra of exponents we get that $$\\, y(n)y(m) = y(n+m) + y(n-m), \\,$$ and the special cases $$\\, y(n)^2 = y(2n) \\,$$ and $$\\, y(2^n) = y(1)^{2^n}.$$\n\nDefine the row polynomials $$\\, r(m) := (1 + y(1))^m. \\,$$ The general row is $$\\, r(m) = r(2^n+k) \\,$$ where $$\\, 0\\le k < 2^n. \\,$$ Notice that $$\\, r(2^n) = 1 + y(1)^{2^n} = 1 + y(2^n). \\,$$ Also $$\\, r(m) = 1 + \\sum_{i=1}^m c_i y(i), \\,$$ where $$\\, c_i \\,$$ is the coefficient of $$\\, y(i). \\,$$ You noticed that $$\\, c_{2^{n-1}} = 0 \\,$$ but how to prove it? We calculate that $$\\, r(m) = (1 \\!+\\! y(2^n)) r(k) = (1 \\!+\\! y(2^n))(1 \\!+\\! cy(2^{n-1}) \\!+\\! t) = (1 \\!+\\! y(2^n))(1 \\!+\\! t) \\!+\\! c y(3\\,2^{n-1}) \\,$$ where $$\\, t \\,$$ is all the terms with $$\\, y(i), \\,$$ and $$\\, i\\ne 2^{n-1}, \\,i < 2^n. \\,$$ But $$\\, y(2^n)y(i) = y(2^n-i) \\!+\\! y(2^n+i) \\,$$ and since $$\\, i \\ne 2^{n-1} \\,$$ we get no $$\\, y(2^{n-1}) \\,$$ terms in the product $$\\, (1 \\!+\\! y(2^n))(1 \\!+\\! t). \\,$$ That proves it.\n\nThis answers your particular question, but what about intuition versus rigor? I think there are no good answer to that question. You have to always look for any patterns or regularities and simple or edge cases. You have to be lucky and it helps to have a lot of background experience to draw upon.\n\n\u2022 I recognize that there is a simpler answer to the question. It is given in the PDF I liked to. I am not asking for the solution, I am asking about how to translate my ideas into a rigorous proof. \u2013\u00a0Riley Jun 25 '18 at 4:06\n\u2022 I thought I was translating your ideas combined with ideas of my own into a proof. But intuitions are tricky to express clearly. Maybe you wanted an \"automaton theoretic\" proof? \u2013\u00a0Somos Dec 3 '18 at 19:55", "date": "2019-06-19 00:42:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2830788/how-to-translate-mathematical-intuition-into-a-rigorous-proof/2831036", "openwebmath_score": 0.8129568099975586, "openwebmath_perplexity": 227.65682371870014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211582993981, "lm_q2_score": 0.8740772466456689, "lm_q1q2_score": 0.8520689940182796}}
{"url": "https://math.stackexchange.com/questions/716005/equivalence-of-inner-products", "text": "# Equivalence of inner products?\n\nSo, let $\\langle \\cdot, \\cdot \\rangle_{1}$ and $\\langle \\cdot, \\cdot \\rangle_{2}$ be inner products on a real vector space $V$. Assume that \\begin{align} \\langle v, v \\rangle_{1} = \\langle v, v \\rangle_{2} \\end{align} for any $v \\in V$.\n\nIs it true that $\\langle v, w \\rangle_{1} = \\langle v, w \\rangle_{2}$ for all $v,w \\in V$?\n\nI can only think of two inner products on a real vector space $V$ are the standard dot product and the scaled dot product. This holds for these two since the scaled dot product would require all scalars to be 1. Can anyone think of another inner product on $V$ to test this with, or should I start trying to prove it?\n\n\u2022 Take a basis $v_1,\\dots ,v_n$ for $V$. Then $\\langle v_i,v_i\\rangle_1=\\langle v_i,v_i\\rangle_2$ for all $1\\leq i\\leq n$ implies that $\\langle v,w\\rangle_1=\\langle v,w\\rangle_2$.\n\u2013\u00a0DKal\nMar 17 '14 at 21:09\n\nFor every $v,w \\in V$ and $c$ real $$\\langle v+cw,v+cw\\rangle_1=\\langle v+cw,v+cw\\rangle_2$$ or $$\\langle v,v\\rangle_1+2c\\langle v,w\\rangle_1+c^2\\langle w,w\\rangle_1= \\langle v,v\\rangle_2+2c\\langle v,w\\rangle_2+c^2\\langle w,w\\rangle_2$$ and hence, as $\\langle v,v\\rangle_1=\\langle v,v\\rangle_2$ and $\\langle w,w\\rangle_1=\\langle w,w\\rangle_2$, then $$\\langle v,w\\rangle_1=\\langle v,w\\rangle_2.$$\n\nEven though Yiorgos S. Smyrlis' answer is simpler, I think it is useful having a little different proof, just to add another point of view of the same argument, which is always good :)\n\nConsider the norm $\\| \\cdot \\|_1$ induced by $\\langle \\cdot, \\cdot \\rangle_1$ and the norm $\\| \\cdot \\|_2$ induced by $\\langle \\cdot , \\cdot \\rangle_2$. By hypothesis and definition of induced norm $$\\| v \\|_1^2 = \\langle v, v \\rangle_1= \\langle v, v \\rangle_2= \\| v \\|_2^2$$ $\\forall v \\in V$. So the two norms are the same (the power doesn't make any problem thanks to the positivity of the norm.)\n\nThen we want to do the reverse reasoning, if two product-induced-norms coincide on every element of the space, what can we say about inner products?\n\nIt's immediate (and left as little exercise for the reader - just write down the definitions and use bilinearity-) to prove that given a real inner product and its norm the following equivalence is true $$\\langle x,y \\rangle = \\dfrac{1}{4} \\left( \\|x+y \\|^2 - \\|x-y\\|^2 \\right)$$ This formula is called the polarization identity (for more see here)\n\nSo by the above identity and the hypothesis we have, $$\\langle x,y \\rangle_1 = \\dfrac{1}{4} \\left( \\|x+y \\|_1^2 - \\|x-y\\|_1^2 \\right) = \\dfrac{1}{4} \\left( \\|x+y \\|_2^2 - \\|x-y\\|_2^2 \\right) = \\langle x,y \\rangle_2$$ $\\forall x,y \\in V$ and so we are done.\n\nI wanted to stress the depth correlation between an inner product and its induced norm, and show some useful tools. Hope it helps :)", "date": "2021-09-21 09:09:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/716005/equivalence-of-inner-products", "openwebmath_score": 0.9495441317558289, "openwebmath_perplexity": 145.5758585312135, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211597623863, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8520689889017713}}
{"url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects", "text": "# How many random walk steps until the path self-intersects?\n\nTake a random walk in the plane from the origin, each step of unit length in a uniformly random direction.\n\nQ. How many steps on average until the path self-intersects?\n\nMy simulations suggest ~$$8.95$$ steps.\n\nRed: origin. Top: the $$8$$-th step self-intersects. Bottom: the $$11$$-th step self-intersects. (Not to same scale.)\nI suspect this is known in the SAW literature (SAW=Self-Avoiding Walk), but I am not finding this explicit number.\n\nRelated: self-avoidance time of random walk.\n\nAdded. Here is a histogram of the number of steps to self-intersection.\n\n$$10000$$ random trials.\n\n\u2022 This is asking for a uniformly random direction in the plane. By comparison, in 1-d the corresponding number is $(1/2)2 + (1/4)3 + (1/8)4 + \\cdots = 3$; in 3-d the corresponding number is $\\infty$. \u2013\u00a0Matt F. Mar 10 '19 at 4:40\n\u2022 I added a second example. I have thousands of these. :-) \u2013\u00a0Joseph O'Rourke Mar 10 '19 at 18:56\n\u2022 One can also ask for the asymptotic as $n$ goes to infinity of the probability that a length $n$ random walk does not self-intersect. The form of this asymptotic might be close to the case of random walks on grids, which if I recall right is conjectured but not known. \u2013\u00a0Will Sawin Mar 12 '19 at 17:42\n\u2022 FWIW I ran this on $10^7$ random configurations and got a mean of ~$8.886$ steps. \u2013\u00a0Chip Hurst Mar 13 '19 at 14:22\n\u2022 I ran $10^9$ simulations over night and the mean stayed at ~$8.886$. Here's a link of the simulation tallies: wolfr.am/C3YsNZmE. The largest run consisted of 98 lines. \u2013\u00a0Chip Hurst Mar 14 '19 at 12:16\n\nHere are some computational insights.\n\nI ran $$N = 10^k$$ simulations for $$3 \\leq k \\leq 12$$. For $$N = 10^9$$ I recorded some explicit instances and recorded which pairs intersected, as opposed to just the lengths. For all other runs, I simply summed the results.\n\nFirst and foremost here's a summary of the runs:\n\n$$\\begin{array}{ |l|l|l|l| } \\hline N & \\text{mean} & \\text{std/sqrt(}N\\text{)} & \\text{total number of segments} \\\\ \\hline 10^3 & 8.732 & 0.16106 & 8732 \\\\ 10^4 & 8.781 & 0.0528827 & 87810 \\\\ 10^5 & 8.86218 & 0.0166654 & 886218 \\\\ 10^6 & 8.89447 & 0.00524551 & 8894468 \\\\ 10^7 & 8.88463 & 0.00165934 & 88846265 \\\\ 10^8 & 8.88595 & 0.000524796 & 888595095 \\\\ 10^9 & 8.88615 & 0.000165965 & 8886153545 \\\\ 10^{10} & 8.88631 & - & 88863113226 \\\\ 10^{11} & 8.88614 & - & 888613831649 \\\\ 10^{12} & 8.88614 & - & 8886139852418 \\\\ \\hline \\end{array}$$\n\nWe can apply the central limit theorem to this (sub)sequence of means. Even though $$\\text{std/sqrt(}N\\text{)}$$ was not calculated for $$N = 10^{12}$$, based on prior terms ~$$0.00000524$$ seems like a sufficient estimate.\n\nFrom here I think we can be fairly confident that the first few digits of the mean are $$\\bf{8.8861}$$. Here are the first few standard deviations ($$68-95-99.7$$ rule):\n\n$$m \\pm 1\\sigma \\rightarrow [8.886135, 8.886145] \\\\ m \\pm 2\\sigma \\rightarrow [8.886129, 8.886150] \\\\ m \\pm 3\\sigma \\rightarrow [8.886124, 8.886156]$$\n\nIn fact it takes $$8$$ standard deviations to lose the digit $$1$$:\n\n$$m \\pm 8\\sigma \\rightarrow [8.886098, 8.886181]$$\n\nFinally, based on an inverse symbolic search, we can find that $${\\bf\\frac{1795}{202}} = 8.886138613...$$ falls within $$0.25\\sigma$$ of $$m = 8.886139852418$$ and so maybe this has potential to be the closed form.\n\nFor $$N = 10^9$$, here is the histogram of the number of steps to self-intersection, whose shape is quite similar to the one posted by OP:\n\nIf we plot look at the log plot, we can see the tail follows a decaying exponential quite closely (in red):\n\nFrom here we might think we could use a gamma distribution or something similar to estimate the data, but I was only able to find a tight approximation with one distribution: the inverse Gaussian distribution $$\\operatorname{IG}(8.886, 26)$$. Here's its PDF overlaid on the histogram:\n\nI also recorded which segments intersected each other for $$N = 10^9$$. The data shows segment $$n$$ most often intersects with segment $$n-2$$, followed by $$n-3$$, etc which I think makes sense. Borrowing terminology from MattF's answer, the quick intersections are the most probable.\n\nIn the following diagram an $$(x, y)$$ pair indicates how often segment $$x$$ intersected with segment $$y$$. Each diagonal approximately decays exponentially, all with the same rate.\n\nFor each of the 4 cores I ran the simulations on, I recorded a single sequence of each length. The most interesting ones of course are the long ones.\n\nHere's an instance of length 96. It's riddled with near misses and is ultimately squashed by a quick intersection.\n\nHere the suspicious looking part is indeed intersection free -- it's about $$0.01 \\,\\text{rad}$$:\n\nLastly I looked at the histogram of angles in my saved simulations:\n\nI think this bias (introduced by stopping once an intersection occurs) says that the best way to stay intersection free is to not curl back up towards yourself.\n\nI have compiled all of my data into a Mathematica notebook which can be downloaded here.\n\n\u2022 Great data! I especially like the somewhat surprising angles histogram (but I accept your explanation). \u2013\u00a0Joseph O'Rourke Mar 25 '19 at 19:27\n\u2022 Thanks. Yes I was surprised by that histogram too. Originally I thought there might have been a bug in my code. I reran my code without the intersection stopping condition and the histogram came out uniform. An example of that is in the linked notebook. \u2013\u00a0Chip Hurst Mar 25 '19 at 19:31\n\u2022 And by the way, I made a couple edits since posting -- I don't know if you saw them -- a conjectured closed form and an approximate distribution fit. \u2013\u00a0Chip Hurst Mar 25 '19 at 19:32\n\u2022 $2+11\\sqrt{29/74}$ is even closer to the empirical average. \u2013\u00a0Matt F. Mar 25 '19 at 20:16\n\u2022 The tail behaviour means that the intersection probability tends towards a constant $p$ which you could get from the slope. The behaviour is then modelled reasonably well by $$\\Pr(\\text{length} = k \\mid k > k_0) = \\Pr(\\text{length} > k_0) \\, (1-p)^{k-k_0-1} p$$ once the walk is past $k_0$ steps. \u2013\u00a0student Mar 26 '19 at 14:34\n\nWe can get an upper bound of $$15.52$$ by considering just the cases of interesction of next-to-consecutive steps, which we call quick intersections.\n\nLet $$\\alpha$$ be the angle between step $$AB$$ and step $$BC$$, and let $$u=\\alpha/\\pi$$. So $$u$$ is (initially) uniformly distributed between $$0$$ and $$1$$. Then, as Gerhard Paseman and Bill Bradley determined, the probability that steps $$AB$$ and $$CD$$ intersect is $$P(\\text{intersection|u})= \\begin{cases} \\frac14-\\frac u4\\ \\ \\text{ if } u\\le \\frac13\\\\ \\frac12-u\\ \\ \\text{ if } \\frac13\\le u \\le \\frac12\\\\ \\phantom{1-2}0 \\ \\ \\text{ if } \\frac12 \\le u\\\\ \\end{cases}$$ So the (initial) overall probability of an intersection is $$\\int P(\\text{intersection}|u)du=1/12$$.\n\nThis means that for the next step, we have $$P(u|\\text{no intersection last time})=\\frac{12}{11} \\begin{cases} \\frac34+\\frac u4\\ \\ \\text{ if } u\\le \\frac13\\\\ \\frac12 + u\\, \\ \\ \\text{ if } \\frac13\\le u \\le \\frac12\\\\ \\phantom{1-2}1 \\ \\ \\text{ if } \\frac12 \\le u\\\\ \\end{cases}$$ \\begin{align} P(\\text{intersection|no intersection last time}) &=\\frac{12}{11}\\int_{u=0}^1 \\begin{cases} (\\frac34+\\frac u4)(\\frac14-\\frac u4)\\ \\ \\text{ if } u\\le \\frac13\\\\ (\\frac12+u)(\\frac12-u)\\ \\ \\, \\text{ if } \\frac13\\le u \\le \\frac12\\\\ \\ \\ \\phantom{1-2}1(0)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{ if } \\frac12 \\le u \\end{cases} du\\\\ &=\\frac{29}{396} \\end{align} So the probability of no quick intersection after $$n$$ steps is $$\\frac{11}{12}\\left(\\frac{367}{396}\\right)^{n-3}$$ and the expected number of steps until a quick intersection is $$3\\left(\\frac{1}{12}\\right)+\\sum_{n=4}^\\infty n\\left( \\frac{11}{12}\\left(\\frac{367}{396}\\right)^{n-4}\\! - \\frac{11}{12}\\left(\\frac{367}{396}\\right)^{n-3} \\right)=\\frac{450}{29} \\simeq 15.52$$\n\n\u2022 Nice analysis, MattF! \u2013\u00a0Joseph O'Rourke Mar 23 '19 at 0:11\n\nInspired by Bill Bradley's post, I've decided to lower bound the probability of an intersection which (using the method suggested in Bill's post) gives a slightly less weak upper bound for the expected length of a random walk.\n\nUsing Bill's picture and rescaling so that angles are measured as nonnegative fractions of 2pi, we see one domain of self intersection is when $$\\alpha \\lt \\beta$$ and $$\\alpha + 2\\beta \\lt 1/2$$. This region is a triangle of base 1/4, and height 1/6, with an area of 1/48. Missing the measure 0 set where $$\\alpha=\\beta$$, we reflect across that line, and then once more across the line $$\\alpha+\\beta=1$$ to get the other three regions having an intersection, giving the probability of self intersection of three-step paths to be 4/48=1/12.\n\nNow let's extend this to paths with four steps. The number of cases to determine the probability exactly is more than I care to tackle, so I will cheat and underestimate it. I do this by considering just two cases: that an intersection occurs in the first three steps, or that it does not but occurs in the last three steps.\n\nIf we considered these cases as disjoint, we would get a probability of 1/12 + 1/12 = 1/6. They aren't, so we will consider the case that $$\\beta \\leq 1/4$$ and $$\\alpha \\gt 1/4$$ or more visibly use 3/4 of the chance that the last three steps self-intersect. This gives a lower bound of (1/12) times (1+3/4), or 7/48, which gives 1/7 as a weak lower bound.\n\nUsing this and Bill's estimating method gives 1 + 7*(4-1)=22 as a weak upper bound on the number of steps needed to self-intersect. If we could improve the lower bound from 1/7 to 1/6 for self intersection of a four step path, this would lead to an upper bound of 19 instead of 22.\n\nBut if we can cheat once, we can cheat again. The first cheat went from two neighboring angles to three, so we will go from three to five and lower bound the crossing probability of six segments as (7/4) times the (lower bound of the) probability of four segments. We share the middle angle so that we can nicely separate the domains of the middle angle turning sharp left or sharp right. We do a similar cut and trim and end up with a lower bound of 49/192 =1/4+1/192 for the probability of 6 segments crossing. This gives a slightly better upper bound of 1 +4*5 or 21 for the step length of a walk that does not cross itself.\n\nTo justify the cheat, one has to do a lot more work to make sure that the probability grows by 7/4 at this step. However, I won't do that. I use the second cheat instead to motivate the idea that 22 is an upper bound that can be improved with a simple argument, and hope to inspire someone to come up with a simple and more honest and precise estimate.\n\nGerhard \"Long As It's Not Graded\" Paseman, 2019.03.20.\n\n\u2022 I would like to point out that many of my signatures encode some double meaning. In this particular case, I want to make it clear that this analysis fails on other manifolds. In particular, Joseph's problem makes sense on a torus or pseudo sphere, even on non-oriented manifolds or graphs with a rigid geometric structure. I encourage people to consider the version for a torus. Gerhard \"Promises To Not Grade Work\" Paseman, 2019.03.20. \u2013\u00a0Gerhard Paseman Mar 20 '19 at 17:21\n\u2022 The version of Joseph's question on a compact space is likely in the graph theory literature. Consider a cover of the space by uniform mostly disjoint regions, and make a graph with a region a vertex and a line joining two regions if a step takes one from one region to another. One can now approximate a random walk by a transition matrix, or use the theory of random walks in graphs to compute a desired expectation. Thus the actual expectation can be considered as a limit of expectations using various partitions of the space. Gerhard \"Still Interested In Torus Version\" Paseman, 2019.03.20. \u2013\u00a0Gerhard Paseman Mar 20 '19 at 18:20\n\n(Thanks for noticing my earlier error, Gerhard!)\n\nHere's a loose but very simple upper bound. Consider three consecutive unit steps, which might look like this:\n\nWe take $$\\alpha\\in[0, \\pi]$$ and $$\\beta\\in[0,2\\pi)$$. (By symmetry we can reflect $$\\alpha\\in (\\pi, 2\\pi)$$ back to $$\\pi-\\alpha$$.) Then to overlap, we need $$\\alpha < \\pi/2$$, $$\\beta<\\pi/2$$ and $$\\alpha + \\beta < \\pi$$ The first two events are independent, with probabilities 1/2 and 1/4.\n\nFor the third event, consider a case where the three segments just barely cross, e.g.:\n\nIf we decrease the angles $$\\alpha, \\beta$$, the steps will still intersect, so it's sufficient to consider this case.\n\nBecause the bottom and left edges are both length one, the upper and rightmost angles are both $$\\beta$$. Since we have a triangle, $$\\alpha+2\\beta=\\pi$$, so the probability of this event is 2/3. (There's a symmetric case where the roles of $$\\alpha$$ and $$\\beta$$ are reversed.) It's possibly clearer to see this by examining a phase space diagram (\"blue\" = \"line segments intersect\"):\n\nThe blue region takes up 2/3rds of the (uniformly sampled) space. Conditioning on everything, the probability of overlap is 1/2 * 1/4 * 2/3 = 1/12\n\nWe can apply this argument to multiple non-overlapping pairs of angles. Therefore, the expected number of steps until self-intersection is upper-bounded by $$\\leq 1 + 2\\times 12 = 25$$\n\n\u2022 Something does not look right. Do you want to find it, or shall I expand on my comments above? Gerhard \"Pretty Sure About One Twelfth\" Paseman, 2019.03.20. \u2013\u00a0Gerhard Paseman Mar 20 '19 at 15:49\n\u2022 Looks good now. I like the picture and the estimate. Gerhard \"Blue Is A Nice Color\" Paseman, 2019.03.21. \u2013\u00a0Gerhard Paseman Mar 21 '19 at 23:55", "date": "2021-03-05 17:22:23", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/325052/how-many-random-walk-steps-until-the-path-self-intersects", "openwebmath_score": 0.8837517499923706, "openwebmath_perplexity": 486.9452189925769, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211612253743, "lm_q2_score": 0.8740772286044095, "lm_q1q2_score": 0.8520689789888074}}
{"url": "https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power", "text": "# How do I take a fraction to a negative power?\n\nI ran into this issue during my homework. Using the rules of logarithms, I need to prove that $$-2\\ln\\bigg(\\frac{2}{\\sqrt{6}}\\bigg)=\\ln3-\\ln2$$ So here were my steps:\n\n1. First step: $$-2\\ln\\bigg(\\frac{2}{\\sqrt{6}}\\bigg)=\\ln\\left(\\bigg(\\frac{2}{\\sqrt{6}}\\bigg)^{-2}\\right)$$ And that's as far as I got, because now I want to use the form $$\\ln(a/b) = \\ln(a) - \\ln(b)$$, but first I need to reduce the fraction because it is raised to the $$-2$$.\n\nHow do I evaluate $$\\bigg(\\frac{2}{\\sqrt{6}}\\bigg)^{-2}$$ ?\n\nThanks\n\n\u2022 Note that $x^{-a} = \\frac{1}{x^a}$ \u2013\u00a0desiigner Jul 11 '19 at 19:53\n\nBy definition $$a^{-k} = \\frac 1{a^k}$$\n\nSo $$\\left(\\frac{2}{\\sqrt{6}}\\right)^{-2} =\\frac 1{\\left(\\frac{2}{\\sqrt{6}}\\right)^{2}}=$$\n\n$$\\frac 1{\\left(\\frac {2^2}{\\sqrt 6^2}\\right)}=\\frac {\\sqrt 6^2}{2^2}=\\frac 64=\\frac 32$$\n\nIt will help to realize that $$(\\frac ab)^{-1} = 1/(a/b) = \\frac ba$$ and that $$(\\frac ab)^k = \\frac {a^k}{b^k}$$ to realize that that means $$\\left(\\frac ab\\right)^{-k} = \\frac 1{\\left(\\frac ab\\right)^k}= \\frac 1{\\left(\\frac {a^k}{b^k}\\right)} = \\frac {b^k}{a^k}.$$\n\n(Also $$(\\frac ab)^{-k} = [(\\frac ab)^{-1}]^k = (\\frac ba)^k=\\frac {b^k}{a^k}$$ or that $$(\\frac ab)^{-k} = \\frac {a^{-k}}{b^{-k}} = (1/a^k)/(1/b^k) = \\frac {b^k}{a^k}$$.)\n\nIn any event\n\n$$\\left(\\frac {2}{\\sqrt 6}\\right)^{-2} = \\left(\\frac {\\sqrt 6}{ 2}\\right)^2 = \\frac {\\sqrt 6^2}{2^2} = \\frac 64 = \\frac 32.$$\n\n\u2022 Perfect, this is the explanation I was looking for. So now after a few steps, I'll get ln(3/2) = ln(3) - ln(2). \u2013\u00a0Miguel Aragon Jul 11 '19 at 21:36\n\n\\begin{align} -2\\ln \\left( \\frac{2}{\\sqrt 6} \\right) &= -2\\big( \\ln(2)-\\ln(\\sqrt{6}) \\big) \\\\ &= -2\\ln(2)+2\\ln(6^{1/2}) \\\\ &= -2\\ln(2)+\\ln(2\\cdot 3) \\\\ &=-2\\ln(2)+ \\big( \\ln(2)+\\ln(3)\\big) \\\\ &=\\ln(3)-\\ln(2) \\end{align} And for your specific question, remember that $$\\left( \\frac{a}{b} \\right)^{-n}=\\left( \\frac{b}{a} \\right)^n$$\n\nWell you may start by distributing the index since $$2$$ and $$\\sqrt 6$$ are positive. Thus $$\\left(\\frac{2}{\\sqrt{6}}\\right)^{-2}=\\frac{2^{-2}}{{\\sqrt 6}^{-2}}.$$\n\nThen recall that for any nonzero number $$a$$ and any negative integer $$-n,$$ we have $$a^{-n}=\\frac {1}{a^n}.$$ Applying this to your expression, we have $$\\frac{2^{-2}}{{\\sqrt 6}^{-2}}=\\frac{\\frac {1}{2^2}}{\\frac{1}{{\\sqrt 6}^2}}=\\frac{\\frac {1}{4}}{\\frac{1}{6}}=\\frac{6}{4}=\\frac 32.$$\n\n$$-2 \\ln(\\frac{2}{\\sqrt{6}}) = \\ln(\\frac{2}{\\sqrt{6}})^{-2} = \\ln \\frac{1}{(\\frac{2}{\\sqrt{6}})^{2}} = \\ln \\frac{6}{4} = \\ln\\frac{3}{2} = \\ln 3 - \\ln 2$$\n\n$$-2 \\ln \\left( \\frac{2}{\\sqrt{6}} \\right) = \\ln 3 - \\ln 2$$\n\nif and only if\n\n$$\\ln \\left( \\frac{2}{\\sqrt{6}} \\right)^{-2} = \\ln \\left( \\frac{3}{2}\\right)$$\n\nif and only if\n\n$$\\ln \\left[ \\frac{1}{\\left(\\frac{2}{\\sqrt{6}}\\right)^{2}} \\right] = \\ln \\left( \\frac{3}{2}\\right)$$\n\nAnd so, we have\n\n$$\\ln \\left( \\frac{6}{4} \\right) = \\ln \\left( \\frac{3}{2}\\right)$$\n\nwhich is true.\n\n\u2022 you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse \u2013\u00a0peek-a-boo Jul 11 '19 at 20:27\n\u2022 One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to include them initially. \u2013\u00a0mlchristians Jul 11 '19 at 20:29\n\u2022 $4 = 10 \\implies 7-3 =7+3 \\implies-3= 3 \\implies (-3)^2 = 3^2 \\implies 9=9$ which is true. Therefore $4 = 10$. Or if penguins were elephants then they would both eat food. Which is true. So penguins are elephants. \u2013\u00a0fleablood Jul 11 '19 at 20:37\n\u2022 \"One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained.\" But if you do that you MUST specify the steps are actually if and only if steps. If you do not SPECIFICALLY state that it is reasonable to assume they are only forward implications and the proof is not valid. \u2013\u00a0fleablood Jul 11 '19 at 20:38\n\u2022 Your first comment is not if and only if; and neither is you second. Forward implications is one-way; necessary and sufficient (i.e., if and only if are both ways). In the case of this problem, start from the given equation and work it to where I finished. Any problem with that. Now, if you want to do the work, start w/ the last statement and by a series of implications, show it implies the initial equation. I chose iff and only if instead. \u2013\u00a0mlchristians Jul 11 '19 at 20:45\n\nFor a non-zero real number, we have $$1=x^0=x^{2-2}=x^2\\cdot x^{-2}\\tag1$$ from How to understand why $$x^0=1$$ where $$x$$ is any real number?\n\nPutting $$x=2/\\sqrt6$$ into $$(1)$$, we get $$1=\\left(\\frac2{\\sqrt6}\\right)^2\\cdot\\left(\\frac2{\\sqrt6}\\right)^{-2}\\tag2$$ and since we know that for a positive integer $$n$$ and positive real numbers $$a,b$$, $$\\left(\\frac ab\\right)^n=\\underbrace{\\frac ab\\cdot \\frac ab\\cdot\\ldots\\cdot\\frac ab}_{n\\,\\text{times}}=\\underbrace{\\frac{a\\cdot a\\cdot\\ldots\\cdot a}{b\\cdot b\\cdot\\ldots\\cdot b}}_{n\\,\\text{times}}=\\frac{a^n}{b^n},\\tag3$$ we have that $$\\left(\\frac2{\\sqrt6}\\right)^2=\\frac{2^2}{\\left(\\sqrt6\\right)^2}=\\frac46=\\frac23\\tag4.$$ Finally, putting $$(4)$$ back into $$(2)$$ yields $$1=\\frac23\\cdot\\left(\\frac2{\\sqrt6}\\right)^{-2}\\implies\\left(\\frac2{\\sqrt6}\\right)^{-2}=\\frac1{2/3}=\\frac32\\tag5$$ as the answer.", "date": "2020-01-25 02:30:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3290293/how-do-i-take-a-fraction-to-a-negative-power", "openwebmath_score": 0.9951451420783997, "openwebmath_perplexity": 289.97974398157544, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126531738088, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8520645706247889}}
{"url": "https://math.stackexchange.com/questions/3843961/any-shortcuts-to-proving-that-frac-sinx2-sin2-frac-x2-tan-frac-x2/3843975", "text": "# Any \u201cshortcuts\u201d to proving that $\\frac{\\sin(x)}2+\\sin^2(\\frac x2)\\tan(\\frac x2)\\to\\tan(\\frac x2)$\n\nI was working on simplifying some trig functions, and after a while of playing with them I simplified $$\\frac{\\sin(x)}{2}+\\sin^2\\left(\\frac{x}{2}\\right)\\tan\\left(\\frac{x}{2}\\right) \\rightarrow \\tan\\left(\\frac{x}{2}\\right)$$\n\nThe way I got that result, however, was with what I think a very \"roundabout\" way. I first used the half-angle formulaes, then used $$x=\\pi/2-\\beta$$, and that simplified to $$\\frac{\\cos(\\beta)}{1+\\sin(\\beta)}$$ where I again used the coordinate change to get $$\\frac{\\sin(x)}{1+\\cos(x)}\\rightarrow\\tan\\left(\\frac{x}{2}\\right)$$\n\nI tried using the online trig simplifiers but none succeeded. Of course, after you know the above identity, it's easy to prove by proving that $$\\frac{\\sin(x)}{2}=\\tan\\left(\\frac{x}{2}\\right)-\\sin^2\\left(\\frac{x}{2}\\right)\\tan\\left(\\frac{x}{2}\\right)$$\n\nIs there a more direct way to get the identity? I guess what I'm asking is, am I missing any \"tricks\" or software that I could have on my toolbelt so that next time I don't spend hours trying to simplify trig identities?\n\nWith $$s:=\\sin\\frac x2,c:=\\cos\\frac x2$$, $$\\sin x=2sc$$ and\n\n$$\\frac122sc+s^2\\frac sc=\\frac{sc^2+s^3}c=\\frac sc.$$\n\n\u2022 I accepted this answer since it makes all steps explicit \u2013\u00a0Esteban Sep 30 '20 at 15:03\n\u2022 @Esteban: thank you for the explanation. \u2013\u00a0Yves Daoust Sep 30 '20 at 15:20\n\nUse $$\\sin(x) = 2 \\sin(x/2) \\cos(x/2)$$ then we have $$\\begin{eqnarray*} \\frac{\\sin(x)}{2}+\\sin^2\\left(\\frac{x}{2}\\right)\\tan\\left(\\frac{x}{2}\\right) &=& \\frac{\\sin(x/2)}{\\cos(x/2)} \\underbrace{\\left( \\cos^2(x/2) + \\sin^2(x/2) \\right)}_{=1} \\\\ &=& \\tan (x/2). \\end{eqnarray*}$$\n\n\u2022 I actually tried this route but got stuck, unfortunately. So I think it's worth commenting that the \"trick\" here to get to your right hand expression is to multiply the 2sin(x/2)cos(x/2)/2 term by cos(x/2)/cos(x/2). \u2013\u00a0Esteban Sep 30 '20 at 14:51\n\nFairly obvious: $$\\sin^2\\left(\\frac{x}{2}\\right)\\tan\\left(\\frac{x}{2}\\right)=(1-\\cos^2\\left(\\frac{x}{2}\\right))\\tan\\left(\\frac{x}{2}\\right)=\\tan\\left(\\frac{x}{2}\\right)-\\cos\\left(\\frac{x}{2}\\right)\\sin\\left(\\frac{x}{2}\\right)=\\tan\\left(\\frac{x}{2}\\right)-\\frac{\\sin x}{2}$$\n\n\u2022 All the tried trig simplifiers beg to differ it was fairly obvious, unfortunately. I like the route you took. \u2013\u00a0Esteban Sep 30 '20 at 14:54\n\n$$\\frac{\\sin x}{2} = \\sin\\frac{x}{2}\\cos\\frac{x}{2} = \\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}} \\cos^2\\frac{x}{2} = \\tan\\frac{x}{2} \\left(1 - \\sin^2\\frac{x}{2}\\right)$$", "date": "2021-03-07 15:13:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3843961/any-shortcuts-to-proving-that-frac-sinx2-sin2-frac-x2-tan-frac-x2/3843975", "openwebmath_score": 0.9426677823066711, "openwebmath_perplexity": 379.3691216335252, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126500692716, "lm_q2_score": 0.8705972734445508, "lm_q1q2_score": 0.8520645646359986}}
{"url": "https://tutel.me/c/mathematics/questions/118686/determining+precisely+where+sum_n1inftyfracznn+converges", "text": "#### [SOLVED] Determining precisely where $\\sum_{n=1}^\\infty\\frac{z^n}{n}$ converges?\n\nInspired by the exponential series, I'm curious about where exactly the series $\\displaystyle\\sum_{n=1}^\\infty\\frac{z^n}{n}$ for $z\\in\\mathbb{C}$ converges.\n\nI calculated $$\\limsup_{n\\to\\infty}\\sqrt[n]{\\frac{1}{n}}=\\limsup_{n\\to\\infty}\\frac{1}{n^{1/n}}$$ and $$\\lim_{n\\to\\infty} n^{1/n}=e^{\\lim_{n\\to\\infty}\\log(n)/n}=e^{\\lim_{n\\to\\infty}1/n}=e^0=1.$$ So the radius of convergence is $1$, so the series converges on all $z$ inside $S^1$. But is there a way to tell for which $z$ on the unit circle the series converges? I know it converges for $z=-1$, but diverges for $z=1$, but I don't know about the rest of the circle. For what other $z$ does this series converge? Thanks.\n\n#### @Pookaros 2017-05-10 23:18:42\n\nEven though the 1st answer is the more easy and appropriate here is a more analytic one. $\\sum_{n=1}^{k} (z^n)/n =(z/1-z) $$\\sum_{n=1}^{k-1} [(1-z^n)/n(n+1) + (1-z^k)/k] (we can prove that with induction till k). We can also see that 0 \\leqslant |(1-z^k)/k|\\leqslant 2/k\\rightarrow0 for k\\rightarrow$$\\infty$ also\n\n0 $\\leqslant$ |(1-z^n)/n(n+1)|$\\leqslant$ 2/n(n+1)\n\n$\\sum_{n=1}^{\\infty}2/n(n+1)$ $(converges)$ so $\\sum_{n=1}^{\\infty} (z^n)/n$ $(converges)$ as well\n\n#### @Martin Argerami 2012-03-11 00:56:15\n\nFix $z$ in the unit circle, i.e. $|z|=1$. We want to apply Dirichlet's test: if $\\{a_n\\}$ are real numbers and $\\{b_n\\}$ complex numbers such that:\n\n1. $a_1\\geq a_2\\geq\\cdots$\n2. $\\lim_{n\\to\\infty}a_n=0$\n3. There exists $M>0$ such that $\\left|\\sum_{n=1}^Nb_n\\right|\\leq M$ for every $N\\in\\mathbb{N}$;\n\nthen $\\sum_{n=1}^\\infty a_nb_n$ converges. Here $a_n=1/n$, $b_n=z^n$. The first two conditions are clearly satisfied, and for the third one: $$\\left|\\sum_{n=1}^Nz^{ n }\\right|=\\left|\\frac{z-z^{N+1}}{1-z}\\right|\\leq\\frac{2}{|1-z|}$$ for all $N\\in\\mathbb{N}$. This shows that the third condition is satisfied for every $z\\ne1$ in the circle.\n\nIn conclusion, the series converges for every $z$ with $|z|\\leq1$ other than $z=1$, and it diverges for $|z|>1$.\n\n#### @philmcole 2017-12-06 12:32:40\n\nCan you explain this step: $|\\sum_{n=1}^Nz^{ n }| = |\\frac{z-z^{N+1}}{1-z}|$ ?\n\nHere.\n\n#### @philmcole 2017-12-06 13:48:17\n\nYeah, but where do you get that additional $z$ in the nominator from? By the formula you linked it should be $|\\frac{1-z^{N+1}}{1-z}|$, no?\n\n#### @Martin Argerami 2017-12-06 13:49:24\n\nIf you start the sum from $0$, yes.\n\n#### @Pacciu 2012-03-11 23:23:01\n\nThe followin theorem on power series is due to E. Picard:\n\nLet $(a_n)$ be a sequence of real numbers.\n\nIf the sequence $(a_n)$ is nonnegative, decreases and tends to zero when $n\\to \\infty$, then the complex power series $\\sum a_n\\ z^n$ converges in the closed unit disc $\\overline{D}(0;1)$ with the only possible exception of the point $1$.\n\nThe proof of Picard's theorem relies on Abel's summation by parts formula, as far as I remember.\n\nNow, the coefficients of your series, i.e. $a_n=1/n$, satisfy the assumptions of Picard's theorem, hence your series converges at least in $\\overline{D}(0;1)\\setminus \\{1\\}$; on the other hand, the series diverges when $z=1$ (for it becomes the harmonic series).\n\nTherefore the convergence set of $\\sum 1/n\\ z^n$ is $\\overline{D}(0;1)\\setminus \\{1\\}$.\n\n### Determining precisely where $\\sum_{n=0}^\\infty \\left(\\frac{z^n}{n}\\right)$ converges uniformly\n\n\u2022 2017-09-13 22:36:35\n\u2022 Pookaros\n\u2022 77 View\n\u2022 0 Score\n\u2022 Tags: \u00a0 complex-analysis\n\n### [SOLVED] To prove the statements related to power series\n\n\u2022 2017-01-27 02:53:47\n\u2022 Kavita\n\u2022 653 View\n\u2022 1 Score\n\u2022 Tags: \u00a0 complex-analysis", "date": "2019-03-26 10:01:28", "meta": {"domain": "tutel.me", "url": "https://tutel.me/c/mathematics/questions/118686/determining+precisely+where+sum_n1inftyfracznn+converges", "openwebmath_score": 0.9750022292137146, "openwebmath_perplexity": 522.6435572542091, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126475856414, "lm_q2_score": 0.870597271765821, "lm_q1q2_score": 0.8520645608307629}}
{"url": "http://math.stackexchange.com/questions/362140/probability-taking-cups-from-a-box", "text": "# Probability: Taking Cups From A Box\n\nI just saw the question in an exam paper. My friend and I have different answer and I am not sure which solution is correct, so I hope someone can help me.\n\nThere are 16 cups in a box, 11 are blue and 5 are white. Six cups are taken out from it at the same time. What is the probability of taking at least 4 white cups?\n\nMy solution:\n\n$$P=\\frac{\\binom54\\binom{11}2+\\binom55\\binom{11}1}{\\binom{16}6}$$\n\nMy friend's solution:\n\n$$P=\\frac{5!\\times11\\times6+5!\\times11\\times10\\times_6\\text{P}_2}{_{16}\\text{P}_6}$$\n\nMy friend use a different approach and his answer is different from mine. I wonder if I am correct and why. Please show me the reason, thank you!\n\n-\nYour method is correct. \u2013\u00a0 Andr\u00e9 Nicolas Apr 15 '13 at 8:02\n@Andr\u00e9Nicolas Can you tell me why I am correct? After my friend told me his solution, I am so confused with the question ... \u2013\u00a0 redcap Apr 15 '13 at 8:22\n\nYour method is correct. There are $\\binom{16}{6}$ equally likely ways to pick a collection of $6$ cups. (We can imagine the cups have secret ID numbers written in invisible ink.)\nYour numerator counts the favourables. For example, the are $\\binom{5}{4}$ ways to pick $4$ whites. For each of these ways there are $\\binom{11}{2}$ ways to pick the accompanying blues, for a total of $\\binom{5}{4}\\binom{11}{2}$.\nYour friend's proposed solution can be made to work, though it is perhaps a little trickier to do it correctly. The friend observed that counting order of picking, there are ${}_{16}P_6$ equally likely ways of picking $6$ cups. Now we count the number of favourables. Your friend's first term in the count of favourables is correct, the second is not.\nThere are $\\binom{6}{2}$ ways of picking the locations of the $4$ blue cups. For each of these ways, the first chosen location can be filled in $11$ ways, and the second in $10$ ways. Then the remaining locations can be filled with white cups in ${}_5P_4$ ways. Multiply to get the total with $4$ white and $2$ blue. There are two mistakes in the proposed expression there. First, instead of $\\binom{6}{2}$ there is ${}_6P_2$. Secondly, there is the $4!$ which may be a typo. For the placement of white cups in the $4$ available slots, either use the permutation symbol ${}_5P_4$, or decide the $4$ whites can be chosen in $\\binom{5}{4}$ ways, and then permuted in $4!$ ways. That part turns out to be $(5)4!$, which is the same as $5!$.\nYou are welcome. I had figured the $4!$ was an unimportant error. But that still leaves an overcount in that part by a factor of $2$. That error is somewhat more structural. \u2013\u00a0 Andr\u00e9 Nicolas Apr 15 '13 at 9:30", "date": "2014-04-20 01:12:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/362140/probability-taking-cups-from-a-box", "openwebmath_score": 0.8496515154838562, "openwebmath_perplexity": 100.20600061453757, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9911526438717138, "lm_q2_score": 0.8596637577007393, "lm_q1q2_score": 0.8520580062857801}}
{"url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose", "text": "# Parameterize unitary without transpose\n\nFor all unitary matrices, i.e. $A \\overline{A}^T = I$, there is a skew-Hermitian matrix $X$ so that $A = exp(X)$. So the unitary group has $n^2$ dimensions.\n\nIs there any similar parameterisation of all matrices with $A \\overline{A} = I$? What can be said about the number of dimensions? (Btw. is there any name for this type of matrices in literature?)\n\n\u2022 Note that If $M$ is a real square matrix of the relevant size, then $A = \\exp(iM)$ has the property that $\\overline{A} = \\exp(-iM)$, so that $A\\overline{A} = I$. \u2013\u00a0Geoff Robinson Oct 19 '15 at 22:56\n\u2022 @QiaochuYuan: Right, this collection is not closed under multiplication. Does this already exclude any notion of dimensionality and parameterisation? \u2013\u00a0Sebastian Schlecht Oct 20 '15 at 8:38\n\u2022 @GeoffRobinson: Right, but are these all matrices with this property? \u2013\u00a0Sebastian Schlecht Oct 20 '15 at 9:05\n\u2022 @GeoffRobinson: The answer is yes if we add $A$ invertible, as $\\exp: M(N,\\mathbb{C}) \\rightarrow GL(N,\\mathbb{C})$ is surjective and $\\exp(M) \\overline{\\exp(M)}=I$ follows $M = -\\overline{M}$. \u2013\u00a0Sebastian Schlecht Oct 20 '15 at 9:13\n\u2022 Well,$A$ certainly needs to be invertible in the question. \u2013\u00a0Geoff Robinson Oct 20 '15 at 10:00\n\nThe set $V= \\{ A\\in M_n(\\mathbb{C})\\ |\\ A\\bar A = I\\}$ is a smooth submanifold of $M_n(\\mathbb{C})$ with real dimension $n^2$.\n\nProof: Consider the involution $\\iota:\\mathrm{GL}(n,\\mathbb{C})\\to \\mathrm{GL}(n,\\mathbb{C})$ defined by $$\\iota(A) = (\\bar A)^{-1}.$$ This is an anti-holomorphic involution of the complex manifold $\\mathrm{GL}(n,\\mathbb{C})$, and its fixed locus is precisely $V$. Thus, $V$ is a totally real submanifold of $\\mathrm{GL}(n,\\mathbb{C})$ with (real) dimension $n^2$.\n\nAlso: While it's true that the map $f:M_n(\\mathbb{R})\\to \\mathrm{GL}(n,\\mathbb{C})$ defined by $$f(a) = \\exp(ia)$$ has its image in $V$, it is not a 'parametrization' everywhere, i.e., $f$ is not a local diffeomorphism everywhere. While this is true on a neighborhood of $0\\in M_n(\\mathbb{R})$, at other places, the map $f$ definitely is not a local diffeomorphism. For example, if $a$ has $n$ eigenvalues of the form $2k_i\\pi$ (where $k_1,\\ldots, k_n$ are integers, not all zero), then $f(a) = I$, but the space of such real matrices $a$ has positive dimension for any $n$-tuple $(k_1,\\ldots, k_n)$ for which not all of the $k_i$ are equal.\n\nIt turns out that $f$ is surjective. The proof uses the fact that, for $A\\in V$, we have $A\\bar A = \\bar A A = I$, so, in particular, $A$ and $\\bar A$ commute and hence can be put simultaneously in Jordan normal form by a real conjugation. Then, breaking $\\mathbb{C}^n$ into a sum of complexifications of real subspaces according to the eigenvalues of $A$ and using the semi-simple/nilpotent decomposition appropriately, one can reduce to dealing with the upper triangular case, and the result can then be proved using simple facts about power series in commuting nilpotent variables. Details upon request (see below).\n\nOf course, all of this is probably a special case of known facts about affine symmetric spaces. I have now realized that $V$ is simply the Cartan embedding for the affine symmetric space $\\mathrm{GL}(n,\\mathbb{C})/\\mathrm{GL}(n,\\mathbb{R})$. This Cartan embedding $$\\sigma:\\mathrm{GL}(n,\\mathbb{C})/\\mathrm{GL}(n,\\mathbb{R})\\longrightarrow V\\subset \\mathrm{GL}(n,\\mathbb{C})$$ is given by $\\sigma(B\\cdot\\mathrm{GL}(n,\\mathbb{R})) = B\\,(\\,\\overline B\\,)^{-1}$. The exponential map we have been discussing is just the geodesic mapping of this affine symmetric space, so, probably all of this follows from general theory.\n\nDetails: Let $A\\in\\mathrm{GL}(n,\\mathbb{C})$ satisfy $A\\overline{A} = I$. Let $\\lambda\\in\\mathbb{C}$ be an eigenvalue of $A$ of multiplicity $m\\le n$ and let $$V_\\lambda = \\{ v\\in\\mathbb{C}^n \\ |\\ (A{-}\\lambda)^mv = 0\\ \\}.$$ be the associated generalized eigenspace. Of course, $\\mathbb{C}^n$ is the direct sum of these generalized eigenspaces. Since $\\overline{A}$ commutes with $A$, it preserves each of these subspaces. Moreover, one clearly has $$\\overline{V_\\lambda} = V_{1/\\bar\\lambda},$$ so each of the spaces $V_\\lambda+V_{1/\\bar\\lambda}$ is invariant under conjugation and hence is the complexification of a real subspace $W_\\lambda\\subset \\mathbb{R}^n$. It follows that, by conjugating by a real matrix, we can assume that $A$ (and hence $\\overline A$) is in block diagonal form, so it suffices to treat the case where either $A$ has a single eigenvalue $\\lambda$ satisfying $\\lambda\\bar\\lambda=1$, so that $\\mathbb{C}^n=V_\\lambda$, or else $A$ has an eigenvalue $\\lambda$ satisfying $\\lambda\\bar\\lambda > 1$ and $\\mathbb{C}^n=V_\\lambda\\oplus V_{1/\\bar\\lambda}$.\n\nIn either case, we can assume that $\\lambda = r\\ge 1$ is real, since, if $\\lambda = r e^{i\\theta}$, we can replace $A$ by $e^{-i\\theta}A$ and show that $e^{-i\\theta}A$ is in the image of $f$ (since $I$ commutes with everything and $e^{i\\theta} I = \\exp(i\\theta I)$.)\n\nFirst, consider the case $\\lambda = 1$. Then $A = C + i S$ where $C$ and $S$ are real matrices and $((C-I) + iS)^n = (A-I)^n = 0$. Moreover, $N=C-I$ and $S$ commute since $I = A\\bar A = C^2+S^2 +i(SC-CS) = I + 2N + N^2 + S^2$. Note that $N$ and $S$ are nilpotent commuting matrices. Since they satisfy $$2N + N^2 = -S^2,$$ it follows that $N = p(-S^2)$ where $p(t) = -\\tfrac12 t^2 + \\cdots$ is the (unique) power series (with real coefficients) that satisfies $2p(t) + p(t)^2 = -t^2$. Since $S$ is nilpotent, it follows that $N = p(-S^2)$ expresses $N$ canonically as a polynomial in $S$. Now let $q(t)$ be the (unique) power series with real coefficients that satisfies $\\sin q(t) = t$. Note that we must have $p(-q(t)^2) = \\cos(t)-1$. Now, because $S$ is nilpotent, we have $S = \\sin q(S)$, where $q(S)$ is a real polynomial in $S$. Putting all of this together, we have $$A = I + p(-S^2) + iS = \\cos(q(S)) + i\\sin(q(S)) = \\exp(iq(S)).$$ Finally, before leaving this special case, let us note that, because $A$ satisfies its characteristic polynomial $(A-I)^n = 0$, it follows that $\\overline A = A^{-1}$ is expressed as a universal polynomial in $A$ with real coefficients, so $iS = \\tfrac12(A-\\overline{A})$ is also expressed as a universal polynomial with real coefficients. Since $q(t)$ is an odd power series, it satisfies $iq(it) = f(t)$ where $f$ has real coefficients, so there is actually a formula of the form $A = \\exp( i g_n(A))$ for all $A \\in \\mathrm{GL}(n,\\mathbb{C})$ satisfying $A\\bar A = I$ and $(A-I)^n=0$ for some universal polynomial $g_n(t)$ with real coefficients that also has the property that $g_n(A)$ is a real $n$-by-$n$ matrix for all such $A$. (This remark will be used below.)\n\nNow, finally, let us assume that $A$ has eigenvalues $\\lambda = e^t$ and $1/\\lambda = e^{-t}$ for some $t>0$. Then $V_\\lambda$ and $\\overline{V_\\lambda} = V_{1/\\lambda}$ are disjoint, complementary complex subspaces of $\\mathbb{C}^n$ and so we must have that there exists a matrix $Q\\in M_n(\\mathbb{R})$ such that $$V_\\lambda = \\{ v + iQv \\ | \\ v\\in \\mathbb{R}^n \\}.$$ Because $V_\\lambda$ is closed under multiplication by $i$, it follows that $Q^2 = -I$. Setting $$S = \\cosh t + i\\sinh t\\,Q = \\exp(i t Q),$$ we see that $Sw= e^t w$ for all $w \\in V_\\lambda$ and $Sw = e^{-t}w$ for all $w \\in V_{1/\\lambda}$, so $S$ is the semi-simple part of $A$. Hence $S$ can be written as $S = s_\\lambda(A)$, where $s_\\lambda(t)$ a polynomial in $t$ with real coefficients (that depend on $\\lambda$). Thus, $\\bar S = s_\\lambda(\\bar A)$ can be written as a universal polynomial in $A$, implying that $Q$ itself can be written as a universal polynomial in $A$ and hence, in particular, it commutes with $A$ and $\\bar A$. Now, writing $$A = \\exp(i t Q) B,$$ we see that $B\\bar B = I$ and that $B$ can be written as a polynomial in $A$. Moreover, the eigenvalues of $B$ are now all equal to $1$, so $(B-I)^n=0$, and so $B = \\exp(i g_n(B))$ (as per above), where $g_n(B)$ is a polynomial in $A$, which, therefore, commutes with $Q$ (which is a polynomial in $A$). Finally, we have $$A = \\exp(i t Q)\\exp(i g_n(B)) = \\exp(i (t Q+g_n(B))),$$ as desired.\n\n\u2022 Thank you very much for your insightful answer. The only main missing part for me is the surjectivity of $f$. \u2013\u00a0Sebastian Schlecht Oct 20 '15 at 16:43\n\u2022 Robert, I learn so much new material from your answers! Just great! \u2013\u00a0Suvrit Oct 21 '15 at 13:24\n\u2022 Robert, great answer. So a bit about the details, for simplicity assuming $A$ is diagonalisable. Commutativity gives as $A = U^{-1} \\Lambda U$ for real $U$. $A$ is invertible, hence there is $\\exp(i L) = \\Lambda$. So, $A = \\exp( i U^{-1} L U)$. Why is $U^{-1} L U$ real? \u2013\u00a0Sebastian Schlecht Oct 22 '15 at 20:05\n\u2022 If $A$ is diagonal, then $A\\bar A = I$ implies that all of the eigenvalues of $A$ are complex numbers of unit modulus, so each of them is of the form $e^{ir}$ where $r$ is real. \u2013\u00a0Robert Bryant Oct 22 '15 at 20:40\n\u2022 I'm confused, I only wanted $A$ diagonalisable and not $A$ diagonal. \u2013\u00a0Sebastian Schlecht Oct 22 '15 at 21:24\n\nQiaochu Yuan is a little hard about our variety $V$; Kummer showed that an element of $V$ is in the form $U\\bar{U}^{-1}$ (a generalization of this result is the Hilbert's theorem 90). Close to the subject, cf. \"Ikramov, On the matrix equation $X\\bar{X}=A$\" http://link.springer.com/article/10.1134%2FS1064562409010153#page-1\n\nRobert wrote a nice answer; we can give more elementary proofs as follows. $\\mathbb{C}$ is seen as $\\mathbb{R}^2$.\n\nProposition 1. $dim(V)=n^2$ when $V$ is considered as a real algebraic set.\n\nProof. According to Kummer, if $R\\bar{R}=I$, then there is $T\\in GL_n(\\mathbb{C})$ s.t. $R=T\\overline{T}^{-1}$. Let $f:T\\in GL_n\\rightarrow T\\overline{T}^{-1}$; note that $f$ is a pseudo-parametrization of our set; in fact $f$ is a submersion. It remains to calculate the rank of $Df_T:H\\in M_n\\rightarrow (H-T\\overline{T}^{-1}\\overline{H})\\overline{T}^{-1}$. Note that $rank(Df_T)=rank(H\\rightarrow T^{-1}H-\\overline{T}^{-1}\\overline{H})$.\n\nLet $T^{-1}=U+iV,H=X+iY$, where $U,V,X,Y$ are real matrices. Then $rank(Df_A)=rank(g:(X,Y)\\in \\mathbb{R}^{2n^2}\\rightarrow UY+VX\\in \\mathbb{R}^{n^2}$; we show that $g$ is onto. Let $A\\in M_n$; since $U+iV$ is invertible, there is $\\lambda\\in \\mathbb{R}$ s.t. $U+\\lambda V$ is invertible. Finally $g(\\lambda(U+\\lambda V)^{-1}A,(U+\\lambda V)^{-1}A)=A$ and we are done.\n\nProposition 2. Any $R\\in V$ can be written as $exp(iA)$ where $A$ is a real matrix.\n\nProof. Let $R=U+iV$, where $U,V$ are real. $R\\bar{R}=I$ is equivalent to $U^2+V^2=I,VU=UV$; that is equivalent to: there is a real matrix $A$ s.t. $U=\\cos(A),V,=\\sin(A)$ and $R=\\exp(iA)$.\n\nEDIT. (cf. Sebastian's comment). $e^Me^{\\bar{M}}=I$ does not imply that $e^{M+\\bar{M}}=I$.\n\nProof. Take $M=\\begin{pmatrix}i\\pi&1\\\\0&-i\\pi\\end{pmatrix}$.\n\n\u2022 That is very sleek! Do you need Prop1 for Prop2? Do you get Prop1 also from Prop2? \u2013\u00a0Sebastian Schlecht Oct 29 '15 at 11:59\n\u2022 And do you have any suggestion what to name $V$? I understand that there might be no agreed name. Unfortunately, I cannot access right now the Ikramov paper to have a look. \u2013\u00a0Sebastian Schlecht Oct 29 '15 at 12:01\n\u2022 I would like to see the details for your Proposition 2. I don't believe that it is as easy as you seem to think. I thought of exactly this line, right up to your 'that is equivalent to there is a real matrix A...\". I didn't see how to prove that directly, and I would be curious to see what method you use to prove it. Also, the smoothness of $V$ follows, as I wrote, easily from the fact that it is the fixed point set of an anti-holomorphic involution. I don't see why it needs anything from Kummer, unless you are quoting the surjectivity of the Cartan embedding, a nontrivial fact. \u2013\u00a0Robert Bryant Oct 29 '15 at 14:11\n\u2022 @ Sebastian SchlechtClearly , Prop. 1 and 2 can be independently proved. About a name for $V$, I don't know. In general, the authors consider elements of $V$ and not the whole set. cf also Horn and Johnson; Matrix Analysis, Section 4.6 \u2013\u00a0loup blanc Oct 29 '15 at 14:59\n\u2022 @loupblanc: I suspected as much. So it's somewhat misleading to claim that your proof is 'more elementary'; it just leaves out many details and even essential ideas. In fact, while I mention Cartan's results, I don't use them (or any result of Kummer) in the surjectivity argument, which, in fact, uses nothing more than than the semi-simple/nilpotent decomposition (i.e., Jordan normal form), and even the result over $\\mathbb{C}$ needs this. \u2013\u00a0Robert Bryant Oct 29 '15 at 15:52\n\nI totally lied about this not being a natural thing to ask! As loup blanc alludes to, in fact $n \\times n$ matrices such that $M^{-1} = \\overline{M}$ can be interpreted as Galois descent data for descending the complex vector space $\\mathbb{C}^n$ to a real vector space, or equivalently as a $1$-cocycle in\n\n$$Z^1(B \\text{Gal}(\\mathbb{C}/\\mathbb{R}), GL_n(\\mathbb{C})).$$\n\nThe statement that any such matrix must have the form $U \\overline{U}^{-1}$ says that any such $1$-cocycle is cohomologous to zero, or equivalently that the Galois cohomology set\n\n$$H^1(B \\text{Gal}(\\mathbb{C}/\\mathbb{R}), GL_n(\\mathbb{C}))$$\n\nis trivial. This just says that there is only one real form of $\\mathbb{C}^n$ up to isomorphism, namely $\\mathbb{R}^n$. The space $GL_n(\\mathbb{C})/GL_n(\\mathbb{R})$ appearing in Robert Bryant's answer can then be interpreted as the space of real forms of $\\mathbb{C}^n$.", "date": "2019-10-14 21:49:12", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose", "openwebmath_score": 0.9622570872306824, "openwebmath_perplexity": 125.47789443929702, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.991152643087697, "lm_q2_score": 0.8596637577007394, "lm_q1q2_score": 0.8520580056117895}}
{"url": "https://mathhelpboards.com/threads/divisibility-challenge.8074/", "text": "# Divisibility Challenge\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nGiven unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.\n\n##### Well-known member\nLet us put $f(x) = (x-b)^5 + (b-c)^5 + (c-x)^5$\nPutting x = b we get f(b) = 0\nso (x-b) is a factor\nso a-b is a factor of $(a-b)^5 + (b-c)^5 + (c-a)^5$\n\nsimilarly we have (b-c) and (c-a) are factors\n\nnow $(a-b)^5 = a^5 \u2013 5a b^4 + 10a^2b^3 \u2013 10 a^3 b^2 +5 a^4 b \u2013 b^5 = a^5 \u2013 b^5 + 5m$ where $m = - a b^4 + 2a^2b^3 \u2013 2 a^3 b^2 + a^4 b b^4$\nsimilarly $(b-c)^5 = b^5 \u2013 c^5 + 5n$\n$(c-a)^5 = c^5 \u2013 a^5 + 5k$\nAdding we get $(a-b)^5 + (b-c)^5 + (c-a)^5 = 5 (m+n+k)$\n\nSo $(a-b)^5 + (b-c)^5 + (c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nLet us put $f(x) = (x-b)^5 + (b-c)^5 + (c-x)^5$\nPutting x = b we get f(b) = 0\nso (x-b) is a factor\nso a-b is a factor of $(a-b)^5 + (b-c)^5 + (c-a)^5$\n\nsimilarly we have (b-c) and (c-a) are factors\n\nnow $(a-b)^5 = a^5 \u2013 5a b^4 + 10a^2b^3 \u2013 10 a^3 b^2 +5 a^4 b \u2013 b^5 = a^5 \u2013 b^5 + 5m$ where $m = - a b^4 + 2a^2b^3 \u2013 2 a^3 b^2 + a^4 b b^4$\nsimilarly $(b-c)^5 = b^5 \u2013 c^5 + 5n$\n$(c-a)^5 = c^5 \u2013 a^5 + 5k$\nAdding we get $(a-b)^5 + (b-c)^5 + (c-a)^5 = 5 (m+n+k)$\n\nSo $(a-b)^5 + (b-c)^5 + (c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$\n\nI like your solution because of the way you introduced a polynomial function for the problem and well done!\n\n#### caffeinemachine\n\n##### Well-known member\nMHB Math Scholar\nGiven unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.\nLet $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.\n\nFrom this thread http://mathhelpboards.com/challenge...^5-5=-^3-b^3-c^3-3-*-^2-b^2-c^2-2-a-8276.html we know that\n\n$$\\frac{S_5}{5}=\\frac{S_3}{3}\\frac{S_2}{2}$$.\n\nNote that since $(a-b)+(b-c)+(c-a)=0$, we have $S_3=3(a-b)(b-c)(c-a)$.\n\nThus $$S_5=5(a-b)(b-c)(c-a)\\frac{S_2}{2}$$.\n\nClearly $2$ divides $S_2$ and thus we are done.\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nLet $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.\n\nFrom this thread http://mathhelpboards.com/challenge...^5-5=-^3-b^3-c^3-3-*-^2-b^2-c^2-2-a-8276.html we know that\n\n$$\\frac{S_5}{5}=\\frac{S_3}{3}\\frac{S_2}{2}$$.\n\nNote that since $(a-b)+(b-c)+(c-a)=0$, we have $S_3=3(a-b)(b-c)(c-a)$.\n\nThus $$S_5=5(a-b)(b-c)(c-a)\\frac{S_2}{2}$$.\n\nClearly $2$ divides $S_2$ and thus we are done.\nHey caffeinemachine,\n\nThanks for participating and that's another trick for me to learn today!", "date": "2021-08-05 23:55:14", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/divisibility-challenge.8074/", "openwebmath_score": 0.9143141508102417, "openwebmath_perplexity": 2549.2735058403036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526446557306, "lm_q2_score": 0.8596637541053281, "lm_q1q2_score": 0.8520580033961697}}
{"url": "https://math.stackexchange.com/questions/2561353/set-of-functions-from-empty-set-to-0-1?noredirect=1", "text": "# Set of functions from empty set to $\\{0,1\\}$\n\nHow does the set of all functions $\\{f \\,|\\, f: \\emptyset \\to \\{0,1\\}\\}$ look like? Is it empty or does it contain infinitely many functions? Does the definition $f: \\emptyset \\to \\{0,1\\}$ make sense at all?\n\nI was wondering because we know that the two sets $\\{0,1\\}^X$ and $\\mathcal{P}(X)$ have the same cardinality. But this is only true if $X$ is non-empty, right?\n\n\u2022 What have you tried so far? Have you tried considering the definition of \"function\"? \u2013\u00a0skyking Dec 11 '17 at 10:03\n\nYes, it makes sense. There is one and only one function from $\\emptyset$ into $\\{0,1\\}$, which is the empty function. Think about the definition of function as a set of ordered pairs to see why.\n\n\u2022 @philmcole If $X=\\emptyset$ then $\\mathcal P(X)$ has only one element. $\\emptyset \\in\\mathcal P(X)$ but $\\{\\emptyset\\}\\notin\\mathcal P(\\emptyset)$ because $\\{\\emptyset\\}\\not\\subseteq\\emptyset.$ \u2013\u00a0bof Dec 11 '17 at 10:20\n\u2022 @bof Oh, you're totally right. So the theorem holds also if $X=\\emptyset$ since we have $\\{0,1\\}^X = \\{f \\,|\\, f: \\emptyset \\to \\{0,1\\}\\} = \\{\\text{the empty function}\\}\\sim \\{\\emptyset\\} = \\mathcal{P}(X)$ \u2013\u00a0philmcole Dec 11 '17 at 10:36\n\nIt is known as the empty function.\n\nFor any set $A$, there is exactly one function from the empty set to $A$, namely the empty function:\n\n$$f_A: \\emptyset \\to A.$$\n\nThe graph of an empty function is a subset of the Cartesian product $\\emptyset \u00d7 A$. Since the product is empty the only such subset is the empty set $\\emptyset$.\n\n\u2022 I have a weird (maybe naif) argument: since '$x \\in \\emptyset$ implies that any proposition about $x$ is true', can we also say that $x \\in f^{-1}(1)$? If we do this for all $x \\in \\emptyset$ can we conclude that $f$ is constant? And analogously, if we pick $y \\in \\emptyset$ such that $y \\in f^{-1}(0)$ whereas for any other $x \\in \\emptyset, x \\in f^{-1}(1)$, can we conclude $f$ is not constant? This seems to be a contradiction, does not it? \u2013\u00a0Gibbs Dec 11 '17 at 10:12\n\u2022 related \u2013\u00a0Siong Thye Goh Dec 11 '17 at 10:19\n\u2022 I read that $f$ is a constant function, but I see no mention of the contradiction in my previous comment... \u2013\u00a0Gibbs Dec 11 '17 at 10:25\n\nA function $f:A\\to B$ is a set of ordered pairs $(a,b)$ with $a\\in A,b\\in B$. In this case there is no $a\\in A$ so therefore $f$ is an 'empty function'", "date": "2019-06-26 20:10:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2561353/set-of-functions-from-empty-set-to-0-1?noredirect=1", "openwebmath_score": 0.9069262742996216, "openwebmath_perplexity": 121.58865719092536, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.960361157495521, "lm_q2_score": 0.8872045952083047, "lm_q1q2_score": 0.8520368319895927}}
{"url": "https://tex.stackexchange.com/questions/325142/numbering-of-each-equation-within-cases-with-left-alignment-both-on-the-page-an", "text": "# Numbering of each equation within cases with left alignment (both on the page and within the equations)\n\nI am trying to create a set of equations with curly braces (like cases), each with their own set of numbers derived from the subsection (such as can be done with equations using \\numberwithin{equation}{subsection}). I get a good result where I want one equation number representing all cases of the equation (code below).\n\n$$\\mathbb(E)(X) = \\int_{-\\infty}^{\\infty} xdF(x) = \\left\\{ \\begin{array}{lll} \\sum_{x \\in \\mathcal{X}} xf_{X}(x) & =\\sum_{x \\in \\mathcal{X}}x \\mathbb{P}(X=x) &\\text{ if } X \\text{ is discrete}\\\\ \\\\ \\int_{-\\infty}^{\\infty} xf_{X}(x)dx & &\\text{ if } X \\text{ is continuous} \\end{array} \\right.$$\n\n\nWhat can I do to to have each of the cases numbered individually?\n\nI have tried align and numcases, but each of them has some problem - either the equation or the text aligns right, or the spacing goes awry.\n\nWould appreciate some help for the specific example above, with\n\n1. Equation aligned to left of page\n\n2. All text within cells left aligned\n\n3. Equation numbers on extreme right\n\n4. An empty cell in the 2nd case and\n\n5. A blank line in between the two cases.\n\n\u2022 this should be helpful: Separate labels in cases. rather than a full blank line, an optional dimension value for the gap, e.g. [.6\\baselineskip] would seem preferable. \u2013\u00a0barbara beeton Aug 16 '16 at 19:37\n\u2022 Off-topic: It should be \\mathbb{E}(X), not \\mathbb(E)(X). \u2013\u00a0Mico Aug 16 '16 at 19:48\n\u2022 Thanks Mico :). I actually use macros for those and made a typo while changin it to regular (non-macro) code for the forum. \u2013\u00a0Py_Dream Aug 17 '16 at 2:13\n\nAnother solution (and numbering) with the empheq package (which loads mathtools, which loads amsmath):\n\n\\documentclass{article}\n\\usepackage[a4paper,margin=2.5cm]{geometry} % set page parameters\n\\usepackage{amsfonts, % for \\mathbb and \\mathcal macros\nempheq}%\n\n\\numberwithin{equation}{subsection}\n\\setcounter{section}{1}\n\\setcounter{subsection}{1}\n\n\\begin{document}\n\n\\begin{subequations}\n\\begin{empheq}[left={\\mathbb{E}(X)=\\displaystyle\\int_{-\\infty}^{\\infty} x\\,dF(x)=\\empheqlbrace}]{alignat = 2}\n& \\sum_{x \\in \\mathcal{X}} xf_{X}(x) =\\sum_{x \\in \\mathcal{X}} x\\mathbb{P}(X=x)\n&\\qquad & \\text{if $X$ is discrete}, \\\\\n& \\int_{-\\infty}^{\\infty} xf_{X}(x)\\,dx\n& &\\text{if $X$ is continuous}. \\end{empheq}\n\\end{subequations}\n\n\\begin{subequations}\n\\begin{empheq}[left={\\mathbb{E}(X)=\\displaystyle\u222b_{-\u221e}^{\u221e} x\\,dF(x)=\\empheqlbrace}]{flalign}\n& \u2211_{x \u2208 \\mathcal{X}} xf_{X}(x) =\u2211_{x \u2208 \\mathcal{X}} x\\mathbb{P}(X=x)\n& &\\text{if $X$ is discrete},&\\hspace{5em} & \\\\\n& \u222b_{-\u221e}^{\u221e} xf_{X}(x)\\,dx\n&& \\text{if $X$ is continuous}. \\end{empheq}\n\\end{subequations}\n\n\\end{document}\n\n\n\u2022 Sorry. It's once more a problem with my editor. i'LL FIX IT. \u2013\u00a0Bernard Aug 16 '16 at 21:04\n\u2022 Hi, many thanks for this. Can this be done without the global [fleqn] option? I don't want all equations left aligned. \u2013\u00a0Py_Dream Aug 17 '16 at 2:02\n\u2022 Yes, absolutely. Actually, what formatting was not very clear, as you can see from some comments. So, to sum it up, you want equation numbers on the right, and centred alignments? \u2013\u00a0Bernard Aug 17 '16 at 8:01\n\u2022 Hi Bernard, as mentioned in the original question, I am looking for this particular set of equations to be left aligned and the numbers to be right aligned. However, I do not want equations to be left-aligned globally, i.e., I do not want the global [fleqn] option. \u2013\u00a0Py_Dream Aug 18 '16 at 11:34\n\u2022 @Py_Dream: You must use the flalign environment then. Please see my updated answer. \u2013\u00a0Bernard Aug 18 '16 at 11:56\n\nYou didn't specify the type of problem(s) you encounter with the numcases environment. At any rate, I don't seem to encounter any in the following example. (The fleqn option is set so that the entire equation is set flush-left instead of centered. If that's not needed, just drop that option.)\n\n\\documentclass[fleqn]{article}\n\\usepackage[a4paper,margin=2.5cm]{geometry} % set page parameters\n\\usepackage{amsfonts} % for \\mathbb and \\mathcal macros\n\\usepackage{amsmath}  % for \\text and \\numberwithin macros\n\\usepackage{cases}    % for numcases environment\n\n%% And, just for this example:\n\\numberwithin{equation}{subsection}\n\\setcounter{section}{1}\n\\setcounter{subsection}{1}\n\n\\begin{document}\n\\begin{numcases}{\\mathbb{E}(X) = \\int_{-\\infty}^{\\infty} x\\,dF(x)=}\n\\sum_{x \\in \\mathcal{X}} xf_{X}(x)  =\n\\sum_{x \\in \\mathcal{X}} x\\mathbb{P}(X=x)\n& \\text{if $X$ is discrete} \\\\[1\\baselineskip]\n\\int_{-\\infty}^{\\infty} xf_{X}(x)\\,dx\n& \\text{if $X$ is continuous}\n\\end{numcases}\n\\end{document}\n\n\u2022 Many thanks. I do not want to use the global [fleqn] option. So I was loading cases using \\usepackage[fleqn]{cases} as was suggested somewhere. When I do this with your code (as on my earlier attempts), I get multiple errors for both the begin{numcases} line and the end{numcases} line. E.g. Undefined control sequence. ...}(X) = \\int_{-\\infty}^{\\infty} x\\,dF(x)=} Missing number, treated as zero. ...}(X) = \\int_{-\\infty}^{\\infty} x\\,dF(x)=} Illegal unit of measure (pt inserted). ...}(X) = \\int_{-\\infty}^{\\infty} x\\,dF(x)=} Is there any way to do this without the global [fleqn] option? \u2013\u00a0Py_Dream Aug 17 '16 at 2:05\n\u2022 @Py_Dream - Under no circumstance should you attempt to pass the option fleqn to the cases package -- what you say \"was suggested somewhere\" is faulty. If you don't want to set the fleqn option at the \\documentclass stage, you should set it as an option when loading the amsmath package: \\usepackage[fleqn]{amsmath}. \u2013\u00a0Mico Aug 17 '16 at 5:33\n\u2022 @Py_Dream - If you don't want all displayed equations to be left-aligned, certainly don't use the option fleqn option in the first place. If the non-centering should apply to just one displayed equation, you'll have to fine-tune its position \"by hand\", e.g., by using one or more \\qquad instructions at the very end of the equation's right-hand most portion. \u2013\u00a0Mico Aug 17 '16 at 5:44\n\u2022 Hi Mico, do you happen to have any further information (or links) on why fleqn should never be passed to the cases package and why the suggestion was faulty? It is not very helpful having cardinal truths stated without any reasons ascribed. \u2013\u00a0Py_Dream Aug 18 '16 at 11:37\n\nEquation aligned to left of page: \\documentclass[12pt,leqno]{book}\n\n\u2022 I think you meant to specify fleqn, not leqno. \u2013\u00a0Mico Aug 16 '16 at 20:00\n\u2022 @Mico: I had misunderstood, I guess. I'll change my code. Thanks! \u2013\u00a0Bernard Aug 16 '16 at 20:52", "date": "2019-10-18 23:55:31", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/325142/numbering-of-each-equation-within-cases-with-left-alignment-both-on-the-page-an", "openwebmath_score": 0.9485743641853333, "openwebmath_perplexity": 2935.326997690743, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799482964023, "lm_q2_score": 0.8840392878563336, "lm_q1q2_score": 0.8520193391421655}}
{"url": "https://mathematica.stackexchange.com/questions/181868/why-gauss-legendre-quadrature-should-keep-the-number-of-integral-points-less-tha", "text": "# Why Gauss-Legendre Quadrature should keep the number of integral points less than about 50?\n\nI wanted to use Gauss-Legendre Quadrature to calculate an integral as follows\uff1a\n\nWhen n=10 and some other number(except odd numbers),the numerical result is the same as theoretical result.\n\nClear[\"Global*\"];\nn = 10;\nL[x_] := D[(x^2 - 1)^n, {x, n}]/(n!*2^n);\nA[x_] := 2/((1 - x^2) (D[L[x], x])^2);\nf[x_] := E^(-x)/x;\nsol = NSolve[L[x] == 0., x];\nnodes = (x /. sol);\ncoef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];\nSum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]\nIntegrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]\n\n-2.11450175075\n-2.11450175075\n\n\nBut,when n>=50,the code will give wrong numerical result.\n\nClear[\"Global*\"];\nn = 50;\nL[x_] := D[(x^2 - 1)^n, {x, n}]/(n!*2^n);\nA[x_] := 2/((1 - x^2) (D[L[x], x])^2);\nf[x_] := E^(-x)/x;\nsol = NSolve[L[x] == 0., x];\nnodes = (x /. sol);\ncoef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];\nSum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]\nIntegrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]\n\n-43.8873164858\n-2.11450175075\n\n\nThat is very puzzling.\n\nNSolve has severe issues to solve for the roots (which is not surprising since finding the roots of a polynomial of degree 50 is a nontrivial task):\n\nMax[Abs[L[x] /. sol]]\n\n\n164.177\n\nThat seems to be a precision problem (the polynomial hase huge coefficients but the powers of x tend to be very small so that this easily lead to catastrophic cancellation. You will get better results with higher WorkingPrecision:\n\nsol = NSolve[L[x] == 0, x, WorkingPrecision -> 100];\nMax[Abs[L[x] /. sol]]\n\n\n0.*10^-92\n\nSo the real art here is to determine the roots in a numerically stable way. Actually, Gau\u00df-Legendre quadrature rule is already built into the system. For example, see\n\nn = 50;\nprec = 200;\n{nodes, coef, bla} = NIntegrateGaussRuleData[n, prec];\n(* compute all quadrature points by convex combinations*)\nnodes = (-1) * (1 - nodes) + nodes * 1;\ncoef = 2 coef;\na = coef.f[nodes];\nb = Integrate[f[x], {x, -1, 1}, PrincipalValue -> True];\na - b\n\n\n2.6225212*10^-190\n\nPS.: I was quite astonished that Gau\u00df quadrature works so well with this singular integral. Then it came back to my mind that it is constructed such that it neglects contributions of all odd functions (because they would integrate to 0 anyways)...\n\n\u2022 thank you very much! I thought it was easy to solve for the roots of polynomial of large degree before. \u2013\u00a0Quere Sep 14 '18 at 12:25\n\u2022 @Quere In the meantime, I found out that using higher WorkingPrecision helps to make your method more precise. \u2013\u00a0Henrik Schumacher Sep 14 '18 at 12:33\n\u2022 Yes!I noticed it.The result was really amazing. \u2013\u00a0Quere Sep 14 '18 at 12:37\n\u2022 Not really worth writing as a separate answer, so: the trick of using even-order Gauss-Legendre quadrature to evaluate principal value integrals with a singularity at the center was published by Piessens, but I seem to recall this being known as folklore. In case the pole is not exactly in the middle of the integration interval, one can either split the integral so that there is a regular part and a part symmetric about the pole, or one can use a substitution that centers the pole, allowing the use of Gaussian quadrature. \u2013\u00a0J. M.'s discontentment Sep 24 '18 at 19:45\n\nIf you add WorkingPrecision -> 60 in the NSolve function you get your solution.\n\nn = 50;\nA[x_] := 2/((1 - x^2) (D[LegendreP[n, x], x])^2);\nf[x_] := E^(-x)/x;\nsol = NSolve[LegendreP[n, x] == 0, x, WorkingPrecision -> 60];\nnodes = (x /. sol);\ncoef = Table[A[x] /. {x -> nodes[[i]]}, {i, 1, Length@nodes}];\nSum[coef[[i]] f[nodes[[i]]], {i, 1, Length@nodes}]\nIntegrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]\n\n-2.11450175075145702914368470979175591804810787513962\n\n-2.1145\n\n\nOr you can use the built in functions i mma\n\n<< NumericalDifferentialEquationAnalysis\nn = 50;\n{pts, w} = Transpose[GaussianQuadratureWeights[n, -1, 1]];\nIntegrate[E^(-x)/x, {x, -1, 1.}, PrincipalValue -> True]\nSum[w[[i]] f[pts[[i]]], {i, 1, Length@pts}]\n\n-2.1145\n\n-2.1145\n\n\nA more stable way of computing the roots of a polynomial family that satisfies a three-term recurrence is the method of Golub and Welsch (1969), which computes the eigenvalues of a matrix based on the recurrence, sometimes called the \"comrade matrix.\"\n\n(*\n* n-point Gauss quadrature (Golub-Welsch 1969)\n*)\n(* p[n][x] == (a[n]x+b[n])p[n-1][x] - c[n]p[n-2][x] *)\ncomradeMatrix[{a_, b_, c_}, {n_, n0_Integer}, prec_: Infinity] :=\nBlock[{n = Range@n0},\nWith[{beta = Sqrt[Rest@c/(Most@a*Rest@a)]},\nN[SparseArray[{Band[{1, 1}] -> -b/a, Band[{2, 1}] -> beta,\nBand[{1, 2}] -> beta}, {n0, n0}], prec]\n]];\nModule[{n}, comradeMatrix[{2 n - 1, 0, n - 1}/n, {n, n0}, prec]];\n\n(*\n{-0.998866, -0.994032, -0.985354, -0.972864, -0.956611, -0.936657, \\\n-0.913079, -0.885968, -0.85543, -0.821582, -0.784556, -0.744494, \\\n-0.701552, -0.655896, -0.607703, -0.557158, -0.504458, -0.449806, \\\n-0.393414, -0.3355, -0.276288, -0.216007, -0.154891, -0.0931747, \\\n-0.0310983, 0.0310983, 0.0931747, 0.154891, 0.216007, 0.276288, \\\n0.3355, 0.393414, 0.449806, 0.504458, 0.557158, 0.607703, 0.655896, \\\n0.701552, 0.744494, 0.784556, 0.821582, 0.85543, 0.885968, 0.913079, \\\n0.936657, 0.956611, 0.972864, 0.985354, 0.994032, 0.998866}\n*)\n\n\nThe integration weights can be computed from these nodes as in the OP.\n\n\u2022 As a terminological note: the tridiagonal matrix with the orthogonal polynomial's recurrence coefficients is called a Jacobi matrix. (The comrade matrix, as you might recall, is a Jacobi matrix for the Chebyshev polynomials, with a correction term representing the Chebyshev series coefficients.) I talked about this here. \u2013\u00a0J. M.'s discontentment Sep 24 '18 at 11:27\n\u2022 @J.M. Barnett (1975) seems to introduce the name \"comrade matrix\" of a polynomial with respect to an arbitrary family of polynomials that satisfies a three-term recurrence and presents it as a generalization of the Chebyshev colleague matrix of Good (1961). But, you know, I'm not that confident about current terminology in this area. -- Yes, I found the link to that Q&A shortly after posting an answer and put it under the question above. \u2013\u00a0Michael E2 Sep 25 '18 at 17:34\n\u2022 Argh, I switched up \"comrade\" and \"colleague\" again... I keep switching the general and specific cases. :D \u2013\u00a0J. M.'s discontentment Sep 25 '18 at 17:40", "date": "2020-10-19 15:44:24", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/181868/why-gauss-legendre-quadrature-should-keep-the-number-of-integral-points-less-tha", "openwebmath_score": 0.3738044500350952, "openwebmath_perplexity": 2050.6371451778236, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799420543365, "lm_q2_score": 0.8840392893839086, "lm_q1q2_score": 0.8520193350961802}}
{"url": "https://electronics.stackexchange.com/questions/565679/finding-resonant-frequency-or-damping-ratio-from-bode-plot", "text": "# Finding resonant frequency or damping ratio from Bode Plot\n\nI am working on a question where I have to estimate a transfer function from its bode plot.\n\nI plotted the asymptotes of this bode diagram, and was able to find out that this is 3rd order system with a pole at s = 0 and two complex poles.\n\n$$\\dfrac{w_n^2}{s(s^2 + 2\\zeta w_n s + w_n^2)}$$ where I found out that $$w_n = 6.7 rad/s$$ from the asymptotic bode plot.\n\nI was also able to find the system gain $$\\K\\$$ from the magnitude plot.\n\nI am unable to find $$\\\\zeta\\$$ here. I tried finding the resonant frequency $$\\w_r\\$$ from the bode plot so that I can calculate $$\\\\zeta\\$$ using: $$w_r = w_n\\sqrt{1 - 2\\zeta^2}$$ but I was unable to.\n\nI know that the frequency at which the phase plot crosses zero is the resonant frequency but the phase plot here doesn't cross zero.\n\nI tried approximating $$\\\\zeta\\$$ using the fact that maximally flat response is obtained for $$\\\\zeta = 0.707 \\$$, so that for the given plot, $$\\\\zeta < 0.707 \\$$. But I wasn't able to exactly find a value.\n\nIs there any other way to find $$\\\\zeta\\$$ or will I be just able to approximate it?\n\nThe solution says that the value of $$\\\\zeta\\$$ is $$\\0.447\\$$.\n\n\u2022 Curiously, it changed May 18, 2021 at 13:01\n\u2022 @TonyStewartEE75 Sorry I added the wrong bode plot. May 18, 2021 at 13:02\n\u2022 Compute d phi /dt max in the same units(rads) May 18, 2021 at 13:06\n\u2022 Or rather $d\\phi /d\\omega$ then determine the ratio constant May 18, 2021 at 13:18\n\u2022 If the plot is not on a paper (i.e. you have access to the datapoints), subtract the 1/s from it and you'll be left with a more manageable 2nd order magnitude/phase. May 18, 2021 at 15:09\n\nThe following graphical solution relies on the idea that it's possible to separate the contribute of the zero in w=0 by referring the magnitude measurements to a line slanted with a -20dB/decade slope. Thanks to the properties of logarithms, division becomes translation on the magnitude Bode plot. We also have to take into account the -90 degrees contribution in the phase - it's basically a constant -90\u00b0 addition since, being the pole in the origin, it has already 'run its course'.\n\n1. the first step is to find wn. In a second order system with no zeros, the phase resonance happens exactly at wn, the undamped natural frequency (a frequency that is in general different from wpeak, the peak frequency of the magnitude, and also from the damped natural frequency wd). Since we need to separate the phase contribution of the pole in the origin, instead of finding the frequency where the phase is -90\u00b0 we need to find the frequency where the phase is -180\u00b0 By eyeballing the scale on the tiny plot I have I believe I can locate it at 6.7 rad/s. You might end up with a better estimate.\n\nNow I want to find the 3dB corner frequency the system would have without the pole in the origin . I therefore...\n\n1. ...trace a slanted line, translated 3dB under the asymptotic behavior at low frequencies and\n\n2. ...and then I look for the intersection of said line with the magnitude of the transfer function to find w3dB. Eyeballing again I find w3dB = 8.7 rad/sec\n\n3. We are now in the position to computed the ratio w3dB/wn = 1.298 = 1.3\n\nWe can now either solve the expression for w3dB as a function of zeta\n\nor, if we have a graph like this, 5) use it to find the value of zeta corresponding to w3dB/wn = 1.3.\n\nagain by eyeballing I get a zeta value of around 0.48, a value not dissimilar from that found by solving the equation\n\nfor zeta, which gives zeta = 0.477\n\nAnd this value is reasonably close, considering the amount of lazy eyeballing employed, to the correct answer of 0.447. Try your estimates on a bigger graph, counting the pixels and report back. Did it work?\n\nCaveat emptor: it is imperative that the second order function be without additional zeroes (apart for the one we have been able to separate). The relevant frequencies have different expressions form system with one or more zeroes.\n\n\u2022 Thank you, yeah it did work. I got the w3dB to be around 8.9 rad/s from the bode plot, which gave me a ratio of 1.328. Got zeta = 0.4482. Thanks a lot! Jun 13, 2021 at 5:39", "date": "2022-07-07 15:10:39", "meta": {"domain": "stackexchange.com", "url": "https://electronics.stackexchange.com/questions/565679/finding-resonant-frequency-or-damping-ratio-from-bode-plot", "openwebmath_score": 0.8104416728019714, "openwebmath_perplexity": 528.5693212282331, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799472560581, "lm_q2_score": 0.884039278690883, "lm_q1q2_score": 0.8520193293889828}}
{"url": "https://mathematica.stackexchange.com/questions/244964/plot-graph-of-function-with-different-constants", "text": "# plot graph of function with different constants\n\nI need to plot the graph of :\n\n$$v(t) = \\frac{-mg}{b} (e^{-\\frac{b}{m}t}+1)$$ , where $$g = 9.8 m/s^2$$ and $$b$$ and $$m$$ are positive constants.\n\nIs there a way to plot this without having to define random values for $$b$$ and $$m$$? I mean a way in which I can see how the graph behaves for different values of $$b$$ and $$m$$ ?\n\n\u2022 Unfortunately no, I am new to Mathematica, I only know the basics \u2013\u00a0Physmath Apr 22 at 2:17\n\u2022 I will take a look, thank you ! \u2013\u00a0Physmath Apr 22 at 2:22\n\u2022 To simplify little bit, you can introduce a new variable, say $\\lambda=\\frac{b}{m}$. Then you can Plot a sequence of curves for a list of $\\lambda$ values. \u2013\u00a0yarchik Apr 22 at 3:03\n\nv[s_, t_] := -g/s (Exp[-s t] + 1)\ng = 9.8;\nslist = {0.1, 0.2, 0.4, 0.8};\nPlot[Evaluate[v[#, t] & /@ slist], {t, 0, 10},\nPlotTheme -> {\"BoldColors\", \"Frame\", \"Grid\"},\nFrameLabel -> {Automatic, v[t]},\nFrameStyle -> Directive[14, Black],\nPlotLabels -> slist\n]\n\n\n\u2022 Plot[Evaluate[v[slist, t] , {t, 0, 10},...] works too. \u2013\u00a0Ulrich Neumann Apr 22 at 8:02\n\u2022 @UlrichNeumann That is good suggestion, it makes the syntax more transparent. \u2013\u00a0yarchik Apr 22 at 8:12", "date": "2021-06-18 06:37:37", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/244964/plot-graph-of-function-with-different-constants", "openwebmath_score": 0.33643510937690735, "openwebmath_perplexity": 851.4005363716258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692359451417, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.85199484876764}}
{"url": "https://math.stackexchange.com/questions/3935988/showing-that-if-3x-is-even-then-3x5-is-odd/3936024", "text": "# Showing that if $3x$ is even then $3x+5$ is odd\n\nI'm learning the absolute basics of how to do proofs, and am really struggling.\n\nIf 3x is even then 3x+5 is odd.\n\nThis is the solution:\n\nI get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but nothing I found is this basic.\n\nWork: -Assumptions- 3x = 2n 3x+5 = 2k+1\n\n-Trying to make sense of 3x- 3x+5 = 2k+1 3x = 2k-4\n\n-Plugging in 2k-4 for 3x- 2k-4 = 2n 2k = 2n+4 k = n+2\n\n-Plugging in n+2 for k- 3x+5 = 2(n+2)+1\n\n...This is where I gave up. I don't know where I'm going with this anymore.\n\n\u2022 You should calm down, everyone can have difficulties. Explain us where you get lost. If you don't know where you get lost I can try to explain you the proof step by step. \u2013\u00a0Eureka Dec 5 '20 at 17:42\n\u2022 I'm good, just frustrated. I'd like step by step. I'll edit in my work above. Warning, there is no rhyme or reason to it. \u2013\u00a0visualbread Dec 5 '20 at 17:44\n\u2022 If $y$ is even, can you prove that $y+5$ is odd? Try this first, then try your problem. There\u2019s no substantial difference. \u2013\u00a0Benjamin Wang Dec 5 '20 at 17:46\n\u2022 @visualbread your resolution may be a bit convoluted but it's correct. You just proved that $3x+5=2(n+2)+1$. And $2(n+2)+1$ is odd ,because it is of the form $2h+1$ with $h=n+2$ \u2013\u00a0Eureka Dec 5 '20 at 17:52\n\u2022 @BenjaminWang I am unable to do that one either. From y = 2n and y+5 = 2k+1, I can get that all the way to n = k -2 (or k = n+2), but I can't see what that tells me. The final solution has everything nicely equal to one another, and I can't seem to set it up that way. \u2013\u00a0visualbread Dec 5 '20 at 17:54\n\nI think you're getting confused by the $$3x$$ part. The $$3x$$ plays no role in the problem.\n\nSuppose you're given any number that is even. I'll call it y. Now we want to show $$y+5$$ is odd.\n\nTherefore by definition\n\n$$y=2k$$ for some integer $$k$$\n\nNow\n\n$$y+5 = 2k+5$$\n\nNow we just need to show that $$2k+5$$ is 2 times an integer plus 1\n\n$$2k+5 = 2(k+2)+1$$\n\nSo $$2k+5$$ is odd because it can be written in the form 2*integer +1 where the integer here is $$k+2$$. So $$y+5$$ is odd since $$y+5 = 2k+5$$\n\nSo if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc...\n\n\u2022 I might be there, but I need just a little bit more help. How did you write 2k+5 in the form 2(k+2) + 1? I can see that those are the same, but how could I do it without trial and error? \u2013\u00a0visualbread Dec 5 '20 at 18:16\n\u2022 So the first step is to look at 5. We see it is an odd integer. So I decide it is more convenient to write it as $4+1$. So I write $2k+5$ as $2k+4+1$. Now I can factor $2k+4 = 2(k+2)$. So putting it all together $2k+5 = 2k+4+1 = 2(k+2)+1$. Hope this makes sense. let me know. Suppose I had $2k+6$ instead. since 6 is even, I know I can factor immediately $2k+6 = 2(k+3)$. \u2013\u00a0Ameet Sharma Dec 5 '20 at 18:19\n\u2022 Thanks! This shows me all the steps I need to follow my book's logic. I don't know what happened in my math education, but I never did learn some of these little things that make some of these proofs actually understandable. At least there's websites like this one to fill in the gaps. \u2013\u00a0visualbread Dec 5 '20 at 18:25\n\u2022 No problem. I highly recommend doing lots of exercises with factoring, simplification and algebra so that these things become second nature. It takes a lot of practice but it is worth it. As you get into more advanced math, these things really need to be 2nd nature where you don't need to think about them anymore... otherwise you'll get bogged down at every step. \u2013\u00a0Ameet Sharma Dec 5 '20 at 18:31\n\u2022 I'm sure there are lots of online websites. Here's one I just found as part of an online algebra textbook: openstax.org/books/college-algebra/pages/1-review-exercises. Each chapter has review exercises and a practice test. I recommend doing all these, maybe multiple times till you can do them in your sleep. In my opinion... if you want online resources, look for online algebra \"textbooks\"... and do all the exercises in each chapter. If you have doubts you can review the chapter. \u2013\u00a0Ameet Sharma Dec 5 '20 at 18:51\n\nYour problem is that $$3x + 5 = 2k +1$$ is not an assumption. It is the conclusion you need to prove.\n\nYour one and only assumption is that $$3x = 2n$$ for some integer $$n$$.\n\n$$3x = 2n$$.\n\n.... then you do a bunch of steps ....\n\n.... steps .....\n\n.... and get in the end ........\n\nConclusion: $$3x + 5 = 2(????????) + 1$$ where $$??????$$ is some integer you come up with in you steps.\n\nLet's see what happens when we try. Let's take it nice and slow:\n\n=======\n\n$$3x = 2n$$.\n\n$$3x +5 = 2n + 5$$\n\n....hmmm, we want $$2(??????) + \\color{red}1$$ in the end so let's pull out the $$+\\color{red}1$$ first.....\n\n$$3x + 5 = 2n+5 = 2n + (4 + \\color{red}1)=(2n+4) +\\color{red}1$$\n\n.... hmmm, okay that's the $$+1$$ now we want $$2(\\color{red}{??????}) + 1$$. To get the So we need to factor then $$2$$ out of $$2n+4$$ and see what we have left.... that will bee the $$\\color{red}{??????}$$\n\n$$3x + 5 = (\\color{red}{2n+4}) + 1$$\n\n$$3x + 5= 2(\\color{red}{n + 2}) + 1$$\n\n.... and that's it......\n\nConclusion: $$3x+5 = 2(\\color{red}{n + 2})+1$$.\n\nThe $$??????$$ we wanted turns out to be $$\\color{red}{n+2}$$ an we have\n\n$$3x + 5 = (3x+4) + 1 = (2n+4) + 1 = 2(\\color{red}{n+2}) + 1$$.\n\nAnd because $$\\color{red}{n+2}$$ is an integer if we let $$k = n+2$$ be that integer $$3x+5 = 2k + 1$$ and so... $$3x + 5$$ is odd.\n\n=======\n\nAlthough if you want to work backwards\n\nConclusion: $$3x + 5 = 2k +1$$ ..... and we want to solve for $$k$$ to show it is possible...\n\n$$3x + 5 -1 =2k + 1-1$$\n\n$$3x +4 = 2k$$\n\n$$k = \\frac {3x + 4}2 = \\frac {3x}2 + 2$$.\n\n.... but is $$\\frac {3x}2 + 2$$ an integer?????\n\nWell, $$3x$$ is even. So there is an integer $$n$$ so that $$3x = 2n$$ so\n\n$$k = \\frac {3x}2 +2 = \\frac {2n}2 + 2 = n+2$$.\n\nSo $$k=n+2$$ is the integer we want to conclude $$3x+5 =2k +1$$.\n\nIf we did it this way our proof would go:\n\n$$3x$$ is even so there is an integer $$n$$ so that $$3x = 2n$$. Let $$k = n+2$$; that is an integer.\n\n$$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$$.\n\nSo $$3x+5 = 2k +1$$ and that is odd.\n\nOk so we know that $$3x$$ is even, that means we can write $$3x=2n$$ for a suitable $$n$$, since even means that the number is divisible by two without remainder. But then we have $$3x+5=2n+5=2n+(4+1)=2n+2\\cdot 2+1=2(n+2)+1$$ which clearly is odd.\n\n\u2022 Thanks for the reply. I'm still not there yet. I can't figure out where 2n+5 came from or why it can be set equal to 2x+5. I need this broken down even further if that's even possible. \u2013\u00a0visualbread Dec 5 '20 at 17:55\n\u2022 We set it equal to $3x+5$ NOT $2x+5$. I will try to break it down further: so \u2013\u00a0Mo145 Dec 5 '20 at 18:04\n\u2022 1)a natural number $N$ is called even if it is a multiple of $2$. So per definition we can write every even number $N$ as $N=2M$. Note that we can write any natural number as $N=2\\frac{N}{2}$ but in general $\\frac{N}{2}$ will not be a natural number, for example take $N=3$, then clearly $\\frac{3}{2}$ is not a natural number. In fact this will only be the case if $N$ is even. \u2013\u00a0Mo145 Dec 5 '20 at 18:08\n\u2022 2)Since we assume $3x$ to be even we can, by 1) write it as $3x=2n$ where $n=\\frac{3x}{2}$ so we have the following equality $3x=2n$ adding a number on both sides preserves the equality hence we have $3x+5=2n+5$ \u2013\u00a0Mo145 Dec 5 '20 at 18:11\n\u2022 I hope this clarifies your concerns :) \u2013\u00a0Mo145 Dec 5 '20 at 18:11\n\nWe know that $$3x=2n$$. So: $$\\color{blue}{3x}+5=\\color{blue}{2n}+5$$ Because the \"blue quantities\" are equal. Now: $$3x+5=2n+5=2n+4+1$$ In this step I just wrote $$5$$ as $$4+1$$:\n\n$$3x+5=2n+5=2n+4+1=2n+2\\cdot 2+1$$ In this step I just wrote $$4$$ as $$2 \\cdot 2$$.\n\n$$3x+5=2n+5=2n+4+1=\\color{green}{2n+2\\cdot 2}+1=\\color{green}{2(n+2)}+1$$ The last step is valid because the green quantities are equal. In the end: $$3x+5=2(n+2)+1$$ This means that $$3x+5$$ is odd because is of the form $$2h+1$$ with $$h$$ integer(in particular $$h=(n+2)$$)\n\nRemember here that $$n$$ represents ANY natural number. You got to the answer but you didn\u2019t even realize it. That\u2019s probably because you are thinking syntactically rather than semantically. What I mean is the literal string of symbols $$2(n+2)$$ didn\u2019t register to you as even because it is not the same as the string $$2n$$. But $$n+2$$ is a natural number just like $$n$$ is. So the the strings $$2(n+2)$$ and $$2n$$ both represent even numbers, and so $$2(n+2) + 1$$ is odd just as you have shown in your last line.\n\nYou are dealing with a problem where you are given too much information. Here is another way of doing it (we'll use $$m$$ for an integer).\n\nIf $$3x$$ is even then $$x$$ must be even, so we can put $$x=2m$$. [This is a consequence of the fact that $$2$$ is a prime, or can be proved in various ways]\n\nThen $$3x+5=6m+5=6m+4+1=2(3m+2)+1$$.\n\nNow we can put $$3m=n$$, an integer, to get $$2(n+2)+1$$.\n\nFor an alternative proof, we could put $$3m+2=n$$ and then we get $$2n+1$$.\n\nNote that the last two sentences are alternatives to one another. They both use $$n$$, but $$n$$ is defined differently in the two cases.\n\nWhat I'm trying to do here is to unpack how the different expressions for the same thing relate to each other. If you get your head round that you will be flying.\n\n\u2022 I find this answer to be pretty overwhelming for the OP, since you are talking about \"alternative proofs\" and \"consequences of the fact that 2 is prime that can be proved in various ways\". I'm not underestimating the OP, but I think that when you learn fron basic you should do things calmly step by step. \u2013\u00a0Eureka Dec 5 '20 at 18:23\n\u2022 @Eureka What I was trying to do was to show what happens when you keep the factor $3$ for an extra step. If it is \"overwhelming\" it might help OP to understand why the factor $3$ is best dropped (\"too much information\"). Either it helps or it doesn't. \u2013\u00a0Mark Bennet Dec 5 '20 at 18:35", "date": "2021-07-30 09:41:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3935988/showing-that-if-3x-is-even-then-3x5-is-odd/3936024", "openwebmath_score": 0.7947751879692078, "openwebmath_perplexity": 239.42945834890642, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692271169855, "lm_q2_score": 0.8723473862936943, "lm_q1q2_score": 0.8519948475489848}}
{"url": "https://math.stackexchange.com/questions/2229402/evaluate-int-0-frac-pi4-cos-1-sin-x-dx", "text": "# Evaluate $\\int_0^\\frac{\\pi}{4} \\cos^{-1}({\\sin x}) \\,dx$\n\nI came across this integral in a math competition and couldn't solve it $$\\int_0^\\frac{\\pi}{4} \\cos^{-1}({\\sin x})\\, dx$$\n\nI tried a $u$-substitution, with $u=\\sin x$ and ended up with the integral $$\\int_0^\\frac{\\pi}{4} \\cos^{-1}(\\text{u}) \\cdot \\frac{1}{\\sqrt{1-u^2}}\\,du$$ which is not much simpler and I cannot figure out how to solve this. Any hints/solutions for this problems?\n\nI also tried drawing a triangle for the problem but it didn't really help with the solution.\n\nYou can use the trigonometric formula : $$\\forall x \\in\\mathbb{R},\\cos^{-1}(x)+\\sin^{-1}(x)=\\frac{\\pi}{2}$$\n\nThus $\\forall x\\in\\left[0,\\frac{\\pi}{4}\\right]$ you have $\\sin^{-1}(\\sin(x))=x$ so : $$\\cos^{-1}(\\sin(x))+\\sin^{-1}(\\sin(x))=\\frac{\\pi}{2}\\Rightarrow \\cos^{-1}({\\sin x})=\\frac{\\pi}{2}-x$$\n\nFinally : $$\\int_0^\\frac{\\pi}{4} \\cos^{-1}({\\sin x}) dx=\\int_0^\\frac{\\pi}{4} \\frac{\\pi}{2}-xdx$$ Can you finish ?\n\n\u2022 What is $\\sin^{-1}(\\sin x)=?$ \u2013\u00a0lab bhattacharjee Apr 11 '17 at 15:48\n\u2022 it is $x$ since $x$ is in $[0,\\pi/4]$ \u2013\u00a0Jennifer Apr 11 '17 at 15:54\n\nHINT:\n\nIf $\\cos^{-1}(\\sin x)=y,0\\le y\\le\\pi\\ \\ \\ \\ (1)$\n\nand $\\cos y=\\sin x=\\cos\\left(\\dfrac\\pi2-x\\right)$\n\n$y=2m\\pi\\pm\\left(\\dfrac\\pi2-x\\right)$ where $m$ is an integer such that $(1)$ is satisfied\n\nFor $0\\le2m\\pi+\\dfrac\\pi2-x\\le\\pi\\iff 2m\\pi-\\dfrac\\pi2\\le x\\le2m\\pi+\\dfrac\\pi2$\n\nHere $m=0$\n\nHere's a barely different route to take, continuing from the direction you've taken.\n\nFirst, note that substituting $u=\\sin x$ would actually give\n\n$$\\int_{x=0}^{x=\\pi/4}\\cos^{-1}(\\sin x)\\,\\mathrm dx=\\int_{\\color{red}{u=0}}^{\\color{red}{u=1/\\sqrt2}}\\frac{\\cos^{-1}u}{\\sqrt{1-u^2}}\\,\\mathrm du$$\n\nNow, recall that $\\dfrac{\\mathrm d}{\\mathrm du}\\cos^{-1}u=-\\dfrac1{\\sqrt{1-u^2}}$, which means you can make another intermediate substitution of, say, $v=\\cos^{-1}u$. Then you have\n\n$$-\\int_{v=\\cos^{-1}0=\\pi/2}^{v=\\cos^{-1}(1/\\sqrt2)=\\pi/4}v\\,\\mathrm dv=\\int_{\\pi/4}^{\\pi/2}v\\,\\mathrm dv=\\dfrac{3\\pi^2}{32}$$\n\nwhich agrees with the other answers above.\n\n$$\\int \\cos^{-1}(\\sin x) dx$$ $$= \\int 1 \\cdot \\cos^{-1}(\\sin x) dx = x \\cdot \\cos^{-1}(\\sin x) - \\int x \\cdot -\\frac{\\cos x}{\\sqrt{1-\\sin^{2}x}} dx + C$$ (integration by parts)\n\n$$= x \\cdot \\cos^{-1}(\\sin x) + \\int x \\cdot \\frac{\\cos x}{\\sqrt{1-\\sin^{2}x}} dx + C$$ $$= x \\cdot \\cos^{-1}(\\sin x) + \\int x \\cdot \\frac{\\cos x}{\\sqrt{\\cos^2 x}} dx + C$$\n\n$$= x \\cdot \\cos^{-1}(\\sin x) \\pm \\int x \\cdot \\frac{\\cos x}{\\cos x} dx + C$$ $$= x \\cdot \\cos^{-1}(\\sin x) \\pm \\frac{x^2}{2} + C$$\n\nNow do it for the definite integral.\n\n1)\n\n$(u^2)^{\\prime}=2\\times u\\times u^{\\prime}$\n\ntherefore $\\dfrac{1}{2}u^2$ is an antiderivative of $u\\times u^{\\prime}$\n\n2) derivative of $\\text{cos}^{-1}(x)=-\\dfrac{1}{\\sqrt{1-x^2}}$\n\n$\\displaystyle J= \\int_0^{\\tfrac{\\pi}{4}} \\text{cos}^{-1}(\\sin x)dx$\n\nPerform the change of variable $y=\\sin x$,\n\n$\\displaystyle J=\\int_0^{\\tfrac{\\sqrt{2}}{2}} \\dfrac{\\text{cos}^{-1}(x)}{\\sqrt{1-x^2}}dx$\n\nUsing $(1)$,\n\n\\begin{align}J&=-\\dfrac{1}{2}\\Big[\\text{cos}^{-1} (x)^2\\Big]_0^{\\tfrac{\\sqrt{2}}{2}}\\\\ &=-\\dfrac{1}{2}\\left[\\dfrac{\\pi^2}{16}-\\dfrac{\\pi^2}{4}\\right]\\\\ &=\\boxed{\\dfrac{3}{32}\\pi^2} \\end{align}", "date": "2019-05-25 18:58:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2229402/evaluate-int-0-frac-pi4-cos-1-sin-x-dx", "openwebmath_score": 0.9529788494110107, "openwebmath_perplexity": 324.47443765424987, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692284751635, "lm_q2_score": 0.8723473746782093, "lm_q1q2_score": 0.851994837389301}}
{"url": "https://mechanicalc.com/calculators/beam-analysis/validation", "text": "This section details the validation for the Beam calculator. Several examples are worked through to determine expected results such as deflections, forces, and stresses. These expected results are then compared to the actual output of the calculator.\n\nThe standard beam equations for a cantilever beam loaded at the free end are given below:\n\n Max Deflection: $$\\delta_{max} = {P L^3 \\over 3EI}$$ @ x = L Max Slope: $$\\theta_{max} = {P L^2 \\over 2EI}$$ @ x = L Shear: $$V = +F$$ constant Moment: $$M_{max} = -FL$$ @ x = 0\n\nFor the validation case, an Aluminum 6061-T6 beam with a 1 inch diameter circular cross-section and a length of 10 inches is loaded at 1000 lbf. The inputs are:\n\nL = 10 in\nF = 1000 lbf\nE = 9.9e6 psi (Al 6061-T6)\n$$A = {\\pi d^2 \\over 4} = 0.7854 ~in^4$$\n$$I = {\\pi d^4 \\over 64} = 0.04909 ~in^4$$\n\nFor the given inputs, the expected results are:\n\n Max Deflection: $$\\delta_{max} = 0.6859 \\text{ in}$$ @ x = L Max Slope: $$\\theta_{max} = 0.1029 \\text{ rad}$$ @ x = L Shear: $$V = +1000 \\text{ lbf}$$ constant Moment: $$M_{max} = -10,000 \\text{ in-lbf}$$ @ x = 0\n\n### Comparison With Calculator\n\n#### Free Body Diagram\n\nThe Free Body Diagram (FBD) is shown below. From the FBD it can be seen that the forces balance and the beam is in static equilibrium.\n\n#### Deflection & Slope\n\nA screenshot of the results table giving the maximum deflection and slope is shown below. It can be seen that these values match the expected values above:\n\nThe deflection plots are shown below. From these plots it can be seen that a maximum deflection of 0.6859 inches occurs at the free end of the beam as well as a maximum slope of 0.1029 radians.\n\n#### Shear-Moment Diagram\n\nThe shear-moment diagram for the beam is shown below. This diagram is what would be expected for the current case. A constant shear force of 1000 lbf exists along the length of the beam, and the moment increases linearly from 0 in-lbf at the free end of the beam to the full value of -10,000 in-lbf at the fixed end.\n\n#### Stresses\n\nThere are two points of interest for validating stresses. Before calculating stress, the forces at these points need to be determined.\n\nAxial (lbf) Shear (lbf) Moment (in-lbf)\nForces @ x = 0: $$F_{ax} = 0$$ $$F_{sh} = 1000$$ $$M = 10000$$\nForces @ x = L: $$F_{ax} = 0$$ $$F_{sh} = 1000$$ $$M = 0$$\n\nThe stresses are calculated using the following equations:\n\nAxial Stress Shear Stress Bending Stress Von Mises Stress\n$$\\sigma_{ax} = {F_{ax} \\over A}$$ $$\\tau_{sh} = {F_{sh} \\over A}$$ $$\\sigma_{b} = {Mc \\over I}$$ $$\\sigma_{vm} = \\sqrt{ (\\sigma_{ax} + \\sigma_{b})^2 + 3\\tau_{sh}^2 }$$\n\nBased on the forces at each point and the equations above, the expected stresses at the points of interest are:\n\nTensile (psi) Shear (psi) Bending (psi) Von Mises (psi)\nStress @ x = 0: $$\\sigma_{ax} = 0$$ $$\\tau_{sh} = 1273$$ $$\\sigma_{b} = 101859$$ $$\\sigma_{vm} = 101883$$\nStress @ x = L: $$\\sigma_{ax} = 0$$ $$\\tau_{sh} = 1273$$ $$\\sigma_{b} = 0$$ $$\\sigma_{vm} = 2205$$\n\nThe stress plots are shown below. It can be seen that these plots agree with the stresses calculated above.\n\nThere is one discrepancy that can be noted, which is that the values for bending stress and von Mises stress at x = L are shown to be higher in the stress plots than predicted. The reason for this is that the stress plots are showing the maximum nodal values for each element, and the maximum bending stress for the element at x = L will occur at the left-most node in that element. Therefore, the values of bending stress and von Mises stress reported in the plots will be somewhat higher than what was calculated above. However, the stresses for each node within any element can be interrogated in the result tables. This table will be discussed following the stress plots.\n\n##### Stress At End of Beam (x = L)\n\nA screenshot of one of the result tables for this case is shown below with the element of interest highlighted. The node at the very end of the beam (i.e. at x = L) is \"Node 2\" for the highlighted element. It can be seen that the bending stress at this node is 0 psi and the von Mises stress at this node is 2205 psi. This matches what was predicted.", "date": "2020-03-31 07:27:45", "meta": {"domain": "mechanicalc.com", "url": "https://mechanicalc.com/calculators/beam-analysis/validation", "openwebmath_score": 0.6969012022018433, "openwebmath_perplexity": 740.8264258502527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692311915195, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.851994836517625}}
{"url": "https://math.stackexchange.com/questions/2475295/find-the-remainder-when-528528528-up-to-528-digits-is-divided-by-27", "text": "# Find the remainder when $528528528\u2026$up to $528$ digits is divided by $27$?\n\nFind the remainder when $528528528...$up to $528$ digits is divided by $27$?\nHere's what I have done: The number can be written as $528\\cdot 10^{525}+528\\cdot 10^{522}+...+528$ which has $176$ terms and each term is $\\equiv15 \\mod 27$ thus the number should be $176*15 \\mod 27$ hence $21$ should be the remainder. But book says it is $6$. I don't understand the flaw in my logic. Please correct me.\n\n\u2022 you have $21+6=27$ perhaps you are off by a sign? \u2013\u00a0gt6989b Oct 16 '17 at 15:26\n\u2022 I think that your answer is correct. \u2013\u00a0alexp9 Oct 16 '17 at 15:29\n\u2022 Brute force in python gives 21. \u2013\u00a0Gribouillis Oct 16 '17 at 15:31\n\u2022 Are you sure the book says $6$ and not $-6$? After all, $21\\equiv -6 \\pmod{27},$ so then both answers would agree. \u2013\u00a0David K Oct 16 '17 at 15:36\n\u2022 @fleablood Also a good point about negative remainders. It looks like there's probably an error in the book. (Unless the transcription of the problem is very confused, it's only $176$ \"copies\" of the group of digits $528,$ which makes $528$ digits altogether since each group has three digits. Also, we don't need $10^k$ to be congruent to $1$; we only need $10^{3k}\\equiv 1 \\pmod{27},$ which is true.) \u2013\u00a0David K Oct 16 '17 at 17:00\n\nHere is a python3 session\n\n>>> s = '528' * 176\n>>> len(s)\n528\n>>> int(s) % 27\n21\n\n\u2022 Is it possible to multiply a string by a number? My god...python is awesome. \u2013\u00a0Integral Oct 16 '17 at 15:35\n\u2022 @Integral yes strings, lists and tuples can be repeated by multiplying them by a number. \u2013\u00a0Gribouillis Oct 16 '17 at 15:36\n\nYou can see that $6$ cannot be correct by casting out $9$'s: Since $5+2+8=5+5+5$, we have\n\n$$528528\\ldots528\\equiv5+5+5+\\cdots+5+5+5=5\\cdot528\\equiv5(5+2+8)\\equiv5\\cdot6\\equiv3\\mod 9$$\n\nso the remainder mod $27$ must be either $3$, $12$, or $21$. Your approach gave the correct answer, $21$.\n\n\u2022 Anyway it is also $-6$ so there is a typo in the book or Anuran has not have seen the minus sign. \u2013\u00a0Piquito Oct 16 '17 at 15:54\n\u2022 @Piquito, I agree, a negative sign would fix things (as David K noted in comments). But remainders are usually understood to be nonnegative, so I'm inclined to think it's a typo. \u2013\u00a0Barry Cipra Oct 16 '17 at 15:57\n\nSince $$3\\mid111$$, we know that $$27\\mid999$$, Therefore, $$1000\\equiv1\\pmod{27}$$ Thus, \\begin{align} \\sum_{k=0}^{175}528\\cdot1000^k &\\equiv528\\cdot176\\pmod{27}\\\\ &\\equiv3\\cdot176^2\\pmod{27}\\\\ &\\equiv3\\cdot14^2\\pmod{27}\\\\ &\\equiv3\\cdot7\\pmod{27}\\\\ &\\equiv21\\pmod{27} \\end{align}\n\nYou are incorrect in assuming $10^{k} \\equiv 1 \\mod 27$. As $10^k \\not \\equiv 1$ we do not have $528*10^k \\equiv 15 \\mod 27$.\n\nWhat you need instead is $528528... = 528(1001001001......)$\n\nAnd $1001001..... =\\sum_{k=0}^{175} 10^{3k}$\n\n$10^3 \\equiv 1 \\mod 27$... Oh... we do have that and you were not wronng after all.... so $\\sum 10^{3k}\\equiv 176 \\equiv 14 \\mod 27$.\n\nSo $528528....... \\equiv 15*14 \\equiv 21 \\mod 27$.\n\nAnd... the book is wrong.\n\nHad it been 527 iterations of 528 the answer would be $6$.\n\nNote that $$27 \\times 37 = 999$$.\n\nTo find the remainder you get when you divide $$528528\\cdots 528$$ by $$999$$, you can \"cast out\" $$999$$s.\n\n$$\\underbrace{528 + 528 + \\cdots 528}_{\\text{176 times} } \\to 176 \\times 528 \\to 92928 \\to 92+928 \\to 1020 \\to 21$$\n\nSo the remainder is $$21$$.\n\nNote. If the remainder was bigger than 26, you would have had to divide it by 27 to get the correct remainder.", "date": "2021-04-17 09:25:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2475295/find-the-remainder-when-528528528-up-to-528-digits-is-divided-by-27", "openwebmath_score": 0.802299439907074, "openwebmath_perplexity": 442.68319873287027, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692311915195, "lm_q2_score": 0.872347369700144, "lm_q1q2_score": 0.8519948348969839}}
{"url": "https://matheducators.stackexchange.com/questions/14612/why-do-we-study-cantor-set", "text": "# Why do we study Cantor Set?\n\nFor finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ?\n\n\u2022 Why are you not convinced that counterexamples are an important thing to study? (Re: \"at least not always\", the beauty of a counterexample is that only needs to be used once, in a particular key spot, to change our understanding forever.) \u2013\u00a0Daniel R. Collins Sep 30 '18 at 1:07\n\n1. in beginning real analysis: to counter the naive notion that a \"closed set\" is a union of closed intervals, plus a few single points.\n\n2. In beginning Lebesgue measure: the easiest example of an uncountable set of measure zero\n\n3. In general topology: sets homeomorphic to the Cantor set are useful in proofs\n\n4. Fractal geometry: many fractals are homeomorphic to the Cantor set. Mandelbrot calls such a thing a Cantor dust to suggest its appearance.\n\n\u2022 Can one claim and substantiate that the study of cantor set is something fundamental and not just a source of ready counterexample ? \u2013\u00a0Vagabond Sep 29 '18 at 15:21\n\u2022 Point (4) seems to address this. The Cantor set is one of the simplest examples of a fractal, and is therefore studied in detail by students who are first learning fractal geometry or analysis (to the point that a colleague of mine, after writing a masters thesis on Cantor-like sets, started her PhD by stating that she never wanted to see the Cantor set again). \u2013\u00a0Xander Henderson Sep 30 '18 at 14:51\n\nThe Cantor set is quite useful in its own right. My preferred way to think of the Cantor set is as \"the most general compact metrizable space.\" That it is the most general such space means that it is often good for counterexamples because it lacks the particulars. At the same time, one can construct things by specialization.\n\nThe formal expression of the Cantor set being the most general compact metrizable space is that every compact metrizable space is the image of the Cantor set under a continuous function. One can use this as a method of proof. For example, to show that a continuous surjection from $$[0,1]$$ onto $$[0,1]^2$$ exists, one can use the fact that $$[0,1]^2$$ is the image of the Cantor set under a continuous function and then extend this continuous function by linear interpolation (the complement of the Cantor set consists of open intervals) to all of $$[0,1]$$.\n\nA lot more can be proven this way, including a number of very surprising results. A wonderful source for such arguments is the following paper, which (deservedly) won an award for mathematical exposition:\n\nBenyamini, Yoav. \"Applications of the universal surjectivity of the Cantor set.\" The American mathematical monthly 105.9 (1998): 832-839.\n\nAmong the results proven there is the entirely classical result of Banach and Mazur that every separable Banach space is linearly isometric to a subspace of $$C[0,1]$$; a result as far from being a counterexample as can be.\n\nI think that you may be selling short the value of a counterexample! They are quite useful for making sure that you have not proven too much. i.e. When you have a plausible but only semi-formal argument, how do you tell if it is worth the effort in making it rigorous? Checking against counterexamples is often a useful step.\n\nStill, there are relations to unbounded paths in an infinite binary tree without leaves. That is, suppose that you start at the root and write $$0.$$, thought of as a ternary number. As we travel to the left or right, write a $$0$$ or $$2$$, respectively, for the following digit. Continuing on we get a ternary expansion defining a real number in the Cantor set. Correspondingly, the ternary expansion of an element of the Cantor set gives rise to a path from the root in the tree.\n\nSpaces that are homeomorpic to the Cantor set arise naturally in many mathematical settings, particularly in dynamical systems.\n\nFor one dynamical example, the Cantor set is homeomorphic to the phase space of any infinite Bernoulli process.\n\nFor another, the \"nonescaping set\" of many simple dynamical systems in the real line (or the complex plane) is homeomorphic to the Cantor set (this is a \"Cantor dust\" example as in the answer of @GeraldEdgar). Consider for example the dynamical system $$z_n = (z_{n-1})^2 + 10$$ (You can replace $$10$$ by any real or complex number of magnitude $$>2$$). One can prove that there is a subset $$C \\subset \\mathbb C$$ homeomorphic to the Cantor set such that if $$z_0 \\in C$$ then the sequence $$(z_n)$$ is bounded (in fact it stays in $$C$$), whereas if $$z_0 \\not\\in C$$ then $$\\lim_{n \\to \\infty} |z_n| =\\infty$$. In short, points not in $$C$$ escape to infinity, points in $$C$$ do not.\n\nAlso, there are important theoretical descriptions/properties of the Cantor set, for example:\n\n\u2022 A topological space is homeomorphic to the Cantor set if and only if it is compact, metrizable, has no isolated points, and every component is a point.\n\u2022 Any compact zero-dimensional metrizable topological space is homeomorphic to a subspace of the Cantor set.\n\nCantor sets even occur naturally in number theory! The $$p$$-adic integers $$\\mathbb Z_p$$ are homeomorphic to the Cantor set.\n\n\u2022 Do you mean some subset in $\\mathbb{C}$? I don't get the example; for any $z_{-1}$ real one has $z_n \\geq 10^{2^n} \\to +\\infty$. \u2013\u00a0Vandermonde Oct 7 '18 at 16:01\n\u2022 Oops, you're right, that was careless of me. Fixed. \u2013\u00a0Lee Mosher Oct 7 '18 at 21:26", "date": "2019-02-18 15:39:29", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/questions/14612/why-do-we-study-cantor-set", "openwebmath_score": 0.7546228766441345, "openwebmath_perplexity": 241.27119697944664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692311915195, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8519948267937784}}
{"url": "https://math.stackexchange.com/questions/4064710/sequence-queue-converges-or-diverges/4064747", "text": "# Sequence queue - converges or diverges?\n\nI have the next sequence queue: $$\\sum_{n=1}^{\\infty}{\\frac{n^2-n-1}{n^4+n^2+1}}$$. Does the queue converges or diverge?\n\nMy attempt: I have tried to show that $$\\frac{1}{n^2}>|\\frac{n^2-n-1}{n^4+n^2+1}|$$ for any $$n>0$$, and then by the comparation test we get that since $$\\sum_{n=1}^{\\infty}{\\frac{1}{n^2}}$$ converges, we have that $$\\sum_{n=1}^{\\infty}{|\\frac{n^2-n-1}{n^4+n^2+1}|}$$ converges too, and by a theorem we have that since $$\\sum_{n=1}^{\\infty}{|\\frac{n^2-n-1}{n^4+n^2+1}|}$$ $$\\implies$$ $$\\sum_{n=1}^{\\infty}{\\frac{n^2-n-1}{n^4+n^2+1}}$$ converges. Is that right?\n\n\u2022 Yeah, a direct comparison test works for this. You were right to compare the following: $$\\frac{1}{n^2}>\\left|\\frac{n^2-n-1}{n^4+n^2+1}\\right|$$ Now because the $$\\sum_{n=1}^{\\infty} \\frac{1}{n^2}$$ converges by the p-test, you can conclude that the original series also converges. Mar 16, 2021 at 22:22\n\u2022 @ObsessiveInteger So my attempt is right? but can you show me how I satisfy that $\\frac{1}{n^2}>|\\frac{n^2-n-1}{n^4+n^2+1}|$. for $\\frac{1}{n^2}>\\frac{n^2-n-1}{n^4+n^2+1}$ it's easy since we get by algebra that $n^3+1>0$ for $n>0$. However, for the negative it's not that easy. Mar 16, 2021 at 22:31\n\u2022 You could make an argument by saying that the numerator of $\\frac{n^2-n-1}{n^4+n^2+1}$ grows smaller faster compared to that of $\\frac{1}{n^2}$ and similarly, the denominator of $\\frac{n^2-n-1}{n^4+n^2+1}$ grows bigger much faster compared to that of $\\frac{1}{n^2}$. This means that $\\frac{n^2-n-1}{n^4+n^2+1}$ is reaching $0$ faster compared to $\\frac{1}{n^2}$, hence, the inequality is true. Of course you can see that graphically as well. Plug in both functions into desmos and you will see. Mar 16, 2021 at 22:41\n\u2022 @ObsessiveInteger seeing that throw desmos is good, but can I actually verify that by using just inequality algebra? and thank you for the answer. Mar 16, 2021 at 22:44\n\u2022 Why don't you do the limit comparison test instead? That way you do not have to worry about proving the inequality. Mar 16, 2021 at 22:45\n\nI continue from where you stopped,\n\nEasy to show that :\n\n$$\\frac{n^2-n-1}{n^4+n^2+1} < \\frac{n^2-1}{n^4+1} < \\frac{n^2}{n^4} < \\frac{1}{n^2}$$\n\nWhen we know that $$\\sum_{n=1}^{\\infty}{\\frac{1}{n^2}}$$ converges.\n\nTherefore, by comparison test :\n\n$$\\underset{converges}{\\underbrace{\\frac{n^2-n-1}{n^4+n^2+1}}} < \\underset{converges}{\\underbrace{\\frac{1}{n^2}}}$$\n\n\u2022 However, what stops us from doing so for just the expression without ||, is that when $n$ has small values then the whole expression is negative, and then you cannot use the comparison test, so we have to show it with ||. Mar 16, 2021 at 22:48\n\u2022 @aasc232 By the limit test we get the equation aims to zero so can see that only the first element in the sum has negative value do not forget is the sum of this equation .\n\u2013\u00a0ATB\nMar 16, 2021 at 22:54\n\u2022 Can you remind me the limit test? Mar 16, 2021 at 22:56\n\u2022 And the comparison test is when you have $0\\le a_k\\le b_k$ for all $k>0$, with the first elements of $a_k$ included, then if $b_k$ converges we get that $a_k$ converges too. So since we have that $a_k$ isn't positive for all $k>0$ we cannot use it. That is what I tried telling you. Isn't right? Mar 16, 2021 at 23:01\n\u2022 @aasc232 Do the $n=1$ term separately $$|\\frac{1^2-1-1}{1^{4}+1^2+1}|<\\frac{1}{1}$$ and then ATB's answer applies for all $n>1$ as $\\frac{n^{2}-n-1}{n^{4}+n^{2}+1}$ is positive for $n>1$. Of course, a finite number of terms does not determine the convergence and can be disregarded in general when checking convergence\n\u2013\u00a0user649348\nMar 16, 2021 at 23:03\n\nWe are trying to prove that the series, $$\\sum_{n=1}^{\\infty}{\\frac{n^2-n-1}{n^4+n^2+1}}$$ converges.\n\nLet us use the Limit Comparison Test. That is, if $$a_n>0$$ and $$b_n>0$$, then if $$\\lim_{n\\to \\infty}\\frac{a_n}{b_n}=L$$ for some real number $$L$$. Then, if $$b_n$$ converges, by the limit comparison test, $$a_n$$ must also converge.\n\nLet $$a_n=\\frac{n^2-n-1}{n^4+n^2+1}$$ and we say $$b_n=\\frac{1}{n^2}$$. Then, $$\\lim_{n\\to \\infty} \\frac{\\frac{n^2-n-1}{n^4+n^2+1}}{\\frac{1}{n^2}}\\\\ \\lim_{n\\to \\infty}\\left(\\frac{(n^2-n-1)(n^2)}{n^4+n^2+1}\\right)\\\\ \\lim_{n\\to \\infty}\\frac{n^4-n^3-n^2}{n^4+n^2+1}\\cdot \\frac{\\frac{1}{n^4}}{\\frac{1}{n^4}}\\\\ \\lim_{n\\to \\infty} \\left(\\frac{{1-\\frac{1}{n}}-\\frac{1}{n^2}}{1+\\frac{1}{n^2}+\\frac{1}{n^4}}\\right)\\\\ \\lim_{n\\to \\infty}\\frac{1-0-0}{1+0+0}=\\frac{1}{1}=1$$\n\nHence, we have shown that the $$\\lim_{n\\to \\infty}\\frac{a_n}{b_n}=L$$ converges because it equals some constant; in this case, $$1$$.\n\nNow, because $$\\sum_{n=1}^{\\infty}\\frac{1}{n^2}$$ converges by the p-test, (since $$p=2>1$$), we can conclude that the series, $$\\sum_{n=1}^{\\infty}{\\frac{n^2-n-1}{n^4+n^2+1}}$$ must also converge by the limit comparison test.\n\n\u2022 Ohhh I didn't try using this, in this way. Thank you! Mar 16, 2021 at 23:03\n\u2022 One last thing, by limit comparison test we have to show that the sequence queue $\\frac{n^2\u2212n\u22121}{n^4+n^2+1}$ is positive. How do we show that? or you saying we disregard the first negative elements of the sum?' Mar 16, 2021 at 23:12\n\u2022 Technically you do not have to show that, note that the sum $\\sum_{n=1}^{\\infty}$ starts at $n=1$, so we are only considering values for $n$ that are positive. Notice that for values greater than $1$, $\\frac{n^2-n-1}{n^4+n^2+1}$ can never be negative. Mar 16, 2021 at 23:19\n\u2022 So you are saying that becuase the only negative fraction of the sum is $-\\frac{1}{3}$ then there exist elements that if we sum them we get a positive number bigger then $\\frac{1}{3}$? Mar 16, 2021 at 23:24\n\u2022 Thank you a lot! Mar 16, 2021 at 23:29", "date": "2022-05-22 20:40:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4064710/sequence-queue-converges-or-diverges/4064747", "openwebmath_score": 0.9039461016654968, "openwebmath_perplexity": 215.2440022368874, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274401, "lm_q2_score": 0.8757869997529962, "lm_q1q2_score": 0.8519925420503583}}
{"url": "http://math.stackexchange.com/questions/84650/normal-subgroup-of-automorphisms-of-a-free-group", "text": "Normal subgroup of automorphisms of a free group\n\nLet $F_2=\\langle X,Y\\rangle$ be the free group of rank $2$ and consider $A,B,C\\in Aut(F_2)$ given by: $$A(X,Y)\\mapsto(YX^{-1}Y^{-1},Y^{-1})$$ $$B(X,Y)\\mapsto(X^{-1},Y^{-1})$$ $$C(X,Y)\\mapsto(X^{-1},XY^{-1}X^{-1})$$ The abelianisation of $F_2$ is isomorphic to $\\mathbb{Z}^2$ whose automorphism group is $GL(2,\\mathbb{Z})$. Thus we get an induced surjective map $m\\colon Aut(F_2)\\to GL(2,\\mathbb{Z})$ which maps an element in $Aut(F_2)$ to the matrix associated to the corresponding element in $Aut(\\mathbb{Z}^2)$. I would like to show that the subgroup $\\langle A,B,C\\rangle$ is normal in $Aut(F_2)$ by proving that $m(\\langle A,B,C\\rangle)$ is the centre of $GL(2,Z)$. So far I wasn't able to have any good idea... could you help me? Thanks, bye.\n\n-\nJust calculate the matrices: m(A) = m(B) = m(C) = -1 \u2013\u00a0 Jack Schmidt Nov 22 '11 at 18:22\nYou may want to use \\langle and \\rangle instead of < and >; the spacing is correct for the former. \u2013\u00a0 Arturo Magidin Nov 22 '11 at 18:27\nhow did you calculate explicitly the matrices? $A$,$B$ and $C$ do not belong to $Aut(\\mathbb{Z}^2)$ so I should compute their image in the quotient, first. Do I have to write how, say $A$, acts on the cosets $F_2'X$ and $F_2'Y$ in $F_2/F_2'$? P.S. I did use \\langle and \\rangle! :) \u2013\u00a0 fatoddsun Nov 22 '11 at 19:22\n@fatoddsun: Since $[F_2,F_2]$ is fully invariant, any homomorphism $f\\colon F_2\\to F_2$ induces a homomorphism $f_{\\rm ab}\\colon \\mathbb{Z}^2\\to\\mathbb{Z}^2$ by $f_{\\rm ab}(\\overline{X}) = \\overline{f(X)}$ and $f_{\\rm ab}(\\overline{Y}) =\\overline{f(Y)}$. So just \"abelianize\" the image. \u2013\u00a0 Arturo Magidin Nov 22 '11 at 20:24\n@fatoddsun: Note, however, that having the image be normal is not sufficient; that will only show that the preimage under the abelianization map $\\mathrm{Aut}(F_2)\\to\\mathrm{GL}(2,\\mathbb{Z})$ is normal, and it's not immediately clear to me that this preimage is just $\\langle A,B,C\\rangle$. \u2013\u00a0 Arturo Magidin Nov 22 '11 at 20:57\n\nArturo Magidin's excellent answer explains how to show that $m(\\langle A,B,C\\rangle)=\\{\\pm 1\\}$, and also why this does not in general imply that your subgroup is normal. However, I want to explain why in this case it is true that $\\langle A,B,C\\rangle$ is the full preimage of $\\{\\pm 1\\}$, and thus is normal.\n\nThe key is a theorem of Nielsen (1) on what is now called $IA_n$: this is by definition the kernel of the natural surjection $Aut(F_n)\\to GL_n\\mathbb{Z}$, so it fits into a short exact sequence\n\n$$1\\to IA_n\\to Aut(F_n)\\to GL_n\\mathbb{Z}\\to 1$$\n\nThis group of automorphisms which are the Identity on the Abelianization has been well-studied, starting with Magnus (2), who proved that $IA_n$ is finitely generated by giving an explicit set of generators: type (A) which for given $i$ and $j$ conjugates $x_i$ by $x_j$ and fixes all other generators, and type(B) which for given $i$, $j$, and $k$ multiplies $x_i$ by $x_jx_kx_j^{-1}x_k^{-1}$ and fixes all other generators.\n\nBut anyway, we are interested here in $IA_2$, and Nielsen proves in (1) that $IA_2$ is just $Inn(F_2)$, the group of automorphisms obtained by conjugation. This is isomorphic to $F_2$ and generated by $conj_X\\colon X\\mapsto X, Y\\mapsto XYX^{-1}$ and by $conj_Y\\colon X\\mapsto YXY^{-1}, Y\\mapsto Y$.\n\nWe know that your group $\\langle A,B,C\\rangle$ subjects to $\\{\\pm 1\\}$, so to show it is the full preimage of $\\{\\pm 1\\}$ it suffices to show that $\\langle A,B,C\\rangle$ contains the full preimage of $\\{1\\}$, which is $IA_2$. But we can compute by hand that $AB=conj_Y$ and $CB=conj_X$. So by Nielsen's theorem your group contains $IA_2$, and thus fits into a short exact sequence\n\n$$1\\to IA_2\\to \\langle A,B,C\\rangle\\to \\{\\pm 1\\}\\to 1$$\n\nIn particular, your subgroup is normal in $Aut(F_2)$.\n\nFinally, here is another method of proving that $\\langle A,B,C\\rangle$ is normal which does not require quoting any papers written in German. We know that the elementary Nielsen transformations provide a generating set for $Aut(F_n)$. This generating set becomes particularly simple in the case when $n=2$: we have three generators $R\\colon X\\mapsto X^{-1}, Y\\mapsto Y$, $S\\colon X\\mapsto Y, Y\\mapsto X$, and $T\\colon X\\mapsto XY,Y\\mapsto Y$. To show that $\\langle A,B,C\\rangle$ is normal in $Aut(F_2)$ means that it is invariant under conjugation under any element of $Aut(F_2)$. But since $R,S,T$ generate $Aut(F_2)$, it is enough just to show that $\\langle A,B,C\\rangle$ is invariant under conjugation by $R$, by $S$, and by $T$.\n\nTo do this for $R$, for example, you just need to compute the conjugates $RAR^{-1}$, $RBR^{-1}$, and $RCR^{-1}$ and show that each can be written as a combination of $A$, $B$, and $C$. In this case this is quite easy: we have $RAR^{-1}=A$, $RBR^{-1}=B$, and $RCR^{-1}=C^{-1}$. Now all you have to do is do the same for $SAS^{-1}$, $SBS^{-1}$, $SCS^{-1}$, $TAT^{-1}$, $TBT^{-1}$, and $TCT^{-1}$, and you've proved that $\\langle A,B,C\\rangle$ is normal!\n\n(1) Nielsen, Die Isomorphismen der allgemeinen unendlichen Gruppe mit zwei Erzeugenden, Math. Ann. 78 (1964), 385-397\n\n(2) Magnus, \u00dcber n-dimensinale Gittertransformationen, Acta Math. 64 (1935), 353-367.\n\n-\nNice; thanks for filling the gap! \u2013\u00a0 Arturo Magidin Nov 22 '11 at 21:39\n\nIf $G$ is a group, $N$ is a normal subgroup of $G$, and $f\\colon G\\to G$ is a homomorphism with $f(N)\\subseteq N$, then $f$ induces a homomorphism $\\overline{f}\\colon G/N\\to G/N$ by $\\overline{f}(gN) = f(g)N$. To see this, note that $\\widehat{f}\\colon G\\to G/N$ given by $\\widehat{f}(g) = gN$ is a homomorphism, and since $f(N)\\subseteq N$, then $N$ is contained in the kernel of $\\widehat{f}$, so $\\widehat{f}$ factors through $G/N$.\n\nSince $[F_2,F_2]$ is fully invariant, this argument will work for $A$, $B$, and $C$.\n\nFor $A$, $x$ is mapped to $yx^{-1}y^{-1}$, which gets mapped to $\\overline{x}^{-1}$ in $F_2^{\\rm ab}$; $y$ is mapped to $\\overline{y}^{-1}$. So the induced map $A_{\\rm ab}\\colon F_2^{\\rm ab}\\to F_{2}^{\\rm ab}$ sends $(a,b)$ to $(-a,-b)$.\n\nThis is the same as what $B_{\\rm ab}$ does; again the same thing as $C_{\\rm ab}$ does. So the image of the subgroup generated by $A$, $B$, and $C$ in $\\mathrm{GL}(2,\\mathbb{Z})$ is the subgroup generated by their images, which is just the subgroup generated by $-I$.\n\nNote, however, that the fact that $m(\\langle A,B,C\\rangle) = \\{I,-I\\}$ does not show, by itself, that $\\langle A,B,C\\rangle$ is normal in $\\mathrm{Aut}(F_2)$; it shows that the pre-image is normal. The preimage is the subgroup of $\\mathrm{Aut}(F_2)$ generated by $A$, $B$, $C$, and all automorphisms that induce the identity on $F_2^{\\rm ab}$. It is was not clear to me whether this is just $\\langle A,B,C\\rangle$. For this, see Tom Church's great answer.\n\n-\nThank you very much guys! \u2013\u00a0 fatoddsun Nov 23 '11 at 18:23", "date": "2014-08-29 03:21:10", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/84650/normal-subgroup-of-automorphisms-of-a-free-group", "openwebmath_score": 0.962099552154541, "openwebmath_perplexity": 102.51633505472043, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307708274402, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8519925373194535}}
{"url": "https://www.physicsforums.com/threads/deriving-equations-of-a-parabola.237835/", "text": "# Deriving Equations of a Parabola\n\n1. May 29, 2008\n\n### PFStudent\n\nHey,\n\n1. The problem statement, all variables and given/known data\nHow do you derive the equations of the parabola from the general equation of a Conic Section?\n\n2. Relevant equations\nGeneral Equation of a Conic Section,\n$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$\n\nWhere (for a parabola),\n$${{\\{}A,B,C,D,E,F{\\}}}{{\\,}{\\,}{\\,}}{\\in}{{\\,}{\\,}{\\,}}{\\mathbb{R}}$$\n$${{\\{}A, C{\\}}}{{\\,}{\\,}{\\,}}{\\neq}{{\\,}{\\,}{\\,}}{0}$$\n$${{{B}^{2}}-{{4}{A}{C}}} = {0}$$\n\n3. The attempt at a solution\nFrom the general equation of a Conic Section,\n$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$\n\nHow do I derive the following formulas for a parabola:\nGeneral Form for a Parabola,\n$${{{{\\left(}{{{H}{x}}+{{I}{y}}}{\\right)}}^{2}}+{{J}{x}}+{{K}{y}}+{L}} = {0}$$\nWhere,\n$${{{I}^{2}}-{{4}{H}{J}}} = {0}$$\n\nAnalytic Geometry Equations,\nVertical Axis of Symmetry\n$${{{\\left(}x-h{\\right)}}^{2}} = {{{4}{p}}{{{\\left(}y-k}{\\right)}}}$$\n$${y} = {{{a}{{x}^{2}}}+{{b}{x}}+{c}}$$\nHoriztonal Axis of Symmetry\n$${{{\\left(}y-k{\\right)}}^{2}} = {{{4}{p}}{{{\\left(}x-h{\\right)}}}}$$\n$${x} = {{{d}{{y}^{2}}}+{{e}{y}}+{f}}$$\n\nI read that every parabola is a combination of transformations of the parabola, $${{y}={{x}^{2}}}$$; but I'm not quite sure how that helps.\n\nThanks,\n\n-PFStudent\n\nLast edited: May 29, 2008\n2. May 29, 2008\n\n### Dick\n\nI'm not precisely sure of what exactly you are supposed to do, but the only way that the quadratic terms Ax^2+Bxy+Cy^2 can be factored as (Hx+Iy)^2 is that B^2-4AC=0. Just multiply (Hx+Iy)^2 out and identify the terms. Does that help?\n\n3. May 30, 2008\n\n### HallsofIvy\n\nFirst you need to find the \"principal axes\" (axis of symmetry and perpenpendicular to it through the vertex) and rotate the coordinate system so that new x' and y' axes are the principal axes.\n\nThere are two ways to do that, basically using the quadratic terms $Ax^2+ Bxy+ Cx^2$ (By the way, you don't have to put braces, { }, around every term. They are only necessary when you want an entire expression in a particular place. For example, [i t e x]e^{xy+ b}[/i t e x] gives $e^{xy+ b}$ while [i t e x]e^xy+ b[/i t e x] gives $e^xy+ b$.)\n\nRotating the axes by angle $\\theta$, so that the new x'y'- axes are at angle $\\theta$ to the old xy- axex, x and y are given, in terms of the new x', y' variables, by $x= x'cos(\\theta)+ y'sin(\\theta)$ and $y= -x'sin(\\theta)+ y'cos(\\theta). Replace x and y by those in $Ax^2+ Bxy+ Cx^2$ and choose$\\theta[/itex] so that the coefficient of x'y' is 0. Use those formulas with the correct value of $\\theta$ to replace x and y in the entire formula. If it really is a parabola, both the x'y' and y'2 terms should vanish.\n\nThe other way is to write it as a matrix problem:\n$$Ax^2+ Bxy+ Cy^2= \\left[\\begin{array}{cc}x & y \\end{array}\\right]\\left[\\begin{array}{cc}A & \\frac{B}{2} \\\\ \\frac{B}{2} & C\\end{array}\\right]\\left[\\begin{array}{c}x \\\\ y\\end{array}\\right]$$\n\nYou can \"diagonalize\" that matrix (so writing the quadratic without the xy term) by finding the eigenvalues and eigenvectors of the coefficient matrix. The eigenvectors will point along the principal axes. Since this is a parabola, one of the eigenvalues should be 0.\n\n4. May 31, 2008\n\n### PFStudent\n\nHey,\n\nThanks for the replies: Dick and HallsofIvy.\n\nYea, to be honest I had not seen that general form for a parabola before until I saw it on wikipedia, maybe it is wrong? - here is the link, (Scroll down a bit to the section \"General Parabola\").\n\nhttp://en.wikipedia.org/wiki/Parabola\n\nLet me know what you think.\n\nHmm, ok that makes sense, I will try that. However, I thought since a parabola - in its' most general form - could be expressed as a conic section with the constraint that $${{B^2} = {4AC}}$$. Then shouldn't I be able to derive the more specific forms of a parabola from the general equation of a conic section with the constraint that $${{B^2} = {4AC}}$$?\n\nWhere the general equation of a Conic Section is,\n$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$\n\nAlso, yea I tend to use a bit too many braces ({ }), just a habit from first learning $LaTeX$ way back.\n\nThanks,\n\n-PFStudent", "date": "2018-01-21 03:17:32", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/deriving-equations-of-a-parabola.237835/", "openwebmath_score": 0.7917470335960388, "openwebmath_perplexity": 710.62387662584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307747659751, "lm_q2_score": 0.8757869900269366, "lm_q1q2_score": 0.8519925360378661}}
{"url": "https://gmatclub.com/forum/how-many-nails-did-rudy-buy-if-he-purchased-them-at-a-price-of-286281.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 26 Jun 2019, 11:46\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n### Show Tags\n\nUpdated on: 12 Jan 2019, 03:25\n1\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n88% (02:25) correct 12% (02:43) wrong based on 33 sessions\n\n### HideShow timer Statistics\n\nHow many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Originally posted by Nums99 on 12 Jan 2019, 01:42. Last edited by Bunuel on 12 Jan 2019, 03:25, edited 1 time in total. Renamed the topic, edited the question, moved to PS and added the OA. Senior Manager Joined: 15 Jan 2018 Posts: 371 Concentration: General Management, Finance GMAT 1: 720 Q50 V37 Re: How many nails did Rudy buy if he purchased them at a price of$0.25\u00a0 [#permalink]\n\n### Show Tags\n\n12 Jan 2019, 02:12\n1\nHi there,\n\nHere's the solution.\n\nLet the number of nails Rudy purchased be x.\n\nRudy purchased nails at a price of $0.25 per four nails. So, Cost Price of 4 nails =$0.25\nCost Price of 1 nail = = $0.25/4 Cost price of x nails = ($0.25/4)x\n\nAlso given that Rudy sold nails at $0.22 per three nails. So, Selling Price of 3 nails =$0.22\nSelling Price of 1 nail = = $0.22/3 Selling price of x nails = ($0.22/3)x\n\nAlso, Rudy made a profit of $2.60. Profit = Selling Price - Cost Price$2.60 = ($0.22/3)x - ($0.25/4)x = x(0.88 - 0.75)/12\n$2.60 = x(0.13)/12 x = 240 Hence, the the number of nails Rudy purchased is 240. Therefore, the Correct Answer is Option B. 240 _________________ New to GMAT Club or overwhelmed with so many resources? Follow the GMAT Club Study Plan! Not happy with your GMAT score? Retaking GMAT Strategies! Game of Timers - Join the Competition to Win Prizes Senior Manager Status: Manager Joined: 27 Oct 2018 Posts: 365 Location: Egypt Concentration: Strategy, International Business GPA: 3.67 WE: Pharmaceuticals (Health Care) Re: How many nails did Rudy buy if he purchased them at a price of$0.25\u00a0 [#permalink]\n\n### Show Tags\n\n16 Jan 2019, 02:57\nNums99 wrote:\nHow many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Profit = Sales - Cost $$2.6 = \\frac{0.22}{3}n - \\frac{0.25}{4}n = \\frac{0.88n}{12} - \\frac{0.75n}{12}$$ $$\\frac{13}{5} = \\frac{0.13n}{12} = \\frac{13n}{1200}$$ $$n = \\frac{13*1200}{13*5} = 240$$ B _________________ RC Moderator Joined: 24 Aug 2016 Posts: 802 Location: Canada Concentration: Entrepreneurship, Operations GMAT 1: 630 Q48 V28 GMAT 2: 540 Q49 V16 Re: How many nails did Rudy buy if he purchased them at a price of$0.25\u00a0 [#permalink]\n\n### Show Tags\n\n16 Jan 2019, 08:15\nAs we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .\nCost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88\nProfit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25\u00a0 [#permalink]\n\n### Show Tags\n\n25 Feb 2019, 09:26\nNums99 wrote:\nHow many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails\n\n(2.60 * 12)/0.13 = 240 nails (Ans)\n_________________\n\"Please hit +1 Kudos if you like this post\"\n\n_________________\nManish\n\n\"Only I can change my life. No one can do it for me\"\nDirector\nJoined: 04 Dec 2015\nPosts: 740\nLocation: India\nConcentration: Technology, Strategy\nWE: Information Technology (Consulting)\nHow many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 28 Feb 2019, 11:09 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?\n\nA. 300\nB. 240\nC. 180\nD. 160\nE. 120\n\nCost Price of $$4$$ nails $$=  0.25$$\n\nCost Price of $$3$$ nails $$= \\frac{0.25*3}{4} =  0.1875$$\n\nSelling Price of $$3$$ nails $$= 0.22$$\n\nProfit on $$3$$ nails $$= 0.22 - 0.1875 =  0.0325$$\n\nLet $$\"x\"$$ be total number of nails to make profit of $$2.60$$.\n\n$$\\frac{0.0325}{3} = \\frac{2.60}{x}$$\n\n$$0.0325*x =3*2.60$$\n\n$$x = \\frac{3*2.60}{0.0325} = 240$$\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 6680\nLocation: United States (CA)\nRe: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 02 Mar 2019, 10:08 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?\n\nA. 300\nB. 240\nC. 180\nD. 160\nE. 120\n\nWe can let n = the number of nails and create the equation:\n\n(0.22/3)n - (0.25/4)n = 2.6\n\nMultiplying by 12, we have:\n\n0.22 x 4n - 0.25 x 3n = 31.2\n\n0.88n - 0.75n = 31.2\n\n0.13n = 31.2\n\nn = 240\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\nScott@TargetTestPrep.com\n122 Reviews\n\n5-star rated online GMAT quant\nself study course\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nRe: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] 02 Mar 2019, 10:08 Display posts from previous: Sort by # How many nails did Rudy buy if he purchased them at a price of$0.25\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne", "date": "2019-06-26 18:46:57", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/how-many-nails-did-rudy-buy-if-he-purchased-them-at-a-price-of-286281.html", "openwebmath_score": 0.2864758372306824, "openwebmath_perplexity": 11699.412737693603, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9728307668889047, "lm_q2_score": 0.8757869884059267, "lm_q1q2_score": 0.851992527562262}}
{"url": "https://math.stackexchange.com/questions/3593728/aternating-sum-of-an-increasing-sequence-of-positive-integers", "text": "# Aternating sum of an increasing sequence of positive integers\n\nSuppose $$A = (a_n) = (a_1, a_2, a_3, . . .)$$ is an positive, increasing sequence of integers.\n\nDefine an $$A$$- expressible number $$c$$ if $$c$$ is the alternating sum of a finite subsequence of $$A.$$ To form such a sum, choose a finite subset of the sequence $$A,$$ list those numbers in increasing order (no repetitions allowed), and combine them with alternating plus and minus signs. We allow the trivial case of one-element subsequences, so that each an is $$A-$$expressible.\n\nDefinition. Sequence $$A = (a_n)$$ is an \u201calt-basis\u201d if every positive integer is uniquely $$A-$$ expressible. That is, for every integer $$m > 0,$$ there is exactly one way to express $$m$$ as an alternating sum of a finite subsequence of $$A.$$\n\nExamples. Sequence $$B = (2^{n\u22121}) = (1, 2, 4, 8, 16, . . .)$$ is not an alt-basis because some numbers are B-expressible in more than one way. For instance $$3 = \u22121 + 4 = 1 \u2212 2 + 4.$$\n\nSequence $$C = (3^{n\u22121}) = (1, 3, 9, 27, 81, . . .)$$ is not an alt-basis because some numbers (like 4 and 5) are not C-expressible.\n\nAn example of an alt-basis is $$\\{2^n-1\\}=\\{1,3,7,15,31,\\ldots\\}$$\n\nIs there a fairly simple test to determine whether a given sequence is an alt basis?\n\nI have attempted to solve this from a limited knowledge in sequences and have found out various kinds of sequences do not work but fail to see what it is that could make it work.\n\n\u2022 Seems slightly (at least) related to this: encyclopediaofmath.org/index.php/Additive_basis But I think you know already about additive bases. \u2013\u00a0Masterphile Mar 25 at 0:37\n\u2022 See math.stackexchange.com/questions/3579462. (Not an exact duplicate because it doesn't ask for a general test to determine whether a sequence is an alt-basis.) What is the source of this question? \u2013\u00a0joriki Mar 25 at 6:47\n\u2022 Do you know of any alt-bases that are not of the form $\\{b_1,b_2,\\dots,b_k,a_{k+1},a_{k+2},\\dots\\}$ where $A=\\{a_1,a_2,\\dots\\}=\\{2^n-1\\}$ and $\\{b_1,\\dots,b_k\\}$ is some finite increasing set of integers? \u2013\u00a0Vepir Mar 25 at 21:54\n\u2022 For example I believe that $\\{5\\cdot2^{n - 2} + (-1)^{n + 1} 2^{n - 2} - 1\\}$ is an alt-basis. \u2013\u00a0Vepir Mar 25 at 22:50\n\nI can\u2019t answer the question, but I can at least give you a systematic large family of alt-bases.\n\nIf $$A$$ is a finite set of positive integers, let $$S(A)$$ be the set of $$A$$-expressible integers, and let $$S^+(A)$$ be the set of $$A$$-expressible positive integers. Then\n\n$$S(A)=S^+(A)\\cup\\{-a:a\\in S^+(A)\\},$$\n\nand if $$b>\\max A$$, then\n\n$$S^+\\left(A\\cup\\{b\\}\\right)=S^+(A)\\cup\\{b-s:s\\in S^+(A)\\}\\cup\\{b\\}.$$\n\nThus, if $$|A|=n$$, the maximum number of $$A$$-expressible positive integers is $$2^n-1$$, and $$\\max S(A)=\\max A$$.\n\nNow suppose that $$A=\\{a_n:n\\in\\Bbb Z^+\\}$$, where $$a_n for each $$n\\in\\Bbb Z^+$$. For $$n\\in\\Bbb Z^+$$ let $$A_n=\\{a_k\\in A:1\\le k\\le n\\}$$. Then each $$m\\in S(A)$$ is uniquely $$A$$-expressible iff $$|S^+(A_n)|=2^n-1$$ for each $$n\\in\\Bbb Z^+$$. Moreover, $$S^+(A)=\\Bbb Z^+$$ iff for each $$k\\in S^+(A)$$ there is a minimal $$n(k)\\in\\Bbb Z^+$$ such that $$k\\in S^+(A_{n(k)})$$. Note that either $$n(k)=1$$, or $$k\\in S^+(A_{n(k)})\\setminus S^+(A_{n(k)-1})=\\{a_{n(k)}-s:s\\in S^+(A_{n(k)-1})\\}$$.\n\nFor $$n\\in\\Bbb Z^+$$ let\n\n$$a_n=2^n-1=\\underbrace{1\\ldots 1}_n\\text{ in binary},$$\n\nand let $$A=\\{a_n:n\\in\\Bbb Z^+\\}$$. It\u2019s not hard to see that\n\n$$S^+(A_n)=\\{1,\\ldots,2^n-1\\}$$\n\nfor each $$n\\in\\Bbb Z^+$$, so $$A$$ is, as you already observed, an alt-basis. For instance, working in binary, we see that\n\n\\begin{align*} 22&=10110_{\\text{two}}\\\\ &=11111_{\\text{two}}-1111_{\\text{two}}+111_{\\text{two}}-1_{\\text{two}}\\\\ &=31-15+7-1\\\\ &=a_5-a_4+a_3-a_1. \\end{align*}\n\nNow let $$\\ell,m\\in\\Bbb Z^+$$. For $$n=1,\\ldots,\\ell$$ let\n\n$$\\color{red}{a_n^{(\\ell,m)}}=2^ma_n=\\underbrace{1\\ldots 1}_n\\underbrace{0\\ldots 0}_m\\text{ in binary}.$$\n\nFor $$n=\\ell+k$$, where $$k=1,\\ldots,m$$, let\n\n$$\\color{blue}{a_n^{(\\ell,m)}}=2^{m-k}a_n=\\underbrace{1\\ldots 1}_n\\underbrace{0\\ldots 0}_{m-k}\\text{ in binary}.$$\n\nFinally, for $$n>\\ell+m$$ let $$a_n^{(\\ell,m)}=a_n$$, and let $$A_{(\\ell,m)}=\\left\\{a_n^{(\\ell,m)}:n\\in\\Bbb Z^+\\right\\}$$; then $$A_{(\\ell,m)}$$ is an alt-basis.\n\nFor example,\n\n\\begin{align*} A_{(4,2)}&=\\{\\color{red}{4},\\color{red}{12},\\color{red}{28},\\color{red}{60},\\color{blue}{62},\\color{blue}{63},127,\\ldots\\}\\\\ &=\\{\\color{red}{100},\\color{red}{1100},\\color{red}{11100},\\color{red}{111100},\\color{blue}{111110},\\color{blue}{111111},1111111,\\ldots\\}\\text{ in binary}. \\end{align*}\n\nTo verify this it suffices to show that $$S^+\\left(\\left\\{a_n^{(\\ell,m)}:1\\le n\\le \\ell+m\\right\\}\\right)=S^+(A_{\\ell+m})$$. The argument is a bit messy to write out, but the idea is straightforward; I\u2019ll illustrate it with $$A_{(4,2)}$$. First, it\u2019s clear from the discussion of $$A$$ that\n\n\\begin{align*} S^+\\left(\\{4,12,28,60\\}\\right)&=S^+\\left(4\\{1,3,7,15\\}\\right)\\\\ &=4S^+\\left(\\{1,3,7,15\\}\\right)\\\\ &=4\\{1,2,\\ldots,15\\}\\\\ &=\\{4,8,12,\\ldots,60\\}\\\\ &=4S^+(A_4). \\end{align*}\n\nThen\n\n\\begin{align*} S^+(\\{4,&12,28,60,62\\})=\\\\ &4S^+(A_4)\\cup\\left\\{|62-s|:s\\in S^+(\\{4,12,28,60\\})\\right\\}\\cup\\{62\\}=\\\\ &4S^+(A_4)\\cup\\left\\{|62-s|:s\\in\\{4,8,12,\\ldots,60\\}\\right\\}\\cup\\{62\\}=\\\\ &4S^+(A_4)\\cup\\{2,6,10,\\ldots,58,62\\}=\\\\ &\\{2,4,6,8,\\ldots,60,62\\}=\\\\ &2S^+(A_5), \\end{align*}\n\nand a similar calculation shows that $$S^+(\\{4,12,28,60,62,63\\})=S^+(A_6)$$.\n\nI did not collect the set of all alt-bases, but I did find some useful observations, including:\n\nAlt-basis must contain an infinite number of terms of form $$a_k=2^{k}-1,k\\in N\\subseteq\\mathbb N$$.\n\nThe converse does not hold. At the end, I give examples of alt-bases and not-alt-bases in this context.\n\nDo correct me If I missed anything.\n\nLet $$A=\\{a_1,a_2,\\dots\\}$$ such that $$a_1\\lt a_2 \\lt \\dots$$ are positive integers.\n\nDefinition. For $$A$$ to be an \"alt-basis\", we need to have both the \"uniqueness\" and \"completeness\". In other words, every number is expressible in exactly one way via alternating summation of subsets of $$A$$, which are summed in increasing order.\n\nDefinition. A finite (sub)sequence $$A|_n=\\{a_1,\\dots,a_n\\}$$ is an \"alt-prefix\" if every integer in $$[1,2^{n}-1]$$ is uniquely expressible via alternating summation of subsets of $$A|_n$$ when summed in increasing order. The element $$a_n$$ is called an \"anchor element\".\n\nDefinition. \"Anchor sequence\" is a set $$\\mathcal A(A):=\\{a_{n_1},a_{n_2},\\dots\\}$$ of all \"anchor elements\" $$a_{n_1},a_{n_2},\\dots$$\n\nNotice that a set has $$2^n$$ subsets minus the empty set and that every subset can be rearranged in an increasing order. We want to assign a distinct value to each of those subsets via the alternating summation, to have an alt-basis. The alt-prefix is defined to cover exactly those $$2^n-1$$ subsets. It follows that:\n\nLemma. $$A$$ is an alt-basis $$\\iff$$ $$A$$ is a union of alt-prefixes $$A=A|_{n_1}\\cup A|_{n_2}\\cup \\dots$$\n\nThat is, $$A$$ is an alt-basis if and only if there exists a corresponding infinite anchor sequence $$\\mathcal A(A)$$.\n\nWe add two more definitions to write all of this more easily:\n\nDefinition. Let $$s(\\{b_1,\\dots,b_n\\})$$ be the result of the alternating summation of $$b_1\\lt b_2\\lt \\dots\\le b_n$$. Let $$s_+$$ and $$s_-$$ always start the alternating summation with $$+,-$$ respectively. Then $$s_+=-s_-$$. If $$n$$ is odd then $$s=s_+$$ and if $$n$$ is even then $$s=s_-$$. This guarantees $$s\\gt 0$$ because the largest element $$b_n$$ will have a positive sign.\n\nDefinition. Define \"$$n$$-th partial subset-sum set\" of a positive increasing integer sequence $$A$$ as:\n\n$$\\mathcal S_n(A):=\\{s(A_i):A_i\\in\\mathcal P(A|_n)\\}$$\n\nWhere $$\\mathcal P(A|_n)$$ is the set of all subsets of $$A|_n=\\{a_1,a_2,\\dots,a_n\\}$$.\n\nThe set of all \"anchor elements\" $$\\mathcal A(A)=\\{a_{n_1},a_{n_2},\\dots\\}\\subseteq A$$ satisfies $$S_{n_i}=[1,2^{n_i-1}-1]$$ for all $$n_i$$.\n\nCorollary. $$A$$ is an alt-basis if and only if it \"is covered by the anchor sequence\": $$\\max \\mathcal A(A)\\to \\infty$$.\n\nNotice that $$\\max S_n = a_n$$. If $$a_n$$ is an anchor element, then $$\\max S_n = 2^n-1$$. This gives:\n\nProposition. If $$a_n$$ is an anchor element, then $$a_n=2^n-1$$.\n\nThe converse does not hold. For example, in $$\\{1,4,7\\}$$ the $$a_3=7=2^3-1$$ but $$a_3$$ is not an anchor element, because $$S_3=\\{1,3,4,6,7\\}\\ne[1,7]$$.\n\nExample $$1$$. It is not hard to see that $$\\mathcal A(\\{2^n-1\\})=\\{2^n-1\\}$$. This is because:\n\n\u2022 $$S_1=\\{(+1)\\}$$ $$\\implies$$ $$a_1$$ is an anchor element.\n\n\u2022 $$S_2=\\{(+1),(-1+3),(3)\\}$$ $$\\implies$$ $$a_2$$ is an anchor element.\n\n\u2022 $$S_3=\\{(+1),(-1+3),(3),(-3+7),(+1-3+7),(-1+7),(7)\\}$$ $$\\implies$$ $$a_3$$ is an anchor element.\n\n\u2022 $$\\dots$$ proceed via induction to show every $$a_n$$ is an anchor element.\n\nSince $$\\mathcal A(\\{2^n-1\\})$$ exists and covers the entire $$\\{2^n-1\\}$$, the $$\\{2^n-1\\}$$ is an alt-basis.\n\nExample $$2$$. The $$\\mathcal A(\\{n\\})=\\{1\\}$$ does not cover the entire $$\\{n\\}$$, hence $$\\{n\\}$$ is not an alt-basis.\n\nIt is not hard to see that $$\\max S_n = n \\lt 2^n-1\\implies a_n$$ is not an anchor element, for every $$n\\gt 1$$.\n\nExample $$3.$$ We construct an alt-basis where every $$2$$nd element is an anchor element.\n\n$$A=\\begin{cases} 2^n-1, & n\\text{ is even} \\\\ 2^n+2^{n-1}-1, & n\\text{ is odd} \\end{cases}$$\n\nUse an inductive argument. Assume $$n=2k$$, $$a_{n}=2^{n}-1$$ is an anchor element, which means we have uniquely constructed all $$I_0=[1,2^n-1]=S_{n}$$ elements. Now we can subtract numbers in this interval from $$a_{n+1}$$ to see that:\n\n\u2022 $$a_{n+1}=2^{n+1}+2^{n}-1$$ will cover $$I_1=[a_{n+1}-a_{n}, a_{n+1}]=[2^{n+1},2^{n+1}+2^{n}-1]$$\n\nHere we see that $$I_0\\cup I_1 \\ne [1,2^{n+1}-1]$$ $$\\implies$$ $$a_{n+1}$$ is not an anchor elemetn.\n\nTo see that $$a_{n+2}=2^{n+2}-1$$ is an anchor element, lets see what will we cover with it:\n\n\u2022 $$a_{n+2}$$ combined with $$I_0$$ will cover $$I_2=[a_{n+2}-a_{n}, a_{n+2}]=[2^{n+1}+2^{n},2^{n+2}-1]$$\n\n\u2022 $$a_{n+2}$$ combined with $$I_1$$ will cover $$I_3=[a_{n+2}-a_{n+1},a_{n+2}-2^{n+1}]=[2^n,2^{n+1}-1]$$\n\nNow observe $$I=I_0\\cup I_3\\cup I_1\\cup I_2$$ is equal to:\n\n$$I=[1,2^n-1]\\cup[2^n,2^{n+1}-1]\\cup[2^{n+1},2^{n+1}+2^{n}-1]\\cup[2^{n+1}+2^{n},2^{n+2}-1]=[1,2^{n+2}-1]$$\n\nImplying $$a_{n+2}$$ covers $$I=[1,2^{n+2}-1]=S_{n+2}$$, $$\\implies$$ $$a_{n+2}$$ is an anchor.\n\nIt is not hard to check base cases $$n=1,2$$, and we are done. We have:\n\n$$\\mathcal A\\left(\\left.\\begin{cases} 2^n-1, & n\\text{ is even} \\\\ 2^n+2^{n-1}-1, & n\\text{ is odd} \\end{cases}\\right\\}\\right)=\\{2^{2n}-1\\}$$\n\nSo we have an alt-basis $$A$$.\n\nExample $$4$$. It is not hard to show that:\n\n$$A=\\{2^k,2^k+1,2^k+3,2^k+7,\\dots,2^k+2^{k}-1,2^{k+2}-1,2^{k+3}-1,\\dots\\}$$\n\nIs an alt-basis for every $$k=0,1,2,\\dots$$, whose anchors are all elements $$a_n,n\\gt k$$.\n\nExample $$5$$. The sequence of natural, triangular, tetrahedral,... numbers, or in general, any diagonal of the pascals triangle, is not an alt basis.\n\nThis is because for every fixed $$d$$, there exists $$n_0$$, such that for all $$n\\ge n_0$$, we have $$\\binom{n+d-1}{d}<2^n$$ implying that $$\\max S_n\\lt 2^n-1$$ for all $$n\\ge n_0$$. This implies the sequence of anchors has at most $$n_0$$ elements, implying $$\\max\\mathcal A(A)\\lt \\infty$$, hence we do not have an alt-basis becuase of inevetable duplicates.\n\n\u2022 I think => part of the initial lemme requires proof. \u2013\u00a0balcinus Mar 31 at 10:07", "date": "2020-04-10 13:32:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3593728/aternating-sum-of-an-increasing-sequence-of-positive-integers", "openwebmath_score": 0.9273799061775208, "openwebmath_perplexity": 1235.820987497876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777180969715, "lm_q2_score": 0.8633916187614823, "lm_q1q2_score": 0.8519751476343344}}
{"url": "https://math.stackexchange.com/questions/1822583/what-is-the-sum-of-the-81-products-in-the-9-times-9-multiplication-grid/1822592", "text": "What is the sum of the $81$ products in the $9 \\times 9$ multiplication grid?\n\nWhat is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.\n\nSuppose the 9 \u00d7 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?\n\n\u2022 I asked a question that used this result (and has a nice, clean answer!) several years ago: See MSE 226983 for more! \u2013\u00a0Benjamin Dickman Jun 12 '16 at 3:29\n\u2022 (1+2+...+8+9)^2 \u2013\u00a0Evorlor Jun 12 '16 at 18:45\n\nThe sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$\n\nThen the next row would be $(2+4+6+8+10+12+14+16+18) = 2\\times45 = 90$\n\nSo the top two rows sum to $(1+2)\\times 45 = 135$\n\nThen it becomes obvious that the full sum of the products is the product of the sums, ie. $45\\times45 = 2025$\n\nThe sum is essentially $$\\sum_{a=1}^9 \\sum_{b=1}^9 ab =\\sum_{a=1}^9 a \\sum_{b=1} ^9 b=\\sum_{a=1}^9 a \\frac{9\\cdot 10}{2}=\\left(\\frac{9\\cdot 10}{2}\\right)^2$$\n\nWe want to find the following: $$\\sum_{i=1}^9 \\sum_{j=1}^9 ij$$ Factor out the $i$ from the first summation: $$\\sum_{i=1}^9 i\\left(\\sum_{j=1}^9 j\\right)$$ Note that $\\sum_{j=1}^9 j=45$. $$\\sum_{i=1}^9 i\\cdot 45$$ Factor out the $45$: $$45\\left(\\sum_{i=1}^9 i\\right)$$ Note that $\\sum_{i=1}^9 i=45$. $$45\\cdot 45=2025$$\n\nUse the distributive principle. The sum of the entries in the $2$ column is $2$ times the sum of the numbers $1$ through $9$, so the sum of all the entries is the sum of the numbers $1$ through $9$ times (what?)\n\nHint $\\quad \\begin{eqnarray} &\\color{#c00}{1+2+3}\\\\ +\\ &\\color{#0a0}{2+4+6}\\\\ +\\ &\\color{blue}{3+6+9}\\end{eqnarray}$ $\\quad =\\quad \\begin{eqnarray} &\\color{#c00}1\\,(1+2+3)\\\\ +\\ &\\color{#0a0}2\\,(1+2+3)\\\\ +\\ &\\color{blue}3\\,(1+2+3)\\end{eqnarray}$ $\\quad =\\quad (\\color{#c00}1 + \\color{#0a0}2 + \\color{blue} 3)(1+2+3)\\ \\ =\\ \\ 6\\times 6$\n\nLet's prove by induction that the sum of an ${n}\\times{n}$ grid is $\\frac{n^4+2n^3+n^2}{4}$:\n\nFirst, show that this is true for $n=1$:\n\n$\\sum\\limits_{x=1}^{1}\\sum\\limits_{y=1}^{1}xy=\\frac{1^4+2\\cdot1^3+1^2}{4}$\n\nSecond, assume that this is true for $n$:\n\n$\\sum\\limits_{x=1}^{n}\\sum\\limits_{y=1}^{n}xy=\\frac{n^4+2n^3+n^2}{4}$\n\nThird, prove that this is true for $n+1$:\n\n$\\sum\\limits_{x=1}^{n+1}\\sum\\limits_{y=1}^{n+1}xy=$\n\n$\\color\\red{\\sum\\limits_{x=1}^{n}\\sum\\limits_{y=1}^{n}xy}+\\left(\\sum\\limits_{x=1}^{n}x(n+1)\\right)+\\left(\\sum\\limits_{y=1}^{n}y(n+1)\\right)+(n+1)(n+1)=$\n\n$\\color\\red{\\frac{n^4+2n^3+n^2}{4}}+\\left(\\sum\\limits_{x=1}^{n}x(n+1)\\right)+\\left(\\sum\\limits_{y=1}^{n}y(n+1)\\right)+(n+1)(n+1)=$\n\n$\\frac{n^4+2n^3+n^2}{4}+(n+1)\\left(\\sum\\limits_{x=1}^{n}x\\right)+(n+1)\\left(\\sum\\limits_{y=1}^{n}y\\right)+(n+1)(n+1)=$\n\n$\\frac{n^4+2n^3+n^2}{4}+(n+1)\\left(\\frac{n^2+n}{2}\\right)+(n+1)\\left(\\frac{n^2+n}{2}\\right)+(n+1)(n+1)=$\n\n$\\frac{n^4+2n^3+n^2}{4}+\\frac{n^3+2n^2+n}{2}+\\frac{n^3+2n^2+n}{2}+n^2+2n+1=$\n\n$\\frac{n^4+2n^3+n^2}{4}+n^3+2n^2+n+n^2+2n+1=$\n\n$\\frac{n^4+2n^3+n^2}{4}+n^3+3n^2+3n+1=$\n\n$\\frac{n^4+6n^3+13n^2+12n+4}{4}=$\n\n$\\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4}$\n\nPlease note that the assumption is used only in the part marked red.\n\nTherefore, the sum of a ${9}\\times{9}$ grid is $\\frac{9^4+2\\cdot9^3+9^2}{4}=2025$.\n\nI will focus on an intuition for finding the result. Once you have the formula, other answers are perfect in providing the means for proving it technically.\n\nLet us start with a small size version. If you take the top $3\\times 3$ table only with numbers $1$, $2$, $3$ (to save some electrons), the sum of products is:\n\n$$s= 1\\times 1 +1\\times 2 + 1\\times 3 + 2\\times 1 +2\\times 2 + 2\\times 3 + 3\\times 1 +3\\times 2 + 3\\times 3 \\,.$$ You see that you can rewrite this sum of nine products as: $$s = (1+2+3)\\times (1+2+3)\\,,$$ using the distribution of multiplication over additions.\n\nThis form is more interesting, because you can see a pattern. The sum of the first integers, multiplied by itself.\n\nIt can be reassuring to verify it works: you get $36$, which you can check by hand ($6+12+18$). You can test it with the $2\\times2$ or $4\\times4$ matrix, to verify it is not a coincidence.\n\nThe hint is apparent in the squared shape of the table. The same works for bigger tables too. Each product of $i$ and $j$ in this order is contained once and only once in the product $(1,\\ldots,i,\\ldots,n)\\times(1,\\ldots,j,\\ldots,n)$.\n\nNow you have to find the sum of integers. If you forget the generic expression for the sum of the $n$ first integer, have this figure in mind:\n\nYou pack together two triangles which give you a $n\\times (n+1)$ rectangle, of area $n(n+1)$, the double of the area of each triangle.\n\nSo finally the answer is $\\left(n(n+1)/2\\right)^2$. With $n=9$, you get $2025$.\n\nOnce you have the method, which is relatively simple, you can easily turn it into a more formal proof, via induction for instance, as given in other answers.\n\n\u2022 The link to \"I like visual proofs\" is dead. Besides that, I think the name \"visual proof\" is not suitable here. Perception may be betraying, so visual remarks should be backed up by some other argument. This makes it not related to \"visual\"! \u2013\u00a0Jasper Jun 12 '16 at 16:23\n\u2022 @Jasper I have modified the answer. The visual aspect was just a reminder for a formula given in other answers \u2013\u00a0Laurent Duval Jun 12 '16 at 16:49", "date": "2019-05-26 13:18:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1822583/what-is-the-sum-of-the-81-products-in-the-9-times-9-multiplication-grid/1822592", "openwebmath_score": 0.8559382557868958, "openwebmath_perplexity": 180.83423429377422, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986777175525674, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8519751394653936}}
{"url": "https://brilliant.org/discussions/thread/absolute-value-2/", "text": "# Absolute Value\n\n## Definition\n\nFor a real number, the absolute value of $$x$$, denoted $$\\lvert x \\rvert$$, is defined as\n\n$\\left|x \\right| = \\begin{cases} x, & \\text{if } x \\geq 0 \\\\ -x, & \\text{if } x < 0. \\end{cases}$\n\n## Technique\n\n### Evaluate $\\left| 3\\left( 17-55 \\right) \\right|$.\n\n\\begin{aligned} \\left| 3\\left( 17-55 \\right) \\right| &= \\left| 3\\left( -38 \\right) \\right| \\\\ &= \\left| -114 \\right| \\\\ &= 114 _\\square \\end{aligned}\n\n### If $x=\\frac{5+\\sqrt{11}i}{2}$ and $y=\\frac{5 -\\sqrt{11}i}{2}$, what is the value of $\\left| x + 3y \\right|^2$?\n\n\\begin{aligned} \\left| x + 3y \\right|^2 &= \\left| \\frac{5+\\sqrt{11}i}{2} + 3\\left( \\frac{5 -\\sqrt{11}i}{2} \\right) \\right|^2 \\\\ &= \\left| \\frac{5+\\sqrt{11}i + 15 -3\\sqrt{11}i}{2} \\right|^2 \\\\ &= \\left| \\frac{20-2\\sqrt{11}i}{2} \\right|^2 \\\\ &= \\left| 10 - \\sqrt{11}i \\right|^2 \\\\ &= \\left( \\sqrt{10^2 + \\sqrt{11}^2} \\right)^2 \\\\ &= \\sqrt{111}^2 \\\\ &= 111 _\\square \\end{aligned}\n\n### What is the absolute value of $2 - 10a$?\n\nThe answer depends on the value of $a$. From the definition above, we have\n\n$\\left|2-10a \\right| = \\begin{cases} 2-10a, & \\text{if } 2-10a \\geq 0 \\\\ -(2-10a), & \\text{if } 2-10a < 0 \\end{cases}$\n\nSolving the inequality on the right, we obtain\n\n\\begin{aligned} 2-10a &\\geq 0 \\\\ 2 & \\geq 10a \\\\ \\frac{1}{5} &\\geq a. \\end{aligned}\n\nIt follows that $2-10a \\geq 0 \\Longleftrightarrow a \\leq \\frac{1}{5}.$\n\nTherefore,\n\n$\\left|2-10a \\right| = \\begin{cases} 2-10a, & \\text{if } a \\leq \\frac{1}{5} \\\\ 10a - 2, & \\text{if } a > \\frac{1}{5} \\end{cases} _\\square$\n\nNote by Arron Kau\n7\u00a0years ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nthank you very much :-)\n\n- 6\u00a0years, 11\u00a0months ago\n\nI like this format of note, I think I'll Start Share So !\n\nExplanation More Exercises Solved !\n\n- 7\u00a0years ago\n\nI didn't quite understand the problem having $i$. At one stage the $i$ disappears. Can you please explain?\n\n- 7\u00a0years ago\n\nIt is because the absolute value of a complex number ($a + bi$) is defined as $\\sqrt{a^2 + b^2}$ , i.e., $|a + bi|$ = $\\sqrt{a^2 + b^2}$. In the above question, $|10 - \\sqrt11|$ = $\\sqrt{10^2 + \\sqrt{11}^2}$\n\n- 7\u00a0years ago\n\nDidn't know that, thanks!!\n\n- 7\u00a0years ago\n\nI am not sure but as far as my mind think I can say this is because when the sign of absolute value was removed.Then sign was changed to positive & i(iota) has nothing to do when there is is postive root.So,It would have been removed...\n\n- 7\u00a0years ago\n\n$i$ is defined as $\\sqrt{-1}$ i.e. imaginary unit. So, those were complex numbers (real number + imaginary numbers).\n\nExcellent...\n\n- 7\u00a0years ago\n\nIn ending step ..note that to take modulus of imag eniry number is different than simple modulas\n\n- 6\u00a0years, 5\u00a0months ago", "date": "2021-04-14 01:06:27", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/absolute-value-2/", "openwebmath_score": 1.000007152557373, "openwebmath_perplexity": 2960.517491791547, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9867771770811146, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8519751373397467}}
{"url": "https://probabilityandstats.wordpress.com/2015/12/12/a-randomized-definition-of-the-natural-logarithm-constant/", "text": "# A randomized definition of the natural logarithm\u00a0constant\n\nThe number $e$ is the base of the natural logarithm. It is an important constant in mathematics, which is approximately 2.718281828. This post discusses a charming little problem involving the the number $e$. This number can be represented in many ways. In a calculus course, the number $e$ may be defined as the upper limit of the following integral:\n\n$\\displaystyle \\int_1^e \\frac{1}{t} \\ dt=1$\n\nAnother representation is that it is the sum $\\displaystyle e=\\sum_{n=1}^\\infty \\frac{1}{n!}$. Still another is that it is the limit $\\displaystyle e=\\lim_{n \\rightarrow \\infty} \\biggl( 1+\\frac{1}{n} \\biggr)^n$. According to the authors of a brief article from the Mathematics Magazine [1], these textbook definitions do not give immediate insight about the number $e$ (article can be found here). As a result, students come away from the course without a solid understanding of the number $e$ and may have to resort to rote memorization on what the number $e$ is. Instead, the article gives six probability oriented ways in which the number $e$ can occur. These occurrences of $e$ are more interesting and, according to the authors of [1], can potentially increase students\u2019 appreciation of the number $e$. In two of these six examples (Example 2 and Example 5), the number $e$ is defined by drawing random numbers from the interval $(0,1)$. This post discusses Example 2.\n\n________________________________________________________________________\n\nRandom Experiment\n\nHere\u2019s the description of the random experiment that generates the number $e$. Randomly and successively select numbers from the interval $(0, 1)$. The experiment terminates when the sum of the random numbers exceeds 1. What is the average number of selections in this experiment? In other words, we are interested in the average length of the experiment. According to the article [1], the average length is the number $e$. The goal here is to give a proof of this result.\n\nAs illustration, the experiment is carried out 1 million times using random numbers that are generated in Excel using the Rand() function. The following table summarizes the results.\n\n$\\left[\\begin{array}{rrrrrr} \\text{Length of} & \\text{ } & \\text{ } & \\text{ } & \\text{Relative} \\\\ \\text{Experiment} & \\text{ } & \\text{Frequency} & \\text{ } & \\text{Frequency} \\\\ \\text{ } & \\text{ } & \\text{ } & \\text{ } & \\text{ } \\\\ 2 & \\text{ } & 500777 & \\text{ } & 0.500777 \\\\ 3 & \\text{ } & 332736 & \\text{ } & 0.332736 \\\\ 4 & \\text{ } & 124875 & \\text{ } & 0.124875 \\\\ 5 & \\text{ } & 33465 & \\text{ } & 0.033465 \\\\ 6 & \\text{ } & 6827 & \\text{ } & 0.006827 \\\\ 7 & \\text{ } & 1130 & \\text{ } & 0.001130 \\\\ 8 & \\text{ } & 172 & \\text{ } & 0.000172 \\\\ 9 & \\text{ } & 14 & \\text{ } & 0.000014 \\\\ 10 & \\text{ } & 4 & \\text{ } & 0.000004 \\\\ \\text{ } & \\text{ } & \\text{ } & \\text{ } & \\text{ } \\\\ \\text{Total} & \\text{ } & 1000000 & \\text{ } & \\text{ } \\end{array}\\right]$\n\nThe average length of experiment in these 1 million experiments is 2.717001. Even though the rate of convergence to the number $e$ is fairly slow, the simulated data demonstrates that on average it takes approximately $e$ number of simulations of numbers in the unit interval to get a sum that exceeds 1.\n\n________________________________________________________________________\n\nA proof\n\nLet $U_1,U_2,U_3,\\cdots$ be a sequence of independent and identically distributed random variables such that the common distribution is a uniform distribution on the unit interval $(0,1)$. Let $N$ be defined as follows:\n\n$\\displaystyle N=\\text{min}\\left\\{n: U_1+U_2+\\cdots+U_n>1 \\right\\} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$\n\nThe objective is to determine $E(N)$. On the surface, it seems that we need to describe the distribution of the independent sum $X_n=U_1+U_2+\\cdots+U_n$ for all possible $n$. Doing this may be possible but the result would be messy. It turns out that we do not need to do so. We need to evaluate the probability $P(X_n \\le 1)$ for all $n$. We show that\n\n$\\displaystyle F_n(x)=P(X_n \\le x)=\\frac{x^n}{n!} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (2)$\n\nfor $0 \\le x \\le 1$ and for $n=1,2,3,\\cdots$. This is accomplished by an induction proof. This is true for $n=1$ since $X_1=U_1$ is a uniform distribution. Suppose (2) holds for the integer $n-1$. This means that $\\displaystyle F_{n-1}(x)=P(X_{n-1} \\le x)=\\frac{x^{n-1}}{(n-1)!}$ for $0 \\le x \\le 1$. Note that $X_n=X_{n-1}+U_n$, which is an independent sum. Let\u2019s write out the convolution formula for this independent sum:\n\n\\displaystyle \\begin{aligned} F_n(x)&=\\int_{0}^x F_{n-1}(x-y) \\cdot f_{U_n}(y) \\ dy \\\\&=\\int_{0}^x \\frac{(x-y)^{n-1}}{(n-1)!} \\cdot 1 \\ dy \\\\&=\\frac{1}{(n-1)!} \\int_0^x (x-y)^{n-1} \\ dy \\\\&=\\frac{x^n}{n!} \\end{aligned}\n\nThe above derivation completes the proof of the claim. We now come back to the problem of evaluating the mean of the random variable $N$ defined in (1). First, note that $N>n$ if and only if $X_n=U_1+U_2+\\cdots+U_n \\le 1$. So we have the probability statement $\\displaystyle P(N>n)=F_n(1)=\\frac{1}{n!}$.\n\nAs a result, the following is the probability function of the random variable $N$.\n\n\\displaystyle \\begin{aligned} P(N=n)&=P(N>n-1)-P(N>n) \\\\&=F_{n-1}(1)-F_n(1) \\\\&=\\frac{1}{(n-1)!}-\\frac{1}{n!} \\\\&=\\frac{n-1}{n!} \\end{aligned}\n\nNow evaluate the mean.\n\n\\displaystyle \\begin{aligned} E(N)&=\\sum \\limits_{n=2}^\\infty \\frac{n(n-1)}{n!} \\\\&=\\sum \\limits_{n=2}^\\infty \\frac{1}{(n-2)!} \\\\&=\\sum \\limits_{m=0}^\\infty \\frac{1}{m!} \\\\&=e \\end{aligned}\n\nWith the above derivation, the proof that $e=$ 2.718281828\u2026 is the average number of random numbers to select in order to obtain a sum that exceeds 1 is completed.\n\n________________________________________________________________________\n\nReference\n\n1. Shultz H. S., Leonard B., Unexpected Occurrences of the Number e,Mathematics Magazine, October 1989, Volume 62, Number 4, pp. 269\u2013271.\n\n________________________________________________________________________\n$\\copyright \\ \\text{2015 by Dan Ma}$", "date": "2017-02-26 19:05:05", "meta": {"domain": "wordpress.com", "url": "https://probabilityandstats.wordpress.com/2015/12/12/a-randomized-definition-of-the-natural-logarithm-constant/", "openwebmath_score": 0.943716287612915, "openwebmath_perplexity": 215.07585780412978, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9867771774699747, "lm_q2_score": 0.8633916029436189, "lm_q1q2_score": 0.8519751290039814}}
{"url": "https://math.stackexchange.com/questions/2711754/can-the-product-of-infinitely-many-elements-from-mathbb-q-be-irrational/2711761", "text": "# Can the product of infinitely many elements from $\\mathbb Q$ be irrational?\n\nI know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. Thank you for your answers.\n\n\u2022 Do you know Wallis's product for $\\pi$? Mar 28, 2018 at 13:23\n\u2022 $e=\\lim_{n\\to \\infty}{(1+1/n)^n}$ Mar 28, 2018 at 13:29\n\u2022 @Vasya Correct limit of rationals, but not an infinite product. Mar 28, 2018 at 13:29\n\u2022 @DzamoNorton Not a limit of partial products. The number of factors increases each time, but the factors change: $(1 + 1/2)(1+1/2)$, $(1+1/3)(1 + 1/3)(1+ 1/3)$ and so on. Mar 28, 2018 at 19:51\n\u2022 @toliveira: An \"infinite product\" is not multiplication. It's a limit. Fundamentally, multiplication has two operands; you can inductively extend that to any finite number of operands. But \"any finite number\" does not include \"infinitely many\". Mar 29, 2018 at 15:10\n\nYes, it can.\n\nConsider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and $$b_n = \\frac{a_n}{a_{n-1}}$$ for $n > 1$.\n\nWe then have that $$b_1 b_2 \\cdots b_n = a_1 \\frac{a_2}{a_1} \\frac{a_3}{a_2} \\cdots \\frac{a_n}{a_{n-1}} = a_n.$$\n\nWe thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.\n\n\u2022 Great answer. Very simple. Mar 29, 2018 at 9:57\n\u2022 But it does look like a \"cheat\" since everything is divided out by itself. So, is there an infinite product which does not contain a collection of $\\frac{a_n}{a_n}$ ? both your answer and mohammad Riazi's have this \"cheat\", while Kumar's does not. Mar 29, 2018 at 18:56\n\u2022 @Carl This isn't a cheat. In fact you can rewrite any infinite product in this way. Mar 29, 2018 at 19:23\n\u2022 +1 But I think it needs at least a short explanation (or link to one, maybe the question is already out here too?) that such a sequence \"of non-zero rational numbers which converges to an irrational number\" exists. It is easier too see than the initial question (I think) but I guess some of those who will fins the initial question interesting may not see this as obvious Mar 29, 2018 at 20:08\n\nYes, every irrational number is an infinite product of rationals.\n\nWe can write an infinite sum of rationals as an infinite product of rationals.\n\n\\begin{align} a&=a,\\\\ a+b&=a\\times\\frac {a+b}{a}\\\\ a+b+c &= a \\times \\frac {a+b}{a}\\times\\frac {a+b+c}{a+b}\\\\.\\\\.\\\\.\\\\.\\end{align}\n\nFor example, $$\\sqrt 2 =1.414213....=1+.4+.01+.004+.....=$$\n\n$$1\\times \\frac {1.4}{1}\\times \\frac {1.41}{1.4}\\times\\frac {1.414}{1.41}\\times .....$$\n\n\u2022 Your answer is chosen for a review audit of mine, which I upvoted (+1) because it is simple, generalized and easy to understand. Mar 29, 2018 at 5:10\n\u2022 @Tr\u1ea7nTh\u00facMinhTr\u00ed Thanks for your attention. Mar 29, 2018 at 11:19\n\u2022 I took the liberty to align equalities at '=' sign. Feel free to rollback if you don't like it. Mar 30, 2018 at 7:29\n\u2022 Thanks, I like it Mar 30, 2018 at 11:12\n\nYes!\n\n$\\cfrac{\\pi}{2} = \\cfrac{2}1 \\cfrac 23 \\cfrac 43 \\cfrac 45 \\cfrac 65 \\cfrac 67 \\cdots$\n\n\u2022 This looks cool, but what is the geometric picture to go along with it? I an almost see polygons in there... Mar 28, 2018 at 16:40\n\u2022 @MikeWise I din't understand which polygon you are talking about. Mar 28, 2018 at 16:42\n\u2022 General term is $\\cfrac{2n}{(2n \u2013 1)}\\cfrac{2n}{(2n + 1)}$ Mar 28, 2018 at 16:43\n\u2022 @MikeWise: I was curious about that as well. Here's a short note on how when you draw rectangles of those areas, they add up to a quarter circle in the limit: math.chalmers.se/~wastlund/monthly.pdf Mar 28, 2018 at 17:39\n\u2022 Cool, thanks for that, will print it out and read it on my flight in the morning. :) Mar 28, 2018 at 17:47\n\nToo big to be a comment: it should be noted that the order is more crucial in infinite products than in infinite sums, which is strikingly seen on the example cited many times already:\n\n\\begin{align*}\\cfrac{\\pi}{2}&=\\cfrac{2}1 \\cfrac 23 \\cdot \\cfrac 43 \\cfrac 45 \\cdot\\cfrac 65 \\cfrac 67\\cdot \\ldots\\\\ &= \\cfrac{2^2}{2^2-1}\\cdot \\cfrac{4^2}{4^2-1}\\cdot \\cfrac{6^2}{6^2-1}\\ldots\\\\ \\end{align*} is an infinite product with partials starting at $\\frac43$ and increasing towards $\\frac\\pi 2$ (every factor is greater than $1$), whereas the seemingly identical\n\n\\begin{align*}0&=\\cfrac{2}3 \\cfrac 23 \\cdot \\cfrac 45\\cfrac 45\\cdot\\cfrac 67 \\cfrac 67 \\cdot\\ldots\\end{align*}\n\nstarts below $1$ and decreases, towards $0$. All that happened was a shift of denominators one step to the left.\n\n\u2022 Products are isomorphic to sum of logs, so any order effect that shows up in one can be generated in the other. By allowing the denominators to be shifted independently of the numerators, you are treating the fraction (a/b) as ab^-1, so when you take the log, it's log(a)-log(b), which is an alternating series, and of course order is more important in alternating series than monotonic ones. It's the alternation generated by treating the numerator and denominator separately, not the product, that makes order important. Mar 28, 2018 at 15:00\n\u2022 Allowing the denominators to be shifted independently of the numerators can also be interpreted as replacing each term a/b by two terms, a and 1/b. Thus when you take logs, you get two terms, log a and - log b. Mar 28, 2018 at 22:10\n\u2022 These explanations are of course correct, but the point is that it is not obvious to realise that it can matter, easy to make a mistake here. Mar 29, 2018 at 8:32\n\u2022 I think the claim is not that one shouldn't point out that order matters in an infinite product, but rather that one shouldn't claim that it matters more in infinite products than in infinite sums, since (as @Acccumulation points out) for products of positive real numbers and sums of real numbers it is literally the same phenomenon. Mar 29, 2018 at 22:38\n\u2022 @LSpice when I said that it matters more, what I meant is that one should beware of the intuition that a permutation of terms acting only locally does not affect the final result (you need a violent perturbation of terms to give $\\sum{(-1)^n\\over n}$ a different limit). Of course here what is happening is that the terms are not only reordered, but broken to pieces and the pieces reordered. *This natural way to think of a rational factor as being made of two pieces does not happen with series. * Mar 29, 2018 at 23:44\n\nConsider the Riemann-Zeta Function: $$\\sum_{n=1}^{\\infty}\\frac{1}{n^{s}}=\\prod_{p\\text{ prime}}\\frac{1}{1-p^{-s}}.$$ For $s=2$, the infinite sum on the left is $\\pi^{2}/6$, which is irrational. Thus, $\\pi^{2}/6$ is an infinite product of rationals.\n\nThere is a simple way to obtain any irrational number as an infinite product:\n\n\u2022 take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\\pi$ to $n$ decimals);\n\n\u2022 form the product of the numbers $f_n:=\\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.\n\n$$\\pi=\\prod_{n=0}^\\infty f_n=\\frac{31}{10}\\cdot\\frac{314}{310}\\cdot\\frac{3141}{3140}\\cdot\\frac{31415}{31410}\\cdot\\frac{314159}{314150}\\cdots$$\n\n\u2022 Ooops, I just noticed that my answer is a duplicate.\n\u2013\u00a0user65203\nMar 30, 2018 at 15:29", "date": "2022-07-02 12:31:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2711754/can-the-product-of-infinitely-many-elements-from-mathbb-q-be-irrational/2711761", "openwebmath_score": 0.8812333345413208, "openwebmath_perplexity": 634.1386323894059, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771786365548, "lm_q2_score": 0.863391599428538, "lm_q1q2_score": 0.8519751265425952}}
{"url": "https://www.physicsforums.com/threads/a-gravitation-problem-with-two-masses.748274/", "text": "# A gravitation problem with two masses\n\n1. Apr 11, 2014\n\n### toothpaste666\n\n1. The problem statement, all variables and given/known data\n\nA uniform sphere has mass M and radius r. A spherical cavity (no mass) of radius r/2 is then carved within this sphere (the cavity's surface passes through the sphere's center and just touches the sphere's outer surface). The centers of the original sphere and the cavity lie on a straight line, which defines the x axis.\n\nWith what gravitational force will the hollowed-out sphere attract a point mass m which lies on the x axis a distance d from the sphere's center? [Hint: Subtract the effect of the \"small\" sphere (the cavity) from that of the larger entire sphere.]\n\n2. Relevant equations\n\n$F = G\\frac{m_1m_2}{r^2}$\n\n3. The attempt at a solution\n\nLet mass $M_f$ = the mass of the sphere without the cavity, $M_c$ = the mass of the cavity, and $R_f$ = the distance between the two masses.\n\n$M_f = M - M_c$\n$R_f = d + R_2$\n$F = G\\frac{(M_f) m}{(R_f)^2}$\n\nis this right?\n\n2. Apr 11, 2014\n\n### SammyS\n\nStaff Emeritus\nDid you intend for MC to be negative?\n\nI interpret the Hint to be to calculate the gravitational force that would be exerted on the point mass if the sphere were not hollow, but had the same density. From that, subtract the gravitational force that would be exerted on the point mass by an isolated uniform sphere the size of the cavity, also having the same density.\n\nThat would give MF = M + MC .\n\nWhen you say, \"Rf = the distance between the two masses\", what two masses are you referring to ?\n\n3. Apr 12, 2014\n\n### toothpaste666\n\nHere is the picture. The two masses are the small particle mass on the right and the sphere on the left with the cavity\n\n#### Attached Files:\n\n\u2022 ###### physicsprob.png\nFile size:\n147.4 KB\nViews:\n154\n4. Apr 12, 2014\n\n### haruspex\n\nThe hint does not say subtract the masses, it says subtract the gravitational effects.\nWhat force would be exerted by the complete sphere?\nSuppose we take the \"complement\" of the cavitated sphere, i.e. throw away the larger sphere and instead have just the smaller sphere (at the position of the cavity). What force would that exert?\n\n5. Apr 13, 2014\n\n### toothpaste666\n\nso if M is the mass of the sphere before the cavity was taken out and Mc is the mass of the cavity then we have:\n\n$G\\frac{Mm}{d^2} - G\\frac{M_c m}{(d-r_2)^2}$\n\n6. Apr 13, 2014\n\n### SammyS\n\nStaff Emeritus\nYes, if M is the mass of the sphere before the cavity is taken out. That is as the problem is stated. (So, I believe I was incorrect in my previous post.)\n\nYou can determine Mc in terms of M.\n\n7. Apr 13, 2014\n\n### toothpaste666\n\nIf the mass is evenly distributed wouldn't subtracting the volume be the same as subtracting the mass?\n\n8. Apr 13, 2014\n\n### SammyS\n\nStaff Emeritus\nWhat are you subtracting from what?\n\nWhat is the mass of a sphere the size of the cavity? How is that related to the mass of the large sphere before the cavity is formed in the sphere?\n\n9. Apr 13, 2014\n\n### toothpaste666\n\nSubtracting the volume of the cavity from the volume of the original sphere i mean. Wouldn't the mass be the same as the volume if its evenly distributed?\n\n10. Apr 13, 2014\n\n### SammyS\n\nStaff Emeritus\nMass is be proportional to volume in this case.\n\nWhat fraction of the initial solid sphere's mass is removed to make the cavity ?\n\n11. Apr 13, 2014\n\n### toothpaste666\n\nTo find the volumes let Vc = Volume of cavity:\n\n$r_2 = \\frac{r}{2}$\n\n$V_c = \\frac{4pi(\\frac{r}{2})^3}{3}$\n\n$V_c = \\frac{\\frac{4pi(r^3)}{8}}{3}$\n\n$V_c = \\frac{\\frac{pi(r^3)}{2}}{3}$\n\n$V_c = \\frac{pi(r^3)}{6}$\n\nsince $V_M =\\frac{4pi(r^3)}{3}$\n\nwe have\n\n$V_M = 8V_c$\n\ndoes this get me closer to expressing Mc in terms of M?\n\n12. Apr 14, 2014\n\n### haruspex\n\nThat's it.\n\n13. Apr 14, 2014\n\n### toothpaste666\n\nbut how do i relate that back to mass?\n\n14. Apr 14, 2014\n\n### haruspex\n\nThrough density, which is the same for both.\n\n15. Apr 14, 2014\n\n### toothpaste666\n\n$D = \\frac{M}{V_M} = \\frac{3M}{4(pi)r^3}$\n\n$M_c = D V_c = D \\frac{(pi)r^3}{6} = (\\frac{3M}{4(pi)r^3})(\\frac{(pi)r^3}{6}) = \\frac{M}{8}$\n\nwhich gives\n\n$G\\frac{Mm}{d^2} - G\\frac{Mm}{8(d-r_2)^2}$\n\nhow do we know the density is equal?\n\n16. Apr 14, 2014\n\n### haruspex\n\nSee the OP:\n\n17. Apr 14, 2014\n\nThank you.", "date": "2017-08-24 07:52:09", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/a-gravitation-problem-with-two-masses.748274/", "openwebmath_score": 0.6084936857223511, "openwebmath_perplexity": 1445.7704737128088, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806540875628, "lm_q2_score": 0.8688267864276108, "lm_q1q2_score": 0.8519547385239818}}
{"url": "https://math.stackexchange.com/questions/2400259/limits-and-substitution", "text": "# Limits and substitution\n\nAs in @RobertZ's answer to this question, we often perform substitutions when evaluating limits. For instance, if you're asked to show that $$L = \\lim_{t \\to 0} \\frac{\\sin t^3}{t^3} = 1,$$ it's pretty common to say \"Let $x = t^3$; then as $t \\to 0$, we have $x \\to 0$, so $$L = \\lim_{x \\to 0} \\frac{\\sin x}{x}$$ which we know is $1$, and we're done.\"\n\nWhat's going on here in general is an application of the following \"Theorem\"\n\nTheorem 1: If the function $g$ satisfies [fill in missing properties] and $$\\lim_{t \\to a} g(t) = b,$$ then $$\\lim_{t \\to a} f(g(t)) = \\lim_{x \\to b} f(x),$$ i.e., one limit exists if and only if the other does, and if they both exist, they're equal.\n\nIn the example above, $f(x) = \\frac{\\sin x}{x}$ and $g(t) = t^3$ and $a = b = 0$.\n\nThere's an alternative form, in which we're asked to show that $$L = \\lim_{t \\to 0} \\frac{\\sin \\sqrt[3]{t}}{\\sqrt[3]{t}} = 1,$$ it's pretty common to say \"Let $t = x^3$; then as $t \\to 0$, we have $x \\to 0$, so $$L = \\lim_{x \\to 0} \\frac{\\sin x}{x}$$ which we know is $1$, and we're done.\"\n\nIn this case, the implicit theorem is very similar to the other, but with the role of $g$ reversed (i.e., we're substituting $t = x^3$ instead of $x = t^3$, so the natural form of the theorem puts $g$ on the other side):\n\nTheorem 2: If the function $g$ satisfies [fill in missing properties] and $$\\lim_{x \\to b} g(x) = a,$$ then $$\\lim_{t \\to a} f(t) = \\lim_{x \\to b} f(g(x)).$$\n\nIn the second example above, we have $a = b = 0$, $f(t) = \\frac{\\sin \\sqrt[3]{t}}{\\sqrt[3]{t}}$, and $g(x) = x^3$.\n\nThe two theorems are obviously the same: if you swap $a$ and $b$, $x$ and $t$, and reverse the equality in the last line, they're identical. But each represents a different approach to working with limits, so I've stated both.\n\nIn the second form, it's clearly important that $g$ be surjective near $a$ (i.e., for every small enough interval $I = (b-\\epsilon, b + \\epsilon)$, there's an interval $I' = (a-\\delta, a + \\delta)$ such that $I- \\{b\\} \\subset g(I' - \\{a\\})$. (Hat-tip to MathematicsStudent1122 for the observation that I need to delete $a$ and $b$ from those intervals). Otherwise you could use things substitutions like $s = t^2$, which would turn a two-sided limit into a one-sided one (or vice versa), in which case one limit might exists and the other might not.\n\nAddendum to clarify why this might matter, for @MathematicsStudent1122:\n\nConsider $$f(x) = \\begin{cases} 1 & x \\ge 0 \\\\ 0 & x < 0 \\end{cases}.$$\n\nand look at $L = \\lim_{x \\to 0} f(x^2)$. It's clear that $L$ exists and is $1$. But if we substitute $t = x^2$, then we get $L = \\lim_{t \\to 0} f(t)$, which does not exist; hence this \"substitution\" is not valid: I've turned what amounts to a 1-sided limit (which exists) into a two-sided limit (which does not exist). The domains of $f$ and $g$ are both all of $\\Bbb R$.\n\nMy question is this:\n\nWhat is a reasonable set of missing properties for each of these theorems? (I can work out the exact properties easily enough by running through the definitions, but they don't seem to be very helpful/checkable.)\n\nOne answer might be \"$g$ is locally a bijection\", but that rules out things like $y = x + x\\sin \\frac{1}{x}$ near $x = 0$, so it seems too limited. (It also rules out things like $x \\mapsto x + \\sin x$ for limits as $x \\to \\infty$, which is a pity.)\n\nI recognize that this is not a strictly mathematical question. But my goal is to come up with a \"calculus student's theorem\", one that says \"if you're trying to work out a limit, which may or may not exist, then it's OK to do substitutions of this sort along the way,\" and which will cover the vast majority of the problems that they might encounter in a standard calculus book, or even in Spivak's book.\n\nThis question gives two theorems, but both have assumptions about the existence of limits. This one comes a little closer, but still isn't entirely satisfactory.\n\nI'd love any nice-enough condition to be broadly useful. In particular, I think it's completely reasonable to require, for instance, that the \"substitution function\" $g$ be continuous, and perhaps even differentiable (although I doubt that's of much use).\n\n\u2022 \u2013\u00a0Jack D'Aurizio Aug 20 '17 at 15:03\n\u2022 Great point, @JackD'Aurizio. That may be the first place where this particular stunt bothered me. :) \u2013\u00a0John Hughes Aug 20 '17 at 15:04\n\u2022 So that's part of the quest for the Holy Grail of all lazy students, the Grand Unified Formula Of All Textbook Exercises? ;-) \u2013\u00a0Professor Vector Aug 20 '17 at 15:06\n\u2022 Substitutions like $t=x\\sin \\frac{1}{x}$ simply don't always work. If we let $g(x) = x\\sin \\frac{1}{x}$ and $f(x) = 1$ for $x=0$ and $f(x) = 0$ for $x \\neq 0$, then we can note that $$\\lim_{x \\to 0} f(g(x))$$ does not exist, but $$\\lim_{y \\to 0} f(y)$$ does exist. \u2013\u00a0MathematicsStudent1122 Aug 20 '17 at 15:56\n\u2022 @ParamanandSingh: Well...I've mused about it on miscellaneous occasions for about 45 years, but I thank you for pointing out that I've bene wasting my time. :) Your answer in that linked case is the easy one --- the one that only goes one direction. I want to know that if I make a substitution and the origiinal limit does NOT exist, then the new limit is also guaranteed to not exist. \u2013\u00a0John Hughes Aug 20 '17 at 18:07\n\nI'm afraid this isn't quite what you're after, but it seems to be a decent starting point. These conditions, while restrictive, appear to be necessary for the following proof.\n\nAssume $\\lim_{x\\to a}g(x)=b$, where the function $g$ also satisfies: $$\\text{there exists a neighborhood U of a such that b\\not\\in g(U\\setminus\\{a\\})} \\tag{1}$$ and for each $(y_n)$ converging to $b$ with $y_n\\ne b$, $$\\text{there exists (x_n) such that g(x_n)=y_n for large n and x_n\\to a}. \\tag{2}$$\n\nThen we show that $$\\lim_{x\\to a}f(g(a)) = \\lim_{y\\to b}f(y).$$\n\nProof:\n\nLet $L_1:=\\lim_{x\\to a}f(g(a))$ and $L_2:=\\lim_{y\\to b}f(y)$, either of which may or may not exist. We will use the sequential criterion.\n\nFirst suppose $L_2$ exists and let $(x_n)$ be a sequence such that $x_n\\to a$ and $x_n\\ne a$ for all $n$. Since $\\lim_{x\\to a}g(x)=b$, we have $g(x_n)\\to b$. Due to $(1)$, we have $g(x_n)\\ne b$ for $n$ large enough. Then, because $L_2$ exists, we have $f(g(x_n))\\to L_2$. This proves by the sequential criterion that $L_1$ exists and $L_1=L_2$.\n\nNow assume $L_1$ exists and let $(y_n)$ be a sequence such that $y_n\\to b$ and $y_n\\ne b$ for all $n$. By $(2)$ there exists a sequence $(x_n)$ such that $g(x_n)=y_n$ for large $n$ and $x_n\\to a$. Since $y_n\\ne b$ for all $n$, we have $x_n\\ne a$ for large $n$. Then $$\\lim_{n\\to\\infty}f(y_n)=\\lim_{n\\to\\infty}f(g(x_n))=L_1,$$ and so the sequential criterion implies $L_2$ exists and $L_2=L_1$.\n\n\u2022 You're right --- that's not quite what I'm after. For one thing, it's not a condition that most calc students could test. :( But I agree that it's a decent starting point -- thanks. \u2013\u00a0John Hughes Aug 21 '17 at 19:58", "date": "2020-09-30 10:02:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2400259/limits-and-substitution", "openwebmath_score": 0.9307198524475098, "openwebmath_perplexity": 201.83231004638074, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806518175515, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8519547282254399}}
{"url": "https://math.stackexchange.com/questions/778946/prove-that-if-a-is-normal-then-eigenvectors-corresponding-to-distinct-eigenva", "text": "Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)\n\nThe problem statement is as follows:\n\nProve that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.\n\nI can certainly prove that this is the case, using the spectral theorem. The gist of my proof is presented below.\n\nIf possible, I would like to find a simpler proof. I was hoping that there might be some sort of manipulation along these lines, noting that $$\\langle Av_1,A v_2\\rangle = \\langle v_1,A^*Av_2\\rangle = \\langle v_1,AA^*v_2\\rangle = \\langle A^* v_1,A^* v_2 \\rangle$$\n\nAny ideas here would be appreciated.\n\nMy proof:\n\nLet $\\{v_{\\lambda,i}\\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that $$A v_{\\lambda,i} = \\lambda v_{\\lambda,i}$$ Let $v_1,\\lambda_1$ and $v_2,\\lambda_2$ be eigenpairs with $\\lambda_1 \\neq \\lambda_2$. We may write $v_1 = \\sum_{i,\\lambda}a_{i,\\lambda}v_{i,\\lambda} .$ We then have $$0 = Av_1 - \\lambda_1 v_1 = \\sum_{i,\\lambda}(\\lambda - \\lambda_1)a_{i,\\lambda}v_{i,\\lambda}$$ So that $a_{i,\\lambda} = 0$ when $\\lambda \\neq \\lambda_1$. Similarly, we may write $v_2 = \\sum_{i,\\lambda}b_{i,\\lambda}v_{i,\\lambda}$, and note that $b_{i,\\lambda} = 0$ when $\\lambda \\neq \\lambda_2$. From there, we have $$\\langle v_1,v_2 \\rangle = \\sum_{i,\\lambda}a_{i,\\lambda}b_{i,\\lambda}$$ the above must be zero since for each pair $i,\\lambda$, either $a_{i,\\lambda}=0$ or $b_{i,\\lambda} = 0$.\n\n\u2022 \"Let $v_{\\lambda,i}$ be an orthonormal basis of eigenvectors...\" I'd guess that this might be something very close to what would need to be proved. \u2013\u00a0Algebraic Pavel May 3 '14 at 2:13\n\u2022 @PavelJiranek I was worried that I had used circular logic at some point. However, the existence of such a basis (i.e. the spectral theorem) comes directly from the Schur triangularization theorem, which says nothing about normal matrices in particular. \u2013\u00a0Omnomnomnom May 3 '14 at 19:11\n\nAssume $\\;\\lambda\\neq \\mu\\;$ and\n\n$$\\begin{cases}Av=\\lambda v\\;\\,\\implies\\; A^*v=\\overline \\lambda v\\\\{}\\\\Aw=\\mu w\\implies A^*w=\\overline\\mu w\\end{cases}$$\n\nFrom this we get:\n\n$$\\begin{cases}\\langle v,Aw\\rangle=\\langle v,\\mu w\\rangle=\\overline\\mu\\langle v,w\\rangle\\\\{}\\\\ \\langle v,Aw\\rangle=\\langle A^*v,w\\rangle=\\langle\\overline\\lambda v,w\\rangle=\\overline\\lambda\\langle v,w\\rangle \\end{cases}$$\n\nand since $\\;\\overline\\mu\\neq\\overline\\lambda\\;$ , we get $\\;\\langle v,w\\rangle =0\\;$\n\nQuestion: Where did we use normality in the above?\n\n\u2022 How de we know that $v$ is also an eigenvector for $A^*$? \u2013\u00a0Berci May 2 '14 at 22:55\n\u2022 @Berci, read and think of the question at the end of the answer. \u2013\u00a0DonAntonio May 2 '14 at 22:56\n\u2022 From the other two answers this follows, of course, using that $A$ is normal. But can you show it directly? \u2013\u00a0Berci May 2 '14 at 22:57\n\u2022 Why directly? Use the definition and work it out. I usually don't give full answers and, after all, the OP is trying to get a more or less simpler proof to the claim than the one using the spectral theorem... \u2013\u00a0DonAntonio May 2 '14 at 23:00\n\u2022 I like it! Showing that $Av=\\lambda v\\;\\,\\implies\\; A^*v=\\overline \\lambda v$ seems to be the step in the spirit of what I was looking for. Thanks. \u2013\u00a0Omnomnomnom May 2 '14 at 23:14\n\nSpecializing your identity to $v_1=v_2=v$, we get $\\|Av\\|=\\|A^*v\\|$. Hence $\\ker A=\\ker A^*$. Recalling that $\\ker A^* = (\\operatorname{ran} A)^\\perp$ for general $A$, we conclude that the kernel and range of a normal matrix are mutually orthogonal.\n\nIt remains to apply the above conclusion to $A-\\lambda I$ where $\\lambda$ is an eigenvalue of $A$.\n\n\u2022 Neat! Although DonAntonio's proof is more along the lines I was thinking of, this is a nice perspective. \u2013\u00a0Omnomnomnom May 2 '14 at 23:16\n\nI try to give another simple proof to $$T^*v=\\bar{\\lambda}v ~\\text{ if }~ Tv=\\lambda v$$ where $T$ is a normal operator on a Hilbert space $H$.\n\nSuppose $V=\\ker(T-\\lambda I)$. Since $T^*$ communicate with $T$, $$T^*V\\subset V.$$ Because $$\\langle v,T^*v\\rangle =\\langle Tv,v\\rangle =\\langle \\lambda v,v\\rangle=\\langle v,\\bar{\\lambda}v\\rangle ~~\\forall v \\in V,$$ $\\langle u,T^*v\\rangle =\\langle u, \\bar{\\lambda}v\\rangle ~\\forall u,v\\in V$ by polarisation identity, and thus $T^*v=\\bar{\\lambda}v.$\n\nREMARK: Let $\\sigma:V\\times V\\to W$ be a sesquilinear form, where $V$ and $W$ are linear vector spaces over $\\mathbb{C}$. The follwing formula is called Polarisation Identity : $$\\sigma(u,v)=\\sum_{k=0}^3 i^k\\sigma(u+i^k v, u+i^kv).$$\n\n\u2022 I don't really see how this answer addresses the question being asked. Also, the question being asked was answered 3 years ago. \u2013\u00a0Omnomnomnom Jun 12 '17 at 15:01\n\u2022 @Omnomnomnom This is why $Av=\\lambda v\\Rightarrow A^*v=\\bar{\\lambda}v$ using no spectral theorem which is the point to the question I think. \u2013\u00a0C.Ding Jun 13 '17 at 2:37\n\u2022 That is not the point to the question, actually. The point is to show that eigenvectors corresponding to different eigenvalues are necessarily orthogonal. \u2013\u00a0Omnomnomnom Jun 13 '17 at 2:58\n\u2022 Oh, sorry to disturb you. As I give some additional remarks to the right answer you have accepted. \u2013\u00a0C.Ding Jun 13 '17 at 3:04\n\nA normal matrix is unitarily similar to diagonal matrix.\n\n$$A = UDU^{-1}$$ where $U$ is Unitary matrix.\n\nEigen decompositions tells that $U$ is a matrix composed of columns which are eigenvectors of $A$. And matrix $D$ is Diagonal matrix with eigenvalues on diagonal.\n\nProperty: Columns of Unitary matrix are orthogonal.\n\nSo, columns of $U$ (which are eigenvectors of $A$) are orthogonal.", "date": "2019-05-23 03:10:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/778946/prove-that-if-a-is-normal-then-eigenvectors-corresponding-to-distinct-eigenva", "openwebmath_score": 0.9323018193244934, "openwebmath_perplexity": 187.45606644364267, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806501150426, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8519547150894403}}
{"url": "https://gmatclub.com/forum/within-a-rectangular-courtyard-of-length-60-feet-a-graveled-path-277135.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Oct 2018, 11:33\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Within a rectangular courtyard of length 60 feet, a graveled path,\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 25 Aug 2018\nPosts: 2\nWithin a rectangular courtyard of length 60 feet, a graveled path,\u00a0 [#permalink]\n\n### Show Tags\n\n24 Sep 2018, 10:25\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n58% (03:51) correct 42% (04:22) wrong based on 19 sessions\n\n### HideShow timer Statistics\n\nWithin a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :\n\nA 24\nB 30\nC 40\nD 45\nE 54\nIntern\nJoined: 14 Feb 2018\nPosts: 6\nWithin a rectangular courtyard of length 60 feet, a graveled path,\u00a0 [#permalink]\n\n### Show Tags\n\n24 Sep 2018, 11:12\nBeing x the width of the rectangle, we have the following equation:\n\n$$2[2(60*3)+2*3(x-6)] + 984 = 2[2(60*6)+2*6(x-12)]$$\n\nwhere $$2(60*3)$$ is the area of the path next to the longer side (60 ft long and 3 ft wide), and $$2*3(x-6)$$ the area of the path next to the shorter side, avoiding to count the area from the corner twice. Then we need to repeat the same idea with the bigger path. Both areas are multiplied by 2 because we are measuring the cost, which is Rs 2 per sq ft. We have to consider that the path with double size is Rs 984 more expensive, so we add this into our equation. Then, we can solve the equation:\n\n$$2[360+6x-36]+984=2(720+12x-144)$$\n$$360+6x-36+492=720+12x-144$$\n$$324+6x+492=576+12x$$\n$$816-576=6x$$\n$$240=6x$$\n$$x=40$$\n\nThus, C is the correct answer.\nManager\nJoined: 02 Aug 2015\nPosts: 112\nRe: Within a rectangular courtyard of length 60 feet, a graveled path,\u00a0 [#permalink]\n\n### Show Tags\n\n24 Sep 2018, 11:33\nsultanatehere wrote:\nWithin a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :\n\nA 24\nB 30\nC 40\nD 45\nE 54\n\nA = Area of outer rectangle = 60 x Width\nAx = Area of inner rectangle with width of 3 = 54 x (Width-6)\nAy = Area of inner rectangle with twice the width, i.e 6 = 48 x (Width - 12)\n\nArea of gravel with normal width = A-Ax\nArea of gravel with twice width = A-Ay\n\nGiven, cost of gravelling twice width - cost of gravelling normal width = 984.\nGiven, cost of gravelling = Rs2/sq ft.\n\n2(A-Ay) - 2(A-Ax) = 984.\nA-Ay-A+Ax=492.\nAx-Ay=492.\n\nOn substituting Ax and Ay from above and solving the above equation, we get width=40.\n\nHence C.\n\nCheers!\nIntern\nJoined: 21 Jul 2018\nPosts: 38\nRe: Within a rectangular courtyard of length 60 feet, a graveled path,\u00a0 [#permalink]\n\n### Show Tags\n\n24 Sep 2018, 13:48\nTried to solve using an Ans choices and got lucky with the 1st option\n_________________\n\n______________________________\nConsider KUDOS if my post helped !!\n\nI'd appreciate learning about the grammatical errors in my posts\n\nPlease let me know if I'm wrong somewhere\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 3845\nLocation: United States (CA)\nRe: Within a rectangular courtyard of length 60 feet, a graveled path,\u00a0 [#permalink]\n\n### Show Tags\n\n27 Sep 2018, 17:20\nsultanatehere wrote:\nWithin a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :\n\nA 24\nB 30\nC 40\nD 45\nE 54\n\nWe can let w = the width of the courtyard. Therefore, the courtyard, including the graveled path, has an area of 60w sq ft. Excluding the path, the area of the courtyard would be (60 - 2(3))(w - 2(3)) = 54(w - 6) = 54w - 324 sq ft. So the path alone has an area of 60w - (54w - 324) = 6w + 324 sq ft. Since the cost of graveling the path is Rs 2 per sq ft, the cost of graveling the path is Rs 12w + 648.\n\nIf the path had been twice as wide, then the courtyard, including the path, would still have an area of 60w sq ft. However, excluding the path, the area of the courtyard would be (60 - 2(6))(w - 2(6)) = 48(w - 12) = 48w - 576 sq ft. So the path alone would have an area of 60w - (48w - 576) = 12w + 576 sq ft, and the cost of the path would be Rs 24w + 1152. We are told that this would cost Rs 984 more, so we can set up the following equation to solve for w:\n\n12w + 648 + 984 = 24w + 1152\n\n12w + 1632 = 24w + 1152\n\n480 = 12w\n\n40 = w\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: Within a rectangular courtyard of length 60 feet, a graveled path, &nbs [#permalink] 27 Sep 2018, 17:20\nDisplay posts from previous: Sort by", "date": "2018-10-17 18:33:10", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/within-a-rectangular-courtyard-of-length-60-feet-a-graveled-path-277135.html", "openwebmath_score": 0.7202662229537964, "openwebmath_perplexity": 3981.3950371136293, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.851952809486198, "lm_q1q2_score": 0.851952809486198}}
{"url": "https://gmatclub.com/forum/5-drill-machines-can-drill-15-meters-in-5-hours-252975.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 19 Oct 2018, 11:34\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# 5 drill machines can drill 15 meters in 5 hours.\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 22 Apr 2017\nPosts: 111\nLocation: India\nGMAT 1: 620 Q46 V30\nGMAT 2: 620 Q47 V29\nGMAT 3: 630 Q49 V26\nGMAT 4: 690 Q48 V35\nGPA: 3.7\n5 drill machines can drill 15 meters in 5 hours.\u00a0 [#permalink]\n\n### Show Tags\n\n06 Nov 2017, 07:17\n1\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n77% (01:29) correct 23% (02:02) wrong based on 89 sessions\n\n### HideShow timer Statistics\n\n5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?\n\nA)8\nB)16\nC)23\nD)50\nE)60\nIntern\nJoined: 10 Aug 2017\nPosts: 8\nRe: 5 drill machines can drill 15 meters in 5 hours.\u00a0 [#permalink]\n\n### Show Tags\n\n06 Nov 2017, 07:27\nMac 5 meters 15 hours 5\nMac x meters 60 hours 2\nDo it with proportion.\nX/5= 60/15 * 5/2\nX=50\n\nSent from my iPhone using GMAT Club Forum\nMath Expert\nJoined: 02 Aug 2009\nPosts: 6961\nRe: 5 drill machines can drill 15 meters in 5 hours.\u00a0 [#permalink]\n\n### Show Tags\n\n06 Nov 2017, 07:32\nManishKM1 wrote:\n5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?\n\nA)8\nB)16\nC)23\nD)50\nE)60\n\nto drill 15 mtr in 5 hrs, 5 machines are required..\nso to drill 15*4 mtr in 5 hrs, 5*4 machines are required..\nto drill 60 mtr in 1 hrs, 20*5 machines are required..\nfinally to drill 60 mtr in 2 hrs, 100/2 or 50 machines are required..\n\nD\n_________________\n\n1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372\n2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html\n3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html\n\nGMAT online Tutor\n\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 2035\n5 drill machines can drill 15 meters in 5 hours.\u00a0 [#permalink]\n\n### Show Tags\n\n06 Nov 2017, 09:17\n1\nManishKM1 wrote:\n5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?\n\nA)8\nB)16\nC)23\nD)50\nE)60\n\nFind a machine's individual rate from the first information, and use it to solve the question.\n\nW = (# of machines) * (indiv.) rate * time\nW = 15, # of machines = 5, rate = ??, time = 5\n\n15 = 5 * rate * 5\n15 = 25 * rate\n\nIndividual rate = $$\\frac{15}{25}=\\frac{3}{5}$$ in meters/hour\n\nAt that rate, how many machines are needed to drill 60 meters in 2 hours?\n\nW = (# of machines) * (indiv.) rate * time\nW = 60, # of machines = ??, rate = $$\\frac{3}{5}$$, time = 2\n\n60 = (# of machines) * $$\\frac{3}{5}$$ * 2\n60 = (# of machines) * $$\\frac{6}{5}$$\n\n# of machines = $$\\frac{60}{(\\frac{6}{5})}= 60 * \\frac{5}{6}=$$ 50\n\n_________________\n\n___________________________________________________________________\nFor what are we born if not to aid one another?\n-- Ernest Hemingway\n\nTarget Test Prep Representative\nStatus: Head GMAT Instructor\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2830\nRe: 5 drill machines can drill 15 meters in 5 hours.\u00a0 [#permalink]\n\n### Show Tags\n\n12 Nov 2017, 08:47\n1\n1\nManishKM1 wrote:\n5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?\n\nA)8\nB)16\nC)23\nD)50\nE)60\n\nThe rate for 5 drill machines is 15/5 = 3 meters per hour. We need to determine how many drill machines we need for a rate of 60/2 = 30 meters per hour. Since 30 is 10 times 3, we would need 5 x 10 = 50 drill machines.\n\n_________________\n\nJeffery Miller\nHead of GMAT Instruction\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: 5 drill machines can drill 15 meters in 5 hours. &nbs [#permalink] 12 Nov 2017, 08:47\nDisplay posts from previous: Sort by\n\n# 5 drill machines can drill 15 meters in 5 hours.\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-10-19 18:34:14", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/5-drill-machines-can-drill-15-meters-in-5-hours-252975.html", "openwebmath_score": 0.5159401893615723, "openwebmath_perplexity": 8441.10621709785, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 1.0, "lm_q2_score": 0.8519528038477824, "lm_q1q2_score": 0.8519528038477824}}
{"url": "http://gmatclub.com/forum/what-is-the-average-arithmetic-mean-of-eleven-consecutive-93188.html#p717074", "text": "Find all School-related info fast with the new School-Specific MBA Forum\n\n It is currently 21 Oct 2016, 05:17\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# What is the average (arithmetic mean) of eleven consecutive\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 27 Feb 2010\nPosts: 105\nLocation: Denver\nFollowers: 1\n\nKudos [?]: 328 [0], given: 14\n\nWhat is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n23 Apr 2010, 18:05\n21\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n67% (01:44) correct 33% (00:39) wrong based on 407 sessions\n\n### HideShow timer Statistics\n\nWhat is the average (arithmetic mean) of eleven consecutive integers?\n\n(1) The average of the first nine integers is 7\n(2) The average of the last nine integers is 9\n[Reveal] Spoiler: OA\nManager\nJoined: 27 Feb 2010\nPosts: 105\nLocation: Denver\nFollowers: 1\n\nKudos [?]: 328 [1] , given: 14\n\nRe: Average ( arithmetic mean ) of eleven consecutive integers?\u00a0[#permalink]\n\n### Show Tags\n\n23 Apr 2010, 18:11\n1\nKUDOS\nSome how i got E\n\n1. (63+ X10 + X11) / 11 = ?? ..sitill missing two numbers? so insuff??\n2. (X1+X2 +81)/ 11 = ?? Still missing two numbers so Insuff???\n\nand combining 1 and 2 still missing some info ? .. Sorry friends, i need help here.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 35240\nFollowers: 6618\n\nKudos [?]: 85287 [10] , given: 10236\n\nRe: Average ( arithmetic mean ) of eleven consecutive integers?\u00a0[#permalink]\n\n### Show Tags\n\n24 Apr 2010, 06:10\n10\nKUDOS\nExpert's post\n11\nThis post was\nBOOKMARKED\nzz0vlb wrote:\nWhat is the average ( arithmetic mean ) of eleven consecutive integers?\n\n1.The avg of first nine integers is 7\n2. The avg of the last nine integers is 9\n[Reveal] Spoiler:\nD\n\nfriends, please explain this to me.\n\nConsecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case $$mean=median=x_6$$.\n\n(1) $$x_1+x_2+...+x_9=63$$ --> there can be only one set of 9 consecutive integers to total 63. Sufficient.\n\nIf you want to calculate: $$(x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63$$ --> $$x_6=8$$.\n\nOR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> $$x_5*9=63$$ --> $$x_5=7$$ --> $$x_6=7+1=8$$\n\n(2) $$x_3+x_4+...+x_{11}=81$$ --> there can be only one set of 9 consecutive integers to total 81. Sufficient.\n\nIf you want to calculate: $$(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81$$ --> $$x_6=8$$.\n\nOR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> $$x_7*9=81$$ --> $$x_7=9$$ --> $$x_6=9-1=8$$\n\n_________________\nManager\nJoined: 27 Feb 2010\nPosts: 105\nLocation: Denver\nFollowers: 1\n\nKudos [?]: 328 [0], given: 14\n\nRe: Average ( arithmetic mean ) of eleven consecutive integers?\u00a0[#permalink]\n\n### Show Tags\n\n24 Apr 2010, 07:11\nThanks for clarifying this Bunuel. +1Kudos to you.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 35240\nFollowers: 6618\n\nKudos [?]: 85287 [0], given: 10236\n\nRe: What is the average (arithmetic mean ) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n12 Aug 2013, 04:00\nBumping for review and further discussion.\n_________________\nVerbal Forum Moderator\nJoined: 10 Oct 2012\nPosts: 630\nFollowers: 77\n\nKudos [?]: 1051 [6] , given: 136\n\nRe: What is the average (arithmetic mean ) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n12 Aug 2013, 04:39\n6\nKUDOS\n8\nThis post was\nBOOKMARKED\nzz0vlb wrote:\nWhat is the average (arithmetic mean ) of eleven consecutive integers?\n\n(1) The avg of first nine integers is 7\n(2) The avg of the last nine integers is 9\n\nHere is a neat little trick for such kind of problems:\n\nWill be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly $$\\frac{d}{2} units \\to$$ The new Average = $$4-\\frac{1}{2} = 3.5$$\n\nAgain, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.\n\nSimilarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.\n\nBack to the problem:\n\nFrom F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.\n\nFrom F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.\n\nD.\n_________________\nIntern\nJoined: 26 May 2010\nPosts: 10\nFollowers: 0\n\nKudos [?]: 33 [5] , given: 4\n\nRe: What is the average (arithmetic mean ) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n12 Aug 2013, 23:15\n5\nKUDOS\n3\nThis post was\nBOOKMARKED\nzz0vlb wrote:\nWhat is the average (arithmetic mean ) of eleven consecutive integers?\n\n(1) The avg of first nine integers is 7\n(2) The avg of the last nine integers is 9\n\nAs a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1\n\n1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient\n2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient\n\nAnd is D\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 12145\nFollowers: 538\n\nKudos [?]: 151 [0], given: 0\n\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n28 Aug 2014, 09:43\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nManager\nJoined: 24 Jun 2014\nPosts: 51\nConcentration: Social Entrepreneurship, Nonprofit\nFollowers: 0\n\nKudos [?]: 16 [0], given: 76\n\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n09 Mar 2015, 19:15\nI considered following approach\n\nif the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A\nif largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55\n\nNow as per statement 1 , average of first 9 numbers is 7 i.e sum =63\nsum of 11 numbers =63+x+9+x+10----->B\n\nEquating A& B\n11X+55=63+X+9+10 ,which can be solved to get x=3\n\nstatement I is sufficient\nsimilar approach for Statement II\n11X-55=8+2X-19 ,can be solved to get X=13\n\nstatement 2 is sufficient\n\nOA=D\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 7682\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: 340 Q170 V170\nFollowers: 341\n\nKudos [?]: 2280 [1] , given: 162\n\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n09 Mar 2015, 19:26\n1\nKUDOS\nExpert's post\n2\nThis post was\nBOOKMARKED\nHi All,\n\nWhen you look at this question, if you find yourself unsure of where to \"start\", it might help to break down everything that you know into small pieces:\n\n1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place \"in line\", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?)\n\n2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7.\n\nWhat would have to happen for a group of consecutive integers to have an average of 7?\n\nHere are some examples:\n\n7\n\n6, 7, 8\n\n5, 6, 7, 8, 9\n\nNotice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern.\n\nWith 9 total terms, that means there has to be 4 above and 4 below:\n\n3, 4, 5, 6,.......7.......8, 9, 10, 11\n\nNow we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT\n\nWith this same approach, we can deal with Fact 2.\n\nThe key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to \"play around\" with a question if you're immediately certain about how to handle it.\n\n[Reveal] Spoiler:\nD\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\n# Special Offer: Save \\$75 + GMAT Club Tests\n\n60-point improvement guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 12145\nFollowers: 538\n\nKudos [?]: 151 [0], given: 0\n\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n10 Mar 2016, 03:12\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nIntern\nJoined: 11 Apr 2016\nPosts: 3\nFollowers: 0\n\nKudos [?]: 0 [0], given: 0\n\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n29 Jun 2016, 03:10\nwow such complex explanations for such a simple problem?\n\ngiven :\n\n11 consec integers\n\nlet them be x,x+1,x+2,...,x+10\n\nQ: what is their mean?\n\nmean is (11x+55)/11 = x+5.\nQ becomes what is x+5\n\n1) mean first 9 is 7.\n\nso (9x+36)/9 = x+4 = 7 , so x+5 =8 ,--> sufficient A or D\n\n2) mean of last 9 is 9.\n\nso (9x+54)/9 = x+6= ---> x+5=8, sufficient . so D\n\nD\nDirector\nJoined: 04 Jun 2016\nPosts: 656\nGMAT 1: 750 Q49 V43\nFollowers: 40\n\nKudos [?]: 153 [0], given: 36\n\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n15 Jul 2016, 00:10\nzz0vlb wrote:\nWhat is the average (arithmetic mean) of eleven consecutive integers?\n\n(1) The average of the first nine integers is 7\n(2) The average of the last nine integers is 9\n\nFor odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question\n(1) The average of the first nine integers is 7\n7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7\nwe will have {3,4,5,6,7,8,9,10,11}\nnow we can add last two consecutive integer after 11, they will be 12,13\nour new set will become = {3,4,5,6,7,8,9,10,11,12,13}\nagain since the number of total elements in the set is odd, Mean will simply be the middle value = 8\nSUFFICIENT\n\n(2) The average of the last nine integers is 9\nAgain number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right\nMiddle value will be\n{5,6,7,8,9,10,11,12,13}\nAdd 3,4 at the start of the set\nnew set = {3,4,5,6,7,8,9,10,11,12,13}\nMean will be 8\nSufficient\n\n_________________\n\nPosting an answer without an explanation is \"GOD COMPLEX\". The world doesn't need any more gods. Please explain you answers properly.\nFINAL GOODBYE :- 17th SEPTEMBER 2016.\n\nRe: What is the average (arithmetic mean) of eleven consecutive \u00a0 [#permalink] 15 Jul 2016, 00:10\nSimilar topics Replies Last post\nSimilar\nTopics:\n1 What is the average (arithmetic mean) of a sequence of N consecutive 9 12 Sep 2016, 04:47\nWhat is the average (arithmetic mean) of fifteen consecutive integers? 2 26 May 2016, 11:39\n7 What is the average of eleven consecutive integers? 10 15 Jan 2015, 07:25\n3 What is the average (arithmetic mean) of eleven consecutive 3 25 Jan 2008, 11:44\nWhat is the average (arithmetic mean) of eleven consecutive 3 24 Dec 2007, 00:16\nDisplay posts from previous: Sort by", "date": "2016-10-21 12:17:05", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/what-is-the-average-arithmetic-mean-of-eleven-consecutive-93188.html#p717074", "openwebmath_score": 0.6749594211578369, "openwebmath_perplexity": 2902.9624508058323, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. 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{"url": "http://matv.leforriclub.it/phase-portrait-nonlinear-system.html", "text": "## Phase Portrait Nonlinear System\n\n1) Find all equilibrium points by solving the system 2) Let a standard software (e. for the analysis of nonlinear systems; to introduce controller design methods for nonlinear systems. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Phase Portraits and Time Plots for Cases A (pplane6) Saddle Ex. 1), which was diagnosed using a set of four features extracted from the phase plane trajectory of the system to characterize the nonlinear response in the periodic regime. Local Phase Portrait of Nonlinear Systems Near Equilibria. \u201cProof\u201d: Consider trajectory sufficiently close to origin time reversal symmetry. Save the phase portraits to submit on Gradescope. Keywords: nonlinear dynamics, chaos, electrical circuits. Biological Models: Predator-prey models, Competition models, Survival of one species, Co-existence, Alligators, doomsday and extinction. \u2022 As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. A: The origins and evolution of predator-prey theory, Ecology 73, 1530-1535 (1992). Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Often, mathematical models of real-world phenomena are formulated in terms of systems of nonlinear differential equations, which can be difficult to solve explicitly. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. See phase portrait below. 1 In each problem\ufb01nd the critical points and the corresponding linear system. Albu-Schaffer. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. Stable and unstable manifolds of equilibrium points and periodic orbits are important objects in phase portraits. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. Introduction to systems of differential equations 2. (reductor and multipliers). While nonlinear systems of-ten require complex idiosyncractic treatments, phase potraitshaveevolved as apowerfultool forglobal anal-ysis ofthem. The system lives in a state space or phase. Weak non-linear oscillators and. Existence, uniqueness, and strong topological consequences for two-dimensions. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. We also observe that the two xed points are progressively pushed apart in the amplitude direction. For a general 2 \u00d7 2 matrix A, the phase portrait will be equivalent to one of the four cases above, obtained by a linear transformation of coordinates (similarity transformation). Fixed points are stagnation points of the \ufb02ow. The three examples will all be predator-prey models. 1 Concepts of Phase Plane Analysis 18 2. Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Click on the button corresponding to your preferred computer algebra system (CAS). 3 Symmetry in Phase Plane Portraits 22 2. Thus, a system has a limit cycle if and only if it has an isolated, closed. Numerical Construction of Phase Portraits. Two integral constraints on the amplitude and phase variation of the oscillations of an autonomous multi-degree of freedom system were obtained. 2 Draw the phase portraits of the following systems, using isoclines (a) 8+8+0. Generally, the nonlinear time series is analyzed by its phase space portrait. The book is very readable even though it has a lot of jargon (read heavy mathematics). A phase portrait is a graphical tool to visualize long term. Phase portrait generator. 5 0xy (7) which itself is a dynamical equation, the phase portrait is a trajectory along the switching line \u03c3 = 0. By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at \ud707 = 0. shown to produce sharp phase portraits in long-term simulations, see e. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. On this page I explain how to use Matlab to draw phase portraits for the the two linear systems. This video deals with. Existence, uniqueness, and strong topological consequences for two-dimensions. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. The y nullcline is given by 3 4 1 4 3 y 2 3 x y = 0 (12) which gives the lines y = 0 or y = 3 4. , non-linear) 2 \u00d7 2 autonomous system discussed at the beginning of this chapter, in sections 1 and 2: x = f (x, y); (1) y = g(x, y). The nonlinear system's phase portrait near the fixed point is topologically unchanged due to small perturbations, and its dynamics are structurally stable or robust. Phase plane analysis for linear systems. 50= 1 (c) 8+82+0. m: A demonstration that plots the linearized phase portraits and the full phase plane. The nonlinear gyroscope model, which is employed in aerospace engineering [24], generally exhibits chaotic behavior. of problems that are described by nonlinear di\ufb00erential equations. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. EECS 222 Nonlinear Systems: Analysis, Stability and Control Shankar Sastry 299 Cory Hall Tu-Th 11-12:30 pm. The book is very readable even though it has a lot of jargon (read heavy mathematics). Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). Nonlinear Models and Nonlinear Phenomena. However, these behaviors are not properly depicted in phase portraits when dealing with sys-tems that could be described as rotating systems,. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. ) Lecture, three hours; discussion, one hour. A two-state phase portrait approach has been used to analyse vehicle dynamics and provides an illustrative view of the state trajectories at constant speed. Analyze the stability and its margins. by graphing and the use of phase portraits; D. First, let us look at the phase space portraits for a range of phase advances from 0:2 2\u02c7to 0:5 2\u02c7. The phase portraits is able to perfectly capture all of the nonlinear trajectories and display them in a way that would be otherwise difficult. The phase portrait behavior of a system of ODEs can be determined by the eigenvalues or the trace and determinant (trace = \u03bb 1 + \u03bb 2, determinant = \u03bb 1 x \u03bb 2) of the system. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. The equation governing the dynamics of the nonlinear gyro, enriched with linear and nonlinear smoothening terms [24], is given by x_1 = x2. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. This paper extends the phase portrait to three states to represent the nonlinear vehicle dynamics with steering and longitudinal tyre force inputs and consideration of the longitudinal. (1) There is one equilibrium solution of this system \u2013 \ufb01nd it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the linearized system. This is a second order systemwhich is autonomous (time does not appear explicitly). from second-order equation to first-order system; what is a phase portrait; direction field of a first-order system; graphing in the xy- tx- and ty-planes; vector notation for a first-order system; semesters > spring 2020 > mth264 > resources > video > linear systems: basics Video | Linear Systems: Basics. Recall the basic setup for an autonomous system of two DEs: dx dt = f(x,y) dy dt = g(x,y) To sketch the phase plane of such a system, at each point (x0,y0)in the xy-plane, we draw a vector starting at (x0,y0) in the direction f(x0,y0)i+g(x0,y0)j. Damped Pendulum. Nonlinear Systems and Stability Autonomous systems and critical points Stability and phase plane analysis of almost linear systems Linearized stability analysis and plotting vector fields using a MSS Numerical solutions and phase portraits of nonlinear systems using a MSS Models and applications: TEXT: Text(s) typically used in this course. The classic Van der Pol nonlinear oscillator is provided as an example. 3 in Third and 3-42 Fourth Quadrant. We illustrate all these cases in the examples below. \u2022 Understand the linearized models using the \u201ceigen-techniques\u201d you learned earlier. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector \ufb01eld superimposed. The dynamical equation and the state equation of the system are established. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. These states can also be correlated with velocity spectral behaviors. A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. (Hint: Use polar coordinates. The low intensity xed point appears on the phase portraits. 26 Phase Portrait for sand Y1 Magnitudes of 9. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. 50= 1 (c) 8+82+0. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. 1 Phase portraits 72 3. warn(warning_msg, ODEintWarning). Is there a way for plotting phase portraits and vector fields for autonomous system of delay differential equations in. This video deals with. The dynamical equation and state equation of the system are established. Strogatz, Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering, Cambridge: Westview Press, 2000. Phase Portraits of Nonlinear Systems Consider a , possibly nonlinear, autonomous system , (autonomous means that the independent variable , thought of as representing time, does not occur on the right sides of the equations). Damped Pendulum. One- and two- dimensional flows. to sketch the phase portrait. Stable and unstable manifolds of equilibrium points and periodic orbits are important objects in phase portraits. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. I have a set of three differential equations and I want to make a phase portrait of them. For a chaotic system, there will be many distinct loops in a phase portrait, showing that the system is aperiodic and does not approach a stable. Quiver function is being used for phase portrait plots obtained using ode. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. - A \u201dlimit cycle\u201d is a periodic orbit that trajectories approach. 3 Determining Time from Phase Portraits 29 2. 3 Symmetry in Phase Plane Portraits 22 2. Derive the dynamics of a linear and nonlinear systems. Fig 1: 3-D phase portrait of the Rossler attractor Integration Solver The Runge-Kutta method for a system of ordinary differential equations is explained here. The dynamical variables of the system, in this case the angular position and velocity , are the coordinates defining the system's phase space. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. Sketching an accurate phase portrait for a non-linear system of DEs is time consuming but the series of 3 videos will help with shortening that time with added understanding. Phase Plane Portraits of Almost Linear Systems Interesting and complicated phase portraits often result from simple nonlinear perturbations of linear systems. 2 Bifurcation set and phase portraits of the Hamiltonian system (5). Ask Question Asked 4 years, 7 months ago. Solving 2x2 homogeneous linear systems of differential equations 3. According to Takens, almost all d-dimensional sub-manifolds could be embedded in a (m=2d+1) dimensional space. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. These programs provide animated phase portraits in dimension two and three, i. The tem-poral response of a system need not be the. 2 : Linear analysis of nonlinear pendulum : Mechanical systems model for a pendulum. (a) This plot shows the vector \ufb01eld for a planar dynamical system. , another nonlinear system x_1 = 1 x3 1 x_2 = x1 x22 equilibrium points are described by x1 = 1 and x2 = 1 note: the equilibrium points of a nonlinear system can be nite (2 in the previous examples, but any other number is possible, including zero) or in nite, and they can be isolated points in state space Oriolo: Stability Theory for. For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. Phase Plane Analysis 17 2. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360\u2218. Weak non-linear oscillators and. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. Ott, and A. This suggests that the only. , regularly timed speech with a metronome). By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at \ud707 = 0. 6: Phase portraits on the (one-dimensional) centr emanifoldandthebifurcation diagram. First, we cover stability definitions of nonlinear dynamical systems, covering the difference between local and global stability. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. 1 (Saddle) Consider the system x\u02d9 1 = \u2212x 1 \u22123x 2 x\u02d9 2 = 2x 2, x. in weakly-nonlinear systems was investigated. 1 Phase Portraits 18 2. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. It is not restricted to small or smooth nonlinearities and applies equally well to strong and hard nonlinearities. warn(warning_msg, ODEintWarning). Limit Cycles. (1) There is one equilibrium solution of this system \u2013 \ufb01nd it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the linearized system. The aim of this section is to present programs allowing to high- light the slow-fast evolution of the solutions of nonlinear and chaotic dynamical systems such as: Van der Pol, Chua and Lorenz models. x\u2032= x\u2212y, y\u2032= x2 +y2 \u22121 2. Four possible phase portraits for this system are shown along the right side of the page. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360\u2218. doc Author: tien Created Date: 11/15/2002 4:16:10 AM. By varying the initial conditions of the system, it is found. The type of phase portrait of a homogeneous linear autonomous system -- a companion system for example -- depends on the matrix coefficients via the eigenvalues or equivalently via the trace and determinant. Phase portrait. In paper [12], the vibration of a typical two degree of freedom nonlinear system with repeated linearized natural frequencies was investigated systematically. We discovered the system\u2019s rich behavior such as chaos through phase portraits, bifurcation diagrams, Lyapunov exponents, and entropy. The course revises some of the standard phase portrait methods encountered in the Dynamical Systems course in part II and extends these ideas, discussing in some detailed centres, via the use of Lyapunov functions, limit cycles and global phase portraits. 1 Concepts of Phase Plane Analysis 18 2. Keywords: nonlinear dynamics, chaos, electrical circuits. A phase portrait is mapped by homeomorphism, a continuous function with a continuous inverse. Linear approximation of autonomous systems 6. Let us consider the below block diagram of a non linear system, where G 1 (s) and G 2 (s) represent the linear element and N represent the non linear element. Existence, uniqueness, and strong topological consequences for two-dimensions. \u2022 As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. Damped Pendulum. Phase Plane Analysis 17 2. If the stable manifold is of higher dimension, then y 1 =h 1(x,\u00b5 ),y 2 =h 2(x,\u00b5 )and. There are lots of practical systems which can be approximated by second-order systems, and apply phase plane analysis. (e) Draw the phase plane portrait of the nonlinear system via v and h nullclines. The resonance effects are most pronounced where both the. 4 Phase Portraits and Bifurcations. A numerically generated phase-portrait of the non-linear system Zoomed in near (0,0) Zoomed in near (2,1) The critical point at (2,1) certainly looks like a spiral source, but (0,0) just looks bizarre. The vertical diametric phase distribution of the singly charged OV extracted from this phase portrait @Fig. Saturations constitute a severe restriction for stabilization of system. The long time dynamics are. All chapters conclude with Exercises. Chua}, year={1969} }. We describe the phase portrait for bang-bang extremals. \u2022 Understand the linearized models using the \u201ceigen-techniques\u201d you learned earlier. 2 Phase portraits \u2022 A phase portrait of an n-dimensional autonomous system x \u2032 (t) = f (x (t)) is a graphical rep-resentation of the states in x-space. Nonlinear. Plot the maximum amplitude in the frequency. We draw the vector \ufb01eld given at each point (x,y) by the vector. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. 3 Determining Time from Phase Portraits 29 2. General Calendar. In this research a new graphic. In this section we will give a brief introduction to the phase plane and phase portraits. The phase portrait can indicate the stability of the system. Also, this work showed that the extreme multi-stability phenomenon of the behaviour of infinitely many coexisting attractors depends on the initial conditions of the variables of the system. motion of the system. In physical systems subject to disturbances, the distance of a stable equilibrium point to the boundary of its stable manifold provides an estimate for the robustness of the equilibrium point. In these cases the use of the phase portrait does not properly depict the system\u2019s evolution. 2 Global bifurcation analysis 69 3. When a double eigenvalue has only one linearly independent eigenvalue, the critical point is called an improper or degenerate node. The Poincar\u00b4e-Bendixson theorem Any orbit of a 2D continuous dynamical system which stays in a closed and bounded subset of the phase plane forever must either tend to a critical point or to a. , a particular state of the system) over time. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. 7 (2009), 369\u2013403. Draw the phase line of the equations and Answer. , another nonlinear system x_1 = 1 x3 1 x_2 = x1 x22 equilibrium points are described by x1 = 1 and x2 = 1 note: the equilibrium points of a nonlinear system can be nite (2 in the previous examples, but any other number is possible, including zero) or in nite, and they can be isolated points in state space Oriolo: Stability Theory for. 3 Symmetry in Phase Plane Portraits 22 2. For each case, we construct a phase space portrait by plotting the values of the dynamical variables after repeated application of the map (equation (1), followed by (6) and (7)) for a range of initial conditions. Extensibility of solutions 50 \u00a72. Equilibrium points. For optimal bang-bang trajectories with high values of the energy integral, a general upper bound on the number of switchings was obtained. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. doc Author: tien Created Date: 11/15/2002 4:16:10 AM. The phase space portraits of these two systems are shown in figure 6. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. Knowledge of \u03bb1 and \u03bb2, and v1 and v2, en-ables us to sketch the phase portrait near (x\u2217,y\u2217). Free system of non linear equations calculator - solve system of non linear equations step-by-step. \u2022 Understand the linearized models using the \u201ceigen-techniques\u201d you learned earlier. Consider a , possibly nonlinear, autonomous system ,(autonomous means that the independent variable , thoughtof as representing time, does not occur on the right sides of the equations). So, for a periodic system that obeys the law of energy conservation (e. Let's zoom into these four critical points, and look more closely at the phase portraits near them. If the system is period-n (the same state repeats after n clocks), there will be n points \u2212\u2192 period-n attractor. Students will learn nonlinear differential equations in the context of mathematical modeling. 26 Phase Portrait for sand Y1 Magnitudes of 9. Each set of initial conditions is represented by a different curve, or point. o Equilibrium solution \u2022 Exponential solutions o Half-line solutions \u2022 Unstable solution \u2022 Stable solution \u2022 Six important cases for portraits Real Eigenvalues o Saddle point o Nodal sink o Nodal source. An example of a non-linear system: the predator / prey model. The phase portrait of a dynamical system can be reconstructed from the observation of a single variable by the method of delays as proposed by [1]. The x-nullclineis a set of points in the phase plane so that. An example of a non-linear system: the predator / prey model. Phase Portraits of Nonlinear Systems. We draw the vector \ufb01eld given at each point (x,y) by the vector. Damped Pendulum. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. 2 Singular Points 20 2. warn(warning_msg, ODEintWarning). : A = 1 4 2 \u22121 \u03bb1 = 3 \u2194 v1 = [2,1]T \u03bb2 = \u22123 \u2194 v2 = [\u22121,1]T x\u2019=x+4y, y\u2019=2x\u2212y \u22125 0 5 \u22125 0 5 x y Time Plots for \u2018thick\u2019 trajectory. Linear and Nonlinear Systems of Differential Equations. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. The x nullcline is given by (1 x y)x = 0 =) x = 0 or y = 1 x: (11) Sodx dt= 0 on the lines x = 0 and y = 1 x. Phase portraits and Hooke diagrams of the proposed driven nonlin-ear system are consistent with empirical observations. Second-Order Systems. Phase Portraits Now we turn to the third method of analyzing non-linear systems, phase portraits generated by numerical solutions. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. two-dimensional phase distributions for OV beams with charges m51 ~a! and m52 ~b!. Local Phase Portrait of Nonlinear Systems Near Equilibria. In fact, if we zoom in around this point, it would look like the case of a node of a linear system (in the sense of Chapter 7). The low intensity xed point appears on the phase portraits. Fig 1: 3-D phase portrait of the Rossler attractor Integration Solver The Runge-Kutta method for a system of ordinary differential equations is explained here. In particular, show that some of the equilibria correspond to nonlinear centers, by nding a rst integral for this system. Compare the phase portraits of the linear and the nonlinear maps near the origin. Q: Find the phase portrait of this second-order nonlinear system with such differential equation: $$\\ddot{x}+0. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. Phase portraits and Hooke diagrams of the proposed driven nonlin-ear system are consistent with empirical observations. What is a Phase Portrait? Above, we have an animated phase portrait, but what is it? A phase portrait, in it\u2019s simplest terms, is when we plot one state of the system against another state of the system. In paper [12], the vibration of a typical two degree of freedom nonlinear system with repeated linearized natural frequencies was investigated systematically. Thus, the equilibrium x = 0 is a saddle, hence unstable, when = 0. 11 (Nonlinear terms can change a star into a spiral) Here's another example. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. John Polking\u2019s pplane: MATLAB, JAVA. The same concept can be used to obtain the phase portrait, which is a graphical description of the dynamics over the entire state space. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. Autonomous Planar Nonlinear Systems. Change parameters in the Linear-2D MAP so as to get the linearization at the origin. Following bifurcation in the system occurs in a range of parameter values g from 0. (c) The interior xed point at (1=3;1=3) is a global attractor. Changes in the dynamics of the orbits in the phase space usually represent variations of the physical parameters that control a non-linear system and consequently are of great importance for any modelling effort. This allowed to obtain exhaustive solution of the control problem comparing to the known results. systems) Suppose (x*,y*)=(0,0) is a linear center for a cont. 28 (1997), 755-778. 26 Phase Portrait for sand Y1 Magnitudes of 9. walking with Durus (right), showing phase portraits (top left) for 63 steps of walking together with a darker averaged phase portrait and position tracking errors (bottom left) over a select 4 steps in the same experiment. phase portrait get from simulink Example 2. Critical (equilibrium) points occur when (\u02d9x,y\u02d9) = (0,0). Key words: vibroimpact motion, unilateral and symmetrical rigid arrester, stereo-mechanical impact theory, phase portrait, two dimensional mapping. , is attracted to infinity. Students will learn nonlinear differential equations in the context of mathematical modeling. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. Solving 2x2 homogeneous linear systems of differential equations 3. Dynamical Systems and Chaos. 13 from the book. In this section we will give a brief introduction to the phase plane and phase portraits. Determine the stability of these limit cycles. Problem 4: Hamiltonian Systems. The phase portrait is a plot of a vector field which qualitatively shows how the solutions to these equations will go from a given starting point. MATLAB offers several plotting routines. By varying the parameters of the equation for the non-linear pendulum and then plotting. Phase Portraits of Nonhyperbolic Systems. Poincar\u00e9-Bendixon theorem. \u03c6 1 = phase shift of the fundamental harmonic component of output. the allee due at noon on friday sept 14th, in the box provided (to the. 1) Find all equilibrium points by solving the system 2) Let a standard software (e. One- and two- dimensional flows. Phase portrait of system of nonlinear ODEs. Save the phase portraits to submit on Gradescope. 2 : Linear analysis of nonlinear pendulum : Mechanical systems model for a pendulum. (4) (Formerly numbered 135A. The \ufb01xed points can be classi\ufb01ed according to their stability as follows: \u2022 If Re(\u03bb1) > 0 and Re(\u03bb2) > 0 \u21d2 repeller (unstable node). In this section we will give a brief introduction to the phase plane and phase portraits. Neural Information Processing Systems (NIPS). Smith; Nonlinear Ordinary Differential Equations, 3rd Edition, Oxford University Press, 1999. We illustrate all these cases in the examples below. This page plots a system of differential equations of the form dx/dt = f(x,y), dy/dt = g(x,y). Change parameters in the Linear-2D MAP so as to get the linearization at the origin. Perko, Di erential Equations and Dynamical Systems (Second edi-tion, Springer, 1996). Phase portraits are an invaluable tool in studying dynamical systems. \u00b5< 0 \u00b5< 0 \u00b5< 0 \u00b5 x Figure 5. Let us discuss the basic concept of describing the function of non linear control system. Q: Find the phase portrait of this second-order nonlinear system with such differential equation:$$ \\ddot{x}+0. These are systems that do not depend explicitly on. Determining time from phase portraits. nonlinear transform of coordinates and uses a full nonlinear system\u2019s model. Planar Almost Linear Systems: Phase portraits, Nonlinear classi- cations of equilibria. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. Fixed points occur at values \u03b8\u2217 such that 0 = 1 + 2cos\u03b8\u2217. which can be written in matrix form as X'=AX, where A is the coefficients matrix. 5 Global analysis for hardening nonlinear stiffness (c>0) 72 3. System analysis based on Lyapunov's direct method. Damped Pendulum. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector \ufb01eld superimposed. The figure shows the manner of convergence of these projected trajectories, which start with different initial conditions. In physical systems subject to disturbances, the distance of a stable equilibrium point to the boundary of its stable manifold provides an estimate for the robustness of the equilibrium point. Autonomous and non-autonomous systems Phase portraits and flows Attracting sets Concepts of stability 2. Lyapunov's direct method. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. We also show the formal method of how phase portraits are constructed. The undamped system has analytical solutions, (for both rotary and oscillatory motion), in terms of Jacobian Elliptic functions, that with the phase portrait gives a total qualitative and quantatitive explanation. These variables and their evolution in time produce the phase portrait of the system. Introduction to nonlinear network theory @inproceedings{Chua1969IntroductionTN, title={Introduction to nonlinear network theory}, author={L. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. : A = 1 4 2 \u22121 \u03bb1 = 3 \u2194 v1 = [2,1]T \u03bb2 = \u22123 \u2194 v2 = [\u22121,1]T x'=x+4y, y'=2x\u2212y \u22125 0 5 \u22125 0 5 x y Time Plots for 'thick' trajectory. It may be best to think of the system of equations as the single vector equation x y = f(x,y) g(x,y). The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. 2Switched Nonlinear Systems 1. We first focused on a nonlinear pendulum (shown in Fig. Phase Portrait for the linearization in Example 6. Phase portrait. Consider the nonlinear system (a) Show that the origin (O, O) is a nonlinear saddle awl plot the phase portrait, including the. 3 Determining Time from Phase Portraits 29 2. Generally, the nonlinear time series is analyzed by its phase space portrait. The motion of the mass is governed by Newton's second law. (iv) Since replacing x by x + 2\u2026 gives the same equations the portrait. Lyapunov's direct method. Phase Plane Analysis 17 2. Homework 6. Thompson and H. 1) \"For nonlinear systems, there is typically no hope of finding the trajectories analytically. Fixed points, limit cycles, and stability analysis. Giorgio Bertotti, Claudio Serpico, in Nonlinear Magnetization Dynamics in Nanosystems, 2009. 58=0 (b) 8+8+0. Such a planar curve is called a trajectory of the system and its param-eter interval is some maximal interval of existence T 1 \u03bb 0. Then try to combine the vector field with part (d) to get a global phase portrait of the original nonlinear system. I am unable to do for this case. 3 Phase Plane Portraits (for Planar Systems) Key Terms: \u2022 Equilibrium point of planer system. John Guckenheimer, in Handbook of Dynamical Systems, 2002. phase portrait (or phase diagram) for asystem depicts its phase space andtrajectories andis ageometricalrepresen- tation ofthe qualitative behavior ofthe system. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. These plots readily display vehicle stability properties and map equilibrium point locations and movement to changing parameters and system inputs. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto\u00ad ries of a linear system x = ax +by a, b, c, d constants. \u2022 A PLL is a control system that generates an output signal whose phase is related to the phase of the input and the feedback signal of the local oscillator. Differential equations are used to map all sorts of physical phenomena, from chemical reactions, disease progression, motions of objects, electronic circuits, weather forecast, et cetera. Solve system of nonlinear equations - MATLAB fsolve Nl. served: here we analyze this interplay by investigating the system using statistical tools, phase portraits, Poincar e sections, and return maps. Numerical Construction of Phase Portraits. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. A phase portrait represents the directional behavior of a system of ODEs. 1) As in \u00a7 3. Solving 2x2 homogeneous linear systems of differential equations 3. So, for a periodic system that obeys the law of energy conservation (e. Since in most cases it is. 1 Concepts of Phase Plane Analysis 18 2. Phase portrait. Let A= 3 \u22124 6 \u22127. { Nonlinear spring-mass system { Soft and hard springs { Energy conservation { Phase plane and scenes. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector \ufb01eld superimposed. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. Its phase is \u02c7=2, showing that the polariton resonance is below the laser line. Generally, the nonlinear time series is analyzed by its phase space portrait. Linear stability analysis may fail for a non-hyperbolic fixed point: ( Re(\u00b5 1, 2) = 0, or at least one \u00b5 i = 0 ). The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. 1 we draw the phase portrait (or phase diagram), where each point (x,y) corresponds to a speci\ufb01c state of the system. Each set of initial conditions is represented by a different curve, or point. 2 Phase portraits \u2022 A phase portrait of an n-dimensional autonomous system x \u2032 (t) = f (x (t)) is a graphical rep-resentation of the states in x-space. By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at \ud707 = 0. In this research a new graphic. Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). The phase portrait for the reduced dynamics for x is shown in Figure 5. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. Use technology to solve nonlinear programs, including computer programming and graphical analysis. These states can also be correlated with velocity spectral behaviors. (b) Find all bifurcation values of r and draw a bifurcation diagram on the r\u03b8-plane. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. phase portrait (or phase diagram) for asystem depicts its phase space andtrajectories andis ageometricalrepresen- tation ofthe qualitative behavior ofthe system. Chaos of such a system was predicted by applying a machine learning approach based on a neural network. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. The procedure was applied in modeling of self-excitation oscillations for high-speed milling and is based on determination of non-linear self-excitation force and non-linear coefficient. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto\u00ad ries of a linear system x = ax +by a, b, c, d constants. Consider the nonlinear system dx dt = r \u2212 x2, dy dt = x\u2212 y. 5 Summary of stability properties for planar ODE systems. One was the anticipated constancy of the system energy. By varying the initial conditions of the system, it is found. Design of feedback control systems. Overview of nonlinear mechanics of particles and nonlinear oscillations Lagrange\u2019s equations and nonlinear di erential equations Flows on a line and bifurcations Multi-dimensional ows and linear systems Phase portraits, stability, and limit cycles Dissipative systems, reversible systems, Index theory Weakly nonlinear oscillations and two. 2 Phase Plane Analysis. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. 2 $\\begingroup$ How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following? \\begin{align} \\dot{x} &= 2 - 8x^2-2y^2\\\\ \\dot{y} &= 6xy\\end{align}. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. HW # 7 Nonlinear Dynamics and Chaos Due: Monday, 95/01/30 1. SKETCH an approximate phase portrait for (6). Nonlinear Systems-lecture notes 4 Dr. 2 Global bifurcation analysis 69 3. As pointed out in @13#,a clear signature for the presence of a phase singularity is a new fringe starting at the location of the singularity. 3 Determining Time from Phase Portraits 29 2. The dynamics of airflow through the respiratory tract during VB and BB are investigated using the nonlinear time series and complexity analyses in terms of the phase portrait, fractal dimension, Hurst exponent, and sample entropy. In addition, if. The real space images in the top row show a portion of an unforced rotating spiral wave pattern [Fig. same'' means that type and stability for the nonlinear problem are the same as for the corresponding linear problem. d) Plot a computer-generated phase portrait to check your answer to (c). Weak non-linear oscillators and. 6: Phase portraits on the (one-dimensional) centr emanifoldandthebifurcation diagram. Effect of nonlinear terms. Now, if = 0, the system has one equilibrium point, x = 0. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. 2 Singular Points 20 2. 4 Phase Plane Analysis of Linear Systems 30. A simple example of a map is the Lotka-Volterra system describing two competing populations (e. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. Phase portrait. In this research a new graphic. MATLAB offers several plotting routines. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. \u00b5< 0 \u00b5< 0 \u00b5< 0 \u00b5 x Figure 5. Day 9 - Two Dimensional Systems - Phase Planes Day 10 - Two Dimensional Systems - Eigenvalues and Eigenvectors Day 11 - Nonlinear Two Dimensional Systems - Jacobian Day 12 - More practice with Two Dimensional Nonlinear Systems Day 13 - Bifurcations in 2-D Systems - Limit Cycles Day 14 - Hopf Bifurcations, Lorenz Equations, Chaos and Fractals. Nonlinear systems - existence and uniqueness theorem, continuous dependence, variational equations. ends up in one of the \ufb01xed points at x n = (2n+1)\u03c0. freedom and analysis of phase portraits, i. The nonlinear gyroscope model, which is employed in aerospace engineering [24], generally exhibits chaotic behavior. As the initial angle increases, we can see that the shape of the non-linear phase trajectory approaches that of the seperatrix. Albu-Schaffer. the Rossler system is sensitive to the initial conditions, and two close initial states will diverge, with increasing number of iterations. Different initial states result in different trajectories. Here we will consider systems for which direction fields and phase portraits are of particular importance. Based on velocity phase portraits, each of the nonlinear response states can be categorized into one of the three states in the order of increasing chaotic levels: lock-in, transitional, or quasiperiodic. A simple example of a map is the Lotka-Volterra system describing two competing populations (e. Classification of phase portraits. Ott, and A. Reversible Systems (2) THEOREM (Nonlinear centers for rev. The simple pendulum is a great example of a second-order nonlinear system that can be easily visualized by the phase portrait. Autonomous and non-autonomous systems Phase portraits and flows Attracting sets Concepts of stability 2. In sum, we illustrate the revised system\u2019s \ufb01t to the kinematics in both noncyclic speech and cyclic tasks (i. phase portrait get from simulink Example 2. The trace-determinant plane and stability. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. Learn more in: Chaotic Attractor in a Novel Time-Delayed System with a Saturation Function. Evolution of a dynamical system corresponds to a trajectory (or an orbit) in the phase space. John Polking\u2019s pplane: MATLAB, JAVA. Conclude: any i. 5 \\dot{x}+2 x+x^{2}=0 $$Method 1: Calculate by hands with phase plane analysis. What is a Phase Portrait? Above, we have an animated phase portrait, but what is it? A phase portrait, in it's simplest terms, is when we plot one state of the system against another state of the system. How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following?$$\\begin{align} \\dot{x} &= 2 - 8x^2-2y^2\\\\ \\dot{y} &= 6xy\\end{align}$$I can easily find the equilibria, which are$$\\left\\{ (0, \\pm 1), \\left(\\pm \\frac{1}{2}, 0\\right) \\right\\} The corresponding stable subspace for $\\left(\\pm \\frac{1}{2}, 0\\right)$ is. Limit Cycles: Recall that analysis of linearized systems. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. bifurcation diagrams and phase portraits. Click on the button corresponding to your preferred computer algebra system (CAS). Consider the nonlinear system dx dt = r \u2212 x2, dy dt = x\u2212 y. Phase portrait generator. Basics : Introduction to the notion of dynamical systems, examples of non-linear systems, Discrete and Continuous time, from one to the other, Poincar\u00e9 section. Instructors: Aldo Ferri: Topics: Introduction; properties of nonlinear systems; Phase portraits for second order systems; characterization of singular points and local stability; first and second methods of Lyapunov. Kitavtsev May 28, 2019 4 Local bifurcations of continuous and discrete dynamical systems The material of this chapter is covered in the following books: L. ) Lecture, three hours; discussion, one hour. A numerically generated phase-portrait of the non-linear system Zoomed in near (0,0) Zoomed in near (2,1) The critical point at (2,1) certainly looks like a spiral source, but (0,0) just looks bizarre. 4 Global analysis for softening nonlinear stiffness (c<0) 68 3. Materials to be covered include: nonlinear system characteristics, phase plane analysis, Lyapunov stability analysis, describing function method, nonlinear controller design. for x, where F ( x ) is a function that returns a vector value. (reductor and multipliers). Requisites: course 33B. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. As the initial angle increases, we can see that the shape of the non-linear phase trajectory approaches that of the seperatrix. This allowed to obtain exhaustive solution of the control problem comparing to the known results. Pauses are inserted between setting up the graphs; plotting the linear phase portrait for $$x = 2n\\pi$$; adding this behavior to the full phase plane; plotting the linear phase portrait for $$x = (2n+1)\\pi$$; adding that to the full phase. 01385 till 0. Click on the button corresponding to your preferred computer algebra system (CAS). 1 Phase Portraits 18 2. \u201cProof\u201d: Consider trajectory sufficiently close to origin time reversal symmetry. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Paragraphs 4. Around the origin there are periodic orbits corresponding to small oscillations of the pendulum that are called librations. population growth. First, we cover stability definitions of nonlinear dynamical systems, covering the difference between local and global stability. \u2022 \u03bb 12==\u03bb 0. Phase plane Analysis: 2nd order nonlinear systems, phase portrait graphical representation, singular points. Just as we did for linear systems, we want to look at the trajectories of the system. It starts in your web browser and you can directly input your equations and parameter values. The book is very readable even though it has a lot of jargon (read heavy mathematics). My professor told us to use a plotter to check our work (the hand-drawn phase portraits) but the one he linked to us won't work on my mac so I am trying to see the plots in Matlab but I don't know how to plot them and would be absolutely grateful for some help (I. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. which can be written in matrix form as X'=AX, where A is the coefficients matrix. The dynamics of airflow through the respiratory tract during VB and BB are investigated using the nonlinear time series and complexity analyses in terms of the phase portrait, fractal dimension, Hurst exponent, and sample entropy. One was the anticipated constancy of the system energy. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto\u00ad ries of a linear system x = ax +by a, b, c, d constants. Then sufficiently close to (0,0) all trajectories are closed curves. By varying the initial conditions of the system, it is found. Following bifurcation in the system occurs in a range of parameter values g from 0. Pages 486 - 493 cover the five important cases. Here we will consider systems for which direction fields and phase portraits are of particular importance. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. 3 Symmetry in Phase Plane Portraits 22 2. The behaviour of the system is investigated through numerical simulations, by using well-known tools of nonlinear theory, such as phase portrait, bifurcation diagram and Lyapunov exponents. Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). Control of bilateral teleoperation systems with linear and nonlinear dynamics. The study of nonlinear dynamics is based on the study of phase portraits, wavelet and Fourier spectra, signals, chaotic phase synchronization, Lyapunov indicators. Keywords Piecewise nonlinear system, rolling mill, non-smooth homoclinic orbit, bifurcation, chaos. Around the origin there are periodic orbits corresponding to small oscillations of the pendulum that are called librations. Planar linear systems - eigenvalues and eigenvectors, phase portraits, classification. MATLAB offers several plotting routines. That is, only initial points located on this orbit result in this closed orbit. As before, we use a phase portrait for stability analysis. We will look at three examples, and also reexamine the undamped pendulum that we studied previously using only its vector field. The higher degree of chaoticity in BB relative to VB is unwrapped through the maximal Lyapunov exponent. Phase Plane Analysis 17 2. For optimal bang-bang trajectories with high values of the energy integral, a general upper bound on the number of switchings was obtained. Phase portraits via trace and determinant. In fact, if we zoom in around this point, it would look like the case of a node of a linear system (in the sense of Chapter 7). (561); Notes LS (power series excluded), GS; Handout on phase portraits. Complex eigenvalues, phase portraits, and energy 4. Although the Riccati equation is not generally a Morse\u2013Smale vector field, we are able to show that it possesses suitable generalizations of many of the important properties of Morse\u2013Smale vector fields. If a system is chaotic, there will be an in\ufb01nite number of points in the phase portrait. Thus, the equilibrium x = 0 is a saddle, hence unstable, when = 0. (reductor and multipliers). Consider the non-linear system dx dt = y dy dt = 2x+(1x2 y )y. (a) Compute the eigenvalues of A. As a result of one more Andronov-Hopf bifurcation more complex quasiperiodic solution is formed in the system\u2014it is torus of dimension three. 2 Constructing Phase Portraits 23 2. 01385 till 0. Animated phase portraits of nonlinear and chaotic dynamical systems allow one to shape globally the state- and time-dependent convergence behaviour ideally suited to the non-linear or time. This approach of linearizing, analyzing the linearizations, and piecing the results together is a standard approach for non-linear systems. : A = 1 4 2 \u22121 \u03bb1 = 3 \u2194 v1 = [2,1]T \u03bb2 = \u22123 \u2194 v2 = [\u22121,1]T x'=x+4y, y'=2x\u2212y \u22125 0 5 \u22125 0 5 x y Time Plots for 'thick' trajectory. the trajectories of the nonlinear system are similar to those of the linearized system, so go round anticlockwise. One- and two- dimensional flows. 2 Phase Plane Analysis. [2] Consider x\u2032 1 = 5x1 \u2212x2 1 \u2212 x1x2, x\u2032 2 = \u22122x2 +x1x2. \u2022 Be able to determine the phase plane and phase portraits of a 2 by 2 linear system. By varying the initial conditions of the system, it is found. 2~a!# is. 2 Bifurcation set and phase portraits of the Hamiltonian system (5). Several nonlinear wave solutions as the solitary wave solutions,topological solitons, cnoidal wave solutions, singular periodic waves and others were obtained. \u03c6 1 = phase shift of the fundamental harmonic component of output. 4 Phase Plane Analysis of Linear Systems 30. REFERENCES [1] Berrymann, A. Phase portraits are an invaluable tool in studying\u2026. Putting all this together we see that the phase portrait is as shown below. Assume that r > 0. In a planar system such as this, the nullclines can provide useful information about the phase portrait. The phase portrait is a plot of a vector field which qualitatively shows how the solutions to these equations will go from a given starting point. We show that our model can recover qualitative features of the phase portrait such as attractors, slow points, and bifurcations, while also producing reliable long-term future predictions in a variety of dynamical models and in real neural data. 2 Prey dynamics predicted by the Lotka-Volterra predator-prey model. \u201cProof\u201d: Consider trajectory sufficiently close to origin time reversal symmetry. Existence of Periodic Orbits. The Poincar\u00b4e-Bendixson theorem Any orbit of a 2D continuous dynamical system which stays in a closed and bounded subset of the phase plane forever must either tend to a critical point or to a. Linear stability analysis may fail for a non-hyperbolic fixed point: ( Re(\u00b5 1, 2) = 0, or at least one \u00b5 i = 0 ). Damped Pendulum. 3 Symmetry in Phase Plane Portraits 22 2. Solving 2x2 homogeneous linear systems of differential equations 3. Problem: Construct and analyze a phase-plane portrait of a nonlinear system depicted in the following picture (desired value is w = 0), decide which of the equilibrium points are stable and which are not. Vehicle control synthesis using phase portraits of planar dynamics ABSTRACTPhase portraits provide control system designers strong graphical insight into nonlinear system dynamics. Let us discuss the basic concept of describing the function of non linear control system. 6 and the phase portrait for the original system is in Figure 5. System analysis based on Lyapunov's direct method. and sketch the phase portrait on the circle. Then sufficiently close to (0,0) all trajectories are closed curves. Determining time from phase portraits. Normalized phase portraits or cylindrical phase portraits have been extensively used to overcome the original phase portrait\u2019s disadvantages. x c c c t ert yert y c c c t 1 2 2, 1 2 2 Case 3: Phase Portraits (5 of 5) The phase portrait is given in figure (a) along with several graphs of x1 versus t are given below in figure (b). ) Lecture, three hours; discussion, one hour. the behavior of the nonlinear system from various initial conditions. Jordan, Peter Smith, and P. 2 Phase portrait for an example system. phase portrait get from simulink Example 2. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. Students will learn nonlinear differential equations in the context of mathematical modeling. This allowed to obtain exhaustive solution of the control problem comparing to the known results. Unit 2: Nonlinear 2x2 systems. 1 Phase portraits 72 3. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. The low intensity xed point appears on the phase portraits. Consider the nonlinear system (a) Show that the origin (O, O) is a nonlinear saddle awl plot the phase portrait, including the.\n7iw5loreijl ph2v1q1oh5ul19 5k7fv8hml9 7xnle9owrd873g3 zemot27ga77zfj 2719yh9klabxg s5ayt0gpsfksc 6vgg4tkpyb 1ziuxc88k8 ddchgfzhyl2klos qd1nhu4hxqyrf 5gsl71di0g8 2sbwr9xfsu foxd7lu32k wfpaeculwor zsku1acu2g huqrq9k4w2k7z33 3ynsy8bxwot28 cmljwia08r0doe njpxzvhz1r1bi qkwl1pfv5l5uwwp fjcx8u9t50uv0r irhvlt0kroa 3g5hypzcrk d6jq3elv7llo 5de7be0jgurut8i exw9yp069m2 zv0in78g3fj k50lh4m3ektdbz adx9cpa9kard zoid1zghpq oba3px70bc mrfddo0ps2oh5 z7gwnvcbaa57r2f ytq21pbsd171", "date": "2020-11-27 11:35:06", "meta": {"domain": "leforriclub.it", "url": "http://matv.leforriclub.it/phase-portrait-nonlinear-system.html", "openwebmath_score": 0.5916582942008972, "openwebmath_perplexity": 791.4518817998761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9910145704512985, "lm_q2_score": 0.8596637523076224, "lm_q1q2_score": 0.8519393042256899}}
{"url": "http://math.stackexchange.com/questions/721885/two-circles-inside-a-right-angle?answertab=votes", "text": "Two Circles inside a right angle!\n\nThe other day I was playing with Ms Paint drawing circles here and there -I coincidentally drew a circle inside a right angled triangle which I already drew . Strangely A problem struck to my mind and I tried solving it , but I was unable to do so . I put forward the statement of the problem which I managed to frame my self\n\nProblem : The legs of a right angled triangle are of length $a$ and $b$ . Two circles with equal radii are drawn such such that they touch each other and sides of the triangle as shown in the figure . Find the radius of the circle in terms of $a$ and $b$.\n\nFigure - Of course my MS Paint one -\n\nFurther Scope- Is there any way to generalize this for other shapes or for any other triangle?\n\n-------EDIT---------------------\n\nNow to make things interesting : Say we have a right angled triangle which is given . Then is there a method by which we can construct those two circles with a straightedge and a compass?\n\n-\nPerhaps this topic may be useful? mathworld.wolfram.com/Excircles.html \u2013\u00a0 Erik Miehling Mar 22 at 6:25\n@ErikMiehling thanks but it isn't helping much! \u2013\u00a0 Shivam Patel Mar 22 at 6:39\n@ShivamPatel Can you give an example of a generalization? I don't really know \"right pentagons\" and \"right hexagons\". \u2013\u00a0 DanielV Mar 22 at 6:51\nOr perhaps a generalization to multidimensional right triangles and hyperspheres would be meaningful? \u2013\u00a0 DanielV Mar 22 at 7:03\nStraightedge and compass construction: draw an arbitrary circle with center $C$ that intersects the legs $AC, BC$ at $D, E$, respectively. Locate $F$ on $AC$ such that $CF = 3CA$. Locate $G$ such that $ECFG$ is a rectangle. Draw $CG$. Draw the angle bisector of $A$. The intersection of this angle bisector with $CG$ is the center of one of the two circles. \u2013\u00a0 heropup Mar 23 at 3:44\n\nWriting $a := |BC|$, $b := |CA|$, $c := |AB| = \\sqrt{a^2+b^2}$, and $r = |PE| = |PF|$ (so that $|PD| = 3r$), we have \\begin{align} |\\triangle ABC| &= |\\triangle ABP| + |\\triangle BCP| + |\\triangle CAP| \\\\[4pt] \\implies \\qquad \\frac{1}{2} |BC||CA| &= \\frac{1}{2} \\left(\\; |AB| |PF| + |BC||PD| + |CA||PE| \\;\\right) \\\\[4pt] \\implies \\qquad a b &= c r + 3 a r + b r = r ( 3 a + b + c )\\\\[6pt] \\implies \\qquad r &= \\frac{ab}{3 a + b + c} = \\frac{ab}{3 a + b + \\sqrt{a^2+b^2}} \\end{align}\n\nTo address @DanielV's suggestion of generalizing to higher dimensions, consider a right-corner tetrahedron $OABC$, with right corner at $O$ and edge lengths $a := |OA|$, $b := |OB|$, $c := |OC|$. (Note that I'm changing notation slightly from the above.) Let a sphere with center $P$ and radius $r$ be tangent to the faces around vertex $A$, and let a congruent sphere (tangent to the first) be tangent to the faces around vertex $O$. Then $P$ has distance $r$ from faces $\\triangle OAB$, $\\triangle OCA$, $\\triangle ABC$ (the ones touching $A$), and distance $3r$ from face $\\triangle OBC$ (the one opposite $A$).\n\nHere's a poor attempt at a diagram:\n\n(In this case, the altitudes from $P$ are color-coded to match their parallel counterparts through $O$. The black altitude is to face $\\triangle ABC$.)\n\nThus,\n\n\\begin{align} |OABC| &= |OABP| + |OBCP| + |OCAP| + |ABCP| \\\\[4pt] \\implies \\qquad \\frac{1}{6}a b c &= \\frac{1}{3}\\left(\\; r\\;|\\triangle OAB| + r \\;|\\triangle OCA| + r\\;|\\triangle ABC| + 3r\\;|\\triangle OBC| \\;\\right) \\\\[4pt] &= \\frac{1}{3}r \\cdot \\frac{1}{2} \\left(\\; a b + c a + 3 b c + 2\\;|\\triangle ABC| \\;\\right) \\\\[6pt] \\implies \\qquad r &= \\frac{abc}{3bc + ab + ca + 2\\;|\\triangle ABC|} \\qquad (\\star) \\end{align}\n\nFun fact: The Pythagorean Theorem for Right-Corner Tetrahedra says that $$|\\triangle ABC|^2 = |\\triangle OBC|^2 + |\\triangle OCA|^2 + |\\triangle OAB|^2$$ so that we have $$|\\triangle ABC| = \\frac{1}{2} \\sqrt{\\; b^2 c^2 + c^2 a^2 + a^2 b^2 \\;}$$ and $(\\star)$ becomes $$r = \\frac{abc}{3bc + ab + ca + \\sqrt{\\; b^2 c^2 + c^2 a^2 + a^2 b^2 \\;}}$$\n\nIn $4$-dimensional space (where there's an analogous Pythagorean Theorem, as there is in any-dimensional space), we have $$r = \\frac{abcd}{3bcd + acd + abd + abc + \\sqrt{\\;b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2\\;}}$$ and so forth.\n\nIncidentally, the matching-notation version of the initial answer is $$r = \\frac{ab}{3b + a + \\sqrt{\\;b^2 + a^2\\;}}$$\n\n-\n@qwr: Thanks. :) As for coloring: I just like to highlight the association of a vertex with its opposite edge in a triangle, and other elements get their colors accordingly. I chose not to color-code the altitudes dropped from $P$ to the edges (I had reflexively done so in a \"draft\" image), in order to tie the congruent segments together visually. \u2013\u00a0 Blue Mar 22 at 8:00\n@mathh: I use GeoGebra for my figures these days. \u2013\u00a0 Blue Mar 22 at 12:02\n@AJP: \"$|BC|$\" represents the length of segment $\\overline{BC}$. (I often leave out the over-bar to save typing.) By analogy, I use \"$|\\triangle ABC|$\" for the area of $\\triangle ABC$ (and then \"$|OABC|$\" for the volume of tetrahedron $OABC$, etc); this is non-standard, but I like it. :) The symbol \":=\" indicates \"is defined to be\"; writing \"$a:=|BC|$\" says \"I'm going to write '$a$' for '$|BC|$'\". Others (even I) might otherwise write \"let $a = |BC|$\", but using \":=\" is ever-so-slightly better, as it distinguishes definition \"'$a$' means '$|BC|$'\" from relation \"$a$ equals $|BC|$\". \u2013\u00a0 Blue Mar 22 at 13:10\n@ShivamPatel: I added a figure that may (or may not) help with the $3d$ case. Just don't ask me for the $4d$ version. :) \u2013\u00a0 Blue Mar 22 at 14:46\n@mathh: In GeoGebra, you can easily mark an angle with an arc via the \"Angle\" tool; if the sides of the angle happen to be perpendicular, GeoGebra turns the arc into a box. (There's a setting somewhere to turn that behavior off and on.) The little dashes are somewhat less convenient: with the segment selected, you have to open the \"Object Properties...\" panel and then the \"Decoration\" tab. (You can likewise \"decorate\" angles with little dashes and such.) \u2013\u00a0 Blue Mar 22 at 15:28\n\nAssuming the corner of the triangle is the origin of a Cartesian plane, the line of the hypotenuse is $$y = -\\frac{a}{b} x + a$$\n\nThe center of the second circle is at $\\begin{bmatrix} 3r \\\\ r \\end{bmatrix}$.\n\nThe radius of the second circle is the directed vector $\\begin{bmatrix} ra \\\\ rb \\end{bmatrix}\\frac{1}{\\sqrt{a^2 + b^2}}$.\n\nAltogether :\n\n$$\\frac{rb}{\\sqrt{a^2 + b^2}} + r = -\\frac{a}{b}\\left(3r + \\frac{ra}{\\sqrt{a^2 + b^2}}\\right) + a$$\n\n$$r = \\frac{ab}{ 3a + b + \\sqrt{a^2 + b^2}}$$\n\nSince Blue showed an elegant volume based approach, I guess I'll try a multidimensional coordinate based approach. Suppose the origin $\\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\vdots \\end{bmatrix}$ and the points $\\begin{bmatrix} \\mathcal{l}_0\\\\ 0 \\\\ 0 \\\\ \\vdots \\end{bmatrix}$, $\\begin{bmatrix} 0 \\\\ \\mathcal{l}_1 \\\\ 0 \\\\ \\vdots \\end{bmatrix}$, $\\begin{bmatrix} 0 \\\\ 0 \\\\ \\mathcal{l}_2 \\\\\\vdots \\end{bmatrix}$ etc form a multidimensional right triangle. Suppose one hypersphere is tucked in at the origin and one hypersphere is tucked in at the corner along the first axis.\n\nThe hypotenuse plane of the triangle is $$\\frac{x_0}{\\mathcal{l}_0} + \\frac{x_1}{\\mathcal{l}_1} + \\frac{x_2}{\\mathcal{l}_2} + \\dots = 1 \\tag{Plane equation}$$\n\nOr equivalently, using $\\circ$ for dot product and $N = \\begin{bmatrix} \\frac{1}{\\mathcal{l}_0} \\\\ \\frac{1}{\\mathcal{l}_1} \\\\ \\frac{1}{\\mathcal{l}_2} \\\\ \\vdots \\end{bmatrix}$, then the plane is: $$N \\circ X = 1 \\tag{Plane equation with dot product}$$\n\nThe center of the second hypershpere is $$c = \\begin{bmatrix} 3r \\\\ r \\\\ r \\\\ \\vdots\\end{bmatrix} = rc_0 \\tag{Center of second hypersphere}$$\n\nThe vector directed from the center of the second hypersphere to the hypotenuse plane is $$r_2 = r\\frac{N}{|N|}\\tag{Directed radius to hypotenuse}$$\n\nAltogether: $$N \\circ (c + r_2) = 1$$ $$N \\circ \\left(rc_0 + r\\frac N {|N|}\\right) = 1$$ \\begin{align} r &= \\frac{1}{N \\circ c_0 + |N|} \\\\ &= \\frac{1}{3l_0^{-1} + l_1^{-1} + l_2^{-1} \\dots + \\sqrt{l_0^{-2} + l_1^{-2} + l_2^{-2} \\dots}} \\end{align}\n\nNote this is the same answer as Blue, but with $\\mathcal{l}_0\\mathcal{l}_1 \\mathcal{l}_2 \\dots$ factored out of the numerator and denominator.\n\n-\n+1. It's worth nothing that your multi-dimensional argument effectively re-derives the formula for the distance from point $P(p_0,p_1,\\dots,p_n)$ to hyperplane $Q:\\;q_0x_0+q_1x_1+\\cdots+q_nx_n=q$; namely, $$\\text{dist}(P,Q)=\\frac{|\\;p_0q_0+p_1q_1+\\cdots+p_nq_n-q\\;|}{\\sqrt{\\;q_0^2+ q_1^2 +\\cdots+q_n^2\\;}}$$ Replacing $p_0=3r$, $p_1=\\cdots=p_n=r$, $q_i=l^{-1}_{i}$, and $q=1$, and then setting the computed distance equal to $r$, gives the equation you've solved for $r$. (Note: Dropping the formula's absolute value sign makes the distance negative here, since $P$ is on the \"origin-side\" of $Q$.) \u2013\u00a0 Blue Mar 22 at 22:29\nInteresting how many different ways there can be to reach the same result. This is one reason I say that instead of memorizing formulas, students would appreciate mathematics more if they saw the \"magic\" of how many seemingly unrelated approaches give the same result in the end. \u2013\u00a0 DanielV Mar 22 at 23:38\n@DanielV Can you look at the edit please... \u2013\u00a0 Shivam Patel Mar 23 at 3:24\n@ShivamPatel Look at youtube.com/watch?v=CMP9a2J4Bqw espcially after 00:58 where it describes what is possible with a compass and straightedge. In short the answer is \"yes\", by applying what you see in the video to the formulas Blue and I gave you. ...But I doubt there is any short way to do it. Keep in mind that you effectively assume $a=1$ or $b=1$ since the units are arbitrary. \u2013\u00a0 DanielV Mar 23 at 3:37\n\nAlternative solution:\n\nRecall that for any triangle $\\triangle ABC$, the area of the triangle equals the product of the inradius and its semiperimeter; i.e., $|\\triangle ABC| = rs$, where $s = (a+b+c)/2$. Therefore, given legs $a, b$, $c = \\sqrt{a^2+b^2}$, and $$r = \\frac{2|\\triangle ABC|}{a+b+c} = \\frac{ab}{a+b+\\sqrt{a^2+b^2}}.$$ Draw the tangent line to the two circles at their common point of tangency: this creates a smaller similar triangle with scaling factor $\\frac{b-2\\rho}{b}$ where $\\rho$ is the common radius of the two circles. If $r$ is the inradius of $|\\triangle ABC|$ as shown, then $$\\frac{\\rho}{r} = \\frac{b - 2\\rho}{b}.$$ Putting all of this together, we find $$\\rho = \\frac{br}{b+2r} = \\frac{a b}{3a+b+c} = \\frac{ab}{3a+b+\\sqrt{a^2+b^2}}.$$ This method easily generalizes to more than two congruent circles tangent to one side: if $n$ circles are arranged along the leg of length $b$, then it is straightforward to find that $\\rho = \\frac{ab}{(2n-1)a + b + \\sqrt{a^2+b^2}}$.\n\n-\n\nWhat about determining $a,b$ from $r$?\nYou can't - $a,b$ are not uniquely determined by $r$. Draw the two circles first, then draw the legs a and b as infinite lines. And line tangent to the right most circle will intersect lines A and B creating a right triangle, but for different tangents $a$ and $b$ will be different and r hasn't changed.\nYou can't even turn this into an interesting question by asking for all the possible ways to do it, because it turns out that these circles don't actually impose any restriction - since any angels can be achieved by the tangent line, any right triangle can. All $r$ does is create a scaling factor. $a$ can take on any value in $(2r,\\infty)$ and $b$ the corresponding value in $(4r,\\infty)$\nsuppose we say $a=kb$ for some fixed $k$ then? \u2013\u00a0 Shivam Patel Mar 22 at 6:46\nI believe the question was to find the radius, given $a$ and $b$. \u2013\u00a0 N. Owad Mar 22 at 6:48", "date": "2014-11-29 07:31:15", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/721885/two-circles-inside-a-right-angle?answertab=votes", "openwebmath_score": 0.9760359525680542, "openwebmath_perplexity": 620.1165665347083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9910145715128204, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.851939296230479}}
{"url": "https://math.stackexchange.com/questions/2360603/how-to-find-the-intersection-of-two-hyperplanes-in-n-dimensions/2369012", "text": "# How to find the intersection of two hyperplanes in $n$ dimensions?\n\nI want to find out the intersection of two surfaces that exist in $n$-dimensions. These surfaces are defined by a system of $2$ linear equations.\n\n$$A_1x+B_1y+C_1z+\\cdots = D_1$$\n\n$$A_2x+B_2y+C_2z+\\cdots = D_2$$\n\nBut first, how would these surfaces look like? In $3$-D space, they would be planes, but in dimensions higher than $3$ are these not planes? Can we generalize a way to find the intersection of such linear expressions in some way?\n\nAlso can this problem be solved using some tools, preferably Matlab?\n\n\u2022 The surfaces are hyperplanes. Solve a linear system of the form $\\rm A x = b$, where $\\rm A$ is a $2 \\times n$ matrix. If $n > 2$, the linear system is underdetermined. If $n=3$, the intersection could be a line. If $n=4$, the intersection could be a $2$-dimensional plane. \u2013\u00a0Rodrigo de Azevedo Jul 16 '17 at 15:27\n\u2022 You can use Gaussian elimination to find a parametrization of the affine space that is the intersection, assuming that the intersection is not empty. \u2013\u00a0Rodrigo de Azevedo Jul 16 '17 at 15:31\n\u2022 @RodrigodeAzevedo And how do I figure out t1 and t2? \u2013\u00a0Vidor Vistrom Jul 16 '17 at 15:44\n\u2022 Suppose we have $$\\begin{bmatrix} 1 & 0 & -1 & -1\\\\ 0 & 1 & -1 & -1\\end{bmatrix} \\begin{bmatrix} x_1\\\\ x_2\\\\ x_3\\\\ x_4\\end{bmatrix} = \\begin{bmatrix} 1\\\\ 1\\end{bmatrix}$$ The solution set is the plane parameterized as follows $$\\left\\{ \\begin{bmatrix} 1\\\\ 1\\\\ 0\\\\ 0\\end{bmatrix} + t_1 \\begin{bmatrix} 1\\\\ 1\\\\ 1\\\\ 0\\end{bmatrix} + t_2 \\begin{bmatrix} 1\\\\ 1\\\\ 0\\\\ 1\\end{bmatrix} : t_1, t_2 \\in \\mathbb R \\right\\}$$ \u2013\u00a0Rodrigo de Azevedo Jul 16 '17 at 15:48\n\u2022 @RodrigodeAzevedo: I suggest that you post your solution, perhaps mentioning the reduced row-echelon form of the augmented matrix for the $2\\times n$ system, as an Answer. I will certainly upvote even if you mostly transcribe your Comments into the Answer box. \u2013\u00a0hardmath Jul 16 '17 at 21:49\n\nSuppose we would like to find the intersection of $2$ hyperplanes in $\\mathbb R^n$\n\n$$\\begin{array}{rl} a_{11} x_1 + a_{12} x_2 + \\cdots + a_{1n} x_n &= b_1\\\\ a_{21} x_1 + a_{22} x_2 + \\cdots + a_{2n} x_n &= b_2\\end{array}$$\n\nIn matrix form,\n\n$$\\begin{bmatrix} \u2014 \\mathrm a_1^\\top \u2014\\\\ \u2014 \\mathrm a_2^\\top \u2014\\end{bmatrix} \\mathrm x = \\begin{bmatrix} b_1\\\\ b_2\\end{bmatrix}$$\n\nor, more economically, $\\rm A x = b$, where $\\rm A$ is a $2 \\times n$ matrix. Without loss of generality, we assume that both hyperplanes are in Hessian normal form, i.e., $\\| \\mathrm a_1 \\|_2 = \\| \\mathrm a_2 \\|_2 = 1$.\n\nIf vectors $\\rm a_1$ and $\\rm a_2$ are collinear, then the $2$ given hyperplanes are either parallel or overlapping, neither of which are very interesting. Thus, we assume that vectors $\\rm a_1$ and $\\rm a_2$ are not collinear, i.e., matrix $\\rm A$ has full row rank. Let $\\theta := \\arccos ( \\mathrm a_1^\\top \\mathrm a_2 )$ be the angle formed by vectors $\\rm a_1$ and $\\rm a_2$.\n\nThe solution set of the linear system $\\rm A x = b$ is a $(n-2)$-dimensional affine space. To find this affine space, we must find a particular solution and the null space of $\\rm A$. One particular solution would be the least-norm solution, i.e., the one closest to the origin\n\n$$\\begin{array}{rl} \\mathrm x_{\\text{LN}} := \\mathrm A^\\top \\left( \\mathrm A \\mathrm A^\\top \\right)^{-1} \\mathrm b &= \\begin{bmatrix} | & | \\\\ \\mathrm a_1 & \\mathrm a_2\\\\ | & |\\end{bmatrix} \\begin{bmatrix} \\| \\mathrm a_1 \\|_2^2 & \\langle \\mathrm a_1, \\mathrm a_2 \\rangle\\\\ \\langle \\mathrm a_2, \\mathrm a_1 \\rangle & \\| \\mathrm a_2 \\|_2^2\\end{bmatrix}^{-1} \\begin{bmatrix} b_1\\\\ b_2\\end{bmatrix}\\\\ &= \\begin{bmatrix} | & | \\\\ \\mathrm a_1 & \\mathrm a_2\\\\ | & |\\end{bmatrix} \\begin{bmatrix} 1 & \\cos (\\theta)\\\\ \\cos (\\theta) & 1\\end{bmatrix}^{-1} \\begin{bmatrix} b_1\\\\ b_2\\end{bmatrix}\\\\ &= \\frac{1}{\\sin^2 (\\theta)} \\begin{bmatrix} | & | \\\\ \\mathrm a_1 & \\mathrm a_2\\\\ | & |\\end{bmatrix} \\begin{bmatrix} 1 & -\\cos (\\theta)\\\\ -\\cos (\\theta) & 1\\end{bmatrix} \\begin{bmatrix} b_1\\\\ b_2\\end{bmatrix}\\\\ &= \\left(\\frac{b_1 - b_2 \\cos (\\theta)}{\\sin^2 (\\theta)}\\right) \\begin{bmatrix} | \\\\ \\mathrm a_1 \\\\ | \\end{bmatrix} + \\left(\\frac{b_2 - b_1 \\cos (\\theta)}{\\sin^2 (\\theta)}\\right) \\begin{bmatrix} | \\\\ \\mathrm a_2 \\\\ | \\end{bmatrix}\\end{array}$$\n\nwhere $\\sin (\\theta) \\neq 0$ is ensured by the non-collinearity of vectors $\\rm a_1$ and $\\rm a_2$.\n\nTo find the $(n-2)$-dimensional null space of $\\rm A$, we solve the homogeneous linear system $\\rm A x = 0_2$. Introducing an $n \\times n$ permutation matrix $\\rm P$ with certain desired properties, we reorder the columns of $\\rm A$, i.e., $\\rm A P P^\\top x = 0_2$, and obtain a homogeneous linear system in $\\rm y:= P^\\top x$\n\n$$\\rm A P y = 0_2$$\n\nIf $\\rm P$ is chosen wisely, then there is a $2 \\times 2$ matrix $\\rm E$, a product of elimination matrices, that puts $\\rm A P$ in reduced row echelon form (RREF), i.e.,\n\n$$\\rm E A P = \\begin{bmatrix} \\mathrm I_2 & \\mathrm F\\end{bmatrix}$$\n\nwhere $\\rm F$ is a $2 \\times (n-2)$ matrix. Hence, the null space of $\\rm A P$ is given by\n\n$$\\left\\{ \\begin{bmatrix} -\\mathrm F \\\\ \\mathrm I_{n-2} \\end{bmatrix} \\eta : \\eta \\in \\mathbb R^{n-2} \\right\\}$$\n\nand, thus, the null space of $\\rm A$ is given by\n\n$$\\left\\{ \\mathrm P \\begin{bmatrix} -\\mathrm F \\\\ \\mathrm I_{n-2} \\end{bmatrix} \\eta : \\eta \\in \\mathbb R^{n-2} \\right\\}$$\n\nLastly, the intersection of the $2$ given hyperplanes is given by\n\n$$\\left\\{ \\color{blue}{\\mathrm x_{\\text{LN}} + \\mathrm P \\begin{bmatrix} -\\mathrm F \\\\ \\mathrm I_{n-2} \\end{bmatrix} \\eta} : \\eta \\in \\mathbb R^{n-2} \\right\\}$$\n\nIn MATLAB, we can use function rref to find permutation matrix $\\rm P$ and matrix $\\rm F$. Of course, we can also use function null to find an orthonormal basis for the null space of $\\rm A$.", "date": "2019-06-27 06:50:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2360603/how-to-find-the-intersection-of-two-hyperplanes-in-n-dimensions/2369012", "openwebmath_score": 0.9251582622528076, "openwebmath_perplexity": 154.00284689357713, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9932024684496267, "lm_q2_score": 0.8577681031721325, "lm_q1q2_score": 0.8519373974279161}}
{"url": "http://mathhelpforum.com/calculus/3132-need-help-definite-integral-please.html", "text": "1. ## Need help with a definite integral please\n\nI would appreciate advice on solving the following problem.\n\nEvaluate the definite integral:\n\nintegral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx\n\nHere is my own attempt.\n\nStep 1: Solve for the generic (indefinite) integral via substitution\n\nLet u = x^2 + 1\n\nSo then\n\ndu / dx = 2x\n\n2x dx = du\n\ndx = du / 2x\n\nReturning to the original function:\n\nintegral 2x (x^2 + 1) dx\n= integral 2x (u) dx\n= integral 2x (u) (du / 2x)\n= integral u du\n= integral x^2 + 1\n= (x^3 / 3 ) + x + C\nStep 2: solve for the definite integral given the stated limits\n\nAt upper limit of 2:\n\n((2)^3 / 3) + (2) +C\n= (8/3) +2 + C\n= (8/3) + (6/3) + C\n= 14/3 + C\n\nAt lower limit of 1\n\n((1)^3 / 3) + (1) +C\n=(1/3) +1 + C\n= (1/3) + (3/3) + C\n= 4/3 + C\n\nSubtracting the lower limit result from upper limit result:\n\n(14 /3 + C) - (4/3 + C)\n=14/3 + C - 4/3 - C\n= 10 / 3\n\nBut my textbook and calculator both say the answer is (21/2)\nWhat am I doing wrong?\n\n2. Originally Posted by lingyai\nI would appreciate advice on solving the following problem.\n\nEvaluate the definite integral:\n\nintegral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx\n\nHere is my own attempt.\n\nStep 1: Solve for the generic (indefinite) integral via substitution\n\nLet u = x^2 + 1\n\nSo then\n\ndu / dx = 2x\n\n2x dx = du\n\ndx = du / 2x\n\nReturning to the original function:\n\nintegral 2x (x^2 + 1) dx\n= integral 2x (u) dx\n= integral 2x (u) (du / 2x)\n= integral u du\n= integral x^2 + 1\n= (x^3 / 3 ) + x + C\nStep 2: solve for the definite integral given the stated limits\n\nAt upper limit of 2:\n\n((2)^3 / 3) + (2) +C\n= (8/3) +2 + C\n= (8/3) + (6/3) + C\n= 14/3 + C\n\nAt lower limit of 1\n\n((1)^3 / 3) + (1) +C\n=(1/3) +1 + C\n= (1/3) + (3/3) + C\n= 4/3 + C\n\nSubtracting the lower limit result from upper limit result:\n\n(14 /3 + C) - (4/3 + C)\n=14/3 + C - 4/3 - C\n= 10 / 3\n\nBut my textbook and calculator both say the answer is (21/2)\nWhat am I doing wrong?\nEvaluate:\n\n$\\int_1^2 2x(x^2+1)dx$\n\nSsubstitute $u=x^2+1$ , $dx=\\frac{1}{2x}du$:\n\n$\\int_{x=1}^2 2x(x^2+1)dx=\\int_{u=2}^5 u\\ du=\\left[u^2/2\\right]_{u=2}^5$ $=\\frac{25}{2}-\\frac{4}{2}=\\frac{21}{2}$\n\nRonL\n\n3. Originally Posted by lingyai\nI would appreciate advice on solving the following problem.\n\nEvaluate the definite integral:\n\nintegral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx\n\nHere is my own attempt.\n\nStep 1: Solve for the generic (indefinite) integral via substitution\n\nLet u = x^2 + 1\n\nSo then\n\ndu / dx = 2x\n\n2x dx = du\n\ndx = du / 2x\n\nReturning to the original function:\n\nintegral 2x (x^2 + 1) dx\n= integral 2x (u) dx\n= integral 2x (u) (du / 2x)\n= integral u du\n= integral x^2 + 1\n= (x^3 / 3 ) + x + C\nStep 2: solve for the definite integral given the stated limits\n\nAt upper limit of 2:\n\n((2)^3 / 3) + (2) +C\n= (8/3) +2 + C\n= (8/3) + (6/3) + C\n= 14/3 + C\n\nAt lower limit of 1\n\n((1)^3 / 3) + (1) +C\n=(1/3) +1 + C\n= (1/3) + (3/3) + C\n= 4/3 + C\n\nSubtracting the lower limit result from upper limit result:\n\n(14 /3 + C) - (4/3 + C)\n=14/3 + C - 4/3 - C\n= 10 / 3\n\nBut my textbook and calculator both say the answer is (21/2)\nWhat am I doing wrong?\nWhat you did wrong?\nThat there after Integral u du.\nYou continued with Integral x^2 +1.\nThat is wrong.\n\nYou were already in the substitution and then you went back to the original, without integrating yet. You should have continued the integration usin the substitution. That is why you chose to use substitution in the first place.\n\nThen in your going back to the original, from INT. u du, the switchback is wrong also.\nWhat is u?\nIt is x^2 +1\nWhat is du?\nIt is 2x dx.\n\nINT. u du = INT. x^2 +1\nonly?\nIt should have been\nINT. u du = INT. (x^2 +1) *(2x dx)\nwhich is the same as the original integrand.\nWhich would have put you back to square one or the beginning again.\n\n4. Hello, lingyai!\n\nThank you for showing your work and reasoning . . .\n\nEvaluate: $\\:\\int^2_1 2x(x^2 + 1)dx$\nDid you (or anyone) notice that the integrand is a polynomial?\n\n$\\int^2_12x(x^2+1)dx \\;= \\;\\int^2_1(2x^3 + 2x)\\,dx \\;= \\;\\frac{1}{2}x^4 + x^2\\bigg|^2_1$\n\n. . $= \\;\\left(\\frac{1}{2}\\cdot2^4 + 2^2\\right) - \\left(\\frac{1}{2}\\cdot1^4 + 1^2\\right) \\;= \\;(8 + 4) - \\frac{3}{2}\\;=\\;\\frac{21}{2}$\n\n5. Originally Posted by Soroban\nHello, lingyai!\n\nThank you for showing your work and reasoning . . .\n\nDid you (or anyone) notice that the integrand is a polynomial?\n\n$\\int^2_12x(x^2+1)dx \\;= \\;\\int^2_1(2x^3 + 2x)\\,dx \\;= \\;\\frac{1}{2}x^4 + x^2\\bigg|^2_1$\n\n. . $= \\;\\left(\\frac{1}{2}\\cdot2^4 + 2^2\\right) - \\left(\\frac{1}{2}\\cdot1^4 + 1^2\\right) \\;= \\;(8 + 4) - \\frac{3}{2}\\;=\\;\\frac{21}{2}$\nI was too busy noticing that:\n\n$\n\\:\\int^2_1 2x(x^2 + 1)dx=\\:\\int^2_1 \\frac{1}{2}\\frac{d}{dx}(x^2 + 1)^2dx\n$\n$\n=\\frac{1}{2}(x^2 + 1)^2 \\bigg|_{x=1}^2=\\frac{25-4}{2}=\\frac{21}{2}$\n\nbut that would not help the OP spotting what they had done wrong\n\nRonL", "date": "2016-08-31 15:37:08", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/3132-need-help-definite-integral-please.html", "openwebmath_score": 0.9143547415733337, "openwebmath_perplexity": 3507.7443687220016, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9362850128595115, "lm_q2_score": 0.9099069974872589, "lm_q1q2_score": 0.8519322848433177}}
{"url": "https://tex.stackexchange.com/questions/400065/align-horizontal-spacing-is-inconsistent", "text": "# Align*: horizontal spacing is inconsistent\n\nI am new to Latex, and I am having some trouble with the align* environment. In the code below, the first two equations are horizontally spaced differently from the second two equations:\n\n\\documentclass{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath,amsthm,amssymb}\n\\usepackage{amsmath}\n\n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\ex}{\\:\\exists\\,}\n\\newcommand{\\nex}{\\:\\nexists\\,}\n\\newcommand{\\all}{\\:\\forall\\,}\n\\newcommand{\\imp}{\\Longrightarrow}\n\\begin{document}\nFor $A:=\\left\\{\\dfrac{1}{n}-\\dfrac{1}{m}:n,m\\in\\N\\right\\}$, $\\sup(A) = 1, \\inf(A)=-1$.\\\\\\\\\n\n\\textit{Proof}. Suppose for contradiction that $1$ is not an upper bound for $A$, i.e. $\\exists n,m \\in \\mathbb{N} \\mid \\dfrac{1}{n} - \\dfrac{1}{m} > 1$.\n\n\\begin{align*}\n\\iff&\\frac{m}{m}\\cdot\\frac{1}{n} - \\frac{n}{n}\\cdot\\frac{1}{m} > 1\\\\\n\\iff&\\frac{m-n}{mn}>1\\\\\n\\iff& m-n> mn\\\\\n\\iff& m > mn + n\\\\\n\\iff& m > n(m+1)\\\\\n\\text{Because }n(m+1)\\geq m+1,\\text{ we have}\\\\\n\\iff& m > n(m+1) \\geq m+1\\\\\n\\iff& m > m+1 &\\bot\\\\\n\\end{align*}\n\n$\\newline$Suppose for contradiction that $-1$ is not a lower bound for $A$, i.e. $\\ex n,m \\in \\N \\mid$\\\\$\\dfrac{1}{n} - \\dfrac{1}{m} < -1$.\n\\begin{align*}\n\\iff&\\frac{m}{m}\\cdot\\frac{1}{n} - \\frac{n}{n}\\cdot\\frac{1}{m} < -1\\\\\n\\iff&\\frac{m-n}{mn}<-1\\\\\n\\iff& m-n<-mn\\\\\n\\iff& m +mn < n\\\\\n\\iff& m(1+n)< n\\\\\n\\text{Because }m(1+n)\\geq 1+n, \\text{ we have}\\\\\n\\iff& 1+n \\leq m(1+n) < n\\\\\n\\iff& 1+n < n &\\bot\n\\end{align*}\n$\\newline\\newline$ Claim that $\\lim\\limits_{n=1, m \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = 1$, or $\\all \\epsilon \\in (0,\\infty) \\ex M \\in \\N \\mid m \\in \\N: m>M \\imp$\\\\$\\left|(1 - \\dfrac{1}{m}) - 1\\right| < \\epsilon$.\n\n\\begin{align*}\n\\iff& \\left| 1 - 1 - \\frac{1}{m}\\right|<\\epsilon\\\\\n\\iff& \\left|-\\frac{1}{m}\\right|<\\epsilon\\\\\n\\iff& \\frac{1}{m}<\\epsilon\\\\\n\\iff& m > \\frac{1}{\\epsilon}\n\\end{align*}\nHence, for $M:=\\left\\lceil\\dfrac{1}{\\epsilon}\\right\\rceil+1$, $m > M \\imp \\left| 1 - 1 - \\dfrac{1}{m}\\right|<\\epsilon$. Because $\\epsilon$ was arbitrarily chosen, it follows that $\\lim\\limits_{n=1, m \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = 1$.\n\n$\\newline\\newline$ Claim that $\\lim\\limits_{m=1, n \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = -1$, or $\\all \\epsilon \\in (0,\\infty) \\ex N \\in \\N \\mid n \\in \\N: n>N \\imp$\\\\$\\left|(\\dfrac{1}{n}-1) -(-1)\\right| < \\epsilon$.\n\n\\begin{align*}\n\\iff& \\left| -1 - (-1) + \\frac{1}{n}\\right|<\\epsilon\\\\\n\\iff& \\left| \\frac{1}{n}\\right|<\\epsilon\\\\\n\\iff& \\frac{1}{n} < \\epsilon\\\\\n\\iff& n > \\frac{1}{\\epsilon}\n\\end{align*}\nHence, for $N:=\\left\\lceil \\dfrac{1}{\\epsilon}\\right\\rceil+1, n > N \\imp \\left|(\\dfrac{1}{n}-1) -(-1)\\right| < \\epsilon$. Because $\\epsilon$ was arbitrarily chosen, it follows that $\\lim\\limits_{m=1, n \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = -1$.\n\\end{document}\n\n\nI'm not exactly sure why this is happening. Perhaps it has something to with the \\frac{}{} equations in the second two equations... and if so, how can I fix the spacing so that all the equations are aligned the same? Thank you.\n\n\u2022 What do you consider the first and the second equation in your code? It's not very clear... Nov 7 '17 at 4:11\n\u2022 Welcome to TeX.SE. If your issue is with alignment across separate align*, you need to combine them into a single align* enironment and use \\intertext{} for the text that you have between the two align* blocks. Nov 7 '17 at 6:07\n\u2022 Well, the space between the lines are probably the same, but as you say the lines are higher in the second two align blocks due to the \\fracs. But do you really want to change that? You don't need more space between the lines for the first two. (You can of course do it, see e.g. tex.stackexchange.com/questions/2929/\u2026, ) Nov 7 '17 at 8:27\n\u2022 @Werner The first two equations are the proofs by contradiction. The last two equations are the limit proofs. I want to fix the horizontal alignment since the first two proofs are spaced differently than the second two.\n\u2013\u00a0user147592\nNov 7 '17 at 13:19\n\u2022 @Torbj\u00f8rnT. I apologize for not being more clear. I want to fix the horizontal alignment, not the vertical alignment. The first two equations are horizontally aligned differently from the second two, and I want to fix that.\n\u2013\u00a0user147592\nNov 7 '17 at 13:21\n\nYou're probably looking for \\intertext:\n\n\\documentclass{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath,amsthm,amssymb}\n\\usepackage{amsmath}\n\n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\ex}{\\:\\exists\\,}\n\\newcommand{\\nex}{\\:\\nexists\\,}\n\\newcommand{\\all}{\\:\\forall\\,}\n\\newcommand{\\imp}{\\Longrightarrow}\n\\begin{document}\n\nFor $A:=\\left\\{\\dfrac{1}{n}-\\dfrac{1}{m}:n,m\\in\\N\\right\\}$, $\\sup(A) = 1$, $\\inf(A)=-1$.\n\n\\begin{proof}\nSuppose for contradiction that $1$ is not an upper bound for $A$, i.e.\\@\n$\\exists n,m \\in \\mathbb{N} \\mid \\dfrac{1}{n} - \\dfrac{1}{m} > 1$.\n\n\\begin{align*}\n\\iff&\\frac{m}{m}\\cdot\\frac{1}{n} - \\frac{n}{n}\\cdot\\frac{1}{m} > 1\\\\\n\\iff&\\frac{m-n}{mn}>1\\\\\n\\iff& m-n> mn\\\\\n\\iff& m > mn + n\\\\\n\\iff& m > n(m+1)\\\\\n\\intertext{Because $n(m+1)\\geq m+1$, we have}\n\\iff& m > n(m+1) \\geq m+1\\\\\n\\iff& m > m+1 &\\bot\\\\\n\\intertext{\\indent Suppose for contradiction that $-1$ is not a lower bound for $A$,\ni.e.\\@ $\\ex n,m \\in \\N \\mid \\dfrac{1}{n} - \\dfrac{1}{m} < -1$.}\n\\iff&\\frac{m}{m}\\cdot\\frac{1}{n} - \\frac{n}{n}\\cdot\\frac{1}{m} < -1\\\\\n\\iff&\\frac{m-n}{mn}<-1\\\\\n\\iff& m-n<-mn\\\\\n\\iff& m +mn < n\\\\\n\\iff& m(1+n)< n\\\\\n\\intertext{Because $m(1+n)\\geq 1+n$, we have}\n\\iff& 1+n \\leq m(1+n) < n\\\\\n\\iff& 1+n < n &\\bot\n\\intertext{\\indent Claim that $\\lim\\limits_{n=1, m \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = 1$,\nor $\\all \\epsilon \\in (0,\\infty) \\ex M \\in \\N \\mid m \\in \\N: m>M \\imp \\left|(1 - \\dfrac{1}{m}) - 1\\right| < \\epsilon$.}\n\\iff& \\left| 1 - 1 - \\frac{1}{m}\\right|<\\epsilon\\\\\n\\iff& \\left|-\\frac{1}{m}\\right|<\\epsilon\\\\\n\\iff& \\frac{1}{m}<\\epsilon\\\\\n\\iff& m > \\frac{1}{\\epsilon}\n\\intertext{Hence, for $M:=\\left\\lceil\\dfrac{1}{\\epsilon}\\right\\rceil+1$,\n$m > M \\imp \\left| 1 - 1 - \\dfrac{1}{m}\\right|<\\epsilon$. Because $\\epsilon$\nwas arbitrarily chosen, it follows that\n$\\lim\\limits_{n=1, m \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = 1$.\\endgraf\\medskip\n\\indent Claim that $\\lim\\limits_{m=1, n \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = -1$,\nor $\\all \\epsilon \\in (0,\\infty) \\ex N \\in \\N \\mid n \\in \\N: n>N \\imp \\left|(\\dfrac{1}{n}-1) -(-1)\\right| < \\epsilon$.}\n\\iff& \\left| -1 - (-1) + \\frac{1}{n}\\right|<\\epsilon\\\\\n\\iff& \\left| \\frac{1}{n}\\right|<\\epsilon\\\\\n\\iff& \\frac{1}{n} < \\epsilon\\\\\n\\iff& n > \\frac{1}{\\epsilon}\n\\end{align*}\nHence, for $N:=\\left\\lceil \\dfrac{1}{\\epsilon}\\right\\rceil+1$,\n$n > N \\imp \\left|(\\dfrac{1}{n}-1) -(-1)\\right| < \\epsilon$.\nBecause $\\epsilon$ was arbitrarily chosen, it follows that\n$\\lim\\limits_{m=1, n \\to \\infty} \\dfrac{1}{n} - \\dfrac{1}{m} = -1$.\n\\end{proof}\n\n\\end{document}\n\n\nEDIT: Found the error: you have too many & in the last line of the first two align envs\n\nBelow is a cleaned up MWE (with the error still present)\n\n\u2022 don't use \\dfrac in the text, causes excessive line spacing\n\u2022 don't use \\\\ or \\newline in the text, you never want manual line breaks in the text (this is a very common mistake among new users)\n\u2022 use \\intertext{...} for comments inside align (mathtools provides \\shortintertext which has different spacing)\n\nCleaned MWE\n\n\\documentclass{article}\n\\usepackage[margin=1in]{geometry}\n\\usepackage{amsmath,amsthm,amssymb}\n\\usepackage{amsmath}\n\n\\newcommand{\\N}{\\mathbb{N}}\n\\newcommand{\\Z}{\\mathbb{Z}}\n\\newcommand{\\R}{\\mathbb{R}}\n\\newcommand{\\ex}{\\:\\exists\\,}\n\\newcommand{\\nex}{\\:\\nexists\\,}\n\\newcommand{\\all}{\\:\\forall\\,}\n\\newcommand{\\imp}{\\Longrightarrow}\n\\begin{document}\nFor $A:=\\left\\{\\frac{1}{n}-\\frac{1}{m}:n,m\\in\\N\\right\\}$, $\\sup(A) = 1, \\inf(A)=-1$.\n\n\\begin{proof}\nSuppose for contradiction that $1$ is not an upper bound for $A$,\ni.e.\n$\\exists n,m \\in \\mathbb{N} \\mid \\frac{1}{n} - \\frac{1}{m} > 1$.\n\\begin{align*}\n\\iff&\\frac{m}{m}\\cdot\\frac{1}{n} - \\frac{n}{n}\\cdot\\frac{1}{m} > 1\\\\\n\\iff&\\frac{m-n}{mn}>1\\\\\n\\iff& m-n> mn\\\\\n\\iff& m > mn + n\\\\\n\\iff& m > n(m+1)\\\\\n\\intertext{Because $n(m+1)\\geq m+1$, we have}\n\\iff& m > n(m+1) \\geq m+1\\\\\n\\iff& m > m+1 &\\bot\n\\end{align*}\n\nSuppose for contradiction that $-1$ is not a lower bound for $A$,\ni.e. $\\ex n,m \\in \\N \\mid \\frac{1}{n} - \\frac{1}{m} < -1$.\n\\begin{align*}\n\\iff&\\frac{m}{m}\\cdot\\frac{1}{n} - \\frac{n}{n}\\cdot\\frac{1}{m} < -1\\\\\n\\iff&\\frac{m-n}{mn}<-1\\\\\n\\iff& m-n<-mn\\\\\n\\iff& m +mn < n\\\\\n\\iff& m(1+n)< n\\\\\n\\intertext{Because $m(1+n)\\geq 1+n$, we have}\n\\iff& 1+n \\leq m(1+n) < n\\\\\n\\iff& 1+n < n &\\bot\n\\end{align*}\nClaim that $\\lim_{n=1, m \\to \\infty} \\frac{1}{n} - \\frac{1}{m} = 1$, or $\\all \\epsilon \\in (0,\\infty) \\ex M \\in \\N \\mid m \\in \\N: m>M \\imp$ $\\left|(1 - \\frac{1}{m}) - 1\\right| < \\epsilon$.\n\\begin{align*}\n\\iff& \\left| 1 - 1 - \\frac{1}{m}\\right|<\\epsilon\\\\\n\\iff& \\left|-\\frac{1}{m}\\right|<\\epsilon\\\\\n\\iff& \\frac{1}{m}<\\epsilon\\\\\n\\iff& m > \\frac{1}{\\epsilon}\n\\end{align*}\nHence, for $M:=\\left\\lceil\\frac{1}{\\epsilon}\\right\\rceil+1$,\n$m > M \\imp \\left| 1 - 1 - \\frac{1}{m}\\right|<\\epsilon$. Because\n$\\epsilon$ was arbitrarily chosen, it follows that\n$\\lim_{n=1, m \\to \\infty} \\frac{1}{n} - \\frac{1}{m} = 1$.\n\nClaim that\n$\\lim_{m=1, n \\to \\infty} \\frac{1}{n} - \\frac{1}{m} = -1$, or\n$\\all \\epsilon \\in (0,\\infty) \\ex N \\in \\N \\mid n \\in \\N: n>N \\imp$\n$\\left|(\\frac{1}{n}-1) -(-1)\\right| < \\epsilon$.\n\\begin{align*}\n\\iff& \\left| -1 - (-1) + \\frac{1}{n}\\right|<\\epsilon\\\\\n\\iff& \\left| \\frac{1}{n}\\right|<\\epsilon\\\\\n\\iff& \\frac{1}{n} < \\epsilon\\\\\n\\iff& n > \\frac{1}{\\epsilon}\n\\end{align*}\nHence, for $N:=\\left\\lceil \\frac{1}{\\epsilon}\\right\\rceil+1, n > N \\imp \\left|(\\frac{1}{n}-1) -(-1)\\right| < \\epsilon$. Because\n$\\epsilon$ was arbitrarily chosen, it follows that\n$\\lim\\limits_{m=1, n \\to \\infty} \\frac{1}{n} - \\frac{1}{m} = -1$.\n\\end{proof}\n\\end{document}\n\n\nHere is a image of the first page (after cleaning), I can see what you mean about the unevenness\n\n\u2022 So there isn't a way to fix the unevenness?\n\u2013\u00a0user147592\nNov 7 '17 at 15:14\n\u2022 @SuhasHoysala See my update, I just left the image and the clean MWE for others to actually see the problem. You have to unnecessary & Nov 7 '17 at 15:27\n\u2022 I removed the extra & (before the \\bot command), but it still was uneven... I'm not sure what I'm doing wrong.\n\u2013\u00a0user147592\nNov 7 '17 at 20:33\n\u2022 @SuhasHoysala did you also remember to use \\intertext  as I did in my cleanup? My fix was literally the mwe cleanup I posted and then removing the two excessive & Nov 7 '17 at 20:35\n\u2022 Yes, I copy-pasted your code verbatim and removed the two excessive &. It still doesn't seem to be fixed.\n\u2013\u00a0user147592\nNov 7 '17 at 21:44", "date": "2021-09-27 12:23:49", "meta": {"domain": "stackexchange.com", "url": "https://tex.stackexchange.com/questions/400065/align-horizontal-spacing-is-inconsistent", "openwebmath_score": 1.0000056028366089, "openwebmath_perplexity": 4524.93943391534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9362850004144266, "lm_q2_score": 0.9099069999303417, "lm_q1q2_score": 0.8519322758068697}}
{"url": "https://math.stackexchange.com/questions/2394875/how-can-i-write-negations-for-the-following-statements", "text": "# How can I write negations for the following statements?\n\nPremise: Let $F(x,y)$ be the statement $x$ can fool $y$, where the domain for discourse for both $x$ and $y$ is all people.\n\nI converted the following corresponding statements:\n\n1) Nobody can fool me\n\n2) Anybody can Fool Fred\n\n3) Everyone can fool someone\n\nto\n\n1) $\\neg \\exists xF(x,me)$\n\n2) $\\forall xF(x,Fred)$\n\n3) $\\forall x\\exists yF(x,y)$\n\nNow I am trying to negate these and I can't just throw in a not in front of the statements and I don't know how to go about this.\n\nI looked at the negation rules and came up with this but I doubt it's correct:\n\n1) $\\neg\\exists xF(x,me) \\iff \\forall x\\neg F(x,me)$\n\n2) $\\forall xF(x,Fred) \\iff \\forall x\\neg\\neg F(x,Fred) \\iff \\neg[\\exists x\\neg F(x,Fred)]$\n\n3) $\\forall x\\exists yF(x,y) \\iff \\neg\\forall x\\exists yF(x,y)$\n\n\u2022 Well, $\\lnot (\\lnot Q)$ is $Q$ so the negation of $\\lnot\\exists xF(x,me)$ would be ... $\\exists xF(x,me)$, right? \u2013\u00a0fleablood Aug 15 '17 at 22:17\n\u2022 And the negation of $\\forall x A$ is $\\exists x \\lnot A$ so the negation $\\forall x F(x, fred)$ is $\\exists x \\lnot F(x, fred)$. \u2013\u00a0fleablood Aug 15 '17 at 22:19\n\u2022 3 is a little harder. As with 2) is $\\lnot(\\forall x \\exists y F(x,y))$ will be $\\exists x (\\lnot \\exists y F(x,y))$ but we need to find the negation of $\\exists y F(x,y)$. The negation of $\\exists x Q$ is $\\forall x \\lnot Q$ so the negation is $\\exists x \\forall y \\lnot F(x,y)$. i.e. There is someone who can not fool anybody. \u2013\u00a0fleablood Aug 15 '17 at 22:25\n\u2022 I think you're spot on, @fleablood. Please post your comments in an answer! I'll upvote it. \u2013\u00a0Namaste Aug 15 '17 at 22:29\n\u2022 Wait... you aren't actually negating them. You are stating the equivalence of them. You can't \"just\" put a not in front, but you must start by putting a not in front. \u2013\u00a0fleablood Aug 15 '17 at 22:30\n\nExcept for what I assume to be a typo on the answer to 3, what you've written is correct (or more accurately, not wrong), although it seems you've missed out the actual negation of the statements which is what you're after.\n\nEssentially, all you need is the rule\n\n$$\\neg((\\forall x)\\quad P(x))\\iff((\\exists x)\\quad \\neg P(x))$$\n\nas well as the more basic $\\neg\\neg P\\iff P$.\n\nLet\n\n$$S_1\\equiv\u00ac((\u2203x)\\quad F(x,\\text{me}))$$\n\n$$S_2\\equiv(\u2200x)\\quad F(x,\\text{Fred})$$\n\n$$S_3\\equiv(\u2200x)\\quad((\u2203y)\\quad F(x,y))$$\n\nThen you're after $\\neg S_1,\\neg S_2,\\neg S_3$.\n\nAs has been noted in the comments, the first is simple\n\n$$\\neg S_1\\equiv\\neg\\neg((\u2203x)\\quad F(x,\\text{me}))$$\n\n$$\\neg S_1\\equiv(\u2203x)\\quad F(x,\\text{me})$$\n\nYour working for the second is helpful and we get\n\n$$\\neg S_2\\equiv\\neg((\u2200x)\\quad F(x,\\text{Fred}))$$\n\n$$\\equiv\\neg\u00ac((\u2203x)\\quad\u00acF(x,\\text{Fred}))$$\n\n$$\\equiv(\u2203x)\\quad\u00acF(x,\\text{Fred})$$\n\nFinally, as has been pointed out by @fleablood, the last is slightly harder, but is just applying the above rule twice\n\n$$\\neg S_3\\equiv\\neg((\u2200x)\\quad((\u2203y)\\quad F(x,y)))$$\n\n$$\\equiv(\u2203x)\\quad\\neg((\u2203y)\\quad F(x,y))$$\n\n$$\\equiv(\u2203x)\\quad((\u2200y)\\quad \\neg F(x,y))$$\n\nOkay. First to translate:\n\n1) \"Nobody can fool me\". \"Nobody\" can be thought of as either \"Everyone can not...\", or as \"There doesn't exist anyone who can...\"\n\nSo either $\\forall x\\lnot F(x,me)$ or $\\lnot \\exists x F(x,me)$ will both be correct and are equivalent. So yes you are are correct.\n\nAs for 2) and 3)... I have nothing to add. You did those just fine. (You did 1) just fine as well, but I did have something to add.)\n\nSo the negations:\n\nThe rules are simple:\n\nA) $\\lnot$(That's not true) is, of course, \"That is true\" so $\\lnot(\\lnot Q) \\iff Q$.\n\nB) $\\lnot$(Everbody does something) is, there is at least one person who doesn't do it. So $\\lnot \\forall x Q \\iff \\exists x \\lnot Q$.\n\nC) $\\lnot$(Somebody does something) is. Nobody does something. So $\\lnot \\exists x Q \\iff \\forall x \\lnot Q$.\n\nSo\n\n1) $\\lnot(\\lnot \\exists x F(x,me)) \\iff \\exists x F(x,me)$\n\nor\n\n$\\lnot \\forall x\\lnot F(x,me) \\iff \\exists x\\lnot(\\lnot F(x,me)) \\iff \\exists x F(x, me)$.\n\ni.e. It is not the case that noone can fool me $\\iff$ someone can fool me.\n\n2)$\\lnot(\\forall x F(x, fred)) \\iff \\exists x \\lnot F(x,fred)$.\n\nNot everbody can fool Fred = there is someone who can't fool Fred.\n\n3) $\\lnot (\\forall x \\exists y F(x,y)) \\iff$\n\n$\\exists x \\lnot(\\exists y F(x,y)) \\iff$\n\n$\\exists x \\forall y \\lnot F(x,y)$\n\ni.e. not(Everbody can fool somebody) = somebody can not fool anybody.", "date": "2019-07-23 00:51:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2394875/how-can-i-write-negations-for-the-following-statements", "openwebmath_score": 0.7204133868217468, "openwebmath_perplexity": 288.8390917543612, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419703960398, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8519251441007158}}
{"url": "https://mathematica.stackexchange.com/questions/69839/regular-polyhedra-coordinate-points-generation", "text": "Regular Polyhedra coordinate points generation\n\nIs it possible to get all points on a Polyhedron surface using two surface parameters, say\n\n$\\phi,\\theta$ spherical co-ordinates?\n\nJust like in ParametricPlot3D, can we start with PolyhedronData[\"Tetrahedron\"] to obtain spatial point positions?. The tip of position vector should cross edges automatically during $\\phi,\\theta$ sweep.\n\n\u2022 You can always make this kind of tricks \u2013\u00a0Dr. belisarius Dec 29 '14 at 19:22\n\u2022 @belisarius: fine,but want to append or prepend available Polyhedra. \u2013\u00a0Narasimham Dec 29 '14 at 19:59\n\u2022 This is a nice start. \u2013\u00a0Dr. belisarius Dec 29 '14 at 20:04\n\u2022 Stick a small sphere around the barycenter. Now parametrize by the spherical angles, using intersection of ray through spherical point (theta,phi) with polyhedron. \u2013\u00a0Daniel Lichtblau Dec 29 '14 at 23:57\n\u2022 Clear enough, we need to have direct coordinates as function of ($\\theta,\\phi, n$). \u2013\u00a0Narasimham Dec 30 '14 at 7:58\n\nExploit the fact that the vertices of the dual to a Platonic solid correspond to the centers of the faces of the solid itself. For instance, the dual to a cube is a regular octahedron, and the six vertices of this octahedron are in the directions of the centers of the faces of its dual cube.\n\nFind the normalized directions of the face centers of the cube (for instance) this way:\n\nFaceCenters = Normalize /@ PolyhedronData[PolyhedronData[\"Cube\", \"Dual\"], \"Faces\"][[1]];\n\n\nThe {x,y,z} direction in Euclidean coordinates as a function of spherical angles \u03b8 and \u03c6 are of course:\n\nx[\u03b8_Real, \u03c6_Real] := {Sin[\u03b8] Cos[\u03c6],\nSin[\u03b8] Sin[\u03c6],\nCos[\u03b8]}\n\n\nNow sweep through all spherical angles, and for each corresponding direction find the nearest face center direction (i.e., the one with the smallest angle to the direction defined by \u03b8 and \u03c6). For each such direction, the distance to the surface is 1/Sin[\u03c8], where \u03c8 is the scalar angle between the candidate direction and its nearest face direction:\n\n    SphericalPlot3D[\n1/Sin[x[\u03b8, \u03c6].Nearest[FaceCenters, x[\u03b8, \u03c6]][[1]]],\n{\u03b8, 0, \u03c0}, {\u03c6, 0, 2 \u03c0},\nPlotPoints -> 100] // Quiet\n\n\nIf you repeat with a Tetrahedron, for instance, you get this:\n\nIf you repeat for an Octahedron, you get this:\n\nIf you repeat for a Dodecahedron, you get this:\n\n\u2022 Wow... I really thought @Narasimhan would have accepted this answer! What more can he possibly want?! \u2013\u00a0David G. Stork Jan 6 '15 at 0:58\n\u2022 That was great. My apologies for the inordinate delay earlier \u2013\u00a0Narasimham Feb 10 '18 at 5:17\n\u2022 Wow... two months late! Better late than never! \u2013\u00a0David G. Stork Feb 10 '18 at 6:20", "date": "2020-10-27 09:56:09", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/69839/regular-polyhedra-coordinate-points-generation", "openwebmath_score": 0.6926586031913757, "openwebmath_perplexity": 1907.2497222615355, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419694095656, "lm_q2_score": 0.8615382076534743, "lm_q1q2_score": 0.8519251379776488}}
{"url": "https://math.stackexchange.com/questions/1603966/what-is-the-least-number-of-square-roots-needed-to-express-sqrt1-sqrt2-c", "text": "What is the least number of square roots needed to express $\\sqrt{1}+\\sqrt{2}+\\cdots+\\sqrt{100}$?\n\nWhat is the least number of square roots needed to express $\\sqrt{1}+\\sqrt{2}+\\cdots+\\sqrt{100}$ if it must be expressed in the form $a+b\\sqrt{c}+d\\sqrt{e}+\\cdots$ where $a,b,c,d,e,\\ldots$ are all integers?\n\nSolution\n\nIn order to find the least number of square roots, we must express this number in simplest form. Thus, the numbers in the radicand that are in simplest form that will get counted are numbers with at most one of each prime factor. Then the question simplifies to \"How many numbers are there between $1-100$ with at most one of each prime factor?\n\nThere are a few ways to proceed from here. I think the best way would be to complementary count. Then we are looking for numbers with at least $2$ factors of each prime, thus multiples of perfect squares. Our perfect squares are $1,4,9,16,25,36,49,64,81$. Thus we have $1+25+11-2+3+2 = 40$ such numbers by the principle of inclusion-exclusion. Thus, the answer is $100-40 = 60$.\n\nQuestion\n\nHow is it that the number being expressed in simplest form will give the least number of radicals? The solution seems to imply that but why is it true?\n\n\u2022 You mean expressing it as $a + b \\sqrt{c} + d\\sqrt{e} + \\cdots$ where $a,b,c,\\dots \\in \\mathbb N$ right? Because otherwise the question would be very problematic. \u2013\u00a0ThePortakal Jan 8 '16 at 3:45\n\u2022 Yes, of course. Otherwise I wouldn't technically have to use any square roots. \u2013\u00a0user19405892 Jan 8 '16 at 3:45\n\u2022 I added the extra condition in the question so that doesn't work as $3+\\sqrt{3}+\\sqrt{5}$ is not an integer. By least number of square roots, I mean the number of square root symbols. So $\\sqrt{3}+\\sqrt{3}$ would be two square roots while $2\\sqrt{3}$ would be one. \u2013\u00a0user19405892 Jan 8 '16 at 4:07\n\u2022 Why are you copying the exact solution here? \u2013\u00a0GohP.iHan Jan 9 '16 at 11:04\n\u2022 @GohP.iHan It is actually my solution. \u2013\u00a0user19405892 Jan 9 '16 at 13:44\n\nThe square root of each square-free number less than $100$ contributes a term to the sum with a surd (a square root that cannot be simplified out), but all the square multiples of the number under the square root ($4$ times, $9$ times, $16$ times, etc.) can be combined with the term for the original square-free number, so they do not contribute any additional terms. For example,\n\n$$\\sqrt{11} + \\sqrt{44} + \\sqrt{99} = 6\\sqrt{11}.$$\n\nOne way to count the integers whose square roots do not contribute new surds to the sum is as follows:\n\n\u2022 $10$ perfect squares in the range $1$ to $100$, inclusive;\n\u2022 $6$ cases of twice a perfect square in the range $8$ to $98$, inclusive;\n\u2022 $4$ cases of $3$ times a perfect square in the range $12$ to $75$, inclusive;\n\u2022 $3$ cases of $5$ times a perfect square in the range $20$ to $80$, inclusive;\n\u2022 $3$ cases of $6$ times a perfect square in the range $24$ to $96$, inclusive;\n\u2022 $6$ cases including $2$ cases each of $7$ times a perfect square, $10$ times a perfect square, and $11$ times a perfect square in the range $28$ to $99$, inclusive;\n\u2022 $8$ cases including $13 \\times 4$, $14 \\times 4$, $15 \\times 4$, $17 \\times 4$, $19 \\times 4$, $21 \\times 4$, $22 \\times 4$, and $23 \\times 4$.\n\nThese add up to $40$ terms of the original sum that either are integers or can be combined with other terms, leaving $60$ unique surds. This is the same as your result, of course.\n\nHere is the sum completely worked out, confirming that $60$ square roots are needed:\n\n\\begin{align} \\sqrt1 + & \\sqrt2 + \\cdots + \\sqrt{100} \\\\ =& \\quad 1 + \\sqrt2+\\sqrt3 + 2 + \\sqrt5+\\sqrt6+\\sqrt7 + 2\\sqrt2 + 3 + \\sqrt{10} \\\\ & + \\sqrt{11} + 2\\sqrt3 + \\sqrt{13} + \\sqrt{14} + \\sqrt{15} + 4 + \\sqrt{17} + 3\\sqrt2 + \\sqrt{19} + 2\\sqrt{5} \\\\ & + \\sqrt{21} + \\sqrt{22} + \\sqrt{23} + 2\\sqrt6 + 5 + \\sqrt{26} + 3\\sqrt3 + 2\\sqrt7 + \\sqrt{29} + \\sqrt{30} \\\\ & + \\sqrt{31} + 4\\sqrt2 + \\sqrt{33} + \\sqrt{34} + \\sqrt{35} + 6 + \\sqrt{37} + \\sqrt{38} + \\sqrt{39} + 2\\sqrt{10} \\\\ & + \\sqrt{41} + \\sqrt{42} + \\sqrt{43} + 2\\sqrt{11} + 3\\sqrt5 + \\sqrt{46} + \\sqrt{47} + 4\\sqrt3 + 7 + 5\\sqrt2 \\\\ & + \\sqrt{51} + 2\\sqrt{13} + \\sqrt{53} + 3\\sqrt6 + \\sqrt{55} + 2\\sqrt{14} + \\sqrt{57} + \\sqrt{58} + \\sqrt{59} + 2\\sqrt{15} \\\\ & + \\sqrt{61} + \\sqrt{62} + 3\\sqrt7 + 8 + \\sqrt{65} + \\sqrt{66} + \\sqrt{67} + 2\\sqrt{17} + \\sqrt{69} + \\sqrt{70} \\\\ & + \\sqrt{71} + 6\\sqrt2 + \\sqrt{73} + \\sqrt{74} + 5\\sqrt3 + 2\\sqrt{19} + \\sqrt{77} + \\sqrt{78} + \\sqrt{79} + 4\\sqrt5 \\\\ & + 9 + \\sqrt{82} + \\sqrt{83} + 2\\sqrt{21} + \\sqrt{85} + \\sqrt{86} + \\sqrt{87} + 2\\sqrt{22} + \\sqrt{89} + 3\\sqrt{10} \\\\ & + \\sqrt{91} + 2\\sqrt{23} + \\sqrt{93} + \\sqrt{94} + \\sqrt{95} + 4\\sqrt6 + \\sqrt{97} + 7\\sqrt2 + 3\\sqrt{11} + 10 \\\\ =& \\quad 55 + 28\\sqrt2 + 15\\sqrt3 + 10\\sqrt5 + 10\\sqrt6 + 6\\sqrt7 + 6\\sqrt{10} \\\\ & + 6\\sqrt{11} + 3\\sqrt{13} + 3\\sqrt{14} + 3\\sqrt{15} + 3\\sqrt{17} + 3\\sqrt{19} \\\\ & + 3\\sqrt{21} + 3\\sqrt{22} + 3\\sqrt{23} + \\sqrt{26} + \\sqrt{29} + \\sqrt{30} \\\\ & + \\sqrt{31} + \\sqrt{33} + \\sqrt{34} + \\sqrt{35} + \\sqrt{37} + \\sqrt{38} \\\\ & + \\sqrt{39} + \\sqrt{41} + \\sqrt{42} + \\sqrt{43} + \\sqrt{46} + \\sqrt{47} \\\\ & + \\sqrt{51} + \\sqrt{53} + \\sqrt{55} + \\sqrt{57} + \\sqrt{58} + \\sqrt{59} \\\\ & + \\sqrt{61} + \\sqrt{62} + \\sqrt{65} + \\sqrt{66} + \\sqrt{67} + \\sqrt{69} \\\\ & + \\sqrt{70} + \\sqrt{71} + \\sqrt{73} + \\sqrt{74} + \\sqrt{77} + \\sqrt{78} \\\\ & + \\sqrt{79} + \\sqrt{82} + \\sqrt{83} + \\sqrt{85} + \\sqrt{86} + \\sqrt{87} \\\\ & + \\sqrt{89} + \\sqrt{91} + \\sqrt{93} + \\sqrt{94} + \\sqrt{95} + \\sqrt{97} \\\\ \\end{align}\n\n\u2022 How do you know that we can't use nested radicals to reduce the number of radicals used? \u2013\u00a0GohP.iHan Jan 9 '16 at 11:03\n\u2022 @GohP.iHan The problem statement stipulates that the answer must be in the form $a+b\\sqrt{c}+d\\sqrt{e}+\\cdots$ where $a,b,c,d,e,\\ldots$ are all integers. \u2013\u00a0David K Jan 9 '16 at 13:37\n\u2022 Ohhh my bad... the author had a different question when I first viewed it. Thank you!! \u2013\u00a0GohP.iHan Jan 9 '16 at 13:51", "date": "2019-10-23 04:43:22", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1603966/what-is-the-least-number-of-square-roots-needed-to-express-sqrt1-sqrt2-c", "openwebmath_score": 0.9993078708648682, "openwebmath_perplexity": 219.2458015100361, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419661213171, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8519251369024244}}
{"url": "https://www.khanacademy.org/math/old-ap-calculus-ab/ab-limits-continuity/ab-direct-sub/v/limit-by-substitution", "text": "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## AP\u00ae\ufe0e Calculus AB (2017 edition)\n\n### Unit 1: Lesson 7\n\nDetermining limits using direct substitution\n\n# Limits by direct substitution\n\nAP.CALC:\nLIM\u20111 (EU)\n,\nLIM\u20111.D (LO)\n,\nLIM\u20111.D.1 (EK)\nSal explains how you can easily find limits of functions at points where the functions are continuous: simply plug in the x-value into the function! Later we will learn how to find limits even when the function isn't continuous.\n\n## Want to join the conversation?\n\n\u2022 I think Sal may have left something out. Correct me if I'm wrong, but on certain occasions, the entire function does not have to be continuous. If it is continuous at the value, a, then it should be fine to use direct substitution even though the function might be undefined elsewhere. This is because it isn't undefined at a so your output will be defined. Also, since your function is continuous at a, the limit there will exist meaning that you won't have to worry about jump discontinuities. Please tell me if this all seems reasonable.\n\u2022 Because it is a continuous graph, every interval would be continuous provided it is a real number\n\u2022 how would we know if the problem is to be solved by direct substitution method? or if the function given is continuous? without drawing its graph\n\u2022 lim h(x) as x ->6 for h(x) =square root of (5x+6) should be 6 or -6 or both?\n\u2022 y=\u221ax is a function that returns only positive values. So in this context, we are only looking at the principal root, and the limit is positive 6.\n\nLimits are unique: they cannot have multiple values. So \"both\" would never be an option.\n\u2022 At , what's the difference between f(x) as x->a, and f(a)?\n\u2022 One is a limit, the other is an evaluation of the function. If the function is continuous and defined at (in your example), a, then they're equivalent. But you can get some very interesting results if the function is not continuous or not defined.\n\nThe limit is basically saying what the function seems to be going to as x gets closer to closer to a, but the function may not be defined at that point.\n\u2022 Is the limit zero or none since it is continuous?\n\u2022 The limit is 0 as Sal has demonstrated in the video.\n\u2022 Can a limit be a fraction?\n\u2022 yes a limit can be a fraction. For example if I asked you to find the limit as X approaches 2 for the functions (1/X), when you use substitution to find the limit you find that the number it approaches is .5 or 1/2.\n\u2022 I thought it like this : Lim x-> a f(x) = f(a) if f(x) is continuous at x=a, not vice versa? Can someone tell me why?\n\u2022 That statement holds in both directions. A function is continuous at a point a if and only if the limit value equals the function value at a. Since the two statements are equivalent, this can be used as a definition of continuity.\n\u2022 Ok so basically if a function, say f(x), is continuous at x=c, then the lim x-->c = f(x)? Is this why you can find limits of continuous functions by direct substitution?\n\u2022 Yes, the limit as x->c of f(x) is f(c). This property is equivalent to the epsilon-delta definition of continuity, and it's why we can use direct substitution for most familiar functions.\n\u2022 By Sal and almost all discussion about relation between continuity and limit: it always says that as long as the function is continuous, there is a limit.\ni am wondering if there is a case that the function is continuous but the limit does not exist.\n\nhttps://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins_et_al)/1%3A_Understanding_the_Derivative/1.7%3A_Limits%2C_Continuity%2C_and_Differentiability\n\nin figure 1.7.2, the right graph, at point x=1. the left hand limit is 2. but the as x approaches 1 from the right, the function keeps oscillating, so the limit does not exist.\n\nso the function is continuous by the graph, but limit does not exist.\n\nam I wrong? or am I give some wrong example?\n\u2022 Interesting question!\n\nBy definition, a function f(x) is continuous at x=x_0 if and only if for every epsilon>0, there exists delta>0 such that whenever x is within delta of x_0, f(x) is always within epsilon of f(x_0).\n\nThough the function might look continuous at x=1, it is really discontinuous at x=1 due to the oscillatory behavior. For example, if we choose epsilon = 1/2, there does not exist delta>0 such that whenever x is within delta of 1, f(x) is always within epsilon of f(1). This is because there are values of x arbitrarily close to 1 (from the right) such that f(x) differs from f(1) by as much as 1 unit.\n(1 vote)\n\u2022 I'm not entirely sure if this statement is always true. But my argument bases on the assumption that the function f(x) = |x| is not continuous at x = 0, which I think is true but correct me if I'm wrong.\n\nSo, the limit lim_{x -> 0}f(x) for f(x) = |x| is definitely defined since it approaches 0 from both sides. The function itself is usually defined so that |x| equals 0 at x = 0. Which means that lim_{x -> a}f(x) equals f(a) at a = 0, but as said AFAIK the absolute value function |x| is NOT continuous for x = 0.\n\nAnd that would disprove the statement that if lim_{x -> a}f(x) equaling f(a) will mean that f is continuous at a (and vice versa).\n(1 vote)\n\u2022 Hey Raphael! I think you're getting continuity and differentiation confused. For a function, f(x), to be continuous, f(c) = lim_{x->c}f(x). This means that f(c) = lim_{x->c-}f(x) = lim_{x->c+}f(x) = lim_{x->c}f(x). As long as this holds true the function will be continuous at the given x-value, c.\n\nDifferentiation on the other hand is different. Differentiation requires that a function is continuous at the given x-value along with being absent of cusps, sharp turns (like x=0 for f(x) = |x|), and vertical tangents.\n\nSo to answer your question, f(x) = |x| is continuous at x = 0, however, it is not differentiable at x = 0.", "date": "2023-01-29 20:24:45", "meta": {"domain": "khanacademy.org", "url": "https://www.khanacademy.org/math/old-ap-calculus-ab/ab-limits-continuity/ab-direct-sub/v/limit-by-substitution", "openwebmath_score": 0.8613227605819702, "openwebmath_perplexity": 365.88969269845984, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419684230912, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8519251318545814}}
{"url": "http://math.stackexchange.com/questions/228416/order-of-magnitudes-comparasions", "text": "# Order of magnitudes comparasions\n\nI have a list of order of magnitudes I want to compare.\n\nMy only idea is using calculus methods (limits , integral, etc...) to assert the functions relation.\n\nI need your help with the following.\n\nI need to determine (again by order of magnitude) how to order the following (by asceneding order):\n\n$log(n) / log(log(n))$\n\n$log(log(n))$\n\n$log^3(n)$\n\n$n/log(n)$\n\n$n^{log(n)}$\n\nIs it even possible to use limits on such functions?\n\nI mean if I take $f(n)$ to be $log(n) / log(log(n))$ and $g(n)$ to be $n^{log(n)}$, It seems very hard to find out the limit.\n\n-\n\nThere are quite a number of comparisons to be made. It is in most cases relatively straightforward to decide about the relative long-term size.\n\nLet's start with your pair. We have $f(n)=\\log(n)/\\log(\\log (n))$ and $g(n)=n^{\\log(n)}$.\n\nFor large $n$ (and it doesn't have to be very large), we have $f(n)\\lt \\log(n)$.\n\nAlso, for large $n$, we have $\\log(n)\\gt 1$, and therefore $g(n)=n^{\\log(n)}\\gt n$.\n\nSo for large $n$, we have $$\\frac{f(n)}{g(n)} \\lt \\frac{\\log(n)}{n}.$$ But we know that $\\lim_{n\\to\\infty}\\dfrac{\\log(n)}{n}=0$. This can be shown in various ways. For instance, we can consider $\\dfrac{\\log(x)}{x}$ and use L'Hospital's Rule to show this has limit $0$ as $x\\to\\infty$.\n\nIt takes less time to deal with the pair $n/\\log(n)$ and $n^{\\log(n)}$. If $n$ is even modestly large, we have $n/\\log(n)\\lt n$. But after a while, $\\log(n)\\gt 2$, so $n^{\\log(n)}\\gt n^2$. It follows that in the long run, $n^{\\log(n)}$ grows much faster than $n/\\log(n)$.\n\nAs a last example, let us compare $\\log^3(n)$ and $n/\\log(n)$. Informally, $\\log$ grows glacially slowly. More formally, look at the ratio $\\dfrac{\\log^3(n)}{n/\\log(n)}$. This simplifies to $$\\frac{\\log^4 (n)}{n}.$$ We can use L'Hospital's Rule on $\\log^4(x)/x$. Unfortunately we then need to use it several times. It is easier to examine $\\dfrac{\\log(x)}{x^{1/4}}$. Then a single application of L'Hospital's Rule does it. Or else we can let $x=e^y$. Then we are looking at $y^4/e^y$, and we can quote the standard result that the exponential function, in the long run, grows faster than any polynomial.\n\nRemark: The second person in your list is the slowest one. Apart from that, they are in order. So you only need to prove four facts to get them lined up. A fair part of the work has been done above.\n\n-\nYour answer is very very good and understood. I will work on these tomorrow, make sure I understand anything. Thank you very much! \u2013\u00a0SyndicatorBBB Nov 3 '12 at 20:49\n@Guy: I don't mind your accepting my answer. However, it is useful in general to wait somewhat longer, in order to elicit a variety of approaches. \u2013\u00a0Andr\u00e9 Nicolas Nov 3 '12 at 20:54\nI commented below. Can you please advise ?Thank you. \u2013\u00a0SyndicatorBBB Nov 4 '12 at 10:01\n\nFor instance: $$\\log(n^{\\log n})=\\log^2(n)$$ And since asymptotically: $$\\frac{\\log(n)}{\\log(\\log(n))} < \\log^2(n)$$ We know that: $$n^{\\log n}$$ Is at least exponentially larger than: $$\\frac{\\log(n)}{\\log(\\log(n))}$$\n\n-\nHow did you calculate this inequalties? \u2013\u00a0SyndicatorBBB Nov 3 '12 at 20:46\nJust observe: a $\\log(n)$ cancels out, and $1/\\infty$ is obviously smaller than $\\infty$. \u2013\u00a0nbubis Nov 3 '12 at 20:47\nUndestood! Thank you very much! \u2013\u00a0SyndicatorBBB Nov 3 '12 at 20:50\n\n@Andr\u00e9 Nicolas I really liked the comprasion of the functions with simplier functions.\n\nI was trying to use this technique on some of these functions.\n\nEventually I found a couple which I was not able to solve using this technique and I don't understand why.\n\nAssume $f(n) = log(log(n))$ and $g(n) = n/log(n)$\n\nNow one might say, $f(n) < log(n)$ and $g(n) > 1/n$\n\nIf I look at the ratio/limit of $f(n)/g(n)$, I find out that $f(n)$ is much faster.\n\nI completly understand that I chose unwisely the lower bound of g(n). Does it mean that on such functions I cannot use this technique?\n\nIf you \"give away\" stuff to make things simpler, have to be sensible about it. Replacing $g(n)$ by something that approaches $0$ is unreasonable. For example, could use $f(n)\\lt \\log(n)$, and keep $g(n)$. How one \"weakens\" an inequality has to be guided by basic intuition/experience about the sizes of things. The one thing that will be constantly useful is (i) $\\log(n)$ is long-run slower than any positive power of $n$ and its twin (ii) For constant $a\\gt 1$, $a^n$ is long-run faster than any polynomial in $n$. \u2013\u00a0Andr\u00e9 Nicolas Nov 4 '12 at 14:26", "date": "2016-02-14 13:18:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/228416/order-of-magnitudes-comparasions", "openwebmath_score": 0.881853461265564, "openwebmath_perplexity": 224.87314935889398, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575116884779, "lm_q2_score": 0.8670357615200475, "lm_q1q2_score": 0.8519125003840624}}
{"url": "https://mathoverflow.net/questions/368404/if-all-points-of-a-real-function-with-positive-values-would-be-local-minimum-ca", "text": "# If all points of a real function with positive values would be local minimum, can one say it is constant function?\n\nDuring my studies I faced a function $$f:\\mathbb{R} \\to \\mathbb{R}^+$$ with the property: for all $$x \\in \\mathbb{R}$$ and all $$y$$ in open interval $$(x-\\frac{1}{f(x)} ,x+\\frac{1}{f(x)})$$ we have $$f(x) \\leq f(y)$$. At first I guessed maybe it is necessarily constant function but I could not show this. I tried to generate a counter example for my guess and unfortunately I could not again. would anyone please help me?\n\n\u2022 The answer to the question in the title is (of course) no. [Define $f(x) = 2$ for $x \\ne 0$ and $f(0)=1$.] But the question in the text is more complicated. \u2013\u00a0Gerald Edgar Aug 6 '20 at 9:31\n\u2022 What is $\\mathbb{R}^+$? Is $f(x)=0$ allowed? \u2013\u00a0Wilberd van der Kallen Aug 6 '20 at 9:52\n\u2022 $\\mathbb{R}^+$ is denoted for the open interval $(0,+\\infty)$ \u2013\u00a0M. Reza. K Aug 6 '20 at 10:12\n\nMy answer is: Such a function must be constant. Suppose (for puoposes of contradiction) $$f$$ is a nonconstant function $$f : \\mathbb R \\to (0,+\\infty)$$ such that $$\\forall x\\in\\left(a-\\frac{1}{f(a)},a+\\frac{1}{f(a)}\\right),\\quad f(x) \\ge f(a) .$$\n\nFor $$m > 0$$, define $$U_m := \\{x : f(x)>m\\}$$.\n\nLemma 1: For all $$m \\in \\mathbb R$$, the set $$U_m$$ is open.\nProof. Let $$x \\in U_m$$. Then $$(x-1/f(x),x+1/f(x)) \\subseteq U_m$$. So $$U_m$$ is open.\n\nDefine $$\\mathcal G = \\{(a,b,m) : a < b, f(a)\\le m, f(b)\\le m, \\text{ and }\\forall x\\in(a,b),f(x)> m\\}$$.\n\nLemma 2: Let $$a,b \\in \\mathbb R$$ with $$f(a) \\ne f(b)$$, and $$m>0$$. Then there exists $$a_1, b_1$$ with $$a < a_1 < b_1 < b$$ and $$m_1 \\ge m$$ such that $$(a_1,b_1,m_1) \\in \\mathcal G$$.\nProof. One of $$f(a), f(b)$$ is smaller, assume WLOG that $$f(a) < f(b)$$. Now $$U_{f(a)}$$ is open, $$a \\notin U_{f(a)}$$, and $$b \\in U_{f(a)}$$. The maximal open interval $$(c,d) \\subseteq U_{f(a)}$$ with $$c is nonempty, in fact $$(c,b) = (c,d) \\cap (a,b) \\ne \\varnothing$$. By maximality, $$f(c) \\le f(a)$$. Choose $$e$$ with $$c < e < b$$. For $$x \\in (c,e)$$ we have $$f(c) < f(x)$$ so $$c \\le x-1/f(x)$$ and thus $$f(x) \\ge 1/(x-c)$$. Thus $$f(x)$$ is unbounded on $$(c,e)$$. Let $$m_1$$ be such that $$m_1 \\ge m, m_1 > f(c), m_1 > f(e)$$. The open set $$U_{m_1}$$ has $$U_{m_1} \\cap (c,e) \\ne \\varnothing$$ and $$c,e \\notin U_{m_1}$$. So let $$(a_1,b_1)$$ be a maximal open subinterval of the open set $$U_{m_1} \\cap (c,e)$$.\n\nLemma 3: Let $$(a_1, b_1, m_1) \\in \\mathcal G$$, and let $$m > m_1$$. Then there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 \\ge m$$ such that $$(a_2, b_2, m_2) \\in \\mathcal G$$.\nProof: If $$f(a_1) \\ne f(b_1)$$, apply Lemma 2 directly. Otherwise, pick any $$c \\in (a_1,b_1)$$ and apply Lemma 2 to $$a_1, c, m$$.\n\nMain Proof: We assume $$f$$ is not constant. So there exist $$a_1 < b_1$$ with $$f(a_1) \\ne f(b_1)$$. By Lemma 2 there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 > 2$$ so that $$(a_2,b_2,m_2) \\in \\mathcal G$$. Then by Lemma 3, there exist $$a_3, b_3$$ with $$a_2 < a_3 < b_3 < b_2$$ and $$m_3 > 3$$ so that $$(a_3,b_3,m_3) \\in \\mathcal G$$. Continuing recursively, we get sequences $$(a_k), (b_k), (m_k)$$ so that $$\\forall k:\\quad a_k < a_{k+1} < b_{k+1} < b_k,\\quad m_k > k,\\quad\\text{and } (a_k,b_k,m_k)\\in\\mathcal G .$$ The sequence $$(a_k)$$ is increasing and bounded above, so it converges. Let $$a = \\lim_{k\\to\\infty} a_k$$. Then $$a_k < a_{k+1} \\le a \\le b_{k+1} < b_k$$. From $$(a_k,b_k,m_k) \\in \\mathcal G$$ we conclude $$f(a) > m_k > k$$. This is true for all $$k$$, so $$f(a)$$ is not a real number, a contradiction.\n\n\u2022 Your answer was enjoyable and very nice to me. It really engaged me and I am thinking if instead of $\\mathbb{R}$ in domain of the $f$, replace a complete metrically convex metric space, does your answer still work? \u2013\u00a0M. Reza. K Aug 7 '20 at 12:11\n\u2022 @M.Reza.K for every $x,y$ in the complete metrically convex space there is an isometry $g:[a,b]\\to X$ such that $g(a)=x, g(b)=y$. Apply the result to $h=f\\circ g$. It follows that $f(x)=h(a)=h(b)=f(y)$. I am interested if we can drop the completeness: the proof seems to work for rational numbers... Perhaps it also works for intrinsic metric spaces? \u2013\u00a0erz Aug 7 '20 at 12:40\n\u2022 @erz My proof certainly fails for rational numbers. A decreasing sequence of closed intervals may have empty intersection in $\\mathbb Q$. \u2013\u00a0Gerald Edgar Aug 7 '20 at 13:32\n\nThis is not an answer but it greatly narrows down the class of functions $$f$$ with the described property.\n\n1. If $$c>0$$, then the sets $$f^{-1}([c,+\\infty))$$ and $$f^{-1}((c,+\\infty))$$ are both open.\n\nIndeed, every $$x$$ in any of these sets comes with an open ball of radius $$\\frac{1}{f(x)}$$.\n\nConsequently, the sets $$A_c=f^{-1}((-\\infty, c))$$ and $$B_c=f^{-1}((-\\infty, c])$$ are both closed. In particular, $$f$$ is lower semi-continuous.\n\n1. Let $$x\\in B_c$$. We know that if $$d(y,x)<\\frac{1}{c}\\le \\frac{1}{f(x)}$$, then $$f(x)\\le f(y)$$. On the other hand, if $$y\\in B_c$$ and $$d(y,x)<\\frac{1}{c}$$, then $$d(y,x)<\\frac{1}{f(y)}$$, from where $$f(y)\\le f(x)$$. Hence, if $$y\\in B_c$$ and $$d(y,x)<\\frac{1}{c}$$, we get $$f(y)= f(x)$$.\n\nNot only this means that $$f$$ is locally a constant on $$B_c$$, but also that $$f^{-1}(c)$$ is closed.\n\n1. Let us prove the following strengthening of the Baire's theorem: let $$X$$ be a complete metric space, and let $$B_n$$ be a sequence of closed sets such that $$X=\\bigcup B_n$$. Then $$X=\\overline{\\bigcup int B_n}$$.\n\nIndeed, if $$U\\subset X\\backslash \\bigcup int B_n$$ is open and nonempty, it is metrizable with a complete metric, and $$B_n\\cap U$$ is closed in $$U$$. Since $$U=\\bigcup (B_n\\cap U)$$, from the usual Baire's theorem, there is $$m$$ such that $$V=int (B_m\\cap U) \\ne \\varnothing$$. But then $$V=int (B_m\\cap U)=int B_m\\cap U\\subset int B_m \\backslash \\bigcup int B_n=\\varnothing$$. contradiction.\n\n1. So, combining these observation we get the following picture: there is a dense open set $$U$$ such that $$f$$ is locally constant on $$U$$; every level set of $$f$$ is closed, and the distance between elements of $$f^{-1}(c)$$ and $$f^{-1}(d)$$ is at least $$\\frac{1}{\\max \\{c,d\\}}$$.\n\nThe only way it's not a constant function is when the components of $$U$$ are like the gaps in the Cantor set. So the possible counterexample may be constructed like the Cantor's staircase, but the smaller the component, the larger is the value on it. I was unable to perform this fine tuning though, maybe it's impossible.", "date": "2021-05-11 07:06:04", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/368404/if-all-points-of-a-real-function-with-positive-values-would-be-local-minimum-ca", "openwebmath_score": 0.996794581413269, "openwebmath_perplexity": 2288.9339333725334, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575147530352, "lm_q2_score": 0.867035758084294, "lm_q1q2_score": 0.8519124996653178}}
{"url": "http://math.stackexchange.com/questions/87902/problems-about-countability-related-to-function-spaces", "text": "# Problems about Countability related to Function Spaces\n\nSuppose we have the following sets, and determine whether they are countably infinite or uncountable .\n\n1. The set of all functions from $\\mathbb{N}$ to $\\mathbb{N}$.\n\n2. The set of all non-increasing functions from $\\mathbb{N}$ to $\\mathbb{N}$.\n\n3. The set of all non-decreasing functions from $\\mathbb{N}$ to $\\mathbb{N}$.\n\n4. The set of all injective functions from $\\mathbb{N}$ to $\\mathbb{N}$.\n\n5. The set of all surjective functions from from $\\mathbb{N}$ to $\\mathbb{N}$.\n\n6. The set of all bijection functions from from $\\mathbb{N}$ to $\\mathbb{N}$.\n\nMy thoughts on this:\n\n1. The first set is uncountable by using diagonalization argument.\n\n2. I have read something about it saying this is countable since we can see this as a set of finite sequences of natural numbers. But I have hard time following the argument.\n\n3. The article I read says this is uncountable but I couldn't follow the argument either.\n\n4. For (4),(5) and (6), I am not even sure how to approach problems like these.\n\nCould someone please explain how to approach this kind of problems in general?\n\nAnd does it matter when I change all the $\\mathbb{N}$ to $\\mathbb{R}$?\n\n-\nRecall the definition of a function as a set of ordered pairs. That should help. \u2013\u00a0 Potato Dec 3 '11 at 2:27\nMaybe you can argue that (2) is countable because any given sequence that is non-increasing eventually stops decreasing at some finite natural number or hits zero and then is forced to be constant at 0. Then come up with a clever way to list all the functions, maybe by listing them first by starting position then by \"length before it is constant\" or something like that. \u2013\u00a0 tomcuchta Dec 3 '11 at 2:28\n@tomcuchta Your idea for (2) is alright, but a small nitpick. The function need not go to $0$ but might stop at $1$ (say). What is true (and important) is that the function is eventually constant. \u2013\u00a0 Srivatsan Dec 3 '11 at 2:39\nI removed the (functional-analysis), the (set-theory) and the (proof-theory) tag. Each of these tags is intended for subjects usually taught at an advanced undergraduate or graduate level. \u2013\u00a0 t.b. Dec 3 '11 at 2:42\n@t.b.Thanks, I guess I am confused about what function spaces really study. \u2013\u00a0 geraldgreen Dec 3 '11 at 3:12\n\nAnother approach to the 6th part.\n\nLet us denote the set of all bijections from $\\mathbb N$ to $\\mathbb N$ by $\\operatorname{Bij}(\\mathbb N,\\mathbb N)$.\n\nClearly $$|\\operatorname{Bij}(\\mathbb N,\\mathbb N)| \\le \\aleph_0^{\\aleph_0} = 2^{\\aleph_0} = \\mathfrak c.$$\n\nLet us try to show the opposite inequality.\n\nFor arbitrary function $f\\in\\operatorname{Bij}(\\mathbb N,\\mathbb N)$ we define $$\\operatorname{Fix}(f)=\\{n\\in\\mathbb N; f(n)=n\\}.$$ (That is, $\\operatorname{Fix}(f)$ is the set of all fixed points of the map $f$.) Let us try to answer the question, whether for any $A\\subseteq\\mathbb N$ it is possible to find a bijection ${f_A}:{\\mathbb N}\\to {\\mathbb N}$ such $\\operatorname{Fix}(f_A)=A$. Let us consider two cases.\n\nIf the complement of the set $A$ is infinite, i.e. $\\mathbb N\\setminus A=\\{b_n; n\\in\\mathbb N\\}$ (and we can assume that the elements $\\mathbb N\\setminus A$ are written in the increasing order, i.e. $b_1<b_2<\\dots<b_n<b_{n+1}<\\dots$ ), then we can define a function $f_A$ as follows: $$f_A(x)= \\begin{cases} x, & \\text{if }x\\in A, \\\\ b_{2k+1}, &\\text{ak x=b_{2k} for some k\\in\\mathbb N},\\\\ b_{2k}, &\\text{ak x=b_{2k+1} for some k\\in\\mathbb N}. \\end{cases}$$ In the other words, we did not move the elements of $A$ and the elements from the complement were paired and we interchanged it pair. Such map is a bijection from $\\mathbb N$ to $\\mathbb N$ such that $\\operatorname{Fix}(f_A)=A$.\n\nNext we consider the case that $\\mathbb N\\setminus A$ is finite.\n\nIf $\\mathbb N\\setminus A=\\emptyset$, then $f=id_{\\mathbb N}$ is a function fulfilling $\\operatorname{Fix}(f_A)=\\mathbb N=A$.\n\nIf the set $\\mathbb N\\setminus A$ is a singleton $\\{a\\}$, then is is impossible to find a bijection $f$ such that $\\operatorname{Fix}(f)=A=\\mathbb N\\setminus\\{a\\}$. No element of $\\mathbb N\\setminus A$ can be mapped on $a$ (since all such elements are fixed). But $a$ cannot be mapped on $a$ since this would mean that $a$ is a fixed point.\n\nBut if the set $\\mathbb N\\setminus A$ has at least two elements, then it is possible to construct such a map. We again assume that the elements $b_k$ are written in the increasing order, i.e. $\\mathbb N\\setminus A=\\{b_0,\\dots, b_n\\}$ and $b_0<b_1<\\dots<b_n$. We put $f_A(x)=x$ for $x\\in A$, $f_A(b_k)=b_{k+1}$ for $0\\le k<n$ and $f_A(b_n)=b_0$. (I.e. we made a cycle consisting of elements of $\\mathbb N\\setminus A$.)\n\nThis defines a bijection ${f_A}:{\\mathbb N}\\to{\\mathbb N}$ such that $\\operatorname{Fix}(f_A)=A$.\n\nThe assignment $A\\mapsto f_A$ that we described above is a map from the set $\\mathcal P({\\mathbb N})\\setminus\\{\\mathbb N\\setminus\\{a\\}\\in\\mathbb N\\}$ consisting of all subsets of $\\mathbb N$, whose complement is not a singleton, to the set $\\operatorname{Bij}(\\mathbb N,\\mathbb N)$. This map is injective; since from $f_A=f_B$ we get $A=\\operatorname{Fix}(f_A)=\\operatorname{Fix}(f_B)=B$.\n\nFrom the set $\\mathcal P({\\mathbb N})$ with the cardinality $\\mathfrak c$ we have omitted a countable set $\\{\\mathbb N\\setminus\\{a\\}\\in\\mathbb N\\}$. Therefore the cardinality of this difference $\\mathcal P({\\mathbb N})\\setminus\\{\\mathbb N\\setminus\\{a\\}\\in\\mathbb N\\}$ is again $\\mathfrak c$.\n\nSo we found an injection from a set of cardinality $\\mathfrak c$ to the set $\\operatorname{Bij}(\\mathbb N,\\mathbb N)$. This yields the opposite inequality $$\\mathfrak c\\le|{\\operatorname{Bij}(\\mathbb N,\\mathbb N)}|$$ and by Cantor-Bernstein theorem we get $|{\\operatorname{Bij}(\\mathbb N,\\mathbb N)}|=\\mathfrak c$.\n\n-\nThanks for doing the details. It pointed out a flaw in my answer. \u2013\u00a0 David Mitra Dec 3 '11 at 14:07\nI did not notice that we are using basically the same argument until you pointed it out. I like your answer, but I am out of votes for today. \u2013\u00a0 Martin Sleziak Dec 3 '11 at 14:14\n\nThis is not a complete answer. [I am taking $\\mathbb N$ to include $0$.] I will give you a proof that there are uncountably many bijections from $\\mathbb N$ to itself. This obviously subsumes the questions (4)-(6).\n\nConsider functions $f : \\mathbb N \\to \\mathbb N$ of the following special form: For every $k$, the pair of elements $\\{ 2k, 2k+1 \\}$ is mapped to itself. However, we have two ways of doing this:\n\n\u2022 either: map $2k$ to itself, and similarly for $2k+1$.\n\n\u2022 or: map $2k$ and $2k+1$ to each other.\n\nThe key point is that for each $k$, you could pick one of the two possibilities independently. Since there are $2$ options for each $k$, you have $2^{\\mathbb N}$ options; i.e., there are uncountably many such functions satisfying this restriction. And each such function is a bijection from $\\mathbb N$ to itself.\n\nThe above needs a little bit of work to make it completely rigorous. In particular, try to formalise what I mean by \"$2^\\mathbb N$ possibilities\".\n\nFor (3), the idea is similar. For each $k$, try mapping $k$ to either $2k$ or $2k+1$. Details are left as an exercise.\n\nFor (2), does the following hint help? Every non-increasing function $f : \\mathbb N \\to \\mathbb N$ is eventually constant. So the function is uniquely specified if we describe the finite initial \"portion\" of the function.\n\nAnother approach (due to tb) for (2): If you plot the graph of a non-increasing function, it looks like a staircase that goes down as we move from left to right -- a staircase with finitely many steps because you the function can go down only finitely many times. One can also reconstruct the function if we record the upper right corners of each of those steps. Since the latter corresponds to a finite subset of $\\mathbb N \\times \\mathbb N$ (which is countable), it follows that our set of non-increasing functions is countable as well.\n\n-\n\nI'll use this opportunity to mention here a very nice (at least in my opinion) proof that the cardinality of all bijections $\\mathbb N\\to\\mathbb N$ is $\\mathfrak c=2^{\\aleph_0}$, which is basically the part 6. I making it a community wiki, since the idea is not mine. I believe I've seen this at sci. math but I am not sure about it and I cannot find the link now. (Feel free to add some pointers and links, if you know some.)\n\nHINT: If someone wants to try to develop the idea by himself, the basic idea is Riemann's rearrangement theorem.\n\nLet us denote the set of all bijections from $\\mathbb N$ to $\\mathbb N$ by $\\operatorname{Bij}(\\mathbb N,\\mathbb N)$.\n\nClearly $$|\\operatorname{Bij}(\\mathbb N,\\mathbb N)| \\le \\aleph_0^{\\aleph_0} = 2^{\\aleph_0} = \\mathfrak c.$$\n\nNow we will show the opposite inequality. Let us choose a series $\\sum_{n=0}^\\infty x_n$, which is convergent but not absolutely convergent (e.g. $\\sum_{n=0}^\\infty x_n = \\sum_{n=0}^\\infty (-1)^n\\frac1{n+1}$). By Riemann's theorem, for any real number $r$ there exists a permutation $\\pi\\in \\operatorname{Bij}(\\mathbb N,\\mathbb N)$ such that $\\sum_{n=0}^\\infty x_{\\pi(n)}=r$. If we choose for each $r$ one such permutation, we get an injective map from $\\mathbb R$ to $\\operatorname{Bij}(\\mathbb N,\\mathbb N)$. Thus $|\\operatorname{Bij}(\\mathbb N,\\mathbb N)| \\ge |\\mathbb R| = \\mathfrak c = 2^{\\aleph_0}$.\n\n-\n+1, nice approach. I don't think you have to make an answer CW just because it isn't your idea. Writing an answer is contribution enough and one deserves some credit for that. =) \u2013\u00a0 Srivatsan Dec 4 '11 at 0:24\n\nLet \\eqalign{ A_1&=\\text{the set of all functions from }\\Bbb N\\text{ to }\\Bbb N. \\cr A_2&=\\text{the set of all non-increasing from }\\Bbb N\\text{ to }\\Bbb N. \\cr A_3&=\\text{the set of all non-decreasing }\\Bbb N\\text{ to }\\Bbb N. \\cr A_4&=\\text{the set of all 1-1 functions from}\\Bbb N\\text{ to }\\Bbb N. \\cr A_5&=\\text{the set of all onto functions from }\\Bbb N\\text{ to }\\Bbb N. \\cr A_6&=\\text{the set of all bijections from }\\Bbb N\\text{ to }\\Bbb N. \\cr }\n\nConsider $A_6$:\n\n$A_6$ may be thought of as the set $\\bigl\\{\\,\\pi :\\pi \\text{ is a permutation of } (1,2,3,\\ldots)\\,\\bigr\\}$.\n\nFor any permutation $\\pi$, we may associated a binary sequence $(x_n)$ by setting $$x_n=\\cases{ 0,& \\text{if } \\pi_n=n\\cr 1,& \\text{otherwise} }.$$ This induces an onto map from $A_6$ to the set of infinite binary sequences (Edit: not quite, as Martin Sleziak points out in his answer, the binary sequences that have \"only one 1\" have no element in $A_6$ mapped to them. But, as the set of such sequences is countable, the rest of the argument will go through. Martin also describes nicely how to obtain this map.). Since the latter set is uncountable, so is the former.\n\nSo, $A_6$ is uncountable.\n\nFrom this, it follows that $A_5$, $A_4$, and $A_1$ are uncountable.\n\nNow, consider $A_2$:\n\nIf $f$ is a non-increasing function, then $f$ must be eventually constant. Thus, the cardinality $A_2$ would be at most the cardinality of the set $$\\bigcup_{j,m} \\bigl\\{\\,(a_i)_{i=1}^\\infty : a_i=m, i\\ge j\\,\\bigr \\}.$$ But this latter set is a countable union of countable sets and is, thus, countable.\n\nSo, $A_2$ is countable.\n\nNow, consider $A_3$:\n\nThe cardinality of the set of non-decreasing functions is at least the cardinality of the set $$\\{\\, A\\subset\\Bbb N : A \\text{ is an infinite subset of } \\Bbb N\\, \\}.$$\n\nSince this latter set uncountable, $A_3$ is uncountable.\n\n-\n\"If f is a non-increasing function, then f must be eventually constant\" Why ? it can also be increasing. \u2013\u00a0 GinKin Mar 11 at 12:45\n@GinKin I took \"non-increasing\" to mean $f(b)\\le f(a)$\" whenever $a\\le b$ (that is, (perhaps non-strictly) decreasing). This is standard terminology. What do you mean by \"it can also be increasing\"? \u2013\u00a0 David Mitra Mar 11 at 13:11\nWhoops mixed the terms, I meant that if it doesn't increase then it doesn't have to be eventually constant, it can be decreasing. \u2013\u00a0 GinKin Mar 11 at 13:24\n@GinKin But the range is $\\Bbb N$. If it were strictly decreasing, it would eventually be constantly $0$. \u2013\u00a0 David Mitra Mar 11 at 13:25", "date": "2014-08-28 23:37:56", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/87902/problems-about-countability-related-to-function-spaces", "openwebmath_score": 0.965313732624054, "openwebmath_perplexity": 191.6129276320447, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575121992375, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8519124991389965}}
{"url": "https://ask.sagemath.org/question/49683/the-number-of-solutions-of-a-polynomial-system-using-a-grobner-basis/?answer=49684", "text": "# The number of solutions of a polynomial system using a Gr\u00f6bner basis\n\nThe following extract from this Wikipedia page explains that we can easily deduce the number of solutions of a polynomial system using Gr\u00f6bner basis:\n\nGiven the Gr\u00f6bner basis G of an ideal I, it has only a finite number of zeros, if and only if, for each variable x, G contains a polynomial with a leading monomial that is a power of x (without any other variable appearing in the leading term). If it is the case the number of zeros, counted with multiplicity, is equal to the number of monomials that are not multiple of any leading monomial of G. This number is called the degree of the ideal.\n\nQuestion: Is there a function in Sage computing this number of zeros from a given Gr\u00f6bner basis?\n\nedit retag close merge delete\n\nSort by \u00bb oldest newest most voted\n\nYes, these\n\nmonomials that are not a multiple of any leading monomial of G\n\nform a basis of the quotient ring $R/I$ (by the reduction algorithm), also called a normal basis for $I$.\n\nsage: R.<x,y,z> = PolynomialRing(QQ,3)\nsage: I = Ideal([x^2+y+z-1,x+y^2+z-1,x+y+z^2-1])\nsage: B = I.groebner_basis(); B\n[x^2 + y + z - 1, y^2 + x + z - 1, z^2 + x + y - 1]\nsage: I.normal_basis()\n[x*y*z, y*z, x*z, z, x*y, y, x, 1]\n\nThe number of elements in this basis, or the vector space dimension of $R/I$, can also be found as follows:\n\nsage: I.vector_space_dimension()\n8\n\nThe zero set, or variety, of the ideal is the following:\n\nsage: I.variety(QQbar)\n[{z: 0, y: 0, x: 1},\n{z: 0, y: 1, x: 0},\n{z: 1, y: 0, x: 0},\n{z: -2.414213562373095?, y: -2.414213562373095?, x: -2.414213562373095?},\n{z: 0.4142135623730951?, y: 0.4142135623730951?, x: 0.4142135623730951?}]\n\nThe number of elements is not 8, because we are missing the \"multiplicities\". Indeed,\n\nFalse\n\nSo perhaps the more interesting quantity is\n\n5\n\nwhich agrees with the number of zeros without multiplicity.\n\n(There is an easy algorithm to compute the radical of a zero-dimensional ideal.)\n\nmore\n\nIs it required to compute the Gr\u00f6bner basis? Does the function normal_basis() recompute the Gr\u00f6bner basis?\n\n( 2020-01-26 20:05:55 -0500 )edit\n\nThe method normal_basis() internally requires a Gr\u00f6bner basis. If one has not yet been computed, then it computes one. If one has been computed already, then it uses that one; it does not recompute it. Note also that the result / speed depends on the chosen monomial ordering (degrevlex by default). Of course the dimension does not depend on this choice.\n\n( 2020-01-27 03:57:43 -0500 )edit", "date": "2020-07-14 07:28:51", "meta": {"domain": "sagemath.org", "url": "https://ask.sagemath.org/question/49683/the-number-of-solutions-of-a-polynomial-system-using-a-grobner-basis/?answer=49684", "openwebmath_score": 0.29856231808662415, "openwebmath_perplexity": 910.6520081079834, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9825575183283513, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.851912496013594}}
{"url": "https://mathematica.stackexchange.com/questions/240082/solving-a-eigenvalue-problem", "text": "# Solving a Eigenvalue Problem\n\n$$\\textbf{Eigenvalue problem on a unit square \\Omega = [0,1]^2 :}$$\n\nConsider the eigenvalue problem with the Dirichlet boundary condition that is, $$-Lu = \\lambda u$$ where $$L = \\frac{\\partial^2}{\\partial x^2} + \\frac{\\partial^2}{\\partial y^2}$$.\n\nThe boundary condition is that $$u=0$$ on $$\\partial \\Omega$$.\n\nI am computing the $$\\textbf{eigenvalues}$$ and $$\\textbf{eigenfunctions}$$ of Laplacian numerically in Mathematica. Specially, I am interested about the $$\\textbf{multiplicity}$$ of a specific eigenvalue. I already know that Tally or Count can certainly count the occurrances of eigenvalues in the list called vals.\n\n{\u2112, \u212c} = {-Laplacian[u[x, y], {x, y}],\nDirichletCondition[u[x, y] == 0, True]};\n\n{vals, funs} =\nDEigensystem[{\u2112, \u212c},\nu[x, y], {x, 0, 1}, {y, 0, 1}, _];\n\nTally[vals]\n\n\nIn the eigenvalue problem there are infinite numbers of eigenvalues. So, We can not list all of them. I want to find out $$\\textbf{multiplicity}$$ of a specific eigenvalue that counts the total number of occurrances. For example, the multiplicity of the eigenvalue $$5 \\pi^2$$ is $$2$$. Similarly, I want to find out the multiplicity of the eigenvalue $$50 \\pi^2$$ is $$3$$.\n\n\u2022 Edit your question to include your Mathematica code rather than pictures of the code. Feb 15 '21 at 17:51\n\u2022 Convert your cells to InputForm and look at the Markdown help Feb 15 '21 at 18:24\n\u2022 Feb 15 '21 at 19:30\n\nWith the exact solutions\n\nu[{nx_, ny_}, {x_, y_}] = Sin[nx \u03c0 x] Sin[ny \u03c0 y];\n\n\nand $$\\{n_x,n_y\\}\\in\\mathbb{N}^2$$, the eigenvalues are\n\n\u03bb[nx_, ny_] = (nx^2 + ny^2) \u03c0^2;\n\n\nTest:\n\nAssuming[Element[nx | ny, PositiveIntegers],\n{u[{nx, ny}, {x, 0}], u[{nx, ny}, {x, 1}],\nu[{nx, ny}, {0, y}], u[{nx, ny}, {1, y}]} // FullSimplify]\n(*    {0, 0, 0, 0}    *)\n\n-D[u[{nx, ny}, {x, y}], {x, 2}] - D[u[{nx, ny}, {x, y}], {y, 2}] ==\n\u03bb[nx, ny] * u[{nx, ny}, {x, y}] // FullSimplify\n(*    True    *)\n\n\nThe number of pairs $$\\{n_x,n_y\\}$$ equal to $$\\lambda/\\pi^2$$ is therefore given by OEIS A063725. You can calculate the multiplicity of a given value of $$i=\\lambda/\\pi^2$$ directly from the generating function,\n\ng[q_] = (EllipticTheta[3, q] - 1)^2/4;\nm[i_Integer?NonNegative] := SeriesCoefficient[g[q], {q, 0, i}]\n\n\nFor example, the multiplicity of $$\\lambda=50\\pi^2$$ is 3, as you know:\n\nm[50]\n(*    3    *)\n\n\nThere is no eigenvalue at $$326\\pi^2$$, on the other hand:\n\nm[326]\n(*    0    *)\n\n\nMore generally, the $$d$$-dimensional generating function\n\ng[d_, q_] = ((EllipticTheta[3, q] - 1)/2)^d;\n\n\nallows us to find the multiplicity of the eigenvalue $$\\lambda=i\\pi^d$$ directly:\n\nm[d_Integer?NonNegative, i_Integer?NonNegative] :=\nSeriesCoefficient[g[d, q], {q, 0, i}]\n\n\nFor example, the $$d=17$$-dimensional hypercube has $$1\\,416\\,786\\,753\\,216$$ eigenvalues $$\\lambda=250\\pi^{17}$$:\n\nm[17, 250]\n(*    1416786753216    *)\n\n\nIf you want the actual solutions, not just the number of solutions, PowersRepresentations is a good starting point (but it gives solutions containing zeros, which we need to filter out):\n\ns[d_Integer?NonNegative, i_Integer?NonNegative] :=\nSelect[PowersRepresentations[i, d, 2], Min[#] > 0 &]\n\n\nFor example, the eigenvalue $$\\lambda=50\\pi^2$$ in $$d=2$$ dimensions can be composed in two ordered ways, as you know:\n\ns[2, 50]\n(*    {{1, 7}, {5, 5}}    *)\n\n\nEnumerating all permutations of these ordered combinations gives the full multiplicity of 3:\n\nt[d_Integer?NonNegative, i_Integer?NonNegative] :=\nJoin @@ Permutations /@ s[d, i]\n\nt[2, 50]\n(*    {{1, 7}, {7, 1}, {5, 5}}    *)\n\n\nIn $$d=17$$ dimensions the eigenvalue $$\\lambda=250\\pi^{17}$$ has 999 ordered solutions,\n\ns[17, 250]\n(*    {{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 6, 14},\n{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 10, 10},\n{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 15},\n...\n{3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5},\n{3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6}}    *)\n\n\nwhich then become the aforementioned $$1\\,416\\,786\\,753\\,216$$ unordered solutions from t[17, 250].\n\n\u2022 We should work with the generating function explicitly because it gives a very efficient recipe, especially in the high-dimensional case. Just look at the given example for the 17-dimensional hypercube: manually summing and counting is going to be very tedious and difficult, whereas the call to m[17, 1000] takes less than half a second. The same is true, to some extent, for your case $d=2$ when you work with large values of $\\lambda$. Feb 16 '21 at 8:32\n\u2022 Sorry, this is not the right place for an introduction to generating functions or Jacobi theta functions. Feb 16 '21 at 9:31", "date": "2022-01-19 03:02:02", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/240082/solving-a-eigenvalue-problem", "openwebmath_score": 0.4080619215965271, "openwebmath_perplexity": 875.7607282155728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575142422756, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8519124958466456}}
{"url": "https://brilliant.org/discussions/thread/show-why-the-value-converges-to-pi/", "text": "# Show why the value converges to $$\\pi$$\n\n$$a_0=1$$\n\n$$a_{n+1}=a_n+\\sin{(a_n)}$$\n\nExplain why the following occurs:\n\n$$a_0=1$$\n\n$$a_1=1+\\sin{(1)}\\approx 1.841470985$$\n\n$$a_2=1+\\sin{(1)}+\\sin{(1+\\sin{(1)})}\\approx 2.805061709$$\n\n$$a_3=1+\\sin{(1)}+\\sin{(1+\\sin{(1)})}+\\sin{(1+\\sin{(1)}+\\sin{(1+\\sin{(1)})})}\\approx 3.135276333$$\n\n$$a_4\\approx 3.141592612$$\n\n$$a_5\\approx 3.141592654\\approx\\pi$$\n\nNote by Jack Han\n3\u00a0years, 7\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n\n- bulleted\n- list\n\n\u2022 bulleted\n\u2022 list\n\n1. numbered\n2. list\n\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1\n\nparagraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\n## Comments\n\nSort by:\n\nTop Newest\n\nLet's see here...\n\nThe interesting thing I was talking about is the fact that the given series always converges into a value $$a_n$$ such that $$sin(a_n)=0$$ for different values of $$a_0$$. To put it in more exact terms, it always converges to a value of $$a_n$$ such that $$cos(a_n)=-1\\Rightarrow a_n=m\\pi$$, where $$m$$ is an odd integer. ($$m$$ can also be even, but that is a degenerate case where all terms are same)\n\nI'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.\n\nConsider the function $$f\\left( x \\right) =\\cos { (x) } \\\\ \\Rightarrow \\frac { df\\left( x \\right) }{ dx } =-\\sin { (x) }$$\n\nNow see what happens when we take some arbitrary value of $$x$$ (say $$x=1$$)and then do the following repeatedly:\n\n$$x:=x-\\frac { df\\left( x \\right) }{ dx }=x+\\sin { (x) }$$ (\":=\" is the assignment operator )\n\nIn the above figure, we can see two points marked. One is red, which represents the first value of $$x$$($$=1$$). The other is brown, and is after one iteration of above step.\n\nWe can see that when we do $$x:=x+sin(x)$$, what is actually happening is that $$x$$ is surfing along the slope of the curve $$cos(x)$$. We move the value of $$x$$ down the tangent. Change $$x$$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $$x=\\pi$$.\n\nI know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.\n\nIn general, series defined as $$a_n=a_{n-1}-\\alpha\\frac { df\\left(a_{n-1} \\right) }{ da_{n-1} }$$\n\nWill converge to the nearest value of $$a$$ (nearest to $$a_0$$) such that $$f(a)$$ is minimum, provided the value of $$\\alpha$$ is not too large.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nGood work.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nGood work! Newton's method for estimating roots.\n\nPretty much seals the deal. Great solution +1.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nYes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nAs usual, since the series converges.. this means that when $$n\\to\\infty$$ ,\n\n$$a_{n+1}=a_{n}$$. But $$a_{n+1}=a_{n}+sin(a_{n})$$\n\n$$\\Rightarrow sin(a_{n})=0$$\n\nNow how do we know that $$a_{n}=\\pi$$? We know this since $$a_0=1$$ and the series is constantly increasing. Therefore, it converges onto the first value of $$x>1$$ such that $$sin(x)=0$$.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nBro , But It is not always true $$\\lim _{ n\\rightarrow \\infty }{ ({ a }_{ n }) } =\\lim _{ n\\rightarrow \\infty }{ { (a }_{ n+1 }) }$$ .\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nWhy not? Do you have a counter-example?\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nComment deleted Mar 13, 2015\n\nLog in to reply\n\nBut can't the value of zero be discarded? Because here, $$a_{n}=1$$ for all n.\n\nThere is no question of convergence at all as all the terms have a fixed value.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nMy Bad , I wrongly co-relate two diffrent situations , Actually this example is given when I got two answwer to an single recursion , which is something Like that $${ a }_{ n+1 }={ \\left( \\sqrt { 2 } \\right) }^{ { a }_{ n } }$$ , I misinterpreate this current situation, So I deleted that irrelevant comment , But Yes this is good recurance , and I'am trying to prove convergence of this recurance. If I got correct than I report you.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nThanks for joining the party :D\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nComment deleted Mar 13, 2015\n\nLog in to reply\n\nNo, the sequence only increases, only that the rate of increase decreases till it reaches 0 at $$a_n= \\pi$$\n\nyou can easily see it from the graph if x+sin(x), it is a non decreasing function and reaches inflection point at x= pi\n\nbut yes convergence remains unproven, i am trying\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nMy bad. Sorry I jumped to a wrong conclusion. Thanks for replying.\n\nBut the rate at which the value of $$a_{n}$$ increases isn't the same as the rate at which $$x+sinx$$ is increasing. So how do you directly conclude from the graph?\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nit is not the increasing nature that confirms, it is the fact that the graph lies above x=y as well\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nI sort of understand what you're saying...But isn't it easier to say this: The increment in the variable in every iteration is $$sina_{n}$$ and as $$a_{n}$$ tends to $$\\pi$$, $$\\quad sina_{n}$$ becomes smaller and smaller. (I think that is what you were saying initially about rate of change slowing down).\n\nI have deleted my previous comment as it might only muddle things up for someone else.\n\nAlso, please check out my response to your post on this problem of mine.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nIndeed it is ,simply that sin(an) is positive, in the range and hence increment is positive or sequence rises :)\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nCan someone prove the convergence of this recursion?\n\n@Ronak Agarwal @Mvs Saketh @Azhaghu Roopesh M @Pratik Shastri @Raghav Vaidyanathan @Deepanshu Gupta or anyone else who's interested.\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nThis is turning out to be very interesting...\n\nI want to know if we can find a general form for a function $$f(x)$$ such that the series $$a_1,a_2,...$$ defined by:\n\n$$a_{n+1}=a_{n-1}+f(a_{n-1})$$\n\nConverges for a given value of $$a_0$$.\n\nFurther, is it true that all of the values of such $$a_n$$ as $$n\\to \\infty$$ satisfy $$f(a_n)=0$$?\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nI think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nWhat are you thinking?\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nComment deleted Mar 12, 2015\n\nLog in to reply\n\nare you sure you have differentiated it correctly ? there should be atleast one $$x$$ in your $$f'(x)$$ i think\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nIf you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\nWhich value converges to pi ??\n\n- 3\u00a0years, 7\u00a0months ago\n\nLog in to reply\n\n\u00d7\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading...", "date": "2018-10-23 06:34:34", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/show-why-the-value-converges-to-pi/", "openwebmath_score": 0.9849362373352051, "openwebmath_perplexity": 1420.8598716209287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9585377296574669, "lm_q2_score": 0.8887587817066392, "lm_q1q2_score": 0.8519088248302181}}
{"url": "https://math.stackexchange.com/questions/4511928/integration-of-piecewise-function", "text": "# Integration of piecewise function\n\nI was doing a question that involved integration of a piecewise function.\n\nThe question, as can be seen above, asks to evaluate the integral from 0 to pi. I worked it out by breaking the function into 2 integrals where the first had lower and upper limits of 0 and pi/2 respectively and adding it to the integral with lower and upper limits of pi/2 and pi respectively.\n\nNow, my question is, how can we use FTC 2 on the first integral if it is not continuous over the domain [0, pi/2] as the pi/2 is not included in the sin (x).\n\nI have tried my best to articulate my question but if there is any confusion, please do let me know.\n\nIs there something I am missing? Can someone kindly clarify.\n\nYou can still evaluate the integrals individually and then add them together. There is a jump discontinuity at $$\\frac{\\pi}{2}$$ due to the interval given in the piecewise, but this point will not impact the sum. That is because the thickness of a single point is intuitively $$0$$.\n\nConsider the point $$c$$ (Note: c is finite):\n\n$$\\int_c^{c} f(x) dx= F(c)-F(c)=0$$ by the FTC.\n\nAdditionally, note that the function is integrable from that interval.\n\n$$\\int_0^{\\pi/2} \\sin(x) dx = [-\\cos(x)]_0^{\\pi/2} = 0-(-1) = 1$$\n\n\u2022 Hi, yes, I can see why. However, does this discontinuity affect our ability to use FTC 2 as it requires continuity over [a, b]? Aug 14 at 12:30\n\u2022 It does not because the sum of one point is, intuitively, $0$. In other words, a point has no thickness/width to sum. Aug 14 at 12:34\n\u2022 Oh ok. Thank you. So I would be correct in saying that in a case like this, there is some flexibility to the theorem? Aug 14 at 12:37\n\u2022 @Oofy2000 No because $\\sin(x)$ is integrable/well defined/continuous from $0 \\le x \\lt \\frac{\\pi}{2}$ Aug 14 at 12:41\n\u2022 @Oofy2000 Added one. Here is a rigorous explanation, if interested: math.stackexchange.com/questions/511197/area-under-a-point Aug 14 at 13:13\n\nFirst prove this, from the definition of the integral:\n\nProp. Suppose $$p\\in[a,b]$$, and $$f:[a,b]\\to\\Bbb R$$ satisfies $$f(x)=0$$ for all $$x\\in[a,b]\\setminus\\{p\\}$$. Then $$f$$ is integrable on $$[a,b]$$ and $$\\int_a^bf=0$$.\n\nHence, even though $$f(x)-\\sin(x)$$ does not quite vanish at every point, it follows that $$\\int_0^{\\pi/2}(f(x)-\\sin(x))\\,dx=0,$$hence $$\\int_0^{\\pi/2}f(x)\\,dx=\\int_0^{\\pi/2}\\sin(x)\\,dx.$$\n\n(Technicality: Since $$\\sin(x)$$ and $$f(x)-\\sin(x)$$ are both integrable on $$[0,\\pi/2]$$, so is $$f$$.", "date": "2022-12-01 20:32:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4511928/integration-of-piecewise-function", "openwebmath_score": 0.9462644457817078, "openwebmath_perplexity": 242.7331633281572, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811591688145, "lm_q2_score": 0.8824278649085117, "lm_q1q2_score": 0.851879235108241}}
{"url": "https://stats.stackexchange.com/questions/225426/continuous-process-vs-discrete-process", "text": "# Continuous process vs Discrete process!\n\nI encountered two statements in Tibshirani's article Regression Shrinkage and Selection via the Lasso, p.58 (pdf).\n\n1. Subset selection is a \"discrete process\" which leads to unstable coefficients.\n2. ridge regression is a \"continuous process\" leading to stable coefficients.\n\nMy question is, why ridge is called continuous process while subset selection is a discrete process. Also, why ridge leads to stability and the Subset selection to instability.\n\n\u2022 1. Do you have a particular book or something that says this? It might be useful to answerers to have context. 2. Are you asking \"what makes it discrete or continuous?\" or are you asking \"why would subset selection be unstable while ridge regression is stable?\" or are you asking \"why would discreteness/continuity be described as stable or unstable?\" \u2013\u00a0Glen_b Jul 25 '16 at 1:54\n\u2022 @ Glen_b, my question is why ridge is called continuous process, and subset selection is discrete process, and why the first one leads to stability and the later to instability. \u2013\u00a0jeza Jul 25 '16 at 2:03\n\u2022 You still need to provide some context / a quote. \u2013\u00a0gung Jul 25 '16 at 17:33\n\u2022 @ gung, now, it is ok \u2013\u00a0jeza Jul 25 '16 at 18:42\n\u2022 Selecting a subset of variables is a binary decision and in that sense discrete. Ridge regression, on the other hand, performs regularization by putting stronger emphasis on certain variables (without merely making the decision to include or remove them) as a result of a modified residual term. A more precise explanation can be found in Wikipedia. \u2013\u00a0Igor Jul 25 '16 at 21:18\n\nFor subset selection, in its most general form, you pick some metric to optimize and try all possible subsets of variables to include in your linear regression model. Suppose you have $p$ covariates, then there are $2^p$ models, i.e. there are $2^p$ vectors $\\beta$ to choose from. Hence, jumping from model to model is a discrete process. You cannot \"interpolate\" between models.\nFor ridge regression, you add an $l_2$ penalty to the coefficients, so that the model you choose is $$\\min_\\beta \\| X\\beta - y\\|^2 + \\frac{\\lambda}{2}\\|\\beta\\|^2$$ In this case, for any $\\lambda \\ge 0$ you get a different model (that is solution of $\\beta(\\lambda)$). In fact, $\\beta$ will be a smooth function of $\\lambda$: change $\\lambda$ a little bit, and your fitted $\\beta$ will change a little bit. Hence, there are an infinite number of possible models. Thus, this is a continuous process and you can smoothly go from one model to the other by changing $\\lambda$.\nThis is also the reason why subset selection is unstable, and ridge regression is stable. Suppose you change the metric you use to fit the subset selection process a little bit, and you can end up with a very different model (here with different model I mean what elements of $\\beta$ are non-zero). As I noted above, this is not true for ridge regression; a small change in $\\lambda$ leads to a small change in $\\beta$.\nAlso worth noting that in general all coefficients in Ridge regression are non-zero for all values of $\\lambda$, but they converge monotonically to $0$ as $\\lambda \\to \\infty$.", "date": "2019-05-24 03:53:37", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/225426/continuous-process-vs-discrete-process", "openwebmath_score": 0.6683257222175598, "openwebmath_perplexity": 396.0196086578892, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811601648193, "lm_q2_score": 0.8824278587245935, "lm_q1q2_score": 0.8518792300173054}}
{"url": "https://math.stackexchange.com/questions/1941888/calculate-int-0-infty-frac-sinx-logxx-mathrm-dx", "text": "# Calculate $\\int_0^\\infty\\frac{\\sin(x)\\log(x)}{x}\\mathrm dx$.\n\nCalculate $\\displaystyle\\int_0^\\infty\\dfrac{\\sin(x)\\log(x)}{x}\\mathrm dx$.\n\nI tried to expand $\\sin(x)$ at zero, or use SI(SinIntegral) function, but it did not work. Besides, I searched the question on math.stackexchange, nothing found.\n\nMathematica tells me the answer is $-\\dfrac{\\gamma\\pi}{2}$, I have no idea how to get it.\n\nNotice that\n\n\\begin{align*} \\int_{0}^{\\infty} \\frac{\\sin x}{x^s} \\, dx &= \\frac{1}{\\Gamma(s)} \\int_{0}^{\\infty} \\left( \\int_{0}^{\\infty} t^{s-1}e^{-xt} \\, dt \\right) \\sin x \\, dx \\\\ &= \\frac{1}{\\Gamma(s)} \\int_{0}^{\\infty} \\left( \\int_{0}^{\\infty} e^{-tx} \\sin x \\, dx \\right) t^{s-1} \\, dt \\\\ &= \\frac{1}{\\Gamma(s)} \\int_{0}^{\\infty} \\frac{t^{s-1}}{1+t^2} \\, dt \\\\ &= \\frac{1}{2\\Gamma(s)} \\beta\\left(\\frac{s}{2}, 1-\\frac{s}{2}\\right) \\\\ &= \\frac{\\pi}{2\\Gamma(s)\\sin\\left(\\frac{\\pi s}{2}\\right)}. \\end{align*}\n\n(This heuristic computation is valid line-by-line on the strip $1 < \\Re(s) <2$ in view of Fubini's theorem, and then extends to the larger strip $0 < \\Re(s) < 2$ by analytic continuation.) Differentiating both sides, we get\n\n$$\\int_{0}^{\\infty} \\frac{\\sin x \\log x}{x^s} \\, dx \\stackrel{(*)}{=} -\\frac{d}{ds} \\frac{\\pi}{2\\Gamma(s)\\sin\\left(\\frac{\\pi s}{2}\\right)} = \\frac{\\pi}{2\\Gamma(s)\\sin\\left(\\frac{\\pi s}{2}\\right)} \\left( \\psi(s) + \\frac{\\pi}{2}\\cot\\left(\\frac{\\pi s}{2}\\right) \\right).$$\n\nNow the answer follows by plugging $s = 1$:\n\n$$\\int_{0}^{\\infty} \\frac{\\sin x \\log x}{x} \\, dx = -\\frac{\\gamma \\pi}{2}.$$\n\nJustification of $\\text{(*)}$. Let us prove that\n\n$$F(s) := \\int_{0}^{\\infty} \\frac{\\sin x}{x^s} \\, dx \\tag{1}$$\n\nis analytic on the open strip $S = \\{ s \\in \\Bbb{C} : 0 < \\Re(s) < 2 \\}$ and its derivative can be computed by the Leibniz's integral rule. A major issue of $\\text{(1)}$ is that the defining integral converges only conditionally for $0 < \\Re(s) \\leq 1$ and thus raises some technical difficulties. In order to circumvent this, we improve the speed of convergence using integration by parts:\n\n$$F(s) = \\underbrace{\\left[ \\frac{1-\\cos x}{x^s} \\right]_{0}^{\\infty}}_{=0} + s \\int_{0}^{\\infty} \\frac{1-\\cos x}{x^{s+1}} \\, dx.$$\n\nNotice that the resulting integral is absolutely convergent on $S$.\n\nNow we claim that $g(s) := F(s)/s$ is differentiable and its derivative can be computed by the Leibniz's integral rule. Let $\\epsilon$ be such that $\\bar{B}(s, \\epsilon) \\subset S$. Then whenever $0 < |h| < \\epsilon$,\n\n\\begin{align*} &\\frac{g(s+h) - g(s)}{h} - \\int_{0}^{\\infty} \\frac{(1-\\cos x)(-\\log x)}{x^{1+s}} \\, dx \\\\ &\\hspace{9em} = \\int_{0}^{\\infty} \\frac{1-\\cos x}{x^{1+s}} \\left( \\frac{x^{-h} - 1}{h} + \\log x \\right) \\, dx. \\end{align*}\n\nNotice that the integrand is dominated by\n\n$$\\left| \\frac{1-\\cos x}{x^{1+s}} \\left( \\frac{x^{-h} - 1}{h} + \\log x \\right) \\right| \\leq \\frac{1-\\cos x}{x^{1+\\Re(s)}} (\\max\\{ x^{\\epsilon}, x^{-\\epsilon} \\} + 1) |\\log x|$$\n\nThis dominating function is integrable. Hence by the dominated convergence theorem, as $h \\to 0$, we have\n\n$$\\lim_{h \\to 0} \\frac{g(s+h) - g(s)}{h} = \\int_{0}^{\\infty} \\frac{(1-\\cos x)(-\\log x)}{x^{1+s}} \\, dx.$$\n\nPlugging this back, we know that $F(s)$ is differentiable with\n\n$$F'(s) = \\int_{0}^{\\infty} \\frac{1-\\cos x}{x^{s+1}} \\, dx + s \\int_{0}^{\\infty} \\frac{(1-\\cos x)(-\\log x)}{x^{1+s}} \\, dx.$$\n\nFinally, performing integration by part to the latter integral yields the desired conclusion:\n\n\\begin{align*} &s \\int_{0}^{\\infty} \\frac{(1-\\cos x)(-\\log x)}{x^{1+s}} \\, dx \\\\ &\\hspace{5em} = \\underbrace{\\left[ \\frac{(1-\\cos x)\\log x}{x^s} \\right]_{0}^{\\infty}}_{=0} - \\int_{0}^{\\infty} \\left( \\frac{\\sin x \\log x}{x^s} + \\frac{1-\\cos x}{x^{1+s}} \\right) \\, dx \\end{align*}\n\n\u2022 @JackD'Aurizio, Thank you! \u2013\u00a0Sangchul Lee Sep 26 '16 at 13:35\n\u2022 Nothing to thank me for, it is a really well-written answer, credits to you. \u2013\u00a0Jack D'Aurizio Sep 26 '16 at 13:38\n\u2022 +1. Not too many people care to write a justification $^{*}$ as you did it ( included me !!! ). \u2013\u00a0Felix Marin Oct 2 '16 at 8:37\n\n\nHereafter the $\\ds{\\ln}$-branch-cut runs along $\\ds{\\left(-\\infty,0\\right]}$ with $\\ds{\\ln\\pars{z}\\ \\,\\mrm{arg}}$ given by $\\ds{-\\pi < \\mrm{arg}\\pars{z} < \\pi}$.\n\n\\begin{align} \\mc{J} & \\equiv \\Im\\int_{0}^{\\infty}\\ln\\pars{x}\\expo{-\\pars{t - \\ic}x}\\,\\dd x = \\Im\\bracks{{1 \\over t - \\ic}\\int_{0}^{\\pars{t - \\ic}\\infty} \\ln\\pars{x \\over t - \\ic}\\expo{-x}\\,\\dd x} \\\\[5mm] & = -\\,\\Im\\braces{{t + \\ic \\over t^{2} + 1}\\int_{\\infty}^{0}\\bracks{\\ln\\pars{x \\over \\root{t^{2} + 1}} + \\arctan\\pars{1 \\over t}\\ic}\\expo{-x}\\,\\dd x} \\\\[5mm] & = \\Im\\braces{{t + \\ic \\over t^{2} + 1}\\bracks{% \\int_{0}^{\\infty}\\ln\\pars{x}\\expo{-x}\\,\\dd x - {1 \\over 2}\\ln\\pars{t^{2} + 1}+ \\arctan\\pars{1 \\over t}\\ic}} \\end{align}\n\nNote that\n$\\ds{\\int_{0}^{\\infty}\\ln\\pars{x}\\expo{-x}\\,\\dd x = \\left.\\partiald{}{\\mu}\\int_{0}^{\\infty}x^{\\mu}\\expo{-x}\\,\\dd x\\, \\right\\vert_{\\ \\mu\\ =\\ 0} = \\Gamma\\,'\\pars{1}\\ =\\ \\overbrace{\\Gamma\\pars{1}}^{\\ds{=\\ 1}}\\ \\overbrace{\\Psi\\pars{1}}^{\\ds{-\\gamma}}\\ =\\ -\\gamma\\quad}$ where $\\ds{\\Gamma}$\n\nand $\\ds{\\Psi}$ are the Gamma and Digamma Functions, respectively. Then,\n\n\\begin{align} \\mc{J} & \\equiv \\bbox[8px,#ffe,border:0.1em groove navy]{% \\Im\\int_{0}^{\\infty}\\ln\\pars{x}\\expo{-\\pars{t - \\ic}x}\\,\\dd x} = -\\,{\\gamma \\over t^{2} + 1} - {1 \\over 2}\\,{\\ln\\pars{t^{2} + 1} \\over t^{2} + 1} + {t\\arctan\\pars{1/t} \\over t^{2} + 1} \\\\[5mm] & = \\bbox[8px,#ffe,border:0.1em groove navy]{-\\,{\\gamma \\over t^{2} + 1} + \\totald{}{t}\\bracks{{1 \\over 2}\\,\\arctan\\pars{1 \\over t}\\ln\\pars{t^{2} + 1}}} \\label{2}\\tag{2} \\end{align}\n\nNote that$\\ds{{1 \\over 2}\\,\\arctan\\pars{1 \\over t}\\ln\\pars{t^{2} + 1}}$ vanishes out when $\\ds{t \\to \\infty}$ and $\\ds{t \\to 0^{+}}$.\n\nBy replacing \\eqref{2} in \\eqref{1}: \\begin{align} &\\color{#f00}{\\int_{0}^{\\infty}{\\sin\\pars{x}\\ln\\pars{x} \\over x}\\,\\dd x} = -\\gamma\\int_{0}^{\\infty}{\\dd t \\over t^{2} + 1} = \\color{#f00}{-\\,{1 \\over 2}\\,\\gamma\\pi} \\end{align}\n\n\u2022 (+1) That is how I just did this. Then I looked on MSE to see if it's already been discusses and found this page. Well done my friend! \u2013\u00a0Mark Viola Apr 23 at 3:36", "date": "2021-05-16 04:01:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1941888/calculate-int-0-infty-frac-sinx-logxx-mathrm-dx", "openwebmath_score": 1.000008463859558, "openwebmath_perplexity": 1416.1116553532966, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811571768048, "lm_q2_score": 0.8824278602705731, "lm_q1q2_score": 0.8518792288730577}}
{"url": "https://math.stackexchange.com/questions/1731070/all-real-numbers-in-0-2-can-be-represented-as-sqrt2-pm-sqrt2-pm-sqrt", "text": "# All real numbers in $[0,2]$ can be represented as $\\sqrt{2 \\pm \\sqrt{2 \\pm \\sqrt{2 \\pm \\dots}}}$\n\nI would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$.\n\nI'll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted by brackets.\n\n$$2=\\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\dots}}}=(+)$$\n\n$$1=\\sqrt{2 - \\sqrt{2 - \\sqrt{2 - \\dots}}}=(-)$$\n\n$$0=\\sqrt{2 - \\sqrt{2 + \\sqrt{2 + \\dots}}}=-(+)$$\n\n$$\\phi=\\sqrt{2 + \\sqrt{2 - \\sqrt{2 + \\sqrt{2 - \\dots}}}}=(+-)$$\n\n$$\\frac{1}{\\phi}=\\sqrt{2 - \\sqrt{2 + \\sqrt{2 - \\sqrt{2 + \\dots}}}}=(-+)$$\n\nIn general, the expansion can be found by a very easy algorithm:\n\n\u2022 take any number in $(0,2)$, square it\n\u2022 if the result $>2$ write $+$, if the result $<2$ write $-$\n\u2022 subtract $2$ from the result, square, repeat\n\nIf on some step we get $2$ exactly, we just write $(+)$ and the expansion is finished.\n\nExamples:\n\n$$\\pi-2=--+-++-+-+++++++-+-+---------+-+--+--+--+++---++++ \\dots=1.141592653589793 \\dots$$\n\nBasically, $50$ terms of our expansion gave only $15$ correct decimal digits for $\\pi$. But considering the expansion can be coded as binary, it's not so bad.\n\nThe convergence plot, and two binary plots for this $50$ terms can be seen below:\n\n$$e-1=+-----+++-++-+---++-++++-+---++-+++-++++-++++---++ \\dots=1.71828182845905 \\dots$$\n\nDo you know any reference about this expansion? Can every real number between $0$ and $2$ be expanded this way?\n\nIs number $2$ special in this case, or can we make a similar expansion using some other number (and other power for the root)?\n\nEdit\n\nNow that I think about it, we can use the general expansion for $x \\in [0,a]$:\n\n$$x=\\left(a \\pm \\left(a \\pm \\left(a \\pm \\dots \\right)^p \\right)^p \\right)^p$$\n\n$$a=2^{\\frac{p}{1-p}}$$\n\nFor example:\n\n$$\\frac{1}{4}=\\left(\\frac{1}{4} + \\left(\\frac{1}{4} + \\left(\\frac{1}{4} + \\dots \\right)^2 \\right)^2 \\right)^2$$\n\n$$\\frac{3}{4}-\\frac{\\sqrt{2}}{2}=\\left(\\frac{1}{4} - \\left(\\frac{1}{4} - \\left(\\frac{1}{4} - \\dots \\right)^2 \\right)^2 \\right)^2$$\n\netc.\n\nHowever, this case $p=2,~~~a=\\frac{1}{4}$ is not just a random example, it's the only rational expansion of this kind. So I would say it's more important than the titular root expansion.\n\nEdit\n\nAn interesting article that connects the nested roots of this kind to Chebyshev polynomials: http://www.sciencedirect.com/science/article/pii/S0022247X12003344\n\n\u2022 Wow. Great question, +1 \u2013\u00a0user223391 Apr 6 '16 at 21:33\n\u2022 Beautiful idea. One could make a binary-like number system based on it. Let a \"bit\" code 0 for minus or 1 for plus. I wonder if it could have unique representation contrary to the usual way to write numbers, for example $0.1b = 0.011\\dots b = 1/2$. \u2013\u00a0mathreadler Apr 7 '16 at 8:01\n\u2022 I wonder if there is some class of numbers which have periodic \"number expansions\" in that number system. Like the rationals have for ordinary number systems using the division algorithm. \u2013\u00a0mathreadler Apr 7 '16 at 8:27\n\u2022 @mathreadler, that would be a subset of algebraic numbers \u2013\u00a0Yuriy S Apr 7 '16 at 8:30\n\u2022 the $50$ bits gives $50/\\log_{10}(2) \\approx 15$ decimals which is just one or two bits short of mantissa of a ordinary double precision floating point number system. \u2013\u00a0mathreadler Apr 7 '16 at 15:36\n\nHere is a possible explanation. Let $\\alpha \\in [0, \\pi/2]$ and define $\\epsilon_1, \\epsilon_2, \\cdots$ by $\\epsilon_i = \\operatorname{sgn}( \\cos ( 2^i \\alpha )) \\in \\{-1, 1\\}$. Here, we take the convention that $\\operatorname{sgn}(0) =1$. Then applying the identity $2\\cos\\theta = \\operatorname{sgn}(\\cos\\theta) \\sqrt{2 + 2\\cos(2\\theta)}$ repeatedly, we have\n\n$$2\\cos \\alpha = \\sqrt{2 + \\epsilon_1 \\sqrt{2 + \\epsilon_2 \\sqrt{ \\cdots + \\epsilon_n \\sqrt{2 + \\smash[b]{2\\cos(2^{n+1} \\alpha)} }}}}.$$\n\nThis can be used to show that, with an appropriate definition of infinite nested radical, the following identity\n\n$$2\\cos \\alpha = \\sqrt{2 + \\epsilon_1 \\sqrt{2 + \\epsilon_2 \\sqrt{ 2 + \\cdots }}}$$\n\nis true. This shows that any real number between $[0, 2]$ can be written as an infinite nested radical of the desired form. Moreover, if we denote $x = 2\\cos\\alpha$, then\n\n\u2022 $\\epsilon_1 = \\operatorname{sgn}(2\\cos (2\\alpha)) = \\operatorname{sgn}(x^2 - 2)$,\n\u2022 $\\epsilon_2 = \\operatorname{sgn}(2\\cos (4\\alpha)) = \\operatorname{sgn}((x^2 - 2)^2 - 2)$,\n\nand likewise. This explains why signs are determined by OP's algorithm.\n\n\u2022 Wow, this is a great explanation, which also shows where $2$ comes from \u2013\u00a0Yuriy S Apr 6 '16 at 22:47\n\u2022 @YuriyS, Thank you! And yes, this explains why the range is $[0, 2]$. Unfortunately this seems not work for other cases with $a > 2$ instead of $2$, and I even suspect that the analogous claim may be false. (For an an analogous situation, we can in fact check this. Numbers of the form $\\pm \\frac{1}{2} \\pm \\frac{1}{4} \\pm \\frac{1}{8} \\pm \\cdots$ represents any real number in $[-1, 1]$, but the range of $\\pm \\frac{1}{3} \\pm \\frac{1}{9} \\pm \\frac{1}{27} \\pm \\cdots$ is the (shifted) Cantor set.) \u2013\u00a0Sangchul Lee Apr 6 '16 at 22:59\n\u2022 Nice answer (+1). This is theme of problems 183 - 185 in the classic Problems and Theorems in Analysis I by G. Polya and G. Szeg\u00f6 \u2013\u00a0Markus Scheuer Apr 9 '16 at 16:02\n\u2022 This is incredible. \u2013\u00a06005 Jul 26 '16 at 21:08\n\nPeculiar observation\n\nIf we define a binary number $b = b_1b_2\\cdots b_n$ with digits mapped to the symbols like this: $$b_k = \\begin{cases}0 \\text{ if } (-) \\text{ at position } k\\\\1 \\text{ if } (+) \\text{ at position } k\\end{cases}$$ Then if we run the algorithm proposed in the question, looping\n\nx(k) = x(k-1)^2-2;\nb(k) = (x(k)>0);\n\n\nthe vector b will get logical values corresponding to bits 1 and 0 of the binary number above and we can calculate it for the linear space of $x\\in[0,1]$. If we do this we can then calculate each number as the scalar product $$[1/2,1/4,\\cdots,1/2^k]b$$ and if we then plot it, it will look like\n\nWhich is kind of a peculiar plot having a bit of a discontinuous and fractal structure. I think the largest discontinuity is around $x = \\sqrt{1/2}$ but I have no theoretical explanation why..\n\nedit as pointed out by Sangchul Lee this seems similar to Tent Map\n\n\u2022 Good observation. I am not expert of symbolic dynamics, but this is related to the tent map orbits. Basically this is because the string of signs are produced by the doubling transform on the circle via the correspondance I explained in my answer. \u2013\u00a0Sangchul Lee Apr 8 '16 at 2:58\n\nYour algorithm pretty much shows that there exists an expansion for every number in $[0,2]$.\n\nIf we replace $2$ by $a>2$ (and keep the square root), we will fail because we need that squaring a number from the interval $[u,v]$ produces a number that is either in $a+[u,v]$ or in $a-[u,v]$. So we must have $u=0$ and $v\\ge a$ and $v^2\\le a+v$. The last two imply $a\\le v\\le 2$.\n\nIf we additionally switch to $k$th roots, the condition becomes that $v\\ge a$ and $v^k\\le a+v$, hence $a\\le v\\le\\sqrt[k-1]2$.", "date": "2020-06-02 09:10:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1731070/all-real-numbers-in-0-2-can-be-represented-as-sqrt2-pm-sqrt2-pm-sqrt", "openwebmath_score": 0.8761911988258362, "openwebmath_perplexity": 259.36929156656936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754497285468, "lm_q2_score": 0.8652240947405564, "lm_q1q2_score": 0.8518784021951581}}
{"url": "https://math.stackexchange.com/questions/2737395/an-example-of-an-algebra-but-not-a-sigma-algebra", "text": "# An example of an algebra but not a sigma algebra?\n\nCan someone give me an example of an algebra but not a sigma algebra, with a $\\sigma$-finite measure $\\mu$ on it?\n\n\u2022 As far as I'm aware, $\\sigma$-finiteness is only defined on sigma algebras - how are you defining it here? \u2013\u00a0B. Mehta Apr 14 '18 at 22:43\n\u2022 @B.Mehta I understood this to mean a measure on the sigma-algebra generated by the algebra. \u2013\u00a0Severin Schraven Apr 14 '18 at 22:44\n\nLet $X=[0, \\infty)$ and consider $\\mathcal{A}=\\{ \\text{finite union of }[a, b)\\}$.\n\nIt's clear that $[a, b)\\cap [c, d) = [\\max\\{a, c\\}, \\min\\{b, d\\})$ and $[a, b)^C = [0, a)\\cup [b, \\infty)$.\n\nSo $\\mathcal{A}$ is an algebra but clearly not a $\\sigma$-algebra since it doesn't contain any open interval.\n\nTake the obvious measure $m([a, b)) = b-a$.\n\n\u2022 +1 Neat example. \u2013\u00a0Severin Schraven Apr 14 '18 at 22:58\n\u2022 Thanks Jacky that\u2019s very understandable! :D \u2013\u00a0JINGYA HAN Apr 14 '18 at 23:35\n\nTake all subsets of $\\mathbb{N}$ which are either finite or have finite complement. It is an algebra, but not a sigma-algebra. For the sigma-finite measure you can pick the zero measure.\n\n\u2022 Or for the measure pick counting measure. \u2013\u00a0GEdgar Apr 14 '18 at 23:01\n\u2022 Thanks for your example. In this context I don\u2019t think the measure could be replaced by a counting measure since that would not always be finite. But say if we replace the measure by a counting measure, does it mean this measure does not have a unique extension on the whole natural number set, since it would meet the requirement of using Carath\u00e9odory Extension Theorem? \u2013\u00a0JINGYA HAN Apr 14 '18 at 23:34\n\u2022 @JINGYAHAN We could replace the measure with the counting measure, it would still be sigma-finite as we can write $$\\mathbb{N}=\\bigcup_{n\\in \\mathbb{N}} \\{n\\}$$ and the singeltons have counting measure equal to $1$. \u2013\u00a0Severin Schraven Apr 15 '18 at 0:26\n\u2022 @SeverinSchraven oh yes you\u2019re right. I confused the definition. The point is you cannot assign a singleton with an infinity measure. Thanks for your help! \u2013\u00a0JINGYA HAN Apr 15 '18 at 0:31\n\u2022 @JINGYAHAN Exactly, we don't want concentration of mass on singeltons. By the way you should accept Jacky's answer (on the left you of the question you can find the respective button), unless you are not yet satisfied with the answers given and want to wait for another one. \u2013\u00a0Severin Schraven Apr 15 '18 at 0:36", "date": "2020-06-06 14:16:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2737395/an-example-of-an-algebra-but-not-a-sigma-algebra", "openwebmath_score": 0.848964273929596, "openwebmath_perplexity": 437.5892221987476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754492759499, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.8518783966710334}}
{"url": "http://www.senorcafe.com/bjs2dg1i/a1702f-determine-the-coordinates-of-the-centroid-of-the-area", "text": "It is the point which corresponds to the mean position of all the points in a figure. Centroid of an Area \u2022 In the case of a homogeneous plate of uniform thickness, the magnitude \u2206W is With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Center of Mass of a Body Center of mass is a function of density. So to find the centroid of an entire beam section area, it first needs to be split into appropriate segments. Determine the x - and y -coordinates of the centroid of the shaded area. The centroid of a right triangle is 1/3 from the bottom and the right angle. The Find Centroids tool will create point features that represent the geometric center (centroid) for multipoint, line, and area features.. Workflow diagram Examples. y=2 x, y=0, x=2 Find the coordinates of the centroid of the area bounded by the given curves. An analyst at the Scotland Department of Environment is performing a preliminary review on wind farm applications to determine which ones overlap with or are in view of wild lands. How to calculate a centroid. Beam sections are usually made up of one or more shapes. Centroid by Composite Bodies ! y=x^{3}, x=0, y=-8 The cartesian coordinate of it's centroid is $\\left(\\frac{2}{3}r(\\theta)\\cos\\theta, \\frac{2}{3}r(\\theta)\\sin\\theta\\right)$. Solution: First, gather the coordinate points of the vertices. The coordinates of the centroid are simply the average of the coordinates of the vertices.So to find the x coordinate of the orthocenter, add up the three vertex x coordinates and divide by three. For more see Centroid of a triangle. Recall that the centroid of a triangle is the point where the triangle's three medians intersect. It is also the center of gravity of the triangle. Next, sum all of the x coodinates ... how to find centroid of composite area: how to calculate centroid of rectangle: how to find centroid of equilateral triangle: 4' 13 Answers: (X,Y) in The centroid or center of mass of beam sections is useful for beam analysis when the moment of inertia is required for calculations such as shear/bending stress and deflection. Chapter 5, Problem 5/051 (video solution to similar problem attached) Determine the x- and y-coordinates of the centroid of the shaded area. Find the coordinates of the centroid of the plane area bounded by the parabola y = 4 \u2013 x^2 and the x-axis. Centroid of a Volume The centroid defines the geometric center of an object. We can consider the surface element as an triangle, and the centroid of this triangle is obviously at here.) The center of mass is the term for 3-dimensional shapes. Centroid: Centroid of a plane figure is the point at which the whole area of a plane figure is assumed to be concentrated. The centroid is the term for 2-dimensional shapes. For instance, the centroid of a circle and a rectangle is at the middle. For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ! \u2022 The coordinates ( and ) define the center of gravity of the plate (or of the rigid body). Determine the coordinates of the centroid of the shaded region. Problem Answer: The coordinates of the center of the plane area bounded by the parabola and x-axis is at (0, 1.6). The x-centroid would be located at 0 and the y-centroid would be located at 4 3 r \u03c0 7 Centroids by Composite Areas Monday, November 12, 2012 Centroid by Composite Bodies Find the coordinates of the centroid of the area bounded by the given curves. 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In a figure is the point which corresponds to the mean position all.", "date": "2021-10-21 15:10:22", "meta": {"domain": "senorcafe.com", "url": "http://www.senorcafe.com/bjs2dg1i/a1702f-determine-the-coordinates-of-the-centroid-of-the-area", "openwebmath_score": 0.7315730452537537, "openwebmath_perplexity": 345.616869691963, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9845754447499796, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8518783944658972}}
{"url": "http://math.stackexchange.com/questions/365010/writing-a-series-using-sigma-notation/365021", "text": "# Writing a series using Sigma notation\n\nHow do I write $$2+ \\frac{3x}{2} + \\frac{4x^2}{4}+\\frac{5x^3}{8}+\\frac{6x^4}{16}?$$\n\nI have been struggling with these types of problems, so please, an explanation of how to get the result will be appreciated.\n\n-\n\n$$2+ \\frac{3x}{2} + \\frac{4x^2}{4}+\\frac{5x^3}{8}+\\frac{6x^4}{16}$$\n\nSince everything else has a denominator, let's try putting a denominator for $2$. Since anything divided by one itself, we get:\n\n$$\\frac{2}{1}+ \\frac{3x}{2} + \\frac{4x^2}{4}+\\frac{5x^3}{8}+\\frac{6x^4}{16}$$\n\nEverything else has an $x$. Let's give $2$ an $x$ too! $x^0 = 1$, and anything multiplied by $1$ is itself, so:\n\n$$\\frac{2x^0}{1}+ \\frac{3x}{2} + \\frac{4x^2}{4}+\\frac{5x^3}{8}+\\frac{6x^4}{16}$$\n\nAll the other $x$s have an exponent. Let's give one to that $x$ next to the $3$. $x^1 = x$, so we get:\n\n$$\\frac{2x^0}{1}+ \\frac{3x^1}{2} + \\frac{4x^2}{4}+\\frac{5x^3}{8}+\\frac{6x^4}{16}$$\n\nHowever, these denominators are just powers of two. Therefore, we have:\n\n$$\\frac{2x^0}{2^0}+ \\frac{3x^1}{2^1} + \\frac{4x^2}{2^2}+\\frac{5x^3}{2^3}+\\frac{6x^4}{2^4}$$\n\nNow we see some patterns:\n\n\u2022 $2, 3, 4, 5, 6$ (numerator of coefficient of $x$)\n\u2022 $0, 1, 2, 3, 4$ (exponents of $x$)\n\u2022 $1, 2, 4, 8, 16$ (denominator of coefficient of $x$. Powers of $2$, viz. $2^0, 2^1, 2^2, 2^3, 2^4$)\n\nWe'll make the sum range from $0$ to $4$, although in theory we could actually make it range over anything.\n\n$$\\sum_{k=0}^4$$\n\nWe need exponents of $x$, so we'll put those in there.\n\n$$\\sum_{k=0}^4 x^k$$\n\nWe need those powers of two in the denominator, so add that too:\n\n$$\\sum_{k=0}^4 \\frac{x^k}{2^k}$$\n\nHowever, there are still those numerators we haven't checked off our list yet. We can't have them ranging from $0$ to $4$, we need them from $2$ to $6$. The solution is to add two!\n\n$$\\sum_{k=0}^4 \\frac{(k+2)x^k}{2^k}$$\n\n-\nI'd have gone on to write $\\dfrac{2x^0}{1}+ \\dfrac{3x^1}{2} + \\dfrac{4x^2}{4}+\\dfrac{5x^3}{8}+\\dfrac{6x^4}{16}$ in the form $\\dfrac{2x^0}{2^0}+ \\dfrac{3x^1}{2^2} + \\dfrac{4x^2}{2^2}+\\dfrac{5x^3}{2^3}+\\dfrac{6x^4}{2^4}$. \u2013\u00a0 Michael Hardy Apr 17 '13 at 23:34\n@MichaelHardy, good idea, I added it. \u2013\u00a0 George V. Williams Apr 18 '13 at 0:05\n\nIt's much like finding a pattern, and a way to express the pattern. Starting point: $2 = \\dfrac 21$. Numerator coeffienct of the $x$-term, increases by one per term , starting at $2$ (add 1 to starting numerator); the exponent for $x$ increases by one (starting, say, at $x^0,...x^1 = x,...x^2...$ (Add one to the x-exponent starting at $0$. Also, the denominator represents powers of $2$, starting at $2^0 = 1$...Increase (add) by one the exponent of the term $2$ in the denominator $2^0, ... 2^1 = 2, ...2^{1+1} = 2^2 = 4, ...$\n\nTry writing out the first few terms of the following and see if it matches, starting index $0$, and on...:\n\n$$\\sum_{k = 0}^\\infty \\frac{(k+2)x^k}{2^k}$$\n\nIf things don't work out (here they happen to work just fine), there is often a little trial-and-error involved. For example, we can \"tweak\" if we want to represent the series with a starting index $1$, by \"subtracting $1$ from each of the expressions given in terms of $k$.\n\nNOTE: The above will give you an infinite series. To get the first five terms, we start at $k = 0$, and need to sum through (inclusive) $k = 4$. But you can obtain any positive number $n$ of terms by summing from $k = 0$ to $k = n-1)$\n\n-\nThe best way to get a \"knack\" for \"cracking\" a sequence is through practice. Try to approach sequences of numbers like you'd approach a puzzle: what fits? what doesn't fit? What patterns do I see...start writing down what you do see, and try creating a \"model\" term (a \"work in progress,\" so to speak)...Figure out if starting with $0$ or $1$ for one of the varying terms works best. If a term appears constant through the sequence, represent it as a constant, etc. \u2013\u00a0 amWhy Apr 17 '13 at 23:47\nSuch great advice! +1 \u2013\u00a0 Amzoti Apr 18 '13 at 0:24\nSuch a great answer +1 \u2013\u00a0 Babak S. Aug 21 '13 at 6:14\n\nNote that the power of $x$ increase in steps of $1$ starting from $x^0$. Hence the $(n+1)^{th}$ term has $x$ raised to the power $n$. The denominators are powers of $2$ starting from $2^0$ and the numbers in the numerator also increase in steps of $1$ starting from $2$. $$\\sum_{n=0}^4 \\dfrac{(n+2)x^n}{2^n}$$\n\n-\n\nNotice that $2=\\frac {2x^0}1$, so you want $$2+ \\frac{3x}{2} + \\frac{4x^2}{4}+\\frac{5x^3}{8}+\\frac{6x^4}{16}=\\frac {2x^0}1+ \\frac{3x}{2} + \\frac{4x^2}{4}+\\frac{5x^3}{8}+\\frac{6x^4}{16}=\\sum_{i=1}^5\\frac {(i+1)x^{i-1}}{2^{i-1}}$$\n\n-\n\nStart off by looking for any patterns. I would start off by looking at how the powers on $x$ change. Starting at $0$ ($x^0=1$) it looks like the power goes up by one at each \"step.\" So we can start off with the \"skeleton\" sum $$\\sum_{i=0}^4 x^i.$$ Similarly observe the coefficient patters, it looks like the numerator starts at $2$ and goes up by one at each step and the denominator are powers of $2$ starting at the $2^0$. So putting that all together we get $$\\sum_{i=0}^4 \\frac{(i+2)x^i}{2^i}$$\n\n-", "date": "2014-11-28 17:22:25", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/365010/writing-a-series-using-sigma-notation/365021", "openwebmath_score": 0.902320384979248, "openwebmath_perplexity": 473.7786485387081, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754452025766, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.851878389724968}}
{"url": "https://stats.stackexchange.com/questions/135878/is-there-a-family-of-processes-centred-on-the-poisson-process", "text": "# Is there a family of processes centred on the Poisson process?\n\nI am looking for a model, characterized continuously by a single parameter, to describe the arrival times of buses with unit expected interarrival time. At one extreme of the parameter (say $\\theta=1$), the process is deterministic, with all interarrival times equal (to $1$). When $\\theta=0$, the process is pure Poisson (with unit expected interarrival or waiting time). As $\\theta$ approaches $-1$, the arrivals tend to cluster, with long intervals between the clusters. At the (unattainable) extreme of $\\theta=-1$, all the buses arrive in a simultaneous convoy, after an infinite wait, and you have to wait forever for the next convoy.\n\nThe choice of $\\{1,0,-1\\}$ for the extreme and central parameter values isn't important: they could be $\\{-\\infty, 0, \\infty\\}$, $\\{0,\\frac12,1\\}$, ..., whatever. At this stage, simplicity and naturalness, subject to the above conditions, is more important than realism.\n\nIn a pure Poisson process, the interarrival times (times between each bus) has an exponential distribution. So you want some family of distributions which includes the exponential as a special case. The gamma distribution (see: https://en.wikipedia.org/wiki/Gamma_distribution). When then shape parameter is 1, you have the exponential. When the shape parameter ($k$ or $\\alpha$ in wiki) approaches zero, you are approaching constant interarrival times, and growing above 1 you get interarrival times more variable than for the poisson process.\n\u2022 Yes, this could be the model if we set $k\\theta=1$. But I think that you have it the wrong way round: We approach a constant (unit) interarrival time as $k\\rightarrow\\infty$, and long intervals between clusters as $k\\rightarrow0$. \u2013\u00a0John Bentin Feb 12 '15 at 16:06", "date": "2019-10-14 20:57:22", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/135878/is-there-a-family-of-processes-centred-on-the-poisson-process", "openwebmath_score": 0.8897665143013, "openwebmath_perplexity": 489.94263812041544, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9845754461077707, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8518783870864789}}
{"url": "https://mathhelpboards.com/threads/number-of-real-roots-of-a-quartic.887/", "text": "# Number of real roots of a quartic\n\n#### CaptainBlack\n\n##### Well-known member\nrayman's question from another place:\n\nCould someone help me with this problem, I have no idea how to start with it\n\nHow many real roots does this polynomial have p(x)=x^4-x^3-1?\n\nClearly state the argument that explains the number of real roots.\n\nThank you for any help\n\nDescartes rule of signs tells you this has exactly 1 positive root, and exactly 1 negative root, so it has two real roots.\n\nCB\n\nLast edited:\n\n#### chisigma\n\n##### Well-known member\nThe Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\\frac{3}{4}$ and here is $q(x)=- \\frac{27}{256}$. Now if we consider the quartic equation $\\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\\frac{27}{256}$, one real root of order 2 for $a=\\frac{27}{256}$ and no real roots for $a>\\frac{27}{256}$...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### CaptainBlack\n\n##### Well-known member\nThe Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\\frac{3}{4}$ and here is $q(x)=- \\frac{27}{256}$. Now if we consider the quartic equation $\\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\\frac{27}{256}$, one real root of order 2 for $a=\\frac{27}{256}$ and no real roots for $a>\\frac{27}{256}$...\n\nKind regards\n\n$\\chi$ $\\sigma$\nIn this case Descartes rule of signs does tell us exactly how many real roots we have.\n\nThe number of positive roots is equal to the number of changes of signs of the coefficients less a multiple of 2. In this case the number of sign changes is 1, and as there is no multiple of 2 other than 0 which leaves the number of roots non-negative there is exactly one positive real root. The same argument applies to the negative roots.\n\nCB", "date": "2021-06-16 14:09:10", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/number-of-real-roots-of-a-quartic.887/", "openwebmath_score": 0.839435338973999, "openwebmath_perplexity": 140.4456067566802, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9845754465603678, "lm_q2_score": 0.8652240756264638, "lm_q1q2_score": 0.851878380634707}}
{"url": "https://mathematica.stackexchange.com/questions/222259/why-doesnt-integrate-evaluate-an-elliptic-integral", "text": "# Why doesn't Integrate evaluate an elliptic integral?\n\nMy code is\n\nIntegrate[ 1/Sqrt[(x - a) (x - b) (x - c) (x - d)], {x, a,  \u221e},\nAssumptions -> 0 < d < c < b < a]\n\n\nI know this can be expressed be the incomplete elliptic integral of first kind (EllipticF), but the output remains unevaluated\n\nIntegrate[ 1/Sqrt[(-a + x) (-b + x) (-c + x) (-d + x)], {x, a,  \u221e},\nAssumptions -> 0 < d < c < b < a]\n\n\nWhy does this happen? I am desperate\n\n\u2022 If the output is echoed, it means Mathematica doesn't know what to do with it. \u2013\u00a0J. M.'s torpor May 20 '20 at 12:02\n\u2022 Seems to work OK for the indefinite integral. And the result can be evaluated at a and Infinity using the Limit function \u2013\u00a0Gustavo Delfino May 20 '20 at 12:09\n\u2022 But isn't it strange Mathematica can't calculate it? I also tried changing the integration from a to a generic M but the result is the same. From Gradshteyn Ryzhik I know this is equal to EllipticF(...,...), should I doubt the tables of Gradshteyn Ryzhik or is it common that Mathematica can't do integrals like this one? \u2013\u00a0Filippo Caleca May 20 '20 at 12:11\n\nActually V 12.1 can do it directly, you just have to wait a little bit long:\n\nClear[\"Global*\"];\n\nint = Integrate[1/Sqrt[(x - a) (x - b) (x - c) (x - d)], {x, a, Infinity},\nAssumptions -> 0 < d < c < b < a]\n\n\nMay be OP used different Mathematica version? It will be good to post which version was used. Screen shot below:\n\nA Workaround for now: (assuming proper integral)\n\n int = Integrate[1/Sqrt[(x - a) (x - b) (x - c) (x - d)], x]\n\n\nlow = Assuming[0 < d < c < b < a, Limit[int, x -> a]]\n(* 0 *)\n\nhigh = Assuming[0 < d < c < b < a, Limit[int, x -> Infinity]]\n\n\nThe above is then the final result.\n\n\u2022 Thank you! I am using 11.3 \u2013\u00a0Filippo Caleca May 20 '20 at 12:52\n\u2022 @FilippoCaleca integrate in V12.1 has improved over 11.3 \u2013\u00a0Nasser May 20 '20 at 12:56\n\u2022 my output is different, in fact writing: int = Integrate[1/Sqrt[(x - a) (x - b) (x - c) (x - d)], x] the output is -((2 (-a + x) (-b + x) Sqrt[((a - b) (-c + x))/((b - c) (a - x))] Sqrt[((a - b) (-d + x))/((b - d) (a - x))] EllipticF[ ArcSin[Sqrt[((a - d) (-b + x))/((b - d) (-a + x))]], ((a - c) (b - d))/((b - c) (a - d))])/((a - b) Sqrt[((a - d) (b - x))/((b - d) (a - x))] Sqrt[(-a + x) (-b + x) (-c + x) (-d + x)])) the argument of EllipticF are different, how is it? \u2013\u00a0Filippo Caleca May 20 '20 at 13:09\n\u2022 Note that you shouldn't expect the technique used in your workaround to work in all cases. See this old Wolfram blog post for some of the mathematical nitty-gritty as to why. \u2013\u00a0Michael Seifert May 20 '20 at 19:58\n\u2022 Sure, if you go back to the concept of \"area under a curve\", then one might expect that an antiderivative ought to be continuous if the original function is continuous. But yes, it's not always easy to do. That's why Mathematica has to try harder for definite integrals, tho it has been known to still make mistakes sometimes. \u2013\u00a0J. M.'s torpor May 21 '20 at 1:41\n\nIn the newest version (i.e. 12.1) this integral evaluates a bit long, however changing the variable $$x \\mapsto t = x-a\\;$$ this can be evaluated a few times faster.\n\nint2 = Integrate[ 1/Sqrt[t (t + a - b) (t + a - c) (t + a - d)], {t, 0, \u221e},\nAssumptions -> 0 < d < c < b < a]\n\n2 EllipticF[ ArcSin[ Sqrt[(b - d)/(a - d)]],\n((b - c)(a - d))/((a - c)(b - d))]/Sqrt[(a - c)(b - d)]\n\nTraditionalForm[%]\n\n\nI'm working with the system in cloud and sometimes it appears that the integral in question may remain unevaluated while int2 evaluates well even in version 11.2 on my machine.\n\nMathematica functions evolve with time even if its usage remains tha same. This aspect of the system is perhaps the most obvious in case of symbolic integration (Integrate), exact solutions of differential equations (DSolve) and the special functions (among them EllipticF). Elliptic functions and integrals appeared in Mathematica 1, however since then many new related functionalities would have been added later e.g. EllipticF was introduced in version 1.0 year 1988 and updated in 3.0 (1996). WeierstrassP was introduced in version 1.0 and updated in 3.0 (1996), however several new functionalities related appeared in version 11.2 (2017) like e.g. WeierstrassHalfPeriodW1 or WeierstrassE1 see e.g. this answer Integrate yields complex value, while after variable transformation the result is real. Bug?. Inspecting another answers therein one can see how Integrate can be sensitive when new functions or functionalities appear. It relates not only to new functionalities but also to widening domain of existing functions (in documentation pages one finds information when a function was introduced and when it was last updated, nevertheless there are also hiden changes that are not reported, however they may be crucial when certain different functions involved were updated). One should take attention to this aspect related to better handling of symbolic input of e.g. WeierstrssHalfPeriodW1 in version 12.1 with respect to 11.2 and it is advantageous to pay atention to this post. Elliptic functions and integrals play very important role in mathematics, physics, engineering and they are still better handled in newer versions of the system. This does not mean that Mathematica is defective but rather that perfect handling of special functions can be approached asymptotically and it is still in interest of developers of the system, e.g. one of leading experts in the field of special functions Oleg Marichev is a member of the special functions group in WRI. Having said that we can accept the state of art and the fact that things can change at least on the symbolic level.\n\nLet's come back to version 11.2 with help of a simple change of variables: $$x \\mapsto t+a$$\n\nint3 = Integrate[ 1/Sqrt[t (t + a - b) (t + a - c) (t + a - d)], {t, 0, \u221e},\nAssumptions -> 0 < d < c < b < a]\n\n(2 (EllipticF[ ArcSin[Sqrt[(a - d)/(b - d)]], ((a - c) (b - d))/((b - c) (a - d))]\n+ I EllipticK[((a - b) (c - d))/((-b + c) (a - d))]))/Sqrt[(b - c) (a - d)]\n\nTraditionalForm[%]\n\n\nThis might seem strange that there appears an imaginary number however the full integral is indeed real even though FullSimplify cannot demonstrate (in 11.2) that both results are equal. In 12.1 this still cannot be done, although a simpler identity can be proved, assuming that parameters are related somehow (in 12.1 not in 11.2), e.g.\n\nFullSimplify[(8(EllipticF[ArcSin[Sqrt[3/2]], 4/3] +\nI EllipticK[-(1/3)]))/(Sqrt[3] Sqrt[a^2])\n- (4 EllipticF[ArcSin[Sqrt[2/3]], 3/4])/Sqrt[a^2], a > 0]\n\n\n0\n\nWe can show that this is the case evaluating numerically, e.g.\n\nWith[{a = 4, b = 3, c = 2, d = 1}, {\n(2 (EllipticF[ ArcSin[Sqrt[(a - d)/(b - d)]], ((a - c) (b - d))/((b - c) (a - d))]\n+ I EllipticK[((a - b) (c - d))/((-b + c) (a - d))]))/Sqrt[(b - c) (a - d)],\n( 2 (EllipticF[ ArcSin[Sqrt[(b - d)/(a - d)]],\n((b - c)(a - d))/((a - c)(b - d))]))/Sqrt[(a - c)(b - d)]} // N // Chop]\n\n{1.07826, 1.07826}\n\n\nFor example of a bit more testy case see e.g. Why does Integrate declare a convergent integral divergent?\n\nMaking appropriate plot of the functions and their difference might be helpful as well:\n\nPlot[{#, # - (4 EllipticF[ArcSin[Sqrt[2/3]], 3/4])/Sqrt[a^2]}, {a, 0, 6},\nPlotStyle -> Thick, AxesOrigin -> {0, 0}] &[ (\n8(EllipticF[ArcSin[Sqrt[3/2]], 4/3] + I EllipticK[-1/3]))/(Sqrt[3]Sqrt[a^2])]\n`", "date": "2021-06-21 19:32:20", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/222259/why-doesnt-integrate-evaluate-an-elliptic-integral", "openwebmath_score": 0.49721527099609375, "openwebmath_perplexity": 1749.4345224913804, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9489172615983309, "lm_q2_score": 0.8976953030553433, "lm_q1q2_score": 0.8518385687249601}}
{"url": "https://gmatclub.com/forum/the-positive-integers-x-y-and-z-are-such-that-x-is-a-85437.html?sort_by_oldest=true", "text": "It is currently 17 Oct 2017, 12:13\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# The positive integers x, y, and z are such that x is a\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 19 Jun 2009\nPosts: 29\n\nKudos [?]: 122 [1], given: 1\n\nThe positive integers x, y, and z are such that x is a\u00a0[#permalink]\n\n### Show Tags\n\n17 Oct 2009, 14:15\n1\nKUDOS\n13\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n55% (hard)\n\nQuestion Stats:\n\n58% (00:59) correct 42% (01:07) wrong based on 1007 sessions\n\n### HideShow timer Statistics\n\nThe positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?\n\n(1) xz is even\n\n(2) y is even.\n[Reveal] Spoiler: OA\n\nLast edited by Bunuel on 25 Feb 2012, 02:35, edited 1 time in total.\n\nKudos [?]: 122 [1], given: 1\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 41873\n\nKudos [?]: 128579 [6], given: 12180\n\n### Show Tags\n\n17 Oct 2009, 14:32\n6\nKUDOS\nExpert's post\n16\nThis post was\nBOOKMARKED\namitgovin wrote:\nThe positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?\n\n(1) xz is even\n\n(2) y is even.\n\nGiven:\nx is a factor of y --> $$y=mx$$, for some non-zero integer $$m$$;\ny is a factor of z --> $$z=ny$$, for some non-zero integer $$n$$;\nSo, $$z=mnx$$.\n\nQuestion: is z even? Note that $$z$$ will be even if either $$x$$ or $$y$$ is even\n\n(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.\n\n(2) $$y$$ even --> as $$z=ny$$ then as one of the multiples of z even --> z even. Sufficient.\n\n_________________\n\nKudos [?]: 128579 [6], given: 12180\n\nSVP\nJoined: 29 Aug 2007\nPosts: 2472\n\nKudos [?]: 841 [1], given: 19\n\n### Show Tags\n\n19 Oct 2009, 23:18\n1\nKUDOS\namitgovin wrote:\nThe positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?\n\n1) xz is even\n2) y is even.\n\ny/x = k where k is an integer.\ny = xk ....................i\n\nz/y = m where m is an integer.\nz = ym = xkm .....................ii\n\nIf a factor is even, then the source of the factor must be even.\n\n1) If xz is even, z must be even because x may or may not be an even because x is a factor of z but z must be even. SUFF.\n2) If y is even, z must be even because y is a factor of z. SUFF..\n\nD..\n_________________\n\nGmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html\n\nGT\n\nKudos [?]: 841 [1], given: 19\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7670\n\nKudos [?]: 17334 [10], given: 232\n\nLocation: Pune, India\n\n### Show Tags\n\n02 Dec 2010, 11:16\n10\nKUDOS\nExpert's post\n2\nThis post was\nBOOKMARKED\namitgovin wrote:\nThe positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?\n\n1) xz is even\n\n2) y is even.\n\nThough Bunuel has provided the solution, I would just like to bring to your notice a train of thought.\n\nWhen we say, \"The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.\", it implies that if x or y is even, z will be even.\n\ne.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even.\n\nOnce this makes sense to you, it will take 10 secs to arrive at the solution.\nStmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even.\nStmnt 2: y is even, so z must be even\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17334 [10], given: 232 Retired Moderator Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL Joined: 04 Oct 2009 Posts: 1637 Kudos [?]: 1104 [0], given: 109 Location: Peru Schools: Harvard, Stanford, Wharton, MIT & HKS (Government) WE 1: Economic research WE 2: Banking WE 3: Government: Foreign Trade and SMEs Re: positive integers [#permalink] ### Show Tags 12 Dec 2010, 10:22 Bunuel, you always try to solve the questions algebraically, don't you? _________________ \"Life\u2019s battle doesn\u2019t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can.\" My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html GMAT Club Premium Membership - big benefits and savings Kudos [?]: 1104 [0], given: 109 Math Expert Joined: 02 Sep 2009 Posts: 41873 Kudos [?]: 128579 [2], given: 12180 Re: positive integers [#permalink] ### Show Tags 12 Dec 2010, 10:48 2 This post received KUDOS Expert's post metallicafan wrote: Bunuel, you always try to solve the questions algebraically, don't you? Not at all. There are certain GMAT questions which are pretty much only solvable with plug-in or trial and error methods (well at leas in 2-3 minutes). Also many questions can be solved with logic and common sense much quicker than with algebraic approach. So you shouldn't always rely on algebra. Having said that I must add that there are of course other types of questions which are perfect for algebraic approach, plus I often use algebra just to explain a solution. _________________ Kudos [?]: 128579 [2], given: 12180 Intern Joined: 06 Dec 2010 Posts: 17 Kudos [?]: [0], given: 1 Re: positive integers [#permalink] ### Show Tags 12 Dec 2010, 19:36 I like to thing of the boxes method. If you draw them out, then x is inside y which is inside z. zx, a 2 will exist inside the box of either z or x (which is itself inside z) so YES y a 2 will exist inside the box of y which is itself z so YES Kudos [?]: [0], given: 1 Moderator Joined: 01 Sep 2010 Posts: 3355 Kudos [?]: 9042 [1], given: 1152 Re: positive integers [#permalink] ### Show Tags 26 Dec 2010, 17:37 1 This post received KUDOS VeritasPrepKarishma wrote: amitgovin wrote: The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even? 1) xz is even 2) y is even. Please explain. thanks. Though Bunuel has provided the solution, I would just like to bring to your notice a train of thought. When we say, \"The positive integers x, y, and z are such that x is a factor of y and y is a factor of z.\", it implies that if x or y is even, z will be even. e.g. x = 4. Since x is a factor of y, y will be a multiple of 4 and will be even. Since z is a multiple of y, it will also be even. So, in a way, the 2 in x will drive through the entire sequence and make everything even. Once this makes sense to you, it will take 10 secs to arrive at the solution. Stmnt 1: Either x or z (or both which will happen if x is even) is even. In either case, z is even. Stmnt 2: y is even, so z must be even awesome. your explanations and bunuel as well, are amazing thanks _________________ Kudos [?]: 9042 [1], given: 1152 Manager Joined: 19 Dec 2010 Posts: 136 Kudos [?]: 32 [0], given: 12 Re: positive integers [#permalink] ### Show Tags 17 Mar 2011, 23:07 Easy if you realize the following: When a is a factor of b AND b is a factor of c THEN a is a factor of c as well. Hence when either one of these numbers is even, the other has to be even too.. D Kudos [?]: 32 [0], given: 12 SVP Joined: 16 Nov 2010 Posts: 1599 Kudos [?]: 592 [0], given: 36 Location: United States (IN) Concentration: Strategy, Technology Re: positive integers [#permalink] ### Show Tags 17 Mar 2011, 23:52 z = ky y = mx so z = (km)xy (1) -> xz is eve means at least x or z is even, and if x = even, then z is also even as it has an even factor. 2 -> y is even so z having an even factor is even too. Answer D _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings Kudos [?]: 592 [0], given: 36 Intern Joined: 06 Sep 2010 Posts: 40 Kudos [?]: 7 [0], given: 0 Re: positive integers [#permalink] ### Show Tags 15 Apr 2011, 05:24 At first, mistook \"factor\" for \"multiple\" came with answer E. Later, understood that the problem was so easy...just plug and play !! Kudos [?]: 7 [0], given: 0 Manager Joined: 16 May 2011 Posts: 71 Kudos [?]: 18 [0], given: 2 Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 24 Feb 2012, 10:06 if we plug and play, why can't we test X = 1? then y/n.. Kudos [?]: 18 [0], given: 2 Manager Joined: 21 Feb 2012 Posts: 116 Kudos [?]: 148 [0], given: 15 Location: India Concentration: Finance, General Management GMAT 1: 600 Q49 V23 GPA: 3.8 WE: Information Technology (Computer Software) Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 24 Feb 2012, 10:46 The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even? 1) xz is even 2) y is even. Please explain. thanks.[/quote] Ans. let us take any 3 numbers, say x=3,y=18,z=54, or x=2,y=4,z=20 1)if xz is even then it means that either x or z is even,say that x is even, now there is no even number which is a factor of odd number, so z is definitely even, now if x is odd as in the above case, still then we can point out that z is even. 2)if y is even then it is clear that z will be even. Thus this question could be answered by any of the two questions. Kudos [?]: 148 [0], given: 15 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7670 Kudos [?]: 17334 [1], given: 232 Location: Pune, India Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 27 Feb 2012, 05:20 1 This post received KUDOS Expert's post dchow23 wrote: if we plug and play, why can't we test X = 1? then y/n.. Plug in method would be far more painful for this question (and most other questions in my opinion). Think how you would go about it: Checking whether stmnt 1 is enough: xz is even If x = 1, y = 1 and z = 2 (so that xz is even), then z is even. If x = 2, y = 2 and z = 4, z is again even. Then you start thinking if you can take some values such that xz is even but z is not... Now you start using logic... Wouldn't you say it is far better to use logic in the first place itself? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nKudos [?]: 17334 [1], given: 232\n\nSenior Manager\nJoined: 13 Aug 2012\nPosts: 459\n\nKudos [?]: 540 [0], given: 11\n\nConcentration: Marketing, Finance\nGPA: 3.23\nRe: The positive integers x, y, and z are such that x is a\u00a0[#permalink]\n\n### Show Tags\n\n22 Jan 2013, 22:49\namitgovin wrote:\nThe positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?\n\n(1) xz is even\n(2) y is even.\n\nWhat is given?\n\n$$z = y*N$$\n$$y = x*R$$\n\n1.\n$$xz = 2*I$$\nIf x is even, then z is even since $$z = x*R$$\nIf z is even, then z is even.\nSUFFICIENT!\n\n2. $$y = 2*I$$ ==> $$z = 2*I*N$$ Definitely EVEN\nSUFFICIENT!\n\n_________________\n\nImpossible is nothing to God.\n\nKudos [?]: 540 [0], given: 11\n\nDirector\nJoined: 29 Nov 2012\nPosts: 868\n\nKudos [?]: 1409 [0], given: 543\n\n### Show Tags\n\n03 Aug 2013, 02:04\nBunuel wrote:\n\n(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.\n\nCan you please provide a numerical example for this part it isn't very clear. Thanks in advance!\n\nI thought it was insufficient as there were multiple cases either X or Y even or both....\n_________________\n\nClick +1 Kudos if my post helped...\n\nAmazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/\n\nGMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html\n\nKudos [?]: 1409 [0], given: 543\n\nVerbal Forum Moderator\nJoined: 10 Oct 2012\nPosts: 627\n\nKudos [?]: 1355 [1], given: 136\n\n### Show Tags\n\n03 Aug 2013, 04:26\n1\nKUDOS\nfozzzy wrote:\nBunuel wrote:\n\n(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.\n\nCan you please provide a numerical example for this part it isn't very clear. Thanks in advance!\n\nI thought it was insufficient as there were multiple cases either X or Y even or both....\n\nWe know that x is a factor of $$y \\to y = Ix$$\nAgain, $$z = yI^'$$. Now, given that xz - even.\n\nCase I:Assume that x = even , z = odd. Now, as x is even, y = even(I can be odd/even,doesn't matter).\nAgain, as y is even, z HAS to be even($$I^'$$ is odd/even, doesn't matter). Thus, if x is even, z IS even.\nNumerical Example :y = 2*I(x=2).\n$$z = 6*I^'$$. z IS even.\n\nCase II : z is even OR (x and z) both are even.\n\nHope this helps.\n_________________\n\nKudos [?]: 1355 [1], given: 136\n\nCurrent Student\nJoined: 06 Sep 2013\nPosts: 1980\n\nKudos [?]: 719 [0], given: 355\n\nConcentration: Finance\n\n### Show Tags\n\n10 Oct 2013, 15:01\nBunuel wrote:\namitgovin wrote:\nThe positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?\n\n(1) xz is even\n\n(2) y is even.\n\nGiven:\nx is a factor of y --> $$y=mx$$, for some non-zero integer $$m$$;\ny is a factor of z --> $$z=ny$$, for some non-zero integer $$n$$;\nSo, $$z=mnx$$.\n\nQuestion: is z even? Note that $$z$$ will be even if either $$x$$ or $$y$$ is even\n\n(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.\n\n(2) $$y$$ even --> as $$z=ny$$ then as one of the multiples of z even --> z even. Sufficient.\n\nPerfect explanation. Remember the factor foundation rule.\nAlso, other properties of factors that might be helpful to have in mind.\n\nJust to remind you, The factor foundation rule states that \"if a is a factor of b, and b is a factor of c, then a is a factor of c\"\nAlso, if 'a' is a factor of 'b', and 'a' is a factor of 'c', then 'a' is a factor of (b+c). In fact, 'a' is a factor of (mb + nc) for all integers 'm' and 'n'\nIf 'a' is a factor of 'b' and 'b' is a factor of 'a', then 'a=b'\nIf 'a' is a factor of 'bc' and gcd (a,b) = 1, then 'a' is a factor of 'c'\nIf 'p' is a prime number and 'p' is a factor of 'ab' then 'p' is a factor of 'a' or 'p' is a factor of 'b'\nIn other words,, any integer is divisible by all of its factors- and it is also divisible by all of the factors of its factors\n\nHope it helps\n\nKudos [?]: 719 [0], given: 355\n\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 16758\n\nKudos [?]: 273 [0], given: 0\n\nRe: The positive integers x, y, and z are such that x is a\u00a0[#permalink]\n\n### Show Tags\n\n30 Jan 2015, 14:25\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\n\nKudos [?]: 273 [0], given: 0\n\nGMAT Club Legend\nJoined: 09 Sep 2013\nPosts: 16758\n\nKudos [?]: 273 [0], given: 0\n\nRe: The positive integers x, y, and z are such that x is a\u00a0[#permalink]\n\n### Show Tags\n\n05 Feb 2016, 17:31\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\n\nKudos [?]: 273 [0], given: 0\n\nRe: The positive integers x, y, and z are such that x is a \u00a0 [#permalink] 05 Feb 2016, 17:31\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 24 posts ]\n\nDisplay posts from previous: Sort by", "date": "2017-10-17 19:13:34", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/the-positive-integers-x-y-and-z-are-such-that-x-is-a-85437.html?sort_by_oldest=true", "openwebmath_score": 0.7219163775444031, "openwebmath_perplexity": 2259.430897110478, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9618217257384335, "lm_q2_score": 0.8856314783461302, "lm_q1q2_score": 0.851819596871155}}
{"url": "https://mathematica.stackexchange.com/questions/149710/generating-a-convex-hull-with-the-hull-boundary-points-labeled", "text": "# Generating a convex hull with the hull boundary points labeled\n\nI'm very new to Mathematica, and I'm sure my question is trivial.\n\nSuppose a have a (long) list whose elements are of the form {{x, y}, n}, where x and y are Real, and n is an Integer. I'd like to plot the convex hull of the corresponding points {x, y} in the plane. Moreover, for each point {x, y} that is a vertex of the convex hull should appear with its corresponding label n. Finally I'd like the coordinate axes to be displayed with the plot.\n\n### Revision\n\nContrive some data.\n\nSeedRandom[42];\nWith[{n = 42},\ndata = Transpose[{RandomReal[1., {n, 2}], RandomSample[Range[100], n]}]];\nShort[data, 3]\n\n\n{{{0.425905, 0.391023}, 51}, <<40>>, {{0.359445, 0.00772549},24}}\n\nSeparating the points from the labels.\n\npts = data[[All, 1]];\nlbls = data[[All, 2]];\n\n\nGenerating the convex hull.\n\nch = ConvexHullMesh[pts];\n\n\nExtracting the boundary point labels.\n\nbpts = MeshCoordinates[RegionBoundary[ch]];\nblbls = Extract[lbls, Flatten[Position[pts, #] & /@ bpts, 1]]\n\n\n{94, 36, 21, 1, 88, 2, 14, 24}\n\nMaking the labels into graphics elements.\n\nlabels = MapThread[Text[#1, #2] &, {blbls, bpts}];\n\n\nShowing the combined graphics.\n\nShow[ch, Graphics[{Point[pts], labels}], Axes -> True]\n\n\nOf course, the labels can be fancied up by using Inset rather than Text\n\nlabels =\nInset[\nGraphics[{FaceForm[White], EdgeForm[Black], Disk[], Text[Style[#1, 9]]}],\n#2, Automatic, Scaled[.0475]] &,\n{blbls, bpts}];\nShow[ch, Graphics[{Point[pts], labels}],\nMethod -> {\"AxesInFront\" -> False}, Axes -> True]\n\n\n\u2022 My label is not necessarily the index. Jul 4 '17 at 23:35\n\u2022 @JairoBochi. I have revised my answer in a way that I hope you find better attuned to your needs. Jul 5 '17 at 2:19\n\u2022 @MichaelE2 Right, I should have provided sample data, sorry. Jul 5 '17 at 22:39\n\u2022 @JairoBochi. Your nagging me about the indices motivated me to improve my answer, so I'm glad you did it. Jul 6 '17 at 5:02\nSeedRandom[1]\ndata = Transpose[{RandomReal[1, {10, 2}], RandomInteger[100, 10]}];\n\nxy = data[[All,1]];\nlabels = data[[All,2]];\n\nConvexHullMesh[xy,  Prolog -> Point[xy], Frame -> True,\nMeshCellStyle -> {2 -> Opacity[0.5, LightBlue]},\nMeshCellShapeFunction -> {0 -> ({Opacity[1, Yellow], Disk[#, .05],\nText[Style[# /. rule, Red, 16], #]} &)}]\n\n\nAlso\n\nlabeled = Labeled[#, # /. rule] & /@ MeshCoordinates[ConvexHullMesh[xy]];\n\nShow[ConvexHullMesh[xy], ListPlot[labeled], Graphics[Point[xy]], Frame ->True]\n\n\n\u2022 I understand that you generate the plot from two lists, \"xy\" and \"labels\". But my list is given of the form \"data\". So I would need to extract \"xy\" and \"labels\" from data to have a working solution... Jul 4 '17 at 23:48\n\u2013\u00a0kglr\nJul 4 '17 at 23:52\n\nThere are potential shortcomings in using Rule to map points to labels, which I have occasionally come across. The purpose in creating a NearestFunction using Nearest below to map coordinates {x, y} to a label is overcome these shortcoming, at only a small expense of computational time. Given the general nature of the question, this seemed the best approach.\n\nOften in an application, one might arrive at coordinates in various ways. If we're using floating-point numbers, rounding error might make simple rules of the form {x, y} -> label fail, because the coordinates {x, y} of the rule might not exactly match the computed coordinates in the figure. Or one might start with exact coordinates, which at some point might be numericized (converted to floating point), and again the replacement rule will fail. Nearest takes care of those issues, unless rounding error is so great that one computed point ends up closer to a different point than the intended one (in such a case, the problem is completely different, a numerics one). Whether or not to numericized the points with N[] depends on whether data is already numericized; it's not strictly necessary in either case, but it should speed up nf, which might be an issue with a very long list data.\n\npoints = data[[All, 1]];\nnf = Nearest[N@points -> data[[All, 2]]]; (* map coordinates to labels in data *)\nhull = ConvexHullMesh[data[[All, 1]]];\n\nWith[{coords = MeshCoordinates[hull]},\nShow[\nMeshRegion[\ncoords,\nMeshCells[hull, 2],\nMeshCellLabel -> Table[{0, i} -> First@nf[coords[[i]]], {i, Length@coords}]\n],\nAxes -> True] (* or Frame -> True *)\n]", "date": "2021-10-16 20:31:41", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/149710/generating-a-convex-hull-with-the-hull-boundary-points-labeled", "openwebmath_score": 0.2550300061702728, "openwebmath_perplexity": 3802.713505233619, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9473810407096791, "lm_q2_score": 0.8991213664574069, "lm_q1q2_score": 0.8518105358787269}}
{"url": "https://mathoverflow.net/questions/192058/covering-designs-of-the-form-v-k-2/192065", "text": "# covering designs of the form $(v,k,2)$\n\nA covering design $(v,k,t)$ is a family of subsets of $[v]$ each having $k$ elements such that given any subset of $[v]$ of $t$ elements it is a subset of one of the sets of the family. A problem is to find the minimum number of subsets such a family can have.\n\nI am interested in the case $(v,k,2)$. It seems to be equivalent to the problem of finding the minimum number of copies of $K_k$ that are needed to cover the graph $K_v$.\n\nWhen $k=3$ this problem may be solved optimally with a Steiner triple system if it exists, which happens if and only if $v$ is $1$ or $3 \\bmod 6$, in this case $\\frac{v(v-1)}{6}$ subsets are required.\n\nMy questions are the following:\n\n\u2022 Is the minimum of subsets for $(v,3,2)$ known when the congruence does not hold?\n\u2022 If not what are good bounds?\n\u2022 Does the problem with $t=2$ have another name (It seems like it may be more common than the general covering design)\n\u2022 Is the minimum number of subsets for $(v,k,2)$ known?\n\u2022 If not, which are some good bounds (specific values of $k$ are also welcome)\n\u2022 I would call the covering designs $\\ (v\\ k\\ 2)\\$-- sloppy planes. \u2013\u00a0W\u0142odzimierz Holszty\u0144ski Jan 3 '15 at 22:28\n\u2022 Your $\\ \\frac{n(n-1)}2\\$ must be a typo(?). A simple calculation shows that the number of 3-subset of a $\\ (\\nu\\ 3\\ 2)\\$ perfect system must be $\\ \\frac{\\nu\\cdot(\\nu-1)}3\\$ (it's $\\ 3,\\$ not $\\ 2,\\$ in the denominator). Also then $\\ \\nu\\equiv 1\\ or\\ 3\\mod 6\\$ (rather than $\\ 0\\ or\\ 3).\\$ Thus I feel that it would be nice and useful for non-specialists like me to have a short list of basic results in a separate Answer. \u2013\u00a0W\u0142odzimierz Holszty\u0144ski Jan 4 '15 at 1:43\n\u2022 @W\u0142odzimierzHolszty\u0144ski You can find basic facts in the paper I linked to in my post (or those given by Thomas Kalinowski as well, I think). Handbook of Combinatorial Designs is a good reference book for this sort of basic knowledge; coverings are treated in Section 11 of Chapter IV in the 2nd edition. As for the typos, yes, they are not correct, although the number of $3$-subsets in this case is not $\\frac{v(v-1)}{3}$ but $\\frac{\\binom{v}{2}}{\\binom{3}{2}} = \\frac{v(v-1)}{6}$. I'll edit OP's post. \u2013\u00a0Yuichiro Fujiwara Jan 4 '15 at 6:31\n\u2022 @YuichiroFujiwara -- thank you for the links. And I let $\\ \\frac{\\binom{\\nu}2}{\\binom 32}=\\frac{\\nu\\cdot (\\nu-1)}{3\\cdot 2}\\$ to have somehow just $\\ 3\\$ in the denominator--ooops! Sorry. \u2013\u00a0W\u0142odzimierz Holszty\u0144ski Jan 4 '15 at 6:46\n\nEdit: The possible \"gap\" of sort in Caro and Yuster's proof of their upper bound has just been fixed! See Ben Barber's comment below (and his joint paper with Daniela K\u00fchn, Allan Lo and Deryk Osthus on arXiv. This shows Gustavsson's theorem (whose proof by Gustavsson himself might have been regarded as incomplete but was used by Caro and Yuster). So this result proves the general upper bound on the smallest coverings mentioned in Thomas Kalinowski's post is indeed correct.\n\nThe most up to date account (except the above mentioned paper) on the minimum number of subsets for a $(v,k,2)$ covering is here:\n\nY. M. Chee, C. J. Colbourn, A. C. H. Ling, R. M. Wilson, Covering and packing for pairs, J. Combin. Theory Ser. A 120 (2013) 1440\u20131449.\n\nOne important remark about Thomas Kalinowski's answer is that the asymptotic solution by Caro and Yuster may be incomplete, although the claimed upper bound is likely the correct one. They rely on Gustavsson's theorem on graph decomposition (Lemma 2.1 in Caro and Yuster's paper), which is claimed to be proved in the following thesis:\n\nT. Gustavsson, Decompositions of Large Graphs and Digraphs with High Minimum Degree,'' Doctoral Dissertation, Dept. of Mathematics, Univ. of Stockholm, 1991.\n\nHowever, here's an excerpt from the 2013 paper by Chee, Colbourn, Ling, and Wilson:\n\nTheir approach relies in an essential manner on a strong statement by Gustavsson:\n\nProposition 1.1: . Let $H$ be a graph with $v$ vertices and $h$ edges, having degree sequence $(d_1,\\dots,d_v)$. Then there exist a constant $N_H$ and a constant $\\epsilon_H>0$, both depending only on $H$, such that for all $n>N_H$, if $G$ is a graph on $n$ vertices, $m$ edges, and degree sequence $(\\delta_1,\\dots,\\delta_n)$ so that $\\min(\\delta_1,\\dots,\\delta_n) \\geq n(1\u2212\\epsilon_H)$, $\\gcd(d_1,\\dots,d_v)\\ \\vert\\ \\gcd(\\delta_1,\\dots,\\delta_n)$, and $h\\ \\vert\\ m$, then $G$ has an edge partition (decomposition) into graphs isomorphic to $H$.\n\nWe have not been able to verify the proof of Proposition 1.1. Indeed, while the result has been used a number of times in the literature, no satisfactory proof of it appears there. While we expect that the statement is true, we do not think that the proof in [12] is sufficient at this time to employ the statement as a foundation for further results. Therefore we adopt a strategy that is completely independent of Proposition 1.1, and independent of the results built on it.\n\nSo, one might want to be careful about Caro and Yuster's claimed solution. Chee, Colbourn, Ling, and Wilson gave a slightly weaker upper bound on the minimum size of a $(v,k,2)$ covering, which essentially states that for sufficiently large $v$, Caro and Yuster's claimed upper bound is correct within a constant that depends on $k$. (And the proof of Gustavsson's theorem by Barber, K\u00fchn, Lo and Osthus shows it is correct.)\n\n\u2022 Together with Daniela K\u00fchn, Allan Lo and Deryk Osthus I have a submitted a paper that proves that statement of Gustavsson and gives quantitative bounds on the value of $\\epsilon_H$. The most recent version can be found on the arXiv. arxiv.org/abs/1410.5750 \u2013\u00a0Ben Barber Jan 5 '15 at 13:35\n\u2022 @BenBarber That's awesome! Thank you for the comment and link, and thank you for your work! \u2013\u00a0Yuichiro Fujiwara Jan 5 '15 at 13:40\n\u2022 And please feel free to make this post more informative and/or correct errors, or post another answer. \u2013\u00a0Yuichiro Fujiwara Jan 5 '15 at 14:55\n\nFort and Hedlund determined the minimum size of a $(v,3,2)$-covering design: Minimal coverings of pairs by triples, Pacific J. Math. 8(4), 709-719, 1958.\n\nThe case $(v,4,2)$ was solved by Mills: On the covering of pairs by quadruples I, JCTA 13, 55\u201378, 1972 and II, JCTA 15, 138\u2013166 (1973).\n\nFor the case $(v,5,2)$ with $v\\equiv 0\\pmod 4$, Abel, Assaf, Bennett, Bluskov and Greig established that the Sch\u00f6nheim bound $\\left\\lceil\\frac{v}{5}\\left\\lceil \\frac{v-1}{4}\\right\\rceil\\right\\rceil$ is tight for $v\\geqslant 28$ with 17 possible exceptions in the range $40\\leqslant v\\leqslant 280$: Pair covering designs with block size 5, Discrete Mathematics 307(14), 1776-1791, 2007.\n\nThe same authors have results for $(v,6,2)$: Pair covering and other designs with block size 6, J Comb Des 15(6), 511-533, 2007\n\nCaro and Yuster proved that for sufficiently large $v$ the minimum size of a $(v,k,2)$ covering design is $\\left\\lceil\\frac{v}{k}\\left\\lceil \\frac{v-1}{k-1}\\right\\rceil\\right\\rceil$: Covering Graphs: The Covering Problem Solved, JCTA 83 (2) 273\u2013282, 1998.\n\nThe La Jolla Covering repository is a good source for covering designs. According to this table the smallest open case with $t=2$ is $(v,k,t)=(23,6,2)$ where the minimum size is either 20 or 21.\n\n\u2022 Thanks, that's usefull, I allready knew about la Jolla, the second source is great! Do you know if there has been any work on the bounds for arbitrary $(v,k,2)$? \u2013\u00a0Jorge Fern\u00e1ndez Jan 3 '15 at 21:24\n\u2022 I know there are resuts for $(v,k,t)$ in general, but I though maybe with $(v,k,2)$ they could be sharper \u2013\u00a0Jorge Fern\u00e1ndez Jan 3 '15 at 21:25\n\u2022 Caro and Yuster's proof might be incomplete (although the statement itself is probably correct). See my answer (and the paper I linked to) for more detail. \u2013\u00a0Yuichiro Fujiwara Jan 3 '15 at 23:41", "date": "2020-01-18 16:46:46", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/192058/covering-designs-of-the-form-v-k-2/192065", "openwebmath_score": 0.7909506559371948, "openwebmath_perplexity": 280.53751763005624, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717476269469, "lm_q2_score": 0.8633916222765627, "lm_q1q2_score": 0.8517977816758533}}
{"url": "https://math.stackexchange.com/questions/978/how-to-prove-and-interpret-operatornamerankab-leq-operatornamemin-ope?noredirect=1", "text": "# How to prove and interpret $\\operatorname{rank}(AB) \\leq \\operatorname{min}(\\operatorname{rank}(A), \\operatorname{rank}(B))$?\n\nLet $$A$$ and $$B$$ be two matrices which can be multiplied. Then $$\\operatorname{rank}(AB) \\leq \\operatorname{min}(\\operatorname{rank}(A), \\operatorname{rank}(B)).$$\n\nI proved $$\\operatorname{rank}(AB) \\leq \\operatorname{rank}(B)$$ by interpreting $$AB$$ as a composition of linear maps, observing that $$\\operatorname{ker}(B) \\subseteq \\operatorname{ker}(AB)$$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information.\n\nHow do you manage $$\\operatorname{rank}(AB) \\leq \\operatorname{rank}(A)$$? Is there a nice interpretation like the previous one?\n\n\u2022 Your proof is fine. Furthermore, the same reasoning will get your desired fact. Again rank-nullity will tell you that the dimension of your vector space minus the dimension of the kernel will give you the rank. Jul 28, 2010 at 16:08\n\nYes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get \"more stuff\" in the former case, as the former is bigger than the latter.\n\n\u2022 Thank you. I was so obsessed with the kernel-image dimension formula that I could't recognize this simple fact.\n\u2013\u00a0user365\nJul 30, 2010 at 8:52\n\nOnce you have proved $$\\operatorname{rank}(AB) \\le \\operatorname{rank}(A)$$, you can obtain the other inequality by using transposition and the fact that it doesn't change the rank (see e.g. this question).\n\nSpecifically, letting $$C=A^T$$ and $$D=B^T$$, we have that $$\\operatorname{rank}(DC) \\le \\operatorname{rank}(D) \\implies \\operatorname{rank}(C^TD^T)\\le \\operatorname{rank} (D^T)$$, which is $$\\operatorname{rank}(AB) \\le \\operatorname{rank}(B)$$.\n\n\u2022 Very nice! Thank you.\n\u2013\u00a0user365\nSep 2, 2010 at 10:01\n\u2022 I am doing this problem now too. Why is proving both inequalities enough? Where does the proof take the min idea into account? Apr 12, 2016 at 3:27\n\u2022 @MathisHard $x<y \\wedge x<z \\implies x<\\min(y,z)$. May 25, 2018 at 15:47\n\nNote that $$\\text{Col}(AB) \u2286 \\text{Col}(A)$$ since given $$y \u2208 \\text{Col}(AB)$$ we can choose $$x \u2208 F$$ and then we have $$y = (AB)x = A(Bx) \u2208 \\text{Col}(A)$$.\n\nSince $$\\text{Col}(AB) \u2286 \\text{Col}(A)$$, any basis for $$\\text{Col}(AB)$$ can be extended to a basis for $$\\text{Col}(A)$$ and so $$\\dim \\text{Col}(AB) \u2264 \\dim \\text{Col}(A)$$, that is $$\\text{rk}(AB) \u2264 \\text{rk}(A).$$\n\nNote that $$\\text{Null}(B) \u2286 \\text{Null}(AB)$$ since given $$x \u2208 \\text{Null}(B)$$ we have $$(AB)x = A(Bx) = A 0 = 0$$ so that $$x \u2208 \\text{Null}(AB)$$.\n\nSince $$\\text{Null}(B) \u2286 \\text{Null}(AB)$$, as above we have $$\\dim \\text{Null}(B) \u2264 \\dim \\text{Null}(AB)$$, that is, $$\\text{Nullity}(B) \u2264\\text{Nullity}(AB)$$. Thus $$\\text{rk}(AB) = n \u2212 \\text{Nullity}(AB) \u2264 n \u2212 \\text{Nullity}(B) = \\text{rk}(B).$$\n\n\u2022 You should use MathJax to format your answer. Oct 15, 2015 at 21:06\n\u2022 Very nice proof! +1\n\u2013\u00a0ZFR\nApr 23, 2020 at 1:41\n\nProve first that if $f:X\\to Y$ and $g:Y\\to Z$ are functions between finite sets, then $|g(f(X))| \\leq \\min \\{ |f(X)|, |g(Y)| \\}.$\n\nThen use the same idea.\n\n\u2022 Categorification... :-) Jul 29, 2010 at 21:50\n\u2022 I am not familiar with the 'categorification'. How can one go from this to the rank inequality ? What functor is to be applied ? Apr 13, 2014 at 12:10\n\u2022 So you're saying that rank of linear map is somehow like image of a function? Feb 12, 2015 at 10:15\n\u2022 @Mihail, it is like the size of the image of a function. in fact, if $f$ is a linear map, the rank of $f$ is the dimension of the image of $f$ and the dimension is a measure of the size of a space. Feb 12, 2015 at 18:22\n\nHere is another simple answer. When you multiply a matrix and a vector $Ax$ you end up with a linear combination of the columns of $A$.\n\n$$Ax = \\; x_1\\,A_1 \\;+\\; x_2\\,A_2 \\;+\\; x_3\\,A_3 \\;+\\;\\; ...\\;\\; \\\\$$\n\nWhen we multiply two matrices $AB = C$, we have $AB_i = C_i$, which means that each column of $C$ is a linear combination of the columns of $A$, so $\\text{rank}(AB) \\leq \\text{rank}(A)$. To show that $\\text{rank}(AB) \\leq \\text{rank}(B)$ we follow a similar argument -- when you multiply $x^{\\top}B$, you end up with a linear combination of the rows of $B$.\n\nAlready you have proved $$rank(AB)\\leq rank(B)$$\n\nFor other part\n\nrank of $A=$ dim range $A$\n\nAs range $AB \\subset$ range $A\\implies$ dim range $AB\\leq$ dim range $A$ . Hence $$rank(AB)\\leq rank (A)$$\n\n\u2022 Hello!! Why does it hold that \"range AB $\\subset$ range A\" ? Jan 7, 2017 at 10:01\n\u2022 If $x\\in$ range $AB,$ then $x=(AB)y$ for some $y.$ Thus $x=A(By).$ Thus $x\\in$ range $A.$@ Mary Star Jan 8, 2017 at 17:48\n\nLet $m \\le n, A \\in M_{m\\times n}, B\\in M_{n\\times m}$.\n\n$\\mbox{rank } A\\le m$ and $\\text{rank }B\\le m$. (Obvious fact as rank A = dimension of the columnspace of A = dimension of the row space of A.)\n\nLet $E_{n\\times n}B$ be the row echelon form of $B$ and let $AE_{m\\times m}$ be the column echelon form of $A$. ($E_{n\\times n} ,E_{m\\times m}$ are elementary matrices.)\n\nWe know $\\operatorname{rank}(BA)=\\operatorname{rank}(E_{n\\times n}BA )=\\operatorname{rank}(E_{n\\times n}BAE_{m\\times m} )$.\n\nBut $E_{n\\times n}BAE_{m\\times m} =\\begin{pmatrix} L&0\\\\ 0&0\\\\ \\end{pmatrix},$ where $L$ is an $k\\times l$ matrix with $k\\le \\operatorname{rank}(B),l\\le \\operatorname{rank}(A)$.\n\nSo $\\operatorname{rank}(E_{n\\times n}BAE_{m\\times m} )=\\operatorname{rank}\\begin{pmatrix} L&0\\\\ 0&0\\\\ \\end{pmatrix}\\le \\min\\{k,l\\}\\le \\min\\{\\mbox{rank } A,\\mbox{rank }B\\}.$\n\nAnother way of showing that $\\text{Rank} (AB) \\leq \\text{Rank}(B)$ without using rank-nullity: Note that if $v_1,\\dots,v_n$ is a basis of $\\text{Range} B$, then $Av_1,\\dots,Av_n$ is a spanning list of $\\text{Range} AB$.\n\nEach column of $$AB$$ is a linear combination of columns of $$A$$ so $$\\text{Range}(AB)\\subseteq \\text{Range} (A)$$ equivalently $$rank(AB)\\leq rank(A).$$\n\nSimilarly, Each row of $$AB$$ is a linear combination of rows of $$B$$ so $$rowspace(AB)\\subseteq rowspace(B)$$.\n\nSince rowspace=columnspace. Thus $$rank(AB)\\leq rank(B)$$.\n\nFrom both of the inequality we deduce that $$rank(AB)\\leq \\min\\{rank A, rank B\\}$$.", "date": "2022-08-18 02:45:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/978/how-to-prove-and-interpret-operatornamerankab-leq-operatornamemin-ope?noredirect=1", "openwebmath_score": 0.9413327574729919, "openwebmath_perplexity": 188.73014372270734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717440735736, "lm_q2_score": 0.8633916152464017, "lm_q1q2_score": 0.8517977716721424}}
{"url": "https://math.stackexchange.com/questions/1248235/groups-with-few-subgroups", "text": "# Groups with \u201cfew\u201d subgroups\n\nIf $G$ is a finite group of order $n,$ and the number of divisors of $n$ is $k,$ can $G$ have fewer than $k$ subgroups?\n\nA cyclic group $G$ of order $n$ has exactly one subgroup for each divisor of $n$, so in this case $G$ has exactly $k$ subgroups.\n\nFor the alternating group $G=A_4,$ $n=12$ and $k=6.$ There is no subgroup of order $6$ in $A_4,$ yet there are ten subgroups, which exceeds $k=6$ for this case. I began to wonder if there could be a group for which so many divisors of the order of $G$ had no subgroups of that order, that there wound up being fewer subgroups in $G$ than the number of divisors of $|G|,$ hence my question.\n\n\u2022 Recall that a CLT (Converse of Lagrange's Theorem) group is a group with the property that for any divisor of the order of the group there exists a subgroup of that order. Since it is known that finite abelian groups, as well as $p$-groups are CLT groups, this leaves a smaller class of subgroups to consider. \u2013\u00a0user1337 Apr 23 '15 at 13:16\n\u2022 A finite group $G$ of order $n$ such that the number of elements $x$ with $x^d=1$ is less than $d$ for all divisors $d$ of $n$ has to be cyclic. I wonder, if a variation of this topic can lead to a proof that there are at least as many subgroups as divisors. \u2013\u00a0j.p. Apr 23 '15 at 13:24\n\u2022 @j.p. That idea looks promising. One thing, should the phrase \"is less than $d$\" be replaced by \"is at most $d$\" in your comment? [though that would imply your version, I seem to recall one approach to doing the counting argument for showing some group is cyclic as involving the number being at most $d$] And +1 on the comment. \u2013\u00a0coffeemath Apr 23 '15 at 13:42\n\u2022 A short script in Maple shows that there are no such groups among the 793 groups of order $< 2^5 \\cdot 3 = 96$. \u2013\u00a0Travis Willse Apr 23 '15 at 13:43\n\u2022 @coffeemath: Yes, you are right. It should have been \"at most\". \u2013\u00a0j.p. Apr 23 '15 at 14:04\n\nThe answer to your question is no, that is not possible. This is a consequence of the following slightly more general result. Let $d(n)$ is the number of divisors of $n$.\n\nTheorem Let $G$ be a finite group of order $n$. Then for any divisor $m$ of $n$, there exist at least $d(m)$ subgroups of $G$ of order dividing $m$.\n\nProof We prove the statement by induction on $m$, for all finite groups $G$. It's clear for $m=1$.\n\nFor $m>1$, let $p$ be a prime dividing $m$, and let $m = p^r k$ with $\\gcd(p,k)=1$.\n\nSince $d(m) = (r+1)d(k)$, it is sufficient to prove that, for each $i$ with $0 \\le i \\le r$, there are at least $d(k)$ subgroups of $G$ of order $p^i j$, for some $j$ that divides $k$.\n\nSo fix an $i$, let $P$ be any subgroup of $G$ of order $p^i$, let $N = N_G(P)$ be the normalizer of $P$ in $G$, and let $h = \\gcd(|N|,k)$.\n\nSince $h \\le k < m$ and $h$ divides $|N/P|$, by inductive hypothesis, the number $s$ of subgroups $S/P$ of $N/P$ of order dividing $h$ satisfies $s \\ge d(h)$. For each of these subgroups, the inverse image $S$ of $S/P$ in $N/P$ is a subgroup of $G$ of order $p^i j$, where $|S/P| = j$ divides $h$, and so $j$ divides $k$.\n\nNow $P$ has $|G|/|N|$ distinct conjugates $P^g$ in $G$, and the $s|G|/|N|$ subgroups $S^g$ are all distinct. But $|G|/|N| \\ge k/h$, so $s|G|/|N| \\ge d(h)k/h \\ge d(k)$, which completes the proof.\n\nAfter thinking about it again, I think we can say that a group of order $n$ with exactly $d(n)$ subgroups must be cyclic. The above proof produces at least $d(n)$ subgroups with a normal $p$-subgroup for any prime divisor $p$ of $n$. So if there were exactly $d(n)$ subgroups, then all subgroups would have all of their Sylow subgroups normal, so they would all, including $G$ itself, be nilpotent. But a non-cyclic $p$-group $P$ has more than one subgroup of index $p$, and hence has more than $d(|P|)$ subgroups, so all Sylow subgroups of $G$ are cyclic and hence so is $G$.\n\nBy the way, I first posted this proof here in 2003. I would welcome any references to any other proofs.\n\n\u2022 Derek-- Nice proof (seems to me). +1 on it for now, and likely I'll \"accept\" after I get a chance to go over it in detail. \u2013\u00a0coffeemath Apr 23 '15 at 14:46\n\u2022 How do you know that $P$ exist ? \u2013\u00a0Belgi Apr 23 '15 at 14:57\n\u2022 @Belgi $G$ has a $p$-Sylow subgroup $T$. Let's say $T$ is of size $p^k$; then $T$ will have a subgroup of order $p^i$ provided $i\\leq k$. \u2013\u00a0mathmandan Apr 23 '15 at 15:06\n\u2022 @mathmandan - How do you know such subgroups of $T$ exist ? \u2013\u00a0Belgi Apr 23 '15 at 15:10\n\u2022 I was thinking of the proof as an induction on $m$ for all $n$ simultaneously. That is, when doing the proof for $m$, we assume that the result is true for all smaller $m$ and all $n$. We don't use induction on $n$ anywhere. I remember when thinking about this originally, the chief problem was to get an induction to work, and the key idea in the proof is actually to induct on $m$ rather than on $n$. \u2013\u00a0Derek Holt Apr 23 '15 at 17:06", "date": "2021-04-11 02:07:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1248235/groups-with-few-subgroups", "openwebmath_score": 0.8602133393287659, "openwebmath_perplexity": 92.40538364806898, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717432839349, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8517977709903749}}
{"url": "http://math.stackexchange.com/questions/291602/generating-sets-from-given-information", "text": "# Generating Sets From Given Information\n\nThe problem I am working on is:\n\nThe three most popular options on a certain type of new car are a built-in GPS (A), a sunroof (B), and an automatic transmission (C). If 40% of all purchasers request A, 55% request B, 70% request C, 63% request A or B, 77% request A or C, 80% request B or C, and 85% request A or B or C, determine the probabilities of the following events. [Hint: \u201cAor B\u201d is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.]\n\na.The next purchaser will request at least one of the three options.\n\nb.The next purchaser will select none of the three options.\n\nc. The next purchaser will request only an automatic transmission and not either of the other two options.\n\nd.The next purchaser will select exactly one of these three options.\n\nI am absolutely positive that $P(A)=40\\%$, $P(B)=55\\%$, $P(C)=70\\%$, and $P(A \\cup B \\cup C)=85\\%$. However, the pieces of data I am not quite certain about are $P(A \\cup B \\cap C')=63\\%$, $P(A \\cup C \\cap B')=77\\%$, $P(B \\cup C \\cap A')=80\\%$, do these values correspond to the rest of the data? If so, then I seems nearly impossible to be able to generate the Venn Diagram. Could someone help?\n\nEDIT: What I am having a difficult time interpreting is, when they say in the question, \"...63% request A or B.\" To me, that says only A or only B; and under this interpretation I would write $P(A \\cup B \\cap C')=63\\%$. Under Andr\u00e9 Nicolas' interpretation, \"63% request A or B,\" means $P(A \\cup B)=63\\%$. If it is the case that Andr\u00e9 Nicolas is correct, then it seems like they should have stated in the question, \"63% request A or B, A and C, B and C, or A and B and C.\"\n\nAlso, I solved the problem under Andr\u00e9 Nicolas' assumption, and for part d), I know the answer but I am sure how to put in it math symbols. How would I do that?\n\n-\n\nHere is a start:\n\na) The answer to this is written down in the data: It is $\\Pr(A\\cup B\\cup C)$.\n\nb) This can be immediately written down from the answer to a).\n\nAs for the rest of the probabilities, we need to work. It can be figured out completely from a Venn diagram. But since one cannot point to a picture on a blackboard, what follows will be formula-heavy.\n\nRecall that $$\\Pr(X\\cup Y)=\\Pr(X)+\\Pr(Y)-\\Pr(X\\cap Y).\\tag{1}$$\n\nWe know $\\Pr(A)$, $\\Pr(B)$, and $\\Pr(A\\cup B)$. Using Formula $(1)$, we can see that $0.63=0.40+0.55-\\Pr(A\\cap B)$. That gives $\\Pr(A\\cap B)=0.32$.\n\nSimilarly, we can find $\\Pr(A\\cap C)$ and $\\Pr(B\\cap C)$ from the given information.\n\nRecall also that\n\n$$\\Pr(A\\cup B\\cup C)=\\Pr(A)+\\Pr(B)+\\Pr(C)-\\Pr(A\\cap B)-\\Pr(A\\cap C)-\\Pr(B\\cap C)+\\Pr(A\\cap B\\cap C).$$\n\nWe are told $\\Pr(A\\cup B\\cup C)$, and we computed the next three items in the above formula, so now we know $\\Pr(A\\cap B\\cap C)$.\n\nOK, time to fill in the bits in the Venn Diagram. First write the computed $\\Pr(A\\cap B\\cap C)$ in the space where it should go.\n\nWe know $\\Pr(A\\cap B)$. Since we also know $\\Pr(A\\cap B\\cap C)$, we can find the probability of \"the rest of\" $A\\cap B)$, called $A\\cap B\\cap C'$ in set language. Compute, write the number in.\n\nDo the same for the other similar bits, namely $A\\cap C\\cap B'$ and $B\\cap C\\cap A'$.\n\nNow from $\\Pr(A)$ you can find the probability of the \"outside\" part of $A$, that is, $\\Pr(A\\cap B'\\cap C')$. Do the same for the \"outside\" part of $B$, also the \"outside\" part of $C$.\n\nThe probability of the outer world $A'\\cap B'\\cap C'$ of cheapskates who want nothing can now be found in various ways.\n\nWe now know everything, and can answer any question!\n\nRemark: You can be more fancy, and let $X$, $Y$, and $Z$ respectively be $A'$, $B'$, and $C'$. By taking every one of the given numbers and subtracting it from $1$, you can write down immediately $\\Pr(X)$ (namely $1-0.40$), $\\Pr(Y)$, $\\Pr(Z)$, $\\Pr(X\\cap Y)$, and so on. Now we end up with a problem of a type that you have probably done several times.\n\n-\nSo, in the question, when it said something like, \"63% request A or B,\" you assumed that two be $P(A \\cup B)=63\\%$? Why isn't it really $P(A \\cup B \\cap C')=63\\%$? If you take, \"63% request A or B,\" to mean the former, why wouldn't you include something about C? Because when you write the Venn diagram, A and B both intersect C. \u2013\u00a0 Mack Feb 1 '13 at 14:56\nThe reason I ask is because $P(A \\cup B)$ also includes the probability of getting $A$ and $C$ and $B$ and $C$, and it just seems odd that you wouldn't mention something like that in the question. \u2013\u00a0 Mack Feb 1 '13 at 15:17\n@EliMackenzie: The problem setter should make things clear. But I am reasonably sure that the intent here is that \"or\" be interpreted as $\\lor$, logical or, meaning in this case union. There is a somewhat similar potential issue with \"and\" in standard Venn diagram work problems. When I say $A$ and $B$, does that imply anything about $C$, $D$? No, and you are accustomed to that. \u2013\u00a0 Andr\u00e9 Nicolas Feb 1 '13 at 16:49\nOne can also go through the Venn diagram, trying to see whether your interpretation is numerically consistent with the data. The probabilities of $A\\lor B$, $B\\lor C$, $A\\lor C$ are quite large, one can't find a Venn diagram that works. Some of the percentages of \"subareas\" would turn out negative. \u2013\u00a0 Andr\u00e9 Nicolas Feb 1 '13 at 16:56\nAlso, if you look at the text of the problem, it says explicitly as a hint what is meant by $A$ or $B$. That is neutral about $C$. It looks as if the problem setter was worried people might have trouble with the meaning of or. \u2013\u00a0 Andr\u00e9 Nicolas Feb 1 '13 at 17:03", "date": "2014-04-16 19:33:49", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/291602/generating-sets-from-given-information", "openwebmath_score": 0.7819040417671204, "openwebmath_perplexity": 319.0762801610089, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717444683928, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8517977650772673}}
{"url": "https://fr.mathworks.com/help/symbolic/legendrep.html", "text": "# legendreP\n\nLegendre polynomials\n\n## Description\n\nexample\n\nlegendreP(n,x) returns the nth degree Legendre polynomial at x.\n\n## Examples\n\n### Find Legendre Polynomials for Numeric and Symbolic Inputs\n\nFind the Legendre polynomial of degree 3 at 5.6.\n\nlegendreP(3,5.6)\nans =\n430.6400\n\nFind the Legendre polynomial of degree 2 at x.\n\nsyms x\nlegendreP(2,x)\nans =\n(3*x^2)/2 - 1/2\n\nIf you do not specify a numerical value for the degree n, the legendreP function cannot find the explicit form of the polynomial and returns the function call.\n\nsyms n\nlegendreP(n,x)\nans =\nlegendreP(n, x)\n\n### Find Legendre Polynomial with Vector and Matrix Inputs\n\nFind the Legendre polynomials of degrees 1 and 2 by setting n = [1 2].\n\nsyms x\nlegendreP([1 2],x)\nans =\n[ x, (3*x^2)/2 - 1/2]\n\nlegendreP acts element-wise on n to return a vector with two elements.\n\nIf multiple inputs are specified as a vector, matrix, or multidimensional array, the inputs must be the same size. Find the Legendre polynomials where input arguments n and x are matrices.\n\nn = [2 3; 1 2];\nxM = [x^2 11/7; -3.2 -x];\nlegendreP(n,xM)\nans =\n[ (3*x^4)/2 - 1/2,        2519/343]\n[           -16/5, (3*x^2)/2 - 1/2]\n\nlegendreP acts element-wise on n and x to return a matrix of the same size as n and x.\n\n### Differentiate and Find Limits of Legendre Polynomials\n\nUse limit to find the limit of a Legendre polynomial of degree 3 as x tends to -\u221e.\n\nsyms x\nexpr = legendreP(4,x);\nlimit(expr,x,-Inf)\nans =\nInf\n\nUse diff to find the third derivative of the Legendre polynomial of degree 5.\n\nsyms n\nexpr = legendreP(5,x);\ndiff(expr,x,3)\nans =\n(945*x^2)/2 - 105/2\n\n### Find Taylor Series Expansion of Legendre Polynomial\n\nUse taylor to find the Taylor series expansion of the Legendre polynomial of degree 2 at x = 0.\n\nsyms x\nexpr = legendreP(2,x);\ntaylor(expr,x)\nans =\n(3*x^2)/2 - 1/2\n\n### Plot Legendre Polynomials\n\nPlot Legendre polynomials of orders 1 through 4.\n\nsyms x y\nfplot(legendreP(1:4, x))\naxis([-1.5 1.5 -1 1])\ngrid on\n\nylabel('P_n(x)')\ntitle('Legendre polynomials of degrees 1 through 4')\nlegend('1','2','3','4','Location','best')\n\n### Find Roots of Legendre Polynomial\n\nUse vpasolve to find the roots of the Legendre polynomial of degree 7.\n\nsyms x\nroots = vpasolve(legendreP(7,x) == 0)\nroots =\n-0.94910791234275852452618968404785\n-0.74153118559939443986386477328079\n-0.40584515137739716690660641207696\n0\n0.40584515137739716690660641207696\n0.74153118559939443986386477328079\n0.94910791234275852452618968404785\n\n## Input Arguments\n\ncollapse all\n\nDegree of polynomial, specified as a nonnegative number, vector, matrix, multidimensional array, or a symbolic number, vector, matrix, function, or multidimensional array. All elements of nonscalar inputs should be nonnegative integers or symbols.\n\nInput, specified as a number, vector, matrix, multidimensional array, or a symbolic number, vector, matrix, function, or multidimensional array.\n\ncollapse all\n\n### Legendre Polynomial\n\n\u2022 The Legendre polynomials are defined as\n\n$P\\left(n,x\\right)=\\frac{1}{{2}^{n}n!}\\frac{{d}^{n}}{d{x}^{n}}{\\left({x}^{2}-1\\right)}^{n}.$\n\n\u2022 The Legendre polynomials satisfy the recursion formula\n\n$\\begin{array}{l}P\\left(n,x\\right)=\\frac{2n-1}{n}xP\\left(n-1,x\\right)-\\frac{n-1}{n}P\\left(n-2,x\\right),\\\\ \\text{where}\\\\ P\\left(0,x\\right)=1\\\\ P\\left(1,x\\right)=x.\\end{array}$\n\n\u2022 The Legendre polynomials are orthogonal on the interval [-1,1] with respect to the weight function w(x)\u00a0=\u00a01, where\n\n\u2022 The relation with Gegenbauer polynomials G(n,a,x) is\n\n$P\\left(n,x\\right)=G\\left(n,\\frac{1}{2},x\\right).$\n\n\u2022 The relation with Jacobi polynomials P(n,a,b,x) is\n\n$P\\left(n,x\\right)=P\\left(n,0,0,x\\right).$", "date": "2020-12-04 21:34:12", "meta": {"domain": "mathworks.com", "url": "https://fr.mathworks.com/help/symbolic/legendrep.html", "openwebmath_score": 0.937826931476593, "openwebmath_perplexity": 2556.7520994579972, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9865717424942961, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8517977633728488}}
{"url": "https://www.physicsforums.com/threads/meaning-of-a-subspace.884758/", "text": "# Meaning of a Subspace\n\n## Main Question or Discussion Point\n\nHi PF!\n\nI want to make sure I understand the notion of a subspace. Our professor gave an example of one: the set of degree n polynomials is a subspace of continuous functions. This is because a) a polynomial is intrinsically a subset of continuous functions and b) summing any polynomials yields another polynomial (closed under addition) and c) any polynomial multiplied by a constant is still a polynomial (closed under scalar multiplication).\n\nIf my reasoning is correct, then I think the set of discontinuous functions is NOT a subset of all real functions. To see why, consider $f(x) = 1$ everywhere except non-existent at $x=1$. Then take the function $g(x)=1$ when $x=1$ and non-existent everywhere else. Then the sum is clearly continuous. (How would this change though if I changed non-existent to zero? I know the result is the same, but is $g(x)$ as I have defined it even discontinuous--I realize this is a real analysis question.)\n\nRelated Linear and Abstract Algebra News on Phys.org\nPeroK\nHomework Helper\nGold Member\nHi PF!\n\nI want to make sure I understand the notion of a subspace. Our professor gave an example of one: the set of degree n polynomials is a subspace of continuous functions. This is because a) a polynomial is intrinsically a subset of continuous functions and b) summing any polynomials yields another polynomial (closed under addition) and c) any polynomial multiplied by a constant is still a polynomial (closed under scalar multiplication).\n\nIf my reasoning is correct, then I think the set of discontinuous functions is NOT a subset of all real functions. To see why, consider $f(x) = 1$ everywhere except non-existent at $x=1$. Then take the function $g(x)=1$ when $x=1$ and non-existent everywhere else. Then the sum is clearly continuous. (How would this change though if I changed non-existent to zero? I know the result is the same, but is $g(x)$ as I have defined it even discontinuous--I realize this is a real analysis question.)\nYou can't choose functions that are not defined at some point, as functions that have different domains can't be added so they don't form a vector space. The space of all real-valued functions would imply that they all all defined on some fixed domain.\n\nAlso, functions that are not defined at a point are not necessarily discontinuous. They may be continuous on their domain. A good example is the function $1/x$, which is a continuous function on its domain.\n\nInstead, you need to think of two genuinely discontinuous functions that can be added to form a contiuous function.\n\nOr, perhaps you could think of a simpler example using a subset of the polynomials that does not form a subspace?\n\nS.G. Janssens\nI would say: \"The set of polynomials of degree at most $n$\". (Here you regard the zero polynomial as having degree $-\\infty$ or you should stipulate that this set includes the zero polynomial.)", "date": "2020-01-18 22:25:27", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/meaning-of-a-subspace.884758/", "openwebmath_score": 0.8620683550834656, "openwebmath_perplexity": 209.91472519802872, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9783846672373523, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.8517790203919058}}
{"url": "http://math.stackexchange.com/questions/124277/how-to-find-the-partial-sum-of-a-given-series", "text": "# How to find the partial sum of a given series?\n\nOn my last exam there was the question if the series $\\sum_{n=2}^{\\infty}\\frac{1}{(n-1)n(n+1)}$ converges and which limit it has. During the exam and until now, I am not able to solve it. I tried partial fraction decomposition, telescoping sum, etc. But I am not able to find the partial sum formula (Wolfram|Alpha):\n\n$$\\sum_{n=2}^{m}\\frac{1}{(n-1)n(n+1)} = \\frac{m^2+m-2}{4m(m+1)}.$$\n\nCould somebody push me in the right direction? Is there any trick or scheme how to find partial sum formulas for given series?\n\n-\nEach term is less than $1/n^2$, so it converges. You don't need to know what a series converges to to know that it converges. \u2013\u00a0 Thomas Andrews Mar 25 '12 at 14:46\nTwo things to observe(Unrelated to the Math): Please do not sign your posts with signature as faq explicitly lists it! Accepting answers is a sign of appreciation for someone who put in effort compiling an answer for you. Please accept answers which you think helped you a lot in solving that problem or cleared up your concepts and whatever. It is done by clicking on the tick mark besides every answer. \u2013\u00a0 user21436 Mar 25 '12 at 14:51\nThe ratio test with $1/n^3$ proves it converges. As often happens, finding the sum takes more work than that. \u2013\u00a0 Michael Hardy Mar 25 '12 at 16:29\nmartini's answer is the right thing to do. Just wanted to add that if you know the answer (given by wolfram) you can just prove the it with recursion arguments \u2013\u00a0 Thomas Dec 15 '13 at 15:11\n\nSo let's try partial fraction decomposition. Writing $$\\frac 1{(n-1)n(n+1)} = \\frac a{n-1} + \\frac bn + \\frac c{n+1}$$ we obtain $$1 = a(n^2 + n) + b(n^2 - 1) + c(n^2 - n)$$ and therefore \\begin{align*} 1 &= -b\\\\ 0 &= a - c\\\\ 0 &= a + b + c. \\end{align*} This gives $b = -1$, $a = c = \\frac 12$. Hence \\begin{align*} \\sum_{n=2}^m \\frac 1{(n-1)n(n+1)} &= \\sum_{n =2}^m \\frac 1{2(n-1)} - \\sum_{n=2}^m \\frac 1n + \\sum_{n=2}^m \\frac 1{2(n+1)}\\\\ &= \\frac 12 + \\sum_{n=2}^{m-1} \\frac 1{2n} - \\sum_{n=2}^m \\frac 1n + \\sum_{n=3}^m \\frac 1{2n} + \\frac 1{2(m+1)}\\\\ &= \\frac 12 + \\frac 14 - \\frac 12 - \\frac 1m + \\frac 1{2m} + \\frac 1{2m+2}\\\\ &= \\frac 14 + \\frac{-2(m+1) + m+1 + m}{2m(m+1)}\\\\ &= \\frac 14 + \\frac{-1}{2m(m+1)}\\\\ &= \\frac{m(m+1) - 2}{4m(m+1)}. \\end{align*}\nI got little confuse here, why the term $\\frac{1}{4}$ doesnt vanish? In my calculation the third step from below, I got $$=\\frac{1}{2}+ (\\frac{1}{4}+ \\frac{1}{6}+\\dots+ \\frac{1}{2(m-1)})-( \\frac{1}{2}+ \\frac{1}{3}+\\dots+ \\frac{1}{m})+( \\frac{1}{6}+ \\frac{1}{8}+\\dots+ \\frac{1}{2m})+ \\frac{1}{2(m+1)} = \\\\ =-( \\frac{1}{3}+ \\frac{1}{5}+\\dots+ \\frac{1}{m})+(\\frac{1}{6}+\\frac{1}{8}+\\dots+ \\frac{1}{2m})+\\frac{1}{2(m+1)}$$ How can I get to the next step? \u2013\u00a0 DadangAH Feb 20 '14 at 5:41\nWhat did you try for a telescoping sum? Your denominator here is the product of three successive terms (this is often called a rising or falling factorial, depending on which side you take as your baseline); this points to looking at a difference of terms that are of the same form but with denominators one degree less. In particular, looking at $t_n=\\dfrac{1}{n(n+1)}$ then $t_n-t_{n-1}$ $=\\dfrac{1}{n(n+1)}-\\dfrac{1}{(n-1)n}$ $=\\dfrac1n\\left(\\dfrac1{n+1}-\\dfrac1{n-1}\\right)$ $=\\dfrac1n\\left(\\dfrac{(n-1)-(n+1)}{(n-1)(n+1)}\\right)$ $=\\dfrac{-2}{(n-1)n(n+1)}$; in other words, $\\dfrac{1}{(n-1)n(n+1)} = -\\dfrac12(t_n-t_{n-1})$, and from here the telescopy should be fairly clear.", "date": "2015-04-27 21:23:32", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/124277/how-to-find-the-partial-sum-of-a-given-series", "openwebmath_score": 0.9995575547218323, "openwebmath_perplexity": 497.2603487002978, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384668497878, "lm_q2_score": 0.8705972600147106, "lm_q1q2_score": 0.8517790116346535}}
{"url": "https://math.stackexchange.com/questions/2042140/precise-proof-of-s-open-set-iff-contains-only-interior-points-complex-analysis", "text": "Precise proof of S Open set iff contains only interior points - Complex Analysis\n\nIn \"Complex Variables and Applications\" by Brown and Churchill (McGraw-Hill) the proof of: $$\\textit{S is an open set \\implies each of its points is an interior point}$$ is left as an exercise. Here are the important definitions that one can use: $$\\textbf{\\epsilon Neighbourhood}: |z-z_0|<\\epsilon$$ $$\\textbf{z_0 Interior point}: \\exists\\,\\,\\ \\text{an}\\,\\,\\, \\epsilon\\text{-neighbourhood containing only}\\, z\\in S$$ $$\\textbf{z_0 Exterior point}: \\exists\\,\\,\\ \\text{an}\\,\\,\\, \\epsilon\\text{-neighbourhood containing no}\\, z\\in S$$ $$\\textbf{z_0 Boundary point}: \\text{all neighbourhood of z_0 have at least a point in S and one not in S}$$ $$\\textbf{S Open}: \\text{does not contain any of its boundary points}$$\n\nI've tried to prove this in the following way, however I got stuck. Can someone tell me how to make it mathematically rigorous, using the above definitions?\n\n$S$ is an Open Set $\\implies$ $S$ does not contain any any of its boundary points. So $$\\text{if} \\,\\, [\\forall \\epsilon>0, (\\exists z_1\\in S\\,\\,\\ \\text{and}\\,\\,\\ \\exists z_2\\notin S ):(|z_1-z_0|<\\epsilon \\,\\,\\text{and} \\,\\,\\ |z_2-z_0|<\\epsilon)] \\implies z_0\\notin S$$ Hence taking the converse of the above we have $$\\text{if}\\,\\, z_0\\in S \\implies \\exists \\epsilon>0, (\\forall z_1\\in S\\,\\,\\ \\text{or}\\,\\,\\ \\forall z_2\\notin S ):|z_1-z_0|\\geq \\epsilon \\,\\,\\text{or} \\,\\,\\ |z_2-z_0|\\geq \\epsilon)$$\n\nWhich doesn't really tell me much. I am really not sure I've even negated it correctly or that my definition with the quantifiers is correct.\n\n\u2022 This are the exact words in the book \"A point $z_0$ is said to be an interior point of a set $S$ whenever there is some neighborhood of $z_0$ that contains only points of $S$\". Is it different from what I wrote? Or is it different from what you wrote? They look pretty similar to me, please tell me if there's some gap between them \u2013\u00a0Euler_Salter Dec 3 '16 at 18:48\n\u2022 Forgive me, I misunderstood you definition. However it is a bit confusing because it seems that you are referring to a specific point $z$. Specifically, if I say that $A$ contains only $z\\in S$ what I immediately think is that $A=\\{z\\}$, whatever $z$ is, while you only meant $A\\subseteq S$. \u2013\u00a0Caligula Dec 3 '16 at 18:54\n\u2022 I removed the \"logic\" tag - that's for questions within the specific field of mathematical logic. \u2013\u00a0Noah Schweber Dec 3 '16 at 19:00\n\nUsing your definitions (of which some are not very well stated), we can proceed this way.\n\nYou want to prove that each point $z$ of an open set $S$ is an interior point. Now, since $S$ is open, $z$ can't be a boundary point, so negating the definition of boundary point, it follows that there exists a neighborhood $U$ of $z$ such that two things can happen:\n\n\u2022 or $U$ contains only point of $S$;\n\n\u2022 or $U$ contains only point outside $S$.\n\nSince $U$ intersect $S$ at least in $z$, the latter is impossible, so you have proved that $U\\subseteq S$. This means that $z$ is interior in $S$.\n\nFollowing the comment, I'll try to rephrase the proof in more logically strict terms - but note that using words instead of quantifiers is not something to despise, as it increases greatly the clearity of proofs.\n\nThe fact $S$ is open means that\n\n$\\forall\\,z\\in S, z\\notin \\partial S$\n\nwhere $\\partial S$ is the boundary. Now, $z\\in \\partial S$ if and only if\n\n$\\forall \\,U\\in\\mathscr{U}_z,\\,U\\cap S\\neq \\varnothing \\text{ and } U\\cap (S^c)\\neq \\varnothing$\n\nwhere $(-)^c$ is the complement and $\\mathscr{U}_z$ denotes the set of all open neighborhoods for $z$ (it has the structure of a filter if you know what it means). Hence, negating the latter will imply that\n\n$\\exists \\,U\\in \\mathscr{U}_z$ such that $U\\cap S = \\varnothing$ or $U\\cap (S^c)=\\varnothing$\n\nThen you conclude as before: the only possible conclusion is that $U\\cap (S^c)=\\varnothing$ and that is equivalent to $U\\subseteq S$.\n\n\u2022 I like your wordy proof, I'm quite a fan of these, as in most of my homework I exactly do them. However I wanted to train myself in proving things in a \"tedious\" way, i.e. using a logic definition and using either negations of it or contrapositives to get to the statement we want to prove. Would you know how I could rephrase those definitions or actually going from there to the wanted result? Thanks \u2013\u00a0Euler_Salter Dec 3 '16 at 18:58\n\u2022 I edited my answer with some more logic inside it. \u2013\u00a0Caligula Dec 3 '16 at 22:12\n\u2022 Thank you! It makes sense! It's a nice proof, based a lot on sets! I have no idea what a filter is in mathematics, however the set $U_z$ makes sense \u2013\u00a0Euler_Salter Dec 4 '16 at 2:13", "date": "2019-05-23 14:47:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2042140/precise-proof-of-s-open-set-iff-contains-only-interior-points-complex-analysis", "openwebmath_score": 0.8620448708534241, "openwebmath_perplexity": 198.39184553187513, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846666070896, "lm_q2_score": 0.8705972549785201, "lm_q1q2_score": 0.8517790050612067}}
{"url": "http://mathhelpforum.com/calculus/15611-area.html", "text": "1. ## Area\n\nFind c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.\nc =?\n\n2. Originally Posted by qbkr21\nFind c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.\nc =?\nyou are integrating with respect to x, so you cannot integrate the c terms like that. you would just attach x to them. by the way, when i did it your way i got 0 = 230, which means you probably made another mistake somewhere\n\n3. ## Re:\n\nCould you trying working it out? or at least some of it... so that I can get a feel for what you mean...\n\nSure appreciate it....\n\n-qbkr21\n\n4. Originally Posted by qbkr21\nCould you trying working it out? or at least some of it... so that I can get a feel for what you mean...\n\nSure appreciate it....\n\n-qbkr21\n$\\int_{-c}^{c}\\left[ c^2 - x^2 - \\left( x^2 - c^2 \\right) \\right]dx$\n\n$=2 \\int_{0}^{c} \\left( 2c^2 - 2x^2 \\right) dx$\n\n$= 4 \\int_{0}^{c} \\left( c^2 - x^2 \\right)dx$\n\n$= 4 \\left[ c^2 x - \\frac {1}{3}x^3 \\right]_{0}^{c}$\n\n$= \\frac {8}{3}c^3$\n\ncheck my computation, i was in a rush. i have to leave for a little while\n\ndid you see the difference with how i integrated c? i treated it as a constant--which it is\n\n5. ## Re:\n\nYes you were right on the money as always!!\n\nc = 4.41828\n\nThanks so much!!!\n\n-qbkr21\n\n6. ## find c\n\nOriginally Posted by Jhevon\n$\\int_{-c}^{c}\\left[ c^2 - x^2 - \\left( x^2 - c^2 \\right) \\right]dx$\n\n$=2 \\int_{0}^{c} \\left( 2c^2 - 2x^2 \\right) dx$\n\n$= 4 \\int_{0}^{c} \\left( c^2 - x^2 \\right)dx$\n\n$= 4 \\left[ c^2 x - \\frac {1}{3}x^3 \\right]_{0}^{c}$\n\n$= \\frac {8}{3}c^3$\n\ncheck my computation, i was in a rush. i have to leave for a little while\n\ndid you see the difference with how i integrated c? i treated it as a constant--which it is\nI think you are right\n\n7. Originally Posted by curvature\nI think you are right\ni know, qbkr21 already confirmed it. i believe he has the answer to the question in the back of his text. i hope he gets why i could have integrated between 0 and c though and then multiply by two, furthermore, i hope he gets why i would want to do that in the first place\n\n8. Originally Posted by Jhevon\n$\\int_{-c}^{c}\\left[ c^2 - x^2 - \\left( x^2 - c^2 \\right) \\right]dx$\n\n$=2 \\int_{0}^{c} \\left( 2c^2 - 2x^2 \\right) dx$\nhi\nwhat did you do to the 2 outside to make the lower integral -c 0?\ncan it be any constants?\n\nhi\nwhat did you do to the 2 outside to make the lower integral -c 0?\ncan it be any constants?\nNote that your integrand is an even function: f(-x) = f(x). In this case\n$\\int_{-c}^c dx \\, f(x) = \\int_{-c}^0 dx \\, f(x) + \\int_0^c dx \\, f(x)$\n\nUse the substitution y = -x in the first integral:\n\n$= \\int_{c}^0 (-dy) \\, f(-y) + \\int_0^c dx \\, f(x)$\n\n$= -\\left ( -\\int_0^c dy \\, f(-y) \\right ) + \\int_0^c dx \\, f(x)$\n\nBut f(-y) = f(y) when f is an even function:\n\n$= \\int_0^c dy \\, f(y) + \\int_0^c dx \\, f(x)$\n\nand now just replace the dummy variable y in the first integration with x:\n\n$= \\int_0^c dx \\, f(x) + \\int_0^c dx \\, f(x)$\n\n$= 2 \\int_0^c dx \\, f(x)$\n\n-Dan", "date": "2017-10-21 16:27:49", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/15611-area.html", "openwebmath_score": 0.8487441539764404, "openwebmath_perplexity": 799.5774073770895, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682481380669, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8517754757370258}}
{"url": "https://tri7bet.co/e5dmwu/4c23b1-circumference-definition-math", "text": "If the circle has radius $$r$$ then the circumference is $$2\\pi r$$. It gives the ratio of circumference to the diameter of a circle or the ratio of circumference to twice the radius. At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! This section of Revision Maths defines many terms in relation to circles, including: Circumference, Diameter, Radius, Chord, Segment, Tangent, Point of contact, Arc, Angles on major and minor arcs, Angle of Centre and Sectors. Meaning of circumference. There is an error: \u201cArea of the segment = ( \u03b8 /360) x \u03c0 r 2 + ( 1 /2) x sin\u03b8 x r 2\u201d \u2026should be \u201cArea of the segment = ( \u03b8 /360) x \u03c0 r 2 \u2013 ( 1 /2) x sin\u03b8 x r 2\u201d sivaalluri (January 22, 2019 - 3:31 pm) Reply. More About Circumference. 3.14 x 100 yards = 314 yards ; Circumference = 314 yards ; Your walk around the lake is 314 yards. Hollow Cylinder Calculator . I consulted a standard dictionary, which said it was the length of a closed curve enclosing an area, so it would seem to apply under that definition. 9 thoughts on \u201c Circle formulas in math | Area, Circumference, Sector, Chord, Arc of Circle \u201d Steve LeVine (December 31, 2018 - 10:05 pm) Reply. Capsule Calculator . For example, the Latin root of the word circumference is circum, meaning around. The circumference of a circle is the distance around the circle. Related Calculators: Helix Curve Calculator . Definition: Pi is a number - approximately 3.142 It is the circumference of any circle divided by its diameter. Circumference is the linear distance around a circle. If you're seeing this message, it means we're having trouble loading external resources on our website. Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The boundary line of a circle. We will be listening the word \u201cperimeter\u201d frequently in the problems which are based on geometry. The circumference of a circle is the distance around the circle. Learn about terms radius, diameter, circumference, chord, pi, etc.. ... what is the official definition of a circle? What does circumference mean? A circle has many different radii and many different diameters, each passing through the center. Download FREE Right circumference of a Circle Worksheets where, R is the radius of the circle. All Free. Circumference, diameter and radii are measured in linear units, such as inches and centimeters. Perimeter is the distance around a two-dimensional object. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Learn more. By Mary Jane Sterling . The circumference of a circle of radius 3 is $$2\\cdot\\pi\\cdot 3$$, which is $$6\\pi$$, or about 18.85. View our Lesson on Circumference of a Circle. The circumference of a circle is the distance around it, but if, as in many elementary treatments, distance is defined in terms of straight lines, this cannot be used as a definition. Radius definition, a straight line extending from the center of a circle or sphere to the circumference or surface: The radius of a circle is half the diameter. The complete distance around a circle or a closed curve is called its Circumference. Math Goodies Glossary. What is circumference? Learn and know what is the perimeter meaning in geometry chapter. See more. These are the two important words for which we have to know the meanings in geometry. Below are our grade 5 geometry worksheets on determining the circumference of circles.Students are provided the radius or the diameter in customary units (worksheets 1-3) or metric units (worksheets 4-6). The Circumference (or) perimeter of a circle = 2\u03c0R. Circumference: The circumference of a circle is the distance around it. If you imagine the circle as a piece of string, it is the length of the string. * By signing up, you agree to receive useful information and to our privacy policy. Definition: Circumference. What does that mean? Top Calculators. Formally, a circumference is defined as the locus of points from the plane equidistant to another point, called the center of the circumference. Circumference to diameter. You can also measure the perimeter of three-dimensional objects like houses, stadiums, buildings and similar shapes. It includes unlimited math lessons on number counting, addition, subtraction etc. The distance around a circle is called the circumference. Courses. Math Made Easy! The length of such a line. Also find the definition and meaning for various math words from this math dictionary. Now you will be able to easily solve problems on circumference definition, circumference geometry definition, c ircumference of a circle formula, circumference equation, and circumference of circle. Definition of circumference in the Definitions.net dictionary. c or circ. Math; Trigonometry; Radius, Diameter, Circumference, and Area of Circles; Radius, Diameter, Circumference, and Area of Circles. What is a circle? the length of a perimeter. While I would never actually refer to the circumference of a circle as its \"perimeter,\" a discussion recently arose about whether the term \"perimeter\" can even APPLY to a circle. Learn area of different shapes here. Circle Geometry math for kids. Jan 31, 2013 - go to this link for a math dictionary tool to use at school or at home (Circumference - the distance around the edge of a circle) A real-life example of a radius is the spoke of a bicycle wheel. Learn the relationship between the radius, diameter, and circumference of a circle. Worksheets > Math > Grade 5 > Geometry > Circumference of circles. Along with the perimeter, we should also know the meaning of \u201cArea\u201d. M Another name for circumference is perimeter. Again, Pi (\u03c0) is a special mathematical constant; it is the ratio of circumference to diameter of any circle. Circumference definition, the outer boundary, especially of a circular area; perimeter: the circumference of a circle. b. Abbr. Definition Of Circumference. perimeter: [noun] the boundary of a closed plane figure. Related Topics: Math Worksheets Objective: I know how to calculate the circumference of a circle. For any given circle, this ratio will remain \u03c0. A circle is a geometric figure that needs only two parts to identify it and classify it: its center (or middle) and its radius (the distance from the center to any point on the circle). The margin or area surrounding something. circumference definition: 1. the line surrounding a circular space, or the length of this line: 2. the outside edge of an\u2026. circumference - WordReference English dictionary, questions, discussion and forums. E-Mail Address * Featured \u2026 Search for courses, skills, and videos. Learn what is circumference. Area and Perimeter is an important topic in Mathematics, which is used in everyday life. If you're seeing this message, it means we're having trouble loading external resources on our website. Learn the relationship between the radius, diameter, and circumference of a circle. Search. Circumference. 2. a. How to use circumference in a sentence. Did You Know? Circumference definition is - the perimeter of a circle. The boundary line of an area or object. ence (s\u0259r-k\u016dm\u2032f\u0259r-\u0259ns) n. 1. Circumference may also refer to the circle itself, that is, the locus corresponding to the edge of a disk Circle. A circle is a shape that is made up of all the points on a plane (a flat surface) that are the same distance from a given point. Perimeter defines the distance of the boundary of the shape whereas area explains the region occupied by it. Area and perimeter, in Maths, are the two important properties of two-dimensional figures. Circum is now considered a prefix also meaning around or round about. 3. The formula for the circumference of a circle, C = 2 \u00d7 \u03c0 \u00d7 r = \u03c0 \u00d7 d, where r is the radius and d is the diameter of the circle.\u03c0 =22/7 = 3.141. Donate Login Sign up. A circle is a two-dimensional shape that is created by drawing a line that is curved so that its ends meet and every point on the line is the same distance from the center. \u03c0 is the mathematical constant with an approximate (up to two decimal points) value of 3.14. Search . The number Pi, denoted by the Greek letter \u03c0 - pronounced 'pie', is one of the most common constants in all of mathematics. Learn More at mathantics.comVisit http://www.mathantics.com for more Free math videos and additional subscription based content! Search form. where C = \u03c0 D. C is the circumference of the circle Sign Up For Our FREE Newsletter! Definition of Perimeter explained with real life illustrated examples. Circumference = 3.14 x 100 yards ; Do the Math. It is the circumference of any circle, divided by its diameter. Make your child a Math Thinker, the Cuemath way. Sign Up For Our FREE Newsletter! It is a constant so whatever be the circumference and the radius the ratio will be equal to \u03c0. \u03c0 is an irrational number hence it cannot be written down as a finite decimal number. Tip. You have probably noticed that, since diameter is twice the radius, the proportion between the circumference and the diameter is equal to \u03c0: C/D = 2\u03c0R / 2R = \u03c0. Okay, so let's start out with the \"given point\". See more. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. The formula is \u03c0d or 2\u03c0r. Sal uses formulas and a specific example to see how area and circumference relate. Learn the circumference of a circle with Definition, Solved examples, and Facts. Circumference: The distance around a circle; the perimeter of a circle. Read the lesson on circumference of circle if you need to learn how to calculate the circumference of a circle. The following is a list of definitions relating to the circumference of a circle. About Cuemath. Information and translations of circumference in the most comprehensive dictionary definitions resource on the web. Circle worksheets: Finding the circumference. It is used in many areas, such as physics and mathematics. If you have trouble remembering geometry terms, it helps to think of other words from the same root with which you may be more familiar. We must never confuse the concept of a circle with the concept of circumference, circumference is actually a curve that encloses a circle (the circumference is a curve, the circle is an area.) This proportion (circumference to diameter) is the definition of the constant pi. Main content. Perimeter Definition; Perimeter of a Triangle; Perimeter of a Square or Rhombus; Perimeter of a Rectangle or Parallelogram; Perimeter of an N-gon; Perimeter Definition. Calculators and Converters \u21b3 Math Dictionary \u21b3 C \u21b3 Circumference ; Ask a Question . Home \u00bb Glossary \u00bb Term \u00bb Math Goodies Glossary.\n\nDublin To Westport, Jason Pierre-paul Hand Video, Venom Vs Superman, Marco Reus Fifa 13, Purple Tier Covid, Easyjet Flights From Luton To Isle Of Man, South Mayo Family Research Centre, Stainless Steel Meaning In Kannada, What Channel Is The Washington Redskins Game On Today, Venom Vs Superman, Elon Women's Soccer Id Camp 2020, St Vincent Dental Clinic Cleveland Ohio, Blue Anodized Ar-15 Grip,", "date": "2021-04-21 04:31:22", "meta": {"domain": "tri7bet.co", "url": "https://tri7bet.co/e5dmwu/4c23b1-circumference-definition-math", "openwebmath_score": 0.6719912886619568, "openwebmath_perplexity": 1044.1904283378854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9886682488058381, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.8517754745549179}}
{"url": "http://mathhelpforum.com/calculus/111456-quick-question-about-position-due-gravity.html", "text": "# Math Help - Quick question about position due to gravity.\n\n1. ## Quick question about position due to gravity.\n\nI have a problem where I'm given a bunch of info. about a ballon dropping a bomb, etc. The thing is that I am not given any functions but I know that I have to use the function for position of falling bodies with respect to time.\nIt's been a while since I saw this formula, so i'm not sure if this is right:\n\nIn feet: $s(t)=-16t^2 + v_0 t + s_0$\nwhere $v_0$ is initial velocity and $s_0$ is starting position.\n\nIs any of this right?\n\n2. I'm not sure what you are trying to do.\n\nYou have a bomb floating to the earth on a balloon, and you have to construct the equation which describes the displacement the balloon as fallen from its original position after time t?\n\n3. Originally Posted by Arturo_026\nI have a problem where I'm given a bunch of info. about a ballon dropping a bomb, etc. The thing is that I am not given any functions but I know that I have to use the function for position of falling bodies with respect to time.\nIt's been a while since I saw this formula, so i'm not sure if this is right:\n\nIn feet: $s(t)=-16t^2 + v_0 t + s_0$\nwhere $v_0$ is initial velocity and $s_0$ is starting position.\n\nIs any of this right?\nNone of what you're asking can be answered without knowing the question. It's always better to post the original question instead of hoping that our crystal balls are in good working order.\n\n4. Try one (or more) of the kinematic equations:\n\n$v = v_0 + at$\n\n$y - y_0 = v_0t + \\frac{1}{2}t^2$\n\n$v^2 = v_0^2 + 2a(y - y_0)$\n\n$y - y_0 = \\frac{1}{2}(v_0 + v)t$\n\nWhere v is the final velocity, $v_0$ is the initial velocity, y is the final height, $y_0$ is the initial height, a is the acceleration due to gravity near Earth's surface ( $g = 9.8 \\frac{m}{s^2}$), and t is the time.\n\nPatrick\n\n5. Originally Posted by mr fantastic\nNone of what you're asking can be answered without knowing the question. It's always better to post the original question instead of hoping that our crystal balls are in good working order.\nOk, here is the problem:\n\nA bomb is dropped from a ballon hovering at an altitude of 800 ft. From directly below the balloon, a projectile is fired directly upward toward the bomb exactly 2 seconds after the bomb is released. With what initial speed should the projectile be fired in order to hit the bomb at an altitude of exactly 400 ft.?\n\n6. The problem is for some reason in the Standard system (feet) instead of the Metric (meters), so be sure to consider that when solving the problem (This is where the -16 comes from in your original equation).\n\nThe bomb accelerates downward at a rate of 9.8 $\\frac{m}{s^2}$ or just over 32 $\\frac{ft}{s^2}$. After t seconds, it will be halfway down, from 800 ft to 400ft. At this precise moment, the projectile from the surface will intercept it. Therefore, you need to solve for t, the time it takes the bomb to drop 400 ft, from an initial velocity of zero.\n\nThen, when you have t, you must subtract 2 seconds to get the time for the projectile's flight from the surface to the bomb, over a distance of 400 ft. When you have this time, you can then use the kinematic equations to solve for the initial velocity, knowing that the acceleration due to gravity will slow the projectile so that the final velocity is less than the initial velocity.\n\nYou will have the variables of t, a, and y. Using the second equation listed (a is not needed), you can solve for the initial velocity. Let me know what the answer is.\n\nPatrick\n\n7. I get t=5 for the bomb to be at an altitude of 400 ft. and when t=3 the projectile needs to be launched with an initial velocity of 181.33 ft. per second.", "date": "2015-05-23 13:48:18", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/111456-quick-question-about-position-due-gravity.html", "openwebmath_score": 0.8240126967430115, "openwebmath_perplexity": 279.96072984209405, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682441314385, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.8517754740425808}}
{"url": "https://physics.stackexchange.com/questions/652716/two-masses-connected-by-a-spring", "text": "# Two masses connected by a spring\n\nI have a question about a problem I saw on a website\n\nA mass is attached to one end of a spring, and the other end of the spring is attached to an immovable wall. The system oscillates with period T. If the wall is removed and replaced with a second mass identical to the first, what is the new period of oscillation of the system?\n\nMy answer: $$\\frac{T}{\\sqrt{2}}$$\n\nWebsite's answer (Incorrect thanks to trula): $$T\\sqrt{2}$$\n\nMy thought process (pretty hand-wavy I know):\n\nBecause the spring constant is inversely proportional (I think) to length of the spring, the spring constant will be doubled if you halve the length of the spring, which is what is essentially happening with two identical massess. Because $$T=2\\pi\\sqrt{\\frac{m}{k}}$$, if you double $$k$$, you divide $$T$$ by $$\\sqrt{2}$$. I don't know why they got $$T\\sqrt{2}$$, but if I am wrong, please correct my misunderstandings.\n\nThanks for reading if you made it this far, and if my question is unclear, please tell me what I can fix about it.\n\n\u2022 Your argument is right, and the published answer probably just a typing error. Jul 21 at 21:17\n\u2022 Ok, thanks for the clarification!\n\u2013\u00a0JD12\nJul 21 at 21:24\n\u2022 the spring constant will be doubled if you halve the length of the spring, which is what is essentially happening with two identical massess. I really don't see why that would be true. If instead $m\\to 2m$ then $T\\to\\sqrt{2}T$, as stated in the answer key.\n\u2013\u00a0Gert\nJul 22 at 0:08\n\nThe approach you have used appears to be correct. The correct answer should be $$\\frac{T}{\\sqrt 2}$$. As you have noted, halving the length of spring with spring constant $$k$$ results in a doubling of the spring constant.\nSince we know that $$T=2\\pi\\sqrt{\\frac{m}{k}}$$ and if we let $$k\\rightarrow 2k$$, then $$T=2\\pi\\sqrt{\\frac{m}{k}} \\rightarrow 2\\pi\\sqrt{\\frac{m}{2k}}=\\frac{2\\pi}{\\sqrt 2}\\sqrt{\\frac{m}{k}}=\\frac{1}{\\sqrt 2} 2\\pi\\sqrt{\\frac{m}{k}}=\\frac{1}{\\sqrt 2}T$$\nThe answer posted - $$T\\sqrt 2$$ - could very well be a typo as also pointed out by @trula in the comments above.\nEqual masses, centre of mass does not move, stretch $$x$$ relative to centre of mass position for both masses, force exerted by spring $$k\\cdot 2x$$, equation of motion for each of the masses of the form $$-2kx = m\\ddot x$$, period is reduced by factor $$1/\\sqrt 2$$.", "date": "2021-09-17 13:06:59", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/652716/two-masses-connected-by-a-spring", "openwebmath_score": 0.7063555717468262, "openwebmath_perplexity": 194.44732809922567, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682444653242, "lm_q2_score": 0.8615382076534742, "lm_q1q2_score": 0.8517754673005623}}
{"url": "https://math.stackexchange.com/questions/1415237/is-there-a-continuous-function-from-0-1-to-mathbb-r-that-satisfies/1415256", "text": "# Is there a continuous function from $[0,1]$ to $\\mathbb R$ that satisfies\n\nIs there a continuous function $f:[0,1] \\to \\mathbb R$ such that $f(x) = 0$ uncountably often and, for every $x$ such that $f(x) = 0$, in any neighbourhood of $x$ there are $a$ and $b$ such that $f(a) > 0$ and $f(b) < 0$?\n\nLet $K$ be the Cantor set; recall that $K$ is uncountable. Then the function $x\\to d(x,K),$ where $d(x,K)$ denotes the distance from $x$ to $K,$ is continuous on $[0,1].$ Define\n\n$$f(x) = \\begin{cases} 0, x \\in K\\\\ d(x,K)\\sin (1/d(x,K)), x \\in [0,1]\\setminus K. \\end{cases}$$\n\nLet $I_1, I_2, \\dots$ be the open intervals thrown out in the construction of $K.$ In each $I_n,$ $f$ is continuous. Hence $f$ is continuous on $[0,1]\\setminus K.$ If $x_n \\to x \\in K,$ then $d(x_n,K)\\to 0,$ hence so does $f(x_n),$ which implies $f$ is continuous at $x.$ Thus $f$ is continuous on $[0,1].$\n\nNow in any neighborhood of a point in $K,$ one of the $I_n$ will be contained in that neighborhood, hence $f$ will be both positive and negative in that neighborhood because of the $\\sin (1/d(x,K))$ term.\n\nAdded: I just noticed that I omitted discussing points in $[0,1]\\setminus K$ where $f=0.$ That happens at an $x$ such that $d(x,K)=1/(m\\pi)$ for some $m\\in \\mathbb {N}.$ Because each $I_n$ has length $1/3^k$ for some $k,$ such an $x$ could not be the midpoint of any $I_n.$ So then $x$ is to the right or left of the midpoint of whatever $I_n$ it belongs to, and thus $d(\\cdot,K)$ will be strictly increasing or decreasing in a neighborhood of $x,$ which implies a change in the sign of $f$ at that $x$ as desired.\n\n\u2022 Very clever! How'd you come up with that? \u2013\u00a0Moya Aug 30 '15 at 23:24\n\u2022 $x\\sin(1/x)$ and its relatives are the gift that keeps on giving. \u2013\u00a0zhw. Aug 30 '15 at 23:28\n\nIn addition to the carefully constructed examples others have given, I just want to note that with probability $1$ one-dimensional Brownian motion starting at $0$ has these properties.\n\nAbsolutely. For example, we can proceed in a similar fashion to the definition of the Cantor function, using the \"middle third construction of the Cantor set. In the first stage, we remove the \"middle third\" $(\\frac13,\\frac23)$ of $[0,1],$ and on this interval, we define $f(x)=(x-\\frac12)^2+c_1$, where $c_1$ is the unique number such that $f(x)\\to 0$ as $x\\searrow\\frac13$ and $x\\nearrow\\frac23.$ (I leave its determination to you.)\n\nIn the second stage, we remove the two \"middle thirds\" $(\\frac19,\\frac29)$ and $(\\frac79,\\frac89)$. On the first of these, we define $f(x)=-(x-\\frac16)^2+c_2,$ where $c_2$ is the unique number so that $f(x)\\to 0$ as $x\\searrow\\frac19$ and $x\\nearrow\\frac29.$ on the second, we define it by $f(x)=-(x-\\frac56)^2+c_2,$ and we can show that $f(x)\\to 0$ as $x\\searrow\\frac79$ and $x\\nearrow\\frac89.$ (I leave it to you to find $c_2$ and verify this.\n\nWe proceed in much the same manner. At the $n$th stage, we remove $2^{n-1}$ open intervals of the form $\\left(m-\\frac1{2\\cdot3^n},m+\\frac1{2\\cdot3^n}\\right),$ and there is some $c_n$ such that for any such \"middle third\" interval, we define the function on that interval by $f(x)=(-1)^{n-1}(x-m)^2+c_n,$ and we have the property that $f(x)\\to0$ as $x\\searrow m-\\frac1{2\\cdot3^n}$ and $x\\nearrow m+\\frac1{2\\cdot3^n}.$\n\nFinally, we define $f(x)=0$ on the Cantor set, itself. Then $f$ has exactly the properties you need, and is in fact differentiable everywhere but at the endpoints of the \"middle third\" intervals, I believe.\n\nYes.\n\nOur function will be $0$ on the Cantor set and nowhere else. For convenience we define a spike function $S(a, b)(x) = \\mathbb{1}_{[a, b]}(x) \\cdot \\min\\{|a-x|, |b-x|\\}$ which is a continuous, positive, compact support function on the interval $[a, b]$. The idea will be to add a countable collection of spike functions to eventually satisfy your conditions.\n\nSo, lets approximate our function in a series of countable steps. Firstly, let $f_1 = S(1/3, 2/3)$. The greatest distance from a point with positive function value is now $1/3 = \\delta_1$. Let $f_2 = f_1 - (S(1/9, 2/9) + S(7/9, 8/9))$. Now, the greatest distance from a point in $x\\in [0, 1]$ with $f_2(x) = 0$ to a point with $y\\in [0, 1]$ with $f_2(y)<0$ is now $1/9 = \\delta_2$. Proceeding in this way, we would next add $4$ spikes to get $f_3 = f_2 + \\sum_{i = 1, 7, 19, 25} S(i/27, (i+1)/27)$, and $\\delta_3 = 1/27$, subtract $8$ spikes to get $f_4$\n\nI claim that the limit of the $\\{f_n\\}$ exists, is well defined, and satisfies your conditions because the $\\delta_n\\to 0$.\n\nCompleted as a pennance for not reading the question properly.\n\n\u2022 The question doesn't ask for the function to be surjective, just that it takes real values. \u2013\u00a0Mark Bennet Aug 30 '15 at 22:50\n\u2022 +1: Nice fix. That's pretty much the direction I was considering, initially, but I decided to try for more differentiability (for no particular reason). \u2013\u00a0Cameron Buie Aug 30 '15 at 23:41\n\u2022 Its the natural solution, though Robert Israel has a nice example/class of examples too. \u2013\u00a0user24142 Aug 31 '15 at 5:49", "date": "2019-07-23 00:55:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1415237/is-there-a-continuous-function-from-0-1-to-mathbb-r-that-satisfies/1415256", "openwebmath_score": 0.9193094968795776, "openwebmath_perplexity": 120.38020162360674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682478041813, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8517754649048597}}
{"url": "https://math.stackexchange.com/questions/2730961/the-difference-between-inverse-function-and-a-function-that-is-invertible", "text": "# The difference between inverse function and a function that is invertible?\n\nMy discrete mathematics book says:\n\nBut I read an answer, https://math.stackexchange.com/a/2415543/390226, said:\n\n[\"...]There are invertible functions which are not bijective,[...\"]\n\n[\"]A function is invertible if and only if it is injective[.\"]\n\nSo for a function to have a inverse, it must be bijective. But any function that is injective is invertible, as long as such inverse defined on a subset of the codomain of original one, i.e. the image of the original function?\n\n\u2022 See Inverse function : \"To be invertible a function must be both an injection and a surjection. Such functions are called bijections. The inverse of an injection $f : X \u2192 Y$ that is not a bijection, that is, a function that is not a surjection, is only a partial function on $Y$, which means that for some $y \u2208 Y, f^{ \u22121}(y)$ is undefined.\" This id the case with $\\exp$ in the post you have linked. \u2013\u00a0Mauro ALLEGRANZA Apr 10 '18 at 14:10\n\u2022 The issue is simply with $A,B$ of $f : A \\to B$. If we \"build in\" domain and co-domain in the def of $f$ we have that the function $\\exp : \\mathbb R \\to \\mathbb R$ is not invertible. But there is an old tradition of calling $\\log : \\mathbb R^+ \\to \\mathbb R$ its inverse. As per Gentlemen Prefer Blondes: \"nobody is perfect\". \u2013\u00a0Mauro ALLEGRANZA Apr 10 '18 at 14:28\n\u2022 @MauroALLEGRANZA: Thank you so much, now it's clear to me... Because the same book says to define a function, the domain, co-domain, image/range are needed. I didn't know the tradition. \u2013\u00a0linear_combinatori_probabi Apr 10 '18 at 14:33\n\nIt all depends on the co-domain of your function.\n\nWhen you have a function $$f:A\\to B$$ which is one-to-one but not onto $B$, you may restrict your co-domain to a subset of $B'\\subset B$ which is the range of $f$.\n\nFor example $$f:N \\to N$$ defined by $$f(n)=2n$$ is not onto but it is one-to-one.\n\nIf we define, $$f^* : N\\to 2N$$ with the same definition $f^*(n)=2n$\n\nWe have an inverse function, $(f^*)^{-1} (n) = n/2.$\n\n\u2022 So when I see \"f is invertible\", I should connect this idea to \"f must be bijective\"? If f is not bijective, then I must change its co-domain? \u2013\u00a0linear_combinatori_probabi Apr 10 '18 at 14:16\n\u2022 Yes you need a one-to-one correspondence which means bijective. \u2013\u00a0Mohammad Riazi-Kermani Apr 10 '18 at 14:17\n\nWhen people said a function is \"invertible\", they mean it can be made invertible. And the rigorous definition of inverse function of $f$ in my book is:", "date": "2021-06-25 00:44:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2730961/the-difference-between-inverse-function-and-a-function-that-is-invertible", "openwebmath_score": 0.8479015231132507, "openwebmath_perplexity": 346.46248806369863, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561703644736, "lm_q2_score": 0.8791467785920306, "lm_q1q2_score": 0.8517667810949385}}
{"url": "https://math.stackexchange.com/questions/806811/smallest-prime-factor-why-does-this-algorithm-find-prime-numbers", "text": "# Smallest Prime Factor - Why does this algorithm find prime numbers?\n\nI have been looking at the problems on Project Euler and a number of them have required me to be able to find the prime factorisation of a given number.\n\nWhile looking for quick ways to do this, I came across a website that could perform this calculation and had it's source code available too.\n\nThe crux of the algorithm was this method: (I hope java code on this site is OK)\n\npublic static long smallestPrimeFactor( final long n ) {\nif (n==0 || n==1) return n;\nif (n%2==0) return 2;\nif (n%3==0) return 3;\nif (n%5==0) return 5;\n\nfor (int i = 7; (i*i) <= n; i += 30 ) {\nif ( n % i == 0 ) {\nreturn i;\n}\nif ( n % ( i+4 ) == 0) {\nreturn i+4;\n}\nif ( n % ( i+6 ) == 0) {\nreturn i+6;\n}\nif ( n % ( i+10 ) == 0) {\nreturn i+10;\n}\nif ( n % ( i+12 ) == 0) {\nreturn i+12;\n}\nif ( n % ( i+16 ) == 0) {\nreturn i+16;\n}\nif ( n % ( i+22 ) == 0) {\nreturn i+22;\n}\nif ( n % ( i+24 ) == 0) {\nreturn i+24;\n}\n}\nreturn n;\n}\n\n\nThe part that really fascinates me is the loop that starts at 7, then considers some gaps and then increments by 30 before carrying on. This is the list of the first few numbers from the sequence:\n\n  7,  11,  13,  17,  19,  23,  29,  31,\n37,  41,  43,  47,  49,  53,  59,  61,\n67,  71,  73,  77,  79,  83,  89,  91,\n97, 101, 103, 107, 109, 113, 119, 121,\n127, 131, 133, 137, 139, 143, 149, 151,\n157, 161, 163, 167, 169, 173, 179, 181,\n187, 191, 193, 197, 199, 203, 209, 211\n\n\nObviously this loop generates some numbers that are not prime numbers (49, 77, 91 for example) but it does appear to produce every possible prime number less than the square root of the given number.\n\nAm I correct in believing that this loop will produce every prime number? And if so, is there a proof or some mathematical reasoning behind why that is the case?\n\n\u2022 See prime sieves and esp. wheel sieves. \u2013\u00a0Bill Dubuque May 23 '14 at 15:57\n\u2022 The loop looks for divisors without checking if they are primes. It provides all possible prime divisors. The reason to exclude some possible divisors is that they are not prime. For example: $7+30k+1=30k+8$ is even and $2$ was checked, $7+30k+2=30k+9$ is multiple of $3$ (also checked), $7+30k+3=30k+10$ which is a multiple of $10$, $7+30k+5=30k+12$ which is a multiple of $3,$ and so on. \u2013\u00a0mfl May 23 '14 at 16:00\n\nIn the loop $i$ is equal to $30k+7$, where $k$ is a non-negative integer. Let's consider what values can't be prime. I'll list the offsets and remove them as they're eliminated. $$0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29$$\n\nWe start with $7$, which is odd. Adding another odd number would give us an even number, which is divisible by $2$. Eliminate those. $$0,2,4,6,8,10,12,14,16,18,20,22,24,26,28$$\n\nWe're adding multiples of $30$, which are divisible by $3$. The last digit is $7$. Adding $2$ would make it $9$, making the whole number divisible by $3$. So $2$ gets skipped, so well as $2+3k$. Eliminate those. $$0,4,6,10,12,16,18,22,24,28$$\n\nSame logic for $5$. Since the last digit is $7$ without an offset, then adding $3,8,13$, etc, would make it a multiple of $5$. Eliminate those. $$0,4,6,10,12,16,22,24$$\n\nAnd you're left with the numbers that are checked. We increment by $30$ since $2\\cdot3\\cdot5=30$, and we checked those at the very beginning. Adding $30$ produces the same pattern of divisibility by those numbers at the same offsets. So from then on we just need to dodge those offsets.\n\n\u2022 That explanation is super clear! \u2013\u00a0Dan Temple May 23 '14 at 16:21\n\nThis should produce all primes. First, $i=30j+7$ for some $j$. Now with this in mind, take a look at the values that get skipped. Obviously, first thing, it skips all even numbers. As an example though, look at the first odd number it skips, $i+2=30j+9=3(10j+3)$. You will find that all the numbers skipped are similarly multiples of $3$ or $5$. In other words, everything it skips is composite.\n\nThe code you posted in fact is not very efficient. The way it works is this:\n\nThe purpose is ro return the smallest prime factor of number n.\n\nAt first it does checks for trivial/simple cases (whether n is divisable by 2,3,5, the first 3 prime numbers)\n\nThen the loop starts fromt the next prime number (=7) and checks up to sqrt(n) (which is enough, but not the most efficient check for factoring n)\n\nThe checks inside the loop check for square-free factors and not multiples of 2,3,5 (limited Eratosthenes' sieve) (thus primes)", "date": "2019-12-09 11:15:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/806811/smallest-prime-factor-why-does-this-algorithm-find-prime-numbers", "openwebmath_score": 0.40097570419311523, "openwebmath_perplexity": 264.441570591761, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9688561730622296, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8517667711954051}}
{"url": "http://terrysylvester.com/in-hr-eipwkv/4e83f0-bio-bidet-bb-600-troubleshooting", "text": "Thank you! View all posts by Whit Ford, am very thankful 2 the information above.it is very helpful to me. The notation that follows a capital Pi describes only the term that is to be multiplied. A \u201cseries\u201d is the sum of the first N terms of a sequence. Chapters & Colonies . Two of these hybrids from each C atom overlap with H 1s orbitals, while the third overlaps with an sp 2 hybrid on the other C atom. (mathematics) Braid group algebra. and therefore I will always be checking this site. A description of the double bond is the sigma-pi model shown in Figure 1. Search for \"sigma symbol\" in these categories . These symbols can be copy and pasted anywhere you want. All that matters in this case is the difference between the starting and ending term numbers\u2026 that will determine how many twos we are being asked to add, one two for each term number. Yes No. (Physics, scattering) Cross_section_(physics). i=2: (14/12)(8/12)(6/12) The pi bond is the \"second\" bond of the double bonds between the carbon atoms and is shown as an elongated green lobe that extends both above and below the plane of the molecule. Differentiating this would turn the right side into the reciprocal of the original sum times its derivative = a mess. The Sigma symbol, , is a capital letter in the Greek alphabet. So, my opinion would be: sure! I will modify my response shortly. The Pi symbol, , is a capital letter in the Greek alphabet call \u201cPi\u201d, and corresponds to \u201cP\u201d in our alphabet. then there is no need for a notation to represent repeated exponentiation, since exponents that are products already represent repeated exponentiation. Pi (capitale \u03a0, minuscule \u03c0 ou parfois \u03d6 ; en grec \u03c0\u03b9) est la 16 e lettre de l'alphabet grec, pr\u00e9c\u00e9d\u00e9e par omicron et suivie par rh\u00f4. Thanks , Very useful post. Therefore, additional notations are not needed to describe repeated subtraction or division\u2026 Which is quite convenient. Ways to make pi symbol, HTML unicode entities and more. Character Palette allows you to view and use all characters and symbols, including pi, available in all fonts (some examples of fonts are \"Arial\", \"Times New Roman\", \"Webdings\") installed on your computer. Post was not sent - check your email addresses! Choose from 500 different sets of phi sigma rho flashcards on Quizlet. The convention is to increase it, just like with Sigma and Pi notation, but they also support decreasing indeces. I have a student asking whether there is a symbol for exponentiation of a sequence? You can put it in Facebook, Youtube or Instagram. it would appear as though the quantity in parentheses is becoming increasingly negative (a sum of growing negative numbers), and therefore the value probably goes to negative infinity. Sigma Symbol, The Eighteenth Letter in Greek Alphabet \u2013 \u00a9Niakris6 at ShutterStock The sigma sign has been used in mathematics, chemistry, and other fields of study for a very long time. Equation for Xn in terms of P1,P2,\u2026\u2026Pn. Sigma Pi Fraternity, International has chartered over 230 chapters with 123 currently active plus 6 additional colonies in the United States and Canada and is headquartered in Lebanon, Tennessee. In my mind, this rounds up each time the value is divided by (1-r). Summation notation does not provide an easy way that I can think of to do what you describe. I like it \u2026 And I hope it will help other students too to acheive their goals \u2026. Sigma Symbol in Greek Alphabet. sigma and pi? Those are only the first numbers of pi, there's a symbol for pi precisely because you can't precisely specify pi with any decimal number, only to a certain degree of accuracy. Excellent blog and makes maths symbols and operations simple to understand. for i = 0 to 32. please help improve this site's prominence in search results by including a link to this site in appropriate places elsewhere on the internet - perhaps in a response to a math question, or in a comment on a math blog, etc. The \u201csquare root of negative one\u201d is not i ; it is represented by i . Acest articol despre o liter\u0103 greceasc\u0103 este deocamdat\u0103 un ciot. A shielding constant. Interesting question! Configure your keyboard layout in Windows so that you can type all additional symbols you want as easy as any other text. 117. If the index limit above the Pi symbol is a variable, as in the example you gave: I am not aware of such notation, and furthermore, I am not aware of situations where such notation would be needed. If I have interpreted the expression you show correctly, it is neither an arithmetic nor a geometric sequence. To facilitate this, a variable is usually listed below the Sigma with an equal sign between it and the starting term number. Quick Help to Type the Pi symbol. Is there a way to rewrite the following expression using both sigma and pi? Sigma notation provides a compact way to represent many sums, and is used extensively when working with Arithmetic or Geometric Series. Sir,If I have equation like this : \u03a3 \"sigma\" Ce symbole math\u00e9matiques permet d'exprimer plus simplement certaine somme et donc certaines expressions Exemples : Exemple 1 : la somme de 1 \u00e0 100 peut s'\u00e9crire : Exemple 2 : la somme des nombres impairs de 1 \u00e0 13 peut s'\u00e9crire : Un nombre pair est de la forme 2j o\u00f9 j est un entier naturel , 2j + 1 est donc l'\u00e9criture d'un nombre impair. Do you know of situations that require repeating exponentiation to model them? It has been represented by \u2026 Since its inception, the fraternity has initiated more than 100,000 men and has 5,300 undergraduate members. It\u2019s basically a for loop in scripting, makes so much sense. I want to do that to signify that the matrices do not commute. If you can illustrate it please. However, your expression leaves me uncertain as to whether you are analyzing the situation correctly or not. But if you are trying to give a general answer, you should show each term individually so that the person reading your answer can see any pattern that is developing, and understand how to fill in the \u201c\u2026\u201d used to represent all the terms that are not shown. In Gematria it has the value of 200. 1 Translingual. Raccourci clavier pour Pi \u03c0. Alt + 227. You are correct \u2013 this can be represented using a combination of Sigma and Pi notation: In the above notation, i is the index variable for the Sum, and provides the starting number for each product. can there be a bigger value at the base and smaller value at top of the PI operator? Good suggestion \u2013 it makes more sense to keep the expression the same as the previous example! When you don't have time in the morning to piece together the perfect outfit, a t-shirt and jeans will always do the trick. Copy and paste pi symbol or look below to find out how to type pi symbol on keyboard. hello sir, Sigma \u03a3 is one of the most popular mathematic signs which means a summation of something. So for instance, if I wanted to round the above to the nearest whole after each division (or multiplication) step I think I could write: \u2308\u2308\u2308I\u00f7(1-r)\u2309\u00f7(1-r)\u2309\u00f7(1-r)\u2309\u2026. In this tutorial, you\u2019ll learn 4 different ways you can insert the sigma symbol in Word. Our men strive for excellence by living our core values to promote fellowship, develop character and leadership, advance heightened moral awareness, enable academic achievement, and inspire service. You can add the Greek language on any iOS devices. 2) The values grow in magnitude linearly by 2 each time. Comment taper tous les symboles et signes sur votre clavier. The post was tremendously helpful. of 14. Sigma notation is most useful when the \u201cterm number\u201d can be used in some way to calculate each term. See more ideas about sigma pi, sigma, delta. This would be easy to do in a computer program, but not so much using summation notation. Sigma Pi Letter T-Shirt comes with Sigma Pi in 4-inch Greek Twill letters sewn on the front. What I am wondering about is this. thank you for the amazing and very helpful post. multiply until n reaches 143 (i.e. Subtraction can be rewritten as the addition of a negative. Putting the three thoughts above together, I get: using Sigma or Pi notation, or possibly both. Make your mail stand out from the crowd with our wonderful selection of designs made with love, just for you! The difficulty you describe is that you wish to specify what happens to the result of that product, and capital Pi notation does not provide any means to do that. Xi Sigma Pi was originally founded at the University of Washington on November 24, 1908, and is now comprised of over 40 chapters across the USA and Canada. For repeated exponentiation I would assume that form rather than (x^a)^b. Achetez AZSTEEL Sigma Pi T-Shirt 3.14 Symbol Math Science | for Men Women Comfortable Fit Wearable Anywhere, White and Black in Sizes S-3xl livraison gratuite retours gratuits selon \u2026 Raccourci clavier pour Tau \u03c4. Alt + 231. You can input pi symbol using it. I\u2019ll challenge him to find a need for it and maybe he can create his own notation. Hello, I am trying to utilize the Pi notation to represent a repeating multiplication, but one that rounds up to the nearest whole after each time there is a multiplication(or division). A factor of (2n) will produce such numbers, but when n=1 this will have a value of 2, not 3\u2026 so I need to add 1 to each value: . To type the Pi (\u03c0) Symbol anywhere (like in Word or Excel), press Option + Z shortcut for Mac.If you are on Windows, simply press down the Alt key and type 227 using the numeric keypad on the right side of your keyboard.For Microsoft Word, just type 03C0 and then press Alt + X to get the symbol. La manipulation de sommes, via le symbole (sigma), repose sur un petit nombre de r\u00e8gles. Sigma Pi Fraternity was founded in 1897 at Vincennes University, in Vincennes, Indiana. Notation is a convention, a commonly shared interpretation of some symbols. Indeed a very lucid exposition of Sigma and Pi notations! I suppose a problem could be posed this way if you are being asked to come up with an expression for such a product that does not involve Pi notation: is there some closed form expression involving \u201cn\u201d that represents this product? For Phi Sigma Pi, Steven has served as National Secretary/Treasurer, VP of Membership Development and National President. Pi Symbol in Greek Alphabet. Before I continue please forgive my mathematical illiteracy, I am taking an amateur interest in this. You earned the right to be called a Phi Sigma Pi by getting it done in the classroom. Este utilizat\u0103 \u00een matematic \u0103 pentru a desemna o sum\u0103 de elemente. Good point, however, x^a^b is not the same as x^ab. We work hard daily to ensure that you have a diverse and complete selection of Sigma Pi T-Shirts to choose from. 1,375 sigma symbol stock photos, vectors, and illustrations are available royalty-free. The alternative form of sigma (\u03c2) must be used in word-final position. Election is a lifelong membership and includes a once-year complimentary membership in the Society of Physics Students (SPS). If I wanted to represent something being rounded up I think I could use ceiling function brackets: \u2308\u2309. does work? In Javascript you should write like a = \"this \\u2669 symbol\" if you want to include a special symbol in a string. (-1)^n will change sign every time \u201cn\u201d grows by one, but when n=1 it is negative \u2013 which is the wrong sign for the first time. Change\u00a0), You are commenting using your Facebook account. The Sigma Pi fellowships strive for excellence and moral and character development by living by their e\u2026 1.1.1 Usage notes; 1.2 See also; 2 Greek. GREEK CAPITAL LETTER PI \u2190 \u039f [U+039F] Greek and Coptic: \u03a1 \u2192 [U+03A1] Contents. Jan 6, 2017 - Explore Something Greek's board \"Delta Sigma Pi\", followed by 11296 people on Pinterest. Note that \u201ctwo cubed\u201d and \u201csigma pi\u201d are puns \u2014 they depend upon the sound of what they are. Tabs Dropdowns Accordions Side Navigation Top Navigation Modal Boxes Progress Bars Parallax Login Form HTML Includes Google Maps Range Sliders Tooltips Slideshow Filter List Sort List. Sigma Pi has chosen for its mascot, the owl, a ubiquitous symbol of wisdom that well-suits its members devotion to the alma mater and to high academic achievement, but the values of the Fraternity are not for college days alone. The letter is pronounced as \"p\". The table below is a Gem in The Rough. Basically this is where k = n. It\u2019s important to emphasize that. And all of them can produce pi text symbol. So maybe we do still need something? I simply cannot figure out how to represent that using big Pi \u03a0. Once that has been evaluated, you can evaluate the next sigma to the left. Annexe E Liste des symboles math\u00e9matiques usuels (LATEX) Vous trouverez ci-dessous la liste des commandes LATEX permettant de produire les symboles math\u00e9matiques les plus courants. Symbolscopyandpaste.com consists of combination of discrete math symbols and latex math symbols like root symbol, sigma symbol, square root symbol, pi symbol text, under root symbol, infinity text symbol, sigma text symbol, square root of pie, square root of infinity, pi emoji, sigma sign and much more. U\u017eit\u00ed ve v\u011bd\u011b. An online LaTeX editor that's easy to use. This plane contains the six atoms and all of the sigma bonds. Using Pi notation in the exponent achieves the desired purpose. Misconception: many students in the Pacific may have this worng notion that a sigma . But, perhaps I do not understand the situation you seek to describe. If you're using it often you can add a shortcut in Settings \u279c General \u279c Keyboard \u279c Text Replacement. If the sum of a bunch of terms in known as a \u201csummation of a series\u201d, then what is the product of a bunch of terms known as in mathematics? The Stefan\u2013Boltzmann constant. HTML CSS JavaScript Python SQL \u2026 The Steven A. DiGuiseppe Excellence in Administration Award was created in his honor. Then, when you need to type it, hold down \u2026 \u03c0\u213c Pi symbol sign Find out how to type Pi sign \u03c0 directly from your keyboard. LIKE US. Try these curated collections. It can also help you lookup Unicode codes for entering symbols with keyboard. My answer to that would be: I probably would not use Sigma Notation to write such a simple expression. Sizing example: 3 \u2026 Moreover, \u03c0 is a transcendental number \u2013 a number that is not the root of any nonzero polynomial having rational coefficients. GeoGebra Applets However, I have never worked with infinite products. It has been represented by the Greek letter \"\u03c0\" since the mid-18th century. An upper bound would be provided by an infinite geometric sequence, but I am uncertain what might best provide a lower bound. The Sigma and Pi expression I used to answer the previous question did not have a value specified for \u201cN\u201d, so any value given for the expression will have to be in terms of \u201cN\u201d\u2026 as your question is. So like E(x+n) for n=1 to 3 would produce (x+1)^(x+2)^(x+3)\u2026 Or maybe((x+1)^(x+2))^(x+3). Reading this post it seems like this would be easy to use the big Pi \u03a0 notation. I was finding how to use Sigma notation, and finally found such a good one. expression is So, the sum of the first two terms would indeed be 30. This site is valid XHTML 1.0 Strict, CSS | Get Buzz via RSS or Atom | Colophon | LegalXHTML 1.0 Strict, CSS | Get Buzz via RSS or Atom | Colophon | Legal It is a weaker type of covalent bond as compared to sigma bonds (\u03c3 bonds). (\u00a0Log\u00a0Out\u00a0/\u00a0 The Pi symbol, , is a capital letter in the Greek alphabet call \u201cPi\u201d, and corresponds to \u201cP\u201d in our alphabet. No new vocabulary is needed. and if you leave the final index as \u201cn\u201d becomes: Is there some closed form expression that represents this product? For example, X1 means we have One term say P3 and rest two are (1-P) and summation of such product terms for 3 values(P1,P2 and P3). \u03a0 \u03c0 Pi: \u0395 \u03b5 Epsilon: \u03a1 \u03c1 Ro: \u0396 ... \u0370 \u0371 Heta: \u03f7 \u03f8 \u0218o: \u03fa \u03fb San: diacritice: Litera \u03a3 (sigma mare) este o liter\u0103 din alfabetul grec (corespondent pentru litera S \u00een limba rom\u00e2n\u0103). n=112, n=113 etc. Gargi, Loops in programming languages can be written to decrease the index each time just as easily as they can increase it. As n grows, the constant power of 2 in the expression will dominate the initial results a lot more, but the infinite number of subtractions from it will eventually catch up to its value, no matter how large it is. I have modified the post. https://mythologian.net/greek-alphabet-greek-letters-symbols-and-meanings Our Executive Office is located in Nashville, Tennessee. I still cannot think of either an application for such an expression or a notation for it. 3) There are seven terms, so n will need a starting value of 1, and an ending value of 7. One other thought\u2026 if It is used in the same way as the Sigma symbol described above, except that succeeding terms are multiplied instead of added: Sigma (summation) and Pi (product) notation are used in mathematics to indicate repeated addition or multiplication. The coefficients will be \u201c32 Choose i\u201d, or. Since So, if n=3, then It came from the Phoenician letter p\u0113, which meant mouth. X2=(1-P1)P2.P3+(1-P2)P2P3+(1-P3)P2P3 The student in question is actually only 11 years old and somehow I don\u2019t think that he will accept the \u201cnot needed\u201d reason! If it ends with, or continues beyond tan(np/2n), which will always be undefined, then my first impression is that there would be no limit to the product. Sigma Pi Phi (\u03a3\u03a0\u03a6) is the first successful and oldest Black Greek-lettered organization.It has always been non-collegiate, designed for professionals at mid-career or older. In Modern Greek, this sound is voiced to /z/ before /m/, /n/, /v/, /\u00f0/ or /\u0263/. So Sigma notation describes repeated subtraction when its argument is a negative quantity. Three sigma bonds are formed from each carbon atom for a total of six sigma bonds total in the molecule. Pi sign is one of the most popular mathematic constants and means a ratio of a circle perimeter to its diameter. If j went from one to three each time, the expression on the right would have to be (i + j \u2013 1). Atomic Term Symbol (Explained in detail) | All types of problems asked in CSIR - Duration: 23:52. However, this particular example can be \u201csimplified\u201d by collecting like terms to become Section: Internet Tutorial: Greek Letters Fabulous Code Chart for Greek Letters & Symbols \u2026 With more than 140 collegiate chapters on campuses throughout the U.S., Phi Sigma Pi is one of the largest and most active fraternities out there. Thanks for your clear explanations. If I wanted to take, let\u2019s say \u201cI\u201d, and multiply \u201cI\u201d by a repeating multiple, let\u2019s say \u201c1/(1-r)\u201d. 1) your expansion of the problem using square brackets Go to Settings, General, Keyboard, and Keyboard again. How can I type a pi symbol on an iPad? There are several in the posting\u2026 Ooops \u2013 I just realized you were asking about my reply to the comment. Sigma Pi Sigma exists to honor outstanding scholarship in physics, to encourage interest in physics among students at all levels, to promote an attitude of service, and to provide a fellowship of persons who have excelled in physics. Sorry, your blog cannot share posts by email. i=1: (14/12)(8/12) Name Unicode Glyph Unicode Name Description Aliases; alefsym: 02135: ALEF SYMBOL : Alpha: 00391: GREEK CAPITAL LETTER ALPHA : alpha: 003B1: GREEK SMALL LETTER ALPHA How would I derive the polynomial for the following expression: n = 112 which would raise the question: \u201cwhy write it using Sigma Notation when you could just as easily write ?\u201d. A more typical use of Sigma notation will include an integer below the Sigma (the \u201cstarting term number\u201d), and an integer above the Sigma (the \u201cending term number\u201d). To make use of them you will need a \u201cclosed form\u201d expression (one that allows you to describe each term\u2019s value using the term number) that describes all terms in the sum or product (just as you often do when working with sequences and series). Your answer options suggest that there is some expansion of a a logarithm that results in an infinite product of tangent functions, however I am not familiar with that. Thank you, this was very helpful. ), I\u2019m interested in simplifying the polynomial to 32 terms and determine the exponents of y, Using Pi notation, I interpret your question to be, Using a binomial expansion, the terms will be Symbol . Math teacher, substitute teacher, and tutor (along with other avocations) GREEK PI SYMBOL: \u03d7 : 983: 03D7 : GREEK KAI SYMBOL ... GREEK CAPITAL REVERSED DOTTED LUNATE SIGMA SYMBOL Previous Next COLOR PICKER. Pi \u03c0 is an irrational number, which means that it cannot be expressed exactly as a ratio of two integers; consequently, its decimal representation never ends and never repeats. These Sigma Pi fraternity stickers measure Approx 2\" tall Greek Letter Sticker. I was just practicing the question wanted to know can 30\u2026.. n(n+1)(n+2) be the ans to the above sigma and product equation given by you. A \u201cproduct\u201d is the result of multiplying two or more \u201cfactors\u201d. 2.1 Letter; 2.2 See also; Translingual Symbol \u03a0. English Wikipedia has an article on: Capital pi notation. This site uses Akismet to reduce spam. In Gematria it has the value of 200. As is the case with most of the other alphabets, even English for that matter, two different symbols are used to represent sigma in uppercase and lowercase respectively. Furthermore is there a way of simplifying the notation and finding a result that is a function of n? But what if the Pi notation is not in closed form, such as. . I do not follow your thinking though when you say you wish to use a descending index value to indicate that matrices do not commute\u2026 I would not perceive a descending index value, or an ascending one, to indicate anything about the commutative property\u2019s applicability to the resulting expression. (mathematics) Sum of divisors. How can u write this using summation notation: 3 -5+7 -9+11-13+15? You'll have to copy paste pi symbol \u03c0 if you're on iOS or iPad OS. All rights to text and images hosted on this blog site are reserved by the author. what is the relation between the two when they are logarithmic differentiated? You can make frequently used technical non-fancy symbols like \"\u221a \u2211 \u03c0 \u221e \u2206 \u2122 \u00a9 \u00e6 \u00a3 \u00a2\" and \u00e5cc\u00e9nted letters on Mac using [Option] key. You press Alt and, while holding it, type a code on Num Pad while it's turned on. So you should wear lettered apparel that gets it done too. For example, if n=1, then the expression would be: Certificates. and if n=4, then So will provide the correct sign for the nth term. It's a text symbol. There actually are 3 different ways to type symbols on Linux with a keyboard. How should I proceed if I want to get it for n instead of 3. 3d sigma sigma logo sigma icon sigma sigma vector sigma letter 6 steps flow greece graffiti omega letter greek lettering. (\u00a0Log\u00a0Out\u00a0/\u00a0 There is no emoji for pi at the moment. Division can be rewritten as multiplication by the reciprocal. M\u1ed9t b\u1ed9 s\u01b0u t\u1eadp c\u00e1c bi\u1ec3u t\u01b0\u1ee3ng to\u00e1n h\u1ecdc, ch\u1eb3ng h\u1ea1n nh\u01b0: Pi, Infinity, Sum, Sigma, Square Root, Integral ... c\u00e1c d\u1ea5u hi\u1ec7u & k\u00fd hi\u1ec7u. I have a question, please. The letter is pronounced as \"p\". Welcome them to your fraternity house properly with Sigma Pi Signs, Flags and Banners. Pi symbol \u03c0 is important for calculation of circular and spherical figures and stands for value of 3.141592653589793238.. . Actually n SHOULD be squared in his reply since he\u2019s saying that that\u2019s the LAST TERM in the product. Character map allows you to view and use all characters and symbols available in all fonts (some examples of fonts are \"Arial\", \"Times New Roman\", \"Webdings\") installed on your computer. Question. How to find the derivative of the pi notation. (\u00a0Log\u00a0Out\u00a0/\u00a0 hope to receive your reply as soon as possible. Sigma Pi (\u03a3\u03a0) is an international social collegiate fraternity founded in 1897 at Vincennes University. \u2022 In both Ancient and Modern Greek, the sigma represents the voiceless alveolar fricative /s/. We have an awesome collection of mathematical symbols for you. Please, read a guide if you're running a laptop. However, while it is common to see uppercase and lowercase sigma signs in science textbooks and the likes, what many people are completely unaware of is the fact that the sigma sign has a third form as well. The symbol of Phi Beta Sigma Fraternity Incorporation is the White Dove What is the name of our publication? what can be the correct answer this equation? In such cases, just as in the example that resulted in a bunch of twos above, the term being added never changes: The \u201cstarting term number\u201d need not be 1. It corresponds to \u201cS\u201d in our alphabet, and is used in mathematics to describe \u201csummation\u201d,\u00a0the addition or sum\u00a0of a bunch of terms (think of the starting sound of the word \u201csum\u201d: Sssigma = Sssum). i.e. Had always skipped these symbols in technical papers and today is the first time i get to understand what they mean. Pi notation provides a compact way to represent many products. The Fraternity currently has over 285,000 total initiated members. Hi Mr. Ford, See more ideas about Phi sigma pi, Sigma pi, New mexico state university. Does Multiplication operator always increment? Change\u00a0). Cet article a pour objet de les \u00e9num\u00e9rer et d\u2019en donner des exemples d\u2019utilisation, sans aucune pr\u00e9tention \u00e0 l\u2019originalit\u00e9. There is often a pattern to them, a formula that can be used to determine the value of the next term in the sequence. For example, $\\prod_{i = 3}^6 (x + 2^i) = (x + 2^3) (x + 2^4) (x + 2^5) (x + 2^6)$ In general, we have $\\boxed{\\prod_{i = a}^b f(i) = f(a) \\cdot f(a+1) \\cdot f(a+2) \\cdot \u2026 \\cdot f(b)}$ Properties of Pi Notation. So, even if it is not commonly used in a particular way, there is no strong reason I can think of why you couldn\u2019t use it that way (if necessary, including a note or example describing how you intend the notation to be interpreted). You can keep the one you order from Greek Gear in the basement or closet of your chapter house for years of use. Series definitions almost always rely on summation notation. Raccourci clavier pour Theta \u0398. Alt + 233 Thanks\u2026. then there are an indeterminate number of factors in the product until such time as \u201cn\u201d is specified. Minususcule : Alt + 229. Let\u2019s list the first few terms of this sequence individually to get a sense of how this series behaves: If sigma is for summation, and pi is for multiplication, are there any notations for division and subtraction? In this case only two of the p orbitals on each C atom are involved in the formation of hybrids. Learn phi sigma rho with free interactive flashcards. It came from the Phoenician letter p\u0113, which meant mouth. The symbol for uppercase sigma is \u03a3 Ch\u1ec9 c\u1ea7n nh\u1ea5p v\u00e0o m\u1ed9t d\u1ea5u hi\u1ec7u ho\u1eb7c bi\u1ec3u t\u01b0\u1ee3ng \u0111\u1ec3 sao ch\u00e9p n\u00f3 v\u00e0o b\u1ea3ng t\u1ea1m v\u00e0 d\u00e1n n\u00f3 \u1edf b\u1ea5t c\u1ee9 \u0111\u00e2u. Thanks! Understanding math topics without memorization. Which \u201ck\u201d are you referring to? I've compiled a list of shortcuts in my article and explained how to open keyboard viewer. This vector image was created with a text editor. If you need to differentiate a sum, I would not expect logarithmic differentiation to be very useful, as the laws of logarithms do not allow us to do anything with something like Pi Bond are formed by the overlapping of atomic orbital\u2019s sidewise. If I were to see an upper index value that is smaller than the lower one, my first assumption would be that I would need to decrease the index by 1 for each iteration \u2013 which seems to be what you intend. HOW TO. Ooops \u2013 didn\u2019t think of CharMap allows you to view and use all characters and symbols available in all fonts (some examples of fonts are \"Arial\", \"Times New Roman\", \"Webdings\") installed on your computer. Phi Sigma Pi Necklace - Phi Sigma Pi Gift - Big Gift - Little Gift - Gift For Big - Gift For Little - Wholesale Sorority - Sorority Gift FelicityAndBliss. The pi symbol will be shown on the virtual keyboard. The Sigma symbol can be used all by itself to represent a generic sum\u2026 the general idea of a sum, of an unspecified number of unspecified terms: But this is not something that can be evaluated to produce a specific answer, as we have not been told how many terms to include in the sum, nor have we been told how to determine the value of each term. Is represented by the Greek letter Sticker, perhaps I do not understand the situation you seek to describe Greek! The Greek letter  \u03c0 '' since the mid-18th century please, read a guide if you 're iOS. Simply can not Figure out how to find a need for summation notation: 3 -9+11-13+15. Convention is to be a leader for others closet of your Chapter house for years use... \u2026 and I hope it will help other students too to acheive their goals \u2026 and!, which meant mouth will be negative v\u00e0o b\u1ea3ng t\u1ea1m v\u00e0 d\u00e1n n\u00f3 \u1edf b\u1ea5t c\u1ee9 \u0111\u00e2u would that! Usually have no need for it and maybe he can create his own notation at New Mexico University... \u03a0. English Wikipedia has an article on: capital Pi notation provides compact... Can be rewritten as multiplication by the Greek alphabet assume that form rather than x^a... Can I type a code on Num Pad while it 's turned on sense to keep the expression same..., such as a member of the p orbitals on each C atom are involved in the Formation of.... Using a backwards sigma symbol list of HTML and JavaScript entities for Pi at the moment we grant this... \\U2669 symbol '' if you could provide an example of what you commenting! Also be considered as math symbols can also be used in word-final position deocamdat\u0103 un ciot sum... Sp Sp2 Sp3, Organic Chemistry, bonding - Duration: 36:31 Arithmetic... Please reply in my mind, this sound is voiced to /z/ before /m/ /n/. I \u201d, or possibly both should wear lettered apparel that gets it done in product. X^A ) ^b, version control, hundreds of LaTeX templates, and found... The exponent achieves the desired purpose initiated members in terms of a circle to! Este deocamdat\u0103 un ciot Alt + 228 an ending value of 1, and finally found such good. Matematic \u0103 pentru a desemna o sum\u0103 de elemente a single \u201c term.... V\u00e0o b\u1ea3ng t\u1ea1m v\u00e0 d\u00e1n n\u00f3 \u1edf b\u1ea5t c\u1ee9 \u0111\u00e2u sigma and Pi is an social! Do with his investigation into combination formulae\u2026 he \u2019 s important to emphasize that s important emphasize... To my list atom for a while scholarship, leadership, and Pi notations (... \u201c series \u201d is the relation between the two when they are I was finding how to interpret notation! Finding how to type sum symbol with keyboard using this technique that has been evaluated, are... ) 1 ( b ) 2-log2 ( C ) 3 -log4 total of six sigma bonds also decreasing... What is the sigma-pi model shown in Figure 1 sigma Rho flashcards Quizlet... Six sigma bonds are formed by the Greek alphabet standard deviation symbol are used interchangeably for this character sigma pi symbol sur. = 112 expression is multiply until n reaches 143 ( i.e sequence \u201d is not ;! Appeared at that time ) \u2026\u2026 } ^ ( 1/n ) on Desktops and Laptops! To open keyboard viewer it done in the molecule symbol '' in these categories and subtraction JavaScript should. 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A transcendental number \u2013 a number that is a convention, a variable is usually listed the! Which are not needed to describe repeated subtraction or division\u2026 which is quite convenient alternative of... V\u011bdn\u00edch oborech ust\u00e1len\u00fd v\u00fdznam, v n\u011bm\u017e se pou\u017e\u00edv\u00e1 sigma pi symbol ho\u1eb7c bi\u1ec3u t\u01b0\u1ee3ng \u0111\u1ec3 sao ch\u00e9p n\u00f3 b\u1ea3ng... Latex editor that 's easy to use it, how about expressing thing one +... And an ending value of 1, and is followed sigma pi symbol Tau just like sigma... The other iPhone languages iOS devices very useful terms, so n will a... \u03a0 directly from your keyboard using this technique Pi bonds, sp Sp2 Sp3, Organic Chemistry bonding... Could use ceiling function brackets: \u2308\u2309 expression: n = 112 is. Solution of following infinite series will appretiated desemna o sum\u0103 de elemente honor fraternity with chapters throughout the.! Not use sigma notation get: using sigma or Pi notation orbital s! Summation, and more sp Sp2 Sp3, Organic Chemistry, bonding - Duration: 36:31 ( \u03c2 must! \u279c text Replacement the first n terms of a negative the entries posted in WaSP Buzz express the of. Values of I will not skip them anymore when I come across them in papers could provide an way... I am not aware of such notation, and keyboard again \u2013 I just realized you asking. Can not share posts by email Physics ) membership and includes a sigma pi symbol. Be provided by an angle of 120\u00b0 selection of designs made with love just. Represent something being rounded up I think I could use ceiling function brackets \u2308\u2309... Used in word-final position variable start at zero, the fraternity currently has over 285,000 total initiated.... Such a good question for a total of six sigma bonds are formed by the letter. And yes, a commonly shared interpretation of some symbols carbon atom for a total of six sigma bonds formed. What you need to type sum symbol with keyboard using different techniques depending on your system,... The addition of a sequence viewer as an alternative to my list voiced! A code on Num Pad while it 's turned on, your expression me! That \u201c two cubed \u201d and \u201c sigma Pi 's colors and symbols an icon Log. On any iOS devices you \u2019 ll challenge him to find out how to type symbol... Type of covalent bond as compared to sigma bonds are formed, separated by an angle 120\u00b0! Decreasing indeces a boss them lasting for a while understand the situation you seek to describe and has 5,300 members! Add New keyboard and find Greek in the Society of Physics students ( SPS ) my to! A list of shortcuts in my mind, this sound is voiced to /z/ before,! Was created with a text editor 5,300 undergraduate members angle of 120\u00b0, separated by an sigma pi symbol geometric.! Sp Sp2 Sp3, Organic Chemistry, bonding - Duration: 36:31 or geometric series University was established 1943. Out to clipboard automatically \u201d is the first n terms of a sequence 143... Has 5,300 undergraduate members be shown on the front bonds are formed separated... Linearly by 2 each time just as easily as they can increase it make the of... Collegiate fraternity founded in 1897 at Vincennes University, in Vincennes, Indiana \u201c two cubed and! Polynomial ( such as, I am uncertain what might best provide a lower bound is voiced /z/... And all of the Modern Greek alphabet value is sigma pi symbol capital Pi describes only the that. Makes so much using summation notation once that has been evaluated, can! Type Pi sign \u03c0 directly from your keyboard layout in Windows so that you can add a shortcut in \u279c! An iPad from your keyboard viewer of any nonzero polynomial having rational coefficients ) there several! Can evaluate the next sigma to the comment the many ways that it can be written to decrease the each! One you order from Greek Gear in the Pacific may have this worng that... Hundreds of LaTeX templates, and keyboard again the polynomial for the solution of following infinite series will.! Formed from each carbon atom for a forum like http: //math.stackexchange.com/questions an nor. And maybe he can create his own notation of your Chapter house for years of use 3p/2n \u2026\u2026! \u2026.Will it be a bigger value at the base and smaller value at top of the sigma,. - Phi sigma Pi is dedicated to supporting its members in scholarship leadership. Find a need for it and maybe he can create his own notation find need! The name of our publication summation of something ust\u00e1len\u00fd v\u00fdznam, v n\u011bm\u017e se pou\u017e\u00edv\u00e1 les et. Our site, you can keep the one you order from Greek Gear in Rough! Voiced to /z/ before /m/, /n/, /v/, /\u00f0/ or /\u0263/ \u03a1 \u2192 U+03A1! Images hosted on this blog site are reserved by the author this blog and receive notifications of New posts email. Shared interpretation of some symbols atomic orbital \u2019 s saying that that \u2019 s basically for... It, hold down \u2026 Pi symbol, HTML Unicode entities and more I knew. This \\u2669 symbol '' if you could provide an easy way that I can think of do. Its members in scholarship, leadership, and more Phoenician letter also gave rise to \u2026 Pi. Calculation of circular and spherical figures and stands for value of 7 ways you can insert the sigma the... Very very useful which meant mouth convention is to increase it oborech ust\u00e1len\u00fd v\u00fdznam, v n\u011bm\u017e se pou\u017e\u00edv\u00e1 Greek!\n\nByron Bay Beach Resort, Call Of Duty: Strike Team Apk Obb 2020, Gta 5 Anime Body Pillow Car, New Orleans Homes For Sale Garden District, New Look High Waisted Trousers, National Arts Club Brunch, Engine Control Module Autozone, United Pentecostal Church Preachers,", "date": "2021-05-17 12:57:24", "meta": {"domain": "terrysylvester.com", "url": "http://terrysylvester.com/in-hr-eipwkv/4e83f0-bio-bidet-bb-600-troubleshooting", "openwebmath_score": 0.7027240991592407, "openwebmath_perplexity": 1894.3436320270612, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9273632976542184, "lm_q2_score": 0.9184802462567087, "lm_q1q2_score": 0.8517648699988799}}
{"url": "https://math.stackexchange.com/questions/508791/number-of-lists-in-which-an-element-is-repeated-consecutively-exactly-twice", "text": "# Number of lists in which an element is repeated consecutively exactly twice\n\nI have an integer list that is n long and each value can be ranging from 1 .. n.\n\nI need a formula that tells me how many of all possible lists for a given n, that have one or more consecutive sequences of a length of exactly 2 of the same number and no other consecutive sequences that are longer than 2.\n\nFor example for n=5:\n\nThese two should count:\n\n{ 1, 1, 5, 3, 3 }\n{ 2, 3, 2, 5, 5 }\n\n\nWhere as these should not:\n\n{ 1, 1, 1, 2, 2 }\n{ 1, 3, 2, 5, 4 }\n\n\nI've been attempting to do this by looking at the possible sequences using the following formula where x = n-1:\n\n(n) x n n = x * n^3\nx (n) x n = x^2 * n^2\nn x (n) x = x^2 * n^2\nn n x (n) = x * n^3\n\n\nAnd sum these four up.\n\nHowever, these also needs to take overlaps between the four into account. This is where I could use a bit of help..? What would the formulas be for excluding the overlaps?\n\nAn alternative approach would also be welcome.\n\nGoing trough all sequences counting manually is not an option - I need this to work for for n larger than what makes that approach computationally feasible.\n\nIf it helps anyone, I've written a little program that runs trough all the sequences and counts have the following results:\n\nn = 2\nL[2]: 2\nL[1]: 2\n\nn = 3\nL[3]: 3\nL[2]: 12\nL[1]: 12\n\nn = 4\nL[4]: 4\nL[3]: 24\nL[2]: 120\nL[1]: 108\n\nn = 5\nL[5]: 5\nL[4]: 40\nL[3]: 280\nL[2]: 1520\nL[1]: 1280\n\n\nWhere n = 5, L[2]: 1520 is the result I've asked for a formula to in the above question.\n\n\u2022 You are using the word permutation to mean something different from what the rest of us mean when we use that word. \u2013\u00a0Gerry Myerson Sep 29 '13 at 12:59\n\u2022 I rephrased it a bit - I hope it makes better sense. \u2013\u00a0Kasper Middelboe Petersen Sep 29 '13 at 13:10\n\u2022 Are ${ 2, 3, 2, 5, 5 }$ and ${ 3, 2, 2, 5, 5 }$ different sequences that both count? \u2013\u00a0miracle173 Sep 29 '13 at 13:37\n\u2022 @miracle173 yes \u2013\u00a0Kasper Middelboe Petersen Sep 29 '13 at 13:40\n\u2022 Does $1, 2, 3, 1$ count? Ie, can the doubled elements \"wrap around\"? \u2013\u00a0Jack M Sep 29 '13 at 13:58\n\nHow many sequences of length $q$ of numbers from $\\{1,...,p\\}$ are there such that consecutive elements are always different? For the first element of such a sequence we can selcet one of the $p$ different values of $\\{1,...,p\\}$. For the following $q-1$ positions we can select always all values from $\\{1,...,p\\}$ except the value of its predecessor in the sequence. These are $p-1$ possible values. So we have\n\n$$p(p-1)^{q-1}$$\n\ndifferent sequences.\n\n$\\Omega_n$ is the number of sequences of length $n$ with elements from $\\{1,\\ldots,n\\}$ such that no three consecutive elements have the same value but at least one pair of consecutive elements have the same value.\n\n$\\Omega_{n,k}$ is the number of sequences of length $n$ with elements from $\\{1,\\ldots,n\\}$ such that no three consecutive elements have the same value but exactly $k$ pairs of consecutive elements have the same value.\n\nFor arbitrary $n$ there are $k \\le \\frac{n}{2}$ different possible values for the number of consecutive element pairs that have equal values.\n\nWe have\n\n$$|\\Omega_{n}|=\\sum_{k=1}^{\\lfloor \\frac{n}{2}\\rfloor }|\\Omega_{n,k}|$$\n\nNow we select a $k$. Choose a sequence of $n-k$ values from $\\{1,\\ldots,n\\}$ such that two consecutive values always differ. There are $n (n-1)^{n-k-1}$ such sequences. For such a sequence we select $k$ of its elements (there are $\\binom{n-k}{k}$ such possibilities) and insert an element with the same value after each selected value. So $$|\\Omega_{n,k}|=\\binom{n-k}{k}n(n-1)^{n-k-1}$$ and $$|\\Omega_{n}|=\\sum_{k=1}^{\\lfloor \\frac{n}{2}\\rfloor }\\binom{n-k}{k}n(n-1)^{n-k-1}$$\n\n\u2022 For $n=3$ there are $12$ sequences.\n\n\u2022 For $n=4$ there are t $120$ sequences.\n\n\u2022 For $n=5$ there are $1520$ sequences.\n\nThe answer to the original question is deleted\n\n\u2022 @KasperMiddelboePetersen The formula would be $n\\sum_{k=1}^{\\lfloor \\frac{n}{2}\\rfloor }\\frac{(n-k)(n-k-1)\\ldots(n-2k+1)}{k!}(n-1)^{n-k-1}$. If you are doing more combinatorics problems, it would be a worth-while investment of time to study binomial coefficients, just a bit. \u2013\u00a0Marc van Leeuwen Sep 29 '13 at 20:27\n\u2022 @MarcvanLeeuwen thanks - I'd just figured it out and deleted the comment 30 seconds before you answered :) It will be a challenge to calculate this for a large value of n though. \u2013\u00a0Kasper Middelboe Petersen Sep 29 '13 at 20:32\n\u2022 I think the only challenge is computing with large integers. For $n=69$, I get 4676495092345476699420593108779018542679373291685221536304769813013106968507850431624264673416798415845073262231682605593395200, no sweat. \u2013\u00a0Marc van Leeuwen Sep 29 '13 at 20:43\n\nHere is a different solution that may interest you. Introduce three sequences $a_{n,k}$, $b_{n,k}$ and $c_{n,k}$ that count the number of strings over $\\Sigma^k$ where $|\\Sigma|=n,$ that end in a digit that is not repeated, a digit that is repeated twice and a digit that is repeated at least three times. In fact we take these sequences to be generating functions in two variables, where the variable $v$ counts occurrences of subsequences of length exactly two and $w$ counts occurrences of subsequences of length at least three.\n\nThis gives the following set of recurrences: $$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v b_{n,k-1} + (n-1) c_{n,k-1},$$ $$b_{n,k} = a_{n,k-1} \\quad\\text{and}\\quad c_{n,k} = w b_{n,k-1} + c_{n,k-1}.$$\n\nWith these settings the generating function of all elements of $\\Sigma^k$ classified according to the number of length 2 and length more than two subsequences is given by $$a_{n,k} + v b_{n,k} + w c_{n,k}.$$\n\nObserve that $a_{n,1} = 1$ and $b_{n,1} = c_{n,1} = 0.$ Furthermore the last recurrence implies that $$c_{n,k} = w \\sum_{q=1}^{k-1} b_{n,q} = w \\sum_{q=1}^{k-2} a_{n,q}.$$\n\nTaken together this yields a recurrence for $a_{n,k}:$ $$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v a_{n,k-2} + (n-1) w \\sum_{q=1}^{k-3} a_{n,q}.$$\n\nIntroduce the trivariate generating function $$G_n(z) = \\sum_{k\\ge 1} a_{n,k} z^k.$$\n\nMultiply the recurrence by $z^k$ and sum for $k\\ge 4:$ $$\\sum_{k\\ge 4} a_{n,k} z^k = (n-1) z \\sum_{k\\ge 4} a_{n,k-1} z^{k-1} + (n-1) vz^2 \\sum_{k\\ge 4} a_{n,k-2} z^{k-2} \\\\+ \\sum_{k\\ge 4} (n-1) w z^k [z^{k-3}] \\frac{1}{1-z} G_n(z).$$ Now $a_{n,1} = n, \\; a_{n,2} = n(n-1)$ and $$a_{n,3} = n(n-1)^2 + n(n-1)v.$$ The equation derived from the recurrence now becomes $$G_n(z) - (n(n-1)^2 + n(n-1)v)z^3 - n(n-1)z^2 -nz \\\\= (n-1)z (G_n(z) - n(n-1)z^2 -nz) + (n-1)vz^2 (G_n(z) - nz) \\\\ + (n-1)w z^3 \\sum_{k\\ge 4} z^{k-3} [z^{k-3}] \\frac{1}{1-z} G_n(z).$$ The sum term simplifies to $$(n-1)w z^3 \\frac{1}{1-z} G_n(z).$$ We may now solve for $G_n(z),$ getting $$G_n(z) = -{\\frac {nz \\left( -1+z \\right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{ z}^{2}-v{z}^{3}-w{z}^{3}n+w{z}^{3}+1}}.$$ Now the generating function for $b_{n,k}$ is $$z G_n(z) \\quad\\text{and the one for}\\; c_{n,k} \\; \\text{is}\\quad \\frac{wz^2}{1-z} G_n(z).$$ It follows that the generating function $H_n(z)$ of $a_{n,k} + v b_{n,k} + w c_{n,k}$ is $$\\left(1 + vz + \\frac{w z^2}{1-z}\\right) G_n(z)$$ or equivalently $$H_n(z) = -{\\frac {zn \\left( -1+z-vz+v{z}^{2}-w{z}^{2} \\right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{z}^{2}-v{z}^{3}-w{ z}^{3}n+w{z}^{3}+1}}.$$ Now we do not permit sequences of length at least three, so we take $$[w^0] H_n(z) = -{\\frac { \\left( vz+1 \\right) nz}{v{z}^{2}n-v{z}^{2}+nz-z-1}}.$$ In fact there must be at least one two-sequence, so we first calculate $$[w^0] H_n(z) - [v^0] [w^0] H_n(z) = {\\frac {v{z}^{2}n}{ \\left( nz-z-1 \\right) \\left( v{z}^{2}n-v{z}^{2}+nz-z-1 \\right) }}$$ and put $v=1$, finally obtaining the generating function $$M_n(z) = {\\frac {n{z}^{2}}{ \\left( nz-z-1 \\right) \\left( n{z}^{2}-{z}^{2}+nz-z-1 \\right) }}.$$ The partial fraction decomposition of $M_n(z)$ is given by $$-{\\frac {n}{ \\left( n-1 \\right) \\left( n{z}^{2}-{z}^{2}+nz-z-1 \\right) }}+{ \\frac {n}{ \\left( n-1 \\right) \\left( nz-z-1 \\right) }}.$$\n\nNow the singularities are at $$\\rho_0 = \\frac{1}{n-1} \\quad\\text{and}\\quad \\rho_{1,2} = -\\frac{1}{2} \\pm \\frac{\\sqrt{n^2+2n-3}}{2(n-1)}.$$ Expanding $M_n(z)$ into series we obtain the following closed form result for the number of admissible strings of length $k$ and an alphabet of $n$ symbols: $$[z^k] M_n(z) = n\\frac{\\rho_0^2 \\rho_1 \\rho_2}{(\\rho_0-\\rho_1)(\\rho_0-\\rho_2)} \\rho_0^{-k}\\\\ + n\\frac{\\rho_0\\rho_1^2 \\rho_2}{(\\rho_1-\\rho_0)(\\rho_1-\\rho_2)} \\rho_1^{-k} + n\\frac{\\rho_0\\rho_1 \\rho_2^2}{(\\rho_2-\\rho_0)(\\rho_2-\\rho_1)} \\rho_2^{-k}.$$ The reader is asked to verify that indeed $$[z^n] M_n(z) = \\sum_{k=1}^{\\lfloor n/2 \\rfloor} {n-k\\choose k} n (n - 1)^{n-k-1}.$$ The above formula for $[z^k] M_n(z)$ is quite powerful. It gives the exact value of the number of admissible strings with any alphabet of size $n$ and of length $k$. For example, when there are $n=2$ symbols, the sequence starting at length $k=1$ is: $$0, 2, 4, 8, 14, 24, 40, 66, 108, 176.$$ With $n=7$ we get the following sequence: $$0, 7, 84, 798, 6804, 54684, 423360, 3194856, 23668848, 172939536.$$\n\nThe generating function $H_n(z)$ encapsulates the complete distribution of the $n^k$ strings classified according to the number of two-sequences (counted by $v$) and sequences of length at least three (counted by $w$).\n\nHere is an example: $$[z^4] H_3(z) = 6\\,{v}^{2}+36\\,v+15\\,w+24$$ for a total of $3^4=81$ terms (strings of length $4$ over $3$ symbols). Another example is: $$[z^6] H_5(z) = 80\\,{v}^{3}+1920\\,{v}^{2}+520\\,vw+20\\,{w}^{2}+6400\\,v+1565\\,w+5120$$ for a total of $5^6=15625$ terms (strings of length $6$ over $5$ symbols). It is a useful combinatorics excercise to verify some of these values with pen and paper. For example, a string of length four over three symbols containing a sequence of length at least three can consist of four equal symbols, giving a contribution of three, or a length three sequence at the front for a contribution of three times two or a length three sequence at the end, again for three times two, for a total of $3+6+6 = 15$ which is indeed the value from the generating function.", "date": "2019-08-22 13:52:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/508791/number-of-lists-in-which-an-element-is-repeated-consecutively-exactly-twice", "openwebmath_score": 0.7548503875732422, "openwebmath_perplexity": 264.7836395677748, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877726405082, "lm_q2_score": 0.8774767954920547, "lm_q1q2_score": 0.8517559961599133}}
{"url": "https://math.stackexchange.com/questions/328071/assigning-values-to-permutations", "text": "# Assigning values to permutations\n\n$N$ objects can be arranged in $N!$ different orders. For example, $10$ playing cards can be stacked $10! = 3,628,800$ different ways. Is there a way to assign a numerical value to each permutation so that every integer from $1$ to $N!$ corresponds to exactly one permutation? Is there a way to derive the permutation from the corresponding numerical value?\n\n\u2022 Well, there are many ways: just ennumerate arbitrarily the permutations from $1$ to $\\,N!\\,$ Mar 12 '13 at 3:05\n\u2022 This is discussed in detail in Higher-Order Perl, section 4.3.1, pp. 128\u2013135, which is available online. It is also discussed in exhaustive detail in volum IV of Knuth's The Art of Computer programming.\n\u2013\u00a0MJD\nMar 12 '13 at 3:41\n\u2022 Could you not number each card and then name permutations according to the order? Mar 12 '13 at 4:59\n\nI don't know if there is a standard way of doing this but you could do something like this:\n\nLets take 4 playing cards b/c it is more managable. Order them some way, say label them a,b,c,d and put them in the order (a,b,c,d). Then\n\n$$1:(a,b,c,d)\\\\ 2:(a,b,d,c)\\\\ 3:(a,c,b,d)\\\\ 4:(a,c,d,b)\\\\ 5:(a,d,b,c)\\\\ 6:(a,d,c,b)$$ So fixing \"a\" as the first entry gives 6 possible permutations, repeating this for b,c,d in the first entry will give you the other 18, for a total of 24. The method is to first fix an ordering and then permute only the last 2, then once thats done permute the last 3 and so on.\n\n\u2022 No help so far. Mar 12 '13 at 15:47\n\u2022 Im not sure what your asking then, you asked if there was a way to assign a number to each permutation. The answer is yes, and here is one way of doing it systematically.\n\u2013\u00a0mv3\nMar 12 '13 at 16:03\n\u2022 And if you know the original ordering of the objects, (a,b,c,d) then you can reconstruct the permutation from the value\n\u2013\u00a0mv3\nMar 12 '13 at 16:08\n\nYes. The easiest way is to order them lexicographically. So for $\\{0,1,2,3,4\\}$ there are $120$ permutations, from $01234$ to $43210$. It is easiest if our permutation numbers run from $0$ to $119$. Of these $4!=24$ have each number first, so if you want permuation $n$, the first number is $a_0=\\lfloor \\frac n{24} \\rfloor$. Then of those, there are $3!=6$ that have each of the remaining numbers first. To find it, compute $a_1=\\lfloor \\frac {n-24a_0}6 \\rfloor$, then increment by $1$ if $a_0 \\le a_1$ because you want the $a_1$st of what is left. Now recurse.\n\n\u2022 I'm looking for a function. Suppose ten cards are in the order Ace-2-3-4-5-6-7-8-9-10 when arranged in permutation #1. What would permutation #1,000,000 look like? Suppose the cards are in the order 5-9-2-4-10-Ace-3-8-6-7. Which permutation # (an integer between 1 and 3,628,800) would that be? Maybe what I'm asking just isn't possible, but it would be useful to me if it can be done. Mar 12 '13 at 16:09\n\u2022 @LeroyNimka: It is quite possible. For permutation 1,000,000 the first card is $\\lfloor\\frac {999,999}{9!}\\rfloor+1=3$ where the -1 and +1 come because you are counting from 1. The ones starting with Ace and 2 use up $2\\cdot 9!=725760$ so now we want permutation 274240 of the rest. $\\lfloor\\frac {274239}{8!}\\rfloor+1=7$ so we want the seventh card, the $8$. That has used up $6 \\cdot 8!=241920$, so we need the $32320$ of the rest, the seventh one, the 9. So we have started with 389. This can be written in a few tens of lines with a loop. Mar 12 '13 at 16:27\n\u2022 @LeroyNimka: for the other direction, the 5 leads off starting with permutation $4 \\cdot 9!+1=1451521. Having the 9 next adds$7 \\cdot 8!=282240. The 2 next adds $7!=5040$, so we are up to 1738801. Again, keep going, and again a few tens of lines. Mar 12 '13 at 16:30\n\nNotice that for one and the same letter kept, out of $$n$$, in one position, you have $$(n-1)!$$ possible choices. And that is all that you need.\n\n$$4231, n=4$$\n\nBefore we reached $$4$$ we had $$(4-1)(n-1)!$$ permutations. With $$4$$ fixed we had $$(2-1)(n-2)!$$ options. With $$42$$ fixed we had $$(3-1-1)(n-3)!$$ options as $$2$$ is preceding $$3$$ and is to the left of it. With $$423$$ fixed we had $$(1-1)(n-4)!$$ options. It is $$18$$th permutation.\n\nFormula is then:\n\n$$P(a_1a_2...a_m)=\\sum_{k=1}^{m}(\\alpha_r(a_k)-1)(m-k)!$$\n\nwhere $$\\alpha_r(a_k)$$ is $$1$$-based alphabetical position of symbol $$a_k$$ reduced by the number of symbols that appear earlier in the alphabet and appear to the left of it.\n\nInverse is more or less obvious. Essentially you write N, the position, in factorial positional system, and read the permutation from the result.\n\n$$18=3110_!$$\n\nWe start\n\n$$1234$$ $$3110$$\n\nPick the one at position $$3$$, remove the position handled and the element ($$4$$).\n\n$$4$$ $$123$$ $$110$$\n\nPick the next one at position $$1$$, remove the position handled and the element ($$2$$).\n\n$$42$$ $$13$$ $$10$$\n\nPick the next one at position $$1$$, remove the position handled and the element ($$3$$).\n\n$$423$$ $$1$$ $$0$$\n\nFinally take the last element to have $$4231$$ as expected.\n\n(You can read the order directly. Start from the left and use order in the alphabet, if you find the same position twice, just increment it. Therefore\n\n$$3110_!$$ is $$3, 31, 311=312, 3120 \\to 4231$$\n\n$$1210_!$$ would be $$1,12,121=122=123,1230 \\to 2341$$)", "date": "2021-09-28 05:18:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/328071/assigning-values-to-permutations", "openwebmath_score": 0.7215588092803955, "openwebmath_perplexity": 273.8530953934899, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877760323724, "lm_q2_score": 0.8774767906859264, "lm_q1q2_score": 0.8517559944709455}}
{"url": "http://stats.stackexchange.com/questions/30842/what-is-the-probability-that-this-person-is-female/30850", "text": "# What is the probability that this person is female?\n\nThere is a person behind a curtain - I do not know whether the person is female or male.\n\nI know the person has long hair, and that 90% of all people with long hair are female\n\nI know the person has a rare blood type AX3, and that 80% of all people with this blood type are female.\n\nWhat is the probability the person is female?\n\nNOTE: this original formulation has been expanded with two further assumptions: 1. Blood type and hair length are independent 2. The ratio male:female in the population at large is 50:50\n\n(The specific scenario here is not so relevant - rather, I have an urgent project that requires I get my mind around the correct approach for answering this. My gut feel is that it's a question of simple probability, with a simple definitive answer, rather than something with multiple debatable answers according to different statistical theories.)\n\n-\nThere are not multiple theories of probability, but it is notoriously true that people have difficulties thinking correctly about probabilities. (Augustus DeMorgan, a good mathematician, gave up the study of probability due to its difficulties.) Don't look at debates: look for appeals to principles of probability (such as the Kolmogorov axioms). Don't let this be resolved democratically: your question is attracting many ill-conceived answers which, even if some of them happen to agree, are merely collectively wrong. @Michael C gives good guidance; my reply tries to show you why he's right. \u2013\u00a0 whuber Jun 22 '12 at 3:46\n@Whuber, if independence is assumed, would you agree that 0.97297 is the correct answer? (I believe that the answer might be anywhere between 0% and 100% without this assumption - your diagrams show this nicely). \u2013\u00a0 ProbablyWrong Jun 22 '12 at 4:53\nIndependence of what, precisely? Are you suggesting that female and male hairstyles are the same? As you say in your question, this particular scenario involving gender/hair/blood type may not be relevant: that tells me you seek to understand how to solve problems like this in general. To do that you will need to know which assumptions imply which conclusions. Thus you need to focus very carefully on the assumptions you are willing to make and determine exactly how much they allow you to conclude. \u2013\u00a0 whuber Jun 22 '12 at 13:07\nThe kind of independence to explore concerns the combination of all three characteristics. E.g., if AX3 is a marker for a syndrome that includes baldness in females (but not in males), then any long-haired person with AX3 is necessarily male, making the probability of being female 0%, not 97.3%. I hope this makes it obvious that anybody producing a definite answer to this question must be making additional assumptions, even if they do not explicitly acknowledge them. The truly useful answers, IMHO, would be those that show directly how different assumptions lead to different results. \u2013\u00a0 whuber Jun 22 '12 at 14:31\nYou're missing the probability that a female doesn't have long hair. That's a critical measure. \u2013\u00a0 Daniel R Hicks Jul 3 '12 at 19:54\n\nMany people find it helpful to think in terms of a \"population,\" subgroups within it, and proportions (rather than probabilities). This lends itself to visual reasoning.\n\nI will explain the figures in detail, but the intention is that a quick comparison of the two figures should immediately and convincingly indicate how and why no specific answer to the question can be given. A slightly longer examination will suggest what additional information would be useful for determining an answer or at least obtaining bounds on the answers.\n\n### Legend\n\nCross-hatching: female / Solid background: male.\n\nTop: long-haired / Bottom: short-haired.\n\nRight (and colored): AX3 / Left (uncolored): non-AX3.\n\n### Data\n\nTop cross-hatching is 90% of the top rectangle (\"90% of all people with long hair are female\").\n\nTotal cross-hatching in the right colored rectangle is 80% of that rectangle (\"80% of all people with this blood type are female.\")\n\n### Explanation\n\nThis diagram shows schematically how the population (of all females and non-females under consideration) can simultaneously be partitioned into females/non-females, AX3/non-AX3, and long haired/non-long haired (\"short\"). It uses area, at least approximately, to represent proportions (there's some exaggeration to make the picture clearer).\n\nIt is evident that these three binary classifications create eight possible groups. Each group appears here.\n\nThe information given states that the upper cross-hatched rectangle (long-haired females) comprises 90% of the upper rectangle (all long-haired people). It also states that the combined cross-hatched parts of the colored rectangles (long-haired females with AX3 and short-haired females with AX3) comprise 80% of the colored region at the right (all people with AX3). We are told that someone lies in the upper right corner (arrow): long-haired people with AX3. What proportion of this rectangle is cross-hatched (female)?\n\nI have also (implicitly) assumed that blood type and hair length are independent: the proportion of the upper rectangle (long hair) that is colored (AX3) equals the proportion of the lower rectangle (short hair) that is colored (AX3). That's what independence means. It is a fair and natural assumption to make when addressing such questions like this, but of course it needs to be stated.\n\nThe position of the upper cross-hatched rectangle (long-haired females)is unknown. We can imagine sliding the top cross-hatched rectangle side-to-side and sliding the bottom cross-hatched rectangle side-to-side and possibly changing its width. If we do this so that 80% of the colored rectangle remains cross-hatched, such an alteration will change none of the stated information, yet it can alter the proportion of females in the upper right rectangle. Evidently the proportion could be anywhere between 0% and 100% and still be consistent with the information given, as in this image:\n\nOne strength of this method is it establishes the existence of multiple answers to the question. One could translate all this algebraically and, by means of stipulating probabilities, offer specific situations as possible examples, but then the question would arise whether such examples are really consistent with the data. For instance, if someone were to suggest that perhaps 50% of long-haired people are AX3, at the outset it is not evident that this is even possible given all the information available. These (Venn) diagrams of the population and its subgroups make such things clear.\n\n-\nWhuber, assuming that blood type and hair length are independent, then surely the portion of long haired women with type AX3 should be the same as the portion of short haired women with AX3? I.e. you don't have flexibility to shift rectangles in the way you propose... If we assume also that men and women are 50:50 in the whole population, doesn't that give us enough info to solve this question with a single indisputable answer? \u2013\u00a0 ProbablyWrong Jun 21 '12 at 6:40\n@whuber +1 very nice. \u2013\u00a0 Michael Chernick Jun 21 '12 at 10:42\nProbablyWrong, take a close look at the question in your comment: because it deals with women, it is making an additional assumption about independence conditional on gender. The assumption of (unconditional) independence of hair and blood type does not mention gender at all, so to understand what it means, erase the cross-hatching from the figures. This, I hope, indicates why we have the flexibility to situate the cross-hatching wherever we like within the upper and lower rectangles. \u2013\u00a0 whuber Jun 21 '12 at 14:46\n@whuber, I like this. However, I have 2 questions / clarifications: 1. the figures seem to assume population proportions for long vs short hair (about 6:4) & ~AX3 vs AX3 (about 85:15), but this is not mentioned in the original question nor discussed in your explanations of the figures. I suspect the pop proportions are not relevant. Am I right / could you clarify that in the explanations? 2. I think this situation is ultimately working w/ the same phenomenon as Simpson's Paradox, only framed differently (coming at the issue from the other direction, as it were). Is that a fair assessment? \u2013\u00a0 gung Jun 27 '12 at 16:25\n@gung, thank you for making those clarifications. The figures of course must represent some proportions in order to work at all, but any proportions not specifically pinned down in the problem statement are free to vary. (I did construct the figure so that about 50% of the population appears female, anticipating a later edit in which this was assumed.) The idea of applying this graphical representation to understanding Simpson's Paradox is intriguing; I think it has merit. \u2013\u00a0 whuber Jun 27 '12 at 16:34\n\nThis is a question of conditional probability. You know that the person has long hair and blood type Ax3 . Let$$\\ \\ \\ \\ \\ A =\\{\\text{'The person has long hair'}\\}\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ B = \\{\\text{'The person has blood type Ax3'}\\} \\\\ C =\\{\\text{'The person is female'}\\}.$$\nSo you seek $P(C|A\\ \\text{and}\\ B)$. You know that $P(C|A)=0.9$ and $P(C|B)=0.8$.\nIs that enough to calculate $P(C|A\\ \\text{and}\\ B)$? Suppose $P(A\\ \\text{and}\\ B\\ \\text{and}\\ C)=0.7$. Then $$P(C|A\\ \\text{and}\\ B)=P(A\\ \\text{and}\\ B\\ \\text{and}\\ C)/ P(A\\ \\text{and}\\ B)=0.7/P(A\\ \\text{and}\\ B).$$ Suppose $P(A\\ \\text{and}\\ B)=0.8$. Then, by the above, $P(C|A\\ \\text{and}\\ B)=0.875$. On the other hand if $P(A\\ \\text{and}\\ B)=0.9$ we would then have $P(C|A\\ \\text{and}\\ B)$=0.78.\n\nNow both are possible when $P(C|A)=0.9$ and $P(C|B)=0.8$. So we can't tell for sure what $P(C|A\\ \\text{and}\\ B)$ is.\n\n-\nHi Michael, If I read you correctly, you're saying the question as posed can't be answered, is that right? Or to put it another way, you'd need more information to answer this question? 1. Let's assume that the rare blood type in my original question doesn't have any impact on a person's desire or ability to grow their hair long. Can the question now be answered? 2. Would you agree that the answer must be GREATER than 0.9? (Because you have a second piece of independent information - blood type - that reinforces the hypothesis that the person is a female) \u2013\u00a0 ProbablyWrong Jun 21 '12 at 3:30\nIf $P(A\\text{ and }B)$ is independant, then $P(A\\text{ and }B)=P(A)P(B)$ and you'll need to specify what fraction of persons have long hair, i.e., $P(A)$ and what fraction of persons have blood type Ax3, i.e., $P(B)$. Also, you can't say that the answer must be greather than 0.9, which is equivalent to stating that $P(C|A\\text{ and }B)>0.9$ (I really don't see why). \u2013\u00a0 N\u00e9stor Jun 21 '12 at 7:39\nThank you Nestor. \u2013\u00a0 Michael Chernick Jun 21 '12 at 10:00\n@ProbablyWrong. Yes the problem as initially stated has insufficient information for a unique answer. \u2013\u00a0 Michael Chernick Jun 21 '12 at 10:04\nWhy does $$P(C|A\\ \\text{and}\\ B)=P(A\\ \\text{and}\\ B\\ \\text{and}\\ C)\\times P(A\\ \\text{and}\\ B)??$$ I thought that $$P(C|A\\cap B)=\\frac{P(C \\cap (A \\cap B))}{P(A \\cap B)}=\\frac{P(A\\cap B\\cap C)}{P(A\\cap B)}$$ using the definition of conditional probability. \u2013\u00a0 Dilip Sarwate Jun 21 '12 at 11:13\n\nFascinating discussion ! I am wondering if we specified P(A) and P(B) as well whether the ranges of P(C| A,B) will not be much narrower than the full interval [0,1], simply because of the many constraints we have.\n\nSticking to the notation introduced above:\n\nA = the event that the person has long hair\n\nB = the event that the person has blood type AX3\n\nC = the event that person is female\n\nP(C|A) = 0.9\n\nP(C|B) = 0.8\n\nP(C) = 0.5 (i.e. let's assume an equal ratio of men and women in the population at large)\n\nit does not seem possible to assume that events A and B are conditionally independent given C ! That leads directly to a contradiction: if $P(A \\wedge B | C) = P(A| C) \\cdot P(B| C) = P(C| A) \\frac{P(A)}{P(C)} \\cdot P(C| B) \\frac{P(B)}{P(C)}$\n\nthen\n\n$P(C| A \\wedge B ) = P(A \\wedge B | C) \\cdot \\left( \\frac{P(C)}{P(A \\wedge B)} \\right) = P(C| A) \\frac{P(A)}{P(C)} \\cdot P(C| B) \\frac{P(B)}{P(C)} \\cdot \\left( \\frac{P(C)}{P(A \\wedge B)} \\right)$\n\nIf we now assume that A and B are independent as well: $P(A \\wedge B) = P(A) P(B)$ most terms cancel and we end up with\n\n$P(C| A \\wedge B ) = \\frac{P(C| A) \\cdot P(C| B)}{P(C)} = \\frac{0.9 \\cdot 0.8}{0.5} > 1$\n\nFollowing up on whuber's wonderful geometric representation of the problem: While it is true that generally speaking $P(C | A \\wedge B)$ can assume any value in the interval $[0,1]$ the geometric constraints do narrow the range of possible values significantly for values of $P(A)$ and $P(B)$ that are not \"too small\". (Though we can also upper bound the marginals: $P(A)$ and $P(B)$)\n\nLet us compute the {\\bf smallest possible value} for $P(C | A \\wedge B)$ under the following geometric constraints:\n\n1. The fraction of the upper area (A TRUE) covered by the upper rectangle must be equal to $P(C|A)=0.9$\n\n2. The sum of the areas of the two rectangles must be equal to $P(C)=0.5$\n\n3. The sum of the fraction of the areas of the two colored rectangles (i.e. their overlap with event B) must be equal to $P(C|B)=0.8$\n\n4. (trivial) The upper rectangle cannot be moved beyond the left boundary and should not be moved beyond its minimum overlap to the left.\n\n5. (trivial) The lower rectangle cannot be moved beyond the right boundary and should not be moved beyond its maximum overlap to the right.\n\nThese constraints limit how freely we can slide the hashed rectangles and in turn generate lower bounds for $P(C | A \\wedge B)$. The figure below (created with this R script ) shows two examples\n\nRunning through a range of possible values for P(A) and P(B) (R script) generates this graph\n\nIn conclusion, we can lower bound the conditional probability P(c|A,B) for given P(A), P(B)\n\n-\nMarkus, the first paragraph belongs as a separate question rather than within an answer. The subsequent material looks like a good observation but it is hard to follow without being told what $A, B,$ and $C$ represent. Please bear in mind that different users will see the answers in different sequences, according to their preferences and when the answers were last edited, so each answer has to be readable independently of the others (although of course you can link to other answers). \u2013\u00a0 whuber Jul 3 '12 at 14:34\n@whuber: thanks for the useful comment ! I hope the new edits make it more readable and clear. \u2013\u00a0 Markus Loecher Jul 3 '12 at 18:36\n@whuber and others: I had hoped to reignite the discussion but the thread seems to have gone inactive ? No more comments by anyone ? \u2013\u00a0 Markus Loecher Jul 8 '12 at 20:36\n\nMake the hypotheses is that the person behind a curtain is a woman.\n\nWe area given 2 pieces of evidence, namely:\n\nEvidence 1: We know the person has long hair (and we're told that 90% of all people with long hair are female)\n\nEvidence 2: We know the person has a rare blood type AX3 (and we're told that 80% of all people with this blood type are female)\n\nGiven just Evidence 1, we can state that the person behind a curtain has a 0.9 probability value of being a woman (assuming 50:50 split between men and women).\n\nRegarding the question posed earlier in the thread, namely \"Would you agree that the answer must be GREATER than 0.9?\", without doing any Math, I would say intuitively, the answer must be \"yes\" (it is GREATER than 0.9). The logic is that Evidence 2 is supporting evidence (again, assuming a 50:50 split for the number of men and women in the world). If we were told that 50% of all people with AX3 type blood were female, then Evidence 2 would be neutral and have no bearing. But since we're told that 80% of all people with this blood type are female, Evidence 2 is supporting evidence and logically should push the final probability of a woman above 0.9.\n\nTo calculate a specific probability, we can apply Bayes' rule for Evidence 1 and then use Bayesian updating to apply Evidence 2 to the new hypothesis.\n\nSuppose:\n\nA = the event that the person has long hair\n\nB = the event that the person has blood type AX3\n\nC = the event that person is female (assume 50%)\n\nApplying Bayes rule to Evidence 1:\n\nP(C|A) = (P(A|C) * P(C)) / P(A)\n\nIn this case, again if we assume 50:50 split between men and women:\n\nP(A) = (0.5 * 0.9) + (0.5 * 0.1) = 0.5\n\nSo, P(C|A) = (0.9 * 0.5) / 0.5 = 0.9 (Not surprising, but it would be different if we didn't have 50:50 split between men and women)\n\nUsing Bayesian updating to apply Evidence 2 and plugging in 0.9 as the new prior probability, we have:\n\nP(C|A AND B) = (P(B|C) * 0.9) / P(E)\n\nHere, P(E) is the probability of Evidence 2, given the hypotheses that the person already has a 90% chance of being female.\n\nP(E) = (0.9 * 0.8) + (0.1 * 0.2) [this is law of total probability: (P(woman)*P(AX3|woman) + P(man)*P(AX3|man)] So, P(E) = 0.74\n\nSo, P(C|A AND B) = (0.8 * 0.9) / 0.74 = 0.97297\n\n-\nThere are a few statements in your answer that do not make sense to me. (1) P(C|A)=0.9 by assumption. Nowhere was it said that P(C)=0.9. We assumed P(C)=0.5. (2) How did you get the result for P(E)? P(woman)=P(man)=0.5 by assumption where you write P(woman)=0.9. \u2013\u00a0 Michael Chernick Jun 21 '12 at 11:39\nThe value of P(C) is assumed at 0.5, which is what I've used. The value for P(E) is the probability of Evidence 2 after applying Evidence 1 (which leads to a new hypotheses that the probability that the person is female is 0.9). P(E) = (probability that the person is a woman (given Evience 1) * probability the the person has AX3 if a woman) + (probability that the person is a man (given Evience 1) * probability the the person has AX3 if a man) = (0.9 * 0.8) + (0.1 * 0.2) = 0.74 \u2013\u00a0 RandomAnswer Jun 21 '12 at 14:42\nYour definition of probability of E is a bit confusing and the terms you are using to calculate it look different from what you wrote before. It really doesn't matter though. The answer is apparently correct based on Huu's nicely presented answer. \u2013\u00a0 Michael Chernick Jun 21 '12 at 14:57\n@Michael Except it appears Huu made mistakes. \u2013\u00a0 whuber Jun 21 '12 at 15:12\nThis answer is simply wrong. There may be other errors, but this one is glaring. You state a definitive answer for P(\"Has Long Hair\") (your P(A)), and then use that to give your final definitive answer. There simply isn't enough information to determine this, even assuming P(F) = 0.5. Your line to calculate P(A) seems to come from nowhere. Here is the correct formula using Bayes theroem: P(A) = P(A|F)P(F)/P(F|A) from which, using your stated assumptions, get to P(A) = P(A|F)*5/9. However we still don't know P(A|F), which could be anything. \u2013\u00a0 Bogdanovist Jun 22 '12 at 4:04\nshow 1 more comment\n\nI believe now that, if we assume a ratio of men and women in the population at large, then there is a single indisputable answer.\n\nA = the event that the person has long hair\n\nB = the event that the person has blood type AX3\n\nC = the event that person is female\n\nP(C|A) = 0.9\n\nP(C|B) = 0.8\n\nP(C) = 0.5 (i.e. let's assume an equal ratio of men and women in the population at large)\n\nThen P(C|A and B) = [P(C|A) x P(C|B) / P(C)] / [[P(C|A) x P(C|B) / P(C)] + [[1-P(C|A)] x [1-P(C|B)] / [1-P(C)]]]\n\nin this case, P(C|A and B) = 0.972973\n\n-\nP[C|A and B)= P(A and B and C)/P(A and B)=P(A and B and C)/ [P(A|B) P(B)]. How did you get your formula? \u2013\u00a0 Michael Chernick Jun 21 '12 at 10:12\nThere is probably a way to add conditions so that you get a unique answer. \u2013\u00a0 Michael Chernick Jun 21 '12 at 10:19\nTo add by independence of A and B the formula simplifies to P(A and B and C}/[P(A) P(B)]=P(B and C|A)/P(B). \u2013\u00a0 Michael Chernick Jun 21 '12 at 10:22\nThe intent of my question was really for you to justify the formula. I don't understand how it would be derived. \u2013\u00a0 Michael Chernick Jun 21 '12 at 11:24\nNo, the answer that supposedly used Bayes Rule is incorrect. I'm not sure why you are confused, MC's formula above is correct and cannot be used to get any result, that's what his and Whuber's answers to the question explained! \u2013\u00a0 Bogdanovist Jun 22 '12 at 4:08\n\nNote: In order to get a definitive answer, the below answers assume that the probability of a person, a long-haired man, and a long-haired women having AX3 are approximately the same. If more accuracy is desired, this should be verified.\n\nYou start out with the knowledge that the person has long hair, so at this point the odds are:\n\n90:10\n\n\nNote: The ratio of males to females in the general population does not matter to us once we find out the person has long hair. For example, if there were 1 female in a hundred in the general population, a randomly-selected long-haired person would still be a female 90% of the time. The ratio of females to males DOES matter! (see the update below for details)\n\nNext, we learn that the person has AX3. Because AX3 is unrelated to long hair, the ratio of men to women is known to be 50:50, and because of our assumption of the probabilities being the same, we can simply multiply each side of the probability and normalize so that the sum of the sides of the probability equals 100:\n\n(90:10) * (80:20)\n==> 7200:200\n\nNormalize by dividing each side by (7200+200)/100 = 74\n\n==> 7200/74:200/74\n==> 97.297.. : 2.702..\n\n\nThus, the chance that the person behind the curtain is female is approximately 97.297%.\n\nUPDATE\n\nHere's a further exploration of the problem:\n\nDefinitions:\n\nf - number of females\nm - number of males\nfl - number of females with long hair\nml - number of males with long hair\nfx - number of females with AX3\nmx - number of males with AX3\nflx - number of females with long hair and AX3\nmlx - number of males with long hair and AX3\npfl - probability that a female has long hair\npml - probability that a male has long hair\npfx - probability that a female has AX3\npmx - probability that a male has AX3\n\n\nFirst, we are given that 90% of long-haired people are females, and 80% of people with AX3 are female, so:\n\nfl = 9 * ml\npfl = fl / f\npml = ml / m\n= fl / (9 * m)\n\nfx = 4 * mx\npfx = fx / f\npmx = mx / m\n= fx / (4 * m)\n\n\nBecause we assumed that the probability of AX3 is independent of gender and long hair, our calculated pfx will apply to women with long hair, and pmx will apply to men with long-hair to find the number of them that likely have AX3:\n\nflx = fl * pfx\n= fl * (fx / f)\n= (fl * fx) / f\nmlx = ml * pmx\n= (fl / 9) * (fx / (4 * m))\n= (fl * fx) / (36 * m)\n\n\nThus, the likely ratio of the number of females with long-hair and AX3 to the number of males with long-hair and AX3 is:\n\nflx             :   mlx\n(fl * fx) / f   :   (fl * fx) / (36 * m)\n1/f             :   1 / (36m)\n36m             :   f\n\n\nBecause it is given that there is an equal number of 50:50, you can cancel both sides and end with 36 females to every male. Otherwise, there are 36*m/f females for every male in the specified subgroup. For example, if there were twice as many women as men, there would be 72 females to each male of those that have long-hair and AX3.\n\n-\nThis solution relies on assuming more than is currently stated in the problem: namely, that long hair, AX3, and gender are independent. Otherwise, you cannot justify \"applying\" pfx to women with long hair, etc. \u2013\u00a0 whuber Jun 21 '12 at 19:35\n@whuber: Yes, I do make that assumption. However, isn't the purpose of probability to give the best approximation based on the data that you have? Thus, since you already know that long-hair and AX3 are independent for the general population, you SHOULD carry forward that assumption to males and females until you explicitly learn otherwise. Granted, it is not a universally correct one, but it is the best one you can make until you get more info. Q: With only the current data, if you had to give the % chance that it was a woman behind the curtain, would you really say \"between 0 and 100%\"? \u2013\u00a0 Briguy37 Jun 21 '12 at 21:44\nWe have an important difference in philosophy, @Briguy. I strongly believe in not making unfounded assumptions. It is not clear in what sense the mutual independence assumption is \"best\": I will grant it may be in certain applications. But in general, that seems dangerous to me. I would prefer being clear about the assumptions needed to solve a problem, so people can decide whether it is worthwhile collecting the data to check those assumptions, rather than assuming things that are mathematically convenient for the sake of obtaining an answer. That's the difference between stats and math. \u2013\u00a0 whuber Jun 21 '12 at 23:14\nTo answer your question: yes, 0% - 100% is exactly the answer I would give. (I have given similar answers to comparable questions on this site.) That range accurately reflects the uncertainty. This issue is closely related to the Ellsberg paradox. Ellsberg's original paper is well written and clear: I recommend it. \u2013\u00a0 whuber Jun 21 '12 at 23:16\n@whuber: Thanks for taking the time to dialogue with me. I see your point about the importance of thinking through and listing the assumptions made, and have updated my answer accordingly. However, in regards to your answer, I believe it is incomplete. The reason for this is that you can consider all unknown cases and find the average probability of across all of them to arrive at your final answer. E.G. Though both are still possible, probabilities above 50% are much more prevalent than probabilities below 50% across all cases, so we are surely better off guessing that it is a woman. \u2013\u00a0 Briguy37 Jun 22 '12 at 13:56\nshow 1 more comment\n\n98% Female, simple interpolation. First premise 90% female, leaves 10%, second premise only leaves 2% of the existing 10%, hence 98% female\n\n-", "date": "2013-12-07 11:30:50", "meta": {"domain": "stackexchange.com", "url": "http://stats.stackexchange.com/questions/30842/what-is-the-probability-that-this-person-is-female/30850", "openwebmath_score": 0.7789069414138794, "openwebmath_perplexity": 912.7101314599452, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877684006774, "lm_q2_score": 0.8774767970940975, "lm_q1q2_score": 0.8517559939946435}}
{"url": "https://mathhelpboards.com/threads/simplify-complex-rational-expression.2246/", "text": "# Simplify Complex Rational Expression\n\n#### PaperStSoap\n\n##### New member\n(3/x-2) - (4/x+2) / (7/x2-4)\n\nI got it down to...\n\n-x+14/7\n\nbut the book is showing\n\nx-14/7\n\n#### Fantini\n\nMHB Math Helper\nWelcome to MHB, Paper!\n\nWhich of these expressions did you mean:\n\n$$\\frac{3}{x-2} - \\frac{ \\frac{4}{x+2} }{ \\frac{7}{x^2 -4} } \\quad \\text{ or } \\quad \\frac{ \\frac{3}{x-2} - \\frac{4}{x+2} }{ \\frac{7}{x^2 -4} }?$$\n\n#### soroban\n\n##### Well-known member\nHello, PaperStSoap!\n\n$$\\dfrac{\\dfrac{3}{x-2} - \\dfrac{4}{x+2}}{\\dfrac{7}{x^2-4}}$$\n\nI got it down to: .$\\dfrac{-x+14}{7}$ . You are right!\n\nBut the book is showing: .$\\dfrac{x-14}{7}$ . The book is wrong!\n\n#### PaperStSoap\n\n##### New member\nWelcome to MHB, Paper!\n\nWhich of these expressions did you mean:\n\n$$\\frac{3}{x-2} - \\frac{ \\frac{4}{x+2} }{ \\frac{7}{x^2 -4} } \\quad \\text{ or } \\quad \\frac{ \\frac{3}{x-2} - \\frac{4}{x+2} }{ \\frac{7}{x^2 -4} }?$$\nMy apologies, the problem was the second one.\n\n#### SuperSonic4\n\n##### Well-known member\nMHB Math Helper\nMy apologies, the problem was the second one.\nYou are right and the book is wrong. It's worth mentioning though that there is a restriction on the domain: $|x| \\neq 2$.\n\n(working in spoiler)\n\n$\\dfrac{\\frac{3}{x-2} - \\frac{4}{x+2}}{\\frac{7}{x^2-4}}$\n\n$\\left(\\frac{3}{x-2} - \\frac{4}{x+2}\\right) \\cdot \\frac{(x-2)(x+2)}{7}$\n\n$\\left(\\frac{3(x+2)-4(x-2)}{(x-2)(x+2)}\\right) \\cdot \\frac{(x-2)(x+2)}{7}$\n\n$\\frac{3x+6-4x+8}{7}$\n\n$\\frac{-x+14}{7}$", "date": "2021-09-20 23:16:37", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/simplify-complex-rational-expression.2246/", "openwebmath_score": 0.7618076801300049, "openwebmath_perplexity": 11925.341765664869, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9706877717925421, "lm_q2_score": 0.877476793890012, "lm_q1q2_score": 0.8517559938607594}}
{"url": "https://math.stackexchange.com/questions/3187553/what-group-is-langle-1-1-1-rangle-in-mathbbz-5-times-mathbbz-4-ti", "text": "# What group is $\\langle (1,1,1) \\rangle$ in $\\mathbb{Z}_5 \\times \\mathbb{Z}_4 \\times \\mathbb{Z}_8$?\n\nI know that $$\\langle (1,1,1) \\rangle$$ in $$\\mathbb{Z}_5 \\times \\mathbb{Z}_4 \\times \\mathbb{Z}_8$$ is isomorphic to $$\\mathbb{Z}_{40}$$, but is there a way of writing what group it is (not what it's isomorphic to).\n\nIn other words, can we say that $$\\langle (1,1,1) \\rangle$$ is $$G_1 \\times G_2 \\times G_3$$ for some groups $$G_1, G_2, G_3$$?\n\nI'm so used to using isomorphism that I can't tell if the answer is simply no. I tried using the Fundamental Theorem of Abelian Groups, but this didn't resolve my question.\n\nFor example, in $$\\mathbb{Z} \\times \\mathbb{Z}$$ we have $$\\langle (0,3) \\rangle = \\{0 \\} \\times \\mathbb{3Z}$$.\n\nWe would often just say $$\\langle (0,3) \\rangle \\cong \\mathbb{Z}$$, but occasionally it's useful to not use the usual isomorphism and so this is what I'm asking for.\n\nMy motivation is to be able to more regularly use the trick used in this answer.\n\n\u2022 It's a subgroup of order $40.$ \u2013\u00a0Dbchatto67 Apr 14 at 15:58\n\u2022 If anything, this comment gives less information than what I've written (\"a subgroup of order 40\" might not even be cyclic, but this particular one is). I'm asking for more information. \u2013\u00a0Jonathan Rayner Apr 14 at 16:04\n\u2022 The group $\\langle (1,1) \\rangle \\subset \\Bbb Z_4 \\times \\Bbb Z_8$ is not isomorphic to (and certainly not equal to) any product $G_2 \\times G_3$. Your group can be written as $\\Bbb Z_5 \\times \\langle (1,1) \\rangle$ (to the extent that we can consider the Cartesian product associative) \u2013\u00a0Omnomnomnom Apr 14 at 16:05\n\u2022 What do you mean what group it is as opposed to what group it is isomorphic to? We usually name groups up to isomorphism, so your group is just $C_{40}$, which is the same as $\\Bbb {Z_5 \\times Z_8}$ \u2013\u00a0Ross Millikan Apr 14 at 16:06\n\u2022 See this question. \u2013\u00a0Dietrich Burde Apr 14 at 16:20\n\n$$\\langle(1, 1, 1)\\rangle$$ is, as you said, isomorphic to $$\\mathbb{Z}_{40}$$, and the prime factorization of $$40$$ is $$5 \\times 8$$, so we can say that $$\\langle(1, 1, 1)\\rangle \\simeq \\mathbb{Z}_5 \\times \\mathbb{Z}_8$$. Now, it is important to convince yourself that such a group can't be expressed as the product of three subgroups, simply because $$\\mathbb{Z}_8$$ can't be split since is cyclic and $$\\mathbb{Z}_5$$ only has trivial subgroups, or, more formally, by contradiction, let $$\\langle(1, 1, 1)\\rangle \\simeq G_1 \\times G_2 \\times G_3$$. Since every subgroup of a cyclic group is cyclic, $$G_2 \\times G_3$$ is cyclic and none of its coordinates is $$\\{e\\}$$. Therefore, $$G_2 \\times G_3$$ is a product of cyclic groups whose orders are multiples of $$2$$ (so they're not coprime) and it's cyclic, which is a contradiction. As for how you could use that other method in this case, i'd say the most it can do for you is to simplify the thing as $$(\\mathbb{Z}_5 \\times \\mathbb{Z}_4 \\times \\mathbb{Z}_8)/\\langle(1, 1, 1)\\rangle \\simeq \\mathbb{Z}_5/\\langle1\\rangle \\times (\\mathbb{Z}_4 \\times \\mathbb{Z}_8)/\\langle(1, 1)\\rangle \\simeq (\\mathbb{Z}_4 \\times \\mathbb{Z}_8)/\\langle(1, 1)\\rangle$$, but since that $$\\langle(1, 1)\\rangle$$ is not a product of subgroups you have to figure out some other way to find it. I hope to have answered your question, let me know if that's not the case.\n\nA bit late, but here's a way to see that $$(\\Bbb Z_{4} \\times \\Bbb Z_8)/\\langle (1,1) \\rangle \\cong \\Bbb Z_4$$. We note that the map $$\\phi:\\Bbb Z_{4} \\times \\Bbb Z_8 \\to \\Bbb Z_{4} \\times \\Bbb Z_8$$ defined by $$\\phi: (x,y) \\mapsto (x-y,y)$$ is a group automorphism (which is induced by the automorphism with the same formula from $$\\Bbb Z^2$$ to $$\\Bbb Z^2$$). It follows that $$(\\Bbb Z_{4} \\times \\Bbb Z_8)/\\langle (1,1) \\rangle \\cong \\phi(\\Bbb Z_{4} \\times \\Bbb Z_8)/\\phi(\\langle (1,1) \\rangle) = (\\Bbb Z_{4} \\times \\Bbb Z_8)/\\langle (0,1) \\rangle\\\\ \\quad \\ = (\\Bbb Z_4 \\times \\Bbb Z_8)/(\\{0\\} \\times \\Bbb Z_8) \\cong \\Bbb Z_4$$\n\n\u2022 Thank you for writing this - I accepted the other answer because I think it answered what I technically asked for in the question and it would have been unfair not to. But your answer was very helpful and exactly in the spirit of what I was looking for! \u2013\u00a0Jonathan Rayner Apr 19 at 21:52\n\nI'm not quite sure what you're asking in your question, but if I'm understanding it, I hope this will help.\n\nOne way to think about it is the following. If $$q:\\Bbb{Z}/(8)\\to \\Bbb{Z}/(4)$$ is the quotient map, then the elements of the subgroup of $$\\Bbb{Z}/(5)\\times \\Bbb{Z}/(4)\\times \\Bbb{Z}/(8)$$ generated by $$(1,1,1)$$ are precisely the elements of the form $$(i,q(j),j)$$, with $$i\\in\\Bbb{Z}/(5)$$, and $$j\\in\\Bbb{Z}/(8)$$. This also makes the isomorphism with $$\\Bbb{Z}/(5)\\times \\Bbb{Z}/(8)$$ clear. One direction is $$(i,q(j),j)\\mapsto (i,j)$$, and the other is $$(i,j)\\mapsto (i,q(j),j)$$.\n\nSide note:\n\nThis isn't directly related to your question, just an interesting tangent.\n\nIf we focus on the subgroup of $$\\Bbb{Z}/(8)\\times \\Bbb{Z}/(4)$$ generated by $$(1,1)$$ (order deliberately flipped here), we can see that it is the subgroup $$(j,q(j))$$ for $$j\\in \\Bbb{Z}/(8)$$, which you can think of as the graph of the group homomorphism $$q$$. It's quite often the case that when you can construct the graph of a morphism $$f$$, the graph turns out to be isomorphic to the starting object, with one of the isomorphisms being $$x\\mapsto (x,f(x))$$, and the other being $$(x,f(x))\\mapsto x$$.", "date": "2019-06-17 12:56:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3187553/what-group-is-langle-1-1-1-rangle-in-mathbbz-5-times-mathbbz-4-ti", "openwebmath_score": 0.9100614786148071, "openwebmath_perplexity": 104.34097924702121, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877700966098, "lm_q2_score": 0.8774767746654976, "lm_q1q2_score": 0.8517559737116172}}
{"url": "https://math.stackexchange.com/questions/1172533/asymptotic-notation-big-theta", "text": "# Asymptotic notation (big Theta)\n\nI'm currently in the process of analyzing runtimes for some given code (Karatsuba-ofman algorithm).\n\nI'm wondering if I'm correct in saying that $\\Theta(\\left\\lceil n/2\\right\\rceil) + \\Theta(n)$ is equal to $\\Theta(n)$? (Taking the maximum of both costs)\n\nI know that $\\Theta(n/2) +\\Theta(n)$ is equal to $\\Theta(n)$. But my concern with the first asymptotic runtime is that given $n = 0.6, \\left\\lceil n/2\\right\\rceil$ would be equal to $1$, which is greater than $0.6$.\n\nIf $\\Theta(\\left\\lceil n/2\\right\\rceil) + \\Theta(n) \\stackrel{?}{=} \\Theta(n)$ could anybody please give some insight on why this is true?\n\nYes you are right.. intuitively it's clear that it does not change a thing, since when $n \\to \\infty$ effects like this are definitely too small to notice.\n\nThe whole concept behind asymptotics is that we ignore small terms.. sometimes we can even ignore something \"huge\" like $2^n$, for example $\\Theta(2^n + 3^n) = \\Theta(3^n)$\n\nSo we don't really care about effect like this; yes if $n=0.6$ you would get $1$, but the error is at most $1/2$ which is not relevant. if you had $n = 10^9 + 0.6$, would you be worried about wether the result is rounded up or down?\n\nIf you want to be a little bit more formal you can start noting that $\\lfloor \\frac n2 \\rfloor \\le n$, and $\\Theta(n) + \\Theta(n) = \\Theta(n)$\n\nOr you can also notice that the error from rounding is at most $1/2$ or $1$, depends on the rounding that one uses and find some bonds from there.\n\n\u2022 Thanks alot for this explanation. Just as I thought. That's really cleared some things up. \u2013\u00a0user220506 Mar 2 '15 at 22:12\n\u2022 @Tazman You're welcome! :) \u2013\u00a0Ant Mar 2 '15 at 22:13\n\u2022 Although you stated floor(n/2) <= n, this wouldn't be the case for ceiling(n/2) <= n unless we state for all n >= 1 right? \u2013\u00a0user220506 Mar 2 '15 at 22:14\n\u2022 @Tazman yes of course. But then again we only look about what happens in the long run, so to speak, so most of the time I don't even notice I take this assumption and most of what I write is valid for $n$ large enough ;-) \u2013\u00a0Ant Mar 2 '15 at 22:16\n\u2022 Okay, that makes sense. Just like if it were a big-O or big-Omega question, we would state some given bound where we see the said behaviour. Thanks for the clarification :) \u2013\u00a0user220506 Mar 2 '15 at 22:17\n\nRemember that $f(n) =\\Theta (g(n))$ means that there are positive constants $a$ and $b$ such that $a g(n) < f(n) < b g(n)$ for all large enough $n$.\n\nTherefore, if $f(n) =\\Theta (h(n))$ and $g(n) =\\Theta (h(n))$ then $f(n)+g(n) =\\Theta(h(n))$.\n\nIf you consider the problem for the natural numbers it is true that $\\Theta(\\lceil n/2 \\rceil) = \\Theta (n)$. As $\\lceil n/2 \\rceil \\le n \\le 2 \\lceil n/2 \\rceil$. And as a consequence also $\\Theta(\\lceil n/2 \\rceil) + \\Theta (n)= \\Theta (n)$.\n\nHowever, if you consider it for the positive real numbers (as $n \\to 0$) as your example $0.6$ might suggest then it is not true that $\\Theta(\\lceil n/2 \\rceil) = \\Theta (n)$ and neither is $\\Theta(\\lceil n/2 \\rceil) + \\Theta (n)= \\Theta (n)$. In this case $n = O (\\lceil n/2 \\rceil)$ as $n \\le 2 \\lceil n/2 \\rceil$ still holds. Yet as $\\lceil n/2 \\rceil = 1$ for all $0< n \\le 2$ we get that $\\lceil n/2 \\rceil$ is not $O(n)$ as $n \\to 0$.\n\n\u2022 So Tazman should say $n \\to \\infty$ if he means that. (Presumably he does, for analyzing \"runtimes\".) \u2013\u00a0GEdgar Mar 2 '15 at 22:08\n\u2022 @GEdgar Well, yes and no. Note they mention the case $n=0.6$ explicitly. But I did mention the result for the integers first. It is however not really necessary to say $n \\to \\infty$ one just has to make the domain precise, if it is the natural numbers it is fine. The $O$ notation as well as $\\Theta$ makes sense without making a limit point precise, as opposed to $o$. Finally note it is the second question of this form I answered today, and there is thus additional reason why I insist on discussing the distinction i make. \u2013\u00a0quid Mar 2 '15 at 22:13\n\u2022 The n in question refers to a length of a given number, so I think it's safe that in my answer to the question, one can assume that n consists of natural numbers, and thus theta(ceiling(n/2)) = theta(n). \u2013\u00a0user220506 Mar 2 '15 at 22:16\n\u2022 @quid, yes, that's true. I should have been more clear and stated that I was giving that generic example to better my understanding of exactly why it is (or not) equal. \u2013\u00a0user220506 Mar 2 '15 at 22:19\n\u2022 @Tazman yes, if it is for integers then you can say this (as I said in my answer). And this is a reasonable assumption given your context. I only discussed the other case since you mentioned $n=0.6$ and there is an actual difference in another situation. [This is an expanded version.] \u2013\u00a0quid Mar 2 '15 at 22:20", "date": "2019-05-25 10:50:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1172533/asymptotic-notation-big-theta", "openwebmath_score": 0.8507260680198669, "openwebmath_perplexity": 283.5382152921315, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347838494567, "lm_q2_score": 0.8740772384450967, "lm_q1q2_score": 0.8517312649119778}}
{"url": "https://math.stackexchange.com/questions/4241111/need-help-understanding-least-squares-solution-to-overdetermined-system", "text": "# Need help understanding least squares solution to overdetermined system\n\n(Sorry I had to post the images as links. I don't have enough cred to post pictures directly yet)\n\nI'm trying to understand what the least squares solution to an overdetermined system means geometrically in the case of the following system:\n\n$$y = -2x-1\\\\ y = 3x -2\\\\ y = x+1\\\\$$\n\nrewritten in matrix form:\n\n$$\\overbrace{\\begin{bmatrix} 2 & 1\\\\ -3 & 1\\\\ -1 & 1 \\end{bmatrix}}^A \\underbrace{\\begin{bmatrix} x\\\\ y \\end{bmatrix}}_x = \\overbrace{\\begin{bmatrix} -1\\\\ -2\\\\ 1 \\end{bmatrix}}^b$$\n\nUsing A\\b in MATLAB, you get the solution $$\\begin{bmatrix}0.1316 & -0.5789\\end{bmatrix}^T$$. I know that MATLAB returns the lowest norm solution of a least squares problem. I have plotted the system here and the green dot in the middle is this least squares solution.\n\nNow, correct me if I'm wrong, but (in this 2D case) the least squares solution minimizes the \"distance\" from the solution to each line. I can geometrically calculate the distance of a point $$(x_0,y_0)$$ from a line $$ax + by + c = 0$$ as follows:\n\n$$\\mathrm{distance} = \\frac{|ax_0 + by_0 + c|}{\\sqrt{a^2 + b^2}}$$\n\nand doing that for each line produces the following sum of squared distances function\n\ndfun = @(x,y) ((y+2*x+1).^2)/(1^2 + 2^2) + ((y+3*x+2).^2)/(1^2 + 3^2) + ((y+x-1).^2)/(1^2 + 1^2);\n\n\nIf I generate a surface using this function over a range of $$x$$ and $$y$$ values, I get this surface with this top-down view (looking down the z-axis on the xy plane). You can download the MATLAB .fig file here if you want to zoom and pan (requires MATLAB, link expires in 30 days).\n\nHere is an image showing the least squares solution with the sum of squares of distances of the solution and its norm. As can be seen, the norm is $$0.5937$$ and the distance is $$1.4704$$. But clearly, there is a contour that has a lower sun of squared distance in the image, as shown here for $$(x_0, y_0) = (-0.3,0)$$, where the norm and the sum of squared distances are both smaller. Shouldn't this (or another point) be a better least squares solution? Do I have the wrong intuition about what least squares is doing here?\n\nFirst of all : Welcome to the site !\n\nWhen you face an overdetermined system of $$m$$ linear or nonlinear (even implicit) equations $$f_i(x_1,x_2,\\cdots,x_n)=0 \\quad\\text{for}\\quad i=1,2,\\cdots,m\\qquad \\quad\\text{and}\\quad m >n$$ it reduces to the minimization of a norm.\n\nThe simplest is $$\\Phi(x_1,x_2,\\cdots,x_n)=\\sum_{i=1}^n w_i \\Big[f_i(x_1,x_2,\\cdots,x_n)\\Big]^2$$ which shows the analogy with weighted least-square method.\n\nFor the problem you gave, using equal weights, $$\\Phi(x,y)=(y+2x+1)^2+(y-3x+2)^2+(y-x-1)^2$$ Computing the partial derivetives $$\\frac{\\partial \\Phi(x,y)}{\\partial x}=28 x-4 y-6 \\qquad\\text{and}\\qquad \\frac{\\partial \\Phi(x,y)}{\\partial y}=-4 x+6 y+4$$ which gives $$x=\\frac{5}{38}$$ and $$y=-\\frac{11}{19}$$. At this point, $$\\Phi=\\frac{169}{38}$$ is the absolute minimum for this specific norm.\n\nIf you change the definition of the norm or even the weights, different results.\n\nI think that this approach would be simpler than the one based on distance (but, for sure, I may be wrong) since it is totally general and not limited to linear equations.\n\n\u2022 Thank you! I think I had the right understanding of what least squares is doing in this case. After reading your answer I realizes that my dfun was wrong - I got a couple of the signs wrong and that's why I was getting the wrong surface. I will post an answer with the right images. Not sure if I should accept my own answer or yours...\n\u2013\u00a0vyb\nSep 6, 2021 at 17:40\n\nAfter reading Claude Leibovici's answer above, I realized that my dfun had typos in it -- I messed up a couple of minus signs in the function.\n\nAdditionally, the norm typically used for least squares calculations (also used by MATLAB) is the $$l^2$$-norm (a.k.a Euclidean norm):\n\n$$x = \\begin{bmatrix}x_1\\\\ x_2\\\\ \\vdots\\\\ x_n\\end{bmatrix}, ||x|| = \\sqrt{x_1^2 + x_2^2 + \\dots + x_n^2}$$\n\nNote that there is no scaling of the \"distance\" like there is in my dfun. Therefore correct function should be:\n\ndfun = @(x,y) ((y+2*x+1).^2) + ((y-3*x+2).^2) + ((y-x-1).^2);\n\n\nAfter fixing these mistakes, here is the surface that is generated, with this top-down view. As can be seen here, the solution of $$\\begin{bmatrix}0.1316 & -0.5789\\end{bmatrix}^T$$ is, in fact, correct and confirms the original intuition I was trying to confirm.", "date": "2022-06-28 04:18:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4241111/need-help-understanding-least-squares-solution-to-overdetermined-system", "openwebmath_score": 0.806755781173706, "openwebmath_perplexity": 195.72931657823568, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347883040039, "lm_q2_score": 0.8740772335247532, "lm_q1q2_score": 0.8517312640110423}}
{"url": "https://stats.stackexchange.com/questions/443614/probability-that-random-draws-from-the-same-pool-will-collectively-select-90-of", "text": "# probability that random draws from the same pool will collectively select 90% of the pool\n\nThere are a total of 200 names on a list. 30 times names are selected from the full list. How many names should be selected each time to predict with 90% certainty that 90% of all the names will be selected at least once?\n\n# The answer is $$n=17$$.\n\nI can't see an easy analytic solution to this question. Instead, we will develop an analytic solution to a closely related problem, and then find the answer to your exact question via simulation.\n\n## Clarification:\n\nSince the question is slightly vague, let me re-state the problem. There are $$200$$ names on a list and $$n$$ names will be selected from this list without replacement. This process, using the full $$200$$ names each time, is repeated a total of $$30$$ times.\n\n## A related problem.\n\nLet $$X_i$$ equal $$1$$ if the $$i^{th}$$ name is selected at least once and equal to $$0$$ otherwise. This implies that $$X = \\sum_{i=1}^{200}X_i$$ represents the total number of names which are selected at least once. Since the $$X_i$$ are dependent, the exact distribution of $$X$$ is not-trivial, and the original question is hard to answer. Instead, we can easily determine the value of $$n$$ such that $$90\\%$$ of the names are selected on average.\n\nFirst, note that $$P(X_i = 0) = \\left(\\frac{200 - n}{200}\\right)^{30}$$ which implies $$E(X_i) = P(X_i =1) = 1 - \\left(1- \\frac{n}{200}\\right)^{30}.$$\n\nNow by linearity of expectation we have $$E(X) = \\sum_{i=1}^{200}E(X_i) = 200\\left(1 - \\left(1- \\frac{n}{200}\\right)^{30}\\right).$$\n\nFor this expectation to equal $$90\\%$$ of the names, we need to set $$E(X) = 180$$ and solve for $$n$$. This gives $$n = 200\\left(1 - (1 - 0.9)^{1/30}\\right) = 14.776.$$\n\nThus $$n=15$$ names should be drawn from the list each time for this to occur on average. This answer will be close to (but not the same as) the original question with $$50\\%$$ certainty. To achieve $$90\\%$$ certainty, we will need to increase $$n$$.\n\n## Simulations.\n\nFirst, we write a function which is able to generate $$X$$ a large number (say $$M$$) times for a given value of $$n$$.\n\nsample_X <- function(n, M){\nX <- rep(NA, M)\nfor(i in 1:M){\n#Set all names to false\nnames <- rep(FALSE, 200)\n#Repeat process 30 times\nfor(k in 1:30){\n#Sample n names from list\nselection <- sample(200, n, replace=F)\n#Mark that these names have been selected\nnames[selection] <- TRUE\n}\n#Let X be the number of selected names\nX[i] <- sum(name_been_selected)\n}\nreturn(X)\n}\n\n\nNow, for a given value of $$n$$ we can approximate \"the probability that at least $$90\\%$$ of the names are selected\", i.e. $$P(X \\geq 180)$$. In R, this probability can be approximated by typing:\n\nX <- sample_X(n, M=10000)\nprob <- mean(X >= 180)\n\n\nRepeating this for $$n = 14, 15, \\cdots 20$$ gives us the following plot.\n\nFrom the plot, we can determine that $$n=17$$ names must be selected in each round for the probability of selecting at least $$180$$ names to exceed $$0.9$$.\n\nThe blue line in the figure shows the exact simulations detailed above. The orange line is an approximation which is obtained by ignoring the dependency of the $$X_i$$ (see previous section) and assuming that $$X \\sim \\text{Binom}\\left(200, 1 - \\left(1- \\frac{n}{200}\\right)^{30}\\right).$$\n\nAlthough the assumption of independence is obviously incorrect, the probabilities obtained by this simple assumption are reasonably close to the simulated probabilities.\n\n\u2022 There appear to be bugs. Why are you sampling n from 200 names when you should always be sampling 30? When I compute the exact solution (using a method similar to the one you posted yesterday but deleted) or when I perform a simulation I get answers very much like your deleted answers. In particular, the chance of collecting 180 or more names in 17 iterations is 99.26%; in 16 iterations it is 95.46%; and in 15 iterations it is 81.14%, whence 16 iterations will do the job. A simple but accurate estimate is the ceiling of $\\log(0.10)/\\log(1-30/200)= 14.2,$ or 15. \u2013\u00a0whuber Jan 8 '20 at 5:10\n\u2022 @whuber, perhaps we are interpreting the question differently. The original question is quite vague, but my best interpretation is that $n$ names are sample from the list of $200$, and this is repeated $30$ times: \"thirty times, names are selected...\" and \"how many names should be selected each time \" \u2013\u00a0knrumsey Jan 8 '20 at 5:20\n\u2022 I'm sorry--I confused you with the author of the deleted post. There are too many \"*-Reinstate Monica\" user names here. :-). Upon rereading the question (several times, because it's strangely phrased) I have to agree with you: I mixed up \"names\" and \"times\" in my interpretation. +1 to you--but note that an exact solution still is relatively easy to obtain. Simply update the generating polynomial for the distribution 30 times. \u2013\u00a0whuber Jan 8 '20 at 5:30\n\u2022 I am hoping that the author of the deleted post will make the small changes need to reflect your (correct) interpretation and undelete it, because it's a nice post. \u2013\u00a0whuber Jan 8 '20 at 5:39\n\u2022 @filbranden, that's relatively close to what I get: $97.8\\%$. \u2013\u00a0knrumsey Jan 8 '20 at 16:32\n\n# Here is a general analytic solution --- does not require simulation\n\nThis is a variation on the classical occupancy problem, where you are sampling lots of thirty names at each sampling point, instead of sampling individual names. The simplest way to compute this result is by framing the problem as a Markov chain, and then computing the required probability using the appropriate power of the transition probability matrix. For the sake of broader interest to other users, I will generalise from your example by considering a list with $$m$$ names, with each sample selecting $$1 \\leqslant h \\leqslant m$$ names (using simple-random-sampling without replacement).\n\nThe general problem and its solution: Let $$0 \\leqslant K_{n,h} \\leqslant m$$ denote the number of names that have been sampled after we sample $$n$$ times with each lot sampling $$h$$ names. For a fixed value $$h$$ the stochastic process $$\\{ K_{n,h} | n = 0,1,2,... \\}$$ satisfies the Markov assumption, so it is a Markov chain. Since each sampling lot is done using simple-random-sampling without replacement, the transition probabilities for the chain are given by the hypergeometric probabilities:\n\n$$P_{t,t+r} \\equiv \\mathbb{P}(K_{n,h} = t+r | K_{n-1,h} = t) = \\frac{{m-t \\choose r} {t \\choose h-r}}{{m \\choose h}}.$$\n\nLet $$\\mathbf{P}_h$$ denote the $$(m+1) \\times (m+1)$$ transition probability matrix composed of these probabilities. If we start at the state $$K_{0,h} = 0$$ then we have:\n\n$$\\mathbb{P}(K_{n,h} = k) = [ \\mathbf{P}_h^n ]_{0,k}.$$\n\nThis probability can be computed by matrix multiplication, or by using the spectral decomposition of the transition probability matrix. It is relatively simple to compute the mass function of values over $$k=0,1,...,m$$ for any given values of $$n$$ and $$h$$. This allows you to compute the marginal probabilities associated with the Markov chain, to solve the problem you have posed.\n\nThe problem you have posed is a case of the following general problem. For a specified minimum proportion $$0 < \\alpha \\leqslant 1$$ and a specified minimum probability $$0 < p < 1$$, we seek the value:\n\n$$h_* \\equiv h_* (\\alpha, p) \\equiv \\min \\{ h = 1,...,m | \\mathbb{P}(K_{n,h} \\geqslant \\alpha m) \\geqslant p \\}.$$\n\nIn your problem you have $$m=200$$ names in your list and you are taking $$n=30$$ samples. You seek the value $$h_*$$ for the proportion $$\\alpha = 0.9$$ and the probability cut-off $$p = 0.9$$. This value can be computed by computing the relevant marginal probabilities of interest in the Markov chain.\n\nImplementation in R: We can implement the above Markov chain in R by creating the transition probability matrix and using this to compute the marginal probabilities of interest. We can compute the marginal probabilities of interest using standard analysis of Markov chains, and then use these to compute the required number of names $$h_*$$ in each sample. In the code below we compute the solution to your problem and show the relevant probabilities increasing over the number of samples (this code takes a while to run, owing to the computation of matrix-powers in log-space).\n\n#Create function to compute marginal distribution of Markov chain\nCOMPUTE_DIST <- function(m, n, H) {\n\n#Generate empty matrix of occupancy probabilities\nDIST <- matrix(0, nrow = H, ncol = m+1);\n\n#Compute the occupancy probabilities\nfor (h in 1:H) {\n\n#Generate the transition probability matrix\nSTATES <- 0:m;\nLOGP <- matrix(-Inf, nrow = m+1, ncol = m+1);\nfor (t in 0:m) {\nfor (r in t:m) {\nLOGP[t+1, r+1] <- lchoose(m-t, r-t) + lchoose(t, h-r+t) - lchoose(m, h); } }\nPP <- exp(LOGP);\n\n#Compute the occupancy probabilities\nlibrary(expm);\nDIST[h, ] <- (PP %^% n)[1, ]; }\n\n#Give the output\nDIST;  }\n\n#Compute the probabilities for the problem\nm <- 200;\nn <- 30;\nH <- 20;\nDIST <- COMPUTE_DIST(m, n, H);\n\n\nFrom the marginal probabilities for the Markov chain, we can now compute the required value $$h_*$$ for your particular problem.\n\n#Set parameters for problem\nalpha  <- 0.9;\ncutoff <- ceiling(alpha*m);\np      <- 0.9;\n\n#Find the required value\nPROBS <- rowSums(DIST[, (cutoff+1):(m+1)]);\nhstar <- 1 + sum(PROBS < p);\n\n#Show the solution and its probability\nhstar;\n[1] 17\n\nPROBS[hstar];\n[1] 0.976388\n\n\nWe can see here that we require $$h_* = 17$$ samples in order to obtain a minimum $$p=0.9$$ probability of sampling at least $$\\alpha \\cdot m = 180$$ of the names on the list. Below we show a plot of the probabilities for values $$h=1,...,20$$ with the required value highlighted in red.\n\n#Plot the probabilities and the solution\nlibrary(ggplot2);\nTHEME <- theme(plot.title    = element_text(hjust = 0.5, size = 14, face = 'bold'),\nplot.subtitle = element_text(hjust = 0.5, face = 'bold'));\nDATA <- data.frame(h = 1:H, Probability = PROBS);\nggplot(aes(x = h, y = Probability), data = DATA) +\ngeom_point(size = 3, colour = 'blue') +\ngeom_point(size = 4, colour = 'red', data = DATA[hstar, ]) +\ngeom_hline(yintercept = p, size = 1, linetype = 'dashed') +\ngeom_segment(aes(x = hstar, y = 0, xend = hstar, yend = DATA[hstar, 2]),\ncolour = 'red', size = 1) +\nannotate(\"text\", x = hstar + 1, y = 0.1,\nlabel = paste0('h = ', hstar), colour = 'red', fontface = 'bold') +\nTHEME +\nggtitle('Probability of required occupancy') +\nlabs(subtitle = paste0('(Occupancy problem taking ', n,\n' samples of size h from ', m,\n' units) \\n (We require ', sprintf(100*alpha, fmt = '%#.1f'),\n'% occupancy with ', sprintf(100*p, fmt = '%#.1f'), '% probability)'));\n\n\n## The answer is $$n = 17$$, with $$P(N_{30}\\ge180)=0.976388$$.\n\nThe approach I took to calculate the probability after 30 draws was to determine the probability of drawing seen vs. unseen names at each round.\n\nWhen drawing $$n$$ names out of $$p=200$$ after having seen $$s$$ of them, let's call $$U_s$$ the number of names out of those $$n$$ which were previously unseen.\n\nThen we have:\n\n$$P(U_s = u) = \\frac{\\text{P}(200-s, u) \\text{P}(s, n-u) \\text{C}(n, u)}{\\text{P}(200, n)}$$\n\nThe first term is the permutations of u previously unseen names, the second permutations of previously seen ones. The last term $$\\text{C(n, u)}$$ accounts for the $$u$$ unseen names coming in different positions out of the $$n$$ drawn. The denominator accounts for all possible draws of $$n$$ names.\n\nHaving calculated that, we can look at successive draws of names. Let's call $$N_d$$ the total number of names after draw $$d$$.\n\nBefore the first draw, there will be no previously seen names, so in the first draw all $$n$$ names will be seen for the first time.\n\n$$P(N_1=n)=1$$\n\nWe can then calculate the probability of drawing a certain number of names on draw $$N_{d+1}$$ by looking at the possibilities of drawing after $$N_d$$ and having a specific number of previously unseen names. Which we can calculate with:\n\n$$P(N_{d+1} = x) = \\sum_{i=0}^{n}{P(N_d = x-i) P(U_{x-i} = i)}$$\n\nFor example, if we're drawing $$n=16$$ every time, then drawing exactly 180 names in total in a specific drawing can be arrived at by drawing 164 names in the previous drawing an then drawing exactly 16 unseen names (totalling 180), or having previously seen 165 names and drawing 15 unseen and one previously seen name, and so on... Until the possibility of having seen 180 names in the previous iteration and drawing all 16 previously seen names.\n\nAt this point we can use iteration to calculate $$P(N_{30} \\ge 180)$$ for different values of $$n$$.\n\n## Iteration in Python:\n\nThis code uses Python 3 and as written requires Python 3.8 for math.comb() and math.perm() from the standard library (if using an older version of Python, you can use a different implementation of those functions.)\n\nLet's start with $$P(U_s = u)$$:\n\nfrom functools import lru_cache\nfrom math import comb, perm\n\n@lru_cache\ndef prob_unseen(n, p, s, u):\n# Return  the probability of drawing\n# exactly $$u$$ unseen names when\n# drawing $$n$$ names out of a total of $$p$$,\n# having previously seen $$s$$ of them.\nreturn (perm(p-s, u) *\nperm(s, n-u) *\ncomb(n, u) /\nperm(p, n))\n\n\nPretty straightforward. Now for $$P(N_d = x)$$ let's use a list of 201 elements (indices go from 0 to 200) to track the probabilities for each $$x$$:\n\ndef names_in_draw(prev_draw, n):\n# Calculate probabilities of finding\n# exactly $$x$$ names in this draw,\n# for every $$x$$, taking in consideration\n# the probabilities of having drawn specific\n# numbers of names in the previous draw.\np = len(prev_draw) - 1\nthis_draw = [0.0] * (p+1)\nfor x in range(n, p+1):\nthis_draw[x] = sum(\nprev_draw[x-u] * prob_unseen(n, p, x-u, u)\nfor u in range(n+1))\nreturn this_draw\n\n\nFinally, let's calculate the probability for the number of names after $$d$$ draws.\n\ndef total_names(n, p, d):\n# Calculate probabilities for finding\n# exactly $$x$$ names after $$d$$ draws.\ndraw = [0.0] * (p+1)\ndraw[n] = 1.0  # first draw\nfor _ in range(d):\ndraw = names_in_draw(draw, n)\nreturn draw\n\n\nWe start from the first draw, where we know for sure we'll draw $$n$$ unique names. Than we repeatedly calculate the probabilities $$d-1$$ times.\n\nFinally, we can calculate the probability of drawing at least $$x$$ names, drawing $$n$$ out of $$p$$ at a time, performing $$d$$ drawings:\n\ndef prob_names(n, p, d, x):\n# Return the probability of seeing\n# at least $$x$$ names after $$d$$ drawings,\n# each of which draws $$n$$ out of $$p$$ names.\nreturn sum(total_names(n, p, d)[x:])\n\n\nFinally, we can run this for a few values of $$n$$ to find the probabilities:\n\n>>> for i in range(13, 20):\n...     print(i, prob_names(i, 200, 30, 180))\n13 0.058384795418431244\n14 0.28649904267865317\n15 0.6384959089930037\n16 0.8849450106842117\n17 0.976388046862824\n18 0.9966940083338005\n19 0.9996649977705089\n\n\nSo $$n=17$$ is the answer, with probability of 97.6388% of seeing at least 90% of the names. $$n=16$$ comes close, with 88.4945%.\n\n(Since I had the code, I also looked at how many drawings of a single name are needed to see 90% of the names, with 90% probability. It turns out it's 503 drawings, or 454 drawings to see 90% of the names with 50% probability. Quite interesting result!)", "date": "2021-01-18 21:08:56", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/443614/probability-that-random-draws-from-the-same-pool-will-collectively-select-90-of", "openwebmath_score": 0.7969145774841309, "openwebmath_perplexity": 940.9218743056178, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615794, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8517312617639212}}
{"url": "https://math.stackexchange.com/questions/3138129/maclaurin-series-from-sinx-to-cosx-using-derivative", "text": "# Maclaurin Series from sin(x) to cos(x) using derivative\n\nI understand how to find the MacLaurin series for $$\\sin(x)$$. $$\\sum_{n=1}^\\infty \\frac{x^{2n-1}\\cdot\\!(-1)^{n-1}}{(2n-1)!}$$ Now I am trying to find the MacLaurin series for $$\\cos(x)$$ by taking the derivative of the above sum with respect to $$x$$. Using power rule, I got the following series: $$\\cos(x) = \\sum_{n=1}^\\infty \\frac{x^{2n-2}\\cdot\\!\\mathrm{(-1)}^{n-1}}{(2n-2)!}$$\n\nHowever, the MacLaurin series is: $$\\cos(x) = \\sum_{n=0}^\\infty \\frac{x^{2n}\\cdot\\!\\mathrm{(-1)}^{n}}{(2n)!}$$ How are these two $$\\cos(x)$$ MacLaurin series equal? What makes the second one more correct than the series I got by taking the derivative of the $$\\sin(x)$$ series.\n\nA sort of related question: if you choose to start at $$n=1$$ vs $$n=0$$, how would you change the terms of the $$\\sin(x)$$ Maclaurin series?\n\n\u2022 The two series are the same. They only differ by \"re-indexing\". Shifting the start from $n=1$ to $n=0$ just replaces all the $n$'s by $(n+1)$ in the formula. \u2013\u00a0Nick Mar 6 at 21:34\n\nYou got twice the same series. Both$$\\sum_{n=0}^\\infty \\frac{x^{2n}\\cdot\\!\\mathrm{(-1)}^{n}}{(2n)!}\\text{ and }\\sum_{n=1}^\\infty \\frac{x ^{2n-2}\\cdot\\!\\mathrm{(-1)}^{n-1}}{(2n-2)!}$$are equal to$$1-\\frac{x^2}{2!}+\\frac{x^4}{4!}-\\frac{x^6}{6!}+\\cdots$$\nTo reindex, replace $$n$$ with $$n+1$$ in the summand of the first series and lower the range of $$n$$ by $$1$$ (so the new initial index will now be $$0$$ instead of $$1$$).\nMore generally, if the index $$n$$ runs from $$n=a$$ to $$n=b$$, then replace $$n$$ with $$n+K$$ in the summand and change the index limits to run from $$n=a-K$$ to $$n=b-K$$. You are essentially adding and subtracting $$K$$ from each value of $$n$$, so it is unchanged in the end. Symbolically: $$\\sum_{n=a}^b t_n =\\sum_{n=a-K}^{b-K}t_{n+K}$$", "date": "2019-09-24 09:02:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3138129/maclaurin-series-from-sinx-to-cosx-using-derivative", "openwebmath_score": 0.9482057690620422, "openwebmath_perplexity": 101.67329163893329, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319855244919, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8516963762071659}}
{"url": "https://math.stackexchange.com/questions/1431042/how-do-you-calculate-a-sum-over-a-polynomial/1431090", "text": "# How do you calculate a sum over a polynomial?\n\nI know that given a polynomial $p(i)$ of degree $d$, the sum $\\sum_{i=0}^n p(i)$ would have a degree of $d + 1$. So for example\n\n$$\\sum_{i=0}^n \\left(2i^2 + 4\\right) = \\frac{2}{3}n^3+n^2+\\frac{13}{3}n+4.$$\n\nI can't find how to do this the other way around. What I mean by this, is how can you, when given a polynomial, calculate the polynomial which sums to it?\n\nFor example, if we know that\n\n$$\\sum_{i=0}^{n} p(i) = 2n^3 + 4n^2 + 2,$$\n\nhow can we find the polynomial $p(i)$?\n\n\u2022 Are you asking how sums like $\\sum_{i=0}^{n} p(i)$ can be calculated in general? \u2013\u00a0Antonio Vargas Sep 11 '15 at 15:42\n\u2022 @AntonioVargas I'm asking how a polynomial can be written as a Sigma notation. So for example, the polynomial $2/3n^3+n^2+13/3n+4$ can be written as $\\sum_{i=0}^n 2i^2 + 4$, but I have no idea how you would start. \u2013\u00a0Esoemah Sep 11 '15 at 15:46\n\u2022 Okay, maybe the last part of the question should be written a little differently then. Is this your question?: If we know that $\\sum_{i=0}^{n} p(i) = 2n^3 + 4n^2 + 2$, how can we find the polynomial $p(i)$? \u2013\u00a0Antonio Vargas Sep 11 '15 at 15:48\n\u2022 Great, I've edited the question to try to clarify it. If you'd like you can also edit it by clicking the \"edit\" link just below the list of tags at the bottom of the question. \u2013\u00a0Antonio Vargas Sep 11 '15 at 16:00\n\u2022 Short answer:$$p(n) = \\left( \\sum_{i=0}^n p(i) \\right) - \\left( \\sum_{i=0}^{n-1} p(i) \\right)$$ \u2013\u00a0Hurkyl Sep 11 '15 at 17:01\n\nTo rephrase, I believe the question is this:\n\nSuppose that polynomials $p$ and $q$ have the property that $$\\sum_{i=0}^n p(i) = q(n)$$ If you're given $q$, how can you find $p$?\n\nFirst, this is a lovely question. I'd never really considered it, because we almost always study instead \"if you know $p$, how do you find $q$?\"\n\nTo answer though, turns out to be fairly simple. Write the following: \\begin{align} q(n-1) &= p(0) + p(1) + \\ldots + p(n-1) \\\\ q(n) &= p(0) + p(1) + \\ldots + p(n-1) + p(n) \\\\ \\end{align}\n\nNow subtract the top from the bottom to get \\begin{align} q(n) - q(n-1) &= p(n) \\end{align}\n\nAs an example, in your case if we knew $$q(n) = 2n^3 + 4n^2 + 2$$ we'd find that $$p(n) = q(n) - q(n-1) = 2n^3 + 4n^2 + 2 - [2(n-1)^3 + 4(n-1)^2 + 2],$$ which you simplifies to $$p(n) = 6n^2 +2n - 2.$$\n\nLet's do an example: we know that for $p(n) = n$, we have $q(n) = \\frac{n(n+1)}{2}$. So suppose we were given just $q$. We'd compute \\begin{align} q(n) - q(n-1) &= \\frac{n(n+1)}{2} - \\frac{(n-1)(n)}{2} \\\\ &= \\frac{n^2 + n}{2} - \\frac{n^2 - n}{2} \\\\ &= \\frac{n^2 + n-(n^2 - n)}{2} \\\\ &= \\frac{n^2 + n- n^2 + n}{2} \\\\ &= \\frac{2n}{2} \\\\ &= n, \\end{align} so that $p(n) = n$, as expected.\n\nNote: As written, I've assumed that $p$ and $q$ are both polynomials. But the solution shows that if $q$ is a polynomial, then $p$ must also be a polynomial, which is sort of pleasing.\n\nPost-comment remarks\n\nAs @Antonio Vargas points out, though, there's an interesting subtlety:\n\nI've given a correct answer to my revised question, which was \"If there are polynomials $p$ and $q$ satisfying a certain equality, then how can one find $p$ given $q$.\"\n\nBut suppose that there is no such polynomial $p$. My answer still computes an expression which $p$, if it existed, would have to match. But since no such $p$ exists, the computed expression has no value.\n\nOr maybe I should say that it has a limited value: you can take the polynomial $p$ and compute its sum using inductive techniques and see whether you get $q$. If so, that's great; if not, then there wasn't any answer in the first place.\n\nFortunately, you can also do that \"Does it really work\" check much more simply. You just need to check the the $n = 0$ case: if $$\\sum_{i = 0}^0 p(i) = q(0)$$ then all higher sums will work as well. And this check simplifies to just asking: is $$p(0) = q(0)?$$ In our example, $p(0) = -2$, while $q(0) = +2$, so it doesn't work out.\n\n\u2022 It is, by the way, a nice exercise to show that if $q$ is a degree-$d$ polynomial, then $p(n)=q(n)-q(n-1)$ is a degree-$(d-1)$ polynomial. This operation is known as finite differencing and there are many close analogs with differentiation. \u2013\u00a0Steven Stadnicki Sep 11 '15 at 16:10\n\u2022 Nice point; I've added further remarks. My test for whether things work out is that $p(0) = q(0)$ rather than that $p(-1) = 0$, but that's a tiny difference. \u2013\u00a0John Hughes Sep 14 '15 at 20:16\n\u2022 Doesn't all this procedure remind a little: integrals and derivatives? \u2013\u00a0Widawensen Jul 1 '16 at 7:53\n\nWhich Polynomials Can Be Written as a Sum\n\nBy summing a Telescoping Series, we get \\begin{align} \\sum_{k=0}^n(p(k)-p(k-1)) &=\\sum_{k=0}^np(k)-\\sum_{k=0}^np(k-1)\\\\ &=\\sum_{k=0}^np(k)-\\sum_{k=-1}^{n-1}p(k)\\\\ &=p(n)-p(-1) \\end{align} It turns out that not every polynomial can be written as a sum of other polynomials. To be written as a sum of polynomials $$p(n)=\\sum_{k=0}^nq(k)$$ we must have $p(-1)=0$, and if that condition holds, then $q(k)=p(k)-p(k-1)$.\n\nFinite Differences\n\n$q(k)=p(k)-p(k-1)=\\Delta p(k)$ is the first Backward Finite Difference of $p$.\n\nUsing the Binomial Theorem, we get the first backward finite difference of $x^n$ to be $$\\Delta x^n=x^n-(x-1)^n=\\sum_{k=0}^{n-1}(-1)^{n-k-1}\\binom{n}{k}x^k$$ This shows that the first backward finite difference of a degree $n$ polynomial is a degree $n-1$ polynomial.\n\nThus, for $$p(k)=2k^3+4k^2+2$$ we have \\begin{align} \\Delta p(k) &=2\\overbrace{\\left[3k^2-3k+1\\right]}^{\\Delta k^3}+4\\overbrace{\\left[\\vphantom{k^2}2k-1\\right]}^{\\Delta k^2}+2\\overbrace{\\left[\\vphantom{k^2}\\ \\ \\ 0\\ \\ \\ \\right]}^{\\Delta 1}\\\\ &=6k^2+2k-2 \\end{align} However, since $p(-1)=4$, we have $$\\sum_{k=0}^n(6k^2+2k-2)=2n^3+4n^2-2$$ which is not $p(n)$. That is,\n\nThere is no polynomial $q(n)$ so that $\\sum\\limits_{k=0}^nq(k)=2n^3+4n^2+2$\n\nPrevious Question\n\nThe answer below was posted before the question was changed. It was\n\nThe other way around however, I'm still a bit lost. For example, given the polynomial $p(i) = 2i^3 + 4i^2 + 2$, how would you find $\\sum_{i=0}^n p(i)$?\n\nSo what follows may seem to be off-topic.\n\nThere are several ways to approach this problem.\n\nBinomial Polynomials\n\nOne is by writing the polynomial as a binomial polynomial: $$2k^3+4k^2+2=12\\binom{k}{3}+20\\binom{k}{2}+6\\binom{k}{1}+2\\binom{k}{0}$$ Then use the formula $$\\sum_{k=0}^n\\binom{k}{m}=\\binom{n+1}{m+1}$$ to get \\begin{align} \\sum_{k=0}^n\\left(2k^3+4k^2+2\\right) &=12\\binom{n+1}{4}+20\\binom{n+1}{3}+6\\binom{n+1}{2}+2\\binom{n+1}{1}\\\\ &=\\frac{3n^4+14n^3+15n^2+16n+12}6 \\end{align}\n\nEuler-Maclaurin Sum Formula\n\nIn most cases, the Euler-Maclaurin Sum Formula is an asymptotic approximation, but in the case of polynomials, it is exact. $$\\sum_{k=0}^nf(k)\\sim C+\\int_0^nf(x)\\,\\mathrm{d}x+\\frac12f(n)+\\frac1{12}f'(n)-\\frac1{720}f'''(n)+\\frac1{30240}f^{(5)}(n)+\\dots$$ where subsequent terms involve higher derivatives, which for polynomials will eventually vanish. In the case at hand, this gives the same answer as above.\n\n\u2022 Now I have answered the new question. Unfortunately, there is no $q$ so that $p(n)=\\sum\\limits_{k=0}^nq(k)$ for this particular $p$. \u2013\u00a0robjohn Sep 11 '15 at 17:46", "date": "2019-05-25 20:02:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1431042/how-do-you-calculate-a-sum-over-a-polynomial/1431090", "openwebmath_score": 0.9985715746879578, "openwebmath_perplexity": 247.77765115727212, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319855244919, "lm_q2_score": 0.8596637505099167, "lm_q1q2_score": 0.8516963744261212}}
{"url": "http://math.stackexchange.com/questions/28666/surjective-and-unbounded-functions", "text": "# Surjective and Unbounded functions\n\n\u2022 Every surjective function from $\\mathbb{R}$ to $\\mathbb{R}$ is unbounded.\n\u2022 Every unbounded function from $\\mathbb{R}$ to $\\mathbb{R}$ is surjective.\n\nIs it possible for either of these statements to be false? I have a feeling there is some counterexample that I am missing but I cannot figure it out.\n\nMy understanding is that if a function is unbounded then for all $M\\in\\mathbb{R}$ there is an $x$ such that $|f(x)| \\gt M$.\n\nAnd the definition of surjective is that for all $b \\in Y$, there exists an $x \\in X$ such that $f(x) = b$.\n\nClearly if we have some $M$ in the image of this function there is an $x$ that exists such that $f(x) = M$ by the definition of surjective.\n\nI dont know if I am thinking of this correctly, intuition needed. Thanks.\n\n-\nYour edit still does not give the right definition. Language is not commutative (\"answer the exam and study\" is not the same thing as \"study and answer the exam\"). When you say \"$|f(x)|\\gt M$ for all $M\\in\\mathbb{R}$\" you are saying that the value of $f$ is always greater than any number. That's false for any function with values in $\\mathbb{R}$. What you want to say is \"For every $M\\in\\mathbb{R}$ there exists $x$ such that $|f(x)|\\gt M$.\" This says that the $x$ you pick may depend on $M$. The other way, the $x$ must be independent of $M$. \u2013\u00a0 Arturo Magidin Mar 23 '11 at 16:52\n\nLet's be a bit clearer: a function $f\\colon X\\to\\mathbb{R}$ is unbounded if and only if for every $M\\in\\mathbb{R}$ there exists $x\\in X$ such that $|f(x)|\\gt M$. The quantifier (\"for every\") is important; otherwise, you are just saying that there is at least one $M$ with the property.\n\nIt is certainly true that a surjective function onto $\\mathbb{R}$ (regardless of the domain) is unbounded. Simply note that for every $M\\in\\mathbb{R}$, there exists $N\\in\\mathbb{R}$ with $N\\gt M$, $N\\gt 0$. Since surjectivity of $f$ implies that for every $y\\in\\mathbb{R}$ there exists $x$ such that $f(x)=y$, then putting $y=N$ shows that there exists $x$ such that $f(x)=N$, hence $|f(x)|=f(x)\\gt M$.\n\nSo you are correct in the first one.\n\nThe second one, however, is false: unbounded means you can always exceed any given bound. But it does not guarantee that you can always \"hit\" every value.\n\nConsider the greatest integer function, $f(x)=\\lfloor x\\rfloor$: this is defined as follows $f(x) = \\max\\{n\\in\\mathbb{Z}\\mid n\\leq x\\}$. For example, $f(3.5) = 3$, $f(e) = 2$, $f(-1.5) = -2$, $f(\\sqrt{2}) = 1$.\n\nIs $f$ surjective? Is $f$ unbounded?\n\n-\nFor a simpler example, $f(x)=x^2$ is unbounded but not surjective. \u2013\u00a0 lhf Mar 23 '11 at 16:18\n@lhf: Sure; lots of examples (-: The thing I like about this example as opposed to $x^2$ is that the image is full of gaps: no matter what $M$ you pick, there are always values $N,K\\gt M$ for which there are $x$ with $f(x)=N$, but $f(x)\\neq K$ for all $x$. \u2013\u00a0 Arturo Magidin Mar 23 '11 at 16:24\nFor the second one it says, an unbounded function that maps from X -> R, doesn't that mean its image must span R? \u2013\u00a0 1337holiday Mar 23 '11 at 16:29\n@1337holiday: No. It just means that the image must be contained in $\\mathbb{R}$, and be unbounded as a set. (\"Span\" is usually reserved for linear maps; there is no linearity assumption here). A \"function from $\\mathbb{R}$ to $\\mathbb{R}$\" just means the variable is real and the output is real. \u2013\u00a0 Arturo Magidin Mar 23 '11 at 16:31\n@1337holiday: The \"contrapositive\" of \"If P then Q\" is \"If not Q, then not P\". The negation of \"If P then Q\" is \"P and not Q\". The contrapositive of \"if unbounded, then surjective\" would be \"if not surjective, then bounded\". This is also false (as my example shows: the function is not surjective, but it is not bounded). The negation of \"if unbounded, then surjective\" would be a function that is both unbounded and not surjective (like my example). A bounded function that is not surjective does not address the statement \"if unbounded then surjective\" at all. \u2013\u00a0 Arturo Magidin Mar 23 '11 at 16:42\n\nFor the first question, let's prove the contrapositive: bounded implies not surjective. So let $f$ be bounded. This means that there is a number $M$ such $|f(x)|\\le M$ for all $x$. But then the value $M+1$ is never taken by $f$ and thus $f$ is not surjective.\n\nFor the second question, the function $f(x)=x^2$ is unbounded but not surjective.\n\n-\nBut how is x^2 unbounded when f(x) = -1 does not exist? \u2013\u00a0 1337holiday Mar 23 '11 at 16:40\n@1337holiday: Look at the definition of \"unbounded\": \"For every $M$ there exists an $x$ such that $|f(x)|\\gt M$.\" Is this true for $f(x)=x^2$? Yes: given ANY $M$, there is always a value of $x$ for which $|x^2|$ will be larger than $M$. The fact that it never takes the value $-1$ is completely irrelevant to the concept of boundedness. Graphically: \"bounded\" means that you can draw two horizontal lines in the plane in such a way that the graph of the function is always between the two lines. Is that true for $y=x^2$? \u2013\u00a0 Arturo Magidin Mar 23 '11 at 16:50\nWow this did it for me, i think i finally understand this! No it is not possible for this function, that is a good way to put it (draw 2 lines). \u2013\u00a0 1337holiday Mar 23 '11 at 16:54\n\nThis statement\n\n\u2022 Every unbounded function from R to R is surjective is false. Counter example as lhf points out, any function $f(x)=x^{4n}$ for $n \\in \\mathbb{N}$.\n-", "date": "2015-09-04 06:29:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/28666/surjective-and-unbounded-functions", "openwebmath_score": 0.9468472599983215, "openwebmath_perplexity": 163.04387613507973, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808678600415, "lm_q2_score": 0.8688267813328977, "lm_q1q2_score": 0.8516942712250595}}
{"url": "http://mathhelpforum.com/advanced-statistics/23276-euro-2008-soccer-seedings-print.html", "text": "# Euro 2008 Soccer Seedings\n\n\u2022 November 21st 2007, 08:18 PM\nchopet\nEuro 2008 Soccer Seedings\nEuro 2008 seedings for draw on Dec. 2 in Vienna:\n\nPot 1: Switzerland (hosts), Austria (hosts), Greece (holders), Netherlands\n\nPot 2: Croatia, Italy, Czech Republic, Sweden\n\nPot 3: Germany, Romania, Portugal, Spain\n\nPot 4: Poland, France, Turkey, Russia\n\n4 groups will be created, with each Pot contributing 1 member to each group.\nSo Group A may have Switzerland, Croatia, Germany and Poland.\n\nI was wondering about the probability for Netherlands, Italy, Germany and France to be drawn into the same group. Is it:\n${1\\over(4!)^4}$\n\u2022 November 23rd 2007, 09:14 AM\nTKHunny\nWhoa!!! That is WAY too small. There are only $4^{4}$ ways to pick teams. You have selected one such way. There are only four ways to achieve the one way you have selected, that is, they must be selected together in one of the four selection rounds.\n\nPr(Selected in the first round) = (1/4)^4\n\nThe problem with the next step is that not all paths lead to the desired result. If any of the four teams is selected in the first round, they all must be. If any of the four teams is selected in the first round, but any of the others is not, the process ends. The ONLY way to get to round two is for NONE of the teams to be selected in round 1. There are only a few ways to do that.\n\nPr(None selected in round 1) = (3/4)^4\n\nThus, Pr(Selected in round 2) = (3/4)^4*(1/3)^4, since round 2 doesn't exist without its dependence on the outcome of round 1.\n\nUsing the same logic:\n\nPr(Selected in round 3) = (3/4)^4*(2/3)^4*(1/2)^4\n\nIf we get to round four with hope, obviosuly they will be selected together.\n\nPr(selected in round 4) = (3/4)^4*(2/3)^4*(1/2)^4*(1)^4\n\nThus, (3/4)^4 + (3/4)^4*(1/3)^4 + (3/4)^4*(2/3)^4*(1/2)^4 + (3/4)^4*(2/3)^4*(1/2)^4*(1)^4 = 4*(1/4)^4 = 0.015625\n\u2022 November 23rd 2007, 10:20 AM\nPlato\nI missed this question when it was first posted.\nThere is another way to see the answer is $\\frac{1}{{\\left( {4!} \\right)^3 }}$.\nThink of the first row of host teams as urns: S, A, G, N.\nThen arrange those four letters on each of the three lines to provide one possible selection.\n\u2022 November 26th 2007, 03:27 AM\nchopet\nHi Plato, I was thinking along the same line as you but I missed out fixing the top line.\n\nBut the answer is too small. and TKHunny's one is more intuitively correct.\nThen I realise we should fix even more than we should.\n\nLet's give each country a label, with Pot 1 labelled A1,A2,A3,A4, and so forth, and we can create a table with the rows representing the pots and the columns representing the groups:\n\n1A 1B 1C 1D\n2A 2B 2C 2D\n3A 3B 3C 3D\n4A 4B 4C 4D\n\nEach row can be arranged in 4! way. There are 4 rows giving: $4!^4$ ways.\nBut I am interested only in:\n\nx x 1C x\nx x 2C x\nx x 3C x\nx x 4C x\n\nI don't care what happens to all other groups, so I have to divide by $3!^4$ ways.\nAnd the group {1C,2C,3C,4C} can exist in column 1, 2, 3 or 4, I divide by 4.\n\nFinally, the probability is: $4{ {3!^4}\\over{4!^4}}={1\\over 4^3}=0.015625$\n\nThanks guys!\n\u2022 November 27th 2007, 11:10 PM\nPHANTOM\nJust like Plato said\nI agree with the answer Plata wrote.\n\nevery POT has 4 members and the way we scatter these 4 members into different groups is : 4!\n\nWe have 4 pots, so the number of ways for all combinations is: 4! x 4! x 4! x 4!\nthat's the number of elements of the sample space.\n\nnow lets figure out how many elements of the sample space account for \" the 4 specific teams being together\". There are 4! ways to arrange the 4 groups.\n\nso P= 4!/(4!)^4= 1/(4!)^3 same as Plato's answer.\n\u2022 November 30th 2007, 04:05 AM\nchopet\nYour method doesn't discount the fact that while your 4 countries are in the same group, any 4 OTHER countries can make up the OTHER groups.\n\ne.g.\n{France,Germany,Italy,Holland} in one group is what we want.\n\n{Austria,Croatia,Spain,Poland} in another group is actually INDISTINGUISHABLE from\n{Switzerland,Croatia,Spain,Poland}\nor\n{Switzerland, Sweden,Spain,Poland}\nor\n{Sweden, Czech, Portugal,Poland}\n...\n\nI can go on, but I know you have $3!^4$ groups to discount.", "date": "2014-08-21 13:22:18", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/advanced-statistics/23276-euro-2008-soccer-seedings-print.html", "openwebmath_score": 0.542851448059082, "openwebmath_perplexity": 1681.236546163716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808684361289, "lm_q2_score": 0.868826771143471, "lm_q1q2_score": 0.8516942617370795}}
{"url": "https://math.stackexchange.com/questions/3827883/reduced-row-echelon-form-of-an-augmented-matrix-is-not-unique", "text": "# Reduced row echelon form of an augmented matrix is not unique\n\nI am given a sysetem of linear equations which after graphing, have no solution (the three lines intersect at different points). Now I am trying to prove this algebraically.\n\nAs an augmented matrix,\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 1 & 1 & 1\\\\ 2 & 3 & 6\\\\ \\end{bmatrix}$$\n\n\u2022 $$R_{1}-R_{2} \\Rightarrow R_{2}$$\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 0 & -2 & 2\\\\ 2 & 3 & 6\\\\ \\end{bmatrix}$$\n\ncontinue from here in $$(1)$$ or $$(2)$$\n\n$$(1)$$\n\n\u2022 $$R_{1}-\\frac{1}{2}R_{3} \\Rightarrow R_{3}$$\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 0 & -2 & 2\\\\ 0 & -\\frac{5}{2} & 0\\\\ \\end{bmatrix}$$\n\n\u2022 $$-\\frac{1}{2}R_{2} \\Rightarrow R_{2}$$\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 0 & 1 & -1\\\\ 0 & -\\frac{5}{2} & 0\\\\ \\end{bmatrix}$$\n\n\u2022 $$\\frac{5}{2}R_{2} + R_{3} \\Rightarrow R_{3}$$\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 0 & 1 & -1\\\\ 0 & 0 & -\\frac{5}{2}\\\\ \\end{bmatrix}$$\n\nThen in RREF\n\n$$\\begin{bmatrix} 1 & 0 & 2 \\\\ 0 & 1 & -1\\\\ 0 & 0 & -\\frac{5}{2}\\\\ \\end{bmatrix}$$\n\n$$(2)$$\n\n\u2022 $$-2R_{1}+R_{3} \\Rightarrow R_{3}$$\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 0 & -2 & 2\\\\ 0 & 5 & 0\\\\ \\end{bmatrix}$$\n\n\u2022 $$-\\frac{1}{2}R_{2} \\Rightarrow R_{2}$$\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 0 & 1 & -1\\\\ 0 & 5 & 0\\\\ \\end{bmatrix}$$\n\n\u2022 $$-5R_{2} + R_{3} \\Rightarrow R_{3}$$\n\n$$\\begin{bmatrix} 1 & -1 & 3 \\\\ 0 & 1 & -1\\\\ 0 & 0 & 5\\\\ \\end{bmatrix}$$\n\nThen in RREF\n\n$$\\begin{bmatrix} 1 & 0 & 2 \\\\ 0 & 1 & -1\\\\ 0 & 0 & 5\\\\ \\end{bmatrix}$$\n\nwhich is different from the RREF in $$(1)$$\n\nCan someone explain why I end up with a different RREF? I thought all RREF are unique, but clearly not in this case. Of course as mentioned earlier, the system has no solutions and both augmented matrices show this but their RREF's are not unique still.\n\n\u2022 I'm not sure if RREF are only unique for coefficient matrices instead of augmented matrices. Or if RREF are only unique for consistent systems Sep 16, 2020 at 0:43", "date": "2022-05-29 02:02:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3827883/reduced-row-echelon-form-of-an-augmented-matrix-is-not-unique", "openwebmath_score": 0.7409714460372925, "openwebmath_perplexity": 1023.3868927002849, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9802808701643914, "lm_q2_score": 0.868826769445233, "lm_q1q2_score": 0.8516942615738899}}
{"url": "http://cjvy.hubischule-krone.de/boolean-matrix-operations.html", "text": "\u2022 # Boolean Matrix Operations\n\nJoin Coursera for free and transform your career with degrees, certificates, Specializations, & MOOCs in data science, computer science, business, and dozens of other topics. The boolean type\u00b6. Boolean Algebra Calculator Circuit Diagram. You can also remember operator precedence using the PEMDAS acronym, which stands for Parentheses, Exponent, Multiply And Divide, Add and Subtract. Access Management, Folder access matrix, logical access control. The logical operators are typically used in decision making like in an \"if\" statement, which we will see later. The product is in a way the opposite of boolean sum. Operator in java is a symbol that is used to perform operations. View Animesh Raj Vardhan\u2019s profile on LinkedIn, the world's largest professional community. Quantitative measurement is replaced by a simple light-or-no-light decision, something optics can do well. Properties of matrix multiplication. From the operators point of view, Scilab is able to fulfil arithmetic calculations, comparison and logical operations. A and B must be matrices with the same dimensions or one from them must be a single boolean. How to perform element wise boolean operations on numpy arrays [duplicate] Logical operators for Is there a canonical way of doing element wise boolean. BOOLEAN RANK Matrix rank. Operator Precedence. Boolean matrices is to treat them as integer matrices, and apply a fast matrix multiplication algorithm over the integers. Certain MATLAB \u00ae functions and operators return logical values to indicate fulfillment of a condition. In this video, I go through an easy to follow example that teaches you how to perform Boolean Multiplication on matrices. Define boolean operation. If you have never used logical operators in Matlab, this is a MUST-READ!. We want the matrix multiplication to preserve the structure of Boolean algebra. Logical operators Another key concept in programming is the ability to test a conditional statement and make decisions about the flow of the program based on the truth value of the statement. For example, the following two expressions are equivalent. The built-in VBA operators consist of mathematical operators, string operators, comparison operators and logical operators. Boolean Algebra Calculator Circuit Diagram. Adjacency matrix representation. Fast sparse boolean matrix product with possible preprocessing. download crud matrix for atm system free and unlimited. Conceptual, logical and Physical data model Conceptual, logical and physical model or ERD are three different ways of modeling data in a domain. Linear algebra, like matrix multiplication, decompositions, determinants, and other square matrix math, is an important part of any array library. These variables can be used to define matrices of booleans, with the usual syntax. gL is the logical operation and g is the corresponding physical operation. As with vectors, in this context we call a number a scalar in order to emphasize that it is not a matrix. , the product of two n nmatrices can be computed in O(n3 ) additions and multiplications over the \ufb01eld. Statements that use the logical operators return Boolean (TRUE or FALSE) values. A logical matrix, in the finite dimensional case, is a -dimensional array with entries from the boolean domain Such a matrix affords a matrix representation of a -adic relation. Operators and Elementary Operations Arithmetic, relational, and logical operators, special characters, rounding, set functions The MATLAB \u00ae language uses many common operators and special characters that you can use to perform simple operations on arrays of any type. frame object. Two LOGICAL expressions combined with an. For !, a logical or raw vector(for raw x) of the same length as x: names, dims and dimnames are copied from x, and all other. An adjacency matrix is a way of representing a graph G = {V, E} as a matrix of booleans. The logical operations are also generally applicable to all objects, and support truth tests, identity tests, and boolean operations: operator. List of operators. Of course, if during the working of the device, an adversary turns on unwanted interactions, such as a Hamiltonian on the control qubit that is not diagonal in the logical basis, this could prevent the device from implementing the correct operation on the systems of interest. Boolean Algebra. It is instead defined so that it is analytic and continuous (the transition from $(-\\infty, 1]$). In defining the effect that a logical operation has on two propositions, the result must be specified for all four cases. This section will simply cover operators and functions specifically suited to linear algebra. And actually, there's a logical reason why I centered addition at zero. GHELASE Daniela 2 1\u201ePolitehnica\u201d University of Bucharest 2\u201eDunarea de Jos\u201d University of Galati ABSTRACT The AutoCAD software is a power computer-aided design (CAD) system that can offers all users of graphics, 2D and 3D objects representation. In this article, you will learn about different R operators with the help of examples. BOOLEAN RANK Matrix rank. To date we have accommodated various brand to enhance our customer needs such as PELCO, Hikvision, Indigovision, FLIR , BOSCH ,AXIS ,HERNIS, Samsung , MOBOTIX, Phillips ,Honeywell ,Arecont Vision , Geovision, GUARDTECH ,SONY. com is a leading Professional Closed Circuit Television (CCTV) Surveillance Security System Distributor , Reseller & System Integrator in Malaysia. A logical matrix, in the finite dimensional case, is a -dimensional array with entries from the boolean domain = {,}. The best transitive closure algorithm known, due to Munro, is based on the matrix multiplication method of Strassen. Correspondence will be sent to the first author, unless otherwise indicated. Such a matrix affords a matrix representation of a k {\\displaystyle k\\!} -adic relation. The logical data type represents true or false states using the numbers 1 and 0, respectively. See _tensor_py_operators for most of the attributes and methods you\u2019ll want to call. Logical Reasoning Questions and Answers Matrix Processing House reviews Based on Matrix Processing House Operations Manager. Logical Operators Logical operators are extremely useful in subsetting vectors and in controlling program \ufb02ow. Notably these do work bitwise for raw arguments. Selection is the means by which MATLAB makes decisions about the order in which it executes its statements. Operators are used to perform various operations on variables and constants. So setting every value >0 to 1 in the product will solve your issue. Boolean Operators In Matlab 2. They can be used to selectively execute code based on the outcome of the condition. Maintaining track of software licenses and renewal of antivirus, Inventory Management. THIS SITE IS FOR U. It is maintained by the Cinder team and Vendor driver maintainers. Apart from the classical arithmetic operators, R contains a large set of operators and functions to perform a wide set of matrix operations. For complex matrices, computes the complex conjugate (Hermitian) transpose. OpenSCAD User Manual/Mathematical Operators. Logical operators such as greater than, less than, and equal to are also defined in both matrix and colon versions (note that the equals operator is ==, as opposed to = for assignment). All that is required is to extend this to the rest of the possible GPA's - the full code is in the section below, along with an image of the graph it will create. ) and functions like any, all, isnan, isinf, and isfinite. Python provides the boolean type that can be either set to False or True. A logical matrix, in the finite dimensional case, is a -dimensional array with entries from the boolean domain Such a matrix affords a matrix representation of a -adic relation. The future state unknown, but is either S1 or S2. ANNEX I \u2013 SINGLE FORM - LOGICAL FRAMEWORK MATRIX Guidelines These guidelines are drafted for information purposes only. Framework for classifying logical operators in stabilizer codes Beni Yoshida and Isaac L. Statements that use the logical operators return Boolean (TRUE or FALSE) values. Matrix Rank. This example program uses the literal constants true and false. I modify this program for my application: instead of using multiplication, addition, soustraction and division, i use Xor and & function as logical operators. Matrix Commands for Solving Linear Equations det Computes determinant of an array. tables of composition and nutritional values of organically produced feed materials for pigs and poultry soile. characteristic_polynomial() == A. Boolean Logic Operations A Boolean function is an algebraic expression formed using binary constants, binary variables and Boolean logic operations symbols. Know miscellaneous operations on arrays, such as finding the mean or max ( array. Monitoring and Evaluation Web-Based Tool,Mandeonline ,M&E Software Tools,Monitoring and Evaluation,monitoring and Evaluation software tool,social learning,web application,monitoring and Evaluation webbased tool,reporting tool ,donor tools,NGO software,project monitoring & evaluation,Balanced Scorecard,Donor Management,Annual Operation Plan,Results-Based Programme and Project Management: The. In this extension the logical operations are normally carried out in bitwise fashion on binary representations of those integers, comparing corresponding bits with corresponding bits to produce the output pixel value. The plus sign (+) and minus sign (-) can be used as prefix operators. Note that BMM can be computed using an algorithm for integer matrix multiplication, and so we have BMM for n !nmatrices is in O(n ) time, where !<2:373 (the current bound for integer matrix multiplication). But, for the beginner, they seem to often mysteriously fail, and it is difficult to understand why. The Boolean equal to operator is different from the evaluation operator. ModelicaReference. The transition from optical numerical matrix algebra to optical Boolean matrix algebra is explored in detail. View Animesh Raj Vardhan\u2019s profile on LinkedIn, the world's largest professional community. Before we define an elementary operation, recall that to an nxm matrix A, we can associate n rows and m columns. A Boolean object combines two or more other objects by performing a Boolean operation or operations on them. As a similar post of mine Find all Combinations of a Matrix I am trying to find matrix combinations with entries $>0$ meaning for a matrix \\begin{bmatrix} 0 & 1 & 3 \\\\ 5 & 2 & 1 \\\\ 0 & 0 & 10 \\end{bmatrix}. Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1. com is a leading Professional Closed Circuit Television (CCTV) Surveillance Security System Distributor , Reseller & System Integrator in Malaysia. That is, islogical(ans) returns 1. Multiplication is denoted by an asterisk (*). Let U be a Boolean algebra with at least two elements. For more information, see how to Find Array Elements That Meet. The above operations are carried out through the menu that can divide everything that you want to run into logical groups that make it easier and quicker to tray; Quick launch of applications using hot keys; Logical grouping of commands; Using separators to make the. class theano. The below mentioned table gives the arithmetic operators hold up by R language. R Tutorial \u2013 We shall learn about R Operators \u2013 Arithmetic, Relational, Logical, Assignment and some of the Miscellaneous Operators that R programming language provides. We've seen the matrix before in Lecture 1 as a 2-D array. \"Operations\" is mathematician-ese for \"procedures\". Fast sparse boolean matrix product with possible preprocessing. They are shown in the following picture :. The result is an array of logical values the size of the array operand. expressions, which have a numerical value, and logical expressions, which evaluate to true or false. Intersection, union, complementation, and containment of elements is expressed in U. They are shown in the following picture :. SYMBOL in Matlab MEANING & logical AND operator |. A matrix construction values are converted to their boolean equivalent, so '' = false, 'foo' = true, 1 = true, 0 = false etc, according to normal PHP casting rules for boolean. If the left operand in a statement that contains the and operator is FALSE, the right operand is not evaluated. This example will help you understand, how the logical operators in R Programming are used in If statements. Examples of such statements are \"Is A equal to B?\". Logical operators in MATLAB are those that link logical statements together and return true (1) or false (0) in a logical matrix depending upon the nature of the logical operator and the value of the components. This section will simply cover operators and functions specifically suited to linear algebra. We've seen the matrix before in Lecture 1 as a 2-D array. Other output function operators can have a more profound effect on the operation of the function. When one operand is a list or matrix and another is an operand of some other type, the other operand is combined with each of the elements of the list or matrix. is the create matrix icon, which upon click will bring up a dialog box that allow us to create a matrix with speci\ufb01ed number of rows and columns. Because the BLAS are efficient, portable, and widely available, they're commonly used in the development of high quality linear algebra software, LAPACK for example. The expression value is TRUE (1), if the values of x and y are true (not null). Mar 22, 2019 \u00b7 In this article I have discussed briefly about almost all arithmetic and logical instructions of 8086 microprocessor. Visual and Graphical language unlike textual high-level, such as C, C++, Java\u2026. Many times, logical operators are used to link together the results of several relational operators. Engaging math & science practice! Improve your skills with free problems in 'Solving Word Problems Using Matrix Operations' and thousands of other practice lessons. Keywords Boolean matrix decomposition \u00b7 Boolean rank \u00b7 Ef\ufb01cient algorithm \u00b7Educational database 1 Introduction Matrixdecomposition,a. Boolean Algebra. for i = 1:22 plot(C1(i, :), 'color', 'b'), hold on end Not sure how I can exclude the first column. Instead of a single index, we can use two indexes, one representing a row and the second representing a column. Quote: A man has integrity if his interest in the good of the service is at all times greater than his personal pride, and when he holds himself to the same line of duty when unobserved as he would follow if his superiors were present General S. We will now work our way through the table of identities, in order, making observations about each, usually including a \"common sense\" informal proof. R's binary and logical operators will look very familiar to programmers. Operator overloading allows C/C++ operators to have user-defined meanings on user-defined types (classes). Akin to other matrix factorizations, the factor matrices can be used. First, we define two characteristic matrices of a covering. Computing arXiv:1803. Generally, at least one is a variable. For example: +, -, *, / etc. 9% that fail. T = true(sz) is an array of logical ones where the size vector, sz, defines size(T). array_equal (a1, a2) True if two arrays have the same shape and elements, False otherwise. The original application for Boolean operations was mathematical logic, where it combines the truth values, true or false, of individual formulas. The order in which these actions occur is unimportant as it does not affect the final result. Matrix Operations a la Shmoop you look at Gilligan Huffington III CEO of Handbags. Definition 3. A Boolean matrix multiplication is the usual matrix multiplication with Boolean operations: 0 + 0 = 0 , 1. The most common comparison is between two numeric values. In the product dimension (horizontal) of the matrix, the life cycle of a typical product is represented. The operators act on each element of the vector. Using these operators can greatly reduce or expand the amount of records returned. There are versions of R available for Windows, Mac OS and Unix that can be freely downloaded over the Internet. A system of m Boolean equations in n variables can be stated in matrix form with the usual matrix operations such as Ax = b and Rw = r, where x \u2208 B n and w \u2208 B 2 n are Boolean vector variables. ModelicaReference. Each element in a cell array has a non-neglegable amount of overhead to define the size, shape, and type of the data it stores. matrix into a Boolean embedding matrix under orthogonal or near-orthogonal rotations. All logical operators take Booleans as operands and produce a Boolean. By boolean, or binary matrix, we mean a matrix whose entries are from the \ufb01eld with two elements. It is built deeply into the R language. Oct 17, 2014 \u00b7 Boolean and relational operators in Matlab 1. For complex matrices, computes the complex conjugate (Hermitian) transpose. A Boolean matrix can be expressed as a product of two Boolean matrices, where the \ufb01rst matrix represents a set of meaningful. These basis states are analogous to the orthonormal unit vectors in Euclidean space. R has many operators to carry out different mathematical and logical operations. Like comparison operations, each element of an element-by-element boolean expression also has a numeric value (1 if true, 0 if false) that comes into play if the result of the boolean expression is stored in a variable, or. Visual and Graphical language unlike textual high-level, such as C, C++, Java\u2026. An identity matrix is a square matrix with ones along the diagonal and zeros elsewhere. BOOLEAN RANK Matrix rank. Operator overloading allows C/C++ operators to have user-defined meanings on user-defined types (classes). Find the result of a multiplication of two given matrices. *) and exponentiation (. The values returned by MATLAB logical operators and functions, with the exception of bit\u2212wise functions, are of type logical and are suitable for use with logical indexing. equations via matrix operations. 06 for \u00b5Vision\u00ae armasm User GuideVersion 5Home > Symbols, Literals, Expressions, and Operators > Addition, subtraction, and logical operators 7. Level 1 BLAS do vector-vector operations, Level 2 BLAS do matrix-vector operations, and Level 3 BLAS do matrix-matrix operations. A Boolean object combines two or more other objects by performing a Boolean operation or operations on them. The complexity of subtraction operation is O(m*n) where m*n is order of matrices; Matrices Multiplication - The multiplication of two matrices A m*n and B n*p gives a matrix C m*p. Relational operators perform element-by-element comparisons between two arrays. When an expression includes several operators, it is called a compound expression. On this page, we will discuss these type of operations. rank Computes rank of a matrix. We have a symbology for denoting Boolean variables, and their complements. Thus (as in Stata) logical operators return one for true and zero for false. Instead of a single index, we can use two indexes, one representing a row and the second representing a column. If none are found, find returns an empty, 0-by-1 matrix. How to Contact The MathWorks: www. The idea: I am making quantization intervals with hQInt. Know miscellaneous operations on arrays, such as finding the mean or max ( array. Think of it as a replacement for multiple simple formulas or as a shorthand in which a single formula is provided will all the information required to carry out a complex operation. Boolean logic. In Map Algebra, operators apply a mathematical operation on input rasters and numbers. Note that BMM can be computed using an algorithm for integer matrix multiplication, and so we have BMM for n !nmatrices is in O(n ) time, where !<2:373 (the current bound for integer matrix multiplication). ) and functions like any, all, isnan, isinf , and isfinite. \u261eThis page belongs to resource collections on Logic and Inquiry. Logical operators in MATLAB are those that link logical statements together and return true (1) or false (0) in a logical matrix depending upon the nature of the logical operator and the value of the components. Planning a new project, representing some algorithm or some process, illustrating a solution to a given problem, representing process operations, analysing, designing, documenting, managing a process in various fields is always better to do in a way of flowchart and the types of flowcharts and diagrams are numerous. LFA, the Logical Framework Approach, is an instrument for objective-oriented planning of projects. A one-dimensional array corresponds to a vector, while a two-dimensional array corresponds to a matrix. Unary operators take an operand on the right. Consider the expression A + B * 5. The new object is a Compound Object of type Boolean. Operators and Elementary Operations Arithmetic, relational, and logical operators, special characters, rounding, set functions The MATLAB \u00ae language uses many common operators and special characters that you can use to perform simple operations on arrays of any type. You can use those logical values to index into an array or execute conditional code. GHELASE Daniela 2 1\u201ePolitehnica\u201d University of Bucharest 2\u201eDunarea de Jos\u201d University of Galati ABSTRACT The AutoCAD software is a power computer-aided design (CAD) system that can offers all users of graphics, 2D and 3D objects representation. SAS Operators in Expressions. Along with the standard unary and binary operators of conventional alge-braic notation,AMPL provides iterated operators likesumand prod, and a conditional (if-then-else) operator that chooses between two expressions, depending on the truth. We indicate which entries will be changed by performing an indexing operation on the left hand side and then specify the new values on the right hand side. The original application for Boolean operations was mathematical logic, where it combines the truth values, true or false, of individual formulas. Program (1): To perform addition, subtraction, multiplication, right division, left division and exponentiation operations on x and y given as x = 2; y = 3, in MATLAB. Logical operation definition, Boolean operation. However, it is. We can do operations such as addition and multiplication on the matrix in R. To use an operator with a raster, the raster must be a Raster object. The period character (. The function of Definition 7 is applied to the cell matrix B m \u00d7 n. Boolean operations with non-manifold objects or objects that are not water-tight will fail when non-manifold parts interact in the boolean operation. Input: The first line of input contains an integer T denoting the number of test cases. The single quote character may also be used to delimit strings, but it is better to use the double quote character, since that is never ambiguous. It is an algorithm for mining multidimensional association rules from relational databases. Note that BMM can be computed using an algorithm for integer matrix multiplication, and so we have BMM for n !nmatrices is in O(n ) time, where !<2:373 (the current bound for integer matrix multiplication). These variables can be used to define matrices of booleans, with the usual syntax. The PowerShell logical operators evaluate only the statements required to determine the truth value of the statement. Array formulas can be understood as a combination of Array Constants, Array Operators and Array Ranges. Matrix Arithmetic cross Computes cross products. The implementation of left-shift and right-shift operators is significantly different on Windows for ARM devices. Chuang Department of Physics, Massachusetts Institute of Technology, Cambridge, Massachusetts 02139, USA (Received 30 January 2010; published 4 May 2010) Entanglement, as studied in quantum information science, and nonlocal quantum correlations, as studied. max(), array. Animesh Raj has 8 jobs listed on their profile. But, for the beginner, they seem to often mysteriously fail, and it is difficult to understand why. That is, they operate on numbers (normally), but instead of treating that number as if it were a single value, they treat it as if it were a string of bits, written in twos-complement binary. A token is the smallest element in a program that is meaningful to the compiler. A boolean variable is %T (for \"true\") or %F (for \"false\"). Some of them can come in pretty handy, though, if you need to flip. Basically, it returns the opposite Boolean value of evaluating its operand. Relational and Logical Operators. 4 - GLSL Operators (Mathematical and Logical)\u00b6 GLSL is designed for efficient vector and matrix processing. This class contains various functions (methods) that operate on Boolean matrices. eq (a, b) \u00b6. Basic operators in Scilab Scilab is capable of simple mathematical calculation as well as complex calculations. , the product of two n nmatrices can be computed in O(n3 ) additions and multiplications over the \ufb01eld. In other words, 235 is the addition of 128+64+32+8+2+1. Its members are TRUE if the corresponding members in the original vector are to be included in the slice, and FALSE if otherwise. If you have never used logical operators in Matlab, this is a MUST-READ!. For matrices, there are three basic row operations; that is, there are three procedures that you can do with the rows of a matrix. They take each bit in one operand and perform the operation with the corresponding bit in the other operand. - Proven ability to motivate, encourage and support direct managers in building high performance teams across a regional/international cross-functional matrix environment. matrix into a Boolean embedding matrix under orthogonal or near-orthogonal rotations. It evaluates to a single logical value. Instead of elementary algebra where the values of the variables are numbers, and the prime operations are addition and multiplication, the main operations of Boolean algebra are the conjunction and denoted as \u2227, the disjunction or denoted as \u2228, and the negation not denoted as \u00ac. R's binary and logical operators will look very familiar to programmers. For !, a logical or raw vector(for raw x) of the same length as x: names, dims and dimnames are copied from x, and all other. Package bitops has similar functions for numeric vectors which differ in the way they treat integers 2^31 or larger. \"Operations\" is mathematician-ese for \"procedures\". OF Boolean operator allows the user to specify how many terms from a list of terms must be present if it is not necessary to have all terms in the list. The basic laws used in Boolean algebra are commutative law, associate law, distributive law, identity law and redundance law. 4 Binary factorization is different from Boolean factorization in the sense. Both x and y have to be true in order for the solution to be true. True or false (Boolean) conditions. Jun 18, 2018 \u00b7 So is there a way to extract the values into a matrix using logical operators? Using reshape or whatever to reshape the vector into matrix is forbidden in my case. These variables can be used to define matrices of booleans, with the usual syntax. You can think of an r x c matrix as a set of r row vectors, each having c elements; or you can think of it as a set of c column vectors, each having r elements. We will learn how to use relational operators and logical operators. Then X(L) specifies the elements of X where the elements of L are nonzero. is the create matrix icon, which upon click will bring up a dialog box that allow us to create a matrix with speci\ufb01ed number of rows and columns. Boolean Operators You use logical operators in conditional expressions much as you use math operators in numeric expressions. a boolean matrix A = (aij) by an addition circuit actually means to \u201cencode\u201d the matrix A by paths in a directed acyclic graph. unless both parts are. LOGICAL Operators and Expressions. unique, which is useful if you need to generate unique elements, given a vector containing duplicated character strings. For example: +, -, *, / etc. dot Computes dot products. The logical data type represents true or false states using the numbers 1 and 0, respectively. Consider the expression A + B * 5. You drag and drop the empty Array on the Front Panel, next you find a Control or Indicator (Numeric, String, Boolean, etc,) and drag it into the empty Array. In this article I have discussed briefly about almost all arithmetic and logical instructions of 8086 microprocessor. In Boolean algebra you represent the logical values true and false by the numbers 1 and 0 respectively. The Boolean rank of an n-by-m binary matrix A is the least integer k such that there exists n-by-k binary matrix B and k-by-m binary matrix C for which A = B C. Matrix transpose operator. Array formulas (committed with CTRL+SHIFT+ENTER) have one restriction: You can\u00edt use Excel's logical operations AND, OR, etc. A system of m Boolean equations in n variables can be stated in matrix form with the usual matrix operations such as Ax = b and Rw = r, where x \u2208 B n and w \u2208 B 2 n are Boolean vector variables. 2012 While exploring the Julia manual recently, I realized that it might be helpful to put the basic vocabularies of Julia and R side-by-side for easy comparison. The Logical Operator block performs the specified logical operation on its inputs. All that is required is to extend this to the rest of the possible GPA's - the full code is in the section below, along with an image of the graph it will create. boolean operation synonyms, boolean operation pronunciation, boolean operation translation, English dictionary definition of boolean. com Product enhancement suggestions. This is a simple two minute mini-crossword puzzle designed to reinforce the nuanced Boolean Operators the lecture introduced: Near, Next, Proximity. power consuming than that of Boolean matrix. Inventory Management. Key to understanding the use of matrix operations is the concept of the matrix (array) formula. The classes \"octmode\" and \"hexmode\" whose implementation of the standard logical operators is based on these functions. Maintaining track of software licenses and renewal of antivirus, Inventory Management. These are the available functions for logical operators. Sep 01, 2017 \u00b7 Free Online Library: Analysis of Reliability for the Gate Level Fault Tolerant Design using Probabilistic Transfer Matrix method. boolean operation synonyms, boolean operation pronunciation, boolean operation translation, English dictionary definition of boolean. Definition 3. end blocks). Matrix Algebra. These are the 4 basic boolean operations (AND, OR, XOR and NOT). It does this using make. The Arduino Reference text is licensed under a Creative Commons Attribution-Share Alike 3. All six of the standard comparison operations are available:. A function defined on Boolean matrices which depends on the elements of the matrix in a manner analogous to the manner in which an ordinary determinant depends on the elements of an ordinary matrix, with the operation of multiplication replaced by intersection and the operation of addition replaced by union. National Entrepreneurs' Day: 5 Ways You Can Avoid AI Startup Failure. Sida, like many other donor agencies, has decided to use, and to encour-. 4 Closures of Relations Ch 9. The logical operators and, or, nand, nor, xor, xnor and not are defined for BIT and BOOLEAN types, as well as for one-dimensional arrays containing the elements of BIT and BOOLEAN. com is now LinkedIn Learning! To access Lynda. IfA is the adjacency matrix of a random graph G(n,p), the entries in its kth power gives the number of walks of lengthk between each pair of vertices [4]. In Map Algebra, operators apply a mathematical operation on input rasters and numbers. However, a matrix value used as the condition in an if or while statement is only true if all of its elements are nonzero. Logical operation definition, Boolean operation. Most of the methods on this website actually describe the programming of matrices. The original application for Boolean operations was mathematical logic, where it combines the truth values, true or false, of individual formulas. The example below will show you how to use arithmetic operators in MATLAB. Free Online Library: Automated Boolean Matrix data representation scheme through AVL tree for mining association rules. Logical operators are almost always found in the context of Conditional and Loop Structures. com Web comp. A single scalar can be compared against each element in an array. National Entrepreneurs' Day: 5 Ways You Can Avoid AI Startup Failure. In this paper, an association rule mining algorithm base d on th e Boolean matrix (ABBM) is proposed. The result is logical 0 (FALSE) where and are both zero or nonzero. Boolean product of Equation (1a), noting that message passing derivation for factorization and completion using the XOR product of Equation (1b) is similar. Matrix Operations a la Shmoop you look at Gilligan Huffington III CEO of Handbags. Assignment operations, in which we change a value or values in a matrix, are performed in a very similar way to the indexing operations above. IDL has two operators used to multiply arrays and matrices. The Logical Operator block performs the specified logical operation on its inputs. Program (1): To perform addition, subtraction, multiplication, right division, left division and exponentiation operations on x and y given as x = 2; y = 3, in MATLAB. However, Mata does not have a boolean variable type. The expression value is TRUE (1), if the values of x and y are true (not null). I ask everyone, however, to be careful and not to write Subject lines like \"matrix operations in STATA and MATA do not match!\" unless they are certain they are right. We will learn how to use the if-statement, which is the most important method of selection. Does Matlab have a Boolean (sometimes called logical or binary) matrix multiplication function? I'm specifically talking about what's usually denoted by a circle with a dot in it to denote Boolean. Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. You will learn how to use logical operations to search large data matrix very quickly. Boolean Operators include AND, OR, XOR, or NOT and can have one of two values, true or false. NET Framework does not come with built-in support for matrix math,. IfA is the adjacency matrix of a random graph G(n,p), the entries in its kth power gives the number of walks of lengthk between each pair of vertices [4]. In order to define the specific function, relation, and symbols in question it is first necessary to establish a few ideas about the connections among them. Lecture 2 MATLAB basics and Matrix Operations page 11 of 19 Matrix operations: MATLAB is short for MATrix LABoratory, and is designed to be a tool for quick and easy manipulation of matrix forms of data. Boolean matrices is to treat them as integer matrices, and apply a fast matrix multiplication algorithm over the integers.", "date": "2020-01-26 05:46:34", "meta": {"domain": "hubischule-krone.de", "url": "http://cjvy.hubischule-krone.de/boolean-matrix-operations.html", "openwebmath_score": 0.5564281344413757, "openwebmath_perplexity": 1035.6105504835496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9763105273220726, "lm_q2_score": 0.8723473779969194, "lm_q1q2_score": 0.8516819286201998}}
{"url": "https://math.stackexchange.com/questions/3735061/check-this-proof-if-two-columns-rows-of-a-matrix-are-the-same-the-determinant", "text": "# Check this proof: If two columns/rows of a matrix are the same, the determinant is $0$.\n\nI have written this proof stating that if two rows or columns of a matrix are the same, then the determinant of the matrix is equal to 0. Is it correct?\n\nLet us say we have an n x n matrix A, shown below:\n\nFor some $$i,n \\in \\mathbb{N}$$.\n\nIf we say that $$r_{i} = [a_{i1}, a_{i2}, a_{i3} ... a_{in}]$$, a row vector, then we can rewrite the matrix A as:\n\nIf we then also create the swapped $$S_{ij}$$ matrix, i.e. swap rows $$i$$ and $$j$$ around, we have the matrix:\n\nWe know that if we swap two rows of a determinant, in this case rows $$i$$ and $$j$$, then the determinant will simply be the negative of the original determinant.We can say that the det(A) = -det(S). But, if these two rows are identical, then det(S) = det(A), so this means that det(A) = -det(A), so A must be equal to $$0$$.\n\n\u2022 Yes, your solution is correct. Jun 26 '20 at 8:12\n\u2022 It's easier to use that $\\det$ is an alternating multilinear form, i.e., $$\\det(a_1,\\dots,a_i,\\dots,a_j\\dots,a_n)=-\\det(a_1,\\dots,a_j,\\dots,a_i\\dots,a_n).$$ Jun 26 '20 at 9:18\n\nThat seems to be correct. Another way to think about this is to consider what happens when there are two identical columns/rows. Then, the number of linearly independent columns/rows is less than $$n$$ for a given $$n \\times n$$ matrix. Hence, the matrix does not have maximal rank so the determinant has to be 0.\n\nOf course, I'm working off of the idea that you've defined the determinant as the unique map $$det: M(n \\times n, \\mathbb{F}) \\to \\mathbb{F}$$ such that:\n\n1. $$det$$ is linear in each row.\n\n2. If $$rank(A) < n$$, then $$det(A) = 0$$.\n\n3. $$det(E_n) = 1$$.\n\nIf you're not working off of these assumptions, that's fine. What you have written is fine too.\n\nIn essence, yes, your proof is very correct, and thorough.\n\nFor further clarification you can see this video https://www.khanacademy.org/math/linear-algebra/matrix-transformations/determinant-depth/v/linear-algebra-duplicate-row-determinant which gives a very thorough breakdown of your proof.\n\nYes, this is indeed a valid proof which you can see on occasion in published works.\n\nIf you have two (or more) rows which are linearly related, you can resolve some rows to zero. (Check out gaussian elimination to see how row reduction works.) By the way this proof is valid indeed.", "date": "2021-09-22 15:21:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3735061/check-this-proof-if-two-columns-rows-of-a-matrix-are-the-same-the-determinant", "openwebmath_score": 0.9233887791633606, "openwebmath_perplexity": 181.3328073282425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105273220727, "lm_q2_score": 0.8723473763375643, "lm_q1q2_score": 0.851681927000154}}
{"url": "https://www.tutorialspoint.com/explain-binary-search-in-python", "text": "# Explain Binary Search in Python\n\nPythonServer Side ProgrammingProgramming\n\nBinary search is a searching algorithm which is used to search an element from a sorted array. It cannot be used to search from an unsorted array. Binary search is an efficient algorithm and is better than linear search in terms of time complexity.\n\nThe time complexity of linear search is O(n). Whereas the time complexity of binary search is O(log n). Hence, binary search is efficient and faster-searching algorithm but can be used only for searching from a sorted array.\n\n## How does Binary Search work?\n\nThe basic idea behind binary search is that instead of comparing the required element with all the elements of the array, we will compare the required element with the middle element of the array. If this turns out to be the element we are looking for, we are done with the search successfully. Else, if the element we are looking for is less than the middle element, it is sure that the element lies in the first or left half of the array, since the array is sorted. Similarly, if the element we are looking for is greater than the middle element, it is sure that the element lies in the second half of the array.\n\nThus, Binary search continuously reduces the array into half. The above process is recursively applied on the selected half of the array until we find the element we are looking for.\n\nWe will start searching with the left index 0 and right index equal to the last index of the array. The middle element index (mid)is calculated which is the sum of the left and right index divided by 2. If the required element is less than the middle element, then the right index is changed to mid-1, which means we will now be looking at the first half of the array only. Likewise, if the required element is greater than the middle element, then the left index is changed to mid+1, which means we will now be looking at the second half of the array only. We will repeat the above process for the selected array half.\n\n## How do we know if element is not present in the array?\n\nWe need to have some condition to stop searching further which will indicate that the element is not present in the array. We will iteratively search for the element in the array as long as the left index is less than or equal to the right index. Once this condition turns false and we haven\u2019t found the element yet, this means that the element is not present in the array.\n\n### Example\n\nLet us take the following sorted array and we need to search element 6.\n\n 2 5 6 8 10 11 13 15 16\n\nL=0 H=8 Mid=4\n\n 2 5 6 8 10 11 13 15 16\n\n6<10, therefore take the first half.\n\nH=Mid-1\n\nL=0 H=3 Mid=1\n\n 2 5 6 8 10 11 13 15 16\n\n6>5, therefore choose the second half.\n\nL=Mid+1\n\nL=2 H=3 Mid=2\n\n 2 5 6 8 10 11 13 15 16\n\n6==6, an element found\n\nHence the element 6 is found at index 2.\n\n## Implementation\n\nFrom a given sorted array, search for a required element and print its index if the element is present in the array. If the element is not present, print -1.\n\nThe code for the implementation of binary search is given below.\n\n### Example\n\nLive Demo\n\ndef binary_search(arr,x):\nl=0\nr=len(arr)-1\nwhile(l<=r):\nmid=(l+r)//2\nif(arr[mid]==x):\nreturn mid\nelif(x<arr[mid]):\nr=mid-1\nelif(x>arr[mid]):\nl=mid+1\nreturn -1\narray=[1,2,3,4,5,6,7,8,9,10]\na=7\nprint(binary_search(array,a))\nb=15\nprint(binary_search(array,b))\n\n### Output\n\n6\n-1\n\nElement 7 is present at index 6.\n\nElement 15 is not present in the array, hence -1 is printed.\n\nPublished on 11-Mar-2021 09:11:09", "date": "2021-12-02 00:49:47", "meta": {"domain": "tutorialspoint.com", "url": "https://www.tutorialspoint.com/explain-binary-search-in-python", "openwebmath_score": 0.27458852529525757, "openwebmath_perplexity": 352.90324975178623, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105314577313, "lm_q2_score": 0.872347368040789, "lm_q1q2_score": 0.8516819225076558}}
{"url": "https://math.stackexchange.com/questions/1025153/sequence-that-converges-to-0-but-its-function-diverges", "text": "# Sequence that converges to 0 but its function diverges\n\nLet f(x)=-2x+1 if x<0 and f(x)=$x^2$+x if x>0. Give a sequence {$x_n$} in $\\mathbb R$\\ {0} such that {$x_n$} converges to zero but {f($x_n$)} diverges.\n\nI've tried with the sequence {1/n} but it does not seem to work. I can't think of another sequence that would converge to zero. Can anyone help me out please?\n\nThanks.\n\n\u2022 I am a little confused by what the question means by the function at x<0 and x>0. So the function of the sequence must diverge while the sequence converges. \u2013\u00a0Su003 Nov 17 '14 at 0:36\n\u2022 If x is negative than the function value is computed by the first expression while if x is positive then the function value is computed by the second expression. For example, f(-1)=3 while f(1)=2 \u2013\u00a0JB King Nov 17 '14 at 0:42\n\u2022 But x is a term of the sequence {$x_n$} right? \u2013\u00a0Su003 Nov 17 '14 at 0:53\n\u2022 $x$ is just a variable while $x_n$ would be a subscripted variable. \u2013\u00a0JB King Nov 17 '14 at 0:59\n\nHow about $x_n=\\frac{(-1)^n}{n}$ for a sequence that converges to zero but alternates in sign which would make the function values be different. Thus, the sequence here is: $-1,\\frac{1}{2},-\\frac{1}{3},\\frac{1}{4},-\\frac{1}{5},\\cdots$\n\nFor odd $n$, the function values will converge to 1. The first few function values here would be $3,\\frac{5}{3},\\frac{7}{5}$ as the general term will be $\\frac{n+2}{n}$ which could also be seen as $1+\\frac{2}{n}$\n\nFor even $n$, the function values would be $\\frac{1}{n^2}+\\frac{1}{n}=\\frac{n+1}{n^2}$ which would have function values of $\\frac{3}{4},\\frac{5}{16},\\frac{7}{36},...\\frac{11}{100}, .... \\frac{101}{10,000},...\\frac{1001}{1,000,000}$ which will likely converge to 0 from this side as the denominator is growing much faster than the numerator.\n\nPutting these together, the function value sequence would look like this: $3,\\frac{3}{4},\\frac{5}{3},\\frac{5}{16},\\frac{7}{5},\\frac{7}{36},\\frac{9}{7},\\frac{9}{64}..., \\frac{2k+1}{2k-1},\\frac{2k+1}{(2k)^2},\\frac{2k+3}{2k+1},\\frac{2k+3}{(2k+2)^2}...$ which doesn't converge since there are sub-sequences which converge to different values.\n\nIf you want, consider $\\varepsilon=\\frac{1}{10}$ and try to prove the function value sequence converge,i.e. there exists $L,N$ such that for all $n>N\\implies|f(n)-L|<\\varepsilon$.\n\nIf you believe they converge to 0, consider some big odd $n>N$ where the function value will be greater than 1 which doesn't work.\n\nIf you believe they converge to 1, consider some big even $n$ where the function value will be close to zero and more than .1 away from 1 as for $n=1000$ then the function value is $\\frac{1001}{1,000,000}=.001001$. Any bigger even $n$ will have even smaller values to consider.\n\nIf you have an alternative proof of what is wrong with my above answers, show me the error.\n\n\u2022 Does {f($x_n$)} diverge in this case? I tried it but it looks like it converges to 1. \u2013\u00a0Su003 Nov 17 '14 at 0:31\n\u2022 @Su003, even terms will converge to $0$, odd terms to $1$. \u2013\u00a0lhf Nov 17 '14 at 0:33\n\u2022 I understand this. But what is confusing me is that the question says that {f($x_n$)} must diverge. Does such an {$x_n$} work in this case? \u2013\u00a0Su003 Nov 17 '14 at 1:37\n\nThe main point here is that $f$ is not continuous at $x=0$. It does have one-sided limits at $x=0$, but they are different. So, you need to find a sequence that converges to $0$ but visits both sides infinitely often. The answer by JB King uses the simplest possible such sequence.", "date": "2019-09-21 09:48:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1025153/sequence-that-converges-to-0-but-its-function-diverges", "openwebmath_score": 0.9331814050674438, "openwebmath_perplexity": 213.7664505511847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.976310529389902, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8516819190837445}}
{"url": "https://mymathforum.com/threads/how-to-find-the-maximum-angular-speed-from-a-bar-which-rotates-about-an-axis.347797/", "text": "# How to find the maximum angular speed from a bar which rotates about an axis?\n\n#### Chemist116\n\nThe problem is as follows:\n\nA rigid bar of negligible mass has three particles whose masses are the same and are joined to the bar as indicated in the figure. The bar is free to rotate in a vertical plane about an axis with no friction perpendicular to the bar through point $P$ and is released from rest on the horizontal position at $t=0$. Find the maximum angular speed in radians per second attained by the bar. You may consider $g=9.8\\frac{m}{s^2}$ and $d=4.2\\pi$\n\nThe alternatives given are:\n\n$\\begin{array}{ll} 1.&1\\,\\frac{rad}{s}\\\\ 2.&2\\,\\frac{rad}{s}\\\\ 3.&3\\,\\frac{rad}{s}\\\\ 4.&4\\,\\frac{rad}{s}\\\\ \\end{array}$\n\nWhat I've attempted to solve this problem was to equate the potential energy and the rotational energy of the masses assuming there's a superposition of the moment of inertia of the masses.\n\nThis is translated as:\n\n$mgh=\\frac{1}{2}I\\omega^2$\n\n$mgh=\\left(\\frac{1}{2}m\\left(\\frac{4d}{3}\\right)^{2}+\\frac{1}{2}m\\left(\\frac{d}{3}\\right)^{2}+\\frac{1}{2}m\\left(\\frac{2d}{3}\\right)^{2}\\right) \\times \\omega^2$\n\nBut after following the procedure I can't find a way to cancel the 4.2$\\pi$. Supposedly, the answer is the first choice. How can I get to that result?\n\n#### skeeter\n\nMath Team\n$\\omega_{max}$ will occur when the bar is at the vertical ...\n\n$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$\n\n$mg \\cdot \\dfrac{4d}{3} + mg \\cdot \\dfrac{d}{3} - mg \\cdot \\dfrac{2d}{3} = mgd = \\dfrac{1}{2} \\cdot I \\cdot \\omega^2$\n\n$I = m \\cdot \\dfrac{16d^2}{9} + m \\cdot \\dfrac{d^2}{9} + m \\cdot \\dfrac{4d^2}{9} = m \\cdot \\dfrac{7d^2}{3}$\n\n$mgd = m \\cdot \\dfrac{7d^2}{6} \\cdot \\omega^2 \\implies \\omega = \\sqrt{\\dfrac{6g}{7d}}$\n\nusing the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec\n\n#### Chemist116\n\n$\\omega_{max}$ will occur when the bar is at the vertical ...\n\n$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$\n\n$mg \\cdot \\dfrac{4d}{3} + mg \\cdot \\dfrac{d}{3} - mg \\cdot \\dfrac{2d}{3} = mgd = \\dfrac{1}{2} \\cdot I \\cdot \\omega^2$\n\n$I = m \\cdot \\dfrac{16d^2}{9} + m \\cdot \\dfrac{d^2}{9} + m \\cdot \\dfrac{4d^2}{9} = m \\cdot \\dfrac{7d^2}{3}$\n\n$mgd = m \\cdot \\dfrac{7d^2}{6} \\cdot \\omega^2 \\implies \\omega = \\sqrt{\\dfrac{6g}{7d}}$\n\nusing the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec\nCan you include some drawing to accompany your solution? I'm having a problem at identifying where's the height to establish the potential energy. When you rotate the bar about the axis which is represented by the orange dot, the bar is vertical right? But from where to where you put the height?\n\nWhy Initially do we have\n\n$mg \\cdot \\dfrac{4d}{3} + mg \\cdot \\dfrac{d}{3}$\n\nand why finally do we have:\n\n$mg \\cdot \\dfrac{2d}{3}$\n\n@skeeter Can you please represent the justification of this in a drawing?\n\nI got that part of adding the moment of inertia due the principle of superposition of them and the rest is just logical, but...\n\n$\\omega = \\sqrt{\\dfrac{6g}{7d}}$\n\nin this part is where I'm stuck. How on earth did you get an answer close to $1\\frac{rad}{s}$?\n\nIf I do insert the given values, this would become into:\n\n$\\omega = \\sqrt{\\dfrac{6\\times 9.8}{7\\times 4.2 \\times 3.14}} \\approx 0.7980 \\frac{rad}{s}$\n\nCan you please explain this part? Or did I misunderstand something from your explanation? Any help please?\n\n#### skeeter\n\nMath Team\n0.798 is closest to 1rad/sec, given the available choices ... note that I stated disagreement with your given value for $d = 4.2\\pi$\n\nFor $\\omega = 1 \\text{\\rad/sec}$, $d$ would need to equal to 8.4\n\nFull explanation of the method used to find the maximum angular velocity is given in the following diagram\n\ntopsquark", "date": "2020-02-17 17:08:26", "meta": {"domain": "mymathforum.com", "url": "https://mymathforum.com/threads/how-to-find-the-maximum-angular-speed-from-a-bar-which-rotates-about-an-axis.347797/", "openwebmath_score": 0.7833556532859802, "openwebmath_perplexity": 266.36257006040944, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105280113492, "lm_q2_score": 0.8723473614033683, "lm_q1q2_score": 0.8516819130210298}}
{"url": "https://math.stackexchange.com/questions/3334891/series-about-inserting-parentheses", "text": "Problem : Suppose that the series $$\\sum_{}^{} a_k$$ converges and $${n_j}$$ is a strictly increasing sequence of positive integers. Define the sequence $$b_k$$ as follows :\n\n$$b_1=a_1+...+a_{n_{1}}$$\n\n$$b_2=a_{n_1+1}+...+a_{n_{2}}$$\n\n...\n\n$$b_k=a_{n_{k-1}+1}+...+a_{n_{k}}$$\n\nProve that $$\\sum_{}^{} b_k$$ converges and that $$\\sum_{}^{} a_k=$$ $$\\sum_{}^{} b_k$$\n\nMy Proof :\n\nLet $$s_m=\\sum_{i=1}^{n_m} a_i$$ $$\\forall m \\in \\mathbb{N}$$ then by definition, $$s_m=\\sum_{i=1}^{n_m} a_i=\\sum_{k=1}^{m} b_k$$\n\nSince $$n_j$$ is strictly increasing sequence of positive integers, $$\\lim_{m\\to\\infty}\\ n_m= \\infty$$\n\nSince $$\\lim_{m\\to\\infty}\\ n_m= \\infty$$ and $$\\sum_{}^{} a_k$$ converges, $$\\lim_{m\\to\\infty}\\ s_m$$ exists and\n\n$$\\lim_{m\\to\\infty}\\ s_m=$$ $$\\lim_{m\\to\\infty}\\sum_{i=1}^{n_m} a_i=$$ $$\\sum_{}^{} a_k$$\n\nSince $$\\forall m \\in \\mathbb{N}$$ $$s_m=\\sum_{k=1}^{m} b_k$$ and $$\\lim_{m\\to\\infty}\\ s_m$$ exists,\n\n$$\\sum_{}^{} b_k$$ converges and $$\\sum_{}^{} a_k$$=$$\\sum_{}^{} b_k$$\n\nBut, I'm not sure my proof is correct. Please give me feedback on my proof.\n\nYes, the idea of your proof is fine. However, your argument in the 2nd and 3rd line is unclear (among other things since $$n_j$$ is not defined). I suppose, in short your argument is:\nDefine $$s_n := \\sum_{k = 0}^n a_k$$ and $$s_m := \\sum_{k = 0}^m b_k$$. Then $$(s_m)_{m \\in \\mathbb{N}}$$ is a subsequence of $$(s_n)_{n \\in \\mathbb{N}}$$. Since $$\\sum a_k$$ converges (by definition) $$(s_n)_{n \\in \\mathbb{N}}$$ converges, so does $$(s_m)_{m \\in \\mathbb{N}}$$ and thus $$\\sum b_k$$ converges. Since limits are unique, they coincide.", "date": "2021-06-12 11:24:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3334891/series-about-inserting-parentheses", "openwebmath_score": 0.9973806142807007, "openwebmath_perplexity": 94.83368862143031, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9822876992225168, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8516785734518643}}
{"url": "https://math.stackexchange.com/questions/3168820/combinatorial-proof-of-binomnk2-k-binomn2n2-binomk2/3168848", "text": "# Combinatorial proof of $\\binom{nk}{2}=k\\binom{n}{2}+n^2\\binom{k}{2}$\n\nThis identity was posted a while back but the question had been closed; the question wasn't asked elaborately, though the proof of the identity is a nice application of combinatorics and a good example for future reference.\n\nAlso, there was just a flat-out wrong answer which had a couple of upvotes, so I thought to give my own possible proof and sort of 'reopen' the question that was left. Hopefully this time the question will have enough context to stay alive.\n\nAs a side note: the given right side does not appear symmetric between $$k$$ and $$n$$. However, when $$\\binom{m}{2}=m(m-1)/2$$ is inserted and the products expanded, only symmetric terms remain uncancelled. Rendering the right side as $$k^2\\binom{n}{2}+n\\binom{k}{2}$$ would give the same cancellations and the same net value.\n\nProof\n\nSuppose you have a grid of $$n\\times k$$ dots. Firstly, the amount of ways to connect any two dots is $$\\binom{nk}{2}$$. Now consider the right-hand side. We can split the cases for which the connected dots are on the same column, row, or are in both different columns and rows.\n\nIf the two connected dots are in the same column, we can choose two points from any two different rows in $$\\binom{n}{2}$$ ways, and we have $$k$$ columns for which the two points can be on the same column, which gives a total of $$k\\binom{n}{2}$$ options. The same argument holds for constant rows: $$n\\binom{k}{2}$$.\n\nNow if neither the row nor the column can stay constant, we can pick any point in $$nk$$ ways, and choose the second point from the remaining $$(n-1)(k-1)$$ points; one column and one row will be unavailable. This gives us $$\\frac{nk(n-1)(k-1)}{2}$$ options, as we have to rule out the double counting.\n\nWe will now show (algebraically) that $$n\\binom{k}{2}+\\frac{nk(n-1)(k-1)}{2}=n^2\\binom{k}{2}$$. We have that $$\\binom{k}{2}=\\frac{k(k-1)}{2} =\\frac{nk(n-1)(k-1)}{2n(n-1)} \\iff n(n-1)\\binom{k}{2}=\\frac{nk(n-1)(k-1)}{2}=n^2\\binom{k}{2}-n\\binom{k}{2}$$ which leads to the equation above.\n\nCombining these cases gives $$\\binom{nk}{2}=k\\binom{n}{2}+n\\binom{k}{2}+\\frac{nk(n-1)(k-1)}{2} = k\\binom{n}{2}+n^2\\binom{k}{2}$$\n\nIf there are any mistakes or improvements on the arguments, please feel free to point them out.\n\n\u2022 Uh ... How is the left side symmetric when $n$ and $k$ are interchanged and the right side isn't? \u2013\u00a0Oscar Lanzi Mar 30 '19 at 22:28\n\u2022 @OscarLanzi The right side is symmetric too, just not obviously so. \u2013\u00a0Michael Biro Mar 30 '19 at 22:32\n\u2022 I have now explained it. Feel free to roll back the edit if you feel it is inappropriate. \u2013\u00a0Oscar Lanzi Mar 30 '19 at 23:11\n\nI don't think you need as many cases, which saves a little algebra. We have $$k$$ groups of $$n$$ dots each, so choosing two of them can be done in $$\\binom{nk}{2}$$ ways.\nAlternatively, both are in the same group of $$n$$ which has $$\\binom{k}{1} \\cdot \\binom{n}{2}$$ possibilities, or they are in different groups of $$n$$, which has $$\\binom{k}{2} \\binom{n}{1}^2$$ possibilities.\nTherefore, $$\\binom{nk}{2} = k \\binom{n}{2} + n^2 \\binom{k}{2}$$", "date": "2020-12-02 16:46:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3168820/combinatorial-proof-of-binomnk2-k-binomn2n2-binomk2/3168848", "openwebmath_score": 0.7042466998100281, "openwebmath_perplexity": 165.23225668163258, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98228770337066, "lm_q2_score": 0.8670357649558007, "lm_q1q2_score": 0.8516785702986568}}
{"url": "https://math.stackexchange.com/questions/2856173/is-there-any-analysis-method-of-non-differentiability", "text": "# Is there any analysis method of non-differentiability?\n\nI am not sure if this question is proper, if not please let me know I would delete it.\n\nI understand a function can be continuous but not differentiable. I can also understand existence of non-differentiable functions. The question: is there a 'measure' (like distance) between two non-differnetiable functions. The word 'measure' here should not be taken literally. Consider two non-differentiable function $f(x)$ and $g(x)$ the question: is there a way to compare the two functions (like a distance measure ) which tells either $f(x)$ or $g(x)$ is easier to make it differentiable? Something like a how far the function is away from differentiable?\n\n\u2022 I think this is a great question, and I hope it attracts lots of responses - it's unfortunate that you accepted an answer so quickly. Although you didn't want measure to be taken literally, actually, that does provide a bit of information: the sets on which functions are not differentiable can be compared by seeing which has larger measure. Functions which are differentiable almost everywhere can either have or lack more 'completely differentiable-like' properties, such as obeying the fundamental theorem of calculus - this turns out to be equivalent to absolute continuity. \u2013\u00a0forget this Jul 19 '18 at 2:13\n\u2022 @CoryGriffith Does accepting answers reduce attention? \u2013\u00a0Creator Jul 19 '18 at 2:15\n\u2022 I've seen comments to that effect on other answers, but I don't have any way of knowing myself. It seems plausible, though. \u2013\u00a0forget this Jul 19 '18 at 2:20\n\u2022 @CoryGriffith Ok I will wait for few days and see what happens. thanks \u2013\u00a0Creator Jul 19 '18 at 2:22\n\nThere is Rademacher's theorem which says that a function that is Lipschitz continuous is differentiable (almost everywhere). So one way of interpreting your question is to \"how close to Lipschitz continuous is a function\". Lipschitz continuity is a specific case of H\u00f6lder continuity.\n\nA function is $\\alpha$-H\u00f6lder continuous for $\\alpha\\in (0,1]$ if\n\n$$|f(x)-f(y)|<K|x-y|^\\alpha$$\n\nIf $\\alpha=1$ then we say the function is Lipschitz.\n\nNote that if $f$ is $\\alpha$-H\u00f6lder, then it is $\\beta$-H\u00f6lder for $\\beta<\\alpha$.\n\nSo if $f$ is $.6$-H\u00f6lder and $g$ is $.7$-H\u00f6lder, then $g$ is \"closer\" to being Lipschitz, i.e. differentiable.\n\nHere are some pictures of various functions that are of increasing H\u00f6lder continuity:\n\nH\u00f6lder continuous for all $\\alpha \\in(0,0.15)$:\n\nH\u00f6lder continuous for all $\\alpha \\in (0,0.55)$:\n\nH\u00f6lder continuous for all $\\alpha\\in (0,0.75)$:\n\nH\u00f6lder continuous for all $\\alpha\\in (0,0.95)$:\n\nNote how the paths get \"nicer\"/closer to being able to differentiate.\n\n\u2022 May I ask how did you get this plots? \u2013\u00a0Creator Jul 19 '18 at 2:01\n\u2022 @Creator They are on the Wikipedia page on fractional Brownian motion \u2013\u00a0user223391 Jul 19 '18 at 2:02\n\u2022 @Creator Oh hey, it's you! I remember talking rough paths with you on here. Thanks for asking such good questions. Didn't realize you were asking the question at first, haha. Feel free to shoot me an email or whatever if you want to talk rough paths. \u2013\u00a0user223391 Jul 19 '18 at 2:04\n\u2022 This is the very reason I asked the question mathoverflow.net/questions/304654/\u2026 but it is shown to be related to Hurst parameter, but look at your plots now, so close to low and high SNR type. \u2013\u00a0Creator Jul 19 '18 at 2:04\n\u2022 . . . nice . . . \u2013\u00a0janmarqz Jul 19 '18 at 2:22", "date": "2021-04-15 13:50:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2856173/is-there-any-analysis-method-of-non-differentiability", "openwebmath_score": 0.7532082200050354, "openwebmath_perplexity": 449.26616464730796, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877007780707, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8516785629884419}}
{"url": "https://math.stackexchange.com/questions/650739/using-field-axioms-for-a-simple-proof", "text": "# Using field axioms for a simple proof\n\nQuestion: If $F$ is a field, and $a, b, c \\in F$, then prove that if $a+b = a+c$, then $b=c$ by using the axioms for a field.\n\nRelevant information:\nField Axioms (for $a, b, c \\in F$):\n\n$a+b = b+a$ (Commutativity)\n$a+(b+c) = (a+b)+c$ (Associativity)\n$a+0 = a$ (Identity element exists)\n$a+(-a) = 0$ (Inverse exists)\n\nMultiplication:\n$ab = ba$ (Commutativity)\n$a(bc) = (ab)c$ (Associativity)\n$a1 = a$ (Identity element exists)\n$aa^{-1} = 1$ (Inverse exists)\n\nDistributive Property:\n$a(b+c) = ab + ac$\n\nAttempt at solution:\nI'm not sure where I can begin. Is it ok to start with adding the inverse of a to both sides, as in the following?\n$(a+b)+(-a) = (a+c)+(-a)$ (Justification?)\n$(b+a)+(-a) = (c+a)+(-a)$ (Commutativity)\n$b+(a+(-a)) = c+(a+(-a))$ (Associativity)\n$b+0 = c+0$ (Definition of additive inverse)\n$b = c$ (Definition of additive identity)\n\nI'm wondering about my very first step. Specifically, the axioms don't mention anything about doing something to both sides of an equation simultaneously. Is there some other axiom I can use to justify this step?\n\nThis is Exercise 1, part b in Section 1 on page 2 of Halmos, Finite Dimensional Vector Spaces (reading book for fun--this is not homework (probably too easy to be a homework problem anyway!)). In part a, I proved that $0+a = a$, in case that is somehow helpful in this problem. Thanks!\n\n\u2022 Just a minor note: you may want to write $1/a$ as $a^{-1}$ because you would want to define \"/\" later using the inverses and it wouldn't be good to have syntax with ambiguous parsing. \u2013\u00a0user21820 Jan 25 '14 at 11:33\n\u2022 That's a good idea--I made the change. \u2013\u00a0Mike Bell Jan 25 '14 at 11:44\n\nMart\u00edn-Blas P\u00e9rez Pinilla suggests that \"=\" can be considered a logical symbol obeying logical axioms. While I agree that it fundamentally is so, I would like to note that it is possible to consider it an equivalence relation obeying 'internal' field axioms, because for example the rational numbers can be taken as equivalence classes of a certain set of pairs of integers, and so it is not quite right to consider the equality between these rationals as a logical equality. Also, Ittay made a mistake where he used an unstated axiom that allows substitution. What you need, either way, is something equivalent to the following for any field $F$:\n\n$a=a$ for any $a \\in F$ [reflexivity of =]\n\n$a=b \\Rightarrow b=a$ for any $a,b \\in F$ [commutativity of =]\n\n$a=b \\wedge b=c \\Rightarrow a=c$ for any $a,b,c \\in F$ [transitivity of =]\n\n(These describe \"=\" as an equivalence relation on $F$)\n\n$a=b \\Rightarrow P(a)=P(b)$ for any $a,b \\in F$ and predicate $P$ [substitution]\n\n(This describes substitution, which can be used to replace separate axioms governing how \"=\" and the field operations interact. Ittay used this in one of his steps.)\n\nThese allow us to \"do the same thing to both sides\", for example:\n\nFor any $a,b,c \\in F$ such that $a=b$,\n\nLet $d=a+c$ [closure under +]\n\n$a+c=a+c$ [transitivity of =; $a+c=d=a+c$]\n\n$a+c=b+c$ [substitution; where the predicate is given by $P(x) \\equiv (a+c=x+c)$]\n\nNote that to prove that something is a field, we will have to prove the substitution axiom, which boils down to proving the following equivalent set of axioms:\n\n$a=b \\Rightarrow a+c=b+c$ for any $a,b,c \\in F$\n\n$a=b \\Rightarrow ac=bc$ for any $a,b,c \\in F$\n\nThe original problem can then be proven as follows:\n\nFor any $a,b,c \\in F$ such that $a+b=a+c$,\n\n$b = 0+b = (-a+a)+b = (-a)+(a+b) = (-a)+(a+c) = (-a+a)+c = 0+c = c$\n\n\u2022 Hurkyl had given a comment indicating that we could use substitution of an argument of $+$ considered as a 2-argument function. As I noted, fields constructed through equivalence relations will still require one to prove the substitution rule before using, so it amounts to the same two more basic axioms that I gave. \u2013\u00a0user21820 Jan 25 '14 at 11:48\n\nThe first step is justified by the existence of $-a$. No further axiom is required in order to deduce that $(a+b)+(-a)=(a+c)+(-a)$. To see that, just write $a+b=y=a+c$ (after all, it is given that $a+b=a+c$). Now, one of the information defining a field is the function of addition: $(u,v)\\mapsto u+v$. Applying it to $u=y$ and $v=(-a)$, yields the element $y+(-a)$. But, since $y=a+b$, $y+(-a)=(a+b)+(-a)$. Similarly, since $y=a+c$, $y+(-a)=(a+c)+(-a)$. By transitivity of equality, it follows that $(a+b)+(-a)=(a+c)+(-a)$.\n\nNote that it would be a lot easier to add $(-a)$ on the left rather than on the right. It will save you using commutativity.\n\n\u2022 The step where you substitute $(a+b)$ into $y$ requires another axiom such as substitution. (See my answer.) With substitution or the alternative, you can just do it in one line. And that very axiom is the most important part of the problem! \u2013\u00a0user21820 Jan 25 '14 at 11:19\n\u2022 Yes, you can put $(-a)$ on the left. But the axiom says $a + (-a) = 0$, not $(-a) + a = 0$. So you have to use Commutativity anyway! \u2013\u00a0TonyK Jan 25 '14 at 11:26\n\n\"Specifically, the axioms don't mention anything about doing something to both sides of an equation simultaneously.\" Because is a logical axiom. If you have two equal things and do the same with both things the results are equal. Search \"first order logic with identity\".\n\n\u2022 See also en.wikipedia.org/wiki/\u2026, for when we don't have two equal things but have two things that behave the same way under some operations, and we still want to have 'substitution'. \u2013\u00a0user21820 Jan 25 '14 at 11:29", "date": "2019-09-22 16:45:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/650739/using-field-axioms-for-a-simple-proof", "openwebmath_score": 0.9354100823402405, "openwebmath_perplexity": 380.9813622254612, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9822877023336244, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.8516785575873659}}
{"url": "https://math.stackexchange.com/questions/3484838/counting-problem-checking-9-squares-out-of-3-times-5-board", "text": "# Counting problem: Checking 9 squares out of $3\\times 5$ board\n\nI'm self learning some combinatorics and I encountered the following counting problem:\n\nHow many ways are there to check 9 squares out of $$3\\times5$$ boards such that in every column there's at least one checked square?\n\n(To be more precise, the board has 3 rows and 5 columns)\n\nI think I know the outline of the solution:\n\nLet $$C$$ be the set of all possible checking of $$9$$ squares off the greed and let $$C_i$$ be the checking of the board where the $$i^{th}$$ column has no square checked- I then proceed by using the Inclusion - Exclusion principle and so the solution is $$|C|-|C_1\\cup C_2\\cup C_3\\cup C_4\\cup C_5|$$\n\nSo I have two questions:\n\n1. How many ways there are actually to check the board without restrictions? When I try to think about it I think about selecting a subset of 9 squares out of 15 squares so $$\\binom{15}{9}$$, is this correct? Somehow it does'nt feel like it's the right number ;\n2. Is the outline to the solution I wrote above the right approach to this problem?\n\nI know this is very elementary but I'm really confused by all the counting arguments and most of the time my initial intuition turns out to be wrong so any help would be much appreciated!\n\nUpdate with solution\n\nFor each $$C_i$$ we restrict our board to be one column less now, so it's actually checking $$9$$ squares on a $$3\\times 4$$ board- there are $$\\binom{12}{9}$$ ways to do so. Furthermore, up to renaming the columns this procedure is symmetric so there are $$5$$ ways to do that.\n\nFor any intersection of the form $$C_i\\cap C_j$$ (for $$i\\neq j$$) we restrict our board to be $$3\\times 3$$, and now there's a single way to check the board. There are $$\\binom{5}{2}$$ such intersections. Any bigger intersection would be empty.\n\nFrom the Inclusion - Exclusion principle we get: \\begin{aligned}\\left|\\bigcup_{i=1}^{5} C_{i}\\right| &=5 C_{i}-\\left(\\begin{array}{c}{5} \\\\ {2}\\end{array}\\right)\\left|C_{i} \\cap C_{j}\\right| \\\\ &=5\\left(\\begin{array}{c}{12} \\\\ {9}\\end{array}\\right)-\\left(\\begin{array}{c}{5} \\\\ {2}\\end{array}\\right) \\end{aligned}\n\nAnd so the number of possible checking that fit the decrepstion of the exercise is:\n\n$$|C|-\\left|\\bigcup_{i=1}^{5} C_{i}\\right|=\\left(\\begin{array}{c}{15} \\\\ {9}\\end{array}\\right)-\\left(5\\left(\\begin{array}{c}{12} \\\\ {9}\\end{array}\\right)-\\left(\\begin{array}{c}{5} \\\\ {2}\\end{array}\\right)\\right)=3915$$\n\n\u2022 A 3 by 5 board has 15 squares. There are 15 ways to select the first square, 14 ways to select the second square down to 15- 8= 7 ways to select the 9th square. There are $15*14*13*...*8*7= \\frac{15!}{6!}$ to select 9 squares out of the 15 in that particular order. If we consider the same 9 square, no matter what the order in which order they are selected, we need to divide by 9!, the number or different ways 9 things can be selected. That gives $\\frac{15!}{6!9!}$, the binomial coefficient. Dec 23, 2019 at 5:00\n\nI think you mean $$|\\color{red}C|-|C_1\\cup C_2\\cup C_3\\cup C_4\\cup C_5|$$\n\nWhere $$C$$ is all possible ways to pick $$9$$ squares ... which is indeed $${15} \\choose 9$$\n\nOK, but you still need to calculate $$|C_1\\cup C_2\\cup C_3\\cup C_4\\cup C_5|$$ .... easier said than done\n\n\u2022 Thank you very much! I fixed the typo and updated the solution. Does it look fine?\n\u2013\u00a0omer\nDec 22, 2019 at 20:56\n\u2022 @omer Yes, well done! Dec 22, 2019 at 20:59\n\nI have found a different solution method which gives the same answer as the one you have, so your method must be sound!\n\nWe examine partitions of $$9$$ into $$5$$ positive parts, each part less than $$4$$.\n\nThese are $$22221, 32211, 33111$$.\n\nThese are our columns, and the number of permutations of each is:\n\n$$\\binom{5}{4}=5, \\binom{5}{1,2}=\\frac{5!}{1!2!2!}=30, \\binom{5}{2}=10$$ respectively.\n\nEach column value of $$k$$ can be displayed in $$\\binom{3}{k}$$ ways.\n\nAll patterns have now been achieved once and once only (by inspection).\n\nThe total number is therefore $$5\\cdot 3^5 + 30\\cdot 3^4 + 10\\cdot 3^3 = 1215+2430+270=3915$$.\n\n\u2022 Thank you very much for the time and effort! Would it be a bother for you to elaborate on the the relation between partitions of 9 into 5 positive parts, each part less than 4- to the problem? I can't see it :( It's a really interesting solution and I'd like to understand it better!\n\u2013\u00a0omer\nDec 23, 2019 at 12:47\n\u2022 5 parts is one for each column, hence positive because each column must be contain at least one filled cell, and each column can have a maximum height of 3, hence less than 4.\n\u2013\u00a0JMP\nDec 23, 2019 at 12:48\n\u2022 Thank you again! This is a great new way to see this problem!\n\u2013\u00a0omer\nDec 23, 2019 at 12:50", "date": "2023-02-03 16:47:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3484838/counting-problem-checking-9-squares-out-of-3-times-5-board", "openwebmath_score": 0.9147350788116455, "openwebmath_perplexity": 203.60480277865648, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363545048393, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8516715226945346}}
{"url": "https://math.stackexchange.com/questions/3033825/injective-homomorphism-from-mathbbr-times-mathbbr-to-the-ring-of-continuo", "text": "# Injective Homomorphism from $\\mathbb{R}\\times\\mathbb{R}$ to the ring of Continuous functions\n\nDoes there exist an injective ring homomorphism from the ring $$\\mathbb{R}\\times\\mathbb{R}$$ to the ring of continuous functions over $$\\mathbb{R}$$?\n\nI know that $$\\mathbb{R}\\times\\mathbb{R}$$ is a field. So, any ring homomrphism should have either identity or the whole field as a kernel. So, does it imply that there is such a homomorphism? I am unable to think of any examples. Any hints. Thanks beforehand.\n\n\u2022 $\\mathbb R \\times \\mathbb R$ is not a field. Remember, the direct product of fields is not going to be a field in general, because, for example, $(1,0) \\cdot (0,1) = (0,0)$. \u2013\u00a0\u0430\u0441\u0442\u043e\u043d \u0432\u0456\u043b\u043b\u0430 \u043e\u043b\u043e\u0444 \u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 Dec 10 '18 at 12:10\n\u2022 @\u0430\u0441\u0442\u043e\u043d\u0432\u0456\u043b\u043b\u0430\u043e\u043b\u043e\u0444\u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 oh, thanks for that fact. So, then, it is not even an integral domain! \u2013\u00a0vidyarthi Dec 10 '18 at 12:11\n\u2022 By the definition of the direct product, two homomorphisms from $\\mathbb R$ to the ring of continuous functions over $\\mathbb R$ give rise to a homomorphism from $\\mathbb R \\times \\mathbb R$ to the latter space, with the two maps as components/projections. However, the pair of maps being injective does not imply that the combined map will be injective. \u2013\u00a0\u0430\u0441\u0442\u043e\u043d \u0432\u0456\u043b\u043b\u0430 \u043e\u043b\u043e\u0444 \u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 Dec 10 '18 at 12:18\n\u2022 @\u0430\u0441\u0442\u043e\u043d\u0432\u0456\u043b\u043b\u0430\u043e\u043b\u043e\u0444\u043c\u044d\u043b\u043b\u0431\u044d\u0440\u0433 ok, any good examples? \u2013\u00a0vidyarthi Dec 10 '18 at 12:23\n\nLet us make a general statement, and then apply it to this case.\n\nDefinition : Given a ring $$R$$, an element $$e \\in R$$ is said to be idempotent, if $$e^2 = e$$. Note that $$0,1$$ are idempotent. Any other idempotent will be referred to as non-trivial.\n\nLet $$R,S$$ be rings such that $$R$$ has a non-trivial idempotent but $$S$$ does not have any non-trivial idempotent. Then, any homomorphism $$\\phi : R \\to S$$ is not injective.\n\nProof : Let $$e \\in R$$ be a non-trivial idempotent, which is to say that $$e^2 = e$$, and $$e \\neq 0,1$$. Let us look at $$\\phi(e)$$. Then, $$(\\phi(e))^2 = \\phi(e^2)= \\phi(e)$$, so $$\\phi(e)$$ is idempotent in $$S$$, hence equals zero or one. if $$\\phi(e) = 0$$ then $$\\phi$$ is not injective as $$e \\neq 0$$. If $$\\phi(e) =1$$, then $$\\phi(1-e) = 0$$, and again injectivity is contradicted.\n\nYou can use the general statement in lots of places. Try to find candidates for $$R$$ and $$S$$.\n\nIn our case, $$S$$ , as the ring of continuous functions over $$\\mathbb R$$,contains no non-trivial idempotent, since if $$f^2 = f$$, then for each $$x$$ we have $$f(x) = 0$$ or $$1$$. However, as $$f$$ is continuous, $$f(\\mathbb R)$$ is connected, hence an interval, but has to be a subset of $$\\{0,1\\}$$, which is discrete. Consequently, $$f$$ is either identically zero or identically one, hence is a trivial idempotent.\n\nHowever, $$\\mathbb R \\times \\mathbb R$$ contains the non-trivial idempotent $$(0,1)$$. Hence, the general statement gives the result you have desired.\n\nAs to non-injective maps, there are many of them. As mentioned earlier, consider $$f,g \\in S$$ not necessarily distinct. Then, $$(a,b) \\to af + bg$$ is a homomorphism. It is not difficult to see that every homomorphism is of this form.\n\nAlso, think about what happens if you treat $$R$$ and $$S$$ as just abelian groups under addition. In that case, can you find an injective group homomorphism between the two? Note that the concept of idempotent cannot be defined without the multiplication, so this might be an interesting side question.\n\nSuppose there is an injective ring homomorphism. The unit element $$(1,1)$$ of $$\\mathbb{R} \\times \\mathbb{R}$$ maps to the unit element of $$\\mathcal{C} (\\mathbb{R} )$$, i.e. the constant function $$1_{\\mathbb{R}}$$. If $$f,g$$ are the images of $$(1,0) , (0,1)$$ respectively, then $$f+g = 1_{\\mathbb{R}}$$ and $$fg = 0$$. This shows that $$f(1_{\\mathbb{R}}-f) = 0 \\implies f = f^2$$. Since $$f$$ is continuous, $$f$$ is identically $$0$$ or identically $$1$$. The first case contradicts injectivity and so does the second because then $$g = 0$$. So there does not exist any injective ring map.\n\n\u2022 +1: Very nice argument! And from this we readily find that the only ring homomorphisms $\\Bbb R\\times\\Bbb R\\to\\mathcal C(\\Bbb R)$ are $(x,y)\\mapsto x\\cdot 1_{\\Bbb R}$ and $(x,y)\\mapsto y\\cdot 1_{\\Bbb R}.$ \u2013\u00a0Cameron Buie Dec 10 '18 at 18:36", "date": "2019-12-09 05:32:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3033825/injective-homomorphism-from-mathbbr-times-mathbbr-to-the-ring-of-continuo", "openwebmath_score": 0.9559090733528137, "openwebmath_perplexity": 86.33953727749726, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984336353126336, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8516715197913934}}
{"url": "http://mathhelpforum.com/algebra/148707-algebra-question-print.html", "text": "# An algebra question\n\n\u2022 June 16th 2010, 11:12 AM\neconlover\nAn algebra question\nQuestion: Simplify the following expression.\n\n(2x5^n)^2 + 25^n\n\nthis whole equation is divided by\n\n5^n\n\nRemarks: I have tried doing this question twice but both attempts were unsuccesful. Please help.\n\u2022 June 16th 2010, 11:28 AM\nearboth\nQuote:\n\nOriginally Posted by econlover\nQuestion: Simplify the following expression.\n\n(2x5^n)^2 + 25^n\n\nthis whole equation is divided by\n\n5^n\n\nRemarks: I have tried doing this question twice but both attempts were unsuccesful. Please help.\n\n1. Use the laws of powers:\n\n$\\frac{(2 \\cdot 5^n)^2 + 25^n}{5^n} = \\frac{4 \\cdot 5^{2n} + 5^{2n}}{5^n}= \\frac{5 \\cdot 5^{2n}}{5^n}=5 \\cdot 5^n = 5^{n+1}$\n\u2022 June 16th 2010, 11:33 AM\neconlover\nslight question\nThanks.\n\nI understand most of it but how did you move along from the 1st step of your answer to the second one\n\nthat is, in specific terms, where did the 4 go?\n\u2022 June 16th 2010, 11:39 AM\nSpringFan25\nconsidering the numerator (top of fraction) only:\n$4 . 5^{2n} + 5^{2n} = (4+1).5^{2n} = 5 \\times 5^{2n} = 5^{2n+1}$\n\u2022 June 16th 2010, 11:41 AM\neconlover\nThanks for the message. But I guess you misinterpret the sign of this \".\"\n\nIt is not 4.5 (4 decimal 5) but 4 . 5 meaning 4 x 5 = 4 times 5\n\nHope this clears up the misconception\n\u2022 June 16th 2010, 11:47 AM\neconlover\nStill don't understand this step. Where did the 1 come from ? What about the positive sign?\n\nCan someone explain? Thanks!\n\u2022 June 16th 2010, 12:18 PM\nSpringFan25\nit is only factorising. i was not interpreting your dot as a decimal.\n\nin general:\n4x + x = (4+1)x = 5x\n\nYou have\n$4 \\times 5^{2n} + 1 \\times 5^{2n}$ = $5 \\times 5^{2n}$\n\nWhich says: \"4 lots of $5^{2n}$ plus 1 lot of $5^{2n}$\" equals \"5 lots of $5^{2n}$\"\n\u2022 June 17th 2010, 02:03 PM\nbjhopper\nQuote:\n\nOriginally Posted by econlover\nQuestion: Simplify the following expression.\n\n(2x5^n)^2 + 25^n\n\nthis whole equation is divided by\n\n5^n", "date": "2015-01-25 22:36:56", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/148707-algebra-question-print.html", "openwebmath_score": 0.8806184530258179, "openwebmath_perplexity": 3356.930983199174, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363503693294, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8516715191163919}}
{"url": "https://math.stackexchange.com/questions/935490/the-limit-of-n1-n-n-as-n-to-infty/935538", "text": "# The limit of $(n!)^{1/n}/n$ as $n\\to\\infty$ [duplicate]\n\n(Proof necessary)\n\n$$\\lim_{n \\to \\infty} \\frac{(n!)^{\\frac{1}{n}}}{n}$$\n\nI don't have an answer yet, but I know it exists, and is less than $1$.\n\nEdit. Winther's answer is the most correct I don't understand how he is jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). Don't presume, it's wrong, I need to go, and I'll keep looking at it when I get back\n\nAny help is appreciated\n\n\u2022 Is this your limit: $\\lim_{n\\rightarrow\\infty} \\frac{(n!)^{\\frac{1}{n}}}{n}$? \u2013\u00a0Jared Sep 17 '14 at 20:15\n\u2022 Hint: If the limit exists, let's say its $L$, then $$\\lim_{n \\to \\infty} \\frac{n!}{n^n}=L^n$$ \u2013\u00a0Oria Gruber Sep 17 '14 at 20:18\n\u2022 Stirling's formula will help you \u2013\u00a0ThePortakal Sep 17 '14 at 20:18\n\u2022 See this question for something very similar. It can be done without Strilings formula. \u2013\u00a0Winther Sep 17 '14 at 20:32\n\u2022 @Oria: The value of a limit (if it exists) cannot depend on the limiting variable. So \"$\\displaystyle\\lim_{n \\to \\infty}\\frac{n!}{n^n} = L^n$\" is definitely incorrect. \u2013\u00a0JimmyK4542 Sep 17 '14 at 20:51\n\nPut $$a_n = \\frac{n!^{1/n}}{n}$$\n\nthen $$\\log a_n = \\frac{1}{n}\\left(\\log n! - n\\log n\\right)= \\frac{1}{n}\\sum_{k=1}^n\\log\\left(\\frac{k}{n}\\right)$$\n\nwhere we have used $\\log n! = \\log 1 + \\log 2 + \\ldots + \\log n$. The sum above is a Riemann sum for the integral $\\int_0^1\\log x dx$ so\n\n$$\\lim_{n\\to\\infty} \\log a_n = \\int_0^1\\log x dx = [x\\log x - x]_0^1 = -1$$\n\nand it follows that $$\\lim_{n\\to\\infty}a_n = \\frac{1}{e}$$\n\n\u2022 Hey winther I don't understand how you're jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). \u2013\u00a0Calvin Hammond Sep 17 '14 at 21:28\n\u2022 @CalvinHammond $$\\log n! - n \\log n = \\log 1 + \\log 2 + \\cdots + \\log n - (\\log n + \\log n + \\cdots + \\log n),$$ where the number of terms in the parentheses is $n$. So $$\\log n! - n \\log n = \\log \\frac{1}{n} + \\log \\frac{2}{n} + \\cdots + \\log \\frac{n}{n}$$ using the identity $\\log x - \\log y = \\log \\frac{x}{y}$. The rest follows. \u2013\u00a0heropup Sep 17 '14 at 21:42\n\u2022 Yes that is. Thanks for explaining it @heropup \u2013\u00a0Winther Sep 17 '14 at 21:46\n\u2022 Could you explain the remain sum more please. \u2013\u00a0Calvin Hammond Sep 17 '14 at 21:56\n\u2022 Notice that you have to be a little careful in applying this technique to improper integrals: math.stackexchange.com/questions/449103/\u2026 \u2013\u00a0user84413 Sep 18 '14 at 16:51\n\nHere is a simple elementary proof I found, but first of all, some lemmas:\n\n\u2022 This one could easily be proven by induction: $\\displaystyle \\prod_{k=1}^{n}\\left(1+\\frac{1}{k} \\right) = n+1$\n\n\u2022 You can try to prove this inequality yourself since it's not difficult: $\\displaystyle \\left (1+\\frac{1}{k} \\right )^k\\leq e \\leq \\left (1+\\frac{1}{k} \\right)^{k+1}\\\\$\n\n\u2022 This inequality is the one I'm going to use though because it gives a much tighter bound on our sequence and that's just more fun, though you could use the second inequality without change in proof. I could give you the proof if needed:$\\displaystyle \\left (1+\\frac{1}{k} \\right )^k\\leq e \\leq \\left (1+\\frac{1}{k} \\right)^{k+1/2}\\\\$\n\nOk, so here is the proof:\n\n\u2022 We first write $n^n/n!$ in a better way: $\\displaystyle \\frac{n^n}{n!}=\\prod_{k=1}^{n-1}\\left(1+\\frac{1}{k} \\right)^n \\cdot \\prod_{i=1}^{n-1}\\prod_{k=1}^{i} \\left(1+\\frac{1}{k} \\right)^{-1}=\\prod_{k=1}^{n-1}\\left(1+\\frac{1}{k} \\right)^n\\cdot \\prod_{i=1}^{n-1} \\left(1+\\frac{1}{i} \\right)^{-(n-i)}=\\prod_{i=1}^{n-1} \\left(1+\\frac{1}{i} \\right)^{i}$\n\u2022 Now, we use our inequalities to bound our sequence. First, an upper bound: $\\displaystyle \\prod_{i=1}^{n-1} \\left(1+\\frac{1}{i} \\right)^{i} \\leq \\prod_{i=1}^{n-1}e^{1}=e^{n-1}$\n\u2022 Then, a lower bound: $\\displaystyle \\prod_{i=1}^{n-1} \\left(1+\\frac{1}{i} \\right)^{i} \\geq \\prod_{i=1}^{n-1}\\left (e^{1} \\cdot \\left(1+\\frac{1}{i}\\right)^{-1/2} \\right )=\\frac{e^{n-1}}{\\sqrt[2]{n}}$\n\u2022 Now, since $\\frac{n!^{1/n}}{n}=(\\frac{n^n}{n!})^{-1/n}$, we get: $\\displaystyle e^{\\frac{1}{n}-1} \\leq \\frac{(n!)^{1/n}}{n} \\leq e^{\\frac{1}{n}-1} \\cdot \\sqrt[2n]{n}$\n\u2022 Finally, by the squeeze theorem, we get $\\displaystyle e^{0-1} \\leq \\lim_{n \\to \\infty} \\frac{(n!)^{1/n}}{n} \\leq e^{0-1} \\cdot 1$\n\u2022 Hence, $\\displaystyle \\lim_{n \\to \\infty} \\frac{(n!)^{1/n}}{n}=e^{-1}$\n\nI know there are simpler proofs, but this one is elementary and I feel like it gives you the direct intuition as to why it's true.\n\nStirling's Approximation $$n!\\sim \\sqrt{2\\pi n}\\left(\\frac{n}{e}\\right)^n$$ Which means that $n!$ is asymptotically equivalent to $\\sqrt{2\\pi n}\\left(\\frac{n}{e}\\right)^n$ as $n$ approaches $\\infty$. If your familiar with asymptotic formulas, then you'd also know that this implies that $$\\lim_{n\\to\\infty} \\frac{n!}{\\sqrt{2\\pi n}\\left(\\frac{n}{e}\\right)^n} =1$$ Now, using the algebraic laws of limits, we have $$\\lim_{n\\to\\infty} n!=\\lim_{n\\to\\infty} \\sqrt{2\\pi n}\\left(\\frac{n}{e}\\right)^n$$ $$\\lim_{n\\to\\infty} e^n=\\lim_{n\\to\\infty} \\sqrt{2\\pi n}\\left(\\frac{n^n}{n!}\\right)$$ $$\\lim_{n\\to\\infty} e=\\lim_{n\\to\\infty} (2\\pi n)^{\\frac{1}{2n}}\\left(\\frac{n}{(n!)^{\\frac{1}{n}}}\\right)$$ So now $$e=\\left[\\lim_{n\\to\\infty} e^{\\ln(2\\pi n)^{\\frac{1}{2n}}}\\right]\\left[\\lim_{n\\to\\infty} \\frac{n}{(n!)^{\\frac{1}{n}}}\\right] =\\left[\\lim_{n\\to\\infty} e^{\\frac{\\ln(2\\pi n)}{2n}}\\right]\\left[\\lim_{n\\to\\infty} \\frac{n}{(n!)^{\\frac{1}{n}}}\\right]$$ $$=\\left[\\lim_{n\\to\\infty} e^{\\frac{\\frac{d}{dn}\\ln(2\\pi n)}{\\frac{d}{dn}2n}}\\right]\\left[\\lim_{n\\to\\infty} \\frac{n}{(n!)^{\\frac{1}{n}}}\\right] =\\left[\\lim_{n\\to\\infty} e^{\\frac{1}{2n}}\\right]\\left[\\lim_{n\\to\\infty} \\frac{n}{(n!)^{\\frac{1}{n}}}\\right]$$ $$=e^0\\left[\\lim_{n\\to\\infty} \\frac{n}{(n!)^{\\frac{1}{n}}}\\right]=1\\cdot \\lim_{n\\to\\infty} \\frac{n}{(n!)^{\\frac{1}{n}}}$$ Thus $$e=\\lim_{n\\to\\infty} \\frac{n}{(n!)^{\\frac{1}{n}}}$$ And now we can easily see that $$\\lim_{n\\to \\infty} \\frac{(n!)^{\\frac{1}{n}}}{n}= \\lim_{n\\to \\infty} \\frac{1}{\\left(\\frac{n}{(n!)^{\\frac{1}{n}}}\\right)}=\\frac{1}{e}$$ Let me know if you have any questions.\n\nLet $\\displaystyle a_n=\\frac{n!}{n^n}$.\n\nThen $\\displaystyle\\lim_{n\\to\\infty}\\frac{a_{n+1}}{a_n}=\\lim_{n\\to\\infty}\\frac{(n+1)!}{(n+1)^{n+1}}\\cdot\\frac{n^n}{n!}=\\lim_{n\\to\\infty}\\frac{n^n}{(n+1)^n}=\\lim_{n\\to\\infty}\\frac{1}{(1+\\frac{1}{n})^n}=\\frac{1}{e}$,\n\nso $\\;\\;\\;\\;\\displaystyle\\lim_{n\\to\\infty}(a_n)^{\\frac{1}{n}}=\\lim_{n\\to\\infty}\\frac{(n!)^{\\frac{1}{n}}}{n}=\\frac{1}{e}$ $\\hspace{.2 in}$ (since $\\frac{a_{n+1}}{a_n}\\rightarrow L\\implies(a_n)^{\\frac{1}{n}}\\rightarrow L)$.\n\n\u2022 This is a great proof, but you would need to prove that the two forms of the limit are equivalent. \u2013\u00a0Fujoyaki Sep 17 '14 at 22:46\n\u2022 Thanks; you are right that I am using the result that if $\\lim_{n\\to\\infty}\\frac{a_{n+1}}{a_n}=L$, then $\\lim_{n\\to\\infty}(a_n)^{\\frac{1}{n}}=L$. See math.stackexchange.com/questions/69386/\u2026 \u2013\u00a0user84413 Sep 17 '14 at 22:49\n\nHint. Take the natural logarithm of $\\dfrac{(n!)^{1/n}}{n}$ and obtain $$\\frac{\\log 1+\\log 2+\\cdots+\\log n}{n}-\\log n=\\frac{\\log (1/n)+\\log(2/n)+\\cdots+\\log(n/n)}{n} \\\\=\\frac{1}{n}\\sum_{k=1}^n\\log\\left(\\frac{k}{n}\\right) \\to \\int_0^1 \\log x\\,dx=-1.$$\n\nHint (Sterling): $$n! \\sim \\sqrt{2\\pi n}(\\frac{n}{e})^n$$\n\n\u2022 Yes that is my limit. I'm using this to help a friend but I don't know much about to how to solve this myself. So I'm asking for a full answer. \u2013\u00a0Calvin Hammond Sep 17 '14 at 20:28", "date": "2021-03-06 12:29:49", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/935490/the-limit-of-n1-n-n-as-n-to-infty/935538", "openwebmath_score": 0.9115509986877441, "openwebmath_perplexity": 348.2621632471697, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363526668351, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8516715176833952}}
{"url": "https://math.stackexchange.com/questions/3221910/how-to-minimise-the-cost-of-guessing-a-number-in-a-high-low-guess-game", "text": "# How to minimise the cost of guessing a number in a high/low guess game?\n\nIn a high/low guess game, the \"true\" number is one of $$\\{1,\\cdots,1000\\}$$. You'll be told if your guess is $$<,>$$ or $$=$$ the true number for each guess you make, and the game terminates when you guess out the true number. Suppose the following three scenarios:\n\n1. If your guess is either lower or higher, you pay $$\\1$$ in both cases. If your guess is correct, you pay nothing and the game ends.\n\n2. If your guess is higher, you pay $$\\1$$; if your guess is lower you pay $$\\2$$. If your guess is correct, you pay nothing and the game ends.\n\n3. If your guess is higher, you pay $$\\1$$; if your guess is lower you pay $$\\1.5$$. If your guess is correct, you pay nothing and the game ends.\n\nIn these three scenarios, what is, respectively, the minimum number of $$\\$$ you must have to make sure you can find the true number?\n\nFormally, define a space of all guess strategies $$\\Sigma$$. We can then identify the cost $$C$$ as a function $$C:\\Sigma\\times \\{1,\\cdots,1000\\}\\to\\Bbb N_+,\\quad v\\mapsto C(S,v)$$ that maps the pair $$(S,v)$$ (where $$v$$ is the \"true\" number) to the cost it's going to take under this arrangement. Our problem is then to find $$\\min_{S} \\max_{v} C(S,v).$$\n\nCan this min-max problem be analytically solved? What would be the corresponding optimal strategy then?\n\nFor scenario 1, my intuition is to use bisection search, which, at worst, costs $$\\10$$ ($$2^{10}=1024$$). But I cannot prove this is indeed optimal. For the second scenario, since making an under-guess means a higher cost, maybe it's more optimal to use a \"right-skewed\" bisection, i.e. you keep making guesses at the $$2/3$$-quantile point. But this is just (rather wild) intuition, not anything close to a proof.\n\nFor this game, the most natural representation of a strategy is a rooted binary tree, with vertices labeled by guesses (i.e. numbers from $$\\{1, \\ldots, N=1000\\}$$), and each vertex having at most two edges (connecting it to the guess to make next, for \"lower\" and \"higher\" cases).\n\nLet the penalties you pay for \"lower/higher\" cases be $$L$$ and $$H$$, respectively. Now consider an arbitrary $$N$$, and consider an optimal tree $$T(N)$$ (that reaches the minimum worst-case total penalty). Its root is labeled with some guess $$R$$, $$1\\leqslant R\\leqslant N$$, its left subtree can be chosen to be [isomorphic to] $$T(R-1)$$, and its right subtree isomorphic to $$T(N-R)$$ (assuming $$T(0)$$ is empty).\n\nSo if $$P(N,L,H)$$ is the optimal penalty, then $$P(0,L,H)=P(1,L,H)=0,\\\\ P(N,L,H)=\\min_{1\\leqslant R\\leqslant N}\\max\\begin{cases}L+P(N-R,L,H)\\\\ H+P(R-1,L,H)\\end{cases}\\quad(N>1)$$ which allows one to compute $$P(1000,1,1)=\\color{red}{9},\\quad P(1000,2,1)=14,\\quad P(1000,1.5,1)=11.5$$ (note $$9$$ but not $$10$$ as you've suggested; the decision process is not just a bisection but has extra \"$$=$$\" outcomes). The trees can be computed along the way.\n\nRegarding the asymptotic behavior (with $$N\\to\\infty$$), if we let $$R(N,L,H)$$ be the \"argmin\" (say, the smallest root of optimal trees), one can show that the limits $$\\alpha=\\alpha(L,H)=\\lim_{N\\to\\infty}\\frac{P(N,L,H)}{\\ln N},\\quad\\beta=\\beta(L,H)=\\lim_{N\\to\\infty}\\frac{R(N,L,H)}{N}$$ exist and satisfy $$H+\\alpha\\ln\\beta=L+\\alpha\\ln(1-\\beta)=0.$$ In particular, $$\\beta(2,1)=(\\sqrt{5}-1)/2$$, and $$\\beta(1.5,1)$$ is the root of $$\\beta^3=(1-\\beta)^2$$.\n\n\u2022 Thanks. This makes sense. There seems to be no way to compute such a complicated recursion analytically though. Did you use DP for this? \u2013\u00a0Vim May 11 at 12:30\n\u2022 Yes, I did. Probably this still can be analyzed further, but I didn't do that. \u2013\u00a0metamorphy May 11 at 12:38\n\u2022 IIRC, your recursion also gives the optimal strategy when you keep track of the partitions chosen at each recursion node. \u2013\u00a0Vim May 11 at 13:41\n\nFor the (2,1) case, I found the worst-case cost increases by 1 at every Fibonacci number, so $$P(F_n,2,1)=1+P(F_n-1,2,1)=1+P(F_{n-1},2,1)$$. These key values $$F_n$$ increase by ratio $$F_n/F_{n-1}\\to\\phi=(1+\\sqrt{5})/2$$.\nThey grow exponentially; conversely, the maximum cost for $$P(N,2,1)$$ grows logarithmically.\n\nFor (1.5,1) the worst-case cost increases by $$0.5$$ at these values: $$P(N,2,1)>P(N-1,1.5,1)\\,if\\, N=2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,\\ldots$$\n\nThe recursion in this sequence is $$a_{n+1}=a_{n-1}+a_{n-2}$$. Their ratio, $$a_{n+1}/a_n$$, approaches 1.3246, which is the root of $$P^3=P+1$$, just as $$\\phi$$ for the (2,1) case is the root of $$P^2=P+1$$.\n\n\u2022 Would you elaborate on how the ratios arise? \u2013\u00a0Vim May 11 at 12:32\n\u2022 For $(4/3,1)$, the recursion is $a_{n+1}=a_{n-2}+a_{n-3}$, and for $(5/3,1)$, the recursion is $a_{n+1}=a_{n-2}+a_{n-4}$ \u2013\u00a0Empy2 May 11 at 13:00\n\nHere are some worked-out cases for smaller lists and variable penalties.\n\nSuppose you pay $$a$$ for a guess that is too low and $$b$$ for a guess that is too high.\n\nGiven a list such as $$\\{1,\\ldots,1000\\}$$, the cost $$C$$ of the optimal strategy is a function only of the number of items $$n$$ in the list. Without loss of generality, assume that $$a\\leq b$$. (If you want the case $$a\\geq b$$, just reverse the order of the list.)\n\nWe can determine the cost of the optimal strategy in a few simple cases:\n\n\u2022 $$C(1) = 0$$. There is only one guess, and it is correct.\n\u2022 $$C(2) = a$$. High guesses are penalized more, so guess the lower element.\n\u2022 $$C(3) = b$$. Guess the middle element. If you're wrong, you know what the correct element is and it costs at worst $$b$$ total.\n\nFor larger values of $$C(n)$$, consider what the optimal first move. Suppose there are $$n$$ elements in the list and you choose index $$1\\leq k\\leq n$$. In the worst case, you are wrong. If the true answer is lower, you must pay $$b$$ and search through a list of $$k-1$$ elements. If the true answer is higher, you must pay $$a$$ and search through a list of $$n-k$$ elements. So in the worst case, by choosing $$k$$ in the first round, you'll incur a cost of $$\\max\\left(b + C(k-1),\\; a + C(n-k)\\right).$$ We can search for an optimal $$k$$ which minimizes this cost in the first move, in which case the recursive solution tells us what the rest of the strategy (and its corresponding cost) is.\n\nNote that the optimal strategy will never be in the latter half of the list. This is because $$C$$ is monotonic. When $$k$$ is in the right half of the list (so that $$n-k \\leq k-1$$), we can always shrink the worst-case cost $$\\max(b+C(k-1), a+C(n-k))$$ by exchanging $$k-1$$ and $$n-k$$, putting $$k$$ into the symmetric position on the left half of the list.\n\nBy this reasoning, we find:\n\n\u2022 $$C(4) = \\max(2a,b).$$ Choose the second element. If you're too high, you'll pay a cost of $$b$$ but know the true answer. If you're too low, you'll pay a cost of $$a$$, then search through the remaining two elements for a worst-case cost of $$C(2)=a$$.\n\n\u2022 $$C(5) = a+b$$. You can choose either the second or third (middle) elements. In the worst case if you choose the middle element, it's too high so you pay $$b+C(2) = a+b$$ to search the two remaining. If you choose the second element, in the worst case it's too low so you pay $$a+C(3) = a+b$$ to search the three remaining.\n\n\u2022 $$C(6) = a+b$$. Choose the third element. In the worst case, it's either too high so you pay $$b+C(2) = a+b$$, or it's too low so you pay $$a+C(3)=a+b$$.\n\n\u2022 $$C(7) = \\min(a+C(4), b+C(3)).$$ As with the case $$n=4$$, the optimal strategy depends on the relative magnitudes of $$a$$ and $$b$$. If $$2b > 3a$$, then choose the third element in the list. Otherwise, choose the fourth (middle).", "date": "2019-06-20 06:52:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3221910/how-to-minimise-the-cost-of-guessing-a-number-in-a-high-low-guess-game", "openwebmath_score": 0.8868332505226135, "openwebmath_perplexity": 310.330390901041, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363480718237, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8516715154181077}}
{"url": "http://ac-immacolata.it/csxr/bisection-method-algorithm-matlab.html", "text": "### Bisection Method Algorithm Matlab\n\nUpdated to reflect the latest version of MATLAB, the second edition of this title introduces the theory and applications of the most commonly used techniques for solving numerical problems on a computer. Secant method D. This method requires two initial guesses satisfying. Numerical Methods with Matlab The Bisection Method 4 1. k 1 is the slope at the beginning of the time step (this is the same as k 1 in the first and second order methods). Design and simulation of three phase induction motor at different load conditions in matlab/simulink. The Bisection Method is a numerical method for estimating the roots of a polynomial f(x). m code; Worksheet 04; 10th - 14th September : Tu: Bisection Method (continued) and order of convergence ; Th: Newton's Method and order of convergence; Homework 01 is due on 09/13! Worksheet 05; Code for Newton's Method; 17th - 21st September : Tu: Secant Method with order of. A few steps of the bisection method applied over the starting range [a 1;b 1]. Here is a Matlab function that carries out the bisection algorithm for our cosmx function. It allows the code writer to focus on the logic of the algorithm without being distracted by details of. 2) using the bisection method. 1 Bisection Method In bisection method we reduce begin with an interval so that 0 2[a;b] and divide the interval in two halves,i. It\u2019s very intuitive and easy to implement in any programming language (I was using MATLAB at the time). They announced that the genetic algorithm is better than classical methods. where the value of the function. 1 The Bisection Method to Solve g(x)=0 Many mathematical problems involve solving one or more equations. On the other hand, the only difference between the false position method and the bisection method is that the latter uses ck = (ak + bk) / 2. The bisection algorithm should be: Save the interval boundaries; Look if [a,b] has a root. In python (with matplotlib), use the savefig function. Provide the function, 'f' and provide two guesses. Students will simultaneously be trained in the theory and practice involved in solving large systems of equations and understand and interpret the quality of such solutions. Bisection Method: Develop a MATLAB program to find the root of the following function using the bisection method gm gc tan g-9. Im studying for a math test and on a old test there is a task about bisection. The standard technique is something like Brent's method (see Numerical Recipes in C, Section 9. From these algorithms, the developer has to explore and exploit the algorithm suitable under specified constraints on the function and the domain. Application Of Bisection Method In The bisection method is an iterative algorithm used to find roots of continuous functions. - Bisection method for bounded searching. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week\u2019s lab problem set involves a numerical method which is very important in applied mathematics. Here's the code:. It shows with bold stripes the length of the bracketed region. These concepts form the foundation for writing full applications, developing algorithms, and extending built-in MATLAB capabilities. 2d Truss Analysis Matlab Program. Matlab has the function \u2018fzero\u2019 to find function zeros. The simplest root-finding algorithm is the bisection method: we start with two points a and b which bracket a root, and at every iteration we pick either the subinterval or, where is the midpoint between a and b. LAB 1: the bisection and secant methods In this section you will learn how to implement and analyse the Bisection and Secant methods in MAT-LAB. Like the others methods, we approach this problem by writing the equation in the form f(x) = 0 for some function f(x). Learn more about bisection, bisection method, algorithm. newton raphson method matlab pdf. We have provided MATLAB program for Bisection Method along with its flowchart and algorithm. The Bisection Method & Intermediate Value Theorem. m (after that day assignments should be put into the mbox of Yinglun ZHU) (1)The original demonstration of Newton\u2019s method was done by Newton almost 350 years ago. The bisection search. f(c)<0 then let b=c, else let a=c. The task is to solve x^2=2 with the bisection method and the precision should be with 10 decimals. What are the applications of the bisection. Fixed Point Iteration 8 1. The setup of the bisection method is about doing a specific task in Excel. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign. 6524; m = 73. The Steepest Descent Algorithm for Unconstrained Optimization and a Bisection Line-search Method Robert M. Problems 197. So let's take a look at how we can implement this. Essentially, the root is being approximated by replacing the. ) (Use your computer code) I have no idea how to write this code. Using this simple rule, the bisection method decreases the interval size iteration by iteration and reaches close to the real root. Learn how genetic algorithms are used to solve optimization problems. In mathematics, the bisection method is a root-finding method that applies to any continuous functions for which one knows two values with opposite signs. In this method, there is no need to find the. Learn more about matlab. A next search interval is chosen by comparing and nding which one has zero. where the value of the function. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. All the Fortran 90 programs listed here are corresponding to the Fortran 77. The equation is of form, f(x) = 0. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. % % OUTPUT: approximate solution p or. Again, as before, Newton\u2019s method does not always converge, but when it does, it does so faster (p = 2) than the bisection method (p = 1) and the secant method (\ud835\udc5d= (1+\u221a5)\u20442). 2D 3D Algorithms ASCII C# C++ Cellular Automata Clustering Cryptography Design Patterns Electronics game Image Processing Integral Approximation Java JavaFX Javascript LED Logic Gates Matlab Numerical Methods Path Finding Pygame Python R Random Root Finding R Shiny Sound UI Unity. It was observed that the Bisection method converges at the 14th iteration while Newton methods. 5 Secant Method 189. The bisection method depends on the Intermediate Value Theorem. Bisection Algorithm Let , - i. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. First I plot the function and then I try to find a domain such that I can see the curve cut through the x -axis. The algorithm applies to any continuous function. B The comparative results are shown in table 3. GitHub Gist: instantly share code, notes, and snippets. This scheme is based on the intermediate value theorem for continuous functions. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. In general, Bisection method is used to get an initial rough approximation of solution. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Bisection method in MATLAB A video of programming the code in realtime. Consider a root finding method called Bisection Bracketing Methods \u2022 If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). % Newton's Method algorithm! n = 2;! nfinal = N + 1; % Store final iteration if tol is reached before N iterations! Newton's Method MATLAB Implementation. The algorithm, created by T. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further. 2 Bisection method 2. Comparative Study Of Bisection, Newton-Raphson And Secant Methods Of Root- Finding Problems International organization of Scientific Research 2 | P a g e Given a function f x 0, continuous on a closed interval a,b , such that a f b 0, then, the function f x 0 has at least a root or zero in the interval. Root approximation through bisection is a simple method for determining the root of a function. Freund February, 2004 1 2004 Massachusetts Institute of Technology. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. Before you start, review the \\Introduction to MATLAB\" notes. The chance of convergence with such a small precision depends on the calculatord: in particular, with Octave, the machine precision is roughly \u22c5 \u2212. Numerical rate of convergence of root has been found in each calculation. Learn how genetic algorithms are used to solve optimization problems. Earlier we discussed a C program and algorithm/flowchart of bisection method. The Bisection Method The Bisection Method at the same time gives a proof of the Intermediate Value Theorem and provides a practical method to find roots of equations. m needed for Homework 4. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. Essentially, the root is being approximated by replacing the. 6: Calculate the ground track of a satellite from its orbital elements. 15625 (you need a few extra steps for \u03b5 abs) Applications to Engineering. I wrote his code as part of an article, How to solve equations using python. The bisection method is one of the simplest and most reliable of iterative methods for the solution of nonlinear equations. Need help with this bisection method code!. Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet; Class vs Instance Methods; PHP5 Method Chaining Example. he gave us this template but is not working. Description. The problem is that it seems like the teachers recommended solution to the task isn't quite right. This Demonstration shows the steps of the bisection root-finding method for a set of functions. f = @(x) (cos(x)); a = input( 'Please enter lower. Above given Algorithm and Flowchart of Bisection Methods Root computation is a simple and easier way of understanding how the bracketing system works, algorithm and flowchart may not follow same procedure, yet they give the same outputs. This is a very simple and powerful method, but it is also relatively slow. 2 (Bisection Method). Bisection Method C Program Bisection Method MATLAB Program. If a change of sign is found, then the root is calculated using the Bisection algorithm (also known as the Half-interval Search). t is the root of the given function if f (t) = 0; else follow the next step. The method is also called the interval halving method. This method will divide the interval until the resulting interval is found, which is extremely small. Powered by Create your own unique website with customizable templates. 000013273393044 9 1. 21 dcm_to_ypr. The following Matlab project contains the source code and Matlab examples used for bisection method. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen appropriately, and that it is relatively. Remark: \ud835\udc5d\ud835\udc5d. So if you need MATLAB programming homework help, feel free to ask for a quote. Set 1: The Bisection Method Set 2: The Method Of False Position. X +4 X i mark the bracket. Bisection Method of Solving a Nonlinear Equation. % Using the. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week\u2019s lab problem set involves a numerical method which is very important in applied mathematics. The bisection search. 4 Basis of Bisection. 1shows the several \ufb01rst iterations of the bisection algorithm. Bayen Bisection algorithm to aproximate an equation result. BISECTION_RC, a MATLAB library which demonstrates the simple bisection method for solving a scalar nonlinear equation in a change of sign interval, using reverse communication (RC). Bisection method. 1 The Bisection Method to Solve g(x)=0 Many mathematical problems involve solving one or more equations. The Secant Method One drawback of Newton's method is that it is necessary to evaluate f0(x) at various points, which may not be practical for some choices of f. Divide the interval [a, b]. 3 Algorithms and convergence 2. Assumption: The function is continuous and continuously differentiable in the given range where we see the sign change. The algorithm for this is given as follows: Choose a;b so that f(a)f(b) <0 1. Bisection Method The Bisection method is a root finding algorithm. Essentially, the root is being approximated by replacing the. The problem I encounter is that the same computation must be performed on each element of a large array(~1. where the value of the function. By the Nested Interval Property of Real Numbers the sequence of Nested Intervals converge to a unique point, which should therefore be r. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use \u03b5 step = 0. MATLAB Tutorial \u2013 Roots of Equations ES 111 1/13 FINDING ROOTS OF EQUATIONS Root finding is a skill that is particularly well suited for computer programming. In these lectures details about how to use Matlab are detailed (but not verbose) and. m This program will implement Euler\u2019s method to solve the di\ufb00erential equation dy dt = f(t,y) y(a) = y 0 (1) The solution is returned in an array y. Suppose we want to solve the equation. These concepts form the foundation for writing full applications, developing algorithms, and extending built-in MATLAB capabilities. Learn more about bisection method, homework. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. 00064404356011 -0. Chapter 6 Finding the Roots of Equations The Bisection Method Copyright \u00a9 The McGraw-Hill Companies, Inc. 11) uses fzero to calculate the root for this heat capacity example. \" by Timmy Siauw and Alexandre M. The software, matlab 2009a was used to find the root of the function for the interval [0,1]. Suppose that f(\u00a2) is a continuous function de\ufb01ned over an interval [a;b] and f(a) and f(b) have opposite signs. This is a quick way to do bisection method in python. function p_min=bisection(func,int,iter,tol_x,tol_f) % It calculates the zero of a regular real function with one variable. The convergence rate of the bisection method could possibly be improved by using a different solution estimate. Bisection method- code stops after one iteration. So the abscissa of point where the chords cuts the x-axis (y=0) is given by,. Because of this, it is often used to roughly sum up a solution that is used as a starting point for a more rapid conversion. Bisection Method Issues/Help. % p_min is the solution and represents the abscissa's value of the zero. What's great about the Bisection Method is that provided the conditions above are satisfied (and hence a root $\\alpha$ exists in the interval $[a, b]$ by the Intermediate Value Theorem), then this method is guaranteed to zone into our root with better and better approximations. Given these facts. Blog Archive. 6 in the text. It\u2019s take a first approximation by apply two times the Bisection method and complete a correct approximation by use the Newton-Raphson method. Freund February, 2004 1 2004 Massachusetts Institute of Technology. Noanyother restrictionsapplied. Additional optional inputs and outputs for more control and capabilities that don't exist in other implementations of the bisection method or other root finding functions like fzero. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. What are the applications of the bisection. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. The x-coordinate of this point is the average of the positive and negative guesses. Newest vertex bisection. However, if there are several solutions present, it finds only one of them, just as Newton's method and the secant method. to a bug in. \"In mathematics, the bisection method is a root-finding algorithm which works by repeatedly dividing an interval in half and then selecting the subinterval in which a root exists. There are classical root-finding algorithms: bisection, false position, Newton-Raphson, modified Newton-Raphson, secant and modified secant method, for finding roots of a non-linear equation f(x) = 0 [7,8,9,10,11]. Freund February, 2004 1 2004 Massachusetts Institute of Technology. Later you will learn how to add it to the calling sequence. Introduction. Bisection Method of finding the roots of an equation is both simple and straight forward - I really enjoyed playing with bisection back in college (oooh yeah ES84 days) and I decided to make a post and implement bisection in scilab. algorithm apply axis bisection method boundary condition button Chebyshev Chebyshev nodes clicking coef\ufb01cient computation constraints converge curve data points de\ufb01ned depicted in Fig derivative diagonal dialog box difference approximation differential equation eigenvalues eigenvectors element end end Example Figure \ufb01nd \ufb01rst flops. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week\u2019s lab problem set involves a numerical method which is very important in applied mathematics. It will helpful for engineering students to learn Bisection method MATLAB program easily. Students will simultaneously be trained in the theory and practice involved in solving large systems of equations and understand and interpret the quality of such solutions. The Algorithm The bisection method is an algorithm, and we will explain it in terms of its steps. MATLAB coding of all methods. - Bisection method for bounded searching. Examples illustrate important concepts such as selection, crossover, and mutation. The points marked as X i are positions of the negative( )andpositive(+)endsoftherootenclosingbracket. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. 6524; m = 73. It is a very simple and robust method, but it is also relatively slow. 1 and \u03b5 abs = 0. Here we are required an initial guess value of root. The Algorithm for The Bisection Method for Approximating Roots Fold Unfold. The Bisection Method looks to find the value c for which the plot of the function f crosses the x-axis. This method is also known as Regula Falsi or The Method of Chords. Python Bisection Method. This is a quick way to do bisection method in python. The number of iterations we will use, n, must satisfy the following formula:. Wednesday, September 28, 11. It requires two initial guesses and is a closed bracket method. Bisection method add iteration table into my code. We are going to find the root of a given function, with bisection method. m with contents. The basic idea is a follows. Matlab will spit out that the root in this interval = '6'. The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. In this course, three methods are reviewed and implemented using Python and MATLAB from scratch. The help page states the following about the algorithm: Algorithms The [code ]fzero[/code] command is a function file. Algorithm of Bisection Method [YOUTUBE 9:47] Example of Bisection Method [YOUTUBE 9:53] Advantages & Drawbacks of Bisection Method [YOUTUBE 8:31] MULTIPLE CHOICE TEST : Test Your Knowledge of Bisection Method PRESENTATIONS. For searching a finite sorted array, see binary search algorithm. The problem I encounter is that the same computation must be performed on each element of a large array(~1. m - matlab file that defines Equation (2. Numerical rate of convergence of root has been found in each calculation. This algorithm is a new approach to compute the roots of nonlinear equations f(x)=0, by propose hybrid algorithm between the Bisection algorithm and Newton-Raphson algorithm. The method is based on the following algorithm: Initialization: The bisection method is initialized by specifying the function , the interval [a,b], and the tolerance > 0. Step 3: If f(a). In this tutorial we are going to develop pseudocode for Bisection Method so that it will be easy while implementing using programming language. Dekker's zeroin algorithm from 1969 is one of my favorite algorithms. Numerical Integration: Rectangle Method. Secant method d. The convergence to the root is slow, but is assured. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Newton's Method in Matlab. Online calculator. The task is to solve x^2=2 with the bisection method and the precision should be with 10 decimals. Bisection method In short, the bisection method will divide one triangle into two children triangles by connecting one vertex to the middle point of its opposite edge. Write a program that calculates the root(s) of a non-linear equation using the bisection method and also the secant method. The Algorithm The bisection method is an algorithm, and we will explain it in terms of its steps. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. This book serves as a textbook for a first course in numerical methods using MATLAB to solve problems in mechanical, civil, aeronautical, and electrical engineering. Algorithm for Bisection Method: Input function and limits. X +1 X-1 X +2 X-2-4 X-3 X +3 f(x) x Figure 1. The problem is that it seems like the teachers recommended solution to the task isn't quite right. 6 Newton Method for a System of Nonlinear Equations 191. Find more Mathematics widgets in Wolfram|Alpha. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. \"Bisection Method and Algorithm for Solving The Electrical Circuits\" Iteration,Bisection Method, Fortran, C, MatLab. This scheme is based on the intermediate value theorem for continuous functions. So, it has a. Also, Newton\u2019s method can be used to approximate complex roots, as well, if the initial value 0 is a complex number satisfying the conditions above. in the case of MATLAB\u00ae, 16 digits). 000003315132176 10 1. The idea is simple: divide the interval in two, a solution must exist within one subinterval, select the subinterval where the sign of. 2 Newton's Method and the Secant Method The bisection method is a very intuitive method for finding a root but there are other ways that are more efficient (find the root in fewer iterations). Im studying for a math test and on a old test there is a task about bisection. Richard Brent's improvements to Dekker's zeroin algorithm, published in 1971, made it faster, safer in floating point arithmetic, and guaranteed not to fail. Using the Bisection method to find the negative root (only) of the equation ,3C2 \u2014 ex \u2014 O (a) [5 points) Choose an initial interval [a, b]. 000000828382262 11 1. Also, Newton\u2019s method can be used to approximate complex roots, as well, if the initial value 0 is a complex number satisfying the conditions above. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign. Online calculator. 6524; m = 73. Figure 3: Synthetic seismic section displayed in \u201ctime from datum\u201d and computed from the log section shown above. The points marked as X i are positions of the negative( )andpositive(+)endsoftherootenclosingbracket. Unless the roots of an equation are easy to find, iterative methods that can evaluate a function hundreds, thousands, or millions of times will be required. 2d Truss Analysis Matlab Program. ^3 - 2; exists. Then faster converging methods are used to find the solution. The basic idea is to switch between inverse quadratic interpolation and bisection based on the step performed in the previous iteration and based on inequalities gauging the difference between guesses:. The algorithm of bisection method is such that it can only find one root between a defined interval. Learn more about bisection method loop. Coding a bisection algorithm using matlab (numerical. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. Bisection \u2013 separate \ufb01les and embedded functions c. then there is at least one real root in the interval. Access Ebooks on other topics. So let's take a look at how we can implement this. Need help with this bisection method code!. It is a very simple and robust method, but it is also relatively slow. 4 Basis of Bisection. m Algorithm 4. As we point out in the introduction, we will mainly discuss newest vertex bisection and include longest edge bisection as a variant of it. The number of iterations we will use, n, must satisfy the following formula:. Another was to say \u201croot. 21 dcm_to_ypr. Bisection method in MATLAB A video of programming the code in realtime. Dekker's zeroin algorithm from 1969 is one of my favorite algorithms. Learn the algorithm of the bisection method of solving nonlinear equations of the form f(x)=0. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function, and that function crosses a horizontal line. Introduction a. k 1 is the slope at the beginning of the time step (this is the same as k 1 in the first and second order methods). Solutions to the Exercises from \"An Introduction to MATLAB and Numerical Methods for Engineers. Find the midpoint of a and b, say \"t\". Bisection method b. , Vasiliou, P. Learn more about bisection. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use \u03b5 step = 0. m, instructions how to run it, an example of a file myfunction. Lab 9 - Bisection Method Introduction In this lab, we will explore a method that we have considered in class for solving nonlinear equations, the bisection method. Initialization: nd [a 1;b 1] \u02c6[a;b], with f(a 1)f(b 1) <0, set i= 1. X +1 X-1 X +2 X-2-4 X-3 X +3 f(x) x Figure 1. A next search interval is chosen by comparing and nding which one has zero. Blog Archive. The first algorithm that I learned for root-finding in my undergraduate numerical analysis class (MACM 316 at Simon Fraser University) was the bisection method. So let's take a look at how we can implement this. 1 Polynomial Interpolation: Method of undetermined coefficients (Vandermonde. In mathematics, the bisection method is a root-finding algorithm which repeatedly bisects an interval then selects a subinterval in which a root must lie for further processing. Hi, I need help solving the function 600x^4-550x^3+200x^2-20x-1=0 using the Bisection and Secant method in MATLAB. At first, two interval-based methods, namely Bisection method and Secant method, are reviewed and implemented. MATLAB Programming Assignment Help, Root ?nding using the bisection method, In many applications, including ?nancial mathematics, ?nding zeros of a function f(x) = 0 (4) is paramount. The bisection search. This is achieved by selecting two points A and B on that interval. The convergence to the root is slow, but is assured. Analysis of the Problem. He used it for \ufb01nding roots of cubic polynomials. m Algorithm 4. Golden section search Fibonacci search Bisection method Secant method Bracketing Chapter 5 Gradient Methods. Input, output. The c value is in this case is an approximation of the root of the function f (x). They are the secant method, bisection method, and newton's method. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. The problem is that it seems like the teachers recommended solution to the task isn't quite right. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. In our improved hybrid algorithm, we compute the x-intercept x using the Newton-Raphson method at the midpoint of the previous interval. It is a very simple and robust method, but it is also relatively slow. Open methods: Newton-Raphson method, Secant method. MATLAB has the function fzero which performs this bisection algorithm. Numerical Methods using MATLAB, 3e, is an extensive reference offering hundreds of useful and important numerical algorithms that can be implemented into MATLAB for a graphical interpretation to help researchers analyze a particular outcome. If a function is continuous between the two initial guesses, the bisection method is guaranteed to converge. Midpoint Method. 000000828382262 11 1. Fixed Point Method Using Matlab Huda Alsaud King Saud University Huda Alsaud Fixed Point Method Using Matlab. Analysis of the Problem. If the guesses are not according to bisection rule a message will be displayed on the screen. We set [a 0;b 0] = [a;b]. Blog Archive. Lecture 31-33 - Rootfinding Table of Contents 31. A few steps of the bisection method applied over the starting range [a 1;b 1]. m and newton. The setup of the bisection method is about doing a specific task in Excel. Bisection Methods: We can pursuse the above idea a little further by narrowing the interval until the interval within which the root lies is small enough. Application Of Bisection Method In The bisection method is an iterative algorithm used to find roots of continuous functions. Implement the Bisection algorithm elegantly and easily (3 answers) Closed 3 years ago. \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014. This method, also known as binary chopping or half-interval method, relies on the fact that if f(x) is real and continuous in the interval a < x < b, and f(a) and f(b) are of opposite signs, that is,. Root Search with the bisection method. The bisection method in Matlab is quite straight-forward. Joseph DeSimone, Applied Mathematics Graduate Student. The previous two methods are guaranteed to converge, Newton Rahhson may not converge in some cases. Essentially, the root is being approximated by replacing the. The instructions of the problem are: Use bisection method to find a root of the function $$\\sin x + x \\cos x = 0$$ Indicate your initial condition and how many steps it requires to reach the tole. Maple f\u00fcr Akademiker. 23 ground_track. Pseudo-code is a simple way to represent an algorithm in a logical and readable form. 5: Calculation of the state vector from the orbital elements. 6 in the text. Suppose we want to solve the equation. 2) using the bisection method. The Bisection Method looks to find the value c for which the plot of the function f crosses the x-axis. Bisection method add iteration table into my code. But they're not live. If (b i+1 a i+1)=2 > , set i= i+1 and go to step 1 4. GitHub Gist: instantly share code, notes, and snippets. We may minimize a convex f : \u2192 by finding a point at which f \u2032 = 0. (original given interval) Thanks for contributing an answer to Code Review Stack Exchange! Bisection method for finding the root of a function. Repeat steps 3 and 4 100 times. 5 Single Variable Newton- Raphson Method 9. m) and see how we can compute the root to a polynomial using this method. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. This tutorial covers in depth algorithm for Bisection Method. Bisection Method is one of the simplest, reliable, easy to implement and convergence guaranteed method for finding real root of non-linear equations. The bisection method is guaranteed to converge to a root of the function f, if the function is continuous between the lower and upper bounds. It will helpful for engineering students to learn Bisection method MATLAB program easily. In simple terms, these methods begin by attempting to evaluate a problem using test (\u201cfalse\u201d) values for the variables, and then adjust the. Title: Microsoft Word - nle_03_bisection_advantages Author: sotirioschat Created Date: 1/15/2013 11:41:47 AM. 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10 \u22124. Bisection Method Example. Lab 9 - Bisection Method Introduction In this lab, we will explore a method that we have considered in class for solving nonlinear equations, the bisection method. First I plot the function and then I try to find a domain such that I can see the curve cut through the x -axis. Another class of mesh refinement method, known as regular refinement, which divide one triangle into 4 similar small triangles, is implemented in uniformrefine. Bisection Method // C++ code Posted: January 31, 2012 by muhammadakif in Algorithms Tags: bisection method , C# code , numerical analysis , numerical computing , numerical methods. Also, a good intermediate approximation may be discarded. It is the slowest algorithm provided by the library, with linear convergence. Initialization: nd [a 1;b 1] \u02c6[a;b], with f(a 1)f(b 1) <0, set i= 1. Designing Robot Manipulator Algorithms. The help page states the following about the algorithm: Algorithms The [code ]fzero[/code] command is a function file. m Algorithm 4. Graphical method useful for getting an idea of what's going on in a problem, but depends on eyeball. You may not use MATLAB's built-in functions for finding roots -- instead, please implement two different algorithms. Algorithms in this toolbox can be used to solve general problems All algorithms are derivative-free methods Direct search: patternsearch Genetic algorithm: ga Simulated annealing/threshold acceptance: simulannealbnd, threshacceptbnd Genetic Algorithm for multiobjective optimization: gamultiobj Kevin Carlberg Optimization in Matlab. For more videos and resources on this topic, please visit http. Let m = (L+H)/2. Roots (Bisection Method) : FP1 Edexcel January 2012 Q2(a)(b) : ExamSolutions Maths Tutorials - youtube Video. I am trying to solve the equation f(x) = x^3 + x - 1 , by using Bisection method within the interval [ 0 , 1] , i have succeeded to generate a code to solve this equation but by using \" while \" function for looping , i need some one to help me to solve it by using \" for \" function , could any one help me to do that ? the code is :. \"In mathematics, the bisection method is a root-finding algorithm which works by repeatedly dividing an interval in half and then selecting the subinterval in which a root exists. This is calculator which finds function root using bisection method or interval halving method. If the guesses are not according to bisection rule a message will be displayed on the screen. function [ r ] = bisection( f, a, b, N, eps_step, eps_abs ) % Check that that neither end-point is a root % and if f(a) and f(b) have the same sign, throw an exception. The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. In this project we use MATLAB to analyze some of the numerical techniques. Im studying for a math test and on a old test there is a task about bisection. This scheme is based on the intermediate value theorem for continuous functions. It was observed that the Bisection method converges at the 14th iteration while Newton methods. In the secant method, it is not necessary that two starting points to be in opposite sign. 5 Single Variable Newton- Raphson Method 9. Next, we study some known numerical algorithms those can be used to find the approximate solutions (roots) for non-linear equations, which are Bisection algorithm, Newton\u2013Raphson algorithm and fixed point algorithm. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen appropriately, and that it is relatively. t is the root of the given function if f (t) = 0; else follow the next step. Step 2: Let c=(a+b)/2. The program assumes that the provided points produce a change of sign on the function under study. B The comparative results are shown in table 3. The proof that the binary search method works is provided by Bolzano's Bisection Theorem. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. The root is then approximately equal to any value in the final (very small) interval. At least one root exists between the two points if the function is real, continuous, and changes sign. 3 Newton's and secant methods 2. 2) using the bisection method. But they're not live. Introduction a. Plot error. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a sub-interval in which a root must lie for further processing. Download MatLab Programming App from Play store. Bisection Method // C# code Posted: January 31, 2012 by Shahzaib Ali Khan in Algorithms Tags: bisection method , C# code , numerical analysis , numerical computing , numerical methods. Useful Computational Methods: The Bisection Method - Finding roots by binary search - Unlike the guess-and-check method, we start with two initial values - one value a below \u221aQ and another value b above \u221aQ, where Q is a positive real number. ContentsDirk DekkerZeroin in AlgolThe test functionBisectionSecant methodZeroin algorithmZeroin in MATLABReferencesDirk DekkerI. As a starting point, let's fix to be the function cosmx that you just wrote. 1997 CREWES software release CREWES Research Report \u2014 Volume 9 (1997) 18-3 Figure 2: The same cross section as above showing the result of the synthesis of an ensemble of new logs. 3 Limits of Accuracy 1. In this course, three methods are reviewed and implemented using Python and MATLAB from scratch. 1shows the several \ufb01rst iterations of the bisection algorithm. Designing Robot Manipulator Algorithms. It is a very simple and robust method, but it is also relatively slow. 1 Polynomial Interpolation: Method of undetermined coefficients (Vandermonde. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. The c value is in this case is an approximation of the root of the function f (x). It requires two initial guesses and is a closed bracket method. Always Converges: like Bisection, it. In this project we use MATLAB to analyze some of the numerical techniques. It shows with bold stripes the length of the bracketed region. Newton's method is an iterative method. Error = x- (a+b)/2. The Bisection Method is used to find the zero of a function. The x-coordinate of this point is the average of the positive and negative guesses. Download MatLab Programming App from Play store. function [ r ] = bisection( f, a, b, N, eps_step, eps_abs ) % Check that that neither end-point is a root % and if f(a) and f(b) have the same sign, throw an exception. com 9/27/01. To code the bisection algorithm. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. The method is also called the interval halving method. Basic Bisection Algorithm: 1. We also check whether f(a) = 0 or f(b) = 0, and if so return the value of a or b and exit. Image: The Bisection Method explained. The problem is that it seems like the teachers recommended solution to the task isn't quite right. Here\u2019s what we do: As with the bisection algorithm, start by choosing an interval [a,b] in which we. The bisection method can be easily adapted for optimizing 1-dimensional functions with a slight but intuitive. 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10 \u22124. f = @(x) (cos(x)); a = input( 'Please enter lower. Use The Following Pseudocode For The Bisection Method To Write MATLAB Code To Approximate The Root Of F(x) = Pt - X - 2, Interval (0,2), Tolerance 10-3, Maximum Number Of Iterations 50. m to determine the root of Equation (2. The convergence to the root is slow, but is assured. Repeat steps 3 and 4 100 times. Analysis of the Problem. The Bisection Method will cut the interval into 2 halves and check which. f(c)<0 then let b=c, else let a=c. Data scientists use a bisection search algorithm as a numerical approach to find a quick approximation of a solution. Steven Chapra's Applied Numerical Methods with MATLAB, third edition, is written for engineering and science students who need to learn numerical problem solving. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use \u03b5 step = 0. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function, and that function crosses a horizontal line. Considering that Scala is similar to the Java programming language, if anyone else needs the Interval-Halving method in Java, this code can easily be adapted to Java as well. The brief algorithm of the bisection method is as follows: Step 1: Choose a and b so that f(a). m - matlab file to determine the root of Equation (2. This method will divide the interval until the resulting interval is found, which is extremely small. (b) [5 points) Determine the number of steps N you should take to find the answer with tolerance 0. You have seen how Matlab functions can return several results (the root and the number of iterations, for example). Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. At first, two interval-based methods, namely Bisection method and Secant method, are reviewed and implemented. For others, an algorithm of Alefeld, Potra, and Shi is used. So in order to use live solutions, we're going to look at the Bisection Method and then the Golden Section Search Method. Before you start, review the \\Introduction to MATLAB\" notes. the Matlab code bisection. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. The \ufb01le EULER. Implement the bisection method to find a zero of the function over [0,1]; Implement the Newton's method to find a zero of the function over [0,1]; Implement the secant method to find a zero of the function over [0,1]. ^3 - 2; exists. This process involves \ufb01nding a root, or solution, of an equation of the form f(x) = 0 for a given function f. Consult the MATLAB TA's if you have any questions. Powered by Create your own unique website with customizable templates. I found where it was in the directory and added the folder to the path so when I entered it again I now get: C:\\Users\\Lulu\\Documents\\MATLAB\\Numerical Optimisation\\bisection. Viewed 601 times 0 $\\begingroup$ To code the bisection algorithm. This method is suitable for finding the initial values of the Newton and Halley's methods. 5: Calculation of the state vector from the orbital elements. Design and simulation of three phase induction motor at different load conditions in matlab/simulink. ) (Use your computer code) I have no idea how to write this code. Root Search with the bisection method. Context Bisection Method Example Theoretical Result The Root-Finding Problem A Zero of function f(x) We now consider one of the most basic problems of numerical approximation, namely the root-\ufb01nding problem. 7 Symbolic Solution for Equations 193. The bisection method is an iterative algorithm used to find roots of continuous functions. Open methods: Newton-Raphson method, Secant method. Analysis of the Problem. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a sub-interval in which a root must lie for further processing. Readers can code the algorithms in the programs of their choice. The bisection method is an enclosure type method for finding roots of a polynomial f(x), i. - Matlab: function end -vs- SciLab: function endfunction - Matlab: [first, second, third] -vs- Scilab: [third, second, first] Lectures for Spring 2009 Intro to Matlab+ Bisection + Newton Method. MATLAB M-files for implementation of the discussed theory and algorithms (available via the book's website) Introduction to Optimization, Fourth Edition is an ideal textbook for courses on optimization theory and methods. The bisection method in math is the key finding method that continually intersect the interval and then selects a sub interval where a root must lie in order to perform the more original process. BISECTION_RC, a MATLAB library which demonstrates the simple bisection method for solving a scalar nonlinear equation in a change of sign interval, using reverse communication (RC). If the guesses are not according to bisection rule a message will be displayed on the screen. As a starting point, let's fix to be the function cosmx that you just wrote. - Use abstracted Autonomous Guided Vehicles (AGVs) system with P controller and supervisory controller as a case study to validate the conflict-driven fault detection method in Matlab. Bisection Method MATLAB Program Note: Bisection method guarantees the convergence of a function f(x) if it is continuous on the interval [a,b] (denoted by x1 and x2 in the above algorithm. The program assumes that the provided points produce a change of sign on the function under study. This is a repository where i put all of the implementation that i have done in numerical analysis. A simple improvement to the bisection method is the false position method, or regula falsi. We start from the 2D case sketched in [3] and the approximation scheme presented in [3, 6], and then we extend the reconstruction scheme of separatrices in the. 000013273393044 9 1. Bisection Method Example. The equation is of form, f(x) = 0. We have developed such an algorithm and it is given in the M-file regfals. f(a)*f(b) < 0. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. He used it for \ufb01nding roots of cubic polynomials. In simple terms, these methods begin by attempting to evaluate a problem using test (\u201cfalse\u201d) values for the variables, and then adjust the. Algorithm To find a solution. Bisection Method C Program Bisection Method MATLAB Program. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. Bisection Algorithm Input: computable f(x) and [a;b], accuracy level. Title: Microsoft Word - nle_03_bisection_advantages Author: sotirioschat Created Date: 1/15/2013 11:41:47 AM. Includes methods used in MATLAB, Mathcad, Mathematica, and various software libraries. Shown here, it is a function, and it crosses the X-axis at just before 2. In order to avoid the shortcoming of the hybrid algorithm[1], we suggest an improved hybrid algorithm. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. 6 was used to nd the root of the function, f(x) = cosx xexp(x) on a close interval [0;1] using the Bisection method and Newton's method the result was compared. There is the graphical interface too. Bisection Method Bisection Method for finding roots of. \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014. This example shows how to generate HDL code from MATLAB\u00ae design implementing an bisection algorithm to calculate the square root of a number in fixed point notation. Bisection Method. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet; Class vs Instance Methods; PHP5 Method Chaining Example. raphson method. This is the same as the slope, k 2 , from the second order midpoint method. It is a very simple and robust method, but it is also rather slow. This is a very simple and powerful method, but it is also relatively slow. Experiment 1. Bisection Method http//numericalmethods. What are the applications of the bisection. This is the first of a three part series. Numerical analysis I 1. Bisection method- code stops after one iteration. Calculates the root of the given equation f(x)=0 using Bisection method. Set r i= (a i+ b i)=2; 2. We won't dwell into explaining what each of them does but will jump straight into explaining the secant method. The basic idea is a follows. Methods by Young and Mohlenkamp, 2018 Bisection Method-- 4 Iterations by Hand (example) Bisection Method-- 4 Iterations by Hand (example) MATLAB Tutorial Part 6 Bisection Method Root finding matlab4engineers. Additional optional inputs and outputs for more control and capabilities that don't exist in other implementations of the bisection method or other root finding functions like fzero. The bisect algorithm is used to find the position in the list, where the data can be inserted to keep the list sorted. Method of Steepest Descent Analysis of Optimization Algorithms Analysis of Gradient Methods. Hi, I need help solving the function 600x^4-550x^3+200x^2-20x-1=0 using the Bisection and Secant method in MATLAB. The algorithm always selects a subinterval which contains a root. The bisection method is an application of the Intermediate Value Theorem (IVT). It is called the bisection method, and it is used for finding roots of a function f (that is, points c where f(c)=. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. Bisection Method http//numericalmethods. - Bisection method for bounded searching. Let us say; f(x,y) = 0 with degree eight and g(x,y) = 0 with degree six; I need a matlab code for 2D Bisection Method to solve f(x,y) = 0 and g(x,y) = 0 and find all possible roots. m; the Matlab code fixedpoint. Matlab will spit out that the root in this interval = '6'. Also use Euler's method for the same problem, and compare your results. Interpolation and approximation (12 lecture hours) 3. We almost have all the tools we need to build a basic and powerful root-finding algorithm, Newton's method*. In mathematics, the bisection method is a root-finding method that applies to any. The algorithm is iterative. bisect(list, element, begin, end). The Bisection Method will cut the interval into 2 halves and check which half interval contains a root of the function. 23 ground_track. bisection method using log10(x)-cos(x) Program to read a Non-Linear equation in one variable, then evaluate it using Bisection Method and display its kD accurate root Basic GAUSS ELIMINATION METHOD, GAUSS ELIMINATION WITH PIVOTING, GAUSS JACOBI METHOD, GAUSS SEIDEL METHOD. Next, we study some known numerical algorithms those can be used to find the approximate solutions (roots) for non-linear equations, which are Bisection algorithm, Newton\u2013Raphson algorithm and fixed point algorithm. Above given Algorithm and Flowchart of Bisection Methods Root computation is a simple and easier way of understanding how the bracketing system works, algorithm and flowchart may not follow same procedure, yet they give the same outputs. GitHub Gist: instantly share code, notes, and snippets. m, instructions how to run it, an example of a file myfunction. Also, this method closely resembles with Bisection method. I found it was useful to try writing out each method to practice working with MatLab. Copy to clipboard. changes sign from. Newton's method requires both the function value and its derivative, unlike the bisection method that requires only the function value. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen. How can we qualify more generally which method. What's great about the Bisection Method is that provided the conditions above are satisfied (and hence a root $\\alpha$ exists in the interval $[a, b]$ by the Intermediate Value Theorem), then this method is guaranteed to zone into our root with better and better approximations. The algorithm of bisection method is such that it can only find one root between a defined interval. It is also known as Binary Search or Half Interval or Bolzano Method. Here we consider a set of methods that find the solution of a single-variable equation , by searching iteratively through a neighborhood of the domain, in which is known to be located. jf(r i)j<. This is more a problem of the algorithm than a MATLAB problem. In Matlab help functions written: The algorithm, created by T. It is a very simple and robust method, but it is also relatively slow. Let us say; f(x,y) = 0 with degree eight and g(x,y) = 0 with degree six; I need a matlab code for 2D Bisection Method to solve f(x,y) = 0 and g(x,y) = 0 and find all possible roots. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. Considering that Scala is similar to the Java programming language, if anyone else needs the Interval-Halving method in Java, this code can easily be adapted to Java as well. Find more Mathematics widgets in Wolfram|Alpha. Bisection algorithm The algorithm itself is fairly straightforward and \"fast\" in some sense: the number of iterations is roughly Log2 of the ratio of the initial interval length and the desired accuracy.\n\nkmrlx2mi1b5na, g8640sik2k85bq2, qxb6kio490lx0s, o1dk1lw8cbg1vij, n1n12g24o9yr, 7nt9hve8ef5it, e2rrjz8kbopebl, ih91ivykr94e, jr79d4l12w, df9ar6amif, sxk2k6p2d2b, lid1boa7hp, guthzsw13jn9esw, rk5cc2ucfco3j, 31bg1hw01snq, fqchs6lh9o02, ph2y08yoiul, m6ndewpv59j, y99w34lzbezim61, vkzsd54vecwijw, 3hxkf8rmyovyvcm, ht59b0zzwo9, hug36h9jtso, qtozz2prn4, 7axno7xf4tek64y, vl7fqen7ka, 5o6iq1q0rn3rn, av73aodlg75zk2l, cydysbtgb3ifakx, txb5upbjhp1nlb6, qbjxjj7apu1k2t, im817raozo9, i7btp0tl9ry, qyzb36emjj", "date": "2020-12-04 19:39:53", "meta": {"domain": "ac-immacolata.it", "url": "http://ac-immacolata.it/csxr/bisection-method-algorithm-matlab.html", "openwebmath_score": 0.4754257798194885, "openwebmath_perplexity": 622.7896160604779, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. 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{"url": "https://marc-petit.com/wwhy41x0/276140-degree-of-expression", "text": "It is also known as an order of the polynomial. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.For a univariate polynomial, the degree of the polynomial is simply the highest exponent occurring in the polynomial. \u2022 If the variable does not have an exponent, the degree is 1. Second Degree Polynomial Function. Trot transitions on a circle: Do transitions between collected and medium trot on a circle, being careful not to let the tempo change. The computer is able to calculate online the degree of a polynomial. Revise the sentences if necessary. The degree of a polynomial with a single variable (in our case, ), simply find the largest exponent of that variable within the expression. A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n \u2026 Expression of Interest Form To undertake a Higher Degree by Research in the Faculty of Science, Engineering and Built Environment. Second degree polynomials have at least one second degree term in the expression (e.g. With thousands of questions available, you can generate as many Degree of expression Worksheets as you want. abstract-algebra polynomials rational-functions. To identify the kind of algebraic expression and determine the degree, variables and constant in. Degree words are traditionally classified as adverbs, but actually behave differently syntactically, always modifying adverbs or adjectives and expressing a degree\u2026 \u2022 If a term does not contain any variable, the degree is 0. See more. The negative and weak positive staining rates in normal appearing mucosa, adenoma, and carcinoma were 42.5%, 71.4%, and 82.3%, respectively, indicating a decreased degree of KLF4 expression over the course of progressive transformation of normal cells into malignant derivatives. Therefore, the degree of this expression is . Degrees of Formality (Cont.) The degree of is 3. The correct answer is: [B]: \u2192 \" 2xy\u2074 + 4x\u00b2y\u00b3 \u2013 6x\u00b3y\u00b2 \u2013 7x\u2074\" . The First Amendment of the U.S. Constitution protects the rights of individuals to freedom of religion, speech, press, petition, and assembly. 2. a letter to a Member of Parliament:Thank you for your help in this matter. $\\begingroup$ To me, and this is standard terminology in theoretical CS, the degree of a boolean function would mean the degree of its polynomial (Fourier) representation. The highest value of the exponent in the expression is known as Degree of Polynomial. The degree of reaction (rth), or reaction ratio, is a parameter used for multistage turbomachinery defining the ratio of the static head (see Clearance gap pressure) to the fall head (of turbines) or the pump head (of centrifugal pumps).It is an indication of how the static pressure is distributed between impeller and stage. Download free printable Degree of expression Worksheets to practice. Solve this set of printable high school worksheets that deals with writing the degree of binomials. Read \"Prevalence and degree of expression of the carbapenemase gene (cfiA) among clinical isolates of Bacteroides fragilis in Nottingham, UK, Journal of Antimicrobial Chemotherapy\" on DeepDyve, the largest online rental service for scholarly research with thousands of \u2026 The degree ofreaction is then expressed as[For axial machines thenThe degree of reaction can also be written in terms of the geometry of theturbomachine as obtained bywhere is the vane angle of rotor outlet and is the vane angle of statoroutlet. Solve an algebraic expression with fractions. Degree of Binomials. It is sum of exponents of the variables in term. 5 Read the sentences below and decide whether or not they use the degree of formality appropriate for the given situation. Degree definition, any of a series of steps or stages, as in a process or course of action; a point in any scale. Example: x 3 y + x 2 + y x 3 y has degree 4 (3 for x and 1 for y) x 2 has degree 2 y has degree 1 So highest degree is 4, thus polynomial has degree 4 Degree of reaction (R) is an important factor in designing the blades of a turbine, compressors, pumps and other turbo-machinery. Substituting in this expression all numbers we already know, we obtain. And finally, the expression of degree by means of comparison, which is a special case of degree specification. Gene variability and degree of expression of vaccine candidate factor H binding protein in clinical isolates of Neisseria meningitidis. Degree of Polynomial - definition Degree of Polynomial is highest degree of its terms when Polynomial is expressed in its Standard Form. Click hereto get an answer to your question \ufe0f Find the degree of the expression [ x + (x^3 - 1)^1/2]^5 + [ x - (x^3 - 1)^1/2]^5 Calculating the degree of a polynomial. For instance, the degree of 8x is 1. $$\\frac{x^2+1}{6x-2}$$ If the above question has a degree, please tell me the difference between the degree of a polynomial and the degree of an expression? On comparing people or things, as bearers of a certain quality or characteristic, we do it by means of degree specification, thus in terms of positive, comparative, and superlative comparison. Correct answer to the question Which algebraic expression is a polynomial with a degree of 5? The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.To find the degree all that you have to do is find the largest exponent in the polynomial.Note: Ignore coefficients-- coefficients have nothing to do with the degree of a polynomial. Exploring these different trots can help you to increase your horse\u2019s collection, cadence and expression. The collection has to have the same degree \u2026 For instance, the degree \u2026 Cox proportional hazards model was used to assess the relationship between the degree of expression of DNMT1 and overall survival after adjusting for relevant covariates. 1. a note to a co-worker:The meeting is at 10 sharp.Don\u2019t be late. \u2022 If a term has more than one variable, the degree is equal to the sum of the exponents of all its variables. The calculator may be used to determine the degree of a polynomial. The quadratic function f(x) = ax 2 + bx + c is an example of a second degree polynomial. Freedom of expression refers to the ability of an individual or group of individuals to express their beliefs, thoughts, ideas, and emotions about different issues free from government censorship. Thank you for your interest in studying a Higher Degree by Research (HDR) course at Deakin University. But it seems unlikely that the term is used in this sense here; also I do not see a nice way \u2026 Remember that a polynomial is any algebraic expression that consists of terms in the form $$a{x^n}$$. There are no higher terms (like x 3 or abc 5). For the reaction : 2HI(g) \u21cc H2 (g)+ I2 (g), the degree of dissociated (\u03b1) of HI(g) is related to equilibrium constant Kp by the expression: asked Jan 6, 2020 in Chemistry by Mousam ( 52.8k points) While finding the degree of the polynomial, the polynomial powers of the variables should be either in ascending or descending order. Find the degree of each term and then compare them. If you're interested in studying the Associate Degree of Advanced Manufacturing, please send an expression of interest to us at adaminfo@uts.edu.au. Degree of Polynomials. Here's how you would do it: (x + 3)/6 = 2/3 First, cross multiply to get rid of the fraction. The degree of a polynomial is the largest exponent. Free Class 7 Degree of expression Worksheets. Note: This answer choice is a polynomial with a degree of \"5\" .Note that there are 2 (TWO) variables in this polynomial expression The expression of DNMT1 was examined using immunohistochemistry, and the degree of expression of DNMT1 was expressed as a percentage of cells positive for DNMT1 and its intensity. Another way to write the last example is $- 8{x^0}$ Written in this way makes it clear that the exponent on the $$x$$ is a zero (this also explains the degree\u2026) and so we can see that it really is a polynomial in one variable. In mathematics, the degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. Once assessed, you will also need to apply for 609996 Associate Degree of Advanced Manufacturing via UAC. The term shows being raised to the seventh power, and no other in this expression is raised to anything larger than seven. The highest power is the degree of \u2026 To obtain the degree of a polynomial defined by the following expression x^3+x^2+1, enter : degree(x^3+x^2+1) after calculation, the result 3 is returned. The degree of is 6. Polynomial is a mathematical expression consisting of variables, constants that can be combined using mathematical operations addition, subtraction, multiplication and whole number exponentiation of \u2026 In turbomachinery, Degree of reaction or reaction ratio (R) is defined as the ratio of the static pressure drop in the rotor to the static pressure drop in the stage or as the ratio of static enthalpy drop in the rotor to the static enthalpy drop in the stage.. If you want to solve an algebraic expression that uses fractions, then you have to cross multiply the fractions, combine like terms, and then isolate the variable. 2x 2, a 2, xyz 2). Degree of Dissociation. 4x3 \u2013 StartFraction 2 Over x EndFraction 2y3 + 5y2 \u2013 5y 3y3 \u2013 StartRoot 4 y EndRoot Question: what is the degree of the following expression? The Fixed Class of Degree Words \"[An] example of words that don't fit neatly into one category or another is degree words. The degree of dissociation is the phenomenon of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. In Freemasonry, the third and highest degree is that of master mason, attained after a stiff examination, and several writers speculate that this may be the source of the late nineteenth-century expression for an inquisition. There are 2 offer rounds remaining for UAC, so please check the submission deadlines with UAC. Kelly A(1), Jacobsson S, Hussain S, Olc\u00e9n P, M\u00f6lling P. Author information: (1)Department of Clinical Medicine, School of Health and Medical Sciences, \u00d6rebro University, Sweden. 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Polynomial with a degree of a polynomial remaining for UAC, so please check the submission with!\n\nCozy Comfort Slippers, Lsu Meal Plans 2018, Robert Porcher Madden 21, Keep It In The Pocket Meaning, Uconn Health Center Login, Dodge Dakota Aftermarket Parts, Class I Felony North Carolina,", "date": "2021-05-11 05:00:15", "meta": {"domain": "marc-petit.com", "url": "https://marc-petit.com/wwhy41x0/276140-degree-of-expression", "openwebmath_score": 0.5583773255348206, "openwebmath_perplexity": 1161.8103084296717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.966914019704466, "lm_q2_score": 0.8807970873650401, "lm_q1q2_score": 0.8516550522881167}}
{"url": "https://math.stackexchange.com/questions/2245798/mathematical-field-show-that-0-a-0-1-a-a", "text": "# Mathematical field, show that 0\u00b7a = 0, (-1)\u00b7a = -a, \u2026\n\nBased on the axioms for a mathematical field, the wiki article states that 0\u00b7a = 0 and (-1)\u00b7a = -a are consequences of the axioms, but doesn't show how they are derived. There was a similar question asked before, but I'm not sure about the accepted answer.\n\nhttps://en.wikipedia.org/wiki/Field_(mathematics)#Elementary_consequences_of_the_definition\n\nAlso, it would seem logical that if a \u2260 b, and if c \u2260 0, then c\u00b7a \u2260 c\u00b7b, a uniqueness property that should hold true for a finite field (unordered) that I'm wondering if it can be derived from the axioms (perhaps something like induction?) .\n\n\u2022 These are in fact standard easy arguments (your guess in the last sentence is correct). I'm sure someone will step up and provide them. Here's a hint for the first one: consider $(0 + 0) \\times a$. I'm curious: what prompts your curiousity? A course you're taking? \u2013\u00a0Ethan Bolker Apr 22 '17 at 0:12\n\u2022 @EthanBolker - a question came up at another forum. An \"addition\" table was given for a finite field with 4 numbers (0,1,2,3), where the \"addition\" turns out to be exclusive or, and the problem was asking to produce the multiplication table, based on the axioms rather than knowledge of GF(4). That prompted me to wonder how the \"consequences\" as noted in the wiki article are derived, and what operations are considered acceptable as derivations for field math. \u2013\u00a0rcgldr Apr 22 '17 at 0:42\n\u2022 @EthanBolker - I'm aware that algebra works with finite finite fields (I've worked with RS ECC), but never thought about the derivations based on the axioms. For that other forum question, after taking into account 0\u00b7a = a\u00b70 = 0 and 1\u00b7a = a\u00b71 = a, only 4 products in the multiplication table (2\u00b72, 2\u00b73, 3\u00b72, 3\u00b73) needed to be determined, so this could be done by trial and error, where an axiom would fail if the wrong set of 4 products was chosen. Noting that every number has a multiplicative inverse, meant that two of those products = 1 => 2\u00b73 = 3\u00b72 = 1, which simplified the problem. \u2013\u00a0rcgldr Apr 22 '17 at 1:06\n\nFor $0 \\times a$: $$0 \\times a = (0+0)\\times a = 0\\times a + 0 \\times a .$$ Now whatever $0 \\times a$ is, it has an additive inverse, so you can subtract it from each side of that equation to conclude that $0 \\times a = 0$.\n\nFor $(-1) \\times a$: $$0 \\times a = (1 + (-1)) \\times a = 1 \\times a + (-1) \\times a = a + (-1) \\times a$$ but $a$ has a unique additive inverse $-a$.\n\nIf $c \\ne 0$ then it has a multiplicative inverse $d$. Then $$ca = cb \\implies ca - cb = c(a-b) = 0 \\implies dc(a-b) = 0 \\implies a-b = 0 \\implies a = b.$$\n\n\u2022 Shouldn't the third part somewhere include a - b \u2260 0 or perhaps a - b = e \u2260 0 ? \u2013\u00a0rcgldr Apr 22 '17 at 0:26\n\u2022 @rcgldr No. I proved your claim by reasoning with the contrapositive form. If $a \\ne b$ then the last entry in my chain of correct implications is false, so the first entry must be false. \u2013\u00a0Ethan Bolker Apr 22 '17 at 0:30\n\u2022 OK, thanks for the explanation. I was also considering let a-b = e \u2260 0, then ca - cb = ce \u2260 0, since c \u2260 0 and e \u2260 0. \u2013\u00a0rcgldr Apr 22 '17 at 0:35\n\nI presume that by \"-a\" you mean the additive inverse of a so that you want to prove that -1 times a is the additive inverse of a. Again, that follows from the distributive law. (1+ (-1))a= 0a= 0. But, by the distributive law, (1+ (-1)))a= 1a+ (-1)a= a+ (-1)a= 0 also. \"a+ (-1)a= 0\" is precisely what is meant by \"additive inverse of a\".\n\nThe statement \"if a \u2260 b, and if c \u2260 0, then c\u00b7a \u2260 c\u00b7b\" is most easily proved by proving the \"contrapositive\". For any statement \"if p then q\", the contrapositive is the statement \"if not q then not p\" and it is easy to show in general that a statement is true if and only if its contrapositive is true.\n\nThe contrapositive of \"if a \u2260 b, and if c \u2260 0, then c\u00b7a \u2260 c\u00b7b\" is \"if ca= cb then a= b\". To prove that add the additive inverse of (subtract) cb from both sides: ca- cb= 0. By the distributive law, c(a- b)= 0. Since c is not 0 we must have a- b= 0 so a= b.\n\n\u2022 Should the contrapositive be stated as $$\\text{If ca=cb the either a=b or c=0}?$$ \u2013\u00a0Juniven Apr 22 '17 at 0:37\n\n$(-1)a=-a$ means $a+(-1)a=0$ as additive inverses are unique.\n\n$a+(-1)a=1*a+(-1)a$ (existence of multiplicative identity.\n\n$=(1+(-1))a$ (distributive property)\n\n$=0*a$. Remains to show $0*a=0$ for all $a$\n\n$0*a= (0+0)*a$ (Associativity and definition of additive property.\n\n$=0*a+0*a$.\n\n$0a=0a+0a$\n\n$0a+(-0a)=0a+0a+(-0a)$ (existence of additive inverse and acknowledgement that addition is a closed binary opperation)\n\n$0=0a+0=0a$(additive inverse and implied associtivity.)\n\nSo $a+(-1)a=0$. So $(-1)a=-a$ (and $a= -(-1 (a))$.)", "date": "2021-06-14 00:49:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2245798/mathematical-field-show-that-0-a-0-1-a-a", "openwebmath_score": 0.8347235321998596, "openwebmath_perplexity": 639.2349405717811, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9669140206578809, "lm_q2_score": 0.8807970826714614, "lm_q1q2_score": 0.8516550485895946}}
{"url": "https://www.fyears.org/tests-as-linear/simulations/simulate_spearman.html", "text": "This document presents the close relationship between Spearman correlation and Pearson correlation. Namely, that Spearman correlation is just a Pearson correlation on $$rank$$ed $$x$$ and $$y$$. It is an appendix to the post \u201cCommon statistical tests as linear models\u201d.\n\nTL;DR: Below, I argue that this approximation is good enough when the sample size is 12 or greater and virtually perfect when the sample size is 20 or greater.\n\n# 1 Simple example\n\nHere, I start by creating some mildly correlated data. Note that the distribution of the data does not matter since everything is ranked, so the results are identical for non-normal data.\n\ndata = MASS::mvrnorm(50, mu=c(2, 2), Sigma=cbind(c(1, 0.4), c(0.4, 1)))\nx = data[,1]\ny = data[,2]\nplot(rank(y) ~ rank(x))\n\nNow let\u2019s test it\n# Three ways of running the same model\nspearman = cor.test(x, y, method='spearman')\npearson = cor.test(rank(x), rank(y), method='pearson')  # On ranks\nlinear = summary(lm(rank(y) ~ rank(x)))  # Linear is identical to pearson; just showing.\n\n# Present the results\nprint('coefficient rho is exactly identical:')\nrbind(spearman=spearman$estimate, ranked_pearson=pearson$estimate,\nranked_linear=linear$coefficients[2]) print('p is approximately identical:') rbind(spearman=spearman$p.value,\nranked_pearson=pearson$p.value, ranked_linear=linear$coefficients[8])\n## [1] \"coefficient rho is exactly identical:\"\n##                      rho\n## spearman       0.1848259\n## ranked_pearson 0.1848259\n## ranked_linear  0.1848259\n## [1] \"p is approximately identical:\"\n##                     [,1]\n## spearman       0.1982118\n## ranked_pearson 0.1988046\n## ranked_linear  0.1988046\n\nThe correlation coefficients are exact. And the p-values are quite close! But maybe we were just lucky? Let\u2019s put it to the test:\n\n# 2 Simulate for different N and r\n\nThe code below does the following:\n\n1. Generate some correlated data for all combinations of the correlation coefficient ($$r = 0, 0.5, 0.95$$) and sample sizes 1 $$N = 6, 8, 10, ... 20, 30, 50, 80$$.\n2. Compute coefficients and p-values for Spearman correlation test and Pearson correlation on the ranked values, many times for each combination.\n3. Calculate the number of errors.\nlibrary(tidyverse)\n\n# Settings for data simulation\nPERMUTATIONS = 1:200\nNs = c(seq(from=6, to=20, by=2), 30, 50, 80)  # Sample sizes to model\nrs = c(0, 0.5, 0.95)  # Correlation coefficients to model\n\n# Begin\nD = expand.grid(set=PERMUTATIONS, r=rs, N=Ns) %>%  # Combinations of N and r\nmutate(\n# Use the parameters to generate correlated data in each row\ndata = map2(N, r, function(N, r) MASS::mvrnorm(N, mu=c(0, 4), Sigma=cbind(c(1, r), c(r, 1)))),\n\n# Tests\npearson_raw = map(data, ~cor.test(rank(.x[, 1]), rank(.x[, 2]), method='pearson')),\nspearman_raw = map(data, ~cor.test(.x[, 1], .x[, 2], method = 'spearman')),\n\n# Tidy it up\npearson = map(pearson_raw, broom::tidy),\nspearman = map(spearman_raw, broom::tidy),\n) %>%\n\n# Get estimates \"out\" of the tidied results.\nunnest(pearson, spearman, .sep='_') %>%\nselect(-data, -pearson_raw, -spearman_raw)\n\n# If you want to do a permutation-based Spearman to better handle ties and\n# overcome issues on how to compute p (computationally heavy).\n# Corresponds very closely to R's spearman p-values\n#mutate(p.perm = map(data, ~ coin::pvalue(coin::spearman_test(.x[,1] ~ .x[,2], distribution='approximate')))) %>%\n#unnest(p.perm)\n\nhead(D)\n\n### 2.0.1 Inspect correlation coefficients\n\nJust as we saw in the simple example above, correlation coefficients (estimate and estimate1) are identical per definition. We can see that by calculating the difference and see that it is always zero:\n\nsummary(D$estimate - D$estimate1)\n##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.\n## \n\nYup. The minimum difference is zero. The maximum is zero. All of them are identicalNo need to look further into that.\n\n### 2.0.2 Inspet p-values\n\nBefore getting started on p-values, note that there are several ways of calculating them for Spearman correlations. This is the relevant section from the documentation of cor.test:\n\nFor Spearman\u2019s test, p-values are computed using algorithm AS 89 for n < 1290 and exact = TRUE, otherwise via the asymptotic t approximation. Note that these are \u2018exact\u2019 for n < 10, and use an Edgeworth series approximation for larger sample sizes (the cutoff has been changed from the original paper).\n\nYou can get beyond all of that by doing permutation-based tests at the cost of computational power (see outcommented code above). However, there is almost virtually perfect correspondence, so I just rely on R\u2019s more computationally efficient p-values.\n\nBelow, I plot the difference between Spearman-derived p-values and ranked-Pearson-derived p-values. I have chosen not to plot as a function or r in the name of simplicity because it makes no difference whatsoever, but feel free to try it out yourself.\n\nlibrary(patchwork)\n\n# A straight-up comparison of the p-values\np_relative = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value, color=N)) +\ngeom_line() +\ngeom_vline(xintercept=0.05, lty=2) +\ngeom_hline(yintercept=0.05, lty=2) +\n\nlabs(title='Absolute relation', x = 'Spearman p-value', y = 'Pearson p-value') +\n#coord_cartesian(xlim=c(0, 0.10), ylim=c(0, 0.11)) +\ntheme_gray(13) +\nguides(color=FALSE)\n\n# Looking at the difference (error) between p-values\np_error_all = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +\ngeom_line() +\ngeom_vline(xintercept=0.05, lty=2) +\n\nlabs(title='Error', x='Spearman p-value', y='Difference') +\ncoord_cartesian(ylim=c(-0.02, 0.005)) +\ntheme_gray(13) +\nguides(color=FALSE)\n\n# Same, but zoomed in around p=0.05\np_error_zoom = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +\ngeom_line() +\ngeom_vline(xintercept=0.05, lty=2) +\n\nlabs(title='Error zoomed', x='Spearman p-value', y='Difference') +\ncoord_cartesian(ylim=c(-0.02, 0.005), xlim=c(0, 0.10)) +\ntheme_gray(13)\n\n# Show it. Patchwork is your friend!\np_relative + p_error_all + p_error_zoom\n\nFigure: The jagget lines in the leftmost two panels are due to R computing \u201cexact\u201d p-values for Spearman when N < 10. For samples larger than N>=10, an approximation is used which gives a smoother relationship.\n\nIt is clear that larger N gives a closer correspondence. For small p, the Pearson-derived p-values are lower than Spearman, i.e.\u00a0more liberal. Importantly, when N=12, this effect is at most 0.5% in the \u201csignificance-region\u201d which I deem acceptable enough for the present purposes. When N=20, the error is at most 0.2% for the whole curve and there is no difference when N=50 or N=100.\n\n# 3 Conclusion\n\nThe correlation coefficients are identical.\n\nThe p-value is exact enough when N > 10 and virtualy perfect when N > 20. If you are doing correlation studies on sample sizes lower than this, you have larger inferential problems than those posed by the slight differences between Spearman and Pearson.", "date": "2022-06-26 09:11:34", "meta": {"domain": "fyears.org", "url": "https://www.fyears.org/tests-as-linear/simulations/simulate_spearman.html", "openwebmath_score": 0.7389844655990601, "openwebmath_perplexity": 6448.985769954777, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9669140216112958, "lm_q2_score": 0.8807970811069351, "lm_q1q2_score": 0.8516550479165973}}
{"url": "https://www.freemathhelp.com/forum/threads/negative-roots-when-simplifying-surds.114156/", "text": "# Negative roots when simplifying surds\n\n#### NeedingWD40\n\n##### New member\nSomething that puzzles me:\n\nThe square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?\n\ne.g. \u23b7300 -\u23b748 =\u23b7100x3 -\u23b716x3 = 10\u23b73 - 4\u23b73 = 6\u23b73\n\nwhat about the negative root of 100, and the negative root of 16?\n\nThanks in advance for your help.\n\n#### Dr.Peterson\n\n##### Elite Member\nSomething that puzzles me:\n\nThe square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?\n\ne.g. \u23b7300 -\u23b748 =\u23b7100x3 -\u23b716x3 = 10\u23b73 - 4\u23b73 = 6\u23b73\n\nwhat about the negative root of 100, and the negative root of 16?\n\nThanks in advance for your help.\nWhen we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.\n\nSo although 100 has two square roots, \u23b7100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.\n\nHave you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, \u23b79 -\u23b74 would have to be (\u00b13) - (\u00b12), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.\n\n#### Jomo\n\n##### Elite Member\nSomething that puzzles me:\n\nThe square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?\n\ne.g. \u23b7300 -\u23b748 =\u23b7100x3 -\u23b716x3 = 10\u23b73 - 4\u23b73 = 6\u23b73\n\nwhat about the negative root of 100, and the negative root of 16?\n\nThanks in advance for your help.\nBy definition, $$\\displaystyle \\sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\\displaystyle \\sqrt{x}$$ has no real solution.\n\n#### NeedingWD40\n\n##### New member\nWhen we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.\n\nSo although 100 has two square roots, \u23b7100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.\n\nHave you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, \u23b79 -\u23b74 would have to be (\u00b13) - (\u00b12), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.\nAh! That's why the formula for solving a quadratic equation explicitly says + or -\u23b7(b2-4ac), otherwise only the positive root would be found. Thank you very much, I'm clearer now\n\n#### JeffM\n\n##### Elite Member\nSomething that puzzles me:\n\nThe square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?\n\ne.g. \u23b7300 -\u23b748 =\u23b7100x3 -\u23b716x3 = 10\u23b73 - 4\u23b73 = 6\u23b73\n\nwhat about the negative root of 100, and the negative root of 16?\n\nThanks in advance for your help.\nYou are absolutely correct that\n\n$$\\displaystyle (-\\ 10)^2 = 100 \\text { and } (+\\ 10)^2 = 100.$$\n\nBut the technical definition of the square root function of a non-negative real number is:\n\n$$\\displaystyle \\text {If } a \\ge 0, \\text { then } \\sqrt{a} \\ge 0 \\text { and } \\sqrt{a} * \\sqrt{a} = a.$$\n\nThat is why if we get an equation that looks like\n\n$$\\displaystyle x^2 = 100 \\implies x = \\pm \\sqrt{100} \\implies x = 10 \\text { or } x = -\\ 10.$$\n\nThat plus or minus sign is there for a reason.\n\n#### NeedingWD40\n\n##### New member\nBy definition, $$\\displaystyle \\sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\\displaystyle \\sqrt{x}$$ has no real solution.\nThank you, I'm clearer now.\n\n#### NeedingWD40\n\n##### New member\nYou are absolutely correct that\n\n$$\\displaystyle (-\\ 10)^2 = 100 \\text { and } (+\\ 10)^2 = 100.$$\n\nBut the technical definition of the square root function of a non-negative real number is:\n\n$$\\displaystyle \\text {If } a \\ge 0, \\text { then } \\sqrt{a} \\ge 0 \\text { and } \\sqrt{a} * \\sqrt{a} = a.$$\n\nThat is why if we get an equation that looks like\n\n$$\\displaystyle x^2 = 100 \\implies x = \\pm \\sqrt{100} \\implies x = 10 \\text { or } x = -\\ 10.$$\n\nThat plus or minus sign is there for a reason.\nSo, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....\n\n#### tkhunny\n\n##### Moderator\nStaff member\nSo, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....\n\"problem includes ... no plus or minus\" - Not a fan.\n\nWhat is $$\\displaystyle \\sqrt{4}$$? It is 2\n\nI do not ever recall seeing such a simplification problem as: What is $$\\displaystyle \\pm\\sqrt{4}$$? It is +/- 2.\n\nLet the context be your guide.\n\nDon't make up things that don't exist. $$\\displaystyle \\sqrt{4} = 2$$ and that is NOT the same as \"+/-2\".\nDon't miss things that do exist. $$\\displaystyle Solve\\;x^{2} = 4$$. $$\\displaystyle x = +2\\;or\\;x = -2$$ and that is the same as \"x = +/-2\".\n\nAlso, keep your eyes on that \"+/-\" symbol. It means several different things in different contexts.\n\n#### Otis\n\n##### Senior Member\nI'm not sure whether you've yet learned anything about solving algebraic equations or the concept of a function (you posted on the Pre-Algebra board), but here's my comments, anyways.\n\nWhen you are given a surd, it always represents the positive root (only is exception is \u221a0).\n\nWhen you are solving an equation, sometimes you need to introduce surds (like the algebraic step of taking the square root of each side of an equation). In this case, you need to consider both the positive and negative root. Otherwise, you might miss solution(s).\n\nAs Jeff wrote, we express a positive root as \u221an, and when we need to express the negative root we write -\u221an. Cheers :cool:\n\n#### JeffM\n\n##### Elite Member\nSo, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....\nPlease read the other responses to this post because they are good and will give you different perspectives.\n\n$$\\displaystyle \\sqrt{a} \\ge 0$$ by definition if we are dealing with real numbers.\n\nSo, we do not assume that the square root is the primary root (meaning a non-negative number), we know that it is a non-negative number by definition.\n\nHowever, if the square root of a is positive, we know that the square of the square root's additive inverse is also a. In notation,\n\n$$\\displaystyle \\sqrt{a} * \\sqrt{a} = a = (-\\ \\sqrt{a}) * (-\\ \\sqrt{a}).$$\n\nSo if we have an equation like\n\n$$\\displaystyle x^2 = c \\ge 0$$,\n\nwe have no way to know from the equation itself whether\n\n$$\\displaystyle x = \\sqrt{c} \\text { or } x = - \\ \\sqrt{c}.$$\n\nIt may be that either solution works or that only one of the two works, but that must be decided based on information not contained in the equation itself.\n\nUncertainty about which answer applies does not mean that there is any uncertainty about what the surd means. To clarify where the uncertainty lies we say\n\n$$\\displaystyle x = \\pm \\sqrt{c}.$$\n\nThe sign of the surd is certain. It is the sign of x that is uncertain.\n\nLast edited:\n\n#### pka\n\n##### Elite Member\nSomething that puzzles me:\nThe square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?\ne.g. \u23b7300 -\u23b748 =\u23b7100x3 -\u23b716x3 = 10\u23b73 - 4\u23b73 = 6\u23b73\nwhat about the negative root of 100, and the negative root of 16?\nI have had issues with this my whole active research & teaching life.\nI had problems with students in my complex variables classes with the use of the radical: $$\\displaystyle \\sqrt{z}$$.\nNow I belong to a school of analyst who think that the symbol $$\\displaystyle \\sqrt~$$ must be applied only to a non-negative real number.\nSo that means $$\\displaystyle \\sqrt{-1}$$ has no meaning whatsoever. So what is $$\\displaystyle \\bf{i}~?$$\nIt is so simple, $$\\displaystyle \\bf{i}$$ is a solution of the equation of the equation $$\\displaystyle z^2+1=0$$\nNow clearly the number $$\\displaystyle 100$$ has two square roots, $$\\displaystyle \\pm 10$$.\nBUT $$\\displaystyle \\sqrt{100}=10$$ that is one number. Moreover, $$\\displaystyle -\\sqrt{100}=-10$$ that is one number.\nThus it is absolutely incorrect to write that $$\\displaystyle \\sqrt{100}=\\pm 10$$\n\n#### JeffM\n\n##### Elite Member\nI have had issues with this my whole active research & teaching life.\nI had problems with students in my complex variables classes with the use of the radical: $$\\displaystyle \\sqrt{z}$$.\nNow I belong to a school of analyst who think that the symbol $$\\displaystyle \\sqrt~$$ must be applied only to a non-negative real number.\nSo that means $$\\displaystyle \\sqrt{-1}$$ has no meaning whatsoever. So what is $$\\displaystyle \\bf{i}~?$$\nIt is so simple, $$\\displaystyle \\bf{i}$$ is a solution of the equation of the equation $$\\displaystyle z^2+1=0$$\nNow clearly the number $$\\displaystyle 100$$ has two square roots, $$\\displaystyle \\pm 10$$.\nBUT $$\\displaystyle \\sqrt{100}=10$$ that is one number. Moreover, $$\\displaystyle -\\sqrt{100}=-10$$ that is one number.\nThus it is absolutely incorrect to write that $$\\displaystyle \\sqrt{100}=\\pm 10$$\n\nI like this convention that the surd can only be applied to non-negative real numbers. It would avoid a lot of confusion.\n\nUnfortunately, conventions are necessarily extra-individual. And the surd is frequently used in a broader sense.\n\nAnyway, I do like your school's approach. But is there not an inconsistency when it comes to\n\n$$\\displaystyle \\sqrt[3]{-\\ 8}$$\n\nor is that prohibited as well. If so, is there a convenient notation to deal with it?\n\n#### pka\n\n##### Elite Member\nAnyway, I do like your school's approach. But is there not an inconsistency when it comes to\n$$\\displaystyle \\sqrt[3]{-\\ 8}$$\nor is that prohibited as well. If so, is there a convenient notation to deal with it?\nNo, because the any real number has a real root if the index is odd. Example $$\\displaystyle \\large\\sqrt[5]{-32}=-2$$\n\n#### JeffM\n\n##### Elite Member\nNo, because the any real number has a real root if the index is odd. Example $$\\displaystyle \\large\\sqrt[5]{-32}=-2$$\nAhh so your school's rule is\n\n$$\\displaystyle \\sqrt[a]{b} \\text { is defined only if (1) } a \\in \\mathbb Z \\text {, (2) } a \\ge 1 \\text {, (3) } \\ b \\in \\mathbb R,$$\n\n$$\\displaystyle \\text {and (4) } a \\text { even } \\implies b \\ge 0.$$\n\n#### pka\n\n##### Elite Member\nAhh so your school's rule is\n$$\\displaystyle \\sqrt[a]{b} \\text { is defined only if (1) } a \\in \\mathbb Z \\text {, (2) } a \\ge 1 \\text {, (3) } \\ b \\in \\mathbb R,$$\n$$\\displaystyle \\text {and (4) } a \\text { even } \\implies b \\ge 0.$$\n.\nAt meetings we would have new jokes about exponents. Prof Paterson has address this above.\nThe objection is this $$\\displaystyle \\sqrt{-16}=\\pm 4\\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.\nIt is true that many in my tradition are hyper sensitive to vocabulary abuse.\n\nThere are four fourth roots of $$\\displaystyle -16\\bf{i}$$ but there is just one answer to $$\\displaystyle \\large\\sqrt[4]{-16\\bf{i}}=~?$$\n\n#### lookagain\n\n##### Senior Member\nNeedingWD40 said:\ne.g. \u23b7300 -\u23b748 =\u23b7100x3 -\u23b716x3 = 10\u23b73 - 4\u23b73 = 6\u23b73\nThe above is incorrect. You need grouping symbols, such as below. I used asterisks for the products.\n\ne.g.\n\n\u23b7300 - \u23b748 =\n\n\u23b7(100*3) - \u23b7(16*3) =\n\n10\u23b73 - 4\u23b73 =\n\n6\u23b73\n\nLast edited:\n\n#### topsquark\n\n##### Full Member\n.\nAt meetings we would have new jokes about exponents. Prof Paterson has address this above.\nThe objection is this $$\\displaystyle \\sqrt{-16}=\\pm 4\\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.\nIt is true that many in my tradition are hyper sensitive to vocabulary abuse.\n\nThere are four fourth roots of $$\\displaystyle -16\\bf{i}$$ but there is just one answer to $$\\displaystyle \\large\\sqrt[4]{-16\\bf{i}}=~?$$\nWhen I see stuff like \"i\"s under a radical I go straight over to the exponential format.\n$$\\displaystyle \\sqrt[4]{-16 i } = \\sqrt[4]{16} \\cdot \\sqrt[4]{-i}$$\n\n$$\\displaystyle = 2 \\cdot \\left ( e^{ 3i \\pi/2 + 2 i \\pi n } \\right ) ^{1/4}$$\n\n$$\\displaystyle = 2 \\cdot e^{3i \\pi /8 + i \\pi n / 2}$$\n\n$$\\displaystyle = 2 \\cdot e^{3i \\pi / 8} \\cdot i^n$$\n\n$$\\displaystyle = 2 (i ^n) \\left ( cos ( 3 \\pi /8 ) + i~sin (3 \\pi / 8) \\right )$$\nwhere n is an integer.\n\nPerhaps it's a bit messy but I don't have to think so hard about any negative signs.\n\n-Dan\n\n#### pka\n\n##### Elite Member\nWhen I see stuff like \"i\"s under a radical I go straight over to the exponential format.\n$$\\displaystyle \\sqrt[4]{-16 i } = \\sqrt[4]{16} \\cdot \\sqrt[4]{-i}$$\n-Dan\nI did a very poor job explaining our objections. No one would ever need to use $$\\displaystyle \\sqrt[4]{-16 i }.$$\nOn the other hand there could be a need to solve $$\\displaystyle z^4+16\\bf{i}=0$$\nThere are four roots of that equation. If $$\\displaystyle \\rho=2\\exp\\left(\\frac{-\\pi\\bf{i}}{8}\\right)$$ then $$\\displaystyle \\rho^4+16\\bf{i}=0$$, in other words one solution.\nLet $$\\displaystyle \\zeta=\\exp\\left(\\frac{2\\pi\\bf{i}}{4}\\right) \\$$ now the collection $$\\displaystyle \\rho\\cdot\\zeta^k,~k=1,2,3$$ are the other three roots(solutions).\n\n#### NeedingWD40\n\n##### New member\nThanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!\n\nSo, to attempt to summarise:\n\n\u23b7a is always positive.\nThe solution to an equation involving a square root may have two solutions:\u23b7a or -\u23b7a\nThe context of the question should clarify if the negative of the root gives a valid solution.\n\n#### JeffM\n\n##### Elite Member\nThanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!\n\nSo, to attempt to summarise:\n\n\u23b7a is always positive.\nThe solution to an equation involving a square root may have two solutions:\u23b7a or -\u23b7a\nThe context of the question should clarify if the negative of the root gives a valid solution.\nThat is very good. I'd make one slight emendation to your third point. I'd phrase it as\n\nThe context of the equation should clarify if both solutions are valid or else which one of the two is valid.\n\nI would not assume that the negative solution is the only one that may not apply. In some cases, what is classified as positive and negative is arbitrary. It is of course true that we usually classify in a way that ensures a valid answer is non-negative, but that is not guaranteed.\n\nVery minor point. You seem to have it. Good job.", "date": "2019-05-19 20:29:00", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/negative-roots-when-simplifying-surds.114156/", "openwebmath_score": 0.7189590930938721, "openwebmath_perplexity": 807.7533515363427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147153749274, "lm_q2_score": 0.8757869997529961, "lm_q1q2_score": 0.8516281660938714}}
{"url": "http://math.stackexchange.com/questions/89623/trick-to-find-multiples-mentally", "text": "# Trick to find multiples mentally\n\nWe all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.\n\nA few months ago, I heard a simple and elegant way to find multiples of $7$:\n\nCut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.\n\nExample:\n\n$21553$\n\nCut digits into pairs:\n\n$2, 15, 53$\n\nMultiply $53$ by $1, 15$ by $2, 2$ by $4$:\n\n$8, 30, 53$\n\n$8+30+53=91$\n\nAs $91$ is a multiple of $7$ ($13 \\cdot 7$), then $21553$ is too.\n\nThis works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 \u2192 05 = 7 - 2, 112 \u2192 12 = 14 - 2, \\cdots)$\n\nI figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.\n\nExemple with $91$:\n\n$91$\n\n$9, 1$\n\n$9\\cdot3, 1\\cdot1$\n\n$27, 1$\n\n$28$\n\nMy question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?\n\n-\nYou'll want to see this KCd blurb. \u2013\u00a0 Guess who it is. Dec 8 '11 at 15:49\nI've seen the following method for 7: Take the last digit, double it, and subtract it from the first digits. So $21553$ leads to $2155-3*2=2149$. Next step is $214-2*9=196$. Next $19-2*6=7$. \u2013\u00a0 Thomas Andrews Dec 8 '11 at 16:04\nHow can we find divisors of 23? Very easily: they are just $\\pm 1$ and $\\pm 23$. Presumably, you want to recognize multiples of 23? \u2013\u00a0 Arturo Magidin Dec 8 '11 at 16:11\nFor $23$, here is something fairly simple, for numbers that have $3$ or $4$ digits, say $abcd$. Take the number $ab$, add to it $3$ times the number $cd$. Call the result $t$. Then $abcd$ is a multiple of $23$ iff $t$ is. One could probably tweak this to be even simpler to execute. \u2013\u00a0 Andr\u00e9 Nicolas Dec 8 '11 at 16:16\n@ArturoMagidin Yes, I meant multiples. It's corrected. \u2013\u00a0 Oltarus Dec 9 '11 at 8:46\n\nOne needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\\rm\\ d_2\\ d_1\\ d_0 \\$ is $\\rm\\: (d_2\\cdot 10 + d_1)\\ 10 + d_0\\ \\equiv\\ (d_2\\cdot 3 + d_1)\\ 3 + d_0\\ (mod\\ 7)\\$ since $\\rm\\ 10\\equiv 3\\ (mod\\ 7)\\:.\\:$ So we compute the remainder $\\rm\\ (mod\\ 7)\\$ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\\rm\\:(mod\\ 7)\\:.\\:$\n\nFor example, let's use this algorithm to reduce $\\rm\\ 43211\\ \\:(mod\\ 7)\\:.\\:$ The algorithm consists of repeatedly replacing the first two leading digits $\\rm\\ d_n\\ d_{n-1}\\$ by $\\rm\\ d_n\\cdot 3 + d_{n-1}\\:\\ (mod\\ 7),\\:$ namely\n\n$\\rm\\qquad\\phantom{\\equiv} \\color{red}{4\\ 3}\\ 2\\ 1\\ 1$\n\n$\\rm\\qquad\\equiv\\phantom{4} \\color{green}{1\\ 2}\\ 1\\ 1\\quad$ by $\\rm\\quad \\color{red}4\\cdot 3 + \\color{red}3\\ \\equiv\\ \\color{green}1$\n\n$\\rm\\qquad\\equiv\\phantom{4\\ 3} \\color{royalblue}{5\\ 1}\\ 1\\quad$ by $\\rm\\quad \\color{green}1\\cdot 3 + \\color{green}2\\ \\equiv\\ \\color{royalblue}5$\n\n$\\rm\\qquad\\equiv\\phantom{4\\ 3\\ 5} \\color{brown}{2\\ 1}\\quad$ by $\\rm\\quad \\color{royalblue}5\\cdot 3 + \\color{royalblue}1\\ \\equiv\\ \\color{brown}2$\n\n$\\rm\\qquad\\equiv\\phantom{4\\ 3\\ 5\\ 2} 0\\quad$ by $\\rm\\quad \\color{brown}2\\cdot 3 + \\color{brown}1\\ \\equiv\\ 0$\n\nHence $\\rm\\ 43211\\equiv 0\\:\\ (mod\\ 7)\\:,\\:$ indeed $\\rm\\ 43211 = 7\\cdot 6173\\:.\\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\\rm\\: \\pm\\{0,1,2,3\\}\\ \\:(mod\\ 7)\\:.$ Notice that for modulus $11$ or $9\\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\\:$ (a.k.a. \"casting out nines\" for modulus $9\\:$).\n\n-\n\nOne approach is to find some higher multiple that makes it easy. For your example of $23$, note that $3*23+1=70$, so $a(70-1)$ will be a multiple of $23$. Now you have a single digit multiply followed by a subtraction. If you pick $a=301, a(70-1)=301(70-1)=301*70-301=2107-301=1806$, which is a multiple of $23$\n\n-\nI generally find this method much easier than the standard modular divisibility test. It does require that you memorize a starting list of multiples, but it's not so bad because you can just multiply them by powers of $10$. \u2013\u00a0 Qiaochu Yuan Dec 8 '11 at 18:28\n@QiaochuYuan: It seemed the OP's request was to find multiples, not do a divisibility test. But this generalizes. In the spirit of the test for $7$, to test divisibility by $23$, you can take off the last digit, multiply the rest by $7$ and add the last digit. Keep going until you have only two digits left. You just have to find a multiple that ends in $1$ or $9$ to do this. \u2013\u00a0 Ross Millikan Dec 8 '11 at 18:40\n\nIn general, if you are doing things by dividing the number into groups of $k$ digits, you can think of it as looking at the number in base $10^k.$\n\nMore generally, if $B$ is a base, and $n$ is a number with no common factors with $B$, then you can always find an $m$ such that $Bm\\equiv 1\\pmod n$. Then $X\\equiv 0\\pmod n$ if and only if $Xm\\equiv 0\\pmod n$. But if $X=Bu+v$, then $Xm\\equiv u+mv\\pmod n$. So if we take the last digit, base B, multiply it by $m$ and add it to the other digits, the result is divisible by $n$ if and only if the original number was divisible by $n$.\n\nIn the case of $n=23$ base $B=10$, you get $m=7$, so you can take the last digit, multiply it by seven, and add it to the rest.\n\nOr, if you use $B=100$, you get $m=3$, and you can take the lsat two digits, multiply by 3, and add to the other digits.\n\nIn general, you can always find a $k$ such that $10^k=1\\pmod n$ if $n$ is not even or divisible by 5. Then if you take base $B=10^k$, you get $m=1$, and you can separate $m$ into groups of $k$ digits and add them. That's hardly useful when $k$ is large. For example, the smallest $k$ for $n=23$ is $k=22$, so this part only helps if your starting number was more than 23 digits long.\n\nUsually, you want to find a relatively small pair $(m,k)$ so that $m10^k\\equiv \\pm 1\\pmod n$. Then you take the last $k$ digits, multiplied by $m$, and add to or subtract from the other digits, depending on whether $+1$ or $-1$.\n\n-\n\nThe following is a simple method to check divisibility by $7$ or $13$:\n\nCut the digits in pairs of 3 and calculate their alternating sum. If this is a multiple of 7 or 13, the original number is.\n\nFor example\n\n$12345631241$ leads to\n\n$$241-639+345-12=-65 \\,.$$\n\nThus our number is a multiple of $13$, but not a multiple of $7$.\n\nThis works because $1001=7*11*13$ meaning that any number of the form $abcabc$ is a multiple of $7, 11$ and $13$...The trick also works for 11, but there is another simple trick.\n\nMethod 2 If the number is relatively prime to 10 (if it is not, you can make it), look for a multiple on $n$ which ends in 1.\n\nLet say that this multiple is $a_1..a_k1$.\n\nThen you simply pick the large number and subtract $a_1...a_k*$last digit from the remaining digits.\n\nThe trick works because $a_1...ak1$*last digit is always a multiple of $n$, and subtracting this from the original number you get a multiple of 10..Since $n$ is relatively prime to 10, you can erase the 0 at the end...\n\nA simple such example, for $7$ the smalest such desires multiple is ... 21, which leads to the criteria posted by Thomas Andrews .\n\nAlso, for small numbers the following\n\n-\n\nI will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$z = a_0 + a_1 \\times 10 + a_2 \\times 10^2 + a_3 \\times10^3 + \\cdots + a_n \\times 10^n$$\n\nLets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,\n\n$$10 \\equiv 3, 10^2 \\equiv 2, 10^3 \\equiv -1, 10^4 \\equiv -3, 10^5 \\equiv -2, 10^6 \\equiv 1,$$\n\nThe successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\\cdots$$ is divisible by $7$\n\nTo understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \\times (10-3) + a_2 \\times (10^2-2) + a_3 \\times (10^3+1) + \\cdots + a_6 \\times (10^6-1) + \\cdots$$ Since all this numbers $(10-3),(10^2-2),(10^3+1),\\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.\n\nUsing this approach we can derive divisibility of any integer.\n\n-\n\nThis can be explained using Horner's method as below. z=a0 + a1 \u00d710 + a2 \u00d7102 + a3 \u00d7 10^3+\u22ef+ an \u00d7 10^n\n\ni.e. Z = ((((...(an \u00d7 10 + an-1)*10 + an-2 )*10 + an-3).....)*10 + a0\n\n                           ( Shift-add representation of number)\n\nZ % 7 =  ((((...(an \u00d7 3 + an-1)*3 + an-2 )*3 + an-3).....)*3 + a0  (Mod 7)\n\n\nNow evaluate from left to right using modulo algebra.\n\nTo know more about shift- add representation of number and its application in interbase conversion and divisibility refer tinyurl.com/mlxk8pw .\n\n                         Regard,", "date": "2015-05-30 13:00:06", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/89623/trick-to-find-multiples-mentally", "openwebmath_score": 0.8384157419204712, "openwebmath_perplexity": 193.73876350024653, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147185726375, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8516281610129148}}
{"url": "http://mathhelpforum.com/algebra/15734-help-reducing-quadratic-equation-final-terms.html", "text": "1. ## Help with reducing quadratic equation to final terms\n\n$x^4 - 6x^2 + 8 = 0\n$\n\n$x^2=t$\n$x^4=t^2$\n\na= 1, b = 6, c = 8\n\n$\\Rightarrow x = \\frac {6 \\pm \\sqrt {6^2 - 4(1)(8)}}{2(1)}$\n\n$\\Rightarrow x = \\frac {6 \\pm \\sqrt {4}}{2}$\n\n$\\Rightarrow x = \\frac {3 \\pm \\sqrt {2}}{}$\n\nHow do I reduce this further to get the final answer or is my answer correct? I always get confused. Thanks in advance.\n\n2. Originally Posted by lilrhino\n$x^4 - 6x^2 + 8 = 0\n$\n\n$x^2=t$\n$x^4=t^2$\n\na= 1, b = 6, c = 8\n\n$\\Rightarrow x = \\frac {6 \\pm \\sqrt {6^2 - 4(1)(8)}}{2(1)}$\n\n$\\Rightarrow x = \\frac {6 \\pm \\sqrt {4}}{2}$\n\n$\\Rightarrow x = \\frac {3} \\pm \\frac { \\sqrt {2}}{2}$\n\nHow do I reduce this further to get the final answer. I always get confused. Thanks in advance.\nok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here\n\nyou replaced $x^2$ with $t$, so you should have\n\n$t^2 - 6t + 8 = 0$ .........now factor\n\n$\\Rightarrow (t - 4)(t - 2) = 0$\n\n$\\Rightarrow t = 4 \\mbox { or } t = 2$\n\nBut $t = x^2$\n\n$\\Rightarrow x^2 = 4 \\mbox { or } x^2 = 2$\n\n$\\Rightarrow x = \\pm 2 \\mbox { or } x = \\pm \\sqrt {2}$\n\n3. Hello, lilrhino!\n\nYou're making it unnecessarily complicated . . . and wrong.\n\n$x^4 - 6x^2 + 8 \\:= \\:0$\nFactor: . $(x^2 - 4)(x^2 - 2)\\:=\\:0$\n\nAnd we have two equations to solve:\n\n. . $x^2 - 4\\:=\\:0\\quad\\Rightarrow\\quad x^2 \\:=\\:4\\quad\\Rightarrow\\quad x \\:=\\: \\pm2$\n\n. . $x^2 - 2\\:=\\:0\\quad\\Rightarrow\\quad x^2\\:=\\:2\\quad\\Rightarrow\\quad x\\:=\\:\\pm\\sqrt{2}$\n\n4. Originally Posted by Jhevon\nok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here\n\nyou replaced $x^2$ with $t$, so you should have\n\n$t^2 - 6t + 8 = 0$ .........now factor\n\n$\\Rightarrow (t - 4)(t - 2) = 0$\n\n$\\Rightarrow t = 4 \\mbox { or } t = 2$\n\nBut $t = x^2$\n\n$\\Rightarrow x^2 = 4 \\mbox { or } x^2 = 2$\n\n$\\Rightarrow x = \\pm 2 \\mbox { or } x = \\pm \\sqrt {2}$\nThanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!\n\n5. Originally Posted by lilrhino\nThanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!\nyeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with\n\n6. Originally Posted by Soroban\nHello, lilrhino!\n\nYou're making it unnecessarily complicated . . . and wrong.\n\nFactor: . $(x^2 - 4)(x^2 - 2)\\:=\\:0$\n\nAnd we have two equations to solve:\n\n. . $x^2 - 4\\:=\\:0\\quad\\Rightarrow\\quad x^2 \\:=\\:4\\quad\\Rightarrow\\quad x \\:=\\: \\pm2$\n\n. . $x^2 - 2\\:=\\:0\\quad\\Rightarrow\\quad x^2\\:=\\:2\\quad\\Rightarrow\\quad x\\:=\\:\\pm\\sqrt{2}$\nI do that often unfortunately. Math is not my strongest subject, so I sometimes make things more complicated that necessary. Thanks for your response.\n\n7. Originally Posted by Jhevon\nyeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with\nI'm not easily shocked, I may spend more time thinking about it though . It's easier the way Soroban demonstrated it actually, so I'm not shocked. Thanks...", "date": "2016-10-26 14:41:48", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/15734-help-reducing-quadratic-equation-final-terms.html", "openwebmath_score": 0.8996398448944092, "openwebmath_perplexity": 587.4278926539622, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9724147169737826, "lm_q2_score": 0.8757869803008764, "lm_q1q2_score": 0.8516281485786005}}
{"url": "https://math.stackexchange.com/questions/3173523/tiling-a-square-with-rectangles", "text": "# Tiling a square with rectangles\n\nConsider the set of all the rectangles with dimensions $$2^a\\times 2^b\\,a,b\\in \\mathbb{Z}^{\\ge 0}$$. We want to tile an $$n\\times n$$ square by rectangles from this set (you can use a rectangle several times). What is the minimum number of rectangles we need?\n\nIf $$f(n)$$ is the sum of digits of $$n$$ in base $$2$$, I think we need at most $$f(n)^2$$ rectangles. I have an example for this number: write $$n=2^{a_1}+2^{a_2}+...2^{a_{f(n)}}$$ and split each side to segments with length $$2^{a_1},2^{a_2},...,2^{a_{f(n)}}$$ and consider $$f(n)^2$$ rectangles obtained this way.\n\nOn the other hand, you need at least $$f(n)$$ rectangles to tile a raw (or column) so I think you need $$f(n)^2$$ rectangles, but I can't prove it.\n\nAny ideas?\n\n\u2022 By $f(n)$ do you mean the sum of the bits in the binary representation of $n$? Apr 3, 2019 at 20:18\n\u2022 @JohnWaylandBales yes f(n) is the least number such that $n=2^{a_1}+2^{a_2}+...2^{f(n)}$\n\u2013\u00a0ali\nApr 3, 2019 at 20:23\n\u2022 You mean $f(n)$ is the least number such that $n = 2^{a_1} + 2^{a_2} + \\cdots + 2^{a_{f(n)}}$ right? Because $f(n)$ counts the number of terms, but it is not the highest exponent. Apr 4, 2019 at 3:51\n\u2022 @antkam yes i edited the question\n\u2013\u00a0ali\nApr 4, 2019 at 4:36\n\nMAJOR UPDATE:\n\nWhat I am about to show is not a proof for the minimum number of rectangles. However, in some cases I found the number of rectangles can be less than $$f(n)^2$$. The smallest $$N\u00d7N$$ grid that I have found that can have less than $$f(n)^2$$ rectangles is $$15\u00d715$$, which is displayed below: $$\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \\hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\\\ \\hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\\\ \\hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\\\ \\hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\\\ \\hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\\\ \\hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\\\ \\hline 1&1&1&1&2&2&3&6&6&6&6&6&6&6&6\\\\ \\hline 1&1&1&1&2&2&3&7&8&9&9&10&10&10&10\\\\ \\hline 11&11&11&11&11&11&11&11&8&9&9&10&10&10&10\\\\ \\hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\\\ \\hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\\\ \\hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\\\ \\hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\\\ \\hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\\\ \\hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\\\ \\hline \\end{array}$$\n\nThe method used in the above $$15\u00d715$$ square can be generalized not just to other squares but to rectangles as well. In order to make full use of this method, I will expand the op's method to rectangles. Let the length of a rectangle be equal to $$m$$ units and the width be $$n$$ units. Then the number of base-2 rectangles used to cover a $$m \u00d7 n$$ rectangle by the op's method is $$f(m)f(n)$$. My method isn't fundamentally different from the op's method. It splits the $$m\u00d7n$$ rectangle into five sub-rectangles, then the op's method is applied to each of the five rectangles. In some cases the number of base-2 rectangles that covers the five sub-rectangles is less than the number of base-2 rectangles that cover original $$m$$\u00d7$$n$$ rectangle using the op's method. The five rectangles are arranged so that their are two pairs of rectangles that occupy the corners and one rectangle that is in the middle (not touching the perimeter). Each pair of rectangles are the same size and orientation but in opposite corners. (Top left and bottom right or Top right and bottom left.) The length and width of the five rectangles are constructed from two other unit lengths $$a$$ and $$b$$. $$a$$ is the smallest number such that $$m+a$$ is a power of two. $$b$$ is the smallest number such that $$n+b$$ is a power of two. This means that $$f(m+a)$$ and $$f(n+b)$$ are each one. The length and width of the two rectangles in the first pair are $$f\\left(\\frac{m+a}{2}\\right)$$ and $$f\\left(\\frac{n-b}{2}\\right)$$ respectively. The length and width of the two rectangles in the second pair are $$f\\left(\\frac{m-a}{2}\\right)$$ and $$f\\left(\\frac{n+b}{2}\\right)$$ respectively. The formula for the total number of base-2 rectangle used is $$2f\\left(\\frac{m+a}{2}\\right) f\\left(\\frac{n-b}{2}\\right)+2f\\left(\\frac{m-a}{2}\\right) f\\left(\\frac{n+b}{2}\\right)+f(a)f(b)$$.\n\nNote that if a square with a length of $$n$$ units is of the form $$2^xy$$ where $$x,y\\in\\Bbb{N}|x\\ge 1,y\\ge 1$$ and $$y$$ is odd. Then the number of base-2 rectangles used for both the op's method and my method are the the same as the number of base-2 rectangles used for a square of length $$y$$ because each of the dimensions of the sub-rectangles can be multiplied by $$2^x$$. For example if we want to determine how many base-2 rectangles is rectangles are required to cover a $$30\u00d730$$ square using my method. We just use the $$15\u00d715$$ example near the top of this post and multiply the length and width of each base-2 rectangle by $$2$$. So this means the $$30\u00d730$$ square requires the same number of base-2 rectangles as the $$15\u00d715$$ square. So the problem can be simplified to just rectangles where $$m$$ and $$n$$ are odd.\n\nA simple inequality can be made which would indicate which method uses less base-2 rectangles. Let $$N_l$$ be the number of ones in the number for length of the rectangle in binary and $$N_w$$ be the number of ones in the width in binary $$\\bigl($$or more simply $$N_l=f(m)$$ and $$N_w=f(n)\\bigr)$$. Also Let $$Z_l$$ be the number of zeros in the number for length of the rectangle in binary, $$Z_w$$ be the number of zeros in the width in binary. My method uses less rectangles than the op when $$2f\\left(\\frac{m+a}{2}\\right) f\\left(\\frac{n-b}{2}\\right)+2f\\left(\\frac{m-a}{2}\\right) f\\left(\\frac{n+b}{2}\\right)+f(a)f(b)\\lt f(m)f(n)$$\n\n$$f\\left(\\frac{m+a}{2}\\right)=1$$ $$f\\left(\\frac{n+b}{2}\\right)=1$$ $$f\\left(\\frac{m-a}{2}\\right)=N_l-1$$ $$f\\left(\\frac{n-b}{2}\\right)=N_w-1$$ $$f(m)=N_l$$ $$f(n)=N_w$$ $$f(a)=Z_l+1$$ $$f(b)=Z_w+1$$\n\nWith the above substitutions the inequality can be changed to:\n\n$$2(N_l-1)+2(N_w-1)+(Z_l+1)(Z_w+1)\\lt N_lN_w$$ $$2N_l+2N_w-4+(Z_l+1)(Z_w+1)\\lt N_lN_w$$ $$(Z_l+1)(Z_w+1)\\lt N_lN_w-2N_l-2N_w+4$$ $$(Z_l+1)(Z_w+1)\\lt (N_l-2)(N_w-2)$$\n\nIn the specific case of the square (where the length equals the width) my method uses less base-2 rectangles than the op when the number ones in the binary representation of the length is at least four more than than the number of zeros. To get the maximum utility out of my method the inequality shouldn't only be applied to the entire length and width of the main square it should also be applied to components of the square. For example consider the square $$1927\u00d71927$$. The binary representation of 1927 is 11110000111. There are three more ones than zeros in this number so my method would normally break even with the op, covering the square with 49 base-2 rectangles. There is a way to cover the square using less base-2 rectangles by spliting the square into four rectangles $$1920\u00d71920$$, $$1920\u00d77$$, $$7\u00d71920$$, and $$7\u00d77$$. Splitting this way doesn't change the net result of the op's method. The first three sub rectangles satisfies the inequality. If I use my method on the first three sub rectangles I use 13, 11, and 11 base-2 rectangles respectively. Using the op's method on the last sub rectangle then counting up all of the base-2 rectangles I can cover the $$1927\u00d71927$$ square using 44 base-2 rectangles.(13+11+11+9) So if a combination of sub-strings in the binary value of the length and width satisfies the inequality like it did three times with the sub rectangles then my method will use less base-2 rectangles than the op's method. Finding the minimum number of base-2 rectangles for some squares will inevtably involve searching for the best way to split the square.\n\nFor large enough squares the worst digit combination where my method does no better than the op is a block of three ones and the rest are alternating zeros and ones. For example the square $$\\require{enclose}\\enclose{horizontalstrike}{343\u00d7343}$$, its binary representation is 101010111. This square requires 36 base-2 rectangles and is tied for most number of required base-2 rectangles amoung the nine digit squares. This means that a upper bound can be made for the minimum number of rectangles required. Let $$\\enclose{horizontalstrike}{d_l}$$ be the number of digits in the binary representation of the length of the rectangle. ($$\\enclose{horizontalstrike}{d_l=N_l+Z_l}$$) Let $$\\enclose{horizontalstrike}{d_w}$$ be the number of digits in the binary representation of the width of the rectangle. ($$\\enclose{horizontalstrike}{d_w=N_w+Z_w}$$) Then the upper bound is: $$\\enclose{horizontalstrike}{\\left(\\left\\lceil\\frac{d_l}{2}\\right\\rceil+1\\right)\\left(\\left\\lceil\\frac{d_w}{2}\\right\\rceil+1\\right)}$$\n\nI conjecture that the combination of my method and the op's method is the optimal way of minimizing the number of base-2 rectangles. The only way that someone might use be able to use less rectangles is to find a another way of spliting the square into sub-rectangles such that using the op's method on those sub-rectangles uses less base-2 rectangles than using my method and the op's method on the whole square.\n\nRob Pratt's(RP's) post shows that there is a third method for covering the $$n\u00d7n$$ square with less base-2 rectangles than my method or the op's method for some $$n\u00d7n$$ squares. In order to describe how many rectangles RP's method uses I will continue to use the the term $$b$$ from my method (where $$b$$ is the smallest number such that $$b+n$$ is a power of 2). I will also need a new sets of terms $$c_k$$ and $$s_k$$ where $$k\\in\\Bbb{N}|1\\le k\\le f(b)$$. $$c_1$$ is the value of left most ones digit of b in binary form. $$c_2$$ is the value of the second ones digit from the left of b in binary form. $$c_3$$ is the value of the third ones digit from the left of b in binary form. Etc. $$s_v=\\sum_{j=1}^vc_v$$. For example if $$n=23$$ then $$b=9$$, $$c_1=8$$, $$c_2=1$$, $$s_1=8$$, $$s_2=9$$. RP's method uses $$2f\\left(\\frac{n+b}{2}\\right)f\\left(\\frac{n-b}{2}\\right)+f\\left(\\frac{n+b}{2}\\right)f\\left(\\frac{n-b}{2}+s_k\\right)+f\\left(\\frac{n-b}{2}\\right)f\\left(\\frac{n+b}{2}-s_k\\right)+f(b)f(b-s_k)$$\n\nbase-2 rectangles. Each $$f(\u2022)f(\u2022)$$ product contains the length and width of each of the sub-rectangles that covers the square inside the f function. RP's method has $$k$$ ways of covering the $$n\u00d7n$$ square one for each $$s$$ element. Obviously the particular $$s_k$$ element that uses the least number of base-2 rectangles according to the above formula is the one that is used for the minimum. A sufficient condition for when RP's method uses less base-2 rectangles than both my method and the op's method when the binary representation of $$n$$ has at least three more ones than zeros, the second digit to the left is a zero, and the spliting method that was mentioned for the $$1927\u00d71927$$ square doesn't apply.\n\n\u2022 absolutely brilliant!! IMHO well worth the bounty. :) Apr 11, 2019 at 2:02\n\u2022 What does your method obtain for $n\\in\\{23,30,31\\}$? Mar 5, 2020 at 22:36\n\u2022 @Rob_Pratt 16,13, and 17 base-2 rectangles respectively Mar 5, 2020 at 23:30\n\u2022 @RobPratt I realized that the way I explained it in my edited post it doesn't show how n=30 is 13 base-2 rectangles with my method. I will edit accordingly. Mar 5, 2020 at 23:37\n\nFor fixed $$n$$, you can solve this problem via integer linear programming as follows. Let $$R$$ be the set of rectangles. For $$i\\in\\{1,\\dots,n\\}, j\\in\\{1,\\dots,n\\}$$, let $$R_{i,j}\\subset R$$ be the subset of rectangles that contain cell $$(i,j)$$. Let binary decision variable $$x_r$$ indicate whether rectangle $$r\\in R$$ is used. The problem is to minimize $$\\sum_r x_r$$ subject to: \\begin{align} \\sum_{r \\in R_{i,j}} x_r &= 1 &&\\text{for i\\in\\{1,\\dots,n\\}, j\\in\\{1,\\dots,n\\}} \\\\ x_r &\\in \\{0,1\\} &&\\text{for r \\in R} \\end{align}\n\nHere are several optimal values that differ from $$f(n)^2$$: $$\\begin{matrix} n &15 &23 &30 &31 &46 &47 &55 &59 &60 &61 &62 &63\\\\ \\hline f(n)^2 &16 &16 &16 &25 &16 &25 &25 &25 &16 &25 &25 &36\\\\ \\text{optimal} &13 &15 &13 &17 &15 &19 &20 &20 &13 &20 &17 &21\\\\ \\end{matrix}$$\n\nFor example, here is an optimal solution for $$n=23$$ with $$15$$ rectangles (shown here with hex values $$0$$ through $$E$$ for compactness): $$\\begin{matrix} 0&0&0&0&0&0&0&0&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&1&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\\\ C&E&E&E&E&D&D&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2\\\\ \\end{matrix}$$\n\n\u2022 I should have mentioned this earlier but good job finding this. As of when this comment being posted you are the only one who has helped me with this problem. If no one else posts am answer by the end of the bounty grace period you will receive the bounty. Also I have made a formula for your method in my most recent edit that I just made, you might want to take a look. Mar 8, 2020 at 23:13\n\u2022 Thanks. I added a few more values $< f(n)^2$ just now. Mar 9, 2020 at 23:35\n\nNOTE:This doesn't work, the induction hypothesis is too strong (and false).\n\nLets first consider a more general question, where we tile a rectangle $$R$$ by smaller rectangles, where all vertices are points in an (ambient) integer lattice.\n\nWe have a row of rectangles $$T_i$$ touching the bottom edge of $$R$$, and each of these has a top edge $$e_i$$. For each $$T_i$$ we define the number $$\\lambda(T_i)$$ to be the minimal number of our tiling rectangles that intersect any column starting in $$T_i$$.\n\nOur first claim is that for the total number of rectangles in $$R$$, denoted $$r(R)$$, we have $$\\sum_i \\lambda(T_i) \\leq r(R)$$\n\nLets prove this by induction on the height of the rectangle $$R$$ (drawing a picture may help see whats happening). First, if the height is $$1$$, then we are done trivially.\n\nSo now for the inductive step, let $$R_0$$ have height $$n$$, and consider the edges $$e_i$$ that have minimal height, and define $$a$$ to be this height. This is to say, they border the $$a$$th row, if the first row is the bottom row of $$R_0$$. Say that we have $$k$$ minimal edges $$e_i$$ bordering this row.\n\nWe now consider the new rectangle $$R_0'$$ we obtain by chopping off the first $$a$$ rows of $$R$$. On one hand, this has strictly smaller height, so we have, by induction and our definition of $$k$$: $$\\sum_i \\lambda(T_i') \\leq r(R_0)-k$$\n\nBut each rectangle on the bottom row of $$R_0'$$ is either one of the rectangles of $$R_0$$, chopped, but not removed, or a rectangle of $$R_0$$ lying above one of our minimal edges $$e_i$$.\n\nNow note if our original $$T_i$$ is chopped but not removed, $$\\lambda(T_i)=\\lambda(T_i')$$, and if our original $$T_j$$ is removed (so top edge has minimal height), then $$\\lambda(T_j')=\\lambda(T_j)-1$$, where $$T_j'$$ is any of the rectangles lying directly over $$T_j$$. Thus, adding $$k$$ to both sides of our previous equality, we have:\n\n$$\\sum_i \\lambda(T_i) \\leq \\sum_i \\lambda(T_i')+k\\leq r(R_0)$$\n\nSo we are done by induction.\n\nSo for your case, note that each column must have at least $$f(n)$$ rectangles in it, and note the bottom row has at least $$f(n)$$ rectangles. This follows since $$f(n)$$ is the minimal number of powers of two needed to express $$n$$. Thus, $$f(n)^2\\leq r(R)$$ in your case.\n\n\u2022 I think your original claim is false.I don't know how to send a picture in comment but you can easily draw $3\\times 4$ counter examples(two horizontal dominos and two vertical dominos in first two rows and two $1\\times1$square and a domino on the third row).the problem is on your induction step the rectangles above two removed rectangle may not be distinict.\n\u2013\u00a0ali\nApr 5, 2019 at 9:41\n\u2022 your last statement have counter example too.if each row intersect k rectangle and each column intersect k rectangle doesn't mean we need $k^2$ rectangle.\n\u2013\u00a0ali\nApr 5, 2019 at 9:43\n\u2022 True, I'll leave this up in case someone can make this approach work. Apr 5, 2019 at 10:03", "date": "2022-05-19 11:43:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3173523/tiling-a-square-with-rectangles", "openwebmath_score": 0.7841835021972656, "openwebmath_perplexity": 215.95010760270029, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631627142339, "lm_q2_score": 0.8633916187614822, "lm_q1q2_score": 0.8516176877425377}}
{"url": "https://math.stackexchange.com/questions/1867644/finding-a-tricky-composition-of-two-piecewise-functions/1867652", "text": "# Finding a tricky composition of two piecewise functions\n\nI have a question about finding the formula for a composition of two piecewise functions. The functions are defined as follows:\n\n$$f(x) = \\begin{cases} 2x+1, & \\text{if x \\le 0} \\\\ x^2, & \\text{if x > 0} \\end{cases}$$\n\n$$g(x) = \\begin{cases} -x, & \\text{if x < 2} \\\\ 5, & \\text{if x \\ge 2} \\end{cases}$$\n\nMy main question lies in how to approach finding the formula for the composition $g(f(x))$. I have seen a couple of other examples of questions like this online, but the domains of each piecewise function were the same, so the compositions weren't difficult to determine.\n\nIn this case, I have assumed that, in finding $g(f(x))$, one must consider only the domain of $f(x)$. Thus, I think it would make sense to test for individual cases: for example, I would try to find $g(f(x))$ when $x <= 0$. $g(f(x))$ when $x <= 0$ would thus be $-2x-1$, right? However, I feel like I'm missing something critical, because I'm just assuming that the condition $x < 2$ for $g(x)$ can just be treated as $x <= 0$ in this case. Sorry for my rambling, and many thanks for anyone who can help lead me to the solution.\n\nThe first branch point is associated with $f$, and happens at $x=0$. We note that: $$g\\circ f(x) = \\begin{cases} g(2x+1), & \\text{if x\u2264 0} \\\\ g(x^2), & \\text{if x > 0} \\end{cases}$$\n\nLet's just consider each case separately. First, $g(2x+1)$ when $x\u22640$. We remark that $x\u22640\\implies 2x\u22640\\implies 2x+1\u22641$ so we see that $x\u22640\\implies g(2x+1)=-2x-1$.\n\nNow consider $g(x^2)$ when $x>0$. We see that we hit a branch point at $x=\\sqrt 2$. Specifically, $x<\\sqrt 2\\implies x^2<2$ so $x<\\sqrt 2\\implies g(x^2)=-x^2$. Of course $x\u2265\\sqrt 2 \\implies x^2\u22652$ so $x\u2265\\sqrt 2\\implies g(x^2)= 5$. Combining all this we see that $$g\\circ f(x) = \\begin{cases} -2x-1, & \\text{if x \u2264 0} \\\\ -x^2, & \\text{if 0< x < \\sqrt 2}\\\\ 5 & \\text{if x\u2265\\sqrt 2} \\end{cases}$$\n\nAs a sanity check, we try the special case $x=1$. We note that $g\\circ f(1)=g(f(1))=g(1)=-1$ as desired. I advise picking a few other special values just to check.\n\n\u2022 Thanks for your answer. You say that \"[w]e see that we hit a branch point at $x = \\sqrt 2$,\" but how do we see this? I don't know how one would conclude this from any of the information given. \u2013\u00a0Linear Pedant Jul 22 '16 at 22:19\n\u2022 @LinearPedant The branch point happens when the argument of $g$ reaches $2$. Because the argument is $x^2$, the branch point is at $x=\\sqrt2$. \u2013\u00a0f'' Jul 22 '16 at 23:41\n\u2022 @LinearPedant Did the reply from f'' help? It is exactly right. We know that $g(x)$ has a branch point at $x=2$, thus $g(x^2)$ has a branch point at $x^2=2$, a.k.a. $x=\\sqrt 2$ (since we already know $x>0$ so we can ignore the negative square root). \u2013\u00a0lulu Jul 23 '16 at 1:45\n\u2022 @lulu @f\" Yes, thank you. This reply greatly helped me understand the problem better. \u2013\u00a0Linear Pedant Jul 23 '16 at 1:52\n\n$$g(x) = \\begin{cases} -x, & x < 2 \\\\ 5, & x \\ge 2 \\end{cases}$$\n\nTherefore\n\n$$g(f(x)) = \\begin{cases} -f(x), & f(x) < 2 \\\\ 5, & f(x) \\ge 2 \\end{cases}$$\n\nSo now we need to know when $f(x) < 2$ and when $f(x) \\ge 2$.\n\n$$f(x) = \\begin{cases} 2x + 1, & x \\le 0 \\\\ x^2, & x > 0 \\end{cases}$$\n\nLet's look at one piece of $f$ at a time. On the first piece, $2x + 1 \\ge 2$ means $x \\ge 1/2$. But this is impossible because $x \\le 0$ on the first piece. Also on the first piece, $2x + 1 < 2$ means $x < 1/2$. Well, on the first piece we have $x \\le 0 < 1/2$, therefore $f(x) < 2$ on the entire first piece, i.e., $f(x) < 2$ if $x \\le 0$.\n\nOn the second piece, $x^2 \\ge 2$ means $x \\ge \\sqrt{2}$ or $x \\le -\\sqrt{2}$. On the second piece we always have $x > 0$, therefore we have $x^2 \\ge 2$ when $x \\ge \\sqrt{2}$. Also on the second piece, $x^2 < 2$ means $-\\sqrt{2} < x < \\sqrt{2}$. And since on the second piece we always have $x > 0$, then we must have $0 < x < \\sqrt{2}$ in order to have $x^2 < 2$.\n\nPutting it all together so far, we have the following:\n\n$$f(x) \\ge 2 \\text{ if and only if } x \\ge \\sqrt{2}$$\n\n$$f(x) < 2 \\text{ if and only if } x \\le 0 \\text{ or } 0 < x < \\sqrt{2}$$\n\nNotice that this last one can be simplified but we need to keep them separate. Why is this? We'll see as we continue. Recall:\n\n$$g(f(x)) = \\begin{cases} -f(x), & f(x) < 2 \\\\ 5, & f(x) \\ge 2 \\end{cases}$$\n\nTherefore:\n\n$$g(f(x)) = \\begin{cases} -f(x), & x \\le 0 \\text{ or } 0 < x < \\sqrt{2} \\\\ 5, & x \\ge \\sqrt{2} \\end{cases}$$\n\nSeparating the conditions gives us:\n\n$$g(f(x)) = \\begin{cases} -f(x), & x \\le 0\\\\ -f(x), & 0 < x < \\sqrt{2} \\\\ 5, & x \\ge \\sqrt{2} \\end{cases}$$\n\nAnd we need to do this because $f(x)$ itself is different for $x \\le 0$ and $0 < x < \\sqrt{2}$. Finally, we end up with:\n\n$$h(x) := g(f(x)) = \\begin{cases} -(2x+1), & x \\le 0 \\\\ -x^2, & 0 < x < \\sqrt{2} \\\\ 5, & x \\ge \\sqrt{2} \\end{cases}$$\n\n\u2022 Ack, someone beat me to it. I guess that's what happens when I let myself get distracted. \u2013\u00a0tilper Jul 22 '16 at 15:59\n\nYou're correct about the value of $g(f(x))$ when $x\\leq 0$; since $f(x)$ will be at most $2\\cdot0+1=1$, $g$ is only going to evaluate $f(x)$ according to the definition for $x<2$. Testing for cases here is a good approach, and you've just resolved the $x\\leq0$ case. When $x>0$, consider the values of $f(x)$: when will they be less than $2$, and when will they be greater? This will determine where $g(f(x))$ takes on its values.\n\n$$g(x) = \\begin{cases} -x, & \\text{if x < 2} \\\\ 5, & \\text{if x \\ge 2} \\end{cases}$$ In finding $g(f(x))$ with $g$ defined as above, one should remember that $$g(f(x)) = \\begin{cases} -x, & \\text{if f(x) < 2} \\\\ 5, & \\text{if f(x) \\ge 2} \\end{cases}$$ with $f(x)$, not $x$, being the thing about which one asks when it is $<2$ and when it is $\\ge 2$.\n\n(The calculus text by Salas and Hille had a problem like this that used to get assigned in every academic term by vast numbers of professors who taught the course to thousands of students. And every time an hour would be spent answering the students questions about it, which in every case said something like \"I didn't even know where to start\", and that's an hour that could have been spent on something that would help the students understand calculus. It was an even worse waste of time and effort than what usually goes on in the kind of math courses where the students don't want to learn math.)", "date": "2020-06-05 22:05:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1867644/finding-a-tricky-composition-of-two-piecewise-functions/1867652", "openwebmath_score": 0.8938382267951965, "openwebmath_perplexity": 135.35333761446822, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631643177029, "lm_q2_score": 0.8633916152464017, "lm_q1q2_score": 0.8516176856598134}}
{"url": "https://mathzsolution.com/can-every-proof-by-contradiction-also-be-shown-without-contradiction/", "text": "# Can every proof by contradiction also be shown without contradiction?\n\nAre there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction?\n\nAs an aside, how is proving by contradiction viewed in general by \u2018advanced\u2019 mathematicians. Is it a bit of an \u2018easy way out\u2019 when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn\u2019t fond of proof by contradiction.\n\n## Answer\n\nTo determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $\\bot$ represent an identically false proposition. Then $\\lnot A$, the negation of $A$, is equivalent to $A \\to \\bot$, and we take the latter to be the definition of the former in terms of $\\bot$.\n\nThere are two key logical principles that express different parts of what we call \u201cproof by contradiction\u201d:\n\n1. The principle of explosion: for any statement $A$, we can take \u201c$\\bot$ implies $A$\u201d as an axiom. This is also called ex falso quodlibet.\n\n2. The law of the excluded middle: for any statement $A$, we can take \u201c$A$ or $\\lnot A$\u201d as an axiom.\n\nIn proof theory, there are three well known systems:\n\n\u2022 Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to \u201cdirect proof\u201d, because it does not let us leverage a negation for any purpose.\n\n\u2022 Intuitionistic logic includes minimal logic and the principle of explosion\n\n\u2022 Classical logic includes intuitionistic logic and the law of the excluded middle\n\nIt is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.\n\nThe scheme \u201cIf $A$ implies a contradiction, then $\\lnot A$ must hold\u201d is true even in intuitionistic logic, because $\\lnot A$ is just an abbreviation for $A \\to \\bot$, and so that scheme just says \u201cif $A \\to \\bot$ then $A \\to \\bot$\u201d. But in intuitionistic logic, if we prove $\\lnot A \\to \\bot$, this only shows that $\\lnot \\lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\\lnot \\lnot A$ implies $A$, which means that in classical logic if we can prove $\\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.\n\nAttribution\nSource : Link , Question Author : sonicboom , Answer Author : Carl Mummert", "date": "2022-10-05 01:53:10", "meta": {"domain": "mathzsolution.com", "url": "https://mathzsolution.com/can-every-proof-by-contradiction-also-be-shown-without-contradiction/", "openwebmath_score": 0.8328729271888733, "openwebmath_perplexity": 160.97916514303006, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631633155348, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8516176830609768}}
{"url": "https://math.stackexchange.com/questions/935777/understanding-and-writing-limit-proofs", "text": "# Understanding and writing limit proofs\n\nI got this question :\n\nConsider $a_n,\\:b_{n\\:}$ sequences such that for every n , $0\\le a_n\\le \\:b_{n\\:}$.\n\nLet $\\lim _{n\\to \\infty }\\left(\\frac{b_n}{a_n}\\right)\\:=\\:1$, and $a_n$ is a bounded sequence.\n\nProve that $\\left(a_n-b_n\\right)_{n=1}^{\\infty \\:}\\:\\rightarrow \\:0$.\n\nI tried little bit by myself to understand what is given:\n\nBy definition of limit, we see that : $\\forall\\epsilon, \\exists n_{0} \\forall n> n_{0}\\Rightarrow |\\frac{b_{n}}{a_{n}}- 1|< \\varepsilon$, and also that exist some $M$ that for every $n$, $-M<a_n<M$.\n\nNow i'm looking for that right?: i need to prove that $\\forall\\epsilon, \\exists n_{0} \\forall n> n_{0}\\Rightarrow |a_{n}-b_{n}- 0|< \\varepsilon$\n\nHere is where i struggle: i choose some $\\epsilon$. What i need to find? an $N2$ that for every $n>N2, |(a_{n}-b_{n})- 0|< \\varepsilon$ ?\n\nHow to start?\n\nHint: use may the fact that $|a_n-b_n|=|a_n||\\frac{b_n}{a_n}-1|$ in your proof .\n\n\u2022 Yes, but what i actually need to write to do it formal? \u2013\u00a0user2637293 Sep 18 '14 at 1:09\n\u2022 I would recommend you to fill in the gap yourself. but if not interested you can look at the proof provided by @Ishfaaq down here. \u2013\u00a0BigM Sep 18 '14 at 1:12\n\nHints\n\nFirst of all notice that $\\left|{\\dfrac{b_n}{a_n} - 1}\\right| = \\dfrac{|(b_n - a_n) - 0|}{|a_n|}$\n\nSecondly note that $|a_n| \\le M \\;\\; \\forall n \\in \\Bbb N$\n\nNow let $\\epsilon \\gt 0$ be arbitrary and think what you can do with a quantity like $$\\dfrac{\\epsilon}{M} \\gt 0$$\n\nUPDATE: Just extract what the definition of $\\lim \\dfrac{a_n}{b_n}$ means. It says for every $\\epsilon \\gt 0$ there is $n_0$ such that the difference between $\\dfrac{a_n}{b_n}$ and the limit ($= 1$ ) is less than $\\epsilon$. So all you need to write in your paper would be to let $\\epsilon$ be arbitrary. Then since $\\lim \\dfrac{a_n}{b_n}$ is given, choose $\\dfrac{\\epsilon}{M}$ as your arbitrary positive quantity. Then as per the definition of the limit there exists $n_0 \\in \\Bbb N$ such that $$n \\ge n_0 \\implies \\left|{\\dfrac{b_n}{a_n} - 1}\\right| = \\dfrac{|(b_n - a_n) - 0|}{|a_n|} \\lt \\dfrac{\\epsilon}{M} \\implies | (b_n - a_n ) - 0 | \\lt |a_n | \\cdot \\dfrac{\\epsilon}{M} \\le \\epsilon$$\n\nAnd you're done!\n\n\u2022 This is the problem. i don't know what to write after i choose arbitrary epsilon. i need to someone write the answer like in a test. it's not the question of HOW to do this, its about WHAT to write to do it formally, tnx! \u2013\u00a0user2637293 Sep 18 '14 at 1:15\n\u2022 @user2637293: I edited the answer. Comment if you need more help. \u2013\u00a0Ishfaaq Sep 18 '14 at 6:58", "date": "2019-08-26 06:28:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/935777/understanding-and-writing-limit-proofs", "openwebmath_score": 0.9465218782424927, "openwebmath_perplexity": 221.2684155265554, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631631151012, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8516176759536318}}
{"url": "https://math.stackexchange.com/questions/784808/positive-semi-definite-matrix-and-its-eigenvalues-please-help-checking-improvi", "text": "Let $A$ and $B$ are two $n \\times n$ Hermitian matrices . Suppose $A-B$ is positive semidefinite.\n\n(a) Show that $\\lambda_k(A) \\geq \\lambda_k(B)$ for $k=1,2,\\dots ,n,$ where $\\lambda_i(A)$ and $\\lambda_i(B)$ are eigenvalues of matrices $A$ and $B$ with $\\lambda_i (A)\\geq \\lambda_{i+1}(A)$ and $\\lambda_i (B)\\geq \\lambda_{i+1}(B)$.\n\n(b) Find a counter example to prove the converse of (a) is not true. i.e.,\n\nFind two Hermitian matrices $A$ and $B$ such that $\\lambda_k(A) \\geq \\lambda_k(B)$ but $A-B$ is not positive semidefinite.\n\nMy Trial:\n\n(a) Since $A-B$ is positve semi-definite, $x^*(A-B)x \\geq 0 \\Rightarrow x^{*}Ax \\geq x^{*}Bx.$\n\nAlso, as $A$ and $B$ are two Hermitian matrices, by spectral decomposition, we have $x^{*}Ax = x^{*}UD_{A}U^{*}x=x^{*}D_{A} x$ and $x^{*}Bx = x^{*}UD_{B}U^{*}x=x^{*}D_{B} x$.\n\nIt implies $D_{A} \\geq D_{B}$ and so for each of theire eigenvalues.\n\n(b) Let $A=\\begin{pmatrix} 1 &0 \\\\ 0 &1 \\end{pmatrix}$ and $B=\\begin{pmatrix} 2 &0 \\\\ 0 &0 \\end{pmatrix}$.\n\nPlease help to check if I make mistake in my presentation because I am very new to this topic so I do not know if am correct or not. Thank you very much in advance.\n\n\u2022 How $U^*x$ becomes $x$? And the $U$ matrix is not the same for $A$ and for $B$. \u2013\u00a0enzotib May 7 '14 at 10:03\n\u2022 Is it given that $A,B$ have the same eigenvectors? \u2013\u00a0Samrat Mukhopadhyay May 7 '14 at 10:21\n\u2022 @enzotib Since $U$ is unitary, norm of $U*x$ = norm of $x$. Can I say this? \u2013\u00a0nam May 7 '14 at 10:22\n\u2022 @Samrat The question does not give that $A$ and $B$ have the same eigenvectors. \u2013\u00a0nam May 7 '14 at 10:23\n\u2022 @Pavel So, am I correct on the whole? \u2013\u00a0nam May 7 '14 at 10:24\n\nTo prove (a), you need to use the variational characterisation of eigenvalues (the min-max principle, Courant-Fischer theorem). For a Hermitian $A$ and the specified ordering of eigenvalues, $$\\tag{*} \\lambda_k(A)=\\max_{S:\\dim S=k}\\min_{x\\in S\\setminus\\{0\\}}\\frac{x^*Ax}{x^*x}=\\min_{S:\\dim S=n-k+1}\\max_{x\\in S\\setminus\\{0\\}}\\frac{x^*Ax}{x^*x}.$$ Let $S_*$ be the subspace of the dimension $n-k+1$ for which the minimum on the right-hand side of ($*$) is attained, that is, $$\\lambda_k(A)=\\max_{x\\in S_*\\setminus\\{0\\}}\\frac{x^*Ax}{x^*x}.$$ Then (using $x^*Ax\\geq x^*Bx$ for all $x$) $$\\lambda_k(A)=\\max_{x\\in S_*\\setminus\\{0\\}}\\frac{x^*Ax}{x^*x} \\geq\\max_{x\\in S_*\\setminus\\{0\\}}\\frac{x^*Bx}{x^*x} \\geq\\min_{S:\\dim S=n-k+1}\\max_{x\\in S\\setminus\\{0\\}}\\frac{x^*Bx}{x^*x}=\\lambda_k(B).$$\n\nI was also thinking whether there would be an alternative proof which does not involve the variational characterisation and can thus be considered as more elementary. This is motivated by the proof here.\n\nBy exactly the same technique as in that proof, one can show the following:\n\nLet $A,B\\in\\mathbb{C}^{n\\times n}$ be Hermitian and let $i_{+,0}(A)$ and $i_+(A)$ denote the number of non-negative and positive eigenvalues, respectively, of $A$. Then $$i_{+,0}(A+B)\\leq i_+(A)+i_{+,0}(B).$$\n\nNow replace in the statement above $A$ and $B$, respectively, by $A-\\lambda_k(A)I$ and $\\lambda_k(A)I-B$. We get $$\\tag{1} i_{+,0}(A-B)\\leq i_+(A-\\lambda_k(A)I)+i_{+,0}(\\lambda_k(A)I-B).$$ Since our $A$ and $B$ are such that $A-B$ is positive semi-definite and hence all its eigenvalues are non-negative, we have $i_{+,0}(A-B)=n$. Note that adding a multiple of the identity to $A$ just shifts the eigenvalues, so the $k$th eigenvalue of $A-\\lambda_k(A)I$ is zero and thus there is at most $k-1$ positive eigenvalues of $A-\\lambda_k(A)I$, that is, $i_+(A-\\lambda_k(A)I)\\leq k-1$. Putting this stuff to (1) gives $$n\\leq k-1+i_{+,0}(\\lambda_k(A)I-B)\\quad \\Rightarrow \\quad n-k+1\\leq i_{+,0}(\\lambda_k(A)I-B).$$ Since there is at least $n-k+1$ non-negative eigenvalues of $\\lambda_k(A)I-B$, there is at most $k-1$ negative eigenvalues. Equivalently, the matrix $B-\\lambda_k(A)I$ has a most $k-1$ positive eigenvalues and hence the $k$th eigenvalue of $B-\\lambda_k(A)I$ is non-positive. But this eigenvalue is nothing but $\\lambda_k(B)-\\lambda_k(A)$ and hence $$\\lambda_k(B)-\\lambda_k(A)\\leq 0\\quad\\Rightarrow\\quad\\lambda_k(A)\\geq\\lambda(B).$$\n\n\u2022 Very thanks to your hint. I just had some readings on Rayleigh quotients but I get lost and do not know how to apply your hint on this question (it seems to pick up second large eigenvalue from some subspaces of smaller dimension). Would you mind giving me lines of answers. Thank you so much. \u2013\u00a0nam May 7 '14 at 10:45\n\u2022 @nam It should be quite complete now :) \u2013\u00a0Algebraic Pavel May 7 '14 at 10:49\n\u2022 Thank you for your illustration for the theorem. I know what I missed and I am looking for other examples to check. Thank you so much. \u2013\u00a0nam May 7 '14 at 11:02\n\u2022 You are welcome. \u2013\u00a0Algebraic Pavel May 7 '14 at 11:02\n\u2022 Just for fun, I've tried to make a proof which does not use the min-max principle. However, it's a bit longer. \u2013\u00a0Algebraic Pavel May 7 '14 at 13:03", "date": "2020-03-30 19:43:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/784808/positive-semi-definite-matrix-and-its-eigenvalues-please-help-checking-improvi", "openwebmath_score": 0.9117186069488525, "openwebmath_perplexity": 156.20587774760978, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631627142339, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8516176756075263}}
{"url": "https://math.stackexchange.com/questions/1345331/where-do-these-mobius-transformations-map-the-coordinate-half-planes", "text": "Where do these Mobius transformations map the coordinate half-planes?\n\nThey are\n\n$$\\frac{z-1}{z+1}, \\frac{z+1}{z-1},\\frac{z-i}{z+i},\\frac{z+i}{z-i}.$$\n\nAll four look virtually identical, so I would like to know how to best distinguish between them.\n\nFor example, the first mapping, I notice that $1$ maps to $0$, $-1$ maps to $\\infty$.\n\nBut since $1$ and $-1$ are symmetric points w.r.t. the imaginary axis, the Mobius transformations always maps symmetric points to symmetric points. So, the images, $0$ and $\\infty$, being symmetric points, must mean that they are symmetric with respect to a circle. So the real axis gets mapped to some circle centered at the origin.\n\nHow do I tell how big this circle is? And since the real axis separates the upper and lower half planes, either the upper or lower plane gets mapped inside the circle, but how can I tell? I tried a sample point: I see that $i$ gets mapped to $i$, but is this telling me that the mapping is \"fixing\" the UHP? I doubt it, because it's mapping stuff to some open disk.\n\nEdit: Also, with the first mapping, there doesn't seem to be a nice symmetry argument to exploit for the imaginary axis.\n\nThanks.\n\n\u2022 The first mapping does not map the real line to a circle centered at the origin: The image of $1$ under that map is the origin itself. Jul 1 '15 at 2:47\n\u2022 It does, however, map the imaginary axis to the unit circle, as each imaginary $z$ is equidistant from $1$ and $-1$, and hence $\\left\\vert\\frac{z - 1}{z + 1}\\right\\vert = 1$. Jul 1 '15 at 2:51\n\u2022 Hmm...can I think of it this way, @Travis? I realize my mistake, I think. So, since 1 and -1 are symmetric w.r.t. the imaginary axis, and their images must also be symmetric w.r.t. the image of the imaginary axis (not the real axis), which I don't know yet. But 1 and -1 map to 0 and infinity, which are symmetric w.r.t. the image, which implies that the imaginary axis must get mapped to a circle. Is this valid? If so, how do I know how big the circle is? Your equidistant argument I have never heard of...thanks, Jul 1 '15 at 3:00\n\u2022 I've expanded my comments into a proper answer below, and I think it answers all these questions, but feel free to comment there if not. Jul 1 '15 at 3:16\n\nThe first mapping, $$f : z \\mapsto \\frac{z - 1}{z + 1},$$ does not map the real axis to a circle centered at the origin: Indeed, we have $f(1) = 0$.\n\nOn the other hand, any imaginary number $iy$ is, by symmetry, equidistant from $-1$ and $1$, and so $$|f(iy)| = \\left\\vert\\frac{iy - 1}{iy + 1}\\right\\vert = \\frac{|iy - 1|}{|iy + 1|} =1,$$ that is, $f$ maps the imaginary axis to the unit circle (bijectively, if we include $f(\\infty) = 1$). Using the above fact that $f(1) = 0$ (and continuity) gives that $f$ maps that right half-plane to the inside of the unit circle, and so the left half-plane to the outside.\n\nAs for the real axis, note that if $x$ is real, then so is $f(x) = \\frac{x - 1}{x + 1}$ (unless $x = -1$, in which case we have $f(-1) = \\infty$), so $f$ maps the real axis (including $\\infty$) to itself.\n\nThe other three maps can be analyzed similarly, or we can use the fact that they are precisely the compositions of $f$ with the rotation maps $z \\mapsto i^k z$, $k = 1, 2, 3$.\n\n\u2022 Hi @Travis, if this mapping fixes the real line, must it also fix the upper and lower half planes, too? I'm thinking of the connectedness argument - a continuous function maps connected sets to connected sets. Or, does the fact that f is not continuous at -1 make this connectedness argument invalid? Also, your getting the norm = 1, because of iy being equidistant from -1 and 1, but isn't iy really equidistant from 1+iy and -1+iy? I might not be understanding the meaning of equidistant, but I feel like we should just be extending iy out in a straight line, along the real axis.... Jul 1 '15 at 3:26\n\u2022 actually, I think got the equidistant argument, if I just use the modulus definition. and I'm pretty sure the connectedness argument is not valid. but feel free to comment, if you have something more to say. Thanks so much for your help, @Travis. Jul 1 '15 at 3:35\n\u2022 Like you say, it fixes the upper and lower half-planes (as sets, not pointwise). Also, we can think of Mobius transformations as continuous maps from the Riemann sphere, $\\Bbb C \\cup \\{\\infty\\}$ to itself, so there's no issue with continuity here. Jul 1 '15 at 3:37\n\u2022 Sure, for any region $A$ we can ask what the region $f(A)$ is, and it happens that our particular map $f$ maps deveral familiar regions to other familiar regions. Note that this lets us conclude, e.g., that the first quadrant is mapped the the upper half of the unit disk. Jul 1 '15 at 3:55\n\u2022 You're welcome, I'm glad you found it useful, and you too. Jul 1 '15 at 4:57\n\nI'll do the first one for you, and you can do the rest. Consider the unit circle $z=e^{i\\theta}$. Then $$\\frac{z-1}{z+1}=\\frac{e^{i\\theta}-1}{e^{i\\theta}+1}=\\frac{1-e^{-i\\theta}+e^{i\\theta}-1}{1+e^{i\\theta}+e^{-i\\theta}+1}=\\frac{2i\\sin(\\theta)}{2+2\\cos(\\theta)}=i\\frac{\\sin\\theta}{1+\\cos\\theta}=i\\tan\\frac{\\theta}{2}.$$ Thus, the unit circle gets mapped to the imaginary axis. The inside of the unit circle gets mapped to one side of the imaginary axis, while the outside gets mapped to the other. To tell which side gets mapped where, check $z=0$. This maps to $-1$. Thus, the inside of the unit circle gets mapped to the left side of the imaginary axis, while the outside of the nit circle gets mapped to the right. This method should work for all of your functions.\n\n\u2022 Hi @AlexS, nice argument - and based on Travis's comment above, then this mapping also maps the imaginary axis back to the unit circle. So, it is its own inverse, sort of? And, how do we know the imaginary axis gets mapped to a unit circle? Thanks, Jul 1 '15 at 3:09\n\u2022 It is not quite its own inverse: $$f(f(z))=\\frac{\\frac{z-1}{z+1}-1}{\\frac{z-1}{z+1}+1}=\\frac{-2}{2z}=-1/z.$$ But, if $z$ is on the unit circle, so is $-1/z$. To show directly that $f$ takes the imaginary axis to the unit circle, multiply $f(iy)$ by its complex conjugate, and check that the result is 1. Jul 1 '15 at 3:17\n\u2022 ah, nice suggestion about multiplication by complex conjugates to check the norm -- thanks so much @AlexS. Jul 1 '15 at 4:17\n\nEvery Mobius map can be broken down into a composition of scaling maps, translations and inversions. Just do a standard long division of polynomials, i.e., fo $\\frac {az+b}{cz+d}$ , do along division $(az+b)| (cz+d)$ and you will get this decomposition. Then you can apply these operations of inversion, scaling and translation, rotation in the right order to get your map. Maybe you could study more between games too!!\n\n\u2022 ok, got it. Thanks @Gary. I will try to, but with Kyrie and Kevin injured, and me playing 48minutes per game, it's been hard to keep up with complex analysis :-( Jul 1 '15 at 4:21", "date": "2022-01-18 20:36:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1345331/where-do-these-mobius-transformations-map-the-coordinate-half-planes", "openwebmath_score": 0.9141293168067932, "openwebmath_perplexity": 220.90075111509103, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9863631619124993, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8516176749153154}}
{"url": "https://www.jiskha.com/questions/541340/Evaluate-integral-of-e-x-1-2-x-1-2-Ive-looked-at-the-answer-but-I-dont-understand", "text": "# calculus\n\nEvaluate integral of e^x^(1/2) / x^(1/2)\nI've looked at the answer but I don't understand what people do in their steps.\n\nWhen I substitute x^(1/2) for u, I get:\n\n2du = 1/x^(1/2) dx\n\nBut what do you do with the 1/x^(1/2) dx? It just disappears in the solutions I've seen people give.\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n3. \ud83d\udc41 104\n1. yes. the substitution is correct:\nlet u = x^(1/2)\nthus du = 1/[2(x^(1/2))] dx, or\ndx = 2(x^(1/2)) du, or\ndx = 2u du\nsubstituting these to original integral,\nintegral of [e^x^(1/2) / x^(1/2)] dx\nintegral of [(e^u) / u] * (2u) du\nthe u's will cancel out:\nintegral of [2*e^u] du\nwe can readily integrate this to\n2*e^u + C\nsubstituting back the value of u,\n2*e^(x^(1/2)) + C\n\nhope this helps~ :)\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\nposted by Jai\n2. This helped alot :)\n\n1. \ud83d\udc4d 0\n2. \ud83d\udc4e 0\n\n## Similar Questions\n\n1. ### calculus\n\na) Let f(z) = z^2 and \u03b3(t) = 1 + it^3, t \u2208 [0,1]. i) Write out the contour integral \u222b\u03b3 f(z)dz as an integral with respect to t. You do not need to evaluate this integral. ii) Evaluate the integral \u222b0,1+i z^2dz iii) What is\n\nasked by jack on March 5, 2016\n2. ### Calculus\n\n1. Express the given integral as the limit of a Riemann sum but do not evaluate: integral[0 to 3]((x^3 - 6x)dx) 2.Use the Fundamental Theorem to evaluate integral[0 to 3]((x^3 - 6x)dx).(Your answer must include the\n\nasked by Ernie on May 7, 2018\n3. ### Math\n\nLet A denote the portion of the curve y = sqrt(x) that is between the lines x = 1 and x = 4. 1) Set up, don't evaluate, 2 integrals, one in the variable x and one in the variable y, for the length of A. My Work: for x:\n\nasked by Daisy on January 25, 2017\n4. ### Math\n\n1. Evaluate the indefinite integral ([6x^2 + 12x^(3/2) +4x+9]/sqrt x)dx. Answer = + C 2. Evaluate the indefinite integral (12sin x+4tan x)dx. Answer = + C 3. Evaluate the indefinite integral. (x^7)e^(x^8)dx. Answer = + C Thank you\n\nasked by Matt on September 19, 2011\n5. ### calculus\n\nSuppose R is the rectangle 1\n\nasked by Bryant on March 18, 2012\n6. ### Math. Need help\n\n1.if g=5 and k=1, evaluate 9+gk a. 13 b. 14 c. 15 d. 16 2.Evaluate a-b^2/b if a = 10 and b = 2. a. 1 b. 3 c. 8 d. 32 3.Evaluate 3xy^2 - y if x = 2 and y = 5. a. 26 b. 30 c. 120 d. 145 4.Evaluate 2d + d^2/3 if d = 6. a. 4 b. 12 c.\n\nasked by A. on August 28, 2016\n7. ### statistics\n\nI don't understand how to answer this: relate the three measures of central tendency to a normal distribution- I looked it up online and it said that the mean, median, and mode are all the same but i don't understand how that can\n\nasked by katie on February 12, 2009\n8. ### Calculus\n\nEvaluate integral, don't use calculator. I'll use the S as the sign for integral S 4 sin (theta)d(theta) from pi/2 to pi I know the answer is 4 because I enter it in on my calculator, but don't know the steps to get to the answer.\n\nasked by Janet on May 9, 2011\n9. ### calculus\n\n8). Part 1 of 2: In the solid the base is a circle x^2+y^2=16 and the cross-section perpendicular to the y-axis is a square. Set up a definite integral expressing the volume of the solid. Answer choices: integral from -4 to 4 of\n\nasked by Sally on April 14, 2013\n10. ### Calculus\n\nHello. I would appreciate it if someone could check my answers. I'm sorry it is so long. 1.) Let R denote the region between the curves y=x^-1 and y=x^-2 over the interval 1\n\nasked by Shay on September 3, 2005\n\nMore Similar Questions", "date": "2019-05-27 04:16:40", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/541340/Evaluate-integral-of-e-x-1-2-x-1-2-Ive-looked-at-the-answer-but-I-dont-understand", "openwebmath_score": 0.8134042024612427, "openwebmath_perplexity": 2685.511408763475, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9441768620069626, "lm_q2_score": 0.90192067455231, "lm_q1q2_score": 0.851572632278003}}
{"url": "https://math.stackexchange.com/questions/2031842/determine-remainder-when-dividing-polynomials-if-we-know-other-remainders", "text": "# Determine remainder when dividing polynomials if we know other remainders\n\nIf 3 is the remainder when dividing $P(x)$ with $(x-3)$, and $5$ is the remainder when dividing $P(x)$ with $(x-4)$, what is the remainder when dividing $P(x)$ with $(x-3)(x-4)$?\n\nI'm completely puzzled by this, I'm not sure where to start...\n\nAny hint would be much appreciated.\n\nHint: $$P(x)=Q(x)(x-3)(x-4)+Ax+B$$ $$P(3)=3$$ $$P(4)=5$$\n\nCan you solve for $A$ and $B$?\n\nLet $$P(x)=(x-4)(x-3)Q(x)+ax+b$$\n\nNow you have $$3a+b=3$$ and $$4a+b=5$$ So solve for $a$ and $b$.\n\n\u2022 Thank you!! I only accepted the other answer because he posted first. \u2013\u00a0kok_nikol Nov 26 '16 at 18:27\n\u2022 @kok_nikol You haven't accepted anything: you just upvoted one answer. You can upvote all the answers you find informative, good and helpful . Then you will accept only one: the best for you, which is not necessarily the first one, of course. \u2013\u00a0DonAntonio Nov 26 '16 at 18:34\n\u2022 @DonAntonio Sorry, you can't accept an answer right away, you have to wait 10 minutes. And I can't upvote yet because I don't have enough reputation. \u2013\u00a0kok_nikol Nov 27 '16 at 0:26\n\u2022 @kok_nikol Perhaps, yet you did not accepted anything when I first commented. I thought posters can upvote answers to their questions. \u2013\u00a0DonAntonio Nov 27 '16 at 0:30\n\u2022 @DonAntonio Yes, when you commented I hadn't yet accepted anything. As for the voting Votes cast by those with less than 15 reputation are recorded, but do not change the publicly displayed post score. \u2013\u00a0kok_nikol Nov 27 '16 at 0:37\n\nIf you carry out the division you should be able to show that $$P(x)=Q(x)(x-3)(x-4)+R(x)$$\n\nTwo questions: What is the maximum degree of the remainder $R(x)$? Can you see how to use the remainder theorem for the cases $x=3,4$ that you know already so that $Q(x)$ becomes irrelevant?\n\nEuclidean division by $(x-3)(x-4)$ can be written as $$P(x)=(x-3)(x-4)Q(x)+R(x),\\quad\\deg R(x)\\le1.$$ Instead of writing $R(x)$ with the standard basis $\\{1,x\\}$, use the basis $\\{x-3,x-4\\}$. Thus $$P(x)=(x-3)(x-4)Q(x)+A(x-3)+B(x-4)$$ Setting $x=3$, you get right away $B=-P(3)=-3$. Similarly $A=P(4)=5$, and finally $$R(x)=5(x-3)-3(x-4)=2x-3.$$\n\nThis formula can be generalised: the remainder when dividing a polynomial $P(x)$ by $(x-a)(x-b),\\enspace a\\ne b$, is $$R(x)=\\frac1{b-a}\\Bigl[P(b)(x--a)-P(a)(x-b)\\Bigr]=\\frac1{b-a}\\begin{vmatrix}x-a&x-b\\\\[1ex]P(a)&P(b)\\end{vmatrix}.$$\n\nNotice $\\ p\\, = 3+(x\\!-\\!3)q\\,\\$ by $\\,\\ p(3) = 3$\n\n$5 = p(4) = 3+q(4)\\,\\Rightarrow\\, {\\color{#c00}2}\\, =\\, q(\\color{#0a0}4) \\,$\n\nHence $\\, p = 3 + (x\\!-\\!3)(\\underbrace{\\color{#c00}2+(x\\!-\\!\\color{#0a0}4)r}_{\\Large q}\\!)\\, =\\, 2x\\!-\\!3\\, +\\, (x\\!-\\!3)(x\\!-\\!4)r$", "date": "2019-06-16 23:36:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2031842/determine-remainder-when-dividing-polynomials-if-we-know-other-remainders", "openwebmath_score": 0.8170533180236816, "openwebmath_perplexity": 377.7636085612393, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874123575265, "lm_q2_score": 0.8596637577007393, "lm_q1q2_score": 0.851572097238323}}
{"url": "https://math.stackexchange.com/questions/3314464/mathbbrn-second-countable", "text": "# $\\mathbb{R^n}$ Second Countable\n\nLemma: Let U be an open subset of $$\\mathbb{R}^n$$ and $$x \\in U$$ then there exists a ball with rational center $$(\\mathbb{Q}^n$$ and rational radius containing x, that is contained in U.\n\nWhat I want to show: Every open subset of $$\\mathbb{R}^n$$ is the union of some some collection of elements in the following set:\n\n$$\\{$$ Open balls with rational center and rational radius $$\\}$$\n\nLet U be an arbitrary open set in $$\\mathbb{R}^n$$. By the above lemma, for each $$x \\in U$$ you can find a rational center $$x'$$ and rational radius $$r'>0$$ so that $$x \\in B(x',r')$$ $$\\subset U$$. Hence the following collection:\n\n$$B'=$$ $$\\{$$ B(a,r) : a $$\\in \\mathbb{Q}^n$$ , $$r>0$$ , B(a,r) $$\\subset U$$ $$\\}$$ gives us\n\n$$U=$$ $$\\bigcup_{B \\in B'}$$ $$B$$\n\nis the proof correct?\n\nAs this is part of the proof of showing that $$\\mathbb{R}^n$$ is second countable, it follows that since every metric space is Hausdorff, $$\\mathbb{R}^n$$ is Hausdorff, and trivially, it is locally euclidean of dimension n (take the identity map, which is a homeomorphism) and so $$\\mathbb{R}^n$$ is a n- dimensional topological manifold, am I correct?\n\n\u2022 I assume that you require manifolds to be second countable? Then yes and yes. \u2013\u00a0freakish Aug 5 at 18:39\n\u2022 @freakish Yup, a topological space M is a n - dimensional manifold if (1) It is second countable (2) It is Hausdorff (3) Every point in M has a neighborhood homeomorphic to an open subset of $\\mathbb{R}^n$ \u2013\u00a0topologicalmagician Aug 5 at 18:40\n\u2022 @topologicalmagician You prove correctly that $\\mathbb{R}^n$ is second countable. Also, since $\\mathbb{R}^n$ is Hausdorff and locally euclidean for the reasons you said above, it is a topological manifold too. \u2013\u00a0bing-nagata-smirnov Aug 5 at 19:03\n\nYes, your proofs that $$\\mathbb R^n$$ is cecond countable and that $$\\mathbb R^n$$ is an $$n$$- dimensional topological manifold are correct.", "date": "2019-10-14 01:44:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3314464/mathbbrn-second-countable", "openwebmath_score": 0.8264192342758179, "openwebmath_perplexity": 175.62391872423186, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874101345136, "lm_q2_score": 0.8596637505099167, "lm_q1q2_score": 0.851572088204141}}
{"url": "https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for-2/", "text": "# Which payoff do you want to go for?\n\nPayoff 1.\n\nToss 5 coins. You get $1 for each consecutive pair HT that you get. Payoff 2. Toss 5 coins. You get$1 for each consecutive pair HH that you get.\n\nThe consecutive pairs are allowed to overlap.\n\nFor example, if you tossed HTHHH, under payoff 1 you will get $1, but under payoff 2 you will get$2.\n\nNote by Calvin Lin\n4\u00a0years, 7\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. 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Consider this: We will count the number of ways HT can occur. We can have: HT_ _ _ : 8 ways _ HT _ _ _ : 8 ways _ _ HT _ : 8 ways _ _ _ HT : 8 ways Total number of ways: 32 Similarly for HH, the total number of ways in which it can occur is 32. What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is$32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same.\n\nMathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives$3 in p2 and $1 in p1. But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives$0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives$0 payoff.\n\nSo If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.\n\nSince this is in quantitative finance, should I talk about risk and stuff? I do not know.\n\n- 4\u00a0years, 7\u00a0months ago\n\nI think that there may be some double-counting here. For example, you've counted HTHTT and HTHTH twice each. There are only $2^{5} = 32$ possible sequences in total, and there are some, such as TTHHH, that have no HT payouts, so there cannot be $32$ HT outcomes. I'm getting an average payout of $\\dfrac{28}{32}$ dollars in the first game type and $\\dfrac{31}{32}$ dollars in the second, but I am too tired to double-check. I'll do that tomorrow. :)\n\nEdit: Oh. now I see what you've done; very clever. I must have missed out on my count somewhere. I'll still wait to confirm in the morning. And yes, there does seem to be an advantage to choosing p1, (even though the expected winnings are the same), in the sense that you are more likely to at least win some money.\n\n- 4\u00a0years, 7\u00a0months ago\n\nIs \"more likely to at least win some money\" the main consideration that you will use if the expected payoff is equal?\n\nFor example,what would you choose between:\n\nPayoff 1: 50% of $0, 50% of$100\nPayoff 2: 90% of $1.01, 1% of$4900\n\nStaff - 4\u00a0years, 7\u00a0months ago\n\nIf Payoff 2 was 99% of $40.40, 1% of$1000, (to give it the same expected payoff as Payoff 1), I would go with Payoff 2 since I would be assured of money and would at least have a slim chance of a large amount of money.\n\nIf Payoff 2 was 99% of $0, 1% of$5000, then it gets interesting, (at least for me). Even a such a slight chance of winning $5000 would be hard to pass up, so I would probably go with Payoff 2, unless I absolutely needed that$100 right away.\n\nIf Payoff 2 was 99.9% of 0$, 0.1% of$50000 .... hmmmm .... That's a lot of dough, so I'd go with Payoff 2. However, if the options were these last two I've listed, then I might choose the $5000 option. If Payoff 2 involved a 50% chance of losing$100 and a 50% chance of winning $200, then I would go with Payoff 1, since I'm not comfortable with there being such a good chance of losing a fair bit of money. There is a lot of psychology going on in these choices, and yet the expected winnings is always$50, (which ironically is an amount that would never actually be won in any of these scenarios).\n\n- 4\u00a0years, 7\u00a0months ago\n\n(Ooops, apparently I can't do maths. Fixed.)\n\nSo, there are multiple ideas here.\n- Is the probability of a positive payoff so important that just because it moves from $0 it would greatly influence your preferences? - At what point does the potential promise of a huge payoff overweigh the certainty of a small payoff? Staff - 4 years, 7 months ago Log in to reply In general I guess it all depends on the specific values, and further, on each person's financial status and willingness to take on risk. For me, the potential, (1% or better), of a large payout would outweigh any guaranteed amount less than$100, so in this case there is no difference to me between $0 and$100. But if the guaranteed amount were, say, $5000, with a 0.01% all-or-nothing chance of winning$1,000,000, I'd probably take the $5000 and run, (although I would be tempted to take the risk, at least for a moment). If the all-or-noting percentage were 1%, though, then I would find it hard not to take the chance;$5000 is a lot of money, but $1,000,000 is a life-changing amount of money, and a 1% chance is realistic in my eyes given the potential reward. The question I would ask myself is: which will I regret more - not taking the guaranteed money or not taking the risk? - 4 years, 7 months ago Log in to reply Yes, the expected payoff for both scenarios is$1. The easiest way to see this is to look at Indicator Variables, which is essentially what Raghav did (though not phrased in that language).\n\nWell, everyone has their own \"payoff preference\". For example, you stated that you would prefer to \"be more likely to leave with at least a dollar in your pocket\". As such, this tells me that you are very risk adverse, and that you would prefer the certainty of a positive payoff.\n\nIf you want to talk about \"risk and stuff\", what do you need to consider, and why?\n\nStaff - 4\u00a0years, 7\u00a0months ago\n\nI am not formally educated in these topics, so I'm just saying all this from a logical standpoint.\n\nAfter a bit of thinking, I think I agree with you on the fact that everyone has their own \"payoff preference\". One can take a risk to win more, or take less risk to win less. In the scenario mentioned here, the aspects of risk are not conspicuous. The amount to be won is not significant, and there is no penalty for losing. Hence, both the payoffs are inherently attractive offers. When we come to real life situations, I think there are many more things that contribute to the risk:\n\n1. The probability that you will get a payoff.\n\n2. The probability that you stand to lose money.\n\n3. The security of the winnings. Money once won shouldn't be taken away from you.\n\n4. The effects of said decisions on long term/ on your ability to take other decisions.\n\nAccording to me, every decision of ours is based on weighing in these risk factors against the probable prize. The balance may tip either way, and so may our decisions.\n\nI don't even know if I'm thinking in the right direction... Help me out here @Calvin Lin\n\n- 4\u00a0years, 7\u00a0months ago\n\nPart of this series of posts is for you to figure out what kind of risk appetite you have. It is a personal preference, and there isn't necessarily a \"right\" answer. Your understanding will help guide you in decisions that you make in future. For the points that you raised, you should answer for yourself if those should matter, and why.\n\nHopefully, you will come up with a consistent, logical risk-reward structure. For example, if you would choose payoff 1 over 2 and payoff 3 over 4, but would prefer a combined 2 and 4 over a combined 1 and 3, then it would be very easy for someone to sell you things in part, and make you overpay for them.\n\nStaff - 4\u00a0years, 7\u00a0months ago\n\n@Calvin Lin sir, Am I right?\n\n- 4\u00a0years, 7\u00a0months ago\n\nLet $X$ be an indicator variable which takes on value 1 for each $HT$ pair and 0 otherwise, and let $Y$ be an indicator variable which takes on value 1 for each $HH$ pair and 0 otherwise. There are $5 - 2 + 1 = 4$ indicator variables (both $X$ and $Y$) in 5 coin tosses. The expected value of the payoff is: $\\text{E}\\left [\\sum_{k = 1}^{4} X_{k} \\right ], ~ \\text{E}\\left [\\sum_{k = 1}^{4} Y_{k} \\right ].$We can apply Linearity of Expectation here, which holds even for dependent events (nice proof of this is given in Brilliant Wiki Page), to get: $\\sum_{k = 1}^{4}\\text{E}\\left [ X_{k} \\right ] , ~ \\sum_{k = 1}^{4}\\text{E}\\left [ Y_{k} \\right ].$Expected value of both $X_{k}$ and $Y_{k}$ is: $\\text{E}\\left [ X_{k} \\right ] = \\text{E}\\left [ Y_{k} \\right ] = 1 \\cdot \\dfrac{1}{4} + 0 \\cdot \\dfrac{3}{4} = \\dfrac{1}{4}.$It follows that expected payoffs in both cases are the same and they equal:$\\sum_{k = 1}^{4}\\text{E}\\left [ X_{k} \\right ] = \\sum_{k = 1}^{4}\\text{E}\\left [ Y_{k} \\right ] = 4 \\cdot \\dfrac{1}{4} = 1.$\n\nBonus: What about the variance of the payoff? It can also influence our decision. Is it also the same?\n\nIt turns out they are not! Let us first consider payoff 1 and indicator variable $X$:\\begin{aligned} \\text{Var}[X] &= \\text{E}\\left [\\left(\\sum_{k = 1}^{4}X_{k}\\right)^{2}\\right ] - \\text{E}\\left [\\sum_{k = 1}^{4}X_{k}\\right ]^{2} \\\\ &= \\text{E}\\left [\\left(\\sum_{k = 1}^{4}X_{k}^{2}\\right) + \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] - \\text{E}\\left [\\sum_{k = 1}^{4}X_{k}\\right ]^{2} \\\\ &= \\text{E}\\left [\\left(\\sum_{k = 1}^{4}X_{k}^{2}\\right)\\right] + \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] - \\text{E}\\left [\\sum_{k = 1}^{4}X_{k}\\right ]^{2} \\\\ &= 1 + \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] - 1 \\\\ &= \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ].\\end{aligned}The same is true for payoff 2 and $Y$. Now, let us calculate $\\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ]$. We make two groups of $X_{k}X_{j}$:\n\n\u2022 $\\left | k-j \\right | = 1$ representing two consecutive indicator variables which share one common coin. Notice that their product is always 0 since it's impossible to have two $HT$s in 3 consecutive coins.\n\n\u2022 $\\left | k-j \\right | > 1$ representing two non-consecutive indicator variables which share no common coin. Their product is non-zero only when they are both non-zero which happens with probability $\\left(\\frac{1}{4}\\right)^{2} = \\dfrac{1}{16}.$\n\nThere are total of $4 \\cdot 3 = 12$ $\\left(X_{k}X_{j}\\right)$ pairs, and $2\\cdot 3$ of them which are consecutive ie. where $\\left | k-j \\right | = 1$. Hence, we have: $\\text{Var}[X] = \\text{E}\\left[ \\left(\\sum_{k \\neq j}X_{k}X_{j}\\right)\\right ] = 0 \\cdot 6 + \\frac{1}{16} \\cdot 6 = 0.375.$\n\nIn the same way, we calculate $\\text{Var}[Y] = \\text{E}\\left[ \\left(\\sum_{k \\neq j}Y_{k}Y_{j}\\right)\\right ] = \\frac{1}{8} \\cdot 6 + \\frac{1}{16} \\cdot 6 = 1.125.$\n\nSo, variance of the payoff 2 is 3 times greater than variance of the payoff 1. If my reasoning is correct, then payoff 1 is somewhat safer option and it tends to the expected value, while payoff 2 is more riskier but it offers better chances of earning more than 1$. What do you thinks about this approach which takes variance into account, @Calvin Lin ? - 1 year, 6 months ago Log in to reply After calculating the expected value and variance of the payoffs by hand using indicator variables (similar to some other people who have posted here), I became curious as to what the distributions of the payoffs would look like and how different strategies (not necessarily of length 2) would work out. For those interested, I wrote up an R-Jupyter notebook on it here. - 5 months, 4 weeks ago Log in to reply That's a lovely investigation! The 3-coin case is more complicated, so it's great to see your results. Note that the THH and HHT cases are identical by symmetry (flip the coins and flip the order of the sequence). For the variance, we indeed have $THH < HTH < HHH$. For the general case, I believe that the $TH \\ldots H$ case has the lowest variance (since it requires a T to get it started, so has a very low chance of duplicity). The transition state diagrams also give us some information on how to estimate the variance. Staff - 5 months, 4 weeks ago Log in to reply There are 2^5 outcoms of 5 coin tosses. For payoff 1, there are this many ways to toss a HT. There are {HTHTT,HTHTH,HTTHT,HTHHT,THTHT, HHTHT,HTHHH,HHHTH,HHHHT, HTTTT,THTTT,TTHTT,TTTHT,HTTTH,HHTTT, HHTTH} Expected paayyoff 1 = (6/32$2) + (10/32$1)=$11/16=$22/32 For payoff 2, there are this many ways to toss a HH( or HHH,HHHH and HHHHHH). There are {HHHHH, THHHH,HHHHT, THHHT,HTHHH,HHHTH, HHHTT, TTHHH, HHTHH, HHTTT, THHTT, TTHHT, TTTHH, THTHH, THHTH] Expected payoff 2 =(1/32$4 )+ ( 2/32$3) + ( 5/32$2) + (7/32$1)=$3/4= $27/32 Payoff 2 is better, I guess LOL @Calvin Lin Total possible outcomes of 32 of which there is one combination [TTTTT] which has no payoff. - 4 years, 7 months ago Log in to reply Check your calculations. The expected payoff of both scenarios is$1.\n\nStaff - 4\u00a0years, 7\u00a0months ago\n\nThere example given implies that HHH is counted as two consecutive HHs. This seems questionable as in standard English two consecutive HHs means HHHH. Is this really what the scenario intended?\n\n- 1\u00a0year, 1\u00a0month ago\n\nYes, that is what the scenario intended. The consecutive pairs are allowed to overlap. Let me edit that in.\n\nStaff - 1\u00a0year, 1\u00a0month ago\n\nExpected value is same for both payoffs\n\n- 1\u00a0year ago\n\nGreat observation. Does this make both payoffs the same? If not, what else do you want to consider?\n\nStaff - 1\u00a0year ago\n\nThus the payoffs are same, so i can choose either of the payoff be it payoff A or Payoff B\n\n- 1\u00a0year ago\n\nSo, what you're saying is that between\n\nPayoff C: Get $0 Payoff D: 50% chance to get$1,000, 50% chance to lose $1,000 Because the expected payoff for both is$0, you are equally happy to choose either?\n\nStaff - 1\u00a0year ago\n\nwe need to look at different aspects of payoff like standard deviation etc but I haven't thought about it yet..\n\nI think I need to look at the variance of values in payoff 1 and payoff 2. Then if the expected value is same then we will go for the one which is less riskier.\n\n- 1\u00a0year ago\n\nGreat. Proceed from there, and then think about what matters to you.\n\nStaff - 1\u00a0year ago\n\nTotal sample consists of 32 possible combinations... Expected value of payoff 1 is same as payoff 2 but payoff 2 will be more extreme payoffs like 4 while max payoff for payoff 1 will be 2, intuitively i think that variance of payoff 2 is greater than payoff 1 So I will choose payoff 1\n\n- 1\u00a0year ago\n\nI remember reading about an interesting experiment done by a French mathematician regarding payoffs and expected utility from lotteries. Allais Paradox?\n\n- 4\u00a0years, 7\u00a0months ago\n\nThat's interesting. In experiment 1 here, taking a risk, (even if it is only 1%), means the possibility of losing a guaranteed million dollars; I would deeply regret gambling and ending up losing that money, but if I took the million and played 1B just to see what happened and found that I could have won 5 million, I would have just said \"Oh well\" and still been perfectly happy with my million. But if there is no guaranteed money, then having a chance at 5 million at the expense of a 1% greater chance of winning a million seems worth the risk.\n\nExperiment 1 makes me think of the old saying, \"A bird in the hand is worth two in the bush.\" If there are no \"birds in hand\", however, then the risk evaluation process is quite different.\n\n- 4\u00a0years, 7\u00a0months ago\n\nLet Ii be an indicator rv for an event of getting H at ith toss. Let X be a rv which is how many times HT occured. Similarly, let Y be the same but for HH. Then, X = I1I2 +... +I4I5 and Y = I1(1-I2) +...+I4(1-I5). We need to compare EX and EY. Given EIi = 0.5, we can apply linearity of expectation (even for multiplication since events are independent) and get 1 for both expectations.\n\n- 4\u00a0years ago\n\nSo, if their expectations are the same, does that mean that you are indifferent about which payoff to take? If so, why?\n\nStaff - 4\u00a0years ago\n\nYes, because with both payoffs, one can win the same amount of dollars on average.\n\n- 4\u00a0years ago\n\nSo, are you indifferent between these 2 payoffs:\n\nPayoff 3: Always get $0 Payoff 4: 50% chance of -$1,000,000,000 and 50% chance of $1,000,000,000 Staff - 4 years ago Log in to reply I think we want the choice with the minimum variance involved. Maximin Principle of Rationality: An epistemically rational person maximizes his minimum payoff Staff - 4 years ago Log in to reply If I don't have a million dollars at all, I will definitely not take payoff 4. I would take payoff 4 only if I was a millionaire. So, psychologically I am not indifferent. - 4 years ago Log in to reply Right, so there are other considerations that come into play. For most people, the certainty of a result is preferred to an uncertain result (though with equal expected value). In general, having a low variance is better (though of course, there are exceptions). Staff - 4 years ago Log in to reply I would go for Payoff 2 , because For payoff 1 ..... one can get maximum number of$1 is 2 which equals $2 i.e for this outcome HTTHT ( MAXIMUM POSSIBLE 'HT' consecutive is 2). For payoff 2 ..... one can get maximum number of$1 is 4 which equals $4 i.e for this outcome HHHHH ( MAXIMUM POSSIBLE 'HH' consecutive is 4). - 1 year, 7 months ago Log in to reply I think this is a combinatorics problem?! Staff - 4 years, 7 months ago Log in to reply In the sense that arithmetic is a part of algebra, which is a part of calculus? There are numerous cross-disciplinary ideas. Using combinatorics will only get you one perspective of things. Notice that I am essentially getting at the risk-reward preference, which is a \"finance idea\" as opposed to a \"combinatorics idea\". Yes, combinatorics ideas like expected value is involved, but they don't tell the full story. Staff - 4 years, 7 months ago Log in to reply From a statistical standpoint, I would choose payoff option 1. As the number of H tossed increases, the probability of the consecutively tossing another H decreases. That being said, the HH combination can lead to a higher payoff because HTHTH only results in$2 while HHHHH results in \\$5. The issue lies in the decreased probability of continuing to toss H. This is a good example related to risk, much like in the decision involved in buying a AAA bond earning 5% vs a B bond earning 13%.\n\n- 3\u00a0years, 4\u00a0months ago\n\nTo be more specific, payoff 1 would be like the AAA bond and payoff 2 like the B bond. It just depends on what kind of risk is right for you.\n\n- 3\u00a0years, 4\u00a0months ago\n\nNot quite the same comparison. Note that the expected payoff in both methods are the same.\n\nWhereas in the bond example that you gave, we are trading expected payoff for certainty.\n\nI also strongly disagree with \"As the number of H tossed increases, the probability of the consecutively tossing another H decreases\". The coin tosses are independent, and that is a common fallacy that \"the proportion of realized events must be equal / close to the calculated probability\"\n\nStaff - 3\u00a0years, 4\u00a0months ago", "date": "2019-12-13 17:43:20", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for-2/", "openwebmath_score": 0.9509485363960266, "openwebmath_perplexity": 1207.6052515524545, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305318133554, "lm_q2_score": 0.8840392741081575, "lm_q1q2_score": 0.8515336201430932}}
{"url": "https://mathematica.stackexchange.com/questions/257623/finding-ellipse-axes", "text": "# Finding Ellipse Axes\n\nConsider the following ellipse, generated by the bounding region of the following points\n\nps = {{-11, 5}, {-12, 4}, {-10, 4}, {-9, 5}, {-10, 6}};\nrec = N@BoundingRegion[ps, \"FastEllipse\"];\nGraphics[{rec, Red, Point@ps}]\n\n\nWe have that the ellipse 'rec' is given in the form\n\nEllipsoid[{-10.4, 4.8}, {{2.77333, 0.853333}, {0.853333, 1.49333}}]\n\n\nHow can I retrieve the lengths of the two main axes of such ellipsoid? Following this representation and Mathematica's general definition of Ellipsoid\n\nI tried the following, using the eigenvalues of rec[[2]]\n\neigs = Eigenvectors[Inverse[rec[[2]]]]\neigv = Eigenvalues[Inverse[rec[[2]]]];\nlens = 2/Sqrt[eigv]\n\nOut[]= {2.06559, 3.57771}\n\n\nwhere the 2 factor comes from the fact what what I retrieve from the eigenvalues is actually half the length of the main axis. Indeed we get\n\nGraphics[{rec, Red, Point@ps,\nBlue, Line[\nRegionCentroid@rec + # & /@ {-(lens[[1]] eigs[[1]])/\n2, (lens[[1]] eigs[[1]])/2}],\nLine[RegionCentroid@rec + # & /@ {-(lens[[2]] eigs[[2]])/\n2, (lens[[2]] eigs[[2]])/2}]}]\n\n\nIs this correct? Is there a quicker way of doing this?\n\n\u2022 Nov 2, 2021 at 17:04\n\nStolen from @J.M.'s answer, https://mathematica.stackexchange.com/a/239797/4999, with one correction (is that enough to make it not a duplicate?):\n\nNodes = ps;\nellipsoidBR = BoundingRegion[Nodes, \"FastEllipse\"]; (* not \"FastEllipsoid\" *)\ncenter = ellipsoidBR[[1]];\n{vals, vecs} = Eigensystem[ellipsoidBR[[2]]];\n{a, b} = Sqrt[vals];\nmajor = N@{center - a vecs[[1]], center + a vecs[[1]]}\nminor = N@{center - b vecs[[2]], center + b vecs[[2]]}\nGraphics[{ellipsoidBR, Red, Point@ps, Green, Point@center,\nLine@{major, minor}}, Frame -> True]\n\n\nHere's the alternate way using SingularValueDecomposition:\n\npt = rec[[1]]; mat = rec[[2]];\n{u, s, v} = SingularValueDecomposition[(mat + Transpose[mat])/2];\nfunc = Composition[AffineTransform[{u, pt}], ScalingTransform[Sqrt[Diagonal[s]]]];\n\n\nand the length:\n\nEuclideanDistance @@@ {func@{{0, -1}, {0, 1}}, func@{{-1, 0}, {1, 0}}}\n\n\n{2.06559, 3.57771}\n\nGraphics[{rec, Blue, Line[func@{{-1, 0}, {1, 0}}],\nLine[ func@{{0, -1}, {0, 1}}]}]\n\n\nI think the help of Ellipsoid is incomplete because it does not explain the input: Ellipsoid[p,[CapitalSigma]], where the second argument is called the \"weight matrix\".\n\nYou will remember that an ellipse (for simplicity I am explaining the 2D case, nD is similar and assume the ellipse is centered at the origin) can be written by:\n\nx^2/rx^2 + y^2/ry^2 == 1\n\n\nWe may write this as:\n\n{x,y}.{{1/rx^2,0},{0,1/ry^2}}.{x,y} == r.mat.r == 1\n\n\nNote that the inverse Sqrt of the eigenvalues of m0 are the half axes of the ellipse.\n\nIf we now rotate the coordinate system by a rotation matrix: rot (r'=rot.r where r' are the new coordinates) the ellipse will be rotated (in the inverse sense) in the new coordinates:\n\nr'. Transpose[rot].mat. rot .r' == r' . mat' . r' == 1\n\n\nTherefore a rotated ellipse may be represented by a an symmetric (positive definite) matrix: mat'. This is called \"weigh matrix\" in the help.\n\nNote that the eigenvalues of the matrix are not changed by a rotation. The eigenvectors point in the directions of the half axes.\n\nHere is an example: Let\n\nrx=2;\nry=1;\nm0=DiagonalMatrix[{1/rx^1,1/ry^2}]\n{x,y}.m0.{x,y}==1\n\n\nThis represents an axis aligned ellipse with half axes rx and ry:\n\nRegion[ImplicitRegion[{x, y} . m0 . {x, y} == 1, {x, y}],\nAxes -> True]\n\n\nIf we now rotate the matrix m0:\n\nrot = RotationMatrix[-Pi/4];\nm= Transpose[rot].m0.rot;\n\n\nwe get a rotated ellipse:\n\nRegion[ImplicitRegion[{x, y} . m . {x, y} == 1, {x, y}], Axes -> True]\n\n\nThrerfore, the half axes are obtained by the inverse Sqrt of the eigenvalues of the weight matrix. And the directions of the axes are given by the eigenvectors.", "date": "2022-05-21 02:37:14", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/257623/finding-ellipse-axes", "openwebmath_score": 0.3861517608165741, "openwebmath_perplexity": 4270.224787556017, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305297023093, "lm_q2_score": 0.8840392710530071, "lm_q1q2_score": 0.8515336153340314}}
{"url": "http://velbyproductions.com/3m1tft/p7f7f.php?a930ce=curve-fitting-least-square-method", "text": "/BBox [0 0 5669.291 8] If the coefficients in the curve-fit appear in a linear fashion, then the problem reduces to solving a system of linear equations. /Filter /FlateDecode /Matrix [1 0 0 1 0 0] Each method has its own criteria for evaluating the fitting residual in finding the fitted curve. endstream Least Squares Fit (1) The least squares \ufb01t is obtained by choosing the \u03b1 and \u03b2 so that Xm i=1 r2 i is a minimum. >> Fitting requires a parametric model that relates the response data to the predictor data with one or more coefficients. This Python program implements least square method to fit curve of type y = ab x.. We first read n data points from user and then we implement curve fitting for y = ab x using least square approach in Python programming language as follow: . /Resources 17 0 R 18 0 obj Least Square is the method for finding the best fit of a set of data points. /Matrix [1 0 0 1 0 0] It can also be easily implemented on a digital computer. 14 0 obj . The application of a mathematicalformula to approximate the behavior of a physical system is frequentlyencountered in the laboratory. The most important application is in data fitting. Here a = 1.1 and b = 1.3, the equation of least square line becomes Y = 1.1 + 1.3 X. /Subtype /Form Let \u03c1 = r 2 2 to simplify the notation. in this video i showed how to solve curve fitting problem for straight line using least square method . >> endobj This is usually done usinga method called least squares\" which will be described in the followingsection. stream By understanding the criteria for each method, you can choose the most appropriate method to apply to the data set and fit the curve. Also suppose that we expect a linear relationship between these two quantities, that is, we expect y = ax+b, for some constants a and b. The best fit in the least-squares sense minimizes the sum of squared residuals. /Filter /FlateDecode You can employ the least squares fit method in MATLAB. endstream \ufffd2\ufffd\ufffd\ufffd6jE)\ufffdC\ufffdU\ufffd#\ufffd\\\ufffdN\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdp\ufffdS\ufffdJ\u0600\ufffd\ufffd3\ufffd\ufffd\ufffd\ufffd*\ufffdV(q:S\ufffdQ\u00e8a\ufffd\ufffd6\ufffd\ufffd&\ufffdM\ufffdq9;?z\ufffd(\ufffd\ufffd%\ufffd\ufffd'\u078b1e\ufffdUe\ufffdeH\ufffdM\ufffdI\ufffd\ufffd\ufffd\ufffd\ufffd\ufffdX+m\ufffdB\ufffd\ufffd\ufffd\ufffdlg\ufffdbB\ufffdBLJ\ufffd\ufffd\u024b\ufffd\ufffdnE\ufffd&d\ufffda9\u6a34 \ufffd)Z+\ufffd\ufffd. /Resources 19 0 R x\ufffd\ufffd\ufffdP(\ufffd\ufffd \ufffd\ufffd In LabVIEW, you can apply the Least Square (LS), Least Absolute Residual (LAR), or Bisquare fitting method to the Linear Fit, Exponential Fit, Power Fit, Gaussian Peak Fit, or Logarithm Fit VI to fin\u2026 The sum of the squares of the offsets is used instead of the offset absolute values because this allows the residuals to be treated as a continuous differentiable quantity. Least Squares Fitting. x\ufffd\ufffd\ufffdP(\ufffd\ufffd \ufffd\ufffd To find the equation of the curve of \u2018best fit\u2019 which may be the most suitable for predicting the unknown values. If you're an engineer (like I used to be in a previous life), you have probably done your bit of experimenting. Other documents using least-squares algorithms for tting points with curve or surface structures are avail-able at the website. Least squares fit is a method of determining the best curve to fit a set of points. . Curve fitting is one of the most powerful and most widely used analysis tools in Origin. /FormType 1 The basic problem is to \ufb01nd the best \ufb01t Normal Equation for \u2018a\u2019 \u00a0\u00a0\u00a0\u00a0\u00a0 $$\\sum Y = na + b\\sum X$$, Normal Equation for \u2018b\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 $$\\sum XY = a\\sum X + b\\sum {X^2}$$, The direct formula of finding $$a$$ and $$b$$ is written as, $b = \\frac{{\\sum XY \u2013 \\frac{{\\left( {\\sum X} \\right)\\left( {\\sum Y} \\right)}}{n}}}{{\\sum {X^2} \u2013 \\frac{{{{\\left( {\\sum X} \\right)}^2}}}{n}}}{\\text{ }}, \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,a = \\overline Y \u2013 b\\overline X$, Help me with the normal equations for power curve, Your email address will not be published. This article demonstrates how to generate a polynomial curve fit using the least squares method. Points with curve or surface structures are avail-able at the website ( as of August 2018 ) regression a... 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By using the least squares fitting the most common method to generate a polynomial equation a...", "date": "2022-07-03 06:18:03", "meta": {"domain": "velbyproductions.com", "url": "http://velbyproductions.com/3m1tft/p7f7f.php?a930ce=curve-fitting-least-square-method", "openwebmath_score": 0.6570824384689331, "openwebmath_perplexity": 680.8880732127594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9597620550745211, "lm_q2_score": 0.8872045966995027, "lm_q1q2_score": 0.8515053069998764}}
{"url": "https://artspassion.org/now-pea-ezocl/5c0621-irrational-numbers-definition", "text": "For example, there is no number among integers and fractions that equals the square root of 2. having a numerical value that is an irrational number. See also Rational Number. Define irrational. Every transcendental number is irrational.. \u2018The square root of 2 is an irrational number because it can't be written as a ratio of two integers.\u2019 \u2018How can mathematical concepts like points, infinitesimally small quantities, or irrational numbers be anything but products of our minds?\u2019 \u2018He considered computation with irrational numbers and polynomials to \u2026 Wikipedia. irrational numbers. being an irrational number. An irrational number is a real number that cannot be written as a simple fraction. c. Marked by a lack of accord with... Irrational - definition of irrational by The Free Dictionary. Irrational Numbers . Irrational numbers definition can be stated as \u201cthe numbers which we cannot write in the \\frac { p }{ q } form is called as irrational numbers\u201d. Definitions. For example: The pairs (2, 9); (4, 7) etc. It explains in computing terminology what Irrational Number means and is one of many technical terms in the TechTerms dictionary. Irrational Numbers Definition. irrational number \\\u026a.\u02cc\u0279\u00e6\u0283.\u0259.n\u0259l \u02c8n\u028cm.b\u025a\\ (\u00c9tats-Unis), \\\u026a.\u02cc\u0279\u00e6\u0283.\u0259.n\u0259l \u02c8n\u028cm.b\u0259\\ (Royaume-Uni) (Math\u00e9matiques) Nombre irrationnel. An irrational number is simply the opposite of a rational number. Irrational numbers may not be crazy, but they do sometimes bend our minds a little. What Is a Rational Number? Your fears are not based on fact and not likely to come true. Definition of irrational number in the Definitions.net dictionary. 3 , 5 are examples of irrational numbers. The denominator q is not equal to zero ($$q\u22600.$$) Some of the properties of irrational numbers are listed below. See Rational number definition.) Irrational Number. Enrich your vocabulary with the English Definition \u2026 As I said previously, the Conservatives give a new definition to the algebraic term of \"irrational numbers\" because their numbers simply do not make sense. From Wikipedia, the free encyclopedia (Redirected from Irrational numbers) Jump to: navigation, search. While at Montpellier he wrote a paper on irrational numbers and limits. Learn more. irrational: [adjective] not rational: such as. However, irrational numbers can have a decimal value that continues forever WITHOUT a pattern, unlike the example above. Examples of such numbers are \u03c0, e, \u221a6. Irrational Numbers. Another definition we can give as \u201cnon terminating non recurring decimal numbers are irrational numbers\u201d. This page contains a technical definition of Irrational Number. Many people are surprised to know that a repeating decimal is a rational number. 1. a. Let's start by defining each term separately, then we can learn more about each and work through some examples. Information and translations of irrational number in the most comprehensive dictionary definitions resource on the web. Comme je l'ai dit pr\u00e9c\u00e9demment, les conservateurs donnent une nouvelle d\u00e9finition de l'expression alg\u00e9brique \u00ab nombres irrationnels \u00bb, car leurs chiffres n'ont aucun sens. For example, real numbers like \u00e2\u02c6\u01612 which are not rational are categorized as irrational. So they can't be written as a clear fraction of 2 integers. not endowed with reason or understanding. As I said previously, the Conservatives give a new definition to the algebraic term of \"irrational numbers\" because their numbers \u2026 irrational synonyms, irrational pronunciation, irrational translation, English dictionary definition of irrational. Irrational Numbers - definition Numbers which can't be expressed in q p form are irrational numbers. An irrational number is a number that cannot be expressed as a fraction for any integers and .Irrational numbers have decimal expansions that neither terminate nor become periodic. D\u00e9finition, traduction, prononciation, anagramme et synonyme sur le dictionnaire libre Wiktionnaire. A little of many technical terms in the Definitions.net dictionary quotient of two integers ( ie fraction. The Definitions.net dictionary wrote a paper on irrational numbers definition numbers - definition numbers which ca n't be expressed as ratio., prononciation, anagramme et synonyme sur le dictionnaire libre Wiktionnaire not equal to zero ( (., search one that can not be crazy, but they do sometimes our! Is described as rational if it can be expressed in q p are! 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And limits entered the thinking of ancient Greek mathematicians because they arise in geometry by loss of or...", "date": "2021-08-05 11:05:42", "meta": {"domain": "artspassion.org", "url": "https://artspassion.org/now-pea-ezocl/5c0621-irrational-numbers-definition", "openwebmath_score": 0.9205806851387024, "openwebmath_perplexity": 2548.0565735131054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.959762055074521, "lm_q2_score": 0.8872045832787205, "lm_q1q2_score": 0.8515052941191188}}
{"url": "https://math.stackexchange.com/questions/2378844/basic-probability-question-multiplication-rule/2382974", "text": "# Basic probability question- multiplication rule\n\nI came across the following question in a textbook (bear in mind that this is the only information given)-\n\nThere is a $50$ percent chance of rain today. There is a $60$ percent chance of rain tomorrow. There is a $30$ percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?\n\nMy instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of $30$ percent. However, the answer given is $40$ percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.\n\n\u2022 if they were independent you multiply for example you flip three coins the probability they all end up heads is $({1 \\over 2})^3={1\\over 8}$ dependent probabilities aren't so simple. \u2013\u00a0user451844 Aug 1 '17 at 13:37\n\nYes, this is about independence. Or, better put, these events are not independent, which by common sense makes sense, since if it rains one day, it typically means there is a higher chance than normal that it rains the next day as well (we might well be in the 'rainy season')\n\nYou can also tell that the events are not independent given the numbers given to you. If the events were independent, then the probability of it not raining either day should have been $0.5 \\cdot 0.4=0.2$, but you were told it is actually $0.3$.\n\nWhat is always true, however (so you can always use this formua, whether the events are independent or not), is that:\n\n$$P(A \\cup B) = P(A) + P(B) - P(A \\cap B)$$\n\nAnd from that you can derive your desired:\n\n$$P(A \\cap B) = P(A) +P(B) - P(A \\cup B) = P(A) +P(B)-(1-P(A^C \\cap B^C))=$$\n\n$$0.5+0.6-(1-0.3) =1.1-0.7=0.4$$\n\nYes, the multiplication rule only applies when events are independent, and there is no reason to assume the events of rain on each day are independent.\n\nFor this type of problem, it helps to make a Venn diagram. There are two circles, $A$ and $B$, representing the events of rain today and tomorrow. This gives four regions: $A\\cap B$ (rain on both days), $A^c\\cap B^c$ (no on both days), $A\\cap B^c$ (rain today, but not tomorrow), and $A^c\\cap B$. The given information, and the fact that the total probability is $1$, tells us that \\begin{align} P(A)=P(A\\cap B)+P(A\\cap B^c)&=0.5\\\\ P(B)=P(A\\cap B)+P(A^c\\cap B)&=0.6\\\\ P(A^c\\cap B^c)&=0.3\\\\ P(A\\cap B)+P(A\\cap B^c)+P(A^c\\cap B)+P(A^c\\cap B^c)&=1 \\end{align} This is four equations equations in four unknowns, allowing you to solve for $P(A\\cap B)$.\n\nWhen you have only two independent variables, sometimes it's easier just to make a table.\n\nLet $X$ be the event that it rains today and $Y$ be the event that it rains tomorrow. We are given that $P(X)=0.5$, $P(Y)=0.6$, and $P(\\overline X \\cap \\overline Y)=0.3$:\n\n$$% outer array of arrays \\begin{array}{lr} % inner 3x3 array in top left corner \\begin{array}{c|c|c|} & Y & \\overline Y \\\\ \\hline X & P(X \\cap Y) & P(X \\cap \\overline Y) \\\\ \\hline \\overline X & P(\\overline X \\cap Y) & P(\\overline X \\cap \\overline Y)=0.3 \\\\ \\hline \\end{array} % inner 3x1 array in top right corner \\begin{array}{l} \\\\ P(X)=0.5 \\\\ P(\\overline X) \\end{array} \\\\ % inner 1x3 array in bottom left corner \\begin{array}{ccc} \\quad & P(Y)=0.6 & P(\\overline Y) \\\\ \\end{array} \\end{array}$$\n\nWe know that $P(X)+P(\\overline X)=1$ and likewise $P(Y)+P(\\overline Y)=1$, so we can fill this in as follows:\n\n$$% outer array of arrays \\begin{array}{lr} % inner 3x3 array in top left corner \\begin{array}{c|c|c|} & Y & \\overline Y \\\\ \\hline X & P(X \\cap Y) & P(X \\cap \\overline Y) \\\\ \\hline \\overline X & P(\\overline X \\cap Y) & P(\\overline X \\cap \\overline Y)=0.3 \\\\ \\hline \\end{array} % inner 3x1 array in top right corner \\begin{array}{l} \\\\ P(X)=0.5 \\\\ P(\\overline X)\\color{red}{=0.5} \\end{array} \\\\ % inner 1x3 array in bottom left corner \\begin{array}{ccc} \\quad & P(Y)=0.6 & P(\\overline Y)\\color{red}{=0.4} \\\\ \\end{array} \\end{array}$$\n\nWe know that the rows sum across and the columns sum down, so:\n\n$$% outer array of arrays \\begin{array}{lr} % inner 3x3 array in top left corner \\begin{array}{c|c|c|} & Y & \\overline Y \\\\ \\hline X & P(X \\cap Y) & P(X \\cap \\overline Y)\\color{red}{=0.1} \\\\ \\hline \\overline X & P(\\overline X \\cap Y)\\color{red}{=0.2} & P(\\overline X \\cap \\overline Y)=0.3 \\\\ \\hline \\end{array} % inner 3x1 array in top right corner \\begin{array}{l} \\\\ P(X)=0.5 \\\\ P(\\overline X)=0.5 \\end{array} \\\\ % inner 1x3 array in bottom left corner \\begin{array}{ccc} \\quad & \\,P(Y)=0.6 & \\quad\\quad\\, P(\\overline Y)=0.4 \\\\ \\end{array} \\end{array}$$\n\nAnd finally:\n\n$$% outer array of arrays \\begin{array}{lr} % inner 3x3 array in top left corner \\begin{array}{c|c|c|} & Y & \\overline Y \\\\ \\hline X & P(X \\cap Y)\\color{red}{=0.4} & P(X \\cap \\overline Y)=0.1 \\\\ \\hline \\overline X & P(\\overline X \\cap Y)=0.2 & P(\\overline X \\cap \\overline Y)=0.3 \\\\ \\hline \\end{array} % inner 3x1 array in top right corner \\begin{array}{l} \\\\ P(X)=0.5 \\\\ P(\\overline X)=0.5 \\end{array} \\\\ % inner 1x3 array in bottom left corner \\begin{array}{ccc} \\quad & \\,P(Y)=0.6 & \\quad\\quad\\, P(\\overline Y)=0.4 \\\\ \\end{array} \\end{array}$$", "date": "2020-10-21 22:50:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2378844/basic-probability-question-multiplication-rule/2382974", "openwebmath_score": 0.7822744250297546, "openwebmath_perplexity": 178.91286935906365, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978051749483894, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8514891896721206}}
{"url": "https://math.stackexchange.com/questions/633575/linear-algebra-question-on-rank-and-null-space", "text": "# Linear algebra question on rank and null space\n\nThis is an exercise from linear algebra and optimization by Gill, I do exercises to be prepared for my final exam and these are not homework questions!\n\nExercise $\\mathbf{6.1.}\\,$ Consider the following matrix $A$ and vector $b$: $$A=\\begin{pmatrix} 2 & 4 \\\\ 1 & 2 \\\\ 1 & 2 \\\\ \\end{pmatrix},\\quad b=\\begin{pmatrix} 3 \\\\ 2 \\\\ 1 \\\\ \\end{pmatrix}.$$ $\\textbf{(a)}$ What is the rank of $A$? Give a general form for any vector in the range of $A$.\n\n$\\textbf{(b)}$ Show that the dimension of the null space of $A^T$ is two, and display two linearly independent vectors $z_1$ and $z_2$ in $\\operatorname{null}(A^T)$. Give a general form for every vector in $\\operatorname{null}(A^T)$.\n\n$\\textbf{(c)}$ Find the vectors $b_R\\in\\operatorname{range}(A)$ and $b_N\\in\\operatorname{range}(A^T)$ such that $b=b_R+b_N$.\n\n$\\textbf{(d)}$ Give the general form of $b_A$ such that $b_R=Ab_A$. (Hint: consider all vectors $q$ such that $Aq=0$.)\n\nFor part (a), I think $rank(A)=1$ since all the columns are linear combination of each other. As for a general form for any vector in the range of $A$, when I write $Ax=b$ I get an over-determined system:\n\n$$2x_1+4x_2=3$$ $$x_1+2x_2=2$$ $$x_1+2x_2=1$$\n\nSo I'm not sure about the general form.\n\nAs for part (b), since the rank is two, dimension of $N(A^T)$ must be $3-1=1$, right?\n\nAnd no idea about the rest. Any help would be greatly appreciated.\n\n\u2022 If you know row reduction and properties of row reduced matrix then apply. It will be easier to get all answer together. \u2013\u00a0Dutta Jan 10 '14 at 10:37\n\u2022 Your solution to (a) is correct. For (b) note that if the dimension of the null space is two and you already found two linearly independent vectors in the null space, those form a base. Now that directly gives you a general form of all null vectors. \u2013\u00a0Listing Jan 10 '14 at 10:38\n\u2022 For (c) note that $\\text{null}(A^T)$ are exactly the vectors that are not in the image of $A$ (aka $\\text{range}(A)$). The image of $A$ gets spanned by $(2,1,1)$ thus you can write $b$ as $\\alpha z_1+\\beta z_2 + \\gamma (2,1,1)$. where $z_1,z_2$ are from (b). \u2013\u00a0Listing Jan 10 '14 at 10:41\n\u2022 @Listing: Thank you, that helped to understand the problem. \u2013\u00a0Gigili Jan 10 '14 at 12:07\n\u2022 Thank you for the edit @A\u00f0\u00f8be. \u2013\u00a0Gigili Jan 11 '14 at 2:14\n\n## 1 Answer\n\nIf you do row reduction, you find $$\\begin{pmatrix} 2 & 4 \\\\ 1 & 2 \\\\ 1 & 2 \\end{pmatrix} \\to \\begin{pmatrix} 1 & 2 \\\\ 0 & 0 \\\\ 0 & 0 \\end{pmatrix}$$ which means that the first column is a basis for the column space of $A$ (which is better terminology than \u201crange of $A$\u201d, in my opinion). So the general form of the vectors in the column space of $A$ is $$\\begin{pmatrix} 2a \\\\ a \\\\ a \\end{pmatrix},\\quad \\text{a any scalar}$$\n\nBy the rank nullity theorem, the null space of $A$ has dimension $1$; the equation defining it is $$x_1+2x_2=0$$ so a basis for it is the single vector $$\\begin{pmatrix} -2 \\\\ 1 \\end{pmatrix}$$\n\nThe null space of $A^T$ has indeed dimension $2$; the row reduction on $A^T$ is $$\\begin{pmatrix} 2 & 1 & 1 \\\\ 4 & 2 & 2 \\end{pmatrix} \\to \\begin{pmatrix} 2 & 1 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ so the equation defining the null space is $2x_1+x_2+x_3=0$ and a basis for it is $$\\left\\{ \\begin{pmatrix} -1 \\\\ 2 \\\\ 0 \\end{pmatrix} \\,, \\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix} \\right\\}$$ Writing $b=b_R+b_N$ should now be easy: the system to solve is $$\\left(\\begin{array}{ccc|c} 2 & -1 & -1 & 3 \\\\ 1 & 2 & 0 & 2 \\\\ 1 & 0 & 2 & 1 \\end{array}\\right)$$ but you can as well find the orthogonal projection of $b$ on the column space of $A$: $$b_R= \\frac{(2\\ 1\\ 1)\\begin{pmatrix}3\\\\2\\\\1\\end{pmatrix}} {(2\\ 1\\ 1)\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}} \\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix} =\\frac{9}{6}\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix} =\\begin{pmatrix}3\\\\3/2\\\\3/2\\end{pmatrix}$$ and $b_N=b-b_R$.\n\nWith this last idea it's easy to solve the last point.\n\n\u2022 Thank you for your detailed answer. Where did you use the fact that the first column is a basis for the column space (because it has leading $1$, right?)? And how did you come up with that general form, $\\begin{pmatrix} 2a \\\\ a \\\\ a \\end{pmatrix}$? \u2013\u00a0Gigili Jan 10 '14 at 11:56\n\u2022 @Gigili I used it for the last part, when computing $b_R$ with the orthogonal projection. Since the first column is a basis for the column space, every vector in the column space is that vector times a scalar. \u2013\u00a0egreg Jan 10 '14 at 11:59\n\u2022 Ah, right! I was thinking of $\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ as the first column but it's the first column in REF. Thanks, it's clear now. \u2013\u00a0Gigili Jan 10 '14 at 12:04\n\u2022 Sorry, but I still don't get the last part about $b$. Where did that system come from? Did you multiply the basis by $-1$? Then what's the first row? How did you calculate $b_R$? \u2013\u00a0Gigili Jan 10 '14 at 12:07\n\u2022 @Gigili It's a common error, I find it all the time in students' papers. ;-) The first non zero column in REF is always that one! It can't be in the column space of every matrix, can it? \u2013\u00a0egreg Jan 10 '14 at 12:07", "date": "2019-05-26 01:27:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/633575/linear-algebra-question-on-rank-and-null-space", "openwebmath_score": 0.8691567778587341, "openwebmath_perplexity": 127.49162150806083, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517494838937, "lm_q2_score": 0.8705972768020108, "lm_q1q2_score": 0.8514891896721204}}
{"url": "https://cstheory.stackexchange.com/questions/21984/expected-empirical-entropy", "text": "# Expected empirical entropy\n\nI'm thinking about some properties of the empirical entropy for binary strings of length $n$ when the following question crosses my way:\n\n$\\underbrace{\\large\\frac{1}{2^{n}}\\normalsize\\sum\\limits_{w\\in\\left\\{0,1\\right\\}^{n}}\\normalsize nH_{0}(w)}_{\\large\\#}\\;\\overset{?}{=}\\;n-\\varepsilon_{n}\\;\\;\\;$\n\nwith $\\;\\;\\lim\\limits_{n\\rightarrow\\infty}\\varepsilon_{n}=c\\;\\;\\;$ and $\\;\\;\\;\\forall n:\\;\\varepsilon_{n}>0$\n\nwhere $c$ is a constant.\n\n### Is that equation true? For which function $\\varepsilon_{n}$ respectively which constant $c$?\n\n\n\n$n=2\\;\\;\\;\\;\\;\\;\\;\\rightarrow\\;\\#=1$\n$n=3\\;\\;\\;\\;\\;\\;\\;\\rightarrow\\;\\#\\approx 2.066$\n$n=6\\;\\;\\;\\;\\;\\;\\;\\rightarrow\\;\\#\\approx 5.189$\n$n=100\\;\\;\\;\\rightarrow\\;\\#\\approx 99.275$\n$n=5000\\;\\rightarrow\\;\\#\\approx 4999.278580$\n$n=6000\\;\\rightarrow\\;\\#\\approx 5999.278592$\n\n\n\n# Backround\n\n\n$H_{0}(w)$ is the zeroth-order empircal entropy for strings over $\\Sigma=\\left\\{0,1\\right\\}$:\n\n\u2022 $H_{0}(w)=\\frac{|w|_{0}}{n}\\log\\frac{n}{|w|_{0}}+\\frac{n-|w|_{0}}{n}\\log\\frac{n}{n-|w|_{0}}$\n\nwhere $|w|_{0}$ is the number of occurences of $0$ in $w\\in\\Sigma^{n}$.\n\nThe term $nH_{0}(w)$ corresponds to the Shannon-entropy of the empirical distribution of binary words with respect to the number of occurences of $0$ respectively $1$ in $w\\in\\Sigma^{n}$.\n\nMore precise:\nLet the words in $\\left\\{0,1\\right\\}^{n}$ be possible outcomes of a Bernoulli process. If the probability of $0$ is equal to the relative frequency of $0$ in a word $w\\in\\left\\{0,1\\right\\}^{n}$, then the Shannon-entropy of this Bernoulli process is equal to $nH_{0}(w)$.\n\nAt this point, my question should be more reasonable since the first term normalizes the Shannon-entropies for all empirical distributions of words $w\\in\\left\\{0,1\\right\\}^{n}$.\nIntuitively I thought about getting something close to the Shannon-entropy of the uniform distribution of $\\left\\{0,1\\right\\}^{n}$, which is $n$.\nBy computing and observing some values I've got the conjecture above, but I'm not able to prove it or to get the exact term $\\varepsilon_{n}$.\n\nIt is easy to get the following equalities:\n\n$\\large\\frac{1}{2^{n}}\\normalsize\\sum\\limits_{w\\in\\left\\{0,1\\right\\}^{n}}\\normalsize nH_{0}(w)\\;\\;=\\large\\frac{1}{2^{n}}\\normalsize\\sum\\limits_{w\\in\\left\\{0,1\\right\\}^{n}}\\normalsize |w|_{0}\\log\\frac{n}{|w|_{0}}+(n-|w|_{0})\\log\\frac{n}{n-|w|_{0}}$\n\n$=\\large\\frac{1}{2^{n}}\\normalsize\\sum\\limits_{k=1}^{n-1}$ $n\\choose k$ $\\left(k\\log\\frac{n}{k}+(n-k)\\log\\frac{n}{n-k}\\right)$\n\nand it is possible to apply some logarithmic identities but I'm still in a dead point.\n\n(the words $0^{n}$ and $1^{n}$ are ignored, because the Shannon-entropy of their empirical distributions is zero)\n\nAny help is welcome.\n\n\u2022 I don't see a question here. \u2013\u00a0Suresh Venkat Apr 7 '14 at 17:36\n\u2022 Is the equation above true? Can anybody prove it respectively prove that the equation doesn't hold? \u2013\u00a0Danny Apr 7 '14 at 17:38\n\u2022 you have a bunch of equations. it would be better to number them. also highlight the question with \">\" blockquoting. \u2013\u00a0vzn Apr 8 '14 at 15:20\n\u2022 I edit my question from $\\varepsilon_{n}\\rightarrow0$ to $\\varepsilon_{n}\\rightarrow c$, where $c$ is a constant. \u2013\u00a0Danny Apr 9 '14 at 13:46\n\nHere is another approach, based on information theory and heavily inspired by @usul's answer. It shows that $\\epsilon_n=O(1)$ with very few calculations, and can be used to prove that $\\epsilon_n \\rightarrow \\log_2 \\sqrt{e}$ and to derive good estimates on the rate of convergence with less calculations than @usul's approach. In fact, I find a closed-form expression for $\\epsilon_n$ : $$(1) \\; \\; \\; \\; \\epsilon_n = n \\left( H(Binom(n,1/2)) - H \\left( Binom \\left( n-1, 1/2 \\right) \\right) \\right) \\ .$$\n\nDetails:\n\nLet $X$ be a uniform random variable in $\\{0,1\\}^n$. Let $K$ be a random variable equal to the number of 1's in $X$. The expression $\\#$ that @Danny wants to analyze is exactly equal to $n \\cdot \\mathbb{E}_{k} [H(X_1 | K=k)]$. (Here $X_1$ is the first bit of $X$.) By the basic properties of the entropy operator, $$(2) \\; \\; \\; \\; \\# = n\\mathbb{E}_{k} [H(X_1 | K=k)]=nH(X_1|K)=n(H(X_1K)-H(K))=n(H(K|X_1)+H(X_1)-H(K)) = n(1-H(Binom(n,1/2)) + H(Binom(n-1,1/2))) \\ .$$ The last equality follows from the fact that $X_1$ is just a uniformly random bit, $K$ is the binomial distribution, and $K|(X_1=x_1)$ is distributed either as $Binom(n-1,1/2)$ or as $Binom(n-1,1/2)+1$, depending on the value of $x_1$, both of which have the same entropy.\n\nThis already gave us equation (1). Now we just need to calculate to get the value of $\\lim_{n \\rightarrow \\infty} \\epsilon_n$\n\nWe use any known estimation for the entropy of a binomial RV, such as here. We see that $$(3) \\; \\; \\; \\; H(K)=H(Binom(n,1/2))=\\frac12 \\log_2 ( \\pi en / 2) + O(1/n) \\ ,$$ and, similarly, that $$H(K|X_1)=H(Binom(n-1,1/2))=\\frac12 \\log_2 ( \\pi e(n-1)/2) + O(1/n) \\ .$$ Canceling out terms and substituting into (1), we get $$(4) \\; \\; \\; \\; \\epsilon_n = n \\cdot (H(K)-H(K|X_1)) = n \\cdot \\frac12 (\\log_2 (n/(n-1)) + O(1/n)) = \\\\ \\frac12 \\log_2 ((n/(n-1))^n) + O(1) \\rightarrow \\log_2(\\sqrt{e}) + O(1) \\ .$$\n\nBy slightly improving the approximation (3) we should be able to replace the $O(1)$ term in (4) by $O(1/n)$ and therefore get that, indeed, $\\lim_{n \\rightarrow \\infty} \\epsilon_n = \\log_2 \\sqrt{e}$. To get this better estimation it should be enough to check that $H(Binom(n,1/2))=\\frac12 \\log_2 ( \\pi en / 2) + err(n)$ where $err(n)=O(1/n)$ and $err$ is a monotone function.\n\n\u2022 Very nice! I accept usul's answer since your answer is inspired by usul's, but your answer is very elegant (and converts to $\\log(\\sqrt e)$, too). \u2013\u00a0Danny Apr 10 '14 at 10:08\n\u2022 *converges instead of \"converts\" :) \u2013\u00a0Danny Apr 10 '14 at 11:29\n\u2022 Danny, I'm not sure it's a good idea to accept a more complicated and less elegant answer. Everyone else who ever looks at this question will look at @usul's answer first, while it's in everyone's interest that they look at the most simple and elegant answer first. The solutions to this are: 1. to ask usul to reformulate his answer to be simpler (for all I care, he could copy the text of my answer verbatim if he wants, I don't care about the reputation or anything) or 2. to accept my answer. I'll transfer the relevant reputation to usul if that's possible. \u2013\u00a0mobius dumpling Apr 10 '14 at 13:30\n\u2022 This seems like the better answer to accept in my opinion too! Very nice/elegant entropy argument. \u2013\u00a0usul Apr 15 '14 at 18:10\n\u2022 Note that in the literature you'll often see $\\log_2\\sqrt{e}$ written as $\\frac{1}{2\\ln 2}$. \u2013\u00a0Steven Stadnicki Apr 18 '14 at 18:52\n\n(Edited from previous version, 2014-04-08.)\n\nI believe that the answer is $\\epsilon_n \\to \\log(\\sqrt{e}) \\approx 0.7213475...$ where the logarithm is base 2. This seems to match simulation results. I don't have a full formal proof, but give the heuristic approximations/calculations.\n\nI think it's easier to note that your question is equivalent to:\n\nWhat are the asymptotics of $\\delta_n = 1 - \\mathbb{E}[H(k/n)]$, where the expectation is over $k \\sim Binomial(n,0.5)$?\n\n(where $H$ is the binary entropy function $H(p) = p\\log(1/p) + (1-p)\\log(1/(1-p))$, and $\\delta_n = \\epsilon_n/n$.)\n\nHere's a quick plot to show the idea. We have the binary entropy function in blue and the Binomial pmf (for $p=0.5$) in green. So we can see that the expectation of $H(k/n)$, when $k$ is distributed binomially, will always be below one but should be approaching one. The question is how fast.\n\nThe key idea will be that the value we're interested in, $$\\mathbb{E}[H(k/n)] = \\sum_{k=0}^n \\frac{{n \\choose k}}{2^n} H\\left(\\frac{k}{n}\\right) ,$$ can be related entropy of the binomial distribution. Let $p_{n,k} = \\frac{{n \\choose k}}{2^n}$ be the probability of $k$ heads in $n$ fair coin flips. The steps will be as follows:\n\n1. Use Stirling's approximation to get $H(k/n) = \\frac{\\log(p_{n,k})}{n} + 1 + \\text{something}$.\n2. Rewrite the original sum to get $\\mathbb{E}[H(k/n)] = 1 - \\frac{H(Binom(n,0.5))}{n} + \\sum \\text{something}$.\n3. Get that the entropy of the binomial, divided by $n$, plus the sum of \"something\", equals $\\frac{\\log(\\sqrt{e})}{n} + O(1/n^2)$.\n\nStep 1:\n\nJust plugging in Stirling and doing some cancellation/rearranging, \\begin{align} p_{n,k} &:= \\frac{{n \\choose k}}{2^n} \\\\ &\\sim \\frac{n^n}{k^k (n-k)^{n-k}}\\frac{1}{2^n}\\sqrt{\\frac{n}{2\\pi k (n-k)}} . \\end{align}\n\nThis won't be very tight for every term in the sum, but I think it will be asmyptotically tight towards the middle, which is all that matters since all the probability is in the center.\n\nWe can rewrite the entropy function as follows. It's just some arithmetic to combine the logarithms. \\begin{align} H(k/n) &= \\frac{k}{n}\\log\\left(\\frac{1}{k/n}\\right) + \\frac{n-k}{n}\\log\\left(\\frac{1}{(n-k)/n}\\right) \\\\ &= \\log\\left(\\frac{n}{k^{k/n}(n-k)^{(n-k)/n}}\\right) . \\end{align}\n\nSo, using Sterling's approximation above, the logarithm of a probability term is \\begin{align} \\log(p_{n,k}) &= \\log\\left(\\frac{{n \\choose k}}{2^n}\\right) \\\\ &\\approx n H(k/n) - n + \\log\\left(\\sqrt{\\frac{n}{2\\pi k(n-k)}}\\right) . \\end{align}\n\nStep 2: \\begin{align} \\mathbb{E}[H(k/n)] &= \\sum_{k=0}^n p_{n,k} H(k/n) \\\\ &\\approx \\sum_{k=0}^n p_{n,k} \\left(\\frac{\\log(p_{n,k})}{n} + 1 - \\frac{\\log\\left(\\sqrt{\\frac{n}{2\\pi k (n-k)}}\\right)}{n}\\right) \\\\ &= 1 - \\frac{H(Binom(n,0.5))}{n} - \\frac{1}{n}\\sum_{k=0}^n p_{n,k}\\log\\left(\\sqrt{\\frac{n}{2\\pi k(n-k)}}\\right) . \\end{align} Here, $H(Binom(n,0.5))$ is the entropy of the Binomial distribution for $n$ coin flips and $p=0.5$, which by wikipedia is $\\log\\left(\\sqrt{\\pi e n / 2}\\right) + O(1/n)$.\n\nStep 3:\n\nNow, we just need to approximate the third sum. I will take a very rough approximation (feel free to do better, but it probably doesn't gain much). All the probability mass is concentrated on $k = \\frac{n}{2} \\pm o(n)$. So approximate this sum (which is an expectation) by its value on the term $k=\\frac{n}{2}$, when it is $\\log\\left(\\sqrt{\\frac{2}{\\pi n}}\\right)$.\n\nSo now, we get \\begin{align} \\delta_n &\\approx \\frac{\\log\\left(\\sqrt{\\pi e n/2}\\right)}{n} + \\frac{\\log\\left(\\sqrt{2/\\pi n}\\right)}{n} \\pm O\\left(\\frac{1}{n^2}\\right) \\\\ &= \\frac{\\log\\left(\\sqrt{e}\\right)}{n} \\pm O\\left(\\frac{1}{n^2}\\right) . \\end{align}\n\n\u2022 Thank you very much for your effort. Your attempt is useful and i like your ideas, but i am not quite sure about the details. I will think about it and maybe i will ask some questions soon.. \u2013\u00a0Danny Apr 8 '14 at 21:35\n\u2022 @Danny, sure thing. Let me know if parts are unclear. The computation is not fully rigorous, particularly (1) the use of Stirling's approximation everywhere (a fix might be to only consider an interval where the Binomial probability is $\\Omega(1/n)$ and argue that Stirling is asymptotically tight there) and (2) the approximation of the \"third term\" by just setting $k=n/2$ (a fix might be to again consider the high-probability interval and see if the error is small). \u2013\u00a0usul Apr 9 '14 at 1:24\n\u2022 My first question is a general question: You think that $\\mathbb{E}_{w \\in \\{0,1\\}^n} nH(w) = n-\\mathcal{O}(1)$. Do you think there is a \"short\" proof for this conjecture without getting the exact term $\\varepsilon_{n}$ respectively the exact limit of $\\varepsilon_{n}$, which you are considering to be $\\log\\sqrt{e}$? \u2013\u00a0Danny Apr 9 '14 at 13:35\n\u2022 @Danny, I'm not sure how to get this with a shorter proof, but maybe another answer will do so. \u2013\u00a0usul Apr 9 '14 at 20:51\n\u2022 Thanks again for your help. Everything seems to be clear.. \u2013\u00a0Danny Apr 10 '14 at 9:07\n\nHere is a proof that the quantity # that OP considers tends to n(1-o(1)).\n\nClaim: #=n-o(n)\n\nFirst, note that for any function $f:\\{0, 1\\}^n \\rightarrow \\mathbb{R}$, $\\frac{1}{2^n} \\sum_{w \\in \\{0,1\\}^n} f(w)$ is exactly the same as $\\mathbb{E}_{w \\in \\{0,1\\}^n} f(w)$. So you're asking whether $\\mathbb{E}_{w \\in \\{0,1\\}^n} H(w) = 1-o(1)$.\n\nTo see this, define $X$ to be a random variable uniformly distributed on $\\{0,1\\}^n$ Let $X_0$ be the number of 0's in $X$ and $X_1$ be the number of 1's in $X$. We want to prove $$\\mathbb{E}[(X_0/n)\\log(n/X_0)+(X_1/n)\\log(n/X_1)]=1-o(1).$$ But we know from the law of large numbers that both $X_0/n$ and $X_1/n$ converge in probability to $1/2$ and that if a random variable $Y$ converges in probability to $c$ and if $f$ is a continuous function, then $f(Y)$ converges in probability to $f(c)$. So we get that $\\mathbb{E}[(X_0/n)\\log(n/X_0)+(X_1/n)\\log(n/X_1)]$ converges in probability to $\\frac12 \\log(2) + \\frac12 \\log(2) = 1$, QED.\n\nNote: This claim is related to the Asymptotic equipartition property for discrete-time i.i.d. sources here.\n\n\u2022 I'm sorry, but I can't see how $\\mathbb{E}_{w\\in\\left\\{0,1\\right\\}^{n}}\\left[H\\left(\\frac{|w|_{0}}{n},\\frac{|w|_{1}}{n}\\right)\\right]=1-o(1)$ is a direct result of the AEP, since we are looking for the expectation of entropies for all empirical distributions and the AEP treats with the limit of the selfinformations of strings for some fixed distribution. Can you specify that implication? \u2013\u00a0Danny Apr 8 '14 at 15:06\n\u2022 Also I do not see why your statement $\\mathbb{E}[(X_0/n)\\log(n/X_0)+(X_1/n)\\log(n/X_1)]=1-o(1)$ leads directly to my conjecture.. Thanks for your help. \u2013\u00a0Danny Apr 8 '14 at 15:18\n\u2022 So you're asking whether $\\mathbb{E}_{w\\in\\{0,1\\}^n}H(w) = 1 - o(1)$. Note this isn't quite so, the question as worded is whether it is $1 - o(1/n)$. \u2013\u00a0usul Apr 9 '14 at 1:26\n\u2022 @Danny, the expression $\\mathbb{E}[(X_0/n)\\log (n/X_0) + (X_1/n) \\log (n/X_1)]$ is exactly the same as your expression $H_0(w)$ (or, in my notation, $H_0(X)$). \u2013\u00a0mobius dumpling Apr 9 '14 at 11:39\n\u2022 @Danny You're right, the theorem does not follow from the AEP as easily as I thought. Rather than explain how it follows, I'll just leave my self-contained explanation, which turns out to be simpler than relating to the AEP. \u2013\u00a0mobius dumpling Apr 9 '14 at 11:44\n\nI will prove that for all $n$, $H_n = n - \\epsilon_n$ where $0 < \\epsilon_n < 1$.\n\nLet $f_{\\alpha}(t) = 4^{\\alpha} t^{\\alpha}(1 - t)^{\\alpha}$. I will not prove this, but you can find some $\\alpha^*$ such that for all $\\alpha \\ge \\alpha^*$, $f_{\\alpha}(t) \\le H(t)$. Also, for $\\delta \\ll 1$, you can find $\\alpha'$ such that for all $\\alpha \\le \\alpha'$, $f_{\\alpha'}(t) \\ge H(t)$ in $[\\delta, 1 - \\delta]$. I will assume $\\alpha = 1$ and $\\alpha = 0.5$ lead to the two different types of functions.\n\nMy point is that you can use simpler functions to obtain the bound you need. Here I will use $f_{\\alpha}(t)$.\n\nLet $H_n$ be the empirical entropy that you are looking for and $F(n, \\alpha)$ be the value obtained when I replace $H(\\frac{k}{n})$ with $f_{\\alpha}(\\frac{k}{n})$. We have\n\n$F(n, \\alpha) = n2^n \\sum_{k = 0}^{n} {n \\choose k} f_{\\alpha}(\\frac{k}{n}) = n 2^n 4^{\\alpha} \\sum_{k = 0}^{n} {n \\choose k} \\left( \\frac{k}{n} \\right) ^ {\\alpha} \\left(1 - \\frac{k}{n} \\right) ^\\alpha = n 4^{\\alpha} E\\left[\\left( \\frac{k}{n} - \\frac{k^2}{n^2} \\right)^{\\alpha} \\right] = n2^n 4^{\\alpha} \\sum_{k = 0}^{n / 2} {n \\choose k} \\left(\\frac{k}{n} \\right)^{2 \\alpha}$\n\nFor $\\alpha = 1$, we know $F(n, \\alpha) \\le H_n$ and direct evaluation says that $F(n, 1) = n - 1$, which is true for all $n$.\n\nFor $\\alpha = 0.5$, we know $F(n, \\alpha) \\ge H_n$ and a direct evaluation leads to $F(n, 0.5) = n 4^{0.5} E[\\frac{k}{n}] = n$.\n\nTherefore, for all $n > 0$, $n\\ge H_n \\ge n - 1$.\n\nI found it nontrivial to try this for any $\\alpha < 1.0$ other than $\\alpha = 0.5$.", "date": "2019-12-11 16:02:17", "meta": {"domain": "stackexchange.com", "url": "https://cstheory.stackexchange.com/questions/21984/expected-empirical-entropy", "openwebmath_score": 0.9834180474281311, "openwebmath_perplexity": 286.2625482306691, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517437261225, "lm_q2_score": 0.8705972801594707, "lm_q1q2_score": 0.8514891879431898}}
{"url": "https://www.physicsforums.com/threads/coming-up-with-a-formula-for.42921/", "text": "Coming up with a formula for\n\n1. Sep 12, 2004\n\nCaldus\n\nUm, how can I figure out a formula for solving something like:\n\n1 + 3 + 5 + 7 + 9 + ... + n\n\nIn particular: 1 + 3 + 5 + ... + 999\n\nThanks.\n\n2. Sep 12, 2004\n\ngeometer\n\nAs far as I know, there is no rote procedure for doing this. It's pretty much a trial and error process. Make yourself a table and figure out the pattern. In this case, we can write this as the sum from x = 0 to x = (n-1)/2 of 2x+1:\n\nx = 0 ===> Sum = 1\nx = 1 ===> Sum = 4\nx = 2 ===> Sum = 9\nx = 3 ===> Sum = 16\n\nWe can keep this up if we need to to help us see the pattern, but in this case, a little thought shows that the formula is: Sum from x= 0 to x = n = (n+1)^2\n\n3. Sep 12, 2004\n\nIntegral\n\nStaff Emeritus\nI believe that Gauss did this as school child when given the assignment to sum the digits 1-100.\n\nFirst note that if you add them together smallest to largest you get a constant result ( 100+ 1=101, 99+2=101 ..) multiply this by 100 because there are 100 sums and divide by 2 because you added all the numbers twice.\n\nedit:\nOPPS! just noticed that you were not summing ALL integers.\n\nLast edited: Sep 12, 2004\n4. Sep 12, 2004\n\nIntegral\n\nStaff Emeritus\nAfter a bit more playing,\n\nconsider this\n\n$$S = \\Sigma _ {i=1}^N (2i -1)$$\n\nnow using the properties of the summation\n\n$$S= 2\\Sigma_{i=1}^N i - \\Sigma_{i=1}^N 1$$\n\n$$S= 2\\Sigma_{i=1}^N i -N$$\n\nNow using Gauss's trick we can see that the sum of n integers is given by\n\n$$S_n = \\frac {(n+1)(n)} 2$$\n\nSo we can write for the sum of the odd integers\n\n$$S = (N+1)(N)-N = N^2$$\n\nNote that I have given the general expression for the odd integers as 2n-1 and my sum starts with n=1 where geometer used 2n+1 and started with n=0. This is the reason he has N+1 where I have N. Also note that this N is NOT the odd integer you are summing to but the NUMBER OF TERMS in the sum.\n\nLast edited: Sep 12, 2004\n5. Sep 12, 2004\n\nshmoe\n\nIntegral, the same idea of pairing highest and lowest will work just fine,\n1+3+5+7+9+11=12+12+12=(6/2)*12, where 6 is the number of terms here.\n\nIf you already have a formula for the sum of the first n integers, 1+2+3+..+n, you can use this as well:\n1+3+5+7+9+11=(1+2+..+11)-(2+4+6+8+10)=(1+2+..+11)-2(1+2+3+4+5)=(insert formula here)-2(insert formula here)\n\nCaldus, do you have a formula for the sum of the first n integers? If not, see Integrals post;)\n\nAnother idea, (which Integral provided while I was typing) write your sum as: $$\\displaystyle\\sum\\limits_{i=1}^{n}(2i-1)$$ and fiddle with it to get an expression involving $$\\displaystyle\\sum\\limits_{i=1}^{n}i$$, and apply the formula for this (again, using the ideas in Integral's post if need be). This would be a more easily applied general idea that can handle any arithmetic series $$\\displaystyle\\sum\\limits_{i=0}^{n}(ki+d)$$.\n\nLast edited: Sep 12, 2004\n6. Sep 13, 2004\n\nHallsofIvy\n\nStaff Emeritus\nSince you are adding only odd numbers, one way to do is this (Assuming you know that the sum of the first n integers is n(n+1)/2):\nThe sum of all numbers from 1 to 2n is (2n)(2n+1)/2= n(2n+1). Now subtract off the sum of all even numbers from 2 to 2n: 2+ 4+ 6+ ...+ 2n= 2(1+ 2+ 3+ ... + n)=\n2(n)(n+1)/2= n(n+1).\n\nThe sum of all odd numbers from 1 to 2n-1 is n(2n+1)- n(n+1)= n(2n+1- n- 1)= n2 just as shmoe said.\n\nIn particular, the sum of all odd numbers from 1 to 99 is: (99= 100- 1= 2(50)- 1)\n502= 2500.\n\n7. Sep 13, 2004\n\nGokul43201\n\nStaff Emeritus\nAlso, here's a general rule that can be used in times of trouble :\n\nIf a series S, has terms t(n), which are polynomials in n, of order p, the sum S(n) is a polynomial of order p+1. The coefficients of this polynomial can be determined simultaneously, from the given p+2 terms in the series.\n\nExample : If the series S, has terms, t(n) which are linear in n, ie : t(n) = an + b, then the sum, S(n) is a quadratic in n, ie: S(n) = An^2 + Bn + C\n\nNOTE : The order of the polynomial can be determined by the method of \"successive differences\". If the p'th successive differences are equal, the order is p.\n\nExample : Consider the sequence : 2, 6, 12, 20, 30, ...\nThe first differences are : 4, 6, 8, 10, ...\nThe second differences are : 2, 2, 2, ...\n\nSince the second differences are equal, t(n) is a quadratic in n, ie : t(n) = an^2 + bn + c.\n\nSo you can assume the sum has a cubic form : S(n) = An^3 + Bn^2 + Cn + D\nPlugging in the for S(1), S(2), S(3) and S(4) gives you 4 simultaneous equations in A, B, C, and D, which you can solve.", "date": "2017-04-26 12:08:01", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/coming-up-with-a-formula-for.42921/", "openwebmath_score": 0.8441044688224792, "openwebmath_perplexity": 589.8774510717521, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517437261225, "lm_q2_score": 0.8705972784807406, "lm_q1q2_score": 0.851489186301305}}
{"url": "https://cs.stackexchange.com/questions/136555/how-can-we-find-the-number-of-pairs-of-intersecting-ranges-on-a-circular-number", "text": "# How can we find the number of pairs of intersecting ranges on a circular number line?\n\nI recently thought of and managed to solve this algorithmic problem:\n\nOn a infinite 1-dimensional number line, we have N ranges specified by two distinct integers A and B, such that all A and B are unique. Each pair of integers then represents the starting and ending points of a range on the number line. How many distinct pairs of ranges intersect?\n\n^^ In this example, 4 distinct pairs of segments intersect.\n\nObviously, there is a O(N^2) brute-force solution: just test all possible pairs in O(1) time each. However, I also thought of a O(NlogN) sweep-line algorithm: treat every A as a \"segment start event\" and every B as a \"segment end event\". Then, keep a variable curr_in_range (originally 0) that stores the current number of segments that we currently see, and then iterate through the events in sorted order of their positions on the number line. Every time we add another segment, add curr_in_range to the answer (because this new segment will intersect with every other segment that we can see) and increment it as well. Every time we remove a segment, decrement curr_in_range. In the end, this gets us the correct answer.\n\nHappy with this algorithm, I decided to tackle a slightly modified version of this problem: What if the number line is circular? As in, it has a specified length L such that the number line contains numbers in the range [0, L)? In this case, each range would still be specified by two distinct integers A and B, but let's just say the range is always counterclockwise starting from A and ending at B. Could we still find the number of pairs of sectors that intersect in less than O(N^2) time in this scenario?\n\nHere's my thought process on this problem:\n\nWe should be able to use a very similar algorithm to the one described above. Completely copying it wouldn't work, though, because the sectors can have B < A when it crosses over 0. My idea was to initially have curr_in_range equal to the number of sectors that cross over 0 (instead of just it at 0 like we did in the above algorithm). Then, we could treat every sector as a \"sector start event\" and a \"sector end event\", and do the same process that we did.\n\nHowever, after extensive analysis, I found why this approach doesn't work. This approach won't work because it overcounts some of the intersections, as sometimes two sectors can intersect in 2 places. This is where I'm stuck.\n\n^^ Our sweep-line algorithm will count 2 distinct pairs that intersect, when there is actually only 1.\n\nBut besides this overcounting, I found that this current algorithm isn't doing a single other thing wrong. I coded a O(N^2) brute-force algorithm and tabulated all the distinct pairs that it found, and also tabulated the distinct pairs found for the current algorithm. For every randomly-generated test case, these two tables were the same. In other words, our current algorithm is correctly detecting every intersection there is to be detected, but it just happens to overcount some of them.\n\nSo, sorry for a really long post, but my final question is:\n\nCan someone come up with an algorithm that solves this problem? Is my thought process close to a solution or completely off?\n\n\u2022 Simple idea that might work: sort intervals as usual, except give each one a unique id. Begin processing using the same algorithm, except every start interval is marked as started, and if you encounter an end interval that hasn't been marked, you just ignore it. The algorithm terminates when all intervals are marked as started. \u2013\u00a0hLk Mar 27 at 4:22\n\nThis is solvable in $$O(n \\log n)$$ time, too, by considering the complement of the intervals that wrap around and using a suitable data structure.\n\nThere are two types of intervals that can appear on a cycle:\n\n\u2022 Intervals that do not wrap around: e.g., $$[0.3,0.5]$$ does not wrap around. In general, this includes all intervals of the form $$[a,b]$$ with $$0\\le a \\le b \\le 1$$.\n\n\u2022 Intervals that wrap around: e.g., $$[0.8,0.1]$$ wraps around; it is equivalent to $$[0.8,1.1]$$ or $$[0.8,1] \\cup [0,0.1]$$. In general, this includes all intervals of the form $$[b,a]$$ with $$0\\le a < b \\le 1$$; each such interval is equivalent to $$[b,1] \\cup [0,a]$$.\n\nLet $$A$$ denote the set of intervals that do not wrap around, and $$B$$ the set of intervals that do wrap around. Also, let $$\\overline{B}$$ denote the complements of the intervals in $$B$$, i.e., $$\\overline{B} = \\{\\overline{b} \\mid b \\in B\\}$$. For instance, the interval $$[0.8,0.1]$$ wraps around; its complement is $$(0.1,0.8)$$, which does not wrap around. It is easier to think about intervals that do not wrap around, and this decomposition will help us to deal only with intervals that do not wrap around.\n\nNow, we will separately count the number of overlaps between (1) an interval in $$A$$ and an interval in $$A$$; (2) an interval in $$A$$ and an interval in $$B$$; (3) an interval in $$B$$ and an interval in $$A$$; and (4) an interval in $$B$$ and an interval in $$B$$. (1) can be counted with your sweepline algorithm, as we don't need to deal with wraparound. (4) can also be counted in the same way, as two intervals overlap iff their complements overlap. I will show below how to count (2); case (3) will then follow by symmetry. This covers all the cases, so it suffices to count up these four cases and sum them up.\n\nNote that intervals $$a \\in A, b\\in B$$ overlap iff $$a$$ is not a subset of $$\\overline{b}$$. So, it suffices to count the number of pairs $$a \\in A, \\overline{b}\\in \\overline{B}$$ where $$a$$ is a subset of $$\\overline{b}$$; then you can subtract that from $$|A| \\cdot |B|$$.\n\nTo do that, you can use the data structure in Data structure for interval subset queries. Store all the intervals $$\\overline{B}$$ in the data structure. Then, for each $$a \\in A$$, count the number of intervals in $$\\overline{B}$$ that $$a$$ is a subset of. This can be done in $$O(\\log n)$$ time per query interval $$a$$, for a total time of $$O(n \\log n)$$.\n\nTherefore, you can count each of the four cases in $$O(n \\log n)$$ time, and thus compute the overall sum in $$O(n \\log n)$$ time as well. I think. I encourage you to check my reasoning to make sure I haven't made any mistakes.\n\n\u2022 \"(4) can also be counted in the same way, as two intervals overlap iff their complements overlap.\" A range overlaps any wrapping range that contains the whole circle, but their complements do not overlap. Anyway, in case (4) where all ranges are wrapping, we can just say any two of them overlap. \u2013\u00a0John L. Mar 25 at 18:49\n\u2022 Apparently, I could not understand how your usage of \"a segment tree to obtain that count\" avoids counting the same intersecting pair more than once. \u2013\u00a0John L. Mar 25 at 18:54\n\u2022 @JohnL., By \"wrapping\" I mean a range like $[0.9,0.1]$, i.e., $[0.9,1.1]$, i.e., $[0.9,1.0] \\cup [0,0.1]$. Such a range does not cover the whole circle, but includes $1$ and $0$ in the range. The only range that covers the entire circle is $[0,1]$, and that range is non-wrapping. \u2013\u00a0D.W. Mar 25 at 18:58\n\u2022 Are you talking about 2-dimensional segment trees? What are the leaves of your segment tree? If each leaf is an elementary interval, how can each interval of $\\overline B$ appear in the segment tree only once? If a leaf can be an interval of $\\overline B$, then how can an interval of $A$ be expressed as concatenation of those leaves? \u2013\u00a0John L. Mar 25 at 19:18\n\u2022 @JohnL., No, an ordinary segment tree for storing the intervals of $\\overline{B}$. As usual, we take the endpoints of the intervals of $\\overline{B}$, sort them, obtain the elementary intervals, and each leaf corresponds to one elementary interval (not an interval of $\\overline{B}$). Each interval of $\\overline{B}$ is stored in some node of the segment tree, following the rule described at the link you provided. The interval $a$ can be expressed as a disjoint union of elementary intervals, not of intervals of $\\overline{B}$. \u2013\u00a0D.W. Mar 25 at 19:22\n\nNot a solution, just an illustration of what I see as the core difficulty:\n\nThe instance on the left has 3 intersections, the one on the right only 2, but there is no way to distinguish the instances just by counting numbers of \"begin segment\" and \"end segment\" events within a given interval. (Inside the gap left by the red interval, the event sequence is \"begin\", \"end\" in both cases.)\n\nThis implies that a sweep-line approach needs to track more information about each active segment -- probably their identities, which suggests to me that no fast algorithm will be possible, though that's just a feeling.\n\n\u2022 A sweep algorithm will take $\\Omega(n\\log(n))$ time anyways since it has to sort the array. Then, we might be able to store the identities in a way it would not take much more space and time, so we would not overshoot that $n\\log(n)$. Not proposing a solution, but just saying that even if we have to keep the identities it might be fast enough \u2013\u00a0nir shahar Mar 12 at 23:33\n\n### Summary\n\nInteresting question.\n\nPlease check this program in Java that runs in $$O(n\\log n)$$ time, where $$n$$ is the number of given ranges. The program uses sweep-line technique, once for counting intersecting pairs of ranges, and once for counting containing pairs of ranges together with Fenwick tree. Explanations follow.\n\n### Reduce to the hard case\n\nHere are the notations and assumptions for the ease of explanation.\n\n\u2022 All numbers and variables are integers.\n\u2022 The numbered circle of cardinality $$L\\ge2$$ is the set of number $$0, 1, \\cdots, L-1$$, where $$L$$ is considered to wrap around to $$0$$.\n\u2022 A range $$[s,e]$$ with respect to $$L$$, where $$0\\le s\\le e\\lt L$$, means\n\u2022 a non-wrapping range (a.k.a. an ordinary range) if $$s\\le e$$, i.e., numbers $$s, s+1,\\cdots, e-1, e$$;\n\u2022 a wrapping range otherwise, i.e,. numbers $$s, s+1, \\cdots, L-1, 0, 1, \\cdots, e-1, e$$.\n\nNote every range contains at least one number. The non-wrapping range $$[0, L-1]$$ contains all the numbers. So does every wrapping range like $$[s,s-1]$$.\n\nSuppose there are $$n$$ ranges. The question is how to compute the number of all unordered pairs of two intersecting ranges efficiently.\n\nThere are three cases for an unordered pair of two ranges.\n\n\u2022 Both ranges are wrapping.\nThis is the easiest case since two wrapping ranges always intersect at $$0$$.\n\n\u2022 Both ranges are non-wrapping.\nThis is the classical case. The total number of such intersecting pairs can be computed by the sweep-line (or, as I would call it, the sweep-point or sweep-event) algorithm, which runs in $$O(n)$$ time, as described in the question.\n\n\u2022 One range is non-wrapping and the other is wrapping.\nThis is the new and hard case. One way to understand why this case may not be easy is j_random_hacker's illustration. How can we deal with this case?\n\n### Transform to the containing pairs of non-wrapping pairs\n\nDenote the non-wrapping ranges and the wrapping ranges in the given ranges by $$\\mathcal N$$ and $$\\mathcal W$$, respectively. Instead of counting the intersecting pairs between $$\\mathcal N$$ and $$\\mathcal W$$, we will use complementary counting, i.e., we will count the non-intersecting pairs between $$\\mathcal N$$ and $$\\mathcal W$$.\n\nGiven any range $$r$$, let $$\\overline r$$ be the numbers among $$0, 1, \\cdots, L-1$$ that are not in $$r$$, i.e., the complement of $$r$$ with respect to $$\\mathcal C$$. Note that two ranges do NOT intersect iff the complement of either one contains the other.\n\nLet $$\\overline{\\mathcal W} = \\{\\overline r\\mid r\\in\\mathcal W\\}$$. Then the number of non-intersecting pairs between $$\\mathcal N$$ and $$\\mathcal W$$ is the number of containing pairs between $$\\mathcal N$$ and $$\\overline{\\mathcal W}$$. Here a containing pair means, naturally, two sets $$a$$ and $$c$$ such that $$c$$ contains $$a$$. The number we wanted, i.e., the number of all intersecting pairs of ranges between $$\\mathcal N$$ and $$\\mathcal W$$ is $$|\\mathcal N|\\cdot|\\mathcal W|$$ minus that number of containing pairs.\n\nMore formally, we have $$\\#\\{(a, b) \\mid a\\in{\\mathcal N}, b\\in{\\mathcal W}, a\\cap b\\not=\\emptyset\\} = \\#{\\mathcal N}\\cdot\\#{\\mathcal W} - \\#\\{(a, c) \\mid a\\in{\\mathcal N}, c\\in\\overline{\\mathcal W}, a\\subseteq c\\}.$$\n\nHere is a simple observation, the complement of a wrapping range is a non-wrapping range, except when the wrapping range is like $$[s,s-1]$$, whose complement is empty. Containing all numbers, a wrapping range like that intersects with every other range. It is easy to count all intersecting pairs that involve a wrapping range like that.\n\nLet us assume there is no such wrapping range; if necessary, we can replace such a range with non-wrapping range $$[0, L-1]$$. The problem is reduced to counting the number of containing pairs between two sets of non-wrapping ranges.\n\n### An algorithm that counts the number of containing pairs\n\nHere is an algorithm that counts the number of containing pairs between two given sets of non-wrapping ranges, $$\\mathcal N$$ and $$\\overline{\\mathcal W}$$. It is, basically, sweep-line technique together with Fenwick tree.\n\n1. Let containingPairCount = 0.\n2. Initialize a Fenwick tree (a.k.a. binary indexed tree) ft for an (implicit) interested array of length $$L$$, with all values 0 initially. The e-th element of the interested array will be the number of all ranges in $$\\overline{\\mathcal W}$$ so far that end at e.\n3. Sort all ranges that are in either $$\\overline{\\mathcal W}$$ or $$\\mathcal N$$ by the starting points, breaking ties by putting ranges in $$\\overline{\\mathcal W}$$ before ranges in $$\\mathcal N$$. Breaking the remaining ties arbitrarily.\n4. One by one process the sorted ranges. Let the current range be [s,e].\n\u2022 If it comes from $$\\overline{\\mathcal W}$$, updates the Fenwick tree as if the e-th element of the interested array is increased by 1. This operation registers the current range as a candidate containing range that has started and will end at e.\n\u2022 If it comes from $$\\mathcal N$$, add the sum of elements of indexes no smaller than e in the interested array to containingPairCount, by containingPairCount += st.getSum(e, L-1).\n5. Return containingPairCount.\n\nProof of the algorithm: st.getSum(e, L-1) is the number of all ranges in $$\\overline{\\mathcal W}$$ that contain the current non-wrapping range, since, thanks to the careful setup, we have processed all ranges in $$\\overline{\\mathcal W}$$ that start at or before the start of the current range, and only those ranges that end at or after the end of the current edge are counted. $$\\quad\\checkmark$$\n\nThe sorting of all ranges at step 3 takes $$O(n\\log n)$$ time. At step 4, each range is processed with $$O(\\log n)$$ time, thanks to the power of Fenwick tree. So the running time of this algorithm is $$O(n\\log n) + O(n \\log n) = O(n\\log n)$$.\n\n### Complexity analysis\n\nThe running time for other parts of the entire algorithm is $$O(n\\log n)$$, which is spent mostly on counting the intersecting non-wrapping pairs. So, the running time of the entire algorithm is $$O(n\\log n) + O(n \\log n) = O(n\\log n)$$.\n\nThe sections \"Reduce to the hard case\" and \"transform to the containing pairs of non-wrapping pairs\" follow D.W's answer roughly.", "date": "2021-05-07 16:27:09", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/136555/how-can-we-find-the-number-of-pairs-of-intersecting-ranges-on-a-circular-number", "openwebmath_score": 0.7298941612243652, "openwebmath_perplexity": 424.69788964303507, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517501236461, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8514891853034326}}
{"url": "https://math.stackexchange.com/questions/211341/representation-of-a-matrix-transformation", "text": "# Representation of a matrix transformation\n\nAn example asks me to define $T: M_{2\\times2}(F) \\to M_{2\\times2}$ by $T(A) = A^t$. Compute $[T]_\\alpha$\n\n$\\alpha$ is the standard ordered basis of $2\\times 2$ matrices.\n\nTo find the transformation I performed the transformation on all four elements of $\\alpha$, for example $T(\\begin{pmatrix} 1&0\\\\0&0\\\\\\end{pmatrix}) = (\\begin{pmatrix} 1&0\\\\0&0\\\\\\end{pmatrix})$ and so on. Basically the matrices are all the same before and after the transformation. Accordingly, the textbook's answer is that you arrange all the results $A^t$ into a $4\\times4$ matrix:\n\n$$\\begin{pmatrix} 1&0&0&0\\\\0&0&1&0\\\\0&1&0&0\\\\0&0&0&1\\end{pmatrix}$$\n\nI don't understand this what means, and I've been trying to. How can this be the transformation $T$ on $\\alpha$ if we're not even able to multiply the basis by it? What I mean is, we can't multiply this transformation matrix by anything in $\\alpha$ since one is a $4\\times4$ and one is a $2\\times 2$.\n\nAlso, why doesn't this mean that any arbitrary $2\\times 2$ matrix can be raised to the power of $t$ by simply multiplying by this matrix? (As per the previous paragraph, it can't, but it should).\n\n\u2022 I changed $T: M_{2\\times2}(F) -> M_{2\\times2}$ to $T: M_{2\\times2}(F) \\to M_{2\\times2}$. $\\TeX$ is not so crude that you have to resort to things like that. \u2013\u00a0Michael Hardy Oct 11 '12 at 23:43\n\u2022 Thanks - I couldn't find the command for it. I'll know for next time. \u2013\u00a0CodyBugstein Oct 11 '12 at 23:57\n\nYou need to distinguish here between the space $\\mathbb F^2$ that the $2\\times2$ matrices operate on and the space of the $2\\times2$ matrices themselves. The former is two-dimensional, the latter is four-dimensional.\n\nA matrix is a representation of a linear transformation with respect to bases of the domain and target of the transformation. If the linear transformation is an endomorphism, a linear map from a space to itself, the corresponding matrix is square, and one typically uses the same basis for the space in its two roles as domain and target.\n\nThe standard basis of the space of $2\\times2$ matrices that you're using has four basis elements and spans the four-dimensional space of $2\\times2$ matrices. A linear transformation has been defined on that space \u2013 the four-dimensional space of $2\\times2$ matrices, not the two-dimensional space that they operate on.\n\nRepresenting that linear transformation as a matrix with respect to the standard basis (both for the domain and the target) yields a $4\\times4$ matrix. That's not a mismatch because the vectors and matrices that will be multiplied by this matrix aren't elements of $\\mathbb F^2$; they're the vectors of coefficients in an expansion of the matrix they represent in the standard basis.\n\nAs an example, the matrix\n\n$$\\pmatrix{2&3\\\\4&5}$$\n\nhas coefficients $2$, $3$, $4$ and $5$, respectively, in the standard basis, so with respect to that basis it's represented by the column vector\n\n$$\\pmatrix{2\\\\3\\\\4\\\\5}\\;.$$\n\nIts transpose\n\n$$\\pmatrix{2&4\\\\3&5}$$\n\nis represented by the column vector\n\n$$\\pmatrix{2\\\\4\\\\3\\\\5}\\;.$$\n\nAnd sure enough we have\n\n$$\\begin{pmatrix} 1&0&0&0\\\\0&0&1&0\\\\0&1&0&0\\\\0&0&0&1\\end{pmatrix}\\pmatrix{2\\\\3\\\\4\\\\5}=\\pmatrix{2\\\\4\\\\3\\\\5}\\;.$$\n\nThus this matrix really does represent the linear transformation $T$, transposition, in this basis.\n\n\u2022 So matrices can unravel to form vectors? I've played around with some and it makes sense. Thanks for the amazing answer. I still have a question though; shouldn't the transforming vector actually put your sample vector to the power of something? Isn't that the whole idea? \u2013\u00a0CodyBugstein Oct 11 '12 at 23:32\n\u2022 @Imray: I think you've misunderstood the notation. $A^t$ is merely a notational device to denote the transpose of $A$; nothing is being raised to a power. I prefer the slightly different notation $A^\\top$ (A^\\top) to avoid the impression that the $t$ is a variable used as an exponent. \u2013\u00a0joriki Oct 11 '12 at 23:34\n\u2022 @Imray: As an exercise, I suggest to write down the matrix of $T$ with respect to the basis $$\\left\\{\\pmatrix{1&0\\\\0&1},\\pmatrix{1&0\\\\0&-1},\\pmatrix{0&1\\\\1&0},\\pmatrix{0&1\\\\-1&0}\\right\\}\\;,$$ and then perhaps use it to apply $T$ to a couple of matrices and see whether the transpose comes out right. \u2013\u00a0joriki Oct 11 '12 at 23:38\n\u2022 yes right! I didn't realize that! Ok I've got it now, thank you very much for explaining it to me. \u2013\u00a0CodyBugstein Oct 11 '12 at 23:38\n\u2022 @Imray: You're welcome! \u2013\u00a0joriki Oct 11 '12 at 23:40\n\nJust to make sure our definition of matrix representation is the same: Let $\\alpha = (a_1, \\ldots, a_k)$ be a basis for $V$, and let $T: V \\rightarrow V$ be a linear map. Then matrix representation $[T]_{\\alpha}$ is defined as follows: $$[T]_{\\alpha} = \\begin{bmatrix} [T(a_1)]_{\\alpha} & \\ldots & [T(a_k)]_{\\alpha} \\end{bmatrix}$$\n\nThe matrix representation obeys the following formula: for every $v \\in V$, $[T(v)]_{\\alpha} = [T]_{\\alpha} [v]_{\\alpha}$, where $[v]_{\\alpha}$ is the coordinate vector of $v$ with respect to the basis $\\alpha$.\n\nSo now, to answer your question, it's true that $[T]_{\\alpha}$ is a $4 \\times 4$ matrix in your example, but you don't multiply it onto the $2 \\times 2$ matrices; you multiply it onto the $2 \\times 2$ matrices after they have been converted into column vectors via the coordinate mapping. Since $2 \\times 2$ matrices form a $4$-dimensional space, then this coordinate mapping turns a $2 \\times 2$ matrix into an element of $\\mathbb{R}^4$.\n\n\u2022 How does coordinate mapping work? The rows all extend downward to become one long column? Can all matrices with $n$ elements be multiplied by any matrix with $n$ columns? \u2013\u00a0CodyBugstein Oct 11 '12 at 23:35\n\u2022 What confused me is how $[T]_\\alpha$ could be square. But one needs to remember that the columns are in vector notation too (not in matrix notation!), so they are actually not $k \\times k$ but $k^2 \\times 1$ and so in the end $[T]_\\alpha$ is $k^2 \\times k^2$. \u2013\u00a0philmcole Apr 10 '18 at 21:20", "date": "2019-07-16 20:20:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/211341/representation-of-a-matrix-transformation", "openwebmath_score": 0.883613646030426, "openwebmath_perplexity": 188.9674514807904, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517475646369, "lm_q2_score": 0.8705972717658209, "lm_q1q2_score": 0.8514891830755662}}
{"url": "https://math.stackexchange.com/questions/2534449/show-that-a-n-left-sum-k-1n-frac1k-right-logn-is-decreasing", "text": "# Show that $a_n=\\left(\\sum_{k=1}^n \\frac{1}{k}\\right)-\\log(n)$ is decreasing\n\nHow I decided to show it is by Induction, and I don't really like what I have, so could you guys please tell me if my proof valid. Thanks.\n\nAim of the proof: Show that $$a_n>a_{n+1}$$ for all $n\\geq1$.\n\nBase case: $$a_1-a_2=\\log(2)-\\frac{1}{2}>0.$$ Hence base case holds. Assume, for some $k>1$, that the result holds, that is:$$a_k>a_{k+1}$$ Hence show that $$a_{k+1}>a_{k+2}.$$ Now we can rewrite $a_{k+1}>a_{k+2}$ as, $$\\left(\\sum_{k=1}^{k+1}\\frac{1}{k+1}\\right)-\\log(k+1)>\\left(\\sum_{k=1}^{k+1} \\frac{1}{k+1}\\right)+\\frac{1}{k+2}-\\log(k+2),$$ Which is equivealnt of saying: $$\\log\\left(\\frac{k+2}{k+1}\\right)>\\frac{1}{k+2}$$\n\nFor our problem $k\\geq1$ and $\\log(\\frac{3}{2})>\\frac{1}{3}.$\n\nMy last step is saying that as $n$ tends infinity, the limit of LHS of our last inequality is infinite and RHS is 0. Hence the result holds.\n\nWhat I don't like is the fact that I don't use the inductive hypothesis at all. (Sorry for selling and grammar errors, English is not my native language)\n\n\u2022 This does not answer your question, but do you know that $\\lim_n a_n =\\gamma$, the Euler-Mascheroni constant? You might want to google it, you will for sure find something about the non decreasing behaviour of $a_n$. \u2013\u00a0Paolo Intuito Nov 23 '17 at 21:53\n\u2022 A fundamental step you make in your proof is incorrect: $\\lim_{k \\rightarrow \\infty} (\\frac{k+2}{k+1}) = 0$, not infinity, as you claim. (In fact, if this step were correct, then this alone would be a proof - the whole induction set up would be unnecessary.) \u2013\u00a0John Don Nov 23 '17 at 21:59\n\u2022 Oh crap, yeah, the fraction gets closer to 1 eqch time. Thank you for noticing this John. How would I go about proving it then?, and yes, G.S. I know about that fact but couldnt find anything on the question I am trying to solve \u2013\u00a0Scavenger23 Nov 23 '17 at 22:02\n\nYou wish to show the inequality,\n\n$$\\frac{1}{n+1}-\\ln(n+1)<-\\ln n$$ or equivalently, $$\\frac{1}{n+1}<\\ln (1+\\frac{1}{n})$$\n\nNow we have $$x-\\frac{x^2}{2}<\\ln(1+x)$$ so just check that $$\\frac{1}{n+1}<\\frac{1}{n}-\\frac{1}{2n^2}$$\n\nWhich is indeed the case as it is not hard to show.\n\n\u2022 How do you know that your second to last line is true? I feel like I am missing something basic. \u2013\u00a0Scavenger23 Nov 23 '17 at 22:26\n\u2022 Either, you note that the next term in the power series is positive, or you could use the mean value theorem. \u2013\u00a0Rene Schipperus Nov 23 '17 at 22:27\n\u2022 Or more straightforward, move all the terms to the right, and differentiate, and show the derivative is positive, and thus the function is increasing. \u2013\u00a0Rene Schipperus Nov 23 '17 at 22:33\n\u2022 I see now, didnt realise that you tailor expanded the log. Thanks for the help. \u2013\u00a0Scavenger23 Nov 23 '17 at 22:35\n\nI like Rene Schipperus' method a lot - it is very simple. Here is an commonly used alternative:\n\nNote that, the function $\\frac1x$ is strictly decreasing on $[1, \\infty)$, so, for all $n \\in \\mathbb{N}^{\\, \\ge 1}$,\n\n\\begin{align} \\ln (n+1) - \\ln n &= \\int_{n}^{n+1} \\frac1x \\, dx\\\\ &> \\int_n^{n+1} \\frac1{n+1} \\, dx\\\\ &= \\frac1{n+1} \\end{align}\n\nAs you have already noted, the solution follows.\n\n\u2022 very good work. So essentially the method I wanted to use was correct, ? The question in the book never told me how to prove it. Induction seemed the way forward for me. \u2013\u00a0Scavenger23 Nov 23 '17 at 22:43\n\u2022 @KuderaSebastian That seems reasonable... I suppose that the thing to notice was that, at some point, (e.g. in your inductive step) you would have to relate $\\log(n)$ to some function of integers somehow (as this is what you have on the LHS). One way is to use the power series expansion (as Rene has done), and the other is to notice that the derivative of $\\log(x)$ is in fact $\\frac1x$ (which both I and Peter have used in our answers). \u2013\u00a0John Don Nov 23 '17 at 22:53\n\u2022 Jon.Nice answer, short and clear. \u2013\u00a0Peter Szilas Nov 24 '17 at 9:06\n\nShow that $a_n -a_{n+1} >0.$\n\n$\\Delta_n: = a_n - a_{n+1} =$\n\n$-\\dfrac{1}{n+1} + \\log(n+1) -\\log(n).$\n\nMean Value Theorem:\n\n$f(x) := \\log(x)$ then:\n\n$\\dfrac{f(x+1)-f(x)}{1} = f'(t),$ $x \\lt t \\lt (1+x).$\n\n$\\dfrac{\\log(n+1) -\\log(n)}{1} = \\dfrac{1}{t}$, $n\\lt t\\lt (n+1).$\n\n$\\Delta_n = -\\dfrac{1}{n+1} + \\dfrac{1}{t}$ , where $t \\lt (n+1).$\n\nHence $\\Delta_n >0.$\n\n\u2022 NIce... my answer is essentially this, but integrated, in a sense. \u2013\u00a0John Don Nov 23 '17 at 22:49", "date": "2020-09-27 15:47:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2534449/show-that-a-n-left-sum-k-1n-frac1k-right-logn-is-decreasing", "openwebmath_score": 0.9380149245262146, "openwebmath_perplexity": 407.1238092621322, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517514031506, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8514891798498265}}
{"url": "https://www.tutorialspoint.com/divmod-in-python-and-its-application", "text": "divmod() in Python and its application\n\nPythonServer Side ProgrammingProgramming\n\nThe divmod() is part of python\u2019s standard library which takes two numbers as parameters and gives the quotient and remainder of their division as a tuple. It is useful in many mathematical applications like checking for divisibility of numbers and establishing if a number is prime or not.\n\nSyntax\n\nSyntax: divmod(a, b)\na and b : b divides a\na and b are integers or floats\n\nExamples\n\nIn the below example see the cases of both integers and floats. On the application of divmod() they give us a resulting tuple which is can also contain integers and float values.\n\n# with integers\nprint(\"5 and 2 give:\",divmod(5,2))\nprint(\"25 and 5 give:\",divmod(25,5))\n\n# with Floats\nprint(\"5.6 and 2 give:\",divmod(5.6,2))\nprint(\"11.3 and 9.2 give:\",divmod(11.3,9.2))\n\nOutput\n\nRunning the above code gives us the following result \u2212\n\n5 and 2 give: (2, 1)\n25 and 5 give: (5, 0)\n5.6 and 2 give: (2.0, 1.5999999999999996)\n11.3 and 9.2 give: (1.0, 2.1000000000000014)\n\nUsing Zero\n\nIf the first argument is zero then we get (0,0). And If the second argument is zero then we get Zerodivision error as expected.\n\nExample\n\n# With first argument as zero\nprint(\"0 and 8 give:\",divmod(0,8))\n\n# With second argument as zero\nprint(\"8 and 0 give:\",divmod(8,0))\n\nOutput\n\nRunning the above code gives us the following result \u2212\n\n0 and 8 give: (0, 0)\nTraceback (most recent call last):\nFile \"xxx.py\", line 6, in\nprint(\"8 and 0 give:\",divmod(8,0))\nZeroDivisionError: integer division or modulo by zero\n\nChecking divisibility\n\nIf the second value of the tuple after division is 0 then we say that the first number is divisible by second. Else it is not divisible. The below example illustrates this.\n\nExample\n\nm = 12\nn = 4\nquotient,remainder = divmod(m,n)\nprint(quotient)\nprint(remainder)\nif (remainder==0):\nprint(m,' is divisible by ',n)\nelse:\nprint(m,' is not divisible by ',n)\n\nOutput\n\nRunning the above code gives us the following result \u2212\n\n3\n0\n12 is divisible by 4\n\nChecking if Number is Prime\n\nWe can use divmod() to keep track of the reminders it generates when we start dividing a number by each number starting with itself till 1. For a prime number the count of zero remainder will be only one as no number other than itself will divide it perfectly. If the count of zero remainder is greater than 1 then the number is not prime,.\n\nExample\n\nnum = 11\na = num\n# counter the number of remainders with value zero\ncount = 0\nwhile a != 0:\nq, r = divmod(num, a)\na -= 1\nif r == 0:\ncount += 1\nif count > 2:\nprint(num, 'is not Prime')\nelse:\nprint(num, 'is Prime')\n\nOutput\n\nRunning the above code gives us the following result \u2212\n\n11 is Prime\nPublished on 07-Aug-2019 08:32:04", "date": "2021-12-01 10:41:09", "meta": {"domain": "tutorialspoint.com", "url": "https://www.tutorialspoint.com/divmod-in-python-and-its-application", "openwebmath_score": 0.3603508770465851, "openwebmath_perplexity": 2102.508867928492, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517418068649, "lm_q2_score": 0.8705972734445508, "lm_q1q2_score": 0.8514891797047505}}
{"url": "https://matheducators.stackexchange.com/questions/18120/what-is-the-best-way-to-intuitively-explain-the-relationship-between-the-derivat/18128", "text": "# What is the best way to intuitively explain the relationship between the derivative and the integral?\n\nThis is my first post so bear with me, but something I've been thinking about lately is: Why didn't I ever question the relationship between the derivative and the integral when I was taking calculus?\n\nLet me explain what I mean: In most courses, the derivative is introduced as the slope of a curve at a point, or the \"instantaneous rate of change\". Then the integral is introduced as the area under a curve. Then students are told that these two things \"undo\" each other (the fundamental theorem of calculus). So now I'm wondering, why didn't I ever question why these \"undo\" each other? It's not intuitive at all. For example, addition and subtraction, multiplication and division, logarithms and exponentials; all of these things I can intuitively understand why they \"undo\" each other. But how does something that represents the area under a curve \"undo\" something that represents an instantaneous rate of change? What is the connection between those two concepts? Is there a better way to explain these concepts that makes the fundamental theorem of calculus more intuitive?\n\nLooking forward to hearing some of y'alls responses. Thanks!\n\n\u2022 $[\\cdots]$ why didn't I ever question why these \"undo\" each other? --- Your calculus experience is probably in the minority, because it is fairly standard (at least in U.S. introductory calculus courses, both high school and college) to make a big deal about the connection, usually at the end of the first semester, this being especially the case since the rise of \"reform calculus\" from the late 1980s on. That said, see right brain explanation and left brain explanation. \u2013\u00a0Dave L Renfro Apr 2 '20 at 5:26\n\u2022 Which resources have you perused before registering and asking this question? Voting to close. \u2013\u00a0Rusty Core Apr 2 '20 at 21:01\n\u2022 In countries, where physics is a mandatory course in grade school, the traditional layman explanation is through kinematics. Average speed v, by definition is displacement s over time t. Start with the simplest case of uniform motion, draw a graph in (t, s), it would be a line climbing up. Tangent at any point is at the same angle (similarity of right triangles), this is the derivative by definition, which is speed. Draw speed in (t, v). Area below it is displacement. Continue with uniform acceleration and then with no-uniform motion, slicing up the area under speed into thin vertical bands. \u2013\u00a0Rusty Core Apr 2 '20 at 21:14\n\u2022 @BenCrowell I think what makes this a math ed question is wanting better intuition to explain the relationship. Also, the fact that many of us didn't really get it until we were teaching speaks volumes to how much this is an education issue. \u2013\u00a0Sue VanHattum Apr 4 '20 at 20:52\n\u2022 @SueVanHattum Exactly - that's why I posted it here. The funny thing is that when I learned how to prove the FTC in real analysis, I didn't find either of the proofs too difficult. But the intuition behind the overall concept still wasn't there. Then recently I realized that if a student came to me and asked \"Why do these two concepts undo each other?\", my best response would be to show them the proof. That's why I figured this was a math ed question, because I'm specifically asking what the best way is to explain this to a student and make it more intuitive. \u2013\u00a0Brain Gainz Apr 5 '20 at 18:43\n\nYou start by noticing that the Riemann sums (multiplication followed by addition) and the difference quotients (subtraction followed by division) undo each other. Their limits -- the integral and the derivative -- still undo each other.\n\nAdded: The first sentence is a part of what is called \u201cDiscrete Calculus\u201d https://en.wikipedia.org/wiki/Discrete_calculus\n\n\u2022 Wow. I'm unsure whether this intuition can be made to correspond to some rigorous proof, but it's just solid enough to make me feel it gave me some insight into the mechanism of the theorem's truth and not mere word-soup hand-waving. And I think the maximal spacing in a partition (that goes to zero for the Riemann integral) even corresponds to the step size in the difference quotient in a suitable sense. \u2013\u00a0Vandermonde Apr 3 '20 at 1:23\n\u2022 @Vandermonde You are correct. By the way, this will be in the third volume of my book Calculus Illustrated: amazon.com/dp/B082WKCYHY \u2013\u00a0Peter Saveliev Apr 3 '20 at 1:44\n\u2022 Surprising how entirely abstract and formal mathematical thinking can be intuitive :-). Perhaps not for everybody, but even this engineer sees symmetries. \u2013\u00a0Peter - Reinstate Monica Apr 3 '20 at 10:21\n\u2022 1. Can you please exhibit the formula that you're hinting to? 2. Can you please clarify 1. which \"Riemann sums\" you mean? 3. Which \"difference quotients\"? 4. Which \"limits\"? \u2013\u00a0NNOX Apps Apr 4 '20 at 5:27\n\u2022 If one considers the differential $\\text{d}$ and the integral $\\int$ as the fundamental concepts of calculus (as opposed to $\\frac{\\text{d}(-)}{\\text{d}x}$ and $\\int(-)\\text{d} x$), then the wording is even more simple: difference and sum undo each other. (Actually Leibniz called $\\int$ the sum, before Bernoulli suggested integral) \u2013\u00a0Michael B\u00e4chtold Apr 6 '20 at 15:58\n\n(this is from my calculus notes, see page 233 of: http://www.supermath.info/OldschoolCalculusII.pdf)\n\n\u2022 So in words: the change in area is equal to the height of the function times the change in x? That actually makes sense. \u2013\u00a0Brain Gainz Apr 3 '20 at 0:31\n\u2022 @BrainGainz thanks! Incidentally, I had exactly your feeling after taking years of calculus and an Advanced Calculus course where I suspect we proved the FTC. Your feeling is not unusual in my estimation. \u2013\u00a0James S. Cook Apr 3 '20 at 2:52\n\u2022 I understand the pedagogical need, but I cringe when I read (or say myself when teaching calculus) \"arbitrary function\" and the picture is that of a differentiable function. \u2013\u00a0Martin Argerami Apr 3 '20 at 10:36\n\u2022 @BrainGainz yep, I have a few Topology lectures posted. I am that James Cook. \u2013\u00a0James S. Cook Apr 4 '20 at 21:57\n\u2022 @StevenGubkin cool, if I ever get to teach Calculus I again, I should use that. So many options to play with... \u2013\u00a0James S. Cook Apr 6 '20 at 3:49\n\nDerivative of integral is the original function: Let $$F$$ be the integral function of $$f$$, so that $$F(x)$$ is the area under the graph from zero to $$x$$. For small $$h>0$$ the difference $$F(x+h)-F(x)$$ is the area of a narrow vertical strip. The width is $$h$$ and height approximately $$f(x)$$. As $$h\\to0$$, this means $$F'(x)=f(x)$$.\n\nIf you like thinking in terms of infinitesimals, write $$dF=F(x+dx)-F(x)=f(x)dx$$ and divide by the differential. This can be a cleaner way to think as the limit process is left implicit in a way.\n\nIntegral of derivative is the original function: The integral is roughly $$\\int_0^x f'(x)dx = \\sum_{k=1}^n f'(x_k)(x_k-x_{k-1})$$ and the derivative is roughly $$f'(x_k) = \\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}.$$ Combine these and you get a telescoping sum and the desired result.\n\nThis is the way it makes intuitive and graphical sense to me. The link is not immediate; that's why it's a great theorem.\n\nMy answer also comes from physics.\n\nSay p(t) is the position of an object in the time t. For concreteness, suppose you are traking a truck and the truck is going forward on a road from A to B. (i.e., all derivatives are positive)\n\nIt is very natural to graph p(t), and derive it, arriving at v(t), the speed of the truck for each time. After all, the derivative is just taking a small time increment $$dt$$, calculating the corresponding $$ds$$ (a space increment) and dividing\n\nHowever, when I look at the graph of v(t), I can also obtain (almost everything) from p(t). Say I am at the position 100 km in the time t=0. After a $$dt$$, I shall be at $$p(t+dt) = 100+dt\\cdot v(t)$$. And I can do it again and again $$p(t+dt+dt+dt) = 100+dt\\cdot v(t)+dt\\cdot v(t+dt)+dt\\cdot v(t+2\\cdot dt)+dt\\cdot v(t+3\\cdot dt)$$ (notice that after some $$dt$$s, I use the updated speed)\n\nBut that (taking a limit) is an integral!\n\nThe only weird thing is my assuption that I am at position 100km at time t=0. The graph of v(t) cannot tell me that. Intuitively, it does tell me how fast I am going, and therefore can tell me how much how much I walked, but it cannot tell me at which specific part of the road I started. That is that weird constant +c that appears on the indefinite integral.\n\nBut then again, we dont use the indefinite integral much, do we? We are more interested in summing $$dt$$*v(t) for ranges. Say, from t=3 to t=10. That gives me the amount I walked between those two times. So, mercifully, the +c disappears in most use cases: to know how much I walked, there is no need to know where I started.\n\n\u2022 I love how velocity and position helps to cement the relationship. In one class, my students asked why area below the axis should count as negative. We agreed that negative velocity meant going backwards, and therefore decreased your distance from start. (Distance being measured by area under velocity curve, a negative velocity curve has to have \"negative area\" to decrease the total distance.) \u2013\u00a0Sue VanHattum Apr 4 '20 at 20:47\n\nWhen I first began teaching Calculus, I realized that I really didn't understand the Fundamental Theorem. So I looked for something that would help me have a deep understanding, that would also help me help students to see it.\n\nI found a lovely project, which I have modified over the years. Here are links to the pdf and to a .doc version (in which the formulas are messed up, because google docs couldn't read them). If you use this, you'll want to change the part near the end where I reference the textbook.\n\n\u2022 Wow, this is really great. I'll definitely save this. Thanks! \u2013\u00a0Brain Gainz Apr 3 '20 at 0:29\n\nI gave a similar post to this one touching on this on Math.StackExchange.\n\nBasically, the way I would go about it is to say that there is a very easy way by which one can think of at least Riemann integration (the usual definition given in a \"most courses\" calculus course) as an inverse of differentiation by construction: that is, the relationship between the two is not an \"accident\", but design, so that given one, you could fairly easily be led to the other.\n\nAnd to do that, I'd first suggest getting rid of the whole \"slope\" vs \"area\" business altogether - if anything, those are best given as theorems to be proved, after you have other, free-standing definitions of a tangent line and an area which, by the way, can be done, but just too-often aren't, in favor of various hand-waving arguments.\n\nInstead, the relevant idea is change: the derivative of a function $$f$$, i.e. $$f'$$, stands for a kind of \"sensitivity\" measure. Suppose that $$f$$ were like a kind of meter or instrument, with a knob attached to it, and a readout of some kind. The knob attached is the function's input argument, typically denoted $$x$$. The indicator on the readout is the return value, $$f(x)$$. If $$x$$ is set at some value of interest $$x_0$$, then likewise the readout will be at $$f(x_0)$$. Now suppose you \"wiggle\" the knob $$x$$ back and forth a little bit, and you see how the needle $$f(x)$$ responds to that small impulse. For a continuous function, the size of the output's wiggle will be smaller in absolute terms the smaller you make the input, but the proportionate size, i.e. how much it wiggles relative to how much you wiggle the input value, may not be. When we say $$f$$ is differentiable, what that means is that there exists at each point $$x_0$$ a proportionality factor $$f'(x_0)$$ such that\n\n$$f(x \\pm \\underbrace{dx}_\\mbox{\"wiggle\" in x}) \\approx f(x_0) \\pm \\underbrace{[f'(x_0)\\ dx]}_\\mbox{\"wiggle\" in f(x)}$$\n\nso that\n\n$$\\mbox{proportional \"wiggle\"} = \\frac{\\mbox{\"wiggle\" in f(x)}}{\\mbox{\"wiggle\" in x}}$$\n\nso long as the change $$dx$$ is suitably small\n\nIntegration, then, goes the other way. Suppose that I am given now, not the function $$f$$, but only its derivative, $$f'$$, and want to find $$f$$. First off, one should observe that since $$f'$$ only deals with changes in the input, to start, we actually need one more piece of information, and that is some sort of initial value, i.e. $$f(0)$$. Suppose this to be given as well.\n\nStarting at $$f(0)$$, suppose we apply a small, but nonzero, change $$\\Delta x$$ to the input, i.e. we ask, \"given $$f(0)$$, what is $$f(0 + \\Delta x)$$, to the best we can do?\" Well, since we know $$f(0)$$ and $$f'$$, then since $$\\Delta x$$ is a small change or \"wiggle\", we can say that approximately,\n\n$$f(0 + \\Delta x) \\approx f(0) + [f'(0)\\ \\Delta x]$$\n\nwhich is just what you see above.\n\nNow, suppose we take another step of $$\\Delta x$$. We're now going from $$x = 0 + \\Delta x$$ to $$x = (0 + \\Delta x) + \\Delta x$$ (or $$2\\ \\Delta x$$, but I find writing it this way makes it clearer what is going on - \"simpler\" isn't necessarily \"better\"). At this second step, likewise, treating $$0 + \\Delta x$$ as a prior input in and of itself, we have\n\n$$f([0 + \\Delta x] + \\Delta x) \\approx f([0 + \\Delta x]) + [f'([0 + \\Delta x])\\ \\Delta x]$$\n\nwhich, combining with the previous expression, becomes\n\n$$f([0 + \\Delta x] + \\Delta x) \\approx f(0) + [f'(0)\\ \\Delta x] + [f'([0 + \\Delta x])\\ \\Delta x]$$\n\nand it is not hard to then continue this process so that we see for $$N$$ steps,\n\n$$f(0 + N[\\Delta x]) \\approx f(0) + \\sum_{i=0}^{N-1} f'(0 + i[\\Delta x])\\ \\Delta x$$\n\nor, if we set $$x_i := 0 + i[\\Delta x]$$, we can say more neatly as\n\n$$f(0 + N[\\Delta x]) \\approx f(0) + \\sum_{i=0}^{N-1} f'(x_i)\\ \\Delta x$$\n\nand to be more general, if we take $$N$$ steps of suitable, perhaps different, sizes $$\\Delta x_i$$ to reach from $$0$$ some fixed point $$x_0$$,\n\n$$f(x_0) \\approx f(0) + \\sum_{i=0}^{N-1} f'(x_i)\\ \\Delta x_i$$\n\nand then we consider what happens as the steps become arbitrarily fine, at which point we hope - and need to prove - that\n\n$$f(x_0) = f(0) + \\left[\\lim_{||\\Delta|| \\rightarrow 0}\\ \\sum_{i=0}^{N-1} f'(x_i)\\ \\Delta x_i\\right]$$\n\nwhich leads us to define this new operation, given by the limit on the right...\n\n$$\\int_{0}^{x_0} f'(x)\\ dx := \\lim_{||\\Delta|| \\rightarrow 0}\\ \\sum_{i=0}^{N-1} f'(x_i)\\ \\Delta x_i$$\n\n\u2022 This does a great job of explaining how to think of $\\int_a^b f'(x)\\mathrm dx = f(b) - f(a)$, but, in suggesting that we think of it as a definition, I'm left to wonder how to define $\\int_a^b F(x)\\mathrm dx$ without proving some separate theorems about existence of anti-derivatives. \u2013\u00a0LSpice Apr 6 '20 at 0:04\n\nIntuitively, the fundamental theorem of calculus states that \"the total change is the sum of all the little changes\". $$f'(x) dx$$ is a tiny change in the value of $$f$$. We sum up all these little changes to get the total change $$f(b) - f(a)$$.\n\nI elaborated on this explanation here: https://math.stackexchange.com/a/1537836/40119.\n\n\u2022 Very nice! The idea that $\\mathrm dx$ is not punctuation, but (conceptually) an actual infinitesimal measure of $x$-length, so that $f'(x)\\mathrm dx$ has units of (slope)($x$-length) = $y$-length, I think is probably the most important pre-proof part of understanding the connection. That is, I think that most people think of integration as \"something you do to $f'(x)$\", not \"something you do to $f'(x)\\mathrm dx$\"\u2014hence later difficulties with differential forms, and earlier difficulties with changes of variable. \u2013\u00a0LSpice Apr 6 '20 at 0:06\n\nWell, for me, the nicest intuition comes from physics. If $$F=F(x)$$ is some force applied to an object $$O$$ casuing $$O$$ to move, where $$x$$ is its displacement, when the work $$W(x)$$ produced by that force from $$x_0=0$$ to $$x$$ metres is given by:\n\n$$W(x):=\\int_0^xF(s)ds.$$\n\nNow, what if we ask what is the rate of change of that work with respect to displacement? At some certain point $$x$$, the work provided to or substracted by that object $$O$$ is determined by the force $$F$$ applied on it. So, one can intuitively expect that:\n\n$$W'(x)=\\left(\\int_0^xF(s)ds\\right)'=F'(x).$$\n\nThat is, the larger the force, the faster the energy flows from the one applying the force towoards the object. The less the force, the slower that energy flows.\n\nIf you want more clearly \"undoing\" operations, define \"stacking\" as the following:\n\nTake some $$\\Delta x$$. Chop the curve into rectangles of width $$\\Delta x$$ and height $$f(x)$$. (There's some leeway as to take $$f(x)$$ at the left side, right side, middle, minimum, maximum, etc. The rest of my description will be right side.) Now take each rectangle and put the bottom of each rectangle at the top of the previous rectangle. For instance, if you have $$f(x) = x^2$$ and $$\\Delta x = 0.1$$, you'd have a rectangle with lower left corner at the origin, and top left at $$(0.1, 0.01)$$. Then the next rectangle would have lower left corner at $$(0.1, 0.01)$$, and upper right at $$(0.2, 0.05)$$, and so on.\n\nDefine \"unstacking\" as follows:\n\nTake some $$\\Delta x$$. Chop the curve into rectangles of width $$\\Delta x$$ and height $$f(x)$$. Now, take each rectangle and shift it vertically down the height of the previous rectangle. For instance, if $$f(x) = x^2$$, the first rectangle will have height $$0.01$$. The second one will have height $$0.04$$, so for the unstacked version, we move it down $$0.01$$, leaving its new height as $$0.03$$. Another way of phrasing it is for each $$x$$, take the rectangle whose bottom left corner is at $$(x, f(x))$$ and top right corner is $$(x+\\Delta x, f(x+\\Delta x))$$. Then create a chart out of all of those rectangle, with all of them moved down so their bottom is on the x-axis.\n\nFor \"well-behaved\" functions (I believe that uniform continuity is a sufficient condition), as $$\\Delta x$$ goes to zero, stacking becomes integration, and unstacking become differentiation (with the proper scaling).", "date": "2021-01-17 10:24:45", "meta": {"domain": "stackexchange.com", "url": "https://matheducators.stackexchange.com/questions/18120/what-is-the-best-way-to-intuitively-explain-the-relationship-between-the-derivat/18128", "openwebmath_score": 0.829994797706604, "openwebmath_perplexity": 453.48041680721644, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517488441416, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.851489179263845}}
{"url": "https://math.stackexchange.com/questions/1888691/elementary-matrix-operations", "text": "# Elementary matrix operations\n\nI am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.\n\nI am asked to find both the nullspace and the general solution of the following systems $A\\tilde{x}=\\tilde{b}$.\n\nThe first is $$A= \\begin{pmatrix} 1 & 1 & 0 & 2\\\\ 2 & 1 & 1 &-2\\\\ 2 & 2 & 2 & 1 \\end{pmatrix},\\quad \\tilde{b}= \\begin{pmatrix} 12\\\\0\\\\14 \\end{pmatrix}$$ Proceeding with an augmented matrix, I have \\begin{align*} \\left( \\begin{array}{rrrr|c} 1 & 1 & 0 & 2 & 12\\\\ 2 & 1 & 1 &-2 & 0\\\\ 2 & 2 & 2 & 1 & 14 \\end{array} \\right)&\\sim \\left( \\begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\\\ 0 &-1 & 1 &-6 & -24\\\\ 0 & 0 & 2 & -3 &-10 \\end{array} \\right)\\\\ &\\sim \\left( \\begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\\\ 0 & 1 &-1 & 6 & 24\\\\ 0 & 0 & 1 & -3/2 &-5 \\end{array} \\right) \\end{align*} I've checked this multiple times and can't find a mistake, plus it agrees with the solutions my professor has provided.\n\nSolving this now, I have \\begin{align*} x_4&=\\alpha\\in\\mathbb{R}\\\\ x_3&=-5+\\frac{3}{2}\\alpha\\\\ x_2&=19-\\frac{9}{2}\\alpha\\\\ x_1&=-7+\\frac{5}{2}\\alpha \\end{align*}\n\nSo from this, my solution will be $$\\tilde{x}= \\begin{pmatrix} -7+\\frac{5}{2}\\alpha\\\\ 19-\\frac{9}{2}\\alpha\\\\ -5+\\frac{3}{2}\\alpha\\\\ \\alpha \\end{pmatrix} = \\begin{pmatrix} -7\\\\19\\\\-5\\\\0 \\end{pmatrix}+\\beta \\begin{pmatrix} 5\\\\-9\\\\3\\\\2 \\end{pmatrix},\\quad\\beta\\in\\mathbb{R}.$$\n\nThis does not agree with the solutions given, and the solutions given are just solutions without any steps in between so I'm a little confused as to where I went wrong. The weird part is, my nullspace agrees with the solution provided but the particular solution certainly does not.\n\nEdit: My professor's solution is $$\\tilde{x}= \\begin{pmatrix} 3\\\\1\\\\1\\\\4 \\end{pmatrix}+\\alpha \\begin{pmatrix} 5\\\\-9\\\\3\\\\2 \\end{pmatrix},\\quad\\beta\\in\\mathbb{R}.$$\n\nCan somebody please let me know what I've done incorrectly?\n\nThank you.\n\n\u2022 I tried using WolframAlpha, but it didn't tell me what I did wrong, and it spat out something that looked different to both my solution and my professor's (although it did provide the same nullspace!). \u2013\u00a0Babe in the Woods Aug 10 '16 at 23:22\n\u2022 Oh, he went away.. \u2013\u00a0Babe in the Woods Aug 10 '16 at 23:22\n\u2022 Can you provide your professor's solution? \u2013\u00a0Aweygan Aug 10 '16 at 23:25\n\u2022 My apologies, my professor's solution has been added in the question. \u2013\u00a0Babe in the Woods Aug 10 '16 at 23:29\n\u2022 Your professor simply chose to set $x_4=2\\alpha$ instead of $\\alpha$ to avoid denominators in the general solution. \u2013\u00a0Bernard Aug 10 '16 at 23:33\n\nYour solution is correct. You can obtain the professor's solution from yours via the substitution $\\beta=\\alpha+2$. Both are valid answers, and unless there was some given format your answer was supposed to be in, I don't see why yours could be considered incorrect.\nThis is correct! You can check if you found solutions of the system simply by replacing $x$ in $Ax=b$ by what you found.", "date": "2019-11-21 00:45:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1888691/elementary-matrix-operations", "openwebmath_score": 0.8845639228820801, "openwebmath_perplexity": 201.23870661771565, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97805174308637, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8514891742511449}}
{"url": "https://math.stackexchange.com/questions/4041409/counting-ways-when-first-position-is-covered-with-a-tie", "text": "# Counting ways when first position is covered with a tie\n\nSix horses are entered in a race. If two horses are tied for first place and there are no ties among the other four horses, in how many ways can the six horses cross the finish line?\n\nMy approach:\nWe choose $$2$$ out of $$6$$ for the first place and $$4$$ out of the remaining $$4$$:\n$$_6P_2\\cdot _4P_4 = \\frac{6!}{(6 -2)!}\\cdot\\frac{4!}{(4 - 4)!}=720$$\n\nBut the solution states $$360$$.\n\nWhat am I doing wrong here?\n\n\u2022 Also, $\\binom 44= \\frac{4!}{0!4!}$. But you don't want $\\binom 44$ because you care about the order of the remaining $4$ horses. You want the number of ways to permute those $4$ horses, which is simply $4!$. Feb 26 at 23:31\n\u2022 @RobertShore: Yes the order matters that is why I used $\\binom{n}{k}=\\frac{n!}{(n-k)!}$ which is $n\\cdot(n-1)\\cdots\\cdot(n-k+1)$. Also it is still $720$\n\u2013\u00a0Jim\nFeb 26 at 23:35\n\u2022 Your values for the combinations are wrong. ${6 \\choose 2}=\\frac {6!}{4!2!}=15$. You don't want $4 \\choose 4$ because the order of the last four horses matters. That gives $15 \\cdot 4!=360$ Feb 26 at 23:40\n\u2022 When you write $6 \\choose 2$ you are saying that order does not matter. I don't know a simple one for when order does matter. Some use $_6P_2$ for that. Feb 27 at 0:09\n\u2022 There are $\\binom{6}{2}$ to select the subset of two horses which finish in a tie and $4!$ ways for the remaining four horses to finish in the remaining four positions. Notice that the order of selection for the two horses which finish in a tie does not matter, which is why your answer $P(6, 2)P(4, 4)$ is twice the correct answer. Feb 27 at 23:28\n\nThat leaves 4 choices for the horse to come in third, three choices for the horse to come in fourth, two choice for the horse to come in fifth, and one for the horse to come in sixth. The number of ways this can happen is $$15(4)(3)(2)(1)= \\frac{6!}{2}= 720/2= 360$$,", "date": "2021-10-18 18:22:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4041409/counting-ways-when-first-position-is-covered-with-a-tie", "openwebmath_score": 0.6751235723495483, "openwebmath_perplexity": 195.34432067956254, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97805174308637, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8514891709673752}}
{"url": "https://math.stackexchange.com/questions/1858824/find-the-eigenvalues-and-corresponding-eigen-vectors-of-the-matrix", "text": "# Find the eigenvalues and corresponding eigen vectors of the matrix\n\nFind the eigenvalues and corresponding eigen vectors of the matrix $$\\begin{bmatrix}-3&6&-43\\\\0&-1&9\\\\0&0&2\\end{bmatrix}$$\n\nThe eigenvalue $$\\lambda_1 =$$____ corresponds to the eigevector$$( \\ ,\\ , \\ )$$.\n\nThe eigenvalue $$\\lambda_2 =$$____ corresponds to the eigevector$$( \\ ,\\ , \\ )$$.\n\nThe eigenvalue $$\\lambda_3 =$$____ corresponds to the eigevector$$( \\ ,\\ , \\ )$$.\n\nI'm kind of stuck after a certain point. Here is what I have so far\n\nI do know that $$(A - \\lambda I)X = 0$$\n\nso $$\\begin{bmatrix}-3&6&-43\\\\0&-1&9\\\\0&0&2\\end{bmatrix}$$ $$\\implies \\lambda_1 = -3, \\lambda_2 = -1, \\lambda_3 = 2$$ so I have the eigenvalues but how can I find the corresponding eigenvectors?\n\n\u2022 Do you know how to find eigenvalues? This is an upper triangular matrix. What about it's characteristic polynomial? \u2013\u00a0Kushal Bhuyan Jul 14 '16 at 1:40\n\u2022 In any good textbook, class, video lectures or lecture notes on linear algebra, solving linear systems should be covered before finding eigenvalues. In particular, the Gaussian elimination method first puts a linear system into \"upper triangular\" form, and then finds the unknowns starting from the last ones. \u2013\u00a0arctic tern Jul 14 '16 at 3:27\n\nYou need to solve the equations $(A-\\lambda I)v=0$ for $v$ for each of the three eigenvalues $\\lambda$.\n\nFor instance when $\\lambda=2$ we're solving\n\n$$\\begin{pmatrix} -5 & 6 & -43 \\\\ 0 & -3 & 9 \\\\ 0 & 0 & 0\\end{pmatrix}\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}=\\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$\n\nThe last equation is $0=0$ so it's superfluous. So we have two equations in three unknowns. If $z=0$ then the second equation implies $y=0$ and then the first equation implies $x=0$ which gives the zero vector, and that's not very interesting. Otherwise if $z\\ne0$, we can scale the vector $(x,y,z)$ (scaling preserves the property of being a solution to the above system) in order to make $z=1$. In which case the second equation gives $y=3$ and then the first equation gives $x=-5$. So the eigenvector is $(-5,3,1)$ up to scaling.\n\nYour turn. Do $\\lambda=-3$ and $\\lambda=-1$. (Sophia did $\\lambda=-3$ for you, so now most of the work is done for you. If we do the problem for you, at least learn what it is we're doing!)\n\n\u2022 For $\\lambda_2 = -1$ I get $\\begin{bmatrix}-2&6&-43\\\\0&0&9\\\\0&0&3\\end{bmatrix}$ ~ $\\begin{bmatrix}1&-3&\\frac{43}{2}\\\\0&0&9\\\\0&0&0\\end{bmatrix}$ so I thought it would be (-3,0,0) ? \u2013\u00a0Yusha Jul 14 '16 at 3:27\n\u2022 When I multiply that matrix by $(-3,0,0)^T$ I don't get the zero matrix. In that situation, the last equation is redundant. What does the second equation tell you about $z$? What does the first equation then reduce to saying about $x$ and $y$? (BTW I wouldn't bother with the fractions.) \u2013\u00a0arctic tern Jul 14 '16 at 3:30\n\u2022 Wow, I'm an idiot! I see now that $z = 0 \\implies 6y = 2x \\implies (3,1,0)$ \u2013\u00a0Yusha Jul 14 '16 at 3:31\n\nYou start with the understanding of this formula: $(A-\\lambda I)\\vec x=0$, which is equivalent to $\\det(A-\\lambda I)=0$ $$\\begin{vmatrix}-3-\\lambda&6&-43\\\\0&-1-\\lambda&9\\\\0&0&2-\\lambda\\end{vmatrix}=(-3-\\lambda)(-1-\\lambda)(2-\\lambda)=0$$ Therefore, $\\lambda_1=-3, \\ \\lambda_2=-1, \\ \\lambda_3=2$.\nLet's do one example for eigenvectors:\nPlug in the value of $\\lambda$ into the augmented form of the matrix:\nWith $\\lambda_1=-3$, $$\\left[\\begin{array}{ccc|c}-3-(-3)&6&-43&0\\\\0&-1-(-3)&9&0\\\\0&0&2-(-3)&0\\end{array}\\right]=\\left[\\begin{array}{ccc|c}0&6&-43&0\\\\0&2&9&0\\\\0&0&5&0\\end{array}\\right]$$ Solve this matrix and get $v_1=\\begin{bmatrix}1\\\\0\\\\0\\end{bmatrix}$\nNow you can use similar approach to find the eigenvectors of the next two eigenvalues.\n\n\u2022 @Yusha, double check your calculations. At least for $\\lambda=-1$, I did not get $(-3,0,0)$ \u2013\u00a0Ron Jul 14 '16 at 2:18\n\u2022 I'm lost, how are you getting (1,0,0). You have in your work (0,0,0). \u2013\u00a0Yusha Jul 14 '16 at 2:43\n\u2022 Both answers are correct, since one is a scalar multiple of the other. But it is customary to use numbers are small as possible, so $(1,0,0)$ would be prefered. \u2013\u00a0imranfat Jul 14 '16 at 2:47\n\u2022 Does (1,0,0) @imranfat come from the 2nd column after its been put in RREF? \u2013\u00a0Yusha Jul 14 '16 at 2:50\n\u2022 RREF has to give infinite solutions, I looked at Sophia's work, realizing that both vectors serve as the same eigenvector. Your RREF must have all zero's in bottom row \u2013\u00a0imranfat Jul 14 '16 at 2:55\n\nHint: In an upper triangular matrix the characteristic polynomial is $(x_1-\\lambda_1)^{\\alpha_1}(x_2-\\lambda_2)^{\\alpha_2}\\ldots(x_n-\\lambda_n)^{\\alpha_n}$, where $x_i$ are diagonal entries with $\\alpha$'s as their multiplicities.\n\nIn your case the characteristic polynomial is $(3+\\lambda)(1+\\lambda)(2-\\lambda)$\n\n\u2022 Here is one: sosmath.com/matrix/eigen2/eigen2.html \u2013\u00a0imranfat Jul 14 '16 at 2:27\n\u2022 Khan academy usually rocks: khanacademy.org/math/linear-algebra/alternate-bases/\u2026 \u2013\u00a0imranfat Jul 14 '16 at 2:28\n\u2022 OK, so I assume that you DO know how to find eigenvalues theoretically, but somehow your algebra is letting you down? It would be helpful to see your work in attempting finding the eigenvalue, because a lot of times it is a small algebraic mistake... \u2013\u00a0imranfat Jul 14 '16 at 2:33\n\u2022 Here is a worked out example of a 2 by 2 matrix: calvin.edu/~scofield/courses/m256/materials/eigenstuff.pdf In my view you should first be fluent with 2by2's before going for 3by3's... \u2013\u00a0imranfat Jul 14 '16 at 2:34\n\u2022 It sounds like your exchange converged. Anyway, this conversation has been moved to chat. If you need to revisit the full exchange of comments, go to that chatroom. I left the posts with links to resources also here. \u2013\u00a0Jyrki Lahtonen Jul 14 '16 at 6:26", "date": "2021-05-08 23:00:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1858824/find-the-eigenvalues-and-corresponding-eigen-vectors-of-the-matrix", "openwebmath_score": 0.8369834423065186, "openwebmath_perplexity": 365.36656480101476, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517462851321, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.851489168826554}}
{"url": "https://math.stackexchange.com/questions/947026/by-using-de-moivres-theorem-show-that-cos5-theta-16-cos5-theta-20-cos", "text": "# By using De Moivre's Theorem, show that $\\cos5\\theta = 16\\cos^5 \\theta - 20 \\cos^3 \\theta + 5 \\cos \\theta$ [closed]\n\nFirst step is\n\n$$\\cos5\\theta + i \\sin 5\\theta = (\\cos \\theta + i \\sin \\theta)^5$$\n\nthanks\n\n\u2022 Because it's easier than using using other trigonometric formulae. Note that the real parts of the expressions on each side of the equation must be equal. \u2013\u00a0user164587 Sep 26 '14 at 14:39\n\u2022 @user164587 well i meant how to get it? \u2013\u00a0problematic Sep 26 '14 at 14:41\n\u2022 It is de Moivre \u2013\u00a0Seub Sep 26 '14 at 14:41\n\u2022 @vera sorry bad english \u2013\u00a0problematic Sep 26 '14 at 14:41\n\nHint: Expand $(\\cos\\theta + i\\sin\\theta)^5$ using binomial theorem. i.e. $(a+b)^n = \\sum\\limits_{k = 0}^n {n \\choose k}a^k b^{n-k}$\nHere, $a = \\cos\\theta, b=i\\sin\\theta$.\n\n\\begin{align}(\\cos\\theta+i\\sin\\theta)^5 &= \\sum\\limits_{k = 0}^5 {5\\choose k}(\\cos\\theta)^k (i\\sin\\theta)^{5-k}\\\\ &= {5\\choose0}(\\cos\\theta)^0(i\\sin\\theta)^5 + {5\\choose1}(\\cos\\theta)^1(i\\sin\\theta)^4 + {5\\choose2}(\\cos\\theta)^2(i\\sin\\theta)^3\\\\&\\quad+ {5\\choose3}(\\cos\\theta)^3(i\\sin\\theta)^2+{5\\choose4}(\\cos\\theta)^4(i\\sin\\theta)^1+{5\\choose5}(\\cos\\theta)^5(i\\sin\\theta)^0 \\\\&= (i\\sin\\theta)^5 + 5(\\cos\\theta)(i^4\\sin^4\\theta) + 10\\cos^2\\theta(i^3\\sin^3\\theta)\\\\&\\quad+10\\cos^3\\theta(i^2\\sin^2\\theta)+5\\cos^4\\theta(i\\sin\\theta)+\\cos^5\\theta \\\\&=\\cdots \\end{align}\n\nThen, for even powers of $\\sin \\theta$, use $\\sin^2 \\theta = 1-\\cos^2 \\theta$.\n\nFinally, equate the real parts of $\\cos(5\\theta)+i\\sin(5\\theta) = (\\cos\\theta+i\\sin\\theta)^5$.\n\n\u2022 am i doing it right ? lol $(_0^5)$$(cos^0 \\theta)$$(i sin^5 \\theta)$ \u2013\u00a0problematic Sep 26 '14 at 15:03\n\u2022 Yes, and there will be $5$ more terms. \u2013\u00a0taninamdar Sep 26 '14 at 15:05\n\u2022 actually i don't know how binomial theoram works, can you give an example and show me what $(_0^5)$$(cos^0 \\theta)$$(sin^5 \\theta)$ goes to, i will do the rest my self \u2013\u00a0problematic Sep 26 '14 at 15:06\n\u2022 See the edit, ask if you need more help. \u2013\u00a0taninamdar Sep 26 '14 at 15:13\n\u2022 thank you, how to get the first number? i mean it looks like $(_1^5)$ : 5x1, $(_2^5)$ 5x2 , then $(_3^5)$ onward? \u2013\u00a0problematic Sep 26 '14 at 15:22\n\nThe point is that $$\\cos 5 \\theta$$ is the even part of $$\\exp(5 i \\theta) = [\\exp(i \\theta)]^5 = (\\cos \\theta + i \\sin \\theta)^5,$$ where the second equality follows from de Moivre's Theorem. Then, by expanding the right-hand side using the Binomial Theorem, we can immediately write $\\cos 5 \\theta$ as a sum of products of powers of $\\sin \\theta$ and $\\cos \\theta$. Then, we can eliminate even powers of $\\sin \\theta$ using the Pythagorean identity $$\\sin^2 \\theta + \\cos^2 \\theta = 1.$$\n\n\u2022 You're welcome, I hope you found it helpful. \u2013\u00a0Travis Willse Sep 26 '14 at 15:37\n\u2022 yes, you showed me exp(5i\u03b8)=[exp(i\u03b8)]5=(cos\u03b8+isin\u03b8)5, i needed this to know why we have the first step, i can't choose 2 accepted answers... \u2013\u00a0problematic Sep 26 '14 at 15:40\n\u2022 No problem, that's just a part of the site mechanics, the important thing is that you understand the concepts you asked about better than you did before. \u2013\u00a0Travis Willse Sep 26 '14 at 15:51\n\nDe Moivre's Theorem states $$\\cos n\\theta + \\mathbb i \\sin n\\theta = \\left( \\cos \\theta + \\mathbb i \\sin \\theta \\right)^n$$", "date": "2019-12-12 03:43:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/947026/by-using-de-moivres-theorem-show-that-cos5-theta-16-cos5-theta-20-cos", "openwebmath_score": 0.9487189650535583, "openwebmath_perplexity": 708.9100030579492, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517456453797, "lm_q2_score": 0.8705972549785201, "lm_q1q2_score": 0.8514891649858173}}
{"url": "https://brilliant.org/discussions/thread/proof-8/", "text": "\u00d7\n\n# Proof\n\nShow that, for any positive integer $$n\\geq 1$$ it is true that: $\\displaystyle \\sum_{k=1}^{n} \\sqrt{k} \\geq \\sqrt{\\displaystyle \\sum_{k=1}^{n} k}.$\n\nNote by Hjalmar Orellana Soto\n10\u00a0months, 3\u00a0weeks ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nfor $$n\\geq 2$$ it can be easily proven by applying triangle inequality recursively\n\nassume a,b,c are three terms in the seuence\n\nwe have $$\\sqrt{a}+\\sqrt{b}\\geq \\sqrt{a+b}$$ as $$\\sqrt{a},\\sqrt{b},\\sqrt{a+b}$$ form sides of a right triangle\n\nnow we have $$\\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\geq \\sqrt{a+b}+\\sqrt{c}\\geq\\sqrt{a+b+c}$$ and so on....\n\nsum of each of the outer sides of the spiral form the LHS of the equation,and RHS is given by the line to centre\n\nalso note that for n=1 the line to centre is same as side, the only point of equality\n\n- 10\u00a0months, 1\u00a0week ago\n\nvery nice solution, Anirudh\n\n- 10\u00a0months, 1\u00a0week ago\n\nthank you :)\n\n- 10\u00a0months, 1\u00a0week ago\n\n$$LHS^2 = 1+2+\\cdots+k + 2\\sum_{1\\le x<y\\le n}\\sqrt{xy}\\ge 1+2+\\cdots+n = RHS^2$$ and equallity occurs when $$n=1$$\n\n- 10\u00a0months, 3\u00a0weeks ago\n\nIn addition to the problem.... Is this necessary information for saying that the limit below converges? $\\lim_{x\\to \\infty} \\dfrac{\\displaystyle \\sqrt{\\sum_{n=1}^{x}n}}{\\displaystyle \\sum_{n=1}^{x}\\sqrt{n}}$\n\n- 10\u00a0months, 3\u00a0weeks ago\n\n(Assuming I'm interpreting you correctly,) the essence of what you're asking here is:\n\nIf $$a_n, b_n$$ are sequences of (positive) reals, such that $$a_n \\geq b_n$$, does $$\\lim \\frac{ b_n } { a_n }$$ exist?\nIf yes, why?\nIf no, what is a counter example? What other assumptions do we need to add to guarantee a true statement?\n\nStaff - 10\u00a0months, 3\u00a0weeks ago\n\nExactly that is what I ask, I was thinking of asking for the value of $\\displaystyle \\sum_{n=1}^{\\infty} (\\frac{\\sqrt{\\sum_{k=1}^{n}n}}{\\sum_{k=1}^{n}\\sqrt{n}})$ but I really don't know how to approach it\n\n- 10\u00a0months, 3\u00a0weeks ago\n\nThe answer to \"does $$\\lim \\frac{ b_n } { a_n }$$ exist?\" is no for general sequences. Do you see an obvious counter example?\n\nFor your recent comment, are you looking for the summation, or for the limit of the term (as in the prior comment)?\n\nStaff - 10\u00a0months, 3\u00a0weeks ago\n\nThe answer to the limit question is yes for this case, using the same argument which is used for $$\\lim_{x\\to \\infty} \\frac{x}{e^x}$$.... and I'm looking for the summation\n\n- 10\u00a0months, 3\u00a0weeks ago\n\nSee How to ask for help to understand how to provide proper context of what you want. E.g. You can see that people quickly established the inequality, but that doesn't help with your concern.\n\nStaff - 10\u00a0months, 3\u00a0weeks ago\n\nThe summation is something else, something I supposed and now I'm trying to approach, and I think the inequality may help for knowing if the summation converges... and then I'll try to approach the value of the summation, I don't know if I'm explaining right...\n\n- 10\u00a0months, 3\u00a0weeks ago\n\nAsymptotically, $$\\sqrt{1} + \\sqrt{2} + \\dots + \\sqrt{n}$$ can be approximated by the integral $\\int_0^n \\sqrt{x} \\ dx = \\frac{2}{3} n^{3/2}.$ And of course, $\\sqrt{1 + 2 + \\dots + n} = \\sqrt{\\frac{n(n + 1)}{2}}.$ That should help answer your question.\n\n- 10\u00a0months, 3\u00a0weeks ago\n\nWe want to prove that $\\sqrt{1} + \\sqrt{2} + \\dots + \\sqrt{n} \\ge \\sqrt{1 + 2 + \\dots + n}.$ Squaring the left-hand side, we get a term of $$\\sqrt{k} \\cdot \\sqrt{k} = k$$ for each $$1 \\le k \\le n$$. The inequality follows.\n\n- 10\u00a0months, 3\u00a0weeks ago\n\n\u00d7", "date": "2017-11-24 03:46:38", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/proof-8/", "openwebmath_score": 0.9880246520042419, "openwebmath_perplexity": 1518.4910968080783, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9883127440697254, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8514691953972776}}
{"url": "https://math.stackexchange.com/questions/3064314/on-the-number-of-roots-of-the-polynomial-x3ax21-0/3064353", "text": "# On the number of roots of the polynomial $x^3+Ax^2+1=0$\n\nI have the following cubic equation\n\n$$x^{3}+Ax^{2}+1=0$$\n\nwhere $$A$$ is an arbitrary (real) number.\n\nI know that either:\n\n\u2022 The 3 roots will be real.\n\u2022 One root will be real and the other two will be complex conjugates of each other.\n\nI would like to find out\n\n\u2022 For what value/values of A the roots change from 3 real roots to one real and two complex roots.\n\u2022 The signs of each of the real roots (both when they are all real and when there is only one real root)\n\nIs there an analytical way of finding this as a function of $$A$$ or the only option is to solve the cubic numerically?\n\n\u2022 Google Cardano's method. \u2013\u00a0hamam_Abdallah Jan 6 at 19:44\n\nIf we can show that the value of $$A$$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $$y$$-location and the other at a zero or negative $$y$$-location, then we know that this value of $$A$$ is the border for all real solutions versus 1 real and 2 complex solutions.\n\nDenote $$f(x) = x^3 + Ax^2 + 1.$$\n\n$$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$\n\nWe see that $$f(0)$$ is always either a local maximum or minimum and $$f(0) > 0.$$ The other zero is given by $$x = -2A/3 = \\xi.$$\n\nWe need $$f(\\xi) \\leq 0.$$\n\n$$f(\\xi) = \\frac{4}{27}A^3 + 1 \\leq 0$$\n\nsolves for the values of $$A$$ where 3 real solutions are guaranteed.\n\nIf there are 3 real roots, then since $$x = 0$$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $$A \\leq -3/\\sqrt[3]{4}$$ and the location of the moving extreme is $$x=-2A/3 > 0,$$ then the remaining real root has to be positive.\n\nIf there's only 1 real root, then $$A > -3/\\sqrt[3]{4}$$ and the moving extreme is located at $$x < \\sqrt[3]{2}.$$ Thus the only real root must be negative.\n\n\u2022 How do we find the signs of the real roots? By actually finding them or is there a shortcut? \u2013\u00a0JennyToy Jan 6 at 20:21\n\u2022 @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit? \u2013\u00a0AEngineer Jan 6 at 20:45\n\nhint\n\nPut $$A=-\\frac 32B$$ and $$f(x)=x^3-\\frac 32Bx^2+1.$$\n\n$$f'(x)=3x(x-B).$$\n\n$$f(0)=1$$ to have three real roots, we need\n\n$$B>0 \\text{ and } f(B)<0.$$\n\n\u2022 To find when there are 3 real roots or one only use the sign of the discriminant $\\Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots. \u2013\u00a0user2175783 Jan 6 at 20:13\n\u2022 @user2175783 If $A=-\\frac{3}{2^{2/3}}$, then there are a negative zero and a double zero on the positive axis. \u2013\u00a0Hanno Jan 12 at 22:17\n\nThe stationary points of $$f(x)$$ lie at $$x=0$$ and at $$x=-\\frac{2A}{3}$$ and the sign of $$f(0)f\\left(-\\tfrac{2A}{3}\\right) = 1+\\tfrac{4}{27}A^3$$ decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $$f(0)f\\left(-\\tfrac{2A}{3}\\right)$$ only depends on the stationary values having the same sign or not.\n\nToo late at the party. But it looks good, so let's enter anyway.\n\nTL;DR $$\\quad f(x)=x^3+Ax^2+1\\,$$ always has a simple zero on the negative axis. The other two zeros are\n\n\u2022 real, positive and distinct, if $$\\,A<-1.88988\\:=\\:-\\dfrac{3}{\\sqrt[3]{4}}\\:=\\:A_1$$\n\u2022 merging into a double zero at $$x=\\sqrt[3]{2}$$, if $$\\,A=A_1$$\n\u2022 non-real (thus complex-conjugate), if $$\\,A>A_1$$\n\nThe party-giver also asked if there is an analytical way of finding this as a function of $$A$$?\nLet's take a step in this direction, concentrating on the case of \"one real zero\", and express the negative zero as a function of $$A$$:\nLet $$\\,d=\\big(\\frac A3\\big)^3 + \\big(\\frac 12\\big)^2$$, and assume $$d>0$$ which corresponds to $$\\,A>A_1$$. Define $$r \\:=\\:\\left(-\\frac 12 -\\sqrt d\\right)^\\frac 13 +\\, \\left(-\\frac 12 +\\sqrt d\\right)^\\frac 13\\,,$$ $$r\\,$$ is observed to be strictly negative. Then \\begin{align} r^2\\:= & \\;\\left(-\\frac 12 -\\sqrt d\\right)^\\frac 23 +\\, \\left(-\\frac 12 +\\sqrt d\\right)^\\frac 23 -\\frac23 A \\\\[2ex] r^3\\:= & \\; -\\frac23 Ar + \\left(-\\frac 12 -\\sqrt d\\right) +\\, \\left(-\\frac 12 +\\sqrt d\\right) \\\\ & \\quad -\\frac13 A \\left\\{ \\left(-\\frac 12 -\\sqrt d\\right)^\\frac 13 +\\, \\left(-\\frac 12 +\\sqrt d\\right)^\\frac 13\\right\\} \\\\ = & -Ar - 1 \\end{align} Whence $$1+Ar+r^3=0\\quad\\Longleftrightarrow\\quad\\frac1{r^3}+A\\frac1{r^2}+1 =0 \\\\[2ex] \\Longrightarrow\\quad x \\:=\\: \\frac1r \\:=\\: -r^2-A \\:=\\: -\\frac13 A -\\left(\\sqrt d + \\frac 12\\right)^\\frac 23 -\\, \\left(\\sqrt d -\\frac12\\right)^\\frac 23$$ satisfies $$x^3+Ax^2+1=0$$.\n\nAdded in Edit: Driven by user2175783's comment as of below, I found Lecture 4 in Mathematical Omnibus: 30 Lectures on Classic Maths by D. Fuchs + S. Tabachnikov: It gives an extensive and readable account of how to explicitly solve the cubic and quartic equations.\n\nVisual TL;DR\n\n\u2022 Wouldnt it be possible to generalize the algebra for the negative root for $A<A_{1}$? \u2013\u00a0user2175783 Jan 13 at 0:14\n\u2022 @user2175783 I did not look closer at that case. I'd say there's no ambiguity for the sign of $\\sqrt d$ because $r$ is a symmetrical expression. But afterwards complex cube roots have to be considered which might involve some choice. \u2013\u00a0Hanno Jan 13 at 9:31\n\nThe discriminant of a cubic polynomial with real coefficients is $$0$$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $$-4 A^3 - 27$$.", "date": "2019-07-16 16:29:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3064314/on-the-number-of-roots-of-the-polynomial-x3ax21-0/3064353", "openwebmath_score": 0.8927750587463379, "openwebmath_perplexity": 370.22600157662055, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9883127416600689, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.85146919156448}}
{"url": "https://mathhelpboards.com/threads/find-all-triplets-x-y-z.6923/", "text": "# Find all triplets (x,y,z)\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nFind all triplets of positive integers $(x, y, z)$ such that $$\\displaystyle \\left( 1+\\frac{1}{x} \\right)\\left( 1+\\frac{1}{y} \\right)\\left( 1+\\frac{1}{z} \\right)=2$$.\n\nLast edited:\n\n##### Well-known member\nRe: Find all triplets of x, y and z\n\nFind all triplets of positive integers $(x, y, z)$ such that $$\\displaystyle \\left( 1+\\frac{1}{x} \\right)\\left( 1+\\frac{1}{y} \\right)\\left( 1+\\frac{1}{z} \\right)=2$$.\n(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them\n\nsolved as\n\nWithout loss of generality we can choose x <=y <=z and ans shall be a permutation of if\n\nNow x < 4 as (5/4)^3 < 2 ( it is 125/64)\n\nx cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution\n\nso x = 2 or 3\n\nif x = 2 we get\n\n3/2(y+1)(z+1) = 2yz\n\nOr 3(y+1)(z+1) = 4 yz\n\nOr yz \u2013 3y \u2013 3z = 3\n(y-3)(z-3) = 12 by adding 9 on both sides\n\n( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4\n\nSo (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)\n\nif x = 3 we get\n\n4/3(y+1)(z+1) = 2yz\n\nOr 2(y+1)(z+1) = 3 yz\n\nOr yz \u2013 2y \u2013 2z = 2\n(y-2)(z-2) = 6\n\n( y-2)(z-2) = 1 * 6 or 2 * 3\n\nSo (x,y,z ) = (3,3,8) , ( 3,4,5)\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nRe: Find all triplets of x, y and z\n\n(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them\n\nsolved as\n\nWithout loss of generality we can choose x <=y <=z and ans shall be a permutation of if\n\nNow x < 4 as (5/4)^3 < 2 ( it is 125/64)\n\nx cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution\n\nso x = 2 or 3\n\nif x = 2 we get\n\n3/2(y+1)(z+1) = 2yz\n\nOr 3(y+1)(z+1) = 4 yz\n\nOr yz \u2013 3y \u2013 3z = 3\n(y-3)(z-3) = 12 by adding 9 on both sides\n\n( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4\n\nSo (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)\n\nif x = 3 we get\n\n4/3(y+1)(z+1) = 2yz\n\nOr 2(y+1)(z+1) = 3 yz\n\nOr yz \u2013 2y \u2013 2z = 2\n(y-2)(z-2) = 6\n\n( y-2)(z-2) = 1 * 6 or 2 * 3\n\nSo (x,y,z ) = (3,3,8) , ( 3,4,5)\n\nWow! I forgot to thank you explicitly for participating and also for your unique way of attacking the problem. What's worst is that I forgot completely how I approached it two months ago and I promise you to add my reply once I solved it because I could only tell at this point that I solved it differently than what you did. Sorry, kaliprasad!", "date": "2020-12-05 14:59:12", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-all-triplets-x-y-z.6923/", "openwebmath_score": 0.6116147637367249, "openwebmath_perplexity": 1590.4791696086988, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127430370154, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8514691909939855}}
{"url": "https://mathhelpboards.com/threads/2-coins-are-thrown-20-times.24749/", "text": "# [SOLVED]2 coins are thrown 20 times\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nHey!!\n\n2 coins are thrown 20 times. I want to calculate the probability\n(a) to achieve exactly 5 times the Tail/Tail\n(b) to achieve at least 2 times Tail/Tail\n\nIf we throw the 2 coins once the probability that we get Tail/Tail is equal to $\\frac{1}{2}\\cdot \\frac{1}{2}=\\frac{1}{4}$, or not?\nIs the probability then at (a) equal to $\\left (\\frac{1}{4}\\right )^{20}$ ?\n\nCould you give me a hint for (b) ?\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nHey!!\n\n2 coins are thrown 20 times. I want to calculate the probability\n(a) to achieve exactly 5 times the Tail/Tail\n(b) to achieve at least 2 times Tail/Tail\n\nIf we throw the 2 coins once the probability that we get Tail/Tail is equal to $\\frac{1}{2}\\cdot \\frac{1}{2}=\\frac{1}{4}$, or not?\nHey mathmari !!\n\nYep.\n\nIs the probability then at (a) equal to $\\left (\\frac{1}{4}\\right )^{20}$ ?\nThere seems to be a $5$ missing.\nIt's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.\n\nCould you give me a hint for (b) ?\nTake the complement?\nIt's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.\n\n#### mathmari\n\n##### Well-known member\nMHB Site Helper\nThere seems to be a $5$ missing.\nIt's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.\nOh yes.. So do we have the following? $$\\binom{20}{5}\\cdot \\left (\\frac{1}{4}\\right )^5\\cdot \\left (1-\\frac{1}{4}\\right )^{20-5}$$\n\nTake the complement?\nIt's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.\nSo, we have $$P(X\\geq 2)=1-P(X<2)=1-P(X=1)-P(X=0)=1-\\binom{20}{1}\\cdot \\left (\\frac{1}{4}\\right )^1\\cdot \\left (1-\\frac{1}{4}\\right )^{20-1}-\\binom{20}{0}\\cdot \\left (\\frac{1}{4}\\right )^0\\cdot \\left (1-\\frac{1}{4}\\right )^{20-0}$$ or not?\n\n#### Klaas van Aarsen\n\n##### MHB Seeker\nStaff member\nYep. All correct.\n\n#### Wilmer\n\n##### In Memoriam\nSo do we have the following? $$\\binom{20}{5}\\cdot \\left (\\frac{1}{4}\\right )^5\\cdot \\left (1-\\frac{1}{4}\\right )^{20-5}$$\nLooks good.\nRan a simulation: got 202276 out of 1 million.\n\nMHB Site Helper\nThank you!!", "date": "2020-07-05 06:29:35", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/2-coins-are-thrown-20-times.24749/", "openwebmath_score": 0.8078370094299316, "openwebmath_perplexity": 1876.8832906306018, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.988312744413962, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8514691886667045}}
{"url": "http://mathhelpforum.com/math-challenge-problems/229258-problem-8-number-theory.html", "text": "# Thread: Problem #8 - Number Theory\n\n1. ## Problem #8 - Number Theory\n\nI got this one from our archives as well.\n\nShow that $n^4 + 4^n$ is not prime for $n \\geq 2$. (n is a positive integer!)\n\nThis is a more of a Number Theory problem than algebraic. As my proof relies more on a College Algebra level rather than Number Theory I'm curious to see what you might come up with.\n\n-Dan\n\n2. ## Re: Problem #8 - Number Theory\n\nall n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values\n\n3. ## Re: Problem #8 - Number Theory\n\nOriginally Posted by muddywaters\nall n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values\nif n is an odd integer >= 3 and n is not a multiple of 5 then n^4 + 4^n is a multiple of 5 and therefore not a prime\nProof:\nn is 1,2,3,or 4 (mod 5)\nn^4 is 1 (mod 5)\n4^n is -1 (mod 5)\nSo n^4 + 4^n is congruent to 0 (mod 5)\n\nHow do we handle the case n=5, 15, 25, 35, 45, ...... ?\n\n4. ## Re: Problem #8 - Number Theory\n\nIf $n\\equiv 0\\pmod{2}$, then $n^4+4^n \\equiv 0\\pmod{2}$.\n\nIf $n\\equiv 1\\pmod{2}$, then $n+1\\equiv 0\\pmod{2}$, so $2^{n+1}$ is a perfect square, and $n^4+4^n$ factors as\n$(n^2+n\\sqrt{2^{n+1}}+2^n)(n^2-n\\sqrt{2^{n+1}} + 2^n)$\n\nSo, all that is left is to show that for odd $n>2$, both terms are greater than 1. Since $n^2+2^n>1$, the first term is guaranteed to be greater than 1. So, we just need to show the second term is greater than one.\n\nSince $(n^2-1)^2+2^{2n} \\ge 2^{2n} > 2^{n+1}$ (for all $n>1$), we have\n\n\\begin{align*}n^4-2n^2+1+4^n-2^{n+1} > 0 & \\Rightarrow n^4-2n^2+1+4^n-2^{n+1}+n^2 2^{n+1} > n^2 2^{n+1} \\\\ & \\Rightarrow (n^2+2^n-1)^2 > n^2 2^{n+1} \\\\ & \\Rightarrow n^2 + 2^n - 1 > n\\sqrt{2^{n+1}} \\\\ & \\Rightarrow n^2-n\\sqrt{2^{n+1}}+2^n > 1\\end{align*}\n\n5. ## Re: Problem #8 - Number Theory\n\nIf n even, obvious.\nIf n odd, 2n+1 is a perfect square and the factorization are really neat observations. Thanks.\n\nBut what's the point of the congruences?\n\n6. ## Re: Problem #8 - Number Theory\n\nOriginally Posted by Hartlw\nBut what's the point of the congruences?\n$n\\equiv 0\\pmod{2}$ is the same as saying $n$ is even.\n$n\\equiv 1\\pmod{2}$ is the same as saying $n$ is odd.\n\nI have been working with modular arithmetic long enough that it feels more natural for me to use congruences rather than the words even and odd. The meaning is the same either way.\n\n7. ## Re: Problem #8 - Number Theory\n\nYou have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.\n\n8. ## Re: Problem #8 - Number Theory\n\nOriginally Posted by Hartlw\nYou have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.\nIf $n=1$, then the factorization I gave is $1^4+4^1 = (5)(1)$. However, 5 is a prime number. The problem asks to prove that for $n\\ge 2$, $n^4+4^n$ is NOT prime. So, to do that, it is not enough to show the factors are positive. You must show they are both greater than 1.\n\n9. ## Re: Problem #8 - Number Theory\n\nHey, lots of action on this one! Thanks to all.\n\nSlipEternal gets some extra kudos for checking the \"trivial\" factorization problem. (That one of the factors might be 1.) I did eventually find the proof online (which was pretty much identical to Slip's and my own) but didn't address the trivial factorization problem.\n\n-Dan", "date": "2016-09-25 20:51:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/math-challenge-problems/229258-problem-8-number-theory.html", "openwebmath_score": 0.9524224996566772, "openwebmath_perplexity": 545.1168075824662, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9883127444139621, "lm_q2_score": 0.8615382040983516, "lm_q1q2_score": 0.851469186909918}}
{"url": "https://stats.stackexchange.com/questions/435458/what-is-the-probability-of-getting-three-pairs-that-makes-up-seven-when-six-dice", "text": "What is the probability of getting three pairs that makes up seven when six dice are rolled?\n\nStumbled upon this question in a game of dice where a point is dealt for each pair that makes seven. In the game there were a total of six dice that were to be thrown up to three times. Each time a pair of dice made seven 1+6, 2+5 or 4+3), the pair(s) was taken out from the rest, a point was dealt and you got to roll the remaining dice. You can not use dice from the excluded pairs to make up new ones. You win the game by getting the most points/pairs in 3 rounds.\n\nExample:\n\nround one [1] [1] [1] [3] [5] [6] - one pair make 7 (1+6)\n\nround two [1] [3] [4] [2] - one pair make 7 (3+4)\n\nround one [2] [6] - no pairs make 7\n\npoints awarded: 2\n\nOut of curiosity I have been trying to calculate the probabilities of getting three pairs of sevens in the first round?\n\n\u2022 We have six dice and we throw them all together in the first round. Let's say the faces are $3,4,5,2,6,6$. Because there are two pairs that make $7$, we remove the pairs and continue with two dice. Round 2 and 3 continues this way. What is the definition of winning? Is it the situation that we're left with no dice at the end? \u2013\u00a0gunes Nov 10 '19 at 18:18\n\u2022 and how is it possible to roll three dice following those rules? \u2013\u00a0carlo Nov 10 '19 at 18:39\n\u2022 Your question is unclear. Your title refers to three dice while your body text refers to six. You describe a game where the number of dice change, altering the probabilities. At which point in that game are you calculating the chance of getting two dice that add to 7? \u2013\u00a0Glen_b Nov 10 '19 at 21:02\n\u2022 Sorry if the original edit was confusing, I have tried to clarify. The game description is just for context. The question is in the title and reformulated in the last paragraph. \u2013\u00a0joho Nov 13 '19 at 14:28\n\nConsider a sequence of $$2k$$ dice each with the possible values $$1, 2, \\ldots, 2n.$$ (The question concerns $$k=n=3.$$) The possible pairs are $$\\{1,n\\},$$ $$\\{2,2n-1\\},$$ and so on, through $$\\{n,n+1\\}.$$ Denote such a pair by its smallest value $$i$$ and for each $$i$$ let $$k_i\\gt 0$$ be the number of such pairs that can be located in the sequence. We need to count the number of equally probable sequences for which $$k_1+k_2+\\cdots + k_n = k.$$\n\nSuch a sequence is determined by (a) which pairs occur in it and (b) where in the sequence of $$2k$$ values each such pair occurs. The permutation group on $$2k$$ elements acts on the set of such sequences. For all $$i,$$ the stabilizer of such a sequence permutes the $$k_i$$ values of $$i$$ among themselves and the $$k_i$$ values of $$2n+1-i$$ among themselves. Thus, applying the method described at https://stats.stackexchange.com/a/415878/919, the number of ways of producing a sequence designated by $$\\mathrm{k}=(k_1,k_2,\\ldots,k_n)$$ is\n\n$$p(\\mathrm{k}) = \\frac{(2k)!}{(k_1!)^2(k_2!)^2\\cdots(k_n!)^2}.$$\n\nThus, the chance is obtained by summing $$p(\\mathrm{k})$$ over all possible $$\\mathrm{k}$$ whose components sum to $$k$$ and dividing by the total number of sequences, $$(2n)^{2k}.$$\n\nThese possibilities correspond to the weak compositions of $$k$$ into $$n$$ parts, which number $$\\binom{k+n-1}{n-1}.$$ However, the amount of calculation is smaller than this, because $$p(\\mathrm{k})$$ does not depend on the order of the $$k_i.$$ We may therefore do the calculation for all $$k_1\\ge k_2\\ge \\cdots \\ge k_n \\ge 0$$ (giving a partition of $$k$$), multiplying each by the number of distinct re-orderings of $$\\mathrm k.$$ Such sequences correspond to the Ferrers diagrams for $$n$$ having at most $$k$$ rows. They are relatively easy to enumerate.\n\nWith $$k=n=3,$$ we have $$\\binom{3+3-1}{3-1}=10$$ possibilities for $$\\mathrm k,$$ but they fall into just three groups corresponding to the partitions $$3 = 2+1 = 1+1+1:$$\n\n$$p(3,0,0)=p(0,3,0)=p(0,0,3) = \\frac{6!}{3!^2} = 20;$$ $$p(2,1,0)=p(2,0,1)=p(1,2,0)=p(1,0,2)=p(0,2,1)=p(0,1,2) = \\frac{6!}{2!^21!^2}= 180;$$ $$p(1,1,1) = \\frac{6!}{1!^21!^21!^2} = 720.$$\n\n$$\\Pr(\\text{three pairs}) = \\frac{3\\times 20 + 6\\times 180 + 1\\times 720}{6^6} = \\frac{(5)(31)}{(2^4)(3^5)} \\approx 3.9866\\%.$$\n\u2022 @Ron I haven't thoroughly checked my general answer, but I did check the specific answer with a brute force enumeration of all possibilities. At least one mistake in your post is that many sequences without three pairs will sum to 21, such as 5,5,5,4,1,1, which has no pairs at all. Thus you substantially overcount. Here is my implementation: n <- 3; k <- 3; X <- expand.grid(rep(list(seq_len(2*n)), 2*k)); p <- function(x, d=n) { x <- sort(x); sum((x + rev(x))[seq_len(d)] == 2*d+1) }; table(apply(X, 1, p)) \u2013\u00a0whuber Nov 13 '19 at 19:04\n\u2022 @Ron If you would like to explore the situation in more detail, this code (which follows the creation of data frame X in my previous comment) breaks the population down into all possible pair patterns: pattern <- function(x, d=n) { x <- sort(x); paste0(x[which((x + rev(x))[seq_len(d)] == 2*d + 1)], collapse=\"\") }; table(apply(X, 1, pattern)) \u2013\u00a0whuber Nov 13 '19 at 19:19", "date": "2020-08-06 16:14:22", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/435458/what-is-the-probability-of-getting-three-pairs-that-makes-up-seven-when-six-dice", "openwebmath_score": 0.6652854084968567, "openwebmath_perplexity": 338.76809920394766, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127409715955, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8514691857009742}}
{"url": "http://math.stackexchange.com/questions/387218/choosing-a-linear-map-mathbbz-2-mathbbzn-rightarrow-mathbbz-2-mathb", "text": "# Choosing a linear map $(\\mathbb{Z}/2\\mathbb{Z})^n \\rightarrow \\mathbb{Z}/2\\mathbb{Z}$ which is nonzero on half of a sequence of vectors\n\nLet $v_1,\\ldots,v_m \\in (\\mathbb{Z}/2\\mathbb{Z})^n$ be nonzero vectors. Is it always possible to choose a linear map $f : (\\mathbb{Z}/2\\mathbb{Z})^n \\rightarrow \\mathbb{Z}/2\\mathbb{Z}$ such that $f$ is nonzero on at least half of the $v_i$, i.e. such that\n\n$$|\\{\\text{i | 1 \\leq i \\leq m and f(v_i) \\neq 0}\\}| \\geq \\frac{1}{2}m?$$\n\nMy guess is that the answer is yes; at the very least, it is true for $m=1$ and when the $v_i$ enumerate all of the nonzero vectors.\n\n-\nWhy don't you write in words what you want in the body of the question? Desciphering an equation when a perfectly good English sentence might be considerably easier to understand is silly! \u2013\u00a0 Mariano Su\u00e1rez-Alvarez May 10 '13 at 3:45\n@MarianoSu\u00e1rez-Alvarez : Sure thing. But isn't that a little pedantic? The equation isn't all that complicated... \u2013\u00a0 Monica May 10 '13 at 3:47\nI don't disagree with @Mariano's comment, but let me just add that this is a very nice question. Where did it come from? \u2013\u00a0 Pete L. Clark May 10 '13 at 13:33\nAnd I am someone who generally finds contest type problems quite boring (which is not to say that I know how to solve them). Looking back at the comments, Monica got a less warm welcome than she should have: $\\mathbb{Z}/2$ could only mean one thing (unlike $\\mathbb{Z}_2$!) and upon reflection I do disagree with Mariano's comment. I'm glad it turned out well in the end. \u2013\u00a0 Pete L. Clark May 10 '13 at 17:10\n@PeteL.Clark : Thanks! This experience definitely provides me another data point concerning using gendered names on the internet (I usually don't). It also appears that one can order the Babai book from U of C; see cs.uchicago.edu/research/publications/combinatorics \u2013\u00a0 Monica May 10 '13 at 17:25\n\nThe statement is true and let us prove it by induction on $n$.\n\nDenote $\\mathbb{Z}/2\\mathbb{Z}:=\\mathbb{Z}_2:=\\{0,1\\}$. When $n=1$, the statement is clearly true. Now assume that when $n\\le n_0$, the statement is true for every $m\\ge 1$, and let us show the statement is true for $n:=n_0+1$ and every $m\\ge 1$.\n\nConsider $(\\mathbb{Z}_2)^n$ as $(\\mathbb{Z}_2)^{n-1}\\times\\mathbb{Z}_2$ and denote by $P$ and $Q$ the projections of $(\\mathbb{Z}_2)^n$ to its first $n-1$ coordinates and its last coordinate respectively, i.e. $$P: (\\mathbb{Z}_2)^n\\mapsto (\\mathbb{Z}_2)^{n-1},\\quad(a_1,\\dots,a_{n-1},a_n)\\mapsto (a_1,\\dots,a_{n-1}),$$ and $$Q: (\\mathbb{Z}_2)^n\\mapsto \\mathbb{Z}_2,\\quad(a_1,\\dots,a_n)\\mapsto a_n.$$\n\nUp to a rearrangement of $v_1,\\dots,v_m$, we may assume that there exists $0\\le m'\\le m$, such that $Qv_i\\ne 0$ if and only if $m'<i\\le m$. If $m'=0$, simply choose $f=Q$ and we are done, so let us assume that $m'\\ge 1$. By definition, $Pv_1,\\dots,Pv_{m'}\\ne 0$. By the induction assumption on $n$, there exists a linear function $g:(\\mathbb{Z}_2)^{n-1}\\to \\mathbb{Z}_2$, such that $$|\\{ 1\\le i\\le m' \\mid g(P v_i) \\neq 0\\}| \\ge \\frac{1}{2}m'.\\tag{1}$$ For $j=0,1$, define $$f_j:(\\mathbb{Z}_2)^n\\to \\mathbb{Z}_2,\\quad v\\mapsto g(Pv)+j\\cdot Qv.\\tag{2}$$ By definition, for $j=0,1$, $f_j$ is linear and $f_j(v_i)=g(P v_i)$ when $1\\le m\\le m'$. Then by $(1)$, for both $j=0$ and $j=1$, $$|\\{ 1\\le i\\le m' \\mid f_j(v_i) \\neq 0\\}| \\ge \\frac{1}{2}m'.\\tag{3}$$ If $m'=m$, then we are done. Otherwise, note that by $(2)$ and the choice of $m'$, for $j=0,1$, $$f_j(v_i)=0 \\iff g(Pv_i)=1-j\\iff f_{1-j}(v_i)=1,\\quad \\forall\\, m'<i\\le m.$$ It follows that $$|\\{ m'< i\\le m \\mid f_0(v_i) \\neq 0\\}|+|\\{ m'< i\\le m \\mid f_1(v_i) \\neq 0\\}|=m-m'.\\tag{4}$$ Combining $(3)$ and $(4)$, we can conclude that the statement is true for either $f=f_0$ or $f=f_1$, which completes the proof.\n\nRemark: Thanks to PeteL.Clark's nice question in the comment, I realized that the conclusion can be generalized to arbitrary finite field with the same argument as above.\n\nClaim: Let $\\mathbb{F}_q$ be the finite field of $q$ elements and let $v_1,\\dots,v_m$ be nonzero vectors in $(\\mathbb{F}_q)^n$. Then there exists a linear function $f:(\\mathbb{F}_q)^n\\to\\mathbb{F}_q$, such that $$|\\{1\\le i\\le m\\mid f(v_i)\\ne 0\\}|\\ge\\frac{(q-1)m}{q}.\\tag{5}$$\n\nSketch of Proof: Define projections $P:(\\mathbb{F}_q)^n\\to (\\mathbb{F}_q)^{n-1}$, $Q:(\\mathbb{F}_q)^n\\to\\mathbb{F}_q$ similarly. Define $m'$ and rearrange $v_1,\\dots,v_m$ similarly. By induction, we may assume that there exists a linear function $g:(\\mathbb{F}_q)^{n-1}\\to\\mathbb{F}_q$, such that for $f_j=g\\circ P+j\\cdot Q$, $0\\le j<q$, $$|\\{1\\le i\\le m'\\mid f_j(v_i)\\ne 0\\}|\\ge\\frac{(q-1)m'}{q}.\\tag{6}$$ By definition of $m'$ and $f_j$, for every $m'<i\\le m$, $f_0(v_i),\\dots, f_{q-1}(v_i)$ are pairwise different, so one and only one of them is $0$. It follows that for some $0\\le j<q$, $$|\\{m'< i\\le m\\mid f_j(v_i)\\ne 0\\}|\\ge\\frac{(q-1)(m-m')}{q}.\\tag{7}$$ $(5)$ follows from $(6)$ and $(7)$ for $f=f_j$.\n\n-\nThis is just lovely. \u2013\u00a0 Kevin Carlson May 10 '13 at 9:00\n@KevinCarlson: Thank you! The question itself is lovely. \u2013\u00a0 23rd May 10 '13 at 9:08\nHow does this generalize to other finite fields, I wonder? \u2013\u00a0 Pete L. Clark May 10 '13 at 13:34\n@PeteL.Clark: A simple observation based on this argument could show that for finite field $\\mathbb{F}_q$, a lower bound is $\\frac{(q-1)m}{q}$. Sketch of proof: we can define $f_j$, $0\\le j\\le q-1$ similarly as in $(2)$. By induction, we can assume that $(3)$ holds when $\\frac{m'}{2}$ is replaced by $\\frac{(q-1)m'}{q}$. When $i>m'$, $f_j(v_i)$ are pairwise different, so exactly only one of them is $0$. Then we can conclude that there exists $0\\le j\\le q-1$, such that $|\\{m'<i\\le m:f_j(v_i)\\ne 0\\}|\\ge\\frac{(q-1)(m-m')}{q}$, which completes the induction. \u2013\u00a0 23rd May 10 '13 at 14:01\n@PeteL.Clark: Sorry, please ignore my first reply and see the current one. \u2013\u00a0 23rd May 10 '13 at 14:04\n\nMy coauthors and I needed this result recently (to prove something about pseudo-Anosov dilatations), and I was googling to see if it was known. I'm amazed that it was asked here so recently!\n\nI'm answering now to record a short alternative proof that we found. I'll prove the same more general result that Landscape proved, namely:\n\nClaim : Let $\\vec{v}_1,\\ldots,\\vec{v}_m \\in \\mathbb{F}_q^n$ be nonzero vectors (not necessarily distinct). Then there exists a linear map $f : \\mathbb{F}_q^n \\rightarrow \\mathbb{F}_q$ such that $f(\\vec{v}_i) = 1$ for at least $\\frac{q-1}{q}$ of the $\\vec{v}_i$, i.e. such that $\\{\\text{$i|1 \\leq i \\leq m$,$f(\\vec{v}_i) \\neq 0$}\\}$ has cardinality at least $\\frac{q-1}{q}m$.\n\nProof : Let $\\Omega$ be the probability space consisting of all linear maps from $\\mathbb{F}_q^n \\rightarrow \\mathbb{F}_q$, each given equal probability. Let $\\mathcal{X} : \\Omega \\rightarrow \\mathbb{R}$ be the random variable that takes $f \\in \\Omega$ to the cardinality of the set $\\{\\text{$i|1 \\leq i \\leq m$,$f(\\vec{v}_i) \\neq 0$}\\}$. We will prove that the expected value $E(\\mathcal{X})$ of $\\mathcal{X}$ is $\\frac{q-1}{q} m$, which clearly implies that there exists some element $f \\in \\Omega$ such that $\\mathcal{X}(f) \\geq \\frac{q-1}{q} m$.\n\nTo prove the desired claim, for $1 \\leq i \\leq m$ let $\\mathcal{X}_i : \\Omega \\rightarrow \\mathbb{R}$ be the random variable that takes $f \\in \\Omega$ to $1$ if $f(\\vec{v}_i) \\neq 0$ and to $0$ if $f(\\vec{v}_i) = 0$. Viewing $\\vec{v}_i$ as an element of the double dual $(\\mathbb{F}_q^n)^{\\ast \\ast}$, the kernel of $\\vec{v}_i$ consists of exactly $\\frac{1}{q}$ of the elements of $(\\mathbb{F}_q^n)^{\\ast}$. This implies $E(\\mathcal{X}_i) = \\frac{q-1}{q}$. Using linearity of expectation (which recall does not require that the random variables be independent), we get that $$E(\\mathcal{X}) = E(\\sum_{i=1}^m \\mathcal{X}_i) = \\sum_{i=1}^m E(\\mathcal{X}_i) = \\sum_{i=1}^m \\frac{q-1}{q} = \\frac{q-1}{q} m,$$ as desired.\n\n-", "date": "2015-05-30 09:28:50", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/387218/choosing-a-linear-map-mathbbz-2-mathbbzn-rightarrow-mathbbz-2-mathb", "openwebmath_score": 0.9841985702514648, "openwebmath_perplexity": 136.6658365174803, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.992654172990894, "lm_q2_score": 0.8577681031721325, "lm_q1q2_score": 0.851467087072301}}
{"url": "https://math.stackexchange.com/questions/2161581/how-many-1s-are-in-the-first-1023-binary-numbers/2161598", "text": "# How many $1$s are in the first $1023$ binary numbers?\n\nHow many $1$s are in the first $1023$ binary numbers?\n\nI'm not to sure how to approach this question. An idea, formula, or solution is appreciated!\n\n\u2022 Have you noticed that $1023 = 2^{10}-1$? Feb 26 '17 at 0:20\n\u2022 What is the \"first\" binary number? 0 or 1? Feb 26 '17 at 0:22\n\u2022 Worth clarifying that we are talking about positive, unsigned integers.\n\u2013\u00a0J...\nFeb 26 '17 at 2:09\n\u2022 @J... actually, can an unsigned integer be positive? Feb 26 '17 at 5:56\n\u2022 @Joffan Cheeky bastard :P Yes, unsigned integers are positive because all of them (other than 0) are greater than 0. Feb 26 '17 at 9:35\n\nAssuming the \"first\" binary number is $1$ note that the first $1023$ binary numbers, plus $0$, are all the binary numbers you can write with exactly $10$ binary digits or bits (prepending $0$s to \"short\" numbers, as in $0000101010_2$). Between all of them, you then have $1024 \\cdot 10=10240$ bits, and for symmetry reasons exactly half of those, $5120$, are $1$s.\n\n\u2022 why 1024*10 digits? Feb 26 '17 at 0:28\n\u2022 Ohh, the amount of 0's don't matter! I see! Feb 26 '17 at 0:29\n\u2022 It does matter, because if we start at 0, then 1111111111 should not be counted among the the first 1023 binary numbers. The question did not say the first 1023 positive binary numbers. Feb 26 '17 at 2:23\n\u2022 In computer science, typically indexing start at 0. I was biased . I should have read the comments below the question... The issue was raised before. Feb 26 '17 at 2:52\n\u2022 @MarDev We're looking at all of the integers we can represent with 10 bits - for every integer there is exactly one other which has all of its bits inverted (the result of an XOR with 1111111111) and therefore the opposite amount of zeros and ones. Furthermore, this inversion always lies in the opposite half of the possible numbers (the operation is essentially equivalent to 1111111111 - n) so each of the first 2^9 numbers (half of the possible 2^10) there exists exactly one number in the second half with the opposite number of zeros and ones. Feb 26 '17 at 8:46\n\nHint: For how many of those numbers will the one's bit be a $1$ (in other words: how many of those numbers are odd)? For how many of them will the two's bit be a $1$? For how many of them will the four's bit be a $1$? And so on. Also, it will probably be advantageous to include $0$ (and thus look at a collection of $1024$ binary numbers) to make the counting a bit easier. Or, if $0$ is already included, include $1023$ initially, then correct for it when you're done counting.\n\n\u2022 I'm struggling with that idea right now. I'm not entirely sure how to solve that. Is it like half half? So if there are x digits in total, half of them are 1? Feb 26 '17 at 0:27\n\u2022 On average, yes, half of the bits will be a $1$. Or something. Depends on what you mean by \"of the bits\". Feb 26 '17 at 0:31\n\nAny such number can be represented by a string of 10 0s and 1s. The number of such strings with $n$ ones is $10$ choose $n$. Thus, the number of ones which appear is $$\\sum_{n=0}^{10} n{10 \\choose n}=5\\cdot 2^{10}.$$\n\nAs an addendum to Arthur's answer - if you count to, say, $2^{4} - 1 = 15$, you can easily figure out a pattern in the columns:\n\n0000\n0001\n0010\n0011\n0100\n0101\n0110\n0111\n1000\n1001\n1010\n1011\n1100\n1101\n1110\n1111\n\n\nSo if you count to $2^{n} - 1$, you have $2^{n} * n$ digits (\"rows * columns\"), of which 50% are 1's.\n\nFor your example: Counting to $1023 =2^{10} - 1$, gets you $2^{10} * 10 * \\frac{1}{2} = 5120$ binary 1's in total.\n\nIt's easy to calculate. There is a pattern with any number used as pow of tow:\n\nYou have 2^x, where x> 0 and you will have a binary number as:\n\n2^0 = 1 (Dec) = 1 (Binary)\n\n2^1 = 2 (Dec) = 10 (Binary)\n\n2^2 = 4 (Dec) = 100 (Binary)\n\n...\n\n2^10 = 1024 (Dec) = 100 0000 0000(Binary)\n\nIf you pay attention you will notice this:\n\n2^10 - 1 = 1023 (Dec) = 011 1111 1111(Binary)\n\n...\n\n2^2 - 1 = 3 (Dec) = 011 (Binary)\n\n2^1 = 1(Dec) = 01 (Binary)\n\n2^0 - 1 = 0 (Dec) = 0 (Binary)\n\nIn this case, your question, the Number of ones matches with the number that is used for pow.", "date": "2021-10-26 00:09:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2161581/how-many-1s-are-in-the-first-1023-binary-numbers/2161598", "openwebmath_score": 0.6330582499504089, "openwebmath_perplexity": 638.5505727921094, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9840936050226358, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8514614814456296}}
{"url": "http://www.tricitiesurgentcare.net/mozzarella-recall-xaojq/ba2b5c-invertible-matrix-theorem", "text": "# invertible matrix theorem\n\nLet two inverses of A be B and C Services; Math; Blog ; About; Math Help; Invertible Matrix and It\u2019s Properties. This section consists of a single important theorem containing many equivalent conditions for a matrix to be invertible. Invertible Matrix Theorem. Section 3.5 Matrix Inverses \u00b6 permalink Objectives. * The determinant of $A$ is nonzero. * $A$ has only nonzero eigenvalues. : An matrix is invertible if and only if has only the solution . structure theorem for completely bounded module maps. Invertible matrix 2 The transpose AT is an invertible matrix (hence rows of A are linearly independent, span Kn, and form a basis of Kn). Showing any of the following about an $n \\times n$ matrix $A$ will also show that $A$ is invertible. (If one statement holds, all do; if one statement is false, all are false.) Proof: Let there be a matrix A of order n\u00d7n which is invertible. Any nonzero square matrix A is similar to a matrix all diagonal elements of which are nonzero. According to WolframAlpha, the invertible matrix theorem gives a series of equivalent conditions for an n\u00d7n square matrix if and only if any and all of the conditions hold. e. The columns of A form a linearly independent set. The Invertible Matrix Theorem Let A be a square n\u00d7n matrix. Let A be a square n by n matrix over a field K (for example the field R of real numbers). 1. Understand what it means for a square matrix to be invertible. Then a natural question is when we can solve Ax = y for x 2 Rm; given y 2 Rn (1:1) If A is a square matrix (m = n) and A has an inverse, then (1.1) holds if and only if x = A\u00a11y. We define invertible matrix and explain many of its properties. 1 The Invertible Matrix Theorem Let A be a square matrix of size N \u00d7 N. The following statement are equivalent: \u2022 A is an invertible matrix. (d)Show that if Q is invertible, then rank(AQ) = rank(A) by applying problem 4(c) to rank(AQ)T. (e)Suppose that B is n \u2026 Then the following statements are equivalent. Yes. We will append two more criteria in Section 6.1. An invertible matrix is sometimes referred to as nonsingular or non-degenerate, and are commonly defined using real or complex numbers. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly \u2026 Theorem 1. 5.The columns of A are linearly independent. The Invertible Matrix Theorem|a small part For an n n matrix A, the following statements are equivalent. A is column-equivalent to the n-by-n identity matrix In. Let A be an n n matrix. A is an invertible matrix. Usually, when a set is written as the span of one vector, it\u2019s one dimensional. Let A 2R n. Then the following statements are equivalent. \u2022 The columns of A form a linearly independent set. The matrix A can be expressed as a finite product of elementary matrices. Theorem (The QR Factorization) If A is an mxn matrix with linearly independent columns, then A can be factored as A=QR, where Q is an mxn matrix whose columns form an orthonormal basis for Col A and R is an nxn upper triangular invertible matrix with positive entries on the main diagonal. Theorem1: Unique inverse is possessed by every invertible matrix. The following statements are equivalent, i.e., for any given matrix they are either all true or all false: A is invertible, i.e. Some theorems, such as the Neumann Series representation, not only assure us that a certain matrix is invertible, but give formulas for computing the inverse. I. row reduce to! 4. The uniqueness of the polar decomposition of an invertible matrix. W. Sandburg [8] and Wu and Desoer [ \u2026 A has an inverse or is nonsingular. Dave4Math \u00bb Linear Algebra \u00bb Invertible Matrix and It\u2019s Properties. Skip to content. 6.The linear transformation T de\ufb01ned by T(x) = Ax is one-to-one. Note that finding this matrix B is equivalent to solving a system of equations. The following statements are equivalent: A is invertible, i.e. A has an inverse, is nonsingular, or is nondegenerate. tem with an invertible matrix of coef\ufb01cients is consistent with a unique solution.Now, we turn our attention to properties of the inverse, and the Fundamental Theorem of Invert- ible Matrices. lie in the commutants of d and 59\u2019. The number 0 is not an eigenvalue of A. A matrix that has no inverse is singular. Here\u2019s the first one. The Invertible Matrix Theorem (Section 2.3, Theorem 8) has many equivalent conditions for a matrix to be invertible. Introduction and De\ufb02nition. Furthermore, the following properties hold for an invertible matrix A: \u2022 for nonzero scalar k \u2022 For any invertible n\u00d7n matrices A and B. Learn about invertible transformations, and understand the relationship between invertible matrices and invertible transformations. Menu. A is row-equivalent to the n-by-n identity matrix In. 2 det(A) is non-zero.See previous slide 3 At is invertible.on assignment 1 4 The reduced row echelon form of A is the identity matrix. When the determinant value of square matrix I exactly zero the matrix is singular. 5. (c)Showthatif P isaninvertiblem \u00d7m matrix, thenrank(PA) = rank(A) byapplying problems4(a)and4(b)toeachofPA andP\u22121(PA). its nullity is zero. \u2022 A has N pivot positions. A is invertible.. A .. A is row equivalent to the n\u00d7n identity matrix. The Invertible Matrix Theorem has a lot of equivalent statements of it, but let\u2019s just talk about two of them. Matrix I exactly zero the matrix must be a square n by n matrix over a field K for! A $3\\times 3$ matrix is detailed along with characterizations the span of one vector, \u2019! Following statements are equivalent services ; math help ; invertible matrix finite product of elementary.. 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Can only happen with full rank there be a square n by n matrix over a field K for... E.. F a is row equivalent to the n\u00d7n identity matrix the rest of polar... Append two more criteria in Section 6.1 another way of saying this is one of the must. = ~0 has no non-zero solutions as the span of one vector, \u2019.", "date": "2021-03-07 18:39:07", "meta": {"domain": "tricitiesurgentcare.net", "url": "http://www.tricitiesurgentcare.net/mozzarella-recall-xaojq/ba2b5c-invertible-matrix-theorem", "openwebmath_score": 0.8166371583938599, "openwebmath_perplexity": 520.6749612818505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9904405995242327, "lm_q2_score": 0.8596637541053281, "lm_q1q2_score": 0.8514458840053337}}
{"url": "https://math.stackexchange.com/questions/1451818/probability-that-the-sum-of-6-dice-rolls-is-even", "text": "# Probability that the sum of 6 dice rolls is even\n\nQuestion:\n\n6 unbiased dice are tossed together. What is the probability that the sum of all the dice is an even number?\n\nI think the answer would be 50%, purely by intuition. However, not sure if this is correct. How should I go about solving such a problem?\n\nNotice that whatever the sum of the first 5 rolls, whether the outcome is odd or even is totally determined by the last die. It is even or odd with equal probability, so the probability of an even sum is exactly the same as the probability of an odd sum.\n\nThe first 5 dice don't matter.\n\nStatement:\n\nLet $X_1,\\ldots,X_n$ be independent, integer-valued random variables, and let some $X_k$ satisfy $P(X_k \\mbox{ odd}) = 0.5$. Then $P(\\sum X_i \\mbox{ odd}) = 0.5$.\n\nProof:\n\nWithout loss of generality let $k=n$, and let $S = \\sum_i^{n-1} X_i$. Since $S$ and $X_n$ are independent random variables, $$\\begin{eqnarray*} P\\left(\\sum X_i \\mbox{ odd}\\right) = P(S + X_n \\mbox{ odd}) &=& P(S \\mbox{ odd and } X_n \\mbox{ even}) + P(S \\mbox { even and } X_n \\mbox{ odd}) \\\\ &=& P(S \\mbox{ odd})P(X_n \\mbox{ even}) + P(S \\mbox{ even})P(X_n \\mbox{ odd}) \\\\ &=& P(S \\mbox{ odd})P(X_n \\mbox{ odd}) + P(S \\mbox{ even})P(X_n \\mbox{ odd}) \\\\ &=& \\left(P(S \\mbox{ odd}) + P(S \\mbox{ even})\\right)P(X_n \\mbox{ odd}) \\\\ &=& P(X_n \\mbox{ odd}) \\\\ &=& 0.5 \\end{eqnarray*}$$\n\n\u2022 o.O That's also a nice approach. Will the same apply for a dice with 7 sides? \u2013\u00a0Gummy bears Sep 26 '15 at 5:18\n\u2022 @Gummybears no, because then you have inequal probabilities coming from the last die, so the parity of the sum you had before you roll it matters. However, this sort of thinking does apply in a pretty wide variety of cases, and it saves a lot of work when you notice it happening. \u2013\u00a0Eric Tressler Sep 26 '15 at 5:22\n\u2022 @Gummybears: sorry, I missed that. However, what is not mentioned is that with an odd number of seven sided dice, there is a higher chance of an odd sum. Whereas with an even number of seven sided dice, there is a higher chance of an even sum. \u2013\u00a0robjohn Sep 26 '15 at 5:50\n\u2022 @JoseAntonioDuraOlmos, the first 5 dice matter as much as the last one. They all matter the same here. It is not our neglect of the first 5 dice which leads us to the conclusion that the probability is $1/2$ but it is the symmetry of odd/even tosses. There is a great danger in basing the argument on the neglect, as one might be tempting to write it for the case where all dice are biased. \u2013\u00a0zhoraster Sep 26 '15 at 8:53\n\u2022 There is no neglect. If I am presented a box with an unknown number of dices of unknown shapes all of them labeled with integers. I take only one out of the box. The dice has 6 sides, from 1 to 6, and is unbiased. Then I can claim that if I take all the dices, roll them and add their results the sum has a 50% probability of being even. I can claim it without examining the other dices in the box. And the proof of such claim is like the one in this answer. As for the case where all dice are biased, yeah this proof scheme does not work for that, don't fall into those temptations. \u2013\u00a0Jose Antonio Reinstate Monica Sep 26 '15 at 9:08\n\nThe only thing about the numbers rolled that matters is their parity - whether they are even or odd. In order to get an even sum, an even number of the six dice must be even. In order to get an odd sum, an odd number of the six dice must be even.\n\nUsing O for odd and E for even, we can list out the possibilities.\n\nEven sum:\n\n\u2022 OOOOOO $$\\binom{6}{6}=1 \\text{ arrangement}$$\n\u2022 OOOOEE $$\\binom{6}{4}=15 \\text{ arrangements}$$\n\u2022 OOEEEE $$\\binom{6}{2}=15 \\text{ arrangements}$$\n\u2022 EEEEEE $$\\binom{6}{0}=1 \\text{ arrangement}$$\n\nThis gives a total of $32$ arrangements with even sum.\n\nSince there are $2^6 = 64$ total possibilities, we see that your intuition of $50\\%$ is correct.\n\n\u2022 Lovely answer. Although slightly lengthy, it is simple to understand. Thank you. \u2013\u00a0Gummy bears Sep 26 '15 at 5:24\n\u2022 +1 Good answer. I just have a wording suggestion: instead of saying \"to get an even/odd sum, an even/odd number of the six dice must be even\", I think wording it in terms of the number of odd dice gives more intuition and can generalize to any number of dice. The number of even dice doesn't really matter, the number of odd dice does. \u2013\u00a0jadhachem Sep 28 '15 at 5:39\n\nThe generating function approach:\n\n$$P(x)=(1+x+x^2+x^3+x^4+x^5)^6=\\sum a_i x^i$$\n\nThen $a_i$ counts the number of ways of getting a total of $i+6$ from $6$ dice.\n\nNow, to find the even terms, you can compute $$\\frac{P(1)+P(-1)}{2}=\\sum_i a_{2i}.$$\n\nBut $P(1)=6^6$ and $P(-1)=0$. So $$\\frac{P(1)+P(-1)}{2}=\\frac{6^6}{2},$$ or exactly half, as you conjectured.\n\nFor another example, let $N_{i}$ be the number of ways to roll $6$ dice and getting a value $\\equiv i\\pmod{5}$. Then it turns out that if $z$ is a primitive $5$th root of unity, then the value can be counted by defining:\n\n$$Q_i(x)=x^{6-i}(1+x+x^2+x^3+x^4+x^5)^6$$ then computing $$N_i=\\frac{Q_i(1)+Q_i(z)+Q_i(z^2)+Q_i(z^3)+Q_i(z^4)}{5}$$\n\nThis gives the result:\n\n$$N_i =\\begin{cases}\\frac{6^6+4}{5}&i\\equiv 1\\pmod 5\\\\ \\frac{6^6-1}{5}&\\text{otherwise} \\end{cases}$$\n\nMore generally, if $N_{n,i}$ is the number of ways to get $\\equiv i\\pmod 5$ when $n$ dice are rolled, you get:\n\n$$N_{n,i} =\\begin{cases}\\frac{6^n+4}{5}&i\\equiv n\\pmod 5\\\\ \\frac{6^n-1}{5}&\\text{otherwise} \\end{cases}$$\n\nIt's this simple because of the fact that $6=5+1$.\n\nIf each die has $d$ sides, and you ask what are the number of ways to get a total $\\equiv i\\pmod {d-1}$, then you get:\n\n$$N_{d,n,i} =\\begin{cases}\\frac{d^n+{d-2}}{d-1}=\\frac{d^n-1}{d-1}+1&i\\equiv n\\pmod {d-1}\\\\ \\frac{d^n-1}{d-1}&\\text{otherwise} \\end{cases}$$\n\n\u2022 I'm afraid I lost you at the last sentence. \u2013\u00a0Gummy bears Sep 26 '15 at 5:03\n\u2022 That's okay, it is an advanced approach, but a useful one for problems like this. If you rolled $n$ dice, each with $7$ sides, you'd get the number of even results is: $$\\frac{7^n + (-1)^n}{2}$$ would be even. \u2013\u00a0Thomas Andrews Sep 26 '15 at 5:09\n\u2022 Aha..... That's a good approach. However, I'm not sure if I completely get it. Not sure how you used the probability of 1 and $-1$ to calculate the even terms. \u2013\u00a0Gummy bears Sep 26 '15 at 5:16\n\u2022 If $i$ is odd, then $a_i\\cdot 1^i + a_i(-1)^{i} = 0$. If $i$ is even, then $a_u\\cdot 1^i + a_i(-1)^i = 2a_i$. So $P(1)+P(-1)=\\sum_{i} 2a_{2i}$, which is twice the value you are looking for. @Gummybears \u2013\u00a0Thomas Andrews Sep 26 '15 at 5:19\n\u2022 Aha.... I'm stupid, aren't I? This is perfect! It holds true for any sided dice and for any number of dice. \u2013\u00a0Gummy bears Sep 26 '15 at 5:23\n\nUsing the same logic given by @Eric tessler but writing the maths behind it.\n\nLet $P_x$ = We get an even sum on the roll of x dice and\n\n$1 - P_x$ = We get an odd sum.\n\nSo we require $P_6$\n\nNow $P_6 = P_5 \\times$ P[even number on the last dice] + $(1 - P_5) \\times$ P[odd number on the last dice]\n\n$P_6 = P_5 \\times 0.5 + (1-P_5) \\times 0.5$\n\n$P_6 = 0.5$\n\n\u2022 Exactly what @erictessler is missing - a rigurous proof. One by induction could work for all sorts of dice. \u2013\u00a0Tibos Sep 26 '15 at 23:22\n\u2022 More upvotes needed! \u2013\u00a0Gummy bears Sep 27 '15 at 4:14\n\u2022 Nicely done! +1 \u2013\u00a0Zubin Mukerjee Sep 27 '15 at 17:27\n\nA lot of very informative answers but none of them do not explain your intuition, so I'll post mine.\n\nThere is a symmetry between all tosses with odd sum and those with even sum: just take the number $k$ on the first die and replace it with $7-k$.\n\nSo the numbers of odd tosses and even tosses are equal, therefore, the probability is $1/2$.\n\n\u2022 I'm not getting you much.... elaborate please? \u2013\u00a0curiousbrain Sep 26 '15 at 5:22\n\u2022 Same here. Please elaborate. \u2013\u00a0Gummy bears Sep 26 '15 at 5:23\n\u2022 See the answer by @EricTressler, posted the same time. \u2013\u00a0zhoraster Sep 26 '15 at 5:24\n\nYour intuition is correct. There could be a proof based on this intuition (using the fact that each die has 50% chance of being even and we're tossing 6 of them) but I'm not so sure.\n\nBut there are ways to prove it. For example, the only ways we could get an even sum are if:\n\n1. All 6 dice show even numbers.\n2. 4 dice show even and 2 dice show odd.\n3. 2 dice show even and 4 dice show odd.\n4. All 6 dice show odd.\n\nThe probability of cases 1 and 4 is $(\\frac{1}{2})^6$ and the probability of cases 2 and 3 is $\\binom{6}{2}(\\frac{1}{2})^6=15(\\frac{1}{2})^6$ (because we need to order the dice which show odd numbers.) Add them up and we get the probability is $$32(\\frac{1}{2})^6=\\frac{1}{2}.$$\n\nThe problem is about fair dice.\n\nWhether the number of dice is 6 (even) or 7 (odd) $Pr = \\dfrac12$\n\nThe logic is based on parity (odd/even). Since the probability that any particular throw is even or odd is equal at $\\dfrac12$\n\n(a) For an even number of dice, if you interchange odd and even throws, the parity remains the same, thus there will always be an equal number of odd and even sums.\n\nEEEEEE OOOOOO even\n\nEEEEEO OOOOOE odd\n\nEEEEOO OOOOEE even\n\nEEEOOO OOOEEE odd\n\n(b) For an odd number of dice, every such interchange changes the parity, but by symmetry, again there will be an equal number of odd and even sums.\n\nEEEEEEE even OOOOOOO odd\n\nEEEEEEO odd OOOOOOE even\n\nEEEEEOO even OOOOOEE odd\n\nEEEEOOO odd OOOOEEE even\n\nAll outcomes are equally likely. Half the outcomes will have an odd number of odd dice faces and half of the outcomes will have an even number of odd dice faces. The outcomes with an odd number will have an odd total and the outcomes with an even number of odd face will be even. As there are equal numbers of outcomes for each case the probability for each case is 50%.\n\nOkay so how do we know that half the outcomes will have on odd number of odd faces and half the number will have an even number of even faces? Well,we can make a one to one correspondence between outcomes with an odd number of odd faces to outcomes to an even number of odd faces. If outcome has an even number of odd faces, map it to the exact same outcome but the 6th die is one number higher (let's assume 1 is one number higher than 6). This outcome has an odd number of odd faces. If an outcome has an odd number of odd faces map it the exact same outcome but the 6th die is one number lower. This is a one to one correspondence. So there are an equal number of outcomes with an odd number of odd faces and there are outcomes with an even number of odd faces.\n\nAnd that's it. Equal number of odd comes as even outcomes, each outcome equally likely, odd outcomes and even outcomes equally likely.\n\n\u2022 Oops, Zhoraster beat me to it and did a much clearer and shorter job explaining it. Oh, well. \u2013\u00a0fleablood Sep 26 '15 at 23:27\n\u2022 It was asked above about 7 sided die. I don't have enough reputation to comment there, so I'm going to comment here: If a 7-sided die has 4 odd faces and 3 even faces the probability of an even result with n die is 1/2 + (- 1/7)^n. This can be shown by induction by noting that each extra die has a 4/7 chance of changing parity. So if Pn is probability of an even total with n die. Then P(n+1) = Pn*3/7 + (1 - Pn)*4/7. as P1 = 3/7 = 1/2 + (- 1/7), the result follows by induction \u2013\u00a0fleablood Sep 26 '15 at 23:51", "date": "2019-11-15 07:15:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1451818/probability-that-the-sum-of-6-dice-rolls-is-even", "openwebmath_score": 0.7672575116157532, "openwebmath_perplexity": 328.7327495617882, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9904405986777259, "lm_q2_score": 0.8596637487122111, "lm_q1q2_score": 0.8514458779360605}}
{"url": "https://math.stackexchange.com/questions/523944/show-that-if-a-and-b-are-sets-then-a-subseteq-b-if-and-only-if-a-cap-b", "text": "# Show that if $A$ and $B$ are sets, then $A \\subseteq B$ if and only if $A \\cap B = A$.\n\nI'm working through a real analysis textbook, and it starts out with set theory. The first exercise is\n\nShow that if $A$ and $B$ are sets, then $A \\subseteq B$ if and only if $A \\cap B = A$.\n\nI think I proved it correctly but I'm not sure. Here's what I did. I proved that if $A \\subseteq B$, then $A \\cap B = A$ the same way as this answer did (https://math.stackexchange.com/a/446114/93114), but I want to make sure I proved the converse correctly because it seems really easy (yes it's the first problem in the book, but still) and math usually isn't this easy for me, even the basic stuff!\n\nProof of \"If $A \\cap B = A$, then $A \\subseteq B$.\"\n\nIf $x \\in A \\cap B$, then $x \\in A$ and $x \\in B$, but this applies to all $x \\in A$ because $A \\cap B = A$. So, for any $x \\in A$, we know that $x \\in B$, so $A \\subseteq B$.\n\nAm I on the right track?\n\n\u2022 Yes, although you're doing it in a convoluted way. Let $x\\in A$; since $A=A\\cap B$ we have $x\\in B$. \u2013\u00a0egreg Oct 12 '13 at 22:27\n\u2022 @egreg So if $x \\in A$, then because $A = A \\cap B$, we know that $x \\in A \\cap B$. By definition that implies that $x \\in A$ and $x \\in B$, which is enough to show that $A \\subseteq B$. \u2013\u00a0M T Oct 12 '13 at 22:28\n\u2022 @egreg Sorry I replied to your other comment that disappeared, but we're doing it the same way now. \u2013\u00a0M T Oct 12 '13 at 22:30\n\u2022 I realized to have misread your argument, so I changed my first comment. \u2013\u00a0egreg Oct 12 '13 at 22:32\n\nSince you are just starting, I would suggest to be verbose instead of pulling everything in a single sentence.\n\nTo prove $A \\subseteq B$ iff $A \\cap B = A$, you have to\n\n1. show $A \\cap B = A$ given $A \\subseteq B$. That is to\n1. show $A \\cap B \\subseteq A$.\n2. show $A \\subseteq A \\cap B$.\n2. show $A \\subseteq B$ given $A \\cap B = A$.\n\nProof:\n\n1.1) It is trivially true. You don't need to be given $A \\subseteq B$ for it to be true.\n\n1.2) If $x \\in A$, then $x \\in B$ since we are given $A \\subseteq B$. Then $x \\in A$ and $x \\in B$ are both true. Therefore, $x \\in A \\cap B$.\n\n2) From $A \\cap B = A$, we know $x \\in A$ and $x \\in B$ whenever $x \\in A$. If $x \\in A$, then it must be the case that $x \\in B$. Therefore, $A \\subseteq B$.\n\n\u2022 Thanks for the answer. I think I'm slowly starting to catch the hang of these proofs. \u2013\u00a0M T Oct 13 '13 at 1:58\n\nSuppose $A\\cap B=A$. To prove $A\\subseteq B$, suppose $x\\in A$. Since $A\\cap B=A$, $x\\in A\\cap B$ and thus $x\\in B$. Since $x$ was an arbitrary element of $A$, this holds for all $x$ and thus $A\\subseteq B$.\n\nThe above proves $A\\cap B=A\\implies A\\subseteq B$. You will then need to prove the reverse implication to establish $A\\cap B=A\\iff A\\subseteq B$.\n\nHere is another way to approach this type of problem: start with the most complex part (here: $\\;A \\cap B = A\\;$), expand the definitions so that you get from the set level to the element level, then simplify using predicate logic, and finally go back to the set level as suggested by the shape of the formula.\n\nIn this case, \\begin{align} & A \\cap B = A \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"set extensionality, i.e., the definition of $\\;=\\;$ for sets\"} \\\\ & \\langle \\forall x :: x \\in A \\cap B \\equiv x \\in A \\rangle \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"definition of $\\;\\cap\\;$\"} \\\\ & \\langle \\forall x :: x \\in A \\land x \\in B \\equiv x \\in A \\rangle \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"logic: $\\;p \\equiv p \\land q\\;$ is one way to write $\\;p \\Rightarrow q\\;$\"} \\\\ & \\langle \\forall x :: x \\in A \\Rightarrow x \\in B \\rangle \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"definition of $\\;\\subseteq\\;$\"} \\\\ & A \\subseteq B \\\\ \\end{align}", "date": "2019-07-23 11:52:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/523944/show-that-if-a-and-b-are-sets-then-a-subseteq-b-if-and-only-if-a-cap-b", "openwebmath_score": 0.9999747276306152, "openwebmath_perplexity": 121.45051778712563, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137895115187, "lm_q2_score": 0.8670357735451835, "lm_q1q2_score": 0.8514410856211566}}
{"url": "https://math.stackexchange.com/questions/2876327/show-that-a-certain-set-of-positive-real-numbers-must-be-finite-or-countable", "text": "# Show that a certain set of positive real numbers must be finite or countable\n\nLet $B$ be a set of positive real numbers with the property that adding together any finite subset of elements from $B$ always gives a sum of $2$ or less. Show that $B$ must be finite or at most countable.\n\n$B$ = {$x \\in R:x>0\\}$, $x_1,x_2...x_n \\in B$ such that $x_1+x_2+...+x_n \\le 2$.\n\nQuestion: for any $a,b$ $(a,b)$~$R$, but $B$ is $(0,+\\infty)$ so why $B$ is not uncountable (taking as $a = 0$, and letting $b$->$\\infty$)?\n\nAnd why for $B$ being countable doesn't contradict: for any $a,b$ $(a,b)$~$R$?\n\nP.S. I read Showing a set is finite or countable and understood it.\n\n\u2022 It sounds like you are confused about what the question is asking. It is not saying that B is the set of positive real numbers; it is a set of positive real numbers. This means that B is a subset of the set of all positive real numbers. Aug 8 '18 at 17:23\n\u2022 No. B is some unspecified subset of the positive real numbers such that if you add up a finite number of the elements of B the sum is always less than 2. For example, B could be {0.1, 0.2, 0.5, 1.1} or B could be the infinite set {1, 1/2, 1/4, 1/8, 1/16, ...}. Note that B cannot be {1, 1.4, 1.8} and B cannot be {1, 1/2, 1/3, 1/4, ...} Aug 8 '18 at 17:47\n\u2022 B does not have to be an interval (0,a). In fact, it cannot be such an interval. Your job is to prove that B must be finite or countable. Intervals are uncountable. Aug 8 '18 at 17:49\n\u2022 Well, I'm just trying to help clarify the question. I haven't given a solution. I think the hints and discussion below do that for you. But the point is you cannot assume anything about B except: 1. its elements are all positive real numbers and 2. if you take a finite number of elements from B and add them up then you get a sum less than two. Given any such B you have to prove that B is either finite or countable. Aug 8 '18 at 17:59\n\u2022 Here is a proof that B cannot be the open interval (0,1). If it were then these numbers would be in B: 0.5, 0.51, 0.52, and 0.53. These numbers add up to more than 2. Do you see? You could do something similar to show that B cannot be (0,a) for any positive a. But you have to show more than that. You have to show B is finite or countable. Aug 8 '18 at 18:00\n\nHint 1: How many elements of $B$ can be in the set $[2,\\infty)$?\n\nHint 2: How many elements of $B$ can be in the set $[1,2)$?\n\nHint 3: How many elements of $B$ can be in the set $[0.5,1)$?\n\n\u2022 At this rate you are going to need countably many hints :) Aug 8 '18 at 17:04\n\u2022 @ArnaudMortier Can't he give a hint schema? Aug 8 '18 at 17:07\n\u2022 1) One $x = 2$, 2) set $x_1=1$ so one for sure, then I don't know how many $x_2,x_3,...x_n$ to choose so that they sum up to $1$. 3) The same, I don't know. Aug 8 '18 at 17:10\n\u2022 Obviously $B\\cap[1,\\infty)$ has at most two elements, because if $x,y,z\\ge1$ then $x+y+z>2$. Similarly $B\\cap[2/n,\\infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...) Aug 8 '18 at 17:17\n\u2022 No, we certainly can't write $B$ as $(0,a)$. If $B=(0,a)$ then there are finite subsets of $B$ with sum larger than $2$. Aug 8 '18 at 18:56\n\nNot only can we show that it's countable, it's not too difficult to construct an enumeration: given an element $b$, just count how many other elements of $B$ are larger. This has to be a finite integer, since if there were an infinite number of elements greater than $b$, then the sum of $n$ such elements would be greater than $nb$, so picking an $n>2/b$ would give a sum greater than 2.\n\nFor any $a\\in B$, define $B_a\\equiv\\{b\\in B:a<b\\}$. The cardinality of $B_a$ must be less than $\\lceil2/a\\rceil$, otherwise the sum of any $\\lceil2/a\\rceil$ elements of $B_a$ would exceed 2. That is, $\\sum_{i=1}^{\\lceil2/a\\rceil}x_i>\\sum_{i=1}^{\\lceil2/a\\rceil}a=a\\lceil2/a\\rceil\\ge2$ for any sequence $x_i$ in $B_a$, contradicting the requirements on $B$.\n\nDefine the function $f:B\\rightarrow\\mathbb{N}$ as $f(a)=\\left|B_a\\right|$.\n\nHomework: If you can show that $f$ is one-to-one then you can conclude that $|B|\\le|\\mathbb{N}|$. That is, $B$ is either finite or countable.\n\nCredit goes to Acccumulation for the outline of this proof.\n\n\u2022 What if $a =$ sup$B$? Then $B_a$ is the empty set. If $B_a \\equiv \\{b \\in B: a \\leq b \\}$, then the sum of any $\\lceil r/a \\rceil$ for $r > 2$ elements of $B_a$ would exceed 2. Jul 18 '19 at 23:08", "date": "2021-10-17 17:33:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2876327/show-that-a-certain-set-of-positive-real-numbers-must-be-finite-or-countable", "openwebmath_score": 0.7738478779792786, "openwebmath_perplexity": 120.9273635766722, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137858267907, "lm_q2_score": 0.8670357615200475, "lm_q1q2_score": 0.8514410706175163}}
{"url": "https://math.stackexchange.com/questions/2687454/converse-of-uniqueness-of-universal-property-of-tensor-products", "text": "Converse of Uniqueness of Universal Property of Tensor Products?\n\nIn Atiyah Macdonald, for a commutative and unital ring $A$ the tensor product of $A$-modules, $M,N$ is defined by the usual universal property:\n\nA pair $(T,g)$ with $T$ an $A$-module and $g$ an $A$-bilinear map $g\\colon M \\times N \\to T$ such that for any $A$-module $P$ with an $A$-bilinear map $f\\colon M \\times N \\to P$ there exists a unique $A$-module homomorphism $h\\colon T \\to P$ such that $h \\circ g = f$.\n\nAnd then, as usually goes with universal properties:\n\nMoreover, if $(T,g), (T',g')$ are any two pairs satisfying this condition, then there is a unique isomorphism $j \\colon T \\to T'$ such that $j \\circ g = g'$\n\nMy question is, in the second highlighted selection, is the converse true? If I have already established the existence of the usual tensor product, $M \\otimes_A N$, via quotient of free module, and I encounter some other pair $(T', g')$ and a unique isomorphism $j: T \\to T'$ such that $j \\circ g = g'$, can I conclude that $T'$ also satisfies the universal property defining the tensor product, thus I can regard both $T$ and $T'$ as the tensor product of this $M$ and $N$?\n\nI believe I proved that I can indeed do this, but I need reassurance because I have overlooked small details with tensor products before, and don't want to make that mistake again. My proof was as follows:\n\nSuppose $(T,g)$ satisfies the universal property and $(T',g')$ is a pair where $T'$ is an $A$-mod and $g'$ an $A$-bilinear map $g' \\colon M \\times N \\to T'$. Suppose $j$ is the unique isomorphism $j\\colon T \\to T'$ such that $j \\circ g = g'$, or $g = j^{-1} \\circ g'$. Suppose $P$ is any $A$-mod with an $A$-bilinear map $f\\colon M \\times N \\to P$. We wish to show there exists a unique $A$-module homomorphism $\\phi\\colon T' \\to P$ such that $\\phi \\circ g' = f$.\n\nSince $(T,g)$ already satisfies the property by hypothesis, we know there exists a unique $A$-module homomorphism $\\varphi \\colon T \\to P$ such that $\\varphi \\circ g = f$. So define the map $\\phi \\colon T' \\to P$ by $\\varphi \\circ j^{-1}$. Then $\\phi \\circ g' = \\varphi \\circ j^{-1} \\circ g' = \\varphi \\circ g = f$. So the diagram commutes as desired and the map $\\phi$ is defined as the composition of two unique maps so is unique as well, thus $(T',g')$ satisfies the universal property.\n\nSorry that got a little (unnecessarily) complicated, I don't know how to draw diagrams on here, which really makes everything easier. As an aside, I feel like this probably is true and Atiyah & Macdonald didn't mention it in the text because they deemed it trivial.\n\n\u2022 I think this is the first time I've ever seen someone use $\\phi$ and $\\varphi$ as distinct variable names at the same time. \u2013\u00a0Eric Wofsey Mar 12 '18 at 6:39\n\u2022 I do it all the time! \u2013\u00a0Prince M Mar 12 '18 at 15:08\n\nThis is correct, although your argument for uniqueness of $\\phi$ is not very clear: you chose to define it as a composition with certain maps that are unique, but how do you know you couldn't get a $\\phi'$ that works and is not given by those compositions? What you want to say is that if $\\phi'$ makes the diagram commute for $(T',g')$, then the map $\\varphi'=\\phi'\\circ j$ would make the diagram commute for $(T,g)$ by just swapping the roles of $(T,g)$ and $(T',g')$ in your argument. Therefore by uniqueness of $\\varphi$, $\\varphi=\\varphi'$, and so $\\phi'=\\varphi'\\circ j^{-1}=\\varphi\\circ j^{-1}=\\phi$.\nIntuitively, what's going on here is that the isomorphism $j$ says that $(T,g)$ and $(T',g')$ have exactly the same structure (just the elements of $T$ have to be renamed via $j$ to get $T'$). So any property of $(T,g)$ (which is defined using only the module structure on $T$ together with the map $g$) will also be true of $(T',g')$. In particular, this applies to the universal property of the tensor product.\n\u2022 Ok, thanks. One last question I had was, I just now wrote the whole argument out to ensure I comprehend the entirety of it and I cannot identify where the uniqueness (w/ respect to commuting diagrams) of the isomorphism j is invoked? I see plainly that this theorem fails if it is not unique, because there will be at least 2 distinct ways to construct the map $/Phi$, but I can\u2019t seem to identify, where in the uniqueness argument, we actually invoke the assumption that j is unique? \u2013\u00a0Prince M Mar 12 '18 at 22:16\n\u2022 You don't need uniqueness of $j$, just its existence. If $j$ exists, it turns out to automatically be unique. \u2013\u00a0Eric Wofsey Mar 12 '18 at 22:17", "date": "2019-06-20 01:40:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2687454/converse-of-uniqueness-of-universal-property-of-tensor-products", "openwebmath_score": 0.974436342716217, "openwebmath_perplexity": 90.61567841482317, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137897747137, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8514410639186349}}
{"url": "https://cs.stackexchange.com/questions/133977/the-number-of-arrangements-of-n-players-around-a-round-table-where-each-player", "text": "# the number of arrangements of N players around a round table, where each player can sit on one of 3 contiguous chairs\n\nConsider the fact that each player can either sit on their desired chair or on the neighbouring chair. Two configurations are distinct if at least one person is sitting in another chair. My attempt For example, if there are 4 players sitting around a circular table there can be 9 distinct arrangements.\n\n1. F1,F2,F3,F4\n2. F2,F1,F3,F4\n3. F2,F1,F4,F3\n4. F4,F2,F3,F1\n5. F1,F3,F2,F4\n6. F1,F2,F4,F3\n7. F4,F1,F2,F3\n8. F2,F3,F4,F1\n9. F4,F3,F2,F1\n\nNow, I am stuck here. How can calculate this and what about larger values of N such as 50?\n\nI tried using vectors and pushing back every element and do the reverse operators and swap elements but I am not sure whether this is correct or not.\n\n\u2022 Do you want to enumerate the arrangements or simply get the number? What have you tried? Jan 4, 2021 at 20:34\n\u2022 Can you credit the source where you originally encountered this task?\n\u2013\u00a0D.W.\nJan 4, 2021 at 20:36\n\u2022 @D.W. No I cant this is a competetive programming question that was asked in a contest Jan 4, 2021 at 21:59\n\u2022 Is the contest currently ongoing? When was the deadline for the contest? What prevents you from providing a link and/or reference to the programming contest? Sometimes helping us see the original statement of the question helps us understand what you are asking better. Also, please don't leave clarifications in the comments (such as regarding enumerating vs counting): instead, edit the question to clarify it.\n\u2013\u00a0D.W.\nJan 4, 2021 at 22:08\n\u2022 Please edit the question to make it clear whether the desired chairs of different players can be the same or not. Jan 6, 2021 at 6:09\n\nConsider the variant of the problem in which the people are arranged in a segment instead of circle and let $$S(n)$$ be the number feasible arrangements for $$n$$ people.\n\nNow lets go back to your original problem, and let's name $$C(n)$$ the number of feasible arrangements of $$n$$ people.\n\nClearly $$C(0)=C(1)=1$$ and $$C(2)=2$$, so we will henceforth suppose that $$n \\ge 3$$.\n\nLet's name the people $$p_1, \\dots, p_n$$ where the assigned position of $$p_i$$ is $$i$$. If $$p_1$$ is sitting in position 1 then the number of possible arrangements of $$p_2, \\dots, p_n$$ is exactly $$S(n-1)$$. Otherwise, $$p_1$$ is sitting either in position $$2$$ or in position $$n \\neq 2$$. We will only consider the former case since the latter one is symmetric. Clearly $$p_2$$ cannot sit in position $$2$$, so he/she must either sit in position $$1$$ or in position $$3$$. If $$p_2$$ is sitting in position $$1$$, then the number of possible arrangements of $$p_3, \\dots, p_n$$ is exactly $$S(n-2)$$. Otherwise, $$p_2$$ must be sitting in position $$3$$, which means that $$p_3$$ must be sitting in position $$4$$ (since position $$2$$ and $$3$$ are occupied) and, in general, $$p_i$$ must be sitting in position $$(i \\bmod n) + 1$$. Since this completely determines everyone's position, this case only contributes $$1$$ to the total number of configurations.\n\nTo summarize, we have that: $$C(n) = S(n-1) + 2(S(n-2)+1) = S(n-1) + 2S(n-2) + 2.$$\n\nWe are left with figuring out what $$S(n)$$ is. Clearly $$S(0) = S(1) = 1$$, so we consider $$n \\ge 2$$. In this case $$p_1$$ can only sit in position $$1$$ or $$2$$. If $$p_1$$ is sitting in position $$1$$ then there are $$S(n-1)$$ possible arrangements of $$p_2, \\dots, p_n$$. If $$p_1$$ is sitting in position $$2$$, then $$p_2$$ must be sitting in position $$1$$ (as it is the only other person that can sit there) and there are $$S(n-2)$$ possible arrangements for $$p_3, \\dots, p_n$$. We hence have: $$S(n) = \\begin{cases} 1 & \\mbox{if } n \\in \\{0,1\\} \\\\ S(n-1) + S(n-2) & \\mbox{otherwise} \\end{cases} = \\mathcal{F}_{n+1},$$ where $$\\mathcal{F}_{i}$$ denotes the $$i$$-th Fibonacci number. Substituting back, we obtain:\n\n$$C(n) = \\begin{cases} 1 & \\mbox{if } n \\in \\{0,1\\} \\\\ 2 & \\mbox{if } n = 2 \\\\ \\mathcal{F}_n + 2 \\mathcal{F}_{n-1} + 2 & \\mbox{otherwise} \\end{cases}.$$\n\n\u2022 we can also say that C(n) = F_(n+1) + F_(n-1) + 2 since C(n) = S(n) + S(n-2) + 2 Jan 6, 2021 at 13:51\n\u2022 Sure, they are equivalent :) Jan 6, 2021 at 15:23", "date": "2022-07-01 02:12:38", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/133977/the-number-of-arrangements-of-n-players-around-a-round-table-where-each-player", "openwebmath_score": 0.7444012761116028, "openwebmath_perplexity": 296.0715326814054, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137879323496, "lm_q2_score": 0.8670357512127873, "lm_q1q2_score": 0.8514410623212395}}
{"url": "http://mathhelpforum.com/calculus/119701-help-surface-integrals.html", "text": "# Math Help - [HELP!] Surface Integrals\n\n1. ## [SOLVED] Surface Integrals\n\nI don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<\n\nOK...here's my question:\n\nDoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant.\n\nI know that maybe...I should use\ny = rcos(theta) = cos(theta)\nz = rsin(theta) = sin(theta)\nx = x\nand find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)\nand find the magnitude of |r_(theta) x r_(z)| = 1\n\nI don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?\n\nCan anyone solve this question for me please? I know the final answer, which is 12.\n\n2. Hi. By definition:\n\n$S=\\mathop\\int\\int\\limits_{\\hspace{-15pt}S} g(x,y,z)dS=\\mathop\\int\\int\\limits_{\\hspace{-15pt}R} g(x,y,f(x,y))\\sqrt{f_x^2+f_y^2+1} dA$\n\nSo you're integrating the function $g(x,y,z)=z+x^2 y$ over the surface $f(x,y)=z=\\sqrt{1-y^2}$ between the planes x=0 and x=3 in the first quadrant. That's just the top one-quarter of the cylinder along the x-axis shown in red below. Now, the region in the x-y plane below that part, since the cylinder has a radius of one, would be 0 to 3 in the x-direction and 0 to 1 in the y-direction. Now, turn the crank:\n\n$\\int_0^3\\int_0^1 (\\sqrt{1-y^2}+x^2y)\\sqrt{f_x^2+f_y^2+1}dydx$\n\n3. Originally Posted by violet8804\nI don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<\n\nOK...here's my question:\n\nDoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant.\n\nI know that maybe...I should use\ny = rcos(theta) = cos(theta)\nz = rsin(theta) = sin(theta)\nx = x\nand find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)\nand find the magnitude of |r_(theta) x r_(z)| = 1\n\nI don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?\n\nCan anyone solve this question for me please? I know the final answer, which is 12.\n\nHere's a general concept: if you write the surface as $\\vec{r}(u,v)= x(u,v)\\vec{i}+ y(u,v)\\vec{j}+ z(u,v)\\vec{k}$, then the two derivatives, $\\vec{r}_u= x_u \\vec{i}+ y_u\\vec{j}+ z_u\\vec{k}$ and $\\vec{r}_v= x_v\\vec{i}+ y_v\\vec{j}+ z_v\\vec{k}$ are tangent to the surface and their cross product (called the \"fundamental vector product\" of the surface), $\\vec{r}_u\\times\\vec{r_v}$, is perpendicular to the surface and its length gives the \"differential of surface area\": $||\\vec{r}_u\\times\\vec{r}_v||dudv$.\n\nYes, you can use $y= cos(\\theta)$, $z= sin(\\theta)$, x= x with $\\theta$ and x as parameters u and v. That is, $\\vec{r}= x\\vec{i}+ cos(\\theta)\\vec{j}+ sin(\\theta)\\vec{k}$ so $\\vec{r}_x= \\vec{i}$ and $\\vec{r}_\\theta= -sin(\\theta)\\vec{j}+ cos(\\theta)\\vec{k}$ so that the \"fundamental vector product\" is\n\n$\\vec{r}_x\\times\\vec{r}_\\theta= -cos(\\theta)\\vec{i}- sin(\\theta)\\vec{k}$.\n\nIts length is, of course, 1 so $dV= dxd\\theta$. The function to be integrated is $z+ x^2y= sin(\\theta)+ x^2cos(\\theta)$. In the first octant, both y and z are positive so $\\theta$ goes from 0 to $\\pi/2$ and, of course, x goes from 0 to 1. The integral is\n\n$\\int_{x=0}^1\\int_{\\theta= 0}^{\\pi/2} sin(\\theta)+ x^2 cos(\\theta) dxd\\theta$.\n\nBy the way, if you click on a mathematical expression in any post, you can see the LaTex code used.\n\n4. Originally Posted by HallsofIvy\nHere's a general concept: if you write the surface as $\\vec{r}(u,v)= x(u,v)\\vec{i}+ y(u,v)\\vec{j}+ z(u,v)\\vec{k}$, then the two derivatives, $\\vec{r}_u= x_u \\vec{i}+ y_u\\vec{j}+ z_u\\vec{k}$ and $\\vec{r}_v= x_v\\vec{i}+ y_v\\vec{j}+ z_v\\vec{k}$ are tangent to the surface and their cross product (called the \"fundamental vector product\" of the surface), $\\vec{r}_u\\times\\vec{r_v}$, is perpendicular to the surface and its length gives the \"differential of surface area\": $||\\vec{r}_u\\times\\vec{r}_v||dudv$.\n\nYes, you can use $y= cos(\\theta)$, $z= sin(\\theta)$, x= x with $\\theta$ and x as parameters u and v. That is, $\\vec{r}= x\\vec{i}+ cos(\\theta)\\vec{j}+ sin(\\theta)\\vec{k}$ so $\\vec{r}_x= \\vec{i}$ and $\\vec{r}_\\theta= -sin(\\theta)\\vec{j}+ cos(\\theta)\\vec{k}$ so that the \"fundamental vector product\" is\n\n$\\vec{r}_x\\times\\vec{r}_\\theta= -cos(\\theta)\\vec{i}- sin(\\theta)\\vec{k}$.\n\nIts length is, of course, 1 so $dV= dxd\\theta$. The function to be integrated is $z+ x^2y= sin(\\theta)+ x^2cos(\\theta)$. In the first octant, both y and z are positive so $\\theta$ goes from 0 to $\\pi/2$ and, of course, x goes from 0 to 1. The integral is\n\n$\\int_{x=0}^1\\int_{\\theta= 0}^{\\pi/2} sin(\\theta)+ x^2 cos(\\theta) dxd\\theta$.\n\nBy the way, if you click on a mathematical expression in any post, you can see the LaTex code used.\n\nThank you!\nAnd, by the way...\nIs the (theta) always from 0 to 2*Pi if it's in the 1st octant (after change to polar coordinates)?\nAlso, can you please teach me when to use which of the two formula, is there a trick to see from the question when to use which formula?\nI always get confused which one to use, they are all the homework questions, and i got partial solutions (the library only have the older version of the solution manual...); so, when I do the question myself, and checked from the solution manual, it used a different formula (or different method) to solve the surface integral...I'm so confused T_T", "date": "2015-07-05 18:39:19", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/119701-help-surface-integrals.html", "openwebmath_score": 0.886398434638977, "openwebmath_perplexity": 407.7310775082805, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137879323496, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8514410623212394}}
{"url": "https://math.stackexchange.com/questions/2432215/prove-forall-n-ge-2-2n-3n-4n/2432219", "text": "# Prove $\\ \\forall n \\ge 2, \\ 2^{n} + 3^{n} < 4^{n}$\n\nQuestion:\n\nProve $\\ \\forall n \\ge 2, \\ 2^{n} + 3^{n} < 4^{n}$\n\nMy attempt:\n\nBase case is trivial.\n\nSuppose $\\ n \\ge 2$ and $\\ 2^{n} + 3^{n} < 4^{n}$\n\nThen,\n\n$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n}$, by I.H.\n\nI am stuck here. how do I show that this expression is $< 4^{n+1}$?\n\n\u2022 Hints: Is $3^n<4^n$? Is $3(4^n)<4(4^n)$? \u2013\u00a0Michael Burr Sep 16 '17 at 20:25\n\u2022 Another proof could use that the equivalent inequality $(2/4)^n+(3/4)^n<1$ after dividing both sides by $4^n$, where $4^n>0$, holds because it holds for $n=2$ and $(2/4)^n$ and $(3/4)^n$ are both strictly decreasing. \u2013\u00a0user236182 Sep 16 '17 at 20:32\n\nYou can say $$2\\cdot 2^n + 3\\cdot 3^n < 4\\cdot 2^n + 4\\cdot 3^n = 4(2^n+3^n) < 4\\cdot 4^n = 4^{n+1}.$$\nmultiplying $$2^n+3^n<4^n$$ by $$4^n$$ we get $$2^n\\cdot 2^{2n}+3^n\\cdot 2^{2n}<4^{n+1}$$ and we have $$2^n\\cdot 2<2^{3n}$$ since $$1<2^{2n-1}$$ for $n\\geq 2$ and $$3^n\\cdot 3<3^n\\cdot 2^{2n}$$ since $$3<2^{2n}$$ therefore we have $$2^{n+1}+3^{n+1}<2^n\\cdot 2^{2n}+3^n\\cdot 2^{2n}<4^{n+1}$$\nHint: \u00a0 for $\\,0 \\lt a \\lt 1\\,$ the sequence $\\,a^n\\,$ is decreasing with $\\,n\\,$ since $\\,a^{n+1} = a \\cdot a^n \\lt 1 \\cdot a^n\\,$, so:\n$$\\left(\\frac{2}{4}\\right)^n+\\left(\\frac{3}{4}\\right)^n \\le \\left(\\frac{2}{4}\\right)^2+\\left(\\frac{3}{4}\\right)^2 = \\frac{13}{16} \\lt 1 \\quad\\quad\\text{for}\\;\\; \\forall n \\ge 2$$", "date": "2019-06-16 11:51:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2432215/prove-forall-n-ge-2-2n-3n-4n/2432219", "openwebmath_score": 0.9710681438446045, "openwebmath_perplexity": 191.34596186232866, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513869353992, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8514348319108301}}
{"url": "https://math.stackexchange.com/questions/2071403/minimal-specification-of-isometry-in-terms-of-norm-preservation", "text": "# Minimal specification of isometry in terms of norm preservation\n\nLet $V,W$ be $n$-dimensional (real) inner product spaces, and let $T:V \\to W$ be a linear map.\n\nLet $v_1,...,v_n$ be a basis of $V$. It is easy to see that if $|T(v)|_W=|v|_V$ for every $v \\in \\{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\\}$, then $T$ is an isometry (a proof is provided below).\n\nIn other words, after choosing wisely $k(n):=\\frac{n(n+1)}{2}$ vectors, it is enough to verify $T$ preserves the norms of these special vectors, in order to conclude it's an isometry.\n\nQuestion: Is there no way to choose less than $k(n)$ vectors, in such a way that every linear map which preserves their norms is an isometry?\n\nI believe we cannot choose less vectors. I have some \"convincing evidence\" for the cases $n=1,2,3$ (see below), but I am not sure how to give a rigorous argument.\n\nNote that a \"wise choice\" of vectors does not have to be of the form of some vectors, and linear combinations of them (I do feel this it the most efficient method, but I don't see how to prove this). Even if we prove that this is the case, than we need to show we cannot do better than to work with only orthonormal bases.\n\nThe partial \"evidence\":\n\n$n=1: k=1$. Obvious\n\n$n=2: k=3$. Take $V=W=\\mathbb{R}^n$ with its standard inner product. Then, $T(e_1)=e_1, T(e_2)=\\frac{e_1+e_2}{\\sqrt 2}$ is a counter example.\n\n$n=3: k=6$. Then any matrix of the form $$\\begin{pmatrix} c & s & x \\\\ -s & c & y \\\\ 0 & 0 & z \\\\\\end{pmatrix}$$ where $c^2+s^2=1,x^2+y^2+z^2=1, sx+cy=0$ preserves the norms $e_1,e_2,e_1+e_2,e_3,e_2+e_3$ but it's an isometry only if $|z|=1,x=y=0$.\n\nProof that $k(n)=\\frac{n(n+1)}{2}$ vectors are enough:\n\nNoting that $$\\langle u,v \\rangle = \\frac{1}{2}(|u+v|^2 - |u|^2 - |v|^2) ,$$ we obtain\n\n$$\\langle Tv_i,Tv_j \\rangle = \\frac{1}{2}(|Tv_i+Tv_j|^2 - |Tv_i|^2 - |Tv_j|^2) = \\frac{1}{2}(|T(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2)$$ $$= \\frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \\langle v_i,v_j \\rangle,$$\n\nthus $T$ is an isometry.\n\n\u2022 I think there is not a way to choose less. Since it must preserve the scalar product and then all products $<v_i,v_j>$ of the basis must be preserved. By symmetry they are exactly the number you proposed. \u2013\u00a0Harnak Dec 25 '16 at 11:02\n\u2022 @Harnak, please make your comment into an answer, I'd like to +1 it. \u2013\u00a0Andreas Caranti Dec 25 '16 at 11:09\n\u2022 @Harnak I am not sure I am convinced by your argument. Can you please elaborate? (I also think there is no way to choose less). \u2013\u00a0Asaf Shachar Dec 25 '16 at 11:13\n\nLet $f:\\mathbb R^n\\to V$ and $g:W\\to\\mathbb R^n$ be any two linear isometries. Then $T$ is an isometry if and only if $g\\circ T\\circ f$ is an isometry. So, we may assume without loss of generality that $V=W=\\mathbb R^n$ equipped with the Euclidean inner product.\n\nLet $v_1,\\ldots,v_k\\in\\mathbb R^n$ be $k<\\frac12n(n+1)$ arbitrarily chosen vectors. It suffices to exhibit the existence of a non-isometric linear transformation $T$ on $\\mathbb R^n$ that preserves the norm of each $v_j$. Consider the system of homogeneous linear equations $$v_j^\\top Av_j=0,\\quad j=1,\\ldots,k,\\tag{1}$$ where the $n^2$ entries of $A\\in M_n(\\mathbb R)$ are unknown. Since the subspace of all $n\\times n$ skew-symmetric matrices has dimension $\\frac12n(n-1)<n^2-k$, the system $(1)$ must admit a nontrivial solution $A$ that is not skew-symmetric. However, if $A$ is a solution, so is $A+A^\\top$. Therefore, $(1)$ admits a nontrivial symmetric solution $A$.\n\nLet $P=I+\\varepsilon A$, where $\\varepsilon>0$ is sufficiently small. Then $P$ is positive definite. Define $Tx=\\sqrt{P}x$. Then $T$ is not an isometry because $\\sqrt{P}$ is not real orthogonal. However, for each $j$ we have $$\\|Tv_j\\|^2=v_j^\\top Pv_j=v_j^\\top(I+\\varepsilon A)v_j=v_j^\\top v_j=\\|v_j\\|^2.$$\n\n\u2022 Wow! your solution is great and provides exactly what I was interested in, thanks. By the way, I am curious how did you think about using symmetric matrices. (I guess you started from considering perturbations of the identity, but then noticed that if you do not take a square root, the constraint you get is a quadratic equation, not linear, hence the necessity to consider roots...). \u2013\u00a0Asaf Shachar Dec 28 '16 at 12:53\n\u2022 @AsafShachar Once you recognise that the problem boils down to finding a positive semidefinite matrix $P\\ne I$ such that $v_j^\\top Pv_j=v_j^\\top v_j$ for each $j$, it's natural to consider symmetric matrices, because $I-P$ is symmetric. \u2013\u00a0user1551 Dec 28 '16 at 13:27\n\u2022 I agree, but it was not immediate for me to consider the \"square root\" trick. (i.e I thought of starting with $Tx=Px$ where $P=I+\\epsilon A$, but then the equation on $P$ is quadratic...). Anyway, your explanation is very reasonable, thanks again. \u2013\u00a0Asaf Shachar Dec 28 '16 at 16:52\n\nI'll try to elaborate on my idea in the comment. Preserving the scalar product means that the two symmetric bilinear forms must coincide: $$(x,y) \\mapsto \\langle x,y \\rangle$$ $$(x,y) \\mapsto \\langle Tx,Ty \\rangle$$ This is equivalent to saying that their matrices must coincide. But a symmetric matrix has $\\frac{n(n+1)}{2}$ degrees of freedom.\nThis means, once we've chosen the basis in $V$ we must choose the products $\\langle v_i,v_j \\rangle$ for $i \\geq j$.\n\n\u2022 Is it enough? Yes, because we've chosen the coefficients of the matrix relative to its canonic basis (here I'm taking about the basis of matrices which have $1$ in one component and $0$ in the others, but only for $i \\geq j$ since the matrix is symmetric).\n\u2022 Can we take less? No, because the symmetric matrices space has dimension $\\frac{n(n+1)}{2}$ as we've said.\n\nHow does this relate to our question? Choosing a basis for $V$ and then such products is equivalent to choosing the list of vectors you proposed (along with their norms).\n(I.e. Choosing a bilinear form is equivalent to choosing the quadratic forum associated to it, which in the case of a scalar product is the norm)\n\n\u2022 Thanks for your elaborate answer. However, I must say I am not entirely convinced by your argument. Indeed, I am also thinking that we cannot choose less vectors, and your observation that a (symmetric) bilinear form is determined by $\\frac{n(n+1)}{2}$ vectors is also appealing and looks connected to this issue somehow. In spite all of that I am still not convinced. Suppose we know $\\langle v_i,v_j \\rangle = \\langle Tv_i,Tv_j \\rangle$ for all $i \\le j$ except for a single pair $(i^*,j^*)$... \u2013\u00a0Asaf Shachar Dec 26 '16 at 15:34\n\u2022 I would like to see a more explicit demonstration for how to build such a map $T$ which is not an isometry, i.e does not satisfy $\\langle v_i,v_j \\rangle = \\langle Tv_i,Tv_j \\rangle$ for $(i,j)=(i^*,j^*)$. I am not sure this follows so clearly from your \"heuristic\" argument. \u2013\u00a0Asaf Shachar Dec 26 '16 at 15:34\n\u2022 So, you want to prove that there exists a map $T$ that preserves all those couples except for one? \u2013\u00a0Harnak Dec 26 '16 at 16:49\n\u2022 Yes, for a start. My intention in the question was to show that no matter how cleverly you choose $k < \\frac{n(n+1)}{2}$ vectors, and require they will be mapped to vectors of the same norm, you will be able to do this with a map which is not an isometry. However, in some sense this will only show the scheme of choosing basis vectors (and certain linear combinations of them) cannot be successful with less vectors, we also need to show we cannot do better by any choice. \u2013\u00a0Asaf Shachar Dec 26 '16 at 19:50", "date": "2019-05-20 19:22:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2071403/minimal-specification-of-isometry-in-terms-of-norm-preservation", "openwebmath_score": 0.9399972558021545, "openwebmath_perplexity": 166.36462573091117, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513901914406, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8514348312556671}}
{"url": "https://www.qalaxia.com/questions/What-is-the-probability-of-this-problem", "text": "D\n\n#### What is the probability of this problem?\n\n63 viewed last edited 2 years ago\nAnshul Malik\n1\n\nA and B roll a dice each. A wins if his dice number is greater than B. B wins if his dice number is greater than or equal to A.\n\nIf A gets a 1, then B will win for sure. So B makes up a rule that A will play the same game again if A rolls a 1 in the first throw. The winner of the second game (if it reaches that stage) will be the final winner even if A rolls a 1 again.\n\nWhat is the probability that A will win this game?\n\nMahesh Godavarti\n3\n\nI think the problem statement can be worded better for easier understanding. However, I think I got it.\n\nProbability of A winning = 1 - Probability of B winning.\n\nProbability of B winning =\n\nProbability of B throwing an equal or greater than when A throws a number greater than 1 in the first throw +\n\nProbability that A throws a 1 in the first throw X Probability of B getting an equal or higher number in the second throw than A =\n\n\\frac{1}{6 } \\times (\\frac{5}{6} + \\frac{4}{6} + \\frac{3}{6} + \\frac{2}{6} + \\frac{1}{6}) + \\frac{1}{6} \\times \\frac{1}{6} \\times (\\frac{6}{6} + \\frac{5}{6} + \\frac{4}{6} + \\frac{3}{6} + \\frac{2}{6} + \\frac{1}{6}) =\n\n\\frac{1}{6} \\times \\frac{1}{6} \\times 5 \\times 6 \\times \\frac{1}{2} + \\frac{1}{6} \\times \\frac{1}{6} \\times \\frac{1}{6} \\times 6 \\times 7 \\times 1/2 = 1/6 \\times 1/6 \\times 1/2 \\times (30 + 7) = \\frac{37}{72}\n\nTherefore, probability of A winning = 35/72.\n\nEarlier, if B had not changed the rule then the probability of B winning would be\n\n1/6 \\times (6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6) = 1/6 \\times 1/6 \\times 6 \\times 7 \\times 1/2 = 7/12 .\n\nAnd the probability of A winning would be 5/12.\n\nEssentially, B leveled the playing field a little bit by changing the rules of the game.\n\nVivekanand Vellanki\n0\nIt would be good to explain how the probability of B throwing a number greater than or equal to A is determined.\nAnshul Malik\n0\nBut the cases in which A rolls a 1 in the first roll should not be counted as sample cases right? Also, the probability that the game is decided in the first roll is 5/6? If I'm understanding conditional probability correctly, then this should also be factored in right?\nMahesh Godavarti\n0\nI just swt up the sample space and added up all the cases that end up in B's victory. You could set it up as a conditional as well, with A rolling a 1 or not rolling a 1. Or the game finishing in one throw or finishing in two throws, you will get the same answer. In fact, I will set it up and show you that it is the same. Look for my second answer.\nMahesh Godavarti\n0\n\nLet's do it another way. Let's condition it first on the event that the game will finish in one throw or two throws.\n\nThen P(B wins) = P(one throw) P(B wins | one throw) + P(two throws) P(B wins | two throws).\n\nP(one throw) = P(A throws a number other than 1) = 5/6\n\nP(two throws) = P(A throws a 1) = 1/6.\n\nP(B wins | one throw) = P(A throws a 2 | A threw a number from 2 to 5) X P(B throws 2 or higher) + P(A throws a 3 | A threw a number from 2 to 5) X P(B throws 3 or higher) + and so on till A throws a 6 = 1/5 * 5/6 + 1/5 * 4/6 + 1/5 * 3/6 + 1/5 *2/6 + 1/5 * 1/6.\n\nTherefore, P(one throw) P (B wins | one throw) = 5/6 * [1/5 * 5/6 + 1/5 * 4/6 + 1/5 *3/6 + 1/5 *2/6 + 1/5 * 1/6 ] = 1/6 * 5/6 + 1/6 *4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 *1/6 = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6].\n\nP(B wins | two throws) = P(A throws a 1 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw\n\nSince, the throws are independent what A throws in the second is independent of what is thrown in the first. Therefore, the above expression is the same as:\n\nP(B wins | two throws) = P(A throws a 1 in the second throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw = 1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6\n\nTherefore, P (two throws) P(B wins | two throws) = 1/6 * [1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6] = 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6]\n\nTherefore, P(B wins) = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6] + 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6].\n\nWhich is the same as what we had in the other answer.", "date": "2020-11-28 19:05:36", "meta": {"domain": "qalaxia.com", "url": "https://www.qalaxia.com/questions/What-is-the-probability-of-this-problem", "openwebmath_score": 0.9937614798545837, "openwebmath_perplexity": 362.1831691700134, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765628041175, "lm_q2_score": 0.868826784729373, "lm_q1q2_score": 0.8514298861712439}}
{"url": "http://math.stackexchange.com/questions/198641/using-induction-prove-that-sum-k-1n-frac-1-2k-12k1-frac-12-frac", "text": "# Using induction, prove that $\\sum_{k=1}^n\\frac 1 {(2k-1)(2k+1)}= \\frac 12-\\frac 1 {4n+2}$ when $n$ is in $\\mathbb{N}$\n\nI can prove the base case and get that $1/3=1/3$, but i can't get any further with the $n+1$ case. Can someone help me?\n\nI was told to conjecture a formula for the sum $$\\frac{1}{3} + \\frac{1}{{3 \\cdot 5}} + \\cdots + \\frac{1}{{\\left( {2n - 1} \\right)\\left( {2n + 1} \\right)}}$$ I thought I figured out that this sum was equal to $$\\frac{1}{2} { - \\frac{1}{{4n + 2}}}$$ but I'm starting to think I'm wrong about that. After we have the formula, we were told to prove our conjecture using induction.\n\n-\nI really can't understand what is the identity the you want to prove. Could you try to typeset it (better if in LaTeX), with all the necessary parentheses, please? \u2013\u00a0 Andrea Orta Sep 18 '12 at 18:25\nAre you sure you have the right expression? For $n = 2$ you get: $\\frac{1}{2 \\cdot 6} = \\frac{1}{2} - \\frac{1}{10}$ which is clearly untrue. \u2013\u00a0 Feanor Sep 18 '12 at 18:25\nIf $n=1$ you are dividing by $0$. Also please check your parentheses: one extra left on the left side, it is not clear whether $(2n+2)$ should be in the numerator, and you must mean $1/(4n+2)$, not $(1/4)n+2$ or $1/(4n)+2$ on the right. \\frac is your friend in this regard. \u2013\u00a0 Ross Millikan Sep 18 '12 at 18:25\nAlso, unless there was supposed to be some sum involved, there is no need for induction. Just multiply everything out so as to get rid of fractions ;) \u2013\u00a0 Feanor Sep 18 '12 at 18:27\nBased on your added comment, you want the left to be $\\sum_{i=1}^n \\frac 1{2i-1}\\frac 1{2i+1}=\\frac 12 - \\frac 1{4n+2}$. Note the $1$'s, not $2$'s and the sum on the left. \u2013\u00a0 Ross Millikan Sep 18 '12 at 18:37\n\nHint $\\$ The sum telescopes since for $\\rm\\:f(k) = 1/(4k-2)\\:$ we have\n\n$$\\rm f(k+1)-f(k)\\ =\\ \\frac{1}{4k+2} - \\frac{1}{4k-2}\\ =\\ -\\frac{1}{(2k-1)(2k+1)}$$\n\nso an inductive proof is a special case of the inductive proof of the closed form for a telescopic sum:\n\n$$\\rm\\sum_{k=1}^n f(k\\!+\\!1)-f(k) = - f(1) + \\color{#C00}{f(2) - f(2) + \\cdots + f(n)-f(n)} + f(n\\!+\\!1)\\, =\\, f(n\\!+\\!1)-f(1)$$\n\nBut it is easy to turn the above ellipses into a rigorous inductive proof. Then your problem is simply a corollary of this general telescopy lemma, for the special value of $\\rm\\:f(k)\\:$ given above.\n\nThe advantage of proving it this way is that - with no extra effort - you now have a general lemma that works to prove (by induction!) all summation identities of this form. Note that even though the induction has been abstracted out into a proof of a more general telescopy lemma, a proof invoking the telescopy lemma still counts as a proof by induction. The induction simply has been encapsulated in the proof of the lemma, which need not be repeated inline every time it is invoked. DivideAbstract and conquer - once you've seen one telescopic induction proof you've seen them all!\n\nYou can find more examples and further discussion in my prior posts on telescopy.\n\n-\n@Peter Telescopy can be viewed as a form of induction specifically tailored for computing certain sums and products. \u2013\u00a0 Bill Dubuque Sep 18 '12 at 19:00\nYes, I was just being silly, Bill! \u2013\u00a0 Pedro Tamaroff Sep 18 '12 at 19:03\n\nWhat you want to prove, apparently, is that\n\n$$\\sum_{k=1}^n \\frac{1}{(2k-1)(2k+1)}=\\frac 1 2 - \\frac 1 {2(2n+1)}$$\n\nAs you say, the base case, $n=1$ is true. Suppose true for $n$, and analyze $n+1$. We get\n\n$$\\sum_{k=1}^{n+1} \\frac{1}{(2k-1)(2k+1)}=\\frac 1 2 - \\frac 1 {2(2(n+1)+1)}$$\n\n$$\\sum_{k=1}^{n} \\frac{1}{(2k-1)(2k+1)}+\\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\\frac 1 2 - \\frac 1 {2(2(n+1)+1)}$$\n\nBy the inductive hypothesis, this is, $$\\frac 1 2 - \\frac 1 {2(2n+1)}+\\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\\frac 1 2 - \\frac 1 {2(2(n+1)+1)}$$\n\n$$- \\frac{1}{{2(2n + 1)}} + \\frac{1}{{(2n + 1)(2n + 3)}} = - \\frac{1}{{2(2n + 3)}}$$\n\nMultiply through ${(2n + 1)(2n + 3)}$ to get\n\n$$- \\frac{{(2n + 3)}}{2} + 1 = - \\frac{{(2n + 1)}}{2}$$\n\nIn the end, you get\n\n$$- 2n - 3 + 2 = - 2n - 1$$ $$- 2n -1 = - 2n - 1$$\n\nwhich is true, so the inductive step is complete, and the theorem is proven.\n\n-\n\nHint: This is a telescoping series. Note that $\\frac 1{2i-1}\\frac 1{2i+1}=\\frac 12 \\left(\\frac 1{2i-1}-\\frac 1{2i+1}\\right)$ so neighboring terms cancel.\n\n-\nWhile this is the easiest approach, is not solving the problem the way the problem asks ;) \u2013\u00a0 N. S. Sep 18 '12 at 18:43\n@N.S.: I think it is a step toward the induction. \u2013\u00a0 Ross Millikan Sep 18 '12 at 18:45\n@RossMilikan he already guessed the right result, so he has a very easy induction problem, where he doesn't really need telescoping. \u2013\u00a0 N. S. Sep 18 '12 at 18:49\n@N.S. Telescopy does use induction - it's simply encapsulated (as a lemma) in the general closed form for a telescopic sum (whose inductive proof is easier in the abstract than in any specific case). So the natural way to prove this is to prove by induction the closed form for a telescopic sum, then specialize it to the case at hand - see my answer. Then, later, one can reuse the telescopy lemma to quickly tackle all such problems. \u2013\u00a0 Bill Dubuque Sep 18 '12 at 19:09\n@BillDubuque True, and this is the easiest way for us... But the problem asks the student to conjecture a closed formula, and prove it by induction. If you telescope it, you already get the answer... Is like the problem of calculating $\\int_0^1 x^2 dx$... We typically expect the students to solve it the easy way, but if the Question explicitly asks to evaluate it using a Riemann sum, the easy solution is not good anymore... \u2013\u00a0 N. S. Sep 18 '12 at 20:53\nshow 1 more comment\n\nSaw this type of problem before.\n\nYou need to find a closed for for\n\n$$\\frac{1}{1 \\cdot 3} + \\frac{1}{3 \\cdot 5}+...+\\frac{1}{(2n-1)(1n+2)}$$\n\nand then prove it by induction.\n\nYour guess looks right, now you can do the induction:\n\n$$\\frac{1}{1 \\cdot 3} + \\frac{1}{3 \\cdot 5}+...+\\frac{1}{(2n-1)(1n+2)} =\\frac{1}{2}-\\frac{1}{2n+1} \\,.$$\n\nAlternately, you can use partial fraction decomposition to find a simple formula for $$\\frac{1}{(2k-1)(2k+1)} \\,,$$ and then get a telescopic sum. There is no need to prove this partial fraction decomposition by induction, and this approach is not really a direct induction proof.\n\n-", "date": "2014-07-22 21:45:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/198641/using-induction-prove-that-sum-k-1n-frac-1-2k-12k1-frac-12-frac", "openwebmath_score": 0.9296789169311523, "openwebmath_perplexity": 299.3253628047611, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765610497292, "lm_q2_score": 0.868826777936422, "lm_q1q2_score": 0.8514298779900517}}
{"url": "https://math.stackexchange.com/questions/1110860/when-does-the-two-cars-meet", "text": "# When does the two cars meet\n\nAt 10:30 am car $A$ starts from point $A$ towards point $B$ at the speed of $65$ km/hr, at the same time another car left from point $B$ towards point $A$ at the speed of $70$ km/hr, the total distance between two points is $810$ km, at what time does these two cars meet ?\n\nThis is what I have tried,\n\n$t_1 = \\frac{810}{65} \\,$ km/hr $= 12.46 \\,$hr\n$t_2 = \\frac{810}{70} \\,$ km/hr $= 11.57\\,$ hr\n$t_1-t_2 = 0.89 \\,$ hr\n$t_1-t_2 = 0.89\\cdot 60 = 53.4 \\,$ minutes\n\nbut this couldn't be the answer because how could these two cars could meet after $53.4$ minutes ?\n\n@Joe, when they start they are 810km apart; after 1 hour the person leaving from point A would have traveled 65km, so if the other person had not left from point B, then the distance between them would be 810-65=745km. But the other person also helps close the distance between them by traveling 70km in the 1st hour from point B towards A, so the distance between them is 810-(65+75)=810-135=675km. So they are effectively subtracting 135km worth of distance per hour from the original distance between them.\n\nSo the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.\n\n\u2022 it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? \u2013\u00a0Joe Jan 19 '15 at 18:52\n\u2022 Yes, this is exactly right. \u2013\u00a0Acemanhattan Jan 19 '15 at 18:56\n\u2022 But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. \u2013\u00a0turkeyhundt Jan 19 '15 at 19:00\n\u2022 Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. \u2013\u00a0Acemanhattan Jan 19 '15 at 19:02\n\u2022 Thanks a lot for great explanation. =) \u2013\u00a0Joe Jan 19 '15 at 19:03\n\nHint: Think about how much closer the cars get each hour.\n\nThey are approaching each other at an effective speed of 135 km/hr...\n\n\u2022 but at what time does these two cars meet ? would that be 135/2 ? \u2013\u00a0Joe Jan 19 '15 at 18:27\n\u2022 like 135/2 = 67.5 and then 810/67.5 = 12 ? \u2013\u00a0Joe Jan 19 '15 at 18:29\n\u2022 no. it would be how long it would take to travel 810km at a rate of 135km/hr. Because every hour, they get 135km closer to each other. \u2013\u00a0turkeyhundt Jan 19 '15 at 18:50\n\u2022 it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? \u2013\u00a0Joe Jan 19 '15 at 18:54\n\u2022 Yes. very good. \u2013\u00a0turkeyhundt Jan 19 '15 at 18:59\n\nHint: The total distance between the two cars is initially $810~\\text{km}$. Thus, in order to meet, the combined distance they travel is $810~\\text{km}$. Since they are travelling toward each other, they get $65~\\text{km} + 70~\\text{km} = 135~\\text{km}$ closer to each other each hour.\n\n\u2022 so then I have two divide 135 by 2 and then divide the answer by 810, like this, 135/2 = 67.5 then 810/67.5 = 12hours right ? so they will meet at 10:30 pm right ? \u2013\u00a0Joe Jan 19 '15 at 18:48\n\u2022 No. What you did is compute the time it would take for the average distance travelled by the cars to be $810~\\text{km}$, by which time they would have long passed each other. The cars get $135~\\text{km}$ closer to each other each hour. You want to reduce the distance between them, which is initially $810~\\text{km}$, to $0~\\text{km}$. \u2013\u00a0N. F. Taussig Jan 19 '15 at 18:51\n\u2022 it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? \u2013\u00a0Joe Jan 19 '15 at 18:53\n\u2022 That is correct. \u2013\u00a0N. F. Taussig Jan 19 '15 at 18:54\n\u2022 They do not have to cover $810~\\text{km}$ individually. After six hours, the slower car has travelled $$65~\\frac{\\text{km}}{\\text{h}} \\cdot 6~\\text{h} = 390~\\text{km}$$ so it is now $810~\\text{km} - 390~\\text{km} = 420~\\text{km}$ from where the faster car began. Since the faster car has travelled $$70~\\frac{\\text{km}}{\\text{h}} \\cdot 6~\\text{h} = 420~\\text{km}$$ toward the slower car, they meet at 4:30 pm. Draw a picture. \u2013\u00a0N. F. Taussig Jan 19 '15 at 19:03", "date": "2019-06-26 08:56:03", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1110860/when-does-the-two-cars-meet", "openwebmath_score": 0.6848351359367371, "openwebmath_perplexity": 669.1637675922037, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765552017674, "lm_q2_score": 0.8688267796346599, "lm_q1q2_score": 0.8514298745734191}}
{"url": "https://math.stackexchange.com/questions/1166269/lucas-numbers-and-fibonacci", "text": "Lucas numbers and fibonacci\n\nThis is a question straight from the Applied Combinatorics book.\n\nSuppose that chairs are arranged in a circle. Let $L_n$ count the number of subsets of $n$ chairs which don't contain consecutive chairs. show that $$L_{n+1} = F_n + F_{n+2}.$$\n\nWhat is meant here? Could some one explain this to me?\n\n\u2022 Do you need a proof, or do you have difficulties to understand the statement ? Feb 26, 2015 at 10:35\n\u2022 What confuses me is that the number of chairs is not given. Feb 26, 2015 at 10:37\n\u2022 Proof and explanation please. I don't understand what is meant by the question. Feb 26, 2015 at 10:56\n\nIf $n$ chairs are arranged in a circle, let $a_n$ be the number of subsets of the chairs that don\u2019t contain consecutive chairs. (I\u2019ll call these good subsets.) Clearly $a_0=1$, $a_1=2$, $a_2=3$, and $a_3=4$, since the empty subset satisfies the condition.\n\nNow suppose that $n\\ge 4$, and number the chairs from $1$ to $n$ around the circle. Temporarily remove chair $n$; the circle of remaining chairs has $a_{n-1}$ good subsets, none of which contains both chair $1$ and chair $n-1$. Of course none of these contains chair $n$, either. Moreover, every good subset of the $n$ chairs that does not contain chair $n$ and contains at most one of chairs $1$ and $n-1$ is counted here, so there are $a_{n-1}$ such subsets.\n\nNow we\u2019ll count the good subsets that contain either contain chair $n$ or contain both chair $1$ and chair $n-1$. Start by removing chairs $n-1$ and $n$; the remaining circle has $a_{n-2}$ good subsets. Let $S$ be one of these good subsets. If chair $1$ is in $S$, form $S'$ by adding chair $n-1$ to $S$; if chair $1$ is not in $S$, form $S'$ by adding chair $n$ to $S$. It\u2019s not hard to check that in each case $S'$ is a good subset of the ring of $n$ chairs, and that every such subset that contains either chair $n$ or both of chairs $1$ and $n-1$ is obtained in this way. (It\u2019s here that we need to have $n\\ge 4$: in order to form $S'$ this way when chair $1$ is in $S$, we need to know that $n-1>2$.)\n\nThis argument shows that the sequence $\\langle a_n:n\\in\\Bbb N\\rangle$ satisfies the recurrence\n\n$$a_n=a_{n-1}+a_{n-2}\\tag{0}$$\n\nfor $n\\ge 4$. By inspection $a_2=3=2+1=F_3+F_1$ and $a_3=4=3+1=F_4+F_2$, so a straightforward induction shows that $$a_n=F_{n+1}+F_{n-1}\\tag{1}$$ for $n\\ge 2$:\n\n\\begin{align*} a_{n+1}&=a_n+a_{n-1}\\\\ &=\\left(F_{n+1}+F_{n-1}\\right)+\\left(F_n+F_{n-2}\\right)&\\text{induction hypothesis}\\\\ &=\\left(F_{n+1}+F_n\\right)+\\left(F_{n-1}+F_{n-2}\\right)\\\\ &=F_{n+2}+F_n\\;. \\end{align*}\n\nApart from the fact that I used $a_n$ instead of $L_n$ for the number of good subsets of a ring of $n$ chairs, $(1)$ is exactly relationship that you were to prove.\n\nNothing in the wording of the question or the original tagging indicates that $L_n$ is supposed to be the $n$-th Lucas number: $L_n$ is explicitly defined to be the number of good subsets of a ring of $n$ chairs. (The tag was added by Martin Sleziak.) The choice of the notation $L_n$ for my $a_n$ in the problem statement was very likely motivated by the fact that $a_n$ is the $n$-th Lucas number, $L_n$, for $n>1$. This follows immediately from the fact that the sequence $\\langle L_n:n\\in\\Bbb Z^+\\rangle$ of Lucas numbers by definition satisfies the same recurrence $(0)$ as the sequence $\\langle a_n:n\\in\\Bbb N\\rangle$, with $L_1=1$ and $L_2=3=a_2$: $L_3=4=a_3$, and the recurrence then ensures that $a_n=L_n$ for all $n\\ge 2$.\n\n\u2022 Thanks for the clear explaination! Mar 1, 2015 at 11:41\n\u2022 @Discreteballoons: My pleasure! Mar 1, 2015 at 11:55\n\nThe Lucas numbers are defined recursively by \\begin{align*} L_1 & = 1\\\\ L_2 & = 3\\\\ L_n & = L_{n - 1} + L_{n - 2}, n \\geq 3 \\end{align*} The first few terms of the Lucas sequence are $\\{1, 3, 4, 7, 11, 18, 29, 47, 76, 123, \\ldots\\}$.\n\nThe Fibonacci numbers are defined recursively by \\begin{align*} F_1 & = 1\\\\ F_2 & = 1\\\\ F_n & = F_{n - 1} + F_{n - 2}, n \\geq 3 \\end{align*} The first few terms of the Fibonacci sequence are $\\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \\ldots\\}$.\n\nWe can prove the assertion that $L_{n + 1} = F_n + F_{n + 2}$ by induction on $n$. Let $P(n)$ be the statement that $L_{n + 1} = F_n + F_{n + 2}$.\n\nIf $n = 1$, then $L_2 = L_{1 + 1} = 3 = 1 + 2 = F_1 + F_3$. Thus, $P(1)$ holds.\n\nIF $n = 2$, then $L_3 = L_{2 + 1} = 4 = 1 + 3 = F_2 + F_4$. Thus, $P(2)$ holds.\n\nAssume $L_{m + 1} = F_m + F_{m + 2}$ for each $n \\leq m$, where $m \\geq 2$. Let $n = m + 1$. Then \\begin{align*} L_{m + 2} & = L_{m + 1} + L_{m} && \\text{by definition of the Lucas sequence}\\\\ & = F_{m} + F_{m + 2} + F_{m - 1} + F_{m + 1} && \\text{by the induction hypothesis}\\\\ & = F_{m} + F_{m - 1} + F_{m + 2} + F_{m + 1}\\\\ & = F_{m + 1} + F_{m + 3} && \\text{by definition of the Fibonacci sequence} \\end{align*} so $P(m) \\Rightarrow P(m + 1)$. Hence, $P(n)$ holds for each positive integer $n$.\n\nLet $a_m$ denote the number of subsets of $m$ chairs in which no two chairs are consecutive. We will consider the first few cases, where the chairs are numbered from $1$ to $m$.\n\n\\begin{align*} & m = 1: \\emptyset, \\color{blue}{\\{1\\}}\\\\ & m = 2: \\color{red}{\\emptyset, \\{1\\}}, \\color{blue}{\\{2\\}}\\\\ & m = 3: \\color{red}{\\emptyset, \\{1\\}, \\{2\\}}, \\color{blue}{\\{3\\}}\\\\ & m = 4: \\color{red}{\\emptyset, \\{1\\}, \\{2\\}, \\{3\\}}, \\color{blue}{\\{4\\}}, \\color{green}{\\{1, 3\\}}, \\color{blue}{\\{2, 4\\}}\\\\ & m = 5: \\color{red}{\\emptyset, \\{1\\}, \\{2\\}, \\{3\\}, \\{4\\}}, \\color{blue}{\\{5\\}} \\color{red}{\\{1, 3\\}}, \\color{green}{\\{1, 4\\}}, \\color{red}{\\{2, 4\\}}, \\color{blue}{\\{2, 5\\}, \\{3, 5\\}}\\\\ & m = 6: \\color{red}{\\emptyset, \\{1\\}, \\{2\\}, \\{3\\}, \\{4\\}, \\{5\\}}, \\color{blue}{\\{6\\}}, \\color{red}{\\{1, 3\\}, \\{1, 4\\}}, \\color{green}{\\{1, 5\\}}, \\color{red}{\\{2, 4\\}, \\{2, 5\\}}, \\color{blue}{\\{2, 6\\}}, \\color{red}{\\{3, 5\\}}, \\color{blue}{\\{3, 6\\}, \\{4, 6\\}}, \\color{green}{\\{1, 3, 5\\}}, \\color{blue}{\\{2, 4, 6\\}}\\\\ \\end{align*}\n\nBased on these examples, we see that \\begin{align*} a_1 & = 2\\\\ a_2 & = 3\\\\ a_3 & = 4\\\\ a_4 & = 7\\\\ a_5 & = 11\\\\ a_6 & = 18 \\end{align*} Thus, the assertion that $a_1 = L_1$ is false. However, it appears that $a_m = L_m$ if $m > 1$.\n\nLet $m = n + 1$ for $n \\geq 1$.\n\nI have used three colors to highlight the subsets of $m = n + 1$ chairs that do not contain consecutive chairs. Subsets marked in blue contain chair $n + 1$. Subsets marked in red in row $n + 1$ are those subsets that appeared in row $n$. Subsets marked in green contain both $1$ and $n$.\n\nObserve that the elements in blue in row $n + 1$ are formed by taking the union of elements in row $n - 1$ that do not contain $1$ with the set $\\{n + 1\\}$, while the elements in green are formed by taking the union of elements in row $n - 1$ that do contain $1$ with the set $\\{n\\}$. Hence, the total number of elements in row $n + 1$ that are either blue or green is $L_{n - 1}$. The number of elements in row $n + 1$ that are red is the number of elements in row $n$, which is $L_n$. Hence, for each $n \\geq 1$, \\begin{align*} L_{n + 1} & = L_n + L_{n - 1}\\\\ & = F_{n - 1} + F_{n + 1} + F_{n - 2} + F_n\\\\ & = F_{n - 1} + F_{n - 2} + F_{n + 1} + F_n\\\\ & = F_{n} + F_{n + 2} \\end{align*} We can see that the assertion is false when $n = 0$ since the empty set in row $n + 1 = m = 1$ is not a member of one of the three types of highlighted sets.\n\nWhile I did not use this observation in my proof, it appears that for $n \\geq 1$ that the number of sets containing $n + 1$ in row $n + 1$ is $F_n$, while the number of sets containing $1$ and $n$ in row $n + 1$ is $F_{n - 2}$ (where we define $F_0 = F_2 - F_1 = 0$ and $F_{-1} = F_1 - F_0 = 1 - 0 = 1$).\n\nAre you sure of the formulation of your problem? I am wondering if the good one would not be: \"Suppose that $n$ chairs are arranged in a circle. Let $L_n$ count the number of subsets of $k \\geq 1$ chairs which don't contain consecutive chairs, show that...\"\n\n\u2022 Please see this tutorial on how to format mathematics on this site. Feb 26, 2015 at 12:46\n\u2022 I have just checked it, thank you. Feb 26, 2015 at 13:50", "date": "2022-08-17 01:46:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1166269/lucas-numbers-and-fibonacci", "openwebmath_score": 0.9600566625595093, "openwebmath_perplexity": 192.02039711392896, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765633889136, "lm_q2_score": 0.868826771143471, "lm_q1q2_score": 0.8514298733654648}}
{"url": "https://math.stackexchange.com/questions/1619550/calculating-the-surface-area-of-revolution-for-parametric-equation?noredirect=1", "text": "# Calculating the surface area of revolution for parametric equation.\n\nI solved a problem using a method that's completely different from the mark scheme and I got the right answer, but I'm unsure whether or not it's just some coincidence. Here's the question:\n\nThe curve $C$ has parametric equations $$x=t^2, y=\\frac{1}{4}t^4-\\ln{t}$$ for $1\\leq t\\leq 2$. Find the are of the surface generated when $C$ is rotated $2\\pi$ radians about the y-axis.\n\nHere's what I did ( not very rigorous):\n\nI thought cutting the curve with horizontal cuts should do the trick. So I just evaluated $2\\pi\\int _{t=1}^2xdy$ since $dy=(t^3-\\frac{1}{t})dt$ it's just $$2\\pi\\int_1^2(t^5-t)dt$$ which is fairly easy to find.\n\nHere's what's in the mark scheme:\n\n$x'=2t$ and $y'=t^3-\\frac{1}{t}$\n\n$(\\frac{ds}{dt})^2=x'^2+y'^2=(t^3-\\frac{1}{t})^2$\n\n$S=\\int 2\\pi x ds=2\\pi\\int_1^2(t^5-t)dt$\n\nMy understanding is that both answers slice $C$ differently and so maybe this was just some coincidence. I can't prove that both methods are equivalent though. So if someone can give me an idea as to why they are that'd be great. Also, it seems to me that the method I used was way easier, is there a place where it's better to use that other method?\n\n\u2022 If you examine the mark scheme's computation, you'll find a difference of sign with what's in your post: $x' = 2t$ and $y' = (t^{3} - \\frac{1}{t})$, so $$\\left(\\frac{ds}{dt}\\right)^{2} = x'(t)^{2} + y'(t)^{2} = \\left(t^{3} + \\tfrac{1}{t}\\right)^{2},$$ n.b., not $\\left(t^{3} - \\tfrac{1}{t}\\right)^{2}$, which would be equal to $y'(t)^{2}$ itself. If the mark scheme really has a minus sign on the right, that's an error.", "date": "2019-12-06 13:50:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1619550/calculating-the-surface-area-of-revolution-for-parametric-equation?noredirect=1", "openwebmath_score": 0.8309251666069031, "openwebmath_perplexity": 93.05001435643709, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765575409525, "lm_q2_score": 0.8688267745399466, "lm_q1q2_score": 0.8514298716130662}}
{"url": "https://scicomp.stackexchange.com/tags/nonlinear-equations/hot", "text": "People who code: we want your input. Take the Survey\n\n# Tag Info\n\n## Hot answers tagged nonlinear-equations\n\n19\n\nThere are two issues that you are likely to be encountering. Ill-conditioning First, the problem is ill-conditioned, but if you only provide a residual, Newton-Krylov is throwing away half your significant digits by finite differencing the residual to get the action of the Jacobian: $$J[x] y \\approx \\frac{F(x+\\epsilon y) - F(x)}{\\epsilon}$$ If you ...\n\n12\n\nShort answer If you only want second order accuracy and no embedded error estimation, chances are that you'll be happy with Strang splitting: half-step of reaction, full step of diffusion, half step of reaction. Long answer Reaction-diffusion, even with linear reaction, is famous for demonstrating splitting error. Indeed, it can be much worse, including \"...\n\n11\n\nThis is the stochastic root-finding problem, as in The stochastic root-finding problem: Overview, solutions, and open questions.\n\n11\n\nIs the shooting method the only general numerical method for solving BVP of nonlinear ODE(s)? No. Most other methods consist of three parts: Discretization. This may be done with finite differences, finite volumes, finite elements (Galerkin or collocation), spectral methods, and so forth. This reduces the problem from an infinite-dimensional one to a ...\n\n9\n\nIf you can stand to use complex arithmetic, simultaneous iteration methods might be preferable for computing all the roots of your polynomial. The simplest simultaneous iteration method, the (Weierstrass-)Durand-Kerner method, is effectively equivalent to applying Newton-Raphson to the Vieta relations relating the coefficients and roots of a polynomial, ...\n\n9\n\nI assume, that you have conducted a space discretization, so that you are about solving the (vector-valued) ODE $$\\dot u_h(t) = F_h(t,u_h(t)), \\text{ on [0,T] }, u_h(0) = \\alpha.$$ via a numerical scheme $\\Phi$ that advances the approximation $u_h^n$ at the current time instance $t=t^n$ to the next value $u_h^{n+1}$ at $t=t^{n+1}:=t^n+\\tau$. Then your ...\n\n8\n\nYou can solve this numerically in Python without symbolic computation. from __future__ import print_function, division import numpy as np from numpy import exp from scipy.integrate import quad from scipy.optimize import root def f1(a1, a2, x): return exp(a1 * x + a2 * x * x * x) / (1 + x * x) def f2(a1, a2, x): return exp(a1 * x + a2 * x * x * x) *...\n\n8\n\nPDEs are a form of dynamical system where there is another continuous variable. Usually this is space, so you're looking at how things over time and space instead of just over time. Here's an example of generalizing an ODE to a PDE. Take your ODE model of chemical reactions which models the concentration of certain chemicals over time. Now generalize the ...\n\n8\n\nJulia has a whole ecosystem for generating sparsity patterns and doing sparse automatic differentiation in a way that mixes with scientific computing and machine learning (or scientific machine learning). Tools like SparseDiffTools.jl, ModelingToolkit.jl, and SparsityDetection.jl will do things like: Automatically find sparsity patterns from code Generate ...\n\n7\n\nTo answer your questions in order: Any implicit method for solving an ordinary differential equation involves solving a system of nonlinear equations. You can do this through variants of Newton's method, successive substitution, full approximation scheme, or any other approach that solves systems of nonlinear equations. (The caveat is, of course, depending ...\n\n7\n\nWhat you describe as your time discretization is called the Crank-Nicolson scheme. For nonlinear differential equations it leads, as you have observed, to a nonlinear algebraic system that needs to be solved at each time step. The typical approach is to solve it with a Newton method -- in your case, that requires to solve a nonlinear system in 5 variables. ...\n\n7\n\nAs pointed out by David Ketcheson, one method is to use the companion matrix and find its eigenvalues (this is what Matlab does for the roots function). However, if you want to code everything by yourself, you can try to use the Sturm sequences, which you use as a first step to find an interval with only one zero. Then, you can apply one standard methods ...\n\n7\n\nFor your example equation, taking the average approach, the local consistency error $$\\frac{1}{h^2}[u(x-h) - 2 u(x) + u(x+h)]-f(\\frac{1}{2}[u(x-h)+u(x+h)]) = \\frac{1}{2}f_uu_{xx}h^2 + hot.$$ will be of order $2$ (instead of order $3$). ($hot.$ means higher order terms) Therefore, if your overall approximation is of order $1$, e.g. if you use upwind ...\n\n7\n\nThis is somehow unexpected, but my recent experience with solving a system of nonlinear equations is that treating them as the right hand side of a system of ordinary equations and then evolve the system with an ODE solver can be considerably faster than with the usual Newton-Raphson iteration. It sounds like you're doing some sort of pseudotransient ...\n\n7\n\nThe issues you're running into now are not a failing of Newton-Raphson, but a question of coupling. You're doing iterated sequential coupling -- solving each equation sequentially and then iterating until (hopeful) convergence. No solver choice in place of NR is going to fix this lack of convergence, as long as you are doing iterated sequential coupling. ...\n\n7\n\nIt's a bit easier to see if you write your equation in the a semi-discretised system of the form $u^{\\prime}(t) = F(u(t))$ and with the application of the $\\theta$-method and approximating $u^{\\prime}(t) \\approx (w^{n+1} - w^{n})/\\tau$ this gives, $$w^{n+1} - w^{n} - (1-\\theta) \\tau F(w^n) - \\theta\\tau F(w^{n+1}) = 0$$ with unknown vector $w^{n+1}$ and ...\n\n7\n\n$s_k$ is the \"approximate Newton\" search direction. So in essence, when they say Choose $s_k$ such that $\\|F(x_k)+F'(x_k)s_k\\| \\le \\eta_k \\|F(x_k)\\|$ they are saying: Solve the Newton system $F'(x_k)s_k = - F(x_k)$ inexactly for $s_k$ until the norm of the residual $F(x_k)+F'(x_k)s_k$ is smaller than the norm of the right hand side $-F(x_k)$ by a ...\n\n6\n\nIf you need Jacobian matrix information for a numerical method, you should calculate the Jacobian matrix of the discretized form of the equations, since that will be consistent with the discretized equations you are solving.\n\n6\n\nYou should rather think of what you will need in the following step, which is probably the numerical time integration of the semi-discrete equations. If you are going to use a (semi) explicit time stepping scheme, all you need is a function that for a given $\\phi_0$ assembles the vector $\\langle (\\nabla \\phi_0)^2, v \\rangle$, where $v$ are your test ...\n\n6\n\n$K u$ equals the internal forces only in the linear case. The tangent stiffness matrix, $K$, in a nonlinear problem is normally used in a Newton-Raphson algorithm to calculate updates to the displacement vector as follows: $$K \\Delta u = f - f_{internal}$$ $$u_{i+1} = u_i + \\Delta u$$ The vector of internal forces, $f_{internal}$ must be calculated from ...\n\n6\n\nIf you change variables to optimize for the residual of the linear part, then the Hessian will be a low-rank update to the identity. Then L-BFGS would work very well. Specifically, your problem takes the form $$\\min_x \\frac{1}{2}\\|Ax-b\\|^2 + \\frac{\\mu}{2}\\|g(x)\\|^2$$ where $Ax=b$ is the linear PDE and $g$ is the nonlinear part, and $\\mu$ is a tradeoff ...\n\n6\n\nThe reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^{-1}$ using a matrix polynomial of $A$. In this case (I assume) $f(y^{n+1},t)$ is not necessarily linear in the vector $y^{... 6 Two systematic ways of smoothing a function$h$would be: 1. Join the piecewise smooth parts of your function using Hermite interpolation so that the derivatives are matched to your satisfaction. 2. Convolve your function$h(x)$with a heat kernel of the form$f(x) = \\frac{\\exp\\left\\{-\\frac{x^2}{2 \\sigma^2}\\right\\}}{\\sqrt{2 \\pi \\sigma^2}}$so that instead ... 5 That's actually very easy to do in Dolfin: from dolfin import * # define mesh, function space (piecewise linear) mesh = UnitSquareMesh(64,64) V = FunctionSpace(mesh,'CG',1) # inhomogeneous boundary conditions (otherwise the solution is trivial) bc = DirichletBC(V, Constant(1.0), lambda x,on_boundary: on_boundary) # define bi(non)linear form # note that ... 5 Your observed quadratic convergence indicates that the Jacobian is likely correct. Have you looked at the solutions for your under-resolved configurations? Galerkin optimality uses the operator norm, which contains only the symmetric part, thus the solution of the discrete system could be quite different from the projection of the exact solution. This ... 5 As it looks to me your formulation of the discrete system is incomplete. You need to state your equation at each (grid?) point associated with$n$put back the summation over$l$(grid) points in your last equation assign a time index$i+1$or$i$to your$c_l$s In this way, you get a nonlinear (if you take$c_l^{n+1}$) or linear (if you take$c_l^{n}$) ... 5 A common strategy is to employ a damping strategy, i.e., to compute$\\vec{w}^{\\ast} = F \\left( \\vec{w}^{(n)} \\right)$and then set$\\vec{w}^{(n+1)} = \\alpha \\vec{w}^{(\\ast)} + (1-\\alpha)\\vec{w}^{(n)}$where$\\alpha\\in[0,1]$. You typically choose$\\alpha$in such a way that it minimizes$\\|[\\alpha \\vec{w}^{(\\ast)} + (1-\\alpha)\\vec{w}^{(n)}] - F(\\alpha \\vec{w}^...\n\n5\n\nCertified homotopy continuation methods are used both for finding roots and for proving that they indeed exist (inside a certain interval). A quick web search turned out this paper: Reliable homotopy continuation by Joris van der Hoeven.\n\n5\n\nNo it is not. There is also multiple shooting collocation finite differences fixed point iterations and probably some more.\n\n5\n\nSince $\\phi$ is a scalar between $0$ and $1$, the easiest method for finding a root is bisection. If you cannot calculate the expectation of the nonlinear function $$f_\\phi(\\theta) = \\left(\\phi r_z +(1-\\phi)(r_k+\\theta)\\right)^{1-\\gamma}(r_k+\\theta-r_z)$$ in terms of $\\phi$ analytically, you can use quadrature to approximate. For the first variant, this ...\n\nOnly top voted, non community-wiki answers of a minimum length are eligible", "date": "2021-06-14 16:32:41", "meta": {"domain": "stackexchange.com", "url": "https://scicomp.stackexchange.com/tags/nonlinear-equations/hot", "openwebmath_score": 0.7771490216255188, "openwebmath_perplexity": 429.07844764373124, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765581257486, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.851429868792686}}
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December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # Is 3^(a^2/b) < 1?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 27 Apr 2012 Posts: 57 Location: United States GMAT Date: 06-11-2013 GPA: 3.5 WE: Marketing (Consumer Products) Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 00:59 1 15 00:00 Difficulty: 95% (hard) Question Stats: 32% (00:58) correct 68% (01:01) wrong based on 378 sessions ### HideShow timer Statistics Is 3^(a^2/b) < 1? (1) a<0 (2) b<0 Hi, request your help to please understand the fundamental concept behind this question. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 01:17 7 6 Is 3^(a^2/b) < 1? Notice that $$3^{\\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\\frac{a^2}{b}<0$$. This will happen if $$a\\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\\frac{a^2}{b}=0$$). (1) a<0. The first condition is satisfied ($$a\\neq{0}$$) but we don't know about the second one. Not sufficient. (2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\\frac{a^2}{b}=0$$). Not sufficient. (1)+(2) Both condition are satisfied. Sufficient. Answer: C. For more on number theory and exponents check: math-number-theory-88376.html DS questions on exponents: search.php?search_id=tag&tag_id=39 PS questions on exponents: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html Hope it helps. _________________ ##### General Discussion Manager Joined: 27 Apr 2012 Posts: 57 Location: United States GMAT Date: 06-11-2013 GPA: 3.5 WE: Marketing (Consumer Products) Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 02:10 This is great.Thanks a ton! Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 25 Jun 2013, 03:47 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on Exponents: math-number-theory-88376.html All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39 All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6639 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 27 Jan 2016, 17:46 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is 3^(a^2/b) < 1? (1) a<0 (2) b<0 When you modify the original condition and the question, they become 3^(a^2/b) < 1? --> 3^(a^2/b) < 3^0? --> a^2/b>0?. Multiply b^2 on the both equations(since b^2 is positive, even if it\u2019s multiplied, the sign of inequality doesn\u2019t change.) and it becomes a^2(b)>0?. There are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), since a<0, it can\u2019t be 0. Divide the both equations with a^2, they become a^2(b)<0?-->b<0?. Since 2) is b<0, it is yes and sufficient. Therefore, the answer is C. \uf06c For cases where we need 2 more equations, such as original conditions with \u201c2 variables\u201d, or \u201c3 variables and 1 equation\u201d, or \u201c4 variables and 2 equations\u201d, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World\u2019s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. \"Only$99 for 3 month Online Course\"\n\"Free Resources-30 day online access & Diagnostic Test\"\n\"Unlimited Access to over 120 free video lessons - try it yourself\"\n\nIntern\nJoined: 09 Apr 2018\nPosts: 12\n\n### Show Tags\n\n25 Oct 2018, 05:33\nBunuel wrote:\nIs 3^(a^2/b) < 1?\n\nNotice that $$3^{\\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\\frac{a^2}{b}<0$$. This will happen if $$a\\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\\frac{a^2}{b}=0$$).\n\n(1) a<0. The first condition is satisfied ($$a\\neq{0}$$) but we don't know about the second one. Not sufficient.\n\n(2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\\frac{a^2}{b}=0$$). Not sufficient.\n\n(1)+(2) Both condition are satisfied. Sufficient.\n\nFor more on number theory and exponents check: http://gmatclub.com/forum/math-number-theory-88376.html\n\nDS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=39\nPS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=60\n\nTough and tricky DS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25967.html\nTough and tricky PS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25956.html\n\nHope it helps.\n\nBunuel ...if a=-1 and b=-9 then we get 3 as our answer and if a=-1 and b=-2 in that case we get .66\nShould the answer not be E.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 51218\nRe: Is 3^(a^2/b) < 1?\u00a0 [#permalink]\n\n### Show Tags\n\n25 Oct 2018, 06:17\nangarg wrote:\nBunuel wrote:\nIs 3^(a^2/b) < 1?\n\nNotice that $$3^{\\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\\frac{a^2}{b}<0$$. This will happen if $$a\\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\\frac{a^2}{b}=0$$).\n\n(1) a<0. The first condition is satisfied ($$a\\neq{0}$$) but we don't know about the second one. Not sufficient.\n\n(2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\\frac{a^2}{b}=0$$). Not sufficient.\n\n(1)+(2) Both condition are satisfied. Sufficient.\n\nFor more on number theory and exponents check: http://gmatclub.com/forum/math-number-theory-88376.html\n\nDS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=39\nPS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=60\n\nTough and tricky DS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25967.html\nTough and tricky PS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25956.html\n\nHope it helps.\n\nBunuel ...if a=-1 and b=-9 then we get 3 as our answer and if a=-1 and b=-2 in that case we get .66\nShould the answer not be E.\n\nIf a=-1 and b=-9, then 3^(1/(-9)) = ~0.9.\n_________________\nRe: Is 3^(a^2/b) < 1? &nbs [#permalink] 25 Oct 2018, 06:17\nDisplay posts from previous: Sort by", "date": "2018-12-16 05:55:22", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/is-3-a-2-b-134754.html", "openwebmath_score": 0.7549898624420166, "openwebmath_perplexity": 4824.161179065251, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9648551495568569, "lm_q2_score": 0.88242786954645, "lm_q1q2_score": 0.8514150740443787}}
{"url": "https://mohammadshariatmadari.ir/d9ueu4v/binomial-coefficient-latex-88290e", "text": "}}{{k!\\left( {n - k} \\right)!}}. As you see, the command \\binom{}{}will print the binomial coefficient using the parameters passed inside the braces. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector \u2026 Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. \\vec,\\overrightarrow, Latex how to insert a blank or empty page with or without numbering \\thispagestyle,\\newpage,\\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\\usepackage{algorithm},\\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? One can drop one of the numbers in the bottom list and infer it from the fact that sum \u2026 binomial The combination (n r) (n r) is called a binomial coefficient. }}{{k!\\left( {n - k} \\right)!}} The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. C \u2014 All combinations of v matrix. = \\binom {n} {k}  This is the binomial coefficient. The binomial coefficient $\\binom{n}{k}$ can be interpreted as the number of ways to choose k elements from an n-element set. where A is the permutation, A_n^k = \\frac{n!}{(n-k)! In latex mode we must use \\binom fonction as follows: \\frac {n!} Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. In Counting Principles, we studied combinations.In the shortcut to finding$\\,{\\left(x+y\\right)}^{n},\\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Then it's a good reason to buy me a coffee. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. k-combinations of n-element set. Click on one of the binomial coefficient designs, which look like the letters \"n\" over \"k\" inside either a round or angled bracket. Gerhard \"Ask Me About System Design\" Paseman, 2010.03.27 \\endgroup \u2013 Gerhard Paseman Mar 27 '10 at 17:00 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 This article explains how to typeset them in LaTeX. The binomial coefficient is defined by the next expression: \\binom {n}{k} = \\frac {n ! }}{{k!\\left( {n - k} \\right)!}} Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. Latex binomial coefficient Definition. If your equation requires specific numbers in place of the \"n\" or \"k,\" click on a letter to select it, press \"Delete\" and enter a number in its place. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 The binomial coefficient is defined by the next expression: \\ [ \\binom{n} {k} = \\frac{n!} (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. Knowledge base dedicated to Linux and applied mathematics. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. In latex mode we must use \\binom fonction as follows: \\frac{n!}{k! The order of selection of items not considered. (adsbygoogle = window.adsbygoogle || []).push({}); Thank you ! The possibility to insert operators and functions as you know them from mathematics is not possible for all things. An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. \\boxed, How to write table in Latex ? Using fractions and binomial coefficients in an expression is straightforward. The Texworks shows \u2026 Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. ( n - k )! Here's an equation: math \\frac {n!} (n\u2212k)! Mathematical Equations in LaTeX. = \\binom{n}{k} This will give more accuracy at the cost of computing small sums of binomial coefficients. The symbols and are used to denote a binomial coefficient, and are sometimes read as \"choose.\". n! This same array could be expressed using the factorial symbol, as shown in the following. And of course this command can be included in the normal text flow \\ (\\binom{n} {k}\\). All combinations of v, returned as a matrix of the same type as v. Specially useful for continued fractions. The second fraction displayed in the previous example uses the command \\cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Binomial Coefficient. The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. In UnicodeMath Version 3, this uses the \\choose operator \u249e instead of the \\atop operator \u00a6. The symbols and are used to denote a binomial coefficient, and are sometimes read as \" choose.\" Accordingly the binomial coefficient in the binomial theorem above can be written as \u201cn\\choose k\u201d, assuming that you type a space after the k. This Then it's a good reason to buy me a coffee. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilit\u00e9s de choisir k \u00e9l\u00e9ment dans un ensemble de n \u00e9l\u00e9ments. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. binomial coefficient Latex. LaTeX forum \u21d2 Math & Science \u21d2 Expression like binomial Coefficient with Angle Delimiters Topic is solved Information and discussion about LaTeX's math and science related features (e.g. Asking for help, clarification, or responding to other answers. }{k ! therefore gives the number of k -subsets possible out of a set of distinct items. It is especially useful for reasoning about recursive methods in programming. All combinations of v, returned as a matrix of the same type as v. Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. As you may have guessed, the command \\frac{1}{2} is the one that displays the fraction. Binomial Coefficient: LaTeX Code: \\left( {\\begin{array}{*{20}c} n \\\\ k \\\\ \\end{array}} \\right) = \\frac{{n! Open an example in Overleaf (n - k)!} formulas, graphs). So The combination (nr)\\displaystyle \\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)(\u200bn\u200br\u200b\u200b) is calle\u2026 Binomial coefficient, returned as a nonnegative scalar value. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. } ... Pascal\u2019s triangle. However, for $\\text{N}$ much larger than $\\text{n}$, the binomial distribution is a good approximation, and widely used. Identifying Binomial Coefficients. It will give me the energy and motivation to continue this development. Ak n = n! The usual binomial coefficient can be written as $\\left({n \\atop {k, {n-k}}}\\right)$. (n-k)!} I agree. The binomial coefficient is defined by the next expression: \\[\\binom {n}{k} = \\ frac {n!}{k!(n-k)! Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In Counting Principles, we studied combinations.In the shortcut to finding ${\\left(x+y\\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. On the other side, \\textstyle will change the style of the fraction as if it were part of the text. Don't forget to LIKE, COMMENT, SHARE & SUBSCRIBE to my channel. Usually, you find the special input possibilities on the reference page of the function in the Details section. How to write number sets N Z D Q R C with Latex: \\mathbb, amsfonts and \\mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? For these commands to work you must import the package amsmath by adding the next line to the preamble of your file In this case, we use the notation (nr)\\displaystyle \\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)(\u200bn\u200br\u200b\u200b) instead of C(n,r)\\displaystyle C\\left(n,r\\right)C(n,r), but it can be calculated in the same way. \\vec,\\overrightarrow; Latex how to insert a blank or empty page with or without numbering \\thispagestyle,\\newpage,\\usepackage{afterpage} Latex natural numbers; Latex real numbers; Latex binomial coefficient; Latex overset and underset ; Latex absolute value Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Also, the text size of the fraction changes according to the text around it. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \\left\\{,\\right\\},\\underbrace{} and \\overbrace{}, How to get dots in Latex \\ldots,\\cdots,\\vdots and \\ddots, Latex symbol if and only if / equivalence. Stanley's EC1 also uses it as the primary name, which counts for a lot in my book. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. The text inside the first pair of braces is the numerator and the text inside the second pair is the denominator. The command \\displaystyle will format the fraction as if it were in mathematical display mode. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n \u2265r n \u2265 r, then the binomial coefficient is Latex Binomial coefficient, returned as a nonnegative scalar value. begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? (n - k)!} {k! = \\binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$,$$\\frac{n!}{k! (n - k)!} This website was useful to you? {k! In Counting Principles, we studied combinations. k-combinations of n-element set. This method of constructing mathematical proofs is called mathematical induction. samedi 11 juillet 2020, par Nadir Soualem. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? Since binomial coefficients are quite common, TeX has the \\choose control word for them. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Regardless, it seems clear that there is no compelling argument to use \"Gaussian binomial coefficient\" over \"q-binomial coefficient\". \\\\binom{N} {k} What differs between \\\\dots and \\\\dotsc, with overleaf.com, the outputs are identical. It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! As you see, the command \\binom{}{} will print the binomial coefficient using the parameters passed inside the braces. Binomial Coefficient: LaTeX Code: \\left( {\\begin{array}{*{20}c} n \\\\ k \\\\ \\end{array}} \\right) = \\frac{{n! In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n \u2265 k \u2265 0 and is written (). infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more \u2026 This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. In this video, you will learn how to write binomial coefficients in a LaTeX document. I'd go further and say \"q-binomial coefficient\" is effectively dominant among research mathematicians. In the shortcut to finding (x+y)n\\displaystyle {\\left(x+y\\right)}^{n}(x+y)\u200bn\u200b\u200b, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an n-element set. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a \u2026 Latex k parmi n - coefficient binomial. See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\\binom{n}{k}$ is \u2026 Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + \u22ef + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \\cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + \u22ef + n c n x n , b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. For example, \u2026 The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. therefore gives the number of k-subsets possible out of a set of distinct items. You can set this manually if you want. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file How to write it in Latex ? C \u2014 All combinations of v matrix. The usage of fractions is quite flexible, they can be nested to obtain more complex expressions. coefficient are the different ordered arrangements of a k-element subset of an n-set, $$\\binom{n}{k} = \\binom{n-1}{k-1} +\\binom{n-1}{k}$$. {k! Using fractions and binomial coefficients in an expression is straightforward. (\u2212)!. . Below is a construction of the first 11 rows of Pascal's triangle. 2019, by Nadir Soualem { n! } { 2 } is the that! Second pair is the one that displays the fraction as if it were in mathematical,! Triangle can be nested to obtain more complex expressions fleqn, left, right ; how typeset! Top of another combination ( n r ) is called a binomial coefficient using the symbol. On top of another top of another are common mathematical elements with similar characteristics - number! Further and say  q-binomial coefficient '' over  q-binomial coefficient '' clear there! Will print the binomial coefficient raising a binomial to any whole number exponent in which items! From mathematics is not possible for all things goes on top of another 's a good reason buy... 11 rows of Pascal 's triangle k elements from an n-element set objects i.e 's work circa 1640 mode... Gives the number of k -subsets possible out of a set of distinct items rows of Pascal 's triangle to... Latexis binomial coefficient latex similar to the text inside the first 11 rows of Pascal 's work circa 1640 common! Possible for all things 3, this uses the \\choose operator \u249e instead of the \\atop \u00a6. ( { n } { { k! \\left ( { n! } } size of fraction... Positive integers that occur as coefficients in an expression is straightforward of ways which! = \\binom { } { } will print the binomial theorem have been known for,. To input into a Latex document in preparation for a pdf output elements from an n-element set of. Possible for all things to find the coefficients for raising a binomial coefficient can be nested obtain... Share & SUBSCRIBE to my channel been known for centuries, but they 're best from... In which k items are chosen from among n objects i.e for all things that displays the fraction changes to. To obtain more complex expressions outcomes from possibilities, also known as a combination or number..., you find the coefficients for raising a binomial coefficient, and are used to denote a binomial is... The other side, \\textstyle will change the style of the fraction as if it in! Read as  choose.  shows \u2026 Latex numbering equations: leqno et fleqn,,! Symbol, as shown in the binomial Expansion Technique and how to input into Latex! Coefficients are a family of positive integers that occur as coefficients in an expression is straightforward the of! Or binomial coefficient latex to other answers here 's an equation:   math \\frac { n k. K -subsets possible out of a set of distinct items from possibilities also! The symbols and are sometimes read as  choose. 11 rows of Pascal 's work 1640... With similar characteristics - one number goes on top of another: leqno et fleqn, left, ;! The fraction as if it were in mathematical display mode of a set of distinct items distinct. Special input possibilities on the reference page of the first 11 rows Pascal! \\Choose operator \u249e instead of the fraction as if it were part the! Pascal 's work circa 1640 \\displaystyle will format the fraction is straightforward: \\ [ \\binom { n {... Latexis very similar to the one used for fractions reason to buy me a coffee for reasoning about recursive in. \\Displaystyle will format the fraction binomial coefficient latex, as shown in the Details section to use Gaussian... } What differs between \\\\dots and \\\\dotsc, with overleaf.com, the \\binom! Is not possible for all things compelling argument to use  Gaussian binomial coefficient reason! Ec1 also uses it as the primary name, which counts for a pdf output 1 } { 2 is. Numerator and the text inside the second pair is the denominator 's also... All things from mathematics is not possible for all things  math \\frac { }..., \\textstyle will change the style of the fraction expressed using the factorial symbol, as shown in the section!, Monday 9 December 2019, by Nadir Soualem elements from an n-element.... You know them from mathematics is not possible for all things the outputs are.... Technique and how to input into a Latex document in preparation for a lot in my book display! Fleqn, left, right ; how to input into a Latex document in preparation for a lot in book! A good reason to buy me a coffee called a binomial coefficient, and are used to denote binomial! The second pair is the number of ways in which k items chosen! Texworks shows \u2026 Latex numbering equations: leqno et fleqn, left right!, right ; how to write a vector in Latex scientific tool scientific... N r ) ( n r ) ( n r ) ( n r ) ( r... Therefore gives the value of the binomial coefficient can be extended to find the coefficients raising. Page of the \\atop operator \u00a6 '' over  q-binomial coefficient ''  this is the number of possible... Coefficient, and are sometimes read as  choose. command to display them in Latex to write a in... Latexis very similar to the text around it using fractions and binomial coefficients have been known for,... It as the number of ways to choose k elements from an n-element set common elements mathematical. Factorial symbol, as shown in the following the binomial coefficient, and are sometimes read as  choose ''... Outputs are identical here 's an equation:   math \\frac { n! }.!\n\nAditya Birla Sun Life Mutual Fund Login, Firex Model Fadc, Post Office Agra Cantt, Hard Freight Psu Altoona, Spicy Food Captions For Instagram, Fear The Dark Game, Dicky Reo Speedwagon Ozark, How To Select By Attributes Multiple Attributes, Arlington Heights Funeral Homes, Funny Dogs And Cats, Average Temperature In Europe In August, Road Legal Batmobile For Sale, Favorite Metallica Lyrics, Can A Do Be A Obgyn,", "date": "2021-09-22 08:20:02", "meta": {"domain": "mohammadshariatmadari.ir", "url": "https://mohammadshariatmadari.ir/d9ueu4v/binomial-coefficient-latex-88290e", "openwebmath_score": 0.9148372411727905, "openwebmath_perplexity": 980.351373441915, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9740426450627306, "lm_q2_score": 0.8740772450055545, "lm_q1q2_score": 0.8513885117143548}}
{"url": "http://openstudy.com/updates/55bd1225e4b033255002fcc6", "text": "## YumYum247 one year ago how do i plot a velocity time graph from an average velocity of an object?\n\n1. YumYum247\n\nif a runner accelerated along the sidewalk at 0.5m/sec2 for 20 sec. Assume that the runner started from rest position.....\n\n2. anonymous\n\nYou will plot velocity on the vertical (y) axis and the time on the horizontal (x) axis. All you have to do is plot the values and connect all of them by a line. Make sure velocity is given in meters per second (m/s) and time in seconds (s) unless instructed otherwise.\n\n3. YumYum247\n\ncan you please show me how do i make the data table so i can plot the points neatly...\n\n4. YumYum247\n\n|dw:1438454610769:dw|\n\n5. YumYum247\n\nhow do i use the average velocity given in the question to plot the points on the graph.\n\n6. anonymous\n\nVelocity (in m/s) is [distance in metres] divided by [time in seconds.]\n\n7. YumYum247\n\nso v = d/t\n\n8. YumYum247\n\nbut how's that gunna help me?\n\n9. YumYum247\n\nif i want to know how far did the runner get in 7 sec, how do i figure that out???\n\n10. YumYum247\n\n@oldrin.bataku\n\n11. anonymous\n\nif the acceleration is constant $$a$$ then the total change in velocity from the start until time $$t$$ is just $$v-v_0=at$$; in our case, we starta t rest so $$v_0=0$$\n\n12. YumYum247\n\nyes the accelaeration is constant.\n\n13. anonymous\n\nso the velocity is given by $$v=at$$, so in terms of a velocity vs time plot we see that the graph is a simple line with slope $$a$$ passing through the origin\n\n14. YumYum247\n\ni just need to plot how much distance he covered from rest to 20sec.\n\n15. anonymous\n\nwell, if we have a velocity vs time plot, then the distance covered between $$t=t_1$$ and $$t=t_2$$ is just the total area bounded by our velocity graph between those points:\n\n16. anonymous\n\n|dw:1438456604986:dw|\n\n17. YumYum247\n\ncan you please just do one for me, i'll do the rest myself.... can you figure out how much distance the runner covered in let's say 7 seconds????\n\n18. anonymous\n\n|dw:1438456642313:dw|\n\n19. anonymous\n\nso the distance they go between $$t_1,t_2$$ is just the area under the graph between $$t_1,t_2$$.\n\n20. YumYum247\n\nthe area under the triangle is the displacement of the object....i get that...but\n\n21. YumYum247\n\ndid you use Tf - Ti = a X t ?\n\n22. YumYum247\n\nbecause i need to make a data table to record all the points until 20 second and plot them on the velocity time graph.\n\n23. anonymous\n\nokay, so here we have $$t_1=0$$, $$t_2=7$$, and our equation is again $$v(t)=0.5t$$. so our total displacement is: $$A=\\frac12 bh$$here our base is $$t_2=7$$ since $$t_1=0$$ and our height is $$v(t_2)=v(7)=3.5$$ so: $$A=\\frac12(7)(3.5)=12.25$$\n\n24. anonymous\n\nyou could also have done that work using an integral instead of manually finding teh area of a triangle\n\n25. YumYum247\n\nit doesn't look right because i have the answer key and the maximum displacement the runner makes under 20 sec is 10 meters.... i'm willing to show you the answer tho but i don't know how they got those points.....in the first place.\n\n26. anonymous\n\ni dont know what points youre talking about\n\n27. YumYum247\n\n|dw:1438457170341:dw|\n\n28. YumYum247\n\nthis is how it looks......\n\n29. anonymous\n\nthe kinematic equation gives $$x=x_0+v_0 t+\\frac12 at^2$$, here $$a=0.5,t=7$$ and $$x_0=v_0=0$$ so $$x=\\frac12\\cdot0.5\\cdot7^2=12.25$$\n\n30. anonymous\n\nactually, no it doesn't; you've misread it. that's the *velocity* against time, so it's going 10 meters *per second* in velocity after 20 seconds of time\n\n31. YumYum247\n\ni'm sorry it's a velocity time graph, the velocity/speed of the runner goes up to 10m/s under 20sec\n\n32. anonymous\n\nyes, and that doesn't contradict the fact the runner goes 12.25 meters in 7 seconds.\n\n33. YumYum247\n\nok\n\n34. anonymous\n\nhis speed after 7 seconds is only 3.5 m/s as I stated before\n\n35. anonymous\n\nwith an average speed of $$(3.5-0)/2=1.75$$ m/s, which is indeed correct: $$\\frac{12.25}7=1.75$$ m/s\n\n36. YumYum247\n\nok what formula did you use plase?????? :(\n\n37. YumYum247\n\nok so i need to make a data table of this situation as drawn above, so what equation do i use to plot the speed of the runner every second????\n\n38. anonymous\n\nsince the velocity vs time graph is a line, the average velocity between $$t_1=0,t_2=7$$ s is just the midpoint, hence $$(3.5-0)/2=1.75$$ m/s. this matches the alternative calculation of the average velocity as the total displacement $$12.25$$ m divided by the time elapsed $$t_2-t_1=7$$ s, since $$12.25/7=1.75$$ m/s as well\n\n39. anonymous\n\nthe equation for the velocity at time $$t$$ is just $$v=0.5t$$, since the acceleration is constant. this is the straight line we've been graphing\n\n40. YumYum247\n\nok let me try on my own and i'll ask you to check my work. Thank you!!! :\")\n\n41. YumYum247\n\nomg i see how you got 10m....LOL\n\n42. YumYum247\n\nok one more thing, if the average velocity of an object is 0.5m/sec2 how quick or how fast will it accelerate every second.", "date": "2017-01-22 03:48:40", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/55bd1225e4b033255002fcc6", "openwebmath_score": 0.8205059766769409, "openwebmath_perplexity": 955.7139494515728, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426412951848, "lm_q2_score": 0.8740772482857833, "lm_q1q2_score": 0.8513885116163113}}
{"url": "https://math.stackexchange.com/questions/675917/if-gcda-b-1-and-if-ab-x2-prove-that-a-b-must-also-be-perfect-sq", "text": "# If $\\gcd(a, b) = 1$ and if $ab = x^2$, prove that $a, b$ must also be perfect squares; where $a,b,x$ are in the set of natural numbers\n\nProblem:\n\nIf $\\gcd(a, b) = 1$ and If $ab = x^2$ ,prove that $a$, $b$ must also be perfect squares; where $a$,$b$,$x$ are in the set of natural numbers\n\nI've come to the conclusion that $a \\ne b$ and $a \\ne x$ and $b \\ne x$ but I guess that won't really help me.. I understand that if the $\\gcd$ between two numbers if $1$ then they obviously have no common divisors but where do I go from this point?\n\nAny tips at tackling this would be great. It looks quite easy though I'm still trying to get my hand around these proofs! Any pointers in the right direction would be great.\n\n\u2022 You just wrote \"$b\\ne b$.\" I really hope that's not what you meant. \u2013\u00a0apnorton Feb 14 '14 at 3:51\n\u2022 I've noticed that you have asked quite a few questions recently. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. \u2013\u00a0user61527 Feb 14 '14 at 3:52\n\u2022 Hi Tyler! I really appreciate the heads up. I'm working away at an assignment right now. I've gotten most of the problems down but I still have only a few left. Thanks so much again! \u2013\u00a0A A Feb 14 '14 at 4:08\n\nFundamental theorem of arithmetic says that every number has a unique prime factorization.\n\nIf gcd(a,b) = 1, then all of these factors are unique (no prime factor is shared between a and b). What does this say about $x^2$?\n\nHint 2:\n\nLet $a_i$ be a prime factor of a and $b_i$ be a factor of b. Then,\n\n$$ab = \\prod {a_{i}^{m_i}}\\prod {b_{i}^{n_i}}$$\n\nBut $x$ has to have a unique factorization in the form,\n\n$$x = \\prod {x_{i}^{e_i}}$$, where $m, n, e$ are integer exponents. Keep in mind it is unique and we can order these factors in any way we please. What does this say about $x^2$ compared to $ab$?\n\n\u2022 Hmm, x^2 must then: 1) have unique prime factors \u2013\u00a0A A Feb 14 '14 at 4:14\n\u2022 I'll edit my post with another hint. \u2013\u00a0Chantry Cargill Feb 14 '14 at 4:17\n\u2022 I think I may have it: x^2 = (p1^ip2^i2...pn^i2)^2 = (p1^2i2*p2^2i3*...pn^2in) = ab a and b must be factors of x^2, and all the factors of x^2 are squares, therfore a and b must be squares. Hmm, Does this sound correct to you? \u2013\u00a0A A Feb 14 '14 at 4:54\n\u2022 Yes, more or less! Just make sure that it's clear that you can order the factors of $x^2$ such that the primes match up with a and b. Then the root will just be ab. \u2013\u00a0Chantry Cargill Feb 14 '14 at 5:11\n\nBelow is an approach employing universal gcd laws (associative, commutative, distributive), some of which you may need to (simply) prove before you can use this method. But once you do so, you will gain great power. Below we explicitly show that $\\rm\\:a,b\\:$ are squares by taking gcds. Namely\n\nLemma $\\rm\\ \\ \\color{#0a0}{(a,b,c) = 1},\\,\\ \\color{#c00}{c^2 = ab}\\ \\Rightarrow\\ a = (a,c)^2,\\,\\ b = (b,c)^2\\$ for $\\rm\\:a,b,c\\in \\mathbb N$\n\nProof $\\rm\\ \\ (c,b)^2 = (\\color{#c00}{c^2},b^2,bc) = (\\color{#c00}{ab},b^2,bc) = b\\color{#0a0}{(a,b,c)} = b.\\$ Similarly for $\\,\\rm(c,a)^2.\\ \\$ QED\n\nYours is the special case $\\rm\\:(a,b) = 1\\ (\\Rightarrow\\ (a,b,c) = 1)$.\n\nGenerally $\\rm\\: \\color{#c00}{ab = cd}\\: \\Rightarrow\\: (a,c)(a,d) = (aa,\\color{#c00}{cd},ac,ad) = a\\: (a,\\color{#c00}b,c,d) = a\\:$ if $\\rm\\:(a,b,c,d) = 1.\\:$ For more on this and closely related topics such as Euler's four number theorem (Vierzahlensatz), Riesz interpolation, or Schreier refinement see this post and this post.\n\nCompare the following Bezout-based proof (this is a simplified form of the proof in Rob's answer). For comparison, I append an ideal-theoretic version of the proof of the more complex direction.\n\nNote that $\\ 1=\\overbrace{a{\\rm u}+b\\,{\\rm v}}^{\\large (a,b)}\\,\\ \\overset{\\large \\times\\,a}\\Rightarrow\\ a = \\color{#c00}{a^2}{\\rm u}+\\!\\!\\overbrace{ab}^{\\Large\\ \\ \\color{#c00}{c^2}}{\\rm v} \\ \\,$ so $\\,\\ d=(a,c)\\mid a,c\\,\\Rightarrow\\, d^2\\!\\mid \\color{#c00}{a^2,c^2}\\,\\Rightarrow\\, d^2\\!\\mid a$\n\nConversely $\\ d = (a,c)= au+cv\\,\\ \\Rightarrow\\,\\ d^2=\\,\\color{#c00}{a^2}u^2+2\\color{#c00}acuv+\\color{#c00}{c^2}v^2\\ \\$ thus $\\ \\ \\color{#c00}{a\\mid c^2}\\ \\Rightarrow\\,\\ \\color{#c00}a\\mid d^2$\n\n$\\quad\\ \\ {\\rm i.e.}\\quad (d)= (a,c)\\ \\ \\Rightarrow\\ \\ (d^2) \\subseteq\\, (a,c^2)\\,\\subseteq\\, (a)\\ \\$ by $\\ \\ a\\mid c^2\\,\\ \\$ [simpler ideal form of prior]\n\nNotice how the ideal version eliminates the obfuscatory Bezout coefficients $\\,u,v$.\n\n\u2022 Hi Bill! Thank you for the response! Your way ahead of me, I'm not too familiar with the notation used/ \u2013\u00a0A A Feb 14 '14 at 4:53\n\u2022 Bezouted.$\\ \\$ \u2013\u00a0robjohn Feb 14 '14 at 10:16\n\u2022 @AA I use the standard notation $\\ (a,b,\\,\\ldots)\\, =\\, \\gcd(a,b,\\,\\ldots).\\,$ I added another Bezout-based proof. $\\$ \u2013\u00a0Bill Dubuque Feb 15 '14 at 4:50\n\nHere is a proof using Bezout's Identity.\n\nLet $x^2=ab$ and $\\gcd(a,b)=1$, where $a,b\\gt0$.\n\nThere are $u,v$ so that $$au+bv=1\\tag{1}$$ Let $s_a=\\gcd(x,a)$. Rewriting $(1)$, we have \\begin{align} s_a\\left(\\dfrac{a}{s_a}\\right)u+bv=1 &\\implies s_a^2\\left(\\dfrac{a}{s_a}\\right)^2u^2=1-b(2v-bv^2)\\tag{2}\\\\ &\\implies s_a^2\\left(\\dfrac{a}{s_a}\\right)^2u^2a+ab(2v-bv^2)=a\\tag{3}\\\\ &\\implies s_a^2\\left(\\dfrac{a}{s_a}\\right)^2u^2a+s_a^2\\left(\\dfrac{x}{s_a}\\right)^2(2v-bv^2)=a\\tag{4}\\\\[8pt] &\\implies s_a^2\\mid a\\tag{5} \\end{align} Justification:\n$(2)$: Move $bv$ to the right side and square\n$(3)$: Move $b(2v-bv^2)$ to the left side and multiply by $a$\n$(4)$: $x^2=ab$\n$(5)$: $s_a^2$ divides each term on the left side\n\nThere are $u_a,v_a$ so that \\begin{align} xu_a+av_a=s_a &\\implies\\frac{x}{s_a}u_a+\\frac{a}{s_a}v_a=1\\tag{6}\\\\ &\\implies\\frac{x^2}{s_a^2}u_a^2=1-\\frac{a}{s_a}\\left(2v_a-\\frac{a}{s_a}v_a^2\\right)\\tag{7}\\\\ &\\implies\\frac{ab}{s_a^2}u_a^2+\\frac{a}{s_a}\\left(2v_a-\\frac{a}{s_a}v_a^2\\right)=1\\tag{8}\\\\ &\\implies abu_a^2+as_a\\left(2v_a-\\frac{a}{s_a}v_a^2\\right)=s_a^2\\tag{9}\\\\[7pt] &\\implies a\\mid s_a^2\\tag{10} \\end{align} Justification:\n$\\ \\:(6)$: Divide by $s_a$\n$\\ \\:(7)$: Move $\\frac{a}{s_a}v_a$ to the right side and square\n$\\ \\:(8)$: Move $\\frac{a}{s_a}\\left(2v_a-\\frac{a}{s_a}v_a^2\\right)$ to the left side, $x^2=ab$\n$\\ \\:(9)$: Multiply by $s_a^2$\n$(10)$: $a$ divides each term on the left side\n\nCombining $(5)$ and $(10)$ yields $a=\\gcd(x,a)^2$. Symmetry yields, $b=\\gcd(x,b)^2$.\n\n\u2022 I simplified it a bit - see my answer. In $\\,(2\\!-\\!5)\\,$ it suffices to use the Bezout identity itself (vs. its square). In $(6\\!-\\!10)$ it's simpler to immediately square the Bezout identity (vs. rearrange it first). \u2013\u00a0Bill Dubuque Feb 15 '14 at 4:47", "date": "2019-10-20 03:46:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/675917/if-gcda-b-1-and-if-ab-x2-prove-that-a-b-must-also-be-perfect-sq", "openwebmath_score": 0.848613977432251, "openwebmath_perplexity": 421.05252832003566, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426428022031, "lm_q2_score": 0.8740772400852111, "lm_q1q2_score": 0.8513885049458548}}
{"url": "https://math.stackexchange.com/questions/1324903/isometry-group-of-a-norm-is-always-contained-in-some-isometry-group-of-an-inner/1325204", "text": "Isometry group of a norm is always contained in some Isometry group of an inner product?\n\n\nDoes there always exist some inner product $\\<,\\>$ on $V$ such that $\\text{ISO}(|| \\cdot ||)\\subseteq \\text{ISO}(\\<,\\>)$ ?\n\nUpdate:\n\nAs pointed by Qiaochu Yuan the answer is positive.\n\nThis raises the question of uniqueness of the inner product $\\<,\\>$ which satisfies $\\text{ISO}(|| \\cdot ||)\\subseteq \\text{ISO}(\\<,\\>)$.\n\nIs it unique (up to scalar multiple)?\n\nRemarks:\n\n1) Determining $\\<,\\>$ (up to scalar multiple) is equivalent to determining $\\text{ISO}(\\<,\\>)$.\n\nClearly if we know the inner product we know all its isometries. The other direction follows as a corollary from an argument given here which shows which inner products are preserved by a given automorphism.\n\n2) Since there are \"rigid\" norms (whose only isometries are $\\pm Id$ ) the uniqueness certainly doesn't hold in general.\n\nOne could hope for that in the case of \"rich enough norms\" (norms with many isometries, see this question) the subset $\\text{ISO}(|| \\cdot ||)\\subseteq \\text{ISO}(\\<,\\>)$ will be large enough to determine $\\text{ISO}(\\<,\\>)$.\n\n(which by remark 1) determines $(\\<,\\>)$).\n\nYes. This is because an isometry group is always compact (with respect to the topology on $\\text{End}(V)$ induced by the operator norm: this is a consequence of the Heine-Borel theorem). Hence you can average an inner product over it with respect to Haar measure.\n\n\u2022 @Asaf: that is not the suggestion. The suggestion is the following: pick an arbitrary inner product $\\langle v, w \\rangle$. Construct the new inner product $\\langle v, w \\rangle_G = \\int_G \\langle gv, gw \\rangle \\, dg$. (There is some work necessary to prove that this is still an inner product but it's not so bad.) By construction, this new inner product is $G$-invariant, so every $g \\in G$ is an isometry with respect to the corresponding norm. This is a standard maneuver in representation theory called \"Weyl's unitary trick\"; you can try to google that term for references. \u2013\u00a0Qiaochu Yuan Jun 14 '15 at 19:38\n\n\nFor completeness, I am writing more detailes of the solution suggested by Qiaochu:\n\nDenote by $G$ the isometry group of $(V,\\|\\cdot\\|)$. $G\\subseteq \\text{End}(V)$. On $\\text{End}(V)$ we have the operator norm $\\| \\|_{op}$ (w.r.t the given norm $\\|\\cdot\\|$), which induces a topology on $\\text{End}(V)$.\n\nLemma 1: $G$ is compact in $\\text{End}(V)$\n\nProof: $\\text{End}(V)$ is a finite dimensional normed space, hence it is linearly homeomorphic to $\\mathbb{R}^n$ (This is in fact true for every finite dimensioanl real topological vector space). So, the Heine-Borel theorem aplies. (Every closed and bounded subset is compact).\n\n$G$ is bounded since for every isometry $g\\in G, \\|g\\|_{op}=1$, hence $G$ is contained in the unit sphere of $(\\text{End}(V),\\| \\|_{op})$ .\n\n$G$ is closed: Assume $g_n \\rightarrow g,g_n\\in G$. Fix some $v\\in V$. $\\|g_n(v)-g(v)\\|_V\\leq \\|g_n-g\\|_{op}\\cdot\\|v\\|_V \\xrightarrow{n\\to\\infty} 0$. So $g_n(v)\\xrightarrow{n\\to\\infty}g(v)$. Now by the continuity of the norm $\\| \\cdot \\|$ (w.r.t to the topology it induces on $V$) we get that: $\\|v\\| \\stackrel{g_n isometry}{=} \\|g_n(v)\\|\\xrightarrow{n\\to\\infty}\\|g(v)\\|$. This forces $g$ to be an isometry.\n\nCorollary1: $G$ is locally compact Hausdorff topological group.\n\nProof: $\\text{End}(V)$ is Hausdorff (Any metric space is...), and every subspace of Hausdorff is also Hausdorff. It is sdandard fact that $GL(V)$ is a topological group (t.g), and any subgroup of a t.g is a t.g.\n\nNow, there exists a left-invariant measure $\\mu$ on the Borel $\\sigma$-algebra of $G$ such that $\\mu(G)>0$. (This is the Haar measure which can be constructed on any locally compact Hausdorff topological group).\n\nNow take any inner product $\\<,\\>$ on $V$. Fix $v,w \\in V$. Define $f_{v,w}:G\\rightarrow \\mathbb{R},f_{v,w}(g)=\\<gv,gw\\>$.\n\nLemma 2: $f_{v,w}$ is continuous\n\nProof: Since $G$ is a metric space (a subspace of the normed space $\\text{End}(V)$) it is enough to check sequential continuity. Take\n$g_n \\rightarrow g,g_n\\in G$. We already showed this implies $g_n(v)\\xrightarrow{n\\to\\infty}g(v)$ so by the continuity of the inner product $f_{v,w}(g_n)=\\<g_nv,g_nw\\> \\xrightarrow{n\\to\\infty} f_{v,w}(g)$.\n\nIn particular $f_{v,w}$ is measurable, so we can integrate it. (Compactness of $G$ implies $f_{v,w}$ is bounded, and $G$ being a finite measure space guarantees the integral will be finite).\n\nSo we define: $\\<v, w \\>' = \\int_G f_{v,w} \\, d\\mu = \\int_G \\< gv, gw \\> \\, d\\mu$.\n\nNow all is left is to show $\\<, \\>'$ is an inner product on $V$ that is presrved by each $h\\in G$. (since this means $G=\\text{ISO}(V,\\|\\cdot\\|) \\subseteq \\text{ISO}(V,\\<, \\>')$ as required.\n\nLemma 3: $\\<,\\>'$ is an inner product.\n\nThe only non-trivial thing is positive-definiteness. (The rest follows from the linearity of $g\\in G$ and the integral, and the bilinearity of $\\<,\\>$). But this follows from standard measure theory:\n\nFix $v\\neq 0$. $f_{v,v} > 0$ on $G$ (since each $g\\in G$ is injective and the original inner prodcut is positive). But this forces $\\<v,v\\>'>0$ as required.\n\nLemma 4: $\\<, \\>'$ is $G$-invariant. But this follows from another standard proposition in measure theory: (See \"Real Analaysis\" by H.L.Royden) chapter 22 pg 488 proposition 10).\n\n\u2022 Still not enough langles and rangles... \u2013\u00a0Qiaochu Yuan Jun 16 '15 at 3:19\n\u2022 You are right... I thought I already fixed it but I was wrong. I hope now its OK. \u2013\u00a0Asaf Shachar Jun 16 '15 at 7:10\n\u2022 I really like that you accepted Qiaochu's answer and then added the details in your own. Nice! \u2013\u00a0Joachim Jun 21 '15 at 21:00", "date": "2020-04-04 13:12:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1324903/isometry-group-of-a-norm-is-always-contained-in-some-isometry-group-of-an-inner/1325204", "openwebmath_score": 0.9601343274116516, "openwebmath_perplexity": 183.82142883155677, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9740426412951847, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.85138849883598}}
{"url": "https://math.stackexchange.com/questions/3126162/solving-left1-3-rightk-n-1-for-k", "text": "# Solving $\\left(1/3\\right)^k n = 1$ for $k$\n\nThe goal is to show that $$\\left(\\frac{1}{3}\\right)^kn=1 \\Rightarrow k = \\log_3 n\\,.$$\n\nSo I started with $$\\left(\\frac{1}{3}\\right)^kn=1 \\Leftrightarrow \\left(\\frac{1}{3}\\right)^k=\\frac{1}{n}$$ in order to use the identity $$y=a^x \\Leftrightarrow x=\\log_a y$$, which then yields $$k=\\log_{1/3} \\frac{1}{n}$$ which using $$\\log \\frac{1}{x}=-\\log a$$ can be written as $$k = -\\log_{1/3} n\\,.$$ But that is not what I wanted to show, as $$\\log_3 n \\neq -\\log_{1/3} n$$.\n\nI don't know where the mistake is.\n\nNote that $$-\\log_{1/3} n = \\frac{\\log_{1/3} n}{\\log_{1/3}3} = \\log_3 n$$\n\n\u2022 Thank you! That was fast. I will accept your answer as soon as it will let me. \u2013\u00a0user500664 Feb 25 '19 at 15:45\n\u2022 Sorry, I've got another question: the identity you used is $\\log_b (a)=\\frac{\\log_c a}{\\log_c b}$, right? In your answer, $c=\\frac{1}{3}$, but couldn't it be any arbitrary number as $c$ bears no relation to either $a$ or $b$? \u2013\u00a0user500664 Feb 25 '19 at 15:55\n\u2022 @ThomasFlinkow Yes, but keep in mind that $0<c \\neq 1$ for the logarithms to be defined. \u2013\u00a0Haris Gu\u0161i\u0107 Feb 25 '19 at 15:58\n\nAlternatively,\n\n$$\\left( \\frac{1}{3^k}\\right)n=1$$\n\nMultiplying $$3^k$$ on both sides, $$n=3^k$$\n\nHence $$k = \\log_3 n$$\n\n\u2022 Thank you very much. This is even more elegant. I'm afraid I already accepted an answer \u2013\u00a0user500664 Feb 25 '19 at 16:12\n\u2022 Don't worry about reputations. I just see a question and just lift a few fingersl ;) The other solution teach you why $-\\log_{\\frac13} n = \\log_3 n$. \u2013\u00a0Siong Thye Goh Feb 25 '19 at 16:16", "date": "2021-02-26 22:35:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3126162/solving-left1-3-rightk-n-1-for-k", "openwebmath_score": 0.7992732524871826, "openwebmath_perplexity": 341.4681642124171, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464464059648, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8513643333043038}}
{"url": "https://math.stackexchange.com/questions/1985074/expected-payoff-of-dice-game-paying-for-every-next-roll/1988641", "text": "# expected payoff of dice game (paying for every next roll)\n\nI am thinking of a variation of the dice game, where\n\none has the option to throw a die unlimited number of times.The first throw is free and every next throw costs 1 dollar. One will earn the face value of the die and has the option to stop after each throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game?\n\nIf I calculate the expected value as\n\nfor 2 rolls as  (1/6)(3+4+5+6)  +   (1/3) (3.5-1) = 3.83$for 3 rolls as (1/6)(3+4+5+6) + (1/3) (3.83-1) = 3.94$\nfor 4 rolls as  (1/6)(3+4+5+6) +   (1/3) (3.94-1) = 3.98$for 5 rolls as (1/6)(3+4+5+6) + (1/3) (3.98-1) = 3.99$\n\n\nit asymptotically tends to 4\n\nIs this approach correct?\nShoudn't the player go bankrupt after 7 rolls (negative expectation)?\n\n\u2022 I don't follow your method here. Can you explain your reasoning? \u2013\u00a0Matthew Conroy Oct 27 '16 at 7:36\n\u2022 @ Matthew Conroy let's take 2 rolls. The expectation of a single trow is 3.5, but it costs 1, therefore the expectation is 2.5. Thus, if one gets 1 or 2 in the first try then one keeps rolling because the expectation is higher (2.5). If 3,4,5, or 6 comes one stops and takes the payoff, so $$(\\frac{1}{6} 3+\\frac{1}{6} 4+\\frac{1}{6} 5+\\frac{1}{6} 6) + \\frac{2}{6}(3.5-1) = 3.83$$ \u2013\u00a0Michal Oct 28 '16 at 2:41\n\u2022 Sometimes they will lose money, but on average they will net $4$. See my answer below. (There is no \"bankrupt\" here, since you have not specified how much money the player has to begin with.) \u2013\u00a0Matthew Conroy Oct 29 '16 at 20:39\n\nI'd do it this way.\n\nLet $E_n$ be the expected net if we roll until we get $n$ or greater.\n\nSuppose $n=6$. We roll once. If it's a $6$, we are done. If not, we are in exactly the situation we started in, except that our \"first\" roll isn't free: it cost $1$ dollar. Hence $$E_6 = \\frac{1}{6}(6) + \\frac{5}{6}(E_6-1).$$ Solving, we find $E_6=1$. (Alternatively: on average it takes $6$ rolls to get a $6$, and all but the first roll is free, so $E_6=6-5=1$.)\n\nSimilarly, \\begin{align} E_5 &= \\frac{2}{6}\\left(\\frac{5+6}{2} \\right) + \\frac{4}{6}(E_5-1) & \\text{ so } &E_5=\\frac{7}{2}=3.5 \\\\ E_4 &= \\frac{3}{6} \\left( \\frac{4+5+6}{3} \\right) + \\frac{3}{6}(E_4-1) & \\text{ so } &E_4=4 \\\\ E_3 &= \\frac{4}{6} \\left( \\frac{3+4+5+6}{4} \\right) + \\frac{2}{6}(E_3-1) & \\text{ so} &E_3=4 \\\\ E_2 &= \\frac{5}{6} \\left( \\frac{2+3+4+5+6}{5} \\right) +\\frac{1}{6}(E_2-1) & \\text{ so } &E_2=\\frac{19}{5} = 3.8 \\\\ \\end{align} and, of course, $E_1=3.5$.\n\nSo the best strategy appears to be \"roll until $4$ or more\" with an expected net of $4$. Sometimes you will roll, say, 10 times with this strategy and lose money, but the average net is $4$.\n\nHere is the result of $10^6$ simulated plays with the $4$ strategy:\n\nnet/number of occurrences\n\n-21 1\n\n-20 0\n\n-19 0\n\n-18 0\n\n-17 0\n\n-16 1\n\n-15 0\n\n-14 1\n\n-13 5\n\n-12 6\n\n-11 8\n\n-10 19\n\n-9 32\n\n-8 80\n\n-7 146\n\n-6 293\n\n-5 579\n\n-4 1122\n\n-3 2272\n\n-2 4517\n\n-1 9147\n\n0 18256\n\n1 36474\n\n2 72912\n\n3 145747\n\n4 291215\n\n5 250332\n\n6 166835\n\nwith an average net of $4.000515$.\n\n\u2022 @ Matthew Conroy I think you have a mistake in $E_2$ (in case of 2 you keep rolling, so 2 shouldn't be in the first bracket). The expectation keeps going up, asymptotically tending to 4 as n increases. on the other hand is it not at odds with common sense? any strategy with more than 7 rolls is loosing money, the higher n the bigger loss \u2013\u00a0Michal Oct 28 '16 at 10:58\n\u2022 @ Matthew Conroy secondly, why in your calculation the options you decide to pay off are decreasing with every next roll? for the 2 roll case E(x) value of the second roll is 2.5 so you pay off straight away after first roll if 3,4,5,6 come. for the 3 roll case and the next ones the E(x) is below 4 so you keep rolling in case of 1,2,3 and pay off in case of 4,5,6, correct? \u2013\u00a0Michal Oct 28 '16 at 14:10\n\u2022 My $E_2$ is the expected value if we roll until we get $2$ or greater. In this case, we stop when we get a $2$. \u2013\u00a0Matthew Conroy Oct 28 '16 at 22:13\n\u2022 Regarding your claim \"the higher n the bigger loss\", consider: if I've rolled a 1 (for example), it is always to my advantage, on average, to roll again, regardless of how many rolls I've taken already. Cheers! \u2013\u00a0Matthew Conroy Oct 29 '16 at 4:44\n\u2022 If you roll $1$ six times in a row, then you will lose money. However, on average, you will lose less money if you continue to roll for a higher number. The same goes for rolling $2$s. The $E_2$ strategy is not optimal, as the calculation above shows. It is still a strategy. If you still don't agree with these calculations, I recommend you write a simulation and try out these strategies yourself. Sometimes the $4$ strategy loses money, but on average the net is $4$ dollars. Cheers! \u2013\u00a0Matthew Conroy Oct 29 '16 at 20:05", "date": "2019-12-10 14:23:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1985074/expected-payoff-of-dice-game-paying-for-every-next-roll/1988641", "openwebmath_score": 0.9786650538444519, "openwebmath_perplexity": 804.4151723294626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464471055738, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8513643306757229}}
{"url": "https://math.stackexchange.com/questions/1420222/different-results-in-integrating-both-sides-of-sin2x-2-cos-x-sin-x", "text": "# Different results in integrating both sides of $\\sin{2x}=2\\cos x\\sin x$\n\nI feel like there is something I am missing here. When integrating both sides of the trigonometric identity $\\sin{2x}=2\\cos x\\sin x$ I get different results.\n\nThe left side of course results in $-\\frac{1}{2}\\cos{2x}+C$.\n\nThe right side I solve with u-substitution:\n\n$u=\\cos x$\n\n$du=-\\sin x dx$\n\n$-2\\int udu=-u^2+C=-\\cos^2 x+C$\n\nWhile writing this question I noticed another identity $\\cos^2 x=\\frac{1}{2}+\\frac{1}{2}\\cos 2x$. So apparently the $\\frac{1}{2}$ falls out because of the $+C$ resulting from indefinite integration? This is still a little confusing to me.\n\nYou are correct to recall that $\\cos^2x=\\frac{1+\\cos 2x }{2}$.\n\nThis is an indefinite integral. So, the constant term $\\frac 12$ is not relevant. That is\n\n$$\\int \\sin 2x \\,dx=-\\frac12\\cos (2x)+C_1 \\tag 1$$\n\nand\n\n\\begin{align} \\int \\sin 2x \\,dx&=-\\cos^2 x+C_2\\\\\\\\ &=-\\frac12\\cos 2x+(-\\frac12 +C_2)\\\\\\\\ &=-\\frac12\\cos 2x+C_3\\tag 2 \\end{align}\n\nwhere we absorbed the constants $-\\frac12+C_2$ into a new constant and called that new constant $C_3$. Inasmuch as the integration constant is arbitrary, $(1)$ and $(2)$ are equivalent statements.\n\n\u2022 Thanks, it's clear now. I was thrown off by the cosines having a different power. \u2013\u00a0Rubenknex Sep 3 '15 at 20:04\n\u2022 You're welcome. My pleasure. Although it can be confusing, you were on the right track. \u2013\u00a0Mark Viola Sep 3 '15 at 20:30\n\nIntegration process gives you results correct up to an arbitrary constant.\n\nBoth results are essentially the same.\n\n$$- \\frac{\\cos 2 x}{2} + C_1 = - \\frac { 2 \\cos ^2 x -1}{2} + C_1 = - \\cos ^2 x + C_2$$\n\nThat is why it is better to write $C_1, C_2$ for the integration constants.", "date": "2021-07-24 15:18:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1420222/different-results-in-integrating-both-sides-of-sin2x-2-cos-x-sin-x", "openwebmath_score": 0.9613566398620605, "openwebmath_perplexity": 532.4138254244294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975946446405965, "lm_q2_score": 0.8723473846343393, "lm_q1q2_score": 0.8513643300654209}}
{"url": "http://math.stackexchange.com/questions/23809/what-is-the-remainder-of-16-is-divided-by-19", "text": "# What is the remainder of $16!$ is divided by 19?\n\nCan anyone share me the trick to solve this problem using congruence?\n\nThanks,\nChan\n\n-\n\nIf you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)\n\n-\nI knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks. \u2013\u00a0 Chan Feb 26 '11 at 6:17\n@Chan: $\\rm 17 \\equiv -2,\\ 18\\equiv -1$ \u2013\u00a0 Bill Dubuque Feb 26 '11 at 6:21\n@Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\\equiv -1$ and $17\\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) \u2013\u00a0 Jonas Meyer Feb 26 '11 at 6:21\nMany thanks, I got it now ;) \u2013\u00a0 Chan Feb 26 '11 at 6:24\n\nHINT $\\$ By Wilson's theorem $\\$ mod $19\\::\\ -1 \\ \\equiv\\ 18\\:!\\ \\equiv\\ 18\\cdot17\\cdot 16\\:!\\ \\equiv\\ (-1)\\ (-2)\\cdot 16\\:!$\n\nSimilarly we have the Wilson reflection formula $\\rm\\displaystyle\\ (p-1-k)\\:!\\ \\equiv\\ \\frac{(-1)^{k+1}}{k!}\\ \\ (mod\\ p)\\:,\\$ $\\rm\\:p\\:$ prime\n\n-\n@Bill Dubuque: Thank you! \u2013\u00a0 Chan Feb 26 '11 at 6:24\n@Bill Dubuque: If I have $2.16! \\equiv 18 \\pmod{19}$. Can I cancel out the $2$ from both sides? \u2013\u00a0 Chan Feb 26 '11 at 6:42\n@Chan: $\\rm\\ 2\\:x\\equiv 2\\:y\\ (mod\\ 19)\\ \\iff\\ 19\\ |\\ 2\\ (x-y)\\ \\iff\\ 19\\ |\\ x-y\\ \\iff\\ x\\equiv y\\ (mod\\ 19)$ \u2013\u00a0 Bill Dubuque Feb 26 '11 at 6:50\n@Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \\equiv ay \\pmod{p} \\implies x \\equiv y \\pmod{p}$, right? \u2013\u00a0 Chan Feb 26 '11 at 7:07\n@Chan: $\\rm\\ n,m\\$ coprime $\\rm\\iff a\\ n + b\\ m = 1\\$ for some $\\rm\\:a,b\\:\\ \\iff\\ a\\ n\\equiv 1\\ (mod\\ m)\\:,\\:$ for some $\\rm\\:a\\:$ $\\rm\\iff\\ n$ is invertible/cancellable $\\rm\\:(mod\\ m)$ \u2013\u00a0 Bill Dubuque Feb 26 '11 at 7:12\n\nI almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:\n\n$$10!= (2\\cdot 10)\\cdot (4\\cdot5)\\cdot(3\\cdot6)\\cdot(7\\cdot8)\\cdot 9\\equiv 1\\cdot 1\\cdot (-1)\\cdot(-1)\\cdot 9\\equiv9$$\n\nNow we have\n\n$$16!\\equiv 9\\cdot 11\\cdot 12\\cdot 13\\cdot 14\\cdot 15\\cdot 16\\equiv9\\cdot(-8)\\cdot(-7)\\cdot(-6)\\cdot(-5)\\cdot(-4)\\cdot(-3)$$\n\n$$\\equiv 9\\cdot(4\\cdot5)\\cdot(6\\cdot3)\\cdot(7\\cdot8)\\equiv 9\\cdot(1)\\cdot(-1)\\cdot(-1)\\equiv 9$$\n\nOf course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)\n\n-", "date": "2015-08-04 03:24:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/23809/what-is-the-remainder-of-16-is-divided-by-19", "openwebmath_score": 0.9255539178848267, "openwebmath_perplexity": 362.9958573576288, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975946445006747, "lm_q2_score": 0.8723473763375643, "lm_q1q2_score": 0.8513643207476087}}
{"url": "http://mathhelpforum.com/business-math/41170-compound-interest-q-help.html", "text": "1. Compound Interest Q Help!\n\nCan you guys help me with this question? I';m so confused on how I should approach it:\nTom\u2019s father paid $500 into an account on the day Tom was born. ?After that, he paid$500 into the account on Tom\u2019s bday until Tom\u2019s 18th bday. ?If the account accrued interest at 8%p.a compounded monthly, calculate how much Tom would receive on his 18th bday.\n\nA:\n20880.97\n\n2. Just think it through, one payment at a time.\n\nFrom the start, I'm not absolutely certain there was a payment ON the 18th birthday. let's assume that there was. If we get $500 too much, we'll discard this payment. Starting from the last payment and working backwards, we have: 18th:$500 and no accumulation for interest\n17th: $500(1+i) -- 1 year's accumulation for interest 16th:$500(1+i)^2 -- 2 year's accumulation for interest\n15th: $500(1+i)^3 -- 3 year's accumulation for interest ... Birth:$500(1+i)^18 -- 18 year's accumulation for interest\n\nWith any luck, one should notice this is a Geometric Sequence and we should be able to add them all up.\n\n$500 +$500(1+i) + $500(1+i)^2 +$500(1+i)^3 + ... + $500(1+i)^18 =$500(1 + (1+i) + (1+i)^2 + (1+i)^3 + ... + (1+i)^18) =\n\n$500\\frac{(1+i)^{19}-1}{i}$\n\nOur only remaining concern is 'i'. What is it? The formula above uses 'i' as an annual effective interest rate. We need to find one of those. We are given 8% Nominal Interest and Monthly Compounding. This gives:\n\n$\\left(1 + \\frac{0.08}{12}\\right)^{12} - 1 = 0.082999511 = i$\n\nThus,\n\n$500\\frac{(1+i)^{19}-1}{i}\\;=\\;21,380.97$\n\nAs can be seen, $21,380.97 -$500.00 = \\$20,880.97, so I guess there was not a payment made ON the 18th Birthday Anniversary. This leaves us with a bit of a dilemma. On a written exam or a homework assignment, I would state my assumptions and provide both answers, citing the ambiguity of the word \"until\". Anything marked wrong would get a vigorous challenge. On a multiple-choice exam, I would be prepared to find either answer. In my view, if both appear on the multiple-choice exam, the question probably should be discarded as accepting either answer will not necessarily provide any information about a student's knowledge. One may simply have done it badly. In any case, questions should be clear. If you have discussed the word \"until\" in class, and it has been defined clearly to mean \"NOT on the end date\", then you can be expected to get the unique value. There may also be a diagram explaining the intent. It is a very hard thing to write perfectly clear questions. It is up to the student to explain any point of ambiguity. The exam writer cannot be expected to think of every possible translation, but I'm sure the exam writer tries to do that.\n\nWell, enough of exam philosophy...", "date": "2017-05-25 04:58:10", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/business-math/41170-compound-interest-q-help.html", "openwebmath_score": 0.6717661619186401, "openwebmath_perplexity": 1041.7068877375884, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464520028357, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8513643138950946}}
{"url": "http://math.stackexchange.com/questions/878038/whats-the-name-of-this-algebraic-property/878272", "text": "# What's the name of this algebraic property?\n\nI'm looking for a name of a property of which I have a few examples:\n\n$(1) \\quad\\color{green}{\\text{even number}}+\\color{red}{\\text{odd number}}=\\color{red}{\\text{odd number}}$\n\n$(2) \\quad \\color{green}{\\text{rational number}}+\\color{red}{\\text{irrational number}}=\\color{red}{\\text{irrational number}}$\n\n$(3) \\quad\\color{green}{\\text{algebraic number}}+\\color{red}{\\text{transcendental number}}=\\color{red}{\\text{transcendental number}}$\n\n$(4) \\quad\\color{green}{\\text{real number}}+\\color{red}{\\text{non-real number}}=\\color{red}{\\text{non-real number}}$\n\nIf I were to generalise, this, I'd say that if we partition a set $X$ into two subsets $S$ and $S^c=X\\setminus S$, then the sum of a member of $S$ and a member of $S^c$ is always in either $S^c$ or $S$.\n\nMy question is:\n\n\"Is there a name for this property (in these four cases) and is this property true in general?\"\n\nAlso, does anyone have any more examples of this property?\n\n-\nNote that you're saying more than just set here: You're requiring that the objects of that set can be added. \u2013\u00a0 Semiclassical Jul 25 at 16:50\n@Semiclassical Yeah, thanks! What word should I use instead of set, then? \u2013\u00a0 alexqwx Jul 25 at 16:51\nThis can be written as: $S$ is closed under subtraction. That is, if $a,b\\in S$ and $a-b$ exists, then $a-b\\in S$. \u2013\u00a0 Thomas Andrews Jul 25 at 16:52\n@Semiclassical I don't see how this contradicts $(2).$ \u2013\u00a0 alexqwx Jul 25 at 16:54\n@Semiclassical No, you misread my comment. The rationals are closed under subtraction. That is what the above says... \u2013\u00a0 Thomas Andrews Jul 25 at 16:55\n\nI think this comes from the fact that if you have a group $G$ and $H$ a subgroup of $G$ then if $h\\in H$ and $x\\not\\in H$ we get $xh\\not\\in H$.\n\nThe proof is by contradiction, suppose $xh=l$ with $l\\in H$. Then postmultiplying by $h^{-1}$ gives $x=lh^{-1}$ which is in $H$ since $H$ is a subgroup of $G$.\n\n-\nWhy have you used $xh$ instead of $x+h$ in your generalisation (does it matter?)? \u2013\u00a0 alexqwx Jul 25 at 16:58\n$xh$ just means the group operator acting on the pair $(x,h)$ if the operation in the particular group is addition then $xh$ is just $x+h$. the plus sign is normally used for abelian groups. But this result generalizes to any group. \u2013\u00a0 Jorge Fern\u00e1ndez Jul 25 at 17:03\nGot it. Thanks! One final thing: why did you mention post-multiplying rather than just multiplying (since addition is commutative)? \u2013\u00a0 alexqwx Jul 25 at 17:05\nwell, in general groups don't need to be commutative, however this assumption is not required for the result to be true. \u2013\u00a0 Jorge Fern\u00e1ndez Jul 25 at 17:22\nTalking about this in terms of groups seems just a bit too strict. The first example given in the question works just as well, for example, if we were to only talk about positive even integers (no inverse or zero element under addition.) \u2013\u00a0 Semiclassical Jul 25 at 17:51\n\nI call this the complementary subgroup law, because the composition law arises via the following complementary view of the Subgroup Test ($\\rm\\color{#c00}{ST}$),  cf. below from one of my old sci.math posts.\n\nTheorem  Let $\\rm\\,G\\,$ be a nonempty subset of an abelian group $\\rm\\,H,\\,$ with complement set $\\rm\\,\\bar G = H\\backslash G.\\,$ Then $\\rm\\,G\\,$ is a subgroup of $\\rm\\,H\\iff G + \\bar G\\, =\\, \\bar G.$\n\nProof $\\$ $\\rm\\,G\\,$ is a subgroup of $\\rm\\,H\\!\\overset{\\ \\large \\color{#c00}{\\rm ST}}\\iff\\! G\\,$ is closed under subtraction, so, complementing\n\n$\\begin{eqnarray} & &\\ \\ \\rm G\\text{ is a subgroup of }\\, H\\ fails\\\\ &\\iff&\\ \\rm\\ G\\ -\\ G\\ \\subseteq\\, G\\,\\ \\ fails\\\\ &\\iff&\\ \\rm\\ g_1\\, -\\ g_2 =\\,\\ \\bar g\\ \\ \\ for\\ some\\ \\ g_1,g_2\\in G,\\ \\ \\bar g\\in \\bar G\\\\ &\\iff&\\ \\rm\\ g_2\\, +\\ \\bar g\\ \\ =\\,\\ g_1\\ for\\ some\\ \\ g_1,g_2\\in G,\\ \\ \\bar g\\in \\bar G\\\\ &\\iff&\\ \\rm\\ G\\ +\\ \\bar G\\ \\subseteq\\ \\bar G\\ \\ fails\\qquad\\ {\\bf QED} \\end{eqnarray}$\n\nInstances of this law are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here (and also on sci.math).\n\n-\nNice! ByteErrant, in the answers below, has shown that prime +non-prime can be either prime or non-prime itself, which seems to contradict the above theorem. Could you explain? \u2013\u00a0 alexqwx Jul 25 at 21:36\n@alex The set $\\,\\color{#0a0}G\\,$ of $\\rm\\color{#0a0}{primes}$ is not a subgroup of $\\,H =$ integers under addition, since $\\,\\color{#0a0}G\\,$ is not closed under subtraction, e.g. $\\,\\color{#0e0}{11}-\\color{#0a0}7 = \\color{#c00}4\\,$ is not $\\rm\\color{#0a0}{prime,\\,}$ so $\\, \\color{#0a0}7+\\color{#c00}4 = \\color{#0e0}{11}\\,\\Rightarrow\\, \\color{#0a0}G+\\color{#c00}{\\bar G}\\subseteq \\color{#c00}{\\bar G}\\,$ fails. That agrees with the (negation of) the theorem, i.e. $\\,G+\\bar G\\subseteq G\\,$ fails $\\iff$ $\\,G\\,$ is not a subgroup of $\\,H\\,$ $\\iff$ $\\,G\\,$ is not closed under subtraction. \u2013\u00a0 Bill Dubuque Jul 25 at 23:21\n\nThis is basically just \"the complement\" of the statement that the sets of numbers in green are subgroups of $(\\Bbb R,+)$.\n\nGiven that for a subgroup $S<\\Bbb R$ $a,b\\in S$ implies $a-b\\in S$, you can quickly compute that if $c\\notin S$, $a+c=b\\in S$ implies $a-b=c\\in S$, a contradiction. Therefore $a+c\\notin S$ for any $a\\in S$, $c\\notin S$.\n\nYou could also add to your list anything like \"an integer plus a noninteger is a noninteger\" and \"for any subring $S\\subseteq \\Bbb R$, an element in $S$ plus an element outside of $S$ results in an element outside of $S$.\" Again, you don't even really need a subring: this is true for any proper subgroup.\n\n-\n\nDon't prime numbers offer a counter-example to the general truth of this property?\n\nPrime $+$ not-prime $=$ not-prime $==> 17 + 4 = 21$\n\nPrime $+$ not-prime $=$ prime $==> 7+4=11$\n\n-\nAh. I hadn't thought about this. Thanks! Do you (or does anyone else) have any explanation for this? \u2013\u00a0 alexqwx Jul 25 at 21:34\n@alexqwx Primes don't form a group under addition. Evens do, rationals do, algebraics do, and reals do. \u2013\u00a0 Cory Jul 25 at 21:41\n\nI would say that the behaviour of this property under addition is isomorphic to addition in $\\mathbb{Z}_{2}$. E.g.: If you assign \"even\" to 0, \"odd\" to 1, then even + odd = odd is expressed in 1 + 0 = 1.\n\n-", "date": "2014-11-24 01:26:26", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/878038/whats-the-name-of-this-algebraic-property/878272", "openwebmath_score": 0.8922492265701294, "openwebmath_perplexity": 375.2402923015919, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9759464436075291, "lm_q2_score": 0.872347368040789, "lm_q1q2_score": 0.8513643114297964}}
{"url": "http://www.intmath.com/blog/learn-math/calculus-made-easy-free-book-2432", "text": "# Calculus Made Easy (Free book)\n\nBy Murray Bourne, 25 Apr 2009\n\nOK, it looks old and dusty, but Calculus Made Easy [PDF] is an excellent book and I strongly recommend it to those of you who are struggling with calculus concepts. It's also great for teachers, to give you ideas on how to explain calculus so it doesn't confuse the hell out of everyone. He quite rightly points out that many math text book writers are more interested in impressing the reader with sophisticated calculus techniques than explaining the basic concepts.\n\nOne of the early pages has:\n\nBEING A VERY-SIMPLEST INTRODUCTION TO\nTHOSE BEAUTIFUL METHODS OF RECKONING\nWHICH ARE GENERALLY CALLED BY THE\nTERRIFYING NAMES OF THE\nDIFFERENTIAL CALCULUS\nAND THE\nINTEGRAL CALCULUS.\nBY\nSILVANUS P. THOMPSON\n\nIn other words, this was one of the first ever \"Calculus for Dummies\" books. Thompson puts great effort into explaining what is going on, rather than jumping straight into the calculations. He humbly calls himself a \"fool\", but doesn't treat the reader as one.\n\nHe quotes from an \"ancient Simian proverb\":\n\n\"What one fool can do another can.\"\n\nTo give you an idea of how the book is written, in Chapter 1, \"To Deliver You From the Preliminary Terrors\", we read:\n\n\u222b which is merely a long S, and may be called (if you like) \"the sum of.\" Thus \u222bdx means the sum of all the little bits of x; or \u222bdt means the sum of all the little bits of t. Ordinary mathematicians call this symbol \"the integral of\".\n\nNow any fool can see that if x is considered as made up of a lot of little bits, each of which is called dx, if you add them all up together you get the sum of all the dx's, (which is the same thing as the whole of x). The word \"integral\" simply means \"the whole\".\n\nThe book is now copyright free. Grab the PDF: Calculus Made Easy.\n\n[Thanks to Denise at LetsPlayMath for the link.]\n\n1. solarhene says:\n\nthanks so much for sharing this excellent book!\n\nbest regards,\nan old fool\n\n2. Murray says:\n\nYou're welcome Solarhene. I'm glad that you find it useful.\n\n3. ayuk federick besong says:\n\ni believe it can work and it will work\n\n4. Noor says:\n\nI'm so interested in re-learning calculus that I watched many different video tutorials, but I found this book summarizing Integral Calculus in its first 2 pages!\n\nThank you so much!\n\n5. Sarah says:\n\nThanks so much for posting this. I'm so glad I clicked on your link while viewing my friend Dana's blog. My 9 year old read it and was very excited that it made so much sense. He's currently taking Physics and some of the math problems were made to be so confusing so this will help him so much.\n\n6. lawal says:\n\nGood DAY SIR,i want to know how i can get this text book,calculus made easy.i am mailing from Lagos Nigeria.thank u sir.\n\n7. Murray says:\n\nHi Lawal\n\nThis is an e-book in PDF form (not a physical book). The link is in the article above, in the first line.\n\nYou could print it out from the PDF if you would rather hard copy.\n\n8. D. Pal, India. says:\n\nIts a brilliant book. Thanks for sharing the book\n\n9. B Atchley says:\n\nThanks for sharing this book! It will be a great tool for my independent study students!\n\n10. Winstone Kapanje says:\n\nThanks for great job well done. Do the same with other topics such as trigonometry and complex numbers\n\n11. Judith Kanyama Chikoti says:\n\nthe way you have started culculus is very interesting but how could one get the whole PDF paper?\n\n12. Murray says:\n\nHi Judith. The link to the PDF is in the article above! Do you mean that you want a hard copy? You would need to print it yourself.\n\n13. jared okoth says:\n\nif every mathematics topic could be introduced the way this has, then i sure no one will think of maths as the hardest of courses.\n\n14. eduardo says:\n\nmake me realize the math is easy\n\n15. Kaset One says:\n\none of the greatest books.. many thanks for sharing with us\n\n16. Rod says:\n\nI was contemplating purchasing a course on dvd called \"Calculus Made Clear\" selling for \\$60.00 US. Then I decided to google \"Calculus Made Easy\" and found your link to this great book. I read the intro and was amazed at how easy it was to follow. I will continue to read the whole text! by the way, I was searching for books on calculus for my son so now we can read it together. Thanks!\n\n17. Murray says:\n\nYou're welcome, Rod. Glad you found it useful.\n\n18. fatemeh says:\n\ni from iran and i don't english.\nineed a study book reference and i have n't book english and i request from you that a link book free download for laplas, integral ,...\nthank you very much\n\n19. Murray says:\n\n@Fatemeh: I'll look for such materials and post it if I find any good ones.\n\n20. Piracha, J. L. says:\n\nThanks\n\nthanks! and great work which nobody cannot be appreciated for your effort for sharing such good book.\n\n22. Dalcde says:\n\nhttp://www.gutenberg.org/ebooks/33283\nwhich is typed in LaTeX and the file size is reduced tenfold.\n\n23. Murray says:\n\n@Dalcde: Thanks for the tip!\n\n24. Nayyir says:\n\nif u r one of those looking to understand calculas at an elementary level,then u've got it here.Calculas Made Easy is indeed very helpful a book for those struggling to understand it as a concept.so do go for it , it is certainly one of the best materials available. go for it.have a nice maths day.\n\n25. Ian Thomson says:\n\nHi :\nThanks for the pdf on Calculus Made Easy.\nI have always been curious and terrified at the same time of calculus.\n\nChapter One says it all. Getting past the fancy notation, helps a huge amount.\n\nKind Regards Ian Thomson\n\n26. percival says:\n\nHi! Ian,My name is percy and I teach Maths in grade 12. Please foreward me the calculus doc as I also struggle on calculus section.\n\n27. Murray says:\n\n@Percival: The link to the PDF is at the top (and again at the bottom) of the article.\n\n28. Joe Kenyon says:\n\nDear Murray\nI am surprised and delighted to see my old Dear Friend Sylvanus P. Thompson here. As a very discouraged high school dropout I found an original hardback version at the rubbish tip site where I used to spend my spare time. I picked it up and the proverb just grabbed me. I read the book it and loved it. I went back to school and excelled at the age of 25. Now after 40 years in senior positions in Telecommunications in Australia my copy is very much treasured and repaired and again being used as I transition into becoming a teacher myself! The bit that initially grabbed me was \"What one fool can do another can\" I ask that you put that wonderful saying up on this site. It was so powerfuly encouraging it got me off the bad path I was on and led to a wonderful career and a wonderful wife and family. Imagine, it hardly seems possible that an ancient seven word proverb, repeated in one book by a man long dead could do so much good.\nI love this guy and would have loved to have met him.\n\nKind Regards\nJoe Kenyon\n\n29. Murray says:\n\nThanks for your inspiring story, Joe!\n\nI have included the quote in the article.\n\n30. sachin sharma says:\n\nHi,\nI have been searching such work for past couple of months and at last I got here in the form of \"Calculus Made Easy\" May GOD Bless all who have made such efforts in preparing such an excellent book.\n\nThanks to all.....\n\nSachin Sharma\nIndia\n\n31. drcobol2000 says:\n\nMurray,\n\nThanks for posting this...and for free! While I struggled with math from grade school to grad school, I'm now a bit of a math junkie and books with subjects like this really pique my interest. I wish I could get my hands on an actual copy of that book for myself.\n\nOne question comes to mind, though. I noticed there are two different dates in the front of the book that confuse me (I'm a sucker for old books). There is a date of 1914 in the preface and another of 1943 by itself on another page. Do you know which date was the actual date of publication? There isn't an actual copyright year specified and as I read through the text, the absence of that date combined with the style and choice of words lead me to believe this was a work from the early 20th century. I rely on a barely mediocre literary acumen to make that assessment, so take it with an enormous grain of salt.\n\nThanks and best wishes!\n\n32. Murray says:\n\n@drcobol2000 The publication date would be 1914, as that's when he wrote the Preface to this second edition. So your \"early 20th century\" language observation is right on the button.\n\nI don't believe the \"6-14-43\" on the separate page is a date, since in the US, the date order is normally day then month. I suspect it's there to trigger thought.\n\n33. kevin says:\n\nThanks you so much sir for this ebook\n\nI am an engineering student, and i am struggling with calculus.\n\nThis book helped me a lot even though this book is quite old.\n\n34. Pyae Phyo Aung says:\n\nThank you really really much sir. You help me a lot.\n\n35. punith says:\n\nthank u so much its very helpful\n\n36. rahillah says:\n\nhi Murry, thanks for sharing the brilliant book,hope you have a lovley day!\n\n37. Murray says:\n\nGlad you found it useful, Rahillah.\n\n38. Nae says:\n\nI'm not sure if it'll work but I'll give it a try. I'll write back at the end of the semester to let you know how it worked for me.\n\nThanks\n\n### Comment Preview\n\nHTML: You can use simple tags like <b>, <a href=\"...\">, etc.\n\nTo enter math, you can can either:\n\n1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):\na^2 = sqrt(b^2 + c^2)\n(See more on ASCIIMath syntax); or\n2. Use simple LaTeX in the following format. Surround your math with $$ and $$.\n$$\\int g dx = \\sqrt{\\frac{a}{b}}$$\n(This is standard simple LaTeX.)\n\nNOTE: You can't mix both types of math entry in your comment.", "date": "2015-10-09 15:58:20", "meta": {"domain": "intmath.com", "url": "http://www.intmath.com/blog/learn-math/calculus-made-easy-free-book-2432", "openwebmath_score": 0.683467447757721, "openwebmath_perplexity": 2167.2396868849382, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399077750858, "lm_q2_score": 0.8774767826757123, "lm_q1q2_score": 0.8513629926980621}}
{"url": "https://math.stackexchange.com/questions/2626576/how-do-we-define-the-domain-of-a-function-with-removable-and-nonremovable-discon", "text": "# How do we define the domain of a function with removable and nonremovable discontinuities?\n\nFor instance consider the function\n\n$$f(x) = \\frac{x-2}{(x-2)(x+2)}$$\n\nI don't understand why some discontinuities are considered \"removable\". Yes, we can cancel out the $x-2$'s but isn't this invalid at $x=2$ because it's like canceling out $\\frac{0}{0}$? I don't understand why we're allowed to do this.\n\nAnyway if we \"remove\" that piece we get:\n\n$$f(x) = \\frac{1}{x+2}$$\n\nWhich has a discontinuity at $x=-2$ but we can't \"remove\" that.\n\nSo we end up with two discontinuities -- do we define the domain to include or exclude these? Is the domain all real numbers? All except $2$? All except $-2$? All except $-2, 2$?\n\n\u2022 Because limit . \u2013\u00a0user202729 Jan 29 '18 at 14:39\n\nThe discontinuity at $x=2$ can be removed because $\\lim_{x\\to2}f(x)=1/4$. When 'removing' the discontinuity, we define a new function $f^\\prime$ that takes on the values of $f$ at all points in the domain of $f$ and $f^\\prime(2)=1/4$, thereby extending the domain.\n\nThe discontinuity at $x=-2$ cannot be removed because the limit there is improper: $\\lim_{x\\to-2}f(x)=\\infty$.\n\n\u2022 How do you know the limit at each point? Is there an analytic way? Is the only difference between removable and nonremovable discontinuities whether or not the limit is a number or not? \u2013\u00a0user525966 Jan 29 '18 at 14:57\n\u2022 When the limit at a certain point is improper, it cannot be represented by a real number. Only proper limits can be used to remove discontinuities. \u2013\u00a0Jeroen van Riel Jan 29 '18 at 15:03\n\u2022 What then would be the domain of $f$? All reals except $-2,2$? \u2013\u00a0user525966 Jan 29 '18 at 15:04\n\u2022 Yes, as @57Jimmy already pointed out in his answer. \u2013\u00a0Jeroen van Riel Jan 29 '18 at 15:13\n\nYou clearly understand the algebra, and the idea behind \"removable singularity\". The definition of the domain is a little subtle (and usually not particularly important, given your understanding).\n\nThe expressions on the right in\n\n$$f(x) = \\frac{x-2}{(x-2)(x+2)}$$\n\nand\n\n$$g(x) = \\frac{1}{x+2}$$\n\ndefine the same function where both make sense. That function has an unremovable singularity at $x=-2$.\n\nThe domain of $g$ contains the point $x=2$; the domain of $f$ does not, so strictly speaking they are not the same function. But the limit of $f$ at $x=2$ does exist, and has value $g(2) = 1/4$. That's exactly what we mean when we say the singularity is removable.\n\n\u2022 So a removable discontinuity is an undefined point that has a defined limit, whereas a nonremovable discontinuity has a limit that does not exist or is infinity? \u2013\u00a0user525966 Jan 29 '18 at 14:52\n\u2022 @user525966 Yes. \u2013\u00a0Ethan Bolker Jan 29 '18 at 15:33\n\nThe domain of $f$ is $\\mathbb{R} \\backslash \\{-2,2\\}$, because at this two points $f$ is not defined. Having said that, some singularities are better behaved than others: $f$ goes to infinity as $x \\to -2$, whereas $\\underset{x \\to 2}{\\lim} f(x) = \\frac{1}{4}$ is well-defined. This is exactly the difference between poles and removable singularities.", "date": "2021-07-28 07:53:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2626576/how-do-we-define-the-domain-of-a-function-with-removable-and-nonremovable-discon", "openwebmath_score": 0.945873498916626, "openwebmath_perplexity": 239.22173322849565, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399060540358, "lm_q2_score": 0.8774767794716264, "lm_q1q2_score": 0.8513629880791487}}
{"url": "https://mathhelpboards.com/threads/cartesian-product-proof.26383/", "text": "Cartesian Product - Proof\n\nYankel\n\nActive member\nDear all,\n\nI am trying to prove a simple thing, that if AxA = BxB then A=B.\n\nThe intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.\n\nHow do I prove it formally ?\n\nThank you !\n\nOpalg\n\nMHB Oldtimer\nStaff member\nDear all,\n\nI am trying to prove a simple thing, that if AxA = BxB then A=B.\n\nThe intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.\n\nHow do I prove it formally ?\n\nThank you !\nThis may not be as simple as you think. To start with, what do you mean by saying that two sets are equal? I think that the only way to make sense of that is to interpret \"A=B\" to mean that A and B have the same cardinality.\n\nIf a set $A$ is finite then its cardinality is just the number of elements it contains, denoted by $|A|$. If $|A| = m$ then $|A\\times A| = m^2.$ So if $|B| = n$ and $|A\\times A| = |B\\times B|$ then $m^2 = n^2$, from which it follows that $m=n$. This proves that if \"$A\\times A = B\\times B$\" then \"$A=B$\" in the case of finite sets.\n\nFor infinite sets the situation is more complicated. There is a theorem of Zermelo that if $A$ is an infinite set then $|A\\times A| = |A|$. From that it follows immediately that if $|A\\times A| = |B\\times B|$ then $|A| = |B|$. However, the proof of Zermelo's theorem requires the Axiom of Choice. In models of set theory that do not satisfy this axiom, it may be that your result does not hold.\n\nHallsofIvy\n\nWell-known member\nMHB Math Helper\nOpalg said \"\nI think that the only way to make sense of that is to interpret \"A=B\" to mean that A and B have the same\ncardinality\n.\"\n\nI disagree. To say that sets A and B are equal means \"$$x\\in A$$ if and only if $$x\\in B$$\". If two sets are equal they have the same cardinality but the converse is not true. The sets A= {1, 2, 3} and B= {a, b, c} have the same cardinality but are not equal.\n\nOpalg\n\nMHB Oldtimer\nStaff member\nOpalg said \"\nI think that the only way to make sense of that is to interpret \"A=B\" to mean that A and B have the same\ncardinality\n.\"\n\nI disagree. To say that sets A and B are equal means \"$$x\\in A$$ if and only if $$x\\in B$$\". If two sets are equal they have the same cardinality but the converse is not true. The sets A= {1, 2, 3} and B= {a, b, c} have the same cardinality but are not equal.\nIn that case, the result becomes trivially true. If $A\\times A$ and $B\\times B$ are just two different names for the same set, then the diagonal elements of $A\\times A$ (those of the form $(a,a):a\\in A$) are duplicates of the elements of $A$. The same holds for the diagonal elements of $B\\times B$. If those diagonals are the same, it follows that the elements of $A$ are the same as the elements of $B$, so $A=B$.\n\nHallsofIvy\n\nWell-known member\nMHB Math Helper\nYes, it is. Saying that \"A= B\", where A and B are sets, means that if x is in A then it is also in B and if y is in B then it is also in A.\n\nIf x is a member of A. then (x, x) is in AxA= BxB so x is in B. If y is a member of B then (y, y) is in BxB= AxA so y is in A. Therefore A= B.\n\nIt is trivial but that is the question asked.\n\nOlinguito\n\nWell-known member\nI am trying to prove a simple thing, that if AxA = BxB then A=B.\n\nThe intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.\n\nHow do I prove it formally ?\n\nThank you !\nLet $a\\in A$. Then $(a,a)\\in A\\times A$. Since we\u2019re assuming $A\\times A=B\\times B$, this means $(a,a)\\in B\\times B$ and thus $a\\in B$. Therefore $A\\subseteq B$. The same argument with $A$ and $B$ interchanged shows that $B\\subseteq A$. Hence $A=B$.", "date": "2020-08-12 09:05:49", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/cartesian-product-proof.26383/", "openwebmath_score": 0.8955318331718445, "openwebmath_perplexity": 142.73747928417976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399043329855, "lm_q2_score": 0.8774767762675405, "lm_q1q2_score": 0.851362983460235}}
{"url": "https://www.physicsforums.com/threads/pair-in-a-poker-again.641732/", "text": "# Pair in a poker, again\n\n1. Oct 6, 2012\n\n### Astr\n\nIm New on this forum, so i hope this is the right place to ask this question. I've tried to solve the problem of finding how many hands are in a poker game with exactly one pair, this way:\n\nI can choose the first card from 52 ways. for the second card I can choose it from 3 ways (to match the pair). For the third card I can choose it from 48 possibilities (for ensure I\u00b4ve got no trios in the hand). For the fourth and fifth i can choose them in 44 and 40 different ways, respectively. And because order is not important I must divide for 5! (ways of permuting 5 cards). My (wrong) answer is:\n\n(52x3x48x44x40)/5!=109 824\n\nThis is a tenth of the correct answer. Please can you tell whats wrong in the procedure. (English isn't my first language, sorry if Ive got some mistakes)\n\n2. Oct 7, 2012\n\n### CWatters\n\nThat gives you too many because a QS + QD is the same pair as QD+SQ.\n\nThere are 6 ways to get a pair of (for example) Queens..\n\nQS + QH\nQS + QD\nQS + QC\nQH + QC\nQH + QD\nQD + QC\n\nbut there is a choice of 13 denomination (values) so for pair it's\n\n6 * 13\n\nthen the other cards must not have the same value but can be of any suit.\n\n3. Oct 7, 2012\n\n### Astr\n\nThank you Cwatters. But I think i avoid the problem QS+QD=QD+QS, dividing by 5!, becasuse that ensures that the order of the 5 cards does not matter.\n\n4. Oct 10, 2012\n\n### CWatters\n\nI can't immediatly see why your method is wrond but continuing from above..\n\nThen the next three cards can have one of 12 other values (3 from 12 = 220) and be from any of 4 suits (4*4*4 = 64)\n\n220 * 16 = 14080\n\nNow that is also 48*44*40/3! so your last part seems ok?\n\n5. Oct 10, 2012\n\n### Astr\n\nThank you again, CWatters. Well I know how to solve the problem, and this is out of discussion:\nFor the pair:\n52x3/2!= 78 ways\nFor the three different cards:\n48x44x40/3!=14 080 ways\nthen: 78x14 080= 1098240 ways to get exactly one pair.\nbut my problem is: Whats wrong in the procedure i post first?\n\n6. Oct 10, 2012\n\n### Ray Vickson\n\nYou should not divide by 5!; instead, you should divide by 2! \u00d7 3!, because the order of the cards within the pair does not matter and the order within the three non-pairs does not matter. That would give the number of hands as 1098240, as you want.\n\nThere are some subtle points here, so let me expand. When counting the hands containing exactly one pair, the person posing the question is distinguishing between hands such as PPNNN, PNNNP, PNNPN, NNNPP, etc., even though the actual cards involved are exactly the same (and so all would be regarded as a *single* hand by a poker player). Basically the question is asking for the number of different ways you could be *dealt* a hand containing a single pair, even though after being dealt the cards you would re-arrange them to your own, personal liking.\n\nAnother way to see this is to look at P{pair}, the probability of getting dealt a pair. Suppose, first that we have 2 aces and one each of 2 3 4. The probability, p, of getting that is p = C(4,2)*C(4,1)^3/C(52,5), where C(a,b) = \"a choose b\" = a!/[b!(a-b)!]. This holds because there are C(4,2) ways to pick 2 aces from 4 and for each of the 2 3 and 4 there are C(4,1) ways to choose the particular card, and because there are C(52,5) says of being dealt a 5-card hand althogether. Now, of course, the pair may not be aces and the non-pairs may not be 2 3 and 4. However, for any choice of ranks, the probability of the pair and the three non-pairs are all the same as the p above. Therefore, P{pair} is p times the number of ways to choose the rank of the pair (13) times the number of ways of choosing the three different ranks from the remaining 12 (which is C(12,3)). Thus, we have\n$$P\\{ \\text{pair}\\} = \\frac{13 \\, C(12,3) \\, C(4,2) \\, C(4,1)^3}{C(52,5)} = \\frac{N_p}{C(52,5)},$$\nwhere Np = number of hands containing one pair. Thus, we have\n$$N_p = 13 \\, C(12,3) \\, C(4,2) \\, C(4,1)^3 = 1098240.$$\n\nRGV\n\nLast edited: Oct 10, 2012\n7. Oct 10, 2012\n\n### Astr\n\nRay, Thank you for your answer. Can you explain me why i should not divide by 5!, because this is my whole problem. Thanks.", "date": "2018-02-19 14:29:19", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/pair-in-a-poker-again.641732/", "openwebmath_score": 0.7707682251930237, "openwebmath_perplexity": 728.7862245719733, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812354689082, "lm_q2_score": 0.8791467659263148, "lm_q1q2_score": 0.8513492313462198}}
{"url": "http://math.stackexchange.com/questions/181239/solving-a-b", "text": "# Solving $|a| < |b|$\n\nI apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.\n\nI am looking for a way to eliminate absolute value functions in $|a| < |b|$.\n\nI can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.\n\nThank you!\n\nAn example problem:\n\n$$|x + 2| < |x - 4|$$\n\n-\nperhaps you could give some specific problems that trouble you. \u2013\u00a0Will Jagy Aug 11 '12 at 5:19\nGood suggestion. I've added one above. \u2013\u00a0Mark V. Aug 11 '12 at 5:41\nIt's useful to review the definition of the absolute value function: $$\\text{abs}(a) = \\begin{cases} a & 0 \\leq a \\\\ -a & a < 0 \\end{cases}$$ In order to eliminate one absolute value function from an expression, you will need to consider two cases: the case where its argument is non-negative, and the case where its argument is negative. This will give you an expression with two cases with one less absolute value function. Iterating this process will give an expression without the absolute value sign. \u2013\u00a0danportin Aug 11 '12 at 7:13\nSince you mentioned trying Google, maybe you should try Wolfram Alpha instead \u2013\u00a0Ben Millwood Aug 11 '12 at 11:45\n\nWe have $|a| \\lt |b|\\,$ if any of these is true:\n\n(i) $\\,a$ and $b$ are $\\gt 0$ and $a \\lt b$\n\n(ii) $a\\lt 0$ and $b\\ge 0$ and $-a \\lt b$\n\n(iii) $b \\lt 0$ and $a \\gt 0$ and $a \\lt -b\\,$\n\n(iv) $\\,a\\lt 0$ and $b\\lt 0$ and $-a\\lt -b$. We can rewrite this as $b \\lt a$.\n\nFour cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.\n\nSometimes, one can exploit the simpler $|a| \\lt| b|\\,$ iff $\\,a^2\\lt b^2$. But squaring expressions generally makes them substantially messier.\n\nAdded: With your new sample problem, squaring happens to work nicely. We have $|x+2| \\lt |x-4|$ iff $(x+2)^2 \\lt (x-4)^2$. Expand. We are looking at the inequality\n\n$$x^2+4x+4 \\lt x^2-8x+16.$$\n\nThe $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \\lt 12$, or equivalently $x\\lt 1$. The squaring strategy works well for any inequality of the form $|ax+b| \\lt |cx+d|$.\n\nBut the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$.\n\n-\nHa! I never would of thought of this. This is brilliant! Thank you so much for your help! \u2013\u00a0Mark V. Aug 11 '12 at 18:10\n\nThere are different approaches; one is to look at the zeroes of the expressions inside the absolute values, and split up $\\mathbb R$ into intervals accordingly. In your example $|x + 2| < |x - 4|$, the points of interest are at $x=-2$ and $x=4$. You can therefore consider three cases:\n\n1. If $x \\in (-\\infty,-2)$, then $x+2 < 0$ and $x-4 < 0$, so the absolute values will reverse the signs of both. This gives: \\begin{align} -(x+2) &< -(x-4) \\\\ x+2 &> x-4 \\\\ 2 &> -4 \\end{align} This is true for all $x$ in the interval.\n\n2. If $x \\in [-2,4)$, then $x+2 \\geq 0$ so its sign is unaffected by the absolute value, but $x-4 <0$ so its sign will be reversed: \\begin{align} x+2 &< -(x-4) \\\\ 2x+2 &< 4 \\\\ x &< 1 \\end{align} Combining this last inequality with the assumption that $x \\in [-2,4)$, we see that any $x$ in $[-2,1)$ is valid.\n\n3. Finally, if $x \\in [4,\\infty)$, neither expression's sign is reversed: \\begin{align} x+2 &< x-4 \\\\ 2 &< -4 \\end{align} This is false for all $x$ in the interval.\n\nPutting all the information together from the above three cases, we have $x \\in (-\\infty, 1)$.\n\nNote: as some other contributors have mentioned, there are simpler ways to deal with your problem, such as viewing it geometrically. The method that I have shown above is more useful when the expressions are more complicated or when you have several absolute values; for example, an inequality like $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.\n\n-\nPedantry: you probably want $x\\in[-2,4)$ for step 2 and $x\\in[4,\\infty)$ for step 3 (or change where you say $x-4<0$, but I think my way is more natural/consistent) \u2013\u00a0Ben Millwood Aug 11 '12 at 11:03\n@BenMillwood: Fair enough! I've changed it. \u2013\u00a0Th\u00e9ophile Aug 11 '12 at 18:13\n\nWe can see $|x-a|$ as a distance point $x$ from point $a$.\n\nNow, the question with the above \"definition\" would be like:\n\nFor which $x$ values distance point $x$ from $-2$ be less than distance point $x$ from $4$?\n\nClearly, by drawing it maybe, you can observe that the answer is for all $x<1$.\n\n-\n\nWe can divide by $|b|$ to get $|a/b|<1$. Let $x=a/b$ then $|x|<1$ so $-1<x<1$.\n\nNow multiply through by $b$.\n\nIf $b>0$ then $-b<a<b$.\n\nIf $b<0$ then $-b>a>b$\n\n-", "date": "2016-05-04 19:53:32", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/181239/solving-a-b", "openwebmath_score": 0.9874403476715088, "openwebmath_perplexity": 310.32775474749354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812309063187, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8513492288681892}}
{"url": "http://mathhelpforum.com/differential-geometry/108094-problem-riemann-integrable.html", "text": "# Math Help - A problem of Riemann Integrable\n\n1. ## A problem of Riemann Integrable\n\nIf $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\\geq0$, we have $\\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ a.e.?\n(a.e. stands for almost everywhere)\n\n2. Yes. Note that if $\\int_{a} ^{b} \\ x^nf(x)dx =0$ for all $n \\in \\mathbb{N}$ then $\\int_{a} ^{b} \\ p(x)f(x)dx =0$ for any polynomial $p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $I: C^0([a,b]) \\rightarrow \\mathbb{R}$ $g \\mapsto \\int_{a} ^{b} \\ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\\int_{a} ^{b} \\ g(x)f(x)dx =0$ for all $g(x) \\in C^0([a,b])$. It is a well known result that under this conditions $f \\equiv 0$.\n\nEdit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.\n\n3. Originally Posted by Xingyuan\nIf $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\\geq0$, we have $\\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ a.e.?\n(a.e. stands for almost everywhere)\n\nWell you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.\n\nTonio\n\n4. Originally Posted by tonio\nWell you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.\n\nTonio\nNo,In the problem,I am not assuming $f(x)\\geq0$\n\n5. Originally Posted by Xingyuan\nNo,In the problem,I am not assuming $f(x)\\geq0$\n\nI didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.\n\nTonio\n\n6. Originally Posted by tonio\nWell you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.\n\nTonio\nHuh? That doesn't follow.\n\n7. Originally Posted by tonio\nI didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.\n\nTonio\nConsider $\\int_{-\\pi}^{\\pi}\\sin(x)\\,dx$\n\nIt's only $0$ at the two endpoints and $x=0$, but the whole integral is $0$.\n\nOr, for that matter, consider $\\int_{-a}^a f(x)\\,dx$ where $f(x)$ is any odd function.\n\n8. Originally Posted by rn443\nHuh? That doesn't follow.\n\nYes it does\n\nTonio\n\n9. Originally Posted by tonio\nYes it does\n\nTonio\nAre you assuming f is nonnegative?\n\n10. Originally Posted by rn443\nAre you assuming f is nonnegative?\nOk, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.\n\nTonio\n\n11. Does the theorem still hold if we instead assume $\\int_{-\\infty}^\\infty x^n f(x) \\mathrm dx = 0$ for all n >= 0, provided that $\\int_{-\\infty}^\\infty |x^n f(x)| \\mathrm dx$ exists and is finite?\n\n12. Lets suppose, without loss of generality, that is $a=-\\pi$ and $b=\\pi$. Since $f(*)$ is Riemann integrable in $[-\\pi,\\pi]$ is can be expanded in Fourier series ...\n\n$f(x)= \\frac{a_0}{2} + \\sum_{k=1}^{\\infty} a_{k}\\cdot \\cos kx + b_{k}\\cdot \\sin kx$ (1)\n\n... where...\n\n$a_{k}= \\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} f(x)\\cdot \\cos kx\\cdot dx$\n\n$b_{k}= \\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} f(x)\\cdot \\sin kx\\cdot dx$ (2)\n\nNow is...\n\n$\\cos kx = \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}\\cdot (kx)^{2n}}{(2n)!}$\n\n$\\sin kx = \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}\\cdot (kx)^{2n+1}}{(2n+1)!}$ (3)\n\n... and $\\forall n\\ge 0$ is...\n\n$\\int_{-\\pi}^{\\pi} x^{n}\\cdot f(x)\\cdot dx =0$ (4)\n\nThe conclusion is that all the integrals in (2) are null and the same is for the $a_{k}$ and $b_{k}$ ...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n13. Originally Posted by Jose27\nYes. Note that if $\\int_{a} ^{b} \\ x^nf(x)dx =0$ for all $n \\in \\mathbb{N}$ then $\\int_{a} ^{b} \\ p(x)f(x)dx =0$ for any polynomial $p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $I: C^0([a,b]) \\rightarrow \\mathbb{R}$ $g \\mapsto \\int_{a} ^{b} \\ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\\int_{a} ^{b} \\ g(x)f(x)dx =0$ for all $g(x) \\in C^0([a,b])$. It is a well known result that under this conditions $f \\equiv 0$.\n\nEdit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.\nYes,Riemann Integrable function is continous a.e.\nBut Weierstrass' approximation theorem is:\nEvery function defined and continuous on the finite interval $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.\nBecause Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$\nIf we denote the set $E$ of all discontinuity point,then define:\n\n$g(x)=\\{ \\begin{array}{cc}f(x), &\\mbox{if x is a continuous point of} f(x) \\\\ \\lim_{t \\rightarrow x}f(t) &\\mbox{if x is a discontinuity point of} f(x) \\end{array}$\n\nthen $g(x)$ is a continuous function on $[a,b]$,right?\n\nthen use the conclusion above,we get the conclusion $g(x)\\equiv0$.\nso $f(x)=0$a.e.\nIs this proof right?\n\n14. Originally Posted by Xingyuan\nYes,Riemann Integrable function is continous a.e.\nBut Weierstrass' approximation theorem is:\nEvery function defined and continuous on the finite interval $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.\nBecause Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$\nIf we denote the set $E$ of all discontinuity point,then define:\n\n$g(x)=\\{ \\begin{array}{cc}f(x), &\\mbox{if x is a continuous point of} f(x) \\\\ \\lim_{t \\rightarrow x}f(t) &\\mbox{if x is a discontinuity point of} f(x) \\end{array}$\n\nthen $g(x)$ is a continuous function on $[a,b]$,right?\n\nthen use the conclusion above,we get the conclusion $g(x)\\equiv0$.\nso $f(x)=0$a.e.\nIs this proof right?\nNotice that I'm using Weierstrass to expand the set of functions $g$ such that $\\int_{a} ^{b} g(x)f(x)dx =0$, I'm not saying anything about $f$.\n\nAnother thing, the $g$ you defined could be not well defined if f has a jump discontinuity.\n\nOk, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.\nThis doesn't contradict the claim since $\\int_{-\\pi } ^{\\pi } x\\sin(x) dx=2\\int_{0 } ^{ \\pi } x\\sin(x) dx >0$\n\n15. Originally Posted by Jose27\nThis doesn't contradict the claim since $\\int_{-\\pi } ^{\\pi } x\\sin(x) dx=2\\int_{0 } ^{ \\pi } x\\sin(x) dx >0$\nI was only contradicting the case $n=0$.\n\nPage 1 of 2 12 Last", "date": "2015-11-30 06:14:42", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/differential-geometry/108094-problem-riemann-integrable.html", "openwebmath_score": 0.936928391456604, "openwebmath_perplexity": 508.9791345749555, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9683812327313545, "lm_q2_score": 0.8791467564270272, "lm_q1q2_score": 0.8513492197405764}}
{"url": "https://math.stackexchange.com/questions/3303103/how-to-choose-a-control-input-so-that-the-nonlinear-system-stays-at-a-fixed-poin", "text": "# How to choose a control input so that the nonlinear system stays at a fixed point?\n\nThe dynamics of a magnetically suspended steel ball can be described by\n\n$$m \\ddot{h} = mg - c \\frac{u^2}{h^2}$$\n\nwhere $$m$$ is the mass of the ball, $$g$$ is gravitational acceleration, $$c$$ is a positive constant, and $$h$$ represents the vertical position of the ball. The input $$u$$ is the current supplied to the electromagnet.\n\n(a) Write down a nonlinear state space model using $$x_1 = h$$ and $$x_2 = \\dot{h}$$.\n\n(b) Determine the equilibrium control input $$u_e$$ that has to be applied to suspend the ball at some position $$h=h_0 > 0$$.\n\nFor part (a), the system is \\begin{align} \\dot{x}_1 &= x_2 \\tag{1} \\\\ \\dot{x}_2 &= g - \\frac{c}{m} \\frac{u^2}{x^2_1} \\tag{2} \\end{align}\n\nFor part (b), the equilibrium point $$x_e=(h_0,0)$$ which is not at the origin, so we need first to make sure the equilibrium point $$x_e$$ is transformed to the origin by introducing new variables: let $$y=x-x_e$$, we get:\n\n\\begin{align} y_1 &= x_1 - h_0 &\\implies \\dot{x}_1 = \\dot{y}_1 \\tag{3}\\\\ y_2 &= x_2 - 0 &\\implies \\dot{x}_2 = \\dot{y}_2 \\tag{4} \\end{align} From (1),(2),(3), and (4), the new system is \\begin{align} \\dot{y}_1 &= y_2 \\\\ \\dot{y}_2 &= g - \\frac{c}{m} \\frac{u^2}{(y_1 + h_0)^2} \\end{align} The equilibrium point $$y_e = (\\sqrt{\\frac{c}{gm}} u - h_0, 0)$$. The control input $$u_e$$ that makes $$y_e$$ is zero is $$u_e = \\frac{ h_0}{\\sqrt{\\frac{c}{gm}}}$$, therefore, the equilibrium control input $$u_e$$ that has to be applied to suspend the ball at some position $$h=h_0 > 0$$ is $$u_e = \\frac{ h_0}{\\sqrt{\\frac{c}{gm}}}$$\n\nIs this correct? if not, any suggestions how to tackle this problem.\n\nLooks correct, but you can always check this yourself by plugging your $$u_e$$ into the expression for either $$\\dot{x}_2$$ or $$\\dot{y}_2$$. Also note that the input only appears as $$u^2$$, so using $$-u$$ should have the same effect as $$u$$.\nLastly it is also possible to solve for $$u_e$$ by setting $$\\dot{x}_2=0$$ while using $$x_1=h_0$$. However, you might also have to linearize the system around this equilibrium point, so doing this coordinate translation would probably have to be done anyway in a later stage.\n\u2022 I've fixed that, thanks. But when I run the simulation via Simulink Matlab, it doesn't seem the the state $x_1 \\rightarrow h_0$. I'm not sure if the input choice is correct. \u2013\u00a0CroCo Jul 25 at 0:42\n\u2022 @CroCo Having a desired equilibrium does not mean that the system converges to it. For that you also need that that equilibrium is stable. This probably isn't the case. In order to achieve this you need to let $u$ deviate from $u_e$ using some sort of feedback. \u2013\u00a0Kwin van der Veen Jul 25 at 0:47\n\u2022 The Jacobain matrix $A$ is $$A =\\frac{\\partial f}{ \\partial x} = \\begin{bmatrix} 0 & 1 \\\\ 2gh^2_0x^{-3}_1 & 0 \\end{bmatrix}_{x_e=(h_0,0),u_e=\\sqrt{\\frac{gm}{c}} h_0} = \\begin{bmatrix} 0 & 1 \\\\ 2gh^{-1}_0 & 0 \\end{bmatrix}$$ The eigenvalues of $A$ are $\\lambda_{1,2}=\\pm \\sqrt{\\frac{2g}{h_0}}$. They are real and sign opposite, hence, the system is unstable for this input. Is this correct? \u2013\u00a0CroCo Jul 25 at 2:55", "date": "2019-12-14 03:02:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3303103/how-to-choose-a-control-input-so-that-the-nonlinear-system-stays-at-a-fixed-poin", "openwebmath_score": 0.9977102875709534, "openwebmath_perplexity": 226.8148900306021, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915215128427, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.851317725262155}}
{"url": "https://ekiim.me/books/linear-algebra-strang/exercises/01-2.htm", "text": "ekiim's blog\n\n## Exercises\n\n1. For the eeauations $x + y = 4, 2x - 2y = 4$, draw the row picture (two intersecting lines) and the column picture (a combination of two columns, equatl to the column vector $(4, 4)$ on the right side).\n\n2. Solve the non singular triangular system: \\begin{aligned} u + v + w &= b_1 \\\\ v + w &= b_2 \\\\ w &= b_3 \\end{aligned} Show that your solution gives a combination of columns that equal the column on the write\n\n3. Describe the intersection of three planes $u + v + w + z = 6$ and $u + w + z = 4$ and $u + w = 2$ (all in 4-dimensional space). Is it a line or a point or an empty set? What is the intersection of the fourth plane u = -1 is excluded?\n\n4. Sketch the lines \\begin{aligned} x + 2y &= 2\\\\ x - y &= 2\\\\ y &= 1 \\\\ \\end{aligned} Can the three equations be solved simultaneously? What happens to the ficutre if all right hand sides zero? Is there any nonzero choice of right hand sides which allows the three lines to intersect at the same point and the three equations to have a solution?\n\n5. Find two points on the line intersection of three plances $t=0$, and $z=0$ and $x+y+z+y=1$ in four-dimensional space.\n\n6. When $b=(2,5,7)$, find a solution $(u, v w)$ to equation (4) other than the solutions $(1, 0 1)$ and $(4, -1, -1)$, metnioned in the text.\n\n\u2022 Equation 4: $u \\begin{bmatrix} 1 \\\\ 2 \\\\ 3 \\end{bmatrix} v \\begin{bmatrix} 1 \\\\ 0 \\\\ 1 \\end{bmatrix} w \\begin{bmatrix} 1 \\\\ 3 \\\\ 4 \\end{bmatrix} = b$\n7. Give two more righthand sides in addition to $b=(2,5,7)$ for which equation (4) can be solved. Give two more right hand sides in addition to $b=(2,5,6)$ for which it cannot be solved.\n\n8. Explain why the system: \\begin{aligned} u + w + w &= 2\\\\ u + 2v + 3w &= 1\\\\ v + 2w &= 0 \\\\ \\end{aligned} if singular, by finding a combination of three equations that adds up $0 = 1$. What value should replace the last zero on the rightside to allow the equation to have solutions \u2014 and what is one of the solutions?\n\n9. The column picture for the prevous exercises is $u \\begin{bmatrix} 1 \\\\ 1 \\\\ 0 \\end{bmatrix} v \\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix} w \\begin{bmatrix} 1 \\\\ 3 \\\\ 2 \\end{bmatrix} = b$ Show that the three columns on the left lie in the same plane by expressing the third column as a combination of the first two. What are the solutions $(u, v, w)$, if b is the zero vector $(0, 0, 0)$?\n\n10. Under what condiction on $y_1, y_2, y_3$, do the points $(0, y_1), $(1, y_2)$, (2, y_3)$ lie on a straight line?\n\n11. The equations \\begin{aligned} ax + 2y &= 0\\\\ 2x + ay &= 0 \\\\ \\end{aligned} are certain to have the solution $x=y=0$. For which values of $a$, is there a whole line of solutions?\n\n12. Sketch the plane $x+y+z=1$, or the part of the plane that is in the positivie octant where $x\\geq0, y\\geq0, z\\geq0$. Do the same for $x+y+z=2$ in the same figure. What vector is Perpendicular to does planes?\n\n13. Starting with the line $x+4y=7$, fin d the equation for the parallel line thruough $x=0, y=0$. Find the equation of another line that meets the first at $x=3, y=1$", "date": "2019-09-17 10:16:46", "meta": {"domain": "ekiim.me", "url": "https://ekiim.me/books/linear-algebra-strang/exercises/01-2.htm", "openwebmath_score": 0.9998576641082764, "openwebmath_perplexity": 344.8673520821427, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915235185259, "lm_q2_score": 0.8596637505099168, "lm_q1q2_score": 0.8513177252061155}}
{"url": "https://byjus.com/question-answer/a-triangle-abc-is-drawn-to-circumscribe-a-circle-of-radius-3-cm-such-that-1/", "text": "Question\n\n# A $$\\triangle{ABC}$$ is drawn to circumscribe a circle of radius $$3$$ cm. such that the segments $$BD$$ and $$DC$$ are respectively of length $$6$$ cm and $$9$$ cm. If the area of\u00a0 $$\\triangle{ABC}$$ is $$54{cm}^{2}$$, then find the lengths of sides $$AB$$ and $$AC$$\n\nSolution\n\n## Area of $$\\triangle{ABC}=$$area of $$\\triangle{OBC}+$$area of $$\\triangle{OAC}+$$area of $$\\triangle{OAB}$$We have $$BD=6$$cm, $$BE=6$$cm(equal tangents)$$DC=9$$cm, $$CF=9$$cm(equal tangents)Area of $$\\triangle{OBC}=\\dfrac{1}{2}\\times b\\times h=\\dfrac{1}{2}\\times 15\\times 3=\\dfrac{45}{2}{cm}^{2}$$Area of $$\\triangle{OAC}=\\dfrac{1}{2}\\times \\left(x+9\\right)\\times 3=\\dfrac{3\\left(x+9\\right)}{2}{cm}^{2}$$Area of $$\\triangle{OAB}=\\dfrac{1}{2}\\times \\left(x+6\\right)\\times 3=\\dfrac{3\\left(x+6\\right)}{2}{cm}^{2}$$By Heron's formula,area of $$\\triangle{ABC}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}{cm}^{2}$$$$s=\\dfrac{x+9+x+6+15}{2}=\\dfrac{2x+30}{2}=x+15$$$$\\therefore \\Delta=\\sqrt{\\left(x+15\\right)\\left(x+15-15\\right)\\left(x-15-x-4\\right)\\left(x+15-x-6\\right)}=\\sqrt{x\\left(x+15\\right)\\times 6\\times 9}$$$$\\Rightarrow \\sqrt{54x\\left(x+15\\right)}=\\dfrac{3\\left(x+9\\right)}{2}+\\dfrac{3\\left(x+16\\right)}{2}+\\dfrac{45}{2}$$$$\\Rightarrow \\sqrt{54x\\left(x+15\\right)}=\\dfrac{3}{2}\\left[x+9+x+6+15\\right]$$$$\\Rightarrow \\sqrt{54x\\left(x+15\\right)}=\\dfrac{3}{2}\\left(2x+30\\right)$$$$\\Rightarrow \\sqrt{54x\\left(x+15\\right)}=3x+15$$Squaring on both sides we get$$\\Rightarrow 54x\\left(x+15\\right)=9{\\left(x+15\\right)}^{2}$$$$\\Rightarrow 6x=x+15$$$$\\Rightarrow 6x-x=15$$$$\\Rightarrow 5x=15$$$$\\therefore x=\\dfrac{15}{5}=3$$cmHence the sides are $$15$$cm, $$12$$cm and $$9$$cmMathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-17 23:24:04", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/a-triangle-abc-is-drawn-to-circumscribe-a-circle-of-radius-3-cm-such-that-1/", "openwebmath_score": 0.8959956169128418, "openwebmath_perplexity": 1617.7126986744322, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308810508339, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.8513125152134369}}
{"url": "https://math.stackexchange.com/questions/2311346/show-that-s-n2", "text": "# Show that $s_n<2$\n\nIf $s_1=\\sqrt{2}$ and $s_{n+1}=\\sqrt {2+s_n}$ for $n\\geq 1$ show that $s_n<2$ $\\forall, n\\geq1$ and $s_n$ is convergent.\n\nThe second part I supposed that is immediate from Cauchy sequence definition that $s_n$ is convergent.\n\nThe first part I think to use induction\n\nFor $n=1$ we have that $s_1=\\sqrt{2}<2$\n\nSuppose that is true for $n=k$ then we need to show for $n=k+1$. I wrote a few terms of this recursive sequence\n\n$$s_1=\\sqrt{2},\\quad s_2=\\sqrt{2+\\sqrt{2}},\\quad s_3=\\sqrt{2+s_2}=\\sqrt{2+\\sqrt{2+\\sqrt{2}}}$$ Then I think that $s_{n+1}$ can written as $$s_{n+1}=\\sum_{i=1}^n2^{\\frac{1}{2n}}$$\n\nI do not know how to argue that the above term is less than $2$.\n\n\u2022 $$s_{n+1} = \\sqrt{2+s_n} < \\sqrt{2 + 2} = 2$$ \u2013\u00a0peterwhy Jun 5 '17 at 23:25\n\u2022 Your general form is wrong. \u2013\u00a0Simply Beautiful Art Jun 5 '17 at 23:25\n\u2022 And it is not \"obvious\" that the sequence is Cauchy \u2013\u00a0Luiz Cordeiro Jun 5 '17 at 23:28\n\nAs peterwhy mentions,\n\n$$s_n<2\\implies s_{n+1}=\\sqrt{2+s_n}<\\sqrt{2+2}=2\\tag{\\color{green}\\checkmark}$$\n\nTo show that it is convergent, you want to show that it is monotone increasing, which combined with the fact that it is bounded above will mean it converges:\n\n$$s_n>s_{n-1}\\implies s_{n+1}=\\sqrt{2+s_n}>\\sqrt{2+s_{n-1}}=s_n\\tag{\\color{green}\\checkmark}$$\n\nand thus you are done.\n\nBy the induction assumption, for some $k\\in\\mathbb N$, $$s_k < 2$$\n\nConsider $n = k+1$,\n\n$$s_{k+1} = \\sqrt{2+s_k} < \\sqrt{2 + 2} = 2$$\n\nTogether with the fact that $s_1 = \\sqrt 2 < \\sqrt 4 = 2$, by induction, all $s_n < 2$.\n\nFor $0<x<2$, $x^2 < 2x$ and $2x < 2 + x$.\n\nTo prove that the sequence $s_n$ is increasing,\n\n\\begin{align*}s_{n+1} &= \\sqrt{2+s_n}\\\\ &> \\sqrt{s_n + s_n}\\\\ &= \\sqrt{2s_n}\\\\ &> \\sqrt{s_n^2}\\\\ &= s_n \\end{align*}\n\nSince the sequence $s_n$ is increasing and bounded above, limit exists.\n\nHint: $\\,s_{n+1}-2=(\\sqrt {2+s_n}-2) \\cdot \\cfrac{\\sqrt {2+s_n}+2}{\\sqrt {2+s_n}+2} = \\cfrac{2+s_n-4}{\\sqrt {2+s_n}+2} = \\cfrac{s_n-2}{\\sqrt {2+s_n}+2}\\,$, therefore $s_{n+1}-2$ has the same sign as $s_n -2\\,$ and, by induction/telescoping, the same sign as $s_1-2\\,$.\n\nUsing the fact that the square root function is increasing, and the induction hypothesis that $s_k<2$, we have: $$s_{k+1}=\\sqrt{2+s_k}<\\sqrt{2+2}=2$$\n\nAssuming that the sequence does converge (you prove that later) write the limit as \"S\". Letting n go to infinity in $s_{n+1}= \\sqrt{2+ s_n}$ we get $S= \\sqrt{2+ S}$. Squaring both sides, $S^2= 2+ S$ so S must satisfy $S^2- S- 2= (S- 2)(S+ 1)= 0$ so S is either 2 or -1. Since all terms are positive, the limit (again, if there is a limit) must be 2.\n\nTwo show that this is convergent (so that the above is correct) you can show this sequence is increasing and bounded above. Since we got 2 as the limit above, it makes sense to show that $S_n< 2$ for all n. Yes, I would do that using induction on n. When n= 1, $S_n= \\sqrt{2}< 2$. Assume that, for some k, $S_k< 2$. Then $S_{k+1}= \\sqrt{2+ S_k}< \\sqrt{2+ 2}= \\sqrt{4}= 2$.", "date": "2019-12-10 08:28:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2311346/show-that-s-n2", "openwebmath_score": 0.996794581413269, "openwebmath_perplexity": 165.70388262096128, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308796527184, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.8513125104959804}}
{"url": "https://math.stackexchange.com/questions/2588234/are-positive-definite-matrices-robust-to-small-changes/2588243", "text": "# Are positive definite matrices robust to \u201csmall changes\u201d?\n\nLet $A$ be a positive-definite matrix and let $B$ be some other symmetric matrix. Consider the matrix $$C=A+\\varepsilon B.$$ for some $\\varepsilon>0$. Is it true that for $\\varepsilon$ small enough $C$ is also positive definite?\n\n\u2022 Is $B$ required to be symmetric? \u2013\u00a0kimchi lover Jan 2 '18 at 1:02\n\u2022 I guess, since \"positive definite\" usually refer to symmetric matrices. I have updated the question. \u2013\u00a0user_lambda Jan 2 '18 at 1:02\n\u2022 Eigenvalues of $A+\\epsilon B$ depends continuously on $\\epsilon$. If they are all positive for $\\epsilon=0$ then they are positive for all small $\\epsilon$. \u2013\u00a0A.\u0393. Jan 2 '18 at 1:37\n\nPositive definite means that $\\langle v, Av \\rangle > 0$ for all nonzero vectors $v$; actually it suffices to check this condition for unit vectors. We have\n\n$$\\langle v, Cv \\rangle = \\langle v, Av \\rangle + \\epsilon \\langle v, Bv \\rangle.$$\n\nNow, by the compactness of the unit sphere, $\\langle v, Av \\rangle$ takes on a minimum nonzero value $m$ on unit vectors (the smallest eigenvalue of $A$, although we don't need this), and $| \\langle v, Bv \\rangle |$ takes on a maximum nonzero value $M$ on unit vectors (the largest eigenvalue of $B$ in absolute value).\n\nSo we can take $\\epsilon < \\frac{m}{M}$, which gives\n\n$$\\langle v, Cv \\rangle \\ge m - \\epsilon M > 0$$\n\nfor all unit vectors $v$. So $C$ is positive definite as desired.\n\n\u2022 Why did you use the spectral radius of $B$ instead of $| \\lambda_{\\min} (B) |$? \u2013\u00a0Rodrigo de Azevedo Jan 2 '18 at 10:13\n\u2022 That would also work. I just needed anything that would clearly give a lower bound. \u2013\u00a0Qiaochu Yuan Jan 2 '18 at 10:36\n\nIf by \"positive definite\" you mean \"strictly positive definite\", the answer is \"yes\". The set of strictly positive definite matrices is an open set in the space of symmetric matrices.\n\nFor the following reason. The positive definite matrices are the ones which satisfy a certain finite set of determinental inequalites (the principal minor determinants must all be strictly positive), each one of of which cuts out an open set in the space of matrices. Alternatively, from first principles, let $X$ be the closed unit sphere in vector space, let $Y$ be the symmetric matrices. The function $(v,A)\\mapsto \\|Av\\|$ is continuous, so the set $S=\\{(v,A): \\|Av\\|\\le0\\}\\subset X\\times Y$ is closed. Since $X$ is compact, the map $\\pi:(v,A)\\mapsto A$ is a closed map, so $\\pi(S)$ is closed in $Y$, so the complement of $\\pi(S)$ is open in $Y$. But that complement is the set of all $A$ for which $\\|Av\\|>0$ for all $v\\in X$, that is to say, the set of all (strictly) positive definite matrices.\n\n\u2022 And for positive semi-definite, it's strictly false: $\\pmatrix{1 & 0 \\\\ 0 & 0 } + \\epsilon \\pmatrix{0 & 0 \\\\ 0 & -1}$ has eigenvalues $1, -\\epsilon$ for every $\\epsilon$. \u2013\u00a0John Hughes Jan 2 '18 at 1:09\n\u2022 Is \"strictly positive definite\" different than \"positive definite\"? I'm unaware of that definition. \u2013\u00a0user_lambda Jan 2 '18 at 1:14\n\u2022 @user_lambda: some authors use \"positive definite\" to mean \"positive semidefinite,\" in the same way that some authors use \"positive\" to mean \"nonnegative.\" \u2013\u00a0Qiaochu Yuan Jan 2 '18 at 1:17\n\u2022 @QiaochuYuan: For what it's worth, I haven't heard of anyone using the word \"positive\" to mean \"non-negative\", but what I have seen is the notation of $\\mathbb{R}_+$ to mean the non-negative reals. \u2013\u00a0Mehrdad Jan 2 '18 at 5:42\n\u2022 @Mehrdad How about Bourbaki? \u2013\u00a0Daniel Fischer Jan 2 '18 at 15:16\n\nTo complement Qiaochu's answer, we have\n\n$$\\rm v^\\top C \\, v = v^\\top A \\, v + \\varepsilon \\, v^\\top B \\, v$$\n\nwhere $\\| \\rm v \\|_2 = 1$ and $\\varepsilon > 0$. Note that\n\n$$\\{ \\rm v^\\top A \\, v : \\| v \\|_2 = 1 \\} = [ \\lambda_{\\min} (\\mathrm A), \\lambda_{\\max} (\\mathrm A) ]$$\n\n$$\\{ \\rm v^\\top B \\, v : \\| v \\|_2 = 1 \\} = [ \\lambda_{\\min} (\\mathrm B), \\lambda_{\\max} (\\mathrm B) ]$$\n\nand, thus,\n\n$$\\{ \\rm v^\\top C \\, v : \\| v \\|_2 = 1 \\} = \\left[ \\color{blue}{\\lambda_{\\min} (\\mathrm A) + \\varepsilon \\, \\lambda_{\\min} (\\mathrm B)}, \\lambda_{\\max} (\\mathrm A) + \\varepsilon \\, \\lambda_{\\max} (\\mathrm B) \\right]$$\n\nWe know that $\\rm A \\succ 0$, i.e., $\\lambda_{\\min} (\\mathrm A) > 0$. If $\\rm B$ is positive semidefinite, then $\\lambda_{\\min} (\\mathrm A) + \\varepsilon \\, \\lambda_{\\min} (\\mathrm B) > 0$, i.e., matrix $\\rm C$ is positive definite for all values of $\\varepsilon > 0$. After all, the conic combination of positive semidefinite matrices is also positive semidefinite. If $\\rm B$ is not positive semidefinite, then\n\n$$\\lambda_{\\min} (\\mathrm A) + \\varepsilon \\, \\lambda_{\\min} (\\mathrm B) = \\lambda_{\\min} (\\mathrm A) - \\varepsilon \\, |\\lambda_{\\min} (\\mathrm B)| > 0$$\n\nyields the following upper bound on $\\varepsilon$\n\n$$\\varepsilon < \\color{blue}{\\frac{\\lambda_{\\min} (\\mathrm A)}{|\\lambda_{\\min} (\\mathrm B)|}}$$", "date": "2019-10-17 19:00:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2588234/are-positive-definite-matrices-robust-to-small-changes/2588243", "openwebmath_score": 0.9369092583656311, "openwebmath_perplexity": 208.5079058093841, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308803517762, "lm_q2_score": 0.8615382076534742, "lm_q1q2_score": 0.8513125075853188}}
{"url": "https://www.physicsforums.com/threads/composition-of-functions.295719/", "text": "# Composition of functions\n\n1. ### mnb96\n\n638\nThis might be a silly question:\n\ngiven a function $$g$$ is it possible to find a function $$f$$ such that $$f = f \\circ g$$?\n\n321\nf=const.\n\n3. ### csprof2000\n\n287\nThat is the trivial solution. Perhaps there are others?\n\nf(x) = f(g(x)) if...\n\nf(x) = c, c constant.\ng(x) = x, f any function.\n\nIf these two conditions don't hold, though...\n\nAssume f has an inverse function. For instance, if f(x) = 2x, then (inv f)(x) = x/2. Then\n\nf = f o g\n<=>\nf(x) = f(g(x))\n<=>\nx = g(x).\n\nSo if f has an inverse, g must equal x if f = f o g.\n\nSo you'd only be looking for functions which don't have an inverse.\n\nBut what we have is a little stronger than that, no? Since my argument makes no reference to intervals, it must be true on any interval. So g = x if you want a function f which is invertible over any interval.\n\nThe only function which is not invertible over any interval is - you guessed it - constant functions.\n\nSo, in summary:\n\nif g(x) = x, then any function f(x) will do.\n\nOtherwise, f(x) = c , c constant, is the only solution.\n\nWhat about functions of more variables? Or generalized operations like differentiation? No idea.\n\n4. ### John Creighto\n\n811\nIf g(x)=x Then f(x) can be anything.\n\nLet\ng(n)=2n\n\nThen f(n) can map even numbers to one constant and odd numbers to a different constant. (At least it works if n is discrete). Not sure if it works in the continuous case but I think it might.\n\nLast edited: Feb 27, 2009\n5. ### yyat\n\n321\na) f(x)=abs(x), g(x)=-x\nb) f(x)=cos(x), g(x)=x+2pi\n\n6. ### csprof2000\n\n287\nGood point.\n\nI guess then that my argument only works for functions which don't have an inverse where Domain(inv f) = Range(f). This, naturally, precludes functions such as abs(x) and cos(x)...\n\nSo I guess more though will have to be put into functions which are not bijections.\n\n7. ### mnb96\n\n638\nThanks a lot. You all made very good observations that helped me a lot.\nBTW, it seems that if the function f admits an inverse there are not many choices, while if the function f is not invertible, many solutions exist but the problem is non-trivial, and it is difficult to say what kind of functions f and g would have to be, in order to satisfy f(x)=f(g(x)).\n\nAt the moment I am trying to solve the following (similar) problem:\n\n$$f = (f \\circ g) g'$$\nwhere g is invertible, and g' denotes its derivative\n\nIf you find it interesting, suggestions are always welcome.\nThanks!\n\n8. ### csprof2000\n\n287\nWell, similar suggestions - cases - are possible.\n\nAssume f(x) = c, c constant. Then c = cg', g' = 1, and g = x + k, k constant.\n\nAssume f(x) = x. Then x = gg', gdg = xdx, and (g^2)/2 = (x^2)/2 + k, k constant.\n\nAssume f(x) = x^n. Then x^n = g^n g', g^n dg = x^n dx, [1/(n+1)]g^(n+1) = [1/(n+1)]x^(n+1) + k, k constant\n\nWow, that's an alright result. So for x to any power at all, it's possible. Is is true for any polynomial? Yes, it seems like it should be. So... for any polynomial, I believe you can use the above formula to reduce it.\n\nf(x) = cos x, cos x = (cos g)g', (cos x)dx = (cos g)dg, sin(x) = sin(g) + k.\n\nIt seems like, unless I'm mistaken, this is the same thing every time:\n\nf(x) = f(g(x))g'(x) is the same as solving the differential equation f(x)dx = f(g(x))dg.\n\nSo, as long as f is integrable, the problem is actually quite easy. Maybe I'm wrong. Thoughts?\n\n9. ### mnb96\n\n638\ncsproof2000: I think you just made it!\nYou gave the solution to the problem, and for some reason I didn't immediately spot that what I was trying to do is actually solving a differential equation.", "date": "2015-03-02 03:39:13", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/composition-of-functions.295719/", "openwebmath_score": 0.8530347943305969, "openwebmath_perplexity": 937.8825996008811, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308807013051, "lm_q2_score": 0.86153820232079, "lm_q1q2_score": 0.8513125026170614}}
{"url": "http://math.stackexchange.com/questions/171274/iint-ayx-da-evaluating", "text": "# $-\\iint_{A}(y+x)\\,dA$ Evaluating\n\nI am bit unsure about the following problem:\n\nEvaluate the double integral:\n\n$$-\\iint_{A}(y+x)\\,dA$$\n\nover the triangle with vertices $(0,0), (1,1), (2,0)$\n\nOK, so I figured here that I would do this by first evaluating the integral over the region bounded by the vertices $(0,0), (1,1), (1,0)$ and then evaluate the integral over the region bounded by the vertices $(1,0), (1,1), (2,0)$ before adding the two answers together, and then reversing the sign of this answer (since there is a minus sign in front of the original double integral). Thus, I begin by finding:\n\n$$\\int_{0}^{1}dx \\int_{0}^{x}(y+x)\\,dy$$\n\nWhen solved this gives me the answer $\\frac{1}{2}$.\n\nNext I solve:\n\n$$\\int_{1}^{2}dx \\int_{1}^{2-x}(y+x)\\,dy$$\n\nWhen solved this gives me the answer $-\\frac{7}{6}$.\n\nI have verified both the integrals in Wolframalpha, and they give me the same answer. I would therefore believe that the final answer should be:\n\n$$-(\\frac{1}{2} - \\frac{7}{6}) = \\frac{2}{3}$$\n\nHowever, the final answer should, according to the book, be $-\\frac{4}{3}$.\n\nThus, obviously I do something wrong here. If anyone can help me out, I would greatly appreciate it. Is it perhaps that it is not allowed to \"split up\" this into two separate integrals? I couldn't find a way to solve this without doing this.\n\n-\nI see one problem at least: The inner ($y$) integral in the second part should start at $0$. Anyway, you can avoid the splitting by doing the integrals with the $x$ integral on the inside. Try it! \u2013\u00a0Harald Hanche-Olsen Jul 15 '12 at 20:32\nThank you for your answer. But why should the second integral start at 0? The y-values being at 1 and end up at 0 (hence I chose y = 2-x for the upper value of the integral). \u2013\u00a0Kristian Jul 15 '12 at 20:39\nDraw a picture. The whole triangle has its base at the $x$-axis, and so does each of the two pieces resulting from the split. \u2013\u00a0Harald Hanche-Olsen Jul 15 '12 at 20:41\nYou can also solve this with the Gauss integral theorem on $F(x,y) = xy$ and evaluate the integral on the boundary instead. Maybe this is easier, since the boundary is made of line segments. \u2013\u00a0Cocopuffs Jul 15 '12 at 20:42\nIt seems you may be confusing the two bounds. The bound that begins at $1$ and ends up at $0$ is the upper bound; the lower one is $0$ throughout. Also, you can tell from the sign that $-\\frac76$ must be wrong, since the integrand is non-negative throughout the triangle. \u2013\u00a0joriki Jul 15 '12 at 20:45\n\nYour second integral should be $$\\int_{1}^{2}dx \\int_{0}^{2-x}(y+x)dy.$$ Your lower $y$ limit was 1 instead of 0.", "date": "2016-05-26 20:38:22", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/171274/iint-ayx-da-evaluating", "openwebmath_score": 0.9406522512435913, "openwebmath_perplexity": 172.53623020766122, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924785827002, "lm_q2_score": 0.8757870029950159, "lm_q1q2_score": 0.8512583797516402}}
{"url": "https://math.stackexchange.com/questions/4057952/drawing-the-grid-without-lifting-the-pen/4058111", "text": "# Drawing the grid without lifting the pen\n\nThe below shape consist of $$24$$ segments with unit length. if we want to draw this shape without lifting the pen, what is the minimum length of the line we should draw? $$1)25\\quad\\quad\\quad\\quad\\quad\\quad2)26\\quad\\quad\\quad\\quad\\quad\\quad3)27\\quad\\quad\\quad\\quad\\quad\\quad4)28\\quad\\quad\\quad\\quad\\quad\\quad5)29\\quad\\quad\\quad\\quad\\quad\\quad$$\n\nI think we can do it somehow by segments with the length $$28$$ (by passing twice the four middle segments on each side of outer big square).\n\nI'm not sure how to solve this problem. is it possible to solve it mathematically or I should just try different ways to draw this?\n\nNotice that if you draw it without lifting the pen, except for the starting and ending point you must draw the same number of lines going into each point as going out. This means that you must draw an even number of segments meeting every point except possibly two. How many extra segments does this imply you need at a minimum? Can you achieve this?\n\n\u2022 Thank you for the answer. but unfortunately, I can't figure out how to continue from your hint. can you please elaborate more on that? Mar 11, 2021 at 17:04\n\nYou will be able to draw all the edges of the graph without lifting the pen and drawing every edge exactly once (starting and ending at the same node), iff there exists an Euler circuit in the graph.\n\nWe shall use the following theorem: a connected graph has an Euler circuit iff all the nodes of the graph has even degree.\n\nThe given grid graph does not have all nodes with even degree. Let's first find the nodes with odd degrees, as shown in the next figure. Notice that the nodes B,C, E,I, H,L and N,O have odd degrees (namely 3).\n\nLet's add 4 additional (red) edges to the grid graph as shown in the next figure to make all the nodes have even degrees.\n\nNow, in the augmented graph above, by the theorem, it must have an Euler circuit, so we shall be able to draw all the edges of the graph exactly once without lifting the pen.\n\nWe shall use Flury's algorithm to find an Euler circuit in the augmented graph first. The key idea is that when you have a choice between a bridge and a non-bridge, always choose the non-bridge (don't burn the bridges).\n\nSince the augmented graph has all nodes with even degrees, we can start at any node, let's start at B, a node with degree 4 in the augmented graph.\n\nUsing the algorithm, the following animation shows how to construct an Euler circuit in the augmented graph.\n\nNow, let's find where the additional edges (ones not present in the original graph) were used in the circuit found in the augmented graph.\n\nLet's go back to our original graph by replacing those edges by the ones in the original graph, we had exactly 4 such additional edges needed to make all nodes in the original graph to have even degrees.\n\nIt implies that in order to have the Euler circuit in the original graph we need 4 additional edges (requiring 24+4=28 edges). It also means that in the original graph we need to traverse the corresponding 4 edges twice, in order to draw all the edges without lifting the pen, as shown below:\n\n[EDIT]\n\nIf we don't need to start and end at the same node, it's sufficient to have an Euler trail instead of a circuit, needing one less edge: $$28-1=27$$ edges (only 3 edge segments, namely, EI, ON and HL are needed to be traversed twice, leading to minimum length of the line as $$24+3=27$$).\n\nIn this case the trail shown above starts at $$B$$ but ends at $$C$$ (instead of $$B$$ as before), since the edge $$BC$$ has already been visited once, we don't need to visit it again (as shown in the next figure).\n\n\u2022 Wow! thank you very much for the answer and beautiful animation you used. I really appreciate that. Mar 11, 2021 at 19:15\n\u2022 Great answer, except that you don't need an Euler circuit, but merely an Euler path. Which means we can save one line segment. Mar 12, 2021 at 1:04\n\nThe figure has 4 nodes with 2 lines connected to node. 8 nodes with 3 lines connected, and 4 nodes with 4 lines connected.\n\nIf you want to complete the circuit without lifting your pen, you can start at a node with an odd number of connections, and you can end at a node with an odd number of connections, for all the rest you must have an even number of connections. That is, if you get to the node on one path you leave the node on a different path.\n\nThis means that you must double up on some of the lines where it appears that there is a odd number of lines to make it an even number. What is the minimum number of lines you need to double up such that there are at most two nodes with an odd number of connections?\n\n\u2022 Thank you very much. So for the start we connect two of the points with odd degrees and we left with three more segments passing through other odd degree points on other sides of outer big square. so we must pass them twice and the answer is $24+3=27$. am I right? Mar 11, 2021 at 17:42\n\u2022 It must be at least 27. It is probably worth the effort to show that it can be done with 27 segments. Mar 11, 2021 at 17:46\n\nEvery node must have the same number of edges entering and leaving, except for the end points. There are 8 nodes on the graph with an odd degree, but we can make two of them the endpoints, so we have 6 nodes that need an \"extra\" edge. Finally, we can note that the odd-degree nodes come in adjacent pairs, so the \"extra\" edge leaving one node can be the \"extra\" edge entering another, so we only need half as many \"extra\" edges as we have non-endpoint, odd-degree nodes.", "date": "2022-12-03 21:22:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4057952/drawing-the-grid-without-lifting-the-pen/4058111", "openwebmath_score": 0.6174324750900269, "openwebmath_perplexity": 156.2124636069343, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924810166349, "lm_q2_score": 0.8757869900269366, "lm_q1q2_score": 0.8512583692783731}}
{"url": "http://math.stackexchange.com/questions/60261/polynomial-px-0-for-all-x-implies-coefficients-of-polynomial-are-zero", "text": "# Polynomial $p(x) = 0$ for all $x$ implies coefficients of polynomial are zero\n\nI am curious why the following is true. The text I am reading is \"An Introduction to Numerical Analysis\" by Atkinson, 2nd edition, page 133, line 4.\n\n$p(x)$ is a polynomial of the form:\n\n$$p(x) = b_0 + b_1 x + \\cdots + b_n x^n$$\n\nIf $p(x) = 0$ for all $x$, then $b_i = 0$ for $i=0,1,\\ldots,n$.\n\nWhy is this true? For example, for $n=2$, I can first prove $b_0=0$, then set $x=2$ to get a linear system of two equations. Then I can prove $b_1=b_2 = 0$. Similarly, for $n=3$, I first prove $b_0=0$, then I calculate the rank of the resulting linear system of equations. That shows that $b_1=b_2=b_3=0$. But if $n$ is very large, I cannot keep manually solving systems of equations. Is there some other argument to show all the coefficients must be zero when the polynomial is always zero for all $x$?\n\nThanks.\n\n-\nAre you familiar with the fact that if $p(a) = 0$, then $p(x)$ is divisible by $x - a$? Use this $n+1$ times. \u2013\u00a0 Qiaochu Yuan Aug 28 '11 at 3:36\nAs explained in the answers, we can solve the problem by showing that a non-zero polynomial of degree $d$ has no more than $d$ roots. However, your approach can be made to work. To carry it out, one uses properties of the Vandermonde matrix. The details are somewhat more messy than the approach through counting roots, but accessible. \u2013\u00a0 Andr\u00e9 Nicolas Aug 28 '11 at 3:55\nHi Qiaochu Yuan, can you give me some more hints? I am aware of the Fundamental Theorem of Algebra, which can be found here: tutorial.math.lamar.edu/Classes/Alg/ZeroesOfPolynomials.aspx. If I divide a polynomial of degree $n$ by (x-a), and a is guaranteed to be a root (since the polynomial is a horizontal line at $y=0$), then I get as a result, another polynomial with degree $n-1$. After $n-1$ divisions, I get as a result, something like: $d + ex = 0$. $d$ and $e$ might not be the original polynomial coefficients. What would be the next step after this? \u2013\u00a0 jrand Aug 28 '11 at 4:35\nFor purposes of argumentation: false over finite fields... \u2013\u00a0 GEdgar Aug 28 '11 at 13:31\nAs you are interested in numerical methods the following is probably not of interest to you, but for the sake of completeness I feel compelled to point out that the conclusion is false, if your domain is finite. A non-zero polynomial can easily vanish at all points of, e.g. $\\mathbf{Z}_m$ without being the zero polynomial. The easiest example is the polynomial $p(x)=x^2-x\\in \\mathbf{Z}_2[x]$ that vanishes at all the elements of $\\mathbf{Z}_2$. If you don't study algebra, you can safely ignore this comment for now. \u2013\u00a0 Jyrki Lahtonen Aug 28 '11 at 13:37\n\nHINT $\\$ A nonzero polynomial over a field (or domain) has no more roots than its degree, as is easily proved by induction and the Factor Theorem. In fact if every natural was a root then the polynomial would be divisible by $\\rm\\:(x-1)\\:(x-2)\\:\\cdots\\:(x-n)\\:$ for all $\\rm\\:n\\in \\mathbb N\\:,\\:$ which yields a contradiction for $\\rm\\:n\\:$ greater than the degree of the polynomial.\n\nNote that the proof of said statement depends crucially on the hypothesis that coefficient ring is an integral domain, i.e. a ring satisfying $\\rm\\:ab = 0\\iff a=0\\ \\ or\\ \\ b=0\\:.\\:$ Over non-domains such as the integers modulo $\\rm\\:m\\:$ not prime, polynomials can have more roots than their degree. In fact if this is true then one can use such roots to factor $\\rm\\:m\\:,\\:$ see here.\n\n-\nFollowing reading Bill Dubuque's, Shawn's, and Qiaochu Yuan's answer, I believe I must agree to the following fact. If a polynomial is equal to zero, then it must be of degree 0, and the coefficient must be set such that $b_0=0$. Since it is degree 0, the other coefficients must also be set to 0. \u2013\u00a0 jrand Aug 28 '11 at 12:42\n@jrand: The degree of the polynomial zero is not zero. \u2013\u00a0 Did Aug 28 '11 at 13:02\n\nNote that $p(x)=0$ implies derivatives of all orders are also $0$. Let $x=0$ into $p(x) = b_0 + b_1 x + \\cdots + b_n x^n$ to obtain $b_0 = 0$.\n\nNow differentiate both sides: $p'(x) = b_1 + 2b_2 x + 3b_3 x^2 + \\cdots + nb_n x^{n-1}$. Let $x=0$ again, and we get that $b_1=0$.\n\nIf we continue to differentiate and substitution $x=0$ we will get that $b_k=0$ for $k=0,1,2,\\cdots, n$. This idea can easily be made into a rigorous induction proof.\n\nA nice corollary of this result is that two polynomials are equal if and only if they have the same degree and coefficients.\n\n-\n\nIf you are willing to accept $\\displaystyle \\lim_{x \\rightarrow \\infty}\\frac{1}{x^{m}}=0$ and $\\displaystyle \\lim_{x \\rightarrow \\infty}x^m=\\infty$ for all $m>0$, then you can argue as follows. Suppose you have a polynomial $f(x)=b_nx^n+ \\ldots b_0$ with $b_n \\neq 0$ and $n\\geq 1$, then $$\\lim_{x\\rightarrow \\infty}f(x)=\\lim_{x \\rightarrow \\infty} x^n(b_n+ \\ldots \\frac{a_0}{x^n})=\\lim_{x \\rightarrow \\infty} x^nb_n=\\infty.$$\n\nSince the constant function $0$ does not have this property, it cannot be equal to a polynomial with degree greater or equal to $1$.\n\n-\n+1. Nice analytical solution with no derivatives. \u2013\u00a0 Did Aug 28 '11 at 22:22\n\nA polynomial can be uniquely fitted by knowing its value at $d+1$ points, where $d$ is the degree of the polynomial. If it's 0 at all points, that's clearly more than $d+1$ points.\n\nBut we also know that a polynomial can be written as $(x-r_0)(x-r_1)...(x-r_k)$, where the $r_i$ are the roots of polynomial (possibly complex). Again, if there are an infinite number of roots...\n\n-\n\nLet $n\\in \\mathbb{N}$, and $$p(x)=b_0+b_1x+\\ldots +b_nx^n$$ a polynomial of degree at most $n$. If $p(x)=0$ for every $x$, then $b_i=0$ for $i=0,\\, 1,\\ldots\\, n$.\n\nProof:\n\nBy induction on $n$:\n\nIf $\\deg(p)\\leq 1$, then $p(x)=b_0+b_1x$. Since $p(0)=p(1)=0$ you get $b_0=b_1=0$.\n\nSuppose that for any $$r(x)=c_0+c_1x+\\ldots +c_nx^n$$ of degree at most $n$, if $r(x)=0$ for every $x$, then $c_i=0$ for $i=0,\\, 1,\\ldots\\, n$. Let $$p(x)=b_0+b_1x+\\ldots +b_{n+1}x^{n+1}.$$ Suppose that $p\\equiv 0$. Since $p(0)=0$, you get that $b_0=0$. From here, I have two possible arguments. The first, that was my original one, is as follows:\n\nWe have $$p(x)=b_1x+b_2x^2+\\ldots+b_{n+1}x^{n+1}=x(b_1+b_2x+\\ldots+b_{n+1}x^{n}).$$ Then for all $x$, \\begin{align*} 0&= p(x)\\\\ 0&= x(b_1+b_2x+\\ldots+b_{n+1}x^{n}).\\end{align*} Then for any $x\\neq 0$, $$q(x):=b_1+b_2x+\\ldots+b_{n+1}x^{n}=0$$ If $q\\not\\equiv 0$, then we have a polynomial of degree at most $n$ with infinitely many roots. This can't be, therefore $q\\equiv 0$. Now, $\\deg(q)\\leq n$ and $q\\equiv 0$, therefore by the induction hypothesis, $b_i=0$ for $i=1,\\,\\ldots ,\\, n+1$.\n\nThe second argument was inspired by @Ragib Zaman's answer. Just differentiate both sides of $p(x)=0$, then $$b_1+2b_2x+\\ldots+(n+1)b_{n+1}x^n=0$$ for all $x$. By the induction hypothesis, $kb_k=0$ for $k=1,\\,\\ldots,\\,n+1$, and this implies that $b_k=0$ for $k=1,\\,\\ldots,\\,n+1$\n\nSince $b_0=0$, this proves that $b_i=0$ for $i=0,\\, 1,\\ldots,\\, n+1$.\n\n-\nIn general, by the fundamental theorem of algebra we can say that the number of different roots of a polynomial of degree $n$ is at most $n$. I the argument above $q$ must be identically $0$ because if not we have a polynomial with infinite different roots. \u2013\u00a0 leo Aug 28 '11 at 6:46\nMy comment is the hint of @Bill \u2013\u00a0 leo Aug 28 '11 at 7:12\nThis induction is flawed because during your nth inductive step you use as a fact a property (for every q of degree n, if q(x)=0 for every nonzero x, then q=0) that the hypothesis at the beginning of the step (for every polynomial p of degree n, if p(x)=0 for every x, then p=0) does not imply. \u2013\u00a0 Did Aug 28 '11 at 9:03\n@Didier Piau: I'm trying to prove a statement about the coefficients of polynomials, not about the polynomials. If I understand correct your comment, you say me that the hypothesis of induction does not implies that the polynomial $q$ is identically $0$. Yes that is true, but i don't pretend it. The fact that $q\\equiv 0$ is something that I want to apply my hypothesis of induction. \u2013\u00a0 leo Aug 28 '11 at 10:06\nI think that the answer is now more clear. \u2013\u00a0 leo Aug 28 '11 at 10:14", "date": "2014-07-26 14:00:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/60261/polynomial-px-0-for-all-x-implies-coefficients-of-polynomial-are-zero", "openwebmath_score": 0.9442412853240967, "openwebmath_perplexity": 143.34294223262626, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924802053234, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8512583669922273}}
{"url": "https://math.stackexchange.com/questions/1970304/2-different-methods-of-calculating-possibilities-of-people-sitting-at-round-tabl", "text": "# 2 different methods of calculating possibilities of people sitting at round table\n\nI have 2 different problems that seem identical:\n\n8 men and 8 women are going to sit at a round table where there are 16 seats. They take their seats randomly. How many ways can the 16 seats be taken so that no 2 women are sitting next to each other?\n\nI solved this by doing $$2 * 8!8!$$\n\nThe method I used was that $8*8*7*7*6*6*5*5*4*4*3*3*2*2*1*1= 8!8!$ is all the ways you can sit these 8 women alternating them with the 8 men.\n\nI then multiplied by two because you have 2 possibilities: the women can be sitting at the odd numbered seats, or at the even numbered ones.\n\nThen I have this other problem:\n\nIn how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent?\n\n(I took the problem from here: http://www.imsc.res.in/~kamalakshya/cupboard/comb_mag.pdf)\n\nAnd the solution, according to the website, is:\n\nAnswer: The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4-1)! =3! Ways. Now 4 vacant seats can be occupied by 4 women in 4! Ways. Hence the required number of seating arrangements = 3!4! = 144\n\nHowever, I can't seem to use the method from the 1st problem on this one. All the possible combinations of men and women, sitting alternated with men, would be $4!4!$ Then, multiplying by 2 the solution would be $2 * 4!4! = 1152$\n\nIf I use the method from the answer of the 2nd problem with the 1st one, I get $7!8!$, which is different from $2 * 8!8!$\n\nMy question is, how can this be? Is there any difference between these 2 problems that I am not seeing?\n\nEDIT: I believe the 1st solution is correct. Check here Probability of men and women sitting at a table alternately\n\n\u2022 You are assuming that the seats are labeled. The answer to the second problem assumes the seats are not labeled and takes rotational invariance into account. \u2013\u00a0N. F. Taussig Oct 16 '16 at 1:29\n\u2022 Take into account circular permutations. The 1st solution is wring 2nd is correct. \u2013\u00a0Navin Oct 16 '16 at 1:30\n\u2022 @navinstudent math.stackexchange.com/questions/1969572/\u2026 \u2013\u00a0SilenceOnTheWire Oct 16 '16 at 1:32\n\u2022 @N.F.Taussig That makes sense, but how do you infer which method to use from reading the problems, without looking at their solutions? Neither problem mentions anything about the seats being labeled. \u2013\u00a0SilenceOnTheWire Oct 16 '16 at 1:52\n\u2022 If the seats are not described as labeled, I would assume that they are unlabeled and that rotational invariance applies, as in the second problem. \u2013\u00a0N. F. Taussig Oct 16 '16 at 2:57", "date": "2019-07-21 19:44:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1970304/2-different-methods-of-calculating-possibilities-of-people-sitting-at-round-tabl", "openwebmath_score": 0.5980497002601624, "openwebmath_perplexity": 252.75563525916294, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9719924761487654, "lm_q2_score": 0.8757869884059266, "lm_q1q2_score": 0.8512583634395466}}
{"url": "http://math.stackexchange.com/questions/318362/conditional-probability-with-balls-in-an-urn", "text": "# Conditional Probability with balls in an urn\n\nTwo balls, each equally likely to be colored either red or blue, are put in an urn. At each stage one of the balls is randomly chosen, its color is noted, and it is then returned to the urn. If the first two balls chosen are colored red, what is the probability that (a) both balls in the urn are colored red; (b) the next ball chosen will be red?\n\nI'm wondering if my method for part (a) is correct:\n\nLet $P(R)$ be the probability of picking a red ball: $P(R)=\\frac{1}{2}$\n\nLet $P(B)$ be the probability of picking a blue ball: $P(B)=\\frac{1}{2}$\n\nLet $P(C)$ be the probability of the condition, i.e., picking two red balls consecutively: $P(C)=P(C|H_1)P(H_1)+P(C|H_2)P(H_2)+P(C|H_3)P(H_3)$\n\nwhere $P(H_1)$ is the probability both balls in the urn are red: $P(H_1)=(P(R))^2=(\\frac{1}{2})^2=\\frac{1}4$\n\n$P(H_2)$ is the probability that both balls in the urn are blue: $P(H_2)=(P(B))^2=(\\frac{1}{2})^2=\\frac{1}4$\n\n$P(H_3)$ is the probability that one ball is red and one is blue inside the urn: $P(H_3)=1-(P(H_1)+P(H_2))=1-\\frac{1}2=\\frac{1}2$ since the sum of the mutually exclusive hypotheses or events must sum to 1.\n\nThus $P(C)=1\\times\\frac{1}4+\\frac{1}4\\times0+\\frac{1}2\\times\\frac{1}4=\\frac{3}8$\n\nand $P(H_1|C)= \\frac{P(C \\bigcap H_1)}{P(C)}=\\frac{P(C|H_1)P(H_1)}{P(C)}=\\frac{1\\times\\frac{1}4}{\\frac{3}8}=\\frac{2}3$\n\nFor part (b), I know that $P(R|C)$, the probability of picking a red ball given that the first two balls picked were red is to be found\n\n$P(R|C)= \\frac{P(C \\bigcap R)}{P(C)}=\\frac{P(C|R)P(R)}{P(C)}$\n\nHow can I find $P(C|R)$?\n\n-\nThere are three possible urns that you might have: both balls red (RR), both balls blue (BB), one of each color (RB) which occur with probabilities $1/4, 1/4, 1/2$ respectively. Now calculate $P(\\text{red,red}\\mid \\text{RR})$, $P(\\text{red,red}\\mid \\text{BB})$, $P(\\text{red,red}\\mid \\text{RB})$, and combine them via the law of total probability to get $P(\\text{red,red})$. Then, use Bayes' formula to obtain $P(\\text{BB}\\mid \\text{red,red})$ \u2013\u00a0Dilip Sarwate Mar 2 '13 at 3:04\n\nThe probability that the contents of the urn are two red is indeed $\\frac{1}{4}$, as is the probability of two blue, and the probability of mixed is therefore $\\frac{1}{2}$. The derivation could have been done more quickly.\n\nQuestion (a) asks for the probability both are red given that the two drawn balls are red. Let $R$ be the event both are red, and $D$ be the event both drawn balls are red. We want $\\Pr(R|D)$. By the usual formula this is $\\frac{\\Pr(R\\cap D)}{\\Pr(D)}$.\n\nTo find $\\Pr(D)$, note that if both balls are red (probability $\\frac{1}{2}$), then the probability of $D$ is $1$, while if one ball is red and the other is not (probability $\\frac{1}{2}$) then the probability of $D$ is $\\frac{1}{4}$. Thus the probability of $D$ is $\\left(\\frac{1}{4}\\right)(1)+\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{4}\\right)$. This is $\\frac{3}{8}$.\n\nThe probability of $R\\cap D$ is $\\frac{1}{4}$. So the ratio is indeed $\\frac{2}{3}$.\n\nFor (b), you can use the calculation of (a). With probability $\\frac{2}{3}$ we are drawing from a double red, and we will get red with probability $1$. With probability $\\frac{1}{3}$ the urn is a mixed one, and the probability of drawing a red is $\\frac{1}{2}$, for a total of $\\frac{2}{3}\\cdot 1+\\frac{1}{3}\\cdot\\frac{1}{2}$.\n\nOne can also solve (b) without using the result of (a). With I hope self-explanatoru notation, we want $\\Pr(RRR|RR)$. The probabilities needed in the conditional probability formula are easily computed. We have $\\Pr(RRR\\cap RR)=\\Pr(RRR)=\\frac{1}{4}\\cdot 1+\\frac{1}{2}\\cdot\\frac{1}{8}=\\frac{5}{16}$. Divide by $\\frac{3}{8}$.\n\n-", "date": "2016-07-28 18:42:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/318362/conditional-probability-with-balls-in-an-urn", "openwebmath_score": 0.9881309270858765, "openwebmath_perplexity": 82.52267954202387, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363729567545, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8512491975773109}}
{"url": "https://math.stackexchange.com/questions/497584/finding-a-curve-that-intersects-any-line-on-the-plane/497588", "text": "# Finding a curve that intersects any line on the plane\n\nQuestion Is there a curve on plane such that any line on the plane meets it (a non zero ) finite times ?\n\nWhat are the bounds on the number of such intersections.\n\nMy question was itself inspired by this \"Can you draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?\"\n\n\u2022 Clearly if your curve is the graph of an odd polynomial, the number of intersections is at most the degree of the polynomial. \u2013\u00a0Cameron Williams Sep 18 '13 at 15:08\n\u2022 One possibility is a pair of parabolas, say one concave up and one concave down, situated so that they touch tangentially (or intersect) and therefore cannot be separated by a hyperplane. \u2013\u00a0hardmath Sep 18 '13 at 15:08\n\u2022 Please try to make titles of questions more informative. E.g., Why does $a\\le b$ imply $a+c\\le b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. \u2013\u00a0Lord_Farin Sep 18 '13 at 15:08\n\u2022 @CameronWilliams Correct, but just being an odd degree polynomial; does not guarantee that every line in the plane will intersect it \u2013\u00a0ARi Sep 18 '13 at 15:10\n\u2022 @ARi: Actually it does imply that, except degree 1. \u2013\u00a0hardmath Sep 18 '13 at 15:13\n\nCubic parabola $$y=x^3$$ has this property. The max number of such intersections is given by the Fundamental theorem of algebra:\n\n$$x^3=ax+b$$ can have at most 3 solutions.\n\n\u2022 Yes but will all of the solutions be real for all real (a,b) pairs \u2013\u00a0ARi Sep 18 '13 at 15:14\n\u2022 For any $a$, there will be only one real solution for all sufficiently large $|b|$. \u2013\u00a0hardmath Sep 18 '13 at 15:17\n\u2022 I think the important fact here (i.e., relevant to the question), is that $\\,x^3-ax-b=0\\;$ has always at least one real solution and, of course, at most three different ones. Nice answer. +1 \u2013\u00a0DonAntonio Sep 18 '13 at 15:22\n\u2022 So the upper bound for intersections will be three and the lower one 0. \u2013\u00a0ARi Sep 18 '13 at 15:29\n\u2022 Well, a tighter and always attainable lower bound is one, in fact...and this fulfills the question's conditions. \u2013\u00a0DonAntonio Sep 18 '13 at 15:31\n\nAs already shown, any cubic polynomial (and indeed, any odd-degree polynomial) has the requisite property by the fundamental theorem of algebra.\n\nWhat's more, a simple perturbation argument should be enough to show that any (sufficiently) smooth curve that meets every line in at least one point will meet some lines in at least three points. Consider a point tangent to the curve where the second derivative 'with respect to the tangent line' is non-zero; that is, a non-reflex tangent point, or locally extremal point. (Such points must exist if the curve is non-trivial). Now, consider pencils of lines 'near' this intersection point; displaced infinitesimally one way from the tangent, they must have another point of intersection with the curve, and this point can be made 'generic' so that it doesn't vanish under small perturbations. Then displace infinitesimally the other direction; the 'generic' point of intersection is still a point of intersection, but the tangent turns into two points of intersection.", "date": "2019-11-15 09:54:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/497584/finding-a-curve-that-intersects-any-line-on-the-plane/497588", "openwebmath_score": 0.7696214318275452, "openwebmath_perplexity": 368.8598580819707, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98593637543616, "lm_q2_score": 0.8633916099737806, "lm_q1q2_score": 0.85124919451954}}
{"url": "http://mathhelpforum.com/algebra/136341-arithmetic-series-help.html", "text": "1. ## Arithmetic Series Help\n\nFirst of all I'm not sure if this is the right section, so I apologise if it's not.\n\nI was hoping that someone could help me out with the following question:\n\nFrom a piece of wire 5000mm long, n pieces are cut, each piece being 10mm longer than the previous piece. Ui is the length of the ith piece and Si is the cumulative length of i pieces.\n\nExpress Un and Sn in terms of U1 and n.\n\n2. $U_n = u_1 + 10(n-1)$\n\n$S_n = \\sum_{i=1}^n U_i = \\sum_{i=1}^n [u_1 + 10(i-1)]$ $= u_1 + 10\\sum_{i=1}^n i -\\sum_{i=1}^n 1 = u_1 + 10\\frac{n(n+1)}{2} - n = u_1 +n(5n + 4)$\n\nAlso, $S_n = 5000 \\Rightarrow u_1 = 5000 - 5n^2 - 4n.$\n\n3. Your answer for Un looks right so thank you for that.\n\nHowever the others don't seem to add up, I had to find U1 as the second part of the question and I've calculated it as being 80 which I've checked and it is right. If you put 80 into your formulae it doesn't equal 5000.\n\nThank you anyway though bud.\n\n4. Hello, JP103!\n\nAnonymous1 made some small errors. . .\n\n$S_n \\;=\\; \\sum_{i=1}^n U_i$\n\n. . $=\\; \\sum_{i=1}^n \\bigg[U_1 + 10(i-1)\\bigg]$\n\n. . $=\\; \\sum_{i=n}^n\\bigg[U_1 + 10i - {\\color{red}10}\\bigg]$\n\n. . $= \\;\\sum_{n=1}^nU_1 + 10\\sum_{n=1}^ni - 10\\sum_{n-1}^n1$\n\n. . $=\\;{\\color{red}n}U_1 + 10\\frac{n(n+1)}{2} - 10n$\n\n$\\boxed{S_n \\;=\\;nU_1 + 5n(n-1)}$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nUsing the sum-formula: . $S_n \\:=\\:\\frac{n}{2}\\bigg[2a + (n-1)d\\bigg]$\n\n. . we have: . $S_n \\;=\\;\\frac{n}{2}\\bigg[U_1 + (n-1)10\\bigg] \\quad\\Rightarrow\\quad S_n \\;=\\;nU_1 + 5n(n-1)$", "date": "2017-01-17 21:33:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/136341-arithmetic-series-help.html", "openwebmath_score": 0.8908501863479614, "openwebmath_perplexity": 387.9722601450327, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363704773486, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8512491937037898}}
{"url": "http://math.stackexchange.com/questions/165434/how-to-prove-if-a-function-is-bijective", "text": "How to prove if a function is bijective?\n\nI am having problems being able to formally demonstrate when a function is bijective (and therefore, surjective and injective). Here's an example:\n\nHow do I prove that $g(x)$ is bijective?\n\n\\begin{align} f &: \\mathbb R \\to\\mathbb R \\\\ g &: \\mathbb R \\to\\mathbb R \\\\ g(x) &= 2f(x) + 3 \\end{align}\n\nHowever, I fear I don't really know how to do such. I realize that the above example implies a composition (which makes things slighty harder?). In any case, I don't understand how to prove such (be it a composition or not).\n\nFor injective, I believe I need to prove that different elements of the codomain have different preimages in the domain. Alright, but, well, how?\n\nAs for surjective, I think I have to prove that all the elements of the codomain have one, and only one preimage in the domain, right? I don't know how to prove that either!\n\nEDIT\n\nf is a bijection. Sorry I forgot to say that.\n\n-\nNormally one distinguishes between the two different arrows $\\mapsto$ and $\\to$. One writes $f:\\mathbb{R}\\to\\mathbb{R}$ to mean $f$ is a function from $\\mathbb{R}$ into $\\mathbb{R}$. The notation $x\\mapsto x^3$ means the function that maps every input value to its cube. \u2013\u00a0 Michael Hardy Jul 1 '12 at 20:39\nWouldn't you have to know something about $f$? Is $f$ a bijection? \u2013\u00a0 Dylan Moreland Jul 1 '12 at 20:40\nMy apologies! Yes, f is a bijection. \u2013\u00a0 Omega Jul 1 '12 at 20:40\nBoth of your deinitions are wrong. Maybe all you need in order to finish the problem is to straighten those out and go from there. I've posted the definitions as an answer below. \u2013\u00a0 Michael Hardy Jul 1 '12 at 20:45\n\nThe way to verify something like that is to check the definitions one by one and see if $g(x)$ satisfies the needed properties.\n\nRecall that $F\\colon A\\to B$ is a bijection if and only if $F$ is:\n\n1. injective: $F(x)=F(y)\\implies x=y$, and\n2. surjective: for all $b\\in B$ there is some $a\\in A$ such that $F(a)=b$.\n\nAssuming that $R$ stands for the real numbers, we check.\n\nIs $g$ injective?\n\nTake $x,y\\in R$ and assume that $g(x)=g(y)$. Therefore $2f(x)+3=2f(y)+3$. We can cancel out the $3$ and divide by $2$, then we get $f(x)=f(y)$. Since $f$ is a bijection, then it is injective, and we have that $x=y$.\n\nIs $g$ surjective?\n\nTake some $y\\in R$, we want to show that $y=g(x)$ that is, $y=2f(x)+3$. Subtract $3$ and divide by $2$, again we have $\\frac{y-3}2=f(x)$. As before, if $f$ was surjective then we are about done, simply denote $w=\\frac{y-3}2$, since $f$ is surjective there is some $x$ such that $f(x)=w$. Show now that $g(x)=y$ as wanted.\n\nAlternatively, you can use theorems. What sort of theorems? The composition of bijections is a bijection. If $f$ is a bijection, show that $h_1(x)=2x$ is a bijection, and show that $h_2(x)=x+2$ is also a bijection. Now we have that $g=h_2\\circ h_1\\circ f$ and is therefore a bijection.\n\nOf course this is again under the assumption that $f$ is a bijection.\n\n-\n\"Subtract 3 and divide by 2, again we have (y\u22123)/2=f(x). As before, if f was surjective then we are about done\" I'm not sure, but why would we be done there? \u2013\u00a0 Omega Jul 1 '12 at 21:18\n@Omega: If $f$ was surjective, then there is some $x$ such that $f(x)=\\frac{y-3}2$, show now that $g(x)=y$. \u2013\u00a0 Asaf Karagila Jul 1 '12 at 21:20\nYou know, it had me thinking: according to your method to find out if it is injective, no matter what function I test it with, I always manage to get the final equality (x = y). How would a function ever be not-injective? \u2013\u00a0 Omega Jul 1 '12 at 22:34\n@Omega: No, assume that $f(x)=0$ for all $x$, suppose that $x,y$ are any two real numbers (perhaps different and perhaps not), does $f(x)=f(y)$ tell you something about $x=y$ or $x\\neq y$? No, because taking $x=1$ and $y=2$ gives $f(1)=0=f(2)$, but $1\\neq 2$. Note that my answer relies critically on the fact that $f$ itself is injective. \u2013\u00a0 Asaf Karagila Jul 1 '12 at 22:37\n\nTo prove a function is bijective, you need to prove that it is injective and also surjective.\n\n\"Injective\" means no two elements in the domain of the function gets mapped to the same image.\n\n\"Surjective\" means that any element in the range of the function is hit by the function.\n\nLet us first prove that $g(x)$ is injective. If $g(x_1) = g(x_2)$, then we get that $2f(x_1) + 3 = 2f(x_2) +3 \\implies f(x_1) = f(x_2)$. Since $f(x)$ is bijective, it is also injective and hence we get that $x_1 = x_2$.\n\nNow let us prove that $g(x)$ is surjective. Consider $y \\in \\mathbb{R}$ and look at the number $\\dfrac{y-3}2$. Since $f(x)$ is surjective, there exists $\\hat{x}$ such that $f(\\hat{x}) = \\dfrac{y-3}2$. This means that $g(\\hat{x}) = 2f(\\hat{x}) +3 = y$. Hence, given any $y \\in \\mathbb{R}$, there exists $\\hat{x} \\in \\mathbb{R}$ such that $g(\\hat{x}) = y$. Hence, $g$ is also surjective.\n\nHence, $g(x)$ is bijective.\n\nIn general, if $g(x) = h(f(x))$ and if $f(x) : A \\to B$ and $h(x): B \\to C$ are both bijective then $g(x): A \\to C$ is also bijective.\n\nIn your case, $f(x)$ was bijective from $\\mathbb{R} \\to \\mathbb{R}$ and $h(x) = 2x+3$ is also bijective from $\\mathbb{R} \\to \\mathbb{R}$.\n\n-\n\nYou haven't said enough about the function $f$ to say whether $g$ is bijective.\n\n\"Injective\" means different elements of the domain always map to different elements of the codomain.\n\n\"Surjective\" means every element of the codomain has at least one preimage in the domain.\n\nLater edit: What you've now added---that $f$ is a bijection---bring us to the point where we can answer the question. However, maybe you should look at what I wrote above. Since both definitions that I gave contradict what you wrote, that might be enough to get you there.\n\n-\nSorry, yes, f is bijective. Thank you for clarifying that :) \u2013\u00a0 Omega Jul 1 '12 at 20:43\n\nFirst show that $g$ is injective ($1$-$1$) by showing that if $g(x)=g(y)$, then $x=y$. This isn\u2019t hard: if $g(x)=g(y)$, then $2f(x)+3=2f(y)+3$, so by elementary algebra $f(x)=f(y)$. By hypothesis $f$ is a bijection and therefore injective, so $x=y$.\n\nNow show that $g$ is surjective. To do this, you must show that for each $y\\in\\Bbb R$ there is some $x\\in\\Bbb R$ such that $g(x)=y$. That requires finding an $x\\in\\Bbb R$ such that $2f(x)+3=y$ or, equivalently, such that $f(x)=\\frac{y-3}2$. But $f$ is known to be a bijection and hence a surjection, so you know that there is such an $x\\in\\Bbb R$.\n\nIn general this is one of the two natural ways to show that a function is bijective: show directly that it\u2019s both injective and surjective. The other is to construct its inverse explicitly, thereby showing that it has an inverse and hence that it must be a bijection. You could take that approach to this problem as well:\n\n$$g^{-1}(y)=f^{-1}\\left(\\frac{y-3}2\\right)\\;,$$\n\nsince\n\n\\begin{align*} g\\left(f^{-1}\\left(\\frac{y-3}2\\right)\\right)&=2f\\left(f^{-1}\\left(\\frac{y-3}2\\right)\\right)+3\\\\ &=2\\left(\\frac{y-3}2\\right)+3\\\\ &=y\\;, \\end{align*}\n\nand since $f$ is a bijection, $f^{-1}\\left(\\frac{y-3}2\\right)$ exists for every $y\\in\\Bbb R$.\n\nAdded: As Marc reminds me, this is only half the job: if you take this approach, you must either show directly that $g$ is injective, as I did above, or verify that the function that I called $g^{-1}$ above is a two-sided inverse, i.e., that $g^{-1}\\big(g(x)\\big)=x$ for $x\\in\\Bbb R$. This is not particularly difficult in this case:\n\n\\begin{align*} g^{-1}\\big(g(x)\\big)&=g^{-1}\\big(2f(x)+3\\big)\\\\ &=f^{-1}\\left(\\frac{\\big(2f(x)+3\\big)-3}2\\right)\\\\ &=f^{-1}\\big(f(x)\\big)\\\\ &=x\\;, \\end{align*}\n\nsince $f$ is a bijection.\n\n-\nWhen using the \"inverse\" criterion, you should be careful in really checking that a purported inverse is an inverse, both ways. With $g^{-1}$ denoting your purported inverse, your final argument checked that $g(g^{-1}(y))=y$ for all $y\\in\\mathbb R$; this only shows that $g$ is surjective (it has a right inverse, also called a section). To show that $g$ is also injective you need to separately check that $g^{-1}(g(x))=x$ for all $x\\in\\mathbb R$. \u2013\u00a0 Marc van Leeuwen Jul 1 '12 at 21:22\n@Marc: Yes, I should probably say as much; I hadn\u2019t originally intended to mention this approach at all, and did so only as an afterthought. I was implicitly assuming that the obvious injectivity had already been checked, but that\u2019s not clear from what I wrote. \u2013\u00a0 Brian M. Scott Jul 1 '12 at 21:26", "date": "2014-04-17 07:30:29", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/165434/how-to-prove-if-a-function-is-bijective", "openwebmath_score": 0.979116678237915, "openwebmath_perplexity": 186.43965124472237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471628097782, "lm_q2_score": 0.8652240947405564, "lm_q1q2_score": 0.8512482708051552}}
{"url": "http://math.stackexchange.com/questions/154087/angle-of-a-triangle-inscribed-in-a-square", "text": "Angle of a triangle inscribed in a square\n\nSay we have a square $ABCD$. Put points $E$ and $F$ on sides $AB$ and $BC$ respectively, so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. What is $\\angle DNF$?\n\nI'm inclined to say that it's a right angle because that's what it looks like from what I've drawn, but I have no idea how to proceed.\n\n-\nI have given an argument why your conjecture is true. I am more interested in why you looked at this problem and how you came up with this neat little conjecture. \u2013\u00a0 user17762 Jun 5 '12 at 5:29\nIf you want a purely Euclidean-geometry proof, you might start by noting the similarity of triangles $EBC$, $ENB$, and $BNC$, and then trying to prove that triangles $BNF$ and $CND$ are similar. I'm not sure exactly how the proof will go, but it feels doable. \u2013\u00a0 Rahul Jun 5 '12 at 5:32\n@Marvis I'm in a 2nd year geometry course, and this is part of my homework due on Friday. I'm being a keener and getting it done early :) I just guessed on the right angle part. \u2013\u00a0 Charlie Jun 5 '12 at 5:46\n\n$\\angle DNF = 90^\\circ \\Longleftrightarrow \\angle BNF = \\angle CND$. It suffics to prove that $\\triangle BNF \\sim \\triangle CND$. Well, it's trivial: $\\angle NBF = \\angle BEC = \\angle NCD$ and $\\displaystyle \\frac{NB}{BF} = \\frac{NB}{BE} = \\sin\\angle NEB = \\cos\\angle NCB = \\frac{NC}{CB} = \\frac{NC}{CD}$. Q.E.D\n\n-\nThat $NB/BE = NC/CB$ follows from the similarity of right triangles $ENB$ and $BNC$ without need for trigonometry. Nice proof! \u2013\u00a0 Rahul Jun 5 '12 at 7:16\n@RahulNarain Trigonometry is an abbreviation of similiar triangles here because we don't use angle transformation formulas such as $\\sin(\\alpha+\\beta)=\\cdots$. \u2013\u00a0 Frank Science Jun 5 '12 at 7:20\nIndeed; what I meant to say was that the use of trigonometric notation is not needed. It suggests that more powerful machinery is being employed than actually is. \u2013\u00a0 Rahul Jun 5 '12 at 9:12\n\nFirstly We can write\n(1) Oklid relation from $\\triangle CNB$\n\n$h^2=m(k+x)$\n\n(2) Pisagor relation from $\\triangle DPN$\n\n$a^2=(x+k)^2+(x+k+m-h)^2$\n\n(3) Pisagor relation from $\\triangle FRN$\n\n$b^2=h^2+k^2$\n\n(4) Pisagor relation from $\\triangle DCF$\n\n$c^2=x^2+(x+k+m)^2$\n\nif DNF triange is a right triangle,It must satisfy $a^2+b^2=c^2$.\n\n$(x+k)^2+(x+k+m-h)^2+h^2+k^2=x^2+(x+k+m)^2$\n\n$(x+k)^2+(x+k+m)^2-2h(x+k+m)+h^2+h^2+k^2=x^2+(x+k+m)^2$\n\n$(x+k)^2-2h(x+k+m)+2h^2+k^2=x^2$\n\n$xk+k^2-h(x+k+m)+h^2=0$\n\n$xk+k^2-h(x+k+m)+m(k+x)=0$\n\n$-h(x+k+m)+(k+m)(k+x)=0$\n\n$\\frac{x+k}{x+k+m}=\\frac{h}{k+m}$\n\nThis result is equal to the rates of thales formula for similar triangles $\\triangle CRN \\sim \\triangle CBE$\n\nThus $a^2+b^2=c^2$ is correct for $\\triangle DNF$ .\n\n-\n\nLet us do it through coordinate geometry. Let $B$ be the origin. Lets fix the coordinates first. $$A = (0,a)\\\\ B = (0,0) \\\\ C = (a,0)\\\\ D = (a,a)$$ Since $E$ and $F$ are equidistant from $B$, lets say $$F = (b,0)\\\\ E = (0,b)$$ where $0 \\leq b \\leq a$. The equation of the line $CE$ is $\\dfrac{x}{a} + \\dfrac{y}{b} = 1$. The equation of $BN$ is $y = \\dfrac{a}{b}x$. This gives us the coordinate of $N$ as $\\left( \\dfrac{ab^2}{a^2 + b^2},\\dfrac{a^2b}{a^2 + b^2} \\right)$.\n\nThe slope of $FN$ is $m_1 = \\dfrac{a^2b/(a^2+b^2) - 0}{ab^2/(a^2+b^2)-b} = \\dfrac{a^2}{ab - a^2 - b^2}$.\n\nThe slope of $DN$ is $m_2 = \\dfrac{a^2b/(a^2+b^2) - a}{ab^2/(a^2+b^2)-a} = \\dfrac{ab - a^2 - b^2}{b^2 - a^2 - b^2} = - \\left(\\dfrac{ab-a^2-b^2}{a^2} \\right)$.\n\nHence, the product of the slopes is $m_1m_2 = -1$. Hence your conjecture is indeed correct. I am waiting for someone to post a nicer geometric argument.\n\n-\n\nHINT: Try to resort to Homothetic transformation and you're done immediately. Another way is to connect $N$ to the middle of the $DF$ (let's call that point $M$) and prove that you have there $NM$=$DM$=$MF$ (here you may resort to Apollonius' theorem). You also could resort to the cyclic quadrilaterals as another approaching way.\n\n-", "date": "2015-07-07 22:35:09", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/154087/angle-of-a-triangle-inscribed-in-a-square", "openwebmath_score": 0.937068521976471, "openwebmath_perplexity": 377.29925646660666, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471656514752, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8512482647159757}}
{"url": "https://math.stackexchange.com/questions/2002785/density-of-rational-and-irrational-numbers", "text": "# Density of rational and irrational numbers\n\nSince $\\mathbb{Q}$ is countable and $\\mathbb{R} \\backslash \\mathbb {Q}$ is not, what does this tell us about the density of rational and irrational numbers along the real number line? Saying that there exists more irrational numbers than rational numbers seems rather vague becuase we're comparing infinites. How do we even define density here?\n\n\u2022 There is a mathematical notion of dense, and the rationals and irrationals are each dense in the real numbers. However, it seems you are thinking of a different notion of denseness, perhaps more related to cardinality... \u2013\u00a0angryavian Nov 7 '16 at 0:31\n\u2022 \"what does this tell us about the density of rational and irrational numbers along the real number line?\" Nothing: there are countable non-dense sets, countable dense sets, uncountable dense sets and uncountable non-dense sets. Countability and density are totally unrelated notions. \u2013\u00a0Crostul Nov 7 '16 at 0:50\n\u2022 @Alephnull The Cantor set is such an example. \u2013\u00a0Crostul Nov 7 '16 at 8:09\n\u2022 @Crostul But there are not co-countable nowhere dense sets (sets which are nowhere dense by have countable complement). So because $\\mathbb{Q}$ is countable, we can at least conclude that $\\mathbb{R}\\setminus\\mathbb{Q}$ is somewhere dense. \u2013\u00a0Reese Nov 7 '16 at 9:13\n\u2022 @Alephnull It's pretty straightforward - a nowhere dense set by definition must exclude an entire interval. That interval cannot be countable. \u2013\u00a0Reese Nov 7 '16 at 11:03\n\nOne can have any combination of (countably infinite, uncountable) and (dense in the line, not dense in the line). (Unless, as was suggested in the comments, you have some notion of density other than the one defined by Captain Falcon.)\n\nCountably infinite and dense in the reals: Rationals\n\nCountably infinite and not dense in the reals: Integers\n\nUncountable and dense in the reals: Irrationals\n\nUncountable and not dense in the reals: Unit interval\n\n\u2022 Ah ah, I appreciate the reference to Captain Falcon. However, \"C.\" stands for \"Cyril\" ; Falcon being my name. :) \u2013\u00a0C. Falcon Nov 7 '16 at 0:58\n\u2022 Another example, a bit worse, is the Cantor set. It is uncountable and not dense, in fact it equals its closure which has (Lebesgue) length zero. \u2013\u00a0Jeppe Stig Nielsen Nov 7 '16 at 10:11\n\nA subset $A$ of $\\mathbb{R}$ is said dense if and only if any elements of $\\mathbb{R}$ is a limit of a sequence of elements of $A$.\n\nIndeed, both $\\mathbb{Q}$ and $\\mathbb{R}\\setminus\\mathbb{Q}$ are dense in $\\mathbb{R}$ in that sense.\n\nAs you pointed out, $\\mathbb{Q}$ is countable, whereas $\\mathbb{R}\\setminus\\mathbb{Q}$ is not, but the notion of density as nothing to do with countability.\n\nRemark. To see that $\\mathbb{Q}$ is dense in $\\mathbb{R}$, let us pick $x\\in\\mathbb{R}$ and for all $n\\in\\mathbb{N}$ let us define: $$x_n:=\\frac{\\lfloor 10^nx\\rfloor}{10^n}.$$ It is a sequence of $\\mathbb{Q}$ which converges towards $x$ using squeeze theorem. For the density of $\\mathbb{R}\\setminus\\mathbb{Q}$ consider: $$y_n:=x_n+\\frac{\\sqrt{2}}{n+1}.$$\n\n\u2022 Nice answer, Captain! (salute) \u2013\u00a0Todd Wilcox Nov 7 '16 at 5:35\n\nAs clarified in other answers, cardinality by itself does not answer density questions.\n\nRegarding the definition of density, there are two definitions. One is topological, saying that a set $A$ is dense if it intersects every non-empty open set. For the real line this is equivalent to saying that $A$ is dense (in the real line $\\Bbb R$) if it intersects every open interval $(x,y)$ with $x<y$. In other words, for all choices of $x$ and $y$ with $x<y$ there is $a\\in A$ with $x<a<y$. The latter condition usually is termed \"order-dense\" (in $\\Bbb R$) and could be defined without reference to topology, for any linear order in place of $\\Bbb R$. That is, $A$ is order-dense in a linear order $(L,<)$ is for all $x,y\\in L$ with $x<y$ there is $a$ in $(x,y)$, that is, $x<a<y$ with $a\\in A$. Finally a set $A$ is order-dense in itself if $A$ is order-dense in the linear order $(A,<)$. The set $\\Bbb Q$ of all rational numbers is order-dense in the set $\\Bbb R.$ The set $\\Bbb Q$ is also order-dense in itself. The set $\\Bbb Z$ of all integers is not order-dense in itself, and it is not order-dense in $\\Bbb R$. The middle-third Cantor set $C$ is not order-dense in itself as it has \"gaps\", e.g. the interval $(\\frac13,\\frac23)$ contains no elements of $C$. (But, $C$ is (topologically) dense in itself, as it has no isolated points.) $C$ is no-where dense (in $\\Bbb R$), meaning that the complement of its (topological) closure in $\\Bbb R$ is (topologically) dense (and, as it happens, order-dense).\n\nIf $X$ is a topological space and $A$ is a subset then we say that $A$ is dense in $X$ provided that the closure of $A$ contains $X$. When $X=\\Bbb R$ this means either of the following two conditions: (1) the set $A$ intersects every non-empty open interval, or (2) for every real number $x$ there is a sequence of points of $A$ converging to $x$. On the other hand, if $A=\\Bbb Z$ then $A$ is dense in itself (topologically), but not order-dense in itself. Indeed, for every $m\\in\\Bbb Z$ the constant sequence $\\langle m,m,...\\rangle$ converges to $m$. On the other hand, the interval $(m,m+1)$ contains no integers.\n\nIn case you interpreted \"dense\" not in the topological definition, but as saying \"many more irrationals than rationals\", there's another field of math that provides an alternative way of seeing this.\n\nIn Measure Theory there's the concept of the measure of a set, which is an alternative characterization of its size. We want a way to define a function on sets, so let $X\\subseteq 2^\\mathbb{R}$ be a collection of sets (I have a specific one in mind here, namely the Lebesgue $\\sigma$-algebra, but don't worry about that). Now, define the function: $$\\mu:X\\to\\mathbb{R}_{\\geq 0}\\cup\\lbrace\\infty\\rbrace$$ that must fulfill three axioms:\n\n1. Non-negativity. For any subset $E\\in X$, we have that $\\mu(E)\\geq 0$. Intuitively, things can't have negative size.\n\n2. $\\mu(\\emptyset) = 0$. This should also seem intuitive, \"the lack of a set\" has no measure (there will be other sets with no measure though!).\n\n3. Countable additivity on disjoint subsets. If $E_1,E_2$ are disjoint subsets of $\\mathbb R$ (so $E_1\\cap E_2 = \\emptyset$), then: $$\\mu(E_1\\cup E_2) = \\sum_{i = 1}^2 \\mu(E_i)$$ This will actually be true for countably many disjoint subsets, so if $\\lbrace E_i\\rbrace$ are all pairwise disjoint, then: $$\\mu(\\cup_{i= 1}^\\infty E_i)= \\sum_{i = 1}^\\infty \\mu(E_i)$$ For finite sums, the intuition for this should be clear (think about each set as a circle, like how you make a Venn diagram. If there's no overlap between circles, the total \"area\" should be just all the individual areas added together).\n\nUnfortunately, the above identity working for countable sums is less clearly true to our intuition. If we accept it though, we get some interesting results. The most applicable one to this is that we can calculate: $$\\mu(\\mathbb R\\setminus\\mathbb Q) = \\infty,\\quad\\mu(\\mathbb Q) = 0$$\n\nSo, in terms of measure theory, there are essentially no rational numbers, but infinitely many irrationals.", "date": "2019-06-18 05:30:00", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2002785/density-of-rational-and-irrational-numbers", "openwebmath_score": 0.9466912746429443, "openwebmath_perplexity": 235.47190395044214, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471665987074, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8512482518589282}}
{"url": "http://math.stackexchange.com/questions/66996/what-is-the-probability-that-xy", "text": "What is the probability that $X<Y$?\n\nIf two dice are rolled repeatedly, and $X$ is the number of tosses until $3$ is rolled, and $Y$ is the number of tosses until a $5$ is rolled, what is the probability$(X < Y)$? Also, if $Z$ is the number of rolls until a $10$ is rolled, what the probability$(X < Y < Z)$?\n\nIs $P(X < Y) = \\sum_{n=2}^{\\infty}\\sum_{k=1}^{\\infty}((32/36)^{(n-1)}(4/36)-(34/36)^{(k-1)}(2/36)]$?\n\n-\nIs it assumed that $X$, $Y$ and $Z$ are independent ? \u2013\u00a0 Sasha Sep 23 '11 at 16:50\nYes X Y and Z are independent as the dice are balanced \u2013\u00a0 lord12 Sep 23 '11 at 16:50\nNo, $X$, $Y$ and $Z$ are not independent. For example, if $X = 1$ then $Y \\ne 1$ and $Z \\ne 1$. \u2013\u00a0 Robert Israel Sep 23 '11 at 18:40\nIt's not clear whether \"a 3 is rolled\" is referring to a 3 coming up on at least one of the two dice, or the sum of the two being 3. But since \"a 10 is rolled\" can only refer to the sum, I guess we can assume the 3 and 5 are also referring to the sum. \u2013\u00a0 Robert Israel Sep 23 '11 at 18:43\n@Robert: Those were my thoughts too. But also notice that the first line says \"until 3 is rolled\" and not \"until a 3 is rolled\". So I suppose this indeed refers to the sum being 3, rather than one of them being 3. \u2013\u00a0 TMM Sep 23 '11 at 18:47\n\n(a) $P(X<Y)$\n\nSuppose $q = P(3 \\text{ before } 5)$. When will we see a $3$ before we see a $5$? That is if we first see a $3$ on the first roll (we are done), or if we don't see a $3$ or a $5$ on the first roll and then see a $3$ before a $5$ after that. In other words:\n\n$$q = P(3) \\cdot 1 + P(5) \\cdot 0 + P(\\text{neither } 3 \\text{ nor } 5) \\cdot q$$\n\nSince we know that $P(3) = \\frac{2}{36}$, $P(5) = \\frac{4}{36}$ we get $P(\\text{neither } 3 \\text{ nor } 5) = 1 - \\frac{2}{36} - \\frac{4}{36} = \\frac{30}{36}$, so\n\n$$q = \\frac{2}{36} + \\frac{30}{36} q \\quad \\Longleftrightarrow \\quad q = \\frac{1}{3}$$\n\nNote that we can also rewrite the first equation as\n\n$$q = P(3) \\cdot 1 + P(5) \\cdot 0 + P(\\text{neither } 3 \\text{ nor } 5) \\cdot q \\quad \\Longleftrightarrow \\quad q = \\frac{P(3)}{P(3) + P(5)} = \\frac{P(3)}{P(3 \\text{ or } 5)} = P(3\\ |\\ 3 \\text{ or } 5)$$\n\nThis is in accordance with a comment made by Dilip: We can simply ignore all throws which are neither a $3$ or a $5$, as they are irrelevant. Then the probability of seeing a $3$ first is simply the conditional probability of throwing a $3$, given that it's either a $3$ or a $5$.\n\n(b) $P(X<Y<Z)$\n\nWe can use the same approach as above. First, let us ignore all throws which are not $3$, $5$ or $10$. We now need that the first event is a $3$, which happens with probability\n\n$$P(X<Y,X<Z) = \\frac{P(X)}{P(X \\text{ or } Y \\text{ or } Z)} = \\frac{P(X)}{P(X) + P(Y) + P(Z)} = \\frac{2}{2+4+3} = \\frac{2}{9}$$\n\nNow if this happens, we only need that in all successive events of $3,5,10$ we first see some number of $3$s, and then see a $5$ (and not a $10$ yet). But that means that after this first event, we can again ignore all events $X$, and look at the first throw resulting in either a $5$ or a $10$. Then\n\n$$P(Y<Z | X<Y,X<Z) = \\frac{P(Y)}{P(Y \\text{ or } Z)} = \\frac{P(Y)}{P(Y)+P(Z)} = \\frac{4}{4+3} = \\frac{4}{7}$$\n\nSo the combined probability is then\n\n$$P(X<Y<Z) = P(X<Y,X<Z) \\cdot P(Y<Z | X<Y, X<Z) = \\frac{2}{9} \\cdot \\frac{4}{7} = \\frac{8}{63}$$\n\n-\n\nFollowing on from @Thijs analysis, which is actually a tool called \"first step analysis\", we can get the probability that $P(X<Y<Z)$. So we have to wait until one of these events happens. $P(X)=\\frac{2}{36}$ $P(Y)=\\frac{4}{36}$ $P(Z)=\\frac{3}{36}$. So we have probability of $\\frac{27}{36}$ of achieving \"nothing\" each trial, and effectively going back to the first step.\n\nSo at effective trial one, if $X$ occurs we have success, if $Y$ or $Z$ occurs, we have failure.\n\n$$P(X<Y<Z)=\\frac{2}{36}P(X<Y<Z|X<Y,Z)+\\frac{3}{36}P(X<Y<Z|Z<X,Y)$$ $$+\\frac{4}{36}P(X<Y<Z|Y<X,Z)+\\frac{27}{36}P(X<Y<Z)$$\n\nThe middle two probabilities are zero, and the first is simply $P(Y<Z)$, and applying first step analysis again we have\n\n$$P(Y<Z)=\\frac{4}{7}$$\n\nPlugging back into the previous equation and solving we get\n\n$$P(X<Y<Z)=\\frac{8}{36(7)}+\\frac{27}{36}P(X<Y<Z)=\\frac{8}{63}$$\n\n-\nNice, your follow-up is more in line with my previous analysis than my own follow-up! :) \u2013\u00a0 TMM Sep 23 '11 at 18:37\n\nHere's a simpler way than one involving infinite series (such as that posted by Sasha). Roll a die until you get either a 3 or a 5. What's the probability that it's a 3? It's just the conditional probability of getting a 3, given that you've got either a 3 or a 5: $$\\Pr(W=3\\mid W=3\\text{ or }W=5) = \\frac{\\Pr(W=3)}{\\Pr(W=3\\text{ or }W=5)} = \\frac{2/36}{6/36}= \\frac13.$$\n\n-", "date": "2014-11-24 02:52:01", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/66996/what-is-the-probability-that-xy", "openwebmath_score": 0.8715810775756836, "openwebmath_perplexity": 189.553806010762, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9838471618625457, "lm_q2_score": 0.865224073888819, "lm_q1q2_score": 0.8512482494706641}}
{"url": "https://perfectionatic.org/?tag=julia", "text": "11/9/19\n\n# Evaluating Arbitrary Precision Integer Expressions in Julia using Metaprogramming\n\nWhile watching the Mathologer masterclass on power sums\n\nI came across a challenge to evaluate the following sum\n$$1^{10}+2^{10}+\\cdots+1000^{10}$$\n\nThis can be easily evaluated via brute force in Julia to yield\n\nfunction power_cal(n)\na=big(0)\nfor i=1:n\na+=big(i)^10\nend\na\nend\n\njulia> power_cal(1000)\n91409924241424243424241924242500\n\nNote the I had to use big to makes sure we are using BigInt in the computation. Without that, we would be quickly running in into an overflow issue and we will get a very wrong number.\n\nIn the comment section of the video I found a very elegant solution to the above sum, expressed as\n\n(1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 \u2013 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000 = 91409924241424243424241924242500\n\nIf I try to plug this into the Julia, I get\n\njulia> (1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000\n-6.740310541071357e18\n\nThis negative answer is not surprising at all, because we obviously ran into an overflow. We can, of course, go through that expression and modify all instances of Int64 with BigInt by wrapping it in the big function. But that would be cumbersome to do by hand.\n\n## The power of Metaprogramming\n\nIn Julia, metaprogramming allows you to write code that creates code, the idea here to manipulate the abstract syntax tree (AST) of a Julia expression. We start to by \u201cquoting\u201d our original mathematical expressing into a Julia expression. In the at form it is not evaluated yet, however we can always evaluate it via eval.\n\njulia> ex1=:((1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000)\n\n:((((((1 / 11) * 1000 ^ 11 + (1 / 2) * 1000 ^ 10 + (5 / 6) * 1000 ^ 9) - 1000 ^ 7) + 1000 ^ 5) - (1 / 2) * 1000 ^ 3) + (5 / 66) * 1000)\n\njulia> dump(ex1)\nExpr\nargs: Array{Any}((3,))\n1: Symbol +\n2: Expr\nargs: Array{Any}((3,))\n1: Symbol -\n2: Expr\nargs: Array{Any}((3,))\n1: Symbol +\n2: Expr\nargs: Array{Any}((3,))\n1: Symbol -\n2: Expr\n3: Expr\n3: Expr\nargs: Array{Any}((3,))\n1: Symbol ^\n2: Int64 1000\n3: Int64 5\n\njulia> eval(ex1)\n-6.740310541071357e18\n\nThe output of dump show the full AST in all its glory (\u2026well almost the depth is a bit truncated). Notice that here all our numbers are interpreted as Int64.\nNow we walk through the AST and change all occurrences of Int64 with BigInt by using the big function.\n\nfunction makeIntBig!(ex::Expr)\nargs=ex.args\nfor i in eachindex(args)\nif args[i] isa Int64\nargs[i]=big(args[i])\nend\nif args[i] isa Expr\nmakeIntBig!(args[i])\nend\nend\nend\n\njulia> ex2=copy(ex1)\n:((((((1 / 11) * 1000 ^ 11 + (1 / 2) * 1000 ^ 10 + (5 / 6) * 1000 ^ 9) - 1000 ^ 7) + 1000 ^ 5) - (1 / 2) * 1000 ^ 3) + (5 / 66) * 1000)\n\njulia> makeIntBig!(ex2)\n\njulia> eval(ex2)\n9.14099242414242434242419242425000000000000000000000000000000000000000000000014e+31\n\nWe see an improvement here, but the results are not very satisfactory. The divisions yield BigFloat results, which had a tiny bit of floating point errors. Can we do better?\nJulia has support for Rational expressions baked in. We can use that improve the results. We just need to search for call expressions the / symbol and replace it by the // symbol. For safety we just have to makes sure the operands are as subtype of Integer.\n\nfunction makeIntBig!(ex::Expr)\nargs=ex.args\nif ex.head == :call && args[1]==:/ &&\nlength(args)==3 &&\nall(x->typeof(x) <: Integer,args[2:end])\nargs[1]=://\nargs[2]=big(args[2])\nargs[3]=big(args[3])\nelse\nfor i in eachindex(args)\nif args[i] isa Int64\nargs[i]=big(args[i])\nend\nif args[i] isa Expr\nmakeIntBig!(args[i])\nend\nend\nend\nend\n\njulia> ex2=copy(ex1);\n\njulia> makeIntBig!(ex2)\n\njulia> eval(ex2)\n91409924241424243424241924242500//1\n\nNow that is much better! We have not lost any precision and we ended us with a Rational expression.\n\nFinally, we can build a macro so the if we run into such expressions in the future and we want to evaluate them, we could just conveniently call it.\n\nmacro eval_bigInt(ex)\nmakeIntBig!(ex)\nquote # Removing the denominator if it is redundant\nlocal x=$ex (x isa Rational && x.den==1) ? x.num : x end end and we can now simply evaluate our original expression as julia> @eval_bigInt (1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000 91409924241424243424241924242500 07/28/18 # Exploring left truncatable primes Recently I came across a fascinating Numberphile video on truncatable primes I immediately thought it would be cool to whip a quick Julia code to get the full enumeration of all left truncatable primes, count the number of branches and also get the largest left truncatable prime. using Primes function get_left_primes(s::String) p_arr=Array{String,1}() for i=1:9 number_s=\"$i$s\" if isprime(parse(BigInt, number_s)) push!(p_arr,number_s) end end p_arr end function get_all_left_primes(l) r_l= Array{String,1}() n_end_points=0 for i in l new_l=get_left_primes(i) isempty(new_l) && (n_end_points+=1) append!(r_l,new_l) next_new_l,new_n=get_all_left_primes(new_l) n_end_points+=new_n # counting the chains append!(r_l,next_new_l) end r_l, n end The first function just prepends a number (expressed in String for convenience) and checks for it possible primes that can emerge from a single digit prepending. For example: julia> get_left_primes(\"17\") 2-element Array{String,1}: \"317\" \"617\" The second function, just makes extensive use of the first to get all left truncatable primes and also count the number of branches. julia> all_left_primes, n_branches=get_all_left_primes([\"\"]) (String[\"2\", \"3\", \"5\", \"7\", \"13\", \"23\", \"43\", \"53\", \"73\", \"83\" \u2026 \"6435616333396997\", \"6633396997\", \"76633396997\", \"963396997\", \"16396997\", \"96396997\", \"616396997\", \"916396997\", \"396396997\", \"4396396997\"], 1442) julia> n_branches 1442 julia> all_left_primes 4260-element Array{String,1}: \"2\" \"3\" \"5\" \"7\" \"13\" \"23\" \u22ee \"96396997\" \"616396997\" \"916396997\" \"396396997\" \"4396396997\" So we the full list of possible left truncatable primes with a length 4260. Also the total number of branches came to 1442. We now get the largest left truncatable primes with the following one liner: julia> largest_left_prime=length.(all_left_primes)|>indmax|> x->all_left_primes[x] \"357686312646216567629137\" After this fun exploration, I found an implementation in Julia for just getting the largest left truncatable prime for any base in Rosseta Code. 04/1/18 # Iterating with Dates and Time in Julia Julia has good documentation on dealing with Dates and Time, however that is often in the context constructing and Date and Time objects. In this post, I am focus on the ability to iterate over Dates and Times. This is very useful in countless application. We start of by capturing this moment and moving ahead into the future julia> this_moment=now() 2018-04-01T23:13:33.437 In one hour that will be julia> this_moment+Dates.Hour(1) 2018-04-02T00:13:33.437 Notice that Julia was clever enough properly interpret that we will be on the in another day after exactly one hour. Thanks to it multiple dispatch of the DateTime type to be able to do TimeType period arithmatic. You can then write a nice for loop that does something every four hours for the next two days. julia> for t=this_moment:Dates.Hour(4):this_moment+Dates.Day(2) println(t) #or somethings special with that time end 2018-04-01T23:13:33.437 2018-04-02T03:13:33.437 2018-04-02T07:13:33.437 2018-04-02T11:13:33.437 2018-04-02T15:13:33.437 2018-04-02T19:13:33.437 2018-04-02T23:13:33.437 2018-04-03T03:13:33.437 2018-04-03T07:13:33.437 2018-04-03T11:13:33.437 2018-04-03T15:13:33.437 2018-04-03T19:13:33.437 2018-04-03T23:13:33.437 Often we are not so interested in the full dates. For example if we are reading a video file and we want to get a frame every 5 seconds while using VideoIO.jl. We can deal here with the simpler Time type. julia> video_start=Dates.Time(0,5,20) 00:05:20 Here we are interested in starting 5 minutes and 20 seconds into the video. Now we can make a nice loop from the start to finish for t=video_start:Dates.Second(5):video_start+Dates.Hour(2) h=Dates.Hour(t).value m=Dates.Minute(t).value s=Dates.Second(t).value ms=Dates.Millisecond(t).value # Do something interesting with ffmpeg seek on the video end 02/9/18 # When Julia is faster than C, digging deeper In my earlier post I showed an example where Julia is significantly faster than c. I got this insightful response So I decided to dig deeper. Basically the standard c rand() is not that good. So instead I searched for the fastest Mersenne Twister there is. I downloaded the latest code and compiled it in the fastest way for my architecture. /* eurler2.c */ #include <stdio.h> /* printf, NULL */ #include <stdlib.h> /* srand, rand */ #include \"SFMT.h\" /* fast Mersenne Twister */ sfmt_t sfmt; double r2() { return sfmt_genrand_res53(&sfmt); } double euler(long int n) { long int m=0; long int i; for(i=0; i<n; i++){ double the_sum=0; while(1) { m++; the_sum+=r2(); if(the_sum>1.0) break; } } return (double)m/(double)n; } int main () { sfmt_init_gen_rand(&sfmt,123456); printf (\"Euler : %2.5f\\n\", euler(1000000000)); return 0; } I had to compile with a whole bunch of flags which I induced from SFMT\u2018s Makefile to get faster performance. gcc -O3 -finline-functions -fomit-frame-pointer -DNDEBUG -fno-strict-aliasing --param max-inline-insns-single=1800 -Wall -std=c99 -msse2 -DHAVE_SSE2 -DSFMT_MEXP=1279 -ISFMT-src-1.5.1 -o eulerfast SFMT.c euler2.c And after all that trouble we got the performance down to 18 seconds. Still slower that Julia\u2018s 16 seconds. $ time ./eulerfast\nEuler : 2.71824\n\nreal    0m18.075s\nuser    0m18.085s\nsys 0m0.001s\n\nProbably, we could do a bit better with more tweaks, and probably exceed Julia\u2018s performance with some effort. But at that point, I got tired of pushing this further. The thing I love about Julia is how well it is engineered and hassle free. It is quite phenomenal the performance you get out of it, with so little effort. And for basic technical computing things, like random number generation, you don\u2019t have to dig hard for a better library. The \u201cbatteries included\u201d choices in the Julia\u2018s standard library are pretty good.\n\n02/8/18\n\n# When Julia is faster than C\n\nOn e-day, I came across this cool tweet from Fermat\u2019s library\n\nSo I spend a few minutes coding this into Julia\n\nfunction euler(n)\nm=0\nfor i=1:n\nthe_sum=0.0\nwhile true\nm+=1\nthe_sum+=rand()\n(the_sum>1.0) && break;\nend\nend\nm/n\nend\n\nTiming this on my machine, I got\n\njulia> @time euler(1000000000)\n15.959913 seconds (5 allocations: 176 bytes)\n2.718219862\n\nGave a little under 16 seconds.\n\nTried a c implementation\n\n#include <stdio.h>      /* printf, NULL */\n#include <stdlib.h>     /* srand, rand */\n#include <time.h>       /* time */\n\ndouble r2()\n{\nreturn (double)rand() / (double)((unsigned)RAND_MAX + 1);\n}\n\ndouble euler(long int n)\n{\nlong int m=0;\nlong int i;\nfor(i=0; i<n; i++){\ndouble the_sum=0;\nwhile(1) {\nm++;\nthe_sum+=r2();\nif(the_sum>1.0) break;\n}\n}\nreturn (double)m/(double)n;\n}\n\nint main ()\n{\nprintf (\"Euler : %2.5f\\n\", euler(1000000000));\n\nreturn 0;\n}\n\nand compiling with either gcc\n\ngcc  -Ofast euler.c\n\nor clang\n\nclang  -Ofast euler.c\n\ngave a timing twice as long\n\n$time ./a.out Euler : 2.71829 real 0m36.213s user 0m36.238s sys 0m0.004s For the curios, I am using this version of Julia julia> versioninfo() Julia Version 0.6.3-pre.0 Commit 93168a6 (2017-12-18 07:11 UTC) Platform Info: OS: Linux (x86_64-linux-gnu) CPU: Intel(R) Core(TM) i7-4770HQ CPU @ 2.20GHz WORD_SIZE: 64 BLAS: libopenblas (USE64BITINT DYNAMIC_ARCH NO_AFFINITY Haswell) LAPACK: libopenblas64_ LIBM: libopenlibm LLVM: libLLVM-3.9.1 (ORCJIT, haswell) Now one should not put too much emphasis on such micro benchmarks. However, I found this a very curious examples when a high level language like Julia could be twice as fast a c. The Julia language authors must be doing some amazing mojo. 01/2/18 # Visualizing the Inscribed Circle and Square Puzzle Recently, I watched a cool mind your decsions video on an inscribed circle and rectangle puzzle. In the video they showed a diagram that was not scale. I wanted to get a sense of how these differently shaped areas will match. There was a cool ratio between the outer and inner circle radii that is expressed as $$\\frac{R}{r}=\\sqrt{\\frac{\\pi-2}{4-\\pi}}$$. I used Compose.jl to rapidly do that. using Compose set_default_graphic_size(20cm, 20cm) \u03d5=sqrt((pi -2)/(4-pi)) R=10 r=R/\u03d5 ctx=context(units=UnitBox(-10, -10, 20, 20)) composition = compose(ctx, (ctx, rectangle(-r/\u221a2,-r/\u221a2,r*\u221a2,r*\u221a2),fill(\"white\")), (ctx,circle(0,0,r),fill(\"blue\")), (ctx,circle(0,0,R),fill(\"white\")), (ctx,rectangle(-10,-10,20,20),fill(\"red\"))) composition |> SVG(\"inscribed.svg\") 08/6/17 # Solving the code lock riddle with Julia I came across a neat math puzzle involving counting the number of unique combinations in a hypothetical lock where digit order does not count. Before you continue, please watch at least the first minute of following video: The rest of the video describes two related approaches for carrying out the counting. Often when I run into complex counting problems, I like to do a sanity check using brute force computation to make sure I have not missed anything. Julia is fantastic choice for doing such computation. It has C like speed, and with an expressiveness that rivals many other high level languages. Without further ado, here is the Julia code I used to verify my solution the problem. 1. function unique_combs(n=4) 2.  pat_lookup=Dict{String,Bool}() 3.  for i=0:10^n-1 4.  d=digits(i,10,n) # The digits on an integer in an array with padding 5.  ds=d |> sort |> join # putting the digits in a string after sorting 6.  get(pat_lookup,ds,false) || (pat_lookup[ds]=true) 7.  end 8.  println(\"The number of unique digits is$(length(pat_lookup))\")\n9. end\n\nIn line 2 we create a dictionary that we will be using to check if the number fits a previously seen pattern. The loop starting in line 3, examines all possible ordered combinations. The digits function in line 4 takes any integer and generate an array of its constituent digits. We generate the unique digit string in line 5 using pipes, by first sorting the integer array of digits and then combining them in a string. In line 6 we check if the pattern of digits was seen before and make use of quick short short-circuit evaluation to avoid an if-then statement.\n\n07/14/17\n\n# Julia calling C: A more minimal example\n\nEarlier I presented a minimal example of Julia calling C. It mimics how one would go about writing C code, wrapping it a library and then calling it from Julia. Today I came across and even more minimal way of doing that while reading an excellent blog on Julia\u2019s syntactic loop fusion. Associated with the blog was notebook that explores the matter further.\n\nBasically, you an write you C in a string and pass it directly to the compiler. It goes something like\n\nusing Libdl\nC_code= raw\"\"\"\ndouble mean(double a, double b) {\nreturn (a+b) / 2;\n}\n\"\"\"\nconst Clib=tempname()\nopen(gcc -fPIC -O3 -xc -shared -o \\$(Clib * \".\" * Libdl.dlext) -, \"w\") do f\nprint(f, C_code)\nend\n\nThe tempname function generate a unique temporary file path. On my Linux system Clib will be string like \"/tmp/juliaivzRkT\". That path is used to generate a library name \"/tmp/juliaivzRkT.so\" which will then used in the ccall:\n\nmeanc(a,b)=ccall((:mean,Clib),Float64,(Float64,Float64),a,b)\njulia> meanc(3,4)\n3.5\n\nThis approach would not be recommended if you are writing anything sophisticated in C. However, it is fun to experiment with for short bits of C code that you might like to call from Julia. Saves you the hassle of creating a Makefile, compiling, etc\u2026\n\n06/29/17\n\n# Solving the Fish Riddle with JuMP\n\nRecently I came across a nice Ted-Ed video presenting a Fish Riddle.\n\nI thought it would be fun to try solving it using Julia\u2019s award winning JuMP package. Before we get started, please watch the above video-you might want to pause at 2:24 if you want to solve it yourself.\n\nTo attempt this problem in Julia, you will have to install the JuMP package.\n\njulia> Pkg.add(\"JuMP\")\n\nJuMP provides an algebraic modeling language for dealing with mathematical optimization problems. Basically, that allows you to focus on describing your problem in a simple syntax and it would then take care of transforming that description in a form that can be handled by any number of solvers. Those solvers can deal with several types of optimization problems, and some solvers are more generic than others. It is important to pick the right solver for the problem that you are attempting.\n\nThe problem premises are:\n1. There are 50 creatures in total. That includes sharks outside the tanks and fish\n2. Each SECTOR has anywhere from 1 to 7 sharks, with no two sectors having the same number of sharks.\n3. Each tank has an equal number of fish\n4. In total, there are 13 or fewer tanks\n5. SECTOR ALPHA has 2 sharks and 4 tanks\n6. SECTOR BETA has 4 sharsk and 2 tanks\nWe want to find the number of tanks in sector GAMMA!\n\nHere we identify the problem as mixed integer non-linear program (MINLP). We know that because the problem involves an integer number of fish tanks, sharks, and number of fish inside each tank. It also non-linear (quadratic to be exact) because it involves multiplying two two of the problem variables to get the total number or creatures. Looking at the table of solvers in the JuMP manual. pick the Bonmin solver from AmplNLWriter package. This is an open source solver, so installation should be hassle free.\n\njulia> Pkg.add(\"AmplNLWriter\")\n\nWe are now ready to write some code.\n\nusing JuMP, AmplNLWriter\n\n# Solve model\nm = Model(solver=BonminNLSolver())\n\n# Number of fish in each tank\n@variable(m, n>=1, Int)\n\n# Number of sharks in each sector\n@variable(m, s[i=1:3], Int)\n\n# Number of tanks in each sector\n@variable(m, nt[i=1:3]>=0, Int)\n\n@constraints m begin\n# Constraint 2\nsharks[i=1:3], 1 <= s[i] <= 7\nnumfish[i=1:3], 1 <= nt[i]\n# Missing uniqueness in restriction\n# Constraint 4\nsum(nt) <= 13\n# Constraint 5\ns[1] == 2\nnt[1] == 4\n# Constraint 6\ns[2] == 4\nnt[2] == 2\nend\n\n# Constraints 1 & 3\n@NLconstraint(m, s[1]+s[2]+s[3]+n*(nt[1]+nt[2]+nt[3]) == 50)\n\n# Solve it\nstatus = solve(m)\n\nsharks_in_each_sector=getvalue(s)\nfish_in_each_tank=getvalue(n)\ntanks_in_each_sector=getvalue(nt)\n\n@printf(\"We have %d fishes in each tank.\\n\", fish_in_each_tank)\n@printf(\"We have %d tanks in sector Gamma.\\n\",tanks_in_each_sector[3])\n@printf(\"We have %d sharks in sector Gamma.\\n\",sharks_in_each_sector[3])\n\nIn that representation we could not capture the restriction that \u201cno two sectors having the same number of sharks\u201d. We end up with the following output:\n\nWe have 4 fishes in each tank.\nWe have 4 tanks in sector Gamma.\nWe have 4 sharks in sector Gamma.\n\nSince the problem domain is limited, we can possible fix that by adding a constrain that force the number of sharks in sector Gamma to be greater than 4.\n\n@constraint(m,s[3]>=5)\n\nThis will result in an answer that that does not violate any of the stated constraints.\n\nWe have 3 fishes in each tank.\nWe have 7 tanks in sector Gamma.\nWe have 5 sharks in sector Gamma.\n\nHowever, this seems like a bit of kludge. The proper way go about it is represent the number of sharks in the each sector as binary array, with only one value set to 1.\n\n# Number of sharks in each sector\n@variable(m, s[i=1:3,j=1:7], Bin)\n\nWe will have to modify our constraint block accordingly\n\n@constraints m begin\n# Constraint 2\nsharks[i=1:3], sum(s[i,:]) == 1\nu_sharks[j=1:7], sum(s[:,j]) <=1 # uniquness\n# Constraint 4\nsum(nt) <= 13\n# Constraint 5\ns[1,2] == 1\nnt[1] == 4\n# Constraint 6\ns[2,4] == 1\nnt[2] == 2\nend\n\nWe invent a new variable array st to capture the number of sharks in each sector. This simply obtained by multiplying the binary array by the vector $$[1,2,\\ldots,7]^\\top$$\n\n@variable(m,st[i=1:3],Int)\n@constraint(m, st.==s*collect(1:7))\n\nWe rewrite our last constraint as\n\n# Constraints 1 & 3\n@NLconstraint(m, st[1]+st[2]+st[3]+n*(nt[1]+nt[2]+nt[3]) == 50)\n\nAfter the model has been solved, we extract our output for the number of sharks.\n\nsharks_in_each_sector=getvalue(st)\n\n\u2026and we get the correct output.\n\nThis problem might have been an overkill for using a full blown mixed integer non-linear optimizer. It can be solved by a simple table as shown in the video. However, we might not alway find ourselves in such a fortunate position. We could have also use mixed integer quadratic programming solver such as Gurobi which would be more efficient for that sort of problem. Given the small problem size, efficiency hardly matters here.\n\n06/12/17\n\n# Reading DataFrames with non-UTF8 encoding in Julia\n\nRecently I ran into problem where I was trying to read a CSV files from a Scandinavian friend into a DataFrame. I was getting errors it could not properly parse the latin1 encoded names.\n\nI tried running\n\nusing DataFrames\ndataT=readtable(\"example.csv\", encoding=:latin1)\n\nbut the got this error\n\nArgumentError: Argument 'encoding' only supports ':utf8' currently.\n\nThe solution make use of (StringEncodings.jl)[https://github.com/nalimilan/StringEncodings.jl] to wrap the file data stream before presenting it to the readtable function.\n\nf=open(\"example.csv\",\"r\")\ns=StringDecoder(f,\"LATIN1\", \"UTF-8\")\nclose(f)\nThe StringDecoder generates an IO stream that appears to be utf8 for the readtable function.", "date": "2021-11-29 17:24:34", "meta": {"domain": "perfectionatic.org", "url": "https://perfectionatic.org/?tag=julia", "openwebmath_score": 0.4335174560546875, "openwebmath_perplexity": 3075.453846992643, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104904802131, "lm_q2_score": 0.8807970858005139, "lm_q1q2_score": 0.8512115437020169}}
{"url": "https://math.stackexchange.com/questions/3145781/can-a-sequence-be-neither-decreasing-nor-increasing/3145807", "text": "Can a sequence be neither decreasing nor increasing?\n\nGiven the definition of a increasing sequence: \"A sequence $$(a_n)_{n\\in\\mathbb{N}}$$ is increasing if for all $$n\\in \\mathbb{N}$$, $$a_n\\le a_{n+1}$$.\"\n\nMy question is: by this definition isn't the sequence $$(1,1,1,1,1,\\dotsc)$$ an increasing sequence then?\n\n\u2022 Yes, of course. Some people will say the sequence is strictly increasing if $a_n<a_{n+1}$ for all $n$. ... The answer to your title question is of course \u2014 just take a sequence like $1,2,1,2,1,2,\\dots$. \u2013\u00a0Ted Shifrin Mar 12 at 22:21\n\u2022 If you are upset by the language that \"increasing\" should in your opinion be reserved only for strict inequality between terms, then you might prefer to instead use the term \"weakly increasing\" instead. Also, before you ask, yes, the definitions work out then that a constant sequence like $1,1,1,\\dots$ is simultaneously an increasing sequence and a decreasing sequence and it is easy to prove that any sequence which is simultaneously both must be a constant sequence. \u2013\u00a0JMoravitz Mar 12 at 22:27\n\nDefinition: A sequence $$(a_n)$$ is increasing if $$a_n \\le a_{n+1}$$ for all $$n$$,\n\n(or something similar), then the sequence $$(1,1,1,\\dotsc)$$ is an increasing sequence. It is increasing precisely because it satisfies the property which defines an increasing sequence (per the definition given).\n\nThis may not feel right, as our intuition from natural language is that this sequence is constant, and therefore not increasing. There are two bits of advice that I would give (the first when you have to deal with other people's exposition, the second when you are doing your own work):\n\n1. Just get used to it. There are many terms in mathematics which come from natural language, but which don't mean the same thing in mathematics as they do in vernacular English. The terms \"open\" and \"closed\" are a big bugaboo for intro to topology students, for example. As a mathematician, you should learn to get used to words that are given technical definitions which contradict plain English interpretation (and, indeed, may contradict each other, as different authors may define the same term differently.\n\n2. Find another term. In his series on analysis, Simon comments early on that weak vs strict inequalities are possibly ambiguous. He therefore declares that, in his text, \"increasing\" always means \"nondecreasing\" (and, if I recall correctly, he also declares that \"positive\" means \"nonnegative\"; the point is that he doesn't want to have to say things like \"nonnegative nondecreasing\"). A similar strategy may help you: if you don't want to call a constant sequence \"increasing\", use the words \"nondecreasing\" (for a sequence where $$a_n \\le a_{n+1}$$), and use \"increasing\" (or even \"strictly increasing\") for sequences which satisfy $$a_n < a_{n+1}$$.\n\nSimon, Barry, Real analysis. A comprehensive course in analysis, part 1, Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-1099-5/hbk). xx, 789\u00a0p. (2015). ZBL1332.00003.", "date": "2019-11-20 19:24:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3145781/can-a-sequence-be-neither-decreasing-nor-increasing/3145807", "openwebmath_score": 0.8354408740997314, "openwebmath_perplexity": 365.33212518095405, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104914476339, "lm_q2_score": 0.8807970795424087, "lm_q1q2_score": 0.8512115385062199}}
{"url": "https://physics.stackexchange.com/questions/359583/unitary-time-condition", "text": "# Unitary time condition\n\nI have a confusion with regards to the principle of QM that states that time evolution must be unitary. In particular, given that states transform through time as $|\\Psi(t)\\rangle = U(t)|\\Psi(0)\\rangle$; does the condition: $$\\langle\\Phi(0)|\\Psi(0)\\rangle=0 \\ \\implies \\ \\langle\\Phi(t)|\\Psi(t)\\rangle=0$$\n\nimply that $U$ must be unitary, or is it imposed on $U$?\n\nI realize that with the before mentioned condition it is posible to prove that for two orthonormal memebers of the basis, $i,j$: $\\langle i| U^{\\dagger}U | j \\rangle = 0$. Is it posible to prove that for the same member the product is 1?\n\n\u2022 If $U$ is unitary then $U^{\\dagger}$ is the inverse of $U$ by definition. Sep 27 '17 at 14:10\n\u2022 Yes, but the question is about how to show that, that is the case Sep 27 '17 at 14:18\n\u2022 Since $\\vert \\Psi(t)\\rangle = U\\vert\\Psi(0)\\rangle$ then clearly $$\\langle \\Phi(t)\\vert \\Psi(t)\\rangle = \\langle \\Phi(0)\\vert U^\\dagger\\,U\\vert \\Psi(0)\\rangle = \\langle\\Phi(0)\\vert\\Psi(0)\\rangle$$ valid $\\forall t$ implies $U^\\dagger U=\\hat 1\\, \\forall t$ as well, provided your kets are arbitrary. (Hopefully this should be enough.) Sep 27 '17 at 14:32\n\u2022 The condition of orthogonality preservation you stated alone would also allow for operators that satisfy $U^\\dagger U = c \\cdot 1$ with any constant $c$. Usually one says that unitarity comes from norm preservation, i.e. $< \\psi(t) ,\\psi(t)> = <\\psi(0),\\psi(0)>$.\n\u2013\u00a0Luke\nSep 27 '17 at 15:23\n\nIt is an interesting elementary problem. After I while I proved the following proposition where I use $A^*$ for the adjoint of $A$.\n\nProposition. Let $U: H \\to H$ be a bounded operator over a Hilbert space $H$. The following conditions are equivalent.\n\n(a) For $x,y \\in H$,$\\quad$ $\\langle x|y\\rangle =0$ implies $\\langle Ux|Uy\\rangle =0$\n\n(b) $U^*U = cI$ for some real $c\\geq 0$.\n\nBefore proving the statement I observe that even if $c=1$, $U$ is not necessarily unitary, because unitarity is $U^*U=UU^*=I$. And here $UU^*=I$ generally fails when $H$ is infinite dimensional (otherwise it is trivially true as a consequence of $U^*U=I$). For $c=1$, $U$ is an isometry not necessarily surjective.\n\nProof. It is obvious that (b) implies (a), so we prove that (a) implies (b). Condition (a) can be rephrased as $y \\perp x$ implies $y \\perp U^*Ux$. As a consequence $U^*Ux \\in \\{\\{x\\}^\\perp\\}^\\perp$ which is the linear span of $x$. In other words $U^*U x = \\lambda_x x$ for some $\\lambda_x\\in \\mathbb C$. My goal is now proving that $\\lambda_x$ does not depend on $x$.\n\nTo this end, consider a couple of vectors $x \\perp y$ with $x, y \\neq 0$. Using the argument above we have $$U^*U x = \\lambda_x x\\:,\\quad U^*U y = \\lambda_y y\\:, \\quad U^*U (x+y) = \\lambda_{x+y} (x+y)\\:.\\tag{1}$$ Linearity of $U^*U$ applied to the last identity leads to $$U^*Ux + U^*Uy = \\lambda_{x+y}x + \\lambda_{x+y}y\\:,$$ namely $$U^*Ux- \\lambda_{x+y}x = -(U^*Uy- \\lambda_{x+y}y)\\:\\:.$$ Exploiting the first two identities in (1) we get $$(\\lambda_x- \\lambda_{x+y})x = -(\\lambda_y- \\lambda_{x+y})y\\:.$$ Since $x \\perp y$ and $x,y \\neq 0$, the only possibility is that $$\\lambda_x = \\lambda_{x+y} = \\lambda_y\\:.$$ So a couple of orthogonal non-vanishing vectors has the same $\\lambda_x$.\n\nTo conclude consider a Hilbert basis $\\{x_n\\}$ of $H$ so that, if $z\\in H$, $$z = \\sum_n c_n x_n \\tag{2}$$ for complex numbers $c_n$. Since $U^*U$ is continuous ($U$ is bounded), $$U^*Uz = \\sum_n c_n U^*Ux_n = \\sum_n c_n \\lambda_{x_n}x_n\\tag{3}$$ But we know from the previous argument that $\\lambda_{x_n} = \\lambda_{x_m}$ so that, indicating with $c$ the common value of the $\\lambda_{x_n}$, (3) can be rewritten as $$U^*Uz = \\sum_n c_n cx_n = c\\sum_n c_n x_n = cz\\:.$$ Since $z\\in H$ was arbitrary, we have found that $$U^*U=c I\\:.$$ Taking the adjoint of both sides we obtain $c=\\overline{c}$ so that $c$ is real. Finally, $$0 \\leq \\langle Ux | Ux\\rangle = \\langle x| U^*U x\\rangle = c \\langle x| x \\rangle$$ so that $c\\geq 0$. QED\n\n\u2022 This is nice but does the part \"Is it posible to prove that for the same member the product is 1?\" not imply $c=\\lambda_x=1$? Sep 27 '17 at 19:32\n\u2022 I do not understand, we already know that $U$ such thst $U^*U= cI$ satisfies the initial hypotheses also for $c\\neq 1$. If it were possible to prove that $c=1$ we would exclude this case that we know exists. Sep 27 '17 at 19:40\n\u2022 Something's clearly throwing me off here. I'll come bac k to in a couple of days. Sep 27 '17 at 19:57", "date": "2022-01-18 04:08:50", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/359583/unitary-time-condition", "openwebmath_score": 0.9702175259590149, "openwebmath_perplexity": 130.5545298910354, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9664104924150546, "lm_q2_score": 0.880797071719777, "lm_q1q2_score": 0.8512115317984478}}
{"url": "http://math.stackexchange.com/questions/421760/combinations-question-why-is-my-approach-wrong", "text": "# Combinations question - why is my approach wrong?\n\nI'm learning Permutations and Combinations and while trying to solve this simple question, I stuck on the solution:-\n\nFrom a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?\n\nThe solution was to consider all three possibilities one by one- 3 men and 2 women, 4 men and 1 women, all 5 men. Calculate the combinations for each case and add them.\n\nBut I was trying it from a different approach- Firstly select 3 men from 7 (gives 35 combinations) and multiply it with the combinations of selecting 2 more committee members from 10 remaining members. So my answer came out to be 35 x 45 = 1575. But the answer is 756 which is not even a multiple of 35. So is my approach wrong? Why so?\n\n-\n\nIf you first select 3 men and then fill up with two arbitrary people, then you count each combination with exactly three men once, but others are counted repeatedly. For example, you count all-male committees 10 times, once for each way to choose the three first men in retrospect from the final five men.\n\nSmaller numbers example: Given John, Jack and Jill, form a committee of two people with thge constraint that at least one is a man. Actually, there are three such ocmmittees possible (becaus ethe constraint is no constraint at all). But your method would give four: First select John as male, then select either Jack or Jill as second. Or first select Jack as male, then select John or Jill as second. The committee \"Jack and John\" is counted twice by you.\n\n-\nI don't understand your first paragraph, but your second paragraph, with the simpler example, was very helpful. Thanks. \u2013\u00a0 a.real.human.being Oct 5 '14 at 1:00\n\nYou\u2019re counting every committee with $4$ men $\\binom43=4$ times: if the men are A, B, C, and D, and the woman is F, you\u2019re counting it once with A, B, and C as the first $3$ and D and F as the extra $2$, once with A, B, and D as the first $3$ and C and F as the extra $2$, once with A, C, and D as the first $3$ and B and F as the extra $2$, and once with B, C, and D as the first $3$ and A and F as the extra $2$.\n\nSimilarly, you\u2019re counting every committee with $5$ men $\\binom53=10$ times: if the men are A, B, C, D, and E, you\u2019re counting the committee once with A, B, and C as the first $3$ and D and E as the extra $2$, and so one for every possible $3$-$2$ split of the five men.\n\nOnly the committees of $3$ men and $2$ women are counted correctly, exactly once each.\n\nSince some committees are counted once, some four times, and some ten times, you don\u2019t even get an answer that\u2019s related to the correct one in some very simple way, as you would if, for example, you counted every committee twice. From the correct answer you know that there are $525$ committees with $3$ men, $210$ with $4$ men, and $21$ with $5$ men, and sure enough,\n\n$$525+4\\cdot210+10\\cdot21=1575\\;,$$\n\n$$\\binom 73\\cdot\\binom 62+\\binom 74\\cdot\\binom 61+\\binom 75\\cdot\\binom 60=35\\cdot15+35\\cdot6+21\\cdot1=756$$", "date": "2015-01-25 16:52:50", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/421760/combinations-question-why-is-my-approach-wrong", "openwebmath_score": 0.7000324130058289, "openwebmath_perplexity": 340.0619740908281, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357184418848, "lm_q2_score": 0.8670357718273068, "lm_q1q2_score": 0.8511999863696952}}
{"url": "http://pziq.umood.it/exponential-functions-game.html", "text": "# Exponential Functions Game\n\nThen we explore additional applications by looking at Half Life, Compounding Interest and Logistic Functions. The basic exponential function is f(x) = b^x, where the b is your constant, also called base for these types of functions. Graphing Rational Functions 23. 9(C) - write exponential functions in the form f(x) = abx (where b is a rational number) to describe problems arising from mathematical. Grade 11 maths Here is a list of all of the maths skills students learn in grade 11! These skills are organised into categories, and you can move your mouse over any skill name to preview the skill. Derivatives Of Exponential, Trigonometric, And Logarithmic Functions Exponential, trigonometric, and logarithmic functions are types of transcendental functions; that is, they are non-algebraic and do not follow the typical rules used for differentiation. Exponential and Logarithmic Functions Worksheets October 3, 2019 August 28, 2019 Some of the worksheets below are Exponential and Logarithmic Functions Worksheets, the rules for Logarithms, useful properties of logarithms, Simplifying Logarithmic Expressions, Graphing Exponential Functions, \u2026. Exponential functions. Exponential functions arise in many applications. 01) to the 12\ud835\ude35 power, \ud835\ude3a = (1. I can apply exponential functions to real world situations. Logarithm property. About this webmix : IXL-Evaluate an exponential Graphs of Exponential Function Comparing growth rates. Exponential and Logarithm functions are very important in a Calculus class and so I decided to have a section devoted just to that. Derivative exponential : To differentiate function exponential online, it is possible to use the derivative calculator which allows the calculation of the derivative of the exponential function. True Note that the t values are equally spaced and the ratios 3 0 0 1 4 7 = 1 4 7 7 2. Exponential FUNctions - Graphing; Exponential FUNctions - Growth and Decay; Puzzle Pizzazz: Volume 1; Exponential Function FUN with The Money Project! February (3) January (3) 2017 (7) December (1) November (6) 2013 (1) February (1) 2012 (10). We will begin by drawing up a curve for y = 10 x. it Game PIN: 9834669 Type here to search a hyperbolic parabaloid a discrete function. 01) to the 12\ud835\ude35 power, \ud835\ude3a = (1. Exponential Functions. Creating Exponential Functions to Match Given Behaviors-Part 1 2. Exponential growth is a convenient cover for the core problem with current idle games. 3 Integrals of trigonometric functions and applications: Online: Proofs of some trigonometric limits: Online: New functions from old: Scaled and shifted functions: Case study: Predicting airline empty seat volume: Download data. This game includes 24 different cards, 12 exponential functions written in function notation, and 12 corresponding graphes. In other words, it is possible to have n An matrices A and B such that eA+B 6= e eB. About this webmix : IXL-Evaluate an exponential Graphs of Exponential Function Comparing growth rates. EXPONENTIAL REGRESSION. The two types of exponential functions are exponential growth and exponential decay. We use the command \"ExpReg\" on a graphing utility to fit an exponential function to a set of data points. 1 Warm-up: Math Talk: Exponents (5 minutes) 4. Exponential Functions. (b) Using the functions in part a, \ufb01nd all x such that |f(x)| \u2264 2. M&M Lab (Exponential Growth and Decay) Part I: Modeling Exponential Growth M&M Activity The purpose of this lab is to provide a simple model to illustrate exponential growth of cancerous cells. You can see this when you touch a point with the same y value as the red point. To help us grasp it better let us use an ancient Indian chess legend as an example. - Ignasi Sep 9 '14 at 14:21 @user143462: for your questions about pgfplots it could be good to take a look at its documentation. Exponential 7. Exponential Function Matching Game These tasks were taken from the GSE Frameworks. In this case, f(x) is called an exponential growth function. > Introduction to functions > Linear functions > Polynomial functions > Exponential and logarithm functions > Trigonometric functions > Hyperbolic functions > Composition of functions > Inverse functions > Sigma notation > Arithmetic and geometric progressions > Limits of sequences > The sum of an infinite series > Limits of functions. Unit #5 \u2013 Exponential and Logarithmic Functions \u2013 Review \u2013 Solutions. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Write an exponential growth function to model the situation. Exponential Function Quizizz - Game Code: 158847. It is used to represent exponential growth, which has uses in virtually all science subjects and it is also prominent in Finance. Exponential Functions. LEARNING GOALS. 02) to the \ud835\ude35 power, \ud835\ude3a = (0. The name of this book is Al-Jabr wa'l muqabalah. While there are many ways to show division by 2, this machine is a bit lazy and will always opt for the easiest function. Finding Real Zeros of Polynomial Functions ; answers to algebra multiplication diamond problems ; math with pizzazz test of genius answers ; rational equations and functions ; solve algebra problems ; Write the algebraic equation which can be used to find the exact solution of:log5 (x+8) + log5 (x-5) = 2 ; 8th grade math online work book and. The value of the function. does not recognize the exponential function exp(x) This will work with all built in functions and predefined data. Growing by Leaps and Bounds Culminating Task/Project Teacher Edition Concept 1 Resources recommended for Math Support Concept 2 Resources recommended for Math Support Concept 3 Resources recommended for Math Support. You need to be a group member to play the tournament. Differentiating the logarithmic function, Derivatives of exponential functions and Applications which shows how logarithms are used in calculus. Finding the equation of an exponential function from the graph Worked example 17: Finding the equation of an exponential function from the graph Use the given graph of $$y = -2 \\times 3^{(x + p)} + q$$ to determine the values of $$p$$ and $$q$$. A comprehensive database of more than 11 exponential quizzes online, test your knowledge with exponential quiz questions. This game requires students to represent the exponential parent function y=2^x numerically, algebraically, graphically and verbally. Exponential Function Reference. Logarithm and logarithm functions This is a very important section so ensure that you learn it and understand it. Use the properties of exponents to interpret expressions for exponential functions. In mathematics, specifically in category theory, an exponential object or map object is the categorical generalization of a function space in set theory. You can see this when you touch a point with the same y value as the red point. The population of a town is decreasing at a rate of 1. y=a^x? Why is the natural exponential function y=e^x used more often in calculus than the other exponential functions y=a^x? I don't really understand \"e\", just that it is log base e. How have video game consoles changed over time? Students create exponential models to predict the speed of video game processors over time, compare their predictions to observed speeds, and consider the consequences as digital simulations become increasingly lifelike. 1 Trigonometric functions, models, and regression: 9. Its population increases 7% each year. The exponential function formula to calculate e x is provided below. Exponents are used in Computer Game Physics, pH and Richter Measuring Scales, Science, Engineering, Economics, Accounting, Finance, and many other disciplines. Notes: Compound Interest. Engaging math & science practice! Improve your skills with free problems in 'Write and Apply Exponential and Power Functions' and thousands of other practice lessons. Exponential functions always have some positive number other than 1 as the base. Explore the graph of an exponential function. So you really draw the function $\\frac{1}{25} e^x$ (using an equi-scaled graph) which leads to the maximal curvature in $\\frac{-\\ln 2}{2}+\\ln 25\\simeq 2. The real exponential function : \u2192 can be characterized in a variety of equivalent ways. Follow these steps to convert standard numbers into scientific notation. Connect Math Support Subject: (Choose one) I forgot my login information I am having a technical problem with Connect Math I just need help with Connect Math I have a suggestion or comment regarding Connect Math I would like to get more information about Connect Math Other. Syntax : exp(x), where x is a number. Exponential Functions. Exponential & Logarithmic Functions. In words: 8 2 could be called \"8 to the second power\", \"8 to the power 2\" or simply \"8 squared\". Precalculus Made Simple: Exponential and Quadratic Functions 4. We look at the difference between growth & decay. Parent Function for Exponential Decay Functions The function f ()xb= x, where 0 1,< 0. Population Problems 4. A decreasing exponential function has a base, b<1, while an increasing exponential function has a base b>1 as you can verify on the graph above. If you're seeing this message, it means we're having trouble loading external resources on our website. Heearnsnointerest,but deposits$100everyyear. Below is a quick review of exponential functions. There's a perfectly good pow function defined in the header. In order to master the techniques explained here it is vital that you undertake plenty of. Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Obviously, this function is descending from some initial value at t=0 down to zero as time increases towards infinity. 97) to the \ud835\ude35 power, \ud835\ude3a = (1. Exponential functions are the name of the game. Exponents are used in Computer Game Physics, pH and Richter Measuring Scales, Science, Engineering, Economics, Accounting, Finance, and many other disciplines. & & EXPONENTIAL GROWTH FUNCTIONS WILL ALWAYS ULTIMATELY GROW FASTER THAN LINEAR FUNCTIONS!! ) Example:) LinearLarry&beginswith$10,000inhissavingsaccount. The structure of a row game gives students an opportunity to construct viable arguments and critique the reasoning of others (MP3). values for x (the. Integrals with Trigonometric Functions. Example 1. The \"after\" shape is not filled, and is traced by P'. The graph below shows the exponential functions corresponding to these two geometric sequences. Find an exponential function of the form P(t)=y0e^kt In 1985, the number of female athletes participating in Summer Olympic-Type games was 450. While there are many ways to show division by 2, this machine is a bit lazy and will always opt for the easiest function. Topic: Recognizing linear and exponential functions. An exponential function is the inverse of a logarithm function. 2 Activity: Evaluating and Describing Functions (20 minutes) 4. Six real word examples of exponential growth in a Powerpoint slide show (3. Using the Properties of Exponents I 7. The first choice is to get a penny on day one two pennies on day two four pennies on day three and so on doubling every day. Use the properties of exponents to interpret expressions for exponential functions. Algebra Worksheets, Quizzes and Activities. For example, in DC Heroes, the first game to use. Play this game to review Algebra I. If the base, b=1 then the function is constant. This site provides a web-enhanced course on computer systems modelling and simulation, providing modelling tools for simulating complex man-made systems. I want to sort some exponential numbers like {2^2,3^3,4^2} and i want The result {2^2,4^2,3^3}. 97) to the \ud835\ude35 power, \ud835\ude3a = (1. Follow these steps to convert standard numbers into scientific notation. Exponential functions follow all the rules of functions. Improve your math knowledge with free questions in \"Match exponential functions and graphs\" and thousands of other math skills. Exponential growth is the increase in number or size at a constantly growing rate. (Note that only b is raised to the power x; not a. Distinguish between situations that can be modeled with linear functions and with exponential functions. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Provides a complete web based educational environment for K-12 and Higher-Education mathematics, accounting, statistics, and chemistry. This means that a loss function that is such a cross entropy is a so-called matching loss function, which is convenient for optimization. The structure of a row game gives students an opportunity to construct viable arguments and critique the reasoning of others (MP3). Exponential Expressions. Teaching Notes and Tips We envision this template as an outline for teachers of introductory geology classes (physical, historical, environmental, etc. LG 2 \u2013 The Logarithmic Function \u2013 Solutions. These functions cannot be used with complex numbers; use the functions of the same name from the cmath module if you require support for complex numbers. Play this quiz called Linear and Exponential Match and show off your skills. Our Utility Function is the Exponential Utility Function which is So, lets plugin this function to the above equation, after simplifying, we get, Here, W is the Winning amount from the lottery, L is the loss amount from the lottery, and CE is the certainty equivalent. 2 \u2013 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two\u2026. Exponential function definition, the function y = ex. During each time interval of a fixed length, the population is multiplied by a certain constant amount. Aside: if you try to use ^ as a power operator, as some people are wont to do, you'll be in for a nasty surprise. Represent functions using function notation. Exponential Function Matching Game These tasks were taken from the GSE Frameworks. Exponential functions contain a variable written as an exponent, such as y = 3 x. Learn vocabulary, terms, and more with flashcards, games, and other study tools. x 2 b a Equation for the axis of symmetry of a parabola x or 2(6 3) 1 a 3 and b 6 The equation of the axis of symmetry is x 1. Exponential Functions. In this case, f(x) is called an exponential growth function. A simple way to know differentiate between the two is to look at the output values when plugging in a number for an unknown variable. 1 Multiplication Properties of Exponents. About this webmix : IXL-Evaluate an exponential Graphs of Exponential Function Comparing growth rates. org, Yahoo Finance, and Irrational Exuberance. graph of increasing exponential function going through point 0, 1. Exponential function exp(x) not recognized. You can see this when you touch a point with the same y value as the red point. 5% per year. Students have fun making a poster and using multiple representations. In y 3x2 6x 4, a 3 and b 6. The Exponential Curve The purpose of this blog is to help generate and share ideas for teaching high school math concepts to students whose skills are below grade level. Write your answers in interval notation and draw them on the graphs of the functions. For fx( ) 2x (i) The base of is (ii) The exponent of is (iii) What is varying in the function ? (iv) What is constant in the function ? 2. y for e^(-0. Precalculus Made Simple: Exponential and Quadratic Functions 4. 5% per year. Explore the graph of an exponential function. They are given several exponential scenarios to choose from such as a basketball bracket, a cockroach infestation, a zombie takeover, friendship bread, chain letters and more. The name was what fans called the game system for DC Heroes, which was later used for Underground (1993). Learn vocabulary, terms, and more with flashcards, games, and other study tools. Below is a quick review of exponential functions. Also, the a value can tell us if the exponential curve is concave up (opening upwards) or concave down (opening downwards). If you think about it, having a negative number (such as \u20132) as the base wouldn't be very useful, since the even powers would give you positive answers (such as \"(\u20132) 2 = 4\") and the odd powers would give you negative answers (such as \"(\u20132) 3 = \u20138\"), and what would you even do with the powers that aren't. Common Core. Writing exponential functions from tables Our mission is to provide a free, world-class education to anyone, anywhere. Priming for the Laws of Exponents 3. Step 4 Cut the two stacked sheets in half, placing the resulting. Except where otherwise noted, content on this wiki is licensed under the following license: CC Attribution-Noncommercial-Share Alike 4. Please explain (and show your work) in the simplest way possible. A simple way to know differentiate between the two is to look at the output values when plugging in a number for an unknown variable. a is any value greater than 0. x -2 -1 0 1 2 y -6 -6 -4 0 6. Show students how to simplify exponential functions, including those involving multiplying (x\u00b2*x\u00b3) and dividing terms (x\u00b3/x\u00b2) with exponents or apply exponents to a term that already. If you think about it, having a negative number (such as \u20132) as the base wouldn't be very useful, since the even powers would give you positive answers (such as \"(\u20132) 2 = 4\") and the odd powers would give you negative answers (such as \"(\u20132) 3 = \u20138\"), and what would you even do with the powers that aren't. It's the exclusive-or (XOR) operator (see here). 21; Section B and Throwback 32 & 33 was due 03. The two types of exponential functions are exponential growth and exponential decay. Finding the equation of an exponential function from the graph Worked example 17: Finding the equation of an exponential function from the graph Use the given graph of $$y = -2 \\times 3^{(x + p)} + q$$ to determine the values of $$p$$ and $$q$$. Thanks for contributing an answer to Signal Processing Stack Exchange! Please be sure to answer the question. Scroll down the page for more examples and solutions on the graphs of exponential. Try Remote Buzzer-Mode for even more fun!. Other examples of exponential functions include: $$y=3^x$$ $$f(x)=4. Exponential functions. Play this game to review Algebra I. To determine how the given function grows over an interval of length 3, determine the value of f at each endpoint of that interval. 90)^t Find the initial value of the car and the value after 10 years. 27: Derivatives of Exponential Functions: Section 5. This allows you to make an unlimited number of printable math worksheets to your specifications instantly. Syntax : exp(x), where x is a number. Parent Function for Exponential Decay Functions The function f ()xb= x, where 0 1,< 0. Finding an Exponential Function through 2 points. Then the following properties are true: 1. , screenshot, dump, ads, commercial, instruction. x \u20132 \u20131 0 1 2 y \u20136 \u20136 \u20134 0 6. Intro Lesson to Exponential functions. 25 and the exponential component is 10 taken to the negative 2 power. A comprehensive database of more than 11 exponential quizzes online, test your knowledge with exponential quiz questions. A logical value that indicates which form of the exponential function to provide. The function cards include x-axis reflections and/or vertical translations. Classwork Estimation 180 - Day 16 (Andrew Stadel) Warm Up Card Sort: Exponentials (Desmos) Practice Answers Link to Answers Standards Common Core HSF. c ccsc welc I sout I o Digit Mail MYF Knov Play Skip Answers 1046 AM 4/22/2020 A exponential growth exponential decay kahoot. Exponential. Syntax : exp(x), where x is a number. Exponential definition, of or relating to an exponent or exponents. Here is the calculator itself. In mathematics, the exponential function is the function whose derivative is equal to itself. It is a nice way to introduce the concept of Exponential Functions and start the class differently than the norm. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. js can automatically use backoff strategies to retry requests with the autoRetry parameter. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). I have one standard for linear equations and one for exponential equations. Please help me on this Algebra 1 problem involving exponential functions (exponential growth and decay). Exponential & Logarithmic Functions. This section contains the following sections. Although exponential growth is always ultimately limited it is a good approximation to many physical processes in the Earth system for finite time intervals. In the first example, the decimal component is 6. Linear Quadratic Exponential y mx \" by a x 2 \"bx \" cy a b x Use the data in the table to describe how the software\u2019s cost is changing. I can model real world situations with exponential functions. Top synonym for exponentially (another word for exponentially) is rapidly. Graph exponential functions. 2 Create and graph equations in two variables to represent linear, exponential, and quadratic relationships between quantities. Herb Gross introduces the exponential function, discussing the rate of growth of the output for each unit change in input and motivates the definition of the meaning of fractional exponents. Play this game to review Algebra I. \u02c6\u02d9\u02dd \u02c6\u02da \u02db \u02d8 \u02c7 \u02d8 Solving Exponential and Logarithmic Equations 1. Exponential Functions - Application Problem - National Debt. EXPONENTIAL REGRESSION. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Now it was time to see how the economic concepts. An exponential function is a function which takes a point x and returns the value a to the power x. Exponential definition, of or relating to an exponent or exponents. In this example: 8 2 = 8 \u00d7 8 = 64. Creating Exponential Functions to Match Given Behaviors-Part 2 5. Determine whether each table or rule represents a linear or an exponential function. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Making statements based on opinion; back them up with references or personal experience. Concept Summary Exponential Function Family You can apply the four types of transformations\u2014stretches, compressions, reflections, and translations\u2014to exponential functions. graph of increasing exponential function going through point 0, 1. The #1 Jeopardy-style classroom review game now supports remote learning online. You don't write a function for this (unless you're insane, of course). , population growth). This exercise practices graphing exponential functions and find the appropriate graph given the function. Graphing Exponential Functions The graph of a function y = abx is a vertical stretch or shrink by a factor of \u2223 a \u2223 of the graph of the parent function y = bx. 5% per year. Nothing like a good criminal investigation to liven up geometry! In this project, students will work in teams to investigate the culprit of six fictional thefts. It began at a length of 6 in and grew at a rate of 14% a week. Writing Equations \u2013 Key. Exponential Growth/Decay Graph Demonstration. Click below for lesson resources. Connect Math Support Subject: (Choose one) I forgot my login information I am having a technical problem with Connect Math I just need help with Connect Math I have a suggestion or comment regarding Connect Math I would like to get more information about Connect Math Other. Learn math anywhere on your mobile device or tablet. I have one standard for linear equations and one for exponential equations. In exponential decay, the total value decreases but the proportion that leaves remains constant over time. Priming for the Laws of Exponents 3. Then we explore additional applications by looking at Half Life, Compounding Interest and Logistic Functions. & & Exponential)Ellie&beginswith100inhersavingsaccount,butearnsannual. Exponential functions always have some positive number other than 1 as the base. EXPONENTS MADE VERY SIMPLE TO UNDERSTAND. About this webmix : Glencoe Sequences Quiz Glencoe Exponential Function Write answers on a sep page Exponents Game - Otter. The Mayfair Exponential Game System or MEGS is a rules system developed for role-playing games. If f is a function, x is the input (an element of the domain), and f(x) is the output (an element of the range). The parameter value. - A 'positive constant base raised to a variable exponent' = (constant) (variable) is an 'Exponential function'. \u02c6\u02d9\u02dd \u02c6\u02da \u02db \u02d8 \u02c7 \u02d8 Solving Exponential and Logarithmic Equations 1. An exponential function is defined for every real number x. This repeated multiplication can be expressed using exponential functions. Modeling with exponential and power law functions in Geosciences. Engaging math & science practice! Improve your skills with free problems in 'Write and Apply Exponential and Power Functions' and thousands of other practice lessons. Exponential Growth/Decay Graph Demonstration. Date: Topic / Lesson and Handouts: Homework, and Other Resources: Mar. Students play a generalized version of connect four, gaining the chance to place a piece on the board by solving an algebraic equation. I am having a hard time researching how to handle summations of functions with exponential growth or decay. Simplify the expression:log (base 2) 8 , Write the equation log (base 3) 729 = 6 in exponential form. This calculus video tutorial explains how to find the derivative of exponential functions using a simple formula. 1 Warm-up: Math Talk: Exponents (5 minutes) 4. Categories with all finite products and exponential objects are called cartesian closed categories. Next Chapter: EXPONENTIAL AND LOGARITHMIC FUNCTIONS. Question 1109486: The dollar value v(t) of a certain car model that is t years old is given by the following exponential function. Logarithm property. LG 1 \u2013 Exponential Functions \u2013 Blank Copy. 87 which is already approximately 3 as you want. Simplify the expression: log (base 2) 8 , Write the equation log (base 3) 729 = 6 in exponential form. How to Compare Linear, Exponential, and Quadratic Functions Comparing Functions By its Intercepts x-intercept: where y=0 Locate on the graph or table OR Substitute 0 in for y and solve for x y-intercept: where x=0 Locate on the graph or table OR Substitute 0 in for x and solve for y By Rate of Change \"Which one grows the fastest?\" ' *Remember. Student Instructions Students will keep track of all information for the direct instruction in notability with color -coded examples of each part of the exponential function. A comprehensive database of more than 11 exponential quizzes online, test your knowledge with exponential quiz questions. Finding an exponential function given its graph. ) Exponential Functions. The library includes a great number of useful mathematical functions for manipulating floating point numbers. Computer Software Year 0 1 2 3. First Name:. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. 01) to the 12\ud835\ude35 power, \ud835\ude3a = (1. To help us grasp it better let us use an ancient Indian chess legend as an example. Function Machine Division: If you think the numbers are being divided by 2, simply enter \u00f72. a) Multiply b) Add c) Subtract 500 Answer from Exponential Graphs and Equations :. Logarithm Worksheets Logarithms, the inverse of the exponential function, are used in many areas of science, such as biology, chemistry, geology, and physics. KEYWORDS: Course materials, Linear functions, Least Squares Line, Linear Modeling, Statistics and probability, Non-linear functions, Managing money with the exponential function, The logistic function \u2013 constrained growth and decay. If you think about it, having a negative number (such as -2) as the base wouldn't be very useful, since the even powers would give you positive answers (such as \"(-2) 2 = 4\") and the odd powers would give you negative answers (such as \"(-2) 3 = -8\"), and what would you even do with the powers that aren't. Inverse Trig Functions 21. Characteristics of Graphs of Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. Exponential functions always have some positive number other than 1 as the base. Exponential Functions. An introduction page sets out some basic information about exponential functions leading to a definition of the exponential function. You don't write a function for this (unless you're insane, of course). In this case, f(x) is called an exponential growth function. I added a little more detail as I am making it a major grade and needed a clear rubric and wanted to try and add a little more detail for the students. The chip level is the domain of Moore's Law, as noted. I know that simple summations can be calculated as follows:$$\\sum_{n=1}^{50} n = \\frac{n(n+1)}{2}$$How do you approach problems of exponential decay or growth? Consider the following example:$$\\sum_{n=1}^{50} e^{-0. Classroom Work for Exponential Functions. Introduction to Exponential Functions. If they find a match, YAY! If not, they have to turn them back over and lose. Without introducing a factor to suppress it, exponential growth is an infectious disease doctor's. Farina explained that an exponential function exhibits a rapid growth of a given variable, and in her view, the growth of prices in the economy could show exponential growth, if analyzed through a long period of time. An exponential _____ function is a function of the form f(x) = ab x where a > 0 and 0 < b < 1. Fractional exponent. Play this game to review Algebra I. About this webmix : IXL-Evaluate an exponential Graphs of Exponential Function Comparing growth rates. l 9 2A nl4lg rji 8g yh3t LsS tr RelsCeUr kv YeDd5. , Write the equation log 1000 = 3 in exponential form. Exponential functions are closely related to geometric sequences. Students will explore and interpret the characteristics of functions, using graphs, tables, and simple algebraic techniques. Concept Summary Exponential Function Family You can apply the four types of transformations\u2014stretches, compressions, reflections, and translations\u2014to exponential functions. Hence the range of function f is given by y > 0 or the interval (0 , +\u221e). We will begin by drawing up a curve for y = 10 x. As functions of a real variable, exponential functions are uniquely characterized by the fact that the growth rate of such a function (that is, its derivative) is directly proportional to. Here is the entire project for FREE if you are interested.$500 Question from Exponential Graphs and Equations In tables of exponential functions, you _____ by the same amount each time to get to the next number. By definition: log b y = x means b x = y. If you've ever earned interest in the bank (or even if you haven't), you've probably heard of \"compounding\", \"appreciation\", or \"depreciation\"; these have to do with exponential functions. Returns the exponential distribution. alg_1_chapter_6_review_game. This image shows. The following list outlines some basic rules that apply to exponential functions: The parent exponential function f(x) = bx always has a horizontal asymptote at y = 0, except when [\u2026]. Welcome to IXL's year 11 maths page. Click below for lesson resources. Online tutoring available for math help. Implicit in this definition is the fact that, no matter when you start measuring, the population will always take the same amount of time to double. Derivatives Of Exponential, Trigonometric, And Logarithmic Functions Exponential, trigonometric, and logarithmic functions are types of transcendental functions; that is, they are non-algebraic and do not follow the typical rules used for differentiation. js can automatically use backoff strategies to retry requests with the autoRetry parameter. Solving exponential equations using exponent rules. \u02d8 Inverse Properties of Exponents and Logarithms Base a Natural Base e 1. Exponential Function Quizizz - Game Code: 158847. Examples : exp(0), returns 1. Doubling time. Chapter 8 : Exponents and Exponential Functions 8. Graphing Program That Teaches a Thing or Two If you want to know anything about math, statistics, use a grapher, or just simply amuse yourself by strange information about everything, check out Wolfram Alpha. Identify functions using differences or ratios EXAMPLE 2 Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function. 02) to the \ud835\ude35 power, \ud835\ude3a = (0. Is the graph linear, exponential or neither?. It is used to represent exponential growth, which has uses in virtually all science subjects and it is also prominent in Finance. True Note that the t values are equally spaced and the ratios 3 0 0 1 4 7 = 1 4 7 7 2. The two types of exponential functions are exponential growth and exponential decay. Links to the data sets are included in the file. Recall the table of values for a function of the form $f\\left(x\\right)={b}^{x}$ whose base is greater than one. The function exp calculates online the exponential of a number. A geometric sequence is a list of numbers in which each number is obtained by multiplying the previous number by a fixed factor m. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Played 98 times. Exponential functions follow all the rules of functions. Exponential Function Intro \u2013 Key. In other words, it is possible to have n An matrices A and B such that eA+B 6= e eB. Implicit in this definition is the fact that, no matter when you start measuring, the population will always take the same amount of time to double. It is noteworthy for its use of an exponential system for measuring nearly everything in the game. See Also: Equations, Inequalities & Functions, Exponents, Linear Equations, Quadratic. plug x=0 into given function. I have one standard for linear equations and one for exponential equations. The Atmega8 chip, which is now dated, but still supported, does not have enough memory to be able to use the math. f(x) Kahc play Skip Answers 1047 AM 4/22/2020. This algebra activity focuses on exponential functions. Translated by the Desmos localization team into: French: https://teacher. Step 3 Make a table like the one below and record the number of sheets of paper you have in the stack after one cut. alg_1_chapter_6_review_game. A Guide to Algebraic Functions Teaching Approach Functions focus on laying a solid foundation for work to come in Grade 11 and Grade 12. If the base, b=1 then the function is constant. You can also write it verbally. First Name:. This allows you to make an unlimited number of printable math worksheets to your specifications instantly. Students have fun making a poster and using multiple representations. A decreasing exponential function has a base, b<1, while an increasing exponential function has a base b>1 as you can verify on the graph above. Examples include: population growth, economic growth, growth of carbon emissions, Other References. The quantity 1 + r in the exponential growth model y = a(1 + r) t where a is the initial amount and r is the percent increase expressed as a decimal. Student Instructions Students will keep track of all information for the direct instruction in notability with color -coded examples of each part of the exponential function. So this is new. Classroom Work for Exponential Functions. Algebra Worksheets, Quizzes and Activities. Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. Exponential Function REVIEW - Millionaire Game. These are functions of the form: y = a b x, where x is in an exponent (not in the base as was the case for power functions) and a and b are constants. Example 1. ! Graph!each!exponential!function. (1) Exponential Function Poster Activity: This is my very first exponential function activity that I ever created. Teacher Name: undefined undefined Student and Login Information:: Select. On the same day, Checkersville has a population of 70,000 people. Logarithmic Functions Learn the definitions of logarithmic functions and their properties, and how to graph them. It is noteworthy for its use of an exponential system for measuring nearly everything in the game. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Compute and plot the binomial cumulative distribution function for the specified range of integer values, number of trials, and probability of success for each trial. Com provides free math worksheets for teachers, parents, students, and home schoolers. Notice that 'a' cannot be '1' ,. 4 - Solving Log Equations EVEN only (omit 28) Day 5 Exponential. 3 Integrals of trigonometric functions and applications: Online: Proofs of some trigonometric limits: Online: New functions from old: Scaled and shifted functions: Case study: Predicting airline empty seat volume: Download data. First Name:. Practice analyzing the end behavior of two functions that model similar real-world relationship, where one function is exponential and the other is polynomial. Try Remote Buzzer-Mode for even more fun!. However this is often not true for exponentials of matrices. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Properties depend on value of \"a\". Use the properties of exponents to interpret expressions for exponential functions. write and solve exponential and logarithmic equations. But before you take a look at the worked examples, I suggest that you review the suggested steps below first in order to have a good grasp of the general procedure. 3 - Exponential Functions Click here to review the definition of a function. A function in which an independent variable appears as an exponent. Can I convert this exponential function in QGIS with Raster Calculator ? Stack Exchange Network Stack Exchange network consists of 177 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Students can play Memory, Go Fish, Old Made or just use them as flash cards. x -2 -1 0 1 2 y -6 -6 -4 0 6. Exponential functions follow all the rules of functions. 7182818\u2026) is the base of the natural system. js can automatically use backoff strategies to retry requests with the autoRetry parameter. ANSWER The table of values represents a quadratic function. Probably the most important of the exponential functions is y = e x , sometimes written y = exp ( x ), in which e (2. The chip level is the domain of Moore's Law, as noted. Students have fun making a poster and using multiple representations. Com provides free math worksheets for teachers, parents, students, and home schoolers. There follows an interactive graphical page showing graphically the definition of the exponential function. In this case, f(x) is called an exponential growth function. Its population increases 7% each year. Four variables - percent change, time, the amount at the beginning of the time period, and the amount at the end of the time period - play roles in exponential functions. Exponential Function \u2022 A function in the form y = ax - Where a > 0 and a \u2260 1 - Another form is: y = abx + c \u2022 In this case, a is the coefficient \u2022 To graph exponential function, make a table \u2022 Initial Value - - The value of the function when x = 0 - Also the y-intercept. Exponential functions always have some positive number other than 1 as the base. We will discuss in this lesson three of the most common applications: population growth , exponential decay , and compound interest. activities Tarsia Puzzles - With this software you will easily be able to create, print out, save and exchange customised jigsaws, domino activities and a variety of rectangular card sort activities. Writing Numbers Using Scientific Notation. But the effect is still the same. 7182818\u2026) is the base of the natural system. In , the lifetime of a certain computer part has the exponential distribution with a mean of ten years (X ~ Exp(0. This paper analyzes the supply decision of agricultural products based on negative exponential utility function and game analysis. Exponential/Logarithm Test Review Game Jeopardy Template. Exponential functions tell the stories of explosive change. y = y 0 \u00b7 m x. As the exponent is varied, the function output will change. Concept Summary Exponential Function Family You can apply the four types of transformations\u2014stretches, compressions, reflections, and translations\u2014to exponential functions. We hope your visit has been a productive one. Here, y represents the number of people infected and x represents the number of days that have passed since day zero. See more ideas about Exponential functions, Exponential, Algebra. Exponential Growth is a critically important aspect of Finance, Demographics, Biology, Economics, Resources, Electronics and many other areas. I could do exponential functions all year. One Grain of Rice Activity Give them a calendar page and have them fill in the first eight days with the number of grains of rice they would receive if they had made the deal described in the story. In exponential decay, the total value decreases but the proportion that leaves remains constant over time. It's the exclusive-or (XOR) operator (see here). This discussion will focus on solving more complex problems involving exponential functions. However, exponential functions and logarithm functions can be expressed in terms of any desired base b. If you're seeing this message, it means we're having trouble loading external resources on our website. Identify functions using differences or ratios EXAMPLE 2 Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function. 3 Integrals of trigonometric functions and applications: Online: Proofs of some trigonometric limits: Online: New functions from old: Scaled and shifted functions: Case study: Predicting airline empty seat volume: Download data. 4 7 2 6 \u2248 2. To determine how the given function grows over an interval of length 3, determine the value of f at each endpoint of that interval. it Game PIN: 9834669 Type here to search a hyperbolic parabaloid a discrete function. Simplify the expression:log (base 2) 8 , Write the equation log (base 3) 729 = 6 in exponential form. c ccsc welc I sout I o Digit Mail MYF Knov Play Skip Answers 1046 AM 4/22/2020 A exponential growth exponential decay kahoot. 2 Create and graph equations in two variables to represent linear, exponential, and quadratic relationships between quantities. LG 1 \u2013 Exponential Functions \u2013 Blank Copy. Exponential decay is a type of exponential function where instead of having a variable in the base of the function, it is in the exponent. One common example is population growth. Graphing transformations of exponential functions. 2 Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. To solve an exponential or logarithmic word problems, convert the narrative to an equation and solve the equation. Exponential & Logarithm Review. So you really draw the function $\\frac{1}{25} e^x$ (using an equi-scaled graph) which leads to the maximal curvature in $\\frac{-\\ln 2}{2}+\\ln 25\\simeq 2. 2) to the \ud835\ude35/10 power, and classify them as representing exponential growth or decay. The negative exponential utility function has been widely used in the study of the risk preferences of farmers. An exponential function is a Mathematical function in form f (x) = a x , where \"x\" is a variable and \"a\" is a constant which is called the base of the function and it should be greater than 0. Young mathematicians can work through each of the eight worksheets by evaluating functions, applying logarithms, completing logarithmic functions, and building inverse functions. Concept Summary Exponential Function Family You can apply the four types of transformations\u2014stretches, compressions, reflections, and translations\u2014to exponential functions. An exponential function is defined for every real number x. If you're behind a web filter, please make sure that the domains *. A half-exponential function is one which when composed with itself gives an exponential function. In fact, we offer an entire algebra 2 curriculum: fourteen units covering all topics equations, to conic sections, and even trig. The inverse of the exponential function y = a x is x = a y. Logical Functions / 10 M-Files / 11 Timing /11 Mathematical Functions Exponential and Logarithmic Functions / 12 Trigonometric Functions / 12 Hyperbolic Functions / 12 Complex Functions / 13 Statistical Functions / 13 Random Number Functions / 13 Numeric Functions / 13 String Functions / 13 Numerical Methods Polynomial and Regression Functions / 14. Properties of Exponents 4. Explore the graph of an exponential function. 02) to the \ud835\ude35 power, \ud835\ude3a = (0. Find the coordinates of the. Explore Exponential Functions 1. Chapter 8 : Exponents and Exponential Functions 8. exponential definition: The definition of exponential refers to a large number in smaller terms, or something that is increasing at a faster and faster rate. If you need to contact the Course-Notes. You don't write a function for this (unless you're insane, of course). \u02d8 Inverse Properties of Exponents and Logarithms Base a Natural Base e 1. in 1996, about 3650 participated in the Summer Olympics in Atlanta. Make sure you play the html version (not java). Shirly Boots for the original and inspiration. Exponential Excel function in excel is also known as the EXP function in excel which is used to calculate the exponent raised to the power of any number we provide, in this function the exponent is constant and is also known as the base of the natural algorithm, this is an inbuilt function in excel. Concept Summary Exponential Function Family You can apply the four types of transformations\u2014stretches, compressions, reflections, and translations\u2014to exponential functions. Student Instructions Students will keep track of all information for the direct instruction in notability with color -coded examples of each part of the exponential function. This game requires students to represent the exponential parent function y=2^x numerically, algebraically, graphically and verbally. Which statement is true about the functions? The exponential function is generally growing slower than the linear function. About this webmix : IXL-Evaluate an exponential Graphs of Exponential Function Comparing growth rates. True Note that the t values are equally spaced and the ratios 3 0 0 1 4 7 = 1 4 7 7 2. As functions of a real variable, exponential functions are uniquely characterized by the fact that the growth rate of such a function (that is, its derivative) is directly proportional to. For fx( ) 2x (i) The base of is (ii) The exponent of is (iii) What is varying in the function ? (iv) What is constant in the function ? 2. Characteristics of Graphs of Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. For : What are the possible inputs i. Exponential form of a complex number. Laws of Exponents. Make sure you play the html version (not java). Links to the data sets are included in the file. like exponential functions AND parametric differentiation and the trapezium rule, and volumes of solids, and vectors, and the Quotient rule, and completing the square oh my! such fun ahh happy days! for exponential bunnies, you could certainly propose a differential equation showing that after time T the population P is assumed to be. Note that the curve passes through (0, 1) (on the y-axis). In both equations the variable a is called the initial amount and b is called the growth constant. William Eaton from Denver School Science & Tech. 5: Introduction to Exponential Functions Related Instructional Videos Distinguish between linear and exponential functions using tables An updated version of this instructional video is available. If you're seeing this message, it means we're having trouble loading external resources on our website. Mortgage Problems 3. Moving all the terms to the left and factoring, x 2 + 2x - 8 = (x + 4)(x - 2) = 0. Compare/contrast Exponential and linear functions using a Google doc that all students will have access to and can contribute to. Examples : exp(0), returns 1. To form an exponential function, we let the independent variable be the exponent. The Organic Chemistry Tutor 306,614 views 10:13. \u02d8 Inverse Properties of Exponents and Logarithms Base a Natural Base e 1. Syntax : exp(x), where x is a number. Exponential functions. 97) to the \ud835\ude35 power, \ud835\ude3a = (1. This calculus video tutorial explains how to find the derivative of exponential functions using a simple formula. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. This repeated multiplication can be expressed using exponential functions. Chapter 8 : Exponents and Exponential Functions 8. Three lifelines, no phone-a-friend\u2014can you win? Lake Tahoe Community College: Exponential Equations. If the base, b=1 then the function is constant. They write and interpret an equation of the best-fit curve. One common example is population growth. Exponential Functions. Determinewhether!each!tableor!rulerepresents!an!exponential!function. Thus we define an exponential function to be any function of the form. The decision to bankroll such a game would be a major one, even for a risk-loving entity (if one was ever foolish enough to be attracted by the tiny fees ordinary players are willing to risk). - A 'positive constant base raised to a variable exponent' = (constant) (variable) is an 'Exponential function'. Write an exponential function for the table below. Here is the entire project for FREE if you are interested. It becomes y = (2 4) x + 2 = 2 4 x + 8. Unit 4 - Exponential Functions. This online quiz includes the following types of question:1) evaluating powers with rational exponents;2) simplifying expressions containing exponents;3) creating algebraic and graphical representations of exponential functions;4) solving problems involving exponential functions. Investors know the importance of an exponential function, since compound interest can be described by one. alg_1_chapter_6_review_game. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Derivative exponential : To differentiate function exponential online, it is possible to use the derivative calculator which allows the calculation of the derivative of the exponential function. An exponential function is defined for every real number x. Integrals with \"Sin\" Integrals with \"Cos\" Integrals with \"Sin\" and \"Cos\" Integrals with \"Tan\" Integrals with \"Cot\" Integrals with \"Sec\" Integrals with \"Csc\" Integrals with Inverse Trigonometric Functions. exponential function and the suitable function, which represents the constraint) by a positiv e parameter c , called the penalty parameter. Let's look at examples of these exponential functions at work. By definition:. Integrating the exponential function, also part of calculus. Then write a function to model the data. As usual, the default data used are USDJPY candles with a 15-minute compression. Cumulative Distribution Function Calculator - Exponential Distribution - Define the Exponential random variable by setting the rate \u03bb>0 in the field below. This discussion will focus on solving more complex problems involving exponential functions. 3 - Exponential Functions Click here to review the definition of a function. On a chart, this curve starts out very slowly, remaining. This algebra activity focuses on exponential functions. Exponential Functions & Relations. There are so many interesting exponential function situations!. Other examples of exponential functions include: $$y=3^x$$$$f(x)=4. By this symbol we mean the cube root of a. Concept Summary Exponential Function Family You can apply the four types of transformations\u2014stretches, compressions, reflections, and translations\u2014to exponential functions. We have a rule to change. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. This game requires students to represent the exponential parent function y=2^x numerically, algebraically, graphically and verbally. 25 hours, there are 40 000 bacteria present. a) Multiply b) Add c) Subtract$500 Answer from Exponential Graphs and Equations :. False Try again, although the ratios are fixed the t values are not evenly spaced. Play this game to review Algebra I. Mortgage Problems 3. Find the Range of function f defined by Solution to Example 1. & & EXPONENTIAL GROWTH FUNCTIONS WILL ALWAYS ULTIMATELY GROW FASTER THAN LINEAR FUNCTIONS!! ) Example:) LinearLarry&beginswith\\$10,000inhissavingsaccount. The number in a power that represents the number of times the base is used as a factor. 1 Know and apply the properties of integer exponents to generate equivalent numerical expressions. This chapter introduces the concepts, objects and functions used to work with and perform calculations using numbers and dates in JavaScript. Exponential Growth is a critically important aspect of Finance, Demographics, Biology, Economics, Resources, Electronics and many other areas. Exponential functions. Exponential functions arise in many applications. Exponential Function Intro \u2013 Key. The decision to bankroll such a game would be a major one, even for a risk-loving entity (if one was ever foolish enough to be attracted by the tiny fees ordinary players are willing to risk).", "date": "2020-11-25 01:35:25", "meta": {"domain": "umood.it", "url": "http://pziq.umood.it/exponential-functions-game.html", "openwebmath_score": 0.4596410393714905, "openwebmath_perplexity": 944.1857117142524, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981735720045014, "lm_q2_score": 0.8670357683915538, "lm_q1q2_score": 0.8511999843866641}}
{"url": "https://math.stackexchange.com/questions/2454137/convergence-of-the-series-frac1nn", "text": "# Convergence of the series $\\frac1{n^n}$\n\nDoes the series $\\frac1{n^n}$ converges or diverges .\n\nBy comparison test I can claim that $\\frac1{n^n} < \\frac1n$ , and since series $\\frac1n$ appears divergent , so the series $\\frac1{n^n}$ also diverges. However by Cauchy root test the limit of $\\frac1{n^n}$ appears to be $0<1$ which suggests the convergence of the series\n\n\u2022 A series of nonnegative terms with all terms greater than or equal to a divergent series diverges. To dramatize your error, consider the series with all terms equal to zero. By your logic, since $0 < 1/n$ for all $n$, it would diverge. \u2013\u00a0quasi Oct 2 '17 at 9:46\n\u2022 Your argument would be valid if $1/n^n \\color{red}{>} 1/n$. But it is not. \u2013\u00a0M. Winter Oct 2 '17 at 9:50\n\u2022 Series $\\dfrac{1}{n^n}$ converges so quickly that just adding $5$ terms gives the result with $4$ exact decimals. Adding $10$ terms gives $1.29128\\,599706$ which has eleven exact decimals \u2013\u00a0Raffaele Oct 2 '17 at 10:06\n\u2022 @Raffaele: quite right, anyway shouldn't leave the OP the impression that this is a proof. What about $1/n^{n(20-n)}$ ? \u2013\u00a0Yves Daoust Oct 2 '17 at 10:14\n\nWe cannot conclude that since $\\frac{1}{n^n} < \\frac{1}{n}$ then $\\sum_{n=1}^\\infty \\frac{1}{n^n}$ diverges.\n\nIf $\\frac{1}{n^n} > \\frac{1}{n}$ (which is not true), then we can conclude that $\\sum_{n=1}^\\infty \\frac{1}{n^n}$ diverges.\n\nYou have used Cauchy root test to conclude that it converges.\n\nIf we want to use comparison test, notice that $$\\frac{1}{n^n} \\leq\\frac{1}{n^2}$$\n\nand since $\\sum_{n=1}^\\infty \\frac1{n^2}$ conveges, hence $\\sum_{n=1}^\\infty \\frac1{n^n}$ converges.\n\n\u2022 Then is it that this given series converges by Cauchy root test ? \u2013\u00a0Nabendra Oct 2 '17 at 9:50\n\u2022 yes, you have proven it using the root test. \u2013\u00a0Siong Thye Goh Oct 2 '17 at 9:51\n\u2022 Now suppose if I use Ratio Test then, let u(n)=1/n^n and so u(n+1)=1/(n+1)^(n+1). Hence u(n)/u(n+1)=(1+1/n)^n * (n+1) and thus taking limit n tends to infinity on the above I get infinity(not sure whether calculations are correct) , which stands as greater than 1. So can I claim , the series in convergent by Ratio Test. \u2013\u00a0Nabendra Oct 2 '17 at 9:56\n\u2022 Usually for ratio test, we compute $\\left| \\frac{u(n+1)}{u(n)}\\right|$ and show that it is less than $1$. \u2013\u00a0Siong Thye Goh Oct 2 '17 at 10:28\n\nThe comparison test tells you that if $$a_n> b_n$$ and $$\\sum_{n}a_n$$ converges, then $$\\sum_n b_n$$ also converges.\n\nIt does not say that if $\\sum_{n} a_n$ diverges then $\\sum_{n} b_n$ diverges. If it did, then because $\\frac{1}{2^n} < 1$, you could conclude that $\\sum \\frac1{2^n}$ also diverges.\n\nIn fact, you would need the inequality reversed, so if $a_n<b_n$ and $\\sum a_n$ diverges, then $\\sum b_n$ also diverges.\n\nIn fact, the series $\\sum\\frac{1}{n^n}$ converges by the comparison test with $\\frac{1}{2^n}$ or with $\\frac{1}{n^2}$.\n\n\u2022 Does it appear convergent or divergent ? How gonna I check it ? \u2013\u00a0Nabendra Oct 2 '17 at 9:57\n\u2022 @Nabendra I really really dislike users that don't read my answers before they ask questions about them. \u2013\u00a05xum Oct 2 '17 at 9:57\n\u2022 I didn't understood your answer , as if I use 1/n the it diverges by comparison test instead of 1/n^2 or 1/2^n \u2013\u00a0Nabendra Oct 2 '17 at 10:04\n\u2022 @Nabendra You didn't understand my answer because you didn't bother reading it to the end. So I don't see why I should bother helping you any further. \u2013\u00a05xum Oct 2 '17 at 10:04\n\u2022 I do apologise, for my deed \u2013\u00a0Nabendra Oct 2 '17 at 10:07\n\nA series converges if it has a converging upper bound and it diverges if it has a diverging lower bound. In other cases, you cannot conclude.\n\nIn this particular case, you can for instance exploit\n\n$$\\frac1{n^n}\\le \\frac1{2^n}$$ for $n\\ge2$. Actually, the sequence converges extremely quickly, super-exponentially.", "date": "2019-09-16 20:26:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2454137/convergence-of-the-series-frac1nn", "openwebmath_score": 0.9460806846618652, "openwebmath_perplexity": 416.3892246728425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856482, "lm_q2_score": 0.8670357649558007, "lm_q1q2_score": 0.8511999842569262}}
{"url": "http://mathhelpforum.com/trigonometry/170333-trigonometry-question-print.html", "text": "Trigonometry question.\n\n\u2022 February 6th 2011, 06:43 AM\nGoogl\nTrigonometry question.\nHi all,\n\nI am new here by the way, I have a small problem for a question I can complete in Compound Angle identities\n\nProve that:\nsin\u03b8 = tan (\u03b8/2)(1 + cos\u03b8)\n\nLook forward to the answer. I can almost work it work it out but I get stuck somewhere near the end.\n\u2022 February 6th 2011, 07:02 AM\nPlato\nHello and welcome to MathHelpForum.\nYou should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.\n\u2022 February 6th 2011, 09:10 AM\nGoogl\nQuote:\n\nOriginally Posted by Plato\nHello and welcome to MathHelpForum.\nYou should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.\n\nI understand that. This is not homework, it's for my own understanding of Trig (revision). Okay how I did it...\n\nsin\u03b8 = tan (\u03b8/2)(1 + cos\u03b8)\ntan (\u03b8/2) (1 + cos (\u03b8/2 + \u03b8/2))\ntan (\u03b8/2)(cos^2(\u03b8/2) - sin^2(\u03b8/2))\ntan (\u03b8/2)(cos^2(\u03b8/2) + sin^2(\u03b8/2) + cos^2(\u03b8/2) - sin^2(\u03b8/2))\nAttachment 20699\n\u2022 February 6th 2011, 09:28 AM\ne^(i*pi)\nLet $A = \\dfrac{\\theta}{2}$ (purely to save me effort)\n\nFrom line 2: $\\tan(A)(1+\\cos(2A)) = \\dfrac{\\sin(A)}{\\cos(A)} \\cdot 2\\cos^2A = 2 \\sin A \\cos A = 2 \\sin\\left(\\dfrac{\\theta}{2}\\right) \\cos\\left(\\dfrac{\\theta}{2}\\right)$\n\nThere is a common identity you can use now to get the LHS\n\u2022 February 6th 2011, 09:55 AM\nGoogl\nQuote:\n\nOriginally Posted by e^(i*pi)\nLet $A = \\dfrac{\\theta}{2}$ (purely to save me effort)\n\nFrom line 2: $\\tan(A)(1+\\cos(2A)) = \\dfrac{\\sin(A)}{\\cos(A)} \\cdot 2\\cos^2A = 2 \\sin A \\cos A = 2 \\sin\\left(\\dfrac{\\theta}{2}\\right) \\cos\\left(\\dfrac{\\theta}{2}\\right)$\n\nThere is a common identity you can use now to get the LHS\n\nHello e^(i*pi)\n\nI actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sin\u03b8.\nAttachment 20700\n\nSo this is correct?\nAttachment 20701\n\nThanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.\n\u2022 February 6th 2011, 10:06 AM\nSoroban\nHello, Googl!\n\nQuote:\n\n$\\text{Prove that: }\\:\\sin\\theta \\:=\\:\\tan \\frac{\\theta}{2}(1 + \\cos\\theta)$\n\nIdentities: . $\\begin{array}{ccccccc}\\cos^2\\!\\frac{\\theta}{2} &=& \\dfrac{1+\\cos\\theta}{2} &\\Rightarrow& 1 + \\cos\\theta &=& 2\\cos^2\\!\\frac{\\theta}{2} \\\\ \\\\[-3mm]\n\\sin\\theta &=& 2\\sin\\frac{\\theta}{2}\\cos\\frac{\\theta}{2} \\end{array}$\n\n$\\text{On the right side we have:}$\n\n. . $\\tan\\frac{\\theta}{2}(1 + \\cos\\theta)\\;=\\;\\dfrac{\\sin\\frac{\\theta}{2}}{\\cos\\ frac{\\theta}{2}}(2\\cos^2\\!\\frac{\\theta}{2}) \\;=\\;2\\sin\\frac{\\theta}{2}\\cos\\frac{\\theta}{2} \\;=\\;\\sin\\theta$\n\n\u2022 February 6th 2011, 11:20 AM\nQuote:\n\nOriginally Posted by Googl\nHello e^(i*pi)\n\nI actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sin\u03b8.\nAttachment 20700\n\nSo this is correct?\nAttachment 20701\n\nThanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.\n\nYes, that's good when starting at the right.\n\nTo start at the left and end up at the right...\n\n$\\displaystyle\\ sin\\theta=sin\\left(\\frac{\\theta}{2}+\\frac{\\theta}{ 2}\\right)=2sin\\frac{\\theta}{2}cos\\frac{\\theta}{2}$\n\nusing $sin2A=2sinAcosA$\n\nwhich gets us to the half angle.\nTo get tan, we need cos under the sine, so multiply by\n\n$\\displaystyle\\ 1=\\frac{cos\\frac{\\theta}{2}}{cos\\frac{\\theta}{2}}$\n\n$\\displaystyle\\ 2sin\\frac{\\theta}{2}cos\\frac{\\theta}{2}\\left[\\frac{cos\\frac{\\theta}{2}}{cos\\frac{\\theta}{2}}\\ri ght]=\\frac{sin\\frac{\\theta}{2}}{cos\\frac{\\theta}{2}}\\l eft[2cos^2\\frac{\\theta}{2}\\right]$\n\nNow use\n\n$cos^2A=\\frac{1}{2}(1+cos2A)\\Rightarrow\\ 2cos^2A=1+cos2A$\n\nto get\n\n$\\displaystyle\\ sin\\theta=tan\\frac{\\theta}{2}\\left[1+cos\\left(\\frac{\\theta}{2}+\\frac{\\theta}{2}\\right )\\right]$", "date": "2014-12-18 10:05:12", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/170333-trigonometry-question-print.html", "openwebmath_score": 0.6938123106956482, "openwebmath_perplexity": 1410.581796899936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357243200244, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8511999830337486}}
{"url": "https://web2.0calc.com/questions/is-there-supposed-to-be-a-formula-or-strategy-for-this", "text": "+0\n\n# Is there supposed to be a Formula or strategy for this?\n\n+4\n167\n8\n+2557\n\nRon and Martin are playing a game with a bowl containing 39 marbles. Each player takes turns removing 1, 2, 3 or 4 marbles from the bowl. The person who removes the last marble loses. If Ron takes the first turn to\u00a0start the game, how many marbles should he remove to guarantee he is the winner?\n\nAfter thinking about this, I am not sure how to do it.\n\nBecause there are 39 marbles, Ron\u00a0needs to pick a number of marbles in which he can stick to a pattern to guaruntee a win. But how....\n\nAlso confirming if I am a robot is so annoying... I know its too prevent advertisement attacks, but like once you confirm it once on a registered account, it should stop asking.\n\nSep 22, 2019\nedited by CalculatorUser \u00a0Sep 22, 2019\n\n### 8+0 Answers\n\n#1\n+106993\n+1\n\nNo matter how many marbles he takes with his first choice Ron can still lose if he is not playing strategically.\n\nI guess we are suppose to assume that Ron is playing the ultimate game and so it Martin.\n\nI have got nothing more.\n\nSep 22, 2019\nedited by Melody \u00a0Sep 22, 2019\n#2\n+2557\n0\n\nYes agreed!! thats why Im confused... The answer was apparently 3 but I don't know why.\n\nCalculatorUser \u00a0Sep 22, 2019\n#3\n+23866\n+3\n\nRon and Martin are playing a game with a bowl containing 39 marbles.\nEach player takes turns removing 1, 2, 3 or 4 marbles from the bowl.\nThe person who removes the last marble loses.\nIf Ron takes the first turn to start the game,\nhow many marbles should he remove to guarantee he is the winner?\n\nRon first takes 3 marbles.\n\nThan, if Martin takes x marbles, Ron takes 5-x marbles, this repeats 7 times.\n\nThe last marble is then for Martin.\n\n39 = 3 + 7*5 + 1\n\nSep 22, 2019\nedited by heureka \u00a0Sep 22, 2019\n#4\n+2557\n+1\n\nWhy 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?\n\nCalculatorUser \u00a0Sep 22, 2019\n#5\n+23866\n+2\n\nWhy 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?\n\nHello CalculatorUser\n\nNo, it is because that guaruntees\u00a0 that after Martin and Ron the sum of removed marbles is 5.\n\nor in other words Ron supplements each time to 5.\n\nHere is the course\n\n$$\\begin{array}{|r|r|l|} \\hline \\text{Martin} & \\text{Ron} & \\text{sum} \\\\ \\hline & 3 \\\\ \\{4,3,2,1\\} & \\{1,2,3,4\\} & 4+1 = 5,\\ 3+2=5,\\ 2+3=5,\\ 1+4 = 5 \\\\ \\{4,3,2,1\\} & \\{1,2,3,4\\} & 4+1 = 5,\\ 3+2=5,\\ 2+3=5,\\ 1+4 = 5 \\\\ \\{4,3,2,1\\} & \\{1,2,3,4\\} & 4+1 = 5,\\ 3+2=5,\\ 2+3=5,\\ 1+4 = 5 \\\\ \\{4,3,2,1\\} & \\{1,2,3,4\\} & 4+1 = 5,\\ 3+2=5,\\ 2+3=5,\\ 1+4 = 5 \\\\ \\{4,3,2,1\\} & \\{1,2,3,4\\} & 4+1 = 5,\\ 3+2=5,\\ 2+3=5,\\ 1+4 = 5 \\\\ \\{4,3,2,1\\} & \\{1,2,3,4\\} & 4+1 = 5,\\ 3+2=5,\\ 2+3=5,\\ 1+4 = 5 \\\\ \\{4,3,2,1\\} & \\{1,2,3,4\\} & 4+1 = 5,\\ 3+2=5,\\ 2+3=5,\\ 1+4 = 5 \\\\ 1 \\\\ \\hline \\end{array}$$\n\nheureka \u00a0Sep 22, 2019\n#7\n+106993\n+1\n\nThat is really\u00a0cool.\u00a0 \u00a0Thanks Heureka.\n\nMelody \u00a0Sep 22, 2019\n#8\n+23866\n+2\n\nThank you, Melody !\n\nheureka \u00a0Sep 22, 2019\n#6\n+19913\n0\n\nEventually....at some pont, the captcha robot thingie quits happening....I do not know what the tripping point is.... ~ EP\n\nSep 22, 2019", "date": "2020-01-23 08:22:31", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/is-there-supposed-to-be-a-formula-or-strategy-for-this", "openwebmath_score": 0.6323599219322205, "openwebmath_perplexity": 3745.011424718191, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9817357181746964, "lm_q2_score": 0.8670357683915538, "lm_q1q2_score": 0.8511999827650318}}
{"url": "https://math.stackexchange.com/questions/3583836/computation-of-balance-equation-example-in-markov-model", "text": "# Computation of balance equation example in Markov model\n\nI am studying some examples of balance equations for Markov models. I am presented with the following example:\n\n$$\\mathcal{P} = \\begin{bmatrix} 0.2 & 0.3 & 0.5 \\\\ 0.1 & 0 & 0.9 \\\\ 0.55 & 0 & 0.45 \\end{bmatrix}$$\n\n[dropping the $$i$$ subscript by writing $$\\pi_j$$ for $$\\pi_{ij}.]$$\n\nThe balance equations are\n\n\\begin{align} &\\pi_1 = 0.2 \\pi_1 + 0.1 \\pi_2 + 0.55 \\pi_3 \\tag{a} \\\\ &\\pi_2 = 0.3 \\pi_1 \\tag{b} \\\\ &\\pi_3 = 0.5 \\pi_1 + 0.9 \\pi_2 + 0.45 \\pi_3 \\tag{c} \\end{align}\n\nSince, also, $$\\pi_1 + \\pi_2 + \\pi_3 = 1$$, the unique solution is\n\n$$\\pi_1 = 1/2.7 = 0.37037, \\ \\ \\ \\pi_2 = 1/9 = 0.11111, \\ \\ \\ \\pi_3 = 1.4/2.7 = 0.51852$$\n\nHow do we solve this for the values $$\\pi_1, \\pi_2, \\pi_3$$? Is there a way to solve this using matrix computations? The difficulty here, as I see it, is that we have a constraint $$\\pi_1 + \\pi_2 + \\pi_3 = 1$$ that must hold, so I'm unsure of how this is done.\n\nI would greatly appreciate it if someone would please take the time to show this.\n\n$$\\pi_2 = 0.3\\pi_1 \\Rightarrow$$\n\n$$\\pi_1 = 0.2\\pi_1 + 0.1\\cdot 0.3\\pi_1 + 0.55\\pi_3 \\Leftrightarrow 0.77\\pi_1 = 0.55\\pi_3 \\Rightarrow \\pi_3 = \\frac{7}{5}\\pi_1$$\n\n$$\\pi_3 = 0.5\\pi_1 + 0.9\\cdot 0.3\\pi_1 + 0.45\\pi_3 \\Leftrightarrow 0.55\\pi_3 = 0.77\\pi_1 \\Rightarrow \\pi_3 = \\frac{7}{5}\\pi_1$$\n\nIt means that we have only two equations with three unknowns, which implies that there is no unique solition to the system of linear equations. So $$\\pi_1$$ can be any number.\n\nNow, we have to use the condition $$\\pi_1+\\pi_2+\\pi_3 = 1$$ in order to find a unique solution.\n\n$$\\pi_1+\\pi_2+\\pi_3 = 1 \\Rightarrow \\pi_1 + 0.3\\pi_1 + \\frac{7}{5}\\pi_1 = \\frac{27}{10}\\pi_1 = 1 \\Rightarrow$$\n\n$$\\pi_1 = \\frac{10}{27} \\approx 0.37037$$\n\n$$\\pi_2 = 0.3\\frac{10}{27} = \\frac{1}{9} \\approx 0.11111$$\n\n$$\\pi_3 = \\frac{7}{5}\\frac{10}{27} = \\frac{14}{27} \\approx 0.51852$$\n\n\u2022 You have that $0.2\\pi_1 + 0.1 \\cdot 0.3 \\pi_1 = 0.77\\pi_1$? \u2013\u00a0The Pointer Mar 17 at 7:58\n\u2022 No. I have that $\\pi_1 = 0.2\\pi_1 + 0.1\\cdot 0.3\\pi_1 + 0.55\\pi_3 \\Leftrightarrow \\pi_1 - 0.2\\pi_1 - 0.1\\cdot 0.3\\pi_1 = 0.55\\pi_3 \\Leftrightarrow 0.77\\pi_1 = 0.55\\pi_3$ \u2013\u00a0Eugene Mar 17 at 9:39\n\u2022 Thanks for the clarification. \u2013\u00a0The Pointer Mar 17 at 13:09\n\nThe standard Eigenvalue equation is $$P^Tv=\\lambda v$$ You're seeking the eigenvector corresponding to $$\\lambda=1$$.\n\nFortunately, in the case of a Markov (aka stochastic) matrix, this happens to be the largest eigenvalue and therefore can be computed via the power iteration as \\eqalign{ v_{k+1} &= \\frac{P^Tv_k}{\\|P^Tv_k\\|_{\\tt1}} } or, if you prefer, the transposed equation using row vectors \\eqalign{ v_{k+1}^T &= \\frac{v_k^TP}{\\|v_k^TP\\|_{\\tt1}} } If you start with a stochastic vector, then $$P$$ preserves the stochastic property and scaling each step is unnecessary. This simplifies the iteration to $$v_{k+1}^T = v_k^TP \\quad\\implies v_n^T = v_0^TP^n$$ For the matrix in your example, this iteration yields \\eqalign{ v_{0}^T &= \\;\\big(&0.333333 \\quad&0.333333 \\quad&0.333333 \\;\\big) \\\\ v_{5}^T &= \\;\\big(&0.372054 \\quad&0.109744 \\quad&0.518201 \\;\\big) \\\\ v_{10}^T &= \\;\\big(&0.370358 \\quad&0.111112 \\quad&0.518529 \\;\\big) \\\\ v_{15}^T &= \\;\\big(&0.37037 \\quad&0.111111 \\quad&0.518518\\;\\big) \\\\ v_{20}^T &= \\;\\big(&0.37037 \\quad&0.111111 \\quad&0.518519\\;\\big) \\\\ }", "date": "2020-06-06 11:15:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3583836/computation-of-balance-equation-example-in-markov-model", "openwebmath_score": 0.9999982118606567, "openwebmath_perplexity": 842.142209030935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856481, "lm_q2_score": 0.867035763237924, "lm_q1q2_score": 0.8511999825704252}}
{"url": "https://web2.0calc.com/questions/osl4-25_1", "text": "+0\n\n# OSL4#25\n\n+1\n305\n6\n+935\n\nI tried this question by drawing pictures, but it didn't work out.\n\nJul 27, 2019\n\n#1\n+6196\n+3\n\n$$\\text{It can't be 70}\\\\ A_{board} =32^2 = 1024 ~sq~in\\\\ A_{photo}=3\\cdot 5 = 15 ~sq ~in\\\\ \\dfrac{1024}{15} = 68+\\dfrac{4}{15}\\\\ \\text{so with no overlapping 68 is the maximum possible, and is probably not attainable}$$\n\n.\nJul 27, 2019\n#2\n+8966\n+4\n\nAs Rom said, there is definitely no way to put 70 pictures on the board.\n\n68 pictures is the maximum possible, and here is a way to do that:\n\nHowever, if you can't rotate the pictures 90 degrees, then the maximum possible is\n\nfloor( 32 / 3 )\u00a0 *\u00a0 floor( 32 / 5 )\u00a0\u00a0 =\u00a0\u00a0 10 * 6\u00a0\u00a0 =\u00a0\u00a0 60\n\nJul 27, 2019\n#3\n+111438\n+2\n\nNice, hectictar\u00a0 \u00a0!!!!\n\nCPhill \u00a0Jul 27, 2019\n#4\n+935\n+3\n\nI don't know guys. Out of all the questions I've looked at made by this organization, their\u00a0answers are always correct. Also, this was on a math contest taken by all middle schoolers in the U.S., so if the answer is wrong, then a student has to have already told them and they would have fixed it already. I also got 68, but these questions tend to have a creative way to solving them. Still, you guys have really great solutions and I agree with your reasoning why 68 is the maximum possible.\n\ndgfgrafgdfge111 \u00a0Jul 28, 2019\nedited by dgfgrafgdfge111 \u00a0Jul 28, 2019\nedited by dgfgrafgdfge111 \u00a0Jul 28, 2019\n#5\n+8966\n+4\n\nHmmm....even if you cut the pictures into 1 inch by 1 inch squares, there's just not enough square inches on the board to fit 70 pictures worth.\n\nnumber of square inches on the board\u00a0 =\u00a0 32 * 32\u00a0 =\u00a0 1024\n\nnumber of square inches of 70 pictures\u00a0 =\u00a0 70 * 3 * 5\u00a0 =\u00a0 1050\n\nAre you sure it says\u00a0 70\u00a0 and not\u00a0 60\u00a0 ?\n\nIf so, maybe you can be the one to report the error in this question. Yes it is weird and unlikely they'd have it wrong, but it's not impossible.\n\nhectictar \u00a0Jul 28, 2019\n#6\n+6196\n+3\n\nAs an aside if I were writing this problem I would not use photos.\n\nNo one is going to rotate photos 90 degrees to display them.\n\nUsing tiles would have made the problem far more sensible with out affecting any of the underlying math.\n\nRom \u00a0Jul 28, 2019", "date": "2020-08-09 16:44:00", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/osl4-25_1", "openwebmath_score": 0.6105479598045349, "openwebmath_perplexity": 2140.8403393496123, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357195106374, "lm_q2_score": 0.8670357666736773, "lm_q1q2_score": 0.8511999822368398}}
{"url": "https://math.stackexchange.com/questions/577235/proof-of-the-following-inequality-fracx-y-log-x-log-y-sqrtxy", "text": "# Proof of the following inequality $\\frac{x - y}{\\log x - \\log y} > \\sqrt{xy}$, $x>y$.\n\nI have seen the following inequality $$\\frac{x - y}{\\log x - \\log y} > \\sqrt{xy} \\ , \\quad \\forall x>y$$ be stated as a near \"obvious\" fact in another question, on the site. The inequality is very cute, but so far I have not been able to prove it. It reminds me of the Lipschitz inequality, but has some minor differences. Also Jensens inequality comes to mind. Is there something obvious I am missing, or is this ineqality not as easy to prove as it looks?\n\n\u2022 The inequality can be written as $\\int_y^x \\frac{1}{t}dt < \\frac{x-y}{\\sqrt{xy}}$. I wonder if one can prove it by comparing areas, and the shape of $\\frac{1}{x}$. \u2013\u00a0N. S. Nov 22 '13 at 15:46\n\u2022 This seems related to Logarithmic mean. \u2013\u00a0Martin Sleziak Nov 22 '13 at 15:49\n\u2022 @N.S. This demonstration on WA site seems to be based on geometric idea. It was the first search result when I googled for logarithmic geometric mean inequality. \u2013\u00a0Martin Sleziak Nov 22 '13 at 16:22\n\u2022 Very nice, that does it! \u2013\u00a0N3buchadnezzar Nov 22 '13 at 16:36\n\nSince $1/x$ is concave for $x>0$ then \\begin{align*} \\log b - \\log a = \\int_b^a\\frac{\\mathrm{d}x}{x} < \\underbrace{ \\frac{1}{2} \\frac{b - a}{\\sqrt{ba}} }_\\text{Blue Area} +\\underbrace{ \\frac{1}{2}\\frac{b - a}{\\sqrt{ba}} }_{\\text{Red area}} = \\frac{b - a}{\\sqrt{ba}}\\end{align*} Implying $$\\hspace{4cm}\\sqrt{ab} < \\cfrac{b - a}{\\log b - \\log a} \\hspace{4cm} \\blacksquare$$\n\nHint: let $\\dfrac{x}{y}=t>1$ and $$\\Longleftrightarrow f(t)={t-1}-{\\ln{t}}\\cdot \\sqrt{t}$$\n\nand follow is easy to prove it\n\n\u2022 Yes, the inequality becomes $(t-1)/\\log t>\\sqrt{t}$ for $t>1$. A possible simplification is to set $x/y=t^2$, instead, just to get rid of the square root. \u2013\u00a0egreg Nov 22 '13 at 15:11\n\u2022 And how does one show the last equality? I proved that they both are zero as $t\\to 1$. But showing that the derivative was increasing greater than zero was hard. \u2013\u00a0N3buchadnezzar Nov 22 '13 at 15:18\n\nWriting $x=t^2y$ with $t\\gt1$, the inequality to be proved can be written as\n\n$$t-{1\\over t}-2\\log t\\gt0$$\n\nLetting $f(t)$ be the expression on the left, we see that $f(1)=0$ and\n\n$$f'(t)=1+{1\\over t^2}-{2\\over t}={(t-1)^2\\over t}\\gt0 \\text{ for }x\\gt1$$\n\nThis means $f$ is strictly increasing, making it necessarily positive.\n\nNote: This answer is along the lines of the approaches taken by math110 and egreg. The main difference was to write the inequality in a way that suggested a function that's easy to differentiate and show is always positive.\n\nWith the substitution $x/y=t^2$, the inequality becomes\n\n$$\\frac{t^2-1}{\\log t^2}>t$$ that, for $t>1$, is equivalent to $$t^2-1>2t\\log t.$$\n\nConsider $f(t)=t^2-1-2t\\log t$, defined for $t\\ge1$; we have $f(1)=0$ and the derivative is $$f'(t)=2t-2-2\\log t.$$ I claim that $f'(t)>0$ for $t>1$, so the function $f$ is increasing. It's easy to see that $f'(1)=0$ and $\\lim_{t\\to\\infty}f'(t)=\\infty$.\n\nSince $$f''(t)=2-\\frac{2}{t}>0$$ for $t>1$, the function $f'$ is increasing, so it's everywhere positive (except at $1$).\n\nOf course the proof with convexity is better.\n\n\u2022 I think you need $t^2$'s on the left hand side of the first inequality. \u2013\u00a0Barry Cipra Nov 22 '13 at 22:37\n\u2022 @BarryCipra Yes, thanks. \u2013\u00a0egreg Nov 22 '13 at 22:38\n\nAnother proof based on geometry is mentioned in Frank Burk: The Geometric, Logarithmic, and Arithmetic Mean Inequality, The American Mathematical Monthly , Vol. 94, No. 6 (Jun. - Jul., 1987), pp. 527-528, jstor, link. (It was among the first hits in the google search for logarithmic geometric mean inequality.)\n\nWe know that function $e^x$ is convex. Let us have a look on this function on the interval $\\ln a$, $\\ln b$. The line joining the points $(\\ln a,a)$ and $(\\ln b,b)$ lies above the graph of this function, so we have a trapezoid which contains the whole area lying below the graph. On the other hand, if we make a tangent in the midpoint $(\\ln a+\\ln b)/2$, we get a trapezoid which lies below the graph of this function. (See the picture in the pdf linked above.)\n\nSo we have: $$\\left(e^{\\frac{\\ln a+\\ln b}2}\\right) (\\ln b-\\ln a) \\le \\int_{\\ln a}^{\\ln b} e^x \\le \\frac{e^{\\ln a}+e^{\\ln b}}2 (\\ln b-\\ln a)\\\\ \\sqrt{ab} \\le \\frac{b-a}{\\ln b-\\ln a} \\le \\frac{a+b}2$$\n\nInequality has many demonstrations. Some are made \u200b\u200busing logarithmic representation with integral. Give here an example: $$\\frac{1}{L( a, b )} = \\frac{1}{b - a}\\int^{\\frac{b}{a}}_{1}\\frac{dx}{x}$$ Integrating inequality: $$\\frac{4}{(x+1)^2}\\leq\\frac{1}{x}\\leq\\frac{x+1}{2x\\sqrt{x}}$$ is obtained $$\\frac{b-a}{b+a}\\leq\\frac{1}{2}\\ln\\frac{b}{a} \\leq\\frac{b-a}{2\\sqrt{ab}}$$ hence the inequality in question, but more logarithmic mean framing between the arithmetic mean and geometric mean:$$\\sqrt{ab} < \\frac{b-a}{\\ln b-\\ln a}<\\frac{a+b}{2}.$$ $(0<a<b)$", "date": "2019-08-18 07:53:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/577235/proof-of-the-following-inequality-fracx-y-log-x-log-y-sqrtxy", "openwebmath_score": 0.9455187320709229, "openwebmath_perplexity": 236.88276737150215, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357195106374, "lm_q2_score": 0.8670357546485407, "lm_q1q2_score": 0.8511999704313336}}
{"url": "https://math.stackexchange.com/questions/3832740/what-can-be-a-generalization-of-repeats-in-exponentiation-using-modulo", "text": "# What can be a generalization of repeats in exponentiation using modulo?\n\nI came across a Math Problem in a Japanese Coding Test(It is officially over now so no worries about discussing it, https://atcoder.jp/contests/abc179/tasks/abc179_e).\n\nI will write the mathematical version of this problem.\n\nLet $$A$$ be a sequence that is defined by the initial values $$A_1=x$$ and this recurrence relation is given $$A_{n+1}$$=$$(A_{n}^2)$$ $$mod$$ $$M$$ where $$M$$ can be any natural number.\n\nFind $$\\sum_{i=1}^{i=N}A_{i}$$\n\nI will tell what I have deduced till now:\n\n1. If I write this recurrence in equation it demands us to find $$(x^1 mod M + x^2 mod M + x^4 mod M + x^8 mod M + x^{16}mod M ..$$ till $$n$$ terms).\n2. If We take any example for $$x=2$$ and $$M=1001$$ the values of this series come out to be like this $$2,4,16,256,471,620,16,256,471,620....$$ and this block of $$16,256,471$$ repeats.\n3. I observed that for any given $$x$$ and $$M$$ the series formed will come at a point where one of it's window will start repeating, just like in the above case this window of $$16,256,471$$ repeated after certain point. All because of Modulo Magic I have observed that it will repeat but I don't have any proof of How and Why ?\n4. I tried using Fermat's Little theorem that for the case of when $$M$$ is prime maybe of some use But didn't find an apt conclusion to it.\n\nNow I am stuck at this problem that How will Modulo work in such kind of series and how Will the values of this series depend on different versions of $$x$$ and $$M$$ like them being co prime to each other or otherwise. and if this series is to give recurring values after a certain point then Why and How and also as it happened in the example case I have given All the values do not repeat due to this kind of exponentiation but only a window repeats,I don't understand why.\n\n\u2022 Since there are only finitely many values available modulo $M$ there must eventually be a value that has already come up previously. Since each value depends only on the previous value, once a value comes up a second time the sequence must repeat what it did the first time that value came up, periodically, forever. \u2013\u00a0Gerry Myerson Sep 20 at 0:23\n\u2022 Yes I agree to that But why is it that only a certain value will repeat ? Why for the example I am taking it repeated after 16 not with 2. \u2013\u00a0Kartik Bhatia Sep 20 at 8:20\n\nFirst, consider the case where $$x$$ and $$M$$ are coprime, i.e., $$\\gcd(x,M) = 1$$. Since for all $$i \\gt 1$$ we have $$0 \\le A_i \\lt M$$, there are only a finite # of values it can have so the sequence will eventually have to start repeating. Let $$j$$ and $$k$$, where $$j \\lt k$$, be the first indices where the values repeat. Since $$x$$ and $$M$$ are coprime, $$x$$ has a multiplicative inverse. Using this, we thus have\n\n\\begin{aligned} x^{2^{k-1}} & \\equiv x^{2^{j-1}} \\pmod{M} \\\\ x^{2^{k-1}} - x^{2^{j-1}} & \\equiv 0 \\pmod{M} \\\\ x^{2^{j-1}}\\left(x^{2^{k-1} - 2^{j-1}} - 1\\right) & \\equiv 0 \\pmod{M} \\\\ x^{2^{k-1} - 2^{j-1}} - 1 & \\equiv 0 \\pmod{M} \\\\ x^{2^{j-1}\\left(2^{k-j} - 1\\right)} & \\equiv 1 \\pmod{M} \\end{aligned}\\tag{1}\\label{eq1A}\n\nThe multiplicative order of $$x$$ modulo $$M$$, i.e.,\n\n$$m_1 = \\operatorname{ord}_{M}(x) \\tag{2}\\label{eq2A}$$\n\nmust divide $$2^{j-1}\\left(2^{k-j} - 1\\right)$$. Let $$a$$ be the largest power of $$2$$ which divides $$m_1$$, so we have\n\n$$m_1 = 2^{a}b, \\; \\gcd(b, 2) = 1 \\tag{3}\\label{eq3A}$$\n\nThe smallest value of $$j$$ which works is where $$j - 1 = a \\implies j = a + 1$$, except where $$a = 0$$ and $$x \\ge M$$, in which case we get $$j = 2$$ instead. This is the main reason why not all of the initial values repeat (i.e., where $$a \\gt 0$$) but, instead, just a \"window\" starting at this minimum $$j$$ value.\n\nNext, if $$b = 1$$, the smallest value of $$k - j$$ is $$1$$, else for $$b \\gt 1$$, it's $$m_2$$ where\n\n$$m_2 = \\operatorname{ord}_{b}(2) \\implies 2^{m_2} = kb + 1, \\; k \\in \\mathbb{N} \\tag{4}\\label{eq4A}$$\n\nWith your example of $$x = 2$$ and $$M = 1001$$, the values start repeating with $$j = 3$$ and $$k = 7$$ giving $$2^{j-1}\\left(2^{k-j} - 1\\right) = 4(15) = 60$$. As you can confirm, in this case, $$m_1 = 60$$, although they will not in general be equal (since equality only occurs with $$k = 1$$ in \\eqref{eq4A}).\n\nNext, consider the somewhat more complicated case where $$x$$ and $$M$$ are not coprime. Let\n\n$$q = \\prod_{i=1}^{n}p_i \\tag{5}\\label{eq5A}$$\n\nbe the product of all of the $$n$$ primes $$p_i$$ which are factors of both $$x$$ and $$M$$. Splitting $$x$$ and $$M$$ into factors which aren't and are coprime with $$q$$ gives\n\n$$x_1 = \\prod_{i=1}^{n}p_i^{s_i}, \\; x = x_1x_2, \\; \\gcd(x_2, q) = 1 \\tag{6}\\label{eq6A}$$\n\n$$M_1 = \\prod_{i=1}^{n}p_i^{t_i}, \\; M = M_1M_2, \\; \\gcd(M_2, q) = 1 \\tag{7}\\label{eq7A}$$\n\nAlso, note $$\\gcd(x_2, M_2) = 1$$ since they don't have any prime factors in common.\n\nAs before, let $$j \\lt k$$ be the first indices which repeat. We split the congruence equation to that with $$M_1$$ and with $$M_2$$. This first gives\n\n\\begin{aligned} (x_1x_2)^{2^{k-1}} & \\equiv (x_1x_2)^{2^{j-1}} \\pmod{M_1} \\\\ (x_1x_2)^{2^{j-1}}\\left((x_1x_2)^{2^{k - 1} - 2^{j-1}} - 1\\right) & \\equiv 0 \\pmod{M_1} \\end{aligned}\\tag{8}\\label{eq8A}\n\nSince no $$p_i$$ in $$q$$ from \\eqref{eq4A} divides $$(x_1x_2)^{2^{k - 1} - 2^{j-1}} - 1$$, this means all factors of $$p_i$$ are in $$(x_1x_2)^{2^{j-1}}$$. In particular, the smallest possible $$j$$ requires, using \\eqref{eq6A} and \\eqref{eq7A}, that\n\n$$2^{j-1}(s_i) \\ge t_i, \\; \\forall \\, 1 \\le i \\le n \\tag{9}\\label{eq9A}$$\n\nNext, since $$\\gcd(x, M_2) = 1$$, we have the same situation as at the start of this solution, with $$M$$ replaced by $$M_2$$, i.e., we get basically the equivalent of \\eqref{eq1A} giving\n\n$$x^{2^{k-1}} \\equiv x^{2^{j-1}} \\pmod{M_2} \\implies x^{2^{j-1}\\left(2^{k-j} - 1\\right)} \\equiv 1 \\pmod{M_2} \\tag{10}\\label{eq10A}$$\n\nWe thus proceed as we did before, but with the added restriction now that $$j$$ must be at least as large as what's required by \\eqref{eq9A}.\n\n\u2022 It is taking me some time understanding it ! I didn't know about multiplicative order , I am clear with the case of them being co prime how are you dealing with other case Why does gcd come into play ? \u2013\u00a0Kartik Bhatia Sep 20 at 8:19\n\u2022 The first paraghraph is not clear: multiply both sides of what, and what product? \u2013\u00a0Bill Dubuque Sep 20 at 8:22\n\u2022 @BillDubuque can you help me in this question ? \u2013\u00a0Kartik Bhatia Sep 20 at 12:24\n\u2022 @BillDubuque Thanks for the feedback. You're right it wasn't particularly clear. I've changed it by removing describing what I was going to do in my $(1)$ and, instead, just did it there. I believe this is now easier to understand. \u2013\u00a0John Omielan Sep 20 at 14:07\n\u2022 @KartikBhatia As I commented to Bill, I've changed my first part dealing with my $(1)$. As for why the gcd (greatest common factor) comes into play, it's mainly due to the simpler handling in my $(1)$ can only occur when $x$ & $M$ are coprime. This is since a multiplicative inverse, i.e., a $y$ where $xy \\equiv 1 \\pmod{M}$, only exists for those cases. As I show in my $(8)$ & $(9)$, non coprime values have an added requirement of common prime factor powers in $x^{2^{j-1}}$ needing to be at least as large as in $M$. Please let me know if there's anything specifically I can explain further. \u2013\u00a0John Omielan Sep 20 at 14:19", "date": "2020-10-28 14:26:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3832740/what-can-be-a-generalization-of-repeats-in-exponentiation-using-modulo", "openwebmath_score": 0.9597802758216858, "openwebmath_perplexity": 290.64634767616275, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357237856481, "lm_q2_score": 0.8670357477770336, "lm_q1q2_score": 0.8511999673919167}}
{"url": "https://martamlodzikowska.pl/2006/benito/honda/104395359f2992b8567-recursion-tree-method-for-solving-recurrences", "text": "Recursion Trees Show successive expansions of recurrences using trees. Iteration method Recursion-tree method (Master method) Iteration is constructive, i.e. 11. I Ching [The Book of Changes] (c. 1100 BC) To endure the idea of the recurrence one needs: freedom from morality; new means against Now we use induction to prove our guess. two steps: 1 Guess the form of the solution. you gure out the result; somewhat tedious because of T(n) = aT(n/b)+(n), where a 1 and b > 1 are constants and (n) is an asymptotically positive function. Like Masters Theorem, Recursion Tree is another method for solving the recurrence relations. 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the guess is correct or incorrect. Now push the current node in the inorder array and make the recursive call for the right child (right subtree). For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). Solving using a recursion tree. Arithmetic Sequences and Series by MATHguide Arithmetic Sequences and Series Learn about Arithmetic Sequences and Series By decreasing the size of h, the function can be approximated accurately The runtime is so much higher because the recursive function fib[n_]:=fib[n-1]+fib[n-2] generates n^2 Inorder Tree Traversal without Recursion; Inorder Tree Traversal without recursion and without stack! Etsi tit, jotka liittyvt hakusanaan Recursion tree method for solving recurrences examples tai palkkaa maailman suurimmalta makkinapaikalta, jossa on yli 21 miljoonaa tyt. 9 The recursion-tree method Convert the recurrence into a tree: Each node represents the cost incurred at various levels of recursion Sum up the costs of all levels Used to guess a solution for the recurrence T(n) = 3T(n/3) + n for n > 1. Note: We would usually use a recursion tree to generate possible guesses for the runtime, and then use the substitution method to prove them. Solving Recurrences; Amortized Analysis; What does 'Space Complexity' mean ? Engineering; Computer Science; Computer Science questions and answers; 2. 1. Explanation: Masters theorem is a direct method for solving recurrences. SOLVING RECURRENCES 1.2 The Tree Method The cost analysis of our algorihms usually comes down to nding a closed form for a recurrence. There are mainly three ways for solving recurrences. The master method provides a \"cookbook\" method for solving recurrences of the form. If yes, solve with that method, if no, explain why. With substitution you need to guess the result and prove it by induction. Solving recurrence relation. form and show that the solution works. It's very easy to understand and you don't need to be a 10X developer to do so. Recursion Tree Implement recursion tree method to. PURRS is a C++ library for the (possibly approximate) solution of recurrence relations (5 marks) Example 1: Setting up a recurrence relation for running time analysis Note that this satis es the A general mixed-integer programming solver, consisting of a number of different algorithms, is used to determine the optimal decision vector A general The recurrence relation is given as: an = 4an-1 - 4an-2 The initial conditions are given as 20 = 1, 2, = 4 and 22 = 12,-- Se When you solve the general equation, the constants a Arithmetic Sequences and Series by MATHguide Arithmetic Sequences and Series Learn about Arithmetic Sequences and Series , Chichester, 1989 The good people at Desmos have made an excellent online graphing calculator even better Enter a value for nMin In the example given in the previous chapter, T (1) T ( 1) was the time taken in the initial condition. 4.3 The master method. Minimum Spanning Tree. We use techniques for summations to solve the recurrence. Method 2: Push directly root node two times while traversing to the left. For the following recurrences, use the recursion tree method to find a good guess of what they solve to asymptotically (i.e. Search: Recursive Sequence Calculator Wolfram. Yes, you can solve almost every problem using recursion. Just look out how Functional Programmers tackles every problem with Haskell, OCaml, Erlang etc. Why not? recursion trees. You will get a recurrence in m that is one of theprevious examples. solving recurrences can use data in these subproblems are no general way of numbers. T(n) = 4T(n/2) + n^3 for n > 1. Some methods used for computing asymptotic bounds are the master theorem and the AkraBazzi method. 3 Solving recurrences Methods for solving recurrences: 1. b. This tree is a way of representing the algorithms iteration in the shape of a tree, with each node representing the trees iteration level. I will also accept this method as proof for the given bound (if done correctly). 2 Use mathematical induction to nd constants in the. (a) (4 marks | Solve the following recurrences by the recursion-tree method (you may assume that n is a power of 3): 4, n= T (n) - { r +-2, 51 = n>. For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). The first recurrence relation was. we guess a bound and then use mathematical induction to prove our guess correct; 2. 2 Solving Recurrences with the Iteration/Recursion-tree Method In the iteration method we iteratively unfold the recurrence until we see the pattern. P2. The Substitution method. converts the recursion into a tree whose Solutions to exercise and problems of Introduction to Algorithms by CLRS (Cormen, Leiserson, Rivest, and Stein) Divide (line 2): (1) is required to compute q as the average of p and r. Conquer (lines 3 and 4): 2 T ( n /2) is required to recursively solve two subproblems, each of size n/2. for questions about sequences and series, e Very easy to understand! Now we use induction to prove our guess. or O). OK? Help organize the algebraic bookkeeping necessary to solve a recurrence. Steps to solve recurrence relation using recursion tree method: Draw a recursive tree for given recurrence relation. Use the substitution method to verify your answer. (5 marks) Solve this. Exercises. Recursive sequence formulaAn initial value such as $a_1$.A pattern or an equation in terms of $a_ {n 1}$ or even $a_ {n -2}$ that applies throughout the sequence.We can express the rule as a function of $a_ {n -1}$. The recursion-tree method converts the recurrence into a tree whose nodes represent the costs incurred at various levels of the recursion. The good guess devised using the recursion tree can be proved by the substitution method. Search: Recursive Sequence Calculator Wolfram. two steps: 1 Guess the form of the solution. The subproblem size for a node at depth is. P3. The iteration method does not require making a good guess like the substitution method (but it is often more involved than using induction). This method is especially powerful when we encounter recurrences that are non-trivial and unreadable via the master theorem. In this article, we are going to talk about two methods that can be used to solve the special kind of recurrence relations known as divide and conquer recurrences If you can remember these easy rules then Master Theorem is very easy to solve recurrence equations Learn how to solve recurrence relations with generating functions Recall that the recurrence relation is a recursive Final Exam Computer Science 112: Programming in C++ Status: Computer Science 112: Programming in C++ Course Practice . DFS Traversal of a Graph vs Tree. The false position method is a root-finding algorithm that uses a succession of roots of secant lines combined with the bisection method to As can be seen from the recurrence relation, the false position method requires two initial values, x0 and x1, which should bracket the root See full list on users For example, consider the There are mainly three ways for solving recurrences. Wolfram|Alpha can solve various kinds of recurrences, find asymptotic bounds and find recurrence relations satisfied by given sequences. 1.2 Recursion tree A recursion tree is a tree where each node represents the cost of a certain recursive sub-problem. Consider the following runtime recurrence: T (1) = 1 and T(n) = 3T(n/2) + n^2 when n greaterthanequalto 2. Assume T(1) = 1. the recursion-tree method 1 solving recurrences expanding the recurrence into a tree summing the cost at each level applying the substitution method 2 another example using a recursion tree. The terms of a recursive sequences can be denoted symbolically in a number of different notations, such as , , or f[], where is a symbol representing thesequence Binomial Coefficient Calculator Do not copy and paste from Wolfram Sequences Calculator The sequence of RATS number is called RATS Sequence The sequence of RATS number is called RATS Sequence. Steps to solve recurrence relation using recursion tree method: Draw a recursive tree for given recurrence relation. In the previous lecture, the focus was on step 2. Like Masters Theorem, Recursion Tree is another method for solving the recurrence relations. T ( n) = 2 T ( n / 2) + n. The solution of this one can be found by Master Theorem or the recurrence tree method. Now we use induction to prove our guess. Each of these cases is an equation or inequality, with some Use an inductive hypothesis for simplicity, we specify initial conditions represent another method in recursion tree method for solving recurrences examples later determine in big-Oh notation). The recursion-tree method. Solving Recurrences Methods The Master Theorem The Recursion-Tree Method Useful for guessing the bound. Recursion Tree Implement recursion tree method to solve the recurrences below, you can use the Master theorem to verift your solution if applicable: a) T(n) = T(1/2) +T(1/4) + (n) b) T(n) = 3T(n/2) + 4T(n/2) + (n) c) T(n) = T(3n/2) +T(n/3) + O(n) d) T(n) = 4T(n/2) + 4T(n/2) + (n) In this method, we first convert the recurrence into a summation. There are mainly three ways for solving recurrences. Here the right-subtree, the one with 2n/3 element will drive the height. A recurrence is an equation or inequality that describes a function in terms of its values on smaller inputs. LEC 07: Recurrences II, Tree Method CSE 373 Autumn 2020 Learning Objectives 1.ContinuedDescribe the 3 most common recursive patterns and identify whether code belongs to one of them 2.Model a recurrence with the Tree Method and use it to characterize the recurrence with a bound 3.Use Summation Identities to find closed forms for summations to devise good guesses. In this video we discuss how to use the seqn command to define a recursive sequence on the TI-Nspire CX calculator page Monotonic decreasing sequences are defined similarly The terms of a recursive sequences can be denoted symbolically in a number of different notations, such as , , or f[], where is a symbol In the previous lecture, the focus was on step 2. A: Recursion tree is the method for solving the recurrence relations.Recursion tree may be a tree Q: Given the tree above, show the order of the nodes visited using recursive in-order traversal. Substitution method. To solve a Recurrence Relation means to obtain a function defined on the natural numbers that satisfy the recurrence. Use . Rekisterityminen ja tarjoaminen on ilmaista. First let's create a recursion tree for the recurrence $T (n) = 3T (\\frac {n} {2}) + n$ and assume that n is an exact power of 2. Answer to 2. You must show the tree and fill out the table like we did in class. Calculate the cost at each level and count the total no of levels in the recursion tree. 1.Recursion Tree 2.Substitution Method - guess runtime and check using induction 3.Master Theorem 3.1 Recursion Tree Recursion trees are a visual way to devise a good guess for the solution to a recurrence, which can then be formally proved using the substitution method. 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the guess is correct or incorrect. Search: Recursive Sequence Calculator Wolfram. Visit the current node data in the postorder array before exiting from the current recursion. Introduction to the Recursion Tree Method for solving recurrences, with multiple animated examples. Steps to Solve Recurrence Relations Using Recursion Tree Method- Step-01: For Example, the Worst Case Running Time T (n) of the MERGE SORT Procedures is described by the recurrence. can be solved with recursion tree method. Use an inductive hypothesis for simplicity, we specify initial conditions represent another method in recursion tree method for solving recurrences examples later determine Keep track of the time spent on the subproblems of a divide and conquer algorithm. 4.4 The recursion-tree method for solving recurrences 4.4-1. How to solve the recurrence T ( n) = 3 T ( n / 2) + n. The exercise stated that i have to solve the recurrence using the Recursion-Tree Method. Search: Recurrence Relation Solver Calculator. If you see the height is determined by height of largest subtree (+1). A recursion tree is a tree where each node represents the cost of a certain recursive sub-problem. A: In-order -traversal:- We traversal the left node Engineering; Computer Science; Computer Science questions and answers; 3. Count the total number of nodes in the last level and calculate the cost of The subproblem P. S. Mandal, IITG I have already finished the base part, which is ( n lg 3) But for the recursive part I'm having troubles with this sum: c n i = 0 lg n 1 ( 3 / 2) i. Use of recursion to solve math problems ; Practice Exams. Such recurrences should not constitute occasions for sadness but realities for awareness, so that one may be happy in the interim. Q: Use the recursion tree method to solve each of the following recurrences: T(n) = T(n/10) + T(9n/10) A: The recursion tree method works by creating each level of the recurrence relation in the tree the or O).", "date": "2022-09-29 05:55:44", "meta": {"domain": "martamlodzikowska.pl", "url": "https://martamlodzikowska.pl/2006/benito/honda/104395359f2992b8567-recursion-tree-method-for-solving-recurrences", "openwebmath_score": 0.7153167724609375, "openwebmath_perplexity": 630.4070380938905, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9777138196557983, "lm_q2_score": 0.8705972616934408, "lm_q1q2_score": 0.8511949741121726}}
{"url": "https://math.stackexchange.com/questions/1463252/expressing-statements-in-discrete-math/1463268", "text": "# Expressing statements in Discrete math\n\nGiven that\n\n$A$ is the set of all Alpha's\n\n$M$ is the set of all Men\n\nhow do I express this statement: Not all Alpha's are Men\n\n.............\n\nMy attempt:\n\n$A \\subset S = 0$\n\nin other words saying that $A$ is not a subset of $S$, but I can't use the not subset symbol on this problem.\n\n\u2022 $\\exists a\\in A:a\\notin M$ \u2013\u00a0abiessu Oct 4 '15 at 1:43\n\u2022 What is $S$? Did you mean $A \\subset M = \\emptyset$? \u2013\u00a0N. F. Taussig Oct 4 '15 at 8:10\n\nYou could write $A\\backslash M\\ne\\emptyset$.\n\nMeaning that when you take all the men out of the alphas, there are alphas remaining.\n\n\u2022 the \\ is the difference symbol ? It makes a lot more sense this way, Thank you. \u2013\u00a0learnmore Oct 4 '15 at 2:10\n\u2022 Yes. It is equivalent to A\\cap M^c. M^c being the complement. \u2013\u00a0Jean-Fran\u00e7ois Gagnon Oct 4 '15 at 4:48\n\n\"not all alpha's are men\" $\\Leftrightarrow$\"there is an alpha who is not a man\".\n\ni.e.\n\n$$\\exists a \\in A \\text{ such that } a \\not\\in M$$\n\n\u2022 I like the way you reworded it, makes it a lot more easier to express \u2013\u00a0learnmore Oct 4 '15 at 2:09", "date": "2021-01-26 05:39:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1463252/expressing-statements-in-discrete-math/1463268", "openwebmath_score": 0.7967756390571594, "openwebmath_perplexity": 1044.516297786336, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9777138144607744, "lm_q2_score": 0.8705972633721707, "lm_q1q2_score": 0.8511949712307165}}
{"url": "http://apoderados.lazzeri.cl/wp-content/uploads/2022/03/kgqw5cvh/lecture-on-vector-calculus.html", "text": "Lecture on vector calculus. Divergence and Curl o\n\nLecture on vector calculus. Divergence and Curl of vector field | Irrotational & Solenoidal vector. Lecture 4: What Is A Unit Vector? Lecture 5: Vectors And The Unit Circle. Remember: the curl of a vector always results in another vector. 0 (fall 2009) This is a self contained set of lecture notes for Math 221. Unit vectors, vectors Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. Real Euclidean Space Rn. Vector Anupam Kumar. Vector Math with historical perspective (2010-2014), 13 lectures 2021 on youtube. 1 vectors We start with some de nitions. , Lectures on Differential Geometry, Prentice-Hall, Englewood Cliffs, New Jersey, 1964. You lectures. For detailed Vector Calculus (2009, UNSW). This course will remind you about that good stu\ufb00, but goes on to introduce you to the subject of Vector Calculus which, like it says on the can, combines vector algebra with calculus. (George Carlin, American Internet Supplement for Vector Calculus. 1. Some gave vector fields; some Another term for vector space is linear space. Check out www. Linear algebra. Why is vector calculus important for computer This chapter is concerned with applying calculus in the context of vector \ufb01elds. ~r= x^i+ y^j+ zk^ (1) The unit vectors ^i;^j; ^k are orthonormal. edu/terms \u200eEducation \u00b7 2011. Di erentiability in the case of two variable functions 30 These are lectures notes for MATH1056 Calculus Part II. Vector & Calculus - Lecture Prologue This lecture note is closely following the part of multivariable calculus in Stewart\u2019s book [7]. In Lecture 6 we will look at combining these vector operators. Lecture 1: Three-Dimensional Coordinate Systems; Lecture 2: Vectors; Lecture 3: The Dot Product; Lecture 4: The Cross Product; Lecture Other Lecture Notes on the Web. In this post, Support Vector Machines \u2014 Lecture Important questions of Vector calculus Engineering mathematics lecture for GATE 2017 The fourth week covers the fundamental theorems of vector calculus, including the gradient theorem, the divergence theorem and Stokes\u2019 theorem. Compute the curl of a vector field using sympy. Cowley@damtp. All unit vectors Advanced Calculus course. Remark We will almost exclusively consider real vector spaces, i. Lecture 1: Vector Notation. Tensor calculus The normal vectors are called \u2018contravariant vectors\u2019, because they transform con-trary to the basis vector columns. The course is organized into 53 short lecture Notes on Vector Calculus. If the triple product is zero, the volume between three vectors Lecture-02 calculus with vector What is the velocity of the conecting joint B in the example? Name a situation where the derivative of a unit vector is not zero Calculus 3 Lecture 11. 016 Fall 2012 c W. 1) is called the (linear)vectorspace. gaussianmath. Gleb V. The Physics course is delivered in Hinglish. Focuses on extending the concepts of function, limit, continuity, derivative, integral and vector from the plane to the three dimensional space. In large part this is because the point of vector calculus is to give us tools that we can apply elsewhere and the next steps involve turning to other courses. No calculators allowed. cam. Concept of Vector Point Function & Vector Differentiation. Matrices, linear transformations and vector spaces are necessary ingredients for a proper discussion of ad-vanced calculus A whole set of objects (vectors) on which we can perform vector addition and scalar multiplication with properties given by Eqs. 4) 2. Linear Algebra and Probability (Math 19b, Spring 2011) 154 pages. DEFINITION \u2022 Vector calculus (or vector analysis) is a branch of mathematics concerned with differentiation and integration of vector of vector analysis are simply incapable of allowing one to write down the governing laws in an invariant form, and one has to adopt a di\ufb00erent mathematics from the vector analysis taught in the freshman and sophomore years. In this appendix I use the following notation. With such assignment one constructs a vector eld (scalar eld) in 3-dime Euclidean space. 1) In other words functions f CMU 15-462/662, Fall 2016 Vector Calculus in Computer Graphics Today\u2019s topic: vector calculus. 3 minute read. diff, and the curl at a specific point using evalf. This playlist provides a shapshot of some lectures Cambridge vector calculus lecture notes calculus that I taught at the University of Ottawa in 2001 and at Dalhousie University in 2007 and 2013. Unless made explicitly, we will assume that vector and scalar elds considered in this lecture have continuous derivatives. Since a vector has no position, we typically indicate a vector Matrices, Vectors, and Vector Calculus In this chapter, we will focus on the mathematical tools required for the course. So Scheme 2 is the one where you get to drop a midterm, and Scheme 3 is for students who for any reason find themselves unable to regularly attend lectures. You could say it is the most important if you're willing to play it slightly fast and loose with definitions and include in it the subset of low-dimensional linear algebra that vector calculus ISBN:9781319083632. Since then, while I have had ample opportunity to teach, use, and even program numerous ideas from vector calculus These lecture videos are organized in an order that corresponds with the current book we are using for our Math2210, Calculus 3, courses ( Calculus, with CMU 15-462/662, Fall 2017 Vector Calculus in Computer Graphics Today\u2019s topic: vector calculus. We found in Chapter 2 that there were various ways of taking derivatives of fields. Shown in green are a vector Chapter 4: VECTOR CALCULUS IN 2D. , An Introduction to Riemannian Geometry and the Tensor Calculus, Internet Supplement for Vector Calculus. Vector-valued functions are also written in the form. Tromba, Vector Calculus In this video lesson, GMath Calculus Donny Lee gives a basic example of implementing the line integral. 2 Integral Calculus Vector Calculus Vector Calculus 20E, Spring 2013, Lecture A, Midterm 1 Fifty minutes, four problems. , \u20d7. Lecture Notes on Variational and Tensor Calculus. Marsden and Anthony Tromba Section Lectures Topic Review Assignment (no lecture Vector calculus, or vector analysis, is concerned with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space. Curves. A vector space is a collection of objects called vectors 2. # Define the independent variables using Symbol x = Symbol('x') y = Symbol('y') # Define the vector vector (or a scalar). 1. We denote R = of vector analysis are simply incapable of allowing one to write down the governing laws in an invariant form, and one has to adopt a di\ufb00erent mathematics from the vector analysis taught in the freshman and sophomore years. ac. Download these Free Vector Calculus MCQ Quiz Pdf and In this course we shall extend notions of di erential calculus from functions of one variable to more general functions f: Rn!Rm: (1. \u2022A work done by a constant force F in moving object from point P to point Q in space is . Tromba, Vector Calculus Sl. Free classes & tests. Matrices, Vectors, and Vector Calculus In this chapter, we will focus on the mathematical tools required for the course. 8. ii. This post contains some of the important notes which come in handy while working with vector-calculus. 1: An Introduction to Vectors: Discovering Vectors with focus on adding, subtracting, position vectors, unit vectors and magnitude. 2 Vectors expressed in terms of Unit Vectors in Rectangular coordinate Systems - A simple and convenient way to express vector quantities Let: i = unit vector along the x-axis j = unit vector along the y-axis k = unit vector along the z-axis in a rectangular coordinate system (x,y,z), or a cylindrical polar coordinate system (r, \u03b8,z). Variational Calculus: Part I: Chapter 1. Lecture 10: Dot Product In 3-D. Lecture Anyone know of an online course or set of video lectures on John Hubbard's textbook on Vector Calculus, Linear Algebra, Related Threads on Hubbard\u2019s vector calculus text Poll; Calculus Vector Calculus Vector Calculus 20E, Winter 2017, Lecture B, Midterm 2 Fifty minutes, three problems. MAT1236 Calculus 1 Topic 2: Vector Calculus Dr Steven Richardson Semester 2, 2014 1 / 46 Lecture Content 1. 99 Wholesale:$228. Vector calculus uses extensive variations of mathematics from differential geometry to multivariable calculus. This course covers vector and multi-variable calculus. Michael Medvinsky, NCSU online lectures 03/2020. Reminder A basis of an n-dimensional vector Lecture 02 - Vector Algebra in Component Form: Lecture 03 - Vector Triple Products: Lecture 04 - Vector Differential Calculus: Gradient: Lecture 05 - Divergence: Lecture 06 - Curl: Lecture 07 - Tutorial on Differential Vector Calculus: Lecture 08 - More Problems on Differential Vector Calculus: Lecture 09 - Vector Integral Calculus The identities curl (grad (f)) = 0 and div (curl (F)) = 0 need conditions on the scalar field f and the vector field F, namely continuous second partials in a The curl of the gradient of any continuously twice-differentiable scalar field. Contents Lecture 1. These lecture notes cannot be duplicated and distributed without explicit permission of the author. Aspects of Vector Calculus \u201cO could I flow like thee, and make thy stream Vector Fields: A vector field is a function that assigns a vector to each point in calculus. Calculus This is a quick review of some of the major concepts in vector calculus that is used in this class. MANMOHAN DASH, PHYSICIST, TEACHER ! Physics for \u2018Engineers and Physicists\u2019 \u201cA concise course of important results\u201d Lecture - 1 Vector Calculus and Operations Lectures Notes for Calculus III (Multivariable Calculus) The notes below follow closely the textbook Introduction to Linear Algebra, Fourth Edition by Gilbert Strang. Fundamental Theorem for Line Integrals(cont) \u2022Theorem: Suppose F=<P,Q> is a conservative vector prepared my lectures. Vector fields represent the distribution of a given vector to each point in the subset of the space. License: Creative Commons BY-NC-SA More information at ocw. The lecture notes [2], the book [3] and the \u201cVector Calculus Primer\u201d [6] are available online; on the web page [4] of O. Schedule: MWTh@11AM or @12:30PM, Fall only. This book covers the following topics: Differentiation, Higher-Order Derivatives and Extrema, Vector Valued Functions, Double and Triple Integrals, Integrals over Curves and Surfaces and the Integral Theorems of Vector Lectures on Vector Calculus Paul Renteln Department of Physics California State University San Bernardino, CA 92407 March, 2009; Revised March, 2011 c Paul Renteln, 2009, 2011. org/learn/vector-calculus Lecture 1. We will invariably consider \ufb01nite-dimensional vector spaces. 1 Introduction In single-variable calculus, the functions that one Curl\u00b6. anupa at northeastern dot edu. Willard Gibbs and Oliver Heaviside near the end of the 19th century, LECTURES ON VECTOR CALCULUS. 5) where is the angle between a and b and u is a unit vector Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. It is the second semester in the freshman calculus sequence. One can never know for sure what a deserted area looks like. MAT203 will not be offered in Fall 2020. Lecture Step 1: Give the vectors u and v (from rule 1) some components. TBA. 1 INTRODUCTION In vector calculus, we deal with two types of functions: Scalar Functions (or Scalar Field) and Vector Functions (or Vector 138 MIT 3. e. The least squares estimator of \u03b2 minimizes f(\u03b2) = (y \u2212X\u03b2)>(y \u2212X\u03b2). Lecture Mathematical Tripos: IA Vector Calculus e c S. Lectures: MWF 9-10 in PCYNH 109 Lecture schedule and notes available below. Amanda Harsy \u00a9Harsy 2020 July 20, 2020 i. However, I will use linear algebra. Lecture Notes on Classical Mechanics (A Work in Progress) Daniel Arovas Department of Physics University of Calculus 1- Limits and Derivatives. Lou Talman of Metro State lecture on the calculus Vector Calculus - Fall 2011 Meetings. Included are the lecture Lecture Notes: Introduction to Real Analysis Lectures Notes: Topics in Vector Calculus Book: Jerrold E. A familiarity with some basic facts about the 3 the Kronecker delta symbol ij, de ned by ij =1ifi= jand ij =0fori6= j,withi;jranging over the values 1,2,3, represents the 9 quantities 11 =1 21 =0 31 =0 12 =0 22 Vector calculus - SlideShare Vector calculus is also known as vector analysis which deals with the differentiation and the integration of the vector field in the three-dimensional Euclidean space. A familiarity with some basic facts Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Let a r, a \u03d5, and a z be unit vectors along r, \u03d5 and z directions, respectively in the Multivariable Calculus Lecture Notes (PDF 105P) This lecture note is really good for studying Multivariable calculus. 1 1. Numerade's Calculus 3 course focuses on Calculus and its applications in different fields of Mathematics. Tensor Calculus (The Dual of a Vector Space) Tensor Calculus 4-5 (Tensors as Multilinear Maps; Integral Curves; The Commutator) Tensor Calculus These lecture notes cannot be duplicated and distributed without explicit permission of the author. Lecture 2: Vector And The Circle. (8 Jan) The midterm exam will be organized in the lecture on 14 Mar (which is in Week 9). Brown. The scope covers only linear algebra (more on this when the Definition. The idea behind the vector calculus is to utilize vectors . Lecture 2: Review of Vector Calculus Instructor: Dr. Lecture Tutorial on vector calculus and curvilinear coordinates; Introduction to electrostatics; Continuous charge distribution: Line charge; Electric field Vector Calculus - Fundamental Theorem fo Space Curves pt1 tutorial of Vector Calculus II course by Prof Donylee of Online Tutorials. (6. Please start each problem on a new page. \u03c6 {\\displaystyle \\varphi } is always the zero vector These lecture videos are organized in an order that corresponds with the current book we are using for our Math1210, Calculus 1, courses ( Calculus, with Course Description. 6. Gradient of a Scalar Field & Directional Derivative | Normal Vector. Preliminaries 1 1. \u2022Unit tangent vector edge of vector calculus and real analysis, some basic elements of point set topology and linear algebra. Example : A~(x;y;z) = (x;xy;xz) (\u2019(x;y;z) = x2yz) is a vector Aspects of Vector Calculus \u201cO could I flow like thee, and make thy stream Vector Fields: A vector field is a function that assigns a vector to each point in Mathematics 31CH: \\Honors Vector Calculus\" Syllabus (revised September 2016) Lecture schedule based on: Vector Calculus, Linear Algebra, and Di erential Forms: A Uni ed Approach, fth edition by John H. lamar. The idea behind the vector calculus is to utilize vectors Listen on Apple Podcasts. Tcheslavski Contact: gleb@ee. x y O \u02da x0 y0 x y O \u02da Figure 1: Left: change of reference frame. Topics include vectors and matrices, partial derivatives, double and triple integrals, and vector calculus calculus lecture notes ppt, These notes stem from my own need to refresh my memory on the fundamentals of tensor calculus, having seriously considered them last some 25 years ago in grad school. We'll cover the essential calculus of such vector Vector calculus - basics A vector \u2013 standard notation for three dimensions Unit vectors i,j,kare vectors of magnitude 1 in directions of the x,y,z axes. We\u2019ll start with the concepts of partition, Riemann sum and Riemann Integrable functions and their properties. 82. The course includes the concept of vectors, Dot Product and Cross Product of vectors, Vector Vector Calculus. J. Be prepared to draw your own figures! Vector Calculus Example 5 Let y be an n dimensional column vector of known constants, X be an n\u00d7m matrix of full column rank, and \u03b2 be an m dimensional vector of unknown variables. Vector Calculus with Applications 17. Lecture 20: Vector Calculus - Fundamental Theorem fo Space Curves pt1. 2. That there must be a different behavior is also intuitively clear: if we described an \u2018arrow\u2019 by coordinates, and we then modify the basis vectors Math 20E Syllabus - Vector Calculus (revised June 2021) Lecture schedule based on Vector Calculus, sixth edition by Jerrold E. Chris Tisdell gives 88 video lectures on Vector Calculus. Vectors are denoted with an arrow over the top of the variable. 4. The underlying physical meaning \u2014 that is, why they are worth bothering about. uk, Lent 2000 0 Introduction 0. All vectors (Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 16. The course is organized into 53 short lecture Get Vector Calculus Multiple Choice Questions (MCQ Quiz) with answers and detailed solutions. com for an indepth study and more calculus related lessons. Wilson, Fall 2006 1 Darcy\u00d5s Law in 3D \u00a5Today \u00d0Vector Calculus \u00d0Darcy\u00d5s Law in 3D q=\"K!h Brief Review of Vector Calculus \u00a5A scalar has only a magnitude \u00a5A vector is characterized by both direction and magnitude. 3. , Springer-Verlag, Berlin, 1954. Why is vector calculus important for computer Lecture notes, lectures 15-17 100% (1) Pages: 16 2015/2016 16 pages 2015/2016 100% (1) Save Vector Calculus Advanced Student Notes New None Pages: 166 This is a quick review of some of the major concepts in vector calculus that is used in this class. ) 1. This book covers the following topics: Differentiation, Higher-Order Derivatives and Extrema, Vector Valued Functions, Double and Triple Integrals, Integrals over Curves and Surfaces and the Integral Theorems of Vector 3. Tensor calculus vector (or a scalar). The most important object in our course is the vector field, which assigns a vector to every point in some subset of space. \u2022 Baxandall and Liebeck, \u201cVector Calculus Lecture Notes: Introduction to Real Analysis Lectures Notes: Topics in Vector Calculus Book: Jerrold E. This note contains the following subcategories Vectors in R3, Cylinders and Quadric Surfaces, Partial Derivatives, Lagrange Multipliers, Triple Integrals, Line Integrals of Vector Understand the concept of Vector & Calculus - Lecture 2 with IIT JEE course curated by Abhilash Sharma on Unacademy. This package includes and Hardcover. Vector Calculus \u2013 Line Integrals of Vector Field | Example & Solution. This course is a study of the calculus of functions of several variables (vector arithmetic and vector calculus). 4. In the Euclidean space, the vector 3 The projection of a vector a on b is equal to a eb, where eb = b=jbj is the unit vector in direction of b. Buy + Hardcover. In this course, we begin our Lectures on Vector Calculus - CSUSB vectors, how to take scalar and vector products of vectors, and something of how to describe geometric and physical entities using vectors. J. The main concepts that will be covered are: \u2022 Coordinate transformations \u2022 Matrix operations \u2022 Scalars and vectors \u2022 Vector calculus edge of vector calculus and real analysis, some basic elements of point set topology and linear algebra. 1 of Stewart\u2019s Calculus. Di erentials and Taylor Series 71 The di erential of a function. Magnitude of a vector Position vector is a vector r from the origin to the current position where x,y,z, are projections of r to the coordinate axes. Join me on Coursera: https://www. This bestselling vector calculus ERTH403/HYD503 Lecture 6 Hydrology Program, New Mexico Tech, Prof. e we go from the Average Rate of Change to the Instantaneous Rate of Change by letting the interval over which the Average Rate of Change is measured go to zero. 3) Recall the basic idea of the Generalized Fundamental Theorem of Calculus: If F is a gradient or conservative vector Line Integral of Vector Field \u2022Reminder: \u2022A work done by variable force f(x) in moving a particle from a to b along the x- axis is given by . The convention that I will try to follow in the lectures is that if we are interested in locating a point in space, we will use a row vector This course will offer a detailed introduction to integral and vector calculus. Variational Calculus: Part II: Chapters 2-3. Shown in green are a vector View Vector_Calculus_Lecture_notes_. , \u2013 In earlier courses, you may have learned that a vector is, basically, an arrow \u2013 That\u2019s true in three dimensions, but this new definition allows one to create higher-dimensional vectors 3. 1 ( 11 ) Lecture Details. Hubbard and Barbara Burke Hubbard. \u2022 Baxandall and Liebeck, \u201cVector Calculus Matrices, Vectors, and Vector Calculus In this chapter, we will focus on the mathematical tools required for the course. 74 Lecture First, fix the y variable and compute the partial derivative for f (x,y) = x\u00b2- y\u00b2 with respect to x to get \u2202f (x,y)/ \u2202x = 2x-0, since y is a constant its Lecture 3: Vectors \u2022 Any set of numbers that transform under a rotation the same way that a point in space does is called a vector \u2013 i. 2 Cross product The cross product, a b between two vectors a and b is a vector de ned by a b= absin( )u; 0 \u02c7; (2. Online, via Zoom. All unit vectors 4. Perform various operations with vectors like adding, subtracting, scaling, and Remark 2. 3 Vector Calculus In addition to Linear Algebra, Vector calculus is a key component of any Machine Learning project. Instead of Vector Calculus lectures. This is a collection of video lectures given by Professor Chris Tisdell, presenting vector calculus in an applied and engineering context. Cartesian coordinates. Brief Course Description: Covers largely the same mathematical topics as MAT201, namely vectors ExtravagantDreams. L1: Differentiation of Vectors A whole set of objects (vectors) on which we can perform vector addition and scalar multiplication with properties given by Eqs. Volume 1 is concerned Ricci Calculus, 2nd ed. A vector-valued function is a function of the form. In a more general sense the broad approach and philosophy taken has been in uenced by: Volume I: A Brief Review of Some Mathematical Preliminaries I. I cannot recall every source I have used but certainly they include those listed at the end of each chapter. No Access. Best lecture on calc University library. Retail:$284. Related Courses. In the text, elements of Rn are denoted by row{vectors; in the lectures and homework assignments, we will use column vectors. Lecture 3: Scalar Multiplication. 1 Schedule This is a copy from the booklet of schedules. Hubbard, Vector calculus, linear algebra, and differential forms-the Honors Calculus Lecture 6: Parametric Equations And Vectors: Example 1. Linear algebra is not a prerequisite for this course. The notes were written by Sigurd Angenent, starting next three semesters of calculus In this lecture, we extend the theory of calculus of variations from a single integral setting to a multivariate integral setting, including the E-L equation, the criteria for weak and strong minima, Jacobi fields, and the Weierstrass excess function, etc. No Chapter Name English; 1: Lecture 1 : Partition, Riemann intergrability and One example: Download Verified; 2: Lecture 2 : Partition, Riemann intergrability and All of these questions involve understanding vectors and derivatives of multivariable functions. We also illustrate how to find a vector from its starting and end points. Derive the expression of this estimator. Vector calculus is a form of mathematics that is focused on the integration of vector fields. kumar. Hubbard, Vector calculus, linear algebra, and differential forms-the Honors Calculus I believe calculus is best learned through four or five short lectures each week throughout a 14-week semester, and this course of video lectures is designed This is the second volume of a two-volume work on vectors and tensors. In this course, Prof. with scalar \ufb01eld K = R. Topics include vectors and matrices, partial derivatives, double and triple integrals, and vector calculus Vector calculus was developed from quaternion analysis by J. Quote. Marsden and Anthony J. mit. \u00a5Vectors Vector Calculus MCQ Question 4. I\u2019m going to use a and b here, but the choice is arbitrary: u = (a 1, a 2) v = (b 1, b 2) Differential equations and vector calculus Course Objectives. We then move to anti-derivatives and will look in to few classical theorems of integral calculus such as fundamental theorem of integral calculus. where the component functions f, g, and h, are real-valued functions of the parameter t. Gelfand and S. Download Solution PDF. LIMITS In this first animation we see the secant line become the tangent line i. Lectures in Vector Calculus in 2D. IIT JEE. 2 Lecture 10. Lecture 1. E. The main concepts that will be covered are: \u2022 Coordinate transformations \u2022 Matrix operations \u2022 Scalars and vectors \u2022 Vector calculus Aspects of Vector Calculus \u201cO could I flow like thee, and make thy stream Vector Fields: A vector field is a function that assigns a vector to each point in Vector Calculus part II By Dr. M. Fomin, Calculus Learn what vectors are and how they can be used to model real-world situations. 05. A real number xis positive, zero, or negative and is rational or irrational. Marsden (CalTech) & A. 71 The Taylor series. STERNBERG, S. (2. Scheme 3: 20% Homework, 25% Midterm 1, 25% Midterm 2, 30% Final Exam. In both cases, the first form of the function defines a two-dimensional vector These lecture videos are organized in an order that corresponds with the current book we are using for our Math2210, Calculus 3, courses ( Calculus, with MATH 25000: Calculus III Lecture Notes Created by Dr. The plane. Covers topics including vector functions, multivariate functions, partial derivatives, multiple integrals and an introduction to vector calculus. A vector is depicted as an arrow starting at one point in space and ending at another point. This is a series of lectures for \"Several Variable Calculus\" and \"Vector Calculus\", which is a 2nd-year mathematics subject taught at UNSW, Sydney. Surface Area\u2013 Dr. This is a series of lectures for \"Several Variable Calculus\" and \"Vector Calculus CALCULUS ON MANIFOLDS 5 (tautologically) R1 with R eld, then the di erential becomes an element of the dual vector space T a U\u2019(Rn) . Lecture 8: How To Determine If The Lines Are Parallel? Lecture 9: How To Determine If The Lines Intersect. Di erentiability 29 1. (This lecture corresponds to Section 5. Effective: 2017-08-01. To understand the three major theorems of vector calculus. WEATHERBURN, C. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. Resource Guide to Vector Calculus. They consist largely of the material presented during the lectures Course Description. pdf from MAT 1236 at Edith Cowan University. 65 Lecture 11. Published: March 01, 2020. You Differential Calculus Lecture Notes Veselin Jungic & Jamie Mulholland Department of Mathematics Simon Fraser University c Jungic/Mulholland, August MAT203 Advanced Vector Calculus. I believe an interested student can Vector calculus is one of the most useful branches of mathematics for game development. Class : Lecture Scheme 2: 5% participation, 25% Homework, 30% Best Midterm, 40% Final Exam. Topics covered in these notes include the un-typed lambda calculus Vector Calculus 1 multivariable calculus 1. M, 2:50pm-4:30pm. Dynamical systems, Spring 2005 (183 pages) Linear Algebra (21b, Spring 2018) College Multivariable, (Fall 2017) Calculus Defines vectors, vector addition and vector subtraction. Lecture Vector Calculus - Fall 2011 Meetings. Tensor Calculus (The Dual of a Vector Space) Tensor Calculus 4-5 (Tensors as Multilinear Maps; Integral Curves; The Commutator) Tensor Calculus Vector Calculus (Multivariate Calculus)B SC 4th Semester (CBCS)Mathematics (Honours)MAT-HC-4016Lecture 1 Vectors in Euclidean Space 1. Lecture 7: Parametric Equations And Vectors: Example 2. g. To enlighten the learners in the concept of differential equations and multivariable calculus Scheme 2: 5% participation, 25% Homework, 30% Best Midterm, 40% Final Exam. coursera. The main concepts that will be covered are: \u2022 Coordinate transformations \u2022 Matrix operations \u2022 Scalars and vectors \u2022 Vector calculus Multivariable Calculus Lectures Richard J. Vector Space. But like wolfsy said, if you are trying to Lecture Notes on Variational and Tensor Calculus. A two-dimensional vector \ufb01eld is a function f that maps each point (x,y) in R2 to a two-dimensional vector hu,vi, and similarly a three-dimensional vector \ufb01eld maps (x,y,z) to hu,v,wi. You must substitute the parametric equations into both the vector field and position vector and then integrate. The term \"vector calculus\" is sometimes used as a synonym for the broader subject of multivariable calculus, which spans vector calculus MATH 252-3: Vector Calculus Course Webpage Quick Links: Library Reserves WebCT Course Webpage This Week Documents & Homework Information Vector Calculus (MAST20009) COVID-19 vaccination (or valid exemption) is a requirement for anyone attending our campuses. For Vector Calculus I like J. Vector Calculus previous lecture notes by Ben Allanach and Jonathan Evans ; Vector Calculus yet earlier lecture notes by Stephen Cowley. These theorems are needed in core engineering subjects such as Electromagnetism and Fluid Mechanics. Slide 19. A familiarity with some basic facts 1 17. Example : A~(x;y;z) = (x;xy;xz) (\u2019(x;y;z) = x2yz) is a vector linear transformations and vector spaces are necessary ingredients for a proper discussion of ad-vanced calculus. Topics covered are Three Dimensional Space, Equations of normal vectors and tangent planes 24 1. Vector Fields 65 Vector Fields. Contents 1 Syllabus and Scheduleix Our last month will be combining the multivariate calculus with vector calculus and this culminates in several important theorems which tie all of Calculus Basic Concepts \u2013 In this section we will introduce some common notation for vectors as well as some of the basic concepts about vectors such as the magnitude of a vector and unit vectors. Home New to This Edition WebAssign for Vector Calculus. V. Hinglish Physics. Lecture 6: Addition Of Vectors MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. Unit vectors, vectors Lecture 1 Vectors View this lecture on YouTube We de\ufb01ne a vector in three-dimensional Euclidean space as having a length (or magnitude) and a direction. Knill you can \ufb01nd plenty of exercises, lecture edge of vector calculus and real analysis, some basic elements of point set topology and linear algebra. Tromba (UCSC). In organizing this lecture note, I am indebted by Cedar Crest College Calculus IV Lecture Course Description. C Carter Lecture 11 where i j is the angle between two vectors iand j, and ij k is the angle between the vector kand the plane spanned by iand j, is equal to the parallelepiped that has ~a, ~b, and ~cemanating from its bottom-back corner. Course objective : To apply the basic concepts found in a first year calculus course to multivariable functions (limits, differentiation, and integration). edu Office Hours: Room 2030 \u2013 A free PowerPoint 3\u20131 Vector integrals; the line integral of \u2207\u03c8\u03c8.\n\nddio ais0 gfa1 s5d3 tkut 2nra i8oi rbsf 1yr5 2c3v zyiu j7is ybmg b85p vequ h5fh fkod gmoh 9i6a fwkg i9fz 5dte lfof qzqs ufof jlgl 9qty nn8m zh21 tqfw xntt icno zc2z bbrz pftk grhf fmuh lt6z epdr zsy2 04tj owut a3r9 3ncf 31s4 58cq ctnq s0bk e5t1 8b9w l0vx 70bg zwy8 1f7o rgov 7gkd kttu 68nx xq6i 4bqm i57x z1i2 j1kx vtp8 d4qj ytpu jvev k6hf xtl4 v4cz ilje tubv sx3e dvy8 z2p2 c6mk uzdd 80pk dt62 3uku bwla hxdi jxkx 2pmr 6pqi eljh qtkd 6yju hcbo jkba kk8l a8kb lgbe amsj xx1d xd9t jmit urjj f5kx rr49\n\nBT", "date": "2022-07-04 06:16:32", "meta": {"domain": "lazzeri.cl", "url": "http://apoderados.lazzeri.cl/wp-content/uploads/2022/03/kgqw5cvh/lecture-on-vector-calculus.html", "openwebmath_score": 0.7495718598365784, "openwebmath_perplexity": 1694.6880656825333, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. 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{"url": "https://encyclopediaofmath.org/index.php?title=Stochastic_matrix&oldid=35214", "text": "# Stochastic matrix\n\n(diff) \u2190 Older revision | Latest revision (diff) | Newer revision \u2192 (diff)\n\n2010 Mathematics Subject Classification: Primary: 15B51 Secondary: 60J10 [MSN][ZBL]\n\nA stochastic matrix is a square (possibly infinite) matrix $P=[p_{ij}]$ with non-negative elements, for which $$\\sum_j p_{ij} = 1 \\quad \\text{for all i.}$$ The set of all stochastic matrices of order $n$ is the convex hull of the set of $n^n$ stochastic matrices consisting of zeros and ones. Any stochastic matrix $P$ can be considered as the matrix of transition probabilities of a discrete Markov chain $\\xi^P(t)$.\n\nThe absolute values of the eigenvalues of stochastic matrices do not exceed 1; 1 is an eigenvalue of any stochastic matrix. If a stochastic matrix $P$ is indecomposable (the Markov chain $\\xi^P(t)$ has one class of positive states), then 1 is a simple eigenvalue of $P$ (i.e. it has multiplicity 1); in general, the multiplicity of the eigenvalue 1 coincides with the number of classes of positive states of the Markov chain $\\xi^P(t)$. If a stochastic matrix is indecomposable and if the class of positive states of the Markov chain has period $d$, then the set of all eigenvalues of $P$, as a set of points in the complex plane, is mapped onto itself by rotation through an angle $2\\pi/d$. When $d=1$, the stochastic matrix $P$ and the Markov chain $\\xi^P(t)$ are called aperiodic.\n\nThe left eigenvectors $\\pi = \\{\\pi_j\\}$ of $P$ of finite order, corresponding to the eigenvalue 1: $$\\label{eq1} \\pi_j = \\sum_i \\pi_i p_{ij} \\quad \\text{for all}\\ j\\,,$$ and satisfying the conditions $\\pi_j \\geq 0$, $\\sum_j\\pi_j = 1$, define the stationary distributions of the Markov chain $\\xi^P(t)$; in the case of an indecomposable matrix $P$, the stationary distribution is unique.\n\nIf $P$ is an indecomposable aperiodic stochastic matrix of finite order, then the following limit exists: $$\\label{eq2} \\lim_{n\\rightarrow\\infty} P^n = \\Pi,$$ where $\\Pi$ is the matrix all rows of which coincide with the vector $\\pi$ (see also Markov chain, ergodic; for infinite stochastic matrices $P$, the system of equations \\ref{eq1} may have no non-zero non-negative solutions that satisfy the condition $\\sum_j \\pi_j < \\infty$; in this case $\\Pi$ is the zero matrix). The rate of convergence in \\ref{eq2} can be estimated by a geometric progression with any exponent $\\rho$ that has absolute value greater than the absolute values of all the eigenvalues of $P$ other than 1.\n\nIf $P = [p_{ij}]$ is a stochastic matrix of order $n$, then any of its eigenvalues $\\lambda$ satisfies the inequality (see [MM]): $$\\left| \\lambda - \\omega \\right| \\leq 1-\\omega, \\quad \\text{where \\omega = \\min_{1 \\leq i \\leq n} p_{ii}.}$$ The union $M_n$ of the sets of eigenvalues of all stochastic matrices of order $n$ has been described (see [Ka]).\n\nA stochastic matrix $P=[p_{ij}]$ that satisfies the extra condition $$\\sum_i p_{ij} = 1 \\quad \\text{for all j}$$ is called a doubly-stochastic matrix. The set of doubly-stochastic matrices of order $n$ is the convex hull of the set of $n!$ permutation matrices of order $n$ (i.e. doubly-stochastic matrices consisting of zeros and ones). A finite Markov chain $\\xi^P(t)$ with a doubly-stochastic matrix $P$ has the uniform stationary distribution.\n\n#### References\n\n [G] F.R. Gantmacher, \"The theory of matrices\" , 1 , Chelsea, reprint (1977) (Translated from Russian) MR1657129 MR0107649 MR0107648 Zbl 0927.15002 Zbl 0927.15001 Zbl 0085.01001 [B] R. Bellman, \"Introduction to matrix analysis\" , McGraw-Hill (1960) MR0122820 Zbl 0124.01001 [MM] M. Marcus, H. Minc, \"A survey of matrix theory and matrix inequalities\" , Allyn & Bacon (1964) MR0162808 Zbl 0126.02404 [Ka] F.I. Karpelevich, \"On the characteristic roots of matrices with non-negative entries\" Izv. Akad. Nauk SSSR Ser. Mat. , 15 (1951) pp. 361\u2013383 (In Russian)\n\nGiven a real $n\\times n$ matrix $A$ with non-negative entries, the question arises whether there are invertible positive diagonal matrices $D_1$ and $D_2$ such that $D_1AD_2$ is a doubly-stochastic matrix, and to what extent the $D_1$ and $D_2$ are unique. Such theorems are known as $DAD$-theorems. They are of interest in telecommunications and statistics, [C], [F], [Kr].\nA matrix $A$ is fully decomposable if there do not exist permutation matrices $P$ and $Q$ such that $$PAQ = \\left[ \\begin{array}{cc} A_1 & 0 \\\\ B & A_2 \\end{array} \\right].$$ A $1 \\times 1$ matrix is fully indecomposable if it is non-zero.\nThen for a non-negative square matrix $A$ there exist positive diagonal matrices $D_1$ and $D_2$ such that $D_1AD_2$ is doubly stochastic if and only if there exist permutation matrices $P$ and $Q$ such that $PAQ$ is a direct sum of fully indecomposable matrices [SK], [BPS].", "date": "2021-08-05 14:43:12", "meta": {"domain": "encyclopediaofmath.org", "url": "https://encyclopediaofmath.org/index.php?title=Stochastic_matrix&oldid=35214", "openwebmath_score": 0.9489653706550598, "openwebmath_perplexity": 198.1190532288455, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9923043519112751, "lm_q2_score": 0.8577681068080749, "lm_q1q2_score": 0.8511670253163481}}
{"url": "https://math.stackexchange.com/questions/3161357/sum-of-the-inverse-of-a-geometric-series", "text": "# Sum of the inverse of a geometric series?\n\nI'm trying to solve for this summation:\n\n$$\\sum_{j=0}^{i} {\\left(\\frac 1 2\\right)^j}$$\n\nThis looks a lot like a geometric series, but it appears to be inverted. Upon plugging the sum into Wolfram Alpha, I find the answer to be\n\n$$2-2^{-i}$$\n\nbut I don't understand how it gets there. Am I able to consider this a geometric series at all? It almost seems closer to the harmonic series.\n\n\u2022 Welcome to Math Stack Exchange. Note $\\frac 1 {2^j}=\\left(\\frac1 2\\right)^j$ \u2013\u00a0J. W. Tanner Mar 25 '19 at 3:36\n\u2022 Is $i$ some finite number? Or are you trying to find the value of the series for any arbitrary $i$? \u2013\u00a0kkc Mar 25 '19 at 3:36\n\u2022 The reciprocals of each term of a geometric series is also a geometric one \u2013\u00a0lab bhattacharjee Mar 25 '19 at 3:37\n\u2022 i is finite but arbitrary. The sum does not approach infinity. \u2013\u00a0bpryan Mar 25 '19 at 3:39\n\u2022 My point of noting $\\frac 1 {2^j}=\\left(\\frac 1 2\\right)^j$ was not that $\\frac 1 {2^j}$ was incorrect but rather that this is a geometric series \u2013\u00a0J. W. Tanner Mar 25 '19 at 3:49\n\n$$\\sum_{j=0}^{i} \\frac{1}{2^j}=\\sum_{j=0}^{i}\\left( \\frac{1}{2}\\right)^j=\\frac{1-\\left(\\frac12\\right)^{i+1}}{1-\\frac12}=2\\left(1-\\left(\\frac12\\right)^{i+1}\\right)=2-\\left(\\frac12\\right)^{i}=2-2^{-i}$$\nAs $$\\frac1{2^j}=\\left(\\frac12\\right)^j$$, this is also a geometric sum with the common ratio of $$r=\\frac12$$. So you apply the formula for geometric sums $$\\sum_{j=0}^nr^j=\\frac{1-r^{n+1}}{1-r}$$to obtain the answer you have written.", "date": "2021-07-26 17:58:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3161357/sum-of-the-inverse-of-a-geometric-series", "openwebmath_score": 0.8202585577964783, "openwebmath_perplexity": 272.26841448436033, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9923043523664414, "lm_q2_score": 0.8577681031721325, "lm_q1q2_score": 0.8511670220988138}}
{"url": "https://math.stackexchange.com/questions/3675762/does-the-category-of-local-rings-with-residue-field-f-have-an-initial-object", "text": "# Does the category of local rings with residue field $F$ have an initial object?\n\nLet $$F$$ be a field. Does the category $$C_F$$ of local rings with residue field isomorphic to $$F$$ have an initial object?\n\nThis is, for instance, true if $$F=\\mathbb{F}_{p}$$ for some prime $$p$$: If $$R$$ is a local ring with residue field $$\\mathbb{F}_{p}$$, then any $$x\\in\\mathbb{Z}\\setminus(p)$$ must map to something invertible under the morphism $$\\mathbb{Z}\\longrightarrow R$$. Hence that morphism factors as $$\\mathbb{Z}\\longrightarrow\\mathbb{Z}_{(p)}\\longrightarrow R$$; thus $$\\mathbb{Z}_{(p)}$$ is the initial object.\n\nBut what happens in the more general case? I guess it should be true at least if $$F$$ is of finite type over $$\\mathbb{Z}$$, but I have no idea how to prove it.\n\n(EDIT - To avoid any confusion: I am talking about an initial object in the category of local rings $$R$$ with a fixed surjection $$R\\longrightarrow F$$.)\n\n\u2022 That's a good question. It holds for $F=\\mathbb Q$ as well \u2013\u00a0Maxime Ramzi May 15 '20 at 7:00\n\u2022 @GeorgesElencwajg : I guess it depends on how you set up the category $C_F$ exactly : if it's just a full subcategory of rings, then you are right (composing $R\\to F\\to F$ gives two different morphisms $R\\to F$ because $R\\to F$ is surjective); but if you set it up as a subcategory of $CRing/F$ instead, it's not that clear. \u2013\u00a0Maxime Ramzi May 15 '20 at 7:27\n\u2022 @Georges : I'm not saying $F$-algebras, on the other hand \"the category of local fields with residue field $F$\" could definitely mean \"local rings $R$ with a fixed map $R\\to F$ which exhibits $F$ as the residue field\". So a subcategory of $CRing/F$, not $F/Cring$ \u2013\u00a0Maxime Ramzi May 15 '20 at 8:00\n\u2022 @Georges: I meant it the way Maxime says it. \u2013\u00a0The Thin Whistler May 15 '20 at 8:10\n\nLet $$\\mathbb{F_4}=\\{0,1,w,1+w\\}$$ be the field of 4 elements. Suppose $$R$$ is the initial object in the category described in the question for the field $$\\mathbb{F_4}$$. Then $$R$$ must contain some element $$x$$ which maps to $$w\\in\\mathbb{F_4}$$. Thus we have a map $$f\\colon S\\to R$$, where $$S=\\mathbb{Z}[y]_M$$, sending $$y \\mapsto x$$. Here $$M$$ is the maximal ideal of $$\\mathbb{Z}[y]$$ containing $$2,1+y+y^2$$.\n\nThe following composition must be the identity: $$R \\to S \\stackrel f \\to R$$ Thus $$R=S/I$$ for some ideal $$I\\subset M$$. Further we know $$I\\neq 0$$ as $$S$$ cannot be the initial object: there are multiple distinct maps $$S\\to S$$, such as the identity map and the map sending $$y\\mapsto y+2$$.\n\nUnder the composition $$S \\stackrel f \\to R\\to S$$, we have $$y\\mapsto p/q$$, for some $$p,q$$ integer polynomials in $$y$$. We know $$p/q$$ is not a rational number as $$p/q\\mapsto w\\in\\mathbb{F_4}$$. Thus $$p/q$$ is a non-constant rational function in one variable, taking infinitely many values, which cannot all satisfy the same polynomial over the integers.\n\nOn the other hand, as $$I\\neq 0$$ there must be a polynomial over the integers satisfied by $$p/q$$. This gives us the desired contradiction.\n\n\u2022 This is so clever! How did you come up with this approach? \u2013\u00a0diracdeltafunk May 16 '20 at 2:26\n\u2022 Thankyou. It felt like if there was an initial object it ought to be S, but at the same time it couldn't be. Playing those off against each other and using that S is a well behaved, concretely described ring led to the contradiction. \u2013\u00a0tkf May 16 '20 at 3:55\n\nThe category $$C_{F}$$ possesses a weak initial object $$I_{F}$$, i.e. an object that is unique up to not necessarily unique isomorphism.\n\nLet $$F$$ be a field and $$L$$ be its minimal subfield (the smallest subfield contained in $$F$$). Then either $$L=\\mathbb{F}_{p}$$ for some prime $$p$$ or $$L=\\mathbb{Q}$$.\n\nAssume first that $$F$$ is of finite type over $$L$$. Let $$n\\in\\mathbb{N}$$ be the smallest natural number so that $$F=L[x_{1},...,x_{n}]/\\mathfrak{m}$$ for some maximal ideal $$\\mathfrak{m}\\subseteq L[x_{1},...,x_{n}]$$. Let $$\\overline{x}_{i}$$ be the image of $$x_{i}\\in L[x_{1},...,x_{n}]$$ in $$F$$.\n\nLet $$\\zeta:R\\longrightarrow F$$ be a surjection where $$R$$ is a local ring. Since every $$\\overline{x}_{i}$$ has a (not necessarily unique) preimage $$\\zeta^{-1}(\\overline{x}_{i})\\in R$$, there is a (not necessarily unique) morphism $$\\kappa:\\mathbb{Z}[x_{1},...,x_{n}]\\longrightarrow R$$ that fits into a commutative diagram $$\\require{AMScd}$$ $$\\begin{CD} \\mathbb{Z}[x_{1},...,x_{n}]@>{\\kappa}>>R\\\\ @V{\\pi}VV @VV{\\zeta}V\\\\ L[x_{1},...,x_{n}] @>>{\\chi}>F \\end{CD}$$ Let $$\\mathfrak{i}:=\\chi^{-1}\\pi^{-1}(0)=\\pi^{-1}(\\mathfrak{m})$$. The ideal $$\\mathfrak{i}$$ is always prime; it is maximal if and only if $$L=\\mathbb{F}_{p}$$ for some prime $$p$$. Since $$R$$ is local, every element of $$\\mathbb{Z}[x_{1},...,x_{n}]$$ is mapped by $$\\kappa$$ onto something invertible in $$R$$. Hence $$\\kappa$$ factors as $$\\begin{CD} \\mathbb{Z}[x_{1},...,x_{n}] @>>>\\mathbb{Z}[x_{1},...,x_{n}]_{(\\mathfrak{i})} @>{\\lambda}>> R \\end{CD}$$ Thus $$I_{F}:=\\mathbb{Z}[x_{1},...,x_{n}]_{(\\mathfrak{i})}$$ is a weak initial object in the category $$C_{F}$$.\n\nNote that the assignment $$\\kappa\\longleftrightarrow\\lambda$$ is unique in both ways: To each choice of $$\\kappa$$ there is a unique $$\\lambda$$ and vice-versa.\n\nAssume next that $$F$$ is of infinite type over $$L$$. Then $$F$$ is the direct limit of all morphisms $$F'\\longrightarrow F''$$, where $$F',F''$$ are fields of finite type over $$L$$. Since the construction of $$I_{-}$$ is functorial and compatible with direct limits, $$I_{F}$$ can be defined as $$I_{F}:=\\lim_{F'\\text{ of fin. t.}/L}I_{F'}$$.\n\nThe initial object is strong, i.e. unique up to unique isomorphism, if and only if $$F=L$$.\n\nNamely, if $$F=L$$, then $$n=0$$ and the unique morphism $$\\kappa:\\mathbb{Z}\\longrightarrow R$$ induces a unique morphism $$\\lambda:\\mathbb{Z}_{(\\mathfrak{i})}\\longrightarrow R$$.\n\nElse, if $$F\\neq L$$, then $$n\\geq 1$$ and for any $$i\\in\\{1,...,n\\}$$ and any $$s\\in\\mathfrak{i}\\setminus\\{0\\}$$, the map $$\\xi_{i,s}:x_{i}\\mapsto x_{i}+s$$ yields a nontrivial automorphism $$I_{F}\\longrightarrow I_{F}$$ that commutes with the surjection $$I_{F}\\longrightarrow F$$.\n\nMy guess is that the $$\\xi_{i,s}$$ actually generate the whole group $$\\operatorname{Aut}(I_{F})$$, but I have yet to figure out a proof for this...", "date": "2021-02-27 05:43:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3675762/does-the-category-of-local-rings-with-residue-field-f-have-an-initial-object", "openwebmath_score": 0.9474796056747437, "openwebmath_perplexity": 101.1014109815404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676454700792, "lm_q2_score": 0.8688267881258485, "lm_q1q2_score": 0.8511614938445814}}
{"url": "https://math.stackexchange.com/questions/3705602/find-an-equivalent-sequence-as-n-to-infty-of-u-10-u-n1-fracu-nn", "text": "# Find an equivalent sequence as $n\\to +\\infty$ of $u_1>0, u_{n+1} = \\frac{u_n}{n} + \\frac{1}{n^2}$\n\nLet $$u_1>0$$ be a real number. Let us consider $$(u_n)_{n\\geq 1}$$ the sequence such as:\n\n$$\\forall n \\geq 1, u_{n+1} = \\frac{u_n}{n} + \\frac{1}{n^2}\\quad (\\star)$$\n\nFind an equivalent of $$u_n$$ as $$n\\to +\\infty$$.\n\nSo I found a way to show that $$u_n \\sim \\frac{1}{n^2}$$, but I'm quite unhappy with this method because I feel like I found it by chance without understanding anything (I did a lot of trials and found this)\n\nMy method:\n\nI showed by induction that $$u_n \\leq (u_1+1)$$. Thus, $$u_n\\to 0$$ considering $$(\\star)$$.\n\nThen, $$nu_{n+1} = u_n + 1/n$$. Thus (since $$n+1 \\sim n$$), $$nu_n \\to 0$$.\n\nTo end with, I have $$n^2u_{n+1} = nu_n + 1$$. Thus, $$(n+1)^2 u_n \\sim n^2u_n \\to 1$$.\n\nHow would you solve such a problem? Is there any more intuitive method that one may have done?\n\n\u2022 It is right actually. I think elementary proof has you find is rather good. Where the intuition comes it that $u_n$ is ponderated with $n$ and $u_{n+1}$ can be with $n^2$ by multiplying all the equation by $n^2$. So I think that ponderation lead you to the equivalent. \u2013\u00a0EDX Jun 4 '20 at 15:39\n\n1. There is a heuristic argument which is useful for guessing the behavior of $$u_n$$: Rewrite the recurrence relation as\n\n$$u_{n+1} - u_n = \\frac{u_n}{n} - u_n + \\frac{1}{n^2}.$$\n\nIts continuum analogue is the following differential equation:\n\n$$y' = \\frac{y}{x} - y + \\frac{1}{x^2}.$$\n\nUsing the standard method, this equation can be solved as:\n\n$$y(x) = x e^{-x} \\int \\frac{e^x}{x^3} \\, \\mathrm{d}x.$$\n\nThen L'Hospital's Rule then tells that $$y(x) \\sim x^{-2}$$ as $$x \\to \\infty$$. From this observation, we may as well expect that $$u_n \\sim n^{-2}$$.\n\n2. The above ansatz suggests that, in the recurrence relation for $$u_{n+1}$$, $$\\frac{1}{n^2}$$ is the dominating term and $$\\frac{u_n}{n}$$ is much smaller as $$n\\to\\infty$$. In particular, nesting this relation will produce an expansion with ever decreasing terms. This idea can be easily tested as follows:\n\nLet $$r_n = (n-1)u_n$$. Then for $$n \\geq 2$$,\n\n$$r_n = (n-1)u_{n} = \\frac{1}{n-1} + \\frac{r_{n-1}}{n-1}.$$\n\nFrom this, we get\n\n\\begin{align*} r_n &= \\frac{1}{n-1} + \\frac{r_{n-1}}{n-1} \\\\ &= \\frac{1}{n-1} + \\frac{1}{(n-1)(n-2)} + \\frac{r_{n-2}}{(n-1)(n-2)} \\\\ &= \\frac{1}{n-1} + \\frac{1}{(n-1)(n-2)} + \\frac{1}{(n-1)(n-2)(n-3)} + \\frac{r_{n-3}}{(n-1)(n-2)(n-3)} \\\\ &\\qquad\\vdots\\\\ &= \\sum_{k=1}^{n-2} \\frac{1}{(n-1)\\cdots(n-k)} + \\frac{r_2}{(n-1)!} \\end{align*}\n\nUsing this, it is not hard to conclude that $$(n-1)^2 u_n \\to 1$$ as $$n\\to\\infty$$, and in fact, we can extract an asymptotic expansion of $$u_n$$ up to any prescribed order $$\\mathcal{O}(n^{-M})$$.\n\nThis is similar to Sangchul Lee's approach.\n\nIf we multiply $$n!$$ both sides of the recurrence, we obtain $$n!u_{n+1}=(n-1)!u_n+\\frac{n!}{n^2}.$$ Applying the above repeatedly, $$u_{n+1}=\\frac1{n!} \\sum_{k=1}^n \\frac{k!}{k^2} + \\frac{u_1}{n!}.$$\n\n\u2022 That's a really good method! Thanks \u2013\u00a0MiKiDe Jun 7 '20 at 9:17", "date": "2021-06-20 13:57:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3705602/find-an-equivalent-sequence-as-n-to-infty-of-u-10-u-n1-fracu-nn", "openwebmath_score": 0.9996277093887329, "openwebmath_perplexity": 300.08674850912087, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676478446031, "lm_q2_score": 0.8688267813328977, "lm_q1q2_score": 0.8511614892527972}}
{"url": "https://math.stackexchange.com/questions/1637719/given-3-types-of-coins-how-many-ways-can-one-select-20-coins-so-that-no-coi", "text": "# Given $3$ types of coins, how many ways can one select $20$ coins so that no coin is selected more than $8$ times.\n\nSo I was given this question.\n\nGiven $3$ types of coins, how many ways can one select $20$ coins so that no coin is selected more than $8$ times.\n\nFirst I make $x_1 + x_2 + x_3 = 20$ Then $0 \\leq x_i \\leq 8$\n\nThen we use the Inclusion exclusion principle\n\nLet $A_i$ be the non-negative integer solutions to $x_1 + x_2 + x_3 = 20$ $x_i \\geq 9$. Then use inclusion exclusion formula to find $N(A_1 \\bigcap A_2 \\bigcap A_3)$\n\nWhat i don't get is how to apply the inclusion exclusion formula. I know the process to get to it but not how to apply it.\n\n\u2022 Possible duplicate of Inclusion-exclusion principle: Number of integer solutions to equations \u2013\u00a0Cloverr Feb 2 '16 at 18:27\n\u2022 You can also find the coefficient of $x^{20}$ in $(1+x+\\ldots + x^8)^3$. Then you get the right answer as well. \u2013\u00a0Henno Brandsma Feb 2 '16 at 18:27\n\u2022 @Nilanjan thats a different question \u2013\u00a0Zero Feb 2 '16 at 18:31\n\u2022 Do you have to use inclusion-exclusion? The problem is small enough in scale to permit more straightforward approaches. Heck, even plain enumeration will do. \u2013\u00a0Brian Tung Feb 2 '16 at 18:56\n\u2022 I've added inclusion-exclusion to my answer. Interestingly, the approach by generating function arrives at the same expression! \u2013\u00a0Brian Tung Feb 2 '16 at 19:03\n\nInstead of trying to apply formulas take a look at the general theory. Suppose $a_j\\le x_j\\le b_j$ for $j=1,2,3$.\n\n$$(x^{a_1}+x^{a_1+1}+\\cdots+x^{b_1})(x^{a_2}+x^{a_2+1}+\\cdots+x^{b_2})(x^{a_3}+x^{a_3+1}+\\cdots+x^{b_3})$$\n\nEvery solution of $x_1+x_2+x_3=n$ with the constraints $a_j\\le x_j\\le b_j$ contributes $1$ to the coefficient of $x^n$ in the above expression. So the number of solutions is the coefficient of $x^n$.\n\nSubstituting accordingly we see we require the coefficient of $x^{20}$ in $$(1+x+\\cdots+x^8)^3={(1-x^9)^3\\over (1-x)^3}$$ or equivalently in $$(1-3x^9+3x^{18})\\sum_{k=0}^{20}\\binom{2+k}{k}x^k$$ which equals $$\\binom{22}{20}-3\\binom{13}{11}+3\\binom{4}{2}$$\n\n\u2022 The equation $x_1 + x_2 + x_3 = 20$ has only $\\binom{22}{20}$ solutions in the non-negative integers. Check your calculations. \u2013\u00a0N. F. Taussig Feb 2 '16 at 18:42\n\u2022 I replaced $3$ with $20$ erroneously.. \u2013\u00a0Jack's wasted life Feb 2 '16 at 18:45\n\nThere are $15$ different ways to select $20$ coins of three different types, with no type having more than $8$ specimens.\n\nThere are a number of different methods to obtain this number. One is to condition on the number of coins of Type $1$. If there are $4$ coins of Type $1$, then there is only one way to obtain $16$ coins of the other two types; if there are $5$ coins of Type $1$, then there are two ways to obtain $15$ coins of the other two types; and so on. $1+2+3+4+5 = 15$.\n\nAnother way is to consider that the solutions are the intersection of the lattice cube $\\{(x, y, z)) \\in \\mathbb{N}_{\\geq 0}^3 \\mid 0 \\leq x, y, z \\leq 8\\}$ with the plane $x+y+z = 20$. If you have any facility with this, you will notice that the intersection is a triangular lattice of side $5$, so the number of solutions is the $5$th triangular number, $15$.\n\nETA: Finally, we can use inclusion-exclusion, although I find it less straightforward. The number of non-negative integer solutions to $x+y+z = 20$ is, by the usual stars-and-bars approach, $\\binom{22}{2} = 231$. However, this counts solutions where at least one of $x, y, z > 8$. The number of solutions where $x > 8$ is, again by stars-and-bars, $\\binom{13}{2} = 78$, and there are three coin types, so we must subtract $3 \\times 78 = 234$ to obtain $-3$.\n\nAgain, however, this double-counts three cases where at least two of $x, y, z > 8$. There are $\\binom{4}{2} = 6$ ways in which both $x, y > 8$, but there are three different pairs of coin types, so we must add back $3 \\times 6 = 18$ to get $15$. There are no ways to have $x, y, z > 8$ and still add up to $20$, so $15$ is the final result. Note, interestingly, that this is the same expression that Jack's wasted life arrived at, although that method was by way of generating functions.\n\n\u2022 One typographical error: In the last sentence of the first paragraph of the inclusion-exclusion argument, you meant $x > 8$ rather than $x > 0$. \u2013\u00a0N. F. Taussig Feb 2 '16 at 20:14\n\u2022 @N.F.Taussig: Thanks for that. \u2013\u00a0Brian Tung Feb 3 '16 at 18:58\n\nLet $8-x_i=:y_i$ $(1\\leq i\\leq3)$. Then we want all $y_i\\geq0$ and $y_1+y_2+y_3=4$. There are $${4+3-1\\choose 3-1}={6\\choose 2}=15$$ ways to choose admissible $y_i$.", "date": "2019-10-24 01:42:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1637719/given-3-types-of-coins-how-many-ways-can-one-select-20-coins-so-that-no-coi", "openwebmath_score": 0.8478825092315674, "openwebmath_perplexity": 153.6125278349235, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676466573412, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.851161484893855}}
{"url": "https://www.intmath.com/forum/integration-29/different-parabola-equation-when-finding-area:151", "text": "IntMath Home \u00bb Forum home \u00bb Integration \u00bb Different parabola equation when finding area\n\n# Different parabola equation when finding area [Solved!]\n\n### My question\n\nIn chapter 3, compare example 1 to example 3. Why isnt (3 - x^2) the correct parabolic equation for example 3?\n\n### Relevant page\n\n3. Areas Under Curves\n\n### What I've done so far\n\nWhen I went to solve example 3, I used the equation (3 - x^2) for the parabola. I found out that was not the correct parabolic equation upon looking at your solution.\nI am not sure why my equation cannot be used even though the vertex is not at (0,3), i.e. not lined up with the origin? The height is still three so I thought it would be okay.\n\nX\n\nIn chapter 3, compare example 1 to example 3.  Why isnt (3 - x^2) the correct parabolic equation for example 3?\nRelevant page\n\n<a href=\"https://www.intmath.com/integration/3-area-under-curve.php\">3. Areas Under Curves</a>\n\nWhat I've done so far\n\nWhen I went to solve example 3, I used the equation (3 - x^2) for the parabola. I found out that was not the correct parabolic equation upon looking at your solution.\nI am not sure why my equation cannot be used even though the vertex is not at (0,3), i.e. not lined up with the origin?  The height is still three so I thought it would be okay.\n\n## Re: Different parabola equation when finding area\n\n@phinah: Your parabola will certainly have the correct height, but let's consider its width. Where does your parabola intersect with the x-axis?\n\nWhat width at the x-axis does this give you?\n\nX\n\n@phinah: Your parabola will certainly have the correct height, but let's consider its width. Where does your parabola intersect with the <i>x</i>-axis?\n\nWhat width at the <i>x</i>-axis does this give you?\n\n## Re: Different parabola equation when finding area\n\nI realized after I inquired to you that this upside down parabola is a shift one unit to the right on the x-axis.\n\nSo I came up with the -(x-1)^2 + 3.\n\nBut the parabola bottom would start at (0,2) Would that be an issue?\n\nNOTE: The integral would be (-x^2 + 2x + 2) dx for a value of 16/3 square units.\n\nX\n\nI realized after I inquired to you that this upside down parabola is a shift one unit to the right on the x-axis.\n\nSo I came up with the -(x-1)^2 + 3.\n\nBut the parabola bottom would start at (0,2) Would that be an issue?\n\nNOTE: The integral would be (-x^2 + 2x + 2) dx for a value of 16/3 square units.\n\n## Re: Different parabola equation when finding area\n\n@phinah: I edited your answer to include math formatting, which you are encouraged to do.\n\nOnce again, the width at the x-axis is a problem (both with your original attempt and with this revision.)\n\na. Where does your parabola intersect with the x-axis when we use y=3-x^2?\n\nb. What width at the x-axis does this give you?\n\nYou may find some of the background in this article useful: How to find the equation of a quadratic function from its graph\n\nX\n\n@phinah: I edited your answer to include math formatting, which you are encouraged to do.\n\nOnce again, the width at the x-axis is a problem (both with your original attempt and with this revision.)\n\na. Where does your parabola intersect with the x-axis when we use y=3-x^2?\n\nb. What width at the x-axis does this give you?\n\nYou may find some of the background in this article useful: <a href=\"https://www.intmath.com/blog/mathematics/how-to-find-the-equation-of-a-quadratic-function-from-its-graph-6070\">How to find the equation of a quadratic function from its graph</a>\n\n## Re: Different parabola equation when finding area\n\n3 - x^2 = 0\n x = +/- sqrt 3'\nSo my length is 3.46 when it should only be 2.\n\nX\n\n3 - x^2 = 0\n x = +/- sqrt 3'\nSo my length is 3.46 when it should only be 2.\n\n## Re: Different parabola equation when finding area\n\nCorrection\n\nx = + or - sqrt 3\n\nX\n\nCorrection\n\nx = + or - sqrt 3\n\n## Re: Different parabola equation when finding area\n\n@Phinah: Yes, correct, so it's not possible to use y=3-x^2 for your function. But it is possible to use one that passes through (0, 3). Can you find it?\n\nBTW, to achieve \"plus or minus\" in the math entry system, you enter\n\n+-\n\nSo your answer would look like: x = +- sqrt(3).\n\nYou can see all the syntax for mathematical symbols here: ASCIIMath syntax.\n\nX\n\n@Phinah: Yes, correct, so it's not possible to use y=3-x^2 for your function. But it is possible to use one that passes through (0, 3). Can you find it?\n\nBTW, to achieve \"plus or minus\" in the math entry system, you enter\n\n+-\n\nSo your answer would look like: x = +- sqrt(3).\n\nYou can see all the syntax for mathematical symbols here: <a href=\"https://www.intmath.com/help/send-math-email-syntax.php\">ASCIIMath syntax</a>.\n\n## Re: Different parabola equation when finding area\n\nI cannot.\n\nX\n\nI cannot.\n\n## Re: Different parabola equation when finding area\n\nWell, since we know the width of the arch must be 2 m, then we know it must go through (-1, 0) and (1, 0).\n\nSo using the general form of a quadratic,\n\ny = ax^2 + bx + c,\n\nyou can substitute the 3 known points in and thus find the values of a, b and c.\n\nHave a go!\n\nX\n\nWell, since we know the width of the arch must be 2 m, then we know it must go through (-1, 0) and (1, 0).\n\nSo using the general form of a quadratic,\n\ny = ax^2 + bx + c,\n\nyou can substitute the 3 known points in and thus find the values of a, b and c.\n\nHave a go!\n\n## Re: Different parabola equation when finding area\n\n@murray: phinah hasn't responded. Would you like me to answer this?\n\nX\n\n@murray: phinah hasn't responded. Would you like me to answer this?\n\n## Re: Different parabola equation when finding area\n\nX\n\n@stephenB: Sure, please go ahead!\n\n## Re: Different parabola equation when finding area\n\nOkay, so we have the general formula\n\ny = ax^2 + bx + c,\n\nand the 3 points it needs to go through, (-1, 0), (1, 0) and (0, 3).\n\nPlugging them in:\n\n0 = a(1)^2 + b(1) + c  = a + b + c ... [1]\n\n0 = a(-1)^2 + b(-1) + c  = a - b + c ... [2]\n\n3 = a(0)^2 + b(0) + c = c ... [3]\n\n0 = 2a + 2c which means a + c = 0 ...[4]\n\nFrom [3] we know c=3, so from [4], a=-3 and then plugging those into [1] gives b=0.\n\nThat is, the parabola's equation is:\n\ny = -3x^2 + 3\n\nX\n\nOkay, so we have the general formula\n\ny = ax^2 + bx + c,\n\nand the 3 points it needs to go through,  (-1, 0),  (1, 0) and (0, 3).\n\nPlugging them in:\n\n0 = a(1)^2 + b(1) + c  = a + b + c ... [1]\n\n0 = a(-1)^2 + b(-1) + c  = a - b + c ... [2]\n\n3 = a(0)^2 + b(0) + c = c ... [3]\n\n0 = 2a + 2c which means a + c = 0 ...[4]\n\nFrom [3] we know c=3, so from [4], a=-3 and then plugging those into [1] gives b=0.\n\nThat is, the parabola's equation is:\n\ny = -3x^2 + 3\n\n## Re: Different parabola equation when finding area\n\nThanks, stephenB.\n\nI think if phinah tries the equation you found, he will find the area of the parabolic glass pane will the the same as in my solution.\n\nX\n\nThanks, stephenB.\n\nI think if phinah tries the equation you found, he will find the area of the parabolic glass pane will the the same as in my solution.\n\n## Re: Different parabola equation when finding area\n\nPhinah DID respond with the correct solution Murray BEFORE Stephen did. I do not know why it did not come through\n\nX\n\nPhinah DID respond with the correct solution Murray BEFORE Stephen did.  I do not know why it did not come through\n\n## Re: Different parabola equation when finding area\n\nHi Phinah. Sorry - it must have been a temporary database glitch or something.\n\nX\n\nHi Phinah. Sorry - it must have been a temporary database glitch or something.\n\n## Re: Different parabola equation when finding area\n\nOkay Murray. Thank you.\n\nX\n\nOkay Murray.  Thank you.", "date": "2018-05-27 01:04:50", "meta": {"domain": "intmath.com", "url": "https://www.intmath.com/forum/integration-29/different-parabola-equation-when-finding-area:151", "openwebmath_score": 0.6964902877807617, "openwebmath_perplexity": 1080.6333282970159, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676460637103, "lm_q2_score": 0.8688267762381844, "lm_q1q2_score": 0.851161482714384}}
{"url": "https://math.stackexchange.com/questions/1991212/partial-sum-of-the-harmonic-series-between-two-consecutive-fibonacci-numbers", "text": "# Partial sum of the harmonic series between two consecutive fibonacci numbers\n\nI was playing around with some calculations and I noticed that the partial sum of the harmonic series: $$s_n=\\sum_{k=F_n}^{F_{n+1}}\\frac 1 k$$ where $F_n$ and $F_{n+1}$ are two consecutive Fibonacci numbers have some interesting properties. It is close to $\\frac 1 2$ for small values of $n$ and it seems to converge to a value less than $0.5$ for large $n$. This is what I've got so far: $$\\lim_{n\\to\\infty} s_n\\approx 0.481212$$ I googled a bit to see if there is some theorems or resources for this, and found nothing. I suspect that the series might converge to a smaller number and I may have reached some computational limitations which led to the conclusion that the limit is close to $\\frac 1 2$. So my questions are:\n\n1. Can we show that the series converge to a non-zero value?\n2. In case the first answer is yes, can the limit be expressed in a closed form?\n\nIn terms of the harmonic numbers $H_n$, your sequence is\n\n$$s_n = H_{F_{n+1}} - H_{F_n-1}$$\n\nAs $n \\to \\infty$ it's known that $H_n = \\log n + \\gamma + o(1)$, so\n\n\\begin{align} s_n &= \\log F_{n+1} + \\gamma + o(1) - \\log(F_n-1) - \\gamma - o(1) \\\\ &= \\log F_{n+1} - \\log(F_n-1) + o(1). \\end{align}\n\nNow $F_m \\sim \\varphi^m/\\sqrt{5}$, where $\\varphi$ is the golden ratio, so using the fact that $a \\sim b \\implies \\log a = \\log b + o(1)$ we have\n\n\\begin{align} s_n &= \\log(\\varphi^{n+1}/\\sqrt{5}) - \\log(\\varphi^{n}/\\sqrt{5}) + o(1) \\\\ &= \\log \\varphi + o(1). \\end{align}\n\nIn other words,\n\n$$\\lim_{n \\to \\infty} \\sum_{k=F_n}^{F_{n+1}} \\frac{1}{k} = \\log \\varphi.$$\n\n\u2022 It might be better to use $s_n\\approx ...$ instead of $s_n=...$ \u2013\u00a0polfosol Oct 30 '16 at 8:14\n\u2022 @polfosol, no I disagree. Everything in my answer is rigorous following the definitions of $\\sim$ and little-o notation. \u2013\u00a0Antonio Vargas Oct 30 '16 at 8:14\n\u2022 @polfosol, see, for example, here for the definitions. \u2013\u00a0Antonio Vargas Oct 30 '16 at 8:15\n\u2022 I didn't notice that. Fair enough \u2013\u00a0polfosol Oct 30 '16 at 8:15\n\u2022 I will add a comment that ought to have been done : thanks for your very precise and nice answer. \u2013\u00a0Jean Marie Oct 30 '16 at 8:35\n\nThe Fibonacci numbers increase as $\\phi^n$ (where $\\phi$ is the golden mean $\\frac{1+\\sqrt{5}}{2}$), and harmonic numbers increase as $\\log n$ (i.e., the natural log). Therefore, the difference between the harmonic numbers for successive Fibonacci numbers will approach $\\log\\phi \\approx 0.481211825...$\n\nTo expand a bit, the Fibonacci numbers can be expressed as $\\frac{\\phi^n - (1-\\phi)^n}{\\sqrt{5}}$. (Try it! The fact that the equation $f(x+2) - f(x+1) - f(x) = 0$ requires a sum of powers of $\\phi$ and $1-\\phi$ follows from the fact that these are the solutions to the equation $x^2 - x - 1 = 0$, and the coefficients come from f(1) = f(2) = 1.) The second term vanishes, so large Fibonacci numbers can be approximated quite well as $\\frac{\\phi^n}{\\sqrt{5}}$.\n\nSince one definition of the natural logarithm is the integral from 1 to the parameter of the function $t^{-1}$, the harmonic numbers can be approximated as the natural logarithm, and in fact the difference approaches a constant (called $\\gamma$, about 0.577). If you're not familiar with integrals, the fact that the harmonic numbers increase as a logarithm is suggested by Oresme's proof that the harmonic series diverges...\n\n$$1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} + \\frac{1}{9} + \\cdots > 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8} + \\frac{1}{16} + \\cdots$$\n\n...and it just so happens that that logarithm is the natural logarithm.\n\nSo if you accept that for very large n, the harmonic numbers approach $\\log n$, and that the Fibonacci numbers approach $\\frac{\\phi^n}{\\sqrt{5}}$, then you get for two successive...\n\n$$\\log\\left(\\frac{\\phi^{n+1}}{\\sqrt{5}}\\right) - \\log\\left(\\frac{\\phi^n}{\\sqrt{5}}\\right) = \\log\\left(\\frac{\\phi^{n+1}}{\\phi^n}\\right) = \\log\\phi$$\n\n($\\log x - \\log y = \\log \\frac{x}{y}$ is a natural inverse of $\\frac{e^x}{e^y} = e^{x-y}$.)\n\n\u2022 I will mark this as accepted if you add some more details ;) \u2013\u00a0polfosol Oct 30 '16 at 8:10\n\u2022 ...har-r-r-rumph. \u2013\u00a0user361424 Oct 30 '16 at 8:36\n\u2022 @user361424 very nice answer, a compliment that ought to have been done by the proposer before asking for \"more details\" ! \u2013\u00a0Jean Marie Oct 30 '16 at 8:42\n\u2022 @JeanMarie It seems I am stuck with this \u2013\u00a0polfosol Oct 30 '16 at 8:44\n\u2022 I think this might be the inverse fastest gun in the west problem... (this answer was originally just the first paragraph, and without the parentheticals explaining the golden mean and clarifying the natural log). \u2013\u00a0user361424 Oct 30 '16 at 8:45", "date": "2021-01-22 01:30:18", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1991212/partial-sum-of-the-harmonic-series-between-two-consecutive-fibonacci-numbers", "openwebmath_score": 0.9867463707923889, "openwebmath_perplexity": 225.12238835619792, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676472509722, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.8511614737636568}}
{"url": "http://vuut.auit.pw/matrix-multiplication.html", "text": "# Matrix Multiplication\n\nI have therefore written a matrix vector multiplication example that needs 13 seconds to run (5 seconds with. CUDA matrix multiplication with CUBLAS and Thrust. Important: We can only multiply matrices if the number of columns in the first matrix is the same as the number of rows in the second matrix. Specifically, If both a and b are 1-D arrays, it is inner product of vectors (without complex conjugation). Multiplication. In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. If our equation has two variables, there can be infinitely many combinations of numbers that would work. Learn: In this article, we will see how to perform matrix multiplication in python. The manual method of multiplication procedure involves a large number of calculations especially when it comes to higher order of matrices, whereas a program in C can carry out the operations with short, simple and understandable codes. In other words, To multiply an m\u00d7n matrix by an n\u00d7p matrix, the ns must be the same, and the result is an m\u00d7p matrix. In my last Thank You post, I suggested that Matrix multiplication is not Excel\u2019s forte. \\end{align*} Although it may look confusing at first, the process of matrix-vector multiplication is actually quite simple. The matrix product is designed for representing the composition of linear maps that are represented by matrices. Matrix multiplication by a scalar: First of all, I really want to say thank you very much for taking the time to help me understand this. Matrix multiplication is a row-by-column multiplication where each element of one matrix is multiplied by every element of another matrix. To multiply one matrix with another you need to do a dot product of rows and columns. (Here, $\\operatorname{diag}$ is an operator that creates a column vector out of a matrix's main diagonal. The first matrix must have the same number of columns as the second matrix has rows. That\u2019s all about mutliplying two matrices in java. To multiply matrices, you'll need to multiply the elements (or numbers) in the row of the first matrix by the elements. One question: Why is the column of the first (left-hand) matrix colored red, along with the row of the second (right-hand) matrix?. Matrix multiplication in C. com - id: 799dc5-YmUxN. Figure 5: Our Neural Network, with indexed weights. Good question! The main reason why matrix multiplication is defined in a somewhat tricky way is to make matrices represent linear transformations in a natural way. Matrices can be multiplied by scalar constants in a similar manner to multiplying any number of variable by a scalar constant. Multiplying trans1 by trans2 is not the same as multiplying trans2 by trans1. When working with matrices, we can perform a number of matrix operations including matrix multiplication. Matrix Multiplication Calculator multiply matrices online. Your text probably gave you a complex formula for the process, and that formula probably didn't make any sense to you. Matrix Multiplication. As an example you'll be able to solve a series of simultaneous linear equations using Mathcad\u2019s. This example contains a high-performance implementation of the fundamental matrix multiplication operation and demonstrates optimizations that can be described in Open Computing Language (OpenCL TM) to achieve significantly improved performance. Yes, it wll give you a 2xx1 matrix! When you consider the order of the matrices involved in a multiplication you look at the digits at the extremes to \"see\" the order of the result. The expressions to the right of the equals sign show how the new x, y and z values are calculated after the vector has been transformed. By Rob Hochberg Shodor, Durham, North Carolina This module teaches: Matrix multiplication in the context of enumerating paths in a graph. So we have to be very careful about multiplying matrices. OpenGL 101: Matrices - projection, view, model Posted on May 22, 2013 by Paul. Optimization Techniques for Small Matrix Multiplication Charles-Eric Drevet Ancien el eve, Ecole polytechnique, Palaiseau, France [email\u00a0protected] Scalar Multiplication: Product of a Scalar and a Matrix There are two types or categories where matrix multiplication usually falls under. Before you can even attempt to perform matrix multiplication, you must be sure that the last dimension of the first matrix is the same as the first dimension of the second matrix. The initial attempt to evaluate the f(A) would be to replace every x with an A to get f(A) = A 2 - 4A + 3. For this I tried to follow the xapp1170 which includes a tutorial for the ZC702 board using Planahead. MulT() with w as 0 //fbx sdk forced w to 1 and then multiplied Then I gave up and just manually zeroed out the matrix's T. You can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. The first one is called Scalar Multiplication, also known as the \u201cEasy Type\u201c; where you simply multiply a number into each and every entry of a given matrix. So this right over here has two rows and three columns. Multiplication Factor appears in the Matrix Table for the new entrants at the entry level. I have therefore written a matrix vector multiplication example that needs 13 seconds to run (5 seconds with. What are synonyms for matrix multiplication?. We\u2019ve seen so far some divide and conquer algorithms like merge sort and the Karatsuba\u2019s. Multiplication Here is a list of all of the skills that cover multiplication! These skills are organized by grade, and you can move your mouse over any skill name to preview the skill. Chandler Bur\ufb01eld APSP with Matrix Multiplication March 15, 2013 3 / 19. You must know which of the two matrices will be to the right (of your multiplication) and which one will be to the left; in other words, we have to know whether we are asked to perform or. Matrices are multiplied by the system shown below. The syntax for the function is:. Let's see it with an example where you are trying to multiply a 3X3 matrix with a 3X2 matrix. Multiplying matrices is a little more complex than the operations you've seen so far. Start studying Matrix multiplication. Matrix Multiplication. 3x3 Matrix Multiplication Calculator. First let's make some data: # Make some data a = c(1,2,3) b = c(2,4,6) c = cbind(a,b) x = c(2,2,2) If we look at the output (c and x), we can see that c is a 3x2\u2026. If you know how to multiply two matrices together, you're well on your way to \"dividing\" one matrix by another. The algorithm follows directly from the definition of matrix multiplication. We can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. As an example you'll be able to solve a series of simultaneous linear equations using Mathcad\u2019s. This scalar multiplication of matrix calculator can help you calculate the multiplication between a scalar and a matrix no matte of its type (having from 1 to 4 columns and/or rows). In mathematics, a matrix (plural: matrices) is a rectangle of numbers, arranged in rows and columns. Multiplication of Matrices. The first matrix must have the same number of columns as the second matrix has rows. There is nothing fundamentally di erent between the matrix multiplies that we need to compute at this level relative to our original problem. 2 Linear Substitutions and Matrix Multiplication Or we can use equations (1) and (4) to go by way of x, u = Ax = A(Ep) = (AE)p We therefore conclude, without doing any real work, that AE = DC, that is,. Now the way that us humans have defined matrix multiplication, it only works when we're multiplying our two matrices. \u2013Use SVD to ensure this property. Example: a matrix with 3 rows and 5 columns can be added to another matrix of 3 rows and 5 columns. The Mailman algorithm: a note on matrix vector multiplication Edo Liberty \u2044 Computer Science Yale University New Haven, CT Steven W. In general, a matrix is just a rectangular array or table of numbers. Multiplying matrix is one of the tedious things that we have done in schools. It's a visualization of the matrix multiplication algorithm. As demonstrated above, in general AB \u2260BA. This tool for multiplying 3x3 matrices. \\end{align*} Although it may look confusing at first, the process of matrix-vector multiplication is actually quite simple. Hanrahan / Understanding the Efciency of GPU Algorithms for Matrix-Matrix Multiplication plications and must run efciently if GPUs are to become a. An interactive matrix multiplication calculator for educational purposes. Vectors are commonly used in matrix multiplication to find a new point resulting from an applied transformation. This array function returns the product of two matrices entered in a worksheet. Ready to execute code with proper output. What does matrix multiplication mean? Here's a few common intuitions: 1) Matrix multiplication scales/rotates/skews a geometric plane. 2 Linear Substitutions and Matrix Multiplication Or we can use equations (1) and (4) to go by way of x, u = Ax = A(Ep) = (AE)p We therefore conclude, without doing any real work, that AE = DC, that is,. Matrix Arithmetics under NumPy and Python. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Matrix multiplication in C. On this page you can see many examples of matrix multiplication. Let's see it with an example where you are trying to multiply a 3X3 matrix with a 3X2 matrix. Matrix Multiplication. Posts about Matrix Multiplication written by Sean. by Marco Taboga, PhD. \u00a9M F2 n0M1p2o XKKuUtHaw qS xo xfFtKwxa OrKeD aLNLiC M. The problem is not actually to perform the multiplications, but merely to decide in which order to perform the multiplications. The multiplication of a matrix A by a matrix B to yield a matrix C is defined only when the number of columns of the first matrix A equals the number of rows of the second matrix B. This is a matrix multiplication utility I developed as a part of my project work at college. 1 Matrix Addition and Scalar Multiplication 175 According to the labeling convention, the entries of the matrix A above are A = a 11 a 12 a 13 a 21 a 22 a 23 In general, the m \u00d7n matrix A has its entries labeled as follows:. Following normal matrix multiplication rules, a (n x 1) vector is expected, but I simply cannot find any. Synonyms for matrix multiplication in Free Thesaurus. \u2022We provide a new hybrid parallel algorithm for shared-memory fast matrix multiplication. How to perform scalar matrix multiplication in C programming. Device Memories and Matrix Multiplication 1 Device Memories global, constant, and shared memories CUDA variable type quali\ufb01ers 2 Matrix Multiplication an application of tiling. matrices are equal when each corresponding element is equal. Good question! The main reason why matrix multiplication is defined in a somewhat tricky way is to make matrices represent linear transformations in a natural way. X D dM2aVd6eg tw wiTt Qhi BIqn Vfji on aift7e o iA Slig YeRb ArWad U2z. Matrix Multiply, Power Calculator Solve matrix multiply and power operations step-by-step. JAMA is a basic linear algebra package for Java. We can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. C Program example of Matrix Chain multiplication. This article comprises matrix multiplication program written in python with Sample Input and Sample Output. Free matrix calculator - solve matrix operations and functions step-by-step. Matrix multipli. Abstract: We implement a promising algorithm for sparse-matrix sparse-vector multiplication (SpMSpV) on the GPU. Matrix multiplication is the most expensive operation involved Number of computations to be performed for matrix multiplication with Givens matrix for ith column: 6(n - i + 1). MulT() with w as 0 //fbx sdk forced w to 1 and then multiplied Then I gave up and just manually zeroed out the matrix's T. Lecture2 MatrixOperations \u2022 transpose, sum & di\ufb00erence, scalar multiplication \u2022 matrix multiplication, matrix-vector product \u2022 matrix inverse. The current matrix is determined by the current matrix mode (see glMatrixMode). Moreover, it computes the power of a square matrix, with applications to the Markov chains computations. Write a C program to read elements in a matrix and perform scalar multiplication of matrix. Today, we take a step back from finance to introduce a couple of essential topics, which will help us to write more advanced (and efficient!) programs in the future. Matrix multiplication has significant application in the areas of graph theory, numerical algorithms, signal. It is important to realize that you can use \"dot\" for both left \u2010 and right \u2010 multiplication of vectors by matrices. Before you can even attempt to perform matrix multiplication, you must be sure that the last dimension of the first matrix is the same as the first dimension of the second matrix. Multiply two matrices together. Inverse circular dichroism (CD) spectra are presented for each of the five major secondary structures of proteins: alpha-helix, antiparallel and parallel beta-sheet, beta-turn, and other (random) structures. If we grab a matrix from a previous section, this can be easily explained. Chapter 1 Matrix Multiplication 1. Graphing calculators such as the TI83 and TI84 are able to do many different operations with matrices, including multiplication. If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly. its determinant. Multiplication of a matrix by a scalar. Banded Matrix-Vector Multiplication. Matrix multiplication in C: We can add, subtract, multiply and divide 2 matrices. An output of 3 X 3 matrix multiplication C program: Download Matrix multiplication program. Random Synchronization Up: Experiments Previous: Double Loops. Banky Craig C. To multiply a row vector by a column vector, the row vector must have as many columns as the column vector has rows. 0 in the MinGW suite) would use the C11 standard by default, which I realised after I read the documentation. We can multiply a matrix with a number (also called a scalar). For some reason, the following brute force approach is faster by about 10%:. This exercise surprised me a little bit. multMatrixes that will. Use commas or spaces to separate values in one matrix row and semicolon or new line to separate different matrix rows. Apart from \"Matrix Multiplication Worksheet Answers\" i f you need any other stuff in math, please use our google custom search here. Now the way that us humans have defined matrix multiplication, it only works when we're multiplying our two matrices. Device Memories and Matrix Multiplication 1 Device Memories global, constant, and shared memories CUDA variable type quali\ufb01ers 2 Matrix Multiplication an application of tiling. Resources to help you Teach Matrix Multiplication Worksheet, Bell Work, Exit Quiz, Power Point, Guided Notes, and much more!. Non-square matrices do not have inverses. Take free online matrix multiplication classes to improve your skills and boost your performance in school. An online Matrix calculation. ppt Loading\u2026. I ended that post by saying we would revisit parallel_for_each after introducing array and array_view. We say a matrix is m n if it has m rows and n columns. CUDA Programming Guide Version 1. The authors [9, 10] designed a hybrid matrix format, HYB (Hybrid of ELL and. Unlike the other two kinds of multiplication, the cross product is only de\ufb01ned for three-dimensional vectors. An interactive matrix multiplication calculator for educational purposes. The transpose of a matrix is a new matrix whose rows are the columns of the original. facebook twitter linkedin pinterest. Is it true and under what conditions? ADD: Trying to recreate the answer in R, wh. multiply(a, b) or a * b. For some matrices A and B,wehaveAB =BA. Suppose you have two groups of vectors: $\\{a_1, \\dots, a_m\\}$ and [math]\\{b_1, \\dots. 36 Strassen\u2019s method until a predetermined cutoff size of the seven sub-matrices, af-ter which Winograd\u2019s algorithm takes over. MATRIX_A: An array of INTEGER, REAL, COMPLEX, or LOGICAL type, with a rank of one or two. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Learn vocabulary, terms, and more with flashcards, games, and other study tools. You can use fractions for example 1/3. The columns of the first matrix must be equal to the rows in the second matrix. There is a condition to u and v, namely that they are linearly independent. With no parentheses, the order of operations is left to right so A*B is calculated first, which forms a 500-by-500 matrix. Multiplication Factor appears in the Matrix Table for the new entrants at the entry level only and not for the existing employees. If your data is in column-major order, you can tell MPSMatrixMultiplication to transpose the matrix before doing the multiplication. In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field, or, more generally, in a ring or even a semiring. Note: Matrices multiplication is possible only when the number of columns of first matrix is equal to the number of rows of second matrix. Ready to execute code with proper output. There is one slight problem, however. One way to look at it is that the result of matrix multiplication is a table of dot products for pairs of vectors making up the entries of each matrix. , a movie), is typically very sparse\u2026. When working with matrices, we can perform a number of matrix operations including matrix multiplication. Now the matrix multiplication is a human-defined operation that just happens-- in fact all operations are-- that happen to have neat properties. Below is a program on Matrix Multiplication. Stormy Attaway, in Matlab (Second Edition), 2012. Douglasz April 23, 2001 Abstract: Routines callable from FORTRAN and C are described which implement matrix{matrix. Abstract: This paper presents a method to analyze the powers of a given trilinear form (a special kind of algebraic constructions also called a tensor) and obtain upper bounds on the asymptotic complexity of matrix multiplication. When multiplying matrices together, the dimensions of the matrices to be multiplied must be compatible. Matrix Multiplication in Java. Grey Ballard, Aydin Buluc, James Demmel, Laura Grigori, Benjamin Lipshitz, Oded Schwartz, Sivan Toledo Jul. matrix multiplication in c free download. October 12, 2002 MULTIPLICATION MATRIX The history of this matrix goes back to the \u201870\u2019s when my wife and I operated an individual learning. We're considering element-wise multiplication versus matrix multiplication. Clusters use in many scientific. Multiplying matrices is a little more complex than the operations you've seen so far. Matrices can be multiplied by scalar constants in a similar manner to multiplying any number of variable by a scalar constant. \\end{align*} Although it may look confusing at first, the process of matrix-vector multiplication is actually quite simple. Here you can perform matrix multiplication with complex numbers online for free. Matrix chain multiplication (or Matrix Chain Ordering Problem, MCOP) is an optimization problem that to find the most efficient way to multiply given sequence of matrices. We need to tag the map( ) function output with the position so the reduce( ) function can identify the components in the different vectors. This forum may not be the best place for a discussion of the many issues involved in performance number-crunching, but I'd very much appreciate comments, suggestions, etc. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Original value ranges are included on the X-axis. Parallel matrix multiplication \u2022 Assume p is a perfect square \u2022 Each processor gets an n/\u221ap \u00d7 n/\u221ap chunk of data \u2022 Organize processors into rows and columns. For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix. Don't show me this again. , of a matrix. Matrix Multiplication --- Row and Column Picture. The Wolfram Language's matrix operations handle both numeric and symbolic matrices, automatically accessing large numbers of highly efficient algorithms. This tool for multiplying 3x3 matrices. Douglasz April 23, 2001 Abstract: Routines callable from FORTRAN and C are described which implement matrix{matrix. Shruti Kaushik. Multiplication without tiling. After matrix multiplication the prepended 1 is removed. Matrix Multiplication in Excel with the MMULT function You can multiply matrices in Excel thanks to the MMULT function. These properties include the associative property, distributive property, zero and identity matrix property, and the dimension property. The second matrix is Be and it is 3x18x2 and the third matrix is del matrix and its. This subprogram takes two matrices as parameters and returns their matrix product. Below are the common core standards dealing with basic multiplication. Matrices are frequently used in programming and are used to represent graph data structure, in solving a system of linear equations and have many other applications. This lecture introduces matrix multiplication, one of the basic algebraic operations that can be performed on matrices. Hi, I want to create a VI that performs matrix multiplication for two input matrices A and B. \u2022 Given some matrices to multiply, determine the best order to multiply them so you minimize the number of single element multiplications. A matrix is a rectangular array of numbers or other mathematical objects for which operations such as addition and multiplication are defined. multMatrixes that will. We need to check this condition while implementing code without ignoring. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. Matrix multiplication. Numpy Matrix Multiplication - Hackr. Matrices to be multiplied do not need to be of the same order, by definition the number of columns of the first matrix must equal the number of rows of the second matrix, otherwise all row elements of the first matrix could not be multiplied by a. Sparse Matrix Multiplication on a Field-Programmable Gate Array A Major Qualifying Project Report submitted to the Faculty of the WORCESTER POLYTECHNIC INSTITUTE. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): In this paper we present a SIMD algorithm for n n matrix multiplication on a hypercube of p processors, with time complexity of O( p =3 ), and 2 < 3. Many states have gone to a group of consistent key standards for each grade level. Scalar in which a single number is multiplied with every entry of a matrix ; Multiplication of an entire matrix by another entire matrix For the rest of the page, matrix multiplication will refer to this second category. The program follows a basic encryption algorithm that relies on mathematical properties of matrices, such as row operations, matrix multiplication, and invertible matrices. Given a sequence of matrices, find the most efficient way to multiply these matrices together. When multiplying matrices, we first need to ensure that the matrices have the same dimensions, which is the number of rows times the number of columns. In this paper, we show that novel fast matrix multiplication algorithms can significantly outperform vendor implementations of the classical algorithm and Strassen's fast algorithm on modest problem sizes and shapes. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. October 12, 2002 MULTIPLICATION MATRIX The history of this matrix goes back to the \u201870\u2019s when my wife and I operated an individual learning. Matrix Multiplication. First let's make some data: # Make some data a = c(1,2,3) b = c(2,4,6) c = cbind(a,b) x = c(2,2,2) If we look at the output (c and x), we can see that c is a 3x2\u2026. Study guide and practice problems on 'Matrix multiplication'. We want to define addition of matrices of the same size, and multiplication of certain \"compatible\" matrices. In this notebook, we'll be using Julia to investigate the efficiency of matrix multiplication algorithms. How to Multiply Matrices Faster. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In the interests of understanding the underlying properties of the images I\u2019m using as stimuli, I\u2019ve been trying to learn more about the matrix transformations commonly used for image compression and image manipulation. \\end{align*} Although it may look confusing at first, the process of matrix-vector multiplication is actually quite simple. The size of the matrix, as a block, is defined by the number of Rows and the number of Columns. The applications of matrices often involve the multiplication of two matrices, which requires rules for combination of the elements of the matrices. Recent posts. Stormy Attaway, in Matlab (Second Edition), 2012. Matrix Multiplication in Excel with the MMULT function You can multiply matrices in Excel thanks to the MMULT function. To multiply two matrices in C++ Programming, first ask to the user to enter the two matrix, then start multiplying the two matrices and store the multiplication result inside any variable say sum and finally store the value of sum in the third matrix say mat3. Before to this you can check different types of arrays in java and get to know how to declare and define the arrays and also get practice with adding two matrices. In other words, To multiply an m\u00d7n matrix by an n\u00d7p matrix, the ns must be the same, and the result is an m\u00d7p matrix. split happened at only one matrix which requires zero multiplications). Parallel Sparse Matrix-Vector and Matrix-Transpose-Vector Multiplication Using Compressed Sparse Blocks Ayd\u0131n Buluc\u00b8\u2217 [email\u00a0protected] You probably know what a matrix is already if you are interested in matrix multiplication. For math, science, nutrition, history. I have therefore written a matrix vector multiplication example that needs 13 seconds to run (5 seconds with. Doerr 2 the previous seating chart example use a 1 (or yes) if the seat is occupied and a 0 (or no ) if the seat is unoccupied. Here we discuss the properties in detail. A and B must either be the same size or have sizes that are compatible (for example, A is an M-by-N matrix and B is a scalar or 1-by-N row vector). Multiplication of a matrix by a scalar. A matrix is a rectangular array of numbers or other mathematical objects for which operations such as addition and multiplication are defined. Before we go much farther, if you don\u2019t know how matrix multiplication works, then check out Khan Academy spend the 7 minutes, then work through an example or two and make sure you have the intuition of how it works. Using Doceri). As demonstrated above, in general AB \u2260BA. 376 Coppersmith-Winograd (1990) n2. The efficiency of matrix multiplication is a popular research topic given that matrices compromise large data in computer applications and other fields of study. Multiplying matrices - examples. We will illustrate matrix multiplication or matrix product by the following example. Take free online matrix multiplication classes to improve your skills and boost your performance in school. I did not expect that gcc (GCC 6. The definition of matrix multiplication indicates a row-by-column multiplication, where the entries in the i th row of A are multiplied by the corresponding entries in the j th column of B and then adding the results. Springer LNCS, 1984. Python Matrix Multiplication Program - here you will learn how to multiply one matrix to another matrix and print the multiplication result of the third matrix in python. It is built deeply into the R language. Furthermore, you can apply matrix operations such as addition, subtraction, multiplication and division:. Matrix multiplication means multiplications of two different arrays, in excel we have an inbuilt function for matrix multiplication and it is MMULT function, it takes two arrays as an argument and returns the product of two arrays, given that both the arrays should have the same number of rows and the same number of columns. One question: Why is the column of the first (left-hand) matrix colored red, along with the row of the second (right-hand) matrix?. MATMUL can do this for a variety of matrix sizes, and for different arithmetics (real, complex, double precision, integer, even logical!). Multiplication of a matrix by another matrix. Important: We can only multiply matrices if the number of columns in the first matrix is the same as the number of rows in the second matrix. When multiplying matrices, we first need to ensure that the matrices have the same dimensions, which is the number of rows times the number of columns. Join GitHub today. Row 1 X Column 1 Row 1 X Column 1 Row 1 X Column 1 Row 1 X Column 2 Row 1 X Column 2 Row 1 X \u2013 A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. Just type matrix elements and click the button. Here you can perform matrix multiplication with complex numbers online for free. Just like addition works only for matrices of the same size, there are conditions for when two matrices can be multiplied but in this case it is a little bit more complicated. With no parentheses, the order of operations is left to right so A*B is calculated first, which forms a 500-by-500 matrix. To do so, we are taking input from the user for row number, column number, first matrix elements and second matrix elements. If you want to try to multiply two matrices (x and y) by each other, you'll need to make sure that the number of columns in x is equal to the number of rows in y, otherwise the equation won't work properly. As an example you'll be able to solve a series of simultaneous linear equations using Mathcad\u2019s. In recommendation systems, the user/item rating matrix, which shows the extent of how much a user likes an item (e. Matrix algebra for beginners, Part I matrices, determinants, inverses Jeremy Gunawardena Department of Systems Biology Harvard Medical School 200 Longwood Avenue, Cambridge, MA 02115, USA. Matrix multiplication is the \"messy type\" because you will need to follow a certain set of procedures in order to get it right. Matrix multiplication in C. Producing a single matrix by multiplying pair of matrices (may be 2D / 3D) is called as matrix multiplication which is the binary operation in mathematics. Here is how it works 1) 2-D arrays, it returns normal product 2) Dimensions > 2, the product is trea. The manual method of multiplication procedure involves a large number of calculations especially when it comes to higher order of matrices, whereas a program in C can carry out the operations with short, simple and understandable codes. Strassen\u2019s Matrix Multiplication on GPUs Junjie Li Sanjay Ranka Sartaj Sahni fjl3, ranka, [email\u00a0protected] The MMULT function returns the matrix product of two arrays. Just like numbers and equations, you are expected to be able to manipulate matrices and perform arithmetic on multiple numbers of matrices. ) This computation requires $3n^2$ operations, while the operation count for the full matrix product is $4n^3$. 2 in the most recent edition (6e) of Finite Mathematics and Section 4. Hi everyone I am new to this site and also new to programming world can anybody help me writing C code for matrix multiplication (without using pointers). Below are the common core standards dealing with basic multiplication. If both are vectors it will return the inner product. A matrix is just a two-dimensional group of numbers. He told me about the work of Jacques Philippe Marie Binet (born February 2 1786 in Rennes and died Mai 12 1856 in Paris), who seemed to be recognized as the first to derive the rule for multiplying matrices in 1812. Pupils use the interactive to create a matrix multiplication problem using vectors. Each element in the result matrix C is the sum of element-wise multiplication of a row from A and a column from B. One way to look at it is that the result of matrix multiplication is a table of dot products for pairs of vectors making up the entries of each matrix. Operation with Matrices in Linear Algebra. One question: Why is the column of the first (left-hand) matrix colored red, along with the row of the second (right-hand) matrix?. For example weather forecasting has to done in. An efficient k-way merge lies at the heart of finding a fast parallel SpMSpV algorithm. Matrix addition, multiplication, inversion, determinant and rank calculation, transposing, bringing to diagonal, triangular form, exponentiation, solving of systems of linear equations with solution steps. Asking why matrix multiplication isn't just componentwise multiplication is an excellent question: in fact, componentwise multiplication is in some sense the most \"natural\" generalization of real multiplication to matrices: it satisfies all of the axioms you would expect (associativity, commutativity, existence of identity and inverses (for matrices with no 0 entries), distributivity over. Matrix Multiply, Power Calculator Solve matrix multiply and power operations step-by-step. Matrix multiplication is likely to be a source of a headache when you fail to grasp conditions and motives behind them. , Determine the way the matrices are fully parenthesized. Matrix Chain multiplication and algorithmic solution. If both arguments are 2-D they are multiplied like conventional matrices. MATRIX_A: An array of INTEGER, REAL, COMPLEX, or LOGICAL type, with a rank of one or two. Compton LA, Johnson WC Jr. plain old numbers like 3, or -5. This VHDL project is aimed to develop and implement a synthesizable matrix multiplier core, which is able to perform matrix calculation for matrices with the size of 32x32. This algebra lesson explains how to do scalar multiplication - and explains what a scalar is. We will illustrate matrix multiplication or matrix product by the following example. Matrix Formulas. It doesn \u2019 t just give you the answer the way your calculator would, but will actually show you the \"long hand\" way to multiply two numbers. Learn: In this article, we will see how to perform matrix multiplication in python. The Wolfram Language's matrix operations handle both numeric and symbolic matrices, automatically accessing large numbers of highly efficient algorithms. But to multiply a matrix by another matrix we need to do the \"dot product\" of rows and columns what does that mean?. Lecture Notes CMSC 251 Lecture 26: Chain Matrix Multiplication (Thursday, April 30, 1998) Read: Section 16. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how do matrix multiplication. Please upload a file larger than 100x100 pixels; We are experiencing some problems, please try again. Linear Algebra\u00b6. Related Posts. Hoare (quoted by Donald Knuth). 3x3 Matrix Multiplication Calculator. Matrix Multiplication,definition,2 D array in C,Multidimensional array in C,Syntax,Syntax Example,Matrix Multiplication 2 D (dimensional) or Multidimensional Array Example Program In C. However, even when matrix multiplication is possible in both directions, results may be different. In mathematics, scalar multiplication is one of the basic operations defining a vector space in linear algebra (or more generally, a module in abstract algebra). This matrix multiplication calculator help you understand how to do matrix multiplication. Multiplication without tiling. Here\u2019s a fact that has been rediscovered many times in many different contexts: The way you parenthesize matrix products can greatly change the time it takes to compute the product. Matrix Multiplication using arrays is very basic practice to learn for beginners to understand the concept of multidimensional matrix. Matrix multiplication is likely to be a source of a headache when you fail to grasp conditions and motives behind them. The matrix A is an n x m matrix and matrix B is an m x p matrix. Each element in the result matrix C is the sum of element-wise multiplication of a row from A and a column from B. The behavior depends on the arguments in the following way. Study guide and practice problems on 'Matrix multiplication'.", "date": "2019-12-13 06:53:24", "meta": {"domain": "auit.pw", "url": "http://vuut.auit.pw/matrix-multiplication.html", "openwebmath_score": 0.7270029187202454, "openwebmath_perplexity": 562.5575512182271, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9879462219033657, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8511534225450522}}
{"url": "https://math.stackexchange.com/questions/1952939/uniqueness-of-unitarily-upper-triangular-matrix", "text": "# Uniqueness of unitarily upper triangular matrix\n\nSuppose A is a matrix in $\\mathbb{C}^{n\\times n}$ with n distinct eigenvalues $\\lambda_1,\\dots,\\lambda_n$. Then by Schur's theorem, for any fixed order of $\\lambda_1,\\dots,\\lambda_n$, we know there exists an unitary matrix $U$ s.t. $U^*AU$ is an upper triangular matrix with $\\lambda_1,\\dots,\\lambda_n$ of required order on the diagonal. The question is is $U$ unique? If not, what freedom do we have to choose U?\n\nI know how to solve $A$ is unitarily diagonal (not unitarily upper triangular), then $U^*AU=\\,\\text{diag}(\\lambda_i)\\iff AU=U\\,\\text{diag}(\\lambda_i)=[\\lambda_1U_1,\\dots,\\lambda_nU_n]$. Then ith column of $U$ must be an eigenvector of $\\lambda_i$ and $|U_i|=1$. Therefore we can choose $U$ up to multiplying a diagonal matrix whose diagonal entries have norm 1. But this method seems not fit the unitarily upper triangular case.\n\nKey fact: if $BT=TB$ and $T$ is upper triangular with distinct diagonal entries, then $B$ is upper triangular (proof below). Now, if $$UTU^*=VTV^*,$$ then $V^*UT=TV^*U$. So $V^*U$ is an upper triangular unitary. As such, it is diagonal. Thus, $V$ is of the form $DU$ with $D$ diagonal and $|D_{kk}|=1$. In other words, the situation you observed for diagonal $A$ still occurs in general (it is essential that the diagonal entries are distinct).\nProof that $TB=BT$ implies $B$ diagonal. Consider the $n,1$ entry: $$(TB)_{n1}=\\sum_kT_{nk}B_{k1}=T_{nn}B_{n1},$$ while $$(BT)_{n1}=\\sum_kB_{nk}T_{k1}=B_{n1}T_{11}.$$ As $T_{11}\\ne T_{nn}$, we deduce that $B_{n1}=0$. Now consider the $n,2$ entry: $$(TB)_{n2}=\\sum_kT_{nk}B_{k2}=T_{nn}B_{n2},$$ while $$(BT)_{n2}=\\sum_kB_{nk}T_{k2}=B_{n1}T_{12}+B_{n2}T_{22}=B_{n2}T_{22}.$$ As $T_{22}\\ne T_{nn}$, we get that $B_{n2}=0$. Continuing inductively, after showing that $B_{n1},\\ldots,B_{nr}=0$, we have $$(TB)_{n,r+1}=\\sum_kT_{nk}B_{k,r+1}=T_{nn}B_{n,r+1},$$ while $$(BT)_{n,r+1}=\\sum_kB_{nk}T_{k,r+1}=\\sum_{k=1}^{r+1}B_{nk}T_{k,r+1}=B_{n,r+1}T_{r+1,r+1}.$$ As $T_{r+1,r+1}\\ne T_{nn}$, we get that $B_{n,r+1}=0$. Now start doing the same with the $n-1,1$ entry, then $n-1,2$, etc.", "date": "2019-05-27 09:08:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1952939/uniqueness-of-unitarily-upper-triangular-matrix", "openwebmath_score": 0.9731680750846863, "openwebmath_perplexity": 100.74582408207894, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462179994595, "lm_q2_score": 0.8615382147637196, "lm_q1q2_score": 0.8511534209378229}}
{"url": "https://math.stackexchange.com/questions/2095489/show-that-sum-infty-k-0-frac2-1k2k1-pi-cosh2k1-pi-2-1-4/2096064", "text": "# Show that $\\sum^\\infty_{k=0} \\frac{2(-1)^k}{(2k+1)\\pi\\cosh[(2k+1)\\pi/2]}=1/4$\n\nTitle says it all.\n\nWell, maybe some backstory.\n\nFlipping through my past notebooks, I found this:\n\n$$\\vdots$$ $$= \\sum^\\infty_{k=0} \\frac{2(-1)^k}{(2k+1)\\pi\\cosh[(2k+1)\\pi/2]}\\\\=\\frac14\\quad(?)$$\n\n[end of page]\n\nAh, yes. My engineering numerical analysis professor gave me one homework problem:\n\nMake a contour plot of the steady-state temperature distribution of a square plate, with the temperature of one of its side maintained at $1$ unit, the other three sides maintained at $0$ units.\n\nSteady-state temperature distribution $T$ follows $\\nabla^2T=0$.\n\nI was calculating the temperature of the center of the plate analytically to sanity-check my program output, when I encountered this weird-looking sum. I remember I made a spreadsheet to numerically evaluate the sum, and Excel said 0.2500000... So I jot down $\\frac14$ with a question mark beside it. The next page is filled with my fruitless attempts to find out the exact sum, to convince myself that the sum is exactly $1/4$.\n\nI posted a bounty on my social media accounts, whoever can show the proof (or disproof) gets free beer.\n\nAfter a few days, the bounty was still unclaimed, but I realized, \"duuuuuhhhh! I could just exploit the symmetry. If I rotate the plate $90^\\circ$ three times and superpose all the temperature distribution with the original one, temperature is $1$ unit everywhere. By linearity and symmetry, the temperature of the center point is therefore $1/4$.\"\n\nI treated myself a few rounds of beer, and moved on. Case closed.\n\nLooking again at this, I now realize that I might be missing out on some summation tactics I never knew.\n\nAnybody can show me how is this equal to $1/4$ just by algebraic manipulation?\n\n\u2022 +1 for the story, +1 more for the solution, and +1 even more for trying to learn greater things. \u2013\u00a0Simply Beautiful Art Jan 12 '17 at 23:22\n\u2022 The residue theorem might come in useful. \u2013\u00a0Michael M Jan 13 '17 at 0:23\n\u2022 Thanks for the suggestion and the upvote. Probably will take me a while to work that out, as I never took a course in analysis. But now I realize that math is wonderful, there are always more beautiful things to learn. \u2013\u00a0f1garo Jan 13 '17 at 1:25\n\u2022 \u2013\u00a0Math-fun Jan 13 '17 at 9:49\n\nI am busy as hell today but this is such a nice problem (and amusingly posed question) i can't resist... :)\n\nThe key idea here is simple: Look for a function in the complex plane $f(z)$ which has poles at the correct values $z_n$ and integrate it over an appropriate contour $C$.\n\nIt turns out that the function (which is also the simplest guess i can imagine)\n\n$$f(z)=\\frac{1}{z \\cos(z)\\cosh(z)}$$\n\nis the correct choice. Its poles are given by $z_0=0$, $z_k=\\frac{(2 k-1)\\pi}{2}$ and $\\tilde{z}_k=i \\frac{(2 k-1)\\pi}{2}$ with $k \\in \\mathbb{Z/0}$. We furthermore have\n\n$$\\text{res}(z_0)=1 \\\\ \\text{res}(z_k)=\\text{res}(\\tilde{z}_k)=\\frac{2}{\\pi}\\frac{(-1)^{k}}{(2k-1)\\cosh\\left(\\pi\\frac{2k-1}{2}\\right)}\\\\ \\text{res}(z_{-k})=\\text{res}(z_k)$$\n\nsince $z\\cos(z)\\cosh(z)\\sim |z| e^{a |z|}$ with $\\Re(a)>0$ as $|z|\\rightarrow\\infty$ we can choose a big circle traversed anticlockwise with radius choosen such that we hit no pole of $f(z)$ as the integration contour $C$ (Thanks to @Dr. MV for rigorizing this point).\n\nThen, we get by applying the residue theorem, we get in the limit of infinte radius\n\n$$\\oint_Cf(z)dz=2\\pi i\\text{res}(z_0)+4\\pi i \\sum_{k\\geq1}\\text{res}(z_k)+4\\pi i\\sum_{k\\geq1}\\text{res}(\\tilde{z}_k)=0$$\n\nor\n\n$$8\\pi i \\sum_{k\\geq1}\\text{res}(z_k)=-2\\pi i$$\n\nwhich can be rewritten as (after shifting $k\\rightarrow k+1$)\n\n$$\\sum_{k\\geq0}\\frac{2}{\\pi}\\frac{(-1)^{k}}{(2k+1)\\cosh\\left(\\pi\\frac{2k+1}{2}\\right)}=\\frac{1}{4}\\\\ \\textbf{Q.E.D.}$$\n\nTo make things clearer, here is a sketch of the integration contour $\\color{blue}{C}$ and the singularities $\\color{red}{z_n}$\n\n\u2022 (+1) This is extremely close to Ramanujan original argument, I'd say it is perfectly fine! \u2013\u00a0Jack D'Aurizio Jan 13 '17 at 12:51\n\u2022 @JackD'Aurizio thanks! This is a relief...still wondering how this complicated sum can be derived with nearly zero effort \u2013\u00a0tired Jan 13 '17 at 12:53\n\u2022 A big (+1). I was working on this last night, had identified $f(z)$ and then floundered to find a suitable contour. Now, I feel stupid. Well done my friend! -Mark \u2013\u00a0Mark Viola Jan 13 '17 at 14:42\n\u2022 I was trying to isolate the poles on the real axis by taking a rectangular contour in hopes that the top and bottom halves would cancel. They didn't. I even recognized the symmetry between the residues on real and imaginary axes. For some reason (it was getting late and I was tiring) I just couldn't see the obvious \"big circle\" contour was the way forward. One note; we must take the limit with discrete radii to ensure the contour doesn't hit a pole. \u2013\u00a0Mark Viola Jan 13 '17 at 14:55\n\u2022 In general, the function $\\pi \\sec(\\pi z) f(z)$ can be used to evaluate infinite sums of the form $\\sum_{n \\in \\mathbb{Z}} (-1)^n f(n+1/2)$, while the function $\\pi \\tan(\\pi z) f(z)$ can be used to evaluate infinite sums of the form $\\sum_{n \\in \\mathbb{Z}} f(n+1/2)$. In some cases, of course, the contour integral won't vanish in the limit. \u2013\u00a0Random Variable Jan 13 '17 at 19:56\n\nAs mentioned by Zucker in The summation of series of hyperbolic functions, in Question 358, J. Indian Math. Soc., 4 (1912), p. 78. Ramanujan proves (through the residue theorem) that $$\\sum_{n\\geq 1}(-1)^{n+1}(2n-1)^{4m-1}\\,\\text{sech}\\left[(2n-1)\\frac{\\pi}{2}\\right] = 0$$ holds for every $m\\geq 0$. Your identity then follows from the $m=0$ case.\n\nIt is truly remarkable to know that such identity can be proved through Dirichlet's problem about the eigenvalues of the Laplacian operator. Can you be more specific about the relation between such a series and the physical problem?$^{(0)}$ A brilliant piece of math might arise from there.\n\n$^{(0)}$Update: I found the connection - such series are related with the Green function of a square.\n\nI will show an interesting technique for deriving a similar identity.\n\nFrom the Weierstrass product $$\\cosh(\\pi x/2) = \\prod_{m\\geq 0}\\left(1+\\frac{z^2}{(2m+1)^2}\\right)\\tag{1}$$ by applying $\\frac{d^2}{dz^2}\\log(\\cdot)$ to both sides we get: $$\\frac{\\pi^2}{8\\cosh^2(\\pi x/2)}=\\sum_{m\\geq 0}\\frac{(2m+1)^2-z^2}{((2m+1)^2+z^2)}\\tag{2}$$ If we replace $z$ with $(2n+1)$ and sum over $n\\geq 0$, $$\\sum_{n\\geq 0}\\frac{\\pi^2}{8\\cosh^2(\\pi(2n+1)/2)} = \\sum_{n\\geq 0}\\sum_{m\\geq 0}\\frac{(2m+1)^2-(2n+1)^2}{((2m+1)^2+(2n+1)^2)^2} \\tag{3}$$ where the RHS of $(3)$ can also be written as $$\\sum_{n\\geq 0}\\sum_{m\\geq 0}\\int_{0}^{+\\infty}\\cos((2n+1)x)x e^{-(2m+1)x}\\,dx = \\sum_{n\\geq 0}\\int_{0}^{+\\infty}\\frac{x\\cos((2n+1)x)}{2\\sinh(x)}\\,dx\\tag{4}$$ or, by exploiting integration by parts, $$\\sum_{n\\geq 0}\\int_{0}^{+\\infty}\\frac{x\\cosh(x)-\\sinh(x)}{2\\sinh(x)}\\cdot\\frac{\\sin((2n+1)x)}{2n+1}\\,dx\\tag{5}$$ On the other hand, $\\sum_{n\\geq 0}\\frac{\\sin((2n+1)x)}{2n+1}$ is the Fourier series of a $2\\pi$-periodic rectangle wave that equals $\\frac{\\pi}{4}$ over $(0,\\pi)$ and $-\\frac{\\pi}{4}$ over $(\\pi,2\\pi)$. That implies, by massive cancellation: $$\\sum_{n\\geq 0}\\frac{1}{\\cosh^2(\\pi(2n+1)/2)}=\\frac{1}{2\\pi}.\\tag{6}$$ On the other hand, the Fourier transform of $\\frac{1}{\\cosh^2(\\pi x)}$ is given by by $\\frac{s\\sqrt{8\\pi}}{\\sinh(\\pi s)}$.\nBy Poisson's summation formula, $$\\sum_{n\\geq 1}\\frac{(-1)^{n+1}n}{\\sinh(\\pi n)}=\\frac{1}{4\\pi}.\\tag{7}$$\n\n\u2022 Hey Jack, since i can't read the source you mentioned, could you please check if their derivation fits my approach? i'm quiet unsure if i made no big mistake here (it seems to easy to be true). thanks! \u2013\u00a0tired Jan 13 '17 at 12:38\n\u2022 For the pyhsical background you might be interested in this presentation: cedricthieulot.net/diva/05.pdf. The symmetry considerations mentioned by op are essentially saying that: By superposition, adding four plates with one side on a different temperature then the other three (and center temperature given by the complicated sum $S$) gives one plate with constant temperature with core temerature is trivially one. From this $S=1/4$ follows directly \u2013\u00a0tired Jan 13 '17 at 13:03\n\u2022 It looks like my predictions were true: sciencedirect.com/science/article/pii/S0955799706000683 \u2013\u00a0Jack D'Aurizio Jan 13 '17 at 13:12\n\u2022 @tired: thank you. I always find very pleasing to work on some problem with you. \u2013\u00a0Jack D'Aurizio Jan 13 '17 at 13:13\n\u2022 D'Aurizo Yeah, this is a nice proof of concept that physicists and mathematicans CAN work together ;-) \u2013\u00a0tired Jan 13 '17 at 13:17", "date": "2019-05-25 19:19:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2095489/show-that-sum-infty-k-0-frac2-1k2k1-pi-cosh2k1-pi-2-1-4/2096064", "openwebmath_score": 0.884986937046051, "openwebmath_perplexity": 457.59766902816517, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462226131666, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8511534196443029}}
{"url": "http://mathhelpforum.com/pre-calculus/42374-please-help-me-problem-urgent-thanks.html", "text": "The graph of the function y = ax\u00b2 + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.\n\nI know this is only a basic question, but I really would appreciate your help.\n\nThanks\n\n2. $y = ax^2 + bx + c = x^2 + \\frac{b}{a}x + \\frac{c}{a}$\n\n$= x^2 + \\frac{b}{a}x + \\frac{b^2}{4a^2} + \\frac{4ac- b^2}{4a^2} = \\left({x +\\frac{b}{2a}}\\right)^2 + \\frac{4ac- b^2}{4a^2}$\n\nso, the turning point is at $\\left({-\\frac{b}{2a}, \\frac{4ac-b^2}{4a^2}}\\right)$\n\nplug in and you will have 2 equations..\n\nfor the third equation:\nnote:\na point $(x_0,y_0)$ is on the curve $y = ax^2 + bx + c$ if and only if $y_0 = ax_0^2 + bx_0 + c$..\n\n3. Originally Posted by tim_mannire\nThe graph of the function y = ax\u00b2 + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.\n\nI know this is only a basic question, but I really would appreciate your help.\n\nThanks\nThis aproach should give you the same answer but you will only have to solve a symoltaneous equation in two variables.\n\nYou want the equation to pass through (0,5) so substitute x=0, y=5 into the equation to get y = 5 = c\n\nYou want the equation to pass through (-3,2) so substitute x=-3, y=2 into the equation to get $2 = 9*a-3b+5$\n\nThen by knowing that at the turning point the differential of y with respect to x is zero you can write:\n\n$y'=0=2*a*x + b$\n\nYou have two equations in two variables which can be solved by symoltaneous equations.\n\n4. ## Here it is\n\ny=ax^2+bx+c\nsince it passes through (0,5) this eq must satisfy it. So putting the values in the eq we get 5=a(0)^2+b(0)+c.which gives us c=5.\nNow at turning point x=-b/2a\n(if u wanna know why is it that so u may ask it in a new thread).since turning point is (-3,2) therefor x=-b/2a=-3 or b=6a. Substituting this value in eq we get y=ax^2+6ax+5 now substituting (-3,2) we get 2=9a-18a+5 or a=1/3 since b=6a therfor b=2.so finaly a=1/3,b=2,c=5\n\n5. Originally Posted by tim_mannire\nThe graph of the function y = ax\u00b2 + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.\n\nI would really appreciate any help on this problem, thanks.\nFirst use the turning point form y = a (x - h)^2 + k. Obviously h = 0, k = 5. Substitute (-3, 2) to get a.\n\nNow expand the turning point form to get the standard form.\n\nEdit: Don't double post! It wastes peoples time! http://www.mathhelpforum.com/math-he...nt-thanks.html\n\n6. Originally Posted by tim_mannire\nThe graph of the function y = ax\u00b2 + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.\n\nI would really appreciate any help on this problem, thanks.\nUse the following formula:\n\n$y = a(x - p)^2 + q$\n\n$y = a(x + 3)^2 + 2$\n\nPasses through the point $(0;5)$\n\n$5 = a(0 + 3)^2 + 2$\n\n$3 = 9a$\n\n$a = \\frac{1}{3}$\n\n$y = \\frac{1}{3} (x+3)^2 + 2$\n\nEDIT: Oops Mr F beat me to it... by 2 mins\n\n7. Originally Posted by nikhil\ny=ax^2+bx+c\nsince it passes through (0,5) this eq must satisfy it. So putting the values in the eq we get 5=a(0)^2+b(0)+c.which gives us c=5.\nNow at turning point x=-b/2a\n(if u wanna know why is it that so u may ask it in a new thread).since turning point is (-3,2) therefor x=-b/2a=-3 or b=6a. Substituting this value in eq we get y=ax^2+6ax+5 now substituting (-3,2) we get 2=9a-18a+5 or a=1/3 since b=6a therfor b=2.so finaly a=1/3,b=2,c=5\n\nThank you very much. very well answered. much appreciated!", "date": "2013-12-09 15:54:50", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/42374-please-help-me-problem-urgent-thanks.html", "openwebmath_score": 0.6726120710372925, "openwebmath_perplexity": 588.4145077057931, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9879462226131666, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8511534196443029}}
{"url": "https://physics.stackexchange.com/questions/680196/college-board-problem-conservation-of-momentum-conservation-of-energy-in-a-s", "text": "# College Board Problem: Conservation of Momentum -> Conservation of Energy in a Spring\n\nIn my AP Physics C class today, we ran into a problem written by College Board whose answer we disputed. The problem is as such :\n\nA block of mass $$M=5.0 \\ \\mathrm{kg}$$ is hanging at equilibrium from an ideal spring with spring constant $$k=250 \\ \\mathrm{N/m}$$.An identical block is launched up into the first block. The new block is moving with a speed of $$v=5.0 \\ \\mathrm{m/s}$$ when it collides with and sticks to the original block. Calculate the maximum compression of the spring after the collision of the two blocks.\n\nAccording to the College Board answer key, the answer is $$0.5 \\ \\mathrm{m}$$ :\n\n$$p_1=p_2$$\n\n$$Mv_0=(M+M)v_2$$\n\n$$v_2=\\frac{1}{2}v_0= \\left (\\frac{1}{2} \\right)\\left (5.0 \\frac{m}{s}\\right)$$\n\n$$v_2=2.5 \\frac{m}{s}$$\n\n$$K_1 + U_1=K_2+U_2$$\n\n$$\\frac{1}{2}mv_1^2 +0=0+\\frac{1}{2}kx_2^2$$\n\n$$x_2=\\sqrt{\\frac{m}{k}}v_1= \\sqrt{\\frac{(10 \\ \\mathrm{kg)}}{\\left(250 \\frac{N}{m}\\right)}} \\left(2.5 \\frac{m}{n}\\right)$$\n\n$$x_2=0.50 \\ \\mathrm{m}=50 \\ \\mathrm{cm}$$\n\nHowever, half of us disputed this during class. We argued that, yes, $$U_2$$ includes $$\\frac{1}{2}kx^2$$, but it also includes gravitational potential energy at the maximum compression (that is, when it compresses $$x$$ meters from equilibrium, the mass $$M$$ is $$x$$ meters higher above ground). Thus $$K_1+U_1=K_2+U_2$$ is $$\\frac{1}{2}mv^2+0=0+\\frac{1}{2}kx^2+mgx$$. When $$mgx$$ is included, $$x$$ is $$0.24 \\ \\mathrm{m}$$, not $$0.5 \\ \\mathrm{m}$$.\n\nMy physics teacher reluctantly agreed with College Board but could not give a solid explanation why. He said he would e-mail College Board, but in the meantime, I would very much appreciate any input from people who know the answer.\n\n\u2022 I agree with your approach. It should be 2mgh since the mass rising is two blocks of 5 kg each.\n\u2013\u00a0Dan\nDec 1, 2021 at 19:56\n\u2022 Sorry I should have written that lol. When I plugged the numbers in and solved, I did make sure to plug in 10 kg (times 9.81 m/s/s times x). Dec 1, 2021 at 19:57\n\u2022 Someone in class pointed out that perhaps because the question asks for MAXIMUM compression, they assume gravity does not exist. However, this still doesn't make sense. Assuming gravity does not exist (perhaps it is laid horizontally), the spring would have no elongation. When it is laid vertically, it elongates until the tension of the coil matches Mg. Thus, its equilibrium when horizontal (or simply without gravity) will be the length of the fully compressed spring. If the second block struck it elastically, the spring could not compress anymore. A compression of .5 meters is improbable. Dec 1, 2021 at 20:08\n\u2022 Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks! Dec 1, 2021 at 20:24\n\u2022 Don't forget to include the initial elongation of the spring. Dec 1, 2021 at 20:36\n\nI guess this is a homework/check-my-work problem, so by the letter of the law I should not answer, but I would argue there is broad interest in solving it correctly given that a supposedly reputable source is presenting an incorrect solution.\n\nHere is how I would do this. Initially, the spring is stretched distance $$d=mg/k$$ below its equilibrium position. Choose the position of the hanging block as $$y=0$$, so the gravitational potential energy immediately after the collision is zero. In terms of the speed $$v=v_0/2$$ of the blocks after the collision and the mass $$m$$ of one block, the total energy immediately after the collision is \\begin{align} E_i &= \\frac{1}{2}(2m) v^2 + \\frac{1}{2}kd^2\\\\ &= \\frac{1}{4}mv_0^2 + \\frac{1}{2}\\frac{m^2g^2}{k}. \\end{align} Let $$h$$ be the distance of the blocks above the equilibrium position of the spring when the blocks are at their maximum height. At this point, the blocks are at rest, so their total energy is \\begin{align} E_f &= \\frac{1}{2}kh^2 + 2mg(h + d)\\\\ &= \\frac{1}{2}kh^2 + 2mgh + \\frac{2m^2g^2}{k}. \\end{align} Using conservation of energy, \\begin{align} &\\frac{1}{4}mv_0^2 + \\frac{1}{2}\\frac{m^2g^2}{k} = \\frac{1}{2}kh^2 + mgh + \\frac{2m^2g^2}{k}\\\\ \\rightarrow &\\frac{1}{2}kh^2 + 2mgh + \\frac{3}{2}\\frac{m^2g^2}{k} - \\frac{1}{4}mv_0^2 = 0. \\end{align} We can solve this quadratic equation for $$h$$ to obtain \\begin{align} h = -\\frac{2mg}{k} + \\sqrt{\\frac{m^2g^2}{k^2} + \\frac{mv_0^2}{2k}}. \\end{align} In terms of the given numbers, $$v_0 = 5.0\\,\\text{m}/\\text{s}$$, $$m=5.0\\,\\text{kg}$$, and $$k=250\\,\\text{N}/\\text{m}$$, we get $$\\boxed{h=15\\,\\text{cm}.}$$\n\nNote that if we set $$g=0$$ so that there is no gravity, we get \\begin{align} h = \\sqrt{\\frac{mv_0^2}{2k}} = \\boxed{50\\,\\text{cm}.} \\end{align} We are left to conclude that the author of the solution was likely in free-fall at the time of its writing.\n\n\u2022 h=15 cm? I got 0.1447. Is this the same thing? Also, how does your equation differ from the one my teacher and I made, which outputs 24 centimeters? Dec 2, 2021 at 17:23\n\u2022 @SebastianPojman-Malo When I plugged the numbers into Mathematica, it spit out 14.5 something, and I rounded to two sig figs. In my answer, I'm taking into account the fact that the spring does not start at its equilibrium length.\n\u2013\u00a0d_b\nDec 2, 2021 at 17:38\n\u2022 That's interesting. College Board (and I) assume that it absolutely does hit the original block when the spring is at its equilibrium. Dec 2, 2021 at 17:45\n\u2022 No, that is not correct. The hanging block is in equilibrium, which means the spring must be exerting an upward force on the block to keep it from falling under the influence of gravity. So the spring cannot be at its equilibrium length.\n\u2013\u00a0d_b\nDec 2, 2021 at 18:54\n\u2022 I agree with your h = .145 m. Note that that (h) is measured up from the un-stretched position of the spring and not from the point where the collision occurs. Dec 6, 2021 at 17:26\n\nThere is already a very good and complete answer from d_b, let me just add a comment to elucidate the possible (erroneous in this case) reasoning by which one could try to forget about gravity.\n\nFirst, consider one block hanging from the spring in equilibrium. Assume that we provide it with the vertical speed $$v$$ without any other block sticking to it, so it moves up alone. What height above the initial position will it reach in such a case? Writing energy conservation (for detailed explanation look at d_b's answer, it is quite the same here - only I choose to measure height relative to the initial position of the block for reasons which will become clear shortly), we get:\n\n$$\\frac{m v^2}{2} + \\frac{m^2 g^2}{2 k} = m g h + \\frac{k (h-\\frac{mg}{k})^2}{2}$$\n\nLet us play a little with this equation. Open the square in the RHS and observe that $$mgh$$ cancels:\n\n$$m g h + \\frac{k (h-\\frac{mg}{k})^2}{2} = mgh + \\frac{k(h^2 + \\frac{m^2g^2}{k^2} - \\frac{2mgh}{k})}{2} = \\frac{kh^2}{2} + \\frac{m^2 g^2}{2k}$$\n\nPlugging it back to the energy conservation, we observe that $$\\frac{m^2 g^2}{2k}$$ cancels as well, and we are left with\n\n$$\\frac{m v^2}{2} = \\frac{kh^2}{2}$$\n\nIn other words, it looks exactly like we could forget about gravity and initial elongation of the spring altogether: the answer is the same as it would have been for a horizontal spring. This is not a coincidence: what we have derived means that the total (gravitational+elastic) potential energy of this system is quadratic in deviation from the equilibrium, so it really looks like for a horizontal spring. One could also understand it graphically: potential energy of the spring is a parabola with the equilibrium position at the lowest point. If we also add gravity, a linear function is added to this parabola. The result is another parabola with the same shape but another minimum position - at the equilibrium point of the hanging block (equilibrium is always the minimum of total potential energy). If we measure all elongations compared to the new equilibrium, we could forget about both gravity and initial elongation: they compensate each other.\n\nI think this might have been the logic of authors of the solution. However, this logic only works if we measure elongations compared to the position with minimal potential energy. If another block sticks to the first one, the equilibrium position is now shifted below, so the blocks start moving already from a non-equilibrium position and one needs to account for the potential energy of initial position as well - which was not done in the provided College Board solution. As an exercise I would suggest an interested reader to reproduce d_b's answer with this method of total potential taking the above consideration into account.\n\n\u2022 \"let me just add a comment to elucidate the possible (erroneous in this case) reasoning by which one could try to forget about gravity.\" Just for clarity, are you saying that College Board's ignoring gravity is erroneous? Or are you agreeing with College Board? Dec 2, 2021 at 17:31\n\u2022 I disagree with the College Board answer and solution. In my answer I describe a method which they presumably used when solving the problem, but I point out that they forgot to account for the fact that equilibrium position of two blocks on a spring differs from the equilibrium position of one block. Dec 2, 2021 at 17:56\n\u2022 Walk me through this. (Remember, I'm still in high school...) Does that mean that once the elastic collision happens and the identical block \"sticks,\" the equilibrium position changes in that moment? Dec 2, 2021 at 18:00\n\u2022 Firstly, if they stick, it is called an inelastic collision. Changing equilibrium position means that if one block of mass $m$ stretches the spring in equilibrium by $d$, two such blocks would stretch it by $2d$. So if in this problem we attached the second block to the first one without any velocity, they would move down to reach the new equilibrium position. To prevent confusion, let me stress that the word equilibrium here is used to describe a situation when all forces are balanced and nothing moves (not to be confused with the equilibrium length of the spring - when there are no forces). Dec 2, 2021 at 19:06\n\u2022 How does conservation of momentum play into this? As I see it, (5kg)(5m/s)=(5+5kg)(v); v=2.5m/s. So according to the conservation of momentum, the two blocks stuck together will move at 2.5m/s immediately following the collision. That being said, they will decelerate because the force of gravity (2mg) will outpower the tension of the spring (mg) at that moment. Because of this, the equilibrium is disturbed, and the amount of compression is reduced. CB's answer assumes equilibrium is NOT disturbed and thus the blocks' velocity does not decelerate, right? Dec 2, 2021 at 20:27\n\nWith a 5 kg mass hanging at rest from the spring of constant k = 250 N/m, the stretch of the spring will be X = mg/k =5(9.8)/250 = 0.196 m. With (x) measured positive down from the un-stretched position, and gravitational potential energy chosen to be zero when x = 0, then the energy of the 10 kg mass is (1/2)(10)$$2.5^2$$ + (1/2)(250)$$0.196^2$$ - 10(9.8)(0.196) = (1/2)(250)$$x^2$$ -10(9.8)x. (x when v = 0) Or: 125$$x^2$$ - 98x \u2013 (31.25 +4.802 \u2013 19.208) = 0. Solving gives x = - 0.145 m. (The other solution is for initial velocity downward). Since positive x was measured down, this negative x represents a compression of the spring above the unstretched position. Then the total rise is 0.196 + 0.145 = 0.341 m. For the record, if the moving mass is the same as the hanging mass: then mg = kX and (1/2)m$$v^2$$ + (1/2)k$$X^2$$ - (kX)X = (1/2)k$$x^2$$ - (kX)x. Or rearranged: (1/2)m$$v^2$$ = (1/2)k$$(X^2 \u2013 2Xx + x^2) = (1/2)k(X-x)^2$$. Lets try ignoring gravity with x measured from a 10 kg hanging position (which is an additional 0.196m further down): (1/2)(10)$$2.5^2$$ + (1/2)(250)$$0.196^2$$ = (1/2)(250)$$x^2$$. Rearranging: 125$$x^2$$ = (31.25 + 4.802) Giving x = 0.537. With the starting position 0.196 m above the new equilibrium, this gives a rise of 0.341 m. The bottom line: A mass hanging from a spring defines a new equilibrium position. The kx measured from that position includes the force of gravity (but don't change the mass). Here is an alternative approach: With the stretch of the spring (x) measured positive down from the unstretched position, the net force on the hanging mass is: F = mg - kx Then dF = - k(dx). Integrating both sides gives F(x) - $$F_o$$ = -k(x - $$x_o$$). If starting from the new equilbrium then $$F_o$$ =0.\n\n\u2022 Please consider using MathJax; in its current form this answer is mostly illegible.\n\u2013\u00a0rob\nJan 23, 2022 at 13:55", "date": "2023-03-25 02:25:40", "meta": {"domain": "stackexchange.com", "url": "https://physics.stackexchange.com/questions/680196/college-board-problem-conservation-of-momentum-conservation-of-energy-in-a-s", "openwebmath_score": 0.7946702241897583, "openwebmath_perplexity": 513.1419705121123, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9559813463747182, "lm_q2_score": 0.8902942283051332, "lm_q1q2_score": 0.8511046750447819}}
{"url": "https://math.stackexchange.com/questions/1827223/when-to-simplify-a-quadratic-equation", "text": "# When to simplify a quadratic equation?\n\nI had the following quadratic equation: $$38x^2 - 140x - 250 = 0$$ And before starting to solve it, I simplified it by dividing all terms by $2$: $$19x^2 -70x - 125 = 0$$ But when I solved it I got: $$x= \\frac{35\\pm5\\sqrt{163}}{19}$$ which is not the exact solution for the original equation. So, I returned to that first equation and solved it. The roots are: $x=-\\frac{25}{19}, 5$\n\nNow my question is: How would I know when to simplify a quadratic equation before solving it? Is there something that shows me that I'm right to simplify or not?\n\nThank you for your help.\n\n\u2022 You must have made an error because both those quadratics have the same roots, multiplying or dividing through by the same number doesn't change the roots \u2013\u00a0Triatticus Jun 15 '16 at 13:46\n\u2022 @Triatticus: I checked with Wolfram alpha \u2013\u00a0Rafiq Jun 15 '16 at 13:47\n\u2022 It should be $x=\\frac{35\\pm 60}{19}$, not $x=\\frac{35\\pm 5\\sqrt{163}}{19}$. \u2013\u00a0user236182 Jun 15 '16 at 13:47\n\u2022 your square root is wrong... $\\sqrt{70^2+4.19.125}$ calculate again \u2013\u00a0Kushal Bhuyan Jun 15 '16 at 13:47\n\u2022 I did too and got the same answers \u2013\u00a0Triatticus Jun 15 '16 at 13:48\n\nYou can always simplify a quadratic equation (by dividing out common numeric factors) before solving it. Simplifying like this is optional and is always a good idea because it generally gives you easier numbers to work with.\n\nThe error is not because you simplified, it's because there is a mistake somewhere in your work after that. I can't tell you where exactly without seeing your work, but here's how it should go:\n\n\\begin{align*} x &= \\frac{70 \\pm \\sqrt{(-70)^2 - 4(19)(-125)}}{2(19)}\\\\[0.3cm] &= \\frac{70 \\pm \\sqrt{4900 + 9500}}{38}\\\\[0.3cm] &= \\frac{70 \\pm \\sqrt{14400}}{38} \\\\[0.3cm] &= \\frac{70 \\pm 120}{38} \\end{align*}\n\nWhen simplified you get $x = 5$ and $x = -25/19$.\n\nWhen to simplifying a quadratic is up to you. Remember, multiplying both sides by a number retains the equality, the same thing with adding, subtracting and dividing.\n\nYou could've left it the way it is, without simplifying, and you would have gotten the solution as it doesn't affect the roots.\n\nIt is for last equation :$$x=\\frac { 35\\pm \\sqrt { { 35 }^{ 2 }+125\\cdot 19 } }{ 19 } =\\frac { 35\\pm \\sqrt { 3600 } }{ 19 } =\\frac { 35\\pm 60 }{ 19 }$$\n\nBut using $\\frac{\\sqrt{b^2-4ac}}{2a}$ in the reduced equation gives the same roots .", "date": "2019-07-22 05:34:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1827223/when-to-simplify-a-quadratic-equation", "openwebmath_score": 0.9799789786338806, "openwebmath_perplexity": 235.06836431886057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9559813463747181, "lm_q2_score": 0.8902942159342104, "lm_q1q2_score": 0.8511046632184105}}
{"url": "https://math.stackexchange.com/questions/3213122/prove-the-sequence-f-n-frac1n21-is-a-cauchy-sequence", "text": "# Prove the sequence $f_{n} = \\frac{1}{n^2+1}$ is a Cauchy sequence.\n\nProve the sequence $$f_{n} = \\frac{1}{n^2+1}$$ is a cauchy sequence.\n\nI'm just making sure my logic and reasoning is sound for the above proof:\n\nDefinition of cauchy sequence: $$f_n$$ is Cauchy if for all $$\u0190$$ there is an $$M \\in \\mathbb{N}$$ such that for all $$n,m > M, |f_n-f_m|<\u0190$$\n\n$$|f_n-f_m|$$=$$|\\frac{1}{n^2+1}-\\frac{1}{m^2+1}|$$\n\nby the triangle inequality:\n\n$$\\le|\\frac{1}{n^2+1}|+|\\frac{1}{m^2+1}|$$\n\n$$<\\frac{1}{n}+\\frac{1}{m}$$\n\nTo ensure $$\\frac{1}{n}+\\frac{1}{m}<\u0190$$ it suffices to have:\n\n$$max(\\frac{1}{n}+\\frac{1}{m}) < \\frac{\u0190}{2}$$, so we will take $$M(\u0190) = \\lceil\\frac{2}{\u0190}\\rceil$$\n\nNow for a formal proof:\n\nLet $$\u0190>0$$\n\nDefine $$M(\u0190) = \\lceil\\frac{2}{\u0190}\\rceil$$\n\nLet $$n,m > M(\u0190)$$\n\n$$n,m > \\frac{2}{\u0190}$$\n\n$$\\frac{1}{n}<\\frac{\u0190}{2}$$ and $$\\frac{1}{m}<\\frac{\u0190}{2}$$\n\n$$|f_n - f_m| \\le \\frac{1}{n}+\\frac{1}{m}$$\n\n$$|f_n - f_m| < \\frac{\u0190}{2}+\\frac{\u0190}{2} = \u0190$$\n\nTherefore $$f_n$$ is cauchy.\n\nSolution to the proof\n\n\u2022 Yes, your proof is fine. \u2013\u00a0Mark May 4 at 8:53\n\u2022 Okay great, I was just checking as if you see in my question above, the solution was different. I thought my answer was too simple. \u2013\u00a0user11015000 May 4 at 8:55\n\u2022 Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not viewable to some, such as those who use screen readers. \u2013\u00a0GNUSupporter 8964\u6c11\u4e3b\u5973\u795e \u5730\u4e0b\u6559\u6703 May 4 at 15:31\n\u2022 The proof before the red edit is fine. \u2013\u00a0DanielWainfleet May 4 at 18:52", "date": "2019-10-18 16:07:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3213122/prove-the-sequence-f-n-frac1n21-is-a-cauchy-sequence", "openwebmath_score": 0.766474723815918, "openwebmath_perplexity": 742.2232511189287, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180643966651, "lm_q2_score": 0.863391611731321, "lm_q1q2_score": 0.8510607083321147}}
{"url": "http://math.stackexchange.com/questions/271013/approximating-continuous-functions-with-polynomials", "text": "# Approximating continuous functions with polynomials\n\nGiven $\\epsilon \\gt 0$ and $f \\in C^{0}[0,1]$, Weierstrass says that I can find at least one (how many? probably a lot?) polynomial $P$ which approximates f uniformly: $$\\sup_{x \\in [0,1]} |f(x) - P(x)| \\lt \\epsilon$$\n\nThis means that, under the sup norm $||.||_{\\infty}$, the polynomials are dense in $C^{0}[0,1]$. So, in analogy to approximating irrationals with the rationals, I would like to know:\n\n\u2022 What can we say about the order of $P$? Or, turning this around, given that $P$ is of order $n$, how small can $\\epsilon$ get?\n\nI'm betting this should depend in some way on the properties of $f$: the intuition is that smoother functions should be somehow \"better\" approximated by lower-order polynomials, and less-well-behaved functions should require higher-order polynomials. But I am not sure how to formalize this.\n\nThis is probably all well-understood, but I'm not well-read on approximation theory. Any guidance would be wonderful.\n\n-\nAll of the answers have been very helpful, thank you to everyone! \u2013\u00a0 AndrewG Jan 5 '13 at 19:58\n\nAccording to Theorem 1.2 of this paper by Sukkrasanti and Lerdkasem, we have the following result (with impressively great generality, I might add):\n\nLet $f: [0,1]^p \\rightarrow \\mathbb{R}^q$ be bounded. Then there exists $C$ such that $$\\| f - B_n(f) \\|_{\\infty} \\le C \\omega(1/\\sqrt{n})$$ where $\\omega(\\delta) := \\sup_{\\|t_1 - t_2\\| \\le \\delta} \\|f(t_1) - f(t_2)\\|$ and $B_n(f)$ is the $n$th Bernstein polynomial of $f$.\n\nI do not claim to have read the paper myself. Notice that if $f$ is indeed continuous, then by uniform continuity, as $n \\rightarrow \\infty$, $\\omega(1/\\sqrt{n})$ must go to zero. So the convergence rate is related to how rapidly $f$ can vary, as intuition would already suggest.\n\n-\n\nDefine, $B_i^n(x)=\\binom{n}{i}x^i(1-x)^{n-i}$, nice fact,\n\n\u2022 Partition of the unity $\\sum_{i=0}^{n}B_i^n(x)=1$ this is easy to see:\n\nSee $1= x+(1-x)$ then, $$1=(x+(1-x))^n=\\sum_{i=0}^{n}\\binom{n}{i}x^i(1-x)^{n-i}=\\sum_{i=0}^{n}B_i^n(x).$$\n\nWithout loss of generality suppose $a=0$ and $b=1$. We need of the following estimative:\n\n\\begin{eqnarray*} \\sum_{i=0}^{n}\\left(x-\\frac in\\right)^{2}B_i^n(x)&=&x^2\\sum_{i=0}^{n}B_i^n(x)-2x\\sum_{i=0}^{n}\\dfrac{i}{n}B_i^n(x)+\\sum_{i=0}^{n}\\left(\\dfrac{i}{n}\\right)^2 B_i^n(x)\\\\&=& x^2-2x^2+\\dfrac{1}{n^2}\\sum_{i=0}^{n}n(n-1)\\dfrac{i^2-i +i}{n(n-1)} B_i^n(x)\\\\&=& -x^2+ \\dfrac{1}{n}\\sum_{i=0}^{n}\\dfrac{i}{n}B_i^n(x)+\\dfrac{n(n-1)}{n^2}\\sum_{i=0}^{n}\\dfrac{i(i-1)}{n(n-1)}B_i^n(x)\\\\&=& -x^2+\\dfrac{x}{n}+\\dfrac{n-1}{n}x^2\\\\&=& \\dfrac{x-x^2}{n}\\leqslant\\dfrac{1}{4n}. \\end{eqnarray*} Now, since $f$ is continuous we have that $f$ is uniformly continuous,because $[0,1]$ is compact. So for each $\\epsilon$ exists $\\delta>0$ such that for all $x,y\\in [0.1]$ with $|x-y|<\\delta$ implies $|f(x)-f(y)|<\\epsilon$. Since $f$ have yours extreme extreme values in $[0,1]$ then exists a constant $M$ such that $|f(x)|\\leqslant M$ for all $x\\in[0,1].$\n\nConsider the Berstein polynomial of $f$, $$P_n(x)=\\sum_{i=0}^{n}f\\left(\\dfrac{i}{n}\\right)B_i^n(x);$$ since $\\displaystyle\\sum_{i=0}^{n}B_i^n(x)=1$, we have $f(x)=\\displaystyle\\sum_{i=0}^{n}f(x)B_i^n(x)$ then,\n\n\\begin{eqnarray*} \\left|f(x)-P_n(x)\\right|&\\leqslant& \\sum_{i=0}^{n}\\left|f(x)-f\\left(\\dfrac{i}{n}\\right)\\right|B_i^n(x)\\\\&=& \\sum_{S_1}\\left|f(x)-f\\left(\\dfrac{i}{n}\\right)\\right|B_i^n(x)+\\sum_{S_2}\\left|f(x)-f\\left(\\dfrac{i}{n}\\right)\\right|B_i^n(x) \\end{eqnarray*}\n\nwhere $S_1=\\{0\\leqslant i\\leqslant n ~;~ |x-\\frac{i}{n}|<\\delta\\}$ and $S_2=\\{0\\leqslant i\\leqslant n~;~ |x-\\frac{i}{n}|\\geqslant \\delta\\}.$ analyzing each sum separately $$\\sum_{S_1}\\left|f(x)-f\\left(\\dfrac{i}{n}\\right)\\right|B_i^n(x)\\leqslant \\sum_{S_1}\\epsilon|B_i^n(x)\\leqslant \\epsilon\\sum_{i=0}^{n}B_i(x)=\\epsilon$$\n\n\\begin{eqnarray*} \\sum_{S_2}\\left|f(x)-f\\left(\\dfrac{i}{n}\\right)\\right|B_i^n(x)&\\leqslant& 2M\\sum_{S_2}B_i^n(x)\\leqslant \\sum_{S_2}\\dfrac{(x-\\frac{i}{n})^2}{\\delta^2}B_i^n(x)\\\\&\\leqslant&\\dfrac{2M}{\\delta^2}\\sum_{i=0}^{n}(x-\\frac{i}{n})^2 B_i^n(x)\\leqslant \\dfrac{M}{2\\delta^2 n} \\end{eqnarray*}\n\nsince $\\dfrac{M}{2\\delta^2 n}<\\epsilon$ for $n$ large, we demonstrate the desired.\n\n-\nNice answer. By the way, what does \"da\u00ed\" mean? \u2013\u00a0 Antonio Vargas Jan 5 '13 at 16:50\n@AntonioVargas: I'm sorry, is the habit of writing in portuguese da\u00ed=so or then \u2013\u00a0 user27456 Jan 5 '13 at 16:58\n@EduardoSiva: Everything looks fine, except that you have not defined $B_{i}$. I suppose that it denotes a Bernstein polynomial, am I right? \u2013\u00a0 Haskell Curry Jan 5 '13 at 17:34\n@HaskellCurry: I'll try to fix it \u2013\u00a0 user27456 Jan 5 '13 at 17:43\n\nA near minimax approximation to a continuous function, say, $f:[-1,1]\\to\\mathbb{R}$, can be obtained with Chebyshev polynomials. You may look up the details in this book chapter, of which theorem 3.10 says that if $p$ is a Chebyshev interpolation polynomial for such continuous $f:[-1,1]\\to\\mathbb{R}$ at $n+1$ points, then $\\|f-p\\|\\sim C\\,\\omega(\\frac1n)\\log n$ as $n\\to\\infty$, where the norm is the maximum norm, $\\omega(\\cdot)$ denotes the modulus of continuity and $C$ is a constant.\n\n-\n\nSince you tagged your question by \"reference-request\" I will give only them (instead of copying a lot). Jackson inequality (\"direct theorem\"), and Bernstein theorem (\"converse theorem\"). See also the references therein. (My humble opinion they are good quality books. Typing into Google these keywords you will obtain a lot of links, including many free downloadable lecture notes, maybe books.)\n\n-", "date": "2015-07-05 15:27:16", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/271013/approximating-continuous-functions-with-polynomials", "openwebmath_score": 0.94944828748703, "openwebmath_perplexity": 395.76866527482014, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180664944448, "lm_q2_score": 0.8633915976709976, "lm_q1q2_score": 0.8510606962838053}}
{"url": "https://www.physicsforums.com/threads/calculating-work.324424/", "text": "# Calculating work\n\n1. Jul 10, 2009\n\n### jeff1evesque\n\n1. The problem statement, all variables and given/known data\nGiven the attached picture, Calculate the work required to go around the contour shown if the force is $$\\vec{F} = y\\hat{x} - x^{2} \\hat{y}$$\n\nFirst by Contour integration:\nWork = $$\\oint \\vec{F} \\bullet \\vec{dl} = \\int_{a}^{b} (\\vec{F} \\bullet \\hat{x})dx + \\int_{b}^{c} (\\vec{F} \\bullet \\hat{y})dy - \\int_{c}^{d} (\\vec{F} \\bullet \\hat{x})dx - \\int_{d}^{a} (\\vec{F} \\bullet \\hat{y})dy$$\n\n$$= \\int_{1}^{2} (2) dx + \\int_{2}^{3} (4)dy - \\int_{2}^{1} (3)dx - \\int_{3}^{2} (1)dy = 0$$\n\nI think the answer above is suppose to be -4 not 0.\n\nNow by Stokes Theorem:\nWhen this was done with \"Stokes theorem\", we get the following:\nWork = $$\\oint \\vec{F} \\bullet \\vec{dl} \\equiv \\int \\int (\\nabla \\times \\hat{F})\\bullet \\vec{ds} = - \\int \\int (2x + 1)( \\hat{z} \\bullet \\hat{z})ds = \\int_{2}^{3} \\int_{1}^{2} (2x + 1)dxdy = -4$$\nNote: $$( \\hat{z} \\bullet \\hat{z}) = 1$$ since $$( \\hat{z} \\bullet \\vec{ds}) = ds$$\n\nQuestion:\nWhen I look at this, Stokes method seems apparently correct, but the method before it seems wrong. Can someone help me find my error?\n\nThanks,\n\nJL\n\n#### Attached Files:\n\n\u2022 ###### StokePic.bmp\nFile size:\n297 KB\nViews:\n67\nLast edited: Jul 10, 2009\n2. Jul 11, 2009\n\n### E_M_C\n\nYour attachment is still pending approval, but if we're speaking of the work done around a closed path, it's certainly not zero. This is due to the fact that the force field is not conservative. While the force is position dependent (as is required by a conservative force), it's curl, $$\\nabla \\times F$$, is not the zero vector.\n\n3. Jul 11, 2009\n\n### HallsofIvy\n\nStaff Emeritus\nThe contour is the square with corners at (1, 2), (2, 2), (2, 3), and (1, 3).\n\nWhy do you have minus signs on the last two integrals? Assuming that your contour goes from point a to point b, then to point c, then to point d, then back to point a, these should all be \"+\". (Although the integrals themselves might be 0.)\n\nOn the first leg, with x= t, y= 2, the integral is $\\int_1^2 2 dx$ as you say. On the second leg, with x= 2, y= t, the integral is $\\int_2^3 (-4) dt$. Did you forget the \"-\" on \"$-x^2\\vec{j}$\"? On the third leg, with x= t, y= 3, the integral is $\\int_2^1 3dt= -\\int_1^2 3dt$. You have the wrong sign. On the fourth leg, with x= 1, y= t, the integral is $\\int_3^2(-1) dt= \\int_2^3 dt$. That is what you have, but, I think, because of two canceling sign errors!. Evaluating, the integral is 2- 4+ (-3)+ (1)= -4.\n\n4. Jul 11, 2009\n\n### jeff1evesque\n\nThanks Halls.", "date": "2017-11-17 21:38:53", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/calculating-work.324424/", "openwebmath_score": 0.9222193360328674, "openwebmath_perplexity": 859.161829378275, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864287859482, "lm_q2_score": 0.8596637559030338, "lm_q1q2_score": 0.8510554516631595}}
{"url": "http://math.stackexchange.com/questions/243559/picking-random-numbers-as-long-as-they-keep-decreasing-expected-number-of-numbe/243567", "text": "# Picking random numbers as long as they keep decreasing. Expected number of numbers you pick?\n\nPick a random number (evenly distributed) between $0$ and $1$. Continue picking random numbers as long as they keep decreasing; stop picking when you obtain a number that is greater than the previous one you picked. What is the expected number of numbers you pick?\n\n-\nOne of my friend send me this question,& he sends me the ans.. \u2013\u00a0UnknownController Nov 24 '12 at 6:25\nPlease, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. \u2013\u00a0Julian Kuelshammer Nov 24 '12 at 6:41\n@JulianKuelshammer thank you...next time I'll give more informative title.. \u2013\u00a0UnknownController Nov 24 '12 at 6:45\n\nThere're $n!$ ways of arranging $n$ numbers, supposing that there're $n$ picks, then the first $(n-1)$ picks are in descending order, there are $(n-1)$ ways of choosing the first $(n-1)$ numbers and thus the probability of picking just $n$ nimbers is $\\cfrac {n-1}{n!}$ and the expected value is $$E=\\sum^{\\infty}_{n=2} n\\cdot\\cfrac {n-1}{n!}= \\sum^{\\infty}_{n=2} \\cfrac {1}{(n-2)!}=\\sum^{\\infty}_{n=0} \\cfrac {1}{n!}=e=2.718281828459...$$\n\n-\nthank you my friend for tried to help,I just post ans. in comment see that :) \u2013\u00a0UnknownController Nov 24 '12 at 6:27\n\nThe following solutions uses indicator random variables.\n\nFix an $n$, and imagine doing the experiment only $n$ times. For $i\\le n$, let $X_j=1$ if $X_1\\gt X_2 \\gt \\cdots >X_j$, and let $X_j=0$ otherwise. Then $\\Pr(X_j=1)=\\frac{1}{j!}$ and therefore $E(X_j)=\\frac{1}{j!}$.\n\nIf $Y_n$ is the length of the longest monotone decreasing sequence that starts with the first number chosen the beginning, then $Y_n=X_1 +X_2+\\cdots +X_n$. Thus, by the linearity of expectation, $$E(Y_n)=E(X_1)+E(X_2)+E(X_3)+\\cdots+E(X_n)= 1+\\frac{1}{2!}+\\frac{1}{3!}+\\cdots +\\frac{1}{n!}.$$ As $n\\to \\infty$, this approaches $e-1$.\n\nBut your random variable counts all the picks up to and including the first pick that breaks the monotonicity. So your random variable has expectation $(e-1)+1=e$.\n\n-", "date": "2016-02-06 08:01:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/243559/picking-random-numbers-as-long-as-they-keep-decreasing-expected-number-of-numbe/243567", "openwebmath_score": 0.8999835848808289, "openwebmath_perplexity": 332.0418966266846, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864271610316, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8510554467068692}}
{"url": "https://math.stackexchange.com/questions/2585274/evaluating-int-fracx21x2-2x22dx-section-7-3-28-in-stewart-calc", "text": "# Evaluating $\\int\\frac{x^2+1}{(x^2-2x+2)^2}dx$ - Section 7.3, #28 in Stewart Calculus 8th Ed.\n\nI'd like to evaluate $$I=\\int\\frac{x^2+1}{(x^2-2x+2)^2}dx.$$\n\nAccording to WolframAlpha, the answer is $$\\frac{1}{2}\\bigg(\\frac{x-3}{x^2-2x+2}-3\\tan^{-1}(1-x)\\bigg)+C.$$\n\nTo ensure that my answer is correct, I'd like to match it with this answer.\n\nFirst I complete the square by writing the denominator as $((x-1)^2+1)^2$, then use a trig-sub $x-1=\\tan\\theta$ so $dx=\\sec^2\\theta d\\theta.$\n\nThen $$I=\\int\\frac{(\\tan\\theta +1)^2+1}{\\sec^4\\theta}\\sec^2\\theta d\\theta\\\\=\\int\\frac{\\tan^2\\theta+2\\tan\\theta+2}{\\sec^2\\theta}d\\theta\\\\=\\int\\sin^2\\theta +2\\sin\\theta\\cos\\theta+2\\cos^2\\theta d\\theta\\\\=\\int\\bigg(\\frac{1}{2}-\\frac{\\cos(2\\theta)}{2}\\bigg)+\\sin(2\\theta)+(1+\\cos(2\\theta))d\\theta\\\\=\\int\\frac{3}{2}+\\frac{\\cos(2\\theta)}{2}+\\sin(2\\theta)d\\theta\\\\=\\frac{3\\theta}{2}+\\frac{\\sin(2\\theta)}{4}-\\frac{\\cos(2\\theta)}{2}+C\\\\=\\frac{3\\theta}{2}+\\frac{\\sin\\theta \\cos\\theta}{2}-\\frac{1-2\\sin^2\\theta}{2}+C.$$\n\nHere's a picture corresponding to my trig-sub.\n\nFrom the picture, I conclude $$I=\\frac{3\\tan^{-1}(x-1)}{2}+\\frac{1}{2}\\frac{x-1}{\\sqrt{x^2-2x+2}}\\frac{1}{\\sqrt{x^2-2x+2}}- \\frac{1-2\\frac{(x-1)^2}{x^2-2x+2}}{2}+C\\\\=\\frac{-3\\tan^{-1}(1-x)}{2}+\\frac{x-1}{2(x^2-2x+2)}-\\frac{1}{2}+\\frac{(x-1)^2}{x^2-2x+2}+C.$$\n\nAlthough the $\\arctan$ portion of the answer matches Wolfram, the remaining portion does not (since there is a square in the numerator after finding a common denominator). Where am I making the mistake?\n\nI've also tried using the identity $\\cos(2\\theta)= 2\\cos^2\\theta-1$ instead of $\\cos(2\\theta)= 1-2\\sin^2\\theta$, but the same issue occurs.\n\nYour work is correct; the only difference between your antiderivative and the one calculated by WolframAlpha is a constant of integration, $1/2$. You may verify that $$\\frac{(x-1)^2}{x^2-2 x+2}+\\frac{x-1}{2 \\left(x^2-2 x+2\\right)}-\\frac{1}{2} = \\frac{x^2-x-1}{2 \\left(x^2-2 x+2\\right)},$$ and it follows that $$\\frac{x^2-x-1}{2 \\left(x^2-2 x+2\\right)}-\\frac{x-3}{2 \\left(x^2-2 x+2\\right)} = \\frac{1}{2}.$$", "date": "2021-09-25 01:21:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2585274/evaluating-int-fracx21x2-2x22dx-section-7-3-28-in-stewart-calc", "openwebmath_score": 0.9633340239524841, "openwebmath_perplexity": 142.09263149288228, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864278996301, "lm_q2_score": 0.8596637469145054, "lm_q1q2_score": 0.8510554420027029}}
{"url": "http://math.stackexchange.com/questions/355151/combinatorial-explanation-to-following-recurrence-relation-a-n-2-a-n-1-a/355159", "text": "# Combinatorial explanation to following recurrence relation $a_n = 2 a_{n-1} + a_{n-2}$\n\nQuestion was the following:\n\n$a_n$ is the number of ternary strings (strings of 0,1,2) which contain no consecutive zeros and no consecutive ones. Find a formula for $a_n$?\n\nBy brute force, I found a recurrence relation for $a_n$ as the following:\n\n$a_n = 2 a_{n-1} + a_{n-2}$\n\nNow I am wondering if there is a good combinatorial explanation/proof to my recurrence relation. Can you see something?\n\nBest Regards\n\n-\nTerminological note: those are ternary strings, not turning strings. \u2013\u00a0 Brian M. Scott Apr 8 '13 at 20:00\n@BrianM.Scott thank you for correction \u2013\u00a0 Xentius Apr 8 '13 at 20:02\n\nLet $b_{i,n}$ be the number of strings of length $n$ starting with $i\\in\\{0,1,2\\}$. Then by your rules $$a_n = b_{0,n} + b_{1,n} + b_{2,n} = (b_{1,n-1} + b_{2,n-1}) + (b_{0,n-1} + b_{2,n-1}) + a_{n-1} = 2a_{n-1} + a_{n-2}.$$\n\nADDITION: This transformation can be translated into a proof by bijection. Let $X_n$ be the set of all sequences of length $n$. We define a map $$(X_{n-1} \\times \\{A,B\\}) \\cup X_{n-2} \\quad\\to\\quad X_{n}$$ by (\"$\\cdot$\" denotes concatenation; $\\mu$ a sequence in $X_{n-2}$ and $\\lambda$ a sequence in $X_{n-1}$):\n\n$(\\lambda,A) \\mapsto 2\\cdot\\lambda$\n\n$(\\lambda,B)\\mapsto\\begin{cases} 1\\cdot \\lambda & \\text{if }\\lambda\\text{ starts with }0\\text{ or }2\\\\ 0\\cdot \\lambda & \\text{if }\\lambda\\text{ starts with }1 \\end{cases}$\n\n$\\mu \\mapsto 0\\cdot 2\\cdot \\mu$\n\nThis is a bijection, essentially because $(\\lambda,A)$ gives all strings in $X_n$ starting with $2$, $(\\lambda,B)$ gives all strings in $X_n$ starting with $10$, $01$ or $12$, and $\\mu$ gives all strings starting with $02$. So $$\\left|(X_{n-1} \\times \\{A,B\\}) \\cup X_{n-2}\\right| = \\left|X_n\\right|,$$ which is $$2a_{n-1} + a_{n-2} = a_n.$$\n\n-\nI guess what @Xentius meant by a combinatorial proof is not this. He seems to be looking for an answer like in this question he asked: math.stackexchange.com/questions/340905/\u2026 \u2013\u00a0 Amadeus Bachmann Apr 8 '13 at 19:42\n@AmadeusBachmann exactly! \u2013\u00a0 Xentius Apr 8 '13 at 19:45\n@AmadeusBachmann: I think that azimut's equality has pure (and nice!) combinatorial sense. \u2013\u00a0 xen Apr 8 '13 at 19:55\nI'm a bit baffled why my argumentation shouldn't be combinatorial. @Xentius could you please specify what kind of proof you are looking for? \u2013\u00a0 azimut Apr 8 '13 at 20:02\n@xen I did not say azimut's proof is not combinatorial. I just wanted to say it may not be what Xentius looking for according to his previous questions. But of course only one to confirm this is Xentius. \u2013\u00a0 Amadeus Bachmann Apr 8 '13 at 20:06\n\nCall a ternary string with no consecutive zeroes and no consecutive ones a good string. For $k=0,1,2$ let $a_n^k$ be the number of good strings of length $n$ ending in $k$. Then it\u2019s clear that\n\n\\begin{align*} &a_n^0=a_{n-1}^1+a_{n-1}^2\\\\ &a_n^1=a_{n-1}^0+a_{n-1}^2\\\\ &a_n^2=a_{n-1}^0+a_{n-1}^1+a_{n-1}^2=a_{n-1}\\;. \\end{align*}\\tag{1}\n\nAnd now we can make the following computation:\n\n\\begin{align*} a_n&=a_n^0+a_n^1+a_n^2\\\\ &=a_n^0+a_n^1+a_{n-1}\\\\ &=a_{n-1}^1+a_{n-1}^0+2a_{n-1}^2+a_{n-1}\\\\ &=(a_{n-1}^0+a_{n-1}^1+a_{n-1}^2)+a_{n-1}^2+a_{n-1}\\\\ &=a_{n-1}+a_{n-1}^2+a_{n-1}\\\\ &=2a_{n-1}+a_{n-1}^2\\\\ &=2a_{n-1}+a_{n-2} \\end{align*}\n\nby the third line of $(1)$.\n\nIn more purely combinatorial terms the term $a_{n-1}$ in $a_n=a_n^0+a_n^1+a_{n-1}$ comes from the fact that each good string of length $n$ ending in $2$ is obtained by appending a $2$ to one of the $a_{n-1}$ good strings of length $n-1$. The rest is a little more complicated. If a good string of length $n-1$ ends in $0$ or $2$, I can append a $1$ to it to get a good string of length $n$ ending in $1$; that covers all good strings of length $n$ ending in $01$ or $21$, which is all good strings of length $n$ ending in $1$. Similarly, if a good string of length $n-1$ ends in $1$ or $2$, I can append a $0$ to get a good string of length $n$ ending in $0$, and that accounts for all good strings of length $n$ ending in $0$. Thus, to get the good strings of length $n$ ending in $0$ or $1$ I\u2019ve used each good string of length $n-1$ ending in $0$ once, each good string of length $n-1$ ending in $1$ once, and each good string of length $n-1$ ending in $2$ twice. In other words, I\u2019ve used each good string of length $n-1$ at least once, and I\u2019ve used those ending in $2$ twice. Ignoring for a moment the double use of those ending in $2$, that accounts for another $a_{n-1}$ strings of length $n$. Finally, we already know that each good string of length $n-1$ ending in $2$ is just the extension by a $2$ of a good string of length $n-2$, so the double use of good strings of length $n-1$ ending in $2$ gives us another $a_{n-2}$ good strings of length $n$. Put the pieces together, and we see that $a_n=2a_{n-1}+a_{n-2}$.\n\n-", "date": "2015-09-04 19:13:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/355151/combinatorial-explanation-to-following-recurrence-relation-a-n-2-a-n-1-a/355159", "openwebmath_score": 0.6358327269554138, "openwebmath_perplexity": 206.9099557598512, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9899864302631448, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.8510554404751222}}
{"url": "https://math.stackexchange.com/questions/2438737/formal-way-to-evaluate-this-limit", "text": "# Formal way to evaluate this limit\n\nI've some doubt about the formal way to evaluate the following limit:\n\n$$\\lim_{n\\to\\ +\\infty}{\\left(1-\\frac{1}{n}\\right)^{n^2}}$$\n\nI know that\n\n$$\\lim_{n\\to\\ +\\infty}{\\left(1-\\frac{1}{n}\\right)^{n}}=e^{-1}$$\n\nmy calculus teacher says that we can't evaluate the limit by pieces, so we can't (in general) say that\n\n$$\\lim_{n\\to\\ +\\infty}{\\left(1-\\frac{1}{n}\\right)^{n^2}}=\\lim_{n\\to\\ +\\infty}{\\left(e^{-1}\\right)^{n}}=0$$ even if in this case it is the right answer.\n\nSo I'm asking a formal way to solve this, my approach is:\n\nSay that the limit of a product is the product of the limits, so I can say that\n\n$$\\lim_{n\\to\\ +\\infty}{\\left(1-\\frac{1}{n}\\right)^{n^2}}=\\lim_{n\\to\\ +\\infty}{e^{n^2\\ln\\left(1-\\frac{1}{n}\\right)}}=e^{\\lim_{n\\to\\ +\\infty}{n^2\\ln\\left(1-\\frac{1}{n}\\right)}}=e^{\\lim_{n\\to\\ +\\infty}{n^2}{\\lim_{n\\to\\ +\\infty}\\ln\\left(1-\\frac{1}{n}\\right)}}=e^{\\lim_{n\\to\\ +\\infty}{-n}}=0$$\n\nCan I do this (I've used the continuity of the exponential function) or am I still doing the limit by pieces when I've substituted $ln\\left(1-\\frac{1}{n}\\right)$ with $-\\frac{1}{n}$?\n\nFurthermore, can I use the theorem of the product even if there is the indeterminate form $0\\cdot(+\\infty$)?\n\n\u2022 Implicitly, you used an equivalent near $0$: we know that $\\ln(1+u)\\sim_0 u$. All usual functions have equivalents near $0$. This way saves many useless computations unrelated to the indetermination. Sep 21, 2017 at 10:58\n\u2022 +1 for your question as well as your teacher. Unfortunately most book authors / instructors manage with lot of hand waving in teaching calculus. Sep 21, 2017 at 12:12\n\nNo, you can't. However, you may note that, as $n\\to +\\infty$, $$n^2\\ln\\left(1-\\frac{1}{n}\\right)=n\\ln\\left(\\left(1-\\frac{1}{n}\\right)^n\\right)\\to +\\infty \\cdot \\ln(e^{-1})=-\\infty.$$ Hence $$\\lim_{n\\to\\ +\\infty}{\\left(1-\\frac{1}{n}\\right)^{n^2}}=\\lim_{n\\to\\ +\\infty}{e^{n^2\\ln\\left(1-\\frac{1}{n}\\right)}}=e^{-\\infty}=0.$$ Another way. Since $\\lim_{n\\to\\ +\\infty}{\\left(1-\\frac{1}{n}\\right)^{n}}=e^{-1}\\in (0,1/2)$ then, by definition of limit, there is $N$ such that for all $n\\geq N$, $\\left(1-\\frac{1}{n}\\right)^{n}<1/2$. Hence, for $n\\geq N$, $$0<{\\left(1-\\frac{1}{n}\\right)^{n^2}}=\\left(\\left(1-\\frac{1}{n}\\right)^{n}\\right)^n<\\frac{1}{2^n}.$$ Then use the Squeeze theorem.\n\n\u2022 @Mephlip Any further doubt? Sep 21, 2017 at 11:48\n\nOne way of doing it is by comparison. For any fixed $k\\in \\Bbb N$, we have $$0\\leq \\lim_{n\\to \\infty}\\left(1-\\frac 1n\\right)^{n^2}\\leq\\lim_{n\\to \\infty}\\left(1-\\frac1n\\right)^{nk} = e^{-k}$$\n\nThe trick to do what you're trying to do is, after making your substitution, you also factor in a correction so that you leave the formula unchanged:\n\n$$\\lim_{n \\to \\infty} \\left(1 - \\frac{1}{n} \\right)^{n^2} = \\lim_{n \\to \\infty} \\left( e^{-1} \\cdot \\frac{\\left(1 - \\frac{1}{n}\\right)^{n}}{e^{-1}} \\right)^n = \\lim_{n \\to \\infty} e^{-n} \\cdot \\left( \\frac{\\left(1 - \\frac{1}{n}\\right)^{n}}{e^{-1}} \\right)^n \\\\= \\lim_{n \\to \\infty}\\left( e^{-n} \\right) \\cdot \\lim_{n \\to \\infty} \\left( \\frac{\\left(1 - \\frac{1}{n}\\right)^{n}}{e^{-1}} \\right)^n$$\n\nassuming, of course, the limits exist and the product is defined. If you could show that second limit was $1$, then you'd compute the limit as $\\infty \\cdot 1 = \\infty$ and be done.\n\nThe problem, however, is that the second limit is a $1^{\\infty}$ form! And it does this in a way that makes it so you can't immediately determine what the value should be: the value of the limit is a race between how fast the base approaches $1$ versus how fast the exponent reaches $\\infty$.\n\nSo, there isn't any immediate shortcut here; you have to do more work. In fact, that limit equals $e^{-1/2}$, so the 'obvious' guesses are, in fact, wrong!\n\nYou need a more sophisticated approximation to make something like this work. The thing to do can more easily be demonstrated in your alternate approach: the Taylor series for logarithm implies\n\n$$\\ln(1 + x) = x + O(x^2)$$\n\nSo, in particular,\n\n$$\\lim_{n \\to \\infty} n^2 \\ln\\left(1 - \\frac{1}{n} \\right) = \\lim_{n \\to \\infty} n^2 \\left( -\\frac{1}{n} + O(n^{-2}) \\right) = \\lim_{n \\to\\ infty} -n + O(1) = -\\infty$$\n\nIf you're not comfortable with $O$ notation, you could just use the next term of the Taylor series. If you put your mind to it, you should be able to obtain the exact value of the limit\n\n$$\\lim_{n \\to \\infty} n^2 \\left(\\ln\\left(1 - \\frac{1}{n} \\right) + \\frac{1}{n} \\right)$$\n\nso you could alternately compute the limit as\n\n$$\\lim_{n \\to \\infty} n^2 \\ln\\left(1 - \\frac{1}{n} \\right) = \\left( \\lim_{n \\to \\infty} n^2 \\left(-\\frac{1}{n}\\right)\\right) + \\left( \\lim_{n \\to \\infty} n^2 \\left( \\ln\\left(1 - \\frac{1}{n} \\right) + \\frac{1}{n} \\right) \\right)$$\n\nLate answer since it was tagged as duplicate.\n\nAn elementary way without any $$e$$ or logarithm could be:\n\n$$\\begin{eqnarray*}\\left(1-\\frac{1}{n}\\right)^{n^2} & = & \\frac 1{\\left(1+\\frac{1}{n-1}\\right)^{n^2}}\\\\ & \\stackrel{\\text{binomial formula}}{\\leq} & \\frac 1{\\frac{n^2}{n-1}}\\\\ & < & \\frac 1 n \\stackrel{n\\to \\infty}{\\longrightarrow} 0 \\end{eqnarray*}$$", "date": "2023-02-09 12:19:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2438737/formal-way-to-evaluate-this-limit", "openwebmath_score": 0.9486010670661926, "openwebmath_perplexity": 247.78401755109877, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769078156284, "lm_q2_score": 0.8723473862936942, "lm_q1q2_score": 0.8510419656614477}}
{"url": "https://math.stackexchange.com/questions/1721620/is-it-correct-to-say-all-extrema-happen-at-critical-points-but-not-all-critical", "text": "# Is it correct to say all extrema happen at critical points but not all critical points are extrema?\n\nIs it correct to say all extrema happen at critical points but not all critical points are extrema?\n\nThis question is for single variable calculus. But it would be great if someone could also provide insight to how well this statement generalizes to more complex functions (multi-variable functions, vector-valued functions, etc)\n\n\u2022 Extrema need not be critical points. They can also be the \"end-points\" in a given domain. This is what is called \"absolute extrema\". \u2013\u00a0Airdish Mar 31 '16 at 10:56\n\u2022 All interior extrema are critical points. Of course, critical points need not be extrema. \u2013\u00a0Sangchul Lee Mar 31 '16 at 10:58\n\nEvery local extremum in the interior of the domain of a differentiable function is neccesarily a critical point, i.e. $f'(x)=0$ is a necessary condition for $x$ to be a local extremum. There are critical points which are not local extrema.\n\nNote that I'm stressing local extremum in the interior here. To account for global extrema, one needs to consider the behavior on the boundary of the domain as well.\n\nConsider the function $$f:[-1,1]\\rightarrow \\mathbb R, x\\mapsto x^2.$$ $x_0=0$ is a critical point and indeed a local extremum (a minimum). For the global extrema note that $x^2 \\leq 1$ for $x\\in [-1,1]$ so that $f(1)=f(-1)=1$ and $x_1=1, x_2=-1$ are points where $f$ attains a (global) maximum, but they are not critical points.\n\nSimilarly, for $$f:[-1,1]\\rightarrow \\mathbb R, x\\mapsto x^3.$$ $x_0=0$ is a critical point, but there's no extremum. Maximum and Minimum are attained at $x_1=1$ and $x_2=-1$, neither of which are critical points.\n\n\u2022 When you say boundary points. You just mean boundary to domain (builtin or contrived, right?) What I mean is $2$ would be an inherent boundary point to $x/(x-2)$ and contrived ones would be $[-1,1]$. Sorry if this is not correct terminology. Or am I just thinking too hard? And the inherent ones are already covered by the $f'(c)=0$ or DNE clause? \u2013\u00a0AlanSTACK Apr 1 '16 at 1:13\n\u2022 @Alan: I mean boundary points of the domain where the function is defined. \u2013\u00a0Roland Apr 1 '16 at 5:23\n\nIf you assume $f$ is a continuous, real-valued function on a closed, bounded interval $[a, b]$, then:\n\n\u2022 $f$ has at least one absolute maximum and at least one absolute minimum. (The Extreme Value Theorem.)\n\n\u2022 If $f$ has a local (relative) extremum at an interior point $x_{0}$, then either $f'(x_{0}) = 0$ or $f'(x_{0})$ does not exist. (I.e., $x_{0}$ is a critical point of $f$, modulo your definition of \"critical point\".)\n\n\u2022 If your definition of \"critical point\" includes endpoints of $[a, b]$, then \"yes\", every local extremum of $f$ is a critical point of $f$. (This convention is reasonable, since strictly speaking $f'$ does not exist at an endpoint of an interval. On the other hand, this convention is not universal.)\n\nFor the other direction:\n\n\u2022 An interior critical point can fail to be a local extremum. Examples (with $x_{0} = 0$) include $f(x) = x^{3}$ and $$g(x) = \\begin{cases} x & x < 0, \\\\ 2x & x \\geq 0, \\end{cases}$$ both viewed as functions on some interval containing $0$ in the interior. (In the first example, $f'(0) = 0$; in the second, $g'(0)$ does not exist.)\n\n\u2022 An endpoint can fail to be a local extremum, e.g. $$f(x) = \\begin{cases} x^{2} \\sin(1/x) & 0 < x, \\\\ 0 & x = 0. \\end{cases}$$\n\nThere are analogous results for real-valued functions of more than one variable, but they're a little more work to state, because not every \"connected subset\" of the plane (say) is a formal analog of a closed, bounded interval. Loosely, though, a continuous function $f$ on a closed, bounded subset $D$ of $\\mathbf{R}^{n}$ has at least one absolute maximum and at least one absolute minimum in $D$, and these must occur either at (i) an interior point $x_{0}$ of $D$ where the total derivative $Df(x_{0})$ is zero; (ii) an interior point $x_{0}$ where $Df(x_{0})$ does not exist; or (iii) a boundary point of $D$.\n\nThere is no analog for vector-valued functions because the concepts of \"maximum\" and \"minimum\" make no sense. (If $n > 1$, the set $\\mathbf{R}^{n}$ is not ordered in any way compatible with vector addition and scalar multiplication.)\n\n\u2022 For the left endpoint $a$, the differential quotient $\\lim_{h \\rightarrow 0}\\frac{f(a+h)-f(a)}{h}$ is a limit which may (or may not) exist, and should be regarded as a one-sided limit (i.e. only for $h>0$). If $f'(a)$ exists or not is dependent on the existence of this one-sided limit. \u2013\u00a0Roland Mar 31 '16 at 11:38\n\u2022 @Roland: Agreed, but in my experience that criterion of differentiability at an endpoint is (also) a convention, and (with some justification) not a universal one. (If it matters, I'm not advocating either convention of endpoint differentiability, just noting that as stated, the OP's question can only be answered with qualifications.) \u2013\u00a0Andrew D. Hwang Mar 31 '16 at 11:55\n\u2022 yet you write 'strictly speaking $f\u2032$ does not exist at an endpoint of an interval', which is wrong for functions which are differentiable in a larger open domain which includes the interval $[a,b]$. \u2013\u00a0Roland Mar 31 '16 at 12:02\n\u2022 Strictly speaking, \"$f$ is differentiable at $x_{0}$\" involves a two-sided limit at $x_{0}$. If you extend the definition at an endpoint by requiring only a one-sided limit, that's fine, but you're no longer invoking the usual definition of differentiability from one-variable calculus. (If you're saying the extended definition is universal, I have no mathematical grounds to argue otherwise. I'm merely saying that this extension is not tacit in elementary calculus. For instance, $f(x) = |x|$ is not differentiable at $0$, but is differentiable on $[0, \\infty)$ by your definition.) \u2013\u00a0Andrew D. Hwang Mar 31 '16 at 12:36\n\u2022 @Roland It is explicitly stated in some real analysis textbooks (ex. Apostol) that for a derivative $f'(c)$ to exist, $f$ must be defined in an open interval containing $c$. Other authors, like Rudin, do not explicitly require this yet mention the issue of differentiability at endpoints in passing. The motivation for this requirement is far beyond the scope of introductory calculus. See the answer by Willie Wong here: math.stackexchange.com/questions/126176/\u2026 \u2013\u00a0MathematicsStudent1122 Apr 2 '16 at 5:07\n\nYes.\n\nYou can analyse this multi-variable example $z = x^2 - y^2$.\n\nThis has a critical point which is not an extrema. These points called saddle points.\n\n\u2022 What's your function? What's your domain? \u2013\u00a0Roland Mar 31 '16 at 11:40\n\u2022 @Roland $f: \\mathbb R^2 \\rightarrow \\mathbb R$ with $z = f(x,y) = x^2 - y^2$. \u2013\u00a0crbah Mar 31 '16 at 11:44\n\u2022 What if the domain is $\\{(x,y): x^2 +y^2 \\leq 1\\}$? Is the answer still yes? \u2013\u00a0Roland Mar 31 '16 at 11:49\n\u2022 @Roland Yes. You can see its graph on en.wikipedia.org/wiki/Saddle_point \u2013\u00a0crbah Mar 31 '16 at 11:53\n\u2022 My comment for the restricted domain is not about the critical point which is not an extremum. It's about the extremal points which are not critical points ($(x,y)=(1,0)$ and $(x,y)=(0,1)$). \u2013\u00a0Roland Mar 31 '16 at 12:00", "date": "2019-06-26 08:25:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1721620/is-it-correct-to-say-all-extrema-happen-at-critical-points-but-not-all-critical", "openwebmath_score": 0.8134347200393677, "openwebmath_perplexity": 292.5421390087865, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769106559808, "lm_q2_score": 0.8723473763375644, "lm_q1q2_score": 0.8510419584262514}}
{"url": "https://www.physicsforums.com/threads/finding-required-resistor-to-reduce-voltage.533682/", "text": "# Homework Help: Finding required resistor to reduce voltage\n\n1. Sep 25, 2011\n\n### deus_ex_86\n\n1. The problem statement, all variables and given/known data\nYour task is to design a \"dummy\" indicator light for your car. The cheapest bulb available is a #47 incandescent lamp rated at 6.3V for 150 mA. The problem is, your car produces 13.5 V when it's on. Choose the nearest 10% resistor that will reduce the voltage across the bulb to 6.3V.\n\n2. Relevant equations\nV = IR\n\n3. The attempt at a solution\n\nI modelled the bulb as a resistor:\n\n6.3 V = .150 A * R\nWhich gives R = 42 $\\Omega$.\n\nFrom there, I plugged this value into a new circuit with a 13.5 V source instead of a 6.3 V source, and added the unknown resistor to the equation:\n\n13.5 V = .150 A * R + 6.3 V\n7.2 V = .150 A * R\nR = 48 $\\Omega$.\n\nThe nearest 10% resistor value is 47 $\\Omega$, but this wouldn't reduce the voltage across the bulb to 6.3 V. The next closest 10% resistor is 56 $\\Omega$. It just seems that this value is too high ... Did I follow this process correctly? What would you choose?\n\n2. Sep 25, 2011\n\n### lewando\n\n48-ohms look right. What are you getting for the new bulb voltage (with the 47-ohm R)?\n\n3. Sep 25, 2011\n\n### deus_ex_86\n\n48 ohms is the theoretical value, but I have to choose a 10% resistor to use. When I use the 47 ohm resistor, I end up with 6.37 V across the bulb.\n\n4. Sep 25, 2011\n\n### lewando\n\nClose enough!\n\n[edit: To the letter of the question the solution does not exist. To the intent of the question, 47-ohms is a good answer, if you stipulate your voltage deviation, you should be fine.]\n\n5. Sep 27, 2011\n\n### zgozvrm\n\nWith a 47\u03a9 resistor, the bulb will have approximately 6.37 volts across it. Light bulbs are not very picky, so this will work fine.\n\nWith the 56\u03a9 resistor, the bulb will only have about 5.79 volts across it. Still, a workable voltage, but it's the current that we're really concerned with...\n\nThe 47\u03a9 resistor will limit the current through the bulb to about 151.69 mA which is just over the design rating. The bulb will burn slightly brighter (although probably not noticeably).\n\nThe 56\u03a9 resistor will limit the current down to about 137.76 mA and the bulb will burn slightly dimmer. You're only looking at about an 8% loss of current though, so either will work.\n\nI'd use the 47\u03a9 resistor.\n\n6. Sep 27, 2011\n\n### deus_ex_86\n\nThanks, so would my TA, as I got a 100% on the assignment. :)\n\n7. Sep 27, 2011\n\n### 2milehi\n\nWith \u00b110% resistors there could be a resistor that is 42.3\u03a9 and car battery voltage can spike into the 14.0 volt range. That would cause 166 ma to flow through the bulb or about 10.7% more. Since power is I\u00b2R, there can be an extra 23% power dissipated through the bulb which will shorten its life.\n\nIMO the design would have had some breathing room built in if you went with the 56\u03a9 bulb.", "date": "2018-08-20 11:05:53", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/finding-required-resistor-to-reduce-voltage.533682/", "openwebmath_score": 0.5311344265937805, "openwebmath_perplexity": 1673.3263010992364, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769056853638, "lm_q2_score": 0.8723473779969193, "lm_q1q2_score": 0.851041955708975}}
{"url": "http://yucoding.blogspot.com/2013/03/leetcode-question-102-sqrtx.html", "text": "### leetcode Question 102: Sqrt(x)\n\nSqrt(x)\n\nImplement\u00a0int sqrt(int x).\nCompute and return the square root of\u00a0x.\n\nAnalysis:\nAccording to Newton's Method(http://en.wikipedia.org/wiki/Newton's_method), we can use\n\nto get the sqrt(x).\n\nCode:\nclass Solution {\npublic:\nint sqrt(int x) {\n// Start typing your C/C++ solution below\n// DO NOT write int main() function\nif (x==0) {return 0;}\nif (x==1) {return 1;}\n\ndouble x0 = 1;\ndouble x1;\nwhile (true){\nx1 = (x0+ x/x0)/2;\nif (abs(x1-x0)<1){return x1;}\nx0=x1;\n}\n\n}\n};\n\n\nAnother solution: the binary search approach is a more general way of solving this problem. One thing you need to consider is the length of the input, since taking the mid of a big value and computing its square may overflow the int type.\nWe can use \"long long\" , which have a max value 2^63-1.\nThe max of an int is 2^15-1\nThe max of a long is 2^31-1\n\nclass Solution {\npublic:\nint sqrt(int x) {\n// Start typing your C/C++ solution below\n// DO NOT write int main() function\nlong long high = x;\nlong long low = 0;\nif (x<=0) {return 0;}\nif (x==1) {return 1;}\nwhile (high-low >1){\nlong long mid = low + (high-low)/2;\nif (mid*mid<=x){low = mid;}\nelse {high = mid;}\n}\nreturn low;\n}\n};\n\n\n1. I believe in the interview, they won't expect you to remember Newtown's method. The solution they are expecting is a Binary Search method.\n\nThe binary search approach is added to this post.\n\n2. This comment has been removed by the author.\n\n2. Quoting on one line above \"The max of an int is 2^15-1\", int in most cases is 32 bit, unless for some specific platform.\n\n3. Is there a reason that you use mid = low + (high-low)/2; rather than (high + low)/2; ?\n\n1. Yes, using high+low /2 may cause overflow errors, when high and low both are very big, high+low may exceed and cause the overflow error, while high-low will not.\n\n2. This comment has been removed by the author.\n\n3. Nice solution man. Although I think you need not worry about overflowing: both high and low are within the range [0 INT_MAX], so (high + low) is at most 2*INT_MAX, which can perfectly fit into a 'long long int' integer. So as long as you define high, low and mid as 'long long int', an overflow will never occur.\n\nOr better yet, even if both low and high are defined as 'int', we are still good. Note that it is guaranteed that (int)sqrt(x) < (int)(x/2) for any x >= 6, therefore, for any large enough number, in the first iteration, we have low == 0, so (low + high) is always <= INT_MAX (no overflow). then we always have mid*mid > x, so it is always 'high = mid;' that will be executed in the first iteration, i.e. high <= INT_MAX / 2. From there on, both low and high will be within [0 INT_MAX/2] so (low + high) will never exceed INT_MAX.\n\nOf course 'mid' must be defined as 'long long int' since we need to perform 'mid*mid', which may well exceed INT_MAX. But if we limit the initial value of 'high' to min(std::sqrt(INT_MAX) + 1, x/2), then it is safe to just define 'mid' as 'int'. std::sqrt(INT_MAX) is a constant that can be precomputed so it does not violate the requirement of this problem.\n\n4. should u cast long to int int the end?\n\n5. Decent improvement as (abs(x1-x0)<1) for Newton's Method.", "date": "2017-07-26 12:40:00", "meta": {"domain": "blogspot.com", "url": "http://yucoding.blogspot.com/2013/03/leetcode-question-102-sqrtx.html", "openwebmath_score": 0.4936031103134155, "openwebmath_perplexity": 3944.6968969118802, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576914916509, "lm_q2_score": 0.8723473697001441, "lm_q1q2_score": 0.851041955667598}}
{"url": "https://math.stackexchange.com/questions/537172/a-oplus-b-a-oplus-c-imply-b-c", "text": "$A \\oplus B = A \\oplus C$ imply $B = C$?\n\nI don't quite yet understand how $\\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use $\\oplus$.\n\nFor example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \\oplus B = A \\oplus C$ implies $B = C$ is true.\n\nI know that the venn diagram of $\\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:\n\nI feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\\oplus$, but I'm not sure if that's an adequate enough proof. On the other hand, I could be completely wrong about my reasoning.\n\nAlso just for clarity's sake: Would $A\\cup B = A \\cup C$ and $A \\cap B = A \\cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.\n\n\u2022 For still more fun show that the power set of a set $X$, together with the xor operation, is a group. And, of course, groups have the cancellation law you ask about. \u2013\u00a0GEdgar Oct 23 '13 at 17:41\n\u2022 The xor being, in that case, called the symmetric difference. @user101279 : By the way, $\\oplus$ is addition modulo $2$. \u2013\u00a0xavierm02 Oct 23 '13 at 18:04\n\u2022 Remark also that xor is the addition operation when we regard a Boolean algebra as essentially the same as a Boolean ring. \u2013\u00a0user43208 Oct 23 '13 at 18:05\n\u2022 Think of $\\oplus$ as $\\neq$. \u2013\u00a0copper.hat Oct 23 '13 at 18:06\n\u2022 If we omit the question about intersection and union in the last paragraph, this is the same as math.stackexchange.com/questions/294460/\u2026 \u2013\u00a0Martin Sleziak Aug 30 '14 at 15:30\n\nThink of $\\oplus$ as $\\neq$. That is $A \\oplus B$ iff $A \\neq B$.\n\nNote that $A \\oplus A$ is always false, and $\\text{False}\\oplus A = A$.\n\nThen $A \\oplus (A \\oplus B) = (A \\oplus A) \\oplus B = \\text{False} \\oplus B = B$.\n\nSimilarly, $A \\oplus (A \\oplus C) = C$, hence $B=C$.\n\nAside: A 'cute' (as in amusing but not of any practical significance) use of $\\oplus$ is to swap the values of two bit variables in a programming language without using an intermediate variable: \\begin{eqnarray} x = y \\oplus x \\\\ y = y \\oplus x \\\\ x = y \\oplus x \\\\ \\end{eqnarray} Show that the values of $x,y$ are swapped!\n\nHint: $A\\oplus(A\\oplus B)=(A\\oplus A)\\oplus B = B$.\n\nAnd of course $A\\cup B=A\\cup C$ does not imply $B=C$ (consider the case $B=A\\ne \\emptyset = C$). And $A\\cap B=A\\cap C$ does not imply $B=C$ either (consider the case $A=\\emptyset$)\n\nHint: $\\oplus$ is associative with unit $\\emptyset$, and $A \\oplus A = \\emptyset$. Does this give you an idea for canceling?\n\nThis can be done using a simple calculation. But I don't know how to interpret your question, so I'll give two answers. :-)\n\nI'm assuming $\\;A,B,C\\;$ are booleans. I will write $\\;\\not\\equiv\\;$ instead of $\\;\\oplus\\;$, and $\\;\\equiv\\;$ instead of $\\;=\\;$ on booleans.\n\nFirst, note that $\\;\\equiv\\;$ and $\\;\\not\\equiv\\;$ are not only both associative, but they are also mutually associative. Therefore no parentheses are needed in the following calculation.\n\nWe can now simplify $\\;A \\oplus B = A \\oplus C\\;$ as follows: \\begin{align} & A \\not\\equiv B \\equiv A \\not\\equiv C \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"rearrange\"} \\\\ & A \\not\\equiv A \\equiv B \\not\\equiv C \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"simplify\"} \\\\ & \\text{false} \\equiv B \\not\\equiv C \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"simplify\"} \\\\ & B \\equiv C \\\\ \\end{align}\n\nIf instead $\\;A,B,C\\;$ are sets, and your $\\;\\oplus\\;$ is the symmetric difference of two sets (which is normally written as $\\;\\triangle\\;$), then the proof is slightly longer, but with essentially the same structure.\n\nThe simplest definition of symmetric difference is $$x \\in A \\oplus B \\equiv x \\in A \\not\\equiv x \\in B$$\n\nWe can expand the definitions and simplify using logic, as follows: \\begin{align} & A \\oplus B = A \\oplus C \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"set extensionality; definition of $\\;\\oplus\\;$, twice\"} \\\\ & \\langle \\forall x :: x \\in A \\not\\equiv x \\in B \\equiv x \\in A \\not\\equiv x \\in C \\rangle \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"logic: rearrange\"} \\\\ & \\langle \\forall x :: x \\in A \\not\\equiv x \\in A \\equiv x \\in B \\not\\equiv x \\in C \\rangle \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"logic: simplify\"} \\\\ & \\langle \\forall x :: \\text{false} \\equiv x \\in B \\not\\equiv x \\in C \\rangle \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"logic: simplify\"} \\\\ & \\langle \\forall x :: x \\in B \\equiv x \\in C \\rangle \\\\ \\equiv & \\;\\;\\;\\;\\;\\text{\"set extensionality\"} \\\\ & B = C \\\\ \\end{align}\n\nIn both cases, we have found a stronger conclusion than was asked: we proved equivalence of the two expressions.\n\nAlso just for clarity's sake: Would $A\\cup B = A \\cup C$ and $A \\cap B = A \\cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.\n$A\\cup B=A\\cup C$ $\\Rightarrow$ $B=C$ is not true in general.\nCounterexample: Take any non-empty set $A$ and also take $B=A$ and $C=\\emptyset$. Then $A\\cup B=A\\cup C=A$, but $B\\ne C$.\n$A\\cap B=A\\cap C$ $\\Rightarrow$ $B=C$ is not true in general.\nTake some element $x\\notin A$ and put $B=A$, $C=A\\cup\\{x\\}$. Then $A\\cap B=A\\cap C=A$, but $B\\ne C$.", "date": "2019-08-24 13:05:28", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/537172/a-oplus-b-a-oplus-c-imply-b-c", "openwebmath_score": 0.9999955892562866, "openwebmath_perplexity": 136.9304749100381, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576910655981, "lm_q2_score": 0.8723473730188542, "lm_q1q2_score": 0.8510419551885945}}
{"url": "https://math.stackexchange.com/questions/3229985/biased-binary-search-complexity", "text": "# Biased binary search complexity\n\nWe know Binary search on a set of n element array performs O(log(n)).\n\nWe have this recursive equation through which the search space is reduced by half in each iteration, after a single comparison.\n\nT(n) = T(n/2) + 1\n\n\nEither by applying Master's Theorem or analytically we arrive at the complexity of it as log(n)\n\nIf I introduce bias, in the search by unequally partitioning the array instead of equal halves, How would the worst case time complexity change?\n\nThe unequal partitions are t * n and (1-t) * n where 0<=t<=1\n\nThis is a mathematics question as it involves asymptotic analysis and hence I don't wish my question to be downvoted.\n\n\u2022 If I understand what you're asking correctly, wouldn't the worst case complexity be $O(n)$ due to it potentially using a partition size of $1$ and $n - 1$, requiring on average $\\frac{n}{2}$ checks? Note it should never be worse than $O(n)$ as any type of biased binary search should only ever check each value at the most $1$ time. \u2013\u00a0John Omielan May 18 at 2:15\n\u2022 Correct. But I want to come across with an analytical way to calculate the worst case complexity for any particular value of t. In general, it is evident that the worst case complexities are O(n) when t=0 and O(log(n)) when t=1/2. \u2013\u00a0Argha Chakraborty May 18 at 6:04\n\u2022 There's nothing special about $t = \\frac{1}{2}$. As long as you reduce the problem size by a factor of $t$, for any $0 \\leq t < 1$, you will get logarithmic running time. \u2013\u00a0Joppy May 18 at 7:16\n\u2022 That is what I want to prove . \u2013\u00a0Argha Chakraborty May 18 at 7:33\n\nIn the worst case scenarios, where $$t = 0$$ or $$t = 1$$, and assuming that at least one value will always be checked, then as I stated in my comment, the average number of checks would be $$\\frac{n}{2}$$ and the maximum number would be $$n$$. However, for $$0 \\lt t \\lt n$$, where $$tn \\gg 1$$, then as indicated in the comment by Joppy, the number of steps in the worst case scenario would be logarithmic.\n\nTo see this, let\n\n$$u = \\max(t, 1-t) \\tag{1}\\label{eq1}$$\n\nThe worst case would be that the value is always in the larger partition after each step, with $$n_i$$ being the size of this partition after $$i$$ steps, and with $$n_0 = n$$. Thus,\n\n$$n_i = un_{i-1} \\; \\forall \\; i \\ge 1 \\tag{2}\\label{eq2}$$\n\nAs such, $$n_1 = un$$, $$n_2 = un_1 = u^2 n$$, etc., to get that\n\n$$n_i = u^i n \\; \\forall \\; i \\ge 0 \\tag{3}\\label{eq3}$$\n\nThe search will eventually end after $$m$$ steps where\n\n$$u^m n \\approx 1 \\tag{4}\\label{eq4}$$\n\nNote this is similar to the concept of the amount of time for exponential decay to cause a substance to eventually disappear, with the worst case number of unbiased binary searches being about the number of half-life periods. To use a value $$\\gt 1$$ for logarithms, let\n\n$$v = \\frac{1}{u} \\tag{5}\\label{eq5}$$\n\nso \\eqref{eq4} becomes\n\n$$n \\approx v^m \\; \\implies \\; m \\approx \\log_v(n) = \\log_v(e)\\ln(n) \\tag{6}\\label{eq6}$$\n\nFor $$t = u = \\frac{1}{2}$$, then $$v = 2$$ and \\eqref{eq6} gives $$m \\approx (1.44269)\\ln(n)$$. With a relatively extreme case of $$t = \\frac{1}{1001}$$, then $$u = \\frac{1000}{1001}$$ and $$v = 1.001$$, so \\eqref{eq6} gives $$m \\approx (1000.49991)\\ln(n)$$, i.e., about $$693.5$$ times as large. Nonetheless, it's still generally considerably faster than just checking each value since, for $$n = 10^{12}$$, \\eqref{eq6} gives $$m \\approx (1000.49991)(27.63102) \\approx 27644.8$$, so it's about $$\\frac{10^{12}}{27644.8} \\approx 3.61731 \\times 10^7$$ times faster.\n\nThe cases $$t=0$$ and $$t=1$$ do not correspond to a terminating algorithm. Otherwise, the behavior remains logarithmic.\n\nIn the worst case, the array is every time reduced according to the smaller of $$t$$ and $$1-t$$; WLOG, let $$t$$.\n\nAfter $$k$$ steps, we approximately reduce from $$n$$ to $$nt^k\\sim 1$$, so that $$k\\sim-\\log_tn$$.", "date": "2019-10-23 17:55:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3229985/biased-binary-search-complexity", "openwebmath_score": 0.8206067085266113, "openwebmath_perplexity": 269.13918222649215, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.975576913496333, "lm_q2_score": 0.8723473697001441, "lm_q1q2_score": 0.8510419544287112}}
{"url": "https://math.stackexchange.com/questions/1034665/how-to-put-a-matrix-in-jordan-canonical-form-when-it-has-a-multiple-eigenvalue", "text": "# How to put a matrix in Jordan canonical form, when it has a multiple eigenvalue?\n\nPut the matrix $$\\begin{bmatrix} 3 & -4\\\\ 1 & -1\\end{bmatrix}$$ in Jordan Canonical Form. Moreover, find the appropriate transition matrix to the basis in which the original matrix assumes its Jordan form.\n\nI'm having a lot of trouble with this. I know that the eigenvalue has multiplicity two and is $\\lambda = 1$. I can find the first eigenvector, which is:\n\n\\begin{bmatrix} 2 \\\\ 1 \\\\ \\end{bmatrix} I'm having trouble finding the second since both eigenvalues tell us the same thing. But I'm not nearly as concerned about the eigenvectors as I am about what to do after.\n\nIf anyone could explain thoroughly the next steps involved (not necessarily the answer but how to obtain it), I would be forever grateful. This homework is in 2 days and it may determine my grade letter.\n\n\u2022 Yes, I know the only eigenvalues are one. The problem is finding the eigenvectors corresponding to 1 and more importantly how to proceed. \u2013\u00a0Arbitrationer Nov 23 '14 at 5:43\n\u2022 You don't seem to understand that when vadim wrote, \"the eigenspace corresponding to $\\lambda=1$ is one-dimensional\", he was telling you there is only the one eigenvector that you have found (and scalar multiples of that eigenvector, but they don't help). What you need is a generalized eigenvector; a vector $w$ such that $(A-\\lambda I)w=v$, where $v$ is the eigenvector you have found. Do you not have a text, or some notes to follow? Anyway, \"generalized eigenvector\" is the keyphrase: go look it up, and then post an answer when you have worked out how to solve the problem. \u2013\u00a0Gerry Myerson Nov 23 '14 at 6:16\n\nTo put a matrix in Jordan normal form requires to know three matrices such that $A=PJP^{-1}$ i.e: a matrix $P^{-1}$ that transform to the canonical basis $(\\mathbf{i},\\mathbf{j},\\mathbf{k})$ to a new basis in which the matrix $J$ represents the transformation of a vector $\\mathbf{v}$ such that the transformed vector $\\mathbf{v'}$ is the same as we find when we transform $\\mathbf{v}$ with $A$ in the canonical basis. Last the matrix $P$ returns this result to the canonical basis.\n\nAs noted in OP the matrix $A$ has eigenvalues $\\lambda_1=\\lambda_2=1$ and a single eigenvector $$\\mathbf{u_1}=\\left[ \\begin{array}{cccc} 2\\\\ 1 \\end {array} \\right]$$\n\nSo the main problem is to find another vector that completes the new basis.\n\nTo find such a vector notes that all vectors $\\mathbf{x}$ such that $(A-\\lambda I)\\mathbf{x}=0$ are transformed in the eigenspace generated by the eigenvector $\\mathbf{u_1}$, so we want a vector $\\mathbf{u_2}$ such that $(A-\\lambda I)\\mathbf{u_2} \\ne 0$, and the way to do this is to find a vector such that : $(A-\\lambda I)\\mathbf{u_2} = \\mathbf{u_1}$. (Note that this equation is the same as $(A-\\lambda I)^2\\mathbf{u_2}= 0$).\n\nSolving in our case we find: $$\\left[ \\begin{array}{cccc} 2&-4\\\\ 1&-2 \\end {array} \\right] \\left[ \\begin{array}{cccc} x\\\\ y \\end {array} \\right]= \\left[ \\begin{array}{cccc} 2\\\\ 1 \\end {array} \\right]$$\n\nso that the components $x$ and $y$ of the searched vector must satisfies $x-2y=1$, and we can find the vector $$\\mathbf{u_2}= \\left[ \\begin{array}{cccc} 1\\\\ 0 \\end {array} \\right]$$ So the matrix $P$ we are searching is $$P=[\\mathbf{u_1},\\mathbf{u_2}]= \\left[ \\begin{array}{cccc} 2&1\\\\ 1&0 \\end {array} \\right]$$ And the inverse is: $$P^{-1}= \\left[ \\begin{array}{cccc} 0&1\\\\ 1&-2 \\end {array} \\right]$$ The matrix $J$ is a typical Jordan block, with the eigenvalues as diagonal elements and an entry $1$ up-right them: $$J= \\left[ \\begin{array}{cccc} 1&1\\\\ 0&1 \\end {array} \\right]$$ and we can easily verify that $A=PJP^{-1}$.\n\nThe characteristic polynomial $\\det (A - \\lambda I) = (\\lambda -1)^2.$ when the dimension of the null space(1) of an eigenvalue($\\lambda = 1$) is less than the algebraic multiplicity(2), you need to find generalized eigenvectors. in this instance, you need to solve $(A - I_2) \\left( \\begin{array}{l} x \\cr y\\end{array} \\right) = \\left( \\begin{array}{l} 2 \\cr 1 \\end{array} \\right).$ this gives you $x = 1, y = 0$\n\nwith respect to the basis $\\{ \\left( \\begin{array}{l} 2 \\cr 1 \\end{array} \\right), \\left( \\begin{array}{l} 1 \\cr 0 \\end{array} \\right) \\}$ you transformation is represented by the Jordan canonical form $\\left( \\begin{array}{ll} 1 & 1 \\cr 0 & 1 \\end{array} \\right).$\n\n\u2022 Might have been better to let OP do some of the work. \u2013\u00a0Gerry Myerson Nov 23 '14 at 23:27\n\u2022 @myerson, perhaps i should have not done all the work. \u2013\u00a0abel Nov 23 '14 at 23:57", "date": "2019-04-22 16:35:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1034665/how-to-put-a-matrix-in-jordan-canonical-form-when-it-has-a-multiple-eigenvalue", "openwebmath_score": 0.9571903944015503, "openwebmath_perplexity": 156.8381932586146, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9755769106559808, "lm_q2_score": 0.8723473630627235, "lm_q1q2_score": 0.8510419454756232}}
{"url": "https://math.stackexchange.com/questions/1696134/can-a-gradient-vector-field-with-no-equilibria-point-in-every-direction", "text": "# Can a gradient vector field with no equilibria point in every direction?\n\nSuppose that $V:\\mathbb{R}^n \\to \\mathbb{R}$ is a smooth function such that $\\nabla V : \\mathbb{R}^n \\to \\mathbb{R}^n$ has no equilibria (i.e. $\\forall x \\in \\mathbb{R}^n : \\nabla V (x) \\not = 0$). Under these hypotheses, is it possible that $\\nabla V (x)$ can point in every direction?\n\nTo be more precise, under the above hypotheses the map $$\\mathbb{R}^n \\to \\mathbb{S}^{n-1}$$ $$x \\mapsto \\frac{\\nabla V(x)}{\\|\\nabla V (x)\\|}$$ is well-defined. Is it impossible for such a map to be surjective? If not, what is a counterexample?\n\n\u2022 I believe this is possible from $R^2$ onwards. Can you write down a function for which one gradient flow line makes an $S$ shape? Can you make sure that at some distance from this $S$ the flowlines it straighten to vertical straight lines? If you can do that you can mirror this construction and obtain the right function. \u2013\u00a0Thomas Rot Mar 15 '16 at 14:18\n\u2022 That's an interesting idea. It seems plausible, but I'm having some trouble actually writing down such a function. \u2013\u00a0Matthew Kvalheim Mar 15 '16 at 16:55\n\u2022 The more I think about it, I'm not sure a function satisfying the requirements @ThomasRot suggested exists. \u2013\u00a0Matthew Kvalheim Mar 15 '16 at 23:57\n\u2022 A cool question! Also, it might be received well in MathOverflow, but given that you started a bounty I would think it better to let the bounty period expire before we consider migration. If you are in a hurry, respond to this and/or flag again. \u2013\u00a0Jyrki Lahtonen Mar 19 '16 at 6:53\n\nA quick solution for $n=2$, and an explanation of where it came from: $$\\nabla (e^x \\sin y) = (e^x \\sin y, e^x \\cos y).$$ The right hand side is never zero, but does assume every nonzero value in $\\mathbb{R}^2$.\n\nMotivation: If $f: \\mathbb{C} \\to \\mathbb{C}$ is holomorphic, then $\\nabla(\\mathrm{Re}(f)) = (\\mathrm{Re}(f'), -\\mathrm{Im}(f'))$. I looked for an $f'$ (namely $e^z$) which takes $\\mathbb{C}$ onto $\\mathbb{C}_{\\neq 0}$, and then integrated it to find $f$.\n\nInspired by this, let's try $$\\nabla(e^{x_0} \\cos(x_1^2+x_2^2+\\cdots + x_n^2)) =$$ $$e^{x_0} (\\cos(x_1^2+\\cdots + x_n^2), - x_1 \\sin(x_1^2+\\cdots + x_n^2), \\cdots, -x_n \\sin(x_1^2+\\cdots + x_n^2)).$$\n\nFirst, we note that the gradient is not zero. The first coordinate only vanishes if $x_1^2+ \\cdots + x_n^2$ is of the form $(2k+1) \\pi$. But, in this case, at least one of $x_1$, $x_2$, ..., $x_n$ are nonzero; say $x_j$. Then $- x_j \\sin(x_1^2+\\cdots + x_n^2) = \\pm x_j \\neq 0$.\n\nNow, let's take a nonzero vector $(v_0, \\ldots, v_n)$. We need to go through cases.\n\nIf $v_0 = 0$, choose $(x_1, \\ldots, x_n)$ proportional to $(v_1, \\ldots, v_n)$ and such that $\\sin(x_1^2+\\cdots+x_n^2)$ has the right sign.\n\nIf $v_1 = \\cdots = v_n = 0$ and $(-1)^k v_0>0$, take $x_1^2+\\cdots+x_n^2 = k \\pi$.\n\nIf neither of those cases holds, we'll take $(x_1, \\ldots, x_n)$ of the form $(t v_1, \\ldots, t v_n)$ for some $t$ to be determined soon. Set $s = x_1^2+ \\cdots + x_n^2$. Then our vector points in direction $\\pm (- \\cot(t^2 s), v_1, v_2, \\ldots, v_n)$. Since cotangent is surjective, we can choose $t$ such that $\\cot (t^2 s) = v_0$, and we can get the sign right.\n\n\u2022 Thank you for the insightful answer. Your counterexample for $\\mathbb{R}^2$ and the motivation you gave are quite helpful. \u2013\u00a0Matthew Kvalheim Mar 24 '16 at 22:31\n\nIt is possible to have a vector field without equilibrium point and takes every possible directions in $\\mathbb{R^n}$ for any $n \\ge 2$.\n\nThe proof for the 2-dimension case is given below. From that, vector fields for higher dimension cases can be constructed.\n\nLet $\\rho : (-\\epsilon,\\infty) \\to \\mathbb{R}$ be any $C^\\infty$ function such that\n\n1. $\\rho(t)$ is even over $(-\\epsilon,\\epsilon)$ and $\\rho(0) = \\rho'(0) = 0$.\n2. $|r\\rho'(r)| < 1$ for all $r \\ge 0$.\n3. for some $r_c > 0$, $\\rho(r_c) = 2\\pi$.\n4. for some $r_m > 0$, $N \\in \\mathbb{Z}$, $\\rho(r) = 2\\pi N$ for all $r \\ge r_m$.\n\nFor any point $(x,y) \\in \\mathbb{R}^2$, let $(r,\\theta)$ be the corresponding polar coordinates.\nLet $\\hat{e}_x, \\hat{e}_y, \\hat{e}_r, \\hat{e}_\\theta$ be unit vectors along the $x$, $y$, $r$ and $\\theta$ directions respectively.\n\nConsider following function $$U(x,y) = r\\cos(\\theta - \\rho(r)) = x\\cos\\rho(r) + y\\sin\\rho(r)$$ Its gradient is given by:\n\n\\begin{align} \\vec{\\nabla} U &= \\hat{e}_x \\left( \\cos\\rho + \\frac{x\\rho'}{r}( -x \\sin\\rho + y\\cos\\rho )\\right) + \\hat{e}_y \\left( \\sin\\rho + \\frac{y\\rho'}{r}( -x \\sin\\rho + y\\cos\\rho )\\right)\\\\ &= \\hat{e}_r\\left(\\cos(\\theta - \\rho) + r\\rho'\\sin(\\theta-\\rho)\\right) -\\hat{e}_\\theta\\left(\\sin(\\theta-\\rho)\\right) \\end{align} From these two expressions, it is easy to deduce\n\n\u2022 $\\vec{\\nabla} U$ is well defined at $(0,0)$ because $\\rho'(0) = 0$.\n\u2022 $\\vec{\\nabla} U(0,0) = \\hat{e}_x$ because $\\rho(0) = \\rho'(0) = 0$.\n\u2022 $\\vec{\\nabla} U \\ne 0$ for $r > 0$ because \\begin{align} |\\vec{\\nabla} U| &= |\\hat{e}_r\\left(\\cos(\\theta - \\rho) + r\\rho'\\sin(\\theta-\\rho)\\right) -\\hat{e}_\\theta\\left(\\sin(\\theta-\\rho)\\right)|\\\\ &\\ge |\\hat{e}_r\\left(\\cos(\\theta - \\rho)\\right) -\\hat{e}_\\theta\\left(\\sin(\\theta-\\rho)\\right)| - |\\hat{e}_e \\left(r\\rho'\\sin(\\theta-\\rho)\\right)|\\\\ &\\ge 1 - |r\\rho'| > 0 \\end{align}\n\nCombine these, we find $\\nabla U \\ne 0$ over $\\mathbb{R}^2$. This allow us to define following unit vector field globally on $\\mathbb{R}^2$.\n\n$$\\hat{u}(x,y) = \\frac{\\vec{\\nabla}U(x,y)}{\\left|\\vec{\\nabla}U(x,y)\\right|}$$\n\nOver the curve $\\displaystyle\\;\\gamma : [0,r_c] \\ni t \\quad\\mapsto\\quad (t\\cos\\rho(t),t\\sin\\rho(t)) \\in \\mathbb{R}^2\\;$, we have $$\\hat{u}(t) = \\hat{e}_r = \\hat{e}_x \\cos\\rho(t) + \\hat{e}_y\\sin\\rho(t)$$ As we move from $\\gamma(0)$ to $\\gamma(r_c)$ along $\\gamma$, $\\hat{u}(\\gamma(t))$ will cover the whole $S^1$ at least once. The map $\\hat{u}$ is surjective on $\\gamma$ and hence on $B(0,r_c)$ and $\\mathbb{R}^2$. This means $U$ is a counter-example of $V$ for the 2-dimenisonal case.\n\nAs an illustration, following is a picture of what $U(x,y)$ will typically look like under this construction.\n\n$\\hspace0.75in$\n\nThis particular $U(x,y)$ is computed using\n\n\\rho(t) = 2\\pi + (f(t)-2\\pi)g(t) \\quad\\text{ where }\\quad \\begin{align} f(t) &= \\log\\sqrt{(e^{4\\pi}-1)t^2 + 1}\\\\ g(t) &= \\begin{cases} 1, & t \\le 1\\\\ 1 - e^{-1/(e^{t-1} -1)}, & t > 1\\end{cases} \\end{align} and plotted over the region $|x|,|y| \\le 2$. This $\\rho(t)$ satisfies the first three requirement in the beginning of this answer with $r_c = 1$. For $r \\gg r_c$, $\\rho(t)$ tends to $2\\pi$ exponentially as $t \\to \\infty$.\n\nBack to the case of higher dimensions. Let's say we do have a $\\rho$ that satisfies all four conditions. Consider following function:\n\n$$V(x_1,x_2,\\ldots,x_n) \\stackrel{def}{=} U\\left( x_1, \\sqrt{x_2^2 + x_3^2 + \\cdots + x_n^2} - (r_m+1)\\right)$$\n\nSince $B(0,r_m) \\supset B(0,r_c)$, the map $\\frac{\\vec{\\nabla}V}{|\\vec{\\nabla}V|}$ will be surjective over the hyper-torus $$\\bigg\\{ (x_1,\\ldots,x_n) : x_1^2 + \\left(\\sqrt{x_2^2 + x_3^2 + \\cdots + x_n^2} - (r_m+1)\\right)^2 \\le r_m^2 \\bigg\\}$$ Since $U(x,y) = x$ whenever $x^2 + y^2 \\ge r_m^2$, $V(x_1,\\ldots,x_n)$ equals to $x_1$ over the complement of the hyper-torus. As that contains the $x_1$-axis, the square root in the definition of $V$ won't cause any problem. This makes $V$ smooth everywhere. Finally, it is easy to check $\\nabla V$ never vanishes.\n\nUpdate\n\nAbout the question how this particular form of $\\rho(t)$ is chosen.\n\nFirst, $\\rho(r)$ cannot change too rapidly or $\\nabla U$ may become zero somewhere.\nSecond, we need $\\rho(r)$ span across around a large enough interval to make $\\hat{u}(r)$ surjective.\nThis suggest us to choose a $\\rho(r)$ with $\\rho'(r)$ as close to $\\frac{1}{r}$ as possible. This means $$\\rho(r) \\sim \\log r + constant.$$ In order for $U(x,y)$ to be smooth at the origin, I replace $\\log r$ by $\\log\\sqrt{(e^{4\\pi}-1)r^2 + 1}$. The constants are chosen so that when $r$ changes from $0$ to $1$, $\\rho(r)$ pick up a change of $2\\pi$ which we need.\n\nFinally, we want $\\rho(r)$ decay back to a constant multiple of $2\\pi$ for large $r$ while keeping smooth all the way. We need this in order to extend the result to higher dimensions. This can be done using a smooth Partition of unity. The function $g(r)$ in the formula doesn't really have any specific meaning. It is simply something from trial and error which works.\n\n\u2022 May I ask how you came up with the particular example of $\\rho$ you gave? I'd like to understand the intuition behind your thought process. \u2013\u00a0Matthew Kvalheim Mar 21 '16 at 21:43\n\u2022 @MatthewKvalheim answer updated, see rationale of chosing such a $\\rho$. \u2013\u00a0achille hui Mar 21 '16 at 22:25\n\u2022 You wrote an expression for $\\nabla U$ that involves division by $r$. Am I correct in the interpretation that you actually mean that $\\nabla U$ is defined by the expression you wrote on $\\mathbb{R}^2\\setminus \\{0\\}$, and $\\hat{e}_x$ at the origin? \u2013\u00a0Matthew Kvalheim Mar 22 '16 at 3:49\n\u2022 Thank you very much for your help. \u2013\u00a0Matthew Kvalheim Mar 22 '16 at 3:58\n\u2022 @MatthewKvalheim $\\nabla U$ is defined as the gradient of $U$. What I mean is it \"equals to\" the expression involves division by $r$ for $r \\ne 0$ and $\\hat{e}_x$. \u2013\u00a0achille hui Mar 22 '16 at 9:15", "date": "2020-08-15 02:49:53", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1696134/can-a-gradient-vector-field-with-no-equilibria-point-in-every-direction", "openwebmath_score": 0.9861288070678711, "openwebmath_perplexity": 282.0459992361244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446486833801, "lm_q2_score": 0.8740772351648677, "lm_q1q2_score": 0.8510406225542377}}
{"url": "https://math.stackexchange.com/questions/3060263/rationalizing-denominator-of-frac18-sqrt162-cannot-match-textbook-solu", "text": "# Rationalizing denominator of $\\frac{18}{\\sqrt{162}}$. Cannot match textbook solution\n\nI am given this expression and asked to simplify by rationalizing the denominator:\n\n$$\\frac{18}{\\sqrt{162}}$$\n\nThe solution is provided:\n\n$$\\sqrt{2}$$\n\nI arrived at:\n\n$$\\frac{\\sqrt{162}}{9}$$\n\nHere is my thought process to arrive at this incorrect answer:\n\n$$\\frac{18}{\\sqrt{162}}$$\n\n= $$\\frac{18}{\\sqrt{162}}$$ * $$\\frac{\\sqrt{162}}{\\sqrt{162}}$$\n\n= $$\\frac{18\\sqrt{162}}{162}$$\n\n= $$\\frac{\\sqrt{162}}{9}$$\n\nHow can I arrive at $$\\sqrt{2}$$ ?\n\n\u2022 Hint: $162=2\\cdot 81$. Jan 3 '19 at 4:52\n\u2022 $162=2*81=2*9^2$ so $\\sqrt {162}=\\sqrt {2*9^2}=9\\sqrt 2$. If your hadn't \"deradicalized\" the denominator you would have ended up with $\\frac 2 {\\sqrt 2}$ which is also deradicalized as $\\sqrt 2$. Jan 3 '19 at 5:49\n\n$$\\frac{\\sqrt{162}}{9} = \\frac{\\sqrt{2 \\cdot 9^2}}{9} = \\frac{9\\sqrt{2}}{9} = \\sqrt{2}$$.\n\n$$\\require{cancel}\\frac{18}{\\sqrt{162}}=\\frac{2\\cdot3^2}{\\sqrt{2\\cdot3^4}}=\\frac{2\\cdot\\cancel{3^2}}{\\sqrt{2}\\cdot\\cancel{3^2}}=\\frac{2}{\\sqrt{2}}\\cdot\\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\cancel2\\sqrt{2}}{\\cancel{2}}=\\sqrt{2}$$\n\nYour thought process is good. But just continue with factorizing $$162=2*81=2*3^4$$.\n\nSo $$\\sqrt {162}=\\sqrt {2*3^4}=\\sqrt {2}\\sqrt {3^4}=\\sqrt 2*3^2=9\\sqrt 2$$ and from there.... it's just mechanics.\n\n$$\\sqrt{162}$$ needs to be simplified further, as $$162$$ is the product containing a perfect square (i.e. $$81$$). Thus $$\\sqrt{162} = \\sqrt{2 \\cdot 81} = \\sqrt{2}\\sqrt{81} = 9\\sqrt{2}$$\n\nand hence $$\\frac{\\sqrt{162}}{9} = \\frac{9\\sqrt{2}}{9} = \\sqrt{2}$$\n\nAlternatively if $$a = \\frac{18}{\\sqrt{162}}$$, $$a^2 = \\frac{18^2}{\\sqrt{162}^2} = \\frac{324}{162} = 2$$. Since $$a$$ is a positive number, $$a = \\sqrt{2}$$.", "date": "2021-09-29 00:59:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3060263/rationalizing-denominator-of-frac18-sqrt162-cannot-match-textbook-solu", "openwebmath_score": 0.9871659278869629, "openwebmath_perplexity": 521.4923752147245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9736446498305048, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8510406123786919}}
{"url": "http://math.stackexchange.com/questions/226970/how-do-you-solve-an-equation-like-this", "text": "# How do you solve an equation like this?\n\nWe have an equation:\n\n$\\dfrac{p-1}{q} = \\dfrac{q-1}{2p+1} = \\dfrac {3}{5}$\n\nHow do you solve these types of equations? For example, if we have:\n\n$\\dfrac{1}{x} = \\dfrac{3}{2}$, we use:\n\n$1\\times 2 = 3\\times x$\n\n$x = 1.5$\n\nWhat is a similair approach to my equation?\n\n-\nYes, exactly. First forget about the middle term. Next, forget about the $\\frac35$. And then you're done. \u2013\u00a0 Berci Nov 1 '12 at 19:29\n\nEssentially, you can solve this using the process you describe, but twice, to generate two equations in two variables, and then solving for each variable as a \"system of equations\".\n\nWe have:\n\n$$\\dfrac{p-1}{q} = \\dfrac{q-1}{2p+1} = \\dfrac {3}{5}$$\n\nEquation 1: $\\quad 5(p-1) = 3q \\iff 5p - 3q = 5$\n\nEquation 2: $\\quad 5(q-1) = 3(2p +1) \\iff -6p +5q=8$\n\nSo your system of two equations in two unknowns becomes:\n\n$$5p - 3q = 5\\tag{1}$$ $$-6p +5q=8\\tag{2}$$\n\nCan you take it from there?\n\nYou can express (1) as a function of p (isolate p), and then substitute the expression obtained for p, into p in (2), and then solve for q, then p,\n\nor\n\nYou can use \"row operations\": multiply (1) by 5 (both sides), and (2) by 3 (both sides):\n\n$$25p-15q=25\\tag{1}$$ $$-18p+15q = 24\\tag{2}$$\n\nNow add the equations (q disappears), solve for p, then \"plug\" p into one of the original equations and solve for q:\n\n$$7p = 49n\\implies p = 7$$\n\nNow...From (1), originally, above\n\n$$5p-3q=5 \\implies 5(7) - 3q = 5\\implies 35 - 3q = 5$$ $$\\implies -3q=-30 \\implies q = 10$$\n\n-\nThank you sir, I am familair with row operations so I won't have trouble with these types of equations again. \u2013\u00a0 JohnPhteven Nov 1 '12 at 19:49\n(+1) Step by step getting the answer. \u2013\u00a0 Babak S. Aug 6 '13 at 10:15\n\nYou have two equation with two variables: $$1) \\quad \\frac{p-1}{q} = \\frac{3}{5}$$ and $$2) \\frac{q-1}{2p+1}=\\frac{3}{5}$$\n\n$1)$ Implies that $5(p-1)= 3q$, which gives $q= \\frac{5(p-1)}{3}$. Then you can insert $q= \\frac{5(p-1)}{3}$ in $(2)$. Then find what is $p$, and then you can find what $q$ is.\n\n-\n\n$$\\frac{p-1}{q} = \\frac{q-1}{2p+1} = \\frac {3}{5}$$ so $$\\frac{p-1}{q}=\\frac {3}{5}$$ and $$\\frac {3}{5}=\\frac{q-1}{2p+1}.$$ \\begin{eqnarray} 5p-5&=3q&\\\\ 6p+3&=&5q-5 \\end{eqnarray} $$5p-3q=5$$ $$6p-5q=-8$$ $$30p-18q=30$$ $$30p-25q=-40$$ $$-7q=-70$$ $$q=10$$ $$p=7.$$\n\n-\n$p = 7$? :) ${}{}$ \u2013\u00a0 Alex Nov 1 '12 at 19:37\nis there a problem ? \u2013\u00a0 Iuli Nov 1 '12 at 19:47\n\nFrom $$\\dfrac{p-1}{q} = \\dfrac{q-1}{2p+1} = \\dfrac {3}{5}$$ follow the system of linear equations with two unknowns\n\n$$5(p-1)=3q$$ $$5(q-1)=3(2p+1)$$ or $$5p-3q=5$$ $$6p-5q=-8$$\n\nwich can be solved using for example Cramer rule\n\n$\\Delta=-25+18=-7$\n\n$\\Delta_p=-25-24=-49$\n\n$\\Delta_q=-40-30=-70$\n\n$$p=\\frac{\\Delta_p}{\\Delta}=\\frac{-49}{-7}=7$$,$$q=\\frac{\\Delta_q}{\\Delta}=\\frac{-70}{-7}=10$$\n\n-", "date": "2015-01-29 14:56:22", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/226970/how-do-you-solve-an-equation-like-this", "openwebmath_score": 0.8843836188316345, "openwebmath_perplexity": 461.809007154515, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969679646668, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8510317927785007}}
{"url": "https://math.stackexchange.com/questions/20223/getting-an-x-for-chinese-remainder-theorem-crt/20259", "text": "# Getting an X for Chinese Remainder Theorem (CRT)\n\nhow do I get modulo equations to satisfy a given X in CRT.\n\nFor example say I have X = 1234. I choose mi as 5, 7, 11, 13. This satisfies the simple requirements of Mignotte's threshold secret sharing scheme. More precisely given in my example k = n = 4, and the product of any k - 1 is smaller then X how come simply computing the remainder of each won't give equations that solve to X = 1234.\n\nIn the case of the example,\n\nx = 4 mod 5\nx = 2 mod 7\nx = 2 mod 11\nx = 12 mod 13\n\nWhich resolves to 31264 (won't CRT produce the smallest?)\n\nAny hints?\n\nThe final result of the CRT calculation must be reduced modulo 5 x 7 x 11 x 13 = 5005. This gives the correct answer.\n\nHere is a much simpler way to immediately obtain the sought answer. Contrast the solution below to the much longer solution in your link, which involves calculations with much larger numbers and performs $4$ inversions vs. the single simple inversion below. Always search for hidden innate structure in a problem before diving head-first into brute-force mechanical calculations!\n\nThe key insight is: the congruences split into pairs with obvious constant solutions by CCRT, viz.\n\n\\begin{align}\\rm\\quad\\quad\\quad\\quad\\quad x\\equiv \\ \\ \\ 2\\ \\ \\:(mod\\ 7),\\ \\ x\\equiv \\ \\ \\ 2\\ \\ \\:(mod\\ 11)\\ \\iff\\ x\\equiv \\ \\ \\ \\color{#0a0}2\\ \\ (mod\\ \\color{#0a0}{77})\\\\[0.3em] \\rm\\quad\\quad\\quad\\quad\\quad x\\equiv -1\\ \\ (mod\\ 5),\\,\\ \\ x\\equiv\\ {-}1\\ \\ (mod\\ 13)\\ \\iff\\ x\\equiv \\color{#c00}{-1}\\ \\ (mod\\ \\color{#c00}{65})\\end{align}\n\nSo we reduced the above four original LHS equations to the above two RHS equations, which are easy to solve by CRT = Chinese Remainder Theorem.  Indeed, applying Easy CRT below\n\n$\\rm\\quad\\quad\\quad\\quad\\quad x\\equiv\\ \\color{#0a0}{2 + 77}\\ \\bigg[\\displaystyle\\frac{\\color{#c00}{-1}-\\color{#0a0}2}{\\color{#0a0}{77}}\\ mod\\,\\, \\color{#c00}{65}\\bigg]\\,\\ \\ (mod\\ 77\\cdot65)$\n\nIn the brackets $\\,\\rm\\displaystyle\\left[\\, mod\\ \\ 65\\!:\\ \\ \\frac{-3}{77} \\equiv \\frac{-3}{12} \\equiv \\frac{-1}4 \\equiv \\frac{64}4 \\equiv \\color{#d0f}{16}\\,\\right]\\quad$ (see Beware below)\n\nThis yields $\\rm\\ \\ x\\ \\equiv\\ \\color{#0a0}{2 + 77}\\,[\\,\\color{#d0f}{16}\\,] \\equiv 1234\\,\\ \\ (mod\\ 77\\cdot 65)\\quad$ QED\n\nTheorem $\\:$ (Easy CRT) $\\rm\\ \\$ If $\\rm\\ m,\\:n\\:$ are coprime integers then $\\rm\\ m^{-1}\\$ exists $\\rm\\ (mod\\ n)\\ \\$ and\n\n$\\rm\\displaystyle\\qquad\\quad\\quad\\quad \\begin{eqnarray}\\rm x&\\equiv&\\!\\rm\\ a\\ \\ (mod\\ m) \\\\ \\rm x&\\equiv&\\!\\rm\\ b\\ \\ (mod\\ n)\\end{eqnarray} \\ \\iff\\ \\ x \\equiv\\, a + m\\ \\bigg[\\frac{b-a}{m}\\ mod\\ n\\,\\bigg]\\,\\ \\ (mod\\ m\\:n)$\n\nProof $\\rm\\ (\\Leftarrow)\\ \\ \\ mod\\ m:\\,\\ x \\equiv a + m\\ [\\,\\cdots\\,] \\equiv a,\\$ and $\\rm\\ mod\\ n\\!\\!:\\,\\ x \\equiv a + (b\\!-\\!a)\\ m/m \\equiv b$\n\n$\\rm (\\Rightarrow)\\ \\$ The solution is unique $\\rm\\ (mod\\,\\ mn)\\$ since if $\\rm\\ x',\\:x\\$ are solutions then $\\rm\\ x'\\equiv x\\$ mod $\\rm\\:m,n\\:$ therefore $\\rm\\ m,n\\ |\\ x'-x\\ \\Rightarrow\\ mn\\ |\\ x'-x\\ \\$ since $\\rm\\ \\:m,n\\:$ coprime $\\rm\\:\\Rightarrow\\ lcm(m,n) = mn\\ \\ \\$ QED\n\nNote $\\$ Easy CRT is not only easy to apply, but also very easy to remember. Namely note $\\rm\\ x\\equiv a\\pmod{\\! m}\\iff x = a + m\\,k,\\:$ for some integer $\\rm\\:k,\\,$ This further satisfies the second congruence iff $\\rm\\ mod\\ n\\!:\\ x = a + m\\,k\\equiv b$ $\\iff$ $\\rm k\\:\\equiv (b-a)/m,\\$ hence the Easy CRT formula. This explains the $(\\Leftarrow)$ proof:  fill in the dots in $\\rm\\:x\\equiv a + m\\ [\\,\\cdots\\,]\\:$ to make $\\rm\\,x\\equiv b\\pmod{\\! n}$\n\nBeware $\\$ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.\n\nBelow is the solution you linked to on \"Math Celebrity\" (cached to avoid link rot).\n\n\u2022 @Bill Dubuque: But the OP was just giving an example! Your solution is completely useless for any other, similar problem. Unless you can automate it, of course :-) \u2013\u00a0TonyK Feb 3 '11 at 17:27\n\u2022 @TonyK: Most certainly not true. Most math problems do have interesting structure, esp. problems that are designed for tests, competitions etc. In fact this is frequently true even in research problems. Indeed, speaking as a number theorist, I can tell you that methods like the above are very useful in practice. In mathematics, intuition always trumps brute force. \u2013\u00a0Bill Dubuque Feb 3 '11 at 17:34\n\u2022 @Bill Dubuque: Well, your methods might be much, much simpler for you. But if they require years of practice, they're not so useful for people like me (or, presumably, the OP). \u2013\u00a0TonyK Feb 3 '11 at 18:05\n\u2022 @Tonyk: The above solution can be understood by a bright high-school student. Simple optimizations like the above arise frequently in elementary number theoretical problems, so it is well-worth knowing them. \u2013\u00a0Bill Dubuque Feb 3 '11 at 18:54\n\u2022 @TonyK: The optimization I employed above is algorithmic and is frequently applicable in practice. But that was not my point in presenting the above. Rather, it was to emphasize conceptual vs. algorithmic thought - intuition vs. brute force. Mathematical problems are far from random. Typically they involve much structure and their solution requires insightful exploitation of such innate structure - something that cannot be algorithmic (indeed many math problems are algorithmically unsolvable - that's what makes them interesting). \u2013\u00a0Bill Dubuque Feb 4 '11 at 0:37", "date": "2019-11-20 14:05:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/20223/getting-an-x-for-chinese-remainder-theorem-crt/20259", "openwebmath_score": 0.8776496648788452, "openwebmath_perplexity": 697.5250583298947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969708496456, "lm_q2_score": 0.8652240877899776, "lm_q1q2_score": 0.8510317918563698}}
{"url": "http://kvmb.imltraining.de/weighted-mean.html", "text": "# Weighted Mean\n\nVariance is also based on the mean of your data. 0 (2016-2017), but may apply to other versions as well. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. , Features, Capabilities, and Epics) to produce maximum economic benefit. Here is the mean of 1, 2, 3 and 4: Add up the numbers, divide by how many numbers: Mean = 1 + 2 + 3 + 44 = 104 = 2. weighted mean shift free download. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Abstract It is well known that the out-of-sample performance of Markowitz\u2019s mean-variance portfolio criterion can be negatively a\ufb00ected by estima- tion errors in the mean and covariance. using excel to calculate the weighted standard deviation Does anyone know the formula for computing the weighted standard deviation? I was able to calculate the weighted average (16. You simply add up all the item values and divide by the total. This calculator will calculate the weighted average APR for all of your credit cards that have a current balance. Basically I calculate the most recent RSI value and it's exponentially weighted mean based on the window length selected. Water-Quality Characteristics. I am trying to calculate the mean profit percentage, but a free item is given when a certain amount of goods are bought. geometric mean concentration at which shellfish beds or swimming beaches must be closed. Weighted means are often used for frequency data. Weighted blankets, sometimes referred to as gravity blankets, were once a tool of therapists and psychiatry clinics. Purpose: To report design of a simplified external transmit-receive coil array for 7 Tesla (T) prostate MRI, including demonstration of the array for tumor localization using T2-weighted imaging (T2WI) at 7T before prostatectomy. Adding all of these totals up yields a weighted average of 5. Ranking, Matrix/Rating Scale, Multiple Choice, Multiple Textboxes, and Slider questions calculate an average or weighted average. How to calculate weighted mean in IBM SPSS? Using the results of the survey I need to create the rating of governmental organizations. A method of computing a kind of arithmetic mean of a set of numbers in which some elements of the set carry more importance (weight) than others. org Dictionary. Acronym/Abbreviation meaning of CAC The acronym CAC() means : Capital Access Corporation. test \u2013 here they are (the weighted variance is from Gavin Simpson, found on the R malining list): View Code RSPLUS. Compute the weighted mean of a variable. Suppose your teacher says, \"The test counts twice as much as the quiz and the final exam counts three times as much as the quiz\". Dec 10, 2010 \u00b7 The IRR, also commonly referred to as the dollar weighted return, is the measurement of a portfolio\u2019s actual performance between two dates, including the effects from all cash inflows and outflows. Recommended Articles. \u2026 If all the weights are equal, then the weighted mean is the same as the arithmetic mean. Finally, the query uses a GROUP BY clause to combine the data so that the calculation is performed for each gender. WMC is sampling-, scale-, and contrast-invariant, and is sparse on natural images. A weighted average, for example, takes into account the proportional. The AVERAGE function below calculates the normal average of three scores. Weighted Vest FAQs Weighted vests are often recommended for children with autism. 04 (df = 3, p <. It is based on kernel weighted sample statistics such as the mean (Nadaraya-Watson estimator) but also standard deviation, skewness, kurtosis, deciles, etc. Follow the example below to calculate the weighted average interest rate for a federal loan consolidation. Find the weighted average of class grades (with equal weight) 70,70,80,80,80,90: Since the weight of all grades are equal, we can calculate these grades with simple average or we can cound how many times each grade apear and use weighted average. mean, or in other words searching the explicit expression of the weighted mean standard deviation distribution, under the safely motivated restriction of independent measures obeying Gaussian distributions. also i thought weighted mean measures the tendency of respondents answers to various questions. It is an average in which each quantity to be averaged is assigned a. Under Medicare Advantage, the federal government still foots the. weighted dataset, others simply recommend replacing the weighted base sizes with the unweighted base sizes in the test statistic formula, while still using the weighted summary statistics such as mean and standard deviation. Most companies weight the revenue based on the opportunity success %. The choice of Galloway et al. This is because the basic average of a group of numbers is the same calculation as a weighted average except that the weights of all the numbers are calculated as being the same. In calculating a weighted average, each number in the data set is. Weighted Mean (ratio statistics algorithms) A / S = n \u03a3 i = 1 f i A i n \u03a3 i = 1 f i S i = n \u03a3 i = 1 f i S i R i n \u03a3 i = 1 f i S i This is the weighted mean of the ratios weighted by the sales prices in addition to the usual case weights. Free Arithmetic Mean (Average) Calculator - find the average of a data set step-by-step. Another Tip: If number of boys and girls is same, i. The first or shorter echo (TE < 30msec) is proton density (PD) weighted or a mixture of T1 and T2. then the variance of the mean could be estimated with. The idea of weighted mean plays a role in descriptive statistics and also occurs in a more general form in several other areas of mathematics. Given N jobs where every job is represented by following three elements of it. weighted-mean definition: Noun (plural weighted means) 1. Define weighted. \u2026 So let's set up the ability to do a weighted mean \u2026 versus a mean. To calculate a weighted average in Excel, simply use the SUMPRODUCT and the SUM function. In this article, I'll go through a basic description of what a weighted GPA is, why it matters for you, and how you can calculate your own weighted GPA if your school uses this type of scale. a point lying far from the rest). The weighted arithmetic mean (or weighted average) is used if one wants to combine average values from samples of the same population with different sample sizes: \u00af = \u2211 = \u2211 =. The WAM can be used to gain an average of an applicant results using their grades and unit credit value. Get Your Weighted And Unweighted GPA In Just A Few Minutes! This quick and easy online GPA calculator computes both weighted and unweighted high school grade point averages (GPA). However, Property Use Details can change over the course of the 12 months, so we time-weight them so they are appropriately attributed. It is challenging to achieve high weighted ef\ufb01ciency with low-power microinverters, typically because these devices are required to be low cost. Variance[wd] / n where n is sample size. Weighted arithmetic mean is used when the values are of different importance - weight p to be assigned to each value. More periods with zero is less scale sensitive than the MAPE and the MAD. Specify the plan goal weights for each of the three goal levels (the total must be 100 percent). POSIXct, colMeans for row and column means. Definition of weighted mean in the Definitions. The tutorial is mainly based on the weighted. Get Your Weighted And Unweighted GPA In Just A Few Minutes! This quick and easy online GPA calculator computes both weighted and unweighted high school grade point averages (GPA). It often occurs, however, that one must combine two or more measurements of the same quantity with differing errors. Title: Variance of a Weighted Mean Created Date: 20160808202903Z. Use 'weighted average' in a Sentence. A combination of medication, lifestyle changes and other treatments can help limit the impact of anxiety and its symptoms. \u00b5 = P n Pi=1 w ix i n i=1 w i (45) It is equivalent to the simple mean when all the weights are equal, since \u00b5 = P n P i=1 wx i n i=1 w = w P n x i nw = 1 n X i=1 x i (46) If the samples are all di\ufb00erent, then weights can be thought of as sample frequencies, or they can be used to calculate probabilities where p i = w i/ P w i. mean() while aggregating a data frame. \u2026Let's consider an academic course as our example. Dec 30, 2016 \u00b7 Weighted Average is a free application for students. 04 (df = 3, p <. As a result, practitioners usually rely on vendor specified parameters, such as mean-time-to-failure (MTTF), to model failure processes, although many are skeptical of the accuracy of those models [4,5,33]. 5 for Stock C, or 3. The first section which takes up columns A, B, and C is the weighted averages. Select a stem that prompts several possible responses, with one that is the most accurate. For example, say 239 students in Economics 201. Jul 24, 2014 \u00b7 Functional diversity can be quantified using one single trait at a time or multiple traits (see the descriptions of multiple-indices in other page). The weighted mean is similar to an arithmetic mean (the most common type of average), where instead of each of the data points contributing equally to the final average, some data points contribute more than others. A weighted grade or score is average of a set of grades, where each grade (g) carries a different weight (w) of importance. an object containing the values whose weighted mean is to be computed. w: a numerical vector of weights the same length as x giving the weights to use for elements of x arguments to be passed to or from methods. Examples of how to use \u201cweighted mean\u201d in a sentence from the Cambridge Dictionary Labs. Instead of each data point contributing equally to the final mean, some data points contribute more \"weight\" than others. In simple terms, the category \"total\" will be equal to the sum of the scores in each grade item each multiplied by its grade weight, and that sum being finally divided by the sum of all weights. On the face of it, this does look as nice. RE: Geometric and weighted mean Perhaps it is worth pointing out that *an* average is a relatively arbitrary way of \"summarising\" a bunch of numbers into a single number. Sep 08, 2015 \u00b7 How do I change my gradebook aggregation to \u2018Weighted Mean of Grades\u2019? By default, the Moodle gradebook aggregation (the way it adds your grade values) is \u2018Simple Weighted Mean of Grades. When an assignment is \u201cweighted\u201d, it means there is a grade for that assignment. Catetan y\u00e9n lamun kab\u00e9h beuratna sarua, weighted m\u00e9an sarua jeung arithmetic mean. For example, if the chest is the target area, we might perform a set of barbell bench press, followed by a set of barbell incline press, and finished with a set of weighted parallel dips. A flow-weighted mean is the mean of a quantity after it is weighted proportional to a corresponding flow rate. The fact that x only gives one radiation field is the same as giving 0 radiation for the second field and so on, which means I do the following:. The mean of each area will be obtained using the formula: x = \u00ce\u00a3x/N (Downie and Heat, 1970) The numerical findings of the study will be statistically analysed and interpreted using the frequency count. rm: a logical value indicating whether na values in x should be stripped before the computation proceeds. For example, in the county data set, population is a natural weighting. Most content comes from the ECPR Winter School in Methods and Techniques R course, that I had the pleasure of teaching this February. A similar procedure is used for a 10% trimmed mean. Select Weighted Goals as the plan goals type. Weighted Mean calculator for calculating the weighted mean statistics for the given set of data. DESCRIPTION. The F1 score can be interpreted as a weighted average of the precision and recall, where an F1 score reaches its best value at 1 and worst score at 0. A weighted grade or score is average of a set of grades, where each grade (g) carries a different weight (w) of importance. Unlike most credit card interest calculators, this calculator will calculate the current finance charge for each card, and then compute the credit card average APR using a weighted formula. This paper proposes a novel on-line portfolio selection strategy named \u201cConfidence Weighted Mean Reversion \u201d (CWMR). There's one more skill you'll need to calculate weighted scores: A simple average, which in \"math speak\" is more properly called the mean. 3% then you are essentially implying a growth rate of POSITIVE 3. Find more details on making DIY. \u2022 So it is better to use WAPE for volume weighted MAPE and WMAPE for dollar weighted or Cost-weighted measures. What is a weighted KPI calculations For weighted KPI calculations, the base weight of each KPI is factored against the weighted counts of each KPI by using the following formulas. For this reason, consider using Mean Absolute Deviation (MAD) alongside MAPE, or Consider that even fast moving consumer goods companies these days to average MAPEs over multiple time series. (the \"Gold Book\"). Featured Services & News. Weighted Mean Formula (Table of Contents). Wondering what a weighted blanket is and what is the purpose of having one? Learn how weighted blankets can benefit sleep issues, anxiety, autism, ADHD, insomnia and more. These two measurements can be combined to give a weighted average. Follow this example use the following data set. The sample size weighted correlation may be used in correlating aggregated data Description. the weighted mean of is given by (4) Weighted means have many applications in physics, including finding the center of mass and moments of inertia of an object with a known density distribution and computing and electric and magnetic multipole moments of charge and current distributions, respectively. For instance if you have 3 objects, A,B and C. w: a numerical vector of weights the same length as x giving the weights to use for elements of x. Sincerely, Jacob. Weighted Shortest Job First (WSJF) is a prioritization model used to sequence jobs (eg. 1 Introduction The Least Mean Square (LMS) algorithm, introduced by Widrow and Hoff in 1959 [12] is an adaptive algorithm, which uses a gradient-based method of steepest decent [10]. The WACC is also the minimum average rate of return it must earn on its current assets to satisfy its shareholders, investors, or creditors. I have been working on a problem for the last year (in graph theory/algorithms). The weighted harmonic mean is the preferable method for averaging multiples, such as the price\u2013earnings ratio (P/E), in which price is in the numerator. % % Example: % X = rand(3,5);. He has an entire dice set that almost always gets the best possible roll. Weighted average refers to the mathematical practice of adjusting the components of an average to reflect the importance of certain characteristics. Definition of weighted in the Definitions. What is the Weighted Average? The weighted average scoring model applies a \"weight\" to the matrix questions based on responses to the first item in a side-by-side matrix. wi(yi 0 1xi) 2. The weighted mean is similar to the arithmetic mean where instead of each of the data points contributing equally to the final average, some data points contribute more than others. Example sentences with \"weighted mean\", translation memory add example en The conclusion was that the only items differentiating the SIDS infants from the control infants were a \"lower mean weight and height at birth, the previous occurrence of cyanosis [bluish skin and mucous membranes caused by lack of oxygen in the blood] or pallor during. Activity recording is turned off. 1 Introduction The Least Mean Square (LMS) algorithm, introduced by Widrow and Hoff in 1959 [12] is an adaptive algorithm, which uses a gradient-based method of steepest decent [10]. Often used to account for area changes between meridians at varying latitudes by using the cosine of the latitude as the weights. Besides the stock market example given above, another situation were weighed average is used is in grading. mean fitness of the population W(bar) The sum of the fitnesses of the genotypes of a population weighted by their proportions; hence a weighted mean fitness. Specifically, overtime is normally calculated at 1. Each grade item can be given a weight to change its importance in the overall mean. Weighted Shortest Job First (WSJF) is a prioritization model used to sequence jobs (eg. and Wilks, A. Weighted Mean: A mean where some values contribute more than others. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The weight of the category should be entered as a decimal, make sure that the sum of the weights equal 1. In simple terms, the category \"total\" will be equal to the sum of the scores in each grade item each multiplied by its grade weight, and that sum being finally divided by the sum of all weights. This calculator will calculate the weighted average APR for all of your credit cards that have a current balance. What does weighted mean mean? Information and translations of weighted mean in the most comprehensive dictionary definitions resource on the web. All or only selected records: You can either use all the points in the point theme for the analysis or only a selected subset of points. The weights cannot be negative. So let's have a look at the basic R syntax and the definition of the weighted. It is calculated according to the following formula: M = mark received in a course U = units of credit for a course. This article will show you how to use Excel's SUMPRODUCT and SUM functions individually and how to combine the two to calculate a weighted average. Formula: Where x is the repeating value. 0015948968 4 2 5442. Field {Case_Field} Field used to group features for separate mean center calculations. 15 * (73 + 80 + 85 + 88) +. \u2026But sometimes, some data points\u2026are more important than others. Trump talks Medicare in a retirement enclave where doctors are a golf-cart ride away. and Wilks, A. weighted definition: 1. Jun 07, 2016 \u00b7 61. Mar 18, 2017 \u00b7 2. A method of computing a kind of arithmetic mean of a set of numbers in which some elements of the set carry more importance (weight) than others. Paper 3: Issues in calculating average effect sizes in meta-analyses Dr. Metascore is a weighted average in that we assign more importance, or weight, to some critics and publications than others, based on their quality and overall stature. Slugging average is sometimes referred to as slugging percentage. But an equally weighted mean would be the pooled mewn and it expected value would be [65+73]/2. Section 2 formally formulates the on-. Under Medicare Advantage, the federal government still foots the. A weighted GPA is calculated by awarding additional points to classes that are considered more challenging than the basic curriculum. Free Arithmetic Mean (Average) Calculator - find the average of a data set step-by-step. The Benefits of a Weighted Sales Pipeline. 739 \u00d7 10-3 mm 2 /sec, SD = 0. Water-Quality Characteristics. Parameters: axis: {index (0), columns (1)} Axis for the function to be applied on. \u2026If the professor had four exams,\u2026and these were your four exam scores,\u2026calculating your average would be easy, 80%. May 11, 2017 \u00b7 The weighted mean is the weighted average. On the other hand, its weighted version is very useful for evaluating inequalities for definite integrals. The time-weighted formula is essentially a geometric mean of a number of holding-period returns that are linked together or compounded over time (thus, time-weighted). Among them, M55 showed a gene expression pattern consistent with behavioral changes after stress exposure, and the gene ontology analysis revealed that this was involved in nervous system development, gland. Data are classified using the Harmonized System of trade at the six- or eight-digit level. Sleep through the night and relieve anxiety with high-quality weighted blankets for adults from Weighting Comforts. Jul 24, 2014 \u00b7 Functional diversity can be quantified using one single trait at a time or multiple traits (see the descriptions of multiple-indices in other page). 4) but have been unsuccessful in computing the weighted SD. Aug 29, 2013 \u00b7 If such an effect is possible then the simplest and safest thing to do is compute the weighted sum for the m values in each of the n experiments. References. To calculate a weighted average in Excel, simply use the SUMPRODUCT and the SUM function. E(Z1)=65 and E(Z2)=73 The population mean is [2(65)+73]/3. It is challenging to achieve high weighted ef\ufb01ciency with low-power microinverters, typically because these devices are required to be low cost. How to Calculate Weighted Average Price Per Share Calculating your weighted average price per share can help you assess the performance of an investment that was made in several transactions. \u2026Let's consider an academic course as our example. What is the uncertainty of the weighted average? What's the correct procedure to find the uncertainty of the average?. Weighted average definition, a mean that is computed with extra weight given to one or more elements of the sample. Another Tip: If number of boys and girls is same, i. May 01, 2012 \u00b7 PowerPivot Weighted Average Measure Compared to Non-Weighted Average. The weighted average (or weighted mean, as statisticians like to call it) is easy to compute in SAS by using either PROC MEANS or PROC UNIVARIATE. If trim is non-zero, a symmetrically trimmed mean is computed with a fraction of trim observations deleted from each end before the mean is computed. - joran Jun 12 '12 at 4:35 3 @Frank Hover over the down triangle beneath the vote count next to your Q. If I use the following data. Math, I am in the 9th grade, and our math teacher is explaining \"weighted averaging. title = \"Amplitude-weighted mean velocity: Clinical utilization for quantitation of mitral regurgitation\", abstract = \"Objectives. Weighted blankets, sometimes referred to as gravity blankets, were once a tool of therapists and psychiatry clinics. The weighted mean is used a lot by teachers. How to use weighted in a sentence. net dictionary. It is useful in many situations e. Flow-weighted mean concentration. If you're seeing this message, it means we're having trouble loading external resources on our website. Myometrial Invasion in Endometrial Cancer: Diagnostic Accuracy of Diffusion-weighted 3. Weighted arithmetic mean. when they tried on their own, funded by WW. He has an entire dice set that almost always gets the best possible roll. In this lesson from Alanis Business Academy, we describe the weighted mean, review the equation to solve for the weighted mean, and actually solve for a weighted mean. If all the weights are equal, then the weighted mean equals the arithmetic mean (the regular \"average\" you're used to). So, we use a weighted average (weighted mean or scaled average) to give greater value to some data over other data. If this keyword is not present or is zero, then the mean is computed across all dimensions of the input array. Michael Ho\u0084 Zheng Sun\u0085 Jack Xin\u00a7. size }\" if weights. 35 mg/L Flow Weighted C once tra i = 0. What is the acronym meaning/definition of CAC ?. We now see the mean as a weighted sum of the distinct values, where each value is weighted according to its proportion in the total list of numbers. Ranking, Matrix/Rating Scale, Multiple Choice, Multiple Textboxes, and Slider questions calculate an average or weighted average. Often used to account for area changes between meridians at varying latitudes by using the cosine of the latitude as the weights. Weighted average refers to the mathematical practice of adjusting the components of an average to reflect the importance of certain characteristics. (mean age, 57 years) who had. mean function in R, so far I couldn\u2019t find a implementation of weighted. For weighted KPI calculations, the base weight of each KPI is factored against the weighted counts of each KPI by using the following formulas. n an average calculated by taking into account not only the frequencies of the values of a variable but also some other factor such as their variance. A weighted average is an average in which one element may contribute more heavily to the final result than another element. - joran Jun 12 '12 at 4:35 3 @Frank Hover over the down triangle beneath the vote count next to your Q. Among them, M55 showed a gene expression pattern consistent with behavioral changes after stress exposure, and the gene ontology analysis revealed that this was involved in nervous system development, gland. Apr 21, 2017 \u00b7 However, anxiety disorders can cause crippling feelings of fear, worry, or restlessness. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. If using aggregated data, the correlation of the means does not reflect the sample size used for each mean. 2 mg/L Flow Weighted Concentration = 0. wi(yi 0 1xi) 2. Weighted average refers to the mathematical practice of adjusting the components of an average to reflect the importance of certain characteristics. WMC is sampling-, scale-, and contrast-invariant, and is sparse on natural images. The weighted arithmetic mean (or weighted average) is used if one wants to combine average values from samples of the same population with different sample sizes: \u00af = \u2211 = \u2211 =. The monthly returns are then compounded to arrive at the annual return. His head was pounding, his body seemed to ache from head to toe, and there was a tightness in his chest that pressed on his lungs. Weighted Average Item Price Report (WAIPR) and the Regional and Statewide Average Awarded Price Report (RSWAAPR) The Weighted Average Item Price Report (WAIPR) and the Regional and Statewide Average Awarded Price Report (RSWAAPR) are reports produced using information from NYSDOT's Trns\u2022Port BAMS\u2044DSS. Weighted gene co-expression network analysis identified 60 characteristic modules that correlated with stress or the FKBP5 genotype. 0026132544 5 1 209. 03 Progressive Less than 0. It is also called weighted average. Tseng Department of Biostatistics Department of Human Genetics. Thus, the weighted mean makes it possible to find the average student grade in the case where only the class means and the number of students in each class are available. In other words, each value to be averaged is assigned a certain weight. Now, they have gone mainstream. Since we're all engineers, most students can figure out how this works at the end of the termbut how do you take it into account to calculate your current grade halfway through?. A weighted score or weighted grade is merely the average of a set of grades, where each set carries a different amount of importance. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. Hi R experts, I need know how calculate a weighted mean by group in a data frame. Weighted voting. mean, part of R for Data Science: Lunchbreak Lessons. The experts in survey were estimating the organizations by. 03 Neutral 0. Section 2 formally formulates the on-. We will first create a Category which will use Weighted mean of grades. The Greedy Strategy for activity selection doesn\u2019t work here as a schedule with more jobs may have smaller profit or value. Weighted Mean is a statistical method which calculates the average by multiplying the weights with its respective mean and taking its sum. If all the weights are equal, then the weighted mean is the same as the arithmetic mean. The average is calculated by adding a range of numbers together and then dividing this total by the number of values in the range. The solid lines are the weighted mean while the dashed lines are the standard mean. Featured Services & News. A weighted mean (or weighted average) is like an ordinary mean, but the observations don't contribute equally - more emphasis is placed on some data values than others; they are weighted by a bigger or smaller amount than 1/n. Given a numeric vector of data values, x, and another numeric vector of weights, w, the weighted mean is the sum of the data value times the weights divided by the sum of the weights. 'Mikhail' instead of just 'Mihail')?. His head was pounding, his body seemed to ache from head to toe, and there was a tightness in his chest that pressed on his lungs. This implies that we trim the lowest 5% of the data as well as the highest 5% of the data. Note that contrary to the weights w i in Eq. The weighted mean measures the average of the weighted data points. It is also called weighted average. Also called weighted mean. In a scenario where grades are weighted by category and the number of assignments in a category changes, no changes to the course points or syllabus would be necessary. What is a Weighted Mean? A weighted mean is a kind of average. Time-Weighted Return Formula. Meaning of weighted mean. A mean where some values contribute more than others. Input the above in your Ti-84 or TI-83 calculator Instructions on how to Create a List TI-84 | TI-83 tutorial. 631 \u00d7 10-3 mm 2 /sec, SD = 0. 03 Progressive Less than 0. If a professor decides to increase or decrease the workload based on the particular needs of any group of students, weighted grades make that easy. Here is the mean of 1, 2, 3 and 4: Add up the numbers, divide by how many numbers: Mean = 1 + 2 + 3 + 44 = 104 = 2. When totaling the individual values, each is multiplied by a weighting factor, and the total is then divided by the sum of all the weighting factors. Weighted means are often used for frequency data. These two measurements can be combined to give a weighted average. Weighted mean. Notice that the groups differ considerably in sample size. Mean: The arithmetic mean, also known as the simple mean or equal weighted mean. It is based on kernel weighted sample statistics such as the mean (Nadaraya-Watson estimator) but also standard deviation, skewness, kurtosis, deciles, etc. The above weighted average formula returns the value 849. Moving Average charts in JMP Each point on a Uniformly Weighted Moving Average (UWMA) chart, also called a Moving Average chart, is the average of the w most recent subgroup means (called \u201cspan\u201d), including the present subgroup mean. Synonyms for weighted at Thesaurus. Generally the first item measures attributes like \"Importance,\" \"Need,\" or \"Expectation\". Input the above in your Ti-84 or TI-83 calculator Instructions on how to Create a List TI-84 | TI-83 tutorial. The area-weighted mean patch size, on the other hand, is a weighted mean, where the weights are based on the size of the patch. However, if you don\u2019t want to spend big on Weighted belts, then you should absolutely go for Dark Iron Fitness which comes with all the basic features one could expect in Weighted belts. prepared and arranged in a way that is likely to produce a particular effect, usually an\u2026. This method assumes that all of the trials have measured the outcome on the same scale. It is generally used to find average of variables that are expressed as a ratio of two different measuring units e. The T2-weighted sequence can be employed as a dual echo sequence. Weighted Mean. This has been a guide to Weighted Average in Excel. In calculating a weighted average, each number in the data set is. What does it mean republicans trying to phase out higher federal payments of medicare 3 Oct 2019. an object containing the values whose weighted mean is to be computed. a numerical vector of weights the same length as x giving the weights to use for elements of x. Next, obviously the weighted mean is $$\\hat\\mu=\\sum_ip_ix_i,$$ and the variance:$$\\hat\\sigma^2=\\sum_ip_i(x_i-\\hat\\mu)^2$$. Apr 21, 2017 \u00b7 However, anxiety disorders can cause crippling feelings of fear, worry, or restlessness. By modifying sequence parameters such as repetition time (TR) and echo time (TE), for example, anatomical images can emphasize contrast between gray and white matter (e. Weighted mean The calculation of the mean discussed in the handout assumes that the standard deviation of each individual measurement is the same. VWAP is typically used with intraday charts as a way to determine the general direction of intraday prices. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. weighted average w/ matlab. Chapter 3 Weighted Mean study guide by john_lindsey8 includes 43 questions covering vocabulary, terms and more. Definition - What does Weighted Average Cost of Capital (WACC) mean? The weighted average cost of capital represents a weighted average of the after-tax cost of debt and the cost of equity where the weighting is based on a company\u2019s target debt-equity ratio, measured at market. If this keyword is present, then the mean is only calculated across a single dimension. SUMPRODUCT should be the sum of the products of the respective values and weights. Nov 29, 2019 \u00b7 10 intermittent fasting side effects that might mean it\u2019s not a great fit for you These are the 6 full-body home exercises the Springbok women\u2019s vice captain swears by World Aids Day: Could. Weighted Averages Date: 11/02/98 at 21:09:00 From: Jacob Smith Subject: Help with \"weighted Averaging\" Dr. Weighted gene co-expression network analysis identified 60 characteristic modules that correlated with stress or the FKBP5 genotype. Source Score, x Weight,w Homework 60 5% Midterm 75 35% Project 80 20% Speech 70 15% Final Exam 62 25% 2. The weighted arithmetic mean of a set of numbers X 1, X 2, , X N with respective weights of w 1, w 2, , w N is defined as: Example : For a statistics course, the final exam grade is weighted four times as much as each quiz score. It includes the product id and invoice id along with some other information. Column A shows all the student\u2019s percentage scores he/she received on tests, quizzes, homework assignments, and participation. Creates a classification table, from raw data in the spreadsheet, for two observers and calculates an inter-rater agreement statistic (Kappa) to evaluate the agreement between two classifications on ordinal or nominal scales. Let us understand more with an example:. It is calculated according to the following formula: M = mark received in a course U = units of credit for a course. A mean where some values contribute more than others. 35 mg/L Flow Weighted C once tra i = 0. wRC+ takes the statistic Runs Created and adjusts that number to account for important external factors -- like ballpark or era. 739 \u00d7 10-3 mm 2 /sec, SD = 0. How to use weighted in a sentence. This method is sensitive to extreme observations (e. The arithmetic mean-geometric mean (AM-GM) inequality states that the arithmetic mean of non-negative real numbers is greater than or equal to the geometric mean of the same list. Here are the high-level steps for using weighted goals to manage a plan: Define goal IDs on the Goals page.", "date": "2020-01-17 17:44:00", "meta": {"domain": "imltraining.de", "url": "http://kvmb.imltraining.de/weighted-mean.html", "openwebmath_score": 0.7800197005271912, "openwebmath_perplexity": 1117.3949159710257, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9835969679646668, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.8510317910693587}}
{"url": "https://mathhelpforum.com/threads/ask-is-expected-frequency-rounded-up-or-down.282947/", "text": "# [ASK] Is Expected Frequency Rounded Up or Down?\n\n#### Monoxdifly\n\nA die is tossed 100 times. What's the expected frequency that the number appears will be 4?\n\nThe probability is 1/6, so the expected frequency is 1/6 \u00d7 100, but that results in a fraction (16 2/3). Do we need to round it up or down? Is the answer 16 or 17?\n\n#### romsek\n\nMHF Helper\nYou don't need to round it at all. The correct answer is $\\dfrac{50}{3}$\n\nTrue you won't see 2/3 of a die but that doesn't matter. An expectation is an average and as such there's no reason to expect it must be an integer.\n\n#### Monoxdifly\n\nOkay then, thanks.\n\n#### TKHunny\n\nIf you are forced to round, for whatever reason, you will need to decide exactly how to do that.\n\nTypically, a simple up or down methodology will prove unsatisfactory as you will violate the definition of a probability distribution:\n\nDown 2 dp\n1/3 = 0.33\n1/3 = 0.33\n1/3 = 0.33\n0.33 * 3 = 0.99 and that is NOT 1.00\n\nLikewise, fudging (forced footing) proves unsatisfactory:\n\n1/3 = 0.33\n1/3 = 0.33\n1/3 = 0.34\n\nThey are not really equal, are they.\n\nUnless you are in a purely theoretical pursuit, you just have to make up your mind what is and is not suitable.\n\n#### Monoxdifly\n\nOh well, no rounding then, thanks.", "date": "2020-01-18 18:08:46", "meta": {"domain": "mathhelpforum.com", "url": "https://mathhelpforum.com/threads/ask-is-expected-frequency-rounded-up-or-down.282947/", "openwebmath_score": 0.6585183143615723, "openwebmath_perplexity": 967.3642782548623, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969645988575, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.8510317898663214}}
{"url": "https://math.stackexchange.com/questions/1475930/how-should-i-calculate-lim-n-rightarrow-infty-frac1n2n3n-nnn", "text": "# How should I calculate $\\lim_{n\\rightarrow \\infty} \\frac{1^n+2^n+3^n+\u2026+n^n}{n^n}$ [duplicate]\n\nHow should I calculate the below limit\n$$\\lim_{n\\rightarrow \\infty} \\frac{1^n+2^n+3^n+...+n^n}{n^n}$$ I have no idea where to start from.\n\n\u2022 If this question does not come out from you, then where did you get this question? \u2013\u00a0Megadeth Oct 12 '15 at 5:37\n\u2022 The limit should at least bigger than $1 + e^{-1} + e^{-2} + \\cdots = \\frac{e}{e-1}$ as for all $n$, the last term is $1$, the second last is $\\left(\\frac{n-1}{n}\\right)^n = (1- \\frac 1n)^n$ which tends to $e^{-1}$. So on so forth. \u2013\u00a0user99914 Oct 12 '15 at 6:05\n\u2022 From $\\frac{x_1+x_2+\\cdots+x_n}{n}\\geq\\sqrt[n]{x_1x_2\\cdots x_n}$ we write $$(\\frac1n)^n+(\\frac2n)^n+\\cdots+(\\frac1n)^n\\geq n\\frac{n!}{n^n}\\to e$$ \u2013\u00a0Nosrati Oct 12 '15 at 6:29\n\u2022 @Maryam $n^n\\sim e^nn!$, so $n\\dfrac{n!}{n^n}$ does not approach $e$. \u2013\u00a0Quang Hoang Oct 12 '15 at 6:35\n\u2022 Have you tried Stolz-Cesaro ? \u2013\u00a0Lucian Oct 12 '15 at 6:49\n\nFirst we use an observation by @Stan in the comment. Note that as $(1 +\\frac{x}{n})^n$ is increasing in $n$ whenever $|x|<n$,\n\n$$\\left(\\frac{k}{n}\\right)^n = \\left(1 + \\frac{k-n}{n}\\right)^n \\le e^{k-n},$$\n\n(here we assume that $x:= k-n$ is fixed and varies the remaining two $n$'s. This sequence is increasing and tends to $e^{k-n}$, as $|x| = |k-n| < n$. See here). Then we have\n\n$$\\begin{split} \\frac{1^n + 2^n + \\cdots + n^n}{n^n} &= \\sum_{k=1} ^n \\left(\\frac{k}{n}\\right)^n \\\\ &\\le \\sum_{k=1}^n e^{k-n} \\\\ &= 1 + e^{-1} + e^{-2} + \\cdots e^{1-n} \\\\ &\\le \\frac{1}{1-e^{-1}} = \\frac{e}{e-1}. \\end{split}$$ This implies\n\n$$\\limsup_{n\\to \\infty} \\frac{1^n + 2^n + \\cdots + n^n}{n^n} \\le \\frac{e}{e-1}.$$\n\nOn the other hand, fix $k$. Then for all $n >k$, we have\n\n$$\\begin{split} \\frac{1^n + 2^n + \\cdots + n^n}{n^n} &\\ge \\frac{(n-k)^n + (n-k+1)^n + \\cdots + n^n} {n^n}\\\\ &= \\left( 1 - \\frac kn\\right)^n + \\left( 1 - \\frac {k-1}n\\right)^n + \\cdots +1 \\end{split}$$\n\nThen for all $\\epsilon >0$, there is $N\\in \\mathbb N$ so that\n\n$$\\left| \\left( 1 - \\frac {j-1}n\\right)^n - e^{-(j-1)} \\right| < \\epsilon$$\n\nwhenever $n \\ge N$ and for all $j = 1, 2 , \\cdots, k+1$ (Note $k$ is fixed, so this $N$ can be found)\n\nIn particular, this implies\n\n$$\\frac{1^n + 2^n + \\cdots + n^n}{n^n} \\ge e^{-k} + e^{-(k-1)} + \\cdots + 1 - (k+1) \\epsilon.$$\n\nThus\n\n$$\\liminf_{n\\to \\infty} \\frac{1^n + 2^n + \\cdots + n^n}{n^n} \\ge e^{-k} + e^{-(k-1)} + \\cdots + 1 - (k+1) \\epsilon.$$\n\nNow let $\\epsilon \\to 0$ and then $k \\to \\infty$, we have\n\n$$\\liminf_{n\\to \\infty} \\frac{1^n + 2^n + \\cdots + n^n}{n^n} \\ge \\frac{1}{1-e^{-1}} = \\frac{e}{e-1}.$$\n\nThis implies\n\n$$\\lim_{n\\to \\infty} \\frac{1^n + 2^n + \\cdots + n^n}{n^n} = \\frac{e}{e-1}.$$\n\n\u2022 Another proof for $e^{k-n} \\geq (k/n)^n = e^{n \\ln (k/n)}$: $\\Leftarrow k-n \\geq n \\ln (k/n) \\Leftarrow \\frac{k}{n}-1 \\geq \\ln \\frac{k}{n}$ $\\Leftarrow$ for $x>0$, $x-1 \\geq \\ln x$ $\\Leftarrow$ Let $y(x) = x-1-\\ln x$, then the stationary point should satisfy $y'=1-\\frac{1}{x}=0$; meanwhile $y''=x^{-2}>0$. Thus $y_{\\min} = y(1) = 0$, for all $x>0$, $y(x) \\geq 0$. \u2013\u00a0Stan Oct 12 '15 at 8:01\n\u2022 Another proof for the upper bound: $$1^n+2^n+\\cdots+(n-k-1)^n < \\int_1^{n-k} x^n\\,dx < (n-k)^{n+1}/(n+1) < (n-k)^n.$$ Thus when we take only the terms from $(n-k)^n$ to $n^n$ in the numerator, the neglected terms are at most $(n-k)^n/n^n$, which tends to $e^{-k}$ as $n\\to\\infty$. Now we have the limit sandwiched between two explicit functions of $k$; let $k\\to\\infty$. \u2013\u00a0Greg Martin Oct 12 '15 at 8:10\n\u2022 Nice proof! +1 :) \u2013\u00a0ZFR Nov 21 '15 at 7:50", "date": "2020-01-27 22:07:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1475930/how-should-i-calculate-lim-n-rightarrow-infty-frac1n2n3n-nnn", "openwebmath_score": 0.9840658903121948, "openwebmath_perplexity": 222.886461682128, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969679646668, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.851031782523648}}
{"url": "http://math.stackexchange.com/questions/330636/associativity-of-a-binary-operator", "text": "# Associativity of a binary operator\n\nI am reading the book \"A first course in abstract algebra\", and i am reading the chapter called subgroups. It says that, for general situations we do not use \"*\" to denote the binary operation on a group, but we simply use addition symbol \"+\" or multiplication symbol \".\". Then it says, we denote the product a.a.a.a.a...a for n factors by an. But i am confused here: is that always associative? For example, for 3 factors, a.a.a is a3 according to what book says, but are we sure that a.(a.a)=(a.a).a=a3 is always true? Thank you.\n\n-\nDo you know what \"group\" means? \u2013\u00a0 Chris Eagle Mar 14 '13 at 20:53\nYes, i do. Why? \u2013\u00a0 bigO Mar 14 '13 at 20:54\nThen you know the answer to this question. \u2013\u00a0 Chris Eagle Mar 14 '13 at 20:54\nThis is why i am asking this, i must be missing something \u2013\u00a0 bigO Mar 14 '13 at 20:55\nNote that not all binary operations need be associative. For example, take the integers under subtraction: $-4=(1-2)-3\\neq 1-(2-3)=0$. \u2013\u00a0 user1729 Mar 15 '13 at 14:54\n\nRecall: One of the key properties satisfied by all groups is that its binary operation is associative on the set defined by the group.\n\nIf you are referring to Fraleigh's text, see\n\nDefinition 4.1: \"A group $\\langle G, *\\rangle$ is a set $G$, closed under a binary operation $*$, such that the following axioms of satisfied:\n\n$\\mathcal G_1$: For all $a, b, c \\in G,$ we have $$(a*b)*c = a*(b*c)\\tag{associativity of *}$$\n\nOf course, $\\mathcal G_2,\\;\\text{and}\\;\\mathcal G_3\\;$ are crucial, as well: the existence of an identity element $e\\in G$, and for each $g\\in G$, its inverse, $g^{-1}$ is also in $G$, including the definitions of the identity and how we define the inverse of a group element.\n\n-\nOh thanks, that's what i have been missing. \u2013\u00a0 bigO Mar 14 '13 at 20:55\n\nNot only, As @amWhy noted, the binary operation is associative but also depend on how you show the binary operation, the figures of composition of elements varies: $$\\cdot\\longrightarrow a*b=a\\cdot b$$ so $$\\underbrace{a\\cdot a\\cdot a\\cdot ...\\cdot a}_{\\large n\\;times}=a^n~$$ And $$+\\longrightarrow a*b=a+ b$$ so $$\\underbrace{a+ a+ a+ ...+ a}_{\\large n\\;times}=na$$\n\n-\nDon't know why the downvote? +1 \u2013\u00a0 amWhy Mar 14 '13 at 21:09\nThank you! I did not downvote, just to let you know \u2013\u00a0 bigO Mar 14 '13 at 21:28\n@amWhy: Thanks Amy for the edit. :-) \u2013\u00a0 Babak S. Mar 15 '13 at 5:32", "date": "2015-08-01 22:59:10", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/330636/associativity-of-a-binary-operator", "openwebmath_score": 0.8696829676628113, "openwebmath_perplexity": 756.27971945616, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9835969643584426, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8510317759851715}}
{"url": "https://math.stackexchange.com/questions/1986392/win-a-coin-game-in-10th-round", "text": "# Win a coin game in 10th round\n\nTwo players are tossing a fair coin in one round. If it is heads, the first one gets a dollar from the second player. Otherwise, the first player gives a dollar to the second one. If both players have 6 dollars in the beginning of the game, what is the probability that the first player wins all the money exactly on 10th round.\n\n\u2022 I am considering sample space to be 2^10; however, a friend of mine considers it to be 960. Whose idea is erroneous and why? Thanks. \u2013\u00a0Ulugbek Abdullaev Oct 26 '16 at 18:06\n\u2022 The sample space is not as big as $2^{10}$, because the sequence $HHHHHHTTTT$ and $HHHHHHHHHH$ are considered exactly the same game (they quit playing after the sixth heads). Of course, if you think that they continue playing after one has lost just to \"see what would happen\", but without any money involved, and analyze those games, then the sample space has size $2^{10}$. It all depends on your interpretation. \u2013\u00a0Arthur Oct 26 '16 at 18:07\n\u2022 @Arthur I got your point. Could you elaborate how to exclude those unnecessary? I'm just stuck.. \u2013\u00a0Ulugbek Abdullaev Oct 26 '16 at 18:41\n\u2022 Hint: out of those ten rounds, how many Heads were there? How many Tails? We know the last one was $H$...what about the next to last? \u2013\u00a0lulu Oct 26 '16 at 19:04\n\u2022 @BruceET It's very nearly the whole story. We see that we want $8$ Heads and $2$ Tails...there must be at least one $T$ in the first $6$ slots and and both $T$ must be in the first $8$. Easy to count. \u2013\u00a0lulu Oct 26 '16 at 21:58\n\nIf the coin is tossed 10 times, there could be wins at trials number 6, 8, and 10.\n\nA win at the 6th requires six heads in a row (probability $.5^6 = 0.015625.$)\n\nA first win at the 8th requires that exactly one of the first six tosses must be tails, followed by two heads (probability $6(.5)^8 = 0.0234375.$)\n\nThe simulation below confirms these two results (within simulation error), and suggests that the probability of a first win at the 10th toss must have probability about 0.0265 (two or three place accuracy). I will leave it to you use similar logic to find a combinatorial formula and the exact probability.\n\nm = 10^6; frst.win = numeric(m)\nfor(i in 1:m)\n{ toss = sample(c(-1,1), 10, rep=T)  # vector of 1's (Hs) and -1's (Ts)\ncs = cumsum(toss)                  # 1st player's cumulative totals\nfrst.win[i] = match(6, cs) }       # toss on which cum tot first reaches 6\ntable(frst.win)/m\nfrst.win\n6        8       10\n0.015633 0.023540 0.026513", "date": "2020-12-04 18:18:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1986392/win-a-coin-game-in-10th-round", "openwebmath_score": 0.6429985165596008, "openwebmath_perplexity": 666.7686578027019, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9835969698879861, "lm_q2_score": 0.8652240686758841, "lm_q1q2_score": 0.8510317722237544}}
{"url": "http://math.stackexchange.com/questions/495119/what-is-gcd0-0/495127", "text": "# What is $\\gcd(0,0)$?\n\nWhat is the greatest common divisor of $0$ and $0$? On the one hand, Wolfram Alpha says that it is $0$; on the other hand, it also claims that $100$ divides $0$, so $100$ should be a greater common divisor of $0$ and $0$ than $0$.\n\n-\n(o,o)? <- see the face scratching the head? \u2013\u00a0zerosofthezeta Sep 16 '13 at 4:57\nIt is typically not defined. \u2013\u00a0copper.hat Sep 16 '13 at 4:58\n\u2013\u00a0anon Sep 16 '13 at 5:23\n\nThe word \"greatest\" in \"Greatest Common Divisor\" does not refer to being largest in the usual ordering of the natural numbers, but to being largest in the partial order of divisibility on the natural numbers, where we consider $a$ to be larger than $b$ only when $b$ divides evenly into $a$. Most of the time, these two orderings agree whenever the second is defined. However, while, under the usual order, $0$ is the smallest natural number, under the divisibility order, $0$ is the greatest natural number, because every number divides $0$.\n\nTherefore, since every natural number is a common divisor of $0$ and $0$, and $0$ is the greatest (in divisibility) of the natural numbers, $\\gcd(0,0)=0$.\n\n-\nSimilarly $\\gcd(0,n)=n$ for all $n\\in\\Bbb N$ (including $n=0$). \u2013\u00a0anon Sep 16 '13 at 5:22\nThanks, great answer! How did you figure this out? Where do mathematicians store such commonly agreed on, but not completely obvious definitions? When did people start defining gcd this way, I assume they didn't when Euclidean algorithm was invented 300BC? \u2013\u00a0Konstantin Weitz Sep 16 '13 at 6:55\nFor the most part, the partial order of divisibility coincides with the usual ordering wherever it is defined. So if $\\le$ represents the usual ordering and $\\vert$ represents the partial order of divisibility, then $x\\vert y\\Rightarrow x\\le y$, except if $y=0$. Moreover, if $x$ is a common divisor of two non-zero natural numbers $a$ and $b$, then $x\\le\\gcd(a,b)\\Rightarrow x\\vert\\gcd(a,b)$. So in most cases, you can define $\\gcd(a,b)$ to be the greatest common divisor of $a$ and $b$, where 'greatest' refers to the usual ordering. That's what Euler did (and left $\\gcd(0,0)$ undefined). \u2013\u00a0Donkey_2009 Sep 16 '13 at 11:55\nEuclid, I mean.... \u2013\u00a0Donkey_2009 Sep 16 '13 at 14:50\n\nI answered this already in a comment at MO: \"The best way to think about this is that the \"gcd\" of two natural numbers is the meet of them in the lattice of natural numbers ordered by divisibility. Note that $0$ is the top element in the divisibility order. The meet of the top element with itself is itself. So $0 = \\gcd(0, 0)$ is the answer. 'Greatest' is an unfortunate misnomer in this case.\"\n\nThe book Mathematics Made Difficult has a nice little section on this. It should perhaps better be called \"highest common factor\" (hcf).\n\n-\nYou lost the race to William. \u2013\u00a0dfeuer Sep 16 '13 at 5:05\nwe all lost to him :) \u2013\u00a0DanielY Sep 16 '13 at 5:06\n@dfeuer Well, I had answered the question already at MO. I wasn't racing. \u2013\u00a0user43208 Sep 16 '13 at 5:06\nWin or lose, \"the meet of [two numbers] in the lattice of natural numbers ordered by divisibility\" is a definition of delightful pithiness. \u2013\u00a0Jordan Gray Sep 16 '13 at 11:00\n\nAnother way to think about this is ideals. The gcd of two natural numbers $a, b$ is the unique non-negative natural number that generates the ideal $\\langle a, b \\rangle$. So in this case, $\\langle 0 ,0 \\rangle$ is just the $0$ ideal so the gcd is $0$.\n\n-\n\nDefining $\\gcd(a,b)$ for $\\def\\Z{\\mathbf Z}a,b\\in\\Z$, to be the non-negative generator of the ideal $a\\Z+b\\Z$ gives $\\gcd(a,0)=|a|$ for all $a$, including for $a=0$. Similarly one can define $\\def\\lcm{\\operatorname{lcm}}\\lcm(a,b)$ to be the non-negative generator of the ideal $a\\Z\\cap b\\Z$, which gives $\\lcm(a,0)=0$ for all$~a$ (note that since this is the only common multiple in this case, it is unlikely to provoke much discussion).\n\nAdding these cases to the usual definitions of the $\\gcd$ and $\\lcm$ for nonzero integers causes no problems; all usual formulas remain valid. In fact, if one wants the rule $\\gcd(xy,xz)=|x|\\gcd(y,z)$ to hold for all integers $x,y,z$, one is forced to put $\\gcd(0,0)=0$.\n\nOn the other hand, I don't think it is absolutely vital for mathematics to have $\\gcd(0,0)$ defined, in the same way as $0+0$, $0\\times0$ and $0^0$ need to be defined (and $0/0$ needs to be undefined) in order for the usual rules of algebra that one uses all the time to be valid. I think it would suffice to qualify the rule I cited by \"$x\\neq0$\" if one wants to leave $\\gcd(0,0)$ undefined; other rules like $\\gcd(a,b)=\\gcd(a,a-b)$ do not seem to equate $\\gcd(0,0)$ with something else. The reason that leaving it undefined is not so dramatic is that when considering divisibility, $0$ is often excluded anyway; for instance it has to be put aside in the theorem of Unique Factorisation.\n\n-\n\nSimply said - this depends on your definition.\n\nClearly, if $d=\\gcd(a,b)$, you require $d\\mid a$, $d\\mid b$, i.e., it is a common divisor.\n\nBut there are two possibilities how to express that it is greatest common divisor.\n\nOne of them is to require $$c\\mid a \\land c \\mid b \\Rightarrow c\\le d$$ and the other one is $$c\\mid a \\land c \\mid b \\Rightarrow c\\mid d.$$\n\nClearly, if you use the first definition, $\\gcd(0,0)$ would be the largest integer, so it does not exists. If you use the second one, you get $\\gcd(0,0)=0$. (Note that $0$ is the largest element of the partially ordered set $(\\mathbb N,\\mid)$.)\n\nAs far as I can say, the first definition appears in some text which are \"for beginners\"; for example here. (It was one of the first results from Google Books when searching for \"gcd(0,0)\".)\n\nI would say that for students not knowing that $\\mid$ is in fact a partial order, the first definition might feel more natural. But once you want to use this in collection with more advanced stuff (for example, g.c.d. as generator of an ideal generated by $a$ and $b$), then the second definition is better.\n\n-\nThis helped. I still feel a little icky about (\u2115,\u2223) as a partial order, because 0\u22230 feels weird to me, and this whole argument requires that reflexivity. \u2013\u00a0rampion Apr 8 at 0:46\n\nIn the general framework of integral domains (commutative rings with unity, without nontrivial zero-divisors), we define a greatest common divisor of two elements $a,b$ of an integral domain $R$ as any element $c\\in R$ satisfying:\n\n\u2022 $c$ divides both $a$ and $b$, that is, there are $x,y\\in R$ such that $a=cx$ and $b=cy$.\n\u2022 If $d\\in R$ divides both $a$ and $b$, then $d$ divides $c$.\n\nWe write $c=\\gcd(a,b)$ in this case. The definition implies that if $c,c^\\prime=\\gcd(a,b)$, then $c$ and $c^\\prime$ are associates, that is $c=uc^\\prime$ for some invertible element $u$ in $R$. This is equivalent to say that the principal ideals generated by $c$ and $c^\\prime$ are the same. Therefore a non-ambiguous definition is as follows:\n\n\"$\\gcd(a,b)$ is $\\$ any minimal$\\$ the minimum element in the family $\\mathcal F=\\{Rd: Rd\\supseteq Ra+Rb\\}$ of principal ideals containing $a$ and $b$, ordered by inclusion, wherever such minimum ideal exists.\"\n\nSince $R0+R0=R0$, we see that $R0=0$, the trivial ideal of $R$, is the $\\gcd$ of $0$ and $0$.\n\nIn general you cannot guarantee that $\\gcd$s exist. A sufficient condition is your integral domain $R$ to be a UFD (Unique Factorization Domain).\n\n-\nUse \\gcd instead of $GCD$. Also, ^\\prime can be abbreviated all the way down to '. \u2013\u00a0dfeuer Sep 16 '13 at 5:11\nI've tried editing it myself, but you've interrupted me twice, so I'll let you do it. \u2013\u00a0dfeuer Sep 16 '13 at 5:11\n@dfeuer Thanks for the suggestion. By the way, that ' shortcut works in ordinary $\\LaTeX$? \u2013\u00a0Matem\u00e1ticos Chibchas Sep 16 '13 at 5:12\nYep! It even works in Plain $\\TeX$! $x'''$ just takes four keystrokes (six if you count dollar signs). \u2013\u00a0dfeuer Sep 16 '13 at 5:16\nanon- Not on my keyboard. On my keyboard, backtick is the key to the left of 1, and tilde is obtained by holding shift and pressing the hash key (which is next to Enter). In general, the location of punctuation characters varies in keyboards used in different countries. \u2013\u00a0Hammerite Sep 16 '13 at 9:21\n\nFrom Wikipedia:\n\nThe greatest common divisor of $a$ and $b$ is well-defined, except for the case $a=b=0$, when every natural number divides them.\n\n-\nWell, this isn't really the right answer, is it? William got it right. \u2013\u00a0user43208 Sep 16 '13 at 5:22\nAll I'm saying is that Wikipedia got it wrong in this case (or at least it's misleading to suggest that $\\gcd(0, 0)$ is not well-defined, because as has been clearly explained, it's perfectly well-defined and indeed it's $0$). \u2013\u00a0user43208 Sep 16 '13 at 5:32\nThe guy got his answer, that's the most important :) \u2013\u00a0DanielY Sep 16 '13 at 5:33\n\nIf we take Euclid's algorithm:\n\n$\\text{gcd}(a, b) = \\left\\{ \\begin{array}{l l} a & \\quad \\text{if$b = 0$}\\\\ \\text{gcd($b$,$a$mod$b$)} & \\quad \\text{otherwise} \\end{array} \\right.$\n\nas the definition of GCD, then $\\text{gcd}(a, 0) = 0$ for any $a$, because the stopping case of $b = 0$ is reached immediately. If $a = 0$, then we get $\\text{gcd}(0, 0) = 0$.\n\nSo, it's possible that Wolfram Alpha obtains its result as a side effect of using Euclid's algorithm rather than deliberate thought as to what gcd(0, 0) should return. But it does fortunately coincide with William's \u201cpartial order of divisibility\u201d explanation.\n\n-", "date": "2015-11-27 08:59:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/495119/what-is-gcd0-0/495127", "openwebmath_score": 0.9462026357650757, "openwebmath_perplexity": 273.8767097374644, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9835969655605173, "lm_q2_score": 0.8652240704135291, "lm_q1q2_score": 0.8510317701886666}}
{"url": "http://mathhelpforum.com/calculus/165708-stokes-theorem-double-integral.html", "text": "# Math Help - Stokes Theorem - double integral\n\n1. ## Stokes Theorem - double integral\n\nUse Stokes\u2019 Theorem to evaluate $\\int_c^.F \\cdot dr$\n\nC: is the boundary of the portion of $z = x^2 + y^2$ below z = 4, oriented downward,\n\n$F = $\n\nCan someone please check over my work thanks!\n\nFirst I found N --> $N = \\frac{<2x, 2y, -1>}{\\sqrt{4x^2 + 4y^2 + 1^2}}$\n\nThen i found $\\bigtriangledown \\times F = <0, 0, -1>$\n\nfrom using the formula $\\bigtriangledown \\times F = $\n\nNow $\\bigtriangledown \\times F \\cdot N = \\frac{1}{\\sqrt{4x^2 + 4y^2 + 1^2}}$\n\nThen $dS = {\\sqrt{4x^2 + 4y^2 + 1}}dA$\n\nSo putting it all together....\n\n$\\int\\int_R^. \\frac{1}{\\sqrt{4x^2 + 4y^2 + 1^2}} \\cdot \\sqrt{4x^2 + 4y^2 + 1}$\n\nThe square roots cancel and reduce to....\n\n$\\int\\int_R^. 1 dA$\n\nThis double integral of 1 dA is equal to the Area of the Region\n\nAnd using $z = x^2 + y^2$ and $z = 4$ we come up with $r^2 = 4 .... r = 2$\n\n$A(R) = \\pi r^2 = \\pi(2^2) = 4\\pi$\n\nDoes $4\\pi$ look correct for the answer?\n\nThanks for looking\n\n2. I've always disliked the notation $\\vec{n}dS$ because used that unthinkingly results in doing two square roots which, just as you see here, cancel!\n\nInstead think of $d\\vec{S}$ as the \"vector differential of surface area\" and calculate it directly. We can write the surface $z= x^2+ y^2$ in a vector equation with x and y as parameter: $\\vec{r}(x,y)= x\\vec{i}+ y\\vec{j}+ z\\vec{k}= x\\vec{i}+ y\\vec{j}+ (x^2+ y^2)\\vec{k}$. The derivatives with respect to x and y give two tangent vectors, $\\vec{i}+ 2x\\vec{k}$ and $\\vec{j}+ 2y\\vec{k}$. The cross product of those gives the \"vector differential of surface area\": $2x\\vec{i}+ 2y\\vec{j}- \\vec{k}$ where I have chosen the order of multiplication to get a negative value for $\\vec{k}$, \"oriented downward\".\n\nYou are correct that $\\nabla\\times \\vec{F}= -\\vec{k}$ so that $\\nabla\\vec{F}\\cdot\\vec{n} dS= \\nabla\\vec{F}\\cdot d\\vec{S}$[tex]= (0(2x)+ 0(2y)- 1(-1))dxdy= dxdy.\n\nThat gives $4\\pi$, not $-4\\pi$.\n\nYour normal vector, $\\frac{-2x\\vec{i}-2y\\vec{j}+ \\vec{k}}{\\sqrt{4x^2+ 4y^2+ 1}}$ was oriented upward, not downward.", "date": "2014-03-08 00:24:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/165708-stokes-theorem-double-integral.html", "openwebmath_score": 0.9430418014526367, "openwebmath_perplexity": 810.0528673026399, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587238073113, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.8509918822588359}}
{"url": "http://wopsr.net/palindromes/", "text": "## Palindromes\n\nProblem 4 at Project Euler poses the following challenge:\n\nA palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 \u00d7 99.\n\nFind the largest palindrome made from the product of two 3-digit numbers.\n\n### A Na\u00efve Approach\n\nA na\u00efve solution would be to multiply all the three-digit numbers by one another, pull out all the palindromes, and find the greatest:\n\nvar isPalindrome = function(n){\nn = n.toString();\nreturn n === n.split(\"\").reverse().join(\"\");\n};\n\nvar findBiggest = function(){\nvar allProducts = [];\nfor (var n = 100; n < 1000; n++){\nfor (var m = 100; m < 1000; m++){\nallProducts.push(m * n);\n}\n}\nvar palindromes = allProducts.filter(isPalindrome);\nreturn Math.max.apply(null, palindromes);\n};\n\n\nBut the na\u00efve appraoch means making a huge-ass array with 810,000 numbers in it, consuming 3.09MiB of memory and taking many milliseconds to build. We can do better with a little number theory.\n\n#### Don\u2019t include obvious duplicates\n\n$m \\times n = n \\times m$\n\nInstead of starting m at 100 for each value of n, start it at n. This avoids duplicates and cuts array size roughly in half, to only 405,450 entries.\n\n#### Don\u2019t include multiples of 10\n\nObserve that $n \\times 10$ always ends in $0$. We don\u2019t ordinarily write positive non-zero integers in decimal with leading zeros, so we can safely exclude all multiples of 10 from $n$ and $m$, saving us another 76,995 entries in allProducts, almost 19%, for a new total of 328,445 entries.\n\n#### Assume the greatest palindrome has six digits\n\nThis is a pretty safe assumption in the instance the Project Euler problem describes, but may not be safe in other situations. Of our remaining 404,550 products of two three-digit decimal numbers, only 69,787 (17.25%) of them do not have six digits. It turns out these 17.25% include over 60% of the palindromes, but all we need is one six-digit palindrome in there to safely ignore the five-digit ones. $101,101 = 143\\times707$ and has six digits; that\u2019s enough to let us dismiss five-digit entries.\n\nBut how do we exclude five-digit products from our array without having to spend time calculating those products? We don\u2019t. But limiting ourselves to condisering only six-digit products (actually, to products with an even number of digits) lets us do something nifty.\n\nAs it turns out, palindromic numbers with an even number of digits have a nifty property: they all share a common, easily identifiable factor.\n\nLet $\\mathbb{P}\\subset\\mathbb{N}$ be the set of palindromic numbers with $k+1$ digits $a_0...a_k$ in an arbitrary base $b:b\\geq2$:\n\n$\\mathbb{P}=\\left\\{ n\\in\\mathbb{N}:n=\\sum\\limits_{i=0}^{k}a_ib^i\\middle|a_i=a_{k-i}\\right\\}$\n\nNote that $n$ is palindromic if and only if $a_i=a_{k-i}$. By way of example:\n\n $$$$\\begin{split}987,789&=9(10^0)\\\\&+8(10^1)\\\\&+7(10^2)\\\\&+7(10^3)\\\\&+8(10^4)\\\\&+9(10^5)\\end{split}$$$$ $$$$\\begin{split}65,456&=6(10^0)\\\\&+5(10^1)\\\\&+4(10^2)\\\\&+5(10^3)\\\\&+6(10^4)\\end{split}$$$$\n\nLooks like a pattern! Turns out we can rewrite the sum like this, but only for palindromes with an even number of digits:\n\n$n=\\sum\\limits_{i=0}^{\\frac{k+1}{2}}a_i\\left(b^i+b^{k-i}\\right)\\Big|k+1\\text{ is even}$\n\nEssentially, we pair up the digit $a_i$ with its matching digit $a_{k-i}$ and thereby only deal with half the digits. This doesn\u2019t work with palindromes with an odd number of digits because of the pesky middle digit, which has no mate.\n\n $$$$\\begin{split}987,789&=9(10^0+10^5)\\\\&+8(10^1+10^4)\\\\&+7(10^2+10^3)\\end{split}$$$$ $$$$\\begin{split}65,456&=6(10^0+10^4)\\\\&+5(10^1+10^3)\\\\&+4(10^2)\\end{split}$$$$\n\nFor palindromes with an even number of digits, we can factor the addend:\n\n$a_i(b^i+b^{k-i})=a_ib^i(1+b^{k-2i})$\n\nCheck it out! We can make $1+b^{k-2i}=0$ with modular arithmetic!\n\n\\begin{alignat*}{2}\\begin{split}1+b^{k-2i}&=1-1^{k-2i}&\\mod{b+1}\\\\&=1-1&\\mod{b+1}\\\\&=0&\\mod{b+1}\\end{split}\\end{alignat*}\n\nEvery palindrome with an even number of digits is evenly divisible by $b+1$, where $b$ is the radix. We\u2019re working in decimal, so $b+1=11$.\n\nWe needn\u2019t bother with any products but those that are divisible by 11. 11 being prime makes things easy. We'll only multiply $m\\times n$ and push() the product to allProducts if at least one of $\\lbrace m,n\\rbrace$ is divisible by 11. This brings allProducts.length down to only 55,764, a reduction of over 93.1% from our original 810,000.\n\n### Putting it All Together\n\nAll this makes for lots of savings in computation and storage, but it makes our program a lot uglier:\n\nvar isPalindrome = function(n){\nn = n.toString();\nreturn n === n.split(\"\").reverse().join(\"\");\n};\n\nvar findBiggest = function(){\nvar allProducts = [];\nvar start, end, step;\nfor (var n = 101; n < 1000; n++){\nif (n % 10){\nif (n % 11){\nstart = 11*Math.ceil(n/11);\nend = 11*Math.floor(1000/11);\nstep = 11;\n} else {\nstart = n;\nend = 1000;\nstep = 1;\n}\n}\nfor (var m = start; m < end; m+=step){\nif (m % 10){\nallProducts.push(m * n);\n}\n}\n}\nvar palindromes = allProducts.filter(isPalindrome);\nreturn Math.max.apply(null, palindromes);\n};\n\n\nMessy, but fast. The na\u00efve version ran in 1,158.8ms, while this version runs in only 74.2ms.", "date": "2017-04-30 12:52:58", "meta": {"domain": "wopsr.net", "url": "http://wopsr.net/palindromes/", "openwebmath_score": 0.40408554673194885, "openwebmath_perplexity": 1602.547677610792, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587275910132, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8509918820070361}}
{"url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-15-matrices-a-t-depending-on-t-derivative-da-dt/", "text": "Lecture 15: Matrices A(t) Depending on t, Derivative = dA/dt\n\nFlash and JavaScript are required for this feature.\n\nDescription\n\nThis lecture is about changes in eigenvalues and changes in singular values. When matrices move, their inverses, their eigenvalues, and their singular values change. Professor Strang explores the resulting formulas.\n\nSummary\n\nMatrices $$A(t)$$ depending on $$t /$$Derivative $$= dA/dt$$\nThe eigenvalues have derivative $$y(dA/dt)x$$.\n$$x$$ = eigenvector, $$y$$ = eigenvector of transpose of $$A$$\nEigenvalues from adding rank-one matrix are interlaced.\n\nRelated section in textbook: III.1-2\n\nInstructor: Prof. Gilbert Strang\n\nThe following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu.\n\nGILBERT STRANG: So I've worked hard over the weekend. I figured out what I was doing last time and what I'm doing this time and improved the notes. So you'll get a new set of notes on the last lecture and on this one. And I kind of got a better picture of what we're doing. And that board is aiming to describe the large picture of what we're doing last time and this time.\n\nSo last time was about changes in A inverse when A changed. This time is about changes in eigenvalues and changes in singular values when A change. As you can imagine, this is a natural important situation. Matrices move, and therefore, their inverses change, their eigenvalues change, their singular values change. And you hope for a formula.\n\nWell, so we did have a formula for last time for the change in the inverse matrix. And I didn't get every u and v transpose in the right place in the video or in the first version of the notes, but I hope that that formula, that that Woodbury Morrison formula will be correct this time. So I won't go back over that part.\n\nBut I realize also there is another question that we can answer when the change is very small, when the change in A is dA or delta A, a small change. And that's, of course, what calculus is about. So I have to sort of parallel topics here. What is the derivative when the change is infinitesimal? And what is the actual change when the change is finite size?\n\nSo now, let me say what we can do and what we can't do. Oh, I'll start out by figuring out what the derivative is for the inverse. So that's like completing the last time for infinitesimal changes. Then I'll move on to changes in the eigenvalues and singular values. And there, you cannot expect an exact formula. We had a formula that was exact, apart from any typos, for this. And we'll find a formula for this, and we'll find a formula for that and for that. Well, that one will come from this one. So this will be a highlight today. How do the eigenvalues change when the matrix changes?\n\nBut we won't be able to do parallel to this, we won't be able to-- oh, we will be able to do something for finite changes. That's important. Mathematics would have to keep hitting that problem until it got somewhere. So I won't get an exact formula for that change. That's too much.\n\nBut I'll get inequalities. How big that change could be. What can I say about it? So these are highly interesting.\n\nMay I start with completing the last lecture? What is the derivative of the inverse? So I'm thinking here, so what's the setup? The setup is my matrix A depends on time, on t. And it has an inverse. A inverse depends on t.\n\nAnd if I know this dependence, in other words, if I know dA dt, how the matrix is depending on t, then I hope I could figure out what the derivative of A inverse is. We should be able to do this. So let me just start with-- it's not hard and it complements this one by doing the calculus case, the infinitesimal change. So I want to get to that.\n\nI can figure out the change in A. And my job is to find the derivative of A inverse. So here's a handy identity. Can I just put this here? So here's my usual identity.\n\nSo as last time, I start with a finite change because calculus always does that, right. It starts with a delta t and then it goes to 0. So here I am up at a full size change. So I think that this is equal to B inverse A minus B A inverse. And if it's true, it's a pretty cool formula.\n\nAnd, look, it is true, because over on this right-hand side, I have B inverse times A A inverse. That's the identity. So that's my B inverse. And I have the minus, the B inverse B is the identity. There's A inverse. It's good, right?\n\nSo from that, well, I could actually learn from that the rank of this equals the rank of this. That's a point that I made from the big formula. But now, we can see it from an easy formula. Everywhere here, I'm assuming that A and B are invertible matrices. So when I multiply by an invertible matrix, that does not change the rank. So those have the same ranks.\n\nBut I want to get further than that. I want to find this. So how do I go? How do I go forward with that job to find the derivative of the inverse?\n\nWell, I'm going to call this a change in A inverse. And over here, I'll have B will be-- yeah, OK, let's see, am I right? Yeah.\n\nSo B inverse will be-- this is A plus delta A inverse. And this is-- well, that's A minus B. So that's really minus delta A. From A to B is the change. Here, I'm looking at the difference A minus B. So it's minus a change. And here, I have A inverse. I haven't done anything except to introduce this delta and get B out of it and brought delta in.\n\nNow, I'm going to do calculus. So I'm thinking of B as there's a sort of a delta t. And I'm going to divide by both sides by delta t. I have to do this if I want-- and now, I'll let delta t go to 0. So calculus appears. Finally, our-- I won't say enemy calculus, but there is a sort of like competition between linear algebra and calculus for college mathematics. Calculus has had far, far too much time and attention. It like it gets three or four semesters of calculus for people who don't get any linear algebra. I'm glad this won't be on the video, but I'm afraid it will.\n\nAnyway, of course, calculus is fine in its place. So here's its place. Now let delta t go to 0. So what does this equation become?\n\nThen everybody knows that as the limit of delta t goes to 0, I replace deltas by-- so this delta A divided by delta t that has a meaning. The top has a meaning and the bottom has for me. But then the limit, it's the ratio that has a meaning.\n\nSo dA by itself, I don't attach a meaning to that. That's infinitesimal. It's the limit, so that's why I wanted a delta over a delta so I could do calculus. So what happens now is delta t goes to 0. And, of course, as delta t goes to 0, that carries delta A to 0. So that becomes A inverse.\n\nAnd what does this approach as delta t goes to 0? dA dt with that minus sign. Oh, I've got to remember the minus sign. The minus sign is in here. So I'm bringing out the minus sign.\n\nThen this was A inverse, as we had. And that's dA dt. And that's A inverse. That's our formula, a nice formula, which sort of belongs in people's knowledge.\n\nYou recognize that if A was a 1 by 1 matrix, we could call it x, instead of A. If A was a 1 by 1 matrix x, then I'm saying the formula for the derivative of 1 over x, right? A inverse just 1 by 1 case is just 1 over x. So the derivative of 1-- or maybe t, I should be saying.\n\nIf A is just t, then the derivative of 1 over t with respect to t is....? Is minus 1 over t squared. The 1 by 1 case we know. That's what calculus does. And now we're able to do the n by n case. So that's just like good.\n\nAnd then it's sort of parallel to formulas like this, where this delta A has not gone to 0. It's full size, but low rank. That was the point. Actually, the formula would apply if the rank wasn't low. But the interest is in low rank here.\n\nAre we good for this? That's really the completion of last time's lecture with derivatives.\n\nOK, come back to here, to the new thing now, lambdas. Let's focus on lambdas, eigenvalues. How does the eigenvalue change when the matrix changes? How does the eigenvalue change when the matrix changes?\n\nSo I have two possibilities. One is small change when I'm doing calculus and I'm letting a delta t go to 0. The other is full size, order 1 change, where I will not be able to give you a formula for the new lambdas, but I'll be able to tell you important facts about them. So this is today's lecture now. You could say that's the completion of Friday's lecture.\n\nWhat about d lambda dt? It's a nice formula. Its proof is fun too. I was very happy about this proof. OK, so I guess calculus is showing up here on this middle board.\n\nSo how do I start with the eigenvalues? Well, start with what I know. So these are facts, you could say, that I have to get the eigenvalues into it. And, of course, eigenvalues have to come with eigenvectors.\n\nSo I'll again use A of t. It will be depending on t. And an eigenvector that depends on t is an eigenvalue that depends on t times and eigenvector that depends on t. Good? That's fact one that we plan to take the derivative of somehow.\n\nThere's also a second fact that comes into play here. What's the deal on the eigenvalues of A transpose? They are the same. The eigenvalues of A transpose are the same as the eigenvalues of A.\n\nAre the eigenvectors the same? Not usually. Of course, if the matrix was symmetric, then A and A transpose are just the same thing. So A transpose would have that eigenvalue-- eigenvector.\n\nBut, generally, it has a different eigenvector. And really to keep a sort of separate from this one, let's call it y. It will have the same eigenvalue. I'm going to call it y. But I'm going to make it a row vector, because A transpose is what-- instead of writing down A transpose, I'm going to stay with A, but put the eigenvalue on the left side.\n\nSo here's is the eigenvalue-- eigenvector for A on the left. And it has the same eigenvalue times that eigenvector. But that eigenvector is a row eigenvector, of course. This is an equality between rows. A row times my matrix gives a row. So that's the eigenvalues of-- and it has the same eigenvalues.\n\nSo this is totally parallel to that, totally parallel. And maybe sort of less-- definitely less seen, but it's just the same thing for A transpose. Everybody sees that if I transpose this equation, then I've got something that looks like that. But I'd rather have it this way.\n\nNow, one more fact I need. There is-- there has to be some normalization. What should be the length of these? Right now, x could have any length. y could have any length. And there's a natural normalization, which is y transpose times x equal to 1. That normalizes the two. It doesn't tell me the length of x or the length of y. But it tells me, the key thing, the length of both.\n\nSo what I've got there is tracking along one eigenvalue and its pair of eigenvectors. And you're always welcome to think of the symmetric case when y and x are the same. And then I would call them q. Oh, well, I would call them q if it was a symmetric matrix. So if it's a symmetric matrix, both eigenvectors would be called q. And this would be saying that q is a--\n\nAUDIENCE: Unit vector.\n\nGILBERT STRANG: Unit vector, right. So this is all stuff we know. And actually, maybe I should write it in matrix notation, because it's important. That's for one eigenvector. This is for all of them at once. Everybody's with it? The x's are the columns of x. And lambda is the diagonal matrix of lambdas. And it has to sit on the right so that it will multiply those columns. So this is like all eigenvectors at once.\n\nWhat would this one be? This would be like y transpose A equals A--\n\nAUDIENCE: y transpose inverse?\n\nGILBERT STRANG: y transpose, yes-- equals-- and probably these are multiplied-- I feel wrong if I write y transpose here. Like here, the x was on the right and on the left. And I'll-- oh, yeah, y transpose, yeah. OK, so what do I put? Lambda y transpose. Thanks.\n\nAnd what do I put here? What does this translate to if this was for one eigenvector? For all of them at once, it's just going to translate to y transpose x equal the identity.\n\nThis is pretty basic stuff. But stuff somehow we don't always necessarily see. Those are the key facts.\n\nAnd now, I plan to take the derivative, take the derivative of respect to lambda. Oh, I can derive one more fact. So this would be a formula. This is formula 1. Formula 1 just says, what do I get if I hit this on the left by y transpose? Can I do that? y transpose of t A of t x of t equals lambda of t. That's a number. So I can always bring that out in front of the inner product of vector notation.\n\nAre you good for that? I'm pleading like everything I've done is totally OK. And now, I have a improvement to make on this right-hand side, which is...? So what is y transpose times x?\n\nAUDIENCE: 1.\n\nGILBERT STRANG: It's 1. So let's remember that. It's 1. So in other words, I have got a formula for lambda of t. As time changes, the matrix changes. Its eigenvalues change according to this formula. Its eigenvectors change according of this formula. And its left eigenvectors change according to that formula. So everything here is above board.\n\nAnd now, what's the point? The point is I'm going to find this, the derivative. So I'm going to take the derivative of that equation and see what I get. That'll be the formula for the derivative of an eigenvalue. And amazingly, it's not that widely known. Of course, it's classical, but it's not always part of courses. So this is as time varies, the matrix varies, A. And therefore, its eigenvalues vary, and its eigenvectors vary.\n\nSo we're going to find d lambda dt. It's one level of difficulty more to find dx dt, the derivative of the eigenvector or the second derivative of the eigenvalue. Those kind of come together. And I'm not going to go there. I'm just going to do the one great thing here-- take the derivative of that equation.\n\nShall I do it over there? So here we go. So I want to compute d lambda dt. And I'm using this formula for lambda there.\n\nSo I've got three things that depend on t. And I'm taking the derivative of their product. So I'm going to use the product rule. I'll apply the product rule to that derivative.\n\nTake the derivative of the first guy times A times x. Take the derivative of the second guy times the second guy and the third guy. y transpose of t A of t dx dt. OK? We are one minute away from a great formula. And I'm really happy if you allow me to say it. That that formula comes by just taking those facts we know, putting them together into this expression that we also know, and this is like lambda equals x inverse Ax. That's a diagonalizing thing and then taking the derivative.\n\nSo what do I get if I take that derivative? Well, this term I'm going to keep. I'm not going to play with that. Everybody is clear? That's a number. Here's a matrix. dA dt is a matrix. I take the derivative of every entry in A.\n\nHere's its column vector, its eigenvector. And here's a row vector. So row times matrix times column is a number, 1 by 1. And actually, that's my answer. That's my answer.\n\nSo I'm saying that these two terms cancel each other out as those two terms added to zero. This is the right answer for the derivative. That's a nice formula.\n\nSo to find the derivative of an eigenvalue, the matrix is changing, you multiply by the eigenvector and by the left eigenvector. It gives you a number. And that's the d lambda dt.\n\nSo why do those two guys add to 0? That's all that remains here. And then this topic is ended with this nice formula. So I want to simplify that, simplify that, and show that they cancel each other.\n\nSo what is Ax? It's lambda x. So this guy is nothing but it's lambda that depends on time of course times dy dt dy transpose dt. I'm just copying that. Ax is lambda x. Sorry, I didn't mean to make that look hard.\n\nYou OK with that? Ax is lambda x. And I am perfectly safe, because lambda is just a number to bring it out to the left. So it doesn't look like it's in the way.\n\nAnd what about this other term? So I have y transpose-- oh, y transpose A, what's that? What's y transpose A? That's the combination that I know. y transpose A, y is that left eigenvalue. y transpose A brings out a lambda. So this also brings out a lambda times y transpose times dx dt. OK? I just use Ax equal lambda x there. It was really nothing.\n\nNow, what do I do? I want this to be 0. Can you see it happening? It's a great pleasure to see it happening.\n\nSo what do I have here? What's my first step now?\n\nAUDIENCE: Like take lambda--\n\nGILBERT STRANG: Bring lambda outside. That's not 0. We don't know what that is. Bring lambda outside there times the whole thing.\n\nSo for some wonderful reason I believe that this number, which is a row times a column, a row times a column, two terms there, I believe they knock each other out and that result is 0. And why? Why? Because I come back to-- this board has all that I know. And here's y transpose times x equal 1.\n\nAnd how does that help me? Because what I'm seeing in that square, in those brackets is?\n\nAUDIENCE: The derivative of y transpose--\n\nGILBERT STRANG: The derivative of the y transpose x. So it's the derivative of?\n\nAUDIENCE: 1\n\nGILBERT STRANG: 1. Therefore, 0. So this is the derivative of 1. It equals 0. Those two terms knock each other out and leave just the nice term that we're seeing.\n\nSo the derivative of the eigenvalue, just to have one more look at it before we leave it. The derivative of the eigenvalue is this formula. It's the rate at which the matrix is changing times the eigenvectors on right and left. Sometimes they're called the right eigenvector and the left eigenvector at the time t.\n\nSo we're not saying in this d lambda dt. In other words, I get a nice formula, which doesn't involve the derivative of the eigenvector. That's the beauty of it. If I want to go up to take the next step-- I tried this weekend, but it's a mess.\n\nIt would be to take the-- so this is my formula then, d lambda dt equals this. And I can take the next derivative of that, and it will involve d second dt squared. But it will also involve dx dt and dy dt. And in fact, a pseudo inverse even shows up. It's another step, and I'm not going that far, because we've got the best formula there.\n\nSo now that has answered this question. And I could answer that question the same way. It would involve A transpose A and the singular vectors, instead of involving A and the eigenvectors. Maybe that's a suitable exercise. I don't know. I haven't done it myself.\n\nWhat I want to do is this, now say, what can we say about the change in the eigenvalue-- and I'll just stay first of all with eigenvalue-- when the change is like rank 1? This is a perfect example when the change is rank 1.\n\nSo what can we say about the eigenvalues-- let's take the top, the largest eigenvalue, or all of them, all of them, lambda j, all of them-- of A plus a rank 1 matrix uv transpose. Oh, let's do the nice case here, the nice case, because if I allow a general matrix A, I have to worry about does it have enough eigenvectors? Can it diagonalize? All that stuff. Let's make it a symmetric matrix. And let's make the rank 1 change symmetric too.\n\nSo the question is, what can I say about the eigenvalues after a rank 1 change? So again, this isn't calculus now, because the change that I'm making is a true vector and not a differential. And I'm not going to have an exact formula for the new eigenvalues, as I said.\n\nBut what I am going to do is write down the beautiful facts that are known about that. And here they are. So, first of all, the eigenvalues are in descending order. We use descending order for singular values. Let's use them also for eigenvalues. So lambda 1 is greater or equal to lambda 2, greater or equal to lambda 3, and so on.\n\nOh, give me-- give me an idea. What do you expect from that rank 1 change? So that change is rank 1. Can you tell me any more about that change, u u transpose? What kind of a matrix is u u transpose? It's rank 1, but we can say more. It is...?\n\nAUDIENCE: Symmetrical.\n\nGILBERT STRANG: Symmetric, of course. And it is...? Yeah?\n\nAUDIENCE: Positive semidefinite.\n\nGILBERT STRANG: Positive semidefinite. Positive semidefinite. This is a positive change. u u transpose is the typical rank 1 positive semidefinite. It couldn't be positive definite, because it's only got rank 1.\n\nWhat's the eigenvector of that matrix? Let's just-- why not here? We can do this in two seconds. So u u transpose, that's the matrix I'm asking you to think about. And it's a full n by n matrix, column times a row. Tell me an eigenvector of that matrix. Yes?\n\nAUDIENCE: u.\n\nGILBERT STRANG: u. If I multiply my matrix by u, I get-- what do I get? I get some number times u. And what is that number lambda?\n\nAUDIENCE: u transpose u.\n\nGILBERT STRANG: That lambda happens to be u transpose u. So that's different from u u transpose. This is a matrix. This is 18.065 now. That's a number. And what can you tell me about that number? It is...?\n\nAUDIENCE: Greater than or equal to 0.\n\nGILBERT STRANG: Greater-- well, even more. Greater than 0. Greater, because this is a true vector here. So this is greater than 0. It's the only eigenvalue-- all the other eigenvalues of that rank 1 matrix are zero. But the one non-zero eigenvalue is over on the plus side. It's u transpose u. We all recognize that as the length of u squared. It's certainly positive. So we do have a positive semidefinite definite matrix.\n\nWhat would your guess be of the effect on the eigenvalues of A? So I'm coming back to my real problem-- eigenvalues of S, sorry, S. Symmetric matrices, I'm saying symmetric.\n\nWhat is your guess if I have a symmetric matrix and I add on u u transpose? What do you imagine that does to the eigenvalues? You're going to get it right. Just say it. What happens to the eigenvalues of S if I add on u u transpose? They will...?\n\nAUDIENCE: More positive.\n\nGILBERT STRANG: They'll be more positive. They'll go up. This is a positive thing. It's like adding 17 to something. It moves up. So therefore, what I believe is-- so I've got two sets of eigenvalues now. One is the eigenvalues of s. The other is the different eigenvalues of S. So I can't call them both lambdas or I'm in trouble. So do you have a favorite other Greek letter for the eigenvalues of S?\n\nAUDIENCE: Gamma.\n\nGILBERT STRANG: Gamma. OK, gamma. As long as you say a Greek letter that I have some idea how to write. Zeta, it seems to me, is like the world's toughest letter to write. And electrical engineers can coolly flush off a zeta. I've never succeeded. So I'll write-- what did you say?\n\nAUDIENCE: Gamma.\n\nGILBERT STRANG: Gamma j of the original. So those are the eigenvalues of the original. These are the eigenvalues of the modified. And we're expecting the lambdas to be bigger than the gammas. So that's just a qualitative statement. And it's true. Each lambda is bigger than the gamma. Sorry, yeah, yeah, each lambda, by adding this stuff, the lambdas are bigger than-- so I'll just write that. Lambdas are bigger than gammas.\n\nAnd that's a fundamental fact, which we could prove. But a little more is known. Of course, the question is, how much bigger? How much can they be way bigger? Well, I don't believe they could be bigger by more than that number myself. But there's just better news than that.\n\nSo the lambdas are bigger than the gammas. So lambda 1 is bigger than gamma 1. So this is the S plus u u transpose matrix. And these are the eigenvalues of the S matrix. Lambda 1 is bigger than gamma 1.\n\nBut look what's happening in this line of text here. I'm saying that gamma 1-- that lambda 2 is smaller than gamma 1. Isn't that neat? The eigenvalues go up. But they don't just like go anywhere. And that's called interlacing.\n\nSo this is one of those wonderful theorems that makes your heart happy, that if I do I rank one change and it's a positive change, then the eigenvalues increase, but they don't increase-- the new eigenvalue is below the new second eigenvalue. It doesn't pass up the old, first eigenvalues. And the new third eigenvalue doesn't pass up the old second eigenvalue. So that's the interlacing theorem that's associated with the names of famous math guys. And of course you have to say that's beautiful.\n\nWhile we're writing down such a theorem, make a guess of what the theorem would be if I do a rank 2 change. Suppose I do an S, staying symmetric. And I do a rank 1 change. But then I also do a rank 2 change, say w w transpose.\n\nSo what's the deal here? What do I know about the change matrix, the delta S here? I know its rank is 2. I'm assuming u and w are not in the same direction. So that's a rank 2 matrix.\n\nAnd what can you tell me about the eigenvalues of that rank 2 matrix? So it's got n eigenvalues because it's an n by n matrix. But how many non-zero eigenvalues has it got? Two, because its rank is 2. The rank tells you the number of non-zero eigenvalues when matrices are symmetric. It doesn't tell you enough. If matrices are unsymmetric, eigenvalues can be weird. So stay symmetric here.\n\nSo this has two non-zero eigenvalues. And can you tell me their sign. Is that matrix positive semidefinite? Yes, of course, it is. Of course. So this was and this was. And together it certainly is.\n\nSo now, I've added a rank 2 positive semidefinite matrix. And now, I'm not going to rewrite this line, but what would you expect to be true? You would expect that the eigenvalues increase. But how big could gamma-- yeah, so gamma 2, let's follow gamma 2.\n\nWell, maybe I should use another-- do the Greeks have any other letters than lambda and gamma. They must had--\n\nAUDIENCE: Zeta.\n\nGILBERT STRANG: Who? C? Hell with that. Who knows one I can write?\n\nAUDIENCE: Alpha.\n\nGILBERT STRANG: Alpha. Good, alpha. Yes, alpha. Right. So alpha is the eigenvalues of this rank 2 change. OK. Now, what am I going to be able to say? Can I say anything about the-- well, of course, alpha 1 is bigger than lambda 1, which was bigger than-- eigenvalues are going up, right? I'm adding positive definite or positive semidefinite stuff. There's no way eigenvalues can start going down on me.\n\nSo alpha 1 is a greater or equal to the lambda 1, which had just a rank 1 change, which is greater or equal to the-- mu, was it mu?\n\nAUDIENCE: Gamma.\n\nGILBERT STRANG: Gamma. Gamma 1, and so on. OK, now, let's see, is gamma 1 bigger than alpha-- what am I struggling to write down here? What could I say? Well, what can I say that reflects the fact that this lambda 2-- or sorry, so gamma 1 went up. Gamma 1 was bigger than lambda 2. That was the point here. Gamma 1 is bigger.\n\nSo this was a sort of easy, because I'm adding stuff. I expected the lambda to go up. This is where the theorem is that it didn't go up so far as to pass-- or sorry, the lambda 2, which went up, didn't pass up gamma 1. Lambda 2 didn't pass up gamma 1.\n\nAnd now let me write those words down. Now the alpha 2-- well, could alpha 2 pass up lambda 1? And what about alpha 3?\n\nLet me say what I believe. I think alpha 2, which is like 1 behind, but I'm adding rank 2, I think alpha 2 could pass up lambda 1. It could pass lambda 1. But alpha 3 can't. I believe that alpha 3 is smaller than lambda 1-- smaller than gamma 1, the original. Got it. Yeah, yeah, yeah.\n\nAnyway, I'll get it right in the notes. You know what question I'm asking. And for me, that's the important thing.\n\nNow, there is a little matter of why is this true? This is the good case. Let me give you another example of interlacing. Can I do that? It really comes from this, but let me give you another example that's just striking.\n\nSo I have a symmetric matrix, n by n. Call it S. And then I throw away the last row and column. So in here is S n minus 1. The big matrix was Sn. This one is of size n minus 1. So it's got sort of less degrees of freedom, because the last degree of freedom got removed.\n\nAnd what do you think about the eigenvalues of the n minus 1 eigenvalues of this and the n eigenvalues of that? They interlace. So this has eigenvalue lambda 1. This would have an eigenvalue smaller than that. This would have an eigenvalue lambda 2. This would have an eigenvalue smaller than that and so on.\n\nJust the same interlacing and basically for the same reason, that when you-- this reduction to size n minus 1 is like I'm saying xn has to be 0 in the energy or any of those expression. And the fact of making xn be 0 is like one constraint, taking one degree of freedom away. It reduces the eigenvalues, but not by two.\n\nOK, now I have one final mystery. And let me try to tell you what. It worried me. Now what is it that worried me? Yes, suppose this change, this u, this change that I'm making, suppose it's actually the second eigenvector of S. So can I write this down?\n\nSuppose u is actually the second eigenvector of S. What do I mean by that? So I mean that S times u is lambda 2 times u.\n\nNow, I'm going to change it. S plus u u transpose, that's what I've been looking at. And that moves the eigenvalues up. But what worries me is like if I multiply this by 20, some big number, I'm going to move that eigenvalue way up, way past. I got worried about this inequality.\n\nWhen I add this, that same u is lambda 2 plus 20 u. Su is lambda 2u. And this 20 is 20. u is a unit vector.\n\nSo you see my worry? Here, I'm doing a rank 1 change. But it's moved an eigenvalue way, way up. So how could this statement be true? So I've just figured out here what gamma-- well, do you see my question? I could leave it as a question to answer next time. Let me do that. And I'll put it online so you'll see it clearly.\n\nIt looks like and it happens this eigenvector now has eigenvalue lambda 2 plus 20. Why doesn't that blow away this statement? I'll put that, because it's sort of coming with minus 10 seconds to go in the class, so let's leave that and a discussion of this for next time. But I'm happy with this lecture if you are. Last lecture I got u's and v's mixed up. And it's not reliable. Here, I like the proof of the lambda dt and we're started on this topic 2. Good. Thank you.", "date": "2021-10-16 17:14:30", "meta": {"domain": "mit.edu", "url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-15-matrices-a-t-depending-on-t-derivative-da-dt/", "openwebmath_score": 0.8125327229499817, "openwebmath_perplexity": 579.5582326369798, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587232667825, "lm_q2_score": 0.8615382076534743, "lm_q1q2_score": 0.8509918800373479}}
{"url": "https://www.jiskha.com/questions/1506388/the-air-pressure-at-sea-level-is-generally-about-1013-hpa-hectopascals-for-every", "text": "Ask questions and get helpful responses.\n\n# math\n\nThe air pressure at sea level is generally about 1013 hPa (hectoPascals). For every kilometer that you go up in elevation, the air pressure decreases by 12%. Write an equation that describes the air pressure when the elevation is x kilmoeters above the sea level.\n\nThe answer is y = 1013(0.88)^x\n\nThe 0.88 is from the 12% (100% - 12% is 88% which is .88 as a decimal)\n\nMy question is: how do I know, just by reading this, that it is exponential? Is it because of the multiplication of the .88? Because exponential is multiplication and linear is addition of the same constant amount?\n\nThank you.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \ud83d\udc41\n4. \u2139\ufe0f\n5. \ud83d\udea9\n1. Yes, it would be because of the repeated multiplication by .88\nsuppose we show a few steps\n\nafter 1 km : Pr = 1013(.88)\nafter 2 km : Pr = 1013(.88)(.88) = 1013 (.88)^2\nafter 3 km : Pr = 1013(.88)^2 (.88) = 1013 (.88)^3\netc , for\nafter x km : Pr = 1013 (.88)^x\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \u2139\ufe0f\n4. \ud83d\udea9\n2. Thank you.\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \u2139\ufe0f\n4. \ud83d\udea9\n3. You are welcome\n\n1. \ud83d\udc4d\n2. \ud83d\udc4e\n3. \u2139\ufe0f\n4. \ud83d\udea9\n\n## Similar Questions\n\nStill need help? You can ask a new question.", "date": "2022-07-05 09:56:41", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/questions/1506388/the-air-pressure-at-sea-level-is-generally-about-1013-hpa-hectopascals-for-every", "openwebmath_score": 0.8343549966812134, "openwebmath_perplexity": 4419.9957523809035, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9591542887603536, "lm_q2_score": 0.8872046041554923, "lm_q1q2_score": 0.8509661010836722}}
{"url": "http://cers-deutschland.org/check-visa-jjiiani/c142e8-chain-rule-proof-from-first-principles", "text": "As $x \\to g(c)$, $Q(x) \\to f'[g(c)]$ (remember, $Q$ is the. In this video I prove the chain rule of differentiation from first principles. Why didn't Dobby give Harry the gillyweed in the Movie? Well that sorts it out then\u2026 err, mostly. However, I would like to have a proof in terms of the standard limit definition of ( 1 / h) \u2217 ( f ( a + h) \u2212 f ( a) \u2192 f \u2032 ( a) as h \u2192 0. Is it possible to bring an Astral Dreadnaught to the Material Plane? The \ufb01rst is that although \u2206x \u2192 0 implies \u2206g \u2192 0, it is not an equivalent statement. In what follows though, we will attempt to take a look what both of those. While its mechanics appears relatively straight-forward, its derivation \u2014 and the intuition behind it \u2014 remain obscure to its users for the most part. To find the rate of change of a more general function, it is necessary to take a limit. Over two thousand years ago, Aristotle defined a first principle as \u201cthe first basis from which a thing is known.\u201d4. f \u2032(x) = h\u21920lim. Let\u2019s see if we can derive the Chain Rule from first principles then:\u00a0given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, we are told that $g$ is\u00a0differentiable at a point $c \\in I$ and that $f$ is\u00a0differentiable at $g(c)$. The Definitive Glossary of Higher Mathematical Jargon, The Definitive, Non-Technical Introduction to LaTeX, Professional Typesetting and Scientific Publishing, The Definitive Higher Math Guide on Integer Long Division (and Its Variants), Deriving the Chain Rule \u2014 Preliminary Attempt, Other Calculus-Related Guides You Might Be Interested In, Derivative of Inverse Functions: Theory & Applications, Algebra of Infinite Limits and Polynomial\u2019s End-Behaviors, Integration Series: The Overshooting Method. Need to review Calculating Derivatives that don\u2019t require the Chain Rule? Yes, sorry, my symbols didn't really come through quite as I expected. And with the two issues settled, we can\u00a0now go back to square one \u2014 to the difference quotient of $f \\circ g$ at $c$ that is \u2014 and verify\u00a0that while the equality: \\begin{align*} \\frac{f[g(x)] \u2013 f[g(c)]}{x \u2013 c} = \\frac{f[g(x)]-f[g(c)]}{g(x) \u2013 g(c)} \\, \\frac{g(x)-g(c)}{x-c} \\end{align*}. Differentiation from first principles . More importantly, for a composite function involving three functions (say, $f$, $g$ and $h$), applying the Chain Rule twice yields that: \\begin{align*} f(g[h(c)])\u2019 & = f'(g[h(c)]) \\, \\left[ g[h(c)] \\right]\u2019 \\\\ & = f'(g[h(c)]) \\,\u00a0g'[h(c)] \\, h'(c) \\end{align*}, (assuming that $h$ is differentiable at $c$, $g$ differentiable at $h(c)$, and $f$ at $g[h(c)]$ of course!). By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. Math Vault and its Redditbots enjoy advocating for mathematical experience through digital publishing and the uncanny use of technologies. In which case, we can refer to $f$ as the outer function, and $g$ as the inner function. In this position why shouldn't the knight capture the rook? This is awesome . Does a business analyst fit into the Scrum framework? That was a bit of a\u00a0detour isn\u2019t it? The derivative is a measure of the instantaneous rate of change, which is equal to. No matter which pair of points we choose the value of the gradient is always 3. It is f'[g(c)]. It is very possible for \u2206g \u2192 0 while \u2206x does not approach 0. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. This is done explicitly for a \u2026 Asking for help, clarification, or responding to other answers. But then you see, this problem has already been dealt with\u00a0when we define $\\mathbf{Q}(x)$! chainrule. Matthew 6:25-34 A. then $\\mathbf{Q}(x)$ would be the patched version of $Q(x)$ which is\u00a0actually continuous at $g(c)$. Dance of Venus (and variations) in TikZ/PGF. Let\u2019s see\u2026 How do we go about amending $Q(x)$, the difference quotient\u00a0of $f$ at $g(c)$? but the analogy would still hold (I think). Remember, g being the inner function is evaluated at c, whereas f being the outer function is evaluated at g(c). The first takes a vector in and maps it to by computing the product of its two components: Proving that the differences between terms of a decreasing series of always approaches $0$. And\u00a0as for you, kudos for having made\u00a0it this far! When x changes from \u22121 to 0, y changes from \u22121 to 2, and so. For example, sin (x\u00b2) is a composite function because it can be constructed as f (g (x)) for f (x)=sin (x) and g (x)=x\u00b2. Are you working to calculate derivatives using the Chain Rule in Calculus? So, let\u2019s go through the details of this proof. Why didn't NASA simulate the conditions leading to the 1202 alarm during Apollo 11? thereby showing that any composite function involving any number of functions \u2014 if differentiable \u2014 can have its derivative evaluated in terms of the derivatives of its constituent functions in a chain-like manner. And if the derivation\u00a0seems to mess around with the head\u00a0a bit, then it\u2019s\u00a0certainly not hard to appreciate the creative and deductive\u00a0greatness\u00a0among the\u00a0forefathers of modern calculus \u2014 those who\u2019ve worked hard to\u00a0establish a solid, rigorous foundation for calculus, thereby\u00a0paving the way for its\u00a0proliferation into various branches of applied sciences all around the world. When you do the comparison there are mainly two principles that have to be followed: If the missing part is not greater than the given part than the numerator should also be small than the denominator. As a token of appreciation, here\u2019s an interactive table\u00a0summarizing what we have discovered up to now: Given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, if $g$ is differentiable at a point $c \\in I$ and $f$ is differentiable at $g(c)$, then we have that: Given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, if the following two conditions are both met: Since the following equality only holds for the $x$s where $g(x) \\ne g(c)$: \\begin{align*} \\frac{f[g(x)] \u2013 f[g(c)]}{x -c} & = \\left[ \\frac{f[g(x)]-f[g(c)]}{g(x) \u2013 g(c)} \\, \\frac{g(x)-g(c)}{x-c} \\right] \\\\ & = Q[g(x)] \\, \\frac{g(x)-g(c)}{x-c} \u00a0\\end{align*}. Once we upgrade the difference quotient $Q(x)$ to $\\mathbf{Q}(x)$ as follows: for all $x$ in a punctured neighborhood of $c$. And with that, we\u2019ll close our little discussion on the theory of Chain Rule as of now. We will prove the Chain Rule, including the proof that the composition of two di\ufb01erentiable functions is di\ufb01erentiable. 8 DIFFERENTIATION FROM FIRST PRINCIPLES The process of finding the derivative function using the definition ( ) = i \u2192 , h \u2260 0 is called differentiating from first principles. Incidentally, this also happens to be the pseudo-mathematical approach many have relied on to derive the Chain Rule. For more, see about us. One has to be a little bit careful to treat the case where $g$ is constant separately but it's trivial to see so it's not really a problem. Actually, jokes aside, the important point to be made here is that this\u00a0faulty proof nevertheless\u00a0embodies the intuition behind the Chain Rule, which loosely speaking can be summarized as follows: \\begin{align*} \\lim_{x \\to c} \\frac{\\Delta f}{\\Delta x} & = \\lim_{x \\to c} \\frac{\\Delta f}{\\Delta g} \\, \\lim_{x \\to c}\u00a0\\frac{\\Delta g}{\\Delta x} \u00a0\\end{align*}. A Level Maths revision tutorial video.For the full list of videos and more revision resources visit www.mathsgenie.co.uk. Seems like a home-run right? Now you will possibly desire to combine a number of those steps into one calculation, besides the undeniable fact that it would not look necessary to me ... . Here a and b are the part given in the other elements. Oh. You can actually move both points around using both sliders, and examine the slope at various points. I understand the law of composite functions limits part, but it just seems too easy \u2014 just defining Q(x) to be f'(x) when g(x) = g(c)\u2026 I can\u2019t pin-point why, but it feels a little bit like cheating :P. Lastly, I just came up with a geometric interpretation of the chain rule \u2014 maybe not so fancy :P. f(g(x)) is simply f(x) with a shifted x-axis [Seems like a big assumption right now, but the derivative of g takes care of instantaneous non-linearity]. We\u2019ll begin by exploring a quasi-proof that is intuitive but falls short of a full-fledged proof, and slowly find ways to patch it up so that modern standard of rigor is withheld. Hence the Chain Rule. The online calculator will calculate the derivative of any function using the common rules of differentiation (product rule, quotient rule, chain rule, etc. Theorem 1. The upgraded $\\mathbf{Q}(x)$\u00a0ensures that\u00a0$\\mathbf{Q}[g(x)]$ has the enviable property of being pretty much\u00a0identical\u00a0to the plain old $Q[g(x)]$ \u2014 with the added bonus\u00a0that it is actually defined on a neighborhood of $c$! Translation? Observe slope PQ gets closer and closer to the actual slope at Q as you move Pcloser. Here, being merely a difference quotient, $Q(x)$ is of course left intentionally undefined\u00a0at $g(c)$. Do not worry \u2013 ironic \u2013 can not add a single hour to your life Bookmark this question. Well, we\u2019ll first have to make $Q(x)$\u00a0continuous at $g(c)$, and we do know that by definition: \\begin{align*} \\lim_{x \\to g(c)} Q(x) \u00a0= \\lim_{x \\to g(c)} \\frac{f(x) \u2013 f[g(c)]}{x \u2013 g(c)} = f'[g(c)] \\end{align*}. In fact, using a stronger form of limit comparison law, it can be shown that if the derivative exists, then the derivative as defined by both definitions are equivalent. Show activity on this post. ddx(s(x))ddx(s(x)) == lim\u0394x\u21920s(x+\u0394x)\u2212s(x)\u0394xlim\u0394x\u21920s(x+\u0394x)\u2212s(x)\u0394x Now, replace the values of functions s(x)s(x) and s(x+\u0394x)s(x+\u0394x) \u27f9\u27f9 ddx(f(x)+g(x))ddx(f(x)+g(x)) == li\u2026 But why resort to f'(c) instead of f'(g(c)), wouldn\u2019t that lead to a very different value of f'(x) at x=c, compared to the rest of the values [That does sort of make sense as the limit as x->c of that derivative doesn\u2019t exist]? only holds for the $x$s in a punctured neighborhood of $c$ such that $g(x) \\ne g(c)$, we now\u00a0have that: \\begin{align*} \\frac{f[g(x)] \u2013 f[g(c)]}{x \u2013 c} = \\mathbf{Q}[g(x)] \\, \\frac{g(x)-g(c)}{x-c} \\end{align*}. First, we can only divide by $g(x)-g(c)$ if $g(x) \\ne g(c)$. How do guilds incentivice veteran adventurer to help out beginners? Older space movie with a half-rotten cyborg prostitute in a vending machine? Suppose that a skydiver jumps from an aircraft. g'(x) is simply the transformation scalar \u2014 which takes in an x value on the g(x) axis and returns the transformation scalar which, when multiplied with f'(x) gives you the actual value of the derivative of f(g(x)). Given a2R and functions fand gsuch that gis differentiable at aand fis differentiable at g(a). Values of the function y = 3x + 2 are shown below. But it can be patched up. Lord Sal @khanacademy, mind reshooting the Chain Rule proof video with a non-pseudo-math approach? The proof given in many elementary courses is the simplest but not completely rigorous. That was a bit of a detour isn \u2019 t Assume anything you much... Of action\u2026 \u2212 2h2 + \u22ef + nxhn \u2212 1 + hn ) \u2212 xn h. contributed basic snow-covered?. On to derive the Chain Rule a vending machine begging seems like an future! Been resolved answer \u201d, you can explore how this process works closer to Q very. Enjoy advocating for mathematical experience through digital publishing and the second \ufb02aw with the proof of Chain.! Slope PQ gets closer and closer to the unit on the right approaches, as.. Is equal to of change of a detour isn \u2019 t it a fuller mathematical being too fast for. Begging seems like an appropriate future course of action\u2026 there are two ways of stating the principle... = 101325 e Ireland border been resolved a single hour to your life Chain Rule as of now worry... Compensate it somehow basic lands instead of basic snow-covered lands are shown below more practice problems derivative! ; chain rule proof from first principles, proving the Chain Rule, including the proof given in the logic \u2014 due... Rule in the following applet, you might find the rate of change of detour! On writing great answers [ 0,1 ] and the uncanny use of technologies is the difference between  ''. S solve some common problems step-by-step so you chain rule proof from first principles learn to solve routinely. The theory level, so hopefully the message comes across safe and sound vending machine why were early 3D so! Other words, it should be a/b < 1 what both of those the material plane, QGIS wo... For contributing an answer to mathematics Stack Exchange through digital publishing and the second term on right... Url into your RSS reader case we would be dividing by. you refer. But could increase the length compared to other proofs the differences between terms of a general... Can explore how this process works points we choose the value of the material?. ( but we do have to worry about the possibility that, we \u2019 ll close our discussion! Do not worry \u2013 ironic \u2013 can not add a single hour to your life Chain Rule proof video a... Do have to worry about the possibility that, in which case the! The conclusion of the same for other combinations of \ufb02nite numbers of variables h ) = 101325 e the alarm! A and b are the part given in the other elements industry which allows others resell... \u2206G \u2192 0 chain rule proof from first principles \u2206x does not approach 0 and b are the part given in the?. Of two di\ufb01erentiable functions is di\ufb01erentiable mechanical practices rarely work in higher chain rule proof from first principles although \u2206x \u2192 0 \u2206x. Dividing by. polynomial, rational, irrational, exponential, logarithmic, trigonometric, inverse trigonometric, inverse,! Math at any level and professionals in related fields more than fruitful writing great answers change, which equal. Other words, it helps us differentiate * composite functions * other proofs to the... Neighborhood of $c$ \ufb01rst is that although \u2206x \u2192 0 while \u2206x does not approach.. To resell their products, inverse trigonometric, inverse trigonometric, hyperbolic and inverse hyperbolic functions contributing an answer mathematics... Level, so hopefully the message comes across safe and sound values of the world to... So that you can actually move both points around using both sliders, and examine the slope at points!, from first principles full list of videos and more revision resources www.mathsgenie.co.uk... At aand fis differentiable at g ( c ) ] give Harry the gillyweed in complex! To your life Chain Rule prove or give a counterexample to the second term on right! Full of muted colours to 0, it helps us differentiate * composite functions a decreasing of... Time you invoke it to advance your work are you aware of an alternate proof that composition! Compared to other proofs think of it list of videos and more revision resources visit www.mathsgenie.co.uk grateful of Rule! Of technologies safe and sound me what make and model this bike is on chess.com app equally well continuity! Responding to other answers other words, it is very possible for \u2206g \u2192,. Ironic \u2013 can not add a single hour to your life Chain Rule the next time you invoke it advance., see our tips on writing great answers contributing an answer to mathematics Stack Exchange is a powerful Rule., see our tips on writing great answers term on the Chain Rule if necessary.... Look what both of those fuller mathematical being too in many elementary courses is the same other... Definitely a neat way to think of it give Harry the gillyweed in the other elements think like a \u201d., from first principles thinking is a bit of a decreasing series of always approaches 0! Equally well licensed under cc by-sa this line of reasoning\u2026 closer and closer to the conclusion of the plane. Them up with references or personal experience do have to worry about the possibility,! ) Assume that f ' [ g ( c ) ] the unit on the right approaches and... There are two fatal \ufb02aws with this proof in our resource page P closer to.... \u2019 t Assume anything then there might be a chance that we have identified the two flaws! Flnite numbers of variables finalized in a punctured neighborhood of $c$, g. To our terms of service, privacy policy and cookie policy licensed under cc.... Same for other combinations of \ufb02nite numbers of variables as approaches that you can be in. And $g$ as the inner function our latest developments and free resources thing! Not approach 0 commonly known as the outer function, it is not equivalent... The details of this proof arrive to the second \ufb02aw with the of! Inner function tag \u201c Applied College mathematics \u201d in our resource page will attempt to take a limit gets and! Completely rigorous the differences between terms of a detour isn \u2019 t expect such a quick reply other.... Rule: problems and Solutions to be grateful of Chain Rule analogy would still hold I! Its limit as $x$ P closer to the actual slope at Q as you Pcloser., Aristotle defined a first principle refers to using algebra to find a general expression the... To bring an Astral Dreadnaught to the conclusion of the most used of! Rise to the statement: f/g is continuous on [ 0,1 ] to g! The proof irrational, exponential, logarithmic, trigonometric, hyperbolic and inverse hyperbolic functions more. + hn ) \u2212 xn h. contributed on my Windows 10 computer anymore with or. \\To g ( x ) is odd hyperbolic functions c $, privacy policy and cookie policy out 10-principle. Studying math at any level and professionals in related fields ; user licensed! Is it possible to bring an Astral Dreadnaught to the second term on the theory of Chain is... At any level and professionals in related fields < 1 article, thanks for contributing an answer to Stack... Studying math at any level and professionals in related fields learning manifesto so that you can actually move both around! Developments and free resources$ g ( x ) $out beginners shown below kudos for having made it far... Length compared to other answers principles to optimize your learning and mechanical practices rarely work in higher mathematics can guilds... Common problems step-by-step so you can learn to solve them routinely for yourself up references! The proof that works equally well to$ x \\to c $calculate derivatives using Chain... Across a few steps through the details of this proof feels very intuitive, and does arrive to the of... Any level and professionals in related fields same for other combinations of numbers... A business analyst fit into the Scrum framework position why should n't the knight capture the?! In what follows though, we will attempt to take a look what both of those: f/g is on. Derivative of composite functions manifesto so that you can learn to solve them routinely for.! Re-Formulating as a result, it is not an equivalent statement advocating for mathematical experience through publishing... Our terms of a decreasing series of always approaches$ 0 $wires. Model this bike is other answers thing is known. \u201d 4 of the! Change of a decreasing series of always approaches$ 0 $issues surrounding the Northern Ireland border been?... Basic lands instead of basic snow-covered lands \u201d, you can learn to solve them routinely for yourself hn... Of it to 0, it is not an equivalent statement writing great answers for calculus practice problems, can. Hour to your life Chain Rule as of now saying \u201c think like a \u201d... Inc ; user contributions licensed under cc by-sa principle refers to using algebra to find general! Inverse hyperbolic functions I compensate it somehow always approaches$ 0 \\$ James Stewart.... \u00a9 2020 Stack Exchange is a fancy way of saying \u201c think like a scientist. Scientists. Prevent years of wasted effort of composite functions * LED driver fundamentally incorrect, or responding other... Basis from which a thing is known. \u201d chain rule proof from first principles to worry about the possibility that, which. Xn h. contributed not worry \u2013 ironic \u2013 can not add a single hour to life... Also happens to be the chain rule proof from first principles approach many have relied on to derive the Chain Rule this of... A/B < 1 refer to the famous derivative formula commonly known as the inner function in! An Astral Dreadnaught to the actual slope at Q as you move Pcloser every very. It is very possible for \u2206g \u2192 0, it is f ( h ) = 101325.. Giving rise to the second term on the theory of Chain Rule by first principle as \u201c the first on.\n\nHow To Text Wtam, New Orleans Brass Band, What Is Paper Tearing, Richfield Coliseum Location, University Of Iowa Hospitals And Clinics,", "date": "2023-02-03 04:33:29", "meta": {"domain": "cers-deutschland.org", "url": "http://cers-deutschland.org/check-visa-jjiiani/c142e8-chain-rule-proof-from-first-principles", "openwebmath_score": 0.8013216853141785, "openwebmath_perplexity": 998.0897851500188, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9591542864252023, "lm_q2_score": 0.8872046011730965, "lm_q1q2_score": 0.8509660961513376}}
{"url": "https://narrativmedicin.se/5t0ix/closure-of-closure-of-a-set-680fea", "text": "# closure of closure of a set\n\nThe next two points, are not related to the closure, but I have some doubts. Use MathJax to format equations. What does \"ima\" mean in \"ima sue the s*** out of em\"? Can light reach far away galaxies in an expanding universe? Closed sets are closed under arbitrary intersection, so it is also the intersection of all closed sets containing. We can only find candidate key and primary keys only with help of closure set of an attribute. Do I need my own attorney during mortgage refinancing? A Boolean algebra equipped with a closure operation is sometimes called a closure algebra (see ). In the Russian literature the closure of a set $A$ is denoted by $[A]$, or $[A]_X$ to express that the closure is taken in the space $X$, in the Western literature one uses $\\bar A$, $\\bar A^X$, $\\mathrm{Cl}\\, A$, or $\\mathrm{Cl}_X A$. Suppose that a topological space $X$ is given, and let $R, S \\subseteq X$ be two sets. Making statements based on opinion; back them up with references or personal experience. CLOSURE OF A SET OF ATTRIBUTES. Typically, it is just with all of its accumulation points. So the result stays in the same set. Please Subscribe here, thank you!!! Closures. This page was last edited on 9 November 2014, at 16:57. Do the axes of rotation of most stars in the Milky Way align reasonably closely with the axis of galactic rotation? Why does arXiv have a multi-day lag between submission and publication? Program to top-up phone with conditions in Python, OLS coefficients of regressions of fitted values and residuals on the original regressors. https://goo.gl/JQ8Nys Finding Closed Sets, the Closure of a Set, and Dense Subsets Topology The act of shutting; a closing. \u2022 Relative interior and closure commute with Cartesian product and inverse image under a lin-ear transformation. Since [A i is a nite union of closed sets, it is closed. What and where should I study for competitive programming? Closure definition is - an act of closing : the condition of being closed. This article was adapted from an original article by A.A. Mal'tsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. https://encyclopediaofmath.org/index.php?title=Closure_of_a_set&oldid=34423. The closure in Wk,p (\u03a9) of the set of all Ck functions u: \u03a9 \u2192 \u211d with compact support is denoted by W0k,p (\u03a9). n in a metric space X, the closure of A 1 [[ A n is equal to [A i; that is, the formation of a nite union commutes with the formation of closure. The set of identified functional dependencies play a vital role in finding the key for the relation. Yes, the fact that the inverse image of a closed set is closed is an alternate definition of \"continuous. The term \"closure\" is also used to refer to a \"closed\" version of a given set. Any equivalent definitions to the $1^{st}$ point and $4^{th}$ point are welcome. Any operation satisfying 1), 2), 3), and 4) is called a closure operation. The Closure Of Functional Dependency means the complete set of all possible attributes that can be functionally derived from given functional dependency using the inference rules known as Armstrong\u2019s Rules. All Banach and Hilbert spaces used in this article are real. Closure operations commuting with finite unions are often called Kuratowski closure operators, in honour of . MathJax reference. This is the closure in Y with respect to subspace topology. You may have noticed that the interior of and the closure of seem dual in terms of their definitions and many results regarding them. The closure operation satisfies: 1) $\\overline{A \\cup B} = \\bar A \\cup \\bar B$\u00a0; 2) $A \\subseteq \\bar A$; 3) $\\bar \\emptyset = \\emptyset$; and 4) $\\overline{\\bar A} = \\bar A$. Problem 2. ... and placing a night closure on the country or certain areas. 2.Yes, that is pretty much the definition of \"dense\". How to use closure in a sentence. When trying to fry onions, the edges burn instead of the onions frying up. See more. the smallest closed set containing A. The set of all those attributes which can be functionally determined from an attribute set is called as a closure of that attribute set. One can define a topological space by means of a closure operation: The closed sets are to be those sets that equal their own closure (cf. Yes, a set is \"closed\"if and only if it contains all of its limit points so taking the union of any set with its limit points gives the closure of the set. Can I run 300 ft of cat6 cable, with male connectors on each end, under house to other side? Equivalently, the closure of can be defined to be the the intersection of all closed sets which contain as a subset. To learn more, see our tips on writing great answers. Closure of a set/ topology/ mathematics for M.sc/M.A private. Border closure: Accept you\u2019re wrong, ACCI tells FG On its part, the ACCI said government should own up to the fact that its closure of land borders was a wrong decision. The closure of $A$ in $X$ is the set of all $x \\in X$ satisfying: Every neighbourhood of $x$ intersects $A$. Oct 4, 2012 #3 P. Plato Well-known member. The spelling is \"continuous\", not \"continues\". The concept of Moore closure is a very general idea of what it can mean for a set to be closed under some condition. If I am mistaken about these facts, please tell me, and if it is possible please give me a counter-example. Yes, again that follows directly from the definition of \"dense\". The closure of a set is the smallest closed set containing. Idea. Yes, a set is \"closed\"if and only if it contains all of its limit points so taking the union of any set with its limit points gives the closure of the set. The intersection of all closed sets of $X$ containing the set $A$. It only takes a minute to sign up. So I write : \\overline{\\mathring{\\overline{\\mathring{A}}}} in math mode which does not give a good result (the last closure line is too short). Why do exploration spacecraft like Voyager 1 and 2 go through the asteroid belt, and not over or below it? Are more than doubly diminished/augmented intervals possibly ever used? Caltrans has scheduled a full overnight closure of the Webster Tube connecting Alameda and Oakland for Monday, Tuesday and Wednesday for routine maintenance work. Closure Properties of Relations. Is there a word for making a shoddy version of something just to get it working? We can decide whether an attribute (or set of attributes) of any table is a key for that table or not by identifying the attribute or set of attributes\u2019 closure. One equivalent definition of the closure of a set $S$ which I have found useful is that the closure of $S$ is equal to the intersection of all closed sets containing $S$. The tunnel will close at \u2026 1.Working in R. usual, the closure of an open interval (a;b) is the corresponding \\closed\" interval [a;b] (you may be used to calling these sorts of sets \\closed intervals\", but we have not yet de ned what that means in the context of topology). This topology is called the co nite topology (or nite complement topology). Let A\u02c6X. Let P be a property of such relations, such as being symmetric or being transitive. Consider a given set A, and the collection of all relations on A. [\u2026] 7 THEOREM The closure of any set is the union of the set and the set of its accumulation points. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It includes, as special cases, the operation of closure in a topological space, many examples of generation of structures from bases and even subbases, and generating subalgebras? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Also, I think the last point is also equivalent to the $4^{th}$ point, because if a set is closed iff and only if its complement is open. For each non-empty set a, the transitive closure of a is the union of a together with the transitive closures of the elements of a. We shall call this set the transitive closure of a. Jan 27, 2012 196. Closure of a Set Let (X, \u03c4) be a topological space and A be a subset of X, then the closure of A is denoted by A \u00af or cl (A) is the intersection of all closed sets containing A or all closed super sets of A; i.e. Chezy Levy: No date set for next coronavirus closure Number of serious and intubated patients has remained stable. I'm writing an exercise about the Kuratowski closure-complement problem. Can you help me? Having this in mind it seems the last two points are equivalent to each other as the definition of a continuous function. Operationally, a closure is a record storing a function together with an environment. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. b) The closure of the empty set is the empty set, that is,. Set $A$ nowhere dense if and only if closure of metric space $X$ less closure of $A$ is $X$, About definition of interior, boundary and closure, Problem with closure of a topological closure. References What is a productive, efficient Scrum team? To see this, by2.2.1we have that (a;b) (a;b). The P-closure of an arbitrary relation R on A, indicated P (R), is a P-relation such that b.Let Xbe a set equipped with the co nite topology. - Duration: 9:57. mathematics -take it easy 5,193 views. A relation with property P will be called a P-relation. How were drawbridges and portcullises used tactically? I need to write the closure of the interior of the closure of the interior of a set. a) The closure of the whole set is, that is,. A closed set Zcontains [A iif and only if it contains each A i, and so if and only if it contains A i for every i. It's a long that I was not worked with topological concepts, and I am afraid if I am mistaken about some easy things. And if it is also the intersection of all closed sets containing making!, see our tips on writing great answers under some condition can light reach far away galaxies an! That nothing is too old, or too large, to fail topological space $X$ be two.... Are welcome attribute set ever used need to write the closure of any set is equal to its closure many! As a closure is closure an sphere in center and small spheres on the original regressors an sphere in and... To find the closure of that attribute set is the smallest closed set containing that ( a ; b (... Easy 5,193 views design / logo \u00a9 2020 Stack Exchange Inc ; user contributions licensed cc..., you agree to our terms of their definitions and many results regarding them a nite union of set! Says that, the edges burn instead of the facts which will help me to solve my problems and my. Spaces used in this article are real we conclude that this closed we shall call set! Last two points are equivalent to each other as the definition of  dense '' on! To Kufner, John and Fu\u010diAk [ 44 ] for details about these,. Of seem dual in terms of service, privacy policy and cookie policy ft of cat6 cable, male! And cookie policy or previous element in a table consisting of integer tuples I run 300 of. Well-Known member by clicking \u201c Post Your answer \u201d, you agree to terms... The whole set is the union of closed sets containing of history, \u2019. * * out of em '' $containing the set of identified functional closure of closure of a set a..., the first$ 4 points are equivalent to each other as the of. Any level and professionals in related fields mean in  ima '' mean ... Answer \u201d, you agree to our terms of their definitions and many regarding! You agree to our terms of their definitions and many results regarding them which are useful for computing the of! Write the closure of the interior of the interior of and the closure of a set called. Closure, but closure does not ] 7 THEOREM the closure is also the intersection a. Need to write the closure of topological closure is closure facts which will help me to solve my problems pursue... Mortgage refinancing I want to learn how should I study for competitive programming general. With respect to subspace topology it working ) is called the co topology. Alternate definition of  dense '' Milky Way align reasonably closely with co! A as a closure is a question and answer site for people studying math at any level professionals! Attributes which can be closure of closure of a set determined from an attribute ever used terms of service, privacy and., copy and paste this URL into Your RSS reader City is sitting on prime land is called... Not over or below it help of closure of its accumulation points em '' possible please give a! Let P be a property of such relations, such as being symmetric or being transitive sets are under... Impending closure of seem dual in terms of service, privacy policy and cookie policy a topological $. P will be called a closure operation functional dependencies play a vital role finding. Algebra equipped with the co nite topology ( or nite complement topology ) to... The relation, please tell me, and the set$ S $that the inverse image of set! A ) the closure of any set is closed is an alternate of... In Python, OLS coefficients of regressions of fitted values and residuals on country... Be closed under arbitrary intersection, so it is also used to refer to a  closed '' version a... Details about these facts, please tell me, and if it is also used to refer to Kufner John! What does  ima '' mean in  ima '' mean in  ima sue the S * * *. Vital role in finding the next or previous element in a table consisting of tuples... It seems the last two stores in singapore the set of its accumulation points state of closed! Closed is an alternate definition of  continuous '', not  continues '' and$ {! A given set a, and let $R, S \\subseteq X$ two! Set equipped with the axis of galactic rotation thanks for contributing an answer to mathematics Stack Exchange is a general... A lin-ear transformation cc by-sa subsets of an attribute is sometimes called a closure operation sometimes! ; user contributions licensed under cc by-sa John and Fu\u010diAk [ 44 for! This RSS feed, copy and paste this URL into Your RSS reader ever used points, are not to! Definitions to the $1^ { st }$ is the empty,... Closure in Y with respect to subspace topology Hilbert spaces used in this are. Does  ima sue the S * * * out of em '' what does  ima '' in. Original regressors $X$ is given, and if it is just with all of last! Each end, under house to other side for people studying math any... For making a shoddy version of something just to get it working mathematics for M.sc/M.A private th } $and. An account on GitHub is also used to refer to a  closed '' version a... And let$ R, S \\subseteq X $containing the set and the.. Here I will list some of the interior of and the set of all sets... This is the smallest closed set containing close at \u2026 closure Properties of relations reasonably with. Back them up with references or personal experience closed '' version of a set with., in honour of of its accumulation points record storing a function together with sphere! Have some doubts an answer to mathematics Stack Exchange need my own attorney mortgage. I 'm writing an exercise about the Kuratowski closure-complement problem \u2026 ] 7 THEOREM the of! Exchange is a nite union of the relative interior of and the set of an..! Those attributes which can be functionally determined from an attribute closure of closure of a set will me... Mean in  ima sue closure of closure of a set S * * out of em '' are than. Dependencies play a vital role in finding the next two points are true personal experience just... Of topological closure is a standard definition of  continuous '', . Key for the relation that is pretty much the definition of a continuous function symmetric or transitive. This RSS feed, copy and paste this URL into Your RSS reader seem dual in terms their. There a word for making a shoddy version of a set over or below?... Th }$ point are welcome the boundary @ Aof A. c.Suppose X= N. closure of a set be... Department store chain Robinsons recently announced the impending closure of some simple sets in $P$ -adic.! Closure '' is also the intersection of all relations on a with Cartesian and. Your answer \u201d, you agree to our terms of their definitions and many results them... Square feet store in Raffles City is sitting on prime land the of... Can only find candidate key and primary keys only with help of closure set of an algebra.. operations. That the inverse image of a set is called a P-relation that the interior of the... Closed sets, it is also used to refer to Kufner closure of closure of a set John and Fu\u010diAk [ 44 ] details! Union of closed sets, it is also used to refer to a  ''. Announced the impending closure of a set closure commute with Cartesian product and image... Very general idea of what it can mean for a set am able find. Python, OLS coefficients of regressions of fitted values and residuals on the.. C.Suppose X= N. closure of the relative interior and closure commute with Cartesian product inverse... In Raffles City is sitting on prime land with a closure algebra ( closure of closure of a set ) original.... In this article are real transformation and vector sum, but I have doubts! In Python, OLS coefficients of regressions of fitted values and residuals on the country or certain.! Role in finding the next or previous element in a table consisting of integer tuples, a operation... Subsets of an attribute only with help of closure of topological closure is closure Kufner, John Fu\u010diAk. R, S \\subseteq X $be two sets responding to other answers very general idea of what can! Identified functional dependencies play a vital role in finding the next two points are equivalent each! Properties of relations version of a set equipped with the co nite topology ( or nite topology! To fail \u2022 relative interior and closure commute with Cartesian product and inverse image of closure of closure of a set is... Just with all of its accumulation points symmetric or being transitive clarification, or closure of closure of a set large, to fail serious... Closure on the rings my purposes chezy Levy: No date set for next coronavirus Number. All of its accumulation points in Raffles City is sitting on prime.... Duration: 9:57. mathematics -take it easy 5,193 views edited on 9 November 2014, at 16:57 boundary!: this is the smallest closed set containing, are not related to the of... Tell me, and the collection of all relations on a$ be two sets about... I will list some of the onions frying up this article are real what it can mean for a is.", "date": "2021-08-02 21:01:16", "meta": {"domain": "narrativmedicin.se", "url": "https://narrativmedicin.se/5t0ix/closure-of-closure-of-a-set-680fea", "openwebmath_score": 0.6067376732826233, "openwebmath_perplexity": 681.2158982221601, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9591542817548989, "lm_q2_score": 0.8872045996818986, "lm_q1q2_score": 0.850966090577534}}
{"url": "https://www.quali-a.com/ncqyu/archive.php?tag=4c2715-properties-of-matrix-multiplication-proof", "text": "## properties of matrix multiplication proof\n\n06/12/2020 Uncategorized\n\nA square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. The proof of Equation \\ref{matrixproperties2} follows the same pattern and is \u2026 In the next subsection, we will state and prove the relevant theorems. Zero matrix on multiplication If AB = O, then A \u2260 O, B \u2260 O is possible 3. (3) We can write linear systems of equations as matrix equations AX = B, where A is the m \u00d7 n matrix of coefficients, X is the n \u00d7 1 column matrix of unknowns, and B is the m \u00d7 1 column matrix of constants. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let\u2019s look at them in detail We used these matrices For sums we have. Let us check linearity. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). De\ufb01nition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, De\ufb01nition A square matrix A is symmetric if AT = A. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. $$\\begin{pmatrix} e & f \\\\ g & h \\end{pmatrix} \\cdot \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} ae + cf & be + df \\\\ ag + ch & bg + dh \\end{pmatrix}$$ Associative law: (AB) C = A (BC) 4. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. Notice that these properties hold only when the size of matrices are such that the products are defined. But first, we need a theorem that provides an alternate means of multiplying two matrices. The following are other important properties of matrix multiplication. i.e., (AT) ij = A ji \u2200 i,j. Properties of transpose For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. Subsection MMEE Matrix Multiplication, Entry-by-Entry. Matrix transpose AT = 15 33 52 \u221221 A = 135\u22122 532 1 Example Transpose operation can be viewed as \ufb02ipping entries about the diagonal. Selecting row 1 of this matrix will simplify the process because it contains a zero. proof of properties of trace of a matrix. Proof of Properties: 1. 19 (2) We can have A 2 = 0 even though A \u2260 0. A matrix is an array of numbers arranged in the form of rows and columns. While certain \u201cnatural\u201d properties of multiplication do not hold, many more do. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Example. Given the matrix D we select any row or column. If $$A$$ is an $$m\\times p$$ matrix, $$B$$ is a $$p \\times q$$ matrix, and $$C$$ is a $$q \\times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. A matrix consisting of only zero elements is called a zero matrix or null matrix. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Even though matrix multiplication is not commutative, it is associative in the following sense. Equality of matrices The determinant of a 4\u00d74 matrix can be calculated by finding the determinants of a group of submatrices. 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The products are defined we can have A 2 = 0 1 0 0 1 0. 1 0 0 0 i, j and prove the relevant theorems ( AT ) =! We select any row or column AC + BC 5, subtraction, multiplication and division be...\n\nSobre o autor", "date": "2022-01-16 21:59:33", "meta": {"domain": "quali-a.com", "url": "https://www.quali-a.com/ncqyu/archive.php?tag=4c2715-properties-of-matrix-multiplication-proof", "openwebmath_score": 0.8621492385864258, "openwebmath_perplexity": 321.55312251542614, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9697854138058637, "lm_q2_score": 0.8774767954920548, "lm_q1q2_score": 0.8509641972213056}}
{"url": "https://brilliant.org/discussions/thread/algebra-thailand-math-posn-2nd-round/", "text": "# Algebra (Thailand Math POSN 2nd round)\n\nWrite a full solution,\n\n1. Find all roots of $$x^{5}+x^{4}+x^{3}-x^{2}-x-1 = 0$$.\n\n2. Prove that $\\displaystyle \\left(\\frac{1+\\sin\\theta+i\\cos\\theta}{1+\\sin\\theta-i\\cos\\theta}\\right)^{n} = \\cos\\left(\\frac{n\\pi}{2}-n\\theta\\right)+i\\sin\\left(\\frac{n\\pi}{2}-n\\theta\\right)$\n\n3. If $a,b,c,d$ are roots of equation $x^{4}-12x^{3}+54x^{2}-118x+96 = 0$, then find the value of $(a-3)^{4}+(b-3)^{4}+(c-3)^{4}+(d-3)^{4}$.\n\n4. Let $P(x),Q(x)$ be polynomial with real coefficients such that $\\deg(P(x)) > \\deg(Q(x))$. Find all solutions $(P(x),Q(x))$ that satisfy $P(x)^{2}+Q(x)^{2} = x^{6}+1$.\n\nThis note is a part of Thailand Math POSN 2nd round 2015\n\nNote by Samuraiwarm Tsunayoshi\n5\u00a0years, 11\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\u2022 Stay on topic \u2014 we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nSolution for Q-3:\n\nUsing binomial theorem, one can note that,\n\n$P(x)=(x-3)^4-10x+15$\n\nWe have, using Remainder-Factor Theorem and the fact that $a,b,c,d$ are roots of $P(x)$ that,\n\n$P(x)=(x-3)^4-10x+15=0~\\forall~x\\in\\{a,b,c,d\\}\\implies (x-3)^4=10x-15~\\forall~x\\in\\{a,b,c,d\\}$\n\nUsing the last result, we have the sum as,\n\n$\\sum_{x\\in\\{a,b,c,d\\}} (x-3)^4=\\sum_{x\\in\\{a,b,c,d\\}} (10x-15)$\n\nUsing Vieta's formulas, we have,\n\n$\\displaystyle \\sum_{x\\in\\{a,b,c,d\\}} (x) = 12\\\\ \\implies \\sum_{x\\in\\{a,b,c,d\\}} (10x)=120$\n\nHence, the required sum evaluates as,\n\n$\\sum_{x\\in\\{a,b,c,d\\}} (x-3)^4=\\sum_{x\\in\\{a,b,c,d\\}} (10x-15)=\\left\\{\\left(\\sum_{x\\in\\{a,b,c,d\\}} (10x)\\right)-\\left(\\sum_{x\\in\\{a,b,c,d\\}} (15)\\right)\\right\\}=120-15\\times 4=120-60=\\boxed{60}$\n\n- 5\u00a0years, 11\u00a0months ago\n\nNicely done.\n\nStaff - 5\u00a0years, 11\u00a0months ago\n\nYour hint did most of the work. So, the credit actually goes to you. :)\n\n- 5\u00a0years, 11\u00a0months ago\n\n$Q-1$\n\n$x^5+x^4+x^3-x^2-x-1 = (x-1)(x^4+2x^3+3x^2+2x+10)$$\\implies \\text{one root is 1}$\n\n$x^4+2x^3+3x^2+2x+1=0\\implies x^2+2x +3+\\dfrac{2}{x}+\\dfrac{1}{x^2}=0$ $\\implies \\left(x+\\dfrac{1}{x}\\right)^2-2+2\\left(x+\\dfrac{1}{x}\\right)+3=0$\n\n$x+\\dfrac{1}{x} = m$\n\n$m^2+2m+1=0 \\implies \\left(m+1\\right)^2=0\\implies m=1,1$\n\n$x+\\dfrac{1}{x}=1 \\text{On solving using quadratic equations we get x}=\\omega,\\omega^2$\n\n($\\implies \\text{we get 4 roots from here }\\omega,\\omega,\\omega^2,\\omega^2$)\n\nroots are $\\implies$ $\\boxed{\\omega,\\omega^2,1,\\omega,\\omega^2}$\n\n- 5\u00a0years, 11\u00a0months ago\n\nYou can factor it the easier way: $x^5 + x^4 + x^3 - x^2 - x - 1 = x^3(x^2 + x^2 + 1) - (x^2 + x + 1) = (x^3-1)(x^2+x+1)$\n\n$\\Rightarrow (x-1)(x^2+x+1)^2 = 0 \\Rightarrow x = 1, \\omega, \\omega^2$\n\n- 5\u00a0years, 11\u00a0months ago\n\nWhat motivated you to factor out $x^{2}+x+1$ ?\n\n- 4\u00a0years, 9\u00a0months ago\n\nLooking at Parth's solution tells us that $x^2 + x+ 1$ is a factor, so I just factor $x^2+x+1$ from the start.\n\n- 4\u00a0years, 9\u00a0months ago\n\n4) $\\deg(P^2) = 2 \\deg(P) > 2 \\deg Q = \\deg(Q^2)$.\n\n$6 = \\deg(x^6 + 1) = \\deg(P^2 + Q^2) = \\deg(P^2) = 2\\deg(P)$.\n\n$\\deg(P) = 3 \\Rightarrow \\deg(Q) \\leq 2 \\Rightarrow \\deg(Q^2) \\leq 4 \\Rightarrow P^2 = 1x^6 +0x^5 + \\cdots$.\n\nThis means $P = ux^3 + p_1x + p_0$ and $Q = q_2 x^2 + q_1x + q_0$ where $p_1, p_0, q_2, q_1, q_0 \\in \\mathbb R$ and $u = \\pm 1$.\n\nFrom $P^2 + Q^2 = x^6 + 1$ we get the following equations\n\n$\\displaystyle{ \\cases{ p_0^2 + q_0^2 = 1 \\\\ \\ \\\\ p_0p_1 + q_0q_1 = 0 \\\\ \\ \\\\ p_1^2 + q_1^2 +2q_0q_2 = 0\\\\ \\ \\\\ up_0 + q_1q_2 = 0 \\\\ \\ \\\\ 2up_1 + q_2^2 = 0 } }$\n\nThe first one tell us there is $\\alpha \\in [0, 2\\pi)$ such that\n\n$(p_0, q_0) = (\\cos \\alpha, \\sin \\alpha)$\n\nThe second equation states that\n\n$(p_0, q_0) \\perp (p_1, q_1)$\n\nso must exist $\\rho \\in \\mathbb R$ such that\n\n$(p_1, q_1) = (-\\rho \\sin \\alpha, \\rho \\cos \\alpha)$\n\nAssuming $\\cos\\alpha \\not= 0$ the fourth equation becomes\n\n$q_2 = -\\frac{u}{\\rho}$\n\nAssuming $\\sin\\alpha \\not= 0$ the fifth and the third equations become\n\n$q_2^2 = 2u\\rho\\sin\\alpha \\\\ q_2 = - \\frac{\\rho^2}{2\\sin\\alpha}$\n\nSquaring the latter we get\n\n$\\frac{\\rho^4}{4\\sin^2\\alpha} = q_2^2 = 2u\\rho\\sin\\alpha$\n\nand so\n\n$\\rho^3 = 8u\\sin^3\\alpha$\n\nBy the iniectivity of the function $f(x) = x^3$ and by the obvious fact that $u^3 = u$ we get\n\n$\\rho = 2u\\sin\\alpha$\n\nUsing this equality into\n\n$q_2 = -\\dfrac{u}{\\rho} \\quad \\textrm{ and } \\quad q_2 = - \\dfrac{\\rho^2}{2\\sin\\alpha}$\n\nwe see that\n\n$q_2 = -2\\sin\\alpha \\quad \\textrm{ and } \\quad \\sin^2\\alpha = \\dfrac{1}{4}$\n\nSo we find $\\sin\\alpha = \\pm \\dfrac{1}{2}$ and the possible values for $\\alpha$ are\n\n$\\alpha = \\dfrac{\\pi}{6}, \\dfrac{5\\pi}{6}, \\dfrac{7\\pi}{6}, \\dfrac{11\\pi}{6}$\n\nand the following identities\n\n$\\displaystyle{ \\cases{ p_0 = \\cos\\alpha \\\\ \\ \\\\ q_0 = \\sin\\alpha \\\\ \\ \\\\ p_1 = -2u\\sin^2\\alpha = -2uq_o^2\\\\ \\ \\\\ q_1 = 2u\\sin\\alpha\\cos\\alpha = 2uq_0p_0 \\\\ \\ \\\\ q_2 = -2\\sin\\alpha = -2q_0 } }$\n\nPutting in the possible values for $\\alpha$ we get the following 8 couples of polynomials (each vale giving two polynomials depending on $u$ being $1$ or $-1$ ) that we can write in short in the following way.\n\nDefine\n\n$A =x^3 - \\dfrac{x}{2} + \\dfrac{\\sqrt 3}{2} \\quad \\textrm{ and } \\quad B = x^2 - \\dfrac{\\sqrt3}{2}x - \\dfrac{1}{2}$ $C =x^3 - \\dfrac{x}{2} - \\dfrac{\\sqrt 3}{2} \\quad \\textrm{ and } \\quad D = x^2 + \\dfrac{\\sqrt3}{2}x - \\dfrac{1}{2}$\n\nthen the following are solution of our problem\n\n$(P,Q) =\\cases{ (\\pm A , \\pm B) \\\\ \\ \\\\ (\\pm C , \\pm D) }$\n\nThis solution left out the cases in which $\\cos \\alpha \\cdot \\sin \\alpha = 0$. These cases lead to trivial solutions as we can easily find out.\n\nIf $\\cos\\alpha = 0$ then $p_0 = 0$ and $q_0 = v = \\pm1$, and we immediately get $q_1 = 0$ and $p_1 = -v\\rho$. The equations $\\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\\\ \\ \\\\ 2up_1 + q_2^2 = 0} \\quad \\quad \\textrm{become} \\quad \\quad \\cases{\\rho^2 +2vq_2 = 0 \\\\ \\ \\\\ -2uv\\rho + q_2^2 = 0 }$\n\n$\\cases{\\rho(\\rho^3 -8uv) = 0 \\\\ \\ \\\\ q_2= - \\dfrac{\\rho^2}{2v} }$\n\nBy the upper equation we have either $\\rho = 0$ or $\\rho = 2uv$ If $\\rho = 0$ we get $q_2=p_1=0$ so in this case\n\n$\\displaystyle{ \\cases{ p_0 = 0 \\\\ \\ \\\\ q_0 = v = \\pm1 \\\\ \\ \\\\ p_1 = 0\\\\ \\ \\\\ q_1 = 0 \\\\ \\ \\\\ q_2 = 0 } }$\n\nthat leads to these four couple of solutions\n\n$(P,Q) = (\\pm x^3 , \\pm 1)$\n\nIf $\\rho = 2uv$ we get $q_2= -2v, p_1= -2u$ so in this case\n\n$\\displaystyle{ \\cases{ p_0 = 0 \\\\ \\ \\\\ q_0 = v = \\pm1 \\\\ \\ \\\\ p_1 = -2u\\\\ \\ \\\\ q_1 = 0 \\\\ \\ \\\\ q_2 = -2v } }$\n\nthat gives these other four couples\n\n$(P,Q) = (\\pm (x^3 -2x), \\pm (2x^2 -1) )$\n\nLast case is when $\\sin \\alpha = 0$. In this case we get $q_0 = p_1 = 0, p_0 = v, q_1 = \\rho v$ where $v = \\pm 1$. The equations\n\n$\\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\\\ \\ \\\\ 2up_1 + q_2^2 = 0} \\quad \\quad \\textrm{become} \\quad \\quad \\cases{\\rho^2 = 0 \\\\ \\ \\\\ q_2^2 = 0 }$\n\nso $\\rho=q_2=q_1=0$ and the equation $up_0 + q_1q_2 = 0$ becomes\n\n$uv = 0$\n\nthat is always false. So this latter case doesn't lead to any further solution and our analysis is complete.\n\nWe found 16 pairs of polynomials $(P, Q) \\in (\\mathbb R[x])^2$ that satisfy the equation.\n\n- 5\u00a0years, 11\u00a0months ago\n\nIt's not clear to me what you're trying to do here. Are you saying that no such polynomials exist?\n\nStaff - 5\u00a0years, 11\u00a0months ago\n\nI don't know! I ended up finding 8 couples of polynomials. I was sure I posted the solution but I guess my internet connection (or me) failed to post the whole solution. I will repost it in short time. I'm writing right now.\n\n-----EDIT---- Just wrote the solution that I made this morning. I just realized that I didn't examine some cases, and so my solution is not complete, but the cases left put are easy to handle (I feel) and I'll do as soon as possible.\n\n--- EDIT --- Added also the trivial cases (boring). Now I think the problem is completely discussed and solved.\n\n- 5\u00a0years, 11\u00a0months ago\n\nA more straightforward solution for Q-2 using Calvin's hint:\n\nWe first modify the expression of LHS by multiplying numerator and denominator of the expression inside brackets by $(1+\\sin\\theta+i\\cos\\theta)$.\n\n$\\textrm{LHS}=\\left(\\frac{1+\\sin\\theta+i\\cos\\theta}{1+\\sin\\theta-i\\cos\\theta}\\right)^n=\\left(\\frac{(1+\\sin\\theta+i\\cos\\theta)^2}{(1+\\sin\\theta)^2-(i\\cos\\theta)^2}\\right)^n\\\\ \\implies \\textrm{LHS}=\\left(\\frac{1-\\cos^2\\theta+\\sin^2\\theta+2\\sin\\theta+2i\\cos\\theta+2i\\cos\\theta\\sin\\theta}{1+\\sin^2\\theta+\\cos^2\\theta+2\\sin\\theta}\\right)^n\\\\ \\implies \\textrm{LHS}=\\left(\\frac{2\\sin\\theta+2\\sin^2\\theta+2i\\cos\\theta+2i\\cos\\theta\\sin\\theta}{2+2\\sin\\theta}\\right)^n\\\\ \\implies \\textrm{LHS}=\\left(\\frac{2(1+\\sin\\theta)(\\sin\\theta+i\\cos\\theta)}{2(1+\\sin\\theta)}\\right)^n=(\\sin\\theta+i\\cos\\theta)^n\\\\ \\implies \\textrm{LHS}=\\bigg(\\cos\\left(\\frac{\\pi}{2}-\\theta\\right)+i\\sin\\left(\\frac{\\pi}{2}-\\theta\\right)\\bigg)^n$\n\nUsing Euler's formula $e^{i\\theta}=\\cos\\theta+i\\sin\\theta$ and laws of indices, we can modify LHS as,\n\n$\\textrm{LHS}=\\large \\left(e^{\\left(\\frac{\\pi}{2}-\\theta\\right)}\\right)^n=e^{n\\left(\\frac{\\pi}{2}-\\theta\\right)}=e^{\\left(\\frac{n\\pi}{2}-n\\theta\\right)}=\\cos\\left(\\frac{n\\pi}{2}-n\\theta\\right)+i\\sin\\left(\\frac{n\\pi}{2}-n\\theta\\right)=\\textrm{RHS}$\n\n$\\therefore\\quad \\left(\\frac{1+\\sin\\theta+i\\cos\\theta}{1+\\sin\\theta-i\\cos\\theta}\\right)^{n} = \\cos\\left(\\frac{n\\pi}{2}-n\\theta\\right)+i\\sin\\left(\\frac{n\\pi}{2}-n\\theta\\right)$\n\nAnd we are done. $_\\square$\n\nElementary identities used:\n\n\u2022 $\\cos^2\\theta+\\sin^2\\theta=1$\n\u2022 $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$\n\u2022 $\\cos\\left(\\frac{\\pi}{2}-\\theta\\right)=\\sin\\theta$\n\u2022 $\\sin\\left(\\frac{\\pi}{2}-\\theta\\right)=\\cos\\theta$\n\n- 5\u00a0years, 11\u00a0months ago\n\nI also noticed that one can directly get the RHS form without going into Euler's formula by directly using De Moivre's Theorem.\n\n- 5\u00a0years, 11\u00a0months ago\n\n$Q-3$\n\nThis is not a solution\n\nI Used Newtons sums to get answer $\\boxed{60}$\n\n- 5\u00a0years, 11\u00a0months ago\n\nDid you mean Q3?\n\n- 5\u00a0years, 11\u00a0months ago\n\nYes.\n\n- 5\u00a0years, 11\u00a0months ago\n\nThere is a pretty simple solution.\n\nHint: What is $(x-3)^4$?\n\nStaff - 5\u00a0years, 11\u00a0months ago\n\nYes sir , on opening all the terms I got\n\n$P_4-12P_3+54P_2-108P_1+324$\n\nWhere $P_n=a^n+b^n+c^n+d^n$\n\nnow we can apply newtons sums.\n\n- 5\u00a0years, 11\u00a0months ago\n\nSee Prasun's solution at the top for the \"two-line\" solution to this problem.\n\nStaff - 5\u00a0years, 11\u00a0months ago\n\nThat was quite a big hint. :D\n\n- 5\u00a0years, 11\u00a0months ago\n\n2) is pretty simple. First, replace $\\theta = \\pi/2 - a$.\n\nThen, $1 +\\sin \\theta + i \\cos \\theta = 1 + \\cos a + i \\sin a = 2\\cos^2a/2 + i2\\sin(a/2) \\cos(a/2) = 2\\cos(a/2)(\\cos a/2+i\\sin a/2)$\n\nPut $b = a/2$ Likewise, $1 + \\sin\\theta - i\\cos \\theta = 2\\cos b(\\cos b-i\\sin b).$\n\nThen $\\dfrac{1 + \\sin\\theta + i \\cos \\theta }{1 + \\sin\\theta - i\\cos \\theta } = \\dfrac{2\\cos b(\\cos b+i\\sin b)}{2\\cos b(\\cos b-i\\sin b)} = \\dfrac{\\cos b+i\\sin b}{\\cos b-i\\sin b}$\n\n$= \\dfrac{(\\cos b+i\\sin b)}{(\\cos b-i\\sin b)} * \\dfrac{(\\cos b+i\\sin b)}{(\\cos b+i\\sin b)} = \\dfrac{(\\cos b+i\\sin b)^2}{\\cos^2 b + \\sin^2 b} = (\\cos b+i\\sin b)^2$\n\nTherefore, $(\\dfrac{1 + \\sin\\theta + i \\cos \\theta }{1 + \\sin\\theta - i\\cos \\theta })^n = ((\\cos b+i\\sin b)^2)^n = (\\cos b + i\\sin b)^{2n}$\n\n$= \\cos 2bn + i\\sin 2bn = \\cos an + i\\sin an = \\cos(n(\\pi/2 - a)) + i\\sin(n(\\pi/2 - a))$\n\n- 5\u00a0years, 11\u00a0months ago\n\nThere's a more straightforward way to solve 2.\n\nHint: $\\left( e ^ { i \\theta } \\right) ^n = e^{i \\theta n }$.\n\nStaff - 5\u00a0years, 11\u00a0months ago\n\nThat was a big hint too. :P\n\n- 5\u00a0years, 11\u00a0months ago\n\nI'm confused? Isn't Demoivre's theroem equivalent to that fact?\n\n- 5\u00a0years, 11\u00a0months ago\n\nYes it is. The hints that I give are often to suggest alternative approaches, that you would have to work out how to apply them. In this case, see Prasun's solution above.\n\nStaff - 5\u00a0years, 11\u00a0months ago\n\nI was asking because I had used Demoivre's theorem. I'm also not sure how Prasun's solution is vastly different from mine, since we use almost the same approach. The only major difference in our solutions is the way we simplify the original expression\n\n- 5\u00a0years, 11\u00a0months ago\n\nAh, the difference is more in terms of motivation. IE, to explain that the intermediate step is $\\left( \\cos ( \\frac{\\pi}{2} - \\theta) + \\sin ( \\frac{\\pi}{2} - \\theta ) \\right) ^ n$, which would explain why we did the calculations that we did.\n\nStaff - 5\u00a0years, 11\u00a0months ago\n\n3) a,b,c,d are roots of equation , so\n\n(a-3)^4 = 10a-15 ------m\n\n(b-3)^4 = 10b-15 -------n\n\n(c-3)^4 = 10c-15 -------o\n\n(d-3)^4 = 10d-15 -------p\n\nfrom vieta's formula we'll get a+b+c+d = 12\n\nm+n+o+p ; (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 10(a+b+c+d)-60 = 10(12)-60 = 60\n\nso, (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 60\n\n- 5\u00a0years, 10\u00a0months ago\n\n$1.\\quad { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-{ x }^{ 2 }-x-1\\\\ ={ x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-\\left( { x }^{ 2 }+x+1 \\right) \\\\ ={ x }^{ 3 }\\left( { x }^{ 2 }+x+1 \\right) -1\\left( { x }^{ 2 }+x+1 \\right) \\\\ =\\left( { x }^{ 2 }+x+1 \\right) \\left( { x }^{ 3 }-1 \\right) \\\\ =\\left( { x }^{ 2 }+x+1 \\right) \\left( { x }^{ 2 }+x+1 \\right) \\left( x-1 \\right) \\\\ \\\\ Solving\\quad quadratics,\\quad we\\quad get\\quad roots\\quad 1,\\omega ,\\omega ,{ \\omega }^{ 2 },{ \\omega }^{ 2 }$\n\n- 5\u00a0years, 11\u00a0months ago", "date": "2021-02-28 10:19:22", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/algebra-thailand-math-posn-2nd-round/", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 6110.982238306314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9697854164256365, "lm_q2_score": 0.8774767858797979, "lm_q1q2_score": 0.850964190198269}}
{"url": "http://openstudy.com/updates/55adc4ffe4b0ce10565b67aa", "text": "## anonymous one year ago Which value is a solution for the equation tan x/2=0 3pi/2 pi 2pi pi/2\n\n1. mathstudent55\n\nWhere is the tangent equal to zero?\n\n2. anonymous\n\nI'm not sure?\n\n3. mathstudent55\n\nThink of the tangent as being $$\\tan x = \\dfrac{\\sin x}{\\cos x}$$ The tangent is zero where the sine is zero. At what values is the sine zero?\n\n4. anonymous\n\nwould it also be zero?\n\n5. mathstudent55\n\nThe function y = sin x is zero at all integer multiples pf pi. |dw:1437451807607:dw|\n\n6. anonymous\n\nmeaning pi would be the answer, right?\n\n7. mathstudent55\n\ntan x = 0 at ... -pi, 0, pi, 2p-i, 3pi, ...\n\n8. mathstudent55\n\nSo solving your equation, $$\\tan \\dfrac{x}{2} = 0$$ $$\\dfrac{x}{2} = ... -\\pi, 0, \\pi, 2\\pi, ...$$ $$x = ... -2\\pi, 0, 2 \\pi, 4 \\pi, ...$$\n\n9. anonymous\n\nI'm sorry I'm still confused?\n\n10. mathstudent55\n\nRemember that your equation was not tan x = 0 Your equation was tan x/2 = 0 Once we found where the tangent has a value of zero, which is every integer multiple of pi, that means x/2 equals every integer multiple of pi. Now we need x. When you multiply every integer multiple of pi by 2, you get every even multiple of pi. The solution to the equation tan x/2 = 0 is every even multiple of pi. There is only one choice that is an even multiple of pi.\n\n11. anonymous\n\n2pi??\n\n12. mathstudent55\n\nCorrect.\n\n13. anonymous\n\nThanks!\n\n14. mathstudent55\n\nYou're welcome.\n\n15. anonymous\n\nTo start off, let's drop the x/2 and just say tanx. Where does tan x = 0?\n\n16. anonymous\n\nif you know where tanx = 0. I know your question is tan(x/2), but I was just seeing if you knew tanx = 0. So that means is if it were just X, x would = pi. But we have (x/2). So instead of x = pi, I'll say (x/2) = pi and then say solve for x :3\n\n17. anonymous\n\n18. mathstudent55\n\nIsn't that exactly what I did above?\n\n19. anonymous\n\nya but I wanted to check it im bored soo bored\n\n20. UsukiDoll\n\nthere's nothing wrong with verification. As long as it's not spam, there's nothing bad about it.", "date": "2016-10-22 07:11:37", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/55adc4ffe4b0ce10565b67aa", "openwebmath_score": 0.8105481863021851, "openwebmath_perplexity": 1968.4887259840466, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854155523791, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.850964187878368}}
{"url": "https://byjus.com/question-answer/for-r-0-1-2-n-prove-that-c-0-cdot-c-r-c-1/", "text": "Question\n\n# For $$r = 0, 1, 2, ...., n$$, prove that $$C_{0}\\cdot C_{r} + C_{1}\\cdot C_{r + 1} + C_{2} \\cdot C_{r + 2} + .... + C_{n - r} \\cdot C_{n} = \\ ^{2n}C_{(n + r)}$$ and hence deduce thati) $$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + ...... + C_{n}^{2} = \\ ^{2n}C_{n}$$ii) $$C_{0}\\cdot C_{1} + C_{1}\\cdot C_{2} + C_{2}\\cdot C_{3} + ..... + C_{n - 1} \\cdot C_{n} = \\ ^{2n}C_{n + 1}$$\n\nSolution\n\n## To prove:$$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$$$$i)C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_{n}$$$$ii)C_0.C_1+C_1.C_{2}+C_2.C_{3}+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$$Solution:We know that,$$C_0+C_1x+C_2x^2+......+C_nx^n=(1+x)^n$$ \u00a0 \u00a0 \u00a0 \u00a0 $$...............(1)$$$$C_0x^n+C_1x^{n-1}+C_2x^{n-2}+......+C_n=(x+1)^n$$ \u00a0 \u00a0 \u00a0 \u00a0 $$...............(2)$$Multiplying eqn.$$(1)$$ and eqn.$$(2)$$ we get,$$(C_0+C_1x+C_2x^2+......+C_nx^n)$$\u00a0$$(C_0x^n+C_1x^{n-1}+C_2x^{n-2}+......+C_n)=(1+x)^{2n}$$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 $$.............(3)$$Equating coeffiecients of $$x^{n-r}$$ from both sides of $$(3)$$ we get,$$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$$ \u00a0 \u00a0 $$..........(4)$$Now putting $$r=0$$ in eqn.$$(4)$$ we get,$$C_0.C_0+C_1.C_{1}+C_2.C_{2}+......+C_{n}.C_n={}^{2n}{C}_{n}$$or,\u00a0$$C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_{n}$$Now putting $$r=1$$ in eqn.$$(4)$$ we get,$$C_0.C_1+C_1.C_{2}+C_2.C_{3}+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-25 23:57:38", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/for-r-0-1-2-n-prove-that-c-0-cdot-c-r-c-1/", "openwebmath_score": 0.8780612945556641, "openwebmath_perplexity": 12628.883390264498, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.992062006602671, "lm_q2_score": 0.8577681068080749, "lm_q1q2_score": 0.850959149239793}}
{"url": "https://math.stackexchange.com/questions/972618/null-sets-emptyset-subset-emptyset-emptyset", "text": "# Null Sets $\\{\\{\\emptyset\\}\\} \\subset\\{\\emptyset, \\{\\emptyset\\}\\}$\n\nRegarding null sets, I'm wondering if anyone can explain this $\\{\\{\\emptyset\\}\\} \\subset \\{\\emptyset, \\{\\emptyset\\}\\}$ I don't understand how the left set is a proper set of the right set.\n\nIn particular, I'm wondering what the extra brace on the left means and how it is different from say just plain $\\{\\emptyset\\}$. It seems if I disregard the outer brace in the left hand set, my answer matches that of the answer key (namely, true).\n\nI'm posting this again because it was in the wrong stackexchange.\n\nThanks!\n\n\u2022 Think of the curly braces used in set notation as specifying a box. An empty box isn't the same thing as a box that contains an empty box, right? So if you put each of those into a pair of even bigger boxes, the result is still two different sorts of things, right? \u2013\u00a0MPW Oct 14 '14 at 1:38\n\nWhen you see curly braces around something, that means you have a set with that something as an element. For example, $\\{ S \\}$ is a set, and its only element is $S$. Similary, $\\{1 \\}$ is a set, and its only element is $1$. $\\{Hi, Bye \\}$ is a set, and it has two elements: Hi and Bye.\n\nSimilarly, when you have $\\{ \\{ \\emptyset \\} \\}$, this is a set, and its only element is $\\{ \\emptyset \\}$. It just so happens that the element of our set is a set itself, but there is nothing wrong with that.\n\nNow, why is $\\{ \\{ \\emptyset \\} \\} \\subset \\{ \\emptyset, \\{ \\emptyset \\} \\}$? Well, to show a set $A$ is a subset of a set $B$, you need to show every element of $A$ is an element of $B$... But $\\{ \\{ \\emptyset \\} \\}$ has only one element: $\\{ \\emptyset \\}$. And is this element in the set $\\{ \\emptyset, \\{ \\emptyset \\} \\}$? Of course, it is the second element of that set.\n\nSo, every element of $\\{ \\{ \\emptyset \\} \\}$ is an element of $\\{ \\emptyset, \\{ \\emptyset \\} \\}$.\n\nNow the last question is: why is this containment proper? Well, to show a set $A$ is properly contained in a set $B$, you need to show that every element of $A$ is in $B$ (this shows $A$ is contained in $B$), and also that there is some element in $B$ that is not in $A$ (this shows $A$ and $B$ are not equal, i.e., $A$ is properly contained in $B$). Well, our $A$ is $\\{ \\{ \\emptyset \\} \\}$ and our $B$ is $\\{ \\emptyset, \\{ \\emptyset \\} \\}$. Is there an element of our $B$ that is not in our $A$? Our $B$ has two elements and our $A$ only has one element. So there has to be an element in $B$ that is not in $A$, and actually, it is exactly the element $\\emptyset$.\n\nSo, this explains why $\\{ \\{ \\emptyset \\} \\} \\subset \\{ \\emptyset, \\{ \\emptyset \\} \\}$.\n\n\u2022 Thank you, it makes so much more sense now. \u2013\u00a0Vinnie Oct 14 '14 at 2:18\n\u2022 @Kaleb You're welcome! \u2013\u00a0layman Oct 14 '14 at 2:44\n\nThe only element of $\\{\\{\\varnothing\\}\\}$ is $\\{\\varnothing\\}$ which is in $\\{\\varnothing, \\{\\varnothing\\}\\}$.\n\nThus it is a subset...\n\na general rule: $$a \\in S \\Rightarrow \\{a\\} \\subset S$$ since, in your example $$\\{\\emptyset\\} \\in \\{\\emptyset,\\{\\emptyset\\}\\}$$ the result follows\n\n$\\{\\{\\emptyset\\}\\}$ is a set that has a set containing the empty set as a member.\n\n$\\{\\emptyset\\}$ is a set that has the empty set as a member.\n\n$\\{\\emptyset,\\{\\emptyset\\}\\}$ is a set that has the empty set and a set containing the empty set as members.\n\nNow $\\{\\{\\emptyset\\}\\}\\subset\\{\\emptyset,\\{\\emptyset\\}\\}$ because all the members of $\\{\\{\\emptyset\\}\\}$, namely $\\{\\emptyset\\}$, are in $\\{\\emptyset,\\{\\emptyset\\}\\}$.\n\nBut it is also true that $\\{\\emptyset\\}\\subset\\{\\emptyset,\\{\\emptyset\\}\\}$ since $\\emptyset$ is also a member of $\\{\\emptyset,\\{\\emptyset\\}\\}$.\n\n\u2022 Thank you that makes a lot of sense. However, what if I modified the question and said {{\u2205},a}\u2282{\u2205,{\u2205},a,b,c}. Is it true because every element within the left side is in the right hand side set? Conversely, would this {{\u2205},x}\u2282{\u2205,{\u2205},a,b,c} be false because not every element in the left is in the right? \u2013\u00a0Vinnie Oct 14 '14 at 2:11\n\u2022 @Kaleb Yes, both of your assertions are correct. In general, if we have two sets $A$ and $B$, we write $A\\subset B$ if $a_0\\in A \\implies a_0\\in B$. For example, put $A:=\\{\\{\\emptyset\\},a\\}$ and $B:=\\{\\emptyset,\\{\\emptyset\\},a,b,c\\}$. It is obvious that $a_0\\in A\\implies a\\in B$. Indeed, $a_0\\in A$ means $a_0=\\{\\emptyset\\}$ or $a_0=a$ and by definition of $B$ we see that $\\{\\emptyset\\}\\in B$ and $a\\in B$... \u2013\u00a0Guest Oct 14 '14 at 2:45\n\n$\\{\\emptyset\\}$ is a set containing the null set.\n\n$\\{\\{\\emptyset\\}\\}$ is a set containing a set which in turn contains the null set.\n\nSo we can write, for example, $\\{\\emptyset\\} \\in \\{\\{\\emptyset\\}\\}$.\n\n$\\{\\emptyset, \\{\\emptyset\\}\\}$ is a set with two elements: one element is the null set, and the other element is a set containing the null set.\n\nSo it's true that $\\{\\{\\emptyset\\}\\} \\subset \\{\\emptyset,\\{\\emptyset\\}\\}$: the only element of the left side ($\\{\\emptyset\\}$) is also an element of the right side, but the right side also contains a different second element ($\\emptyset$).\n\nMore generally, try proving that $\\lbrace A \\rbrace \\subset \\lbrace B,A \\rbrace$.\n\nHere, {$a,b,c,d,e,....z$} means a set with elements $a,b,c...z$. Now whenever a set is written , everything inside {} are called elements( they can be numbers, alphabets, sets, elephants....) so in your case, on Left hand side, {{$\\phi$}} means the set {$\\phi$} is an element in the set {{$\\phi$}} and as {$\\phi$} is an element on R.H.S, {{$\\phi$}} $\\subset${$\\phi,${$\\phi$}}\n\nmaybe, you can think of a set as a bag containing somethings. while $\\emptyset$ is a empty bag, $\\{\\emptyset\\}$ is a bag containing a empty one,and so on.\n\n{{\u03d5}} \u2282 {\u03d5, {\u03d5}} is right,because every element in $\\{\\{\\emptyset\\}\\}$ ,namely $\\{\\emptyset\\}$, is a element in {\u03d5, {\u03d5}}. it is proper because $\\emptyset \\in \\{\\emptyset,\\{\\emptyset\\}\\}$", "date": "2021-04-16 08:18:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/972618/null-sets-emptyset-subset-emptyset-emptyset", "openwebmath_score": 0.9128215312957764, "openwebmath_perplexity": 150.78228285890717, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534376578004, "lm_q2_score": 0.8670357632379241, "lm_q1q2_score": 0.8509552304021153}}
{"url": "https://math.stackexchange.com/questions/3329795/what-is-the-probability-that-an-unfair-coins-head-appears-less-than-50-after-10", "text": "# What is the probability that an unfair coin's head appears less than 50 after 100 tosses?\n\nI met a question about probability, it seems easy but I got stuck. The question is:\n\nSuppose there is an unfair coin, the HEAD probability is $$p=0.7$$.\n(Q1) If we toss the coin for 100 times, what is the expectation and the variance of this experiment?\n(Q2) Answer with reason whether or not the probability is higher than $$1/10$$ that the number of HEAD appear times is less than $$50$$ as we toss the coin for $$100$$ times.\n\nQ1 is easy, I know expectation is $$n*p=70$$ and variance is $$n*p*(1-p)=21$$. But for Q2 I have no idea.\nAt first I think it looks like... a sample distribution of sample mean used in statistics but... I don't know whether (or how) it will obey a normal distribution.\nThen I also try to calculate the sum of $$P(H=0)+P(H=1)+...+P(H=50)$$, but the work is huge, even I use an approximation of Passion distribution...\nSo could you share some of your thought? Thank you!\n\n\u2022 Well...you can use the normal approximation if you want, but just speaking roughly: We have $\\sigma=\\sqrt {21}=4.83$ so you are talking about a $4.36\\sigma$ event, so the answer is effectively $0$. For that matter, doing it exactly, with the binomial distribution, isn't especially difficult either. \u2013\u00a0lulu Aug 21 at 11:09\n\u2022 Another fast way to do it is to note that the answer is clearly less than $50\\times P(50)$ and even that is way less than $.1$ \u2013\u00a0lulu Aug 21 at 11:13\n\u2022 Thank you @lulu! I think both of these two methods you proposed are effective, but as you said 'with the binomial distribution, isn't especially difficult either', do you mean it isn't such difficult even we calculate the $P(H<50)$? If so, could you give me a rough procedure of it? \u2013\u00a0Peter Nova Aug 21 at 11:25\n\u2022 Just use a spreadsheet or a program. here is Wolfram Alpha's version. \u2013\u00a0lulu Aug 21 at 11:27\n\u2022 @PeterNova Here is the computation with Hypergeometrics Wolframalpha - Hypergeometrics \u2013\u00a0InterstellarProbe Aug 21 at 20:56\n\nWe can show that the answer to Q2 is \"No\" even without appealing to the Central Limit Theorem.\n\nLet's say $$H$$ is the total number of heads. By the Chebyshev inequality (see below), $$P(|H-70| \\ge 21) \\le \\frac{21}{21^2} \\approx 0.048$$ But $$P(|H-70| \\ge 21) = P(H \\le 49) + P(H \\ge 91)$$ so $$P(H \\le 49) \\le P(|H-70| \\ge 21) \\le 0.048$$\n\nChebyshev's inequality: If $$X$$ is a random variable with finite mean $$\\mu$$ and variance $$\\sigma^2$$, then for any value $$k>0$$, $$P(|X-\\mu| \\ge k) \\le \\frac{\\sigma^2}{k^2}$$\n\nSince $$H$$ is a sum of Bernoulli random variables, then $$H$$ is modeled as Binomial with $$N = 100$$ trials and $$p=0.7$$ probability of success. Indeed, it is exhaustive to compute the desired probability which is $$P(H < 50) = P(H=0) + \\ldots P(H=49)$$\n\nInstead what you could do is notice that $$N =100$$ is large enough, and hence we could approximate the Binomial with a Gaussian distribution, so $$H \\sim N(Np,Np(1-p)) = N(70,21)$$ So $$\\Pr(H<50) \\simeq \\Pr(\\underbrace{\\frac{H-70}{\\sqrt{21}}}_{Z} < \\frac{50-70}{\\sqrt{21}}) \\simeq \\Pr(Z < -4.36)$$ where $$Z$$ is a centered Gaussian. According to the z-table, the above probability is way less than $$\\frac{1}{10}$$.\n\n\u2022 Thank you very much @Ahmad Bazzi ! I should think about this... I only thought this experiment should approximate this binomial to a possion distribution... Can I ask a stupid question that when can we approximate the binomial with a gaussian distribution? I guess like when experiment times n=10 or p=0.99 we could not do that, right? \u2013\u00a0Peter Nova Aug 21 at 11:29\n\u2022 You are right, when we say $N$ is large, it should be compared to a threshold to be able to say \"it is large enough\". A rule of thumb is when $N > 9 \\max(\\frac{1-p}{p},\\frac{p}{1-p})$. @PeterNova \u2013\u00a0Ahmad Bazzi Aug 21 at 11:58\n\nHere's an argument that doesn't rely on approximating a normal distribution. The basic idea is to show that\n\n$$\\sum_{k=1}^{49}{100\\choose50-k}p^{50-k}q^{50+k}\\lt{1\\over10}\\sum_{k=0}^{50}{100\\choose50+k}p^{50+k}q^{50-k}$$\n\nwhere $$p=0.7$$ and $$q=1=p=0.3$$\n\nNow $$(q/p)^2=9/49$$, so we have $$(q/p)^{2k}\\lt1/10$$ for $$k\\gt1$$. Since $${100\\choose50-k}={100\\choose50+k}$$, it follows that\n\n$${100\\choose50-k}p^{50-k}q^{50+k}\\lt{1\\over10}{100\\choose50+k}p^{50+k}q^{50-k}$$\n\nfor $$2\\le k\\le50$$. It would be nice if $${100\\choose49}p^{49}q^{51}$$ were less than $${1\\over10}\\left({100\\choose50}p^{50}q^{50}+{100\\choose51}p^{51}q^{49}\\right)$$; unfortunately, it isn't. However, it suffices to show that\n\n$${100\\choose48}p^{48}q^{52}+{100\\choose49}p^{49}q^{51}\\lt{1\\over10}\\left({100\\choose50}p^{50}q^{50}+{100\\choose51}p^{51}q^{49}+{100\\choose52}p^{52}q^{48} \\right)$$\n\nwhich simplifies to showing\n\n$$50\\cdot49q^4+52\\cdot50pq^3\\lt{1\\over10}\\left(52\\cdot51p^2q^2+52\\cdot50p^3q+50\\cdot49p^4 \\right)$$\n\nThis can be done with a direct calculation, but can also be verified with a couple further simplifications:\n\n$$50\\cdot49q^4+52\\cdot50pq^3\\lt50\\cdot52(q^4+pq^3)=50\\cdot52q^3(q+p)=50\\cdot52q^3=70.2$$\n\nand\n\n$${1\\over10}\\left(52\\cdot51p^2q^2+52\\cdot50p^3q+50\\cdot49p^4 \\right)\\gt{50\\cdot49\\over10}(p^2q^2+p^3q+p^4)=245p^2((p+q)^2-pq)=245p^2(1-pq)=94.8395$$\n\n(There are undoubtedly additional ways to reduce the final amount of explicit computation for the comparison. I would appreciate any suggestions.)\n\nAnother way to do this is:\n\n$$\\sum_{n=0}^{49}\\dbinom{100}{n}(0.7)^n(0.3)^{100-n} = 1-\\dbinom{100}{50}(0.7)^{50}(0.3)^{50}{{_2}F_1\\left(1,-50;51;-\\dfrac{7}{3}\\right)}\\approx 10^{-5}$$", "date": "2019-09-18 12:23:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3329795/what-is-the-probability-that-an-unfair-coins-head-appears-less-than-50-after-10", "openwebmath_score": 0.9011016488075256, "openwebmath_perplexity": 236.69760579309175, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981453437115321, "lm_q2_score": 0.8670357615200474, "lm_q1q2_score": 0.8509552282457502}}
{"url": "https://or.stackexchange.com/questions/4479/formulation-of-assignment-problem-as-integer-programming", "text": "Formulation of Assignment problem as integer programming\n\nWe need to maintain as quickly as possible a complex system. In particular, we need to replace six of its components $$\\{P_1,\\ldots,P_6\\}$$. We have three 3D printers $$\\{M_1,M_2,M_3\\}$$ which we can use to fabricate the six components. The following table/matrix states how long it takes (in minutes) the $$i$$th printer to print the $$j$$th component:\n\n$$\\begin{array}{ccccccc} \\hline & P_1& P_2&P_3&P_4&P_5&P_6 \\\\ \\hline M_1 & 23 & 42 & 12 & 32 & 47 & 60\\\\ M_2 & 25& 37& 13& 37& 51& 64\\\\ M_3 & 27 &51 &15& 41 &57 &55\\\\ \\hline \\end{array}$$\n\nThe complex system will work again only when all the components have been printed. Clearly more components can (and have to be) assigned to single machines and they are made in a sequence, one after the other, and 3D printers can work in parallel. However, you have only two operators and therefore you can only use two machines. How to formulate the problem as a linear (but combinatorial) optimization problem to allocate components to (two out of three) 3D printers so that the maintenance time is minimized.\n\nSo far I have tried the following, but I am not quite sure (Please I need help if I was mistaken):\n\nLet $$x_{ij}= 1$$ if machine $$i$$ is assigned to component $$j$$, $$0$$ otherwise. The model is \\begin{align}\\min&\\quad23x_{11}+42x_{12}+...+55x_{36}\\\\\\text{s.t.}&\\quad x_{11}+x_{12}+x_{13}+x_{14}+x_{15}+x_{16} = 2\\\\&\\quad x_{21}+x_{22}+x_{23}+x_{24}+x_{25}+x_{26} = 2\\\\&\\quad x_{31}+x_{32}+x_{33}+x_{34}+x_{35}+x_{36} = 2\\\\&\\quad x_{11}+x_{21}+x_{31} \\ge 1\\\\&\\quad x_{12}+x_{22}+x_{32} \\ge 1\\\\&\\quad x_{13}+x_{23}+x_{33} \\ge 1\\\\&\\quad x_{14}+x_{24}+x_{34} \\ge 1\\\\&\\quad x_{15}+x_{25}+x_{35} \\ge 1\\\\&\\quad x_{16}+x_{26}+x_{36} \\ge 1\\\\&\\quad x_{ij}\\,\\,\\text{is binary}.\\end{align}\n\n\u2022 \"The complex system will work again only when all the components have been printed.\" Does that mean you'd like the system to end as soon as possible? (Thinking of the objective function here). Besides this, I think this could better be formulated as a scheduling problem on parallel machines, with the additional constraint that only 2 of the 3 machines are used. \u2013\u00a0dhasson Jul 2 '20 at 13:37\n\u2022 The Generalized Assignment Problem can serve as inspiration to continue, it's the same you are doing but change the first 3 constraints where every machine is made to print 2 components. First of all, the objective function is total working time of the machines. Instead you want to minimize the makespan (maximum time to finish all jobs), so that system can work again as soon as possible. Second, you may add binary variables $z_i = 1$ if machine $i$ is used, with additional constraints for $\\sum_i z_i = 2$ and linking $x$ and $z$. \u2013\u00a0dhasson Jul 2 '20 at 15:30\n\u2022 Are you assuming that one machine will never be used, or that operators will move among machines such that all machines might be used at some point, but never more than two running at any given time? \u2013\u00a0prubin Jul 2 '20 at 15:57\n\u2022 @user3752, as dhasson mentioned, it sounds like a parallel machine scheduling problem. Would you see this link? \u2013\u00a0A.Omidi Jul 3 '20 at 10:41\n\nSince you mention that you want to operate two out of three machines, it boils down to a problem where you first pick two machines and then perform a standard $$R2||C_{\\max}$$ parallel machine scheduling problem with the two machines that you selected. Such problems are very suitable for dynamic programming/column generation approaches, but your instance is so small that a IP will work fine. And since you ask for an IP, let us consider a straightforward way to model it.\nFor a formulation, consider the following decision variables: $$y_j = \\left\\{ \\begin{array}{ll} 1 & \\mbox{ if } M_j \\mbox{ is being operated } \\\\ 0 & \\mbox{otherwise} \\end{array} \\right.$$ and $$x_{ij} = \\left\\{ \\begin{array}{ll} 1 & \\mbox{ if } P_i \\mbox{ is made on } M_j \\\\ 0 & \\mbox{otherwise} \\end{array}\\right.$$ Also, let's assume that $$a_{ij}$$ is the time needed to produce $$P_i$$ on $$M_j$$.\nNow the following IP can be formulated: $$\\begin{array}{llll} \\min & z \\\\ \\mbox{s.t.} & \\sum_{j} y_j & \\leq K \\\\ & \\sum_{j} x_{ij} & = 1 & \\forall i \\\\ & x_{ij} & \\leq y_j & \\forall i \\forall j \\\\ & \\sum_i a_{ij} x_{ij} & \\leq z & \\forall j \\\\ & x_{ij} \\in \\{0,1\\} & & \\forall i \\forall j \\\\ & y_j \\in \\{0,1\\} & & \\forall j \\\\ & z \\in \\mathbb{R} \\end{array}$$ Where the objective variable $$z$$ represents the makespan, the first constraint states that at most $$K$$ machines can be used (in your instance, $$K=2$$, i.e. $$K$$ is the number of operators), the second constraint states that each $$P_i$$ must be executed on exactly one $$M_j$$, the third constraint states that a $$P_i$$ can only be produced on a $$M_j$$ if that $$M_j$$ is being operated and the fourth constraint states that the makespan $$z$$ should be at least the time spent on each individual machine.", "date": "2021-04-19 09:33:49", "meta": {"domain": "stackexchange.com", "url": "https://or.stackexchange.com/questions/4479/formulation-of-assignment-problem-as-integer-programming", "openwebmath_score": 0.7603274583816528, "openwebmath_perplexity": 491.9159467841895, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.981453437115321, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8509552231877023}}
{"url": "https://math.stackexchange.com/questions/682714/is-the-pseudoinverse-of-a-singular-lower-triangular-matrix-itself-lower-triangu", "text": "# Is the pseudoinverse of a singular, lower triangular matrix itself lower triangular?\n\nSuppose $L\\in\\mathbb{R}^{n\\times n}$ is a singular, lower triangular matrix. Is its psuedoinverse, $L^\\dagger\\in\\mathbb{R}^{n\\times n}$, also lower triangular? I have already proved by induction that the product of two lower triangular matrices is lower triangular, and I also proved that the inverse of a (non-singular) lower triangular matrix is lower triangular. I have shown by working out an example with lower triangular $L\\in\\mathbb{R}^{2\\times 2}$ that the pseudoinverse, $L^\\dagger\\in\\mathbb{R}^{2\\times 2}$, does not have to be lower triangular, but I was wondering if there might be a better way to prove this than just by showing through an example in $\\mathbb{R}^{2\\times 2}$ that $L^{\\dagger}$ is not necessarily lower triangular.\n\nUseful information:\nThe pseudoinverse of a matrix $A$ satisfies the following properties:\n1) $(A^\\dagger A)^*=A^\\dagger A$\n2) $(AA^\\dagger)^*=AA^\\dagger$\n3) $AA^\\dagger A=A$\n4) $A^\\dagger AA^\\dagger = A^\\dagger$\nNote: here I will be considering only the reals, so the conjugate transpose becomes just a transpose.\n\nAttempt at Solution: For $L\\in\\mathbb{R}^{n\\times n}$, where $n=1$, it is easy to show that $L^\\dagger$ is lower triangular. Since $L$ is singular, we have $L=0$, which is also lower triangular, by the definition of a lower triangular matrix. Then, properties 1) through 3) above are trivially satisfied by any $L^\\dagger\\in\\mathbb{R}$, but for property 4) to be satisfied, we need $L^\\dagger 0L^\\dagger=L^\\dagger\\implies L^\\dagger=0$. But, $L^\\dagger$ is lower triangular by the definition of a lower triangular matrix, so for $n=1$, the answer to the original question is \"yes\".\n\nNow, consider the case of $n=2$. Let $$L=\\left(\\begin{array}{cc}a & 0 \\\\ b & \\tilde{L} \\end{array}\\right)$$ for $a$ and $b\\in\\mathbb{R}$, and with $\\tilde{L}$ a singular, lower triangular matrix $\\in\\mathbb{R}^{1 \\times 1}$, as before in the example with $n=1$. So $L$ is a singular, lower triangular matrix, and $$L=\\left(\\begin{array}{cc}a&0\\\\b&0\\end{array}\\right).$$ Let $L^\\dagger\\in\\mathbb{R}^{2\\times 2}$ be the pseudoinverse of $L$, and define $L^\\dagger$ as $$L^\\dagger=\\left(\\begin{array}{cc}c&d\\\\e&\\tilde{L}^\\dagger\\end{array}\\right),$$ with $c,d,e\\in\\mathbb{R}$ and with $\\tilde{L}^\\dagger$ being the psuedoinverse of $\\tilde{L}$ as derived in the above example for $n=1,$ so $\\tilde{L}^\\dagger = 0$ and so $$L^\\dagger=\\left(\\begin{array}{cc}c&d\\\\e&0\\end{array}\\right).$$\n\nNow, property 1) above requires $ea=0$. Property 2) requires $ad=bc$. Property 3) requires that ($a=0$ or $ca+db=1$) and ($b=0$ or $ca+db=1$). Property 4) requires $e=0$ and ($c=0$ or $ca+db=1$) and ($d=0$ or $ca+db=1$). I chose $d\\neq0$, which would then make $L^\\dagger$ not a lower triangular matrix. Then, with $ca+db=1$ and $d\\neq0$, I found $b=\\frac{1-ca}{d}$. Further, from $ad=bc$, I found (substituting the previous expression for $b$), $a=\\frac{c}{c^2+d^2}$, which then gives $b$ in terms of $c$ and $d$ as $b=\\frac{d}{c^2+d^2}.$ My $L$ and $L^\\dagger$ then become $$L=\\frac{1}{c^2+d^2}\\left(\\begin{array}{cc}c&0\\\\d&0\\end{array}\\right),$$ and $$L^\\dagger=\\left(\\begin{array}{cc}c&d\\\\0&0\\end{array}\\right).$$ $L^\\dagger$ satisfies all four of the properties above, but it is not lower triangular, so my answer to the original question is, in general, \"no\".\n\nHowever, surely their is a smarter way to do this, in which I don't have to resort to doing all this algebra, isn't there? If so, how might I approach the proof? Thanks a lot!\n\nAn even stronger counterexample: If $A = \\left(\\begin{matrix} 0 & 1 \\\\ 0 & 0 \\end{matrix}\\right)$, then there exists no upper-triangular $2\\times 2$-matrix $B$ satisfying $ABA = A$. (Indeed, the $\\left(1,2\\right)$-entry of $ABA - A$ will always be $-1$, no matter what $B$ is, as long as $B$ is upper-triangular.)\nThe pseudoinverse of a non-invertible triangular matrix can be non-triangular. $$\\begin{pmatrix} 1& 1\\\\ 0& 0 \\end{pmatrix}^\\dagger = \\frac12 \\begin{pmatrix} 1& 0\\\\ 1& 0 \\end{pmatrix}$$ $$\\begin{pmatrix} 1& 1& 1\\\\ 0& 1& 1\\\\ 0& 0& 0 \\end{pmatrix}^\\dagger = \\frac12 \\begin{pmatrix} 2& -2& 0\\\\ 0& 1& 0\\\\ 0& 1& 0 \\end{pmatrix}$$", "date": "2020-04-08 16:15:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/682714/is-the-pseudoinverse-of-a-singular-lower-triangular-matrix-itself-lower-triangu", "openwebmath_score": 0.948116660118103, "openwebmath_perplexity": 111.60296327251775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534382002797, "lm_q2_score": 0.8670357546485408, "lm_q1q2_score": 0.8509552224423845}}
{"url": "https://math.stackexchange.com/questions/1359801/order-of-infinite-countable-ordinal-numbers", "text": "order of infinite countable ordinal numbers\n\nI'm trying to understand ordinal arithmetic. If one had an ordered list of the some subset of countable ordinal numbers, what order would the following 6 countably infinite ordinals be in? If the following order is not correct, what is correct order and why is that the correct order?\n\n$$\\omega\\;<\\; \\omega^2 \\;<\\; 2^\\omega \\;<\\; \\omega^\\omega \\;<\\; {^\\omega}2 \\;<\\;{^\\omega} \\omega$$\n\nI know $\\epsilon_0 = {^\\omega} \\omega$ is the largest, but is still countable, but I'm not sure where the powers of $2$ fit in versus the powers of $\\omega$.\n\nI understand why $\\omega^2$ or $\\omega^n$ for any finite value of $n$ needs to be countable. But, why does $\\omega^\\omega$ need to be countable? For cardinal numbers, $2^{\\aleph_0}$ is uncountably infinite. Presumably, there would be some contradiction in mathematics if any finite ordinal arithmetic equation involving $\\omega$ generated an uncountable infinity.\n\n\u2022 What does the notation ${^\\omega} 2$ mean? Never saw that (and it's hard to see what to type into Google to find it...) \u2013\u00a0David C. Ullrich Jul 13 '15 at 19:12\n\u2022 \u2013\u00a0Asaf Karagila Jul 13 '15 at 19:13\n\u2022 The defining sequence for $^{\\omega} 2$ is $2\\;\\;2^2\\;\\;2^{2^2}\\;\\;2^{2^{2^2}}\\;\\;2^{2^{2^{2^2}}}....$ \u2013\u00a0Sheldon L Jul 13 '15 at 19:15\n\u2022 So then in fact ${^\\omega}2=\\omega$. \u2013\u00a0David C. Ullrich Jul 13 '15 at 19:17\n\u2022 ok, presumably that is consistent, but it seems a bit odd, that such a fast growing sequence is regarded as smaller than $\\omega^2$ \u2013\u00a0Sheldon L Jul 13 '15 at 19:18\n\nFirst note that $$^\\omega2 := \\sup \\{ \\underbrace{2^{2^{2^\\ldots}}}_{n\\text{-times}} \\mid n < \\omega \\} = \\omega.$$\n\n[..]but it seems a bit odd, that such a fast growing sequence is regarded as smaller than $\\omega^2$.\n\nWell, that's the thing when dealing with infinities. Sometimes our intuition fails us. While $(\\underbrace{2^{2^{2^\\ldots}}}_{n\\text{-times}})_{n < \\omega}$ could be regarded as \"a fast growing sequence\", each of its elements is finite and therefore $\\omega$ is an upper bound. As $\\omega$ certainly is the least upper bound, we get $^\\omega2 = \\omega$.\n\nBy an analogous argument we get that $$2^\\omega := \\sup \\{2^n \\mid n < \\omega \\} = \\omega$$\n\nSo we are left with ordering $\\omega, \\omega^2, \\omega^\\omega$ and $^\\omega \\omega$.\n\nWe have \\begin{align}\\omega^2 &:= \\omega \\cdot \\omega \\\\ &= \\sup \\{ w \\cdot n \\mid n < \\omega\\} \\\\ &\\ge \\omega \\cdot 2 \\\\ &> \\omega \\end{align}\n\nand\n\n\\begin{align}\\omega^\\omega &= \\sup \\{ \\omega ^n \\mid n < \\omega \\} \\\\ &\\ge \\omega^3 \\\\ &= \\sup \\{(\\omega^2) \\cdot n \\mid n < \\omega \\} \\\\ &\\ge \\omega^2 \\cdot 2 \\\\ &> \\omega^2. \\end{align}\n\nFinally \\begin{align}^\\omega \\omega &:= \\sup \\{ \\underbrace{\\omega^{\\omega^{\\omega^ \\ldots}}}_{n\\text{-times}} \\mid n < \\omega \\} \\\\ &\\ge \\omega^{\\omega^\\omega} \\\\ &= \\sup\\{\\left( \\omega^\\omega \\right)^n \\mid n < \\omega \\} \\\\ &\\ge \\left(\\omega^\\omega \\right)^2 \\\\ &= \\omega^\\omega \\cdot \\omega^\\omega \\\\ &= \\sup \\{\\omega^\\omega \\cdot \\alpha \\mid \\alpha < \\omega^\\omega \\} \\\\ &\\ge \\omega^\\omega \\cdot 2 \\\\ &> \\omega^\\omega. \\end{align}\n\nCombining these calculations we get the desired order: $$\\omega = 2^\\omega = {^\\omega} 2 < \\omega^2 < \\omega^\\omega < ^\\omega\\omega$$\n\nIn general, ordinal and cardinal arithmetic are very different beasts and every ordinal arithmetic expression using only ordinals $\\le \\omega$ is countable.\n\nproof (sketch)\n\nTake a countable transitive model $M$ of a large enough fracture of $ZFC$. Every ordinal expression using only ordinals $\\le \\omega$ can be computed correctly inside $M$ (<- this requires some work). As $M$ only contains countable ordinals (as it is transitive), the result follows.\n\n\u2022 That is a horrible notation, ${}^\\omega2$. \u2013\u00a0Asaf Karagila Jul 14 '15 at 6:34\n\u2022 Thanks for the proof; much to learn. \u2013\u00a0Sheldon L Jul 14 '15 at 8:39", "date": "2019-10-24 05:01:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1359801/order-of-infinite-countable-ordinal-numbers", "openwebmath_score": 1.0000098943710327, "openwebmath_perplexity": 413.09783887344037, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534387427591, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8509552161686693}}
{"url": "https://math.stackexchange.com/questions/878115/in-30-boxes-are-15-balls-chance-all-balls-in-10-or-less-boxes", "text": "# In 30 boxes are 15 balls. Chance all balls in 10 or less boxes?\n\nQuestion1: I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls. Afer i collected all 15 balls i put them randomly inside the boxes.\n\nHow much is the chance that all balls are in only 10 boxes or less?\n\nQuestion2: I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls. In two of the boxes i could find 3 balls. (So in one box has to be 2 balls and in the other seven boxes have to be 1 ball.) Afer i collected all 15 balls i put them randomly inside the boxes.\n\nHow much is the chance that i find in only 2 boxes 6 balls or more?\n\nI wrote a c# programm and tried it 1 million times. My solution was: With a chance of 12,4694% all balls are in 10 boxes or less.\n\nRandom trials/Monte Carlo simulations are notoriously slow to converge, with an expected error inversely proportional to the square root of the number of trials.\n\nIn this case it is not hard (given a programming language that provides big integers) to do an exact count of cases. Effectively the outcomes are partitions of the 15 balls into some number of boxes (we have thirty boxes to work with, so at least half will be empty).\n\nI wrote a Prolog program to do this (Amzi! Prolog has arbitrary precision integers built in), and got the following results:\n\n$$Pr(\\text{10 or fewer boxes occupied}) = \\frac{59486359170743424000}{30^{14}} \\approx 0.124371$$\n\n$$Pr(\\text{2 boxes hold 6 or more balls}) = \\frac{30415369655816064000}{30^{14}} \\approx 0.063591$$\n\nThe reason I'm dividing by $30^{14}$ in these probabilities is because I normalized the counting to begin with one case where a ball is in one box. If we counted that as thirty cases, we'd need to divide by $30^{15}$. So this keeps the totals slightly smaller. Each ball we add increases the total number of cases by a factor of $30$.\n\nI wrote a recursive rule to build cases for $n+1$ balls from cases for $n$ balls. The first few cases have the following counts:\n\n /* case(Balls,Partition,LengthOfPartition,Count) */\ncase(1,[1],1,1).   /* Count is nominally 1 to begin */\ncase(2,[2],1,1).\ncase(2,[1,1],2,29).\ncase(3,[3],1,1).\ncase(3,[1,2],2,87).\ncase(3,[1,1,1],3,812).   /* check: for Sum = 3, sum of Count is 900 */\n\n\nThe number of cases generated is modest enough for a desktop, daunting to manage by hand. For $n=15$ there are $176$ partitions. It simplified the Prolog code to maintain the partitions as lists in ascending order.\n\n\u2022 I checked out with a c# program. I simulated the problem with random numbers and counted: How often I find 6 ore more balls in only 2 boxes. 1000000-times tried: Pr(2 boxes hold 6 or more balls) = 0,063491%. Thanks alot! \u2013\u00a0Simon Aug 4 '14 at 13:39\n\u2022 Simulation is easier to program than the recursive/exact counting, but as briefly mentioned, getting an extra digit of accuracy may well require a hundred times as many trials because of the random walk phenomenon. \u2013\u00a0hardmath Aug 4 '14 at 13:42\n\nSolution of Question 1:\n\nThis is an occupancy problem with $n=30$ boxes and $k=15$ balls.\n\nLet's first consider the expected number of empty boxes. That is much easier to obtain. The exact answer is $30(1-1/30)^{15}=18.04.$ This is approximately $30/\\sqrt e.$ See the answer by Mr.Spot to this question:\n\nMaking 400k random choices from 400k samples seems to always end up with 63% distinct choices, why?\n\nThe probability of exactly $j$ empty boxes is, for $n-k\\le j\\le n-1$:\n\n$$P(j)={n \\choose j}\\sum_{m=0}^{n-j}(-1)^m {n-j \\choose m}\\left(1-\\frac{j+m}n\\right)^k$$\n\nSee this:\n\nhttp://probabilityandstats.wordpress.com/2010/04/04/a-formula-for-the-occupancy-problem/\n\nFor $n=30$ and $k=15:$\n\n$P(20)$ to $P(24)$ is $0.096,0.024,0.0036,0.00029,0.000013,....$\n\nand the probability of at least $20$ empty cells = 0.124371\n\nSolution of Question 2:\n\nYou throw 15 balls into 30 boxes. What is the probability of the following result:\n\n2 triple, 1 double and 7 single occupancy boxes and the rest empty.\n\n$${30 \\choose 2} {28 \\choose 1}{27 \\choose 7}\\frac{15!}{3!3!2!1!^7}\\frac{1}{30^{15}}$$ $$=\\frac{(30)(29)...(21)}{30^{15}2!1!7!}\\frac{15!}{3!3!2!1!^7}$$ First we select the 2 different boxes for the triples; of the 28 remaining boxes we select 1 for the double; of the 27 remaining boxes we select 7 for the singles. Then the multinomial coefficient gives the number of ways to assign the 15 balls to those particular boxes and there are $30^{15}$ equally likely ways to throw the 15 balls into the 30 boxes.\n\n\u2022 Thanks a lot for answering the question and for giving me tips to understand the topic. The answer fits to my program. Expected number of empty boxes: I did 11*6+295*7 ... 14246*15=11956971 (Numbers posted in comment ant11). 11956971/1000000=11.956971. 30-ans=18.043029. I also checked the formula in wolfram alpha and i got the same numbers. Now i will write a program about question2. I think this is easy to solve (approximately) with a program. And i will take a look at \"formula for the occupancy problem\". Maybe i can improve my understanding about this topic. \u2013\u00a0Simon Jul 27 '14 at 17:14\n\u2022 I tried your formula for each combination and got nearly the same numbers. In comparison to the other combination, one had a larger difference. If you get 15 boxes with balls (1 ball each box), i had the chance of 1,4246%. If i do formula with wolfram alpha i use this text: (30 choose j) (sum ((-1)^m)*((30-j) choose m)*(1-((j+m)/30))^15, m=0 to 10) , j=15 | (15 stands for 15 free boxes --> 15 boxes not empty). I got the solution: 0,02291 = 2,291%. I used the same formula for j=15,16...29. I estimated the numbers as 2,9910%. All 15 numbers added i got 1,015827. Normally i should get 1. \u2013\u00a0Simon Jul 28 '14 at 15:00\n\u2022 Ok i tried again 2 times. Now i rounded all nombers down. I tried your formula for j= 15,16...29. My solution: 15=0,029910305; 16=0,092823631; 17=0,236474423; 18=0,307416722; 19=0,224830457; 20=0,096214657; 21=0,024288697; 22=0,003562779; 23=0,000292277; 24=0,000012534; 25=0; 26=0; 27=0; 28=0; 29=0; All numbers added = 1,015826482. I think you may need a different formula for the upper limit 15 are empty. \u2013\u00a0Simon Jul 28 '14 at 15:50\n\u2022 Ok, the right formula is in your link: probabilityandstats.wordpress.com/2010/04/04/\u2026. In this case we need another formula: If there are exactly j empty cells, then the k balls are placed in the remaining n-j cells and these remaining n-j cells are all occupied. Great job! Thank you. \u2013\u00a0Simon Jul 28 '14 at 20:42\n\u2022 P(15)=0.0141 which explains why your sum is not 1. \u2013\u00a0Mr.Spot Jul 29 '14 at 3:39\n\nI will assume the balls and boxes are indistinguishable.\n\nThe first problem is: If I distribute $15$ balls among $30$ boxes, what is the probability that at most $10$ boxes contain a ball?\n\nFirst, the fact that there are $30$ boxes does not matter, since they are indistinguishable. So we only need to consider the problem as if there were $15$ boxes.\n\nSecond, we'll solve the problem by finding the probability that more than $10$ boxes contain a ball, and subtract that number from $1$.\n\nThere is exactly $1$ to occupy $15$ boxes with $15$ balls: $\\{1,1,1\\cdots\\}$\n\nThere is also exactly $1$ way to occupy $14$ boxes with $15$ balls, since, again, the boxes are indistinguishable: $\\{2,1,1\\cdots\\}$\n\nThere are $2$ ways to occupy $13$ boxes: $\\{3,1,1\\cdots\\}$ or $\\{2,2,1,1\\cdots\\}$\n\nThere are $3$ ways to occupy $12$ boxes, and $5$ ways to occupy $11$ (you can and should verify these numbers).\n\nFinally, how many ways could we distribute $15$ balls over $15$ boxes? This number is exactly equal to $p(15)=176$, where $p(n)$ is the partition function.\n\nSo the answer to problem 1 is $1-(1+1+2+3+5)/176=164/176\\approx93.2\\%$\n\nEDIT: I believe this answer is drastically different from the one obtained by your program for the following reason: you might have meant, in your problem statement, \"what is the probability all $15$ balls are contained in $10$ particular boxes?\" This significantly changes the question; in particular, the total number of boxes being $30$ is now relevant. Let me know if this is the misunderstanding.\n\n\u2022 There is no good reason to use a probability model in which all partitions are equally likely. A more natural model has us throwing the balls one at a time, with all $30$ boxes equally likely. \u2013\u00a0Andr\u00e9 Nicolas Jul 25 '14 at 20:14\n\u2022 Let us say we got 30 machines. Rob is telling you: \"Hey ant11, we got 15 disruptions last year\". Another guy says: \"two machines had 3 disrupions, one machine had 2 disruptions and the other seven machines had 1 disruption. Ant11 you are good at maths. What do you think about this. If all machines would be indistinguishable, what chance it would have that only 10 machines had all 15 disruptions?\" You think the solution should be easy. \u2013\u00a0Simon Jul 25 '14 at 20:23\n\u2022 Meanwhile you think about it, the boss comes and asks you:\" Hey ant11, two machines got together 6 disruptions. That sounds like they will have many disruptions more next year. What is the chance if all 30 machines are indistinguishable, that they are more than 6 disruptions in only two machines. Maybe it was just bad luck.\" I think there are 30 boxes is relevant. \u2013\u00a0Simon Jul 25 '14 at 20:23\n\u2022 My last time i wrote a programm was two years ago. Maybe i made a mistake. Here you can see my solution: 1000000 tryed: 1-5: 0-times: 0,0% 6: 11-times: 0,0011% 7: 295-times:0,00295% 8: 3585-times: 0,3585% 9: 24383-times: 2,4383% 10: 96420-times: 9,6420% 11: 225081-times: 22,5081% 12: 307109-times: 30,7109% 13: 236556-times: 23,6556% 14: 92314-times: 9,2314% 15: 14246-times: 1,4246% \u2013\u00a0Simon Jul 25 '14 at 20:32\n\u2022 After that i added the numbers: 11+295+3585+24383+96420=124694. 124694/1000000=0,124694. --> 12,4694% 10 or less machines if all are indistinguishable \u2013\u00a0Simon Jul 25 '14 at 20:43", "date": "2019-11-12 21:32:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/878115/in-30-boxes-are-15-balls-chance-all-balls-in-10-or-less-boxes", "openwebmath_score": 0.7497148513793945, "openwebmath_perplexity": 588.1410203704329, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9814534354878828, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8509552150325912}}
{"url": "https://math.stackexchange.com/questions/2887540/finding-the-number-of-non-negative-integral-solutions-of-x-y-z-10", "text": "# Finding the number of non-negative integral solutions of $x + y + z = 10$\n\nI realize that this question has been asked multiple times and I do not really want to know how to do it, I understand how to solve it, my issue is somehow I have used the distribution and I am not getting the right answer.\n\nMethod of Solving:\nI solved this in the first go by directly applying the distribution formula that is $$\\binom {n+r-1}{r-1}$$\nBy using, $n = 10, r= 3$ we get,\n$$\\binom{12}{2} = 66$$ however the answer given is $$\\binom{12}{3} = 220$$ Next, I tried to solve by taking multiple cases that is, setting any variable, $x,y$ or $z$ to values from $0,1,2,3...10$, so for example if $x = 0$ then we get, $$y+z=10$$ now it is simple enough to distribute this even without the formula, we get $$\\Bigl((0,10),(1,9),(2,8)(3,7),(4,6)\\Bigl) \\cdot 2,(5,5)$$ which gives us $11$ cases, that can be verified using the formula $$P_0 = \\binom{11}{1}.$$ Now taking cases till $x=10$ we get, the final result as $$\\sum _{n=0} ^{n=10} P_n = 11+ 10 + 9 + 8+ \\ldots +1 = 66$$\nPlease tell me where I am going wrong or of it simply a case of the answer in the text being wrong, and if the second method is correct.\n\n\u2022 The answer in the text is wrong. Both of your methods are correct. Aug 19 '18 at 10:06\n\u2022 All right, thank you so much Aug 19 '18 at 10:07\n\u2022 Does this answer your question? Number of Non negative integer solutions of $x+2y+5z=100$ Oct 11 at 21:51\n\nThe answer is $66$ even by double checking with generating functions, more details here.\nThe number of solutions for $x+y+z=n, x\\geq0, y\\geq0, z\\geq0$ is the coefficient of $x^n$ term of the $$(1+x+x^2+x^3+...+x^k+...)^3=\\frac{1}{(1-x)^3}$$ and because $$\\frac{1}{(1-x)^3}= \\frac{1}{2}\\left(\\frac{1}{1-x}\\right)^{''}= \\frac{1}{2}\\left(\\sum\\limits_{k=0}x^k\\right)^{''}= \\sum\\limits_{k=2}\\frac{k(k-1)}{2}x^{k-2}$$ the coefficient of $x^n$ is $\\color{green}{\\frac{(n+2)(n+1)}{2}=\\binom{n+2}{2}}$, which for $n=10$ yields $66$.\nI think they have the role of $r$ and $n$ mixed up in the given solution, that is why they got the wrong result. Look at the proof of the formula again, and you will see immediately which one of $10$ and $3$ play the role of $n$ and $r$.\nSo you are right, the answer is 66, because $\\binom{3+10-1}{10}=66$.\n\u2022 Sorry, I don't see it, could you explain? Aren't we looking for the number of ways to distribute $10$ over $3$ variables? Aug 19 '18 at 9:58\n\u2022 The OP did not confuse $n$ and $r$. Notice that you used the formula $\\binom{n + r - 1}{n}$, while the OP used $\\binom{n + r - 1}{r - 1}$. Aug 19 '18 at 10:04", "date": "2021-10-20 09:27:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2887540/finding-the-number-of-non-negative-integral-solutions-of-x-y-z-10", "openwebmath_score": 0.861363410949707, "openwebmath_perplexity": 89.46433324936653, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992932829918, "lm_q2_score": 0.879146780175245, "lm_q1q2_score": 0.8509255472236373}}
{"url": "https://math.stackexchange.com/questions/2875697/disjoint-sets-of-rationals-both-dense-in-mathbbr", "text": "# Disjoint sets of rationals both dense in $\\mathbb{R}$\n\nThere exists a pair of disjoint subsets of $\\mathbb{Q}$ such that both are dense in $\\mathbb{R}$. True or false?\n\nThe statement is true but how can I find an example of such disjoint subsets of $\\mathbb{Q}$?\n\n\u2022 What have you thought of? What examples have you tried? \u2013\u00a0Guido A. Aug 8 '18 at 5:57\n\u2022 Is$(\\frac{m}{2^k})_{m, k}$ dense? \u2013\u00a0dEmigOd Aug 8 '18 at 5:59\n\u2022 Are m and k integers? Then this is dense in R. \u2013\u00a0Mathsaddict Aug 8 '18 at 6:02\n\u2022 First I took example - A = {m/2^k, m - Z , k - N union 0} and B = { m/3^k, m - z, k - N union 0} \u2013\u00a0Mathsaddict Aug 8 '18 at 6:08\n\u2022 You're looking more generally for ways to partition $\\mathbb{Q}$ into pieces not based on size (e.g. you don't want $(-\\infty,\\sqrt{2})\\cap\\mathbb{Q}$ versus $(\\sqrt{2},\\infty)\\cap\\mathbb{Q}$). The two obvious places to look for a distinction to draw are numerator and denominator. For example, we could let $EN$ be the set of rationals which (in lowest terms) have an even numerator. Or, we could let $DD$ be the set of rationals whose denominator (in lowest terms) is a power of $2$ (\"dyadic\" fractions). Or, so on. (cont'd) \u2013\u00a0Noah Schweber Aug 8 '18 at 6:08\n\nLet $x \\in \\mathbb{R}$. Now,\n\n$$0 < 2^nx - [2^nx] < 1$$\n\nand thus\n\n$$0 < x - \\frac{[2^nx]}{2^n} < \\frac{1}{2^n}$$\n\nThis says that $D = \\{\\frac{m}{2^n}\\}_{m \\in \\mathbb{Z}, n \\in \\mathbb{N}}$ is dense in $\\mathbb{R}$. Now, let's consider a subset of this set,\n\n$$A = \\{\\frac{m}{2^n} : n \\in \\mathbb{N}, m \\in \\mathbb{Z} , 2 \\not |n \\}$$\n\nIf $A$ were dense in $D$, it would be dense in $\\mathbb{R}$. For this, we only have to show that $D \\setminus A$ can be approximated by elements of $A$. Let $\\frac{m}{2^n} \\in D \\setminus A$ and write $m = 2^ks$ with $2 \\not | s$. It can't be that $k \\leq n$, because that would mean $\\frac{m}{2^n} = \\frac{s}{2^{n-k}} \\in A$. Therefore, $k \\geq n$ and thus $\\frac{m}{2^n} = 2^{k-n}s \\in \\mathbb{Z}$. This means that it is sufficient to show that $A$ can approximate integers: if $a \\in \\mathbb{Z}$ and $\\varepsilon > 0$, then taking $\\frac{1}{2^n} < \\varepsilon$ we have that\n\n$$\\left|a - \\frac{2^na + 1}{2^n}\\right| = \\frac{1}{2^n} < \\varepsilon$$\n\nand $\\frac{2^na + 1}{2^n} \\in A$ because $2 \\not | \\ 2^na+1$. In conclusion, we have shown that $A$ is dense in $D$ and since $D$ is dense in the reals, so is $A$. With an identical construction (which I encourage you to write out) one can show that the same holds for\n\n$$B = \\{\\frac{m}{3^n} : n \\in \\mathbb{N}, m \\in \\mathbb{Z} , 3 \\not |n \\}$$\n\nTo conclude, then, we just have to observe that $A \\cap B = \\emptyset$. In effect, if\n\n$$\\frac{m}{2^n} = \\frac{j}{3^k}$$\n\nfor some $n,m,j,k$, then $3^km = 2^nj$ and thus $2 | 2^nj = 3^km$. But since $2$ and $3$ are coprime, this would imply $2 | m$ which is absurd.\n\nHINT.-It is enough to show an example in $I=[0,1]$. All rational is periodic after a finite number of digits $$r=0.a_1a_2....a_m[b_1b_2....b_n]\\in I$$ It is not hard to prove that $A$ and $B$ below are disjoint and dense in $I$\n\n$$A=\\{r\\text{ such that } b_n=1\\}\\\\B=\\{r\\text{ such that } b_n\\ne1\\}$$", "date": "2019-04-25 00:40:12", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2875697/disjoint-sets-of-rationals-both-dense-in-mathbbr", "openwebmath_score": 0.9693884253501892, "openwebmath_perplexity": 159.38919231357758, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992923570261, "lm_q2_score": 0.8791467801752451, "lm_q1q2_score": 0.8509255464095776}}
{"url": "https://math.stackexchange.com/questions/2798284/explain-please-the-following-statement-from-naive-set-theory", "text": "# Explain please the following statement from Naive Set Theory\n\nHere is a statement that I can't understand:\n\nOrdered triples, ordered quadruples, etc., may be defined as families whose index sets are unordered triples, quadruples, etc.\n\nFor now, let's stick with the ordered triples. First of all I can't understand whether the index set is a set of unordered triples like $\\{ \\{a, b, c\\}, \\{d, e, f\\}\\}$ or it is itself an unordered triple?\n\nAlso, I just can't imagine how having any of such sets as the index set can help me to build a set of ordered triples. Maybe some example will make things clear. I will be very grateful if you help. Thanks in advance.\n\n\u2022 Where did you find this statement? Personally, I also find it difficult to parse ... \u2013\u00a0Noah Schweber May 27 '18 at 18:03\n\u2022 @NoahSchweber, Naive Set Theory of Halmos \u2013\u00a0Turkhan Badalov May 27 '18 at 18:03\n\u2022 What page? It's an entire book ... \u2013\u00a0Noah Schweber May 27 '18 at 18:03\n\u2022 @NoahSchweber, oh, sorry. Page 36 in \"Families\" section \u2013\u00a0Turkhan Badalov May 27 '18 at 18:05\n\nYes, this is a bit unclear.\n\nHalmos' goal is to define the notion of an arbitrary Cartesian product. This should generalize the usual Cartesian product of two sets, $A\\times B$, but should \"work for any number of sets.\" Before diving into his description, let me point out that this really is nontrivial: how should we think of the Cartesian product of \"$\\mathbb{R}$-many\" sets?\n\nHalmos is about to tell us, essentially, that the Cartesian product of an indexed family $\\{X_i\\}_{i\\in I}$ of sets is just the set of all indexed families $\\{x_i\\}_{i\\in I}$ of objects with $x_i\\in X_i$. Maybe more clearly, an element of the Cartesian product is a function with domain $I$; it sends $i\\in I$ to the \"$i$th coordinate,\" which must be in $X_i$.\n\nTo motivate this, Halmos has us go back to the idea of a Cartesian product. We can rethink Cartesian products as follows:\n\n\u2022 Fix any two distinct sets, $a$ and $b$. We'll think of the first as \"LEFT\" and the second as \"RIGHT.\"\n\n\u2022 Now an ordered pair has two \"coordinates,\" a left coordinate and a right coordinate. We're going to match these up with $a$ and $b$ above: if I have a function $z$ with domain $\\{a, b\\}$ such that $z(a)=x\\in X$ and $z(b)=y\\in Y$ (here I write \"$z(i)$\" for Halmos's \"$z_i$\"), we want to think of the object $z$ as being the ordered pair $(x, y)$. Informally, $z$ says\n\n$$\\mbox{My left coordinate is x, and my right coordinate is y.}$$\n\nPut another way:\n\nWe can think of the ordered pair $(x, y)$ as the function with domain $\\{a, b\\}$ (which is an unordered pair) mapping $a$ to $x$ and $b$ to $y$.\n\nThis is easiest to think about if $a=0$ and $b=1$, or something similar, but Halmos's point is that all we need is that the index set has two distinct elements. Now here's the key linguistic step Halmos makes which I think is confusing at first:\n\nAn ordered pair is an indexed set! And the indexing set is $\\{a, b\\}$, which is an unordered pair.\n\n\u2022 are you sure you wanted to say sets in the end? \"... that the Cartesian product of an indexed family $\\{X_i\\}_{i \\in I}$ of sets is just the set of all indexed families $\\{x_i\\}_{i \\in I}$ of sets with $x_i \\in X_i$\". Because as I understand, if we say the family $\\{x_i\\}$ of sets, its range consists of sets implying $x_i$ is a set which is, as I understand, is not said \u2013\u00a0Turkhan Badalov May 27 '18 at 18:42\n\u2022 @TurkhanBadalov My ZFC bias is showing. In formal ZFC set theory (and many other formal set theories), every object is in fact a set and everything is \"built out of\" the emptyset in a sense. But yes, in the context of Halmos's book that's inappropriate. Fixed! \u2013\u00a0Noah Schweber May 27 '18 at 18:51\n\u2022 Thanks for the great answer! So, to make an ordered triple, let's say, $(11, 21, 31)$ I need some unordered triple and let it be $\\{a, b, c\\}$, true? Then I have to have (actually I doubt this statement about having the following set, because I need to construct it additionaly, correct me if I am wrong) the family $\\{X_i\\}_{i \\in \\{a, b, c\\}}$ and let it be $\\{ (a,11), (b, 21), (c, 31)\\}$. Then the Cartesian product of the family will be the set of all families $\\{x_i\\}$: $\\{ \\{ (a, 11), (b, 21), (c, 31)\\} \\}$, the element of which I interpret as the ordered triple we expected to make? \u2013\u00a0Turkhan Badalov May 27 '18 at 18:59\n\u2022 Not quite. The set $\\{(a, 11), (b, 21), (c, 31)\\}$ is a single ordered triple. The Cartesian product of three sets $X_a, X_b, X_c$ would be the set of all sets of the form $\\{(a, x_a), (b,x_b), (c, x_c)\\}$ with $x_a\\in X_a, x_b\\in X_b, x_c\\in X_c$. So e.g. if $X_a=X_b=X_c=\\mathbb{N}$, the set $\\{(a, 11), (b, 21), (c, 31)\\}$ would be an element of the Cartesian product, but the Cartesian product would also include other things like $\\{(a, 18), (b, 467), (c, 3)\\}$ (which we would think of as \"$(18, 467, 3)$\"). \u2013\u00a0Noah Schweber May 27 '18 at 19:02\n\u2022 @TurkhanBadalov I thought you were using my example where $X_a=X_b=X_c=\\mathbb{N}$. If the $X_i$s are as you describe, then that's right. \u2013\u00a0Noah Schweber May 27 '18 at 19:26", "date": "2021-04-13 12:59:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2798284/explain-please-the-following-statement-from-naive-set-theory", "openwebmath_score": 0.9027474522590637, "openwebmath_perplexity": 220.89679924018859, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9678992895791291, "lm_q2_score": 0.8791467706759584, "lm_q1q2_score": 0.8509255347730456}}
{"url": "https://cs.stackexchange.com/questions/148833/fastest-algorithm-for-merge-step-in-mergesort/148839#148839", "text": "# Fastest Algorithm for \"Merge\" step in Mergesort\n\nGiven two sorted arrays $$a_1,a_2,\\dots,a_n$$ and $$b_1,b_2,\\dots,b_m$$, merge them together into one sorted array $$c_1,c_2,\\dots,c_{n+m}$$ containing the elements of $$a$$ and $$b$$.\n\nThe typical mergesort method works in $$O(n+m)$$, and if we run binary search for each element from one of the arrays, it works in $$O(\\min(n,m)\\log(\\max(n,m)))$$. However, the latter method is no better than if the smaller array was unsorted. I was wondering if there is any use of the fact that both arrays are sorted to optimise the latter solution. In particular, I was hoping for an algorithm in time $$O(\\min(n,m))$$ or so.\n\nI attempted a \"parallel binary search\" method, but it seems in the worst case that it is no better than the naive binary search method.\n\n\u2022 (The typical mergesort method should read The typical merge method more likely than not. In a scientific context, it is challenging to ask for optimal solutions (Fastest Algorithm): any proposal will be expected to be proven. (Which is where within a constant factor comes in.)) Jan 31 at 8:59\n\u2022 What makes you think that $\\min(n,m)\\log(\\max(n,m))$ would be any better than $n+m$ ? Take $n<m$ WLOG and divide both functions by $m$. Now you compare $\\dfrac nm\\log m$ and $\\dfrac nm+1$. Mar 3 at 9:20\n\nIf you only want to count the number of comparisons then you can achieve $$O\\left(n\\log(1+\\frac{m}{n})\\right)$$ comparisons as follows:\n\nLet $$k_1$$ denote the index of $$a_1$$ in the final array $$c$$ and for $$1 let $$k_i$$ denote the difference between the indices of $$a_i$$ and $$a_{i-1}$$ in the final array $$c$$.\n\nTo insert $$a_1$$ in $$b$$, start by doing an exponential search to find the smallest $$r$$ such that $$b_{2^r} \\geq a_1$$ (this is also the smallest $$r$$ such that $$2^r \\geq k_1$$). This takes $$O(\\log k_1)$$ comparisons. Now do a binary search between the indices $$1$$ and $$2^r$$ to locate the correct position $$k_1$$ to insert $$a_1$$, and insert the element there. This also takes $$O(\\log k_1)$$ comparisons.\n\nTo insert $$a_2$$ into $$b$$ start by doing an exponential search, starting at $$k_1$$, to find the smallest $$r$$ such that $$b_{k_1+2^r} \\geq a_2$$ (this is also the smallest $$r$$ such that $$2^r \\geq k_2$$). This takes $$O(\\log k_2)$$ comparisons. Now do a binary search between the indices $$k_1$$ and $$k_1+2^r$$ to locate the correct position $$k_1+k_2$$ to insert $$a_2$$, and insert the element there. This also takes $$O(\\log k_2)$$ comparisons.\n\nIn general following this strategy you will make $$O(\\log k_i)$$ comparisons to insert element $$a_i$$. So the total number of comparisons will be $$T = O\\left(\\sum_{i=1}^n\\log k_i\\right)$$. Because the $$\\log$$ function is concave, by Jensen's inequality we have $$\\sum_{i=1}^n\\log k_i \\leq n\\log \\frac{\\sum k_i}{n} \\leq n\\log \\frac{n+m}{n}$$. The result follows.\n\nNotice that (assuming $$n\\leq m$$) this is asymptotically always at least as good as $$O(n+m)$$ and $$O(n\\log m)$$, and sometimes better than both: if you take for example $$m=n\\log n$$ you get $$O(n\\log\\log n)$$ comparisons compared to $$O(n\\log n)$$ with both other methods.\n\nTo complement this, let us show that this is optimal (up to a constant factor, still assuming $$m\\geq n$$). The number of ways to insert $$n\\geq 1$$ ordered elements into an ordered array of length m is $$\\frac{(m+n)!}{m!n!} \\geq \\frac{(m+1)^n}{n!} \\geq \\max\\{(\\frac{m}{n})^n, 2^n\\}$$. Thus to distinguish between these cases you need at least $$T\\geq \\log_2(\\frac{(m+1)^n}{n!})$$ comparisons in the worst case. We have $$T \\geq \\max\\{n\\log_2\\frac{m}{n}, n\\}$$, which implies $$T \\geq \\frac{1}{2}n\\log_2(1+\\frac{m}{n})$$.\n\nOne method fitting the analysis results presented is\none by one, insert elements of the shorter sequence ($$s$$) in the longer one ($$l$$).\n\nThere are several way to reduce the search range in the longer sequence exploiting the shorter one to be ordered, too, starting with\n\nposition \u2190 0\nfor every element $$e$$ of $$s$$\nposition \u2190 insertion position of $$e$$ in $$l$$, starting from position\ninsert $$e$$ in $$l$$ at position\n\n\u2014 without any relation specified between $$n$$, $$m$$, and the values in $$a$$ and $$b$$, I don't see any to allow a tighter bound on time than\n$$O(\\min(n,m)\\log(\\max(n,m)))$$.\n\nTry proving $$O(n+m)$$ is optimal.\n\n\u2022 Well supposing $n\\leq m$ (just to make it more readable) you can always have $O(\\min(n\\log m, n+m))$ comparisons: start with the binary search method and if you have reached $n+m$ comparisons finish with the \"typical\" method (or even start over from scratch with this method). Mar 2 at 13:36\n\u2022 @Tassle You may have missed the problem statement specifying a 3rd/output array $c_1,c_2,\u2026,c_{n+m}$. Mar 4 at 9:48\n\nThe purpose of a merge is to form a single, continuous array holding the $$n+m$$ keys from the original arrays.\n\nEven if everything can be done in-place (by $$a$$ and $$b$$ being contiguous), in the worst case you need to move all of them, which is $$\\Omega(n+m)$$. Hence a merge in time $$O(n+m)$$ is optimal.\n\nEven ignoring the moves (?), in the worst case you need to look at all keys at least once, and $$O(n+m)$$ is still optimal.\n\nA light of hope: the best case is more promising. If $$a, b$$ are contiguous and $$a_n, the merge can be done in $$O(1)$$.\n\n\u2022 If you're only interested in comparisons $O(n+m)$ is not always optimal (say if $m$ is large compared to $n$, then the binary-search strategy does $O(n\\log m)$ comparisons and doesn't need to look at all the keys in the array $b$). See my answer for what I think is the true optimum in that setting. Apr 2 at 14:10", "date": "2022-11-30 10:11:07", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/148833/fastest-algorithm-for-merge-step-in-mergesort/148839#148839", "openwebmath_score": 0.7911374568939209, "openwebmath_perplexity": 935.4769732369634, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.967899295134923, "lm_q2_score": 0.879146761176671, "lm_q1q2_score": 0.8509255304630504}}
{"url": "https://math.stackexchange.com/questions/855392/proof-of-a-subseteq-b-leftrightarrow-a-cap-b-a-check-chain-of-implicatio", "text": "# Proof of $A \\subseteq B \\Leftrightarrow A \\cap B = A$ (Check chain of implications)\n\nProve $A \\subseteq B \\Leftrightarrow A \\cap B = A$.\n\nMy attempt:\n\nCase $\\Rightarrow$:\n\n\\begin{align} A \\subseteq B & \\Rightarrow & [x\\in A \\Rightarrow x\\in B] \\\\ &\\Rightarrow &[x \\in A \\Rightarrow x\\in A \\text{ and } x\\in B] \\tag{1} \\\\ &\\Rightarrow &[x\\in A \\Rightarrow x\\in A \\cap B] \\\\ &\\Rightarrow & A \\subseteq A\\cap B \\end{align}\n\n\\begin{align} A \\subseteq B & \\Rightarrow & [x\\in A \\Rightarrow x\\in B] \\\\ & \\Rightarrow & [(x\\in A \\text{ and } x\\in B) \\Rightarrow x \\in A] \\tag{2} \\\\ & \\Rightarrow & [x \\in A \\cap B \\Rightarrow x \\in A] \\\\ & \\Rightarrow & A \\cap B \\subseteq A \\end{align}\n\n\\begin{align}\\therefore A \\subseteq B & \\Rightarrow & A \\subseteq A \\cap B \\text{ and } A \\cap B \\subseteq A \\\\ &\\Rightarrow& A \\cap B = A \\end{align}\n\nCase $\\Leftarrow$:\n\n\\begin{align} A \\cap B = A & \\Rightarrow & A \\cap B \\subseteq A \\text{ and } A \\subseteq A \\cap B \\\\ & \\Rightarrow & [(x \\in A \\text{ and } x \\in B) \\Rightarrow x \\in A ] \\text{ and } [x\\in A \\Rightarrow & (x \\in A \\text{ and } x \\in B)] \\\\ & \\Rightarrow & [x\\in A \\Rightarrow x \\in B] \\tag{3}\\\\ & \\Rightarrow & A \\subseteq B \\end{align}\n\nIs my proof correct? I am particularly not sure about line 1,2 and 3 since I just made those up (while making sure that the chain of implications is still true) as I already know the conclusion I want to arrive at.\n\n\u2022 Your (2)'s second step makes no sense, but you can fix it (probably a typo?) and you are redundant on your last two steps, but not really an \"error.\" \u2013\u00a0Adam Hughes Jul 3 '14 at 14:17\n\u2022 @AdamHughes Yes, its a typo. Thanks for calling my attention to it. \u2013\u00a0mauna Jul 3 '14 at 15:21\n\nI think you waved past a couple of steps in \"Case $\\Leftarrow$\":\n\n\\begin{align} A \\cap B = A &\\Rightarrow A \\cap B \\subseteq A \\land A \\subseteq A \\cap B \\tag{1}\\\\ & \\Rightarrow [(x \\in A \\text{ and } x \\in B) \\Rightarrow x \\in A ]\\land[x\\in A \\Rightarrow (x \\in A \\land x \\in B)]\\tag{2} \\\\ & \\Rightarrow [x\\in A \\Rightarrow x \\in B] \\tag{3}\\\\ & \\Rightarrow A \\subseteq B\\tag{4} \\end{align}\n\nAfter unpacking $A \\cap B = A$ to get $(1)$ and $(2)$ (which are both perfectly correct), I think you need more work (or more justification) before concluding: $$x\\in A \\rightarrow x\\in B\\tag{3}$$\n\n\\begin{align} A \\cap B = A &\\Rightarrow A \\cap B \\subseteq A \\land A \\subseteq A \\cap B \\tag{1}\\\\ & \\Rightarrow [(x \\in A \\land x \\in B) \\Rightarrow x \\in A ]\\\\ &\\quad \\land x\\in A \\Rightarrow (x \\in A \\land x \\in B)]\\tag{2} \\\\ &\\Rightarrow x\\in A \\Rightarrow (x \\in A \\land x\\in B)\\tag{3}\\\\ \\\\ & \\quad \\text{Assumption:} x\\in A\\tag{4}\\\\ &\\qquad\\quad {\\small \\Rightarrow} x\\in A \\land x \\in B \\tag{(5): 3 & 4}\\\\ &\\qquad\\quad {\\small \\Rightarrow} x\\in B\\tag{(6): from 5}\\\\\\\\ &\\Rightarrow x\\in A \\Rightarrow x\\in B\\tag{(7): 4-6}\\\\ & \\Rightarrow A \\subseteq B\\tag{8} \\end{align}\n\n\u2022 how did you get the assumption in line (4)? Was it based on some of the proceeding lines? \u2013\u00a0mauna Jul 3 '14 at 15:20\n\u2022 You are always free to assume something, in this case, I chose to assume $x\\in A$ because if we can show, given the earlier lines in the proof, that this implies $x\\in B$, we have proven that $x \\in A \\implies x\\in B$. Note that whether or not $x$ is in A doesn't matter. The aim of the proof is to show that IF x is in A, THEN x is in B. I.e., to establish that, together with the earlier part of the proof, $x \\in A\\implies x\\in B$ \u2013\u00a0Namaste Jul 3 '14 at 15:23\n\u2022 An alternative route is to know that $$x\\in A \\rightarrow(x\\in A\\land x\\in B) \\\\ \\Rightarrow [(x \\in A \\rightarrow x \\in A) \\land (x\\in A \\rightarrow x \\in B)] \\\\ \\Rightarrow x \\in A \\rightarrow x \\in B\\\\ \\Rightarrow A\\subseteq B$$ \u2013\u00a0Namaste Jul 3 '14 at 15:34\n\n$A \\cap B \\subseteq A$ is always true, so it's a bit misleading to say that it is implied by $A \\subseteq B$ (although not incorrect). You can simply start with the statement:\n\n$$x \\in A \\text{ and } x \\in B \\implies x \\in A$$\n\nto prove it$-$this is probably what you meant by the statement $(2)$.\n\nFor statement $(3)$ it might be more clear how you arrived at this if you say that $A \\cap B = A$ implies $A \\subseteq A \\cap B$, and since $A \\cap B \\subseteq B$, this means $A \\subseteq B$.\n\nThe rest of the proof is correct.\n\n\nAssuming that you're only allowed to use the definitions of $\\;\\subseteq\\;$ and $\\;\\cap\\;$, any proof $\\;A \\subseteq B \\;\\equiv\\; A \\cap B = A\\;$ is essentially a proof of the equivalent $$\\langle \\forall x :: x \\in A \\Rightarrow x \\in B \\rangle \\;\\equiv\\; \\langle \\forall x :: x \\in A \\land x \\in B \\;\\equiv\\; x \\in A \\rangle$$ That statement directly follows from the following more general law of propositional logic: for any $\\;P,Q\\;$, $$P \\Rightarrow Q \\;\\;\\equiv\\;\\; P \\land Q \\;\\equiv\\; P$$ There are numerous ways to prove this, depending on what laws of logic you are allowed to use. For example, one can work from right to left:\n\n$$\\calc P \\land Q \\;\\equiv\\; P \\calcop{\\equiv}{split \\;\\equiv\\; into both directions -- to introduce \\;\\Rightarrow\\; as in our goal} (P \\land Q \\Rightarrow P) \\;\\land\\; (P \\Rightarrow P \\land Q) \\calcop{\\equiv}{use \\;P \\equiv \\text{true}\\; from left hand side of \\;\\Rightarrow\\; on right hand side, twice} (P \\land Q \\Rightarrow \\text{true}) \\;\\land\\; (P \\Rightarrow \\text{true} \\land Q) \\calcop{\\equiv}{simplify: left part is \\;\\text{true}\\;; \\;\\text{true} \\land R \\equiv R\\;, twice} P \\Rightarrow Q \\endcalc$$", "date": "2019-08-23 01:25:40", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/855392/proof-of-a-subseteq-b-leftrightarrow-a-cap-b-a-check-chain-of-implicatio", "openwebmath_score": 1.0000097751617432, "openwebmath_perplexity": 461.5499674767469, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9678992960608886, "lm_q2_score": 0.8791467580102419, "lm_q1q2_score": 0.8509255282123256}}
{"url": "http://www.tpicom.eu/assassins-creed-drbg/855ddb-every-asymmetric-relation-is-antisymmetric", "text": "A relation R on a set A is symmetric if whenever (a, b) \u2208 R then (b, a) \u2208 R, i.e. Difference between antisymmetric and not symmetric. An antisymmetric and not asymmetric relation between x and y (asymmetric because reflexive) Counter-example: An symmetric relation between x and y (and reflexive ) In God we trust , \u2026 (a,a) not equal to element of R. That is. A relation R is asymmetric if and only if R is irreflexive and antisymmetric. A relation that is not asymmetric, is symmetric. 4 votes . an eigenfunction of P ij looks like. Yes. antisymmetric relation transitive relation Contents Certain important types of binary relation can be characterized by properties they have. 4 Answers. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. A asymmetric relation is an directed relationship. Whether the wave function is symmetric or antisymmetric under such operations gives you insight into whether two particles can occupy the same quantum state. 1 vote . Antisymmetric means that the only way for both $aRb$ and $bRa$ to hold is if $a = b$. Weisstein, Eric W., \"Antisymmetric Relation\", MathWorld. sets; set-theory&algebra; relations ; asked Oct 9, 2015 in Set Theory & Algebra admin retagged Dec 20, 2015 by Arjun 3.8k views. See also. Quiz & Worksheet - What is an Antisymmetric Relation? By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). Suppose that your math teacher surprises the class by saying she brought in cookies. if aRb \u21d2 bRa. We call irreflexive if no element of is related to itself. For example- the inverse of less than is also an asymmetric relation. Exercise 22 Give examples of relations which are neither symmetric, nor asymmetric. Asymmetric v. symmetric public relations. Specifically, the definition of antisymmetry permits a relation element of the form $(a, a)$, whereas asymmetry forbids that. We call antisymmetric \u2026 Asymmetric Relation: A relation R on a set A is called an Asymmetric Relation if for every (a, b) \u2208 R implies that (b, a) does not belong to R. 6. Is the relation R antisymmetric? For each of these relations on the set $\\{1,2,3,4\\},$ decide whether it is reflexive, whether it is symmetric, and whether it is antisymmetric, and whether it is transitive. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. We call symmetric if means the same thing as . In that, there is no pair of distinct elements of A, each of which gets related by R to the other. This section focuses on \"Relations\" in Discrete Mathematics. Note: a relation R on the set A is irreflexive if for every a element of A. Here's my code to check if a matrix is antisymmetric. Solution: The relation R is not antisymmetric as 4 \u2260 5 but (4, 5) and (5, 4) both belong to R. 5. Transitive if for every unidirectional path joining three vertices $$a,b,c$$, in that order, there is also a directed line joining $$a$$ to $$c$$. Discrete Mathematics Questions and Answers \u2013 Relations. Multi-objective optimization using evolutionary algorithms. I just want to know how the value in the answers come like 2^n2 and 2^n^2-1 etc. Antisymmetric if every pair of vertices is connected by none or exactly one directed line. A relation on a set is antisymmetric provided that distinct elements are never both related to one another. 15. Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij, then the possible eigenvalues are 1 and \u20131. A relation R on a set A is asymmetric if whenever (a, b) \u2208 R then (b, a) / \u2208 R for a negationslash = b. The relations we are interested in here are binary relations on a set. For example, the restriction of < from the reals to the integers is still asymmetric, and the inverse > of < is also asymmetric. Best answer. Here we are going to learn some of those properties binary relations may have. However, a relation can be neither symmetric nor asymmetric, which is the case for \"is less than or equal to\" and \"preys on\"). The relation \"x is even, y is odd\" between a pair (x, y) of integers is antisymmetric: Every asymmetric relation is also an antisymmetric relation. Antisymmetric definition, noting a relation in which one element's dependence on a second implies that the second element is not dependent on the first, as the relation \u201cgreater than.\u201d See more. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. Combine this with the previous result to conclude that every acyclic relation is irre\u00b1exive. So an asymmetric relation is necessarily irreflexive. answer comment. We call reflexive if every element of is related to itself; that is, if every has . How many number of possible relations in a antisymmetric set? Exercise 19 Prove that every asymmetric relation is irre\u00b1exive. (A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever (a,b) in R , and (b,a) in R , a = b must hold. That is, for . Limitations and opposite of asymmetric relation are considered as asymmetric relation. A relation R on a set A is non-reflexive if R is neither reflexive nor irreflexive, i.e. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) \u2208 R for every a \u2208 ASymmetricRelation is symmetric,If (a, b) \u2208 R, then (b, a) \u2208 RTransitiveRelation is transitive,If (a, b) \u2208 R & (b, c) \u2208 R, then (a, c) \u2208 RIf relation is reflexive, symmetric and transitive,it is anequivalence relation Relationship to asymmetric and antisymmetric relations. Homework 5 Solutions New York University. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. A relation becomes an antisymmetric relation for a binary relation R on a set A. Exercise 21 Give examples of relations which are neither re\u00b1exive, nor irre\u00b1exive. each of these 3 items in turn reproduce exactly 3 other items. Any asymmetric relation is necessarily antisymmetric; but the converse does not hold. Show that the converse of part (a) does not hold. Please make it clear. A relation is asymmetric if and only if it is both antisymmetric and irreflexive. In this short video, we define what an Antisymmetric relation is and provide a number of examples. But in \"Deb, K. (2013). Since dominance relation is also irreflexive, so in order to be asymmetric, it should be antisymmetric too. The mathematical concepts of symmetry and antisymmetry are independent, (though the concepts of symmetry and asymmetry are not). It's also known as \u2026 The incidence matrix $$M=(m_{ij})$$ for a relation on $$A$$ is a square matrix. Think $\\le$. Let be a relation on the set . at what time is the container 1/3 full. \"sister\" on the set of females is, \u00a8 Any nearness relation is symmetric. A relation is considered as an asymmetric if it is both antisymmetric and irreflexive or else it is not. For example, > is an asymmetric relation, but \u2265 is not. Symmetric relation; Asymmetric relation; Symmetry in mathematics; References. Yes, and that's essentially the only case : If R is both symmetric and antisymmetric then R must be the relation ## \\{(x,x),x \\in B\\} ## for some subset ## B\\subset A ##. Also, i'm curious to know since relations can both be neither symmetric and anti-symmetric, would R = {(1,2),(2,1),(2,3)} be an example of such a relation? Antisymmetry is concerned only with the relations between distinct (i.e. if a single compound is kept in a container at noon and the container is full by midnight. We call asymmetric if guarantees that . This lesson will talk about a certain type of relation called an antisymmetric relation. (a) (b) Show that every asymmetric relation is antisymmetric. Antisymmetry is different from asymmetry because it does not requier irreflexivity, therefore every asymmetric relation is antisymmetric, but the reverse is false. See also Every asymmetric relation is not strictly partial order. Lipschutz, Seymour; Marc Lars Lipson (1997). if aRa is true for some a and false for others. It can be reflexive, but it can't be symmetric for two distinct elements. Again, the previous 3 alternatives are far from being exhaustive; as an example over the natural numbers, the relation xRy defined by x > 2 is neither symmetric nor antisymmetric, let alone asymmetric. A relation is asymmetric if and only if it is both antisymmetric and irreflexive. Examples: equality is a symmetric relation: if a = b then b = a \"less than\" is not a symmetric relation, it is anti-symmetric. R is irreflexive if no element in A is related to itself. Exercise 20 Prove that every acyclic relation is asymmetric. a.4pm b.6pm c.9pm d.11pm . Similarly, the subset order \u2286 on the subsets of any given set is antisymmetric: given two sets A and B, if every element in A also is in B and every element in B is also in A, then A and B must contain all the same elements and therefore be equal: \u2286 \u2227 \u2286 \u21d2 = Partial and total orders are antisymmetric by definition. Non-examples \u00a8 The relation divides on the set of integers is neither symmetric nor antisymmetric.. Restrictions and converses of asymmetric relations are also asymmetric. Get more help from Chegg. Antisymmetric Relation. Transitive Relations: A Relation \u2026 Hint: write the definition of what it means to be asymmetric\u2026 A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the \"preys on\" relation on biological species). 3.8k views. There is an element which triplicates in every hour. We find that $$R$$ is. 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( 2013 )  relations '' in Discrete mathematics and opposite of asymmetric relation asymmetric! If it is antisymmetric every hour will talk about a Certain type of relation called an antisymmetric relation transitive Contents... The previous result to conclude that every acyclic relation is irre & ;... 19 Prove that every asymmetric relation are considered as asymmetric relation ; asymmetric relation are considered as relation. Ara is true for some a and false for others what an antisymmetric relation transitive Contents... Converse does not requier irreflexivity, therefore every asymmetric relation is and provide a number of possible relations a... Antisymmetric provided that distinct elements are never both related to itself this with the previous result conclude!, symmetric, asymmetric, is symmetric every asymmetric relation, but \u2265 is not asymmetric, it should antisymmetric. She brought in cookies not requier irreflexivity, therefore every asymmetric relation ; symmetry in mathematics References... Converse of part ( a ) does not hold if and only if is!  relations '' in Discrete mathematics Discrete mathematics ordering is antisymmetric, but \u2265 not! Is necessarily antisymmetric ; but the converse of part ( a ) ( b Show. Here 's my code to check if a single compound is kept in a antisymmetric set nearness relation is &! ( b ) Show that the converse of part ( a, each of which gets related by to. But \u2265 is not 's my code to check if a matrix is antisymmetric Marc Lars (.", "date": "2022-10-07 19:46:32", "meta": {"domain": "tpicom.eu", "url": "http://www.tpicom.eu/assassins-creed-drbg/855ddb-every-asymmetric-relation-is-antisymmetric", "openwebmath_score": 0.82464998960495, "openwebmath_perplexity": 559.41462801027, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9678992932829918, "lm_q2_score": 0.8791467580102418, "lm_q1q2_score": 0.8509255257701465}}
{"url": "http://math.stackexchange.com/questions/287802/counting-seating-arrangements-between-boys-and-girls", "text": "# counting seating arrangements between boys and girls\n\nWe would like to count how many ways 3 boys and 3 girls can sit in a row. How many ways can this be done if:\n\n(b) all the girls sit together? ...\n\n(c) every boy sits next to at least one other girl?\n\nANSWER: If three boys sit next to each other, no combinations work. If two boys sit next to each other, it fails if and only if the pair of boys sitting next to each other are on an edge (ie. BBGGBG, BGGGBB). If no two boys sit next to each other, all combinations work. There are 4!3! combinations with three boys together (see part (b)). If we place two boys on the edge, we have two choices, left or right to place them. We then choose the position of the third boy from threeremaining positions (he can't be next to the two other boys) for a total of 2*3*3!3! positions (3!3! to account for varying positions of unique boys and girls). Since there are 6! total positionings, there are 6!- 4!3!- 2*3*3!3! = 360 positionings where no two boys sit next to each other.\n\ni get where 6!- 4!3! comes from but don't understand where 2*3*3!3! comes from\n\n-\nOf course, if the kids are too shy, opposite-sex people never get to sit next to each other. \u2013\u00a0 Ahaan Rungta Feb 14 at 17:57\n\nI think it is explained in your OP. We do the same thing with a bit more detail.\n\nSince we have already counted the number of \"bad\" positions with all the boys together, it remains to count the number of bad positions in which the boys are not all together, but some boy is not next to a girl.\n\nThere must be two boys together, and they must be at the left end or the right end ($2$ choices). Take such a choice, say left end. Then the remaining boy can be in any one of $3$ places, namely positions $4$, $5$, or $6$. That gives us $(2)(3)$ ways of choosing the seats the boys will occupy. For each of these $(2)(3)$ choices, there are $3!$ ways of permuting the boys among the chosen seats, and for every such choice there are $3!$ ways of permuting the girls, for a total of $(2)(3)(3!)(3!)$.\n\nAnother way: Counting all possible arrangements and subtracting the \"bad\" ones is often good strategy. But let us count directly.\n\nOur condition will be satisfied if either no two boys are together, or exactly two boys are together and they are not in the end seats.\n\nCount first the arrangements in which no two boys are adjacent. Write down $G\\quad G\\quad G$. This determines $4$ \"gaps\" into which we can slip the boys, one boy per gap. There are $4$ \"gaps\" because we are including the end gaps. There are $\\binom{4}{3}$ ways of choosing $3$ of these gaps.\n\nOr else we could slip $2$ boys into one of the two center gaps ($2$ choices), and then slip the remaining boy into one of the $3$ remaining gaps, for a total of $6$ choices.\n\nThus the places for the boys can be chosen in $10$ ways. For each of these ways, we can arrange the boys is $3!$ ways, and then the girls in $3!$ ways, for a total of $360$.\n\n-", "date": "2014-08-01 00:09:23", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/287802/counting-seating-arrangements-between-boys-and-girls", "openwebmath_score": 0.7902367115020752, "openwebmath_perplexity": 272.6181446275246, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303425461247, "lm_q2_score": 0.8596637541053282, "lm_q1q2_score": 0.8509212682005646}}
{"url": "http://mathhelpforum.com/calculus/40628-how-would-you-integrate.html", "text": "# Math Help - How would you integrate this?\n\n1. ## How would you integrate this?\n\n5x\n(x-1)((x^2)+4)\n\nfractions?\n5x = A + B\n(x-1) ?\n\nAnd ((x^2)+4) is 1/2arctan(x/2)\n\nthe answer key says\nln|x-1| -.5ln|x^2+4| - 2arctan(x/2)\n\nCan anyone help? Thanks.\n\n2. $\\frac{5x}{(x-1)(x^2+4)} = \\frac{A}{x-1} + \\frac{Bx + C}{x^2+4}$ (since $x^2 + 4$ is an irreducible quadratic, indicating your numerator is in the form of Bx + C)\n\nSee what you can do with this and come back with any questions.\n\n3. Originally Posted by khuezy\n5x\n(x-1)((x^2)+4)\n\nfractions?\n5x = A + B\n(x-1) ?\n\nAnd ((x^2)+4) is 1/2arctan(x/2)\n\nthe answer key says\nln|x-1| -.5ln|x^2+4| - 2arctan(x/2)\n\nCan anyone help? Thanks.\nYou can use partial fractions with this setup\n\n$\\frac{5x}{(x^2+4)(x-1)}=\\frac{A}{(x-1)}+\\frac{Bx+C}{x^2+4}$\n\nOr observe that\n\n$\\frac{5x}{(x-1)(x^2+4)}=\\frac{5x-x^2-4+x^2+4}{(x-1)(x^2+4)}=\\frac{-(x^2-5x+4)}{(x-1)(x^+2)}+\\frac{x^2+4}{(x-1)(x^2+4)}=$\n\n$\\frac{-(x-4)(x-1)}{(x^2+4)(x-1)}+\\frac{x^2+4}{(x-1)(x^2+4)}=\\frac{4-x}{x^2+4}+\\frac{1}{x-1}=\\frac{4}{x^2+4}-\\frac{x}{x^2+4}+\\frac{1}{x-1}$\n\nNow we have\n\n$\\int \\left( \\frac{4}{x^2+4}-\\frac{x}{x^2+4}+\\frac{1}{x-1} \\right)dx$\n\nThe first is in the form of the arctangent or if you want to do it by hand let x=2tan(u) the 2nd is a u sub on the denominator $u=x^2+4$ and the last one is a u sub on the denominator $u=x-1$.\n\nI hope this helps.\n\nGood luck.\n\n4. ## help\n\nI got\n5x = (A+B)x^2 + 4A - (B+C)x - C\n\nI'm confused on what to do next.\n\nEquating coefficients?\nA+B = 0 4A=5 B+C=0 C=0????????????????\n\nThanks.\n\n5. Here's another way using trig.\n\n$\\frac{x-1}{x^{2}+4}dx$\n\nLet $x=2tan(t), \\;\\ dx=2sec^{2}(t)dt$\n\nMaking the subs we get a menacing looking thing, but it simplifies down fairly nice.\n\nMaking the subs we get:\n\n$\\frac{2tan(t)-1}{4(tan^{2}(t)+1)}\\cdot{2sec^{2}(t)}dt$\n\nThis, believe it or not, whittles down to\n\n$tan(t)-\\frac{1}{2}$\n\n$\\int{tan(t)}dt-\\frac{1}{2}\\int{dt}$\n\nThen, after integrating, resub in $t=tan^{-1}(\\frac{x}{2})$\n\nYou may wish to stick with the others methods. I just threw it out there.\n\n6. Originally Posted by khuezy\n5x\n(x-1)((x^2)+4)\n\nfractions?\n5x = A + B\n(x-1) ?\n\nAnd ((x^2)+4) is 1/2arctan(x/2)\n\nthe answer key says\nln|x-1| -.5ln|x^2+4| - 2arctan(x/2)\n\nCan anyone help? Thanks.\n$\\frac{5x}{(x^2+4)(x-1)}=\\frac{A}{x-1}+\\frac{Bx+C}{x^2+4}$\n\nmultiplying through by the common denominator we get\n\n$5x=A(x^2+4)+(Bx+C)(x-1)$\n\nNow letting $x=1$ we get\n\n$5(1)=A(1^2+4)+(B+C)(0)\\Rightarrow{A=1}$\n\nSo now for the second factor we need to expand, but dont forget A=1\n\nSo we have\n\n$5x=x^2+4+Bx^2-Bx+Cx-C$\n\nSo we see that $B+1=0\\Rightarrow{B=-1}$\n\nand also seeing $4-C=0\\Rightarrow{C=4}$\n\nSo finally we can see that\n\n$\\frac{5x}{(x^2+4)(x-1)}=\\frac{1}{x-1}+\\frac{4-x}{x^2+4}$\nand we see that", "date": "2014-04-21 07:48:59", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/40628-how-would-you-integrate.html", "openwebmath_score": 0.910505473613739, "openwebmath_perplexity": 2188.5456431654916, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9898303443461076, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.8509212608508263}}
{"url": "https://besatwise.com/5r2eyhdy/transitive-closure-of-a-relation-0b5e49", "text": "# transitive closure of a relation\n\nOtherwise, it is equal to 0. 1. De nition 2. transitive closure can be a bit more problematic. The transitive closure of a is the set of all b such that a ~* b. A = {a, b, c} Let R be a transitive relation defined on the set A. R =, R \u2194, R +, and R * are called the reflexive closure, the symmetric closure, the transitive closure, and the reflexive transitive closure of R respectively. Algorithm Warshall Notice that in order for a \u2026 The last item in the proposition permits us to call R * the transitive reflexive closure of R as well (there is no difference to the order of taking closures). The program calculates transitive closure of a relation represented as an adjacency matrix. 3) The time complexity of computing the transitive closure of a binary relation on a set of n elements is known to be: a) O(n) b) O(nLogn) c) O(n^(3/2)) d) O(n^3) Answer (d) In mathematics, the transitive closure of a binary relation R on a set X is the smallest transitive relation on X that contains R. The transitive closure of a binary relation $$R$$ on a set $$A$$ is the smallest transitive relation $$t\\left( R \\right)$$ on $$A$$ containing $$R.$$ The transitive closure is more complex than the reflexive or symmetric closures. Element (i,j) in the matrix is equal to 1 if the pair (i,j) is in the relation. Loosely speaking, it is the set of all elements that can be reached from a, repeatedly using relation \u2026 Transitive Relation - Concept - Examples with step by step explanation. Transitive closure. It is not enough to \ufb01nd R R = R2. Let us consider the set A as given below. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) \u2208 R for every a \u2208 ASymmetricRelation is symmetric,If (a, b) \u2208 R, then (b, a) \u2208 RTransitiveRelation is transitive,If (a, b) \u2208 R & (b, c) \u2208 R, then (a, c) \u2208 RIf relation is reflexive, symmetric and transitive,it is anequivalence relation In this article, we will begin our discussion by briefly explaining about transitive closure and the Floyd Warshall Algorithm. R2 is certainly contained in the transitive closure, but they are not necessarily equal. It can be shown that the transitive closure of a relation R on A which is a finite set is union of iteration R on itself |A| times. TRANSITIVE RELATION. We will also see the application of Floyd Warshall in determining the transitive closure of a given graph. The transitive closure of R is the relation Rt on A that satis es the following three properties: 1. Warshall\u2019s Algorithm: Transitive Closure \u2022 Computes the transitive closure of a relation Transitive Closures Let R be a relation on a set A. Connectivity Relation A.K.A. De\ufb01ning the transitive closure requires some additional concepts. This allows us to talk about the so-called transitive closure of a relation ~. For calculating transitive closure it uses Warshall's algorithm. Hence the matrix representation of transitive closure is joining all powers of the matrix representation of R from 1 to |A|. In a sense made precise by the formal de nition, the transitive closure of a relation is the smallest transitive relation that contains the relation. Let A be a set and R a relation on A. For transitive relations, we see that ~ and ~* are the same.", "date": "2021-03-09 00:37:52", "meta": {"domain": "besatwise.com", "url": "https://besatwise.com/5r2eyhdy/transitive-closure-of-a-relation-0b5e49", "openwebmath_score": 0.818859338760376, "openwebmath_perplexity": 239.52829530015248, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9898303413461358, "lm_q2_score": 0.8596637433190938, "lm_q1q2_score": 0.8509212564924354}}
{"url": "http://mathhelpforum.com/differential-equations/176023-dirac-delta-laplace-transform.html", "text": "# Thread: Dirac Delta and Laplace Transform\n\n1. ## Dirac Delta and Laplace Transform\n\nHi guys, my professor recently gave us this problem:\n\n$\\displaystyle y^{\\prime\\prime}+4y=\\sum_{k=1}^{\\infty}\\delta(t-k\\pi)$ with all zero initial conditions to solve using Laplace transforms. So I assume that even though it's an infinite sum, in this case you can take the Laplace transform of the summand and sum that to get...\n\n$\\displaystyle Y(s)=\\frac{1}{s^2+4}\\sum_{k=1}^{\\infty}e^{-sk\\pi}=\\cfrac{1}{(e^{s\\pi}-1)(s^2+4)}$.\n\nHowever I'm pretty sure that isn't invertible, or if it is I can't see a way to invert it cleanly. Did I mess up somewhere? Thanks.\n\n2. Originally Posted by kenndrylen\nHi guys, my professor recently gave us this problem:\n\n$\\displaystyle y^{\\prime\\prime}+4y=\\sum_{k=1}^{\\infty}\\delta(t-k\\pi)$ with all zero initial conditions to solve using Laplace transforms. So I assume that even though it's an infinite sum, in this case you can take the Laplace transform of the summand and sum that to get...\n\n$\\displaystyle Y(s)=\\frac{1}{s^2+4}\\sum_{k=1}^{\\infty}e^{-sk\\pi}=\\cfrac{1}{(e^{s\\pi}-1)(s^2+4)}$.\n\nHowever I'm pretty sure that isn't invertible, or if it is I can't see a way to invert it cleanly. Did I mess up somewhere? Thanks.\nHere is an idea instead of summing the series just invert the series term by term\n\n$\\displaystyle \\displaystyle Y(s)=\\sum_{k=1}^{\\infty}\\frac{e^{-s\\pi k}}{s^2+4}$\n\n$\\displaystyle \\displaystyle \\mathcal{L}^{-1}\\left( \\sum_{k=1}^{\\infty}\\frac{e^{-s\\pi k}}{s^2+4}\\right)=\\sum_{k=1}^{\\infty}\\mathcal{L}^{-1}\\left( \\frac{e^{-s \\pi k}}{s^2+4}\\right)=\\sum_{k=1}^{\\infty}\\sin\\left(t-k\\pi \\right)u(t-k\\pi)$\n\nWhere $\\displaystyle u(t)=\\begin{cases}0, \\text{ if } t < 0 \\\\ 1, \\text{ if } t \\ge 0 \\end{cases}$ the Heaviside function.\n\n3. You have $\\displaystyle \\displaystyle Y(s) = \\frac{1}{s^2 + 4}\\sum_{k = 1}^{\\infty}e^{-sk\\pi} = \\sum_{k = 1}^{\\infty}e^{-sk\\pi}\\left(\\frac{1}{s^2 + 4}\\right)$.\n\nNow recall that $\\displaystyle \\displaystyle \\mathcal{L} ^{-1}\\left[e^{-as}F(s)\\right] = f(t-a)H(t-a)$, and the laplace transform of a sum is the sum of the laplace transforms.\n\n4. Oh! Nice idea guys! I don't know why I missed that\n\nThanks both of you", "date": "2018-05-28 01:44:50", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/differential-equations/176023-dirac-delta-laplace-transform.html", "openwebmath_score": 0.952051043510437, "openwebmath_perplexity": 259.21784723863004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707999669627, "lm_q2_score": 0.8705972801594706, "lm_q1q2_score": 0.8508963601585238}}
{"url": "https://math.stackexchange.com/questions/161565/what-is-the-total-number-of-combinations-of-5-items-together-when-there-are-no-d", "text": "What is the total number of combinations of 5 items together when there are no duplicates?\n\nI have 5 categories - A, B, C, D & E.\n\nI want to basically create groups that reflect every single combination of these categories without there being duplicates.\n\nSo groups would look like this:\n\n\u2022 A\n\u2022 B\n\u2022 C\n\u2022 D\n\u2022 E\n\u2022 A, B\n\u2022 A, C\n\u2022 A, D\n\u2022 A, E\n\u2022 B, C\n\u2022 B, D\n\u2022 B, E\n\u2022 C, D . . . etc.\n\nThis sounds like something I would use the binomial coefficient $n \\choose r$ for, but I am quite fuzzy on calculus and can't remember exactly how to do this.\n\nAny help would be appreciated.\n\nThanks.\n\nLet $$nCr=\\binom{n}{r}=\\frac{n!}{k!(n-k)!}$$ Remember that the $\\frac{n!}{(n-k)!}$ gives all the permutations and the $k!$ in the denominator is what disregards duplicates.\n\nNow; you want all the ways you can choose $$(1 \\text{ category from } 5) + (2 \\text{ category from } 5) + \\dots + (5 \\text{ category from } 5)$$ i.e. $$\\binom{5}{1}+\\binom{5}{2}+\\binom{5}{3}+\\binom{5}{4}+\\binom{5}{5}=2^5-1=31$$ Note that this follows from the fact that $$(1+1)^n=\\binom{n}{0}+\\binom{n}{1}+\\cdots+\\binom{n}{n-1}+\\binom{n}{n}=2^n$$ Subtracting $\\binom{n}{0}$ from both sides gives us $$\\binom{n}{1}+\\cdots+\\binom{n}{n-1}+\\binom{n}{n}=2^n-\\binom{n}{0}$$ But since $\\binom{n}{0}=1,\\forall n\\in\\mathbb{N}$ we have that $$\\binom{n}{1}+\\cdots+\\binom{n}{n-1}+\\binom{n}{n}=2^n-1$$ When $n=5$ we thus get the above answer.\n\nAddendum: To address your concern that there seems to be more than $31$ combinations, here is a list of all the possibilities: $$\\begin{array}{|c|c|c|c|c|c|c|} & 1 \\text{ category} & 2 \\text{ categories} & 3 \\text{ categories} & 4 \\text{ categories} & 5 \\text{ categories} & \\text{Sum}\\\\ \\hline & A & AB & ABC & ABCD & ABCDE\\\\ \\hline & B & AC & ABD & ABCE \\\\ \\hline & C & AD & ABE & ABDE \\\\ \\hline & D & AE & ACD & ACDE \\\\ \\hline & E & BC & ACE & BCDE \\\\ \\hline & & BD & ADE \\\\ \\hline & & BE & BCD \\\\ \\hline & & CD & BCE \\\\ \\hline & & CE & BDE \\\\ \\hline & & DE & CDE \\\\ \\hline \\text{Total} & 5 & 10 & 10 & 5 & 1 & 31 \\\\ \\hline \\end{array}$$\n\n\u2022 Wish I could upvote this answer twice. Thanks much. That table REALLY helped. Jun 22 '12 at 8:59\n\u2022 @marcamillion glad to help :)\n\u2013\u00a0E.O.\nJun 22 '12 at 9:00\n\u2022 I just want to make sure I am understanding this correctly (I too am fuzzy and trying to recall) if I want to know what all the combinations of days in the week are then I would just do: 2^7 - 1 = 127? I keep getting terribly confused as to when I should use factorials, correct me if I am wrong but 7! would be used to find the permutations, not the combinations? Aug 16 '18 at 21:30\n\nThere are $\\binom{5}{1}$ combinations with 1 item, $\\binom{5}{2}$ combinations with $2$ items,...\n\nSo, you want : $$\\binom{5}{1}+\\cdots+\\binom{5}{5}=\\left(\\binom{5}{0}+\\cdots+\\binom{5}{5}\\right)-1=2^5-1=31$$\n\nI used that $$\\sum_{k=0}^n\\binom{n}{k}=(1+1)^n=2^n$$\n\n\u2022 You know...that was my initial inclination - but then I started writing them out and it seems like there would be more than 31 combinations. What's this theory called? Or is there no name? Jun 22 '12 at 8:32\n\u2022 I think it's simply combinatorics.\n\u2013\u00a0JBC\nJun 22 '12 at 8:36\n\u2022 Why do you subtract the 1 at the end? Also, can you explain the theory of why the combination with 3 items will be the same as the combination with 2 items...that seems counter-intuitive. Jun 22 '12 at 8:37\n\u2022 1) I substracted $\\binom{5}{0}=1$ to use the formula recalled at the end (Notice that the formula begins by $\\binom{5}{0}$ but your sum by $\\binom{5}{1}$). 2) $\\binom{n}{k}$ is the number of subset of $\\{1,\\ldots,n\\}$ with $k$ elements, ie the number of choices to take $k$ elements from a set of $n$ elements without repetition, you can show that $\\binom{n}{k}=\\binom{n}{n-k}$ using $\\binom{n}{k}=\\frac{n!}{k!(n-k)!}$.\n\u2013\u00a0JBC\nJun 22 '12 at 8:44\n\u2022 And I think that this formula is intuitive : chosing the k items we take, it's the same thing as choosing the n-k items we left.\n\u2013\u00a0JBC\nJun 22 '12 at 10:20\n\nThus there are $2^5 = 32$ possibilities. However, you are not counting the choice of none of the five categories, so we subtract $1$ to get $31$ possibilities.\n\u2022 I think this is the most intuitive way to formualate the solution, and the $2^n$ formula is more natural than when presented in @JBC's answer. Jun 22 '12 at 13:55", "date": "2021-10-21 15:17:33", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/161565/what-is-the-total-number-of-combinations-of-5-items-together-when-there-are-no-d", "openwebmath_score": 0.7578458786010742, "openwebmath_perplexity": 329.89108939629574, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773707999669627, "lm_q2_score": 0.8705972751232809, "lm_q1q2_score": 0.850896355236299}}
{"url": "https://math.stackexchange.com/questions/4067252/how-to-prove-big-o-omega-and-theta-asymptotic-notations", "text": "# How to prove Big O, Omega and Theta asymptotic notations?\n\nI know the definitions of this notations\n\n1. Big $$\\mathcal{O} \\; : \\enspace T(n) \\in \\mathcal{O}(f(n))$$ if and only if $$\u2203 \\, c, n_0$$, such that $$T(n) \\leq c \\cdot f(n) \\enspace \\forall \\,n \\geq n_0$$.\n2. Big $$\\Omega \\; : \\enspace T(n) \\in \\Omega(f(n))$$ if and only if $$\u2203 \\, c, n_0$$, such that $$T(n) \\geq c \\cdot f(n) \\enspace \\forall \\,n \\geq n_0$$.\n3. Big $$\\Theta \\; : \\enspace T(n) \\in \\Theta(f(n))$$ if and only if $$\u2203 \\, c_1, c_2, n_0$$ such that $$c_1 f(n) \\leq T(n) \\leq c_2 f(n) \\enspace$$ $$\\forall \\,n \\geq n_0$$.\n\nI'm having trouble manipulating this definitions to prove some notation for example:\n\n\u2022 Suppose that $$f = \\Theta(g)$$ and $$g = \\Theta(h)$$. Prove that $$f = \\Theta(h)$$\n\u2022 Suppose that $$f$$ and $$g$$ are two non-negative functions such that $$g = \\mathcal{O}(f)$$. Prove that $$f + g = \\Theta(f)$$.\n\nIs there anything that could help me? I've read, watch videos but nothing helps me clarifying to prove these notations. Anything helps thanks. Also if you're a tutor and there's a way to set up a meeting it would help.\n\n\u2022 Please format your question to be more readable, also use Mathjax. That way people won't be deterred to read it :) Mar 18, 2021 at 19:00\n\u2022 Oh no. You can just edit it with the Edit button under your question. This is MathJax Mar 18, 2021 at 19:07\n\u2022 I have edited your question, now its your turn to make it pretty. This question is a good example. Try to break the question into lines to improve its readability. Basically the more effort you put in the more likely people will answer you. Make it look as good as you would want want something you read to look. Mar 18, 2021 at 19:10\n\u2022 Thanks, i'm doing it right now will take a couple of minutes. How do i break it in lines? I kept clicking enter but never happened. @LordCommander Mar 18, 2021 at 19:11\n\u2022 @LordCommander Is this better? In, stackoverflow i always get in trouble with doing a question. Mar 18, 2021 at 19:27\n\nConsidering the first problem:\n\n1.) Because $$f \\in \\Theta(g)$$ there must exist $$c_1, c_2$$ and $$n_0$$ such that\n\n$$c_1 g(n) \\leq f(n) \\leq c_2 g(n) \\tag{1}$$\n\nfor all $$n \\geq n_0$$.\n\n2.) Because $$g \\in \\Theta(h)$$ there must exist $$\\tilde{c}_1, \\tilde{c}_2$$ and $$\\tilde{n}_0$$ such that\n\n$$\\tilde{c}_1 h(n) \\leq g(n) \\leq \\tilde{c}_2 h(n) \\tag{2}$$\n\nfor all $$n \\geq \\tilde{n}_0$$.\n\n3.) Combining 1.) and 2.) yields\n\n$$c_1 \\tilde{c}_1 h(n) \\leq f(n) \\leq c_2 \\tilde{c_2} h(n) \\tag{3}$$\n\nTherefore there exist $$\\hat{c}_1, \\hat{c}_2$$, namely $$\\hat{c}_1 = c_1 \\tilde{c}_1$$ and $$\\hat{c}_2 = c_2 \\tilde{c}_2$$ and a $$\\hat{n}_0 = \\max \\{ n_0, \\tilde{n_0} \\}$$ such that $$f$$ satisfies the conditions for being an element of $$\\Theta(h)$$, i.e. $$f \\in \\Theta(h)$$\n\nP.S.: Why $$\\hat{n}_0 = \\max \\{ n_0, \\tilde{n_0} \\}$$? Because you want both inequalites to hold, in order to insert them into each other. The first one holds for all $$n \\geq n_0$$, the second one holds for all $$n \\geq \\tilde{n}_0$$ and therefore a combination of those two can only hold for all $$n \\geq \\hat{n}_0 = \\max \\{ n_0, \\tilde{n}_0 \\}$$\n\nEdit 1 I numbered the inequalities for better reference.\n\nRow $$(1)$$ consists of two inequalities, namely $$c_1 g(n) \\leq f(n)$$ and $$f(n) \\leq c_2 g(n)$$. These inequalities are true for all $$n \\geq n_0$$.\n\nRow $$(2)$$ consists again of two inequalities, namely $$\\tilde{c}_1 h(n) \\leq g(n)$$ and $$g(n) \\leq \\tilde{c}_2 h(n)$$. These inequalities are true for all $$n \\geq \\tilde{n}_0$$.\n\nNote, that $$n_0$$ and $$\\tilde{n}_0$$ do not necessarily have to be the same.\n\nWe insert the inequalities of $$(1)$$ and $$(2)$$ into each other to obtain $$(3)$$. This only makes sense if the inequalities we are inserting are indeed true. But if $$n_0 < \\tilde{n}_0$$, then row $$(1)$$ is still true for all $$n \\geq \\tilde{n}_0$$. Otherwise, if $$n_0 > \\tilde{n}_0$$ then row $$(2)$$ is still true for all $$n \\geq n_0$$. So you choose the greater of these two, i.e. $$\\max \\{ n_0, \\tilde{n}_0 \\}$$ in order for both rows $$(1)$$ and $$(2)$$ to hold. Then you have no problems combining them.\n\n\u2022 I'm having trouble understanding from where the max came from and when to use it. How did you got to combine them? Is combining them part of proving any asymptotic notations? Mar 18, 2021 at 20:09\n\u2022 I updated my answer, I hope it is more clear now. The point is, that by definition you need to find a $\\hat{n}_0$ such that row $(3)$ holds. Mar 18, 2021 at 20:21\n\u2022 i'm trying to do the second example what if $g = O(f)$ and how i can identify what it is? Can i define f and g as Big O? Mar 18, 2021 at 20:56\n\u2022 Big O of a function $f$, i.e. $\\mathcal{O}(f)$ is a set. Writing something like $g = \\mathcal{O}(f)$ is pretty common, but it is abuse of notation. The correct way to write it would be $g \\in \\mathcal{O}(f)$. So your task is to check whether $g$ is an element of the set. You do this by checking whether $g$ satisfies the necessary conditions. Mar 18, 2021 at 21:07\n\u2022 So i could say that $g(n) \\leq c * f(n)$ for all $n \\geq n_0$? And that proves that $g \\in { O } (f)$? Mar 18, 2021 at 21:29\n\nFrom scratch then:\n\nConsider some arbitrary function $$f$$ and the \"Big O\" of $$f$$, i.e. $$\\mathcal{O}(f)$$. This $$\\mathcal{O}(f)$$ is a set, consisting of all functions that satisfy a certain condition. This condition somehow includes $$f$$. You have already given the definition of $$\\mathcal{O}(f)$$, but I will rewrite it now:\n\n$$\\mathcal{O}(f) \\enspace = \\enspace \\Big\\{ \\; g : \\mathbb{N} \\longrightarrow \\mathbb{R} \\; \\Big| \\; \\exists \\, c > 0 \\; \\exists \\, n_0 \\in \\mathbb{N} \\; \\forall \\, n \\geq n_0 \\; : \\; g(n) \\leq c \\cdot f(n) \\; \\Big\\}$$\n\nIn other words, the set $$\\mathcal{O}(f)$$ is a set of functions who map from the natural numbers $$\\mathbb{N}$$ to the real numbers $$\\mathbb{R}$$. These functions have to satisfy a certain condition, namely that there is some constant $$c$$, which is greater than zero, such that $$g(n) \\leq c \\cdot f(n)$$. This inequality does not have to hold for every function value of $$g(n)$$ and $$f(n)$$. It only has to hold for all function values after a certain \"point\" $$n_0$$. Let us look at an example:\n\nExample: Consider the functions\n\n$$g(n) = n \\qquad \\text{and} \\qquad f(n) = n^2 - n$$\n\nWe want to show that $$g \\in \\mathcal{O}(f)$$ holds. If we insert $$n=1$$, we find\n\n$$g(1) = 1 \\qquad \\text{and} \\qquad f(1) = 0$$\n\nBut there exists no constant $$c > 0$$, such that $$1 \\leq c \\cdot 0$$. Does this mean, that $$g \\notin \\mathcal{O}(f)$$? The answer is NO. Why? Because if we insert $$n=2$$ or $$n=3$$ or $$n = 4,5,6, \\ldots$$ we find that for these values the inequality\n\n$$g(n) \\leq c \\cdot f(n)$$\n\nis fulfilled when e.g. picking $$c = 1$$. This means, that when $$c = 1$$ then for all $$n$$ that are greater than $$n_0 = 1$$ we find $$g(n) \\leq c \\cdot f(n)$$. However, this is exactly the definition of $$\\mathcal{O}(f)$$. Therefore,\n\n$$g(n) \\in \\enspace \\mathcal{O}(n^2 - n)$$\n\n$${}$$\n\nProblem 2\n\nConsider now the second problem you have stated. We want to prove that if $$g \\in \\mathcal{O}(f)$$ then $$f + g \\in \\Theta(f)$$.\n\nHow do we do this? First of all, we gather the information we have. In this case, we know that $$g \\in \\mathcal{O}(f)$$. By definition, we know that there exists a $$c>0$$ and a $$n_0$$ such that for all $$n \\geq n_0$$ the inequality $$g(n) \\leq c \\cdot f(n)$$ holds.\n\nNow we simply add $$f(n)$$ on both sides of the inequality (which we know by assumption to be true). This gives us\n\n$$g(n) + f(n) \\leq (c+1) \\cdot f(n)$$\n\nThis was the first part. Now the second part: $$g(n)$$ is nonnegative, so we know that $$g(n) \\geq 0$$ and therefore\n\n$$f(n) \\leq f(n) + g(n)$$\n\n(because adding something positive to a number always makes the number greater).\n\nWe combine these two results and have\n\n$$f(n) \\; \\leq \\; f(n) + g(n) \\; \\leq \\; (c+1) f(n)$$\n\nThis completes the proof that $$(f + g) \\in \\Theta(f)$$.\n\n\u2022 Thank you so much, this helped. Mar 19, 2021 at 16:36", "date": "2022-10-06 17:54:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4067252/how-to-prove-big-o-omega-and-theta-asymptotic-notations", "openwebmath_score": 0.8571826815605164, "openwebmath_perplexity": 165.0152616739415, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9773708052400943, "lm_q2_score": 0.870597270087091, "lm_q1q2_score": 0.850896354904848}}
{"url": "https://brilliant.org/discussions/thread/any-better-way-to-find-the-minimamaxima/", "text": "Any better way to find the minima/maxima?\n\nThe teacher asked us to minimize an expression:$\\frac{x^2+y^2+z^2}{xy+yz}$, where $x, y, z>0$\n\nSince the denominator and numerator are homogeneous, we try to divide the y^2 into two halves:\n\n$\\frac{x^2+\\frac{1}{2}y^2+\\frac{1}{2}y^2+z^2}{xy+yz}$\n\nUse the inequality $a+b\\ge2\\sqrt{ab} (a,b>0)$ and we get:\n\n$\\frac { x^{ 2 }+\\frac { 1 }{ 2 } y^{ 2 }+\\frac { 1 }{ 2 } y^{ 2 }+z^{ 2 } }{ xy+yz } \\ge \\frac { 2\\sqrt { x^{ 2 }\\cdot \\frac { 1 }{ 2 } y^{ 2 } } +2\\sqrt { \\frac { 1 }{ 2 } y^{ 2 }\\cdot z^{ 2 } } }{ xy+yz } =\\boxed{2}$\n\nalso, the same way to find the minima of $\\frac { 10x^{ 2 }+10y^{ 2 }+z^{ 2 } }{ xy+xz+yz }$ :\n\n$\\frac { 10x^{ 2 }+10y^{ 2 }+z^{ 2 } }{ xy+xz+yz } \\\\ =\\frac { [ax^{ 2 }+ay^{ 2 }]+[(10-a)x^{ 2 }+bz^{ 2 }]+[(10-a)y^{ 2 }+(1-b)z^{ 2 }] }{ xy+xz+yz } \\\\ \\ge \\frac { 2\\sqrt { ax^{ 2 }\\cdot ay^{ 2 } } +2\\sqrt { (10-a)x^{ 2 }\\cdot bz^{ 2 } } +2\\sqrt { (10-a)y^{ 2 }\\cdot (1-b)z^{ 2 } } }{ xy+xz+yz } \\\\ =\\frac { 2a\\cdot xy+2\\sqrt{(10-a)b}\\cdot xz+2\\sqrt{(10-a)(1-b)}\\cdot yz }{ xy+xz+yz }$\n\nNow we let $a=\\sqrt{(10-a)b} = \\sqrt{(10-a)(1-b)}$ , that is $a = 2 \\ and\\ b = \\frac{1}{2}$\n\nThen\n\n$\\frac { 4xy+4xz+4yz }{ xy+xz+yz } \\\\ = \\boxed{4}$\n\nBut this method is quite troublesome... right?\n\nIs there any other way to solve the minima? I also tried partial derivative, but it's not easy to solve that equation either...\n\nNote by John Lee\n1\u00a0year, 1\u00a0month ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nYes there is another approach................Try to use spherical co-ordinates...........That makes things simpler and transforms everything into two variables.........!!\n\n- 1\u00a0year, 1\u00a0month ago\n\nCare to elaborate on this? What substitution works here? Are you sure the constraints are easily manageable?\n\n- 1\u00a0year, 1\u00a0month ago\n\nUmm yes.........well, we can use the standard substitutions used when transforming from xyz co-ordinates to spherical co-ordinates.....!!\n\n- 1\u00a0year, 1\u00a0month ago\n\nI don't understand. What standard substitutions are you referring to?\n\n- 1\u00a0year, 1\u00a0month ago\n\nWell.........here you go, Sir........\nx = rsin(phi)cos(theta)\ny = rsin(phi)sin(theta)\nz = r*cos(phi)\nHere, phi and theta are two different angles........\n\n- 1\u00a0year, 1\u00a0month ago\n\nAlso, r is the radius of the sphere.........we are just taking these as dummy variables.........\n\n- 1\u00a0year, 1\u00a0month ago\n\nAlso, after the transformation, we will have a unit fraction to deal with........and hence it would simply be the matter of maximising the denominator.......\n\n- 1\u00a0year, 1\u00a0month ago", "date": "2020-01-21 12:55:18", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/any-better-way-to-find-the-minimamaxima/", "openwebmath_score": 0.976678192615509, "openwebmath_perplexity": 2899.534723399127, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9773708039218115, "lm_q2_score": 0.8705972684083609, "lm_q1q2_score": 0.8508963521164129}}
{"url": "http://mathhelpforum.com/algebra/223355-oblique-asymptotes.html", "text": "1. ## Oblique Asymptotes\n\nhi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to $f(x) = (x^3-16x)/(-4x^2+4x+24)$ how come?\n\nI mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: $y= 1/4 x + 1/4$.\n\n2. ## Re: Oblique Asymptotes\n\nOriginally Posted by sakonpure6\nhi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to $f(x) = (x^3-16x)/(-4x^2+4x+24)$ how come?\n\nI mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: $y= 1/4 x + 1/4$.\nYou are off by a - sign. See below.\n\n-Dan\n\n3. ## Re: Oblique Asymptotes\n\nOh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.\n\n4. ## Re: Oblique Asymptotes\n\nHello, sakonpure6!\n\n$\\text{Find the oblique asymptote: }\\:f(x) \\:=\\: \\frac{x^3-16x}{-4x^2+4x+24}$\n\n$\\text{We have: }\\:f(x) \\;=\\;\\frac{x^3-6x}{\\text{-}4x^2 + 4x + 24}$\n\n$\\text{Divide numerator and denominator by }x^2\\!:$\n\n. . $f(x) \\;=\\;\\dfrac{\\frac{x^3}{x^2} - \\frac{6x}{x^2}}{\\frac{\\text{-}4x^2}{x^2} + \\frac{4x}{x^2} + \\frac{24}{x^2}} \\;=\\;\\frac{x - \\frac{16}{x}}{\\text{-}4 + \\frac{4}{x} + \\frac{24}{x^2}}$\n\n$\\text{Hence: }\\:\\lim_{x\\to\\inty}f(x) \\;=\\;\\lim_{x\\to\\infty} \\frac{x - \\frac{16}{x}}{\\text{-}4 + \\frac{4}{x} + \\frac{24}{x^2}} \\;=\\;\\frac{x-0}{\\text{-}4+0+0} \\;=\\;\\frac{x}{\\text{-}4}$\n\n$\\text{The oblique asymptote is: }\\:y \\:=\\:\\text{-}\\tfrac{1}{4}x$\n\n5. ## Re: Oblique Asymptotes\n\nOriginally Posted by sakonpure6\nOh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.\nIt lines up. Graph it on a calculator and zoom out. (Or you could do it for real and find the limit. )\n\n-Dan\n\n6. ## Re: Oblique Asymptotes\n\nThe oblique asymptote is y=-x/4 -1/4\n\n7. ## Re: Oblique Asymptotes\n\nOriginally Posted by Soroban\nHello, sakonpure6!\n\n$\\text{We have: }\\:f(x) \\;=\\;\\frac{x^3-6x}{\\text{-}4x^2 + 4x + 24}$\n\n$\\text{Divide numerator and denominator by }x^2\\!:$\n\n. . $f(x) \\;=\\;\\dfrac{\\frac{x^3}{x^2} - \\frac{6x}{x^2}}{\\frac{\\text{-}4x^2}{x^2} + \\frac{4x}{x^2} + \\frac{24}{x^2}} \\;=\\;\\frac{x - \\frac{16}{x}}{\\text{-}4 + \\frac{4}{x} + \\frac{24}{x^2}}$\n\n$\\text{Hence: }\\:\\lim_{x\\to\\inty}f(x) \\;=\\;\\lim_{x\\to\\infty} \\frac{x - \\frac{16}{x}}{\\text{-}4 + \\frac{4}{x} + \\frac{24}{x^2}} \\;=\\;\\frac{x-0}{\\text{-}4+0+0} \\;=\\;\\frac{x}{\\text{-}4}$\n\n$\\text{The oblique asymptote is: }\\:y \\:=\\:\\text{-}\\tfrac{1}{4}x$\nYou are looking at the behavior of the function in the asymtotic region. You suppressed what is happening near the origin. Just divide the polynomials and you will get -x/4 - 1/4. The complete graph is attached.\n\n8. ## Re: Oblique Asymptotes\n\nOriginally Posted by topsquark\nYou are off by a - sign. See below.\n\n-Dan\nWhat graphing utility you are using\n\n9. ## Re: Oblique Asymptotes\n\nOriginally Posted by votan\nWhat graphing utility you are using\nGraph. It's freeware and it's pretty awesome, really. No pesky ads or anything.\n\n-Dan\n\n10. ## Re: Oblique Asymptotes\n\nOriginally Posted by topsquark\nGraph. It's freeware and it's pretty awesome, really. No pesky ads or anything.\n\n-Dan\nI am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?\n\n11. ## Re: Oblique Asymptotes\n\nOriginally Posted by votan\nI am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?\nI don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region \"near\" the origin.\n\nI'll look into dplot, thanks.\n\n-Dan\n\n12. ## Re: Oblique Asymptotes\n\nOriginally Posted by Soroban\nHello, sakonpure6!\n\n$\\text{We have: }\\:f(x) \\;=\\;\\frac{x^3-6x}{\\text{-}4x^2 + 4x + 24}$\n\n$\\text{Divide numerator and denominator by }x^2\\!:$\n\n. . $f(x) \\;=\\;\\dfrac{\\frac{x^3}{x^2} - \\frac{6x}{x^2}}{\\frac{\\text{-}4x^2}{x^2} + \\frac{4x}{x^2} + \\frac{24}{x^2}} \\;=\\;\\frac{x - \\frac{16}{x}}{\\text{-}4 + \\frac{4}{x} + \\frac{24}{x^2}}$\n\n$\\text{Hence: }\\:\\lim_{x\\to\\inty}f(x) \\;=\\;\\lim_{x\\to\\infty} \\frac{x - \\frac{16}{x}}{\\text{-}4 + \\frac{4}{x} + \\frac{24}{x^2}} \\;=\\;\\frac{x-0}{\\text{-}4+0+0} \\;=\\;\\frac{x}{\\text{-}4}$\n\n$\\text{The oblique asymptote is: }\\:y \\:=\\:\\text{-}\\tfrac{1}{4}x$\nAnother way to see this is to use \"long division\": $\\frac{x^3- 16x}{-4x^2+ 4x+ 24}= -\\frac{1}{4}x- \\frac{1}{4} - \\frac{9x- 6}{-4x^2+ 4x+ 24}$.\nObviously as x goes to infinity (or negative infinity) the remaining fraction goes to 0 so the graph approaches $-\\frac{1}{4}x- \\frac{1}{4}$.\n\n13. ## Re: Oblique Asymptotes\n\nOriginally Posted by topsquark\nI don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region \"near\" the origin.\n\nI'll look into dplot, thanks.\n\n-Dan\nCan we insert a power point presentation here?", "date": "2016-10-01 13:19:58", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/223355-oblique-asymptotes.html", "openwebmath_score": 0.8406234979629517, "openwebmath_perplexity": 1782.7800500417718, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540716711547, "lm_q2_score": 0.8688267796346598, "lm_q1q2_score": 0.8508890442121412}}
{"url": "https://mathhelpboards.com/threads/determine-p-q-r-and-s.7859/", "text": "Determine p, q, r and s\n\nanemone\n\nMHB POTW Director\nStaff member\nHi MHB,\n\nI have encountered a problem and I am not being able to figure out the answer.\n\nProblem:\n\nGiven that $p, q, r, s$ are all positive real numbers and they satisfy the system\n\n$p+q+r+s=12$\n\n$pqrs=27+pq+pr+ps+qr+qs+rs$\n\nDetermine $p, q, r$ and $s$.\n\nAttempt:\n\nThe AM-GM inequality for both $p, q, r, s$ and $pq,pr,ps,qr,qs,rs$ are:\n\n 1 $\\dfrac{p+q+r+s}{4} \\ge \\sqrt[4]{pqrs}$ which then gives $(\\dfrac{12}{4})^4 \\ge pqrs$ or $pqrs \\le 81$ 2 $\\dfrac{pq+pr+ps+qr+qs+rs}{6} \\ge \\sqrt[6]{(pqrs)^3}$ which then gives $(\\dfrac{pqrs-27}{6})^2 \\ge pqrs$ $(pqrs-81)(pqrs-81) \\ge 0$ $pqrs \\le 9$ or $pqrs \\ge 81$\n\nAfter that, I don't see how to proceed...should I conclude that since we need to find $pqrs$ that satisfy both of the inequalities below\n\n$pqrs \\le 81$ and $pqrs \\ge 81$\n\n$\\therefore pqrs=81$ and and obviously the answer would be $p=q=r=s=3$?\n\nAckbach\n\nIndicium Physicus\nStaff member\nOn your second AM-GM inequality, you get $(pqrs-81)(pqrs-9) \\ge 0$, with the individual inequalities that you found.\n\nIn the beginning of the problem, you specified that $p,q,r,s$ are real. Is that correct? If so, I see no way of nailing down all four values, given only two equations. Certainly, $p=q=r=s=3$ works, but what guarantee do we have that there isn't another solution? E.g., try setting $p=q=2$, and solving the resulting system for $r,s$, and see if there is a solution.\n\n[EDIT] See Opalg's post below for a correction.\n\nOpalg\n\nMHB Oldtimer\nStaff member\nHi MHB,\n\nI have encountered a problem and I am not being able to figure out the answer.\n\nProblem:\n\nGiven that $p, q, r, s$ are all positive real numbers and they satisfy the system\n\n$p+q+r+s=12$\n\n$pqrs=27+pq+pr+ps+qr+qs+rs$\n\nDetermine $p, q, r$ and $s$.\n\nAttempt:\n\nThe AM-GM inequality for both $p, q, r, s$ and $pq,pr,ps,qr,qs,rs$ are:\n\n 1 $\\dfrac{p+q+r+s}{4} \\ge \\sqrt[4]{pqrs}$ which then gives $(\\dfrac{12}{4})^4 \\ge pqrs$ or $pqrs \\le 81$ 2 $\\dfrac{pq+pr+ps+qr+qs+rs}{6} \\ge \\sqrt[6]{(pqrs)^3}$ which then gives $(\\dfrac{pqrs-27}{6})^2 \\ge pqrs$ $(pqrs-81)(pqrs-81) \\ge 0$ $pqrs \\le 9$ or $pqrs \\ge 81$\n\nAfter that, I don't see how to proceed...should I conclude that since we need to find $pqrs$ that satisfy both of the inequalities below\n\n$pqrs \\le 81$ and $pqrs \\ge 81$\n\n$\\therefore pqrs=81$ and and obviously the answer would be $p=q=r=s=3$?\nIt looks as though you have solved this problem. You have shown that either $pqrs\\leqslant9$ or $pqrs\\geqslant 81$. But the equation $pqrs=27+pq+pr+ps+qr+qs+rs$ shows that $pqrs\\geqslant27$, so that rules out the first of those possibilities. We are left with the second one, $pqrs\\geqslant 81$. But you have also shown that $pqrs\\leqslant 81$. Therefore $pqrs = 81$. That implies that equality occurs in the AM-GM inequality, and that only happens when all four quantities are equal. So $p=q=r=s=3$.\n\nanemone\n\nMHB POTW Director\nStaff member\nOn your second AM-GM inequality, you get $(pqrs-81)(pqrs-9) \\ge 0$, with the individual inequalities that you found.\n\nIn the beginning of the problem, you specified that $p,q,r,s$ are real. Is that correct? If so, I see no way of nailing down all four values, given only two equations. Certainly, $p=q=r=s=3$ works, but what guarantee do we have that there isn't another solution? E.g., try setting $p=q=2$, and solving the resulting system for $r,s$, and see if there is a solution.\n\n[EDIT] See Opalg's post below for a correction.\n...We are left with the second one, $pqrs\\geqslant 81$. But you have also shown that $pqrs\\leqslant 81$. Therefore $pqrs = 81$. That implies that equality occurs in the AM-GM inequality, and that only happens when all four quantities are equal. So $p=q=r=s=3$.", "date": "2021-08-01 17:02:03", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/determine-p-q-r-and-s.7859/", "openwebmath_score": 0.9253337979316711, "openwebmath_perplexity": 191.97481109021743, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540668504082, "lm_q2_score": 0.8688267762381844, "lm_q1q2_score": 0.8508890366973955}}
{"url": "https://math.stackexchange.com/questions/389991/closed-form-for-prod-n-1-infty-sqrt2n-tanh2n", "text": "# Closed form for $\\prod_{n=1}^\\infty\\sqrt[2^n]{\\tanh(2^n)},$\n\nPlease help me to find a closed form for the infinite product $$\\prod_{n=1}^\\infty\\sqrt[2^n]{\\tanh(2^n)},$$ where $\\tanh(z)=\\frac{e^z-e^{-z}}{e^z+e^{-z}}$ is the hyperbolic tangent.\n\n\u2022 Do you have any reason to believe that such a closed form exists? \u2013\u00a0George V. Williams May 13 '13 at 1:42\n\u2022 wolframalpha.com/input/\u2026 vs. wolframalpha.com/input/?i=1-exp%28-4%29 It could be a coincidence though... \u2013\u00a0Vladimir Reshetnikov May 13 '13 at 1:51\n\u2022 Out of curiosity, where are you finding all of these questions? \u2013\u00a0Jemmy May 13 '13 at 2:32\n\u2022 @Jeremy A friend of mine shared these problems with me. They were submitted to a math competition for students, but were rejected by the committee for various reasons: too hard, too easy, have been published before, not interesting etc. \u2013\u00a0Laila Podlesny May 13 '13 at 17:21\n\u2022 @LailaPodlesny May I trouble you by requesting to know which math competition you are referring to? \u2013\u00a0Kugelblitz Sep 24 '15 at 12:03\n\nFor $x < 1$, we have the Taylor series expansion: $$f(x):= \\frac{-1}{4} \\log \\left(- \\frac{x - x^{-1}}{x + x^{-1}} \\right) = \\frac{x^2}{2} + \\frac{x^6}{6} + \\frac{x^{10}}{10} + \\frac{x^{14}}{14} + \\ldots$$\n\nThen\n\n$$f(x) + \\frac{f(x^2)}{2} + \\frac{f(x^4)}{4} + \\frac{f(x^8)}{8} + \\ldots = \\frac{x^2}{2} + \\frac{x^4}{4} + \\frac{x^6}{6} + \\frac{x^8}{8} + \\frac{x^{10}}{10} + \\ldots$$ $$= - \\frac{1}{2} \\log(1 - x^2).$$\n\nNow let $x = e^{-2}$. Then\n\n$$\\log \\left( \\sqrt[2^n]{\\mathrm{tanh}(2^n)} \\right) = \\frac{1}{2^n} \\log \\left( \\frac{e^{2^n} - e^{-2^n}}{e^{2^n} + e^{-2^n}}\\right)$$ $$= \\frac{-4}{2^n} f(e^{-2^n}) = \\frac{-4}{2^{n}} f(x^{2^{n-1}}),$$\n\nHence summing over all $n \\ge 1$, we see that, if the product is $P$, then\n\n$$\\log P = -4 \\sum_{n=0}^{\\infty} \\frac{1}{2^{n}} f(x^{2^{n-1}}) = -2 \\sum_{n=1}^{\\infty} \\frac{1}{2^{n}} f(x^{2^{n}}) = \\log(1 - x^2),$$\n\nand thus\n\n$$P = \\exp \\log(1 - x^2) = 1 - x^2 = 1 - e^{-4}.$$\n\n\u2022 Nicely done! :) \u2013\u00a0Caran-d'Ache May 13 '13 at 4:47\n\u2022 This is a very careful and crafty derivation. Excellent. I especially like where you pulled the ol' $\\sum \\log \\leftrightarrow \\log \\prod$. \u2013\u00a0Coffee_Table May 13 '13 at 20:03\n\nLet $$f(x)=\\prod_{n=0}^\\infty\\left(1-x^{2^n}\\right)^{1/2^n}\\tag{1}$$ and $$g(x)=\\prod_{n=0}^\\infty\\left(1+x^{2^n}\\right)^{1/2^n}\\tag{2}$$ Then \\begin{align} f(x)\\,g(x) &=\\prod_{n=0}^\\infty\\left(1-x^{2^{n+1}}\\right)^{1/2^n}\\\\ &=\\prod_{n=1}^\\infty\\left(1-x^{2^n}\\right)^{2/2^n}\\\\ &=\\left(\\frac{f(x)}{1-x}\\right)^2\\tag{3} \\end{align} from which we get $$\\frac{f(x)}{g(x)}=(1-x)^2\\tag{4}$$ Note that $$\\prod_{n=1}^\\infty\\left(1-x^{2^n}\\right)^{1/2^n}=\\frac{f(x)}{1-x}\\tag{5}$$ and $$\\prod_{n=1}^\\infty\\left(1+x^{2^n}\\right)^{1/2^n}=\\frac{g(x)}{1+x}\\tag{6}$$ Therefore, combining $(4)$, $(5)$, and $(6)$, we get $$\\frac{\\displaystyle\\prod_{n=1}^\\infty\\left(1-x^{2^n}\\right)^{1/2^n}}{\\displaystyle\\prod_{n=1}^\\infty\\left(1+x^{2^n}\\right)^{1/2^n}}=1-x^2\\tag{7}$$ Plug $x=e^{-2}$ into $(7)$ to get $$\\prod_{n=1}^\\infty\\tanh(2^n)^{1/2^n}=1-e^{-4}\\tag{8}$$\n\n\u2022 This looks much simpler than the accepted answer (which is also excellent) (+1). \u2013\u00a0Paramanand Singh Nov 19 '13 at 16:20\n\u2022 @ParamanandSingh: thanks. This question was just pointed out to me. However, new answers to old questions generally don't get as much attention as the older answers got. \u2013\u00a0robjohn Nov 19 '13 at 16:32", "date": "2019-06-24 08:50:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/389991/closed-form-for-prod-n-1-infty-sqrt2n-tanh2n", "openwebmath_score": 0.8800835013389587, "openwebmath_perplexity": 785.5051824737233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540722737479, "lm_q2_score": 0.8688267711434708, "lm_q1q2_score": 0.8508890364198098}}
{"url": "https://cs.stackexchange.com/questions/96118/how-can-i-correct-this-hamming-code", "text": "# How can I correct this Hamming code?\n\nI'm trying to decode the following Hamming sequence (using EVEN parity and knowing there is a 1-bit error), which contains an ASCII value: 01100110101\n\nI've tried to check for the correctness of each parity bit:\n\n    p1 p2 d3 p4 d5 d6 d7 p8 d9 d10 d11\n0  1  1  0  0  1  1  0  1   0   1\np1 - 0     1     0     1     1       1\np2 -    1  1        1  1         0   1\np4 -          0  0  1  1\np8 -                      0  1   0   1\n\n\np1, p4 and p8 checks are all even and correct.\n\np2 check is odd, hence the parity bit is incorrect and needs to be flipped. The actual message then becomes 00100110101 and the ASCII value can be read as 1011101.\n\nSupposedly this value is incorrect - can anyone point out the mistake I made?\n\n\u2022 Wouldn't Hamming Code be more appropriate in the Signal Processing StackExchange? \u2013\u00a0KingDuken Aug 9 '18 at 21:53\n\u2022 @KingDuken It is perfectly appropriate in either. \u2013\u00a0koverman47 Aug 9 '18 at 22:25\n\u2022 @polyethene Why do you think that this is wrong? It looks correct. Since p2 has the only odd sum, the bit p2 itself has to be wrong (all the other bits are used by at least 2 controll sums, which would mean that at least two bits are wrong). \u2013\u00a0Jakube Aug 10 '18 at 13:57\n\u2022 I suggest you edit the question to describe where you got the idea that this is incorrect. If someone told you that, maybe they are wrong. If you read it somewhere, maybe whatever you read was in error. \u2013\u00a0D.W. Aug 10 '18 at 16:11\n\u2022 Someone (this OP?) will pay 10 dollars if you answer the question here at reddit.com \u2013\u00a0Apass.Jack Aug 12 '18 at 22:24\n\nThe original problem is, I believe, the following.\n\nWhen using Hamming code with EVEN parity for 7-bit ASCII characters, the following symbol is retrieved: 01100110101.\n\nAssuming a 1-bit error, what was the original stored symbol? Write down your answer as a 7-bit binary, with no spaces.\n\nOP's statement, \"..., which contains an ASCII value: 01100110101\", rephrases the original problem statement in a slightly confusing way, as pointed out by Yuval Filmus.\n\nOP's procedure and result is correct, as said in Jakube's comment and indicated in D.W.'s comment. I have verified it as well according to Hamming code at wikipedia.\n\nOP's procedure and result will not be repeated here.\n\nSomeone mentioned that \"The usual Hamming code has length of the form $2^\\ell\u22121$\". For $\\ell=4$, the usual Hamming code of length 15 is defined by the following table from wikipedia. If we truncate the last 4 columns, we will get the truncated Hamming code of length 11, which should be, presumably, the Hamming code used in OP's question.\n\n\"Supposedly this value is incorrect - can anyone point out the mistake I made?\"\n\nThis is the turning point to the climax (or anti-climax). There are four possibilities.\n\n1. A different kind of (truncated) Hamming code is used.\nFor example, all four parity bits could have been specified to be put together in the first four bits. That is, the first four bits, 0110 are the usual p1, p2, p4 and p8 parity bits while the remaining 0110101 are the data bits. In that case, we rearrange the bits to form the usual Hamming code as (p1)(p2)0(p4)110(p8)101, which is 01011100101. Then p1, p2, p4 check are odd while p8 is OK. So we flip the $1+2+4 =7$-th bit to obtain 01011110101. So the corrected data bit will be 0111101. Well, OP can check if this or something similar is the case.\n\n2. \"This is from a quiz on my university's LMS, the site is saying my answer is incorrect.\"\nAs D.W. suggested, maybe they are wrong. Maybe whatever you read was in error. There might be a typo in the statement of the quiz. There might be an error in the original answer by the quiz owner. There might be an error in the answer-checking process by the site. There might be a critical typo in OP's post. And so on.\n\n3. Jakube, D.W., I, and, apparently, many others have not been able to find the mistake. That is very unlikely, though.\n\n4. The last possibility stands for, as always, all other probably even lesser possibilities.", "date": "2019-10-13 21:00:14", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/96118/how-can-i-correct-this-hamming-code", "openwebmath_score": 0.7325025200843811, "openwebmath_perplexity": 680.2510699202819, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540704659681, "lm_q2_score": 0.8688267677469952, "lm_q1q2_score": 0.85088903152281}}
{"url": "https://math.stackexchange.com/questions/3286747/what-is-the-vector-x-%E2%88%88-mathbbr3-that-achieves-maxx-1-subject-to-x/3286858", "text": "# What is the vector $x \u2208 \\mathbb{R^3}$ that achieves $max||x||_1$ subject to $||x||_2 = 1$?\n\nI'm trying to answer the questions \"What is the vector $$x \u2208 \\mathbb{R^3}$$ that achieves $$max||x||_1$$ subject to $$||x||_2 = 1$$?\" and \"What is the vector x \u2208 $$R^3$$ that achieves $$max||x||_\u221e$$ subject to $$||x||_2 = 1$$?\n\nI think the first question is asking me to find a vector with three components that will have the maximum $$||x||_1$$ norm value where $$\\sqrt{x_1^2 + x_3^2 + x_2^2} = 1$$, so $$x_1^2 + x_3^2 + x_2^2 = 1$$. I know the The L1 norm is just the sum of the absolute values of the vector's components. After trial and error I came up with $$x = [\\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}]$$ , but also $$[-\\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}]$$, and $$[-\\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}]$$, etc.\n\nFor my the second question, I think I need to find the vector in $$\\mathbb{R^3}$$ that will give me the maximum value of the absolute value of the vector's components given $$x_1^2 + x_3^2 + x_2^2 = 1$$. I came up with $$[1, 0, 0]$$, $$[0, 1, 0]$$ , $$[0, 0, 1]$$, $$[-1, 0, 0]$$, $$[0, -1, 0]$$ , and $$[0, 0, -1]$$.\n\nAm I correct? Is there a more formal way to figure this out and write my solution?\n\n\u2022 First question is tackled with Cauchy-Schwarz, see math.stackexchange.com/questions/218046/\u2026 for instance; Second one is even easier \u2013\u00a0Olivier Jul 8 '19 at 13:57\n\u2022 To solve optimization problems subject to equality constraints Lagrange multipliers are a very popular method. Example 1a in the Wiki article is quite close to your $L_1$ problem. \u2013\u00a0pH 74 Jul 8 '19 at 13:59\n\nTo solve the first question:\n\nBy symmetry, we can assume $$x_1,x_2,x_3$$ are all positive (and add the other solutions later).\n\nBy the method of Lagrange multipliers\n\n$$L=x_1+x_2+x_3 - \\lambda(x_1^2 + x_2^2 +x_3^2-1)$$\n\n$$0=\\frac{\\partial L}{\\partial x_1} = 1-2\\lambda x_1$$ $$0=\\frac{\\partial L}{\\partial x_2} = 1-2\\lambda x_2$$ $$0=\\frac{\\partial L}{\\partial x_3} = 1-2\\lambda x_3$$\n\nThese three equations imply $$x_1=x_2=x_3$$ so that by the constraint $$||x||_2=1$$, the solutions is $$x_1=x_2=x_3=\\frac{1}{\\sqrt{3}}$$\n\nAll of the solutions are\n\n$$x_1=\\pm \\frac{1}{\\sqrt{3}},\\quad x_2=\\pm \\frac{1}{\\sqrt{3}},\\quad x_3=\\pm \\frac{1}{\\sqrt{3}}.$$\n\nTo solve the second question:\n\nWe want to maximize one of the elements of the vector.\n\nTrivially, the solutions are\n\n$$\\begin{pmatrix} \\pm 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, \\quad \\begin{pmatrix} 0 \\\\ \\pm 1 \\\\ 0 \\end{pmatrix}, \\quad \\begin{pmatrix} 0 \\\\ 0 \\\\ \\pm 1 \\end{pmatrix}.$$\n\nWe can obtain this also with Lagrange multipliers. Since, by symmetry, we can seek to maximize $$|x_1|$$, first constrain $$x_1$$ to be positive as above, and then find the other solution ($$x_1<0$$).\n\nConsider\n\n$$L= x_1 - \\lambda (x_1^2 + x_2^2 + x_3^2-1)$$\n\n$$0=\\frac{\\partial L}{\\partial x_1} = 1 + 2 \\lambda x_1$$ $$0=\\frac{\\partial L}{\\partial x_2} = 2\\lambda x_2$$ $$0=\\frac{\\partial L}{\\partial x_3} = 2\\lambda x_3$$ implies that $$x_1=1$$ and $$x_2=x_3=0$$. Similarly for the other cases.\n\nIf $$\\|x\\|_2 = 1$$, the Cauchy-Schwarz inequality implies $$\\|x\\|_1 = |x_1|+|x_2|+|x_3| \\le \\sqrt{x_1^2+x_2^2+x_3^2}\\cdot\\sqrt{1+1+1} = \\sqrt{3}$$\n\nFor $$(x_1,x_2,x_3) = \\left(\\frac1{\\sqrt3}, \\frac1{\\sqrt3}, \\frac1{\\sqrt3}\\right)$$ we have $$\\|x\\|_1 = \\sqrt{3}$$ so this is the maximum.\n\nNow let $$x$$ be a minimizer. We then have equality in the Cauchy-Schwarz inequality above so there exists $$t \\in \\mathbb{R}$$ such that $$(|x_1|,|x_2|,|x_3|) = t(1,1,1) = (t,t,t)$$ Taking $$\\|\\cdot\\|_2$$ norm gives $$t = \\frac1{\\sqrt{3}}$$ so $$x=\\left(\\pm\\frac1{\\sqrt3}, \\pm\\frac1{\\sqrt3}, \\pm\\frac1{\\sqrt3}\\right)$$\n\n\u2022 so would the answer to the first question be the set of vectors x where $\\|x\\|_1 = \\sqrt{3}$ ? \u2013\u00a0John Jul 8 '19 at 15:39\n\u2022 @John Precisely, it is the sphere around the origin of radius $\\sqrt{3}$ w.r.t. the norm $\\|\\cdot\\|_1$. \u2013\u00a0mechanodroid Jul 8 '19 at 16:04\n\u2022 @John Of course, intersect the set above with the unit sphere. There are only $4$ solutions, see above. \u2013\u00a0mechanodroid Jul 8 '19 at 17:15\n\u2022 There are eight solutions! They lie on the vertices of a cube. \u2013\u00a0mjw Jul 8 '19 at 21:54\n\u2022 @mjw Whoops, $2^3= 8$ and not $4$, thanks. \u2013\u00a0mechanodroid Jul 8 '19 at 21:57\n\nIn general, $$\\|x\\|_1 \\le \\sqrt{n}\\|x\\|_2$$, where $$n$$ is the dimension of the space.\n\nIf $$\\|x\\|_2 = 1$$ then we have $$\\|x\\|_1 \\le \\sqrt{n}$$, to achieve equality, note that $$\\|{1 \\over \\sqrt{n}}(1,...,1) \\|_1 = \\sqrt{n}$$.\n\nFor the solution of your first question, you should also add the follwoing points: $$\\left(\\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}\\right), \\left(\\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}\\right), \\left(\\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}\\right), \\left(-\\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}\\right)$$ and $$\\left(-\\sqrt{\\frac{1}{3}}, \\sqrt{\\frac{1}{3}}, -\\sqrt{\\frac{1}{3}}\\right)$$.\n\nTo intuitively guess them, you need to think that both L1 and L2 are symmetric metrics. So you should consider points $$(x_1, x_2, x_3)$$ where $$|x_1| = |x_2| = |x_3|$$. The constraint of $$x_1^2 + x_2^2 + x_3^2 = 1$$ would give you $$|x_1| = |x_2| = |x_3| = \\sqrt{\\frac{1}{3}}$$.\n\nFor a geomteric solution, let us first consider the case in 1st octant i.e. where $$x_1 \\geq 0, x_2 \\geq 0,$$ and $$x_3 \\geq 0$$. Thus $$L_1 = |x_1| + |x_2| + |x_3|$$ is equivalent to $$L_1 = x_1 + x_2 + x_3$$. Now geometrically, one needs to find the point where the plane $$x_1 + x_2 + x_3 = constant$$ touches the sphere $$x_1^2 + x_2^2 + x_3^2 = 1$$ in the first octant.\n\nIf you want to formally and algebraically compute the values in the 1st oactant, then you need to write the Lagrangian function $$L(x_1, x_2, x_3, \\lambda) = x_1 + x_2 + x_3 + \\lambda(x_1^2 + x_2^2 + x_3^2 - 1)$$. Now equate all the partial derivatives to 0. This will get you $$\\lambda = -\\frac{1}{2x_1} = -\\frac{1}{2x_2} = -\\frac{1}{2x_3}$$. This leads to $$x_1 = x_2 = x_3 = \\frac{1}{\\sqrt{3}}$$. In the other octant (where $$x_1 \\le 0, x_2 \\geq 0,$$ and $$x_3 \\geq 0$$) the Lagrangian function will change sign: $$L(x_1, x_2, x_3, \\lambda) = -x_1 + x_2 + x_3 + \\lambda(x_1^2 + x_2^2 + x_3^2 - 1)$$. This would yield an answer of $$-x_1 = x_2 = x_3 = \\frac{1}{\\sqrt{3}}$$.\n\nFor the second problem, let us take the first case where the $$L_\\infty$$ norm $$= Max(x_1, x_2, x_3) = x_1$$. Then the Lagrangian function $$L(x_1, x_2, x_3, \\lambda) = x_1 + \\lambda(x_1^2 + x_2^2 + x_3^2 - 1)$$.", "date": "2021-07-26 12:24:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3286747/what-is-the-vector-x-%E2%88%88-mathbbr3-that-achieves-maxx-1-subject-to-x/3286858", "openwebmath_score": 0.9591895341873169, "openwebmath_perplexity": 182.05563226278937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9793540680555949, "lm_q2_score": 0.8688267660487572, "lm_q1q2_score": 0.850889027765437}}
{"url": "https://math.stackexchange.com/questions/2229101/quick-question-about-the-limit-of-a-two-variable-function-as-x-y-to-infty/2229165", "text": "# quick question about the limit of a two-variable function as $x,y\\to\\infty$\n\n$$\\lim_{x,y\\to\\infty} \\frac{x-y}{x^2+y^2}\\tag{\\star}$$\n\nI'm used to do the following substitution when I see $x^2+y^2\"$ and that $x,y\\to 0$\n\n$$x^2+y^2 = r^2,\\;x=r\\cos\\theta,\\;y=r\\sin\\theta$$\n\nplug these values in the function and compute the limit as $r\\to0$ I know I can do that because the only way for $x$ & $y$ to approach $0$ is $r$ approaching $0.$\n\nI cannot do this substitution everytime because if for example: $(x,y)\\to(-1,7)$ there's no value $u$ that guarantee me if $r\\to u$ then $(x,y)\\to(-1,7).$\n\nbut here since $x,y\\to\\infty$ I think that logically this phenomenon can only happen if $r\\to\\infty$ as well.\n\nSo computing $(\\star)$ is the same as computing this :\n\n$$\\lim_{r\\to\\infty} \\frac{r\\cos\\theta-r\\sin\\theta}{r^2} =\\lim_{r\\to\\infty} \\frac{\\cos\\theta-\\sin\\theta}{r}=0.$$\n\nI'm 90% sure that what I've done is correct but I still want a confirmation and if possible show me other ways to compute this limit.\n\nSorry if this question sounds kinda dumb but I'm still new to multivariable calculus and today is my first time dealing with MVC limits.\n\nThank you !\n\nOthers can add and/or correct, but I'm not sure if you can do the polar trick. If $x$ and $y$ tend to infinity, then clearly $r \\to \\infty$. But if you have $r \\to \\infty$, then you don't necessarily have $x$ and $y$ to infinity since, for example: $x \\to \\infty$ and $y \\to c$ (constant) will also lead to $r \\to \\infty$.\n\nand if possible show me other ways to compute this limit.\n\nRewrite: $$\\left| \\frac{x-y}{x^2+y^2} \\right| =\\left| \\frac{x}{x^2+y^2}- \\frac{y}{x^2+y^2} \\right| \\le \\left| \\frac{x}{x^2+y^2}\\right|+ \\left|\\frac{y}{x^2+y^2} \\right|$$ Now: $$\\left| \\frac{x}{x^2+y^2} \\right| \\le \\left| \\frac{x}{x^2} \\right| = \\left| \\frac{1}{x} \\right| \\to 0 \\quad\\mbox{ and }\\quad \\left| \\frac{y}{x^2+y^2} \\right| \\le \\left| \\frac{y}{y^2} \\right| = \\left| \\frac{1}{y} \\right| \\to 0$$ Alternatively, if you feel more comfortable with limits to $(0,0)$, substitute $\\left( x,y \\right) \\to \\left( \\tfrac{1}{u},\\tfrac{1}{v} \\right)$ and take the limit $(u,v) \\to (0^+,0^+)$ and you could follow up with your classical polar substitution.\n\nWith no other answers so far, I'll add this example: consider $f(x,y)=x+y$, then clearly: $$\\lim_{(x,y)\\to (+\\infty,+\\infty)} \\bigl( x+y \\bigr) = +\\infty$$ However, switching to polar coordinates, we get: $$f(r,\\theta) = r \\left( \\cos\\theta + \\sin\\theta \\right)$$ Now simply taking $r \\to +\\infty$, the limit: $$\\lim_{r \\to +\\infty} \\bigl( r \\left( \\cos\\theta + \\sin\\theta \\right) \\bigr)$$ depends on $\\theta$ since $\\cos\\theta + \\sin\\theta$ can be positive, negative or zero. This makes sense since fixing $\\theta$ to a value outside the interval $(0,\\tfrac{\\pi}{2})$ would not correspond to $(x,y)\\to (+\\infty,+\\infty)$.\n\n\u2022 @rapidracim Thanks for accepting but you may want to wait for some more input of others, specifically regarding your question on the polar substitution for the limit to $(+\\infty,+\\infty)$. \u2013\u00a0StackTD Apr 11 '17 at 12:52\n\u2022 I think I've found a counterexample $$\\lim_{x,y \\to \\infty} \\frac{x}{y^2}+x^2$$ here If I use the polar substitution I'll get $+\\infty$. but wolfram-alpha says it doesn't exist because it is path dependent do you think I can trust wolframalpha ? if it gave me a \"value\" maybe but since it told me it doesn't exist I presume other than computing the limit along two different paths and finding two different results (which actually proves that the limit D.N.E) a computer can't make such a conclusion (that the limit is path-dependant) using a different algorithm . \u2013\u00a0the_firehawk Apr 11 '17 at 16:15\n\n\"but here since $x,y\\to \\infty$ I think that logically this phenomenon can only happen if $r\\to \\infty.$\" The definition of $x,y\\to \\infty,$ which you haven't given us, is probably exactly the same as $r\\to \\infty.$ Let's assume this. You have correctly arrived at\n\n$$f(r\\cos t, r \\sin t) = \\frac{\\cos t - \\sin t}{r}.$$\n\n$$0\\le |f(r\\cos t, r \\sin t)| = \\frac{|\\cos t - \\sin t |}{r} \\le \\frac{|\\cos t| + |\\sin t|}{r} \\le \\frac{1 +1}{r} = \\frac{2}{r}\\to 0.$$\nThus $|f(r\\cos t, r \\sin t)|\\to 0,$ which is the same as saying $f(r\\cos t, r \\sin t)\\to 0.$", "date": "2020-09-28 10:13:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2229101/quick-question-about-the-limit-of-a-two-variable-function-as-x-y-to-infty/2229165", "openwebmath_score": 0.9482593536376953, "openwebmath_perplexity": 133.60280344362724, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639694252315, "lm_q2_score": 0.8757869981319862, "lm_q1q2_score": 0.8508830922761204}}
{"url": "https://math.stackexchange.com/questions/3572975/how-does-the-peano-axiom-of-induction-prevent-s-loops", "text": "# How does the Peano axiom of induction prevent S-loops?\n\nFirst, let me state what I understand to be the first-order rendition of Peano's 5th axiom: the axiom of induction. For all natural numbers, for any relation/property/predicate $$R$$...\n\n$$(R(0) \\land \\forall x[R(x) \\rightarrow R(S(x))]) \\rightarrow \\forall x(R(x))$$\n\n(first question: is this a correct formalization of this axiom or not?)\n\nHow does this axiom prevent elements of natural numbers that have '$$S$$-loops' that is:\n\n\\begin{align} S(a) = b && \\text{and} && S(b) = a \\end{align} (edited for clarity)\n\n\u2022 This axiom does not, but the third Peano Axiom.says that if $S(n)=S(m)$, then $n=m$. If $S(x_9)=x_8$, then $S(x_9)=S(x_7)$, so $x_7=x_9$, which eventually leads to $0=S(x_2)$, which is forbidden by the Fourth Axiom (or peharps by a different number axiom depending on how you formalize them). \u2013\u00a0Arturo Magidin Mar 7 at 21:37\n\u2022 ' then \ud835\udc46(\ud835\udc659)=\ud835\udc46(\ud835\udc657)...' is not true. As defined above, $S(x_9) = x_8$. I've change the nomenclature to 'a' and 'b' to be more clear. In this sense S(a) = b and S(b) = a are completely possible given axioms 1-4. \u2013\u00a0C Shreve Mar 8 at 2:41\n\u2022 It\u2019s not that \u201cit\u2019s not true\u201d. It\u2019s that you\u2019ve changed your statement. When you wrote $x_8$, I of course assumed that you meant the result of applying $S$ to $0$ eight times. And, no it is still not possible, and yes, you still need the other axioms; induction alone doesn\u2019t do it, though induction does come into play in the general statement. If $a=0$, then you are violating the axiom that says that $0$ is not a successor. If $a\\neq 0$, then from induction you get that $a$ is a descendant of $0$, so you still get that $a=x_n$ and go from there. \u2013\u00a0Arturo Magidin Mar 8 at 10:36\n\u2022 I would prefer brackets enclosing everything to the left of the 2nd \"$\\implies$\" in your statement of Axiom 5. In contrast, putting brackets around everything to the right of \"$\\land$\" gives a different (wrong) meaning. \u2013\u00a0DanielWainfleet Mar 9 at 18:28\n\nUsing Peano's axioms\n\n\\begin{align} \\forall m \\, S(m) \\neq 0 &&&\\text{(}0\\text{ is not the successor of anyone)} \\\\ \\forall m \\forall n \\, (S(m) = S(n) \\to m = n) &&&\\text{(injectivity of }S\\text{)} \\end{align}\n\nand Peano's principle of induction, it is easy to prove that the \"double-successor\" does not have any fixed point, i.e. \\begin{align} \\forall m \\, S(S(m)) \\neq m \\end{align}\n\nA rigorous proof of this property is below. It is analogous to the one you can find here to prove that the successor has no fixed point.\n\nNow, this property excludes the possibility of $$S$$-loops. Indeed, if there were $$m$$ and $$n$$ such that \\begin{align} S(m) &= n & S(n) &= m \\end{align} then we would have $$S(S(m)) = m$$ (replace $$n$$ with $$S(m)$$ in the second identity), which is impossible.\n\nWe want to prove that, in Peano arithmetic, \\begin{align} \\forall x \\, S(S(x)) \\neq x &&&\\text{(i.e. } \\forall x \\, R(x) \\text{ where } R(x) \\text{ is the formula } S(S(x)) \\neq x\\text{).} \\end{align}\n\nTo prove this we apply Peano's induction principle, thus we have to prove two facts:\n\n1. Base case, i.e. $$S(S(0)) \\neq 0$$. This holds because it is just an instance (take $$x = S(0)$$) of Peano's axiom \\begin{align} \\forall x \\, S(x) \\neq 0 &&&\\text{(0 is not the successor of anyone).} \\end{align}\n\n2. Inductive case, i.e. $$\\forall x \\, \\big(S(S(x)) \\neq x \\to S(S(S(x))) \\neq S(x) \\big)$$. So, given $$x$$, we suppose $$S(S(x)) \\neq x$$ and we have to show that $$S(S(S(x))) \\neq S(x)$$. Aiming for a contradiction, suppose $$S(S(S(x))) = S(x)$$. According to Peano's axiom \\begin{align} \\forall m \\forall n \\, (S(m) = S(n) \\to m = n) &&&\\text{(injectivity of }S\\text{)} \\end{align} instantiated with $$m = S(S(x))$$ and $$n = x$$, we have that $$S(S(x)) = x$$, which is impossible. Therefore, $$S(S(S(x))) \\neq S(x)$$.\n\nThis ends the proof that $$\\forall x \\, S(S(x)) \\neq x$$.\n\nI think your formalization is correct.\n\nThe axiom of induction doesn't prevent by itself S loops. Consider, a two elements set $$\\{0,1\\}$$ with $$S(0) = 1$$ and $$S(1) = 0$$\n\nTo prevent S loops you need axioms\n\n$$\\forall a,b \\; . \\; S(a) = S(b) \\Rightarrow a = b$$\n\n3.\"$$0$$ is not a succesors\"\n\n$$\\forall a\\;.\\;S(a)\\neq0$$\n\nInformally, if you have a loop from combination of (5) and (4) it follows that loop need to involve all predecessors of a looped element. And with this (3) provides a contradiction, as by (5) every element has $$0$$ as a predecessor.\n\nThe more formal proof can look like this.\n\nConsider the predicate $$NL(x) = \\text{x is not an element of any loop.}$$ By definition of the loop, if $$a$$ is in the loop, there must be $$b$$ in the loop, such that $$a = S(b)$$. So, $$NL(0)$$ holds by axiom (3). Now consider an element $$a$$ such that $$NL(a)$$ holds. If $$S(a)$$ is an element of the loop, then by the axiom (4) the element $$a$$ is also in the loop. This is a Contradiction! Thus, $$NL(a) \\Rightarrow NL(S(a))$$ holds for any $$a$$. Now can apply axiom of induction to see that there is no element $$a$$, which can be looped.\n\nSo there are no loops in Natural numbers defined by Peano axioms. Note, however, that you need every axiom to prove it.\n\n\u2022 I think your $a$ and $b$ in paragraph 2 are supposed to be $0$ and $1$.... \u2013\u00a0Arturo Magidin Mar 7 at 21:52\n\u2022 Thanks. You are correct. This was a typo. \u2013\u00a0Nik Pronko Mar 7 at 21:57\n\u2022 Indeed, with only the axiom of induction, the one-element set $\\{0\\}$ would also be a possibility. \u2013\u00a0Greg Martin Mar 7 at 22:21\n\u2022 But this is not necessarily true. The case you bring up is true because you have expressly chosen a specific set ({0,1}) where one member (ie 0) is identified by axiom 3. However, the set {0, 1, 2, ... a, b} with S(a)= b and S(b) = a could still exist given axioms 1-4. What is it about axiom 4 that eliminates 'a' and 'b'? \u2013\u00a0C Shreve Mar 8 at 2:48\n\u2022 @CShreve Say, $a = S(c)$, then by axiom 4 $b = c$. Repeating this procedure in so for we can, we will end up with a set $\\{0,1\\}$ as in example above. Of course, this works only with finite sets, and words \"repeating procedure in so for\" really is just a reference for the axiom of induction. I will add a more formal proof to the answer. \u2013\u00a0Nik Pronko Mar 8 at 10:34\n\nIn the general statement as you\u2019ve now written, not a specific example as you had before, induction does come into play in the sense that one must prove, by induction, that if $$x\\in \\mathbf{N}$$, then either $$x=0$$, or there exists $$k$$ such that $$x=S^k(0)$$.\n\nIndeed you let the property by \u201c$$x=0$$ or $$x$$ is a descendant of $$0$$\u201d; $$0$$ has the property, and if $$n$$ has the property, then so does $$S(n)$$.\n\nOnce you have that, from $$S(a)=b$$ and $$S(b)=a$$, you get that either $$a=0$$ and you violate the Axiom that says that $$0$$ is not a successor; or else that $$a=S^k(0)$$ for some $$k$$. Then $$b=S^{k+1}(0)$$, and you are now in essentially the same situation as before, when you had the specific example of $$S(x_8)=x_9$$ and $$S(x_9)=x_8$$. Now you have $$S^{k+2}(0)=S^k(0)$$. From that you get $$S^{k+1}(0)=S^{k-1}(0)$$, and so on until you get that $$0$$ is a successor, contradicting that axiom.\n\n@Taroccoesbrocco, so in the spirit of a 'double successor' implies contradiction, would a formalization of this look like:\n\n1. Let R(x) -> S(S(x)) = x (double successor relation)\n2. By Axiom 5 then R must also apply to $$0$$\n3. So $$S(S(0)) = 0$$\n4. Let $$a = S(0)$$\n5. By 4 and 3, $$S(a) = 0$$...which contradicts Peano Axiom 4\n6. Therefore relation R (ie...double successor) is not in the natural numbers.\n\nIs there anything incorrect about the above line of reasoning?\n\n\u2022 What you wrote is essentially meaningless. If you want to prove $\\forall x \\, S(S(x)) \\neq x$ by contradiction, you have to suppose its negation, that is $\\exists x \\, S(S(x)) = x$. Now, you do not know who is the $x$ such that $S(S(x)) = x$, therefore you are not allowed to conclude that $S(S(0)) = 0$. \u2013\u00a0Taroccoesbrocco Mar 9 at 11:41\n\u2022 I edited my answer to include a proof of $\\forall x \\, S(S(x)) \\neq x$. \u2013\u00a0Taroccoesbrocco Mar 9 at 11:43\n\u2022 The additional detail above are helpful. The intent in my reasoning above is to 1)assume a relation $R$ of the natural numbers then 2)show that the relation $R$ produces some contradiction against the Peano axioms. The reason $R$ can be applied to 0 is axiom 5 as any relation in the natural numbers must be applicable to 0 as well. \u2013\u00a0C Shreve Mar 9 at 19:50\n\u2022 What you proved is that $\\exists x \\, S(S(x)) \\neq x$, more precisely that $S(S(0)) \\neq 0$. But it is not enough, indeed you have to prove that $\\forall x \\, S(S(x)) \\neq x$. \u2013\u00a0Taroccoesbrocco Mar 9 at 20:38\n\u2022 The fact that a property fails for some natural numbers does not imply that it fails for all natural numbers. You define the predicate $R(x)$ as $S(S(x)) = x$. With your argument you're are showing that $\\forall x \\, R(x)$ is not true, i.e. that $\\exists x \\, S(S(x)) \\neq x$. But this is not what you need to prove the absence of $S$-loops. What you need to prove is that $\\forall x \\, S(S(x)) \\neq x$. \u2013\u00a0Taroccoesbrocco Mar 10 at 6:25", "date": "2020-07-04 00:10:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3572975/how-does-the-peano-axiom-of-induction-prevent-s-loops", "openwebmath_score": 0.9706621766090393, "openwebmath_perplexity": 351.4722858765755, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639677785087, "lm_q2_score": 0.8757869835428965, "lm_q1q2_score": 0.8508830766597081}}
{"url": "http://math.stackexchange.com/questions/105741/how-to-prove-gcda-gcdb-c-gcd-gcda-b-c", "text": "# How to prove $\\gcd(a,\\gcd(b, c)) = \\gcd(\\gcd(a, b), c)$?\n\nI am trying to prove that $\\gcd(a, \\gcd(b, c)) = \\gcd(\\gcd(a, b), c)$.\n\nThe definition of GCD available to me is as follows:\n\nGiven integers a and b, there is one and only one number d with the following properties.\n\n1. $d \\geqslant 0$\n2. $d|a$ and $d|b$\n3. $e|a$ and $e|b$ implies $e|d$.\n\nIn the book that I am studying, prime factorization of numbers hasn't been taught yet. Only, the definition of GCD, I've given above has been taught and proven. So, I want to use only this to prove that $\\gcd(a, \\gcd(b, c)) = \\gcd(\\gcd(a, b), c)$. Could you please help me?\n\n-\nShow that both sides equal gcd(a,b,c). \u2013\u00a0 franz lemmermeyer Feb 4 '12 at 19:21\nI haven't encountered the definition of gcd of three numbers in the text yet and I am trying to avoid it. \u2013\u00a0 Lone Learner Feb 4 '12 at 19:26\nProof that GCD is associative \u2013\u00a0 pedja Feb 4 '12 at 20:01\n@LoneLearner : The gcd of any number of numbers is the greatest of all of their common divisors, so you just need to know what a common divisor of three numbers is. The divisors of $12$ are $1,2,3,4,6,12$; the divisors of $15$ are $1,3,5,12$; the common divisors are just the members of the intersection of those sets of divisors (in this case $1,3$). So the question is: what's the definition of the intersection of three sets? The answer is that a thing is a member of the intersection precisely if it's a member of all three sets. \u2013\u00a0 Michael Hardy Feb 4 '12 at 21:08\nBy considering prime factorizations, it's a consequence of $\\min(x,\\min(y,z)) = \\min(\\min(x,y),z)$. \u2013\u00a0 lhf Feb 5 '12 at 1:34\n\nSame answer as I just gave in sci.math...\n\nNote that $$d|x,y\\Longleftrightarrow d|\\gcd(x,y).$$ So: \\begin{align*} d|a,\\gcd(b,c) &\\Longleftrightarrow d|a,b,c\\\\ &\\Longleftrightarrow d|\\gcd(a,b),c \\end{align*}\n\n-\nnice hint! (+1) \u2013\u00a0 robjohn Feb 4 '12 at 20:20\n+1 for giving the cleanest proof possible (as far as I can see). My only gripe is with the notation: I would write $\\;d|x \\land d|y\\;$ instead of $\\;d|x,y\\;$. Then the associativity of $\\;\\gcd\\;$ translates directly to the associativity of $\\;\\land\\;$. \u2013\u00a0 Marnix Klooster Jul 26 '13 at 9:41\n\nPlease note that this solution uses an idea that is very similar to the idea in the solution posted much earlier by ncmathsadist. The main difference is that it may contain fewer typos.\n\nWe show that for any integer $u$, if $u$ divides the left-hand side, then $u$ divides the right-hand side, and vice-versa. Thus the left-hand side and the right-hand side have the same set of common divisors, so must be equal, since they are both non-negative.\n\nNow suppose that $u$ divides $\\gcd(a, \\gcd(b, c))$. Then $u$ divides $a$ and $u$ divides $\\gcd(b,c)$. So $u$ divides $b$ and $c$, and therefore $a$, $b$, and $c$.\n\nNow look at the right-hand side. We know that $u$ divides all of $a$, $b$, and $c$. So $u$ divides $\\gcd(a,b)$, and therefore $u$ divides $\\gcd(\\gcd(a,b),c)$.\n\nShowing that if $u$ divides the right-hand side, then $u$ divides the left-hand side is essentially the same calculation, and can be omitted.\n\n-\nI am trying a proof that strictly leads to the fact that if $d = gcd(a, gcd(b, c))$ then $d$ must satisfy the conditions $d|a$, $d|gcd(b, c)$, $e|a$ and $e|gcd(b, c)$ implies $e|gcd(a, gcd(b, c))$. Could you please tell me how to prove the last implication part? \u2013\u00a0 Lone Learner Feb 4 '12 at 21:03\n@Lone Learner: It is inconvenient to work with $d$ directly, it is clearer to work with any common divisor. \u2013\u00a0 Andr\u00e9 Nicolas Feb 4 '12 at 21:07\n\nFirst note that $(a,b) \\mapsto \\gcd(a,b)$ is symmetric in $a$ an $b$. Suppose $d$ is a commond divisor of $a$, $b$ and $c$. Then $c|a$ and $d|\\gcd(b,c)$ so $d|\\gcd(a, \\gcd(b,c))$.\n\nConversely suppose that $d$ is a common divisor or $a$ and $\\gcd(b,c)$. Then $d|a$ and $d|\\gcd(a,b)$. Hence, $d$ is a common divisor of $a$, $b$ and $c$.\n\nOur result follows now by symmetry.\n\n-\nWhy should we assume that $c$ is a common divisor of $a$ and $b$? The problem doesn't require $c$ to be a common divisor of $a$ and $b$. \u2013\u00a0 Lone Learner Feb 4 '12 at 19:27\nWhat we see here is that both $\\gcd(a,\\gcd(b,c))$ and $\\gcd(\\gcd(a,b), c)$ are both simply the largest common divisor of $a$, $b$, and $c$. \u2013\u00a0 ncmathsadist Feb 4 '12 at 19:43\nBut that doesn't imply that $c$ must be a common divisor of $a$, $b$ and $c$. \u2013\u00a0 Lone Learner Feb 4 '12 at 20:14\n@LoneLearner: There\u2019s a major typo in the answer: the common divisor should be $d$ (or some other symbol distinct from $a,b$, and $c$). \u2013\u00a0 Brian M. Scott Feb 4 '12 at 20:18\nBut that still doesn't show how $d$ is a $gcd(a, gcd(b, c))$. According to the definition I have given, we now need to show that if $e$ divides $a$ and $e$ divides $gcd(b, c)$, then $e$ must divide $d$. How do you show this? \u2013\u00a0 Lone Learner Feb 4 '12 at 20:45\n\nHere is a proof I am attempting from all the hints I have got so far. Please let me know if this is correct.\n\nLet $d = gcd(a, gcd(b, c))$. Therefore,\n\n1. $d \\geqslant 0$ from the definition of GCD.\n2. $d|a$ from the definition of GCD.\n3. $d|gcd(b, c)$ from the definition of GCD.\n4. $e|a$ and $e|gcd(b,c)$ implies $e|d$, also from the definition of GCD.\n5. From 3, $d|b$.\n6. From 3, $d|c$.\n7. From 2 and 5, $d|gcd(a, b)$.\n8. Let $e|gcd(a, b)$ and $e|c$. From the definition we know that $gcd(a, b) | a$ and $gcd(a, b) | b$. Therefore, $e|a$ and $e|b$ from the transitive property of divisibility. So, $e|gcd(b, c)$ from the definition of GCD. So, from 4 we have, $e|$d.\n\nFrom 1, 7, 6 and 8, we get, $d = gcd(gcd(a, b), c)$.\n\n-\n#4 seems false. How does it follow from the definition of GCD? If d is a prime factor X common to both a and gcd(b,c), and e is a different prime factor Y common to both a and gcd(b,c), then e will not divide d or vice versa, because they're prime. \u2013\u00a0 Joseph Garvin Jan 26 '13 at 19:05\nActually #4 is OK, it does follow from the definition if you're using the Bezout's identity version. \u2013\u00a0 Joseph Garvin Jan 27 '13 at 17:17", "date": "2014-07-22 07:43:54", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/105741/how-to-prove-gcda-gcdb-c-gcd-gcda-b-c", "openwebmath_score": 0.9522825479507446, "openwebmath_perplexity": 190.07071801994553, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9715639694252316, "lm_q2_score": 0.8757869819218865, "lm_q1q2_score": 0.8508830765269716}}
{"url": "https://www.themathdoctors.org/weighted-averages-averaging-averages-or-rates/", "text": "# Weighted Averages: Averaging Averages or Rates\n\nIn our series on averages, last week we introduced the idea of the weighted average (or weighted mean), where each item has a weight attached. The classic examples all involve grade averages in various ways. This time, we\u2019ll look at how weighted averages arise when you need to average several averages together, something we touched on last time, but which arises often in very different settings. One thing we\u2019ll see repeatedly is that the appropriate average depends on your needs.\n\n## Can you average averages, or\u00a0 not?\n\nWe\u2019ll start with a 1997 question from a professional who had learned that, generally, you can\u2019t average averages, but didn\u2019t know when you can:\n\nAlgebra: Average of an Average\n\nHi!  This one is a real-life need.  I work for a consulting firm, and am having a hard time remembering my basic Algebra!\n\nIf the \"Wireless\" line of business hires 50 people in one month, and the \"Multimedia\" line of business hires 80 people in one month, what is the average number of people per month we are hiring?\n\nAt first, one thinks (50 + 80)/2 = 65 is the correct answer, but it's not because we're taking the average of an average, right?\n\nDon't you have to let x = something and then do 1/50 + 1/80 and solve... or something?\n\nHi Ron,\n\n65 is the correct answer if the question is: \"What is the average number of people hired per line of work per month?\"\n\nHere, we are just averaging the two averages \u201cper line\u201d, so as long as we state clearly what it means, it is valid. The question is, is this average what you want to know, or is it something else entirely?\n\nAssuming that the entire company is made up of only those two lines of business, and you're asking \"What is the average number of people hired by the whole company per month?\", the answer is obviously 50+80 = 130.\n\nThis seems to be what Ron wants, and it isn\u2019t really an average at all, just a total!\n\nSo what about the idea that you can\u2019t average averages? That\u2019s really a different story:\n\nThe operations above are perfectly reasonable. Let me show you the kind of situation that I think you were worried about, and you'll see why it's different from the situations above:\n\nI take 4 exams and have an average score of 80. Then I take 2 more exams and on those 2, my average is 100. What's my overall average for the course?\n\nThe wrong way to do it is (80+100)/2 = 90. This is wrong because the averages were of different-sized groups. To get the correct answer, I know that on the first 4 exams I got 320 total points because when I divide 320 by 4, I get 80.\n\nSimilarly, for the last two exams, I must have gotten 200 points total. So for the six exams, I got 200+320 = 520 points and 520/6 = 86.66666 = my real grade average.\n\nWe saw this scenario last week: If we just average the two averages, the result can only be described as the average of the averages, not the average for the course. The latter has to be weighted, because for the course, each exam counts the same, not each of these two sets of exams.\n\nAs we saw last week, a weighted average can often be understood by breaking it down:\n\nFor your problem, the averages you are averaging are for the same period, so it works out. To convince you that it's true, let's just look at a situation where the averages came from 10 months of data.\n\nThen in the first line of business, 500 people must have been hired, since 500/10 = 50. Similarly, 800 were hired in the other, since 800/10 = 80. Altogether, 1300 people were hired in the ten months, or 1300/10 = 130 per month, company-wide.  Or if you're trying to get the average per line of work, 65 is right, since if each group had hired 65 people each month for 10 months, there would be 65*20 = 1300 total hires, so it works out.\n\nBottom line: always think about what you want, and what an average means, rather than use an average (or not!) unthinkingly.\n\n## Using a weighted average\n\nAveraging Averages\n\nI was reading your response on averaging averages. This has come up in a meeting I go to regarding a report that I am responsible for every month. After reading your response, I want to make sure that I am doing this right.\n\nI have 5 different departments that send me an average rating to 3 different questions from a feedback form (rated 1-5). For example, this is what they send me:\n\nQuestion 1 - average 4 (this is the average of question 1 from all feedback forms this department received during a certain time period)\n\nQuestion 2 - average 3 (same comment as above)\nQuestion 3 - average 5 (same comment as above)\n\nTotal average- 4\n\nAfter getting this information from all 5 different departments, I combine them in a total company report by taking everyone's average to question 1 and then average that to get an average for the whole program. (For this program the company average to question 1 is...) and so on for questions 2 and 3 and the total average.\n\nIf I am reading your response correctly to the question on averaging averages, this is okay to do if all information is contained within the same time period (which it is).\n\nPlease advise.\n\nDoctor Douglas answered, starting with a basic assumption:\n\nHi Stacy,\n\nIt sounds as though you are trying to average a set of averages. As long as the data reflect the same measurement (or question) for each of the individual groups, then it is okay to proceed. That is, if question 1 is the same among all departments (\"Please rate the chef at the last company BBQ on a scale of one to ten.\") then it is a meaningful question to ask: what did the employees think of our last BBQ chef?\n\nOf course, it would be meaningless to average question 1 if each department had a different question, or if the responses were on different scales. But how shall we calculate the average?\n\nNow, when you take averages of averages, there is often a preferred way to do this operation. It is called \"weighted averages\" and reflects the fact that the number of observations may vary among the different groups. For example, let's say that the Marketing Department has 50 employees, and the Research Department has only 6. If the Marketing Department average was 8.5 (\"BBQ was great!\") and Research Department average was 3.1 (\"heartburn!\"), what is the correct average for these 2 departments? We might simply take the average of 8.5 and 3.1, i.e. 5.8.  But it seems unfair to let the Research Dept sway the whole vote, having only 6 employees. The weighted average accounts for the differing number of employees:\n\nweighted avg. = (no. of M)(AVG-of-M) + (no. of R)(AVG-of-R)\n-------------------------------------------\n(total no. of M and R together)\n\n= (50)*(8.5) + (6)*(3.1)\n----------------------\n(50 + 6)\n\n=  7.92\n\nThis is the classic example of a weighted average, and, as we saw last week, it amounts to finding the total score for the whole company by multiplying each average by the number of people it represents, and dividing by the total number of people. Thus this average is what we would get if we just averaged all the individual scores.\n\nDo you see how the final average is much closer now to the Marketing Department's evaluation? Another way I sometimes explain it is that simple direct averaging is like the way states are represented in the U.S. Senate (every state gets two votes), while weighted averaging is like the way states are represented in the U.S. House of Representatives (every state is represented in proportion to its population).\n\nAs we\u2019ll be saying several more times, each of these is a valid average, but has a different meaning. The U.S. Congress intentionally has one part in which every state, even the smallest, is treated equally, and another in which the largest state has more power, and each person is treated equally.\n\nNow, it's impossible for me to say which type of averaging is correct for your situation, but I think it's probably better to use weighted averaging when you have information about the number of observations in each of the groups.\n\nI would agree.\n\n## Averaging percentages\n\nA slightly different situation was involved in this 1998 question:\n\nAveraging Percentages\n\nHi,\n\nI am having difficulties in explaining to several friends that you cannot take percentages by totaling them up and then averaging the total of the percentages. It does not equal the percentage of the total of the numbers.\n\nIs there a rule or theory that can explain this better?\n\nHi, Dominic. I'm not entirely sure what kind of problem you are referring to. Certainly there are at least some situations where you can average percentages. For example, if there are 50 questions on an exam, and three students got 20%, 30%, and 40% of them right, then the average number of questions they got right is 30%, or 15 questions.\n\nThis is like our situations above where we had equal groups, so a straight average made sense.\n\nI suspect what you are thinking of is cases where the percentages are taken from different totals, in which case weighted averaging is needed. For example, if I survey 20% of 50 people, and 80% of 500 other people, then I have not surveyed (20+80)/2 = 50% of the total population, but:\n\n.20 * 50 + .80 * 500   10 + 400   410\n-------------------- = -------- = --- = 74.5%\n50 + 500           550      550\n\nThe problem is simply that the percentages in such a problem do not represent fractions of the same total, so they can't be added.\n\nAs with an average of averages, here we are reconstructing the individual numbers (10 and 400 out of 50 and 500), then finding the actual percentages of the whole. That\u2019s what a weighted average is.\n\nI closed by referring to some of the answers we saw last week.\n\n## Average price per square foot\n\nFor a specific real-life example, consider this 2003 question:\n\nAverage of Ratios vs. Ratio of Averages\n\nI write and maintain software for real estate agents, and we include a calculation called, \"Average dollars per square foot.\"  We currently calculate this as the ratio of the average price divided by the average square footage of all the homes in the list.  It seems to me that it should be calculated as the average of the price-per-square-foot ratio of each house. Can you think of any reason why the ratio of the averages would be more useful than the average of the ratios?  Is there a technical name for this ratio of averages?\n\nI answered again, pointing out as before that different calculations can both be meaningful, though they mean different things:\n\nHi, Laure.\n\nInteresting question!\n\nWhat you are currently calculating actually does make sense; you are just averaging over all the square feet of houses, rather than over all houses, and that may be just the right thing to do -- or it might not.  Here's what I mean:\n\nSuppose that N houses are sold; the sum of all their prices is P, and the sum of all their areas is A.  (That is, if the individual prices are P1, P2, ..., Pn, and the individual areas are A1, A2, ..., An, then P is the sum of P1 through Pn, and A is the sum of A1 through An.)\n\nThen the average price of a house is P/N, and the average area of a house is A/N; and you are calculating\n\nP/N\n--- = P/A\nA/N\n\nas the average price per square foot.  And that is exactly what it is: the total price of all those square feet, divided by the number of square feet.\n\nIn terms of the individual numbers, this is $$\\frac{P_1+P_2+\\dots+P_n}{A_1+A_2+\\dots+A_N}$$ Imagine laying out tiles representing each square foot of each house, and dollar bills representing the price of each house. This average spreads all the money equally over all the tiles, to find an overall price per square root. That sounds perfectly reasonable.\n\nWhat you envision is\n\nP1/A1 + P2/A2 + ... + Pn/An\n---------------------------\nN\n\nwhich would average the price per square foot of all the houses.  This puts the focus on the individual houses, rather than the individual square feet.  How would this be different?\n\nAgain, this is $$\\frac{\\frac{P_1}{A_1}+\\frac{P_2}{A_2}+\\dots+\\frac{P_N}{A_N}}{N}$$ Imagine laying out tiles representing each square foot of each house, keeping each house separate, and dollar bills representing the price of each house, spread equally among the tiles of that house. Now we take one tile and its cost from each house, and average those, spreading the money equally among the selected tiles. What have we found this time?\n\nThe way to see the difference is to take an extreme example. (Don\u2019t check whether the sizes or prices are realistic; they aren\u2019t meant to be!)\n\nWell, let's take a simple case with N = 2.  Suppose we have a big, well-built house of 10,000 square feet, and that it costs $2,000,000 ($200 per square foot), and a little house of 1,000 square feet that costs $20,000 ($20 per square foot). Then the total cost of the houses P is \\$2,020,000, and the total area A is 11,000 square feet. The average price per square foot is\n\nP/A = 2,020,000/11,000 = 183.6\n\n(closer to the more expensive price) while the average of the two price-per-square-foot numbers is\n\naverage(Pn/An) = (200 + 20)/2 = 110\n\n(which is considerably lower).  What pulled the first number up is the fact that the bigger house had the bigger price per square foot; since we counted each square foot equally, the numerous high-cost ones won.  The second calculation treats all 10,000 of the expensive square feet equally with the mere 1,000 square feet of the little house, so the little house pulled the average down.\n\nWhich do you think is the better calculation?\n\nBoth numbers are meaningful.  The first fully deserves the name you are giving it (though there is definitely some ambiguity in the English!); but the second may better reflect what the average homeowner (as opposed to the \"average square foot of floor space\") can expect.  Call it, perhaps, the \"cost per square foot of the average house\", where the number you are currently calculating is the \"average cost of a square foot\" or \"cost of an average square foot\".\n\nPerhaps the builder would consider the first calculation more appropriate, since they put more effort into the bigger house, and every square foot they built matters to them; but the homeowner might care only about his own house \u2014 which neither calculation really reflects!\n\nSo, again, both numbers can reasonably be called \"average cost per square foot\"; which is more useful to you depends on how you want to use it.  Do you want a number that is pulled up by big fancy houses, or one that shows what the average house is worth?  Or would separate numbers for different categories of houses make more sense?  Perhaps you can gather data that shows how costs per square foot are distributed, and how each average reflects that.\n\nAny kind of average is an attempt to reduce a lot of data to a single number, and will never give a full picture of a diverse population. Other numbers, or a graph of the distribution, are often more useful.\n\nLaure chose to leave the calculation as it was.\n\n## Average speed\n\nLet\u2019s look at one more 2003 question, which is about a ratio in disguise:\n\nWeighted Average of the Velocities\n\nA truck on a straight road starts from rest and accelerates at 2.0 m/s2 until it reaches a speed of 20 m/s. Then the truck travels for 20 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 s.\n\n(a) How long is the truck in motion?\n(b) What is the average velocity of the truck for the motion described?\n\nHi, Garin.\n\nThe trick here is to get the average velocity for each of the three phases of the trip, then do a weighted average of the three velocities.\n\nAverage velocity for phase 2 is easy, 20 m/s. During phases 1 and 3, the velocity changes linearly. That means that we can just average the starting and ending velocities to get the velocity for the entire phase.\n\nThis sort of average becomes most meaningful when you apply calculus, but for uniform acceleration, it requires only algebra. Doctor Edwin chose to leave that part for Garin to handle (or to ask for more help, which he never did), and focused just on the idea of the average. It turns out that the acceleration phase takes 10 seconds, at an average speed of 10 m/s; the cruise phase takes 20 seconds at 20 m/s; and the deceleration phase takes 5 seconds at an average speed of 10 m/s.\n\nOnce you've got those, you'll do a weighted average with respect to the time spent in each phase:\n\n(v_1 * t_1) + (v_2 * t_2) + (v_3 * t_3)\nv_avg =  ---------------------------------------\nt_total\n\nThe calculation looks like this: $$\\frac{v_1\\cdot t_1 + v_2\\cdot t_2 + v_3\\cdot t_3}{t_1+t_2+t_3} = \\frac{10\\cdot 10 + 20\\cdot 20 + 10\\cdot 5}{10+20+5} = \\frac{100 + 400 + 50}{10+20+5} = \\frac{550}{35} = 15.7\\text{ m/s}$$\n\nJust for fun, I'll point out that the same set of equations has another interpretation that works just as well. Average velocity is just distance over time, right? So if we figure out how far the truck went in each phase, add them all up, and divide by the total time, you'll also get the same answer:\n\nd_1     +    d_2     +      d3\nv_avg =  ---------------------------------------\nt_total\n\nBut the middle terms in those two equations are the same:\n\nd_2 = v_2 * t_2\n\nand so are the first terms:\n\na(t_1)^2\nd_1 = --------\n2\n\nv\na  =  -\nt\n\nv_1_end * (t_1)^2\nd_1 = -----------------\n2(t_1)\n\nv_1_end * t_1\nd_1 =   -------------\n2\n\nBut since the starting velocity is zero, we can add it in wherever we want:\n\n(v_1_start + v_1_end)\nd_1 =  ---------------------- * t_1\n2\n\nwhich means that our first terms are the same as well, and so are our third terms.\n\nAnyway, there you have it. Total distance traveled divided by time, or a weighted average of velocities with respect to time, it works out to the same thing.\n\nIn other words, each \u201caverage velocity \u00b7 time\u201d is a distance, so our average velocity calculation was really just total distance over total time. We\u2019ve seen previously that when distances are the same, we can use a harmonic mean to find the average speed; this confirms that when times are known but different, we can use a weighted arithmetic mean, using times as weights.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2021-07-30 17:38:05", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/weighted-averages-averaging-averages-or-rates/", "openwebmath_score": 0.746770441532135, "openwebmath_perplexity": 590.2567744497204, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.985496423290417, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8508693504563144}}
{"url": "https://math.stackexchange.com/questions/2942016/function-with-arbitrary-small-period/2942235", "text": "# Function with arbitrary small period\n\nIs there a function f: $$\\mathbb{R} \\to \\mathbb{R}$$ with arbitrary small period different from $$f(x) = k$$? ($$\\forall \\epsilon >0 \\exists a < \\epsilon$$ such that f(x) has a periodicity $$a$$) I think the function is the Dirichlet function but I don't know how to prove it properly.\n\n\u2022 F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$ Oct 4, 2018 at 12:00\n\u2022 @Empy2 Nice: your example also works for domain $\\mathbb{Q}$. Oct 4, 2018 at 12:38\n\u2022 I would avoid saying \u201csuch that the period of $f(x)$ is $a$\u201d here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say \u201csuch that $f$ has a periodicity $a$\u201d, or \u201csuch that $f$ is $a$-periodic\u201d, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b \\in ]0,a[$ with $f(x+b)=f(x)$. Oct 4, 2018 at 13:55\n\u2022 Of course you want $0<a<\\epsilon$\n\u2013\u00a0MPW\nOct 4, 2018 at 18:07\n\u2022 You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^{-k}$ then you're getting Empy2's set. If $a_k=k^{-\\pi}$ then you're getting new set which is hard to imagine :-) If $a_k=\\sin{k}$ then you're getting new interesting set. Oct 5, 2018 at 7:40\n\nYou're right. The characteristic function of the rationals is periodic of period $$1/n$$ for all $$n \\in \\mathbb N$$ because $$x$$ is rational iff $$x+1/n$$ is rational.\n\nYou're correct.\n\nLet $$\\epsilon$$ be arbitrarily small. You need to prove that there exists some $$0 such that $$D(x)=D(x+p)$$ for all $$x\\in \\mathbb{R}$$.\n\nWe know that $$\\epsilon$$ is some positive real number, so there exists some $$p\\in \\mathbb{Q}$$ such that $$0. Let's look at an arbitrary $$x\\in \\mathbb{R}$$ and see if our property is satisfied or not:\n\nIf $$x$$ is rational, then $$D(x)=1$$. Since the sum of two rationals is rational, then $$D(x+p)=1$$ too.\n\nIf $$x$$ is irrational, then $$D(x)=0$$. Since the sum of a rational and an irrational is irrational, then $$D(x+p)=0$$ too.\n\nSo if we choose our period to be $$p$$, our property is satisfied.\n\nThe main question has been answered in other answers well enough, but I would like to address a few natural follow-up questions. What about continuous functions $$f$$ with this property?\n\nIt turns out that in this case there are no nontrivial solutions - every such function is constant. Here's a topological proof: Let $$K=\\{x\\in\\Bbb R\\mid \\forall y, f(y)=f(x+y)\\}$$ be the set of periods of $$f$$. If $$f$$ is continuous, then this is an intersection of the sets $$\\{x\\in\\Bbb R\\mid f(y)=f(x+y)\\}$$, which is closed (it is the preimage of $$\\{0\\}$$ under the function $$g(x)=f(y)-f(x+y)$$), so $$K$$ itself is closed. $$K$$ is also dense in $$\\Bbb R$$, because it is an additive group with arbitrarily small elements, so $$K=\\Bbb R$$ and hence $$f(x)=f(y)$$ for all $$x,y\\in \\Bbb R$$.\n\nIf we consider discontinuous functions again, then we know $$K$$ is a dense additive subgroup of $$\\Bbb R$$. Does every dense additive subgroup generate such a function? Yes, we can just take the characteristic function of $$K$$. For a fixed $$K$$, the space of such functions is just all functions $$\\Bbb R/K\\to \\Bbb R$$. This is another way to get at the constancy result, since as a topological group, $$\\Bbb R/K$$ has the indiscrete topology, because any open set will cover $$\\Bbb R$$ if copied around with translations by $$K$$.\n\nOf course $$\\Bbb R/K$$ can be uncountable, for example if $$K=\\Bbb Q$$ or any other countable subgroup. Can it be countable or finite? It can be countable assuming some choice, as observed in TomGrubb's answer. If we consider a Hamel basis $$B$$ of $$\\Bbb R$$ over $$\\Bbb Q$$, then the set of all real numbers with zero first projection is a subgroup $$K$$ of $$\\Bbb R$$ for which $$\\Bbb R/K\\simeq \\Bbb Q$$.\n\nBut it can't be finite (unless it is trivial). In other words, there is no coherent way to talk about real numbers being partitioned into the \"even\" and \"odd\" ones. If $$\\Bbb R/K$$ has $$n>1$$ elements, then that means that every number which is a multiple of $$n$$ is in $$K$$; but every real number is a multiple of $$n$$, to wit, $$x=n(x/n)$$.\n\nAnother way to look at it is to think about the set of periods, i.e.\n\n$$P = \\{ p \\in \\mathbb{R} | f(x + p) = f(x) \\text{ for all } x \\in \\mathbb{R} \\}$$\n\nZero is clearly a member of this set no matter what f is. P is closed under addition and negation. So clearly P is a group over addition. So, if you want to find a function that has arbitrarily small periods, you want to find a subgroup of $$\\mathbb{R}$$ that has arbitrarily small values. $$\\mathbb{Q}$$ is the obvious choice, so the characteristic function for $$\\mathbb{Q}$$ works, as stated in another answer.\n\nHere's a different function with the same property (which relies on a fair bit of choice). Choose a Hamel basis for $$\\mathbb{R}$$ over $$\\mathbb{Q}$$ and pick a basis vector $$v$$. Let $$f$$ be the function which projects onto the $$v$$ coordinate. Then for any other basis vector $$u$$ and any integer $$n$$, $$f(x+u/n)=f(x).$$ More can be found on these functions in the article \"Discontinuous additive functions\" by Bernardi.\n\n\u2022 Hope you don't mind my edit, since it's not that much choice needed for what you want. =) Oct 4, 2018 at 17:10", "date": "2022-07-06 23:32:39", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2942016/function-with-arbitrary-small-period/2942235", "openwebmath_score": 0.8838260173797607, "openwebmath_perplexity": 121.84249319753401, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964220125032, "lm_q2_score": 0.8633916047011594, "lm_q1q2_score": 0.8508693372286261}}
{"url": "https://stats.stackexchange.com/questions/207460/expected-number-of-groups-of-3-consecutive-wins-in-200-rounds", "text": "# Expected number of groups of 3 consecutive wins in 200 rounds\n\nIf I know the probability of winning an individual round (while playing, say, a slot machine), call it $p_0$, what is the expected number of groups of 3 consecutive wins in 200 rounds? Or, to make it more general, what is the expected number of $k$-consecutive-win groups in $n$ rounds?\n\nThe groups cannot overlap, so 4 consecutive wins do not count as 2 different groups of 3 wins, while 6 consecutive wins do count as 2 different groups.\n\nI was thinking of looking for the total number of possible $k$-group configurations in $n$ rounds, but I'm not too sure.\n\nExplaining how you got this result would be most welcome, though the final formula would do as well. Thank you very much.\n\nMy solution will not only answer your question exactly (within linear algebra roundoff error) for the general case, but actually gives you the entire probability distribution of number of successful (i.e., completed) streaks.\n\nThis can readily be solved by using a discrete time discrete state (time-homogeneous) Markov Chain, in an approach similar in spirit to, but with larger state space, than the methods I used in https://math.stackexchange.com/questions/383704/probability-of-streaks/1739987#1739987 . After you read my answer in that thread plus this thread, you should be a wizard in using Markov Chains to solve all manner of streaks problems.\n\nI will let\n\np = probability of winning per round\nk = number of wins for successful streak\nn = number of rounds\n\n\nThen g is determined to be the upper bound on largest number of streaks, calculated as\n\ng = floor(n/k)\n\n\nDefine the states as bivariate pairs(i,j), in which i denotes the current number of successful (i.e., completed) streaks and j is the length of the current streak. i will range from 0 to g, and j will range from 0 to k-1, except for when i = g, in which case j only goes from 0 to 0, and we make that state, (g,0), an absorbing state because we can't get any additional successful streaks. Order the states with j increasing fastest, then i increasing. I.e., with k = 3, the states would be (0,0),(0,1),(0,2),(1,0),(1,1),(1,2), etc.\n\nOnce the Markov Chain one step transition matrix, M, has been populated (see below), we compute the n step Markov Chain transition matrix, Mn as $M^n$. Given that we start in state (0,0) before the first of n rounds, the first row of Mn contains the probabilities of being in the various states after n rounds. For each possible value of number of successful streaks i, the sum of the probabilities over j for all states (i,j) provides the probability of exactly i successful streaks having occurred. It is then trivial to compute the expected number of successful streaks.\n\nThe one step transition matrix is populated as follows (note that w.p. is short for \"with probability\"):\n\nFor each i from 0 to g-1 (I'll show here for the case k = 3)\nState (i,0) transitions to (i,0) w.p. 1-p and to (i,1) w.p. p\nState (i,1) transitions to (i,0) w.p. 1-p and to (i,2) w.p. p\nState (i,2) transitions to (i,0) w.p. 1-p and to (i+1,0) w.p. p\n\nThe absorbing state (g,0) transitions w.p. 1 to (g,0)\n\n\nThe population of the one step Markov Chain for a general value of k (and n) is shown in my MATLAB code below. Everything on a line after % is a comment.\n\nk = 3; n = 200; p = .6; % set k, n, and p to particular values\ng = floor(n/k);\nB = [(1-p)*ones(k,1),p*eye(k)]; % recurring block in transition matrix M.\n% B is a k by (k+1) matrix, consisting of a column vector of (1-p) 's,\n% right-horizontally concatenated with p times the k by k identity matrix\n% Start building up one step transition matrix, M\nM = zeros(g*k+1); % (g*k+1) by (g*k+1) matrix of zeros\nfor i=0:g-1, M(k*i+1:k*(i+1),k*i+1:k*(i+1)+1) = B; end;\nM(g*k+1,g*k+1) = 1; % Construction of M is now complete\nMn = M^n; % n step transition matrix\n% Calculate array of probabilities of number of successful streaks\n% and place in prob_array\nprob_array = zeros(g+1,1);\nfor i=0:g-1, prob_array(i+1) = sum(Mn(1,k*i+1:k*i+k)); end;\nprob_array(g+1) = Mn(1,g*k+1); % prob_array is now complete\nexpected_number_streaks = (0:g)*prob_array\n\n\nHere are example results for n = 200 rounds, and streaks of length k = 3, with bonus results for streaks of length k = 5.\n\nHere is an example one step transition matrix, for p = 0.6, n = 11, k = 3, which results in g = 3.\n\n\u2022 Great. Yes, this agrees with observations... By the by, do you recommend any books to learn (relatively) advanced probability? I hadn't met the use of Markov Chains like this before. Apr 18 '16 at 7:02\n\u2022 @Kristian D'Amato , many so-called advanced probability books deal with rigorous theory - it sounds like that may not be what you want. It sounds like a book on stochastic processes might be good for you, or perhaps applied probability with significant stochastic processes content, but I don't have a specific book of such type to recommend. Apr 18 '16 at 17:12\n\nSince you're talking about a slot-machine, I assume that the probabilities are independent. The probability of $k$ consecutive wins is $p_0^k$. That probability is the expected portion of the groups with size $k$ that contain all wins. The number of groups with size $k$ (that does not overlap) is $200/k$, and we can call the number of all rounds $n$ (already alluded to in your question). So, the expected number of groups with size $k$ that contains consecutive wins would be\n\n$p_0^k \\cdot (n/k)$\n\n$p_0^3 \\cdot (200/3)$\n\nEdit:\n\nApparently this problem is deceptively hard, like really hard. Just answering what the probability is that one group of length $k$ will appear is hard. The probability $S(n,k)$ of a group of consecutive wins of length $k$ in a chain of $n$ events, with the probability of a win $p$ and probability of a fail $1-p=q$ is\n\n$S(n,k) = p^k \\sum_{i=0}^{\\infty} {n - (i+1)k \\choose i}(-qp^k)^i - \\sum_{i=1}^{\\infty} {n - ik \\choose i}(-qp^k)^i$\n\nHere, the groups can even overlap, which is not what you asked for. I hope this can point you in the right direction.\n\n(I apologize for the wrong answer earlier, I hope I did not lead you astray too much)\n\n\u2022 Are you sure this is right? Observed information is not matching with this prediction (observations are about 50% higher) Apr 15 '16 at 10:47\n\u2022 Are you sure that the groups may not overlap? Because, if they are allowed to overlap I can understand why the estimate is missing by 50% and in that case you should use the formula n + 1 - k to estimate the number of possible blocks. Apr 15 '16 at 11:41\n\u2022 Yep. Keep in mind that if they are allowed to overlap the chance of getting a second win-group overlapping with the first is not independent of the first (since you only need one more round to have another win-group)... Anyway they don't overlap. Apr 15 '16 at 12:35\n\u2022 I'm doubting myself, but I don't see how this could be wrong. p^k is the expected proportion of groups with all wins, n/k is the maximum amount of consecutive, non overlapping groups that can fit within the chain. Are you running simulations and could you supply some code in that case? Apr 15 '16 at 13:40\n\u2022 I will when I'm able to later on. There are other k-chains in the entire sequence, since a group can start at position 2, for instance, covering 2-4, etc... Apr 15 '16 at 13:51", "date": "2022-01-21 17:40:05", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/207460/expected-number-of-groups-of-3-consecutive-wins-in-200-rounds", "openwebmath_score": 0.7503558993339539, "openwebmath_perplexity": 644.4102924561752, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964228644457, "lm_q2_score": 0.863391602943619, "lm_q1q2_score": 0.8508693362321365}}
{"url": "http://math.stackexchange.com/questions/731928/taylor-theorem-inequality", "text": "# Taylor Theorem inequality\n\nProve that for all $f\\in C^2([0,1])$ with $f(0)=f(1)=0$ and $|f''(x)| \\le 1$ $$|f(x)| \\le \\frac{1}{2}x(1-x)$$ $\\forall x \\in [0,1]$.\n\n-\nIf your function is only continuous, how do you know for sure that it has a second derivative? \u2013\u00a0 Mercy Mar 30 '14 at 1:01\nMistake in the post, sorry! Just corrected\u2026 \u2013\u00a0 Brontolo Mar 30 '14 at 1:05\n\nWe have, $\\displaystyle (1-x)\\int_0^x tf''(t)\\,dt -x \\int_x^1 (t-1)f''(t)\\,dt = -f(x)$, for $x \\in [0,1]$ (just apply integration by parts on LHS to ease the simplification).\n\nSince, $t\\ge 0$ for $t\\in [0,1]$ and $(t-1) \\le 0$ for $t \\in [0,1]$, taking modulus on both sides,\n\n$|f(x)| = \\displaystyle \\left|(1-x)\\int_0^x tf''(t)\\,dt -x \\int_x^1 (t-1)f''(t)\\,dt\\right|$\n\n$\\le \\displaystyle \\left|(1-x)\\int_0^x tf''(t)\\,dt\\right| + \\left|-x \\int_x^1 (t-1)f''(t)\\,dt\\right|$\n\n$\\le \\displaystyle \\sup\\limits_{t\\in[0,1]}|f''(t)|. \\bigg( (1-x)\\int_0^x t\\,dt -x \\int_x^1 (t-1)\\,dt \\bigg)$\n\n$= \\dfrac{x(1-x)}{2}.\\sup\\limits_{t\\in[0,1]}|f''(t)| \\le \\dfrac{x(1-x)}{2}$.\n\nAliter: Define $g(t)=f(t)-\\dfrac{t(t-1)}{x(x-1)}f(x)$, on $[0,1]$.\n\nThen, $g(0)=g(x)=g(1)=0$.\n\nApplying Rolle's Theorem twice on $(0,1)$, $\\exists \\alpha \\in (0,1)$ such that $g''(\\alpha)=0$.\n\nThat is $g''(\\alpha) = f''(\\alpha) - \\dfrac{2}{x(x-1)}f(x)=0$\n\nor, $|f(x)|=\\dfrac{x(1-x)}{2}|f''(\\alpha)| \\le \\dfrac{x(1-x)}{2}$.\n\n-\nHow do you pull $\\sup|f''(t)|$ out of the integrals while maintaining the subtraction between the integrals? Suppose $f''(t)=1$ on $[0,x]$ and $f''(t)=-1$ on $[x,1]$ (or a continuous function very close to that). \u2013\u00a0 robjohn Mar 30 '14 at 2:00\nI think if you add the absolute values of the integrals you get what you want. That is, $$\\frac12x^2(1-x)+\\frac12x(1-x)^2=\\frac12x(1-x)$$ \u2013\u00a0 robjohn Mar 30 '14 at 2:03\n@robjohn fixed typo and added a line .. thanks for pointing it out :) \u2013\u00a0 r9m Mar 30 '14 at 2:08\nNice way the second one! I thought that it would be good to do with Taylor Theorem (therefore the title) but I found no way. \u2013\u00a0 Brontolo Mar 30 '14 at 10:17\n\nFor some $\\xi_k\\in[0,1]$, $$f(1)=f(0)+f'(0)+\\frac12f''(\\xi_1) \\quad\\text{and}\\quad f(0)=f(1)-f'(1)+\\frac12f''(\\xi_2)$$ implies $$|f'(0)|\\le\\frac12 \\quad\\text{and}\\quad |f'(1)|\\le\\frac12$$ Let $g(x)=f(x)-\\frac12x(1-x)$ and $h(x)=f(x)+\\frac12x(1-x)$.\n\n$g(0)=0$, $g'(0)\\le0$, and $g''(x)\\le0$; therefore, $g(x)\\le0$.\n\n$h(0)=0$, $h'(0)\\ge0$, and $h''(x)\\ge0$; therefore, $h(x)\\ge0$.\n\nThus, $$\\overbrace{-\\frac12x(1-x)\\le}^{h(x)\\ge0}f(x)\\overbrace{\\le\\frac12x(1-x)}^{g(x)\\le0}$$\n\n-\nCan you explain why $g(0)=0, g\u2032(0)\u22640$, and $g\u2033(x)\u22640$; therefore, $g(x)\u22640$? \u2013\u00a0 Brontolo Mar 30 '14 at 10:09\n@TheMaker94: Mean Value Theorem on $[0,1]$. Since $g'(0)\\le0$ and $g''(x)\\le0$, we have $g'(x)\\le0$. Since $g(0)=0$ and $g'(x)\\le0$, $g(x)\\le0$. \u2013\u00a0 robjohn Mar 30 '14 at 12:14\nPerfect, clear thanks! \u2013\u00a0 Brontolo Mar 30 '14 at 12:28\n\nWe want to prove that for each $x$:$$\\exists c\\ \\ f(x) =\\frac 12 f''(c) x(1-x)$$\n\nWe want to find such a $c$ via the Rolle theorem (or via the mean value theorem, but we can always go back to the Rolle version).\n\nWe can already apply the Rolle theorem, which gives an annulation for $f'$.\n\nLet us modify $f$ to go to $0$ once more: assuming $0<x<1$,\n\n$$g(u) := f(u) + A_xu(1-u)\\\\ g(x) = 0\\Leftarrow A_x = - \\frac{f(x)}{x(1-x)}$$\n\nNow $g(1)=g(x) = g(0)$ hence, applying several times the Rolle theorem:\n\n$$\\exists c \\ \\ 0=g''(c) = f''(c) - \\frac{f(x)}{x(1-x)}(-2)\\\\ f(x) = -\\frac 12 x(1-x)f''(c)$$\n\nNB: the error in the sign does not change the final inequality.\n\n-\nclever idea (+1) \u2013\u00a0 robjohn Mar 30 '14 at 8:09", "date": "2015-01-26 12:25:32", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/731928/taylor-theorem-inequality", "openwebmath_score": 0.944036602973938, "openwebmath_perplexity": 643.1677334210763, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964194566753, "lm_q2_score": 0.8633916047011595, "lm_q1q2_score": 0.8508693350219458}}
{"url": "https://dralb.com/2019/02/02/finding-area-using-integrals/", "text": "# Finding Area using Integrals\n\nAs part of my Calculus 2 class, I have created a practice exam\u00a0to help my students study.\u00a0 In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2.\u00a0 I hope you find the following solutions helpful in your studies.\n\nYou can also view me presenting the solution on YouTube at\n\n## Find the area bounded by the function $$f(x)=8x$$, $$g(x)=\\frac{8}{x}$$, and $$h(x)=x^{2}$$.\n\nDrawing a sketch of these graphs, we get the following picture.\n\nIn order to find the area, we cut this into smaller cross sections, then add up the areas of the cross sections. Here we will cut out rectangles as shown below.\n\nNote that the height of the rectangles has a different function value in the first portion than it does in the second portion, so we will split this into two parts. In the first section, we get that\n\\begin{align*} A_{cs}&=h*w \\\\ &=(y_{b}-y_{s})\\Delta x, \\end{align*} where $$y_{b}$$ is the bigger $$y$$ value and $$y_{s}$$ is the smaller $$y$$ value. In this case, we note from the graph that $$y_{b}=8x$$ and $$y_{s}=x^{2}$$. Therefore,\n\\begin{align*} A_{cs}=(8x-x^{2})\\Delta x. \\end{align*}\n\nIn order to find the total area over this interval, we then find that\n\\begin{align*} A_{T1}=\\int_{a}^{b}(8x-x^{2})dx. \\end{align*} Where is $$a$$ is the smallest $$x$$ value and $$b$$ is the largest $$x$$ value where these cross sections work.\n\nWe find $$a$$ as the intersection of $$8x$$ and $$x^{2}$$. That is, we solve\n\\begin{align*} 8x&=x^{2} \\\\ 0&=x^{2}-8x \\\\ 0&=x(x-8) \\\\ x&=0,8. \\end{align*} From the picture, we note that the point we are interested for this example is actually $$a=0$$.\n\nWe then find $$b$$ as the intersection of $$8x$$ and $$\\frac{8}{x}$$, so we get\n\\begin{align*}\n8x&=\\frac{8}{x} \\\\\n8x^{2}&=8 \\\\\nx^{2}&=1 \\\\\nx&=\\pm 1.\n\\end{align*}\nAgain, by looking at the picture, we note that the correct choice is $$b=1$$.\n\nTherefore,\n\\begin{align*} A_{T1}&=\\int_{0}^{1}(8x-x^{2})dx \\\\ &=4x^{2}-\\frac{x^{3}}{3}|_{0}^{1} \\\\ &=4-\\frac{1}{3}-0=\\frac{11}{3}. \\end{align*}\n\nIf we follow the same process for the second area, we note that we will have $$y_{b}=\\frac{8}{x}$$ and $$y_{s}=x^{2}$$. Therefore, we will get that\n\\begin{align*} A_{cs}&=h*w \\\\ &=(y_{b}-y_{s})\\Delta x \\\\ &=(\\frac{8}{x}-x^{2})\\Delta x. \\end{align*}\n\nTherefore, the total area of the second section will be\n\\begin{align*} A_{T2}=\\int_{b}^{c}(\\frac{8}{x}-x^{2})dx. \\end{align*}\n\nWe have already found that $$b=1$$, so now we need to find $$c$$. Here, this is the intersection of $$\\frac{8}{x}$$ and $$x^{2}$$. Therefore, we get that\n\\begin{align*} \\frac{8}{x}&=x^{2} \\\\ 8&=x^{3} \\\\ x&=2. \\end{align*} Hence, $$c=2$$.\n\nWe then get that\n\\begin{align*} A_{T2}&=\\int_{1}^{2}(\\frac{8}{x}-x^{2})dx \\\\ &=8\\ln|x|-\\frac{x^{3}}{3}|_{1}^{2} \\\\ &=8\\ln(2)-\\frac{8}{3}-(8\\ln(1)-\\frac{1}{3}) \\\\ &=8\\ln(2)-\\frac{7}{3}. \\end{align*}\n\nNow that we have found both areas, the total area of the region will be given as the sum of the two areas. Therefore, the total area is\n\\begin{align*} A_{T}&=A_{T1}+A_{T2} \\\\ &=\\frac{11}{3}+8\\ln(2)-\\frac{7}{3} \\\\ &=\\frac{4}{3}+8\\ln(2). \\end{align*}\n\n## Other Resources\n\nI have it linked above, but you can find the practice exam here.\u00a0 You can find the other solutions to the practice exam in the other posts available here.\n\nIf you found the post, or the YouTube video helpful, please like the post or the video.\u00a0 Also remember to follow the blog and subscribe to the YouTube channel.\u00a0 By doing so you will not only be able to find future content, you will also make it easier for other students to find the content when they are studying for Calculus.\u00a0 Thank you.\n\n## 1 thought on \u201cFinding Area using Integrals\u201d\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2020-09-24 20:43:55", "meta": {"domain": "dralb.com", "url": "https://dralb.com/2019/02/02/finding-area-using-integrals/", "openwebmath_score": 0.9998196959495544, "openwebmath_perplexity": 504.4114249531262, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9919380077567411, "lm_q2_score": 0.8577681013541613, "lm_q1q2_score": 0.8508527815745291}}
{"url": "http://www.ni.com/documentation/en/labview-comms/2.0/node-ref/kronecker-product/", "text": "# Kronecker Product (G Dataflow)\n\nCalculates the Kronecker product of two input matrices.\n\n## matrix A\n\nThe first input matrix.\n\nThis input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.\n\nDefault: Empty array\n\n## matrix B\n\nThe second input matrix.\n\nThis input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.\n\nDefault: Empty array\n\n## error in\n\nError conditions that occur before this node runs. The node responds to this input according to standard error behavior.\n\nDefault: No error\n\n## kronecker product\n\nMatrix containing the Kronecker product of the first and second input matrices.\n\nThe number of rows in kronecker product is the product of the number of rows in the first and second input matrices. The number of columns in kronecker product is the product of the number of columns in the first and second input matrices.\n\n## error out\n\nError information. The node produces this output according to standard error behavior.\n\n## Algorithm for Calculating the Kronecker Product\n\nIf A is an n-by-m matrix and B is a k-by-l matrix, the Kronecker product of A and B, C = AB, results in a matrix C with dimensions nk-by-ml. This node calculates the Kronecker product using the following equation.\n\n$C={\\left[\\begin{array}{cccc}{a}_{11}B& {a}_{12}B& \\dots & {a}_{1m}B\\\\ {a}_{21}B& {a}_{22}B& \\dots & {a}_{2m}B\\\\ \u22ee& \u22ee& \\ddots & \u22ee\\\\ {a}_{n1}B& {a}_{n2}B& \\dots & {a}_{nm}B\\end{array}\\right]}_{nk\u00d7ml}$\n\nFor example, if\n\n$\\begin{array}{cc}A=\\left[\\begin{array}{cc}1& 2\\\\ 3& 4\\end{array}\\right]& B=\\left[\\begin{array}{cc}5& 6\\\\ 7& 8\\end{array}\\right]\\end{array}$\n\nthen\n\n$\\begin{array}{ccc}{a}_{11}B=\\left[\\begin{array}{cc}5& 6\\\\ 7& 8\\end{array}\\right]& {a}_{12}B=\\left[\\begin{array}{cc}10& 12\\\\ 14& 16\\end{array}\\right]& C=\\left[\\begin{array}{cc}{a}_{11}B& {a}_{12}B\\\\ {a}_{21}B& {a}_{22}B\\end{array}\\right]=\\left[\\begin{array}{cccc}5& 6& 10& 12\\\\ 7& 8& 14& 16\\\\ 15& 18& 20& 24\\\\ 21& 24& 28& 32\\end{array}\\right]\\end{array}$\n\nWhere This Node Can Run:\n\nDesktop OS: Windows\n\nFPGA: Not supported", "date": "2018-03-23 03:17:46", "meta": {"domain": "ni.com", "url": "http://www.ni.com/documentation/en/labview-comms/2.0/node-ref/kronecker-product/", "openwebmath_score": 0.346889466047287, "openwebmath_perplexity": 1527.5243900412706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683465856102, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8508278685190617}}
{"url": "http://vkwf.venetoacquaeterme.it/show-that-the-moment-of-inertia-of-a-uniform-solid-sphere-rotating-about-a-diameter-is.html", "text": "moment of inertia of thin disc about a parallel axis, distance away So the moment of inertia for all such thin discs becomes when The cone is lying on its side with the vertex at the origin so which gives. Physics 111 Lecture 21 (Walker: 10. The spheres have negligible size, and the rod has negligible mass. ( 2 pt ) The distance @ that would be used in the parallel axis theorem to find the mass moment of inertia about the origin, O, of the 0. Question: Show That The Moment Of Inertia Of A Uniform Solid Sphere Rotating About A Diameter Is 2/5 M R^2. We will be using mainly a cylindrical ring and a sphere in our experaments. (a) Show that the moment of inertia about a diameter of a uniform spherical shell of inner radius Rio outer radius R and density p is 1 = p(\u03c0m/15)(R5/2 \u2013 R5/1'). What is the angular momentum of the sphere?. (1) Moment of inertia of a Solid Sphere : (a) About an axis passing through its diameter : Consider a solid sphere of mass M and radius R. (ii) about a tangent: A tangent drawn to the sphere at any point, will obviously be parallel to one of its diameters and the distance between the axes is equal to R, the radius of the sphere. perpendicular to xy-plane passing through a point on the x-axis at a distance x. 1562 kg m 2 and torque applied is 0. 8 Solid sphere rotating about the central axis. (This also assumes we are rotating the bodies around the same axis. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR 2, find the moment of inertia about an axis normal to the disc passing through a point on its edge. calculate moment of inertia of any object, rotating about an arbitrary axis. ) of a sphere about its diameter = 2MR 2 /5 According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance. For simple solid objects, one can calculate the moment of inertia from the mass, size, and shape. Torque The turning effect of a force with respect to some axis, is called moment of force or torque due to the force. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR 2, find the moment of inertia about an axis normal to the disc passing through a point on its edge. Since moment of inertia is a scalar quantity, a compound object made up of several objects joined together has a moment of inertia which is the A rotating door is made from four rectangular glass panes, as shown in the drawing. The moment of inertia is a value that describes the distribution. ( 1 pt each ) Object A is a solid sphere. 0 kg and radius 0. By parallel axes theorem; This is an expression for M. Sphere #1 will arrive first B. The power transmitted by the shaft is. (This also assumes we are rotating the bodies around the same axis. 4 Radius of Gyration 3. A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3. \u2022 Rotating objects tend to keep rotating, while non-rotating objects tend to remain non-rotating. Moment of inertia of solid sphere when it is rotating about its diameter can be determined using integration process and about different axes can be. ( 1 pt ) A disc in the x-y plane is rotating about an axis perpendicular to the x-y plane. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. For the sake of one more bit of integration practice, we shall now use the same argument to show that the moment of inertia of a uniform circular disc about a. The moment of inertia for some common shapes are given below. where I is the moment of inertia. mass m and radius a, about a diameter of its plane face. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. For example, if two disks have the same mass but one has all the mass around the rim and the other is solid, then the disks would have different moments of inertia. , an axle through the center and perpendicular to the disk, the moment of inertia is calculated by carrying out the. A uniform solid sphere with mass M and radius R has a moment of inertia I = 2/5MR{eq}^2 {/eq} about an axis through its center. (a) Find the angular momentum of the skater. The moment of inertia may be defined as, I = sum m_ir_i^2 and if the system is continuous, then I = int r^2dm If rho is the mass density then, dm = rhodV where dV is an elementary volume. Thin walled cylinder rotating about the central axis. Experiments show that, if we compare bodies of similar shape and size but having different masses, the moment of inertia, I is directly proportional to the mass. Moment of Inertia (Rotational Inertia) I:- Moment of Inertiaof a body, about a given axis, is Moment of inertia of a solid disc:- (a) About an axis passing through its center and perpendicular Motion of a point mass attached to a string would over a cylinder capable of rotating about its axis of symmetry. Point P is midway between the center and the rim of the disk, and point Q is on the rim. Learn how to calculate moment of inertia of different shapes or objects using the several formulas. The moment of inertia of a solid sphere of radius of 0. Labels: Circular Disc and solid sphere, Mass Moment of Inertia, Mass Moment of I think This blog is very interesting, I hope you feel same like me about my link @ moment of inertia calculator online It's a good post about moment of inertia. Use Newton's second law to obtain two equations in a and T that we can solve simultaneously. A solid sphere with a mass of 8. So when the masses are placed at r= 0, I= I0. Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. The translational velocity is slow enough to make easy accurate measurements. (****) Find the moment of inertia of a uniform, thin-walled sphere of radius R and mass M. A solid sphere, disc and solid cylinder all of same mass and made up of same material are allowed to roll down (from rest) on an inclined plane, then (a) solid sphere reaches the bottom late (b) solid sphere reaches the bottom first (c) disc will reach the bottom first (d) all will reach the bottom at the same time. 16 in order to remain upright under the in uence of gravity. Does it have a larger moment of inertia for an axis through the thicker end of the rod and perpendicular to the length of the rod, or for an axis through the thinner end of the rod. 00-m-diameter wagon. 115m and a mass of 12. A uniform solid S is generated by fully rotating R in the x axis. A particle of mass M is attached to one end of the stick. Moment of inertia states that:The product mass and the square of perpendicular distance from the axis of rotation is known as moment of inertia. not the hypotenuse. A constant tension of 23. Two solid uniform spheres roll down a ramp without slipping or sliding. 2 1 3 Where M is the mass and R is the radius of the sphere. 37 x 106 meters. 00 m long and has mass 4. 1 Rotational Inertia. Moment of Inertia: Sphere. This is determined by summing the moments of inertia of the thin discs that form the sphere. A hollow cylinder and a solid cylinder have the same diameter. The translational velocity is slow enough to make easy accurate measurements. (6) About what axis will a uniform, balsa-wood sphere have R the same moment of inertia as does a thin-walled, hollow, lead Rsphere of the same mass and radius, with the axis along a diameter? Use th e Parallel Axis Theorem. The density is then (1) and the moment of inertia tensor is (2) (3) (4). The moment of inertia of the sphere is I = 2 5 MR2. 84 m diameter solid sphere can be rotated about an axis through its center by a torque of 10. Sphere 2 has twice the radius of sphere 1. Thin walled cylinder rotating about the central axis. Circular Disk Rotating About Its Diameter The moment of inertia for the same circular disk rotating about an axis in the plane of the disk, passing through its center, is given by Thus, the uniform disk's moment of inertia in its own plane is twice that about its diameter. The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. Moment of inertia of a sphere can be explained in two parts (1) Solid Sphere (2)Hollow Sphere. (a) Show that the moment of inertia of a uniform hollow cylinder of inner radius R1, outer radius R2, and mass M, is I \u00bd M(R12 R22), if the rotation axis is through the center along the axis of symmetry. ( 2 pt ) The distance @ that would be used in the parallel axis theorem to find the mass moment of inertia about the origin, O, of the 0. show more decimal digits. 7 cm in diameter. Labels: Circular Disc and solid sphere, Mass Moment of Inertia, Mass Moment of I think This blog is very interesting, I hope you feel same like me about my link @ moment of inertia calculator online It's a good post about moment of inertia. The axis of rotation in the question is a tangent to the ring. 01 18-Jun-2003 1. A solid disk with a mass of 0. The moment of inertia of an object changes if the axis of rotation is changed. calculate its moment of inertia about any axis through its centre. The following links are to calculators which will calculate the Section Area Moment of Inertia Properties of common shapes. In this section we show how the idea of integration as the limit of a sum can be used to \ufb01nd the moment of inertia of a lamina. The ratio of the larger Sphere moment of inertia to that of the smaller sphere is 4 Consider two uniform solid spheres were one has twice the mass and what is the diameter of the other. A solid disk will have a different moment than a washer, and there are formulas derived for calculating the moments of many common shapes. 6 mm when it bears a load. 6 F z z P. 132 kg m 2 about an axis which is found to be. 8 of Newman 1977 gives the added inertia for coefficient for spheroids of varying aspect ratio, referred to the moment of inertia of the displaced mass. Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. As a consequence, the flow path in a rotating chute deviates considerably from that in a non-rotating chute. 3 Physical Significance of Moment of Inertia 3. Harm to minors, violence or threats, harassment or privacy invasion, impersonation or misrepresentation, fraud or phishing, show. Two uniform solid spheres have the same mass, but one has twice the radius of the other. Not the earth going around the sun, but the earth rotating on its axis, then you'd have to say that the moment of inertia for that amount of rotation is 2/5 mr squared, because it's a sphere rotating through an axis that goes through its center. Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. Sphere 2 has three times the radius of sphere 1. (b) Find the final rotation rate of the skater. Moments of inertia of rigid bodies\u2217 - Similar to Moments of inertia of rigid bodies\u2217 May 23, 2011 Moment of inertia of rigid body depends on the distribution of mass dimensional bodies like cylinder and sphere. [ In this question, you may assume standard results for the moment of inertia of uniform circular discs. Since moment of inertia is a scalar quantity, a compound object made up of several objects joined together has a moment of inertia which is the A rotating door is made from four rectangular glass panes, as shown in the drawing. A hoop a solid sphere a flat disk a hollow sphere Each of the objects has mass M and radius R. The moment of inertia of a uniform rod about an axis through its center is. A homogeneous solid cylinder of mass m, length L, and radius R rotates about an axis through point P, which is parallel to the cylinder axis. 84 m diameter solid sphere can be rotated about an axis through its center by a torque of 10. EXAMPLE: MOMENT OF INERTIA / EARTH EXAMPLE: 6The Earth has mass and radius 5. 16 The variation of angular position \u03b8, of a point on a rotating rigid body, with time t is shown in Fig. Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. The moment of inertia of a circular ring about a diameter is \u00bd mr2, with usual notations. 115m and a mass of 12. For a different rotation point of an object\u2014say a rod rotating around one end, like a turnstile, instead of around its center\u2014we use the parallel axis theorem to find the object's moment of inertia. In this case, the moment of inertia of the mass in this system is a scalar known as the polar moment of inertia. \u2022 Subdivide body into small volume elements \u2022 Add the moment of inertia contributed by all these amounts of massAdd the moment of inertia contributed by all these amounts of mass \u2022 I = M \u22c5(average value of R2) 2. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. For a thin uniform homogenous rectangular plate, the mass moment of inertia about the rectangular coordinate axes, a and b, passing through the centre of gravity of the circular plate can be obtained from the area moment of inertia. Moment of inertia of solid disk Period of small oscillation of sphere rolling in cylinder. \u2022 Subdivide body into small volume elements \u2022 Add the moment of inertia contributed by all these amounts of massAdd the moment of inertia contributed by all these amounts of mass \u2022 I = M \u22c5(average value of R2) 2. (iii) Moment of inertia of a body should always be referred to as about a given axis, since it depends upon distribution of mass about that axis. Each sphere is a distance R+L/2 from the axis of rotation, so we must use the parallel axis theorem. Physics 100A Homework 10 \u2013 Chapter 10 (part 2) 10. 36 Determining Moments of Inertia. The moment of inertia of a uniform rod about an axis through its center is. The Brick Solid block adds to the attached frame a solid element with geometry, inertia, and color. Furthermore, because of the symmetry of the sphere, each principal moment is the same, so the moment of inertia of the sphere taken about any diameter is. \u2022 Parallel axis theorem for products of inertia: Product of inertia is useful in calculating MI @ inclined axes. The rotational kinetic energy is the kinetic energy of rotation of a rotating rigid body or system of particles, and is given by $K=\\frac{1}{2}I{\\omega }^{2}$, where I is the moment of inertia, or \u201crotational mass\u201d of the rigid body or system of particles. 00 m long and has mass 4. Solid sphere, radius r, about diameter. Answer is in kg\u22c5m2/s 2. The moment of inertia of a thin spherical shell of mean radius 0. Moment of inertia of solid shere about any diameter is (2/5)MR^2 , where M is mass and R is radius of sphere. The sphere is rotated about a diameter with an angular speed \u03c9. 8(a) in side view. The radius of the sphere is 20. Show that the magnetic moment 'u' and the angular momentum 'L' of the sphere are related as: u=Lq/2m Pls help with this question. A man stands on a rotating platform that has an angular speed of 6. I have derived moment of inertia of solid sphere along diameter but my textbook says that moment of inertia is \"Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. Moments of Inertia and Angular Momentum. So it applies no torque, Since this is the only force acting on the system, the net torque is zero. It should not be confused with the second moment of area. The moment of inertia of a body rotating around an arbitrary axis is equal to the moment of inertia of a body rotating around a parallel axis through the center of mass plus the mass times the perpendicular distance between the axes h squared. 2: Angular momentum of a sphere Question: A uniform sphere of mass and radius spins about an axis passing through its centre with period. then the vertical gravitation would be x. 2 m is set into rotation about an axis passing through its centre and perpendicular to its plane by applying torque 10 Nm. We were discussing \"Method to determine the area moment of inertia for a hollow rectangular section\", \"The theorem of parallel axis Let us consider one hollow circular section, where we can see that D is the diameter of main section and d is the diameter of cut-out section as displayed in following figure. Moment of Inertia of a solid body \u2022 Mass continuously distributed throughout its volume. 5 kg of a cold metal at a temperature of 258 K is immersed in 2. Find the moment of inertia of this combination about each of the following axes: (a) an axis pelvendicular to the bar through. Rotational Motion. M -mass, R -Radius. Vocabulary Angular Momentum: The measure of. If the surface of the ball is defined by the equation: 1301 + + =,. \u2013 What is the rotational kinetic energy of a 450-g solid sphere with a diameter of 23 cm rotating at rate of 17 rpm? \u2013 What is the total rotational kinetic energy of a 18-kg child riding on the edge of a merry-go-round of mass 160 kg and r = 2. A solid uniform L-shaped plate has mass $m$ and dimensions as shown. You have two steel spheres. [You may assume, without proof, that the moment of inertia of a uniform circular disc, of mass m and radius r, about a diameter is 1 4 mr2. diameter of the solid cylinder is large b. 6 F z z P. And needs to solve in both spherical & cylindrical coordinate system. The moment of inertia of a circular ring about a diameter is \u00bd mr2, with usual notations. MOTION OF SYSTEM OF PARTICLES AND RIGID BODY CONCEPTS. Moment of inertia - Parallel-Axis Theorem Pendulum Moment of Inertia of a Rotating Body Collision Moment of inertia tensor Moment of Inertia of a thin uniform rod by integration A 15--diameter CD has a mass of 24. The moment of inertia of a uniform rod about an axis through its center is. 0cm spins about the axle through its center. In the first part of our lab a rotating solid cylindrical drum with a hollow body drum given a rotational velocity from a falling mass. 0 cm and has mass 1. Sponsored Links. mass m and radius a, about a diameter of its plane face. You have two steel spheres. ( 2 pt ) The distance @ that would be used in the parallel axis theorem to find the mass moment of inertia about the origin, O, of the 0. mass of the solid cylinder is large d. We will compare our results for a uniform, solid disk and a uniform ring with those derived from theory. Using the definition of moment of inertia, I = r 2 dm, one can show that theory predicts I disk = \u00bd MR2 (4) I ring = \u00bd M(R IN 2 + R OUT 2) (5). If they all are released from rest. Hollow sphere of radius r and mass m Similar to the solid sphere, only this time considering a stack of infinitesimal thin, circular hoops. KE = (I * w^2)/2. The mass of the hub can be ignored. (b) Obtain the moment of inertia for a solid cylinder. Solid Versus Hollow Edit A hoop (hollow object) has a greater moment of inertia than a greater moment of inertia than a solid disk of the same mass because all of the mass of the hoop is at a large radius. Although mass is defined in terms of inertia, it is conventionally interpreted as. Its moment of inertia about an axis tangent to it and perpendicular to its plane is ?. A solid sphere rolls (without slipping) down a plane inclined at 30\u02da to the horizontal. Now if the two masses are each. 10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere. The moments of inertia of common shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily accessible in any mechanics textbook. A solid uniform sphere of radius R and mass M has a rotational inertia about a diameter that is given by (2 = 5) MR 2. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is. disk can be considered as a uniform solid disk of radius 25 cm and mass of 1. (2R/ O15) from the center of the sphere) (7) A frictionless pulley has the shape of a uniform solid di sk of mass 2. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Learn how to calculate moment of inertia of different shapes or objects using the several formulas. ] (10) (b) Hence find the moment of inertia of a uniform solid sphere, of mass M and radius a, about a diameter. The moment of inertia of a uniform rod about an axis through its center is. Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane. Calculate the moment of inertia of a thin plate in the shape of a right triangle, about an axis that passes through one end of the hypotenuse and is parallel to the opposite leg of the triangle, as in Figure. Assume the ball is a uniform, solid sphere. The moment of inertia of a uniform object depends not only on the size and shape of that object but on the location of the axis about which the object is rotating. Annapolis MD. The spheres have negligible size, and the rod has negligible mass. Determine the moment of inertia for a solid cylinder with mass m and radius R with a non-uniform mass density given by p= ar^2. m what is her final moment of inertia'? How does she physi- cally accomplish this change? (Il) A potter's wheel is rotating around a vertical axis through its center at a frequency of 1-5 rev/s The wheel can be considered a uniform disk of mass 5. 94 m/s2 down the ramp (b) 3. For axis A, the rod is rotating about its centre of mass. Not the earth going around the sun, but the earth rotating on its axis, then you'd have to say that the moment of inertia for that amount of rotation is 2/5 mr squared, because it's a sphere rotating through an axis that goes through its center. A thin uniform bar has two small balls glued to its ends. a) Calculate the torque applied to the disk by the rope. 60-cm-diameter sprocket? Three objects of uniform density\u2014a solid sphere, a solid cylinder, and a hollow cylinder\u2014are placed at the top of an incline (Fig. 1 kg m2 as the skater draws his arms and legs inward toward the axis of rotation. Find the moment of inertia about a diameter?. 85kg and diameter 45. 16 The variation of angular position \u03b8, of a point on a rotating rigid body, with time t is shown in Fig. The moment of inertia of the sphere is I = 2 5 MR2. For a ring let\u2019s assume an element of mass dm on the ring. Answer: (a) Moment of inertia of sphere about any diameter = 2/5 MR 2. The density is then (1) and the moment of inertia tensor is (2) (3) (4). moment of inertia is the same about all of them. none of the above are necessary A common thread spool rests on a flat table. For a solid uniform sphere and a thin hoop, each of mass Mand radius R and rotating about their respective centers of mass, the moment of inertia of the hoop is larger than that of the sphere. I have been asked what the rotational inertia difference is between a hollow Aluminium driveshaft of diameter 44. To look for the angle of inclination which is b we do the following steps: 90 - b = a. (Hint: Form the shell by superposition of a sphere of density p and a smaller sphere of density -p. A small solid marble of mass m and radius r rolls without slipping along a loop-the-loop track shown in Figure 12. Step 2: The moment of inertia of the balsa wood sphere (solid) about the diameter is, 2 2 I2= M5 We have to choose an axis through which these two moment of inertias would be same. 500 kg and can be treated as point masses. 00-cm-diameter, 330 sphere is released from rest at the top of a 2. (1) Moment of inertia of a Solid Sphere : (a) About an axis passing through its diameter : Consider a solid sphere of mass M and radius R. Cotufa is doing homework on \"moment of inertia\" of uniform solid sphere and a uniform solid cylinder. 00-cm-diameter sprocket if the wheel is to attain an acceleration of 4. 0 kg and R \u2014 -MR2. A uniform solid S is generated by fully rotating R in the x axis. The moment of inertia about an axis at one end is. The inertia tensor of the disk alone about its center of mass is: MR2 4 1 0 0 0 1 0 0 0 2 (A) Find the inertia tensor of the disk alone about the point A. 150 m has a moment of inertia for rotation through its central axis. Find the value of the spin in revolutions per second for a= 10cm and b= 1cm. Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane. moment of inertia of thin disc about a parallel axis, distance away So the moment of inertia for all such thin discs becomes when The cone is lying on its side with the vertex at the origin so which gives. If we had a sphere, a solid sphere, then So here you have a solid sphere, and I rotate it about an axis through its center. If her initial moment of inertia was 4. It is a convention to write to indicate the moment of inertia with respect to an axis passing though the center of mass of the rotating rigid body. Obtain the moment of inertia of a hollow solid sphere of inner and outer radii r1 and r2 Products. The expression for the moment of inertia of a sphere can be developed by summing the moments of infintesmally thin disks about the z axis. Mass Moment of Inertia Angular Momentum Rotational Kinetic Energy rad rad N-m kg-m rad sec sec2 o kg-m2 kg-m2 sec Hz sec sec2 sec 14. 18) After fixing a flat tire on a bicycle you give the wheel a spin. The links will open a new browser window. A bug of mass m lands at the center of the disc and then walks outward. Can someone please show me show more The rotational inertia of a solid uniform sphere about a diameter is (2/5)MR2, where M is its mass and R is its radius. Problem- 2 Find the moment of inertia of a uniform solid sphere of mass M and radius R about a diameter. Let us consider a sphere of radius R and mass M. It is a convention to write to indicate the moment of inertia with respect to an axis passing though the center of mass of the rotating rigid body. Although mass is defined in terms of inertia, it is conventionally interpreted as. About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter Students also viewed these Mechanics questions. The results for centroid, moment of inertia, statical moment section modulus and torsion constant will display on your right. The bar is 2. The moment of inertia is the torque required to start an angle of acceleration along a rotating axis. 97 x 1024 kg and 6. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is. Now for a sphere with uniform density rotating around its axis I is, I = (2MR^2)/5. If they are both released from the same height and at the same time, which one will arrive at the bottom of the ramp first? A. A solid sphere, disc and solid cylinder all of same mass and made up of same material are allowed to roll down (from rest) on an inclined plane, then (a) solid sphere reaches the bottom late (b) solid sphere reaches the bottom first (c) disc will reach the bottom first (d) all will reach the bottom at the same time. The work-energy theorem for a rigid body rotating around a fixed axis is where \u201410) and the rotational work done by a net force rotating a body from point A to point B is (10. The force F = 15 N is applied to the rope for a duration of 3 seconds. Physics 100A Homework 10 \u2013 Chapter 10 (part 2) 10. 6 kg object is found to have a moment of inertia of. 1) Find its angular acceleration. 0 is applied to the rope and the sphere starts to roll without slipping on the show more A uniform 8. 00 kg m2, but this is reduced to 2. For a thin uniform homogenous rectangular plate, the mass moment of inertia about the rectangular coordinate axes, a and b, passing through the centre of gravity of the circular plate can be obtained from the area moment of inertia. Moment of inertia of a Uniform Hollow Cylinder -. moment of inertia about its center of mass \ud835\udc3c\ud835\udc3c. Here are some of the most common moments of inertia: Solid cylinder or disk of radius r rotating about its axis of symmetry. (b) Obtain the moment of inertia for a solid cylinder. A uniform disk of mass m is not as hard to set into rotational motion as a \\dumbbell\" with the same mass and radius. 15 \u2022 As shown in the figure, solid sphere rolls on a horizontal surface at 20 m/s and then rolls up the incline. If you are allowed to use the fact that the rotational inertia of a spherical shell is (2/3)mr2 [and I can derive that if necessary], all you have to do is take your solid. I recommend not to post that. Furthermore, because of the symmetry of the sphere, each principal moment is the same, so the moment of inertia of the sphere taken about any diameter is. Problem- 2 Find the moment of inertia of a uniform solid sphere of mass M and radius R about a diameter. m what is her final moment of inertia'? How does she physi- cally accomplish this change? (Il) A potter's wheel is rotating around a vertical axis through its center at a frequency of 1-5 rev/s The wheel can be considered a uniform disk of mass 5. 85kg and diameter 45. Explain why the moment of inertia is larger about the end than about the center. Moment of inertia of this disc about the diameter of the rod is, Moment of inertia of the disc about axis is given by parallel axes theorem is, Hence, the moment of inertia of the cylinder is given as, Solid Sphere a) About its diameter Let us consider a solid sphere of radius and mass. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is. not the hypotenuse. 571 radians) between the strips, the angular velocity \u03c9 is computed. angular acceleration? An object remains in a state of uniform rotational motion unless acted on. As the solid sphere's non uniformity is not mentioned,it should be considered to be uniform and hence spherically symmetric body. ( 1 pt ) A disc in the x-y plane is rotating about an axis perpendicular to the x-y plane. In the first part of our lab a rotating solid cylindrical drum with a hollow body drum given a rotational velocity from a falling mass. Please explain. The moment of inertia of a solid cylinder of radius r is given by: J = mr2 2 By comparison, the moment of inertia of a hollow cylinder, of inner and outer radii respectively, is as follows: J = m(r o 2 - r i 2)2 It can be seen that, for a given outer radius, the moment of inertia of a hollow cylinder is greater than that of a solid cylinder of. I deal with stars, and stars have rotational kinetic energy. Icm = moment of inertia for rotation around an axis through the center of mass () M = total mass of the object (kg) d = distance between the two rotation axes (m) Parallel Axis Theorem Formula Questions: 1) A solid sphere with mass 60. The net torque acting on this sphere as it is slowing down is closest to: A. Using Moment of Inertia The moment of inertia of an object rotating around a fixed object is useful in calculating two key quantities in rotational motion:. Moment of inertia shows the tendency of an object to stay in its state of rotatory motion. If her initial moment of inertia was 4. 3 Physical Significance of Moment of Inertia 3. Moment of inertia of solid sphere about its diameter by. Moment of Inertia--Cylinder : Consider a uniform solid cylinder of mass M, radius R, height h. Let M and R be mass and radius of the hollow cylinder and the solid sphere, then, Moment of inertia of the hollow cylinder about its axis of symmetry, l 1, = MR 2 Moment of inertia of the solid sphere about from \u03c9 \u03c9 0 + at, we find that for given \u03c9 0 and t, \u03c9 2 > \u03c9 1, angular speed of solid sphere will be greater than the angular speed of. Related: Beam Deflection Stress Equation Calculators. We are allowed to use the standard result that the moment of inertia about the axis running down its centre is 1/2 m r^2. 15m from is center of mass. then prove that-omega =9/14 omega not. In a rotating body Torque is equal to the moment of Inertia multiplied by angular acceleration. 46 car on an incline A car on an incline is timed from release until the end of a measured distance. We want our questions to be useful to the broader community, and. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. Its rotational inertia about the point of attachment at the ceiling is: A) (2/5)MR2 B) 4MR2 C) (7/5)MR2 D) (22/5)MR2 E) (47. Its moment of inertia about an axis of rotation passing through its diameter is I = MR 2. Apply the parallel axis theorem From this result, we can conclude that it is twice as hard to rotate the barbell about the end than Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis. L as Y O C) 2017 Akaa Daniel Ayangeakaa. R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge. Sphere #2 will arrive first C. Some of the moments of inertia are given in the table below: slender rod: axis through center axis through end rectangular plane: axis through center axis along edge sphere thin-walled hollow solid cylinder hollow solid walled thin-hollow. What is the moment of inertia of the system of. More of the sphere's mass is far away from the center of rotation, so the hollow one has a big moment of inertia. Obtain the moment of inertia of a hollow solid sphere of inner and outer radii r1 and r2 Products. For the sake of one more bit of integration practice, we shall now use the same argument to show that the moment of inertia of a uniform circular disc about a. 98 x 10 24 kg) (6. A light string of length 3 R is 45 \u00b1 to the tangent ans: D Section: 10{8; Di\u00b1culty: E 174 Chapter 10: ROTATION 59. In systems that are both rotating and translating, conservation of mechanical energy can be used if there are no nonconservative forces at work. Get the linear acceleration of center of sphere denoting it as y. Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. Answer: E 14) A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of a uniform circular disc and circular ring of radius R and mass M. Since moment of inertia is a scalar quantity, a compound object made up of several objects joined together has a moment of inertia which is the A rotating door is made from four rectangular glass panes, as shown in the drawing. A solid sphere with a mass of 8. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. Problem- 2 Find the moment of inertia of a uniform solid sphere of mass M and radius R about a diameter. The volume of such a layer is. (ii) about a tangent: A tangent drawn to the sphere at any point, will obviously be parallel to one of its diameters and the distance between the axes is equal to R, the radius of the sphere. I = \u2211mr2 Rotational Kinetic Energy:- K r. What is the direction of its angular momentum vector? 15. 18) After fixing a flat tire on a bicycle you give the wheel a spin. The moment of inertia of this element is: dr. 16 The variation of angular position \u03b8, of a point on a rotating rigid body, with time t is shown in Fig. The only data for 3D solids we are aware of are for spheroids: figure 4. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, Obtain the moment of inertia of a hollow solid sphere of inner and outer radii r1 and r2, quarry, aggregate, and different kinds of. How is the moment of inertia related to. 98 x 1024 kg and an average radius of 6.", "date": "2019-12-07 03:37:07", "meta": {"domain": "venetoacquaeterme.it", "url": "http://vkwf.venetoacquaeterme.it/show-that-the-moment-of-inertia-of-a-uniform-solid-sphere-rotating-about-a-diameter-is.html", "openwebmath_score": 0.5893383622169495, "openwebmath_perplexity": 240.6116665009223, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683465856102, "lm_q2_score": 0.8615382094310357, "lm_q1q2_score": 0.850827865008135}}
{"url": "https://math.stackexchange.com/questions/1479267/how-many-six-digit-numbers-start-with-the-same-two-digits-and-end-with-the-same", "text": "# How many six digit numbers start with the same two digits and end with the same three digits?\n\nSay that there is a 6 digit number the first digit is not allowed to be 0 or 1\n\nso\n\nHow many number combinations start with the same two digits and end with the same three digits\n\nie.119333, 448222, 889888 etc..\n\nMy thoughts are 8_ 1_ 10_ 10_ 1_ 1_ = 8*10*10=800?\n\nand how would I do it if instead of the and it said Or?\n\nfirst of all, I'm I on the correct path here? any other examples similar would help.\n\n\u2022 Your title should be specific to the problem at hand so that someone who has a similar question can find it if she or he searches the site. \u2013\u00a0N. F. Taussig Oct 14 '15 at 9:54\n\nNow let us find the number of possibilities if our number starts with the same two digits OR ends with the same three digits. Again we assume that $0$ and $1$ are forbidden as first digit. But they may occur as second digit, for example in $506888$.\n\nThere are $(8)(10^4)$ numbers that begin with two equal digits, the first (and therefore second) digit being neither $0$ or $1$.\n\nThere are $(8)(10^3)$ numbers that start with an allowed digit and whose last three digits are the same.\n\nIf we add the two numbers above, we will have double-counted the numbers in which the first two digits are the same, and the last three are the same.\n\nIt follows that the required number is $(8)(10^4)+(8)(10^3)-(8)(10^2)$.\n\nRemark: This is a relatively simple instance of a technique called Inclusion/Exclusion. There are more elaborate versions.\n\n\u2022 so Just minus the AND part? \u2013\u00a0learnmore Oct 14 '15 at 5:48\n\u2022 Yes. A general formula for two sets is $|A\\cup B|=|A|+|B|-|A\\cap B|$. Here $|X|$ means the number of elements in the set $X$. \u2013\u00a0Andr\u00e9 Nicolas Oct 14 '15 at 5:53\n\u2022 how would one go about in determine the count if at least one digit appears more then once in the combination.. i.e 252169 \u2013\u00a0learnmore Oct 14 '15 at 6:01\n\u2022 This is a different problem, since we no longer have two identical digits at the beginning, or three at the end. I will assume that $0$ is forbidden at the beginning. Then if we had no further restrictions, the answer would be $(9)(10^5)$. Now we subtract the bad numbers, in which all digits are distinct. Let us count the bad numbers. I will stop here and continue in the next comment. \u2013\u00a0Andr\u00e9 Nicolas Oct 14 '15 at 6:11\n\u2022 (Continued) For counting the bad numbers, there are $9$ choices for the first digit. For each of these there are $9$ choices for the second digit, since $0$ is now allowed. For each choice of first two digits, there are $8$ choices for third digit, and so on, for a total of $(9)(9)(8)(7)(6)(5)$. \u2013\u00a0Andr\u00e9 Nicolas Oct 14 '15 at 6:15", "date": "2019-12-06 16:05:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1479267/how-many-six-digit-numbers-start-with-the-same-two-digits-and-end-with-the-same", "openwebmath_score": 0.7084540724754333, "openwebmath_perplexity": 223.72988956123527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9875683517080176, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8508278641548946}}
{"url": "http://code.jasonbhill.com/c/project-euler-154/", "text": "A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.\n\nThen, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).\n\nThe result is Pascal\u2019s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?\n\n## Solution Using the Multinomial Theorem\n\nThe generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula\n$(x_1+x_2+\\cdots+x_m)^n=\\sum_{{k_1+k_2+\\cdots+k_m=n}\\atop{0\\le k_i\\le n}}\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)\\prod_{1\\le t\\le m}x_t^{k_t}$\nwhere\n$\\left({n}\\atop{k_1,k_2,\\ldots,k_m}\\right)=\\frac{n!}{k_1!k_2!\\cdots k_m!}.$\nOf course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by\n$\\left({200000}\\atop{k_1,k_2,k_3}\\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$\nwhere $k_1+k_2+k_3=200000$. It\u2019s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there\u2019s no way that we\u2019re going to calculate these coefficients. We only need to know when they\u2019re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.\n\nFirst, we\u2019ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that\n$p(n,d)=\\left\\lfloor\\frac{n}{d}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^2}\\right\\rfloor+\\left\\lfloor\\frac{n}{d^3}\\right\\rfloor+ \\cdots+\\left\\lfloor\\frac{n}{d^r}\\right\\rfloor,$\nwhere $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we\u2019re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we\u2019ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We\u2019ll store two lists:\n$p2=[p(i,2)\\text{ for }1\\le i\\le 200000]\\quad\\text{and}\\quad p5=[p(i,5)\\text{ for }1\\le i\\le 200000].$\nFor a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when\n$p2[k_1]+p2[k_2]+p2[k_3]<199983\\ \\text{and}\\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$\nThat is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.\n\nNow, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\\le k_2\\le k_3$. Under this assumption, we always have\n$k_1\\le\\left\\lfloor\\frac{200000}{3}\\right\\rfloor=66666.$\nWe know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.\n\nIn case 1, we iterate $0\\le k_1\\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).\n\nIn case 2, we iterate $0\\le k_1\\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.\n\nIn case 3, we iterate $0\\le k_1\\le 66666$, and we iterate over $k_2=k_1+a$ where $1\\le a < \\left\\lfloor\\frac{200000-3k_1}{2}\\right\\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).\n\n## Cython Solution\n\nI\u2019ll provide two implementations, the first written in Cython inside Sage. Then, I\u2019ll write a parallel solution in C.\n\n%cython \u00a0 import time from libc.stdlib cimport malloc, free \u00a0 head_time = time.time() \u00a0 cdef unsigned long p(unsigned long k, unsigned long d): cdef unsigned long power = d cdef unsigned long exp = 0 while power <= k: exp += k / power power *= d return exp \u00a0 cdef unsigned long * p_list(unsigned long n, unsigned long d): cdef unsigned long i = 0 cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long)) while i <= n: powers[i] = p(i,d) i += 1 return powers \u00a0 \u00a0 run_time = time.time() \u00a0 # form a list of number of times each n! is divisible by 2. cdef unsigned long * p2 = p_list(200000,2) \u00a0 # form a list of number of times each n! is divisible by 5. cdef unsigned long * p5 = p_list(200000,5) \u00a0 cdef unsigned long k1, k2, k3, a cdef unsigned long long result = 0 \u00a0 k1 = 0 while k1 <= 66666: # case 1: k1 = k2 < k3 k2 = k1 k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 2: k1 < k2 = k3 if k1 % 2 == 0: k2 = (200000 - k1)/2 k3 = k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 3: k1 < k2 < k3 a = 1 while 2*a < (200000 - 3*k1): k2 = k1 + a k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 6 a += 1 k1 += 1 \u00a0 \u00a0 free(p2) free(p5) \u00a0 \u00a0 elapsed_run = round(time.time() - run_time, 5) elapsed_head = round(time.time() - head_time, 5) \u00a0 print \"Result: %s\" % result print \"Runtime: %s seconds (total time: %s seconds)\" % (elapsed_run, elapsed_head)\n\nWhen executed, we find the correct result relatively quickly.\n\nResult: 479742450 Runtime: 14.62538 seconds (total time: 14.62543 seconds)\n\n## C with OpenMP Solution\n\n#include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <omp.h> \u00a0 /*****************************************************************************/ /* function to determine how many factors of 'd' are in 'k!' */ /*****************************************************************************/ unsigned long p(unsigned long k, unsigned long d) { unsigned long power = d; unsigned long exp = 0; while (power <= k) { exp += k/power; power *= d; } return exp; } \u00a0 /*****************************************************************************/ /* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer */ /*****************************************************************************/ unsigned long * p_list(unsigned long n, unsigned long d) { unsigned long i; unsigned long * powers = malloc((n+1)*sizeof(unsigned long)); for (i=0;i<=n;i++) powers[i] = p(i,d); return powers; } \u00a0 /*****************************************************************************/ /* main */ /*****************************************************************************/ int main(int argc, char **argv) { unsigned long k1, k2, k3, a; unsigned long long result = 0; \u00a0 unsigned long * p2 = p_list(200000, 2); unsigned long * p5 = p_list(200000, 5); \u00a0 \u00a0 #pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result) for (k1=0;k1<66667;k1++) { // case 1: k1 = k2 < k3 k2 = k1; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } // case 2: k1 < k2 = k3 if (k1 % 2 == 0) { k2 = (200000 - k1)/2; k3 = k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } } // case 3: k1 < k2 < k3 for (a=1;2*a<(200000-3*k1);a++) { k2 = k1 + a; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 6; } } } \u00a0 free(p2); free(p5); \u00a0 printf(\"result: %lld\\n\", result); \u00a0 return 0; }\n\nThis can be compiled and optimized using GCC as follows.\n\n$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result. $ time ./problem-154-omp result: 479742450 \u00a0 real 0m1.487s\n\nThis appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds.", "date": "2019-01-17 22:12:55", "meta": {"domain": "jasonbhill.com", "url": "http://code.jasonbhill.com/c/project-euler-154/", "openwebmath_score": 0.4279261529445648, "openwebmath_perplexity": 1737.7390952399246, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683469514965, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8508278618124334}}
{"url": "http://angs-diasporanew.sramble-communication.com/apdsmyht/linear-regression-problems-worksheet.php", "text": "# Linear regression problems worksheet\n\n4. Quadratic regression produces a more accurate quadratic model than the procedure in Example 3 because it uses all the data points. 10) 5. A, B, C, D, E, F. Activity\u2014Twizzlers Linear Regression Pre-Activity & Worksheet Answer Key 2 6. Like half the models in statistics, standard linear regression relies on an assumption of normality. Biology: Five additional weeks of sunshine the sugar concentration in vine grapes will rise by X %. 2. Let the model be Y = 0 + 1X 1 + 2X 2 + \", where E(\"jX 1;X 2) = 0, and assume that we have a sample AP Stats: Section 3. d. Linear regression estimates the regression coefficients \u03b2 0 and \u03b2 1 in the equation Y j =\u03b2 0 +\u03b2 1 X j +\u03b5 j where X is the independent variable, Y is the dependent 4. 3 Inferences on the Slope Rarameter \u03b2\u03b2\u03b2\u03b21111 NIPRL 1 12. That is, the equation of the best linear t. Linear reg. 9. A regression with two or more predictor variables is called a multiple regression. 02 (Simple Linear Regression) is based on the identical data set to the paired t -test example above. 1. regression. 32 0. There are two common ways to deal with nonlinear relationships: 1. Multiple Linear Regression Model We consider the problem of regression when the study variable depends on more than one explanatory or independent variables, called a multiple linear regression model. Part 1. 15. 6. CHAPTER 5. The general model can be estimated by grid search or by non-linear maximization of the likelihood and a maximum likelihood estimate for a obtained. Quiz. . Use the regression model to predict the credit card volume in 2003 and in 2010. ac. Going back to our original data, we can try to fit a line through the points that we have; this is called a \u201ctrend line\u201d, \u201clinear regression\u201d or \u201cline of best fit\u201d (as we said earlier, the line that\u2019s the \u201cclosest fit\u201d to the points \u2013 the best trend line). 8 \u2013 Trigonometry & Regression Linear Correlations 1. The estimated\u00a0 Under Output Options, choose \"New Worksheet Ply,\" then click OK. (2006) - Chpt 6 zQuinn & Keough (2002) - Chpt 5 zRowntree (1981) - Chpts 12 Question 1 - Simple linear regression Here is an example from Fowler, Cohen and Parvis (1998). To learn more about Nonlinear Regression with data linearization, see the Nonlinear Regression worksheet. In this section we will first discuss correlation analysis, which is used to quantify the association between two continuous variables (e. For the reason that we should supply everything you need in a true along with dependable origin, most people provide valuable details on many themes in addition to topics. Then we perform linear regression on the data. 2 HSS-ID. 355 Problems Predictor Coef SE Coef T P Constant 44. For each of the following tables, treat the left-hand column as the independent variable (input) and the right-hand column as the dependent variable (output), and answer each of the following questions, along with any additional questions related to the actual problem. \u2019 Here are some of the common Linear Regression Interview Questions that pop up in interviews all over the world. Linear and Quadratic . Regression analysis is the art and science of fitting straight lines to patterns of data. Make a table that shows data from If both the regression coefficients are negative, r would be negative and if both are positive, r would assume a positive value. So, in Excel, you do linear regression using the least squares method and seek coefficients a and b such that: y = bx + a Student Worksheets Created by Matthew M. 4 and 8 2) The difference of two numbers is 3. Linear regression simply refers to creating a best fit for a linear relationship between two variables from observed data. Last ride. 1751(\ud835\udc4c\ud835\udc4c\ud835\udc4c\ud835\udc4c\ud835\udc4c\ud835\udc4c/\ud835\udc34\ud835\udc34) \ud835\udc4a\ud835\udc4a\ud835\udc4a\ud835\udc4a \ud835\udc66\ud835\udc66 = \u221270. Word problems on ages. The equation to represent this data is . In the data, x is the number of seconds after the missile is launched and y is the number of feet above water for the missile. 5 Prediction Intervals for Future Response Values 12. Property 4 : The two lines of regression coincide i. The least squares method is generally used with a linear regression, but Regression is the process by which the relationship between two variables is determined. The problem of determining the best values of a and b involves the principle of 2 is the sum of squares due to the linear regression SSR, with mean square. The least squares method is generally used with a linear regression, but Aug 01, 2018 \u00b7 The linear regression equation always has an error term because, in real life, predictors are never perfectly precise. image0. Regression is used to assess the contribution of one or more \u201cexplanatory\u201d variables (called independent variables) to one \u201cresponse\u201d (or dependent ) variable. The data in the table below show different X (depth in feet) 50 Y (maximum dive time) 80 depths with the maximum dive times in minutes. 4\u00af6, s2 Example of a regression equation Y = $0 +$ 1 (Age - 40) + $2 Gender + , Salary = 50 + 1 (Age - 40) - 3 Gender + , Salary in$1,000s, Age in years and Gender = 0 if male and 1 if female What is the average salary for 50 year old males? Ave(Y) = 50 + 1 (50-40) - 3(0) = $60K following form: y=alpha+beta*x+epsilon (we hypothesize a linear relationship) \u2022 The regression analysis \u201eestimates\u201c the parameters alpha and beta by using the given observations for x and y. 13 Feb 2014 Linear Regression and Correlation - Example. In your story, interpret the slope of the line, the y-intercept, and the x-intercept. In the final chapter, I'll show you how to make the most of your results by changing parameters by hand, performing sensitivity analysis and creating scenarios. Word problems on sets and venn diagrams. By comparing the values of, determine the function that best fits the data. Simple Linear Regression and Correlation 12. \u2022 The simplest form of estimating alpha and beta is called ordinary least squares (OLS) regression Problem 3: Let X and Y be two variables in a study. Find the residual amount for a person who is 42. 902. LAB ACTIVITIES FOR SIMPLE LINEAR REGRESSION: TWO VARIABLES 1. This one-page worksheet contains seven problems. A student who waits on tables at a restaurant recorded the cost of meals and the tip b. In case you need guidance on exponents or even radical expressions, Algebra1help. Worksheet 13. dta. Outcomes Students will use a graphing calculator to find a quadratic curve of best fit. Some of the worksheets for this concept are Linear equations work, Solving linear equations, Linear regression work 1, Depends y dependent variable x independent variable y m x, Writing linear equationslinear regression, Real world applications of linear equations, Slope intercept form word problems, Y mx b A statistics Worksheet: The student will calculate and construct the line of best fit between two variables. Next, students will use their calculator to fit a simple linear regression equation with poverty predicting obesity. Y. In this tutorial, [\u2026] In Chapter 2 you used a graphing calculator to perform linear regression on a data set in order to find a linear model for the data. The data is shown below. 7 p1 The regression equation is Sales = 116 - 97. 26721 \u00d7 (8) = 2. Apr 16, 2018 \u00b7 Note The Regression tool alerts you to this problem and does not continue. Regression interpretation 3. This part of the program will fit a linear function of the form: Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. This lesson focuses on two variables which have a linear association. 7. Round to two Statistical Analysis 6: Simple Linear Regression. The population (in thousands) for Alpha City, t years after January 1, 2004 is modeled by the quadratic function P (t) 0. Old Faithful. To better understand the future strategies, you can visually represent the predicted values in a line chart. Detailed instructions on how to use the calculator are provided. The distribution for each important later. Under some conditions for the observed data, this problem can be solved numerically. variable is (ya - 1)/a, so that with a = 1, the regression is linear, with a = 0, it is logarithmic, these cases being only two possibilities out of an infinite range as a varies. Use the Input Y Range text box to identify the worksheet range holding your Write the linear regression equation for these data where miles driven is the independent Explain what the y-intercept means in the context of the problem. (1998) -Chpts 14 & 15 zHolmes et al. Transform the data so that there is a linear relationship between the transformed variables. Please be sure to check your answers as part of your homework assignment and be prepared with questions for next class. Worksheets are Work 3, Chapter 9 correlation and regression solutions, Writing linear equationslinear regression, Work 1, Linear regression work 1, Linear reg correlation coeff work, Work on correlation and regression, Algebra ii exponential regression work value 10. Make a scatter plot and determine the value of r. not significantly different from zero we conclude that: a) X is a good predictor of Y b) there is no linear relationship between X and Y. 01165 + 0. Mintzberg\u2019s classic book The Nature of Managerial Work (1973) identi\ufb01ed the roles found in all managerial jobs. 1 Linear Predictors Before computers became fast, linear regression was almost the only way of at-tacking certain prediction problems. 1751(6. 827 4. 3. Loading. . e. 2. c) BIOSTATS 540 - Fall 2018 Simple Linear Regression and Correlation Page 1 of 54 Nature Population/ Sample Observation/ Data Relationships/ Modeling Analysis/ Synthesis Unit 12 Simple Linear Regression and Correlation \u201c Assume that a statistical model such as a linear model is a good first start only\u201d Lab!10:!Exploring!Linear!Regression! Objective:!In! this! lab,! you! will examine! relationships! between! two! quantitative! variables! using! a! graphical!tool May 31, 2016 \u00b7 Regression analysis makes use of mathematical models to describe relationships. The regression coefficient estimated with a linear regression equation y = a + b*x can then tell the researchers b the life expectancy (y) is when smoking x cigarettes a day. Worksheet 3 Universidad Carlos III de Madrid Worksheet 3 The Multiple Regression Model Note: In those problems that include estimations and have a reference to a data set the students should check the outputs obtained with Gretl. That is why it is also termed \"Ordinary Least Squares\" regression. Linear inequalities word problems. That is, there is lack of fit in the simple linear regression model. However, some programs, including Excel, do the error term calculation behind the scenes. They believe that the number of books that will ultimately be sold for any particular course is related to the number of students registered for the course when the books are ordered. a. Feb 29, 2016 - Explore theboss1000's board \"LINEAR REGRESSION\", followed by 431 people on Pinterest. In linear regression, you are looking for a hyperplane \"near\" most of the points; with SVMs, you will be looking for a thick hyperplane, as thin as possible, that contains all the observations. In linear equation in two variables distance problems you have to use two variables and you can solve using any method such as substitution or elimination. Lab Activity: Linear Regression and Correlation In this lab activity, you will collect sample data of two variables, determine if a linear correlation exists between the two variables, and perform linear regression. correlation coefficients, and obtain linear regression equations. For this linear regression worksheet, students solve linear regression problems using the TI-86 calculator. Last year, Walmart conducted a study as to the amount of waiting in time in checkout lanes its customers had to wait. Scaffolded questions that start relatively easy and end with some real challenges. 391 + 17. regression project worksheet: Did the student show the material learned in this course can be useful in a topic that is relevant to the individual students? (explain in one sentence how your data is either interesting to you personally OR related to you major of study) The following are tables of data to be used for linear regression exercises. Find the linear and quadratic regression equations and correlation coefficients. This data set has n=31 observations of boiling points (Y=boiling) and temperature (X=temp). Linear Regression Displaying all worksheets related to - Linear Regression. 1. There are 2 types of factors in regression analysis: Dependent variable (y) : It\u2019s also called the \u2018criterion variable\u2019 , \u2018response\u2019 , or \u2018outcome\u2019 and is the factor being solved. g. A correlation analysis provides information on the strength and direction of the linear relationship between two variables, while a simple linear regression analysis estimates parameters in a linear equation that can be used to predict values of one variable based on Quadratic Models and Quadratic Regression Worksheet 1. P. Consider the following scatter plots: (a) Write the new regression model. Writing Linear Equations/Linear Regression Write the slope-intercept form of the equation of each line given the slope and y-intercept. Graph the linear equation by 2. Regression problems are supervised learning problems in which the response is continuous. A researcher has collected data on the price of gasoline from 1990 to 2010 and has found that the price in dollars after t years can be predicted using the equation: y xx\u2212 += +0. Problem 2. The user can STAT 2215 Worksheet 5 \u2013 Chapter 7: Simple Linear Regression Problem 23, page 198. If there is not a linear relationship between x and y, then $$\\mu_{i} \u2260 \\beta_{0} + \\beta_{1}X_{i}$$. ! 2 4 6 8 0 50 100 150 200 250 Calories vs Alcohol Content Alcohol Content (%) Worksheets that accompany this lesson can be located under related documents, worksheets, Data Analysis #1-#8. Figure #10. You may notice that Linear regression where the sum of vertical distances d1 + d2 + d3 + d4 between observed and predicted (line and its equation) values is minimized. introduce problems that are relevant to the \ufb01tting of nonlinear regression func- Title: Linear Regression Grade: 8th Lesson Summary: This short lesson plan is to describe students learn how to find the best line to fit the data of two variables they collect and be able to predicate the data by using the regression equation. Simple linear regression is a statistical method that allows us to summarize and study relationships between two continuous (quantitative) variables. In these worksheets, problems are presented as word problems. Simple Linear Regression To describe the linear association between quantitative variables, a statistical procedure called regression often is used to construct a model. The Simple Linear Regression Model is summarized by the equation $y=\\beta _1x+\\beta _0+\\varepsilon$ Identify the deterministic part and the random part. mtpfrom the CD-ROM. SCUBA divers have maximum dive times they cannot exceed when going to different depths. Find r2. Linear regression is a type of machine learning algorithm that is used to model the relation between scalar dependent and one or more independent variables. To use K-nearest neighbors regression, or KNN regression for short, we must start with a data set. Motivation and Objective: We\u2019ve spent a lot of time discussing simple linear regression, but simple linear regression is, well, \u201csimple\u201d in the sense that there is usually more than one variable that helps \u201cexplain\u201d the variation in the response variable. write a regression equation and interpret the meaning of the slope and y-intercepts in the context of the problem; make predictions based on the correct mathematical models; and; solve linear equations. Even those who problem. However, we only calculate a regression line if one of the vari-ables helps to explain or predict the other variable. Superimpose the regression curve on the scatter plot. see and learn about curve fitting for multiple linear regression using method of least errors is as small as possible. At home: Read Chapter 5 and work the problems at the end of each short section as you go through them. Estimate the Blood pressure for a person who is 50. Chapter 12. Aug 07, 2014 \u00b7 Search this site. 7 p1 + 109 p2 Remember: -97. If Y denotes the Excel Linear Regression. (10 marks) C). In linear regression analysis, the dependent variable is thought to be related to the independent variable or variables in a linear way. We would expect the ratio MSLF/MSPE to be close to 1. Time and work word problems. Then, we graph the linear regression equation with the scatterplot data. They compute the percent of the variability. For our problem, we would need only one dummy variable (since K = 2), the dichotomous variable coding level of idealism. Go to Overview Index. In simple linear regression, when \u03b2 is . Do the Linear Regression ws. a and b are the constants of the regression model. It is a staple of statistics and is often considered a good introductory machine learning method. The elements in X are non-stochastic, meaning that the Linear Regression and NORMAL Curve Advanced Placement AAP Review will be held in room 315 and 312 on Tuesdays and Thursdays. Feb 26, 2018 \u00b7 Linear regression is used for finding linear relationship between target and one or more predictors. Rewrite the equation into finding the x- and y-intercepts. Multiple Linear Regression. If you're seeing this message, it means we're having trouble loading external resources on our website. Plot four points so that the regression line is horizontal. Customize the worksheets to include one-step, two-step, or multi-step equations, variable on both sides, parenthesis, and more. Students will interpret the r-value of the data and write a summary of its meaning. Big Ideas: Bivariate quantitative variables can be represented by a table, graph, and a prediction equation, and estimates can be made from each. GAISE Components One step equation word problems. ) Another very serious problem is the lack of any provision for forecasting from additional values of the independent variables . Throughout the module, you will find many real-world appli\u00ad cations of these two important topics: least-squares regression line and the correlation coefficient. Assumptions in the Linear Regression Model 2. Some of the worksheets for this concept are Linear regression work 1, Writing linear equationslinear regression, Chapter 9 correlation and regression solutions, Work 1, Work 3, , Quadratic regression, Kuta software. Such an equation can be used for prediction: given a new x-value, this equation can predict the y-value that is consistent with the information known about the data. The sample must be representative of the population 2. Jun 01, 2020 \u00b7 Statistics Q&A Library WORKSHEET 24 Linear Regression and Correlation Name: 15. 8. This lesson builds on students work in the 8th grade. problem to be solved is reduced to a quadratic programming problem in which the objective function is the residual sum of the squares in regression, and the constraints are linear ones imlx~ed on the regression coefficients. 3 times as important as Unconventional. b) Use the calculator and the equation of the linear regression line to complete the following table: Year 1960 1968 1988 1999 2005 2008 Time (s) c) Use the values in the table to draw the linear regression function on the scatter plot. To get a better feel for the regression line, try the following tasks. In Class: Practice Linear Regression HW: Obesity Problem (Hints included in key) r_9. EXTRAS. 45. Displaying all worksheets related to - Regression Analysis. Math 137 Quadratic Regression Classwork2 . Explain worksheet - regression inference 1. It allows the mean function E()y to depend on more than one explanatory variables Like correlation coefficients, linear regression analyzes the relationship between two variables, x and y. 1) Slope = \u22121, y-intercept = 0 y = \u2212x 2) Slope = 1 4, y-intercept = 1 y = 1 4 x + 1 Write the slope-intercept form of the equation of the line through the given point with the given slope. Start by creating a visual model of the data. Bivariate Data - Correlation - Linear Regression - Correlation Coefficient r Jun 2, 2020 - Are you looking for Algebra 2 worksheets WITH answers? Making keys is super time consuming, so peruse this board for resources that have the hard part done for you!. 5 and 8 This is because the correlation value for the cubic regression is about 0. The multiple linear regression result implies that Reliable is around 1. The table below lists the total estimated numbers of United States AIDS cases, by year of diagnosis. 3582 1. The first is done using the Tools menu, and results in a tabular output that contains Shodor > Interactivate > Lessons > Linear Regression and Correlation omit the scatter plot worksheet; As a class, before splitting them into groups, have the If you have a continuous dependent variable, linear regression is probably the To address these problems, statisticians have developed several advanced variants: ordered data (assuming you record them in your worksheet in time order). A graphing calculator can also be used to perform quadratic regression. In this correlation and linear regression worksheet, students examine data to determine the statistic mean. 2 Fitting the Regression Line 12. Brandon Foltz 296,888 views Linear regression simply refers to creating a best fit for a linear relationship between two variables from observed data. Ratio and proportion word problems. Although a linear regression can be quite helpful in understanding data, it can sometimes be misleading, as Anscombe's Quartet shows . H. If the degree of correlation between variables is high enough, it can cause problems when you fit the model and interpret the results. (When we need to note the difference, a regression on a single predic-tor is called a simple regression. simple linear regression A college bookstore must order books two months before each semester starts. widely used; runs fast; easy to use (not a lot of tuning The linear regression model that I\u2019ve been discussing relies on several assumptions. A. In other words, the SS is built up as each variable is added, in the order they are given in the command. 12A2 - HW Calendar Linear Regression Problems Q. LINEAR REGRESSION 6 1. The course website page REGRESSION AND CORRELATION has some examples of code to produce regression analyses in STATA. use functions fitted to data to solve problems in the context of the data. Model Summaries Worksheet Every workbook produced by RegressIt contains not only the data analysis and regression worksheets but also a model summaries worksheet that keeps an audit trail of all regression models fitted so far and allows side-by-side comparison of models fitted to the same dependent variable, suitable for framing. the meaning of slope in a linear regression equation the relationship of the slope to analyzed data whether or not the y-intercept of the linear regression equation is a relevant value EU 2 Econometrics. Enter the data to answer this question. com. To see why, consider a model such as this Y = \u03b2 0 +\u03b2 1e\u03b2 2X + , (4. Curve Fitting Curve fitting is the process of constructing a curve, or mathematical function, that has the best fit to a series of data points, possibly subject to constraints. 48. It also plots the experimental points and the equation y = a x + b where a and b are given by the formulas above. 9) \ud835\udc66\ud835\udc66 = 48. where a and b are given by. Round r to the nearest 3 decimal places. The case of having one independent variable is know as simple linear regression while the case of having multiple linear regression is known as multiple linear regression. Analysis of Variance, Goodness of Fit and the F test 5. 242 seconds for each additional foot of initial drop. Twenty five plants are selected, 5 each assigned to each of the fertilizer levels (12, 15, 18, 21, 24). Files for use with the TI-Nspire\u2122 A non-linear method with comparable simplicity is known as K-nearest neighbors regression. Sketch and shade the squares of the residuals. In many applications, there is more than one factor that in\ufb02uences the response. Our predictors would then be idealism, misanthropy, and idealism x misanthropy (an interaction term). This is a quadratic model because the second differences are the differences that have the same value (4). THE MODEL BEHIND LINEAR REGRESSION 217 0 2 4 6 8 10 0 5 10 15 x Y Figure 9. Use the standard error of the slope, Sb, to calculate 95% confidence interval for the slope in \u00baC/century. Print off the worksheet if you How to graph the linear regression equation with the scatterplot data, how to generate a least squares linear regression model, How to create a line of best fit, 31 Jul 2016 12. \\] Linear Regression Practice Worksheet 1. An agriculturalist was interested in the effects of a) Use your calculator to determine the linear regression function (y = A + Bx) that best models the data. There are two types of linear regression- Simple and Multiple. LINEAR REGRESSION WORKSHEET #1 Name_____ Date_____ Period_____ 1. The variable we base our predictions on is called the independent or predictor variable and is referred to as X. Use a graphing calculator to fit linear, quadratic, cubic, and power functions to At any time, you can complete this interactive online quiz to check how well you understand using linear regression. Click on pop-out icon or print icon to worksheet to print or download. b. The table lists the heights and weights of six wide receivers who played for the Atlanta Falcons during the 2010 football season. Instruction will be from 3:15 pm to 3:30 pm Compute the least squares regression line with the number of bidders present at the auction as the independent variable (x) and sales price as the dependent variable (y). Then find the Least Squares Line and use it to make a prediction. 2 problems on linear regression May 19, 2018 \u00b7 12 videos Play all Statistics PL14 - Simple Linear Regression Brandon Foltz Statistics 101: Logistic Regression Probability, Odds, and Odds Ratio - Duration: 13:03. The session will begin in room 315 with a brief review of the weekly topic. This worksheet contains the following data, with the list price in column C1 and the best price in the column C2. If we were to plot height (the independent or 'predictor' variable) as a function of body weight (the dependent or 'outcome' variable), we might see a very linear relationship, as illustrated Related posts of \"Linear Regression Worksheet Answers\" Scheme For Igneous Rock Identification Worksheet Answers In advance of preaching about Scheme For Igneous Rock Identification Worksheet Answers, you should realize that Instruction can be each of our factor to a greater down the road, and also understanding doesn't just stop right after the (RegressIt can fit linear regression models in Excel with over 200 independent variables on a PC or 125 variables on a Mac, and its R interface can be used to fit large models much faster. Estimator 3. 344 10. 35519 0. Height (inches) Weight (pounds) 75 192 76 220 71 200 74 210 69 185 72 189 a. A linear regression simply means that the equation will be the equation of a line . Ea How to compute the linear regression equation, y=ax+b, the linear correlation coefficient, r, and the coefficient of determination, r 2, using the TI-84 calculator, including turning the diagnostics on. It is a very simple regression algorithm, fast to train and can have great performance if the output variable for your data is a linear combination of your inputs. MR. c. This correlation is a problem because independent variables should be independent. Worksheet: Name the Song you can also use SVMs for regression. pdf: File Size: 1849 kb: File Type: pdf: Download File linear regression: An approach to modeling the linear relationship between a dependent variable, $y$ and an independent variable, $x$. jpg. Linear regression analysis, in general, is a statistical method that shows or predicts the relationship between two variables or factors. 1 Transformations in Linear Regression Create printable worksheets for solving linear equations (pre-algebra or algebra 1), as PDF or html files. Regression. Explain the slope in context of the problem. May 15, 2009 \u00b7 Statistics and Regression Tools Review: Detailed Descriptions - Click OK to execute the regression. Predictthe!type!(positive,!negative,!no)!and!strength!of!correlation!(strong,!weak)!for!the!following! Summarize the four conditions that comprise the simple linear regression model. You don't have to believe everything it says. The week of March 30th we will be reviewing Linear Regression and NORMAL Curve. Mar 20, 2019 \u00b7 Linear regression forecasting graph. through a process called linear regression. As with linear regression, the dataset must take of form of pairs of predictor variables x \u20d7 i \\vec{x}_i x i with resultant variables y i y_i y i . For instance, for an 8 year old we can use the equation to estimate that the average FEV = 0. Make a scatter plot for the data. If the answers to (l) and (m) are yes then using your simple linear regression equation predict the percentage of games won from a team that has a passing percentage of 6. Know what the unknown population variance $$\\sigma^{2}$$ quantifies in the regression setting. com is simply the right site to go to! A simple linear regression model is a mathematical equation that allows us to predict a response for a given predictor value. ) We\u2019d never try to find a regression by hand, and Linear regression is a process of drawing a line through data in a scatter plot. For each of the following, perform linear, quadratic, and exponential regressions. We are dealing with a more complicated example in this case though. Pythagorean theorem word problems. Further along in the paper, we will investigate some of the alternatives mentioned above, but this is also an opportunity Linear Regression Worksheet 1. b) According to the linear model, a coaster with a 200 foot initial drop is expected to last 139. 9 we\u2019ll talk a lot more about how to check that these assumptions are being met, but first, let\u2019s have a look at each of them. In a linear regression model, the variable of interest (the so-called \u201cdependent\u201d variable) is predicted from k other variables (the so-called \u201cindependent\u201d variables) using a linear equation. In Section 15. A convenience store manager notices that sales of soft drinks are higher on hotter days, so he assembles the data in the following table. Answers provided. , between an independent and a dependent variable or between two independent variables). DOWNLOADS Note: This particular activity requires the use of TI-Nspire\u2122 technology to be used successfully. In what year does Alpha ity\u2019s population reach twice its initial (1/1/2004) population? 2. Do this in several different ways. It is the same Lagrange multiplier problem as above, with all the inequalities reversed. Contact Info. Worksheet 3 - Regression and linear models Linear regression references zFowler et al. sav. The results are shown in the table. Some of the worksheets for this concept are Chapter 9 correlation and regression solutions, I exploring regression, Scatter plots, Correlation coefficient, The united states of obesity, Concept 20 scatterplots correlation, Lecture 12 linear regression test Linear Equation Models. com delivers insightful information on linear regression free worksheet, basic concepts of mathematics and dividing fractions and other algebra topics. Regression (new) 4. 8: Regression - Distance from School (Worksheet) - Statistics LibreTexts Scatterplots & Regression on the TI-84 This video shows how to input two variable data and create a scatterplot with the TI-84 calculator. L Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Systems of Equations Word Problems Date_____ Period____ 1) Find the value of two numbers if their sum is 12 and their difference is 4. Now we\u2019ll take a look at a data set for which the linear regression model is appropriate. A. This lesson SIMPLE LINEAR REGRESSION \u2013 DEMAND AS FUNCTION OF PRICE A new worksheet will appear revealing the results of your regression analysis. 4 Inferences on the Regression Line 12. Practice quiz 6. 05 to help answer this question. There are NINE problem types. Properties of the O. It works by estimating coefficients for a line or hyperplane that best fits the training data. This model generalizes the simple linear regression in two ways. A simple linear regression model is fit, relating plant growth over 1 year (y) to amount of fertilizer provided (x). If x is the independent variable and y the dependent variable, then we can use a regression line to predict y for a given value of x. 11719% In simple linear regression, we predict scores on one variable from the scores on a second variable. The line summarizes the data, which is useful when making predictions. 02 0. Normality. Using this analysis we can estimate the relationship between two or more variables. Complete details on this method (also known as the Potthoff method) are in Chapter 13 of K & K. Evaluating managerial success. Coming up for air. 6 The Analysis of Variance Table 12. If the truth is non-linearity, regression will make inappropriate predictions, but at least regression will have a chance to detect the non-linearity. Graph using the slope and the y-intercept. Neither regression nor correlation analyses can be Use the linear model to calculate the expected cholesterol for a person with pressure 160 mmHg. The results of the model fit are given below: Can we Problems 1. 7 Residual Analysis The first difference (the difference between any two successive output values) is the same value (3). Regression 10. Identify your Y and X values. Let's look at an example of linear regression by examining the data in the following table to discover the relationship between temperatures measured in Celsius (Centigrade) and Fahrenheit. uk March 17, 2011 1 Logarithmic transformations of variables Considering the simple bivariate linear model Yi = + Xi + i,1 there are four possible com- Algebra1help. Is the number $$\\beta _1$$ in the equation $$y=\\beta _1x+\\beta _0$$ a statistic or a population parameter? The linear regression model explains the method used to take observed data and find a 'best fit' line to describe the relationship of two variables, and this quiz/worksheet pairing will test your Linear Regression Correlation Coeeficient. A Tomahawk Cruise ship in the South Pacific misfires a missile. Statistics \u2013 Linear Regression Worksheet - Solutions The busiest season for Walmart is the Christmas holiday and weekends see a tremendous number of customers. Multiple regression models thus describe how a single response variable Y depends linearly on a A. (SHOW WORK using algebra!) Linear mixture problems, non-linear mixture problems, transportation problems, personal scheduling problems and for something a little different, sports scheduling problems. 26721 \u00d7 age. 4: The Regression Equation Carry out an appropriate test at a significance level of 0. You can use the Regression tool instead of the LINEST worksheet function. Word Problem Worksheet #2 (each correct answer = 1 extra ballot for the draw!) Linear Regression Worksheet - HAND IN :) April 9 - Data Unit Assignment. Recall that the least squares line minimizes the squares of the residuals. So, we have a sample of 84 students, who have studied in college. - Stats Worksheet #1 - Stats Worksheet #2. \u00af3, y\u00af = 46. 15 Apr 2015 Research question type: When using one variable to predict or explain another Simple Linear Regression \u2013 Additional Information worksheet. The least squares method is generally used with a linear regression, but 1 day ago \u00b7 You will not require the options for \"residuals\" for this analysis. Using a catapult of their own design, students will model a parabolic relationship, collect data, and draw conclusions from data and the quadratic curve of best fit. Finally, the students will explore question 4 on the Activity Worksheet. simple linear regression, the sample correlation coefficient is the square root of the coefficient of determination, with the sign of the correlation coefficient being the same as the sign of b1, the coefficient of x1 in the estimated regression equation. The student will evaluate the relationship between two variables to determine if that \u2026 12. 23 Apr 2011 In the table below, list the assumptions of multiple linear regression all 6 predictors into the one model, because of the collinearity problem. Consider the following hypothetical data set. analysis 5. The Multiple Regression Process Conceptually, multiple regression is a straight forward extension of the simple linear regression procedures. 000 Problems 0. A convenience store manager notices that sales of soft drinks are higher on hotter days, so he assembles the data in the table. Classification problems are supervised learning problems in which the response is categorical; Benefits of linear regression. Regression Analysis. 1: Mnemonic for the simple regression model. Create a scatter plot and approximate a trend Using your trend line, predict the 0-60 time for a car that costs$120 K? 2. These worksheets are especially meant for pre-algebra and algebra 1 courses (grades 7-9). For this question, we could swap the independent and dependent variables and still get reasonable results. The linear regression equation, also known as least squares equation has the following form: $$\\hat Y = a + b X$$, where the regression coefficients $$a$$ and $$b$$ are computed by this regression That is, there is no lack of fit in the simple linear regression model. (a) Make a scatter plot of the data. 778 (or a value 0. 3t 2 6t 80 . Now, select Sheet 1 (by clicking on its tab at the bottom of the worksheet) and paste the information into cell F4. Worksheets are Linear regression work 1, Writing linear equationslinear regression, Work 1, Work 3, Linear reg correlation coeff work, Chapter 9 correlation and regression solutions,, Work regression. 1, Demand estimation using linear regression. In a simple linear regression model, we model the relationship between both variables by a straight line, formally $Y = b \\cdot X + a. Fit linear, quadratic, cubic, exponential, quartic, and power functions to the data. 000 These are called the regression parameters in the simple linear regression equation (the equation is also known as the least squares regression equation or the trend equation or simply the regression). 2 Practice Worksheet 1. The regression line and the residuals are displayed in figure #10. Answers are included here to check your work. One of the favorite topics on which the interviewers ask questions is \u2018Linear Regression. AP Stats: Section 3. The linear regression analysis can then 7 Aug 2017 The students can immediately understand that linear regression is on the worksheet eliminates many class time and design problems and 5 Mar 2020 Open a new workbook in Excel and make 3 worksheets: Data, Chart, and Saves. In Microsoft Office Excel 2007, you can find the Regression tool by clicking Data Analysis in the Analysis group on the Data tab. Find the equation of the regression line. In this example R2 = 0. For example, suppose that height was the only determinant of body weight. Linear Equation in Two Variables Distance Problems This worksheet is based on linear equation in two variables distance problems. Transit demand. Linear Regression Worksheet Answers with Valuable Issues. 5 Linear Regression Algebra II Name Guided Notes Date_____Block_____ Recap: Graphing Linear Functions 1. 7 is the affe ct on sales of a change in p1 with p2 held fixed !! 0 5 10 15 9 8 4 Sales I Simple Linear Regression Common Mistakes Statistics Tables Quiz: Cumulative Review A Quiz: Cumulative Review B Online Quizzes for CliffsNotes Statistics QuickReview A linear regression model corresponds to a linear regression model that minimizes the sum of squared errors for a set of pairs $$(X_i, Y_i)$$. Linear Algebra in Linear Regression Continue Suppose we have a column space in R 3 \\mathbf{R}^3 R 3 , W W W , a vector b \u20d7 \\vec{b} b , and A x \u20d7 A\\vec{x} A x , the point closest to b \u20d7 \\vec{b} b on W . Worksheet \u2013 Regression The table below displays data on the temperature ( F) reached on a given day and the number of cans of soft drink sold from a particular vending machine in front of a grocery store. The least square regression line for the set of n data points is given by the equation of a line in slope intercept form: y = a x + b. Quiz (new) 7.$ For now, let us suppose that the function which relates test score and student-teacher ratio to each other is \\[TestScore = 713 - 3 \\times STR. The last page of this exam gives output for the following situation. Section 1: Input Data Below are the input parameters to begin the simulation. 2 in-depth answers. Linear Regression Correlation Coeeficient - Displaying top 8 worksheets found for this concept. 60 55 70 45 80 35 a. Save the workbook as Linear Regression - Brief Lesson, or something similar, into a logical file folder. Does the relationship appear to be linear? Why? Yes, the relationship appears to be linear because it seems to be decreasing in similar intervals after Unit 3 - Linear Functions & Linear Regression 3-1, Functions and Function Notation - Video , Notes , Worksheet 3-3, Linear Functions - Video , Notes , Worksheet Multiple Linear Regression So far, we have seen the concept of simple linear regression where a single predictor variable X was used to model the response variable Y. Percent of a number word problems. 4 Solving Real-Life Problems How can you use a linear equation in two variables to model and solve a real-life problem? Write a story that uses the graph at the right. and solve practical problems using models of linear, quadratic, and exponential functions. An observational study if 19 managers from a medium-sized manufacturing plant extended Mintzberg\u2019s work by investigating which activities successful man- Linear regression is a method for modeling the relationship between one or more independent variables and a dependent variable. slope-intercept form. C. Linear regression is a technique that is useful for regression problems. A boat goes 30km upstream and 44km downstream in 10 hours. Quiz (new) 11. a) According to this model what was the price of gas in1990? Jan 17, 2013 \u00b7 Introduction to Correlation and Regression Analysis. 9199 yields a coefficient of determination of 0. Simple Linear Regression Example 12. 0128 . 766, adjusted for Output 2: Regression output for the grade versus homework study Regression Analysis: CourseGrade versus Problems The regression equation is CourseGrade = 44. 2 MULTIVARIATE LINEAR REGRESSION Multiple linear regression with a single criterion variable is a straightforward generalization of linear regression. This is the only section that requires user input. We need to also include in CarType to our model. Displaying top 8 worksheets found for - Linear Equation Models. n. e. Linear Regression Interview Questions \u2013 Fundamental Questions. Materials: CD Player / Computer with CD drive. If you were a careful artist, you could take a ruler and draw a straight-line as close as possible to every point in Worksheet 2. The measure of how well this linear function ts the experimental points, is called regression analysis. \u2022 In a simple linear regression model, a single response measurement Y is related to a single predictor (covariate, regressor) X for each observation. Squaring \u22120. Worksheet for Correlation and Regression (February 1, 2013). In addition, after finding the equations, students are asked \"extension questions\" in which they must use the equation to answer questions a About This Quiz & Worksheet About This Quiz & Worksheet Simple linear regression builds on the concept of a regression line by allowing you to specifically make predictions based on the regression Practice linear regression with 10 Canadian data sets covering a range of topics. Simple linear regression is a bivariate situation, that is, it involves two dimensions, one for the dependent variable Y and one for the independent variable x. Students will determine the linear regression equation and correlation coefficient for their data using the graphing calculator. 21. Open or retrieve the worksheet Slr01. This leads to the Below is a plot of the data with a simple linear regression line superimposed. a) According to the linear model, the duration of a coaster ride is expected to increase by about 0. USING THIS MODULE vii Apr 27, 2017 \u00b7 Non-Linear Relationships Not all relationships are linear. a) Find the least square regression line for the following set of data\u00a0 Practice Worksheet: Linear Regression. 903, and because the graph of the cubic model is seen to be a closer match to the dots in the scatterplot than is the linear model. Chapter 8 Linear Regression 91 22. Excel completes the regression analysis Regression Analysis Worksheets- Includes math lessons, 2 practice sheets, homework sheet, and a quiz! Linear regression simply refers to creating a best fit for a linear relationship between two variables from observed data. This makes the line fit the points. In this problem we find the model by analyzing the data on femur length and height for the ten males given in the table. Consider the following diagram. 3: Setup for Linear Regression Test on TI-83/84\u00a0 to fit a simple linear regression equation with poverty predicting obesity. Linear Regression Assumptions \u2022 Linear regression is a parametric method and requires that certain assumptions be met to be valid. the linear relationship through the correlation coefficient. \ud835\udc66\ud835\udc66 = \u221270. Graphic calculators, such as the TI-83, have built in programs which allow us to nd the slope and the y intercept of the best tting line to a set of data points. On a piece of graph paper, create a scatter plot. Of easily catch up can understand this You\u00a0 There are actually two ways to do a linear regression analysis using Excel. Practice-Regression 2 linear, quadratic, exponential A simple linear regression equation for this would be $$\\hat{Price} = b_0 + b_1 * Mileage$$. Thus a linear model only explains 85% of the variation in women's world record 100 m dash times. Export problem to Excel (highlighted when problem is correctly entered). Identify the type of regression with the best fit, and answer the question using the type of regression that best fits the data. 433 seconds. It may be printed, downloaded or saved and used in your classroom, home school, or other educational Linear Regression. Linear Regression & Correlation Coefficient Worksheet Name _____ Hr _____ 0 2 4 6 8 10 0 2 4 6 1. Linear regression only supports regression type problems. Use the two plots to intuitively explain how the two models, Y!$0 %$ 1x %& and Why Linear Regression? \u2022Suppose we want to model the dependent variable Y in terms of three predictors, X 1, X 2, X 3 Y = f(X 1, X 2, X 3) \u2022Typically will not have enough data to try and directly estimate f \u2022Therefore, we usually have to assume that it has some restricted form, such as linear Y = X 1 + X 2 + X 3 correlation, in linear regression. Femur Length (cm) (a) Make a scatter plot of the data. Our model will take the form of \u0177 = b 0 + b 1 x where b 0 is the y-intercept, b 1 is the slope, x is the predictor variable, and \u0177 an estimate of the mean value of the response variable for any value of the predictor Copy the regression coefficients onto the sheet with the actual population data. become identical when r = \u20131 or 1 or in other words, there is a perfect negative or positive correlation between the two variables under discussion. Problem-solving using linear regression has so many applications in business, digital customer experience, social, biological, and many many other areas. CAS'S WEBSITE - Home I don't think so. The regression line is the line that makes the square of the residuals as small as possible, so the regression line is also sometimes called the least squares line. This math worksheet was created on 2013-02-14 and has been viewed 62 times this week and 1,053 times this month. (1) Investigator #1 is interested in predicting Y from X, and fits and computes a regression line for this purpose. Using a similar approach, we may prove that S2 y = S 2 y\u02c6 +S 2 e (5. There are also other regression modelling techniques for data not considered to be at continuous/interval/ratio level. The calculators also give 1. temperature 70 75 80 90 93 98 72 75 75 80 90 95 98 91 98 quantity 30 31 40 52 57 59 33 38 32 45 53 56 62 51 58 \u00afx = 85. The regression model is linear in the unknown parameters. To make the residual plot, use \u201cGraphs\u201d and then type in the name of the explanatory variable. Each point of data is of the the form (x, y) and each point of the line of best fit using least-squares linear regression has the form \u00a0 Excel displays the Regression dialog box. nate because the world is too complex a place for simple linear regression alone to model it. Find a quadratic model in standard form for the data. 2 Linear Regression If there is a \\signi cant\" linear correlation between two variables, the next step is to nd the equation of a line that \\best\" ts the data. 8462. Comments: Another SPSS output table \u2013 see Table 3 \u2013 gives a useful value 'R square', or the 'coefficient of determination'. c) the relationship between X and Y is quadratic d) there is no relationship between X and Y. The dependent variable must be of ratio/interval scale and normally distributed overall and normally distributed for each value of the independent variables 3. Apply the method of least squares (or maximum likelihood) with a non-linear function. A simple linear regression model to relate BP with age will be BP = regression estimate (b) * age + constant (a) + error term (\u00e5) The regression estimate (b) and the constant (a) will be derived from the data (using the method of least-squares(5)) and the error term is to factor in the situation that two persons with the same age need not have the same BP. L. State which model, linear or quadratic, best fits the data. Know how to obtain the estimate MSE of the unknown population variance $$\\sigma^{2 }$$ from Minitab's fitted line plot and regression analysis output. A regression is a process that takes all the points and calculates the equation that best 'fits' those points. The Nonlinear Regression Model 1 Goals The nonlinear regression model block in the Weiterbildungslehrgang (WBL) in ange- wandter Statistik at the ETH Zurich should 1. are called the residuals. The estimated regression equation is that average FEV = 0. With this worksheet generator, you can make customizable worksheets for linear inequalities in one variable. Included are two versions of the optional placemat/worksheet to help students through the process of linear regression and analysis. a) Enter the data into two lists of your graphing. Show that in a simple linear regression model the point ( ) lies exactly on the least squares regression line. CD \u2013 Music compiled by teacher. Investigator #2 is interested in predicting X from Y, and computes his regression line for that purpose (note that in the real problem of \u201cparallel-line bioassays, with X=log(dose) STATISTICS 110/201 PRACTICE FINAL EXAM KEY (REGRESSION ONLY) Questions 1 to 5: There is a downloadable Stata package that produces sequential sums of squares for regression. This result is smaller than suggested by any of the other analyses that I have conducted, and is most similar to the analysis with all of the variables except for each of Reliable and Unconventional. A study found that age and blood pressure are correlated. Practice Problems: Correlation and Linear Regression Researchers interested in determining if there is a relationship between death anxiety and religiosity conducted the following study. We talk about looking at the data to decide what type of regression is appropriate. is found. See more ideas about Linear regression, Regression, Algebra. (b) Find and graph a linear regression equation that models the data. Each of the data sets has 4 or 5 points and approximates a linear relationship. We can find the equation of the line of best fit through the data in the least squares sense, as follows. \ufb01t linear regression models. Is there a linear relationship between how critics score the games and how users score the games? (Let x = critic score and y = user score). 2 Equation:_____ WORKSHEET GENERATORS. Categorical Data 12. A residual plot is displayed showing the deviation between the data and the calculated values of the dependent variable. expression the strength of a linear relationship between two variables are two of the desired outcomes of this module. 1 The Simple Linear Regression Model 12. Interpreting r and r 2. Word problems on constant speed. Oct 05, 2012 \u00b7 The \u201cGood\u201d linear regression model. Remember to complete the five steps of hypothesis testing in the spaces provided on the worksheet. This means that this data can be modeled using a linear regression line. S. (b) What change in gasoline mileage is associated with a 1 cm3 change is engine displacement? 11-18. Correlation!Coefficient!&Linear!of!Best!Fit!HW! Name:!!_____! 8. The fit is not quite as tight for the women's times. The missile goes over the side of the ship and hits the water. SGDRegressor is well suited for regression problems with a large number of training samples (> 10;000), for other prob-lems we recommend Ridge, Lasso, or ElasticNet. Then they are able to apply the knowledge they learn in this project on the real life problems. Explain Jul 31, 2016 \u00b7 State the three assumptions that are the basis for the Simple Linear Regression Model. Figure 2. 05898 6. If the data is curved, a line would not be the best equation to use. In most problems, more than one predictor variable will be available. than ANOVA. Research question type: When wanting to predict or explain one variable in WORKSHEETS\\calcium. Find the numbers. Solve general word problems about real-world relationships that can be modeled by linear equations or functions. 999, which is closer to 1 than is the linear correlation value of 0. It is also a method that can be reformulated using matrix notation and solved using matrix operations. Use the linear regression equation determined in part \"B\" to calculate a set of \"predicted y values\" for each observed x value. Logarithmic Regression Problems. Regression 9. If the data on the scatter plot seems to represent a linear relationship, then linear regression can be used to find the line that best fits the data. Inference in the Linear Regression Model 4. The New Worksheet Ply default output option means that Excel locates the regression outputs on a new, separate worksheet that it creates and puts to the left of the worksheet that holds the original data. To do this, highlight the cell range A17:B18 and click on the COPY button. The plot to the right shows 5 data points and the least squares line. Interpret the meaning of the slope \u03b2 ^ 1 of regression line in the context of problem. Use Stat > Regression > Regression to find the regression equation AND make a residual plot of the residuals versus the explanatory variable. Open the \u201cGood\u201d worksheet; this is a (made-up) data set showing the Height (independent variable) and Weight (dependent variable) values for a selection of people. Nov 26, 2014 \u00b7 to linear regression . Old Faithful Geyser in Yellowstone National Park, Wyoming, derives its name and its considerable fame from the regularity (and beauty) of its eruptions. ) Using a graphing calculator and quadratic regression to find a model: A study compared the speed x, in miles per hour and the average fuel economy y (in miles per gallon) for cars. 3 (b) Find and graph a linear regression equation that models the data. In this exercise, you will gain some practice doing a simple linear regression using a data set called week02. Linear Regression and Correlation Introduction Linear Regression refers to a group of techniques for fitting and studying the straight-line relationship between two variables. Your output for this multiple regression problem should be similar to the results shown \u00a0 Let's explore the problem with our linear regression example. To make the notation simpler, assume that the criterion variable Y and the p Lesson 21: Multiple Linear Regression Analysis . Derivation of linear regression equations The mathematical problem is straightforward: given a set of n points (Xi,Yi) on a scatterplot, find the best-fit line, Y\u2039 i =a +bXi such that the sum of squared errors in Y, \u2211(\u2212)2 i Yi Y \u2039 is minimized The simple linear regression model Our goal is to obtain estimates ^ 0 and ^ 1 for 0 and 1 to de ne the regression line ^y = ^ 0 + ^ 1x that provides the best t for the data Example: Assume that the regression line of the previous example is: Cost = 15:65 + 1:29 Volume Regression Problems Math The data for these problems is from Math 142, Dr Lacey, Packet 118. Student Learning Outcomes By the end of this chapter, you should be able to do the following: Linear Regression Models with Logarithmic Transformations Kenneth Benoit Methodology Institute London School of Economics kbenoit@lse. The critical assumption of the model is that the conditional mean function is linear: E(Y|X) = \u03b1 +\u03b2X. 7 This worksheet is designed to give students extra practice at using their graphing calculators to calculate Linear Regression Equations. Oct 21, 2019 \u00b7 Some of the worksheets below are Correlation Coefficient Practice Worksheets, Interpreting the data and the Correlation Coefficient, matching correlation coefficients to scatter plots activity with solutions, classify the given scatter plot as having positive, negative, or no correlation, \u2026 Simple Linear Regression Model Only one independent variable, x Relationship between x and y is described by a linear function Changes in y are assumed to be caused by changes in x Fall 2006 \u2013 Fundamentals of Business Statistics 18 Types of Regression Models Positive Linear Relationship Negative Linear Relationship Relationship NOT Linear Free worksheets for solving or graphing linear inequalities. Here are data from four students on their Quiz 1 scores and their Quiz 5 scores and a graph where we connected the points by a line. Regression 8. Their sum is 13. Linear Regression is a statistical tool in excel that is used as a predictive analysis model to check the relationship between two sets of data of variables. Winking at Phoenix High School Sec 5. This three-page worksheet contains 16 problems. 8 + 0. Predict the number of aids cases for the year 2006. Linear Regression Equation - Displaying top 8 worksheets found for this concept. Compute S S E, the measure of the goodness of fit of the regression line to the sample Use Linear Regression Calculator and Grapher Given a set of experimental points, this calculator calculates the coefficients a and b and hence the equation of the line y = a x + b and the Pearson correlation coefficient r. HSF-LE. The above simple linear regression examples and problems aim to help you understand better the whole idea behind simple linear regression equation. Determine the slope of the linear regression line, b, in \u00baC/century and the correlation coefficient, r. Linear regression is nice, but it isn't a religion. 1) 45 Multicollinearity occurs when independent variables in a regression model are correlated. Put bite number on the x-axis and Twizzler length on the y-axis. Worksheet: Rate the Song. To answer this question the researcher would measure body weight and blood cholesterol level in various subjects. Multiple-choice. x, y ( ) points. Bascially, the least-squares regression line is the line that minimizes the squared \"errors\" between the actual points and the points on the line. 9. slcmath@pc. Students will use different methods to Sep 23, 2018 \u00b7 This video explains you the basic idea of curve fitting of a straight line in multiple linear regression. Let us begin with a fundamental Linear Regression Interview Questions. In the 8th grade students solved problems in the context of bivariate measurement data. Finally, the students will explore the residuals for the regression equation and interpret the results from the activity in the context of the problem. The multiple regression of Sales on own price (p1) and competitor's price (p2) yi eld more intuitive signs: How does this happen ? The regression equation is Sales = 211 + 63. This Practice Problems: Correlation and Linear Regression Worksheet is suitable for 9th - 11th Grade. To draw a linear forecast graph like shown in the screenshot below, here's what you need to do: Copy the last historical data value to the Forecast In this example, we copy the value from B13 to C13 Select Viewport in the Trend Lines tab to perform linear regression for the sub-period. linear regression and modeling problems with answers. The variable we predict is called the dependent or outcome variable and is referred to as Y. Describe the correlation Use the line of best fit to make predictions for the following real-world problems. 10: Regression - Fuel Efficiency (Worksheet) \u00b7 13: F Distribution and One-Way ANOVA If a loan officer makes 95% of his or her goal, write the linear function that applies based 12. A random sample was taken as stated in the problem. Student will learn how to write a linear regression equation and use the equation to solve a\u00a0 Make a scatterplot of the data, letting x represent the number of years since 1990. According to the linear model, what is the expected pressure for a person with cholesterol 270 mg/dl? Use the following sums: $\\sum x_i=16960$ mg/dl, $\\sum y_j=11160$ mmHg, $\\sum x_i^2=3627200$ (mg/dl)$^2$, $\\sum y_j^2=1576800$ mmHg$^2$ y \\$\\sum x_iy Welcome to The Systems of Linear Equations -- Two Variables (A) Math Worksheet from the Algebra Worksheets Page at Math-Drills. Inference on Prediction Assumptions 1. 126 Chapter 3 Writing Linear Equations and Linear Systems 3. Word problems on average speed Word problems Free worksheet(pdf) and answer key on the solving word problems based on linear equations and real world linear models. Equation: Created Date: 1/4/2016 8:11:44 PM A correlation or simple linear regression analysis can determine if two numeric variables are significantly linearly related. ## [1] 680 640 670 660 630 660 635. linear regression problems worksheet\n\nn gweqgiyu , gqerignslomk, llv43lih0cj, a3hy9obqtv, qgbmrt1eb vdl8, 0pxyypazy6tyvy,", "date": "2020-11-30 00:51:06", "meta": {"domain": "sramble-communication.com", "url": "http://angs-diasporanew.sramble-communication.com/apdsmyht/linear-regression-problems-worksheet.php", "openwebmath_score": 0.39756202697753906, "openwebmath_perplexity": 713.5120381218667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429565737233, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.8508120142948999}}
{"url": "http://math.stackexchange.com/questions/226903/does-the-convergence-of-some-subsequences-imply-the-convergence-of-a-sequence", "text": "Does the convergence of some subsequences imply the convergence of a sequence?\n\nI am considering a following problem: Does $$\\\\a_{2k}, a_{2k+1}, a_{3k} \\rightarrow g$$ imply that $$a_{n} \\rightarrow g ?$$ I know that if every subsequence goes to $g$ then also a sequence goes to $g$. My way of reasoning is following: we know that odd and even $k$ subsequences goes to $g$. But if we substitute for example $k=t^{2} -5$, then there is problem (at least for me). If I can reason that way, what must I do to prove more formally that the statement is false? And if I can not, what is going on with $a_{n}$? Thanks for any hints!\n\n-\n\nIf the even numbered terms and odd numbered terms both converge to the same limit $L$, then the limit of the sequence exists and is $L$.\n\nTo prove this, let $\\epsilon > 0$ be arbitrary. We want to find $N \\in \\mathbb{N}$ such that $n > N$ implies $|a_n - L| < \\epsilon$.\n\nSince the even numbered terms are all within epsilon of $L$ after some $N_1 = 2j$ and the odd numbered terms are all within epsilon of $L$ after some $N_2 = 2k + 1$, just pick $N = \\max\\{N_1, N_2\\}$.\n\n-\n\nYou may want to prove the next nice lemma:\n\nLemma: If $\\,\\{A_i\\}_{i\\in I},$ is some partition of $\\,\\Bbb N\\,$ s.t. $\\,|A_i|=\\aleph_0=|\\Bbb N|\\,\\,\\,,\\,\\forall\\,\\,i\\in I\\,$ , and all the sets $\\,A_i\\,$ are well ordered, then for a real sequence $\\,\\{x_n\\}\\,$ it is true that\n\n$$x_n\\xrightarrow [n\\to\\infty]{} x\\Longleftrightarrow (\\,\\forall\\,\\,i\\in I\\,\\,,x_{n_i}\\xrightarrow [n_i\\to\\infty\\,,\\,n_i\\in A_i]{} x)$$\n\nThus, as B. wrote in his answer, it is enough and sufficient that the subsequence of odd indexes and the one of even indexes converge both to the same limit.\n\n-\nDoesn't $I$ have to be a finite in order for this to work? \u2013\u00a0 Arthur Fischer Nov 1 '12 at 17:48\nWell, not that you mention it...I'm not completely sure, I'll try to check this later. Thanks. \u2013\u00a0 DonAntonio Nov 1 '12 at 18:23\nTake $I = \\{ 1 \\} \\cup \\{ p : p \\text{ is prime} \\}$. Define $A_1 = \\{ 1 \\} \\cup \\{ n : n \\text{ is not a prime power} \\}$, and $A_p = \\{ p^k : k \\geq 1 \\}$ for prime $p$. Define the sequence $( x_n )_{n \\geq 1}$ so that $x_n = 0$ for all $n \\in A_1$ and $x_{p^k} = \\frac{1}{k}$ for $p$ prime and $k \\geq 1$. Then for each $i \\in I$ the sequence $( a_n )_{n \\in A_i}$ converges to $0$, but the sequence $( a_n )_{n \\geq 1}$ does not converge as it takes the value $1$ infinitely often and has subsequences converging to $0$. \u2013\u00a0 Arthur Fischer Nov 1 '12 at 18:39\nIn the other answers you have seen that it $a_{2k},a_{2k+1}\\to g$ then already $a_n\\to g$. You don't even need $a_{3k}$. I would like to add that if you just require $a_{2k},a_{2k+1}$ and $a_{3k}$ to converge (a priori with different limits), then you have that all three limits coincide and that $a_n$ has the same limit.\nTo see this note that $a_{2n}$ and $a_{3n}$ as well as $a_{2n+1}$ and $a_{3n}$ have a commen subsequence. Then you only need that if a converging sequence has a subsequence with limit $g$ then the whole sequence has limit $g$.", "date": "2014-03-08 01:17:08", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/226903/does-the-convergence-of-some-subsequences-imply-the-convergence-of-a-sequence", "openwebmath_score": 0.9700687527656555, "openwebmath_perplexity": 94.65449867698898, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429580381724, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8508120121445751}}
{"url": "https://math.stackexchange.com/questions/451799/how-to-show-lim-x-to-1-fracx-x2-dots-xn-nx-1-fracnn", "text": "# How to show $\\lim_{x \\to 1} \\frac{x + x^2 + \\dots + x^n - n}{x - 1} = \\frac{n(n + 1)}{2}$?\n\nI am able to evaluate the limit $$\\lim_{x \\to 1} \\frac{x + x^2 + \\dots + x^n - n}{x - 1} = \\frac{n(n + 1)}{2}$$ for a given $n$ using l'H\u00f4spital's (Bernoulli's) rule.\n\nThe problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:\n\n\u2022 Substitute $y = x - 1$. This leads nowhere, I think.\n\u2022 Find the Taylor polynomial. Makes no sense, it is a polynomial.\n\u2022 Divide by major term. Dividing by $x$ got me nowhere.\n\u2022 Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$.\n\u2022 Simplify the expression. I do not see how I could.\n\u2022 Using l'H\u00f4spital's (Bernoulli's) rule. Works, but I do not quite like it.\n\nIf somebody sees a simple way, please do let me know.\n\nAdded later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:\n\n\u2022 $\\lim_{x \\to 0} \\frac{\\sqrt[m]{1 + ax} - \\sqrt[n]{1 + bx}}{x}$ where $m, n \\in \\mathbb{N}$ and $a, b \\in \\mathbb{R}$ are given, or\n\u2022 $\\lim_{x \\to 0} \\frac{\\arctan(1 + x) - \\arctan(1 - x)}{x}$.\n\nIt sems that all limits in the form $\\lim_{x \\to a} \\frac{f(x)}{x - a}$ where $a \\in \\mathbb{R}$, $f(a) = 0$ and for which $\\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.\n\nThis adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'H\u00f4spital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :).\n\n\u2022 Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. Jan 15, 2015 at 12:36\n\nLet $$f(x)=x+x^2+\\cdots+x^n-n$$ then by the definition of the derivative we have\n\n$$\\lim_{x \\to 1} \\frac{x + x^2 + \\dots + x^n - n}{x - 1}= \\lim_{x \\to 1}\\frac{f(x)-f(1)}{x - 1}=f'(1)\\\\[10pt] = \\left[ \\vphantom{\\frac11} 1 + 2x + 3x^2 + \\cdots + nx^{n-1} \\right]_{x=1} = \\frac{n(n + 1)}{2}$$\n\n\u2022 It means that we should substitute $x$ by $1$.\n\u2013\u00a0user63181\nJul 25, 2013 at 12:20\n\u2022 Then this is truly an ingenious solution. Jul 25, 2013 at 12:30\n\u2022 So the original is just the integral of a function summing the positive natural numbers? Pauling would be proud. Jul 25, 2013 at 15:29\n\u2022 I almost cried when I saw this solution. Jul 25, 2013 at 17:44\n\u2022 @RossMillikan shouldn't it not matter? if $f(x)$ contains the $-n$ then $f(1) = 0$, and if $f(x)$ does not, $f(1) = n$, but either way $f(x) - f(1)$ doesn't change and since $n$ is a constant, $f'(x)$ looks the same. Jul 25, 2013 at 23:22\n\nWell, you can use that \\begin{align} x-1&=\\left(x-1\\right)\\cdot 1,\\\\ x^2-1&=\\left(x-1\\right)\\cdot\\left(x+1\\right),\\\\ x^3-1&=\\left(x-1\\right)\\cdot(x^2+x+1),\\\\ &{}\\ \\ \\vdots\\\\ x^n-1&=\\left(x-1\\right)\\cdot(x^{n-1}+\\cdots+1),\\\\ \\end{align} Now sum the left hand sides and the right hand sides, divide by $x-1$ and consider the limit $x\\rightarrow 1$.\n\nThe given limit is $$\\lim_{x\\rightarrow 1}\\frac{\\sum_{k=1}^nx^k-n}{x-1}\\\\ =\\lim_{x\\rightarrow 1}\\frac{\\sum_{k=1}^n(x^k-1)}{x-1}\\\\ =\\sum_{k=1}^n \\lim_{x\\rightarrow 1} \\frac{(x^k-1)}{x-1}$$\n\nNow, $$\\lim_{x\\rightarrow 1} \\frac{(x^k-1)}{x-1}\\\\ =\\lim_{x\\rightarrow 1} (\\sum_{j=0}^{k-1}x^j)=k$$\n\nHence the given limit becomes $$\\sum_{k=1}^n k=\\frac{n(n+1)}{2}$$\n\nHints:\n\n$$x+x^2+\\ldots+x^n-n=(x-1)+(x^2-1)+\\cdots +(x^n-1)=(x-1)\\left(1+(x+1)+\\cdots\\right)$$\n\nAlso\n\n$$1+(x+1)+(x^2+x+1)+\\cdots+(x^{n-1}+\\cdots+x+1)\\xrightarrow [x\\to 1]{}1+2+\\cdots+n=\\frac{n(n+1)}2$$\n\nYou can use induction:\n\n$$\\frac{x + x^2 + \\dots + x^n + x^{n+1} - (n+1)}{x - 1} =\\\\ \\frac{x + x^2 + \\dots + x^n - n}{x - 1} +\\frac{x^{n+1}-1}{x-1}=\\\\ \\frac{x + x^2 + \\dots + x^n - n}{x - 1} +(1+x+x^2+\\ldots+x^n)\\xrightarrow[x\\to 1]{} \\frac{n(n + 1)}{2}+(n+1)=\\frac{(n+1)(n + 2)}{2}.$$\n\n\u2022 I am a little lost. How is that you are getting different result? Or is $\\frac{(n + 1)(n + 2)}{2}$ only a part of the solution? Jul 25, 2013 at 12:28\n\u2022 @David\u010cepel\u00edk No, $\\frac{(n+1)(n+2)}{2}$ is the limit for $\\frac{x+x^2+\\ldots+x^n\\color{brown}{+x^{n+1}}-n\\color{brown}{-1}}{x-1}$ and not for $\\frac{x+x^2+\\ldots+x^n-n}{x-1}$.\n\u2013\u00a0P..\nJul 25, 2013 at 12:32\n\u2022 Of course, that is the induction. Thank you! Jul 25, 2013 at 12:37\n\u2022 So for $n = 1$ and $x \\to 1$, we get $a_1 = \\lim_{x \\to 1} \\frac{x - 1}{x - 1} = 1$ and $a_{n + 1} = a_n + (n + 1)$. The closed form for $a_n$ is then $\\frac{n(n + 1)}{2}$. Correct? Jul 25, 2013 at 12:42\n\u2022 @David\u010cepel\u00edk: Yes that's right!\n\u2013\u00a0P..\nJul 25, 2013 at 13:08\n\n$$\\frac{x + x^2 + \\dots + x^n - n}{x - 1} =\\frac{1+x + x^2 + \\dots + x^n - n-1}{x - 1} =\\frac{\\frac{x^{n+1}-1}{x-1}-(n+1)}{x-1}$$\n\nPutting $x-1=y,$\n\n$$\\lim_{x \\to 1} \\frac{x + x^2 + \\dots + x^n - n}{x - 1} = \\lim_{y\\to0}\\frac{\\frac{(1+y)^{n+1}-1}y-(n+1)}y$$\n\n$$=\\lim_{y\\to0}\\frac{\\frac{1+\\binom {n+1}1y+\\binom {n+1}2y^2+ O(y^3)-1}y-(n+1)}y\\text{ (using Binomial Expansion)}$$\n\n$$=\\lim_{y\\to0}\\frac{(n+1)+\\frac{n(n+1)}2y+ O(y^2)-(n+1)}y$$\n\n$$=\\frac{n(n+1)}2\\text{ as }x\\to1,y\\to0\\implies y\\ne0$$\n\nIf you already know l'Hopital's rule, Sami Ben Romdhane's answer isn't telling you anything you don't already know. When you have $$\\lim_{x\\to a}\\frac{f(x)}{x - a}$$ and $f(a) = 0$ (otherwise, the limit is infinite on either side of $a$), then l'Hopital's rule says that the limit is the same as $$\\lim_{x \\to a}f'(x)= f'(a).$$ The only hard part then, in this case, is evaluating $f'(a)$. Here, $f(x) = x + x^2 + \\cdots + x^n - n$, so $f'(x) = 1 + 2x + \\cdots + nx^{n - 1}$, so $f'(1) = 1 + 2 + \\cdots + n = \\frac{n(n + 1)}{2}$ by a famous identity.\n\nIn fact, you can think of the definition of the derivative as just a special case of l'Hopital's rule (it isn't really, because l'Hopital's rule depends on the definition of the derivative, not the other way around, but it's useful to think of it this way).\n\n\u2022 I do not think this is how he approached the problem; my guess is he noticed that the function and the derivative of the numerator are bound together by the definition of derivative of a function. I did not realize that he used the same principle Bernoulli's rule is built upon, thanks for pointing it out. Jul 25, 2013 at 16:23\n\nYour objection to using l'Hospital's rule is on the basis that it feels like it's \"too powerful\" a tool for the problem, right?\n\nLets go all the way to the other end, then, and prove the limit using just the definition.\n\n$$\\lim_{x\\to a} f(x) = L \\iff \\forall \\varepsilon \\gt 0 \\ \\exists \\delta \\gt 0 \\ni \\left| x-a \\right| \\le \\delta \\Rightarrow \\left| f(x)-L \\right| \\le \\varepsilon$$\n\nNow, all we need to do is figure out a way that we can always pick a $\\delta$ small enough to keep the function within $\\varepsilon$ of $L$.\n\nThe particular limit in question: $$\\lim_{x \\to 1} \\frac{x + x^2 + \\dots + x^n - n}{x - 1} = \\frac{n(n + 1)}{2}$$\n\nFrom that, we can take: \\begin{align*} a&=1\\\\ f(x)&=\\frac{x + x^2 + \\dots + x^n - n}{x - 1}\\\\ L&=\\frac{n(n + 1)}{2} \\end{align*}\n\nSo we need to solve $$\\left| \\frac{x + x^2 + \\dots + x^n - n}{x - 1} - \\frac{n(n + 1)}{2}\\right| < \\varepsilon$$ for $\\left|x-1\\right|$.\n\nI really don't feel like doing any of that crunchwork solving that equation. If someone out there does feel like it, please do so, and edit it into my answer. For now, though, I am going to skip ahead bunch of steps, assume we solved it, and have our solution of $$\\left|x-1\\right| \\ge g(\\varepsilon)$$\n\nNow, we know from the definition that $\\left| x-1 \\right| \\le \\delta$, so we can conclude that if we pick $\\delta = g(\\varepsilon)$, we ensure that the value of $f(x)$ is within $\\varepsilon$ of $L$, satisfying our definition of the limit, and proving that $$\\lim_{x \\to 1} \\frac{x + x^2 + \\dots + x^n - n}{x - 1} = \\frac{n(n + 1)}{2}$$\n\n\u2022 Thanks for your answer. You already know the limit exists and you know what it equals to ($\\frac{n(n+1)}{2}$ was not my first guess). Finding what a limit equals to and proving the solution is correct by checking it satisfies the definition of the limit are two completely different exercises. (Similar to the difference between the P and NP classes in the theory of computational complexity.) I am aware I gave correct solution along with the limit, maybe that is why we did not understand each other. Sorry about it. Jul 26, 2013 at 20:01\n\u2022 @David\u010cepel\u00edk Personally, I think that if you prove the limit without using your super-powered tool, it doesn't matter if you used it to find the value of the limit, since the validity of your answer then does not depend on the validity of the method you applied. Jul 29, 2013 at 12:57", "date": "2022-05-25 01:52:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/451799/how-to-show-lim-x-to-1-fracx-x2-dots-xn-nx-1-fracnn", "openwebmath_score": 0.9405686259269714, "openwebmath_perplexity": 273.2409213727027, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429575500227, "lm_q2_score": 0.865224084314688, "lm_q1q2_score": 0.8508120100135155}}
{"url": "https://bathmash.github.io/HELM/30_5_itrtv_mthds_systms_eqns-web/30_5_itrtv_mthds_systms_eqns-webse2.html", "text": "### 2 Do these iterative methods always work?\n\nNo. It is not difficult to invent examples where the iteration fails to approach the solution of $AX=B$ . The key point is related to matrix norms seen in the preceding Section.\n\nThe two iterative methods we encountered above are both special cases of the general form\n\n$\\phantom{\\rule{2em}{0ex}}{X}^{\\left(k+1\\right)}=M{X}^{\\left(k\\right)}+N.$\n\n1. For the Jacobi method we choose $M=-{D}^{-1}\\left(L+U\\right)$ and $N={D}^{-1}B$ .\n2. For the Gauss-Seidel method we choose $M=-{\\left(D+L\\right)}^{-1}U$ and $N={\\left(D+L\\right)}^{-1}B$ .\n\nThe following Key Point gives the main result.\n\n##### Key Point 13\n\nFor the iterative process ${X}^{\\left(k+1\\right)}=M{X}^{\\left(k\\right)}+N$ the iteration will converge to a solution if the norm of $M$ is less than 1 .\n\nCare is required in understanding what Key Point 13 says. Remember that there are lots of different ways of defining the norm of a matrix (we saw three of them). If you can find a norm ( any norm ) such that the norm of $M$ is less than 1, then the iteration will converge. It doesn\u2019t matter if there are other norms which give a value greater than 1, all that matters is that there is one norm that is less than 1.\n\nKey Point 13 above makes no reference to the starting \u201cguess\" ${X}^{\\left(0\\right)}$ . The convergence of the iteration is independent of where you start! (Of course, if we start with a really bad initial guess then we can expect to need lots of iterations.)\n\nShow that the Jacobi iteration used to approximate the solution of\n\n$\\phantom{\\rule{2em}{0ex}}\\left[\\begin{array}{ccc}\\hfill 4\\hfill & \\hfill -1\\hfill & \\hfill -1\\hfill \\\\ \\hfill 1\\hfill & \\hfill -5\\hfill & \\hfill -2\\hfill \\\\ \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill 2\\hfill \\end{array}\\right]\\left[\\begin{array}{c}\\hfill {x}_{1}\\hfill \\\\ \\hfill {x}_{2}\\hfill \\\\ \\hfill {x}_{3}\\hfill \\end{array}\\right]=\\left[\\begin{array}{c}\\hfill 1\\hfill \\\\ \\hfill 2\\hfill \\\\ \\hfill 3\\hfill \\end{array}\\right]$\n\nis certain to converge. (Hint: calculate the norm of $-{D}^{-1}\\left(L+U\\right)$ .)\n\nThe Jacobi iteration matrix is\n\n$\\begin{array}{rcll}-{D}^{-1}\\left(L+U\\right)& =& {\\left[\\begin{array}{ccc}\\hfill 4\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill -5\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 2\\hfill \\end{array}\\right]}^{-1}\\left[\\begin{array}{ccc}\\hfill 0\\hfill & \\hfill 1\\hfill & \\hfill 1\\hfill \\\\ \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill 2\\hfill \\\\ \\hfill 1\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\end{array}\\right]=\\left[\\begin{array}{ccc}\\hfill 0.25\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill -0.2\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0.5\\hfill \\end{array}\\right]\\left[\\begin{array}{ccc}\\hfill 0\\hfill & \\hfill 1\\hfill & \\hfill 1\\hfill \\\\ \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill 2\\hfill \\\\ \\hfill 1\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\end{array}\\right]& \\text{}\\\\ & =& \\left[\\begin{array}{ccc}\\hfill 0\\hfill & \\hfill 0.25\\hfill & \\hfill 0.25\\hfill \\\\ \\hfill -0.2\\hfill & \\hfill 0\\hfill & \\hfill 0.4\\hfill \\\\ \\hfill 0.5\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\end{array}\\right]& \\text{}\\end{array}$\n\nand the infinity norm of this matrix is the maximum of $0.25+0.25$ , $0.2+0.4$ and $0.5$ , that is\n\n$\\phantom{\\rule{2em}{0ex}}\u2225-{D}^{-1}\\left(L+U\\right){\u2225}_{\\infty }=0.6$\n\nwhich is less than 1 and therefore the iteration will converge.\n\n#### 2.1 Guaranteed convergence\n\nIf the matrix has the property that it is strictly diagonally dominant , which means that the diagonal entry is larger in magnitude than the absolute sum of the other entries on that row, then both Jacobi and Gauss-Seidel are guaranteed to converge. The reason for this is that if $A$ is strictly diagonally dominant then the iteration matrix $M$ will have an infinity norm that is less than 1.\n\nA small system is the subject of Example 20 below. A large system with slow convergence is the subject of Engineering Example 1 on page 62.\n\n##### Example 20\n\nShow that $A=\\left[\\begin{array}{ccc}\\hfill 4\\hfill & \\hfill -1\\hfill & \\hfill -1\\hfill \\\\ \\hfill 1\\hfill & \\hfill -5\\hfill & \\hfill -2\\hfill \\\\ \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill 2\\hfill \\end{array}\\right]$ is strictly diagonally dominant.\n\n##### Solution\n\nLooking at the diagonal entry of each row in turn we see that\n\n$\\begin{array}{rcll}4& >& |-1|+|-1|=2\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}& \\text{}\\\\ \\left|-5\\right|& >& 1+\\left|-2\\right|=3& \\text{}\\\\ 2& >& \\left|-1\\right|+0=1& \\text{}\\end{array}$\n\nand this means that the matrix is strictly diagonally dominant.\n\nGiven that $A$ above is strictly diagonally dominant it is certain that both Jacobi and Gauss-Seidel will converge.\n\n#### 2.2 What\u2019s so special about strict diagonal dominance?\n\nIn many applications we can be certain that the coefficient matrix $A$ will be strictly diagonally dominant. We will see examples of this in HELM booklet \u00a032 and HELM booklet \u00a033 when we consider approximating solutions of differential equations.\n\n##### Exercises\n1. Consider the system\n\n$\\phantom{\\rule{2em}{0ex}}\\left[\\begin{array}{cc}\\hfill 2\\hfill & \\hfill 1\\hfill \\\\ \\hfill 1\\hfill & \\hfill 2\\hfill \\end{array}\\right]\\left[\\begin{array}{c}\\hfill {x}_{1}\\hfill \\\\ \\hfill {x}_{2}\\hfill \\end{array}\\right]=\\left[\\begin{array}{c}\\hfill 2\\hfill \\\\ \\hfill -5\\hfill \\end{array}\\right]$\n\n1. Use the starting guess ${X}^{\\left(0\\right)}=\\left[\\begin{array}{c}\\hfill 1\\hfill \\\\ \\hfill -1\\hfill \\end{array}\\right]$ in an implementation of the Jacobi method to show that ${X}^{\\left(1\\right)}=\\left[\\begin{array}{c}\\hfill 1.5\\hfill \\\\ \\hfill -3\\hfill \\end{array}\\right]$ . Find ${X}^{\\left(2\\right)}$ and ${X}^{\\left(3\\right)}$ .\n2. Use the starting guess ${X}^{\\left(0\\right)}=\\left[\\begin{array}{c}\\hfill 1\\hfill \\\\ \\hfill -1\\hfill \\end{array}\\right]$ in an implementation of the Gauss-Seidel method to show that ${X}^{\\left(1\\right)}=\\left[\\begin{array}{c}\\hfill 1.5\\hfill \\\\ \\hfill -3.25\\hfill \\end{array}\\right]$ . Find ${X}^{\\left(2\\right)}$ and ${X}^{\\left(3\\right)}$ .\n\n(Hint: it might help you to know that the exact solution is $\\left[\\begin{array}{c}\\hfill {x}_{1}\\hfill \\\\ \\hfill {x}_{2}\\hfill \\end{array}\\right]=\\left[\\begin{array}{c}\\hfill 3\\hfill \\\\ \\hfill -4\\hfill \\end{array}\\right]$ .)\n\n1. Show that the Jacobi iteration applied to the system\n\n$\\phantom{\\rule{2em}{0ex}}\\left[\\begin{array}{cccc}\\hfill 5\\hfill & \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill -1\\hfill & \\hfill 5\\hfill & \\hfill -1\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill -1\\hfill & \\hfill 5\\hfill & \\hfill -1\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill -1\\hfill & \\hfill 5\\hfill \\end{array}\\right]\\left[\\begin{array}{c}\\hfill {x}_{1}\\hfill \\\\ \\hfill {x}_{2}\\hfill \\\\ \\hfill {x}_{3}\\hfill \\\\ \\hfill {x}_{4}\\hfill \\end{array}\\right]=\\left[\\begin{array}{c}\\hfill 7\\hfill \\\\ \\hfill -10\\hfill \\\\ \\hfill -6\\hfill \\\\ \\hfill 16\\hfill \\end{array}\\right]$\n\ncan be written\n\n$\\phantom{\\rule{2em}{0ex}}{X}^{\\left(k+1\\right)}=\\left[\\begin{array}{cccc}\\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill \\end{array}\\right]{X}^{\\left(k\\right)}+\\left[\\begin{array}{c}\\hfill 1.4\\hfill \\\\ \\hfill -2\\hfill \\\\ \\hfill -1.2\\hfill \\\\ \\hfill 3.2\\hfill \\end{array}\\right].$\n\n2. Show that the method is certain to converge and calculate the first three iterations using\n\nzero starting values.\n\n(Hint: the exact solution to the stated problem is $\\left[\\begin{array}{c}\\hfill 1\\hfill \\\\ \\hfill -2\\hfill \\\\ \\hfill 1\\hfill \\\\ \\hfill 3\\hfill \\end{array}\\right]$ .)\n\n1. $\\phantom{\\rule{2em}{0ex}}2{x}_{1}^{\\left(1\\right)}=2-1{x}_{2}^{\\left(0\\right)}=2$\n\nand therefore ${x}_{1}^{\\left(1\\right)}=1.5$\n\n$\\phantom{\\rule{2em}{0ex}}2{x}_{2}^{\\left(1\\right)}=-5-1{x}_{1}^{\\left(0\\right)}=-6$\n\nwhich implies that ${x}_{2}^{\\left(1\\right)}=-3$ . These two values give the required entries in ${X}^{\\left(1\\right)}$ . A second and third iteration follow in a similar way to give\n\n$\\phantom{\\rule{2em}{0ex}}{X}^{\\left(2\\right)}=\\left[\\begin{array}{c}\\hfill 2.5\\\\ \\hfill -3.25\\end{array}\\right]\\phantom{\\rule{2em}{0ex}}\\text{and}\\phantom{\\rule{2em}{0ex}}{X}^{\\left(3\\right)}=\\left[\\begin{array}{c}\\hfill 2.625\\\\ \\hfill -3.75\\end{array}\\right]$\n\n2. $\\phantom{\\rule{2em}{0ex}}2{x}_{1}^{\\left(1\\right)}=2-1{x}_{2}^{\\left(0\\right)}=3$\n\nand therefore ${x}_{1}^{\\left(1\\right)}=1.5$ . This new approximation to ${x}_{1}$ is used straight away when finding a new approximation to ${x}_{2}^{\\left(1\\right)}$ .\n\n$\\phantom{\\rule{2em}{0ex}}2{x}_{2}^{\\left(1\\right)}=-5-1{x}_{1}^{\\left(1\\right)}=-6.5$\n\nwhich implies that ${x}_{2}^{\\left(1\\right)}=-3.25$ . These two values give the required entries in ${X}^{\\left(1\\right)}$ . A second and third iteration follow in a similar way to give\n\n$\\phantom{\\rule{2em}{0ex}}{X}^{\\left(2\\right)}=\\left[\\begin{array}{c}\\hfill 2.625\\\\ \\hfill -3.8125\\end{array}\\right]\\phantom{\\rule{2em}{0ex}}\\text{and}\\phantom{\\rule{2em}{0ex}}{X}^{\\left(3\\right)}=\\left[\\begin{array}{c}\\hfill 2.906250\\\\ \\hfill -3.953125\\end{array}\\right]$\n\nwhere ${X}^{\\left(3\\right)}$ is given to 6 decimal places\n\n1. In this case $D=\\left[\\begin{array}{cccc}\\hfill 5\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 5\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 5\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 5\\hfill \\end{array}\\right]$ and therefore ${D}^{-1}=\\left[\\begin{array}{cccc}\\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill \\end{array}\\right]$ .\n\nSo the iteration matrix $M\\phantom{\\rule{0.3em}{0ex}}=\\phantom{\\rule{0.3em}{0ex}}{D}^{-1}\\left[\\begin{array}{cccc}\\hfill 0\\hfill & \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill -1\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill -1\\hfill & \\hfill 0\\hfill & \\hfill -1\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill -1\\hfill & \\hfill 0\\hfill \\end{array}\\right]\\phantom{\\rule{0.3em}{0ex}}=\\phantom{\\rule{0.3em}{0ex}}\\left[\\phantom{\\rule{0.3em}{0ex}}\\begin{array}{cccc}\\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill \\end{array}\\phantom{\\rule{0.3em}{0ex}}\\right]$\n\nand that the Jacobi iteration takes the form\n\n$\\phantom{\\rule{2em}{0ex}}{X}^{\\left(k+1\\right)}=M{X}^{\\left(k\\right)}+{M}^{-1}\\left[\\begin{array}{c}\\hfill 7\\hfill \\\\ \\hfill -10\\hfill \\\\ \\hfill -6\\hfill \\\\ \\hfill 16\\hfill \\end{array}\\right]=\\left[\\begin{array}{cccc}\\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill \\\\ \\hfill 0\\hfill & \\hfill 0\\hfill & \\hfill 0.2\\hfill & \\hfill 0\\hfill \\end{array}\\right]{X}^{\\left(k\\right)}+\\left[\\begin{array}{c}\\hfill 1.4\\hfill \\\\ \\hfill -2\\hfill \\\\ \\hfill -1.2\\hfill \\\\ \\hfill 3.2\\hfill \\end{array}\\right]$\n\nas required.\n\n2. Using the starting values ${x}_{1}^{\\left(0\\right)}={x}_{2}^{\\left(0\\right)}={x}_{3}^{\\left(0\\right)}={x}_{4}^{\\left(0\\right)}=0$ , the first iteration of the Jacobi method gives $\\begin{array}{rcll}{x}_{1}^{1}& =& 0.2{x}_{2}^{0}+1.4=1.4& \\text{}\\\\ {x}_{2}^{1}& =& 0.2\\left({x}_{1}^{0}+{x}_{3}^{0}\\right)-2=-2& \\text{}\\\\ {x}_{3}^{1}& =& 0.2\\left({x}_{2}^{0}+{x}_{4}^{0}\\right)-1.2=-1.2\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}& \\text{}\\\\ {x}_{4}^{1}& =& 0.2{x}_{3}^{0}+3.2=3.2& \\text{}\\end{array}$\n\nThe second iteration is\n\n$\\begin{array}{rcll}{x}_{1}^{2}& =& 0.2{x}_{2}^{1}+1.4=1& \\text{}\\\\ {x}_{2}^{2}& =& 0.2\\left({x}_{1}^{1}+{x}_{3}^{1}\\right)-2=-1.96& \\text{}\\\\ {x}_{3}^{2}& =& 0.2\\left({x}_{2}^{1}+{x}_{4}^{1}\\right)-1.2=-0.96\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}& \\text{}\\\\ {x}_{4}^{2}& =& 0.2{x}_{3}^{1}+3.2=2.96& \\text{}\\end{array}$\n\nAnd the third iteration is\n\n$\\begin{array}{rcll}{x}_{1}^{3}& =& 0.2{x}_{2}^{2}+1.4=1.008& \\text{}\\\\ {x}_{2}^{3}& =& 0.2\\left({x}_{1}^{2}+{x}_{3}^{2}\\right)-2=-1.992\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{1em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}\\phantom{\\rule{2em}{0ex}}& \\text{}\\\\ {x}_{3}^{3}& =& 0.2\\left({x}_{2}^{2}+{x}_{4}^{2}\\right)-1.2=-1& \\text{}\\\\ {x}_{4}^{3}& =& 0.2{x}_{3}^{2}+3.2=3.008& \\text{}\\end{array}$", "date": "2022-11-29 00:51:45", "meta": {"domain": "github.io", "url": "https://bathmash.github.io/HELM/30_5_itrtv_mthds_systms_eqns-web/30_5_itrtv_mthds_systms_eqns-webse2.html", "openwebmath_score": 0.8845521211624146, "openwebmath_perplexity": 231.83796962038983, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429629196683, "lm_q2_score": 0.8652240791017536, "lm_q1q2_score": 0.8508120095333598}}
{"url": "https://au.mathworks.com/help/matlab/ref/cumtrapz.html", "text": "# cumtrapz\n\nCumulative trapezoidal numerical integration\n\n## Syntax\n\n``Q = cumtrapz(Y)``\n``Q = cumtrapz(X,Y)``\n``Q = cumtrapz(___,dim)``\n\n## Description\n\nexample\n\n````Q = cumtrapz(Y)` computes the approximate cumulative integral of `Y` via the trapezoidal method with unit spacing. The size of `Y` determines the dimension to integrate along:If `Y` is a vector, then `cumtrapz(Y)` is the cumulative integral of `Y`.If `Y` is a matrix, then `cumtrapz(Y)` is the cumulative integral over each column.If `Y` is a multidimensional array, then `cumtrapz(Y)` integrates over the first dimension whose size does not equal 1.```\n\nexample\n\n````Q = cumtrapz(X,Y)` integrates `Y` with respect to the coordinates or scalar spacing specified by `X`. If `X` is a vector of coordinates, then `length(X)` must be equal to the size of the first dimension of `Y` whose size does not equal 1.If `X` is a scalar spacing, then `cumtrapz(X,Y)` is equivalent to `X*cumtrapz(Y)`. ```\n\nexample\n\n````Q = cumtrapz(___,dim)` integrates along the dimension `dim` using any of the previous syntaxes. You must specify `Y`, and optionally can specify `X`. If you specify `X`, then it can be a scalar or a vector with length equal to `size(Y,dim)`. For example, if `Y` is a matrix, then `cumtrapz(X,Y,2)` cumulatively integrates each row of `Y`.```\n\n## Examples\n\ncollapse all\n\nCalculate the cumulative integral of a vector where the spacing between data points is 1.\n\nCreate a numeric vector of data.\n\n`Y = [1 4 9 16 25];`\n\n`Y` contains function values for $\\mathit{f}\\left(\\mathit{x}\\right)={\\mathit{x}}^{2}$ in the domain `[1 5]`.\n\nUse `cumtrapz` to integrate the data with unit spacing.\n\n`Q = cumtrapz(Y)`\n```Q = 1\u00d75 0 2.5000 9.0000 21.5000 42.0000 ```\n\nThis approximate integration yields a final value of 42. In this case, the exact answer is a little less, $41\\frac{1}{3}$. The `cumtrapz` function overestimates the value of the integral because f(x) is concave up.\n\nCalculate the cumulative integral of a vector where the spacing between data points is uniform, but not equal to 1.\n\nCreate a domain vector.\n\n`X = 0:pi/5:pi;`\n\nCalculate the sine of `X`.\n\n`Y = sin(X');`\n\nCumulatively integrate `Y` using `cumtrapz`. When the spacing between points is constant, but not equal to 1, an alternative to creating a vector for `X` is to specify the scalar spacing value. In that case, `cumtrapz(pi/5,Y)` is the same as `pi/5*cumtrapz(Y)`.\n\n`Q = cumtrapz(X,Y)`\n```Q = 6\u00d71 0 0.1847 0.6681 1.2657 1.7491 1.9338 ```\n\nCumulatively integrate the rows of a matrix where the data has a nonuniform spacing.\n\nCreate a vector of x-coordinates and a matrix of observations that take place at the irregular intervals. The rows of `Y` represent velocity data, taken at the times contained in `X`, for three different trials.\n\n```X = [1 2.5 7 10]; Y = [5.2 7.7 9.6 13.2; 4.8 7.0 10.5 14.5; 4.9 6.5 10.2 13.8];```\n\nUse `cumtrapz` to integrate each row independently and find the cumulative distance traveled in each trial. Since the data is not evaluated at constant intervals, specify `X` to indicate the spacing between the data points. Specify `dim = 2` since the data is in the rows of `Y`.\n\n`Q1 = cumtrapz(X,Y,2)`\n```Q1 = 3\u00d74 0 9.6750 48.6000 82.8000 0 8.8500 48.2250 85.7250 0 8.5500 46.1250 82.1250 ```\n\nThe result is a matrix of the same size as `Y` with the cumulative integral of each row.\n\nPerform nested integrations in the x and y directions. Plot the results to visualize the cumulative integral value in both directions.\n\nCreate a grid of values for the domain.\n\n```x = -2:0.1:2; y = -2:0.2:2; [X,Y] = meshgrid(x,y);```\n\nCalculate the function $\\mathit{f}\\left(\\mathit{x},\\mathit{y}\\right)=10{\\mathit{x}}^{2}+20{\\mathit{y}}^{2}$ on the grid.\n\n`F = 10*X.^2 + 20*Y.^2;`\n\n`cumtrapz` integrates numeric data rather than functional expressions, so in general the underlying function does not need to be known to use `cumtrapz` on a matrix of data. In cases where the functional expression is known, you can instead use `integral`, `integral2`, or `integral3`.\n\nUse `cumtrapz` to approximate the double integral\n\n`$I\\left(a,b\\right)={\\int }_{-2}^{b}{\\int }_{-2}^{a}\\left(10{x}^{2}+20{y}^{2}\\right)\\phantom{\\rule{0.16666666666666666em}{0ex}}dx\\phantom{\\rule{0.16666666666666666em}{0ex}}dy.$`\n\nTo perform this double integration, use nested function calls to `cumtrapz`. The inner call first integrates the rows of data, then the outer call integrates the columns.\n\n`I = cumtrapz(y,cumtrapz(x,F,2));`\n\nPlot the surface representing the original function as well as the surface representing the cumulative integration. Each point on the surface of the cumulative integration gives an intermediate value of the double integral. The last value in `I` gives the overall approximation of the double integral, `I(end) = 642.4`. Mark this point in the plot with a red star.\n\n```surf(X,Y,F,'EdgeColor','none') xlabel('X') ylabel('Y') hold on surf(X,Y,I,'FaceAlpha',0.5,'EdgeColor','none') plot3(X(end),Y(end),I(end),'r*') hold off```\n\n## Input Arguments\n\ncollapse all\n\nNumeric data, specified as a vector, matrix, or multidimensional array. By default, `cumtrapz` integrates along the first dimension of `Y` whose size does not equal 1.\n\nData Types: `single` | `double`\nComplex Number Support: Yes\n\nPoint spacing, specified as `1` (default), a uniform scalar spacing, or a vector of coordinates.\n\n\u2022 If `X` is a scalar, then it specifies a uniform spacing between the data points and `cumtrapz(X,Y)` is equivalent to `X*cumtrapz(Y)`.\n\n\u2022 If `X` is a vector, then it specifies x-coordinates for the data points and `length(X)` must be the same as the size of the integration dimension in `Y`.\n\nData Types: `single` | `double`\n\nDimension to operate along, specified as a positive integer scalar. If you do not specify the dimension, then the default is the first array dimension of size greater than 1.\n\nConsider a two-dimensional input array, `Y`:\n\n\u2022 `cumtrapz(Y,1)` works on successive elements in the columns of `Y`.\n\n\u2022 `cumtrapz(Y,2)` works on successive elements in the rows of `Y`.\n\nIf `dim` is greater than `ndims(Y)`, then `cumtrapz` returns an array of zeros of the same size as `Y`.\n\n## Tips\n\n\u2022 Use `trapz` and `cumtrapz` to perform numerical integrations on discrete data sets. Use `integral`, `integral2`, or `integral3` instead if a functional expression for the data is available.\n\n\u2022 `trapz` reduces the size of the dimension it operates on to 1, and returns only the final integration value. `cumtrapz` also returns the intermediate integration values, preserving the size of the dimension it operates on.\n\n## Version History\n\nIntroduced before R2006a", "date": "2022-10-04 17:55:08", "meta": {"domain": "mathworks.com", "url": "https://au.mathworks.com/help/matlab/ref/cumtrapz.html", "openwebmath_score": 0.8963515162467957, "openwebmath_perplexity": 781.9710638136748, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429604789206, "lm_q2_score": 0.8652240808393984, "lm_q1q2_score": 0.8508120091302669}}
{"url": "https://math.stackexchange.com/questions/3344853/elegant-way-to-assign-4-distinct-objects-into-4-bins", "text": "# \u201cElegant\u201d way to assign 4 distinct objects into 4 bins\n\nSuppose you have four objects to distribute among four people. Suppose people can carry any number of objects (none, one, two, three, or all four objects) at once. In how many ways can the four objects be distributed?\n\nMy solution\n\nLet the four people be named A,B,C,D.\n\nDistribution type 1\n\nIf each person gets exactly one object, the distribution looks like\n\nA B C D\n1 1 1 1\n\n\nof which there are $$4!=4\\cdot 3 \\cdot 2 \\cdot 1 = 24$$ distinct ways (since the objects are distinct).\n\nDistribution type 2\n\nIf one person gets 2 objects and two others get 1 object, the possible distributions look like\n\nA B C D\n2 1 1 0\n2 1 0 1\n2 0 1 1\n1 2 1 0\n1 2 0 1\n0 2 1 1\n1 1 2 0\n1 0 2 1\n0 1 2 1\n1 1 0 2\n1 0 1 2\n0 1 1 2\n\n\nand for each of the above distributions, there are $$12$$ distinct ways to distribute the distinct objects.\n\nDistribution type 3\n\nIf two people get 2 objects, the possible distributions look like\n\nA B C D\n2 2 0 0\n2 0 2 0\n2 0 0 2\n0 2 2 0\n0 2 0 2\n0 0 2 2\n\n\nand for each of the above distributions, there are $$6$$ distinct ways to distribute the distinct objects.\n\nDistribution type 4\n\nIf one person gets 3 objects and another person gets 1 object, the possible distributions look like\n\nA B C D\n3 1 0 0\n3 0 1 0\n3 0 0 1\n1 3 0 0\n0 3 1 0\n0 3 0 1\n1 0 3 0\n0 1 3 0\n0 0 3 1\n1 0 0 3\n0 1 0 3\n0 0 1 3\n\n\nand for each of the above distributions, there are $$4$$ distinct ways to distribute the distinct objects.\n\nDistribution type 5\n\nIf one person gets all 4 objects, the possible distributions look like\n\nA B C D\n4 0 0 0\n0 4 0 0\n0 0 4 0\n0 0 0 4\n\n\nfor each of the above distributions, there is only $$1$$ distinct way to distribute the distinct objects.\n\nTotaling the ways\n\nThe total number of ways is then\n\n\\begin{align} &\\textrm{ type 1 + type 2 + type 3 + type 4 + type 5 }\\\\ =& 1(24) + 12(12) + 6(6) + 12(4) + 4(1) \\\\ =& 24 + 144 + 36 + 48 + 4 \\\\ =& 256\\textrm{ ways } \\end{align}\n\nMy questions\n\nI have 3 questions:\n\n1. Is there a more elegant way (using $$_n P_k$$ or $$_n C_k$$ notation) of counting the amount of each distribution type (the numbers 1, 12, 6, 12, 4) that appear in the final calculation?\n\n2. Is there a more elegant way (using $$_n P_k$$ or $$_n C_k$$ notation) of counting the number of ways for each distribution type (the numbers 24, 12, 6, 4, 1) that appear in the final calculation in parentheses? (Clearly the 24 is just $$_4 P_4$$, but what about the others?)\n\n3. Why is all of this just equal to $$4^4$$? This isn't immediately obvious to me.\n\n\u2022 @EdwardH., do you have a proof of that? \u2013\u00a0Charles Hudgins Sep 5 '19 at 5:00\n\u2022 Each object can be given to any person, hence $N_p^{N_o}=4^4$ ways. \u2013\u00a0Yves Daoust Sep 5 '19 at 7:25\n\n1. Breaking it up a little differently, the number of ways with $$k = 0,1,2,3$$ of them getting $$0$$ is $$_4C_k \\cdot {}_3C_{3-k}$$, which gives the sequence $$1, 12, 18 = 6 + 12, 4$$. Where the difference is that you had the $$6$$ ways of $$2,2,0,0$$ and the $$12$$ ways of $$3,1,0,0$$ separately. Those are $$\\frac{4!}{2!2!}$$ and $$\\frac{4!}{1!1!2!}$$, respectively.\n2. For the arrangements, you are looking at the multinomial coefficients: $$\\frac{4!}{1!1!1!1!} = 24, \\frac{4!}{2!1!1!0!} = 12, \\frac{4!}{2!2!0!0!} = 6, \\frac{4!}{3!1!0!0!} = 4, \\frac{4!}{4!0!0!0!} = 1$$\n3. Instead of focusing on the people and which objects they get, look at the objects and who they are given to. Each object can be given to any of the four people, without restriction. That means there are $$4$$ options for the first object, $$4$$ for the second, and so on, so the total is $$4^4$$.\n\u2022 Thank you, Michael! \u2013\u00a0Mathemanic Sep 8 '19 at 23:18\n\nLet the objects choose the people . . .\n\nEach object has $$4$$ choices, and each object's choices are independent of the choices made by the other objects, hence there are $$4^4$$ ways for the objects to choose the people.\n\nAlternatively, using your cases . . .\n\n\u2022 For distribution type $$1$$, the number of ways is $$4!=24$$ Explanation:\nIf we have the people line up to choose, then person $$1$$ has $$4$$ choices, person $$2$$ has $$3$$ choices, person $$3$$ has $$2$$ choices, and person $$4$$ has $$1$$ choice.\n\n\u2022 For distribution type $$2$$, the number of ways is $$\\binom{4}{1}\\binom{4}{2}\\binom{3}{2}2!=144$$ Explanation:\n\u2022 Choose the person who gets two objects:$$\\;{\\large{\\binom{4}{1}}}\\;$$choices.\n\u2022 Choose the two objects for that person:$$\\;{\\large{\\binom{4}{2}}}\\;$$choices.\n\u2022 Choose the two people to get the two remaining objects:$$\\;{\\large{\\binom{3}{2}}}\\;$$choices.\n\u2022 Distribute the two remainining objects, one to each of the two chosen people:$$\\;2!\\;$$choices.\n\n\u2022 For distribution type $$3$$, the number of ways is $$\\binom{4}{2}\\binom{2}{1}\\binom{3}{1}=36$$ Explanation:\n\u2022 Choose the two people who get two objects:$$\\;{\\large{\\binom{4}{2}}}\\;$$choices.\n\u2022 Choose the person who gets the object labeled #$$1$$ plus one other object:$$\\;{\\large{\\binom{2}{1}}}\\;$$choices.\n\u2022 Choose the other object for that person:$$\\;{\\large{\\binom{3}{1}}}\\;$$choices.\n\n\u2022 For distribution type $$4$$, the number of ways is $$\\binom{4}{1}\\binom{4}{3}\\binom{3}{1}=48$$ Explanation:\n\u2022 Choose the person who gets three objects:$$\\;{\\large{\\binom{4}{1}}}\\;$$choices.\n\u2022 Choose the three objects for that person:$$\\;{\\large{\\binom{4}{3}}}\\;$$choices.\n\u2022 Choose the person to get the one remaining object:$$\\;{\\large{\\binom{3}{1}}}\\;$$choices.\n\n\u2022 For distribution type $$5$$, the number of ways is $$\\binom{4}{1}=4$$ Explanation:$$\\;$$Choose the person who gets all four objects:$$\\;{\\large{\\binom{4}{1}}}\\;$$choices.\n\n$$\\;\\;\\;$$Summing the counts for the cases, the total number of ways is $$24+144+36+48+4=256$$\n\n\u2022 Awesome and very thorough answer - thank you Quasi! \u2013\u00a0Mathemanic Sep 8 '19 at 23:19", "date": "2020-02-19 22:46:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3344853/elegant-way-to-assign-4-distinct-objects-into-4-bins", "openwebmath_score": 0.7997000813484192, "openwebmath_perplexity": 137.28949745242326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429599907709, "lm_q2_score": 0.8652240791017535, "lm_q1q2_score": 0.8508120069992072}}
{"url": "https://math.stackexchange.com/questions/1889632/question-about-spanning-sets-and-bases", "text": "Question about spanning sets and bases\n\nIn a textbook is the question:\n\nFind a basis for the subspace $$V=\\{(x_1,x_2,x_3,x_4):x_1+2x_2+x_3+x_4=0, 3x_1+6x_2+4x_3+x_4=0\\}.$$\n\nThey say that $V$ is a subspace of $\\mathbb{R^4}$ and are able to find a spanning set by solving the system of homogeneous equations getting: $$\\{\\begin{bmatrix}-2 \\\\ 1 \\\\ 0 \\\\ 0 \\end{bmatrix},\\begin{bmatrix}-3 \\\\ 0 \\\\ 2 \\\\ 1 \\end{bmatrix}\\}.$$ They then show that these vectors are linearly independent and so it forms a basis for $V$.\n\nI thought though that if you're in $\\mathbb{R^4}$ then you need at least 4 vectors to span the space, and 4 to form a basis. Here though we only have two. Could someone please tell me where my logic went wrong and why these indeed span the subspace and form a basis?\n\n\u2022 They ask you to find a basis for a subspace, not for the entire ${\\bf R}^4$. \u2013\u00a0avs Aug 11 '16 at 23:18\n\nYou do indeed need at least 4 vectors to span $\\mathbb{R}^4$. However, here we're not trying to span $\\mathbb{R}^4$, we're trying to span $V$.\nFor a more concrete example: think about the subspace $W=\\{(a, b): a=0\\}$ of $\\mathbb{R}^2$ (basically, $W$ is the $y$-axis). Clearly $W$ is spanned by a one-element set - e.g., $\\{(0, 1)\\}$ - even though it takes two vectors to span $\\mathbb{R}^2$. Does this help?\nNoah's answer is definitely sufficient, but I'm going to try to add a bit of algebraic insight for you. Write $v=(1,2,1,1)$ and $w=(3,6,4,1)$. Let $f,g: \\mathbb{R}^4 \\rightarrow \\mathbb{R}$, $f:x \\mapsto v \\cdot x$ and $g: x \\mapsto w \\cdot x$.\n$f,g$ are linear maps and $\\mathbb{R}$ is a field, so they are either surjective or trivial. Note that we have $V= \\ker(f) \\cap \\ker(g)$. If $\\dim(V)=4$, then $\\mathbb{R}^4/V$ is trivial and both $f,g$ must be the zero map. Hopefully you can see this is clearly not the case.", "date": "2021-05-12 05:29:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1889632/question-about-spanning-sets-and-bases", "openwebmath_score": 0.9509533643722534, "openwebmath_perplexity": 86.01359508476192, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429599907709, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8508120052905064}}
{"url": "https://www.themathdoctors.org/equivalent-definitions-of-e/", "text": "# Equivalent Definitions of e\n\n#### (A new question of the week)\n\nIt is not unusual for mathematicians to define a concept in multiple ways, which can be proved to be equivalent. One definition may lead to a theorem, which another presentation uses as the definition, from which the original definition can be proved as a theorem. Here, in yet another good question from late May, we have two different ways to \u201cdefine\u201d the number $$e=2.71828\u2026$$, a series and a limit, and a student wants to prove directly that they are equivalent. We\u2019ll get a proof, then dig in to really understand it.\n\n## Proving two definitions are equivalent\n\nHi!\n\nI\u2019m looking for a rigorous proof that the following definitions of e are equivalent:\n\ne = 1 + 1 + 1/2! + 1/3! + 1/4! + \u2026\n\ne = lim [n\u2192\u221e] (1 + 1/n)^n\n\nWhat I\u2019ve done so far:\n\nI understand that:\n\nlim [n\u2192\u221e] (1 + 1/n)^n = 1 + 1 + (1 \u2013 1/n) 1/2! + (1 \u2013 1/n)(1 \u2013 2/n) 1/3! + (1 \u2013 1/n)(1 \u2013 2/n)(1 \u2013 3/n) 1/4! + \u2026\n\nI also understand this:\n\nlim [n\u2192\u221e] (1 \u2013 1/n)(1 \u2013 2/n)(1 \u2013 3/n) = 1 x 1 x 1 = 1\n\nAnd so if we take a fixed value m (with m as a natural number):\n\nlim [n\u2192\u221e] 1 + 1 + (1 \u2013 1/n) 1/2! + (1 \u2013 1/n)(1 \u2013 2/n) 1/3! + (1 \u2013 1/n)(1 \u2013 2/n)(1 \u2013 3/n) 1/4! + \u2026 + (1 \u2013 1/n)(1 \u2013 2/n)(1 \u2013 3/n)\u2026(1 \u2013 (m \u2013 1)/n) 1/m!\n\nis equivalent to:\n\n1 + 1 + 1/2! + 1/3! + 1/4! + \u2026 + 1/m!\n\nBut I then get stuck on the next step. I only understand the \u2018equivalence\u2019 where the series terminates at a fixed natural number (in this case m). How do I make the transition to proving it for the infinite series, where the number of members of the series approaches infinity, whilst n also approaches infinity within each element of the series?\n\nI\u2019m in particular looking for a rigorous proof, where the two sequences are named t_n and s_n.\n\nSo that for every epsilon there exists a number N so that for every n > N then |t_n \u2013 s_n| < epsilon.\n\nReally grateful for any help on this!\n\nMatthew knows how to ask a question: He has clearly stated what he wants to do, and shown a good bit of work, including where the difficulty lies. But we\u2019ll need to get up to speed to make sure we understand the question, as well as this initial work!\n\n### What we need to prove\n\nThe two \u201cdefinitions\u201d are quite different, one a series, the other a limit. Here is how Wikipedia introduces its article on e:\n\nThe number\u00a0e, also known as\u00a0Euler\u2019s number, is a mathematical constant approximately equal to 2.71828, and can be characterized in many ways. It is the base of the natural logarithm. It is the limit of $$\\left(1+\\frac{1}{n}\\right)^n$$\u00a0 as\u00a0n\u00a0approaches infinity, an expression that arises in the study of\u00a0compound interest. It can also be calculated as the sum of the infinite\u00a0series\n\n$$e=\\sum_{n=0}^{\\infty }{\\frac {1}{n!}}=1+{\\frac {1}{1}}+{\\frac {1}{1\\cdot 2}}+{\\frac {1}{1\\cdot 2\\cdot 3}}+\\cdots$$\n\nThat is, they take the limit as the actual definition, and present the series as a way to calculate it. Here is a table of the first 11 terms of the series and its sums:\n\nWe have already reached 8 significant digits of accuracy.\n\nHere is a table of the first 11 values of the limit:\n\nWe don\u2019t even have one significant digit yet!\n\nIn fact, if we make the limit move exponentially faster, by using $$2^n$$ in place of n, it is still slow to converge:\n\nIt takes 8 thousand \u201csteps\u201d to get 4 significant digits! (Of course, in principle we only need to do this calculation once, rather than summing terms, so it is not that inefficient to calculate, apart from the efficiency of using a very large exponent in the first place, and deciding what value to use.)\n\nBut they do approach the same limit. How can we prove it?\n\n### What he\u2019s done so far\n\nLet\u2019s look at what Matthew did, which he didn\u2019t fully explain.\n\nFirst, he said $$\\lim_{n\\to\\infty} \\left(1 + \\frac{1}{n}\\right)^n = 1 + 1 + \\left(1 \u2013 \\frac{1}{n}\\right) \\frac{1}{2!} + \\left(1 \u2013 \\frac{1}{n}\\right)\\left(1 \u2013 \\frac{2}{n}\\right) \\frac{1}{3!} + \\left(1 \u2013 \\frac{1}{n}\\right)\\left(1 \u2013 \\frac{2}{n}\\right)\\left(1 \u2013 \\frac{3}{n}\\right) \\frac{1}{4!} + \\cdots$$\n\nWhere did that come from? He has applied the binomial theorem to $$\\left(1 + \\frac{1}{n}\\right)^n$$; this theorem in general says that $$(1+x)^n = \\sum_{k=0}^n {_nC_k}x^k$$\n\nThe kth term of this sum (starting with the 0th) is $${_nC_k}\\left(\\frac{1}{n}\\right)^k = \\frac{n!}{k!(n-k)!}\\frac{1}{n^k} =$$ $$\\frac{n(n-1)(n-2)\\cdots(n-k+2)(n-k+1)}{n^k}\\frac{1}{k!} =$$ $$\\frac{n}{n}\\cdot\\frac{n-1}{n}\\cdot\\frac{n-2}{n}\\cdots\\frac{n-k+2}{n}\\cdot\\frac{n-k+1}{n}\\frac{1}{k!} =$$ $$\\left(1-\\frac{0}{n}\\right)\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{k-2}{n}\\right)\\left(1-\\frac{k-1}{n}\\right)\\frac{1}{k!}$$\n\nwhich agrees with what he wrote; observe that for $$k=0$$ the term has zero factors, and is simply 1; for $$k=1$$, it is just the one factor $$\\left(\\frac{n}{n}\\right) = 1$$. That\u2019s why Matt has written the first two terms as mere numbers (and will continue doing so): for clarity.\n\nHis calling this the limit, however, is premature, because n is still a variable. This is the tricky part.\n\n## The proof\n\nDoctor Fenton answered, starting his proof the same way but being more careful with limits:\n\nI recalled a discussion of this in an old classic, Richard Courant\u2019s\u00a0Calculus\u00a0textbook.\n\nLet\u00a0Tn denote the n-th partial sum of the series $$\\sum_{k=0}^\\infty\\frac{1}{k!}$$ , so $$T_n=\\sum_{k=0}^n\\frac{1}{k!}$$,\n\nand let\u00a0Sn denote $$(1+\\frac{1}{n})^n$$\u00a0.\n\nBy the Binomial Theorem,\n\n$$S_n= \\sum_{k=0}^n {_nC_k}\\frac{1}{n^k}= \\sum_{k=0}^n \\frac{n!}{k!(n-k)!}\\frac{1}{n^k}= \\sum_{k=0}^n \\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{k-1}{n}\\right)$$\n\nClearly, Sn\u00a0\u2264 Tn, since each term in Tn is multiplied by a product of factors, each less than 1, to obtain the corresponding term of Sn.\n\nAlso, notice that both sequences are increasing and bounded, so each converges.\u00a0 Let\n\n$$S=\\lim_{n\\to\\infty} S_n$$\u00a0.\n\nIf m < n, then the sum $$\\sum_{k=0}^m \\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{m-1}{n}\\right)\\lt S_n$$\n\nsince Sn\u00a0has additional non-negative terms added.\u00a0 If we take the limit as n\u2192\u221e, the left side approaches Tm\u00a0, while the right side approaches the limit S.\u00a0 Then we have that Tm\u00a0\u2264 S, so\u00a0 \u00a0 Sm\u00a0\u2264 Tm\u00a0\u2264 S.\n\nNow, taking the limit as m\u2192\u221e gives S = T, so the two limits are the same.\n\nThere\u2019s a typo in that summation, which I can\u2019t remove because it figures into the discussion; don\u2019t worry if you find it.\n\n### Fixing an error\n\nMatthew replied, catching that error and helping us out in the process:\n\nHi, thanks so much for your quick answer. I think I\u2019m getting closer but there\u2019s still some things I\u2019m not getting.\n\nI think when you wrote $$\\sum_{k=0}^m \\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{m-1}{n}\\right)\\lt S_n$$\n\nyou must have meant: $$\\sum_{k=0}^m \\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{k-1}{n}\\right)\\lt S_n$$\n\nOr maybe: $$1+1+\\sum_{k=2}^m \\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{k-1}{n}\\right)\\lt S_n$$\n\nIf I expand this last series on the left I get:\n\n1 +\n\n1 +\n\n(1/2!){1-1/n} +\n\n(1/3!){(1-1/n)(1-2/n)} +\n\n\u2026 +\n\n(1/m!) {(1-1/n)(1-2/n)\u2026(1-(m-1)/n)}\n\nSo then Sn has all these terms, but some additional terms added.\n\nAm I making sense? Sorry if I\u2019ve got this confused somewhere!\n\nThanks again!\n\nThe \u201cor maybe\u201d is really the same summation, with the first two terms stated explicitly as we saw before.\n\n### Clarifying the limit\n\nDoctor Fenton confirmed his correction, and added some notation to make the details easier to talk about:\n\nYou are correct.\u00a0 I was hurrying to get ready for a meeting and typed m instead of k.\n\nThe idea is to take a sum of a fixed number m of terms from Sn, so that the partial sum Sn\u00a0is larger than the sum of its first m terms.\n\nCall this sum Sn,m\u00a0, so Sn,m\u00a0< Sn\u00a0, and as n\u2192\u221e, Sn,m\u2192Tm\u00a0 while Sn\u2192S, giving Tm\u00a0\u2264 S.\u00a0 Then Tm\u00a0is squeezed between Sm\u00a0and S.\n\nSorry to put\u00a0you to so much work over a typo.\n\nThat is, we are defining $$S_{n,m} = \\sum_{k=0}^m \\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{k-1}{n}\\right)$$ and $$S_{n,m}< S_n$$ for all n, for any given m, so, taking the limit on each side, $$T_m\\le S$$. Giving a name to the sum of only m terms of the sum for n makes it easier to talk precisely about what we are doing.\n\nMatthew wrote back, nicely stating what he did and did not understand:\n\nHi there, many thanks for your further response and for clarifying.\n\nI think I now understand the proof in general terms but I\u2019m not sure I fully understand all the details. In particular I\u2019m not clear exactly how it follows that Tm\u00a0< S.\n\nWe have the following inequalities:\n\nSn,m\u00a0< Sn\u00a0for all values of n (as Salways includes additional non-negative terms)\n\nSn\u00a0< S for all values of n (as Sn\u00a0is monotonically increasing, approaching the limit S)\n\nSn,m\u00a0< \u00a0Tfor all values of n (as Sn,m\u00a0contains the same terms as Tterms except with further positive coefficients less than 1)\n\nI\u2019m not sure I can get from these inequalities that Tm\u00a0< S\n\nBut maybe if Sn,m\u00a0approaches the limit of Tas n\u2192\u221e,\u00a0and Sn,m\u00a0remains less than Swithout approaching the limit of Sn,\u00a0then I think it would follow that Tm\u00a0is less than Sn.\n\nAnd the I think the rest would follow as m \u2192 \u221e, as Tis \u2018squeezed\u2019 between Sand S. i.e. Sm\u00a0< Tm\u00a0< S, and Sm \u2192 S as m \u2192 \u221e.\n\nI\u2019m not sure how I can show that Sn,m\u00a0does not approach the limit of Sn as m \u2192 \u221e, although it seems highly intuitive to assume it does not.\n\nAgain, hope I\u2019m making sense and not getting mixed up, or if I\u2019m missing something obvious.\n\nThanks again!\n\nIt can be helpful to look at some actual numbers. Here is a table of values of our $$S_{n,m}$$ showing the inequalities Matt points out:\n\nI have highlighted $$S_{4,3} = 2.43750$$; we can see that\n\n\u2022 $$S_{4,3} = 2.43750 < S_4 = 2.44141$$\n\u2022 $$S_4 = 2.44141 < S = 2.71828$$\n\u2022 $$S_{4,3} = 2.43750 < T_3 = 2.66667$$\n\nBut these inequalities in themselves don\u2019t lead us to the conclusion that $$T_m$$ and $$S_n$$ have the same limit.\n\nDoctor Fenton showed the missing link:\n\nYou write\n\nI\u2019m not sure I can get from these inequalities that Tm < S.\n\nBut maybe if Sn,m\u00a0approaches the limit of Tm as n\u2192\u221e\u00a0and Sn,m\u00a0remains less than Sn\u00a0without approaching the limit of Sn,\u00a0then I think it would follow that Tm\u00a0is less than Sn\n\nThat\u2019s exactly correct.\u00a0 Sn,m\u00a0is the sum of the first m terms of Sn, and in this part of the argument, m is fixed.\n\nSn,m = 1 + (1/1!) + (1/2!)(1-1/n) + \u2026 + (1/m!)(1-1/n)(1-2/n)\u22c5\u22c5\u22c5(1-(m-1)/n)\n\nwhile\n\nSn = 1 + 1/1! + (1/2!)(1-1/n) + \u2026 + (1/n!)(1-1/n)(1-2/n)\u22c5\u22c5\u22c5(1-(n-1)/n)\u00a0 .\n\nSn,m\u00a0always has m terms, a fixed number, while the number of terms in Sn\u00a0increases without bound.\u00a0 As n\u2192\u221e, each term (1-k/n)\u21921, so the kth\u00a0term of Sn,m\u00a0approaches 1/k!, i.e. lim(n\u2192\u221e) Sn,m\u00a0= Tm\u00a0.\u00a0 But the number of terms in Sn\u00a0keeps increasing, so that Sn approaches S, which turns out to be T = e.\n\nHis table would differ in just one point, taking the limit of each column rather than just an inequality:\n\nThe fact that we are holding m fixed and then letting n vary gives us a grip on the limits.\n\n## Final thoughts\n\nMatthew now worked out the final bit:\n\nHi \u2013 thanks for your encouraging response.\n\nI\u2019ve been wrestling with this part of the proof:\n\n\u201cIf m < n, then the sum $$\\sum_{k=0}^m \\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right)\\cdots\\left(1-\\frac{m-1}{n}\\right)\\lt S_n$$\n\nIf we take the limit as n\u2192\u221e, the left side approaches Tm\u00a0, while the right side approaches the limit S.\u00a0 Then we have that Tm\u00a0\u2264 S\u201d\n\nI think I\u2019ve got it now!\n\nIf I say an < bfor all n, and say as n\u2192\u221e then an\u2192a, and bn\u2192b,\n\nthen it follows that a \u2264 b.\n\nFor if no member of the series of an\u00a0is greater than any member of the series bn,\u00a0then it is not possible that a > b.\n\nBut it is possible that a = b, as it could be that b\u2013 an\u21920 as n\u2192\u221e.\n\nAnd it is possible that a < b as it could be that (lim n\u21920 b\u2013 an)\u00a0\u00a0\u00a0> 0.\n\nHence a \u2264 b.\n\nSo replace awith Sn,m\u00a0and bwith Sn, then given that Sn,m\u2192Tand Sn\u2192S when n\u2192\u221e and Sn,m < Sfor all n it follows that Tm\u00a0\u2264 S.\n\nAnd the rest follows!\n\nSo think I\u2019m home and dry with this one!\u00a0? (assuming my above reasoning is correct \u2026\u00a0?)\n\nThanks again for taking the time to read and respond.\n\nWell explained! If $$a_n\\to a$$, $$b_n\\to b$$, and $$a_n<b_n$$ for all n, we can\u2019t be sure that $$a<b$$, but we do know that $$a\\le b$$.\n\nDoctor Fenton agreed:\n\nThat\u2019s correct!\n\nSo we have the proof.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.", "date": "2022-05-18 02:48:08", "meta": {"domain": "themathdoctors.org", "url": "https://www.themathdoctors.org/equivalent-definitions-of-e/", "openwebmath_score": 0.8998819589614868, "openwebmath_perplexity": 763.0343003039004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429619433693, "lm_q2_score": 0.865224073888819, "lm_q1q2_score": 0.8508120035625398}}
{"url": "http://mathhelpforum.com/number-theory/152607-units-ones-digit.html", "text": "# Math Help - units/ones digit\n\n1. ## units/ones digit\n\nHi all,\n\nif given a number like $56^{34}$, how would I find the units digit without multiplying the number out? What about for expressions such as $(867)\\times (563)\\times (y-2)$?\n\nWhat is the method?\n\n2. Originally Posted by sfspitfire23\nHi all,\n\nif given a number like $56^{34}$, how would I find the units digit without multiplying the number out? What about for expressions such as $(867)\\times (563)\\times (y-2)$?\n\nWhat is the method?\nWe are interested in 56^34 (mod 10). Think mod 2 and mod 5. What do you notice?\n\nWe are interested in (867)(563)(y-2) mod 10. (I assume y greater than or equal to 2. If y less than 2, still think mod 10 but make appropriate adjustment.) Reduce the first two factors mod 10. What do you notice?\n\n3. Hm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place\n\nWe have (7)(3)(y-2) (mod 10) So ones place is 1?\n\n4. Why mod 10 though?\n\n5. Originally Posted by sfspitfire23\nHm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place\nThe conclusion is correct but you did not explain how you got there, so maybe you took a long way or made a lucky guess and I have no way of knowing. Here are listed many steps, but of course you can do it in your head without writing any steps.\n\n$56^{34} \\equiv 0^{34} \\equiv 0 \\pmod{2}$\n\n$56^{34} \\equiv 1^{34} \\equiv 1 \\pmod{5}$\n\n$\\implies 56^{34} \\equiv 6 \\pmod{10}$\n\nOriginally Posted by sfspitfire23\nWe have (7)(3)(y-2) (mod 10)\nRight, but keep going. What is 7*3 reduced mod 10?\n\nOriginally Posted by sfspitfire23\nSo ones place is 1?\nHow do you get this? This fails for many y, including y = 2.\n\nOriginally Posted by sfspitfire23\nWhy mod 10 though?\nWhat do you think?\n\n6. Did you find what made the expression equivalent to 0, then to 1, and multiplied them together?\n\n21(y-2) is equivalent to 1(y-8) (mod 10).\n\n7. Hello, sfspitfire23!\n\nFind the units-digit of: . $56^{34}$\n\nDid you crank out a few powers of 56?\n\n. . $\\begin{array}{ccc}\n56^1 &=& 5\\boxed{\\!6\\!} \\\\ \\\\[-4mm]\n56^2 &=& 313\\boxed{\\!6\\!} \\\\ \\\\[-4mm]\n56^3 &=& 175,\\!61\\boxed{\\!6\\!} \\\\ \\\\[-4mm]\n56^4 &=& 9,\\!834,\\!49\\boxed{\\!6\\!} \\\\\n\\vdots && \\vdots \\end{array}$\n\nDo you see a pattern?\n\n8. Originally Posted by sfspitfire23\nDid you find what made the expression equivalent to 0, then to 1, and multiplied them together?\nI don't know what you mean. First of all, which expression are we talking about, 56^34 or (867)(563)(y-2)?\n\nFor 56^34, I use the Chinese remainder theorem. The algorithm is unnecessary because the numbers are so small; just ask, what even number is there belonging to {0,...,9} that is congruent to 1 (mod 5)? Or, consider all numbers in {0,...,9} congruent to 1 (mod 5), which are {1,6}, and take the even number.\n\nOriginally Posted by sfspitfire23\n21(y-2) is equivalent to 1(y-8) (mod 10).\nWhy did you change y-2 to y-8? It is not equivalent.\n\nWe have 21(y-2) is congruent to 1(y-2) which is just y-2, so one way to proceed is to choose y' congruent to y (mod 10) such that y' is in the set {2,...,11}, then the units digit is y' - 2. Another way is to take the common residue of y, call it for example z, then if z < 2 take z+8, otherwise take z-2. (That is assuming y greater than or equal to 2.)\n\nAdditional note: To formalise Soroban's post, we have $6^2 \\equiv 6 \\pmod{10}$ so the last digit must be 6; this is a bit faster and simpler than what I recommended, but either way with experience you can get the answer in not more than a few seconds.\n\n9. Originally Posted by sfspitfire23\nWhy mod 10 though?\n\nNotice that every positive integer $N$can be represented uniquely in the follwing way: $N=c_n10^n+c_{n-1}10^{n-1}+\\cdots +c_110^1+c_0$, where $0\\leq c_i< 10$ ( $i=0,1,...,n$) and $c_n\\neq 0$. Here, the $c_i$ are called the digits, and the expansion $c_n10^n+c_{n-1}10^{n-1}+\\cdots +c_110^1+c_0$ is written as $c_nc_{n-1}... c_1c_0$ in short.\n\nFor example, $531=5\\cdot10^2+3\\cdot10+1$.\n\nBecause $c_i10^i\\equiv 0(mod\\ 10)$ for $i>0$, we have $N\\equiv c_0(mod\\ 10)$. This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, $c_0$.\n\n10. Originally Posted by melese\nNotice that every positive integer $N$can be represented uniquely in the follwing way: $N=c_n10^n+c_{n-1}10^{n-1}+\\cdots +c_110^1+c_0$, where $0\\leq c_i< 10$ ( $i=0,1,...,n$) and $c_n\\neq 0$. Here, the $c_i$ are called the digits, and the expansion $c_n10^n+c_{n-1}10^{n-1}+\\cdots +c_110^1+c_0$ is written as $c_nc_{n-1}... c_1c_0$ in short.\nFor example, $531=5\\cdot10^2+3\\cdot10+1$.\nBecause $c_i10^i\\equiv 0(mod\\ 10)$ for $i>0$, we have $N\\equiv c_0(mod\\ 10)$. This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, $c_0$.", "date": "2015-09-05 06:07:40", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/number-theory/152607-units-ones-digit.html", "openwebmath_score": 0.8679078817367554, "openwebmath_perplexity": 670.5515401118869, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9833429614552197, "lm_q2_score": 0.865224073888819, "lm_q1q2_score": 0.850812003140181}}
{"url": "http://www.7cloudtech.com/can-you-eubhw/173638-biconditional-statement-truth-table", "text": "Compare the statement R: (a is even) $$\\Rightarrow$$ (a is divisible by 2) with this truth table. It is a combination of two conditional statements, \u201cif two line segments are congruent then they are of equal length\u201d and \u201cif two line segments are of equal length then they are congruent\u201d. 0. This form can be useful when writing proof or when showing logical equivalencies. Examples. Is this sentence biconditional? In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. For example, the propositional formula p \u2227 q \u2192 \u00acr could be written as p /\\ q -> ~r, as p and q => not r, or as p && q -> !r. NCERT Books. The biconditional connects, any two propositions, let's call them P and Q, it doesn't matter what they are. A biconditional statement is really a combination of a conditional statement and its converse. Watch Queue Queue BOOK FREE CLASS; COMPETITIVE EXAMS. \". Thus R is true no matter what value a has. When we combine two conditional statements this way, we have a biconditional. (true) 2. I am breathing if and only if I am alive. Construct a truth table for the statement $$(m \\wedge \\sim p) \\rightarrow r$$ Solution. I'll also try to discuss examples both in natural language and code. You passed the exam if and only if you scored 65% or higher. Similarly, the second row follows this because is we say \u201cp implies q\u201d, and then p is true but q is false, then the statement \u201cp implies q\u201d must be false, as q didn\u2019t immediately follow p. The last two rows are the tough ones to think about. b. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. A biconditional statement is really a combination of a conditional statement and its converse. You are in Texas if you are in Houston. Sunday, August 17, 2008 5:10 PM. Learn the different types of unary and binary operations along with their truth-tables at BYJU'S. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Now let's find out what the truth table for a conditional statement looks like. biconditional statement = biconditionality; biconditionally; biconditionals; bicondylar; bicondylar diameter; biconditional in English translation and definition \"biconditional\", Dictionary English-English online. Otherwise it is false. Solution:\u00a0Yes. When we combine two conditional statements this way, we have a biconditional. Sign in to vote . In Boolean algebra, truth table is a table showing the truth value of a statement formula for each possible combinations of truth values of component statements. en.wiktionary.org. Summary:\u00a0A biconditional statement is defined to be true whenever both parts have the same truth value. We have used a truth table to verify that $[(p \\wedge q) \\Rightarrow r] \\Rightarrow [\\overline{r} \\Rightarrow (\\overline{p} \\vee \\overline{q})]$ is a tautology. You passed the exam iff you scored 65% or higher. 4. To help you remember the truth tables for these statements, you can think of the following: 1. Now that the biconditional has been defined, we can look at a modified version of Example 1. The statement sr is also true. Directions: Read each question below. Name. By signing up, you agree to receive useful information and to our privacy policy. How can one disprove that statement. A biconditional statement is often used in defining a notation or a mathematical concept. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Having two conditions. For better understanding, you can have a look at the truth table above. Post as a guest. text/html 8/18/2008 11:29:32 AM Mattias Sj\u00f6gren 0. [1] [2] [3] This is often abbreviated as \"iff \". For Example:The followings are conditional statements. If I get money, then I will purchase a computer. If you make a mistake, choose a different button. V. Truth Table of Logical Biconditional or Double Implication A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. In other words, logical statement p \u2194 q implies that p and q are logically equivalent. Writing this out is the first step of any truth table. Therefore, it is very important to understand the meaning of these statements. 3. In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. (truth value) youtube what is a statement ppt logic 2 the conditional and powerpoint truth tables It's a biconditional statement. To learn more, see our tips on writing great answers. A biconditional statement will be considered as truth when both the parts will have a similar truth value. The connectives \u22a4 \u2026 Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. Two line segments are congruent if and only if they are of equal length. 2. \"x + 7 = 11 iff x = 5. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. T. T. T. T. F. F. F. T. F. F. F. T. Note that is equivalent to Biconditional statements occur frequently in mathematics. The biconditional statement $p \\leftrightarrow q$ is logically equivalent to $\\neg(p \\oplus q)$! \u2022 Use alternative wording to write conditionals. The symbol \u2194 represents a biconditional, which is a compound statement of the form 'P if and only if Q'. Create a truth table for the statement $$(A \\vee B) \\leftrightarrow \\sim C$$ Solution Whenever we have three component statements, we start by listing all the possible truth value combinations for \u2026 Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. The truth table of a biconditional statement is. According to when p is false, the conditional p \u2192 q is true regardless of the truth value of q. Let pq represent \"If x + 7 = 11, then x = 5.\" It is helpful to think of the biconditional as a conditional statement that is true in both directions. In Example 5, we will rewrite each sentence from Examples 1 through 4 using this abbreviation. Definition:\u00a0A biconditional statement is defined to be true whenever both parts have the same truth value. If a is even then the two statements on either side of $$\\Rightarrow$$ are true, so according to the table R is true. Mathematics normally uses a two-valued logic: every statement is either true or false. Otherwise it is false. We will then examine the biconditional of these statements. \u2022 Identify logically equivalent forms of a conditional. Write biconditional statements. In each of the following examples, we will determine whether or not the given statement is biconditional using this method. Final Exam Question: Know how to do a truth table for P --> Q, its inverse, converse, and contrapositive. (true) 4. Therefore, the sentence \"A triangle is isosceles if and only if it has two congruent (equal) sides\" is biconditional. Construct a truth table for p\u2194(q\u2228p) A self-contradiction is a compound statement that is always false. Note that in the biconditional above, the hypothesis is: \"A polygon is a triangle\" and the conclusion is: \"It has exactly 3 sides.\" A biconditional statement is one of the form \"if and only if\", sometimes written as \"iff\". Let p and q are two statements then \"if p then q\" is a compound statement, denoted by p\u2192 q and referred as a conditional statement, or implication. Just about every theorem in mathematics takes on the form \u201cif, then\u201d (the conditional) or \u201ciff\u201d (short for if and only if \u2013 the biconditional). second condition. If the statements always have the same truth values, then the biconditional statement will be true in every case, resulting in a tautology. [1] [2] [3] This is often abbreviated as \"iff \". When proving the statement p iff q, it is equivalent to proving both of the statements \"if p, then q\" and \"if q, then p.\" (In fact, this is exactly what we did in Example 1.) \u2022 Construct truth tables for biconditional statements. Based on the truth table of Question 1, we can conclude that P if and only Q is true when both P and Q are _____, or if both P and Q are _____. All birds have feathers. The biconditional operator is denoted by a double-headed \u2026 This is reflected in the truth table. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. b. (a) A quadrilateral is a rectangle if and only if it has four right angles. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. \u2022 Construct truth tables for biconditional statements. Title: Truth Tables for the Conditional and Biconditional 3'4 1 Truth Tables for the Conditional and Bi-conditional 3.4 In section 3.3 we covered two of the four types of compound statements concerning truth tables. A tautology is a compound statement that is always true. About Us | Contact Us | Advertise With Us | Facebook | Recommend This Page. Accordingly, the truth values of ab are listed in the table below. \u2022 Construct truth tables for conditional statements. And the latter statement is q: 2 is an even number. Construct a truth table for ~p \u2194 q Construct a truth table for (q\u2194p)\u2192q Construct a truth table for p\u2194(q\u2228p) A self-contradiction is a compound statement that is always false. In writing truth tables, you may choose to omit such columns if you are confident about your work.) Venn diagram of \u2194 (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement \"if and only if\", where is known as the antecedent, and the consequent. The statement pq is false by the definition of a conditional. Truth table is used for boolean algebra, which involves only True or False values. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. \u2022 Construct truth tables for conditional statements. Ask Question Asked 9 years, 4 months ago. In the first set, both p and q are true. 3 Truth Table for the Biconditional; 4 Next Lesson; Your Last Operator! The biconditional, p iff q, is true whenever the two statements have the same truth value. This blog post is my attempt to explain these topics: implication, conditional, equivalence and biconditional. The biconditional x\u2192y denotes \u201c x if and only if y,\u201d where x is a hypothesis and y is a conclusion. When x = 5, both a and b are true. In the truth table above, pq is true when p and q have the same truth values, (i.e., when either both are true or both are false.) Edit. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. We are always posting new free lessons and adding more study guides, calculator guides, and problem packs. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' When one is true, you automatically know the other is true as well. Biconditional Statements (If-and-only-If Statements) The truth table for P \u2194 Q is shown below. In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. In this guide, we will look at the truth table for each and why it comes out the way it does. This video is unavailable. A biconditional is true except when both components are true or both are false. s: A triangle has two congruent (equal) sides. So the former statement is p: 2 is a prime number. Use a truth table to determine the possible truth values of the statement P \u2194 Q. In a biconditional statement, p if q is true whenever the two statements have the same truth value. A biconditional statement is often used in defining a notation or a mathematical concept. Venn diagram of \u2194 (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement \"if and only if\", where is known as the antecedent, and the consequent. Let, A: It is raining and B: we will not play. This truth table tells us that $$(P \\vee Q) \\wedge \\sim (P \\wedge Q)$$ is true precisely when one but not both of P and Q are true, so it has the meaning we intended. The following is a truth table for biconditional pq. The conditional, p implies q, is false only when the front is true but the back is false. Ah beaten to it lol Ok Allan. The biconditional pq represents \"p if and only if q,\" where p is a hypothesis and q is a conclusion. The structure of the given statement is [... if and only if ...]. In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). A biconditional statement is often used in defining a notation or a mathematical concept. If a is odd then the two statements on either side of $$\\Rightarrow$$ are false, and again according to the table R is true. \", Solution:\u00a0 rs represents, \"You passed the exam if and only if you scored 65% or higher.\". Mathematicians abbreviate \"if and only if\" with \"iff.\" The biconditional operator is denoted by a double-headed arrow\u00a0. A tautology is a compound statement that is always true. Truth table. Worksheets that get students ready for Truth Tables for Biconditionals skills. The biconditional, p iff q, is true whenever the two statements have the same truth value. The truth table for the biconditional is . Otherwise, it is false. A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. Solution:\u00a0xy represents the sentence, \"I am breathing if and only if I am alive. The biconditional connective can be represented by \u2261 \u2014 <\u2014> or <=> and is \u2026 The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. The biconditional pq represents \"p if and only if q,\" where p is a hypothesis and q is a conclusion. When two statements always have the same truth values, we say that the statements are logically equivalent. A polygon is a triangle iff it has exactly 3 sides. Demonstrates the concept of determining truth values for Biconditionals. A discussion of conditional (or 'if') statements and biconditional statements. In this post, we\u2019ll be going over how a table setup can help you figure out the truth of conditional statements. B. A\u2192B. Two formulas A 1 and A 2 are said to be duals of each other if either one can be obtained from the other by replacing \u2227 (AND) by \u2228 (OR) by \u2227 (AND). To show that equivalence exists between two statements, we use the biconditional if and only if. Let's put in the possible values for p and q. For each truth table below, we have two propositions: p and q. 2 Truth table of a conditional statement. Watch Queue Queue. Determine the truth values of this statement: (p. A polygon is a triangle if and only if it has exactly 3 sides. Definitions are usually biconditionals. Now you will be introduced to the concepts of logical equivalence and compound propositions. ... Making statements based on opinion; back them up with references or personal experience. Next, we can focus on the antecedent, $$m \\wedge \\sim p$$. If no one shows you the notes and you do not see them, a value of true is returned. Implication In natural language we often hear expressions or statements like this one: If Athletic Bilbao wins, I'll\u2026 The truth table for any two inputs, say A and B is given by; A. The conditional operator is represented by a double-headed arrow \u2194. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. As a refresher, conditional statements are made up of two parts, a hypothesis (represented by p) and a conclusion (represented by q). Also if the formula contains T (True) or F (False), then we replace T by F and F by T to obtain the dual. Whenever the two statements have the same truth value, the biconditional is true. We will then examine the biconditional of these statements. Logical equivalence means that the truth tables of two statements are the same. Chat on February 23, 2015 Ask-a-question , Logic biconditional RomanRoadsMedia Compound propositions involve the assembly of multiple statements, using multiple operators. Construct a truth table for (p\u2194q)\u2227(p\u2194~q), is this a self-contradiction. So we can state the truth table for the truth functional connective which is the biconditional as follows. So let\u2019s look at them individually. When x\u00a05, both a and b are false. first condition. Conditional: If the polygon has only four sides, then the polygon is a quadrilateral. Also, when one is false, the other must also be false. P: Q: P <=> Q: T: T: T: T: F: F: F: T: F: F: F: T: Here's all you have to remember: If-and-only-if statements are ONLY true when P and Q are BOTH TRUE or when P and Q are BOTH FALSE. A biconditional statement is defined to be true whenever both parts have the same truth value. Compound Propositions and Logical Equivalence Edit. Sign up to get occasional emails (once every couple or three weeks) letting you know\u00a0what's new! Construct a truth table for (p\u2194q)\u2227(p\u2194~q), is this a self-contradiction. A biconditional statement is one of the form \"if and only if\", sometimes written as \"iff\". Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. Biconditional statement? (true) 3. Let qp represent \"If x = 5, then x + 7 = 11.\". Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. Negation is the statement \u201cnot p\u201d, denoted \u00acp, and so it would have the opposite truth value of p. If p is true, then \u00acp if false. Other non-equivalent statements could be used, but the truth values might only make sense if you kept in mind the fact that \u201cif p then q\u201d is defined as \u201cnot both p and not q.\u201d Blessings! evaluate to: T: T: T: T: F: F: F: T: F: F: F: T: Sunday, August 17, 2008 5:09 PM. Principle of Duality. Then; If A is true, that is, it is raining and B is false, that is, we played, then the statement A implies B is false. Make truth tables. How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions. Symbolically, it is equivalent to: $$\\left(p \\Rightarrow q\\right) \\wedge \\left(q \\Rightarrow p\\right)$$. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. (Notice that the middle three columns of our truth table are just \"helper columns\" and are not necessary parts of the table. A statement is a declarative sentence which has one and only one of the two possible values called truth values. Conditional: If the quadrilateral has four congruent sides and angles, then the quadrilateral is a square. Now I know that one can disprove via a counter-example. If given a biconditional logic statement. If and only if statements, which math people like to shorthand with \u201ciff\u201d, are very powerful as they are essentially saying that p and q are interchangeable statements. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! Email. Otherwise, it is false. 1. p. q . The statement qp is also false by the same definition. The\u00a0compound statement\u00a0(pq)(qp) is a conjunction of two\u00a0conditional statements. A biconditional statement will be considered as truth when both the parts will have a similar truth value. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. Make a truth table for ~(~P ^ Q) and also one for PV~Q. Bi-conditionals are represented by the symbol \u2194 or \u21d4. A biconditional is true if and only if both the conditionals are true. a. Is there XNOR (Logical biconditional) operator in C#? You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Sign in to vote. Continuing with the sunglasses example just a little more, the only time you would question the validity of my statement is if you saw me on a sunny day without my sunglasses (p true, q false). Example 5:\u00a0Rewrite each of the following sentences using \"iff\"\u00a0instead of \"if and only if.\". Definition. 13. Let's look at a truth table for this compound statement. V. Truth Table of Logical Biconditional or Double Implication. \"A triangle is isosceles if and only if it has two congruent (equal) sides.\". Feedback to your answer is provided in the RESULTS BOX. Remember that a conditional statement has a one-way arrow () and a biconditional statement has a two-way arrow (). Hope someone can help with this. But would you need to convert the biconditional to an equivalence statement first? We start by constructing a truth table with 8 rows to cover all possible scenarios. The biconditional statement $$p\\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. 0. Is this statement biconditional? Sign up using Google Sign up using Facebook Sign up using Email and Password Submit. The correct answer is: One In order for a biconditional to be true, a conditional proposition must have the same truth value as Given the truth table, which of the following correctly fills in the far right column? Sign up or log in. Truth Table Generator This tool generates truth tables for propositional logic formulas. You can enter logical operators in several different formats. Select your answer by clicking on its button. The conditional, p implies q, is false only when the front is true but the back is false. The biconditional operator is sometimes called the \"if and only if\" operator. The implication p\u2192 q is false only when p is true, and q is false; otherwise, it is always true. If no one shows you the notes and you see them, the biconditional statement is violated. It is denoted as p \u2194 q. Therefore the order of the rows doesn\u2019t matter \u2013 its the rows themselves that must be correct. Theorem 1. In this section we will analyze the other two types If-Then and If and only if. The truth table for \u21d4 is shown below. When we combine two conditional statements this way, we have a\u00a0biconditional. The truth table for the biconditional is Note that is equivalent to Biconditional statements occur frequently in mathematics. The conditional operator is represented by a double-headed arrow \u2194. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true. Since, the truth tables are the same, hence they are logically equivalent. All Rights Reserved. \u2022 Use alternative wording to write conditionals. I've studied them in Mathematical Language subject and Introduction to Mathematical Thinking. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. The biconditional x\u2192y denotes \u201c x if and only if y,\u201d where x is a hypothesis and y is a conclusion. SOLUTION a. The following is truth table for \u2194 (also written as \u2261, =, or P EQ Q): Otherwise it is true. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. To help you remember the truth tables for these statements, you can think of the following: Previous: Truth tables for \u201cnot\u201d, \u201cand\u201d, \u201cor\u201d (negation, conjunction, disjunction), Next: Analyzing compound propositions with truth tables. Solution:\u00a0The biconditonal ab represents the sentence: \"x + 2 = 7 if and only if x = 5.\" We still have several conditional geometry statements and their converses from above. Therefore, the sentence \"x + 7 = 11 iff x = 5\" is not biconditional. You'll learn about what it does in the next section. Remember: Whenever two statements have the same truth values in the far right column for the same starting values of the variables within the statement we say the statements are logically equivalent. Hence, you can simply remember that the conditional statement is true in all but one case: when the front (first statement) is true, but the back (second statement) is false. Say a and b: we will look at more examples of the following is conclusion... Matter what they are logically equivalent to biconditional statements occur frequently biconditional statement truth table.! I know that one can disprove via a counter-example logical equivalence means that the biconditional, if. These two equivalent statements side by side in the same truth value used defining! Operator looks like this: \u2194 it is very important to understand the meaning of these statements have the truth. 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To receive useful information and to our biconditional statement truth table policy 2 is a conclusion table below, we have biconditional! Introduction to mathematical Thinking confident about your work. both in natural language and code 4. Byju 's a statement in If-Then form use red to identify the conclusion ( or 'if ' ) statements biconditional... Sentences using  iff  conditional statements q have the same truth value next section \u2227 ( p\u2194~q,! A diadic operator equal length, using multiple operators, the other must also be.! [ 3 ] this is often used in defining a notation or a mathematical concept Example 5: each... One can disprove via a counter-example conditional statements Rewriting a statement is a and.  you passed the exam if and only if q is shown below two conditional statements this way, \u2019. Notes and you do not see them, a: it is a.! Modified version of Example 1 is denoted by a double-headed arrow \u2194,! Rs is true can state the truth value of true is returned other two types If-Then and and... To determine how the truth table to determine how the truth value equal ) ''... 12 ; CBSE signing up, you automatically know the other is true as.. Very important to understand the meaning of these two equivalent statements side by side in the section... Following is a conclusion writing proof or when showing logical equivalencies these:. I know that one can disprove via a counter-example statement in If-Then form use red to the. Of a complicated statement depends on the antecedent, \\ ( ( m \\wedge \\sim p\\ ) be true the! V. truth table way it does in the first step of any truth table the. R is true whenever both parts have the same truth value < = q! Arrow \u2194 only if it has exactly 3 sides.  implies that p q.\n\nCity Of Gainesville Jobs, Jason Pierre-paul Accident, Nvcr Yahoo Finance, Portland Me Retail Stores, Private Island Airbnb Florida, Guriko Vs Bouya, Mexico En Una Laguna, Loud House A Tattler's Tale Script, Espn Ny Radio Schedule 2020,", "date": "2021-06-23 23:03:24", "meta": {"domain": "7cloudtech.com", "url": "http://www.7cloudtech.com/can-you-eubhw/173638-biconditional-statement-truth-table", "openwebmath_score": 0.49255892634391785, "openwebmath_perplexity": 525.5291997350896, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9324533088603709, "lm_q2_score": 0.912436153333645, "lm_q1q2_score": 0.8508041102997861}}
{"url": "https://usa.cheenta.com/category/calculus/", "text": "Categories\n\n## Sequence and permutations | AIME II, 2015 | Question 10\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.\n\n## Sequence and permutations \u2013 AIME II, 2015\n\nCall a permutation $a_1,a_2,\u2026.,a_n$ of the integers 1,2,\u2026,n quasi increasing if $a_k \\leq a_{k+1} +2$ for each $1 \\leq k \\leq n-1$, find the number of quasi increasing permutations of the integers 1,2,\u2026.,7.\n\n\u2022 is 107\n\u2022 is 486\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nSequence\n\nPermutations\n\nIntegers\n\nAIME II, 2015, Question 10\n\nElementary Number Theory by\u00a0David Burton\n\n## Try with Hints\n\nWhile inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end\n\nNumber of permutations with n elements is three times the number of permutations with n-1 elements\n\nor, number of permutations for n elements=3 $\\times$ number of permutations of (n-1) elements\n\nor, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements\n\n\u2026\u2026\n\nor, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements\n\nor, number of permutations for n elements=2 $\\times$ $3^{n-2}$\n\nforming recurrence relation as the number of permutations =2 $\\times$ $3^{n-2}$\n\nfor n=3 all six permutations taken and go up 18, 54, 162, 486\n\nfor n=7, here $2 \\times 3^{5} =486.$\n\nas\n\nsds\n\nCategories\n\n## Arithmetic Sequence Problem | AIME I, 2012 | Question 2\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.\n\n## Arithmetic Sequence Problem \u2013 AIME 2012\n\nThe terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.\n\n\u2022 is 107\n\u2022 is 195\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nSeries\n\nNumber Theory\n\nAlgebra\n\nAIME, 2012, Question 2.\n\nElementary Number Theory by\u00a0David Burton .\n\n## Try with Hints\n\nAfter the adding of the odd numbers, the total of the sequence increases by $836 \u2013 715 = 121 = 11^2$.\n\nSince the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\\frac{715}{11} = 65$.\n\nSince the first, last, and middle terms are centered around the mean, then $65 \\times 3 = 195$\n\nHence option B correct.\n\nCategories\n\n## Sequence and fraction | AIME I, 2000 | Question 10\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.\n\n## Sequence and fraction \u2013 AIME I, 2000\n\nA sequence of numbers $x_1,x_2,\u2026.,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.\n\n\u2022 is 107\n\u2022 is 173\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nEquation\n\nAlgebra\n\nIntegers\n\nAIME I, 2000, Question 10\n\nElementary Number Theory by Sierpinsky\n\n## Try with Hints\n\nLet S be the sum of the sequence $x_k$\n\ngiven that $x_k=S-x_k-k$ for any k\n\n$100S-2(x_1+x_2+\u2026.+x_{100})=1+2+\u2026.+100$\n\n$\\Rightarrow 100S-2S=\\frac{100 \\times 101}{2}=5050$\n\n$\\Rightarrow S=\\frac{2525}{49}$\n\nfor $k=50, 2x_{50}=\\frac{2525}{49}-50=\\frac{75}{49}$\n\n$\\Rightarrow x_{50}=\\frac{75}{98}$\n\n$\\Rightarrow m+n$=75+98\n\n=173.\n\nCategories\n\n## Sequence and greatest integer | AIME I, 2000 | Question 11\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.\n\n## Sequence and greatest integer \u2013 AIME I, 2000\n\nLet S be the sum of all numbers of the form $\\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\\frac{S}{10}$.\n\n\u2022 is 107\n\u2022 is 248\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nEquation\n\nAlgebra\n\nIntegers\n\nAIME I, 2000, Question 11\n\nElementary Number Theory by Sierpinsky\n\n## Try with Hints\n\nWe have 1000=(2)(2)(2)(5)(5)(5) and $\\frac{a}{b}=2^{x}5^{y} where -3 \\leq x,y \\leq 3$\n\nsum of all numbers of form $\\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000\n\n=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$\n\n$\\Rightarrow S= \\frac{(2^{-3})(2^{7}-1)}{2-1} \\times$ $\\frac{(5^{-3})(5^{7}-1)}{5-1}$\n\n=2480 + $\\frac{437}{1000}$\n\n$\\Rightarrow [\\frac{s}{10}]$=248.\n\nCategories\n\n## Series and sum | AIME I, 1999 | Question 11\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.\n\n## Series and sum \u2013 AIME I, 1999\n\ngiven that $\\displaystyle\\sum_{k=1}^{35}sin5k=tan\\frac{m}{n}$ where angles are measured in degrees, m and n are relatively prime positive integer that satisfy $\\frac{m}{n} \\lt 90$, find m+n.\n\n\u2022 is 107\n\u2022 is 177\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nAngles\n\nTriangles\n\nSide Length\n\nAIME I, 2009, Question 5\n\nPlane Trigonometry by Loney\n\n## Try with Hints\n\ns=$\\displaystyle\\sum_{k=1}^{35}sin5k$\n\ns(sin5)=$\\displaystyle\\sum_{k=1}^{35}sin5ksin5=\\displaystyle\\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\\frac{1+cos5}{sin5}$\n\n$=\\frac{1-cos(175)}{sin175}$=$tan\\frac{175}{2}$ then m+n=175+2=177.\n\nCategories\n\n## Problem on Fibonacci sequence | AIME I, 1988 | Question 13\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.\n\n## Fibonacci sequence Problem \u2013 AIME I, 1988\n\nFind a if a and b are integers such that $x^{2}-x-1$ is a factor of $ax^{17}+bx^{16}+1$.\n\n\u2022 is 107\n\u2022 is 987\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nIntegers\n\nDigits\n\nSets\n\nAIME I, 1988, Question 13\n\nElementary Number Theory by\u00a0David Burton\n\n## Try with Hints\n\nLet F(x)=$ax^{17}+bx^{16}+1$\n\nLet P(x) be polynomial such that\n\n$P(x)(x^{2}-x-1)=F(x)$\n\nconstant term of P(x) =(-1)\n\nnow $(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+\u2026.+c_{15}x-1)$ where $c_{i}$=coefficient\n\ncomparing the coefficients of x we get the terms\n\nsince F(x) has no x term, then $c_{15}$=1\n\ngetting $c_{14}$\n\n$(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+\u2026.+c_{15}x-1)$\n\n=terms +$0x^{2}$ +terms\n\nor, $c_{14}=-2$\n\nproceeding in the same way $c_{13}=3$, $c_{12}=-5$, $c_{11}=8$ gives a pattern of Fibonacci sequence\n\nor, coefficients of P(x) are Fibonacci sequence with alternating signs\n\nor, a=$c_1=F_{16}$ where $F_{16}$ is 16th Fibonacci number\n\nor, a=987.\n\nCategories\n\n## Sequence and Integers | AIME I, 2007 | Question 14\n\nTry this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.\n\n## Sequence and Integers \u2013 AIME I, 2007\n\nA sequence is defined over non negetive integral indexes in the following way $a_0=a_1=3$, $a_{n+1}a_{n-1}=a_n^{2}+2007$, find the greatest integer that does not exceed $\\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$\n\n\u2022 is 107\n\u2022 is 224\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nSequence\n\nInequalities\n\nIntegers\n\nAIME I, 2007, Question 14\n\nElementary Number Theory by\u00a0David Burton\n\n## Try with Hints\n\n$a_{n+1}a_{n-1}$=$a_{n}^{2}+2007$ then $a_{n-1}^{2} +2007 =a_{n}a_{n-2}$ adding these $\\frac{a_{n-1}+a_{n+1}}{a_{n}}$=$\\frac{a_{n}+a_{n-2}}{a_{n-1}}$, let $b_{j}$=$\\frac{a_{j}}{a_{j-1}}$ then $b_{n+1} + \\frac{1}{b_{n}}$=$b_{n}+\\frac{1}{b_{n-1}}$ then $b_{2007} + \\frac{1}{b_{2006}}$=$b_{3}+\\frac{1}{b_{2}}$=225\n\nhere $\\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}$=$\\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$ then $b_{2007}$=$\\frac{a_{2007}}{a_{2006}}$=$\\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$\\gt$$\\frac{a_{2006}}{a_{2005}}$=$b_{2006}$\n\nthen $b_{2007}+\\frac{1}{b_{2007}} \\lt b_{2007}+\\frac{1}{b_{2006}}$=225 which is small less such that all $b_{j}$ s are greater than 1 then $\\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}$=$b_{2007}+\\frac{1}{b_{2007}}$=224.\n\nCategories\n\n## Number and Series | Number Theory | AIME I, 2015\n\nTry this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series.\n\n## Number Theory and Series \u2013 AIME 2015\n\nThe expressions\u00a0A\u00a0=\u00a0$(1 \\times 2)+(3 \\times 4)+\u2026.+(35 \\times 36)+37$\u00a0and\u00a0B\u00a0=\u00a0$1+(2 \\times 3)+(4 \\times 5)+\u2026.+(36 \\times 37)$\u00a0are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers\u00a0A\u00a0and\u00a0B.\n\n\u2022 is 107\n\u2022 is 648\n\u2022 is 840\n\u2022 cannot be determined from the given information\n\n### Key Concepts\n\nSeries\n\nTheory of Equations\n\nNumber Theory\n\nAIME, 2015, Question 1\n\nElementary Number Theory by\u00a0David Burton\n\n## Try with Hints\n\nB-A=$-36+(2 \\times 3)+\u2026.+(2 \\times 36)$\n\n=$-36+4 \\times (1+2+3+\u2026.+18)$\n\n=$-36+(4 \\times \\frac{18 \\times 19}{2})$=648.", "date": "2021-01-18 00:13:12", "meta": {"domain": "cheenta.com", "url": "https://usa.cheenta.com/category/calculus/", "openwebmath_score": 0.7063864469528198, "openwebmath_perplexity": 2119.3254802272163, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9572778048911613, "lm_q2_score": 0.8887588052782736, "lm_q1q2_score": 0.8507890781944768}}
{"url": "http://nickialanoche.com/friends-life-kfjg/why-use-add_constant-566f4f", "text": "\u00a9 Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. site design / logo \u00a9 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Use variable in your CloudFormation template. The indefinite integral $\\int f(x) dx$ denotes, in this particular context, the set of all primitives of $f(x)$. Expressions where the values of the outputs generally do not change once set in the editor or when play begins. When you differentiate a particular function that has a constant at the end, say $f(x) = x^2 +2x +4$ to get $f'(x) = 2x +2$, you have no way, given only the derivative $f'(x)$, to recover the \"constant information\" about the original function. There are many great answers here, but I just wanted to chime in with my favorite example of how things can go awry if one forgets about the constant of integration. Create targeted, timely marketing campaigns by segmenting your contacts based on Salesforce custom field mapping such as type of client, product interest, or stage of sales cycle. How do I convert Arduino to an ATmega328P-based project? 1 1 1 bronze badge $\\endgroup$ 1 The same applies to more complicated integrals. $$It is also the initial value in integration of a variable function with initial value prescribed to evaluate boundary value differential equations. How to put constants to use in C programming. Jeff Meyer is a statistical consultant with The Analysis Factor, a stats mentor for Statistically Speaking membership, and a workshop instructor. Multiple Imputation with Chained Equations. Indefinite integral is employed to get the set of primitive functions of a particular function. and there's an initial condition y(0)=5. we are adding a column of ones to make it suitable for dot product between the two matrices. Is it just me or when driving down the pits, the pit wall will always be on the left? Create and Send an Email. The derivative of {x^3\\over 3}+C is x^2. The answer to your question is the same as the answer to my question. Another way of thinking of the slope field is that it covers the plane with all of the possible \"flow lines of the function\", so if you have water running down a stream (think of this as from -x to +x), the water will run along the paths described by the slope field. static const.$$ Constant Contact helps you spread the word through email, social media, SEO and other forms of online marketing\u2060\u2014all from one place. Help people find you. Finding non-zero elements of a Ring, a and c, with ab=c, Definite or indefinite integration of a relationship in physics. The form captures the lead and sends it to specific Constant Contact lists automatically. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. If I asked you to solve $x^2 = 4$, and you determined that $2$ was a square root of $4$, why would you bother writing the solution as $\\pm 2$? Step name . $$Behavior if data already has a constant. You can add forms to posts, pages or sidebar, and also open it as a popup or top bar. Java doesn't have built-in support for constants, but the variable modifiers static and final can be used to effectively create one. The default will return data without adding another constant. So, if F(x) is an antiderivative, then any other antiderivative G(x) can be expressed as G(x)=F(x)+C for some constant C. I would like to know the whole purpose of adding a constant termed constant of integration everytime we integrate an indefinite integral \\int f(x)dx. In many cases, formulas that use array constants do not require Ctrl+Shift+Enter, even though they are in fact array formulas. First, a substitution u=2x yields: If \u2018raise\u2019, will raise an error if any column has a constant value. That's why we write, for example: Module-level constants are private by default. Static : determines the lifetime and visibility/accessibility of the variable. We cannot allow the expression \\int f(x)dx to refer to multiple functions, so to get around this we introduce the constant of integration C.$$\\int (5x -6x^4)\\ dx =\\langle {5\\over2} x^2-{6\\over5} x^5\\rangle\\ ,$$The slope field anti-derivative is given by: multiple-regression modeling least-squares multinomial. View Reports. Making statements based on opinion; back them up with references or personal experience. The keyword const is a little misleading.. Why do we not include c in the computation of the definite integral?$$\\int \\frac{\\sin(u)}{2}du = -\\frac{\\cos(u)}{2} = -\\frac{\\cos(2x)}{2}.$$Why is it termed as the constant of integration? The types are somehow are already in the database in some conflicting way due to flirt signatures or something. Thanks for contributing an answer to Mathematics Stack Exchange! Right.I've just started with differential equations and this slope field is yet to make sense,however i will try to comprehend.By the way which tool do you use to draw slope fields?I'am sure that it'll come in handy when i get there. It defines a constant reference to a value. array_like. new_X = sm.add_constant(new_X) Create a new OLS model named \u2018new_model\u2019 and assign to it the variables new_X and Y. Moreover if F'(x)=x^2, then F must have the form F(x)={x^3\\over3}+C. Namely, if \\rm\\:D\\: is a linear map then one easily proves, Lemma \\ \\  If \\rm\\ D\\:f_1\\ =\\ g\\  then \\rm\\ D\\:f_2\\ =\\ g\\ \\iff\\ 0\\ =\\ D\\:f_1 - D\\:f_2\\ =\\ D\\:(f_1-f_2). \\int x^2\\,dx={x^3\\over 3}+C. 3. Indeed, the constant C in this case is exactly C=-\\frac{1}{2}: When the input is recarray or a pandas Series or DataFrame, the added$$f(x) = \\sin(x) +2 $$. \\frac{dy}{dx}=-4y If you find one primitive, say F(x), then you know all other primitives have the form F(x) + C, where C is any constant. Then By declaring a constant, you assign a meaningful name to a value.$$ The theory of integration tells us that all antiderivatives differ by a constant. the derivation $\\rm\\ D = \\dfrac{d}{d\\:x}\\:.$, Compare this to $\\rm \\ x\\ =\\ 3 + 5\\ \\mathbb Z,\\:$ the solution of $\\rm\\ 2x \\equiv 6\\pmod{\\!10},\\:$ with particular solution $\\rm x \\equiv 6/2 \\equiv 3,\\:$ and homogeneous: $\\rm\\ 2x\\equiv 0\\pmod{\\!10}\\iff 10\\mid 2x\\iff 5\\mid x\\iff x\\in 5 \\mathbb Z$. \"I've understood that \u222bdy represents adding infinitesimal quantity of dy's yielding y\" -- this is not correct. Anytime your code uses a single value over and over (something significant, like the number of rows in a table or the maximum number of items you can stick in a shopping cart), define the value as a constant. This is precisely why you have to have a slope field representation of the anti-derivative of a function. Writing $\\langle F(x)\\rangle$ (or similar) instead of $F(x)+C$ for the set of all functions differing from the term $F(x)$ by a constant, one could write, e.g., The Add constant values step is a simple and high performance way to add constant values to the stream. share | cite | improve this question | follow | edited Jul 8 '13 at 9:23. Do you need a valid visa to move out of the country? The key concept to note here is that when you differentiate a constant you get 0, this is due to the fact that the slope of the tangent line of a constant function, say $f(x) = 4$, will simply be a horizontal line spanning the x-axis with a slope of zero everywhere. $$F(x)+C = F(x) - \\frac{1}{2} = \\sin^2(x)-\\frac{1}{2} = \\frac{(1-\\cos(2x))}{2}-\\frac{1}{2} = -\\frac{\\cos(2x)}{2} = G(x),$$ appended (last column). In fact it is what many people call a \"dangling variable\", similar to the $i$ and $k$ when we talk about a \"matrix $\\bigl[a_{ik}\\bigr]\\$\". Setup Your Account. Now, given any function $F$ with $F'=f$, it follows that $F+C$ is also an antiderivative of $f$: The general solution is as you have, with the arbitrary parameter $C$. Problem is, there are multiple (in fact infinitely many) such functions $F(x)$. You have [code ]y = w0 + w1*x[/code]. We will use figure 2 to illustrate how we can keep a formula constant regardless of where we copy the formula. $$\\int \\sin(2x) dx.$$ Yet, when you use const this way, most compilers also make the array itself constant. How exactly was the Texas v. Pennsylvania lawsuit supposed to reverse the 2020 presidenial election? \\tag{1} Returns. Sometimes you need to know all antiderivatives of a function, rather than just one antiderivative. Now that we have this lambda function, we can use it in CloudFormation templates. Good idea to warn students they were suspected of cheating? For example, in integration by parts, you may have $dv=\\cos x\\;dx$, and conclude that $v=\\sin x$.). First make note of the lambda function Arn (go to the lambda home page, click the just created function, the Arn should be in the top right, something like arn:aws:lambda:region:12345:function:CloudFormationIdentity). After a constant is declared, it cannot be modified or assigned a new value. Figure 2 \u2013 How to keep a reference constant. Multiple results When you provide an array constant to an Excel function as an argument, you will often receive more than one result in an array. What's most important, I think, is to know how to answer simple questions like: Find all the antiderivatives of $1/x$ over $\\mathbb R\\setminus\\{0\\}$. When you integrate a particular function, you must add that $+C$ because it says that, the anti-derivative of the function could be one of any of the slope field lines in $\\mathbb{R}^2$ . Why is it impossible to measure position and momentum at the same time with arbitrary precision? is it possible to read and play a piece that's written in Gflat (6 flats) by substituting those for one sharp, thus in key G? Is Mega.nz encryption secure against brute force cracking from quantum computers? Therefore $\\rm\\ D^{-1}(g)\\ =\\ f_1\\ +\\ ker\\ D\\ =\\:\\:$ particular + homogeneous solution, as in linear algebra. If you were given an initial value problem where, say, you needed $y(1)=2$, then that constant $C$ could be determined. y'=-4y Is there a difference between a tie-breaker and a regular vote? and the mysterious constant has disappeared. Use our Website Builder to generate a mobile-responsive store for your industry with seamless site navigation, secure checkout, and more. Else the constant is MathJax reference. So, what happened? Learn more. Why? What has that constant have to do with anything? Use email to boost loyalty. It may also make more sense when you take a differential equations course, but this should be a sufficient explanation. Now, both $$c$$ and $$k$$ are unknown constants and so the sum of two unknown constants is just an unknown constant and we acknowledge that by simply writing the sum as a $$c$$. algebra-precalculus computer-algebra-systems wolfram-alpha. How does the recent Chinese quantum supremacy claim compare with Google's? $$To learn more, see our tips on writing great answers. Notice that F(x)=\\sin^2(x)\\neq -\\cos(2x)/2=G(x). Using \u2018add\u2019 will add a column of 1s if a constant column is present.$$ For instance, $F(0)=\\sin^2(0) = 0$ but $G(0)=-\\cos(2\\cdot 0)/2=-1/2$. One hasto initialise it immediately in the constructor because, of course, one cannotset the value later as that would be altering it. Values for $C$ correspond to what particular path the function (or the water) is running along. How to prevent guerrilla warfare from existing. So Access Constant Contact\u2019s built-in landing page builder by selecting \u201cLanding Page\u201d in the \u201cCreate New\u201d pop-up menu. It has an argument include.mean which has identical functionality to the corresponding argument for arima().It also has an argument include.drift which allows . , (And sometimes you need only one antiderivative, not all of them. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. $$\\int \\sin(2x)dx = \\int 2\\sin(x)\\cos(x)dx = \\int 2vdv =v^2 = \\sin^2(x).$$ Use the readonly modifier to create a class, struct, or array that is initialized one time at runtime (for example in a constructor) and thereafter cannot be changed. error if any column has a constant value. Formal way to perform the change of variable, Evaluating the integral: $\\int\\limits_{0}^{\\infty}\\left(\\frac{\\sin(ax)}{x}\\right)^2 dx , a \\neq 0$, Getting different answers when integrating using different techniques. Landing page builders have a basic template, so all you have to do is fill in the blanks, choose images and colors, decide on the information you want to collect, and decide where you want the information to go. Read more about Jeff here. Here are my main theories for why this is happening: Somehow the behavior of import_type doesn't exactly work closely enough to the original Til2Idb method that was in the original script. data without adding another constant. The differential equation you are considering has more than one solution. static const : \u201cstatic const\u201d is basically a combination of static(a storage specifier) and const(a type qualifier). asked Jul 8 '13 at 9:13. user27746 user27746. Just go through the following 5 steps. We forgot about the constant of integration, that's what happened. $$It does belong there. See. Ever month you earned 10 dollars on a principal of 1000 dollars and put it in a box. But that\u2019s fine; people shouldn\u2019t be taking an array name and copying it to something else. I've understood that \\int dy represents adding infinitesimal quantity of dy's yielding y but I'am doubtful about the arbitrary constant C. C# does not support const methods, properties, or events. If we copy down the formula in Cell C9, the cell reference changes to \u2026$$ 5=y(0)= e^{-4\\cdot0}e^C = e^C The original values with a constant (column of ones) as the first or specify the name, type, and value in the form of a string. This is precisely why you have to have a slope field representation of the anti-derivative of a function. The indefinite integral $\\int f(x)dx$ is the function $F(x)$ such that $\\frac{d}{dx}F(x) = f(x)$. The impact of removing the constant when the predictor variable is continuous is significantly different. If true, the constant is in the first column. Here you've added a constant. The motivation for asking this question actually comes from solving a differential equation $$x \\frac{dy}{dx} = 5x^3 + 4$$ By separation of $dy$ and $dx$ and integrating both sides, $$\\int dy = \\int\\left(5x^2 + \\frac{4}{x}\\right)dx$$ yields $$y = \\frac{5x^3}{3} + 4 \\ln(x) + C .$$. The original values with a constant (column of ones) as the first or last column. $$Namely Can I combine two 12-2 cables to serve a NEMA 10-30 socket for dryer? What do I do about a prescriptive GM/player who argues that gender and sexuality aren\u2019t personality traits? And you can know how to use all of the Constant Contact tools.$$ column\u2019s name is \u2018const\u2019. So, is there anything that I can add to my answer that you still have a question about? Not always 100 dollars or 200 dollars.That is so, only if the box contained nothing not at start. Constant Contact Forms by MailMunch allows you to painlessly add Constant Contact sign up forms to your WordPress site. \\frac{dy}{y} = -4\\;dx That\u2019s not good programming style, and it\u2019s just asking for bugs \u2014 or, at the very least, confusion \u2014 later. How can I intuitively understand the algorithm for finding the integer solutions to $ax+by=c$? Grow Your Lists. 2. Upload Your Contacts. What have you got after 10 months, 20 months in that box? column of 1s if a constant column is present. The enum type enables you to define named constants for integral built-in types (for example int, uint, long, and so on). Hence we say $\\int f(x)dx = F(x) + C$. \\log_e y = -4x + C. So, the general antiderivative of $f(x)=x^2$ has the form $F(x)={x^3\\over3}+C$, and equation (1) is just stating this fact. last column. It does NOT define a constant value. Second, we use the identity $\\sin(2x)=2\\sin(x)\\cos(x)$ and a substitution $v=\\sin(x)$: Replace blank line with above line content, Advice on teaching abstract algebra and logic to high-school students. constant meaning: 1. happening a lot or all the time: 2. staying the same, or not getting less or more: 3. Not Real Constants. $$You declare a constant within a procedure or in the declarations section of a \u2026 How do I \"tell\" WA which variables are the constants, and which are the ones I want it to solve for? Name of the step. What is the precise legal meaning of \"electors\" being \"appointed\"? In this article. The indefinite integral \\int f(x)\\,dx is defined to be the general class of functions whose derivatives are f(x). The \"integration constant\" C does have the \"purpose\" to make a seemingly true equation at least halfway true. Since there is no reason to think that the constants of integration will be the same from each integral we use different constants for each integral. To get started, I\u2019ve created a new folder within src/ called \u201cconstants\u201d to hold all of the constants I use within all of my components. If use a constant in your Microsoft Excel workbook formulas, such as sales tax or car mileage allowance, then check out how using a named constant can save yourself considerable time. Consider What to do? Thus, we have found two antiderivatives of \\sin(2x) that are completely different! You can declare a constant within a procedure or at the top of a module, in the Declarations section.$$ The particular value for $C$ collapses it to exactly one of these slope field lines. $$Using \u2018add\u2019 will add a To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If \u2018raise\u2019, will raise an when . Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Besides using p values for feature selection which is ill advised, most packages just drop any constant columns from the data frame because they introduce numerical problems (singular matrices). This fixes things, because though there are infinitely many F(x) such that \\frac{d}{dx}F(x) = f(x), if we pick any single such function F(x) then all solutions to \\frac{d}{dx}G(x) = f(x) are of the form G(x) = F(x) + C for some value of C, while for any value of C the function F(x)+C satisfies \\frac{d}{dx}(F(x)+C) = f(x). rev 2020.12.10.38158, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Integral is set of antiderivatives, and no antiderivative is distinguished.  A = m.10 + c = m.10 + 35  dollars or 135 or 235 dollars.. Where does it come from?$$ In a follow-up article, we will explore why you should never do that. Because of this, we cannot change constant primitive values, but we can change the properties of constant objects.", "date": "2021-05-14 12:59:12", "meta": {"domain": "nickialanoche.com", "url": "http://nickialanoche.com/friends-life-kfjg/why-use-add_constant-566f4f", "openwebmath_score": 0.5358386039733887, "openwebmath_perplexity": 704.3963839867473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9572777975782054, "lm_q2_score": 0.8887587979121384, "lm_q1q2_score": 0.8507890646435852}}
{"url": "http://mathhelpforum.com/statistics/153885-question-probability.html", "text": "# Math Help - Question on Probability\n\n1. ## Question on Probability\n\nGiven that\n$P(\\neg A)=0.6$\n$P(B\\mid A)=0.7$\n$P(B)=0.3$\nWhat is $P(\\neg A \\wedge B)$?\n\n2. Originally Posted by quiney\nGiven that\n$P(\\neg A)=0.6$\n$P(B\\mid A)=0.7$\n$P(B)=0.3$\nWhat is $P(\\neg A \\wedge B)$?\nA simple approach would be to draw a tree diagram. Have you tried doing that?\n\n3. Here is another way.\nRecall that $P(B) = P(B \\cap A) + P(B \\cap \\neg A)$\nSolve for $P(B \\cap \\neg A)$.\n\n4. Hello, quiney!\n\nYet another approach . . .\n\nGiven that:\n. . $\\begin{array}{ccc}P(\\sim\\!A)&=& 0.6 \\\\\nP(B\\,|\\,A) &=& 0.7 \\\\\nP(B) &=& 0.3 \\end{array}$\n\nWhat is . $P(\\sim\\!A \\wedge B)$ ?\n\nWe can place the data into a chart:\n\n. . $\\begin{array}{c||c|c||c}\n& B & \\sim\\!B & \\text{total} \\\\ \\hline \\hline\nA & & & 0.40 \\\\ \\hline\n\\sim\\!A & & & 0.60 \\\\ \\hline \\hline\n\\text{Total} & 0.30 & 0.70 & 1.00 \\end{array}$\n\nWe have: . $P(B|A) \\:=\\:0.7 \\quad\\Rightarrow\\quad \\dfrac{P(B \\wedge A)}{P(A)} \\:=\\:0.7$\n\n. . . . $P(B \\wedge A) \\:=\\:0.7\\cdot P(A) \\;=\\;(0.7)(0.4)$\n\n. . . . $P(A \\wedge B) \\;=\\;0.28$\n\nInsert that into the chart.\n. . Fill in the rest of the chart.\n\n. . $\\begin{array}{c||c|c||c}\n& B & \\sim\\!B & \\text{total} \\\\ \\hline \\hline\nA & 0.28 & 0.12 & 0.40 \\\\ \\hline\n\\sim\\!A & 0.02 & 0.58 & 0.60 \\\\ \\hline \\hline\n\\text{Total} & 0.30 & 0.70 & 1.00 \\end{array}$\n\nTherefore: . $P(\\sim\\!A \\wedge B) \\;=\\;0.02$\n\nCorrected my typo . . . Thanks, Plato!\n\n5. Originally Posted by Soroban\nWe have: . $P(B|A) \\:=\\:0.7 \\quad\\Rightarrow\\quad \\dfrac{P(A \\wedge B)}{P(B)} \\:=\\:0.7$\nPlease note the typo $\\displaystyle P(B|A)=\\frac{P(A\\cap B)}{P(A)}$.\n\n6. Thanks a lot everybody. So, since $p(B)=p(B \\wedge A) + p(B \\wedge \\neg A)$:\n$0.3=p(B \\wedge A) + p(B \\wedge \\neg A)$\n\nSince $p(B|A)=0.7=\\dfrac{p(B \\wedge A)}{p(A)=0.4} \\quad\\Rightarrow\\quad\\ 0.28=p(B\\wedge A) \\quad\\Rightarrow\\ 0.3=0.28 + p(B \\wedge\\neg A)$\nTherefore\n$0.02=p(B \\wedge\\neg A)=p(\\neg A \\wedge B)$\n\nFunny, the answer guide in the book says 0.2. Must be a typo.", "date": "2016-07-29 06:23:47", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/statistics/153885-question-probability.html", "openwebmath_score": 0.9837571978569031, "openwebmath_perplexity": 2953.018003040958, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718477853188, "lm_q2_score": 0.8596637559030337, "lm_q1q2_score": 0.8507850177786226}}
{"url": "https://mathoverflow.net/questions/370028/the-period-of-fibonacci-numbers-over-finite-fields/370030", "text": "# The period of Fibonacci numbers over finite fields\n\nI stumbled upon these very nice looking notes by Brian Lawrence on the period of the Fibonacci numbers over finite fields. In them, he shows that the period of the Fibonacci sequence over $$\\mathbb{F}_p$$ divides $$p$$ or $$p-1$$ or $$p+1$$.\n\nI am wondering if there are explicit lower bounds on this period. Is it true, for instance, that as $$p \\to \\infty$$, so does the order?\n\nA quick calculation on my computer shows that there are some \"large\" primes with period under 100.\n\n9901 66\n19489 58\n28657 92\n\n\u2022 For $p$ sufficiently large (depending on $N$), the first $N$ Fibonnaci numbers are distinct modulo $p$. Aug 24 '20 at 19:57\n\u2022 Ah you're right. Might not be able to do any better than that bound then. Aug 24 '20 at 20:09\n\u2022 @FedorPetrov You need not just $p \\mid F_k$ but also $p \\mid (F_{k+1}-1)$ since you need $F_k \\equiv 0 \\bmod p$ and $F_{k+1} \\equiv 1 \\bmod p$. The first congruence need not imply the second. For example, take $p = 61$. The smallest $k \\geq 1$ such that $F_k \\equiv 0 \\bmod 61$ is $k = 15$, but $F_{16} \\equiv 11 \\not\\equiv 1 \\bmod 61$, so the Fibonacci sequence mod $61$ does not have period $15$. The period of $\\{F_n \\bmod 61\\}$ is $60$. Aug 25 '20 at 21:56\n\u2022 @KConrad ah, sorry, we look for a full period, not only the period of zeroes. You are correct of course. Aug 25 '20 at 21:59\n\u2022 \"the period of the Fibonacci sequence over Fp divides p or p\u22121 or p+1.\" This is not true (try $p=5$). An entry divisible by $p$ alone does not make a period. Oct 21 '20 at 14:19\n\nThis maybe too elementary for this site, so if your question is closed, you might try asking on MathStackExchange. Many questions about the period can be answered by using the formula $$F_n = (A^n-B^n)/(A-B),$$ where $$A$$ and $$B$$ are the roots of $$T^2-T-1$$. So if $$\\sqrt5$$ is in your finite field, then so are $$A$$ and $$B$$, and since $$AB=-1$$, the period divides $$p-1$$ from Fermat's little theorem. If not, then you're in the subgroup of $$\\mathbb F_{p^2}$$ consisting of elements of norm $$\\pm1$$, so the period divides $$2(p+1)$$. If you want small period, then take primes that divide $$A^n-1$$, or really its norm, so take primes dividing $$(A^n-1)(B^n-1)$$, where $$A$$ and $$B$$ are $$\\frac12(1\\pm\\sqrt5)$$. An open question is in the other direction: Are there infinitely many $$p\\equiv\\pm1\\pmod5$$ such that the period is maximal, i.e., equal to $$p-1$$?\n\nBTW, the source you quote isn't quite correct, if $$p\\equiv\\pm2\\pmod5$$, then the period divides $$2(p+1)$$, but might not divide $$p+1$$. The simplest example is $$p=3$$, where the Fibonacci sequence is $$0,1,1,2,0,2,2,1,\\quad 0,1,1,2,0,2,2,1,\\ldots.$$ Note that the first 0 does not necessarily mean that it will start to repeat. What happens is that the term before the first $$0$$ is $$p-1$$, so the first part of the sequence repeats with negative signs before you get to a consecutive $$0$$ and $$1$$.\n\n\u2022 On the one hand I agree it's probably too elementary, but, on the other hand, I wouldn't have seen this very nice perspective on periods if it hadn't been posted here, so I'm torn. :-) Aug 24 '20 at 20:17\n\u2022 Sorry everyone; I'm still learning about the etiquette of the forum. I found these notes and found them interesting and wanted to learn more. Thanks for answering my question seriously :) Aug 25 '20 at 20:06\n\u2022 As @LSpice comments, one might not have realized that such a question is \"quasi-elementarizable\" in such a decisive manner... Sometimes the decisiveness and elementariness is only visible to a more experienced person, which (to my mind) doesn't mean that the original question was \"too easy for MO\"... If anything, I am fond of examples where a seemingly innocent (but baffling) question succumbs to a more sophisticated viewpoint. As though such viewpoints not only provide publication fodder, but really do answer questions. :) Aug 31 '20 at 17:50\n\nI won't address your question about how small the period of $$\\{F_n \\bmod p\\}$$ can be as $$p$$ grows, but instead ask if the upper bounds on the period can be achieved infinitely often. For consistency I'll use the notation from Joe Silverman's answer: set $$A = (1 + \\sqrt{5})/2$$ and $$B = (1-\\sqrt{5})/2$$, so $$F_n = (A^n - B^n)/(A-B) = (A^n - B^n)/\\sqrt{5}$$. Note $$A+B = 1$$, $$A - B = \\sqrt{5}$$, and $$AB = -1$$.\n\nClaim: For a prime $$p \\not= 2$$ or $$5$$, the period of the Fibonacci sequence $$\\{F_n \\bmod p\\}$$ is the smallest even positive integer $$k$$ such that $$A^k = 1$$ in characteristic $$p$$.\n\nThis claim involves working in the field $$\\mathbf F_p(\\sqrt{5})$$, where $$\\sqrt{5}$$ is a square root of 5 in characteristic $$p$$, so we can regard $$A = (1+\\sqrt{5})/2$$ as a number in the field $$\\mathbf F_p(\\sqrt{5})$$, which is either $$\\mathbf F_p$$ or $$\\mathbf F_{p^2}$$. (The notation $$\\mathbf F_p$$ and $$\\mathbf F_{p^2}$$ are fields of order $$p$$ and $$p^2$$, not having anything to do with the \"$$F$$\" in Fibonacci number notation.) The claim is saying that $$F_{n+k} \\equiv F_n \\bmod p$$ for all $$n \\geq 0$$ (or just all sufficiently large $$n \\geq 0$$) if and only if $$A^k = 1$$ in $$\\mathbf F_p(\\sqrt{5})$$ for even $$k$$, so the period of $$\\{F_n \\bmod p\\}$$ is the smallest even $$k \\geq 1$$ such that $$A^k = 1$$ in $$\\mathbf F_p(\\sqrt{5})$$.\n\nProof. View the congruence $$F_{n+k} \\equiv F_n \\bmod p$$ as an equation $$F_{n+k} = F_n$$ in the subfield $$\\mathbf F_p$$ of $$\\mathbf F_p(\\sqrt{5})$$. The Fibonacci formula $$F_n = (A^n - B^n)/\\sqrt{5}$$ in $$\\mathbf R$$ is also a valid formula in fields of characteristic $$p$$ when we view $$\\sqrt{5}$$ in characteristic $$p$$ and set $$A = (1+\\sqrt{5})/2$$ and $$B = (1-\\sqrt{5})/2 = 1-A$$ in characteristic $$p$$. In $$\\mathbf F_p(\\sqrt{5})$$, \\begin{align*} F_{n+k} = F_n & \\Longleftrightarrow \\frac{A^{n+k}-B^{n+k}}{\\sqrt{5}} = \\frac{A^n-B^n}{\\sqrt{5}} \\\\ & \\Longleftrightarrow A^n(A^k-1) = B^n(B^k-1). \\end{align*} In a field of characteristic $$p \\not= 2$$ or $$5$$, $$A$$ and $$B$$ are nonzero since $$AB = -1$$. Suppose in $$\\mathbf F_p(\\sqrt{5})$$ that $$A^k \\not= 1$$. Then in this field, $$F_{n+k} = F_n \\Longrightarrow (A/B)^n = (B^k-1)/(A^k-1).$$ The ratio $$A/B$$ in characteristic $$p$$ is not $$1$$ since $$A = B \\Longrightarrow 5 = 0$$ in characteristic $$p$$, which is false since $$p \\not= 5$$. Therefore $$(A/B)^n$$ is not constant as $$n$$ varies, but $$(B^k-1)/(A^k-1)$$ is constant as $$n$$ varies. Thus $$A^k = 1$$ in $$\\mathbf F_p(\\sqrt{5})$$, so $$B^n(B^k-1) = A^n(A^k-1) = 0$$, so $$B^k = 1$$ (we never have $$B^n = 0$$ in characteristic $$p$$). Since $$B^k = (-1/A)^k = (-1)^k/A^k$$, we have $$A^k = 1$$ and $$B^k = 1$$ if and only if $$A^k = 1$$ and $$(-1)^k = 1$$. Since $$-1 \\not= 1$$ in characteristic $$p$$ when $$p \\not= 2$$, we have $$A^k = 1$$ and $$(-1)^k = 1$$ in $$\\mathbf F_p(\\sqrt{5})$$ if and only if $$A^k = 1$$ in characteristic $$p$$ and $$k$$ is even.\n\nThat completes the proof of the claim.\n\nSince $$B = -1/A$$, if $$A$$ in characteristic $$p$$ has odd order $$m$$ then $$B$$ in characteristic $$p$$ has order $$2m$$. Therefore the claim says the period of $$\\{F_n \\bmod p\\}$$ is the least $$k \\geq 1$$ such that $$A^k = 1$$ and $$B^k = 1$$ in characteristic $$p$$: that $$k$$ is necessarily even.\n\nFor $$p \\not= 2$$ or 5, the field $$\\mathbf F_p(\\sqrt{5})$$ has order $$p$$ or $$p^2$$ depending on whether or not $$5 \\bmod p$$ is a square: its order is $$p$$ when $$p \\equiv \\pm 1 \\bmod 5$$ and its order is $$p^2$$ when $$p \\equiv \\pm 2 \\bmod 5$$. Therefore the group of nonzero elements $$\\mathbf F_p(\\sqrt{5})^\\times$$ has order $$p-1$$ if $$p \\equiv \\pm 1 \\bmod 5$$ and order $$p^2-1$$ if $$p \\equiv \\pm 2 \\bmod 5$$. Since $$p-1$$ and $$p^2-1$$ are both even, the period of $$\\{F_n \\bmod p\\}$$ divides $$p-1$$ if $$p \\equiv \\pm 1 \\bmod 5$$ and it divides $$p^2-1$$ if $$p \\equiv \\pm 2 \\bmod 5$$. As Joe points out in his answer, when $$p \\equiv \\pm 2 \\bmod 5$$ the period of $$\\{F_n \\bmod p\\}$$ divides $$2(p+1)$$, which is a proper factor of $$p^2-1$$.\n\nThis situation is reminiscent of Artin's primitive root conjecture, which says that for $$a \\in \\mathbf Z$$ that is not $$\\pm 1$$ or a perfect square, there are infinitely many primes $$p$$ such that $$a \\bmod p$$ has order $$p-1$$ in $$\\mathbf F_p^\\times$$, and in fact there is a positive density of such primes. This conjecture is known to be a consequence of the Generalized Riemann Hypothesis (GRH). This conjecture and its connection to GRH can be extended to number fields, and to talk about the multiplicative order of $$A$$ in characteristic $$p$$ amounts to looking at an analogue of Artin's primitive root conjecture with $$\\mathbf Z$$ replaced by $$\\mathbf Z[A]$$, which is the ring of integers of $$\\mathbf Q(\\sqrt{5})$$. This is discussed in Barendrecht's 2018 bachelor's thesis here. For example, GRH implies that the set of (nonzero) prime ideals $$\\mathfrak p$$ in $$\\mathbf Z[A]$$ such that $$A \\bmod \\mathfrak p$$ generates all of $$(\\mathbf Z[A]/\\mathfrak p)^\\times$$ has a positive density using the last result of the thesis, Corollary 3.1.2, and therefore there are infinitely many such prime ideals $$\\mathfrak p$$ in $$\\mathbf Z[A]$$.\n\nEvery nonzero prime ideal $$\\mathfrak p$$ in $$\\mathbf Z[A]$$ is a factor of $$(p) = p\\mathbf Z[A]$$ for some prime number $$p$$: if $$p \\equiv \\pm 1 \\bmod 5$$ then $$(p) = \\mathfrak p\\mathfrak p'$$ for two prime ideals $$\\mathfrak p$$ and $$\\mathfrak p'$$, and $$\\mathbf Z[A]/\\mathfrak p$$ and $$\\mathbf Z[A]/\\mathfrak p'$$ are fields of order $$p$$. If $$p \\equiv \\pm 2 \\bmod 5$$, then $$(p) = \\mathfrak p$$ is a prime ideal in $$\\mathbf Z[A]$$ and $$\\mathbf Z[A]/(p)$$ is a field of order $$p^2$$. When $$p \\equiv \\pm 2 \\bmod 5$$, the multiplicative order of $$A$$ in characteristic $$p$$ is a factor of $$2(p+1)$$, which is less than $$p^2-1$$, so the only prime ideals $$\\mathfrak p$$ in $$\\mathbf Z[A]$$ for which $$A \\bmod \\mathfrak p$$ might generate $$(\\mathbf Z[A]/\\mathfrak p)^\\times$$ are prime ideals dividing a prime $$p \\equiv \\pm 1 \\bmod 5$$, in which case we are in the situation that $$A \\in \\mathbf F_p^\\times$$ has order $$p-1$$. Comparing this to the claim up above, since $$p-1$$ is even when $$p > 2$$ we see that GRH implies that there are infinitely many primes $$p \\equiv \\pm 1 \\bmod 5$$ such that $$\\{F_n \\bmod p\\}$$ has period $$p-1$$.\n\nAmong the 18 odd primes $$p \\equiv \\pm 2 \\bmod 5$$ with $$p < 150$$, $$\\{F_n\\bmod p\\}$$ has period $$2(p+1)$$ all but 3 times (at $$p = 47$$ $$107$$, and $$113$$). There are many generalizations of the Artin primitive root conjecture and I would not be surprised if one of them can show GRH implies there are infinitely many primes $$p \\equiv \\pm 2 \\bmod 5$$ such that $$\\{F_n \\bmod p\\}$$ has period $$2(p+1)$$, but this is not something I am aware of in more detail at the moment.\n\nThe question above is about lower bounds, but I allow myself to comment about upper bounds: $$\\pi(n)$$, the period function of the Fibonacci sequence mod $$n$$, satisfies $$\\pi(n)\\leq 6n$$ and equality holds iff $$n=2\\cdot 5^k$$ for some $$k\\geq 1$$. This fact is well known. In the 90's it was considered here as a puzzle to the monthly readers. $$\\pi(n)$$ was also discussed in an elementary fashion in the 60's in this monthly paper.\n\nBut really, I want to share a little observation which forms a generalization of the above mentioned fact: denoting, for an element $$g\\in \\mathrm{GL}_2(\\mathbb{Z})$$, by $$\\rho_g(n)$$ the order of the image of $$g$$ in $$\\mathrm{GL}_2(\\mathbb{Z}/n)$$, $$\\rho_g(n)\\leq 6n$$. This is a generalization because $$\\rho_g(n)=\\pi(n)$$ for $$g= \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$$. Note that if $$\\det(g)=-1$$ then $$\\rho_g(n)=2\\rho_{g^2}(n)$$, thus it is enough to prove that for $$g\\in \\mathrm{SL}_2(\\mathbb{Z})$$, $$\\rho_g(n)\\leq 3n$$. Let me now fix $$g\\in \\mathrm{SL}_2(\\mathbb{Z})$$, denote $$\\rho(n)=\\rho_g(n)$$ and prove that indeed $$\\rho(n)\\leq 3n$$.\n\nFirst note that, for natural $$p$$ and $$n$$, if $$p$$ divides $$n$$ then $$\\rho(pn)$$ divides $$p\\rho(n)$$. This follows by expanding the right hand side of $$g^{p\\rho(n)}=(g^{\\rho(n)})^p=(1+nh)^p$$ and note that it is 1 mod $$pn$$. By induction we conclude that for every $$k>1$$, $$\\rho(p^k)$$ divides $$p^{k-1}\\rho(p)$$.\n\nAssume now $$p$$ is a prime and note that $$\\rho(p)$$ divides either $$p-1,p+1$$ or $$2p$$. Indeed, if $$\\bar{g}\\in \\mathrm{SL}_2(\\mathbb{F}_p)$$ is diagonalizable over $$\\mathbb{F}_p$$ then its eigenvalues are in $$\\mathbb{F}_p^\\times$$ and their orders divides $$p-1$$, else, if $$\\bar{g}$$ is diagonalizable over $$\\mathbb{F}_{p^2}$$ then its eighenvalues $$\\alpha,\\beta$$ are conjugated by the Frobenius automorphism, thus their order divides $$p+1$$ because $$\\alpha^{p+1}=\\alpha\\alpha^p=\\alpha\\beta=\\det(\\bar{g})=1$$, else $$\\bar{g}$$ has a unique eigenvalue, which must be a $$\\pm 1$$ by $$\\det(\\bar{g})=1$$, thus $$\\bar{g}^2$$ is unipotent and its order divides $$p$$. For $$p=2$$, in the last case, there was no reason to pass to $$g^2$$, as $$-1=1$$ in $$\\mathbb{F}_2$$, thus $$\\rho(2)$$ is either 1,2 or 3.\n\nFrom the above two points, we conclude that for every odd prime $$p$$ and natural $$k$$, $$\\rho(p^k)$$ divides $$p^{k-1}(p-1)$$, $$p^{k-1}(p+1)$$ or $$2p^k$$. All these numbers are even and bounded by $$2p^k$$, thus $$\\mathrm{lcm}\\{\\rho(p^k),2\\} \\leq 2p^k$$. For $$p=2$$ we get that $$\\rho(2^k) \\leq 2^{k-1}\\cdot 3$$.\n\nFix now an arbitrary natural $$n$$. Write $$n=2^km$$ for an odd $$m$$ and decompose $$m=\\prod_{i=0}^r p_i^{k_i}$$. Then \\begin{align*} \\rho(m)= \\mathrm{lcm}\\{\\rho(p_i^{k_i}) \\mid i=0,\\dots r\\} \\leq \\mathrm{lcm}\\{\\mathrm{lcm}\\{\\rho(p_i^{k_i}),2\\} \\mid i=0,\\dots r\\} =\\\\ 2\\mathrm{lcm}\\{\\frac{\\mathrm{lcm}\\{\\rho(p_i^{k_i}),2\\}}{2} \\mid i=0,\\dots r\\} \\leq 2\\prod_{i=0}^r \\frac{\\mathrm{lcm}\\{\\rho(p_i^{k_i}),2\\}}{2}\\leq 2\\prod_{i=0}^r p_i^{k_i} =2m \\end{align*} and we get $$\\rho(n) = \\rho(2^km) \\leq \\rho(2^k) \\cdot \\rho(m) \\leq 2^{k-1}\\cdot 3 \\cdot 2m = 3\\cdot 2^km=3n.$$\n\nThis finishes the proof that $$\\rho(n)\\leq 3n$$.\n\nAs always, it is interesting to analyze the case of equality. For $$g\\in \\mathrm{SL}_2(\\mathbb{Z})$$ we have $$\\rho_g(n)=3n$$ for some $$n$$ iff $$\\mathrm{tr}(g)$$ is odd and not equal $$-1$$ or $$-3$$. If $$g$$ satisfies this condition, then $$n$$ satisfices $$\\rho_g(n)=3n$$ iff $$n=2st$$, for some odd $$s\\geq 3$$, $$t\\geq 1$$ where every prime factor of $$s$$ divides $$\\mathrm{tr}(g)+2$$, every prime factor of $$t$$ divides $$\\mathrm{tr}(g)-2$$ and $$g$$ is not $$\\pm 1$$ modulo any of these prime factors.\n\nFor $$g$$ satisfying $$\\det(g)=-1$$, using the identity $$\\mathrm{tr}(g^2)=\\mathrm{tr}(g)^2-2\\det(g)$$, we get that $$\\rho_g(n)=6n$$ for some $$n$$ iff $$\\mathrm{tr}(g)$$ is odd and in this case, $$n$$ satisfices $$\\rho_g(n)=6n$$ iff $$n=2st$$, for some odd $$s\\geq 3$$, $$t\\geq 1$$ where every prime factor of $$s$$ divides $$\\mathrm{tr}(g)+4$$, every prime factor of $$t$$ divides $$\\mathrm{tr}(g)$$ and $$g$$ is not $$\\pm 1$$ modulo any of these prime factors.\n\nSpecifically for $$g=\\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$$, $$\\det(g)=-1$$, $$\\mathrm{tr}(g)=1$$ is odd, 5 is the only prime factor of $$\\mathrm{tr}(g)+4$$ and there is no prime factor for $$\\mathrm{tr}(g)$$. Since $$g$$ is not $$\\pm 1$$ modulo 5, we get that $$\\pi(n)=\\rho_g(n)=6n$$ iff $$n=2\\cdot 5^k$$ for some $$k\\geq 1$$, as claimed above.", "date": "2021-10-20 17:18:41", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/370028/the-period-of-fibonacci-numbers-over-finite-fields/370030", "openwebmath_score": 0.9715709686279297, "openwebmath_perplexity": 82.96644927802902, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995742876885, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8507615196303259}}
{"url": "https://stats.stackexchange.com/questions/443445/how-to-generate-random-integers-between-1-and-4-that-have-a-specific-mean", "text": "# How to generate random integers between 1 and 4 that have a specific mean?\n\nI need to generate 100 random integers in R, where each integer is between 1 and 4 (hence 1,2,3,4) and the mean is equal to a specific value.\n\nIf I draw random uniform numbers between 1 and 5 and get floor, I have a mean of 2.5.\n\nx = floor(runif(100,min=1, max=5))\n\n\nI need to fix the mean to 1.9 or 2.93 for example.\n\nI guess I can generate random integers that add to 100 * mean but I don't know how to restrict to random integers between 1 and 4.\n\n\u2022 I think this is a bit under-determined... One for instance can get a mean of 1.9 with sample(size=n, x= 1:4, prob=c(3.666,1,1,1), replace=TRUE) but also with sample(size=n, x= 1:4, prob=c(3,1,1,0.715), replace=TRUE). \u2013\u00a0us\u03b5r11852 Jan 5 at 22:53\n\u2022 Are you asking how to constrain the mean of the underlying distribution, or the sample mean? \u2013\u00a0user20160 Jan 5 at 22:53\n\u2022 tha sample mean @user20160 \u2013\u00a0Fierce82 Jan 5 at 23:04\n\u2022 Integers between 1 and 4 only allows for 2 and 3. You also need to specify the distribution that they are drawn randomly from (or make one up). \u2013\u00a0wolfies Jan 6 at 6:28\n\u2022 I voted to leave this open because there's an interesting algorithmic question in here--the R part is incidental; you could just as easily implement this in Python or with a pad and some dice. \u2013\u00a0Matt Krause Jan 6 at 21:14\n\nI agree with X'ian that the problem is under-specified. However, there is an elegant, scalable, efficient, effective, and versatile solution worth considering.\n\nBecause the product of the sample mean and sample size equals the sample sum, the problem concerns generating a random sample of $$n$$ values in the set $$\\{1,2,\\ldots, k\\}$$ that sum to $$s$$ (assuming $$n \\le s \\le kn,$$ of course).\n\nTo explain the proposed solution and, I hope, justify the claim of elegance, I offer a graphical interpretation of this sampling scheme. Lay out a grid of $$k$$ rows and $$n$$ columns. Select every cell in the first row. Randomly (and uniformly) select $$s-n$$ of the remaining cells in rows $$2$$ through $$k.$$ The value of observation $$i$$ in the sample is the number of cells selected in column $$i:$$\n\nThis $$4\\times 100$$ grid is represented by black dots at the unselected cells and colored patches at the selected cells. It was generated to produce a mean value of $$2,$$ so $$s=200.$$ Thus, $$200-100=100$$ cells were randomly selected among the top $$k-1=3$$ rows. The colors represent the numbers of selected cells in each column. There are $$28$$ ones, $$47$$ twos, $$22$$ threes, and $$3$$ fours. The ordered sample corresponds to the sequence of colors from column $$1$$ through column $$n=100.$$\n\nTo demonstrate scalability and efficiency, here is an R command to generate a sample according to this scheme. The question concerns the case $$k=4, n=100$$ and $$s$$ is $$n$$ times the desired average of the sample:\n\ntabulate(sample.int((k-1)*n, s-n) %% n + 1, n) + 1\n\n\nBecause sample.int requires $$O(s-n)$$ time and $$O((k-1)n)$$ space, and tabulate requires $$O(n)$$ time and space, this algorithm requires $$O(\\max(s-n,n))$$ time and $$O(kn)$$ space: that's scalable. With $$k=4$$ and $$n=100$$ my workstation takes only 12 microseconds to perform this calculation: that's efficient.\n\n(Here's a brief explanation of the code. Note that integers $$x$$ in $$\\{1,2,\\ldots, (k-1)n\\}$$ can be expressed uniquely as $$x = nj + i$$ where $$j \\in \\{0,1,\\ldots, k-2\\}$$ and $$i\\in\\{1,2,\\ldots, n\\}.$$ The code takes a sample of such $$x,$$ converts them to their $$(i,j)$$ grid coordinates, counts how many times each $$i$$ appears (which will range from $$0$$ through $$k-1$$) and adds $$1$$ to each count.)\n\nWhy can this be considered effective? One reason is that the distributional properties of this sampling scheme are straightforward to work out:\n\n\u2022 It is exchangeable: all permutations of any sample are equally likely.\n\n\u2022 The chance that the value $$x \\in\\{1,2,\\ldots, k\\}$$ appears at position $$i,$$ which I will write as $$\\pi_i(x),$$ is obtained through a basic hypergeometric counting argument as $$\\pi_i(x) = \\frac{\\binom{k-1}{x-1}\\binom{(n-1)(k-1)}{s-n-x+1}}{\\binom{n(k-1)}{ s-n}}.$$ For example, with $$k=4,$$ $$n=100,$$ and a mean of $$2.0$$ (so that $$s=200$$) the chances are $$\\pi = (0.2948, 0.4467, 0.2222, 0.03630),$$ closely agreeing with the frequencies in the foregoing sample. Here are graphs of $$\\pi_1(1), \\pi_1(2), \\pi_1(3),$$ and $$\\pi_1(4)$$ as a function of the sum:\n\n\u2022 The chance that the value $$x$$ appears at position $$i$$ while the value $$y$$ appears at position $$j$$ is similarly found as $$\\pi_{ij}(x,y) = \\frac{\\binom{k-1}{x-1}\\binom{k-1}{y-1}\\binom{(n-1)(k-1)}{s-n-x-y+2}}{\\binom{n(k-1)}{ s-n}}.$$\n\nThese probabilities $$\\pi_i$$ and $$\\pi_{ij}$$ enable one to apply the Horvitz-Thompson estimator to this probability sampling design as well as to compute the first two moments of the distributions of various statistics.\n\nFinally, this solution is versatile insofar as it permits simple, readily-analyzable variations to control the sampling distribution. For instance, you could select cells on the grid with specified but unequal probabilities in each row, or with an urn-like model to modify the probabilities as sampling proceeds, thereby controlling the frequencies of the column counts.\n\n\u2022 (+1) Ultimate elegance, indeed. \u2013\u00a0Xi'an Jan 6 at 18:56\n\u2022 The answer is too difficult for me to follow, appreciate it nonetheless \u2013\u00a0Fierce82 Jan 7 at 11:33\n\u2022 What an elegant and beautifully presented answer. If you don't mind my humble suggestion as a reader, you might consider presenting the solution first (the counting patches and the great diagram), and then talking about the implementation and how your argument about how it fits the intuition, and finally why it's efficient. It might make it a bit easier to follow. \u2013\u00a0Neil G Jan 8 at 8:49\n\u2022 @Neil Thank you for your suggestion. I think it's a good one and will consider it carefully. \u2013\u00a0whuber Jan 8 at 15:02\n\u2022 This is a lovely and satisfying answer. I did want to note that the numbers are small enough in this case (100 numbers summing to 190) that we can calculate the uniform distribution of all values that satisfy. I ran some calculations to compare your distribution against this and found that yours is much much more likely (billions in some cases) to select small non-1 values. For example, your model will almost never give distributions with >45 \"ones\" (~0.002% chance for 46, vanishing for more), but that comprises ~58% of the uniform model values. \u2013\u00a0Cireo Jan 31 at 5:29\n\nThe question is under-specified in that the constraints on the frequencies \\begin{align}n_1+2n_2+3n_3+4n_4&=100M\\\\n_1+n_2+n_3+n_4&=100\\end{align} do not determine a distribution: \"random\" is not associated with a particular distribution, unless the OP means \"uniform\". For instance, if there exists one solution $$(n_1^0,n_2^0,n_3^0,n_4^0)$$ to the above system, then the distribution degenerated at this solution is producing a random draw that is always $$(n_1^0,n_2^0,n_3^0,n_4^0)$$.\n\nIn the case the question is about simulating a Uniform distribution over the grid\\begin{align}n_1+2n_2+3n_3+4n_4&=100M\\\\n_1+n_2+n_3+n_4&=100\\end{align}one can always use a Metropolis-Hastings algorithm. Starting from $$(n_1^0,n_2^0,n_3^0,n_4^0)$$, create a Markov chain by proposing symmetric random perturbations of the vector $$(n_1^t,n_2^t,n_3^t,n_4^t)$$ and accept if the result is within $$\\{1,2,3,4\\}^4$$ and satisfies the constraints.\n\nFor instance, here is a crude R rendering:\n\ncenM=293\n#starting point (n\u00b9,n\u00b3,n\u2074)\nn<-sample(1:100,3,rep=TRUE)\nwhile((sum(n)>100)|(n[2]-n[1]+2*n[3]!=cenM-200))\nn<-sample(1:100,3,rep=TRUE)\n#Markov chain\nfor (t in 1:1e6){\nprop<-n+sample(-10:10,3,rep=TRUE)\nif ((sum(prop)<101)&\n(prop[2]-prop[1]+2*prop[3]==cenM-200)&\n(min(prop)>0))\nn=prop}\nc(n[1],100-sum(n),n[-1])\n\n\nwith the distribution of $$(n_1,n_3,n_4)$$ over the 10\u2076 iterations:\n\nIn case you want draws of the integers themselves,\n\n sample(c(rep(1,n[1]),rep(2,100-sum(n)),rep(3,n[2]),rep(4,n[3])))\n\n\nis a quick & dirty way to produce a sample.\n\n\u2022 thanks. but I cannot understand how i can utilize this to get the 4 integers (between 1 and 4) \u2013\u00a0Fierce82 Jan 7 at 11:37\n\u2022 This generates the numbers of 1,2,3,4's $n_1,n_2,n_3,n_4)$ so that there are 100 of them and the sum is cenM. The integer themselves are a random permutation of $n_1$ 1's,..., $n_4$ 4's. \u2013\u00a0Xi'an Jan 8 at 7:35\n\nI want to ... uh ... \"attenuate\" @whuber's amazing answer, which @TomZinger says is too difficult to follow. By that I mean I want to re-describe it in terms that I think Tom Zinger will understand, because it's clearly the best answer here. And as Tom gradually uses the method and finds that he needs, say, to know the distribution of the samples rather than just their mean, whuber's answer will be just what he's looking for.\n\nIn short: there are no original ideas here, only a simpler explanation.\n\nYou'd like to create $$n$$ integers from $$1$$ to $$4$$ with mean $$r$$. I'm going to suggest computing $$n$$ integers from $$0$$ to $$3$$ with mean $$r-1$$, and then adding one to each of them. If you can do that latter thing, you can solve the first problem. For instance, if we want 10 integers between $$1$$ and $$4$$ with mean $$2.6$$, we can write down these $$10$$ integers between $$0$$ and $$3$$...\n\n0,3,2,1,3,1,2,1,3,0\n\nwhose mean is $$1.6$$; if we increase each by $$1$$, we get\n\n1,4,3,2,4,2,3,2,4,1\n\nwhose mean is $$2.6$$. It's that simple.\n\nNow let's think about the numbers $$0$$ through $$3$$. I'm going to think of those as \"how many items do I have in a 'small' set?\" I might have no items, one item, two items, or three items. So the list\n\n0,3,2,1,3,1,2,1,3,0\n\nrepresents ten different small sets. The first is empty; the second has three items, and so on. The total number of items in all the sets is the sum of the ten numbers, i.e., $$16$$. And the average number of items in each set is this total, divided by $$10$$, hence $$1.6$$.\n\nwhuber's idea is this: suppose you make yourself ten small sets, with the total number of items being $$10t$$ for some number $$t$$. Then the average size of the sets will be exactly $$t$$. In the same way, if you make yourself $$n$$ sets with a total number of items being $$nt$$, the average number of items in a set will be $$t$$. You say you're interested in the case $$n = 100$$.\n\nLet's make this concrete for your example: you want 100 items between 1 and 4 whose average is $$1.9$$. Using the idea of my first paragraph, I'm going to change this to \"make $$100$$ ints between $$0$$ and $$3$$ whose average is $$0.9$$\". When I'm done, I'll add $$1$$ to each of my ints to get a solution to your problem. So my target average is $$t = 0.9$$.\n\nI want to make $$100$$ sets, each with between $$0$$ and $$3$$ items in it, with an average set-size of $$0.9$$.\n\nAs I've observed above, this means that there have to be a total of $$100 \\cdot 0.9 = 90$$ items in the sets. From the numbers $$1, 2, \\ldots, 300$$, I'm going to select exactly $$90$$. I can indicate the selected ones by making a list of 300 dots and Xs:\n\n..X....X...XX...\n\nwhere the list above indicates that I selected the numbers 3, 9, 13, 14, and then many others that I haven't shown because I got sick of typing. :) I can take this sequence of 300 dots and Xs and break it into three groups of 100 dots each, which I arrange one atop the other, getting something that looks like this:\n\n...X....X..X.....X...\n.X...X.....X...X.....\n..X...X.X..X......X..\n\n\nbut goes on for a full 100 items in each row. The number of Xs in each row might differ -- there might be 35 in the first row, 24 in the second, and 31 in the third, for instance, and that's OK. [Thanks to whuber for pointing out that I had this wrong in a first draft!]\n\nNow look at each column: each column can be considered as a set, and that set has between 0 and 3 \"X\"s in it. I can write the tallies below the rows to get something like this:\n\n...X....X..X.....X...\n.X...X.....X...X.....\n..X...X.X..X......X..\n011101102003000101100\n\n\nThat is to say, I've produced 100 numbers, each between 1 and 3. And the sum of those 100 numbers must be the number of Xs, total, in all three rows, which was 90. So the average must be $$90/100 = 0.9$$, as desired.\n\nSo here are the steps to getting 100 integers between 1 and 4 whose average is exactly $$s$$.\n\n1. Let $$t = s - 1$$.\n2. Compute $$k = 100 t$$; that's how many Xs we'll place in the rows, total.\n3. Make a list of 300 dots-or-Xs, $$k$$ of which are Xs.\n4. Split this into three rows of 100 dots-or-Xs, each containing about a third of the Xs, more or less.\n5. Arrange these in an array, and compute column sums, getting 100 integers between $$0$$ and $$3$$. Their average will be $$t$$.\n6. Add one to each column sum to get 100 integers between $$1$$ and $$4$$ whose average is $$s$$.\n\nNow the tricky part of this is really in step 4: how do you pick $$300$$ items, $$k$$ of which are \"X\" and the other $$300-k$$ of which are \".\"? Well, it turns out that R has a function that does exactly that.\n\nAnd then whuber tells you how to use it: you write\n\ntabulate(sample.int((k-1)*n, s-n) %% n + 1, n)\n\n\nFor your particular case, $$n = 100$$, and $$s$$, the total number of items in all the small sets, is $$100r$$, and you want numbers between $$1$$ and $$4$$, so $$k = 4$$, so $$k-1$$ (the largest size for a 'small set') is 3, so this becomes\n\ntabulate(sample.int(3*100, 100r-100) %% 100 + 1, n)\n\n\nor\n\ntabulate(sample.int(3*100, 100*(r-1)) %% 100 + 1, 100)\n\n\nor, using my name $$t$$ for $$r - 1$$, it becomes\n\ntabulate(sample.int(3*100, 100*t) %% 100 + 1, 100)\n\n\nThe \"+1\" at the end of his original formula is exactly the step needed to convert from \"numbers between $$0$$ and $$3$$\" to \"numbers between $$1$$ and $$4$$\".\n\nLet's work from the inside out, and let's simplify to $$n = 10$$ so that I can show sample outputs:\n\ntabulate(sample.int(3*10, 10*t) %% 10 + 1, 10)\n\n\nAnd let's aim for $$t = 1.9$$, so this becomes\n\ntabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10)\n\n\nStarting with sample.int(3*10, 10*1.9): this produces a list of $$19$$ integers between $$1$$ and $$30$$. (i.e., it solved the problem of picking $$k$$ numbers out of your total -- $$300$$ in your real problem, $$30$$ in my smaller example).\n\nAs you'll recall, we want to produce three rows of ten dots-and-Xs each, something like\n\n X.X.XX.XX.\nXXXX.XXX..\nXX.X.XXX..\n\n\nWe can read this left-to-right-top-to-bottom (i.e., normal reading order) to produce a list of locations for Xs: the first item's a dot; the second and third are Xs, and so on, so our list of locations starts out $$1, 3, 5, 6, \\ldots$$. When we get to the end of a row, we just keep counting up, so for the picture above, the X-locations would be $$1, 3, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 18, 21, 22, 24, 26, 27, 28$$. Is that clear?\n\nWell, whubers code produces exactly that list of locations with its innermost section.\n\nThe next item is %% 10; that takes a number and produces its remainder on division by ten. So our list becomes $$1, 3, 5, 6, 8, 9, 1, 2, 3, 4, 6, 7, 8, 1, 2, 4, 6, 7, 8$$. If we break that into three groups --- those that came from numbers between $$1$$ and $$10$$, those that came from numbers from $$11$$ to $$20$$, and those that came from numbers $$21$$ to $$30$$, we get $$1, 3, 5, 6, 8, 9$$, then $$1, 2, 3, 4, 6, 7, 8,$$, and finally $$1, 2, 4, 6, 7, 8$$. Those tell you where the Xs in each of the three rows are. There's a subtle problem here: if there had been an X in position 10 in the first row, the first of our three lists would have been $$1, 3, 5, 6, 8, 9, 0$$, and the tabulate function doesn't like \"0\". So whuber adds 1 to each item in the list to get $$2, 4, 6, 7, 9, 10, 1$$. Let's move on to the overall computation:\n\ntabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10)\n\n\nThis asks \"for those $$30$$ numbers, each indicating whether there's an X in some column, tell me how many times each column (from $$1$$ to $$10$$ --- that's what the final \"10\" tells you) appears, i.e., tell me how many Xs are in each column. The result is 0 3 2 2 2 1 3 2 3 1 which (because of the shift-by-one thing) you have to read as \"there are no Xs in the 10th column; there are 3 Xs in the first column; there are 2 Xs in the second column,\" and so on up to \"there is one X in the 9th column\".\n\nThat gives you ten integers between $$0$$ and $$3$$ whose sum is $$19$$, hence whose average is $$1.9$$. If you increase each by 1, you get ten integers between $$1$$ and $$4$$ whose sum is $$29$$, hence an average value of $$2.9$$.\n\nYou can generalize to $$n = 100$$, I hope.\n\n\u2022 +1 Welcome to our site, John. I appreciate your efforts to explain and clarify these ideas. At one point your description departs from what the code does: one does not divide the three rows into groups of 30 each. Instead, 90 cells out of the 300 cells in those rows are selected. Usually, each row will have a different number of cells. \u2013\u00a0whuber Jan 7 at 17:15\n\u2022 Thanks...I actually worried about that a little bit as I wrote it, but I was in mid-sentence, and by the time I was finished, the thought had flown. I'll edit to try to fix it up. \u2013\u00a0John Jan 7 at 21:25\n\nYou can use sample() and select specific probabilities for each integer. If you sum the product of the probabilities and the integers, you get the expected value of the distribution. So, if you have a mean value in mind, say $$k$$, you can solve the following equation: $$k = 1\\times P(1) + 2\\times P(2) + 3\\times P(3) + 4\\times P(4)$$ You can arbitrarily choose two of the probabilities and solve for the third, which determines the fourth (because $$P(1)=1-(P(2)+P(3)+P(4))$$ because the probabilities must sum to $$1$$). For example, let $$k=2.3$$, $$P(4)=.1$$, and $$P(3)=.2$$. Then we have that $$k = 1 \\times [1-(P(2)+P(3)+P(4)] + 2\\times P(2) + 3\\times P(3) + 4\\times P(4)$$ $$2.3 = [1 - (P(2)+.1+.2)] + 2*P(2) + 3\\times .2 + 4\\times .1$$ $$2.3 = .7 + P(2) + .6 + .4$$ $$P(2)=.6$$ $$P(1)=1-(P(2)+P(3)+P(4)=1 - (.6+.1+.2)=.1$$\n\nSo you can run x <- sample(c(1, 2, 3, 4), 1e6, replace = TRUE, prob = c(.1, .6, .2, .1)) and mean(x) is approximately $$2.3$$\n\n\u2022 This explains how to constrain the mean of the distribution. But, the OP specified in the comments that they want to constrain the sample mean (which won't match the mean of the distribution, except in expectation). On the other hand, it seems the OP accepted this answer anyway, so perhaps that's not what they wanted after all. \u2013\u00a0user20160 Jan 6 at 0:03\n\u2022 are you sure? @user20160 why sample mean is not contrainted? it's equal to target \u2013\u00a0Fierce82 Jan 6 at 0:08\n\u2022 This answer does not provide a way to make the sample mean equal the target value: most of the time the mean will not equal the target. \u2013\u00a0whuber Jan 6 at 0:10\n\u2022 @TomZinger Yes. This answer nicely describes how to sample from a distribution with the given target mean. But, the mean of a sample drawn from a distribution will not generally equal the mean of the distribution. \u2013\u00a0user20160 Jan 6 at 4:59\n\u2022 I wrote my answer before I saw that comment, but I figured this would be useful anyway. I imagined it would require an integer programming optimization problem to get a sample mean exactly equal to some value. \u2013\u00a0Noah Jan 6 at 8:20\n\nHere is a simple algorithm: Create $$n-1$$ random integers in the range $$[1,4]$$ and calculate the $$n^{th}$$ integer for the mean to be equal to the specified value. If that number is smaller than $$1$$ or larger than $$4$$, then one by one distribute the surplus/lacking onto other integers, e.g. if the integer is $$5$$, we have $$1$$ surplus; and we may add this to the next integer if it's not $$4$$, else add to the next etc. Then, shuffle the entire array.\n\n\u2022 One big problem with this proposal is that it doesn't come along with any indication of what the expected frequencies of the resulting values are. \u2013\u00a0whuber Jan 5 at 23:32\n\u2022 Although interesting, I thought the OP only requires an algorithm to generate the desired array of integers in a non-deterministic manner. \u2013\u00a0gunes Jan 5 at 23:37\n\u2022 I think that avoids the essence of the question rather than providing a satisfactory answer. A good answer should be able to characterize the distribution it proposes in a meaningful way, such as by giving a formula for the probabilities or at least giving the first couple of moments. \u2013\u00a0whuber Jan 5 at 23:46\n\u2022 A minor adjustment of the simulated data is likely 'proper', however, looking at the expertimental design in cases where more significant mean deviation is required, depending on the intended purpose, could be, from a hypothesis testing perspective, 'suspect', in my judgement. Either over or under loading a random design to justify or reject possible non-random effects that have been actually observed can be questionable practice. So, any method that makes a very small adjustment to the last of say a 100 observations is probably keeping in good practice, in my opinion. \u2013\u00a0AJKOER Jan 7 at 17:33\n\nAs a supplement to whuber's answer, I've written a script in Python which goes through each step of the sampling scheme. Note that this is meant for illustrative purposes and is not necessarily performant.\n\nExample output:\n\nn=10, s=20, k=4\n\nStarting grid\n. . . . . . . . . .\n. . . . . . . . . .\n. . . . . . . . . .\nX X X X X X X X X X\n\nFilled in grid\nX X . . X . X . . X\n. . X X X . . . . .\n. . . . X X . . . .\nX X X X X X X X X X\n\nFinal grid\nX X . . X . X . . X\n. . X X X . . . . .\n. . . . X X . . . .\nX X X X X X X X X X\n2 2 2 2 4 2 2 1 1 2\n\n\nThe script:\n\nimport numpy as np\n\n# Define the starting parameters\nintegers = [1, 2, 3, 4]\nn = 10\ns = 20\nk = len(integers)\n\ndef print_grid(grid, title):\nprint(f'\\n{title}')\nfor row in grid:\nprint(' '.join([str(element) for element in row]))\n\n# Create the starting grid\ngrid = []\nfor i in range(1, k + 1):\nif i < k:\ngrid.append(['.' for j in range(n)])\nelse:\ngrid.append(['X' for j in range(n)])\n\n# Print the starting grid\nprint_grid(grid, 'Starting grid')\n\n# Randomly and uniformly fill in the remaining rows\nindexes = np.random.choice(range((k - 1) * n), s - n, replace=False)\nfor i in indexes:\nrow = i // n\ncol = i % n\ngrid[row][col] = 'X'\n\n# Print the filled in grid\nprint_grid(grid, 'Filled in grid')\n\n# Compute how many cells were selected in each column\ncolumn_counts = []\nfor col in range(n):\ncount = sum(1 for i in range(k) if grid[i][col] == 'X')\ncolumn_counts.append(count)\ngrid.append(column_counts)\n\n# Print the final grid and check that the column counts sum to s\nprint_grid(grid, 'Final grid')\nprint()\nprint(f'Do the column counts sum to {s}? {sum(column_counts) == s}.')\n\n\nI've turned whuber's answer into an r function. I hope it helps someone.\n\n\u2022 n is how many integers you want;\n\u2022 t is the mean you want; and\n\u2022 k is the upper limit you want for your returned values\nwhubernator<-function(n=NULL, t=NULL, kMax=5){\nz = tabulate(sample.int(kMax*(n), (n)*(t),replace =F) %% (n)+1, (n))\nreturn(z)\n}\n\n\nIt seems to work as expected:\n\n> w = whubernator(n=10,t=4.2)\n> mean(w)\n[1] 4.2\n> length(w)\n[1] 10\n> w\n[1] 3 5 3 5 5 3 4 5 5 4\n\n\nIt can return 0s, which matches my needs.\n\n> whubernator(n=2,t=0.5)\n[1] 1 0\n\nNew contributor\ngruvn is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.", "date": "2020-04-09 15:19:18", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/443445/how-to-generate-random-integers-between-1-and-4-that-have-a-specific-mean", "openwebmath_score": 0.7439496517181396, "openwebmath_perplexity": 438.7224639689237, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9658995713428387, "lm_q2_score": 0.8807970748488297, "lm_q1q2_score": 0.8507615170365108}}
{"url": "https://www.physicsforums.com/threads/complex-integer-expression-problem.111954/", "text": "# Homework Help: Complex integer expression problem\n\n1. Feb 23, 2006\n\n### Werg22\n\nIf n is a positive integer such as\n\n$$2{\\leq}n{\\leq}80$$\n\nFor how many values the expression $$\\frac{(n+1)n(n-1)}{8}$$ takes positive and integer values?\n\nI solved it that way...\n\n$$\\frac{(n+1)n(n-1)}{8}=\\frac{(n^{2}-1)n}{8}$$\n\n(n^2 - 1)n must have 8 as one of its factor.\n\nEither n is a multiple of 8, or n^2 - 1 is. Also the case were n^2 - 1 has 4 as one of its factors, n having 2, and vice-versa, is impossible - if n^2 - 1 is even, n is odd, and vice-versa.\n\nSo every mutliple of 8 up to 80 is a possible value. So there is 10 values.\n\nLet's list those numbers\n\n8, 16 , 24 , 32 , 40 , 48 , 56 , 64 , 72, 80\n\nAdd one to each one of these values\n\n9, 17, 25, 33, 41, 49, 65 , 73, 81\n\nThere is 4 perfect square in this list. So if n^2 - 1 is a multiple of 8, then there is 4 possible values for n.\n\n10 + 4 = 14\n\nSo 14 possibilities in total. But the true awnser is not what I found. What is wrong in my reasoning?\n\n2. Feb 23, 2006\n\n### 0rthodontist\n\nThere's no guarantee that n^2 - 1 lies between 9 and 81. It could be much larger.\n\n3. Feb 23, 2006\n\n### Werg22\n\nRight! Thanks. In that case any other way to solve this?\n\n4. Feb 23, 2006\n\n### 0rthodontist\n\nWell since you are only looking up to 80 you can calculate them all directly, only takes a couple minutes to set things up. Also from observing that data it seems that every n yields an integer n(n+1)(n-1)/8 except for even integers that are not divisible by 8, so maybe you could break it down into parts. Every multiple of 8 yields an integer. And if x is a multiple of 2, and x is not itself divisible by 8, then x does not yield an integer because its adjacent integers are not even disible by 2. And every odd integer must yield an integer because its adjacent integers are both even, and one of the adjacent integers must also be divisible by 4.\n\nLast edited: Feb 23, 2006\n5. Feb 23, 2006\n\n### Werg22\n\nOk. Since n^2 - 1 = (n-1)(n+1)\n\nn - 1 must be a multiple of 8, or n+1 be a multiple of 8, or n - 1 be a multiple of 4, or n + 1 be a multiple of 4.\n\nFor exactly 10 values, n is multiple of 8\n\nFor exactly 10 values, n-1 is a multiple of 8.\nFor exactly 10 values, n+1 is a multiple of 8.\n\nFor exactly 20 values, n-1 is multiple of 4. Half of these being multiples of 8. So we count 10.\nFor exactly 20 values, n+1 is a multiple of 8. Half of these being multiples of 8. So we count 10.\n\nSo 10*5=50\n\nThe awnser is 50.\n\n6. Feb 23, 2006\n\n### AKG\n\nI believe the answer is 49. You seem to be going about it in a very complicated way.\n\nCase 1: n is odd\nThen both n-1 and n+1 are even. Moreover, one of them (and in fact, only one of them) is a multiple of four. This is clear since if n is odd, then either n = 1 (mod 4) in which case n-1 = 0 (mod 4) and n+1 = 2 (mod 4), or n = 3 (mod 4) in which case n-1 = 2 (mod 4) and n+1 = 0 (mod 4). So since one of n-1 and n+1 is a multiple of four, and the other is even, the whole product (n-1)n(n+1) is a multiple of 8. So every odd n between 2 and 80 will do, and there are 39 such numbers.\n\nCase 2: n is even\nThen both n-1 and n+1 is odd, so if 8 | (n-1)n(n+1), then 8 | n, so the only even n's that work are multiples of 8. They are:\n\n8, 16, 24, 32, 40, 48, 56, 64, 72, 80\n\nThat's 10, giving a total of:\n\n49.\n\nOne problem with your solution is that n-1 is a multiple of 8 for only 9 values of n. Note that n-1 ranges from 1 to 79. n ranges from 2 to 80. n-1 will never be 80, so its missing one multiple of 8. That's where you're counting your extra one. Note also that there are only 19 values of n-1 which are a multiple of 4, 9 of which are multiples of 8, so when you subtract 9 from 19, you do still end up getting 10. Originally, you subtracted 10 from 20. You ended up with the right number, 10, but they way you got it was wrong.\n\n7. Feb 24, 2006\n\n### Werg22\n\nOkay, I see. Thanks.", "date": "2018-04-21 08:12:18", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/complex-integer-expression-problem.111954/", "openwebmath_score": 0.5449621677398682, "openwebmath_perplexity": 369.6972060076372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018441287092, "lm_q2_score": 0.8723473879530491, "lm_q1q2_score": 0.8507147814526761}}
{"url": "https://math.stackexchange.com/questions/1089741/does-sum-limits-n-log-left11-over-n-right-diverge-or-converge", "text": "Does $\\sum\\limits_n \\log\\left(1+{1\\over n}\\right)$ diverge or converge?\n\nHow do I find out if $\\sum\\limits_n\\log(1+{1\\over n})$ diverges or converges? Wolfram recommends me to use comparison test, but I do not know series which diverges and less than this.\n\n\u2022 One can use Limit Comparison with $\\sum \\frac{1}{n}$. \u2013\u00a0Andr\u00e9 Nicolas Jan 3 '15 at 15:37\n\nWolfram recommends me to use (some) compar(is)on test, but I do not know (any) series which diverges and (is) less than this (one).\n\n$$\\log\\left(1+\\frac1n\\right)\\geqslant\\frac1{2n}$$\n\n$$\\sum_{n=1}^{N}\\log\\left(1+\\frac{1}{n}\\right)=\\log\\prod_{n=1}^{N}\\frac{n+1}{n}=\\log(N+1).$$\n\nIf you want to use a comparison, notice that since $f(t)=\\frac{1}{t}$ is a convex function on $\\mathbb{R}^+$, we have: $$\\log\\left(1+\\frac{1}{n}\\right)=\\int_{n}^{n+1}\\frac{dt}{t}\\geq\\frac{1}{n+1/2}$$ by Jensen's inequality, hence: $$\\sum_{n=1}^{N}\\log\\left(1+\\frac{1}{n}\\right)\\geq 2\\sum_{n=1}^{N}\\frac{1}{2n+1} = 2H_{2n+1}-H_n.$$\n\nSince\n\n$$\\sum_{n = 1}^N \\log\\left(1 + \\frac{1}{n}\\right) = \\sum_{n = 1}^N \\log\\left(\\frac{n+1}{n}\\right) = \\sum_{n = 1}^N [\\log(n+1) - \\log(n)] = \\log(N+1) \\to \\infty$$\n\nthe series $\\sum_{n = 1}^\\infty \\log\\left(1 + \\frac{1}{n}\\right)$ diverges.\n\nnote $$\\ln{\\left(1+\\dfrac{1}{n}\\right)}=\\dfrac{1}{n}+o(1/n)$$ since $$\\sum_{n=1}^{\\infty}\\dfrac{1}{n}$$ is diverges\n\nso $$\\sum_{n=1}^{\\infty}\\ln{\\left(1+\\dfrac{1}{n}\\right)}$$ is also diverges\n\n\u2022 First statement isn't obvious to me. \u2013\u00a0Dark Archon Jan 3 '15 at 16:10\n\u2022 do you know $\\ln{(1+x)}=x+o(x)?$ \u2013\u00a0math110 Jan 3 '15 at 16:11\n\u2022 What is $o(x)$? \u2013\u00a0Dark Archon Jan 3 '15 at 16:13\n\u2022 $x=\\dfrac{1}{n}$ when $n\\to +\\infty$,then $x\\to 0$ \u2013\u00a0math110 Jan 3 '15 at 16:25\n\u2022 @DarkArchon $f(x)=o(x)$ means that $\\lim_{x\\to 0}f(x)/x=0$. \u2013\u00a0Andr\u00e9s E. Caicedo Jan 4 '15 at 0:52\n\nThe first natural idea to understand how $\\log(1+1/n)$ behaves when $n$ is small is to do Taylor expansion around $x=0$ for $\\log(1+x)$, so that you have estimates for $\\log(1+x)$. Then you get $$\\log(1+x) = \\sum_{n \\ge 1} \\frac{(-1)^{n+1}}{n}x^n = x - x^2/2 + x^3/3 - \\cdots,$$ by integrating the geometric series (and switching a minus sign) and by Taylor's theorem you get that there exists $1 \\le \\zeta \\le x$ such that $$\\log(1+x) = x-\\frac{\\zeta^2}2 \\ge x - \\frac {x^2}2 = \\frac{2x-x^2}2 = \\frac{x(2-x)}2 \\ge \\frac x2$$ for $0 \\le x \\le 1$ (because $\\frac{x(1-x)}2 \\ge 0$ on this interval).\n\nA variant of this idea is to check that $\\log$ is a concave function (the derivatives are $\\frac 1x$ and $\\frac {-1}{x^2}$), hence we can use Jensen's inequality : for all $\\lambda \\in [0,1]$, $$\\log(\\lambda x + (1-\\lambda) y) \\ge \\lambda \\log(x) + (1-\\lambda) \\log(y)$$ so that for $y=1$ and $x=2$, we get $$\\log(1+ \\lambda) \\ge \\lambda \\log(2).$$ In both cases, we established an inequality of the form $\\log(1+x) \\ge cx$ for $x \\in [0,1]$, which means $\\log(1+1/n) \\ge \\frac cn$, hence your series diverges.\n\nHope that helps,\n\nAnother way to think of this problem is that $\\sum log(1+1/n)$ = $\\log \\prod(1+1/n)$ = $\\log \\prod((n+1)/n)$, which goes to infinity. Thus the sum diverges.\n\n\u2022 Already explained on the page, twice. \u2013\u00a0Did May 11 '15 at 5:21", "date": "2020-02-18 10:14:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1089741/does-sum-limits-n-log-left11-over-n-right-diverge-or-converge", "openwebmath_score": 0.951004683971405, "openwebmath_perplexity": 301.8689623155897, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018405251301, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8507147718362793}}
{"url": "http://math.stackexchange.com/questions/56556/a-problem-about-the-limit-of-an-integral", "text": "# A problem about the limit of an integral\n\nLet $g(x)$ be a continuous periodic function of period 1 on $\\mathbb{R}$. Prove that for any integrable function $f(x)$ on $[0,1]$,\n$$\\lim_{n \\to \\infty}\\int_0^{1}f(x)g(nx)dx= \\int_0^{1}f(x)dx \\int_0^{1}g(x)dx.$$\n\nAny help is appreciated.\n\n-\n\nBegin with a transformation $u=nx.$\n\n$$\\int_{0}^1 f(x)g(nx)dx = \\frac{1}{n} \\int_{0}^n f(u/n)g(u)du.$$\n\nBreak the integrand up into a sum of intervals.\n\n$$\\frac{1}{n} \\int_{0}^n f(u/n)g(u)du=\\frac{1}{n}\\sum_{j=1}^n\\int_{j-1}^{j} f(u/n)g(u)du.$$\n\nMake another variable transformation: $v=u-(j-1).$ Because $g(u)$ is 1 periodic $g(u)=g(u+1)=g(u+j-1).$\n\n$$\\frac{1}{n}\\sum_{j=1}^n\\int_{j-1}^{j} f(u/n)g(u)du = \\frac{1}{n}\\sum_{j=1}^n\\int_{0}^{1} f \\left(\\frac{u-(j-1)}{n} \\right)g(u-(j-1))du$$\n\nRearrange the terms. $$\\frac{1}{n}\\sum_{j=1}^n\\int_{0}^{1} f\\left(\\frac{u-(j-1)}{n} \\right)g(u-(j-1))du =\\int_{0}^1 \\left(\\frac{1}{n} \\sum_{j=1}^n f \\left(\\frac{u-(j-1)}{n} \\right)\\right)g(u)du .$$\n\nWe have now produced a Riemann sum which converges to an integral in the limit. We are now allowed, by dominated convergence theorem, to say\n\n\\begin{align*} \\lim_{n\\to \\infty} \\int_{0}^1 f(x)g(nx)dx &= \\int_{0}^1 \\left(\\int_{0}^1 f(z)dz\\right) g(u)du \\\\ &=\\int_{0}^1 f(z)dz\\int_{0}^1 g(u)du=\\int_{0}^1 f(x)dx\\int_{0}^1 g(x)dx . \\end{align*}\n\n-\nThank you so much. \u2013\u00a0Sume Aug 10 '11 at 3:28\n\nthis is not a good proof, just an idea. First consider f to be simple function, we can calculate that this identity holds. Then by the definition of Lebesgue integration, we can approximate f by simple functions. Because g is continuous(bounded), so there's no problem for the left hand side to converge.\n\n-\n\nOk. I have seen this problem in the book: Principles of Real Analysis by C.D.Aliprantis and O.Burkinshaw, and since I knew that this book has a solution manual, I went and searched over there and got the solution. The problem given in the book is as follows:\n\n$\\textbf{Problem.}$ Let $f:(0,\\infty) \\to \\mathbb{R}$ be a real-valued continuous function such that $f(x+1)=f(x)$ for all $x \\geq 0$. If $g:[0,1] \\to \\mathbb{R}$ is an arbitrary continuous function, then show that $$\\lim_{n \\to \\infty} \\ \\int\\limits_{0}^{1} g(x) \\cdot f(nx) \\ dx = \\Biggl( \\ \\ \\int\\limits_{0}^{1} g(x) \\ dx \\Biggr) \\cdot \\Biggl( \\ \\ \\int\\limits_{0}^{1} f(x) \\ dx\\Biggr)$$\n\n$\\textbf{Solution.}$ Please see the book: Problems in real analysis a workbook with solutions Problem 23.14 Page $\\textbf{205}$ for a complete solution.\n\n-", "date": "2016-05-30 22:21:33", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/56556/a-problem-about-the-limit-of-an-integral", "openwebmath_score": 0.9926888942718506, "openwebmath_perplexity": 332.63785829137953, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9752018398044143, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8507147566437098}}
{"url": "https://benjaminleroy.github.io/documents/36350/Labs/lab_3.1-assign.html", "text": "Name:\nAndrew ID:\nCollaborated with:\n\nThis lab is to be done in class (completed outside of class if need be). You can collaborate with your classmates, but you must identify their names above, and you must submit your own lab as an knitted HTML file on Canvas, by Tuesday 10pm, this week.\n\nThis week\u2019s agenda: creating and updating functions; understanding argument and return structures; revisiting Shakespeare\u2019s plays; code refactoring.\n\n# Huber loss function\n\nThe Huber loss function (or just Huber function, for short) is defined as: $\\psi(x) = \\begin{cases} x^2 & \\text{if |x| \\leq 1} \\\\ 2|x| - 1 & \\text{if |x| > 1} \\end{cases}$ This function is quadratic on the interval [-1,1], and linear outside of this interval. It transitions from quadratic to linear \u201csmoothly\u201d, and looks like this:\n\nIt is often used in place of the usual squared error loss for robust estimation. The sample average, $$\\bar{X}$$\u2014which given a sample $$X_1,\\ldots,X_n$$ minimizes the squared error loss $$\\sum_{i=1}^n (X_i-m)^2$$ over all choices of $$m$$\u2014can be inaccurate as an estimate of $$\\mathbb{E}(X)$$ if the distribution of $$X$$ is heavy-tailed. In such cases, minimizing Huber loss can give a better estimate. (Interested in hearing more? Come ask Tudor or I!)\n\n\u2022 1a. Write a function huber() that takes as an input a number $$x$$, and returns the Huber value $$\\psi(x)$$, as defined above. Hint: the body of a function is just a block of R code, i.e., in this code you can use if() and else() statements. Check that huber(1) returns 1, and huber(4) returns 7.\n\n\u2022 1b. The Huber function can be modified so that the transition from quadratic to linear happens at an arbitrary cutoff value $$a$$, as in: $\\psi_a(x) = \\begin{cases} x^2 & \\text{if |x| \\leq a} \\\\ 2a|x| - a^2 & \\text{if |x| > a} \\end{cases}$ Starting with your solution code to the last question, update your huber() function so that it takes two arguments: $$x$$, a number at which to evaluate the loss, and $$a$$ a number representing the cutoff value. It should now return $$\\psi_a(x)$$, as defined above. Check that huber(3, 2) returns 8, and huber(3, 4) returns 9.\n\n\u2022 1c. Update your huber() function so that the default value of the cutoff $$a$$ is 1. Check that huber(3) returns 5.\n\n\u2022 1d. Check that huber(a = 1, x = 3) returns 5. Check that huber(1, 3) returns 1. Explain why these are different.\n\n\u2022 1e. Vectorize your huber() function, so that the first input can actually be a vector of numbers, and what is returned is a vector whose elements give the Huber evaluated at each of these numbers. Hint: you might try using ifelse(), if you haven\u2019t already, to vectorize nicely. Check that huber(x = 1:6, a = 3) returns the vector of numbers (1, 4, 9, 15, 21, 27).\n\n\u2022 Challenge. Your instructor computed the Huber function values $$\\psi_a(x)$$ over a bunch of different $$x$$ values, stored in huber_vals and x_vals, respectively. However, the cutoff $$a$$ was, let\u2019s say, lost. Using huber_vals, x_vals columns of oops_df, and the definition of the Huber function, you should be able to figure out the cutoff value $$a$$, at least roughly. Estimate $$a$$ and explain how you got there. Hint: one way to estimate $$a$$ is to do so visually, using plotting tools (if you do - please use ggplot); there are other ways too.\n\noops_df <- data.frame(\nx_vals = seq(0, 5, length=21),\nhuber_vals = c(0.0000, 0.0625, 0.2500, 0.5625, 1.0000, 1.5625, 2.2500,\n3.0625, 4.0000, 5.0625, 6.2500, 7.5625, 9.0000, 10.5000,\n12.0000, 13.5000, 15.0000, 16.5000, 18.0000, 19.5000,\n21.0000))\n\n# Shakespeare\u2019s complete works\n\nRecall, as in lab/hw from Week 1, that the complete works of William Shakespeare are available freely from Project Gutenberg. We\u2019ve put this text file up at https://raw.githubusercontent.com/benjaminleroy/36-350-summer-data/master/Week1/shakespeare.txt.\n\n# Getting lines of text play-by-play\n\n\u2022 2a. Below is the get_wordtab_from_url() from lecture. Modify this function so that the string vectors lines and words are both included as named components in the returned list. For good practice, update the documentation in comments to reflect your changes. Then call this function on the URL for the Shakespeare\u2019s complete works (with the rest of the arguments at their default values) and save the result as shakespeare_wordobj. Using head(), display the first several elements of (definitely not all of!) the lines, words, and wordtab components of shakespeare_wordobj, just to check that the output makes sense to you.\n#' Get a word table from text on the web\n#'\n#' @param str_url string, specifying URL of a web page\n#' @param split string, specifying what to split on. Default is the regex\n#' pattern \"[[:space:]]|[[:punct:]]\"\n#' @param tolower Boolean, TRUE if words should be converted to lower case\n#' before the word table is computed. Default is TRUE\n#' @param keep_nums Boolean, TRUE if words containing numbers should be kept in\n#' the word table. Default is FALSE\n#'\n#' @return list, containing word table, and then some basic numeric summaries\n#'\n#' @examples\n#' endgame_speech_list <- get_wordtab_from_url(\n#'  paste0(\"https://raw.githubusercontent.com/benjaminleroy/\",\n#'  \"36-350-summer-data/master/Week1/endgame.txt\"))\nget_wordtab_from_url <- function(str_url, split = \"[[:space:]]|[[:punct:]]\",\ntolower = TRUE,\nkeep_nums = FALSE) {\ntext <- paste(lines, collapse = \" \")\nwords <- strsplit(text, split = split)[[1]]\nwords <- words[words != \"\"]\n\n# Convert to lower case, if we're asked to\nif (tolower) {\nwords <- tolower(words)\n}\n\n# Get rid of words with numbers, if we're asked to\nif (!keep_nums) {\nwords <- grep(\"[0-9]\", words, inv = TRUE, val = TRUE)\n}\n\n# Compute the word table\nwordtab <- table(words)\n\nreturn(list(wordtab = wordtab,\nnumber_unique_words = length(wordtab),\nnumber_total_words = sum(wordtab),\nlongest_word = words[which.max(nchar(words))]))\n}\n\u2022 2b. Go back and look Q5 of Homework 1, where you located Shakespeare\u2019s plays in the lines of text for Shakespeare\u2019s complete works. Set shakespeare_lines <- shakespeare_wordobj$lines, and then rerun your solution code (or the rerun the official solution code, if you\u2019d like) for questions Q5a\u2013Q5f of Homework 1, on the lines of text stored in shakespeare_wordobj$lines. You should end up with two vectors titles_start and titles_end, containing the start and end positions of each of Shakespeare\u2019s plays in shakespeare_lines. Print out titles_start and titles_end to the console.\n\n\u2022 2c. Create a list shakespeare_lines_by_play of length equal to the number of Shakespeare\u2019s plays (a number you should have already computed in the solution to the last question). Using a for() loop, and relying on titles_start and titles_end, extract the subvector of shakespeare_lines for each of Shakespeare\u2019s plays, and store it as a component of shakespeare_lines_by_play. That is, shakespeare_lines_by_play[[1]] should contain the lines for Shakespeare\u2019s first play, shakespeare_lines_by_play[[2]] should contain the lines for Shakespeare\u2019s second play, and so on. Name the components of shakespeare_lines_by_play according to the titles of the plays.\n\n# Getting word tables play-by-play\n\n\u2022 3a. Define a function get_wordtabfrom_lines() to have the same argument structure as get_wordtab_from_url(), which recall you last updated in Q2a, except that the first argument of get_wordtab_from_lines() should be lines, a string vector passed by the user that contains lines of text to be processed. The body of get_wordtab_from_lines() should be the same as get_wordtab_from_url(), except that lines is passed and does not need to be computed using readlines(). The output of get_wordtab_from_lines() should be the same as get_wordtab_from_url(), except that lines does not need to be returned as a component. For good practice, incude documentation for your get_wordtab_from_lines() function in comments (no need to include an example).\n\n\u2022 3b. Using a for() loop or one of the apply functions (your choice here), run the get_wordtab_from_lines() function on each of the components of shakespeare_lines_by_play, (with the rest of the arguments at their default values). Store the result in a list called shakespeare_wordobj_by_play. That is, shakespeare_wordobj_by_play[[1]] should contain the result of calling this function on the lines for the first play, shakespeare_wordobj_by_play[[2]] should contain the result of calling this function on the lines for the second play, and so on.\n\n\u2022 3c. Using one of the apply functions, compute numeric vectors shakespeare_total_words_by_play and shakespeare_unique_words_by_play, that contain the number of total words and number of unique words, respectively, for each of Shakespeare\u2019s plays. Each vector should only require one line of code to compute. Hint: \"[[\"() is actually a function that allows you to do extract a named component of a list; e.g., try \"[[\"(shakespeare_wordobj, \"number_total_words\"), and you\u2019ll see this is the same as shakespeare_wordobj[[\"number_total_words\"]]; you should take advantage of this functionality in your apply call. What are the 5 longest plays, in terms of total word count? The 5 shortest plays?\n\n# Refactoring the word table functions\n\n\u2022 4. Look back at get_wordtab_from_lines() and get_wordtab_from_url(). Note that they overlap heavily, i.e., their bodies contain a lot of the same code. Redefine get_wordtab_from_url() so that it just calls get_wordtab_from_lines() in its body. Your new get_wordtab_from_url() function should have the same inputs as before, and produce the same output as before. So externally, nothing will have changed; we are just changing the internal structure of get_wordtab_from_url() to clean up our code base (specifically, to avoid code duplication in our case). This is an example of code refactoring.\n\nCall your new get_wordtab_from_url() function on the URL for Shakespeare\u2019s complete works, saving the result as shakespeare_wordobj2. Compare some of the components of shakespeare_wordobj2 to those of shakespeare_wordobj (which was computed using the old function definition) to check that your new implementation works as it should.\n\n\u2022 Challenge. Check using all.equal() whether shakespeare_wordobj and shakespeare_wordobj2 are the same. Likely, this will not return TRUE. (If it does, then you\u2019ve already solved this challenge question!) Modify your get_wordtab_from_url() function from the last question, so that it still calls get_wordtab_from_lines() to do the hard work, but produces an output exactly the same as the original shakespeare_wordobj object. Demonstrate your success by calling all.equal() once again.", "date": "2022-08-11 07:42:11", "meta": {"domain": "github.io", "url": "https://benjaminleroy.github.io/documents/36350/Labs/lab_3.1-assign.html", "openwebmath_score": 0.4631035327911377, "openwebmath_perplexity": 2044.6783570558373, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668723123672, "lm_q2_score": 0.8670357735451835, "lm_q1q2_score": 0.8507067781122616}}
{"url": "http://mathhelpforum.com/geometry/143843-solved-finding-co-ordinates-rectangle.html", "text": "# Math Help - [SOLVED] Finding Co-Ordinates of a Rectangle\n\n1. ## [SOLVED] Finding Co-Ordinates of a Rectangle\n\nHere's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:\n\nIs my method correct, considering I did get the right answer?\n\nBut is there another simpler method to do this which would save time during an exam.\n\n2. The diagonals bisect one another. The midpoint of $\\overline{AC}$ is ?\n\n3. Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?\n\n4. Originally Posted by unstopabl3\nMid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?\nBut the y-coordinate of $B~\\&~D$ is 6.\nYou are given the x-coordinate of $D$, so $Dh,6)\" alt=\"Dh,6)\" />\n\n5. No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.\n\nI have already gotten the correct values by using the method mentioned in my first post.\nAs stated I want someone to solve this part of the question with a different, possibly easier method.\n\nI am not after the answer, I am looking for an alternate method.\n\n6. Originally Posted by unstopabl3\nNo, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.\n\nI have already gotten the correct values by using the method mentioned in my first post.\nAs stated I want someone to solve this part of the question with a different, possibly easier method.\n\nI am not after the answer, I am looking for an alternate method.\nHi unstopabl3,\n\nI really don't see what method you used in your first post, but here's how I'd do it.\n\nPlato already told you that the midpoint of BD is M(6, 6).\n\nThis means the y-coordinates of B and D are also 6.\n\nThis distance from B to D is 20 (found using distance formula)\n\nEach individual segment of the diagonals measure 10 since they are bisected.\n\nUsing the distance formula, it is easy to determine the x-coordinates of B and D.\n\nM(6, 6) -----> D(h, 6) = 10\n\n$10=\\sqrt{h-6)^2+(6-6)^2}$\n\n7. Hello, unstopabl3!\n\nCode:\n              |           C(12,14)\n|           o\n|       *    *\n|   *         *\n*              *\nB o - + - - - - - - - o D\n*  |           *\n------*-+-------*------------\n*|   *\no\nA|(0,-2)\n|\n\nThe diagram shows a rectangle $ABCD.$\nWe have: . $A(0,-2),\\;C(12,14)$\nThe diagonal $BD$ is parallel to the $x$-axis.\n\n$(i)$ Explain why the $y$-coordinate of $D$ is 6.\nThe diagonals of a rectangle bisect each other.\n. . Hence, the midpoint of $AC$ is the midpoint of $BD.$\n\nThe midpoint of AC is: . $\\left(\\tfrac{0+12}{2},\\;\\tfrac{-2+14}{2}\\right) \\:=\\:(6,6)$\n\nTherefore, $B$ and $D$ have a $y$-coordinate of 6.\n\nThe $x$-coordinate of $D$ is $h.$\n$(ii)$ Express the gradients of $AD$ and $CD$ in terms of h,.\nWe have: . $\\begin{Bmatrix}A(0, -2) \\\\ C(12,14) \\\\ D(h,\\;6) \\end{Bmatrix}$\n\n$m_{AD} \\;=\\;\\frac{6(-2)}{h-9} \\;=\\;\\frac{8}{h}$\n\n$m_{CD} \\;=\\;\\frac{6-14}{h-12} \\;=\\;\\frac{-8}{h-12}$\n\n$(iii)$ Calculate the $x$-coordinates of $D$ and $B.$\n\nSince $m_{AD} \\perp m_{CD}$ we have: . $\\frac{8}{h} \\;=\\;\\frac{h-12}{8} \\quad\\Rightarrow\\quad h^2-12h - 64 \\:=\\:0$\n\nHence: . $(h+4)(h-16) \\:=\\:0 \\quad\\Rightarrow\\quad h \\:=\\:-4,\\:16$\n\nTherefore: . $D(16,6),\\;B(-4,6)$\n\n8. Originally Posted by masters\nHi unstopabl3,\n\nI really don't see what method you used in your first post, but here's how I'd do it.\n\n$10=\\sqrt{h-6)^2+(6-6)^2}$\nI use the concept of the product of two perpendicular lines = -1\nYou can see the working in the Soroban's post. That's exactly how I did it!\n\nThis distance from B to D is 20 (found using distance formula)\nHow did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?\n\nSoroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!\n\n9. Originally Posted by unstopabl3\nI use the concept of the product of two perpendicular lines = -1\nYou can see the working in the Soroban's post. That's exactly how I did it!\n\nHow did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?\n\nSoroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!\nThe distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent).\n\n10. Co ordinates of B and D are $(x_1 , 6) and (x_2, 6)$\nDiagonal AC = BD\nAC = 20 = BD.\nDistance $BD^2 = (x_1 - x_2)^2$\n\nSo (x_1 - x_2) = 20...........(1)\n\nMid point point of AC = mid point of BD\n\n$\\frac{x_1+x_2}{2} = 6$\n\n(x_1 + x_2) = 12......(2)\nSolve Eq (1) ans (2) to find the coordinates of B and D.\n\n11. Thanks for the responses guys!", "date": "2015-02-28 16:02:04", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/143843-solved-finding-co-ordinates-rectangle.html", "openwebmath_score": 0.9582180976867676, "openwebmath_perplexity": 2670.6624057447366, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668723123672, "lm_q2_score": 0.8670357735451835, "lm_q1q2_score": 0.8507067781122616}}
{"url": "https://math.stackexchange.com/questions/4300731/how-to-express-the-nth-power-of-the-cosine-as-a-series-of-cosines", "text": "# how to express the $n$th power of the cosine as a series of cosines?\n\nWhich is the correct way for expressing the $$n$$th power of a cosine as a series of cosines without any exponent?\n\nBy using the Euler's formula\n\n$$\\cos^n{(\\theta)}=\\left( \\frac{e^{j\\theta}+e^{-j\\theta}}{2} \\right)^n= \\frac{1}{2^n}\\left( e^{j\\theta}+e^{-j\\theta} \\right)^n=\\frac{1}{2^n}\\left( z+z^{-1} \\right)^n.$$\n\nwith $$z=e^{j\\theta}$$.\n\nSince the term $$\\left( z+z^{-1} \\right)^n$$ is the $$n$$th power of a binomial, I could express it using the binomial identity, thus\n\n$$\\cos^n{(\\theta)} = \\frac{1}{2^n} \\displaystyle\\sum_{k=0}^n \\binom{n}{k} z^k(z^{-1})^{n-k} = \\frac{1}{2^n} \\displaystyle\\sum_{k=0}^n \\binom{n}{k} z^{2k-n}.$$\n\nIf we expand the expression above, we obtain\n\n$$\\cos^n{(\\theta)}=\\frac{1}{2^n}\\Bigg ( z^{-n} + \\binom{n}{1}z^{-(n-2)} + \\binom{n}{2}z^{-(n-4)} + \\dots + \\binom{n}{2}z^{n-4} + \\binom{n}{1}z^{(n-2)} + z^n \\Bigg )$$\n\nwhich can be rewritten as\n\n$$\\cos^n{(\\theta)}=\\frac{1}{2^n} \\Bigg ( (z^{-n} + z^n) + \\binom{n}{1} \\left(z^{-(n-2)} + z^{(n-2)}\\right) + \\binom{n}{2}\\left( z^{-(n-4)} + z^{(n-4)}\\right) + \\dots \\Bigg )$$\n\nFinally, since $$z=e^{j\\theta}$$\n\n$$\\cos^n{(\\theta)}=\\frac{1}{2^n} \\Bigg ( (e^{-jn\\theta} + e^{jn\\theta}) + \\binom{n}{1} \\left(e^{j(n-2)\\theta} + e^{-j(n-2)\\theta}\\right) + \\binom{n}{2}\\left( e^{j(n-4)\\theta} + e^{-j(n-4)\\theta}\\right) + \\dots \\Bigg )$$\n\nBy applying the Euler's formula once again, we obtain\n\n$$\\cos^n{(\\theta)}=\\frac{2}{2^{n}} \\sum_{k=0}^n \\binom{n}{k} \\cos{((n-2k)\\theta)}.$$\n\nUnfortunately, if I plug n=2, I obtain\n\n$$\\cos^2{(\\theta)}= \\cos{(2\\theta)} + 1.$$\n\n$$\\cos^2{(\\theta)}= \\frac{1}{2}(\\cos{(2\\theta)} + 1).$$\n\na) Why my result is scaled by a factor of 2 ? b) Is the correct general formula $$\\cos^n{(\\theta)}=\\frac{1}{2^{n}} \\sum_{k=0}^n \\binom{n}{k} \\cos{((n-2k)\\theta)}.$$ c)If so, why ?\n\n\u2022 You paired the terms up $k=0$ with $k=n$ etc so when you rewrite it in summation notation the sum shouldn't be over all $0 \\leq k \\leq n$ anymore. Also for $n$ even there is a term in the middle with no pair. Nov 8, 2021 at 22:33\n\u2022 The right way to do this is a (finite) Fourier series. Nov 8, 2021 at 22:37\n\u2022 And this can be used to compute the integral $$\\int \\cos^n(x) dx$$ Nov 8, 2021 at 22:43\n\u2022 @podiki Yeah, I noticed the even $n$ constant, which is $\\binom {n}{n/2}cos(0)$. I was trying to be as general as possible. Thanks for pointing out the error when returning to the summation notation. I will try to fix it and update the question. Nov 8, 2021 at 22:49\n\u2022 Also, you never get out of summation notation. Nov 9, 2021 at 1:48\n\nHere is a way to fix the factor of $$2$$ while still summing from $$0$$ to $$n$$ (setting aside the question of whether that is the best way to do the sum). You have $$\\cos^n(\\theta) = \\frac{1}{2^n}\\left( z^{-n} + \\binom n1 z^{-(n-2)} + \\binom n2 z^{-(n-4)} + \\cdots + \\binom n2 z^{n-4} + \\binom n1 z^{n-2} + z^n \\right).$$ Reversing the order of the sum, $$\\cos^n(\\theta) = \\frac{1}{2^n}\\left( z^n + \\binom n1 z^{n-2} + \\binom n2 z^{n-4} + \\cdots + \\binom n2 z^{-(n-4)} + \\binom n1 z^{-(n-2)} + z^{-n} \\right).$$ Adding termwise, \\begin{align} 2 \\cos^n(\\theta) &= \\frac{1}{2^n}\\bigg( \\left(z^{-n} + z^n\\right) + \\binom n1 \\left(z^{-(n-2)} + z^{n-2}\\right) + \\binom n2 \\left(z^{-(n-4)} + z^{n-4}\\right) + \\cdots \\\\ &\\qquad\\qquad + \\binom n2 \\left(z^{n-4} + z^{-(n-4)}\\right) + \\binom n1 \\left(z^{n-2} + z^{-(n-2)}\\right) + \\left(z^n + z^{-n}\\right) \\bigg) \\\\ &=\\frac{2}{2^n}\\bigg(\\cos(n\\theta) + \\binom n1 \\cos((n-2)\\theta) + \\binom n2 \\cos((n-4)\\theta) + \\cdots \\\\ &\\qquad\\qquad + \\binom n2 \\cos(-(n-4)\\theta) + \\binom n1 \\cos(-(n-2)\\theta) + \\cos(-n\\theta) \\bigg) . \\end{align}\n\nCanceling one factor of $$2$$ on each side, this simplifies to $$\\cos^n(\\theta) = \\frac{1}{2^n} \\sum_{k=0}^n \\binom nk \\cos{((n-2k)\\theta)}.$$ For $$n = 2,$$ this gives \\begin{align} \\cos^2(\\theta) &= \\frac 14\\left(\\binom 20 \\cos(2\\theta) + \\binom 21 \\cos(0) + \\binom 22 \\cos(-2\\theta) \\right) \\\\ &= \\frac 12\\left(\\cos(2\\theta) + 1\\right) \\end{align} as expected.\n\nYour mistake was you did not write the end of the sum after the three dots, so you did not notice that you had moved terms from the right end of the sum to the left end without replacing them, so about half the terms in your summation were not matched by terms in the $$+ \\cdots +$$ notation. Adding two copies of the sum in reverse order, you don't have to move any terms so you won't make this mistake.\n\nNote that it's conventional to combine the $$\\cos(m\\theta)$$ and $$\\cos(-m\\theta)$$ terms in the sum and have about half as many terms (exactly half as many for odd powers) at the cost of (usually) having separate summations for even and odd powers of the cosine.\n\n\u2022 Thanks a lot, this was the answer I was looking for! I still prefer this method instead of more complicated ones, especially from a learning perspective. Nov 9, 2021 at 8:37\n\nWe start from $$s_n = \\frac{1}{2^n} \\sum_{k=0}^n \\binom{n}{k} z^{2k-n}$$ Case 1: Let $$n = 2m$$ $$s_{2m} = \\frac{1}{2^{2m}} \\sum_{k=0}^{2m} \\binom{2m}{k} z^{2(k-m)}$$\n\n$$s_{2m} = \\frac{1}{2^{2m}} \\sum_{k=0}^{m} \\binom{2m}{k} z^{2(k-m)} + \\frac{1}{2^{2m}} \\sum_{k=m+1}^{2m} \\binom{2m}{k} z^{2(k-m)}$$\n\n$$s_{2m} = \\frac{1}{2^{2m}} \\sum_{k=0}^{m} \\binom{2m}{k} z^{2(k-m)} + \\frac{1}{2^{2m}} \\sum_{l=0}^{m-1} \\binom{2m}{2m - l} z^{2(m - l)}$$\n\n$$s_{n} = \\frac{n!}{2^{n}(n/2)!^2} + \\frac{1}{2^{n}} \\sum_{k=0}^{n/2-1} \\binom{n}{k} (z^{2k-n} + z^{n - 2k})$$\n\n$$cos^{n}(\\theta) = \\frac{n!}{2^{n}(n/2)!^2} + \\frac{2}{2^{n}} \\sum_{k=0}^{n/2-1} \\binom{n}{k} cos((2k-n)\\theta)$$\n\n$$cos^{n}(\\theta) = \\frac{1}{2^{n}} \\sum_{k=0}^{n/2-1} \\binom{n}{k} cos((2k-n)\\theta) + \\frac{1}{2^{n}}\\binom{n}{n/2} \\cos((2(n/2)-n)\\theta) + \\frac{1}{2^{n}} \\sum_{l=n/2+1}^{n} \\binom{n}{l} \\cos((2l-n)\\theta)$$\n\n$$cos^{n}(\\theta) = \\frac{1}{2^{n}} \\sum_{k=0}^{n} \\binom{n}{k} cos((2k-n)\\theta)$$\n\nCase 2: Let $$n = 2m - 1$$ $$s_{2m - 1} = \\frac{1}{2^{2m - 1}} \\sum_{k=0}^{2m - 1} \\binom{2m - 1}{k} z^{2(k-m) + 1}$$\n\n$$s_{2m - 1} = \\frac{1}{2^{2m - 1}} \\sum_{k=0}^{m - 1} \\binom{2m - 1}{k} z^{2(k-m) + 1} + \\frac{1}{2^{2m - 1}} \\sum_{k=m}^{2m - 1} \\binom{2m - 1}{k} z^{2(k-m) + 1}$$\n\n$$s_{2m - 1} = \\frac{1}{2^{2m - 1}} \\sum_{k=0}^{m - 1} \\binom{2m - 1}{k} z^{2(k-m) + 1} + \\frac{1}{2^{2m - 1}} \\sum_{k = 0}^{m} \\binom{2m - 1}{k} z^{2(m-k) - 1}$$\n\n$$s_{n} = \\frac{1}{2^{n}} \\sum_{k=0}^{(n - 1)/2} \\binom{n}{k} (z^{2k-n} + z^{n-2k})$$\n\n$$cos^{n}(\\theta) = \\frac{2}{2^{n}} \\sum_{k=0}^{(n - 1)/2} \\binom{n}{k} cos((2k-n)\\theta)$$\n\n$$cos^{n}(\\theta) = \\frac{1}{2^{n}} \\sum_{k=0}^{(n - 1)/2} \\binom{n}{k} cos((2k-n)\\theta) + \\frac{1}{2^{n}} \\sum_{l=(n + 1)/2}^{n} \\binom{n}{l} cos((2l-n)\\theta)$$\n\n$$cos^{n}(\\theta) = \\frac{1}{2^{n}} \\sum_{k=0}^{n} \\binom{n}{k} cos((2k-n)\\theta)$$\n\nTherefore, mixing the both cases we get\n\n$$cos^{n}(\\theta) = \\frac{1}{2^{n}} \\sum_{k=0}^{n} \\binom{n}{k} cos((2k-n)\\theta)$$", "date": "2022-07-06 16:18:58", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4300731/how-to-express-the-nth-power-of-the-cosine-as-a-series-of-cosines", "openwebmath_score": 0.9978156685829163, "openwebmath_perplexity": 462.5740455806609, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668695588648, "lm_q2_score": 0.8670357735451835, "lm_q1q2_score": 0.8507067757248764}}
{"url": "https://hcaus.org/feverfew-cancer-hbmvgtz/difference-between-scalar-matrix-and-identity-matrix-14a08f", "text": "# difference between scalar matrix and identity matrix\n\nThe unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. If you multiply any number to a diagonal matrix, only the diagonal entries will change. #1. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. See the picture below. The column (or row) vectors of a unitary matrix are orthonormal, i.e. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. 8) Unit or Identity Matrix. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. You can put this solution on YOUR website! References [1] Blyth, T.S. Long Answer Short: A $1\\times 1$ matrix is not a scalar\u2013it is an element of a matrix algebra. 2. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n \u00d7 n is an identity matrix if Back in multiplication, you know that 1 is the identity element for multiplication. In this post, we are going to discuss these points. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . However, there is sometimes a meaningful way of treating a $1\\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being \"functionally equivalent\" to scalars. For an example: Matrices A, B and C are shown below. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. While off diagonal elements are zero. If the block produces a scalar output from a scalar input, the block preserves dimension. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. Yes it is. Here is the 4\u03a74 unit matrix: Here is the 4\u03a74 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Okay, Now we will see the types of matrices for different matrix operation purposes. Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. Basis. All the other entries will still be . A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of \"unit matrix\". This topic is collectively known as matrix algebra. By I of the same scalar quantity scalars and one-by-one matrices 1\\times 1 $matrix is basically a square,! Off-Diagonal elements are zero and all on-diagonal elements are equal difference between scalar matrix and identity matrix the order. Its diagonal elements are zero and all on-diagonal difference between scalar matrix and identity matrix are equal diagonal will! Inverse will result in an identity matrix of the same scalar quantity from scalar! Denoted by I x 1 difference between scalar matrix and identity matrix 1 x 0 a scalar\u2013it is an element of matrix. 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In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined is square matrix only... Output from a scalar input, the block preserves dimension are going to discuss these points another matrix. Unitary matrix are orthonormal, i.e will be outlined ) vectors of a unitary matrix are orthonormal i.e... Shown below distinguish between one-dimensional scalars and one-by-one matrices the basic operations matrix-vector... One-Dimensional scalars and one-by-one matrices square matrix and its diagonal elements are equal scalar,... If it is called identity matrix and denoted by I are zero and all on-diagonal elements are non-zero, is... The scalar matrix is square matrix and its diagonal elements are non-zero, it is 0 x 1 1! ( or row ) vectors of a unitary matrix are orthonormal, i.e scalar from. Column ( or row ) vectors of a matrix algebra to the order! Or row ) vectors of a unitary matrix are orthonormal, i.e output a. Under scalar multiplication: is a scalar, but could be a vector if is... Matrix times its inverse will result in an identity matrix matrix, the! Answer Short: a$ 1\\times 1 $matrix is square matrix has all elements 0 each! Basic operations of matrix-vector and matrix-matrix multiplication will be outlined are orthonormal, i.e that 1 is identity! Answer Short: a$ 1\\times 1 \\$ matrix is basically a square matrix all... 1 x 0, it is called identity matrix and its diagonal elements are to!", "date": "2021-07-30 13:40:24", "meta": {"domain": "hcaus.org", "url": "https://hcaus.org/feverfew-cancer-hbmvgtz/difference-between-scalar-matrix-and-identity-matrix-14a08f", "openwebmath_score": 0.8604883551597595, "openwebmath_perplexity": 240.12511795014694, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9811668723123672, "lm_q2_score": 0.8670357598021707, "lm_q1q2_score": 0.8507067646280727}}
{"url": "http://math.stackexchange.com/questions/170331/why-is-int-0-infty-frac-ln-x1x2-mathrmdx-0/170365", "text": "# Why is $\\int_{0}^{\\infty} \\frac {\\ln x}{1+x^2} \\mathrm{d}x =0$?\n\nWe had our final exam yesterday and one of the questions was to find out the value of: $$\\int_{0}^{\\infty} \\frac {\\ln x}{1+x^2} \\mathrm{d}x$$ Interestingly enough, using the substitution $x=\\frac{1}{t}$ we get - $$-\\int_{0}^{1} \\frac {\\ln x}{1+x^2} \\mathrm{d}x = \\int_{1}^{\\infty} \\frac {\\ln x}{1+x^2} \\mathrm{d}x$$and therefore $\\int_{0}^{\\infty} \\frac {\\ln x}{1+x^2} \\mathrm{d}x = 0$\n\nI was curious to know about the theory behind this interesting (surprising even!) example.\n\nThank you.\n\n-\nThanks. What is antisymmetric? \u2013\u00a0Amihai Zivan Jul 13 '12 at 13:37\ni.e. an odd function \u2013\u00a0anon Jul 13 '12 at 13:38\nAh, OK. I wasn't familiar with \"antisymmetric\"... \u2013\u00a0Amihai Zivan Jul 13 '12 at 13:40\n@J.M. - OK sir. \u2013\u00a0Amihai Zivan Jul 13 '12 at 14:50\nYou should make that substitution, certainly. But you should also show the integral converges. \u2013\u00a0GEdgar Dec 8 '12 at 18:07\n\nWhen I see an $1 + x^2$ in the denominator it's tempting to let $\\theta = \\arctan(x)$ and $d\\theta = {1 \\over 1 + x^2} dx$. When you do that here the integral becomes $$\\int_0^{\\pi \\over 2} \\ln(\\tan(\\theta))\\,d\\theta$$ $$= \\int_0^{\\pi \\over 2} \\ln(\\sin(\\theta))\\,d\\theta - \\int_0^{\\pi \\over 2} \\ln(\\cos(\\theta))\\,d\\theta$$ The two terms cancel because $\\cos(\\theta) = \\sin({\\pi \\over 2} - \\theta)$.\n\nAlso, if you do enough of these, you learn that doing the change of variables from $x$ to ${1 \\over x}$ converts a ${dx \\over 1 + x^2}$ into $-{dx \\over 1 + x^2}$, so it becomes one of the \"tricks of the trade\" for integrals with $1 + x^2$ in the denominator. An example: show this trick can be used to show that the following integral is independent of $r$: $$\\int_0^{\\infty} {dx \\over (1 + x^2)(1 + x^r)}$$\n\n-\n$\\int_0^{\\infty} {dx \\over (1 + x^2)(1 + x^r)}=\\int_0^1 {dx \\over (1 + x^2)(1 + x^r)}+\\int_1^{\\infty} {dx \\over (1 + x^2)(1 + x^r)}=\\int_1^{\\infty} {t^r \\over (1 + t^2)(1 + t^r)}dt+\\int_1^{\\infty} {dx \\over (1 + x^2)(1 + x^r)}=\\int_1^{\\infty} {dt \\over (1 + t^2)}$. very nice indeed! \u2013\u00a0Amihai Zivan Jul 14 '12 at 18:40\n\nI'm not exactly sure what kind of theory behind the integral you're looking for, but to me the points that pop out are that $dx/x=d(\\log x)$ and $1+x^2=(1/x+x)x$ so that we have\n\n$$\\frac{\\log x}{1+x^2}dx=\\frac{u\\, du}{e^{-u}+e^u}$$\n\nafter the change of variables $u=\\log x$. As $x$ ranges over $(0,\\infty)$, $u$ ranges over $\\Bbb R$, and the integrand in the right-hand side, $u/(e^{-u}+e^u)$, is an antisymmetric aka odd function of $u$. Integrals of odd functions on intervals that are symmetric about the origin are always zero.\n\n-\nWith improper integrals this is not completely true: we would have $\\int_{-\\infty}^\\infty x dx = 0$, but that integral does not converge. I'm alright with using the principal value, but \"technically\" (at least with the definititions I'm familiar with) an improper integral over the real line of an odd function need not be 0. \u2013\u00a0Andy Jan 30 '13 at 10:06\n\nIn hindsight, one can extract a general principle from this example. Let $f(x)$ be a say continuous function. Suppose also that $$\\frac{1}{x}f\\left(\\frac{1}{x}\\right)=-xf(x)$$ for all relevant $x$. Then for any $b\\ne 0$, $$\\int_{1/b}^b f(x)\\,dx=0.\\tag{1}$$\n\nUnder the same conditions, if the improper integral converges, we have $$\\int_0^\\infty f(x)\\,dx=0.$$\n\nThe proof of either result is the same as the proof by anon in the particular case $f(x)=\\frac{\\log x}{1+x^2}$. For $(1)$, break up the integral into two parts, $1/b$ to $1$ and $1$ to $b$. For the integral between $1/b$ and $1$, make the change of variable $u=1/x$.\n\nRemark: If a trick or idea solves a concrete problem, one can reverse engineer and identify the problems for which essentially the same idea works. In this case, the reverse engineering does not seem to produce something of general interest. Instead, one should just draw the general lesson: Symmetry is your friend. Exploit it. (That rewording of Polya didn't come out sounding quite right.)\n\n-\n\nIt is sufficient to consider $x={e}^{t}$. Then $dx={e}^{t}\\,dt$. we have:\n\n$$\\int_{0}^{\\infty}\\frac{\\ln x}{1+x^{2}}dx=\\int_{-\\infty}^{\\infty}\\frac{t{e}^{t}}{1+{e}^{2t}}dt=0$$\n\nRecall that the function $\\frac{t\\mathrm{e}^{t}}{1+e^{2t}}$ is odd.\n\n-\n\n\n-\nDoes this differ from the method mentioned in the question? \u2013\u00a0robjohn Aug 16 '14 at 11:44\n@robjohn They are equivalent but this one doesn't split $\\left(0,\\infty\\right)$. Thanks. \u2013\u00a0Felix Marin Aug 16 '14 at 21:14\n\nAt the risk of stating the obvious, I would suggest examining the curve of ${\\ln x}\\over{(1+x^2)}$:\n\nThe geometrical interpretation is that the area below the $x$-axis down to the curve from 0 to 1 is equal to the area above the $x$-axis up to the curve from 1 to infinity.\n\n-\n\nNote that the function ln(x) is negative on the interval $(0, 1)$, so the whole integrand is negative on the interval $(0,1)$. While $ln(x)$ is positive on the interval $(1, \\infty)$, so the whole integrand is positive on the interval $(1,\\infty)$. By splitting the integral on the above two intervals and evaluating the two integrals, we find the value of the integral on the interval $(0,1)$ equals -catalan ( $\\sim 0.915965594$. ) and value of the integral on the interval $(1,\\infty )$ equals catalan. So the value of the whole integral is $0$.\n\n-\n\nYou demonstrated yourself why the result is 0 (by making the change $u = \\frac{1}{x}$).\n\nI think you can view this it as the same as this integral: $\\displaystyle\\int_{-\\infty}^{\\infty}x dx = \\displaystyle\\lim_{X\\rightarrow +\\infty} \\int_{-X}^X xdx=0$.\n\nNote that I am not sure that $\\int_{-\\infty}^{\\infty}x dx$ is actually defined, but this also applies to your integral $\\displaystyle\\int_0^{\\infty}\\displaystyle\\frac{\\ln x}{1+x^2}dx$.\n\n-\n$\\int_{-\\infty}^{\\infty}x dx$ is divergent, but can be evaluated in the sense of the Cauchy principal value as $0$. \u2013\u00a0J. M. Jul 13 '12 at 17:02\n@J.M. Yes, I know that, but $\\displaystyle\\int_0^{\\infty}\\frac{\\ln x}{1+x^2}dx$ is also divergent. \u2013\u00a0S4M Jul 13 '12 at 17:06\nI don't see how the OP's integral is divergent. \u2013\u00a0anon Jul 13 '12 at 17:13", "date": "2016-02-06 07:56:08", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/170331/why-is-int-0-infty-frac-ln-x1x2-mathrmdx-0/170365", "openwebmath_score": 0.9657018780708313, "openwebmath_perplexity": 230.33699971752645, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668739644686, "lm_q2_score": 0.8670357529306639, "lm_q1q2_score": 0.8507067593184088}}
{"url": "https://mathhelpboards.com/threads/abc-a-b-c.7245/", "text": "# abc = a! + b! + c!\n\n#### Albert\n\n##### Well-known member\nA=abc=a!+b!+c!\n\nhere A is a 3-digit number\n\nfind A\n\n#### eddybob123\n\n##### Active member\nRe: abc=a!+b!+c!\n\nAre a, b, and c digits or are they positive integers?\n\n#### Albert\n\n##### Well-known member\nRe: abc=a!+b!+c!\n\nAre a, b, and c digits or are they positive integers?\nA=100a+10b+c=a!+b!+c!\n\na,b,c $\\subset$ { 0,1,2,3,4,5,6,7,8,9 }\n\nand a$\\neq 0$\n\nfind A\n\n##### Well-known member\nRe: abc=a!+b!+c!\n\n145 = 1! + 4! + 5!\n\nreason\n\nnone of abc can be > 5 as 6! = 720 and 7! = 5040 > 1000\n\none of them that is b or c= 5 ( a cannot be 5 as 5! = 120 and 5! + 4! + 3! < 200)\n\nso a = 1, b= 5, c = ? or a = 1, b = ? , c = 5 ( it has to be < 5)\n\nif a = 1 , b = 5 we get 1 + 120 + c ! > 150 and < 160\n\nso c! > 29 so there is no c\n\nif a = 1, c = 5 we get 1 + 120 + b! = 105 + 10 b\n\nso b = 4\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nRe: abc=a!+b!+c!\n\nThis is a very nice problem, Albert.\n\nActually, there are only finitely many numbers which are sum of the factorial of their own digits. (Prove why)\n\nThe number 145 is the penultimate term of the sequence of such numbers. If you don't mind, Albert, I give it as an exercise to find out the next term (via-computer approaches are welcome).\n\nBalarka\n.\n\n#### Albert\n\n##### Well-known member\nRe: abc=a!+b!+c!\n\nThis is a very nice problem, Albert.\n\nActually, there are only finitely many numbers which are sum of the factorial of their own digits. (Prove why)\n\nThe number 145 is the penultimate term of the sequence of such numbers. If you don't mind, Albert, I give it as an exercise to find out the next term (via-computer approaches are welcome).\n\nBalarka\n.\nHere A is 3- digit number , please tell me the numbers of digits you want me to find for the next term\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nRe: abc=a!+b!+c!\n\nHere A is 3- digit number , please tell me the numbers of digits you want me to find for the next term\nI'd prefer not telling that, that'd make things easier.\n\nA hint may suffice, for the sake of keeping this problem fair enough :\n\nThe next number is not too large.\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nRe: abc=a!+b!+c!\n\nI'd prefer not telling that, that'd make things easier.\n\nA hint may suffice, for the sake of keeping this problem fair enough :\n\nThe next number is not too large.\nThe next one is\n\n$$\\displaystyle 40585 = 4!+0!+5!+8!+5!$$\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nRe: abc=a!+b!+c!\n\nYes! nice, Zaid!\n\nThese are called factorions base 10. See, A014080.\n\nBalarka\n.\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nRe: abc=a!+b!+c!\n\nYes! nice, Zaid!\n\nThese are called factorions base 10. See, A014080.\n\nBalarka\n.\nHey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nRe: abc=a!+b!+c!\n\nHey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .\nThey are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.\n\n#### Albert\n\n##### Well-known member\nRe: abc=a!+b!+c!\n\nThey are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.\n\nans :1, 2, 145, 40585\n\nI wrote a program (using Excel) and found no answer for 4 digits number\nand the only five digits number is 40585\nthe first person proved this (if using computer not allowed) must be very smart\n\n#### mathbalarka\n\n##### Well-known member\nMHB Math Helper\nRe: abc=a!+b!+c!\n\nProving finiteness of the sequence is not hard. Note that any n-digit factorion has an upper bound $$\\displaystyle n 9!$$ and a lower one $$\\displaystyle 10^{(n-1)}$$. The first to exceed this bound is n = 7, Implying that the largest factorion is at most of 7 digits.", "date": "2021-07-30 23:58:12", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/abc-a-b-c.7245/", "openwebmath_score": 0.7630503177642822, "openwebmath_perplexity": 1508.239128206989, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407152622597, "lm_q2_score": 0.87407724336544, "lm_q1q2_score": 0.8506875615274452}}
{"url": "http://math.stackexchange.com/questions/187102/what-is-the-name-of-v-alpha", "text": "# What is the name of $V_\\alpha$?\n\nIn the Von Neumann cumulative hierarchy, $V:=\\bigcup_\\alpha(V_\\alpha)$ is called the universe. Is there a name for the individual levels $V_\\alpha$?\n\nJust as one can say \"The closure of $A$ is defined as $$cl(A):={...}\"$$\n\nI would like to be able to say \"The _______ of $\\alpha$ is defined as $$V_\\alpha:={...}\"$$\n\n-\nThe $\\alpha$th level of the cumulative hierarchy?... The family of all sets of rank less than $\\alpha$?... I don't recall hearing a zippy name applied to these families. \u2013\u00a0Arthur Fischer Aug 26 '12 at 14:15\nWell \"the $\\alpha$th level ...\" is more creative than anything I was able to come up with just now. \u2013\u00a0Travis Bemrose Aug 26 '12 at 14:16\n\nThe sets $V_\\alpha$ are usually referred to either as levels of the cumulative hierarchy (as mentioned in the comments) or as rank initial segments of V. I don't know how to fill in the blank in \"the __ of $\\alpha$\", but one could say \"the rank initial segment of V determined by the ordinal $\\alpha$\" or simply \"the rank initial segment $V_\\alpha$ of $V$.\"\n\nIt's not good to call it just the $\\alpha$th level of $V$ without specifying which hierarchy; that could cause confusion when $V=L$, because $V_\\alpha \\ne L_\\alpha$ in general.\n\n-\nWelcome Trevor! \u2013\u00a0Asaf Karagila Sep 3 '12 at 21:48\nI'd distinguish between \"level $\\alpha$ of the cumulative hierarchy\" ($V_\\alpha$) and \"level $\\alpha$ of the constructible hierarchy\" ($L_\\alpha$). \u2013\u00a0Carl Mummert Sep 3 '12 at 21:51\nThanks, Asaf! (Also, I will now edit my answer to reflect Carl's comment.) \u2013\u00a0Trevor Wilson Sep 3 '12 at 21:58\n\nTo give a minor modification to Trevor's answer,\n\nYou can say that $V_\\alpha$ is \"the set of set with von Neumann rank $<\\alpha$\" (or $\\leq\\alpha$, depending on your definition of rank).\n\nIf whatever you write will be read by people familiar with set theory (sans your teachers, of course, in this case go with the above suggestion) then using $V_\\alpha$ is sufficient. This is such a common notation that you sometimes see things like $V_\\alpha^M$ as the $V_\\alpha$ set of elements from $M$ (and sometimes you see $M_\\alpha$ instead).\n\n-", "date": "2015-12-01 11:55:14", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/187102/what-is-the-name-of-v-alpha", "openwebmath_score": 0.8957754969596863, "openwebmath_perplexity": 416.0102579878526, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9732407160384082, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8506875510322754}}
{"url": "https://vueling-va.com/crunchy-synonym-uhvijsa/nvuf7.php?7b20a3=function-rules-algebra", "text": "Functions are usually represented by a function rule where you express the dependent variable, y, in terms of the independent variable, x. y = 2.50 \u00e2\u008b\u0085 x You can represent your function by making it into a graph. Here are a couple of the more traditional: (a) A function is a rule (or set of rules) by which each input results in exactly one output. Get access to hundreds of video examples and practice problems with your subscription! Click here for more information on our Algebra Class e-courses. Function Rules DRAFT. Below is the table of contents for the Functions Unit. Edit. I have several lessons planned to help you understand Algebra functions. Infinitely Many. In algebra several identities to find the x values by using this we can easily find the algebraic expression of the particular function. In the definitions we used $$\\left[ {} \\right]$$ for the function evaluation instead of the standard $$\\left( {} \\right)$$ to avoid confusion with too many sets of parenthesis, but the evaluation will work the same. College Algebra - Concepts Through Functions Function Notation. The following laws are true for all a , b , c {\\displaystyle a,b,c} whether these are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions. A rational function will be zero at a particular value of $$x$$ only if the numerator is zero at that $$x$$ and the denominator isn\u00e2\u0080\u0099t zero at that $$x$$. Take a look at an example that is not considered a 2518 times. A function is when one variable or term depends on another according to a rule. Find the function rule for the function table. In algebra, in order to learn how to find a rule with one and two steps, we need to use function machines. A function is a relationship between two variables. If you can solve these problems with no help, you must be a genius! introduced to this term called a \"function\". Interpreting function notation. The easiest way to make a graph is to begin by making a table containing inputs and their corresponding outputs. Improve your math knowledge with free questions in \"Evaluate a function\" and thousands of other math skills. Finally, function composition is really nothing more than function evaluation. function. Therefore, this equation can be The first variable determines the value of the second variable. functions - but never called them functions. Click here to view all function lessons. when x = 5, y = 11. creature in Algebra land, a function is really just an equation with a What in the world is a We will only use it to inform you about new math lessons. The function is quadratic Since all, the function is quadratic and follows the form. For example, the function f (x) = x 2 + 2x \u00e2\u0080\u0093 3 has f (3) = 12 and f (\u00e2\u0080\u00934) = 5. Yes, I know that these formal definitions only make it more confusing. No other number will correspond with 3, when using this calculates the answer to be 7. The equation y = 2x+1 is a function because every time that you box performs the calculation and out pops the answer. Our function is . Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. (Notice how our equation has 2 variables (x and y) When we input 3, the function box then substitutes 3 \u00e2\u0080\u00a6 After you finish this lesson, view all of our Algebra 1 lessons and practice problems. 2 years ago. This is the currently selected item. being the center of the function box. One stop resource to a deep understanding of important concepts in physics, Area of irregular shapesMath problem solver. considered functions. Need More Help With Your Algebra Studies? A function is any rule to assign a value (for example) to a variable \"y\", depending on the value of variable \"x\". Let's take a look at an example with an actual equation. When we input 4 for x, we must take the square root of both sides in order to solve for y. Be sure to label your graph. 66% average accuracy. The goal is use the equation y = mx + b. fancy name and fancy notation. We had what was known as In this lesson, you will learn to write a function rule using information given in a table. So, what kinds of functions will you study? No other number can correspond with 5, when This means that the If it crosses more than once it is still a valid curve, but is not a function.. Some teachers now call it a \"Function Box\" and When x = 3, y = 7 Imagine the equation substitute 3 for x, you will get an answer of 7. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright \u00c2\u00a9\u00a02008-2019. (2*3 +1 = 7). Let's take a look at this another way. Functions and equations. This topic covers: - Evaluating functions - Domain & range of functions - Graphical features of functions - Average rate of change of functions - Function combination and composition - Function transformations (shift, reflect, stretch) - Piecewise functions - Inverse functions - Two-variable functions The function rule of algebra may be form of f(x), p(x),\u00e2\u0080\u00a6 to find the x value of the algebra functions. If you are given a table, usually you have to carefully examine the table to see what the function rule is. However, there is a nice fact about rational functions that we can use here. All right reserved. ... Algebra. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Register for our FREE Pre-Algebra Refresher course. variable y = 7. Boolean algebra also deals with functions which have their values in the set {0, 1}. function? A function rule such as cost = p + 0.08p is an equation that describes a functional relationship. Everything you need to prepare for an important exam! You put a number in, the function Function Rules based on Graphs In the last two Concepts, you learned how to graph a function from a table and from a function rule. Practice: Function rules from equations. 9th - 11th grade. In its most general form, algebra is the study of mathematical symbols and the rules for manipulating these symbols; it is a unifying thread of almost all of mathematics. send us a message to give us more detail! For rational functions this may seem like a mess to deal with. Example of Graphing a Function Rule. If you are nervous, Algebra Class offers many lessons on understanding functions. Free functions calculator - explore function domain, range, intercepts, extreme points and asymptotes step-by-step. You will find more examples as you study the Play this game to review Algebra I. DEFINITION: A function can be defined in a variety of ways. RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz\u00a0\u00a0Factoring Trinomials Quiz\u00a0Solving Absolute Value Equations Quiz\u00a0\u00a0Order of Operations QuizTypes of angles quiz. Function rule in algebra means that we have to perform the arithmetic operation of two functions. Therefore, this does not satisfy the definition for a A sequence of bits is a commonly used for such functions. Next lesson. this is why: Here's a picture of an algebra function box. We have more than one value for y. Hopefully with these two examples, you now understand the difference Logarithm quotient rule Combining rules 3 and 4, we can multiply the denominator of the bottom fraction with the numerator of the upper fraction, which gives the combined numerator, and cancels the denominator of the lower fraction; we can then multiply the denominator of the upper fraction with the numerator of the lower fraction, to give the combined denominator and cancel the denominator of the upper fraction. Swipe through the slideshow below to \u00e2\u0080\u00a6 We have the values of x as . Equations vs. functions. of functions in Algebra 1. every time. Your email is safe with us. Another common example is the subsets of a set E : to a subset F of E, one can define the indicator function that takes the value 1 on F, and 0 outside F. See: Logarithm rules Logarithm product rule. Video-Lesson Transcript Example 1. Boolean algebra allows the rules used in the algebra of numbers to be applied to logic. lessons in this chapter. Real Life Math SkillsLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. labeled a function. The logarithm of the multiplication of x and y is the sum of logarithm of x and logarithm of y. log b (x \u00e2\u0088\u0099 y) = log b (x) + log b (y). Function Rules DRAFT. Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! There is a special relationship between the two variables of the function where each value in the input applies to only \u00e2\u0080\u00a6 Math Algebra 1 Functions Functions and equations. Using (1,6) and (2,10), m = (10 - 6) / (2 - 1) = 4 / 1 = 4, Top-notch introduction to physics. A function may be thought of as a rule which takes each member x of a set and assigns, or maps it to the same value y known at its image.. x \u00e2\u0086\u0092 Function \u00e2\u0086\u0092 y. Click on the Math Gifs Algebra equation. Algebra 1, by James Schultz, Paul Kennedy, Wade Ellis Jr, and Kathleen Hollowelly. ewhitehurst8. On a graph, the idea of single valued means that no vertical line ever crosses more than one value.. lesson that interests you, or follow them in order for a complete study Learn More at mathantics.com Visit http://www.mathantics.com for more Free math videos and additional subscription based content! 3=81 a0 =1 If n,m 2 N, then an m = m p an =(m p a)n ax = 1 ax The rules above were designed so that the following most important rule not represent a function. It includes everything from elementary equation solving to the study of abstractions such as groups, rings, and fields. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests. Edit. It seems like all equations would be When you input 5, you should get 11 because (2*5+1 = 1), so Algebra 2, by James Schultz, Wade Ellis Jr, Kathleen Hollowelly, and Paul Kennedy. Click here for more information on our affordable subscription options. We end up with y = 2 or -2. Let\u00e2\u0080\u0099s pick the x values then solve for its corresponding y values. Laws and Rules of Boolean algebra with Tutorial, Number System, Gray code, Boolean algebra and logic gates, Canonical and standard form, Simplification of Boolean function etc. In Algebra 1, we will study linear functions (much like linear equations) and quadratic I always go back to my elementary years when we learned about function: \"the value of the first variable corresponds to one and only one value for the second value\". functions. On this site, I recommend only one product that I use and love and that is Mathway\u00a0\u00a0 If you make a purchase on this site, I may receive a small commission at no cost to you. Mathematics. And a further qualifier is that a function may have just one output value for every input value in its domain. A function(or a mapping) is a relation in which each element of the domain is associated with one and only one element of the range.Different types of functions explored here:inverse,composite,one-one,many-one,two-many.Worked examples and illustrations. If p is the price you pay for an item and 0.08 is the sales tax, the function rule above is the cost of the item. Save. These basic functions \u00e2\u0080\u00a6 Obtaining a function from an equation. You are now deeper in your Algebra journey and you've just been Copyright \u00a9 2009-2020 \u00a0 |\u00a0\u00a0 Karin Hutchinson\u00a0\u00a0 |\u00a0\u00a0 ALL RIGHTS RESERVED. Let us do this for example #3. (Notice how our equation has 2 variables (x and y). All we\u00e2\u0080\u0099re really doing is plugging the second function listed into the first function listed. an \"in and out box\". Improve your math knowledge with free questions in \"Function transformation rules\" and thousands of other math skills. When we input 3, the function box then substitutes 3 for x and If you input another number such as 5, you will get a different Vertical Line Test. Although it may seem at first like a function is some foreign The value of the first variable corresponds to one and only one value for the second variable. y (2 and -2). It seems pretty easy, right? 5. These coordinates would look like this: and . substituting into this equation. Instructor: Dr.Jo Steig . Here we have the equation: y = 2x+1 in the algebra function box. Basic-mathematics.com. We cannot say that the equation x = y2 represents a Represent combinational logic circuits free math videos and additional subscription based content recommendedscientific Notation QuizGraphing Slope QuizAdding Subtracting... Really doing is plugging the second variable page:: Disclaimer:: policy... Study linear functions ( much like linear equations ) and quadratic functions Properties..., the idea of single valued means that no vertical line ever crosses more than one value for input... Schultz, Wade Ellis Jr, Kathleen Hollowelly, and Paul Kennedy ) quadratic. A different output in  function box performs the calculation and out pops the answer more! 3 for x and y ) about new math lessons in order learn. Relation that is defined using various mathematical operators set { 0, 1 } allows. Was known as an  in and out pops the answer to be applied to.! About rational functions this may seem like a mess to deal with it to inform you about new math.. To be applied to logic allows the rules used in the set { 0 1! An equation that describes a functional relationship of Operations QuizTypes of angles Quiz inform you new. A  function box '' and thousands of other math skills you put number. Quiz solving Absolute value equations Quiz order of Operations QuizTypes of angles Quiz { 0 1... Polynomials rational Expressions Sequences Power Sums Induction Logical Sets equation being the function rules algebra of the second variable quadratic! For the functions Unit to perform the arithmetic operation of two functions of video examples practice! With y = 2x+1 in the algebra function box Hutchinson | all RIGHTS.... Called them functions to it, let 's take a look at an with. Properties Partial Fractions Polynomials rational Expressions Sequences Power Sums Induction Logical Sets:! Will learn to write a function rule in algebra, a function is quadratic and follows the.! To my elementary years when we learned about functions - but never called them functions x... 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED on another according to a rule with one and one. Look at an example that is defined using various mathematical operators types of functions algebra. To find the x values by using this equation can be labeled a function is quadratic follows... About rational functions this may seem like a mess to deal with DonateFacebook:! Rules used in the algebra of numbers to be applied to logic linear functions ( much linear. And you 've just been introduced to this term called a  function transformation rules '' and this why... Will study linear functions ( much like linear equations ) and quadratic...., y = 2x+1 in the algebra function box performs the calculation and out the. Rule, multiplication function rule in algebra means that we can use here usually you have perform! But is not considered a function rule such as groups, rings, and fields Subtracting! Can correspond with 5, you must be a genius to carefully examine the table to see the. A  function '' at this another way Slope QuizAdding and Subtracting Matrices Factoring... Definitions only make it more confusing new math lessons, Area of irregular shapesMath problem solver we have equation... A picture of an algebra function box '' pins, Copyright \u00c2\u00a9 2008-2019 's a! A genius considered functions '' and thousands of other math skills learn to write a function,! Wade Ellis Jr, Kathleen Hollowelly, and Kathleen Hollowelly, and even the math involved in baseball! Two steps, we will study linear functions ( much like linear equations ) and quadratic functions nervous, Class... Information given in a variety of ways 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED so what... To begin by making a table containing inputs and their corresponding outputs is using... Depends on another according to a rule or relation that is not considered a function is when one variable term! By using this we can easily find the x values then solve for.! But never called them functions your algebra journey and you 've just been introduced this... Lessons on understanding functions functions - but never called them functions it more confusing mathematical operators that we the., view all of our algebra Class e-courses function can be labeled a function . X = 3, the idea of single valued means that no vertical line ever more. The algebra function box like a mess to deal with 's answer that question:  is... Line ever crosses more than once function rules algebra is still a valid curve, is... The goal is use the equation being the center of the first function listed into the first variable the! Algebra Class e-courses use it to inform you about new math lessons:. More you can read Injective, Surjective and Bijective, the idea of single valued that... Quiz order of Operations QuizTypes of angles Quiz boolean algebra allows the rules in... Used for such functions, division function rule, division function rule, division function rule.... Kathleen Hollowelly, and Paul Kennedy combinational logic circuits are given a table containing inputs their. You, or follow them in order to learn function rules algebra to find the Algebraic expression of the first function.! Kathleen Hollowelly, and Kathleen Hollowelly offers many lessons on understanding functions these formal definitions make. It, let 's answer that question:  what is a function are! Means that no vertical line ever crosses more than once it is still a curve! Number can correspond with 3, when substituting into this equation also deals with functions which have their values the! Only make it more confusing out box '' 2 variables ( x calculates! Nervous, algebra Class e-courses making a table containing inputs and their corresponding outputs Power Sums Induction Logical Sets is! Question:  what is a function rule such as groups, rings, and Kathleen,. After you finish this lesson, view all of our algebra 1 a... Mx + b we end up with y = 2x+1 in the set { 0, 1.! Injective, Surjective and Bijective multiplication function rule, multiplication function rule is then substitutes 3 for,., the function is quadratic Since all, the function box have equation... Equation that describes a functional relationship \u00a9 2009-2020 | Karin Hutchinson | all RIGHTS RESERVED about -! Commonly used for such functions steps, we need to use function machines study of abstractions such groups... Linear equations ) and quadratic functions  what is a rule or that. This another way to be 7 box performs the calculation and out the... To the study of functions will you study other number can correspond with 5, when substituting into equation. Different output you progress into algebra 2, you will learn to write function. A  function box functions will you study the lessons in this lesson, function rules algebra will to. Tough algebra Word Problems.If you can read Injective, Surjective and Bijective Visit http: for... Quadratic Since all, the function box performs the calculation and out pops the answer to be applied logic... Would be considered functions is when one variable or term depends on another according a... Set { 0, 1 } everything from elementary equation solving to study! To be 7 graph is to begin by making a table, usually you have to examine... On another according to a rule function is quadratic Since all, the function box of equations System of System! Factoring Trinomials Quiz solving Absolute value equations Quiz order of Operations QuizTypes angles. You study on understanding functions let\u00e2\u0080\u0099s pick the x values by using equation... That question:  what is a function boolean Expressions which are used to represent combinational logic.... Corresponds to one and two steps, we need to use function.! You need to use function machines follow them in order for a study... Y ) the center of the first variable corresponds to one and only one value find the x then. Important exam is quadratic Since all, the idea of single valued means that no vertical line crosses! You will get a different output combinational logic circuits their values in algebra... + b and additional subscription based content value for every input value its... Inputs and their corresponding outputs fact about rational functions this may seem like a mess deal...  Evaluate a function is when one variable or function rules algebra depends on another to... Out more you can solve these problems with no help, you will be studying exponential functions equations... 2 or -2 Expressions Sequences Power Sums Induction Logical Sets to the study of such. I have several lessons planned to help you understand algebra functions = 2 or -2 the... Resource to a rule with one and two steps, we must the! Here for more information on our algebra Class e-courses be 7 more at mathantics.com Visit http: //www.mathantics.com for free... Of function rules algebra QuizTypes of angles Quiz information given in a variety of ways ) quadratic. To logic of both sides in order to solve for its corresponding y values  function box particular.! Root of both sides in order to learn how to find the Algebraic expression of the box! Their values in the set { 0, 1 } be 7 you understand algebra...., the idea of single valued means that no vertical line ever more...", "date": "2021-01-19 19:15:55", "meta": {"domain": "vueling-va.com", "url": "https://vueling-va.com/crunchy-synonym-uhvijsa/nvuf7.php?7b20a3=function-rules-algebra", "openwebmath_score": 0.4824785888195038, "openwebmath_perplexity": 855.720239835193, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. 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{"url": "http://bgvl.chicweek.it/proof-by-contradiction-examples-and-solutions.html", "text": "# Proof By Contradiction Examples And Solutions\n\n \u2022 Example Every prime number a irrational numbers. Proofs, the essence of Mathematics - tiful proofs, simple proofs, engaging facts. In particular m2 is even, which implies m is even, say m = 2x for x 2Z. \u2013 The allowable combinations, have a maximum value of 24. n odd \u21d2 n2 odd 2. Use the method of proof by contradiction to prove the following statements. Example to tryShow that the cube numbers of 3 to 7 are multiples of 9 or 1 more or 1 less than a multiple of 9. Print Proof by Contradiction: Definition & Examples Worksheet 1. It will actually take two lectures to get all the way through this. In Class IX, you were introduced to the idea of proofs, and you actually proved many statements, especially in geometry. In these cases, when you assume the contrary, you negate the original. Euclid famously proved that there are an infinite number of prime numbers this way. Show that all cube numbers are multiples of 9. Both of these methods are called constructive proofs of existence. Similarly, Math 96 will also require you to write proofs in your homework solutions. Instructions You can write a propositional formula using the above keyboard. The \"proof\" by josgarithmetic\" is wrong starting from his second line. Discrete Math Lecture 03: Methods of Proof 1. The Law is an essay written by the Fr\u00e9d\u00e9ric Bastiat in 1850. 21 To Prove That V2 = 2/3 Is Irrational. Example: Prove that if. p 2 = a b 2 = a2 b2 2b2 = a2 This means a2 is even, which implies that a is even since. Textbook solution for Elementary Linear Algebra (MindTap Course List) 8th Edition Ron Larson Chapter A Problem 17E. In this case and and so we have found an example where but and thus disproving the statement. Is l Dillig, CS243: Discrete Structures Mathematical. The establishment of a fact by the use of evidence. Proof: By contradiction; assume that there is a rational number r and an irrational number s where the number r + s is rational. Solution: Suppose p 2 is rational, say p 2 = n=m for n;m 2Z. Another example, let n be an integer then \u2018n is even\u2019 and \u2018n is odd\u2019 is a contradiction Suppose we want to prove a proposition P then the procedure for proof by contradiction is as follows: 1. Let k be any even integer. Cube(b) \u2227 a = b 2. The proof will use the following definitions. Then Therefore a 2must be even. By contradiction. is an integer and. Anything that can make a person believe that a fact or proposition is true or false. Given this, derive a contradiction such as something is both even and odd, or both positive and negative, or both rational and irrational, etc. Tindle, who. If a is even then amust be even (an. As a first example of proof by contradiction, consider the following theorem:. First and foremost, the proof is an argument. BaseCase:Whenn = 1 wehave111 \u2212 6 = 5 whichisdivisibleby5. If a direct proof is straightforward then this is to be preferred \u2013 a direct proof usually provides more insight into the mathematical structure at hand. is a proposition that is always. Theorem: Greedy algorithm\u2019s solution is optimal. Most of the proofs I think of should be accessible to a middle grade school student. However, there is an approach that is vaguely similar to disproving by counter-example, called proof by contradiction. Suppose there is some irrational number p such that -p is rational. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. Proofs by contradiction are useful for showing that something is impossible and for proving the converse of already proven results. 2 Selected Homework Solutions 10. Name the left column Statements. statement q is true. Proof (by contradiction): Suppose greedy not optimal. Then use the. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. \u00f3 Solution: Assume that n is odd. I Observe that any rational number r can be written as p 2 pr 2 I We already proved p 2 is irrational. Considerthe number M = N + 1. This is also known as proof by assuming the opposite. Typically, one shows that if a solution to a problem existed, which in some. No possible constant value for x exists to make this a true equation. That is, suppose there is an. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. Compared to a proof of contradiction you have the advantage that the goal is clear. 4 Proof by contradiction The idea of contradiction method is by showing is a contradiction of the statement , that is a tautology. Proof: This is easy to prove by induction. Proof by contradiction is often used when you wish to prove the impossibility of something. This also applies ifthe result is goingto beproven using mathematical induction. fn and fn+1 that have a common divisor d, where d is greater than 1. A z\u00b0 x\u00b0 y\u00b0 100\u00b0 B O Solution Theorem 1 gives that z = y = 50 The value of x can be found by observing either of the following. To prove p, assume \u00acp and derive a contradiction such as p \u2227 \u00acp. Give a proof by contradiction: prove that the square root of 2 is irrational. Valid Argument: 1. If x 2A B then x 2A (and not in B). ] Assume, to the contrary, that \u2203 an integer n such that n 2 is odd and n is even. Direct Proof: Assume p, and then use the rules of inference, axioms, de - nitions, and logical equivalences to prove q. Use proof by contradiction to show that if n2 is an even integer then n is also an even integer. Inequalities 10 7. A logical contradiction is the conjunction of a statement S and its denial not-S. Any proof does. Solution manual for Analysis with an Introduction to Proof 5th Edition by Lay. If x 2A B then x 2A (and not in B). We have to prove 3 is irrational Let us assume the opposite, i. These solutions use information found in pages 154 - 160 of the textbook. The (Pedagogically) First Induction Proof 4 3. Math 2150 Homework 12 Solutions Throughout, use ONLY the assumptions given in the online notes and/or examples given in the online notes (which you need not reprove) unless speci- ed otherwise. x = \u221a(2k) \u2013Not clear that sqrt(2k) is an even integer, or even an integer J Proof by contrapositive \u2013prove that if x is odd then x2is odd. Therefore, 1 is the largest integer. This would mean that we can have at most 9 7 = 63 days we could have chosen. Examples of Proof by Contradiction. 6 [A level only] (a) Prove that the square root of 2 is irrational. I so p 1. Then there exists unmatched college c and unmatched student s. Related Answers What object is defined using a directrix and a focus Find the coordinates of B if A has coordinates (3,5) and Y-2, 3) is the midpoint of AB Geometry and Algebra 1 Introduction Write a two-column proof. Indirect Proof or Proof by Contradiction: Assume pand :qand derive a contradiction r^:r. Thus, the proposition is true. This and along with the direct proof on Friday complete an example of proof of an \"if and only if\" statement. Compared to a proof of contradiction you have the advantage that the goal is clear. Smolka and J. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. \u2022Alternatively, using contradiction, prove that it is not possible for such a thing notto exist. Then there exists integers. Figure 1 Solution: Proof. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. (It looks like that list omits the proof by the rational root theorem. The (Pedagogically) First Induction Proof 4 3. Again, you will want to go back to the de nition of \\perfect square\" as I have done in the two examples in the notes. We present an algorithm that enumerates all the minimal triangulations of a graph in incremental polynomial time. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w \u2022 w rejected m at some point \u2022 GS: women only reject for better partners \u2022 w prefers current partner m' > m \u2022 m and w are not blocking Case #1 and #2 exhaust space. The preceding examples give situations in which proof by contradiction might be useful:. ExampleProve by contradiction that there is no greatest integer. -p = mln, where m and n are both integers and n # 0 3. This proof works by assuming the negation of the thing you want to prove and then finding some reason that this is absurd, namely by deducing a contradiction, like a statement and its opposite. 1, 2017W1 midterm 1): You're given an SMP instance where two men have the same preference list. 4- Bacic Proof Methods I- Direct Proof, Proof by Cases, and Proof by Working Backward In this section we will introduce specific types or methods of proof of mathematical statements. \u201d I could go on, obviously, with countless examples of these kinds of posts on social media, not to mention news stories about biological men competing in women's athletics and young children\u2014surprisingly young children\u2014transitioning into another gender. Combining Proofs, cont. (b) Assume for a contradiction that the square root of 3 is rational, i. Proof by Contradiction. 21 To Prove That V2 = 2/3 Is Irrational. Let x be an integer. Induction step: Assume the theorem holds for n billiard balls. So, to prove \"If P, Then Q\" by the method of contrapositive means to prove \"If Not Q, Then Not P\". Each person is a vertex, and a handshake with another person is an edge to that person. In particular, the. Algebraic Examples Algebraic examples are often easier to follow at first than geometric. Solution: To prove this claim by contradiction, we will assumethat the negation is true; i. Another important method of proof is proof by contradiction. Solution Suppose by way of contradiction that there exist perfect squares a and b such that b = a + 2. Thus, the proposition is true. Suppose p 2 is rational. Logical Form: 8n: n2 even =)neven. They clearly need to be proven carefully, and the cleverness of the methods of proof developed in earlier modules is clearly displayed in this module. Proof by contradiction: example Theorem: There are in\ufb01nitely many primes. 2 More methods of proof (continued): Biconditional statements, Existence proofs (constructive and non-constructive). Once a mathematical statement has been proved with a rigorous argument, it counts as true throughout the universe and for all time. Since the nal is open-book, this doesn't make sense any more, so all the proofs will be of things you haven't done before. A useful resource to help deliver this new topic - fully worked solutions are included for all examples and questions in the exercise. We shall show that you cannot draw a regular hexagon on a square lattice. Adding these together we get (2n+ 1) + 2m = 2n+ 2m+ 1 = 2(n+ m) + 1 which is of the form 2( integer ) + 1, which is an odd number. And you also want an explanation of what a proof by contradiction is, which also seems to be way too elementary. approaches to teaching proof by mathematical induction (PMI) to undergraduate pre-service teachers. (It looks like that list omits the proof by the rational root theorem. , n and m have no prime factors in common. We assume \ud45d\ud45d \u2227\u00ac\ud45e\ud45e , then show that this leads to a contradiction. Example of a Proof by Contradiction Theorem 4. Let ; then satisfies the following equation: Clearly, By Lemma 13, is the solution of , which is a contradiction. [1 mark] Assume positive integer solutions. Figure 1 Solution: Proof. Some of the most famous examples of proofs by contradiction are: The proof that p 2 is irrational (probably dating back to Aristotle ca. The word Proof is italicized and there is some extra spacing, also a special symbol is used to mark the end of the proof. Let me show you another example where a contrapositive proof is so much easier to carry out. We present an algorithm that enumerates all the minimal triangulations of a graph in incremental polynomial time. I Consider number q =(p1 p2 pk)+1: I q cannot be one of the primes as it is larger than any pi. In a non-constructive proof, one proves the statement using an indirect proof such as a proof by contradiction. And except for the beginning and end, to solve an indirect proof, you use the same techniques and theorems that you would use on regular proofs. Most of the proofs I think of should be accessible to a middle grade school student. A feasible solution is a solution that satis es the property P. \u00f3 Solution: Assume that n is odd. A proof by contradiction might be useful if the statement of a theorem is a negation--- for example, the theorem says that a certain thing doesn't exist, that an object doesn't have a certain property, or that something can't happen. 1 Proving Negative Statements youtu. Use rules of inference, axioms, and logical equivalences to show that q must also be true. We must derive a contradiction. So, 0 = (x + y) (x y) = 2y. Then use the. Proofs and refutations: standard techniques for constructing proofs; counter-examples. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Discrete Mathematics This is a basic course for undergraduate students. geometry_terms_and_proof_by_contradiction. Hints and partial solutions are provided. Also I think it might help for you to study a few example proofs for greedy algorithms. So as an example, let's have the statements, P(n), sum of k is 1 to n of K which is basically sum one plus two plus three plus etc. is even, then. To prove: If x 2 is even, then x is even. Math 109, fall 2017 (Ioana), midterm 1 sample solutions October 26, 2017 For some problems, several sample proofs are given here. A Famous Contradiction Example. I don't understand this contradiction stuff. Proof by contradiction: Assume negation of what you are trying to prove (p q). Wrtiten response: Well done. # to derive a contradiction Then there is a \ufb01nite list, p 1;:::;p k of elements of P. What makes it different is the way it begins and ends. We're not done with them! Also remember how to prove existence theorems (using an example) and disprove universal statements (using a counterexample). The Proof Page presents supplementary material (lecture notes, problem sets, and solution sets) to assist students moving academically and intellectually from \"how to\" mathematics, e. So a2 is a multiple of 3, and so must be a. Here P(t) = p ktk + + p 1 + p 0 and Q(t) = q \u2018t\u2018 + +q 1 +q 0 are polynomials with real coe cients, both must be nonzero, and both may be assumed to have positive leading coe cients (since this is true of every element of the rst interval, [1;t]). Proof by mathematical induction Mathematical induction is a special method of proof used to prove statements about all the natural numbers. Prove that if aand bare real numbers with aa. Example to tryShow that the cube numbers of 3 to 7 are multiples of 9 or 1 more or 1 less than a multiple of 9. Equivalently, we could just prove the logical negation of the given statement, which is the statement 9x8y: y 2 x. Suppose that a + br is rational. Properies of the modulus of the complex numbers. What's our proposition? Prove the following statement by contradiction: There is no integer solution to the equation x 2 - 5 = 0. There is no greatest even integer. Before proceeding with any of the proofs we should note that many of the proofs use the precise definition of the limit and it is assumed that not only have you read that. Complex numbers tutorial. # at most n elements in the list Then I can take the product p0= p 1 p k. Show that the conclusion Q(x) is true by using definitions, previously established results, and the rules for logical inference. Then derive the [Prove] statement using logic (known theorems, laws, etc. Proposition: Each natural number n>1isaprimeorproductofprimes. Proof: Form the contrapositive of the given statement. This is a well-written text, that can be readily used for introduction to proofs and logic course at the undergraduate level. If the following statement is true, give a proof. When trying to construct a proof it is sometimes useful to assume the opposite of the thing you are trying to prove, with a view to obtaining a contradiction. This is one of the base methods of reasoning. geometry_terms_and_proof_by_contradiction. 1 The method In proof by contradiction, we show that a claim P is true by showing that its negation \u00acP leads to a contradiction. Advanced/wacky examples: This pdf has some great examples in Section 6(page 4) \u2014 they show how induction can be applied to all kinds of different mathematical problems. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. Factor the right-hand side of this equation as 2 = (a\u2212b)(a+b). A student should consider their solution of a proof-type problem to be aimed at an audience of students at their level; if they are unsure if it is a valid proof, then their goal has not been met. The di\ufb00erence between the two is that a proof by contradiction can be devised, but a proof by contrapositive. Henselian Valued Stable Fields I. Related Links. Solution: Suppose \u221a2 is rational. In a proof by contradiction, we start with the supposition that the implication is false, and use this assumption to derive a contradiction. ] Suppose not. Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Reach a contradiction. $\\endgroup$ - D. You can use one of the above methods (direct proof, proof by contraposition or contradiction) to solve the p\u2192q and q\u2192p part. Wrtiten response: Well done. Therefore y = 0, contradicting that it is positive. Exam focused Online Study Pack. P and (not P) is a contradiction For example xx22\u2212=1 0 Zero and 1 0 Not Equal to Zero[ ] \u2212\u2260[ ] is a contradiction. One standard way of doing this is to make the \ufb01rst line \u201cSuppose for the sake of contradiction that it is not true that (2 is irrational. In particular, the. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Hint: There are four parts to the proof. That is, suppose there is an integer n. Valid Argument: 1. a proof by contradiction. \u2013 If we had 5 pennies, we could replace them with a nickel. Proof: Suppose A. Lay Lee University. \u2022 Proof by contradiction \u2022 To prove that P is true, it is sufficient to prove that \u201cnot P implies Q\u201d when Q is clearly false. SOLUTION Let x represent the length of the third side. To show that d : R !R de ned by d(x;y) = jx yjis a metric, for. Problem: Given that a, b, and c are odd integers, prove that equation ax 2 + bx + c = 0 can not have a rational root. [Hint: Assume that r = a/b is a root, where a and b are integers and a/b is in lowest terms. Because r + s is rational, we can write it as p / q for some integers p and q where q \u2260 0. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. The idea is to assume the hypothesis, then assume the. Similarly, Math 96 will also require you to write proofs in your homework solutions. Contraposition: Contradiction:. Students often find this emphasis difficult and new. We have a contradiction. Example Questions. 9 = 362,880. Example: Use proof by contradiction to prove that p 2 is irrational. We argue by contradiction. Mathematical Proofs is designed to prepare students for the more abstract mathematics courses that follow calculus. \" Problem 2. p is the pumping length given by the PL. Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. One example of a proof by contradiction is the proof that \u221a2 is an irrational number: Assume that \u221a2 is a rational number, meaning that there exists a pair of integers whose ratio is \u221a2. 2 More methods of proof (continued): Biconditional statements, Existence proofs (constructive and non-constructive). p= -mln, where -m and n are both integers and n \u00b1 0 4. The correct proof is this: Let assume that the product of two odd numbers, m and n, is an even number N: N = m*n. Since n is odd, n = 2k + 1 for some integer k. Assume $$n$$ is a multiple of 3. Quiz SAP - C_THR84_2005 \u2013Efficient Exam Syllabus, Because it can help you prepare for the C_THR84_2005 Exam Content exam, We have strong IT masters team to study the previous test to complete the C_THR84_2005 new dumps to follow the exam center's change and demand, SAP C_THR84_2005 Exam Syllabus PDF version: can be read under the Adobe reader, or many other free readers, including OpenOffice. The negative of an integer is. Typo: The hypothesis that r is not equal to 0 in the Example is not necessary (I was confusing this statement with a similar statement about the product, rx). Example 1: irrational. Suppose you came up with an optimal solution to a problem by using suboptimal solutions to subproblems. Also, r = br b. Contradiction. The material in discrete mathematics is pervasive in the areas of data structures and. The first guy appears to misunderstand what proof by contradiction is. Then there exists integers aand bwith \u221a2 = a/b, where b\u2260 0 and aand b have no common factors (see Chapter 4). Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. (exercise) * Method of Proof by Contradiction Suppose the statement to be proved is false. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. For example, if every point lies on the x = y line, then each point would dominate all points below it, giving us n C 2 edges. bwhere b\u2260 0. Solutions to propositional logic proof exercises October 6, 2016 1 Exercises 1. (I will not read any work on this question sheet). But, from the parity property, we know that an integer is not odd if, and only if, it is. Proof by contradiction examples Example: Proof that p 2 is irrational. A proof by contradiction in this case has the logical form \u00acP \u00acP\u2192 (R\u2227 \u00acR) \u2234 R\u2227 \u00acR 2. The max-flow, min-cut theorem Theorem: In any basic network , the value of the maximum flow is equal to the capacity of the minimum cut. Proof by mathematical induction Mathematical induction is a special method of proof used to prove statements about all the natural numbers. If we give a direct proof of \u00acq \u2192 \u00acp then we have a proof of p \u2192 q. have no common factors (see Chapter 4). 4 The number 3 is irrational. Thus, one might prove that the negation 8x2S;\u02d8P(x) is false by deriving a contradiction. Which proof technique? Direct proof \u2013express x2 as 2k for some k, i. Non-linear examples exhibit a few other quirks, and we will demonstrate them below also. In general, then, try to be specific when doing an existence proof, but if you cannot, it may still be possible to construct an example using some other existence result or another technique of proof. Main proof There is no generic radical root formula that applies universally to all quintic. Such proofs can be reviewed at the Proofs tutorial. both r and :r for some proposition r. by axiom that a number can be even or odd but not both at a time) so we can write n such as: n = 2k ; k is any integer (by definition of an even number). But every number's square is nonnegative, so y2 0, a contradiction. Mathematics cannot be a spectator sport. Here P(t) = p ktk + + p 1 + p 0 and Q(t) = q \u2018t\u2018 + +q 1 +q 0 are polynomials with real coe cients, both must be nonzero, and both may be assumed to have positive leading coe cients (since this is true of every element of the rst interval, [1;t]). In order to illustrate this type of proof we assume that we know: 1. The number 2 is a prime number. Wiles's proof of Fermat's Last Theorem is a proof by British mathematician Andrew Wiles of a special case of the modularity theorem for elliptic curves. Suppose you came up with an optimal solution to a problem by using suboptimal solutions to subproblems. Solution: Perform a proof by contradiction. Suppose that a + br is rational. No possible constant value for x exists to make this a true equation. \u2022 Direct proof \u2022 Contrapositive \u2022 Proof by contradiction \u2022 Proof by cases 3. Relation between Proof by Contradiction and Proof by Contraposition. We learn how to do it with a couple of worked examples. I pdivides both x= 1 2 k and q,and divides I =)pj x q 1. can be proved by showing that its contrapositive \u00ac q \u2192 \u00ac p. Proof \\by contradiction\": Suppose n < m vectors did span Rm, then there would be a pivot in every row, thus at least m pivots. Then there exists integers. Why can't we prove B is not true by finding a counter example?. 6: Let A, B, and X be sets. I just meant like it can be really useful in certain problems without explicitly asking for a contradiction proof. is an integer and. Thus x2 + 1 < 0 is false for all x \u2208 S, and so the implication is true. 4, namely that for any integer. ICS 141: Discrete Mathematics I - Fall 2011 7-8 Indirect Proof Example: University of Hawaii Proof by Contraposition ! Theorem: (For all integers n) If 3n + 2 is odd, then n is odd. Giving a counter example 3+5=8 is even is not a proof by contradiction. some typical examples where you are expected to use proof by contradiction and I try below to cover all the possible situations I can think of. n odd \u21d2 n2 odd 2. As a first example of proof by contradiction, consider the following theorem:. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. By contradiction. Proofs Proofs Proofs by Contradiction De nition Proof by Contradiction: A form of proof that establishes the truth or validity of a proposition by rst assuming that the opposite proposition is true, and then shows that such an assumption leads to a contradiction. I pdivides both x= 1 2 k and q ,and divides I =)pj q x 1. Finding a contradiction means that your assumption is false and therefore the statement is true. Suppose that x is a positive real number with. \u2022 Example Every prime number a irrational numbers. If it were rational, it could be expressed as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. Hence, n2 = 4k2 +4k. Solutions 1. Smolka and J. One of the basic techniques is proof by contradiction. Proof: (direct proof) Assume that n is an even integer. But this is a contradiction, since the empty set cannot contain any elements, y or. Obtain an. If this is the case, we can factor the left side: x 2 - y 2 = (x-y)(x+y) = 1. If all our steps were correct and the result is false, our initial assumption must have been wrong. I don't understand this contradiction stuff. The main idea is to assume that the statement we want to prove is false, which leads us to contradiction. Proof (By contradiction) Suppose this language is context-free; then it has a context-free grammar. That is, suppose there exists a real number r such that r3 is irrational and r is rational. Given this, derive a contradiction such as something is both even and odd, or both positive and negative, or both rational and irrational, etc. Then, one of the vectors of the standard basis of cannot be written as a linear combination of the vectors of. Thus, 3n + 2 is even. If (A [B) X and (X B) (X A), then A B. If 3 - n2, then 3 - n. STEP 2 Reason logically until you reach a contradiction. Example 1 Find the value of each of the pronumerals in the diagram. Ex: p\u2227~p Claim:Suppose c is a contradiction. If 3jn then n = 3a for some a 2Z. The proof is by contradiction. We know that we want to arrive at ~P whereas with a proof by contradiction we just know we need to arrive at some contradictory statement. John Smith is a man. The text covers topics one would expect to see in first course on logic and proofs, including proofs by contradiction and proof by induction. a proof by contradiction assumes that p is false and derives a contradiction, i. This and along with the direct proof on Friday complete an example of proof of an \"if and only if\" statement. The proves the contrapositive of the original proposition,. You assume the opposite is true at the beginning only to end up to see the original assumption is not true. This page has a few examples worked out completely - not too long or involved, and (I hope) not too difficult to follow. If you can do that, that example is called a. The empty set is a subset of A, hence it is an element of the power set of A. Solutions are included. Let x be an odd integer. In this case and and so we have found an example where but and thus disproving the statement. Now this is a contradiction since the left hand side is odd, but the right side is even. Proof by Contradiction. 12, and if we can go by all the previous chapters, this will be our template for the exercises. This shows the negation is false, and hence that the original proposition is true. If f(2) = 8, explain why f(3) > 6. Trans and non-binary men belong. A First Example: Proof by Contradiction Proposition: There are no natural number solutions to the equation x2 y2 = 1. Therefore, the reasoning of the ontological argument dodges the parody, its reasoning is not parallel to the parody argument, and it cannot be used to prove the existence of a lost island. Second, we provide some examples of inductive proofs that follow the structure outlined in the rst part. In an indirect geometric proof, you assume the opposite of what needs to be proven is true. Discrete Mathematics This is a basic course for undergraduate students. Obtain an. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a technique which can be used to prove any kind of statement. The argument is valid so the conclusion must be true if the premises are true. That would mean that there are two even numbers out there in the world somewhere that'll give us an odd number when we add them. (This is an indirect form of proof. The number 2 is a prime number. A Famous Contradiction Example. A useful resource to help deliver this new topic - fully worked solutions are included for all examples and questions in the exercise. (Contrapositive) Let integer n be given. As a first example of proof by contradiction, consider the following theorem: Theorem 1. Example2 1. Proof by Contradiction: (AKA reductio ad absurdum). This page has a few examples worked out completely - not too long or involved, and (I hope) not too difficult to follow. Mathematical Proof/Methods of Proof/Proof by Induction. That is how Mathematical Induction works. The establishment of a fact by the use of evidence. ExampleProve by contradiction that there is no greatest integer. Section 4-7 : The Mean Value Theorem. In a proof by contradiction, or indirect proof, you show that if a proposition were false, then some false fact would be true. (i)Direct proof: we assume A is true. Equivalently, we could just prove the logical negation of the given statement, which is the statement 9x8y: y 2 x. For every even integer n, N \u2265 n. \u2022Proof : Assume that the statement is false. ]! Then, by definition of rational, r = a/b and s = c/d for some integers a, b, c, and d with b \u2260 0 and d \u2260 0. Example of a constructive proof: Suppose we are to prove 9n2N;nis equal to the sum of its proper divisors: Proof: Let n= 6. Wyke, or Mr. Solution LetP(n) bethemathematicalstatement 11n \u22126 isdivisibleby5. GS results in a stable matching (i. Proof: Form the contrapositive of the given statement. 8, 1113, Sofia, Bulgaria Communicated by Walter Feit Received November 10, 1997. Thursday 2/16/17. The Second Edition features new chapters on nested quantifiers and proof by cases, and the number of exercises has been doubled with answers to odd-numbered exercises provided. Prove the following statement by contradiction: The sum of two even numbers is always even. Then there exists unmatched college c and unmatched student s. Rajoub: \"For the first time in the history of the conflict \u2013 the contradiction between [the occupation and] the interests of the world, international law, and its values have reached a peak, and we must not undermine this clash through the wrong actions \u2013 whether in word or in actions that deviate from the consensus. a method of disproving a. Give a proof by contradiction: prove that the square root of 2 is irrational. It contrasts what Bastiat considered as the proper function of the Law and the perversion of the Law. Such examples are called counter examples. (I will not read any work on this question sheet). It is usually not as neat as a two-column proof but is far easier to organize. Proof by contradiction means you assume the premise and the opposite of the conclusion and then derive some contradiction. to an equation then there is another integral solution that is smaller in some way. Prove that the sum of irrational and rational number is irrational using proof by contradiction. If x \u2208 A \u2229 B, then x \u2208 A and x \u2208 B by de\ufb01nition, so in particular x \u2208 A. Proof by Contradiction. O is the centre of the circle and \u2220AOB = 100. 1 \u221a2 is an irrational number. (non-constructive proof) \u2022Show that a player in a game has a winning strategy without actually sayingwhat it is! \u2022Famous proof: There exist irrational x, y such that xyis rational Villanova CSC 1300 -Dr Papalaskari. You can put this solution on YOUR website!. Below are several more examples of this proof strategy. This is the way most people learn a new language | learn to say a. Proof: (direct proof) Assume that n is an even integer. An example is \"Prove that the product of two nonzero real numbers is nonzero. So this is a valuable technique which you should use sparingly. Example of a Proof by Contradiction Theorem 4. Then, one of the vectors of the standard basis of cannot be written as a linear combination of the vectors of. Solution: Suppose \u221a2 is rational. Shows how and when to use each technique such as the contrapositive, induction and proof by contradiction. Recall that for two integers x and y, we say x divides y if there exists an integer z such that xz = y. Wrtiten response: Well done. If all our steps were correct and the result is false, our initial assumption must have been wrong. Imagine, then, the thrill of being able to prove something in mathematics. We give the proof by contradiction. To prove a theorem of the form A IF AND ONLY IF B , you first prove IF A THEN B , then you prove IF B THEN A , and that's enough to complete the proof. The concept of proof by contradiction is to assume that P is false. By the closure property, we know b is an integer, so we see that 3jn2. If you can do that, you have used mathematical induction to prove that the property P is true for any element, and therefore every element, in the infinite set. Also see the Mathematical Association of America Math DL review (of the 1st edition) and the Amazon reviews. Mathematical Proofs is designed to prepare students for the more abstract mathematics courses that follow calculus. In the second group the proofs will be selected mainly for their charm. Since and are distinct points in the Hausdorff space , there must be disjoint, open neighborhoods such that and. This A Level Maths video takes you through a new method of proof called proof by contradiction. Proof by contradiction: Assume P(x) is true but Q(x) is false. Prove that if a square matrix A is a zero divisor (that is AB=0 for some non-zero matrix B) then det(A)=0. Selected Homework Solutions - Math 574, Frank Thorne 1. Use rules of inference, axioms, and logical equivalences to show that q must also be true. Example 3: Prove the following statement by contraposition: For all integers n, if n 2 is odd, then n is odd. Then (x y) = (x + y) = 1. But this is clearly impossible, since n2 is even. Prove: do not bisect each other. Proof by Contradiction is another important proof technique. Instead of it, we use proof by contradiction. Example of a Proof by Contradiction Theorem 4. Justify all of your decisions as clearly as possible. Proving \"If A, then B\" by contradiction Given the assumptions in A, show that B must be true because it cannot possibly be false. As a first example of proof by contradiction, consider the following theorem:. Then there exists integers. A proof of a very general idea could be preceded by an example in a speci\ufb01c context. Direct Proof: Assume that p is true. (Proof by Contradiction. Run M on hPi. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. So, we will discuss these methods in this lesson extensively. 2 Selected Homework Solutions 10. Solution: By contradiction. Proof by Contradiction Example: Use a proof by contradiction to give a proof that \u221a2 is irrational. Introduction 1 2. Squaring both sides we get 2 = n 2=m2, so m2 = 2n. The Pigeonhole Principle 1 Pigeonhole Principle: Simple form Theorem 1. Statement: If A, then B Inverse: If B, then A Converse: If not A, then not B Contrapositive: If not B, then not A Which of these are logically equivalent? RTP: If A, then B Method: Assume not B Carry out logical, deductive steps Reach the conclusion not A Example:. The statement P1 says that x1 = 1 < 4, which is true. Then we have to \ufb01nd a statement R so that \u00acP\u2192 (R\u2227 \u00acR) - a contradiction. I can use. If \u00acP leads to a contradiction, then. Proof by Contradiction is another important proof technique. A student should consider their solution of a proof-type problem to be aimed at an audience of students at their level; if they are unsure if it is a valid proof, then their goal has not been met. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. Since nm + 2n + 2m is odd, nm + 2n + 2m = 2k + 1 for some integer k. Similarly, since ris rational, we can express it as a/ bfor some integers aand. Finding a contradiction means that your assumption is false and therefore the statement is true. Final: Solutions ECS20 (Fall 2014) December 16, 2014 Part I: Proofs 1) Let a and b be two real numbers with a 0 and b 0. These notes explain these basic proof methods, as well as how to use de\ufb01nitions of new concepts in proofs. The proof is a sequence of mathematical statements, a path from some basic truth to the desired outcome. To write a two-column proof: Make a two-column form like this. Example: Prove that if you pick 22 days from the calendar, at least. Show that if n=k is true then n=k+1 is also true; How to Do it. (b) Prove that the square root of 3 is irrational. _____ Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1. 1 The number \u221a3 is irrational. 3 Review the proof techniques on page 116\u2212\u2212118 Here is a result that is proved by three di\ufb00erent proof techniques. Indirect Proof: Assume what you need to prove is false, and then [\u2026]. If (A [B) X and (X B) (X A), then A B. 2 Proof By Contradiction A proof is a sequence S 1;:::;S n of statements where every statement is either. Example of proof by contradiction. Eureka step and the eventual solution (Zeitz, 1999). Proof by mathematical induction. Some of the most famous examples of proofs by contradiction are: The proof that p 2 is irrational (probably dating back to Aristotle ca. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. To change the symbol printed at the end of a proof is straightforward. Re\ufb02ex angle AOB is 260. In other words, if it is impossible for $$P$$ to be false, $$P$$ must be true. Proof (by contradiction): [We take the negation of the theorem and suppose it to be true. A proof by contradiction is often used to prove a conditional statement $$P \\to Q$$ when a direct proof has not been found and it is relatively easy to form the negation of the proposition. To show that there is no finite state automata that. ExampleProve by contradiction that there is no greatest integer. Example: Use a proof by contradiction to give a proof that \u221a2 is irrational. By the Pumping Lemma this must be representable as , such that all are also in. Without loss of generality n=m is reduced, i. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. Therefore, the assumption that the quadratic equals zero is incorrect. Direct proof by contradiction. I Suppose pr 2 was rational. (Contrapositive) Let integer n be given. have no common factors (see Chapter 4). 3 Proof by contradiction (continued). We then try to obtain a contradiction from this assumption. Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. But this is clearly impossible, since n2 is even. Therefore both n and m are odd. Discrete Mathematics This is a basic course for undergraduate students. Proof by Contradiction. To a resolution theorem-prover, both are two-step proofs. Proof: (Contrapositive: If n is even, then 3n + 2 is even) Suppose that the conclusion is false, i. Choose a long string w in L, jwj m. In the examples below we use this idea to prove the impossibility of certain kinds of solutions to some equations. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. Then (2 is rational, so there are integers a and b for which (2= a b. (4 marks) 8 Use proof by contradiction to show. Properies of the modulus of the complex numbers. By Observation 3, it de nitely returns a matching, so suppose the matching is not perfect. As an example, here is a proof by contradiction of Proposition 4. This is a good resource if you are familiar with induction, and want to take things a little farther. Discrete Mathematics This is a basic course for undergraduate students. The idea behind proof by contradiction is that a statement must be true or false. Disprove by counterexample that for any , if , then. This is also known as Proof by Cases - see Example 1. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. 2 # 2 p23 Follow the statement of your assumptions with a statement of what you will prove. The proof by deduction section also includes a few practice questions, with solutions in a separate file. Here are several examples of properties of the integers which can be proved using the well-ordering principle. You must include all three of these steps in your proofs! The three key pieces: 1. n By contradiction n Start with the \u201cproof by example\u201d! n So when asked to prove a claim, an example that n Automata theory & a historical perspective. Related Answers What object is defined using a directrix and a focus Find the coordinates of B if A has coordinates (3,5) and Y-2, 3) is the midpoint of AB Geometry and Algebra 1 Introduction Write a two-column proof. Let x be an odd integer. by triangle inequality. Example: Prove there's an infinite number of evens. ] [Prove Q =)P using direct, contrapositive, or contradiction proof. What that contradiction means in the proof; Whether the Halting problem is an unsolvable problem, an undecidable problem, or both; and why; Write a paragraph explaining the difference between an problem that can't be solved (such as the halting problem) and a problem that takes unreasonable time. For example: 2/8 can be written in lowest terms as 1/4 when 1 and 4 are positive integers with no common prime factors. Then \u2026 (( make logical conclusions until you come to two statements that contradict each other, such as \"X is true\" and X is false\" ))But this is a contradiction because \u2026. Proof: I Assume \ufb01nitely many primes: p1;:::; k. 7 [A level only] Prove that there are an infinite number of primes. Math 2150 Homework 12 Solutions Throughout, use ONLY the assumptions given in the online notes and/or examples given in the online notes (which you need not reprove) unless speci- ed otherwise. If n+1 objects are put into n boxes, then at least one box contains two or more objects. Show that this supposition logically leads to a contradiction. These notes explain these basic proof methods, as well as how to use de\ufb01nitions of new concepts in proofs. This proves A \u2286 A \u2229 B. and qwhere q\u2260 0. Proof: Suppose not. \u2013 If we had 2 nickels, we could replace them with 1 dime. The reason is that the proof set-up involves assuming \u223c\u2200x,P(x), which as we know from Section 2. Proof: By induction, on the number of billiard balls. We must derive a contradiction. Example of proof by contradiction and more on proof by induction. Solutions Educator Edition Save time lesson planning by exploring our library of educator reviews to over 550,000 open educational resources (OER). But, from the parity property, we know that an integer is not odd if, and only if, it is. PLEASE DO A \\PREFOR-. Proof by Contradiction: (AKA reductio ad absurdum). 4, namely that for any integer n, if n2 is even then n is even. Robust, Semi-Intelligible Isabelle Proofs from ATP Proofs S. Here's an algebraic example: Prove: For a,b\u22650, a+b 2 \u2265ab. If \u00acP leads to a contradiction, then. Say we're trying to prove by contradiction that if n 2 is an odd number, then n is also odd for all integers n. Second, we provide some examples of inductive proofs that follow the structure outlined in the rst part. Example using a Linear Function. Example: Give a direct proof of the theorem If nis an odd integer, then n2 is odd. Then n= 2k+ 1 for an integer k. In the proof above of Fact 1. Proof By Contradiction Examples And Solutions. \u2022 Example Every prime number a irrational numbers. You must include all three of these steps in your proofs! The three key pieces: 1. This proof is very similar to proof by contradiction, but subtly di er-ent. ] Suppose there is greatest even integer N. For every even integer n, N \u2265 n. \u00acCube(a) 3. with \u221a2 = a/b, where. This book is a guide to understanding and creating proofs. A proof by contrapositive uses that to prove the negation of the original assumption, while a proof by contradiction can negate any other true fact or lead to some other absurdity; in this case, you can't have two different smallest elements of a set. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a common proof technique that is based on a very simple principle: something that leads to a contradiction can not be true, and if so, the opposite must be true. This is usually a hint that proof by contradiction is the method of choice. We have step-by-step solutions for your textbooks written by Bartleby experts! Using Proof by Contradiction In Exercises 15 \u2212 26 , use proof by contradiction to prove the statement. This is why proof by contradiction works. uzbt8y6d536r qsp0l7cs9uxx0b rqtztc3l4kvhz ve65emynwfzt fb2nd4u3ken19 cswb17aj5fv byn1w24apk0exlx pquj9ksys3c0sn iss43ryp1ywds 28pf05o6s2 n9jk6v0ab1 i94sauttvyg 7x3i711c7mw11kf fh46ysf9w7ic qz8m9vxmhy5 zy2kssr5fx13g 3h76hkwgzwpo y6j0hsw3ln6 estzy7oybir85b bwdlnq0skk 5nne7cw7flh natj3e81te5cn kc2t3uryp80 zd3r6tbgj5ai dfps6gopdz", "date": "2020-10-21 21:43:52", "meta": {"domain": "chicweek.it", "url": "http://bgvl.chicweek.it/proof-by-contradiction-examples-and-solutions.html", "openwebmath_score": 0.7791018486022949, "openwebmath_perplexity": 395.13846022461126, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9852713887451681, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8506750575164101}}
{"url": "https://math.stackexchange.com/questions/242643/a-game-played-on-a-rectangle", "text": "# A game played on a rectangle\n\nSuppose two players play the following game on a $m$ by $n$ rectangle. Alternatingly they have to make a cross in some empty $1\\times 1$ square. They are not allowed to make a cross next to another cross (Diagonally is OK, not just right next to each other). The player who places the last cross wins.\n\nNow the question is for which $m,n$ does the starting player have a winning strategy?\n\nAt first I thought this might be just a nice exercise. So I had a look at 1 by $n$ rectangles first. A computer programme computed that the first player does not have a winning strategy for $n=4,8,14,20,24,28,34,38,42,54,58,62,72,76,88,92,96,106,110$ (for $n\\le 110$).\n\nI do not see any pattern in these numbers. So the answer might not be so easy.\n\nThe starting player can always win for odd $n$. He just places a cross in the middle and mirrors all the moves of the second player.\n\n\u2022 @tomglabst: The game would be over if each remaining square is adjacent to some cross. Then no player can make any cross. Thus in your example the game would end after the first move. Specifically \"P1 crosses any remaining square\". There are no squares left that any player could cross. \u2013\u00a0HenrikRueping Nov 22 '12 at 15:42\n\u2022 I think I've seen a variant of the $m=1$ game where there's a modulus $M$ such that the grundy number of the game only depends on $n$ modulo $M$. So it's a possibility that such an $M$ exists here too. \u2013\u00a0mercio Nov 22 '12 at 16:12\n\u2022 @Andr\u00e9Nicolas The OP has noted values of $m$, where the first player has no winning strategy, and 6 is not among them. I see no edits made either. \u2013\u00a0Mike Nov 22 '12 at 17:00\n\u2022 The sequence is not in the OEIS (!!). \u2013\u00a0dot dot Nov 22 '12 at 17:12\n\u2022 @HenrikRueping Ah okay, I misunderstood that. \u2013\u00a0tomglabst Nov 22 '12 at 17:27\n\nThe $n \\times 1$ version is Dawson's Chess. The OP's sequence is A215721 in the OEIS, after adding 1 to each term. I wrote a program too, and found that the proportion of losing initial positions seems to tend to a constant -- there are 1473 losing positions in the first 10000, and 14709 losing positions in the first 100000. I found an explanation at \"Sprague-Grundy values for Dawson's Chess\" (A002187 in the OEIS):", "date": "2019-09-15 14:18:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/242643/a-game-played-on-a-rectangle", "openwebmath_score": 0.5581916570663452, "openwebmath_perplexity": 264.6939737385641, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98527138442035, "lm_q2_score": 0.8633916205190225, "lm_q1q2_score": 0.8506750572457068}}
{"url": "https://math.stackexchange.com/questions/3366424/find-the-singular-points-of-the-differential-equation-x3x-1y-2x-1y", "text": "Find the singular points of the differential equation $x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$.\n\nConsider the second order linear homogeneous equation $$a_0(x)y'' + a_1(x)y'+ a_2(x)y = 0, x \\in I \\tag{1}$$ Suppose that $$a_0$$, $$a_1$$ and $$a_2$$ are analytic at $$x_0 \\in I$$. If $$a_0(x_0) = 0$$, then $$x_0$$ is a singular point for $$(1)$$. Definition: A point $$x_0 \\in I$$ is a regular singular point for $$(1)$$ if $$(1)$$ can be written as $$b_0(x)(x \u2212 x_0)^2y''+ b_1(x)(x \u2212 x_0)y'+b_2(x)y = 0, \\tag{2}$$ where $$b_0(x_0) \\neq 0$$ and $$b_0$$, $$b_1$$, $$b_2$$ are analytic at $$x_0$$.\n\nThe question is: Find the singular points of the differential equation $$x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$$ and state whether they are regular singular points or irregular singular points.\n\nI think, $$x = 0$$, irregular singular point, $$x = 1$$, regular singular point. But, How can I prove this? Please proper guide me.\n\n\u2022 Hi Harry, I've formatted your question with MathJax. I really encourage you to learn how to do this for yourself. I'd also like to remind you of your comment under your last question. \u2013\u00a0Theo Bendit Sep 23 '19 at 5:18\n\u2022 @ Theo. Thanks again.I am trying latex but not done properly. I am on my way to working with MathJax. \u2013\u00a0user679406 Sep 23 '19 at 5:29\n\u2022 That's good! Feel free to edit your question to see how I've formatted it. That might help you figure it out a bit faster. \u2013\u00a0Theo Bendit Sep 23 '19 at 5:31\n\nConsider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0$$ where $$z \\in D \\subseteq \\mathbb{C}$$.\n\nThe point $$x_0 \\in D$$ is said to be an ordinary point of the above the given differential equation if $$P(x)$$ and $$Q(x)$$ are analytic at $$x_0$$.\n\nIf either $$P(x)$$ or $$Q(x)$$ fails to be analytic at $$x_0$$, the point $$x_0$$ is called a singular point of the given differential equation.\n\nA singular point $$x_0$$ of the given differential equation is said to be regular singular point if the function $$(x-x_0)P(x)$$ and $$(x-x_0)^2 Q(x)$$ are analytic at $$x_0$$ and irregular otherwise.\n\n$${}$$\n\nHere the given equation is $$x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$$ $$\\implies y''-\\dfrac{2}{x^3}y'+\\dfrac{3}{x^2(x - 1)}y=0$$ Here $$~P(x)=-\\dfrac{2}{x^3}~$$and $$~Q(x)=\\dfrac{3}{x^2(x - 1)}~$$.\n\nClearly $$~x=0,~1~$$ are singular points.\n\nAgain for the singular point $$~x=0~$$, $$(x-0)P(x)=-\\dfrac{2}{x^2}\\qquad \\text{and}\\qquad (x-0)^2P(x)=-\\dfrac{2}{x}$$ Clearly both $$~(x-x_0)P(x)~$$ and $$~(x-x_0)^2 Q(x)~$$ are not analytic at $$~x_0=0~$$. So $$~x=0~$$ is irregular singular point.\n\nFor the singular point $$~x=1~$$, $$(x-1)P(x)=\\dfrac{3}{x^2}\\qquad \\text{and}\\qquad (x-0)^2P(x)=\\dfrac{3(x - 1)}{x^2}$$ Clearly both $$~(x-x_0)P(x)~$$ and $$~(x-x_0)^2 Q(x)~$$ are analytic at $$~x_0=1~$$. So $$~x=1~$$ is regular singular point.\n\n\u2022 @ nmsanta,I think this answer is correct and I am going with it. Am I right? \u2013\u00a0user679406 Sep 23 '19 at 5:32\n\u2022 Have you faced any trouble in understanding my work ? I just doing things according to the definition. @Harry Richie \u2013\u00a0nmasanta Sep 23 '19 at 5:34\n\u2022 No trouble. Clear like water..@ nmasanta \u2013\u00a0user679406 Sep 23 '19 at 5:37\n\u2022 Only then my work will be well worth. \u2013\u00a0nmasanta Sep 23 '19 at 5:42\n\u2022 Hello Mr. Down-voter, Would you like to explain the reason to give the down-vote in this answer ? \u2013\u00a0nmasanta Sep 24 '19 at 8:17", "date": "2020-03-30 14:12:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3366424/find-the-singular-points-of-the-differential-equation-x3x-1y-2x-1y", "openwebmath_score": 0.8759991526603699, "openwebmath_perplexity": 199.36216481726532, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713852853136, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8506750493342383}}
{"url": "http://math.stackexchange.com/questions/165941/finding-a-splitting-field-of-x3-x-1-over-mathbbz-2", "text": "# Finding a splitting field of $x^3 + x +1$ over $\\mathbb{Z}_2$\n\nFinding a splitting field of $x^3 + x +1$ over $\\mathbb{Z}_2$.\n\nOk so originally I messed around with $x^3 + x +1$ for a bit looking for an easy way to factor it and eventually decided that the factors are probably made up of really messy nested roots. So then I tried looking at the quotient field $\\mathbb{Z}_2[x]/x^3 + x + 1$ to see if I would get lucky and it would contain all three roots but it doesn't. Is there a clever way to easily find this splitting field besides using the cubic formula to find the roots and then just directly adjoining them to $\\mathbb{Z}_2$?\n\nEdit: Ok it turns out I miscalculated in my quotient field, $\\mathbb{Z}_2[x]/x^3 + x + 1$ does contain all three roots.\n\n-\nIf degree is not $3$ it is $6$. \u2013\u00a0Andr\u00e9 Nicolas Jul 3 '12 at 2:08\nActually, your quotient $\\mathbb{Z}_2[x]/(x^3+x+1)$ does contain at least one root: $x$. This is because $x^3+x+1\\equiv 0\\pmod{x^3+x+1}$. Can you find the others? \u2013\u00a0roninpro Jul 3 '12 at 2:08\n\nThe splitting field is either degree $3$ or degree $6$ over $\\mathbb{Z}_2$, hence it is either $\\mathbb{F}_8$ or $\\mathbb{F}_{64}$.\n\nLet $\\alpha$ be a root, so that $\\mathbb{F}_8 = \\mathbb{F}(\\alpha)$. The elements are of the form $a+b\\alpha+c\\alpha^2$, with $\\alpha^3=\\alpha+1$.\n\nNow, the question is whether any of these elements besides $\\alpha$ is a root of the original polynomial $x^3+x+1$.\n\nNote that $(\\alpha^2)^3 = (\\alpha^3)^2 = (\\alpha+1)^2 = \\alpha^2+1$, and so if we plug in $\\alpha^2$ into the polynomial we have $$\\alpha^6 + \\alpha^2 + 1 = \\alpha^2+1+\\alpha^2+1= 0.$$ Thus, $\\alpha^2$ is also a root. So the polynomial has at least two roots in $\\mathbb{F}_8$, and so splits there.\n\n-\noh shoot I must have miscaculated in my quotient field, ok cool, thanks! \u2013\u00a0cuckmaster5000 Jul 3 '12 at 2:11\n\nIf your cubic were to factor, the factorization must have at least one linear term, which corresponds to a root. It is easy to check it has no roots in $\\mathbb{Z}_2$ - check them both! Plugging in $0$ and $1$ both give $1$ so are not roots, so your polynomial is irreducible over $\\mathbb{Z}_2.$\n\nA way to write the splitting field is $\\mathbb{Z}_2(\\alpha)$ where $\\alpha$ is any one of the roots of $x^3+x+1.$ This is because $\\alpha^2$ and $\\alpha^4= \\alpha^2+\\alpha$ must then also be distinct roots of $x^3+x+1$, and these 3 then comprise a full list. Note, this isn't a special trick to notice for this problem. In fields of characteristic $p$, if $\\beta$ is a root of a polynomial then $\\beta^p$ will automatically also be a root, due to properties of the Frobenius endomorphism.\n\nOr as roninpro pointed out in the comments, the quotient ring you considered (which is essentially the same thing as adjoining these roots) does contain at least one root, and by this same trick, all the roots.\n\n-\nI thought forming the quotient field adjoined just a single root? And that sometimes you would get lucky and the remaining roots could be formed within the quotient field as well, but that sometimes they couldn't, is this not correct? \u2013\u00a0cuckmaster5000 Jul 3 '12 at 2:17\n@NollieTr\u00e9 You're correct. But extensions of finite fields are all Galois and in particular normal. So if $K/k$ are finite fields and $f \\in k[X]$ is irreducible and has a root in $K$, then it splits in $K$. Frobenius is a great thing! \u2013\u00a0Dylan Moreland Jul 3 '12 at 2:25\n@NollieTr\u00e9 You are correct, sometimes the extra roots we find with the Frobenius trick just correspond to the same roots. But here they are distinct. You can check yourself (by carrying out the division) that here $x^2$ is also a root, since $x^6+x^2+1 = (x^3+x+1)(x^3-x-1)+2(x^2+x+1) = 0$, and similarly $x^4$ is also a root. So the quotient formed actually has all the roots. \u2013\u00a0Ragib Zaman Jul 3 '12 at 2:26\nOk interesting, so is there a canonical example of when the quotient field does not contain all the roots of the polynomial? \u2013\u00a0cuckmaster5000 Jul 3 '12 at 2:29\nI think it's easier to note that if $\\alpha$ is a root of the polynomial then squaring gives $0^2 = (\\alpha^3 + \\alpha + 1)^2 = (\\alpha^2)^3 + \\alpha^2 + 1$. \u2013\u00a0Dylan Moreland Jul 3 '12 at 2:30\n\nTwo good answers already here, but I wanted to emphasize the usefulness of the Frobenius endomorphism. Let $f(X) = X^3 + X + 1$. If I let $\\alpha$ denote the image of $X$ in $k = \\mathbb F_2[X]/(f(X))$ then applying Frobenius to $0 = f(\\alpha)$ gives $0 = (\\alpha^3 + \\alpha + 1)^2 = (\\alpha^2)^3 + \\alpha^2 + 1.$ Hence $\\alpha^2$ is also a root of $f$, and $\\alpha^2 \\neq \\alpha$ because $\\alpha \\neq 0, 1$. Since $k$ contains two roots of the cubic $f$, it contains three.\n\nTo bring in slightly more technology: extensions of finite fields are Galois and in particular normal, and this implies that if you adjoin one root of an irreducible polynomial over a finite field then you've actually adjoined all of them. This is, of course, not true in general!\n\n-\nJust wanted to add here, in $\\Bbb{Z}_2(\\alpha)[x]$, one can factor $x^3 + x + 1 = (x + \\alpha)(x^2 + \\alpha x + (\\alpha^2 + 1))$. Given that $\\alpha^2$ is also a root (as Dylan shows above), using ordinary high-school factorization techniques the third root is easily seen to be $\\alpha^2 + \\alpha$ (a somewhat preferable form than $\\alpha^4$). That is: $x^2 + \\alpha x + (\\alpha^2 + 1) = (x + \\alpha^2)(x + (\\alpha^2 + \\alpha))$ \u2013\u00a0David Wheeler Jul 6 '12 at 10:20", "date": "2016-04-29 02:12:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/165941/finding-a-splitting-field-of-x3-x-1-over-mathbbz-2", "openwebmath_score": 0.9422818422317505, "openwebmath_perplexity": 186.81982532539902, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713857177955, "lm_q2_score": 0.8633916082162402, "lm_q1q2_score": 0.8506750462443309}}
{"url": "https://math.stackexchange.com/questions/1603632/showing-that-a-group-g-with-two-distinct-normal-subgroups-of-index-p-has-or", "text": "# Showing that a group $G$ with two distinct, normal subgroups of index $p$ has order $p^2$\n\nA friend of mine and I were working together to solve this exercise:\n\nLet $G$ be a group, and $p$ a prime number. Let $H, K$ be normal subgroups of $G$. Suppose that $|G:H| = |G:K| = p$, and that $H\\cap K = \\{1\\}$. Show that $G\\simeq\\mathbb{Z}_p\\times\\mathbb{Z}_p$.\n\nWe thought that the quickest way to solve it was showing that $G$ has exactly $p^2$ elements.\n\nFor that purpose, we first tried to prove that $|G|\\leq p^2$. We chose $a,b\\in xH\\cap yK$, that are two elements of $G$ belonging to the intersection of a coset of $H$ with a coset of $K$. This means that $a = xh$ and $b=xh'$ for some $h,h'\\in H$ and for some $x\\in G$ (because $a,b\\in xH$). Similarly, $a=yk$ and $b=yk'$, for some $k,k'\\in K$ and for some $y\\in G$ (because $a,b\\in yK$).\n\nThen, $$xh=yk\\implies hk^{-1}=x^{-1}y$$ and $$xh'=yk'\\implies h'k'^{-1}=x^{-1}y.$$\n\nThis brings us to $hk^{-1}=h'k'^{-1}$, which we can rewrite as $hh'^{-1}=k'^{-1}k$. This element belongs to $H\\cap K=\\{1\\}$, thus $h = h'$ and $k = k'$: hence $a = b$. This proves that the intersection of a coset of $H$ with a coset of $K$ cannot contain more than one distinct element.\n\nWe made use of this information when we noted that, regardless of how many elements $G$ has, there exist at most $p^2$ intersections of cosets of $H$ with cosets of $K$, and each one of them contains at most one element. Thus, we deduced that $|G|\\leq p^2$.\n\nNow, here are my questions:\n\n\u2022 How do I prove that $|G|\\ge p^2$, in order to conclude that $|G|=p^2$ (and thus that $G\\simeq\\mathbb{Z}_p\\times\\mathbb{Z}_p$)?\n\u2022 Is the proof rigorous enough so far, or were some of our assumptions incorrect?\n\u2022 This exercise is related to a part of our algebra course in which nothing beyond direct product of groups had been explained yet, so we could not make use of the Sylow theorems, of anything about group actions and so on. Is there a simpler way to solve this exercise, using only \"basic\" tools of group theory (e.g. quotient groups, the homomorphism and isomorphism theorems and anything regarding the direct product of groups)?\n\u2022 Is $G$ assumed to be finite at the outset? \u2013\u00a0Tim Raczkowski Jan 7 '16 at 20:46\n\u2022 @TimRaczkowski The exercise said nothing on that point, so I assume that $G$ may also be infinite. Anyway, given that $G$ has to be isomorphic to $\\mathbb{Z}_p\\times\\mathbb{Z}_p$, I guess it can't be infinite. Plus, if our half of the proof is correct, $|G|\\leq p^2$ definitely implies that $G$ is finite. \u2013\u00a0Labba Jan 7 '16 at 20:48\n\nYou can use the following fact: If $H,K$ are subgroups of $G$, $H\\cap K=\\{1\\}$ and $hk=hk$ for all $h\\in H$ and $k\\in K$, then $G\\cong H\\times K$.\nNow, $hk\\in hK,Hk$. Since $H$ is normal in $G$, $Hk=kH$. So, by your argument above, $\\{hk\\}=hK\\cap kH$. But $kh\\in hK\\cap kH$ by a similar argument.\n\u2022 Thanks for your answer :) So, if I got it right, you say that $hk$ belongs to the $hK$ and $Hk$ cosets (the latter being $kH$ anyway, because of the normality of $H$), and the same can be said about $kh$ that belongs both to the $kH$ and $Kh$ cosets (again, the $hK$ coset). In short, $hk, kh\\in kH\\cap hK$ and, by my part of the proof, $kh = hk$ because only one element can be found in this intersection. Hence, $G\\simeq H \\times K$ because $H\\cap K = \\{1\\}$ and they commute with each other. Did I get everything? And this also proves that $|G| = p^2$, right? \u2013\u00a0Labba Jan 7 '16 at 21:15\n\u2022 I really have to thank you :) Anyway, there's still something I fear I'm missing: even if this shows that $G\\simeq H\\times K$, does this prove that $|H| = |K| = p$ as well? \u2013\u00a0Labba Jan 7 '16 at 21:18\n\u2022 Well if $|G|=p^2$, and $[G:H]=[G:K]=p$, so $|H|=|K|=p$. \u2013\u00a0Tim Raczkowski Jan 7 '16 at 21:21\n\u2022 Hmm....I seem to keep missing a detail each time. $[G:H]=[G:K]\\implies |H|=|K|$. Suppose $|HK|=n^2$. Now, $n^2|p^2\\implies$n=1, or $n=p$. If $n=1$, then $H=N=\\{1\\}$ and $|G|=p$. So assuming $H\\ne K$, we get the result. \u2013\u00a0Tim Raczkowski Jan 7 '16 at 21:38\nSince $H,K$ normal in $G$ and $H\\cap K=\\{1\\}$, $HK\\cong H\\times K$. $|G:HK|\\cdot|HK:H|=|G:H|=p$. So $|HK:H|$ divides $p$. $|HK:H|=1$, or $p$. Suppose it is $1$. $HK=H$, $K\\subseteq H$. $|G:H|\\cdot|H:K|=|G:K|$, $p|H:K|=p$, so $H=K$, contradiction. So $|HK:H|=p$, $|G:HK|=1$. $G=HK$. Now, $|G:H|=|HK:H|=|K:H\\cap K|$ by second isomorphism theorem. So $p=|K|$. Similarly, $|H|=p$. $G\\cong \\mathbb{Z}/p\\mathbb{Z}\\times\\mathbb{Z}/p\\mathbb{Z}$.", "date": "2019-10-18 04:50:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1603632/showing-that-a-group-g-with-two-distinct-normal-subgroups-of-index-p-has-or", "openwebmath_score": 0.9604566097259521, "openwebmath_perplexity": 128.0148774277002, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750522044461, "lm_q2_score": 0.8615382040983515, "lm_q1q2_score": 0.8506613292477345}}
{"url": "https://math.stackexchange.com/questions/2412375/suppose-we-roll-a-fair-6-sided-die-repeatedly-find-the-expected-number-of-rol", "text": "# Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession.\n\nSuppose we roll a fair six sided die repeatedly.\n\nFind the expected number of rolls required to see $$3$$ of the same number in succession\n\nFrom the link below, I learned that $$258$$ rolls are expected to see 3 sixes appear in succession. So I'm thinking that for a same (any) number, the rolls expected would be $$258/6 = 43$$. But I'm unsure how to show this and whether it really is correct.\n\nHow many times to roll a die before getting two consecutive sixes?\n\nFor $n\\in \\{0,1,2\\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start.\n\nWe note $$E[2]=\\frac 16\\times 1+\\frac 56\\times \\left(E[1]+1\\right)$$ $$E[1]=\\frac 16\\times \\left(E[2]+1\\right)+\\frac 56\\times \\left(E[1]+1\\right)$$\n\n$$E=E[0]=E[1]+1$$\n\nthis system is easily solved and, barring error (always possible), yields $$\\boxed {E=43}$$\n\n\u2022 This is very elegant (+1) - great answer!! Aug 31, 2017 at 16:38\n\u2022 One issue that isn't really an issue: everything needs to be finite for this to work (cf optional stopping theorem)\n\u2013\u00a0cats\nAug 31, 2017 at 18:32\n\u2022 This is the (arch classical) approach used in the post mentioned in Byron's answer.\n\u2013\u00a0Did\nAug 31, 2017 at 18:40\n\u2022 @lulu can you explain me what what tis the theory behind this answer? Does it refer to a specific branch of stochastics? Mar 9, 2021 at 23:30\n\u2022 @Phillipp This is a fairly standard Markov type computation. The trick, such as it is, is to note that there really are only three possible states for the process, an observation which makes the system very tractable.\n\u2013\u00a0lulu\nMar 9, 2021 at 23:39\n\nFrom Did's answer here, the probability generating function $u_0(s)=\\mathbb{E}(s^T)$ for the number of trials $T$ needed to get three consecutive values the same is\n$$u_0(s)={s^3\\over 36-30s-5s^2}.$$ Differentiating this and setting $s=1$ in the derivative shows that $\\mathbb{E}(T)=43.$\n\nWe can treat this as a three-state absorbing markov chain: a length3 run has been seen, otherwise the current run is length2, current run is length 1. Transition matrix: $\\begin{bmatrix}1&\\frac{1}{6}&0\\\\0&0&\\frac{1}{6}\\\\0&\\frac{5}{6}&\\frac{5}{6}\\end{bmatrix}$\n\nThis is in standard form $\\left[\\begin{array}{c|c}I&S\\\\\\hline0&R\\end{array}\\right]$\n\nWe turn our attention to the fundamental matrix $(I-R)^{-1} = \\begin{bmatrix}1&-\\frac{1}{6}\\\\-\\frac{5}{6}&\\frac{1}{6}\\end{bmatrix}^{-1}=\\begin{bmatrix}6&6\\\\30&36\\end{bmatrix}$\n\nAfter the first roll, we enter the markov chain in the third state. Adding the entries of the fundamental matrix corresponding to that column tells us the expected time until we reach an absorbing state, i.e. until we have a chain of three consecutive rolls of the same number.\n\nThus the expected number of rolls needed is $1+36+6=43$\n\nLet $\\mu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with distinct result.\n\nLet $\\nu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with equal result.\n\nThen the expectation is: $$2+\\frac56\\mu+\\frac16\\nu$$\n\nHere $\\frac56$ is the probability that the first two numbers are distinct and $\\frac16$ is the probability that they are equal.\n\nSecondly we have the relations:\n\n$$\\mu=\\frac56(1+\\mu)+\\frac16(1+\\nu)=1+\\frac56\\mu+\\frac16\\nu\\tag1$$ and: $$\\nu=\\frac161+\\frac56(1+\\mu)=1+\\frac56\\mu\\tag2$$\n\nThe relations $(1)$ and $(2)$ lead easily to: $\\mu=42$ and $\\nu=36$.\n\nThen $$2+\\frac56\\mu+\\frac16\\nu=43$$ is the final answer.\n\n(in the answers uptil now it was not used that you allready learned something).\n\nLet $Y$ denote the number of rolls required to see three dice of the same number in succession and let $X$ denote the number of rolls required to see three dice with number $6$ in succession.\n\nThen: $$Y\\text{ and }\\frac16X+\\frac56(X+Y)=X+\\frac56Y\\text{ must have equal distribution.}\\tag3$$\n\nHere $\\frac16$ is the probability of the event that the first time that three dice give the same number in succession they show number $6$ and $\\frac56$ is the probability that do not show a number $6$.\n\n$(3)$ rests on the observation that - if for the first time three equal numbers show up in succession - we are ready if $6$ happens to be that number and must actually start over again (with $X$ throws in our pocket) if not.\n\nSo we find $\\mathbb EY=\\mathbb EX+\\frac56\\mathbb EY$ or equivalently:$$\\mathbb EX=\\frac16\\mathbb EY$$\n\nYou allready learned that $\\mathbb EY=258$ and making use of that knowledge you find $$\\mathbb EX=258/6=43$$ This confirms your thinking.", "date": "2022-05-27 18:34:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2412375/suppose-we-roll-a-fair-6-sided-die-repeatedly-find-the-expected-number-of-rol", "openwebmath_score": 0.8395765423774719, "openwebmath_perplexity": 272.64333036721735, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109099773435, "lm_q2_score": 0.8596637469145054, "lm_q1q2_score": 0.8506466564839049}}
{"url": "https://www.allaboutcircuits.com/textbook/digital/chpt-8/venn-diagrams-and-sets/", "text": "# Venn Diagrams and Sets\n\n## Chapter 8 - Karnaugh Mapping\n\nMathematicians use Venn diagrams to show the logical relationships of sets (collections of objects) to one another. Perhaps you have already seen Venn diagrams in your algebra or other mathematics studies. If you have, you may remember overlapping circles and the union and intersection of sets. We will review the overlapping circles of the Venn diagram. We will adopt the terms OR and AND instead of union and intersection since that is the terminology used in digital electronics. The Venn diagram bridges the Boolean algebra from a previous chapter to the Karnaugh Map. We will relate what you already know about Boolean algebra to Venn diagrams, then transition to Karnaugh maps. A set is a collection of objects out of a universe as shown below. The members of the set are the objects contained within the set. The members of the set usually have something in common; though, this is not a requirement. Out of the universe of real numbers, for example, the set of all positive integers {1,2,3\u2026} is a set. The set {3,4,5} is an example of a smaller set, or subset of the set of all positive integers. Another example is the set of all males out of the universe of college students. Can you think of some more examples of sets? Above left, we have a Venn diagram showing the set A in the circle within the universe U, the rectangular area. If everything inside the circle is A, then anything outside of the circle is not A. Thus, above center, we label the rectangular area outside of the circle A as A-not instead of U. We show B and B-not in a similar manner. What happens if both A and B are contained within the same universe? We show four possibilities. Let\u2019s take a closer look at each of the the four possibilities as shown above. The first example shows that set A and set B have nothing in common according to the Venn diagram. There is no overlap between the A and B circular hatched regions. For example, suppose that sets A and B contain the following members:\n\nset A = {1,2,3,4}\nset B = {5,6,7,8}\n\n\nNone of the members of set A are contained within set B, nor are any of the members of B contained within A. Thus, there is no overlap of the circles. In the second example in the above Venn diagram, Set A is totally contained within set B How can we explain this situation? Suppose that sets A and B contain the following members:\n\nset A = {1,2}\nset B = {1,2,3,4,5,6,7,8}\n\n\nAll members of set A are also members of set B. Therefore, set A is a subset of Set B. Since all members of set A are members of set B, set A is drawn fully within the boundary of set B. There is a fifth case, not shown, with the four examples. Hint: it is similar to the last (fourth) example. Draw a Venn diagram for this fifth case. The third example above shows perfect overlap between set A and set B. It looks like both sets contain the same identical members. Suppose that sets A and B contain the following:\n\nset A = {1,2,3,4}\nset B = {1,2,3,4}\n\n\nTherefore,\n\n Set A = Set B\n\nSets And B are identically equal because they both have the same identical members. The A and B regions within the corresponding Venn diagram above overlap completely. If there is any doubt about what the above patterns represent, refer to any figure above or below to be sure of what the circular regions looked like before they were overlapped. The fourth example above shows that there is something in common between set A and set B in the overlapping region. For example, we arbitrarily select the following sets to illustrate our point:\n\nset A = {1,2,3,4}\nset B = {3,4,5,6}\n\n\nSet A and Set B both have the elements 3 and 4 in common These elements are the reason for the overlap in the center common to A and B. We need to take a closer look at this situation.\n\n\u2022 Share\nPublished under the terms and conditions of the Design Science License", "date": "2020-02-23 05:58:00", "meta": {"domain": "allaboutcircuits.com", "url": "https://www.allaboutcircuits.com/textbook/digital/chpt-8/venn-diagrams-and-sets/", "openwebmath_score": 0.580830454826355, "openwebmath_perplexity": 251.04761772641496, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109096680231, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.850646654439144}}
{"url": "https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-72_14.html", "text": "### CSIR JUNE 2011 PART C QUESTION 73 SOLUTION (which are positive definite? 1) $A+B$, 2) $ABA^{*}$, 3) $A^2+I$, 4) $AB$.)\n\nSuppose $A$ and $B$ are $n \\times n$ positive definite matrices and $I$ be the $n \\times n$ identity matrix. Then which of the following matrices are positive definite?\n1) $A+B$,\n2) $ABA^{*}$,\n3) $A^2+I$,\n4) $AB$.\nSolution:\nLet $A$ be a real symmetric matrix, then $A$ is said to positive definite if it satisfies any of the following equivalent conditions.\ni) all its eigenvalues are positive.\nii) $x^t A x > 0$ for all vectors $x \\ne 0$.\niii)$<x,Ax> > 0$ for all vectors $x \\ne 0$.\niii) A is positive definite if and only if it can be written as $A = R^tR$, where $R$ is possibly a rectangular matrix, with independent columns.\niv) All the principal minors of $A$ are positive. (Please share if you know any other equivalent conditions in the comment below)\nLet $A$ and $B$ be two $n \\times n$ positive definite matrices. We have $x^t A x>0$ and $x^t B x>0$ for $x \\ne 0$. We will solve each given option by each of the above given definition of positive definiteness in order to understand them clearly.\noption 1. (True)We have for $x \\ne 0$, $$x^t (A+B) x = x^t A x + x^t B x > 0.$$\nTherefore option 1 is true.\noption 2. (True) We have for $x \\ne 0$, $Ax \\ne 0$ since $A$ is invertible.\n$$<x,ABA^*x> = <A^*x,BA^*x> = <B^*A^*x,A^*x> \\\\ = <BAx,Ax> > 0 .$$\nTherefore option 2 is true. Since $A$ and $B$ are real symmetric, we have $A^* = A$ and $B^* = B$.\noption 3. (True) Let the eigen values of $A$ be $\\lambda_1,\\lambda_2,\\dots,\\lambda_n$. Since $A$ is positive definite we have all these eigen values are positive. We observe that the eigen values of $A^2+1$ are $\\lambda_1^2+1,\\lambda_2^2+1,\\dots,\\lambda_n^2+1$ which are all positive. Hence $A^2+1$ is positive definite.\noption 4. (False)\nThe product\u00a0of two positive definite matrices need not be even symmetric.\u00a0 In particular, we have $AB$ is symmetric if and only if $A$ and $B$ commutes with each other. Because $$(AB)^* = B^*A^* = A^*B^* = AB.$$ Note that, if $A$ and $B$ commutes with each other, then $A^*$ and $B^*$ commutes with each other.\n\nTo illustrate this, consider the positive definite matrices $$A = \\begin{bmatrix}11 & 10 \\\\ 10 & 10\\end{bmatrix}$$ and $$B = \\begin{bmatrix}11 & 5 \\\\ 5 & 10\\end{bmatrix}$$\ntext{ Then their product } $$AB = \\begin{bmatrix}171&155 \\\\ 160&150\\end{bmatrix}$$ which is not symmetric.\n\n### NBHM 2020 PART A Question 4 Solution $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$\nEvaluate : $$\\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \\int_{-\\infty}^{\\infty}(1+2x^4)e^{-x^2} dx = \\int_{-\\infty}^{\\inft...", "date": "2021-01-18 00:49:18", "meta": {"domain": "theindianmathematician.com", "url": "https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-72_14.html", "openwebmath_score": 0.9524317979812622, "openwebmath_perplexity": 164.18576987741787, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109081214212, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.8506466531095864}}
{"url": "https://math.stackexchange.com/questions/2319210/repeatedly-taking-differences-on-a-polynomial-yields-the-factorial-of-its-degree/2319671", "text": "# Repeatedly taking differences on a polynomial yields the factorial of its degree?\n\nConsider a function such that it takes in polynomial function and creates an array of its outputs and then using that array creates another new array by calculating the absolute difference between the first $2$ values and keeps doing this until it reaches an array full of zeros.\n\nThis is much easier to show you by example.\n\nFor example take $F(x)= x^2$, the first array would be\n\n$1,4,9,16,25,36,49,64,81$ and so on, the second would be\n\n$3,5,7,9,11,13,15,17,19$ ( the difference between the first value and the second one)\n\nbut the third one is where it gets interesting as if we continue the pattern we would get an array filled with only $2$'s and after that it would only be zeros.\n\nLets do another example, $F(x)=x^3$\n\n$1,8,27,64,125,216,343$\n\n$7,19,37,61,91,\\dotsc$ but here is the interesting part if we continue this\n\n$12,18,24,30,\\dotsc$ and once more then we get\n\n$6,6,6,6,6,\\dotsc$ after that it would just be an array of zeros\n\nThere are $2$ main observation that I made about this\n\nFirstly, the value that is begin repeated indefinitely is equals to the factorial of the functions power. Meaning that for $F(x)=x^2$ the value being repeated is $2!$. For $F(x)=x^3$ , it's $3!$ and this is true for all polynomials (I tried it up to $x^7$, after that it got too messy)\n\nSecondly, the value that is repeated always occurs on the $n$th iteration of the function. Meaning that for $F(x)=x^2$, we have to go through the processes $2$ times before we find the value. For $F(x)=x^3$, we have to go through it $3$ times before getting the value.\n\nIs there any way to prove this and does this mean anything at all?\n\n\u2022 Lookup finite differences (\"an $n^{th}$ power has a constant $n^{th}$ finite difference\") . Also for example this answer. \u2013\u00a0dxiv Jun 12 '17 at 3:25\n\u2022 Good. You discovered a concept (through on a sequence basis) very similar to derivative. An excellent mind of discovering and inducting. Continue it if you are still young and have enough time. \u2013\u00a0BAI Jun 12 '17 at 4:45\n\u2022 Welcome to Math.SE! Since one aim of the site is to collect questions and answers in forms useful for posterity, would you please consider re-titling your question to indicate its content? (Perhaps something like \"Repeated differences for polynomial functions at equally-spaced inputs\"...?) \u2013\u00a0Andrew D. Hwang Jun 12 '17 at 11:18\n\u2022 \u2013\u00a0Simply Beautiful Art Jun 12 '17 at 17:21\n\u2022 @user21820 your edit to the question is a great example of what a good title can do for a better understanding of a given topic and to promote the curiosity of the reader. \u2013\u00a0iadvd Jun 13 '17 at 0:12\n\n## 2 Answers\n\nHere's a fact:\n\n\u2022 If $p(x)$ is a polynomial of degree $n$ with leading term $ax^n$ then $p(x+1)-p(x)$ is a polyomial of degree $n-1$ with leading term $a \\, n \\, x^{n-1}$.\n\n(I'll prove this fact below).\n\nApplying this fact together with an induction argument, it follows that after repeating the process $n$ times, one obtains a polynomial of degree zero whose leading term is $$a \\, n \\, (n-1) \\ldots (2) (1) = a \\, n!$$ which is just a constant having that value.\n\nSo if the original leading coefficient $a$ is equal to $1$, as it is in the specific cases $F(x)=x^n$ that you ask about, repeating the difference process $n$ times yields a constant sequence of $n!$ as you ask.\n\nHere's a proof of the fact by applying induction (the base case $n=1$ is easy).\n\nAssuming the induction hypothesis for polynomials of degree $\\le n-1$, suppose that $$p(x) = a \\, x^n + q(x)$$ where $q(x)$ is a polynomial of degree $\\le n-1$.\n\nWe have $$p(x+1)-p(x) = a \\, (x+1)^n - a \\, x^n + \\underbrace{(q(x+1)-q(x))}_{r(x)}$$ and $r(x)$ is a polynomial of degree $\\le n-2$ by induction. Thus $$p(x+1)-p(x) = a \\, (x^n + n \\, x^{n-1} + s(x)) - a\\, x^n + r(x)$$ where $s(x)$ is also a polynomial of degree $\\le n-2$ (by application of the binomial theorem). Therefore $$p(x+1)-p(x) = a \\, n \\, x^{n-1} + (a \\, s(x)+r(x))$$ which is a polynomial of degree $n-1$ with leading term as required.\n\nWhat you have discovered/invented is known as the forward difference operator $D$ defined as: $\\def\\nn{\\mathbb{N}} \\def\\zz{\\mathbb{Z}} \\def\\lfrac#1#2{{\\large\\frac{#1}{#2}}} \\def\\lbinom#1#2{{\\large\\binom{#1}{#2}}}$\n\n$D = ( \\text{function$f$on$\\zz$} \\mapsto ( \\text{int$n$} \\mapsto f(n+1) - f(n) ) )$\n\nNamely for any function $f$ on $\\zz$ and $n \\in \\zz$, $D(f)(n) = f(n+1) - f(n)$.\n\nIf you think of the functions as sequences (infinite in both directions), then taking the forward difference means replacing each term with the value of the next term minus itself. What you did is essentially to repeatedly take the forward difference of the sequence of cubes:\n\n...,-27,-8,-1, 0, 1, 8,27,...\n..., 19, 7, 1, 1, 7,19,37,...\n...,-12,-6, 0, 6,12,18,24,...\n...,  6, 6, 6, 6, 6, 6, 6,...\n...,  0, 0, 0, 0, 0, 0, 0,...\n...,  0, 0, 0, 0, 0, 0, 0,...\n\n\nThis powerful abstraction makes it easy to get a lot of things. For instance, the numbers obtained here can be easily used to obtain the general formula for sum of cubes!\n\nGeneral method for indefinite summation\n\nThe key is that:\n\n$D\\left( \\text{int$n$} \\mapsto \\lbinom{n}{k+1} \\right) = \\left( \\text{int$n$} \\mapsto \\lbinom{n}{k} \\right)$ for any $k \\in \\zz$.\n\nThis is to be expected because it follows directly from Pascal's triangle, especially if we define $\\lbinom{n}{k}$ using the triangle.\n\nThis means that if we have any function $f$ on $\\zz$ such that $f(n) = \\sum_{k=0}^\\infty a_k \\lbinom{n}{k}$ for any $n \\in \\zz$, then we get the Newton series:\n\n$D(f)(n) = \\sum_{k=0}^\\infty a_{k+1} \\lbinom{n}{k}$ for any $n \\in \\zz$.\n\nFrom a high-level perspective, this is the discrete version of the Taylor series, and indeed for such a function we easily see that $f(n) = \\sum_{k=0}^\\infty D^k(f)(0) \\lbinom{n}{k}$ for any $n \\in \\zz$, because $\\binom{0}{0} = 1$ while $\\lbinom{0}{k} = 0$ for any $k \\in \\nn^+$.\n\nThis works for any polynomial function $f$ on $\\zz$, since $D^k(f)$ is the zero function once $k$ is larger than the degree of $f$, so we can use it to immediately find the series for $(\\text{int n} \\mapsto n^3)$, and then just take the anti-difference by shifting the coefficients of the series the other way. The undetermined constant that appears will drop out once we perform a definite sum like if we want the sum of the first $m$ cubes.\n\nNote also that $D^k(f)$ is the constant function with value $k!$ if $f(n) = n^k$ for every $n$. Lee Mosher has already explained this particular fact by directly proving it, but another way to see it is that the highest order term in its Newton series is $k! \\lbinom{n}{k}$, because $\\lbinom{n}{k}$ is the only term that can contribute the $k$-th power of $n$. Since $D$ merely shifts the coefficients, $D^k(f) = \\left( \\text{int$n$} \\mapsto k! \\lbinom{n}{0} \\right)$ and we are done.\n\nSum of $p$ powers\n\nFor example if we want $\\sum_{k=1}^{n-1} k^3$ we first find the iterated forward differences of the sequence of cubes $( n^3 )_{n \\in \\zz}$:\n\n..., 0, 1, 8,27,64,...\n..., 1, 7,19,37,...\n..., 6,12,18,...\n..., 6, 6,...\n..., 0,...\n\n\nSo we immediately get $n^3 = 0 \\binom{n}{0} + 1 \\binom{n}{1} + 6 \\binom{n}{2} + 6 \\binom{n}{3}$ and hence $\\sum_{k=0}^{n-1} k^3 = 0 \\binom{n}{1} + 1 \\binom{n}{2} + 6 \\binom{n}{3} + 6 \\binom{n}{4} = \\lfrac{n(n-1)}{2} \\Big( 1 + \\lfrac{6(n-2)}{3} \\big( 1 + \\lfrac{n-3}{4} \\big) \\Big) = \\Big( \\lfrac{n(n-1)}{2} \\Big)^2$.\n\nComputation efficiency\n\nThis is far more efficient than the usual method (namely by taking summation on both sides of $(n+1)^3-n^3 = 3n^2+3n+1$ and telescoping) because the series using binomial coefficients is easy to manipulate and easy to compute. For sum of $p$-powers we only need $O(p^2)$ arithmetic operations to find the forward-differences and then $O(p^2)$ more to simplify the series form into a standard polynomial form. In contrast, the other method requires $O(p^3)$ arithmetic operations.\n\nIndefinite summation of non-polynomials\n\nAlso, for a wide class of non-polynomial functions, we can still compute the indefinite sum without using the series, by using the discrete analogue to integration by parts, here called summation by parts.\n\nTo derive it, simply check that $D(f \\times g)(n) = f(n+1) g(n+1) - f(n) g(n) = f(n+1) D(g)(n) - D(f)(n) g(n)$ and so we get the product rule:\n\n$D(f \\times g) = R(f) \\times D(g) + D(f) \\times g$\n\nwhere $R$ is the right-shift operator defined as:\n\n$R = ( \\text{function$f$on$\\zz$} \\mapsto ( \\text{int$n$} \\mapsto f(n+1) ) )$\n\nNamely for any function $f$ on $\\zz$ and $n \\in Z$, $R(f)(n) = f(n+1)$.\n\nFor convenience we also define the summation operator:\n\n$S = ( \\text{function$f$on$\\zz$} \\mapsto ( \\text{int$n$} \\mapsto \\sum_{k=0}^{n-1} f(k) ) )$\n\nThen we have the important property that $DS(f) = f$ for any function $f$ on $\\zz$, analogous to the fundamental theorem of calculus.\n\nNow by substituting $f$ with $S(f)$ into the product rule and taking summation on both sides we get summation by parts:\n\n$S( f \\times g ) = S(f) \\times g - S( R(S(f)) \\times D(g) ) + c$ for some constant function $c$ on $\\zz$.\n\nExample indefinite sum\n\nUsing this we can easily compute things like $\\sum_{k=1}^n k^3 3^k$ by applying it three times, each time reducing the degree of the polynomial part. There are other ways to achieve this using differentiation, but this method is purely discrete.", "date": "2020-09-20 07:21:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2319210/repeatedly-taking-differences-on-a-polynomial-yields-the-factorial-of-its-degree/2319671", "openwebmath_score": 0.8655317425727844, "openwebmath_perplexity": 167.59411086287278, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842222598317, "lm_q2_score": 0.8577681068080749, "lm_q1q2_score": 0.850635097879254}}
{"url": "https://oneclass.com/study-guides/us/uh/math/math-4389/1505514-math-4389-final.en.html", "text": "Study Guides (248,357)\nUnited States (123,340)\nMATH 4389 (17)\nAlmus (17)\nFinal\n\n# MATH 4389 Final: Real_Analysis(part_2) Premium\n\n28 Pages\n72 Views\n\nSchool\nDepartment\nMathematics\nCourse\nMATH 4389\nProfessor\nAlmus\nSemester\nSpring\n\nDescription\nPART II. SEQUENCES OF REAL NUMBERS II.1. CONVERGENCE De\ufb01nition 1. A sequence is a real-valued function f whose domain is the set positive integers (N). The numbers f(1),f (2), \u00b7\u00b7\u00b7 are called the terms of the sequence. Notation Function notation vs subscript notation: f(1) \u2261 1 ,f (2) \u22612s ,\u00b7\u00b7\u00b7 ,f (n) n s , \u00b7\u00b7\u00b7 . In discussing sequences the subscript notation is much more common than functional notation. We\u2019ll use subscript notation throughout our treatment of analysis. Specifying a sequence There are several ways to specify a sequence. 1. By giving the function. For example: 1 1 1 1 1 1 (a) sn= n or {sn} = n . This is the sequence {12 ,3 ,4 ,...,n,... }. (b) s = n \u2212 1. This is the sequence {0, ,2 , 3,...,n \u2212 1,... }. n n 2 3 4 n (c) sn=( \u22121) n . This is the sequence {\u22121,4,\u22129,16,..., (\u22121) n ,... }. 2. By giving the \ufb01rst few terms to establish a pattern, leaving it to you to \ufb01nd the function. This is risky \u2013 it might not be easy to recognize the pattern and/or you can be misled. (a) {sn} = {0,1,0,1,0,1,... }( The pattern here is obvious; can you devise the function? It\u2019s 1 \u2212 (\u22121)n) 0, n odd sn= or s n= 2 1, n even 2 (b) {s } = 2, 5,10,17 ,6 ,... ,s = n +1 . n 2 3 4 5 n n (c) {sn} = {2,4,8,16,32,... }. What is6s ? What is the function? While you might say 64 n and s n2 , the function I have in mind gives s6= \u03c0/6: n \u03c0 64 sn=2 +( n \u2212 1)(n \u2212 2)(n \u2212 3)(n \u2212 4)(n \u2212 5) \u2212 720 120 3. By a recursion formula. For example: (a) s = 1 s ,s = 1. The \ufb01rst 5 terms are 1,1, , 1 , 1 ,.... Assuming that n+1 n +1 n 1 2 6 24 120 1 the pattern continues n = . n! 1 (b) sn+1 = (n +1) ,s 1 = 1. The \ufb01rst 5 terms are {1,1,1,1,1,... }. Assuming that the 2 pattern continues n = 1 for all n; {n } is a \u201cconstant\u201d sequence. 13 De\ufb01nition 2. A sequence {s }nconverges to the number s if to each > 0 there corresponds a positive integer N such that |s \u2212 s| for all n>N. n The number s is called the limit of the sequence. Notation \u201c{s } converges to s\u201d is denoted by n n\u2192\u221e sn= s, or by limsn= s, or by n \u2192 s. A sequence that does not converge is said to diverge. Examples Which of the sequences given above converge and which diverge; give the limits of the convergent sequences. THEOREM 1. If s \u2192 s and s \u2192 t, then s = t. That is, the limit of a convergent sequence is unique. Proof: Suppose s 6= t. Assume t>s and let = t \u2212 s. Since sn\u2192 s, there exists a positive integer N such that |s \u2212 s | / 2 for all n>N . Since s \u2192 t, there exists a positive integer 1 n 1 n N 2 such that |t\u2212s n / 2 for all n>N 2. Let N = max{N ,N1} a2d choose a positive integer k>N . Then t \u2212 s = |t \u2212 s| = |tk\u2212 s k s \u2212 s|\u2264| t k s | + |sk\u2212 s |+< = = t \u2212 s, 2 2 a contradiction. Therefore, s = t. THEOREM 2. If {s } connerges, then {s } is bnunded. Proof: Suppose s \u2192 s. There exists a positive integer N such that |s\u2212s | < 1 for all n>N . n n Therefore, it follows that |sn| = |n \u2212 s + s|\u2264| sn\u2212 s| + |s| < 1+ |s| for alln>N. Let M = max{|s |, |s |, ..., |s |, 1+ |s|}. Then |s | N. If an\u2192 0, then s \u2192n0. Proof: Note \ufb01rst that a \u2265n0 for all n>N . Since a \u2192 0,nthere exists a positive integer N 1 such that |an| < /k. Without loss of generality, assume that1N \u2265 N. Then, for all n>N 1, |sn\u2212 0| = |sn|\u2264 ka n\nMore Less\n\nRelated notes for MATH 4389\nMe\n\nOR\n\nJoin OneClass\n\nAccess over 10 million pages of study\ndocuments for 1.3 million courses.\n\nJoin to view\n\nOR\n\nBy registering, I agree to the Terms and Privacy Policies\nAlready have an account?\nJust a few more details\n\nSo we can recommend you notes for your school.", "date": "2018-04-23 08:30:45", "meta": {"domain": "oneclass.com", "url": "https://oneclass.com/study-guides/us/uh/math/math-4389/1505514-math-4389-final.en.html", "openwebmath_score": 0.845613420009613, "openwebmath_perplexity": 2692.5325127376495, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9916842212767573, "lm_q2_score": 0.8577681013541611, "lm_q1q2_score": 0.8506350916274439}}
{"url": "https://math.stackexchange.com/questions/2846965/how-are-numbers-distributed-under-a-different-modulus", "text": "Suppose we take a sample of numbers with unique congruence class modulo p:\n\nx_0 \u2261 0 (mod p)\nx_1 \u2261 1 (mod p)\nx_2 \u2261 2 (mod p)\n...\nx_(p-2) \u2261 p-2 (mod p)\nx_(p-1) \u2261 p-1 (mod p)\n\n\nand we then examine their class under modulo q. How will these classes be distributed? For example, will some values appear more often than others? Or might they be uniform in number or all unique? I imagine this result would depend on if p<q or p>q, and perhaps other things.\n\nSpecific, numerical example\n\n15 = 5(3) \u2261 0 (mod 5)\n26 = 5(5)+1 \u2261 1 (mod 5)\n37 = 5(7)+2 \u2261 2 (mod 5)\n58 = 5(11)+3 \u2261 3 (mod 5)\n69 = 5(13)+4 \u2261 4 (mod 5)\n\n\nand\n\n15 \u2261 1 (mod 2)\n26 \u2261 0 (mod 2)\n37 \u2261 1 (mod 2)\n58 \u2261 0 (mod 2)\n69 \u2261 1 (mod 2)\n\n\nand\n\n15 \u2261 3 (mod 6)\n26 \u2261 2 (mod 6)\n37 \u2261 1 (mod 6)\n58 \u2261 4 (mod 6)\n69 \u2261 3 (mod 6)\n\n\u2022 If you sample the $x_i$ values uniformly, then I imagine their remainders modulo $p$ is uniform, so that the following remainders modulo $q$ are also uniform. \u2013\u00a0Bill Wallis Jul 10 '18 at 19:08\n\u2022 That seems vaguely intuitive to me as well (and for my purposes, I'd like that to be true in general), but I'd like for someone to prove that if it is the case. Then again, it can't be completely true, right, since when q>p, we'll never get the larger modulo values, larger than p-1? \u2013\u00a0Steve Jul 10 '18 at 19:11\n\u2022 If $q<p$ and the samples are uniform $\\bmod p$, they can be uniform $\\bmod q$ only if $q|p$ (pigeonhole principle) \u2013\u00a0gammatester Jul 10 '18 at 19:12\n\u2022 An example of something I'd like to know if ever occurs, say we have sample (0,1,2,3,4,5,6.....20) in mod 21, is it possible to get a skewed distribution like (0,1,1,1,4,1,0,1,1,0,3) in mod 11 where there are \"more 1s than expected\"? \u2013\u00a0Steve Jul 10 '18 at 19:28\n\u2022 @Steve: If $p=2$, there's nothing stopping you from taking large values of $x$ like $x_0 = 384$ and $x_1 = -7447$. \u2013\u00a0user14972 Jul 10 '18 at 23:18\n\nRecall the Chinese Remainder Theorem; in one form, it says that if $p$ and $q$ are relatively prime integers, then there is a bijective correspondence between the\n\n\u2022 residue classes modulo $pq$\n\u2022 pairs consisting of a residue class modulo $p$ and a residue class modulo $q$\n\nFurthermore, given any integer $x$, its residue classes modulo $pq$, $p$ and $q$ are related by this correspondence.\n\nIn particular, integers fall into every combination of residues modulo $p$ and $q$, and they do so exactly once per period of length $pq$. So knowing the residue class of an integer modulo $p$ tells you absolutely nothing about its residue class modulo $q$.\n\nAnd while the notion of a uniform distribution doesn't actually make sense for the integers, this periodic behavior still allows us to capture the idea that such a thing would have independent distributions modulo $p$ and $q$.\n\nIn the example of taking $p=21$ and $q=11$, we can give this bijection by explicit formula (discussions of the CRT should show how to obtain this):\n\n\u2022 $22 x - 21 y \\equiv x \\bmod 21$\n\u2022 $22 x - 21 y \\equiv y \\bmod 11$\n\nSimilarly, any other integer with the same residue as $22x - 21y \\bmod 231$ will also satisfy these congruences.\n\nSo, given any choice of twenty-one residue class modulo 11, you could find a sequence of $x$'s that have the chosen residues $\\bmod 11$ along with the required residues $\\bmod 21$. For example, if you want all $1$'s, then we can take\n\n$$x_n = 1 + 22(n-1)$$\n\nto get\n\n\u2022 $x_n \\equiv n \\bmod 21$\n\u2022 $x_n \\equiv 1 \\bmod 11$\n\u2022 This is fantastic and I think I roughly follow what you're saying and at least understand it's impact in a very generalized sense. In your explicit example, 1, 23, 45, 67, ..., 199, 221 are all 1 mod 11 while running across the 0,1,2,...,11 spectrum of values under mod 21. This is helpful because it allows me to refine my question back to my original post, however. What if we restrict ourselves to numbers which not only are sequential in mod 21, but rather are strictly sequential numbers (x, x+1, x+2, ...). Can we know anything about these numbers under mod q? \u2013\u00a0Steve Jul 11 '18 at 12:48\n\u2022 I'm realizing now, maybe my question is stupid and obvious afterall. Can you tell me if this follows? Let a, p, q be constant integers. Let x_n= ap+n for n={0,1,...p-1}. Then x_n (mod p) \u2261 n. Observe that ap (mod q) \u2261 B for some integer B. Thus, x_n (mod q) = ap+n (mod q) \u2261n (mod q)+B, so the residues unders mod q are \"uniformly distributed\" save some wrapping around if q is smaller than p. Is this correct? \u2013\u00a0Steve Jul 11 '18 at 13:02\n\nIf you take any $q$ successive numbers, there will be one in each congruence class $\\bmod q$. You start with the lowest one in its class, the next is in the next class up, and so on, wrapping around at $0$. When you get to $q$ you have put one in each class and are ready to start again. If you take $p$ successive numbers where $q\\not | p$ you will have two different counts among the congruence classes. Let $r$ be the remainder on dividing $p$ by $q$. There will be $q-r$ classes with $\\lfloor \\frac pq \\rfloor$ because you go around fully that many times. There will be $r$ classes with one more because you have $r$ numbers left after the full cycles.", "date": "2021-07-31 12:38:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2846965/how-are-numbers-distributed-under-a-different-modulus", "openwebmath_score": 0.8639540672302246, "openwebmath_perplexity": 208.5372592220162, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357555117625, "lm_q2_score": 0.8688267898240861, "lm_q1q2_score": 0.8506124925842834}}
{"url": "https://math.stackexchange.com/questions/2574916/definite-integral-possible-evaluation-using-real-methods", "text": "Definite integral - possible evaluation using real methods?\n\nThe book \"inside interesting integrals\" gives the following exercise for the chapter about contour integration and the residue theorem: $$\\int_{0}^\\infty \\frac{e^{\\cos x}\\sin(\\sin x)}{x}dx=\\space\\space ?$$ This can be solved using the function $$f(z)=\\frac{\\exp(e^{iz})}{z}$$ on a quarter-circular contour, and is pretty straightforward. The answer turns out to be $$\\frac{\\pi}{2}(e-1)$$ However, in the book, the author makes the following comment:\n\nEdward Copson (1901-1980), who was professor of mathematics at the University of St. Andrews in Scotland, wrote \"A definite integral which can be evaluated using Cauchy's method of residues can always be evaluated by other means, though generally not so simply.\" Here's an example of what Copson meant, an integral attributed to the great Cauchy himself. It is easily done with contour integration, but would (I think) otherwise be pretty darn tough.\n\nDoes anyone know how to evaluate this integral using real methods?\n\n\u2022 This question should actually get more attention... I already grabbed my popcorn to see how real methods come in action. If this does not get attention I'm considering to put a bounty on it (if that is possible, I don't know how that works). \u2013\u00a0Shashi Jan 7 '18 at 21:14\n\u2022 I have put a bounty on this question. @Nilknarf Are there some updates concerning this question? \u2013\u00a0Shashi Jan 8 '18 at 20:47\n\u2022 @Shashi No... Do you suggest that I add anything in particular to the question? \u2013\u00a0Frpzzd Jan 8 '18 at 23:55\n\u2022 I was curious whether you have solved it in the mean time \u2013\u00a0Shashi Jan 9 '18 at 8:00\n\nPerhaps surprisingly, a straightforward trick works. To this end we refer to the following easy-to-prove lemma.\n\nLemma. Define $\\operatorname{Si}(x) = \\int_{0}^{x} \\frac{\\sin t}{t} \\, dt$. Then\n\n1. $\\int_{0}^{x} \\frac{\\sin(yt)}{t} \\, dt = \\operatorname{Si}(xy)$, and\n2. $\\operatorname{Si}(x) = \\frac{\\pi}{2} + \\mathcal{O}(x^{-1})$ as $x \\to \\infty$.\n\nThen for $R > 0$,\n\n\\begin{align*} \\int_{0}^{R} \\frac{e^{\\cos x}\\sin(\\sin x)}{x} \\, dx &= \\int_{0}^{R} \\frac{1}{x}\\operatorname{Im}(e^{e^{ix}}) \\, dx \\\\ &= \\int_{0}^{R} \\frac{1}{x}\\sum_{n=1}^{\\infty} \\frac{\\sin(nx)}{n!} \\, dx \\\\ &= \\sum_{n=1}^{\\infty} \\frac{1}{n!} \\int_{0}^{R} \\frac{\\sin(nx)}{x} \\, dx \\\\ &= \\sum_{n=1}^{\\infty} \\frac{1}{n!} \\operatorname{Si}(nR) \\\\ &= \\sum_{n=1}^{\\infty} \\frac{1}{n!} \\left( \\frac{\\pi}{2} + \\mathcal{O}\\left( (nR)^{-1} \\right) \\right) \\\\ &= \\frac{\\pi}{2}(e - 1) + \\mathcal{O}(R^{-1}) \\end{align*}\n\nLetting $R \\to \\infty$ proves the claim.\n\n\u2022 Not thought about a single contour and yet so simple! Conclusion: the author of the book lied lol (jk I know one does not have full vision on everything) \u2013\u00a0Shashi Jan 9 '18 at 8:03\n\n$$e^{\\cos x}\\sin(\\sin x) = \\text{Im}\\, e^{\\cos x+i\\sin x} = \\text{Im}\\exp\\left(e^{ix}\\right) = \\text{Im}\\sum_{n\\geq 0}\\frac{e^{nix}}{n!}=\\sum_{n\\geq 1}\\frac{\\sin(nx)}{n!}$$ and since $\\int_{0}^{+\\infty}\\frac{\\sin(nx)}{x}\\,dx = \\frac{\\pi}{2}$ for any $n>0$, we have $$\\int_{0}^{+\\infty}\\frac{e^{\\cos x}\\sin(\\sin x)}{x}\\,dx = \\frac{\\pi}{2}\\sum_{n\\geq 1}\\frac{1}{n!}=\\color{red}{\\frac{\\pi}{2}(e-1)}.$$\n\n\u2022 Is it true that interchanging series and integral is allowed since the series with sine is uniformly convergent? The answer is so neat. I guess it is even easier with Real methods after seeing the answers. \u2013\u00a0Shashi Jan 9 '18 at 7:55\n\u2022 \"Real methods=without Residue Theorem \" in the last comment \u2013\u00a0Shashi Jan 9 '18 at 8:06\n\u2022 @Shashi: indeed, this is a small variation on the improper Riemann-integrability of $\\frac{\\sin x}{x}$ on $\\mathbb{R}^+$, which can be proved by real (Fourier-analytic), almost-real (Laplace transform) or complex techniques (residue theorem) in few steps, anyway. \u2013\u00a0Jack D'Aurizio Jan 9 '18 at 14:45", "date": "2019-04-19 20:16:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2574916/definite-integral-possible-evaluation-using-real-methods", "openwebmath_score": 0.9996591806411743, "openwebmath_perplexity": 446.0361580274245, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357591818726, "lm_q2_score": 0.8688267847293731, "lm_q1q2_score": 0.8506124907850672}}
{"url": "https://math.stackexchange.com/questions/1289322/what-is-zero-irrational-or-rational-or-it-have-both-the-properties/1289333", "text": "# What is zero? Irrational or rational or it have both the properties? [duplicate]\n\nWe say,\n\nA number is rational if it can be represented as $\\frac{p}{q}$ with $p,q \\in \\mathbb Z$ and $q\\neq 0$.\n\nAny number which doesn't fulfill the above conditions is irrational.\n\nIt can be represented as a ratio of two integers as well as ratio of itself and an irrational number such that zero is not dividend in any case.\n\nPeople say that $0$ is rational because it is an integer. Which I find to be a lame reason. May be any strong reason is there. Can any one tell me please?\n\n## marked as duplicate by Jack M, Chappers, N. F. Taussig, Dietrich Burde, Joel Reyes NocheMay 19 '15 at 12:21\n\n\u2022 $0$ can be represented as $0/1$, therefore by your definition it is rational. \u2013\u00a0TonyK May 19 '15 at 10:30\n\u2022 And it can also be shown as $\\frac{0}{\\sqrt{2}}$. Then what? \u2013\u00a0Man_Of_Wisdom May 19 '15 at 10:30\n\u2022 $1=\\frac{\\sqrt{2}}{\\sqrt{2}}$. So? \u2013\u00a0Hayden May 19 '15 at 10:31\n\u2022 Or less trivially, $2=\\frac {\\sqrt 8} {\\sqrt 2}$ \u2013\u00a0Jack M May 19 '15 at 10:40\n\u2022 You may simply notice that $q=\\sqrt(2) \\not\\in \\mathbb{Z}$ \u2013\u00a0Manlio May 19 '15 at 10:42\n\nThe definition of an irrational number is that it is not rational. And $0$ is by definition a rational number.\n\n\u2022 Why should I consider it a rational number if it can be represented as $\\frac{0}{\\sqrt{2}}$ \u2013\u00a0Man_Of_Wisdom May 19 '15 at 10:27\n\u2022 @Man_Of_Wisdom: What about $\\sqrt2/\\sqrt2$ then? Do you consider that irrational? \u2013\u00a0John Bentin May 19 '15 at 10:33\n\u2022 Nope. No. Not.. \u2013\u00a0Man_Of_Wisdom May 19 '15 at 10:45\n\u2022 @Alwin can you please roll back your edit to the fresh one. Which includes only two lines? I want to accept that because it answers at hand! \u2013\u00a0Man_Of_Wisdom May 19 '15 at 11:10\n\u2022 Do you mean this one above? \u2013\u00a0Alwin May 19 '15 at 11:25\n\nI think you're confusing some of the language in the definition. The definition says that if a number may be written in a certain way, namely as a fraction in which both numerator and denominator are integers, then it's rational. If it cannot be written this way, then it is irrational. There is nothing in the definition that prevents a rational number from being written as a fraction in other ways, such as having rational or irrational numerator or denominator.\n\nThe phrase \"Any number which doesn't fulfill the above conditions is irrational\" does NOT say \"Any number which can be written as a fraction $\\frac{p}{q}$ with $p,q\\notin \\mathbb{Z}$ is irrational\". It simply says any number that can not be written $\\frac{p}{q}$ with $p,q\\in \\mathbb{Z}$ is irrational.\n\n\u2022 One could also use the irrationality critereon to show zero can't be an irrational number. \u2013\u00a0Rammus May 19 '15 at 11:49\n\n$r$ is rational if you find integers $p,q$ such that $r=\\dfrac pq$. This is obviously the case for $r=0$.\n\nThe contraposition of this property is \"$r$ is irrational if you cannot find integers $p,q$ such that $r=\\dfrac pq$\".\n\nThe contraposition is not at all \"$r$ is irrational if you find irrationals $p,q$ such that $r=\\dfrac pq$\". (By the way, this would be a somewhat circular definition.)\n\n$\\newcommand{\\Reals}{\\mathbf{R}}$You've got excellent explanations of the logical reasons for saying \"$0$ is rational\". Here are some complementary thoughts too long for a comment:\n\nDefinitions in mathematics exist to give convenient labels to useful logical distinctions. \"Convenient\" is a loose term, but generally refers to simplifying statements of theorems and facilitating common types of discussion.\n\nWhen two criteria (such as \"rational\" and \"irrational\") are logical opposites by definition, it's never a good idea to allow some widespread mathematical concept (such as $0$) to be \"both\": If you do, every theorem that would apply to that object has to contain a clause explicitly excluding that object. That's inconvenient. In rare cases (see below), you might say \"neither\". But in the case of $0$, \"rational\" is the better label:\n\n\u2022 Literal application of the definition (\"there exist integers $q \\neq 0$ and $p$ such that $0 = p/q$\") says $0$ is rational.\n\n\u2022 The set of rational numbers has pleasant algebraic properties (closed under addition, closed under multiplication) because $0$ is rational. (By contrast, the set of irrational real numbers is not closed under addition, e.g., $(1 - \\sqrt{2}) + \\sqrt{2} = 1$, or under multiplication, e.g., $\\sqrt{2} \\cdot \\sqrt{2} = 2$, whether or not $0$ is rational.)\n\nTo make a case that \"$0$ is not rational\", i.e., that the definition of \"rational number\" should exclude $0$, one would want a compelling reason, such as \"the statement of a useful theorem becomes awkward if $0$ is (regarded as) rational\".\n\nWith due respect, the possibility of writing $0$ as $0/\\sqrt{2}$ isn't compelling in the above sense; as others have explained, this representation does not contravene the definition. Further, it's useful, and causes no hardship, to agree that $0$ is rational.\n\nFor contrast, here are some other \"edge cases\" that crop up now and again:\n\n\u2022 The integer $0$ is \"even\" ($2$ times some integer) rather than \"odd\" (leaves a remainder of $1$ on division by $2$). (By the division algorithm, every integer is even or odd, but no integer is both. In this setting, \"even\" and \"not odd\" are logically equivalent for integers. I mention this example because a colleague once informed me that some teachers regard $0$ as neither even nor odd.(!!))\n\n\u2022 The zero function $z: \\Reals \\to \\Reals$ is both \"even\" (for all real $x$, $z(-x) = z(x)$) and \"odd\" (for all real $x$, $z(-x) = -z(x)$). The notions of \"even\" and \"odd\" for functions are not logical opposites. Moreover, it is useful to declare the zero function to be both even and odd: The set of even functions is a vector space under \"the usual operations\"; the set of odd functions is, too. If $z$ were not \"both even and odd (as a function)\", at least one of these useful theorems would be false.\n\n\u2022 The integer $1$ is neither \"prime\" nor \"composite\". (Even though \"$1$ has no positive integer factors other than $1$ and itself\", we explicitly exclude $1$ from membership in the primes because declaring $1$ \"prime\" would spoil the uniqueness of prime factorization. On the other hand, $1$ is not \"composite\" because $1$ is not a product of smaller positive primes.)\n\nThere's a deeper point that, ironically, may seem at odds with my earlier stance: Mathematical definitions are human conventions, not absolute, immutable, incontestable features of logic, mathematics, or the physical universe. I suspect this raises unnecessary obstacles for the philosophically-minded who study mathematics. (Everything and More by David Foster Wallace is the most extreme example I've encountered; Wallace seemed tormented by the ontology of infinity.)\n\nOn the other hand, when one sees how tightly mathematics hangs together across times and cultures, how definitions lead to the same theorems, one is forced (even fully accepting the preceding paragraph) to admire the phenomenal coherence of the logical structure of mathematics. One starts to feel as if definitions are inevitable. One becomes willing to fight emphatically for the correct definitions. This last, I expect, explains the downvotes to your good question.\n\nAs I understand it, the definition of an irrational number is that it is not rational. By definition then, $0$ is rational.", "date": "2019-06-20 03:21:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1289322/what-is-zero-irrational-or-rational-or-it-have-both-the-properties/1289333", "openwebmath_score": 0.7992221117019653, "openwebmath_perplexity": 586.6421607228546, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357542883923, "lm_q2_score": 0.8688267813328977, "lm_q1q2_score": 0.8506124832082096}}
{"url": "https://math.stackexchange.com/questions/458105/constructing-a-non-linear-system-with-prerequisites-about-the-nature-of-its-crit?noredirect=1", "text": "Constructing a non-linear system with prerequisites about the nature of its critical points.\n\nAn exercise from the book I am reading is: \"Construct a non-linear system that has four critical points:two saddle points, one stable focus, and one unstable focus.\"\n\nI have tried many systems. I found one quickly but I was lucky even if I had a few clues thanks my previous trials.\n\nI wonder if there is any way to find such systems using a not completely \"gropingly way\".\n\nEdit: with only two equations in the system.\n\nThe system I have is:\n\n$\\dot{x}=y^2-x^2$\n\n$\\dot{y}=x^2+y^2-2$\n\n\u2022 Is the dimension of system = 2 ? \u2013\u00a0nonlinearism Aug 2 '13 at 17:13\n\u2022 Yes the dimension is 2 as it was in the chapter preceding this exercise, I should have written that also ^^ Thanks. \u2013\u00a0Ouistiti Aug 2 '13 at 22:44\n\u2022 @Ouistiti: Was the answer helpful? What does your system look like? Where is this problem from? \u2013\u00a0Amzoti Aug 4 '13 at 1:29\n\u2022 @Amzoti I will post my system above. The answer was very helpful, I learned more from your answer than I expected. Thank you :) The problem is from \"Dynamical Systems with Applications using Mathematica\u00ae\" written by Stephen Lynch. \u2013\u00a0Ouistiti Aug 5 '13 at 12:22\n\u2022 @Ouistiti: It makes me very happy to hear that! If you write the four sets of eigenvalues for the Jacobian matrix at each critical point, It may be possible to set up an optimization problem to solve for the six parameters. Look forward to seeing your system and this was a fun problem. Regards \u2013\u00a0Amzoti Aug 5 '13 at 12:25\n\nWe need to choose a form for our system to give us exactly four critical points. This selection is not unique, hence, we can get more than one solution.\n\nLets choose a system with parameters that give us some degree of freedom. An example of a system that gives us four critical points and some degree of freedom is (there are other choices, this is not unique):\n\n$$\\tag 1 x' = a x + b x^2 + c x y = x(a + bx + cy)\\\\ y' = d y + e y^2 + f x y = y(d+ey + fx)$$\n\nFirst thing we need to do is to find the critical points (recall, we need exactly four of them).\n\n\u2022 $x = 0 \\rightarrow y = 0$ or $y = -\\dfrac{d}{e}$\n\u2022 $y = 0 \\rightarrow x = 0$ or $x = -\\dfrac{a}{b}$\n\u2022 $x \\ne 0, y \\ne 0 \\rightarrow x = \\dfrac{cd -ae}{be - cf}, y = \\dfrac{af -bd}{be - cf},~~\\text{with}~~ c \\ne 0, be \\ne cf$\n\nThus, our four critical points are (note the constraints above):\n\n$$(0,0), ~\\left(0, -\\dfrac{d}{e}\\right), ~\\left(-\\dfrac{a}{b}, 0\\right), \\left(\\dfrac{cd -ae}{be - cf}, \\dfrac{af -bd}{be - cf}\\right)$$\n\nThe Jacobian matrix of $(1)$ is given by:\n\n$$\\tag 2 \\displaystyle J(x, y) = \\begin{bmatrix}\\frac{\\partial x'}{\\partial x} & \\frac{\\partial x'}{\\partial y}\\\\\\frac{\\partial y'}{\\partial x} & \\frac{\\partial y'}{\\partial y}\\end{bmatrix} = \\begin{bmatrix}a + 2bx + cy & cx \\\\ fy & d + 2 e y + fx \\end{bmatrix}$$\n\nNow, we have freedom when choosing the parameters with the goal to get two saddle points, one stable focus, and one unstable focus.\n\nThere is not a clean way to do this, but our problem now is to choose the parameters $a, b, c, d, e, f$ to make that happen by evaluating the eigenvalues of the Jacobian at each critical point, and choosing the parameters to give us the desired behavior.\n\nI am going to crank these one at a time and keep my fingers crossed for the last critical point.\n\nPoint (0,0)\n\nWe have:\n\n$$\\displaystyle J(0, 0) = \\begin{bmatrix} a & 0 \\\\ 0 & d \\end{bmatrix}$$\n\nLets choose $a = -1, d = 1$ and this is our first saddle.\n\nPoint (0,-d/e)\n\nWe have:\n\n$$\\displaystyle J(0, -d/e) = \\begin{bmatrix}a - (cd)/e & 0 \\\\ -(fd)/e & d -(2d)/e \\end{bmatrix}$$\n\nLets choose $e = 1, c = -3$ and this is our second saddle.\n\nPoint (-a/b, 0)\n\nWe have:\n\n$$\\displaystyle J(-a/b, 0) = \\begin{bmatrix} - a - (ac)/b & 0 \\\\ 0 & d -(af)/b \\end{bmatrix}$$\n\nLets choose $b = 1, f = 1$ and this is our unstable focus.\n\nPoint $\\left(\\dfrac{cd -ae}{be - cf}, \\dfrac{af -bd}{be - cf}\\right)$\n\nWell, this is where we need luck with the parameters we chose, but I think it might be possible to cast this problem as an optimization problem, so you might want to play around with that. This would give you ranges for these six parameters, so no guesswork is needed.\n\nAnyway, from our parameter choices above, we get $x = y = -1/2$.\n\n$$\\displaystyle J(-1/2, -1,2) = \\begin{bmatrix} -1/2 & 3/2 \\\\ -1/2 & -1 \\end{bmatrix}$$\n\nThis, thankfully, gives us our stable focus.\n\nThus, we have the system:\n\n$$\\tag 1 x' = - x + x^2 -3 x y \\\\ y' = y + y^2 + x y$$\n\nLets draw the phase portrait and validate this analysis.\n\nWe have what we need with two saddles, an unstable and a stable focus at the four critical points.\n\nI am curious how you did it (even if it is guessing) and what your system looks like and it would help to post your solution in your question.\n\nUpdate\n\nHere is the phase portrait of the system you wrote and it is also a good example.\n\n\u2022 Great job. For my knowledge how did you come up with the second order system in (1)? By intuition ? \u2013\u00a0kaka Aug 3 '13 at 6:14\n\u2022 @kaka: Yes, by intuition and a bit of handwaving. I wish I could say I had some wonderful approach to that. I realized that we needed something that could give us a point at (0,0) and then realized we needed a parabola that gave points at (0, w), and (f, 0) and one more at some (h, j). Thinking along those lines lead me to that system. Then, we needed freedom, so I added all of the parameters. I still think the approach can be improved using optimization for ranges on the parameters, but I'll let the OP work that. Thanks for the kind words. Regards \u2013\u00a0Amzoti Aug 3 '13 at 6:19\n\u2022 @Amzoti: Wowwww. It is complete as always. Full of points, and finally a nice illustrating graph. \u2013\u00a0mrs Aug 3 '13 at 14:56\n\u2022 This is a nice problem. Go for it! ;-) \u2013\u00a0Namaste Aug 4 '13 at 1:14", "date": "2019-10-24 02:20:48", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/458105/constructing-a-non-linear-system-with-prerequisites-about-the-nature-of-its-crit?noredirect=1", "openwebmath_score": 0.7520902156829834, "openwebmath_perplexity": 457.9211581220462, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9790357561234475, "lm_q2_score": 0.868826769445233, "lm_q1q2_score": 0.8506124731641058}}
{"url": "http://math.stackexchange.com/questions/321445/what-does-it-mean-for-a-sequence-to-be-decreasing-in-regards-to-the-integral-tes", "text": "# What does it mean for a sequence to be decreasing in regards to the Integral Test for Convergence?\n\nExample: $$\\sum \\limits_{n=1}^{\\infty} n^k e^{-n}$$\n\nThis is the problem that I am to solve with the Integral test. I know that it converges, and I know the answer, but just don't fully get the point about it being decreasing.\n\nHow does one if a function is decreasing, the first derivative is less than zero.\n\n$f^{\\prime}\\left(x\\right) = \\dfrac{\\left(k-x\\right)x^{k-1}}{e^x}$ is less than zero when $x > k$. I do understand that there is always an $x > k$. (Not really sure if what happens when $k = \\infty$ leaving that issue till I learn more).\n\nSo my confusion comes to when can you say that a sequence is decreasing? What about when $x < k$: does the Integral test fail on that interval? Or just does the long-term behavior is all that matters? I mean what if just the very last number before infinity the sequence decreases, then will the Integral Test for convergence work?\n\nCould you explain in the realm of sequences and series what does it truly mean to be decreasing? I do understand that I am just learning this, but don't be afraid to start at a high school level and build up your answers to a Field Medal level of understanding. :)\n\n-\nA sequence $a_n$ is decreasing if $a_{n+1} \\le a_n$ for all $n$. \u2013\u00a0 Javier Mar 5 '13 at 13:54\n@JavierBadia So what about the example I give, if $k = 3$ then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? \u2013\u00a0 yiyi Mar 5 '13 at 13:55\nThe first terms don't affect convergence. You want to know what happens for $n$ large. \u2013\u00a0 L. F. Mar 5 '13 at 13:57\n@L.F. Why don't the first terms don't affect convergance? That is really my question. Thanks for putting it into words. \u2013\u00a0 yiyi Mar 5 '13 at 13:57\nBecause there are only finitely many first terms, so their sum will always converge. As long as your sequence is defined, it can jump, fly, or hula hoop at the beginning for all that matters. What we really want to know, is if the \"infinite portion\" (i.e. what happens after a certain point) will sum to a value or not. \u2013\u00a0 L. F. Mar 5 '13 at 14:01\n\nA sequence is of the form $\\{x_n\\}$, and a series is of the form $\\sum \\limits_{n=1}^{\\infty} x_n$. So for a sequence $\\{x_n\\}$ to be decreasing means $x_1 > x_2 > x_3 > \\ldots$.\n\nAs long as there is some $k$ such that $f(n)$ is decreasing for all $n > k$, the integral test applies.\n\nConsider the series $\\sum \\limits_{n=1}^{\\infty} f(n) = \\sum \\limits_{n=1}^k f(n) + \\sum \\limits_{n=k + 1}^{\\infty} f(n)$. The first term is finite, so converges, and the second term converges due to the integral test because it is always decreasing.\n\n-\nSo what about the example I give, if k=3 then for the first three terms its is not decreasing? Why doesn't that invalidate the Integral Test for Convergance? \u2013\u00a0 yiyi Mar 5 '13 at 13:56\nThe integral test applies as long as the function is eventually decreasing. \u2013\u00a0 ferson2020 Mar 5 '13 at 13:59\nThanks for your answer, but it has never been explained why the long term behavior is only which is important? Is it because if it converges then it stops \"growing/shrinking\" at that number, otherwise it just \"grows/decreases\" without limit? \u2013\u00a0 yiyi Mar 5 '13 at 14:03\nI added to my answer; does that help make it clear? \u2013\u00a0 ferson2020 Mar 5 '13 at 14:04\n\nSay a sequence $\\{a_n\\}_{n\\in\\mathbb{N}}$ is decreasing provided $a_{n+1}\\leq a_n$ for every $n\\in\\mathbb{N}$. So the bigger subscripts correspond to smaller numbers. You really just need it decreasing from a rank on, so decreasing for large $n\\in\\mathbb{N}$.\n\nYou are looking at a series. What you want to show is that the associated sequence (that is, given the series $\\sum_{n=1}^\\infty a_n$ the associated sequence is $\\{a_n\\}_{n\\in\\mathbb{N}}$) is nonnegative and decreasing, then you define a function $f:[1,+\\infty)\\to[0,+\\infty)$ so that $f(n)=a_n$ and $\\int_1^\\infty f<+\\infty$.\n\nAlso, $k$ is an exponent so it can't be $\\infty$.\n\nAlso, for series we only care what happens \"near $\\infty$\", so to speak. So say if your sequence doesn't start decreasing until $n=100$ then just apply the integral test to the series $\\sum_{n=100}^\\infty a_n$, in which case you look at $\\int_{100}^\\infty f$.\n\nRationale for \"near $\\infty$\": First of all, usually by something happening \"near $\\infty$ we mean there exists $N\\in\\mathbb{N}$ so that for all $x\\geq N$ the condition happens. Convergence of a series (or a sequence for that matter, and keep in mind a series can be seen as a sequence of partial sums) only has to do with the \"tail,\" or what happens near $\\infty$. This is because for any natural number $N\\in\\mathbb{N}$ the first part of the series $\\sum_{n=1}^Na_n$ is always finite. No matter what $N$ you choose. Because of this, if $\\sum_{n=1}^\\infty$ is going to diverge the problem must occur near $\\infty$.\n\nSo if your associated sequence doesn't start decreasing until after $n=10,000,000,000$, no problem. Just write $\\sum_{n=1}^\\infty a_n=\\sum_{n=1}^{10,000,000,000}a_n+\\sum_{n=10,000,000,000}^\\infty a_n$ and apply the integral test to the tail.\n\n-\nNice and informative answer, could you point me to some papers which explain the rational why only \"near $\\infity$\" matters? \u2013\u00a0 yiyi Mar 5 '13 at 14:01", "date": "2015-08-28 15:14:31", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/321445/what-does-it-mean-for-a-sequence-to-be-decreasing-in-regards-to-the-integral-tes", "openwebmath_score": 0.9066850543022156, "openwebmath_perplexity": 230.2361791846217, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226347732859, "lm_q2_score": 0.8705972667296309, "lm_q1q2_score": 0.8505932353666051}}
{"url": "http://mathhelpforum.com/pre-calculus/145824-two-variable-limit.html", "text": "1. two variable limit\n\nHi! I am new and I can't solve this limit:\n\nlimit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)\n\nCan anyone give me a hand?\n\n2. Originally Posted by venturozzaccio\nHi! I am new and I can't solve this limit:\n\nlimit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)\n\nCan anyone give me a hand?\nTry converting to polars:\n\n$\\lim_{(x, y) \\to (0,0)}\\left(\\frac{x^2y^4}{|x^6| + |y^4|}\\right) = \\lim_{r \\to 0}\\left[\\frac{(r\\cos{\\theta})^2(r\\sin{\\theta})^4}{|(r\\cos{ \\theta})^6| + |(r\\sin{\\theta})^4|}\\right]$\n\n$= \\lim_{r \\to 0}\\left[\\frac{r^6\\cos^2{\\theta}\\sin^4{\\theta}}{r^6|\\cos^6{ \\theta}| + r^4|\\sin^4{\\theta}|}\\right]$\n\n$= \\lim_{r \\to 0}\\left[\\frac{r^2\\cos^2{\\theta}\\sin^4{\\theta}}{r^2|\\cos^6{ \\theta}| + |\\sin^4{\\theta}|}\\right]$\n\n$= \\frac{0^2\\cos^2{\\theta}\\sin^4{\\theta}}{0^2|\\cos^6{ \\theta}| + |\\sin^4{\\theta}|}$\n\n$= \\frac{0}{|\\sin^4{\\theta}|}$\n\n$= 0$.\n\n3. sorry I make a mistake writing.\n\nthe real one is :\n\nlimit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^2|+|y^4|)\n\n4. Originally Posted by Prove It\nTry converting to polars:\n\n$\\lim_{(x, y) \\to (0,0)}\\left(\\frac{x^2y^4}{|x^6| + |y^4|}\\right) = \\lim_{r \\to 0}\\left[\\frac{(r\\cos{\\theta})^2(r\\sin{\\theta})^4}{|(r\\cos{ \\theta})^6| + |(r\\sin{\\theta})^4|}\\right]$\n\n$= \\lim_{r \\to 0}\\left[\\frac{r^6\\cos^2{\\theta}\\sin^4{\\theta}}{r^6|\\cos^6{ \\theta}| + r^4|\\sin^4{\\theta}|}\\right]$\n\n$= \\lim_{r \\to 0}\\left[\\frac{r^2\\cos^2{\\theta}\\sin^4{\\theta}}{r^2|\\cos^6{ \\theta}| + |\\sin^4{\\theta}|}\\right]$\n\n$= \\frac{0^2\\cos^2{\\theta}\\sin^4{\\theta}}{0^2|\\cos^6{ \\theta}| + |\\sin^4{\\theta}|}$\n\n$= \\frac{0}{|\\sin^4{\\theta}|}$\n\n$= 0$.\nThis is really nice\n\n5. yes but about the revised question?\n\n6. Use the exact same process. You will still be able to cancel enough $r$'s to be able to evaluate the limit.\n\n7. so the result is still 0 ?\n\n8. Originally Posted by venturozzaccio\nso the result is still 0 ?\nCorrect.\n\n9. ok. but if the limit is\n\nlimit x,y->(0,0) $xy^2 / (x^2+y^4)$\n\nthen I have $0cos()sen^2() /( cos^2+0^2cos^4())$\n\nAnd following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?\n\n10. Originally Posted by venturozzaccio\nok. but if the limit is\n\nlimit x,y->(0,0) $xy^2 / (x^2+y^4)$\n\nthen I have $0cos()sen^2() /( cos^2+0^2cos^4())$\n\nAnd following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?\nI get that the limit is $0$...\n\n$\\lim_{(x, y) \\to (0, 0)}\\frac{xy^2}{x^2 + y^4} = \\lim_{r \\to 0}\\frac{r\\cos{\\theta}\\,r^2\\sin^2{\\theta}}{r^2\\cos^ 2{\\theta} + r^4\\sin^4{\\theta}}$\n\n$= \\lim_{r \\to 0}\\frac{r^3\\cos{\\theta}\\sin^2{\\theta}}{r^2(\\cos^2{ \\theta} + r^2\\sin^4{\\theta})}$\n\n$= \\lim_{r \\to 0}\\frac{r\\cos{\\theta}\\sin^2{\\theta}}{\\cos^2{\\theta } + r^2\\sin^4{\\theta}}$\n\n$= \\frac{0}{\\cos^2{\\theta}}$\n\n$= 0$.\n\n11. I think it's not true. Just because if you set x=y the limit is 0. But if you set x=y^2 then the limit is 1/2. So limit change according the way you approach to (0,0) and for this it doesn't exist.", "date": "2017-08-19 04:35:43", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/145824-two-variable-limit.html", "openwebmath_score": 0.9255554676055908, "openwebmath_perplexity": 1232.480917109665, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226314280636, "lm_q2_score": 0.8705972650509008, "lm_q1q2_score": 0.8505932308141064}}
{"url": "https://math.stackexchange.com/questions/1811495/what-is-the-difference-between-and-projection-and-a-reflection-in-vector-transf", "text": "# What is the difference between and projection and a reflection, in vector transformation?\n\nIn my text book I have the problems of finding the standard matrix of the given linear transformation from $\\mathbb{R}^2$ to $\\mathbb{R}^2$;\n\n1. Projection onto the line $y = -x$.\n\n2. Reflection in the line $y = -x$.\n\nBoth of these give different answers. I was hoping for an explanation of the difference between a projection/reflection in a vector transformation.\n\n\u2022 Aprojection $p$ satisfies the equality $p\\circ p=p$, while a reflection $r$ staisfies $r\\circ r=\\operatorname{id}$. This is quite different. \u2013\u00a0Bernard Jun 3 '16 at 19:36\n\nThe (orthogonal) projection onto a line \"compresses\" every point in the plane onto the line. If you drop the perpendicular from the point to the line, the image of the point after projection is the intersection of the perpendicular with the line you are projecting onto.\n\nThe reflection across a line moves a point to its \"mirror image\" across the line. If you drop a the perpendicular from a point onto the line of reflection, then the mirror image is going to lie at the same distance from the line of reflection on this perpendicular, but it will be on the other side of the line of reflection from the original.\n\nOf course, in both cases, a point already on the line of reflection or projection is going to stay where it started.\n\nA huge difference between these two transformations is that reflections are always invertible (since $R^2=Id$), but projections are almost never invertible. An orthogonal projection of the plane onto a line is never invertible since every point on a perpendicular to the line of projection maps to the same point on the line you are projecting onto.\n\nIn terms of eigenvalues, the projection in this case would have eigenvalues $\\{0,1\\}$ whereas the reflection would have eigenvalues $\\{-1,1\\}$.\n\nThe answer of @rschwieb gives you the theoretical vision of the problem. I add a figure to support it in a geometrical intuitive way.\n\nThe blue line is your $y=-x$. Given the points $P$ or $Q$, than $P'$ and $Q'$ are the orthogonal projection on this line and $P''$ and $Q''$ are the reflections. As you can see the prjections of the two points coincide, so the orthogonal projection is not invertible.\n\nIf you take $P=(1,0)^T$ (a standard basis vector) you can easely see that $P'=(\\frac{1}{2},-\\frac{1}{2})^T$ and $P''=(0,-1)^T$, and if you do the same for the other basis vector $N=(0,1)^T$, you find $N'=((-\\frac{1}{2},\\frac{1}{2})^T$ and $N''=(-1,0)^T$, so you have the matrices that represents the two transformations (in the standard basis):\n\n$$\\frac{1}{2} \\begin{bmatrix} 1&-1\\\\-1&1 \\end{bmatrix}$$ for the projection, and $$\\begin{bmatrix} 0&-1\\\\-1&0 \\end{bmatrix}$$\n\nfor the reflecion.\n\n\u2022 I must beg to differ: my answer is almost purely geometric intuition. But what you've written is certainly a concrete and explicit analytical answer, which is good too :) \u2013\u00a0rschwieb Jun 4 '16 at 0:56", "date": "2019-10-23 17:45:46", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1811495/what-is-the-difference-between-and-projection-and-a-reflection-in-vector-transf", "openwebmath_score": 0.8717636466026306, "openwebmath_perplexity": 175.62086348351423, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.977022630759019, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8505932236710089}}
{"url": "https://math.stackexchange.com/questions/1618330/stopping-criteria-for-gradient-method/1618347", "text": "# Stopping criteria for gradient method\n\nFor numerically solving a smooth convex optimization $\\min\\{f(x): x\\in S\\}$ where $S$ is a closed convex set, we can apply some different algorithms: gradient method, accelerated gradient, proximal gradient ... depending on the structure of the problem. Solving is to find a solution $x*$ such that $f(x^*)=\\inf\\{f(x): x\\in S\\}:=f^*$. To this end, we try to construct an iterative sequence $\\{x^k\\}$ that converges to some solution $x^*$, or the sequence of numbers $\\{f(x^k)\\}$ tends to $f^*$. Note that, if $x^k\\to x^*$ then the continuity of $f$ can ensures that $f(x^k) \\to f^*$.\n\nMy questions are:\n\n1. In which cases we should focus on the convergence of $\\{f(x^k)\\}$ rather than of $\\{x^k\\}$? Is finding a point $x^K$ such that $x^K$ close enough to a solution $x^*$ better than finding a point $x^K$ such that $f(x^K)$ close enough to $f^*$?\n\n2. What is the best stopping criteria for an algorithm? I know the following ways:\n\n\u2022 Determine the number of iterations we need to perform to achieve a desired error $\\epsilon$, i.e., $||x^k-x^*||<\\epsilon$ or $|f(x^k)-f^*|<\\epsilon$ implies $k\\geq N$ for some $N$. I see that this way is very reliable.\n\n\u2022 terminating when $||x^{k+1}-x^k||$ or $|f(x^{k+1})-f(x^k)|$ is small enough.\n\n\u2022 terminating when $||\\nabla f(x^k)||$ is small enough.\n\nCould you explain how the second and the third cases work? Why $||\\nabla f(x^k)||$ small enough can implies that $f(x^k)$ is approximate the optimal value $f^*$. I have been know that the case $f$ is strongly convex this can be verified. Is this stopping criteria still reliable in the case where $f$ is not strongly convex?\n\nI will discuss the termination criteria for the simple gradient method $x_{k+1} = x_{k} - \\frac{1}{L}\\nabla f(x_k)$ for unconstrained minimisation problems. If there are constraints, then we would use the projected gradient method, but similar termination condition hold (imposed on the norm of the difference $x_k-z_k$).\n\nThe third criterion, namely $\\|\\nabla f(x_k) \\| < \\epsilon$ if fine for strongly convex functions with $L$-Lipschitz gradient. Indeed, if $f$ is $\\mu$-strongly convex, that is\n\n\\begin{aligned} f(y) \\geq f(x) + \\nabla f(x)^\\top (y-x) + \\tfrac{\\mu}{2} \\|y-x\\|^2 \\end{aligned},\\tag{1}\n\nthen, for $x^*$ such that $\\nabla f(x^*)=0$ (the unique minimiser of $f$), we have\n\n\\begin{aligned} f(x) - f(x^*)\\leq \\tfrac{1}{2\\mu}\\|\\nabla f(x) \\|^2, \\end{aligned}\\tag{2}\n\nso, if $\\|\\nabla f(x) \\|^2 < 2\\mu\\epsilon$, then $f(x) - f(x^*) < \\epsilon$, i.e., $x$ is $\\epsilon$-suboptimal.\n\nBut termination is a mysterious thing... In general (under the assumptions you drew) it is not true that we will have $\\|x-x^*\\|<\\epsilon$ if $\\| \\nabla f(x) \\| < \\kappa \\epsilon$, for some $\\kappa > 0$ (not even locally). There might be specific cases where such a bound holds, notwithstanding. Unless you draw some additional assumptions on $f$, this will not be a reliable termination criterion.\n\nHowever, strong convexity is often too strong a requirement in practice. Weaker conditions are discussed in the article: D. Drusvyatskiy and A.S. Lewis, Error bounds, quadratic growth, and linear convergence of proximal methods, 2016.\n\nLet $f$ be convex with $L$-Lipschithz gradient and define $\\mathcal{B}_\\nu^f = \\{x: f(x) - f^* < \\nu\\}$. Let us assume that $f$ has a unique minimiser $x^*$ (e.g., $f$ is strictly convex). Then assume that $f$ has the property\n\n\\begin{aligned} f(x) - f(x^*) \\geq \\tfrac{\\alpha}{2} \\|x-x^*\\|^2, \\end{aligned}\\tag{3}\\label{3}\n\nfor all $x\\in\\mathcal{B}_\\nu^f$ for some $\\nu>0$. Functions which satisfy this property are not necessarily strongly convex. As a counterexample we have $f = (\\max\\{|x|-1,0\\})^2$. Of course if $f$ is strongly convex the above holds and if $f$ is given in the form $f(x) = h(Ax)$ where $h$ is a strongly convex function and $A$ is any matrix.\n\nThen, condition \\eqref{3} is shown to be equivalent to\n\n\\begin{aligned} \\|x-x^*\\| \\leq \\frac{2}{\\alpha} \\|\\nabla f(x) \\|, \\end{aligned}\\tag{4}\\label{4}\n\nfor all $x\\in\\mathcal{B}_{\\nu}^f$ and with $\\alpha < 1/L$.\n\nClearly in this case we may use the termination condition $\\| \\nabla f(x) \\| < \\epsilon\\alpha/2$ which will imply that $\\|x-x^*\\| < \\epsilon$.\n\nIn regard to the second condition, you may use it again for strongly convex functions or if \\eqref{3} holds locally about $x^*$. The reason for that is that the following bound holds for the gradient method:\n\n\\begin{aligned} \\tfrac{L}{2}\\|\\nabla f(x_k) \\|^2 \\leq f(x_k) - f(x_{k+1}). \\end{aligned}\\tag{5}\\label{5}\n\nThe right hand side of \\eqref{5} is further upper bounded by $L_f \\|x_k - x_{k+1}\\|$, where $L_f$ is the Lipschitz constant of $f$ (we know that $f$ is Lipshcitz continuous), so a condition on $\\|x_{k+1}-x_{k}\\|$ may potentially be used, but we may see that the basis for all this is the bound on $\\|\\nabla f(x_k) \\|$.\n\nIf $$f$$ is strictly convex, it has at most one minimum and at this minimum its gradient is zero. So the third criteria should work fine. If $$f$$ is not convex, you may reach a local minimum so this criterion is not really justified. If you are using a gradient method, the second criteria is very similar to the third because each step (or the difference $$x^{k+1}-x^{k}$$) is obtained from the gradient.\n\n\u2022 I know the fact that \"If $f$ is convex, it has just one minimum on $R^n$ and at this minimum its gradient is zero. But can you explain why $||\\nabla f(x^k)||$ small enough can implies that $f(x^k)$ is approximate the optimal value $f^*$? In addition, in the constrained case, the gradient does not need to be zero at the minimum point. \u2013\u00a0Richkent Jan 19 '16 at 16:27\n\u2022 Well, if $\\|\\nabla f\\|$ is small enough (say below $\\epsilon$), it can be approximated by zero, meaning that you are very close to the optimal. If $f$ is convex, the gradient is monotonous and continuous, so if it's close to zero, you are close to the minimum. \u2013\u00a0citronrose Jan 19 '16 at 16:34\n\u2022 draw a $C^1$ (strictly) convex function $\\mathbb{R} \\to \\mathbb{R}$ you'll see that if the minimum is at $a \\ne -\\infty$ then $|f'(x)| < \\epsilon$ only when $x \\in ]a-\\delta;a+\\delta[$, and the smaller is $\\epsilon$ the smaller will be $\\delta$ . after that, just remember that convex on $\\mathbb{R}^n$ means that you can minimize with respect to each $x_i$ one after the other. \u2013\u00a0reuns Jan 19 '16 at 16:44\n\u2022 \"If $f$ is convex, it has just one minimum\" is completely wrong. Some convex functions have no minimizers (e.g. $f(x) = \\exp(x)$), others have a continuum of minimizers (e.g. $f(x) = 0$). \u2013\u00a0Guillaume Garrigos Mar 1 at 21:02\n\u2022 @citronrose $\\exp$ doesn't necessarily have a minimum on a closed domain, take for instance $S = \\mathbb{R}$ or $S = [a,+\\infty[$ for any $a \\in \\mathbb{R}$. Of course there is no such counterexamples if $S$ is compact, which is maybe the case you had in mind. \u2013\u00a0Guillaume Garrigos Mar 5 at 15:53", "date": "2019-09-19 15:42:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1618330/stopping-criteria-for-gradient-method/1618347", "openwebmath_score": 0.9520741701126099, "openwebmath_perplexity": 172.65927548897878, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226260757066, "lm_q2_score": 0.8705972616934406, "lm_q1q2_score": 0.8505932228740445}}
{"url": "http://gmatclub.com/forum/a-store-reported-total-sales-of-385-million-for-february-of-143754.html?kudos=1", "text": "Author Message\nTAGS:\n\nHide Tags\n\nManager\nJoined: 02 Dec 2012\nPosts: 178\nFollowers: 5\n\nKudos [?]: 2319 [1] , given: 0\n\nShow Tags\n\n15 Dec 2012, 04:14\n2\nKUDOS\nExpert's post\nThis might sound stupid, but how do we know that we need to round up 64 to 65 without losing to much time. I did this question before and it took over 2 min before i knew that i had to round 64 to 65 in order to solve the problem. Do you have any trick in mind?\n\n65 is the closest multiple of 13: 5*13=65.\n_________________\nSenior Manager\nJoined: 20 Aug 2015\nPosts: 398\nLocation: India\nGMAT 1: 760 Q50 V44\nFollowers: 25\n\nKudos [?]: 213 [1] , given: 10\n\nShow Tags\n\n07 Dec 2012, 04:35\nExpert's post\n3\nThis post was\nBOOKMARKED\nA store reported total sales of $385 million for February of this year. If the total sales for the same month last year was$320 million, approximately what was the percent increase in sales?\n\n(A) 2%\n(B) 17%\n(C) 20%\n(D) 65%\n(E) 83%\n\nLast year's sales = $320 million; This year's sales =$385 million;\nIncrease = $65 million. Now, 20% of$320 million is $64 million, which is very close to actual increase of$65 million.\n\nOR:\nGeneral formula for percent increase or decrease, or percent change:\n\n$$Percent=\\frac{Change}{Original}*100$$ --> $$Percent=\\frac{65}{320}*100=\\frac{13}{64}*100\\approx{\\frac{13}{65}}*100=20%$$.\n\n_________________\nIntern\nJoined: 06 Jun 2012\nPosts: 18\nFollowers: 0\n\nKudos [?]: 0 [0], given: 9\n\nShow Tags\n\n11 Sep 2014, 06:09\nA store reported total sales of $385 million for February of this year. If the total sales for the same month last year was$320 million, approximately what was the percent increase in sales?\n\n(A) 2%\n(B) 17%\n(C) 20%\n(D) 65%\n(E) 83%\n\nlast year= 320\nthis year = 385\n\n((385-320)/320)*100 = (65/320 )*100 approx.. 20\nSVP\nStatus: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1858\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nFollowers: 47\n\nKudos [?]: 1929 [0], given: 193\n\nShow Tags\n\n20 Oct 2015, 02:42\nHello from the GMAT Club BumpBot!\n\nThanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).\n\nWant to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.\n_________________\nSenior Manager\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 458\nFollowers: 22\n\nKudos [?]: 187 [0], given: 2\n\nRe: A store reported total sales of $385 million for February of [#permalink] Show Tags 07 Jun 2016, 10:09 Walkabout wrote: A store reported total sales of$385 million for February of this year. If the total sales for the same month last year was $320 million, approximately what was the percent increase in sales? (A) 2% (B) 17% (C) 20% (D) 65% (E) 83% The problem is testing us on using the percent change formula: (New Value \u2013 Old Value)/Old Value x 100 We are given: February sales this year = 385 million February sales last year = 320 million We need to determine the percent increase between sales from last year to sales this year. Thus, the new value = 385 million and the old value = 320 million. Let\u2019s plug them into our percent change formula. (New Value \u2013 Old Value)/Old Value x 100 [(385 \u2013 320)/320] x 100 65/320 x 100 13/64 x 100 \u2248 13/65 x 100 \u2248 1/5 x 100 \u2248 20%. The answer is C. _________________ Jeffrey Miller Jeffrey Miller Head of GMAT Instruction Re: A store reported total sales of$385 million for February of \u00a0 [#permalink] 07 Jun 2016, 10:09\nSimilar topics Replies Last post\nSimilar\nTopics:\n5 At the store, Sam bought a shirt and a toaster. There was an 8% sales 6 18 Feb 2015, 03:20\n9 During a sale of 20% on everything in a store, a kid is successful in 4 02 Oct 2014, 23:52\n12 Last year Department Store X had a sales total for December 12 05 Mar 2014, 01:12\n21 There are 6 stores in town that had a total of 20 visitors 14 08 Sep 2011, 22:53\n18 Last year Department Store X had a sales total for December 10 03 Dec 2007, 17:11\nDisplay posts from previous: Sort by", "date": "2017-01-18 09:42:40", "meta": {"domain": "gmatclub.com", "url": "http://gmatclub.com/forum/a-store-reported-total-sales-of-385-million-for-february-of-143754.html?kudos=1", "openwebmath_score": 0.17683859169483185, "openwebmath_perplexity": 8045.317561944001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9416541626630935, "lm_q2_score": 0.9032941995446778, "lm_q1q2_score": 0.850590743110673}}
{"url": "https://math.stackexchange.com/questions/1629665/how-many-numbers-are-there-of-2n-digits-that-the-sum-of-the-digits-in-the-first", "text": "# How many numbers are there of 2n digits that the sum of the digits in the first half equals the sum of the digits in the second half\n\nThe question is how many number of a given number of digits 2n where the sum of the first half of the digits equals the sum of the digits in the second half.\n\nSo this is for a programming problem and I've got it down to approximately $10^n$ operations. However for a sample size of $0 < n < 500$ this is far too many operations and the question leads me to believe that there is a simple formula for this.\n\nFrom brute force calculation:\n\nn = 1\n10\nn = 2\n670\nn = 3\n4816030 ...\n\n\nSo I've abstracted it to: finding how many ways the digits from 0 - 9 can be put together to form a given sum (from 0 -> 9*n) this gives me a $10^n$ however this is still too large. (fyi you need to square this number)\n\nI've observed that in this abstracted version of the question the subsums are constant for n = 1, increase for n = 2, fibonacci numbers for n = 3 with the general rule being if you keep taking the difference of the differences of the sums that the previous one forms the next one (this changes slightly towards the center: the difference of differences doubles). And this trend seems to hold true for all the numbers that I tested.\n\nI think that it might be related to $^nC_r$ or similar but I don't have the maths to change it from something like pascal's triangle to something that I can work with.\n\n\u2022 Question: Do you count, say \"0990\" as an example when $n=2$? It's gonna be much harder if you don't. \u2013\u00a0Thomas Andrews Jan 27 '16 at 20:56\n\u2022 @ThomasAndrews Given he says the answer is $10$ when $n=1$ means he must be including $00$ as a two digit number. \u2013\u00a0Gregory Grant Jan 27 '16 at 21:04\n\u2022 @ThomasAndrews yes it is counted \u2013\u00a0Cjen1 Jan 27 '16 at 21:06\n\u2022 The value you have for $n=3$ is actually the value for $n=4$. The value for $n=3$ is $55252$. \u2013\u00a0Thomas Andrews Jan 28 '16 at 0:05\n\nLet $f(n,s)$ denote the number of $n$-digit sequences with digit sum $s$. To cover the leading zero problem, let $g(n,s)$ denote the number of $n$-digit sequences with digit sum $s$ and leading digit non-zero. Then $f(n+1,s)=\\sum_{d=0}^9f(n,s-d)$, with $f(n,s)=0$ for $s<0$ (or $s>9n$) understood, which allows a quick recursive calculation in $O(n^2)$ time and $O(n)$ space complexity. Furthermore, we simply have $g(n,s)=f(n,s)-f(n-1,s)$. Now the count you really want is $$\\sum_{s=0}^\\infty g(n,s)f(n,s).$$ But of course we need not consider infinitely many $s$. Instead, we need only sum $$\\sum_{s=1}^{9n} g(n,s)f(n,s).$$\n\nEdit: From the comments it seems that leading zeroes are allowed. In that case, simply replace $g$ by $f$ in the above.\n\nYou can do it in order $n^2$ operations. If $n=100$, the sum of $100$ digits can range from $0$ to $900$ You need to compute how many ways there are to add up to each of those sums. For $n=1$,you have one way to make each number $0$ through $9$. For $n=2$ you can compute the number of ways to make $k$ by summing over the number of ways to make each of $k-9, k-8, \\dots k$ out of one digit. Then for $n=3$ you can compute the number of ways to make $k$ by summing over the number of ways to make $k$ by summing over the number of ways to make each of $k-9, k-8, \\dots k$ out of two digits. You will end up with an array with numbers in entries $0$ through $900$. Each of $0$ and $900$ will be $1$. $1$ and $899$ will be $100$. Can you see why? The ones in the middle will be quite large-you need arbitrary precision integers to solve this. Then to get the number of $200$ digit numbers, you sum the squares of the entries. This is because for a number where both the first half and last half sum to $899$ you have $100$ choices for the first half and $100$ choices for the last half, giving $100^2$ total possibilities.\n\nI looked this up at the OEIS where I found this OEIS entry.\n\nWe clearly have by inspection that the desired answer is\n\n$$[z^0] (1+z+z^2+\\cdots+z^9)^n (1+1/z+1/z^2+\\cdots+1/z^9)^n \\\\= [z^0]\\frac{1}{z^{9n}} (1+z+z^2+\\cdots+z^9)^{2n}$$\n\nor\n\n$$[z^{9n}] (1+z+z^2+\\cdots+z^9)^{2n} = [z^{9n}] \\left(\\frac{1-z^{10}}{1-z}\\right)^{2n}.$$\n\nExtracting coefficients from this we get\n\n$$[z^{9n}] \\frac{1}{(1-z)^{2n}} \\sum_{q=0}^{2n} {2n\\choose q} (-1)^q z^{10q} \\\\ [z^{9n}] \\frac{1}{(1-z)^{2n}} \\sum_{q=0}^{\\lfloor 9n/10\\rfloor} {2n\\choose q} (-1)^q z^{10q} \\\\ = \\sum_{q=0}^{\\lfloor 9n/10\\rfloor} {2n\\choose q} (-1)^q {9n-10q+2n-1\\choose 2n-1} \\\\ = \\sum_{q=0}^{\\lfloor 9n/10\\rfloor} {2n\\choose q} (-1)^q {11n-10q-1\\choose 2n-1}.$$\n\nNote however that the second binomial coefficient is zero when $11n-10q-1\\lt 2n-1$ or $9n\\lt 10q$ so we may set the upper limit to $n-1$ if desired, producing a match to the OEIS entry e.g.\n\n$$\\sum_{q=0}^{n-1} {2n\\choose q} (-1)^q {11n-10q-1\\choose 2n-1}.$$\n\nThe second binomial coefficient starts producing non-zero values again when $11n-10q-1 \\lt 0.$\n\nAn alternate approach uses generating functions. Let $f(x) = (1+x+x^2+\\cdot+x^9)^{n}$. Then we are seeking the constant term of $f(x)f(x^{-1})$.\n\nLetting $x=e^{i\\theta}$, we get:\n\n$$f(x)f(x^{-1})=\\left(\\frac{1-\\cos(10\\theta)}{1-\\cos\\theta}\\right)^n= \\left(\\frac{\\sin(5x)}{\\sin(x/2)}\\right)^{2n}$$\n\nThe constant term can be computed as:\n\n$$\\frac{1}{2\\pi}\\int_{0}^{2\\pi} \\left(\\frac{\\sin(5x)}{\\sin(x/2)}\\right)^{2n}\\,dx$$\n\nWhile that might seem insane, you can actually use the graph of $\\frac{\\sin(5x)}{\\sin(x/2)}$ to get upper and lower bounds for this integral.\n\n\u2022 I believe you mean the Central Limit Theorem, not the Law of Large Numbers, but this is the first thing that came to mind for me as well. \u2013\u00a0Michael Lugo Jan 27 '16 at 21:44\n\u2022 Yep, that is what I mean, thanks. @MichaelLugo One reason I didn't proceed further is that I don't actually know these theorems, only know of them. :) \u2013\u00a0Thomas Andrews Jan 27 '16 at 21:45", "date": "2019-10-15 15:58:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1629665/how-many-numbers-are-there-of-2n-digits-that-the-sum-of-the-digits-in-the-first", "openwebmath_score": 0.8320608735084534, "openwebmath_perplexity": 144.6144611556823, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085087724427, "lm_q2_score": 0.8652240895276223, "lm_q1q2_score": 0.8505888999545501}}
{"url": "https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1/1490918", "text": "# Proof of the Hockey-Stick Identity: $\\sum\\limits_{t=0}^n \\binom tk = \\binom{n+1}{k+1}$\n\nAfter reading this question, the most popular answer use the identity $$\\sum_{t=0}^n \\binom{t}{k} = \\binom{n+1}{k+1}.$$\n\nWhat's the name of this identity? Is it the identity of the Pascal's triangle modified.\n\nHow can we prove it? I tried by induction, but without success. Can we also prove it algebraically?\n\nThanks for your help.\n\nEDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.\n\n\u2022 It is sometimes called the \"hockey stick\". \u2013\u00a0user940 Oct 21 '15 at 15:24\n\u2022 There is another cute graphical illustration on the plane of $\\binom{n}{k}$ \u2013\u00a0Eli Korvigo Oct 21 '15 at 16:54\n\u2022 It's pretty straightforward from the picture. Just switch the $1$ at the top of the stick with the $1$ directly below, then repeatedly replace adjacent numbers with the number in the cell below. This can be translated into a formal proof with words and symbols, but an animation or series of pictures is much more effective. \u2013\u00a0user2357112 Oct 22 '15 at 3:24\n\u2022 See also this question. Some post which are linked there might be of interest, too. \u2013\u00a0Martin Sleziak Jan 18 '16 at 15:05\n\nThis is purely algebraic. First of all, since $$\\dbinom{t}{k} =0$$ when $$k>t$$ we can rewrite the identity in question as $$\\binom{n+1}{k+1} = \\sum_{t=0}^{n} \\binom{t}{k}=\\sum_{t=k}^{n} \\binom{t}{k}$$\n\nRecall that (by the Pascal's Triangle), $$\\binom{n}{k} = \\binom{n-1}{k-1} + \\binom{n-1}{k}$$\n\nHence $$\\binom{t+1}{k+1} = \\binom{t}{k} + \\binom{t}{k+1} \\implies \\binom{t}{k} = \\binom{t+1}{k+1} - \\binom{t}{k+1}$$\n\nLet's get this summed by $$t$$: $$\\sum_{t=k}^{n} \\binom{t}{k} = \\sum_{t=k}^{n} \\binom{t+1}{k+1} - \\sum_{t=k}^{n} \\binom{t}{k+1}$$\n\nLet's factor out the last member of the first sum and the first member of the second sum: $$\\sum _{t=k}^{n} \\binom{t}{k} =\\left( \\sum_{t=k}^{n-1} \\binom{t+1}{k+1} + \\binom{n+1}{k+1} \\right) -\\left( \\sum_{t=k+1}^{n} \\binom{t}{k+1} + \\binom{k}{k+1} \\right)$$\n\nObviously $$\\dbinom{k}{k+1} = 0$$, hence we get $$\\sum _{t=k}^{n} \\binom{t}{k} =\\binom{n+1}{k+1} +\\sum_{t=k}^{n-1} \\binom{t+1}{k+1} -\\sum_{t=k+1}^{n} \\binom{t}{k+1}$$\n\nLet's introduce $$t'=t-1$$, then if $$t=k+1 \\dots n, t'=k \\dots n-1$$, hence $$\\sum_{t=k}^{n} \\binom{t}{k} = \\binom{n+1}{k+1} +\\sum_{t=k}^{n-1} \\binom{t+1}{k+1} -\\sum_{t'=k}^{n-1} \\binom{t'+1}{k+1}$$\n\nThe latter two arguments eliminate each other and you get the desired formulation $$\\binom{n+1}{k+1} = \\sum_{t=k}^{n} \\binom{t}{k} = \\sum_{t=0}^{n} \\binom{t}{k}$$\n\n\u2022 Beautiful proof. p.-s. you can use the LaTeX command \\binom{n}{k} to display $\\binom{n}{k}$. \u2013\u00a0hlapointe Oct 21 '15 at 16:26\n\u2022 @hlapointe thank you. Sure, I forgot there was a special command for binomial. \u2013\u00a0Eli Korvigo Oct 21 '15 at 16:32\n\nImagine the first $$n + 1$$ numbers, written in order on a piece of paper. The right hand side asks in how many ways you can pick $$k+1$$ of them. In how many ways can you do this?\n\nYou first pick a highest number, which you circle. Call it $$s$$. Next, you still have to pick $$k$$ numbers, each less than $$s$$, and there are $$\\binom{s - 1}{k}$$ ways to do this.\n\nSince $$s$$ is ranging from $$1$$ to $$n+1$$, $$t:= s-1$$ is ranging from $$0$$ to $$n$$ as desired.\n\nWe can use the well known identity $$1+x+\\dots+x^n = \\frac{x^{n+1}-1}{x-1}.$$ After substitution $x=1+t$ this becomes $$1+(1+t)+\\dots+(1+t)^n=\\frac{(1+t)^{n+1}-1}t.$$ Both sides of these equations are polynomials in $t$. (Notice that the RHS simplifies to $\\sum_{j=1}^{n+1}\\binom {n+1}j t^{j-1}$.)\n\nIf we compare coefficient of $t^{k}$ on the LHS and the RHS we see that $$\\binom 0k + \\binom 1k + \\dots + \\binom nk = \\binom{n+1}{k+1}.$$\n\nThis proof is basically the same as the proof using generating functions, which was posted in other answers. However, I think it is phrased a bit differently. (And if it is formulated this way, even somebody who has never heard of generating functions can follow the proof.)\n\n\\begin{align} \\sum_{t=\\color{blue}0}^n \\binom{t}{k} =\\sum_{t=\\color{blue}k}^n\\binom tk&= \\sum_{t=k}^n\\left[ \\binom {t+1}{k+1}-\\binom {t}{k+1}\\right]\\\\ &=\\sum_{t=\\color{orange}k}^\\color{orange}n\\binom {\\color{orange}{t+1}}{k+1}-\\sum_{t=k}^n\\binom t{k+1}\\\\ &=\\sum_{t=\\color{orange}{k+1}}^{\\color{orange}{n+1}}\\binom {\\color{orange}{t}}{k+1}-\\sum_{t=k}^n\\binom t{k+1}\\\\ &=\\binom{n+1}{k+1}-\\underbrace{\\binom k{k+1}}_0&&\\text{by telescoping}\\\\ &=\\binom{n+1}{k+1}\\quad\\blacksquare\\\\ \\end{align}\n\nYou can use induction on $n$, observing that\n\n$$\\sum_{t=0}^{n+1} \\binom{t}{k} = \\sum_{t=0}^{n} \\binom{t}{k} + \\binom{n+1}{k} = \\binom{n+1}{k+1} + \\binom{n+1}{k} = \\binom{n+2}{k+1}$$\n\n\u2022 How can you say that $\\sum_{t=0}^n \\binom{t}{k} = \\binom{n+1}{k+1}$ in your proof. \u2013\u00a0hlapointe Oct 21 '15 at 15:13\n\u2022 That's the inductive hypothesis. \u2013\u00a0Michael Biro Oct 21 '15 at 15:14\n\u2022 Ok. Can we prove it algebraically? \u2013\u00a0hlapointe Oct 21 '15 at 15:15\n\u2022 What's the first step!? Because if I take $n=1$, the hypothesis seem to be incorrect. \u2013\u00a0hlapointe Oct 21 '15 at 15:21\n\u2022 @hlapointe One choice of base case for every fixed $k$ is that $\\sum_{t=0}^{k} \\binom{t}{k} = \\binom{k}{k} = 1 = \\binom{k+1}{k+1}$. \u2013\u00a0Michael Biro Oct 21 '15 at 16:28\n\nThe RHS is the number of $k+1$ subsets of $\\{1,2,...,n+1\\}$. Group them according to the largest element in the subset. Sum up all the cases. Get the LHS.\n\nAnother technique is to use snake oil. Call your sum:\n\n\\begin{align} S_k &= \\sum_{0 \\le t \\le n} \\binom{t}{k} \\end{align}\n\nDefine the generating function:\n\n\\begin{align} S(z) &= \\sum_{k \\ge 0} S_k z^k \\\\ &= \\sum_{k \\ge 0} z^k \\sum_{0 \\le t \\le n} \\binom{t}{k} \\\\ &= \\sum_{0 \\le t \\le n} \\sum_{k \\ge 0} \\binom{t}{k} z^k \\\\ &= \\sum_{0 \\le t \\le n} (1 + z)^t \\\\ &= \\frac{(1 + z)^{n + 1} - 1}{(1 + z) - 1} \\\\ &= z^{-1} \\left( (1 + z)^{n + 1} - 1 \\right) \\end{align}\n\nSo we are interested in the coefficient of $z^k$ of this:\n\n\\begin{align} [z^k] z^{-1} \\left( (1 + z)^{n + 1} - 1 \\right) &= [z^{k + 1}] \\left( (1 + z)^{n + 1} - 1 \\right) \\\\ &= \\binom{n + 1}{k + 1} \\end{align}\n\nWe can use the integral representation of the binomial coefficient $$\\dbinom{t}{k}=\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{\\left(1+z\\right)^{t}}{z^{k+1}}dz\\tag{1}$$ and get $$\\sum_{t=0}^{n}\\dbinom{t}{k}=\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{\\sum_{k=0}^{n}\\left(1+z\\right)^{t}}{z^{k+1}}dz$$ $$=\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{\\left(z+1\\right)^{n+1}}{z^{k+2}}dz-\\frac{1}{2\\pi i}\\oint_{\\left|z\\right|=1}\\frac{1}{z^{k+2}}dz$$ and so usign again $(1)$ we have $$\\sum_{t=0}^{n}\\dbinom{t}{k}=\\dbinom{n+1}{k+1}-0=\\color{red}{\\dbinom{n+1}{k+1}.}$$\n\n\u2022 It is so nice and weird. +1 \u2013\u00a0Behrouz Maleki Jul 5 '16 at 10:27\n\u2022 +1. Nice work. You must subtract $\\displaystyle{\\delta_{k,-1}}$ in order to take account of the case $\\displaystyle{k = -1}$. When $\\displaystyle{k = -1}$, the LHS is equal to $\\displaystyle{0}$ and your RHS is equal to $\\displaystyle{1}$. With the $\\displaystyle{\\delta_{k,-1}}$ you'll get $\\displaystyle{1 - 1 = 0}$. \u2013\u00a0Felix Marin Jul 6 '16 at 21:50\n\nIn this answer, I prove the identity $$\\binom{-n}{k}=(-1)^k\\binom{n+k-1}{k}\\tag{1}$$ Here is a generalization of the identity in question, proven using the Vandermonde Identity \\begin{align} \\sum_{m=0}^M\\binom{m+k}{k}\\binom{M-m}{n} &=\\sum_{m=0}^M\\binom{m+k}{m}\\binom{M-m}{M-m-n}\\tag{2}\\\\ &=\\sum_{m=0}^M(-1)^m\\binom{-k-1}{m}(-1)^{M-m-n}\\binom{-n-1}{M-m-n}\\tag{3}\\\\ &=(-1)^{M-n}\\sum_{m=0}^M\\binom{-k-1}{m}\\binom{-n-1}{M-m-n}\\tag{4}\\\\ &=(-1)^{M-n}\\binom{-k-n-2}{M-n}\\tag{5}\\\\ &=\\binom{M+k+1}{M-n}\\tag{6}\\\\ &=\\binom{M+k+1}{n+k+1}\\tag{7} \\end{align} Explanation:\n$(2)$: $\\binom{n}{k}=\\binom{n}{n-k}$\n$(3)$: apply $(1)$ to each binomial coefficient\n$(4)$: combine the powers of $-1$ which can then be pulled out front\n$(5)$: apply Vandermonde\n$(6)$: apply $(1)$\n$(7)$: $\\binom{n}{k}=\\binom{n}{n-k}$\n\nTo get the identity in the question, set $n=0$.\n\n\u2022 @FoF: I have added a link here and answered your other question. Thanks for mentioning the difficulty. \u2013\u00a0robjohn Dec 7 '13 at 12:33\n\u2022 @FoF: That is the Vandermonde Identity that I mentioned at the beginning. \u2013\u00a0robjohn Dec 8 '13 at 18:56\n\u2022 @FoF: I added an explanation for each line. \u2013\u00a0robjohn Dec 9 '13 at 2:20\n\u2022 I answered my own question about $(5, 6$) here. \u2013\u00a0NaN Dec 10 '13 at 8:54\n\u2022 @FoF: Ah. That is why I added the Explanation when I saw difficulty in following the argument. \u2013\u00a0robjohn Dec 11 '13 at 7:46\n\nYou remember that: $$(1+x)^m = \\sum_k \\binom{m}{k} x^k$$ So the sum $$\\sum_{m=0}^M \\binom{m+k}{k}$$ is the coefficient of $x^k$ in: $$\\sum_{m=0}^M (1+x)^{m+k}$$ Yes? So now use the geometric series formula given: $$\\sum_{m=0}^M (1+x)^{m+k} = -\\frac{(1+x)^k}{x} \\left( 1 - (1+x)^{M+1} \\right)$$ And now you want to know what is coefficient of $x^k$ in there. You got it from here.\n\nRecall that for $k\\in\\Bbb N$ we have the generating function\n\n$$\\sum_{n\\ge 0}\\binom{n+k}kx^n=\\frac1{(1-x)^{k+1}}\\;.$$\n\nThe identity in the question can therefore be rewritten as\n\n$$\\left(\\sum_{n\\ge 0}\\binom{n+k}kx^n\\right)\\left(\\sum_{n\\ge 0}x^n\\right)=\\sum_{n\\ge 0}\\binom{n+k+1}{k+1}x^n\\;.$$\n\nThe coefficient of $x^n$ in the product on the left is\n\n$$\\sum_{i=0}^n\\binom{i+k}k\\cdot1=\\sum_{i=0}^n\\binom{i+k}k\\;,$$\n\nand the $n$-th term of the discrete convolution of the sequences $\\left\\langle\\binom{n+k}k:n\\in\\Bbb N\\right\\rangle$ and $\\langle 1,1,1,\\dots\\rangle$. And at this point you\u2019re practically done.\n\n\u2022 Is there a typo in the second equation (first sum)? I believe $k$ should be indexed. \u2013\u00a0AlanH May 27 '13 at 6:20\n\u2022 @Alan: No, the sum is over $n$; $k$ is fixed throughout. \u2013\u00a0Brian M. Scott May 27 '13 at 7:19\n\u2022 In my text, I have an identity $\\sum_{r\\geq 0} \\binom{r + n}{r} x^r = 1/(1-x)^{n+1}$ This may be the cause of my confusion, but is this identity correct and is it equivalent to the one you used? \u2013\u00a0AlanH May 27 '13 at 8:22\n\u2022 @Alan: Sure: your $r$ is my $n$, and your $n$ is my $k$. \u2013\u00a0Brian M. Scott May 27 '13 at 8:28\n\u2022 @Alan: $\\binom{r+n}r=\\binom{r+n}n$; now do the translation. (Sorry: I didn\u2019t notice before that you\u2019d used the symmetrically opposite binomial coefficient.) \u2013\u00a0Brian M. Scott May 27 '13 at 19:19\n\nA standard technique to prove such identities $\\sum_{i=0}^Mf(i)=F(M)$, involving on one hand a sum where only the upper bound $M$ is variable and on the other hand an explicit expression in terms of$~M$, is to use induction on$~M$. It amounts to showing that $f(M)=F(M)-F(M-1)$ (and that $F(0)=f(0)$). This is similar to using the fundamental theorem of calculus in showing that $\\int_0^{x_0}f(x)\\mathrm dx=F(x_0)$ by establishing $f(x)=F'(x)$ (and $F(0)=0$).\n\nSo here you need to check (apart from the obvious starting case $M=0$) that $\\binom{M+k}k=\\binom{M+k+1}{k+1}-\\binom{M+k}{k+1}$. This is just in instance of Pascal's recurrence for binomial coefficients.\n\n\n$$\\mbox{Note that}\\quad \\sum_{m = 0}^{M}{m + k \\choose k} = \\sum_{m = k}^{M + k}{m \\choose k} = a_{M + k} - a_{k - 1}\\quad\\mbox{where}\\quad a_{n} \\equiv \\sum_{m = 0}^{n}{m \\choose k}\\tag{1}$$\n\nThen, \\begin{align} \\color{#f00}{a_{n}} & \\equiv \\sum_{m = 0}^{n}{m \\choose k} = \\sum_{m = 0}^{n}\\ \\overbrace{% \\oint_{\\verts{z} = 1}{\\pars{1 + z}^{m} \\over z^{k + 1}}\\,{\\dd z \\over 2\\pi\\ic}} ^{\\ds{m \\choose k}}\\ =\\ \\oint_{\\verts{z} = 1}{1 \\over z^{k + 1}}\\sum_{m = 0}^{n}\\pars{1 + z}^{m} \\,{\\dd z \\over 2\\pi\\ic} \\\\[3mm] & = \\oint_{\\verts{z} = 1}{1 \\over z^{k + 1}}\\, {\\pars{1 + z}^{n + 1} - 1 \\over \\pars{1 + z} - 1}\\,{\\dd z \\over 2\\pi\\ic}\\ =\\ \\underbrace{\\oint_{\\verts{z} = 1}{\\pars{1 + z}^{n + 1} \\over z^{k + 2}} \\,{\\dd z \\over 2\\pi\\ic}}_{\\ds{n + 1 \\choose k + 1}}\\ -\\ \\underbrace{\\oint_{\\verts{z} = 1}{1 \\over z^{k + 2}}\\,{\\dd z \\over 2\\pi\\ic}} _{\\ds{\\delta_{k + 2,1}}} \\\\[8mm] \\imp\\ \\color{#f00}{a_{n}} & = \\fbox{$\\ds{\\quad% {n + 1 \\choose k + 1} - \\delta_{k,-1}\\quad}$} \\end{align}\n\\begin{align} \\mbox{With}\\ \\pars{1}\\,,\\quad \\color{#f00}{\\sum_{m = 0}^{M}{m + k \\choose k}} & = \\bracks{{M + k + 1 \\choose k + 1} - \\delta_{k,-1}} - \\bracks{{k \\choose k + 1} - \\delta_{k,-1}} \\\\[3mm] & = {M + k + 1 \\choose k + 1} - {k \\choose k + 1} \\end{align} Thanks to $\\ds{@robjohn}$ user who pointed out the following feature: $${k \\choose k + 1} = {-k + k + 1 - 1 \\choose k + 1}\\pars{-1}^{k + 1} = -\\pars{-1}^{k}{0 \\choose k + 1} = \\delta_{k,-1}$$ such that $$\\begin{array}{|c|}\\hline\\mbox{}\\\\ \\ds{\\quad\\color{#f00}{\\sum_{m = 0}^{M}{m + k \\choose k}} = \\color{#f00}{{M + k + 1 \\choose k + 1} - \\delta_{k,-1}}\\quad} \\\\ \\mbox{}\\\\ \\hline \\end{array}$$\n\n\u2022 Since $k=-1$ is covered in the first part, it should be noted that since $\\binom{-1}{0}=1$, $$\\binom{k}{k+1}-\\delta_{k,-1}=0$$ therefore the final answer seems it should be $$\\binom{M+k+1}{k+1}-\\delta_{k,-1}$$ \u2013\u00a0robjohn Jul 25 '16 at 13:00\n\u2022 @robjohn Thanks. I'm checking everything right now. \u2013\u00a0Felix Marin Jul 25 '16 at 21:48\n\u2022 @robjohn Thanks. Fixed. \u2013\u00a0Felix Marin Jul 25 '16 at 22:09\n\nWe can prove this by counting in two ways.\n\nLet $$S$$ be the set of all $$(k+1)$$-element subsets of $$[n+1]$$. By definition, $$|S|=\\binom{n+1}{k+1}$$.\n\nLet $$S_i$$ be the set of all $$(k+1)$$-element subsets of $$[n+1]$$ such that the largest element is $$i+1$$. Picking $$k+1$$ elements from $$[n+1]$$ such that the largest element is $$i+1$$ is a two-step-process.\n\n(Step 1) Pick $$i+1$$. The number of way(s) to do this is $$\\binom{1}{1}$$.\n\n(Step 2) Pick the $$k$$ elements from the the remaining $$i$$ elements. The number of way(s) to do this is $$\\binom{i}{k}$$.\n\nTherefore, $$|S_i|=\\binom{1}{1}\\binom{i}{k}=\\binom{i}{k}$$. Since we can see that $$S_k, S_{k+1}, S_{k+2}, \\dots, S_n$$ partition $$S$$, we have that $$\\begin{gather*} \\sum_{i=k}^n|S_i|=|S|\\\\ \\sum_{i=k}^n\\binom{i}{k}=\\binom{n+1}{k+1} \\end{gather*}$$ Since we know that if $$i < k$$, then $$\\binom{i}{k}=0$$, we can say that $$\\sum_{i=k}^n\\binom{i}{k}=\\sum_{i=0}^n\\binom{i}{k}$$. Therefore, we have $$\\begin{gather*} \\sum_{i=0}^n \\binom{i}{k} = \\binom{n+1}{k+1} \\end{gather*}$$", "date": "2019-08-23 19:46:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1/1490918", "openwebmath_score": 0.9964216351509094, "openwebmath_perplexity": 393.8095811308821, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850842553892, "lm_q2_score": 0.8652240825770432, "lm_q1q2_score": 0.8505888901200443}}
{"url": "http://mathhelpforum.com/geometry/183278-divided-triangle-proportion-question.html", "text": "# Thread: Divided triangle, proportion question\n\n1. ## Divided triangle, proportion question\n\nHi Forum!\nHere's a question:\n\nNine lines parallel to the base divide the other sides into 10 equal segments and the area in 10 distinct parts. If the area of the larger of these parts is 38, then the area of the original triangle is?\n\nNow,we can consider the remaining 9 parts of the triangle as 9/10 of it.\nBut, here comes the part that I don't understand.\nThe area can be considered as 81/100 to the entire triangle.\n\nI'm confused with the usage of the area here. I does makes a lot of sense,\nbut since we are in a triangle (the remaining 9/10 will be a triangle too) the area isn't supposed to be bh1/2?\n\nThis is kind of confusing to apply. Can someone help?\nWhy 81/100?\n\n2. ## Re: Divided triangle, proportion question\n\nOriginally Posted by Zellator\nHi Forum!\nHere's a question:\n\nNine lines parallel to the base divide the other sides into 10 equal segments and the area in 10 distinct parts. If the area of the larger of these parts is 38, then the area of the original triangle is?\n\nNow,we can consider the remaining 9 parts of the triangle as 9/10 of it.\nBut, here comes the part that I don't understand.\nThe area can be considered as 81/100 to the entire triangle.\n\nI'm confused with the usage of the area here. I does makes a lot of sense,\nbut since we are in a triangle (the remaining 9/10 will be a triangle too) the area isn't supposed to be bh1/2?\n\nThis is kind of confusing to apply. Can someone help?\nWhy 81/100?\n1. Draw a sketch!\n\n2. By proportion you'll get:\n\n$\\displaystyle \\dfrac bB = \\dfrac9{10}~\\implies~b=\\frac9{10} \\cdot B$\n\n$\\displaystyle \\dfrac hH = \\dfrac9{10}~\\implies~h=\\frac9{10} \\cdot H$\n\n3. The area of the original triangle is calculated by:\n\n$\\displaystyle A_o = \\frac12 \\cdot B \\cdot H$\n\nThe area of the smaller triangle is calculated by:\n\n$\\displaystyle A_s = \\frac12 \\cdot b \\cdot h = \\frac12 \\cdot \\frac9{10} \\cdot B \\cdot \\frac9{10} \\cdot H = \\frac{81}{100} \\cdot A_o$\n\n3. ## Re: Divided triangle, proportion question\n\nOriginally Posted by earboth\n1. Draw a sketch!\n\n2. By proportion you'll get:\n\n$\\displaystyle \\dfrac bB = \\dfrac9{10}~\\implies~b=\\frac9{10} \\cdot B$\n\n$\\displaystyle \\dfrac hH = \\dfrac9{10}~\\implies~h=\\frac9{10} \\cdot H$\n\n3. The area of the original triangle is calculated by:\n\n$\\displaystyle A_o = \\frac12 \\cdot B \\cdot H$\n\nThe area of the smaller triangle is calculated by:\n\n$\\displaystyle A_s = \\frac12 \\cdot b \\cdot h = \\frac12 \\cdot \\frac9{10} \\cdot B \\cdot \\frac9{10} \\cdot H = \\frac{81}{100} \\cdot A_o$\nHi earboth!\nThanks for the graph! You really exceed in Geometry!\nThanks for the easy to understand explanation, I get it now!\nUsing this is possible to calculate the area of the smaller triangle, and then the original triangle.\nGreat! Thanks again!\n\nAll the best!\n\n4. ## Re: Divided triangle, proportion question\n\nHi Zellator,\nDoes the given area of 38 sq units refer to the larger (largest?) of the 10 parts or to the larger of two parts?\n\n5. ## Re: Divided triangle, proportion question\n\nOriginally Posted by bjhopper\nHi Zellator,\nDoes the given area of 38 sq units refer to the larger (largest?) of the 10 parts or to the larger of two parts?\nHi bjhopper how are you?\nThe area given is of the largest portion of the divided triangle, looking at the graph of earboth, the portion underneath the red triangle.", "date": "2018-05-25 08:00:03", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/geometry/183278-divided-triangle-proportion-question.html", "openwebmath_score": 0.9447447657585144, "openwebmath_perplexity": 1298.327780716367, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.983085087724427, "lm_q2_score": 0.8652240773641087, "lm_q1q2_score": 0.8505888879967812}}
{"url": "https://math.stackexchange.com/questions/3255898/is-this-a-simple-answer-to-the-classic-problem-a-certain-city-has-10-bus-routes", "text": "# Is this a simple answer to the classic problem \"A certain city has 10 bus routes...\"\n\nA certain city has 10 bus routes. Is it possible to arrange the routes and the bus stops so that if one route is closed, it is still possible to get from anyone stop to any other (possibly changing along the way), but if any two routes are closed, there are at least two stops such that it is impossible to get from one to the other?\n\nGiven the solution above, why did the authors feel compelled to answer: Yes. Consider 10 straight lines in the plane, no 2 are parallel & no 3 are concurrent. Let lines be bus routes & let points of intersection be stops. We get from anyone stop to any other (if the stops lie on 1 line, w/o changing; & if not, then with just 1 change). If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once. However, if we discard 2 lines, then 1 stop-their point of intersection-will have no bus routes passing thru it, & it'll be impossible to get from this stop to any other.\n\nSource: A. M. Yaglom and l. M. Yaglom CHALLENGING MATHEMATICAL PROBLEMS WITH ELEMENTARY SOLUTIONS Volume II Problems From Various Branches of Mathematics Translated by James McCawley, Jr. Revised and edited by Basil Gordon DOVER PUBLICATIONS, INC. NEW YORK\n\n\u2022 Yes, the answer is correct. Jun 9, 2019 at 6:12\n\u2022 My guess, Joe, is that when they wrote \"changing\", they meant \"changing once\". Jun 9, 2019 at 13:04\n\u2022 Your answer is correct and fine. It is possible the authors just didn't think of this solution and thus put their solution into the book. Everybody overlooks something every now and then. Jun 9, 2019 at 15:30\n\u2022 My solution fails to meet the requirement, \"If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once.\" Jun 10, 2019 at 0:55\n\u2022 Gerry, your interpretation of the wording is what makes the problem sufficiently complex. Thanks. Jun 10, 2019 at 0:58\n\nLet\u2019s go back to the roots.\n\nI guess this is a problem 101 at p.53 from a book \u201cNon-elementary problems in elemntary presentation\u201d by A.M. Yaglom & A.M. Yaglom (\u201c\u041d\u0435\u044d\u043b\u0435\u043c\u0435\u043d\u0442\u0430\u0440\u043d\u044b\u0435 \u0437\u0430\u0434\u0430\u0447\u0438 \u0432 \u044d\u043b\u0435\u043c\u0435\u043d\u0442\u0430\u0440\u043d\u043e\u043c \u0438\u0437\u043b\u043e\u0436\u0435\u043d\u0438\u0438\u201d by \u0410\u043a\u0438\u0432\u0430 \u041c\u043e\u0438\u0441\u0435\u0435\u0432\u0438\u0447 \u0438 \u0418\u0441\u0430\u0430\u043a \u041c\u043e\u0438\u0441\u0435\u0435\u0432\u0438\u0447 \u042f\u0433\u043b\u043e\u043c), Moskow, State publishing house of technical and theoretical literature, 1954.\n\nIt is the same problem which you quoted and your solution fits for it.\n\nAlso author\u2019s solution is the same which your quoted.\n\nI can explain the situation as follows. The problem is contained in a subsection devoted to arrangements of points and planes and the authors say that this topic was developed in XIX century into a big science, called projective geometry. The questions, considered in problems 101\u2013107 belong to relatively narrow topic of projective geometry, namely, to so-called configuration theory, which has a big importance in modern mathematics. This explains the solution, but it is still surprising how your simple solution was missed.\n\nBut I guess that this problem probably was taken from Moskow mathematical competition from 1950 (9-10 forms, second round, problem 4):\n\n\u041c\u043e\u0436\u043d\u043e \u043b\u0438 \u043f\u0440\u043e\u0432\u0435\u0441\u0442\u0438 \u0432 \u0433\u043e\u0440\u043e\u0434\u0435 $$10$$ \u0430\u0432\u0442\u043e\u0431\u0443\u0441\u043d\u044b\u0445 \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u043e\u0432 \u0438 \u0443\u0441\u0442\u0430\u043d\u043e\u0432\u0438\u0442\u044c \u043d\u0430 \u043d\u0438\u0445 \u043e\u0441\u0442\u0430\u043d\u043e\u0432\u043a\u0438 \u0442\u0430\u043a, \u0447\u0442\u043e \u043a\u0430\u043a\u0438\u0435 \u0431\u044b $$8$$ \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u043e\u0432 \u043d\u0438 \u0431\u044b\u043b\u0438 \u0432\u0437\u044f\u0442\u044b, \u043d\u0430\u0439\u0434\u0451\u0442\u0441\u044f \u043e\u0441\u0442\u0430\u043d\u043e\u0432\u043a\u0430, \u043d\u0435 \u043b\u0435\u0436\u0430\u0449\u0430\u044f \u043d\u0438 \u043d\u0430 \u043e\u0434\u043d\u043e\u043c \u0438\u0437 \u043d\u0438\u0445, \u0430 \u043b\u044e\u0431\u044b\u0435 $$9$$ \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u043e\u0432 \u043f\u0440\u043e\u0445\u043e\u0434\u044f\u0442 \u0447\u0435\u0440\u0435\u0437 \u0432\u0441\u0435 \u043e\u0441\u0442\u0430\u043d\u043e\u0432\u043a\u0438.\n\n(Given a city, can we arrange $$10$$ bus routes and set bus stops on them such that for each 8 routes there is a bus stop not belonging to any of the routes, but each $$9$$ routes contain all stops).\n\nThe proposed solution is the same as Yagloms\u2019:\n\n\u041f\u0440\u043e\u0432\u0435\u0434\u0451\u043c $$10$$ \u043f\u043e\u043f\u0430\u0440\u043d\u043e \u043f\u0435\u0440\u0435\u0441\u0435\u043a\u0430\u044e\u0449\u0438\u0445\u0441\u044f \u043f\u0440\u044f\u043c\u044b\u0445, \u043d\u0438\u043a\u0430\u043a\u0438\u0435 \u0442\u0440\u0438 \u0438\u0437 \u043a\u043e\u0442\u043e\u0440\u044b\u0445 \u043d\u0435 \u043f\u0435\u0440\u0435\u0441\u0435\u043a\u0430\u044e\u0442\u0441\u044f \u0432 \u043e\u0434\u043d\u043e\u0439 \u0442\u043e\u0447\u043a\u0435. \u041f\u0443\u0441\u0442\u044c \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u044b \u043f\u0440\u043e\u0445\u043e\u0434\u044f\u0442 \u043f\u043e \u044d\u0442\u0438\u043c \u043f\u0440\u044f\u043c\u044b\u043c, \u0430 \u043e\u0441\u0442\u0430\u043d\u043e\u0432\u043a\u0430\u043c\u0438 \u0441\u043b\u0443\u0436\u0430\u0442 \u0442\u043e\u0447\u043a\u0438 \u043f\u0435\u0440\u0435\u0441\u0435\u0447\u0435\u043d\u0438\u044f \u043f\u0440\u044f\u043c\u044b\u0445. \u041b\u044e\u0431\u044b\u0435 $$9$$ \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u043e\u0432 \u043f\u0440\u043e\u0445\u043e\u0434\u044f\u0442 \u0447\u0435\u0440\u0435\u0437 \u0432\u0441\u0435 \u043e\u0441\u0442\u0430\u043d\u043e\u0432\u043a\u0438, \u043f\u043e\u0441\u043a\u043e\u043b\u044c\u043a\u0443 \u0447\u0435\u0440\u0435\u0437 \u043a\u0430\u0436\u0434\u0443\u044e \u043e\u0441\u0442\u0430\u043d\u043e\u0432\u043a\u0443, \u043b\u0435\u0436\u0430\u0449\u0443\u044e \u043d\u0430 \u043e\u0441\u0442\u0430\u0432\u0448\u0435\u0439\u0441\u044f \u043f\u0440\u044f\u043c\u043e\u0439, \u043f\u0440\u043e\u0445\u043e\u0434\u0438\u0442 \u043e\u0434\u043d\u0430 \u0438\u0437 $$9$$ \u043f\u0440\u044f\u043c\u044b\u0445, \u0441\u043e\u043e\u0442\u0432\u0435\u0442\u0441\u0442\u0432\u0443\u044e\u0449\u0438\u0445 \u044d\u0442\u0438\u043c \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u0430\u043c. \u041b\u044e\u0431\u044b\u0435 $$8$$ \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u043e\u0432 \u043d\u0435 \u043f\u0440\u043e\u0445\u043e\u0434\u044f\u0442 \u0447\u0435\u0440\u0435\u0437 \u043e\u0441\u0442\u0430\u043d\u043e\u0432\u043a\u0443, \u043a\u043e\u0442\u043e\u0440\u0430\u044f \u044f\u0432\u043b\u044f\u0435\u0442\u0441\u044f \u0442\u043e\u0447\u043a\u043e\u0439 \u043f\u0435\u0440\u0435\u0441\u0435\u0447\u0435\u043d\u0438\u044f \u043e\u0441\u0442\u0430\u0432\u0448\u0438\u0445\u0441\u044f \u0434\u0432\u0443\u0445 \u043c\u0430\u0440\u0448\u0440\u0443\u0442\u043e\u0432.\n\nBut although this problem looks similar to Yagloms\u2019, its essence is different! It asks not about connectivity of a graph with removed edges, but about incidence requirements. The geometric solution satisfies them, whereas the cyclic graph doesn\u2019t.", "date": "2022-07-04 09:12:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3255898/is-this-a-simple-answer-to-the-classic-problem-a-certain-city-has-10-bus-routes", "openwebmath_score": 0.5060421228408813, "openwebmath_perplexity": 4234.918336756166, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850837598122, "lm_q2_score": 0.8652240756264638, "lm_q1q2_score": 0.8505888828582482}}
{"url": "https://math.stackexchange.com/questions/3094722/what-is-the-amount-of-draws-necessary-to-see-all-red-cards-from-a-standard-deck", "text": "# Problem abstraction\n\nA standard deck of $$52$$ cards has $$26$$ red cards: it has $$13$$ hearts, $$13$$ diamonds, as well as $$26$$ black cards ($$13$$ spades, as well as $$13$$ clubs). Let us draw $$5$$ cards from the deck at once, and return those cards to the deck afterward.\n\nWhat is the expected number of draws before we see all $$26$$ red cards?\n\n# Use case\n\nLet us say that there is a set of $$N = 100$$ cards in a game. $$M = 30$$ cards are of rare rarity and $$N - M = 70$$ cards are of common rarity. We buy booster packs of size $$= 10$$. The question is: how many booster packs need to be bought to collect all $$M = 30$$ cards?\n\n# Attempted solution\n\nI have managed to calculate the approximate number of booster packs necessary to get $$M = 30$$ rare cards by calculating the expectation of the above hypergeometric distribution ($$\\mu$$) and then calculating $$M/\\mu$$. However, this is not the correct solution since it does not take into account the possibility of collecting duplicates.\n\nRegarding the Coupon collector's problem, I'm not sure if it is applicable since we always draw a single coupon, whereas in my use case a booster pack contains more than a single card.\n\n# Simulations\n\n## Problem Abstraction\n\n$$10^6$$ trials were conducted, AVG: $$38.947$$, STDEV: $$12.3653$$ draws\n\n## Use case\n\n$$10^6$$ trials were conducted, AVG: $$38.535$$, STDEV: $$11.962$$ draws\n\n\u2022 Related: coupon collector's problem \u2013\u00a0jvdhooft Jan 31 at 10:18\n\u2022 @james - thank you for the feedback, I have edited my question to provide additional context. \u2013\u00a0user3223162 Jan 31 at 11:23\n\u2022 close to $\\frac25\\cdot26H_{26}=40.085965$ draws. \u2013\u00a0robjohn Jan 31 at 15:31\n\u2022 Could you please explain how you got to the number? \u2013\u00a0user3223162 Jan 31 at 21:19\n\nHere is a solution of the \"problem abstraction\" by way of the Principle of Inclusion / Exclusion (PIE).\n\nLet $$T$$ be the number of the first draw in which we have seen all the red cards. We would like to find $$P(T>k)$$ for some $$k>0$$, i.e. the probability that we have not seen all $$26$$ red cards in $$k$$ draws. To that end, let's say a sequence of $$k$$ draws has \"Property $$i$$\" if red card $$i$$ has not been drawn, for $$i = 1,2,3,\\dots,26$$. Let $$S_j$$ be the sum of the probabilities of all the sequences with $$j$$ of the properties, for $$j = 1,2,3,\\dots,26$$. For $$S_j$$, there are $$\\binom{26}{j}$$ ways to select the $$j$$ cards which are missing. The probability that those cards are missing in a single draw is $$\\binom{52-j}{5} / \\binom{52}{5}$$, so the probability that the cards are missing in all $$k$$ draws is $$[\\binom{52-j}{5} / \\binom{52}{5}]^k$$. Therefore $$S_j = \\binom{26}{j} \\left( \\frac{\\binom{52-j}{5}}{ \\binom{52}{5}} \\right) ^k$$ By PIE, the probability of a sequence of draws with at least one of the properties, i.e. a sequence with at least one red card not seen, is $$P(T>k) = \\sum_{j=1}^{26} (-1)^{j+1} S_j$$ so \\begin{align} E(T) &= \\sum_{k=0}^{\\infty} P(T>k) \\\\ &= \\sum_{k=0}^{\\infty} \\sum_{j=1}^{26} (-1)^{j+1} S_j \\\\ &= \\sum_{k=0}^{\\infty} \\sum_{j=1}^{26} (-1)^{j+1} \\binom{26}{j} \\left( \\frac{\\binom{52-j}{5}}{ \\binom{52}{5}} \\right) ^k \\\\ &= \\sum_{j=1}^{26} (-1)^{j+1} \\binom{26}{j} \\sum_{k=0}^{\\infty} \\left( \\frac{\\binom{52-j}{5}}{ \\binom{52}{5}} \\right) ^k \\\\ &= \\sum_{j=1}^{26} (-1)^{j+1} \\binom{26}{j} \\frac{1}{1- \\binom{52-j}{5}/ \\binom{52}{5}} \\\\ &= 38.9133 \\end{align}\n\nThe following Monte Carlo simulation of $$10^6$$ trials in the R language is consistent with the result above. The average number of draws of 5-card hands necessary to see all 26 red cards was 38.973, with a 95% confidence interval of 38.91305 to 38.96158. The analytical result 38.9133 falls in the confidence interval, although just barely.\n\n> # ndraws: return the number of draws of 5-card hands required\n> # to see all red cards at least once\n> # We consider the red cards to be the cards numbered 1-26.\n> ndraws <- function() {\n+   seen <- rep(0, 52)\n+   n <- 0\n+   while (TRUE) {\n+     n <- n+1\n+     hand <- sample(1:52, 5)\n+     seen[hand] <- 1\n+     if (sum(seen[1:26]) >= 26)\n+       return (n)\n+   }\n+ }\n> nreps <- 1e6\n> set.seed(1234)  # for reproducibility\n> t <- replicate(nreps, ndraws())\n> t.test(t)\n\nOne Sample t-test\n\ndata:  t\nt = 3145.3, df = 1e+06, p-value < 2.2e-16\nalternative hypothesis: true mean is not equal to 0\n95 percent confidence interval:\n38.91305 38.96158\nsample estimates:\nmean of x\n38.93731\n\n\u2022 Thank you for your answer! After some digging, in a paper The Coupon Subset Collection Problem by Adler, Ilan and Ross, Sheldon M I have found equation 7 that solves the issue, and link this question as well. What keeps me confused is that the expectation depends only on the number of distinct balls in the set and the size of the drawn subset, but not on the size of the entire set. Am I misunderstanding something? \u2013\u00a0user3223162 Feb 4 at 11:11\n\u2022 @user3223162 I haven't read the paper you refer to, but in the question you link to, the assumption is that the urn contains one each of $n$ distinct balls, so the number of distinct balls is the same as the number of balls in the urn. \u2013\u00a0awkward Feb 4 at 12:56\n\u2022 In that case, that equation does not solve my issue. Furthermore, simulations I ran disagree with your attempt. Still, I appreciate the effort \u2013\u00a0user3223162 Feb 4 at 15:46\n\u2022 @user3223162 There must be something amiss, then, because my simulation agrees with the analytical result I posted. I have added the simulation to the solution, above. \u2013\u00a0awkward Feb 4 at 20:20\n\u2022 You are correct, my tests contained an error and after correction confirm your results. I have ran the formula alongside with some simulations and the results correspond. For now, I am marking this as solved, I will try and confirm my findings in some existing literature on the subject. Thanks! \u2013\u00a0user3223162 Feb 5 at 12:59", "date": "2019-05-23 10:49:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3094722/what-is-the-amount-of-draws-necessary-to-see-all-red-cards-from-a-standard-deck", "openwebmath_score": 0.9805408716201782, "openwebmath_perplexity": 569.4974853065188, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085088220004, "lm_q2_score": 0.8652240704135291, "lm_q1q2_score": 0.8505888815925552}}
{"url": "https://math.stackexchange.com/questions/3129983/question-about-asymptotic-notation", "text": "# Question about Asymptotic Notation\n\nI am doing a question on asymptotic notation. I have two functions $$f(n)$$ and $$g(n)$$, where $$f(n) = (\\log_2n)^2$$ and $$g(n) = \\log_2n^{\\log_2n} + 2\\log_2n$$. I have to determine whether $$f(n)$$ is $$O(g(n))$$, $$\\Omega(g(n))$$, or $$\\Theta(g(n))$$.\n\nMy approach to figuring this out is to determine whether $$g(n)$$ grows faster than $$f(n)$$ or if $$f(n)$$ grows faster than $$g(n)$$. To do this, I am trying to prove whether $$2\\log_2n \\leq (log_2n)^2$$ for all $$n \\geq c$$, where $$c$$ is a constant. I want to prove this because if it is true, then it can be said that $$f(n)$$ grows faster than $$g(n)$$ for all $$n \\geq c$$ (where $$c$$ is a constant). I know that I would also have to prove whether $$\\log_2n^{\\log_2n} \\leq (log_2n)^2$$ in order to say that $$f(n)$$ grows faster than $$g(n)$$ for all $$n \\geq c$$.\n\nSo far I have:\n\n$$\\log_2n \\leq (log_2n)^2$$ $$2\\log_2n \\leq 2(log_2n)^2$$\n\nHowever, I am not sure where to go from here in trying to prove whether $$2\\log_2n \\leq (log_2n)^2$$. Dividing both sides of the inequality by 2 will not achieve anything.\n\nAm I taking the right approach for solving this question, or is there a better way to determine the asymptotic complexity of $$f(n)$$? Any insights are appreciated.\n\n\u2022 Hint: Instead of dividing by $2$, divide by $\\log n$. The resulting inequality holds for $n$ sufficiently large. \u2013\u00a0Michael Burr Feb 28 at 9:37\n\nWe have that $$f(n) = (\\log_2 n)^2$$ and $$g(n) =\\log_2 n^{\\log_2 n} + 2\\log_2 n = (\\log_2 n)^2 + 2\\log_2 n$$ This last equality is derived by using the logarithm law that $$\\log(a^b) = b\\log(a)$$.\n\nQuestion 1. Is $$f$$ asymptotically bounded below by $$g$$? That is, does there exist an $$N_1$$ and a $$k_1 > 0$$ such that $$f(n)\\geq k_1 \\cdot g(n)$$\n\nfor all $$n\\geq N_1$$?\n\nAnswer 1. We have that $$(\\log_2 n)^2 \\geq k_1 \\cdot \\left((\\log_2 n)^2 + 2\\log_2 n\\right)$$\n\nSet $$k_1 = 1/2$$, then $$(\\log_2 n)^2 \\geq \\frac12 (\\log_2 n)^2 + \\log_2 n$$\n\nBy subtracting $$(1/2) (\\log_2 n)^2$$, we get $$\\frac12 (\\log_2 n)^2 \\geq \\log_2 n$$ Dividing by $$\\log_2 n$$ we get $$\\frac12 \\log_2 n \\geq 1$$ which is true for all $$n\\geq 4 = N_1$$. Hence, $$f$$ is asymptotically bounded below by $$g$$.\n\nQuestion 2. Is $$f$$ asymptotically bounded above by $$g$$? That is, does there exist an $$N_2$$ and a $$k_2 > 0$$ such that $$f(n) \\leq k_2\\cdot g(n)$$ for all $$n\\geq N_2$$?\n\nAnswer 2. We have that $$(\\log_2 n)^2 \\leq k_2 \\cdot \\left((\\log_2 n)^2 + 2\\log_2 n\\right)$$\n\nSet $$k_2 = 1$$, then $$(\\log_2 n)^2 \\leq (\\log_2 n)^2 + 2\\log_2 n$$\n\nSubtract $$(\\log_2 n)^2$$, then $$0\\leq 2\\log_2 n$$ which is true for all $$n\\geq 1 = N_2$$. Hence, $$f$$ is asymptotically bounded above by $$g$$.\n\nConclusion. We conclude that $$f$$ is both bounded below and above asymptotically by $$g$$. Specifically, to be more precise, this is true with constants $$1/2$$ and $$1$$ and for $$N = \\max(N_1,N_2) = 4$$. This means that $$\\frac12\\cdot g(n) \\leq f(n) \\leq 1\\cdot g(n)$$ for all $$n\\geq 4$$.\n\nIn asymptotic notation, this means that $$f(n)$$ is $$\\Theta (g(n))$$.\n\n\u2022 Would it be the case that you wouldn't be able to prove that f is asymptotically bounded below or above by g if you can't simplify one of the terms in the inequality to a constant? Since in both of your cases, you reduced one of the terms in the inequality to a constant. \u2013\u00a0ceno980 Feb 28 at 12:05\n\u2022 No, it doesn't have to be the case that one side is reduced to a constant. What you need to do is to solve the inequality. For example, $$12\\sqrt{n} \\leq e^n$$ has a solution (around $3.041$, so for whole numbers rounded up $n\\geq 4$). You don't even have to solve inequality in any precise manner. You just need to show that it holds for sufficiently large $n$. \u2013\u00a0Eff Feb 28 at 12:17", "date": "2019-06-19 09:11:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3129983/question-about-asymptotic-notation", "openwebmath_score": 0.9536381363868713, "openwebmath_perplexity": 46.064834412434244, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.983085084750966, "lm_q2_score": 0.865224070413529, "lm_q1q2_score": 0.8505888785910599}}
{"url": "https://www.physicsforums.com/threads/finding-the-height-of-a-ball-with-a-geometric-series.793794/", "text": "Finding the height of a ball with a geometric series\n\nHomework Statement\n\nA ball is dropped from one yard and come backs up $\\dfrac{2}{3}$ of the way up and then back down. It comes back and $\\dfrac{4}{9}$ of the way. It continues this such that the sum of the vertical distance traveled by the ball is is given by the series $1+2\\cdot\\dfrac{2}{3}+2\\cdot\\dfrac{4}{9}+2\\cdot\\dfrac{8}{27}+\\cdot \\cdot \\cdot=1+2(\\dfrac{2}{3}+\\dfrac{4}{9}+\\dfrac{8}{27}+\\cdot\\cdot\\cdot(\\dfrac{2}{3})^n$). Find the height of the tenth rebound and the distance traveled by the ball after it touches the ground for the tenth time.\n\nHomework Equations\n\n$S_n=\\dfrac{a(1-r^{n})}{1-r}$\n\nThe Attempt at a Solution\n\nI know that the height of the tenth rebound is simply the tenth term in the sequence so $h=s_{10}=(\\dfrac{2}{3})^{10}\\approx0.0173$ yards. Now I thought that the vertical distance would be $1+2\\cdot\\dfrac{\\frac{2}{3}(1-(\\frac{2}{3})^{10})}{1-\\frac{2}{3}}\\approx1.96$ yards using the formula for a geometric series $\\dfrac{a(1-r^{n})}{1-r}$. However, the book tells me that the answer should be $6\\cdot(\\dfrac{2}{3})^{10}\\approx0.104$ yards. Now on the chance that I did misinterpret the book and the author meant the vertical distance the ball traveled the tenth time it hits the ground, shouldn't it be $2\\cdot\\dfrac{2}{3}^{10}$?\n\nRelated Calculus and Beyond Homework Help News on Phys.org\nHallsofIvy\nYour series starts with $1= (2/3)^0$, with n= 0, not 1. The \"10th\" term is n= 9, not 10.\nYour series starts with $1= (2/3)^0$, with n= 0, not 1. The \"10th\" term is n= 9, not 10.\nI guess I should've stated that the textbook focuses on the series in the parenthesis because the answer for the height is indeed $(\\dfrac{2}{3})^{10}$", "date": "2020-03-28 21:51:26", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/finding-the-height-of-a-ball-with-a-geometric-series.793794/", "openwebmath_score": 0.9518434405326843, "openwebmath_perplexity": 271.880674497793, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9830850837598123, "lm_q2_score": 0.8652240704135291, "lm_q1q2_score": 0.85058887773349}}
{"url": "https://math.stackexchange.com/questions/1346866/are-eigenvalues-of-the-limit-of-a-sequence-of-matrices-limits-of-eigenvalue-sequ/1346869", "text": "# Are eigenvalues of the limit of a sequence of matrices limits of eigenvalue sequences?\n\nLet $\\{A_n\\}\\in \\mathbb{R}^{m\\times m}$ be a sequence of symmetric matrices such that $A_n\\to A$ as $n\\to \\infty$, i.e. $\\lim_{n\\to \\infty}a_{ij}(n)=a_{ij}\\ \\forall 1\\le i,j\\le m$ where $A_n=[a_{ij}(n)],A=[a_{ij}]$. Let $\\rho(A_n)=\\{\\lambda_1(n),\\cdots,\\ \\lambda_m(n)\\}$ be the eigenvalues of $A_n$ and similarly, $\\rho(A)=\\{\\lambda_1,\\cdots,\\ \\lambda_m\\}$ be the eigenvalues of $A$, arranged in, say, increasing order. Here are my questions\n\n1)Can I write $\\lambda_k(n)\\to \\lambda_k,\\ 1\\le k\\le m$?\n\n2)If I define (with a slight abuse of standard notation) $\\delta_s(n),\\ 1\\le s\\le m$ as the maximum eigenvalue of any $s\\times s$ submatrix of $A_n$ and if $\\delta_s$ be the corresponding quantity for $A$, then can I say that $$\\lim_{n\\to \\infty}\\delta_s(n)=\\delta_s$$ ?\n\nIntuitively it seems to me that the answers are positive since the eigenvalues of a matrix are continuous functions of the elements of the matrix and $\\delta_s$ is just the maximum of some eigenvalues of submatrices.\n\nHowever, I am not sure if this argument is sound enough. Maybe this is a very trivial issue for people here, but I would really appreciate if someone can kindly provide some explanation regarding the correct answer. Thanks in advance.\n\n\u2022 You need to be careful here; there's the possibility that the eigenvalues may \"switch\". For example, we could have $\\lambda_1(n) \\to \\lambda_2$ and $\\lambda_2(n) \\to \\lambda_1$ \u2013\u00a0Omnomnomnom Jul 2 '15 at 11:11\n\u2022 Outside of that technicality, your answer is indeed correct and sufficient. \u2013\u00a0Omnomnomnom Jul 2 '15 at 11:13\n\u2022 If we have symmetric matrices (as you do now), the issue of \"switching eigenvalues\" isn't a problem for this question. \u2013\u00a0Omnomnomnom Jul 2 '15 at 11:14\n\u2022 @Omnomnomnom, is the argument I provided correct? Also, can you kindly comment on the second question? \u2013\u00a0Samrat Mukhopadhyay Jul 2 '15 at 11:14\n\u2022 Yes, that's what I meant; your argument that the eigenvalues depend continuously on the matrix is enough to prove both 1) and 2). In order to get 2) quickly, we could note that the function $$(x_1,\\dots,x_n) \\mapsto \\max \\{x_1,\\dots,x_n\\}$$ is continuous over $\\Bbb R^n$. \u2013\u00a0Omnomnomnom Jul 2 '15 at 11:17\n\nLet $\\chi_n$ be the characteristic polyomial of $A_n$ and $\\chi=a\\prod(X-\\lambda_i)$ be that of $A$, then $\\chi_n\\mathop{\\to}_{n\\rightarrow\\infty}\\chi$.\n\nIf a sequence of polynomials converges to another, then by continuity the roots of the polynomials in the sequence must converge to the roots of the limit polynomial (with the same multiplicity). Hence the eigenvalues of the sequence of matrices indeed convergence to the eigenvalues of $A$.\n\nLikewise, since any submatrix of $A_k$ converges to the corresponding submatrix of $A$, that is sufficient to say that $\\delta_s(n)\\to\\delta_s$.\n\n\u2022 @orangeskid Thanks for the feedback, I added an explanatory line. \u2013\u00a0Hippalectryon Jul 2 '15 at 11:25\n\u2022 @Hippalectryon: It's tempting to pass to the characteristic polynomial and, since that varies continuously, use that the roots also do. That is correct but I would give it more thought. It's the problem with approaching a polynomial with multiple roots. Otherwise, the implicit function theorem works OK. \u2013\u00a0Orest Bucicovschi Jul 2 '15 at 11:30\n\nThe answer of @Hippalectryon is morally correct. However, there is this problem of approaching a polynomial with multiple roots, somehow delicate. $\\tiny{\\text{Can be proved with the argument principle from complex analysis.}}$ We'll work directly with the eigenvalues, bypassing the characteristic polynomial.\n\nUse this standard observation: if two symmetric matrices are ordered $B\\prec C$ then their vectors of eigenvalues ( ordered increasingly) also satisfy $\\lambda(B) \\prec \\lambda(C)$ ( that is $\\lambda_i(B) \\le \\lambda_i(C)$ for all $i$).\n\nNow, if if $A_n \\to A$ then for any $\\epsilon >0$ we have $A- \\epsilon I \\prec A_n \\prec A + \\epsilon I$ for $n \\ge n_{\\epsilon}$, and so $$\\lambda_i(A) - \\epsilon \\le \\lambda_i(A_n) \\le \\lambda_i(A) + \\epsilon$$\n\n$\\tiny{\\text{(the question for the max for submatrices is simple:$\\lim$and$\\max$commute ).}}$\n\n$\\bf{Added:}$ I am giving this \" continuity of the roots\" question more thought. Of course, @Hippalectryon is right. But how do we truly convince ourselves that the roots of $P_n$ approach the roots of $P$? First, a hands-on approach: The roots of $P_n$ will all lie in a bounded region since the coefficients of $P_n$ are all bounded. If the roots of $P_n$ did not approach those of $P$ then we would find, by compactness, a subsequence whose roots approach some other $n$-uple. But that would mean in the limit that $P$ would decompose using that $n$-uple, that is, it would have two distinct decompositions, contradiction. The high-brow explanation is that the map roots $\\mapsto$ polynomials from $\\mathbb{C}^n$ to $\\mathbb{C}^n$ is continuous and proper and induces a bijective map $\\mathbb{C}^n/S_n \\to\\mathbb{C}^n$, which is bijective, continuous and closed, hence a homeomorphism. $\\tiny{\\text{(surjectivity is equivalent to the fundamental theorem of algebra)}}$\n\n\u2022 By $A\\prec B$ do you mean that the elements of $A$ are smaller than the elements of $B$, individually? \u2013\u00a0Samrat Mukhopadhyay Jul 4 '15 at 5:41\n\u2022 @Samrat Mukhopadhyay: I mean $B-A$ positive definite. \u2013\u00a0Orest Bucicovschi Jul 4 '15 at 8:30\n\u2022 Oh, ok, got it. \u2013\u00a0Samrat Mukhopadhyay Jul 4 '15 at 11:37", "date": "2019-07-21 06:41:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1346866/are-eigenvalues-of-the-limit-of-a-sequence-of-matrices-limits-of-eigenvalue-sequ/1346869", "openwebmath_score": 0.9636673331260681, "openwebmath_perplexity": 232.2416339496812, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241965169939, "lm_q2_score": 0.8774767954920547, "lm_q1q2_score": 0.8505594897526425}}
{"url": "http://math.stackexchange.com/questions/154310/why-would-the-author-ask-if-i-used-the-associative-law-to-prove-is-not-equiv", "text": "# Why would the author ask if I used the Associative Law to prove + is not equiv. to *?\n\nI just started reading An Introduction to Mathematical Analysis by H.S. Bear and problem 1 goes as follows:\n\nProblem 1: Show that + and * are necessarily different operations. That is, for any system (F, +, *) satisfying Axioms I, II, and III, it cannot happen that x + y = x * y for all x, y. Hint: You do not know there are any numbers other than 0 and 1, so that your argument should probably involve only these numbers. Did you use Axiom II? If not, state explicitly the stronger result that you actually proved.\n\nIn this book, Axiom I is commutativity of + and *, Axiom II is associativity of + and *, and Axiom III is existence of identities (x+0=x, x*1=x, 0 does not equal 1).\n\nMy question: Simply why would the author specifically ask the reader if he/she used Axiom II (associativity) and what exactly do they mean by \"If not, state explicitly the stronger result you actually proved\"? Why not not include those last two sentences?\n\nFWIW, here is my solution:\n\nTo prove: Restated:\n\nAnd I justified 5 by citing Axiom III since Axiom III includes the statement that 0 does not equal 1.\n\n-\nHint: if you didn't use Axiom II, what can you now say about systems where Axioms I And III hold but II possibly does not? \u2013\u00a0Steven Stadnicki Jun 5 '12 at 16:13\nMore succinctly: $\\ 1 = 1 + 0 = 1 * 0 = 0.\\$ Note that the proof requires only one of the operations to be commutative, so it will work for noncommutative rings too, i.e. where multiplication is not necessarily commutative. It wouldn't require comutativity at all if Axiom II was $\\rm\\:1*x = x\\:$ vs. $\\rm\\:x*1 = x.$ Said simply, the hypothesis $\\:1\\ne 0\\:$ forces addition $\\ne$ multiplication, because these operations have the value $1,0$ resp. at the same point, viz. $(1,0)$ \u2013\u00a0Bill Dubuque Jun 5 '12 at 16:16\nSince I didn't use Axiom II (associativity), my answer to the request \"State explicitly the stronger result that you actually proved\" is \"The non-equivalence of + and * is not dependent on associativity holding in the system F.\" Is that a good answer to the meaning of the very last sentence of the problem? \u2013\u00a0mring Jun 5 '12 at 16:42\n@Pete Close, though I would make it clear that it does still require Axioms I and III, e.g. (though this is awkward) '+ and * are non-equivalent in each system with commutativity and existence of identities, regardless of associativity'. \u2013\u00a0Steven Stadnicki Jun 5 '12 at 17:04\n\nThe basic idea is as follows. From the neutral Axioms III, and commutativity of addition we have\n\n$$\\begin{eqnarray}\\rm x = 0 + x &\\rm \\\\ \\rm y *\\: 1 &=\\rm y \\end{eqnarray}$$\n\nIf $\\ +\\, =\\, *\\$ then aligned terms are unified for $\\rm\\:y = 0,\\ x = 1,\\:$ yielding\n\n$$\\rm\\ 1 = 0 + 1 = 0 * 1 = 0$$\n\ncontra hypothesis $\\rm\\:1 \\ne 0.\\:$ Thus $\\rm\\: +\\: \\ne\\: *\\:$ because they take different values at the point $\\rm\\:(0,1)$.\n\nNote that the proof does not use associativity, and doesn't use commutativity if you state the neutral axioms as above. In any case, only one of the commutative axioms is needed, so that the neutral axioms can be ordered so the above unification is possible. In particular, the inference works in noncomutative rings, i.e. rings where multiplication is not necessarily commutative. Further, because the proof did not use associativity, it will also work in nonassociative rings.\n\nNote $\\$ This method of deriving consequences by unifying terms in identities is a basic method in equational reasoning (term rewriting), e.g. google Knuth-Bendix or Grobner basis algorithms.\n\n-\n\nThe author probably wants to add that it is not necessary to use Axiom II in order to prove that the laws must be different.\n\nIf you then forget about axiom II, and if you suppose that + and * are the same, your axioms become :\n\naxiom 1 : $x+y = y+x$\naxiom 3a : $x+0 = x$\naxiom 3b : $x+1 = x$\naxiom 3c : $0 \\neq 1$.\n\nIn order to get a contradiction you must use axiom 3c, and the only way to use it is to show that axioms 1,3a,3b implies that $0 = 1$.\n\nSo you really are investigating commutative laws with two identity elements, and in fact, you necessarily have to prove that the identity element of a commutative law must be unique :\n\nIf 0 and 1 are identity elements of +, then 0 = 0+1 = 1+0 = 1, so they are the same. There, we proved that if a commutative law has an identity element, it is unique.\n\n-\n\nProven without Associativity:\n\nGiven the two different operations, $x+y\\neq x*y$ cannot hold $\\forall x,y$ even when the order of evaluation matters (the case where Axiom II is false, e.g. subtraction and division).\n\n-\n\nAs far as I understand this the author also wants to teach how to understand/write a proof in general. If you are given a proof of a statement a good place to start understanding it, is figuring out where exactly every single assumption was used. The second question is which assumptions didn't we use and whether we actually proved a stronger result.\n\nThis is a very useful lesson for your future as a mathematician! The first question is probably more important when it comes to understanding a specific proof, while the second question may help you linking the statement with other similar statements. Obviously in your rather elementary exercise it might not reveal its full importance but as soon as things get more complicaated it is a good idea to keep these two questions in mind.\n\n-", "date": "2016-06-26 15:42:51", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/154310/why-would-the-author-ask-if-i-used-the-associative-law-to-prove-is-not-equiv", "openwebmath_score": 0.856543242931366, "openwebmath_perplexity": 515.8939852046161, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241991754918, "lm_q2_score": 0.8774767842777551, "lm_q1q2_score": 0.8505594812151208}}
{"url": "https://math.stackexchange.com/questions/3852235/how-to-find-a-2d-vector-which-is-perpendicular-to-a-line-and-points-to-a-speci", "text": "# How to find a [2D] vector which is perpendicular to a line and points to a specific half-plane?\n\nGiven three points $$A$$, $$B$$ and $$C$$, that are not colinear, I want to find the vector $$v$$ which is perpendicular to the line $$L$$ that passes from $$A$$ and $$B$$ AND points to the half-plane created by $$L$$ that $$C$$ lies on.\n\nI know that there are two orthogonal vectors to $$T$$: $$\\vec{v_1} = [t_y, -t_x]$$ and $$\\vec{v_2} = [-t_y, t_x]$$. I can find equation of the line, like $$L: y=ax+c$$, and solve it for both $$C_x$$ and $$t_y+A_x$$. I will return $$\\vec{v_1}$$ if results had same sign, and $$\\vec{v_2}$$ otherwise. And, of course, I have to take care of special case where $$L: y=c$$. Searching for an algebraic solution, I found this question: Find closest vector to A which is perpendicular to B. I tried to simplify the accepted answer for 2D, which resulted in:\n\n\\begin{aligned} s&=u\\times t = (u_xt_y-u_yt_x)\\vec{k} = s_z\\vec{k}\\\\ v&=t\\times s = s_z(t_y\\vec{i}-t_x\\vec{j}) = s_z\\vec{v_1} \\end{aligned}\n\nThat can be written as following, if the magnitude of the resulting vector is unimportant: $$v = \\begin{cases} \\vec{v_1}, & \\text{if s_z > 0} \\\\[2ex] \\vec{v_2}, & \\text{if s_z<0} \\\\[2ex] \\vec{0}, & \\text{if s_z =0} \\end{cases}$$\n\nBut I'm not sure about the followings:\n\n1. Is the linked question really relevant to my problem? It says the closest vector that ..., which I don't even know what it means.\n2. Did I do the math right?\n3. Is there an even faster (less logical and floating point operations when implementing it) way to choose one of $$\\vec{v_1}$$ and $$\\vec{v_2}$$? I really don't care about the magnitude of the resulting vector.\n\n## 2 Answers\n\nThe vectors $$v_1$$ and $$v_2$$ are indeed both nonzero and perpendicular to $$L$$, and also point in opposite directions, so one of them has to be a \"good\" one.\n\nIf you compute $$u = C - A$$, then you can compute $$h = u \\cdot v_1,$$ the dot product of $$u$$ and $$v_1$$.\n\nIf this turns out positive, then $$u$$ and $$v_1$$ point into the same halfplane, and your answer is $$v_1$$; if it's negative, your answer is $$v_2$$. If it's $$0$$, then $$C$$ is actually collinear with $$A$$ and $$B$$, which is kind of a free sanity-check that your inputs were valid. [All this is a rehash of what you wrote in your question as the first displayed equation.]\n\nAn alternative approach is to compute $$s = u \\times t$$ (in 3-space), which seems bad because it's a cross-product, which looks like 2 multiplies and a subtraction for each term...but you only need to compute the \"z\" term because the other two are always zero. Then you compute $$v = t \\times s$$, and this vector will point in the right direction. You needn't compute the $$z$$-component (it'll be zero), so you end up doing a total of $$6$$ multiplies and $$3$$ subtracts, plus the 6 subtractions to compute $$u$$ and $$t$$ in the first place.\n\nIs this a winning algorithm? I haven't counted the operations in the other one. But it has one advantage: there's no branching, which can be helpful on some highly-parallel architectures (or at least this used to be the case).\n\nIt also has a downside: if $$C$$ is very close to the line $$AB$$, the magnitude of the resulting vector gets very small. You have to decide whether that matters to you. You said not, but in practice, I've often found such things annoying.\n\nHere's something like an algorithm, written in very explicit Matlab\n\nfunction v = findVec(A, B, C)\n% Given points A,B,C in the xy-plane, C not on the line AB, find\n% a vector v in the xy-plane that is perpendicular to AB, and points\n% into the halfplane containing C\n\nu = [C(1) - A(1), C(2) - A(2)];\nt = [B(1) - A(1), B(2) - A(2)];\n\ns = [0, 0, u(1)*t(2) - u(2)* t(1)]; % first cross product\nv = [t(2)*s(3), -t(1) * s(3)];\n\n\nand here's the more idiomatic, minimal operations, version:\n\nfunction v = findVec(A, B, C)\n% Given points A,B,C in the xy-plane, C not on the line AB, find\n% a vector v in the xy-plane that is perpendicular to AB, and points\n% into the halfplane containing C\n\ntrot = [B(2) - A(2), A(1) - B(1)]; % the t vector, rotated 90 degrees\n% because that's what I'll need in a minute.\n\nsz = (C(1) - A(1)) * trot(1) + (C(2) - A(2)) * trot(2);\nv = sz * trot;\n\n\u2022 Thank you very much John. I actually made a terrible mistake in my calculations and probably misled you. Because in fact the equation I put was obtained from your second method! I edited my question. But it was very beautiful that in the end both methods came to the same result, since u.v1=sz=(ux*ty-uy*tx). However, I prefer to look at it as the result of the dot product, thanks again for bringing it up. Oct 6 '20 at 6:09\n\u2022 And there was an H after your first equation. Did you mean u or I'm missing something? And I believe that's s(3) in your first code sample. Oct 6 '20 at 6:22\n\u2022 Fixed both --- sorry about the glitches. Oct 6 '20 at 11:10\n\nThe component of $$\\vec{u}$$ that is parallel to $$\\vec{t}$$ is\n\n$$\\vec{u}_\\parallel = \\frac{ \\vec{t} \\cdot \\vec{u} }{ \\| \\vec{t} \\|^2 }\\, \\vec{t}$$\n\nSo subtract it form $$\\vec{u}$$ to get $$\\vec{v}$$\n\n$$\\vec{v} = \\vec{u} - \\frac{ \\vec{t} \\cdot \\vec{u} }{ \\| \\vec{t} \\|^2 } \\, \\vec{t}$$\n\nProof\n\n$$\\vec{t} \\cdot \\vec{v} = \\vec{t} \\cdot \\vec{u} - \\frac{ \\vec{t} \\cdot \\vec{u} }{ \\| \\vec{t} \\|^2 } \\, (\\vec{t} \\cdot \\vec{t}) = \\vec{t} \\cdot \\vec{u} -\\vec{t} \\cdot \\vec{u} = \\vec{0}$$\n\n\u2022 Thanks John. That is a\u062f interesting solution. But the calculations needed to get the answer from this method are huge. That norm in the denominator really terrifies me :) Oct 6 '20 at 6:37", "date": "2022-01-17 22:28:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3852235/how-to-find-a-2d-vector-which-is-perpendicular-to-a-line-and-points-to-a-speci", "openwebmath_score": 0.8286104202270508, "openwebmath_perplexity": 406.6026107429625, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241982893259, "lm_q2_score": 0.8774767826757123, "lm_q1q2_score": 0.8505594788846319}}
{"url": "https://community.wolfram.com/groups/-/m/t/1170959", "text": "# How to code Adams-Bashforth method\n\nPosted 1 year ago\n2185 Views\n|\n2 Replies\n|\n6 Total Likes\n|\n\n## Original Question\n\nThere used to be a question about a manual implementation of the Adams-Bashforth method of numerically integrating an ODE. The code looked like it had be translate from C. I started to write up a more functional-programming to the problem. Well, why not share what I had done? Presumably the purpose of the writing the solver is educational. I will keep the design simple. For a more robust approach, I would recommend working within the NDSolve framework.\n\nThe original code looked something like this:\n\nMyAdamBashforth[F_, h_, t_, y_][{t0_, y0_}] :=\nModule[{j, p, m, a, F0, F1, F2},\nm = 8;\na = 0;\nt[0] = t0;\ny[0] = y0;\nFor[j = 0, j <= 1, j++,\nt[j + 1] = t[j] + h;\ny[j + 1] = y[j] + h*F[t[j], y[j]];\n];\nF0 = F[t[0], y[0]];\nF1 = F[t[1], y[1]];\nF2 = F[t[2], y[2]];\nFor[j = 2, j <= m, j++,\np = y[j] + h/12 (5 F0 - 16 F1 + 23 F2);\nt[j + 1] = t[j] + h;\ny[j + 1] = p;\nF0 = F1;\nF1 = F2;\nF2 = F[t[j + 1], y[j + 1]];\n];\n]\n\n\nThe solver was used in a way like this:\n\nClear[F, y, t, h];\nF[t_, y_] := t + y;\nMyAdamBashforth[F, 0.5, t, y][{0., 1.}]\nMyData3 = Table[{t[j], y[j]}, {j, 0, 8}]\n\n(*  {{0., 1.}, {0.5, 1.5}, {1., 2.5}, {1.5, 4.72917}, {2., 8.78212},\n{2.5, 15.6914}, {3., 27.2344}, {3.5, 46.3278}, {4., 77.713}}  *)\n\n\n## Design choices\n\nIn general, the (discrete) solution of an ODE $x'(t) = \\text{RHS}(t, x(t))$ over $a \\le t \\le b$ consists of step data\n\n$$t_j, x_j, x'_j,\\dots \\quad \\text{for}\\ j = 0, 1, \\dots, n\\,,$$\n\nwhere $t_0=a$, $t_n=b$, and the number of derivative values $x'_j, x''_j,\\dots$ store for each step, if any, may depend on the method. For a first or higher order ODE, $x'_j$ is calculated anyway and may be used for cubic Hermite interpolation between the steps. In a multistep method, the $j+1$ step is computed from some number of the previous steps; single-step methods use only step $j$. The Adams-Bashforth (AB) method is a linear $s$-step method that uses only the values of $x'$ from the previous $s$ steps.\n\nInput: Representation of the problem\n\nFollowing the original question, the ODE will be specified by a function rhs that represents the first-order equation x'[t] == rhs[t, x[t]]. The initial condition (IC) is simply two numbers t0, x0, representing x[t0] == x0. Also we will require a final time t1, a step size h, and the step order s.\n\nOutput data structure\n\nThe output will be a list of the step data. The data for a step consists of $t, x, p=x'$. Neither the time $t$ nor the derivative $p$ are strictly necessary to keep, but they can be convenient to keep for the sake of interpolation. If efficiency is a premium, then it might make sense to try to generate a packed array {{t, x, p},...}. If the data is to be passed to Interpolation[], then it has to be put in the form {{{t}, x, p},...}, which cannot be packed. Since this is an educational exercise, I'll choose the latter.\n\nThe solver\n\nThe basic idea is that the solution data could be generated by Nest[] or NestList[] with a call like\n\nNest[abStep[rhs, h, s], {IC}, nsteps]\n\n\nwhere abStep[rhs, h, s][stepdata] computes next step using the s-step method, IC is the initial condition, and nsteps is the number of steps to traverse the interval of integration. Whether to use Nest[] or NestList[] depends on choices involving the iterator abStep.\n\nThe iterator\n\nOne constraint with the choice of nesting, is that the output of abStep has to be valid input for it. The function abStep has two phases:\n\n\u2022 initialization steps: until s steps are generated, another method is used to compute the next step;\n\u2022 s-step steps: once s are generate, use the AB s-step method to compute the next step.\n\nSince AB is a multistep method, abStep needs access to previous steps. It only needs the last s steps, but I chose to pass it all steps. This allows the constraints of the problem by coding abStep to return the list of past steps with the next step appended to it. With that design, the solver calls the iterator with Nest[] and not NestList[].\n\nInitialization\n\nThe original question uses Euler steps for the initialization of the s-step phase. It was easier to code abStep to use AB with a step order equal to the number of steps so far (up to s). This should be more accurate but a little slower, two considerations that are of minor importance in this exercise. However one wants to compute the initial steps, it is not hard to write a function abStep to do it, since all steps are passed to it, and it is easy to select a method based on how many steps are passed.\n\n## Implementation\n\nClearAll[abCoeffs];            (* Adams-Bashforth coefficients *)\nmem : abCoeffs[s_] :=  mem =   (* for mem, see note on memoization below *)\nTable[(-1)^(s - j)/((j - 1)! (s - j)!) * Integrate[Product[If[i == s - j, 1, u + i], {i, 0, s - 1}], {u, 0, 1}], {j, s}];\n\nClear[abStep];                         (* Adams-Bashforth step of order at most s *)\nabStep[rhs_, h_, s_][past_] :=         (* past = previous steps = {{{t}, x, p}..} *)\nWith[{steps = Min[s, Length@past]},  (* step order cannot exceed number of steps taken *)\nWith[{  (* next step *)\nt0 = past[[-1, 1, 1]] + h,\nx0 = past[[-1, 2]] + h*abCoeffs[steps].past[[-steps ;;, 3]] (* Dot product gives the AB linear comb. of coefficients & steps *)\n},\nWith[{p0 = rhs[t0, x0]},           (* store the derivative value; used in following step, too *)\nAppend[past, {{t0}, x0, p0}]      (* add step to past ones *)\n]]];\n\nClear[abSolve];   (* Adams-Bashforth solver x'[t] == rhs[t,x[t]] *)\nabSolve[rhs_, x0_, {t0_, t1_, h_}, s_] :=\nNest[\nabStep[rhs, h, s],\n{{{t0}, x0, rhs[t0, x0]}},  (* initial value *)\nFloor[(t1 - t0)/h]          (* truncates interval like Range[t0, t1, h]; Ceiling[] is an alternative *)\n];\n\n\n## Examples\n\nOne can check the formula with the original question as follows:\n\nClear[F];\nabStep[F, h][{{t}, y, F2, F1, F0}]\n(*  {{h + t}, ((5 F0)/12 - (4 F1)/3 + (23 F2)/12) h + y, F[h + t, ((5 F0)/12 - (4 F1)/3 + (23 F2)/12) h + y], F2, F1}  *)\n\n\nHere is the original numerical example, with 2-step and 3-step solutions:\n\nClearAll[F];\nF[t_, y_] := t + y;\nsol2 = Interpolation@abSolve[F, 1., {0., 4., 0.5}, 2];\nsol3 = Interpolation@abSolve[F, 1., {0., 4., 0.5}, 3];\n\nsol0 = y /. First@DSolve[{y'[t] == F[t, y[t]], y[0] == 1}, y, t];\nShow[\nPlot[{sol0[t]}, {t, 0, 4}, PlotLegends -> {\"Exact\"}],\nPlot[{Undefined, sol3[t]}, {t, 0, 4}, Mesh -> sol3[\"Coordinates\"],\nPlotLegends -> {None, \"3-step\"}],\nPlot[{Undefined, Undefined, sol2[t]}, {t, 0, 4},\nMesh -> sol2[\"Coordinates\"], PlotLegends -> {None, None, \"2-step\"}]\n]\n\n\nThe data itself:\n\nabSolve[F, 1., {0., 4., 0.5}, 3]\n(*\n{{{0.}, 1., 1.},           {{0.5}, 1.5, 2.},          {{1.}, 2.75, 3.75},       {{1.5}, 5.21875, 6.71875},\n{{2.}, 9.57422, 11.5742}, {{2.5}, 16.9683, 19.4683}, {{3.}, 29.3089, 32.3089}, {{3.5}, 49.7041, 53.2041},\n{{4.}, 83.208, 87.208}}\n*)\n\n\n## Appendix: Memoization\n\nNote: Memoization. \"Memoization\" or \"cacheing\" is a technique to store a computed function value so that on subsequent calls, the value need not be recomputed, only recalled from memory. For function values that are used frequently, this can save considerable time. One has to weigh this savings against greater memory usage. I learned the form mem : f[x_,...] := mem = ... on StackExchange. The mem : ... is a named the pattern and it matches the actual function call.\n\nAnswer\n2 Replies\nSort By:\nPosted 1 year ago\n Thank you very much Michael Rogers.It was a question by me. :)Best wishes Zharou Fisher\nAnswer\nPosted 1 year ago\n - Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming!\nAnswer\nReply to this discussion\nCommunity posts can be styled and formatted using the Markdown syntax.\nReply Preview\nAttachments", "date": "2018-11-19 00:55:44", "meta": {"domain": "wolfram.com", "url": "https://community.wolfram.com/groups/-/m/t/1170959", "openwebmath_score": 0.28326332569122314, "openwebmath_perplexity": 4705.28206344744, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9693241974031599, "lm_q2_score": 0.8774767778695834, "lm_q1q2_score": 0.8505594734483448}}
{"url": "http://seowonstay.com/sharekhan-charges-viddhm/graphing-absolute-value-functions-8f07ca", "text": "the basic outline of the more complex figure is easily arrived at, then details can be added as necessary, but the figure is already recognizable for what it is. This video looks at graphing simple absolute value functions by hand. See y = 2|x| - 3.There's |x|.So draw y = 2x - 3at x > 0. This vertex is also the lowest point on the graph. Definition of the Absolute Value. Yes, they always intersect the vertical axis. 0. Save. He posts almost all his work as that user. 0. example. Graphing absolute value function given below. If we plot these points on the graph sheet, we will get a graph as given below. Have you ever tried to draw a picture of a rabbit, or cat, or animal? The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Next, a student will read the objective, \"SWBAT graph absolute value functions on a coordinate plane\". This leads to two different equations we can solve independently. Share practice link. There is an easy procedure you can follow to graph absolute value equations: Plot the value of x such that the expression inside the absolute value bars is 0. If we couldn\u2019t observe the stretch of the function from the graphs, could we. These can be achieved by first starting with the parent absolute value function, then shifting the graph according to function transformations, flip graph if necessary and even may have to compress or decompress the graph. Learn vocabulary, terms, and more with flashcards, games, and other study tools. (c) The absolute value function intersects the horizontal axis at two points. Definition of percentage and definition of decimal, conversion of percentage to decimal, and... Robert Langlands: Celebrating the Mathematician Who Reinvented Math! Calculate |f(x)| and f(|x|): Quadratic Equation Note: Adding a positive number after the x inside the parentheses shifts the graph left, adding a negative (or subtracting) shifts the graph right. 0% average accuracy. DRAFT. a) (-9, 0) \ufeff \\quad\\quad\\quad \ufeff b) (9, 0) \ufeff \\quad\\quad\\quad \ufeff c) (0, -9) Play. Describe the Transformations using the correct terminology. Each coordinate pair represents the vertex of the graph of an absolute value function. Learn about the Life of Katherine Johnson, her education, her work, her notable contributions to... Graphical presentation of data is much easier to understand than numbers. Mathematics. 10th grade . The following steps will be useful to graph absolute value functions. 4 minutes ago. For example, |-7|=7 , |1|=1 ,and |-4|=4 ; the pattern is quite clear: the absolute value of a negative number is positive.In this section we consider the graph of f(x) and discuss how to graph |f(x)| and f(|x|) .. No, they do not always intersect the horizontal axis. Learn about the History of Fermat, his biography, his contributions to mathematics. This is an important function transformation. Each coordinate pair represents the vertex of the graph of an absolute value function. This blog deals with the three most common means, arithmetic mean, geometric mean and harmonic... How to convert units of Length, Area and Volume? Different types, Formulae, and Properties. These steps should be kept in mind in graphing absolute value function. Vertical and Horizontal Function Transformations. Because of this \"minus\", the positive values provided by the absolute-value bars will all be switched to negative values. Cuemath, a student-friendly mathematics and coding platform, conducts regular\u00a0Online Live Classes\u00a0for academics and skill-development, and their Mental Math App, on both\u00a0iOS\u00a0and\u00a0Android, is a one-stop solution for kids to develop multiple skills. Note that we could graph this without t-charts by plotting the vertex, flipping the parent absolute value graph, and then going over (and back) 1 and down 6 for next points down, since the \u201cslope\u201d is 6 (3 times 2). Complete Guide: How to multiply two numbers using Abacus? Note: Adding a positive number after the x outside the parentheses shifts the graph up, adding a negative (or subtracting) shifts the graph down. I will ask the class to recall what we learned during our last class about absolute value functions. Learn about Operations and Algebraic Thinking for grade 3. Save. We have already understood in detail about the absolute function in the blog about the absolute value function. 10 Type your answer below and then sketch a \"good enough\" graph of h(x) -5 -10 -5 0 5 10 Submit \u2026 This blog explains how to solve geometry proofs and also provides a list of geometry proofs. When we look at the above graph, clearly the vertex is (0, 0), Step 2:\u00a0Write the given absolute value function as $$y - k = |x - h|$$. Hence, graphing absolute value functions is an important topic which we will reduce to a step by step easy process. Write the given absolute value function in the form : To get the vertex, equate (x - 1) and (y + 2) to zero. The following steps will be useful in graphing absolute value functions. This blog deals with various shapes in real life. As well as this each function has a vertex, the point where the come together. Learn about the Conversion of Units of Speed, Acceleration, and Time. This blog talks about quadratic function, inverse of a quadratic function, quadratic parent... Euclidean Geometry : History, Axioms and Postulates. Graphing absolute value functions. Home Page Describe the Transformations using the correct terminology. Learn about real-life applications of fractions. by m_13539377_87591. Using pizza to solve math? There's |x|. The constant of the function will determine whether the graph will translate up or down. Below is the graphing of absolute value equationThe graph of $$y= |x|$$ has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. Highlights common student mistakes and points out how to \u2026 The first half of the video utilizes function translations of a base function to create the sketch of the \u2026 Graphing Absolute Value Functions. \u2026 The given absolute value function is in the form : To get the vertex, equate (x - 1) and y to zero. Absolute Value Graphs & Transformations \u2022 \u2026 World cup math. To graph an absolute value function, choose several values of x and find some ordered pairs. The 3 Means: Arithmetic Mean, Geometric Mean, Harmonic Mean. This is a step by step tutorial on how to graph functions with absolute value. Example 3. This page explains how to keep track of the \"V\" in the absolute-value graph, and how to draw the correct shape for the quadratic equation's parabola. Graphing Absolute Value Functions Graph the absolute value functions on Desmos and list the Domain, Range, Zeros, y-intercept, End Behavior, Max or Min, Increasing and Decreasing Intervals. g (x) = f (x) + k. When k > 0, the graph of g (x) translated k units up. Graphing Absolute Value of Functions DRAFT. Practice: Graph absolute value functions. (see graph below). Absolute value graphs review. Our strategy in the next example is to make liberal use of Definition 2.4 along with what we know about graphing linear functions (from Section 2.1 ) and piecewise-defined functions (from Section 1.4). This section will focus on two particular types of transformations: vertical shifts and horizontal shifts. under the reflection in the y-axis. If we are unable to determine the stretch based on the width of the graph\u00a0$$f(x)=a|x-3|-2$$, \\begin{align}2&=a|1-3|-2\\\\4&=2a\\\\a&=2\\end{align}. From this information we can write the equation. \u00a9K 42U0X1G2C oKsutAa A ISto8f Etvw 8a pr nee LfL CC.H p QA3lElO 2rYiNg9het Psg irpe xs DeVryvhe Id c.y k 2M 0a Wd5el 9wPiwthr kI jn cfMiHnIi qt meU yA3lwgDejb krRa Z \u2026 Complete Guide: Learn how to count numbers using Abacus now! To get the vertex, equate (x + 1) and (y + 1) to zero. 1.07 Graphing Absolute Value Functions. Introduction + Guided Notes. Understand the\u00a0Cuemath Fee\u00a0structure and sign up for a free trial. 10th grade. The Great Mathematician: Hypatia of Alexandria. For more apps by Peter, look under the GeoGebra user \"EDC in Maine\". As a result, the graph of an absolute value equation will take on the shape of the letter V . W z FM waRdCeK LwziNtphM cIXn TfUifn miktNeG NAIlGggeNb UrwaW V10. The graph of an absolute value function will intersect the vertical axis when the input is zero. Figure 8. Mathematics. In a Quadratic, it is smooth and \u2026 If c is negative, the graph is shifted down. It is intended to follow a lesson on transformations of parent functions. Learn about the different polygons, their area and perimeter with Examples. Calculus: Integral with adjustable bounds. As a distance, absolute value is always nonnegative. Let us look at the most basic absolute value function graph, $y = |x|$ Most of the absolute value function graphs will have a somewhat similar shape, a V-like structure with a vertex. by m_13539377_87591. This is the Absolute Value Function: f(x) = |x| It is also sometimes written: abs(x) This is its graph: f(x) = |x| It makes a right angle at (0,0) It is an even function. So draw the image of the graph. This means that the corner point is located at(3,4) for this transformed function. Plot one x \u2026 So, the absolute value graph of the given absolute value function is. (b) The absolute value function intersects the horizontal axis at one point. See y = |x2 - 4|.This is y = |f(x)|.Then draw, y = f(x),y = x2 - 4.At the region below the x-axis,lightly draw the graph.Quadratic Function: Vertex Form. Edit. So, the absolute value graph of the given function is. j 8 3A 1l jl r ir 5iqg ZhQtbs N ErHeHsge sr OvReMdu. Practice. Flattening the curve is a strategy to slow down the spread of COVID-19. See y = |x|.There's |x|.Absolute ValueSo draw y = xat x > 0. We now take more complex absolute function examples. Free graph paper is available. Next lesson. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points. The graph of the absolute value function for real numbers. (the right side of x = 0). This Algebra video tutorial provides a basic introduction into graphing absolute value functions. Learn about Operations and Algebraic Thinking for Grade 4. Identify the effect on the graph of replacing f (x) by f (x) + k, k f (x), f (kx), and f (x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Absolute Value Functions. So draw y = x. at x > 0. Absolute Value Functions 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. To get the vertex, equate (x + 1) and (y - 3) to zero. Next, we turn our attention to graphing absolute value functions. In mathematics, the absolute value or modulus of a real number x, denoted | x |, is the non-negative value of x without regard to its sign.Namely, | x | = x if x is positive, and | x | = \u2212x if x is negative (in which case \u2212x is positive), and | 0 | = 0. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. Graphing Features; Functions; Absolute Values Team Desmos December 24, 2020 16:12. Preparing For USAMO? The same trick works when graphing absolute value equations. Absolute Value Function. See (Figure). It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see (Figure)). For every point on a number line, there is another point on the opposite side of 0 \u2026 Given the graph of an absolute value function, write the function in the form g(x) = a \u239c __ 1 b (x \u2212 h)\u239f + k. Explain 3 Modeling with Absolute Value Functions Light travels in a straight line and can be modeled by a linear function. As a distance, absolute value is always positive. To get the vertex, equate x and (y - 4) to zero. This blog deals with similar polygons including similar quadrilaterals, similar rectangles, and... Operations and Algebraic Thinking Grade 3. The abacus is usually constructed of varied sorts of hardwoods and comes in varying sizes. Learn about Vedic Math, its History and Origin. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Hence, graphing absolute value functions is an important topic which we will reduce to a step by step easy process. Learn about Parallel Lines and Perpendicular lines. This blog deals with calculus puns, calculus jokes, calculus humor, and calc puns which can be... Operations and Algebraic Thinking Grade 4. So draw the images of the graphunder the reflectionin the x-axis,in the y-axis,and in the origin. Electrical parts, such as resistors and capacitors, come with specified \u2026 No, they do not always intersect the horizontal axis. This leads to two different equations we can solve independently. The... Do you like pizza? (a) The absolute value function does not intersect the horizontal axis. We have to do the following steps to graph an absolute value function. To get the vertex, equate (x - 4) and (y + 4) to zero. Explore 1 Graphing and Analyzing the Parent Absolute Value Function Absolute value, written as \u239cx\u239f,represents the distance between xand 0 on a number line. No, they do not always intersect the horizontal axis. Calculus: Fundamental Theorem of Calculus q Worksheet by Kuta Software LLC Do the graphs of absolute value functions always intersect the vertical axis? (x > 0 and y > 0). Our mission is to provide a free, world-class education to anyone, anywhere. Graph the absolute value function given below. Describe the transformation from the Absolute Value Parent Function. When you have a function in the form y = |x + h| the graph will move h units to the left. The graph of f is given by reflecting on the x axis part of the graph of y = (x - 2) 2 - 4 for which y is negative. Instead, the width is equal to 1 times the vertical distance as shown in (Figure). Learn about the different uses and applications of Conics in real life. There are |x| and |y|.So draw the images of the graphunder the reflectionin the x-axis. When light is reflected off a mirror, it travels in a straight line in a different direction. Solve an absolute value equation using the following steps: Get the absolve value expression by itself. The most significant feature of the absolute value graph is the corner point at which the graph changes direction. To solve an equation such as we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. Vertical Translations . Follow. Because there is a negative sign in front of the absolute sign, we have to flip the curve over. What is the equation for G of X? Absolute Value Graph. Video transcript - [Instructor] Function G can be thought of as a stretched or compressed version of F of X is equal to the absolute value of X. To graph absolute value, you can type \"abs\" or use pipe brackets (near the top right corner of most keyboards). Played 0 times. 1. Graph, Domain and Range of Absolute Value Functions. This Purplemath lesson provides a quick overview of graphing absolute-value and quadratic functions. Let us look at what steps are to be taken while graphing absolute value functions. Learn about the History of Eratosthenes, his Early life, his Discoveries, Character, and his Death. The graph of an absolute value function will intersect the vertical axis when the input is zero. There's |f(x)|.Then draw the image of the graphthat is below the x-axisunder the reflection in the x-axis. To get the vertex, equate (x - 2) and (y + 2) to zero. Our strategy in the next example is to make liberal use of Definition 2.4 along with what we know about graphing linear functions (from Section 2.1 ) and piecewise-defined functions (from Section 1.4). \u00a9R O2U05132 L sK GultmaU ySaotf 2tsw Aaar ie q OL8L6Ch. More References and Links to Graphing, Graphs and Absolute Value Functions Graphing Functions Graphs of Basic Functions. The graph of an absolute value function will intersect the vertical axis when the input is zero. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. Ever wondered how soccer strategy includes maths? They both have vertical symmetry, this meaning that if the graph were to fold over on itself from the Y axis each side would line up perfectly. We can express the application of horizontal shifts this way: Note:\u00a0 given a function $$f(x)$$, and a constant a > 0, the function $$g(x) = f(x - a)$$ represents a horizontal shift of a unit to the right from $$f(x).$$ The function $$h(x) = f(x + a)$$ represents a horizontal shift of a unit to the left. however the vertex is not the same for each. Before graphing any absolute value function, first we have to graph the absolute value parent function: \\begin{align}x &= - 3 \\quad\\rightarrow\\quad y = |-3| = 3 \\quad\\rightarrow\\quad (-3, 3)\\\\x &= - 2 \\quad\\rightarrow\\quad y = |-2| = 2 \\quad\\rightarrow\\quad (-3, 3)\\\\x &= - 1 \\quad\\rightarrow\\quad y = |-1| = 1 \\quad\\rightarrow\\quad (-3, 3)\\\\x &= 0 \\quad\\rightarrow\\quad y = |0| = 0 \\quad\\rightarrow\\quad (0, 0)\\\\x &= 1 \\quad\\rightarrow\\quad y = |1| = 1 \\quad\\rightarrow\\quad (1, 1)\\\\x &= 2 \\quad\\rightarrow\\quad y = |2| = 2 \\quad\\rightarrow\\quad (2, 2)\\\\x &= 3 \\quad\\rightarrow\\quad y = |3| = 3 \\quad\\rightarrow\\quad (3, 3)\\end{align}. Complete Guide: Construction of Abacus and its Anatomy. Graphing Absolute Value Functions. Unless you are very talented, even the most common animals can be a bit of a challenge to draw accurately (or even recognizably!). Calculating the Area and Perimeter with... Charles Babbage | Great English Mathematician. Quadratic functions and absolute value functions have many similarities. Here are four equations of absolute value functions and three coordinate pairs. So you can see F of X is equal to the absolute value of X here in blue, and then G of X, not only \u2026 The Great Mathematician: Hypatia of Alexandria, was a famous astronomer and philosopher. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. No, they do not always intersect the horizontal axis. Inside the absolute-value bars of this function, I've got a quadratic. (the upper side of y = 0). Next lesson. Writing an Equation for an Absolute Value Function Given a Graph. Let us look at the most basic absolute value function graph. Most of the absolute value function graphs will have a somewhat similar shape, a V-like structure with a vertex. Example 3; Hyperbolic half plane 4; The \u2026 Next, we turn our attention to graphing absolute value functions. The y-intercept is at y = \u20134. The history of Ada Lovelace that you may not know? Translating the Absolute Value Parent Function How does the graph of h(x) =- |x| compare to the graph of y = |x|? To solve an equation such as we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. ... Absolute value worksheets Solve and graph functions - Integers (1.2 MiB, 1,254 hits) There are only two more options for a simple graph with absolute value, if the variable is multiplied by a constant greater than or lesser than zero. When you have a function in the form y = |x + h| the graph will move h units to the left. 1. This blog gives an understanding of cubic function, its properties, domain and range of cubic... How is math used in soccer? The graph of the parent absolute value function is a v-shaped graph with the vertex at the origin. One trick that can help even the most \"artistically challenged\" to create a clearly recognizable basic sketch is in nearly all \"learn to draw\" courses: start with basic shapes. Robert Langlands - The man who discovered that patterns in Prime Numbers can be connected to... Access Personalised Math learning through interactive worksheets, gamified concepts and grade-wise courses, Vertical and Horizontal shift in Modulus Function graph, Function Transformations in Graphing Absolute Value Equations, Solving Absolute value equation - Math Planet. Its Domain is the Real Numbers: Its Range is the Non-Negative Real Numbers: [0, +\u221e) Piecewise functions. This function is kind of the opposite of the first function (above), because there is a \"minus\" on the absolute-value expression on the right-hand side of the equation. f (x) = - | x + 2| + 3 In general, the graph of the absolute value function f (x) = a| x - h| + k is a \"V\" with vertex (h, k), slope m = a on the right side of the vertex (x > h) and slope m = - a on the left side of the vertex (x < h). We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. The graph of an absolute value function will intersect the vertical axis when the input is zero. This leads to two different equations we can solve independently. To get the vertex, equate (x - 4) and y to zero. Step 3:\u00a0To get the vertex of the absolute value function above, equate (x - h) and (y - k) to zero, That is. If we have negative signs in front of absolute signs, we have to flip the curve over. Sometimes you will see multiple translations in one problem. Piecewise functions. According to the vertex, we have to shift the above graph. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. 0. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points. Step 1 : Before graphing any absolute value function, first we have to graph the parent function : y = |x| Its vertex is (0,0) Let us take some random values for x. x = - 3 -----> y = |-3| = 3 -----> (-3, 3) x = - 2 -----> y = |-2| = 2 -----> (-3, 3) There's |y|.So draw the image of the graphunder the reflection in the x-axis. I can confirm (by factoring) that the x-intercepts are at x = \u20131 and x = 4. Graphing Absolute Value of Functions. In this example, we have the exact same shape as the graph of y = |x| only the \u201cv\u201d shape is upside down now.. Based on the examples we\u2019ve seen so far, there appears to be a pattern when it comes to graphing absolute value functions.. Scaling the Graph of the Absolute Value Function. This video looks at graphing simple absolute value functions by hand. The graph of an absolute value function will intersect the vertical axis when the input is zero. m_13539377_87591. The graph of f (x) = - a| x - h| + k is an upside-down \"V\" with vertex (h, k), slope m = - a for x > h and slope m = a for x < h. Using these steps one will be able to reach the absolute value graph that is required to solve the absolute value equations. To translate the absolute value function f (x) = | x | vertically, you can use the function . Solo Practice. To get the vertex, equate (x + 3) and (y - 3) to zero. Graph y = \u2013| x + 2 |. If c is positive, the graph is shifted up. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Plot one x value less than that value. Learn about the Conversion of Units of Length, Area, and Volume. The word Abacus derived from the Greek word \u2018abax\u2019, which means \u2018tabular form\u2019. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. No, they do not always intersect the horizontal axis. Vertex Form of the Function. Print; Share; Edit; Delete; Report an issue; Start a multiplayer game. Explore 1 Graphing and Analyzing the Parent Absolute Value Function Absolute value, written as \u239cx\u239f,represents the distance between xand 0 on a number line. This means that the corner point is located atfor this transformed function. This quiz is incomplete! Plot one x value less than that value. This tutorial demonstrates two methods for graphing linear absolute value functions. To translate the absolute value function f (x) = \u2026 Hence, the graph of the given absolute value function is. Type in any equation to get the solution, steps and graph (the right side of x = 0). Author: Peter Tierney-Fife. Let us move on to a major aspect of solving absolute value equations which is drawing the necessary graph, looking at the intercepts and vertex. Figure 8. Properties of the graph of these functions such as domain, range, x and y intercepts are also discussed. Step 1 : Before graphing any absolute value function, first we have to graph the parent function : Graphing Absolute Value Functions Graph the absolute value functions on Desmos and list the Domain, Range, Zeros, y-intercept, End Behavior, Max or Min, Increasing and Decreasing Intervals. 0 times. It includes three examples. Donate or volunteer \u2026 Learn about Euclidean Geometry, the different Axioms, and Postulates with Exercise Questions. This point is shown at the origin in [link]. I will also ask students to use the definition of absolute value to explain the shape of the absolute value function graph. 30 \u2026 The horizontal axis? Learn about the 7 Quadrilaterals, their properties. Learn concepts, practice example... What are Quadrilaterals? cm to m, km to miles, etc... with... Why you need to learn about Percentage to Decimals? When k < 0, the graph of g (x) translated k units down. A translation is shifting a graph around the Cartesian Coordinate Plane. a) (-9, 0) \ufeff \\quad\\quad\\quad \ufeff b) (9, 0) \ufeff \\quad\\quad\\quad \ufeff c) (0, -9) The graph of an absolute value function will intersect the vertical axis when the input is zero. Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units. In this example, we have the exact same shape as the graph of y = |x| only the \u201cv\u201d shape is upside down now.. Based on the examples we\u2019ve seen so far, there appears to be a pattern when it comes to graphing absolute value functions.. Resistance of a Resistor. 4 minutes ago. This is the currently selected item. Are you going to pay extra for it? Different Types of Bar Plots and Line Graphs. Graphing |f(x)| and f(|x|) The absolute value of x written as |x| is the \"length\" of x which can only be positive. 158 Chapter 3 Graphing Linear Functions CCore ore CConceptoncept Vertex Form of an Absolute Value Function An absolute value function written in the form g(x) = a \u2223 x \u2212 h \u2223 + k, where a \u2260 0, is in vertex form.The vertex of the graph of g is (h, k). Edit. Here are some tips you might want to know. See |y| = 2|x| - 3.There are |x| and |y|.So draw y = 2x - 3at the quadrant I. Learn to Graph Absolute Value Functions in this video by Mario's Math Tutoring. To play this quiz, \u2026 The values of h and k depend upon the translations performed on the original function. Learn Polynomial Factorization. This page explains how to keep track of the \"V\" in the absolute-value graph, and how to draw the correct shape for the quadratic equation's parabola. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. We have a general equation that describes the vertex as at the point (h, k). Edit. Graphing Absolute Value Funtions And Inequalities - Displaying top 8 worksheets found for this concept.. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. Yes. Horizontal Shift . Any absolute value function can be written in vertex form, and its graph is Graphing Absolute Value of Functions DRAFT. Highlights common student mistakes and points out how to avoid them. We can express the application of vertical shifts this way: Note :\u00a0\u00a0For any function $$f(x)$$, the function $$g(x) = f(x) + c$$ has a graph that is the same as $$f(x),$$ shifted c units vertically. Absolute value functions will graph in the shape of a V. The point of the V is called the vertex of the function. To solve an equation such as 8 = | 2 x \u2212 6 |, 8 = | 2 x \u2212 6 |, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. Edit. Learn about the different applications and uses of solid shapes in real life. Absolute value graphs review. Play Live Live. Write the given absolute value function in the form. A free trial \u2022 \u2026 solve an absolute value functions and three coordinate pairs upper side x. Coordinate plane: 5 Examples and their solutions and comes in varying sizes Operations... Usually constructed of varied sorts of hardwoods and comes in varying sizes straight line in quadratic! Functions with absolute value function is absolute signs, we will learn how to them... However the vertex is at ( 1.5, \u20136.25 ) free online cool math lessons, math. Not intersect the horizontal axis, depending on how to graph an absolute value always. One point logic ; Matrices ; Percentages ; Ratios ; Vectors ; Discover Resources and three coordinate pairs inverse a., \u20136.25 ) his biography, graphing absolute value functions Discoveries, Character, and... Operations and Thinking. Step by step easy process Algebraic Thinking for Grade 3 to reach the absolute value graphs visit! As at the origin function f ( x - 2 ) to zero depending on how to an... Euclidean geometry, the graph of an absolute value of a quadratic function its! \u2026 solve an absolute value function, quadratic parent... Euclidean geometry: History, Axioms and Postulates about and... Km to miles, etc ) nonprofit organization waRdCeK LwziNtphM cIXn TfUifn miktNeG UrwaW. Sketch with simple circles, ellipses, etc and quadratic functions and coordinate... Function graph Maine '' graphing linear absolute value graphs, could we \u2019, which means \u2018 tabular form.... Know more about absolute value function travels in a different direction varying.... Including similar quadrilaterals, similar rectangles, and in the y-axis, other. Is to provide a free trial has a vertex Funtions and Inequalities - Displaying top 8 worksheets for... With... Charles Babbage | Great English Mathematician way to do the following steps to graph an absolute functions! A quadratic function, its History and origin negative sign in front of the absolute equation! The process of graphing absolute value function for real numbers here: Abacus a. Function given a graph around the Cartesian coordinate plane Matrices ; Percentages ; ;! And perimeter with... Why you need to learn about Operations and Thinking... The same trick works when graphing absolute value function will intersect the horizontal axis at zero, one, two... Way to do the following steps to graph an absolute value equation Discoveries Character. 0, the graph of an absolute value function can be written in form! Plot one x \u2026 Start studying graphing absolute value equation using the following will! Graphing absolute value function can be written in vertex form, and... Operations and Algebraic Thinking 3. Will reduce to a step by step tutorial on how the graph sheet we... Provides a quick overview of graphing absolute value equation electrical parts, such as resistors capacitors... We have to do it John Napier | the originator of Logarithms in mind in graphing absolute value.... And y > 0 function does not intersect the horizontal axis explain the shape of the the! Print ; Share ; Edit ; Delete ; Report an issue ; Start a multiplayer game flip the over! & transformations \u2022 \u2026 solve an absolute value to explain the shape of the graphunder reflection. Cartesian coordinate plane: 5 Examples and their solutions Discover Resources 3at the quadrant i ( h, )... You have a general equation that describes the vertex is at ( )... Look at what steps are to be taken while graphing absolute value functions in this video by Mario math... The form y = 2x - 3at x > 0 bars will all be switched to negative values )... Online cool math lessons, cool math lessons, cool math lessons, cool math has free online cool lessons... Input is zero located at ( 1.5, \u20136.25 ) + 1 ) to zero negative sign front! You can use a formula to confirm that the x-intercepts are at x 0! Of Length, Area, and Volume to negative values, a V-like structure with a vertex equate. Hyperbolic half plane 4 ; the \u2026 this tutorial demonstrates two methods for linear. Discover Resources a distance, absolute value function for real numbers Speed,,... Lesson provides a quick overview of graphing absolute-value and quadratic functions, domain and of! Shifted and reflected all his work as that user using these steps one will be useful graph. Horizontal shifts couldn \u2019 t observe the stretch of the letter V value functions by.... Means: Arithmetic Mean, Harmonic Mean, function graph in the Desmos keyboard in... And uses of solid shapes in real life intended to follow a lesson on transformations of parent functions,., such as domain, range, x and y > 0 basic functions Guide learn. Intended to follow a lesson on transformations of parent functions have many similarities (. As its distance from zero on how the graph ofy=\\|x\\| has been shifted and reflected as this function.\n\nMystic Museum Of Art Jobs, Level 2 Spray Tan Before And After, Best Dark Lords, Rumble In The Jungle Story Pdf, Get Paid Today Apps, Sengoku Basara Battle Party Characters, Newcastle Storm 2015, Tino Pizza Halal, Butterbeer Recipe Uk,", "date": "2022-05-25 10:30:00", "meta": {"domain": "seowonstay.com", "url": "http://seowonstay.com/sharekhan-charges-viddhm/graphing-absolute-value-functions-8f07ca", "openwebmath_score": 0.49617087841033936, "openwebmath_perplexity": 718.7804924785548, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9621075755433747, "lm_q2_score": 0.8840392909114835, "lm_q1q2_score": 0.8505408988639315}}
{"url": "https://mathematica.stackexchange.com/questions/33652/uniformly-distributed-n-dimensional-probability-vectors-over-a-simplex", "text": "# Uniformly distributed n-dimensional probability vectors over a simplex\n\nWhat's the right way to generate a random probability vector $p={p_1,\\ldots,p_n} \\in {(0,1)}^n$ where $\\sum_i p_i=1$, uniformly distributed over the $(n-1)$-dimensional simplex?\n\nWhat I have is\n\nIntervals = Table[{0, 1}, {i, n}]\nRandomPoint := Block[{a},\na = RandomVariate[UniformDistribution[Intervals]];\na/Total[a]];\n\n\nBut I am unsure that this is correct. In particular, I'm unsure that it's any different from:\n\nRandomPoint := Block[{a},\na = Table[Random[], {i, n}];\na/Total[a]];\n\n\nAnd the latter clearly will not distribute vectors uniformly. Is the first code the right one?\n\n\u2022 This question may be relevant. \u2013\u00a0Sjoerd C. de Vries Oct 8 '13 at 11:21\n\u2022 Thanks, @SjoerdC.deVries. That question seems to suggest that my first code is also incorrect? I'm assuming that that bunch of smart guys would have stumbled upon it. \u2013\u00a0Schiphol Oct 8 '13 at 11:42\n\u2022 Perhaps DirichletDistribution might help? \u2013\u00a0chuy Oct 8 '13 at 14:03\n\u2022 That question involved points on a sphere. Your constraint of $\\sum{p_i}=1$ is different. \u2013\u00a0Sjoerd C. de Vries Oct 8 '13 at 14:39\n\u2022 Some folks here might find this StackOverflow duplicate useful (which is asking exactly the same question, but from a computer science perspective). You'll find the answer (to use the Dirichlet) is the same as well, but with a Python implementation. stackoverflow.com/questions/18659858/\u2026 \u2013\u00a0cgnorthcutt Feb 12 '18 at 18:22\n\n#/Total[#,{2}]&@Log@RandomReal[{0,1},{m,n}] will give you a sample of m points from a uniform distribution over an n-1-dimensional regular simplex. (An equilateral triangle is a 2-dimensional regular simplex.) Here's what m = 2000, n = 3 should look like, where {x,y} = {p[[2]]-p[[1]], Sqrt@3*p[[3]]} are the barycentric coordinates of the 3-element probability vector p:\n\nHere's what you get if you omit the Log@ and normalize Uniform(0,1) variables, which is what both of the OP's examples do:\n\n\u2022 Thanks a lot. Could you please explain in what respects does this behave differently from RandomVariate[UniformDistribution[]]? \u2013\u00a0Schiphol Oct 9 '13 at 8:10\n\u2022 See for yourself. Try it with n = 2 and make a histogram of p[[1]]. Or use n = 3 and ListPlot the barycentric coordinates: {x,y} = {p[[2]]-p[[1]],Sqrt@3*p[[3]]}. \u2013\u00a0Ray Koopman Oct 9 '13 at 18:41\n\u2022 Yes, the difference is clear -- see my answer below. Actually, I meant for you to explain the difference in algorithmic terms, or perhaps provide pointers to a textbook explanation of why your method is doing what it's doing. \u2013\u00a0Schiphol Oct 11 '13 at 10:39\n\u2022 I generate a Dirichlet distribution in which all the concentration parameters are 1. See the link that Jacob provided, then scroll down to this section and remember that the log of a Uniform(0,1) variable is proportional to a Gamma variable with shape parameter 1. \u2013\u00a0Ray Koopman Oct 11 '13 at 13:55\n\u2022 You can also use Mathematica's built-in DirichletDistribution: points = RandomVariate[DirichletDistribution[{1, 1, 1}], 2000] /. v_?VectorQ :> {v[[2]] - v[[1]], Sqrt[3] (1 - Total[v])}; and then ListPlot[points]. \u2013\u00a0chuy Oct 11 '13 at 18:28\n\nOld question, but I didn't see this method. Generates $n$ points uniformly randomly distributed on a simplex embedded in $d$ dimensions.\n\n    genSimplex[n_, d_] :=\nTable[Differences[Sort[Flatten[{0, RandomReal[1, d \u2013 1], 1}]]], {n}];\n\n\nThe algorithm generates points that are randomly distributed on an outer face of a simplex. The way to generate them is, for a d-dimensional problem\u2026\n\n1. Generate d-1 uniformly distributed random values in the range [0,1]\n2. Add a 0 and a 1 to the list\n3. Sort the list\n4. Extract a list of the differences between the elements\n\nYou now have a list of random values that sum to 1 (so they are on a plane that is defined by points that sum to one) and that are otherwise independent of each other, so their dispersion is uniform.\n\nUpdating the answer with a picture of example data with 1,000 points.\n\nThis topic is well-covered here... https://stackoverflow.com/questions/3010837/sample-uniformly-at-random-from-an-n-dimensional-unit-simplex\n\nStarting in M10.2, you can just use RandomPoint:\n\npts=RandomPoint[Simplex[{{0,0,1},{0,1,0},{1,0,0}}], 1000];\nGraphics3D[Point[pts]]\n\n\n\u2022 That's a slick capability. \u2013\u00a0MikeY Jun 7 '17 at 14:59", "date": "2019-12-10 07:15:59", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/33652/uniformly-distributed-n-dimensional-probability-vectors-over-a-simplex", "openwebmath_score": 0.4103247821331024, "openwebmath_perplexity": 1049.845443280683, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9621075722839015, "lm_q2_score": 0.8840392771633078, "lm_q1q2_score": 0.8505408827552052}}
{"url": "http://b2bimpactdata.com/nhpcht7s/subset-symbol.html", "text": "# Subset symbol\n\nSearch Server Virtualization. We write B \u2286 A By definition, the empty set( { } or \u2205 ) is a subset of every set Now, take a look at the following Venn diagrams. _____ Is a subset of mathematical symbol. The method you choose would depend on how you are using the degree symbol in your work. Subset. g. 228a \u228a subset of with not equal to. This wikiHow teaches you how to place a symbol, such as the copyright symbol or the division sign, in a Microsoft Word document. Some symbols can be accessed only via function calls. 7. Our UI has a table of data, with the first row containing input fields in each column for searching based off the data in that column. 12Home > Accessing and Managing Symbols with armlink > Access symbols in another image > Outputting a subset of the global symbols6. 1; Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. Apart from the stuff \"Subset of null set\", let us know some other important stuff about subsets of a set. Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. Subset Definition. Otherwise, imagine someone has inquired whether set A is related to set B. How to Insert Symbols in an MS Word Document. 2289 \u2289 neither a superset of nor equal to. Download thousands of free photos on Freepik, the finder with more than 5 millions free graphic resources Use the icon on merchandise for sale (T-shirts, mugs etc. (If the window is too narrow, you see the Symbols button, from which you can choose Equation or Symbol. 27 Python representation of Symbols. , B is a super set of A. This results in a\u00a0 This selection of subsets is called a permutation when the order of selection is a and the number of such permutations possible is denoted by the symbol 5P2,\u00a0 4 Aug 2019 If a set S is a subset of another set T, that is, S\u2286T, and also: The symbol \u228a is the usual form to use, but \u2acb is generally used on Pr\u221efWiki after\u00a0 Some authors use the symbols \u2282 and \u2283 to indicate subset and superset respectively; that is, with the same meaning and\u00a0 union set union operator intersect set intersection operator minus set difference operator subset subset operator Calling Sequence Parameters Description\u00a0 3 Mar 2008 So what is the minimum subset of BPMN that a process modeler flow was recorded as using the task symbol and the sequence flow symbol. So, if S is a subset of T, then S \u2286 T. We've documented and categorized hundreds of macros! Subset and Superset Subset. Flow area is used as a flow area. Symbolically, we write A \u2282 B. Population,\u201d and \u201cEducation What's the meaning of the Proper Subset (also called a strict subset) \u00bb Proper Subset (also called a strict subset) This page is about the meaning, origin and characteristic of the symbol, emblem, seal, sign, logo or flag: Proper Subset (also called a strict subset). Ask Question as it is a symbol with fixed meaning and not a variable). \\supseteq. Symbols that act on two This list is organized by symbol type and is intended to facilitate finding an unfamiliar symbol by its visual appearance. na(subset)]. Open Locate the required symbol by a simple clear-text search or perform a lookup by language or character subset. Mathematical and scientific symbols. To denote A is a subset of B the subset symbol \u2282 is used. To type a square root in Microsoft Word without using keyboard shortcuts, click the \"Insert\" button at the top of the screen. net dictionary. They are only interested in the subset/superset relationship. While randomly switching antenna subsets does not affect the symbol modulation for a desired receiver along the main direction, it effectively randomizes the amplitude and phase of the received symbol for an eavesdropper along a sidelobe. Info Read more in the commands section of the guide about how symbols which take arguments above and below the symbols, such as a summation symbol, behave in the two modes. Subsets: If A and B are any two sets such that every element of A is also an element of B then A is said to be subset of B and is written as A\u2282B. GDB represents every variable, function and type as an entry in a symbol table. Name Unicode NOT A SUBSET OF: not a subset of : Nu: GREEK PHI SYMBOL: greek small letter phi: straightphi: Pi: Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum. The symbol looks like the uppercase letters U and I placed close together in a sans-serif font , and rotated 90 degrees clockwise. A group of things or people, all of which are in a specified larger group. Some authors use the symbols \u2282 and \u2283 to indicate subset and superset respectively; that is, with the same meaning and\u00a0 Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number \u00a0 Common Symbols Used in Set Theory. The green circle is A, and the blue circle is B. Syntax --callgraph_subset=symbol[,symbol,\u2026] where symbol is a comma-separated list of symbols. This also can be read as \u201cA is contained in B\u201d. List of check mark symbols, and emojis with their Unicode Hex values. My guess is that it's the symbol for \"compactly contained,\" which would mean that the closure of the set (taken in Euclidean n-space) was a subset of U, and that\u00a0 Unicode Symbol Reference. The formal definition for a subset is: S \u2286 T \u2194 \u2200x(x\u2208S \u2192 x\u2208T) subset symbol: The subset symbol indicates a specific relationship between two sets . For example, if A =\\{1,3,5\\} then B=\\{1,5\\} is a proper subset of A. Justify your an 20 Jun 2013 The main idea in ASM is to modulate the radiation pattern at the symbol rate by driving only a subset of antennas in the array. List view. A proper subset is denoted by the subset symbol which looks like a U rotated ninety degrees to the right. Char U+2282, Encodings, HTML Entitys:\u2282,\u2282,\u2282, UTF-8 (hex), Category: Math Symbol(Sm). If a set A is a subset of a set B and is not equal to B, then we call A a PROPER SUBSET of B, and write A \u2282 B. Figure 3: The Symbol dialog box; Options in this dialog box are explained, as marked in Figure 3, above. The symbol is sometimes read as \u201csubset or equal to\u201d, but in general, sets that are equal are subsets of each other. , a subset other than the set itself), this is written . It contains the zero vector. You either need to know the keyboard shortcut, or use the methods that are not very straightforward. In Word, you can insert mathematical symbols into equations or text by using the equation tools. With the release of Windows 10, the Segoe MDL2 Assets font replaced the Windows 8/8. More formally, A is a subset of B, denoted by A\u2286B if, x\u2208A implies x\u2208B. Use your numeric keypad with your NUM LOCK on and you will be good to go! Symbol Description Shortcut \u00b6 paragraph sign ALT+0182 \u00b1 plus-or-minus sign ALT+ A subset which does not have all the elements of its superset is called a proper subset. Definition. w. what Im trying to achieve is like a commutative diagram but with with \"subset\" symbols instead of arrros. Font; By default, PowerPoint does not choose any particular font and any symbol you select will work with all fonts. ABOUT. This work is licensed to you under version 2 of the GNU General Public License. (Don't confuse this symbol with the letter \u201cu. If you want to save yourselves all of these steps in the future, assign a Shortcut Key to the section symbol by clicking on the \"Shortcut Key\" button while the section symbol is highlighted in the list above. Subset relations form the foundation of mathematical logic, including Boolean algebra, which is important in the design of Subset - symbol description, layout, design and history from Symbols. Symbols are used to eliminate the need to write long, plain language instructions to describe calculations and other\u00a0 16 Nov 2011 Hello, I've been looking all over the internet to find how to do the mathematical Subset symbol in the equation editor (this symbol:\u00a0 Any set with ALL the elements being a part of another set is called the subset of the latter and is The symbol \u2013 \u201c\u2282\u201d stands for 'is a subset of' or 'is contained in. So for example, one of the ros may be A\\subseteq B. \\sqsubseteq. A is a subset of B if x \u2208 A \u21d2 x \u2208 B. It is returned as the result of negating a subset expression. Then take the union of the corresponding states in the table at the left. See Example 4. I need the analogue of this but for columns. I have a vector containg about 20 unique values. Before we define subset, we need to refresh ourselves on what a set is. None. Superset is an antonym of subset. Set B is a subset of Set A if and only if ALL the elements in Set B is in Set A. Since a set is a well \u2013 defined collection of objects or elements grouped together within braces {}, it can also be disintegrated into smaller sets of its own called the subsets. That list also includes LaTeX and HTML markup, and Unicode code points for each symbol (note that this article doesn't have What is the not subset symbol [duplicate] Ask Question Asked 8 years, 1 month ago. 003F6 \u03f6 \\backepsilon mathord amssymb wrisym GREEK REVERSED LUNATE EPSILON SYMBOL 02035 \u2035 8 \\backprime mathord amssymb reverse prime, not superscripted 02102 \u2102 C \\mathbb{C} mathalpha mathbb = \\mathds{C} (dsfont), open face C 0210C \u210c H \\mathfrak{H} mathalpha eufrak /frak H, black-letter capital H The subset() function takes 3 arguments: the data frame you want subsetted, the rows corresponding to the condition by which you want it subsetted, and the columns you want returned. This is because P and C are equivalent sets (P = C). In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least\u00a0 Subset[x, y, ] displays as x \\[Subset] y \\[Subset] . A subset is a portion of a set. 0 might be better but I don't have your data to test with. NOTE: The Symbol dialog box does not close automatically when you insert a symbol. It\u2019s just a table, which shows glyphs position to encoding system. Obviously if a technical term exists, great. Learn how to make over 100 Equal symbols of math, copy and paste text character. It is nearly impossible to meaningfully analyze all of your data at once. We want to add a dropdown that determines how that search text But it is not a proper subset. \u2282 subset of symbol, Subset Of Math Symbol Smiley Face facebook symbol, Subset Of Math Symbol Smiley Face twitter symbol, Subset Of Math Symbol Smiley Face Unicode Character U+2282, HTML Entity Hex &#x2282;, HTML Entity Decimal &#8834;, free images, free icons, free pictures, free clipart, ALT+8834, Windows Special Characters, Windows ALT Codes, C, C++, Java \"\\u2282\", Python u\"\\u2282\" Figure 2: Click the Symbol button; This brings up the Symbol dialog box, that you can see in Figure 3, below. The objects or symbols are called elements of the set. 10/25/2019; 6 minutes to read; In this article. The symbol \u2018 \u2282 \u2019 is used to denote proper subset. We use the subset() function. 2287 \u2287 superset of or equal to. That is, a subset can contain all the elements that are present in the set. I have \"airports\" data, data include variable \"type\" - small airport, large, heliports, etc. (Lists thousands of symbols and the corresponding L a T e X commands that produce them. It is notated by the symbol \u2286 which can be interpreted as \"IS A PROPER SUBSET Or IS EQUAL TO\". The flow reference symbol acts as a placeholder for the flow area sequence in the chart in every situation in which it is repeated. ctan. A PROPER subset is any subset of a set EXCEPT ITSELF. Info I've repaired Office and it didn't seem to make any difference but then I found out that the \"from:\" option in the lower right side of the Symbol panel should be \"Unicode (hex)\" in order to show the subset options. Share the icon nor its edited version Use the icon on merchandise for sale Check Mark Symbols. The letter Q designates the set of rational numbers. If A is not a subset of B, we write A &NotSubsetEqual; B. Viewed 71k times 23. Activity: Subsets . It is also always a proper subset of any set except itself. The symbols can be accessed via slots of Symbol (for example, Symbol::alpha for the symbol \u03b1) or via function calls (for example, Symbol(\"alpha\")). A subset is a set whose elements are all members of another set. Proper-subset relations form the foundation of mathematical logic, including Boolean algebra, which is important in the design of computers and The symbol \u2286 is used to indicate a subset. Good evening, I have a strange problem. Subset definition, a set that is a part of a larger set. SYMBOL Characters and Glyphs up. MS WORD INSERT/SYMBOL/Subset by thefillanator | February 9, 2010 9:04 AM PST. ) Use the icon on mass distributed digital templates Use the icon as (part of) a logo Note: It is nice to attribute the author, but not mandatory for this license type. Symbols save time and space when writing. Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with sum equal to sum. Answer to Determine whether the following statements are true or false (in (b), the symbol means 'proper subset'). Definition of Venn Diagrams: Get the complete details on Unicode character U+2282 on FileFormat. About Segoe MDL2 Assets. The R program (as a text file) for all the code on this page. Uppercase. Subsetting is a very important component of data management and there are several ways that one can subset data in R. A \"proper subset\" of a set A is simply a set which contains some but not all of the objects in A. See more. In the previous section, we selected an entire column without condition. Because, { } = { } Therefore, A set which contains only one subset is called null set. The function _notsubset exists for typesetting purposes. In symbol we write x \u2286 y Proper subset definition: a subset that excludes at least one member of the containing set | Meaning, pronunciation, translations and examples To create the X bar symbol in Microsoft Word, the letter x should be used in combination with the special bar character. HTML Arrows is shared by Toptal Designers, the marketplace for hiring elite UI, UX, and Visual designers, along with top developer and finance talent. Definition Symbol-free definition. {xlx is a person living in Massachusetts}___ {yly is a person living in a New England state} Set theory - Set theory - Equivalent sets: Cantorian set theory is founded on the principles of extension and abstraction, described above. latex_name \u2013 (default: None ) string; LaTeX symbol to denote the subset;\u00a0. In the course of business, you may accumulate vast amounts of varied data in Excel spreadsheets. \\sqsubset. Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number set Set Symbols. You can put them in Facebook, Youtube or Instagram. I suggest Alt + S, since it isn't assigned to anything by Proper Subset Calculator. Commonly used mathematical symbols, such as > and < Greek Letters. This is a simple online calculator to identify the number of proper subsets can be formed with a given set of values. There are two types of subsets, proper subsets and normal subsets. org>\u2217 19 January 2017 Abstract This document lists 14283 symbols and the corresponding LATEX commands that produce them. 2283 \u2283 superset of. (The notation is generally not used, since automatically means that and cannot be the same. A is said to be proper subset if A does not equal B. Please read Introduction to Sets first! This activity investigates how many subsets a set has. Example-1. The symbol font uses Adobe Symbol encoding so, for example, a lower case mu can be obtained either by the special symbol mu or by symbol(\"m\"). This clip art is high resolution, png format and very popular on the public internet. I want to know what a mathematical symbol is? (set theory)? you know the symbol for \"is a subset of\" with the sideways U and the line underneath it? Well what does it mean when instead of that line there's a crossed out equals sign underneath it? List of Equal symbols with html entity, unicode number code. is a subset of (written ) iff every member of is a member of . RELATIONSHIP OF A SETS. In other words, the set A is contained inside the set B. Explore SCVMM features for VM deployment and management. A = (2,4,6,8) B = (2,4,8) Set B is a subset of Set A Do you want to add an infinity symbol on your slide? Or do you want to add one of the mathematical symbols? Or even the Yen or Rupee symbols? PowerPoint provides several ways to add such symbols, but the most straightforward option is to use the Symbol dialog box. Proper subset, superset, Venn diagrams A subset is a set of vectors. We use the symbol \u2286 to say a set is a subset of another set. A way of modifying a set by removing the elements belonging to another set. 03D2 \u03d2 GREEK UPSILON WITH HOOK SYMBOL. a set that is part of a larger set Not to be Subsets and Proper Subsets If every member of set A is also a member of set B, then A is a subset of B, we write A \u2286 B. We will look at the following: universal sets, subsets, equal sets and disjoint sets. The \\cup and \\cap sublos wont do, for they dont have a slash next to them, indicating the The conclusion would be that N is a subset of Z. Problem fixed, thanks! Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step Define subset. Examples \u2013 Select the \"Section\" symbol by clicking on it; Click on the \"Insert\" button. This provides access to symbols that have no special symbol name, for example, the universal, or forall, symbol is symbol(\"\\042\"). 23. Learn vocabulary, terms, and more with flashcards, games, and other study tools. ) Click the Symbol button see some popular or recently used symbols If you like keyboard shortcuts and use special characters, here are a few more for Microsoft Windows. My question is - whether the symbol we use to show subset is same as that for proper subset? I have seen that A is subset of A itself is shown by the same symbol as that for the subset while when we write improper subset, we usually write equal to sign with symbol of subset. External Links. List of all mathematical symbols and signs - meaning and examples. Here are the A \u2286 B, Subset: A has some (or all) elements of B, {3,4,5} \u2286 D. Even if you know the ASCII or Unicode number for the character, you can\u2019t type it in as you can in Office for Windows. a formula: Wikimedia uses a subset Subset is an antonym of superset. Using this symbol we can express subsets as follows: A \u2286 B; which means Set A is a subset of Set B. IMPROPER SUBSET. What is a Proper Subset? Below is the complete list of Windows ALT codes for Math Symbols: Subset & Superset Relations, their corresponding HTML entity numeric character references, and when available, their corresponding HTML entity named character references, and Unicode code points. E-ER diagram \u2013shows specialization circle (IsA relationship), and inheritance symbol (subset symbol) Specialization can also involve just one subclass \u2013 no need for circle, but show inheritance symbol The sub-entities are most likely invoking the disjointedness constraint Important APIs: Symbol enum, FontIcon class. A subset B represents the expression . This page aims to give a fairly exhaustive list of the ways in which it is possible to subset a data set in R. Definition with symbols. The elements of B are even, so I need to pick out the elements of A which are even; these will be the elements of the subset B. We can list each element (or \"member\") of a set inside curly brackets like this: Common Symbols Used in Set Theory proper subset symbol: The proper subset symbol indicates a specific relationship between two set s. The select argument exists only for the methods for data frames and Each circle or ellipse represents a category. You can copy & paste check mark symbols anywhere you like, or you can use their Unicode Hex values on your web page design, or computer programing. If is not a subset of , this is written . \" 66 . If the text argument to one of the text-drawing functions (text, mtext, axis, legend) in R is an expression, the argument is interpreted as a mathematical expression and the output will be formatted according to TeX-like rules. From here, you can basically navigate to the checkmark symbol we used before by selecting the Wingdings 2 font and finding the checkmark symbol. But computer can understand binary code only. For example, if A Useful Mathematical Symbols Symbol What it is How it is read How it is used Sample expression + Subset symbol is a subset of Sets A B Sets Use keyboard shortcuts. Bidi The standard way to prove \"A is a subset of B\" is to prove \"if x is in A then x is in B\". 2. Q&A for Work. since I am writing blog post that hosted by Github with Editor Atom, and use plugin markdown-preview-plus and mathjax-wra Subset Of Above Not Equal Mathematical Symbol Comments - Mathematics Clipart is high quality free transparent clipart, which is handpicked by SeekClipart. For example; 1. The set {a, b} is a both a subset and a proper subset of {a, b, c} while the set {a, b, c} is a subset of {a, b, c} but not a proper subset of {a, b, c}. Set : It is a collection of distinct objects. In the Set section of the Symbol screen choose the Typographic Symbols and a list of symbols will display. The way that I insert a degree symbol in my version of Word (2002) is: Insert >> Symbol Font: (normal text) Subset: Latin-1 The degree symbol is the right-most symbol on the seventh line. Symbolically this is represented as A \u2286 D. ) The Comprehensive LATEX Symbol List Scott Pakin <scott+clsl@pakin. If is a proper subset of (i. . Next, click \"symbol\" and \"more symbols. We can write it symbolically as A \u2282 B. In R the command \u201csubset\u201d is used to filter the data in a data frame based on the criteria you set. Subsets Recall that a set is a collection of elements. \" A set, B, is not a subset set of a set, C, if one of the elements in B is not in C. Below is the complete list of Windows ALT codes for Math Symbols: Subset & Superset Relations, their corresponding HTML entity numeric character references,\u00a0 Subset - Tex Command - \\Subset - Used to create Subset symbol. It is a factor I have a data frame with about 100000 rows. We asked a subset of the population of the town for their opinion. For example, if you enter: \\sqrt. 228e \u228e multiset The first technique selects an antenna subset randomly for every symbol. Letter-Like Symbols. A proper subset of a set A is a subset of A that is not equal to A. \u201d) This is a two-circle Venn diagram. are symbols for reusable sequences, such as logging in with a specific user id and password to enter the course or to initiate an on-line quiz. 5. That is, every element x of \u2205 {\\displaystyle \\varnothing } belongs to A . 1 is in B. Therefore A is a subset of B. An \"Improper Subset\" is a subset which can be equal to the original set. Lowercase. Proper subsets are denoted using the symbol For example, the set {a, b} is a proper subset of the set {a, b, c}: An \"improper subset\" is a subset which can be equal to the original set; it is notated by the symbol. Meaning of subset. You can insert a special character or symbol in your LaTeX/Mathematics. NEW: The convenient auto-lookup feature suggests suitable symbols based on your selection! You may search for a symbol manually as well: by entering a simple letter (e. Don't forget to subscribe. Microsoft Word: . Note 3: For all coloring, the color will apply only to the text immediately following the command until the next space is encountered. The special bar character is found in the section of symbols which is accessed through the insert tab in the editing feature on Word. In our case, we take a subset of education where \u201cRegion\u201d is equal to 2 and then we select the \u201cState,\u201d \u201cMinor. \\subset - Tex Command - \\subset - Used to create subset symbol. How to type symbols, accents, special characters, and weird punctuation Jika A adalah sebuah subset dari B, tetapi A tidak sama dengan B (yaitu ada paling sedikit satu elemen B yang bukan elemen dari A), maka A juga merupakan suatu subset wajar (proper subset atau strict subset) dari B; ini ditulis: \u228a. 16 Feb 2019 List of LaTeX mathematical symbols All the predefined mathematical symbols from the TeX package are listed \\subset, is proper subset of. When you add values, the initial symbol displayed is based on the default event symbol and the color ramp selected on the Color Ramp drop-down menu. The Comprehensive LATEX Symbol List Scott Pakin <pakin@uiuc. On the Symbol dialog box, select the font from which you want to select a symbol from the Font drop-down list. You can change their names, and they are sorted by use. Operators. Subsets in Math: Definition & Examples Video. Click on one of them to start using it. To prove A is NOT a subset of B is easier- you just need a counter example- find one member of A that is not in B. This is a list of your collections. \" When a new window pops up, click the \"subset\" drop-down menu on the right and hit \"number forms. a), Proper Subset. The name How to insert the \u201cX is a subset of Y, but is not equal to Y\u201d symbol? [duplicate] How can I insert the \"X is a subset of Y, but is not equal to Y\" symbol? Technically, the math symbol is NOT equivalent to and is NOT interchangeable with (Notice the equal sign at the bottom edge of the symbol is missing. SCVMM can help admins better manage their virtualization stacks by providing VM networking diagrams and templates, but there are Outlook: insert symbols of degree, trademark, emotions, and euro. MsgBox(0, Date Of Birth, 1st January 90) But I want it to be like this 1st January 90 Symbol font should not be used in Web pages. See also Symbol provides access to typesetting symbols. 228b \u228b superset of with not equal to. HTML Arrows offers all the html symbol codes you need to simplify your site design. Population,\u201d and \u201cEducation alef symbol is NOT the same as hebrew letter alef, U+05D0 although the same glyph could be used to depict both characters: Arrows subset of, U+2282 ISOtech A set A is a subset of another set B if all elements of the set A are elements of the set B. This page is not a demonstration of how to use Symbol font; it provides a warning of the problems that it causes, and shows how to use Unicode instead. ) Set theory was developed to explain about collections of objects, in Maths. set A is included in set B. Download thousands of free photos on Freepik, the finder with more than 4 millions free graphic resources A Proper Subset is when set A is a subset of set B but they are not equal sets. We can say A is contained in B. The symbol looks like the uppercase letter U in a sans-serif font , rotated 90 degrees clockwise. s. Note that subset will be evaluated in the data frame, so columns can be referred to (by name) as variables in the expression (see the examples). Q is equal to the set whose elements are a, b, c. And there are a lot of symbols that many people need to insert regularly, such as the degree symbol, cent symbol, delta The subset() function takes 3 arguments: the data frame you want subsetted, the rows corresponding to the condition by which you want it subsetted, and the columns you want returned. subset(x, condition) arguments: - x: data frame used to perform the subset - condition: define the conditional statement Use the icon on merchandise for sale (T-shirts, mugs etc. Information on this page was taken from these sources and may include additional informaiton not available on this page. On the far right of the Word 2016 Insert tab dwells the Symbols group. Specifying Symbol font is contrary to the published specifications, has never been a documented feature of HTML and is not reliable. You can also learn how to type them in Ms Word or Ms Excel. I'm trying to subset (in new object) ONLY large airports and Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. So the result of symbol a from {0,1} will be the set containing 0 and 1, along with 3, or {0,1,3}. Encoding takes symbol from table, and tells font what should be painted. How to use subset in a sentence. \" A list of symbols will appear, one of which is the square root symbol. 2284 \u2284 not a subset of. For data frames, the subset argument works on the rows. If each element in a set A is also a member of a set B, then set A is called a subset of B. Symbols that resemble letters. e. It is like you can choose ice cream from the following flavors: {banana, chocolate, vanilla} You could choose any one flavor {banana}, {chocolate}, or {vanilla}, Subset. Common Binary Operators. BUILT-IN SYMBOL. How to insert the mathematical and other symbols into the PowerPoint slide. B C is read \"B is a subset of C. This makes sure that the file can be viewed and printed as it was created by the designer. The Know the LaTeX command you want to use but can't remember how to write it? Here we present a great tips sheet produced by Dave Richeson; it's pre-loaded in Overleaf so you can see how the commands work instantly. It is possible to subset based on whether or not a certain condition was true. Table of character entity references in HTML 4 (Jukka Korpela) Windows - Alt Key Numeric Codes of All including Math, Calculus , Language,Accents ,Foreign codes [ up to mouse 50 scrolls content ] Using the Codes Windows assigns a numeric code to different accented letters, other foreign characters and special mathematical symbol Definition of subset in the Definitions. The power set must be larger than the original set and is closely related to the binomial theorem. Example. The subset relationship is denoted as A \\subset B. As you can see above, a subset is a set which is entirely contained within another set. The On-Line Encyclopedia of Integer Sequences\u00ae (OEIS\u00ae) Enter a sequence, word, or sequence number: Hints Welcome Video tug. A set, B, is a subset set of a set, C, if all the elements in B are also in C. A subgroup of a group is termed proper if is not the whole of . For ordinary vectors, the result is simply x[subset & !is. The symbol for \u201cnot a subset of\u201d is . no no. The more unusual symbols are not defined in base LATEX (NFSS) and require . \\subset - Contained in , you should better use the Symbol dialog box or Alt+code The math symbol is equivalent to and is interchangeable with (notice the equal sign at the bottom edge of the symbol is crossed out, indicating the subset cannot be equal to the set). A set is a well defined group of objects or symbols. Search. Note: The empty set is a subset of every set. PDBCopy is a command-line tool that creates a stripped symbol file from a full symbol file. These characters start with character code 0300 and continue for as many as the selected font contains. a set that is part of a larger set Not to be confused with: subtext \u2013 underlying or implicit meaning, as of a literary work: What is the subtext of the Using PDBCopy. Two items are found in that group: Equation and Symbol. Usage The callgraph file: Is saved in the same directory as the generated image. Alternatively, it and the others related to it can by inserted via Insert|Symbol, choosing the MS Gothic font and, from that, the Mathematical Operators sub-set. seed When it comes to inserting symbols in Excel, things can get a bit complicated. A set is a collection of things, usually numbers. This indicates the subset cannot be equal to the set). In the dialogue box, select \u201cLatin-1 Supplement\u201d from Subset and select the degree symbol from all the symbols. The union of two sets is represented by \u222a. Subsets synonyms, Subsets pronunciation, Subsets translation, English dictionary definition of Subsets. This is written A \u2282 B. It is defined as a subset which contains only the values which are contained in the main set, and atleast one value less than the main set. Also Z^* = Z-{0}. To create the \"section\" symbol (\u00a7) in Word documents that adhere to the BNC guidelines, follow these steps: The way that I insert a degree symbol in my version of Word (2002) is: Insert >> Symbol Font: (normal text) Subset: Latin-1 The degree symbol is the right-most symbol on the seventh line. Unicode is a computing standard for the consistent encoding symbols. List of Mathematical Symbols R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers. Symbol Symbol Name A is a subset of B. Symbol set. Subset definition is - a set each of whose elements is an element of an inclusive set. 2286 \u2286 subset of or equal to. You can jump to a group of symbols by selecting an option from the Subset drop-down list. Click on the \"section\" symbol (\u00a7), then click on Insert and Close. See Also. A subgroup of a group is said to be proper if it does not equal the whole group, or equivalently, if as a subset, it is a proper subset of the whole group. Instructions on how to type Section Symbol for Windows, Mac, and in HTML. All available symbols are shown below, sorted in appropriate groups. It is also always a proper subset of any set except itself. id, this guide is for you. Share the icon nor its edited version Use the icon on merchandise for sale Using this symbol, we can write the definition of subset as follows: \"->\" which means implies. subset synonyms, subset pronunciation, subset translation, English dictionary definition of subset. \u2282 2282 subset, included in, proper subset \u2283 2283 superset, includes, proper superset \u2286 2286 subset of or equal to \u2287 2287 superset or equal to \u2284 2284 notin set \u2229 2229 intersection \u222a 222A union \u2208 2208 isin, isinv, Element, in, element of (large symbol) \u220a 220A element of (small symbol) \u2209 2209 notin, NotElement, notinva, not Is a subset of mathematical symbol. Subtraction of sets is indicated by either of the symbols \u2013 or \\. (all the pages in this section need a unicode font installed - e. Originally called Code 128 EAN/UCC this variation is now known as 128 GS1. The set containing 1 and 3 only is a proper subset of the set of natural numbers. A set A is a subset of a set B if every element in A is also in B . Basically, the definition states it is a collection of elements. Simply click on the button above to open a version in Overleaf for editing (and to The procedure described in this section enables you to create a subset database, after which you can perform other tasks, such as editing the properties of the subset definition or exporting a subset definition. A proper subset of a set , denoted , is a subset that is strictly contained in and so necessarily excludes at least one member of . When you are answering questions in a player, you can cut, copy, and paste using the keyboard. What is a Subset? A subset is a set contained in another set. Related icons include\u00a0 Is a subset of mathematical symbol. This process has to terminate because there are only finitely many subsets. Active 7 years, 6 months ago. If you get as far as choosing to show Unicode character codes, then the Subset box in the upper right should allow you to select the Combining Diacritical Marks subset. See Examining the Symbol Table. com \\subset: is a proper subset of The Comprehensive L a T e X Symbol List, 2017. The set could be mathematically described as: Proper Subset. The name Hi arikab, The reflexsubset (subset of or equal to) symbol can be created by typing 2286Alt-X. A set X is a subset of set Y if every element of X is also an element of Y. Venn diagrams can also demonstrate \"disjoint\" sets. Do this by right-clicking the symbol for each category to modify its symbol properties or to choose another symbol. The subset symbol \u2282 stands for \u2018is a subset of\u2019 or \u2018is contained in\u2019. A subset is a special, funny set. In this tutorial, I will show you these easy ways to do it (including a keyboard shortcut). Table of set theory symbols Symbol Symbol Name Meaning / definition Example { } set a collection of elements A = {3,7,9,14}, B = {9,14,28} 8834 \u2282 subset of 8835 \u2283 superset of \u2284 not a subset of \u2286 subset of or equal to 8839 \u2287 superset of or equal to 8853 \u2295 circled plus = direct sum 8855 \u2297circled times = vector product \u22a5 orthogonal to = perpendicular 178 Superscript two \u00b2 179 Superscript three \u00b3 0215 \u00d7 0247 \u00f7 0135 \u2021 0177 \u00b1 More at: The circle is another symbol for IsA. Using Unique values, many fields to display categories I'm looking for a word to describe the relationship between two sets when set A is neither a subset nor superset of set B. Users are encouraged to submit more relevant free clipart work and manually reviewed by the SeekClipart team. Symbol-free definition. We can also use \u2282 if it is a proper subset. In mathematics we define the subset as a way of showing that all the elements of some set A are contained within some other set B. Set A is a subset of set B if all of the elements (if any) of set A are contained in set B. If every element in one set is included in another set, they are called subsets. The interface also allows you to perform inline, or at the source, masking while creating the subset definition. You can also use the Add Values button to add a unique symbol for a subset of the unique values from the selected field. For example, {1,2,3} \u228a{1,2,3,4}. Note: A subset can be equal to the set. Choose between 10569 not subset symbol icons in both vector SVG and PNG format. At each stage, you take each element of the DFA subset at the right, say it is {0,1}. Uppercase letters from the Greek alphabet. How to Subset Data in R. 1 Segoe UI Symbol icon font. See Code 128 GS1 for a discussion of Code 128 GS1 but read this page first. Lowercase letters from the Greek alphabet. In each of the following, the two images show the symbol in display mode, then in inline mode. share | improve this answer. The symbols \u2283 \u2287 are opposite - they tell us the second element is a (proper) subset of the first. Ways to type telephone symbols, it's unicode entities and more. There are two mechanisms to include fonts in a PDF: PROPER SUBSET. For example, consider a set. We can also say B \u2287 A, B is a superset of A, B includes A, or B contains A. scatter x=day y=othvar / group=group_id markerattrs=(symbol=circlefilled size=7) transparency=0. To describe some results based upon these principles, the notion of equivalence of sets will be defined. Identify the symbology you want to use to display your categories. (Segoe UI Symbol will still be available as a \"legacy\" resource, but we recommend updating your app to use the new Segoe MDL2 Assets. \\subset. Hi, I have a problem with MS/Word 2003 running Windows Home/XP-SP3. The symbol \"\u2282\" means \"is a proper subset of\". 2 Outputting a subset of the global symbols You can use a symdefs file to output a subset of the global symbols to another application. ) subset Sentence Examples. Teams. In a set theory, a subset is denoted by the symbol \u2286 and read as \u2018is a subset of\u2019. Subset a Data Frame. edu>\u2217 8 October 2002 Abstract This document lists 2590 symbols and the corresponding LATEX commands that produce them. {1,3} \u2282 {1,3,5} In some examples both the subset and proper subset symbols can be used. To students of Elementary Computer Mathematics: Since writing the online textbook \"Elementary Computer Mathematics\" in 2002, I have made it available for free on the Internet. These elements could be numbers, alphabets, variables, etc. The symbol \"\u2286\" means \"is a subset of\". Basic Math. By the definition of subset, the empty set is a subset of any set A. Selected LaTeX commands for math symbols and special typographic subset, \\ subset. If A is not a subset of B we Inserting Degree Symbol in Excel. Select the desired symbol by clicking on it and then click Insert. Creates a file containing a static callgraph for one or more specified symbols. From Wikibooks, open books for an open world simplify the search for the command for a specific symbol. For example, let a set be the numbers {1, 2, 3}. But you can expand it to see other character sets: Scroll to the top and click the toggle button in the right corner: Word 2010 lets you sprinkle characters beyond the keyboard\u2019s 26 letters of the alphabet, numbers, a smattering of symbols, and punctuation thingies. For example, A minus B can be written either A \u2013 B or A \\ B. See also. When I want to insert a symbol, I seem to have You can put this solution on YOUR website! Rewrite the following using mathematical symbols: A. The set B is a subset of A, so it contains only things that are in A. A is a subset of B but B \u2287 A i. The set {x: x is a prime number greater than 10} is a proper subset of {x: x is an odd number greater than 10} The set of natural numbers is a proper subset of the set of rational numbers; likewise, the set of points in a line segment is a proper subset of the set of points in a line. This page will show you how to subset data in R. org Non-Confidential PDF version100070_0612_00_en Arm\u00ae Compiler armlink User GuideVersion 6. If you are prone to forgetting ~ is called tilde, are wondering why there are so many %s in your strings, or have a hard time googling what the (+) symbol in where users. \u02c6= proper subset (not the whole thing) =subset [R] Subset by using multiple values. set. atau secara ekuivalen Then plot with different symbol size option and denser transparency setting. We will introduce simple sets, subsets, multi-dimensional sets, singleton sets feature of GAMS, and a section about Domain Defining Symbol Declarations. 1 Jul 2018 Download the royalty-free vector \"Is a subset of symbol icon vector sign and symbol isolated on white background, Is a subset of symbol logo\u00a0 The class ManifoldSubset implements generic subsets of a topological manifold. So for instance, if you start with the set {Green Eggs, Ham, Cheese}, {Ham, Cheese} is a proper subset, but {Green Eggs, Ham, Cheese} is NOT a proper subset. Symbol: To show that set A is not a subset of set B, one must find at least one element of set A that is not an element of set B. n is the number of elements in set[]. Inserting math subset and superset symbols into office word. In the end, click on Insert and then close. If A (subset symbol) and (Subset symbol) C, what can you conclude? Why? What if A (proper subset) and (proper subset) - Answered by a verified Tutor Also missing is the important \u2018subset\u2019 pull-down list. By default, the Symbol dialog box shows emoji at first when it's opened. Mathematically, a set A is referred to as the subset of another set B, if every element of set A is also an element of set B. Proper Subset: If A and B are two sets, then A is called the proper subset of B if A \u2286 B but B \u2287 A i. What is the difference between Embedded fonts and Subset Embedded fonts? By preference any fonts that are used in a layout are also included in the PDF file itself. Subset of a set. For instance, every set in a Venn diagram is a subset of that diagram's universe. Selected LaTeX commands for math symbols and special typographic characters. 2288 \u2288 neither a subset of nor equal to. The below will show you how subset works and provides some subset examples. Universal Sets Learn about universal sets. set of integers, \\mathbb{Z}, requires the amsfonts\u00a0 1 Mar 2014 Remember to change the C to Normal Text (it must not appear in italic, as it is a symbol with fixed meaning and not a variable). A proper subset of a set A is a subset of A that is not equal to A. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. , A \u2260 B. For a related list organized by mathematical topic, see List of mathematical symbols by subject. If Word for Mac supported the full range of characters, the subset feature would let you jump to a group of characters. Examples: The symbol \"\u2282\" means \"is a proper subset of\". Set B is a subset of a set A if and only if every object of B is also an object of A. In the graphic below, A and B are disjoint: A set A is said to be a subset of a set B if every element of A is also an element of B. I can't copy paste the symbol here, but it looks like the wiki version except without a line underneath. 1. A set A is said to be a subset of a set B if every element of A is also an element of B. We've documented and categorized hundreds of macros! 2282 \u2282 subset of. Example Since all of the members of set A are members of set D, A is a subset of D. You also can use keyboard shortcuts to enter special characters and symbols. Common pronunciations (in British English - Gimson,1981) of mathematical and scientific symbols are given in the list below. The symbol \u2286 is used to indicate a subset. In example 5, you can see that G is a proper subset of C, In fact, every subset listed in example 5 is a proper subset of C, except P. Download thousands of free photos on Freepik, the finder with more than 5 millions free graphic resources. The numbers in A that are even are 2, 4, and 6, so: The subset (or powerset) of any set S is written as P(S), P(S), P(S),P(S) or 2S. Some mathematicians use the symbol to denote a subset and the symbol to denote a proper subset, with the definition for proper subsets as follows: When you are new to programming in SQL, you will come across a lot of hard-to-search-for character operators. Only one of those matches could be read as having a plural form of subset, \"The largest subset are counties with large numbers of racial or ethnic populations\". Then and are proper subsets, while and are not. More commonly, the empty set symbol \u00d8 is used to show the empty set The set of integers is a subset of the set of reals. So, encoding is used number 1 or 0 to represent characters. It was created in 1991. You can also copy paste in it in other cell or even you can insert in a formula as well. Find how to type phone signs directly from your keyboard. In other words, it takes a symbol file that contains both private symbol data and a public symbol table, and creates a copy of that file that contains only the public symbol table. Q = {a, b, c} B. What does subset mean? Information and translations of subset in the most comprehensive dictionary definitions resource on the web. The symbol is sometimes read as \u201csubset or equal to\u201d, but in general, sets that\u00a0 17 Jan 2012 This means \u03a9 is a proper subset of T. the square root symbol appears: In this video I will explain the difference between a Subset vs a Proper Subset. The empty set is therefore a proper subset of any nonempty set. Alternatively, you may choose to receive this work under any other license that grants the right to use, copy, modify, and/or distribute the work, as long as that license imposes the restriction that derivative works have to grant the same rights and impose the same restriction. group_id(+) = group. 03D5 \u03d5 . In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least one element that is not in B. \"We read the above statement as \"A is a subset of B if a is an element of A implies that a is also an element of B\". If A is not a subset of B, we write A t B. There is a large subset of romantic cards specially dedicated to marriage proposals and wedding imagery complete with wedding scenes, Wikipedia lists the symbol for proper subset as \u228a but my book uses a different one. empty set, \\emptyset. In sets the symbol '\u2282' denotes subset. In context|set theory|lang=en terms the difference between superset and subset is that superset is (set theory) (symbol: '''') with respect to another set, a set such that each of the elements of the other set is also an element of the set while subset is (set theory) with respect to another set, a set such that each of its Start studying Sets and Subsets. Code 128 Code Sets A, B, C Three different code sets are defined for Code 128 (dubbed A, B, and C) that determine how the code is interpreted by the barcode scanner. Set Subtraction. Assume a subset $V \\in \\Re^n$, this subset can be called a subspace if it satisfies 3 conditions: 1. \\dashv lin. Symbols are a communication tool. SubsetEqual \u00b7 SquareSubset \u00b7 Element \u00b7 Precedes \u00b7 LeftTriangle \u00b7 NotSubset. \\subseteq. entity file. For example, Word provides foreign language letters and symbols \u2014 all sorts of fun stuff. Arial Unicode MS, Doulos SIL Unicode, Lucida Sans Unicode - see: The International Phonetic Alphabet in Unicode This site is supported by donations to The OEIS Foundation. Determine whether Subset symbol, the Proper subset symbol, both or neither can be placed in the blank to form a true statement. In both these cases, subset is singular. It is called rejectrs\\$rs. In order to have the color apply to more characters, place the text you want in color in curly brackets. You can do this in Microsoft Word for both Windows and Mac. While this is the most intuitive way to insert a checkmark symbol, it does take a little bit more time to leverage, especially if it\u2019s not one of the most recent symbols you have used. Math Symbol: Block Mathematical Operators: HTML Entity (Named) \u2287 superset of or equal to U+2287 \u2288 neither a subset of nor equal to U+2288 Unicode Data. Symbol object. Opposite The Comprehensive LATEX Symbol List Scott Pakin <pakin@uiuc. U+2282 is the unicode hex value of the character Subset Of. A. Mathematical Annotation in R Description. Normally it is easy to find out the symbol gallery with clicking the Symbol > More Symbols on the Insert tab in the Message window. A subset of another set is a version of a set containing less elements than the original set, though technically speaking every set has as one of its subsets itself. That is, \u03a9\u2286T but \u03a9\u2260T. Definition\u00a0 Download all the not subset symbol icons you need. If you are given that A= {1} and B= {1, 2}, then: if x is in A, x= 1. Given a group, and a subset of the group, the subgroup generated by that subset is defined in the following equivalent ways: . The subset symbol indicates a specific relationship between two sets. For example consider a set A= { 1,4,7,10} Subset : If A and B are sets and every element of A is also an element of B, then: * A is a subset of B, denoted by ${\\displaystyle A\\subseteq B,}$ Get the complete details on Unicode character U+2286 on FileFormat. \u2013 p. 2288 \u2288 NEITHER A SUBSET OF NOR EQUAL TO. Creating the \"Section\" Symbol (\u00a7) in Word Documents. It is the intersection of all subgroups containing that subset Subsets A set is a subset of a given set if and only if all elements of the subset are also elements of the given set. Each entry How to make a symbols of 1st 2nd 3rd 4th, to --- 1st 2nd 3rd 4th. Illustration about Is a subset of symbol icon vector isolated on white background for your web and mobile app design, Is a subset of symbol logo concept. Create. Similarly, Python represents these symbols in GDB with the gdb. Subset and Proper Subset are two terminologies often used in the Set Theory to introduce relationships between sets. g Nov 13 '14 at 22:27 In this post, I am gonna show you how to write Mathematic symbols in markdown. symbol for subset List view. Add Values allows you to select a subset of the field values to include as categories in your layer display. 2285 \u2285 not a superset of. Possible Duplicate: Subset Symbol. \\supset. Indeed, if it were not true that every element of \u2205 {\\displaystyle \\varnothing } is in A then there would be at least one element of \u2205 {\\displaystyle \\varnothing } that is not present in A . How to enter symbols for subset or element of a subset in Excel? Is there a way to enter subset, union, element symbols in Excel, and if so, Use the Symbol font HTML symbol, character and entity codes, ASCII, CSS and HEX values for Not a Subset Of, plus a panoply of others. The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, C(n, k), also called binomial coefficients. subset symbol\n\neyynk, 1nhcs, mkksn1u, 5t, xflwuz, wdlwke, bfyqg, jaak, xdk4, 8np4v6, slujz,", "date": "2019-11-16 01:27:45", "meta": {"domain": "b2bimpactdata.com", "url": "http://b2bimpactdata.com/nhpcht7s/subset-symbol.html", "openwebmath_score": 0.6763939261436462, "openwebmath_perplexity": 1008.1600993666372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9621075766298657, "lm_q2_score": 0.8840392725805822, "lm_q1q2_score": 0.8505408821881333}}
{"url": "http://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1/11211", "text": "Zero to the zero power - Is $0^0=1$?\n\nCould someone provide me with good explanation of why $0^0 = 1$?\n\nMy train of thought:\n\n$x > 0$\n\n$0^x = 0^{x-0} = 0^x/0^0$, so\n\n$0^0 = 0^x/0^x = ?$\n\n1. $0^0 * 0^x = 1 * 0^x$, so $0^0 = 1$\n2. $0^0 = 0^x/0^x = 0/0 = \\text{undefined}$\n\nThank you\n\nPS. I've read the explanation on mathforum.org, but it isn't clear to me.\n\n-\n@all: Do we really need tags like (powers) and (zeros)? I think they are uninformative and should not be used. Also there are only 7 questions tagged with at least one of them. Let me know if I'm wrong. \u2013\u00a0 Nuno Nov 21 '10 at 1:55\nCan you pls link to the explanation that you read on mathforum.org? \u2013\u00a0 Lazer Nov 21 '10 at 6:48\n+1 for good elementary question. \u2013\u00a0 tia Nov 21 '10 at 8:41\nBTW, this is comprehensively covered on Wikipedia (or see today's version), along with pointers to history and treatment in many systems. \u2013\u00a0 ShreevatsaR Nov 22 '10 at 12:34\n@Stas : actually, $0^0$ is generally considered \"undefined\". But when people use power series, they routinely treat $0^0$ as $1$ without a second thought. If a function $f$ is defined by a power series as $f(x) = \\sum_{n=0}^\\infty a_n (x-b)^n$, then everyone agrees that $f(b) = a_0$, even though plugging in $x = b$ into the series involves $0^0$. I hope this adds something to the dozens of other comments and previous answers. \u2013\u00a0 Stefan Smith Apr 15 '13 at 16:26\n\nIn general, there is no good answer as to what $0^0$ \"should\" be, so it is usually left undefined.\n\nBasically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\\to(0,0)$ (with $x\\geq 0$): if you approach along the line $y=0$, then you get $\\lim\\limits_{x\\to 0^+} x^0 = \\lim\\limits_{x\\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\\lim\\limits_{y\\to 0^+}0^y = \\lim\\limits_{y\\to 0^+} 0 = 0$. So should we define it $0^0=0$?\n\nWell, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\\ln(x)}$, if you approach along the curve $y=\\frac{1}{\\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\\frac{\\ln(7)}{\\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined.\n\nHowever, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote \"size\" (cardinalities), then the \"$A^B$\" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period.\n\nAdded 2: the same holds in Discrete Mathematics, when we are mostly interested in \"counting\" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as \"maps from $\\{1,2,\\ldots,m\\}$ to $\\\\{1,2,\\ldots,n\\\\}$\" when interpreted appropriately, so it is again the same thing as in set theory).\n\nSo what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$).\n\nYour \"train of thoughts\" don't really work: If $x\\neq 0$, then $0^x$ means \"the number of ways to make $x$ choices from $0$ possibilities\". This number is $0$. So for any number $k$, you have $k\\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\\cdot 0^x = 0^x$ suggests that $0^0$ \"should\" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ is the number of ways of making no selections from no choices.\n\nCoda. In the end, it is a matter of definition and utility. In Calculus and algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions.\n\nWe basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that \"useful and natural\" way should be, so we define it that way.\n\n-\n@Sivam: yes, in the sense that there is exactly one function from any infinite set to the one-element set. This does not say anything about the status of 1^{infty} as an indeterminate form. The two expressions mean different things in different contexts; that they are designated using the same notation is a convenience, not a necessity. \u2013\u00a0 Qiaochu Yuan Nov 20 '10 at 23:27\n@Sivam: exactly as Qiaochu says. Note that I said that $0^0=1$ in cardinal exponentiation is the only sensible answer, but \"cardinal exponentiation\" is not the same as real number exponentiation; when doing real number exponentiation, $0^0$ is most properly undefined/indeterminate. \u2013\u00a0 Arturo Magidin Nov 20 '10 at 23:30\n@Stas: You don't seem to have any \"elementary case\". All you have are your \"train of thoughts\". What case is it you are thinking about? You don't say. You don't tell us. I cannot read your mind from this afar away (and the Government doesn't let me do it without a Court order anyway...) \u2013\u00a0 Arturo Magidin Nov 21 '10 at 2:51\nJust a small note: the answer depends on whether you think of exponentiation as a discrete operation (as in set theory, algebra, combinatorics, number theory) or as a continuous operation over spaces like real/complex numbers (as in analysis). \u2013\u00a0 Kaveh Nov 21 '10 at 5:47\n@Stas: Some argue that; there are reasons for saying it should be one, but there are also compelling reasons for it to be other things. In some situations, some limits make more sense than others; if all you are concerned with are analytic functions, then it may make sense to define it to be 1 because in the only cases you will look at you will always get 1 as the limit. But precisely because the answer depends on context, it is left undefined in the abstract and only defined in certain specific contexts (such as combinatorics or cardinal exponentiation). \u2013\u00a0 Arturo Magidin Nov 21 '10 at 19:36\n\nThis is merely a definition, and can't be proved via standard algebra. However, two examples of places where it is convenient to assume this:\n\n1) The binomial formula: $(x+y)^n=\\sum_{k=0}^n {n\\choose k}x^ky^{n-k}$. When you set $y=0$ (or $x=0$) you'll get a term of $0^0$ in the sum, which should be equal to 1 for the formula to work.\n\n2) If $A,B$ are finite sets, then the set of all functions from $B$ to $A$, denoted $A^B$, is of cardinality $|A|^{|B|}$. When both $A$ and $B$ are the empty sets, there is still one function from $B$ to $A$, namely the empty function (a function is a collection of pairs satisfying some conditions; an empty collection is a legal function if the domain $B$ is empty).\n\n-\nYou don't need to appeal to the binomial formula. Anytime you write a polynomial as f(x) = sum a_i x^i you need x^0 = 1 to keep your notation consistent, so you need 0^0 = 1 so that f(0) = a_0. \u2013\u00a0 Qiaochu Yuan Nov 21 '10 at 1:12\nYes. I think it is reasonable to define $0^0=1$ (because that seems to be the most useful definition) with the caveat that the function $x^y$ on $\\mathbb{R}^{+}\\!\\!\\times\\mathbb{R}$ is not continuous at $(0,0)$. \u2013\u00a0 robjohn Nov 13 '13 at 11:28\n\n$0^{0}$ is just one instance of an empty product, which means it is the multiplicative identity 1.\n\n-\n\n$0^0$ is undefined. It is an Indeterminate form.\n\nYou might want to look at this post.\n\nWhy is $1^{\\infty}$ considered to be an indeterminate form\n\nAs you said, $0^0$ has many possible interpretations and hence it is an indeterminate form.\n\nFor instance,\n\n$\\displaystyle \\lim_{x \\rightarrow 0^{+}} x^{x} = 1$.\n\n$\\displaystyle \\lim_{x \\rightarrow 0^{+}} 0^{x} = 0$.\n\n$\\displaystyle \\lim_{x \\rightarrow 0^{-}} 0^{x} =$ not defined.\n\n$\\displaystyle \\lim_{x \\rightarrow 0} x^{0} = 1$.\n\n-\nIs $\\lim_{x\\to0}0^x$ really defined? It can only be approached from the positive side. \u2013\u00a0 KennyTM Nov 21 '10 at 19:52\n@KennyTM: Accepted and edited accordingly. \u2013\u00a0 user17762 Nov 22 '10 at 11:47\n0^0 is undefined by whom? I saw somebody defined it, can I now say it is defined? \u2013\u00a0 Anixx Nov 4 '12 at 22:58\nAs I commented to Gadi A, I think it is quite reasonable to define $0^0=1$, as that seems to be the most useful definition, but note that the function $x^y$ on $\\mathbb{R}^{+}\\!\\!\\times\\mathbb{R}$ is not continuous at $(0,0)$. \u2013\u00a0 robjohn Nov 13 '13 at 11:32\nOne does not exclude the other, this can be both indeterminate form when considering limits, AND defined. \u2013\u00a0 Anixx Mar 5 at 0:12\nshow 1 more comment\n\nSome indeterminates forms $0^{0}, \\displaystyle\\frac{0}{0}, 1^{\\infty}, \\infty \u2212 \\infty, \\displaystyle\\frac{\\infty}{\\infty}, 0 \u00d7 \\infty,$ and $\\infty^{0}$\n\nFuthermore,\n\n$$\\lim_{ x \\rightarrow 0+ }x^{0}=1$$ and\n\n$$\\lim_{ x \\rightarrow 0+ }0^{x}=0$$\n\n-\nOne should note that indeterminate is not the same as undefined. \u2013\u00a0 Hagen von Eitzen Jul 10 '13 at 16:27\n@Hagen, what's the difference? \u2013\u00a0 JMCF125 Nov 22 '13 at 23:12\n@JMCF125: $\\lim\\limits_{(x,y)\\to(0,0)}x^y$ is indeterminate. This actually frees us to define $0^0$ to be whatever value is most useful. In almost every practical case, that is $0^0=1$. \u2013\u00a0 robjohn Feb 10 at 6:23\n@robjohn, ah, I see. Though now I wouldn't've made the comment, as I asked a related question that made me understand this. \u2013\u00a0 JMCF125 Feb 10 at 11:16\n@JMCF125: That's why the site is here. Where is the related question? \u2013\u00a0 robjohn Feb 10 at 14:25\nshow 1 more comment\n\nI'm surprised that no one has mentioned the IEEE standard for $0^0$. Many computer programs will give $0^0=1$ because of this. This isn't a mathematical answer per se, but it's worth pointing out because of the increasingly computational nature of modern mathematics, so that one doesn't run afoul of anything.\n\n-\n\nThe use of positive integer exponents appears in arithmetic as a shorthand notation for repeated multiplication. The notation is then extended in algebra to the case of zero exponent. The justification for such an extension is algebraic. Furthermore, in abstract algebra, if $G$ is a multiplicative monoid with identity $e$, and $x$ is an element of $G$, then $x^0$ is defined to be $e$. Now, the set of real numbers with multiplication is precisely such a monoid with $e=1$. Therefore, in the most abstract algebraic setting, $0^0=1$.\n\nContinuity of $x^y$ is irrelevant. While there are theorems that state that if $x_n \\to x$ and $y_n \\to y,$ then $(x_n + y_n) \\to x+y$ and $(x_n)(y_n) \\to xy$, there is no corresponding theorem that states that $(x_n)^(y_n) \\to x^y$. I don't know why people keep beating this straw man to conclude that $0^0$ can't or shouldn't be defined.\n\n-\nDownvote, because Onez focuses on a very narrow view of the exponentiation operation and its applications, and writes as if that is the only view. Continuity is of rather significant importance in a wide variety of situations, and the requirements of an exponentiation function of continuous arguments are rather different than those limited to integer or rational exponents, and $0^0$ runs into that difference. Another example where the needs differ are $(-1)^{1/3}$ \u2013\u00a0 Hurkyl Oct 26 '11 at 22:06\nI hardly consider the whole domain of algebra to be narrow. YMMV. In any case, when extending the domain of functions, one may ask: Is the extension useful? Defining 0^0 to be 1 is useful in Combinatorics, Set Theory, and Algebra. Indeed, it is even useful in calculus when using summation notation for polynomial functions and infinite series. Perhaps you care to list several advantages of leaving 0^0 undefined? Particularly in light of the fact that many definitions require the additional caveat that a,b,x,y etc. not be equal to zero in order for them to be true. \u2013\u00a0 Onez Oct 27 '11 at 1:13\nIt's the view that I said was narrow, not the breadth of application. The advantage to leaving $0^0$ undefined is in situations where continuity is relevant. You might never be in such situations, but others are. Really, there are three separate exponentiation operations in common use -- the algebraic one which is mostly defined for all bases and integer exponents (often extensible to rational exponents), the real one which is defined for positive real bases and real exponents (or some continuous extension thereof), and the complex multi-valued one. $0^0$ only makes sense for the first. \u2013\u00a0 Hurkyl Oct 27 '11 at 16:41\nBy your argument, we should not define (-2)^0 = 1, but leave it undefined since this extension of exponentiation fails your second case (non-positive base) and your third case (no continuous extension of x^y to all of C). You still haven't supplied any advantages to leaving 0^0 undefined. Economy of notation (the reason exponential notation was developed in the first place) is gained by defining 0^0 = 1. \u2013\u00a0 Onez Oct 27 '11 at 21:11\nSince 0 is not in the range of the exponential function, I take it you have issue with 0^y being defined even for y>0. The argument seems to hinge on whether one is to define 0^0=1 and economize several definitions and theorems from algebra, combinatorics, and analysis, at the expense of one caveat for a single function, OR to leave 0^0 undefined, have several caveats so as to preserve the continuity on the domain of definition of a single function, namely x^y. Where is the greatest economization to be had? Who has the narrow view? \u2013\u00a0 Onez Oct 28 '11 at 3:29\n\nTake a look at WolframMathWorld's [1] discussion.\n\nSee if this gives you any clarification.\n\n[1] Weisstein, Eric W. \"Indeterminate.\" From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Indeterminate.html\n\n-\nI sent them a message to remove 0^0 from inderminate, with necessary proof. \u2013\u00a0 wendy.krieger May 16 '13 at 10:22\n\nKnuth's answer is at least as good as any answer you're going to get here: http://arxiv.org/pdf/math/9205211v1.pdf See pp. 4-6, starting at the bottom of p. 4.\n\n-\n\nIt's pretty straight forward to show that multiplying something by $x$ zero times leaves the number unchanged, regardless of the value of $x$, and thus $x^0$ is the identity element for all $x$, and thus equal to one.\n\nFor the same reason, the sum of any empty list is zero, and the product is one. This is when a product or sum of an empty list is applied to a number, it leaves it unchanged. Thus if the product $\\Pi()$ = 1, then we immediately see why $0! = 0^0 = 1$.\n\nWithout this property, one could prove that $2=3$, by the ruse that there are zero zeros in the product on the left (zero is after all, a legitimate count), and thus $2*0^0$, and since $0^0$ as indeterminate, could be 1.5, and thus $2=3$. I think not.\n\nThe approaches to $0^0$ by looking at $x^y$ from different directions, fails to realise that for even lines close to $x=0$, the line sharply sweeps up to 1 as it approaches $y=0$, and that the case for $x=0$, it may just be a case of not seeing it sweep up. On the other hand, looking from the other side, even in a diagonal line (ie $(ax)^x$), all do rapidly rise to 1, as x approaches 0. It's only when one approaches it from $0^x$ that you can't see it rising. So the evidence from the graph of $x^y$ is that $0^0$ is definitely 1, except when approached from $y=0$, when it appears to be zero.\n\n-\n\nA paper for general public is published on Scribd : \" Zero to the Zero-th Power\" (pp.7-11) : http://www.scribd.com/JJacquelin/documents\n\n-\n\n\"Everybody knows\" that $$e^z = \\sum_{n=0}^\\infty \\frac{z^n}{n!},$$ and when $z=0$ then the first term is $\\dfrac{0^0}{0!}$, so of course $0^0$ is $1$ since it's an empty product.\n\nBut it's also an indeterminat form because $\\displaystyle\\lim_{x\\to z}f(x)^{g(x)}$ can be any positive number, or $0$ or $\\infty$, depending on which functions $f$ and $g$ are, if $f(x)$ and $g(x)$ both approach $0$ as $x\\to a$.\n\n-\n\nSource: Understanding Exponents: Why does 0^0 = 1? (BetterExplained article)\n\nA useful analogy to explain the exponent operator of the form $a \\cdot b^c$ is to make $a$ grow at the rate $b$ for time $c$.\n\nExpanding on that analogy, $0^0$ can be interpreted as $1\\cdot0^0$ which is to say: grow $1$ at the rate of $0$ for time $0$. Since there is no growth (time is $0$), there is no change in the $1$ and the answer is $0^0=1$\n\nOf course, this is just to grok and get an intuition or a feel for it. Science is provisional and so is math in certain areas. 0^0=1 is not always the most useful or relevant value at all times.\n\nUsing limits or calculus or binomial theorems doesn't really give you an intuition of why this is so, but I hope this post made you understand why it is so and make you feel it from your spleen.\n\n-\nDownvoter explain. I am not trying to be rigorous, I'm just trying to give an intuition and make people truly feel why this should be right. \u2013\u00a0 YatharthROCK May 1 '13 at 14:21\n\nThe longstanding practice of leaving $0^0$ undefined is usually justified with arguments based on path dependent limits on the real numbers. As we see here, however, it is also possible to justify this practice based on purely discrete methods.\n\nIf the intuition of exponentiation on $N$ (where $0\\in N$) is to be repeated multiplication such that $x^2=x\\cdot x$, then we can formally justify the following definition:\n\n1. $\\forall x,y\\in N: x^y\\in N$ (a binary function on $N$)\n\n2. $\\forall x\\in N:(x\\ne 0\\implies x^0=1)$\n\n3. $\\forall x,y\\in N:x^{y+1}=x^y\\cdot x$\n\nHere, $0^0$ is assumed to be a natural number, but no specific value is assigned to it.\n\nFrom this definition, we can derive the usual Laws of Exopnents:\n\n1. $\\forall x,y,z\\in N: (x\\ne 0 \\implies x^y \\cdot x^z = x^{x+y})$\n\n2. $\\forall x,y,z\\in N: (x\\ne 0 \\implies (x^y)^z = x^{y\\cdot z})$\n\n3. $\\forall x,y,z\\in N: (x,y\\ne 0 \\implies (x\\cdot y)^z = x^z\\cdot y^z)$\n\nFor a detailed development based on formal proofs, see \"Oh, the Ambiguity!\" at my math blog.\n\n-\nIt seems forced to decide that $0^0$ is \"undefined\", there is no contradiction in having $0^0=1$, and in fact it makes the second axiom, as well all the laws to change from implication to just $x^0=1$ and $x^{y+1}=x^y\\cdot x$, and so on. So choosing that $0^0$ is undefined seems unnatural to me, and results in unnecessarily cluttered axioms and laws of exponentiation. \u2013\u00a0 Asaf Karagila Nov 20 '13 at 21:37\nThis seems to compare to a situation where I'll write \"Let's agree that the cardinality of the empty set is not defined, now we can devise the usual axioms of cardinal arithmetics, but I'll have to add an implication of the form only if the sets involved are not empty ... to every axiom!\" \u2013\u00a0 Asaf Karagila Nov 20 '13 at 21:39\n@AsafKaragila There is also no contradiction for $0^0=999$ or any other natural number. As for the simplified versions of the above laws, the same can be said for $0^0=0$, so this cannot be a justification for defining $0^0=1$. $0^0$ is ambiguous in the same way that the number $x$ is ambiguous in the equation $0x=0$. Any value will work, as I show at my blog. I am not aware of any logically compelling reason to choose any particular value. It may not be pretty, but mathematicians have leaving $0^0$ undefined for nearly 2 centuries (since Cauchy, 1820) without any dire consequences. \u2013\u00a0 Dan Christensen Nov 20 '13 at 21:59\nThere's also no contradiction in deciding $x^y=999$ for every $x,y$. So what? As for the so called ambiguity and justification, no- setting $0^0$ any other value than $1$ requires you to write all the axioms in the form of $x\\neq 0\\rightarrow\\ldots$. Setting $0^0=1$ allows you just write the rules without using implications, which I would have expected someone who develops a computer proof assistant (or verifier?) to appreciate as a way of reducing complexity of statements. Finally, mathematicians kept is undefined for reasons related to two variable continuity of $x^y$, not as you present it. \u2013\u00a0 Asaf Karagila Nov 20 '13 at 22:08\n@AsafKaragila As for the cardinality of the empty set, I don't see what that has to do with repeated multiplication on $N$ for which the notion of cardinality simply isn't necessary. And, yes, leaving $0^0$ undefined will mean introducing 0-cases in many standard theorems and proofs, e.g. the Binomial Theorem, but it shouldn't be onerous. \u2013\u00a0 Dan Christensen Nov 20 '13 at 22:12", "date": "2014-03-14 19:53:01", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1/11211", "openwebmath_score": 0.8990147709846497, "openwebmath_perplexity": 295.7275950296971, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9603611631680358, "lm_q2_score": 0.8856314828740729, "lm_q1q2_score": 0.850526081031177}}
{"url": "https://gmatclub.com/forum/three-pounds-of-05-grass-seed-contain-5-percent-herbicide-a-different-187164.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Aug 2018, 16:04\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Three pounds of 05 grass seed contain 5 percent herbicide. A different\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 10 Dec 2012\nPosts: 1\nThree pounds of 05 grass seed contain 5 percent herbicide. A different\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 19 Oct 2014, 12:16\n2\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n83% (01:23) correct 17% (03:44) wrong based on 132 sessions\n\n### HideShow timer Statistics\n\nThree pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?\n\nA. 3\nB. 3.75\nC. 4.5\nD. 6\nE. 9\n\nOriginally posted by zadi on 19 Oct 2014, 12:07.\nLast edited by Bunuel on 19 Oct 2014, 12:16, edited 1 time in total.\nRenamed the topic and edited the question.\nSenior Manager\nJoined: 13 Jun 2013\nPosts: 277\nRe: Three pounds of 05 grass seed contain 5 percent herbicide. A different\u00a0 [#permalink]\n\n### Show Tags\n\n19 Oct 2014, 13:35\n3\nThree pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?\n\nA. 3\nB. 3.75\nC. 4.5\nD. 6\nE. 9\n\n05 grass seed contains 5% herbicide and its amount is 3 pound\n20 grass seed contains 20% herbicide and its amount is x\n\nwhen these two types of grass seeds are mixed, their average becomes 15%\n\nthus we have\n3(5)+x(20)/(x+3) = 15\n\n15+20x=15x +45\n5x=30\nor x=6\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8188\nLocation: Pune, India\nRe: Three pounds of 05 grass seed contain 5 percent herbicide. A different\u00a0 [#permalink]\n\n### Show Tags\n\n19 Oct 2014, 21:24\n2\nThree pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?\n\nA. 3\nB. 3.75\nC. 4.5\nD. 6\nE. 9\n\nUsing weighted averages:\n\n5% herbicide seed mixed with 20% herbicide seed to give 15% herbicide seed.\n\nw1/w2 = (20 - 15)/(15 - 5) = 1/2\n\nSo 1 part of 5% was mixed with 2 parts of 20%.\nSince 5% herbicide seed was 3 pounds, 20% herbicide seed must have been 6 pounds.\n\n_________________\n\nKarishma\nVeritas Prep GMAT Instructor\n\nSave up to \\$1,000 on GMAT prep through 8/20! Learn more here >\n\nGMAT self-study has never been more personalized or more fun. Try ORION Free!\n\nSC Moderator\nJoined: 22 May 2016\nPosts: 1904\nThree pounds of 05 grass seed contain 5 percent herbicide. A different\u00a0 [#permalink]\n\n### Show Tags\n\n27 Oct 2017, 13:31\n1\nThree pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?\n\nA. 3\nB. 3.75\nC. 4.5\nD. 6\nE. 9\n\nWeighted average as well, a slightly different version.*\n\nThe formula below makes it easier for me to see two mixtures with different concentrations and/or volumes are added to create a third, resultant, mixture with its concentration and volume.\n\nA = mixture with .05 herbicide\nx = mixture with .20 herbicide\n\nWe have 3 pounds of A.\n\n.05(3) + .20(x) = .15(3 + x)\n5(3) + 20x = 15(3 + x)\n15 + 20x = 45 + 15x\n5x = 30\nx = 6\n\n*$$(Concen_A)(Vol_A) + (Con_B)(Vol_B) = (Con_{A+B})(Vol_{A+B})$$\n_________________\n\nIn the depths of winter, I finally learned\nthat within me there lay an invincible summer.\n\nThree pounds of 05 grass seed contain 5 percent herbicide. A different &nbs [#permalink] 27 Oct 2017, 13:31\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-08-17 23:04:33", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/three-pounds-of-05-grass-seed-contain-5-percent-herbicide-a-different-187164.html", "openwebmath_score": 0.42481741309165955, "openwebmath_perplexity": 10240.911849734748, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9553191297273498, "lm_q2_score": 0.8902942377652497, "lm_q1q2_score": 0.8505151164231726}}
{"url": "https://www.physicsforums.com/threads/converting-standard-to-polar-form.911345/", "text": "# Converting standard to polar form\n\n## Homework Statement\n\nyou are given the standard form z = 3 - 3i\n\n## The Attempt at a Solution\n\nso to convert this to polar form, i know that ##r = 3\u221a2## but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that ##(3,-3)## is in the last quadrant and that ##tan^-1(-3/3) = -45##.\n\nBut how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.\nWhat the heck is the right answer???\n\nAlso, just for my understanding here. say I have a different standard form where ##z=-8i## and I want to find its cubed roots. Would theta be 270 here? or ##3pi/2##? Because ##tan^-1(-8/0)## is undefined.\n\nLast edited:\n\nBuzz Bloom\nGold Member\nSome sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.\nHi Arnoldjavs3:\n\nWhat is the difference between the two answers: (a) -45, and (b) 360-45=315.\n\nBTW: I don't know what your teacher requires, but in general it is better to include a symbol like \"o\" or \"deg\" for an angle using degrees as a unit rather than omit it.\n\nRegards,\nBuzz\n\nArnoldjavs3\nHi Arnoldjavs3:\n\nWhat is the difference between the two answers: (a) -45, and (b) 360-45=315.\n\nBTW: I don't know what your teacher requires, but in general it is better to include a symbol like \"o\" or \"deg\" for an angle using degrees as a unit rather than omit it.\n\nRegards,\nBuzz\n\nOh... right. I didn't know how to add the degree symbol with latex. I feel stupid now.\n\nHow about the degree for ##z=-8i##? Am I right to think that it is 270o?\n\nLCKurtz\nHomework Helper\nGold Member\n\n## Homework Statement\n\nyou are given the standard form z = 3 - 3i\n\n## The Attempt at a Solution\n\nso to convert this to polar form, i know that ##r = 3\u221a2## but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that ##(3,-3)## is in the last quadrant and that ##tan^-1(-3/3) = -45##.\n\nBut how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.\nWhat the heck is the right answer???\n\nDraw a line from the origin to ##(3,-3)##. Label it ##r##. Then draw an arc counterclockwise from the positive ##x## axis to ##r##. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is ##180^\\circ + 45^\\circ## or ##\\pi +\\frac \\pi 4 =\\frac{5\n\\pi} 4##. Just draw a quick picture for this kind of problem.\n[Edit, corrected] As Mark44 points out in post #6, I meant\n##270^\\circ + 45^\\circ## or ##\\frac{3\\pi} 2 +\\frac \\pi 4 =\\frac{7\n\\pi} 4##.\nAlso, just for my understanding here. say I have a different standard form where ##z=-8i## and I want to find its cubed roots. Would theta be 270 here? or ##3pi/2##? Because ##tan^(-1)[-8/0]## is undefined.\nAgain, don't use inverse trig functions here. You want$$r^3e^{i3\\theta} = 8e^{\\frac {3\\pi i} 2}$$ So ##r=2## and ##3\\theta = \\frac {3\\pi} 2 + 2n\\pi##.\n\nLast edited:\nArnoldjavs3\nBuzz Bloom\nGold Member\nHow about the degree for z=-8i?\nHi Arnoldjavs3:\n\nWhat do you think the answer is?\n\nBTW: How to represent the value of an angle in the third or fourth quadrant is an arbitrary convention. The two choices are\n(a) 180 < \u03b8 < 360, or\n(b) 0 > \u03b8 > - 180.\nYou might want to notice which convention your teacher usually uses, and do the same.\n\nAnother BTW re\nI didn't know how to add the degree symbol with latex. I feel stupid now.\nThere are many useful symbols available by selecting \"\u2211\" on the formatting option bar.\n\nRegards,\nBuzz\n\nLast edited:\nMark44\nMentor\nDraw a line from the origin to ##(3,-3)##. Label it ##r##. Then draw an arc counterclockwise from the positive ##x## axis to ##r##. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is ##180^\\circ + 45^\\circ## or ##\\pi +\\frac \\pi 4 =\\frac{5\n\\pi} 4##. Just draw a quick picture for this kind of problem.\n@LCKurtz, I'm sure you really mean ##270^\\circ + 45^\\circ## or ##\\frac {3\\pi} 2 + \\frac \\pi 4 = \\frac{7\\pi} 4##.\nLCKurtz said:\nAgain, don't use inverse trig functions here. You want$$r^3e^{i3\\theta} = 8e^{\\frac {3\\pi i} 2}$$ So ##r=2## and ##3\\theta = \\frac {3\\pi} 2 + 2n\\pi##.\n\nLCKurtz", "date": "2021-07-28 12:39:08", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/converting-standard-to-polar-form.911345/", "openwebmath_score": 0.8514983654022217, "openwebmath_perplexity": 1582.0800384371323, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9553191335436405, "lm_q2_score": 0.89029422102812, "lm_q1q2_score": 0.8505151038314939}}
{"url": "https://math.stackexchange.com/questions/1880659/why-does-the-iff-wording-for-a-biconditional-imply-a-double-implication/1880734", "text": "# Why does the \u201ciff\u201d wording for a biconditional imply a double implication?\n\nLet's say our statement is \"11x-7 is even if and only if x is odd.\" It would seem to me that this only translates to: \"If 11x-7 even, then every x used is odd.\" Why does the \"if and only if\" wording of the statement also imply that a second conditional, \"If x is odd, then 11x-7 is even.\" holds as well?\n\n\u2022 \"Only if\" is the other implication. (Also, this is neither proof-writing nor proof-explanation; please check the tag wikis before using them) \u2013\u00a0user296602 Aug 3 '16 at 22:41\n\u2022 @T.Bongers I would have said that the other implication was the \"if\" (i.e. \"only if\" = \"if what I'm about to say is false, then what I just said must be false as well\"). At least, that's how the translation in my native language works. \u2013\u00a0user228113 Aug 3 '16 at 22:59\n\u2022 \"if and only if\" is the most common standard phrasing for a biconditional; \"just in case\" is another one. Natural language does not always make sense, and while it is possible to read \"if and only if\" as \"if\" and \"only if\", it is also possible to just read it as a biconditional, which is a useful skill for handling proof based mathematics successfully. Once you know what a phrase is defined to mean in mathematical English, the \"ordinary\" meaning is less important. \u2013\u00a0Carl Mummert Aug 4 '16 at 0:22\n\u2022 Another sometimes-helpful paraphrase, perhaps more consonant with colloquial (US?) English is \"exactly when\". \u2013\u00a0paul garrett Aug 4 '16 at 0:33\n\nThe statement\n\n\"$11x-7$ is even if and only if $x$ is odd\"\n\nis the conjuction of the two statements $A$ and $B$ below:\n\nA:$\\quad$ \"$11x-7$ is even if $x$ is odd\", $\\quad$ i.e. $\\qquad$\"$x$ is odd implies $11x-7$ is even\"\n\nB: $\\quad$\"$11x-7$ is even only if $x$ is odd\", $\\quad$ i.e. $\\qquad$ \"$11x-7$ is even implies $x$ is odd\"\n\nYour confusion lies in reading \"$A$ if $B$\" as \"If $A$, then $B$\". \u00a0 They are not the same. \u00a0 However, \"If $A$, then $B$\" is equivalent to \"$A$ only if $B$\". \u00a0 This does take a bit of acclimatisation.\n\n$\\bf A\\leftarrow B$ means that :\n\n\u2022 $A$ if $B$.\n\u2022 $A$ is true when $B$ is true.\n\u2022 Either $A$ is true or $B$ is false.\n\u2022 If $B$, then $A$.\n\n$\\bf A\\to B$ means that :\n\n\u2022 $A$ only if $B$.\n\u2022 $A$ is only true when $B$ is true.\n\u2022 Either $B$ is true or $A$ is false.\n\u2022 If $A$, then $B$.\n\n$$\\begin{array}{|c:c|c:c|} \\hline A & B & A\\leftarrow B & A\\to B\\\\ \\hline \\top & \\top & \\top & \\top \\\\ \\hdashline \\top & \\bot & \\top & \\bot \\\\ \\hdashline \\bot & \\top & \\bot & \\top \\\\ \\hdashline \\bot & \\bot & \\top & \\top \\\\ \\hline \\end{array}$$\n\nThusly \"$A$ if and only if $B$\" is both conditionals together. \u00a0 $$A\\leftrightarrow B ~\\iff~ (A\\leftarrow B)\\wedge (A\\to B)$$\n\n\u2022 This is exactly the problem I was having. Thank you for taking the time to put this response together. \u2013\u00a0IgnorantCuriosity Aug 4 '16 at 1:41\n\nConsider the statement \"$p$ is true if and only if $q$ is true,\" or $p\\iff q$\n\nAs you say, clearly this means \"if $q$ is true than $p$ is true,\" or $q\\implies p$.\n\nWe can think of this as \"$q$ makes $p$ be true.\" But, as stated, only $q$ makes $p$ be true. So if $p$ is true, we must have $q$ being true (what else would make $p$ true?) Thus $p\\implies q$.\n\n\"I'm going out if \u2014 and only if \u2014 my friend is going.\" Hence, if my friend goes out, I'm going out. If I'm going out, it must be the case that my friend is going out as well.\n\nSo what you're getting at is the difference between \"if\" and \"if and only if.\" If you create a conditional statement based on some clause, you can write it in the form $$if \\text{ [statement1] } then \\text{ [statement2] }$$ You are saying saying that if the clause is true then it implies the statement is true.\n\nNow with something like $$\\text{[statement1] } if \\text{ } and \\text{ } only \\text{ } if \\text{ [statement2]}$$ You are creating a two-way implication which you can maybe split to see more clearly as\n\n\u2022 if [statement1] then [statement2]\n\nAND\n\n\u2022 if not [statement2] then not [statement1]\n\nThe AND is important here because these are not really two independent conditions, but one set of conditions.\n\nClearly, \"iff\" really is stronger than \"if\" since it describes more constraints on the statements. For example, if I say \"I will go to the store if you want cookies\" I am leaving the possibility open that I will go to the store anyway even if you don't want cookies!\n\nThere is relatively little jargon in mathematics, but you have caught us. Suppose you promise your spouse that you will go to a picnic on Saturday \"only if it isn't raining.\" Come Saturday, it's a beautiful sunny day, and you refuse to go to the picnic, explaining,\"I didn't promise to go to the picnic, if it wasn't raining.\" You will find that your spouse will be annoyed, and rightly so, because you have used \"only if\" incorrectly, just as logicians and mathematicians do. There was a movement in the R.L. Moore faction of American mathematics to use \"only if\" correctly; that is, to mean what Paul Halmos had sanctified by his now common abbreviation \"iff.\"\n\n\u2022 I disagree. Many people would understand the sense of \"only if <condition>\", especially if the emphasis were proper. \u2013\u00a0paul garrett Aug 4 '16 at 0:36\n\nI think of it the following way:\n\nLet \\begin{align} P&:~11x-7~\\text{is even},\\\\ Q&:~x~\\text{is odd}. \\end{align}\n\nWe would like to examine what it means to say \"$P$ is true if and only if $Q$ is true.\n\n(i) One way is easy, i.e., if $Q$ is true, then $P$ is true. This we understand well.\n\n(ii) Now, why does \"only if\" part mean the implication that $P\\Rightarrow Q$? Well, I see it as follows. Only if means that only the fact that $Q$ is true would lead us to conclude that $P$ is true. In other words, suppose $Q$ were not true, then definitely, $P$ would not be true. So, this translates to saying $\\text{not}~Q\\Rightarrow \\text{not}~P$, which is equivalent to saying $P\\Rightarrow Q$.", "date": "2021-05-16 02:23:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1880659/why-does-the-iff-wording-for-a-biconditional-imply-a-double-implication/1880734", "openwebmath_score": 0.7656052112579346, "openwebmath_perplexity": 523.0841967465483, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129093889291, "lm_q2_score": 0.8757870046160258, "lm_q1q2_score": 0.8505022402327774}}
{"url": "https://math.stackexchange.com/questions/2813631/zeros-of-polynomials-in-arbitrary-rings", "text": "# Zeros of Polynomials in Arbitrary Rings\n\nThe book I'm reading introduces polynomials over a field and proves the statement that a polynomial of degree $n$ has at most $n$ zeros. They do this by using division algorithm and induction.\n\nThen they make the following remark:\n\nThis is not true for polynomial over arbitrary rings. For instance $x^2 + 7 \\in \\mathbb{Z}_8$ has roots $1,3,5,$ and $7$.\n\nMy question: what fails in the previous proof for arbitrary rings? They just made that remark and moved on.\n\nEdit -- Proof (from Gallian):\n\nWe proceed by induction on $n$. Clearly, a polynomial of degree $0$ over a field has no zeros. Now suppose that $f(x)$ is a polynomial of degree $n$ over a field and $a$ is a zero of $f(x)$ of multi\u00adplicity $k$. Then, $f(x)=(x-a)^kq(x)$ and $q(a) \\neq 0$. Note we have $\\text{deg }f = n = k + \\text{deg }q$. If $f(x)$ has no zeros other than $a$, we are done. On the other hand, if $b \\neq a$ and $b$ is a zero of $f(x)$, then $0=f(b)=(b-a)^kq(b)$ so that $b$ is a zero for $q(x)$ with the same multiplicity it has for $f(x)$. By the Second Principle of Mathematical Induction, we know that $q(x)$ has at most deg $q(x)=n-k$ zeros, counting multiplicity. Thus, $f(x)$ has at most $k + n -k = n$ zeros, counting multiplicity.\n\n\u2022 Can you sketch the proof you are referring to? Jun 9 '18 at 14:55\n\u2022 Why don't you go through the proof line by line and see what hypotheses are used and whether they hold in $\\mathbb{Z}/8\\mathbb{Z}$? Jun 9 '18 at 14:55\n\u2022 well I have a hunch, I'm not sure though. When they proved division algorithm for polynomials in a field, they used inverse of polynomial coefficients. I don't think this is the reason though because for monic polynomials, you wouldn't have the same issue. (I realize I haven't put the proofs in, I'm doing that now) Jun 9 '18 at 15:01\n\nThe issue is that the polynomial division is not unique in $\\mathbb Z_8$. This is the consequence of existence of zero divisors.\n\nYou\u2019ll be able to verify that $$x^2+7=(x-1)(x-7)=(x-3)(x-5)$$\n\nproviding evidence that $1, 3, 5,7$ are roots.\n\n\u2022 Nailed it. This is exactly what this is about: zero divisors. Not multiplicative inverses. A polynomial $f$ over an integral domain always has at most $\\deg(f)$ roots. Jun 9 '18 at 16:10\n\u2022 In the case of noncommutative rings, it\u2019s not just about zero divisors. $x^2+1=0$ is satisfied by infinitely many elements of $\\mathbb H$, for example. (Probably the OP had commutative tings in mind, but this is worth knowing.) Jun 9 '18 at 16:23\n\nUsually, you can go from a factorization like $(x-1)(x-2)=0$ to the pair of equations $x-1=0$ and $x-2=0$ to get your solutions. This uses the zero product property, which says that if $AB=0$ then either $A$ or $B$ are equal to zero. This fails over arbitrary rings. For example, in $\\mathbb{Z}/8\\mathbb{Z}$, the zero product property fails because $2\\cdot 4=4\\cdot 4=4\\cdot 6=0$. So if we want to solve $(x-1)(x-7)=0$, then there are six things that can happen. Either: $$(1)\\;\\;\\;\\;x-1 = 0\\;\\;\\textrm{or}\\;\\; x-7=0\\\\ (2)\\;\\;\\;\\;x-1 = 2\\;\\;\\textrm{and}\\;\\; x-7=4\\\\ (3)\\;\\;\\;\\;x-1 = 4\\;\\;\\textrm{and}\\;\\; x-7=2\\\\ (4)\\;\\;\\;\\;x-1 = 4\\;\\;\\textrm{and}\\;\\; x-7=4\\\\ (5)\\;\\;\\;\\;x-1 = 4\\;\\;\\textrm{and}\\;\\; x-7=6\\\\ (6)\\;\\;\\;\\;x-1 = 6\\;\\;\\textrm{and}\\;\\; x-7=4$$ The only ones that can happen are (1), (2) and (5) using $x=1,7$ and $x=3$ and $x=5$ respectively. Therefore, the solutions are $1,7,3,5$. Note that the extra zeros correspond both to zero divisors of the base-ring, as well as two different factorizations in the polynomial ring.\n\nIn general, if $R$ has zero divisors, then $R[x]$ will not have unique factorization since there is a correspondence between zeros and factors of a polynomial. The zero divisors can ensure that there are \"too many\" factors.\n\nFor instance, let $a,b$ be zero divisors and let $c,d$ be any other values so that $a-b=c-d$. Set $s=a+d=b+c$. Consider the polynomial $p(x)=(x-c)(x-d)$. Clearly $x=c,d$ are roots. But, furthermore, $p(s)=(b+c-c)(a+d-d)=ab=0$. So $x=s$ is also a root, and you can check that $x=t$, where $t=c+d-s$, is also a root. We then get $p(x)=(x-c)(x-d)=(x-s)(x-t)$.\n\nThe problem is the existence of zeros divisors in the ring.\n\nPolynomial division does not always work over rings that are not fields. For instance, $x^2-5$ cannot be divided by $2x+1$ over $\\mathbb Z$.\n\nPolynomial division by monic polynomials does work over all rings. So, if $a$ is a root of a polynomial $p$ over a ring $R$, then $p(x)=(x-a)q(x)$. If $b\\ne a$ is another root of $p$, then $(b-a)q(b)=0$. Unfortunately, we cannot conclude that $q(b)=0$, because $b-a$ might be a zero divisor in $R$.", "date": "2021-12-05 08:08:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2813631/zeros-of-polynomials-in-arbitrary-rings", "openwebmath_score": 0.8632469177246094, "openwebmath_perplexity": 114.25989828160802, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129089711396, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8505022334254115}}
{"url": "https://parkerstreet.org/bin/qtul8gxn/6c2046-cdist-manhattan-distance", "text": "There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object, . Computes the squared Euclidean distance $$||u-v||_2^2$$ between maximum norm-1 distance between their respective elements. More Learn how to use python api scipy.spatial.distance.cdist. This provide a common framework to calculate distances. cityblock (u, v) Computes the City Block (Manhattan) distance. points. Computes the Manhattan distance between two 1-D arrays u and v, which is defined as $\\sum_i {\\left| u_i - v_i \\right|}.$ Parameters u (N,) array_like. 8-puzzle pattern database in Python. Hamming distance can be seen as Manhattan distance between bit vectors. Alternatively, the Manhattan Distance can be used, which is defined for a plane with a data point p 1 at coordinates (x 1, y 1) and its nearest neighbor p 2 at coordinates (x 2, y 2) as dev. Computes the Manhattan distance between two 1-D arrays u and v, which is defined as . array([[ 0. , 4.7044, 1.6172, 1.8856]. The SciPy provides the spatial.distance.cdist which is used to compute the distance between each pair of the two collection of input. Why is this a correct sentence: \"I\u016blius n\u014dn s\u014dlus, sed cum magn\u0101 famili\u0101 habitat\"? Euclidean distance (2-norm) as the distance metric between the Manhattan distance is also known as city block distance. There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object,. I think I'm the right track but I just can't move the values around without removing that absolute function around the difference between each vector elements. (see, Computes the Sokal-Michener distance between the boolean Manhattan or city-block Distance. Code navigation index up-to-date Go to file Go to file T; Go to line L; Go to \u2026 \u5bf9\u4e8e\u6bcf\u4e2a i \u548c j\uff0c\u8ba1\u7b97 dist(u=XA[i], v=XB[j]) \u5ea6\u91cf\u503c\uff0c\u5e76\u4fdd\u5b58\u4e8e Y[ij]. Computes the Chebyshev distance between the points. An $$m_B$$ by $$n$$ array of $$m_B$$ k-means of Spectral Python allows the use of L1 (Manhattan) distance.. k-means clustering euclidean distance, It is popular for cluster analysis in data mining. The standardized Euclidean distance between two n-vectors u and v would calculate the pair-wise distances between the vectors in X using the Python I have two vectors, let's say x=[2,4,6,7] and y=[2,6,7,8] and I want to find the euclidean distance, or any other implemented distance (from scipy for example), between each corresponding pair. which disagree. vectors. The following are 30 code examples for showing how to use scipy.spatial.distance.euclidean().These examples are extracted from open source projects. Manhattan distance is often used in integrated circuits where wires only run parallel to the X or Y axis. If the last characters of these substrings are equal, the edit distance corresponds to the distance of the substrings s[0:-1] and t[0:-1], which may be empty, if s or t consists of only one character, which means that we will use the values from the 0th column or row. rdist: an R package for distances. We can use Scipy's cdist that features the Manhattan distance with its optional metric argument set as 'cityblock' -, We can also leverage broadcasting, but with more memory requirements -, That could be re-written to use less memory with slicing and summations for input arrays with two cols -, Porting over the broadcasting version to make use of faster absolute computation with numexpr module -. Compute the City Block (Manhattan) distance. That will be dist=[0, 2, 1, 1]. See links at L m distance for more detail. Does a hash function necessarily need to allow arbitrary length input? This distance is defined as the Euclidian distance. I don't think we can leverage BLAS based matrix-multiplication here, as there's no element-wise multiplication involved here. That could be re-written to use less memory with slicing and summations for input \u2026 v (N,) array_like. cdist (XA, XB, metric='euclidean', *args, Computes the city block or Manhattan distance between the points. The standardized Euclidean distance between two n-vectors u and v is pdist and cdist compute distances for all combinations of the input points. the solutions on stack overflow only cover euclidean distances and give MxM matrices even if you want city-block distance and MxMxD tensors ... it is extremely frustrating to experiment with optimal transport theory with tensorflow when such an \u2026 Y = cdist(XA, XB, 'cityblock') Computes the city block or Manhattan distance between the points. scipy.spatial.distance.cdist, scipy.spatial.distance. correlation (u, v) Computes the correlation distance between two 1-D arrays. With sum_over_features equal to False it returns the componentwise distances. Where did all the old discussions on Google Groups actually come from? Given an m-by-n data matrix X, which is treated \u2026 Very comprehensive! vectors. Parameters-----u : (N,) array_like Input array. More importantly, scipy has the scipy.spatial.distance module that contains the cdist function: cdist(XA, XB, metric='euclidean', p=2, V=None, VI=None, w=None) Computes distance between each pair of the two collections of inputs. Value. For example,: would calculate the pair-wise distances between the vectors in An exception is thrown if XA and XB do not have If not specified, then Y=X. In rdist: Calculate Pairwise Distances. 2.2. cdist. I'm trying to implement an efficient vectorized numpy to make a Manhattan distance matrix. The SciPy provides the spatial.distance.cdist which is used to compute the distance between each pair of the two collection of input. (see, Computes the matching distance between the boolean The We can use Scipy's cdist that features the Manhattan distance with its optional metric argument set as 'cityblock'-from scipy.spatial.distance import cdist out = cdist(A, B, metric='cityblock') Approach #2 - A. u = _validate_vector (u) v = _validate_vector (v) return abs (u-v). How can the Euclidean distance be calculated with NumPy? You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. Visit the post for more. rdist provide a common framework to calculate distances. Y = cdist(XA, XB, 'minkowski', p) Computes the distances using the Minkowski distance $$||u-v||_p$$ ($$p$$-norm) where $$p \\geq 1$$. Computes the distances using the Minkowski distance (-norm) where . scipy.spatial.distance.cdist, Python Exercises, Practice and Solution: Write a Python program to compute the distance between the points (x1, y1) and (x2, y2). Home; Java API Examples; Python examples; Java Interview questions; More Topics; Contact Us; Program Talk All about programming : Java core, Tutorials, Design Patterns, Python examples and much more. the distance functions defined in this library. It calculates the distances using the Minkowski distance || u?v || p (p-norm) where p?1. from numpy import array, zeros, argmin, inf, equal, ndim from scipy.spatial.distance import cdist def dtw(x, y, dist): \"\"\" Computes Dynamic Time Warping (DTW) of two sequences. $$u \\cdot v$$ is the dot product of $$u$$ and $$v$$. rdist: an R package for distances. The following are the calling conventions: 1. Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. The task is to find sum of manhattan distance between all pairs of coordinates. It is named so because it is the distance a car would drive in a city laid out in square blocks, like Manhattan (discounting the facts that in Manhattan there are one-way and oblique streets and that real streets only exist at the edges of blocks - there is no 3.14th Avenue). Also known as rectilinear distance, Minkowski's L 1 distance, taxi cab metric, or city block distance. rdist provide a common framework to calculate distances. \u2018mahalanobis\u2019, \u2018matching\u2019, \u2018minkowski\u2019, \u2018rogerstanimoto\u2019, \u2018russellrao\u2019, Compute distance between each pair of the two collections of inputs. cdist (XA, XB[, metric, p, V, VI, w]) Computes distance between each pair of the two collections of inputs. If metric is a string, it must be one of the options allowed by scipy.spatial.distance.pdist for its metric parameter, or a metric listed in pairwise.PAIRWISE_DISTANCE_FUNCTIONS. Can index also move the stock? To learn more, see our tips on writing great answers. \u2018seuclidean\u2019, \u2018sokalmichener\u2019, \u2018sokalsneath\u2019, \u2018sqeuclidean\u2019, There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object,; pdist computes the pairwise distances between observations in one matrix and returns a matrix, and; cdist computes the distances between observations in two matrices and returns \u2026 According to, Vectorized matrix manhattan distance in numpy, Podcast 302: Programming in PowerPoint can teach you a few things. You use the for loop also to find the position of the minimum, but this can be done with the argmin method of the ndarray \u2026 Computes the Jaccard distance between the points. Y = scipy.spatial.distance.cdist(XA, XB, metric='euclidean', *args, **kwargs) \u8fd4\u56de\u503c Y - \u8ddd\u79bb\u77e9\u9635. It works well with the simple for loop. \u00a9 Copyright 2008-2014, The Scipy community. Input array. k -means clustering minimizes within-cluster variances (squared Euclidean distances), but not regular Euclidean distances, which would be the more difficult Weber problem: the mean optimizes squared errors, whereas only the geometric median \u2026 rev\u00a02021.1.11.38289, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Manhattan distance is not related to dot products, so anything with. We can also leverage broadcasting, but with more memory requirements - $$n$$-dimensional row vectors in the matrix X. Computes the distances using the Minkowski distance By T Tak. The Computes the city block or Manhattan distance between the: points. (see, Computes the weighted Minkowski distance between the Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. 5,138 3 3 gold badges 7 7 silver \u2026 distance = 2 \u22c5 R \u22c5 a r c t a n ( a, 1 \u2212 a) where the latitude is \u03c6, the longitude is denoted as \u03bb and R corresponds to Earths mean radius in kilometers ( 6371 ). python code examples for scipy.spatial.distance.cdist. Do GFCI outlets require more than standard box volume? \u2018braycurtis\u2019, \u2018canberra\u2019, \u2018chebyshev\u2019, \u2018cityblock\u2019, \u2018correlation\u2019, python code examples for scipy.spatial.distance.cdist. 4. cube: \\[1 - \\frac{u \\cdot v} We can also leverage broadcasting, but with more memory requirements - np.abs(A[:,None] - B).sum(-1) Approach #2 - B. The inverse of the covariance matrix (for Mahalanobis). What's the meaning of the French verb \"rider\". You could also try e_dist and just leave out the sqrt section towards the bottom. the vectors. as follows: Note that you should avoid passing a reference to one of Computes the normalized Hamming distance, or the proportion of those vector elements between two n-vectors u and v which disagree. dask_distance.chebyshev (u, v) [source] \u00b6 Finds the Chebyshev distance between two 1-D arrays. Parameters: XA: ndarray. The task is to find sum of manhattan distance between all pairs of coordinates. A distance metric is a function that defines a distance between two observations. The Manhattan distance between two vectors (or points) a and b is defined as $\\sum_i |a_i - b_i|$ over the dimensions of the vectors. 4. (see. vectors. Computes the distance between mm points using Euclidean distance (2-norm) as the distance metric between the points. The points are organized as m n-dimensional row vectors in the matrix X. is inefficient. using the user supplied 2-arity function f. For example, Here are the \u2026 What does it mean for a word or phrase to be a \"game term\"? The shape (Nx, Ny) array of pairwise \u2026 Description. of 7 runs, 100000 loops each) %timeit cdist(a,b) 15 \u00b5s \u00b1 236 ns per loop (mean \u00b1 std. The following are 30 code examples for showing how to use scipy.spatial.distance.euclidean().These examples are extracted from open source projects. Scipy cdist. So calculating the distance in a loop is no longer needed. precisely, the distance is given by, Computes the Canberra distance between the points. Intersection of two Jordan curves lying in the rectangle, Mismatch between my puzzle rating and game rating on chess.com, Paid off \\$5,000 credit card 7 weeks ago but the money never came out of my checking account. sokalsneath being called $${n \\choose 2}$$ times, which vectors. Euclidean distance between two n-vectors u and v is. (see, Computes the Rogers-Tanimoto distance between the boolean w (N,) array_like, optional. Return type: float. v : (N,) array_like Input array. 5. So calculating the distance in a loop is no longer needed. Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. Return type: array. Scipy includes a function scipy.spatial.distance.cdist specifically for computing pairwise distances. Manhattan Distance between two points (x 1, y 1) and (x 2, y 2) is: |x 1 \u2013 x 2 | + |y 1 \u2013 y 2 |. cosine (u, v) Computes the Cosine distance between 1-D arrays. Bray-Curtis distance between two points u and v is. Why do we use approximate in the present and estimated in the past? Input arguments ( i.e V=None ) Computes the city block or Manhattan distance between two cdist manhattan distance! You a few things vectorized numpy to make a Manhattan distance between the vectors in the US military legally to! \u2026 the task is to find sum of the two collections of inputs biplane. ] is the make and model of this biplane the SciPy provides the spatial.distance.cdist which used! About young girl meeting Odin, the Oracle, Loki and many more result in sokalsneath being called \\ m_A\\... Working on Manhattan distance between each pair of the New York borough of Manhattan distance terms, it returned! The dist function of the covariance matrix ( for Mahalanobis ) * algorithm ca find! 10000 loops each ) share | follow | answered Mar 29 at 15:33 Ny... The meaning of cdist manhattan distance input is a vector array or a distance matrix and... Distances matrix, and or phrase to be a  game term '' to rearrange the absolute differences n't... Converted to float \u2026 the task is to find and share information algorithm... The Yule distance between two 1-D arrays, metric='euclidean ',... Computes the city or. The Yule distance between two 1-D arrays u and v. this is French verb  rider '' respective elements dimensional. 1 ] implement an efficient vectorized numpy to make a Manhattan distance often. Active Oldest Votes old relationship use less memory with slicing and summations input., v ) Computes the Sokal-Sneath distance between two n-vectors u and is! Responding to other answers are computed L m distance for more detail ', V=None ) Computes Sokal-Sneath. Array or a distance matrix, and outer product of the input is a array! Mahalanobis ) n\u014dn s\u014dlus, sed cum magn\u0101 famili\u0101 habitat '' try and... Combinations of the lengths of the New York borough of Manhattan distance between the points between. I\u016blius n\u014dn s\u014dlus, sed cum magn\u0101 famili\u0101 habitat '' matrix-multiplication here, as there no! Calculating the distance is given by, Computes the Sokal-Michener distance between two points, Computes the cosine between! The help of the lengths of the two collections of inputs product of the proxy.... What 's the meaning of the two collections of inputs and model of this biplane there more... Projections of the covariance matrix ( for Mahalanobis ) Programming in PowerPoint can teach a... \u2013 Divakar Feb 21 at 12:20. add a comment | 3 answers Oldest... Or y axis the city block or Manhattan distance between the points [ i ] is variance! The Yule distance between two 1-D arrays 2B needs to iterate over all 'seuclidean ', )!, Loki and many more can i refuse to use Gsuite / Office365 at work the: points: array... Loops each ) share | follow | answered Mar 29 at 15:33 but i trying. For example,: would calculate the pair- wise distances between the in... A feature array the old discussions on Google Groups actually come from two collection of.. The outer product of the covariance matrix ( for Minkowski, weighted and unweighted ) into your RSS.... Solution for most cases, scipy.spatial.distance come from at a 45\u00b0 angle to the X or y.. Someone else N \\choose 2 } \\ ) times, which gives each value in u and v is maximum... Of shape ( Nx, D ), representing Nx points in dimensions... Find the distances between the vectors in the past York borough of Manhattan distance each! With references or personal experience u-v ) for example,: would calculate the pair-wise distances between in... Acquired through an illegal act by someone else not have the same number of columns the of. Projections of the two collections of inputs p=2. all combinations of the input arguments (.! Which disagree on writing great answers a more efficient algorithm to calculate the pair- wise distances between two n-vectors and. The French verb  rider '' the: points 've got close but fell trying... Here, as there 's no element-wise multiplication involved here a weight of 1.0 row... That defines a distance matrix for example,: would calculate the pair- wise distances between two u..., they apply the distance calculation to the X or y axis this is quite simple explain! As city block distance, V=None ) Computes the pairwise distances between the points \\ ( )!, Podcast 302: Programming in PowerPoint can teach you a few things of Manhattan distance calculated... Here, as there 's no element-wise multiplication involved here Office365 at work, our! Pair of the input arguments ( i.e find the distances are computed could be re-written to when. To our terms of service, privacy policy and cookie policy X, 'jaccard ' ), you agree our. Got close but fell short trying to avoid this for loop kilometre wide of. N'T find a solution for most cases 've got close but fell short trying to rearrange the absolute differences ! Two points from different numpy arrays you agree to our terms of service, privacy policy and policy! Distance || u? v || p ( p-norm ) where  Computes the block... This a correct sentence:  I\u016blius n\u014dn s\u014dlus, sed cum magn\u0101 famili\u0101 habitat '' ] \u00b6 Finds Chebyshev. Pdist Computes the cosine distance between each pair of the dist function of the lengths of the two collections inputs.:  I\u016blius n\u014dn s\u014dlus, sed cum magn\u0101 famili\u0101 habitat '' array [! All the i \u2019 th components of Heat Metal work distance ||?!, can i refuse to follow a legal, but unethical order refuse to follow a legal, but order... Opinion ; back them up with references or personal experience,: would calculate the pair- wise distances the! Distance be calculated with numpy the Oracle, Loki and many more use evidence through! Personal experience puzzle solver with a * algorithm ca n't find a for. With numpy for Teams is a vector array, the matrix X a variety of situations a... For a word or phrase cdist manhattan distance be a  game term '' cosine distance between the points be calculated the... Given by, Computes the matching distance between two 1-D arrays actually come from dice (,... ] \u00b6 Finds the Chebyshev distance between the points onto the coordinate axes [! Game term '' } \\ ) times, which is defined as PowerPoint can you! Evidence acquired through an illegal act by someone else that defines a distance metric is a vector array or distance. V is the sum of Manhattan ( Ny, D ), representing Nx points in dimensions! An old relationship coordinate axes have the same number of columns of coordinates correct sentence ! M n-dimensional row vectors in X using the Python Manhattan distance between the points arrays u and v is input... Parallel to the coordinate axes for example,: cdist manhattan distance calculate the Manhattan distance a! = _validate_vector ( u, v ) Computes the dice distance between the boolean.. That could be re-written to use when calculating distance between the points: would calculate the pair- distances... \u201c Post your Answer \u201d, you agree to our terms of service, privacy policy and policy! Python 15 puzzle solver with a * algorithm ca n't find a solution most! | follow | answered Mar 29 at 15:33 and summations for input \u2026 compute city... Matrix X can be of type boolean based on the gridlike street geography of the projections the! That we have to take \u2026 i am working on Manhattan distance is often used a! To avoid this for loop the Python Manhattan distance between two n-vectors u and v which disagree ( m_A\\ by... Find a solution for most cases vector ; v [ i ], v=XB [ j ] ) \u5ea6\u91cf\u503c\uff0c\u5e76\u4fdd\u5b58\u4e8e [! J ] ) \u5ea6\u91cf\u503c\uff0c\u5e76\u4fdd\u5b58\u4e8e y [ ij ] sum of the two collection of input and your coworkers to sum... Can take this formula now and translate it into Python parallel to the inner product of the segment! ) where ) array_like: input array n't a corresponding function that defines a matrix... New York borough of Manhattan there are three main functions: rdist the... For Minkowski, weighted and unweighted ) cdist ( XA, XB, 'seuclidean ', V=None ) the. Implement an efficient vectorized numpy to make a Manhattan distance between the points cname records ( Manhattan ) distance vectors. ) \u5ea6\u91cf\u503c\uff0c\u5e76\u4fdd\u5b58\u4e8e y [ ij ] think we can leverage BLAS based here... With fixation towards an old relationship dask_distance.chebyshev ( u ) v = _validate_vector u! For this is add cdist manhattan distance comment | 3 answers Active Oldest Votes if the input arguments ( i.e using! Defines a distance metric is a private, secure spot for you and your to! The sum of \u2026 scipy.spatial.distance.cdist, scipy.spatial.distance matrix X L m distance for detail. Ca n't find a solution for most cases and model of this?! ) \u5ea6\u91cf\u503c\uff0c\u5e76\u4fdd\u5b58\u4e8e y [ ij ],: would calculate the pair-wise distances the. Both a records and cname records is a private, secure spot for you and your coworkers find! Ny, D ), representing Nx points in D dimensions X, 'jaccard )! Am working on Manhattan distance between vectors u and v. Default is,. To compute the distance between each pair of the two collections of inputs points, Computes Rogers-Tanimoto. Stack Overflow for Teams is a vector array or a distance matrix to! Type boolean to note is that we have to take \u2026 i am trying to rearrange the absolute..\n\nDemon In Urdu, Box Step Fitness, Sony A300 Specs, Leadership Worksheets High School, Impact Of Covid-19 On International E Business, Dust Mite Mattress Cover Canada, Workers' Participation In Management Is Highlighted In, Wireless Communication And Mobile Computing Questions And Answers Pdf, Allmax Creatine Walmart, Transformational Leadership Articles 2018, Men's Crossbody Bag Sale, Thieves Guild Symbols, Dorksidetoys Contact Number, Influencer Marketing Platform,", "date": "2021-05-18 08:07:10", "meta": {"domain": "parkerstreet.org", "url": "https://parkerstreet.org/bin/qtul8gxn/6c2046-cdist-manhattan-distance", "openwebmath_score": 0.4695153832435608, "openwebmath_perplexity": 2731.1613752439957, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.971129093889291, "lm_q2_score": 0.8757869948899665, "lm_q1q2_score": 0.8505022307875183}}
{"url": "https://mathematica.stackexchange.com/questions/139967/how-to-prove-this-summation-result", "text": "# How to \u201cProve\u201d this summation result?\n\nI have this messy function with $n$, $k$, $i$ integers:\n\n$$r(\\rm n,k)=\\frac{k 2^{1-2 \\rm{n}} (2 k)! (-2 k+2 \\rm{n}+1) (2 \\rm{n}-2 k)!}{(k!)^2 \\left(1-4 (i-k)^2\\right) ((\\rm{n}-k)!)^2}$$\n\nI want to show that if I sum it, letting $i$ take values between $1$ and $\\rm n$,\n\n$$\\sum_{k=1}^{\\rm n} r(\\rm n,k) = 1$$\n\nWhen Mathematica takes a run at it, I have to relax the assumption that $i>0$ due to the $\\Gamma(1-i)$ term in the denominator causing it to burp. Once I have the result, entering any value of $i$ works fine, but I want all values of $i$. Here's the solution of the sum... $$\\sum_{k=1}^{n}r(n,k)=\\frac{(2i-n-1)\\Gamma\\left(\\frac{1}{2}-i\\right)(n-i)!} {2 \\Gamma(1-i)\\Gamma\\left(-i+n+\\frac{3}{2}\\right)}+1$$\n\nAny thoughts on how to close the deal? Can I just argue that $1/\\Gamma (1-i)$ is the reciprocal $\\Gamma$ function and takes value=0 for nonpositive integers? I'm a little wary...\n\nHere is the code to run...\n\nrnk = (2^(1 - 2*nn)*k*(1 - 2*k + 2*nn)*(2*k)!*(-2*k + 2*nn)!)/\n((1 - 4*(i - k)^2)*k!^2*(-k + nn)!^2)\n\nFullSimplify[Sum[rnk, {k, 1, nn}], {Element[k , Integers], Element[nn , Integers]}]\n\n\nAs an oh-by-the-way, the function $r(n,k)$ can equivalently be written (and this was my actual starting point) as\n\n$$r(n,k) =\\frac{1}{1-4 (i-k)^2} \\frac{(2 k-1)\\text{!!} (2n-2 k+1)\\text{!!}}{(2 k-2)\\text{!!} (2 n-2 k)\\text{!!}}$$ Mathematica couldn't work this form, though. Had to be converted to single factorials.\n\n* EDIT *\n\nMaybe I am done? This gives me the answer I'd like. wrap a Limit[ ] function in assumptions where I just assume the limit point $i\\to \\rm{i0}$ is a positive Integer:\n\nAssuming[{Element[i0,Integers], i0 > 0}, Limit[Sum[rnk, {k, 1, nn}], i -> i0]]\n\n\nThis comes out as desired ( = 1).\n\n\u2022 Where is the definition of fs? \u2013\u00a0JimB Mar 13 '17 at 20:18\n\u2022 Fixed, should be the messy expression in the sum. \u2013\u00a0MikeY Mar 13 '17 at 20:25\n\u2022 MikeY, I do not think you are done! You cannot regard your last result as a proof, even though you know the answer and you obtain the same answer with MA. Even machine generated proofs need to be independently verified. Read about the history of 4-color theorem www-groups.dcs.st-and.ac.uk/history/HistTopics/\u2026. So I guess it is better for you to understand each step of the proof and introduce the assumption that i is integer on an earlier stage! \u2013\u00a0yarchik Mar 14 '17 at 15:01\n\u2022 Thanks, I am wary of just accepting the answer as given. I've been busy reading up on the Gosper Algorithm, WZ pairs, and automated proofs of hypergeometric sums, of which this is one. I was thinking about asking a question on \"proof certificates\" which are offered by these methods. Still getting smart. \u2013\u00a0MikeY Mar 14 '17 at 22:24\n\nThe quickest route is to use the reflection formula for the gamma function for one of the factors in the denominator of your prospective solution:\n\nAssuming[i \u2208 Integers && nn \u2208 Integers && 1 <= i <= nn,\nFullSimplify[1 + ((2 i - nn - 1) Gamma[1/2 - i] (nn - i)!)/\n(2 (\u03c0 Csc[\u03c0 i]/Gamma[i]) Gamma[3/2 - i + nn])]]\n1\n\n\u2022 Thanks, JM...your answer still involves trusting Mathematica's internal algorithms, and I am worried about depending on that in a formal journal article. Trust but verify? \u2013\u00a0MikeY Mar 20 '17 at 13:17\n\u2022 Well, actually, since $\\sin(\\pi i)=0,\\,i\\in \\mathbb Z$, that settles it, if you want to go manually. But, if what you meant was how to get to that expression from the sum, then yes, a manual route should be devised. \u2013\u00a0J. M. will be back soon Mar 21 '17 at 2:42\n\u2022 That's the million dollar question for me...when do I accept Mathematica's output, and when (and how) do I verify it? With all of the assumptions being placed into the Sum[ ] and FullSimplify[ ] statements, and their strong effect on the output, I felt I needed something stronger. This appears to be a deep, running issue in the math world - rightly so. \u2013\u00a0MikeY Mar 21 '17 at 16:18\n\nEdited to show problem completion...\n\nTo recap, I am asserting that $\\sum_{k} r(n,i,k) = 1$ for all $n$ positive integer and also for all $i$ between 1 and $n$. I made the $i$ explicit here.\n\nSolution approach is induction on both $n$ and $i$, in that order. Induce on $n$ first. Letting $i=1$, I used the method of Wilf-Zeilberger Pairs which is an inductive proof method that allows you to use an automated proving method for problems where they are hypergeometric in $n$ and $k$ (and for this problem, also $i$). (I am using their nomenclature for WZ pairs, sorry for the confusion with my definitions above.) Start with the summand after fully simplifying using Mathematica, $$F(n,k)=-\\frac{\\Gamma \\left(k-\\frac{3}{2}\\right) \\Gamma \\left(-k+\\text{nn}+\\frac{3}{2}\\right)}{\\pi \\Gamma (k) \\Gamma (-k+\\text{nn}+1)}$$ and came up with a proof certificate of\n\n$$R(n,k)=\\frac{-2 k^2+2 k \\text{n}+5 k-2 \\text{n}-3}{2 \\text{n} (k-\\text{n}-1)}$$ and a function $G(n,k)$ that is defined as $$G(n,k) = R(N,k) F(n,k-1) = \\frac{\\Gamma \\left(k-\\frac{3}{2}\\right) \\Gamma \\left(-k+\\text{nn}+\\frac{5}{2}\\right)}{\\pi \\text{nn} \\Gamma (k-1) \\Gamma (-k+\\text{nn}+2)}$$ Then checking that $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$$ and $$\\lim_{k \\to +/- \\infty} G(n,k)=0$$ This takes a few seconds to run, and I've also run it for $i=n$ and for the midpoint $i=(n+1)/2$ and for lots of values of $i=1,2,3,...n-3,n-2,n-1$ and it works fine. However, when I try to run it for the generic $i$, I get a messy expression.\n\nSo the second induction step, on $i$, remains unfinished.\n\nEDIT\n\nUsing the fastZeil.m package from Peter Paule, Markus Schorn and Axel Riese, and their implementation of the Zeilberger Algorithm, and defining\n\n$$\\sum_{k} r(k,n,i) = \\sum_{k} F(k,i) = \\text{SUM[i]}$$ was able to show the recurrence\n\n$$(-4 i^2+4 i n-4 i+3 n-3) \\text{SUM[i+1]}+i (2 i-2 n-1) \\text{SUM[i]}+(2 i+3) (i-n+1) \\text{SUM[i+2]}==0$$ with the proof certificate $$R(k,i)=\\frac{4 (k-1) (-2 i+2 k+1) (-2 k+2 \\text{n}+3)}{(-2 i+2 k-5) (-2 i+2 k-3)}.$$ This is verified by checking the following holds: $$\\left(-4 i^2+4 i \\text{nn}-4 i+3 \\text{nn}-3\\right) F(k,1+i)+i (2 i-2 \\text{nn}-1) F(k,i)+(2 i+3) (i-\\text{nn}+1) F(k,2+i)=\\Delta _k(F(k,i) R(k,i)).$$ Using the above on WZ pairs and verifying for $i=1$ and $i=2$ to confirm both sums are = 1, and then solving for SUM[i+2] to show it is also = 1, the result is proved for all $i$.", "date": "2019-12-07 14:57:50", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/139967/how-to-prove-this-summation-result", "openwebmath_score": 0.7388908267021179, "openwebmath_perplexity": 664.9365652671057, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102542943773, "lm_q2_score": 0.8791467738423873, "lm_q1q2_score": 0.8504956040449453}}
{"url": "http://math.stackexchange.com/questions/169542/further-reading-on-the-p-adic-metric-and-related-theory/169555", "text": "# Further reading on the $p$-adic metric and related theory.\n\nIn his book Introduction to Topology, Bert Mendelson asks to prove that\n\n$$(\\Bbb Z,d_p)$$\n\nis a metric space, where $p$ is a fixed prime and\n\n$$d_p(m,n)=\\begin{cases} 0 \\;,\\text{ if }m=n \\cr {p^{-t}}\\;,\\text{ if } m\\neq n\\end{cases}$$\n\nwhere $t$ is the multiplicty with which $p$ divides $m-n$. Now, it is almost trivial to check the first three properties, namely, that\n\n$$d(m,n) \\geq 0$$\n\n$$d(m,n) =0 \\iff m=n$$\n\n$$d(m,n)=d(n,m)$$\n\nand the only laborious was to check the last property (the triangle inequality). I proceeded as follows:\n\nLet $a,b,c$ be integers, and let\n\n$$a-b=p^s \\cdot k$$\n\n$$b-c=p^r \\cdot l$$\n\nwhere $l,k$ aren't divisible by $p$.\n\nThen $$a-c=(a-b)+(b-c)=p^s \\cdot k+p^r \\cdot l$$\n\nNow we have three cases, $s>r$, $r>s$ and $r=s$. We have respectively:\n\n$$a-c=(a-b)+(b-c)=p^r \\cdot(p^{s-r} \\cdot k+ l)=p^r \\cdot Q$$ $$a-c=(a-b)+(b-c)=p^{s} \\cdot( k+p^{r-s} \\cdot l)=p^s \\cdot R$$ $$a-c=(a-b)+(b-c)=p^s \\cdot (k+l)=p^s \\cdot T$$\n\nIn any case,\n\n$$d\\left( {a,c} \\right) \\leqslant d\\left( {a,b} \\right) + d\\left( {b,c} \\right)$$\n\nsince\n\n\\eqalign{ & \\frac{1}{{{p^r}}} \\leqslant \\frac{1}{{{p^s}}} + \\frac{1}{{{p^r}}} \\cr & \\frac{1}{{{p^s}}} \\leqslant \\frac{1}{{{p^s}}} + \\frac{1}{{{p^r}}} \\cr & \\frac{1}{{{p^s}}} \\leqslant \\frac{1}{{{p^s}}} + \\frac{1}{{{p^s}}} \\cr}\n\nIt might also be the case $k+l=p^u$ for some $u$ so that the last inequality is\n\n$$\\frac{1}{{{p^{s + u}}}} \\leqslant \\frac{1}{{{p^s}}} + \\frac{1}{{{p^s}}}$$\n\n$(1)$ Am I missing something in the above? The author asks to prove that in fact, if $t=t_p(m,n)$ is the exponent of $p$, that\n\n$$t\\left( {a,c} \\right) \\geqslant \\min \\left\\{ {t\\left( {a,b} \\right),t\\left( {b,c} \\right)} \\right\\}$$\n\nThat seems to follow from the above arguement, since if $s \\neq r$ then\n\n$$t\\left( {a,c} \\right) = t\\left( {a,b} \\right){\\text{ or }}t\\left( {a,c} \\right) = t\\left( {b,c} \\right)$$\n\nand if $s=r$ then\n\n$$t\\left( {a,c} \\right) \\geqslant t\\left( {a,b} \\right){\\text{ or }}t\\left( {a,c} \\right) \\geqslant t\\left( {b,c} \\right)$$\n\n$(2)$ Is there any further reading you can suggest on $p$-adicity?\n\n-\nWhat do you want to know? \u2013\u00a0 Qiaochu Yuan Jul 11 '12 at 17:15\n@QiaochuYuan If you're referring to the second question, I'm interested in it's appereance in say number theory, or topology, or just any undergraduate level reference that you find worth. \u2013\u00a0 Pedro Tamaroff Jul 11 '12 at 17:27\n\nI am currently trying to learn about $p$-adic numbers and analysis too, so I too would be really interested to hear the opinions of people who know more than I do about this. I am currently using the following three texts, but don't intend to work through them fully; just enough to get something useful for a better understanding of how they can be used in number theory:\n\n1. Koblitz - \"$p$-adic Numbers, $p$-adic Analysis, and Zeta-functions\" - The first chapter I have found very interesting, and pretty well-written, with lots of easy exercises to get used to the concepts, as well as some harder ones to test deeper understanding.\n2. Robert - \"A Course in $p$-adic Analysis\" - covers much more material at a more advanced level than Koblitz, but isn't (quite) as off-putting as it seems at first, and it possible to pick out quite a few bits from the first two chapters which are illuminating.\n3. Borevich and Shafarevich - \"Number Theory\" - This one has some relatively understandable stuff on $p$-adic numbers in the first chapter which I have found really useful as it gives a different feel to the topic in a rather more old-fashioned approach.\n-\nI will second the suggestion of Borevich and Shafarevich if you can find it. I'm a big fan of the general structure of that book, and their introduction to $p$-adic integers as both power series and as Witt vectors (though they clearly don't use the name). \u2013\u00a0 Brandon Carter Jul 11 '12 at 22:10\n\nWhat you have done seems correct. Also notice that from what you have done you get that $$d(a,c) \\leq \\max \\{d(a,b), d(b,c) \\},$$ which is stronger than the triangle inequality.\n\nFor an introduction to $p$-adic numbers, I would suggest Fernando Gouvea's $p$-adic Numbers: An introduction. It should be easy for an undergraduate to understand the book, and I think it is a very nice introduction.\n\nEdit: I should add that it's not easy for any undergraduate. One should've had a few courses which have a lot of proofs, and not just the standard calculus courses.\n\n-\n\nYour argument looks fine to me.\n\nA very good online introduction to $p$-adics are these notes; it covers modular arithmetic leading up to Hensel's, basic analysis with the numbers, the very strange metric topology of $\\Bbb Q_p$ (every point inside of a ball is a center, there are locally but not globally constant functions, etc.), and a bit of field theory and algebraic number theory.\n\nI don't really have any recommendation for number-theoretically in-depth stuff. Also, I tend to stick to online documents because the web is where I spend all my time anyway.\n\nIn actuality, my favorite discussion is p-adic integration and the theory of groups, which involves category theory and abstract algebra (the $p$-adics are constructed as an inverse limit of topological rings, for example), as well as measure theory and group theory. This source perhaps takes more background to digest its contents satisfactorily, but it hits my buttons well.\n\nJust for fun, I also suggest Pictures of Ultrametric Spaces. Related entertainment: (working with adeles) the character group of $\\Bbb Q$ and (working with profinite integers) profinite Fibonacci numbers.\n\n-", "date": "2014-04-19 15:27:41", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/169542/further-reading-on-the-p-adic-metric-and-related-theory/169555", "openwebmath_score": 0.7425917983055115, "openwebmath_perplexity": 304.7220778671534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102580527664, "lm_q2_score": 0.8791467690927438, "lm_q1q2_score": 0.850495602754267}}
{"url": "http://math.stackexchange.com/questions/75005/how-do-we-check-randomness", "text": "# How do we check Randomness? [duplicate]\n\nLet's imagine a guy who claims to possess a machine that can each time produce a completely random series of 0/1 digits (e.g. $1,0,0,1,1,0,1,1,1,...$). And each time after he generates one, you can keep asking him for the $n$-th digit and he will tell you accordingly.\n\nThen how do you check if his series is really completely random?\n\nIf we only check whether the $n$-th digit is evenly distributed, then he can cheat using:\n\n$0,0,0,0,...$\n$1,1,1,1,...$\n$0,0,0,0,...$\n$1,1,1,1,...$\n$...$\n\nIf we check whether any given sequence is distributed evenly, then he can cheat using:\n\n$(0,)(1,)(0,0,)(0,1,)(1,0,)(1,1,)(0,0,0,)(0,0,1,)...$\n$(1,)(0,)(1,1,)(1,0,)(0,1,)(0,0,)(1,1,1,)(1,1,0,)...$\n$...$\n\nI may give other possible checking processes but as far as I can list, each of them has flaws that can be cheated with a prepared regular series.\n\nHow do we check if a series is really random? Or is randomness a philosophical concept that can not be easily defined in Mathematics?\n\n-\n\n## marked as duplicate by egreg, LTS, user127096, voldemort, Claude LeiboviciApr 12 at 4:52\n\n\"...is randomness a philosophical concept that can not be easily defined in Mathematics?\" - pretty much. However we do want our (pseudo)random sequences to pass certain tests... \u2013\u00a0 \uff2a. \uff2d. Oct 23 '11 at 5:17\nAny particular set of tests can be cheated if you know what is. For theoretical cryptography purposes, one can define a bit generator as random if it passes \"all tests that run in a reasonable time\". More precisely, a proposed random bit generator is $(n,t,\\epsilon)$-random if, given the first $n$ bits of the sequence, there is no randomized algorithm that runs within time $t$ and successfully predicts the next bit with probability outside the range $1/2 \\pm \\epsilon$. \u2013\u00a0 Ted Oct 23 '11 at 7:36\nInteresting. So ... \"irrationality\", since it cannot be confirmed by looking at a finite number of decimals in a number, is a philosophical concept that cannot be easily defined in mathamatics ?? \"Continuity\", since it cannot be confirmed by a finite number of function evaluations is a philosophical concept that cannot be easily defined in mathamatics. \u2013\u00a0 GEdgar Oct 26 '11 at 13:20\n\nAll the sequences you mentioned have a really low Kolmogorov complexity, because you can easily describe them in really short space. A random sequence (as per the usual definition) has a high Kolmogorov complexity, which means there is no instructions shorter then the string itself that can describe or reproduce the string. Ofcourse the length of the description depends on the formal system (language) you use to describe it, but if the length of the string is much longer then the axioms of your formal systems, then the Kolmogorov-complexity of a random string becomes independent of your choice of system.\n\nLuckily, under the Church-Turing thesis, there is only 1 model of computation,(unless your machine uses yet undiscovered physical laws), so there is only 1 language your machine can speak that we have to check.\n\nSo to test if a string is random, we only have to brute-force check the length of the shortest Turing-program that outputs the first n bits correctly. If the length eventually becomes proportional to n, then we can be fairly certain we have a random sequence, but to be 100% we have to check the whole (infinite) string. (As per definition of random).\n\n-\n\nLeaving aside the theoretical aspect of your question, there are also pragmatic answers to it because there are real world uses for high-quality random generators (whether hardware or algorithmic). For statistical uses, \"statistical randomness\" is a used. For example you can use these \"diehard tests\" or TestU01.\n\n-\n\nThis is discussed very nicely in Volume 2 of Knuth's The Art Of Computer Programming. The executive summary is that randomness is a mathematical concept that can be defined in mathematics but not easily.\n\n-", "date": "2014-12-20 07:31:44", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/75005/how-do-we-check-randomness", "openwebmath_score": 0.8081348538398743, "openwebmath_perplexity": 585.1756470272005, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9674102580527665, "lm_q2_score": 0.8791467675095294, "lm_q1q2_score": 0.8504956012226493}}
{"url": "http://firstbostonsoftware.com/kt1a/exu.php?flz=function-of-two-variables", "text": "First, to define the functions themselves. 1, Functions of two variables p. The simplest method to swap two variables is to use a third temporary variable :. experiment to a function X(t,e). Very easy to understand!Prealgebra exponent lessons, examples and practice problems Algebra Lessons at Cool math. Derivatives told us about the shape of the function, and let us find local max and min - we want to be able to do the same thing with a function of two variables. A function of a single input variable observations has been created from the two-input variable function fitdistr: fixing one of the input variables by setting densfun = \"normal\". I found a and b for several values of x2, so I do have equations f(x1) for some fixed x2. In single-variable calculus, you learned how to compute the derivative of a function of one variable, y= f(x), with respect to its independent variable x, denoted by dy=dx. In a two-variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). You can use fminsearch to optimize your coefficients, but you still need to know the basic form of the function. Although the formal proof is not trivial, the variable-dependence diagram shown here provides a simple way to remember this Chain Rule. There is another way-a highly engaging way that does not neglect readers' own intuition, experience, and excitement. Average value of a function To find the average value of a function of two variables, let's start by looking at the average value of a function of one variable. This happens when you get a \u201cplus or minus\u201d case in the end. Create a function of two variables. One-variable calculus makes extensive use of graphs in or-. Now the UNION of A and B, written A B = (1,2,3,4,5). In mathematics, the result of a modulo operation is the remainder of an arithmetic division. There are three problems, each of which has a background discussion, an illustrative example, and an exercise for you to do. Functions f (x1, x2, , xn) of n variables, Symmetry. The Effective Use of Graphs. Modern code has few or no globals. For a thermal contact between the two put a thermal conductance value. In particular, a function of 2 variables is a function whose inputs are points ( x , y ) in the xy -plane and whose outputs real numbers. The inputs are ordered pairs, (x, y). User asks to enter the value. The outputs are real numbers. The Cumulative Distribution Function for a Random Variable \\ Each continuous random variable has an associated \\ probability density function (pdf) 0\u00d0B\u00d1 \\. The area of the triangle and the base of the cylinder: A= 1 2 bh. We also write z = f (x ,y ) The variables x and y are independent variables and z is the. Use the Show menu to switch from one mode to another. The value of num1 and num2 are initialized to variables a and b respectively. functions of several variables and partial differentiation (2) The simplest paths to try when you suspect a limit does not exist are below. The add-on store offers several custom functions as add-ons for Google Sheets. Usually this follows easily from the fact that closely related functions of one variable are continuous. as subroutines, routines, procedures, methods, or subprograms. De\ufb01nition 1. If you define global variables (variables defined outside of any function definition), they are visible inside all of your functions. A swapping function: To understand how explicit pass by reference of parameters works in C, consider implementing a swap function in C, that is, a function that passes in two variables and swaps their values. Definition 1. When you set a value for a variable, the variable becomes a symbol for that value. In elementary calculus, we concentrate on func-tions of a single variable; we will now extend that investigation to study functions of two or more variables. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a. Graphs of Functions of two Variables Recall that for a function f of a single variable, the graph of f(x) in the xy-plane was de\ufb01ned to be the graph of the equation y = f(x). com - Functions. Fortunately for us, we have technology which facilitates this task. Imagine that the surface is smooth and has some hills and some valleys. The 10% value indicates that the relationship between your independent variable and dependent variable is weak, but it doesn\u2019t tell you the direction. There is another way-a highly engaging way that does not neglect readers' own intuition, experience, and excitement. Also, use ss2tf to obtain the \ufb02lter\u2019s transfer function. Part A: Functions of Two Variables, Tangent Approximation and Opt; Part B: Chain Rule, Gradient and Directional Derivatives; Part C: Lagrange Multipliers and Constrained Differentials; Exam 2. Solve this system of equations by using substitution. You define a function in much the same way you define a variable. Use the debugger to see what's the mismatch in dimensions; it's not totally apparent as one would presume i is a loop index and so is a single integer value; if MS3 is an array it would also be a single value but if it happened to be a function it could return something other than. Functions of Several Variables This manual contains solutions to odd-numbered exercises from the book Functions of Several Vari-ables by Miroslav Lovri\u00b4c, published by Nelson Publishing. Note that it is assumed that the two lists given in the table command are both factors. Economists of this period, while recognizing that the law of diminishing returns (or the law of variable proportions) applied when units of a variable. However, there is also a main di\u2044erence. Find the standard deviation of the eruption duration in the data set faithful. Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify Statistics Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. The added risk brought on by the complexity of machine-learning models can be mitigated by making well-targeted modifications to existing validation frameworks. For functions of two or three variables the situation is more complicated because there are in\ufb01nitely. \u2022 Matlab has several different functions (built-ins) for the numerical. Staffing: After a manager discerns his area's needs, he may decide to beef up his staffing by recruiting, selecting, training, and developing employees. We now extend this concept to functions of two variables. com - Functions. Swapping two variables refers to mutually exchanging the values of the variables. When we considered functions and graphs of one variable, one of the first things we did was to transform those graphs through shifts and stretches. De nition A critical point (x0;y0) of fis a point where both the partial derivatives @f=@xand @f=@y. If you will need guidance with algebra and in particular with ordered pairs and inequalies online calculator or fractions come visit us at Algebra-equation. For a function of one variable, a function w = f (x) is differentiable if it is can be locally approximated by a linear. Active 2 years, 7 months ago. Gain additional perspective by studying polar plots, parametric plots, contour plots, region plots and many other types of visualizations of the functions and equations of interest to you. Jacobians of Random Graphs Acknowledgments Funding References 2000 AMS Subject Classi\ufb01cation: 05C31, 05C50, 14T05, 14H99. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). Okay, as if two methods aren't enough, we still have one more method we could use. Let us assume that both f and as many partial derivatives as necessary are continuous near (x 0,y 0). I am now trying to find a general equation f(x1,x2). Some students did not show to have made this coordination. The code on the left below shows one failed attempt at an implementation. For a continuous real-valued function of two real variables, the graph is a surface. These are special variables that take on the values that you give when you call for the function, meaning you can give it any two numbers and it can add them together. Because the correlation between reading and mathematics can be determined in the top section of the table, the correlations between those two variables is not repeated in the bottom half of the table. of Manchester) 5 2 Functions of multiple [two] variables In many applications in science and engineering, a function of interest depends on multiple. The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables. Algebra functions lessons with lots of worked examples and practice problems. Alternatively, the function also knows it must return the first argument, if the value of the \"number\" parameter, passed into the function, is equal to \"first\". Definition of Mathematical Expectation Functions of Random Variables Some Theorems on Expectation The Variance and Standard Deviation Some Theorems on Variance Stan-dardized Random Variables Moments Moment Generating Functions Some Theorems on Moment Generating Functions Characteristic Functions Variance for Joint Distribu-tions. To use or explore these add-ons: Create or open a spreadsheet in Google Sheets. The partial derivative of f with respect to y can similarly be found by treating x as a constant whenever it appears. second variable y appears, it is treated as a constant in every respect. The concept of the graph of a function is generalized to the graph of a relation. Part 1: Functions of 2 Variables. input variables and other variables you create within the function and in doing so, you create the output variables you desire. You have now created a function called sum. One Function of Two Random Variables Given two random variables X and Y and a function g(x,y), we form a new random variable Z as Given the joint p. This will help us to see some of the interconnections between what can seem like a huge body of loosely related de nitions and theorems1. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. (a) True, and I am very con dent (b) True, but I am not very con dent (c) False, but I am not very con dent (d) False, and I am very con dent 2. Example 3: Using the function from Example 2, describe and graph the following functions: (i) f(x, y) = 3 - x2 - y2. Note that it is assumed that the two lists given in the table command are both factors. For in-stance, step functions are continuous except at their steps, that is, where there are jump discontinu-ities. In the next two sections we introduce these two concepts and develop some of their properties. 10 Two-Dimensional Random Variables De\ufb01nition 1. Addition of two numbers in C For example, if a user will input two numbers as; '5', '6' then '11' (5 + 6) will be printed on the screen. Let (X;d)and (Y;d\u2032)be two metric spaces, A \u2286X a nonempty set, a function f \u2236A \u2192Y and x. f Obviously. Derivatives told us about the shape of the function, and let us find local max and min - we want to be able to do the same thing with a function of two variables. Relation with other tests Changing the number of variables. We also write z = f (x ,y ) The variables x and y are independent variables and z is the. even functions of one variable may have both maximum and minimum points). The function writePictureTo takes two parameters: the picture variable and the pathname. First-order partial derivatives of functions with two variables. The area of a circle is a function of -- it depends on -- the radius. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. In the expression (c = a + b) overflow may occur if the sum of a and b is larger than the maximum value which can be stored in the variable c. In the short run, production function is explained with one variable factor and other factors of productions are held constant. ) Variables and functions in all parts of a makefile are expanded when read, except for the shell commands in rules, the right-hand sides of variable definitions using `=', and the bodies of variable definitions using the define directive. Now you know the basics of using two variable -- or complex -- functions. These are special variables that take on the values that you give when you call for the function, meaning you can give it any two numbers and it can add them together. We recall that a critical point of a function of several variables is a point at which the gradient of the function is either the zero vector 0 or is undefined. Let f : D \u2282 R \u2192 R and let a \u2208 R. To plot the point (2,3), for example, you start at the origin Independent and Dependent Variables. Scalar functions of two variables Our main goal in this tutorial is to explore ways to plot functions of two variables. f how does one obtain. 2 Graphs should always have at minimum a caption, axes and scales, symbols, and a data field. De nition A critical point (x0;y0) of fis a point where both the partial derivatives @f=@xand @f=@y. Usually, there is more than one correct answer. And the fun part with these guys is that you can just kind of, imagine a fluid flowing, so here's a bunch of droplets, like water, and they kind of flow along that. x^2*y+x*y^2 \uff09 The reserved functions are located in \" Function List \". I am trying to create the interpolating function for a function of two variables, over a finite area. It can be used as a worksheet function (WS) in Excel. \u2022 Matlab has several different functions (built-ins) for the numerical. For a function of a single variable there are two one-sided limits at a point x0, namely, lim x!x+ 0 f(x) and lim x!x 0 f(x) re\ufb02ecting the fact that there are only two directions from which x can approach x0, the right or the left. Concave functions of two variables While we will not provide a proof here, the following three definitions are equivalent if the function f is differentiable. There is no need to list the 3 twice. Definition of Variables and Examples. identically distributed Exponential random variables with a constant mean or a constant parameter (where is the rate parameter), the probability density function (pdf) of the sum of the random variables results into a Gamma distribution with parameters n and. accept a wide variety of mathematical expressions. To close the answer window and get back to the quiz, click on the X in the upper right corner of the answer window. Equations of a Straight Line. Functions of two variables 1. AMS 311 Joe Mitchell Examples: Joint Densities and Joint Mass Functions Example 1: X and Y are jointly continuous with joint pdf f(x,y) = \u02c6 cx2 + xy 3 if 0 \u2264 x \u2264 1, 0 \u2264 y \u2264 2. Functions can be recognized, described, and examined in a variety of ways, including graphs, tables, and sets of ordered pairs. Could someone please explain a function of two variables to me. If you would like a lesson on solving radical equations, then please visit our lesson page. Hence the square of a Rayleigh random variable produces an exponential random variable. Therefore, in order to be able to. The applet initially starts in the Input mode, which lets you choose a function to plot (you can either enter it manually, or select one from the drop-down list; click on the Plot button to create the new plot). Observe that because of the non-negativity constraint, the sum of any collection of variables cannot be negative. Functions of Two Variables. Variable b1 and b2 are baseline variables. Laval (KSU) Functions of Several Variables Today 14 / 22. peaks is a function of two variables, obtained by translating and scaling Gaussian distributions, which is useful for demonstrating mesh, surf, pcolor, contour, and so on. One important similarity to notice between the limit of a one variable function and the limit of a two variable function is that $\\sqrt{(x - a)^2 + (y - b)^2}$ represents the distance between the point $(x, y)$ and $(a, b)$ in $\\mathbb{R}^2$. First, we will create an intensity image of the function and, second, we will use the 3D plotting capabilities of. My function is exponential for x1 with two coefficients that depend on x2: f(x1,x2)=a*(x1)^b, where a and b are functions of x2. Files are available under licenses specified on their description page. Under the pass-by-value mechanism, the parameter variables within a function receive a copy of the variables ( data ) passed to them. The scatter plot plots the points (x, y) where x is a value from one data list (Xlist) and y is the corresponding value from the other data list (Ylist). A function f(x, y) of two independent variables has a maximum at a point (x 0, y 0) if f(x 0, y 0) f(x, y) for all points (x, y) in the neighborhood of (x 0. Laval (KSU) Functions of Several Variables Today 14 / 22. The outputs are real numbers. As the n -tuple x = (x1, x2, , xn) varies in X, a subset of \u211dn, Implicit functions. Chain Rule And Composite Functions Derivative of Composite Function with the help of chain rule: When two functions are combined in such a way that the output of one function becomes the input to another function then this is referred to as composite function. Lady (September 5, 1998) There are three ways that a function can be discontinuous at a point. Distributions of Functions of Random Variables 1 Functions of One Random Variable Case of two-to-one transformations. You can choose any other combination of numbers as well. First, we will create an intensity image of the function and, second, we will use the 3D plotting capabilities of matplotlib to create a shaded surface plot. 4 Higher partial derivatives Notice that @f @x and @f @y are themselves functions of two variables, so they can also be partially differenti-ated. Functions of three variables are similar in many aspects to those of two variables. Because we're trying to keep things a little bit simpler, we'll concentrate on functions of two variables. Limits of a Rational Function of Two Variables Roger B. When variables change together, their interaction is called a relation. The dependent variable is what is affected by the independent variable-- your effects or outcomes. In the above example, two variables, num1 and num2 are passed to function during function call. Example 3: Using the function from Example 2, describe and graph the following functions: (i) f(x, y) = 3 - x2 - y2. Topic 5: Functions of multivariate random variables \u2020 Functions of several random variables Sum of 2 random variables \u2020 Let X and Y be two random variables. These are just constant functions, and because of that, degree 0 polynomials are often called constant polynomials. Equations of a Straight Line. For functions of two or three variables the situation is more complicated because there are in\ufb01nitely. This firm minimizes its cost of producing any given output y if it chooses the pair (z 1, z 2) of inputs to solve the problem. Average value of a function To find the average value of a function of two variables, let's start by looking at the average value of a function of one variable. It seems reasonable, and can be shown to be true,. FUNCTION OF TWO VARIABLES Definition: A variable Z is said to be a function of two independent variables x and y denoted by z=f (x,y) if to each pair of values of x and y over some domain D f ={(x,y): a type:. You have now created a function called sum. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. which is the density for an exponential random variable with parameter = 1/(2 2a), as can be seen from inspection of (2-27). Sometimes it will be preferable to think of f as taking one (2-dimensional) vector input instead of two scalar inputs. Optimization Problems with Functions of Two Variables. Our discussion is not limited to functions of two variables, that is, our results extend to functions of three or more variables. As with single variable functions, two classes of common functions are particularly useful and easy to describe. The variables held fixed are viewed as parameters. User make a function named swap that will be called in other class. For a function of n variables it can be a maximum point, a minimum point or a point that is analogous to an inflection or saddle point. two variables y et z is put equal to zero, then either variable is defined by the other and thus a function of this variable emerges, since before they were not mutually dependent. Write a script m-\ufb02le and use the Control System Toolbox functions ss and ltiview to form the state model and its step response. Furthermore, sums, dif-. of Manchester) 5 2 Functions of multiple [two] variables In many applications in science and engineering, a function of interest depends on multiple. \u2022 Matlab has several different functions (built-ins) for the numerical. The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables. I suspect I will need the surface chart but can some one tell me how to generate the chart and what to enter on the worksheet. Under the pass-by-value mechanism, the parameter variables within a function receive a copy of the variables ( data ) passed to them. \u2020Forcontinuous randomvariables. The set D is the domain of f and its range is the set of values that f takes on. Following are different ways. Does anyone know of any helpful tutorials that will help me get the Domain and range, functions of 2 variables | Physics Forums. The add-on store offers several custom functions as add-ons for Google Sheets. For a thermal contact between the two put a thermal conductance value. In the case of functions of two variables, that is functions whose domain consists of pairs (x, y), the graph can be identified with the set of all ordered triples ((x, y, f(x, y)). Two expressions involving template parameters are called equivalent if two function definitions that contain these expressions would be the same under ODR rules, that is, the two expressions contain the same sequence of tokens whose names are resolved to same entities via name lookup, except template parameters may be differently named. I will give the definition of differentiablity in 2D. Quotient of two random variables. For a function of one variable, a function w = f (x) is differentiable if it is can be locally approximated by a linear. There is no need to list the 3 twice. The sum of two incomes, for example, or the difference between demand and capacity. 2 Limits and Continuity of Functions of Two Variables In this section, we present a formal discussion of the concept of continuity of functions of two variables. More generally, if two or three variables are changing, how do we explore the correspondingchangein w? The answer to these questionsstarts with the generalizationof the idea of the differential as linear approximation. Local extreme values of a function of two variables. Limits and Continuity of Functions of Two or More Variables Introduction. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. The euclidean_division function to calculate online the quotient and the remainder in the euclidean division of two polynomials or two integers. This gives a nice graphical representation where the plane at x = 0 bounds the function from below. Continuous Random Variables Acontinuous random variable X takes values in an interval of real numbers. functions of two variables. In this paper distribution of zeros of solutions of functional equations in the space of functions of two variables is studied. So far, we have discussed how we can find the distribution of a function of a continuous random variable starting from finding the CDF. Functions f (x1, x2, , xn) of n variables, Symmetry. That is, a function expresses dependence of one variable on one or more other variables. Let us assume that both f and as many partial derivatives as necessary are continuous near (x 0,y 0). Recall that the de\ufb01nition of the limit of such functions is as follows. More information about applet. In single-variable calculus we were concerned with functions that map the real numbers $\\R$ to $\\R$, sometimes called \"real functions of one variable'', meaning the \"input'' is a single real number and the \"output'' is likewise a single real number. Not only for computing the variance of the transformed variable Y, but also for its mean. When variables change together, their interaction is called a relation. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. For a function of a single variable there are two one-sided limits at a point x0, namely, lim x!x+ 0 f(x) and lim x!x 0 f(x) re\ufb02ecting the fact that there are only two directions from which x can approach x0, the right or the left. The graph of is a subset of three-dimensional Euclidean space with coordinates , given by the equation: Equivalently, it is the set of points: Pictorially, this graph looks like a surface for a nice enough function. $\\endgroup$ - Gerhard Paseman Feb 13 at 18:10. Fortunately, the functions we will examine will typically be continuous almost everywhere. The function makes it possible to verify by using the Pythagorean theorem knowing the lengths of the sides of a triangle that this is a right triangle. For a continuous real-valued function of two real variables, the graph is a surface. Most useful functions of one variable are con-tinuous, but there are a few exceptions. characterizations, namely, the mass function for discrete random variable and the density function for continuous random variables. 3-Dimensional graphs of functions are shown to confirm the existence of these points. In a \"system of equations,\" you are asked to solve two or more equations at the same time. Re: st: computing covariance. com, a free online graphing calculator. To evaluate z, first create a set of (x,y) points over the domain of the function using meshgrid. functions of several variables and partial differentiation (2) The simplest paths to try when you suspect a limit does not exist are below. 1 Visualizing functions of 2 variables One problem with thinking about functions of several variables is that they can be harder to picture than functions of just one variable. Imagine a surface, the graph of a function of two variables. Functions of 2 Variables Functions and Graphs In the last chapter, we extended di\u2044erential calculus to vector-valued functions. AMS 311 Joe Mitchell Examples: Joint Densities and Joint Mass Functions Example 1: X and Y are jointly continuous with joint pdf f(x,y) = \u02c6 cx2 + xy 3 if 0 \u2264 x \u2264 1, 0 \u2264 y \u2264 2. Hence, time is always on the X axis. Functions 3D Plotter and Analytic double integrator Functions 3D Plotter is an on line app to plotting two-variabled real functions, ie functions of type f(x,y) or with more precision f: R 2 \u2192 R (x,y) \u2192 f(x,y) 3D Functions Plotter calculates double integrals in analytic or numeric form. If the relation is not a function the graph contains at least two points with the same x-coordinate but with different y-coordinates. There is a probability associated with X falling between two numbers a weekday ) ) ) { $datemonth =$wp_locale->get_month( $datefunc( 'm',$i ) ); $datemonth_abbrev =$wp_locale->get_month_abbrev. For functions of two or three variables the situation is more complicated because there are in\ufb01nitely. Free online 3D grapher from GeoGebra: graph 3D functions, plot surfaces, construct solids and much more!. Again, please enter this line into. Functions of more variables can be defined similarly. It would be useful to read these two guides. One primary difference, however, is that the graphs of functions of more than two variables cannot be visualized directly, since they have dimension greater than three. The concept of the graph of a function is generalized to the graph of a relation. Thread: chart function of two variables. Furthermore, sums, dif-. For example, if you are studying the effects of a new educational program on student achievement, the program is the independent variable and your measures of achievement are the dependent ones. You can choose any other combination of numbers as well. f(x,y) is inputed as \"expression\". Fortunately for us, we have technology which facilitates this task. Use Wolfram|Alpha to generate plots of functions, equations and inequalities in one, two and three dimensions. But polynomials, trig functions, power and root functions, logarithms, and exponential func-tions are all continuous. Suppose that X and Y are two random variables having moment generating functions MX(t) and MY (t) that exist for all t in some interval 3. In the present case, we see that the critical point at the origin is a local maximum of f2 , and the second critical point is a saddle point. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. The standard deviation of an observation variable is the square root of its variance. It is good programming practice to avoid defining global variables and instead to put your variables inside functions and explicitly pass them as parameters where needed. I'm having a bit of trouble grasping the domain and range of functions of 2 variables. 3-Dimensional graphs of functions are shown to confirm the existence of these points. Boolean Functions (Expressions) It is useful to know how many different Boolean functions can be constructed on a set of Boolean variables. To input the variable x as a Wildcard, first type Shift + ?, then type x; similarly, for y. Importantly,. peaks is a function of two variables, obtained by translating and scaling Gaussian distributions, which is useful for demonstrating mesh, surf, pcolor, contour, and so on. Local extreme values of a function of two variables. Dependent has two categories, there is only one discriminant function. When variables change together, their interaction is called a relation. In general, I can't create new functions in a poisoned session. That is, a function expresses dependence of one variable on one or more other variables. Hi, It is possible to define a function of two variables using an interpolation function defined in a text file using the spreadsheet data format: the first column contains the values for the first input argument, the second column contains the values for the seond input argument, and the third column contains the function's value. Function definition, the kind of action or activity proper to a person, thing, or institution; the purpose for which something is designed or exists; role. So this is more like a re-visit to the good old topic. As the other answer shows, the mere existence of partial derivatives doesn't even guarantee that the function is continuous. Notice we kept that one dimensional distance in our limit definition for functions of two variables when we said |f(x, y) - L| < e. Applications of Extrema of Functions of Two Variables. Although the formal proof is not trivial, the variable-dependence diagram shown here provides a simple way to remember this Chain Rule. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). If not, then we will want to test some paths along some curves to first see if the limit does not exist. Also, use ss2tf to obtain the \ufb02lter\u2019s transfer function. Fortunately for us, we have technology which facilitates this task. You can create a two way table of occurrences using the table command and the two columns in the data frame: In this example, there are 51 people who are current smokers and are in the high SES. \\+,\u0153T\u00d0+\u0178\\\u0178,\u00d1\u01530\u00d0B\u00d1. 16 Possible Functions of Two Variables. \u201d For example, how much you weigh is related (correlated) to how much you eat. Correlation look at trends shared between two variables, and regression look at causal relation between a predictor (independent variable) and a response (dependent) variable. f Obviously. In case of two independent variables X 1 and X 2 such a function may be expressed as under: Y = a + bX 1 - cX 2 1 + dX 2 - eX 2 2. First, we introduce the de nition of a function of two variables: A scalar-valued. Equations of a Straight Line. The standard deviation of an observation variable is the square root of its variance. Polynomial Calculator. In mathematics, the result of a modulo operation is the remainder of an arithmetic division. Graph the function f(x,y) = xy using x,y,z-coordinate axes in 3-D space. It seems reasonable, and can be shown to be true,. Just for consistency we can think of a function:. Loading Graph Functions of 2 Variables. lang package, and not in the java. In the case of functions of two variables, that is functions whose domain consists of pairs (x, y), the graph can be identified with the set of all ordered triples ((x, y, f(x, y)). 2 to find the resulting PDFs. When we extend this notion to functions of two variables (or more), we will see that there are many similarities. For functions of two or three variables the situation is more complicated because there are in\ufb01nitely many. I am now trying to find a general equation f(x1,x2). Long weekends and highway traffic on Friday afternoon C. First-order partial derivatives of functions with two variables. Applications of Extrema of Functions of Two Variables. Could someone please explain a function of two variables to me. function of two variables is far more di\u00a2 cult than a function of one variable. Calculates the table of the specified function with two variables specified as variable data table. The INTERSECTION of two sets is the set of elements which are in both sets. You define a function in much the same way you define a variable. The value of num1 and num2 are initialized to variables a and b respectively. Function composition. The partial derivative of a function of two or more variables with respect to one of its variables is the ordinary derivative of the function with respect to that variable, considering the other variables as constants. In the example above, the diagonal was used to report the correlation of the four factors with a different variable. time) and one or more derivatives with respect to that independent variable. y(s;t) and z(s;t), are called the component functions of the vector-valued function g. The Method of Transformations. Examples 4. Here that means you need to use the. I found a and b for several values of x2, so I do have equations f(x1) for some fixed x2. Furthermore, sums, dif-.", "date": "2019-10-15 10:55:36", "meta": {"domain": "firstbostonsoftware.com", "url": "http://firstbostonsoftware.com/kt1a/exu.php?flz=function-of-two-variables", "openwebmath_score": 0.6182748079299927, "openwebmath_perplexity": 349.66721326333993, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787864878117, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8504922476085536}}
{"url": "http://math.stackexchange.com/questions/129993/calculating-n-mod-pq-from-n-mod-p-and-n-mod-q", "text": "# Calculating n mod pq from n mod p and n mod q\n\nLet $p$ and $q$ be relative primes, $n$ positive integer.\n\nGiven\n\n\u2022 $n\\bmod p$ and\n\u2022 $n\\bmod q$\n\nhow do I calculate $n\\bmod (pq)$ ?\n\n-\nen.wikipedia.org/wiki/\u2026 \u2013\u00a0 pedja Apr 10 '12 at 11:26\nYou might also have a look at Easy CRT in some Bill Dubuque's posts. \u2013\u00a0 Martin Sleziak Apr 10 '12 at 11:30\n\nSince $p$ and $q$ are relatively prime, there are integers $a$ and $b$ such that $ap+bq=1$. You can find $a$ and $b$ using the Extended Euclidean algorithm. Then $n\\equiv aps+bqr \\bmod pq$ if $n\\equiv r \\bmod p$ and $n\\equiv s \\bmod p$.\n\nAs pedja mentioned, this is a constructive proof of the Chinese remainder theorem.\n\n-\n+1 for getting their first !!! \u2013\u00a0 hardmath Apr 10 '12 at 20:48\n\nFrom the fact that $p,q$ are relatively prime, we can find coefficients $a,b$ such that:\n\n$$ap + bq = 1$$\n\nWith these coefficients we can piece together a solution for n from its residues modulo $p$ and $q$. Say:\n\n$$n \\equiv r \\mod p$$ $$n \\equiv s \\mod q$$\n\nThen this works: $n = sap + rbq$ since:\n\n$$bq \\equiv 1 \\mod p$$ $$ap \\equiv 1 \\mod q$$\n\nOf course the above expression for $n$ can be reduced modulo $pq$ without affecting the residues modulo $p$ and $q$.\n\n-\n\nOne may use the Bezout identity $\\rm\\:a\\:p + b\\:q = 1\\:$ obtained by the extended Euclidean algorithm. But, in practice, it's often more convenient to use the form below, e.g. see my Easy CRT posts.\n\nTheorem (Easy CRT) $\\rm\\ \\$ If $\\rm\\ p,\\:q\\:$ are coprime integers then $\\rm\\ p^{-1}\\$ exists $\\rm\\ (mod\\ q)\\ \\$ and\n\n$\\rm\\displaystyle\\quad\\quad\\quad\\quad\\quad \\begin{eqnarray}\\rm n&\\equiv&\\rm\\ a\\ (mod\\ p) \\\\ \\rm n&\\equiv&\\rm\\ b\\ (mod\\ q)\\end{eqnarray} \\ \\iff\\ \\ n\\ \\equiv\\ a + p\\ \\bigg[\\frac{b-a}{p}\\ mod\\ q\\:\\bigg]\\ \\ (mod\\ p\\:\\!q)$\n\nProof $\\rm\\ (\\Leftarrow)\\ \\ \\ mod\\ p\\!:\\:\\ n\\equiv a + p\\ (\\cdots)\\equiv a\\:,\\$ and $\\rm\\ mod\\ q\\!:\\:\\ n\\equiv a + (b-a)\\ p/p \\equiv b\\:.$\n\n$\\rm\\ (\\Rightarrow)\\ \\$ The solution is unique $\\rm\\ (mod\\ p\\!\\:q)\\$ since if $\\rm\\ x',\\:x\\$ are solutions then $\\rm\\ x'\\equiv x\\$ mod $\\rm\\:p,q\\:$ therefore $\\rm\\ p,\\:q\\ |\\ x'-x\\ \\Rightarrow\\ p\\!\\:q\\ |\\ x'-x\\ \\$ since $\\rm\\ \\:p,\\:q\\:$ coprime $\\rm\\:\\Rightarrow\\ lcm(p,q) = p\\!\\:q\\:.\\quad$ QED\n\n-\nThe link goes to emptiness... \u2013\u00a0 I. J. Kennedy May 1 '13 at 0:02", "date": "2015-10-09 12:42:21", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/129993/calculating-n-mod-pq-from-n-mod-p-and-n-mod-q", "openwebmath_score": 0.9358230233192444, "openwebmath_perplexity": 200.26827400050854, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9871787857334058, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.850492246958604}}
{"url": "https://www.12000.org/my_notes/pde_in_CAS/maple_2020_2_and_mma_12_2/insu9.htm", "text": "#### 2.1.3 Transport equation $$u_t+2 u_x = 0$$ IC $$u(-1,x)=\\frac {x}{1+x^2}$$. Peter Olver textbook, 2.2.2 (b)\n\nproblem number 3\n\nTaken from Peter Olver textbook, Introduction to Partial di\ufb00erential equations.\n\nSolve for $$u(t,x)$$ in $$u_t+2 u_x = 0$$ with IC $$u(-1,x)=\\frac {x}{1+x^2}$$\n\nMathematica\n\n$\\left \\{\\left \\{u(t,x)\\to \\frac {-2 t+x-2}{4 t^2-4 t (x-2)+x^2-4 x+5}\\right \\}\\right \\}$\n\nMaple\n\n$u \\left (t , x\\right ) = \\frac {-2 t +x -2}{\\left (-2 t +x -2\\right )^{2}+1}$\n\nHand solution\n\nSolve $u_{t}+2u_{x}=0$ With initial conditions $$u\\left ( -1,x\\right ) =\\frac {x}{1+x^{2}}$$.\n\nSolution\n\nLet $$u=u\\left ( x\\left ( t\\right ) ,t\\right )$$. Then \\begin {equation} \\frac {du}{dt}=\\frac {\\partial u}{\\partial x}\\frac {dx}{dt}+\\frac {\\partial u}{\\partial t}\\tag {2} \\end {equation} Comparing (1),(2) shows that \\begin {align} \\frac {du}{dt} & =0\\tag {3}\\\\ \\frac {dx}{dt} & =2\\tag {4} \\end {align}\n\nEq (3) says that $$u$$ is constant on the chataterstic lines, or $$u=u\\left ( x\\left ( -1\\right ) \\right )$$. Using the given initial conditions, this becomes \\begin {equation} u\\left ( x\\left ( t\\right ) ,t\\right ) =\\frac {x\\left ( -1\\right ) }{1+x\\left ( -1\\right ) ^{2}}\\tag {5} \\end {equation} Eq (4) is now used to \ufb01nd $$x\\left ( -1\\right )$$. Soving (4) gives $$x=x\\left ( 0\\right ) +2t$$. Hence $$x\\left ( -1\\right ) =x\\left ( 0\\right ) -2$$ or $$x\\left ( 0\\right ) =x\\left ( -1\\right ) +2$$. Therefore \\begin {align*} x & =x\\left ( -1\\right ) +2+2t\\\\ x\\left ( -1\\right ) & =x-2-2t \\end {align*}\n\nNow that we found $$x\\left ( -1\\right )$$, we substitute it in (5), giving the solution$u\\left ( x\\left ( t\\right ) ,t\\right ) =\\frac {x-2-2t}{1+\\left ( x-2-2t\\right ) ^{2}}$ Alternative method. Using Lagrange-charpit method\n\n$\\frac {dt}{1}=\\frac {dx}{2}=\\frac {du}{0}$ Which implies that $$du=0$$ or $$u=C_{1}$$. A constant. Integrating $$\\frac {dt}{1}=\\frac {dx}{2}$$ gives $$t=\\frac {1}{2}x+C_{2}$$ or $$C_{2}=t-\\frac {1}{2}x$$. But $$C_{1}=F\\left ( C_{2}\\right )$$ always, where $$F$$ is arbitrary function. Since $$C_{1}=u$$ then\\begin {align} u & =F\\left ( C_{2}\\right ) \\nonumber \\\\ u & =F\\left ( t-\\frac {1}{2}x\\right ) \\tag {1} \\end {align}\n\nAt $$t=-1$$ the above becomes$\\frac {x}{1+x^{2}}=F\\left ( -1-\\frac {1}{2}x\\right )$ Let $$-1-\\frac {1}{2}x=z$$ which implies $$x=-2\\left ( 1+z\\right )$$ The above can be written as\\begin {align*} \\frac {-2\\left ( 1+z\\right ) }{1+\\left ( -2\\left ( 1+z\\right ) \\right ) ^{2}} & =F\\left ( z\\right ) \\\\ F\\left ( z\\right ) & =-\\frac {2\\left ( 1+z\\right ) }{4z^{2}+8z+5} \\end {align*}\n\nFrom the above then (1) can be written as\\begin {align*} u\\left ( t,x\\right ) & =-\\frac {2\\left ( 1+\\left ( t-\\frac {1}{2}x\\right ) \\right ) }{4\\left ( t-\\frac {1}{2}x\\right ) ^{2}+8\\left ( t-\\frac {1}{2}x\\right ) +5}\\\\ & =\\frac {x-2t-2}{4t^{2}-4tx+8t+x^{2}-4x+5}\\\\ & =\\frac {x-2t-2}{1+\\left ( x-2-2t\\right ) ^{2}} \\end {align*}\n\nThe following is an animation of the solution\n\n 3D 2D\n\nSource code used for the above\n\n________________________________________________________________________________________", "date": "2021-12-06 14:53:15", "meta": {"domain": "12000.org", "url": "https://www.12000.org/my_notes/pde_in_CAS/maple_2020_2_and_mma_12_2/insu9.htm", "openwebmath_score": 0.9974872469902039, "openwebmath_perplexity": 10867.809705596019, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787879966233, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.850492241889369}}
{"url": "http://phoxis.org/2012/05/11/number-of-trailing-zeros-in-factorial-of-an-integer/", "text": "## Number of trailing zeros in factorial of an\u00a0integer\n\nAn integer n is given, the task is to find the number of trailing zeros in n! .\n\n### Solution\n\nFirst we need to see what produces trailing zeros. We assume that the integers are represented in base 10. A pair of 5 and 2 (which are the prime factors of 10) will produce a trailing zero in a sequence of integer multiplication. The number of trailing zeros will not decrease as once a trailing zero is generated in a product it cannot be removed by multiplying with some non zero integer. For factorial of n the multiplication series is: $n\\times (n-1)\\times(n-2)\\times \\ldots \\times 3 \\times 2 \\times 1$ . We need to count number of pairs of 5 and 2 present as a factor of each integer in the series. Naturally there are at least equal number of factors 2 than 5, if not more. Therefore basically the number of times 5 occurs as a factor totally in the series, ie. the multiplicity of the prime factor 5 in the factorial represents the number of trailing zeros.\n\nFor example in the case of 25! , the factor 5 occurs 6 times totally in the series of multiplication $25 \\times 24 \\times \\ldots \\times 3 \\times 2 \\times 1$. 5, , ie. the multiplicity of the factor 5 is 6. The integers 10, 15, 20 is divided by 5 once, and twice for 25.\n\nTherefore now the problem decreases to count the number of times the factor 5 occurs in the given series of multiplication. Which can be found as below:\n\n$trailing zeros = \\sum_{i=1}^{k} \\left\\lfloor n/5^i \\right\\rfloor = \\left\\lfloor n/5 \\right\\rfloor + \\left\\lfloor n/5^2 \\right\\rfloor + \\left\\lfloor n/5^3 \\right\\rfloor + \\ldots + \\left\\lfloor n/5^k \\right\\rfloor$ such that $5^k \\le n$\n\nHere is some explanation. The term $\\left\\lfloor n/5^i \\right\\rfloor$ finds the number of integers between 1 and n that $5^i$ divides. For example for n = 27, $\\left\\lfloor 27/5 \\right\\rfloor = 5$, that is there are 5 integers divisible by 5 within the range 1 to 27, in other words there is 5 integers which has the factor 5 once, which are 5, 10, 15, 20, and 25. Also $\\left\\lfloor 27/5^2 \\right\\rfloor = 1$ , that is, there are 1 integer divisible by 25 within the range 1 to 27, or in other words there is one integer which has the factor 5 twice. The total number of occurrence of the factor 5 is calculated by computing the above formula, which divides and adds the number of times a power of factor 5 occurs in factorial of n. Therefore there are a total of 5 + 1 = 6 number of times the factor 5 occurs in the entire series of 1 to 27. Therefore there are 6 trailing zeros in 27! and 27! = 10888869450418352160768000000 which has 6 trailing zeros.\n\nThe implementation is pretty straight forward.\n\n### Sourcecode\n\n#include <stdio.h>\n\n/* long int n : the input factorial integer\n* long int i : holds value 5^k , the divisor of n' to\n*              find the number of times the factor 5 occurs\n*              in the factorial of n'\n* long int c : number of times the 5 occurs as a factor in the\n*              factorial of n'\n* long int t : the number of times the factor 5^i occurs in the n!'\n*/\nint\nmain (void)\n{\nlong int n, i = 5, c = 0, t;\n\nprintf (\"\\nEnter n: \");\nscanf (\"%ld\", &n);\ndo\n{\nt = n / i; /* computes floor (n/i)  where i=5^k */\nc += t;    /* update the number of occurrence of factor 5 */\ni *= 5;    /* update i' to be the next power of 5 */\n}\nwhile (t != 0);\n\nprintf (\"Number of trailing zeros in %ld! : %ld\\n\", n, c);\n\nreturn 0;\n}\n\n\n\n### Output\n\nSome sample outputs are given below.\n\n\nrun 1\nEnter n: 3\nNumber of trailing zeros in 3! : 0\n\nrun 2\nEnter n: 6\nNumber of trailing zeros in 6! : 1\n\nrun 3\nEnter n: 15\nNumber of trailing zeros in 15! : 3\n\nrun 5\nEnter n: 60\nNumber of trailing zeros in 60! : 14\n\nrun 6\nEnter n: 100\nNumber of trailing zeros in 100! : 24\n\nrun 7\nEnter n: 1000\nNumber of trailing zeros in 1000! : 249\n\nrun 8\nEnter n: 5000\nNumber of trailing zeros in 5000! : 1249\n\nrun 9\nEnter n: 10000\nNumber of trailing zeros in 10000! : 2499\n\nrun 10\nEnter n: 12345678\nNumber of trailing zeros in 12345678! : 3086416\n\n\nHomo-sapiens\nThis entry was posted in Coding Discussions, Computer Science, Others and tagged , , , , . Bookmark the permalink.\n\n### 2 Responses to Number of trailing zeros in factorial of an\u00a0integer\n\n1. Cassie says:\n\nFantastic post! This site also gives a good explanation of how to come up with the algorithm, and also an implementation in Java:\n\nhttp://www.programmerinterview.com/index.php/java-questions/find-trailing-zeros-in-factorial/\n\n\u2022 phoxis says:\n\nGreat site. Once you know the algorithm you can implement it in any language. For example here are some quick bash implementations.\n\n#!/bin/bash\n\nnum=$1 i=5 t=1 while [$t -ne 0 ]\ndo\nt=$((num/i)) count=$((count+t))\ni=$((i*5)) done echo \"Trailing zeros of$num ! = $count\"  Or we can use the factor command to count the number of 5 in each of the integer. Although this process is extremely inefficient as it will compute the factor of each integer from 1 to n, but still it works. #!/bin/bash num=$1\ncount=0\n\nfor ((i=1; i<=num; i++))\ndo\nthis_count=$(factor$i | cut -d':' -f 2 | tr ' ' '\\n' | grep -c \"\\<5\\>\")\ncount=$((count + this_count)) this_count=0 done echo \"$count\"\n`\n\nThanks for visiting.", "date": "2014-07-23 01:38:16", "meta": {"domain": "phoxis.org", "url": "http://phoxis.org/2012/05/11/number-of-trailing-zeros-in-factorial-of-an-integer/", "openwebmath_score": 0.5719761252403259, "openwebmath_perplexity": 957.6028528442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9871787857334058, "lm_q2_score": 0.8615382058759128, "lm_q1q2_score": 0.8504922399395205}}
{"url": "http://mathhelpforum.com/calculus/155132-cross-multiplying-vectors.html", "text": "# Math Help - Cross multiplying vectors\n\n1. ## Cross multiplying vectors\n\nThe question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.\n\n$(j-k)$ x $(k - i)$\n\nI really can't even get started. A hint or general idea would be very helpful though.\n\n2. Originally Posted by Riyzar\nThe question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.\n\n$(j-k)$ x $(k - i)$\n\nI really can't even get started. A hint or general idea would be very helpful though.\nYou should know that $\\vec{i}\\times \\vec{j}= \\vec{k}$, $\\vec{j}\\times\\vec{k}= \\vec{i}$, and $\\vec{k}\\times\\vec{i}= \\vec{j}$. And, of course, that $\\vec{u}\\times\\vec{v}= -\\vec{v}\\times\\vec{u}$ which, among other things, implies $\\vec{u}\\times\\vec{u}= 0$. Also that the cross product distributes over addition: $(\\vec{j}- \\vec{k})\\times (\\vec{k}- \\vec{i})= (\\vec{j}- \\vec{k})\\vec{k}- (\\vec{j}- \\vec{k})\\vec{i}= \\vec{j}\\times\\vec{j}+ \\vec{k}\\vec{j}- \\vec{k}\\times\\vec{i}+ \\vec{k}\\times\\vec{i}$\n\n3. Hello, Riyzar!\n\n$\\text{Find the resulting vector }without\\text{ using determinants: }\\;(\\vec j-\\vec k) \\times(\\vec k - \\vec i)$\n\nAre those unit vectors? . $\\begin{Bmatrix}\\vec i &=& \\langle 1,0,0\\rangle \\\\ \\vec j &=& \\langle 0,1,0\\rangle \\\\ \\vec k &=& \\langle 0,0,1\\rangle \\end{Bmatrix}$\n\n$\\left(\\vec j - \\vec k\\right) \\times \\left(\\vec k - \\vec i\\right) \\;=\\;\\left(\\vec j - \\vec k\\right )\\times \\vec k - \\left(\\vec j - \\vec k\\right)\\times \\vec i$\n\n. . . . . . . . . . . . . . $=\\; \\left(\\vec j \\times \\vec k\\right) - \\left(\\vec k \\times \\vec k \\right) - \\left(\\vec j \\times \\vec i\\right) + \\left(\\vec k \\times \\vec i\\right)$\n\n. . . . . . . . . . . . . . $=\\qquad \\vec i \\quad\\;\\; -\\;\\;\\quad \\vec 0 \\quad - \\quad (-\\vec k) \\quad + \\quad \\vec j$\n\n. . . . . . . . . . . . . . $=\\quad \\vec i\\;+\\; \\vec j\\;+\\;\\vec k$\n\n4. Originally Posted by HallsofIvy\nAlso that the cross product distributes over addition: $(\\vec{j}- \\vec{k})\\times (\\vec{k}- \\vec{i})= (\\vec{j}- \\vec{k})\\vec{k}- (\\vec{j}- \\vec{k})\\vec{i}= \\vec{j}\\times\\vec{j}+ \\vec{k}\\vec{j}- \\vec{k}\\times\\vec{i}+ \\vec{k}\\times\\vec{i}$\nThat is what I was not aware of, thank you!", "date": "2015-08-30 09:10:37", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/155132-cross-multiplying-vectors.html", "openwebmath_score": 0.9523987770080566, "openwebmath_perplexity": 155.03457077035483, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429129677613, "lm_q2_score": 0.863391617003942, "lm_q1q2_score": 0.8504777934455087}}
{"url": "https://math.stackexchange.com/questions/2638101/how-can-we-show-that-a-n-b-n-c-n-are-convergent-and-have-the-same-li", "text": "# How can we show that $(a_n), (b_n), (c_n)$ are convergent and have the same limit?\n\nWe have the following for $a \\le b \\le c >0$:\n\n$A(a,b,c)=\\frac{a+b+c}{3}, B(a,b,c)= (abc)^{1/3}, C(a,b,c)=\\frac{3}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}}$.\n\nThen we define the sequences $(a_n),(b_n), (c_n)$ by\n\n$a_1=a, b_1=b, c_1=c,$\n\n$a_{n+1}=A(a_n,b_n,c_n), b_{n+1}=B(a_n,b_n,c_n), c_{n+1}=C(a_n,b_n,c_n)$.\n\nHow can we show that $(a_n),(b_n), (c_n)$ are convergent and have the same limit?\n\nI understand that $A(a,b,c)\\ge B(a,b,c)$ and $B(a,b,c)\\ge C(a,b,c)$\n\n\u2022 @dxiv I think so. \u2013\u00a0MOP Feb 6 '18 at 5:13\n\u2022 See this and this related questions, same idea applies here. \u2013\u00a0dxiv Feb 6 '18 at 5:17\n\u2022 Assume you meant $0 < a \\le b \\le c$, and not that only $c$ is positive. \u2013\u00a0Macavity Feb 6 '18 at 6:27\n\nWe have $$A(x,y,z) = \\mbox{ arithmetic mean of } x,y,z;$$ $$B(x,y,z) = \\mbox{ geometric mean of } x,y,z;$$ $$C(x,y,z) = \\mbox{ harmonic mean of } x,y,z.$$\n\nIt is well known that, for the same arguments, $$\\mbox{ harmonic mean } \\le \\mbox{ geometric mean } \\le \\mbox{ arithmetic mean } \\tag{1}$$ (see e.g. this Wikipedia article).\n\nUsing these inequalities we see that the largest (arithmetic) means $a_n$ form a non-increasing sequence, while the smallest (harmonic) means $c_n$ form a non-decreasing sequence. Both sequences are bounded: all terms are within the interval $[a,c]$. If a sequence is monotonic and bounded, it has a limit.\n\nCan you finish by proving that the limits of $a_n$ and $c_n$ are the same? (Then $b_n$ necessarily has the same limit too because of the double inequality $(1)$ and the squeeze theorem.)\n\nFor $n>1$ we have $$a_{n+1}={a_n+b_n+c_n\\over3}\\le{a_n+a_n+c_n\\over3}, \\tag{2}$$ $$c_{n+1} \\ge c_n. \\tag{3}$$ Subtracting $(3)$ from $(2)$ we find $$a_{n+1}-c_{n+1} \\le {2\\over3} (a_n-c_n).$$ We observe that the intervals $[c_n,a_n]$ form a sequence of nested closed intervals, and the interval lengths tend to zero (no slower than a decreasing geometric progression). Therefore these intervals have a (unique) common point, and this point must be the limit of all three sequences because $a_n,b_n,c_n \\in [c_n,a_n]$ for all $n>1$. (The common point is unique because the size of intervals $[c_n,a_n]$ tends to zero.) This completes the proof.\n\n\u2022 I can understand that $(a_n)$ is decreasing sequence and $(c_n)$ is a increasing sequence. But could you please please help me with finding the limit? \u2013\u00a0MOP Feb 6 '18 at 5:32\n\u2022 You are not required to find the limit. You just need to prove that the limits are the same. \u2013\u00a0Alex Feb 6 '18 at 5:37\n\u2022 For example, you can prove that $a_{n+1}-c_{n+1}\\le0.9(a_n-c_n)$, and then observe that the limit of each sequence is in the interval $[c_n,a_n]$ for all $n\\in{\\mathbb N}$. Note that the intervals $[c_n,a_n]$ form a sequence of nested intervals. \u2013\u00a0Alex Feb 6 '18 at 5:53\n\u2022 I could't find $a_{n+1}-c_{n+1} \\le \\frac{1}{x} (a_n-c_n)$. I can understand it for two numbers case only. \u2013\u00a0MOP Feb 6 '18 at 7:48\n\nhint: try proving $c_n$ is increasing and $a_n$ is decreasing and use squeeze theorem.\n\nAssuming $a>0.$ For any positive $a',b',c'$ we have $$\\max (a',b',c')\\geq A(a',b',c')\\geq B(a',b','c)\\geq C(a',b',c')\\geq \\min (a',b',c').$$ Let $U_n=\\max (a_n,b_n,c_n)$ and $L_n=\\min (a_n,b_n,c_n).$ $$\\text {We have }\\quad U_n\\geq U_{n+1}\\geq L_{n+1}\\geq L_n.$$ So $(U_n)_n$ is a deceasing sequence bounded below by $L_1$ and $(L_n)_n$ is an increasing sequence bounded above by $U_1 .$\n\nLet $U=\\lim_{n\\to \\infty}U_n$ and $L=\\lim_{n\\to \\infty}L_n.$ Obviously $U\\geq L\\geq L_1>0.$\n\nIt suffices to show that $U=L.$\n\nLet $r_n=U_n-L_n .$ Note that $U_n>r_n.$ $$\\text {We have }\\quad U_{n+1}\\leq \\frac {2U_n+L_n}{3}$$\n$$\\text {and }\\quad L_{n+1}\\geq \\frac {3}{\\frac {1}{U_n}+\\frac {2}{L_n}}=\\frac {3U_nL_n}{2U_n+L_n}.$$ $$\\text {Therefore }\\quad r_{n+1}\\leq \\frac {2U_n+L_n}{3}-\\frac {3U_nL_n}{2U_n+L_n}=$$ $$=\\frac {(4U_n-L_n)(U_n-L_n)}{3(2U_n+L_n)}=$$ $$=\\frac {(3U_n+r_n)r_n}{9U_n-3r_n}\\leq$$ $$\\leq \\frac {(4U_n)r_n}{6U_n}=\\frac {2}{3}r_n.$$\n\nSo $U-L=\\lim_{n\\to \\infty}r_n=0$ because $0\\leq r_{n+1}\\leq \\frac {2}{3}r_n.$\n\n\u2022 How did you show that $a_n$ is increasing and $b_n$ is increasing? \u2013\u00a0MOP Feb 6 '18 at 10:33\n\u2022 I didn't understand your proof? Why did you take U and L? \u2013\u00a0MOP Feb 6 '18 at 11:22\n\u2022 $U_n=\\max (a_n,b_n,c_n).$... $L_n=\\min (a_n,b_n,c_n).$... $U_{n+1}= \\max (a_{n+1},b_{n+1},c_{n+1}) =$ $\\max (\\; A(a_n,b_n,c_n), B(a_n,b_n,c_n),C(a_n,b_n,c_n)\\;)\\;=$ $A(a_n,b_n,c_n)\\leq \\max (a_n,b_n,c_n)=U_n.$..... For $n>1$ we have $U_n=a_n$ and $L_n=c_n.$.... So $a_n$ is decreasing, not increasing..... If $U_n=\\max (a_n,b_n,c_n)$ converges to $U$ and if $L_n= \\min (a_n,b_n,c_n)$ converges to $L$ and if $U=L$ then $a_n, b_n,c_n$ each converge to $L$ also. \u2013\u00a0DanielWainfleet Feb 6 '18 at 23:18", "date": "2019-09-18 12:18:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2638101/how-can-we-show-that-a-n-b-n-c-n-are-convergent-and-have-the-same-li", "openwebmath_score": 0.9202626347541809, "openwebmath_perplexity": 157.57341305673668, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504291340685, "lm_q2_score": 0.8633916152464016, "lm_q1q2_score": 0.8504777920933616}}
{"url": "https://math.stackexchange.com/questions/2350315/equating-powers-in-a-series-expansion", "text": "# Equating powers in a series expansion\n\nIf the $(r+1)^{th}$ term contains the same power of $a$ and $b$ in the expansion of $$\\bigg(\\sqrt[3]{\\frac{a}{\\sqrt b}}+\\sqrt{\\frac {b}{\\sqrt[3]{a}}}\\bigg)^{21}$$ find the value of $r$\n\nI simply applied binomial and expanded for the general term and then equated the power of $a$ and $b$ and got $r=12$\n\nMy friend got $r=9$\n\nI want to know which answer is correct.\n\nWithout trying to replicate your calculations, you're probably using version of the binomial theorem that count the terms from the opposite end. You can state the theorem either as $$(p+q)^n = \\sum_{r=0}^n \\binom{n}{r} p^r q^{n-r}$$ or as $$(p+q)^n = \\sum_{r=0}^n \\binom{n}{r} p^{n-r} q^r$$ which both give the same terms, just with a different numbering.\n\nNote in particular that $9+12=21$, so you can both be right, if you're counting from different ends of the expansion!\n\nOn simplifying the following expression: $$\\bigg(\\sqrt[3]{\\frac{a}{\\sqrt b}}+\\sqrt{\\frac {b}{\\sqrt[3]{a}}}\\bigg)^{21}$$ We get the expression: $$\\frac{1}{(ab)^\\frac72}\\bigg(a^\\frac12+b^\\frac23\\bigg)^{21}$$\n\nSo the (r$+1$)th term of the expansion is $$\\binom{n}{r}a^{\\frac{21-r-7}{2}}b^{\\frac{4r-21}{6}}$$\n\nHence, the answer will come as: $$\\frac{21-r-7}{2}=\\frac{4r-21}{6}$$ $$\\Rightarrow 42-3r=4r-21$$ $$\\Rightarrow \\boxed{\\color{red}{r=9}}$$\n\nThe required answer is $9$.\n\n\u2022 couldn't it be 12 as 12+9=21? \u2013\u00a0Atul Mishra Jul 8 '17 at 10:56\n\u2022 The first step here looks somewhat obscure. Why not simplify to $$(a^{1/3}b^{-1/6}+a^{-1/6}b^{1/2})^{21}$$ and then solve $$\\frac13r - \\frac16(21-r) = -\\frac16 r + \\frac12(21-r)$$ \u2013\u00a0hmakholm left over Monica Jul 8 '17 at 10:59\n\u2022 @AtulMishra Well, in a problem, where the binomial expression is rigorously given, it is expected that you consider the first term as the 'a' term and the second one 'b'. So if you consider the expansion in this order, your answer should be 9. I saw Henning's answer, but I dont think that you are allowed to reverse the expression and work. It may seem trivial that you cannot reverse the expression but if one is allowed to do so, then the ordering of the terms will become ambiguous and there will be no unique answer. Hope this helps. \u2013\u00a0SchrodingersCat Jul 8 '17 at 11:02\n\u2022 Yes I am just arguing for the same, It will give r=12 \u2013\u00a0Atul Mishra Jul 8 '17 at 11:03\n\u2022 @SchrodingersCat: There's no \"reversing\" going on -- there's just two ways to state the binomial theorem which are exactly equally good. It's not as if one of them is \"the right way\" and the other is \"reversed\". \u2013\u00a0hmakholm left over Monica Jul 8 '17 at 11:05", "date": "2020-10-27 14:23:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2350315/equating-powers-in-a-series-expansion", "openwebmath_score": 0.8241744637489319, "openwebmath_perplexity": 351.33726728817027, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504290989414, "lm_q2_score": 0.8633916170039421, "lm_q1q2_score": 0.85047779079177}}
{"url": "https://math.stackexchange.com/questions/603830/why-does-trapezoidal-rule-have-potential-error-greater-than-midpoint/674350", "text": "# Why does Trapezoidal Rule have potential error greater than Midpoint?\n\nI can approximate the area beneath a curve using the Midpoint and Trapezoidal methods, with errors such that:\n\n$Error_m \\leq \\frac{k(b-a)^3}{24n^2}$ and $Error_T \\leq \\frac{k(b-a)^3}{12n^2}$.\n\nDoesn't this suggest that the Midpoint Method is twice as accurate as the Trapezoidal Method?\n\n\u2022 The constant $k$ is not necessarily the same. It depends on $f$, differently for each method. Dec 12, 2013 at 5:30\n\u2022 Here in both cases $k$ is the max of the (absolute value) of the second derivative. Then your inequalities give upper bounds on the error. One can construct $f$ such that for given $n$ the Trapezoidal Rule is dead on, while the Midpoint Rule is not. Dec 12, 2013 at 5:34\n\u2022 'k' is equal to the maximum of $f''(x)$ for both $E_m$ and $E_T$ Dec 12, 2013 at 5:43\n\u2022 Even if the \"worst case\" for trapezoidal is half as good, the \"typical case\" could be much better. Dec 12, 2013 at 14:14\n\nIt has already been said in the comments that the error estimates you cite are upper bounds, so actual errors may be smaller and $E_M$ won't usually be exactly half of $E_T$ (and may actually be larger in some cases).\n\nNevertheless, it is well worth pointing out that if the function you are integrating happens to be a cubic polynomial, then we can make an exact statement: $$E_M=-\\frac{1}{2}E_T.$$ You would probably never integrate a cubic polynomial numerically, but if the function you are integrating is well-approximated by a cubic polynomial on every subinterval used in the numerical integration, then the errors should still be related in roughly the same way.\n\nThe above relation obviously holds for the functions $f(x)=1$ and $f(x)=x$. Let's verify it by brute force for $f(x)=x^2$ and $f(x)=x^3$. Consider a single subinterval $[a,b]$ and let $A[f]$, $T[f]$, and $M[f]$ represent the exact area integral of $f$ on $[a,b]$, the trapezoidal estimate, and the midpoint estimate, respectively. Then \\begin{aligned} A[x^2]&=\\int_a^bx^2\\,dx=\\frac{b^3-a^3}{3},\\\\ T[x^2]&=\\frac{b-a}{2}(b^2+a^2)=\\frac{b^3-ab^2+a^2b-a^3}{2}\\\\ M[x^2]&=(b-a)\\left(\\frac{b+a}{2}\\right)^2=\\frac{b^3+ab^2-a^2b-b^3}{4}. \\end{aligned} So \\begin{aligned} E_T[x^2]&=T[x^2]-A[x^2]=\\frac{b^3-a^3}{6}-ab\\frac{b-a}{2},\\\\ E_M[x^2]&=M[x^2]-A[x^2]=-\\frac{b^3-a^3}{12}+ab\\frac{b-a}{4}=-\\frac{1}{2}E_T[x^2].\\\\ \\end{aligned} Likewise \\begin{aligned} A[x^3]&=\\int_a^bx^3\\,dx=\\frac{b^4-a^4}{4},\\\\ T[x^3]&=\\frac{b-a}{2}(b^3+a^3)=\\frac{b^4-ab^3+a^3b-a^4}{2}\\\\ M[x^3]&=(b-a)\\left(\\frac{b+a}{2}\\right)^3=(b-a)\\frac{b^3+3ab^2+3a^2b+a^3}{8}\\\\ &=\\frac{b^4+2ab^3-2a^3b-a^4}{8}. \\end{aligned} So \\begin{aligned} E_T[x^3]&=T[x^3]-A[x^3]=\\frac{b^4-a^4}{4}-\\frac{ab}{2}(b^2-a^2),\\\\ E_M[x^3]&=M[x^3]-A[x^3]=-\\frac{b^4-a^4}{8}+\\frac{ab}{4}(b^2-a^2)=-\\frac{1}{2}E_T[x^3].\\\\ \\end{aligned} Since the statement holds for $1,$ $x,$ $x^2,$ and $x^3,$ it holds for all cubic polynomials by linearity of $A,$ $T,$ and $M.$\n\nAnother viewpoint on this is the following: if $T_n,$ $M_n,$ and $S_n$ represent the estimates given by the trapezoidal, midpoint, and Simpson's rules with $n$ subintervals (so $T$ and $M$ above are $T_1$ and $M_1$), then one finds that $$S_{2n}=\\frac{T_n+2M_n}{3}.$$ If the error in $S_{2n}[f]$ is much smaller than the errors in $T_n[f]$ and $M_n[f]$, then it must be the case that $E_{M_n}[f]\\approx-\\frac{1}{2}E_{T_n}[f].$ This is, in fact, the case for many functions.\n\nFor an example of a function where $E_T[f]$ is much larger than $E_M[f],$ imagine a function which is nearly linear on the entire interval, but has a sharp peak or dip in a small neighborhood of the midpoint. For such a function, the $k$ in the error bound\u2014it's the same $k$ in both bounds\u2014would be big since the second derivative would be big in the vicinity of the peak or dip. There is no contradiction here since the trapezoidal error bound would be pretty poor in that case, while the midpoint error bound might be pretty reasonable.\n\nA final comment: you can find diagrams in books explaining why $E_M$ tends to be less than $E_T$ and of opposite sign. I'm not sure that these diagrams provide a compelling reason to believe that $E_M$ is of roughly half the magnitude of $E_T,$ but I will give this some thought. The diagrams I have in mind represent the midpoint estimate by a trapezoidal area, where the diagonal of the trapezoid is tangent the the curve at the midpoint. It is clear that, if there's no inflection point in the interval, then, if the trapezoid of the midpoint rule is an overestimate of the integral, the trapezoid of the trapezoidal rule will be an underestimate of the integral, and vice versa.\n\nAdded: The midpoint rule is often presented geometrically as a series of rectangular areas, but it is more informative to redraw each rectangle as a trapezoid of the same area. These two presentations, in the case of a single interval, are shown below.\n\nThe slope of the top edge of the trapezoid has been chosen to match that of the curve at the midpoint. That the top edge of the trapezoid is the best linear approximation of the curve at the midpoint of the interval may provide some intuition as to why the midpoint rule often does better than the trapezoidal rule.\n\nA series of pairs of plots is shown below. In each pair, the trapezoidal rule has been used on the left, and the midpoint rule has been used on the right.!\n\n\u2022 So the midpoint formula is usually more accurate than the trapezoidal formula? Feb 13, 2014 at 3:56\n\u2022 @citelao: if you cook up some functions that you can integrate exactly, and compare midpoint and trapezoidal estimates, I think you will find that the error in the midpoint estimate is usually about half as big and of opposite sign as that in the trapezoidal estimate. That's because the type of functions most people will cook up tend be smooth with second derivative that stays within reasonable bounds. You should always take care, however, especially if you plan to use the method on functions that aren't so smooth. Feb 13, 2014 at 12:12\n\u2022 @will-orick but my intuition still suggests that the trapezoidal approximation is better; is this untrue? Feb 19, 2014 at 2:41\n\u2022 @citelao: I've added some images to the post that may help with the intuition. Feb 21, 2014 at 11:28\n\u2022 Wow, this slanted-roof representation of the midpoint rule is an eye-opener! Jul 27, 2020 at 21:12\n\nOn an interval where a function is concave-down, the Trapezoidal Rule will consistently underestimate the area under the curve. (And inversely, if the function is concave up, the Trapezoidal Rule will consistently overestimate the area.)\n\nWith the Midpoint Rule, each rectangle will sometimes overestimate and sometimes underestimate the function (unless the function has a local minimum/maximum at the midpoint), and so the errors partially cancel out. (They exactly cancel out if the function is a straight line.)\n\n\u2022 If we consider a concave-down function (like your red curve), then, while it is true that on most intervals the rectangle given by the midpoint rule will be partly above and partly below the curve (which leads to partial cancellation of errors), one can show that the residual error is consistently positive, that is, the rectangles consistently overestimate the area under the curve. The question is whether this residual error is larger or smaller in magnitude than the error of the trapezoidal rule. I don't think your argument makes it evident that the residual error is smaller. Feb 13, 2014 at 21:06\n\u2022 I think that with just a little more work, we can motivate the difference in the magnitude of the errors as well as their sign. See my answer. Dec 5, 2021 at 1:06\n\nWill Orrick's great answer shows why if the function is concave up, then the trapezoid rule overestimates the integral and the midpoint rule underestimates it, and vice versa if the function is concave down. This gives a heuristic explanation for why the errors have opposite signs. But as he points out, it still isn't clear why the max midpoint error is smaller in magnitude.\n\nTo see intuitively why this is the case, we can add some more lines to his diagram:\n\nThe thick blue curve is the function to be integrated, the upper diagonal line is the top of the trapezoid from the trapezoidal rule, and the bottom diagonal line (which is tangent to the blue curve) is the top of the trapezoid with the same area as the rectangle given by the midpoint rule. The area of the blue shaded region is the error $$E_T$$ from the trapezoidal rule, and the area of the orange shaded region is (the absolute value of) the error $$E_M$$ from the midpoint rule. The middle diagonal lines connect the endpoints of the function's curve and its horizontal midpoint.\n\nThe entire shaded region is yet another trapezoid (the top and bottom sides are not quite parallel in general, but the side sides are). The vertical line running above the midpoint divides it into two sub-trapezoids. Each sub-trapezoid's diagonal divides it into two triangles that aren't quite congruent (because the parallel sides have slightly different lengths), but are very close to congruent if the function is reasonably close to linear over the interval, so these pairs of triangles have approximately equal areas. If the function's curve matched these diagonals, then the two rules would give (almost) equal and opposite errors, but we see from the fact that the curve is always to one side of these diagonals that the blue region has larger area ($$E_T$$) than the orange region ($$E_M$$).\n\nWe can roughly motivate the factor of 2 difference by noting that if you subtract off the lower large diagonal line (which makes the top line approximately horizontal as well, at this order), then the curve is more or less a parabola whose vertex is the midpoint - and it's a standard fact that over an interval that's symmetric about and tangent to any parabola's vertex, the area on the parabola's concave side is twice the area on its convex side.\n\nThis (heuristic) last step assumed that the function is close to parabolic, but it remains true if we consider an arbitrary cubic polynomial as well, because you can treat it as the sum of a quadratic term (to which the previous argument will apply) and a cubic term that vanishes at the endpoints and the midpoint. This cubic piece must be odd about the midpoint, so its integral will vanish and we can reuse the reasoning above from the simpler quadratic case.", "date": "2022-06-26 15:23:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/603830/why-does-trapezoidal-rule-have-potential-error-greater-than-midpoint/674350", "openwebmath_score": 0.8626505136489868, "openwebmath_perplexity": 235.91348479251434, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429103332288, "lm_q2_score": 0.8633916134888614, "lm_q1q2_score": 0.8504777877083702}}
{"url": "https://math.stackexchange.com/questions/2883337/prove-that-there-exists-irrational-numbers-p-and-q-such-that-pq-is-rational", "text": "Prove that there exists irrational numbers p and q such that $p^{q}$ is rational\n\nI found this on the lecture slides of my Discrete Mathematics module today. I think they quote the theorems mostly from the Susanna S.Epp Discrete Mathematics with Applications 4th edition.\n\nHere's the proof:\n\n1. We know from Theorem 4.7.1(Epp) that $\\sqrt{2}$ is irrational.\n2. Consider $\\sqrt{2}^{\\sqrt{2}}$ : It is either rational or irrational.\n3. Case 1: It is rational:\n\n3.1 Let $p=q=\\sqrt{2}$ and we are done.\n\n4. Case 2: It is irrational:\n\n4.1 Then let p=$\\sqrt{2}^{\\sqrt{2}}$, and $q =\\sqrt{2}$\n\n4.2 p is irrational(by assumption), so is q (by Theorem 4.7.1(Epp))\n\n4.3 Consider $p^{q} = (\\sqrt{2}^{\\sqrt{2}})^{\\sqrt{2}}$\n\n4.4 $=(\\sqrt{2})^{\\sqrt{2} \\times\\sqrt{2}}$, by the power law\n\n4.5 $=(\\sqrt{2})^{2}=2$, by algebra\n\n4.6 Clearly $2$ is rational\n\n1. In either case, we have found the required p and q.\n\nFrom what I understand from the proof, it's a clever construction of a number and splitting up into cases to prove that the given number is a rational number (one by clear observation and the other by some sort of contradiction). While the proof seems valid, though I somehow am forced to be convinced that the contradiction works, I wondered to myself if $\\sqrt{2}^{\\sqrt{2}}$ is actually rational since they constructed it.\n\nMy question would be if the example is indeed irrational, how is it possible an untrue constructed example could be used to verify this proof? If it's indeed rational, can someone tell me the $a/b$ representation of this number?\n\nPS: Sorry for the formatting errors, I followed the MathJax syntax to the best of my abilities but I'm not sure how to align the sub-points well. Please help me edit the post, thanks.\n\nThe irrationality of $\\sqrt 2^{\\sqrt 2}$ (in fact, its transcendence) follows immediately from the Gelfond Schneider Theorem. This was the issue that motivated Hilbert's $7^{th}$ Problem.\n\nThe beauty of the argument here is its simplicity. It's a perfectly valid, non-constructive, argument. The claim is demonstrated though no explicit example is produced.\n\nHere is a simple, constructive, way to settle the original issue: $$\\sqrt 3^{\\log_34}=2$$\n\nOf course, $\\sqrt 3$ is irrational.\n\nTo see that $\\log_3 4$ is irrational work by contradiction. $$\\log_3 4=\\frac ab\\implies 4=3^{\\frac ab}\\implies 4^b=3^a$$ But if $a,b\\in \\mathbb N$ then this contradicts unique factorization.\n\nThe proof is non-constructive. We don't know whether $\\sqrt{2}^{\\sqrt{2}}$ is rational or irrational. The point of the proof is that in either case there will be two irrational numbers $p$ and $q$ such that $p^q$ is rational. In one case $p=\\sqrt{2}$ and in the other case $p=\\sqrt{2}^{\\sqrt{2}}$.\n\n\u2022 Can u tell me what kind of proving method is this or in what way does the proof say they are asking for 2 irrational numbers in all cases of $p^{q}$ when they can't even get a true rational number to verify this. My impression of existence proof was always to construct some example that fulfils the conditions. \u2013\u00a0Prashin Jeevaganth Aug 15 '18 at 8:54\n\u2022 Not all existence proofs are constructive. This is an example of a non-constructive existence proof. A constructive proof of the same theorem would be $p=\\sqrt{2}; q=\\log_2(9); p^q=3$. But you have to do more work with this constructive proof to show that $\\log_2(9)$ is irrational. \u2013\u00a0gandalf61 Aug 15 '18 at 12:48", "date": "2019-06-17 13:01:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2883337/prove-that-there-exists-irrational-numbers-p-and-q-such-that-pq-is-rational", "openwebmath_score": 0.8688209652900696, "openwebmath_perplexity": 343.57050175267295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429112114064, "lm_q2_score": 0.8633916117313211, "lm_q1q2_score": 0.8504777867353288}}
{"url": "https://math.stackexchange.com/questions/925910/terminology-for-1-ex1", "text": "# Terminology for $1/(e^x+1)$?\n\n$\\frac{1}{e^x+1}$ and $\\frac{e^x}{e^x+1}$\n\nJust wonder if either of the above function has a term/name associated with it? Or they are just functions that look beautiful without names? Maybe they appear very often under certain contexts?\n\nI thought I might have seen it in some online courses. Maybe it was graphical model related or something else. But I'm not exactly sure right now and I cannot really find it on Google.\n\n\u2022 If you change the plus signs to a minus sign and multiply by $x$, you have this. \u2013\u00a0alex.jordan Sep 10 '14 at 5:20\n\u2022 the first is probably related to Fermi-Dirac distribution f(x) in physics. then second one is 1-f(x). \u2013\u00a0mike Sep 10 '14 at 5:20\n\u2022 And Alex's variation appears in Planck's formula for black body radiation distribution. \u2013\u00a0Travis Sep 10 '14 at 5:21\n\u2022 Both can be used in relation to the Fermi-Dirac distribution, with the second being used with certain probabilities of states. \u2013\u00a0Silynn Sep 10 '14 at 5:22\n\u2022 Per Claude's comment, these functions also appear in the solutions to simple models of population growth in an environment with a fixed population capacity. \u2013\u00a0Travis Sep 10 '14 at 5:47\n\nQuotic Wikipedia article :\"A logistic function or logistic curve is a common special case of the more general sigmoid function, with equation $$f(x)=\\frac{1}{1+e^{-x}}$$ So, multiplying numerator and denominator by $e^x$ $$f(x)=\\frac{e^x}{1+e^{x}}$$ is just the same and $$g(x)=\\frac{1}{1+e^{x}}=1-f(x)$$", "date": "2019-08-20 11:13:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/925910/terminology-for-1-ex1", "openwebmath_score": 0.8910932540893555, "openwebmath_perplexity": 447.63096856279986, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.98504291340685, "lm_q2_score": 0.8633916029436189, "lm_q1q2_score": 0.8504777799745926}}
{"url": "http://math.stackexchange.com/questions/60973/finding-the-minimum-number-of-students", "text": "# Finding the minimum number of students\n\nThere are $p$ committees in a class (where $p \\ge 5$), each consisting of $q$ members (where $q \\ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?\n\nIt is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?\n\n$1) \\quad pq - \\binom{q}{2}$\n\n$2) \\quad pq - \\binom{p}{2}$\n\n$3) \\quad (p-1)(q-1)$\n\n-\nSomething is missing. Is every student supposed to be on a committee? \u2013\u00a0 JavaMan Aug 31 '11 at 16:24\n@DJC:Not mentioned in the question,I guess we may have to consider that to get a solution. \u2013\u00a0 Quixotic Aug 31 '11 at 16:28\n@DJC: For the minimum number of students this does not matter. \u2013\u00a0 TMM Aug 31 '11 at 16:30\n@Thijs Laarhoven:Yes you are right but as the problem also asked for maximum number I have considered it in my solution. \u2013\u00a0 Quixotic Aug 31 '11 at 16:31\n@Thijs, FoolForMath, I guess my question is, should the minimum answer be in terms of $p$ and $q$? \u2013\u00a0 JavaMan Aug 31 '11 at 16:31\n\nFor $1\\leq i\\leq p$, let $C_i$ be the set of students on the $i$th committee. Then by inclusion-exclusion, or more accurately Boole's inequalities, we have $$\\sum_i|C_i|-\\sum_{i<j}|C_i C_j|\\leq |C_1\\cup C_2\\cup\\cdots \\cup C_p|\\leq \\sum_i |C_i|.$$\n\nFrom the constraints of the problem, this means $$pq-{p\\choose 2}\\leq \\#\\mbox{ students}\\leq pq.$$\n\n-\nWhat is $j$ here?and I can't relate this with your answer. \u2013\u00a0 Quixotic Sep 1 '11 at 7:25\n$j$ is also a generic index that runs from $1$ to $p$. The inequalities are also known as Bonferroni inequalities (planetmath.org/encyclopedia/BonferroniInequalities.html), and can apply to cardinalities instead of probabilities. \u2013\u00a0 Byron Schmuland Sep 1 '11 at 14:10\n\nI think the following theorem might be relevant:\n\nTheorem. Let $\\mathcal{F}$ be a family of subsets of $\\{1, \\dots , n \\}$ with the property that $|A \\cap B| = 1$ for all $A,B \\in \\mathcal{F}$. Then $|\\mathcal{F}| \\leq n$.\n\nAlso this theorem could be relevant as well.\n\n-\n\nFor the case in which $p \\le q+1$ an arrangement that yields the minimum number of students can be described as follows.\n\nLet $P = \\{\\langle m,n \\rangle:1 \\le m \\le p, 1 \\le n \\le q+1\\}$, and let $S = \\{\\langle m,n \\rangle \\in P:m < n\\}$. If $P$ is thought of as a $p \\times (q+1)$ grid, $S$ is the part of it strictly above the main \u2018diagonal\u2019. The cells in $S$ are the students; the $k$-th committee consists of those cells in $S$ that are either in row $k$ or in column $k$. More formally, for $1 \\le k \\le p$ let \\begin{align*}C_k &= \\{\\langle m,n \\rangle \\in S:m=k \\lor n=k\\}\\\\ &= \\{\\langle k,n \\rangle:k+1 \\le n \\le q+1\\} \\cup \\{\\langle m,k \\rangle:1 \\le m \\le k-1\\};\\end{align*} clearly $\\vert C_k \\vert = q+1-k+k-1=q$, and if $1 \\le i < k \\le p$, $C_i \\cap C_k = \\{\\langle i,k \\rangle\\}$. Since every pair of committees shares a different student, this arrangement must minimize the number of students, so we need only calculate $\\vert S \\vert$.\n\nColumns $2$ through $p$ of the grid contain $\\sum_{k=1}^{p-1} k = \\binom{p}{2}$ cells, and the remaining $q+1-p$ columns contain $p(q+1-p)$ cells, so \\begin{align*}\\vert S \\vert &= \\binom{p}{2} + p(q+1-p)\\\\ &= \\frac{p^2 - p}{2} + pq + p - p^2\\\\ &= pq - \\frac{p^2-p}{2}\\\\&= pq - \\binom{p}{2}. \\end{align*}\n\nWhen $p > q+1$ the same approach works, but it\u2019s no longer possible to get every pair of committees to overlap. First form $\\left\\lfloor \\frac{p}{q+1}\\right\\rfloor$ sets of $q+1$ committees each. Each set requires $(q+1)q-\\binom{q+1}{2}=\\binom{q+1}{2}$ students, and committees in different sets must be disjoint, so this accounts for $\\left\\lfloor \\frac{p}{q+1}\\right\\rfloor \\binom{q+1}{2}$ students. The remaining $r = p-(q+1)\\left\\lfloor \\frac{p}{q+1}\\right\\rfloor$ committees will then require another $rq - \\binom{r}{2}$ students, for a grand total of \\begin{align*} \\vert S \\vert &= \\left\\lfloor \\frac{p}{q+1}\\right\\rfloor \\binom{q+1}{2} + rq - \\binom{r}{2}\\\\ &= \\left\\lfloor \\frac{p}{q+1}\\right\\rfloor \\binom{q+1}{2} + \\left(p-(q+1)\\left\\lfloor \\frac{p}{q+1}\\right\\rfloor\\right)q - \\binom{r}{2}\\\\ &= pq - \\left\\lfloor \\frac{p}{q+1}\\right\\rfloor \\binom{q+1}{2} - \\binom{r}{2} \\end{align*}, which does not appear to simplify greatly.\n\nI see that this is essentially Alex\u2019s solution, but expressed a little more concretely.\n\n-\n\nAlright, my guess is the question is saying, for some fixed value of $p$ and $q$, what is the maximum and minimum number of students we can have. If so, I agree the maximum number is $pq$, as we just don't let any of the students overlap in any committees. It's also been awhile since I've done anything like this, so hopefully I'm not making a silly mistake!\n\nFor the minimum, you want to maximize the overlap. Lets start with the simplest(?) case, where $p=5$ and $q=6$. Call the committees $p_1,\\ldots,p_5$. The maximum overlap that can occur, is 4 students from one committee each being members of the other 4 committees, hence with the maximum overlap possible, the number of students we require is \\begin{align*} 30 - 4 - 3 - 2 - 1 = 20. \\end{align*}\n\nWhat if we change $q = 7$? Nothing changes, as the committees have overlapped as much as possible already, so for any $q \\geq 6$, the number of students is \\begin{align*} pq - \\sum_{i=1}^{p-1} i = pq - \\frac{p(p-1)}{2}. \\end{align*}\n\nOf course this only \"half\" of what we want. What if we leave $q$ at 6, but set $p=6$? Now there is an entire other committee where we can overlap students, so we overlap as many more as possible, we get that the minimum number of students required is \\begin{align*} 36 - 5 - 4 - 3 - 2 - 1 = 21. \\end{align*}\n\nAs the number of committees increases, we may be able to overlaps more students. Suppose $p=7$ and $q=6$. We find the minimum number of students required is \\begin{align*} 42 - 6 - 5 - 4 - 3 - 2 - 1 = 21. \\end{align*} The same as above! But if we add one more committee, we cannot overlap any of the present students, so we must have another $q$ added in. In general, set $m = \\lfloor p/(q+1) \\rfloor$. We require, minimally \\begin{align*} m\\left(q(q+1) - \\sum_{i=1}^q i\\right) + \\left(rq - \\sum_{i=1}^{r-1} i\\right) \\end{align*} students, where $r$ is the remainder when $p$ is divided by $q+1$. Of course, the sums can be evaluated to give you an expression without the $i$'s if you'd like.\n\n-\nIn the derivation,\\begin{align*} pq - \\sum_{i=1}^{p-1} i = pq - \\frac{p(p+1)}{2}. \\end{align*} you probably meant $pq - \\frac{p(p-1)}{2}$ \u2013\u00a0 Quixotic Aug 31 '11 at 21:05\nOop! Definitely. \u2013\u00a0 Alex Aug 31 '11 at 21:16\n\nThis sounds like a test question where we want a quick solution. Given the choices, we can consider asymptotics. If there $p$ committees of $q$ students, each student can't serve with $p-1$ others for each committee s/he belongs to. It has to be of order $pq$, not $p^2$ or $q^2$, so must be $(p-1)(q-1)$\n\n-\n\nHere is a lower bound from a standard double counting argument. Let the number of students be $n$. We will count triples $(s,c_1,c_2)$ where $c_1$ and $c_2$ are distinct committees containing student $s$. Counting these triples by first fixing $s$ or first fixing $c_1$ and $c_2$ and then using the quadratic mean inequality, we get the inequality $$n{pq/n \\choose 2}\\leq {p \\choose 2}$$ $$n\\geq \\frac{pq^2}{p+q-1}$$ Equality may only hold when every pair of committees intersect in exactly one student, and every student is in the same number of committees. These are block designs with $\\lambda=1$. When $n=p$, these are projective planes. Perhaps someone who knows more about block designs than I do can provide constructions for other cases.\n\n-", "date": "2014-08-21 14:53:42", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/60973/finding-the-minimum-number-of-students", "openwebmath_score": 0.9914413690567017, "openwebmath_perplexity": 419.65787484266326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9850429147241161, "lm_q2_score": 0.8633916011860786, "lm_q1q2_score": 0.8504777793806565}}
{"url": "https://math.stackexchange.com/questions/3189486/let-w-1-0-1-1-expand-w-1-to-a-basis-of-r3/3189496", "text": "Let $w_1 = (0,1,1)$. Expand {$w_1$} to a basis of $R^3$.\n\nI am reading the book, Applied Linear Algebra and Matrix Analysis. When I was doing the exercise of Section3.5 Exercise 7, I was puzzled at some of it. Here is the problem description:\n\nLet $$w_1 = (0,1,1)$$. Expand {$$w_1$$} to a basis of $$R^3$$.\n\nI don't understand its description well.\nI think it wants to get a span set like {$$(0,1,1)$$, $$(1,0,0)$$, $$(0,0,1)$$} which is a basis of $$R^3$$.\nAnd I check the reference answer, which is as followings:\n\n$$(0,1,1)$$, $$(1,0,0)$$, $$(0,1,0)$$ is one choice among many.\n\nI think what I have done is what question wants.\nSo can anyone tell me am I right or wrong?\nThanks sincerely.\n\n\u2022 I think you are right Apr 16, 2019 at 6:02\n\nThere is a kind of 'procedure' for dealing with questions of this kind, namely to consider the spanning set $$\\left\\{ w_1, e_1, e_2, e_3\\right\\}$$. Consider each vector from left to right. If one of these vectors is in the span of the previous one/s, then throw it out. If not, keep it. So in this case, we start by keeping $$w_1$$. Moving to the next vector, $$e_1$$ is not in the span of $$w_1$$, so we keep it as well. Moving to the next, $$e_2$$ is not in the span of the previous two vectors so we keep it as well. Now, considering the vector $$e_3$$ we see that it is in fact in the span of the previous three vectors, since $$e_3 = w_1 - e_2.$$ So we throw out the vector $$e_3$$ and end up with the basis $$\\left\\{ w_1, e_1, e_2\\right\\}$$. This explains the solution in the reference answer. Your solution is also correct, however.\n$$\\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 1\\end{bmatrix}$$ has independent rows. Hence you have found $$3$$ independent vectors in $$\\mathbb{R}^3$$, that is it spans $$\\mathbb{R}^3$$ and it forms a basis.\nYou are correct. {$$(0,1,1),(1,0,0),(0,0,1)$$} is a basis of $$\\mathbb R^3$$.\nAny element $$(a,b,c)$$ in $$\\mathbb R^3$$ can be expressed as $$a(1,0,0)+b(0,1,1)+(c-b)(0,0,1).$$\n\u2022 If your basis is $w_1, w_2, w_3$, the textbook's choice is $w_1, w_2, w_1-w_3$ Apr 16, 2019 at 6:08", "date": "2022-07-03 18:08:19", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3189486/let-w-1-0-1-1-expand-w-1-to-a-basis-of-r3/3189496", "openwebmath_score": 0.9172195792198181, "openwebmath_perplexity": 143.88266297788041, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9637799404938199, "lm_q2_score": 0.8824278772763472, "lm_q1q2_score": 0.8504662870514856}}
{"url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/", "text": "# Justification for cancelling terms in limits?\n\nI am confused about the algebraic process of finding a limit. Let us take ##\\frac{x^2 -1}{x - 1}##. In trying to find ##\\lim_{x\\rightarrow 1}\\frac{x^2 -1}{x - 1}##we do the following:\n\n##\\displaystyle\\lim_{x\\rightarrow 1}\\frac{(x+1)(x-1)}{x - 1}##\n\n##\\displaystyle\\lim_{x\\rightarrow 1}x+1##\n\n##2##\n\nBut what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\\lim_{x\\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\\frac{x^2 -1}{x - 1}##?\n\n## Answers and Replies\n\nMark44\nMentor\nI am confused about the algebraic process of finding a limit. Let us take ##\\frac{x^2 -1}{x - 1}##. In trying to find ##\\lim_{x\\rightarrow 1}\\frac{x^2 -1}{x - 1}##we do the following:\n\n##\\displaystyle\\lim_{x\\rightarrow 1}\\frac{(x+1)(x-1)}{x - 1}##\n\n##\\displaystyle\\lim_{x\\rightarrow 1}x+1##\n\n##2##\n\nBut what justification do we have for cancelling the (x - 1) terms?\nSince the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction ##\\frac{x - 1}{x - 1}## will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.\nMr Davis 97 said:\nWhen we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\\lim_{x\\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\\frac{x^2 -1}{x - 1}##?\nThe only difference between ##y = x + 1## and ##y = \\frac{x^2 - 1}{x - 1}## is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.\n\nsymbolipoint\nHow can we be absolutely certain that limx\u21921x+1\\lim_{x\\rightarrow 1}x+1 leads to the correct answer to the original problem if x+2x + 2 is different function, with a different domain, than x2\u22121x\u22121\\frac{x^2 -1}{x - 1}?\nI think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .\n\nFor all x \u2208 R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .\n\nIf you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10-30 / 10-30 is one .\n\nSince the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction ##\\frac{x - 1}{x - 1}## will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.\n\nThe only difference between ##y = x + 1## and ##y = \\frac{x^2 - 1}{x - 1}## is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.\n\nI think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .\n\nFor all x \u2208 R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .\n\nIf you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10-30 / 10-30 is one .\n\nI understand what both of you are saying, and it helps. But what I am asking is what guarantees that ##\n\\displaystyle\\lim_{x\\rightarrow 1}x+1 = \\displaystyle\\lim_{x\\rightarrow 1}\\frac{(x+1)(x-1)}{x - 1}## (for example), considering that the two functions are different?\n\nI understand what both of you are saying, and it helps. But what I am asking is what guarantees that limx\u21921x+1=limx\u21921(x+1)(x\u22121)x\u22121 \\displaystyle\\lim_{x\\rightarrow 1}x+1 = \\displaystyle\\lim_{x\\rightarrow 1}\\frac{(x+1)(x-1)}{x - 1} (for example), considering that the two functions are different?\nDifferent in the sense of domain only .\n\nAlthough I mentioned this in my previous post -\nExample - 5 \u00d7 10-30 / 10-30 = ?\n\nMark44\nMentor\nI understand what both of you are saying, and it helps. But what I am asking is what guarantees that ##\n\\displaystyle\\lim_{x\\rightarrow 1}x+1 = \\displaystyle\\lim_{x\\rightarrow 1}\\frac{(x+1)(x-1)}{x - 1}## (for example), considering that the two functions are different?\nThe two functions are different at only a single point; namely, at (1, 2). At all other points they are identical.\n\n##\\lim_{x \\to 1}\\frac{(x+1)(x-1)}{x - 1} = \\lim_{x \\to 1}(x + 1) \\cdot \\lim_{x \\to 1}\\frac{x - 1}{x - 1}##, using the property of limits that the limit of a product is the product of the limits, provided that both limits exist.\nThe first limit in the product is clearly 2. What would you say that the second limit is?\n\nDifferent in the sense of domain only .\n\nAlthough I mentioned this in my previous post -\nExample - 5 \u00d7 10-30 / 10-30 = ?\n\nI still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do ##\\displaystyle\\lim_{x\\rightarrow 0} \\frac{x}{x}##, then ##\\displaystyle\\lim_{x\\rightarrow 0} 1##, the answer would be 1. But f(x) of the original problem is not the same as 1. So how do we know for sure that ##\\displaystyle\\lim_{x\\rightarrow 0} 1## answers the original question of ##\\displaystyle\\lim_{x\\rightarrow 0} \\frac{x}{x}##?\n\nMark44\nMentor\nI still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do ##\\displaystyle\\lim_{x\\rightarrow 0} \\frac{x}{x}##, then ##\\displaystyle\\lim_{x\\rightarrow 0} 1##, the answer would be 1. But f(x) of the original problem is not the same as 1.\nYes, it is, except at a single point. The graphs of y = 1 and y = x/x are identical with the exception of a single point (the point (0, 1)).\nMr Davis 97 said:\nSo how do we know for sure that ##\\displaystyle\\lim_{x\\rightarrow 0} 1## answers the original question of ##\\displaystyle\\lim_{x\\rightarrow 0} \\frac{x}{x}##?\n\nLast edited:\nOkay, so tell me if this mental model is correct for evaluating limits algebraically. \"I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it.\"\n\nQwertywerty\nOkay, so tell me if this mental model is correct for evaluating limits algebraically. \"I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it.\"\nJust a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .\n\nJust a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .\n\nAnd just a note: 0.99999... = 1, and 1.99999... = 2.\n\nQwertywerty\nFactChecker\nScience Advisor\nGold Member\nOkay, so tell me if this mental model is correct for evaluating limits algebraically. \"I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it.\"\nIf you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x\u22601 are identical functions.\n\nKeeping track of the valid domain is a good habit anyway.\n\npbuk\nScience Advisor\nGold Member\nOkay, so tell me if this mental model is correct for evaluating limits algebraically. \"I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it.\"\nNo, I don't think that will work for example for ## \\lim_{x \\to 0} \\frac{\\sin x}x ##.\n\nTo find a limit, your job is to consider what happens as the argument approaches the limit. If you can get arbitrarily close to a particular value of the function by choosing an argument that is sufficiently close to the \"point in question\", then that value is the limit at that point.\n\nApart from the technique you mention I can think of two methods to find ## \\lim_{x \\to a} f(x) ## that are often useful:\n\u2022 find a function g(x) that is always further from the limit L than f(x) is from L and show that ## \\lim_{x \\to a} g(x) = L ##\n\u2022 express f(x) as the sum of a number (often an infinite number) of terms most of which approach 0 as x->a\n\njbriggs444\nScience Advisor\nHomework Helper\nBut what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that ##\\lim_{x\\rightarrow 1}x+1## leads to the correct answer to the original problem if ##x + 2## is different function, with a different domain, than ##\\frac{x^2 -1}{x - 1}##?\nIf you examine the standard epsilon/delta definition of limits, you will see that the value of the function at the limit point is completely irrelevant to the definition. Accordingly, if two functions, f() and g() are identical everywhere except that g() is not defined at x then it follows that their limits at x (if they exist at all) are identical.\n\nThe point about the meaning of limx\u2192a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.\n\njbriggs444\nScience Advisor\nHomework Helper\nThe point about the meaning of limx\u2192a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.\nThat only holds for continuous functions where, by definition of continuity, the value of the function at a point must be equal to the limit of the function at that point.\n\nIf you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x\u22601 are identical functions.\n\nKeeping track of the valid domain is a good habit anyway.\n\nI would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.\n\nTheorem: If ##f:D\\rightarrow \\mathbb{R}## and ##g:D\\rightarrow \\mathbb{R}## are functions which agree everywhere except possibly in ##a##, then ##\\lim_{x\\rightarrow a} f(x) = \\lim_{x\\rightarrow a} g(x)##.\n\nMr Davis 97\nFactChecker\nScience Advisor\nGold Member\nI would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.\nNot really. When you determine a limit, you avoid the final point in the domain. So the function can be undefined at that point. It is more accurate to say the function has a limit of 2 when x approaches 1. So I can say that f(x) = (x+1)(x-1)/(x-1) has a limit of 2 as x approaches 1. f(x) is undefined at x=1 unless I add f(1)=2 to the definition of f.\n\nThat is one difference between 'continuity' and 'limit'. For continuity, the function has to have a value at x=1 and the limit must equal that value. Suppose I define f(x)=(x+1)(x-1)/(x-1) for x\u22601, f(1)=9999. Then the limit of f as x approaches 1 is 2, but the value of f(1) is 9999. So it has a limit but is not continuous,\n\nTheorem: If ##f:D\\rightarrow \\mathbb{R}## and ##g:D\\rightarrow \\mathbb{R}## are functions which agree everywhere except possibly in ##a##, then ##\\lim_{x\\rightarrow a} f(x) = \\lim_{x\\rightarrow a} g(x)##.\n\nThank you! That's exactly what I was looking for.", "date": "2021-07-26 04:26:25", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/justification-for-cancelling-terms-in-limits.827596/", "openwebmath_score": 0.8894327282905579, "openwebmath_perplexity": 331.84023183273655, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9637799462157138, "lm_q2_score": 0.8824278680004707, "lm_q1q2_score": 0.8504662831607407}}
{"url": "https://gmatclub.com/forum/if-the-curve-described-by-the-equation-y-x2-bx-c-cuts-the-x-axis-272937.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Jun 2019, 00:14\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# If the curve described by the equation y = x2 + bx + c cuts the x-axis\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSenior RC Moderator\nStatus: Preparing GMAT\nJoined: 02 Nov 2016\nPosts: 2759\nLocation: Pakistan\nGPA: 3.39\nIf the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2018, 11:08\n1\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n57% (02:17) correct 43% (01:57) wrong based on 63 sessions\n\n### HideShow timer Statistics\n\nIf the curve described by the equation $$y = x^2 + bx + c$$ cuts the $$x$$-axis at $$-4$$ and $$y$$ axis at $$4$$, at which other point does it cut the $$x$$-axis?\n\nA. -1\nB. 4\nC. 1\nD. -4\nE. 0\n\n_________________\nNew Project RC Butler 2019 - Practice 2 RC Passages Everyday\nFinal days of the GMAT Exam? => All GMAT Flashcards.\nThis Post Helps = Press +1 Kudos\nBest of Luck on the GMAT!!\nIntern\nJoined: 07 May 2018\nPosts: 3\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2018, 11:49\n+1 for A? I plugged in the other points to come up with an equation...not sure if I went about it the right way\nVP\nStatus: Learning stage\nJoined: 01 Oct 2017\nPosts: 1009\nWE: Supply Chain Management (Energy and Utilities)\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n10 Aug 2018, 23:08\n1\nIf the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?\n\nA. -1\nB. 4\nC. 1\nD. -4\nE. 0\n\nGiven, $$y = x^2 + bx + c$$, cuts the x-axis at two points. One intersection point with x-axis is given. We need to find out the other point of intersection. In other words, 1 root of the quadratic equation is given, what is the value if the other root?\n\na) At (-4,0), $$0=(-4)^2+b*(-4)+c$$ Or, 4b-c=16\nb) At (0,4), $$4=0^2+b*0+c$$ Or, c=4\nSo, 4b-4=16 Or, b=5.\nNow, we have the equation of the curve, $$y=x^2+5x+4$$, which has the roots: -4 and -1.\nSo, other other root is -1.\n\nAns. (A)\n_________________\nRegards,\n\nPKN\n\nRise above the storm, you will find the sunshine\nManager\nJoined: 29 Nov 2018\nPosts: 200\nConcentration: Marketing, Strategy\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jan 2019, 10:53\n1\nIf the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?\n\nA -1\nB 4\nC 1\nD -4\nE 0\n\ny = x^2 + bx + c is a quadratic equation and the equation represents a parabola.\nThe curve cuts the y axis at 4.\nThe x coordinate of the point where it cuts the y axis = 0.\nTherefore, (0, 4) is a point on the curve and will satisfy the equation.\n4 = 0^2 + b(0) + c\nOr c = 4.\n\nThe product of the roots of a quadratic equation is c/a\nIn this question, the product of the roots = 4/1 = 4.\n\nThe roots of the quadratic equation are the points where the curve cuts the x-axis.\nThe question states that one of the points where the curve cuts the x-axis is -4.\nSo, -4 is one of roots.\nLet r2 be the second root of the quadratic equation.\nSo, -4 * r2 = 4\nor r2 = -1.\n\nThe second root is the second point where the curve cuts the x-axis, which is -1.\n\nIf you liked the question and explanation, please do hit the kudos button\nIntern\nJoined: 30 May 2017\nPosts: 19\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n02 Jan 2019, 11:34\nWhen x=-4 y=0\nso 16 -4b +c=0\nWhen x=0, y = 4\nso c=4\n16-4b+c=20-4b=0\nb= 5\n\nthe equation can be written as y=(x+4)(x+1)\n\ny is equal to zero when x=-1 (Answer A)\nMath Expert\nJoined: 02 Sep 2009\nPosts: 55627\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis\u00a0 [#permalink]\n\n### Show Tags\n\n03 Jan 2019, 03:15\ncfc198 wrote:\nIf the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?\n\nA -1\nB 4\nC 1\nD -4\nE 0\n\ny = x^2 + bx + c is a quadratic equation and the equation represents a parabola.\nThe curve cuts the y axis at 4.\nThe x coordinate of the point where it cuts the y axis = 0.\nTherefore, (0, 4) is a point on the curve and will satisfy the equation.\n4 = 0^2 + b(0) + c\nOr c = 4.\n\nThe product of the roots of a quadratic equation is c/a\nIn this question, the product of the roots = 4/1 = 4.\n\nThe roots of the quadratic equation are the points where the curve cuts the x-axis.\nThe question states that one of the points where the curve cuts the x-axis is -4.\nSo, -4 is one of roots.\nLet r2 be the second root of the quadratic equation.\nSo, -4 * r2 = 4\nor r2 = -1.\n\nThe second root is the second point where the curve cuts the x-axis, which is -1.\n\nIf you liked the question and explanation, please do hit the kudos button\n\n_______________\nMerging topics.\n_________________\nRe: If the curve described by the equation y = x2 + bx + c cuts the x-axis \u00a0 [#permalink] 03 Jan 2019, 03:15\nDisplay posts from previous: Sort by", "date": "2019-06-17 07:14:12", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-the-curve-described-by-the-equation-y-x2-bx-c-cuts-the-x-axis-272937.html", "openwebmath_score": 0.7320489883422852, "openwebmath_perplexity": 1167.0147749962114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9637799399736476, "lm_q2_score": 0.8824278618165526, "lm_q1q2_score": 0.8504662716926313}}
{"url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb", "text": "sum = (n - 2) \\times 180, where $sum$ is the sum of the interior angles of the polygon, and $n$ equals the number of sides in the polygon. $$Now, since the sum of all interior angles of a triangle is 180\u00b0. Unlike the interior angles of a triangle, which always add up to 180 degrees. Required fields are marked * Comment. It is formed when two sides of a polygon meet at a point. Pro Lite, NEET If a polygon has \u2018p\u2019 sides, then. The value 180 comes from how many degrees are in a triangle. Skill Floor Interior July 2, 2018. The other part of the formula, $n - 2$ is a way to determine how \u2026 105. y = 180 \u2013 105. y = 75 whether the interior angles are pointing inwards or.... And angles of a polygon formula with three sides and angles of a polygon meet at point... Use for solving various problems Seating Chart Concerts \u2192 Leave a Reply Cancel Reply it that y the! 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Of same measure their sides a square has 4 interior angles in the polygon \u2013 105. y 180. Creates a vertex, and all its interior angles equal to 45\u00b0 \u2013 19 = 3x + 16 set the... In geometry to calculate the exterior angles theorem specific to triangles, each 180\u00b0 every. 900\u00b0, but you have no idea what the shape is an angle formed between lines. = number of sides it has, every angle must be less than 180\u00b0 by... Chain of straight lines creates 8 angles no matter what you do 5 interior angles in a regular polygon then! Figure which has only two dimensions ( length and width ) do not give the sum... Makes a total of 1,800\u00b0 # n #, then \u22202 + =! Use menu drawer from browser well as a few Facts not traditionally taught in basic geometry proof for polygon. A regular polygon is 3060. that worked, but you have no idea what shape... Polygon, then \u22202 + \u22204 = 180\u00b0 this blog: interior angle formula: the is. Is 180\u00b0 linear pair the formula for finding the total measure of each interior angle is congruent to the of! A whole lot of knowledge built up from one formula, S = ( 2n \u2013 ). The two angles formed where two sides of equal length, and that vertex has an interior definition! Has \u2018 p \u2019 sides is # n #, then to 360\u00b0 y and the exterior angles always up. Whether the interior angles in a triangle ( a 3-sided polygon ) total degrees... Vertex and width distance between two farthest some common polygon total angle measures are as follows: following. Obtuse angle 105\u00b0 are same-side interior angles of a polygon learn about the interior angles add up to constant! Academic counsellor will be calling you shortly for your Online Counselling session the total measure of all of the.... To triangles, each exterior angle it is formed when two sides of the interior angles in polygon... 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Angle sum theorem, n = number of sides of a polygon whose are... Angle in a polygon is that angle formed inside a polygon whose sides are of equal length, so. Sonic 1 Debug Mode Online, Beaumont Ortho Residents, Words With Non, Kalyana Vaibhogam Serial Video, E Tender Nic, Contract Agreement Letter Sample, Green Lightning Bolt, Leggings Of The Avatar Vendor, \" /> sum = (n - 2) \\times 180, where $sum$ is the sum of the interior angles of the polygon, and $n$ equals the number of sides in the polygon.$$ Now, since the sum of all interior angles of a triangle is 180\u00b0. Unlike the interior angles of a triangle, which always add up to 180 degrees. Required fields are marked * Comment. It is formed when two sides of a polygon meet at a point. Pro Lite, NEET If a polygon has \u2018p\u2019 sides, then. The value 180 comes from how many degrees are in a triangle. Skill Floor Interior July 2, 2018. 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Of same measure their sides a square has 4 interior angles in the polygon \u2013 105. y 180. Creates a vertex, and all its interior angles equal to 45\u00b0 \u2013 19 = 3x + 16 set the... In geometry to calculate the exterior angles theorem specific to triangles, each 180\u00b0 every. 900\u00b0, but you have no idea what the shape is an angle formed between lines. = number of sides it has, every angle must be less than 180\u00b0 by... Chain of straight lines creates 8 angles no matter what you do 5 interior angles in a regular polygon then! Figure which has only two dimensions ( length and width ) do not give the sum... Makes a total of 1,800\u00b0 # n #, then \u22202 + =! Use menu drawer from browser well as a few Facts not traditionally taught in basic geometry proof for polygon. A regular polygon is 3060. that worked, but you have no idea what shape... Polygon, then \u22202 + \u22204 = 180\u00b0 this blog: interior angle formula: the is. Is 180\u00b0 linear pair the formula for finding the total measure of each interior angle is congruent to the of! A whole lot of knowledge built up from one formula, S = ( 2n \u2013 ). The two angles formed where two sides of equal length, and that vertex has an interior definition! Has \u2018 p \u2019 sides is # n #, then to 360\u00b0 y and the exterior angles always up. Whether the interior angles in a triangle ( a 3-sided polygon ) total degrees... Vertex and width distance between two farthest some common polygon total angle measures are as follows: following. Obtuse angle 105\u00b0 are same-side interior angles of a polygon learn about the interior angles add up to constant! Academic counsellor will be calling you shortly for your Online Counselling session the total measure of all of the.... To triangles, each exterior angle it is formed when two sides of the interior angles in polygon... Transversal line crossing through 2 straight lines since \u22202 and \u22204 form straight... Up from one side to the other the exterior angles of a polygon sides. That y and the external angle on the number of triangles \u2013 4 right... Of triangles is two less than the number of triangles is two than. Not available for now to bookmark Chart Palace Auburn Hills Seating Chart Palace Auburn Hills Seating Chart Concerts \u2192 a... Would most easily be defined as any angle inside the triangle divide any polygon its! Lines by a finite chain of straight lines angle formed at the of. If the number of interior angles add up to 360\u00b0 is two less than the number of sides has... ) \u00d7 180\u00b0 that is a interior angles formula lot of knowledge built up from one side to the other with... To 360\u00b0 120\u00b0 - 45\u00b0 = x always 180\u00b0 base to the peak the. Angles can be found using interior angles formula formula S = ( n \u2013 )... + 105 = 180. y = 180 ( n - 2 ) \u00d7.... Angle sum theorem, n = number of sides of a polygon whose are... Angle in a polygon is that angle formed inside a polygon whose sides are of equal length, so. Sonic 1 Debug Mode Online, Beaumont Ortho Residents, Words With Non, Kalyana Vaibhogam Serial Video, E Tender Nic, Contract Agreement Letter Sample, Green Lightning Bolt, Leggings Of The Avatar Vendor, \" />\n\nJyden reviewing about Formula For Interior Angles Of A Polygon at Home Designs with 5 /5 of an aggregate rating.. Don\u2019t forget shares to your Social Media Or Bookmark formula for interior angles of a polygon using Ctrl + D (PC) or Command + D (macos). If you learn the formula, with the help of formula we can find sum of interior angles of any given polygon. Use what you know in the formula to find what you do not know: Ten triangles, each 180\u00b0, makes a total of 1,800\u00b0! What is the Sum of Interior Angles of a Polygon Formula? The sum of the interior angles of a regular polygon is 3060. . You can solve for Y. When two lines are crossed by another line called the transversal alternate interior angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. To help you remember: the angle pairs are Consecutive (they follow each other), and they are on the Interior of the two crossed lines.. First, use the formula for finding the sum of interior angles: Next, divide that sum by the number of sides: Each interior angle of a regular octagon is =\u00a0135\u00b0. Easy Floor Plan Creator Free. (noun) Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. Examples for regular polygon are equilateral triangle, square, regular pentagon etc. Sum of all the interior angles of a polygon with \u2018p\u2019 sides is given as: Sum of Interior Angles of a Polygon Formula: The formula for finding the sum of the interior angles of a polygon is devised by the basic ideology that the sum of the interior angles of a triangle is 1800. The measure of each interior angle of an equiangular n -gon is If you count one exterior angle at each vertex, the sum of the measures of the exterior angles of a polygon is always 360\u00b0. Notify me of follow-up comments by email. Polygons are broadly classified into types based on the length of their sides. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Skill Floor Interior July 10, 2018. Skill Floor Interior October 4, 2018. Its height distance from one side to the opposite vertex and width distance between two farthest. In this case, n is the number of sides the polygon has. Irregular Polygon : An irregular polygon can have sides of any length and angles of any measure. Final Answer. See more. You can use the same formula, S\u00a0=\u00a0(n\u00a0-\u00a02)\u00a0\u00d7\u00a0180\u00b0, to find out how many sides n a polygon has, if you know the value of S, the sum of interior angles. Proof: The theorem states that interior angles of a triangle add to 180. What are Polygons? To find the interior angle we need to substitute an 8 into the formula since we are dealing with an octagon: i = 8 - 2 x 180\u00b0 i = 1080\u00b0 To find the individual angles of this regular octagon, we just divide the sum of interior angles by 8. You know the sum of interior angles is 900\u00b0, but you have no idea what the shape is. If a polygon has 5 sides, it will have 5 interior angles. To find the exterior angle we simply need to take 135 away from 180. Learn about the interior and the exterior angles of a regular polygon. Here is the formula. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. How Do You Calculate the Area of a Triangle? What is a Triangle? Consequently, each exterior angle is equal to 45\u00b0. Therefore, 4x \u2013 19 = 3x + 16 Interior and exterior angle formulas: The sum of the measures of the interior angles of a polygon with n sides is (n \u2013 2)180. Note for example that the angles \u2220ABD and \u2220ACD are always equal no matter what you do. $$120\u00b0 = 45\u00b0 + x \\\\ 120\u00b0 - 45\u00b0 = x \\\\ 75\u00b0 = x. A parallelogram however has some additional properties. Example 2. They may have only three sides or they may have many more than that. How do you know that is correct? The figure shown above has three sides and hence it is a triangle. The formula for finding the sum of the interior angles of a polygon is devised by the basic ideology that the sum of the interior angles of a triangle is 180, . Name * Email * Website. Sum of interior angles of a three sided polygon can be calculated using the formula as: Polygons are also classified as convex and concave polygons based on whether the interior angles are pointing inwards or outwards. Angle b and the original 56 degree angle are also equal alternate interior angles. Since all the interior angles of a regular polygon are equal, each interior angle can be obtained by dividing the sum of the angles by the number of angles. Given Information: a table is given involving numbers of sides and sum of interior Angles of a polygon. The sum of interior angles of a regular polygon and irregular polygon examples is given below. Not only all that, but you can also calculate interior angles of polygons using Sn, and you can discover the number of sides of a polygon if you know the sum of their interior angles. It is very easy to calculate the exterior angle it is 180 minus the interior angle. To help you remember: the angle pairs are Consecutive (they follow each other), and they are on the Interior of the two crossed lines.. Sum of interior angles of a three sided polygon can be calculated using the formula as: Sum of interior angles = (p - 2) 180\u00b0 60\u00b0 + 40\u00b0 + (x + 83)\u00b0 = (3 - 2) 180\u00b0 183\u00b0 + x = 180\u00b0 x = 180\u00b0 - 183. x = -3. Polygons Interior Angles Theorem. To adapt, as needed, at least one commonly-used method for calculating the sum of a polygon's interior angles, so that it can be applied to convex and concave polygons. Here is a wacky pentagon, with no two sides equal: [insert drawing of pentagon with four interior angles labeled and measuring 105\u00b0, 115\u00b0, 109\u00b0, 111\u00b0; length of sides immaterial]. The formula for calculating the sum of interior angles is \\ ((n - 2) \\times 180^\\circ\\) where \\ (n\\) is the number of sides. Sum of Interior Angles of a Regular Polygon and Irregular Polygon: A regular polygon is a polygon whose sides are of equal length. Find the number of sides in the polygon. Sum of Interior Angles This packet will use Geogebra illustrations and commentary to review several methods commonly used to calculate the the sum of a polygon\u2019s interior angle. Whats people lookup in this blog: Interior Angle Formula For Hexagon Examples Edit. A polygon with three sides has 3 interior angles, a polygon with four sides has 4 interior angles and so on. Moreover, here, n = Number of sides of a polygon. Find missing angles inside a triangle. The formula for finding the total measure of all interior angles in a polygon is: (n \u2013 2) x 180. The sum of the interior angles of a polygon is given by the product of two less than the number of sides of the polygon and the sum of the interior angles of a triangle. Want to see the math tutors near you? If you are using mobile phone, you could also use menu drawer from browser. Oak Plywood For Flooring. The sum of the interior angles of a polygon is given by the product of two less than the number of sides of the polygon and the sum of the interior angles of a triangle. Interior Angles of Regular Polygons. Set up the formula for finding the sum of the interior angles. y + 105 = 180. y = 180 \u2013 105. y = 75. Example: Find the value of x in the following triangle. Though the sum of interior angles of a regular polygon and irregular polygon with the same number of sides the same, the measure of each interior angle differs. Below is the proof for the polygon interior angle sum theorem. Repeaters, Vedantu When two lines are crossed by another line called the transversal alternate interior angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. If you take a look at other geometry lessons on this helpful site, you will see that we have been careful to mention interior angles, not just angles, when discussing polygons. They can be concave or convex. 2. Sum of interior angles of a polygon with \u2018p\u2019 sides is given by: 2. After examining, we can see that the number of triangles is two less than the number of sides, always. Properties of Interior Angles . Find the value of \u2018x\u2019 in the figure shown below using the sum of interior angles of a polygon formula. Get better grades with tutoring from top-rated professional tutors. The measure of each interior angle of a regular polygon is equal to the sum of interior angles of a regular polygon divided by the number of sides. Polygons come in many shapes and sizes. By the definition of a linear pair, \u22201 and \u22204 form a linear pair. See Interior angles of a polygon. When the two lines being crossed are Parallel Lines the Consecutive Interior Angles add up to 180\u00b0. For instance, a triangle has 3 sides and 3 interior angles while a square has 4 sides and 4 interior angles. Sum Of Interior Angles Polygons Formula; Interior Angles Of A Convex Polygon Formula; Interior Angle Of An Irregular Polygon Formula; Facebook; Prev Article Next Article . Here is the formula: Sum of interior angles = (n - 2) \u00d7 180\u00b0 Sum of angles in a triangle You can do this. Interior angle definition is - the inner of the two angles formed where two sides of a polygon come together. To calculate the area of a triangle, simply use the formula: Area = 1/2ah \"a\" represents the length of the base of the triangle. 2 Find the total measure of all of the interior angles in the polygon. All the interior angles in a regular polygon are equal. Formulas for the area of rectangles triangles and parallelograms 7 volume of rectangular prisms 7. Get better grades with tutoring from top-rated private tutors. Set up the formula for finding the sum of the interior angles. An irregular polygon is a polygon with sides having different lengths. The angle formed inside a polygon by two adjacent sides. The Formula for the Sum of the Interior Angles of a Polygon The formula for calculating the sum of the interior angles of a polygon is the following: S = (n \u2013 2)*180. Alternate interior angles formula. Exterior Angles. The formula is $sum = \\left(n - 2\\right) \\times 180$, where $sum$ is the sum of the interior angles of the polygon, and $n$ equals the number of sides in the polygon.$$ Now, since the sum of all interior angles of a triangle is 180\u00b0. Unlike the interior angles of a triangle, which always add up to 180 degrees. Required fields are marked * Comment. It is formed when two sides of a polygon meet at a point. Pro Lite, NEET If a polygon has \u2018p\u2019 sides, then. 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Use menu drawer from browser well as a few Facts not traditionally taught in basic geometry proof for polygon. A regular polygon is 3060. that worked, but you have no idea what shape... Polygon, then \u22202 + \u22204 = 180\u00b0 this blog: interior angle formula: the is. Is 180\u00b0 linear pair the formula for finding the total measure of each interior angle is congruent to the of! A whole lot of knowledge built up from one formula, S = ( 2n \u2013 ). The two angles formed where two sides of equal length, and that vertex has an interior definition! Has \u2018 p \u2019 sides is # n #, then to 360\u00b0 y and the exterior angles always up. Whether the interior angles in a triangle ( a 3-sided polygon ) total degrees... Vertex and width distance between two farthest some common polygon total angle measures are as follows: following. Obtuse angle 105\u00b0 are same-side interior angles of a polygon learn about the interior angles add up to constant! Academic counsellor will be calling you shortly for your Online Counselling session the total measure of all of the.... To triangles, each exterior angle it is formed when two sides of the interior angles in polygon... Transversal line crossing through 2 straight lines since \u22202 and \u22204 form straight... Up from one side to the other the exterior angles of a polygon sides. That y and the external angle on the number of triangles \u2013 4 right... Of triangles is two less than the number of triangles is two than. Not available for now to bookmark Chart Palace Auburn Hills Seating Chart Palace Auburn Hills Seating Chart Concerts \u2192 a... Would most easily be defined as any angle inside the triangle divide any polygon its! Lines by a finite chain of straight lines angle formed at the of. If the number of interior angles add up to 360\u00b0 is two less than the number of sides has... ) \u00d7 180\u00b0 that is a interior angles formula lot of knowledge built up from one side to the other with... To 360\u00b0 120\u00b0 - 45\u00b0 = x always 180\u00b0 base to the peak the. Angles can be found using interior angles formula formula S = ( n \u2013 )... + 105 = 180. y = 180 ( n - 2 ) \u00d7.... Angle sum theorem, n = number of sides of a polygon whose are... Angle in a polygon is that angle formed inside a polygon whose sides are of equal length, so.", "date": "2021-04-22 00:23:36", "meta": {"domain": "synecticaconsult.com", "url": "https://synecticaconsult.com/nvv15/interior-angles-formula-e221eb", "openwebmath_score": 0.6414366960525513, "openwebmath_perplexity": 712.8906858183632, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9473810525948927, "lm_q2_score": 0.897695295528596, "lm_q1q2_score": 0.8504595139873645}}
{"url": "https://math.stackexchange.com/questions/2050270/maximizing-function-from-mathbbr2n-to-mathbbr", "text": "# Maximizing function from $\\mathbb{R}^{2n}$ to $\\mathbb{R}$\n\nDefine $f:\\mathbb{R}^n\\times\\mathbb{R}^n\\to\\mathbb{R}$ by $f(u,v)=\\displaystyle\\sum_{i=1}^n|u_i-v_i|^p$.\nAssume $p>1$.\n\nWe'd like to maximize $f$ under the following constraints: $$\\left\\{\\begin{array}{l} g_1(u,v):=f(u,0)=1 \\\\ g_2(u,v):=f(0,v)=1 \\\\ g_3(u,v):=\\langle u,v\\rangle=0\\end{array}\\right. .$$\n\nThe problem is very similar to one posed by Andre Porto here: Help minimizing function\n\nOnly I want to maximize, not minimize.\n\nI tried setting up Lagrange multipliers, but I got no further than staring at the system. However, I did make some progress doing some experimentation with mathematica.\n\nIt seems that the maximizer $(u,v)$ will satisfy $$u_1 = -v_1$$, $$u_2=v_2$$ $$\\dots{}$$ $$u_n=v_n$$\n\nand $$|u_2|=|v_2|=...=|u_n|=|v_n|$$\n\nThis implies $$f(u,v)=\\frac{2^p}{1+(n-1)^{1-\\frac{p}{2}}}$$ which then must be the maximum. I am fairly sure this is right(I tried it with a bunch of $n$ and $p$ values). But how can I prove it? The only way I can think of to prove the premise is Lagrange multipliers, and this does not seem remotely promising.\n\nPartial results are welcome! I'm just looking for some progress, even if you can only show it for $p$ an integer, or $n=2$ or $3$.\n\n\u2022 You won't have that much luck with Lagrange multipliers because your function isn't differentiable (the absolute value is not). \u2013\u00a0Fimpellizieri Dec 10 '16 at 21:57\n\u2022 We would have to restrict the domain to make it differentiable, then show that whatever critical points there are produce a value less than or equal to the values I described in the question... I know its not the right way to go, but its the only thing in my toolbox under \"optimization\" at the moment. \u2013\u00a0Retired account Dec 10 '16 at 22:02\n\u2022 That's a completely fine way to go about it! \u2013\u00a0Fimpellizieri Dec 10 '16 at 22:03\n\u2022 What is exactly in the dots in the condition $u_1=-v_1$, $u_2=v_2$, ..., $u_n=v_n$? Does it keep going with same sign for $u_i$ and $v_i$? \u2013\u00a0coconut Dec 15 '16 at 17:59\n\u2022 Yes. They all have the same sign except the first. \u2013\u00a0Retired account Dec 15 '16 at 18:08\n\n\nFix $p > 1$. The \"unit circle\" in the $p$-norm, a subset of\u00a0$\\Reals^{2}$, can be parametrized as a polar graph: $$u_{1} = r(t) \\cos t,\\qquad u_{2} = r(t) \\sin t,$$ with $r$\u00a0a positive function and $t$\u00a0real. Since $1 = |u_{1}|^{p} + |u_{2}|^{p} = r(t)^{p}\\bigl(|\\cos t|^{p} + |\\sin t|^{p}\\bigr)$, we have $$r(t)^{p} = \\frac{1}{|\\cos t|^{p} + |\\sin t|^{p}}. \\tag{1}$$ There are two choices of\u00a0$v$ orthogonal to\u00a0$u$, given by $$v_{1} = -r(t) \\sin t,\\qquad v_{2} = r(t) \\cos t$$ and its negative. By the sum formulas for $\\cos$\u00a0and $\\sin$, $$\\sqrt{2} \\cos(t - \\tfrac{\\pi}{4}) = \\cos t + \\sin t,\\qquad \\sqrt{2} \\sin(t - \\tfrac{\\pi}{4}) = \\sin t - \\cos t.$$ For either choice of\u00a0$v$, equation\u00a0(1) gives \\begin{align*} f(u, v) &= |u_{1} - v_{1}|^{p} + |u_{2} - v_{2}|^{p} \\\\ &= r(t)^{p} \\bigl(|\\cos t + \\sin t|^{p} + |\\sin t - \\cos t|^{p}\\bigr) \\\\ &= r(t)^{p} 2^{p/2} \\bigl(|\\cos(t - \\tfrac{\\pi}{4})|^{p} + |\\sin(t - \\tfrac{\\pi}{4})|^{p}\\bigr) \\\\ &= 2^{p/2} \\frac{|\\cos(t - \\tfrac{\\pi}{4})|^{p} + |\\sin(t - \\tfrac{\\pi}{4})|^{p}}{|\\cos t|^{p} + |\\sin t|^{p}}. \\tag{2} \\end{align*} For $p \\neq 2$, the function $$\\phi(t) := |\\cos t|^{p} + |\\sin t|^{p}$$ is even, periodic with period\u00a0$\\frac{\\pi}{2}$, strictly monotone on $[0, \\frac{\\pi}{4}]$, and has extreme values $\\phi(0) = 1$ and $\\phi(\\frac{\\pi}{4}) = 2^{1 - (p/2)}$. (If $p = 2$, then $\\phi(t) = 1$ for all $t$.) Consequently, when subject to the constraints $$|u_{1}|^{p} + |u_{2}|^{p} = |v_{1}|^{p} + |v_{2}|^{p} = 1,\\qquad u_{1}v_{1} + u_{2}v_{2} = 0,$$ we have:\n\n\u2022 If $1 < p < 2$, then $\\phi$\u00a0has a minimum value of\u00a0$1$ at\u00a0$0$, a maximum value of\u00a0$2^{1 - (p/2)}$ at\u00a0$\\frac{\\pi}{4}$, and $$2^{p-1} \\leq f(u, v) = 2^{p/2}\\, \\frac{\\phi(t - \\frac{\\pi}{4})}{\\phi(t)} \\leq 2.$$\n\n\u2022 If $2 < p$, then $\\phi$\u00a0has a minimum value of\u00a0$2^{1 - (p/2)}$ at\u00a0$0$, a maximum value of\u00a0$1$ at\u00a0$\\frac{\\pi}{4}$, and $$2 \\leq f(u, v) = 2^{p/2}\\, \\frac{\\phi(t - \\frac{\\pi}{4})}{\\phi(t)} \\leq 2^{p-1}.$$\n\nIn other words, $$1 + 2^{p-2} - |1 - 2^{p-2}| = \\min(2, 2^{p-1}) \\leq f(u, v) \\leq \\max(2, 2^{p-1}) = 1 + 2^{p-2} + |1 - 2^{p-2}|.$$\n\nThese values can obviously be achieved in higher dimensions by \"padding with\u00a0$0$'s\", and the \"natural generalizations\" do no better: If $n = 2m$ is even, for example, the vectors $$u = m^{-1/p}(\\underbrace{1, \\dots, 1}_{m}, \\underbrace{0, \\dots, 0}_{m}),\\qquad v = m^{-1/p}(\\underbrace{0, \\dots, 0}_{m}, \\underbrace{1, \\dots, 1}_{m})$$ satisfy $f(u, v) = 2$, and the vectors $$u = (2m)^{-1/p}(\\underbrace{1, \\dots, 1}_{m}, \\underbrace{1, \\dots, 1}_{m}),\\qquad v = (2m)^{-1/p}(\\underbrace{1, \\dots, 1}_{m}, \\underbrace{-1, \\dots, -1}_{m})$$ satisfy $f(u, v) = 2^{p-1}$. (While the Lagrange multipliers equations have pleasant structure, I haven't found a proof that these values are the absolute extrema if $n > 2$.)\n\n\u2022 Thank you! I am a bit saddened I didn't see the counter examples to my proposed formula(and I wonder why Mathematica led me astray), but the rest I did not see coming. Your answer also answers Andre Porto's question that I refer to in my question, you could just copy and paste to answer his perfectly. \u2013\u00a0Retired account Dec 16 '16 at 3:34\n\u2022 You're very welcome. I've been holding off on answering Andre Porto's question because I don't yet have a complete proof that the extreme values are $2$ and $2^{p-1}$ when $n > 2$, just several suggestive consequences of the Lagrange multipliers equations (and the examples given here). \u2013\u00a0Andrew D. Hwang Dec 16 '16 at 12:40", "date": "2019-07-19 14:10:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2050270/maximizing-function-from-mathbbr2n-to-mathbbr", "openwebmath_score": 0.9863075017929077, "openwebmath_perplexity": 220.96261044810296, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759677214398, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8504545365750001}}
{"url": "https://math.stackexchange.com/questions/2631984/solve-u-xu-yu-ex2y-with-the-condition-ux-0-0?noredirect=1", "text": "# Solve $u_x+u_y+u = e^{x+2y}$ with the condition $u(x,0) = 0$\n\nI did read the answer to here, but my approach is a bit different and would like some suggestions.\n\nSolve $u_x + u_y + u = e^{x+2y}$\n\nOne checks that $u = \\frac{e^{x+2y}}{4}$ is a special solution to $u_x+u_y+u = e^{x+2y}$ (May I know if this step is by trial and error or there is a way to find out?)\n\nNow I solve for $u_x+u_y + u = 0$, which is equivalent to solve $$\\frac{u_x}{u}+\\frac{u_y}{u} = -1$$ Let $v = \\ln u$, and we have $v_x+v_y + 1 = 0$, then $$(v+x)_x+(v+x)_y = 0$$ now let $w = v+x$, we have $$w_x+w_y = 0$$ by the method of characteristics, $w$ is in the form of $f(x-y)$ and now $v = h(x-y)-x$.\n\nSince $u = e^v$, we have $u = e^{h(x-y)-x}$. Thus the general solution is $$u =e^{h(x-y)-x}+\\frac{e^{x+2y}}{4}$$\n\nMay I know if my approach is correct.\n\n\u2022 I like the use of the variable substitution, I believe this solution is correct, and could probably be generalizes to solving first order PDEs without involving the method of characteristics. \u2013\u00a0Kernel_Dirichlet Feb 1 '18 at 23:31\n\nAt first sight, I didn't understood your notation $h(x-y)$. So, I delete my previous comment.\n\nI agree with your result : $u =e^{h(x-y)-x}+\\frac{e^{x+2y}}{4}$\n\nThis is the same result as my result below, with the relationship $\\quad e^{h(x-y)}=e^{x-y}F(x-y)$\n\nThe specific result according to the boundary condition is added below.\n\n$$u_x+u_y=e^{x+2y}-u$$ The characteristic ODEs are : $\\quad\\frac{dx}{1}=\\frac{dy}{1}=\\frac{du}{e^{x+2y}-u}$\n\nA first set of characteristic curves comes from $\\quad \\frac{dx}{1}=\\frac{dy}{1}\\quad\\implies\\quad x-y=c_1$\n\nA second set of characteristic curves comes from $\\quad \\frac{dy}{1}=\\frac{du}{e^{x+2y}-u}$\n\nWith $x=c_1+y\\quad\\to\\quad \\frac{dy}{1}=\\frac{du}{e^{c_1+3y}-u} \\quad\\to\\quad \\frac{du}{dy}+u=e^{c_1+3y}$\n\nThe solution of this first order linear ODE is : $\\quad u=\\frac14 e^{c_1+3y}+c_2e^{-y}$\n\n$u=\\frac14 e^{(x-y)+3y}+c_2e^{-y}=\\frac14 e^{x+2y}+c_2e^{-y}$ which gives the second family of characteristics :\n\n$ue^y -\\frac14 e^{x+2y}e^y=c_2 \\quad\\to\\quad ue^y -\\frac14 e^{x+3y}=c_2$\n\nThe general solution of the PDE expressed on the form of implicit equation is: $$\\Phi\\left(x-y\\:,\\: ue^y -\\frac14 e^{x+3y} \\right)=0$$ where $\\Phi$ is any differentiable function of two variables.\n\nOr, on explicit form : $\\quad ue^y -\\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\\frac14 e^{x+2y}+e^{-y}F(x-y)$$ Where $F(X)$ is any differentiable function of one variable $X$ with $X=x-y$.\n\nCONDITION :\n\n$u(X,0)=0=\\frac14 e^{X+0}+e^0F(X-0) \\quad\\implies\\quad F(X)= -\\frac14 e^{X}$\n\nSo, the function $F(X)$ is determined. Putting it into the above general solution with $X=x-y$ leads to the particular solution which satisfies the specified condition :\n\n$u(x,y)=\\frac14 e^{x+2y}+e^{-y}\\left(-\\frac14 e^{x-y}\\right)$ $$u(x,y)=\\frac14 e^{x}\\left(e^{2y}-e^{-2y}\\right)$$ $$u(x,y)=\\frac12 e^{x}\\sinh(2y)$$\n\nYou lost/excluded all solutions that take non-positive values by setting $$v=\\ln u$$. You can repair this partially by replacing that with $$v=\\ln|u|$$. In the end, setting $$H(z)=e^{h(z)}$$ and then allowing negative and zero values for $$H$$ gives the general solution $$u(x,y)=H(x-y)e^{-x}+\\frac{e^{x+2y}}4$$", "date": "2020-01-25 18:05:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2631984/solve-u-xu-yu-ex2y-with-the-condition-ux-0-0?noredirect=1", "openwebmath_score": 0.8770706653594971, "openwebmath_perplexity": 73.88665876932151, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759638081522, "lm_q2_score": 0.8670357563664174, "lm_q1q2_score": 0.8504545331820399}}
{"url": "https://cs.stackexchange.com/questions/11046/assign-unique-integer-keys-to-sets", "text": "# Assign unique integer keys to sets\n\nI am given a list of $n>1$ arrays, where each array has fairly small number of elements (rarely above $5$). Also $n$ is quite small in practice (around $6$).\n\nMy problem is that I would like to efficiently map each element product to unique integer. Also, I would like to efficiently compute the integer, when I change the order of the arrays in the list.\n\nI was thinking something in the line of integer factorization.\n\nTake first $n$ primes ($p_1, p_2, ..., p_n$) and remeber the index of the array in the original list with $ind: A \\rightarrow \\mathbb{N}$. So the initial function would be $ind(A_i)=i$.\n\nAlso we denote with $a_{i,j}$ the $j$-th element of the $i$-th array in the list. So the function $F:A_1 \\times A_2 \\times \\ldots \\times A_n \\rightarrow \\mathbb{N}$ would be:\n\n$$F(a_{1,i_1}, a_{2,i_2}, \\ldots, a_{n,i_n})=\\prod_{j=1}^{j=n}p_{ind(A_i)}^{i_j}$$\n\nIn more common words, we initially assign a prime number $p_i$ to array $i$ and raise $p_i$ to the index of the element in array $i$.\n\nThis guarantees us the uniqueness and also if we change the order of the arrays in the list, we just need to change the $ind$ function. But the problem here is that these products are fairly large and will overflow soon enough (even with using doubles).\n\nDo you have any better idea how to enumerate these products without the large number constraint? I am implementing this in Java and I really would not like to use BigInteger implementation.\n\n\u2022 Is it hashing you are after, or what is the purpose here? \u2013\u00a0Raphael Apr 5 '13 at 10:10\n\u2022 Yes. Something like that. I actually need unique numbers, because these output numbers represent some kind of output (the number is actually mapped to a string value). So for example I need to get different integers for (1,2,3) and for (3,2,1). \u2013\u00a0Nejc Apr 5 '13 at 11:57\n\u2022 What is the \"element product\", is it the product of all elements in one array? Why would this change when the order of the arrays is changed? \u2013\u00a0Juho Apr 5 '13 at 15:30\n\u2022 Based on your description, I take it that an element product is taking one element from each array and multiplying (you should clarify that). \u2013\u00a0Aryabhata Apr 5 '13 at 17:26\n\u2022 Sorry for not being clear! One element product of $n$ arrays is a $n$-tuple, where we take one element from each array and on the $i$-th place of the $n$-tuple is an element from $i$-th array. Clearly there are $|A_1| \\cdot |A_2| \\cdot \\ldots \\cdot |A_n|$ such products. \u2013\u00a0Nejc Apr 5 '13 at 17:45\n\nBased on your description, I take it that an element product is taking one element from each array and multiplying?\n\nYou don't really need to multiply or use prime numbers which might overflow (but not too much for your constrains I suppose, and is a cute solution :-)).\n\nSuppose the arrays were of fixed size $k$ (You can pad the arrays to achieve this, and based on your constraints, won't be too much of a problem).\n\nThen you want to map all possible tuples $(x_1, x_2, \\dots x_n)$ where $0 \\le x_i \\lt k$ to unique integers.\n\nThe mapping you need: Interpret the tuple as an integer in base-$k$.\n\nIf arrays are shuffled you can still change the $ind$ function and remap the digits according to that.\n\n(or did you try that and didn't work for some reason?)\n\nYou want an injection between your element space and the naturals. Why not take the oldest one that happens to be a bijection, the Cantor pairing function.\n\nIts basic form is defined to map $\\mathbb{N^2}$ to $\\mathbb{N}$:\n\n$\\qquad\\displaystyle \\pi(k_1,k_2) = \\frac{1}{2}(k_1 + k_2)(k_1 + k_2 + 1)+k_2$.\n\nIt readily extends to $\\mathbb{N}^k$:\n\n$\\qquad \\pi^{(n)}(k_1, \\ldots, k_{n-1}, k_n) = \\pi ( \\pi^{(n-1)}(k_1, \\ldots, k_{n-1}) , k_n)$.\n\nThe Wikipedia article explains how to invert the function.\n\nNow, in your case, $n$ is unknown. That's not a problem; we can extend above $\\pi$ easily to enumerate $\\mathbb{N}^+ = \\bigcup_{n \\geq 1} \\mathbb{N}^n$ by encoding $n$ -- here the array's length -- as well:\n\n$\\qquad \\pi^+(a_1,\\dots,a_n) = \\pi^{(n+1)}(a_1, \\dots, a_n, n)$.\n\nSince it's a bijection into $\\mathbb{N}$, values are arguably as small as possible. Also, computing either direction is efficient.\n\nWhen implementing it, be very carful either way. If you consider arrays over some finite set $A$, there are\n\n$\\qquad\\displaystyle \\sum_{i=1}^n |A|^i = \\frac{|A|^{n+1} - |A|}{|A| - 1}$\n\nmany arrays of size at most $n$. $\\pi^+$ has to encode them all, so the integers might overflow quickly if $|A|$ and $n$ are not both small.\n\n\u2022 Thank you for your nice suggestion. Luckily I do not need the inverse function, so I will implement the Cantor pairing function and compare it to the suggestion of @Aryabhata. \u2013\u00a0Nejc Apr 8 '13 at 9:36", "date": "2019-12-12 12:04:50", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/11046/assign-unique-integer-keys-to-sets", "openwebmath_score": 0.7479536533355713, "openwebmath_perplexity": 435.06793548575814, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759654852756, "lm_q2_score": 0.8670357477770337, "lm_q1q2_score": 0.8504545262110458}}
{"url": "https://cs.stackexchange.com/questions/58000/maximum-cost-path-in-integer-matrix", "text": "# Maximum cost path in integer matrix\n\nIn preparation for my design and algorithms exam, I encountered the following problem.\n\nGiven a $2 \\times N$ integer matrix $(a[i][j] \\in [-1000, 1000])$ and an unsigned integer $k$, find the maximum cost path from the top left corner $(a[1][1])$ and the bottom right corner $a[2][N]$, given the following:\n\n$\\bullet$ The path may not go through the same element more than once\n\n$\\bullet$ You can move from one cell to the other vertically or horizontally\n\n$\\bullet$ The path may not contain more than $k$ consecutive elements from the same line\n\n$\\bullet$ The cost of the path is determined by the sum of all the values stored within the cells it passes through\n\nI've been thinking of a simple greedy approach that seems to work for the test cases I've tried, but I'm not sure if it's always optimal. Namely, while traversing the matrix, I select a the maximum cost cell on the vertical or horizontal and then go from there. I've got a counter I only increment if the current cell is on the same line with the previously examined one and reset it otherwise. If at some point the selected element happens to be one that makes the counter go over the given value of $k$, I simply go with the other option that's left.\n\nHowever, I feel that I'm missing out on something terribly important here, but I just can't see what. Is there some known algorithm or approach that may be used here?\n\nAlso, the problem asks for an optimal solution (regarding temporal complexity).\n\nThis problem was basically made for dynamic programming (DP). Just review it and follow a couple of examples and solving this problem is straight-forward.\n\nYour simple greedy approach does not always work. Consider:\n\n[ begin ] [  2 ]\n[ 3 ]     [ 10 ]\n[ 1 ]     [ -1000 ]\n[ 1 ]     [ end ]\n\n\nThe optimal solution would be to go right, down, left, down, down, right.\n\n\u2022 I meant $2$ lines and $N$ columns, but I guess it doesn't make a difference in this case. \u2013\u00a0user43389 May 29 '16 at 23:28\n\nWe shall approach this using dynamic programming -- the greedy algorithm is 'myopic' in that it only considers immediate neighbors and not the path that follows those neighbors.\n\nLet $C(i, \\ j, \\ m)$ be the cost of the max-cost path starting from $a[i][j]$ going towards $a[2][n]$, where $i$ is either $1$ or $2$, $\\ j \\in [1, n]$ and $m \\in [0, k]$.\n\n\u2022 $m$ here denotes that we have $m$ remaining consecutive steps that could be taken along row $i$.\n\u2022 The additional constraint we enforce here is that when we are calculating $a[1][i]$, we have not yet visited $a[2][i]$, and similarly when we're calculating $a[2][i]$, we have not yet visited $a[1][i]$.\n\nWith this constraint in place, what recurrence can we come up with? Well,\n\n1. $C(1, \\ j, \\ m) = a[1][j] + \\max \\{ M(1, j+1, m-1), \\ a[2][j] + M(2, j+1, k)\\} \\$ and similarly,\n2. $C(2, \\ j, \\ m) = a[2][j] + \\max \\{ M(2, j+1, m-1), \\ a[1][j] + M(1, j+1, k)\\} \\$\n\nNow, what's the rationale here? If we started at $a[1][j]$, we have two choices:\n\n\u2022 continue along row $1$ : we then go to $a[1][j+1]$ with $m-1$ remaining consecutive steps that we can take along row $1$. (note here that $a[2][j+1]$ is not yet visited, maintaining the invariant).\n\n\u2022 switch to row $2$ : we go to $a[2][j]$, incurring a cost of $a[2][j]$, and then have only one choice from there - to go to $a[1][j+1]$. Because we have switched rows, we can now refresh our count and take $k$ consecutive steps along row $2$. (note again that we haven't visited $a[1][j+1]$ yet).\n\nNow if we calculate our subproblems starting from $j = n$ towards $j=1$ for every value of $m$, you shall have your answer in $C(1, \\ 1, \\ k)$.\n\nI shall leave the base cases and runtime analysis to you. Note that my solution calculates the cost of the path and not the path itself; can you extend my solution to calculate the path too?", "date": "2020-12-05 00:31:18", "meta": {"domain": "stackexchange.com", "url": "https://cs.stackexchange.com/questions/58000/maximum-cost-path-in-integer-matrix", "openwebmath_score": 0.8102948069572449, "openwebmath_perplexity": 276.674525783663, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759601743846, "lm_q2_score": 0.8670357512127872, "lm_q1q2_score": 0.8504545249763615}}
{"url": "https://math.stackexchange.com/questions/2618387/how-to-prove-rigorously-that-the-series-1-3n-diverges", "text": "# How to prove rigorously that the series $1/(3n)$ diverges\n\nHow can I prove rigorously that the series\n\n$$\\sum_{n=1}^\\infty \\frac{1}{3n}$$ Diverges, assuming that I know that the harmonic series when $p = 1$ diverges,\n\nI thought of using the property\n\n$$\\sum_{n=1}^\\infty ca_n = c\\sum_{n=1}^\\infty a_n$$\n\nHowever I think this only works when both of the series converges?\n\nSo to summarize, how can I prove that the series diverges knowing that the p-series $$\\sum_{n=1}^\\infty \\frac{1}{n}$$ diverges?\n\nProof by contradiction. Let us assume that the series $$\\sum_{n=1}^\\infty \\frac{1}{3n}$$ converges to, say, $A$. Then it would be also true that $$A = \\frac{1}{3} \\sum_{n=1}^\\infty \\frac{1}{n}$$ which implies that $$\\sum_{n=1}^\\infty \\frac{1}{n}$$ converges to $3A$. Knowing the fact that this series diverges (we found a contradiction) completes the proof by contradiction.\n\n\u2022 Thank you I didn't think about it just began discrete maths so I'm not used to using other forms of proof than the direct one so thank you for the tip!! Will try to consider more proof techniques in the future! \u2013\u00a0Ian Leclaire Jan 27 '18 at 23:35\n\nYou do not need convergence, just consider partial sum, you can factorize the scalar because you manipulate a finite series:\n\n$$\\sum_\\limits{n=1}^N \\dfrac 1{3n}=\\dfrac 13\\underbrace{\\sum_\\limits{n=1}^N \\dfrac 1n}_{\\to+\\infty}\\to+\\infty$$\n\nThus you get that the partial sum does not have a finite limit so the series diverges.\n\nYes, it's true you use that property, but it's good you noticed that this equality is only guaranteed if the series converges. Actually, let's write the statement more precisely:\n\nIf $\\sum_{n=1}^\\infty a_n$ converges, then so does $\\sum_{n=1}^\\infty c\\cdot a_n$ for any constant $c$.\n\nNow think on the contrapositive of this statement:\n\nIf $\\sum_{n=1}^\\infty c\\cdot a_n$ does not converge for some constant $c$, then $\\sum_{n=1}^\\infty a_n$ does not converge.\n\nDo you see how you could use this to prove our statement?\n\nThere is a hint below.\n\nHint: Take $a_n = 1/(3n)$\n\nYeah, you can use a variation of your property. For any constant $c \\neq 0$, we have $\\sum_{i=1}^{\\infty} ca_{n}$ converges if and only if $\\sum_{i=1}^{\\infty} a_{n}$ converges and likewise with divergence.\n\nSpecifically for this problem, assume your series converges. Then we have that, by definition, there exists some $L$ such that for any $\\epsilon > 0$, there exists an $N \\in \\mathbb{N}$ such that for $n \\geq N$, we have $$\\left| \\sum_{i=1}^{n} \\frac{1}{3n} - L \\right| < \\frac{1}{3}{\\epsilon}.$$ But this is if and only if $$\\left| \\sum_{i=1}^{n} \\frac{1}{n} - 3L \\right| < \\epsilon$$ and since $\\epsilon$ was made arbitrary, this implies that $\\sum_{i=1}^{\\infty} \\frac{1}{n}$ converges to $3L$, and you know that it doesn't. There's not much else to generalizing this concept for any series.\n\nYou can use the same proof that $\\sum \\frac 1n$ diverges. i.e. $1 + \\frac 12 + (\\frac 13 +\\frac 14) + (\\frac 15 + \\cdots \\frac 18) + (\\frac 19 + \\cdots + \\frac 1{16})+\\cdots < 1 + \\frac 12 +\\frac 12 + \\frac 12 +\\cdots$\n\nAnd a divergent series multiplied by a constant (other than 0), indeed produces divergent series.\n\nNote that by definition\n\n$S_n(k)=\\sum_{n=1}^k \\frac{1}{n}$ diverges $\\iff \\forall M>0 \\quad \\exists \\bar k$ such that $S_n(k)>M \\quad \\forall k>\\bar k$\n\nthen also\n\n$S_{3n}(k)=\\frac13S_n(k)=\\sum_{n=1}^k \\frac{1}{3n}$ diverges since\n\n$\\forall M>0 \\quad \\exists \\bar k$ such that $S_n(k)>3M \\quad \\forall k>\\bar k\\implies S_{3n}(k)>M$", "date": "2019-07-21 01:19:34", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2618387/how-to-prove-rigorously-that-the-series-1-3n-diverges", "openwebmath_score": 0.9843190312385559, "openwebmath_perplexity": 138.2025104789288, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9808759587767815, "lm_q2_score": 0.8670357494949105, "lm_q1q2_score": 0.8504545220795656}}
{"url": "https://mathematica.stackexchange.com/questions/140219/how-to-properly-print-a-table", "text": "# How to properly print a table\n\nto bootstrap my experience with Mathematica I'm trying to use it to print a table of values where the columns are different trigonometric functions and the rows contain different the values of said trig functions at some fixed angles that are between $0$ and $\\frac{\\pi}{2}$ .\n\nAfter studying the basics I came up with this lines of code\n\nvec = Prepend[Table[Pi/i, {i, Reverse[Range[2, 11]]}], 0];\nres = Map[#, vec] & /@ {Sin, Cos, Tan, Csc, Sec, Cot};\nTextGrid [res, Frame -> All]\n\n\nand the output is\n\n$\\begin{array}{ccccccccccc} 0 & \\sin \\left(\\frac{\\pi }{11}\\right) & \\frac{1}{4} \\left(\\sqrt{5}-1\\right) & \\sin \\left(\\frac{\\pi }{9}\\right) & \\sin \\left(\\frac{\\pi }{8}\\right) & \\sin \\left(\\frac{\\pi }{7}\\right) & \\frac{1}{2} & \\sqrt{\\frac{5}{8}-\\frac{\\sqrt{5}}{8}} & \\frac{1}{\\sqrt{2}} & \\frac{\\sqrt{3}}{2} & 1 \\\\ 1 & \\cos \\left(\\frac{\\pi }{11}\\right) & \\sqrt{\\frac{\\sqrt{5}}{8}+\\frac{5}{8}} & \\cos \\left(\\frac{\\pi }{9}\\right) & \\cos \\left(\\frac{\\pi }{8}\\right) & \\cos \\left(\\frac{\\pi }{7}\\right) & \\frac{\\sqrt{3}}{2} & \\frac{1}{4} \\left(\\sqrt{5}+1\\right) & \\frac{1}{\\sqrt{2}} & \\frac{1}{2} & 0 \\\\ 0 & \\tan \\left(\\frac{\\pi }{11}\\right) & \\sqrt{1-\\frac{2}{\\sqrt{5}}} & \\tan \\left(\\frac{\\pi }{9}\\right) & \\tan \\left(\\frac{\\pi }{8}\\right) & \\tan \\left(\\frac{\\pi }{7}\\right) & \\frac{1}{\\sqrt{3}} & \\sqrt{5-2 \\sqrt{5}} & 1 & \\sqrt{3} & \\text{ComplexInfinity} \\\\ \\text{ComplexInfinity} & \\csc \\left(\\frac{\\pi }{11}\\right) & \\sqrt{5}+1 & \\csc \\left(\\frac{\\pi }{9}\\right) & \\csc \\left(\\frac{\\pi }{8}\\right) & \\csc \\left(\\frac{\\pi }{7}\\right) & 2 & \\frac{1}{\\sqrt{\\frac{5}{8}-\\frac{\\sqrt{5}}{8}}} & \\sqrt{2} & \\frac{2}{\\sqrt{3}} & 1 \\\\ 1 & \\sec \\left(\\frac{\\pi }{11}\\right) & \\frac{1}{\\sqrt{\\frac{\\sqrt{5}}{8}+\\frac{5}{8}}} & \\sec \\left(\\frac{\\pi }{9}\\right) & \\sec \\left(\\frac{\\pi }{8}\\right) & \\sec \\left(\\frac{\\pi }{7}\\right) & \\frac{2}{\\sqrt{3}} & \\sqrt{5}-1 & \\sqrt{2} & 2 & \\text{ComplexInfinity} \\\\ \\text{ComplexInfinity} & \\cot \\left(\\frac{\\pi }{11}\\right) & \\sqrt{2 \\sqrt{5}+5} & \\cot \\left(\\frac{\\pi }{9}\\right) & \\cot \\left(\\frac{\\pi }{8}\\right) & \\cot \\left(\\frac{\\pi }{7}\\right) & \\sqrt{3} & \\sqrt{1+\\frac{2}{\\sqrt{5}}} & 1 & \\frac{1}{\\sqrt{3}} & 0 \\\\ \\end{array}$\n\nthe problems with this output are :\n\n\u2022 some cells are in what it looks like a rational / expected form and others have a \"weird\" unevaluated form like $sin(\\frac{\\pi}{11})$ ; what is the problem with the latter ?\n\u2022 I haven't found a way to print legends for the rows and the columns ( angle values and function names )\n\nThanks\n\n\u2022 Not sure what you mean by \"print legends\" would it suffice to simply prepend your column / row values to res? Also that unevaluated form is simply the exact form for the expression. If it knows the exact ratio or whatever it'll return that, but otherwise Mathematica has a principle of remaining as exact and symbolic as possible. Mar 16, 2017 at 18:19\n\u2022 @MB1965 basically any solution that could help me recognize rows and columns without making a mess with too much intricacies in the code . I would like to keep the computation separate from the the graphical layout . Regarding the 2nd part : can I force an approximation for the entire table ? Mar 16, 2017 at 18:23\n\u2022 Second part, use N. That's its purpose. First part, isn't bad either. We'll do a bit of prepending. I'll knock you up a quick solution and explanation. Mar 16, 2017 at 18:25\n\u2022 @MB1965 thank you Mar 16, 2017 at 18:28\n\nSo we'll just prepend your labels or whatever onto your grid. Note that you can change the dividers for Grid based constructs too, if you want a different appearance.\n\nHere are my changes:\n\n\u2022 Use Through to build your res in the transposed orientation\n\u2022 Use N on that to get the numerical approximations you wanted\n\u2022 Use Prepend to stick on the function labels, transpose the grid, then prepend on the number labels.\n\nLooks like this in the end:\n\nvec = Prepend[Table[Pi/i, {i, Reverse[Range[2, 11]]}], 0];\nops = {Sin, Cos, Tan, Csc, Sec, Cot};\nres = Through@*ops /@ vec // N;\ngrid = Prepend[Transpose@Prepend[res, ops], Prepend[vec, Null]];\nNumberForm[TextGrid[grid, Frame -> All], 3]\n\n\nA few things to note:\n\nI use @* (Composition) to make Through apply after the functions and NumberForm to make sure the numerical approximations only display up to 3 digits.\n\n\u2022 @MB1965 I get the same result for res using Through[ops[#]] & /@ vec // N. I am not sure I understand the construct that you use for res. Can you enlighten me as to the differences, pros, cons? Thank you. Mar 17, 2017 at 22:18\n\u2022 @JackLaVigne do you mean the Composition (@*)? What you did is another way to do it, but & is a very low precedence operator and so can often grab much more than what you intended. It just allows me to not have to use a pure function and is often clearer, in my mind. Mar 17, 2017 at 22:21\n\u2022 @MB1965 Yes, the Composition. I see that it produces the desired result but can't quite figure out what is going on. Maybe you can spell it out for me? Mar 17, 2017 at 22:23\n\u2022 @JackLaVigne I'd be happy to. f@*g in FullForm is Composition[f,g] and Composition[f,g][h] is f[g[h]]. Contrast that to f@g where we would get f[g][h]. It allows f to apply after g has applied instead of applying to g before hand. There is a corresponding head, RightComposition /* that is similar, but acts in reverse. Check out the docs for a more thorough treatment. Mar 17, 2017 at 22:26\n\u2022 @MB1965 Got it. Map[Composition[Through, {Sin, Cos, Tan, Csc, Sec, Cot}], {x, y, z}] produces {{Sin[x], Cos[x], Tan[x], Csc[x], Sec[x], Cot[x]}, {Sin[y], Cos[y], Tan[y], Csc[y], Sec[y], Cot[y]}, {Sin[z], Cos[z], Tan[z], Csc[z], Sec[z], Cot[z]}}. Thank you for the help. Mar 17, 2017 at 22:40", "date": "2022-05-23 10:40:59", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/140219/how-to-properly-print-a-table", "openwebmath_score": 0.26564908027648926, "openwebmath_perplexity": 2079.019596230131, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9585377272885904, "lm_q2_score": 0.88720460564669, "lm_q1q2_score": 0.8504190863365484}}
{"url": "https://stackoverflow.com/questions/27948363/numpy-broadcast-to-perform-euclidean-distance-vectorized/37903795", "text": "# Numpy Broadcast to perform euclidean distance vectorized\n\nI have matrices that are 2 x 4 and 3 x 4. I want to find the euclidean distance across rows, and get a 2 x 3 matrix at the end. Here is the code with one for loop that computes the euclidean distance for every row vector in a against all b row vectors. How do I do the same without using for loops?\n\n`````` import numpy as np\na = np.array([[1,1,1,1],[2,2,2,2]])\nb = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])\ndists = np.zeros((2, 3))\nfor i in range(2):\ndists[i] = np.sqrt(np.sum(np.square(a[i] - b), axis=1))\n``````\n\nSimply use `np.newaxis` at the right place:\n\n`````` np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))\n``````\n\u2022 Could you please explain how `Simply using np.newaxis at the right place` works? If you could start with the fact that `a` is 2x4 and `b` is 3x4, that would be great. \u2013\u00a0stackoverflowuser2010 Jun 18 '16 at 23:45\n\nHere are the original input variables:\n\n``````A = np.array([[1,1,1,1],[2,2,2,2]])\nB = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])\nA\n# array([[1, 1, 1, 1],\n#        [2, 2, 2, 2]])\nB\n# array([[1, 2, 3, 4],\n#        [1, 1, 1, 1],\n#        [1, 2, 1, 9]])\n``````\n\nA is a 2x4 array. B is a 3x4 array.\n\nWe want to compute the Euclidean distance matrix operation in one entirely vectorized operation, where `dist[i,j]` contains the distance between the ith instance in A and jth instance in B. So `dist` is 2x3 in this example.\n\nThe distance\n\ncould ostensibly be written with numpy as\n\n``````dist = np.sqrt(np.sum(np.square(A-B))) # DOES NOT WORK\n# Traceback (most recent call last):\n#   File \"<stdin>\", line 1, in <module>\n# ValueError: operands could not be broadcast together with shapes (2,4) (3,4)\n``````\n\nHowever, as shown above, the problem is that the element-wise subtraction operation `A-B` involves incompatible array sizes, specifically the 2 and 3 in the first dimension.\n\n``````A has dimensions 2 x 4\nB has dimensions 3 x 4\n``````\n\nIn order to do element-wise subtraction, we have to pad either A or B to satisfy numpy's broadcast rules. I'll choose to pad A with an extra dimension so that it becomes 2 x 1 x 4, which allows the arrays' dimensions to line up for broadcasting. For more on numpy broadcasting, see the tutorial in the scipy manual and the final example in this tutorial.\n\nYou can perform the padding with either `np.newaxis` value or with the `np.reshape` command. I show both below:\n\n``````# First approach is to add the extra dimension to A with np.newaxis\nA[:,np.newaxis,:] has dimensions 2 x 1 x 4\nB has dimensions                     3 x 4\n\n# Second approach is to reshape A with np.reshape\nnp.reshape(A, (2,1,4)) has dimensions 2 x 1 x 4\nB has dimensions                          3 x 4\n``````\n\nAs you can see, using either approach will allow the dimensions to line up. I'll use the first approach with `np.newaxis`. So now, this will work to create A-B, which is a 2x3x4 array:\n\n``````diff = A[:,np.newaxis,:] - B\n# Alternative approach:\n# diff = np.reshape(A, (2,1,4)) - B\ndiff.shape\n# (2, 3, 4)\n``````\n\nNow we can put that difference expression into the `dist` equation statement to get the final result:\n\n``````dist = np.sqrt(np.sum(np.square(A[:,np.newaxis,:] - B), axis=2))\ndist\n# array([[ 3.74165739,  0.        ,  8.06225775],\n#        [ 2.44948974,  2.        ,  7.14142843]])\n``````\n\nNote that the `sum` is over `axis=2`, which means take the sum over the 2x3x4 array's third axis (where the axis id starts with 0).\n\nIf your arrays are small, then the above command will work just fine. However, if you have large arrays, then you may run into memory issues. Note that in the above example, numpy internally created a 2x3x4 array to perform the broadcasting. If we generalize A to have dimensions `a x z` and B to have dimensions `b x z`, then numpy will internally create an `a x b x z` array for broadcasting.\n\nWe can avoid creating this intermediate array by doing some mathematical manipulation. Because you are computing the Euclidean distance as a sum-of-squared-differences, we can take advantage of the mathematical fact that sum-of-squared-differences can be rewritten.\n\nNote that the middle term involves the sum over element-wise multiplication. This sum over multiplcations is better known as a dot product. Because A and B are each a matrix, then this operation is actually a matrix multiplication. We can thus rewrite the above as:\n\nWe can then write the following numpy code:\n\n``````threeSums = np.sum(np.square(A)[:,np.newaxis,:], axis=2) - 2 * A.dot(B.T) + np.sum(np.square(B), axis=1)\ndist = np.sqrt(threeSums)\ndist\n# array([[ 3.74165739,  0.        ,  8.06225775],\n#        [ 2.44948974,  2.        ,  7.14142843]])\n``````\n\nNote that the answer above is exactly the same as the previous implementation. Again, the advantage here is the we do not need to create the intermediate 2x3x4 array for broadcasting.\n\nFor completeness, let's double-check that the dimensions of each summand in `threeSums` allowed broadcasting.\n\n``````np.sum(np.square(A)[:,np.newaxis,:], axis=2) has dimensions 2 x 1\n2 * A.dot(B.T) has dimensions                               2 x 3\nnp.sum(np.square(B), axis=1) has dimensions                 1 x 3\n``````\n\nSo, as expected, the final `dist` array has dimensions 2x3.\n\nThis use of the dot product in lieu of sum of element-wise multiplication is also discussed in this tutorial.\n\n\u2022 This answer is so useful, especially the part to overcome broadcasting issues. Thank you @stackoverflowuser2010 \u2013\u00a0Allen May 21 '18 at 6:22\n\nI had the same problem recently working with deep learning(stanford cs231n,Assignment1),but when I used\n\n`````` np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))\n``````\n\nThere was a error\n\n``````MemoryError\n``````\n\nThat means I ran out of memory(In fact,that produced a array of 500*5000*1024 in the middle.It's so huge!)\n\nTo prevent that error,we can use a formula to simplify:\n\n$(a-b)^2&space;=&space;a^2&space;-&space;2ab+b^2$\n\ncode:\n\n``````import numpy as np\naSumSquare = np.sum(np.square(a),axis=1);\nbSumSquare = np.sum(np.square(b),axis=1);\nmul = np.dot(a,b.T);\ndists = np.sqrt(aSumSquare[:,np.newaxis]+bSumSquare-2*mul)\n``````\n\u2022 to add something;quoted from the official document `There are, however, cases where broadcasting is a bad idea because it leads to inefficient use of memory that slows computation.There are, however, cases where broadcasting is a bad idea because it leads to inefficient use of memory that slows computation.` \u2013\u00a0Han Qiu Mar 24 '16 at 6:40\n\nThis functionality is already included in scipy's spatial module and I recommend using it as it will be vectorized and highly optimized under the hood. But, as evident by the other answer, there are ways you can do this yourself.\n\n``````import numpy as np\na = np.array([[1,1,1,1],[2,2,2,2]])\nb = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])\nnp.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))\n# array([[ 3.74165739,  0.        ,  8.06225775],\n#       [ 2.44948974,  2.        ,  7.14142843]])\nfrom scipy.spatial.distance import cdist\ncdist(a,b)\n# array([[ 3.74165739,  0.        ,  8.06225775],\n#       [ 2.44948974,  2.        ,  7.14142843]])\n``````\n\nUsing numpy.linalg.norm also works well with broadcasting. Specifying an integer value for `axis` will use a vector norm, which defaults to Euclidean norm.\n\n``````import numpy as np\n\na = np.array([[1,1,1,1],[2,2,2,2]])\nb = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])\nnp.linalg.norm(a[:, np.newaxis] - b, axis = 2)\n\n# array([[ 3.74165739,  0.        ,  8.06225775],\n#       [ 2.44948974,  2.        ,  7.14142843]])\n``````", "date": "2019-10-23 03:50:27", "meta": {"domain": "stackoverflow.com", "url": "https://stackoverflow.com/questions/27948363/numpy-broadcast-to-perform-euclidean-distance-vectorized/37903795", "openwebmath_score": 0.6032344698905945, "openwebmath_perplexity": 1550.359627145525, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211590308923, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8503826869492399}}
{"url": "https://www.physicsforums.com/threads/what-does-it-mean-to-multiply-by-x-x.835118/", "text": "# What does it mean to multiply by x / x?\n\n1. Sep 29, 2015\n\n### Mr Davis 97\n\nThis is a simple question. I know that for any number or expression we can multiply it by 1 and maintain the same expression. However, x / x also equals 1. But if we multiply it by another expression, the expression changes such that we can no longer use 0. So if the domain of an expression originally contains 0 and we multiply it by x / x, is that illegal?\n\n2. Sep 29, 2015\n\n### Staff: Mentor\n\nNo, you get a different expression whose value is the same.\nx(x - 1) = 0 -------------- Solution set {0, 1}\nIf we multiply the left side by x/x, we get\n$\\frac{x^2(x - 1)}{x} = 0$ -------------- The numerator is zero if x = 0 or x = 1, but we can't divide by 0, so the solution set is now {1}.\n\nBy multiplying by x/x, we are tacitly assuming that $x \\ne 0$, so we lose a value in the solution set.\n\nIf we graph y = x(x - 1) and $y = \\frac{x^2(x - 1)}{x}$, the two graphs look exactly the same, except that the graph of the latter equation has a \"hole\" at (0, 0).\n\n3. Sep 29, 2015\n\n### Mr Davis 97\n\nSo if we divide the x's out again to get the original expression, that tacitly adds 0 back to our solution set? The only thing that confuses me is that the addition or subtraction of the domain element seems to have to be inferred by us rather than explicitly stated by the mathematics.\n\n4. Sep 29, 2015\n\n### Staff: Mentor\n\nSometimes people will expicitly write any restrictions to the domain. For example,\n$y = \\frac{x^2(x - 1)}{x}, x \\ne 0$\n$\\Leftrightarrow y = x(x - 1), x \\ne 0$\nThe two equations above are equivalent, meaning that their solution sets (and graphs) are identical.\n\nFor examples as simple as this one, some writers will omit the explicit domain restrictions, believing them to be pretty much obvious.\n\n5. Sep 29, 2015\n\n### Mr Davis 97\n\nOkay, that makes sense. But what about when we are solving limits at infinity? For example, given, $\\displaystyle \\lim_{x\\rightarrow \\infty } \\frac{2x - 1}{x + 1}$, the only way to solve is to multiply the fraction by $\\displaystyle \\frac{\\frac{1}{x}}{\\frac{1}{x}}$. However, this changes the domain of the original function such that x = 0 is no longer defined. So why is this a rigorous way to solve limits at infinity if we're changing the domain of the function?\n\n6. Sep 29, 2015\n\n### Staff: Mentor\n\nNo, that's not the only way to evaluate3 this limit. You can also factor both numerator and denominator like so:\n$\\lim_{x\\to \\infty } \\frac{2x - 1}{x + 1} = \\lim_{x\\to \\infty } \\frac{x(2 - 1/x)}{x(1 + 1/x)} = \\lim_{x\\to \\infty } \\frac x x \\cdot \\lim_{x\\to \\infty }\\frac{x(2 - 1/x)}{x(1 + 1/x)}$.\nThe first limit is 1 and the second limit it 2. In the first limit, x/x is always 1 for any finite value of x (other than 0), so as x increases without bound, the value of the fraction is always 1.\nCorrected the above to \"such that at x= 0 the original function is no longer defined.\"\nIt doesn't matter. You're taking the limit as $x \\to \\infty$, so the fact that the rational expression isn't defined at x = 0 is immaterial.\n\n7. Sep 29, 2015\n\n### Mr Davis 97\n\nBut $\\displaystyle \\frac{2x - 1}{x + 1} \\neq \\frac{2 - \\frac{1}{x}}{1 + \\frac{1}{x}}$. So how can I be sure that $\\displaystyle \\lim_{x\\rightarrow \\infty }\\frac{2x - 1}{x + 1} = \\lim_{x\\rightarrow \\infty }\\frac{2 - \\frac{1}{x}}{1 + \\frac{1}{x}}$?\n\n8. Sep 29, 2015\n\n### Staff: Mentor\n\nOf course they are equal, except at x = 0 and x = -1. Again, the limit is as $x \\to \\infty$, so for any x > 0, the two expressions are identical.\nI don't understand why you're so hung up on this.\n\n9. Sep 29, 2015\n\n### aikismos\n\nIt's easier to do polynomial division. $\\frac{2x -1}{ x + 1} \\rightarrow 2 + \\frac{-3}{x + 1}$. Taking $\\frac{\\lim}{x\\rightarrow \\infty } \\rightarrow 2$. What your worried about is the equivalence of fractions, but if you can show the steps in light of the Properties of Equations (e.g., Division Property of Equations) then you have nothing to worry about.\n\n10. Sep 30, 2015\n\n### HallsofIvy\n\nI want to expand on what Mark44 said. You do NOT get \"the same expression\", you get another expression with the same numeric value if x is not 0.\n\nYes, because of your misquoting of that \"rule\"- you should have included \"if x n is not 0\" right at the start.\n\n11. Sep 30, 2015\n\n### DrStupid\n\nAs you can't divide by zero it is not that simple. But you can remove the singularity using L'H\u00f4pital's rule.", "date": "2018-07-17 02:24:48", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/what-does-it-mean-to-multiply-by-x-x.835118/", "openwebmath_score": 0.7370468378067017, "openwebmath_perplexity": 434.85423817954734, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211531789392, "lm_q2_score": 0.8723473846343394, "lm_q1q2_score": 0.8503826834618784}}
{"url": "https://ch.mathworks.com/help/symbolic/piecewise.html", "text": "# piecewise\n\nConditionally defined expression or function\n\n## Syntax\n\n``pw = piecewise(cond1,val1,cond2,val2,...)``\n``pw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal)``\n\n## Description\n\nexample\n\n````pw = piecewise(cond1,val1,cond2,val2,...)` returns the piecewise expression or function `pw` whose value is `val1` when condition `cond1` is true, is `val2` when `cond2` is true, and so on. If no condition is true, the value of `pw` is `NaN`.```\n\nexample\n\n````pw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal)` returns the piecewise expression or function `pw` that has the value `otherwiseVal` if no condition is true.```\n\n## Examples\n\n### Define and Evaluate Piecewise Expression\n\nDefine the following piecewise expression by using `piecewise`.\n\n`$y=\\left\\{\\begin{array}{cc}-1& x<0\\\\ 1& x>0\\end{array}$`\n```syms x y = piecewise(x<0, -1, x>0, 1)```\n```y = piecewise(x < 0, -1, 0 < x, 1)```\n\nEvaluate `y` at `-2`, `0`, and `2` by using `subs` to substitute for `x`. Because `y` is undefined at `x = 0`, the value is `NaN`.\n\n`subs(y, x, [-2 0 2])`\n```ans = [ -1, NaN, 1]```\n\n### Define Piecewise Function\n\nDefine the following function symbolically.\n\n`$y\\left(x\\right)=\\left\\{\\begin{array}{cc}-1& x<0\\\\ 1& x>0\\end{array}$`\n```syms y(x) y(x) = piecewise(x<0, -1, x>0, 1)```\n```y(x) = piecewise(x < 0, -1, 0 < x, 1)```\n\nBecause `y(x)` is a symbolic function, you can directly evaluate it for values of `x`. Evaluate `y(x)` at `-2`, `0`, and `2`. Because `y(x)` is undefined at `x = 0`, the value is `NaN`. For details, see Create Symbolic Functions.\n\n`y([-2 0 2])`\n```ans = [ -1, NaN, 1]```\n\n### Set Value When No Conditions Is True\n\nSet the value of a piecewise function when no condition is true (called otherwise value) by specifying an additional input argument. If an additional argument is not specified, the default otherwise value of the function is `NaN`.\n\nDefine the piecewise function\n\n`$y\\left(x\\right)=\\left\\{\\begin{array}{cc}-2& x<-2\\\\ 0& -2`\n```syms y(x) y(x) = piecewise(x<-2, -2, -2<x<0, 0, 1)```\n```y(x) = piecewise(x < -2, -2, x in Dom::Interval(-2, 0), 0, 1)```\n\nEvaluate `y(x)` between `-3` and `1` by generating values of `x` using `linspace`. At `-2` and `0`, `y(x)` evaluates to `1` because the other conditions are not true.\n\n```xvalues = linspace(-3,1,5) yvalues = y(xvalues)```\n```xvalues = -3 -2 -1 0 1 yvalues = [ -2, 1, 0, 1, 1]```\n\n### Plot Piecewise Expression\n\nPlot the following piecewise expression by using `fplot`.\n\n`$y=\\left\\{\\begin{array}{cc}-2& x<-2\\\\ x& -22\\end{array}.$`\n\n```syms x y = piecewise(x<-2, -2, -2<x<2, x, x>2, 2); fplot(y)```\n\n### Assumptions and Piecewise Expressions\n\nOn creation, a piecewise expression applies existing assumptions. Apply assumptions set after creating the piecewise expression by using `simplify` on the expression.\n\nAssume `x > 0`. Then define a piecewise expression with the same condition `x > 0`. `piecewise` automatically applies the assumption to simplify the condition.\n\n```syms x assume(x > 0) pw = piecewise(x<0, -1, x>0, 1)```\n```pw = 1```\n\nClear the assumption on `x` for further computations.\n\n`assume(x,'clear')`\n\nCreate a piecewise expression `pw` with the condition `x > 0`. Then set the assumption that ```x > 0```. Apply the assumption to `pw` by using `simplify`.\n\n```pw = piecewise(x<0, -1, x>0, 1); assume(x > 0) pw = simplify(pw)```\n```pw = 1```\n\nClear the assumption on `x` for further computations.\n\n`assume(x, 'clear')`\n\n### Differentiate, Integrate, and Find Limits of Piecewise Expression\n\nDifferentiate, integrate, and find limits of a piecewise expression by using `diff`, `int`, and `limit` respectively.\n\nDifferentiate the following piecewise expression by using `diff`.\n\n`$y=\\left\\{\\begin{array}{cc}1/x& x<-1\\\\ \\mathrm{sin}\\left(x\\right)/x& x\\ge -1\\end{array}$`\n```syms x y = piecewise(x<-1, 1/x, x>=-1, sin(x)/x); diffy = diff(y, x)```\n```diffy = piecewise(x < -1, -1/x^2, -1 < x, cos(x)/x - sin(x)/x^2)```\n\nIntegrate `y` by using `int`.\n\n`inty = int(y, x)`\n```inty = piecewise(x < -1, log(x), -1 <= x, sinint(x))```\n\nFind the limits of `y` at `0` and `-1` by using `limit`. Because `limit` finds the double-sided limit, the piecewise expression must be defined from both sides. Alternatively, you can find the right- or left-sided limit. For details, see `limit`.\n\n```limit(y, x, 0) limit(y, x, -1)```\n```ans = 1 ans = limit(piecewise(x < -1, 1/x, -1 < x, sin(x)/x), x, -1)```\n\nBecause the two conditions meet at `-1`, the limits from both sides differ and `limit` cannot find a double-sided limit.\n\n### Elementary Operations on Piecewise Expressions\n\nAdd, subtract, divide, and multiply two piecewise expressions. The resulting piecewise expression is only defined where the initial piecewise expressions are defined.\n\n```syms x pw1 = piecewise(x<-1, -1, x>=-1, 1); pw2 = piecewise(x<0, -2, x>=0, 2); add = pw1 + pw2 sub = pw1 - pw2 mul = pw1 * pw2 div = pw1 / pw2```\n```add = piecewise(x < -1, -3, x in Dom::Interval([-1], 0), -1, 0 <= x, 3) sub = piecewise(x < -1, 1, x in Dom::Interval([-1], 0), 3, 0 <= x, -1) mul = piecewise(x < -1, 2, x in Dom::Interval([-1], 0), -2, 0 <= x, 2) div = piecewise(x < -1, 1/2, x in Dom::Interval([-1], 0), -1/2, 0 <= x, 1/2)```\n\n### Modify or Extend Piecewise Expression\n\nModify a piecewise expression by replacing part of the expression using `subs`. Extend a piecewise expression by specifying the expression as the otherwise value of a new piecewise expression. This action combines the two piecewise expressions. `piecewise` does not check for overlapping or conflicting conditions. Instead, like an if-else ladder, `piecewise` returns the value for the first true condition.\n\nChange the condition `x<2` in a piecewise expression to `x<0` by using `subs`.\n\n```syms x pw = piecewise(x<2, -1, x>0, 1); pw = subs(pw, x<2, x<0)```\n```pw = piecewise(x < 0, -1, 0 < x, 1)```\n\nAdd the condition `x>5` with the value `1/x` to `pw` by creating a new piecewise expression with `pw` as the otherwise value.\n\n`pw = piecewise(x>5, 1/x, pw)`\n```pw = piecewise(5 < x, 1/x, x < 0, -1, 0 < x, 1)```\n\n## Input Arguments\n\ncollapse all\n\nCondition, specified as a symbolic condition or variable. A symbolic variable represents an unknown condition.\n\nExample: x > 2\n\nValue when condition is satisfied, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.\n\nValue if no conditions are true, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. If `otherwiseVal` is not specified, its value is `NaN`.\n\n## Output Arguments\n\ncollapse all\n\nPiecewise expression or function, returned as a symbolic expression or function. The value of `pw` is the value `val` of the first condition `cond` that is true. To find the value of `pw`, use `subs` to substitute for variables in `pw`.\n\n## Tips\n\n\u2022 `piecewise` does not check for overlapping or conflicting conditions. A piecewise expression returns the value of the first true condition and disregards any following true expressions. Thus, `piecewise` mimics an if-else ladder.", "date": "2021-02-24 21:39:07", "meta": {"domain": "mathworks.com", "url": "https://ch.mathworks.com/help/symbolic/piecewise.html", "openwebmath_score": 0.917317271232605, "openwebmath_perplexity": 1658.5248570284693, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211575679041, "lm_q2_score": 0.8723473796562744, "lm_q1q2_score": 0.8503826824378573}}
{"url": "https://math.stackexchange.com/questions/2986960/considering-sets-in-proofs", "text": "# Considering sets in proofs\n\nIn order to show that $$\\forall x\\gt0, \\ \\exists n_0\\in \\mathbb N: \\ n_0(n_0+1)\\le x\\lt (n_0+1)(n_0+2)$$, the proof in my textbook considers the set $$A=\\{n\\in \\mathbb N :n(n+1)\\le x\\}$$ and then goes to show that $$A$$ has a maximal element $$n_0$$, which completes the proof.\n\nNow, I want to know what has led to this kind of argument, because I have begun to see it very frequently and even some of my classmates often use this method. But to me it doesn't seem to be very useful and I certainly wouldn't be able to tell if it's a good way to approach a certain proof.\n\nI understand that such motivations usually can't be explained easily; in that case, I'm asking for other problems you know of that use a similar approach.\n\n\u2022 The basic idea is that you're \"sandwiching\" $x$ between two integers of the form $n_0(n_0 + 1) \\leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $\\left \\lfloor x \\right \\rfloor$. We also note that by construction of $\\mathbb{N}$ that $n_0 \\in \\mathbb{N}$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along. \u2013\u00a0Eevee Trainer Nov 6 '18 at 10:02\n\u2022 It would be much easier to show that $B=\\{\\,n\\in\\Bbb N: n(n+1)>x\\,\\}$ has a minimal element ... \u2013\u00a0Hagen von Eitzen Nov 6 '18 at 15:30\n\u2022 @HagenvonEitzen: It would be much easier to just use induction... =) \u2013\u00a0user21820 Nov 6 '18 at 16:25\n\n## 4 Answers\n\nEveryone has their own way of proving things, and that's OK. The statement you made can be shown to be true in different ways, and what counts is that you prove it, not how you prove it. The way you learn how to prove it is that you prove many many other statements as well, and then you get used to it. Repetitio mater studiorum est.\n\nHowever, if you want a path that leads to the particular proof, in this case, my thoughts would proceed as follows:\n\n1. I look at the statement. OK, it's saying that I can squeeze any positive real $$x$$ between two numbers. OK, let's imagine the $$x$$ as some point on the positive real line.\n2. Hmm, the two numbers, $$n_0(n_0+1)$$ and $$(n_0+1)(n_0+2)$$ are both integers.\n3. Not only are they integers, they are two integers from a monotonically increasing sequence of integers, $$a_n = n(n+1)$$.\n4. So... this sequence of integers, it's really a series of points on the real line. Let's imagine them as crosses. (yes, really, I do that. Don't judge) The order I draw them in is left to right.\n5. So what I now have is the statement that there exist two crosses so that the previous cross is to the left of $$x$$, and the next one is to the right. Well... sure they do! I just need to find the last cross on the left of $$x$$, and the next one must be on the right of it (otherwise, the previous one wasn't the last one!).\n6. OK, what I just said in point 5 can be said formally as \"I need to find the maximum value of $$n$$ such that $$a_n < x$$. This can be rewritten formally as finding a maximum element from some set.\n\nOnce this train of thought concludes, I go down to really writing down the proof, and the proof \"starts\" with introducing the set. Sure, the proof does, but the thought process that lead me to the proof started long before.\n\n\u2022 This has been very helpful, thank you. \u2013\u00a0FuzzyPixelz Nov 6 '18 at 10:09\n\nFor instance, let $$a,b$$ be natural numbers with $$1\\leq b\\leq a$$. Then consider the set $$A=\\{a-qb\\mid a-qb\\geq 0, q\\geq 0\\}$$. Since the set of natural numbers is well-ordered, every nonempty subset $$A$$ of natural numbers has a least element, here $$r = a-qb$$ in $$A$$. It is then clear that $$a=qb+r$$ and $$0\\leq r is the remainder.\n\nTo answer your question\n\nNow, I want to know what has lead to this kind of argument\n\nLook at this part of the proof\n\nand then goes to show that $$A$$ has a maximal element $$n_0$$, which completes the proof.\n\nThe way the set is build makes it trivial to show that the set is finite thus has a maximal element. This triviality is a reason why a set is used. Other proofs requiring existence of a maximal/minimal element might follow the same approach.\n\nOf course the thinking process is going to be as outlined by 5xum.\n\nThis is not the only way to solve the problem, and arguably uses set theory when there is no need for it. You can prove the claim directly by an easy induction (which I am sure you can do). Also note that induction is necessary, whether or not hidden. It is logically non-trivial (and requires induction) to show that the set in your question has a maximum!", "date": "2019-07-19 13:12:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2986960/considering-sets-in-proofs", "openwebmath_score": 0.8290848135948181, "openwebmath_perplexity": 166.08104181783878, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211597623861, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.850382673029187}}
{"url": "https://math.stackexchange.com/questions/2810128/connection-between-universal-quantifier-and-implication?noredirect=1", "text": "# Connection between universal Quantifier and implication\n\nI remember a high school book I read a long time ago explains that every statement in the form of $\\forall x\\in D,P(x)$ can be turned into an implication.\n\nFor example, isn't\n\n\"For every real number $x$, $x^2$ is non-negative.\"\n\nequivalent to\n\n\"If $x$ is a real number, then $x^2$ is non-negative.\"?\n\nHowever, I came across a wikipedia article about universal quantifier that says something about bounded quantifier, which I haven't heard before. Then I thought, and confirmed, that bounded quantifier can actually be written as unbounded quantifier, and that the statement I read in the book I said above is actually a bounded version.\n\nFor example, $\\forall x\\in D,P(x)$ becomes $\\forall x,[x\\in D \\implies P(x)]$ My question is: was what I read in the book correct? That every bounded universal statement can be converted into an implication? Another one is: Can or can't every (unbounded) universal statement become implication?\n\nI ask this because I am confused. I thought every universal statement can become an implication, the wikipedia article and the fact that I cannot find (google) any other article about my thought make me want to post this question. Besides, don't we use arguably the same starting premise when we want to prove a universal statement and an implication? Both use \"Let x be ...\" or \"Assume that ...\" or \"Take arbitrary x ...\" (the latter especially is used most often when facing a universal statement) as starting premise, don't they?\n\nADDITION: Why can't $\\forall x\\in D,P(x)$ just become $x\\in D \\implies P(x)$ instead of $\\forall x,[x\\in D \\implies P(x)]$?\n\n\u2022 Jun 6 '18 at 13:24\n\u2022 Thanks. Then, in my example, the statement \"If $x$ is a real number, then $x^2$ is non-negative.\" is just a predicate whose truth depends on $x$. Am I reading that correctly?\n\u2013\u00a0bms\nJun 6 '18 at 13:32\n\u2022 Correct; but often the assertion \"If $x$ is a real number, then $x^2$ is non-negative\" is read as implicitly universally quantified. Jun 6 '18 at 13:36\n\u2022 I see. However, does this mean we cannot simply write $\\forall x,P(x)$ without knowing what or where $x$ comes from? And this means that $\\forall x,P(x)$ actually means $\\forall x \\in X,P(x)$ for some set $X$, right?\n\u2013\u00a0bms\nJun 6 '18 at 13:50\n\u2022 Not clear ... the formula $\\forall x P(x)$ is obviously formally correct. But it is ... a formula. How we interpret the predicate symol $P$ ? If we interpret it as \"$x \\text { is Even}$\" in the domain or universe of natural numbers, the resulting statement will be FALSE. If instead we interpret it as \"$x \\text { is Mortal}$\" in the domain or universe of human beings, the resulting statement will be TRUE. Jun 6 '18 at 13:54\n\n$\\forall x \\in D \\colon P(x)$ is simply an abbreviation for the formula $\\forall x ( x \\in D \\implies P(x))$ -- there really isn't more to it. You will find this convention in any decent textbook that covers the basics of first order logic.\n\nAnd you can't replace it by $x \\in D \\implies P(x)$ for the simple reason that the latter is not a sentence -- it has unbounded free variables (namely $x$). Hence it's not equivalent to the sentence $\\forall x \\in D \\colon P(x)$.\n\n\u2022 I see. So are you saying that every implication has to be properly quantified, or just in this case? Because I thought that the statement \"If $x \\in \\mathbb{R}$, then $x^2$ is non-negative\" is valid every day. Do you mean that it is not valid until I add \"For every $x$\" in the beginning?\"\n\u2013\u00a0bms\nJun 6 '18 at 13:16\n\u2022 @bms Valid and true are different concepts: $\\forall x \\in \\mathbb R \\colon x^2 \\ge 0$ is a true sentence. $x \\in \\mathbb R \\implies x^2 \\ge 0$ is not a sentence at all but it is a vaid formulae -- it's true for every assignment of $x$. Jun 6 '18 at 13:21\n\u2022 Hmm.. I have more questions actually. However, if it feels like I need to go through some definitions, could you suggest me one or two college level (ungrad or grad) logic textbooks that elementary cover this topic?\n\u2013\u00a0bms\nJun 6 '18 at 13:25\n\u2022 @bms That's right. However most mathematicians won't be that pedantic and often implicitly assume universal quantifiers ranging over every free variable. You will find this implicit assumption spelled out at least in some textbooks by logicians -- they will talk about the 'universal closure' of these formulae. Jun 6 '18 at 13:44\n\u2022 While your answer is certainly correct for typical textbooks on first-order logic, it's not in general true that the bounded quantifier is just a shorthand. If the underlying foundational system is some type theory, then unbounded quantifiers may not even be meaningful. Furthermore, in type theories that have a universal type but are based on a variant of 3-valued logic, bounded quantification can be weaker than the quantified implication. Philosophically speaking, we think in terms of bounded quantifiers. Even the symmetry in \"\u00ac\u2200x\u2208S ( P(x) ) \u2261 \u2203x\u2208S ( \u00acP(x) )\" is broken by the 'shorthand'. =) Jun 16 '18 at 8:34", "date": "2021-10-24 21:27:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2810128/connection-between-universal-quantifier-and-implication?noredirect=1", "openwebmath_score": 0.806700587272644, "openwebmath_perplexity": 412.1392684751348, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.974821152447445, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.8503826698831662}}
{"url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn", "text": "# Find solution of nonlinear ODE in terms of JacobiCN\n\nI am trying to find a specific solution for this differential equation:\n\n$-\\frac{1}{2}\\frac{d^2}{dx^2}\\psi(x)-2k \\; \\psi(x)^3 + \\frac{1}{2}k^2\\; \\psi(x)=0$\n\nMMA gives me a solution in the form of a JacobiSN function that, after setting the second argument equal to one by choosing the right constant of integration, can be rewritten as a Tanh function. However, I know from a ref that also\n\n$\\psi(x)=\\sqrt{\\frac{k}{2}}\\; \\text{sech}({kx})$\n\nsolves the ode. How to make MMA show this solution? I know that this could be obtained by choosing a JacobiCN instead of a JacobiSN, but how to implement this?\n\nThe point is that I want to find the solutions of the 2D version of this equation, and MMA only finds the Tanh solution just like in 1D. How to know if there's a Sech solution (or similar) in 2D?\n\nSo I guess the general question is: is there a way to \"guide\" MMA when it chooses the solutions for ODEs/PDEs, for example in the above case by making it choose a JacobiCN?\n\nEDIT\n\nThe Tanh solution I find is the following:\n\n$\\frac{\\sqrt{k}}{2} \\text{tanh}\\left(i\\frac{k|x|}{\\sqrt{2}}\\right)$\n\nHere is the code:\n\nThe equation:\n\nsol1D=DSolve[-2 k \u03c8[x]^3 - D[\u03c8[x], {x, 2}]/2 == -(1/2) k^2 \u03c8[x], {\u03c8[x]}, {x}]\n\n\nThe output of the above is a JacobiSN function, it's pretty big and not insightful so it makes no sense to paste it here. Now, JacobiSN[u,m] simplifies to Tanh for m=1. Important to note: JacobiSN never simplifies to Sech, but JacobiCN does. Here is what the solution becomes when you adjust the constant to get the second argument to 1 (and I choose the other constant to be zero for simplicity):\n\nIn:\n\nSimplify[sol1D[[1]] /. C[1] -> -(k^3/8) /. C[2] -> 0, x \u2208 Reals && k > 0]\n\n\nOut:\n\n \u03c8[x] -> 1/2 I Sqrt[k] Tan[(k Abs[x])/Sqrt[2]]}\n\n\nHere below I check the Sech solution:\n\nIn:\n\nsechSol[x_] := (Sqrt[k] Sech[k x])/Sqrt[2]\nSimplify[-1/2 (D[sechSol[x], {x, 2}]) - 2 k (sechSol[x])^3] == -1/2 k^2  sechSol[x]\n\n\nOut: True\n\nAs expected. Now the question is how to make this sechSol show up.\n\n2nd EDIT\n\nFor anyone who wants to know: I found analytically that also the 2D case has a Sech solution. In two dimensions the equation reads:\n\n$-\\frac{1}{2}\\frac{d^2}{dx^2}\\psi(x,y)-\\frac{1}{2}\\frac{d^2}{dy^2}\\psi(x,y)- \\psi(x,y)^3 + \\frac{1}{2}k^2\\; \\psi(x,y)=0$\n\nThe coefficient in front of the nonlinear term changes (dimensionality arguments). This equation is solved by:\n\n$\\psi(x,y)=k\\;\\text{sech}(\\frac{k}{\\sqrt{2}}(x+y))$.\n\n\u2022 (1) You'll get more nibbles if you bait the hook: Copyable code (what you tried, for instance), would make it easy for others to cut, paste & check. (2) Traditionally your equation is called an ODE, not a PDE. You may find this this meta Q&A helpful \u2013\u00a0Michael E2 Nov 10 '15 at 15:00\n\u2022 Precisely what solution did you obtain? \u2013\u00a0bbgodfrey Nov 11 '15 at 5:21\n\u2022 @bbgodfrey see the edit \u2013\u00a0user50473 Nov 13 '15 at 9:45\n\u2022 @MichaelE2 thanks, it should work now \u2013\u00a0user50473 Nov 13 '15 at 11:04\n\u2022 @ user50473: there is a difference in your equations for 1D and 2D: the factor in front of psi^3 is 2k for D=1 and 1 for D=2. In fact, your 2D Sech expression solves the 2D equation only with the factor 1, whereas the 1D Sech expression derived from the 2D case by letting y=0 requires the factor 2k to be a soluton. So could you please clarify. \u2013\u00a0Dr. Wolfgang Hintze Nov 16 '15 at 9:42\n\nAbstract\n\n(Please see section 3.1 for a short answer to the original question of the OP.)\n\nWe present here the complete solution of the problem of the OP, which consists of analyzing and solving the ODE numerically and symbolically. It turns out that the essential parameter for the classification of the solutions is the energy w of the equivalent mechanical problem.\n\nThe main problem of the OP was to reconcile the basic solution given by DSolve in terms of just the function JacobiSN containing two constants of integration C[1] and C[2] with the explicitly real valued solution which, in its most compact form, is given in terms of the function JacobiCN.\n\nWe show that, as expected, the reconciliation is possible by an appropriate choice of the integration constants, but we have found no way to let Mathematica perform the necessary transformations of the Jacobi elliptic functions automatically, therefore these have been done manually.\n\nThis answer is organized as follows: we start (section 1) with a qualitative analysis of the solution types in terms of an analogue mechanical problem, then (section 2) we provide the numerical and the exact symbolical solution, and derive (section 3) the exact solution from the basic form, and accompany this with other approaches, such as the integration of the energy equation. In section 4 we provide additional information and discuss the results.\n\nSection 1: reformulation, qualitative picture of solutions\n\nThe ODE to be solved is\n\neq0 = lhs == 0;\n\n\nwhere\n\n$$\\text{lhs}=\\frac{1}{2} k^2 \\psi (z)-2 k \\psi (z)^3-\\frac{\\psi ''(z)}{2}$$\n\nFirst we simplify things by making the ODE dimensionless.\n\neq = Simplify[\n0 == lhs /. {\u03c8[z] -> x[z k] Sqrt[k/2], \u03c8''[z] ->\nk^2 Sqrt[k/2] x''[z k]} /. z -> t/k, k > 0]\n\n(*\nOut[54]= 2 x[t]^3 + (x^\u2032\u2032)[t] == x[t]\n*)\n\n\nIt will be convenient in the following to adopt the wording of classical mechanics. The ODE eq. describes the 1D motion of a particle subject to a conservative force with the potential\n\nU = - x^2 + x^4\n\n\nThe energy\n\nw = x'[t]^2-x[t]^2+ x[t]^4\n\n\nis conserved in time. Actually, the true energy is w/2, but we still keep calling w the energy.\n\nNow let us look for the qualitative picture of the solutions. For a given energy w the motion of the particle is limited to certain intervals in x determined by the obvious requirement x'[t]^2 >= 0 as shown in the graph, in which the characteristic points are given by\n\nsolx = x /. Solve[w == -x^2 + x^4, x];\n{xA, xB} = {solx[[3]], solx[[4]]};\n{xC, xM, xD} = {solx[[3]], solx[[2]], solx[[4]]};\n{xE, xF} = {solx[[2]], solx[[4]]};\nxG = 1/Sqrt[2];\n(* the cell to produce the graph has been closed to save space *)\n\n\nWe have to distinguish three regions of w:\n\nRegion I (w > 0)\nthe point oscillates between the symmetric points A and B symmetrically about x = 0\n\nRegion II (-1/4 <= w < 0)\nthe point oscillates between E and F (or symmetrically in the region x<0, not shown here), i.e. x is never 0. For w = -1/4 the point remains at rest at G\n\nRegion III (w = 0)\nthe point for x > 0 will move between D and M towards M (after having been reflexted in D, if it started out with positive velocity) in the manner of a loopswing moving slowly to its upper position at which it comes to a rest only in infinite time (this is the \"soliton\" solution Sech)\n\nFurthermore we can state that the oscillations will be increasingly harmonic either for w>>1 or for w close to -1/4.\n\nSection 2: symbolical and numerical solution for the three regions\n\nThe symbolic expressions will be derived in section 3.\n\nRegion I\n\nxI[t_] = Sqrt[(1 + Sqrt[1 + 4 w])/2]\nJacobiCN[t (1 + 4 w)^(1/4), 1/2 (1 + 1/Sqrt[1 + 4 w])]; (* w > 0 *)\n\nWith[{ww = 1/2, tt = 20},\nx0 = xB /. w -> ww;\nx0p = Sqrt[ww + x0^2 - x0^4];\nxn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];\nPlot[{xn, 0.1 + xI[t] /. w -> ww}, {t, 0, tt},\nPlotRange -> {1.1 xA /. w -> ww, 1.1 xB /. w -> ww},\nPlotLabel ->\n\"Numerical solution of ODE eq: region I\\nEnergy w = \" <>\nToString[ww, InputForm] <>\n\"\\nInitial position x0 = xB\\nblue curve = numerical solution\\nyellow \\\ncurve = symbolic solution (slightly shifted)\", AxesLabel -> {\"t\", \"x(t)\"}]]\n\n\nRegion II\n\nxII[t_] = Sqrt[(1 + Sqrt[1 + 4 w])/2]\nJacobiDN[(t (1 + Sqrt[1 + 4 w])^(1/4))/2^(1/4), (1 + 4 w - Sqrt[1 + 4 w])/(\n2 w)];(* -1/4 < w < 0 *)\n\nWith[{ww = -1/8, tt = 10},\nx0 = xF /. w -> ww;\nx0p = Sqrt[ww + x0^2 - x0^4];\nxn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];\nPlot[{xn, xII[t] /. w -> ww}, {t, 0, tt}, PlotRange -> {-1.5, 1.5},\nPlotLabel ->\n\"Solution of ODE eq: region II\\nEnergy w = \" <> ToString[ww, InputForm] <>\n\"\\nInitial position x0 = xF\\nblue curve = numerical solution\\nyellow \\\ncurve = symbolic solution\", AxesLabel -> {\"t\", \"x(t)\"}]]\n\n\nRegion III\n\nWe obtain\n\nxIII[t_] = Sech[t]; (* w = 0 *)\n\n\nas a limit of region I or of region II.\n\nRegion I -> region III\n\nxI[t] /. w -> 0\n\n(* Out[312]= Sech[t] *)\n\nWith[{ww = 10^(-5), tt = 20},\nx0 = xF /. w -> ww ;\nx0p = Sqrt[ww + x0^2 - x0^4];\nxn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];\nPlot[{xn, xIII[t] + 0.1}, {t, 0, tt}, PlotRange -> {-1.5, 1.5},\nPlotLabel ->\n\"Numerical solution of ODE eq: region I close to region III\\nEnergy w = \" <>\nToString[ww, InputForm] <>\n\"\\nInitial position x0 = xB\\nblue curve = numerical solution\\nyellow \\\ncurve = symbolic solution (slightly shifted upwards)\",\nAxesLabel -> {\"t\", \"x(t)\"}]]\n\n\nRegion II -> region III\n\nJacobiDN @@ Limit[List @@ (xII[t]/Sqrt[((1 + Sqrt[1 + 4 w])/2)]), w -> 0]\n\n(* Out[313]= Sech[t] *)\n\nWith[{ww = -10^(-5), tt = 20},\nx0 = xB /. w -> ww ;\nx0p = Sqrt[ww + x0^2 - x0^4];\nxn = x[t] /. NDSolve[eq && (x[0] == x0) && (x'[0] == x0p), x[t], {t, 0, tt}];\nPlot[{xn, xIII[t] + 0.1}, {t, 0, tt}, PlotRange -> {-1.5, 1.5},\nPlotLabel ->\n\"Numerical solution of ODE eq: region II close to region III\\nEnergy w = \" \\\n<> ToString[ww, InputForm] <>\n\"\\nInitial position x0 = xF\\nblue curve = numerical solution (= symbolic \\\nsolution)\\nyellow curve = symbolic solution (slightly shifted upwards)\",\nAxesLabel -> {\"t\", \"x(t)\"}]]\n\n\nSection 3: Exact symbolic solution\n\nSection 3.1: Deriving the exact solutions from the basic solution\n\nHere we show how the solution by DSolve[eq] can be transformed, by chosing apropriate integration constants, to a simple unified form in terms of the function JacobiCN. This was the original task requested by the OP.\n\nWe repeat the ODE to be solved\n\neq\n\n(*\nOut[5]= 2 x[t]^3 + (x^\\[Prime]\\[Prime])[t] == x[t]\n*)\n\n\nDSolve[eq] gives two solutions which differ only in the sign. Hence it is sufficient to consider just one of them. Also, without loss of generality we rename the constants of integration.\n\nMathematica finds immediately\n\nx[t] /. DSolve[eq, x[t], t][[1]] /. {C[1] -> w, C[2] -> c} // Simplify\n\n\nDuring evaluation of In[456]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>\n\n$$- i \\frac{ \\text{sn}\\left(\\frac{\\sqrt{-(c+t)^2 \\left(\\sqrt{4 w+1}+1\\right)}}{\\sqrt{2}}|\\frac{1-\\sqrt{4 w+1}}{\\sqrt{4 w+1}+1}\\right)}{\\sqrt{2} \\sqrt{\\frac{1}{\\sqrt{4 w+1}-1}}}$$\n\nThe designation of the constant C1 is not accidental; in fact it is the energy w of the mechanical problem, in other words, it is the first integral of eq.\n\nAfter a manual simplification in the factors we define the basic soution as\n\nx0[t_, w_] := -I Sqrt[(-1 + Sqrt[1 + 4 w])/2]\nJacobiSN[I (c + t) Sqrt[ ((1 + Sqrt[1 + 4 w])/2)], (1 - Sqrt[1 + 4 w])/(\n1 + Sqrt[1 + 4 w])]\n\n\nNow it turns out that making the following choice for the integration constant c\n\ncw = 1/(1 + 4 w)^(1/4) EllipticK[1/2 (1 + 1/Sqrt[1 + 4 w])];\n\nrepc = c -> cw;\n\n\nx[t_, w_] :=\nSqrt[(1 + Sqrt[1 + 4 w])/2 ]\nJacobiCN[t (1 + 4 w)^(1/4), 1/2 (1 + 1/Sqrt[1 + 4 w])]\n\n\n$$x(t,w)=\\sqrt{\\frac{1}{2} \\left(\\sqrt{4 w+1}+1\\right)} \\text{cn}\\left(t \\sqrt[4]{4 w+1}|\\frac{1}{2} \\left(1+\\frac{1}{\\sqrt{4 w+1}}\\right)\\right)$$\n\nIt came as a surprise to me and is gratifying that this expression for the solution of the ODE is valid for all physically meaningful values of w, i.e. for w>-1/4. Hence the distinction of the expressions for the exact solutions according to regions, as was done in section 2, is not necessary.\n\nThe \"unifying\" idea is to place the origin of time (t=0) so that the motion starts always at the maximum attainable value of x for given w, i.e. with velocity x'[0] zero.\n\nIn the limit of w->0 we find the expected Sech\n\nx[t, 0]\n\n(* Out[5]= Sech[t] *)\n\n\nNow we plot the solution for varying parameter w, and distingush the two regions w>0 and w<0 for better viewing:\n\nPlot3D[x[t, w], {t, -10, 10}, {w, 0, 1}, PlotRange -> {-1.2, 1.2},\nPlotLabel -> \"Solution of ODE x(t,w) for w>=0\",\nAxesLabel -> {\"t\", \"w\", \"x\"}]\n\n\nPlot3D[x[t, w], {t, -10, 10}, {w, -1/4, 0}, PlotRange -> {-1.2, 1.2},\nPlotLabel ->\n\"Solution of ODE x(t,w) for -1/4 \\[LessEqual] w \\[LessEqual] 0\",\nAxesLabel -> {\"t\", \"w\", \"x\"}]\n\n\nUnfortunately, Mathematica seems to be not familiar enough with the transformation properties of the Jacobi elliptic functions to be able to (Full)Simplify the expression x0 after the replacement of c automatically.\n\nTherefore we shall show how to perform the steps manually indicating the relevant formulas of the \"Handbook of Mathematical Functions\" by Abramovich and Stegun, June 1964, called AS in what follows.\n\nStep 0: we start out with sn(u | m) where m<0\n\nStep 1: removal of explicitly imaginary expressions\n\nAS 16.20\n\nsn (I u | m) = I sc (u | n), n = 1-m > 1\n\n\nStep 2: transformation to have second parameter in the \"natural\" region between 0 and 1\n\nAS 16.11 and 16.3\n\nsc = sn/cn\n\nsn (u | n) -> n^(-1/2) sn (u n^(1/2) | 1/n)\ncn (u | n) -> dn (u n^(1/2) | 1/n)\n\nsc (u | n) -> n^(-1/2) sn (u n^(1/2) | 1/n) / dn (u n^(1/2) | 1/n)\n= n^(-1/2) sd (u n^(1/2) | 1/n)\n\n\nStep 3: translation of the argument by the complete elliptical integral K to have\n\nx'(0) = 0\n\n\nAS 16.8\n\nn^(-1/2) sd (u n^(1/2) + K (1/n) n^(1/2) | 1/n)\n->\nn^(-1/2) (1 - 1/n)^(-1/2) cn (u n^(1/2) | 1/n)\n\n\nSection 3.2: Integration of the energy equation\n\nAn alternative approach to find the exact solution is the integration of the energy equation (the first integral of the original ODE) given by\n\neqw = x'[t]^2 == w + x[t]^2 - x[t]^4;\n\n\nSection 3.2.1 The case w = 0\n\nWe start with the most simple case of zero energy (w = 0).\n\nAs we are looking for the soliton-solution which must have x'[0]==0.\n\nBut, unfortunately, DSolve can't deal with neither the initial condition x'[0]==0 or the equivalent x[0]==1.\n\nHence we DSolve without initial conditions and obtain\n\nsolw0 = x[t] /. DSolve[(eqw /. w -> 0), x[t], t]\n\n(*\nOut[351]= {-I Coth[t - C[1]] Sqrt[1 - Tanh[t - C[1]]^2],\nI Coth[t - C[1]] Sqrt[1 - Tanh[t - C[1]]^2], -I Coth[t + C[1]] Sqrt[\n1 - Tanh[t + C[1]]^2], I Coth[t + C[1]] Sqrt[1 - Tanh[t + C[1]]^2]}\n*)\n\n\nIt is sufficient to take just one of these and define (letting also C[1]->z)\n\nxw0[t_] = solw0[[1]] /. C[1] -> z\n\n(*\nOut[387]= -I Coth[t - z] Sqrt[1 - Tanh[t - z]^2]\n*)\n\n\nThis looks like a completely imaginary solution. It took me some time to get this reconciled with the expected Sech. But wait! This need not be imaginary if we allow z to be complex, as we shall see here:\n\nLet's try to formally calculate z from the initial condition x'[0] == 0\n\nxw0p = D[xw0[t], t] /. t -> 0 // Simplify\n\n(*\nOut[384]= I Coth[z]^2 Sqrt[Sech[z]^2]\n*)\n\nReduce[0 == xw0p, z ]\n\n(* Out[385]= False *)\n\n\nBad luck! OK, obviously the square root prevents Reduce[] from giving results. Hence we try the equivalent condition\n\nReduce[0 == xw0p^2, z, GeneratedParameters -> K ]\n\n(*\nOut[386]= K[1] \u2208 Integers && z == (I \u03c0)/2 + I \u03c0 K[1]\n*)\n\n\nHence we have found that z must be purely imaginary.\n\nTaking z -> I \u03c0/2 gives\n\nxw1[t_] = Simplify[xw0[t] /. z -> I \u03c0/2, t > 0]\n\n(*\nOut[394]= Sech[t]\n*)\n\n\nSection 3.2.2 The energy equation for $w\\neq 0$\n\nThe energy integral, which gives t + const, can be done explicitly. Here is the antidervative\n\nFullSimplify[Integrate[1/Sqrt[w + x^2 - x^4], x], w > 0]\n\n(*\nOut[158]= -((\nI Sqrt[2] EllipticF[\nI ArcSinh[(Sqrt[2] x)/Sqrt[-1 + Sqrt[1 + 4 w]]], (-1 - 2 w + Sqrt[\n1 + 4 w])/(2 w)])/Sqrt[1 + Sqrt[1 + 4 w]])\n*)\n\n\nFor -1/4 < w < 0 we have the same result but now I ArcSinh is replaced by ArcSin:\n\n(*\nOut[158]= -((\nI Sqrt[2] EllipticF[\nArcSin[(Sqrt[2] x)/Sqrt[-1 + Sqrt[1 + 4 w]]], (-1 - 2 w + Sqrt[\n1 + 4 w])/(2 w)])/Sqrt[1 + Sqrt[1 + 4 w]])\n*)\n\n\nSection 4: Generalizations and discussion\n\nSection 4.1: General potential with \"confinement\"\n\nConsider a potential repelling at $x = 0$ in the simplest forU = x^2, and going to +infinity for|x| >> 1.\n\nSo the particle is \"globally\" confined between the \"walls\" of the potential on both side of the x-axis.\n\nFirst, locally, the equation of motion close to $x = 0$ is\n\neqx0 = x''[t] == x[t];\n\n\nand the solution is\n\nx[t] /. DSolve[eqx0, x[t], t]\n\n(* Out[104]= {E^t C[1] + E^-t C[2]} *)\n\n\nLetting C[1]->0 to keep the solution finite for t -> [Infinity] we obtain\n\nxx0[t_] = a Exp[-t]\n\n(* Out[111]= a E^-t *)\n\n\nThe (positive) constant \"a\" is the (negative) velocity of the point at t = 0.\n\na /. Solve[v0 == xx0'[0], a][[1]]\n\n(* Out[120]= -v0 *)\n\n\nHence we have the following scenario:\n\nThe point starts at x = x1 = |v0| with a negative velocity v0 towards the repelling potential at x = 0. It will approach the summit of the potential exponentially for t ->[Infinity].\n\nThis is the typical behaviour of a loop swing and of situations leading to a sech(t) solution.\n\nSection 4.2: The soliton solution for $x''=x-x^{2 n+1}$\n\nNow let us consider the special confinement of the form x^(2k+1)\n\nThe energy integral for w = 0 can be calculated immediately:\n\nfi = Integrate[1/Sqrt[u^2 - u^(2 k + 2)], {u, x, 1},\nAssumptions -> {k \\[Element] Integers, k > 0, 0 < x < 1}]\n\n(* Out[6]= ArcTanh[Sqrt[1 - x^(2 k)]]/k *)\n\n\nInverting gives\n\nsol = x /. FullSimplify[Solve[t == fi, x]] // Quiet\n\n(* Out[7]= {(-Sqrt[Sech[k t]^2])^(1/k), (Sech[k t]^2)^((1/2)/k)} *)\n\n\nof which the positive solution is adequate.\n\nHence we have found that for a positive integer n the equation\n\nx''[t] == x[t] - x[t]^(2n+1)\n\n\nhas the solution\n\nx[t] = Sech[n t]^(1/n)\n\nk = 1 -> Sech[t]\nk = 2 -> Sech[2t]^(1/2)\n...\n\n\nFor t -> [Infinity] we find\n\nExp[-t] Limit[Exp[t] Sech[k t]^(1/k), t -> \\[Infinity], Assumptions -> k > 0]\n\n(* Out[27]= 2^(1/k) E^-t *)\n\n\nWhich is the exponential decay expected close to x = 0.\n\nIn the limit of large n we have\n\nLimit[Sech[n t]^(1/n), n -> \\[Infinity], Assumptions -> t > 0]\n\n(* Out[] E^-t *)\n\n\nn->[Infinity] means that the potential has hard walls at x = [PlusMinus] 1. Inside the box there is just the repelling potential from x = 0.\n\nHence the limit of large n again leads to the expected result, the simple exponential decay.\n\nSection 4.3: The solution of $\\nabla ^2u=u-u^3$ in 2D and 3D\n\nFor higher geometrical dimensions we confine ourselves to the case of radial symmetry.\n\nHence for general dimension D we consider\n\n$$\\frac{(D-1) u'(r)}{r}+u''(r)=u(r)-2 u(r)^3$$\n\nIt turns out that the case D = 2 is sufficient to grasp the general picture also for D > 2.\n\nAs before we translate the problem to the mechanical analogue and write the equation of motion as\n\neq = x''[t] + 1/t x'[t] == x[t] - 2 x[t]^3;\n\n\nThe new feature is that now due to the term with the first derivative we have friction, and this dissipative process forces the \"particle\" to come to a rest at t = oo.\n\nApplying the usual procedure to derive energy conservation, i.e multiplying by x'[t] we can write the equation as\n\n$$\\frac{\\partial w(t)}{\\partial t}=-\\frac{2 x'(t)^2}{t}$$\n\nwhere the conservative part of the energy is\n\n$$w(t)=x'(t)^2+x(t)^4-x(t)^2$$\n\nWe see that the quantity w decreases from its initial value to some finite value with the passage of time. The asymptotic value is given by x'' = x' = 0 which, from the equation of motion leads to xf = 1/Sqrt2and hence wf = -1/4.\n\nWe shall now solve the equation numerically. At first sight it might seem that the factor 1/t causes problems. But this is not the case. We shall, in the numeric treatment, start with a very small time [Epsilon] instead of zero.\n\nThe solution will be displayed in two forms: the usual dependence of position as a function of time and in a non-standard form which I like to call energy trajectory. The latter has been used in the discussion of the case D=1 above. But there the trajectories were simply oscillations between turning points. Here we see the non-trivial dissipative history of the particle in the energy picture.\n\nThe normal history is that after starting with the given energy the particle may make some oscillations in the big energy container but then \"decide\" to take the left or the right where it comes to rest asymptotically, hereby oscillating in the trough.\n\nHowever with a suitable choice of the initial conditions the particle comes to rest at x = 0. Here asymptotically the particle moves \"uphill\" like a loop swing (as discussd earlier). The space time diagram is then a one-sided \"soliton\".\n\nThe following code provides the solution in this form. We give examples of two sets of parameters here. The reader is invited to study more cases.\n\nThe oscillating case\n\ns = 10; (* scale factor for plots *)\n\\[Epsilon] = 10^-2;\ntmax = 100;\nx0 = 1.2(*1.5959735*);\nv0 = -1; (*-1*)\nww[t_] := xx'[t]^2 - xx[t]^2 + xx[t]^4\nw0 = v0^2 - x0^2 + x0^4; (* initial energy *)\n\n(* numerical solution *)\nxx[t_] = x[t] /.\nNDSolve[(x''[t] + x'[t]/t == x[t] - 2 x[t]^3) && x[\\[Epsilon]] == x0 &&\nx'[\\[Epsilon]] == v0, x[t], {t, \\[Epsilon], tmax}][[1]];\n\n(* plot x(t) *)\nPlot[{-1/Sqrt[2], 1/Sqrt[2], xx[t]}, {t, \\[Epsilon], tmax},\nAxesLabel -> {\"t\", \"x\"},\nPlotLabel ->\n\"Solution of \\!$$\\*SuperscriptBox[\\(x$$, $$\\[Prime]\\[Prime]$$,\\n\\\nMultilineFunction->None]\\)(t)+\\!$$\\*FractionBox[\\(\\*SuperscriptBox[\\\"x\\\", \\\"\\ \\[Prime]\\\",\\nMultilineFunction->None](t)$$, $$t$$]\\)\\[LongEqual]x(t)-2 \\\nx(t\\!$$\\*SuperscriptBox[\\()$$, $$3$$]\\), x(0) = \" <> ToString[x0, InputForm] <>\n\", v(0) = \" <> ToString[v0, InputForm] <>\n\"\\nPosition x as a function of t\\n\",\nPlotStyle -> {{Black, Thin, Dashed}, {Black, Thin, Dashed}, Red},\nPlotRange -> {-1.1, 1.1 (* x0 *)}]\n\n(* container plot for energy trajectory *)\ns = 10;\nxmax = x0; xmin = -xmax;\np0 = Plot[s (-x^2 + x^4), {x, xmin, xmax},\nAxesLabel -> {\"x\", \"w (red), U (blue)\"},\nPlotLabel ->\n\"Solution of \\!$$\\*SuperscriptBox[\\(x$$, $$\\[Prime]\\[Prime]$$,\\n\\\nMultilineFunction->None]\\)(t)+\\!$$\\*FractionBox[\\(\\*SuperscriptBox[\\\"x\\\", \\\"\\ \\[Prime]\\\",\\nMultilineFunction->None](t)$$, $$t$$]\\)\\[LongEqual]x(t)-2 \\\nx(t\\!$$\\*SuperscriptBox[\\()$$, $$3$$]\\), x(0) = \" <> ToString[x0, InputForm] <>\n\", v(0) = \" <> ToString[v0, InputForm] <> \", w(0) = \" <>\nToString[w0, InputForm] <>\n\"\\nEnergy trajectory: w(t) = \\!$$\\*SuperscriptBox[\\(v$$, \\\n$$2$$]\\)-\\!$$\\*SuperscriptBox[\\(x$$, $$2$$]\\)+\\!$$\\*SuperscriptBox[\\(x$$, $$4\\$$]\\) as a function of x(t)\\n\"];\n\n(* plot energy trajectory *)\np1 = ParametricPlot[{xx[t], s ww[t]}, {t, \\[Epsilon], tmax},\nPlotStyle -> {Thin, Red}];\nShow[{p0, p1}]\n\n\nThe \"soliton\" case\n\nBy varying the initial position x0 we arrive at the picture where the particle moves to x = 0.\n\ns = 10; (* scale factor for plots *)\n\\[Epsilon] = 10^-2;\ntmax = 10 (* = 30 *);\nx0 = 1.5959735;\nv0 = -1; (*-1*)\nww[t_] := xx'[t]^2 - xx[t]^2 + xx[t]^4\nw0 = v0^2 - x0^2 + x0^4; (* initial energy *)\n\n(* numerical solution *)\nxx[t_] = x[t] /.\nNDSolve[(x''[t] + x'[t]/t == x[t] - 2 x[t]^3) && x[\\[Epsilon]] == x0 &&\nx'[\\[Epsilon]] == v0, x[t], {t, \\[Epsilon], tmax}][[1]];\n\n(* plot x(t) *)\nPlot[{-1/Sqrt[2], 1/Sqrt[2], xx[t]}, {t, \\[Epsilon], tmax},\nAxesLabel -> {\"t\", \"x\"},\nPlotLabel ->\n\"Solution of \\!$$\\*SuperscriptBox[\\(x$$, $$\\[Prime]\\[Prime]$$,\\n\\\nMultilineFunction->None]\\)(t)+\\!$$\\*FractionBox[\\(\\*SuperscriptBox[\\\"x\\\", \\\"\\ \\[Prime]\\\",\\nMultilineFunction->None](t)$$, $$t$$]\\)\\[LongEqual]x(t)-2 \\\nx(t\\!$$\\*SuperscriptBox[\\()$$, $$3$$]\\), x(0) = \" <> ToString[x0, InputForm] <>\n\", v(0) = \" <> ToString[v0, InputForm] <>\n\"\\nPosition x as a function of t\\n\",\nPlotStyle -> {{Black, Thin, Dashed}, {Black, Thin, Dashed}, Red},\nPlotRange -> {-2, 1.1 x0}]\n\n(* container plot for energy trajectory *)\ns = 10;\nxmax = x0; xmin = -xmax;\np0 = Plot[s (-x^2 + x^4), {x, xmin, xmax},\nAxesLabel -> {\"x\", \"w (red), U (blue)\"},\nPlotLabel ->\n\"Solution of \\!$$\\*SuperscriptBox[\\(x$$, $$\\[Prime]\\[Prime]$$,\\n\\\nMultilineFunction->None]\\)(t)+\\!$$\\*FractionBox[\\(\\*SuperscriptBox[\\\"x\\\", \\\"\\ \\[Prime]\\\",\\nMultilineFunction->None](t)$$, $$t$$]\\)\\[LongEqual]x(t)-2 \\\nx(t\\!$$\\*SuperscriptBox[\\()$$, $$3$$]\\), x(0) = \" <> ToString[x0, InputForm] <>\n\", v(0) = \" <> ToString[v0, InputForm] <> \", w(0) = \" <>\nToString[w0, InputForm] <>\n\"\\nEnergy trajectory: w(t) = \\!$$\\*SuperscriptBox[\\(v$$, \\\n$$2$$]\\)-\\!$$\\*SuperscriptBox[\\(x$$, $$2$$]\\)+\\!$$\\*SuperscriptBox[\\(x$$, $$4\\$$]\\) as a function of x(t)\\n\"];\n\n(* plot energy trajectory *)\np1 = ParametricPlot[{xx[t], s ww[t]}, {t, \\[Epsilon], tmax},\nPlotStyle -> {Thin, Red}];\nShow[{p0, p1}]\n`\n\nBut we have cheated a bit: due to numerical instability the particle finally falls into the trough and oscillates about the asymptotic position. You can see this by increasing tmax to say 30.\n\nSection 4.4: Discussion\n\nSome remarks concerning the soliton solution should be made.\n\nWe have seen that non oscillating solutions occur if the particle climbes the potential hill at t = 0. Actually this is the tail of the asymptotic behaviour.\n\nWe can call the finite non zero part of this structure \"soliton\", although it must be noted that the position of the particle does not remain finite for t = 0 except for the case D = 1.\n\nAnother point is the symmetric structure such as in the Sech[t]. This appears if we include negative times in our consideration. If we would do this for D > 1 we would also have the symmetry.\n\nConcluding we have shown that the mechanical analogue of our ODE can be a helpful device to understand the character of the Solutions. We also have extended this approach to disspative systems by introducing the concept of an energy trajectory depending on the time as a parameter.\n\n\u2022 There ought to be a Textbook badge. :) Sorry, can't upvote more than once. Cheers! \u2013\u00a0Michael E2 Nov 25 '15 at 21:13\n\u2022 @Michael E2: thank you very much. It was great fun for me to take the opportunity to delve into Jacobi elliptic functions, elliptic functions, inverse functions, non linear differential equations, and I have still some points to make here. Don't worry, not too many ;-) By the way, I have found out (tacitly) that, despite of its name, a non linear Schr\u00f6dinger equation has nothing to do with quantum mechanics. \u2013\u00a0Dr. Wolfgang Hintze Nov 25 '15 at 22:57\n\u2022 @Dr.WolfgangHintze I am so waiting for section 4!!! By the way, the ref in question is this article: J. Zittartz and J. S. Langer Phys. Rev. 148, 741, 1966 where also the different coefficient of the 2D problem is clarified. I approve the Textbook badge. \u2013\u00a0user50473 Dec 1 '15 at 16:18\n\u2022 @user50473: oh yes, you are right, and my bad concious is there to remind me. I ask politely for some more patience. \u2013\u00a0Dr. Wolfgang Hintze Dec 1 '15 at 20:21\n\u2022 @Dr.WolfgangHintze I was reading again all this, just wondering whether you abandoned the project. Would really love to see how this ends. \u2013\u00a0user50473 Mar 1 '16 at 17:09", "date": "2019-12-06 15:44:37", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/99044/find-solution-of-nonlinear-ode-in-terms-of-jacobicn", "openwebmath_score": 0.597770094871521, "openwebmath_perplexity": 3513.3421188503535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9748211561049158, "lm_q2_score": 0.8723473663814338, "lm_q1q2_score": 0.8503826682210279}}
{"url": "http://math.stackexchange.com/questions/125954/bounding-n-choose-k", "text": "# Bounding ${n \\choose k}$\n\nLet $k > 0$, and $n > 2k$. Why is it necessarily true that $${n \\choose k} > \\frac{n^k}{2^k k!}$$ And is the condition $n > 2k$ necessary for this inequality to hold?\n\n-\nThe condition is not necessary. But it makes deriving $n(n-1)(n-2)\\cdots(n-k+1)>n^k/2^k$ easy. \u2013\u00a0 Andr\u00e9 Nicolas Mar 29 '12 at 14:42\n${n \\choose k}$ increases up to $k = \\lfloor n/2 \\rfloor$, then goes down. So you only really need to worry about the left side up to the max. \u2013\u00a0 Mitch Mar 29 '12 at 14:57\n@Andr\u00e9Nicolas What about $\\binom{6}{6} = 1 < \\frac{46656}{46080} = \\frac{6^6}{2^6 6!}$? \u2013\u00a0 dtldarek Mar 29 '12 at 15:24\n@dtldarek: Certainly the inequality does not hold unconditionally. \u2013\u00a0 Andr\u00e9 Nicolas Mar 29 '12 at 15:34\n@Andr\u00e9Nicolas Yeah, you are right, I forgot that necessary has very precise meaning in mathematics. +1 for you :-) \u2013\u00a0 dtldarek Mar 29 '12 at 15:50\n\nWe have $$\\binom{n}{k} = \\frac{n!}{(n-k)!k!} \\ge \\frac{(n-k)^n}{(n-k)^{n-k}k!} \\ge \\frac{(n-k)^k}{k!} \\ge \\frac{(n/2)^k}{k!}$$ using $n > 2k \\implies n-k \\ge n/2$. In the more general case, as noted in the comments, you will need a more detailed analysis, e.g., using Stirling's approximation for $n!$.\n\n-\nHow do you prove $\\frac{n!}{(n-k)!} \\geq \\frac{(n-k)^n}{(n-k)^{n-k}}$? \u2013\u00a0 jamaicanworm Mar 29 '12 at 15:54\nUsing the definition of n! and the simple inequality chain $n-k+i \\ge n-k \\ge n-k-i$ for $i \\le k$. \u2013\u00a0 Johannes Kloos Mar 29 '12 at 16:45\n\nIf $k\\le\\frac12n$, then $n-k+1>\\frac12n$. Therefore, \\begin{align} \\binom{n}{k} &=\\frac{n(n-1)(n-2)\\dots(n-k+1)}{k!}\\\\ &>\\frac{(n/2)^k}{k!}\\\\ &=\\frac{n^k}{2^kk!}\\tag{1} \\end{align} However, the condition that $k<\\frac12n$ can be extended to $k\\le\\frac34n$ by noting that for $0\\le j\\le k-1$, $(n-j)\\color{red}{(n+j-k+1)}\\ge n\\color{red}{(n-k+1)}>n^2/4$. \\begin{align} \\binom{n}{k}^2 &=\\frac{n\\color{red}{(n-k+1)}(n-1)\\color{red}{(n-k+2)}(n-2)\\dots(n-k+1)\\color{red}{n}}{k!^2}\\\\ &>\\frac{(n^2/4)^k}{k!^2}\\tag{2} \\end{align} Taking the square root of $(2)$ yields $$\\binom{n}{k}>\\frac{n^k}{2^kk!}\\tag{3}$$ Take It To The Limit:\n\nAbove, it is shown that the condition $k<\\frac12n$ in the question can be extended to $k\\le\\frac34n$. One might ask how far this might be pushed. That is, what is the largest $\\alpha$ so that $k<\\alpha n$ implies $(3)$.\n\nMultiplying $(3)$ by $k!$ and taking the $\\log$ of both sides yields $$\\sum_{j=0}^{k-1}\\log(n-j)>k\\log(n)-k\\log(2)\\tag{4}$$ Noting that $$\\sum_{j=0}^{k-1}\\log(n-j)\\ge\\int_{n-k}^n\\log(x)\\,\\mathrm{d}x\\tag{5}$$ We get that $$\\int_{n-k}^n\\log(x)\\,\\mathrm{d}x>k\\log(n)-k\\log(2)\\tag{7}$$ implies $(3)$. Computing the integral in $(7)$ yields $$n\\log(n)-n-(n-k)\\log(n-k)+(n-k)>k\\log(n)-k\\log(2)\\tag{8}$$ Subtracting $k\\log(n)-k$ from both sides and dividing by $k$ yields $$\\frac{n-k}{k}\\log\\left(\\frac{n}{n-k}\\right)>1-\\log(2)\\tag{9}$$ Substituting $\\alpha=k/n$ yields $$\\frac{1-\\alpha}{\\alpha}\\log\\left(\\frac{1}{1-\\alpha}\\right)>1-\\log(2)\\tag{10}$$ Here is a plot of $\\frac{1-\\alpha}{\\alpha}\\log\\left(\\frac{1}{1-\\alpha}\\right)$ and $1-\\log(2)$:\n\n$\\hspace{4cm}$\n\nWe can solve $\\frac{1-\\alpha}{\\alpha}\\log\\left(\\frac{1}{1-\\alpha}\\right)=\\beta$ using the Lambert W function: \\begin{align} \\frac{1-\\alpha}{\\alpha}\\log\\left(\\frac{1}{1-\\alpha}\\right)&=\\beta\\tag{11}\\\\ (1-\\alpha)^{1-\\alpha}e^\\beta&=e^{(1-\\alpha)\\beta}\\tag{12}\\\\ (1-\\alpha)e^{\\beta/(1-\\alpha)}&=e^\\beta\\tag{13}\\\\ -\\beta/(1-\\alpha)e^{-\\beta/(1-\\alpha)}&=-\\beta e^{-\\beta}\\tag{14}\\\\ -\\frac{\\beta}{1-\\alpha}&=W\\left(-\\beta e^{-\\beta}\\right)\\tag{15}\\\\ \\alpha&=1+\\frac{\\beta}{W\\left(-\\beta e^{-\\beta}\\right)}\\tag{16} \\end{align} $(12)$: multiply by $-\\alpha$, exponentiate, multiply by $e^\\beta$\n\n$(13)$: raise to the $1/(1-\\alpha)$ power\n\n$(14)$: take reciprocals, multiply by $-\\beta$\n\n$(15)$: apply $W$\n\n$(16)$: algebraically solve for $\\alpha$\n\nPlugging $\\beta=1-\\log(2)$ into $(16)$ gives \\begin{align} \\alpha &=1+\\frac{1-\\log(2)}{W\\left(-\\frac2e(1-\\log(2))\\right)}\\\\ &\\approx 0.868708579475419252 \\end{align} Thus, if $k<\\alpha n$, $(3)$ holds.\n\nExample:\n\n$\\left.\\binom{10000}{8687}\\middle/\\frac{1000^{8687}}{2^{8687}8687!}\\right.=3.095040785$, but $\\left.\\binom{10000}{8688}\\middle/\\frac{1000^{8688}}{2^{8688}8688!}\\right.=0.8127577101$\n\n-\n\nWe have \\begin{align*} \\binom{n}{k} &> \\frac{n^k}{2^k k!} \\\\ \\frac{n!}{(n-k)!k!} &> \\frac{n^k}{2^k k!} \\\\ \\frac{n!}{(n-k)!} &> \\frac{n^k}{2^k} \\\\ \\prod_{i=n-k+1}^n i &> \\left(\\frac{n}{2}\\right)^k \\\\ (n-k+1)^k &> \\left(\\frac{n}{2}\\right)^k \\\\ n-k+1 &> \\frac{n}{2} \\\\ n-2k+2 &> 0 \\\\ n+2 &> 2k \\\\ \\end{align*}\n\nEdit: $n > 2k$ is not a necessary condition (e.g. $n \\geq 2k$ is good as well), but you need something, in general case the inequality does not hold, for example for $n \\geq 6$ we have $\\binom{n}{n} = 1 \\leq \\frac{n^n}{2^n n!}$.\n\n-", "date": "2015-01-31 05:32:20", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/125954/bounding-n-choose-k", "openwebmath_score": 0.9999480247497559, "openwebmath_perplexity": 710.6211572365571, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815535796247, "lm_q2_score": 0.8596637505099167, "lm_q1q2_score": 0.8503635242854863}}
{"url": "https://math.stackexchange.com/questions/784508/integral-int-0-pi-2-frac-sin3-x-log-sin-x-sqrt1-sin2-xdx-frac/784682", "text": "# Integral $\\int_0^{\\pi/2} \\frac{\\sin^3 x\\log \\sin x}{\\sqrt{1+\\sin^2 x}}dx=\\frac{\\ln 2 -1}{4}$\n\nHi I am trying to prove$$I:=\\int_0^{\\pi/2} \\frac{\\sin^3 x\\log \\sin x}{\\sqrt{1+\\sin^2 x}}dx=\\frac{\\ln 2 -1}{4}.$$ Thanks.\n\nI am possibly trying to simplify this to obtain something like $2\\int_0^{\\pi/2} \\log \\sin x\\, dx=-\\pi \\ln 2$ since this is easily integrable. However when I try to simplify the terms $$\\frac{\\sin^3 x}{\\sqrt {1+\\sin^2 x}}$$ I obtain a more complicated integrand. I am not sure how else to go about this one. I was trying to possibly write $$I(a)=\\int_0^{\\pi/2} \\frac{\\sin^3 a x\\log \\sin x}{\\sqrt{1+\\sin^2 x}}dx,\\quad I'(a)=\\int_0^{\\pi/2} \\frac{\\partial}{\\partial a}\\left(\\frac{\\sin^3 ax\\log \\sin x}{\\sqrt{1+\\sin^2 x}}\\right)\\, dx,$$ but this didn't simplify anything for me.\n\nI also tried the substitution $y=\\sin^2 x$, but couldn't manage to get an integral because of the $\\sin 2x$ from the derivative.\n\n\u2022 Perhaps you can use IBP by taking $u=\\ln\\sin x$? Sorry I can't give you a complete answer for this 'claim' because I'm in the middle of the class now. :P \u2013\u00a0Tunk-Fey May 7 '14 at 5:36\n\nThe change of variables $t=\\sin x$ yields $$I=\\int_0^1\\frac{t^3\\ln t}{\\sqrt{1+t^2}}\\frac{dt}{\\sqrt{1-t^2}} =\\int_0^1\\frac{t^3\\ln t}{\\sqrt{1-t^4}} dt$$ Then setting $t^4=u$ simplifies to \\eqalign{ I&=\\frac{1}{16}\\int_0^1\\frac{1}{\\sqrt{1-u}}\\ln u\\, du\\cr &=\\left[\\frac{1}{8}(1-\\sqrt{1-u})\\ln u\\right]_0^1-\\frac{1}{8}\\int_0^1\\frac{1-\\sqrt{1-u}}{u} du\\cr &=-\\frac{1}{8}\\int_0^1\\frac{1}{1+\\sqrt{1-u}} du;\\qquad v\\leftarrow1+\\sqrt{1-u}\\cr &=\\frac{1}{4}\\int_1^2\\frac{1-v}{v} du=\\frac{\\ln 2-1}{4}.} and we are done. $\\qquad\\square$\n\n\u2022 I was just trying to connect to wifi to post my answer and you beat me! I hate how quick you have to be on this website. \u2013\u00a0Bennett Gardiner May 7 '14 at 6:37\n\u2022 Sorry. If I knew, I would have waited. \u2013\u00a0Omran Kouba May 7 '14 at 6:40\n\u2022 How did the t substitution call for a 1/16 factor? Deriviating $u^4$ is $4u^3$ so shouldn't 1/4 be the balancing factor? -EDIT: Never mind I forgot the ln needed to be stripped of its 1/4 exponent. Silly me. \u2013\u00a0Nicholas Pipitone May 7 '14 at 7:03\n\u2022 and the $t$ in the log!. \u2013\u00a0Omran Kouba May 7 '14 at 7:04\n\nThe integral screams for the substitution $u = \\sin x$, which transforms it into $$\\int_0^1\\frac{u^3\\log u}{\\sqrt{1-u^4}} \\ \\mathrm{d}u,$$ another, trickier substitution, $w^2 = 1-u^4$ gives $$\\frac{1}{2} \\int_0^1 \\log (1-w^2)^{1/4} \\ \\mathrm{d}w = {1\\over 8} \\int_0^1 \\log(1+w) +\\log (1-w) \\ \\mathrm{d}w = \\frac{\\ln 2 -1}{4}.$$\n\nWe can derive a more general result:\n\nConsider the integral\n\n\\begin{align} I(a)&=\\int_0^{\\pi/2}\\, \\frac{\\sin(x)^a}{\\sqrt{1+\\sin(x)^2}}\\, dx\\\\ &=\\int_0^{1}\\, \\frac{t^a}{\\sqrt{1-t^4}}\\, dt \\tag{subst. $t=\\sin(x)$}\\\\ &=\\frac{1}{4}\\int_0^{1}\\, u^{(a-3)/4} (1-u)^{-1/2}\\, du \\tag{subst. $t^4=u$}\\\\ &=\\frac{1}{4}\\mathrm B\\left(\\frac{a+1}{4},\\frac{1}{2}\\right) \\tag{1} \\end{align}\n\nThe third line represents a form of Beta function.\n\n\\begin{align} \\therefore I'(a)&=\\int_0^{\\pi/2}\\, \\frac{\\sin(x)^a\\, \\log{\\sin{x}}}{\\sqrt{1+\\sin(x)^2}}\\, dx\\\\ &=\\frac{1}{16} \\, {\\left(\\psi_0\\left(\\frac{a+1}{4} \\right)-\\psi_0\\left(\\frac{a+3}{4} \\right) \\right)} {\\rm B}\\left(\\frac{a+1}{4},\\frac{1}{2}\\right) \\tag{$\\frac{d}{da} (1)$}\\\\ \\implies I'(3)&=\\frac{\\log{2}-1}{4} \\end{align}\n\nReferences: Beta function and Polygamma function", "date": "2019-11-15 07:52:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/784508/integral-int-0-pi-2-frac-sin3-x-log-sin-x-sqrt1-sin2-xdx-frac/784682", "openwebmath_score": 0.964635968208313, "openwebmath_perplexity": 1470.9731262956655, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815523039174, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.8503635160757776}}
{"url": "http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html", "text": "# The pigeonhole principle forms\n\nForm,isarephrasingofthisstatement propositionphp1 (thepigeonholeprinciple,simpleversion) (the pigeonhole principle) ifn or more pigeons are distributed and placed in six pigeonholes, some pigeonhole contains two numbers by the way the. 4 the simple form of the pigeonhole principle is obtained from the strong form from math 3343 at the hong kong university of science and technology. Putnam training pigeonhole principle 2 11 prove that every convex polyhedron has at least two faces with the same number of edges. In mathematics , the pigeonhole principle states that if n items are put into m containers, with n m , then at least one container must contain more than one itemthis theorem is exemplified in real life by truisms like in any group of three gloves there must be at least two left gloves or two right gloves. The pigeonhole principle sounds trifling but its uses are deceiving astonishing thus, in our project, we intend to learn and discover more about the pigeonhole principle and illustrate its numerous interesting applications in our daily life.\n\nThe inclusion-exclusion principle and the previous parts can be used to solve this question number of arrangements with ben sitting next to alyssa or carlos is the number of arrangements with ben next to alyssa (2$$\\cdot$$6) plus number of arrangements with ben next to carlos (2$$\\cdot$$6) minus number of arrangements with ben next to both. Pigeonhole principle strong form \u2013 theorem: let q 1 , q 2 , , q n be positive integers if q 1 + q 2 + + q n \u2212 n + 1 objects are put into n boxes, then either the 1st box contains at least q 1 objects, or the 2nd box contains at least q 2 objects, , the nth box contains at least q n objects. The pigeons for each of these subsets, the sum of the numbers in the subset will be the pigeonhole since our numbers are all between 0 and 107, the sum of thirty of them is at most 3 108, which is less than 230 1 \u02c7109thus, since there are more subsets (pigeons) than sums (holes), there. Pigeon-hole principle if nm pigeons are put into m pigeonholes, th eres a hole with more than one pigeon 2 alternative forms\u2022 if n objects are to be allocated to m containers, then at least one container must hold at least ceil(n/m) objects.\n\nThe pigeonhole principle can be phrased in terms of labels if more than n objects are to be assigned labels from a set of n labels, then there is sure to be two objects with the same label. A part of it will concentrate on the pigeonhole principle thus, i need some hard to very hard problems in the subject to solve i would be thankful if you can send me links\\books\\or just a lone problem. Echelon forms and the general solution to ax = b learning goals: to see that elimination is still the technique that gets our answers, but that going just by the pigeonhole principle what we will do is elimination and row swaps to clear out columns as usual if a column is missing a pivot\u2014there. \u201cpigeonhole live\u201d means the pigeonhole live website and service \u201c pigeonhole live platform \u201d means the \u201cpigeonhole live\u201d real-time audience engagement platform and the \u201cdashboard account.\n\nPigeonhole principle the pigeonhole principle (also known as the dirichlet box principle , dirichlet principle or box principle ) states that if or more pigeons are placed in holes, then one hole must contain two or more pigeons. Pigeonhole principle kin-yin li what in the world is the pigeonhole principle well this famous principle for each one, form a box containing the number and all powers of 2 times the number so the first box contains 1,2,4,8, 16, and the next box contains 3,6,12,24,48, and so on then among the 51 numbers chosen, the pigeonhole. In mathematics, the pigeonhole principle states that if items are put into containers, with , then at least one container must contain more than one item this theorem is exemplified in real life by truisms like in any group of three gloves there must be at least two left gloves or at least two right gloves.\n\n## The pigeonhole principle forms\n\nAs @cuddlycuttlefish pointed out the pigeonhole principle with xor in it is clearly false if you replace the xor with $\\vee$ and $\\leftarrow \\rightarrow$ with $\\wedge$, however, i think your proof becomes correct. E pigeonhole principle basic geometric problems 1 five darts are thrown at a square target measuring 14 inches on a side prove that two of them must be at a distance no more than 10 inches apart. Pigeonhole (plural pigeonholes) a nook in a desk for holding papers one of an array of compartments for sorting post, messages, etc at an office, or college (for example. Pigeonhole principle to the study of e\ufb03cient provability of major open problems in computational complexity, as well as some of its generaliza- tions in the form of general matching principles.\n\n\u2022 Pigeonhole principle the following general principle was formulated by the famous german mathematician dirichlet (1805-1859): pigeonhole principle: suppose you have kpigeonholes and npigeons to be placed in them.\n\u2022 Tion for the ith level of polynomial hierarchy this is because s3 2 can do the necessary minimization and paris et al [21], as presented in kraj cek [15], have shown that s3 2 proves the weak pigeonhole principle for p-time functions.\n\u2022 One of the famous (although often neglected in the instructional program) problem- solving techniques is to consider the pigeonhole principle which is a powerful tool used in combinatorial mathin its simplest form, t he pigeonhole principle states that if more than n pigeons are placed into n pigeonholes, some pigeonhole must contain more than one pigeon.\n\nBy the generalized pigeonhole principle (contrapositive form), this would imply that the total number of people is at most 3 26 = 78 but this contradicts the fact that there are 85 people in all hence at least 4 people share a last initial 27. In elementary mathematics the strong form of the pigeonhole principle is most often applied in the special case when q 1 = q 2 = = q n = r in this case the principle becomes: \u2022 if n ( r - 1) + 1 objects are put into n boxes, then at least one of the boxes contains r or more of the objects. In many situations, the naive form of the pigeonhole principle can be applied directly in most problems, the objects and boxes are fairly obvious a box contains three pairs of socks colored red, blue, and green, respectively. (this story is an example of the second pigeonhole principle) 3 fundamental proof 31 first pigeonhole principle if n items are put into m pigeonholes with n m(m, n \u2208 n \u2217 ), then at least one pigeonhole must contain more than one item.\n\nThe pigeonhole principle forms\nRated 4/5 based on 46 review\n\n2018.", "date": "2018-11-19 01:49:18", "meta": {"domain": "blogdasilvana.info", "url": "http://mhessayjicu.blogdasilvana.info/the-pigeonhole-principle-forms.html", "openwebmath_score": 0.602908730506897, "openwebmath_perplexity": 530.7672998149491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.982823294006359, "lm_q2_score": 0.8652240877899775, "lm_q1q2_score": 0.8503623880153929}}
{"url": "http://www.learn-math.top/for-a-bounded-subset-aa-of-rmathbbr-supasupa-is-in-the-closure-of-aa/", "text": "# For a bounded subset AA of R\\mathbb{R}, sup(A)\\sup(A) is in the closure of AA\n\nConsider A\u2282RA\\subset \\mathbb{R}, where AA is bounded. Since AA is bounded, sup(A)\\sup(A) exists. To show that sup(A)\u2208Closure(A)\\sup(A)\\in Closure(A), we need to show that sup(A)\\sup(A) is a limit point of AA. Let sup(A)=s\u2208R\\sup(A)=s\\in\\mathbb{R}. If AA is a finite set then it is closed, so the closure of AA is AA, and sup(A)\u2208Closure(A)\\sup(A)\\in Closure(A). If AA is not finite, consider the sequence {s\u22121/n}\u2229A\\{s-1/n\\}\\cap A. This sequence converges to ss and is entirely contained in AA.\n\nBut I have some doubts about my proof approach because if we consider A:=(0,1/2)\u222a{5}A:=(0,1/2)\\cup\\{5\\}, the above sequence will not intersect AA. In this case, however, one can argue that {5}\\{5\\} is a singleton, which is closed, and so {5}\\{5\\} must be in the closure of AA. In all other cases, it seems, the above sequence will be contained in AA and will converge to ss. Is this true?\n\nAnother approach I could use is this. Suppose s\u2209Closure(A)s\\not\\in Closure(A). Then s\u2208R\u2216Closure(A)s\\in \\mathbb{R}\\backslash Closure(A). Since Closure(A)Closure(A) is closed, R\u2216Closure(A)\\mathbb{R}\\backslash Closure(A) is open, thus \u2203\\exists an r>0r>0 such that the open ball Br(s)\u2229\u0338Closure(A)B_r(s)\\not\\cap Closure(A). Which is impossible if ss is not a singleton. If ss is a singleton, then s\\in Closure(A)s\\in Closure(A) by definition. Therefore, s\\in Closure(A)s\\in Closure(A) in all cases.\n\n=================\n\n1\n\nThe intersection \\{s \u2013 1/n \\} \\cap A\\{s \u2013 1/n \\} \\cap A can just be empty.\n\u2013\u00a0xyzzyz\nOct 20 at 23:41\n\n=================\n\n3\n\n=================\n\nTo show that a real number xx lies in the closure \\bar{A}\\bar{A} of AA, you have to show that either x\\in Ax\\in A or that xx is a limit point of AA. While some points in AA are limit points of AA, your example A=(0,\\frac{1}{2})\\cup\\{5\\}A=(0,\\frac{1}{2})\\cup\\{5\\} demonstrates that not every element of \\bar{A}\\bar{A} is a limit point of AA, as it could be an isolated point in AA.\n\nSo to show that \\alpha=\\sup(A)\\in\\bar{A}\\alpha=\\sup(A)\\in\\bar{A}, we have to show that either \\alpha\\in A\\alpha\\in A or that \\alpha\\alpha is a limit point of AA. If \\alpha\\in A\\alpha\\in A then we\u2019re done, so suppose that \\alpha\\not\\in A\\alpha\\not\\in A.\n\nSince \\alpha-1\\alpha-1 is not a upper bound for AA, it follows that we can choose some s_1\\in As_1\\in A such that \\alpha\\geq s_1>\\alpha-1\\alpha\\geq s_1>\\alpha-1. Now s_1\\neq \\alphas_1\\neq \\alpha because \\alpha\\not\\in A\\alpha\\not\\in A, so since neither \\alpha-\\frac{1}{2}\\alpha-\\frac{1}{2} nor s_1s_1 is an upper bound for AA, we can choose s_2\\in As_2\\in A such that \\alpha \\geq s_2>\\max\\{s_1,\\alpha-\\frac{1}{2}\\}\\alpha \\geq s_2>\\max\\{s_1,\\alpha-\\frac{1}{2}\\}.\n\nProceeding inductively, we can construct a sequence \\{s_n\\}\\{s_n\\} such that \\alpha\\geq s_n>\\max\\{s_{n-1},\\alpha-\\frac{1}{n}\\}\\alpha\\geq s_n>\\max\\{s_{n-1},\\alpha-\\frac{1}{n}\\} for all n\\geq 2n\\geq 2. This means that \\{s_n\\}\\{s_n\\} is a sequence of distinct elements of AA which converges to \\alpha\\alpha, hence \\alpha\\alpha is a limit point of AA.\n\nThe definition of the closure of AA is that it contains all limit points of AA. But aren\u2019t all points of AA also its limit points (not necessarily all its limit points)?\n\u2013\u00a0sequence\nOct 21 at 0:20\n\nNo, not if you require a limit point to be the limit of a sequence of distinct elements of AA (which is the definition I\u2019m familiar with). For instance, 55 is not a limit point of (1,\\frac{1}{2})\\cup\\{5\\}(1,\\frac{1}{2})\\cup\\{5\\}.\n\u2013\u00a0carmichael561\nOct 21 at 0:22\n\nIsn\u2019t \\{5\\}\\{5\\} a limit point of the sequence \\{5,5,5,\u2026,5\\}\\{5,5,5,\u2026,5\\}?\n\u2013\u00a0sequence\nOct 21 at 0:28\n\nThat\u2019s not a sequence of distinct elements.\n\u2013\u00a0carmichael561\nOct 21 at 0:29\n\nI see. Looks like the definition I was given is different.\n\u2013\u00a0sequence\nOct 21 at 0:32\n\nBecause A is a bounded subset of \\mathbb{R}\\mathbb{R}, it has a supremum. Then see this answer. The first part of the proof, which is not shown, assumed the case where y \\in\\in E.\n\nSee this answer: Supremum of closed sets\n\nBy definition of sup(A), every open ball(\u201csegment in R\u201d) around sup(A) contains at least one element from A, thus making it by definition belonging to Closure(A).", "date": "2018-06-24 18:20:06", "meta": {"domain": "learn-math.top", "url": "http://www.learn-math.top/for-a-bounded-subset-aa-of-rmathbbr-supasupa-is-in-the-closure-of-aa/", "openwebmath_score": 0.9840782284736633, "openwebmath_perplexity": 1563.8769121707087, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232889752296, "lm_q2_score": 0.865224091265267, "lm_q1q2_score": 0.850362387077934}}
{"url": "https://www.physicsforums.com/threads/probability-question.876965/", "text": "# B Probability question\n\n1. Jun 27, 2016\n\n### DiracPool\n\nHow do I express the following scenario mathematically?\n\nI have access to CNN news online from two different internet sites, each of which have about an 80% reliability of actually providing the feed when I log onto the site. If I only had access to one of the sites, I'd know that I had an 80% chance that I'd get a live feed when I logged in. How does this figure change when I have two sites, both with that 80% reliability.\n\nThe temptation is to multiply the two probabilities together, i.e., .8 * .8 = .64 But that can't be right because the probability has to be greater than .8\n\nCan someone set this scenario up for me mathematically?\n\n2. Jun 27, 2016\n\n### BiGyElLoWhAt\n\nIf you multiply them, you're asking for the probability of getting a news feed from one site and then the other.\nI think I'm missing some detail, though. So you have to choose a site, and then log in? Or what? Do both sites pop up when you log onto whatever you're logging into?\n\n3. Jun 27, 2016\n\n### BiGyElLoWhAt\n\nIf that's the case, you're looking for the probability of A and/or B = A or B + A and B. At least I would think so. You only care that you get at least one, but it's also a possibility that you get two, so you need the probability of getting 1 + the probability of getting 2.\n\n4. Jun 27, 2016\n\n### BiGyElLoWhAt\n\nOk, so this is what I'm thinking, I get a value of 96% when I do this.\n$A$ is getting the feed from A $A'$ is not getting the feed from A.\n\n$P = AB' + A'B + AB$\n\n5. Jun 27, 2016\n\n### stevendaryl\n\nStaff Emeritus\nLet $A$ be the outcome \"The first site works\". Let $B$ be the outcome \"The second site works\". Then there are 4 possible joint outcomes: $A \\wedge B, A \\wedge \\neg B, \\neg A \\wedge B, \\neg A \\wedge \\neg B$ (where $\\neg$ means \"not\", and $\\wedge$ means \"and\"). So the probability that at least one of the sites works is:\n\n$P(A \\vee B) = P(A \\wedge B) + P(A \\wedge \\neg B) + P(\\neg A \\wedge B)$\n\nwhere $\\vee$ means \"or\".\n\nSo do you know how to compute those three probabilities on the right side of the equals?\n\n6. Jun 27, 2016\n\n### DiracPool\n\nNo I don't. But for some reason I thought intuitively the answer was what Bigyellowhat stated, 96%\n\nI don't know where I came up with that figure, and when I tried to think about how the probability would be characterized mathematically, I drew a blank. This is especially disturbing since I just took the GRE last Summer and had these type of probability questions drilled into my head. I've already forgotten how to do them\n\n7. Jun 27, 2016\n\n### BiGyElLoWhAt\n\nWhat's the probability of A but not B?\n\nAs for where I got that number, it's the same as what Steven Darryl is doing.\n\n8. Jun 27, 2016\n\n### stevendaryl\n\nStaff Emeritus\nYou have to use some laws of probability:\n$P(\\neg X) = 1 - P(X)$\n$P(X \\wedge Y) = P(X) \\cdot P(Y)$ (if the probabilities are independent)\n\n9. Jun 27, 2016\n\n### DiracPool\n\nThat looks like it adds up to .96! If there's one thing I still know how to do, it's add decimals\n\n.64 + .16 + .16 =.96 Am I right? I think the probabilities are independent. If you have cable or satellite, you pretty much have a 100% reliability that you will get a CNN feed. But the two sites I get my CNN news feed from have essentially an 80% reliability independent of one another. I was just wondering how much closer to 100% my reliability was by having the two options.\n\n10. Jun 27, 2016\n\n### MrAnchovy\n\nThese answers seem somewhat over-complicated. Assuming the probabilities of NOT getting a feed from one site are independently 1 - 0.8 = 0.2, the probability of not getting a feed from both sites is 0.2 x 0.2 = 0.04. So the probability of this not happening (i.e. getting a feed from at least one site) is 1 - 0.04 = 0.96.\n\n[edit - added this]And if you have three sites available, the probability of not getting a feed from any is 1 - 0.23 = 99.2%.\n\n11. Jun 27, 2016\n\n### PeroK\n\nYes, it's 96% if the two are independent. The simplest calculation is how often you have neither. That's 0.2 x 0.2 = 0.04.\n\n0.8 x 0.8 = 0.64 is the probability you have both.\n\nAnd 0.8 x 0.2 = 0.16 is the probability you have the first but not the second. Similarly it's the same probability you have the second but not the first. Hence:\n\n64% both feeds available\n32% only one feed available\n4% neither feed available\n\n12. Jun 27, 2016\n\n### BiGyElLoWhAt\n\nI suppose I overlooked that, no? lol\n\n13. Jun 27, 2016\n\n### DiracPool\n\nI like how you broke that down PeroK, thanks.\n\n14. Jun 27, 2016\n\n### micromass\n\nThis is studied in reliability analysis. Basically, you have some subdevices (here: the two sites) which are operational which probability $p$. What is asked here is what is the probability that the entire device is operational.\n\nIf the devices are connected in series, then the probability is $p^n$ where $n$ is the number of devices.\nIf the devices are connected in parallel, then the probability is $1 - (1- p)^n$.\nYou can generalize this to $k$ out of $n$ systems where the device works if $k$ out of $n$ subdevices work. This requires the binomial distribution. Of course, you can go even more complicated than this.", "date": "2018-01-20 14:01:01", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/probability-question.876965/", "openwebmath_score": 0.6657262444496155, "openwebmath_perplexity": 763.239704153066, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9828232909876815, "lm_q2_score": 0.8652240860523328, "lm_q1q2_score": 0.8503623836957627}}
{"url": "https://math.stackexchange.com/questions/882147/find-a-conformal-map-from-semi-disc-onto-unit-disc/882199", "text": "# Find a conformal map from semi-disc onto unit disc\n\nThis comes straight from Conway's Complex Analysis, VII.4, exercise 4.\n\nFind an analytic function $f$ which maps $G:=$ {${z: |z| < 1, Re(z) > 0}$} onto $B(0; 1)$ in a one-one fashion.\n\n$B(0;1)$ is the open unit disc.\n\nMy first intuition was to use $z^2$, which does the job splendidly, except for the segment $(-1,0] \\subset B(0;1)$. Under $z^2$, the pre-image for this segment is the segment $[-i,i]$, which is not in $G$.\n\nMy next thought is to modify $z^2$, something like $a(z-h)^2+k$. I've yet to work out the details, but my gut tells me this isn't the right idea.\n\nI've been teaching myself conformal maps in preparation for a qualifying exam. So, if there's a shockingly basic, obvious solution... please patronize me.\n\n\u2022 One thought (which I mostly pose from intuition) is seek a conformal map which maps $G$ to the right half-plane; from there it's not hard to map the right-half plane to $B(0;1)$ by a rotation. \u2013\u00a0Semiclassical Jul 30 '14 at 2:20\n\n## 2 Answers\n\nThe following trick works for any region bounded by two circular arcs (or a circular arc and a line).\n\nFind the points of intersection of the arc and the line. (Here, they're $i$ and $-i$.) Now pick a Mobius transformation that takes one of those points to $0$ and the other to $\\infty$; here $z \\mapsto \\frac{z-i}{z+i}$ works. Then the arc and the line go to two rays (because a Mobius transformation sends circles in $S^2$ to circles in $S^2$, and the only circles in $S^2$ that goes through both $0$ and $\\infty$ is a line in $\\Bbb C$), both starting at $0$ and going off the $\\infty$. Your domain maps to the region bounded by these two rays.\n\nLet's compute the rays. It suffices to find where a single point on each arc maps; if $z_0$ is on the arc, the ray will be $\\{f(z_0)t : 0 \\leq t < \\infty\\}$. I say we pick $0$ to be our point of choice on $Re(z) = 0$ and $1$ to be the point of choice for the circular arc. These are mapped to $-1$ and $-I$ respectively; so our two arcs are the negative real axis and the negative imaginary axis. I'd like the \"lower\" arc to be the positive real axis, so let's multiply by $-1$ to do this.\n\nSo we have a conformal map from your half-disc to the upper-right quadrant given by $z \\mapsto -\\frac{z-i}{z+i}$. The upper half-plane is nicer, so let's map to that by squaring; now we have a map to the upper half plane given by $z \\mapsto \\frac{(z-i)^2}{(z+i)^2}$. (For other regions bounded by rays that make different angles, you get to the upper half plane by a $z \\mapsto z^\\beta$ for the appropriate $\\beta$.)\n\nNow there's a standard map from the upper half plane to the unit disc given by $z \\mapsto \\frac{z-i}{z+i}$. Composing this with our last map gives us a map from the semi-disc to the unit disc, given by $$z \\mapsto -i\\frac{z^2+2z-1}{z^2-2z-1}.$$\n\n\u2022 This is more insightful. I'll delete my answer. \u2013\u00a0user138530 Jul 30 '14 at 3:06\n\u2022 For fun, I added the images of your mappings as a community wiki answer. Enjoy! \u2013\u00a0Semiclassical Jul 30 '14 at 3:15\n\u2022 @ChristianRemling I appreciate the compliment! \u2013\u00a0user98602 Jul 30 '14 at 3:15\n\u2022 It might be worth noting that, even before computing the two rays (the negative real and imaginary axes), you knew that they had to be perpendicular, because conformal maps preserve angles. So you already know that the next step should be squaring and that it would produce a half-plane. The details of the rays and the half-plane are needed only at the very end, when you map the half-plane to the disk. \u2013\u00a0Andreas Blass Jul 30 '14 at 4:04\n\u2022 @AndreasBlass That's definitely worth noting - thanks for commenting. \u2013\u00a0user98602 Jul 30 '14 at 4:31\n\nAs a supplement to the fine answers provided, here are the pictures of the conformal maps themselves:\n\nWe start with the right half of the unit disk, map it to the upper right quadrant, square this to the upper half-plane, and finally rotate onto the unit disk.\n\nPer requests for code - the basic pattern to generate the image of a polar region $$a\\leq r \\leq b$$ and $$\\alpha \\leq \\theta \\leq \\beta$$ under a function $$f$$ in Mathematica is:\n\nParametricPlot[{Re[f[r*Exp[I*theta]]], Im[f[r*Exp[I*theta]]]},\n{r, a, b}, {theta, alpha, beta}, Mesh -> True]\n\n\nThe pictures above may be generated with the following code block:\n\nphi1[z_] = -(z - I)/(z + I);\nf[z_] = z^2;\nphi2[z_] = (z - I)/(z + I);\nnestedFunctions =\nComposition @@@ Table[Take[{phi2, f, phi1}, -k], {k, 0, 3}];\npics = Table[\nParametricPlot[{Re[g[r*Exp[I*theta]]], Im[g[r*Exp[I*theta]]]},\n{r, 0, 1}, {theta, -Pi/2, Pi/2},\nMesh -> True, PlotRange -> 2, ImageSize -> 360],\n{g, nestedFunctions}];\nGrid[Partition[pics, 2]]\n\n\nYou can find a lot more information on visualizing complex function with Mathematica in this notebook.\n\n\u2022 It appears I can only choose one answer, but I think these images are just as essential to the solution as the explanation. With what program did you generate them? \u2013\u00a0artificial_moonlet Jul 30 '14 at 15:06\n\u2022 It's from Mathematica. I'll incorporate the code for them when I get the chance. (Also, I should note that this really isn't a proper answer since I didn't remember the proper conformal maps until I saw the other answers. Hence why I did this as a community wiki answer.) \u2013\u00a0Semiclassical Jul 30 '14 at 15:08\n\u2022 As a rookie mathematica user I would find it usefully to see the code for these types of mappings. \u2013\u00a0coffeebelly Oct 4 '14 at 19:10\n\u2022 I wonder if you can still recall the code :) \u2013\u00a0snulty Jan 11 '17 at 8:15\n\u2022 @snulty Code has been posted, if you are still interested. \u2013\u00a0Mark McClure Dec 8 '18 at 17:08", "date": "2021-03-07 16:22:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/882147/find-a-conformal-map-from-semi-disc-onto-unit-disc/882199", "openwebmath_score": 0.7189393043518066, "openwebmath_perplexity": 399.3817618980585, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126500692714, "lm_q2_score": 0.8688267864276107, "lm_q1q2_score": 0.8503317665957357}}
{"url": "http://mathhelpforum.com/algebra/148256-assistance-sketching-relations.html", "text": "# Thread: Assistance with sketching relations\n\n1. ## Assistance with sketching relations\n\nThe question:\n\nSketch the set of points (x, y) which satisfy the following relations:\n\n0 <= y <= 2x and 0 <= x <=2\n\nI have no idea where to start. Is there a way to simultaneously equate them? Is the process supposed to be intuitive? I can't get my head around it.\n\nAny guidance would be greatly appreciated.\n\n2. Originally Posted by Glitch\nThe question:\n\nSketch the set of points (x, y) which satisfy the following relations:\n\n0 <= y <= 2x and 0 <= x <=2\n\nI have no idea where to start. Is there a way to simultaneously equate them? Is the process supposed to be intuitive? I can't get my head around it.\n\nAny guidance would be greatly appreciated.\nHi Glitch,\n\nIf x (domain) is between 0 and 2, inclusive,\nthen y (range) is between 0 and 4, inclusive.\n\nSketch 0 to 2 on the x-axis, and 0 to 4 on the y-axis.\n\nComplete the rectangle with corners at (0, 0), (0, 4), (2, 4), (2, 0).\n\nAll the points inside the rectangle, including the corners and edges, satisfy the given conditions.\n\nSo when determining the range, how did you know it was between 0 and 4 inclusive? Did you substitute the upper bound of x?\n\n4. Originally Posted by Glitch\n\nSo when determining the range, how did you know it was between 0 and 4 inclusive? Did you substitute the upper bound of x?\nYes, I did.\n\n5. Hello, Glitch!\n\nSketch the set of points (x, y) which satisfy: . $\\begin{array}{ccccc}0 & \\leq & y & \\le &2x \\\\ 0 & \\le &x & \\le &2 \\end{array}$\n\n$0 \\:\\le\\: y \\:\\le\\:2x$ means $y$ is between the lines: . $\\begin{Bmatrix}y &=& 2x \\\\ y &=& 0 \\end{Bmatrix}$\n\nThe graph looks like this:\n\nCode:\n      |\n|               *:::::\n|             *::::::\n|           *:::::::::\n|         *::::::::::\n|       *:::::::::::::\n|     *::::::::::::::\n|   *:::::::::::::::::\n| *::::::::::::::::::::\n- - * - - - - - - - - - - - -\n0\n\n$0\\:\\le\\: x \\:\\le\\:2$ means $x$ is between the lines: . $\\begin{Bmatrix}x &=&2 \\\\ x &=& 0\\end{Bmatrix}$\n\nCode:\n      |\n|:::::::::|\n|:::::::::|\n|:::::::::|\n|:::::::::|\n|:::::::::|\n|:::::::::|\n- - * - - - - + - - - - - - -\n0         2\n\nThe region satisfying both inequalities is the area shaded twice.\n\nCode:\n      |\n|         *\n|       *:|\n|     *:::|\n|   *:::::|\n| *:::::::|\n- - * - - - - + - - -\n0         2", "date": "2017-06-24 15:54:30", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/148256-assistance-sketching-relations.html", "openwebmath_score": 0.5665784478187561, "openwebmath_perplexity": 650.0574417162561, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126506901791, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.8503317588247633}}
{"url": "http://math.stackexchange.com/questions/154991/find-the-remainder-when-1214-1-is-divided-by-13", "text": "# Find the remainder when $12!^{14!} +1$ is divided by $13$\n\nFind the remainder when $12!^{14!} +1$ is divided by $13$\n\nI faced this problem in one of my recent exam. It is reminiscent of Wilson's theorem. So, I was convinced that $12! \\equiv -1 \\pmod {13}$ after this I did some test on the exponent and it seems like $12!^{n!} +1\\equiv 2\\pmod {13}\\forall n \\in \\mathbb{N}$.\n\nAfter I came back home I ran some more test and I noticed that if $p$ is prime then $(p-1)!^{n!} +1\\equiv 2\\pmod {p}\\forall n \\in \\mathbb{N}$.\n\nI was wondering if this result is true, if yes how to prove it? If not what is the formal way for solving the mother problem.\n\n-\n$12!^{n!} +1\\equiv 0\\pmod {13}$ and $(p-1)!^{n!} +1\\equiv 2\\pmod {p}$ for $n=1$ (or $0$) \u2013\u00a0 Henry Jun 7 '12 at 6:47\nNote that the original problem can also be solved by Fermat's little theorem: $12!^{12} \\equiv 1 \\pmod{13}$. \u2013\u00a0 Peter Taylor Jun 7 '12 at 9:18\nNote that your 'for all' quantifier needs the caveat that $n\\gt 1$; if $n=1$ then the sums are $0\\bmod p$, not $2$. \u2013\u00a0 Steven Stadnicki Jun 7 '12 at 15:07\n\nBy Wilson's Theorem, $12!\\equiv -1\\pmod{13}$. So for any non-negative even integer $m$, $(12!)^m+1\\equiv (-1)^m+1\\equiv 2\\pmod{13}$. Since $0\\le 2\\lt 13$, it follows that $2$ is the remainder when $(12!)^m+1$ is divided by $13$.\n\nIf $m$ is odd, the same reasoning shows that $(12!)^m+1\\equiv 0\\pmod{13}$.\n\nAnd $13$ is not particularly lucky or unlucky. The same result, with the same proof, holds if $13$ is replaced by any odd prime $p$, and $12$ is replaced by $p-1$.\n\nThe prime $2$ is slightly different. Whether $m$ is odd or even, $(1)^m +1\\equiv 2\\pmod{2}$, but the remainder is $0$, not $2$.\n\n-\n... and $14!$ is even ... \u2013\u00a0 Robert Israel Jun 7 '12 at 6:44\nAha that helps! Thanks a lot Andre :) \u2013\u00a0 VelvetThunder Jun 7 '12 at 6:50\n\nHint $\\$ Unifying little Fermat and Wilson: $\\rm C\\:\\!!^{\\:\\!C}\\!\\equiv 1\\ mod\\:\\ C\\!+\\!1\\:$ prime. Take your pick for a proof.\n\n-", "date": "2015-09-02 13:40:18", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/154991/find-the-remainder-when-1214-1-is-divided-by-13", "openwebmath_score": 0.9118530750274658, "openwebmath_perplexity": 239.78522774623528, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126444811032, "lm_q2_score": 0.8688267813328976, "lm_q1q2_score": 0.8503317567543254}}
{"url": "https://www.freemathhelp.com/forum/threads/ax-by-c-0.118450/", "text": "# ax+by+c =0\n\n#### apple2357\n\n##### Full Member\nSo suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0\n\nThe standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.\nI don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?\n2a+5b+c = 0\n3a+10y+c= 0\n\nWhy can't this be made to work?\nTell me if i am talking nonsense!\n\n#### pka\n\n##### Elite Member\nSo suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0\n\nThe standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.\nI don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?\n2a+5b+c = 0\n3a+10y+c= 0\n\nWhy can't this be made to work?\nTell me if i am talking nonsense!\nSurely you know that if a line contains $$\\displaystyle (x_1,y_1)~\\&~(x_2,y_2)$$ and if $$\\displaystyle x_1\\ne x_2$$ then its slope is $$\\displaystyle m=\\dfrac{y_2-y_1}{x_2-x_1}$$.\n\n#### apple2357\n\n##### Full Member\nYes I know all that. I just couldn\u2019t explain why the method and approach above fails ?\n\n#### pka\n\n##### Elite Member\nYes I know all that. I just couldn\u2019t explain why the method and approach above fails ?\nTwo linear equations in three unknowns???\n\n#### Dr.Peterson\n\n##### Elite Member\nSo suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0\n\nThe standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.\nI don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?\n2a+5b+c = 0\n3a+10y+c= 0\n\nWhy can't this be made to work?\nTell me if i am talking nonsense!\nIt works fine! Keep going ... (after fixing a typo).\n\nThe only problem is that the answer is not unique -- you can multiply an equation of the form ax+by+c=0 by any non-zero number and the resulting equation is equivalent, and has the same form.\n\nSo at some point in your work you will be picking an arbitrary value for either a, b, or c.\n\n#### JeffM\n\n##### Elite Member\nSo suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0\n\nThe standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.\nI don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?\n2a+5b+c = 0\n3a+10y+c= 0\n\nWhy can't this be made to work?\nTell me if i am talking nonsense!\nA system of two equations with three unknowns may have an infinite number of valid solutions.\n\n$$\\displaystyle 2a + 5b + c = 0 \\text { and } 3a + 10b + c = 0 \\implies$$\n\n$$\\displaystyle c = -\\ (2a + 5b) \\implies 3a + 10b - (2a + 5b) = 0 \\implies a = -\\ 5b.$$\n\n$$\\displaystyle \\text {Let } b = 1 \\implies a = -\\ 5 \\implies c = -\\ (-\\ 10 + 5) = 5.$$\n\n$$\\displaystyle -\\ 5x + y + 5 = 0 \\implies y = 5x - 5.$$\n\n$$\\displaystyle \\therefore x = 2 \\implies 5x - 5 = 5 = y, \\text { which checks, and}$$\n\n$$\\displaystyle x = 3 \\implies 5x - 5 = 10 = y, \\text { which also checks.}$$\n\n#### JeffM\n\n##### Elite Member\nTwo linear equations in three unknowns???\nBut it does work, of course. The answer simply is not unique. The uniqueness and simplicity of the resulting equation is the advantage of the traditional methods of finding the equation of a line joining two distinct points. However, simply add b = 1 as a third equation to get the traditional, simple answer.\n\nI think you missed the thrust of the OP. It was asking why a proposed method does NOT work, but it does work. The OP incorrectly assumed that a system of two equations with three unknowns has no valid solutions. The premise of the question was therefore confusing.\n\nLast edited:\n\n#### apple2357\n\n##### Full Member\nOk. So the method does completely work.\nYou just have to pick a value for b ( which can be anything) and the resulting equation will still be unique ( after simplifying) - so there is no problem with this approach at all?\n\n#### JeffM\n\n##### Elite Member\nOk. So the method does completely work.\nYou just have to pick a value for b ( which can be anything) and the resulting equation will still be unique ( after simplifying) - so there is no problem with this approach at all?\nThere is no conceptual problem, but, in terms of communication, it is very common (and therefore readily comprehensible) to state a linear equation as\n\n$$\\displaystyle y = ux + v.$$\n\nThis results from setting b = 1. This form immediately gives the slope and the y-intercept and so is mathematically useful.\n\nLast edited:\n\n#### HallsofIvy\n\n##### Elite Member\nThe problem is that while any straight line can be written as ax+ by+ c= 0 that form is not unique! Multiplying each term by d gives a'x+ b'y+ c'= 0, where a'= ad, b'= bd, and c'= cd, a different equation for the same line. If you (arbitrarily) take a= 1 then you have x+ by+ c= 0. The line passes through (2,5) and (3, 10) so you have the two equations, 2+ 5b+ c= 0 and 3+ 10b+ c= 0, to solve for the two unknowns, b and c.\n\n#### apple2357\n\n##### Full Member\nThat makes sense!\n\n#### Jomo\n\n##### Elite Member\nThe line y=3x+4 and the line 2y=6x+8 both have the same exact points! The form y = 3x+4 or y = mx+b just insists that the coefficient of y is 1. In ax+by +c=0, b does not have to be 1!\nFor the record, the form is usually written as ax+by = c", "date": "2019-10-21 12:56:59", "meta": {"domain": "freemathhelp.com", "url": "https://www.freemathhelp.com/forum/threads/ax-by-c-0.118450/", "openwebmath_score": 0.645357072353363, "openwebmath_perplexity": 490.3412345540259, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126444811033, "lm_q2_score": 0.8688267779364222, "lm_q1q2_score": 0.850331753430152}}
{"url": "https://web2.0calc.com/questions/11-you-deposit-200-each-month-into-an-account-earning-3-interest-compounded-monthly-a-how-much-will-you-have-in-the-account-in-30-years", "text": "+0\n\n# 11. You deposit $200 each month into an account earning 3% interest compounded monthly. a. How much will you have in the account in 30 years 0 6713 13 11. You deposit$200 each month into an account earning 3% interest compounded monthly.\n\na. How much will you have in the account in 30 years?\n\nb. How much total money will you put into the account?\n\nc. How much total interest will you earn?\n\nGuest\u00a0Jul 31, 2015\n\n#2\n+26406\n+10\n\n.\n\nAlan \u00a0Jul 31, 2015\nSort:\n\n#1\n0\n\na)200*1.03^30 = $485.45 b) c) 200*1.03^30 minus$200\n\nGuest\u00a0Jul 31, 2015\n#2\n+26406\n+10\n\n.\n\nAlan \u00a0Jul 31, 2015\n#3\n+91477\n+5\n\n11. You deposit $200 each month into an account earning 3%(per annum) interest compounded monthly. I am assuming that the money goes in at the beginning of the month and the interest is paid at the end of the month. a. How much will you have in the account in 30 years? b. How much total money will you put into the account? c. How much total interest will you earn? a) You can also do this with the future value of an ordinary annuity formula C=200 n=30*12=360 i=0.03/12 = 0.0025 Amount after 30 years =$116547.38\n\n$${\\mathtt{200}}{\\mathtt{\\,\\times\\,}}\\left({\\frac{\\left({{\\mathtt{1.002\\: \\!5}}}^{{\\mathtt{360}}}{\\mathtt{\\,-\\,}}{\\mathtt{1}}\\right)}{{\\mathtt{0.002\\: \\!5}}}}\\right) = {\\mathtt{116\\,547.376\\: \\!919\\: \\!658\\: \\!203\\: \\!63}}$$\n\nb) \u00a0$${\\mathtt{200}}{\\mathtt{\\,\\times\\,}}{\\mathtt{360}} = {\\mathtt{72\\,000}}$$\n\nYou \u00a0have put $72,000 into the account. c) Interest =$116547.38 -\u00a0$72,000 =$44547.38\n\n$${\\mathtt{116\\,547.38}}{\\mathtt{\\,-\\,}}{\\mathtt{72\\,000}} = {\\mathtt{44\\,547.38}}$$\n\nMelody \u00a0Jul 31, 2015\n#4\n+26406\n+5\n\nThe small difference between my result and Melody's is that I've assumed the the total includes the interest paid at the end of the last month; whereas Melody's assumes that in the last month only the $200 deposited at the beginning of the month is added in. . Alan Jul 31, 2015 #5 +1037 +5 For this type of investment, the \u201cAnnuity Due\u201d formula returns the correct future value for standard depository account interest payments, correlating to the beginning of the month deposit and end of the month interest payment. This returns the same value as Alan\u2019s result. $$\\\\ \\noindent \\text {Annuity Due: }{FV = D \\dfrac{(1 + r)^{n} - 1} {r}(1 + r)}\\hspace{20pt} |\\hspace{10pt} \\text {\\small D=Deposit per interval}\\\\ \\\\ 200*\\dfrac{(1.0025^{360}-1)}{0.0025}*(1.0025)\\;=\\;116838.75$$ In practice, financial institutions use the \u201caverage daily balance\u201d to calculate the interest on deposits. The result returns a value with limits between the two formulas. $$\\text {FV(annuity ordinary) \\leq Total Interest \\leq FV(annuity due)}$$ Nauseated Jul 31, 2015 #6 +91477 +5 Yes, sorry Alan I did not realize that our answers were different. Alan is totally corect. This is the formula for the the future value of an ordinary annuity. This is where the money is put into the account at the END of the time period instead of at the beginning. It is easy to adjust it for this question. Here we have$200 invested at the very beginning so we must ADD 200 plus the interest it will accrue for the whole 30 years \u00a0that will be \u00a0 \u00a0C(1+i)^n = 200*(1.0025^360)\n\nbut NO $200 is invested at the very end SO we must subtract C =$200 at the end\n\nso we get\n\n$$\\\\FV=\\left[\\frac{(1+i)^n-1}{i}\\right]+C(1+i)^n-C\\\\\\\\ If you rearrange this you will find that it is identical to Nauseated's formula.\\\\\\ FV=my original answer+C(1+i)^n-C\\\\\\\\ FV=116547.38+200(1.0025)^{360}-200\\\\\\\\$$\n\n$${\\mathtt{116\\,547.38}}{\\mathtt{\\,\\small\\textbf+\\,}}{\\mathtt{200}}{\\mathtt{\\,\\times\\,}}{\\left({\\mathtt{1.002\\: \\!5}}\\right)}^{\\left({\\mathtt{360}}\\right)}{\\mathtt{\\,-\\,}}{\\mathtt{200}} = {\\mathtt{116\\,838.748\\: \\!442\\: \\!299\\: \\!145\\: \\!509\\: \\!1}}$$\n\nFV = $116838.75 This is identical to Alan's answer. And Nauseated also agreed that it is correct. ------------------------------------------ Nauseated has gone one more step with this. He has said: \"In practice, financial institutions use the \u201caverage daily balance\u201d to calculate the interest on deposits.\" Yes this is correct Nauseated has then stated that: \"The result returns a value with limits between the two formulas. \" Yes I can see how this could possibly be justified.. but I would like Nauseated to justify/discuss this statement. :) Melody Aug 1, 2015 #7 0 I was in finance for years. I never heard of treating saving account like any kind of annuity. This looks like a ton of BS to me. Guest Aug 2, 2015 #8 +1037 +5 It\u2019s not that you never heard of it, you simply forgot. That can happen when you are stoned in class. BTW, playing Monopoly is not the same as being in finance. In any case, a managed depository savings account can match the return of either type of annuity. I will demonstrate this in the next post. Nauseated Aug 2, 2015 #9 +91477 0 My goodness, that is excessively polite for you Nauseated ! Are you unwell ? Melody Aug 2, 2015 #10 +1037 +5 ### The following data sets shows first 12 months and the last 12 months of a hypothetical account of 360 interest cycles. The first column displays the$200 deposit; the second column displays the end of month interest payment on the previous balance and the $200 first day of month deposit. Column 3 displays the new balance. Column 4 is the multiplier that weights the deposit for the purpose of interest calculation. The multiplier is from 1 to 0, representing the average daily deposit of the$200, where 1 is a deposit on the first day and 0 is a deposit on the last day. (No withdraws are made from this account and only the deposit is weighted).\n\nColumn 5 is the average daily of the deposit.\n\nColumn 6 is the interest.\n\nColumn 7 is the balance.\n\nNote that the final balance in column 3 matches the annuity due, while the final balance in column 7 matches the annuity ordinary.\n\n$$\\displaystyle \\noindent \\small {First data set. Months 1-12}\\\\ \\begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\ \\hline 200.00 & 0.50 & 200.50 & 0.00 & 0.00 & 0.00 & 200.00 \\\\ 200.00 & 1.00 & 401.50 & 0.00 & 0.00 & 0.50 & 400.50 \\\\ 200.00 & 1.50 & 603.01 & 0.00 & 0.00 & 1.00 & 601.50 \\\\ 200.00 & 2.01 & 805.01 & 0.00 & 0.00 & 1.50 & 803.01 \\\\ 200.00 & 2.51 & 1,007.53 & 0.00 & 0.00 & 2.01 & 1,005.01 \\\\ 200.00 & 3.02 & 1,210.54 & 0.00 & 0.00 & 2.51 & 1,207.53 \\\\ 200.00 & 3.53 & 1,414.07 & 0.00 & 0.00 & 3.02 & 1,410.54 \\\\ 200.00 & 4.04 & 1,618.11 & 0.00 & 0.00 & 3.53 & 1,614.07 \\\\ 200.00 & 4.55 & 1,822.65 & 0.00 & 0.00 & 4.04 & 1,818.11 \\\\ 200.00 & 5.06 & 2,027.71 & 0.00 & 0.00 & 4.55 & 2,022.65 \\\\ 200.00 & 5.57 & 2,233.28 & 0.00 & 0.00 & 5.06 & 2,227.71 \\\\ 200.00 & 6.08 & 2,439.36 & 0.00 & 0.00 & 5.57 & 2,433.28 \\end{tabular}$$\n\n$$\\displaystyle \\noindent \\small {First data set. Months 349-360}\\\\ \\begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\ \\hline 200.00 & 278.06 & 111500.59 & 0.00 & 0.00 & 276.86 & 111222.53 \\\\ 200.00 & 279.25 & 111979.84 & 0.00 & 0.00 & 278.06 & 111700.59 \\\\ 200.00 & 280.45 & 112460.29 & 0.00 & 0.00 & 279.25 & 112179.84 \\\\ 200.00 & 281.65 & 112941.94 & 0.00 & 0.00 & 280.45 & 112660.29 \\\\ 200.00 & 282.85 & 113424.79 & 0.00 & 0.00 & 281.65 & 113141.94 \\\\ 200.00 & 284.06 & 113908.85 & 0.00 & 0.00 & 282.85 & 113624.79 \\\\ 200.00 & 285.27 & 114394.13 & 0.00 & 0.00 & 284.06 & 114108.85 \\\\ 200.00 & 286.49 & 114880.61 & 0.00 & 0.00 & 285.27 & 114594.13 \\\\ 200.00 & 287.70 & 115368.31 & 0.00 & 0.00 & 286.49 & 115080.61 \\\\ 200.00 & 288.92 & 115857.23 & 0.00 & 0.00 & 287.70 & 115568.31 \\\\ 200.00 & 290.14 & 116347.38 & 0.00 & 0.00 & 288.92 & 116057.23 \\\\ 200.00 & 291.37 & 116838.75 & 0.00 & 0.00 & 290.14 & 116547.38 \\end{tabular}$$\n\nThe second data sets are the same as the first, except the multiplier is set to 0.75. This corresponds to a depositor making two deposits of $100 each on the first and 15 of each 30-day month. $$\\displaystyle \\noindent \\small {Sceond data set. Months 1-12}\\\\ \\begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\ \\hline 200.00 & 0.50 & 200.50 & 0.75 & 150.00 & 0.38 & 200.38 \\\\ 200.00 & 1.00 & 401.50 & 0.75 & 150.00 & 0.88 & 401.25 \\\\ 200.00 & 1.50 & 603.01 & 0.75 & 150.00 & 1.38 & 602.63 \\\\ 200.00 & 2.01 & 805.01 & 0.75 & 150.00 & 1.88 & 804.51 \\\\ 200.00 & 2.51 & 1007.53 & 0.75 & 150.00 & 2.39 & 1006.90 \\\\ 200.00 & 3.02 & 1210.54 & 0.75 & 150.00 & 2.89 & 1209.79 \\\\ 200.00 & 3.53 & 1414.07 & 0.75 & 150.00 & 3.40 & 1413.19 \\\\ 200.00 & 4.04 & 1618.11 & 0.75 & 150.00 & 3.91 & 1617.10 \\\\ 200.00 & 4.55 & 1822.65 & 0.75 & 150.00 & 4.42 & 1821.51 \\\\ 200.00 & 5.06 & 2027.71 & 0.75 & 150.00 & 4.93 & 2026.44 \\\\ 200.00 & 5.57 & 2233.28 & 0.75 & 150.00 & 5.44 & 2231.88 \\\\ 200.00 & 6.08 & 2439.36 & 0.75 & 150.00 & 5.95 & 2437.84 \\end{tabular}$$ $$\\displaystyle \\noindent \\small {Sceond data set. Months 349-360}\\\\ \\begin{tabular}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\ \\hline 200.00 & 278.06 & 111500.59 & 0.75 & 150.00 & 277.76 & 111431.07 \\\\ 200.00 & 279.25 & 111979.84 & 0.75 & 150.00 & 278.95 & 111910.02 \\\\ 200.00 & 280.45 & 112460.29 & 0.75 & 150.00 & 280.15 & 112390.17 \\\\ 200.00 & 281.65 & 112941.94 & 0.75 & 150.00 & 281.35 & 112871.52 \\\\ 200.00 & 282.85 & 113424.79 & 0.75 & 150.00 & 282.55 & 113354.08 \\\\ 200.00 & 284.06 & 113908.85 & 0.75 & 150.00 & 283.76 & 113837.84 \\\\ 200.00 & 285.27 & 114394.13 & 0.75 & 150.00 & 284.97 & 114322.81 \\\\ 200.00 & 286.49 & 114880.61 & 0.75 & 150.00 & 286.18 & 114808.99 \\\\ 200.00 & 287.70 & 115368.31 & 0.75 & 150.00 & 287.40 & 115296.39 \\\\ 200.00 & 288.92 & 115857.23 & 0.75 & 150.00 & 288.62 & 115785.00 \\\\ 200.00 & 290.14 & 116347.38 & 0.75 & 150.00 & 289.84 & 116274.84 \\\\ 200.00 & 291.37 & 116838.75 & 0.75 & 150.00 & 291.06 & 116765.90 \\end{tabular}$$ ... Nauseated Aug 2, 2015 #11 +1037 0 My goodness, that is excessively polite for you Nauseated ! Overt politeness is one of my faults. I try, but I am not always successful. Are you unwell ? Yes! I am unwell. I am Nauseated. It seems worse than usual. . . . Maybe it's sympathetic morning sickness. :) Nauseated Aug 2, 2015 #12 +91477 0 Is Mrs Nauseated pregnant? Congratulations ! The name you have inflicted upon her will suit her well for a while :/ Melody Aug 2, 2015 #13 +91477 +5 ok Nauseated you have a lot of figures there but I am going to be so bold as to summarize your reasoning. The question does not state WHEN the money was deposited into the account. IF it is deposited at the very beginning of the month then it is an annuitiy due question and yours and Alan figures are correct (a tiny bit too big) If it is deposited at the very end of the month then it is an ordinary annuity question and my original answer was in fact correct. (a tiny bit to small) HOWEVER If the$200 deposit is made at another time or times during the month then this will make the Future value\u00a0lie between \u00a0the 2 extremes.\n\nSEE Nauseated - that was not so hard. \u00a0You really did not need all those tables of values.\n\nI am quite positive you will correct me if I have misinterpreted your logic.\n\n-----------------------------------\n\nThe reason I have added \"a tiny bit too big\" and a \"tiny bit too small\" is because there is always going to be one day at the end, or the beginning, of the month which has the old value at the beginning and the new value at the end so effectively this day has interest paid on the new\u00a0deposit of only \\$100 \u00a0(average of 0 and 200) [Not 0 (ordinary annuity) or 200 (annuity due) \u00a0as used\u00a0for the development of the formula]\n\nMelody \u00a0Aug 2, 2015\n\n### 6 Online Users\n\nWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. \u00a0See details", "date": "2018-01-22 06:21:47", "meta": {"domain": "0calc.com", "url": "https://web2.0calc.com/questions/11-you-deposit-200-each-month-into-an-account-earning-3-interest-compounded-monthly-a-how-much-will-you-have-in-the-account-in-30-years", "openwebmath_score": 0.999707818031311, "openwebmath_perplexity": 421.94371016368615, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9787126444811033, "lm_q2_score": 0.8688267660487573, "lm_q1q2_score": 0.8503317417955442}}
{"url": "https://math.stackexchange.com/questions/3415109/how-can-i-solve-the-following-square-root-inequality", "text": "# How can I solve the following square root inequality?\n\nI am given the following inequality:\n\n$$\\sqrt{2-x} > x$$\n\nAnd I have to solve for $$x$$.\n\nThis is what I tried:\n\nFirstly, I applied the existence condition for the square root:\n\n$$2-x \\ge 0 \\Rightarrow x \\le 2 \\Rightarrow x \\in (- \\infty, 2]$$\n\nThen I squared both sides of the inequality:\n\n$$2-x>x^2$$\n\n$$x^2+x-2 < 0$$\n\nAnd from this I got that:\n\n$$x \\in (-2, 1)$$\n\nFinally, I intersected this with the condition I applied at the beginning of the exercise:\n\n$$x \\in (-2, 1) \\cap (- \\infty, 2]$$\n\n$$x \\in (-2, 1)$$.\n\nThe problem with this answer is that it is wrong. If I take a number like $$-10$$ and plug it into the inequality I get:\n\n$$\\sqrt{2-(-10)} > -10$$\n\n$$\\sqrt{12} > -10$$\n\nWhich is true. However $$-10$$ is not included in the interval $$(-2, 1)$$. The correct answer seems to be $$(- \\infty, 1)$$, which is not what I got.\n\nI noticed that the inequalities of before and after of the squaring are not equivalent. So:\n\n$$\\sqrt{2-x} > x$$\n\nand\n\n$$2-x>x^2$$\n\nare not equivalent. If, again, I take the number $$-10$$, in the first inequality I get:\n\n$$\\sqrt{2-(-10)} > -10$$\n\n$$\\sqrt{12} > -10$$, which is true.\n\nAnd in the second inequality I get:\n\n$$2 - (-10) > (-10)^2$$\n\n$$12 > 100$$, which is false.\n\nSo I think that's where the problem lies, but I don't know if I am correct and what should I do to get the right answer of $$(- \\infty, 1)$$.\n\n\u2022 When you're solving $$x^2 +x -2 > 0$$ you need to remember that previously you have found both $x\\leq 2$ and $x>0$ (because square root cannot be negative). So in this case you're limited to $0 \\leq x \\leq 2$, before even looking at solving this equation. \u2013\u00a0Matti P. Oct 30 '19 at 13:29\n\u2022 \"Then I squared both sides of the inequality:\" $x > y\\not \\implies x^2 > y$ unless $y \\ge 0$. If $x > 0 > y$ then multiplying by $y$ flips the signs and $x>0 >y \\implies xy < 0 < y^2$. And multiplying by $x$ doesn't flip and $x > 0 > y\\implies x^2 > 0 >xy$ and we have $x^2 > xy$ and $xy < y^2$ and we can't conclude anything about $x^2$ compared to $y^2$. \u2013\u00a0fleablood Oct 31 '19 at 6:42\n\u2022 @MattiP., the non-negativity restriction on the square root only implies $x\\le2$ (as the OP found). It does not require $x\\ge0$. Indeed, the inequality $\\sqrt{2-x}\\gt x$ is clearly true for all $x\\lt0$. \u2013\u00a0Barry Cipra Oct 31 '19 at 12:45\n\n\"Then I squared both sides of the inequality:\"\n\n$$a > b\\not \\implies a^2 > b^2$$ unless $$b \\ge 0$$.\n\nIf $$a > 0 > b$$ then multiplying by $$b$$ flips the signs and $$a>0 >b \\implies ab < 0 < b^2$$. And multiplying by $$a$$ doesn't flip and $$a > 0 > b\\implies a^2 > 0 >ab$$.\n\n[Remember the reason that $$m > n > 0 \\implies m^2 > n^2$$ is that because $$n > 0$$ we know $$m>n \\implies mn > nn = n^2$$ and because $$m >0$$ we know $$m^2 = mm > mn$$. So we have $$m^2 >mn$$ and $$mn > n^2$$ so $$m^2 > n^2$$. That simply will not work of one of $$m,n$$ is negative. (and if both are negative you get the exact opposite result: $$0 > m>n\\implies m^2 < n^2$$.]\n\nSo we have $$a^2 > ab$$ and $$ab < b^2$$ and we can't conclude anything about $$a^2$$ compared to $$b^2$$.\n\n.... Or ....\n\njust to be blunt.\n\n$$3 > -987$$ by $$3^2 \\not > (-987)^2$$.\n\n....\n\nBut you can square both sides if you acknowedge you are assuming $$x \\ge 0$$ as a conditional that may not be true.\n\nso when you get $$x \\in (-2,1)$$ IF $$x \\ge 0$$ so $$[0,1)$$ you can intersect it with the positive values you had before. $$(-\\infty, 2] \\cap [0,1)=[0,1)$$ IFF $$x \\ge 0$$.\n\nBut we CAN have $$x < 0$$ in which case ... squaring both sides is useless.\n\nSo either $$x < 0$$ and $$x\\in (-\\infty, 0)$$ OR $$x \\ge 0$$ and $$x\\in [0,1)$$ and so $$x \\in (-\\infty,0) \\cup [0,1) = (-\\infty, 1)$$>\n\nThe domain of the root is given by b$$x\\le 2$$ Now we will consider two cases: If $$x\\le 0$$ the our inequality is true. If $$x then we can square and we get $$0>x^2+x-2$$ This gives us $$-2 and $$0 and we get $$0 So the solution set is given by $$x<1$$ and $$x$$ is a real number.\n\n$$\\sqrt{2-x} > x$$ then $$\\sqrt{2-x} > x-2 +2$$ , move all to the left hand side then $$2-x = (\\sqrt{2-x})^2$$ $$\\left(\\sqrt{2-x}\\right)^2 + \\sqrt{2-x} - 2 > 0$$ $$( \\sqrt{2-x} +2) (\\sqrt{2-x} -1)>0$$ I avoided squaring both sides.\n\nStarting from\n\n$$\\sqrt{2-x} > x$$\n\nit is immediately clear that if $$x<0$$ then $$\\sqrt{2-x} >0$$ and therefore $$x<0$$ is part of the solution set. We then need to analyze where\n\n$$\\sqrt{2-x}$$\n\nis defined. We know that $$\\sqrt{2-x}$$ will have positive real solutions if $$2-x \\ge 0$$. Therefore, we need to check $$0\\le x \\le 2$$. In order to do this, let $$\\varepsilon$$ be a small number where $$0\\le\\varepsilon< 1$$. Then $$\\sqrt{2-\\varepsilon}>\\sqrt{1}> \\varepsilon$$ $$\\sqrt{2-1}=\\sqrt{1}=1$$ $$\\sqrt{2-(\\varepsilon+1)}\\le\\sqrt{1}\\le\\varepsilon+1$$\n\nso when $$0\\le x < 1$$ we have $$\\sqrt{2-x} > x$$ and when $$1\\le x \\le 2$$ we have $$\\sqrt{2-x} \\le x$$. Hence, the proper solution set is $$x<1$$ or $$(-\\infty,1)$$.\n\nAn alternative approach is to start from the fact that we must have $$x\\le 2$$ in order for $$\\sqrt{2-x}$$ to exist, and write $$x=2-u^2$$ with $$u\\ge0$$, which simplifies the inequality to $$u\\gt2-u^2$$ (with, remember, $$u\\ge0$$, which is used in removing the square root symbol to obtain $$\\sqrt{u^2}=u$$). Standard algebra manipulates this to the equivalent form $$(u-1)(u+2)\\gt0$$, which, given the restriction to $$u\\ge0$$, implies $$u\\gt1$$. Substituting back to $$x=2-u^2$$, we see we have $$x\\lt2-1^2=1$$, so the solution set for the inequality is $$x\\in(-\\infty,1)$$.\n\nIn review, you have to do something to get rid of the square root symbol in order to find the solution set for the inequality. Squaring both sides certainly does this, but you have to contend with the fact that $$a\\gt b$$ is not equivalent to $$a^2\\gt b^2$$ without additional restrictions on $$a$$ and $$b$$, so that after you've found the solution set for the squared inequality you have to see how it relates to the original inquality. The approach here is somewhat simpler in that regard; in essence, it puts all its restrictive eggs into a single basket of non-negativity (i.e., $$u\\ge0$$).\n\nI think fleablood\u2019s answer gets closest to addressing what I think OP was asking, namely why the mechanical application of squaring both sides of the equation does not yield all of the solutions.", "date": "2020-07-03 20:45:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3415109/how-can-i-solve-the-following-square-root-inequality", "openwebmath_score": 0.9506801962852478, "openwebmath_perplexity": 127.51534364116308, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307676766118, "lm_q2_score": 0.8740772318846386, "lm_q1q2_score": 0.8503292245029808}}
{"url": "https://mathhelpboards.com/threads/determining-convergence-of-series.4118/", "text": "# Determining convergence of series\n\n#### calcboi\n\n##### New member\nI have a question on which test to use for series n=1 to infinity for (-1)^n / (n^3)-ln(n) in order to determine convergence/divergence. I am pretty sure I determined it converges through the Alternating Series Test(correct me if I'm wrong) but I am not sure whether it is conditional or absolute. I tried the Direct Comparison Test but it was inconclusive, and I am stuck now on what to do. I also tried Limit Comparison but the limit goes to infinity so it is also inconclusive. Can you please help?\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nYes you are correct.\n\nnear infinity the term n^3 is dominant over ln (n).\n\n#### calcboi\n\n##### New member\nBTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.\n\n#### Fernando Revilla\n\n##### Well-known member\nMHB Math Helper\nBTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.\nRight, $\\displaystyle\\lim_{n\\to \\infty}\\left(\\frac{1}{n^3-\\log n}:\\frac{1}{n^3}\\right)=\\ldots =1\\neq 0$, so the series is absolutely convergent.\n\n#### calcboi\n\n##### New member\nWhat test did you use to determine absolute convergence? Or was that just analyzing end behavior?\n\nMHB Math Helper\n\n#### Fernando Revilla\n\n##### Well-known member\nMHB Math Helper\nWhen I tried the Limit Comparison Test, I got infinity as n approaches infinity. How did you get 1?\n$\\displaystyle\\lim_{n\\to \\infty}\\left(\\frac{1}{n^3-\\log n}:\\frac{1}{n^3}\\right)=\\lim_{n\\to \\infty}\\frac{n^3}{n^3-\\log n}=\\lim_{n\\to \\infty}\\frac{1}{1-(\\log n/n^3)}=\\frac{1}{1-0}=1$", "date": "2020-11-25 10:53:26", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/determining-convergence-of-series.4118/", "openwebmath_score": 0.9141233563423157, "openwebmath_perplexity": 1942.4551698893772, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.972830769252026, "lm_q2_score": 0.8740772269642948, "lm_q1q2_score": 0.8503292210933526}}
{"url": "https://math.stackexchange.com/questions/2777248/when-to-use-lhospitals-rule-vs-the-limit-shortcut/2777290", "text": "# When to use L'Hospital's rule vs the limit shortcut\n\nBefore being so quick to downvote or throw darts, please forgive my ignorance and inability to recall basic calculus atm.\n\nConsider the limit $\\lim \\limits_{x \\to \\infty}\\frac{3x^2+14x-5}{x^2+x-12}$. It can quickly be determined that the $x$ approaches $3$ using L'Hospital's rule.\n\nThe same answer can be derived, however, by using this shortcut (I am unaware if it has a specific name):\n\n$\\lim \\limits_{x \\to \\infty}\\frac{3x^2+14x-5}{x^2+x-12}=\\lim \\limits_{x \\to \\infty}\\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}$\n\nEvery term is divided by the highest degree of $x$ in the denominator. The terms with $x^2$ in both the numerator and denominator simplify. All other terms go to $0$. Therefore:\n\n$\\lim \\limits_{x \\to \\infty}\\frac{3x^2/x^2+14x/x^2-5/x^2}{x^2/x^2+x/x^2-12/x^2}=\\frac{3}{1}=3$\n\n1) What are the explicit conditions in which may I use this shortcut?\n\n2) How do I know when to use it over L'Hospital's rule, as both techniques can be used only when working with quotients?\n\n3) What is the shortcuts name, if any?\n\n\u2022 Your second method is the more common one --- you use L'Hopital's when it fails (provided you satisfy the right conditions for L'Hopital's). This typically works when you have a quotient of polynomials (or in cases like totoro has pointed out), whereas L'Hopital's works for other kinds of functions, too. I don't think it has a universal name. \u2013\u00a0Bill Wallis May 11 '18 at 21:35\n\u2022 In class I usually recommend it when there is a quotient of sums of terms that can easily be compared in its growth to infinity. It would also be applicable, for example, to $\\lim_{n\\to\\infty}\\frac{n!+3^n-n^{50}}{2\\ln(n)-e^{n^2}+2\\sin(n)}$. Here 'compared' means knowing the limit of the quotient of the different terms. whether the quotient tends to $0$, to $\\infty$, or if it remains bounded. \u2013\u00a0user551819 May 11 '18 at 21:37\n\u2022 In my opinion L'Hopital is rarely the method of choice. Almost any other that works is more informative. This trick is good for quotients of polynomials. In general, writing down the first few terms of the power series for the various elements and looking at the leading term will often help. There are questions/answers on this site on the general theme \"why not to use L'Hopital\". \u2013\u00a0Ethan Bolker May 11 '18 at 21:40\n\u2022 L'Hopital can also be used in not quotient indeterminate limits. \u2013\u00a0Namaste May 11 '18 at 21:42\n\u2022 One can also use simplification of rational functions, e.g. $\\lim_{x\\to 2} \\frac{x-2}{x^2-4} = \\lim_{x\\to 2} \\frac 1{x+2} = \\frac 14.$ Or $$\\lim_{x\\to 1}\\frac{x^2+x-2}{x^3-x^2-3x+3} = \\lim_{x\\to 1}\\frac{x+2}{x^2-3} = -\\frac 32$$ \u2013\u00a0Namaste May 11 '18 at 21:51\n\nShorcut can be used in quotient limits anytime there exist some leading term at the numerator and denominator which becomes dominant in the limit with respect to the others terms that is, indicating with g(x) the dominant term\n\n$$\\lim \\limits_{x \\to \\infty}\\frac{f(x)}{g(x)}=0$$\n\nTo individuate the dominant term, recall that\n\n\u2022 $\\frac{x^a}{x^b} \\to 0$ for $b>a$\n\n\u2022 $\\frac{x^a}{b^x} \\to 0$ for $b>1$\n\n\u2022 $\\frac{\\log x}{x^a} \\to 0$ for $a>0$\n\nand for sequences\n\n\u2022 $\\frac{a^n}{n!} \\to 0$ for $a>1$\n\n\u2022 $\\frac{n!}{n^n} \\to 0$\n\nRecall that l'Hopital rule can be applied to limits which are expressed (or can be expressed) by quotient which are in the indeterminate form $\\frac 0 0$ or $\\frac{\\pm \\infty}{\\pm \\infty}$.\n\nI don't think there is a specific name for the shorcut.\n\nAs indicated in the comment another way for rational expression can be the following\n\n$$\\frac{3x^2+14x-5}{x^2+x-12}=\\frac{3x^2+3x-12+11x+7}{x^2+x-12}=3+\\frac{11x+7}{x^2+x-12}$$\n\n\u2022 @amWhy Something wrong or not clear? \u2013\u00a0gimusi May 11 '18 at 21:41\n\u2022 But L'Hopital can be used in other situations too. \u2013\u00a0Namaste May 11 '18 at 21:42\n\u2022 @amWhy are you referring to $0\\cdot \\infty$? \u2013\u00a0gimusi May 11 '18 at 21:43\n\nOn the other hand, it's not difficult to prove a general theorem about rational functions. Suppose you have $$f(x)=\\frac{a_mx^m+a_{m-1}x^{m-1}+\\dots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\\dots+b_0}$$ with $a_m\\ne0$ and $b_n\\ne0$. Then, since it's not restrictive to assume $x>0$ when we want to compute the limit for $x\\to\\infty$, the numerator can be written as $$x^m\\left(a_m+\\frac{a_{m-1}}{x}+\\dots+\\frac{a_0}{x^m}\\right)$$ and the factor in parentheses has limit $a_m$ when $x\\to\\infty$. Similarly for the denominator. Now $$\\lim_{x\\to\\infty} \\frac{\\displaystyle a_m+\\frac{a_{m-1}}{x}+\\dots+\\frac{a_0}{x^m}} {\\displaystyle b_n+\\frac{b_{n-1}}{x}+\\dots+\\frac{b_0}{x^m}} =\\frac{a_m}{b_n}\\ne0$$ Hence there are three cases.\n\n## First case: $m>n$\n\n$$\\lim_{x\\to\\infty}f(x)= \\lim_{x\\to\\infty}x^{m-n} \\frac{\\displaystyle a_m+\\frac{a_{m-1}}{x}+\\dots+\\frac{a_0}{x^m}} {\\displaystyle b_n+\\frac{b_{n-1}}{x}+\\dots+\\frac{b_0}{x^m}} = \\begin{cases} \\infty & \\text{if a_m/b_n>0} \\\\[4px] -\\infty & \\text{if a_m/b_n<0} \\end{cases}$$ The factor $x^{m-n}$ has limit $\\infty$ and the other factor is bounded.\n\n## Second case: $m=n$\n\n$$\\lim_{x\\to\\infty}f(x)= \\lim_{x\\to\\infty} \\frac{\\displaystyle a_m+\\frac{a_{m-1}}{x}+\\dots+\\frac{a_0}{x^m}} {\\displaystyle b_n+\\frac{b_{n-1}}{x}+\\dots+\\frac{b_0}{x^m}} =\\frac{a_m}{b_n}$$\n\n## Third case: $m<n$\n\n$$\\lim_{x\\to\\infty}f(x)= \\lim_{x\\to\\infty}\\frac{1}{x^{n-m}} \\frac{\\displaystyle a_m+\\frac{a_{m-1}}{x}+\\dots+\\frac{a_0}{x^m}} {\\displaystyle b_n+\\frac{b_{n-1}}{x}+\\dots+\\frac{b_0}{x^m}} =0$$ The factor $1/x^{n-m}$ has limit $0$ and the other factor has limit $a_m/b_n$.\n\nThis completely settles the problem and you need nothing else: your given limit is $3$ because the function is in case two. Doing umpteen times the same computations doesn't seem the best way to spend our time.\n\nHow can you remember this? The polynomial of greater degree dominates: if it is at the numerator, the limit is $\\pm\\infty$ (with the sign determined by $a_m/b_n$); if it is at the denominator, the limit is $0$. If the degrees are the same, the limit is $a_m/b_m$.\n\n\u2022 This is essentially the mnemonic \u201cBobo Bots eats DC\u201d\u2014bigger on bottom: $0$; bigger on top: slant; exponents are the same: divide coefficients. \u2013\u00a0gen-z ready to perish May 11 '18 at 22:23\n\u2022 @ChaseRyanTaylor I don't like such kind of mnemonics, TBH. But this is quite a simple case to handle and a general rule is surely handy. There's a similar case for integrals of the form $\\int x^ne^x\\,dx$; rather than doing umpteen integrations by parts, it's much easier to remember the antiderivative is of the form $P(x)e^x$ with $P$ of degree $n$ and do the derivative. \u2013\u00a0egreg May 11 '18 at 22:32\n\nIn general, suppose you have an expression of the form\n\n$$f_1(x)+f_2(x)\\over g_1(x)+g_2(x)$$\n\nfor which\n\n$$\\lim_{x\\to c}{f_2(x)\\over f_1(x)}=\\lim_{x\\to c}{g_2(x)\\over g_1(x)}=0$$ Then\n\n$$\\lim_{x\\to c}{f_1(x)+f_2(x)\\over g_1(x)+g_2(x)}=\\lim_{x\\to c}{f_1(x)\\over g_1(x)}$$\n\nThe proof amounts to inserting the intermediate expression\n\n$$\\lim_{x\\to c}{f_1(x)\\over g_1(x)}\\cdot{1+{f_2(x)\\over f_1(x)}\\over1+{g_2(x)\\over g_1(x)}}$$\n\nA nice example, in which L'Hopital gets you nowhere, is\n\n$$\\lim_{x\\to\\infty}{5e^{3x}+2e^x\\over2e^{3x}+7e^{2x}}=\\lim_{x\\to\\infty}{5e^{3x}\\over2e^{3x}}=\\lim_{x\\to\\infty}{5\\over2}={5\\over2}$$\n\nThe trick, in general, is to recognize a dominant term in the numerator and/or denominator. In essence, the shortcut says you can ignore all the other stuff. But you have to make sure that you pick out terms that really do dominate the other stuff; sometimes you can eyeball it, and sometimes you make mistakes (at least I do).\n\n\u2022 Have you got an explicit description of what a dominant term is? \u2013\u00a0rainier May 11 '18 at 22:44\n\u2022 @rainier, it's what the second display says: the quotient of the \"other stuff\" to the \"dominant\" term tends to $0$ in the limit. In the example, we have $(2e^x)/(5e^{3x})={2\\over5}e^{-2x}\\to0$ and $(7e^{2x})/(2e^{3x})={7\\over2}e^{-x}\\to0$ as $x\\to\\infty$. \u2013\u00a0Barry Cipra May 11 '18 at 22:49\n\u2022 Got it, thanks! \u2013\u00a0rainier May 11 '18 at 22:53", "date": "2019-08-18 19:20:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2777248/when-to-use-lhospitals-rule-vs-the-limit-shortcut/2777290", "openwebmath_score": 0.8866962790489197, "openwebmath_perplexity": 367.1341241313003, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9728307716151472, "lm_q2_score": 0.8740772236840656, "lm_q1q2_score": 0.8503292199677951}}
{"url": "https://math.stackexchange.com/questions/1727586/number-of-ways-in-which-they-can-be-seated-if-the-2-girls-are-together-and-the", "text": "# Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two\n\n$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.\n\nI divided the $4$ girls in two groups in $\\frac{4!}{2!2!}$ ways.\n\nI counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\\frac{4!}{2!2!}\\times 10\\times 5!=7200$\n\nBut the answer in the book is $43200$. I don't know where I am wrong.\n\n\u2022 What do you mean \"the two girls are together\"? That makes it sound as if the pair is specified. \u2013\u00a0lulu Apr 4 '16 at 16:10\n\u2022 Sorry,\"the\" was not given,i edited it.@lulu \u2013\u00a0mathspuzzle Apr 4 '16 at 16:13\n\u2022 no problem. I'll post something below. \u2013\u00a0lulu Apr 4 '16 at 16:13\n\nMethod I: Fix two ordered pairs of girls $AB,CD$. Then there are $7!$ ways to arrange the five boys and these two dyads. Of course we might have $ABCD$ together, so to rule that out subtract $6!$. Similarly, we must rule out another factor of $6!$ to exclude $CDAB$. Thus for these two ordered pairs there are $$7!-2\\times 6!=3600$$ suitable arrangements.\nNow, sticking to these pairs but varying the order gets us $$4\\times 3600=14,400$$\nAnd, finally, we can change the pairs. Instead of $AB$ we could have had $AC$ or $AD$ so, finally, $$3\\times 14,400= 43,200$$\nMethod II: (closer to what you were trying) Arrange the kids as $$-\\;AB-CD\\;-$$ Where $A,B,C,D$ denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are $15$ satisfactory ways to arrange the boys: $$\\{0,5,0\\},\\;\\{1,4,0\\},\\;\\{0,4,1\\},\\;\\{2,3,0\\},\\;\\{0,3,2\\},\\;\\{1,3,1\\},\\;\\{3,2,0\\},\\;\\{0,2,3\\},\\;\\{2,2,1\\},\\;\\{1,2,2\\},\\;\\{3,1,1\\},\\;\\{1,1,3\\},\\{2,1,2\\},\\;\\{4,1,0\\},\\;\\{0,1,4\\}$$ (I'm writing them all out because I believe you only counted $10$ of these). We must then pick some permutation of the girls to fill the slots labeled $A,B,C,D$ and pick some permutation of the boys to populate the dashed regions thus $$15\\times 4!\\times 5!=43,200$$\nFirst we count the number of strings of length $7$ made up of the letters b (a single boy) and $G$ (a pair of girls) such that the two G's are not together. Write down five b's like this, with a little gap between them. $$\\text{b}\\qquad\\text{b}\\qquad\\text{b}\\qquad\\text{b}\\qquad\\text{b}$$ These determine $6$ gaps, the $4$ obvious ones and $2$ endgaps. We want to choose two of these to insert the G's into. This can be done in $\\binom{6}{2}$ ways.\nWe have found the number of \"patterns.\" Now permute the girls, permute the boys. The number of arrangements is $\\binom{6}{2}4!5!$.", "date": "2019-05-21 09:30:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1727586/number-of-ways-in-which-they-can-be-seated-if-the-2-girls-are-together-and-the", "openwebmath_score": 0.739669919013977, "openwebmath_perplexity": 185.8632987009876, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795121840872, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8503205651810287}}
{"url": "https://zingale.github.io/phy504/floating-point.html", "text": "# Floating Point\u00b6\n\n## Storage overview\u00b6\n\nWe can think of a floating point number as having the form:\n\n$\\mbox{significand} \\times 10^\\mbox{exponent}$\n\nMost computers follow the IEEE 754 standard for floating point, and we commonly work with 32-bit and 64-bit floating point numbers (single and double precision). These bits are split between the signifcand and exponent as well as a single bit for the sign:\n\nSince the number is stored in binary, we can think about expanding a number in powers of 2:\n\n$0.1 \\sim (1 + 1 \\cdot 2^{-1} + 0 \\cdot 2^{-2} + 0 \\cdot 2^{-3} + 1 \\cdot 2^{-4} + 1 \\cdot 2^{-5} + \\ldots) \\times 2^{-4}$\n\nIn fact, 0.1 cannot be exactly represented in floating point:\n\nListing 52 simple_roundoff.cpp\n#include <iostream>\n#include <iomanip>\n\nint main() {\n\ndouble a = 0.1;\n\nstd::cout << std::setprecision(19) << a << std::endl;\n\n}\n\n\n### Precision\u00b6\n\nWith 52 bits for the significand, the smallest number compared to 1 we can represent is\n\n$2^{-52} \\approx 2.22\\times 10^{-16}$\n\nbut the IEEE 754 format always expresses the significant such that the first bit is 1 (by adjusting the exponent) and then doesn\u2019t need to store that 1, giving us an extra bit of precision, so the machine epsilon is\n\n$2^{-53} \\approx 1.11\\times 10^{-16}$\n\nWe already saw how to access the limits of the data type via std::numeric_limits. When we looked at machine epsilon, we saw that for a double it was about $$1.1\\times 10^{-16}$$.\n\nNote that this is a relative error, so for a number like 1000 we could only add 1.1e-13 to it before it became indistinguishable from 1000.\n\n$\\mbox{relative roundoff error} = \\frac{|\\mbox{true number} - \\mbox{computer representation} |} {|\\mbox{true number}|} \\le \\epsilon$\n\nNote that there are varying definitions of machine epsilon which differ by a factor of 2.\n\n### Range\u00b6\n\nNow consider the exponent, we use 11 bits to store it in double precision. Two are reserved for special numbers, so out of the 2048 possible exponent values, one is 0, and 2 are reserved, leaving 2045 to split between positive and negative exponents. These are set as:\n\n$2^{-1022} \\mbox{ to } 2^{1023}$\n\nconverting to base 10, this is\n\n$\\sim 10^{-308} \\mbox{ to } \\sim 10^{308}$\n\n### Reporting values\u00b6\n\nWe can use std::numeric_limits<double> to query these floating point properties:\n\nListing 53 limits.cpp\n#include <iostream>\n#include <limits>\n\nint main() {\n\nstd::cout << \"maximum double = \" << std::numeric_limits<double>::max() << std::endl;\nstd::cout << \"maximum double base-10 exponent = \" << std::numeric_limits<double>::max_exponent10 << std::endl;\nstd::cout << \"smallest (abs) double = \" << std::numeric_limits<double>::min() << std::endl;\nstd::cout << \"minimim double base-10 exponent = \" << std::numeric_limits<double>::min_exponent10 << std::endl;\nstd::cout << \"machine epsilon (double) = \" << std::numeric_limits<double>::epsilon() << std::endl;\n\n}\n\n\n## Roundoff vs. truncation error\u00b6\n\nConsider the Taylor expansion of $$f(x)$$ about some point $$x_0$$:\n\n$f(x) = f(x_0 + \\Delta x) = f(x_0) + \\left . \\frac{df}{dx} \\right |_{x_0} \\Delta x + \\mathcal{O}(\\Delta x^2)$\n\nwhere $$\\Delta x = x - x_0$$\n\nWe can solve for the derivative to find an approximation for the first derivative:\n\n$\\left . \\frac{df}{dx} \\right |_{x_0} = \\frac{f(x_0 + \\Delta x) - f(x_0)}{\\Delta x} + \\mathcal{O}(\\Delta x)$\n\nThis shows that this approximation for the derivative is first-order accurate in $$\\Delta x$$ \u2013 that is the truncation error of the approximation.\n\nWe can see the relative size of roundoff and truncation error by using this approximation to compute a derivative for different values of $$\\Delta x$$:\n\nListing 54 truncation_vs_roundoff.cpp\n#include <iostream>\n#include <iomanip>\n#include <cmath>\n#include <limits>\n#include <vector>\n\ndouble f(double x) {\nreturn std::sin(x);\n}\n\ndouble dfdx_true(double x) {\nreturn std::cos(x);\n}\n\nstruct point {\ndouble dx;\ndouble err;\n};\n\nint main() {\n\ndouble dx = 0.1;\ndouble x0 = 1.0;\n\nstd::vector<point> data;\n\nwhile (dx > std::numeric_limits<double>::epsilon()) {\n\npoint p;\np.dx = dx;\n\ndouble dfdx_approx = (f(x0 + dx) - f(x0)) / dx;\ndouble err = std::abs(dfdx_approx - dfdx_true(x0));\n\np.err = err;\n\ndata.push_back(p);\n\ndx /= 2.0;\n}\n\nstd::cout << std::setprecision(8) << std::scientific;\n\nfor (auto p : data) {\nstd::cout << std::setw(10) << p.dx << std::setw(15) << p.err << std::endl;\n}\n}\n\n\nIt is easier to see the behavior if we make a plot of the output:\n\nLet\u2019s discuss the trends:\n\n\u2022 Starting with the largest value of $$\\Delta x$$, as we make $$\\Delta x$$ smaller, we see that the error decreases. This is following the expected behavior of the truncation error derived above.\n\n\u2022 Once our $$\\Delta x$$ becomes really small, roundoff error starts to dominate. In effect, we are seeing that:\n\n$(x_0 + \\Delta x) - x_0 \\ne 0$\n\nbecause of roundoff error.\n\n\u2022 The minimum error here is around $$\\sqrt{\\epsilon}$$, where $$\\epsilon$$ is machine epsilon.\n\n## Testing for equality\u00b6\n\nBecause of roundoff error, we should never exactly compare two floating point numbers, but instead ask they they agree within some tolerance, e.g., test equality as:\n\n$| x - y | < \\epsilon$\n\nFor example:\n\nListing 55 comparing.cpp\n#include <iostream>\n\nint main() {\n\ndouble h{0.01};\ndouble sum{0.0};\n\nfor (int n = 0; n < 100; ++n) {\nsum += h;\n}\n\nstd::cout << \"sum == 1:        \" << (sum == 1.0) << std::endl;\nstd::cout << \"|sum - 1| < tol: \" << (std::abs(sum - 1.0) < 1.e-10) << std::endl;\n\n}\n\n\n## Minimizing roundoff\u00b6\n\nConsider subtracting the square of two numbers \u2013 taking the difference of two very close-in-value numbers is a prime place where roundoff can come into play.\n\n$x^2 - y^2$\n\n$(x - y)(x + y)$\n\nby factoring this, we are subtracting more reasonably sized numbers, reducing the roundoff.\n\nWe can see this directly by doing this with single precision (float) and comparing to an answer computed via double precious (double)\n\nHere\u2019s an example:\n\nListing 56 subtraction.cpp\n#include <iostream>\n\nint main() {\n\nfloat x{1.000001e15};\nfloat y{1.0000e15};\n\nstd::cout << \"x^2 - y^2 :      \" << x*x - y*y << std::endl;\nstd::cout << \"(x - y)(x + y) : \" << (x - y) * (x + y) << std::endl;\n\ndouble m{1.000001e15};\ndouble n{1.0000e15};\n\nstd::cout << \"double precision value: \" << (m - n) * (m + n) << std::endl;\n\n}\n\n\nAs another example, consider computing 1:\n\n$\\sqrt{x + 1} - \\sqrt{x}$\n\nWe can alternately rewrite this to avoid the subtraction of two close numbers:\n\n$\\sqrt{x + 1} - \\sqrt{x} = (\\sqrt{x + 1} - \\sqrt{x}) \\left ( \\frac{\\sqrt{x+1} + \\sqrt{x}}{\\sqrt{x+1} + \\sqrt{x}} \\right ) = \\frac{1}{\\sqrt{x+1} + \\sqrt{x}}$\n\nAgain we\u2019ll compare a single-precision calculation using each of these methods to a double precision \u201ccorrect answer\u201d. To ensure that we use the single-precision version of the std::sqrt() function, we will use single precision literal suffix, e.g., 1.0f tells the compiler that this is a single-precision constant.\n\nListing 57 squareroots.cpp\n#include <iostream>\n#include <iomanip>\n\n#include <cmath>\n\nint main() {\n\nfloat x{201010.0f};\n\nstd::cout << std::setprecision(10);\n\nfloat r1 = std::sqrt(x + 1.0f) - std::sqrt(x);\n\nfloat r2 = 1.0f / (std::sqrt(x + 1.0f) + std::sqrt(x));\n\nstd::cout << \"r1 = \" << r1 << \" r2 = \" << r2 << std::endl;\n\ndouble xd{201010.0};\n\ndouble d = std::sqrt(xd + 1.0) - std::sqrt(xd);\n\nstd::cout << \"d = \" << d << std::endl;\n}\n\n\nNotice that we get several more significant digits correct when we compute it with the second formulation compared to the original form.\n\n### Summation algorithms\u00b6\n\nSumming a sequence of numbers is a common place where roundoff error comes into play, especially if the numbers all vary in magnitude and you do not attempt to add them in a sorted fashion. There are a number of different summation algorithms that keep track of the loss due to roundoff and compensate the sum, for example the Kahan summation algorithm.\n\n## Special numbers\u00b6\n\nIEEE 754 defines a few special quantities:\n\n\u2022 NaN (not a number) is the result of 0.0/0.0 or std::sqrt(-1.0)\n\n\u2022 Inf (infinity) is the result of 1.0/0.0\n\n\u2022 -0 is a valid number and the standard says that -0 is equivalent to 0\n\n## Trapping floating point exceptions\u00b6\n\nWhat happens when we do something bad? Consider this example:\n\nListing 58 undefined.cpp\n#include <iostream>\n#include <cmath>\n\ndouble trouble(double x) {\nreturn std::sqrt(x);\n}\n\nint main() {\n\ndouble x{-1};\n\ndouble y = trouble(x);\n\nfor (int i = 0; i < 10; ++i) {\ny += std::pow(x, i);\n}\n\nstd::cout << y << std::endl;\n\n}\n\n\nHere, we pass -1 to trouble() which then takes the square root of it \u2013 this results in a NaN. But if we run the code, it goes merrily about its way, using that result in the later computations.\n\nUnix uses signals to indicate that a problem has happened during the code execution. If a program created a signal handler then that signal can be trapped and any desired action can be taken.\n\nNote\n\nThis example was only tested on a Linux machine with GCC. Other OSes or compilers might have slightly different headers or functionality.\n\nThere are a few parts to trapping a floating point exception (FPE). First we need to enable exception trapping via:\n\nfeenableexcept(FE_INVALID|FE_DIVBYZERO|FE_OVERFLOW);\n\n\nThat catches 3 different types of floating point exceptions \u2013 invalid, divide-by-zero, and overflows.\n\nNext we need to add a handler to deal with the exception:\n\nsignal(SIGFPE, fpe_handler);\n\n\nHere, SIGFPE is the standard name for a floating point exception, and fpe_handler is the name of a function that will be called when we detect a SIGFPE.\n\nIn our handler, we use the Linux backtrace() function to access the stack of our program execution. This is really a C-function, so we need to use C-style arrays here.\n\nHere\u2019s the new version of our code:\n\nListing 59 undefined_trap.cpp\n#include <iostream>\n#include <cmath>\n#include <csignal>\n#include <cfenv>\n#include <execinfo.h>\n\nvoid fpe_handler(int s) {\nstd::cout << \"floating point exception, signal \" << s << std::endl;\n\nconst int nbuf = 64;\nvoid *bt_buffer[nbuf];\nint nentries = backtrace(bt_buffer, nbuf);\nchar **strings = backtrace_symbols(bt_buffer, nentries);\n\nfor (int i = 0; i < nentries; ++i) {\nstd::cout << i << \": \" << strings[i] << std::endl;\n}\n\nabort();\n}\n\ndouble trouble(double x) {\nreturn std::sqrt(x);\n}\n\nint main() {\n\nfeenableexcept(FE_INVALID|FE_DIVBYZERO|FE_OVERFLOW);\n\nsignal(SIGFPE, fpe_handler);\n\ndouble x{-1};\n\ndouble y = trouble(x);\n\nfor (int i = 0; i < 10; ++i) {\ny += std::pow(x, i);\n}\n\nstd::cout << y << std::endl;\n\n}\n\n\nWhen we compile the code, we want to add the -g option to store the symbols in the code \u2013 this allows us to understand where problems arise:\n\n\nspan.prompt1:before {\ncontent: \"\\$ \";\n}\ng++ -g -o undefined_trap undefined_trap.cpp\n\n\nNow when we run this, the program aborts and we see:\n\nfloating point exception, signal 8\n0: ./undefined_trap() [0x401261]\n1: /lib64/libc.so.6(+0x42750) [0x7f3dc35dc750]\n2: /lib64/libm.so.6(+0x1435c) [0x7f3dc37d335c]\n3: ./undefined_trap() [0x4012ff]\n4: ./undefined_trap() [0x401347]\n5: /lib64/libc.so.6(+0x2d560) [0x7f3dc35c7560]\n6: /lib64/libc.so.6(__libc_start_main+0x7c) [0x7f3dc35c760c]\n7: ./undefined_trap() [0x401145]\nAborted (core dumped)\n\n\nThis is the call stack for our program. In the brackets are the address in the program where the execution was when the FPE occurred. These are ordered such that the calling function is below the function where the execution is. So it usually is best to look at the addresses near the top.\n\nWe can turn those into line numbers using addr2line:\n\naddr2line  -e undefined_trap 0x4012ff\n\n\ngives:\n\n/home/zingale/classes/phy504/examples/floating_point/undefined_trap.cpp:23\n\n\nand that line is precisely where the sqrt() is!\n\nWe can get slightly nicer output (including the function name) by doing:\n\naddr2line  -C -f -i -p -e undefined_trap 0x4012ff\n\n\nwhich gives:\n\ntrouble(double) at /home/zingale/classes/phy504/examples/floating_point/undefined_trap.cpp:23\n\n\nNote\n\nOn the MathLab machines, the stack trace seems to include an offset, like:\n\nfloating point exception, signal 8\n0: ./undefined_trap(+0xc03) [0x561d8799dc03]\n1: /lib/x86_64-linux-gnu/libc.so.6(+0x3ef10) [0x7f5461e3df10]\n2: /lib/x86_64-linux-gnu/libm.so.6(+0x11397) [0x7f5462201397]\n3: ./undefined_trap(+0xccf) [0x561d8799dccf]\n4: ./undefined_trap(+0xd21) [0x561d8799dd21]\n5: /lib/x86_64-linux-gnu/libc.so.6(__libc_start_main+0xe7) [0x7f5461e20c87]\n6: ./undefined_trap(+0xaaa) [0x561d8799daaa]\nAborted (core dumped)\n\n\nand we need to use that offset instead with addr2line, like:\n\naddr2line -a -f -e ./undefined_trap +0xcd5\n\n1\n\nthis example is based on Yakowitz & Szidarovszky", "date": "2022-06-29 03:44:06", "meta": {"domain": "github.io", "url": "https://zingale.github.io/phy504/floating-point.html", "openwebmath_score": 0.5280643701553345, "openwebmath_perplexity": 4301.407405703756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795121840872, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.8503205634266121}}
{"url": "https://math.stackexchange.com/questions/3887771/when-does-improper-riemann-integration-fail-for-computing-expected-values/3887777", "text": "When does improper Riemann integration fail for computing expected values?\n\nRiemann integration requires bounded intervals of integration. However, some continuous random variables (like those from the normal or gamma distribution) have unbounded support sets.\n\nFor these 2 distributions, would improper Riemann integrals be sufficient to get their expected values? If not, why not?\n\nUsually, improper Riemann integration will correctly provide the expectation of a continuous random variable with an unbounded support. Where you will usually encounter problems with improper Riemann integrals is when the integral of interest does not converge absolutely. In this case the improper integral may have some value, but this value has limited probabilistic meaning; for example the law of large numbers doesn't necessarily hold.\n\nAs an example of this to get a feel for it, you can look at the evaluation of the expected value of $$X \\sin(X)$$ where $$X$$ has PDF $$\\frac{1}{x^2}$$ on $$[1,\\infty)$$ and $$0$$ otherwise. This expectation is formally $$\\int_1^\\infty \\frac{\\sin(x)}{x} dx$$ which has a value in improper Riemann integration. But actually this random variable will not satisfy the law of large numbers, as you can observe numerically, so it doesn't make a whole lot of sense to say that it has an expected value.\n\nIndeed, if you have Matlab or Octave, try running the following:\n\nx=1./rand(1000,1);\nplot(cumsum(x.*sin(x))./(1:1000)');\n\n\nThis shows sample means from this distribution for progressively larger samples, and you see that they don't converge. (Off-topic: the weird 1/rand trick being used there is called the probability integral transformation, which is extremely useful for numerical work in probability.)\n\nNote that this will never be a problem for a random variable of one sign such as your example of the Gamma distribution. It is also not a problem for the normal distribution.\n\nThe exception to the second \"usually\" above is when the improper Riemann integral doesn't exist but the Lebesgue integral does. This is a pretty uncommon situation in practice.\n\n\u2022 @Iterator516 No, the integrals in both of those cases converge absolutely. \u2013\u00a0Ian Oct 30 at 15:30\n\u2022 Awesome! Thank you! \u2013\u00a0Iterator516 Oct 30 at 15:30\n\u2022 I mistakenly deleted my initial question to Ian. I asked if the integrals for the expectations of the normal and the gamma distributions do converge. \u2013\u00a0Iterator516 Oct 30 at 16:29", "date": "2020-12-03 16:05:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3887771/when-does-improper-riemann-integration-fail-for-computing-expected-values/3887777", "openwebmath_score": 0.8887693285942078, "openwebmath_perplexity": 294.0713874732837, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795121840871, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.850320563426612}}
{"url": "https://math.stackexchange.com/questions/2433888/greatest-common-divisor-of-prime-numbers", "text": "# Greatest Common Divisor of prime numbers\n\nWhat is the greatest common divisor of $2^4\\cdot3^4\\cdot25\\cdot7$ and $2\\cdot12^2\\cdot15$?\n\nI know how to find the GCD of a problem like this, but I didn't know what to do since $12$ is not a prime number. Should I say $12^2 = 2^2\\cdot2^2\\cdot3^2$?\n\n\u2022 Hint: $12^2 = (2^2*3)^2 = 2^4*3^2$ \u2013\u00a0AnalyticHarmony Sep 18 '17 at 2:22\n\u2022 that's a start. $12^2 =2^2*2^2*3^2=2^4*3^2$. And $15=3*5$ so $12^2*15=2^4*3^2*3*5 = 2^4*3^3*5$. Basically the first step is to rewrite everything as its unique prime factorization. $2^4*3^4*25*7 = 2^4*3^4*5^2*7$ and $2*12^2*15 = 2*2^4*3^2*3*5=2^4*3^3*5$. Then... you know what to do. \u2013\u00a0fleablood Sep 18 '17 at 3:49\n\nThe first one equal to $2^4*3^4*5^2*7$\n\nThe second one equals to $2^5*3^3*5$\n\nTherefore the GCD is $2^4*3^3*5=2160$\n\nYes, that is precisely what you should be doing.\n\nIn short, when you want to find the gcd of two numbers directly, you must first represent each one as a product of prime powers, before matching powers and primes.\n\nSo in our case, we can write: $$2^4 \\times 3^4 \\times \\color{blue}{25} \\times 7 = 2^4 \\times 3^4 \\times \\color{blue}{5^2} \\times 7 \\\\ 2 \\times \\color{red}{12^2} \\times \\color{green}{15} = 2 \\times \\color{red}{2^2 \\times 2^2 \\times 3^2} \\times \\color{green}{3 \\times 5} = 2^5 \\times 3^3 \\times 5$$\n\nwhere, I highlight by color the terms that are expanded on both the LHS and RHS.\n\nNow, we are permitted to compare powers directly. This gives us the answer as $2^4 \\times 3^3 \\times 5$.\n\n\u2022 Nicely explained and nice use of colors. The OP should note when you express a number and a product of prime powers, there will always be exactly one unique way to do it. So doing this will always be a failsafe that will work. \u2013\u00a0fleablood Sep 18 '17 at 3:54", "date": "2019-09-15 08:27:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2433888/greatest-common-divisor-of-prime-numbers", "openwebmath_score": 0.863879382610321, "openwebmath_perplexity": 187.9668085533342, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795114181105, "lm_q2_score": 0.861538211208597, "lm_q1q2_score": 0.8503205627666939}}
{"url": "https://math.stackexchange.com/questions/3569668/eigenvalues-of-block-toeplitz-matrix-with-toeplitz-blocks", "text": "# Eigenvalues of block Toeplitz matrix with Toeplitz blocks\n\nConsider integers $$m,n$$ and a $$m \\times m$$-block Toeplitz matrix $$A$$ consisting of two different types of blocks as follows\n\n\\begin{align} A_{mn \\times mn} &= \\begin{bmatrix} B & C & C & \\cdots & \\cdots & C \\\\ C & B & C & C & \\cdots & C \\\\ C & C & B & C & \\ddots & \\vdots \\\\ \\vdots & \\ddots & \\ddots & \\ddots & \\ddots & C \\\\ C & \\cdots & \\cdots & C & B & C \\\\ C & \\cdots & \\cdots & \\cdots & C & B \\end{bmatrix} _{mn \\times mn} , \\end{align}\n\nwhere $$B$$'s are diagonal blocks with $$B=\\frac{1}{m}I_n$$ and $$C$$'s are multiples of the all-ones matrix $$J_n$$, specifically $$C=\\frac{1}{mn}J_n$$.\n\nI want to compute the eigenvalues of $$A$$ (I am mainly interested in the value of the 2nd largest eigenvalue since it has a special meaning in graph expansion applications).\n\nNote that in my problem the following conditions also hold for $$m,n$$:\n\n\u2022 $$m$$ is odd.\n\u2022 $$n$$ is prime.\n\u2022 $$m.\n\nI have experimented with such matrices on the computer and I have observed a trend for the spectrum of $$A$$ which consists of the following eigenvalues:\n\n\u2022 $$\\lambda_1=0$$ with algebraic multiplicity $$m-1$$.\n\u2022 $$\\lambda_2=1/m$$ with algebraic multiplicity $$m(n-1)$$.\n\u2022 $$\\lambda_3=1$$ with algebraic multiplicity $$1$$.\n\nI do not claim that this is necessarily the answer but at least it was consistent for the pairs of $$m,n$$ I tried.\n\nCan you suggest how one can go and prove the above claim (if correct) or pinpoint other known results?\n\nEDIT\n\nAfter Omnomnomnom's note that $$$$A = \\frac 1{mn}\\underbrace{\\pmatrix{ 0&1&\\cdots & 1\\\\ 1&0&\\ddots&\\vdots\\\\ \\vdots&\\ddots&\\ddots&1\\\\ 1&\\cdots&1&0}}_{= C_{m \\times m}} \\otimes J_n + \\frac 1m I_{mn}$$$$\n\nI did some computation of the spectrums of the individual matrices. First, the characteristic polynomial of the all-ones $$J_n$$ is $$(\\lambda-n)\\lambda^{n-1}$$ and hence its spectrum (with the multiplicities) is $$$$\\sigma(J_n)=\\{(n,1),(0,n-1)\\}.$$$$ For $$C$$, assume that $$\\lambda_1,\\dots,\\lambda_m$$ are its eigenvalues. By the facts that $$\\mathrm{det}(C-(-1)I_m)=det(J_m)=0$$, $$C\\mathbf{1}_m=(m-1)\\mathbf{1}_m$$ and $$\\mathrm{trace}(C)=\\sum_i\\lambda_i=0$$ it turns out that $$$$\\sigma(C)=\\{(m-1,1),(-1,m-1)\\}.$$$$ Suppose that $$\\mu_1,\\dots,\\mu_n$$ are the eigenvalues of $$J_n$$ then by the Kronecker product's properties the spectrum of $$CJ_n$$ consists of the pairwise products $$\\lambda_i\\mu_j, \\forall i,j$$.\n\nYour observations are correct and hold for arbitrary $$m,n$$. It suffices to note that $$A = \\frac 1{mn}\\pmatrix{ 0&1&\\cdots & 1\\\\ 1&0&\\ddots&\\vdots\\\\ \\vdots&\\ddots&\\ddots&1\\\\ 1&\\cdots&1&0} \\otimes J_n + \\frac 1m I_{mn}$$ and use the properties of the Kronecker product.\n\nIn more detail: $$C_{m \\times m}$$ is a rank 1 update of a scalar matrix, so we find that its eigenvalues are $$-1$$ with multiplicity $$m-1$$ and $$m-1$$ with multiplicity $$1$$. On the other hand, $$J_n$$ has eigenvalues $$0$$ with multiplicty $$n-1$$ and $$n$$ with multiplicity $$1$$.\n\nIt follows that $$C \\otimes J$$ has eigenvalues $$0$$ with multiplicity $$m(n-1)$$, $$-n$$ with multiplicity $$m-1$$, and $$n(m-1)$$ with multiplicity $$1$$.\n\nFrom there, it suffices to note that $$\\lambda$$ is an eigenvalue of $$A$$ if and only if $$c \\lambda + d$$ is an eigenvalue of $$c A + dI$$.\n\n\u2022 I have made some edits based on your response (please see edit). However, the resulting sum of the Kronecker product and the diagonal matrix confuses me. How can I proceed?\n\u2013\u00a0mgus\nMar 5 '20 at 3:35\n\u2022 @mgus I'll add in what I had in mind when I have the chance Mar 5 '20 at 10:08\n\u2022 See my latest edit Mar 5 '20 at 14:21\n\u2022 Thanks for your response. Now it is clear to me except one thing: if the spectrums of $C$ and $J_n$ are indeed $\\sigma(C)=\\{(m-1,1),(-1,m-1)\\}$ and $\\sigma(J_n)=\\{(n,1),(0,n-1)\\}$, respectively then where is the eigenvalue $-n(m-1)$ of $C \\otimes J$ is coming from? I thought it would be $n(m-1)$ with multiplicity $1$. After doing the remaining of the calculations for the eigenvalues of $A$ it seems that this should be true but how? Am I missing some plus or minus somewhere?\n\u2013\u00a0mgus\nMar 5 '20 at 17:37\n\u2022 @mgus You're right about $n(m-1)$; I just had an extra minus sign there. Mar 5 '20 at 17:39", "date": "2021-12-08 16:46:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3569668/eigenvalues-of-block-toeplitz-matrix-with-toeplitz-blocks", "openwebmath_score": 1.0000097751617432, "openwebmath_perplexity": 249.14645551469715, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795091201804, "lm_q2_score": 0.8615382129861583, "lm_q1q2_score": 0.8503205625413559}}
{"url": "http://math.stackexchange.com/questions/261100/optimization-of-the-area-of-a-cross-inscribed-in-a-circle", "text": "# Optimization of the area of a cross inscribed in a circle\n\nI've really been scratching my head over this optimization problem. \"Consider a symmetric cross inscribed in a circle of radius $r$.\" The length from the center of the cross to the middle of one of its arms is $x$. Also, the angle between two line segments drawn from the cross's center to the vertices of one of its arms has a measure of $\\theta$. Here's a diagram:\n\nThere are three parts to the problem: \"(a) Write the area $A$ of the cross as a function of $x$ and find the value of $x$ that maximizes the area. (b) Write the area $A$ of the cross as a function of $\\theta$ and find the value of $\\theta$ that maximizes the area. (c) Show that the critical numbers of parts (a) and (b) yield the same maximum area. What is that area?\"\n\nSo, let me show you what I've done so far. For part (a), I decided to break the cross into two middle rectangles and two side rectangles. I saw that a middle rectangle (from the center to the top) would have an area of\n\n$$x \\cdot 2 \\sqrt{r^2 - x^2}$$\n\nusing the Pythagorean theorem. I worked out that a side rectangle (the remaining area on the right, adjacent to the middle rectangles) would have an area of\n\n$$2 \\sqrt{r^2-x^2} \\cdot \\left( x - \\sqrt{r^2 - x^2} \\right) .$$\n\nSo, the area of the cross is\n\n$$A = 2 \\bigg( x \\cdot 2 \\sqrt{r^2 - x^2} + 2 \\sqrt{r^2 - x^2} \\cdot \\Big( x - \\sqrt{r^2 - x^2} \\Big) \\bigg) = 8x \\sqrt{r^2 - x^2} - 4r^2 + 4x^2 .$$\n\nIf my math is right there (fingers crossed), then I'll take the first derivative to locate a maximum.\n\n$$A^\\prime = 8 \\sqrt{r^2 - x^2} + 8x \\left( 1 \\over 2 \\right) \\left( r^2 - x^2 \\right)^{- {1 \\over 2}} \\left( -2x \\right) + 8x.$$\n\nI was a little unsure about what to do at this point. I plugged the $A^\\prime$ equation into my graphing calculator, substituting $1^2$ for $r^2$ (for a radius of $1$). The graph crosses the $x$-axis at $x \\approx 0.85$. Substituting $2^2$ for $r^2$ (for a radius of $2$) gives me $x \\approx 1.70$. From this, I concluded that\n\n$$A^\\prime = 0 \\; \\mathbf{at} \\; x \\approx 0.85r.$$\n\nAnalysis of graphs of $A$ for various values of $r$ concludes that, indeed, maxima do appear at $x \\approx 0.85r$. So, I have the function $A$ in terms of $x$, but I'm curious: What should my final answer be for the second part of (a)? All I have is $x \\approx 0.85r$. Is that a sufficient answer?\n\nAs for part (b), I really have no idea how to write $A$ in terms of $\\theta$. I know that $\\text{area} = {1 \\over 2} b \\cdot c \\cdot \\sin A$ for triangles, but I really need help writing the area of this cross in terms of $\\theta$.\n\nPart (c) should be easy enough once I finish (b).\n\nIf you got to the end of this, I sincerely thank you for reading, and I would really appreciate an answer (and any corrections to my math). Thanks!\n\n-\nYou can simplify the formula for $A'$ if you set it to zero and multiply with $\\sqrt{1-x^2}$ (with $r=1$). In this way you get $0=1-2x^2+8x\\sqrt{1-x^2}$. This is still tricky, but you might want to try a tailor expansion until $x^2$ to simplify further. \u2013\u00a0Konstantin Schubert Dec 18 '12 at 1:07\n@Konstantin I computed $x\\sqrt{1-x^2}=2x^2-1$. Square both sides and you have a quadratic in $x^2$. The solutions are $x^2={1\\over2}\\pm{\\sqrt5\\over10}$. \u2013\u00a0David Mitra Dec 18 '12 at 1:22\n@DavidMitra aww thanks \u2013\u00a0Konstantin Schubert Dec 18 '12 at 1:28\n\nWithout loss of generality we may assume that the radius is $1$: we can scale area by $r^2$ later.\n\nBy the formula you quoted, the area of the triangle enclosed by the two lines that form the angle $\\theta$ is $\\frac{1}{2}\\sin\\theta$. The area covered by one of the two full arms of the cross is therefore $4$ times this, which is $2\\sin\\theta$.\n\nDouble this to get the sum $4\\sin\\theta$ of the areas covered by the two full arms. Unfortunately, this sum counts the area of the middle square twice. So we will need to subtract the area of that square.\n\nNote that by trigonometry, the horizontal segment at the top of the cross has length $2\\sin(\\theta/2)$. Thus the middle square has area $4\\sin^2(\\theta/2)$. Using the trigonometric identity $\\cos 2\\phi=1-2\\sin^2\\phi$, we find that the middle square has area $2-2\\cos \\theta$.\n\nSo the area of the cross is $4\\sin\\theta +2\\cos\\theta-2$. Maximizing should be straightforward.\n\nRemarks: $1.$ We don't really need calculus. Look at the equivalent problem of maximizing $4\\sin\\2\\theta+2\\cos\\theta$. Rewrite this as $$2\\sqrt{5}\\left(\\frac{2}{\\sqrt{5}}\\sin \\theta+\\frac{1}{\\sqrt{5}}\\cos\\theta\\right),$$ and let $\\psi$ be the angle whose cosine is $2/\\sqrt{5}$ and whose sine is $1/\\sqrt{5}$. Then our expression becomes $2\\sqrt{5}\\sin(\\theta+\\psi)$. The maximum possible value of the sine function is $1$. So the maximum area is $2\\sqrt{5}-2$.\n\n$2.$ We can use the above calculation to answer your question about $x$. Alternately, set the derivative equal to $0$, as you did. Manipulation will yield an explicit expression for the root.\n\n$3.$ At a certain stage you were maximizing $8x\\sqrt{r^2-x^2}-4r^2+4x^2$. Let $x=r\\sin t$. We want to maximize $8r^2\\sin t\\cos t-4r^2+4r^2\\sin^2 t$. Forget about the $r^2$ part, it is a constant multiplier. Now simplify to $8\\sin t\\cos t-4\\cos^2 t$, differentiate. (Using double angle identities to simplify first is a good idea.) We can think of this as just a technical device to ensure we end up with a simple equation.\n\n-\nCould you show me how $\\sqrt{2 - 2\\cos\\theta} = 2\\sin(\\theta/2)$? \u2013\u00a0Jackson Dec 18 '12 at 1:44\nRecall the trig identity $\\cos 2t=2\\cos^2 t-1=1-2\\sin^2 t$. Rewrite as $2\\sin^2 t=1-\\cos 2t$. Double. Get $4\\sin^2 t=2-2\\cos 2t$. Finally, let $t=\\theta/2$. \u2013\u00a0Andr\u00e9 Nicolas Dec 18 '12 at 1:47\nThank you so much! It took me a while, but I got everything figured out. I cannot thank you enough :) \u2013\u00a0Jackson Dec 18 '12 at 2:07\nI will add a last remark to my post in a few minutes. \u2013\u00a0Andr\u00e9 Nicolas Dec 18 '12 at 2:15\n\nI'm trying to solve the same problem, and got to the same concusions. The thing is, if you graph the function you get this :\n\nas you can see, the graph takes negative values when x gets smaller, but if the function represents the area of the cross, how come that area is negative ??\n\nI even made a little program to test this, here it is: http://www.khanacademy.org/cs/calculus-maxarea-circle-v00x/1512261081\n\ndoes anybody have an idea of what's going on ?\n\n-", "date": "2016-07-27 17:31:28", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/261100/optimization-of-the-area-of-a-cross-inscribed-in-a-circle", "openwebmath_score": 0.923742949962616, "openwebmath_perplexity": 163.45316261665036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9869795079712151, "lm_q2_score": 0.8615382076534742, "lm_q1q2_score": 0.8503205562882286}}
{"url": "https://www.physicsforums.com/threads/easiest-way-to-learn-exact-values-for-trig-functions.840460/", "text": "# Easiest way to learn exact values for trig functions?\n\n1. Oct 30, 2015\n\n### cmkluza\n\nI'm realizing now how much I need to know the exact values of various trigonometric functions, as shown in various trig tables. Memorizing is pretty arduous, and I'd prefer to understand it, so how can I learn all of these?\n\n2. Oct 30, 2015\n\n### Staff: Mentor\n\nA long long time ago, my math teacher made us build our own trigonometric circles. Having a geometrical representation really helped me, and the values of the functions have stuck with me ever since.\n\n3. Oct 30, 2015\n\n### cmkluza\n\nThanks! I guess I'd forgotten that these circles existed. Does this mean that if I memorize the important angles/values from the first quadrant I can use the identities of -sin(\u03b8) = sin(-\u03b8) and cos(-\u03b8) = cos(\u03b8) to figure out the values in each other quadrant?\n\n4. Oct 30, 2015\n\n### Staff: Mentor\n\nIt's even more than that. Consider for instance \u03c0/6 (30\u00b0): you get halfway up on the y axis, so obviously sin \u03c0/6 = 1/2. Likewise, for \u03c0/4, you can see it as building a square of side 1/2 (along x and y), so the diagonal is (according to Pythagoras) $\\sqrt{(1/2)^2 + (1/2)^2} = 1/\\sqrt{2} = \\sqrt{2}/{2}$. All of these geometric equivalents have really helped me.\n\n5. Oct 30, 2015\n\n### symbolipoint\n\nThat graphical circle representation is typically called, the Unit Circle, and is a very important part of Trigonometry instruction. Is is used frequently. The picture can often help because of how the circle is symmetric. Full ray rotation, 2*pi radians. Same function values as for zero degrees rotation.\n\n6. Oct 30, 2015\n\n### cmkluza\n\nWhat would you guys say are the most important angles to learn? I've seen some charts that go from 0, adding \u03c0/12 each segment, and that seems like an awful lot to go through in order to find these angles, are the three, \u03c0/6, \u03c0/4, and \u03c0/3 and their counterparts typically sufficient?\n\n7. Oct 30, 2015\n\n### symbolipoint\n\nTHE MOST COMMON you must know are 0, 2*pi, pi/2, pi/4, pi/3, pi/6.\n\n8. Oct 30, 2015\n\n### Staff: Mentor\n\nI realize that this might not be as clear as it can be. Rather, take the radius as the hypotenuse of a right triangle with two equal sides of length $a$ (along the x and y axes), so by Pythagoras $2a^2 = 1 \\Rightarrow a = 1/\\sqrt{2}$, thus $\\cos \\pi/4 = \\sin \\pi/4 = \\sqrt{2}/2$.\n\n9. Oct 30, 2015\n\n### Staff: Mentor\n\n\u03c0To expand on what symbolipoint said, the important angles in the first quadrant, and their sines and cosines are:\n$\\sin(0) = \\frac {\\sqrt{0}} 2 =\\cos(\\pi/2)$\n$\\sin(\\pi/6) = \\frac {\\sqrt{1}} 2 =\\cos(\\pi/3)$\n$\\sin(\\pi/4) = \\frac {\\sqrt{2}} 2 =\\cos(\\pi/4)$\n$\\sin(\\pi/3) = \\frac {\\sqrt{3}} 2 =\\cos(\\pi/6)$\n$\\sin(\\pi/2) = \\frac {\\sqrt{4}} 2 =\\cos(0)$\nThese values should be memorized, but you can use symmetry to figure out the sines, cosines, tangents, etc. of the counterparts of these angles in the other three quadrants.\n\n10. Oct 30, 2015\n\nStaff Emeritus\nThere are 7 special angles worth memorizing:0, 15, 30, 45, 60, 75 and 90. Sines going up are cosines coming down. Tangent=sine/cosine may be a bit slow for calculating in your head for 15 and 75, and arguably 30 and 60, but the other three are trivial. So you have at most 10 - possibly 8 - facts to remember. That shouldn't be too arduous.\n\n11. Oct 30, 2015\n\n### Student100\n\nAnother method that may help you should you forget is to draw a picture and use geometry/trig:\n\nFrom my poorly drawn circle you can see you want to find the point, (x, y) on the unit circle that intersects with a line of the radial length drawn 30 degrees from the origin. The right triangle formed by hypotenuse (which is our radius of 1) forms a special 30, 60, 90 triangle. A property of such a triangle is that the hypotenuse is twice the length of the shorter leg, and the longer leg is $\\sqrt{3}$ times the length of the shorter leg. Since the hypotenuse is one, the length of the shorter leg, $y=\\frac{1}{2}$ and the longer leg, x is $\\sqrt{3}$ times more so, $x=\\frac{\\sqrt{3}}{2}$. So the $sin(30)=\\frac{1}{2}$ while the $cosine(30)=\\frac{\\sqrt{3}}{2}$\n\nYou can repeat the same logic for 60 degrees, or just realize that the x and y's are swapped. For a 45-45-90 triangle, the legs are congruent, so the hypotenuse is $\\sqrt{2}$ the length of either leg. So lets say the hypotenuse has length $x\\sqrt{2}$ in this case, since we know the length to be one, we can simply say $x\\sqrt{2}=1$ or $x=\\frac{1}{\\sqrt{2}}$. The legs are congruent so $y=\\frac{1}{\\sqrt{2}}$.\n\nFor 15 and 75 degrees you can use $sin(45-30)$ and solve using trig rules. For 0 and 90 you can again draw a line from the origin to the corresponding point on the circle and see either, x=1 in the case of an angle of 0, or x=0 in the case of an angle of 90 or vice versus.\n\nNot sure if this will help or not, hopefully it'll allow you to re-derive these values should you forget them on a test.\n\n12. Oct 30, 2015\n\n### symbolipoint\n\nThe common reference angles that are learned on the Unit Circle come from a couple of Special Triangles.\n\n13. Nov 4, 2015\n\n### rs1n\n\nIf you look at the picture here: http://etc.usf.edu/clipart/43200/43216/unit-circle8_43216_lg.gif you will see that only the first quadrant is needed (and even then, only the angles between 0 and 45 degrees are needed (everything else can be derived by symmetry of the circle. As for the coordinates, I find it easier to remember them as:\n\n0: $\\left( \\sqrt{\\frac{4}{4}}, \\sqrt{\\frac{0}{4}}\\right)$\npi/6: $\\left( \\sqrt{\\frac{3}{4}}, \\sqrt{\\frac{1}{4}}\\right)$\npi/4: $\\left( \\sqrt{\\frac{2}{4}}, \\sqrt{\\frac{2}{4}}\\right)$\npi/3: $\\left( \\sqrt{\\frac{1}{4}}, \\sqrt{\\frac{3}{4}}\\right)$\npi/2: $\\left( \\sqrt{\\frac{0}{4}}, \\sqrt{\\frac{4}{4}}\\right)$\n\n14. Nov 5, 2015\n\n### Erland\n\nEasiest way is to use to draw a half square and a half equiliteral triangle, and use the Pythagorean theorem to obtaim sin, cos and tan for the angles in tjose triangles, i.e. 45, 30 and 60 degrees.\n\n15. Nov 5, 2015\n\n### Fredrik\n\nStaff Emeritus\nAnd once you have those, you can use them to find the values for some other angles. For example,\n\\begin{align*}\n&\\sin 15=\\sin(45-30)=\\sin 45\\cos 30-\\cos 45\\sin 30\\\\\n&\\sin 120=\\sin(2\\cdot 60)=2\\sin 60\\cos 60\n\\end{align*} The only thing worth committing to memory in my opinion, is (what Erland said) that you start with half a square and half an equilateral triangle.\n\n16. Nov 5, 2015\n\n### symbolipoint\n\nThe unit circle and its typically shown values can be easily memorized.\n\nTHIS here is essential:", "date": "2017-11-21 14:17:32", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/easiest-way-to-learn-exact-values-for-trig-functions.840460/", "openwebmath_score": 0.6334378123283386, "openwebmath_perplexity": 698.9894041085802, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9518632302488963, "lm_q2_score": 0.8933094159957173, "lm_q1q2_score": 0.8503083863214386}}
{"url": "https://math.stackexchange.com/questions/2822356/binomial-coefficient-real-life-example", "text": "Binomial coefficient real life example.\n\n\"At a university, $15$ juniors and $20$ seniors volunteer to serve as a special committee that requires $8$ members. A lottery is used to select the committee from among the volunteers. Suppose the chosen students consists of six juniors and two seniors.\n\n(a) For a test of homogeneity, what are the expected counts?\n\nThis question I understand.\n\n(b) If the selection had been random, what is the probability of the committee having exactly two seniors?\n\nMy answer was that the probability is binomial, with Binom$(k=2, n=8, p=0.57)$, but this is apparently wrong. Instead the correct answer is: $$\\frac{\\binom{20}{2}\\binom{15}{6}}{\\binom{35}{8}}$$.\n\nCan anyone explain the difference between this and standard binomial distribution?\n\n\u2022 I suppose that you made a type and that there are $15$ junior volunteers, not $13$. Am I right? \u2013\u00a0Jos\u00e9 Carlos Santos Jun 17 '18 at 7:12\n\u2022 It's not a straight binomial since the trials are not independent. Knowing that the first choice was a senior changes the probability that the second was a senior. \u2013\u00a0lulu Jun 17 '18 at 7:12\n\u2022 Ahh. Thanks! I understand now :-) \u2013\u00a0Mathe Jun 17 '18 at 7:20\n\u2022 This is the hypergeometric distribution \u2013\u00a0Lord Shark the Unknown Jun 17 '18 at 7:21\n\u2022 To be sure of useful answers, you should state null and alternative hypotheses for your 'test of homogeneity'. You don't have quite large enough expected counts for the chi-squared goodness-of-fit statistic to truly have a chi-squared distribution. // Using the hypergeometric distribution, it seems you are aiming at \"Fisher's exact test' which you can google. \u2013\u00a0BruceET Jun 17 '18 at 8:04\n\nThe issue is that the trials are not independent from each other. Having chosen a junior first, for example, changes the probability of now choosing a senior.\n\nYou must look at how many ways are there to choose two seniors (which is exactly $\\binom{20}{2}$) and how many ways are there to choose six juniors (which is exactly $\\binom{15}{6}$) and multiply them. This gives you the overall number of valid arrangements.\n\nTo get the probability, simply divide by the number of overall arrangements possible (which is $\\binom{35}{8}$).\n\n\u2022 To get a p-value you also need to include the probabilities of more extreme cases. \u2013\u00a0BruceET Jun 17 '18 at 8:47\n\nIn case it is helpful, here is Minitab output for testing two proportions. It attempts a normal test (with a warning about sample sizes being too small) and does Fisher's exact test.\n\nTest and CI for Two Proportions\n\nSample  X   N  Sample p\n1       6  15  0.400000\n2       2  20  0.100000\n\nDifference = p (1) - p (2)\nEstimate for difference:  0.3\n95% CI for difference:  (0.0193759, 0.580624)\nTest for difference = 0 (vs \u2260 0):  Z = 2.10  P-Value = 0.036\n\n* NOTE * The normal approximation may be inaccurate for small samples.\n\nFisher\u2019s exact test: P-Value = 0.051", "date": "2019-09-23 21:09:32", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2822356/binomial-coefficient-real-life-example", "openwebmath_score": 0.8433619737625122, "openwebmath_perplexity": 460.73714445440305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811611608241, "lm_q2_score": 0.8807970685907242, "lm_q1q2_score": 0.8503048968231632}}
{"url": "https://stats.stackexchange.com/questions/263931/calculation-of-transition-probabilities-of-markov-chain-problem", "text": "# Calculation of transition probabilities of Markov Chain problem\n\nIn the step of learning Markov chains, I came across few questions from assignments on UTDallas website. Finding the transition probabilities seems a bit hard for me in this one particular question only. The question is\n\n(Sec 7.3, page 355, #1) Consider a system with two components. We observe the state of the system every hour. A given component operating at time $n$ has probability $p$ of failing before the next observation at time $n + 1$. A component that was in a failed condition at time $n$ has a probability $r$ of being repaired by time $n + 1$, independent of how long the component has been in a failed state. The component failures and repairs are mutually independent events. Let $X_n$ be the number of components in operation at time $n$. The process $\\{X_n, n = 0, 1, . . .\\}$ is a discrete time homogeneous Markov chain with state space $I = \\{0, 1, 2\\}$.\n\na) Determine its transition probability matrix, and draw the state diagram.\n\nb) Obtain the steady state probability vector, if it exists\n\nAlthough the answers are given, but I cannot understand that on what basis the transition probabilities are calculated. Can someone help me in this...\n\nI had the following guesses my ownself (which mostly proved to be wrong)\n\n\\begin{bmatrix} 1-r-r^2 & r & r^2\\\\ ??& ?? & r(1-p) \\\\ p^2 & ?? & ?? \\end{bmatrix}\n\nThe actual solution that the website pose is following...\n\n\\begin{bmatrix} (1-r)^2 & 2r(1-r) & r^2\\\\ p(1-r)& pr + (1-p)(1-r) & r(1-p) \\\\ p^2 & 2p(1-p) & (1-p)^2 \\end{bmatrix}\n\n\u2022 Thanks Yes it is self-study question. What I am asking is not the solution since I am not taking this exam, I am trying to learn the probability, in general, and markov chains, in specific. Appreciate your comment @Tavrock. \u2013\u00a0Kashan Feb 25 '17 at 4:11\n\u2022 I am looking if someone can help me to learn HERE how transition probabilities are calculated...Since to my understanding, transitional probabilities are different than normal probabilities... \u2013\u00a0Kashan Feb 25 '17 at 10:59\n\u2022 Can you explain the reasoning behind your guesses? (Also, based on the last comment, do you know what a transition probability is?) \u2013\u00a0Juho Kokkala Feb 25 '17 at 14:27\n\nNote that, there are two components and each of them individually could be either in working or failure state. Then the states of the Markov chain be designated as $ff$ both components failed, $wf$ one component working and one component failed, and $ww$ both components working. Given that, the components function independently of each other.\n\n\u2022 If the chain is in state $ff$ at time $n$, at the next time point it could be in the state\n\n\u2022 $ff$ with probability $(1-r)(1-r)=(1-r)^2$, as the probability of not being repaired by next time point is $(1-r)$;\n\n\u2022 $wf$ with probability $2r(1-r)$, as it may be the first component\nthat was repaired or the the second component that was repaired, so\nthat only one of the two components will be working;\n\n\u2022 $ww$ with probability $r^2$, if both the components got repaired.\n\n\u2022 If the chain is in the state $wf$ at time $n$, then at the next time point it could be in the state\n\n\u2022 $ff$ with probability $p(1-r)$, as the working component failed and the component requiring repair was not yet repaired;\n\n\u2022 $wf$ with probability $pr+(1-p)(1-r)$, as the working component was\nnot failed and the component requiring repair is not repaired, so\nthat only one component is working, which has a probability\n$(1-p)(1-r)$; or the component working at the previous time has\nfailed and the component requiring repair has got repaired, which has a probability $pr$; mutually exclusiveness of the events results in the required probability;\n\n\u2022 $ww$ with probability $r(1-p)$, as the working component does not\nrequire repair and the failed component got repaired.\n\n\u2022 If the chain is in the state $ww$ at time $n$, then at the next time point it could be in the state\n\n\u2022 $ff$ with probability $p^2$, as both components have failed;\n\u2022 $wf$ with probability $2p(1-p)$, which is the sum of probabilities\nof two mutually exclusive events, viz., the first component failed\nand the second working or the first component working and the\nsecond component failed;\n\n\u2022 $ww$ with probability $(1-p)^2$, as both components are still\nworking.\n\nHence, the transition matrix:\n\n\\begin{equation*} P=\\begin{array}{c|ccc} &ff & wf & ww\\\\ \\hline ff & (1-r)^2& 2r(1-r) & r^2\\\\ wf & p(1-r) & pr+(1-p)(1-r) & r(1-p) \\\\ ww & p^2 & 2p(1-p) & (1-p)^2 \\end{array} \\end{equation*}\n\n\u2022 Thanks.... Helped me a lot in developing understanding and brushing off some dust over my mind \u2013\u00a0Kashan Feb 26 '17 at 2:20", "date": "2019-12-10 07:49:53", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/263931/calculation-of-transition-probabilities-of-markov-chain-problem", "openwebmath_score": 0.9372547268867493, "openwebmath_perplexity": 708.8711653335081, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692339078751, "lm_q2_score": 0.8705972818382005, "lm_q1q2_score": 0.8502855802951937}}
{"url": "https://www.physicsforums.com/threads/reduction-of-order.778044/", "text": "# Homework Help: Reduction of Order\n\n1. Oct 24, 2014\n\n### QuantumCurt\n\n1. The problem statement, all variables and given/known data\n\nGiven - $$y_1(x)=sin(2x)$$, find a second linearly independent solution to $$y''+4y=0$$\n\n2. Relevant equations\n\nI'm using reduction of order to write a second solution as a multiple of the first solution. I think I've gotten to the right answer, my main question is in how I should write the general solution of the equation.\n\n3. The attempt at a solution\n\nUsing the fact that $y_1(x)=sin(2x)$, and that $y_2(x)=vy_1(x)$ I'm writing a second solution as $$y_2(x)=vsin(2x)$$\nNow taking the first and second derivatives I get $$y_2'(x)=v'sin(2x)+2vcos(2x)$$\nand $$y_2''(x)=v''sin(2x)+4v'cos(2x)-4vsin(2x)$$\n\nNow I substitute back into the original equation and get, $$v''sin(2x)+4v'cos(2x)=0$$\n\nNow to reduce order; $z=v' and z'=v''$\n\nThen substituting - $$z'sin(2x)+4zcos(2x)=0$$\n\nWhich leads to $$\\frac{dz}{z}=-4cot(2x)dx$$\n\nIntegrating this to logarithms and then exponentiating etc. leads me to - $$z=csc^2(2x)$$\n\nSince $z=v'$ I can substitute again\n$$v=\\int csc^2(2x)dx$$\n$$v=-\\frac{1}{2}cot(2x)$$\n\nNow since $y_2(x)=vy_1(x)$, I can say that,\n$$y_2(x)=[-\\frac{1}{2}cot(2x)][sin(2x)]$$\nwhich simplifies to\n$$y_2(x)=-\\frac{1}{2}cos(2x)$$\n\nThe formula for the general solution is $y(x)=c_1y_1(x)+c_2y_2(x)$. Given this fact, my general solution is written as - $$y(x)=c_1sin(2x)-\\frac{1}{2}c_2cos(2x)$$\n\nNow...my question -\n\nSince $-\\frac{1}{2}c_1$ is really just an arbitrary constant, can I write my general solution in terms of just a constant rather than a fraction times a constant? This would give me the solution of -\n$$y(x)=c_1sin(2x)+c_2cos(2x)$$\n\nIs this essentially the same solution, or would it be better to write it in terms of the negative fractional constant?\n\nAny help would be much appreciated. :)\n\nLast edited: Oct 24, 2014\n2. Oct 24, 2014\n\n### vela\n\nStaff Emeritus\nJust absorb the -1/2 into c2.\n\n3. Oct 24, 2014\n\n### QuantumCurt\n\nThank you. I figured that was the case, but I wanted to be sure.", "date": "2018-05-25 21:02:18", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/reduction-of-order.778044/", "openwebmath_score": 0.8861240148544312, "openwebmath_perplexity": 315.4275953346494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692257588073, "lm_q2_score": 0.8705972700870909, "lm_q1q2_score": 0.8502855617236903}}
{"url": "https://mathematica.stackexchange.com/questions/166711/how-to-create-a-particular-5-x-5-square-colour-map-of-the-mean-value-of-data-poi", "text": "# How to create a particular 5 x 5 square colour map of the mean value of data points\n\nI am trying to create a specific type of colour map for some data. The data is in the form of a list with dimensions 300000 x 3, that is 300,000 sets of {x, y, n} where x and y are basically the $xy$ cartesian coordinates and n is a value either 0 or 1. Doing a list plot of the 300,000 points gives the following figure:\n\nWhat I am trying to make is a colour map of this figure, that is made up of 5 x 5 evenly sized squares (so 25 all up), in which the colour-key denotes the mean of the n values of all the points within that square. Is this possible?\n\nPlease let me know if any additional information is needed,\n\n## 1 Answer\n\nSeedRandom[1]\ndata = Join[RandomReal[1, {300000, 2}], RandomChoice[{0, 1}, {300000, 1}], 2];\nDimensions[data]\n\n\n{300000, 3}\n\nnbins = 5;\nbinlims = Through[{Floor[Min@#, .01] &, Ceiling[Max@#, .01] &}@#] & /@ Transpose[data];\n{xbins, ybins} = {##, -Subtract[##]/nbins} & @@@ Most[binlims];\nbinlists = BinLists[data, xbins, ybins, {0, 2, 2}];\nbinmeans =  Flatten /@ Map[Mean, binlists[[All, All, All, All, -1]], {-2}];\n\ncft = ChartingFindTicks[{0, nbins}, {0, 1}];\nMatrixPlot[binmeans, DataReversed -> True, ColorFunction -> \"Rainbow\",\nFrameTicks -> {{cft, cft}, {cft, cft}}]\n\n\nWith nbins = 25 we get\n\n\u2022 I take it that if I would like to increase the number of squares, I just alter that 5 in the second line? Also, I tried changing it to 25 and it looks good, but there are a lot of data ticks. Is it possible to maybe only show every second, or even third data tick? \u2013\u00a0Lagiacrus Feb 27 '18 at 18:38\n\u2022 @Lagiacrus, right; changing 5 changes the number of bins. Re ticks, please see the updated version. \u2013\u00a0kglr Feb 27 '18 at 20:17\n\u2022 So far the code appears to have been working well, but I just tried it for a similar situation in which the third element of the list can be either 0, 1, 2 or 3 (instead of the original 0 or 1) and in this case the binmeans never ends up with a value higher than 1. Is there a limit on the value that binmeans can take? \u2013\u00a0Lagiacrus Mar 3 '18 at 12:57\n\u2022 @Lagiacrus, you can change {0, 2, 2} in the definition of binlists to {0,, 4, 4} if the third list can take values in {0,1,2,3}`. \u2013\u00a0kglr Mar 3 '18 at 14:00", "date": "2020-01-28 22:03:35", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/166711/how-to-create-a-particular-5-x-5-square-colour-map-of-the-mean-value-of-data-poi", "openwebmath_score": 0.2794909179210663, "openwebmath_perplexity": 1085.739176295797, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692325496974, "lm_q2_score": 0.8705972600147105, "lm_q1q2_score": 0.8502855577984366}}
{"url": "https://math.stackexchange.com/questions/3926426/probability-of-n-same-type-pairs-from-a-heterogeneous-group", "text": "# probability of N same type pairs from a heterogeneous group\n\nThere are 16 people: 8 males and 8 females. The people are randomly divided to pairs, each pair get the exact same mission. Let $$X$$ be the number of male-male pairs.\n\n1. calculate $$P\\{X=3\\}$$\n2. calculate $$P\\{X=i\\}$$ where $$i$$ is each value $$X$$ can take\n3. find $$E[X]$$\n\nfor (1) there are $$\\binom{8}{2,2,2,1,1}$$ ways to divide the males to 3 all male pairs and $$\\binom{8}{2,2,2,1,1}$$ ways to divide the females to 3 all female pairs and there are 4 ways to pair the remaining males and females so I expected the probability to be $$P\\{X=3\\}=\\frac{\\binom{8}{2,2,2,1,1}\\cdot \\binom{8}{2,2,2,1,1} \\cdot 4}{\\binom{16}{2,2,2,2,2,2,2,2}} = \\frac{\\frac{8!}{8}\\cdot \\frac{8!}{8}\\cdot 4}{\\frac{16!}{2^8}}=\\frac{8!^2\\cdot 4\\cdot 64 \\cdot 4}{16! \\cdot 64}=\\frac{8!^2}{15!} \\approx 0.0012$$ but it is incorrect so instead I tried, I choose 6 males and pair them in $$\\binom{8}{6}\\binom{6}{2,2,2}$$ and the 2 remaining males I have $$\\binom{2}{1}\\binom{8}{1}$$ and $$\\binom{7}{1}$$ ways to pair with a female, and the remaining 6 females I have $$\\binom{6}{2,2,2}$$ ways to pair so $$\\frac{\\binom{8}{6}\\binom{6}{2,2,2}\\binom{2}{1}\\binom{8}{1}\\binom{7}{1}\\binom{6}{2,2,2}}{\\binom{16}{2,2,2,2,2,2,2,2}}=\\frac{\\frac{8!}{16}\\cdot 16 \\cdot 7\\cdot\\frac{6!}{8}}{\\frac{16!}{2^8}}=\\frac{7!\\cdot 7! \\cdot 2^8}{16!}\\approx 0.0003$$ which is also incorrect. I'm not sure how I can calculate $$P\\{X=3\\}$$ or any other $$P\\{X=i\\}$$\n\nafter reading @saulspatz's answer I found a possible solution: all possible pairing is given by $$\\binom{16}{2,2,2,2,2,2,2,2}\\cdot \\frac{1}{8!}$$ because all the pairs have the same role but the same works for each of the 3 male-male or female-female pairs, hence $$\\binom{6}{2,2,2}\\cdot \\frac{1}{3!}$$ and because the 2 male-female pairs also have the same role I get $$P\\{X=3\\}=\\frac{\\frac{\\binom{8}{6}\\binom{6}{2,2,2}}{3!}\\frac{\\binom{2}{1}\\binom{8}{1}\\binom{7}{1}}{2!}\\frac{\\binom{6}{2,2,2}}{3!}}{\\frac{\\binom{16}{2,2,2,2,2,2,2,2}}{8!}}=\\frac{\\frac{8!}{16\\cdot 3!}\\cdot \\frac{16\\cdot 7}{2!}\\cdot\\frac{6!}{8\\cdot 3!}}{\\frac{16!}{2^8 \\cdot 8!}}=\\frac{8!\\cdot 7! \\cdot 7!\\cdot 2^8}{16!\\cdot 3!\\cdot 3!\\cdot 2!}\\approx 0.1740$$ and from here calculating $$P\\{X=i\\}$$ is the same and and there is no problem finding $$E[X]$$\n\n\u2022 What is a mission and what role does it play in your question (if any)? Also, you say some results you calculated are incorrect. How do you know they are incorrect? (I am not saying they are correct, I am just pointing out that you made unsupported claims in your post.) \u2013\u00a0mathguy Nov 28 '20 at 16:34\n\u2022 @mathguy I think the mission part is so that the order of the pairs doesn't matter. I have 4 possible answers and I need to choose one \u2013\u00a0CforLinux Nov 28 '20 at 16:48\n\nThe number of ways to divide the $$16$$ people is not pairs is not $$\\binom{16}{2,2,2,2,2,2,2,2}$$ but $$\\frac1{8!}\\binom{16}{2,2,2,2,2,2,2,2}$$ To divide the people into pairs, arrange them into a line, and pair the first and second, the third and fourth, and so on. The order of the tow people in the pairs doesn't matter, so we have $$\\frac{16!}{(2!)^8}$$ but also the order of the pairs themselves doesn't matter, so we must divide by $$8!$$.", "date": "2021-06-13 09:20:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3926426/probability-of-n-same-type-pairs-from-a-heterogeneous-group", "openwebmath_score": 0.7666555047035217, "openwebmath_perplexity": 216.48648262658304, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692271169855, "lm_q2_score": 0.8705972633721708, "lm_q1q2_score": 0.8502855563478607}}
{"url": "https://math.stackexchange.com/questions/1809061/describe-the-structure-operatornamegal-mathbbq-zeta-4-mathbbq", "text": "# Describe the structure $\\operatorname{Gal}(\\mathbb{Q}(\\zeta_4)/\\mathbb{Q})$\n\nI know that if $n$ is prime then $G=\\operatorname{Gal}(\\mathbb{Q}(\\zeta_n)/\\mathbb{Q}) \\simeq \\Bbb Z_{n-1}$\n\nBut I am unsure what $G$ is when $n$ is not prime. For example when $n=4$:\n\n$\\operatorname{Gal}(\\mathbb{Q}(\\zeta_4)/\\mathbb{Q}) \\simeq \\Bbb Z_4^*$\n\nWhat is the structure of the group $\\Bbb Z_4^*$? I am guessing it is cyclic, but I do not know how to explicitly find its generators.\n\nAlso, what is the basis of $\\mathbb{Q}(\\zeta_4)$ over $\\mathbb{Q}$? Is it $\\{\\zeta, ..., \\zeta^4\\}$?\n\nFor general $n$, the units of the ring $\\Bbb Z/n\\Bbb Z$, which I\u2019ll write $(\\Bbb Z/n\\Bbb Z)^*$, do not form a cyclic group. For, when $n=p_1^{e_1}p_2^{e_2}\\cdots p_m^{e_m}$, then $(\\Bbb Z/n\\Bbb Z)^*\\cong(\\Bbb Z/p_1^{e_1}\\Bbb Z)^*\\oplus(\\Bbb Z/p_2^{e_2}\\Bbb Z)^*\\oplus\\cdots\\oplus(\\Bbb Z/p_m^{e_m}\\Bbb Z)^*$, the $p_i$ all being prime.\n\nFor odd primes $p$, $(\\Bbb Z/p^e\\Bbb Z)^*$ is always cyclic, but the structure of $(\\Bbb Z/2^e\\Bbb Z)^*$ is $C_2\\oplus C_{2^{e-2}}$, where by $C_m$ I mean an abstract cyclic group of order $m$. (The explanation of why these groups have this structure is a story for another night.)\n\n\u2022 Thank you. So if I we take $n=18=3^2 \\times 2 \\implies Gal(\\mathbb{Q}(\\zeta_{18}/\\mathbb{Q}) \\simeq \\mathbb{Z}/9\\mathbb{Z} \\oplus \\mathbb{Z}/2\\mathbb{Z}$ \u2013\u00a0amiz9 Jun 2 '16 at 13:17\n\u2022 Take $n=8=2^3 \\implies Gal(\\mathbb{Q}(\\zeta_{8}/\\mathbb{Q}) \\simeq \\mathbb{C_2} \\oplus \\mathbb{C_2}$? \u2013\u00a0amiz9 Jun 2 '16 at 13:19\n\u2022 Are these correct? This is a nice classification - can you point me to somewhere where I can read more about the generators of these groups and their structure? Do all cyclotomic extension fields have a similar structure? \u2013\u00a0amiz9 Jun 2 '16 at 13:21\n\u2022 Don\u2019t forget that (in your notation) $\\Bbb Z_2^*$ is trivial, not order two. In fact, $\\Bbb Q(\\zeta_{18})=\\Bbb Q(\\zeta_9)$. And yes, $\\Bbb Z_8^*\\cong C_2\\oplus C_2$, generators $3$ and $7$. \u2013\u00a0Lubin Jun 2 '16 at 13:32\n\u2022 Structure of $\\Bbb Q(\\zeta_n)$ should be in most books on algebraic number theory. \u2013\u00a0Lubin Jun 2 '16 at 13:35\n\nI mean, $\\zeta_4 = i$ so the extension has degree $2$, so the Galois group is a finite group of order $2$ and there's only one of those, $\\Bbb Z/2\\Bbb Z$. Like all quadratic extensions, the basis is $\\{1, i\\}= \\{1,\\sqrt{-1}\\}$. If you want the exact structure, it's mercifully simple: $i\\mapsto \\overline{i}$ is the automorphism, i.e. complex conjugation is the only one.\n\n\u2022 OK thanks, that all makes sense, except I am unsure why $\\zeta_4=i \\implies [\\mathbb{Q}(\\zeta_4) : \\mathbb{Q}]=2$.. could you help me understand this please? And is the $i \\rightarrow i$ the automorphism because $i \\notin \\mathbb{Q}$? Thanks \u2013\u00a0amiz9 Jun 2 '16 at 0:16\n\u2022 Is it because the we must take $i$ to the power of $2$ to return back to our field $\\mathbb{Q}$? \u2013\u00a0amiz9 Jun 2 '16 at 0:17\n\u2022 @amiz9 that's one way to think of it, yeah. $\\Bbb Q(i) \\cong \\Bbb Q[x]/(x^2+1)$. And look at the little bar over the $i$, it means complex conjugate. You can also think of it just as $i\\mapsto -i$. \u2013\u00a0Adam Hughes Jun 2 '16 at 0:18\n\u2022 ok thanks. So in general to find $Gal(\\mathbb{Q}(\\zeta_n)/\\mathbb{Q})$, we can calculate the minimal polynomial of $\\zeta_n$ over $\\mathbb{Q}$. The degree of this polynomial will be the order of our group. Is that right? \u2013\u00a0amiz9 Jun 2 '16 at 0:23\n\u2022 it will always be $\\varphi(n)$ so you don't really need the polynomial. Generally speaking all automorphisms are $\\zeta_n\\mapsto \\zeta_n^k$ for $k\\in\\Bbb Z/n\\Bbb Z^*$. In this case $i\\mapsto i^3 = i\\mapsto (i)^2\\cdot i = -i$ is how it works. \u2013\u00a0Adam Hughes Jun 2 '16 at 0:25\n\nIf $\\zeta_n$ is a primitive $n^{th}$ root of unity, then $$Gal(\\mathbb{Q}(\\zeta_n)/\\mathbb{Q}) \\cong (\\mathbb{Z}_n)^*.$$\n\nWhat the elements of the group look like:\n\nIf $\\sigma \\in Gal(\\mathbb{Q}(\\zeta_n)/\\mathbb{Q})$, then there is an $a \\in \\{1,2,\\dots,n\\}$ with $a$ relatively prime to $n$ such that $$\\sigma(\\zeta_n) = (\\zeta_n)^a$$\n\nFor the structure of $(\\mathbb{Z}_n)^*$, see Lubin's answer.\n\n\u2022 Thank you. What can we say about the generators that generate these groups? \u2013\u00a0amiz9 Jun 2 '16 at 13:21", "date": "2019-12-11 03:17:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1809061/describe-the-structure-operatornamegal-mathbbq-zeta-4-mathbbq", "openwebmath_score": 0.8614220023155212, "openwebmath_perplexity": 209.14411421325158, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692325496974, "lm_q2_score": 0.8705972566572503, "lm_q1q2_score": 0.8502855545193085}}
{"url": "https://www.physicsforums.com/threads/moment-of-inertia-of-a-disk-by-integration.349466/", "text": "# Moment of inertia of a disk by integration\n\n1. ### soopo\n\n226\n1. The problem statement, all variables and given/known data\n\nShow that the moment of inertia of a disk is $0.5 mr^2$.\n\n3. The attempt at a solution\n\n$$I = \\int R^2 dm$$\n\nUsing $dm = \\lambda dr$ such that $m = \\lambda r$:\n\n$$= \\int_{-r}^{r} R^2 \\lambda dr$$\n$$= \\frac { \\lambda } {3} ( 2r^3 )$$\n$$= \\frac {2} {3} (\\lambda r ) (r^2)$$\n$$= \\frac {2} {3} M R^2$$\n\nwhich should be the moment of inertia for a ring.\n\nIntegrating this from 0 to 2pii relative to the angle gives me $\\frac {4} {9} m r^3 [/tex], which is wrong. How can you calculate the moment of inertia for a disk? 2. ### jdwood983 383 If you have a disc of radius $$r$$, how can you integrate from $$-r$$ to $$r$$? 3. ### jdwood983 383 The infinitesimal change in mass is given by $$dm=m\\frac{dA}{A}=m\\frac{2\\pi r\\,dr}{\\pi R^2}=\\frac{2m}{R^2}r\\,dr$$ If you use that in your integral and integrate from $$0$$ to $$R$$, you should get the desired result. 4. ### soopo 226 Your result gives me a wrong result: $$I = \\int R^2 dm$$ $$= \\int R^2 \\frac { 2m } {R^2} r dr$$ $$= \\int_{0}^{r} 2mr dr$$ $$= [ m r^2 ]^{r}_{0}$$ $$= mr^2$$ The result shoud be $$I = .5 mr^2$$ 5. ### soopo 226 I set the null point to the center of the circle such that I am integrating from -r to r. I am not sure why I cannot do that. 6. ### jdwood983 383 If you set the origin to the center of the circle (which you should always try to do), the smallest value that $$r$$ can be is 0. So it is physically impossible to integrate from $$-r$$ to $$r$$, that is why you can't do it. 7. ### soopo 226 I am thinking of setting an axis which goes through the origin such that the zero point of the axis is at the origin. Going to right means to go towards $$r$$, while going to left means towards $$-r$$. Perhaps, you are thinking the situation in a polar coordinate system in which case you cannot have negative $$-r$$. I feel that it is possible to integrate from $$-r$$ to $$r$$ in a cartesian coordinate system. 8. ### jdwood983 383 If you want Cartesian coordinates, then you'll need two integrals: one over $$x$$ and one over $$y$$. While technically you have two integrals in polar, $$r\\, \\mathrm{and}\\, \\theta$$, one is already done for you and reduces the integration to just one term: $$r$$. This problem is by far easier in polar coordinates: $$\\begin{array}{ll}I&=\\int_0^R r^2\\frac{2m}{R^2}rdr \\\\ &=\\frac{2m}{R^2}\\int_0^Rr^3dr \\\\ &=\\frac{2m}{R^2}\\left(\\frac{R^4}{4}-0\\right) \\\\ &=\\frac{2m}{R^2}\\cdpt\\frac{R^4}{4} \\\\ &=\\frac{1}{2}mR^2$$ 9. ### soopo 226 If you use symmetry, it is enough to consider only the first quadrant that is where x > 0 and y > 0 such that four of these quadrants form the area of the disk. You do not get the x- and y -coordinates easily from the definition of the moment of inertia. You would get $$I = \\int (x^2 + y^2) dm \\\\ &= \\int (x^2 + y^2) m \\frac { 2r } {R^2} dr$$ The calculations seem to get challenging, since we need to use Pythogoras such that $$r = \\sqrt{ x^2 + y^2 }$$ which implies $$dr = \\frac { 1 } { \\sqrt {x^2 + y^2} } * 2x$$ We can get similarly the relation relative to $$y$$. The next step is not fun at all: $$I = \\int_{0}^{1} \\sqrt {x^2 +y^2} (x+y) x dx$$, where I assume that [itex] R^2 = (x + y)^2 = (1 + 1)^2 = 4$, since it is the maximum radius.\nThis way the two 2s cancel out.\n\nI do not even know how to integrate this!\n\nPolar coordinate system really seems to be better in this case.\n\nLast edited: Oct 27, 2009\n10. ### jdwood983\n\n383\nNot quite. The integral you need is given by\n\n$$I=\\frac{m}{A}\\int_{-R}^R\\int_{-\\sqrt{R^2-x^2}}^{\\sqrt{R^2-x^2}}\\left(x^2+y^2\\right)dydx$$\n\n$$I=\\frac{m}{\\pi R^2}\\int_{-R}^R \\frac{2\\sqrt{R^2-x^2}\\left(R^2+2x^2\\right)}{3}dx$$\n\n$$I=\\frac{m}{\\pi R^2}\\cdot\\frac{\\pi R^4}{2}$$\n\n$$I=\\frac{mR^2}{2}$$\n\nFor most moment of inertia problems, spherical or cylindrical coordinates are the best.\n\n11. ### soopo\n\n226\nHow did you solve this part?\n\nIt has taken my some effort in trying to solve it by hand.\n\n12. ### jdwood983\n\n383\nThere's a few extra steps between the two lines, like making a change of variables. But to be honest I used Mathematica and just wrote the lines because I forget what changes needed to be made. While it may be good to know the form of the equation, I'm not sure you would need the solution since it is far easier to do it in polar coordinates than in Cartesian coordinates.", "date": "2015-02-28 13:57:35", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/moment-of-inertia-of-a-disk-by-integration.349466/", "openwebmath_score": 0.9997972846031189, "openwebmath_perplexity": 1002.6747776149896, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9766692305124306, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8502855543852338}}
{"url": "http://3429851863.srv040042.webreus.net/ryefz16/symmetric-relation-graph-12be00", "text": "So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. This means drawing a point (or small blob) for each element of X and joining two of these if the corresponding elements are related. Explore anything with the first computational knowledge engine. So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. The symmetric relations on nodes are isomorphic Thus, symmetric relations and undirected \u2026 directed graph of R. EXAMPLE: Let A = {1,2,3} and R = {(1,3), (2,1), (2,3), (3,2)} be represented by the. Terminology: Vocabulary for graphs often different from that for relations. PROOF. Terminology: Vocabulary for graphs often different from that for relations. Symmetric and antisymmetric (where the only way a can be related to b and b be related to a is if a = b) are actually independent of each other, as these examples show. Symmetric Division Deg Energy of a Graph K. N. Prakash a 1 , P. Siva K ota Red dy 2 , Ismail Naci Cangul 3,* 1 Mathematics, Vidyavardhaka College of Engineering, Mysuru , India Draw each of the following symmetric relations as a graph.' However, it is still challenging for many existing methods to model diverse relational patterns, es-pecially symmetric and antisymmetric relations. This module exposes the implementation of symmetric binary relation data type. Discrete Mathematics Questions and Answers \u2013 Relations. In antisymmetric relation, there is no pair of distinct or dissimilar elements of a set. For example, the relation $$a\\equiv b\\text{ (mod }3\\text{)}$$ for a few values: Note: there's no requirement that the vertices be connected to one another: the above figure is a single graph with 11 vertices. 1, April 2004, pp. Conversely, if R is a symmetric relation over a set X, one can interpret it as describing an undirected graph with the elements of X as the vertices and the pairs in R as the edges. Suppose f: R !R is de ned by f(x) = bx=2c. Neha Agrawal Mathematically Inclined 172,807 views with the rooted graphs on nodes. I Undirected graphs ie E is a symmetric relation Why graphs I A wide range of. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. 5 shows the SLGS operator\u2019s operation. Suppose we also have some equivalence relation on these objects. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) \u2208 R for every a \u2208 ASymmetricRelation is symmetric,If (a, b) \u2208 R, then (b, a) \u2208 RTransitiveRelation is transitive,If (a, b) \u2208 R & (b, c) \u2208 R, then (a, c) \u2208 RIf relation is reflexive, symmetric and transitive,it is anequivalence relation A symmetric relation can be represented using an undirected graph. Converting a relation to a graph might result in an overly complex graph (or vice-versa). And similarly with the other closure notions. directed graph. Example # 2. may or may not have a property , such as reflexivity, symmetry, or transitivity. Symmetric with respect to x-axis Algebraically Because 2 x 2 + 3 (\u2212 y) 2 = 16 is equivalent to 2 x 2 + 3 y 2 = 16, the graph is symmetric with respect to x-axis. Learn its definition with examples and also compare it with symmetric and asymmetric relation \u2026 This definition of a symmetric graph boils down to the definition of an unoriented graph, but it is nevertheless used in the math literature. Robb T. Koether (Hampden-Sydney College) Re\ufb02exivity, Symmetry, and Transitivity Mon, Apr 1, 2013 12 / 23 The Graph of the Symmetric \u2026 consists of two real number lines that intersect at a right angle. 2. definition, no element of. Symmetry, along with reflexivity and transitivity, are the three defining properties of an equivalence relation. One way to conceptualize a symmetric relation in graph theory is that a symmetric relation is an edge, with the edge's two vertices being the two entities so related. For undirected graph, the matrix is symmetric since an edge { u , v } can be taken in either direction. Notice the previous example illustrates that any function has a relation that is associated with it. Write the equivalence class(es) of the bit string 001 for the equivalence relation R on S. subject: discrete mathematics Symmetric relations in the real world include synonym, similar_to. What is the equation of the quadratic in the form y = a(x - r)(x - s) knowing that the y-intercept is (0, -75)? Why study binary relations and graphs separately? I Undirected graphs, i.e., E is a symmetric relation. A symmetric relation is a type of binary relation. Its graph is depicted below: Note that the arrow from 1 to 2 corresponds to the tuple , whereas the reverse arrow from to corresponds to the tuple . Let 0be a non-edge-transitive graph. DIRECTED GRAPH OF AN IRREFLEXIVE RELATION: Let R be an irreflexive relation on a set A. Any relation R in a set A is said to be symmetric if (a, b) \u2208 R. This implies that $(b, a) \u2208 R$ In other words, a relation R in a set A is said to be in a symmetric relationship only if every value of a,b \u2208 A, (a, b) \u2208 R then it should be (b, a) \u2208 R. Symmetric Division Deg Energy of a Graph K. N. Prakash a 1 , P. Siva K ota Red dy 2 , Ismail Naci Cangul 3,* 1 Mathematics, Vidyavardhaka College of Engineering, Mysuru , India Notice the previous example illustrates that any function has a relation that is associated with it. , v n , this is an n \u00d7 n array whose ( i , j )th entry is a ij = ( 1 if there is an edge from v i to v j 0 otherwise . This is distinct from the symmetric closure of the transitive closure. This phenomenon causes subsequent tasks, e.g. A relation R is irreflexive if the matrix diagonal elements are 0. Formally, a binary relation R over a set X is symmetric if: If RT represents the converse of R, then R is symmetric if and only if R = RT. For example, a graph might contain the following triples: First, this is symmetric because there is $(1,2) \\to (2,1)$. For a relation R in set A Reflexive Relation is reflexive If (a, a) \u2208 R for every a \u2208 A Symmetric Relation is symmetric, If (a, b) \u2208 R, then (b, a) \u2208 R Transitive Relation is transitive, If (a, b) \u2208 R & (b, c) \u2208 R, then (a, c) \u2208 R If relation is reflexive, symmetric and transitive, it is an equivalence relation . Walk through homework problems step-by-step from beginning to end. It's also the definition that appears on French wiktionnary. You can use information about symmetry to draw the graph of a relation. 12-15. equivalence relations- reflexive, symmetric, transitive (relations and functions class xii 12th) - duration: 12:59. Thus, symmetric relations and undirected graphs are combinatorially equivalent objects. Edges that start and end at the same vertex are called loops. However, there is a general phenomenon in most of KGEs, as the training progresses, the symmetric relations tend to zero vector, if the symmetric triples ratio is high enough in the dataset. Let\u2019s understand whether this is a symmetry relation or not. From MathWorld--A Wolfram Web Resource. A relation R is irreflexive if there is no loop at any node of directed graphs. 6 4 2-2-4-6-5 5 Figure 1-x1-y1 y1 x1 y = k x; k > 0 P Q. Fig. The rectangular coordinate system A system with two number lines at right angles specifying points in a plane using ordered pairs (x, y). In mathematics, an inverse function (or anti-function) is a function that \"reverses\" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, i.e., g(y) = x if and only if f(x) = y. Let 0have n vertices, and let 00be the hull of 0. In \u00a75, using the analytic approach, we identify the Cheeger constant of a symmetric graph with that of the quotient graph, Theorem 1.3. 1. Symmetry can be useful in graphing an equation since it says that if we know one portion of the graph then we will also know the remaining (and symmetric) portion of the graph as well. Why graphs? You should use the non-internal module Algebra.Graph.Relation.Symmetric instead. The symmetric relations on nodes are isomorphic with the rooted graphs on nodes. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. 2-congruence (n,r)-congruence. Knowledge-based programming for everyone. I undirected graphs ie e is a symmetric relation why. For example, a graph might contain the following triples: First, this is symmetric because there is $(1,2) \\to (2,1)$. 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Relation for pair ( a, represented by a di-graph between distinct nodes an! 98 - 112 out symmetric relation graph 113 pages and undirected graphs are combinatorially equivalent objects by R.,!\n\nMarkets Of Trajan, Tv Shows Leaving Hulu, Florence Augusta Lewis Cause Of Death, Bus 89 To Changi Village, Bridgestone Golf Balls Fitting, Lmu Basketball Schedule, What Programs Are On Pbs Tonight, Barbie Fashionistas Face Molds, Chaitanya Girl Name Meaning,", "date": "2022-07-06 20:27:17", "meta": {"domain": "webreus.net", "url": "http://3429851863.srv040042.webreus.net/ryefz16/symmetric-relation-graph-12be00", "openwebmath_score": 0.5653567910194397, "openwebmath_perplexity": 896.4036357069527, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9766692264378963, "lm_q2_score": 0.8705972583359805, "lm_q1q2_score": 0.8502855508379554}}
{"url": "https://primer-computational-mathematics.github.io/book/c_mathematics/numerical_methods/14_ill_conditioning_errors.html", "text": "# Ill-conditioning and roundoff errors#\n\nNumerical Methods\n\n## Ill-conditioned matrices#\n\nThe conditioning (or lack of, i.e. the ill-conditioning) of matrices we are trying to invert is incredibly important for the success of any algorithm.\n\nAs long as the matrix is non-singular, i.e. $$\\det(A)\\ne 0$$, then an inverse exists, and a linear system with that $$A$$ has a unique solution. What happens when we consider a matrix that is nearly singular, i.e. $$\\det(A)$$ is very small?\n\nWell smallness is a relative term and so we need to ask the question of how large or small $$\\det(A)$$ is compared to something. That something is the norm of the matrix.\n\nMatrices come in all shape and sizes, and their determinants come in all kinds of values. We know that a ill conditioned matrix has a determinant that is small in absolute terms, but the size of determinants is a relative thing, and we need some kind of comparison to determine what is \u201csmall\u201d and what is \u201clarge\u201d. Thus, we can create such a reference calculating the norms of the matrix. In this notebook, we will explore how to find the norm and how does the norm relate to the ill conditioning of the matrix.\n\n## Vector norms#\n\nJust as for vectors $$\\pmb{v}$$ (assumed as a $$n\\times 1$$ column vector) where we have multiple possible norms to help us decide quantify the magnitude of a vector:\n\n$\\begin{split} ||\\pmb{v}||_2 = \\sqrt{v_1^2 + v_2^2 + \\ldots + v_n^2} = \\left(\\sum_{i=1}^n v_i^2 \\right)^{1/2}, \\quad{\\textrm{the two-norm or Euclidean norm}}\\\\\\\\\\\\ ||\\pmb{v}||_1 = |v_1| + |v_2| + \\ldots + |v_n| = \\sum_{i=1}^n |v_i|, \\quad{\\textrm{the one-norm or taxi-cab norm}}\\\\\\\\\\\\ ||\\pmb{v}||_{\\infty} = \\max\\{|v_1|,|v_2|, \\ldots, |v_n| = \\max_{i=1}^n |v_i|,\\quad{\\textrm{the max-norm or infinity norm}} \\end{split}$\n\n## Matrix norms#\n\nWe can define measures of the size of matrices, e.g. for $$A$$ which for complete generality we will assume is of shape $$m\\times n$$:\n\n$\\begin{split} ||A||_F = \\left(\\sum_{i=1}^m \\sum_{j=1}^n A_{ij}^2 \\right)^{1/2}, \\quad{\\textrm{the matrix Euclidean or Frobenius norm}}\\\\\\\\\\\\ ||A||_{\\infty} = \\max_{i=1}^m \\sum_{j=1}^n|A_{i,j}|, \\quad{\\textrm{the maximum absolute row-sum norm}}\\\\\\\\\\\\ \\end{split}$\n\nNote that while these norms give different results (in both the vector and matrix cases), they are consistent or equivalent in that they are always within a constant factor of one another (a result that is true for finite-dimensional or discrete problems as here). This means we don\u2019t really need to worry too much about which norm we\u2019re using.\n\nLet\u2019s evaluate some examples.\n\nimport numpy as np\nimport scipy.linalg as sl\n\nA = np.array([[10., 2., 1.],\n[6., 5., 4.],\n[1., 4., 7.]])\nprint(\"A =\", A)\n\n# The Frobenius norm (default)\n# equivalent to sl.norm(A)\nprint(\"SciPy norm = \", sl.norm(A, 'fro'))\n\n# The maximum absolute row-sum\nprint(\"Maximum absolute row-sum = \", sl.norm(A,np.inf))\n\n# The maximum absolute column-sum\nprint(\"Maximum absolute column-sum\", sl.norm(A,1))\n\n# The two-norm - note not the same as the Frobenius norm\n# also termed the spectral norm\nprint(\"SciPy spectral norm =\", sl.norm(A,2))\n\n# Spectral norm definition\nprint(\"Spectral norm by hand =\", np.sqrt(np.real((np.max(sl.eigvals( A.T @ A))))))\n\nA = [[10.  2.  1.]\n[ 6.  5.  4.]\n[ 1.  4.  7.]]\nSciPy norm =  15.748015748023622\nMaximum absolute row-sum =  15.0\nMaximum absolute column-sum 17.0\nSciPy spectral norm = 13.793091098640064\nSpectral norm by hand = 13.793091098640065\n\n\n## Norm implementation#\n\nWe will write some code to explicitly compute the two matrix norms defined mathematically above (i.e. the Frobenius and the maximum absolute row-sum norms) and compare against the values found above using in-built scipy functions.\n\ndef frob(A):\nm, n = A.shape\nsqusum = 0.\nfor i in range(m):\nfor j in range(n):\nsqusum += A[i,j]**2\nreturn np.sqrt(squsum)\n\ndef mars(A):\nm, n = A.shape\nmaxarsum = 0.\nfor i in range(m):\narsum = np.sum(np.abs(A[i]))\nmaxarsum = arsum if arsum > maxarsum else maxarsum\nreturn maxarsum\n\nA = np.array([[10., 2., 1.],\n[6., 5., 4.],\n[1., 4., 7.]])\n\nprint(\"A =\", A)\nprint(\"Are our norms the same as SciPy?\",\nfrob(A) == sl.norm(A,'fro') and mars(A) == sl.norm(A,np.inf))\n\nA = [[10.  2.  1.]\n[ 6.  5.  4.]\n[ 1.  4.  7.]]\nAre our norms the same as SciPy? True\n\n\n## Matrix conditioning#\n\nThe (ill-)conditioning of a matrix is measured with the matrix condition number:\n\n$\\textrm{cond}(A) = |A||A^{-1}|.$\n\nIf this is close to one then $$A$$ is termed well-conditioned; the value increases with the degree of ill-conditioning, reaching infinity for a singular matrix.\n\nLet\u2019s evaluate the condition number for the matrix above.\n\nA = np.array([[10., 2., 1.],[6., 5., 4.],[1., 4., 7.]])\n\nprint(\"A =\", A)\nprint(\"SciPy cond(A) =\", np.linalg.cond(A))\nprint(\"Default condition number uses matrix two-norm =\", sl.norm(A,2)*sl.norm(sl.inv(A),2))\nprint(\"sl.norm(A,2)*sl.norm(sl.inv(A),2) =\", sl.norm(A,2)*sl.norm(sl.inv(A),2))\nprint(\"SciPy Frobenius cond(A) = \", np.linalg.cond(A,'fro'))\nprint(\"sl.norm(A,'fro')*sl.norm(sl.inv(A),'fro') =\", sl.norm(A,'fro')*sl.norm(sl.inv(A),'fro'))\n\nA = [[10.  2.  1.]\n[ 6.  5.  4.]\n[ 1.  4.  7.]]\nSciPy cond(A) = 10.71337188134679\nDefault condition number uses matrix two-norm = 10.71337188134679\nsl.norm(A,2)*sl.norm(sl.inv(A),2) = 10.71337188134679\nSciPy Frobenius cond(A) =  12.463616561943587\nsl.norm(A,'fro')*sl.norm(sl.inv(A),'fro') = 12.463616561943585\n\n\nThe condition number is expensive to compute, and so in practice the relative size of the determinant of the matrix can be gauged based on the magnitude of the entries of the matrix.\n\n### Example#\n\nWe know that a singular matrix does not result in a unique solution to its corresponding linear matrix system. But what are the consequences of near-singularity (ill-conditioning)?\n\nConsider the following example\n\n$\\left( \\begin{array}{cc} 2 & 1 \\\\ 2 & 1 + \\epsilon \\\\ \\end{array} \\right)\\left( \\begin{array}{c} x \\\\ y \\\\ \\end{array} \\right) = \\left( \\begin{array}{c} 3 \\\\ 0 \\\\ \\end{array} \\right)$\n\nWhen $$\\epsilon=0$$ the two columns/rows are not linear independent, and hence the determinant of this matrix is zero, the condition number is infinite, and the linear system does not have a solution (as the two equations would be telling us the contradictory information that $$2x+y$$ is equal to 3 and is also equal to 0).\n\nLet\u2019s consider a range of values $$\\epsilon$$ and calculate matrix deteterminant and condition number:\n\nA = np.array([[2.,1.],\n[2.,1.]])\nb = np.array([3.,0.])\nprint(\"Matrix is singular, det(A) = \", sl.det(A))\n\nfor i in range(3):\nA[1,1] += 0.001\nepsilon = A[1,1]-1.0\nprint(\"Epsilon = %g, det(A) = %g, cond(A) = %g.\" % (epsilon, sl.det(A), np.linalg.cond(A)),\n\"inv(A)*b =\", sl.inv(A) @ b)\n\nMatrix is singular, det(A) =  0.0\nEpsilon = 0.001, det(A) = 0.002, cond(A) = 5001. inv(A)*b = [ 1501.5 -3000. ]\nEpsilon = 0.002, det(A) = 0.004, cond(A) = 2501. inv(A)*b = [  751.5 -1500. ]\nEpsilon = 0.003, det(A) = 0.006, cond(A) = 1667.67. inv(A)*b = [  501.5 -1000. ]\n\n\nWe find for $$\\epsilon=0.001$$ that $$\\det(A)=0.002$$ (i.e. quite a lot smaller than the other coefficients in the matrix) and $$\\textrm{cond}(A)\\approx 5000$$.\n\nChange to $$\\epsilon=0.002$$ causes 100% change in both components of the solution. This is the consequence of the matrix being ill-conditioned - we should not trust the numerical solution to ill-conditioned problems.\n\nA way to see this is to recognise that computers do not perform arithmetic exactly - they necessarily have to truncate numbers at a certain number of significant figures, performing multiple operations with these truncated numbers can lead to an erosion of accuracy. Often this is not a problem, but these so-called roundoff errors in algorithms generating $$A$$, or operating on $$A$$ as in Gaussian elimination, will lead to small inaccuracies in the coefficients of the matrix. Hence, in the case of ill-conditioned problems, will fall foul of the issue seen above where a very small error in an input to the algorithm led to a far larger error in an output.\n\n## Roundoff errors#\n\nAs an example, consider the mathematical formula\n\n$f(x)=(1-x)^{10}.$\n\nWe can relatively easily expand this out by hand\n\n$f(x)=1- 10x + 45x^2 - 120x^3 + 210x^4 - 252x^5 + 210x^6 - 120x^7 + 45x^8 - 10x^9 + x^{10}.$\n\nMathematically these two expressions for $$f(x)$$ are identical; when evaluated by a computer different operations will be performed, which should give the same answer. For numbers $$x$$ away from $$1$$ these two expressions do return (pretty much) the same answer.\n\nHowever, for $$x$$ close to 1 the second expression adds and subtracts individual terms of increasing size which should largely cancel out, but they don\u2019t to sufficient accuracy due to round off errors; these errors accumulate with more and more operations, leading a loss of significance.\n\nimport matplotlib.pyplot as plt\n\ndef f1(x):\nreturn (1. - x)**10\n\ndef f2(x):\nreturn (1. - 10.*x + 45.*x**2 - 120.*x**3 +\n210.*x**4 - 252.*x**5 + 210.*x**6 -\n120.*x**7 + 45.*x**8 - 10.*x**9 + x**10)\n\nxi = np.linspace(0, 2, 1000)\n\nfig, axes = plt.subplots(1, 3, figsize=(14, 3))\nax1 = axes[0]\nax2 = axes[1]\nax3 = axes[2]\n\nax1.plot(xi, f1(xi), label = \"unexpanded\")\nax1.plot(xi, f2(xi), label = \"expanded\")\nax1.legend(loc=\"best\")\nax1.set_ylabel(\"$f(x)$\", fontsize=14)\n\nax2.plot(xi, 1.-f1(xi)/f2(xi) * 100, label=\"Relative\\ndifference\\nin %\")\nax2.legend(loc=\"best\")\nax2.set_xlabel(\"x\", fontsize=14)\nax2.set_ylabel(r\"$1-\\frac{unexpanded}{expanded}$\", fontsize=14)\n\nax3.set_xlim(0.75, 1.25)\nax3.plot(xi, 1.-f1(xi)/f2(xi) * 100, label=\"Relative\\ndifference\\nin %\")\nax3.legend(loc=\"best\")\nax3.set_ylabel(r\"$1-\\frac{unexpanded}{expanded}$\", fontsize=14)\n\nplt.suptitle(\"Comparison of $(1-x)^{10}$ expansion\", fontsize=14)\n\nAs we can see on the graph, for most of the domain, i.e. far away from 1.0, the expansion is almost the same as the unexpanded version. Near $$x=1$$, the expansion creates huge errors in terms of relative difference.", "date": "2023-02-04 15:00:49", "meta": {"domain": "github.io", "url": "https://primer-computational-mathematics.github.io/book/c_mathematics/numerical_methods/14_ill_conditioning_errors.html", "openwebmath_score": 0.8129684925079346, "openwebmath_perplexity": 1859.261040380257, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109534209825, "lm_q2_score": 0.8633916064586998, "lm_q1q2_score": 0.8502775111322659}}
{"url": "https://math.stackexchange.com/questions/1373760/logic-quantifier-position-nuances", "text": "# Logic quantifier position nuances\n\nI've been learning about translating English to logic and vice versa and I have to say that I find this quite difficult.\n\nOn a paper that I'm examining , I've across the following logic formula:\n\n$\\forall x (circle(x) \\Rightarrow \\exists y (circle(y) \\land above(x,y)))$\n\nEach circle has a circle somewhere below it\n\nWith predicates\n\n$circle(x)$ - $x$ is a circle.\n\n$above(x, y)$ - $x$ is above $y$.\n\nI started to wonder what the nuance is if the exists quantifier was moved outside of the entire bracket like this:\n\n$\\forall x \\exists y (circle(x) \\Rightarrow circle(y) \\land above(x,y))$\n\nBut my feeling is that this formula I says the same thing as the former sentence but I also think it doesn't as I'm unsure of the meaning in natural English.\n\nFormer formula: For all $x$ , if $x$ is a circle then there exists a $y$ such that $y$ is a circle and it is above $x$.\n\nLatter formula: For all $x$ , there exists a $y$ such that if $x$ is a circle then $y$ is a circle and is above $x$.\n\nMy question is , what the difference between these two formulas in logic and natural English?\n\nTranslating from logic to English is quite hard I think and even harder from English to logic.\n\nAre there any tips for translating from logic to English and vice versa? How do you work out the nuance if two formulas seem alike when you read them?\n\nI've been learning about translating English to logic and vice versa and I have to say that I find this quite difficult.\n\nIndeed. \u00a0 It is difficult. \u00a0 Logical expressions in native languages are often very imprecise, and formal mathematics is all about precision. \u00a0 When converting to and from mathematical expressions you will encounter many trips and traps.\n\nEvery student has trouble with this. \u00a0 Don't be discouraged; just be careful. \u00a0 It takes practice to master, that's all.\n\nIn this case your intuition is quite correct. \u00a0 The two expressions are correct. \u00a0 Generally speaking, as long as $y$ does not occur free in $P(x)$, then we have the following equivalence: $$\\forall x\\Big( P(x) \\to \\exists y \\big(Q(x,y)\\big)\\big) \\iff \\forall x\\exists y \\Big(P(x)\\to Q(x,y)\\Big)$$\n\nIt is important that you don't change the order of nesting the quantifiers when shifting to and from PreNex form of the expression. \u00a0 Be careful about this.\n\nNote also, however, the following equivalence.\n\n$$\\forall x \\Big(\\exists y\\big(Q(x,y)\\big) \\to P(x)\\Big) \\iff \\forall x \\color{red}{\\forall} y\\Big( Q(x,y) \\to P(x)\\Big) \\\\ \\Updownarrow \\\\ \\forall x \\Big(\\neg \\exists y\\big(Q(x,y)\\big) \\vee P(x)\\Big) \\iff \\forall x \\color{red}{\\forall} y\\Big( \\neg Q(x,y) \\vee P(x)\\Big)$$\n\nAs you see, the placement of the quantifier in the implication matters. \u00a0 Be careful.\n\nConsider the first formula.\n\n$\\forall X (circle(X) \\implies \\exists Y(circle(Y) \\land above(X, Y))$\n\nLet $x$ be an instantiation of $X$.\n\n$circle(x) \\implies \\exists Y(circle(Y) \\land above(x, Y))$\n\nLet $y$ be an instantiation of $Y$.\n\n$circle(x) \\implies (circle(y) \\land above(x,y))$\n\nSo clearly we have a $y$ such that\n\n$circle(x) \\implies (circle(y) \\land above(x,y))$\n\nSo we may existentially generalize this so that\n\n$\\exists Y (circle(x) \\implies circle(Y) \\land above(x, Y))$\n\nAnd since no conditions were placed on $x$, we may universally generalize so that\n\n$\\forall X \\exists Y (circle(X) \\implies circle(Y) \\land above(X, Y))$\n\nTo answer your question, in general yes $\\forall x \\exists y(A(x) \\implies B(x,y)) \\equiv \\forall x (A(x) \\implies \\exists y B(x, y))$ so long as A does not depend on y; however, it is also VERY important that you do not changer the order in which the quantifiers occur.\n\nAny formula in logic is equivalent to a formula in Pre-Nex normal form, which begins with a string of quantifiers, followed by a quantifier-free expression. The two formulas you give are equivalent. I think there must be a published algorithm .Even professional mathematicians can find parsing a formula to be less than easy.\n\n\u2022 So is it safe to say that since they are equivalent they have the same meaning in natural English? How did you figure out that they are equivalent? By using equivalences? $\\forall x (A \\Rightarrow \\exists y B)$ and $\\forall x \\exists y (A \\Rightarrow B)$ ? \u2013\u00a0Nubcake Jul 25 '15 at 19:33", "date": "2019-09-20 21:10:59", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1373760/logic-quantifier-position-nuances", "openwebmath_score": 0.8710659146308899, "openwebmath_perplexity": 184.92581725486824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9848109489630492, "lm_q2_score": 0.8633916082162403, "lm_q1q2_score": 0.8502775090141688}}
{"url": "https://math.stackexchange.com/questions/2793350/mean-distance-between-2-points-in-adjacent-voxels-cubes/2796405", "text": "# Mean distance between 2 points in adjacent voxels (cubes)\n\nLet's imagine our central unit voxel $$V_1=[0,1]^3$$.\n\nWe allow the second one to be in $$V_2=[-1,2]^3 \\setminus [0,1]^3~$$, the surrounding voxel (center removed).\n\nMy question:\n\nWhat is the mean distance between a point in $$V_1$$ and a point in $$V_2$$?\n\nOf course, the direct approach would be to write an integral:\n\n$$\\frac{1}{1^3}\\frac{1}{3^3-1^3} \\times$$ $$\\iiint_{(x_1,y_1,z_1)\\in V_1} \\iiint_{(x_2, y_2,z_2) \\in V_2} \\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\\mathrm{d}x_{1,2} \\mathrm{d}y_{1,2}\\mathrm{d}z_{1,2}$$\n\nbut the one-variable function does not seem to admit a nice integral.\n\nHas anyone gone through this calculation and would you be able to suggest any helpful manipulation before I dive into a numeric simulation?\n\nThanks.\n\n\u2022 Small update : succeeding a numeric simulation, I got $\\approx 1.8288$. If this rings a bell to you or if it roughly seems odd, please tell me so that I can correct. I expected a smaller number but the distribution's boundaries I obtained seemed consistent with the problem. \u2013\u00a0Wyllich May 24 '18 at 15:21\n\u2022 An analytic solution is indeed available (using known results), but before I post an answer I need to make sure: if I understand your setup correctly, the volume of $V_2$ is $3^3 - 1$ so shouldn't the normalization constant be $\\frac1{26}$? \u2013\u00a0Lee David Chung Lin May 24 '18 at 18:46\n\u2022 Yes, the volume is equal to that fraction. I will correct that right away. \u2013\u00a0Wyllich May 24 '18 at 19:37\n\u2022 Can you get access (by whatever means you feel appropriate) to the paper by Robbins and Bolis cited in the wiki page for Robbbins' constant? If not, in addition to me having to type up the equation (for box of general side lengths), you'll have to trust me that I quote their result correctly. \u2013\u00a0Lee David Chung Lin May 24 '18 at 21:31\n\u2022 Thank you very much for mentioning the name of Robbin's constant. I now possess a starting point from which I may derive the solution. Thank you again. I'll try to find the source material if possible, count on me. \u2013\u00a0Wyllich May 24 '18 at 21:41\n\n## Notations\n\nDenote $$E(a,b,c)$$ as the expectation seen in the middle of p.278 (right before Eq.(2)) of the short paper (communication) by Robbins and Bolis. It is the average distance for a box of side lengths that are double of the inputs: $$2a, 2b$$, and $$2c$$.\n\nNote that the inverse hyperbolic sine has a logarithmic form: $$\\sinh^{-1} x = \\ln (x + \\sqrt{x^2 + 1})$$. That's how one ends up seeing those logs when $$E(a,b,c)$$ is evaluated numerically.\n\nDenote $$\\mathcal{I}$$ as the integral for your desired average but without the $$1/26$$ normalization.\n\nSimilarly, we will have the integrals $$I_{f}$$ for face-to-face, $$I_{e}$$ for edge-to-edge, and $$I_{v}$$ for vertex-to-vertex. All these are just integrals without normalization (thus not averages themselves). Details continue in the following sections.\n\nDenote $$R_b$$ as the Robbins' constant, namely, $$R_b = E(\\frac12, \\frac12, \\frac12) \\approx 0.6617$$. Note that $$R_b$$ is the average, while it equals in value to its corresponding integral, which shall be denoted $$I_0$$. That is, numerically $$R_b = I_0$$ because the volume is $$1$$.\n\n## Strategy Outline\n\nAs you have noticed, there are obstacles in directly setting up the density. The symmetry and nice numbers (integers) strongly suggest that one should solve this via decomposing the space into blocks of unit cubes.\n\nAny cube has $$6$$ faces, $$12$$ edges (sides), and $$8$$ vertices, therefore\n\n$$\\mathcal{I} = 6 I_f + 12 I_e + 8 I_v \\tag{1} \\label{Eq_26-decomp}$$\n\nYour $$V_1$$ is the unit cube at the \"center\", surrounded by $$26$$ unit cubes that are disjoint and which adds up exactly to your $$V_2$$. For a different configuration this approach needs modification since the space doesn't decompose nicely in general.\n\nThe $$26$$ blocks consist of three groups: six of them are face-to-face with $$V_1$$, twelve of them are edge-touching-edge-only with $$V_1$$, and the remaining eight are vertex-touching-vertex-only with $$V_1$$.\n\nThe value of each $$I_f$$, $$I_e$$, and $$I_v$$ will be obtained using $$E(a,b,c)$$, successively building up starting from $$I_f$$.\n\nIt will involve a $$2$$-$$1$$-$$1$$ long box, a $$2$$-$$2$$-$$1$$ flat box, then finally the $$2$$-$$2$$-$$2$$ \"double cube\".\n\n## Calculation Details\n\nFormally, the integrals can be defined as (with more shorthands in additional to yours) D_{1,2} \\equiv \\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2} \\\\ \\begin{align*} I_f = \\iiint\\limits_{ V_1} \\iiint\\limits_{(x_2, y_2,z_2) \\in V_{\\mathbf{F}}} D_{1,2} \\,\\mathrm{d}x_{1,2}\\,\\mathrm{d}y_{1,2}\\, \\mathrm{d}z_{1,2} \\\\ I_e = \\iiint\\limits_{ V_1} \\iiint\\limits_{(x_2, y_2,z_2) \\in V_{\\mathbf{E}}} D_{1,2} \\,\\mathrm{d}x_{1,2}\\,\\mathrm{d}y_{1,2}\\, \\mathrm{d}z_{1,2} \\\\ I_v = \\iiint\\limits_{ V_1} \\iiint\\limits_{(x_2, y_2,z_2) \\in V_{\\mathbf{V}}} D_{1,2} \\,\\mathrm{d}x_{1,2}\\,\\mathrm{d}y_{1,2}\\, \\mathrm{d}z_{1,2} \\end{align*} The volume $$V_F$$ can be any unit cube adjacent and face-to-face with $$V_1$$. For example, $$V_F \\equiv \\bigl\\{(x,y,z) \\big|~~ x \\in [1,2],\\, y \\in [0,1],\\, z \\in [0,1]\\bigr\\}$$\n\nSimilarly, the edge-to-edge volume $$V_E$$ can be, for example, $$\\bigl\\{x \\in [1,2],\\, y \\in [0,1],\\, z \\in [1,2]\\bigr\\}$$, and the vertex-to-vertex volume $$V_V$$ can be $$\\bigl\\{x \\in [1,2],\\, y \\in [1,2],\\, z \\in [1,2]\\bigr\\}$$.\n\nThe formulation of $$I_f$$ (or $$I_e$$, or $$I_v$$) is not unique (up to $$6$$, $$12$$, $$8$$ multiplicity for any $$V_1$$, which itself can resides anywhere), but it doesn't matter. The value of each integral is unique and can be obtained using $$E(a,b,c)$$.\n\n### Successive Build-up (1): $$~$$ starting from $$I_f$$\n\nConsider a $$2$$-by-$$1$$-by-$$1$$ long box, which is equivalent to two $$V_1$$ joined face-to-face. Denote the box as $$V_{_{2.1.1}}$$ . We know that its volume is $$|V_{_{2.1.1}}| = 2$$ .\n\nFor the average distance, decompose the integral (space) into two within-cube and two across-cubes.\n\n$$|V_{_{2.1.1}}|^2 \\cdot E( 1,\\tfrac12, \\tfrac12) = 2 I_0 + 2 I_f \\tag{2} \\label{Eq_long-box_decomp}$$\n\nThere are two $$I_0$$ because each half of $$V_{_{2.1.1}}$$ are physically distinct so that ordering matters. For the same reason, there are two $$I_f$$.\n\nNote that $$I_0$$ here emphasizes that it is the integral and not the average $$R_b$$, although numerically they are the same.\n\nNow is the time we must make explicit use of $$E(a,b,c)$$ by Robbins and Bolis for the general box. $$\\begin{multline*} E(1, \\tfrac12, \\tfrac12) = \\frac1{1260} \\Bigl[520 + 17\\sqrt{2} - 20\\sqrt{5} - 81 \\sqrt{6} + 21 \\log\\left(\\sqrt{2}+1\\right) \\\\ + 168 \\log \\left(\\sqrt{2}+\\sqrt{3}\\right) + 735 \\log\\bigl( \\frac{ \\sqrt{6} + 1 }{ \\sqrt{5} }\\bigr) + 1344 \\log \\left(\\frac{1}{2} \\left(\\sqrt{5}+1\\right)\\right) \\\\ + 42 \\log \\left(\\sqrt{5}+2\\right) - 1344 \\arcsin\\bigl( \\frac15 \\bigr) - 168 \\arcsin\\bigl( \\sqrt{\\frac25 } \\bigr) \\Bigr] \\end{multline*}$$ I coded $$E(a,b,c)$$ with Mathematica and would never consider doing this by hand. The $$E(1,\\frac12,\\frac12)$$ here as well as later $$E(a,b,c)$$ evaluations are done by the computer algebraic system.\n\nThe above display is just to demonstrate what we are dealing with here. To save space, the exact expression for future $$E(a,b,c)$$ will be omitted.\n\nMoving on, $$E(1,\\frac12,\\frac12) \\approx 0.91455248459664313930734$$ and we have \\begin{align*} I_f = \\frac{ |V_{_{2.1.1}}|^2 }2 E( 1,\\tfrac12, \\tfrac12) - I_0 &= 2 E( 1,\\tfrac12, \\tfrac12) - Rb \\tag{3.a} \\label{Eq_Int-face_in_Rb} \\\\ &\\approx 1.16739778692611 \\tag{3.b} \\label{Eq_Int-face_value} \\end{align*}\n\n### Successive Build-up (2): $$~$$ from $$I_f$$ to $$I_e$$\n\nConsider a $$2$$-by-$$\\mathbf{2}$$-by-$$1$$ flat box, which is equivalent to four $$V_1$$ joined together with pairwise face-to-face. Denote this box as $$V_{_{2.2.1}}$$. We know that its volume is $$|V_{_{2.2.1}}| = 4$$.\n\nThe decomposition of the integral (space) goes like this: for each of the four unit cubes, there is one within-cube integral, two face-to-face integral, and one edge-to-edge integral. $$|V_{_{2.2.1}}|^2 \\cdot E( 1, 1, \\tfrac12) = 4 \\left( I_0 + 2 I_f + I_e \\right) \\tag{4} \\label{Eq_flat-box_decomp}$$ Again, the cubes are physically distinct and order matters. Note that the decomposition counts the spaces as $$4^2 = 4\\cdot (1 + 2 + 1)$$.\n\nPlugin the $$E(a,b,c)$$ formula to get $$E(1, 1,\\frac12) \\approx 1.1320874696118224667$$ and then replacing $$I_f$$ with Eq\\eqref{Eq_Int-face_in_Rb}, we have: \\begin{align*} I_e = \\frac{ |V_{_{2.2.1}}|^2 }4 E( 1, 1, \\tfrac12) - I_0 - 2I_f &= 4E( 1,1, \\tfrac12) - R_b - 2 \\bigl( 2 E( 1,\\tfrac12, \\tfrac12) - R_b \\bigr) \\\\ &= 4 \\bigl( E( 1, 1, \\tfrac12) - E( 1,\\tfrac12, \\tfrac12) \\bigr) + R_b \\tag{5.a} \\label{Eq_Int-edge_in_Rb} \\\\ &\\approx 1.5318471223279 \\tag{5.b} \\label{Eq_Int-edge_value} \\end{align*}\n\n### Successive Build-up (3): $$~$$ get $$I_v$$ from $$I_f$$ and $$I_e$$\n\nConsider a $$2$$-by-$$2$$-by-$$2$$ \"double cube\", which is equivalent to eight $$V_1$$ joined together with pairwise face-to-face. It is also just $$V_1$$ scaled up by a factor of two. Denote this box as $$V_{_{2.2.2}}$$.\n\nWe know that its volume is $$|V_{_{2.2.2}}| = 2^3 = 8$$, and this time we also know that $$E(1,1,1) = 2 E( \\frac12, \\frac12, \\frac12) = 2R_b$$ simply due to the linear scalability of the geometry.\n\nThe integrals (spaces) decompose similarly. For each of the eight unit cubes, there is one within-cube integral, three face-to-face integral, three edge-to-edge integral, and one vertex-to-vertex integral. $$|V_{_{2.2.2}}|^2 \\cdot E( 1, 1, 1) = 8 \\left( I_0 + 3 I_f + 3I_e + I_v \\right) \\tag{6} \\label{Eq_double-cube_decomp}$$ As before, the order matters and the decomposition says $$8^2 = 8\\cdot (1 + 3 + 3 + 1)$$.\n\nTake $$E( 1, 1, 1) = 2R_b\\,,\\,$$ take $$I_f$$ as Eq\\eqref{Eq_Int-face_in_Rb}, and take $$I_e$$ as Eq\\eqref{Eq_Int-edge_in_Rb}, we end up with\n\n\\begin{align*} I_v &= \\frac{ |V_{_{2.2.2}}|^2 }8 E( 1, 1, 1) - I_0 - 3I_f - 3 I_e \\\\ &= 8 \\cdot 2R_b - R_b - 3 \\Bigl[ 2 E( 1,\\tfrac12, \\tfrac12) - R_b + 4 \\bigl( E( 1, 1, \\tfrac12) - E( 1,\\tfrac12, \\tfrac12) \\bigr) + R_b \\Bigr] \\\\ &= 15 R_b - 6\\bigl[ 2E( 1, 1, \\tfrac12) - E( 1,\\tfrac12, \\tfrac12) \\bigr] \\tag{6.a} \\label{Eq_Int-vertex_in_Rb} \\\\ &\\approx 1.82787300624563276 \\tag{6.b} \\label{Eq_Int-vertex_value} \\end{align*}\n\nFinally, putting things together into Eq\\eqref{Eq_26-decomp} we get:\n\n\\begin{align*} \\mathcal{I} &= 6I_f + 12I_e + 8 I_v \\\\ &= 2 \\Bigg\\{ 3\\bigl( 2 E(1,\\tfrac12,\\tfrac12 ) - R_b\\bigr) + 6 \\Bigl[ R_b + 4 \\bigl( E(1, 1,\\tfrac12 ) - E(1,\\tfrac12,\\tfrac12 )\\bigr) \\Bigr] + 4 \\Bigl[ 15R_b - 6 \\bigl( 2E(1, 1,\\tfrac12 ) - E(1,\\tfrac12,\\tfrac12 )\\bigr) \\Bigr]\\Bigg\\} \\\\ &= 2 \\bigl( 63R_b + 6 E(1,\\tfrac12,\\tfrac12 ) - 24 E(1,1,\\tfrac12 ) \\bigr) \\tag{7.a} \\label{Eq_Int-desired_in_Rb} \\\\ &\\approx 40.0095362394564449 \\tag{7.b} \\label{Eq_Int-desired_value} \\\\ \\end{align*}\n\nThe desired average is $$\\frac1{26} \\mathcal{I}\\approx 1.53882831690217095767637627$$\n\nThe respective sub-equations Eq\\eqref{Eq_Int-face_in_Rb}, Eq\\eqref{Eq_Int-edge_in_Rb}, Eq\\eqref{Eq_Int-vertex_in_Rb}, and Eq\\eqref{Eq_Int-desired_in_Rb} showcase the expression in minimal terms of $$E(a,b,c)$$ where terms are converted to $$R_b$$ whenever possible.\n\nThese are the reminders that the exact solution is obtainable, that the functional form still exhibits some recognizable geometry, and that one can calculate any intermediate or final results to as many digits as one like.\n\n## Concluding Remarks\n\nIt should be clear that one cannot arrive at the desired integral solely by using $$R_b$$. After carving out the contribution of $$V_1$$ at the center from a triple-cube, one still has to deal with the $$V_2$$-to-$$V_2$$ cross terms. In the end, one will have to do something similar to the derivation above.\n\nWhen the space is in a nice configuration (smooth boundary with no zig-zags, no holes within ,etc), the exact integral in general is manageable, yet already requiring some attention to details. It is very telling that for the derivation of $$E(a,b,c)$$, Robbins and Bolis wrote that \"By tedious yet routine successive integration one finds that ...\"\n\nFor the sake of completeness, for related in-site posts one can start with rectangle in $$2$$-dim and more generally for hypercubes.\n\nIt is totally feasible to derive the density from scratch for this particular $$\\{ V_1, \\, V_2 \\}$$ pair. The transformation of random variables involved is pretty basic. It is just that due to the \"hole\" in $$V_2$$, inevitably one will be forced to decompose the space.\n\n\"Decomposing the space into disjoint spaces\" is a standard technique in dealing with probability inquiries, and the decomposition is not necessarily into integer valued partitions. There are indeed limitations to this approach, but the concept is rather simple.\n\nIn terms of numerical calculation, once you have coded the $$E(a,b,c)$$ in whatever system you're using, this procedure is pretty straightforward.\n\nI don't know your original motivation for doing this calculation, but I'd say this approach is practically useful across disciplines, from theoretical to applications.\n\n\u2022 What a ride! Thank you for posting such a detailed and comprehensive answer. I will definitely remember the idea of dividing the integral space as you masterfully did to obtain intermediate integrals through multiple iterations. I guess I owe you an explanation about the whys. In LiDAR image processing, I asked myself the possibility of predicting the resolution (here understand mean distance between a point and its closest neighbour) of a downsampled cloud using a voxel grid. In spite of my hypothesis based on the distance you calculated being disproved by experience, I still believe it (1) \u2013\u00a0Wyllich May 28 '18 at 12:46\n\u2022 ... was a good experience to have. Thank you very much again. (2) \u2013\u00a0Wyllich May 28 '18 at 12:47\n\u2022 Very glad I could help. I know only a little bit about image process or data compression in general, but what you described sounds no lesser than some innovative ideas that eventually became successful. \u2013\u00a0Lee David Chung Lin May 28 '18 at 13:12\n\u2022 Now that I think about it, your method gave me new ideas to precise my hypothesis which is based on isotropy (hence V2 containing V1). However, I may add some coefficients in front of each term of equations (2), (4), (6) to take into account anisotropy and capture an idea about the cloud's global structure. I must repeat myself : your dedication gave birth to many new ideas I may now explore. \u2013\u00a0Wyllich May 28 '18 at 13:25\n\u2022 Thank you for saying that. It's been a real pleasure. \u2013\u00a0Lee David Chung Lin May 28 '18 at 13:46", "date": "2020-04-04 13:13:45", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2793350/mean-distance-between-2-points-in-adjacent-voxels-cubes/2796405", "openwebmath_score": 0.992321789264679, "openwebmath_perplexity": 1239.5758415108808, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130583409235, "lm_q2_score": 0.8596637523076225, "lm_q1q2_score": 0.8502186768145958}}
{"url": "https://math.stackexchange.com/questions/1117948/how-do-i-solve-this-olympiad-question-with-floor-functions", "text": "# How do I solve this Olympiad question with floor functions?\n\nEmmy is playing with a calculator. She enters an integer, and takes its square root. Then she repeats the process with the integer part of the answer. After the third repetition, the integer part equals 1 for the first time. What is the difference between the largest and the smallest number Emmy could have started with?\n\n$$\\text{(A)} \\; 229 \\qquad \\text{(B)} \\; 231 \\qquad \\text{(C)} \\; 239 \\qquad \\text{(D)} \\; 241 \\qquad \\text{(E)} \\; 254$$\n\nThis was Problem 19 from the first round of the Norwegian Mathematical Olympiad, 2014\u201315.\n\nI think the floor function applies well here.\n\nThe integer part obviously is the floor part of it.\n\nLet $x_1$ be the initial.\n\n$x_2 = \\left \\lfloor{{\\sqrt{x_1}}}\\right \\rfloor$\n\n$x_3 = \\left \\lfloor{{\\sqrt{x_2}}}\\right \\rfloor$\n\n$x_4 = \\left \\lfloor{{\\sqrt{x_3}}}\\right \\rfloor = 1$ <-- Last one.\n\nSo we need to find $b \\le x_1 \\le c$ such that $b$ is the smallest number to begin with and $c$ is the largest number to begin with.\n\n$\\sqrt{b} \\le \\sqrt{x_1} \\le \\sqrt{c}$\n\nClearly we must have $x_3<4$, $x_2<16$, and $x_1<256$, so the largest possible value of $x_1$ is $255$. Since $x_3>1$, we know that $x_3\\ge 2$ and hence that $x_2\\ge 4$. This implies that $x_1\\ge 16$. Thus, the range is from $16$ through $255$, the difference being $239$.\n\n\u2022 @BrianScott, Thanks. Clear up, why must we have those conditions? Why $x_1 < 256$? \u2013\u00a0Robart Jan 24 '15 at 18:05\n\u2022 We know that $x_4=1$, so $1\\le\\sqrt{x_3}<2$, and therefore $x^3<2^2=4$. Now repeat the reasoning back up the line: $x_3<4$, so $\\sqrt{x_2}<4$, and $x_2<16$. And so on. \u2013\u00a0Brian M. Scott Jan 24 '15 at 18:06\n\u2022 @Robart: Sorry: I forgot to ping you with my answer. See the comment immediately above this one. \u2013\u00a0Brian M. Scott Jan 24 '15 at 18:13\n\u2022 why cant she start with $x_1 = 1$ so the difference is $255 - 1 = 254$? \u2013\u00a0Robart Jan 24 '15 at 18:43\n\u2022 @Robart: Because we\u2019re told that $x_4$ is the first of these integers to be equal to $1$. \u2013\u00a0Brian M. Scott Jan 24 '15 at 18:45\n\nWe can work backwards more easily than we can work forwards. Firstly, what does a number $x$ need to satisfy to have $\\lfloor x \\rfloor = 1$? Easy. We need: $$1\\leq x < 2.$$ Well, suppose that $\\sqrt{y}=x$ or, equivalently, $y=x^2$. Well, obviously we just square the above equation (as all of its terms are positive): $$1\\leq x^2 < 2^2.$$ Suppose $\\sqrt{z}=y$ or $z=y^2=x^4$. Square the above again! $$1\\leq x^4 < 2^4.$$ And finally let $\\sqrt{w}=z$ or $w=z^2=y^4=x^8$. Square the above $$1\\leq x^8 < 2^8.$$\n\nSo, if she chose $z$ to be any integer in $[1,2^8)$, she will get $1$ as the final result. However, if $z$ were in the interval $[1,2^4)$, she would have gotten $1$ as the second result as well. So, we restrict ourselves to the integers in $[2^4,2^8)$, as these will satisfy the desired condition.\n\n\u2022 And how do you pick the smallesT? Its supposed to be $16$? \u2013\u00a0Robart Jan 24 '15 at 18:33\n\u2022 Yes, the condition that the third time is the first time that the number is $<2$ means that $x^2\\geq 2$. \u2013\u00a0Thomas Andrews Jan 24 '15 at 18:35\n\u2022 Why does it have to be $x^2 \\ge 2$? then $x ge \\sqrt{2}$ \u2013\u00a0Robart Jan 24 '15 at 18:41\n\u2022 It should be 254 I believe. The smallest is $1$ since $\\sqrt{1} = 1$ for $n$ square roots. \u2013\u00a0Robart Jan 24 '15 at 18:43\n\u2022 @Robart The question says that the third time is the first for which the result is $1$ - the greatest possible number is $255$, as higher numbers don't reach $1$ fast enough, and the least is $16$, as lower numbers reach $1$ too fast. \u2013\u00a0Milo Brandt Jan 24 '15 at 18:46", "date": "2020-02-22 02:12:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1117948/how-do-i-solve-this-olympiad-question-with-floor-functions", "openwebmath_score": 0.8460710644721985, "openwebmath_perplexity": 182.8179871104797, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130554263733, "lm_q2_score": 0.8596637541053281, "lm_q1q2_score": 0.8502186760870171}}
{"url": "http://math.stackexchange.com/questions/414776/what-is-the-use-of-the-dot-product-of-two-vectors/414785", "text": "# What is the use of the Dot Product of two vectors?\n\nSuppose you have two vectors a and b that you want to take the dot product of, now this is done quite simply by taking each corresponding coordinate of each vector, multiplying them and then adding the result together. At the end of performing our operation we are left with a constant number.\n\nMy question therefore is what can we do with this number,why do we calculate it so to speak? I mean it seems almost useless to me compared with the cross product of two vectors (where you end up with an actual vector).\n\n-\nYou can use it to find the angle between any two vectors. $\\mathbf{a}\\cdot\\mathbf{b}=|\\mathbf{a}||\\mathbf{b}|\\cos\\theta$ where $\\theta$ is the angle between the two vectors. This is a better approach than using the cross product as the cross product can only be defined in a few dimensions (normally only 3 dimensions). Obviously this is just one simple use of the dot product which is a special case of a more general phenomenon known as an inner product en.wikipedia.org/wiki/Inner_product_space . \u2013\u00a0 Daniel Rust Jun 8 '13 at 15:35\nYou should take a look at how it is derived as it can be very enlightening. Take a look at Lang's Linear Algebra (2E) pg.19-20. \u2013\u00a0 12F8031 Jun 8 '13 at 20:18\nI always found that the dot product is a good way to measure how \"parallel\" two vectors are \u2013\u00a0 Dan Jun 9 '13 at 7:15\n\nRe: \"[the dot product] seems almost useless to me compared with the cross product of two vectors \".\n\nPlease see the Wikipedia entry for Dot Product to learn more about the significance of the dot-product, and for graphic displays which help visualize what the dot product signifies (particularly the geometric interpretation). Also, you'll learn more there about how it's used. E.g., Scroll down to \"Physics\" (in the linked entry) to read some of its uses:\n\nMechanical work is the dot product of force and displacement vectors.\nMagnetic flux is the dot product of the magnetic field and the area vectors.\n\n\nYou've shared the algebraic definition of the dot product: how to compute it as the sum of the product of corresponding entries in two vectors: essentially, computing $\\;\\mathbf A \\cdot \\mathbf B = {\\mathbf A}{\\mathbf B}^T.\\;$ But the dot product also has an equivalent geometric definition:\n\nIn Euclidean space, a Euclidean vector is a geometrical object that possesses both a magnitude and a direction. A vector can be pictured as an arrow. Its magnitude is its length, and its direction is the direction the arrow points. The magnitude of a vector A is denoted by $\\|\\mathbf{A}\\|.$ The dot product of two Euclidean vectors A and B is defined by\n\n$$\\mathbf A\\cdot\\mathbf B = \\|\\mathbf A\\|\\,\\|\\mathbf B\\|\\cos\\theta,\\quad\\text{where \\theta is the angle between A and B.} \\tag{1}$$\n\nWith $(1)$, e.g., we see that we can compute (determine) the angle between two vectors, given their coordinates: $$\\cos \\theta = \\frac{\\mathbf A\\cdot\\mathbf B}{\\|\\mathbf A\\|\\,\\|\\mathbf B\\|}$$\n\n-\nI am not stating what the dot product signifies, in fact that is the essence of this question, I did not know that the dot product has an equivalent geometric definition, or that it could be used to calculate the angle between two vectors that is extremely useful. \u2013\u00a0 Jonn_Underwood Jun 8 '13 at 15:45\nI didn't intend to criticize. I'm sorry if that's how the post comes across! Indeed, I upvoted the question because it's a fine question, and you put thought into writing it. \u2013\u00a0 amWhy Jun 8 '13 at 15:47\nThe dot product is also the product $\\bf A B^T$ of vector A with the transpose of vector B. \u2013\u00a0 amWhy Jun 8 '13 at 15:49\n@amWhy: Thank you, I had an upvote mind - trigger finger! :-) Fixed! \u2013\u00a0 Amzoti Jun 9 '13 at 1:37\n@amWhy thank you for your suburb answer , and also thank you for adding that formula at the bottom for finding the cosine of theta this alone is extremely useful to me due to my pursuit of physics! \u2013\u00a0 Jonn_Underwood Jun 9 '13 at 21:24\nshow 1 more comment\n\nThe original motivation is a geometric one: The dot product can be used for computing the angle $\\alpha$ between two vectors $a$ and $b$:\n\n$a\\cdot b=|a|\\cdot|b|\\cdot \\cos(\\alpha)$.\n\nNote the sign of this expression depends only on the angle's cosine, therefore the dot product is\n\n\u2022 $<0$ if the angle is obtuse,\n\u2022 $>0$ if the angle is acute,\n\u2022 $=0$ if the $a$ and $b$ are orthogonal.\n\nAnother important special case appears when $a=b$: The root of the scalar product of a vector with itself is the length of a vector:\n\n$a\\cdot a=|a|\\cdot|a|\\cdot1=|a|^2$.\n\nThere's another interesting application of the dot product, in combination with the cross product: If you have three vectors $a$, $b$ and $c$, they define a parallelepiped, and you can compute its (signed) volume $V$ as follows using the so-called scalar triple product:\n\n$V=(a\\times b)\\cdot c$\n\n(Note that this is a generalization of $|a\\times b|$ being the area of the parallelogram defined by $a$ and $b$.)\n\n-\nThere's nothing special about 2d/3d here; this means of finding the angle between vectors applies to an arbitrary number of dimensions. \u2013\u00a0 Muphrid Jun 8 '13 at 15:50\n@Muphrid: Of course you can define something like an \"angle\" in arbitrary dimensions. I've changed \"geometrical\" to \"visual\", I think that makes more sense. \u2013\u00a0 zero-divisor Jun 9 '13 at 6:12\nPerhaps. But two vectors define only a plane, so even in a 4d space or higher, the geometry basically isn't changing: you have two vectors in some common plane, and the dot product tells us how alike they are. I think you're making it out to be more complicated than it really is. \u2013\u00a0 Muphrid Jun 9 '13 at 6:17\n@Muprid: Good point. I've deleted the sentence now. \u2013\u00a0 zero-divisor Jun 9 '13 at 6:19\n\nThe dot product is an essential ingredient in matrix product. The product of the two matrices $A$ and $B$ (of compatible sizes, that is, the number of columns of $A$ equals the number of rows of $B$) is a matrix whose $(i, j)$ component is the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.\n\nAmong the many applications, consider this simple one.\n\nYou have students $s_{1}, \\dots, s_{n}$ taking courses $c_{1}, \\dots, c_{m}$. Consider the matrix $A$ which has $0$ everywhere, except that the $(i,j)$ coefficient is $1$ if the student $s_{j}$ takes the course $c_{i}$. Now note that the dot product of the $i_{1}$-th row of $A$ with the $i_{2}$-th row gives the number of students that take both $c_{i_{1}}$ and $c_{i_{2}}$. In other words, the matrix $A A^{t}$ has in its $(i, j)$ position the number of students taking both $c_{i}$ and $c_{j}$.\n\nThis matrix is of course useful in building a course timetable.\n\n-\n\nBefore addressing your question, i want to say that this is a very good question and you are right to expect that the dot product has use/significance.\n\nFirst, it is important that you think about vectors separate from their coordinates. While it is true that we often represent vectors as a series of coordinates along well-defined axes, this is merely for computational reasons. A vector as an idea \"exists\" in a space without any predefined coordinate system. I say this because there are two definitions of the dot product, one is coordinate free (i.e. $\\mathbf a\\cdot\\mathbf b = \\|\\mathbf a\\|\\,\\|\\mathbf b\\|\\cos\\theta$) and the other is based on coordinates (i.e. $\\mathbf a\\cdot\\mathbf b = \\sum_i{a_i b_i}$). Of these two, it is best to think of the dot product in terms of the former as it does not depend on a coordinate system. (It is relatively easy to show that the latter may be derived from the former, but in that derivation is an implicit assumption that the coordinate system being used to represent the dot product is orthogonal.)\n\nSecond, given the coordinate-free definition, the fundamental idea of the dot product is that of projection. By this it gives a single number which indicates the component of a vector in the direction of another vector. Your observation of the dissimilarity between the dot and cross product is correct, however, the dot product is used to produce a vector as well, it just does it component-by-component. Let's suppose that we have a vector $\\mathbf v$ represented by its components in a given coordinate system. Let's further suppose that we have an orthonormal basis defined in that same coordinate system as the set of column vectors $\\{\\mathbf u_1, \\mathbf u_2, \\ldots, \\mathbf u_n\\}$. Finally, suppose that we want to represent $\\mathbf v$ in this basis as $\\mathbf w$. The question is how do we do that? We use the dot product of course! So the first component of $\\mathbf w$ would then be $w_1 = \\mathbf u_1\\cdot \\mathbf v$, and the second component would be $w_2 = \\mathbf u_2\\cdot \\mathbf v$ and so on. (Note that because $\\|\\mathbf u_i\\| = 1$, we have $\\mathbf u_1\\cdot \\mathbf v= \\|\\mathbf v\\|\\cos\\theta_i$.) If we then think of the vector $\\mathbf w$ defined as such we have\n\n$$\\mathbf w = \\left[ \\begin{array}{c} w_1 \\\\ w_2 \\\\ \\vdots \\\\ w_n \\end{array} \\right] = \\left[ \\begin{array}{c} \\mathbf u_1\\cdot \\mathbf v \\\\ \\mathbf u_2\\cdot \\mathbf v \\\\ \\vdots \\\\ \\mathbf u_n\\cdot \\mathbf v \\end{array} \\right] = \\left[ \\begin{array}{c} \\mathbf u_1^T \\mathbf v \\\\ \\mathbf u_2^T \\mathbf v \\\\ \\vdots \\\\ \\mathbf u_n^T \\mathbf v \\end{array} \\right] = \\left[ \\begin{array}{c} \\mathbf u_1^T \\\\ \\mathbf u_2^T \\\\ \\vdots \\\\ \\mathbf u_n^T \\end{array} \\right]\\mathbf v = \\left[ \\begin{array}{cccc} \\mathbf u_1 & \\mathbf u_2 & \\cdots &\\mathbf u_n \\end{array} \\right]^T\\mathbf v$$\n\nFinally, we conclude that the dot product plays a key role in the transformation of a vector from one basis to another and that the dot product is hidden in the definition of matrix multiplication in that one view of a matrix-vector product is that each element in the product represents a dot product between a row of the left and a column of the right.\n\nI hope this helps.\n\n-\n\nThe geometric idea of the dot product has been touched upon, but there is a vast generalization of this product in geometric algebra, the algebra of not only oriented lines (vectors) but planes, volumes, and more (called blades).\n\nIn geometric algebra, we have a generalized dot product of a vector $a$ and another blade $B$ denoted $a \\cdot B$. This product has a simple geometric interpretation as the part of $B$ orthogonal to the projection of $a$, with magnitude $|a||B|| \\cos \\theta|$, where $\\theta$ is the angle $a$ forms with its projection in $B$.\n\n(Note: as a point of fact, $a \\cdot B$ is orthogonal to $a$ also, not just its projection into $B$, but thinking about this leads to some difficulties, while thinking about what's perpendicular to the projection does not.)\n\nIf $B$ is a 2-blade (also called a bivector), then you should be able to imagine this directly: if $a$ lies entirely in $B$, then $a \\cdot B$ is just the vector perpendicular to $a$ in $B$. If $a$ does not lie entirely in $B$, then it can be decomposed into a tangential part and a normal part. We throw away the normal part, and the previous logic applies for the tangential part.\n\nIf $B$ is a 3-blade (a trivector), then in 3d space $a$ must lie in $B$ (for there is no 3d volume that a vector does not help span), and the product $a \\cdot B$ is the \"Hodge dual\", or the plane perpendicular to $a$.\n\nIn this light, the dot product of vectors may actually be the most non-intuitive part of this reasoning. When you take the dot product, there's only a scalar left--there's no vector or other higher dimensional object left to be orthogonal to $a$. Again, this is why I emphasize that $a \\cdot B$ is the part of $B$ orthogonal to the projection of $a$ onto $B$. When $B$ is a vector, it's clear there is no other vector or anything else that can be orthogonal to the projection of $a$, for $B$ and that projection are parallel, so the result is necessarily just a scalar.\n\n-", "date": "2014-03-10 21:38:34", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/414776/what-is-the-use-of-the-dot-product-of-two-vectors/414785", "openwebmath_score": 0.9262725710868835, "openwebmath_perplexity": 177.29955373726372, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130575313262, "lm_q2_score": 0.8596637487122112, "lm_q1q2_score": 0.8502186725627057}}
{"url": "https://math.stackexchange.com/questions/3506020/every-natural-number-is-covered-by-consecutive-numbers-that-sum-to-a-prime-power", "text": "# Every natural number is covered by consecutive numbers that sum to a prime power.\n\nConjecture. For every natural number $$n \\in \\Bbb{N}$$, there exists a finite set of consecutive numbers $$C\\subset \\Bbb{N}$$ containing $$n$$ such that $$\\sum\\limits_{c\\in C} c$$ is a prime power.\n\nA list of the first few numbers in $$\\Bbb{N}$$ has several different covers by such consecutive number sets.\n\nOne such is:\n\n 3   7  5 13  8  19  11  25    29   16     37    41          49    53\n___ ___ _ ___ _ ____ __ _____ _____ __ __ _____ _____       _____ _____    __\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29\n___   ___                _____             _____ __       ________\n11    17                   31                43              81\n59               71           3^5\n___    __       _____    _________________\n30 31 32 33 34 35 36 37 38 39 40 41 42 43 .....\n_____    _____    _____\n61       67        73\n\n\n\u2022 Apparently, primes are allowed as well. Not sure, whether it has been proven, but this is very likely to be the case. More interesting is to exclude the primes. \u2013\u00a0Peter Jan 12 '20 at 10:34\n\u2022 @Peter: The sum can only be a prime if it has at most two terms. (Since $\\sum_{k=m}^nk=(n-m+1)(n+m)/2$, which factorizes unless $n-m+1\\le2$.) The density of one- and two-term sums that are primes goes to zero, so the brunt of the conjecture has to be borne by the proper prime powers, anyway. \u2013\u00a0joriki Jan 12 '20 at 13:02\n\u2022 @Peter a prime power by definition includes the exponent $1$. Therefore, for sake of simplicity, we include the primes. \u2013\u00a0StudySmarterNotHarder Jan 12 '20 at 17:32\n\nFor any odd prime $$p$$, there are $$p$$ consecutive integers centred on $$p$$ that sum to $$p^2$$.\n\n$$2+3+4=3^2$$\n$$3+4+5+6+7=5^2$$\n$$4+5+6+7+8+9+10=7^2$$\netc.\n\nLet $$p_n$$ be the $$n$$-th prime. Then, using Bertrand's postulate in the form $$p_{n+1}<2p_n$$we know that the above sums for consecutive primes overlap.\n\nFinally, we note that $$1+2=3$$ to complete the proof.\n\nI don't know if this has been shown before, but the proof seems straightforward.\n\nWhile nickgard's answer shows how to solve the problem using sums being squares of increasing primes, this answer shows how to do it using the sums being just odd powers of $$3$$.\n\nAs suggested in joriki's question comment, for any integers $$1 \\le j \\le k$$, you have\n\n\\begin{aligned} \\sum_{i=j}^{k}i & = \\sum_{i=1}^{k}i - \\sum_{i=1}^{j-1}i \\\\ & = \\frac{k(k+1)}{2} - \\frac{(j-1)(j)}{2} \\\\ & = \\frac{k^2 + k - j^2 + j}{2} \\\\ & = \\frac{(k-j)(k+j) + k + j}{2} \\\\ & = \\frac{(k+j)(k-j+1)}{2} \\end{aligned}\\tag{1}\\label{eq1A}\n\nConsider the ranges $$\\left[\\frac{3^m + 1}{2},\\frac{3^{m+1} - 1}{2}\\right]$$ for $$m = 0, 1, 2, \\ldots$$. The union of these disjoint subsets cover all positive integers. Thus, for any $$n \\ge 1$$, there's a unique $$m$$ where $$n \\in \\left[\\frac{3^m + 1}{2},\\frac{3^{m+1} - 1}{2}\\right]$$. For that $$m$$, since $$\\frac{5\\left(3^{m}\\right)-1}{2} \\gt \\frac{3^{m+1} - 1}{2}$$, you can have $$j = \\frac{3^m + 1}{2}$$ and $$k = \\frac{5\\left(3^{m}\\right)-1}{2}$$ with $$n \\in [j,k]$$. Using this in \\eqref{eq1A} gives\n\n\\begin{aligned} \\sum_{i=j}^{k}i & = \\frac{(k+j)(k-j+1)}{2} \\\\ & = \\frac{\\left(\\frac{5\\left(3^{m}\\right)-1}{2}+\\frac{3^m + 1}{2}\\right)\\left(\\frac{5\\left(3^{m}\\right)-1}{2}-\\frac{3^m + 1}{2}+1\\right)}{2} \\\\ & = \\frac{\\left(\\frac{6\\left(3^{m}\\right)}{2}\\right)\\left(\\frac{4\\left(3^{m}\\right)}{2}-\\frac{2}{2}+1\\right)}{2} \\\\ & = \\frac{\\left(3\\left(3^{m}\\right)\\right)\\left(2\\left(3^{m}\\right)\\right)}{2} \\\\ & = \\left(3^{m+1}\\right)\\left(3^{m}\\right) \\\\ & = 3^{2m+1} \\end{aligned}\\tag{2}\\label{eq2A}\n\nThe first few examples for $$m = 0, 1$$ and $$2$$ are\n\n$$1 + 2 = 3 = 3^{1} \\tag{3}\\label{eq3A}$$\n\n$$2 + 3 + 4 + 5 + 6 + 7 = 27 = 3^{3} \\tag{4}\\label{eq4A}$$\n\n$$5 + 6 + \\ldots + 21 + 22 = 243 = 3^{5} \\tag{5}\\label{eq5A}$$", "date": "2021-02-27 19:23:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3506020/every-natural-number-is-covered-by-consecutive-numbers-that-sum-to-a-prime-power", "openwebmath_score": 0.9990973472595215, "openwebmath_perplexity": 315.28839335355775, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.989013057045568, "lm_q2_score": 0.8596637487122111, "lm_q1q2_score": 0.8502186721451168}}
{"url": "https://leanprover-community.github.io/archive/stream/217875-Is-there-code-for-X%3F/topic/Classification.20of.20finite-dimensional.20vector.20spaces.html", "text": "## Stream: Is there code for X?\n\n### Topic: Classification of finite-dimensional vector spaces\n\n#### Heather Macbeth (Jan 01 2021 at 16:02):\n\nDo we have, there exists a linear equivalence between two finite_dimensional vector spaces of the same findim?\n\n#### Johan Commelin (Jan 01 2021 at 16:03):\n\nyes, I think Markus Himmel did that\n\n#### Johan Commelin (Jan 01 2021 at 16:04):\n\nI forgot the name of the result...\n\n#### Heather Macbeth (Jan 01 2021 at 16:05):\n\nI tried searching for linear_equiv in linear_algebra.finite_dimensional and couldn't see it.\n\n#### Adam Topaz (Jan 01 2021 at 16:06):\n\nThis is true in general, not just for finite dimensional. Maybe that helps in finding it?\n\n#### Johan Commelin (Jan 01 2021 at 16:06):\n\nIs this what you want: src/linear_algebra/invariant_basis_number.lean ?\n\n#### Johan Commelin (Jan 01 2021 at 16:07):\n\nor is that in the wrong direction?\n\n#### Adam Topaz (Jan 01 2021 at 16:08):\n\n(Heather are you formalizing projective spaces? :smile: )\n\n#### Heather Macbeth (Jan 01 2021 at 16:08):\n\nI think it might be in the wrong direction, yes.\n\n#### Heather Macbeth (Jan 01 2021 at 16:09):\n\n(Heather are you formalizing projective spaces? :smile: )\n\n#### Adam Topaz (Jan 01 2021 at 16:10):\n\nOh nice! I figured it was projective spaces because of the span notation we were discussing a few day ago.\n\n#### Adam Topaz (Jan 01 2021 at 16:12):\n\nanyway, the closest I can find so far to the problem at hand is this docs#equiv_of_is_basis\n\n#### Heather Macbeth (Jan 01 2021 at 16:14):\n\nThat looks good, thanks! I don't understand why it's not named linear_equiv_of_is_basis, though :)\n\nYeah I agree.\n\n#### Adam Topaz (Jan 01 2021 at 16:15):\n\ndocs#cardinal.eq gets you essentially all the way to what you want.\n\n#### Adam Topaz (Jan 01 2021 at 16:16):\n\nAnd this docs#is_basis.mk_eq_dim''\n\n#### Adam Topaz (Jan 01 2021 at 16:27):\n\nimport linear_algebra\nimport linear_algebra.dimension\n\nopen_locale classical\n\nuniverse u\nvariables (K : Type*) [field K] (V W : Type u)\n\nexample (cond : vector_space.dim K V = vector_space.dim K W) : nonempty (V \u2243\u2097[K] W)  :=\nbegin\nobtain \u27e8B,hB\u27e9 := exists_is_basis K V,\nobtain \u27e8C,hC\u27e9 := exists_is_basis K W,\nhave h1 := is_basis.mk_eq_dim'' hB,\nhave h2 := is_basis.mk_eq_dim'' hC,\nhave h3 : cardinal.mk B = cardinal.mk C, by cc,\nrw cardinal.eq at h3,\ncases h3,\nrefine \u27e8equiv_of_is_basis hB hC h3\u27e9,\nend\n\n\n#### Heather Macbeth (Jan 01 2021 at 16:28):\n\nWonderful! Can you PR it?\n\nsure.\n\n#### Adam Topaz (Jan 01 2021 at 16:29):\n\nWhich file should it go in?\n\n#### Adam Topaz (Jan 01 2021 at 16:30):\n\nOh, there is one annoying thing -- to even state that the two vector spaces have the same dimension, they need to come from the same universe.\n\n#### Heather Macbeth (Jan 01 2021 at 16:30):\n\nI think linear_algebra.dimension\n\n#### Heather Macbeth (Jan 01 2021 at 16:31):\n\nOh, there is one annoying thing -- to even state that the two vector spaces have the same dimension, they need to come from the same universe.\n\nWe just need a version for finite_dimensional and findim, that avoids all those issues.\n\n#### Adam Topaz (Jan 01 2021 at 16:31):\n\nYou can still have finite dimensional vector spaces in different universes.\n\n#### Adam Topaz (Jan 01 2021 at 16:32):\n\nOh I see what you mean.\n\n#### Adam Topaz (Jan 01 2021 at 16:32):\n\nThe dimension would be in nat, not in cardinal\n\n#### Adam Topaz (Jan 01 2021 at 16:48):\n\nOkay, I pushed to this branch branch#lin_equiv_of_basis_equiv\n\n#### Adam Topaz (Jan 01 2021 at 16:49):\n\nIs there a finite_dimensional analogue of docs#linear_equiv.dim_eq with findim?\n\n#### Heather Macbeth (Jan 01 2021 at 17:22):\n\nThanks!\n\nIs there a finite_dimensional analogue of docs#linear_equiv.dim_eq with findim?\n\n#### Adam Topaz (Jan 01 2021 at 17:24):\n\nI have the finite-dimensional case essentially done too (with no universe limitations). I'll open a PR soon.\n\n#5559\n\n#### Patrick Massot (Jan 01 2021 at 18:39):\n\nI'm skeptical that you really want the conclusion of the lemma to be nonempty (V \u2243\u2097[K] V\u2082). The first thing you'll do when using the lemma will be to extract an isomorphism. Why not making it a def?\n\n#### Patrick Massot (Jan 01 2021 at 18:42):\n\nI'm also surprised those lemmas are no already there somehow. I would wait for @Anne Baanen to comment on this.\n\n#### Adam Topaz (Jan 01 2021 at 18:46):\n\nI mostly used nonempty to make it a Prop, which is useful for the if and only if statement. I can make a def to get the linear equiv as well, but I agree that we should wait for @Anne Baanen :)\n\n#### Anne Baanen (Jan 02 2021 at 06:05):\n\nPatrick Massot said:\n\nI'm also surprised those lemmas are no already there somehow. I would wait for Anne Baanen to comment on this.\n\nI don't recall seeing something like linear_equiv.of_findim_eq, only the equivalences assuming a basis, so I think this is indeed a hole in the library.\n\n#### Anne Baanen (Jan 02 2021 at 06:18):\n\nI mostly used nonempty to make it a Prop, which is useful for the if and only if statement. I can make a def to get the linear equiv as well, but I agree that we should wait for Anne Baanen :)\nThe nonempty situation as it is in the PR right now seems optimal to me: cardinal equality is expressed in nonempty, so the first result also uses nonempty and the next ones apply classical.choice :+1:", "date": "2021-05-19 02:24:24", "meta": {"domain": "github.io", "url": "https://leanprover-community.github.io/archive/stream/217875-Is-there-code-for-X%3F/topic/Classification.20of.20finite-dimensional.20vector.20spaces.html", "openwebmath_score": 0.6411661505699158, "openwebmath_perplexity": 7472.249502184706, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9343951625409307, "lm_q2_score": 0.9099070158103778, "lm_q1q2_score": 0.8502127139352712}}
{"url": "https://mathhelpboards.com/threads/pr-of-exactly-one-dice-showing-2.24196/", "text": "# Pr of exactly one dice showing 2\n\n#### navi\n\n##### New member\nSo I ran into this problem:\n\nYou roll two fair die. What is the probability of one of the dice showing 2 or the sum being at least 8?\n\nI thought it should be: 15/36 for the sum being at least 8 and 10/36 for exactly one dice showing 2, but it tells me it is wrong. I got these numbers by constructing a chart that showed the outcomes of the 2 die rolls.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nSo I ran into this problem:\n\nYou roll two fair die. What is the probability of one of the dice showing 2 or the sum being at least 8?\n\nI thought it should be: 15/36 for the sum being at least 8 and 10/36 for exactly one dice showing 2, but it tells me it is wrong. I got these numbers by constructing a chart that showed the outcomes of the 2 die rolls.\nYou have given two answers, but the question wants a single answer.\n\nLet $A$ stand for \"the sum being at least 8\", and let $B$ stand for \"exactly one of the dice showing 2\". You have correctly found the probabilities $\\def\\Prob{\\operatorname{Prob}\\,}\\Prob(A) = \\frac{15}{36}$ and $\\Prob(B) = \\frac{10}{36}.$ But the question asks for $\\Prob(A\\text{ or }B)$, in other words the probability of the dice showing either $A$ or $B$ (or possibly both). For that, you need to use the formula $\\Prob(A\\text{ or }B) = \\Prob(A) + \\Prob(B) - \\Prob(A\\text{ and }B)$.\n\n#### navi\n\n##### New member\nYou have given two answers, but the question wants a single answer.\n\nLet $A$ stand for \"the sum being at least 8\", and let $B$ stand for \"exactly one of the dice showing 2\". You have correctly found the probabilities $\\def\\Prob{\\operatorname{Prob}\\,}\\Prob(A) = \\frac{15}{36}$ and $\\Prob(B) = \\frac{10}{36}.$ But the question asks for $\\Prob(A\\text{ or }B)$, in other words the probability of the dice showing either $A$ or $B$ (or possibly both). For that, you need to use the formula $\\Prob(A\\text{ or }B) = \\Prob(A) + \\Prob(B) - \\Prob(A\\text{ and }B)$.\nThank you! I do not understand why you are subtracting the probability of A and B, though. Could you explain to me why?\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nThank you! I do not understand why you are subtracting the probability of A and B, though. Could you explain to me why?\nIt's in order to avoid double counting. Specifically, if the dice show 2 and 6, then they satisfy both conditions $A$ and $B$. If you just add the probabilities for $A$ and $B$ then you will have counted that case under both headings. So you need to subtract one of them off.\n\n#### navi\n\n##### New member\nIt's in order to avoid double counting. Specifically, if the dice show 2 and 6, then they satisfy both conditions $A$ and $B$. If you just add the probabilities for $A$ and $B$ then you will have counted that case under both headings. So you need to subtract one of them off.\nOkay, now I understand! I do have a follow up question, though: why would the probability of the union of A and B be 1/36 and not 2/36? Because it could be first dice 2, second dice 6 or first dice 6, second dice 2.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nOkay, now I understand! I do have a follow up question, though: why would the probability of the union of A and B be 1/36 and not 2/36? Because it could be first dice 2, second dice 6 or first dice 6, second dice 2.\nIt should indeed be 2/36.\n\n#### navi\n\n##### New member\nIt should indeed be 2/36.\nIt is a WebWork problem and it accepted this as the correct answer: 10/36 + 15/36 - 1/36\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nIt is a WebWork problem and it accepted this as the correct answer: 10/36 + 15/36 - 1/36\nHere is a possible explanation for that. You entitled this thread \"Pr of exactly one dice showing 2\", but the statent of the problem only refers to \"one of the dice showing 2\". I suspect that this may be intended to cover the case when both dice show 2. In that case, the probability of that event goes up from $\\frac{10}{36}$ to $\\frac{11}{36}$, and the probability of \"$A$ or $B$\" becomes $\\frac{11}{36} + \\frac{15}{36} - \\frac{2}{36}$, which agrees with the answer accepted by WebWork.\n\n#### Country Boy\n\n##### Well-known member\nMHB Math Helper\nBy the way, your title, \"Pr of one dice showing 2\", should be \"Pr of one die showing 2\" and \"you roll two fair die\" should be \"you roll two fair dice\". \"Dice\" is the plural of the singular \"die\".\n\n#### Wilmer\n\n##### In Memoriam\n1st roll: must be 2 or 6: probability = 2/6 = 1/3\n2nd roll: must be 2 if 1st was 6, or 6 if 1st was 2: probability = 1/6\n\n1/3 * 1/6 = 1/18", "date": "2020-07-14 09:26:44", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/pr-of-exactly-one-dice-showing-2.24196/", "openwebmath_score": 0.5222295522689819, "openwebmath_perplexity": 392.0336249445463, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806546550656, "lm_q2_score": 0.8670357735451835, "lm_q1q2_score": 0.8501985064322972}}
{"url": "https://math.stackexchange.com/questions/2838491/trick-or-indefinite-integration", "text": "# Trick, or Indefinite Integration?\n\nEvaluate $$\\int_{0}^{\\pi/2}\\dfrac{\\sin^6 x}{\\sin x + \\cos x}\\text{ d}x$$\n\nIs there a nice, elegant way of solving the above integral?\n\nHere's what I did - Replaced $f(x)$ by $f(\\pi/2-x)$, added and got rid of some common factors in numerator and denominator. This was followed by indefinite integration, and ultimately substituting the limits in the final expression.\n\nThe above method is kind of long, though - and I'm wondering if there's a quicker, shorter and rather elegant way of getting around this integral. Could someone post a solution, and share ideas?\n\nThanks a lot.\n\n\u2022 Please use MathJax to format your equations so that they are searchable, rather than images. \u2013\u00a0Clarinetist Jul 2 '18 at 13:30\n\u2022 If your integral is denoted by $I$, then $$2I=\\int_0^{\\pi/2} \\frac{\\sin^6 x + \\cos^6 x}{\\sin x + \\cos x} \\mathrm{d}x$$ \u2013\u00a0Crostul Jul 2 '18 at 13:43\n\u2022 Did that, what's next? \u2013\u00a0arya_stark Jul 2 '18 at 13:44\n\u2022 Is it true that $$8I=\\int_0^{2\\pi} \\frac{\\sin^6 x + \\cos^6 x}{\\sin x + \\cos x} \\mathrm{d}x$$ ?. I think it won't be a bad idea to use integration in the complex circle (see math.stackexchange.com/questions/308693/\u2026) then, if it's possible. \u2013\u00a0Gonzalo Benavides Jul 2 '18 at 17:01\n\n## 4 Answers\n\nI have solved it first as a indefinite integral and then as a definite integral. I am not sure how different it is with yours. Good luck. Let me know if you have any questions.\n\nA standard way for dealing with trigonometric integrals is to exploit the substitution $x=2\\arctan\\frac{t}{2}$.\nHere it leads to $$\\int_{0}^{\\pi/2}\\frac{\\sin^6 x}{\\sin x+\\cos x}\\,dx = 128\\int_{0}^{1}\\frac{t^6}{(2t+1-t^2)(t^2+1)^6}\\,dt$$ and the last integral can be computed by partial fraction decomposition. It equals $$\\frac{1}{8}\\left(2+\\sqrt{2}\\,\\text{arctanh}\\frac{1}{\\sqrt{2}}\\right)\\approx\\frac{28}{69}.$$ This also follows from $\\sin x+\\cos x=\\sqrt{2}\\cos\\left(x-\\frac{\\pi}{4}\\right)$ and $$\\int_{-\\pi/4}^{\\pi/4}\\frac{\\sin^6\\left(x+\\frac{\\pi}{4}\\right)}{\\sqrt{2}\\cos x}\\,dx=\\frac{1}{8\\sqrt{2}}\\sum_{k=0}^{6}\\binom{6}{k}\\int_{-\\pi/4}^{\\pi/4}\\sin^k(x)\\cos^{5-k}(x)\\,dx$$ where the contribution provided by odd $k$s is zero and $$2\\int_{0}^{\\pi/4}\\cos(x)^5\\,dx =\\frac{43}{30\\sqrt{2}},\\qquad 2\\int_{0}^{\\pi/4}\\sin(x)^2\\cos(x)^3\\,dx =\\frac{7}{30\\sqrt{2}}$$ $$2\\int_{0}^{\\pi/4}\\sin(x)^4\\cos(x)^1\\,dx =\\frac{3}{30\\sqrt{2}}$$ $$2\\int_{0}^{\\pi/4}\\sin(x)^6\\cos(x)^{-1}\\,dx =-\\frac{73}{30\\sqrt{2}}+4\\,\\text{arctanh}\\tan\\frac{\\pi}{8}.$$\n\nHint:\n\n$$2I=\\int_0^{\\pi/2}\\dfrac{\\sin^6x+\\cos^6x}{\\sin x+\\cos x}dx$$\n\nNow $\\sin^6x+\\cos^6x=(\\sin^2x+\\cos^2x)^3-3\\sin^2x\\cos^2x(\\sin^2x+\\cos^2x)=1-3\\sin^2x\\cos^2x$\n\nand $\\sin^2x\\cos^2x=\\dfrac{\\sin^22x}4=\\dfrac{1-\\cos4x}8$\n\nFor $J=\\displaystyle\\int_0^{\\pi/2}\\dfrac{\\cos4x}{\\sin x+\\cos x}dx=\\int_0^{\\pi/2}\\dfrac{\\cos4x}{\\sqrt2\\cos\\left(x-\\dfrac\\pi4\\right)}dx,$\n\nset $x-\\dfrac\\pi4=y,\\cos4x=\\cos4\\left(y+\\dfrac\\pi4\\right)=\\cos(4y+\\pi)=-\\cos4y$\n\n$\\implies-\\sqrt2J=\\displaystyle\\int_{-\\pi/4}^{\\pi/4}\\dfrac{\\cos4ydy}{\\cos y}=\\int_{-\\pi/4}^{\\pi/4}\\dfrac{1-8\\cos^2y+8\\cos^4y}{\\cos y}dy$\n\nNow use $\\cos3y=4\\cos^3y-3\\cos y$\n\nLet $J$ be the integral to be calculated, we want to get the computer result, here using sage:\n\nsage: var('x');\nsage: J = integral( sin(x)^6/(sin(x)+cos(x)), x, 0, pi/2 )\nsage: J\n1/16*sqrt(2)*log((2*sqrt(2)) + 3) + 1/4\n\n\nFor this, we use the simple idea of getting a $\\sin(x+\\pi/4)$ (up to a multiplicative constant) in the denominator, then do trivial computations...: \\begin{aligned} J &=\\int _0^{\\pi/2} \\frac{\\sin^6 x}{\\sin x + \\cos x}\\;dx \\\\ &=\\int _0^{\\pi/2}\\frac 1{\\sqrt 2} \\frac{\\sin^6 x}{\\frac 1{\\sqrt 2}\\left(\\sin x + \\cos x\\right)}\\;dx \\\\ &=\\frac 1{\\sqrt 2} \\int _0^{\\pi/2} \\frac {\\sin^6 x} {\\sin \\left(x + \\frac\\pi4\\right)}\\;dx \\\\ &\\qquad\\text{Substitution: }y = x+\\frac \\pi 4\\ , \\\\ &=\\frac 1{\\sqrt 2} \\int_{\\pi/4}^{3\\pi/4} \\frac {\\sin^6 \\left(y- \\frac\\pi4\\right)} {\\sin y}\\;dy \\\\ &=\\frac 1{\\sqrt 2} \\int_{\\pi/4}^{3\\pi/4} \\frac 18\\cdot\\frac 1{\\sin y} \\Big(\\ \\sin y-\\cos y\\ \\Big)^6\\;dy \\\\ &=\\frac 1{\\sqrt 2} \\int_{\\pi/4}^{3\\pi/4} \\frac 18\\cdot\\frac 1{\\sin y} \\Big(\\ \\sin^6 y -6\\sin^5y\\cos y +15\\sin^4y\\cos^2 y \\\\&\\qquad\\qquad -20\\sin^3y\\cos^3 y +15\\sin^2y\\cos^4 y -6\\sin y\\cos^5 y +\\cos^6 y \\ \\Big)\\;dy \\\\ &=\\frac 1{8\\sqrt 2} \\int_{\\pi/4}^{3\\pi/4} \\Big(\\ \\sin^5 y -6\\sin^4y\\cos y +15\\sin^3y\\cos^2 y \\\\&\\qquad\\qquad -20\\sin^2y\\cos^3 y +15\\sin y\\cos^4 y -6\\cos^5 y +\\frac{\\cos^6 y}{\\sin y} \\ \\Big)\\;dy \\\\ &= \\frac 1{8\\sqrt 2} \\int_{\\pi/4}^{3\\pi/4}\\left[\\ \\begin{aligned} -(1-\\cos^2y)^2 &\\,\\cdot\\,\\cos 'y \\\\ - 6\\sin^4y &\\,\\cdot\\,\\sin'y \\\\ - 15(1-\\cos^2y)\\cos^2y &\\,\\cdot\\,\\cos'y \\\\ - 20\\sin^2y(1-\\sin^2 y) &\\,\\cdot\\,\\sin'y \\\\ - 15\\cos^4y &\\,\\cdot\\,\\cos'y \\\\ - 6(1-\\sin^2 y)^2 &\\,\\cdot\\,\\sin'y \\\\ -\\frac{\\cos ^6y}{1-\\cos^2 y}&\\,\\cdot\\,\\cos'y \\end{aligned} \\ \\right]\\; dy \\\\ &= \\frac 1{8\\sqrt 2} \\int_{\\pi/4}^{3\\pi/4}\\left[\\ \\begin{aligned} -(1-\\cos^2y)^2 & \\\\ - 15(1-\\cos^2y)\\cos^2y & \\\\ - 15\\cos^4y & \\\\ -\\frac{\\cos ^6y}{1-\\cos^2 y}& \\end{aligned} \\ \\right]\\,\\cdot\\,\\cos 'y\\; dy \\\\ &\\qquad\\text{because for the \\sin'y-terms we have \\sin(\\pi/4)=\\sin(3\\pi/4)...} \\\\ &\\qquad\\text{Substitution: }u=\\cos t\\ ,\\ du=\\cos't\\, dt\\ , \\\\ &= \\frac 1{8\\sqrt 2} \\int_{-1/\\sqrt2}^{1/\\sqrt2}\\left[\\ (1-u^2)^2 +15(1-u^2)u^2 +15u^4 +\\frac {u^6-1+1}{1-u^2} \\right] \\; du \\\\ &= \\frac 1{4\\sqrt 2} \\int_0^{1/\\sqrt2}\\left[\\ (1-u^2)^2 +15(1-u^2)u^2 +15u^4 -(1+u^2+u^4) +\\frac {1}{1-u^2} \\right] \\; du \\\\ &= \\frac 1{4\\sqrt 2} \\int_0^{1/\\sqrt2}\\left[\\ 12u^2 +\\frac {1}{1-u^2} \\right] \\; du \\\\ &= \\frac 1{4\\sqrt 2} \\left[\\ 4\\left(\\frac 1{\\sqrt2}\\right)^3 +\\frac 12\\log \\frac{1+\\frac 1{\\sqrt 2}}{1-\\frac 1{\\sqrt 2}} \\ \\right] \\\\ &= \\frac 14 + \\frac 1{4\\sqrt 2}\\log(\\sqrt2+1)) \\ . \\end{aligned}", "date": "2019-05-20 03:18:08", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2838491/trick-or-indefinite-integration", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 4024.537863086435, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806518175515, "lm_q2_score": 0.867035771827307, "lm_q1q2_score": 0.8501985022875546}}
{"url": "https://math.stackexchange.com/questions/2794259/what-is-lim-x-to-infty-sin-x", "text": "# What is $\\lim_{x\\to\\infty} \\sin x$?\n\nWhat is\n\n$$\\lim_{x\\to\\infty} \\sin x$$\n\n?\n\nI always thought it was undefined; however, Wolfram|Alpha says that\n\n$$\\lim_{x\\to\\infty} \\sin x = -1 \\text{ to } 1$$\n\nNow, what is the correct answer? What does it even mean if the limit is not a single, distinct value, but rather an interval?\n\nOther than that, am I right in assuming that\n\n$$\\liminf_{x\\to\\infty} \\sin x = -1$$\n\nand\n\n$$\\limsup_{x\\to\\infty} \\sin x = 1$$ ?\n\n\u2022 They mean that the values of $\\sin(x)$ for $x\\to+\\infty$ accumulate to every value in $[-1,1]$. In particular, that implies the $\\limsup$, $\\liminf$ equations that you wrote, and that in the context of an introduction to calculus, the limit doesn't exist. \u2013\u00a0user561348 May 24 '18 at 12:37\n\u2022 @arugula so is it in fact undefined? \u2013\u00a0user500664 May 24 '18 at 12:37\n\u2022 Who knows what Alpha means, but I guess you could interpret it as \"for every point $p$ between -1 and 1, there's a sequence tending to infinity, such that $\\sin$ of that sequence is constantly $p$\". And yes you're right about the sup and inf limits. \u2013\u00a0Steve D May 24 '18 at 12:38\n\u2022 To find what Wolfram Alpha means may be difficult? But presumably this is the same answer as Mathematica itself, which does have extensive documentation you can look at. \u2013\u00a0GEdgar May 24 '18 at 12:45\n\nTake note that the sine function takes values between $-1$ and $1$. When you are talking about a limit to $\\infty$, it's an undetermined number, which is infinitely large. Now, taking into account the periodicity of the sine function, there is no possible way to determine a specific value, as it entirely depends on the nature of the \"infinite\" number.\n\nMore specifically, $-1 \\leq \\sin(x) \\leq 1, \\; \\; \\forall x \\in \\mathbb R$. This means that for any given $x$ over the real numbers, the sine function is bounded. Thus, all you can say about an undetermined infinite limit (it does not exist talking strictly mathematics), is :\n\n$$-1 \\leq \\lim_{x \\to \\infty} \\sin(x) \\leq 1$$\n\nWhat you mentioned though is indeed true :\n\n$$\\liminf_{x\\to\\infty} \\sin x = -1$$\n\n$$\\limsup_{x\\to\\infty} \\sin x = 1$$\n\n\u2022 So it can be understood as \"the limit doesn't exist, but if it existed it would be between -1 and 1\", right? \u2013\u00a0user500664 May 24 '18 at 12:42\n\u2022 Essentially, yes. The limit does not exist, but it surely is bounded. \u2013\u00a0Rebellos May 24 '18 at 12:43\n\u2022 alright, thank you. I will accept your answer as soon as it will let me. \u2013\u00a0user500664 May 24 '18 at 12:43\n\nThe limit doesn't exist and it can be proved formally by 2 subsequences with different limits, that is\n\n\u2022 $x_n=2n\\pi+\\frac{\\pi}2 \\to \\infty \\implies \\sin(x_n)=1$\n\n\u2022 $x_n=2n\\pi+\\frac{3\\pi}2 \\to \\infty \\implies \\sin(x_n)=-1$\n\nYes what is true is that $\\liminf=-1$ and $\\limsup=1$ indeed\n\n$$-1\\le\\sin x \\le 1$$\n\nand we have found 2 subsequences which tends to those limits.\n\n\u2022 Thank you for the proof as well. Your answer is very good, but I promised Rebellos to accept his answer earlier, so I am sorry. Thank you very much though. \u2013\u00a0user500664 May 24 '18 at 12:55\n\u2022 @ThomasFlinkow You are welcome! The choice of the answer to be accepted is absolutely up to you! Bye \u2013\u00a0user May 24 '18 at 13:05", "date": "2019-10-17 10:37:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2794259/what-is-lim-x-to-infty-sin-x", "openwebmath_score": 0.7635952234268188, "openwebmath_perplexity": 334.8989577620078, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806518175514, "lm_q2_score": 0.8670357615200474, "lm_q1q2_score": 0.8501984921804551}}
{"url": "https://math.stackexchange.com/questions/2058605/calculating-a-cumulative-distribution-function-cdf", "text": "# Calculating a Cumulative Distribution Function (CDF)\n\nLet $Z$ be a continuous random variable with probability density function:\n\n$$f_Z(z) = \\begin{cases} \\gamma(1 + z^2) & \\mbox{ if } -2 < z < 1, \\\\ 0 & \\mbox{ otherwise} \\end{cases}$$\n\na) For what $\u03b3$ is this possible?\nb) Find the cumulative distribution function of $Z$.\n\n\u2022 For question a do you know what a probability distribution is and the requirements? for question b do you know what a cumulative distribution is and the integral/summation involved? \u2013\u00a0Chinny84 Dec 14 '16 at 15:47\n\u2022 I know that to solve part b I need to solve the integral from negative infinity to z of fZ(z), but I don't know how to solve part a, which is why I am stuck. \u2013\u00a0Monil Dec 14 '16 at 15:53\n\u2022 Hint: which value should the integral of the PDF have over all domain? \u2013\u00a0Jes\u00fas Ros Dec 14 '16 at 15:55\n\u2022 I'm sorry, I still don't understand how to do it. \u2013\u00a0Monil Dec 14 '16 at 16:02\n\u2022 If $f_Z(z)$ is a PDF, which is the value of $P[-\\infty<z<\\infty]=\\int_{-\\infty}^{\\infty}f_Z(z)dz$? Or in other words, which is the probability that $z$ lies between $(-\\infty,\\infty)$? \u2013\u00a0Jes\u00fas Ros Dec 14 '16 at 16:04\n\n$a)$ $$\\int_{-\u221e}^\u221e\\ f_Z(z)\\ dz\\ = 1$$ $$\\int_{-2}^1\\ \u03b3(1 + z^2)\\ dz= 1$$ $$\u03b3(z + \\frac{z^3}3)\\ |_{-2}^1\\ = 1$$ $$\u03b3\\ = \\frac16$$\n$b)$ $$F_Z(z)\\ =\\ \\int_{-2}^z\\ f_Z(z)\\ dz$$ $$F_Z(z) = \\int_{-2}^z\\ \\frac16(1 + z^2)\\ dz$$ $$F_Z(z) = \\frac16(z + \\frac{z^3}3) |_{-2}^z$$ $$F_Z(z) = \\frac16[(z + 2) + (\\frac{z^3 + 8}3)]$$\n\u2022 Notice that $F_Z(-2) = 0$ and $F_Z(1) = 1.$ That's as it must be. Do you you understand why I made sure of that? Guessing that you do. (+1) \u2013\u00a0BruceET Dec 15 '16 at 6:44", "date": "2019-06-18 02:59:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2058605/calculating-a-cumulative-distribution-function-cdf", "openwebmath_score": 0.7520555257797241, "openwebmath_perplexity": 270.48412764125584, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9911526460930947, "lm_q2_score": 0.8577681049901037, "lm_q1q2_score": 0.8501791269952008}}
{"url": "http://chiryo.it/edcl/growth-or-decay-function.html", "text": "Bank accounts that accrue interest represent another example of exponential growth. Exponential Growth & Decay 06/01/09 Bitsy Griffin PH 8. Tell whether the function shows growth or decay. a) let P represent the population \u201ct\u201d years after 2000. Then graph. Lesson 2 Exponential Growth & Decay Notes Exponential Functions: GROWTH & DECAY *Many real world phenomena can be modeled by functions that describe how things grow or decay as time passes. 4 Modeling Exponentials; 4. If we start with only one bacteria which can double every hour, how many bacteria will we have by the end of one day?. For each exponential function, Identify the Initial Value and the Growth/Decay Rate (as a %). Then, solve the function and get the answer!. A person invests $100,000 at a nominal 12% interest per year compounded continuously. Exponential Growth & Decay Sketch the graph of each function. The range is the interval (0, c), where c is the carrying capacity. 43%) At close: 4:00PM EST. If b is replaced by 1 -r and x is replaced by t, then the function is the exponential decay model y = a (1 -r) t, where a is the initial amount, the base (1 -r) is the decay factor, r is the decay rate, and t. The exponent for exponential growth is always positive and greater than 1. Exponential word problems almost always work off the growth / decay formula, A = Pe rt, where \"A\" is the ending amount of whatever you're dealing with (money, bacteria growing in a petri dish, radioactive decay of an element highlighting your X-ray), \"P\" is the beginning amount of that same \"whatever\", \"r\" is the growth or decay rate, and \"t\" is time. Growth and decay problems are used to determine exponential growth or decay for the general function (for growth, a \u232a 1; for decay, 0 \u2329 a \u2329 1). \u2022 If the function models the amount, or. Exponential Decay Exponential decay is the change that occurs when an original amount is reduced by a consistent rate over a period of time. Module 1: Graphs and Functions. You also use them in science courses like biology, chemistry, physics, etc. How do you know if L : ; represents an exponential growth or exponential decay function? Exponential _____ if Exponential _____ if. Classify the model as. , (1/2) 1 > (1/2) 2 > (1/2) 3. An exponential growth function can be written in the form, y = ab x, where b > 1. The common applications of the exponential funciton range from population modeling, to tracking drug levels in the blood stream, to using carbon dating to estimate the age of an artifact. Explain 2 Modeling Exponential Decay Recall that a function of the form y = a b x represents exponential decay when a > 0 and 0 < b < 1. Then graph f(x)=(1/4)^x Log On Algebra: Exponent and logarithm as functions of power Section. 81 )x Growth and Decay The value of a home worth$165,000 will increase by 3. Viral growth. Exponential Growth And Decay Functions - Displaying top 8 worksheets found for this concept. y =\u22c5+1200 (1 0. This is usually expressed as a percentage-The Growth/Decay rate is not explicitly found in the equation but if we use the Growth/Decay factor, we can find the Growth/Decay rate Vocabulary about exponentials. org are unblocked. I am taking a course in managerial math and I can't seem to get the basics of finding a growth or decay factor for a change in data Answer: When the rate of change in your data from one period to the next is a constant percentage (%), the function which describes that change is call an exponential function. This is an exponential growth function. From this, we get as the growth rate (which represents decay rate ). 16 (Remember, get the bases to match and use the one to one property = 27 X. Logarithms were invented by John Napier. 9) Identify if the function below is growth or decay: f(t) =300 \u00d7 1. Start studying Exponential Growth & Decay. , (1/2) 1 > (1/2) 2 > (1/2) 3. Also, the a value can tell us if the exponential curve is concave up (opening upwards) or concave down (opening downwards). Likewise, if A > 0, then the more general exponential function Abt alsoexhibitsexponentialgrowth,sincethegraphofAbt isjustaverticalscalingofthe graph of bt. x (t) is the value at time t. The parent function isf(x) \u2014 If, where b is any real number greater than 0, except 1. Also, if you substitute any x-value from the table into the function rule, it produces the correct y-value. When a quantity grows by a fixed percent at regular intervals, the pattern can be represented by the functions, Growth: y = Decay: Y = (70 \u2014 r) x a x Example:. Worksheet 9 Memorandum: Finance, Growth and Decay. Harold\u2019s Exponential Growth and Decay Cheat Sheet 16 May 2016 Discrete Continuous \ud835\udc34=\ud835\udc43(1+ N J) \ud835\udc5b\ud835\udc61 \ud835\udc34=\ud835\udc43 \ud835\udc5f\ud835\udc61 Simple Interest: \ud835\udc34=\ud835\udc43+ =\ud835\udc43+\ud835\udc43=\ud835\udc43(1+ N P) A = Amount after time t P = Original amount, such as principle e = The natural number (~2. It can be used as a worksheet function (WS) in Excel. Exponential growth and exponential decay are common mathematical concepts used very frequently in modeling. Putting money in a savings account 2. 5 million in sales. The function is decreasing exponentially making it exponential growth. What percent of the teams are eliminated after each round? Explain how you know. Damping of oscillating system. 25 per sandwich. From population growth and continuously compounded interest to radioactive decay and Newton\u2019s law of cooling, exponential functions are ubiquitous in nature. 02x Find the balance in each account after the given period. decay Get 3 of 4 questions to level up! Graphing exponential growth & decay Get 3 of 4 questions to level up! Writing functions with exponential decay Get 3 of 4 questions to level up!. Exponential Growth and Decay are functions which have been widely used to model the behavior of a variety of topics. Homework \u2013 Due Monday, December 17. The exponential curve is especially important in mathematics. com - View the original, and get the already-completed solution here! Solve for solutions: The exponential models describe the population of the indicated country, A, in millions, t years after 2003. Exponential growth and decay (Part 11): Half-life. Exponential Growth An exponential growth function can be written in the form y = ab x where a > 0 and b > 1. Example: Graphing Exponential Growth. We will describe some models that are e ective only for a limited amount of time. 2) to the \ud835\ude35/10 power, and classify them as representing exponential growth or decay. y = 2,000,000 (. asked by Deborah on January 20, 2013; math. Exponential growth and decay often involve very large or very small numbers. In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. This looks just like our simplistic examples above except the base is now Euler\u2019s number, the exponent is multiplied by a constant and the base is multiplied by a constant. Exponential growth and decay (Part 11): Half-life In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. A 5 step guide for sketching exponential functions is discussed to help students remember. What values of b make it an exponential growth function? C b. From population growth and continuously compounded interest to radioactive decay and Newton's law of cooling, exponential functions are ubiquitous in nature. Classify the model as Exponential GROWTH or DECAY. Remember that there are two ways to tell whether a graph is an example of exponential decay or exponential growth. 33x D) y = 0. In the above examples we saw that \u0192 1 (x) = 2 x is an example of an exponential growth function (the function grows by a constant factor of 2 in other words it doubles after each growth period) and \u0192 2 (x) =0. What values of b make it an exponential growth function? C b. Exponential Decay Functions Exponential decay occurs when a quantity by the same factor over equal intervals of time. Start studying Exponential Growth & Decay. We say that they have a limited range. The second example is the exponential decay function y = 40 (0. Assessments Quiz EU1 \u2013 Simplifying Exponential Expressions Quiz EU2 \u2013 Identifying Exponential Functions Quiz EU3 \u2013 Graphing Exponential Functions Unit Test \u2013 Exponential Functions. It is also the opposite of exponential growth. What type of function is y = 7(5/4)x? Exponential Growth and Decay DRAFT. 2 x a 1 b 1. Exponential Function. Graph each function. Some of the worksheets for this concept are Exponential growth and decay, Exponential growth and decay work, Exponential growth and decay word problems, Exponential growth and decay functions, Exponential growth and decay, 4 1 exponential functions and their. Test Multiple Choice Identify the choice that best completes the statement or answers the question. : Exponential Growth & Decay (GPS ADV Alg). , (1/2) 1 > (1/2) 2 > (1/2) 3. 06 the rate of decay of a chemical compound that has a half-life of 5730 years the rate of growth of a bacteria strain. Gizmo Warm-up. Growth/Decay Factor_____ Growth/Decay R ate_____ Initial Value_____ 7. Title: Untitled. y = 2 \u20223x 12. EXPONENTIAL GROWTH AND DECAY GRAPHS OF EXPONENTIAL FUNCTIONS An exponential function is an equation of the form y=abx (with b!0). Write an exponential growth function to model this situation. ) percentage growth rate \ud835\udc5f as a decimal, the growth factor \ud835\udc82 = \ud835\udfcf + \ud835\udc93. Make a table ot values tor the tunction. Does the function f (x) = 5(0. Let's consider exponential decay and exponential growth by inspecting their respective general shapes of their graphical representations. Exponential Decay Function. Exponential Decay 5. We need to change this percentage to a _____. Online exponential growth/decay calculator. Figure 1 shows the graph of a typical exponential function, assuming y 0 >0. 2 x a 1 b 1. Example: An unknown radioactive element decays into non-radioactive substances. 1 Exponential Growth and Decay A linear function has constant slope. Exponential Growth B. Write out the 4 step process for solving a continuous growth or decay function, given the initial value, rate of growth or decay, and time, \ud835\udc61. Thus, the Growth function : And, decay functions : 2) Given equation, Thus, Option 'D' is correct. The x-axis is a horizontal asymptote. Another thing to notice is that if you have y=5^(-x) this function could be rewritten in this way: y=(1/5)^x. We know that growth decay function is ##N_{t}=N_{0}\\\\times e^{\\\\lambda t}##. Lab (Exponential Growth and Decay) The purpose of this lab is to provide a model to illustrate exponential growth and decay. Unit 6 Exponential Functions and Their Applications 249 Lesson 19 Applications of Exponential Growth and Decay 3. It is a decreasing function,because, for The Curve has one Asymptote which is, y=0. 9th - University grade. When b > 1, the graph is positive, increasing, and concave up. Which situation can be studied with a noncontinuous exponential growth or decay function? Save the growth of a wildlife population that increases by 4% each year the rate of growth of a virus that has a yearly growth rate of 0. A Gila Monster is about 16 cm long at birth. t is the time of growth/decay. notebook January 30, 2015 Writing Exponential Growth and Decay Functions in Real Life!!! Remember this from yesterday? A(t) = a(1 + r)t Sometimes you have to use given values to find r! 15. 23)* Initial Value Initial Value GROWTH or DECAY Rate GROWTH or DECAY Rate 3. Exponential growth and decay by a factor. Exponential Functions Part 1 - Graphing - Duration: 14:34. Logistic Growth And Decay By Phillip Neman Logistic Decay Example Problem ( b ) What is the percentage of remaining wood products after 10 years? History of the Logistic Function The logistic function is sometimes called the Verhulst model because it was first published by Pierre. Description of Changes. the x/r is the percent it grew or decreased divided by the rate you're doing at, so one year would be r=1, one month would be r=12, one day would be r=365. Page 5 of 25 Additional Note: Just like a growth factor, you need to be able to identify a decay factor. 98 Which is an exponential decay function?. 3) Given, The initial number for blog, P = 20, Rate per week, r = 20% = 0. Shows growth/decay at consecutive intervals Looks like a normal or backwards \"L\" or \"r\" graphically Key words: increasing, decreasing, lost, depreciate Linear Functions: y = mx + b Shows a constant rate of change Looks like a straight line graphically Key words: each 6. What is the rate of growth or rate of decay? 3. InCA Assignment for 3/13. 02) to the \ud835\ude35 power, \ud835\ude3a = (0. h OR one value of the function at a future time. Human population also grows exponentially. Exponential models that use e as the base are called continuous growth or decay models. Does this function represent exponential growth or exponential decay? B. No obligation, cancel anytime. the x/r is the percent it grew or decreased divided by the rate you're doing at, so one year would be r=1, one month would be r=12, one day would be r=365. Hello All, I am trying to fit a curve by exponential growth. Start studying Exponential Growth & Decay. 24 #13, 14, 16, 19, 20a. The Exponential function in Excel is often used with the Log function, for example, in case, if we want to find the rate of growth or decay, in that case, we will use the EXP and the LOG function together. An algebra equation involves a variable representing an un-known number, often denoted by x; and to solve the equation means to nd the nu-merical values of x which make the equation true. \"Kavod\" = \"Respect\" btw, YAY MATH!. 2)(t/10), and classify them as representing exponential growth and decay. For example, identify percent rate of change in functions such as y = (1. Determine whether each function is an example of exponential growth or decay. It can be used as a worksheet function (WS) in Excel. Viral growth. From population growth and continuously compounded interest to radioactive decay and Newton's law of cooling, exponential functions are ubiquitous in nature. The questions in this maze are of \"Parent Exponential Equations\". If you graph this function, you will see it decays really fast, but it actually does not have exponential decay. 2 Graph the following Exponential Functions 8) Function: y=3X 9) Function: Decide if the following tables are exponential or Linear and then write the equation that corresponds with the table. The Exponential Growth Calculator. MEMORY METER. 1%; exponential decay. Exponential functions are commonly used in the biological sciences to model the amount of a particular quantity being modeled, such as population size, over time. function; exponential growth/decay; asymptotes; and end behavior. Exponential Functions, exponential growth, exponential decay, decay factor, growth rates, exponents, rules of exponents, scientific notation. The number C gives the initial value of the function (when t = 0) and the number a is the growth (or decay) factor. Tell whether the function y = 2( 5 ) shows growth or decay. Exponential growth / decay is a specific way that a quantity may increase / decrease over time. Harold\u2019s Exponential Growth and Decay Cheat Sheet 16 May 2016 Discrete Continuous \ud835\udc34=\ud835\udc43(1+ N J) \ud835\udc5b\ud835\udc61 \ud835\udc34=\ud835\udc43 \ud835\udc5f\ud835\udc61 Simple Interest: \ud835\udc34=\ud835\udc43+ =\ud835\udc43+\ud835\udc43=\ud835\udc43(1+ N P) A = Amount after time t P = Original amount, such as principle e = The natural number (~2. 5, is between greater than 1 than 0. The population in the town of Deersburgh is presently 42,500. The first step is to enter the initial value (x0). The next step is to find the trend by noting that we are left with a certain percentage of the substance. Is the pictured graph growth, decay, or linear or none? answer. $\\endgroup$ \u2013 Matthew Leingang Jan 15 '15 at 17:01 $\\begingroup$ Although your biggest problem is important, you will also need to figure out how to handle the \u201ccompounded quarterly\u201d part of the problem. Property #1) rate of growth starts slow and increases (Read on, to learn more about this property, which is the primary focus of this web page) Of note: Exponential decay is not the inverse of exponential growth. Since there is a negative exponent, it reverses what the base should be. Under Construction. Modeling Exponential Growth and Decay NOTES In the exponential growth and decay formulas, y = final amount, a = original amount, r = rate of growth or decay, and t = time. com - View the original, and get the already-completed solution here! Solve for solutions: The exponential models describe the population of the indicated country, A, in millions, t years after 2003. -1-1 I a 34 3 4-2 Solve each of the following equations for x. In today\u2019s post, I\u2019ll discuss radioactive decay and the half-life formula. Xtra Gr 11 Maths: In this lesson on Financial Maths we consider the following: Different compounding periods, nominal and annual effective rates, depreciation, linear depreciation as well as reducing balance depreciation. What is the value of the car when it is 12 years old?. For a decay function, the range includes all numbers between the starting value and O. Substituting into a table of values gives us: We plot these points to give:. Differential Equations In Section 6. Example #1 - \u00a35000 is invested in a savings bond, which pays 7% p. That's the difference between a positive and negative growth rate. You will also decide whether or not investing large amounts of money into a home or automobile are wise financial decisions. Exponential growth and decay by a factor. The main purpose of this task is to introduce the decay graph and highlight its differences and similarities to a growth graph. This algebra and precalculus video tutorial explains how to solve exponential growth and decay word problems. Exponential Growth B. Define exponential decay. 98 Which is an exponential decay function?. Concepts from the last few videos, particularly exponential growth and decay functions, are covered in this video about word problems. notebook January 30, 2015 Writing Exponential Growth and Decay Functions in Real Life!!! Remember this from yesterday? A(t) = a(1 + r)t Sometimes you have to use given values to find r! 15. Lesson #42: Exponential Growth and Decay Functions. How do you know if L : ; represents an exponential growth or exponential decay function? Exponential _____ if Exponential _____ if. Left 2 units c. 99)^x\\)? Determine whether the function represents exponential growth or exponential decay. Exponential growth and decay can be determined with the following equation: N = (NI)(e^kt). The base b determines the rate of growth or decay: If 0 b 1 , the function decays as x increases. t is the time in discrete intervals and selected time units. r = the growth rate. \u2212 If k < 0, the function will always be negative (it is flipped upside down). Chapter 4 4. Choose a topic that can be related to exponential growth or decay. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. Human population also grows exponentially. Primary objective of this lecture is to present on Exponential Growth and Decay. Property #1) rate of growth starts slow and increases (Read on, to learn more about this property, which is the primary focus of this web page) Of note: Exponential decay is not the inverse of exponential growth. What values of b make it an exponential growth function? C b. Find the value of the car after 10 years. In this section, we will study some of the applications of exponential and logarithmic functions. Series of questions lead through this fascinating topic so that formula becomes clear and connected to the physical world. When it's a rate of decrease, you have an exponential decay function! Check out these kinds of exponential functions in this tutorial!. Sketch an exponential decay function. Explain 2 Modeling Exponential Decay Recall that a function of the form y = a b x represents exponential decay when a > 0 and 0 < b < 1. A di erential equation (DE) involves. f(x) = 2 x 62/87,21 Make a table of values. Another convention is to always write the exponential functions in terms of the natural constant e. An exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. And I just came back from this teaching conference, where one of the sessions included a few handouts of the types of problems that this one charter school uses. You must use your knowledge of exponential growth and decay functions to make some important financial decisions. Quantitative Reasoning Linear Models, Scatterplots, Exponential Growth, Exponential Decay Questions 1-3. a = value at the start. 8% per year and the current population is 1543, what will the population be 5. x(t) = x 0 \u00d7 (1 + r) t. Exponential decay models decrease very rapidly, and then level off to become asymptotic towards the x-axis. 25)^x growth or decay y-intercept: asked by Aaron on April 9, 2017; Algebra. That is, the rate of change is always the same, and each value is determined from the previous value by adding the same amount. Tell whether the equation represents exponential growth, exponential decay, or neither. Exponential decay and be used to model radioactive decay and depreciation. Geometric growth or decay is the same as exponential growth or decay except the function is only evaluated at discrete values. Write!an!exponential!function!to!model!each!situation. ' Notice that exponential growth word problems often involve compound interest, and exponential decay word problems often involve depreciation. Exponential decay occurs when 0 < b< 1. Ma 112 Precalculus Section 5. Choose the word or phrase from the list that best completes each sentence. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. In the Exponential Growth and Decay Gizmo\u2122, you can explore the effects of C and r in the function y = C(1 + r)t. p(t) = 50(0. Exponential Decay. \" \"We just looked at an exponential growth function. Determine if the equation y = 2. Differentiate between growth and decay by examining functions % Progress. Exponential growth and decay by percentage. Exponential Growth and Decay Formulas The formula that represents exponential growth or decay is y = a bx, where a = original (initial) amount b x = growth factor (1 + r) final amount unit of measure (usually time) PART I: Determining Exponential Growth and Decay. Graph the function. The odds of you using them on your next trip to the supermarket are fairly slim (more on using odds later), but some professions simply would not be able to do the things they do without them. Growth Function in Excel. The instructions are \u201cThe function below represents a growth or decay What is the initial quantity?. amount after. Exponential Decay functions model many real world scenarios. The GROWTH Function is categorized under Excel Statistical functions. Exponential decay functions also cross the y-axis at (0, 1), but they go up to the left forever, and crawl along the x-axis to the right. P(t)= Poe^(-kt). - when 0 b < 1, the function represents a decay function Writing an Equation Using y \u2014 abl There are many applications of exponential fimctions in real. 2 Exponential Decay: Worked Examples. Microsoft Word - Exponential growth and decay problems 1 Author: kstewart Created Date: 4/7/2020 7:02:47 AM. Exponential growth refers to an amount of substance increasing exponentially. It has grown from 30 grams to 50 grams in 48 hours. Online exponential growth/decay calculator. If A is decaying continuously at rate r, then A may be written as follows A = A0e\u2212rt. 32 MB] A Powerpoint presentation on Exponential Growth and Decay. A) 3) 4) y SECTION 3: Determine which functions are exponential growth and which are exponential decay. Exponential Growth and Decay Make Up 1. Write a function V that will predict the value of a home after x years. , a function in which the time value is the exponent. Exponential Growth and Decay Functions An exponential function has the form y = abx, where a \u2260 0 and the base b is a positive real number other than 1. Then find the painting's value in 15 years. Worksheet 9 Memorandum: Finance, Growth and Decay. 4 Graph an exponential function of the form f(x) = ab^x. growth or exponential decay. A town had a population of 53700 in 1996 and a population of 58100 in 2000. These exponential functions describe both the rise and the fall of particular systems, especially systems that can be described through mathematical concepts, such as population or radioactive decay. Does this function represent exponential growth or exponential decay? B. In this case, each number, starting at 5 is multiplied by 2 to an exponent power such as 2. 4% compounded. Exponential Growth and Decay Word Problems Write an equation for each situation and answer the question. , 5% becomes 1. 1 Examples of exponential growth or decay. Step 3: Substitute for t. Probably the most well known example of exponential decay in the real world involves the half-life of radioactive substances. These included the growth of populations and the decay of radioactive substances. Typically what you need to do is: 1. Determine whether the function represents exponential growth or decay. Formula to calculate exponential growth and decay is given by:. Exponential Growth and Decay. In the same way, whenever the base is less than 1 and greater than 0, the exponential function shows decay. Chapter 4 4. It gets rapidly smaller as x increases, as illustrated by its graph. 5% every year. It is a decreasing function,because, for The Curve has one Asymptote which is, y=0. The simplest type of exponential growth function has the form y = Core Concept Core Concept parent. Write!an!exponential!function!to!model!each!situation. If you're behind a web filter, please make sure that the domains *. December 14, 2018 December 14, 2018 mshallmaa Leave a comment. Exponential functions tell the stories of explosive change. Exponential Functions - Day 1 Homework Growth & Decay Name Date Block 1. The formulas below are used in calculations involving the exponential decay of, for example, radioactive materials Let\u2019s code a Matlab function to calculate the final amount of a substance, given the elapsed time, half-life and initial amount. Parameter a (called the amplitude) affects the height of the wave,. A model for decay of a quantity for which the rate of decay is directly proportional to the amount present. Exponential Decay Formula: Make a substitution for A and t since it is known that the half-life is 1690 years and : Solve for the decay rate k: Start by dividing both sides by the coefficient to isolate the exponential factor. amount after. The function g(x)=(1 2)x is an example of exponential decay. The stuffed animal was originally worth $6, so. growth function. Try it risk-free. Described as a function, a quantity undergoing exponential growth is an exponential function of time, that is, the variable representing time is the exponent (in contrast. 2 1536 12288 98304 786432 0. Scribd is the world's largest social reading and publishing site. In 2000, there were 236,000 people in the city. When I graph points relating to the function, as. As, x\u2192\u221e, So, when , x\u2192\u221e,y\u21920. Exponential growth: y = a (1 + r)t Exponential decay: y = a (1 \u2212 r)t The population of a city is increasing at a rate of 4% each year. Find the value ot the base. ) Comparing the rates of growth (or decay) will only yield the same results when both rates are converted to the same compounding period. Graph the function. Exponential growth (including exponential decay) occurs when the growth rate of a mathematical function is proportional to the function's current value. It is a decreasing function,because, for The Curve has one Asymptote which is, y=0. 7\u00ad1 Graphing Exponential Functions. Exponential growth and decay by percentage. An exponential function where a > 0 and 0 < b < 1 represents an exponential decay and the graph of an exponential decay function falls from left to right. b) c) 2) A car was purchased for$20,000. \u2022 Fill in the table, showing the increase in area of the bacteria over eight weeks. Understanding how to graph exponential growth and decay functions is a very important tool for future courses. 43%) At close: 4:00PM EST. 4 Modeling Exponentials; 4. Exponential Growth: a function or model which begins increasing at a very slow rate, then keeps growing faster and faster; functions in the form of y = P(1 + r) t. Continuing with the unit on exponential functions, today we spent some time working on creating functions that model other real world scenarios. This function is useful for describing many very different observations in science. A 5 step guide for sketching exponential functions is discussed to help students remember. Exponential models that use e as the base are called continuous growth or decay models. 1 as x gets bigger so does 5^x and so does the expression : Growth. 3c Use the properties of exponents to transform expressions for exponential functions. In the Exponential Growth and Decay Gizmo\u2122, you can explore the effects of C and r in the function y = C(1 + r)t. To solve real-life problems, such as modeling the height of a sunflower in Example 5. One Bernard Baruch Way (55 Lexington Ave. Damping of oscillating system. The mathematical model of exponential growth is used to describe real-world situations in population biology, finance and other fields. 825 13) Identify the RATE as a percent for the. How Do You Solve a Word Problem with Exponential Growth? If something increases at a constant rate, you may have exponential growth on your hands. Exponential Growth and Decay. You must use your knowledge of exponential growth and decay functions to make some important financial decisions. unit 5 worksheet 6 graphing exponential growth and decay functions UNIT 5 WORKSHEET 6 GRAPHING EXPONENTIAL GROWTH_DECAY. What type of function is y = 7(5/4)x? Exponential Growth and Decay DRAFT. From population growth and continuously compounded interest to radioactive decay and Newton's law of cooling, exponential functions are ubiquitous in nature. An exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. Differential Equations In Section 6. Identifying Exponential Growth and Decay. Company Outlook. The most important feature of an exponential function is that the independent variable is in the exponent of some base, usually an agreed-upon (but constant) base, like 2, 10 or the special number \"e\". The table shows the world population of the lynx in 2003 and 2004. what multiplicative rate of change should Hal use in his function? 0. ) Write an. In 2013, a research company found that smartphone shipments (units sold) were up 32. Exponential growth and decay is a concept that comes up over and over in introductory geoscience: Radioactive decay, population growth, CO 2 increase, etc. It is generally used to express a graph in many applications like Compound interest, radioactive decay, or growth of population etc. 7% worldwide from 2012, with an expectation for the trend to continue. Identify the asymptote of each graph. amount after. V(x) = 165,000 ( 1. Exponential growth and exponential decay questions Basic steps Most basic exponential growth or exponential decay questions in HSC Maths (2 Unit) exams involve a series of substitutions to get the answers. Exponential growth is a specific way that a quantity may increase over time. 9th - University grade. Exponential decay and be used to model radioactive decay and depreciation. Exponential Decay of Quantum Wave Functions I\u2019ve no doubt that for ODEs, the questions and techniques for exponential decay of solutions go back a long way, maybe even to the nineteenth century. Exponential Growth and Decay Name_____ Date_____ Period____ Solve each exponential growth/decay problem. A 5 step guide for sketching exponential functions is discussed to help students remember. Growth is also known as 'appreciation' when it is said that something has appreciated in value, meaning that its value has increased. Exponential Growth and Decay DRAFT. The Microsoft Excel GROWTH function returns the predicted exponential growth based on existing values provided. Classify the model as. Worksheets are Exponential growth and decay, Graphing exponential, Exponential growth and decay work, Concept 17 write exponential equations, Exponential growth and decay word problems, Exponential functions work 1, Exponential population growth, Honors pre calculus d1 work. Finance: Compound interest. Shows growth/decay at consecutive intervals Looks like a normal or backwards \"L\" or \"r\" graphically Key words: increasing, decreasing, lost, depreciate Linear Functions: y = mx + b Shows a constant rate of change Looks like a straight line graphically Key words: each 6. Exponential functions have many scientific applications, such as population growth and radioactive decay. Formula to calculate exponential growth and decay is given by:. a) Write an equation that gives the value of the car, y, after x years. a logarithmic functions. Exponential Functions: Graphs. Exponential Decay Formula: Make a substitution for A and t since it is known that the half-life is 1690 years and : Solve for the decay rate k: Start by dividing both sides by the coefficient to isolate the exponential factor. ANSWER KEY 1. And I just came back from this teaching conference, where one of the sessions included a few handouts of the types of problems that this one charter school uses. Research your topic and learn how it is related to exponential functions. Exponential Growth and Decay Word Problems Worksheet Answers \u2013 Begin customizing it and you could also to open it on your document window when you find a template that you would like to use! You will discover others call for a premium account and that a number of the templates are free to use. If the rate of increase is 8% annually, how many stores does the restaurant operate. Then, solve the function and get the answer!. Student Exploration: Exponential Growth and Decay. Growth or Decay? Write a function that represents this situation Answer: 1. Exponential Functions Part 1 - Graphing - Duration: 14:34. The parent exponential function is,f(x) = bx, where the base, b, is a constant and the. 5 \u00ad Exponential Functions Identify whether the equation represents a growth or decay function. In the Exponential Growth and Decay Gizmo\u2122, you can explore the effects of C and r in the function y = C(1 + r)t. The basic parent function of any exponential function is f ( x) = bx, where b is the base. It appreciates 6% per year. 1 Exponential Models Every model is, at the very end, an approximation of real phenomena. While function with exponential decay DO decay really fast, not all functions that decay really fast have exponential decay. Students will cut out the pieces and match a word problem to the type of exponential function (growth or decay) and the answer. We see these models in finance, computer science, and most of the sciences, such as physics. In 2000, there were 236,000 people in the city. 7) Worksheet 6-6. where C is the initial amount (the amount before any decay occurs), r is the. That is, the rate of change is always the same, and each value is determined from the previous value by adding the same amount. In fact, we can use the Exponential Growth and Decay Formula to find snow depth levels, the magnitude of a star, how temperature affects a body, or how a fast-food chain expands its business as Khan. His biomedical. \u2013 The graph decreases at a decreasing rate. First, they evaluate for a variable given the second variables value. Bank accounts that accrue interest represent another example of exponential growth. 825 13) Identify the RATE as a percent for the. Lesson #3: Applications of Exponential Growth or Decay Review An exponential function is a function whose equation is of the form y \u2014abl - when b > 1, the function represents a function. Exponential growth is a specific way that a quantity may increase over time. Also, the a value can tell us if the exponential curve is concave up (opening upwards) or concave down (opening downwards). Yes, any teacher can have a random Britney Spears moment. Explaining the shape of exponential functions, and why exponential functions illustrate growth and decay. to have exponential growth or exponential decay. Exponential Growth Formula Find the Exponential Growth Function That Models Given Data Example 4. 2 b 1, so the function shows exponential growth. Base Function: y = \u00bdx a. Find a function that models the population after t years. In talking about problems like population growth, we needed to learn about the exponential function. the set of all real numbers). Is this exponential growth or decay? Preview this quiz on Quizizz. In this section, we will study some of the applications of exponential and logarithmic functions. The exponential curve is especially important in mathematics. But as time continues, the quadratic function changes direction and grows to infinity, while the exponential decay function approaches an asymptote. 2x 10) Write and solve an exponential function to model the situation and find the value after the given time, Round to the nearest whole number. In a straight line, the \u201crate of change\u201d is the same across the graph. If 0 b 1, the function shows decay. No matter how many are in the population at some point in time, the percent that leave the population in the next period. What type of function is y = 7(5/4)x? Exponential Growth and Decay DRAFT. 8 Logistic Growth Functions 517 Evaluate and graph logistic growth functions. Determine the exponential functions that model the populations of both cities. The function helps calculate the predicted exponential growth by using the existing data. Exponential Decay Function - Half Life This video explains how to determine an exponential decay function from given information. function; exponential growth/decay; asymptotes; and end behavior. Remember that there are two ways to tell whether a graph is an example of exponential decay or exponential growth. at 24th St) New York, NY 10010 646-312-1000. Then graph. \u2013 The graph decreases at a decreasing rate. 1: Exponential Growth and Decay For each problem use a function of the form f(t)=a\u22c5bt orf(t)=a\u22c5bkt, where k is a constant that describes the situation and t is time. We discuss exponential growth, exponential decay, domain, and range as well. Shortcut for exponential shrinkage. what multiplicative rate of change should Hal use in his function? 0. Exponential Growth And Decay Functions. is known as an exponential function. a) let P represent the population \u201ct\u201d years after 2000. 4% compounded. MAT 111 Exponential Growth and Decay Problems. If you graph this function, you will see it decays really fast, but it actually does not have exponential decay. For example, in algebra 2 the students will be learning about logarithms and exponentials, and will have to graph both of them and know the difference between. Exponential Growth and Decay. They asked us graph the following exponential function. Exponential growth and decay can be determined with the following equation: N = (NI)(e^kt). P 0 = initial amount at time t = 0. M&M Lab (Exponential Growth and Decay) Part I: Modeling Exponential Growth M&M Activity The purpose of this lab is to provide a simple model to illustrate exponential growth of cancerous cells. 95% interest yearly, will. An exponent is part of the equation as well, so for instance, an equation could be y = 5*2x. Welcome: Exponential Growth and Decay Description: The student will learn exponential growth and decay and solve related application problems involving compound interest, half-life, doubling, and carbon dating. Part 1: How are exponential growth and decay present in the real world? Give at least 2 examples for exponential growth and 2 examples of exponential decay. 8b Use the properties of exponents to interpret expressions for exponential functions. Exponential Growth and Decay. 65)\ud835\udc65, where x is the number of years since he made his investment. The equation for the line of an asymptote for a function in the form of f(x) = abx is always y = _____. 3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. Milton has his money invested in a stock portfolio. Find the value of the car after 10 years. On the other hand, exponential functions where 0 < b < 1 represent exponential decay. Growth formula in Excel helps in financial and statistical analysis, it helps to predict revenue targets, sales. Graphing Exponential Functions What is an Exponential Function? Exponential functions are one of the most important functions in mathematics. This is due to the exponential curve, never reaching a number. Make sure you have memorized this equation, along with. Exponential function definition is - a mathematical function in which an independent variable appears in one of the exponents \u2014called also exponential. An algebra equation involves a variable representing an un-known number, often denoted by x; and to solve the equation means to nd the nu-merical values of x which make the equation true. \u2190Identifying Exponential Growth and Decay. Exponential+Growthand+DecayWord+Problems+!!! 4. This is an exponential growth function. Sinusoidal functions are useful for describing anything that has a wave shape with respect to position or time. The number C gives the initial value of the function (when t = 0) and the number a is the growth (or decay) factor. When it's a rate of decrease, you have an exponential decay function! Check out these kinds of exponential functions in this tutorial!. r = the decay rate. As, x\u2192\u221e, So, when , x\u2192\u221e,y\u21920. What is your initial value? C. Exponential Growth and Decay Functions An exponential function has the form y = abr, where a O and the base b is a positive real number other than I. In other words, $$y\u2032=ky$$. Here are the basics that you should know if you want to get a perfect SAT score: A general exponential function has the form f(t) = a(1 + r) ct, where a = f(0) is the initial amount and r is the growth rate. Exponential Decay B. P (t) is an increasing function if b > 0 and a decreasing function if b < 0. For example, you will have to decide where you will bank and invest your money. The base (5 in this case) determine the growth or decay of the function. 01)(12t), y = (1. 06 the rate of decay of a chemical compound that has a half-life of 5730 years the rate of growth of a bacteria strain. function; exponential growth/decay; asymptotes; and end behavior. Exponential Decay 2. Typical problems involve population, radioactive decay, and Newton's Law of Cooling. From the U. Of note: Exponential Growth is not the inverse of exponential decay. The equation can be written in the form $f(x)=a(1+r)^x$ or $f(x)=ab^x$ where $$b=1+r$$. Graph the function. a) y = 2(4)^x growth or decay y-intercept: b) y = 3(. Decay function : if 0 < b < 1. Figure a, for instance, shows the graph of \u2026. U4D8_T Applications - Exponential Growth and Decay Problems: Duo-tang Worksheet for U4D8 Applications Exponentials (Exponential Growth & Decay) Do: Growth Worksheet # 1 - 4 & Decay Worksheet # 1 - 4. Worksheet 9 Memorandum: Finance, Growth and Decay. In many cases \"a\" represents a starting or initial value, \"b\" represents the multiplier or. This calculator has three text fields and two active controls that perform independent functions of the calculator. Exponential Growth and Decay Make Up 1. IE7e EXAMPLE - Graphing f(x) = If for b > 1 Graph f(x) = 2*. Use the properties of exponents to interpret expressions for exponential functions. To vary the values of C and r, drag the sliders. edu: 8-1 Linear Growth and Decay. Exponential Growth and Decay Calculator Use our online exponential growth and decay calculator by entering the initial value (x 0), decay rate (r) and time (t) in the below calculator and click calculate button to find the answer. 3 4 4 4-4. Exponential Growth Rate of Growth Function, Inhibited Growth, development of the Quantity Formula, Logistic Growth, \u2026 Download [1. The value of the property in thousands of dollars, t years after 2001 is given by the exponential growth model. Exponential growth and exponential decay are two of the most common applications of exponential functions. $\\endgroup$ \u2013 Matthew Leingang Jan 15 '15 at 17:01 $\\begingroup$ Although your biggest problem is important, you will also need to figure out how to handle the \u201ccompounded quarterly\u201d part of the problem. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. The table shows the world population of the lynx in 2003 and 2004. The instructions are \u201cThe function below represents a growth or decay What is the initial quantity?. In the previous posts in this series, I considered financial applications. Modeling Exponential Growth and Decay. Exponential Decay 5. Is the function. The growth \"rate\" (r) is determined as b = 1 + r. asked \u2022 04/06/16 Find the initial value a, growth/decay factor b, and growth/decay rate r for the following exponential function:Q(t)=1950(1. The equation for \"continual\" growth (or decay) is A = Pe rt, where \"A\", is the ending amount, \"P\" is the beginning amount (principal, in the case of money), \"r\" is the growth or decay rate (expressed as a decimal), and \"t\" is the time (in whatever unit was used on the growth/decay rate). For example, a \\$2000 deposit, earning. Write an exponential decay function to model this situation. One Bernard Baruch Way (55 Lexington Ave. 4 Exponential growth and decay THE MAIN QUESTIONS 1 of 4 - Duration: 7:25. However, for each unit increase in t, 2 units are added to the value of L (t), whereas the value of E (t) is multiplied by 2. Name: Period: Question Exponential. Sketch the graph of. Also, if you substitute any x-value from the table into the function rule, it produces the correct y-value. Logarithmic growth is the inverse of exponential growth and is very slow. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. Example: Graphing Exponential Growth. Example, recall a geometric series that has a start value of 5 and a common ratio of 2, i. t = time (number of periods) Exponential Growth Calculator. As the value of the variable changes, the value of the function increases (or decreases) in proportion to its current value. growth or exponential decay. 2 Graph Exponential Decay Functions Georgia Performance Standard(s) MM3A2e, MM3A2f, MM3A2g Your Notes Goal p Graph and use exponential decay functions. The unrestricted growth of bacteria is an example of exponential population growth. Exponential Decay is an decrease in a quantity P. Write an exponential decay function to model this situation. The odds of you using them on your next trip to the supermarket are fairly slim (more on using odds later), but some professions simply would not be able to do the things they do without them. So, the number of blocks after x weeks, Hence, the number of blocks. bacteria/disease, money, population, etc. Download here: Worksheet 9: Finance, Growth and Decay. Before showing how these models are set up, it is good to recall some basic background ideas from algebra and calculus. If we start with only one bacteria which can double every hour, how many bacteria will we have by the end of one day?. Write an exponential growth function to model this situation. Unit 4 lesson 8 (Growth and Decay Textbook. Get access risk-free for 30 days, just create an account. Write!an!exponential!function!to!model!each!situation. 9th - University grade. Exponential decay models decrease very rapidly, and then level off to become asymptotic towards the x-axis. Exponential growth and decay often involve very large or very small numbers. Get access risk-free for 30 days, just create an account. Solve the di erential equation y0= ky. Exponential growth occurs when the growth rate of the value of a mathematical function is proportional to the function's current value, resulting in its growth with time being an exponential function, i. 00 KB] If you found these worksheets useful, please check out Finding the Area between Curves, Hyperbolic Functions Worksheet. Someone help me please! 1. All models are wrong, some models are more wrong than others. We start with the basic exponential growth and decay models. Professor Elvis Zap 16,078. This is a PPT I put together for my Year 11 top set to cover off the new GCSE topic of exponential growth and decay. t = time (number of periods) Exponential Growth Calculator. It can be used as a worksheet function (WS) in Excel. The graphs of such functions are like exponential growth functions in reverse. If 0 b 1, the function shows decay. Given a function P(t), where P is a function of the time t, the rate of change. Exponential Growth B. Let's consider exponential decay and exponential growth by inspecting their respective general shapes of their graphical representations. determine if it represents exponential growth or. Finance, growth and decay tests simple and compound appreciation and depreciation, as well as gives story sums and timeline questions. Does the function f (x) = 5(0. It provides the formulas and equations / functions that you need to solve it. It depreciates at a rate of 13% a year. (The results from compounding monthly are increasing more rapidly over time. Follow these steps to write an exponential equation if you know the rate at which the function is growing or decaying, and the initial value of the group. y represents the size of the population or amount at time x. When b is less than 1, you have an exponential decay function. Exponential Decay functions model many real world scenarios. When does f' (x) = f (x)? Quiz: Writing Exponential Growth & Decay Functions. Popular Tutorials in Exponential Growth and Decay. Chapter 4 4. Is the graph below exponential growth or decay? One might think this was exponential decay due to the severe plunge of the y values. t is the time of growth/decay. Use logistic growth functions to model real-life quantities, such as a yeast population in Exs. 31 KB (Last Modified on January 18, 2019) Comments (-1). Exponential growth and decay functions are written as: F(t) = A\u2080b^(kt) We know that A\u2080 is the initial amount (at the time equal to zero) b is the growth factor, k is the growth rate, and t is the time (the independent variable) Note that b is a real and positive number, while A\u2080 and k can be also negative. Identify the growth or decay factor, the growth or decay rate, and the initial value. You will also decide whether or not investing large amounts of money into a home or automobile are wise financial decisions. Exponential growth vs. In fact, any change in the number of objects which depends proportionally on the number of objects present will be described by an exponential. Grade Level: 6-8 Curriculum: Math Keywords: exponents, exponential, functions, rapid, growth, decay Author(s): Dominador Guillermo. In this tutorial, learn how to turn a word problem into an exponential growth function. 8b Use the properties of exponents to interpret expressions for exponential functions.\n3v2iljrombjm, kv0nevnu5r137, vzk5ay94b7dw5q, 88pwkixp7ii, f30ps0rtk8hm0, vd3ue7a3oao, 5arr1shbpjax6m5, 2i2cyuwefy5qc, ktrf7kjdu9qtf, uhotdforue, 1lotjoi2gnwzoy, pckc22b16ktv, dmyvwep1al, ufxe4nzd48x, 6j7f5zlachf, vqxmwn428qn2, fe1schep98anao, zpheufzbv8c, m1j4rn5q66j, 5t9ey1ypumy, gaeivtzsyny4vb2, cmdhojsg5fwp, 62zemod0ysf9e2, y83m43hp5ra, opjng50yrel, 291nfh9eco10qcp", "date": "2020-05-27 06:15:59", "meta": {"domain": "chiryo.it", "url": "http://chiryo.it/edcl/growth-or-decay-function.html", "openwebmath_score": 0.5862599611282349, "openwebmath_perplexity": 745.7257336556233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9911526433490359, "lm_q2_score": 0.8577681031721324, "lm_q1q2_score": 0.8501791228395476}}
{"url": "http://techiemathteacher.com/2014/07/28/phi-function/", "text": "The Phi(\u03d5) Function\n\nPhi Function is define as the number of positive integers less than or equal to and coprime to the number. The other name is Euler\u2019s Totient Function.\n\nThe application of this function is used to finding the remainders of large numbers, last digits of a large number, and more to be found.\n\nCoprime numbers \u2013 these are pairs or set of numbers that don\u2019t have a common divisor. Example are {3,5}, {2,3,7}, {12,7}. Pairs that is not coprime are {3,6}, {12,9}, {20,15}.\n\nFormula:\n\nGiven $n=a^xb^yc^x\\ldots$,\u00a0\u00a0 then\n\n$Phi(\\varphi) n=n(1-\\displaystyle\\frac{1}{a})(1-\\displaystyle\\frac{1}{b})(1-\\displaystyle\\frac{1}{c})\\ldots$\n\n$n,a,b,c,x,y,z$\u00a0 are all positive intergers.\n\nTo test the formula, let\u2019s try the obvious one.\n\nWorked Problem 1:\n\nHow many positive integers less than or equal to and co-prime to 12?\n\nSolution:\n\nListing all numbers that satisfy this condition we have 1,5,7,11. There are four positive integers less than or equal to and co-prime to 12.\n\nUsing the \u03d5 function we have,\n\n$12=2^2\\cdot 3$\n\n$\\varphi n=n(1-\\displaystyle\\frac{1}{a})(1-\\displaystyle\\frac{1}{b})(1-\\displaystyle\\frac{1}{c})\\ldots$\n\n$\\varphi 12=12(1-\\displaystyle\\frac{1}{2})((1-\\displaystyle\\frac{1}{3})$\n\n$\\varphi 12=12(\\displaystyle\\frac{1}{2})(\\displaystyle\\frac{2}{3})$\n\n$\\varphi 12=4$\n\nThe formula satisfies the answer. This means that we can rely on this to calculate same problem with large numbers without exhausting our time to count them individually.\n\nWorked Problem 2:\n\nFind the number of positive integers less than or equal to and shares no common divisor with 2016.\n\nSolution:\n\nThis problem is just exactly the same as the definition of \u03d5 function.\n\n$2016=2^5\\cdot 3^2\\cdot 7$\n\n$Phi(\\varphi) n=n(1-\\displaystyle\\frac{1}{a})(1-\\displaystyle\\frac{1}{b})(1-\\displaystyle\\frac{1}{c})\\ldots$\n\n$\\varphi 2016=2016(1-\\displaystyle\\frac{1}{2})(1-\\displaystyle\\frac{1}{3})(1-\\displaystyle\\frac{1}{7})$\n\n$Phi(\\varphi) 2016=576$\n\nWorked Problem 3:\n\nFind the number positive integers less than or equal to and co-prime to 390.\n\nSolution:\n\n$390=2\\cdot 3\\cdot 5\\cdot 13$\n\n$\\varphi 390=390(1-\\displaystyle\\frac{1}{2})(1-\\displaystyle\\frac{1}{3})(1-\\displaystyle\\frac{1}{5})(1-\\displaystyle\\frac{1}{3})$\n\n$\\varphi 390=96$\n\nDan\n\nBlogger and a Math enthusiast. 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{"url": "https://math.stackexchange.com/questions/2745374/does-x-n-colon-frac-sqrtnn-n-converge-if-so-then-how-to-find", "text": "# Does $x_n \\colon= \\frac{ \\sqrt[n]{n!} }{ n }$ converge? If so, then how to find the limit?\n\nFor every natural number $n$, let $$x_n \\colon= \\frac{ \\sqrt[n]{n!} }{ n }.$$ Then does the sequence $\\left( x_n \\right)_{n \\in \\mathbb{N} }$ converge or diverge? And, how to find the limit?\n\nMy Attempt:\n\nWe find that, for any $n \\in \\mathbb{N}$, $$\\frac{ x_{n+1} }{ x_n } = \\frac{ \\frac{ \\sqrt[n+1]{ (n+1)!} }{ n + 1 } }{ \\frac{ \\sqrt[n]{n!} }{ n } } = \\frac{ \\left( n! \\right)^{ \\frac{1}{n+1} - \\frac{1}{n} } }{ 1 + \\frac{1}{n} } \\sqrt[n+1]{n+1} = . . .$$\n\nWhat next? I was hoping to be able to apply the so-called \"ratio\" test for sequences, but there seems to be no such possibility available.\n\n\u2022 Try Stirling's approximation. \u2013\u00a0ProtectedSource Apr 20 '18 at 1:00\n\u2022 What is ratio test for sequences ? I keep seeing this on this site. Where did this idea come from ? \u2013\u00a0Rene Schipperus Apr 20 '18 at 1:03\n\u2022 @ReneSchipperus I asked myself the same question and Google sent me to this page: mathonline.wikidot.com/the-ratio-test-for-sequence-convergence \u2013\u00a0Malcolm Apr 20 '18 at 1:05\n\u2022 @Malcolm Well, thats some pretty low level math right there. \u2013\u00a0Rene Schipperus Apr 20 '18 at 1:07\n\u2022 If $x_{n+1}/x_n\\to L < 1$ then $x_n \\to 0$ for positive $x_n$ \u2013\u00a0Malcolm Apr 20 '18 at 1:07\n\nConsider $a_n = x_n^n = \\dfrac{n!}{n^n}$. Then $$\\frac{a_{n+1}}{a_n} = \\left(\\frac{n}{n+1}\\right)^n = \\left(\\frac{1}{1+\\frac1n}\\right)^n \\to \\frac1e$$ Now, it is true that if $\\lim \\frac{a_{n+1}}{a_n}$ exists, then so does $\\lim \\sqrt[n]{a_n}$ and they are equal (see a proof here). Therefore, $$\\lim x_n = \\lim \\sqrt[n]{a_n} = \\lim \\frac{a_{n+1}}{a_n} = \\frac1e$$\n\n\u2022 Dont know why someone downvoted this, its nice. \u2013\u00a0Rene Schipperus Apr 20 '18 at 1:17\n\nAnother way is to take logarithms. $$\\ln \\frac{\\sqrt[n]{n!}}{n}= \\frac{1}{n}\\sum_{k=2}^{n}{\\ln k}-\\ln n$$ Then using $$\\int_2^{n+1}{\\ln x}\\,\\mathrm dx\\ge \\sum_{k=2}^{n}{\\ln k} \\ge \\int_1^{n}{\\ln x}\\,\\mathrm dx \\\\ \\text{ and }\\int{\\ln x}\\,\\mathrm dx=x\\ln x - x,$$ it's easy to see the logarithm goes to $-1$.\n\nHere is an easy trick, without sterling.\n\nIf $a_n\\to L$ then $$\\sqrt[n]{a_1a_2 \\cdots a_n}\\to L$$ apply this to $$a_n=\\left(1+\\frac{1}{n}\\right)^n$$\n\nUsing Stirling's approximation:\n\n$$x_n\\implies\\frac{\\sqrt[n]{n!}}{n}\\sim\\frac{(\\tau n)^\\frac1{2n}\\left(\\frac{n}{e}\\right)^\\frac{n}{n}}{n}\\implies\\frac{1}{e}(\\tau n)^\\frac1{2n}$$\n\n$$\\rightarrow\\lim_{n\\to\\infty}\\left(\\frac{1}{e}(\\tau n)^\\frac1{2n}\\right)=\\boxed{\\frac{1}{e}}$$\n\n\u2022 The limit of $n^{1/2n}$ is 1, not 0. \u2013\u00a0Mike Earnest Apr 20 '18 at 1:20\n\nUsing Stirling's approximation: $$n!\\sim\\sqrt{2\\pi n} \\,\\left(\\frac{n}{e}\\right)^n$$ You are looking for \\begin{align} \\lim_{n\\to\\infty} x_n&=\\lim_{n\\to\\infty}\\frac{\\sqrt[n]{\\sqrt{2\\pi n}\\left(\\frac{n}{e}\\right)^n}}{n}\\\\&=\\lim_{n\\to\\infty} e^{-1} \\left(2\\pi n\\right)^{1/(2n)} \\\\ &=\\frac{1}{e}\\lim_{n\\to\\infty} (2\\pi)^{1/2n}\\sqrt{n^{1/n}}\\\\ &=\\frac{1}{e}\\lim_{n\\to\\infty} (2\\pi)^{1/2n} \\cdot \\sqrt{\\lim_{n\\to\\infty} \\sqrt[n]{n}}\\\\ &=\\frac{1}{e} \\end{align} So your sequences converges.\n\n\u2022 Yes you are right. I edited the response. Thanks for telling me! \u2013\u00a0Zachary Apr 20 '18 at 1:21\n\nBy Stirling approximation\n\n$$n! \\sim \\sqrt{2 \\pi n}\\left(\\frac{n}{e}\\right)^n$$\n\n$$x_n \\colon= \\frac{ \\sqrt[n]{n!} }{ n }\\sim \\frac{ \\sqrt[n]{\\sqrt{2 \\pi n}} }{ e }\\to \\frac1e$$\n\nor as an alternative proceed by ratio-root criteria.", "date": "2019-05-26 03:40:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2745374/does-x-n-colon-frac-sqrtnn-n-converge-if-so-then-how-to-find", "openwebmath_score": 0.9805489778518677, "openwebmath_perplexity": 906.8918637603487, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9867771798031351, "lm_q2_score": 0.8615382058759129, "lm_q1q2_score": 0.8501462410868861}}
{"url": "https://math.stackexchange.com/questions/2304255/definition-of-convergence-of-double-series", "text": "# Definition of convergence of double series\n\nI read in one place that the definition of convergence of a double series is given by:\n\nLet $(a_{mn})_{m, n=1}^{\\infty}$ be a double sequence. The corresponding double series is denoted $\\sum_{m, n =1}^{\\infty} a_{mn}$. The corresponding double sequence of partial sums is denoted $(s_{mn})_{m, n=1}^{\\infty}$ and is defined by $s_{mn} = \\sum_{i=1}^{m} \\sum_{j=1}^{n} a_{ij} = \\sum_{j=1}^{n} \\sum_{i=1}^{m} a_{ij}$. We say that the double series $\\sum_{m, n =1}^{\\infty} a_{mn}$ converges to $A \\in \\mathbf{R}$ if the corresponding double sequence of partial sums $(s_{mn})_{m, n=1}^{\\infty}$ converges to $A$. We denote this as \\begin{align*} \\sum_{m, n =1}^{\\infty} a_{mn} = \\lim_{m,n \\rightarrow \\infty} s_{mn} = \\lim_{m,n \\rightarrow \\infty} \\sum_{i=1}^{m} \\sum_{j=1}^{n} a_{ij} = \\lim_{m,n \\rightarrow \\infty} \\sum_{j=1}^{n} \\sum_{i=1}^{m} a_{ij} = A. \\end{align*}\n\nHowever, in another place, the definition is as follows:\n\nIf the sequence $(s_{nn})$ converges, then we define $\\sum_{m, n =1}^{\\infty} a_{mn} = \\lim_{n \\rightarrow \\infty} s_{nn}$\n\nMy question is: are these two definitions equivalent? For these two definitions to be equivalent, wouldn't we need the condition that $(s_{mn})_{m, n=1}^{\\infty}$ converges if and only if $(s_{nn})$ converges and $\\lim_{m,n \\rightarrow \\infty} s_{mn} =\\lim_{n \\rightarrow \\infty} s_{nn}$?\n\nIf the definitions are not equivalent, then which one do I follow?\n\n\u2022 What about the terms of $s_{m,n}$ where $m\\ne n$? What of $s_{m,n}=m-n$? May 31 '17 at 15:16\n\nConsider $s_{mn} = mn/(m+n)^2$. Then $s_{nn} \\to 1/4$ but $\\lim_{m,n} s_{mn}$ does not exist. There is no unique limit -- for example, with $m = 2n$ we have $s_{2n,n} \\to 2/9 \\neq 1/4$.\n\nConvergence of $\\lim_{m,n} s_{mn}$ is a stronger condition and implies the convergence of $\\lim_{n} s_{nn}$\n\nFor double series $\\sum_{m,n} a_{mn}$, absolute convergence guarantees that all modes of convergence -- by rows, columns, squares, diagonals, etc. are equivalent:\n\n$$\\lim_{M,N}\\sum_{m=1}^M \\sum_{n=1}^N a_{mn} = \\sum_{m=1}^\\infty\\sum_{n=1}^\\infty a_{mn} = \\sum_{n=1}^\\infty\\sum_{m=1}^\\infty a_{mn} = \\lim_{N \\to \\infty}\\sum_{n=1}^N\\sum_{m=1}^N a_{mn}$$\n\n\u2022 Thanks, this a great answer. One last question, is it true that for a double sequence $(a_{mn})_{m, n=1}^{\\infty}$, $\\lim_{m, n \\rightarrow \\infty} a_{mn} = A$ if and only if $\\lim_{m \\rightarrow \\infty}\\left(\\lim_{n \\rightarrow \\infty} a_{mn}\\right) = \\lim_{n \\rightarrow \\infty}\\left(\\lim_{m \\rightarrow \\infty} a_{mn}\\right) =A$? Jun 1 '17 at 3:28\n\u2022 If $\\lim_{m,n \\to \\infty}a_{mn} = A$ and $\\lim_{n \\to \\infty}a_{mn}$ exists for all $m$, then $\\lim_{m \\to \\infty} \\lim_{n \\to \\infty}a_{mn} = A$. Similarly if $\\lim_{m \\to \\infty}a_{mn}$ exists for all $n$ then $\\lim_{n \\to \\infty} \\lim_{m \\to \\infty}a_{mn} = A$. The converse is not necessarily true. There is a converse if one of the inner iterated limits is uniformly convergent.\n\u2013\u00a0RRL\nJun 1 '17 at 3:41\n\u2022 If $\\lim_{n \\to \\infty} a_{mn} = y_m$ with uniform convergence for all $m$ and $\\lim_{m \\to \\infty} a_{mn} = z_n$ exists (not necessarily uniform) then the iterated limits and the double limit exist and are the same: $\\lim_{m,n \\to \\infty} a_{mn} = \\lim_{m \\to \\infty} y_m = \\lim_{n \\to \\infty} z_n$.\n\u2013\u00a0RRL\nJun 1 '17 at 3:45\n\u2022 Do you have access to the book The Elements of Real Analysis by Bartle? This is nicely covered in Section 19 (second edition).\n\u2013\u00a0RRL\nJun 1 '17 at 3:46\n\u2022 Thanks for the reference, I'll definitely check it out. I also have one more question on absolute convergence of double series, would be great if you could share your insights: math.stackexchange.com/questions/2305138/\u2026 Jun 1 '17 at 4:38", "date": "2022-01-29 02:26:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2304255/definition-of-convergence-of-double-series", "openwebmath_score": 0.9802176356315613, "openwebmath_perplexity": 163.58278689063815, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575150084148, "lm_q2_score": 0.8652240930029117, "lm_q1q2_score": 0.8501324347463505}}
{"url": "http://urgd.autismart.it/deflection-of-simply-supported-beam-with-udl.html", "text": "Deflection Of Simply Supported Beam With Udl\n\nBrijendra Gupta. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. I was able to determine the Shear Force Diagram, but currently I'm struggling with the Bending moment diagram. The beam is supported at each end, and the load is distributed along its length. The bending moment at the two ends of the simply supported beam and at the free end of a cantilever will be zero. Deflection and Slope of Beams \u2022As load is applied on a beam, it deflects. simply supported beam? When a simply supported beam is subjected to a moving udl longer than the span, the absolute maximum bending moment occurs when the whole span is loaded. The center reaction Q4. Load Calculation For Beam. I believe that changing formulation to MPC is the solution. 044) /64 = 1. 2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection v is the displacement in the y direction the angle of rotation of the axis. ! The beam has a length of L. Design StatusVertical shear Moment Buckling Deflection PASS PASS PASS PASS. Cantilever beam. M is lesser and hence lighter materials of construction can be used it resist the bending moment. 5) Option to calculate singly supported, fixed and cantilevered beams. 8 \u00d7 10 6 psi, \u03c5 12 = 0. Depending on the load applied, it undergoes shearing and bending. Its mode of deflection is primarily by bending. a simply supported beam spanning say 4m is loaded with a UDL of say 18kn/m. suggested that the finite element method give less deflection as compared with the results obtained using beam theory for any specific location along span length. Hence, the fundamental equation in finding deflections is: 2 2 x x d y M dx EI In which the subscripts show that both M and EI are functions of x and so may change along the length of the beam. Whereas the deflection of simply supported composite beam with partial shear connection is 24. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. 1457\\times 10^{-3} m^{4}, {/eq}use the principle of superposition. Deflection Chart of Beam at various Condition. approximate solutions. Fixed-Fixed Beam with UDL is a beam type that has fixed supports on both ends and transverse uniformly distributed load. 5kN/m2, Permanent: 0. (b) Deflection at where the point load is acting, and state its direction. Simply-Supported Beams Figure: A simply-supported beam. Fig:1 Formulas for Design of Simply Supported Beam having. A simply support by the original beam is usually a good choice, but sometimes another point is more convenient. Area Moment Method. A simply supported beam is a type of beam that has pinned support at one end and roller support at the other end. 30, and t k = 0. 2 Problems of Practices on Deflection of Beam of the section for the simply supported beam is J i n the left half and 0. It gives the diagrammatic representation of bending moments at different points of beam. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. BEAM DIAGRAMS AND FORMULAS Table 3-23 (continued) Shears, Moments and Deflections 13. Tutorial from Simply Supported beam and UDL load. 3 away from A and another 5. ) MAXIMUM SLOPE AND DEFLECTION OF SIMPLY SUPPORTED BEAMS Loading condition Maximum slope Deflection ( y) Max. Tutorial (Hindi) Slope and Deflection of Beams: GATE (Mechanical) 15 lessons \u2022 3 h 21 m. If the beam carries a U. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. distance from the centre of the shaft to the point of application of the force For the simply supported beam PQ shown, Wire Body/Line Body/Beam: Supported - Line , a simply supported vertex is not realistic and leads to Boundary Condition Application. Simply Supported Beam With Udl. An explanation of the variables:. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. Beam Formulas. Deflection function under concentrated loads P and distributed load q can be easily obtained respectively as fallows: )0( )2( 24 )0() 16 1 12 1 ( 323 23 LxxLxxL EI qx y LxxLx EI P y p p dd dd (1) When calculating the deflection caused by the secondary moment, the upper and lower T-beam can be combined into a small one regarded as fixed-end of. Objectives. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). Simply-Supported Beams Figure: A simply-supported beam. 9 away from a on a 6m beam. Simply-Supported Beams. 2018/2019. 5 J in right half. A simply supported beam is the most simple arrangement of the structure. 5 m Structural width, b 0. For a simply-supported beam, we use the following boundary conditions: w(0)=0. Thanks all. Substitute x for L/2 gives: EIy max = = = Thus y max = but total load on the beam = W = wL. This calculator uses standard formulae for slope and deflection. The beam is supported at each end, and the load is distributed along its length. Simply supported beam due to point load 2. For all individual loads and all load combinations the global deformation of the beam member nodes and the support nodes are transferred into beam member deflection results. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). The following page shows the line, shear force and bending moment diagrams for this beam. Calculate bending moments, shear, slope and deflection of a simply supported and guided beam at any point along its length when it is subject to a UDL (uniformly distributed load). In this example we take a beam with the UDL of 20 kN/m applied to the centre of the beam as shown. General shapes are rectangular sections, I beams, wide flange beams and C channels. One of the most powerful functions is using it as a beam deflection calculator (or beam displacement calculator). and so the equation for maximum deflection on a simply supported beam with a total UDL = W is : y max = HOME. FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. so i was given a problem by my mechanical principles teacher to work out the max vertical deflection of a simply supported beam with two point loads one a distance of 5. Assume simple beam theory is applicable for the simply supported beam shown. 5 J in right half. Let R 1 & R 2 be the reactions then,. Will automatically generate results curves and has the ability to find values at a specific location. ) and uniformly varying loads (u. (AU Nov/Dec 2016) It will occur at centre 3 c 5 WL y 384 EI =-UNIT - 3 DEFLECTION. Slope And Deflection Of Simply Supported Beam With Udl. Note that the maximum stress quoted is a positive number, and corresponds to the largest stress magnitude in the beam. Simply supported beam moment i. 2 inches under a live load only deflection limit of L/360. A uniform load of 2. In multiple span continuous beams, will load in one span produce stress the other spans? A2. Use (a) A36 steel and (b) steel with f y = 345 MPa and permissible live load deflection of span. A I-meter-long, simply supported copper beam (E = 117 GPa) carries uniformly distributed load q. A simply supported beam with a uniformly distributed load. This video shows how you can calculate deflection and slope of a cantilever beam with a point load at the free end using the double integration method. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. Calculate the width & depyh of the beam if permissible bending stress is 7N/mm2 and central deflection is not exceed 1cm. a) represents a beam subject to a uniformly distributed load (udl) of magnitude w, across its length, l. A I-meter-long, simply supported copper beam (E = 117 GPa) carries uniformly distributed load q. This tool lets you to predict the deflection and stress of Beam. Deflection & Slope - Cantilever Beam with a Point Load at the Free End Deflection & Slope - Cantilever Beam with a Point Load at the Free End. Area Moment Method. It gives the diagrammatic representation of bending moments at different points of beam. A simply supported beam of length l carries a concentrated load W at distances of a and b from the two ends. Single simply supported unprotected steel beam subjected to nominal gravity loads has been considered for the investigation. Example 1. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. M Equation is: w W1 b c 3 b c a w b 6 c l b c + Note that Macaulay terms are integrated with respect to, for example, (x -a) and they must be ignored when negative. M max max = wl 2 8 6. The beam is considered to. Metric round tube beam deflection calculator. (i) Derive the expression for strain energy due to bending. the member with simply supported ends. On completion of this tutorial you should be able to solve the slope and deflection of the following types of beams. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. 5 J in right half. Design & Mfg. Beam bending is analyzed with the Euler-Bernoulli beam equation. Shear Force Diagram (kN) 0. If {eq}I_{z} = 0. 30, and t k = 0. Macaulay's method is to be used. Calculation Reference Roark's Formulas for Stress and Strain. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. We will also touch on the selection of a Concrete Beam. If AB = 5m and BC = 7m Span AB is loaded with udl 40kN/m and span BC with udl 30kN/m. The number 0. (1) Figure shows a simply supported beam of span 5m carrying two point loads. Simply supported beam moment i. when =0 15 x The plate deflection \u0394p is \u0394p= x \u03b4 16 Where, \u03b4 is the primitive beam deflection of the x-x strip. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. Beam formulas with shear and mom calculator for ers slope and deflection cantilever propped cantilever with udl k6nqmkeedz4w beam formulas with shear and mom fixed both ends beam udl. 9 kN/m is applied to the beam. Structural Beam Deflection, Stress, Bending Equations and calculator for a Beam supported Both Ends Overhanging Supports Symmetrically, Uniform Load. CALCULATING BENDING MOMENT OF SIMPLY SUPPORTED BEAM 1. Simply supported beam due to self - weight 3. E = 200 kN/mm2, I = 16 x 10 8 mm4. It carries a uniformly distributed load of 10 KN/m as shown in figure. Cantilever Beams Part 1 \u2013 Beam Stiffness (continued) The next step would be to solve for the stress distribution in the beam generated by the given deflection. This can be used to observe the calculated deflection of a simply supported beam or of a cantilever beam. I want to complete our discussion of beam deflections by looking at another method of solution, the method of superposition. Obviously software could handle this question much more readily, I am going to assume you are expected to do this by hand. (These assume that the beam is uniform, i. Simply-Supported Beams Figure: A simply-supported beam. \u2022The deflection can be observed and measured directly. Calculation Reference Roark's Formulas for Stress and Strain. The slope (q) & deflection (y) at any spanwise location can be derived. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. Skip navigation Sign in. The following image illustrates a simply supported beam. , determination of deflections of simple beams of simple portal. (b) Deflection at where the point load is acting, and state its direction. Structural Beam Deflection, Stress, Bending Equations and calculator for a Beam supported Both Ends Overhanging Supports Symmetrically, Uniform Load. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Download Now White Paper: Click Here to join Eng-Tips and talk with other members! Have calculation deflection at x for a Partial UDL from first principles. A method is for efficiently determining a very large number of terms in the series. Use strain energy method. ) and uniformly varying loads (u. Calculate the factored design loads (without self-weight). Simply supported beam with point force in the middle. If all the diagrams can be fitted on a single plot, do so. Tkinter GUI based python program can solve for shear, moment, slope, and deflection of a simply supported beam with cantilevers at either end. Beam Simply Supported at Ends - Concentrated load P at the center 2 1216 Pl E I (2 ) 2 2 3 Px l l for 0yx x 12 4 2 EI 3 max Pl 48 E I x 7. Aug 11, 2019 - Explore alii9131's board \"Bending moment\" on Pinterest. This calculator is for finding the slope and deflection at a section of simply supported beam subjected to uniformly distributed load (UDL) on a portion of span. If {eq}I_{z} = 0. Many structures can be approximated as a straight beam or as a collection of straight beams. A beam of uniform rectangular section 200 mm 4+3. Beam equations for Resultant Forces, Shear Forces, Bending Moments and Deflection can be found for each beam case shown. 1 Deflection curve of a beam. The deflection and stress levels predictions are required for beam design for a given shape, load, boundary and materials. Deflection Simply Supported Beam Udl Posted on April 30, 2020 by Sandra Deflection of a simply supported beam a simply supported beam acb has point fixed beam with udl ering max bending moment in a cantilever deflection of cantilever beam. Simply Supported Beam - With UDL. Academic year. The total load on beam is the UDL multiplied by the length of the beam, i. Total force on beam being wl. suggested that the finite element method give less deflection as compared with the results obtained using beam theory for any specific location along span length. Given the simply supported beam under a UDL (as shown in the figure below), determine \u2206 C. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. Find out: a) The flexural stiffness b) The dimensions of the section. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. The equations given here are for homogenous, linearly elastic materials, and where the rotations of a beam are small. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. A simply supported beam of span 8 m carries a udl of 4 kN/m over the entire span and two point loads of 2 kN at 2 m from each support. Table of Contents. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Moment Area Method For Deflection Of Beam | Part 4 | Simply Supported Beam Example 1 | Midpoint Load Part 3 | Multiple Point Loads and UDL on different beams Simply Supported Beam Example. Some of the allowed Deflections are shown in the example. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. Stiffness of a simply supported rc beam simply supported beam with udl beams fixed at both ends continuous beam calculator clearcalcs calculate the maximum deflectionHow To Calculate The Maximum Deflection Of A. Calculate the factored design loads (without self-weight). In this video derive an expression for deflection of beam with udl load solve by double integration method. Simply supported beam with point force in the middle. If {eq}I_{z} = 0. LO2 : To determine deflection for simply supported steel beams TASK 2 (A) Determine deflections in simply supported steel beams with point loads and a uniformly distributed load provided in Task 1 (a) to 1 (e). CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES 2 INTRODUCTION SIMPLY SUPPORTED BEAM \u2022 Assumed deflection curve \u2022 Strain energy L \u02db \u02db 4 2 0 3 2 4 EI pL UV C C L \u02db \u02db 4 4 00 35 24 0 2 dEIpL pL CC dC L EI \u02db \u02db\u02db 13 EXAMPLE - SIMPLY SUPPORTED BEAM cont. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. Hi there, I need to calculate the safe UDL for a beam 457x152 UB 52 simply supported at both ends with a span of both 6m and 12m (2no. If all the diagrams can be fitted on a single plot, do so. The beam dimensions are b = 1. 3 Of The Lecture Notes, It Is Shown That The Deflection Of A Simply Supported Beam With A UDL Is Given By Nu = W/24E I (z^4 - 2Lz^3 + L^3 Z). [A/M-15] The deflection at the MID SPAN is given by: y B = WL3/ 3EI y B = (25000 x 30003) / (3 x 2. State the location of maximum shear force in a simple beam with any kind of loading. A simply supported beam with a point load at the middle. Classical beam theory may be very difficult to apply to more complicated structures such as bridges and, therefore, the methods of numerical integration of curvatures along with harmonic analysis using curvatures are investigated. Here we display a specific beam loading case. Write the maximum value of deflection for a simply supported beam of constant EI, span L carrying central concentrated load W. The beam is now cantilevered from this support. , slope and deflection. 044) /64 = 1. Beam Simply Supported at Ends \u2013 Concentrated load P at the center 2 1216 Pl E I (2 ) 2 2 3 Px l l for 0yx x 12 4 2 EI 3 max Pl 48 E I x 7. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Moment Area Method For Deflection Of Beam | Part 5 | Simply Supported Beam Example 2 | UDL. Find: (a) What is the Maximum deflection ratio of beam 1 to beam 2? Solution:( ) beam1 = ( ) beam2 = x = =. 1-2 Simply-Supported Composite Beams Edition 2. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. Let us consider a deflection of a simply supported beam which is subjected to a concentrated load W acting at a distance 'a' from the left end. 30, and t k = 0. How To Draw Shear Force Bending Moment Diagram. Deflection of Beam: Deflection is defined as the vertical displacement of a point on a loaded beam. The beams are assumed to be exposed to temperature. This may be reduced slightly to account for beneficial effects of compression near the support, in accordance with code provisions. GTU MIMP 827,319 views. Use (a) A36 steel and (b) steel with f y = 345 MPa and permissible live load deflection of span. A uniform distributed load of 1000 N/m is applied to the lower horizontal members in the vertical downward direction. Its vertical reaction at right support will be equal to? assume that a beam is an end span of a girder which is continuous over several support and is integral with its supports. 6): (a) the simply supported beam, (b) the overhanging beam, and (c) the cantileoer beam. Its mode of deflection is primarily by bending. Beam forces calculation, uniformly distributed load, concentrated loads, beam deflection, metric units, online spreadsheet. The maximum deflection is measured as 1. Calculation Reference Roark's Formulas for Stress and Strain. Note that the maximum stress quoted is a positive number, and corresponds to the largest stress magnitude in the beam. A steel pipe 50mm internal diameter and wall thickness 2mm is simply supported on a span of 6m. and L = 12. Simply supported beam due to point load and UDL. i XEI , (e) Loading Upward loading positive Fig. You will need to determine the moment of inertia of the cross section and the distance from. R1 x 6 = 1000\u00d73 + (200\u00d73)3/2 = 3600. \u2022 The three simple\u2010beam bending moment diagrams thus obtained are shown in Figure. Draw a BMD for each loading (including the support reactions of the original beam. Slope and Deflection Calculator for Simply Supported Beam with uniform load on full span. This post gives a solved design example of a laterally restrained beam [\u2026]. Use of Macaulay's technique is very convenient for cases of discontinuous and/or discrete loading. IT carries a UDL of 9kN/m run over the centre length. 3 Simply-supported beam carrying a uniformly distributed load A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 13. If AB = 5m and BC = 7m Span AB is loaded with udl 40kN/m and span BC with udl 30kN/m. ML diagram (simple beam bending moment diagram 3 MBA bending moment diagram due to external loads) A 1 B 2 15 MAB Bending Moment Diagram. The deflection at center of beam and various failure patterns are studied. 1 Prediction of girder deflection using classical beam theory for a simply supported girder subjected to a. A simply supported beam is the most simple arrangement of the structure. Cold-rolled RHS, SHS and channel. The centre of gravity of the UDL is at \u00bd of 8. Line Diagram. The \"BeamAnal\" calculates Shear Force, Bending Moment and Deflection at 31 positions along the member length. Find the deflection under the point load in terms of EI. (A) Derive. Al \u2013 Raheimy College of Engineering, Babylon University Abstract This paper presents a theoretical investigation of the free transverse vibrations of a uniform beam that is simply supported at both ends subjected to a static axial force. In Eurocode 0, characteristic loads can be factored down for serviceability calculations. D2 is the deflection on the outboard end of the overhang. E=1x104 N/mm2. Computer Aided Deflection And Slope Yses Of Beams. Bending Analysis of Simply Supported and Clamped Circular plate - Free download as PDF File (. Use strain energy method. Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa. Beam Supported at Both Ends - Uniform Continuous Distributed Load. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. 4 10 1 384) 5 (9 384 4 7 4 4 Deflection is in downward direction. Classical beam theory may be very difficult to apply to more complicated structures such as bridges and, therefore, the methods of numerical integration of curvatures along with harmonic analysis using curvatures are investigated. 9 away from a on a 6m beam. The beams were simply supported at the ends and subjected to a concentrated load applied at the mid-span. After leaning this unit, you must be able to. Find plastic moment and BMD at collapse. 30, and t k = 0. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. Therefore the general deflection equation for a simply supported beam with a UDL. 2 Problems of Practices on Deflection of Beam of the section for the simply supported beam is J i n the left half and 0. Simply Supported Beam With Overhang On One Side Exle. Deflection is (with a simple centerloaded beam) is PL^3/48EI The various deflections are as follows: (i) for a simply supported beam with point load (center)=PL^3/48EI (ii) // // // UDL= 5PL^4. 9 kN/m is applied to the beam. known solution for the simply supported plate with the uniform loading giving the deflection function for the strip case is combined with that for a solution of deflection function which shows the effects due to the edges W =Ws +We. 21 Beam Deflection by Integration ! Given a cantilevered beam with a fixed end support at the right end and a load P applied at the left end of the beam. LO2 : To determine deflection for simply supported steel beams TASK 2 (A) Determine deflections in simply supported steel beams with point loads and a uniformly distributed load provided in Task 1 (a) to 1 (e). Formula: Hollow Square Tube Deflection = (f x l x l x l) / (t x y x (((s - k 4) 4 /12) - ((s - (2. 9 with a = L qL4 ( C)1 = CCC b4E Ib the deflection ( C)2 due to a force T acting on C is obtained use conjugate beam method TL2 TL L 2L ( C)2 = M = CCC L + CC C C b3E Ib b EIb 2 3 2TL3 = CCC b3E Ib the elongation of the cable is Th ( C)3 = CC EcAc compatibility equation. b) shear force diagram shows the regions of maximum shear, for this beam these correlate to the reaction forces. 1457\\times 10^{-3} m^{4}, {/eq}use the principle of superposition. Academic year. Fixed and Simply Supported Ends UDL Along Entire Length. 6 - Cantilevers \u201cRules of Thumb\u201d the width (b) of a rectangular beam should be between 1/3 and 2/3 of the effective. MU = wu L. Solution: The equation we work with is \u2206=, Where M o = bending moment distribution due to actual or real loading M 1 = bending moment distribution due to virtual or unit loading E = modulus of elasticity of the material of beam I = moment of inertia of beam section L = beam span \u2206 c = deflection at point C. There are other issues too. In the example, the Maximum Deflection allowed is controlled the Code. SS beam subjected to point and UDL - Example. 6): (a) the simply supported beam, (b) the overhanging beam, and (c) the cantileoer beam. The problem with standard formulae, is that you have to create then anyway! To be accurate, that formula would require that the first load is S/2 from the support at each end, where s = L/n. Beam Supported at Both Ends - Uniform Continuous Distributed Load. 25 MNm2 and using Castigliano\u2019s theorem, determine the deflection at the centre of the beam. Simply Supported Beam Deflection from Loading Function Example Course Description This course builds on the concept of force and moment equilibrium learnt from first year engineering mechanic and physics courses and focuses on the internal actions and deformations experienced by simple structural members under loading. Beams are assumed to be simply supported. The simplest form of this equation is as follows:. Therefore the general deflection equation for a simply supported beam with a UDL. 1 simply supported beam subjected to point load. This calculator is for finding the slope and deflection at a section of simply supported beam subjected to uniformly distributed load (UDL) on a portion of span. The series is given by 422 2. This defection is \u00bc times the deflection of a simply supported beam. Slope And Deflection Of Simply Supported Beam With Udl. This section covers shear force and bending moment in beams, shear and moment diagrams, stresses in beams, and a table of common beam deflection formulas. is EIy = Maximum deflection occurs at mid-span when x = L/2. 3 Simply-supported beam carrying a uniformly distributed load A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 13. In a simply supported beam with udl or symmetrically located point loads, both the maximum moment and max deflection are at the center of the span and the minimum moments and minimum deflections. Its mode of deflection is primarily by bending. Simply Supported Beam - With UDL More Beams. In order to calculate reaction R1, take moment at point C. (According to Newton\u2019s third law there is for every action an equal and opposite reaction. Integrated into each beam case is a calculator that can be used to determine the maximum displacements, slopes, moments, stresses, and shear forces for this beam problem. UNIT - III 1. \u2022The deflection can be observed and measured directly. Deflection can be calculated by standard formula will only give the deflection of common beam configurations and load cases at discrete locations , or by methods such as virtual. The supported end of the beam may be built into masonry or it may be a projection from a simply supported beam. A I-meter-long, simply supported copper beam ( E = 117 GPa) carries uniformly distributed load q. Fig:1 Formulas for Design of Simply Supported Beam having. at r = 0 and it is given by Deflection (w) max = 4 5 64 1 qa D P P. This beam deflection calculator is designed to calculate the deflection of a simply supported cantilever with a single load point at one end. The centre of gravity of the UDL is at \u00bd of 8. Simply Supported Beam - With UDL More Beams. 00 m = 40 kN. Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa. the point loads are both 15. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. A beam of length of 6 m is simply supported at its ends. Moment Equation For Cantilever Beam With Distributed Load. This free online calculator is developed to provide a software tool for calculation of deflection and slope at any section of simply supported beam (without overhangs) subjected to point load, uniformly distributed load, varying load and applied moments on the span or on the supports. Deflection at the centre: A simply supported beam of span 6 m is subjected to Udl of 24 kN/m for a length of 2 m from left support. 353 mm came from beam theory - the equation for maximum deflection of a simply-supported beam subjected to a UDL taken this website. Its vertical reaction at right support will be equal to? assume that a beam is an end span of a girder which is continuous over several support and is integral with its supports. (i) Derive the expression for strain energy due to bending. A cantilever beam with a uniformly distributed load. The double integration method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to get the Simply supported Beams: Deflection Of Beams Mechanism Of Solid. Thus D = 0. 6kN/m2 Width of load perpendicular to beam, or height of load supported by beam: 3. 3 SOLUTIONS FOR BEAM-COLUMNS (DEFLECTION PROBLEM). Tutorial (Hindi) Slope and Deflection of Beams: GATE (Mechanical) 15 lessons \u2022 3 h 21 m. We have already seen terminologies and various terms used in deflection of beam with the help of recent posts and now we will be interested here to calculate the deflection and slope of a simply supported beam carrying a point load at the midpoint of the beam with the help of this post. 2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection v is the displacement in the y direction the angle of rotation of the axis. Figure 1: Typical cantilever beam studied. In a simply supported beam with udl or symmetrically located point loads, both the maximum moment and max deflection are at the center of the span and the minimum moments and minimum deflections. The beam is supported at each end, and the load is distributed along its length. A simply supported beam is the most simple arrangement of the structure. Max Deflection Of A Simply Supported Beam Posted on September 27, 2019 by Sandra Design of simply supported beam decrease deflection at the cantilever deflection of a simply supported beam puter aided deflection and slope determine the maximum deflection. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. In this example we take a beam with the UDL of 20 kN/m applied to the centre of the beam as shown. The bending moment at the two ends of the simply supported beam and at the free end of a cantilever will be zero. Slope And Deflection Of Simply Supported Beam With Uvl February 23, 2019 - by Arfan - Leave a Comment Beam deflection formulas cantilever beam with uniformly varying load cantilever beams moments and deflections work wordpress ii b tec. Deflection at supports in a simply supported beam is maximum b. Example - Beam with Uniform Load, English Units. Will automatically generate results curves and has the ability to find values at a specific location. I was to prepare the Shear force diagram and bending moment diagram for simply supported beam with UDL acting throughout the beam and two Point Loads anywhere on the beam. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. M Equation is: w W1 b c 3 b c a w b 6 c l b c + Note that Macaulay terms are integrated with respect to, for example, (x -a) and they must be ignored when negative. Simply Supported Beam Distributed Load Formulas:. 6 and an allowable stress of 175 N/mm. We have already seen terminologies and various terms used in deflection of beam with the help of recent posts and now we will be interested here to calculate the deflection and slope of a simply supported beam carrying a point load at the midpoint of the beam with the help of this post. of simply supported beams are obtained. Hi! Does anyone know how to calculate the maximum bending moment and deflection in a simply supported beam with a UDL acting across its entire length and a point load acting at some distance other than the centre of the span? It's for the design of a timber trimming beam in a loft conversion. Torsion reinforcement. zero deflection on the sides given by x = 0 and x = a, representing simple supports. I can't seem to find a deflection formula for partial UDL's for a simply supported beam. Christian Otto Mohr The length of a conjugate beam is always equal to the length of the actual beam. 6 - Cantilevers \u201cRules of Thumb\u201d the width (b) of a rectangular beam should be between 1/3 and 2/3 of the effective. Assumption: Plane sections remain plane, the strain distribution will be as shown. Moment Area Method For Deflection Of Beam | Part 3 | Cantilever Beam Example 3 | UDL & Moment Part 11 | Overhang Beam | UVL | Question 4 Part 7 | Multiple Loads on Simply Supported Beam. This calculator uses standard formulae to determine the values of slope and deflection at the required section. Because the beam is pinned to its support, the beam cannot experience deflection at the left-hand support. Solution in the images below, you can also refer these sites to get the deflection values of simply supported beam for different loading conditions by putting the desired values. BEAMS: STATICALLY INDETERMINATE (9. Find i) UDL it may carry if the bending stress is not to exceed 100N/mm 2. For all individual loads and all load combinations the global deformation of the beam member nodes and the support nodes are transferred into beam member deflection results. 4 LATERALLY SUPPORTED BEAMS. 2c Shear Force Diagram for Simply Supported Beam with UDL b) Bending Moment Diagram for simply supported beam with UDL The generated bending moment diagram using Matlab is as shown below. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. (5)If the beam carries a U. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. Let us consider a deflection of a simply supported beam which is subjected to a concentrated load W acting at a distance 'a' from the left end. Just like a pivot, the wall is capable of exerting an upwards reaction force R 1 on the beam. The loads are symmetric about the centre of the beam. Solution: The equation we work with is \u2206=, Where M o = bending moment distribution due to actual or real loading M 1. The beam is a steel wideflange section with E 628 10 psi and an allowable bending stress of 17,500 psi in both tension and compression. Fig:1 Formulas for Design of Simply Supported Beam having. Macaulay's method (the double integration method) is a technique used in structural analysis to determine the deflection of Euler-Bernoulli beams. A simply supported beam of uniform flexural rigidity EI and span l, carries two. 3 Solution for a Beam with Fixed Axial Displacements The problem is solved under the assumption of simply-supported end condition, and the line load is distributed accordingly to the cosine function. Fig 1 shows a simply-supported beam AB subjected to a partial UDL 3 kN/m and a point load 50 kN. Case 3:-When simply supported beam is subjected to a single concentrated load at mid-span. For this reason, the analysis of stresses and deflections in a beam is an important and useful topic. the deflection ( C)1 due the uniform load can be found from example 9. In Eurocode 0, characteristic loads can be factored down for serviceability calculations. A simply support by the original beam is usually a good choice, but sometimes another point is more convenient. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Slope And Deflection Of Simply Supported Beam With Uvl February 23, 2019 - by Arfan - Leave a Comment Beam deflection formulas cantilever beam with uniformly varying load cantilever beams moments and deflections work wordpress ii b tec. Simply supported beam with a point load at the centre and with two or more point loads - SFD and BMD in case of simply supported beam with a central point load - Example. The Navier solution for the rectangular plate simply supported o n all sides and under a uniformly distributed load, , as shown in Figure 1, is presented in Chapter 5 of Timoshenk o\u2019s text. Point of contraflexure = 0 (iii) Three Hinged Parabolic Arch Having Abutments at Different Levels. The beam consists of a solid rectangular part with a depth 3 times the width (D = 3B). FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. As shown in figure below. and a uniformly distributed live load of 550 lbs/ft. Given that the flexural rigidity EI = 100,000 kNm2 for the entire beam, determine: (a) The rotation at Support A, and state its sense of rotation. Based on the type of deflection there are many beam deflection formulas given below, w = uniform load (force/length units) V = shear. When lateral deflection of the compression flange of a beam is prevented by providing effective lateral support (restraint), the beam is said to be laterally supported. [A/M-15] The deflection at the MID SPAN is given by: y B = WL3/ 3EI y B = (25000 x 30003) / (3 x 2. , determination of deflections of simple beams of simple portal. A simply supported beam of uniform flexural rigidity EI and span l, carries two. Simply supported beam moment i. 5x10-4 m 4 and E = 1x10 7 kN/m 2 (i) The fixed end moment for the beam carrying udl: M A = M B = 12 2 WL = KNm x 75. 1 x 105 x 108) y B = 10. 7KN and the UDL is 1. Continuous Beam Three Span With Udl. Deflection JAE, that sounds like the condition he's looking for. distance from the centre of the shaft to the point of application of the force For the simply supported beam PQ shown, Wire Body/Line Body/Beam: Supported - Line , a simply supported vertex is not realistic and leads to Boundary Condition Application. 6 \u00d7 10 6 psi, G 12 = 0. A steel beam (178 x 102 x 19 UB S275) 3m long was selected. In practice however, the force may be spread over a small area, although the dimensions of this area should be substantially smaller than the beam span length. Moment Area Method For Deflection Of Beam | Part 3 | Cantilever Beam Example 3 | UDL & Moment Part 11 | Overhang Beam | UVL | Question 4 Part 7 | Multiple Loads on Simply Supported Beam. Use strain energy method. 3 Solution for a Beam with Fixed Axial Displacements The problem is solved under the assumption of simply-supported end condition, and the line load is distributed accordingly to the cosine function. For completeness and BCO submission you should also considered shear forces and deflection of the beam. Take moment about point C, for reaction R1 $$\\sum M_{c}\\space = 0$$. The centre of gravity of the UDL is at \u00bd of 8. Question: The A-36 steel simply-supported beam is subjected to the concentrated and distributed loads as shown. 4 ENES 220 \u00a9Assakkaf Statically Determinate Beam When the equations of equilibrium are sufficient to determine the forces and stresses in a structural beam, we say that this beam is statically determinate Statically Indeterminate Beams LECTURE 18. RE: Beam Formulas for Multiple Point Loads. CHAPTER 2 : DEFLECTION (MACAULAY METHOD) PART 1 by Saffuan Wan Ahmad Let us again consider a simply supported beam AB of length L and carrying concentrated load P at mid span,C as shown below. The beam is supported at each end, and the load is distributed along its length. Solution in the images below, you can also refer these sites to get the deflection values of simply supported beam for different loading conditions by putting the desired values. 1 Simply supported circular plate subjected to uniformly distributed load The deflection, moment and transverse shear are to be finite at the center of the plate (r = 0). I was able to determine the Shear Force Diagram, but currently I'm struggling with the Bending moment diagram. The Young\u2019s Modulus of the beam is 30 x 10^6 Psi. ! The beam has a length of L. shear force and bending moment diagram for simply supported beam with udl - Duration: 19:45. 2d Bending Moment Diagram for Simply Supported. Draw the positive. The tables below give equations for the deflection, slope, shear, and moment along straight beams for different end conditions and loadings. Calculate the flexural stiffness of a simply supported beam which will limit the deflection to 2 mm at the middle. The beam dimensions are b = 1. 01 radians at the ends. The beam is supported at each end, and the load is distributed along its length. (b) Deflection at where the point load is acting, and state its direction. In the above image, I need to find the deflection at point C and D of the simple supported beam. The simplest form of this equation is as follows:. Skip navigation Sign in. Example 1. These programs are of high quality but the approach, and hence the output, is too complex for a quick solution of continuous beam problems. Remember the actual value of v is still intact in the deflection as it is part of the flexural rigidity of the plate. Uniform Load on full span. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. In a coil spring, the stress is distributed evenly along the length of the coil. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. By taking into account the stiffness of the composite joints, the deflection is therefore reduced by roughly 50%. Another method of determining the slopes and deflections in beams is the area-moment method, which. The load on the conjugate beam is the M/EI diagram of the loads on the actual beam. (B) Explain how deflection in beams affects structural stability. The deflection of the beam towards a particular direction when force is applied on it is called Beam deflection. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. 3 away from A and another 5. The Euler-Bernoulli equation describes a relationship between beam deflection and applied external forces. 5 kN/m \u00d7 8. In this post i will briefly explain different types of beams. The ClearCalcs beam calculator allows the user to input the geometry and loading of a beam for analysis in a few simple steps. FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2. shear force and bending moment diagram for simply supported beam with udl - Duration: 19:45. Could anyone please tel me how to calculate the Bending Stress, Compressive stress and the Deflection for any beam (UB 203x102x23) subjected to Uniform Distributed Load of 8063 N/m over a 8 m long beam (simply supported beam). Deflection & Slope - Cantilever Beam with a Point Load at the Free End Deflection & Slope - Cantilever Beam with a Point Load at the Free End. Engineers adopt deflection limits which suit the nature of the building. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. This square tubing deflection calculator calculates tube deflection for square based on length of beam, tube size, decimal gauge and load. D1 is the maximum deflection midspan between supports. Where y is the deflection at the point, and x is the distance of the point along the beam. 9 with a = L qL4 ( C)1 = CCC b4E Ib the deflection ( C)2 due to a force T acting on C is obtained use conjugate beam method TL2 TL L 2L ( C)2 = M = CCC L + CC C C b3E Ib b EIb 2 3 2TL3 = CCC b3E Ib the elongation of the cable is Th ( C)3 = CC EcAc compatibility equation. Use strain energy method. ALL calculators require a Premium Membership. Deflection of Cantilever Beam. All structural/civil formula books, etc seem to neglect this information. This can be used to observe the calculated deflection of a simply supported beam or of a cantilever beam. General shapes are rectangular sections, I beams, wide flange beams and C channels. Remember the actual value of v is still intact in the deflection as it is part of the flexural rigidity of the plate. The beam is also pinned at the right-hand support. EI is constant. Problem 691 Determine the midspan deflection for the beam shown in Fig. For a simply sagging supported structure with distributed mass - or load due to gravitational force - can be estimated as. What is the absolute maximum bending moment due to a moving udl longer than the span of a simply supported beam? 11. deflection curve of beams and finding deflection and slope at specific points along the axis of the beam 9. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. Let say for an example, If we want to calculate the Deflection for Simply Supported Beam having 1m [1000mm] with acting load of 1000N. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. The loads are symmetric about the centre of the beam. 01 radians at the ends. This calculator uses standard formulae to determine the values of slope and deflection at the required section. Introduction Universal beam sections are normally employed in buildings to carry load. 2018/2019. 353 mm came from beam theory - the equation for maximum deflection of a simply-supported beam subjected to a UDL taken this website. The ends A and C are hinged supports and B is a continuous support. (Hint: Apply Case No. Differential Equation of the Deflection Curve of Beam: When a beam is subjected to pure bending couple M, it is bent into a circular arc the radius of curvature. As shown in figure below. It carries a UDL of 9 kN/m over the entire length. The beams used for frame work are selected on the basis of deflection, amongst other factors. For information on beam deflection, see our reference on. Formula: Hollow Square Tube Deflection = (f x l x l x l) / (t x y x (((s - k 4) 4 /12) - ((s - (2. STATIC ANALYSIS OF A SIMPLY SUPPORTED BEAM WITH UNIFORMLY DISTRIBUTED LOAD Figure 1 All dimensions are in mm Objective: To find the deflection, stress, strain, shear force and bending moment diagram of simply supported beam with uniformly distributed load as shown in Figure 1. This beam deflection calculator is designed to calculate the deflection of a simply supported cantilever with a single load point at one end. A simply supported beam is a type of beam that has pinned support at one end and roller support at the other end. Beam bending is analyzed with the Euler-Bernoulli beam equation. (ii) Derive the expression for strain energy due to. The center reaction Q4. y of a simply supported beam under uniformly distributed load (Figure 1) is given by EI qx L x dx d y 2 ( ) 2 2 \u2212 = (3) where. 1 Simply supported circular plate subjected to uniformly distributed load The deflection, moment and transverse shear are to be finite at the center of the plate (r = 0). In multiple span continuous beams, will load in one span produce stress the other spans? A2. I did not have large deflection turned on. Find expressions for the total strain energy of the beam and the deflection under load. Now we have the deflection for a simply supported beam with UDL acting on the whole length. Sign conventions for load, S. Any help would be much appreciated! Thanks. INTRODUCTION (Fig. In this study, Navier's solution for the analysis of simply supported rectangular plates is extended to consider rigid internal supports. 0GPa Tributary width for loading, tw 6. 4 Analysis of Singly Reinforced Beam Bending Consider a simply supported singly reinforced rectangular beam Load causes - deflection (downwards) - bottom of the beam will be in tension while top in compression. From the geometry of the figure, (6. The beam is now cantilevered from this support. 6 and an allowable stress of 175 N/mm. The Navier solution for the rectangular plate simply supported o n all sides and under a uniformly distributed load, , as shown in Figure 1, is presented in Chapter 5 of Timoshenk o\u2019s text. deflection (YId Cantilever with concentrated load Wat end WL2 2EI W 6E1 - ~2~3 - 3 ~2~ + x3~ WL3 3EI. Shear Force diagram for Simply supported Beam with three segments, when third segment start and end is reversed from other 2, then the in-correct shear force diagram generated by Staad pro, Please clarify. Christian Otto Mohr The length of a conjugate beam is always equal to the length of the actual beam. 7KN and the UDL is 1. I believe that changing formulation to MPC is the solution. The centre of gravity of the UDL is at \u00bd of 8. 4 Analysis of Singly Reinforced Beam Bending Consider a simply supported singly reinforced rectangular beam Load causes - deflection (downwards) - bottom of the beam will be in tension while top in compression. Based on the supports of the beam, Following are some of the classification of the beam. The equivalent UDL method is a useful tool for estimating the deflection in a simply supported beam with a complex loading. These programs are of high quality but the approach, and hence the output, is too complex for a quick solution of continuous beam problems. Chapter-5 Deflection of Beam Page- 7 (ix) A simply supported beam with a continuously distributed load the intensity of which at any point 'x' along the beam is x sin x ww L \u239b\u239e\u03c0 = \u239c\u239f \u239d\u23a0 (i) A Cantilever beam with point load at the free end. The calculator produced a report suitable for building regulation approval which shows the bending, shear and deflection for the beam are all within safe limits. Here we display a specific beam loading case. Monash University. Deflection Of Simply Supported Beam With Uniformly Varying Load Posted on May 4, 2020 by Sandra Uniformly varying load 37 review bending moment and shear force moment area theorems civil er uniformly varying load 37 review deflection on a cantilever beam. A simply supported beam cannot have any translational displacements at its support points, but no restriction is placed on rotations at the supports. The force is concentrated in a single point, located in the middle of the beam. when the beam is loaded with a 14KNload at each one third span point, it failed. helping to develop a coherent picture of the deflection behaviour. Deflection of RC (Rectangular Beam) Simply Supported With UDL Elastic modulus, E 7. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. Markings for. Deflection Simply Supported Beam Udl Posted on April 30, 2020 by Sandra Deflection of a simply supported beam a simply supported beam acb has point fixed beam with udl ering max bending moment in a cantilever deflection of cantilever beam. The modulus of elasticity is 205GPa. Beam Deflection Ronni Kaptanband Co. Good question, Thanks For The A2A, OK fine, now why we consider larger side as a depth? Let Us take one example Assume you have one Steel scale which have depth is d and breadth is b. Simple Support Beam with Overhanging. Aug 11, 2019 - Explore alii9131's board \"Bending moment\" on Pinterest. 1 Classification based on supports; 2 Simply Supported Beam : U. See more ideas about Bending moment, Civil engineering construction, Civil engineering design. 044) /64 = 1. Lesson 15 of 15 \u2022 2 upvotes \u2022 12:56 mins. Chapter 9 deflections of beams chapter 9 beams on elastic foundations where is the maximum deflection in a simply supported load quora solved where on the beam below. The shear force diagram indicates the shear force withstood by the beam section along the length of the beam. Total force on beam being wl. The beam is laterally supported. Slope and Deflection for sim. The dead load does not include the self-weight of the beam. Differential Equation of the Deflection Curve of Beam: When a beam is subjected to pure bending couple M, it is bent into a circular arc the radius of curvature. Skip navigation Sign in. 9 away from a on a 6m beam. Beams - Fixed at One End and Supported at the Other - Continuous and Point Loads.\n4wqm9qh2ki0945j 540mmj6nu9t 4ccx8nd6gy4y mip4qag40jqv jib9uh8ne1tn6s2 0gbl4vh4wj d1s2c5evxkjyf jlbax8c46j2xf2 9cwt7mito0b anquvshpp6 925w9babsnq3mwc i9wmrr4w8fercco hipz5bvzvw5uix y36g2dawrsvtf5j e8yz3mnz5ea76hx yrol77n13xvbn w19j1mlvl95bd2 e5nszxy5pqsc3ui 2i7wqx0ccqddd lr7xnx7htc0y 1ym3nift6ij 98o2ter7jo41 at08bqwv8sa sxkdilpoh0 8ogxyeyibgecj", "date": "2020-08-09 15:02:03", "meta": {"domain": "autismart.it", "url": "http://urgd.autismart.it/deflection-of-simply-supported-beam-with-udl.html", "openwebmath_score": 0.556337833404541, "openwebmath_perplexity": 1316.018877203321, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9825575116884779, "lm_q2_score": 0.8652240877899775, "lm_q1q2_score": 0.8501324267518535}}
{"url": "https://math.stackexchange.com/questions/3008575/double-angle-formulas-finding-tan-2-theta", "text": "# Double Angle Formulas: Finding $\\tan 2\\theta$\n\nI am trying to find $$\\tan 2\\theta$$ where $$sin \\theta = \\frac{5}{13}$$ and $$\\theta$$ is in Quadrant One.\n\nAccording to my textbook, $$\\tan 2\\theta = \\frac{120}{119}$$, but I get $$\\frac{-10}{13}$$ instead.\n\nThe Identity I am using:\n\n$$\\tan 2\\theta = \\frac{2 \\tan \\theta}{1 - \\tan^{2}\\theta}$$\n\nMy Process:\n\nSince $$y = 5\\;$$ and $$r = 13,\\; x = 12.$$\n\nApply Tangent Double Angle Formula: $$\\frac{2(\\frac{5}{12})}{1 - (\\frac{5}{12})^2}$$\n\n$$\\frac{\\frac{10}{12}}{1 - \\frac{25}{12}} \\cdot \\frac{12}{12}$$\n\n$$\\frac{10}{12-25}$$\n\n$$\\frac{-10}{13}$$\n\nWhat am I doing wrong?\n\n\u2022 $(5/12)^2$ is not $25/12$. It is $25/144$. \u2013\u00a0Nick Nov 22 '18 at 0:12\n\u2022 Erm, that's probably my issue then. \u2013\u00a0LuminousNutria Nov 22 '18 at 0:12\n\nAlternatively:\n\n$$\\sin\\theta = 5/13$$, implies $$\\cos\\theta = \\sqrt{1-(5/13)^2} = 12/13$$.\n\n$$\\sin 2\\theta = 2\\sin \\theta \\cos \\theta = 120/169$$.\n\n$$\\cos 2\\theta = 2 \\cos^2 \\theta-1 = 1 - 2 \\sin^2 \\theta = 119/169$$.\n\nThis gives $$\\tan 2\\theta = \\sin 2\\theta/ \\cos 2 \\theta = 120/119$$.\n\nI made a mistake solving the problem. $$(\\frac{5}{12})^2 \\neq \\frac{25}{12}$$.\n\nActually, $$(\\frac{5}{12})^2 = \\frac{25}{144}$$.\n\nTaking that into account:\n\n$$\\frac{\\frac{10}{12}}{1 - \\frac{25}{144}} \\cdot \\frac{12}{12}$$\n\n$$\\frac{10}{12-\\frac{300}{144}}$$\n\n$$\\frac{10}{12 - \\frac{25}{12}} \\cdot \\frac{12}{12}$$\n\n$$\\frac{120}{144-25}$$\n\n$$\\frac{120}{119}$$\n\nTherefore, $$\\tan 2\\theta = \\frac{120}{119}$$", "date": "2019-07-21 09:01:37", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3008575/double-angle-formulas-finding-tan-2-theta", "openwebmath_score": 0.8991622924804688, "openwebmath_perplexity": 1058.8126321176442, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575178175919, "lm_q2_score": 0.865224070413529, "lm_q1q2_score": 0.8501324149815505}}
{"url": "https://math.stackexchange.com/questions/1667546/sigma-algebra-generated-by-a-subset", "text": "$\\sigma$-algebra generated by a subset\n\nI'm new to this concept of a $\\sigma$-algebra generated by a collection of subsets.\n\nLet $\\Omega = \\{a, b, c, d\\}$ and \\begin{align} &\\mathcal{F}_1 = \\{\\Omega, \\emptyset, \\{a\\}\\} \\\\ &\\mathcal{F}_2 = \\{\\Omega, \\emptyset, \\{a\\}, \\{b, c, d\\}\\}\\text{.} \\end{align} I wish to show that $$\\sigma\\langle \\mathcal{F}_1 \\rangle = \\bigcap_{\\mathcal{F} \\in \\mathcal{I}(\\mathcal{F}_1)}\\mathcal{F} = \\mathcal{F}_2$$ where $\\mathcal{I}(\\mathcal{F}_1) = \\{\\mathcal{F}: \\mathcal{F}_1 \\subset \\mathcal{F} \\text{ and }\\mathcal{F} \\text{ a }\\sigma\\text{-algebra on }\\Omega \\}$.\n\nI know immediately from this that any $\\sigma$-algebra $\\mathcal{F}$ must, at the very least, contain elements of $\\mathcal{F}_1$: $$\\{\\Omega, \\emptyset, \\{a\\}\\}$$ but other than literally listing every possible collection of subsets of $\\Omega$ and checking which are $\\sigma$-algebras, and then intersecting such sets, I don't see what's an efficient way to do this.\n\nIs my way of thinking correctly or is there a quicker way?\n\n\u2022 Complete the collection of subsets to be closed under the conditions for an sigma algebra, i.e., closed under the operations of countable union or intersection and complementary. This is the minimal sigma algebra that contain the collection. \u2013\u00a0Masacroso Feb 22 '16 at 19:26\n\u2022 @Masacroso Ah, so that's why it's called the \"smallest\" $\\sigma$-algebra generated by a set. Thank you! \u2013\u00a0Clarinetist Feb 22 '16 at 19:29\n\nYes. but one quick shortcut is that the complements of the sets are in the sigma algebra as well, so you can easily see $\\{b,c,d\\}$ must be in $\\sigma\\langle\\mathcal{F}_1\\rangle$.\n\n\u2022 And we know that this $\\{b, c, d\\}$ set with everything else in $\\mathcal{F}_1$ must be the \"smallest\" $\\sigma$-algebra generated by $\\mathcal{F}_1$. Thank you! ... now, the question is, how do I prove this rigorously? I don't see how else to do this other than finding every possible $\\mathcal{F}$ and then intersecting them. \u2013\u00a0Clarinetist Feb 22 '16 at 19:31\n\u2022 @Clarinetist, no, it's the smallest also because it is the minimum extension of the current set to a sigma-algebra. So suffices to find one, and show that none of its subsets are sigma algebras. \u2013\u00a0gt6989b Feb 22 '16 at 19:34\n\u2022 Very good and clear explanation. Thank you! \u2013\u00a0Clarinetist Feb 22 '16 at 19:35\n\nHere are some more general statements that imply the desired equation. We assume the following without proof.\n\n1. The intersection of $\\sigma$-algebras is again a $\\sigma$-algebra.\n2. A set of the form $\\lbrace \\emptyset, \\Omega, A, A^c \\rbrace$ is a $\\sigma$-algebra.\n\nLet $\\mathscr{G}$ (particular case $\\mathcal{F}_1$) be any set of subsets of $\\Omega$. $\\mathscr{A}:=\\bigcap_{\\mathcal{F} \\in \\mathcal{I}(\\mathscr{G})}\\mathcal{F}$ is nonempty because the powerset $\\mathscr{P}(\\Omega)$ has $\\mathscr{P}(\\Omega) \\in \\mathcal{I}(\\mathscr{G})$. It follows that $\\mathscr{G} \\subset \\mathscr{A}$ and that, because of assumption 1, $\\mathscr{A}$ is a $\\sigma$-algebra. If $\\mathscr{A}'$ is another $\\sigma$-algebra with $\\mathscr{G} \\subset \\mathscr{A}'$, then, because $\\mathscr{A}'$ is one of the sets over which the intersection is taken, we have $\\mathscr{A} \\subset \\mathscr{A}'$. So $\\mathscr{A}$ satisfies the definition of the smallest $\\sigma$-algebra generated by $\\mathscr{G}$, i.e. $\\sigma(\\mathscr{G}) = \\mathscr{A}$.\n\nWe conclude $\\sigma(\\mathcal{F}_1) = \\bigcap_{\\mathcal{F} \\in \\mathcal{I}(\\mathscr{\\mathcal{F}_1})} \\mathcal{F}$.\n\nWe now show that $\\sigma(\\mathcal{F}_1) = \\mathcal{F}_2$. Note that $A \\in \\mathcal{F}_1$ implies $A \\in \\sigma(\\mathcal{F}_1)$ and $\\lbrace a \\rbrace \\in \\sigma(\\mathcal{F}_1)$ implies $\\lbrace b,c,d \\rbrace = \\lbrace a \\rbrace^c \\in \\sigma(\\mathcal{F}_1)$, so that $\\mathcal{F}_2 \\subset \\sigma(\\mathcal{F}_1)$. By assumption 2, $\\mathcal{F}_2$ is a $\\sigma$-algebra. We obtain $\\sigma(\\mathcal{F}_1) = \\mathcal{F}_2$.\n\nNote\n\nIt holds in general that $\\sigma(\\lbrace A \\rbrace) = \\sigma(\\lbrace \\emptyset, \\Omega, A \\rbrace) = \\lbrace \\emptyset, \\Omega, A, A^c \\rbrace$, we could have proved this instead of proving this for our particular case in the last step.", "date": "2019-09-22 16:26:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1667546/sigma-algebra-generated-by-a-subset", "openwebmath_score": 0.9799019694328308, "openwebmath_perplexity": 194.6361182394072, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575142422757, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8501324135954369}}
{"url": "https://math.stackexchange.com/questions/3102390/show-that-limit-does-not-exist-two-variables", "text": "# Show that limit does not exist (two variables)\n\nI just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.\n\nI want to answer this question:\n\nShow that the following limit does not exist\n\n$$\\lim_{(x,y)\\to (0,0)}\\frac{x^2y^2}{x^2y^2+(x-y)^2}$$\n\nSo, my working:\n\n$$\\lim_{x\\to 0}\\frac{x^2y^2}{x^2y^2+(x-y)^2}=\\lim_{x\\to 0}\\frac{0}{y^2}=0$$\n\nand\n\n$$\\lim_{y\\to 0}\\frac{x^2y^2}{x^2y^2+(x-y)^2}=\\lim_{y\\to 0}\\frac{0}{x^2}=0$$\n\nI wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $$y=x$$, not sure if that's correct - or my working.\n\nSo, let $$y=x$$, then\n\n$$\\lim_{(x,y)\\to (0,0)}\\frac{x^2y^2}{x^2x^2+(x-y)^2}=\\lim_{x\\to 0}\\frac{x^4}{x^4+(x-x)^2}=1$$\n\nTherefore limit doesn't exist.\n\nIs this somewhat correct? What's the best way to answer a question like this?\n\nAlso, when showing that this limit does not exist, do I need to find different values of limits for both $${x\\to 0}$$ AND $${y\\to 0}$$? Or is one enough, e.g. if I just find two different values for two limits for $${x\\to 0}$$ without using $${y\\to 0}$$ in my calculation at all, is that fine?\n\nThanks!\n\nWhat you have done is correct. The limit exists only if the value of the limit along every direction that leads to $$(0,0)$$ is same.\n\nSo when you calculate $$\\lim_{x\\to 0}\\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $$x=0$$.\n\nSimilarly,\n$$\\lim_{y\\to 0}\\frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $$y=0$$.\n\nAnd the last limit you calculated is along line $$y=x$$.\n\nSo to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $$y=0$$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.\n\n\u2022 \"The limit exists only if the value of the limit along every direction ... is same\" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist. \u2013\u00a0leonbloy Feb 6 at 14:35\n\u2022 @leonbloy Do you have an example of that case? \u2013\u00a0Shufflepants Feb 6 at 15:48\n\u2022 @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist. \u2013\u00a0Swapnil Rustagi Feb 6 at 15:53\n\u2022 If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?) \u2013\u00a0gidds Feb 6 at 16:28\n\u2022 @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/\u2026 (from page 19) \u2013\u00a0leonbloy Feb 6 at 17:11\n\nYour reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.\n\n\u2022 How can I improve my reasoning? \u2013\u00a0Mandingo Feb 6 at 11:48\n\u2022 No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same. \u2013\u00a0PierreCarre Feb 6 at 11:54\n\nWhat you did is completely correct, but perhaps you can ease it a little bit as follows:\n\nFor $$\\;x=0\\;,\\;\\;y\\to0\\;$$ the limit clearly is $$\\;0\\;$$, whereas for $$\\;x=y\\;$$ and $$\\;x\\to0\\;$$ we get\n\n$$\\frac{x^4}{x^4+(x-x)^2}=1\\xrightarrow[x=y\\to0]{}1$$\n\nThus the limit depends on the path chosen $$\\;\\implies\\;$$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.\n\nYou did right.\n\nA possible improvement is to compute the limit along an arbitrary line: along $$y=mx$$ you have to compute $$\\lim_{x\\to0}\\frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}= \\lim_{x\\to0}\\frac{m^2x^4}{m^2x^4+x^2(1-m)^2}= \\lim_{x\\to0}\\frac{m^2}{m^2+(1-m)^2x^{-2}}= \\begin{cases} 0 & m\\ne 1 \\\\[4px] 1 & m=1 \\end{cases}$$ Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.\n\nWhat you did is correct, but note that after proving that the limit is $$0$$ when you take $$y=0$$, the fact that it is also $$0$$ when you take $$x=0$$ is irrelevant. What matters here is that going to $$(0,0)$$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $$(0,0)$$.", "date": "2019-09-16 08:51:30", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3102390/show-that-limit-does-not-exist-two-variables", "openwebmath_score": 0.8499876856803894, "openwebmath_perplexity": 183.92543095249806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575137315161, "lm_q2_score": 0.865224072151174, "lm_q1q2_score": 0.8501324131535155}}
{"url": "http://math.stackexchange.com/questions/250768/is-mathbb-r2-a-field", "text": "# Is $\\mathbb R^2$ a field?\n\nI'm new to this very interesting world of mathematics, and I'm trying to learn some linear algebra from Khan academy.\n\nIn the world of vector spaces and fields, I keep coming across the definition of $\\mathbb R^2$ as a vector space ontop of the field $\\mathbb R$.\n\nThis makes me think, Why can't $\\mathbb R^2$ be a field of its own? Would that make $\\mathbb R^2$ a field and a vector space?\n\nThanks\n\n-\nA field has multiplication. How would you define multiplication on $\\mathbb R^2$ so that it is a field? (There is a way to do so, but it isn't \"obvious\" until you realize that the resulting field is the complex numbers...) \u2013\u00a0Thomas Andrews Dec 4 '12 at 16:07\n$\\mathbb{R}^2$ can be a field but with multiplication defined as follows: $(a,b)(c,d) = (ac - bd, ad + bc)$. Indeed, this is one way of defining the complex numbers. \u2013\u00a0Rankeya Dec 4 '12 at 16:08\nBut, if you want to try to do this for $\\mathbb{R}^n$, for $n \\geq 3$ such that $\\mathbb{R}$ is naturally embedded in $\\mathbb{R}^n$ as a subfield, then it is not possible to do so, and this is a harder fact to prove. \u2013\u00a0Rankeya Dec 4 '12 at 16:11\nok this just answered my follow-up question. Is there any proof of that being true? \u2013\u00a0vondip Dec 4 '12 at 16:11\nVondip - Perhaps this is at a slight tangent, but a significant difference between R and C is that R is an ordered field and C is not. e.g. 5 is larger than 3, but which is \"larger\", 4 + 7i or 6 + 5i ? (Answer: well, defining how \"large\" or \"the length\" a complex number is is not as obvious as for the reals. In fact, there are many different ways of defining the length of a complex number). Just something to think about. \u2013\u00a0Adam Rubinson Dec 4 '12 at 16:37\n\nIf you define:\n\n$$(a,b)+(x,y):=(a+x,b+y)$$\n\n$$(a,b)\\cdot (x,y):=(ax-by,ay+bx)$$\n\nthen the set $\\,\\Bbb R^2=\\Bbb R\\times\\Bbb R\\,$ turns into a field, and a rather well known and important one. Can you identify it?\n\n-\nSpoiler: take a look at one of the comments given. $\\^\\smile\\^$ \u2013\u00a0FrenzY DT. Dec 4 '12 at 16:10\ncomplex numbers indeed! fantastic how it all connects! Would that make R^2 a field and a vector space? \u2013\u00a0vondip Dec 4 '12 at 16:11\nYes, it would, because addition in $\\mathbb{R}^2$, as @DonAntonio defines it, is component wise (the usual way). Remember, the vector space structure depends only on the underlying abelian group. \u2013\u00a0Rankeya Dec 4 '12 at 16:15\nWhat do you mean \"vector space\"? Any field is a vector space over any subfield, so the field $\\,\\Bbb R^2\\cong\\Bbb C\\,$ is a vector field ove an infinite number of subfields, say $\\,\\Bbb C\\,,\\,\\Bbb R\\,,\\,\\Bbb Q\\,,\\,\\Bbb Q(i)\\ldots\\,$ , etc. \u2013\u00a0DonAntonio Dec 4 '12 at 16:15\nSo would that mean that I any vector space could be defined when F -being the field, as : F^n ? \u2013\u00a0vondip Dec 4 '12 at 16:17\n\nIt is important to understand that a set on its own has no algebraic structure. By defining operators on $\\mathbb{R}^2$ you could turn it into (almost) anything you like.\n\nThe natural operators on $\\mathbb{R}^2$, namely $(x, y) + (a, b) \\mapsto (x+a, y+b)$ and $(x, y) \\cdot (a, b) \\mapsto (x\\cdot a, y\\cdot b)$ do not define a field as $(0, 1)$ has no multiplicative inverse.\n\n-\n\nUsually in mathematics one defines these structures as tuples\n\nA field is a triple $(K,+,\\cdot)$ such that $K$ is a set and [...] and $\\cdot:K \\times K \\rightarrow K$\n\nA Vectorspace is a triple $(V,+,\\cdot)$ such that $V$ is a set and [...] and $\\cdot: K \\times V \\rightarrow V$\n\nSo your question is meaningless: A set (say $\\mathbb R^2$) cannot be a field or a vectorspace or a group or anything - only if you add some additional structure (most of the time operations) you can ask this question.\n\nFor example $\\mathbb R^2$ can be the set used in the definition of a field, as well as the underlying set used in the definition of a vectorspace. And we are happy, the addition operation $$+:\\mathbb R^2 \\times \\mathbb R^2 \\rightarrow \\mathbb R^2$$ is the same, and the multiplicaton for \"the\" vectorspace structure $$\\mathbb R \\times \\mathbb R^2 \\rightarrow \\mathbb R^2$$ is \"compatible\" with the multiplication for \"the\" field structure $$\\mathbb R^2 \\times \\mathbb R^2 \\rightarrow \\mathbb R^2$$\n\n-\n\nAdding to the above answer. With the usual exterior multiplication of the $\\mathbb{R}-\\{0\\}$ as a ring with the natural addition and multiplication you can not make a field out of $\\mathbb{R}^{2}-(0,0)$ \\ But there may exist other products such as the one in the answers which can make a field out of ${\\mathbb{R}\\times\\mathbb{R}}-\\{ 0\\}$ \\ According to one of the theorems of Field theory every field is an Integral domain. So by considering : ${\\mathbb{R}\\times\\mathbb{R}}-\\{ 0\\}$ With the following natural product: $(A,B)*(C,D)=(AB,CD)$ We see that $(1,0)*(0,1)=(0,0)$ Which means that $\\mathbb{R}^{2}$is not an integral domain and hence not a field.\n\n-\nWhy are you considering $\\mathbb{R}^*\\times\\mathbb{R}^*$? \u2013\u00a0Tobias Kildetoft Feb 23 '15 at 10:16\nBcoz he is asking that if $\\mathbb{R}^{2}$is a field ... And I think he meant with the exterior multiplication and exterior addition. Otherwise it is obvious we can consider it as field aince it is isomorphism to $\\mathbb{C}$ and hence a field \u2013\u00a0Romel Feb 23 '15 at 10:21\nBut what does that have to do with this? You are removing way more elements than $0$ when you consider this (in fact, it is clear that the usual multiplication does turn this into a group). \u2013\u00a0Tobias Kildetoft Feb 23 '15 at 10:22\nAnd since the natural multiplication is defined on $\\mathbb{R}$ We have to consider $\\mathbb{R}^{*}$ as the set which the natural multiplication works on \u2013\u00a0Romel Feb 23 '15 at 10:24\nNo, to be a field we would need $(\\mathbb{R}\\times\\mathbb{R})\\setminus \\{0\\}$ to be a group, not $\\mathbb{R}^*\\times\\mathbb{R}^*$. \u2013\u00a0Tobias Kildetoft Feb 23 '15 at 10:25", "date": "2016-04-29 10:36:52", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/250768/is-mathbb-r2-a-field", "openwebmath_score": 0.9212859272956848, "openwebmath_perplexity": 279.8928370392323, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.982557512709997, "lm_q2_score": 0.8652240686758841, "lm_q1q2_score": 0.8501324088550003}}
{"url": "https://gmatclub.com/forum/each-term-of-a-certain-sequence-is-calculated-by-adding-a-particular-271963.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Oct 2018, 06:49\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Each term of a certain sequence is calculated by adding a particular\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 50004\nEach term of a certain sequence is calculated by adding a particular\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jul 2018, 00:45\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n86% (01:48) correct 14% (01:40) wrong based on 42 sessions\n\n### HideShow timer Statistics\n\nEach term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?\n\n(A) 20\n(B) 15\n(C) 13\n(D) 12\n(E) 8\n\n_________________\nPS Forum Moderator\nJoined: 25 Feb 2013\nPosts: 1216\nLocation: India\nGPA: 3.82\nEach term of a certain sequence is calculated by adding a particular\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jul 2018, 10:41\nBunuel wrote:\nEach term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?\n\n(A) 20\n(B) 15\n(C) 13\n(D) 12\n(E) 8\n\nlet the constant be $$d$$ and the first term be $$a$$\n\nso 2nd term $$= a+d=27$$ and 5th term will be $$a+4d=84$$. Subtract the two equations to get\n\n$$3d=57 =>d=19$$\n\nHence $$a=27-19=8$$\n\nOption $$E$$\nVP\nJoined: 07 Dec 2014\nPosts: 1104\nEach term of a certain sequence is calculated by adding a particular\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 31 Jul 2018, 17:11\nBunuel wrote:\nEach term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?\n\n(A) 20\n(B) 15\n(C) 13\n(D) 12\n(E) 8\n\nlet d=difference between terms\n84-27=57=3d\nd=19\n27-19=8=1st term\nE\n\nOriginally posted by gracie on 30 Jul 2018, 18:15.\nLast edited by gracie on 31 Jul 2018, 17:11, edited 1 time in total.\nManager\nJoined: 11 Mar 2018\nPosts: 80\nRe: Each term of a certain sequence is calculated by adding a particular\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jul 2018, 19:34\nBunuel wrote:\nEach term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?\n\n(A) 20\n(B) 15\n(C) 13\n(D) 12\n(E) 8\n\nLet the first term be 'a' and the constant being added be d.\n\nSo, 2nd term becomes -\n$$a_2$$ = a + d = 27 ------ (1)\n\nand the 5th term becomes -\n$$a_5$$ = a + d + d + d + d = 84\nHence,\n$$a_5$$ = a + 4d = 84 ------ (2)\n\nNow multiplying equation (1) by 4 to eliminate d\n\nhence (1) becomes -\n\n$$a_2$$ * 4 =>\n4a + 4d = 108 ---- (3)\n\nSubtracting equation (2) from (3)\n\n4a + 4d - a - 4d = 108 - 84\n\n3a = 24\n\na = 8\n\n_________________\n\nRegards\n---------------------------------\nA Kudos is one more question and its answer understood by somebody !!!\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 3896\nLocation: United States (CA)\nRe: Each term of a certain sequence is calculated by adding a particular\u00a0 [#permalink]\n\n### Show Tags\n\n01 Aug 2018, 16:36\nBunuel wrote:\nEach term of a certain sequence is calculated by adding a particular constant to the previous term. The 2nd term of this sequence is 27 and the 5th term is 84. What is the 1st term of this sequence?\n\n(A) 20\n(B) 15\n(C) 13\n(D) 12\n(E) 8\n\nLet n = the constant added to each term to get the next term.\n\nWe have:\n\n2nd term = 27\n\n3rd term = 27 + n\n\n4th term = 27 + 2n\n\n5th term = 27 + 3n\n\nSince we are given that the 5th term is 84, we have:\n\n27 + 3n = 84\n\n3n = 57\n\nn = 19\n\nSince the second term was given as 27, we know the first term is 27 - n. So the first term is 27 - 19 = 8.\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: Each term of a certain sequence is calculated by adding a particular &nbs [#permalink] 01 Aug 2018, 16:36\nDisplay posts from previous: Sort by", "date": "2018-10-20 13:49:12", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/each-term-of-a-certain-sequence-is-calculated-by-adding-a-particular-271963.html", "openwebmath_score": 0.8939218521118164, "openwebmath_perplexity": 1686.1567387233056, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9825575121992375, "lm_q2_score": 0.8652240686758841, "lm_q1q2_score": 0.8501324084130789}}
{"url": "https://www.physicsforums.com/threads/iterated-integrals-setting-up-limits-of-integration.260486/", "text": "# Iterated Integrals - setting up limits of integration\n\n1. Sep 30, 2008\n\n### mirandasatterley\n\n1. The problem statement, all variables and given/known data\n\nFind the volume of the region under the graph of f(x,y) = x+y and above the region y2\u2264x, 0\u2264x\u22649\n\n3. The attempt at a solution\n\nFrom these equations, x will be integrated from 0-9, but i'm not sure about y.\n\nMy thinking is that y will be intgrated from 0-3 because y2\u2264x and the smallest value of x is 0, and the square root of 0 is 0, so that is the smallest y, and the largest x value possible is 9 and the positive quare root of 9 is 3, so this is the largest value of y,\nSo I would integrate:\n\n\u222b0-9\u222b0-3 (x+y)dydx.\n\nIs this correct or am I missing something, where the limits of integration also involve using f(x,y)?\n\n2. Sep 30, 2008\n\n### HallsofIvy\n\nStaff Emeritus\nNo, that is not correct. The region $a\\le x\\le b$, $c\\le y\\le d$, with a, b, c, d numbers is always a rectangle and the figure here is not a rectangle. But the bounds do NOT involve f(x,y)- that is a \"z\" value and goes inside the integral as you have it.\n\nAlways draw a picture for problems like this. y^2= x is a parabola, of course, \"on its side\". The line x= 9 is a vertical line crossing the parabola at (9,3) and at (9, -3). Yes, you can integrate with x going from 0 to 9. On your picture, mark an arbitrary \"x\" by marking a point on the x-axis between 0 and 9. Now draw a vertical line from one boundary to the other. The y bounds, for that x, are y values of those endpoints: $(x, -\\sqrt{x})$, and $(x, -\\sqrt{x})$. Your integral is\n$$\\int{x=0}^9\\int_{y=-\\sqrt{x}}^{\\sqrt{x}} f(x)dy dx= \\int{x=0}^9\\int_{y=-\\sqrt{x}}^{\\sqrt{x}}(x+ y) dy dx$$\n\nOf course, like any double integral, you can reverse the order of integration. If you look at your picture you will see that y ranges, overall, from -3 to 3. Draw a horizontal line across the parabola, representing an arbitrary value of y in that range. It will have left endpoint on the parabola: x= y^2, and right endpoint on the vertical line x= 9. Those will now be the limits of integration for this order:\n$$\\int_{y= -3}^3\\int_{x= y^2}^9 f(x,y)dx dy= \\int_{y= -3}^3\\int_{x= y^2}^9(x+ y)dx dy$$\n\nTry it both ways. You should get the same answer.\n\n3. Sep 30, 2008\n\n### mirandasatterley\n\nIn a similar situation where i have to switch the order of integration from\n\u222b0-3\u222by2-9 f(x,y) dxdy to dydx,\n\nIs this also a parabola on it's side, with intercepts through (3,9) and (0,9), meaning that I would be setting up the new limits of integration, fro the portion of the parabola above y=0?\n\n4. Sep 30, 2008\n\n### mirandasatterley\n\nSo, \u222b0-9\u222bSquare root of x -3 f(x,y) dydx", "date": "2017-05-27 10:22:03", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/iterated-integrals-setting-up-limits-of-integration.260486/", "openwebmath_score": 0.7749534249305725, "openwebmath_perplexity": 406.71554735861827, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877658567787, "lm_q2_score": 0.8757870046160257, "lm_q1q2_score": 0.8501157308771303}}
{"url": "https://www.physicsforums.com/threads/inequality-3-x-3.706872/", "text": "# Inequality (-3/x) < 3\n\n1. Aug 23, 2013\n\n### Qube\n\n1. The problem statement, all variables and given/known data\n\n(-3/x) < 3\n\n2. Relevant equations\n\nDividing/multiplying an inequality causes the inequality sign to change.\n\n3. The attempt at a solution\n\nI keep getting the wrong solution. I tried two methods. I cannot get the textbook solution (x < -1)\n\nMethod one:\n\n-3 < 3x\n-1 < x\n\nMethod two:\n\nDivide both sides by negative one, resulting in:\n\n3/x > -3\n3 > -3x\n-1 < x\n\n2. Aug 23, 2013\n\n### Staff: Mentor\n\nHere you have multiplied by x, which could be positive or negative (or zero). Since you don't know the sign of x, you need two cases - one where x is assumed to be positive and the other where x is assumed to be negative.\nAgain, you've multiplied by x, whose sign you don't know. Same comments as above apply here.\n\n3. Aug 23, 2013\n\n### Qube\n\nAll right! Now I have -1 < x if x is positive and -1 > x if x is negative. How do I determine which is correct? Is the time for trial-and-error?\n\n4. Aug 23, 2013\n\n### Staff: Mentor\n\nThe book's answer is wrong or at least incomplete.\n\nIf x > 0, you get x > - 1. This means that x > 0 AND x > -1. Together, these mean that x > 0.\nIf x < 0, you get x < -1. This means that x < 0 AND x < -1. Together, these mean that x < -1.\n\nIf you look at the graph of y = -3/x, you'll see that it has two parts. For the curve on the left, -3/x < 3 when x < -1. For the curve on the right, -3/x < 3 for any x > 0.\n\n5. Aug 23, 2013\n\n### Qube\n\nSo I guess I'll have to do a bit of trial and error to determine the exact intervals? As in, x > -1 is technically correct, but the best answer is that x is actually greater than 0 or x > 0?\n\nHow do you recommend solving these problems? I'm having massive trouble.\n\n6. Aug 23, 2013\n\n### verty\n\nJust remember the rule, NEVER multiply or divide by x or an expression containing x, it could be negative. Read Mark44's reply again, the answer is: x < -1 OR x > 0.\n\nExample: $\\frac{x+2}{x-3} > 0$. If $x-3 > 0, x+2 > 0$, and if $x-3 < 0, x+2 < 0$. Can you give the answer for this example?\n\n7. Aug 23, 2013\n\n### rock.freak667\n\nNot to detract from what the others are saying, if you multiply by x, you are multiplying by an unknown value which can be +ve or -ve and will thus change the sign of the inequality (i.e. < might become >). So what you can do to prevent this is multiply instead by x2 as this will always be positive.\n\nThen you can solve it as you would using normal algebraic means. You will just need to be careful when doing this as you might pick up an extra solution. But simply analyzing your solution set for the problem would lead you in the correct path.\n\n8. Aug 23, 2013\n\n### Ray Vickson\n\nTurn your inequality into one of the form\n$$\\frac{1}{x} < a$$\nor\n$$\\frac{1}{x} > a$$\nI will let you figure out which inequality applies, and what is the value of $a$.\n\nAnyway, once you have a simple inequality in $1/x$, you can envision the graph $y = 1/x$, and then figure out what the x-region must be for the corresponding inequality in y.\n\nNote: this suggested method is safe; it never has you multiplying or dividing by an x of unknown sign, at least, not until the very end.\n\n9. Aug 24, 2013\n\n### vanhees71\n\nI think it's time to give the solution to clarify the issue:\n\nYou just have to do a case differentiation. First you can simplyfy the inequality by divding through $-3$:\n$$-\\frac{3}{x} < 3 \\Leftrightarrow \\frac{1}{x}>-1.$$\nNow for $x>0$ the inequality is always fulfilled and for $x<0$ you get by multiplying with $(-x)>0$\n$$-1>x \\; \\Leftrightarrow \\; x<-1$$.\nThus the inequality is fulfilled for either $x>0$ or $x<-1$.\n\n10. Aug 24, 2013\n\nYou can also try\n\\begin{align*} \\frac{-3}{x} & < 3 \\tag{Given} \\\\ \\frac{-3}{x} \\cdot x^2 & < 3x^2 \\\\ -3x & < 3^2 \\\\ 0 & < 3x^2 + 3x \\tag{Now factor to obtain}\\\\ 0 & < 3x\\left(x + 1\\right) \\end{align*}\n\nIf you set up a sign table (or graph $3x(x+1)$ you'll find that the inequality is solved\nfor numbers $x < -1$ or $x > 0$ as stated in other places. In short, it is not that one or the other of these two inequalities gives the solution set, is that the solution set consists of any number that satisfies either the first or the second of these two.\n\n11. Aug 24, 2013\n\n### Infrared\n\nI think this problem is being made much harder than it really is.\n\n$\\frac{-3}{x}<3 \\\\ \\frac{-3}{x}-3<0 \\\\ \\frac{1}{x}+1>0 \\\\ \\frac{x+1}{x}>0$\n\nSo $x>0$ or $x<-1$\n\n12. Aug 24, 2013\n\n### Ray Vickson\n\nYou are doing exactly what I suggested the OP do, but he/she never reported back.\n\n13. Aug 25, 2013", "date": "2017-11-24 08:04:21", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/inequality-3-x-3.706872/", "openwebmath_score": 0.8445603251457214, "openwebmath_perplexity": 589.3431137519043, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877675527114, "lm_q2_score": 0.875787001374006, "lm_q1q2_score": 0.8501157292154172}}
{"url": "https://math.stackexchange.com/questions/875627/the-definitions-of-limit-infimum-and-limit-supremum", "text": "# The definitions of limit infimum and limit supremum\n\nI have begun reading Rosenthal's \"A First Look to Rigorous Probability Theory\" and in order to reinforce my calculus background I am studying through the appendix section. Here he defined the limit of a sequence of real numbers as $\\lim_{n \\to \\infty}x_n=x$ if for each $0 <\\epsilon$ there is a natural number $N$ such that it is $|x-x_n| < \\epsilon$ for $n > N$.\n\nThen the limit infimum and supremum are defined as $\\liminf_n x_n = \\lim_{n \\to \\infty} \\inf_{k\\geq n} x_k$ and $\\limsup_n x_n = \\lim_{n \\to \\infty} \\sup_{k\\geq n} x_k$.\n\n1. My first question is about the interpretation of the limit infimum and limit supremum definitions. It looks to me like they are defined as the limits of sequences as well. For example, for the infimum, one can define a sequence $a_n = \\inf_{k \\geq n}x_k$ where $x_n$ was already a sequence of real numbers. Then the limit infimum is the limit of the sequence $a_n$ , $\\lim_{n \\to \\infty}a_n$, so the infimums of the subsequences $x_k$ where $k \\geq n$ are converging if the limit exists. It is the similar for the supremum. Is my interpretation correct here?\n\n2. It is said the limit infimum and supremum do always exist, though they can be infinite. I did not understand this; why it is so?\n\n3. In the book it is stated that the limit $\\lim_{n \\to \\infty} x_n$ exists if and only if it is $\\liminf_n x_n = \\limsup_n x_n$. I could not prove this to myself. How can it be shown that this statement is true?\n\n\u2022 1. ok, 2. they are monotone sequences. \u2013\u00a0enzotib Jul 23 '14 at 7:28\n\u2022 For 3. In general the limit infimum is the least value to which a subsequence of ${a_n}$ can converge and the limit supremum is the greater value to wihch a subsequence of ${a_n}$ can converge \u2013\u00a0Dimitri Jul 23 '14 at 7:44\n\nThese are answers to the secondary questions in the comments after my answer to the original question; I cannot put the answers in comments, because the size of a comment is limited.\n\nThe second question. Let $(x_n)$ be a sequence taking values in the set $\\{0,1,2,\\ldots,9\\}$, and $a:=\\liminf_n x_n$. If $0$ appears infinitely many times in the sequence, then $a=0$. If $0$ appears only finitely many times, but $1$ appears infinitely often, then $a=1$. And so on. This is a slick answer. It is entirely another matter to answer the question with a definite value for a particular given sequence, such as the sequence of digits after the comma in the decimal expansion of $\\pi$. Well, if you can prove that the digit $0$ appears infinitely many times in the decimal expansion of $\\pi$, then you have $a=0$, and so on; in this case it takes a serious theoretical effort to determine the $\\liminf$.\n\nAs for the third question, let $(x_n)$ be a sequence of real numbers such that $a:=\\liminf_n x_n$ is a real number (that is, $-\\infty<a<\\infty$). Since the sequence of infima $a_n:=\\inf_{k\\geq n} x_n$ is increasing (meaning that $m\\leq n$ implies $a_m\\leq a_n$ -- I hate \"nondecreasing\"), the limit $a=\\lim_{n\\to\\infty} a_n$ is actually the supremum: $a=\\sup_n a_n$.\n$\\quad$Let $a_1\\leq a$ and $a'<a_1$; then $a'<a$. We claim that there exists $m$ such that $x_n\\geq a'$ for every $n\\geq m$: since $a=\\sup_n a_n$, there exists $m$ such that $a_m\\geq a'$; but then $x_n\\geq a_m\\geq a'$ for every $n\\geq m$.\n$\\quad$Now let $a_1>a$, and set $a':=\\tfrac{1}{2}(a+a_1)$; we have $a<a'<a_1$. In this case we claim that given any $m$ there exists $n\\geq m$ such that $x_n<a'$: since $a_m=\\inf_{k\\geq m} x_k\\leq a<a'$, there exists $n\\geq m$ such that $x_n<a'$.\n$\\quad$We have proved that $a$ is the largest of all real numbers $a_1$ with the property that for every $a'<a_1$ there are only finitely many $n$ such that $x_n<a'$.\n\n1. Yes, in the book you are reading, $\\liminf_n x_n$ and $\\limsup_n x_n$ are defined as limits of sequences. Note that the sequence $a_n=\\inf_{k\\geq n} x_k$ is increasing (which means that $m\\leq n$ implies $a_m\\leq a_n$), while the sequence $b_n=\\sup_{k\\geq n} x_k$ is decreasing.\n\n2. The $\\liminf$ and $\\limsup$ are taken in the extended real line $\\overline{\\mathbb{R}}=\\{-\\infty\\}\\cup\\mathbb{R}\\cup\\{\\infty\\}$. As an ordered set, $\\overline{\\mathbb{R}}$ is obtained from the ordered set $\\mathbb{R}$ by adding to it the bottom (the least) element $-\\infty$ and the top (the greatest) element $\\infty$. As a topological space, $\\overline{\\mathbb{R}}$ is homeomorphic to any closed interval $[a,b]$ (with $a<b$) in $\\mathbb{R}$, and is therefore compact. For example, $x\\mapsto(2/\\pi)\\arctan x$ is a homeomorphism $\\overline{\\mathbb{R}}\\to[-1,1]$ (where it is understood that ${-}\\infty\\mapsto-1$ and $\\infty\\mapsto1$), and at the same time it is an isomorphism of (totally) ordered sets. All this means that the limits, $\\liminf$'s, and $\\limsup$'s in the extended real line are just `infinitely stretched out' versions of the limits, $\\liminf$'s, and $\\limsup$'s in a closed interval. Since in a closed interval all increasing/decreasing sequences converge, so then do all increasing/decreasing sequences in the extended real line; in particular, $\\liminf_n x_n$ and $\\limsup_n x_n$ always exist. When you are thinking about the extended real line, you can imagine it as a closed interval, you can' go astray with that. If you have to prove some assertion about sequences in the extended real line, formulated in terms of the ordering and the topology (which is, after all, also determined by the ordering), then it suffices to prove the assertion for sequences in a closed interval.\n\n3. You can assume that $x_n$ is a sequence in a closed interval; this will make your life easier, because $\\liminf_n x_n$ and $\\limsup_n x_n$ are real numbers (belonging to the interval). Now prove the following: $a=\\liminf_n x_n$ is the largest real number with the property that for every real number $a'<a$ there are only finitely many $n\\in\\mathbb{N}$ such that $x_n\\leq a'$, and the analogous assertion for $b=\\limsup_n x_n$ (which you will formulate yourself --- just turn everything upside-down). With these characterizations of $\\liminf_n x_n$ and $\\limsup_n x_n$ under the belt, you will easily prove that $\\lim_{n\\to\\infty} x_n$ exists iff $\\liminf_n x_n=\\limsup_n x_n$.\n\nRemark. You did not ask this, but since I have mentioned imagining the extended real line as a closed interval$\\ldots$ $~$You can actually see $\\overline{\\mathbb{R}}$ as the closed interval $[-1,1]$, and transfer anything that's happening in $\\overline{\\mathbb{R}}$ to $[-1,1]$. Let $f(y)=\\tan((\\pi/2)y)$ for $y\\in[-1,1]$, so that $f^{-1}(x)=(2/\\pi)\\arctan(x)$ for $x\\in\\overline{\\mathbb{R}}$. Now define, for all $y,z\\in[-1,1]$, \\begin{aligned} \\mathit{sum}(y,z)~ &:= f^{-1}(f(y)+f(z))~,\\\\ \\mathit{prod}(y,z)~ &:= f^{-1}(f(y)f(z))~. \\end{aligned} Actually the operations $\\mathit{sum}$ and $\\mathit{prod}$ are not defined everywhere on $[-1,1]\\times[-1,1]$; for example, $\\mathit{plus}$ is not defined at the two points $(1,-1)$ and $(-1,1)$ that correspond to the risque limits of the form $\\infty-\\infty$. Write the definitions of $\\mathit{sum}$ and $\\mathit{prod}$ in Mathematica (or a similar program), and then draw their $3\\text{D}$ diagrams, placing on all three axes the labels $-\\infty$, $-1$, $0$, $1$, $\\infty$ at positions, respectively, $-1$, $-1/2$, $0$, $1/2$, $1$. You will be able to observe in the diagrams the behavior of the two operations all the way to the infinity (in either direction); at the points where the operations are not defined you will clearly see the discontinuities (at these points the diagrams contain vertical lines --- more precisely, the closures of the diagrams contain vertical lines).\n\n\u2022 A quick question: What exactly do we mean by a sequence in a closed interval? Do we mean that all values $x_n$ are mapped to values in an interval $[a,b]$ such that it is $a \\leq x_n \\leq b$? (a and b are real numbers) \u2013\u00a0Ufuk Can Bicici Jul 23 '14 at 9:42\n\u2022 Yes, we mean exactly that. \u2013\u00a0chizhek Jul 23 '14 at 9:58\n\u2022 I have another question, it may be weird but still it bugged me. Let's assume that the sequence $x_n$ takes the values of $\\pi$'s digit values after the comma. So each $x_n$ takes a value in the set $(0,1,2,3,4,5,6,7,8,9)$. If we think of the sequence $a_n = \\inf_{k \\geq n} x_k$ this sequence is clearly nondecreasing and bounded in $[0,9]$, but how can we know the value of limit $\\liminf_n x_n$? Since we don't know exactly the values of $x_n$ for very large $n$s, we can be never sure about the value for $a_n$ for these values. \u2013\u00a0Ufuk Can Bicici Jul 23 '14 at 12:36\n\u2022 Continued: The infimum can be zero, one or any value, we cannot observe this exactly for this almost random sequence $x_n$ at large $n$s. So, what can we say about the value $\\liminf_n x_n$ for such sequences? \u2013\u00a0Ufuk Can Bicici Jul 23 '14 at 12:39\n\u2022 For the third question, unfortunately I have failed to find a rigorous proof for that if $a$ is the largest real number such that for every $a' < a$ there is finitely many $x_n < a'$ then $a=\\liminf_n x_n$. But intuitively it is clear that this is valid: If we have a plot of a bounded sequence $x_n$ and visualize a vertical line $y=a'$ then there will be finitely many $x_n$ points under that line and as soon as we increase the line over $a$, the number of $x_n$ points will be infinitely many! But I frustratingly fail to express that visualization as a mathematically valid statement... \u2013\u00a0Ufuk Can Bicici Jul 23 '14 at 15:00", "date": "2019-12-06 18:43:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/875627/the-definitions-of-limit-infimum-and-limit-supremum", "openwebmath_score": 0.967788815498352, "openwebmath_perplexity": 102.1858039005688, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877692486436, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8501157212597258}}
{"url": "https://math.stackexchange.com/questions/496494/smallest-number-with-specific-number-of-divisors", "text": "# Smallest number with specific number of divisors\n\nIs there a general method for finding smallest number of specific number of divisors? I am doing \"Higher Algebra by Barnard JM Child\" and came across a question that \"find the smallest number with 24 divisors\", that's how I tried to solve it, alert me if I am wrong:\n\nSince $24$ can not have more than $4$ prime factors, the number can not have more than 4 prime factors.\n\nAs a single number it is :$2$^${23}$\n\nas product of two numbers: $2^5*3^3$,$2^{11}*3$,$2^7*3^2$, since $5+3<7+2<11+1$, so $2^5*3^3$ is the min of these numbers\n\nas product of three numbers :$2^5*3*5$,$2^3*3^2*5$ since $3+2+1<5+1+1$, hence $2^3*3^2*5$ is the lesser of two\n\nas product of 4 numbers $2^2*3*5*7$\n\n$k=\\min(2^{23},2^5*3^3,2^3*3^2*5,2^2*3*5*7)$\n\n=$2^3*3^2*5=360$\n\nThe above method seems to be fishy and laborious, is there a general approach to find the smallest number with specific number of divisors?\n\n\u2022 For multicharacter exponents, put them in braces. So 2^{11} to get $2^{11}$ Also, \\min gets it set as a function. \u2013\u00a0Ross Millikan Sep 17 '13 at 15:51\n\u2022 oeis.org/A005179 \u2013\u00a0barrycarter Sep 7 '17 at 19:17\n\nThis approach is not fishy at all, but can be laborious. Note that you can't use $5+3 \\lt 7+2$ to conclude that $2^5*3^3 \\lt 2^7*3^2$, although it is true, you have to take the ratio and use $3 \\lt 2^2$. For example, if you were looking for $36$ factors, one comparison would be $2^8*3^3$ versus $2^5*3^5$. Despite the fact that $5+5 \\lt 3+8, 2^5*3^5=7776 \\gt 6912=2^8*3^3$. You can take logs and compare $5 \\log 2 + 3 \\log 3$ with $7 \\log 2 + 2 \\log 3$ to get it right.\n\nI don't know an easier way. Your example shows the failure of a greedy algorithm. Factor the desired number of factors, here as $3*2^3$. Starting from the largest factor, find the cheapest way to get that many. So start with $2^2$, which has $3$ factors. Now you need to double it. You can either multiply by a new prime, clearly $3$, or increase the exponent of $2$ to $5$. Since the first has a factor $3$ and the second $8$, we choose $2^2*3$ Now we want to double again, and our choices are $2^3, 3^2, \\text{ or } 5$ and we take $5$, giving $2^2*3*5$ One more doubling comes from $7$, and we come up with $2^2*3*5*7=420$, not the best.\n\nTake any number $N=p_1^{m_1}...p_k^{m_k}$, where $p_i$ are its prime divisors, and compute how many divisors it has. Each divisor would be a product of the same primes in varying powers, $D=p_1^{d_1}...p_k^{d_k}$, where $0\\leq d_i \\leq m_i$. Different divisors have different collections of powers, so the number of divisors will be $(m_1+1)(m_2+1)...(m_k+1)$.\n\nNow let's find the smallest $N$ such that $(m_1+1)(m_2+1)...(m_k+1)=24$. There are only few possibilities to break 24 into a product of decreasing numbers: $3*2*2*2=4*3*2=6*4=6*2*2=8*3=12*2=24$. The corresponding candidates with the smallest primes chosen for the smallest powers would be these ones:\n\n$2^{3-1}*3^{2-1}*5^{2-1}*7^{2-1}=2^2*3*5*7=420$\n\n$2^{4-1}*3^{3-1}*5^{2-1}=2^3*3^2*5=360$\n\n$2^5*3^3=2592$\n\n$2^5*3*5=480$\n\n$2^{11}*3=6144$\n\n$2^{23}=8388608$\n\nThen just pick the smallest one, which happens to be 360.\n\n\u2022 @ Michael That's what I tried at first,this algorithm becomes laborious if we are dealing with such numbers whose number of divisors has many prime factors,so I want a generalized method in order to avoid comparing numbers>>> \u2013\u00a0Tom Lynd Sep 17 '13 at 17:23\n\nA005179 lists some resources on the problem. In particular:\n\nGrost, M. (1968). The Smallest Number with a Given Number of Divisors. The American Mathematical Monthly, 75(7), 725-729. doi:10.2307/2315183\n\nFrom the introduction:\n\nGiven $h = q_1 q_2 \\dots q_n$, with primes $q_1 \\le q_2 \\le \\dots \\le q_n$, let $A(h)$ be the smallest number with $h$ divisors.\n\nIn many cases, $$A(h) = 2^{q_1-1} 3^{q_2-1} \\dots p_n^{q_n-1}$$\n\nThe primary objective of the paper is to determine the exceptions.\n\nWe call these numbers ordinary, from R. Brown, The minimal number with a given number of divisors, Journal of Number Theory 116 (2006) 150-158. In this paper Brown shows almost all $A(h)$ are ordinary, in particular \"We show here that all square-free numbers are ordinary and that the set of ordinary numbers has natural density one.\"", "date": "2020-01-25 23:36:55", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/496494/smallest-number-with-specific-number-of-divisors", "openwebmath_score": 0.8289042711257935, "openwebmath_perplexity": 297.41830065468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9706877667047451, "lm_q2_score": 0.8757869916479466, "lm_q1q2_score": 0.8501157190318125}}
{"url": "https://stats.stackexchange.com/questions/159905/why-does-drawing-one-card-at-a-time-increase-the-probability-of-choosing-the-ace", "text": "Why does drawing one card at a time increase the probability of choosing the Ace of Spades?\n\nIf you draw 5 cards from a standard deck of 52 cards, then the probability of your hand having the Ace of Spades is: $$\\frac{51\\choose 4}{52\\choose 5} = \\frac{51!5!47!}{4!47!52!} = \\frac{5}{52}$$ If, however, you choose one card a time until you've drawn 5 cards, the probability of having the Ace of Spades is: $$\\frac{1}{52}+\\frac{1}{51}+\\frac{1}{50}+\\frac{1}{49}+\\frac{1}{48}=\\frac{433507}{4331600}\\approx\\frac{1}{10}>\\frac{5}{52}$$ Why does choosing one card a time increase the probability of finding the Ace of Spades, when the resulting hands are equivalently drawn?\n\n\u2022 Hint: instead of drawing 5 cards one at a time, let's say you draw 34. Would you then say your probability of drawing the ace of spades is $1/52 + 1/51 + \\ldots + 1/19 \\approx 1.04 > 1$? \u2013\u00a0Hong Ooi Jul 4 '15 at 13:31\n\u2022 Another hint : What you wanted to do is compute P(X=1)+P(X=2)+..+P(X=5)...with P(x=k) representing the probability of the Ace of Spade drawn exactly at the k th draw.This is one way to find the answer (not the most simple) but one legit way as these are disjunctive cases. The problem comes from how you compute these probabilities. For P(X=2), for example, If you had to answer the question \"what is the probability that the ace of space is drawn exactly at the second draw, would you find 1/51 ? \u2013\u00a0brumar Jul 4 '15 at 13:39\n\u2022 It's not actually homework, just a genuine question. But I think I know what you're saying: I'm conflating the idea of the geometric distribution and dependent draws. So, it'd be better to say $P(X=1)+P(X=2|X\\neq 1)+\\dots+P(X=5|X\\neq 1,2,3,4)$? \u2013\u00a0AJS Jul 4 '15 at 14:26\n\u2022 One thing I can't figure out is, why does it cancel out to $5/52$? $P(X=1) = 1/52$, $P(X=2\\mid X\\neq 1)=P(X=2\\cap X\\neq 1)/P(X\\neq 1)=((51/52)*(1/51)/(51/52))$, giving me the original (incorrect) sum of fractions? \u2013\u00a0AJS Jul 4 '15 at 14:47\n\u2022 Focus on P(X=2\u2229X\u22601) = P(X=2 |X\u22601) * P(X\u22601). What is that? How does the answer to this relate to probability of getting an ace of spaces for the first time on the 2nd card drawn? \u2013\u00a0Mark L. Stone Jul 4 '15 at 15:23\n\nAs @Glen_b suggested, it would be a good idea to summarize the comments part in an answer. I'll do that and also give an alternative for the formula related to the probabilistic point of view at the end of the answer. The apparent contradiction between the two computations came from this line :\n\nthe probability of having the Ace of Spades is: $$\\frac{1}{52}+\\frac{1}{51}+\\frac{1}{50}+\\frac{1}{49}+\\frac{1}{48}=\\frac{433507}{4331600}\\approx\\frac{1}{10}>\\frac{5}{52}$$\n\nThe idea behind this computation was good but the logic was flawed. If we sum the probability of the Ace of Spade drawn exactly at the k th draw with k going from 1 to 5, we have the probability we want. But $\\frac{1}{51}$, for example, does not represent the probability that the ace of spade is exactly drawn at the second attempt but the probability that the ace of space is drawn at the second attempt given that it has not been drawn at the first one.\n\nAJS finally found the right formula\n\nI got it! $1/52+((51/52)\u2217(1/51))+\u22ef+((51/52)\u2217(50/51)\u2217(49/50)\u2217(48/49)\u2217(1/48))=5/52$\n\nWith the idea that the probability to exactly draw the ace of spade at the $k$ th trial is the probability to not draw the ace of spade during previous attempts.... $$(51/52)*(50/51)...(52-k+1)/(52-k+2)$$...multiplied by the probability to draw the card at the $k$ th attempt $$1/(52-k+1)$$\n\nA general rule of thumb that I have to avoid this kind of error is to be cautious when it comes to adding probabilities. If you can turn your problem around to avoid additions to the profit of multiplications, this is less prone to error.\nA more classic approach giving a shorter path would have been to consider that the probability of having the Ace of Spades after 5 draw is 1-(the probability to not draw it with 5 draws) which gives, as you know : $$1-(51/52)*(50/51)*(49/50)*(48/49)*(47/48)=1-47/52=5/52$$\n\n\u2022 Thank you, brumar. I appreciate you taking the time to write it out. \u2013\u00a0AJS Jul 4 '15 at 23:22", "date": "2020-08-06 13:14:58", "meta": {"domain": "stackexchange.com", "url": "https://stats.stackexchange.com/questions/159905/why-does-drawing-one-card-at-a-time-increase-the-probability-of-choosing-the-ace", "openwebmath_score": 0.838527500629425, "openwebmath_perplexity": 310.3443161049843, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754447499796, "lm_q2_score": 0.863391611731321, "lm_q1q2_score": 0.850074180113767}}
{"url": "https://www.grae.io/post/euler_problem_7/", "text": "Project Euler Problem 7: 10001st prime\n\nBack to primes! So far we\u2019ve been able to get away with being a little greedy with our compute when playing with primes. Now Euler is ratcheting up the difficulty and we\u2019ll have to focus on efficiency.\n\nAs usual, if you haven\u2019t spent time with Problem 7 yet, take a chance to play with it on your own and come back.\n\nCounting Primes\n\nLet\u2019s start by looking at each integer, deciding whether it\u2019s prime, and counting it if it is until we get to the 10,001st prime.\n\nOur first stab at an is_prime(n) function will be the simplest and we\u2019ll iterate into more optimized (and complicated) versions after. Here\u2019s the starting point:\n\ndef is_prime(n):\nif n < 2:\nreturn False\nfor x in range(2, n):\nif n % x == 0:\nreturn False\nreturn True\n\n\nThis checks every number less than $n$ to see if it\u2019s a factor of $n$. It\u2019s almost good enough to solve the problem in under a minute. My laptop chugs through the first 10,001 primes in 68 seconds using this version of the is_prime(n) function (full code later). But the rules only give us a minute and we can do better.\n\nCap the Search Space\n\nLooking back at our Problem 3 Solution we optimized our is_prime(n) function by caping the space we search to find factors factors by checking only numbers up to $\\sqrt{n}$. Check out that post if you want to dig deep into why / how that works.\n\ndef is_prime(n):\nif n < 2:\nreturn False\nfor x in range(2, math.floor(math.sqrt(n)) + 1):\nif n % x == 0:\nreturn False\nreturn True\n\n\nThis runs a lot faster. It finds the 10,001st prime in 0.29 seconds on my machine. But can we make it even better?\n\nSkip Through the Search Space\n\nPerhaps the most rediscovered result about primes numbers is the fact that every prime bigger than 3 is \u201cnext\u201d to a multiple of 6. That is, for every prime number starting at 5 you can get a multiple of 6 by adding 1 or subtracting 1.\n\nFor example:\n\n\u2022 5 is prime, add 1 and get 6\n\u2022 13 is prime, subtract 1 and get (6 * 2)\n\u2022 1,361 is prime, add 1 and get (6 * 227)\n\nThis works for every prime number.\n\nWe can use this property to skip potential factors we don\u2019t need to check. When checking to see if a number $n$ has factors we can get away with just looking for the prime factors, we don\u2019t also need to know if it has any factors that are themselves composite. For example, we don\u2019t need to know that 24 is divisible by 8. We can stop as soon as we see it\u2019s divisible by 2. So we can skip every potential factor except for those which might be prime.\n\nIn code:\n\ndef is_prime(n):\nif n < 2:\nreturn False\nif n == 2 or n == 3:\nreturn True\nif n % 2 == 0 or n % 3 == 0:\nreturn False\nfor x in range(6, math.floor(math.sqrt(n)) + 2, 6):\nif n % (x - 1) == 0 or n % (x + 1) == 0:\nreturn False\nreturn True\n\n\nThis version takes advantage of Python\u2019s \u201cstep\u201d argument to range(). We\u2019re looking at every multiple of 6 (below our limit) and checking whether the number before or after it divides our target.\n\nThis optimizes things a bit more and, indeed, finds the 10,001st prime in 0.17 seconds on my machine.\n\nPutting it Together\n\nOnce we have an efficient is_prime() function the solution is a matter of counting primes with a while loop.\n\nseen = 0\nn = 1\nwhile seen < 10001:\nn += 1\nif is_prime(n):\nseen += 1\n\nprint(n)\n\n\nGoing Further\n\nThere are ways to solve this problem even faster. You could use the Prime Number Theorem to approximate an upper bound for a Sieve of Eratosthenes and sieve out the answer. We\u2019ll deal with those concepts in coming problems so for now I\u2019ll leave that as an exercise for the reader.\n\nSee an issue on this page? Report a typo, bug, or give general feedback on GitHub.", "date": "2020-09-29 23:36:16", "meta": {"domain": "grae.io", "url": "https://www.grae.io/post/euler_problem_7/", "openwebmath_score": 0.5414897203445435, "openwebmath_perplexity": 743.9199693840225, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419710536894, "lm_q2_score": 0.8596637559030337, "lm_q1q2_score": 0.8500716028305736}}
{"url": "https://geocabirthve.web.app/1580.html", "text": "# Implicit differentiation examples and solutions pdf\n\nExplicit means fully revealed, expressed without vagueness or ambiguity. Now i will solve an example of the differentiation of an implicit function. Find two explicit functions by solving the equation for y in terms of x. For each problem, use implicit differentiation to find d2222y dx222 in terms of x and y. In this presentation, both the chain rule and implicit differentiation will. Implicit differentiation is as simple as normal differentiation. Implicit differentiation extra practice date period. Implicit di erentiation statement strategy for di erentiating implicitly examples table of contents jj ii j i page2of10 back print version home page method of implicit differentiation. Multivariable calculus implicit differentiation examples. Solutions to implicit differentiation problems uc davis mathematics. They are laying the groundwork for stokestheorem, a. Implicit di erentiation statement strategy for di erentiating implicitly examples table of contents jj ii j i page1of10 back print version home page 23. If we are given the function y fx, where x is a function of time. Since the point 3,4 is on the top half of the circle fig.\n\nThere will also be one or two exercises on material in the next set of notes, which are not taken from the text. Search within a range of numbers put between two numbers. In fact, all you have to do is take the derivative of each and every term of an equation. Implicit differentiation helps us find dydx even for relationships like that. Implicit differentiation practice questions dummies. Implicit differentiation problems are chain rule problems in disguise. Find dydx by implicit differentiation and evaluate the derivative at. For each of the following equations, find dydx by implicit differentiation. Let us remind ourselves of how the chain rule works with two dimensional functionals. Example using the product rule sometimes you will need to use the product rule when differentiating a term. Calculusdifferentiationbasics of differentiationsolutions. Calculus i implicit differentiation practice problems.\n\nEquations inequalities system of equations system of inequalities basic operations algebraic properties partial fractions polynomials rational expressions sequences power sums. How implicit differentiation can be used the find the derivatives of equations that are not functions, calculus lessons, examples and step by step solutions, what is implicit differentiation, find the second derivative using implicit differentiation. In other words, the use of implicit differentiation enables. It is the fact that when you are taking the derivative, there is composite function in there, so you should use the chain rule. Implicit differentiation if a function is described by the equation \\y f\\left x \\right\\ where the variable \\y\\ is on the left side, and the right side depends only on the independent variable \\x\\, then the function is said to be given explicitly. These are functions of the form fx,y gx,yin the first tutorial i show you how to find dydx for such functions. This section contains lecture video excerpts and lecture notes on implicit differentiation, a problem solving video, and a worked example. Check that the derivatives in a and b are the same. Calculus implicit differentiation solutions, examples, videos.\n\nTo generalize the above, comparative statics uses implicit differentiation to study the effect of variable changes in economic models. To perform implicit differentiation on an equation that defines a function \\y\\ implicitly in terms of a variable \\x\\, use the following steps. Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. Differentiation of implicit function theorem and examples.\n\nThen the derivative of y with respect to t is the derivative of y with respect to x multiplied by the derivative of x with respect to t dy dt dy dx dx dt. Implicit differentiation solved practice problems timestamp. This means that when we differentiate terms involving x alone, we can differentiate as usual. Given an equation involving the variables x and y, the derivative of y is found using implicit di erentiation as follows. Implicit differentiation multiple choice07152012104649. For example, according to the chain rule, the derivative of y. Suppose that the nth derivative of a n1th order polynomial is 0. This page was constructed with the help of alexa bosse. To do this, we use a procedure called implicit differentiation. Calculus bc parametric equations, polar coordinates, and vectorvalued functions defining and differentiating parametric equations parametric equations differentiation ap calc.\n\nYou know that the derivative of sin x is cos x, and that according to the chain rule, the derivative of sin x3 is you could finish that problem by doing the derivative of x3, but there is a reason for you to leave. Implicit differentiation mctyimplicit20091 sometimes functions are given not in the form y fx but in a more complicated form in which it is di. Implicit differentiation basic idea and examples youtube. Jul, 2009 implicit differentiation basic idea and examples. Implicit differentiation can help us solve inverse functions. The following problems require the use of implicit differentiation. Multivariable calculus implicit differentiation this video points out a few things to remember about implicit differentiation and then find one partial derivative. Implicit diff free response solutions 07152012145323. The basic idea about using implicit differentiation 1. If a value of x is given, then a corresponding value of y is determined. In such a case we use the concept of implicit function differentiation. Some relationships cannot be represented by an explicit function. Implicit differentiation example walkthrough video khan. There is a subtle detail in implicit differentiation that can be confusing.\n\nThe chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. Here i introduce you to differentiating implicit functions. Jan 22, 2020 implicit differentiation is a technique that we use when a function is not in the form yf x. That is, i discuss notation and mechanics and a little bit of the. Jun 24, 2016 implicit differentiation solved practice problems timestamp. For example, in the equation we just condidered above, we. The majority of differentiation problems in firstyear calculus involve functions y written explicitly as functions of x. Implicit differentiation is a technique that we use when a function is not in the form yf x. Then the derivative of y with respect to t is the derivative of y with respect to x multiplied by the derivative of x with respect to t.\n\nTo make our point more clear let us take some implicit functions and see how they are differentiated. The technique of implicit differentiation allows you to find the derivative of y with respect to x without having to solve the given equation for y. Implicit di erentiation implicit di erentiation is a method for nding the slope of a curve, when the equation of the curve is not given in \\explicit form y fx, but in \\implicit form by an equation gx. Examples of the differentiation of implicit functions.\n\nThis is done using the chain rule, and viewing y as an implicit function of x. Implicit differentiation is a method for finding the slope of a curve, when the equation of the. Calculus implicit differentiation solutions, examples. Implicit di erentiation implicit di erentiation is a method for nding the slope of a curve, when the equation of the curve is not given in \\explicit form y fx, but in \\ implicit form by an equation gx. For difficult implicit differentiation problems, this means that its possible to differentiate different individual pieces of the equation, then piece together the result. Implicit differentiation basic idea and examples what is implicit differentiation. This means that when we differentiate terms involving x. For each problem, use implicit differentiation to find dy dx in terms of x and y. Implicit functions are often not actually functions in the strict definition of the word, because they often have multiple y values for a single x value. Find dydx by implicit differentiation and evaluate the derivative at the given point. The chain rule must be used whenever the function y is being differentiated because of our assumption that y may be expressed as a function of x. However, some functions y are written implicitly as functions of x. The solutions to this equation are a set of points x,y which implicitly define a relation between x and y which we will call an implicit function. Parametric equations differentiation practice khan academy.\n\nDifferentiation of implicit functions engineering math blog. Uc davis accurately states that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve both x and y. Use implicit differentiation directly on the given equation. Implicit differentiation sometimes functions are given not in the form y fx but in a more complicated form in which it is di. Find the equation of the tangent line to the graph of 2. Implicit differentiation mcty implicit 20091 sometimes functions are given not in the form y fx but in a more complicated form in which it is di. In any implicit function, it is not possible to separate the dependent variable from the independent one. Preference bundles, utility and indifference curves. Search for wildcards or unknown words put a in your word or phrase where you want to leave a placeholder.\n\n293 802 632 1207 1048 331 259 386 1012 524 1522 1299 1300 730 1542 1000 938 1032 708 353 1555 1317 459 852 320 880 1127 184 841", "date": "2021-10-20 00:54:47", "meta": {"domain": "web.app", "url": "https://geocabirthve.web.app/1580.html", "openwebmath_score": 0.8376717567443848, "openwebmath_perplexity": 404.06666719092937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419674366167, "lm_q2_score": 0.8596637469145054, "lm_q1q2_score": 0.8500715908328732}}
{"url": "https://www.gradesaver.com/textbooks/math/other-math/discrete-mathematics-with-applications-4th-edition/chapter-11-analysis-of-algorithm-efficiency-exercise-set-11-4-page-764/38", "text": "## Discrete Mathematics with Applications 4th Edition\n\nProve $\\frac{1}{5} + \\frac{4}{5^2} + \\frac{4^2}{5^3} + \\cdots + \\frac{4^n}{5^{n+1}}$ is $\\Theta(1)$. By theorem 5.2.4: $\\frac{1}{5} + \\frac{4}{5^2} + \\frac{4^2}{5^3} + \\cdots + \\frac{4^n}{5^{n+1}} = \\frac{1}{5} (1 + \\frac{4}{5} + \\frac{4^2}{5^2} + \\cdots + \\frac{4^n}{5^{n}}) = \\frac{1}{5}(\\frac{\\frac{4}{5}^{n+1}-1}{\\frac{4}{5}-1}) = -(\\frac{4}{5}^{n+1}-1)=1-\\frac{4}{5}^{n+1}$. For $n>0$, $1-\\frac{4}{5}^{n+1} \\leq 1$. By the transitive property: $\\frac{1}{5} + \\frac{4}{5^2} + \\frac{4^2}{5^3} + \\cdots + \\frac{4^n}{5^{n+1}} \\leq 1$. For $n>0$, $\\frac{1}{5}(1) \\leq \\frac{1}{5} + \\frac{4}{5^2} + \\frac{4^2}{5^3} + \\cdots + \\frac{4^n}{5^{n+1}}$. Since all terms are positive: $\\frac{1}{5}|(1)| \\leq |\\frac{1}{5} + \\frac{4}{5^2} + \\frac{4^2}{5^3} + \\cdots + \\frac{4^n}{5^{n+1}}| \\leq |1|$. Thus for A=1/5, B=1, k=1, $A|(1)| \\leq |\\frac{1}{5} + \\frac{4}{5^2} + \\frac{4^2}{5^3} + \\cdots + \\frac{4^n}{5^{n+1}}| \\leq B|1|$ for all $x>k$. Thus $\\frac{1}{5} + \\frac{4}{5^2} + \\frac{4^2}{5^3} + \\cdots + \\frac{4^n}{5^{n+1}}$ is $\\Theta(1)$.\nRecall the definition of $\\Theta$-notation: $f(x)$ is $\\Theta(g(x))$ iff there exist positive real numbers A, B, k, such that $A|g(x)| \\leq |f(x)| \\leq B|g(x)|$ for all $x>k$. Recall theorem 5.2.3: Sum of a geometric sequence: For any real number $r$ except 1, and any integer $n \\geq 0$, $\\sum_{i=0}^{n}r^i = \\frac{r^{n+1}-1}{r-1}$. In this case $r=\\frac{4}{5}$.", "date": "2019-11-21 23:59:33", "meta": {"domain": "gradesaver.com", "url": "https://www.gradesaver.com/textbooks/math/other-math/discrete-mathematics-with-applications-4th-edition/chapter-11-analysis-of-algorithm-efficiency-exercise-set-11-4-page-764/38", "openwebmath_score": 0.9701132774353027, "openwebmath_perplexity": 184.410606193641, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419677654414, "lm_q2_score": 0.8596637451167997, "lm_q1q2_score": 0.850071589337905}}
{"url": "https://math.stackexchange.com/questions/2257156/prove-sum-k-0n-2n-k-choose-2-n1-choose-3/2257198", "text": "# Prove $\\sum_{k=0}^{n-2}{n-k \\choose 2} = {n+1 \\choose 3}$\n\nProve $$\\sum_{k=0}^{n-2}{n-k \\choose 2} = {n+1 \\choose 3}$$\n\nIs there a relation I can use that easily yields above equation?\n\n\u2022 You could start with $\\;{n-k \\choose 2} = \\frac{(n-k)\\cdot(n-k-1)}{2\\cdot 1}\\,$.\n\u2013\u00a0dxiv\nApr 29, 2017 at 4:59\n\u2022 @mvw not, the same. I edited the problem to consider that. Apr 29, 2017 at 5:00\n\u2022 @dxiv I already knew that. Thank you. What are the next steps? Apr 29, 2017 at 5:01\n\u2022 $\\sum \\frac{(n-k)\\cdot(n-k-1)}{2\\cdot 1} = \\frac{1}{2}\\left(n^2 \\cdot \\sum 1 - n \\cdot \\sum (2k+1) + \\sum k(k+1)\\right)$\n\u2013\u00a0dxiv\nApr 29, 2017 at 5:03\n\u2022 Note: The original question just stated the left hand side\n\u2013\u00a0mvw\nApr 29, 2017 at 11:30\n\nThe right hand side ${n+1 \\choose 3}$ is the number of ways to choose 3 elements from $\\{1,2,\\ldots,n+1\\}$. Let $A$ denote the set of all 3-subsets of $\\{1,2,\\ldots,n+1\\}$. We want to show that $|A|$ is equal to the left hand side.\n\nLet $A_i$ denote the set of all 3-subsets of $\\{1,2,\\ldots,n+1\\}$ which have $i$ as their largest element. For example, if $i=n+1$, then $A_{n+1}$ is the set of all 3-subsets which contain $n+1$, and the number of ways to choose the remaining 2 elements is ${n \\choose 2}$. Hence, $|A_{n+1}| = {n \\choose 2}$. More generally, $|A_i| = {i-1 \\choose 2}$, for $i=3,\\ldots,n+1$, because the remaining 2 elements must be chosen from $\\{1,\\ldots,i-1\\}$. Note that $i$ has to be at least 3 because the largest of 3 elements will be at least 3.\n\nUsing the fact that the $A_i$'s $(i=3,4,\\ldots,n+1)$ are disjoint and their union is all of $A$, we obtain the desired formula.\n\n\u2022 If the largest element in a 3-subset is say 7, then this largest element can't be any of $3, 4, 5, 6, 8, 9,\\ldots,n+1$. So $A_7$ is disjoint from $A_3, A_4, A_5, A_6, A_8, A_9, \\ldots, A_{n+1}$. Similarly for the other $A_i$'s, and so the $A_i$'s are pairwise disjoint. Apr 29, 2017 at 5:56\n\u2022 Thanks for comment. I got it and because of this I deleted my comment. Apr 29, 2017 at 6:10\n\n$$\\sum_{k=0}^{n-2}\\frac{k^2+k(1-2n)+n^2-1}{2}\\tag{Simplify what @dixv said}$$ now, $$\\sum_{k=1}^nk=\\frac{n(n+1)}{2}\\text{ and }\\sum_{k=1}^nk^2=\\frac{n(n+1)(2n+1)}{6}$$\n\nHope it helps?\n\nFor $$S=\\sum_{k=0}^{n-2}(n-k)(n-k-1)$$ set $n-k=r$ to get $$S=\\sum_{r=1}^nr(r-1)=\\sum_{r=1}^nr^2-\\sum_{r=1}^nr$$\n\nUsing index transformation $m = n - k$, reversal of the summation order and the definition of the binomial coefficient in terms of factorials we can write for $n\\ge 2$: \\begin{align} \\sum_{k=0}^{n-2} \\binom{n-k}{2} &= \\sum_{m=2}^{n} \\binom{m}{2} \\\\ &= \\sum_{m=2}^{n} \\frac{m!}{2! (m-2)!} \\\\ &= \\sum_{m=2}^{n} \\frac{m(m-1)}{2} \\\\ &= \\sum_{m=1}^n \\frac{m(m-1)}{2} \\\\ &= \\frac{1}{2} \\sum_{m=1}^{n} m^2 - \\frac{1}{2} \\sum_{m=1}^n m \\\\ &= \\frac{n(n+1)(2n+1)}{12} - \\frac{n(n+1)}{4} \\\\ &= \\frac{1}{6} n^3 + \\frac{1}{4} n^2 + \\frac{1}{12} n - \\left( \\frac{1}{4} n^2 + \\frac{1}{4} n \\right) \\\\ &= \\frac{1}{6} n^3 - \\frac{1}{6} n \\end{align} where we used Faulhaber's formula for the square pyramidal and triangular numbers.\n\nOn the other side of the equation we have: \\begin{align} \\binom{n+1}{3} &= \\frac{(n+1)!}{3!(n+1-3)!} \\\\ &= \\frac{(n+1)n(n-1)}{6} \\\\ &= \\frac{n^3 - n}{6} \\end{align}\n\nYou can prove it using generating functions. We wish to prove that $$\\sum_{m=0}^{n} \\binom{m}{2}=\\binom{n+1}{3}\\tag{1}$$ Use the identity $$\\frac{1}{(1-x)^k}=\\sum_{n=0}^\\infty \\binom{k+n-1}{k-1}x^n.\\tag{2}$$ (which can be obtained by repeatedly differentiating the geometric series) where $k\\geq1$ to get that the generating function whose $m$th coefficent is $\\binom{m+2}{2}$ is $(1-x)^{-3}$ and the generating function whose $m$th coefficent is $\\binom{m}{2}$ is $x^2(1-x)^{-3}$. Thus $$\\sum_{m=0}^{n} \\binom{m}{2}= [x^n]\\left(\\frac{1}{1-x}\\frac{x^2}{(1-x)^{3}}\\right) = [x^n] \\left(\\frac{x^2}{(1-x)^{4}}\\right)\\tag{3} =\\binom{n+1}{3}$$ by (2) as desired where $[x^n]$ extracts the coefficient of $x^n$.\n\n\nWith so many proofs there is no need for one more. But I like using finite differences.\n\nLet $f(n)$ represent the sum.\n\n$\\Delta f(n) = \\binom {n+1}{2} = \\Delta \\binom {n+1}{3}$\n\n$\\implies f(n) = \\binom {n+1}{3} + c$\n\nfor $n=2$, we get $c + 1 = 1 \\implies c=0$", "date": "2023-03-21 04:10:29", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2257156/prove-sum-k-0n-2n-k-choose-2-n1-choose-3/2257198", "openwebmath_score": 1.000009298324585, "openwebmath_perplexity": 244.33631311856286, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419697383903, "lm_q2_score": 0.8596637433190939, "lm_q1q2_score": 0.8500715892563308}}
{"url": "http://vurg.curaben.it/curve-sketching.html", "text": "Sketch the rest of the graph. I now introduce you to plotting the curve r=a sin2\u03b8. 5a limits at infinity 3. If f and g are functions that have derivatives, then the composite function has a derivative given by f. uk A sound understanding of Curve Sketching is essential to ensure exam success. Domain: For what values ofx is f(x) de\ufb01ned? Avoid division by zero and square roots of negative numbers. CURVE SKETCHING EXERCISE 1 Sketch the following Find the gradient of the tangent to the curve x 2 2 + xy 2 + y2 = 14 at (2, 3). MCV4U CURVE SKETCHING QUIZ Name: Give all answers as exact numbers (fractions, terminating decimals, etc. To the left zooms in. Likes scottdave. And the goal here--STUDENT: [INAUDIBLE. Carefully, state L\u2019Hospital\u2019s Rule. 5: Summary of Curve Sketching Last updated; Save as PDF Page ID 4465 If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function. 5 Man vs machine. X \u2013 axis is the tangent at (2,0). Curve sketching lesson plan template and teaching resources. But there were no suitable comparison stars nearby to help today. Calculus plays a much smaller part in curve sketching than is commonly believed; it is just one of the tools at our disposal. Limits and Curve Sketching. This page will take a look at how people have assessed curve sketching in STACK, including some promising projects and alternatives. Instead of focusing on details at the start of a picture, make light sketch lines to capture the posture, proportions, and angles of your subject. Learn More. And so let's get started with that. Each topic builds on the previous one. Assignment. The sketch must include the coordinates of \u2022 \u2026 all the points where the curve meets the coordinate axes. The following incorporates the additional Calculus techniques you have recently learned. Never runs out of questions. Click below to download the free player from the Macromedia site. 5 \u2013 Summary of Curve Sketching Math& 151 Warnock - Class Notes Here are the Guidelines for Sketching a Curve. f 0 (x) > 0). Start your bird sketch by noting the posture of the bird or the angle at which it sits with a single line. No calculator unless otherwise stated. Now after this how do i plot the remaining curve for x >1 (where y become s <0) Can you tell me a smaller approach. There are a number of mathematical curves that produced heart shapes, some of which are illustrated above. AP Calculus. Put the critical numbers in a sign chart to see where the first derivative is positive or negative (plug in the first derivative to get signs). Find the intervals of increase and decrease 8. Well, the free Urban Sketching 101 guide covers everything there is to know, including: what it is, where to go and starter techniques and tips for the urban sketcher on the go. Perhaps someone should change the thread title to \"Not Curve Sketching\". Although these problems are a little more challenging, they can still be solved using the same basic concepts covered in the tutorial and examples. Reading: Curve Sketching Maxima and Minima of Functions Much can be done to sketch the approximate graph of a function without calculus, in fact I strongly encourage you to rely mostly on your pre-calculus skills to sketch graphs. Curve sketching with calculus: logarithm. }\\) Generally, we assume that the domain is the entire real line then find restrictions, such as where a denominator is $$0$$ or where negatives appear under the radical. Sample CHART for Sketching Curves. The ten steps of curve sketching each require a specific tool. We also need to \ufb01nd lim x\u2192c+ g(x) h(x) and lim x\u2192c\u2212 g(x) h(x). How do I tell which ones have or don't have horizontal asymptotes? Please use a simple method to understand! Thank you everyone who answers. The curve does not intersects the y \u2013 axis other than origin. Extrema and Curve Sketching Two Types of Extrema: Absolute (Global) Extremum at x = c: fHxL \u2021 fHcL (respectively, fHxL \u00a3 fHcL) over entire interval of consideration. The graph shown is the DERIVATIVE of f. Keyword-suggest-tool. Math 140: Calculus with Analytic Geometry I Spring 2013 Penn State University Sections 7, 9, 16 Curve Sketching The following is a list topics to consider when drawing the graph of y = f(x). b) horizontal: No horizontal asymptotes because. how to sketch a curve that has asymptotes. Example 3 (f(x,y) = x2 +4y2 \u2212 2x+2) Sketch the level curves of f(x,y) = x2 +4y2 \u22122x+2. The second set of holes uses the sketch to drive a law curve helix (2 turns) that matches the conical shape in that area. C2 Sketching Trigonometric graphs (trigonometric graph shapes) Sketching graphs: the reciprocal graph - C1 Edexcel A Level Maths This video reminds you of a basic index law and then explains the shape of the graph of y=k/x. AP Calculus AB/BC - M. move objects in sketch. While some sketching tools allow 3D drafting, the feature is not universal. The parabola is the envelope of the straight lines. Watch all CBSE Class 5 to 12 Video Lectures here. Curve sketching is a handy tool, used both directly and indrectrly in these examinations. Perhaps someone should change the thread title to \"Not Curve Sketching\". AP Calculus Project 4 - Curve Sketching Name_____ You will be in a group of 2 to 4. The general approach to curve sketching. Find the critical numbers 7. Fix any real number C. Get smarter on Socratic. 4a homework questions 3. STEP 2 Curve Sketching Questions 2. Curve Sketching packet. Heart Curve. Simple Curve Sketching; Higher Derivative and Concavity; Curve Sketching Techniques; Indeterminate Forms; The Integral. Curve Sketching Summary Introduction Now that you have learned how to find relative extrema, intervals where a function is increasing/decreasing, and intervals where a function is concave up/concave down, we will now \"pull it all together\" and work through several AP problems that involve the analysis of functions and curve sketching. While we have been treating the properties of a function separately (increasing and decreasing, concave up and concave down, etc. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. Let\u2019s do another curve sketching example. If you're interested, take a look. While we have been treating the properties of a function separately (increasing and decreasing, concave up and concave down, etc. In the past, one of the important uses of derivatives was as an aid in curve sketching. Now after this how do i plot the remaining curve for x >1 (where y become s <0) Can you tell me a smaller approach. Curve Sketching with Calculus \u2022 First derivative and slope \u2022 Second derivative and concavity. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Active 4 years, 7 months ago. Applications to Curve Sketching. I have also included teacher pages that give: 1) the equation that was actually graphed 2. Intercepts C. Guidelines For Analyzing The Graph Of A Function. MCV4U Unit 3 Test Curve Sketching (V1) Name: Knowledge: /25 Application: /19 Inquiry: /8 Comm. View Homework Help - curve_sketching. a) Domain: Find the domain of the function. L5 \u2013 Curve Sketching Unit 2 MCV4U Jensen Algorithm for Curve Sketching 1. \u201cAlliances\u201d Written by Jeri Taylor Directed by Les Landau Season 2, Episode 14 Production episode 131 Original air date: January 22, 1996 Stardate: 49337. Get smarter on Socratic. In these notes, we will review the critical attributes of the graphs that you studied. Find the - and -intercepts. Here are all the components that we must include: \u00adDomain \u00adIntercepts. Sketching Solution Curves for Autonomous DEs By \ufb01nding and classifying critical points for an Autonomous DE we can greatly simplify the process of sketching a solution curve. All you need is the most minimal of kit, and you're away on a journey of enjoyment and pleasure. Unit 1: Limits & Continuity. Summary of Curve Sketching 1 Domain of f(x) 2 x and y intercepts 1 x-intercepts occur when f(x) = 0 2 y-intercept occurs when x = 0 3 Find the asymptotes (vertical, horizontal / slant). An asymptote is a line that the curve gets very very close to but never intersect. Calculus 3 very difficult curve sketching problem? Sketch the curve traced out by the tip of the radius vector and indicate the direction in which the curve is traversed as t increases. Look at any item sitting around you. In particular, Section 4. If f( x) = f(x), then f(x) is symmetric about. Be sure to nd any horizontal and ver-tical asymptotes, show on a sign chart where the function is increasing/decreasing, concave up/concave down, and identifying (as ordered pairs) all relative extrema and in ection points. Determine the x- and y- intercepts of the function, if possible. A critical point may be a maximum point, minimum point, or neither. However, this equation, y =x / x^2 -1 does not have any horizontal asymptotes. With a model open, click Edit > Project. The curve passes through origin and meets the x \u2013 axis at two coincident points (2,0) and (2,0). This is for a few reasons, but primarily because curve sketching takes a little bit of intuition. XXIII \u2013 Curve Sketching 1. If f and g are functions that have derivatives, then the composite function has a derivative given by f. GraphSketch is provided by Andy Schmitz as a free service. A critical point may be a maximum point, minimum point, or neither. Find more Mathematics widgets in Wolfram|Alpha. Curve Sketching. Link to Binder: Link to Current Tab: Email Embed Facebook Twitter Google+ Classroom Upgrade to Pro Today! The premium Pro 50 GB plan gives you the option to download a copy of your binder to your local machine. 5 Algorithm for Curve Sketching The Algorithm for Curve Sketching provides us with a framework from which we can determine all the key elements of a curve so that we can sketch it relatively accurately. Example 1 Sketch the parametric curve for the following set of parametric equations. Design Sketching Learning Curves (2011, 177 pages, Klara Sj\u00f6l\u00e9n and Allan Macdonald) is a brand new sketch book, aimed at teaching how to really learn to sketch. dominique_cimafranca writes \"The Dynamic Graphics Project of the University of Toronto has released a pretty nifty 3D curve sketching system. Some of the worksheets for this concept are Perspective drawing work, Pencil sketching 2nd edition, Mind map templates, Graphs of trig functions, Drawing basic shapes, The effect of exploratory computer based instruction on, Basic technical mathematics with calculus si version, Using. Curve Sketching - Displaying top 8 worksheets found for this concept. A normal to a curve is a line perpendicular to a tangent to the curve. Textbook Authors: Thomas Jr. curve_sketching_solutions. In this case, it does not have a vertical asymptote. The graph shown is the DERIVATIVE of f. View Homework Help - curve_sketching. We use a multi-stroke pentimenti style curve sketching approach with an ink dry-ing visualization that allows users to sketch uninterrupted. : If your making something that is less than simple it will almost always pay you to do some kind of drawing the try to get things straight in your head before you commit to cutting expensive materials up. Advanced Trigonometry 1 Revision Notes Inverse Trigonometric Function, Stationary Points, Curve Sketching. How to draw curves, pfft! Not so fast! Curves are in a great deal of things you'll want to draw. 3: # 1, 2a-f, 3-9 5. And a lot of people struggle with them even when they don't realise it. Curve Sketching The concepts of domain, limits, derivative, extreme values, monotonicity and concavity have been introduced. Learn more about ferguson curve, curve, draw curve, draw ferguson curve. Sample Problem #1: f(x) = x3 - 6x2 + 9x + 1. From the home tab, select move curve icon command. State any horizontal and vertical asymptotes or holes in the graph. When you exit the sketch, regions are formed by intersecting lines. We begin by making some general remarks about curve sketching, by which we mean more specifically, sketching or drawing the graph of the function y equals f of x in the xy-plane. Sketch the Curve!. While we have been treating the properties of a function separately (increasing and decreasing, concave up and concave down, etc. Instead of focusing on details at the start of a picture, make light sketch lines to capture the posture, proportions, and angles of your subject. 10 determines the solution for given polynomials. unit 4: curve sketching Lesson 1: Increasing/Decreasing Functions. Turning point Axis of Symmetry Mirror point Y intercept X intercepts { the real roots The turning point is always required, and another two points are needed for a rough sketch. Although these problems are a little more challenging, they can still be solved using the same basic concepts covered in the tutorial and examples. WORKSHEETS: Practice-Curve Sketching 1 open ended. Show that, if a > 1, then C has exactly one stationary point. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. If a lack of sketching tips are holding you back from beginning your sketching journey, then we've got you covered. Honors Calculus -e - Asymptotes Plans changed for Date Period Domain: Range: Zeros: y-intercept: HA: c. Tangent lines are useful in calculus because they can magnify the slope of a curve at a single point. Curve sketching, the methods for drawing approximately a curve defined by an equation; Sketch (mathematics), a generalization of algebraic theory A summary of a mathematical proof; Software and computing. This is achieved by adding a sketch relation to your finished curve. how to sketch a curve that has asymptotes. You should be able to quickly sketch straight-line graphs, from your knowledge that in the equation y = mx + c, m is the gradient and. Determine the domain and range. The figure illustrates a means to sketch a sine curve \u2013 identify as many of the following values as you can: asymptotic behaviour,. Get smarter on Socratic. ), we combine them here to produce an accurate graph of the function without plotting lots of extraneous points. Find the location of the x and y intercepts and plot them on the graph. This usually isn\u2019t of help. An asymptote is a line that the curve gets very very close to but never intersect. Detailed Example of Curve Sketching x Example Sketch the graph of f(x) =. Curve Sketching Recipe: 1. When x < 0 then y < 0 so in this case the curve lies in the 3rd quadrant. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. Today, we are going to lay out the principles behind these questions, and explain the methods on how to attack them. Curve Sketching Quiz. 1 $$y=x^5-5x^4+5x^3$$. We will review the main topics that you'll need to know for the AP Calculus exams. Please check your network connection and refresh the page. The system coherently integrates existing techniques of sketch-based interaction with a number of novel and enhanced features. docx: File Size: 255 kb Video - The Mean Value Theorem. Sketching Sketching is useful if you want to create a region that can be pulled into 3D. Theory: This section will review the basic principles and equations that you should know to answer the exam. DUE TUESDAY FEBRUARY 16 AT THE BEGINNING OF CLASS. 1 (i) A curve has equation Find the x-coordinates of the points on the curve where [2] (ii) The curve is translated by Write down an equation for the translated curve. Some of the worksheets for this concept are Perspective drawing work, Pencil sketching 2nd edition, Mind map templates, Graphs of trig functions, Drawing basic shapes, The effect of exploratory computer based instruction on, Basic technical mathematics with calculus si version, Using. Exit Tickets. 5 Man vs machine. Play this game to review Calculus. Madas Created by T. \u00ad10 \u00ad5 5 10 7 \u00ad45. The graph shown is the DERIVATIVE of f. Curve Sketching Example: Sketch 1 Review: nd the domain of the following function. Alternatively, Curve Sketching 1. Turning point Axis of Symmetry Mirror point Y intercept X intercepts { the real roots The turning point is always required, and another two points are needed for a rough sketch. Curve Sketching Calculus, free curve sketching calculus software downloads, Page 2. [Grade 12 Differential Calculus: Curve Sketching] I have ALOT of little questions/confusion about graphing. You should be able to quickly sketch straight-line graphs, from your knowledge that in the equation y = mx + c, m is the gradient and. 4 Curve Sketching V63. Keyword-suggest-tool. org helps support GraphSketch and gets you a neat, high-quality, mathematically-generated poster. the methods for drawing approximately a curve defined by an equation. Curve Sketching The concepts of domain, limits, derivative, extreme values, monotonicity and concavity have been introduced. \u00a92007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Get step-by-step solutions to your Curve sketching problems, with easy to understand explanations of each step. Curve sketching. 5 Curve Sketching \u00b6 permalink. Curve Sketching. A normal to a curve is a line perpendicular to a tangent to the curve. C1 Curve Sketching - Factorising & Sketching Polynomials 2 QP C1 Curve Sketching - Factorising & Sketching Polynomials 3 MS C1 Curve Sketching - Factorising & Sketching Polynomials 3 QP. Label x and y intercepts. A curve with two loops. 5\u2014Curve Sketching Show all work on a separate sheet of paper. Let's put all of our differentiation abilities to use, by analyzing the graphs of various functions. If x is large negative then y is large positive. Concavity/Inflection Points H. The first thing I did to solve this problem was sketch the curve{s}. Leading artists share their sketching tips to help you get started, then take things further. 1 Extreme Values. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. x -Intercept (s) Vertical Asymptote (s) Horizontal and/or Oblique Asymptote (s) First Derivative. Sketch the vector function f(t) = < t 2, t 3 > for -5\u2264 t \u2264 5. Tags: curve sketching. Review of Prerequisite Skills. This is an input location where either f0(x. Unit 1: Limits & Continuity. Created by T. Zeros The zeros of the function f(x) are: x 1 = 3 and x 2 = 2 y-intercept. This usually isn\u2019t of help. Sketch the graph of the curve with equation y x x x= + + \u2212(1 4 2)( )( ), x\u2208. To plot a function just type it into the function box. Because we often represent functions by their graphs, you could say that calculus is all about the analysis of graphs. A full lesson on sketching cubics, quartics and reciprocal functions. is to determine the following: 1) find y(0) 2) find y = 0, if. 2: CURVE SKETCHING POLYNOMIALS Example 3. We have been learning how we can understand the behavior of a function based on its first and second derivatives. Determine asymptotes: a) for vertical asymptotes, check for rational function zero denominators, or unde ned log function points; b) for horizontal asymptotes, consider lim. Let's put all of our differentiation abilities to use, by analyzing the graphs of various functions. We can roughly sketch the graph with stationary point, point of inflection, and y-intercept. Curve Sketching We've done most of the legwork needed for this section. The graph shown is the DERIVATIVE of f. Students describe a curve given the equation of the curve in polar form. The following steps are taken in the process of curve sketching: Find the domain of the function and determine the points of discontinuity (if any). 5 Algorithm for Curve Sketching part 2. So let's hope we can do this. Curve sketching The roots , stationary points , inflection point and concavity of a cubic polynomial x 3 \u2212 3 x 2 \u2212 144 x + 432 (black line) and its first and second derivatives (red and blue). The process of using the first derivative and second derivative to graph a function or relation. It is an application of the theory of curves to find their main features. Fusion 360 Blog. Keyword-suggest-tool. Horizontal and vertical asymptotes may be calculated by taking the appropriate limits of. , ISBN-10: 0-32187-896-5, ISBN-13: 978-0-32187-896-0, Publisher: Pearson. In this calculus worksheet, 12th graders answer questions about derivatives, increasing and decreasing functions, relative maximum and minimum and points of inflection. Prior observations indicate that professionals want full control of the final shapes of 2D/3D curves while leveraging their sketching skills. The course is intended to be challenging and demanding. Concavity/Inflection Points H. Course Description: UCI Math 2A is the first quarter in Single-Variable Calculus and covers the following topics: Introduction to derivatives, calculation of derivatives of algebraic and trigonometric functions; applications including curve sketching, related rates, and optimization; exponential and logarithm functions. Polynomial Curve Sketching - Displaying top 8 worksheets found for this concept. How do I know which equations have horizontal asymptotes? For example, y=x+2 / x-1 has the horizontal asymptote of y=1. GraphSketch is provided by Andy Schmitz as a free service. 10 creates exercises with solutions and graphs in the field of curve sketching of linear, quadratic, cubic, quartic and quintic polynomials. Sketching Sketching is useful if you want to create a region that can be pulled into 3D. Limits by Direct Evaluation. When x < 0 then y < 0 so in this case the curve lies in the 3rd quadrant. Curve sketching (Q350877) From Wikidata. A Rhino curve is similar to a piece of wire. Curve Sketching Summary Introduction Now that you have learned how to find relative extrema, intervals where a function is increasing/decreasing, and intervals where a function is concave up/concave down, we will now \"pull it all together\" and work through several AP problems that involve the analysis of functions and curve sketching. Therefore the domain is D f = R. The curve consist of straight lines and some arces, all tangent to each other. 5 B\u2014Curve Sketching Summary For a function f \u2032, the combined information of the first derivative f and the second derivative f \u2032\u2032 can tell us the shape of a graph. Sketching Polynomials 1 January 16, 2009 Oct 11 \u00ad 9:12 AM Sketching Polynomial Functions Objective \u00ad Sketch the graphs of Polynomial Functions. Solve Curve sketching problems with our Curve sketching calculator and problem solver. Given the information, determine the following about f(x): (Explain each of your answers) 8. Instead, WebAssign will ask limited submission questions about your graphs. Sketching the Curve Summary \u2013 Graphing Ex 2 \u2013 Part 4 of 4. The following problems illustrate detailed graphing of functions of one variable using the first and second derivatives. Curve Sketching. In my opinion, one of the more difficult topics in Extension 1 and 2 has got to be curve sketching. 5 Summary of Curve Sketching(A) GUIDELINES FOR SKETCHING A CURVE \u0629\u0644\u0627\u062f \u0646\u064a\u0628 \u0645\u0633\u0631 \u062a\u0627\u0648\u0637\u062e 1 Find the domain \u0629\u0644\u0627\u062f\u0644\u0627 \u0644 \u062c\u0645 \u062f\u062c\u0648\u0627 1 the domain is, the set of values of for which is defined. However, for sketching \"basic\" cubics, you should be given \"nice\" equations. Explore math with our beautiful, free online graphing calculator. It is recommended that you start with Lesson 1 and progress through the video lessons, working through each problem session and taking each quiz in the order it appears in the table of contents. If f '(x) =0 , then P(x, f (x)) is a local extrema and tangent is horizontal. Welcome to highermathematics. r(t)=(2cost)i+(2sint)j+(2pi-t)k 0 =t=2pi Ok, so I've drawn out the curve, and my curve starts at (2)i+(2pi)k and ends at (2)i. Write NONE if there are none. Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function looks like. How to draw curves, pfft! Not so fast! Curves are in a great deal of things you'll want to draw. Click below to download the free player from the Macromedia site. We will focus on polynomials, but the same methods apply to roughly sketching the graph of any function. Intercepts C. The following techniques may also be of help and they should be employed whenever appropriate. A rational function is looked at as an. We have been learning how we can understand the behavior of a function based on its first and second derivatives. That tells us that our midline drooping down 4. Applications of the Derivative: Curve Sketching and Extrema Practice Problems: The following six pages contain 28 problems to practice curve sketching and extrema problems. In this section, we learn methods of drawing graphs by hand. When f''(x) = 0, there is an inflection point changing the concavity of f(x). Find the - and -intercepts. 6 curve sketching 3. Curve sketching for calculus. Semester Test 1 I Saturday 25 August I D1 Lab 308 I Starts at 09h00. 3 Second Derivative Test. Mathematics / Analysis - Plotter - Calculator 3. Curve Sketching Learning Outcomes Make tables and draw the graphs of various equations to include: Linear Functions Quadratic Functions Cubic Functions - A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. Curve sketching is a handy tool, used both directly and indrectrly in these examinations. there is no y-. Capture drawings on site will forever embed that experience in your mind. \u0629\u0641\u0631\u0639\u0645 \u0644\u0639\u062c\u062a \u064a\u062a\u0644\u0627 \u0645\u064a\u0642 \u0629\u0639\u0648\u0645\u062c\u0645 \u0648\u0647 \u0629\u0644\u0627\u062f\u0644\u0627 \u0644\u062c\u0645. xy\u2013plane where f takes the value C. Curve sketching (Q350877) From Wikidata. Curve Sketching Using Calculus - Part 1of 2. The distance from the starting line of a runner in the 100-meter dash is a. Curve sketching This PowerPoint presentation shows the different stages involved in sketching the graph Sketching the graph Step 1: Find where the graph cuts the axes When x = 0, y = 4/3, so the graph goes through the point (0, 4/3). Incorporates the use of GeoGebra and the Casio fx-991EX Classwiz. In this method, we\u2019ll skip steps 1 to 4 of curve sketching and go straight to steps 5, 6 and 7. y = (sin x)/ x (Pay particular attention to the shape of the curve around x = 0). 2 Sample Problems. Put the critical numbers in a sign chart to see where the first derivative is positive or negative (plug in the first derivative to get signs). Even high school students love to color!. This interactive workshop is designed for students currently enrolled in Math 226 who would like more exposure on understanding the relationships between first derivatives, second derivatives, and curve sketching. Be sure to list the Domain and Range, intercepts, the equation of any asymptotes, intervals of increasing/decrease,. Recall, if they exist, we find the -intercept(s) by setting =0 and. 2 shows graphs A and B. pdf File history uploaded by Paul Kennedy 11 months, 1 week ago No preview is available for MCV 4U Unit 8 Shell-Curve Sketching. Curve Sketching. From the derivative's graph, identify the interval graph where f (the original function) is concave up. The sketch must include the coordinates of \u2022 \u2026 all the points where the curve meets the coordinate axes. edu Abstract Space curve sketching using 2D user interface is a chal-lenging task and forms the foundation for almost all sketch. 2: CURVE SKETCHING RATIONAL FUNCTIONS EXERCISES Give a complete graph of the following functions. And there's relatively little computation. Some of the worksheets for this concept are Perspective drawing work, Pencil sketching 2nd edition, Mind map templates, Graphs of trig functions, Drawing basic shapes, The effect of exploratory computer based instruction on, Basic technical mathematics with calculus si version, Using. 4: # 1 (what you need), 3abc 5. 5 Curve Sketching. Curve Sketching 1. For the sake. Binder ID: 93334. Sketch the graph of the following functions by finding the domain, symmetry, intercepts, asymptotes, intervals of increase and decrease, local maximum and minimum, concavity and points of inflection. Calculus 3 very difficult curve sketching problem? Sketch the curve traced out by the tip of the radius vector and indicate the direction in which the curve is traversed as t increases. And the goal here--STUDENT: [INAUDIBLE. 00 Price per gallon 2. \u2022 The techniques used in algebra for graphing functions do not demonstrate subtle behaviors of curves. 5 Curve Sketching. Polar curves: wrapping a function around the pole. So the next topic is curve sketching. Here input is an equation. XXIII \u2013 Curve Sketching 1. Curve Sketching. This property is called the asymptote. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Curve Sketching 1. Sketching curves the curve r = asin 2\u03b8. Limits and Curve Sketching. 4a concavity and pois 3. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. Curve sketching Graphing The topic menu above allows you to move directly to any of the four sections for each topic. That\u2019s the first step ,in any curve sketching problem. I have a curve sketching assignment and this one question i am having trouble with (this is x2= x squared and 3x2 is 3 x squared. Polar curves: wrapping a function around the pole. f 0 (x) 0). b) horizontal: No horizontal asymptotes because. Passing the fast paced Higher Maths course significantly increases your career opportunities by helping you gain a place on a college/university course, apprenticeship or even landing a job. Find vertical asymptotes and holes. An asymptote is a line that the curve gets very very close to but never intersect. In this calculus worksheet, 12th graders answer questions about derivatives, increasing and decreasing functions, relative maximum and minimum and points of inflection. Voyager \u2026. As $$x$$ increases, the slope of the tangent line increases. The first curve is a rotated cardioid (whose name means \"heart-shaped\") given by the polar equation. That tells us that our midline drooping down 4. Sketching Infinite Lines: the Conic tool applies tangency at each endpoint and selects the top vertex of the curve. The graph shown is the DERIVATIVE of f. CURVE SKETCHING BLAKE FARMAN Lafayette College Name: 1. So even 10 problems you should be able to get through in a few hours. Domain of f(x) 2. y: f 00 ( x ) > 0 ) f ( x up. Recall, if they exist, we find the -intercept(s) by setting =0 and. Topic: Calculus, Derivatives. Curve Sketching Recipe: 1. From the home tab: Direct sketch group -> sketch curve gallery -> edit curve gallery -> move curve. These are potential local extrema. But there were no suitable comparison stars nearby to help today. Although these problems are a little more challenging, they can still be solved using the same basic concepts covered in the tutorial and examples. Convert Entities: Creates one or more entities in a 3D sketch by projecting an edge, loop, face, external curve, external sketch contour, set of edges, or set of external curves onto the sketch plane. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by carefully labeling critical points, intercepts, and inflection. If a graph is given, then simply look at the left side and the right side. The graph of y. The curve will be exactly the same as when you add hydrochloric acid to sodium hydroxide. Because it is a curve in 2d, it is usually easier to sketch than the graph of f. Notes - Curve Sketching (Extrema, Critical Numbers, Intervals of Increase and Decrease, etc. Curve Sketching Date_____ Period____ For each problem, find the: x and y intercepts, x-coordinates of the critical points, open intervals where the function is increasing and decreasing, x-coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. And the goal here--STUDENT: [INAUDIBLE. 5 yesterday, which fit well with the light curve. Let's look at a couple of techniques for making our curve-drawing life a little easier. Oct 5, 2019 #10 Kolika28. 5 How to Find Inflection Points 2 3 2 12 inflection point f(x) =. Curve Sketching. Give a complete graph of f(x) = 1 3 x3 1 2 x2 2x+ 1: Be sure to show on a sign chart where the function is increasing/decreasing, concave up/concave down, and identifying (as ordered pairs) all relative extrema and in ection points. This interactive workshop is designed for students currently enrolled in Math 226 who would like more exposure on understanding the relationships between first derivatives, second derivatives, and curve sketching. Get feedback on your graphs. Further information regarding the Curve Sketching Summer School. Veitch 1 p x 1 = 0 1 p x = 1 1 = p x 1 = x The other critical value is at x = 1. A rational function is looked at as an. We will focus on polynomials, but the same methods apply to roughly sketching the graph of any function. Domain and Range. Curve Sketching Example: Sketch 1 Review: nd the domain of the following function. Right-click the line and select Set as Mirror Line from the context menu. A function f (x) is decreasing on an interval if the values of f decrease as x increases (i. In this method, we\u2019ll skip steps 1 to 4 of curve sketching and go straight to steps 5, 6 and 7. 1 use many of the techniques discussed in this chapter. Take a quick interactive quiz on the concepts in Curve Sketching Derivatives, Intercepts & Asymptotes or print the worksheet to practice offline. MCV4U CURVE SKETCHING QUIZ Name: Give all answers as exact numbers (fractions, terminating decimals, etc. Learn new vocabulary: f is concave up wherever f0 is increasing. Microsoft Word - MCV4U1 - Task - Curve Sketching. This is good because as well as triggering the rebuild it also fully constrains the curve. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. A 3D Curve Sketching System For Tablets 72 Posted by timothy on Saturday October 11, 2008 @10:53PM from the no-mention-of-license-terms dept. No calculator unless otherwise stated. After memorizing the concepts of the second derivative, we move onto the next topic: creating sign charts. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by carefully labeling critical points, intercepts, and inflection. 201-103-RE - Calculus 1 WORKSHEET: CURVE SKETCHING General Guidelines (1) domain of f(x) (2) intercepts (3) asymptotes (a) horizontal asymptotes lim. Curve Sketching Quiz. Y U bA Ql Yl9 irmiwgth1tes m srdeWs3e 0r Vv vebd E. When is the function f(x) concave up? 12. Characteristics of curve. Limits by Direct Evaluation. The basic sine curve has a midline at the x-axis (y = 0). In order to sketch the curve of a function, you need to:. Computer-generated graph of y = x 2 /(x + 3) One of the interesting attributes of curve sketches is that the sketches we make by hand are rarely to scale and can grossly exaggerate features of. DUE TUESDAY FEBRUARY 16 AT THE BEGINNING OF CLASS. You can sketch a curve by listing down a range of values for x, calculating the values for y and from the points drawn on the graph, join the dots. Let\u2019s put it all together; here are some general curve sketching rules: Find critical numbers (numbers that make the first derivative 0 or undefined). CURVE SKETCHING Curve Sketching Steps: for sketching the graph of f(x). Created by T. NOTES: There are now many tools for sketching functions (Mathcad, Scientific Notebook, graphics calculators, etc). 7 Curve Sketching. A full lesson on sketching cubics, quartics and reciprocal functions. And the goal here--STUDENT: [INAUDIBLE. Applet: Curve Sketching: Increasing/Decreasing Try it! The Second Derivative: Concavity and In ection Points Suppose y = f(x) is a given function. 6 # 1-3 SPICY 5. Curve Sketching Introduction Prior to learning calculus, you studied functions of various types, and you learned how to sketch their graphs with and without the support of a calculator. In my opinion, one of the more difficult topics in Extension 1 and 2 has got to be curve sketching. From the derivative's graph, identify the interval graph where f (the original function) is concave up. Polar curves: wrapping a function around the pole. Likes scottdave. Graphs reveal the behaviour of functions and are used for many purposes in mathematics, science and engineering. To zoom, use the zoom slider. This is a graph of the derivative of function h(x). Similarly, we set x = 0 to find the y- intercept. But some of the steps are closely related. (If f \u02dc\u02dcx\u02da \u02da f \u02dcx\u02da, the graph is symmetric with respect to the y-axis; if f \u02dc\u02dcx\u02da \u02da\u02dcf \u02dcx\u02da, the graph is symmetric with respect to the origin). C2 Sketching Trigonometric graphs (trigonometric graph shapes) Sketching graphs: the reciprocal graph - C1 Edexcel A Level Maths This video reminds you of a basic index law and then explains the shape of the graph of y=k/x. Displaying all worksheets related to - Polynomial Curve Sketching. Sample CHART for Sketching Curves. 2 First Derivative Test. The parabola is the envelope of the straight lines. Concavity/Inflection Points H. Sketch the curve using the information for the previous items: Sketch the asymptotes as dashed lines. These are general guidelines for all curves, so each step may not always apply to all functions. org helps support GraphSketch and gets you a neat, high-quality, mathematically-generated poster. These cannot be graded by WebAssign. 201-103-RE - Calculus 1 WORKSHEET: CURVE SKETCHING General Guidelines (1) domain of f(x) (2) intercepts (3) asymptotes (a) horizontal asymptotes lim. Polar curves: wrapping a function around the pole. Created by T. Solution: 1. You can access the circle tools from the Sketch tab of command manager. dvi Created Date. 707\\) and then switch back to concave down at. Let's look at a couple of techniques for making our curve-drawing life a little easier. ) - Domain - Symmetry (Plug in \"-x\" for every \"x\" in the equation, and if the equation doesn't change, then it's an even. Never runs out of questions. (b) Critical Numbers \u2014 numbers a in the domain of f where f\u2032(a) is 0 or unde\ufb01ned. We'll cover two types of curves. how to sketch a curve that has asymptotes. 5 Curve Sketching. 2 First Derivative Test. Andrew Cuomo Saturday said he was signing an executive order allowing the state's roughly 5,000 independent pharmacies to. Add HW points so we can figure out your Unit 3 HW grade. So now, happily in this subject, there are more pictures and it's a little bit more geometric. Let's take a look at an example to see one way of sketching a parametric curve. a) 2f(x) = x3 + x - x + 4 5 b) g(x) = 5x - 3x3 + 3 2. In this video I discuss the following topics to help produce the graph of a function: domain, x-y intercepts, symmetry of the function, intervals of. Title: math142weekinreview6. Powered by Create your own unique website with customizable templates. Some things that might keep your lines from being more fluid loose: using a guide to draw the lines (I would not recommend using a French curve for this reason), drawing the line slowly to maintain precision, pressing too hard (usually goes along with previous). 3 Higher Degree Polynomials and Curve Sketching Name_____ Period____ \u00a9E q2O0e1V7_ jKruStYaB wSuoxfZtGwma^rFe] CLvLeCW. 4 Curve Sketching V63. Curve Sketching. Give x- and y-intercepts. Check your answers with 1t calculator. November 18, 2013. Before we move onto using concavity as a part of curve sketching, we note that using a function\u2019s concavity can be a helpful tool for classifying its extrema. First Derivative Test Find where dy/dx (the deriviative, which is the slope) is zero or undefined; find the critcal numbers for the function. We use a multi-stroke pentimenti style curve sketching approach with an ink dry-ing visualization that allows users to sketch uninterrupted. Let's suppose the function you need to sketch is x^3-8/(x^2-4). 6 A Summary of Curve Sketching 209 Section 3. 147 seventh pages Chapter 3 Curve Sketching How much metal would be required to make a 400-mL soup can? What is the least amount of cardboard needed to build a box that holds 3000. Buying a poster from posters. There are a number of mathematical curves that produced heart shapes, some of which are illustrated above. If you're interested, take a look. Analysis of graphs (or curve sketching) includes finding: Domain and range. pdf: File Size: 12 kb: File Type: pdf: Download File. Plot the intercepts, maximum and minimum points, and in ection points. Okay, so in the last blog, we went over tips for curve-sketching. Brief Notes for STEP Section 06 \u2013 Curve Sketching Curve-sketching is a challenging exercise, one with much variety, and one which represents a good opportunity to display analytical skill. Maximum-minimum by LearnOnline Through OCW. At this point the graph starts to decrease and will continue to decrease until we hit $$x = 1$$. Well, the free Urban Sketching 101 guide covers everything there is to know, including: what it is, where to go and starter techniques and tips for the urban sketcher on the go. We will review the main topics that you'll need to know for the AP Calculus exams. Because it is a curve in 2d, it is usually easier to sketch than the graph of f. That is, this is when f intercepts the x-axis. With a model open, click Edit > Project. y = x sin(1/x) (In particular, what does x sin(1/x) tend to as x tends to 0? The answer is not 1, as a cursory application of might lead you to believe). Multiple-version printing. These are potential local extrema. Give the domain. 1 Increasing and Decreasing Functions. quadratic: 6: PDF: Practice-Curve Sketching 2 open ended. Algebra and pre-calculus. Connecting a function, its first derivative, and its second derivative. Curve Sketching. These are are the sampe problems that we did in class. Determine y-intercept and x-intercepts, if possible. Design Sketching Learning Curves (2011, 177 pages, Klara Sj\u00f6l\u00e9n and Allan Macdonald) is a brand new sketch book, aimed at teaching how to really learn to sketch. Curve sketching, the methods for drawing approximately a curve defined by an equation; Sketch (mathematics), a generalization of algebraic theory A summary of a mathematical proof; Software and computing. How to draw curves, pfft! Not so fast! Curves are in a great deal of things you'll want to draw. Titration curves for weak acid v strong base. On-screen applet instructions: For the function shown, the applet identifies the relationship between the derivative (positive, negative, or zero) and the function (increasing, decreasing, max or min) that can aid in sketching a graph of the function. We now look at an example of sketching curves with asymptotes, i. Curve Sketching: Level 4 Challenges on Brilliant, the largest community of math and science problem solvers. So the next topic is curve sketching. MAXIMUM, MINIMUM, AND INFLECTION POINTS: CURVE SKETCHING - Applications of Differential Calculus - Calculus AB and Calculus BC - is intended for students who are preparing to take either of the two Advanced Placement Examinations in Mathematics offered by the College Entrance Examination Board, and for their teachers - covers the topics listed there for both Calculus AB and Calculus BC. The Reference panel opens. This is the graph of the second derivative of a function. Never runs out of questions. Sketching Solution Curves for Autonomous DEs By \ufb01nding and classifying critical points for an Autonomous DE we can greatly simplify the process of sketching a solution curve. This handout contains three curve sketching problems worked out completely. edu Abstract Space curve sketching using 2D user interface is a chal-lenging task and forms the foundation for almost all sketch. Sketch the Curve!. A function can have two, one, or no asymptotes. If it appears that the curve levels off, then just locate the y-coordinate to which the curve seems to be. Likes scottdave. edu is a platform for academics to share research papers. 2: CURVE SKETCHING POLYNOMIALS Example 3. Determine the x- and y- intercepts of the function, if possible. Concavity and inflection points Critical points (maxima, minima, inflection) Video transcript. Look for any. We can obtain a good picture of the graph using certain crucial information provided by derivatives of the function and certain. Video Lesson Part 2. It cannot have \"no solution\" since a cubic curve has to cross the x-axis at least once. Heart Curve. x 3 - 3 x 2 - 9 x + 5; The sign of the derivative can be used to determine where a function is monotonic, i. Put the critical numbers in a sign chart to see where the first derivative is positive or negative (plug in the first derivative to get signs). Plot the intercepts, maximum and minimum points, and in ection points. These are general guidelines for all curves, so each step may not always apply to all functions. My claim is that we can write $\\cos x + \\sin x$ as $\\sqrt{2}\\cos\\left(x-\\frac{\\pi}{4}\\right)$. Technical Sketching and Drawing. asymptotes: Polynomial functions do not have asymptotes: a) vertical: No vertical asymptotes because f(x) continuous for all x. Math 170 Curve Sketching I Notes All homework problems will require that you create both a sign chart and a graph. Now if g(c) 6= 0, then x= cis a Vertical Asymptote to the curve y= f(x). 1 Crit #s and Abs Extrema 3. com \u00a93 r2I0 E1K3 A YKTurt fa V 9S eo Rfbt NwraWrie A PLyL 5C Q. These are are the sampe problems that we did in class. This is the definitive app for calculus!Simply insert your function into The Calculus. Int: V I ) Max: Min:. geometry, curve sketching (or curve tracing) includes techniques that can be used to produce a rough idea of overall shape of a plane curve given its equation without computing the large numbers of points required for a detailed plot. 2) Curve Sketching Color by Number - In this activity, students practice finding the characteristics of curves. When f''(x) = 0, there is an inflection point changing the concavity of f(x). And there's relatively little computation. Math 170 Curve Sketching II Notes This homework is, once again, mostly about sign charts and graphing. 5 Man vs machine. 5 - Summary of Curve Sketching Math& 151 Warnock - Class Notes Here are the Guidelines for Sketching a Curve. We will focus on polynomials, but the same methods apply to roughly sketching the graph of any function. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. This section deals with recognizing when a curve is symmetric by performing a simple. [Grade 12 Differential Calculus: Curve Sketching] I have ALOT of little questions/confusion about graphing. Mark any asymptotes (if the limit of f(x) (as x approaches positive or negative infinity equals a y-value, then the y-value is a horizontal asymptote). 4 Curve Sketching V63. 5 How to Find Inflection Points 2 3 2 12 inflection point f(x) =. Comet Swan continues to brighten - posted in Sketching: This mornings observation was hampered by windy weather. c O + + +. 8 \u2013 A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. ; Standard Deviants (Performing group); Goldhil Video (Firm);] -- How does art figure into calculus? This program illustrates applications of the derivative through graphing. My claim is that we can write $\\cos x + \\sin x$ as $\\sqrt{2}\\cos\\left(x-\\frac{\\pi}{4}\\right)$. In Curve Sketching 2, we have learned the different properties of quadratic functions that can help in sketching its graphs. 1, Relative Maxima and Minima: Curve Sketching 1 Increasing and Decreasing Functions We say that a function f (x) is increasing on an interval if the values of f increase as x increases (i. If you're interested, take a look. MCV4U CURVE SKETCHING QUIZ Name: Give all answers as exact numbers (fractions, terminating decimals, etc. y \u2260 0 for any values of x, so the graph does not cut the x axis. Convert Entities: Creates one or more entities in a 3D sketch by projecting an edge, loop, face, external curve, external sketch contour, set of edges, or set of external curves onto the sketch plane. Curve Sketching.", "date": "2020-07-04 17:28:29", "meta": {"domain": "curaben.it", "url": "http://vurg.curaben.it/curve-sketching.html", "openwebmath_score": 0.5482892990112305, "openwebmath_perplexity": 1251.2958620737718, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9910145712474399, "lm_q2_score": 0.8577681122619883, "lm_q1q2_score": 0.8500606980030403}}
{"url": "https://davidespataro.it/codinginterviewessentials/18.Generate_points_in_circle_uniformly.html", "text": "# Generate points in circle uniformly\n\n## Introduction\n\nThis chapter\u2019s problem concerns uniformly generating a (potentially large) number of random points in a circle of a certain radius. Despite its simplicity the problem poses some unexpected challenges. We will discuss the best approach to this problem as well as one solution that many candidates provide which, whilst initially appearing correct actually fails in one crucial aspect (spoiler: it does not distribute points uniformly).\n\n## Problem statement\n\nWrite a function that, given a circle of radius $$r$$ and centered at $$(x,y)$$ where $$r,x,y \\in \\mathcal{R}$$ returns a uniformly distributed point in the circle.\n\n## Clarification Questions\n\nWhat exactly does it mean for the point to be uniformly distributed?\n\nIt means that every point of the circle has the same probability of being picked/generated by the function.\n\n## Discussion\n\nBefore discussing solutions it is worth mentioning that the fact that the circle is centered at $$(x,y)$$ makes very little difference and we can continue our discussion as if it were centered at $$(0,0)$$. This is the case because all the points we generate can then be translated to $$(x,y)$$ by simply adding $$x$$ and $$y$$ to the $$x$$-coordinate and $$y$$-coordinate of the generated point.\n\n### Polar Coordinates - The wrong approach\n\nLet\u2019s start by discussing an intuituve, but ultimately incorrect, approach. One might think that in order to pick a point in the circle it is sufficient to\n\n1. Pick a random angle $$\\theta \\in [0, 2\\pi[$$\n\n2. Pick a random radius $$\\overline{r} \\in [0,r]$$\n\n3. Generate the Cartesian coordinates of the point given the radius and the angle (polar coordinates [@cit:wiki:polarcoordinates]) as (see Figure 22.1): $\\begin{gathered} x=\\overline{r}\\sin(\\theta) \\\\ y=\\overline{r}\\cos(\\theta) \\end{gathered}$\n\n[fig:random_points_in_cirle:polar_coordinates]\n\nUnfortunately, despite its appealing simplicity, this approach is wrong as it fails to produce points that are distributed uniformly in the circle. Before examining the mathematical proof it is instructive to have a look at Figure 22.2 which is drawing a large number of points on the circle generated using this incorrect solution. As you can see, the points are not generated uniformly as their density is higher towards the center. The bottom line is, do not use this solution in an interview. A possible matlab implementation of this buggy approach is shown at [list:random_points_in_circle:buggy]\n\nListing 1: Non-uniform random point in a circle generation using Matlab\n\nfunction [px, py] = buggy_random_point(radius, x,y)\ntheta = rand()*2*pi;\npx = r * sin(theta);\npy = r *cos(theta);\nendfunction\n\n[fig:random_points_in_cirle:buggy]\n\n### Loop approach\n\nA good way to ensure that the point density is uniform across on the surface of the circle is to pick a point randomly in an enclosing square and make sure that we discard all the points that lie outside the circle. In other words, we keep asking for a random point $$(p_x=\\text{rand()}, p_y=\\text{rand()})$$ in the enclosing square until the following is true: $$p_x^2 + p_y^2 \\leq r$$. In this way we are guaranteed to generate uniformly distributed points because we pick those points from a set of points that are already uniformly distributed in a square, and we exclude those which are not inside the circle. This method is also known as the exclusion method. Figure 22.3 depicts a large number of points generated with this method.\n\nThe downside here is that we might need to generate a number of points in the square before getting lucky and picking one lying in the circle. We need to make on average $$\\approx 1.2732$$ tries before getting a point in the circle. This number is the ratio between the are of enclosing square and the area of the enclosed circle i.e. $$\\frac{(2r)^2}{\\pi r^2} = \\frac{4}{\\pi}$$.\n\n[fig:random_points_in_cirle:loop]\n\nA Matlab implementation of this approach is shown in Listing [list:random_points_in_circle:loop].\n\nListing 2: Random point in a circle generation using the exclusion method.\n\nfunction [px, py, t] = random_point_loop(radius, x,y)\npx = 100;\npy = 100;\nt=0;\nwhile (px*px + py*py > 1)\nsignx = 2*randi([0,1])-1;\nsigny = 2*randi([0,1])-1;\nt=t+1;\nendwhile\n\nendfunction\n\n### Polar Coordinates - The right approach\n\nIn order for the points to be distributed uniformly it is necessary that the average distance between the points be the same regardless of how far they lie from the center of the circle. This means that, looking at the points generated on a circumference of radius $$2$$, there have to be twice as many points as the the number of points on a circumference of radius $$1$$. A circumference that is twice as long translates to needing twice as many points to maintain the same density. Another intuitive way to understand why simply picking a random angle and a random radius is not enough would be to think about having to distribute $$10$$ points at random on a circle of radius $$1$$ and $$2$$. It is clear that the circumference of radius $$2$$ would look emptier than the one with radius $$1$$ simply because there is more circumference to be filled but a constant amount of points.\n\nThe fundamental problem with the appraoch described in Section 22.3.1 is that the area of the circle is not uniformly covered. The random radius cuts through the area of the circle and this is the only parameter that affects how the points are going to be distributed across the full area of the circle. Therefore we should focus our attention how we can pick a better radius by making sure that larger radii are picked more often to accommodate for the larger area they define. In other words, we need to ensure that our random function for picking the radius takes the area of our circle into account.\n\nConsider the area $$A$$ of a circle of radius $$r$$ i.e.\u00a0$$A = \\pi r^2$$. We can rearrange the formula so that $$r = \\sqrt{\\frac{A}{\\pi}}$$. What this formula is really telling us is that the radius is proportional to $$\\sqrt{A}$$. Now we have a way of choosing the radius that depends on the area of the circle. We can simply pick an area at random and then calculate the radius accordingly. This will make sure that the radius is picked taking into consideration the area of the circle. Figure 22.4 shows many points generated using this method. As you can see the points are generated uniformly across the area of the circle and the picture looks similar to Figure 22.3.\n\nA C++ implementation of this method is shown in Listing [list:random_points_in_circle:sqrtcpp]. Details on the random number generation in Modern C++ can be found in [@cit::std::random].\n\nListing 3: C++ implementation of the function for generating a random point in a circle described in Section \\ref{random_points_in_circle:sec:polar_sqrt}\n\nauto generate_random_point_in_circle()\n{\nstatic std::uniform_real_distribution<double> dist_angle(0, 2 * M_PI);\nconst auto theta = dist_angle(rnd);\nconst auto x     = r * cos(theta);\nconst auto y     = r * sin(theta);\nreturn std::make_pair{x + x_center, y + y_center};\n}\n\n[fig:random_points_in_cirle:polar_sqrt]\n\nA Matlab implementation of this approach is shown in Listing [list:random_points_in_circle:sqrt].\n\nListing 4: Random point in a circle generation using polar coordinates and the $\\approx \\sqrt{A}$ dependency of the radius on the area of the circle.\n\nfunction [px, py] = random_sqrt_area(radius, x,y)\nr = sqrt(area/pi);\ntheta = rand()*2*pi;\npx = r * sin(theta);\npy = r *cos(theta);\nendfunction\n\n### Conclusion\n\nFor both the viable methods for generating random points withing a circle which we have discussed the time and space complexity is constant although the one presented in Section 22.3.3 will probably have better performance when included in a hot path i.e.\u00a0in a loop for the generation of many random points.\n\nAll the code used to generate the Figures in this chapter is shown in Listing [list:random_points_in_circle:drivercode].\n\nListing 5: Matlab driver code for the generation of all figures in Chapter \\ref{ch:random_points_in_circle}\n\nfunction draw_points(n)\nclf(1);\n% n is the total number of points\npx = zeros(1,n);\npy = zeros(1,n);\n\ntries = 0;\nfor i =0:n\n%[x,y] = buggy_random_point(1,0,0);\n% [x,y,t] = random_point_loop(1,0,0);\n[x,y] = random_sqrt_area(1,0,0);\n% tries = tries + t;\npx(i+1) = x;\npy(i+1) = y;\nendfor\n\naverage = tries/n\n\n% Plot a circle.\nangles = linspace(0, 2*pi, n);\nxCenter = 0;\nyCenter = 0;\ncx = radius * cos(angles) + xCenter;\ncy = radius * sin(angles) + yCenter;\n% Plot circle.\nplot(cx, cy, 'b-', 'LineWidth', 2);\n% Plot center.\nhold on;\nplot(xCenter, yCenter, 'k+', 'LineWidth', 2, 'MarkerSize', 16);\ngrid on;\naxis equal;\nxlabel('X', 'FontSize', 16);\nylabel('Y', 'FontSize', 16);\n\n% Plot random points.\nplot(px, py, 'r.', 'MarkerSize', 1);\n\nrectangle('Position',[-1 -1 2 2], 'LineWidth',3,  'EdgeColor' , [0 .5 .5])\n\nendfunction", "date": "2021-09-27 06:30:18", "meta": {"domain": "davidespataro.it", "url": "https://davidespataro.it/codinginterviewessentials/18.Generate_points_in_circle_uniformly.html", "openwebmath_score": 0.750119686126709, "openwebmath_perplexity": 434.14399494217605, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846678676151, "lm_q2_score": 0.8688267830311355, "lm_q1q2_score": 0.850046803550406}}
{"url": "http://math.stackexchange.com/questions/312981/rewrite-so-that-the-denominator-does-not-have-any-root-expressions-frac-sqrt", "text": "# Rewrite so that the denominator does not have any root expressions: $\\frac{\\sqrt[3]{49} +\\sqrt[3]{7x} + \\sqrt[3]{x^2}}{\\sqrt[3]{x} -\\sqrt[3]{7}}$\n\nI am struggling with rewriting the following so that the denominator does not have any root expressions:\n\n$$\\frac{\\sqrt[3]{49} +\\sqrt[3]{7x} + \\sqrt[3]{x^2}}{\\sqrt[3]{x} -\\sqrt[3]{7}}$$\n\nI guess I should start with the denominator and try to get rid of the cube root expressions. But I cannot really get how one would do that easily. Is there another way to solve this problem?\n\nThank you kindly for your help!\n\n-\nRemember $x^3-y^3=(x-y)(x^2+xy+y^2)$. \u2013\u00a0 Maesumi Feb 24 '13 at 15:47\n@Maesumi Post it as an answer? \u2013\u00a0 Git Gud Feb 24 '13 at 15:50\n\nHint:$a^3-b^3=(a-b)(a^2+ab+b^2)$\n\nLet $a=x^{1/3},b=7^{1/3}$\n\n$x-7=a^3-b^3=(a-b)(a^2+ab+b^2)=(x^{1/3}-7^{1/3})(7^{2/3}+(7x)^{1/3}+x^{2/3})$\n\n$\\displaystyle \\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}=\\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}\\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(7^{2/3}+(7x)^{1/3}+x^{2/3})}=\\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})^2}{(x-7)}$\n\n-\nThank you for your answer! I would very much appreciate if you would write the full answer. I want to study in what order you would do the arithmetic. \u2013\u00a0 Lukas Arvidsson Feb 24 '13 at 15:58\nNow is it okay?????? @LukasArvidsson \u2013\u00a0 Abhra Abir Kundu Feb 24 '13 at 16:04\nThank you very much! \u2013\u00a0 Lukas Arvidsson Feb 24 '13 at 16:05\nYou are welcome @LukasArvidsson \u2013\u00a0 Abhra Abir Kundu Feb 24 '13 at 16:06\n\nSince $u^3-v^3=(u-v)(u^2+uv+v^2)$, then $$\\frac{v^2+uv+u^2}{u-v}=\\frac{u^2+uv+v^2}{u-v}=\\frac{\\left(u^2+uv+v^2\\right)^2}{u^3-v^3}.$$ What are $u$ and $v$ in your particular case?\n\n-", "date": "2015-07-31 07:34:52", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/312981/rewrite-so-that-the-denominator-does-not-have-any-root-expressions-frac-sqrt", "openwebmath_score": 0.895392656326294, "openwebmath_perplexity": 546.4246400125473, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846628255123, "lm_q2_score": 0.8688267745399465, "lm_q1q2_score": 0.8500467908620429}}
{"url": "http://math.stackexchange.com/questions/3999/is-a-prime-number-still-a-prime-when-in-a-different-base/198752", "text": "# Is a prime number still a prime when in a different base?\n\nIs a prime number in the decimal system still a prime when converted to a different base?\n\n-\nWhat do you mean exactly? 4 is divisible by 2 independently of base. \u2013\u00a0Il-Bhima Sep 4 '10 at 9:11\nThink of representing a number as a pile of rocks. If a number n can be factored as n = a*b, then we can arrange the pile of rocks into an a by b rectangle. If a number n is prime then it can only take the form of a trivial a 1 by n or n by 1 rectangle. Notice we haven't made any mention of a base. \u2013\u00a0yjj Sep 4 '10 at 9:11\nDeveloping on yjj's answer: n being a prime number is a property of the number in terms of arithmetic operations (e.g. multiplications). A decimal representation is just a way of representing the number: the representation doesn't affect any of its \"arithmetical\" properties. As the Bard said, \"a rose by any other name would smell as sweet.\" \u2013\u00a0Niel de Beaudrap Sep 4 '10 at 10:17\nOn a semi-related note, Mersenne primes have the trivial but fun property that if $2^p -1$ is a Mersenne prime, then it can be represented as $p$ $1$'s in base $2$. \u2013\u00a0Joshua Shane Liberman Sep 4 '10 at 13:53\n\nHere is the same question which has been asked at Mathforum: Here is the link\n\nhttp://mathforum.org/library/drmath/view/55880.html\n\n-\nIt would have been helpful if you quoted the answer here in addition to the link. \u2013\u00a0Sniper Clown May 14 '12 at 4:45\n\nAlas, the accepted answer is misleading (and arguably incorrect). The reason that primality (or any other purely arithmetic property) is preserved in radix representation is simply that such representation faithfully preserves all of the arithmetic operations on integers. More precisely, $\\:$ if $\\rm\\;n\\to r(n)\\;$ is the map from $\\rm\\:n\\:$ to its radix $\\rm\\:d\\;$ representation, then it preserves addition $\\rm\\;r(m+n) = r(m) + r(n),\\;$ and multiplication $\\rm\\;r(mn) = r(m)\\ r(n),\\;$ and $\\rm\\;r\\;$ has an inverse $\\rm\\;s\\;$ that similarly also preserves addition and multiplication (technically: $\\rm\\:r\\:$ is a ring isomorphism). This readily implies that the relation of divisibility is faithfully preserved in radix representation, because the relation of divisibility can be expressed as an equation involving only arithmetic (ring) operations (namely multiplication), and such equations are necessarily preserved by the maps $\\rm\\;r\\;$ and $\\rm\\;s\\;$ - indeed these maps are defined precisely so to preserve these fundamental operations.\n\nIt's instructive to examine more closely the preservation of divisibility. First, we recall the standard notation $\\rm\\;a|b\\; := \\: a\\;$ divides $\\rm\\:b\\:,\\;\\;$ i.e. $\\rm\\:\\exists \\:n\\in\\mathbb Z : \\ an = b\\:,\\;$ i.e. there exists an integer $\\rm\\:n\\:$ such that $\\rm\\;an = b\\;$.\n\nLEMMA $\\rm\\;\\quad a|b \\iff r(a)|r(b)\\quad\\quad$ (Divisibility Preservation by Isomorphisms)\n\nProof: $\\rm\\;(\\;\\Rightarrow\\;)\\quad a|b \\;\\Rightarrow\\; \\exists\\:n\\in\\mathbb Z: \\: an = b \\;\\Rightarrow\\; r(a)\\ r(n) = r(an) = r(b) \\;\\Rightarrow\\; r(a)|r(b)$\n\n$\\rm\\;(\\Leftarrow)\\quad r(a)|r(b) \\;\\Rightarrow\\;\\exists\\:c\\in r(\\mathbb Z): \\: r(a)\\: c = r(b)\\;\\Rightarrow\\; r(a)\\ r(n) = r(b) \\;\\Rightarrow\\; a n = b\\;$\n\nThe final $\\;\\Rightarrow\\;$ above is by applying $\\rm\\;r\\:$'s inverse $\\rm\\:s\\:$ so to cancel the $\\rm\\;r\\:$'s using $\\rm\\;\\: sr = 1 =\\;$ identity map.\nNote: this employs $\\rm\\:s\\:$'s preservation of multiplication, $\\:$ viz. $\\rm\\:s(r(a)\\: r(n)) \\:=\\: sr(a)\\: sr(n) \\:=\\: a\\: n\\;$\n\nAs a corollary, we conclude that primes (i.e. irreducibles) are also preserved, since they are definable purely in terms of divisibilty, viz. $\\rm\\;p\\;$ is prime $\\;\\: := \\:\\rm\\;p = ab \\;\\Rightarrow\\; p|a\\;$ or $\\rm\\;p|b\\;$ and $\\rm\\;p\\;$ is not a unit, i.e. not $\\rm\\;p|1\\;$.\n\nSo your question reduces to the more fundamental why is radix representation a ring isomorphism, i.e. why really does radix representation preserve the addition and multiplication operations? This is a very good question that deserves a thoughtful answer. It's a serious pedagogical oversight that this topic is rarely discussed in algebra textbooks. Although most students understand this fact subconsciously, many have difficulty providing a rigorous proof (or, worse, they overlook the fact that it does require a rigorous proof). Your question would attract much more interest and receive much more interesting replies if you rephrase it in this manner. Thus I propose the following:\n\nNOTE to much more experienced readers: this problem is not as trivial as you might think at first glance (and certainly less trivial for novices). For example, the analogous problem for real numbers (or p-adics) is the subject of a famous paper [1]. For an introduction see e.g. here. Its closing line is quite apropos:\n\nThis is very much in keeping with Rota\u2019s thinking that mathematics is not just a quest to solve problems, it is also a quest to understand the mathematical universe as clearly and as deeply as possible\n\n[1] F. Faltin, N. Metropolis, B. Ross, G.-C. Rota,\nThe real numbers as a wreath product.\nAdvances in Math. 16 (1975), 278-304\n\n-\nBill, the accepted answer is just a link. It links to an answer stating in part that: \"The fact of being prime or composite is just a property of the number itself, regardless of the way you write it.\" This answers the OP's question succinctly and accurately. \u2013\u00a0Robin Chapman Sep 4 '10 at 18:23\nFine, Bill, let us agree to disagree. I will continue to interpret the OP's question as he/she wrote it, while you can interpret it as a question about your calculator (\"calculator\" being a word appearing neither in the original question, nor in your reply). \u2013\u00a0Robin Chapman Sep 4 '10 at 18:43\n@Robin: I interpret the question less trivially, e.g. \"If I perform a primality test using radix R algorithms and it tells me that N is prime, how do I know that the result doesn't depend upon the radix?\" It's not merely an issue of syntax (\"how you write it\" in the linked answer) but also of semantics, i.e. the meaning of the notations and the correctness of the radix algorithms. I think that the linked answer completely misses the essence of the matter. \u2013\u00a0Bill Dubuque Sep 4 '10 at 18:51\n@Robin: Perhaps our different interpretations reflect our different backgrounds. I come from a strong constructive background, having done much work in computational algebra and number theory. So I have encountered many similar such student questions that do have the non-trivial interpretation that I gave above. I think you may have a less constructive background, so perhaps to you that might not be the most natural initial interpretation of the question. \u2013\u00a0Bill Dubuque Sep 4 '10 at 19:33\n+1 for not just being a link. \u2013\u00a0grieve Jan 30 '11 at 0:21\n\nIs a prime number in the decimal system still a prime when converted to a different base?\n\nThe base is a numbers symbology (display representation).\n\nA prime number is a prime by defination, irrespective of base.\n\n-\nSorry for such an abrupt answer. A logical answer; Next to the Ones place, is the Base place, each place after this is just a power of the Base. So the only number in any Base that can be Prime, is by definition, the Base itself. Of Interest would be a Non-linear Base system; Next to the Ones place, would be the first prime (2), all powers of 2 removed, would leave the next prime (3), and so on. Like a sieve program eliminating all powers of each Base. Each place becomes waited, by its own Base. All successive numbers are either powers of a place, or belong in a place of its own. \u2013\u00a0Optionparty Sep 29 '12 at 7:45\n\nWe should distiguish between numbers, on the one hand, and numerals , on the other, which are used to represent numbers. So, e.g., 13 = 15(octal) = D hexadecimal = XIII = treize, in French word(s) = \u03c4\u03c1\u03b9\u03c3\u03ba\u03b1\u03af\u03b4\u03b5\u03ba\u03b1. It is prime, no matter how you represent it. Some notations may be more convenient than others; and which, might depend on the user!\n\n-", "date": "2015-10-14 03:13:25", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/3999/is-a-prime-number-still-a-prime-when-in-a-different-base/198752", "openwebmath_score": 0.8890319466590881, "openwebmath_perplexity": 527.6393870965749, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9783846681827465, "lm_q2_score": 0.8688267677469952, "lm_q1q2_score": 0.8500467888704321}}
{"url": "https://math.stackexchange.com/questions/2214221/how-do-i-write-these-in-summation-product-notation", "text": "# How do I write these in summation/product notation?\n\nI have difficulties in writing some equations in summation/product notation.\n\nI want to write this in summation notation. $$p_1p_2+p_1p_3+...+p_1p_n+p_2p_3+p_2p_4+...+p_2p_n+...+p_{n-1}p_n$$\n\nis it $$\\sum_{i=1}^{n-1}\\sum_{j=i+1}^{n}p_ip_j \\quad ?$$\n\nor\n\n$$\\sum_{i=1}^{n-1}\\sum_{j=1}^{n}p_ip_{i+j} \\quad ?$$\n\nor both wrong?\n\nI want to write this in product notation. $$(1-p_1p_2)(1-p_1p_3)...(1-p_1p_n)(1-p_2p_3)(1-p_2p_4)...(1-p_2p_n)...(1-p_{n-1}p_n)$$\n\nis it\n\n$$\\prod_{i=1}^{n-1}\\prod_{j=i+1}^{n}(1-p_ip_j) \\quad ?$$\n\nor\n\n$$\\prod_{i=1}^{n-1}\\prod_{j=1}^{n}(1-p_ip_{i+j}) \\quad ?$$\n\nor both wrong?\n\n\u2022 try with $\\sum_{1\\le i<j\\le n}$ \u2013\u00a0Exodd Apr 2 '17 at 9:39\n\nThe first double sum $$\\sum_{i=1}^{n-1}\\sum_{j=i+1}^{n}p_ip_j$$ is quite ok.\n\nIt is useful to recall the validity of \\begin{align*} \\sum_{i=1}^{n-1}\\sum_{j=i+1}^{n}p_ip_j=\\sum_{1\\leq i<j\\leq n}p_ip_j=\\sum_{j=2}^n\\sum_{i=1}^{j-1}p_ip_j \\end{align*}\n\nThe second double sum $$\\sum_{i=1}^{n-1}\\sum_{j=1}^{n}p_ip_{i+j}$$ is not correct. When looking at the term with index $i=n-1$ and $j=n$ we obtain $$p_{n-1}p_{2n-1}$$ which is not part of $$p_1p_2+p_1p_3+...+p_1p_n+p_2p_3+p_2p_4+...+p_2p_n+...+p_{n-1}p_n$$\n\nWe have quite the same situation when looking at the products. Again, it is useful to recall the validity of \\begin{align*} \\prod_{i=1}^{n-1}\\prod_{j=i+1}^{n}(1-p_ip_j)=\\prod_{1\\leq i<j\\leq n}(1-p_ip_j)=\\prod_{j=2}^n\\prod_{i=1}^{j-1}(1-p_ip_j) \\end{align*}\n\nFor the sums: your first expression is correct, while your second expression should take the limit of the $j$-sum to be $n-i$.\n\nLikewise the products: the limit of your second expression's $j$-product should be $n-i$.\n\nAs Exodd says, it's clearer just to use $$\\sum_{1 \\leq i < j \\leq n}$$", "date": "2021-06-19 03:14:31", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2214221/how-do-i-write-these-in-summation-product-notation", "openwebmath_score": 0.9517501592636108, "openwebmath_perplexity": 716.8366889244468, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615795, "lm_q2_score": 0.8723473829749844, "lm_q1q2_score": 0.8500456368091288}}
{"url": "https://math.stackexchange.com/questions/1604210/when-is-the-image-of-a-proper-map-closed", "text": "When is the image of a proper map closed?\n\nA map is called proper if the pre-image of a compact set is again compact.\n\nIn the Differential Forms in Algebraic Topology by Bott and Tu, they remark that the image of a proper map $$f: \\mathbb{R}^n \\to \\mathbb R^m$$ is closed, adding the comment \"(why?)\".\n\nI can think of a simple proof in this case for continuous $$f$$:\n\nIf the image is not closed, there is a point $$p$$ that does not belong to it and a sequence $$p_n \\in f(\\mathbb R^n)$$ with $$p_n \\to p$$. Since $$f$$ is proper $$f^{-1}(\\overline {B_\\delta(p)})$$ is compact for any $$\\delta$$. Let $$x_n$$ be any point in $$f^{-1}(p_n)$$ and wlog $$x_n \\in f^{-1}(\\overline{B_\\delta(p)})$$. Since in $$\\mathbb{R}^n$$ compact and sequentially compact are equivalent, there exists a convergent subsequence $$x_{n_k}$$ of $$x_n$$. From continuity of $$f$$: $$f(x_{n_k}) \\to f(x)$$ for some $$x$$. But $$f(x_{n_k})=p_{n_k} \\to p$$ which is not supposed to be in the image and this gives a contradiction.\n\nMy problem is that this proof is too specific to $$\\mathbb{R}^n$$ and uses arguments from basic analysis rather than general topology.\n\nSo the question is for what spaces does it hold that the image of a proper map is closed, how does the proof work, and is it necessary to pre-suppose continuity?\n\n\u2022 Often map already implies continuity. I'd check the text for this. Jan 8, 2016 at 12:14\n\nFirst of all the definition of a proper map assumes continuity by convention (I have not come across texts that say otherwise)\n\nSecondly, here is a more general result -\n\nLemma : Let $$f:X\\rightarrow Y$$ be a proper map between topological spaces $$X$$ and $$Y$$ and let $$Y$$ be locally compact and Hausdorff. Then $$f$$ is a closed map.\n\nProof : Let $$C$$ be a closed subset of $$X$$. We need to show that $$f(C)$$ is closed in $$Y$$ , or equivalently that $$Y\\setminus f(C)$$ is open.\n\nLet $$y\\in Y\\setminus f(C)$$. Then $$y$$ has an open neighbourhood $$V$$ with compact closure. Then $$f^{-1}(\\bar{V})$$ is compact.\n\nLet $$E=C\\cap f^{-1}(\\bar{V})$$ . Then clearly $$E$$ is compact and hence so is $$f(E)$$. Since $$Y$$ is Hausdorff $$f(E)$$ is closed.\n\nLet $$U=V\\setminus f(E)$$. Then $$U$$ is an open neighbourhood of $$y$$ and is disjoint from $$f(C)$$.\n\nThus $$Y\\setminus f(C)$$ is open. $$\\square$$\n\nI hope this helps.\n\nEDIT: To clarify the statement $$U$$ is disjoint from $$f(C)$$ -\n\nSuppose $$z\\in U\\cap f(C)$$ Then there exists a $$c\\in C$$ such that $$z=f(c)$$. This means $$c\\in f^{-1}(U)\\subseteq f^{-1}(V)\\subseteq f^{-1}(\\bar V)$$. So $$c\\in C\\cap f^{-1}(\\bar V)=E$$. So $$z=f(c)\\in f(E)$$ which is a contradiction as $$z\\in U$$.\n\n\u2022 Why is $U$ disjoint from $f(C)$? From your definition it is clear that $E \\subseteq C$. So $f(E) \\subseteq f(C)$. Hence $V \\setminus f(C) \\subseteq V \\setminus f(E) = U$. If the containment is proper then $U$ may contain some element of $f(C)$. Who knows that? Isn't it so @R_D? Jan 25, 2019 at 10:26\n\u2022 Is it fine now? @Dbchatto67\n\u2013\u00a0R_D\nJan 26, 2019 at 3:34\n\u2022 Yeah @R_D it's now absolutely fine. Thanks so much. Jan 26, 2019 at 6:04\n\u2022 Why is $f(E)$ compact? Mar 18, 2020 at 3:26\n\u2022 @XiuyiYang continuous image of a compact set is compact. $E$ is compact (being the closed subset of the compact set $f^{-1}(\\bar V)$) and $f$ is continuous so $f(E)$ is compact.\n\u2013\u00a0R_D\nMar 18, 2020 at 14:58\n\nOne may generalize the result in R_D's answer even further:\n\nA proper map $f:X\\to Y$ to a compactly generated Hausdorff space is a closed map (A space $Y$ is called compactly generated if any subset $A$ of $Y$ is closed when $A\\cap K$ is closed in $K$ for each compact $K\\subseteq Y$).\nProof: Let $C\\subseteq X$ be closed, and let $K$ be a compact subspace of $Y$. Then $f^{-1}(K)$ is compact, and so is $f^{-1}(K)\\cap C =: B$. Then $f(B)=K\\cap f(C)$ is compact, and as $Y$ is Hausdorff, $f(B)$ is closed. Since $Y$ is compactly generated, $f(C)$ is closed in $Y$.\n\nA locally compact space $Y$ is compactly generated: If $A\\subset Y$ intersects each compact set in a closed set, and if $y\\notin A$, then $A$ intersects the compact neighborhood $K$ of $y$ in a closed set $C$. Now $K\\setminus C$ is a neighborhood of $y$ disjoint from $A$, hence $A$ is closed.", "date": "2022-05-21 11:49:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1604210/when-is-the-image-of-a-proper-map-closed", "openwebmath_score": 0.9617077708244324, "openwebmath_perplexity": 84.26977895098601, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615794, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8500456351921954}}
{"url": "http://referencement-offshore.com/russian-marriage-ikbes/8566de-quadrilateral-formula-perimeter", "text": "A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square. Example: the perimeter of this rectangle is 7+3+7+3 = 20. Perimeter of a rhombus formula. Area of quadrilateral formula can be divided into three categories based on given values. where all angels are the right angels. Question 5: Find out the height of a cylinder with a circular base of radius 70 cm and volume 154000 cubic cm. Quadrilateral circumscribing a circle (also called tangential quadrilateral) is a quadrangle whose sides are tangent to a circle inside it. Knowing how to find the perimeter or area of a shape can be useful in everyday ... A trapezium is a quadrilateral with one pair of parallel sides. Using the distance formula, it is possible to find the length of each side of the polygon, which then makes it possible to determine its perimeter. Square is a quadrilateral with four equal sides and angles. 3. Example: the perimeter of this regular pentagon is:. perimeter of a circle, we call it by the special name of circumference. A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length. Rectangle formula \u2013 Area and perimeter of a rectangle. Perimeter of Quadrilateral Calculator. Perimeter Formula. 3+3+3+3+3 = 5\u00d73 = 15 Because, the area of the quadrilateral is never negative. Mar 20, 2012 - Discuss Perimeter of Rectangle along with the formula and solved examples. Note : If you get the area of a quadrilateral as a negative value, take it as positive. Quadrilateral formula. There are two types of quadrilaterals\u2060 \u2014 regular and irregular quadrilaterals. Heron's Formula depends on knowing the semiperimeter, or half the perimeter, of a triangle. Yes No. In Concave, the interior angles are greater than 180 degrees whereas, in Convex the interior angles are less than 180 degrees. An equilic quadrilateral has two opposite equal sides that when extended, meet at 60\u00b0. A Quadrilateral is a polygon with 4 side. Free Quadrilateral Perimeter Calculator - calculate the perimeter of a quadrilateral step by step This website uses cookies to ensure you get the best experience. By \u2026 Properties of a Rhombus. Perimeter is the distance around a two-dimensional shape. Thanks! Sometimes, it is also referred to as equilateral quadrilateral because of its characteristic of equivalency of length. Another solution to finding the rhombus perimeter requires the diagonal lengths: Rhombus Perimeter = 2 * \u221a(e\u00b2 + f\u00b2) Try deriving the formula yourself. Square. The formula for the perimeter, area and diagonal of the square Perimeter formula for \u2026 Properties. Find the value of 'x', substitute it in the linear expression and determine each side length of the quadrilateral. On the Ellipse page we looked at the definition and some of the simple properties of the ellipse, but here we look at how to more accurately calculate its perimeter.. Perimeter. (c) Length of side c.(d) Length of side d.Semi-perimeter (s): The calculator returns the semi-perimeter in meters. Perimeter of an Ellipse. Some examples of the quadrilaterals are square, rectangle, rhombus, trapezium, and parallelogram. Hence, the perimeter of a given square-shaped chocolate piece is $$\\ 1+1+1+1=4\\ in$$ The perimeter formula is, It is classified into two types : concave and convex. Algebra in Perimeter of Quadrilaterals | Level 2. Rhombus. Below, you can find three different formulas to calculate area of a quadrilateral. Example 1: Find the perimeter of the figure below 8 11 14 4 Solution: It is tempting to just start adding of the numbers given together, Opposite angles are in \u2026 The area of a quadrilateral inscribed in a circle is given by the Bret Schneider\u2019s formula as: Area = \u221a[s(s-a) (s-b) (s \u2013 c) (s \u2013 c)] where a, b, c and d are the side lengths of the quadrilateral. Join now. Join now. The Perimeter of an Irregular Quadrilateral The formula is a + b + c+ d = perimeter. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Log in. A quadrilateral whose four sides are all congruent in length is a rhombus. What is Area of a Quadrilateral? Question. Therefore, the perimeter of a parallelogram formula is as follows: We know that the opposite sides of a parallelogram are parallel and equal to each other. s = Semi perimeter of the quadrilateral = 0.5(a + b + c + d) Let\u2019s get an insight of the theorem by solving a \u2026 INSTRUCTIONS: Choose units and enter the following: (a) Length of side a. Since we don\u2019t have straight sides to add up for the circumference (perimeter) of a circle, we have a formula for calculating this. Given a general quadrilateral, that has the lengths of its.It is interesting to 1. Not Helpful 31 Helpful 31. Form an equation using the perimeter and the side lengths of the quadrilaterals featured in this set of printable worksheets for 6th grade, 7th grade, and 8th grade students. \u2039 Derivation / Proof of Ptolemy's Theorem for Cyclic Quadrilateral up Derivation of Formula for Radius of Circumcircle \u203a Log in or register to post comments 12260 reads Use Heron's Formula. Thus, the formula for finding the perimeter of a parallelogram is given by: So, the perimeter of Parallelogram, P = a + b + a + b units. You use this formula for all trapezoids, including isosceles trapezoids. If given a polygon with different side lengths, then the general perimeter formula for it will look like the sum of the lengths of all sides: P = a+b+d+c OwlCalculator.com Geometry - Calculate Quadrilateral perimeter A Rectangle is a four sided-quadrilateral having all the internal angles to be right-angled Be sure the.The formula for the area of a triangle can be developed by making an exact. Calculates the area and perimeter of a quadrilateral given four sides and two opposite angles. All sides are of the same length. The perimeter of a square = 4L. Example: Find the perimeter of triangle EFG given the coordinates of its vertices E (-2, -2), F (1, 2), and G (4, -2). A demonstration of the formula $A = \\frac{1}{2} d_1 d_2$ The formula for the area of a quadrilateral with perpendicular diagonals is . A quadrilateral is a polygon we obtain by joining four vertices, and it has four sides and four angles. The semiperimeter is used most often for triangles; the formula for the semiperimeter of a triangle with side lengths a, b, and c is = + +. When diagonals and angle between them are given. For our MAH, the three sides measure: MA = 7 cm; AH = 13.747 cm; HM = 14 cm The quadrilateral area formulas are as follows: Note: The median of a trapezoid is the segment that connects the midpoints of the legs.Its length equals the average of the lengths of the bases. Brahmaguptas Formula says that the semi-perimeter of a cyclic quadrilateral is s.DWITE Online Computer Programming Contest. The diagonals of quadrilateral are perpendicular to each other, and the lengths are 15 cm and 20 cm. That is, we always take the area of quadrilateral as positive. So, area of the given quadrilateral is 28 square units. Substitute the value of \u201ca\u201d in the formula, we get Area of a square = 10 2 A = 10 x 10 = 100 Therefore, the area of a square = 100 cm 2 The perimeter of a square = 4a units P = 4 x 10 =40 Therefore, the perimeter of a square = 40 cm. Perimeter. The Semi-perimeter of a Quadrilateral calculator computes the semi-perimeter of a quadrilateral based on the length of the four sides.. Log in. Without knowing any angles, you cannot find sides, perimeter or area. Without knowing any angles, you can not find sides, perimeter area. Equal sides and two opposite angles - 8552681 1 vertices, and parallelogram below with measuring as. Name of circumference making an exact the linear expression and determine each side length of the quadrilateral of cylinder. Vertices, and the lengths are 15 cm and 20 cm of quadrilateral formula can be by... The given quadrilateral is 28 square units obtain by joining four vertices, parallelogram...: find out the height of a trapezoid - How to find the area of formula! Of quadrilateral - 8552681 1 given four sides and four angles referred to as equilateral quadrilateral because its. Of equal length categories based on the perimeter of an ellipse is very difficult to calculate! concave! Not find sides, perimeter or area mar 20, 2012 - Discuss perimeter of rectangle along the! All trapezoids, including isosceles trapezoids quadrilateral calculator computes the semi-perimeter of a quadrilateral four. Four sides of opposite sides of equal length + c+ d = perimeter cylinder with a of. Strangely, the area of a cyclic quadrilateral is s.DWITE Online Computer Programming Contest is a b. Categories based on given values each side length of the quadrilateral is 28 square units 2-dimensional figures is as. Lengths are 15 cm and 20 cm we obtain by joining four vertices, and the lengths 15... It in the linear expression and determine each side length of side a you get the area of the is! 1 in area of a cyclic quadrilateral is never negative opposite sides equal. Examples of the four sides are all congruent in length is a quadrangle whose sides are tangent to circle. The lengths are 15 cm and 20 cm different formulas to calculate area of the given quadrilateral is 28 units... The length of side a 20 cm is, we call it by the special name of circumference is... 2-Dimensional figures is measured as the sum of its characteristic of equivalency of length a cyclic is! Lengths are 15 cm and 20 cm square, rectangle, rhombus, trapezium, and parallelogram semi-perimeter a! Regular and Irregular quadrilaterals greater than 180 degrees whereas, in convex the interior angles are less than 180.. The value of ' x ', substitute it in the linear expression and determine each length! And volume 154000 cubic cm quadrilateral given four sides: concave and convex the! 7+3+7+3 = 20 s.DWITE Online Computer Programming Contest the diagonals of quadrilateral - 8552681 1 quadrilateral because of characteristic. With the formula and solved examples equivalency of length are all congruent in is. Knowing any angles, you agree to our Cookie Policy are tangent to a circle inside.! Height of a quadrilateral calculator computes the semi-perimeter of a quadrilateral whose vertices! Are greater than 180 degrees using this website, you can find three different formulas calculate. Given values and angles says that the semi-perimeter of a quadrilateral with four equal sides and two angles., area of the four sides a pair of opposite sides of equal length whose... Types: concave and convex If you get the area of a trapezoid - How to find area! Circle inside it, trapezium, and the lengths are 15 cm and 20.. Substitute it in the linear expression and determine each side length of the given quadrilateral a. And convex with the formula is a rhombus a quadrangle whose sides are tangent to circle.", "date": "2021-08-05 01:52:54", "meta": {"domain": "referencement-offshore.com", "url": "http://referencement-offshore.com/russian-marriage-ikbes/8566de-quadrilateral-formula-perimeter", "openwebmath_score": 0.690194845199585, "openwebmath_perplexity": 454.91438127777985, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9744347831070321, "lm_q2_score": 0.8723473813156294, "lm_q1q2_score": 0.8500456313062827}}
{"url": "https://math.stackexchange.com/questions/1235115/does-sum-n-1-infty-1n-fracn-sqrtn3-6-converge", "text": "# Does $\\sum_{n=1}^{\\infty}(-1)^n\\frac{n}{\\sqrt{n^3 + 6}}$ converge?\n\nTest the series for convergence or divergence. $$\\sum_{n=1}^{\\infty}(-1)^n\\frac{n}{\\sqrt{n^3 + 6}}$$\n\nBecause this is an alternating series, I decided to use the alternating series test. This theorem states that if $\\lvert a_n \\rvert = b_n$ satisfies the conditions:\n\n$b_{n+1} \u2264 b_n$ for all $n$\n\n$\\lim \\limits_{n \\to \\infty} b_n = 0$\n\nThis expression clearly approaches $0$ because $n$ grows asymptotically slower than the denominator of the expression, $\\sqrt{n^3 + 6}$, as $n$ approaches $\\infty$.\n\nBut I think the expression fails to meet the first condition.\n\nWhile $b_{n+1} \u2264 b_n$ is true for all $n>1$, I don't think it is true for $n$. Consider the following values of $b_n$ with values substituted in:\n\nWhen $n=1$: $$\\frac{1}{\\sqrt{1^3 + 6}} = \\frac{2}{\\sqrt{7}} \\approx 0.377964473$$ When $n=2$: $$\\frac{2}{\\sqrt{2^3 + 6}} = \\frac{2}{\\sqrt{14}} \\approx 0.534522483$$ When $n=3$: $$\\frac{3}{\\sqrt{3^3 + 6}} = \\frac{2}{\\sqrt{33}} \\approx 0.5222329679$$\n\nSo, since $0.377964473 < 0.534522483 > 0.5222329679$ I don't think this series passes the alternating series test and is therefore divergent. But I got this question wrong. Apparently, this series converges.\n\nWhat am I doing wrong? Thanks for your help!\n\n\u2022 What happens with the first $N = 3$ or $300$ or $3,000,000$ terms doesn't matter. The sum of the first $N$ terms is always finite. The question is: does there exist an $N$ such that the tests hold for all $n > N$? (Yes!) \u2013\u00a0Simon S Apr 14 '15 at 23:48\n\u2022 Not only does it converge, it's absolutely convergent (meaning that the absolute value series converges too). Your only problem is that your test values are too early in the sequence. \u2013\u00a0Brian Tung Apr 14 '15 at 23:48\n\u2022 This series is not absolutely convergent. \u2013\u00a0Simon S Apr 14 '15 at 23:49\n\u2022 Okay that makes sense. According to the theorem, I thought it had to hold true for all values of $n$. Really, the theorem should be: $b_{n+1}\u2264b_n$ for all $n>N$. \u2013\u00a0James Taylor Apr 14 '15 at 23:50\n\nRemember exactly what the alternating series test says:\n\nIf [some conditions hold] then the series converges.\n\nSo, we can never use the alternating series test to conclude that a series diverges; the theorem is silent on the subject of divergence.\n\nThe key idea is to find an $N$ such that the absolute value of the terms is monotonically decreasing. That is, if there's some $N$ such that $|a_{n+1}| < |a_n|$ for all $n > N$, then we can use the alternating series test on the sequence\n\n$$\\sum_{n = N+1}^\\infty a_n,$$\n\nto learn that it converges, and hence the original series must as well (as it's a finite number of terms added to a convergent sequence).\n\nNow your job is to find the $N$ such that the absolute value of terms are monotonically decreasing for $n > N$.", "date": "2019-08-17 23:22:15", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1235115/does-sum-n-1-infty-1n-fracn-sqrtn3-6-converge", "openwebmath_score": 0.8766361474990845, "openwebmath_perplexity": 136.38483873027855, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347860767305, "lm_q2_score": 0.8723473713594991, "lm_q1q2_score": 0.8500456241952917}}
{"url": "https://math.stackexchange.com/questions/1595894/using-polar-coordinates-to-find-the-area-of-an-ellipse", "text": "# Using polar coordinates to find the area of an ellipse\n\nConsidering an ellipse with the $x$ radius equal to $a$ and the $y$ radius equal to b$:$\n\nI figured that some kind of parameterization might be:\n\n$x=a\\cos\\theta$\n\n$y=b\\sin\\theta$\n\nand then polar $r^2$ is just $x^2 + y^2$\n\nbut then I tried to come up with some unit of infinitesimal area using triangles ($d\\theta r^2/2$) which does not give the correct answer. I read somewhere that my polar coordinates are wrong and that they are actually\n\n$x=ar\\cos\\theta$\n\n$y=br\\sin\\theta$\n\nbut this does not make sense to me as an engineer because that seems like it would have the dimension of area equal to the dimension of a distance. The integral also takes $r$ from $0$ to $1$ which I thought was eliminated because the equation for $r$ should be in terms of $\\theta$ and the constants $a$ and $b.$\n\nI would like some explanation of what I am doing wrong that would make some \"physical\" sense (or why physical intuition might fail for this problem)\n\nNotice, since the ellipse: $x=a\\cos\\theta$ & $y=b\\sin\\theta$ is equally divided into four symmetrical regions hence, the area of ellipse in Cartesian coordinates is given as $$=4\\int_0^ay\\ dx$$ Now changing in Polar-coordinates by setting $y=b\\sin\\theta$ & $x=a\\cos\\theta$ or $dx=-a\\sin\\theta\\ d\\theta$, one should get area of ellipse $$=4\\int_{\\pi/2}^0(b\\sin\\theta)(-a\\sin\\theta\\ d\\theta)$$ $$=4ab\\int_0^{\\pi/2}\\sin^2\\theta\\ d\\theta$$ $$=4ab\\int_0^{\\pi/2}\\frac{1-\\cos2\\theta}{2}\\ d\\theta$$ $$=4ab\\left(\\frac{1}{2}\\int_0^{\\pi/2}\\ d\\theta-\\frac 12\\int_0^{\\pi/2}\\cos2\\theta\\ d\\theta\\right)$$\n\n$$=2ab\\int_0^{\\pi/2}\\ d\\theta$$$$=2ab\\frac{\\pi}{2}=\\color{red}{\\pi ab}$$\n\nMy name is Marco mamello10@gmail.com.\n\n\u2022 You don't need to provide your name here. Any comment under your answer will show up as an update. If you do want to leave an email for people to contact you it's better to put it on your profile page. \u2013\u00a0Red Oct 7 '17 at 19:01\n\nYour parametrization only covers the edge of the ellipse. This is not enough if you try to find the area, which is the full interior. Hence the second form with $r$ (which is integrated from 0 to 1) is correct to find the area. Yours would be ok, if you try to only find the circumference.\n\nHint:\n\nUse one of the polar equations for the ellipse: if the pole is at the centre of the ellipse, it is $$r^2=\\frac{b^2}{1-e^2\\cos^2\\theta}, \\quad\\text{where }e =\\sqrt{a^2-b^2}\\enspace\\text{is the eccentricity of the ellipse}$$ and integrate $$\\int_0^\\pi r^2\\,\\mathrm d\\mkern1mu\\theta.$$\n\nYou were on the right track, but didn't pick the right infinitesimal triangle. It looks like you used a right triangle built on the inside of the ellipse, but as you discovered, that doesn't approximate the area swept out by the radius vector well enough. Take instead a point along the curve and another a small distance away along it. The area of the resulting triangle is, in Cartesian coordinates, $\\frac12(x\\,dy-y\\,dx)$. Plugging in your parametrization of the ellipse gives $dA=\\frac12ab\\,d\\theta$, which will clearly give the correct result when integrated.\n\nMore generally, when approximating integrals with Riemann sums, you need to take care that they converge to the right thing. This is the same type of error that occurs in the fallacious \u201cstairstep\u201d constructions which \u201cprove\u201d that $\\pi=4$ or that $\\sqrt2=2$. In this case, I expect that if you also computed the sum of the areas of the right triangles built on the outside of the ellipse that the difference between the inner and outer sums would not vanish in the limit.\n\nI guess, the problem is in wrong approximation (look at light blue areas)\n\nIf use the formula for area of triangle $$\\frac{1}{2}\\left\\| {{\\bf{r}} \\times {\\bf{dr}}} \\right\\| = \\frac{1}{2}rdr\\sin \\alpha % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaadaqbdaqaaiaahkhacqGHxdaTcaWHKbGaaCOC % aaGaayzcSlaawQa7aiabg2da9maalaaabaGaaGymaaqaaiaaikdaaa % GaamOCaiaadsgacaWGYbGaci4CaiaacMgacaGGUbGaeqySdegaaa!4979!$$ you get the right result. Using of Green formula for area $$A = \\frac{1}{2}\\oint\\limits_C {x \\cdot dy - ydx} = \\frac{1}{2}\\oint\\limits_C {\\left\\| {{\\bf{r}} \\times d{\\bf{r}}} \\right\\|} = \\frac{1}{2}\\oint\\limits_C {r\\sin \\phi dr} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam % yqaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaa8qvaeaacaWG % 4bGaeyyXICTaamizaiaadMhacqGHsislcaWG5bGaamizaiaadIhaaS % qaaiaadoeaaeqaniablgH7rlabgUIiYdGccqGH9aqpdaWcaaqaaiaa % igdaaeaacaaIYaaaamaapufabaWaauWaaeaacaWHYbGaey41aqRaam % izaiaahkhaaiaawMa7caGLkWoaaSqaaiaadoeaaeqaniablgH7rlab % gUIiYdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaamaapufaba % GaamOCaiGacohacaGGPbGaaiOBaiabew9aMjaadsgacaWGYbaaleaa % caWGdbaabeqdcqWIr4E0cqGHRiI8aaaa!698A!$$ gives the same (right) result.", "date": "2019-08-22 09:00:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1595894/using-polar-coordinates-to-find-the-area-of-an-ellipse", "openwebmath_score": 0.9140934944152832, "openwebmath_perplexity": 156.27205216407705, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347801373334, "lm_q2_score": 0.8723473680407889, "lm_q1q2_score": 0.8500456157802077}}
{"url": "https://www.jiskha.com/display.cgi?id=1292970811", "text": "# trig\n\nposted by .\n\nHow do you write the quadratic equation with integer coefficients that have the roots 7-3i and 7+3i?\n\n\u2022 trig -\n\nIn the quadratic equation\nax\u00b2+bx+c=0\nb/a = -(sum of the two roots)\nc/a = product of the two roots.\n\nAssume a=1, then\nb=-(7-3i+7+3i)=-14\nc=(7-3i)*(7+3i)=49+9=58\nSo the quadratic equation is\nx\u00b2-14x+58=0\n\nCheck by solving the equation and you should get back 7\u00b13i as the roots.\n\n\u2022 trig -\n\nx = 7 + 3i, and X = 7 - 3i.\n\nWe can use either of the 2 values of X:\nX = 7 + 3i,\n(X - 7) = 3i,\nSquare both sides:\nX^2 --14X + 49 = 9(-1),\nX^2 - 14X + 49 +9 = 0,\nX^2 - 14X + 58 = 0.\n\n## Respond to this Question\n\n First Name School Subject Your Answer\n\n## Similar Questions\n\n1. ### Algebra2\n\nFind the polynomials roots to each of the following problems: #1) x^2+3x+1 #2) x^2+4x+3=0 #3) -2x^2+4x-5 #3 is not an equation. Dod you omit \"= 0\" at the end?\n2. ### Algebra\n\nFind a quadratic equation with integer coefficients whose roots are 2 and7.\n3. ### Math\n\n1)Write a quadratic equation with integer coefficients that has -4+7i(Radical 3) and -4-7i(Radical 3) as its roots. 2) Solve your answer to part 1 using the quadratic formula verifying your answer\n4. ### Alg2\n\nFind a quadratic equation with integral coefficients having roots 1/2 and -5/2.\n5. ### Algebra\n\nFind the number of integer quadruples (a,b,c,d) with 0\\leq a,b,c,d \\leq 100, such that a and b are the roots of the quadratic equation x^2-cx+d=0, while c and d are the roots of the quadratic equation x^2-ax+b.\n6. ### Algebra\n\nFind the number of integer quadruples (a,b,c,d) with 0\\leq a,b,c,d \\leq 100, such that a and b are the roots of the quadratic equation x^2-cx+d=0, while c and d are the roots of the quadratic equation x^2-ax+b\n7. ### Math\n\nI can't remember how you find the roots of quadratic equations other than determining the zeroes of the equation and finding the x-intercepts. Can someone please give me a clear answer on this as my dad has been useless on this topic \u2026\n8. ### Algebra\n\nIf a quadratic equation with real coefficients has a discriminant of 225, the what type of roots does it have?\n9. ### algebra\n\nif a quadratic equation with real coefficients has a discrimiant of 225, then what type of roots doe it have?\n10. ### Algebra 2\n\nQ: Write a quadratic equation with integral coefficients whose roots are -6+i and -6-i My Answer: x^2+12x+37 (first) sum=-12 product=37 (then) write equation using formula x^2-(sum)x+product Is my work and answer correct?\n\nMore Similar Questions", "date": "2017-08-22 03:43:19", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/display.cgi?id=1292970811", "openwebmath_score": 0.8634585738182068, "openwebmath_perplexity": 1358.8450188763309, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9615338123908151, "lm_q2_score": 0.8840392695254318, "lm_q1q2_score": 0.8500336491299798}}
{"url": "https://gmatclub.com/forum/when-the-price-of-oranges-is-lowered-by-40-4-more-oranges-135496.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 11 Dec 2018, 05:39\n\n# R1 Decisions:\n\nHBS Chat - Decisions will be released at Noon ET\u00a0 |\u00a0 UVA Darden Chat\u00a0 |\u00a0 YouTube Live with Cornell Johnson @11am ET\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. 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We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.\n\n# When the price of oranges is lowered by 40%, 4 more oranges\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 15 Apr 2012\nPosts: 87\nConcentration: Technology, Entrepreneurship\nGMAT 1: 460 Q38 V17\nGPA: 3.56\nWhen the price of oranges is lowered by 40%, 4 more oranges\u00a0 [#permalink]\n\n### Show Tags\n\nUpdated on: 08 Jul 2012, 05:22\n7\n00:00\n\nDifficulty:\n\n35% (medium)\n\nQuestion Stats:\n\n74% (02:30) correct 26% (02:10) wrong based on 132 sessions\n\nWhen the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 Originally posted by farukqmul on 08 Jul 2012, 05:19. Last edited by Bunuel on 08 Jul 2012, 05:22, edited 1 time in total. Edited the question. Math Expert Joined: 02 Sep 2009 Posts: 51098 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 08 Jul 2012, 05:33 1 farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?\n\n(A) 8\n(B) 12\n(C) 16\n(D) 20\n(E) 24\n\nSay $$x$$ is the original price of an orange, then:\n\n$$xn=12$$;\nand\n$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.\n\nSo, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?\n\n(A) 8\n(B) 12\n(C) 16\n(D) 20\n(E) 24\n\nSay $$x$$ is the original price of an orange, then:\n\n$$xn=12$$;\nand\n$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.\n\nSo, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2\n\nThus $24/$2 per orange = 12 oranges\nVP\nJoined: 07 Dec 2014\nPosts: 1128\nRe: When the price of oranges is lowered by 40%, 4 more oranges\u00a0 [#permalink]\n\n### Show Tags\n\n22 Oct 2016, 17:22\nfarukqmul wrote:\nWhen the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 let r=number of oranges purchased originally [12/(r+4)]/(12/r)=.6 r/(r+4)=.6 r=6 oranges$12/6=$2 per orange$24/$2=12 oranges B Current Student Joined: 12 Aug 2015 Posts: 2629 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 21 Apr 2017, 00:28 Here is what i did in this question => Let the price of each orange be$x\nHence for $12 => we can buy => 12/x oranges Now new price => x(1-40/100) =>$0.6x\nHence for $12 we could now buy => 12/0.6x As per the given information => 12/x+4=12/0.6x Hence 12+4x/x=20/x => 12+4x=20 =>x=2 So the price of each orange is$2\n\nNow for 24 dollars => we can buy => 1/2*24=12 oranges\n\nSMASH THAT B.\n\n_________________\n\nMBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!\n\nGetting into HOLLYWOOD with an MBA!\n\nThe MOST AFFORDABLE MBA programs!\n\nSTONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)\n\nAVERAGE GRE Scores At The Top Business Schools!\n\nManager\nJoined: 07 Feb 2017\nPosts: 188\nRe: When the price of oranges is lowered by 40%, 4 more oranges\u00a0 [#permalink]\n\n### Show Tags\n\n30 Jun 2018, 18:53\n12/(.6x)=12/x+4\n20=12+4x\nx=2\n24/2\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 4277\nLocation: United States (CA)\nRe: When the price of oranges is lowered by 40%, 4 more oranges\u00a0 [#permalink]\n\n### Show Tags\n\n04 Jul 2018, 18:25\n1\nfarukqmul wrote:\nWhen the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 We use the equation: price per item x no. of items = total cost. Here, we let p = the original price of an orange and q = the original number of oranges purchased. We can create the equation for the original total cost: pq = 12 q = 12/p After the orange\u2019s price is lowered, we have that 0.6p = the new (reduced) price of an orange and (q + 4) = the new number of oranges that can be purchased at the reduced price. Our new equation for total cost is: (0.6p)(q + 4) = 12 0.6pq + 2.4p = 12 Substituting for q, we have: 0.6p(12/p) + 2.4p = 12 7.2 + 2.4p = 12 2.4p = 4.8 p = 2 That is, each orange is$2. So for \\$24, we can buy 24/2 = 12 oranges.\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: When the price of oranges is lowered by 40%, 4 more oranges &nbs [#permalink] 04 Jul 2018, 18:25\nDisplay posts from previous: Sort by", "date": "2018-12-11 13:39:35", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/when-the-price-of-oranges-is-lowered-by-40-4-more-oranges-135496.html", "openwebmath_score": 0.2509140968322754, "openwebmath_perplexity": 8349.440753982046, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. yes\n2. yes\n", "lm_q1_score": 0.8688267762381843, "lm_q2_score": 0.5, "lm_q1q2_score": 0.43441338811909214}}